SOLUTIONS TO RAINVILLE’S “SPECIAL FUNCTIONS” (1960) SOLUTIONS BY SYLVESTER J. PAGANO AND LEON HALL; EDITED BY TOM CUCHTA
1. Introduction 2. Chapter 1 Solutions Problem 1. Show that the following product converges and find its value: ∞ Y 6 1+ . (n + 1)(2n + 9) n=1 Solution 1. (Solution by Leon Hall) By Theorem 3, page 3, this product converges ∞ X 6 converges absolutely. absolutely because (n + 1)(2n + 9) n=1 1+
6 (n + 1)(2n + 9)
(n + 1)(2n + 9) + 6 (n + 1)(2n + 9) 2n2 + 11n + 15 = (n + 1)(2n + 9) (2n + 5)(n + 3) = . (n + 1)(2n + 9)
n Y 1+
=
So, if Pn
=
k=1 n Y
6 (k + 1)(2k + 9)
(2k + 5)(k + 3) (k + 1)(2k + 9) k=1 [7 · 9 · · · (2n + 5)][4 · 5 · · · (n + 3)] = [2 · 3 · · · (n + 1)][11 · 13 · · · (2n + 9)] [7 · 9][(n + 2) · (n + 3)] = [2 · 3][(2n + 7) · (2n + 9)] 21 (n + 2)(n + 3) = 2 (2n + 7)(2n + 9) =
then
21 n2 + 5n + 6 21 = . n→∞ n→∞ 2 4n2 + 32n + 63 8 Note: The use of Theorem 3 is not needed because finding the value of the infinite product is sufficient itself to show convergence. lim Pn = lim
1
2
SOLUTIONS BY SYLVESTER J. PAGANO AND LEON HALL; EDITED BY TOM CUCHTA
Problem 2. Show that
∞ Y n=2
1 1− 2 n
=
1 . 2
Solution 2. (Solution by Leon Hall) n2 − 1 (n + 1)(n − 1) 1 = = 2 n n2 n2
1− Let Pn
= =
n Y 1 1− 2 k k=2 n Y (k + 1)(k − 1)
k2 [3 · 4 · · · (n + 1)][1 · 2 · 3 · · · (n − 1)] = (2 · 3 · 4 · · · n)2 (n + 1) . = 2n k=2
n+1 = lim n→∞ 2n 1 = 2∞ Y 1 = 1− 2 . n n=2
lim Pn
n→∞
Problem 3. Show that
∞ Y n=2
1−
1 n
diverges to 0.
Solution 3. (Solution by Leon Hall) n Y k−1 1 · 2 · 3 · · · (k − 1) 1 = = Pn = k 2·3·4···n n k=2
Since 1 =0 n the product diverges to 0. [The product does not converge to 0 because none of the terms in the product are 0.] lim Pn = lim
n→∞
n→∞
Problem 4. Investigate the product
∞ Y
n
(1 + z 2 ) in |z| < 1.
n=0
Solution 4. (Solution by Leon Hall) Let Pn =
n Y
k
(1 + z 2 ). Then
k=0
P0 = 1 + z, P1 = (1 + z)(1 + z 2 ) P2 = (1 + z)(1 + z 2 )(1 + z 4 ) = (1 + z)(1 + z 2 + z 4 + z 6 ).
SOLUTIONS TO RAINVILLE’S “SPECIAL FUNCTIONS” (1960) 2(2n −1)
X
Assume Pn = (1 + z)
(z 2 )k =
2n+1 X−2
(z 2 )k . Then
k=0
Pn+1
3
k=0
n+1
= Pn (1 + z 2 ) n+1 = Pn + z 2h Pn
n+1
= (1 + z) 1 + z 2 + · · · + z 2 = (1 + z)
−2
n+1
+ z2
n+1
+ z2
+2
n+2
+ · · · + z2
−2
i
2n+2 X−2
(z 2 )k
k=0
So we have shown by induction that Pn = (1 + z)
2n+1 X−2
z 2k ,
k=0
1 n , for |z| < 1. |1 + z 2 | ≤ 2 1−z ∞ Y 1 2n . and same process works. Thus, (1 + z ) converges absolutely to 1 − z n=0
which is a geometric series converging to (z + 1) 1 + |z|2
n
∞ Y
1 exp Problem 5. Show that diverges. n n=1 n Y
Solution 5. (Solution by Leon Hall) Let Pn =
exp
k=1
Sn = log Pn =
1 and let n
n X 1 . k
k=1
Sn is the nth partial sum of the harmonic series, which diverges. As in the proof of Theorem 2, page 3, Pn = exp Sn and lim Pn = lim exp Sn = exp lim Sn .
n→∞
n→∞
n→∞
Thus, because {Sn } diverges, so does {Pn }. ∞ Y 1 Problem 6. Show that exp − diverges to 0. n n=1 Solution 6. (Solution by Leon Hall) Let log Pn = Sn =
n n X X 1 1 − =− k k
k=1
k=1
as in Problem 5. Then n X 1 Pn = exp Sn = exp − k k=1
! =
1 !. n X 1 exp k k=1
4
SOLUTIONS BY SYLVESTER J. PAGANO AND LEON HALL; EDITED BY TOM CUCHTA
Because
∞ X 1 diverges to ∞ we have k
k=1
lim Pn = 0
n→∞
and so ∞ Y
1 exp − n n=1
diverges to 0. Problem 7. Test
∞ Y n=1
1−
z2 n2
.
Solution 7. (Solution by Leon Hall) The product diverges to 0 for any z such z2 that z = ±m, m a positive integer. For all z such that 1 − 2 6= 0, we have by n ∞ ∞ Y X z2 z2 Theorem 3, page 3, that 1 − 2 is absolutely convergent because − 2 n n n=1 n=1 is absolutely convergent. In fact, ∞ X z2 z2 π2 . − 2 =− n 6 n=1 ∞ Y (−1)n+1 converges to unity. Problem 8. Show that 1+ n n=1 Solution 8. (Solution by Leon Hall) Let Pn =
n Y (−1)k+1 1+ . k
k=1
Case 1: n is even. Then 2−1 3+1 4−1 n n−1 1+1 ··· Pn = 1 2 3 4 n−1 n n! Rearranging, we get Pn = = 1 for even n. n! Case 2: n is odd. Then 1+1 2−1 3+1 4−1 n−1 n−2 n+1 n+1 Pn = ··· = . 1 2 3 4 n−2 n−1 n n In both cases lim Pn = 1. n→∞
∞ Y 1 Problem 9. Test for convergence: 1 − p for real p 6= 0. n n=2 Solution 9. (Solution by Leon Hall) For p > 1 the series of positive numbers ∞ ∞ Y X 1 1 is known to be convergent (e.g., by the Integral Test). Thus, 1− p np n n=2 n=2 is absolutely convergent by Theorem 3.
SOLUTIONS TO RAINVILLE’S “SPECIAL FUNCTIONS” (1960)
5
1 > 0 and so convergence and absolute convergence are the np ∞ X 1 diverges for 0 < p ≤ 1, our product diverges by same. Because the series np n=2 Theorem 3. For p < 0, let p = −q where q > 0. Then 1 1 − p = 1 − nq = 1 + (−nq ). n But lim (−nq ) 6= 0, so in this case our product diverges by Theorem 1. n→∞ ∞ Y 1 Summary: 1 − p diverges when p ≤ 1 (an p 6= 0), and converges when n n=2 p > 1. For 0 < p ≤ 1, 1 −
Problem 10. Show that
∞ Y sin( nz ) z n
n=1
is absolutely convergent for all finite z with
the usual convention at z = 0. Solution 10. (Solution by Leon Hall) Let sin( nz ) z n
Then an (z)
=
= 1 + an (z).
sin( nz ) z n
−1
1 z2 1 z4 1 z6 =− + − + ... 2 4 3!n 2 5! n 7! n6 1 1 z = 2 − +O . n 6 n2 Thus, there exists a constant M such that |an (z)| <
M , n2
∞ X M and because converges, the product n2 n=1 ∞ Y
(1 + an (z)) =
n=1
sin( nz ) z n
converges absolutely and uniformly for all finite z by Theorems 3 and 4. If z = 0 the product is, with the usual convention, ∞ Y
1 = 1.
n=1
Problem 11. Show that if c is not a negative integer, ∞ z Y z 1− exp c+n n n=1 is absolutely convergent for all finite z.
6
SOLUTIONS BY SYLVESTER J. PAGANO AND LEON HALL; EDITED BY TOM CUCHTA
Solution 11. Let z z 1 + an (z) = 1 − exp c+n n 1 z3 1 1 z4 1 z2 z2 1 z3 z + + ... − + + ... z+ + = 1+ + 2 3 c+n n 2! n2 3! n3 n 2! n 3! n 1 1 1 1 1 1 2 =1+ − − − z+ z + z3 + . . . n c+n 2!n2 n(c + n) 3!n3 2!n2 (c + n) ∞ X c c−n c − (k − 1)n k =1+ z+ 2 z2 + z n(c + n) 2n (c + n) k!nk (c + n) k=3 c 1 1 . =1+ z− z2 + O n(c + n) 2n(c + n) n2 Thus, for c not a negative integer and for any finite z, there is a constant M such M that |an (z)| ≤ 2 and so by Theorems 3 and 4 the product converges absolutely and n uniformly. 3. Chapter 2 Solutions ∞ 1 Γ0 (z) 1 X 1 Problem 12. Start with (†) = −γ − − − , prove that Γ(z) z n=1 z + n n 2Γ0 (2z) Γ0 (z) Γ0 (z + 21 ) − − = 2 log 2, Γ(2z) Γ(z) Γ(z + 12 ) and thus derive Legendre’s duplication formula, page 24. Solution 12. Applying (†) three times and simplifying yields 2Γ0 (2z) Γ0 (z) Γ0 (z + 12 ) − − Γ(2z) Γ(z) Γ(z + 12 ) X X ∞ ∞ ∞ X 2 1 1 2 2 1 1 1 1 = −2γ − +γ+ +γ+ − − − + + − 2z z z+n n z + 21 n=1 2z + n n z + 21 + n n n=1 n=1 2n n n X X X 2 2 1 1 2 1 2 − lim − + lim − + lim − = n→∞ n→∞ 2z + 1 n→∞ 2z + k k z+k k 2z + 1 + 2k k k=1 k=1 k=1 " # 2n n n X −2 X X 2 2 2 = + lim + 2H2n + − Hn + − Hn 2z + 1 n→∞ 2z + k 2z + 2k 2z + 2k + 1 k=1 k=1 k=1 " 2n # 2n+1 X −2 X 2 2 = + lim + + 2H2n − 2Hn 2z + 1 n→∞ 2z + k 2z + k k=1 k=2 2 −2 2 = + + lim + lim (2H2n − 2Hn ) 2z + 1 2z + 1 n→∞ 2z + 2n + 1 n→∞ = 0 + 0 + 2 lim [(H2n − log 2n) − (Hn − log n) + log 2n − log n] n→∞
= 2[γ − γ + log 2] = 2 log 2. √ Problem 13. Show that Γ0 ( 21 ) = −(γ + 2 log 2) π. Solution 13. By Problem 12, we know that 2Γ0 (2z) Γ0 (z) Γ0 (z + 21 ) − − = 2 log 2. Γ(2z) Γ(z) Γ(z + 12 )
SOLUTIONS TO RAINVILLE’S “SPECIAL FUNCTIONS” (1960)
Now let z =
7
1 to get 2 Γ0 (1) Γ0 − 2 Γ(1) Γ
1 2 1 2
−
Γ0 (1) = 2 log 2, Γ(1)
and so, algebra yields Γ0 ( 12 ) Γ0 (1) 1 = Γ(1) − 2 log 2. Γ( 2 ) √ But Γ(1) = 1, Γ0 (1) = −γ, Γ( 21 ) = π, hence Γ0 ( 12 ) γ √ = − − 2 log 2, 1 π and by rearrangement, √ 1 Γ0 ( ) = −(γ + 2 log 2) π. 2 Z ∞ Problem 14. Use Euler’s integral form Γ(z) = e−t tz−1 dt to show that 0
Γ(z + 1) = zΓ(z). Z Solution 14. From Γ(z) =
∞
e−t tz−1 dt, for R(z) > 0, integration by parts
0
yields dv = e−t dt
u = tz
Z Γ(z + 1)
∞
=
e−t tz dt
0 z−1
du = zt
v = −e
−t
=
∞ [−tz e−t ]0
Z
∞
+z
e−t tz−1 dt
0
= 0 + zΓ(z), where lim −tz e−t converges for R(z) > 0 t→∞
.
Problem 15. Show that Γ(z) = lim nz B(z, n). n→∞
Solution 15. From page 28 (1), we know (n − 1)!nz , n→∞ (z)n
Γ(z) = lim but B(z, n) =
Γ(z)Γ(n) Γ(z)(n − 1)! (n − 1)! = = . Γ(z + n) (z)n Γ(z) (z)n
Hence Γ(z) = lim nz B(z, n). n→∞
Problem 16. Derive the following properties of the beta function: (a) (b) (c) (d)
pB(p, q + 1) = qB(p + 1, q); B(p, q) = B(p + 1, q) + B(p, q + 1); (p + q)B(p, q + 1) = qB(p, q); B(p, q)B(p + q, r) = B(q, r)B(q + r, p).
8
SOLUTIONS BY SYLVESTER J. PAGANO AND LEON HALL; EDITED BY TOM CUCHTA
Solution 16. (a) We know B(p, q) = pB(p, q + 1) =
Γ(p)Γ(q) , so Γ(p + q)
Γ(p + 1)qΓ(q) pΓ(p)Γ(q + 1) = = qB(p + 1, q). Γ(p + q + 1) Γ(p + 1 + q)
(note: p → q and q → p – is this the symmetric property?) (b) B(p, q)
Γ(p)Γ(q) Γ(p + q) Γ(p)Γ(q) = Γ(p + q + 1) p+q (p + q)Γ(p)Γ(q) = Γ(p + q + 1) qΓ(p)Γ(q) pΓ(p)Γ(q) + = Γ(p + q + 1) Γ(p + q + 1) Γ(p + 1)Γ(q) = Γ(p)Γ(q + 1) Γ(p + q + 1) + Γ(p + q + 1) = B(p + 1, q) + B(p, q + 1). =
(c) (p + q)B(p, q + 1)
(p + q)Γ(p))Γ(q + 1) Γ(p + q + 1 (p + q)Γ(p)Γ(q + 1) = (p + q)Γ(p + q) Γ(p)Γ(q + 1) = Γ(p + q) Γ(p)qΓ(q) = Γ(p + q) = qB(p, q).
=
(d) B(p, q)B(p + q, n)
Γ(p)Γ(q) Γ(p + q)Γ(n) Γ(p + q) Γ(p + q + n) Γ(p)Γ(q)Γ(n) = Γ(p + q + n) Γ(q)Γ(n) Γ(p)Γ(q + n) = Γ(q + n) Γ(p + q + n) = B(q, n)B(q + n, p). =
SOLUTIONS TO RAINVILLE’S “SPECIAL FUNCTIONS” (1960)
Problem 17. Show that for positive integral n, B(p, n + 1) =
9
n! . (p)n+1
Solution 17. For integer n and using Theorem 9 (pg. 23), B(p, n + 1)
= = = = =
Γ(p)Γ(n + 1) Γ(p + n + 1) Γ(p)Γ(n + 1) (p + 1)n Γ(p + 1) Γ(p)n! (p + 1)n pΓ(p) n! p(p + 1)n n! . (p)n+1
1
Z
(1 + x)p−1 (1 − x)q−1 dx.
Problem 18. Evaluate −1
Z
1
Solution 18. Let A =
(1 + x)p−1 (1 − x)q−1 dx. Now let y =
−1
1+x , x = 2
2y − 1, 1 − x = 2 − 2y = 2(1 − y). Hence Z 1 A = 2p−1 y p−1 2q−1 (1 − y)q−1 2dy 0 Z 1 = 2p+q−1 y p−1 (1 − y)q−1 dy 0
= 2p+q−1 B(p, q).
Problem 19. Show that for 0 ≤ k ≤ n, (−1)k (α)n . (1 − α − n)k Note particularly the special case α = 1. (α)n−k =
Solution 19. Consider (α)n−k for 0 ≤ k ≤ n. Then (α)n−k
= α(α + 1) . . . (α + n − k − 1) α(α + 1) . . . (α + n − k − 1)[(α + n − k)(α + n − k + 1) . . . (α + n − 1)] = (α + n − 1)(α + n − 2) . . . (α + n − k) (α)n = (α + n − k)k (α)n = (−1)k (1 − α − n)k (−1)k (α)n = . (1 − α − n)k
Note for α = 1, that (n − k)! =
(−1)k n! . (−n)k
10 SOLUTIONS BY SYLVESTER J. PAGANO AND LEON HALL; EDITED BY TOM CUCHTA
Problem 20. Show that if α is not an integer, Γ(1 − α − n) (−1)n = . Γ(1 − α) (α)n Solution 20. Consider for α not equal to an integer Γ(1 − α − n) Γ(1 − α)
as desired.
Γ(1 − α − n) −αΓ(−α) Γ(1 − α − n) = (−α)(−α − 1)Γ(−α − 1) Γ(1 − α − n) = (−α)(−α − 1) . . . (−α − n + 1)Γ(1 − α − n) 1 = , (−1)n (α)n
=
1 In the following problems, the function P (x) := x − bxc − . 2 Z x Problem 21. Evaluate P (y)dy. 0
Z
x
P (y)dy when P (y) = y − byc −
Solution 21. To evaluate 0
1 . Let m be an 2
integer so that m ≥ 0. If m ≤ x < m + 1, then bxc = m and Z x Z x P (y)dy = P (y)dy 0 Zmx 1 = y−m− dy 2 # m " 2 x 1 1 y−m− = 2 2 " 2 m 2 # 1 1 1 x−m− − = 2 2 2 1 1 = P 2 (x) − 2 4 1 2 1 = P (x) − . 2 8 Problem 22. Use integration by parts and the result of the above exercise to show that Z ∞ P (x)dx 1 ≤ 8(1 + n) . 1 + x n Z ∞ P (x) dx and use integration by parts Solution 22. Consider 1+x n 1 ∞ 1 2 2 Z ∞ Z 1 ∞ P (x) − 4 P (x) 1 P (x) − 4 −1 dv = P (x)dx u = (1 + x) dx = dx. + 1+x 2 1+x 2 n (1 + x)2 n v=
1 2 1 P (x) − 2 8
n
du = −(1 + x)−2 dx
SOLUTIONS TO RAINVILLE’S “SPECIAL FUNCTIONS” (1960)
Now max P 2 (x) − Z
11
1 1 1 = and P 2 (n) = implies 4 4 4 ∞
n
P (x) 1 dx = 0 − 0 + 1+x 2
Z
∞
P 2 (x) −
1 4 dx
(1 + x)2
n
and Z
1 ∞ Z P (x) 1 ∞ 1 1 4 dx = − dx ≤ 1+x 2 n (1 + x)2 8 1+x n
∞
n
or ∞
Z
P (x) 1 1 dx ≤ . 1+x 8 1+n
n
Z
Problem 23. With the aid of the above problem, prove the convergence of 0
Z Solution 23. 0
cise 22,
∞
P (x) dx converges ←→ lim n→∞ 1+x Z lim n→∞
∞
n
Z Hence 0
∞
Z
∞
n
∞
P (x)dx . 1+x
P (x) dx = 0 but from Exer1+x
1 P (x) dx ≤ lim = 0. N →∞ 8(1 + n) 1+x
P (x) dx < ∞. 1+x
Problem 24. Show that Z ∞ ∞ Z ∞ Z P (x)dx X n+1 P (x)dx X 1 (y − 21 )dy = = . 1+x 1+x 1+n+y 0 n=0 n n=0 0 Then prove that lim n
n→∞
Z
∞
and thus conclude that 0
2
Z
1
0
(y − 21 )dy 1 =− 1+n+y 12
P (x)dx is convergent. 1+x
Solution 24. (Solution by Leon Hall) Because P (x) is periodic with period 1, it is clear that Z ∞ ∞ Z n+1 X P (x) P (x) dx = dx. 1+x 1+x 0 n=0 n Let x = n + y. Then Z
n+1
n
P (x) dx 1+x
Z
1
= Z0 1 = Z0 1 = 0
P (n + y) dy 1+n+y P (y) dy 1+n+y y − 21 dy. 1+n+y
12 SOLUTIONS BY SYLVESTER J. PAGANO AND LEON HALL; EDITED BY TOM CUCHTA
This establishes the first set of equalities. Z 1 Z 1 y − 12 3 1 1− n+ dy = dy 2 y+n+1 0 y+n+1 0 1 3 = y− n+ log(y + n + 1) 2 0 3 n+2 =1− n+ log 2 n+1 3 1 =1− n+ log 1 + 2 n+1 1 1 3 1 1 + − + ... =1− n+ − 2 n + 1 2(n + 1)2 3(n + 1)3 4(n + 1)4 ∞ Z 1 X (y − 21 ) To determine the convergence of dy we compare with the known y+n+1 n=0 0 ∞ X 1 convergent series using the limit comparison test. 2 n n=1 Z 1 (y − 12 ) ∞ dy 1 3 X 0 y+n+1 2 2 = n − n n + 1 2 k(n + 1)k n2 k=1 3 1 1 1 1 = n2 − n2 n + − + + O 2 n + 1 2(n + 1)2 3(n + 1)3 n + 1 3 2 2 3 2 6n (n + 1) − n (n + 2 )[6(n + 1) − 3(n + 1) + 2] 1 = +O 6(n + 1)3 n +1 15 2 3 6n5 + 18n4 + 18n3 + 6n2 − [6n5 + 18n4 + 37 1 2 n + 2 n ] +O = 3 n+1 6(n + 1) 1 3 −2n 1 = +O . 6(n + 1)3 n+1 Thus, lim n
n→∞
and
∞ Z X n=0
0
1
2
Z 0
1
(y − 21 ) 1 dy = − , y+n+1 12
(y − 21 ) dy = y+n+1
Z 0
∞
P (x) dx 1+x
converges. Problem 25. Apply Theorem 11, page 27, to the function f (x) = (1 + x)−1 ; let n → ∞ and thus conclude that Z ∞ 1 γ= − y −2 P (y)dy. 2 1 Solution 25. (Solution by Leon Hall) Let f (x) =
1 . Theorem 11, page 27 1+x
gives with p = 0, q = n, Z n Z n n X 1 1 1 1 1 = dx + + + f 0 (x)B1 (x)dx. 1+k 2 2 1+n 0 1+x 0 k=0
SOLUTIONS TO RAINVILLE’S “SPECIAL FUNCTIONS” (1960)
13
So, n X k=0
1 − log(1 + n) 1+k
=
1 1 + 2 2
1 1+n
Z
n
−
+ 0
B1 (x) dx (1 + x)2
Z n+1 1 1 1 B1 (y + 1) y := x + 1 = + dy + − 2 2 1 + n Z1 y2 n+1 B1 (y) 1 1 1 dy = + − 2 2 1+n y2 1 π Problem 26. Use the relation Γ(z)Γ(1 − z) = and the elementary result sin πz 1 sin x sin y = [cos(x − y) − cos(x + y)] 2 to prove that 1−
Γ(2 − c)Γ(c − 1)Γ(c − a − b)Γ(a + b + 1 − c) Γ(c)Γ(1 − c)Γ(c − a − b)Γ(a + b + 1 − c) = . Γ(c − a)Γ(a + 1 − c)Γ(c − b)Γ(b + 1 − c) Γ(a)Γ(1 − a)Γ(b)Γ(1 − b)
Solution 26. Note that 1 − (c − a − b) = a + b + 1 − c, 1 − (c − a) = a + 1 − c, and 1 − (c − b) = b = 1 − c, so we can use the gamma function relation four times to get 1−
Γ(c)Γ(1 − c)Γ(c − a − b)Γ(a + b + 1 − c) Γ(c − a)Γ(a + 1 − c)Γ(c − b)Γ(b + 1 − c)
π 2 sin π(c − a) sin π(c − b) π 2 sin(πc) sin π(c − a − b) sin(πc) sin π(c − a − b) − sin π(c − a) sin π(c − b) = . sin(πc) sin π(c − a − b)
=1−
Now using the given trig identity, we get, continuing the equality: + b) − cos π(2c − a − b)] − 12 [cos π(b − a) − cos π(2c − a − b)] 1 2 [cos π(a + b) − cos π(2c − a − b)] 1 − 2 [cos π(b − a) − cos π(a + b)] = 1 2 [cos π(a + b) − cos π(2c − a − b)] − sin(πa) sin(πb) = sin(πc) sin π(c − a − b) 0 sin(πa) sin(πb) = . − sin π(c − 1) sin π(c − a − b) =
1 2 [cos π(a
Canceling minus signs and multiplying and dividing by π 2 yields =
Γ(c − 1)Γ(2 − c)Γ(c − a − b)Γ(a + b + 1 − c) Γ(a)Γ(1 − a)Γ(b)Γ(1 − b)
as desired. Problem 27. With the assumptions of Watson’s lemma, show, with the aid of ∞ X k the convergence of the series F (t) = an expt − 1 in |t| ≤ a + δ, that for r k=1
14 SOLUTIONS BY SYLVESTER J. PAGANO AND LEON HALL; EDITED BY TOM CUCHTA
0 ≤ t ≤ a, there exists a positive constant c1 such that n X k n+1 ak expt − 1 < c1 expt −1 . F (t) − r r k=1
Solution 27. We wish to show that there exists c1 such that for 0 ≤ t ≤ a (see problem (2) for t > a), n X n+1 k −1 ak t r < c1 t r −1 F (t) − k=1
under the condition of Watson’s lemma. By the convergence of
∞ X
n
an t r −1 = F (t)
n=1
in |t| ≤ a we write n X k −1 ak t r F (t) − k=1
∞ X k −1 ak t r = k=n+1 ∞ X k−n−1 n+1 −1 =t r ak t r k=n+1 ∞ X k−n−1 n+1 ≤ t r −1 ak a r k=n+1
< c1 t
n+1 r −1
, ∞ X k−n−1 ak a r . Remember from Watson’s lemma where c1 > k=n+1
F (t) =
∞ X
n
an t r −1
k=1
for |t| ≤ a + δ where δ > 0.
Problem 28. With the assumptions of Watson’s lemma, page 41, show that for t > a, there exist positive constants c2 and β such that n X k n+1 ak expt − 1 < c2 expt − 1 eβt . F (t) − r r k=1
Solution 28. Under the assumption of Watson’s lemma we wish to show that for t > a, there exist constants c2 , β such that n X n+1 k −1 ak t r < c2 eβt t r −1 . F (t) − k=1
Now for t > 0, we have given |F (t)| < kebt . Hence n n X X k−n−1 k n+1 −1 bt −1 ak t r < ke + t r ak t r F (t) − k=1 k=1 n X k = kebt + ak t r −1 , k=1
SOLUTIONS TO RAINVILLE’S “SPECIAL FUNCTIONS” (1960)
15
but k − n − 1 < 0 and t > a, so n n X X n+1 k−n−1 k bt −1 −1 ak t r < ke + t r ak a r F (t) − k=1 k=1 ∞ X k but ak a r converges. Hence, since t > a, there exist constants M1 , M2 such k=1 that n a n+1 X k n+1 n+1 r −1 ak t r −1 < M1 ebt t r −1 + M2 t r −1 F (t) − t k=1
n+1
< M1 ebt t r −1 + M2 t n+1 < c1 eβt t r −1 .
n+1 r −1
ebt
Problem 29. Derive the asymptotic expansion (6) immediately preceding these exercises by applying Watson’s lemma to the function Z ∞ −xt te f 0 (x) = − dt t + t2 0 and then integrating the resultant expansion term by term. Solution 29. Consider Watson’s lemma with F (t) = F (t) = Choose r = Hence
−t . Hence 1 + t2
∞ ∞ X X −t n+1 2n+1 (−1)n t2n−1 ; 0 ≤ t < 1. (−1) t = = 1 + t2 n=1 n=0
1 1 1 t 1 , a = , δ = . Also for t ≥ 0, et ≥ 1 and < . 2 2 2 3 1+t 2 |F (t)| < 1 · et ; t ≥
1 , 2
satisfying condition (2) of Watson’s Lemma, and F (t) =
∞ X
n
(−1)n t 1/2 −1
n=1
for |t| ≤
5 1 1 + = . 2 3 6
Z
Hence by Watson’s lemma for f 0 (x) =
∞
0
f 0 (x) ∼
n ∞ (−1) Γ X n=1
∞ 0
f (s)ds ∼ x
n 1/2
=
x2n
Then Z
∞ Z X n=1
−te−xt dt we have 1 + t2
∞
x
∞ X (−1)n (2n − 1)! . x2n n=1
(−1)n (2n − 1)! ds. s2n
After integrating, this becomes Z ∞ ∞ X (−1)n (2n − 2)! . (∗) f 0 (s)ds ∼ x2n−1 x n=1
16 SOLUTIONS BY SYLVESTER J. PAGANO AND LEON HALL; EDITED BY TOM CUCHTA
Now ∞
Z
d dx
0
e−xt dt = 1 + t2
∞
Z 0
−te−xt dt ≡ f 0 (x). 1 + t2
Let “A” label the integral in the middle of the last formula. Hence Z ∞ −xt e f (x) = dt, 1 + t2 0 which we label as “B”. Also note that integrals “A” and “B” are uniformly convergent. Hence Z ∞ e−xt dt = 0. lim f (x) = lim x→∞ x→∞ 1 + t2 0 Therefore ∞
Z
∞ f 0 (s)ds = f (s) = 0 − f (x). x
x
By (∗), 0 − f (x) ∼
∞ X (−1)n (2n − 2)! ; |x| → ∞, R(x) > 0. x2n−1 n=1
So ∞
e−xt dt 1 + t2 0 ∞ X (−1)n (2n − 2)! ∼− x2n−1 n=1 ∞ X (−1)n (2n)! ; |x| → ∞, R(x) > 0. ∼ x2n+1 n=0 Z
f (x)
=
Problem 30. Establish (6), page 43, directly, first showing that f (x) −
n X
(−1)k (2k)!x−2k−1 = (−1)n+1
Z
∞
0
k=0
e−xt t2n+2 dt, 1 + t2
and thus obtain not only (6) but also a bound on the error made in computing with the series involved. ∞
Solution 30. (Solution by Leon Hall) Because
X 1 = (−1)k t2k we have 1 + t2 k=0
1 1 + t2
= = =
n X k=0 n X k=0 n X k=0
(−1)k t2k +
∞ X
(−1)k t2k
k=n+1 k 2k
(−1) t
+ (−1)n+1 t2n+2
∞ X k=0
t2n+2 (−1)k t2k + (−1)n+1 . 1 + t2
(−1)k t2k
SOLUTIONS TO RAINVILLE’S “SPECIAL FUNCTIONS” (1960)
17
So f (x)
∞
e−xt dt 1 + t2 Z Z0 ∞ n X = e−xt (−1)k t2k dt + (−1)n+1 Z
=
0
=
n X
(−1)k
k=0 Z ∞
e−xt t2n+2 dt 1 + t2
∞
e−xt t2n+2 dt. 1 + t2
0
e−xt t2k dt + (−1)n+1
Z 0
0
k=0
∞
Integration by parts give the reduction formula (for x as specified) Z
∞
e
2k(2k − 1) t dt = x2
−xt 2k
0
Z
∞
e−xt dt =
This, plus the fact that 0
f (x) =
n X
(−1)k
k=0
Z
∞
e−xt t2k−2 dt.
0
1 , yields x
(2k)! + (−1)n+1 x2k+1
Z
∞
0
e−xt t2n+2 dt. 1 + t2
Then n X (2k)! (−1)k 2k+1 f (x) − x k=0
Z ∞ −xt 2n+2 e t = dt 1 + t2 0 Z ∞ < |e−xt |t2n+2 dt Z0 ∞ < e−Re(x)t t2n+2 dt 0
(2n + 2)! . [Re(x)]2n+3
=
π N − ∆, ∆ > 0, if Re(x) > N , then |x| > and as 2 sin ∆
In the region |arg x| ≤ |x| → ∞
(2n + 2)! = O |x|−2n−2 , 2n+3 [Re(x)] and so f (x) ∼
∞ X
(−1)k
n=0
as |x| → ∞ in the sector |arg x| <
(2k)! x2k+1
π − ∆, ∆ > 0. 2
Problem 31. Use integration by parts to establish that for real x → ∞, Z
∞
x
e−t t−1 dt ∼ e−x
∞ X
(−1)n n!x−n−1 .
n=0
18 SOLUTIONS BY SYLVESTER J. PAGANO AND LEON HALL; EDITED BY TOM CUCHTA ∞
e−t dt as x → ∞. Using integration by parts, t x ∞ Z ∞ 1 −t t−2 e−t dt f (x) = − e − t x ∞ x Z ∞ 1 −t 1 1 = e−x + 2 e−t + 2 e dt 3 x t x t 0 Z ∞ ∞ 1 −t 1 −t 1 −x 1 =e − 2 3e −2·3 e dt − 4 x x2 t t x x Z
Solution 31. Consider f (x) = u=
1 t
u=− u=
dv = e−t dt 1 t2
v = −e−t
1 t2
du = −2
dv = e−t dt 1 t3
v = −e−t = ....
This pattern can clearly continue forever, so we can write Z ∞ n X (−1)k k! n+1 f (x) = e−x + (−1) (n + 1)! t−(n+2) e−t dt. xk+1 x k=0
Now since 0 < x < t, Z ∞ −t n X (−1)k k! e x x dt e f (x) − = (n + 1)!e n+2 xk+1 t k=0 Zx ∞ (n − 1)! x e e−t dt < xn+2 x (n + 1)! < . xn+2 Hence n X (−1)k k! 1 1 x e f (x) − =O =O , xk+1 xn+2 xn+1 k=0
so ex f (x) ∼
∞ X (−1)n n! k=0
xn+1
or Z
∞
x
t−1 e−t dt ∼ e−x
∞ X (−1)n n! as x → ∞. xn+1 n=0
Problem 32. Let the Hermite polynomials Hn (x) be defined by exp(2xt − t2 ) =
∞ X Hn (x)tn n! n=0
for all x and t, as in Chapter 11. Also let the complementary error function erfc x be defined by Z ∞ 2 erfc x = 1 − erf x = √ exp(−β 2 )dβ. π x Apply Watson’s lemma to the function F (t) = exp(2xt − t2 ); obtain 2 Z ∞ ∞ X 1 exp x − s exp(−β 2 )dβ ∼ Hn (x)s−n−1 , s → ∞, 1 2 s−x 2 n=0
SOLUTIONS TO RAINVILLE’S “SPECIAL FUNCTIONS” (1960)
19
and thus arrive at the result " 2 # X ∞ 1 −1 √ 1 −1 1 −1 π exp t t −x t −x ∼ erfc Hn (x)tn , t → 0+ . 2 2 2 n=0 Solution 32. Because expt (n − 1) = tn−1 , and if F (t) = exp(2xt − t2 ) =
∞ X Hn (x) n t , n! n=0
we can write F (t) =
∞ X Hn−1 (x) expt (n − 1) (n − 1)! n=1
so the first condition in Watson’s Lemma is satisfied for any fixed x with r = 1, a = 1, and any δ > 0. Also, 2
F (t) = exp(2xt − t2 ) = e−t e2xt < 2e2xt for t ≥ 0, so the second condition in Watson’s Lemma is satisfied with k = 2 (or anything > 1) and b = 2x. Thus, we get Z ∞ ∞ X Hn−1 (x)Γ(n) e−st F (t)dt ∼ (n − 1)!sn 0 n=1 as s → ∞, or Z
∞
0
∞ X 1 Hn (x)s−n−1 , exp −t2 + 2 x − s t dt ∼ 2 n=0
as s → ∞. Completing the square in the exponent leads to " Z ∞ 2 # ∞ X 1 1 exp x − s dt ∼ Hn (x)s−n−1 exp − t + s − x 2 2 0 n=0 1 as s → ∞, and if we let β = t + s − x we get 2 " 2 # Z ∞ ∞ X 2 1 exp x − s e−β dβ ∼ Hn (x)s−n−1 , 1 2 s−x 2 n=0 as s → ∞. 1 (not the Using the definition of erf c, and then making the substitution t = s inverse Laplace transform) we get " 2 # X √ ∞ π 1 1 exp x − s erfc s−x ∼ Hn (x)s−n−2 , 2 2 2 n=0 as s → ∞ or √
π exp 2t
as t → 0
+
as desired.
"
1 x− 2t
2 #
erfc
1 −x 2t
∼
∞ X n=0
Hn (x)tn ,
20 SOLUTIONS BY SYLVESTER J. PAGANO AND LEON HALL; EDITED BY TOM CUCHTA
Problem 33. Use integration by parts to show that if R(α) > 0, and if x is real, Z ∞ ∞ X (−1)n (α)n e−t t−α dt ∼ x1−α e−x , x → ∞, xn+1 x n=0 of which Problem 31 is the special case α = 1. Solution 33. Integration by parts with u = t−α and dv = e−t dt yields Z ∞ Z ∞ e−t t−α dt = e−x x−α − α e−t t−(α+1) dt. x
x
The same integration by parts with α replaced by α + 1 applied to the last integral gives Z ∞ Z ∞ h αi e−t t−α dt = e−x x−α 1 − + α(α + 1) e−t t−(α+2) dt. x x x Continuing, after n + 1 integrations by parts, we have Z ∞ Z ∞ n X (−1)k (α)k n+1 + (−1) (α) e−t t−(α+n+1) dt e−t t−α dt = e−x x−α+1 n+1 xk+1 x x k=0
or x α−1
Z
∞
e x
−t −α
e t x
dt −
n X (−1)k (α)k k=0
xk+1
x α−1
=e x
n
Z
∞
(−1) (α)n+1
e−t t−(α+n+1) dt.
x
Thus, we have the desired asymptotic series if the right side of the last equation is 1 O + 1 as x → ∞. This is true because xn Z ∞ x α−1 −t −(α+n+1) n e x (−1) (α)n+1 e t dt x x α−1 Z e x (α)n+1 ∞ −t e dt ≤ α+n+1 x x x e (α)n+1 −x e = xn+2 |(α)n+1 | = xn+2 1 =O n+1 x as x → ∞. 4. Chapter 4 Results Cited By definition, F (a, b; c; 1) ≡
∞ X (a)n (b)n . (c)n n! n=0
Theorem 4.1. If R(c − a − b) > 0 and if c is neither zero nor a negative integer, F (a, b; c; 1) =
Γ(c)Γ(c − a − b) . Γ(c − a)Γ(c − b)
SOLUTIONS TO RAINVILLE’S “SPECIAL FUNCTIONS” (1960)
21
pg.66, (2): a a+1 , ; 2 2 −a (1 + z) F 1 + a − b;
4z (1 + z)2
a, a − b + =F b+
1 ; 2
x2
1 ; 2
z < 1, Theorem 4.2. If |z| < 1 and 1 − z
a, b;
a, c − b;
z = (1 − z)−a F
F c;
−z . 1−z
c;
Theorem 4.3. If |z| < 1, F (a, b; c; z) + (1 − z)c−a−b F (c − a, c − b; c; z). 1 Theorem 4.4. If 2b is neither zero nor a negative integer and if |y| < and 2 y 1 − y < 1, a a+1 , ; 2 2 (1 − y)−a F 1 b+ ; 2 Theorem 4.5. Kf a + b + and |4x(1 − x)| < 1,
2
y (1 − y)2
a, b; =F 2b;
2y .
1 is neither zero nor a negative integer, and if |x| < 1 2
F
a, b; 1 a+b+ 2
4x(1 − x) .
Theorem 4.6. If c is neither zero nor a negative integer, and if both |x| < 1 and |4x(1 − x)| < 1, c−a c+a−1 a, 1 − a; , ; 2 x = (1 − z)c−1 F 2 F 4x(1 − x) . c; c; 5. Chapter 4 Solutions Problem 34. Show that a, b; d F dx c;
a + 1, b + 1; ab x = F c c + 1;
x .
22 SOLUTIONS BY SYLVESTER J. PAGANO AND LEON HALL; EDITED BY TOM CUCHTA
Solution 34. From F (a, b; c; 1) ≡
∞ X (a)n (b)n , (c)n n! n=0
we get d F (a, b; c; x) dx
= =
∞ X (a)n (b)n xn−1 (c)n (c − 1)! n=1 ∞ X (a)n+1 (b)n+1 xn
(c)n+1 n! ∞ X (a + 1)n (b + 1)n xn ab = c n=0 (c + 1)n n! ab = F (a + 1, b + 1; c + 1; x). c n=0
Problem 35. Show that 2a, 2b;
F
1 a+b+ ; 2
1 1 Γ Γ a+b+ 1 2 2 . 2 = 1 1 1 1 1 Γ c+ a Γ c− a+ 2 2 2 2 2
Solution 35. Wish to evaluate F
F
2a, 2b; 1 a+b+ 2
2a, 2b;
1 2 . From Theorem 4.5,
1 a+b+ ; 2
x =F
a, b; 1 a+b+ 2
4x(1 − x)
1 1 for |x| < 1, |4x(1−x)| < 1. We need to use x = , but note that R(a+b+ −a−b) > 2 2 0. Hence, by Theorem 4.1, F
2a, 2b;
a+b+
1 2
1 2 =F
a, b; a+b+
1 2
1 1 Γ a+b+ Γ 2 2 1 = . 1 1 Γ b+ Γ a+ 2 2
Problem 36. Show that
a, 1 − a;
F c;
1 2 Γ(c)Γ 2 1 = 1 1 c−a+1 2 Γ c+ a Γ 2 2 2
1−c
SOLUTIONS TO RAINVILLE’S “SPECIAL FUNCTIONS” (1960)
23
c−1 c+a−1 , ; 2 x = (1−x)c−1 F 2
a, 1 − a;
Solution 36. Consider F c;
c; By Theorem 4.6, for |x| < 1, |4x(1 − x)| < 1, c−a c+a−1 a, 1 − a; , ; 2 x = (1 − x)c−1 F 2 F 4x(1 − x) . c; c; c a c a 1 Since R c − + − − + > 0, we may use Theorem 4.1 to conclude that 2 2 2 2 2 c−1 c+a−1 a, 1 − a; c−1 , ; 1 1 2 F F 2 = 1 2 2 c; c; 1 Γ(c)Γ 2 , = c + a c−a+1 2c−1 Γ Γ 2 2 as desired. Problem 37. Obtain the result −n, b; F c;
(c − b)n 1 = . (c)n
Solution 37. Consider F (−n, b; c; 1). At once, if R(c − b) > 0, F (−n, b; c; 1) =
Γ(c)Γ(c − b + n) (c − b)n = . Γ(c + n)Γ(c − b) (c)n
Actually the condition R(c − b) > 0 is not necessary because of the termination of the series involved. Problem 38. Obtain the result −n, a + n; F c;
(−1)n (1 + a − c)n 1 = . (c)n
Solution 38.
−n, a + n;
1 =
F c;
Γ(c)Γ(c − a) . Γ(c + n)Γ(c − a − n)
By Exercise 9, Chapter 2, if (c − a) is nonintegral, Γ(1 − α − n) (−1)n = . Γ(1 − α) (α)n Hence,
−n, a + n;
F a;
(−1)n (1 − c + a)n 1 = , (c)n
4x(1 − x) .
24 SOLUTIONS BY SYLVESTER J. PAGANO AND LEON HALL; EDITED BY TOM CUCHTA
as desired. Problem 39. Show that −n1 − b − n; F a; Solution 39. −n, 1 − b − n; F a;
1 =
1 =
(a + b − 1)2n . (a)n (a + b − 1)n
Γ(a)Γ(a − 1 + b + 2n) (a − 1 + b)2n = . Γ(a + n)Γ(a − 1 + b + n) (a)n (a − 1 + b)n
Of course a 6= nonpositive integer, as usual. Problem 40. Prove that if gn = F (−n, α; 1 + α − n; 1) and α is not an integer, then gn = 0 for n ≥ 1, g0 = 1. Solution 40. Let gn = F (−n, α; 1 + α − n; 1). Then gn =
n X k=0
n
X n!(α)k (−α)k (−n)k (α)k = . k!(1 + α − n)k n!(n − k)!(α)n k=0
Hence, compute the series ∞ X (−α)n gn tn n! n=0
=
n ∞ X X (α)k (−α)n−k tn
k!(n − k)! ! ∞ ! X (−α)n tn (α)n tn = n! n! n=0 n=0 = (1 − t)α (1 − t)−α = 1. n=0 k=0 ∞ X
Therefore, g0 = 1 and gn = 0 for n ≥ 1. (Note: easiest to choose α 6= integer, can actually do better than that probably.) Problem 41. Show that dn a−1+n x F (a, b; c; x) = (a)x xa−1 F (a + n, b; c; x). n dx Solution 41. Consider Dn [xa−1+n F (a, b; c; x)] (D ≡ Dn [xa−1+n F (a, b; c; x)]
= Dn = =
∞ X k=0 ∞ X k=0 ∞ X
d ). We have dx
∞ X (a)k (b)k xn+k+a−1 k=0
(c)k k!
(a)k (n + k + a − 1)(n + k + a − 2) . . . (k + a)xk+1−a (b)k (c)k k! (a)k (a)n+k xk+a−1 (b)k (c)k (a)k k!
(a + k)n (a)n xk+a−1 (b)k k!(c)k k=0 a−1 = (a)n x F (a + n, b; c; x). =
SOLUTIONS TO RAINVILLE’S “SPECIAL FUNCTIONS” (1960)
Problem 42. Use equation (2), page 66, with non-negative integer, to conclude that −n, a; −x = (1 − x)−a F F 1 + a + n;
25
z = −x, b = −n, in which n is a 1 1 1 a, a + ; 2 2 2
−4x . (1 − x)2
1 + a + n; Solution 42. From (2) on page (66) we get a a+1 , ; 2 2 −a −4z (1 + z) F (1 + z)2 1 + a − b; Use z = −x, b = −n to arrive at a a+1 , ; 2 2 −a (1 − x) F 1 + a + n;
a, b; =F 1 + a − b;
z
.
−4x (1 − x)2
a, −n; =F 1 + a + n;
−x ,
as desired. 1 Problem 43. In Theorem 23, page 65, put b = γ, a = γ + , 4x(1 + x)−2 = z and 2 thus prove that 1 γ, γ + ; 2 F
2γ−1 1 2 2 √ . z = (1 − z) 1 + 1 − z
2γ
Solution 43. Theorem 4.4 gives us a, b; 4x (1 + x)−2a F =F (1 + x)2 ab; Put b = γ, a = γ +
1 and 2
1 a, a − b + ; 2 1 b+ ; 2
4x = z. (1 + x)2
Then zx2 + 2(z − 2)x + z = 0 p √ zx = 2 − 3 ± z 2 − 4z + 4 − 32 = 2 − z ± 2 1 − z. Now x = 0 when z = 0, so √ √ zx = 2 − z − 2 1 − z = 1 − z + 1 − 2 1 − z or x=
(1 −
√
√ 1 − z)2 (1 − 1 − z[1 − (1 − z)] √ = . z z(1 + 1 − z)
x2 .
26 SOLUTIONS BY SYLVESTER J. PAGANO AND LEON HALL; EDITED BY TOM CUCHTA
Thus
√ 1− 1−z √ x= 1+ 1−z
and 1+x=
2 √ . 1+ 1−z
Then we obtain √ √ 4x 4(1 − 1 − z) (1 + 1 − z)2 √ = · = z, (1 + x)2 4 1+ 1−z 1 a check. Now with b = γ, a = γ + , Theorem 4.4 yields 2 1 1 γ + , 1; −2γ−1 γ + , γ; 2 2 2 √ x2 F =F z 1+ 1−z 1 2γ; γ+ ; 2 1; x2 =1F0 −; = (1 − x2 )−1 . √ 2 2 1−z √ √ and 1 + x = , Since 1 − x = 1+ 1−z 1+ 1−z √ 4 1−z 2 √ (1 − x ) = . (1 + 1 − z)2 Thus we have 1 γ, γ + ; 2 F
2γ;
as defined. Now we use Theorem 4.3 to see that 1 1 γ, γ − ; γ, γ + ; 1 2 2 −2 F 2 = (1 − z) F 2γ;
z
2γ;
so that we also get 1 γ, γ − ; 2 F
2γ; as desired.
2γ+1 −2 2 2 1 √ √ = (1 − z)− 2 1+ 1−z 1+ 1−z 2γ−1 2 − 21 √ = (1 − z) , 1+ 1−z
z
z =
2 √ 1+ 1−z
2γ−1 ,
SOLUTIONS TO RAINVILLE’S “SPECIAL FUNCTIONS” (1960)
27
Problem 44. Use Theorem 4.6 to show that 1 1 1 1 1 c − a, c − a + ; 2 2 2 2 2 x = (1−2x)a−c F c;
a, 1 − a;
(1−x)1−c F c;
Solution 44. By Theorem 4.6, a, 1 − a; x (1 − x)1−c F a;
c−a c+a−1 , ; 2 2 =F
c; c−a c+a−1 , ; 2 =F 2
4x(x − 1 . (1 − 2x)2
4x(1 − x)
1 − (1 − 2x)2
c; c−a c−a+1 ; a − c 2 ,c − 2 2 = (1 − 2x) 2 F
−1 + (1 − 2x)2 (1 − 2x)2
c; c−a c−a+1 , ; 2 2 = (1 − 2x)a−c F c;
4x(x − 1) , (1 − 2x)2
which we wished to obtain. Problem 45. In the differential equation (3), page 54, for w = F (a, b; c; z) introduce a new dependent variable u by w = (1 − z)−a u, thus obtaining z(1 − z)2 u00 + (1 − z)[c + (a − b − 1)z]u0 + a(c − b)u = 0. −z Next change the independent variable to x by putting x = . Show that the 1−z equation for u in terms of x is d2 u du + [c − (a + c − b + 1)x] − a(c − b)u = 0, dx2 dx and thus derive the solution a, c − b; −z w = (1 − z)−a F . 1−z c; x(1 − x)
Solution 45. We know that w = F (a, b; c; z) is a solution of the equation (1)z(1 − z)w00 + [c − (a + b + 1)z]w0 − abw = 0. In (1) put w = (1 − z)−a u. Then w0 = (1 − z)−a u0 + a(1 − z)−a−1 u, w00 = (1 − z)−a u00 + 2a(1 − z)−a−1 u0 + a(a + 1)(1 − z)−a−2 u.
28 SOLUTIONS BY SYLVESTER J. PAGANO AND LEON HALL; EDITED BY TOM CUCHTA
Hence the new equation is z(1−z)u00 +2azu0 +a(a+1)z(1−z)−1 u+cu0 +ca(1−z)−1 u−(a+b+1)zu0 −a(a+b+1)z(1−z)−1 u−abu = 0, or z(1−z)u00 +[c−(b−a+1)z]u0 +(1−z)−1 [(a2 +a)z+ca−(a2 +ab+a)z−ab(1−z)]u = 0, or (2)z(1 − z)2 u00 + (1 − z)[c + (a − b − 1)z]u0 + a(c − b)u = 0. −z −x 1 . Then z = ,1 − z = , and we use equation (12) 1−z 1−x 1−x dx −1 on page 12 of IDE for the change of variable. First, = −(1 − x)2 : = dz (1 − z)2 −2 d2 x = = −2(1 − x)3 . The old equation (2) above may be written dz 2 (1 − z)3 d2 u a − b − 1 du a(c − b) c + + + u = 0, dz 2 z(1 − z) 1−z dt z(1 − z)2 Now put x =
which then leads to the new equation 2 du a(c − b)(1 − x)3 c(1 − x)2 4d u 3 2 (1−x) + (a − b − 1)(1 − x) − u = 0, + −2(1 − x) − (1 − x) dx2 −x dx x or x(1 − x)
du d2 u + [−2x − {−c(1 − x) + (a − b − 1)x}] − a(c − b)u = 0, dx2 dx
or d2 u du + [x − (a − b + c + 1)x] − a(c − b)u = 0. 2 dx dx Now (3) is a hypergeometric equation with parameters γ = c, α + beta + 1 = a − b + c + 1, αβ = a(c − b). Hence α = a, β = c − b, γ = c. One solution of (3) is (3)x(1 − x)
u = F (a, c − b; c; x), so one solution of equation (1) is
a, c − b;
W = (1 − z)−a F
−z . 1−z
c; Problem 46. Use the result of Exercise 12 and the method of Section 40 to prove Theorem 4.2. z < 1, Solution 46. We know that in the region in common to |z| < 1 and 1 − z there is a relation a, c − b; −z 1−c (1−z)−a F = AF (a, b; c; z)+Bz F (a+1−c, b+1−c; z−c; z). 1−z c;
SOLUTIONS TO RAINVILLE’S “SPECIAL FUNCTIONS” (1960)
29
Since c is neither zero nor a negative integer, the last term is not analytic at z = 0. Hence B = 0. Then use z = 0 to obtain 1 · 1 = A · 1, so A = 1. Hence a, c − b; −z F (a; b; c; z) = (1 − z)−a F . 1−z c; Problem 47. Prove Theorem 4.3 by the method suggested by Exercises 12 and 13. Solution 47. (Solution by Leon Hall) Note that the first two parameters in F (a, b; c; z) are interchangeable, so results involving one of them also apply to the other. By Exercise 12, −z F (a, b; c; z) = (1 − z)−a F a, c − b; c; . 1−z −z Let w = so this becomes 1−z F (a, b; c; z) = (1 − z)−a F (a, c − b; c; w). Again, by Exercise 12, applied to the second parameter, −w −a −(c−b) F (a, b; c; z) = (1 − z) (1 − w) F c − a, c − b; c; . 1−w −w But 1 − w = (1 − z)−1 , and = z, so 1−w F (a, b; c; z)
= (1 − z)−a ((1 − z)−1 )−(c−b) F (c − a, c − b; c, z) = (1 − z)c−a−b F (c − a, c − b; c; z)
as desired. Problem 48. Use the method of Section 39 to prove that if both |z| < 1 and |1 − z| < 1, and if a, b, c are suitably restricted, a, b; a, b; Γ(c)Γ(c − a − b) 1−z z = F F Γ(c − a)Γ(c − b) 1 + b + 1 − c; c; c − a, c − b; c−a−b Γ(c)Γ(a + b − c)(1 − z) 1 − z . + F Γ(a)Γ(b) c − a − b + 1; Solution 48. (Solution by Leon Hall) We denote the hypergeometric differential equation: z(1 − z)w00 (z) + [c − (a + b + 1)z]w0 (z) − abw(z) = 0 by HGDE. If we make the change of variable z = 1 − y, then HGDE becomes y(1 − y)w00 (y) + [c ∗ −(a + b + 1)y]w0 (y) − abw(y) = 0 where c∗ = a + b + 1 − c. Thus, two linearly independent solutions are F (a, b; c∗; y) and y 1−c∗ F (a + 1 − c∗, b + 1 − c∗; 2 − c∗; y). These solutions as function of z are valid in |1 − z| < 1 and are F (a, b; a + b + 1 − c; 1 − z)
30 SOLUTIONS BY SYLVESTER J. PAGANO AND LEON HALL; EDITED BY TOM CUCHTA
and (1 − z)c−a−b F (c − a, c − b; c − a − b + 1; 1 − z). Thus, in the region D where both |z| < 1 and |1 − z| < 1, F (a, b; c; z) = AF (a, b; a+b+1−c; 1−z)+B(1−z)c−a−b F (c−a, c−b; c−a−b+1; 1−z) for some constants A and B. Assume Re(c − a − b) > 0 and c 6= 0 or a negative integer and let z → 1 inside the region D to get F (a, b; c; 1) = A · 1 + B · 0. Thus, by Theorem 18, page 49, we get A=
Γ(c)Γ(c − a − b) . Γ(c − a)Γ(c − b)
Now, let z → 0 inside the region D and assume Re(1−c) > 0 and neither a+b+1−c nor c − a − b + 1 is zero or a negative integer. Then 1 = AF (a, b; a + b + 1 − c; 1) + BF (c − a, c − b; c − a − b + 1; 1) and we get 1 − AF (a, b; a + b + 1 − c; 1) . F (c − a, c − b; c − a − b + 1; 1) Again using Theorem 18, this becomes B=
1− B=
Γ(c)Γ(c−a−b)Γ(a+b+1−c)Γ(1−c) Γ(c−a)Γ(c−b)Γ(b+1−c)Γ(a+1−c) Γ(c−a−b+1)Γ(1−c) Γ(1−b)Γ(1−a)
.
By Exercise 15, page 32 the numerator is equal to Γ(2 − c)|Gamma(c − 1)Γ(c − a − b)Γ(a + b + 1 − c) . Γ(a)Γ(1 − a)Γ(b)Γ(1 − b) Hence, B
= =
Γ(2 − c)Γ(c − 1)Γ(c − a − b)Γ(a + b + 1 − c) Γ(a)Γ(b)Γ(c − a − b + 1)Γ(1 − c) (1 − c)Γ(1 − c) Γ(c) c−1 (a + b − c)Γ(a + b − c)
Γ(a)Γ(b)(c − a − b)Γ(c − a − b)Γ(1 − c) Γ(c)Γ(a + b − c) = . Γ(a)Γ(b) This yields the desired formula for F (a, b; c; z) in terms of the given hypergeometric functions of 1 − z. Problem 49. In a common notation for the Laplace transform Z ∞ L{F (t)} = e−st F (t)dt = f (s); L−1 {f (s)} = F (t). 0
Show that L−1
1 s
a, b;
F s + 1;
a, b; z =F 1;
z(1 − e−t ) .
SOLUTIONS TO RAINVILLE’S “SPECIAL FUNCTIONS” (1960)
Solution 49. Let A = L −1
1 s
31
z . We wish to evaluate A. Now
a, b;
F 1 + s;
∞ X (a)n (b)n z n −1 1 A= L . n! s(1 + s)n n=0 But 1 s(s + 1)n
Γ(1 + s) sΓ(1 + s + n) 1 Γ(1 + s)Γ(1 + n) = sn! Γ(1 + s + n)Γ(1) −n, s; 1 1 = F n!s 1 + s; n X (−n) k (s)k 1 = n!s . k!(1 + s)k =
k=0
Hence n
1 1 X (−n)k . = s(s + 1)n n! k!(s + k) k=0
Then L −1
1 s(s + 1)n
n
=
1 1 X (n)k e−kt = (1 − e−t )n . n! k! n! k=0
Therefore ∞ a, b; n −t n X (a)n (b)n z (1 − e ) A= =F n!n! n=0 1;
z(1 − e−t ) .
There are many other ways of doing Exercise 16. Probably the easiest, but most undesirable, is to work from right to left in the result to be proved. It is hard to see any chance for discovering the relation that way. Problem 50. With that notation of Exercise 16 show that n 3+n 1+ , ; 2 2 aΓ(n + 2) a2 L{tn sin at} = F − 2 n+2 s s 3 ; 2
.
Solution 50. We wish to obtain the Laplace Transform of tn sin at. Now n
t sin at =
∞ X (−1)k a2k+1 tn+2k+1 k=0
(2k + 1)!
and L {tm } =
Γ(m + 1) . sm+1
.
32 SOLUTIONS BY SYLVESTER J. PAGANO AND LEON HALL; EDITED BY TOM CUCHTA
Hence L {tn sin at}
=
∞ X (−1)k a2k+1 Γ(n + 2k + 2)
(2k + 1)!sn+2k+2 ∞ aΓ(n + 2) X (−1)k (n + 2)2k a2k
k=0
=
sn+2
(2)2k s2k n+2 n+3 k (−1) a2k ∞ 2 2 aΓ(n + 2) X k k = 3 sn+2 2k k=0 k! s 2 k n+2 n+3 , ; 2 2 aΓ(n + 2) −a2 . F = sn+2 s2 3 ; 2 k=0
Problem 51. Obtain the results log(1 + x) = xF (1, 1; 2; −x), 1 1 3 2 arcsin x = xF , ; ;x , 2 2 2 1 3 arctan x = xF , 1; ; −x2 . 2 2 Solution 51. Using
(1)n n! = we know that (n + 1)! (2)n log(1 + x)
∞ X (−1)n xn+1 n+1 n=0 ∞ X (−1)n (1)n (1)n xn =x (2)n n! n=0 = xF (1, 1; 2; −x).
=
Next, start with 2 − 12
(1 − y )
=
∞ X n=0
using
y 2n n!
1 2
1 = 2n + 1 n+
1 2 n 3 2 n
1 2 n
1 2
= x
Z
to get 2 − 21
(1 − y )
dy =
0
∞ X n=0
Thus we arrive at arcsin x =
∞ X n=0
= xF
1 2 n
1 2 n
x2n+1 . n!(2n + 1) 1 2n+1 2 nx 3 2 n n!
1 1 3 2 , ; ;x . 2 2 2
SOLUTIONS TO RAINVILLE’S “SPECIAL FUNCTIONS” (1960)
33
Finally form (1 + y 2 )−1 =
∞ X
(−1)n y 2n
n=0
to obtain Z
x 2 −1
(1 + y ) 0
∞ ∞ X X (−1)n (−1)n x2n+1 dy = = 2n + 1 n=0 n=0
or
arctan x = xF
1 2 n 3 2 n
x2n+1
·
(1)n n!
1 3 2 , 1; ; −x . 2 2
Problem 52. The complete elliptic integral of the first kind is Z π2 dφ p . K= 0 1 − k 2 sin2 φ π 1 1 Show that K = F , ; 1; k 2 . 2 2 2 Z π2 dφ p Solution 52. From K = we obtain 0 1 − k 2 sin2 φ 2n 2n Z π2 X ∞ 1 sin φdφ 2 nk . K= n! 0 n=0 But Z
π 2
2n
sin 0
Hence
1 Γ2 12 12 n Γ n + 21 Γ 12 1 1 π 2 n 1 = = = . φdφ = B n + , 2 2 2 2Γ(n + 1) 2n! 2 n! ∞ πX K= 2 n=0
1 2 n
1 2 n
k 2n
n!n!
π = F 2
1 1 2 , ; 1; k . 2 2
Problem 53. The complete elliptic integral of the second kind is Z π2 p 1 − k 2 sin2 θdθ. E= 0
1 1 2 , − ; 1; k . 2 2 Z π2 p Solution 53. From E = 1 − k 2 sin2 θdθ, we get
π Show that E = F 2
0
E
π 2
∞ X − 12
k 2n sin2n φdφ n! 0 n=0 1 2n ∞ 1 X −2 n 2 n k π = . 2 n=0 n!n! Z
=
Hence E=
π F 2
n
1 1 − , ; 1; k 2 . 2 2
Problem 54. From the contiguous function relations 1-5 obtain the relations 6-10.
34 SOLUTIONS BY SYLVESTER J. PAGANO AND LEON HALL; EDITED BY TOM CUCHTA
(1) (2) (3) (4) (5) (6) (7) (8) (9) (10) (11) (12) (13) (14) (15)
(a − b)F = aF (a+) − bF (b+), (a − c + 1)F = aF (a+) − (c − 1)F (c−), [a + (b − c)z]F = a(1 − z)F (a+) − c−1 (c − a)(c − b)zF (c+), (1 − z)F = F (a−) − c−1 (c − b)zF (c+), (1 − z)F = F (b−) − c−1 (c − a)zF (c+), [2a − c + (b − a)z]F = a(1 − z)F (a+) − (c − a)F (a−), (a + b − c)F = a(1 − z)F (a+) − (c − b)F (b−), (c − a − b)F = (c − a)F (a−) − b(1 − z)F (b+), (b − a)(1 − z)F = (c − a)F (a−) − b(1 − z)F (b+), [1 − z + (c − b − 1)z]F = (c − a)F (a−) − (c − 1)(1 − z)F (c−), [2b − c + (a − b)z]F = b(1 − z)F (b+) − (c − b)F (b−), [b + (a − c)z]F = b(1 − z)F (b+) − c−1 (c − a)(c − b)zF (c+), (b − c + 1)F = bF (b+) − (c − 1)F (c−), [1 − b + (c − a − 1)z]F = (c − b)F (b−) − (c − 1)(1 − z)F (c−), [c − 1 + (a + b + 1 − 2c)z]F = (c − 1)(1 − z)F (c−) − c−1 (c − a)(c − b)zF (c+).
Solution 54. From (3) and (4) we get [a + (b − c)z − (c − a)(1 − z)]F = a(1 − z)F (a+) − (c − a)F (a−), or [2a − c + (b − a)z]F = a(1 − z)F (a+) − (c − a)F (a−).
(6)
From (3) and (5) we get [a + (b − c)z − (c − b)(1 − z)]F = a(1 − z)F (a+) − (c − b)F (b−), or (7)
[a + b − c]F = a(1 − z)F (a+) − (c − b)F (b−).
From (1) and (6) we get [(a − b)(1 − z) − 2a + c − (b − a)z]F = (c − a)F (a−) − b(1 − z)F (b+), or (8)
[c − a − b]F = (c − a)F (a−) − b(1 − z)F (b+).
From (6) and (7) we get (9)
(b − a)(1 − z)F = (c − a)F (a−) − (c − b)F (b−).
Use (2) and (6) to obtain [(a − c + 1)(1 − z) − 2a + c − (b − a)z]F = (c − a)F (a−) − (c − 1)(1 − z)F (c−), or [1 − a + (c − b − 1)z]F = (c − a)F (a−) − (c − 1)(1 − z)F (c−).
(10)
From (1) and (7) we get [a + b − c − (a − b)(1 − z)]F = b(1 − z)F (b+) − (c − b)F (b−), or (11)
[2b − c + (1 − b)z]F = b(1 − z)F (b+) − (c − b)F (b−),
which checks with (6). Easier method: in (6) interchange a and b. From (1) and (3) we get [a + (b − c)z − (a − b)(1 − z)]F = b(1 − z)F (b+) − c−1 (c − a)(c − b)zF (c+),
SOLUTIONS TO RAINVILLE’S “SPECIAL FUNCTIONS” (1960)
35
or (12)
[b + (a − c)z]F = b(1 − z)F (b+) − c−1 (c − a)(c − b)zF (c+),
more easily found by changing b to a and a to b in (3). In (2) interchange a and b to get (13) (b − c + 1)F = bF (b+) − (c − 1)F (c−). In (10) interchange a and b to get (14)
[1 − b + (c − a − 1)z]F = (c − b)F (b−) − (c − 1)(1 − z)F (c−).
From (2) and (3) we get [a + (b − c)z − (a − c + 1)(1 − z)]F = (c − 1)(1 − z)F (c−) − c−1 (c − a)(c − b)zF (c+), or (15)
[c−1+(a+b−2c+1)z]F = (c−1)(1−z)F (c−)−c−1 (c−a)(c−b)zF (c+).
Problem 55. The notation used in Exercise 54 and in Section 33 is often extended as in the examples F (a−, b+) = F (a − 1, b + 1; c; z), F (b+, c+) = F (a, b + 1; c + 1; z). Use the relations (4) and (5) of Exercise 54 to obtain F (a−) − F (b−) + c−1 (b − a)zF (c+) = 0 and from it, by changing b to (b + 1) to obtain the relation (c − 1 − b)F = (c − a)F (a−, b+) + (a − 1 − b)(1 − z)F (b+), or (c − 1 − b)F (a, b; c; z) = (c − a)F (a − 1, b + 1; c; z) + (a − 1 − b)(1 − z)F (a, b + 1; c; z), another relation we wish to use in Chapter 16. Solution 55. From Exercise 54 equation (4) and (5) we get (1)(1 − z)F = F (a−) − c−1 (c − b)zF (c+), (2)(1 − z)F = F (b−) − c−1 (c − a)zF (c+). From the above we get F (a−) − F (b−) + c−1 (b − a)zF (c+) = 0. Now replace b by b + 1 to write F (a−, b+) − F + c−1 (b + 1 − a)zF (b+, c+) = 0, or F (a, b; c; z) = F (a − 1, b + 1; c; z) + c−1 (b + 1 − a)zF (a, b + 1; c + 1; z). Problem 56. In equation (9) of Exercise 54 shift b to b + 1 to obtain the relation (c − 1 − b)F = (c − a)F (c−, b+) + (a − 1 − b)(1 − z)F (b+), or (c − 1 − b)F (a, b; c; z) = (c − a)F (a − 1, b + 1; c; z) + (a − 1 − b)(1 − z)F (a, b + 1; c; z), another relation we wish to use in Chapter 16.
36 SOLUTIONS BY SYLVESTER J. PAGANO AND LEON HALL; EDITED BY TOM CUCHTA
Solution 56. Equation (9) of Exercise 54 is (b − a)(1 − z)F = (c − a)F (a−) − (c − b)F (b−) from which we may write (b + 1 − a)(1 − z)F (b+) = (c − a)F (a−, b+) − (c − b − 1)F, or (c − b − 1)F (a, b; c; z) = (c − a)F (a−, b + 1; c; z) + (a − 1 − b)(1 − z)F (a, b + 1; c; z). 6. Results from Chapter 5 used Theorem 6.1. If z is nonintegral, Γ(z)Γ(1 − z) =
π . sin πz
Theorem 6.2. (Dixon’s Theorem) The follow is an identtiy of a, b, and c are so restricted that each of the functions involved exists: 3 F2
a, b, c;
1 =
1 + a − b, 1 + a − c;
Γ(1 + 12 a)Γ(1 + a − b)Γ(1 + a − c)Γ(1 + 21 a − b − c) . Γ(1 + a)Γ(1 + 12 a − b)Γ(1 + 21 a − c)Γ(1 + a − b − c)
Theorem 6.3. If R(α) > 0, R(β) > 0, and if k and s are positive integers, then insdie the region of convergence of the resultant series Z
t
xα−1 (t − x)β−1 p Fq
0
a1 , . . . , ap ;
cxk (t − x)s b1 , . . . , b q ;
α α+1 α+k−1 β β+1 β+s−1 , , ,..., ; a1 , . . . , a p , k , k , . . . , k s s s = B(α, β)α+β−1 p+k+s Fq+k+s α+β α+β+1 α+β+k+s−1 b1 , . . . , bq , , ,..., ; k+s k+s k+s
7. Results from Chapter 5 used Theorem 7.1. If z is nonintegral, Γ(z)Γ(1 − z) =
π . sin πz
Theorem 7.2. (Dixon’s Theorem) The follow is an identtiy of a, b, and c are so restricted that each of the functions involved exists: a, b, c; Γ(1 + 12 a)Γ(1 + a − b)Γ(1 + a − c)Γ(1 + 21 a − b − c) 1 = . 3 F2 Γ(1 + a)Γ(1 + 12 a − b)Γ(1 + 21 a − c)Γ(1 + a − b − c) 1 + a − b, 1 + a − c;
k k ss ctk+s (k + s)k+s
.
SOLUTIONS TO RAINVILLE’S “SPECIAL FUNCTIONS” (1960)
37
Theorem 7.3. If R(α) > 0, R(β) > 0, and if k and s are positive integers, then insdie the region of convergence of the resultant series Z t a1 , . . . , ap ; cxk (t − x)s xα−1 (t − x)β−1 p Fq 0 b1 , . . . , b q ; α α+1 α+k−1 β β+1 β+s−1 , , ,..., ; a1 , . . . , a p , k , k , . . . , k s s s = B(α, β)α+β−1 p+k+s Fq+k+s α+β α+β+1 α+β+k+s−1 b1 , . . . , bq , , ,..., ; k+s k+s k+s 8. Chapter 5 Problem 57. Show that −; −; x 0 F1 0 F1 b; a;
1 1 1 1 1 a + b, a + b − ; 2 2 2 2 2 x = 2 F3
4x .
a, b, a + b − 1;
Solution 57. Consider the product 0 F1 (−; a; x)0 F1 (−; b; x)
= =
∞ X
xn+k (a)k (b)n k!n! n,k=0 n ∞ XX xn n=0 k=0 ∞ X n X
(a)k (b)n−k k!(n − k)!
(1 − b − n)k (−n)k xn · (a)k k! (b)n n! n=0 k=0 ∞ −n, 1 − b − n; X xn 1 . = F (b)n n! n=0 a; =
We then use the result in Exercise 6, page 69, to get ∞ X (a + b − 1)2n xn 0 F1 (−; a; x)0 F1 (−; b; x) = (b) (a)n (a + b − 1)n n! n=0 n a+b−1 ∞ X 2 n = (a)n (b)n (a + b − 1)n n! a+b n=0 22n xn n 2 a+b a+b−1 , ; 2 = 2 F3 F 2 4x . a, b, a + b − 1; Problem 58. Show that 1 3 , ; t 1 1 4 4 − 12 2 2 − 12 2 x (t − x) [1 − x (t − x) ] dx = πt2 F1 2 0 1;
Z
t4 . 16
k k ss ctk+s (k + s)k+s
.
38 SOLUTIONS BY SYLVESTER J. PAGANO AND LEON HALL; EDITED BY TOM CUCHTA
Solution 58. We use theorem 7.3 on the integral Z t 1 1 1 A= x 2 (t − x)− 2 [1 − x2 (t − x)2 ]− 2 dx. 0
Now t
1 ; −; x2 (t − x)2 dx, 2 0 1 1 3 so in Theorem 7.3 we pput |alpha = , β = , p = 1, q = 0, a1 = , c = 1, k = 2 2 2 2, s = 2. The result is 1 3 5 1 3 , , , , ; 2 4 4 4 4 3 1 22 22 t2+2 , , t 5 F4 A=B 2 2 44 2 3 4 5 , , , ; 4 4 4 4 or 1 3 1 3 3 1 , ; , ; Γ Γ 4 4 π 4 4 2 2 t4 t4 A= t 2 F1 = t2 F1 , Γ(2) 2 16 16 1, 1; 1; Z
A=
1
1
x 2 (t − x)− 2 1 F0
as desired. Problem 59. With the aid of Theorem 7.1, show that Γ(1 + 21 a) cos 21 πaΓ(1 − a) = Γ(1 + a) Γ(1 − 21 a and that sin π(b − 12 a)Γ(b − 21 a) Γ(1 + a − b) . = sin π(b − a)Γ(b − a) Γ(1 + 12 a − b) Thus put Dixon’s theorem (Theorem 7.2) in the form a, b, c; cos 21 πa sin π(b − 12 a) Γ(1 − a)Γ(b − 21 a)Γ(1 + a − c)Γ(1 + 21 a − b − c) 1 = · . 3 F2 sin π(b − a) Γ(1 − 21 a)Γ(b − a)Γ(1 + 12 a − c)Γ(1 + a − b − c) 1 + a − b, 1 + a − c; π Solution 59. We first note that, since Γ(z)Γ(1 − z) = , sin πz 1 1 1 1 1 Γ 1+ a Γ 1− a aΓ a γ 1− a 2 2 2 2 2 = Γ (1 + a) Γ(1 − a) aΓ(a)Γ(1 − a) sin πa · π = πa 2π sin π2 π 2 cos a sin a 2 2 = π 2 sin a πa 2 = cos . 2
SOLUTIONS TO RAINVILLE’S “SPECIAL FUNCTIONS” (1960)
3 F2
a, b, c;
1 =
1 + a − b, 1 + a − c;
cos
39
1 1 πa 1 sin π b − a Γ(1 − a)Γ b − a Γ(1 + a − c)Γ 1 + a − b − c 2 2 2 2 , · 1 1 sin π(b − a) Γ 1 − a Γ(b − a)Γ 1 + a − c Γ(1 + a − b − c) 2 2
1 1 so long as a, a, b − a, b − a are not integers and the gamma functions involved 2 2 have no poles. But now we have arrived at an identity (for non-integral values of certain parameters) which has the property that both members are well-behaved if a is a negative integer or zero. It follows that the identity continues to be valid for a = −n, n a πa = 0. nonnegative integer. If a = −(2n + 1), an odd negativer integer, cos 2 Problem 60. Use the result in Exercise 59 to show that if n is a non-negative integer, −2n, α, 1 − β − 2n; (2n)!(α)n (β − α)n 1 = . 3 F2 n!(α)2n (β)n 1 − α − 2n, β; Solution 60. If a = −2n in the identity of Exercise 59 above, and if we chose b = α, c = 1 − β −2n , we obtain −2n, α, 1 − β − 2n; cos(−πn) sin π(α + n) Γ(1 + 2n)Γ(α + n)Γ(1 − 2n − 1 + β + 2n)Γ(1 − n − α − 1 + β + 2n) 1 = 3 F2 sin π(α + 2n) Γ(1 + n)Γ(α + 2n)Γ(1 − n − 1 + β + 2n)Γ(1 − 2n − α − 1 + β + 2n) 1 − α − 2n, β; n n (−1) (−1) sin πα (2n)!(α)n Γ(β)Γ(β − α + n) = sin πα n!(α)2n Γ(β + n)Γ(β − α) (2n)!(α)n (β − α)n = , n!(α)2n (β)n as desired. Problem 61. With the aid of the formula in Exercise 60 prove Ramanujan’s theorem: α, β − α; α; α; x2 x −x . F F = F 1 1 1 2 3 4 1 1 1 β; β; 1 β, β, β + ; 2 2 2 Solution 61. Consider the product ∞ X n X (−1)k (α)k (α)n−k xn F (α; β; x) F (α; β; −x) = 1 1 1 1 (β)k (β)n−k k!(n − k)! n=0 k=0 ∞ X n X (−n)k (α)k (1 − β − n)k (α)n xn = k!(β)k (1 − α − n)k n!(β)n n=0 k=0 ∞ −n, α, 1 − β − n; X (α)n xn 1 = . 3 F2 n!(β)n n=0 β, 1 − α − n; Since the product of the two 1 F10 s is an even function of x, we may conclude that −2n − 1, α, 1 − β − 2n − 1; 1 =0 3 F2 β, 1 − α − 2n − 1;
40 SOLUTIONS BY SYLVESTER J. PAGANO AND LEON HALL; EDITED BY TOM CUCHTA
and that 1 F1 (α; β; x)1 F1 (α; β; −x)
=
∞ X
3 F2
−2n, α, 1 − β − 2n;
β, 1 − α − 2n; (2n)!(α)n (β − α)n (α)2n x2n = , n!(α)2n (β)n (2n)!(β)2n n=0 n=0 ∞ X
by Exercise 60. Hence we get Ramanujan’s theorem
(α)2n x2n 1 (2n)!(β)2n
α, β − α;
2
1 F1 (α; β; x)1 F1 (α; β; −x)
= 2 F3
1 1 1 β, β, β + ; 2 2 2
x 4
.
Problem 62. Let γn =3 F2 (−n, 1 − a − n, 1 − b − n; a, b; 1). Use the result in Exercise 59 to show that γ2n+1 = 0 and γ2n =
(−1)n (2n)!(a + b − 1)3n . n!(a)n (b)n (a + b − 1)2n
Solution 62. From Exercise 59, we get α, β, γ; α α α cos πa 2 sin π(β − 2 ) Γ(1 − α)Γ(β − 2 )Γ(1 + α − γ)Γ(1 + 2 − β − γ) 1 = . 3 F2 α α sin π(β − α) Γ(1 − 2 )Γ(β − α)Γ(1 + 2 − γ)Γ(1 + α − β − γ) 1 + α − β, 1 + α − γ; Consider πn γn = 3 F2 (−n, 1 − a − n, 1 − b − n; a, b; 1). We wish to use α = −n, but cos = 0 for n odd. Hence γ2n+1 = 0 and 2 γ2n = 3 F2 (−2n, 1 − a − 2n, 1 − b − 2n; a, b; 1). Therefore in the result from Exercise ?? we put α = −2n, β = 1 − a − 2n, γ = 1 − b − 2n, and thus obtain γn
cos(nπ) sin π(1 − a − n) Γ(1 + 2n)Γ(1 − a − n)Γ(b)Γ(−1 + a + b + 3n) sin π(1 − a) Γ(1 + n)Γ(1 − a)Γ(b + n)Γ(−1 + a + b + 2n) cos2 (nπ) sin π(1 − a) (2n)!(−1)n (a + b − 1)3n = , sin π(1 − a) n!(a)n (b)n (a + b − 1)2n =
so that γn =
(−1)n (2n)!(a + b − 1)3n . n!(a)n (b)n (a + b − 1)2n
Problem 63. With the aid of the result in Exercise 62 show that 1 1 1 (a + b − 1), (a + b), (a + b + 1); 3 3 3 0 F2 (−; a, b; t)0 F2 (−; a, b; −t) =3 F8 1 1 1 1 1 1 1 a, b, a + , b, b + , (a + b − 1), (a + b); 2 2 2 2 2 2 2
−27t2 64
.
SOLUTIONS TO RAINVILLE’S “SPECIAL FUNCTIONS” (1960)
41
Solution 63. Let us consider the product ψ(t)
= 0 F2 (−; a, b; t)0 F2 (−; a, b; −t) ∞ X n X = n=0 k=0 ∞ X n X
tn (−n)k (1 − a − n)k (1 − b − n)k k!(a)k (b)k n!(a)n (b)n n=0 n=0 ∞ n X t = γn n!(a) n (b)n n=0 ∞ X γ2n t2n = , (2n)!(a)2n (b)2n n=0 =
in terms of the γn of Exercise 62 above. We already knew that γ2n+1 = 0 which checks with the fact that Ψ(t) is an even function of t. Since, by Exercise 62, γn =
(−1)n (2n)!(a + b − 1)3n , n!(a)n (b)n (a + b − 1)2n
we have ψ(t)
∞ X
(−1)n (a + b − 1)3n t2n n!(a)n (b)n (a)2n (a + b − 1)2n n=0 a+b a+b+1 2n ∞ X (−1)n 33n a+b−1 t 3 n 3 n 3 n = a+1 b+1 2n b 2n a+b−1 2n a 2 2 n!(a) (b) 2 n n 2 n 2 2 n 2 2 n=0 n n n =
,
a+b 2 n
or F (−1, a, b; x) F (−; a, b; −x) = F 0 2 0 2 3 8
1 3 (a
1 1 + b − 1), (a + b), (a + b + 1); 3 3
−
a a+1 b b+1 a+b−1 a+b , , , , ; a, b, , 2 2 2 2 2 2
27t2 64
Problem 64. Prove that n X (−1)n−k (γ − b − c)n−k (γ − b)k (γ − c)k xn−k k=0
k!(n − k)!(γ)k
3 F2
−k, b, c;
(γ − b)n (γ − c)n (1 − x)n 3 F2 n!(γ)n
x
1 − γ + b − k, 1 − γ + c − k;
=
1 1 1 − n, − n + , 1 − γ − n; 2 2 2 1 − γ + b − n, 1 − γ + c − n;
and note the special case γ = b + c, Whipple’s theorem.
−4x (1 − x)2
.
42 SOLUTIONS BY SYLVESTER J. PAGANO AND LEON HALL; EDITED BY TOM CUCHTA
γ − b, γ − c;
t(1 − x + xt) . Then
Solution 64. Let ψ = 2 F1 γ; ψ
= = = =
∞ X (γ − b)n (γ − c)n tn [(1 − x) + xt]n (γ)n n! n=0 ∞ X n X (γ − b)n (γ − c)n (1 − x)n−k xk tn+k k=0 k=0 [n ∞ X 2] X n=0 k=0 [n ∞ X 2] X
k!(n − k)!(γ)n (γ − b)n−k (γ − c)n−k (1 − x)n−2k xk tn k!(n − 2k)!(γ)n−k
(γ − b)n (γ − c)n (1 − x)n tn (−n)2n (1 − γ − n)k (−1)k xk , k!(1 − γ + c − n)k (1 − γ + b − n)k (1 − x)2k n!(γ)n n=0 n=0
or ψ=
∞ X
3 F2
n=0
n n−1 − ,− , 1 − γ − n; 2 2
(γ − b) (γ − c) (1 − x)n tn n n −4x . n!(γ) n (1 − x)2
1 − γ + b − n, 1 − γ + c − n; But also, since 1 − t(1 − x + x) = (1 − t)(1 + xt),
b, c;
Ψ = (1 − t)b+c−γ (1 + xt)b+c−γ 2 F1
t(1 − x + xt) . γ;
Hence = (1 − t)b+c−γ (1 + xt)b+c−γ
ψ
= (1 − t)b+c−1 (1 + xt)b+c−γ
∞ X (b)n (c)n tn [1 − x(1 − t)]n n!(γ)n n=0 n ∞ X k X (−1) (b)n (c)n xk (1 − t)k tn n=0 k=0
k!(n − k)!(γ)n
,
or ψ
∞ X (−1)k (b)n+k (c)n+k xk (1 − t)x tn+k k!n!(γ)n+k n,k=0 ∞ b + k, c + k; X 1 = (1 + xt)b+c−γ (1 − t)b+c−γ 2 F1 γ + k; k=0
= (1 − t)b+c−γ (1 + xt)b+c−γ
Now 2 F1
b + k, c + k;
γ − b, γ − c;
t = (1 − t)γ−b−c−k 2 F1
γ + k;
t .
γ + k;
SOLUTIONS TO RAINVILLE’S “SPECIAL FUNCTIONS” (1960)
43
Therefore ψ
= (1 + xt)b+c−γ
∞ X
2 F1
γ − b, γ − c;
(−1)k (b)k (c)k (xt)k t k!(γ)k
γ + k; k k n+k (−1) (b) (c) k k (γ − b)n (γ − c)n x t = (1 + xt)b+c−γ k!n!(γ)n+k n,k=0 ∞ n X X (−1)s (b)s (c)s (γ − b)n−s (γ − c)n−s xs tn = (1 + xt)b+c−γ s!(n − s)!(γ)n n=0 s=0 ! ∞ ∞ −n, b, c; X (−1)n (γ − b − c)n xn tn X = 3 F2 n! n=0 n=0 1 − γ + b − n, 1 − γ + c − n; k=0 ∞ X
x ,
or ψ=
n ∞ X X n=0 k=0
3 F2
−k, b, c;
n−k
(−1) x
1 − γ + c − b, 1 − γ + c − x;
(γ − b − c)n−k (γ − b)k (γ − c)k xn−k n t . k!(γ)k (n − k)!
By equating coefficients of tn in the two expansions we obtain the desired identity Exercises 9-11 below use the notation of the Laplace transform as in Exercise 16, page 71. Problem 65. Show that a1 , . . . , a p ; L tcp Fq b1 , . . . , bq ;
1 + c, a1 , . . . , ap ; Γ(1 + c) zt = Fq s1+c p+1 b1 , . . . , b q ;
Solution 65. We know that L {tm } = a1 , . . . , a p ; L tc p Fq b1 , . . . , bq ;
z . s
Γ(m + 1) . Then, sm+1
∞ X (a1 )n . . . (ap )n z n zt = L {tn+k } (b ) . . . (b ) n! 1 n q n n=0 ∞ X
(a1 )n . . . (ap )n z n (1 + c)n Γ(1 + c) n+k+1 (b ) . . . (b ) n!s 1 1 n p n n=0 1 + c, a1 , . . . , ap ; z F (1 + c) = . p+1 Fq 5 s1+c b1 , . . . , b q ;
=
Problem 66. Show that a1 , . . . , ap ; 1 L−1 Fq+1 sp s + 1, b1 , . . . , bq ;
z =p Fq+1
a1 , . . . , ap ;
z(1 − e−1 ) .
1, b1 , . . . , bq ;
Solution 66. In Chapter 4, Exercise 16 we found that 1 (1 − e−t )n −1 L = . s(s + 1)n n!
44 SOLUTIONS BY SYLVESTER J. PAGANO AND LEON HALL; EDITED BY TOM CUCHTA
It follows that 1 L −1 F s p q+1
z = p Fq+1
a1 , . . . , ap ; 1 + s, b1 , . . . , bq ;
Problem 67. Show that L−1
sk (s − z)k+1
a1 , . . . , ap ;
z(1 − e−t ) . 1, b1 , . . . , bq ;
k + 1;
zt
= 1 F1 1;
Solution 67. Consider k
1 1 1 s = = 1 F0 (s − z)k+1 s (1 − 54 )k+1 s By Exercise 65 with c = 0, k + 1; Γ(1) −1 L 1 F0 s −;
z s
= L −1
k + 1;
Γ(1)
z . 5
−;
1, k + 1;
2 F1 s k + 1; zt = t0 F 1; k + 1; zt . =F 1;
1;
z s
Problem 68. Show that d Fq dz p
p Y
a1 , . . . , ap ;
z = b1 , . . . , bq ;
m=1 q Y
am
p Fq
bj
a1 + 1, . . . , ap + 1;
z .
b1 + 1, . . . , bq + 1;
j=1
Solution 68. a1 , . . . , ap ; d p Fq dz b1 , . . . , b q ;
z
=
∞ X (a1 )n . . . (ap )n z n−1 (b1 )n . . . (bq )n (n − 1)! n=1
∞ X (a1 )n+1 . . . (a0 )n+1 z n (b1 )n+1 . . . (bq )n+1 n! n=0 a1 + 1, . . . , ap + 1; a1 . . . ap = p Fq b − 1 . . . bq b1 + 1, . . . , bq + 1;
=
z .
Problem 69. In Exercise 19 page 71, we found that the complete elliptic integral of the first kind is given by 1 1 1 2 K(k) = π 2 F1 , ; 1; k . 2 2 2
SOLUTIONS TO RAINVILLE’S “SPECIAL FUNCTIONS” (1960)
45
Show that t
Z
p K( x(t − x))dx = π arcsin
0
1 t . 2
Solution 69. We are given that 1 π 2 F1 2
K(k) =
1 1 , ; 1; k 2 . 2 2
Now consider Z A=
t
p K( x(t − x))dx.
0
By the integral of Section 56, (i.e. Theorem 57) with α = 1, β = 1, k = 1, s = 1, etc. t
1 1 , ; 1; x(t − x) dx 2 2 0 1 1 1 1 , , , ; 2 2 1 1 π = B(1, 1)t2−1 4 F3 2 2 3 1, , ; 2 2 π Γ(1)Γ(1) 1 1 3 t2 = t 2 F1 , ; ; 2 Γ(2) 2 2 2 4! 2 t 1 1 3 t =π F , ; ; 2 1 2 2 2 2 2 t , = π arcsin 2
A =
π 2
Z
2 F1
1 · 1t2 1· 22
by Exercise 18, Chapter 4.
9. Chapter 6 Problem 70. By collecting powers of x in the summation on the left, show that ∞ X
1 J2n+1 (x) = 2 n=0
Z
x
J0 (y)dy. 0
46 SOLUTIONS BY SYLVESTER J. PAGANO AND LEON HALL; EDITED BY TOM CUCHTA
Solution 70. ∞ X
J2k+1 (x)
=
k=0
=
∞ X (−1)n ( x2 )2n+2k+1 n!(n + 2k + 1)! k,n=0 ∞ X n X (−1)n−k ( x )2n+1 2
n=0 k=0 ∞ X n X
(n − k)!(n + k + 1)!
(−n)k (−1)n ( x2 )2n+1 (n + 1)k n!(n + 1)! n=0 k=0 ∞ −n, 1; X (−1)n ( x2 )2n+1 1 = 2 F1 n!(n + 1)! n=0 n + 2; ∞ X Γ(n + 2)Γ(2n + 1)(−1)n ( x )2n+1 2 = Γ(2n + 2)Γ(n + 1)n!(n + 1)! n=0 ∞ n x 2n+1 X (−1) ( 2 ) = (2n + 1)n!n! n=0 Z xX ∞ (−1)n ( y2 )2n 1 dy = 2 0 n=0 n!n! Z x 1 J0 (y)dy. = 2 0 =
Problem 71. Put the equation of Theorem 39, page 113, into the form ∞ X 1 (A) exp z(t − t−1 ) = J0 (z) + Jn (z)[tn + (−1)n t−n ]. 2 n=1 Use equation (A) with t = i to conclude that cos z = J0 (z) + 2
∞ X
(−1)k J2k (z),
k=1
sin z = 2
∞ X
(−1)k J2k+1 (z).
k=0
Solution 71. We know that ∞ −1 ∞ X X X 1 −1 n n Jn (z)t = Jn (z)t + J0 (z) + Jn (z)tn . exp z(t − t ) = 2 n=−∞ n=−∞ n=1 Now J−n (z) = (−1)n Jn (z). Hence ∞ ∞ X X z 1 exp (t − = J−n (z)E n + J0 (z) + Jn (z)T tn 2 t n=1 n=1 ∞ X n = J0 (z) + Jn (z)[t + (−1)n t−n ]. n=1
Now use t = i. Then (−1)n i−n = i2n i−n = in and π 1 exp i− = exp(iz) = cos z + i sin z. 2 i
SOLUTIONS TO RAINVILLE’S “SPECIAL FUNCTIONS” (1960)
47
Therefore cos z + i sin z = J0 (z) + 2
∞ X
in Jn (z).
n=1
But Jn (−z) = (−1)n Jn (z), so we have cos z − i sin z = J0 (z) + 2
∞ X
(−1)n in Jn (z).
n=1
(note: above argument can be done more easily by equating even and odd function of z) Thus we get cos z = J0 (z) + 2
∞ X
i2n d2n (z) = J0 (z) + 2
n=1
∞ X
(−1)n J2n (z)
n=1
and i sin z = 2
∞ X
i2n+1 J2n+1 (z),
n=0
or
∞ X
sin z = 2
(−1)n J2n+1 (z).
n=0
Problem 72. Use t = e
iθ
in equation (A) of Exercise 71 to obtain the results
cos(z sin(θ)) = J0 (z) + 2
∞ X
J2k (z) cos(2kθ),
k=1
sin(z sin(θ)) = 2
∞ X
J2k+1 (z) sin(2k + 1)θ.
k=0
Solution 72. Put t = eiθ . Then tn + (−1)n t−n = eniθ + (−1)N e−niθ and 1 z t− = exp(iz sin(θ)) = cos(z sin(θ)) + i sin(z sin(θ)). exp 2 t z 1 niθ n −niθ n n Also e +(−1) e = [1+(−1) ] cos(nθ)+[1−(−1) ] sin(nθ). From exp t− = 2 t ∞ X J0 (z) + Jn (z)[tn + 9 − 1)n t−n ] we thus obtain n=1
cos(z sin θ)+i sin(z sin θ) = J0 (z)+
∞ X
Jn (z)[(1+(−1)n ) cos(nθ)+(1−(−1)n ) sin(nθ)].
n=1
Now equate even function of z on the two sides, then odd functions of z on the two sides to get ∞ X cos(z sin θ) = J0 (z) + 2 J2k (z) cos(2kθ) k=1
and sin(z sin θ) = 2
∞ X k=0
J2k+1 (z) sin(2k + 1)θ.
48 SOLUTIONS BY SYLVESTER J. PAGANO AND LEON HALL; EDITED BY TOM CUCHTA
Problem 73. Use Bessel’s integral, page 114, to obtain for integral n in the relations Z 2 π n cos(nθ) cos(z sin(θ))dθ, (B) [1 + (−1) ]Jn (z) = π 0 Z π 2 (C) [1 − (−1)n ]Jn (z) = sin(nθ) sin(z sin(θ))dθ. π 0 With the aid of (B) and (C) show that for integral k, Z 1 π J2k (z) = cos(2kθ) cos(z sin(thetaA))dθ, π 0 Z 1 π J2k+1 (z) = sin(2k + 1)θ sin(z sin(θ))dθ, π 0 Z π cos(2k + 1)θ cos(z sin(θ)dθ = 0, 0 Z π sin(2kθ) sin(z sin(θ))dθ = 0. 0
Solution 73. We know that 1 Jn (z) = π
Z
π
cos(nθ − z sin θ)dθ. 0
Then Jn (z) =
1 π
Z
π
cos(nθ) cos(z sin θ)dθ + 0
1 π
Z
π
sin(nθ) sin(z sin θ)dθ. 0
Now change z to (−z) to get Z Z 1 π 1 π (−1)n Jn (z) = cos(nθ) cos(z sin θ)dθ − sin(nθ) sin(z sin θ)dθ. π 0 π 0 We then obtain Z 2 π n cos(nθ) cos(z sin θ)dθ (B) [1 + (−1) ]Jn (z) = π 0 Z π 2 (C) [1 − (−1)n ]Jn (z) = sin(nθ) sin(z sin θ)dθ. π 0 Use (B) with n = 2k and (C) with n = 2k + 1 to obtain Z 1 π J2k (z) = cos(2kθ) cos(z sin θ)dθ, π 0 Z 1 π J2k+1 (z) = sin(2k + 1)θ sin(z sin θ)dθ. π 0 Use (B) with n = 2k + 1 and (C) with n = 2k to obtain Z π cos((2k + 1)θ) cos(z sin θ)dθ = 0, 0
Z
π
sin(2kθ) sin(z sin θ)dθ = 0. 0
Problem 74. Expand cos(z sin(θ)) and sin(z sin(θ)) in Fourier series over the interval −π < θ < π. Thus use Exercise 73 to obtain in another way the expansions in Exercise 72.
SOLUTIONS TO RAINVILLE’S “SPECIAL FUNCTIONS” (1960)
49
Solution 74. In the interval −π < θ < π, the Fourier approximation of f (θ) is f (θ) =
∞ X 1 a0 + (an cos(nθ) + bn sin(nθ)), 2 n=1
in which
Z 1 π f (θ) cos(nθ)dθ, π −π Z 1 π bn = f (θ) sin(nθ)dθ. π −π Consider first f (θ) = cos(z sin θ), an even function of θ. For this function Z 2 π an = cos(nθ) cos(z sin θ)dθ, bn = 0. π 0 an =
By the results in Exercise 73 we obtain a2k+1 = 0, a2k = 2J2k (z). Hence cos(z sin θ) = J0 (z) + 2
∞ X
J2k (z) cos(2kθ), −π < θ < π.
k=1
Next, let f (θ) = sin(z sin θ), an odd function of θ. For this function, Z 2 π an = 0, bn = sin(nθ) sin(z sin θ)dθ. π 0 From Exercise 73 we get b2k = 0, b2k+1 = 2J2k+1 (z). Hence sin(z sin θ) = 2
∞ X
J2k+1 (z) sin((2k + 1)θ)), −π < θ < π.
k=0
1 1 x(t − t−1 ) by exp − x(t − t−1 ) , obtain 2 2 the coefficient of t0 and thus show that ∞ X J02 (x) + 2 Jn2 (x) = 1.
Problem 75. In the product of exp
n=1 1
For real x conclude that |J0 (x)| ≤ 1 and |Jn (x)| ≤ 2− 2 for n ≥ 1. Solution 75. We know that Z ∞ γ 1 exp t− = Jn (x)tn 2 t n=−∞ and thus that Z ∞ γ 1 exp − t− = (−1)k Jk (x)Jn (x)tn+k . 2 t n=−∞ The coefficient of t0 on the right is (k = −n) ∞ X
(−1)−n J−n (x)Jn (x) = 0,
n=−∞
from which we obtain J02 (x) + 2
∞ X
(−1)n J−n (x)Jn (x) = 1.
n=1
50 SOLUTIONS BY SYLVESTER J. PAGANO AND LEON HALL; EDITED BY TOM CUCHTA
But J2n (x) = (−1)n Jn (x). Hence J02 (x) + 2
∞ X
Jn2 (x) = 1.
n=1
It follows at once, for real x, that |J0 (x)| ≤ 1, 1 |Fn (x)| ≤ √ , n ≥ 1. 2 Problem 76. Use Bessel’s integral to show that |Jn (x)| ≤ 1 for real x and integral n. Solution 76. Bessel’s integral is 1 Jn (x) = π
π
Z
cos(nθ − x sin θ)dθ. 0
For real x (and n), |cos(nθ − x sin θ)| ≤ 1. Hence Z 1 π |Jn (x)| ≤ dθ = 1. π 0 Problem 77. By iteration of equation (8), page 111, show that 2m
m X dm J (z) = (−1)m−k Cm,k Jn+m−2k (z), n dz m n=0
where Cm,k is the binomial coefficient. Solution 77. We have, with D =
d , from (8), page 100, dz
2DJn (z) = Jn−1 (z) − Jn+1 (z). Then 22 D 2 Jn (z)
= 2DJn−1 (z) − 2DJn+1 (z) = Jn−2 (z) − Jn (z) − Jn (z) + Jn+2 (z) = Jn−2 (z) − 2Jn (z) + Jn+2 (z).
Let us use induction. Assume 2m D m Jn (z) =
m X
(−1)m−k Cm,k Jn+m−2k (z),
k=0
as we k now is true for m = 1, 2. Then 2m+1 D m+1 Jn (z)
= = =
m X
(−1)m−k Cm,k [Jn+m−2k−1 (z) − Jn+m−2k+1 (z)]
k=0 m+1 X
m X
k=1
k=0
(−1)m−k+1 Cm,k−1 Jn+m+1−2k (z) +
m X
(−1)m+1−k Cm,k Jn+m+1−2k (z)
(−1)m+1−k [Cm,k + Cm,k−1 ]Jn+m+1−2k (z) + (−1)m+1 Jn+m+1 (z) + Jn−m−1 (z).
k=1
SOLUTIONS TO RAINVILLE’S “SPECIAL FUNCTIONS” (1960)
51
Now Cm,k + Cm,k−1 = Cm+1,k (Pascal’s triangle), so that (using the last two terms also), m+1 X 2m+1 D m+1 Jn (z) = (−1)m+1−k Cm+1,k Jn+m+1−2k (z), k=0
which completes the induction. Problem 78. Use the result in Exercise 1, page 105, to obtain the probduct of two Bessel functions of equal argument. Solution 78. We know already that a+b a+b−1 , ; 2 2 0 F1 (−; a; x)0 F1 (−; b; x) = 2 F3
4x .
a, b, a + b − 1; Then z n z m Jn (z)Jm (z)
z2 z2 2 2 )0 F1 (−; m + 1; − ) 0 F1 (−; n + 1; − Γ(n + 1)Γ(m + 1) 4 4 z n+m n+m+2 n+m+1 , ; 2 2 2 = 2 . 2 F3 −z Γ(n + 1)Γ(m + 1) n + 1, m + 1, n + m + 1; =
Problem 79. Start with the power series for Jn (z) and use the form (2), page 18, of the Beta function to arrive at the equation Z π2 2( 21 z)n Jn (z) = sin2n φ cos(z cos φ)dφ, Γ( 12 )Γ(n + 21 ) 0 1 for R(n) > − . 2 Solution 79. We know that 1 z 2k+n ∞ ∞ (−1) k 2k+n X X 2 (−1) z k Jn (z) = = . 22k+1 k!Γ(k + n + 1) 2n (2k)!Γ(k + n + 1) k
k=0
k=0
Also 1 2 k Γ(k + n + 1)
1 1 Γ n+ Γ k+ 2 2 = 1 1 Γ Γ(k + n + 1)Γ n + 2 2 1 1 B k + ,n + 2 2 = 1 1 Γ Γ n+ 2 2 Z π2 2 = cos2k φ sin2n φdφ. 1 1 0 Γ Γ n+ 2 2
52 SOLUTIONS BY SYLVESTER J. PAGANO AND LEON HALL; EDITED BY TOM CUCHTA
Therefore z n 2 Jn (z) = 2 1 Γ Γ n+ 2 z n 2 = 2 1 Γ Γ n+ 2
π 2
Z 1 2
1 2
sin2n φ cos(z cos φ)dφ
0
π 2
Z
sin2n φ cos(z cos φ)dφ.
0
Problem 80. Use the property d 1 du 0 F1 (−; a; u) = 0 F1 (−; a + 1; u) dx a dx to obtain the differential recurrence relation (6) of Section 60. Solution 80. We know that
d 1 du 0 F1 (−; a; u) = 0 F1 (−; a + 1; u). Since dx a dx z n
(1)
Jn (z) =
z2 2 0 F1 −; 1 + n; − Γ(1 + n) 4
we obtain d −n [z Jn (z)] dz
1 1 z z2 F − −; 2 + n; − 0 1 2n Γ(1 + n) 1 + n 2 4 z n+1 z2 = −z −n 2 0 F1 −; 2 + n; − Γ(2 + n) 4 = −z −n Jn+1 (z),
=
which yields (6) of Section 60. Problem 81. Expand 0 F1
−;
1 + α;
2xt − t2 4
in a series of powers of x and thus arrive at the result
t − 2x t
− 12 α
∞ p X Jα+n (t)xn Jα ( t2 − 2xt) = . n! n=0
SOLUTIONS TO RAINVILLE’S “SPECIAL FUNCTIONS” (1960)
53
1 Solution 81. Consider 0 F1 −; 1 + α; (2xt − t2 ) . We obtain 4 −; ∞ X tn (2x − t)n 2xt − t2 F = = 0 1 2n 2 (1 + α)n n! 4 n=0 1 + α; ∞ X n X (−1)k (2k)n tn+2k = 2n+2k w (1 + α)n+k k!n! n=0 k=0 ∞ X ∞ X tn (2x)n (−1)k t2k = 22k k!(1 + α + n)k 22n n!(1 + α)n n=0 k=0 n t xn Γ(1 + α) ∞ X t2 2 = . 0 F1 −; 1 + α + n; − 4 n!Γ(1 + α + n) n=0 Now
α+n t t2 2 Jα+n (t) = , 0 F1 −; 1 + α + n; − Γ(α + n + 1) 4
so we obtain −;
0 F1
1 + α;
2xt − t 4
2
= Γ(1 + α)
−α X ∞ Jn+α (t)xn t 2 n! n=0
or √
t2 − 2xt 2
!−α Γ(1 + α)Jα
or
t − 2x t
α2
p
−α X ∞ Jn+α (t)xn t , t2 − 2xt = Γ(1 + α) 2 n! n=0
∞ p X Jn+α (t)xn . Jα ( t2 − 2xt) = n! n=0
Problem 82. Use the realtions (3) and (6) of Section 60 to prove that: For real x, between any two consecutive zeros of x−n Fn (x), there lies one and only one zero of x−n Fn+1 (x). Solution 82. We are given that d n [x Jn (x)] = xn Jn−1 (x), dx d −n x Jn (x) = −x−n Jn+1 (x). dx We know that Jn (x) has exactly n zeros at x = 0. Let the others (we have proved there are any) on the axis of reals be at α1,n , α2,n , . . . . The curve y = x−n Jn (x) has its real zeros only at the α’s. By Rolle’s theorem we see that the zeros β1,n1 , β2,n2 , . . . of y 0 = −x−n Jn+1 (x) are such that an odd number of them lie between each two consecutive α’s. The curve y2 = xn+1 Jn+1 (x)
54 SOLUTIONS BY SYLVESTER J. PAGANO AND LEON HALL; EDITED BY TOM CUCHTA
has its zeros at x = 0 and at the β’s. But y20 = xn+1 Jn (x), so the α’s lie between consecutive β’s. Problem 83. For the function In (z) of Section 65 obtain the following properties by using the methods, but not the results, of this chapter: (1)zIn0 (z) = zIn−1 (z) − nIn (z), (2)zIn0 (z) = zIn+1 (z) + nIn (z), (3)2In0 (z) = In−1 (z) + In+1 (z), (4)2nIn (z) = z[In−1 (z) − In+1 (z)]. Solution 83. (Solution by Leon Hall) = i−n Jn (iz)
In (z)
z n 2
z2 = 0 F1 −; 1 + n; Γ(1 + n) 4 2 k n ∞ z X 1 z 2 = Γ(1 + n) (1 + n)k k! 4 k=0 ∞ X z 2k+n . = 22k+n k!Γ(k + n + 1)
k=0
(1 + n)k = (1 + n)(2 + n) . . . (k + n), so Γ(1 + n)(1 + n)k = Γ(k + n + 1). So, as in the method of Section 60, d n [z In (z)] dz
=
∞ X k=0
= zn =z
n
z 2k+n2n−1 + k)
22k+n−1 k!Γ(n
∞ X
k=0 ∞ X k=0
z 2k+n−1 22k+n−1 k!Γ(n + k) z 2k+n−1 + (n − 1) + 1)
22k+n−1 k!Γ(k
= z n In−1 (z), or z n In0 (z) + nz n−1 In (z) = z n In−1 (z), which is equivalent to zIn (z) = zIn−1 (z) − nIn (z), which is (1).
SOLUTIONS TO RAINVILLE’S “SPECIAL FUNCTIONS” (1960)
55
Similarly, d −n [z In (z)] dz
∞
= =
d X z 2k dz 22k+n k!Γ(k + n + 1) k=0 ∞ X z 2k−1 k=1 ∞ X
22k+n−1 (k − 1)!Γ(k + n + 1)
z 2k+1 22k+n+1 k!Γ(k + n + 1 + 1) k=0 −n = z In+1 (z), =
and z −n In0 (z) − nz −n−1 In (z) = z −n In+1 (z), or zIn0 (z) = nIn (z) + zIn+1 (z), which is (2). Adding (1) and (2): 2zIn0 (z) = zIn−1 (z) + zIn+1 (z) or 2In0 (z) = In−1 (z) + In+1 (z), which is (3). Equating the right sides of (1) and (2): zIn−1 (z) − nIn (z) = zIn+1 (z) + nIn (z), or 2nIn (z) = z[In−1 (z) − In+1 (z)], which is (4). Problem 84. Show that In (z) is one solution of the equation z 2 w00 + zw0 − (z 2 + n2 )w = 0. Solution 84. (Solution by Leon Hall) Because In is a 0 F1 function times z n , we know from Section 46 that u = 0 F1 (−; b; y) is a solution of y
d2 y du +b − u = 0, dy 2 dy
z2 and so 0 F1 −; 1 + n; is a solution of 4 d2 u du − zu = 0. + (2n + 1) dz 2 dz Thus is w = z n u, In (z) will be a solution of z
z −n+1 w00 − 2nz −n w0 + n(n + 1)z −n−1 w + (2n + 1)[z −n w0 − nz −n−1 w] − z −n+1 w = 0 or z −n [zw00 − (2n − 2n − 1)w0 − [−n(n + 1)z −1 + n(2n + 1)z −1 + z]w] = 0 or z 2 w00 + zw0 − (n2 + z 2 )w = 0.
56 SOLUTIONS BY SYLVESTER J. PAGANO AND LEON HALL; EDITED BY TOM CUCHTA
1 Problem 85. Show that, for Re(n) > − , 2 n Z π2 1 2 2z sin2n φ cosh(z cos(φ))dφ. In (z) = Γ 12 Γ n + 12 0 Solution 85. (Solution by Leon Hall) For n not a negative integer, In (z) = i−n Jn (iz), and using the result of Problem 79, In (z) = i−n
2( 12 iz)n 1 Γ( 2 )Γ(n + 12 )
1 2π
Z
sin2n φ cosh(z cos φ)dφ.
0
The powers of i cancel, and because cos(iw) = cosh w we get Z 12 π 2( 21 z)n In (z) = sin2n φ cosh(z cos φ)dφ Γ( 12 )Γ(n + 21 ) 0 for Re(n) > −
1 as desired. 2
Problem 86. For negative integral n define In (z) = (−1)n I−n (z), thus completing the definition in Section 65. Show that In (−z) = (−1)N In (z) and that " # ∞ X 1 n −1 exp z(t + t ) = In (z)t . 2 n=−∞ Solution 86. (Solution by Leon Hall) We have ∞ X
In (z)tn
=
n=−∞
=
−1 X
(−1)n I−n (z)t−n +
n=−∞ ∞ X
∞ X
In (z)tn
n=0
(−1)
n+1
In+1 (z)t
−n−1
+
∞ X
In (z)tn .
n=0
n=0
Now proceed exactly difference beingthat as in the 2proof of Theorem 39, the only z z2 In (z) involves 0 F1 −; 1 + n; whereas Jn (z) involves 0 F1 −; 1 + n; − , to 4 4 get ∞ X 1 In (z)tn = exp z(t + t−1 ) . 2 n=−∞ Problem 87. Use the integral evaluated in Section 56 to show that Z tp p √ n+ 1 t n −n 2 1 [ x(t − x)] Jn ( x(t − x))dx = 2 πt Jn+ 2 . 2 0 Solution 87. Z 0
t
p p [ x(t − x)]n Jn ( x(t − x))dx =
Z
1 2n Γ(1
+ n)
0
t
−;
xn (t−x)n 0 F1
−
1 + n;
x(z − x) dx. 4
SOLUTIONS TO RAINVILLE’S “SPECIAL FUNCTIONS” (1960)
57
Now use Theorem 37 with α = n + 1, β = n + 1, p = 0, q = 1, b1 = 1 + n, c = 1][4 − k = 1, s = 1. The result is , n + 1, n + 1; Z tp t2 p 1 − [ x(t − x)]n Jn ( x(t − x))dx = n B(1 + n, 1 + n)t2n+1 F 4·4 2 Γ(1 + n) 0 2n + 2 2n + 3 1 + n, , ; 2 2 −; t2 Γ(1 + n)Γ(1 + n) 2n+1 − = n t 0 F1 16 2 Γ(1 + n)Γ(2 + 2n) 3 n+ ; 2 −; n+ 12 2 3 t 1 t tn+ 2 Γ(1 + n)Γ n + 2n+1 2 r 2 F = 0 1 − 3 4 Γ(2 + 2n)Γ n + 3 2 n+ ; 2 3 Γ(1 + n)Γ +n t 2 1 = 2n+1 tn+ 2 Jn+ 12 . Γ(2 + 2n) 2 1 2z−1 2 Γ(z)Γ z + 2 √ , we get By Legendre’s duplication formula, Γ(2z) = π 3 1+2n 2 Γ(1 + n)Γ +n 2 √ Γ(2 + 2n) = . π Hence Z
t
p p [ x(t − x)]n Jn ( x(t − x))dx
0
1√ 2n+1 tn+ 2 π t 1 J n+ 2 1+2n 2 2 √ n+ 1 t −n =2 πt 2 Jn+ 12 . 2 =
Problem 88. By the method of Exercise 87 show that Z 1 √ √ 1 − x sin(α x)dx = πα−1 J2 (α), 0
and, in general, that c Z 1 √ 2 c−1 12 n (1 − x) x Jn (α x)dx = Γ(c) Jn+c (α). α 0 Solution 88. Consider Z
1
√
√ 1 − x sin(α x)dx.
0
We know that
3 z2 sin z = z 0 F1 −; ; − . 2 4
58 SOLUTIONS BY SYLVESTER J. PAGANO AND LEON HALL; EDITED BY TOM CUCHTA
Hence 1
Z
√
√ 1 − x sin(α x)dx = α
Z
0
1
1 1 3 α2 x x 2 (1 − x) 2 0 F1 −; ; − dx 2 4
0
We now use Theorem 3 with t = 1, α = −
α2 , k = 1, s = 0. We get 4
3 3 3 , β = , p = 0, q = 1, b1 = , c = 2 2 2
Z
1
√
√
1 − x sin(α x)dx
Now let us turn to Z 1 √ 1 (1 − x)c−1 x 2 n Jn (α x)dx = 0
α2 = αB ·1 F − 4 3 3 , ; 2 1 3 3 Γ Γ α2 2 2 =α 0 F1 −; 3; − Γ(3) 4
0
3 ; 2
( α2 )n Γ(1 + n)
3 3 , 2 2
Z
2
1 c−1 n
(1 − x)
x 0 F1
0
α2 x −; 1 + n; − 4
dx
and use Theorem 37 with α = n + 1, β = c, p = 0, q = 1, b1 = 1 + n, t = 1, c = α2 − , k = 1, s = 0. The result is 4 n + 1; Z 1 α n 2 √ (2) α B(n + 1, c)F (1 − x)t−1 xtn Jn (α x)dx = − ·1 Γ(1 + n) 4 0 1 + n,c + n + 1; ( α2 )n Γ(n + 1)Γ(c) α2 F = −; c + n + 1; − 0 1 Γ(1 +n) Γ(n + c + 1) 4 α −c = Γ(c) Jn+c (α), 2 as desired. Problem 89. Show that Z t Z t exp[−2x(t − x)]I0 [2x(t − x)]dx = exp(−β 2 )dβ. 0
0
Z
t
exp[−2x(t − x)]I0 [2x(t − x)]dx. Now
Solution 89. Consider 0
exp[−2x(t − x)]I0 [2x(t − x)] = exp[−2x(t − x)]0 F1 [−; 1; x2 (1 − x)2 ]. In Kummer’s second formula we have 1 z2 . 1 F1 (a; 2a; 2z) = e 0 F1 −; 1 + ; 2 4 z
Use a =
1 and z = −2x(t − x) to get 2 exp[−2x(t − x)]I0 [2x(t − x)] = 1 F1
1 ; 1; −4x(t − x) . 2
SOLUTIONS TO RAINVILLE’S “SPECIAL FUNCTIONS” (1960)
59
Then, t
Z
Z exp[−2x(t − x)]I0 [2x(t − x)]dx =
1 F1
0
0
t
1 ; 1; −4x(t − x) dx, 2
to which we may apply Theorem 37 with α = 1, β = 1, p = 1, q = 1, a1 = 1, c = −4, k = 1, s = 1. We thus get 1 1 1 , , ; 2 1 1 Z t t2 exp[−2x(t − x)]I0 [2x(t − x)]dx = B(1, 1)tF −4 4 0 2 3 1, , ; 2 2 1 3 2 = t 1 F1 ; ; −t 2 2 1 n t2n+1 ∞ (−1) X 2 n = 3 n=0 n! 2 n ∞ X (−1)n tn+1 = n!(2n + 1) n=0 Z tX ∞ (−1)n β 2n = dβ n! 0 n=0 Z t = exp(−β 2 )dβ.
1 , b1 = 2
0
Problem 90. Show that Z t 1 1 2 1 2 t I0 t . [x(t − s)]− 2 exp[4x(t − x)]dx = π exp 2 2 0 Solution 90. Z t Z t 1 1 − 12 [x(t − x)] exp[4x(t − x)]dx = x− 2 (t − x)− 2 0 F0 (−; −; 4x(t − x))dx. 0
0
1 1 , β = , p = q = 0, k = 1, s = 1, c = 4 : 2 2 1 1 , ; 2 2 1 1 0 4t2 exp[4x(t − x)]dx = B , t F 2 2 4 1 2 , ; 2 2 Γ( 12 )Γ( 12 ) 1 = F ; 1; t2 1 1 Γ(1) 2 1 2 t4 = π exp t 0 F1 −; t; 16 22 2 t t = π exp I0 . 2 2
We use Theorem 37 with α =
Z 0
t
1
[x(t − x)]− 2
60 SOLUTIONS BY SYLVESTER J. PAGANO AND LEON HALL; EDITED BY TOM CUCHTA
Problem 91. Obtain Neumann’s expansion ∞ z n X (n + 2k)(n + k − 1)!Jn+2k (z) = , n ≥ 1. 2 k! k=0
Solution 91. Problem 92. Prove Theorem 39, page 113, by forming the product of the series 1 1 for exp zt and the series for exp − zt−1 . 2 2 Solution 92. 10. Chapter 7 Solutions Problem 93. The function 2 erf (x) = √ π was defined on page 36. Show that
x
Z
exp(−t2 )dt
0
2x erf (x) = √ 1 F1 π
1 3 ; ; −x2 . 2 2
Solution 93. Let 2 erf x = √ π
Z
x
exp(−t2 )dt.
0
Then,
erf x
2 =√ π
∞ X n=0 ∞ X
(−1)n
Z
x
t2n dt
0
n! n 2n+1
(−1) x n!(2n + 1) n=0 ∞ 1 2x X (−1)n 2 n x2n =√ π n=0 n! 23 n 2x 1 3 2 = √ 1 F1 ; ; −x . 2 2 π 2 =√ π
Problem 94. The incomplete Gamma function may be defined by the equation Z x γ(α, x) = e−t tα−1 dt, R(α) > 0. 0
Show that γ(α, x) = α−1 xα 1 F1 (α; α + 1; −x). Solution 94. Let Z γ(α, x) = 0
Then,
x
e−t tα−1 dt; Re(α) > 0.
SOLUTIONS TO RAINVILLE’S “SPECIAL FUNCTIONS” (1960)
∞ xX
Z γ(α, x)
= 0
(−1)n tn+α−1 n! n=0
∞ X (−1)n xn+α . = n!(α + n) n=0
α(α + 1)n . Hence (α)n ∞ X (−1)n (α)n xn γ(α, x) = α−1 xα = α−1 xα 1 F1 (α; α + 1; −x). n!(α + 1) n n=0
Now, (α + n) =
Problem 95. Prove that dk −z e F (a; b; z) = (−1)k (b − a)k e−z 1 F1 (a; b + k; z). 1 1 dz k You may find it helpful to use Kummer’s formula, Theorem 42. (b)k
Solution 95. Let D =
d . Consider D[e−z 1 F1 (a; b; z)]. We know that dx 1 F1 (a; b; z)
= ez 1 F! (b − a; b; −z).
Hence D k [e−z 1 F! (a; b; z)]
We used D 1 F! (a; b; z) =
= D k 1 F1 (b − a; b; −z) (b − a)k (−1)k = 1 F! (b − 1 + k; b + k; −z) (b)k (−1)k (b − a)k −z = e 1 F! (a; b + k; z). (b)k
a 1 F! (a + 1; b + 1; z) k times. b
Problem 96. Show that 1 F1 (a; b; z) =
1 Γ(a)
Z
∞
e−t tα−1 0 F1 (−; b; zt)dt.
0
Solution 96. We know that Z Γ(x) =
∞
e−t tx−1 dt, Re(x) > 0.
0
Then, 1 F1 (a; b; z)
= =
∞ X (a)n z n n!(b)n n=0 ∞ 1 X Γ(a + n)z n
Γ(a) n=0 n!(b)n Z ∞ ∞ a+n−1 n X 1 t z = e−t dt Γ(a) 0 n!(b)n n=0 Z ∞ 1 = e−t ta−1 0 F1 (−; b; tz)dt, Re(a) > 0. Γ(a) 0
61
62 SOLUTIONS BY SYLVESTER J. PAGANO AND LEON HALL; EDITED BY TOM CUCHTA
Problem 97. Show, with the aid of the result in Exercise 96, that ∞
Z
exp(−t2 )t2a−n−1 Jn (zt)dt =
0
Γ(a)z n z2 F a; n + 1; − . 1 1 2n+1 Γ(n + 1) 4
Solution 97. We obtain Z
∞ 2
A=
2a−n−1
exp(−t )t
∞
Z Jn (zt)dt =
0
0
2 z 2 t2 e−t t2a−n−1 z n tn dt. 0 F1 −; 1 + n; − 2n Γ(1 + n) 4
2
Put t = β. Then Z z2β zn 1 ∞ −β a−1 e β A = n dβ 0 F1 −; 1 + n; − 2 Γ(1 + n) 2 0 4 n 2 Γ(a) z z , = n+1 1 F! a; 1 + n; − 2 Γ(1 + n) 1 4 as desired. Problem 98. If k and n are non-negative integers, show that −k, α + n; ;k > n 0 (−n)k 1 = F ; 0 ≤ k ≤ n. α; (α)k Solution 98. Let V (k, n) = F (−k, α + n; α; 1). Then V (k, n) =
k X (−k)s (α + n)s
s!(α)s
s=0
=
k X s=0
(−1)s k!(α)n+s . s!(k − s)!(α)s (α)n
Hence ψ
= =
∞ X V (k, n)tk k=0 ∞ X k X k=0 s=0 ∞ X
k! (−1)s (α)n+s tk s!(k − s)!(α)s (α)n
(−1)s (α)n+s tk+s s!k!(α)s (α)n k,j=0 ∞ X (−1)s (α + n)s ts = et s!(α)s s=0 t = e 1 F! (α + n; α; −t) = et e−t 1 F! (−n; α; t) n X (−n)k tk = . k!(α)k =
k=0
Thus (−n)k V (k, n) = (α)k 0
;0 ≤ k ≤ n ; k > n.
SOLUTIONS TO RAINVILLE’S “SPECIAL FUNCTIONS” (1960)
note: this method requires material in chapter 7 (on line 5) Can do by Chapter 4 if first show that e−t 1 F1 (−n; α; t) = 1 F1 (α + n; α; −t). 11. Chapter 8 Solutions ∞ X Problem 99. From et ψ(xt) = σn (x)tn , show that n=0
σn (xy) =
n X y k (1 − y)n−k σk (x)
(n − k)!
k=0
,
and in particular that
n
2 σn
X n 1 σk (x) x = . 2 (n − k)! k=0
Solution 99. Let t
e ψ(xt) =
∞ X
σn (x)tn .
n=0
Then et ψ(xyt) =
∞ X
σn (xy)tn
n=0
and eyt ψ(xyt) =
∞ X
σn (x)y n tn .
n=0
But et ψ(xyt) = e(1−y)t eyt ψ(xyt) so that ∞ X
σn (xy)t
n=0
n
! ∞ ! ∞ X X (1 − y)n tn n n = σn (x)y t n! n=0 n=0 ∞ X n X (1 − y)n−k y k σk (x) n = t . (n − k)! n=0 k=0
Hence σn (xy) =
n X y k (1 − y)n−k σk (x) k=0
For y =
(n − k)!
1 , we obtain 2 n
2 σn
X n 1 σk (x) x = . 2 (n − k)! k=0
.
63
64 SOLUTIONS BY SYLVESTER J. PAGANO AND LEON HALL; EDITED BY TOM CUCHTA
Problem 100. Consider the set (called Appell polynomials) αn (x) generated by ext A(t) =
∞ X
αn (x)tn .
n=0
Show that α00 (x) = 0, and that for n ≥ 1, αn0 (x) = αn−1 (x). Solution 100. Consider αn (x) defined by xt
(A)
e A(t) =
∞ X
αn (x)tn .
n=0
From (A) we obtain text A(t) =
∞ X
αn0 (x)tn .
n=0
Hence ∞ X
αn (x)tn+1 =
n=0
∞ X
αn0 (x)tn ,
n=0
or ∞ X
αn−1 (x)tn =
n=1
∞ X
αn0 (x)tn .
n=0
Therefore, α00 (x) = 0, and, for n ≥ 1, αn0 (x) = αn−1 (x). Problem 101. Apply Theorem 50, page 141, to the polynomials σn (x) of Section 73 and thus obtain Theorem 45. Solution 101. We have et ψ(xt) =
∞ X
σn (x)tn
n=0
and wish to apply Theorem 50, page 239. In the notation of Theorem 50 we have A(t)et , H(t) = t, ψ(t) =
∞ X
γn tn , γ0 6= 0.
n=0 ∞ n X t 1 Now A(t) = et = ,, so an = ; H(t) = t, so h0 = 1, hn = 0 for n ≥ 1. n! n n=0 Then also ∞ X tA0 (t) tet = t =t= αn tn+1 , A(t) e n=0
so that α0 = 1, αn = 0 for n ≥ 1. Furthermore, ∞ X t·1 tH 0 (t) = =1=1+ βn tn+1 H(t) t n=0
SOLUTIONS TO RAINVILLE’S “SPECIAL FUNCTIONS” (1960)
65
so that βn = 0 for n ≥ 0. Hence by Theorem 50, xσn0 (x) − nσn (x) = −1 · σn−1−0 (x) = −σn−1 (x), which is Theorem 45, page 224. Problem 102. The polynomials σn (x) of Exercise 101 and Section 73 are defined by ∞ X
et ψ(xt) =
(A)
σn (x)tn ,
n=0
but by equation (9), page 134, they also satisfy −c
(1 − t)
(B)
F
xt 1−t
=
∞ X
(c)n σn (x)tn ,
n=0
for a certain function F . By applying Theorem 50, page 141, to (B), conclude that the σn (x) of (A) satisfy the relation (c)n [xσn0 (x) − nσn (x)] = −
n−1 X
(c)k [cσk (x) + xσk0 (x)],
k=0
for arbitrary c. Solution 102. For the σn (x) of Exercise 101 above, we know that −c
(1 − t)
(B)
F
xt 1−t
=
∞ X
(c)n σn (x)tn ,
n=0
where F (u) =
∞ X
(c)n γn un
n=0
in terms of ψ(u) =
∞ X
γn un .
n=0
We now use Theorem 50 on the polynomials fn (x) = (c)n σn (x) of (B). Here A(t) = (1 − t)−c , H(t) =
t . 1−t
Then log A(t) = −c log(1 − t), H(t) = −1 +
1 1 ; H 0 (t) = , 1−t (1 − t)3
A0 (t) c = , A(t) 1−t ∞ X tH 0 (t) t 1−t 1 = = = 1 + tn . H(t) (1 − t)2 t 1−t n=1
66 SOLUTIONS BY SYLVESTER J. PAGANO AND LEON HALL; EDITED BY TOM CUCHTA
Hence ∞ tA0 (t) X n+1 = ct , so αn = c for n ≥ 0; A(t) n=0 ∞ X tH 0 (t) =1+ tn+1 , so βn = 1 for n ≥ 0. H(t) n=0 Thus Theorem 50 yields
x(c)n σn0 (x)−n(c)n σn (x) = −
n−1 X
c(c)n−1−k σn−1−k (x)−x
k=0
n−1 X
0 1·(c)n−1−k σn−1−k (x),
k=0
or, with reversal of order of summation, (c)n [xσn0 (x) − nσn (x)] = −
n−1 X
(c)k [cσk (x) + xσk0 (x)].
k=0
Problem 103. Apply Theorem 50, page 141, to the polynomials yn (x) defined by (1), page 135. You do not, of course, get Theorem 47, since that theorem depended upon the specific character of the exponential. Solution 103. On page 228 we find A(t) exp
−xt 1−t
=
∞ X
yn (x)tn .
n=0
In the notation of Theorem 50 we have A(t) = A(t), H(t) = −
t 1 =1− 1−t 1−t
from which H 0 (t) = −
∞ X −t 1 1 tH 0 (t) 1−t tn . = − = = 1 + , · (1 − t)2 H(t) (1 − t)2 −t 1−t n=1
so ∞ tA0 (t) X = αn tn+1 A(t) n=0
and ∞ X tH 0 (t) =1+ tn+1 , H(t) n=0
so that βn = 1 for n ≥ 0. Therefore, we may conclude that there exist constants αn such that xyn0 (x) − nyn (x) = −
n−1 X k=0
αk yn−1−k (x) − x
n−1 X k=0
0 1 · yn−1−k (x).
SOLUTIONS TO RAINVILLE’S “SPECIAL FUNCTIONS” (1960)
67
Problem 104. Apply Theorem 50 to the Laguerre polynomials through the genrating relation 914), page 135, to get (α) xDL(α) n (x) − nLn (x) = −
n−1 X
(α)
(α)
[(1 + α)Lk (x) + xDLk (x) = 0
k=0
for the Laguerre polynomials. Solution 104. From page 228 we have (1 − t)
−1−α
exp
−xt 1−t
=
∞ X
Ln(α) (x)tn .
n=0
We may therefore use Theorem 50 with A(t) = (1 − t)−1−α , H(t) = −
1 t =1− , 1−t 1−t
log A(t) = −(1 + α) log(1 − t), H 0 (t) = −
1 , (1 − t)2
∞ X tA0 (t) (1 + α)t = = (1 + α) tn+1 , A(t) 1−t n=0 ∞ ∞ X X t 1−t 1 tH 0 (t) n =− = = 1 + t = 1 + tn+1 . H(t) (1 − t)2 −t 1−t n=1 n=0
Hence αn = 1 + α, n ≥ 0; βn = 1, n ≥ 0. Therefore we obtain xDLn(α) (x) − nLn(α) (x) = −(1 + α)
n−1 X
(α)
Ln−1−k (x) − x
k=0
n−1 X
(α)
DLn−1−k (x)
k=0
or
(A)
xDLn(α) (x) − nLn(α) (x) = −
n−1 X
(α)
k=0
On page 230 we had (B)
DLn(α) (x) = −
n−1 X k=0
Using (B) with (A) we obtain
(α)
[(1 + α)Lk (x) + xDLk (x)].
(α)
Lk (x).
68 SOLUTIONS BY SYLVESTER J. PAGANO AND LEON HALL; EDITED BY TOM CUCHTA
xDLn(α) (x) − nLn(α) (x) = (1 + α)DLn(α) (x) + xD 2 Ln(α) (x), or [xD 2 + (1 + α − x)D + n]Ln(α) (x) = 0, as desired. Problem 105. The Humbert polynomials hn (x) are defined by (1 − 3xt + t3 )−p =
∞ X
hn (x)tn .
n=0
Use Theorem 52, page 144, to conclude that xh0n (x) − nhn (x) = h0n−2 (x). Solution 105. For hn (x) we have (1 − 3xt + t3 )−ν =
∞ X
hn (x)tn = 1 F0
n=0
ν;
3xt − t3 . −;
In the notation of Theorem 52 we have A(t) = 1, H(t) = 3t, g(t) = −t3 , ψ(t) = (1 − t)−ν . Then tA0 (t) tH 0 (t) 3t tg 0 (t) t(−3t2 ) = 0, = = 1, = = −t2 . A(t) H(t) 3t H(t) 3t Hence αn = 0, n ≥= 0; βn = 0, n ≥ 0; δ1 = −1, δn = 0 for n = 0, n ≥ 2. Therefore Theorem 52 yields xh0n (x) − nhn (x) = h0n−2 (x). Problem 106. For the yn (x) of Section 74 show that −xt F = A(t) exp 1−t satisfies the equation ∂F ∂F ∂F (1 − t)tA0 (t) −t = −t2 − F ∂x ∂t ∂t A(t) and draw what conclusions you can about yn (x). X ∞ −xt Solution 106. Let F = A(t) exp = yn (x)tn . 1−t n=0 Then ∂F t −xt =− A(t) exp ∂x 1−t 1−t x
SOLUTIONS TO RAINVILLE’S “SPECIAL FUNCTIONS” (1960)
∂F x A(t) exp =− ∂t (1 − t)2 Thus we obtain x
−xt 1−t
+ A0 (t) exp
−xt 1−t
69
.
∂F ∂F A0 (t) − t(1 − t) = −t(1 − t) F, ∂x ∂t A(t)
or x
∂F ∂F tA0 (t) ∂F −t = −t2 − (1 − t) F. ∂x ∂t ∂t A(t)
Now let ∞ tA0 (t) X αn tn+1 . = A(t) n=0
Then ∞ X
[xyn0 (x) n=0
− nyn (x)]t
n
=− =− =−
Hence we get
xyn0 (x)
y00 (x)
∞ X
n+1
−
yn (x)ct
n=1 ∞ X
nyn (x)tn+1 −
∞ X
!
αn t n=0 n ∞ X X
n+1
n=2
n=1 k=0
(n − 1)yn−1 (x)t −
=
yn (x)t
αn−k yk (x)tn+1 +
n=0 k=0 ∞ n−1 X X n
∞ X
!
n=0
n=0 ∞ X
0, xy10 (x)
∞ X
n
+
n ∞ X X
∞ X
! αn t
n+2
n=0
− nyn (x) = −(n − 1)yn−2 (x) −
αn−k yk (x)tn+2
n=0 k=0 ∞ n−2 X X n
αn−1−k yk (x)t +
αn−2−k yk (x)tn .
n=2 k=0
(αn−1−k − αn−2−k )yk (x) − α0 yn−1 (x).
k=0
We may also write x(1 − t)
∂F ∂F ∂F A0 (t) − t(1 − t) = −xt − t(1 − t) F, ∂x ∂t ∂x A(t)
or ∂F ∂F xt ∂F tA0 (t) −t =− − F, ∂x ∂t 1 − t ∂x A(t) from which it follows that x
∞ X
[xyn0 (x) n=0
− nyn (x)]y
n
= −x = −x
∞ X
! n+1
t
n=0 ∞ X n X n=0 k=0
Hence, for n ≥ 1,
∞ X
! yn0 (x)tn
n=0
yk0 (x)tn+1 −
∞ X n X n=0 k=0
−
∞ X
yn (x)t
n=0
− y1 (x) = −α0 y0 (x), and for n ≥ 2, n−2 X
!
! n+1
αn t
n=0
αn−k yk (t)tn=1 .
∞ X n=0
! n
yn (x)t
n
70 SOLUTIONS BY SYLVESTER J. PAGANO AND LEON HALL; EDITED BY TOM CUCHTA
xyn0 (x) − nyn (x) = −
n−1 X
[xyk0 (x) + αn−1−k yk (x)].
k=0
Problem 107. For polynomials an (x) defined by −c
(1 − t)
A
−xt 1−t
∞ X
=
an (x)tn
n=0
obtain what results you can parallel to those of Theorem 48, page 137. Solution 107. Let an (x) be defined by (1 − t)
(1)
−c
A
−xt 1−t
=
∞ X
an (x)tn .
n=0
Let (2)
A(u) =
∞ X
αn un .
n=0
Put −c
F = (1 − t)
A
−xt 1−t
.
Then ∂F = −t(1 − t)−c−1 A0 ∂x ∂F = c(1 − t)−c−1 A − x(1 − t)−c−2 A0 . ∂t Hence x
∂F ∂F − t(1 − t) = −ctF. ∂x ∂t
From which we may write (3)
x
∂F ∂F ∂F −t = −ctF − t2 . ∂x ∂t ∂t
From (3) we obtain ∞ X
[xa0n (x)
− nan (x)]t
n
= −c
n=0
=−
∞ X
n=0 ∞ X
an (x)t
n+1
−
∞ X
ngn (x)tn+1
n=0
[c + n − 1]gn−1 (x)tn .
n=1
Therefore, a00 (x) = 0 and for n ≥ 2, xa0n (x) − nan (x) = −(c + n − 1)an−1 (x). We may rewrite equation (3) in the form
SOLUTIONS TO RAINVILLE’S “SPECIAL FUNCTIONS” (1960)
x(1 − t)
71
∂F ∂F ∂F − t(1 − t) = −ctF − xt . ∂x ∂t ∂x
This leads to the identity
x
∂F ct xt ∂F ∂F −t =− F− ∂x ∂t 1−t 1 − t ∂x
from which we get ∞ X
[xa0n (x) − nan (x)]tn
∞ X
= −c
n=0
∞ X
! tn+1
n=0
= −c =−
∞ X n X
! an (x)tn
n=0 n+1
ak (x)t
−x
n=0 k=0 ∞ n−1 X X
−x
∞ X n X
∞ X
! tn+1
n=0
a0k (x)tn+1
k=0 k=0
[cak (x) + xa0k (x)]tn .
n=0 k=0
Hence we obtain the relation
xa0n (x) − nan (x) = −
n−1 X
[cak (x) + xa0k (x)].
k=0
In (1), let us put v = −
v 1 t . Then t = − and 1 − t = . 1−t 1−t 1−v
Hence (1) becomes
(1 − v)c A(xv) =
∞ X an (x)(−v)n , (1 − v)n n=0
or ∞ X
αn xn v n
=
n=0
= = =
∞ X an (x)(−1)n v n (1 − v)n+c n=0 ∞ X (−1)n an (x)(c)n+k v n+k n,k=0 ∞ X n X n=0 k=0 ∞ X n X n=0 k=0
k!(c)n (c)n (−1)n−k an−k (x) n v k!(c)n−k (−1)k (c)n ak (x) n v , (c)k (n − k)!
which can be put in various other forms. Next, from (1) we get
∞ X n=0
! a0n (x)tn
72 SOLUTIONS BY SYLVESTER J. PAGANO AND LEON HALL; EDITED BY TOM CUCHTA
∞ X
an (x)tn
=
n=0
= =
∞ X αn (−x)n tn (1 − t)n+c n=0 ∞ X αn (−1)n xn (c)n+k tn+k n,k=0 ∞ X n X n=0 k=0
(c)n k! (−1)n−k xn−k αn−k (c)n tn k1(c)n−k
or ∞ X
an (x)tn =
n=0
∞ X n X (−1)k xk αk (c)n tn n=0 k=0
(n − k)!(c)k
.
Therefore, an (x) =
n X (−1)k xk αk (c)n k=0
(n − k)!(c)k
.
12. Chapter 9 Solutions No problems in Chapter 9. 13. Chapter 10 Solutions Problem 108. Start with the defining relation for Pn (x) at the beginning of this chapter. Use the fact that p p 1 1 1 (1 − 2xt + t2 )− 2 = [1 − (x + x2 − 1)t]− 2 [1 − (x − x2 − 1)t]− 2 and thus derive the result √ √ n X ( 12 )k ( 12 )n−k (x + x2 − 1)n−k (x − x2 − 1)k Pn (x) = k!(n − k)! k=0
Solution 108. We know that 1
(1 − 2xt + t2 )− 2 =
∞ X
Pn (x)tn .
n=0
Now [1 − (x +
p
x2 − 1)t][1 − (x −
p
x2 − 1)t] = (1 − xt)2 − (x2 − 1)t2 = 1 − 2xt + t2 .
Hence ∞ X n=0
Pn (x)tn
√ 1 1 x2 − 1)t]− 2 [1 − (x − x2 − 1)t]− 2 ! ∞ ! √ √ ∞ X X ( 1 )n (x − x2 − 1)n tn ( 21 )n (x + x2 − 1)n tn 2 = n! n! n=0 n=0 √ √ ∞ X n 1 1 n−k 2 2 X ( 2 )k ( 2 )n−k (x + x − 1) (x − x − 1)k tn = . k!(n − k)! n=0 = [1 − (x +
k=0
√
SOLUTIONS TO RAINVILLE’S “SPECIAL FUNCTIONS” (1960)
73
Hence we obtain Pn (x) =
∞ X ( 1 )k ( 1 )n−k (x + 2
2
k=0
√
x2 − 1)n−k (x − k!(n − k)!
√
x2 − 1)k
.
Problem 109. Use the result in Exercise 108 to show that
Pn (x) =
( 21 )n (x
1 −n; ; 2 + x2 − 1)n 2 F1 n! 1 − n; 2
√
Solution 109. From Exercise 108 we get, since x +
(x − √
√
x2 − 1)2 .
x2 − 1 = (x −
√
x2 − 1)−1 ,
√ x2 − 1)n (x − x2 − 1)2k Pn (x) = k!n!(1 − 12 − n)k k=0 1 −n, ; √ 1 n 2 2 √ ( )n (x + x − 1) = 2 (x − x2 − 1)2 2 F1 n! 1 − n; 2 n X ( 1 )k (−n)k ( 1 )n (x + 2
√
2
Problem 110. In Section 93, equation (4), page 166, is 1 1 1 − n, − n + ; 2 2 2 ( 21 )n (2x)n 1 . Pn (x) = 2 F1 n! x2 1 − n; 2 We know from Section 34 that the 2 F1 equation has two linearly independent solution: 2 F1 (a, b; c; z)
and z 1−c F (a + 1 − c, b + 1 − c; 2 − c; z). Combine these facts to conclude that the differential equation (1 − t2 )y 00 − 2ty 0 + n(n + 1)y = 0 has the two linearly independent solutions y1 = Pn (t) and y2 = Qn (t), where Qn (t) is as given in equation (4) of Section 102. Solution 110. We know that n −n − 1 ; −2, 2 1 n ( 2 )n (2x) Pn (x) = 2 F1 n! 1 − n; 2
1 x2
,
74 SOLUTIONS BY SYLVESTER J. PAGANO AND LEON HALL; EDITED BY TOM CUCHTA
from which w = x−n Pn (x) is a solution of the 2 F1 differential equation with n n−1 1 a = − ,b = − , c = − n, z = x−2 . 2 2 2 Then consider 1 d2 w 3 dw n(n − 1) z(1 − z) 2 + −n− −n z − w=0 dz 2 2 dz 4 dz dx 1 and put z = x−2 . Then, = −2x−3 , = − x3 , dx dz 2 dw dx dw 1 3 dw = =− x dz dz dx 2 dx 2 2 d w 1 d w 3 1 dw 1 3 = x6 2 − x2 − x3 = x6 w00 + x5 w1 . 2 dz 4 dx 2 2 dx 4 4 The equation in w and x becomes 1 3 1 1 3 n(n − 1) x−2 (1−x−2 ) x6 w00 + x5 x−2 (1−x−2 )w0 − x3 −n− − n x−2 w0 − w = 0, 4 4 2 2 2 4 or x2 (x2 − 1)w00 + 3x(x2 − 1)w0 − x[(1 − 2n)x2 − (3 − 2n)]w0 − n(n − 1)w = 0, or x2 (x2 − 1)w00 + x[(2 + 2n)x2 − 2n]w0 − n(n − 1)w = 0. Now put w = x−n y w0 = x−n y 0 − nx−n−1 y w00 = x−n y 00 − 2nx−n−1 y 0 + n(n + 1)x−n−2 y or xn w = y xn w0 = y 0 − nx−1 y xn w00 = y 00 − 2nx−1 y 1 + n(n + 1)x−2 y and the differential equation becomes x2 (x2 −1)y 00 −2nx(x2 −1)y 0 +n(n+1)(x2 −1)y+x[(2+2n)x2 −2n]y 0 −n[(2+2n)x2 −2n]y−n(n−1)y = 0, or x2 (x2 − 1)y 00 + x[2x2 ]y 0 + [n(n + 1)x2 (1 − 2) − n2 − n + 2n2 − n2 + n]y = 0, or (1)
(1 − x2 )y 00 − 2xy 0 + n(n + 1)y = 0.
Since the differential equation for w has the two linearly independent solutions w1 = x−n Pn (x) and
SOLUTIONS TO RAINVILLE’S “SPECIAL FUNCTIONS” (1960)
n 1 n−1 1 + + n; − + + n; 2 2 2 2
75
1+n 2+n , ; 2 2 1− 12 +n −1−2n z =x w2 = z 2 F1 2 F1 3 3 + n; + n; 2 2 Then the differential equation (1) for y has the linearly independent solutions y1 = Pn (x) and 1+n 2+n , ; 2 2 1 −1−n , y2 = x 2 F1 x2 3 + n; 2 which is a constant multiple of Qn (x) as given on page 314. Note that the whole thing could have been done much more simply by using the properties of the Riemann P-symbol.
−
Problem 111. Show that ∞ X
3
[xPn0 (x) − nPn (x)]tn = t2 (1 − 2xt + t2 )− 2
n=0
and n
[2] ∞ X X
3
(2n − 4k + 1)Pn−2k (x)tn = (1 − 2xt + t2 )− 2 .
n=0 k=0
Thus conclude that [ n−2 2 ]
xPn0 (x)
− nPn (x) =
X
(2n − 4k − 3)Pn−2−2k (x).
k=0
Solution 111. We know that 1
(1 − 2xt + t2 )− 2 =
∞ X
Pn (x)tn .
n=0
Put 1
F = (1 − 2xt + t2 )− 2 . Then 3 ∂F = t(1 − 2xt + t2 )− 2 ∂x
and 3 ∂F = (x − t)(1 − 2xt + t2 )− 2 . ∂t
Hence
x−2 .
76 SOLUTIONS BY SYLVESTER J. PAGANO AND LEON HALL; EDITED BY TOM CUCHTA
3 ∂F ∂F −t = t2 (1 − 2xt + t2 )− 2 , ∂x ∂t from which it follows that
x
∞ X
(1)
3
[xPn0 (x) − nP (x)]tn = t2 (1 − 2xt + t2 )− 2 .
n=0
Next we form the series n
[2] ∞ X X
(2n − 4k + 1)Pn−2k (x)tn
∞ X
=
n=0 k=0
(2n + 1)Pn (x)tn+2k
n,k=0
= (1 − t2 )−1
∞ X
(2n + 1)Pn (x)tn .
n=0
By equation (5), page 271, we get 0 0 (2n + 1)Pn (x) = Pn+1 (x) − Pn−1 (x), n ≥ 1.
Hence n
[2] ∞ X X
∞ X
" (2n − 4k + 1)Pn−2k (x)t
n
0 Pn+1 (x)tn
n=0
n=2
Now
= 1 and
P00 (x)
= 0. Hence we have
n
[2] ∞ X X
" (2n − 4k + 1)Pn−2k (x)t
n
2 −1
= (1 − t )
n=0 k=0
∞ X
Pn0 (x)tn−1
n=0 ∂F 2 −1 1 ∂F −t = (1 − t ) t ∂x ∂x 1 ∂F = t ∂x 3 = (1 − 2xt + t2 )− 2 .
−
∞ X
# Pn0 (x)tn+1
n=0
Now we use (1) to conclude that n
∞ X
[xPn0 (x) n=0
# 0 Pn−1 (x)tn
− = (1 − t2 )−1 1 + n=1 n=1 # " ∞ ∞ X X 0 n+1 0 n−1 2 −1 Pn (x)t . PN (x)t − = (1 − t ) 1+
n=0 k=0
P10 (x)
∞ X
− nPn (x)]t
n
=
=
[2] ∞ X X
(2n − 4k + 1)Pn−2k (x)tn+2
n=0 k=0 n−2 ∞ [X 2 ] X
(2n − 4k − 3)Pn−2−2k (x)tn .
n=2 k=0
Hence, for n ≥ 2, [ n−2 2 ]
xPn0 (x) − nPn (x) =
X k=0
(2n − 4k − 3)Pn−2−2k (x).
SOLUTIONS TO RAINVILLE’S “SPECIAL FUNCTIONS” (1960)
77
There are many other methods of obtaining this result. Problem 112. Use Bateman’s generating function (3), page 163, with x = 0, t = 2y to conclude that 1 1 2 . 0 F1 (−; t; y)0 F1 (−; 1; −y) = 0 F3 −; 1, 1, ; − y 2 4 Solution 112. We have, from Bateman, 0 F1
−; 1;
t(x − 1) 2
0 F1
X ∞ t(x + 1) Pn (x)tn . −; 1; = 2 (n!)2 n=0
Use x = 0, t = 2y to obtain 0 F1 (−; 1; −y)0 F1 (−; 1; y)
=
∞ X 2n Pn (0)y n . (n!)2 n=0
(−1)k ( 12 )k . Therefore k! ∞ X 22k (−1)k ( 21 )k y 2k F (−; 1; −y) F (−; 1; y) = 0 1 0 1 k![(2k)!]2 k=0 ∞ X (−1)k 22k ( 21 )k y 2k = k!22k k!( 12 )k 22k k!( 21 )k k=0 ∞ X (−1)k y 2k = 22k (k!)3 ( 21 )k k=0 1 1 2 = 0 F3 −; 1, 1, ; − y , 2 4
now P2k+1 (0) = 0 and P2k (0) =
as desired. Problem 113. Use Brafman’s generating function, page 168, to conclude that 2 F1
c, 1 − c; 1−t−
√
1 + t2 2 F1
c, 1 − c; 1+t−
2 1; 1 1 1 1 1 1 c, c + , − c, 1 − c; 2 2 2 2 2 2 −t2 = 4 F3 1 1, 1, ; 2 Solution 113. Brafman’s generating relation, page 287, is 2 F1
c, 1 − c;
1;
1−t−p 2 F1 2 1 2
c, 1 − c; 1;
√
1 + t2 2
∞ 1 + t − p X (c)n (1 − c)n pn (x)tn , = (n!)2 2 n=0
with p = (1 − 2xt + t2 ) . We use x = 0. Note that P2n+1 (0) = 0 and P2n (0) = (−1)n ( 12 )n . Then n!
78 SOLUTIONS BY SYLVESTER J. PAGANO AND LEON HALL; EDITED BY TOM CUCHTA
2 F1
c, 1 − c; 1−t−
1;
√
c, 1 − c;
1 + t2 2 F1 2
∞ X (c)2n (1 − c)2n (−1)n ( 21 )n t2n 1+t−p = n![(2n)!]2 2 n=0
1;
∞ 1−c 2−c n 2n X ( 2c )n ( c+1 2 )n ( 2 )n ( 2 )n (−1) t 1 n!n!( 2 )n n! n=0 c c+1 1−c 2−c , , , ; 2 2 2 2 −t2 = 4 F3 , 1 1, 1, ; 2
=
as desired. Problem 114. Use equation (5), page 168, to obtain the results sinn βPn (sin β) =
n X
(−1)k Cn,k cosk (β)Pk (cos β),
k=0
Pn (x) =
n X
(−1)k Cn,k (2x)n−k Pk (x),
k=0
Pn (1 − 2x2 ) =
n X
(−2x)k Cn,k Pk (x).
k=0
Solution 114. We are given (1)
Pn (cos α) =
First use α = β − Thus (1) leads to
sin α sin β
n X n
Cn,k
k=0
sin(β − α) sin α
n−k Pk (cos β).
π . Then cos α = sin β, sin α = − cos β, sin(β − α) = 1. 2
n X n−k n cos β 1 Pn (sin β) = − = Cn,k Pk (cos β), sin β − cos β k=0
or (2)
(sin β)n Pn (sin β) =
n X
(−1)k Cn,k cosk βPk (cos β).
k=0
Next let us use β = −α in (1). We thus obtain n X n−k n sin α −2 sin α cos α Pn (cos α) = Cn,k Pk (cos α) − sin α sin α k=0 n X (−1)n−k Cn,k (2 cos α)n−k Pk (cos α) = (−1)n
=
n X
k=0
(−1)k Cn,k (2 cos α)n−k Pk (cos α).
k=0
SOLUTIONS TO RAINVILLE’S “SPECIAL FUNCTIONS” (1960)
79
With cos α = x we get (3)
Pn (x) =
n X
(−1)k Cn,k (2x)n−k Pk (x).
k=0
Finally put α = 2β in (1). The result is Pn (cos(2β)) =
2 sin β cos β sin β
n X n
Cn,k
k=0
− sin β 2 sin β cos β
n−k Pk (cos β),
or Pn (cos 2β)
= (2 cos β)n =
n X
n X
(−1)n−k Cn,k (2 cos β)k−n Pk (cos β)
k=0
(−1)n−k Cn,k (2 cos β)k Pk (cos β).
k=0
Put cos β = x. Then cos(2β) = 2 cos2 (β) − 1 = 2x2 − 1 and Pn (2x2 − 1) = (−1)n Pn (1 − 2x2 ). Hence we have Pn (1 − 2x2 ) =
(4)
n X
(−1)k Cn,k (2x)k Pk (x).
k=0
Problem 115. Use the technique of Section 96 to derive other generating function relations for Pn (x). For instance, obtain the results ∞ X n=0
1 F2 (−n; 1, 1; y)Pn (x)t
n
−;
−yt(x − t − ρ) 0 F1 2 2ρ
= ρ−1 0 F1 1;
−;
−yt(x − t + ρ) 2ρ2 1;
1 2
in which ρ = (1 − 2xt + t2 ) , and ∞ X
2 F1 (−n, c; 1; y)Pn (x)t
n
n=0
1 1 1 c, c + ; 2 = ρ2c−1 (ρ2 + xyt − yt2 )−c 2 F1 2 2
. y 2 t2 (x2 − 1) 2 2 2 (ρ + xyt − yt )
Also sum the series ∞ X n=0
3 F2 (−n, c, 1
− c; 1, 1; y)Pn (x)tn .
80 SOLUTIONS BY SYLVESTER J. PAGANO AND LEON HALL; EDITED BY TOM CUCHTA
Solution 115. We start with
p−n−1 Pn
(1)
x−t p
=
∞ X (n + k)!Pn+k (x)tk
k!n!
k=0
1
, p = (1 − 2xt + t2 ) 2 .
In the generating relation 0 F1
t(x − 1) −; 1; 2
we replace x by
X ∞ t(x + 1) Pn (x)tn , = 0 F1 −; 1; 2 (n!)2 n=0
x−t −yt and t by to get p p
∞ n n X (−1)n p−n−1 Pn ( x−t −yt x − t −yt x − t p )y t F −1 −; 1; + 1 = p−1 0 F1 −; 1; 0 1 2p p 2p p (n!)2 n=0 ∞ X (−1)n (n + k)!Pn+k (x)y n tn+k = k!n!(n!)2 n,k=0 ∞ n X X n!(−1)n−k Pn (x)y n−k tn = k![(n − k)!]3 n=0 k=0 n ∞ X X (−1)k n!y k = P (x)tn , 3 (n − k)! n (k!) n=0 k=0
or X ∞ −n; −yt(x − t + p) −yt(x − t − p) F −; 1; = F p−1 0 F1 −; 1; 0 1 1 2 2p2 2p2 n=0 1, 1; Next let us apply the same process to the known generating relation c c+1 , ; 2 2 −c (1 − xt) 2 F1
∞ X (c)n Pn (x)tn t2 (x2 − 1) . = n! (1 − xt)2 n=0
1; We replace x by
x−t −yt , t by and note that (1 − xt) becomes p p 1+
yt(x − t) = p−2 [p2 + xyt − yt2 ] p2
while t2 (x2 − 1) becomes y 2 t2 p2
"
x−t p
2
We thus obtain
# −1 =
y 2 t2 2 y 2 t2 (x2 − 1) 2 2 [x − 2xt + t − t + 2xt − t ] = . p4 p4
y Pn (x)tn .
SOLUTIONS TO RAINVILLE’S “SPECIAL FUNCTIONS” (1960)
c c+1 , ; 2 p2c−1 [p2 + xyt − yt2 ]−c 2 F1 2
81
2 2
2
y t (x − 1) (p2 + xyt − yt2 )2
∞ n n X (−1)n (c)n p−n−1 Pn ( x−t p )y t = n! n=0
= = =
∞ X (−1)n (c)n (n + k)!Pn+k (x)y n tn+k k!n!n! n,k=0 ∞ X n n−k X (−1) (c)n−k n!Pn (x)y n−k tn n=0 k=0 ∞ X n X n=0 k=0
k![(n − k)!]2 k
(−1) (c)k n!y k Pn (x)tn . (k!)2 (n − k)!
We have thus obtained c c+1 , ; 2 p2c−1 [p2 +xyt−yt2 ]−c 2 F1 2
∞ −n, c; X = 2 F1 n=0 1;
y 2 t2 (x2 − 1) 2 (p + xyt − yt2 )2
y Pn (x)tn .
Finally let us sum the series ∞ X
3 F2 (−n, c, 1
− c; 1, 1; y)Pn (x)tn
=
n=0
= =
∞ X n X (−1)k n!(c)k (1 − c)k Pn (x)y k tn
(k!)3 (n − k)!
n=0 k=0 ∞ X
(−1)k (n + k)!(c)k (1 − c)k Pn+k (x)y k tn+k (k!)3 n! n,k=0 ∞ ∞ X X (n + k)!Pn+k (x)tn (−1)k (c)k (1 − c)k (yt)k (k!)2
k!n!
k=0 n=0
With the aid of (1) we get ∞ X
3 F2 (−n, c, 1
n=0 ∞ X
=
k=0
= p−1
− c; 1, 1; y)Pn (x)tn
k k p−k−1 Pk ( x−t p )(−1) (c)k (1 − c)k (yt)
(k!)2 yt k ∞ X (c)k (1 − c)k Pk ( x−t p )(− p ) k=0
(k!)2 r 2 2 2y(x−t)t 1+ yt 1+ + yp2t p − p2
= p−1 2 F1
c, 1 − c; 2
1 c, 1 − c; p+yt−
= p−1 2 F1
√
" 2 F1
1−2xt(1−y)+(1−y)2 t2 2p
c, 1 − c; 1 1− yt p −p
1 2
Problem 116. With ρ = (1 − 2xt + t2 ) , show that
p2 +2yt(x−t)+y 2 t2 2
c, 1 − c; p−yt−
2 F1 1;
1;
√
#
√
1−2xt(1−y)+t2 (1−y)2 2
.
82 SOLUTIONS BY SYLVESTER J. PAGANO AND LEON HALL; EDITED BY TOM CUCHTA
ρ n Pn
1 − xt ρ
=
n X
(−1)k Cn,k tk Pk (x).
k=0
Solution 116. Consider ∞ X
n
p Pn
n=0
1 − xt p
yn
1
2 2 −2 = [1 − 2 1−xt p yp + y p ] 1
= [1 − 2y(1 − xt) + y 2 p2 ]− 2 1 = [1 − 2y + 2xyt + y 2 − 2xy 2 t + y 2 t2 ]− 2 1 = [(1 − y)2 + 2xyt(1 − y) + y 2 t2 ]− 2 −yt −yt 2 − 12 = (1 − y)−1 [1 − 2x( 1−y ) + ( 1−y ) ] ∞ k k k X (−1) Pk (x)y t = (1 − y)k+1 k=0 ∞ X (−1)k Pk (x)(n + k)!tk y n+k = n!k! n,k=0 n ∞ X X (−1)k Pk (x)n!tk y n = k!(n − k)! n=0 k=0 ∞ X n X = (−1)k Cn,k Pk (x)tk y n . n=0 k=0
It follows that pn Pn
1 − xt p
=
n X
(−1)k Cn,k tk Pk (x)
k=0
as desired. Problem 117. With the aid of the result in Example 7 pg. 31, show that
1
Z
−1
−n, n + 1, β;
(1 + x)α−1 (1 − x)β−1 Pn (x)dx = 2α+β−1 B(α, β)3 F2
1 . 1, α + β;
Investigate the three special cases α = 1, β = 1, α + β = n + 1. Solution 117. We know that 1−v Pn (x) = 2 F1 −n, n + 1; 1; . 2 Then 1
Z
α−1
(1+x)
β−1
(1−x)
Z
1
Pn (x)dx =
−1
(1+x)α−1 (1−x)β−1
−1
n X (−n)k (n + 1)k (1 − x)k k=0
2k (k!)2
But, by Ex.7, page ? we have Z
1
−1
(1 + x)α−1 (1 − x)β+k−1 dx = 2α+β+k−1 B(α, β + k) =
2α+β+k−1 Γ(α)Γ(β + k) . Γ(α + β + k)
dx.
SOLUTIONS TO RAINVILLE’S “SPECIAL FUNCTIONS” (1960)
83
Hence 1
Z
(1 + x)α−1 (1 − x)β−1 Pn (x)dx
=
−1
n X (−n)k (n + 1)k 2α+β+k−1 Γ(α)Γ(β + k) k=0 α+β−1
=
2
2k (k!)2 P (α + β + k) n Γ(α)Γ(β) X (−n)k (n + 1)k (β)k
Γ(α + β)
k=0
(k!)2 (α + β)k
.
Therefore Z
1
(A)
α−1
(1+x)
β−1
(1−x)
α+β−1
Pn (x)dx = 2
−1
−n, n + 1, β;
1 .
B(α, β)3 F2 1, α + β;
Let use choose α = 1 in (A). We get −n, n + 1, β; β 2 Γ(1)Γ(β) F (1 − x)β−1 Pn (x)dx = Γ(β + 1) −1 1, 1 + β;
Z
1
1 .
We now turn to Theorem 30 with a = 1, b = 1 − β. To see that −n, 1 + n, β; (1 − β)n 1 = . 3 F2 (1 + β)n 1 + β, 1; Thus we get Z
1
(1) −1
(1 − x)β−1 Pn (x)dx =
2β (1 − β)n 2β (1 − β)n − . β (1 + β)n (β)n+1
Next choose β = 1 in equation (A). We thus get Z 1 −n, n + 1, 1; Γ(α)Γ(1) F (1 + x)α−1 Pn (x)dx = 2α Γ(α + 1) −1 1, α +1; −n, n + 1; 2α 1 = 2 F1 α α + 1; 2α Γ(α + 1)Γ(α) = α Γ(α + 1 + n)Γ(α − n) 2α (−1)n (1 − α)n = α(1 + α)n (−1)n 2α (1 − α)n = . (α)n+1 Finally, we choose α + β = n + 1 in (A) and obtain
1
−n, n + 1, n + 1 − α; Γ(α)Γ(n + 1 − α) (1+x)α−1 (1−x)n−α Pn (x)dx = 2n F Γ(n + 1) −1 1, n + 1;
Z
1
Hence
1 .
84 SOLUTIONS BY SYLVESTER J. PAGANO AND LEON HALL; EDITED BY TOM CUCHTA
Z
−n, n + 1 − α; 2n Γ(α)Γ(1 − α)(1 − α)n 1 = 2 F1 n! 1; 2n Γ(α)Γ(1 − α)(1 − α)n (−1)n (1 + 1 − α − 1)n = , n! (1)n
1
(1 + x)α−1 (1 − x)n−α Pn (x)dx
−1
in which we used Ex.5, page 119. Therefore we have Z
1
(1 + x)α−1 (1 − x)n−α Pn (x)dx =
−1
(−1)n 2n π(1 − α)n (1 − α)n . sin(πα)(n!)2
Problem 118. Obtain from equation (5), page 168, the result ! r n X n 1+x −n 2 2 (1 + x) Pn Cn,k Pk (x) =2 2 k=0
and use it to evaluate the integral Z
1
r
n 2
(1 + x) Pn −1
1+x 2
! Pm (x)dx.
Solution 118. We put β = 2α and cos β = x in the known relation (1)
Pn (cos α) =
sin α sin β
n X n
Cn,k
k=0
sin(β − α) sin α
n−k Pk (cos β)
to get Pn
β cos 2
1 2 cos( β2 )
=
!n
n X
Cn,α Pk (cos β),
k=0
or r
n 2
(2)
[1 + x] Pn
1+x 2
! n
= 2− 2
n X
Cn,k Pk (x).
k=0
Then Z
1
B=
r
n 2
(1 + x) Pn −1
1+x 2
! n
Pm (x)dx = 2− 2
n X
Z
k=0
If n < m, then k < m and β = 0. If n ≥ m, B
n
= 21− 2
1
Cn,k
n X (−1)k (−n)k ( 1 )k 2
k!( 32 )k k=0 1 −n, ; 2 n −1 = 21− 2 2 F1 . 3 ; 2
Pk (x)Pm (x)dx. −1
SOLUTIONS TO RAINVILLE’S “SPECIAL FUNCTIONS” (1960)
85
Problem 119. Evaluate 1
Z 1 1 n x Pn−2k (x)dx, 2 −1 0 and check your result by means of Theorem 65, page 181. Thus show that Z
xn Pn−2k (x)dx =
1
Z
xn Pn−2k (x)dx =
0
n! 2n k!( 32 )n−k
and, equivalently, that 1
Z
xn+2k Pn (x)dx =
0
(n + 2k)! 2n+2k k!( 32 )n+k
.
Solution 119. Consider 1
Z
1 x Pn−2k (x)dx = 2 n
A= 0
Z
1
xn Pn−2k (x)dx.
−1
We have Z A= 0
1
−n + 2k, n − 2k + 1;
1−x dx 2
x n 2 F1
1; and we apply Theorem 37 with α = n + 1, β = 1, t = 1, p = 2, q = 1, a1 = 1 −n + 2k, a2 = n − 2k + 1, b1 = 1, c = , k = 0s = 1. The result is 2 −n + 2k, n − 2k + 1, 1; Z 1 1 xn P n − 2k(x)dx = B(n + 1, 1)F . 2 0 1, n + 2; By Example 3, page 39 a, 1 − a; 2 F1
21−c Γ(c)Γ( 12 ) 1 = c−a+1 , Γ( c+a ) 2 2 )Γ( 2
c; which we use with a = −n + 2k, 1 − a = 1 + n − 2k, c = n + 2. Then Γ(n + 1)Γ(1) 2−n−1 Γ(n + 2)Γ( 12 ) 2n−2k+3 Γ(n + 2) Γ( 2k+2 ) 0 2 )Γ( 2 3 Γ( 2 ) n! = n 2 k!Γ(n − k + 32 ) n! = n 3 . 2 k!( 2 )n−k Recall that we found in Theorem 65 that Z
1
xn Pn−2k (x)dx
=
n
[2] (2n + 4s + 1)Pn−2s (x) n! X . x = n 2 s=0 s!( 32 )n−s n
86 SOLUTIONS BY SYLVESTER J. PAGANO AND LEON HALL; EDITED BY TOM CUCHTA
We may then write 1
Z
1 = 2
n
x Pn−2k (x)dx 0
=
Z
1
−1
xn Pn−2k (x)dx n
[2] (2n − 4s + 1) n! X
2n+1
R1 −1
Pn−2k (x)Pn−2s (x)dx
s!( 32 )n−s
s=0
. .
Only the term with s = k remains. We get Z
1
xn Pn−2k (x)dx =
0
n! 2n − 4k + 1 2 n! . = n 3 2n+1 k!( 32 )n−k 2n − 4k + 1 2 k!( 2 )n−k
Thus a check on the earlier result. A simple change from n to (n + 2k) yields 1
Z
xn+2k Pn (x)dx =
0
(n + 2k)! . 2n+2k k!( 32 )n+k
Problem 120. Use the formula (5), page 104, to obtain the result n t 1 xn (1 − x)n dx = Q , n 2 )n+1 (1 − x 2 t 0 where Qn (t) is the function given in (4), page 182. t
Z
Solution 120. Consider Z t xn (t − x)n dx = xn (t − x)n 1 F0 (n + 1; −; x2 )dx. 2 n+1 0 (1 − x ) 0 We may use Theorem 37 with α = n + 1, β = n + 1, p = 1, q = 0, a1 = n + 1, c = 1, k = 2, s = 0 to get Z
1
B1 =
n + 1,
B1 = B(n + 1, n + 1)t2n+1 F
n+1 n+2 , ; 2 2
2n + 2 2n + 3 , ; 2 2
t2 ,
or
Z 0
t
n+1 n+2 , ; 2 2 Γ(n + 1)Γ(n + 1) 2n+1 F = t Γ(2n + 2) 3 n+ ; 2 (n!)2 t2n+1 2n ( 1t )n+1 ( 32 )n 1 = Qn , (2n + 1)! n! t
xn (t − x)n dx (1 − x2 )n+1
t2
in terms of the !n (t) of page 182. We thus obtain, since n!22n ( 32 )n = (2n + 1)!, Z 0
t
xn (t − x)n dx = (1 − x2 )n+1
n t 1 Qn . 2 t
SOLUTIONS TO RAINVILLE’S “SPECIAL FUNCTIONS” (1960)
87
Problem 121. Show that −n, −n;
Pn (x) =
2n ( 12 )n (x
− 1)
n
F
n!
−2n;
2 . 1−x
Solution 121. We know that X n k (−1)k (n + k)!( 1−x 1−x 2 ) = . Pn (x) = 2 F1 −n, n + 1; 1; 2 2 (k!) (n − k)! k=0
Reversing the order of the terms, we get Pn (x)
= =
n X (−1)n−k (2n − k)!( 1−x )n−k 2
k![(n − k)!]2 k=0 n X n 2 k (2n)!(−n)k (−n)k ( 1−x ) x−1
(−2n)k k!(n!)2 −n, −n; 2n ( 21 )n (x − 1)n 2 = . 2 F1 n! 1−x −2n; 2
k=0
Problem 122. Show that −n, −n;
Pn (x) =
2n ( 21 )n (x n!
+ 1)
n
F −2n;
2 . 1+x
Solution 122. Since Pn (x) = (−1)n Pn (−x), it follows from the result in Exercise 121 that −n, −n; 2n ( 21 )n (x + 1)n 2 Pn (x) = . 2 F1 n! 1+x −2n; Problem 123. Show that for |t| sufficiently small ∞ X
3
(2n + 1)Pn (x)tn = (1 − t2 )(1 − 2xt + t2 )− 2 .
n=0
Solution 123. From 1
(1 − 2xt + t2 )− 2 =
∞ X
Pn (x)tn
n=0
we obtain 1
t(1 − 2xt2 + t4 )− 2 =
∞ X
Pn (x)t2n+1 .
n=0
We then differentiate both members with respect to t to get
88 SOLUTIONS BY SYLVESTER J. PAGANO AND LEON HALL; EDITED BY TOM CUCHTA
∞ X
3
1
(2n + 1)Pn (x)t2n = (1 − 2xt2 + t4 )− 2 + t(2xt − 2t3 )(1 − 2xt2 + t4 )− 2 .
n=0
Now replace t by t2 : ∞ X
3
1
(2n + 1)Pn (x)tn
= (1 − 2xt + t2 )− 2 + t(2x − 2t)(1 − 2xt + t2 )− 2
n=0
3
= (1 − 2xt + t2 )− 2 (1 − 2xt + t2 + 2xt − 2t2 ) 3 = (1 − t2 )(1 − 2xt + t2 )− 2 , as desired. Problem 124. Use Theorem 48, page 137, with c = 1, x replaced by γn =
( 12 )n to arrive at n! (1 − x)n = 2n (n!)2
n X (−1)k (2k + 1)Pk (x) k=0
(n − k)!(n + k + 1)!
1 (1 − x) and 2
.
Solution 124. We know the generating relation ∞ X
Pn (x)tn
1
= (1 − 2xt + t2 )− 2
n=0
1
= [(1 − t)2 − 2t(x − 1)]− 2 − 1 2t(x − 1) 2 −1 = (1 − t) 1− 2 (1 − t) 1 2t(x − 1) −1 = (1 − t) 1 F0 ; −; . 2 (1 − t)2 In the above, put x = 1 − 2v to get 1 ; 2 −1 (1 − t) 1 F0 −;
∞ X −4vt = Pn (1 − 2v)tn . 2 (1 − t) n=0
We may now use Theorem 48 with γn =
( 12 )n and c = 1. This yields n!
n
vn =
22n n!( 12 )n n! X (−1)k (2k + 1)Pk (1 − 2v) , 22n ( 12 )n k=0 (n − k)!(n + k + 1)!
or n
n
2
(1 − x) = 2 (n!)
n X (−1)k (2k + 1)Pk (x) k=0
as desired.
(n − k)!(n + k + 1)!
,
SOLUTIONS TO RAINVILLE’S “SPECIAL FUNCTIONS” (1960)
89
Problem 125. Use Theorem 48, page 137, to show that (1 − x)Pn0 (x) + nPn (x)
0 = nPn−1 (x) − (1 − x)Pn−1 (x) n−1 n−1 X X = Pk (x) − 2(1 − x) Pk0 (x)
=
k=0 n−1 X
k=0
(−1)
n−k+1
(1 + 2k)Pk (x).
k=0
Solution 125. Using the transformation in Exercise 124 above we obtain v
d Pn (1 − 2v) − nPn (1 − 2v) dv
d Pn−1 (1 − 2v) dv n−1 X d =− [Pk (1 − 2v) + 2v Pk (1 − 2v)] dv k=0 n−1 X (−1)n−k (2k + 1)Pk (1 − 2v). = = −nPn−1 (1 − 2v) − v
k=0
dy dy Since x = 1 − 2v, we know that v = −(1 − x) . dv dx Hence the above relations become 0 (1 − x)Pn0 (x) + nPn (x) = nPn−1 (x) − (1 − x)Pn−1 (x),
(1 − x)Pn0 (x) + nPn (x) =
n−1 X
[Pk (x) − 2(1 − x)Pk0 (x)],
k=0
(1 − x)Pn0 (x) + nPn (x) =
n−1 X
(−1)n−k+1 (2k + 1)Pk (x).
k=0
Problem 126. Use Rodrigues’ formula, page 162, and successive integrations by parts to derive the orthogonality property for Pn (x) and to show that Z
1
−1
Pn2 (x)dx =
2 . 2n + 1
1 Solution 126. We know Pn (x) = n D n (x2 − 1)n . 2 n! Then Z
1
Pn (x)Pm (x)dx −1
Z 1 1 = n [D n (x2 − 1)n ]Pm (x)dx 2 n! −1 Z 1 1 1 n−1 2 1 = n {D (x − 1)n }Pm (x) −1 − n [D n−1 (x2 − 1)n ][DPm (x)]dx 2 n! 2 n! −1 = ... Z (−1)n 1 0 2 = n [D (x − 1)n ][D n Pm (x)]dx. 2 n! −1
If n > m, D n Pm (x) ≡ 0. Also the original integral is symmetric in n and m. Hence
90 SOLUTIONS BY SYLVESTER J. PAGANO AND LEON HALL; EDITED BY TOM CUCHTA
1
Z
Pn (x)Pm (x)dx = 0, m 6= n. −1
But we also have Z
1
Pn2 (x)dx =
−1
Now Pn (x) =
2n ( 21 )n xn n!
(−1)n 2n n!
Z
1
(x2 − 1)n [D n Pn (x)]dx.
−1
1 . 2 n
+ πn−1 , so D n Pn (x) = 2n
Therefore Z
1
Pn2 (x)dx =
−1
( 21 )n n!
Z
1
(1 − x2 )n dx =
−1
( 12 )n n!
Z
1
(1 − x)n (1 + x)n dx.
−1
By Example 7, page 51, we have Z
1
Pn2 (x)dx =
−1
22n+1 ( 21 )n Γ(n + 1)Γ(n + 1) ( 12 )n n+1+n+1−1 2 B(n + 1, n + 1) = . n! n!Γ(2n + 2)
Hence Z
1
Pn2 (x)dx
−1
22n+1 ( 21 )n n! (2n + 1)! 22n+1 ( 21 )n n! = 2n 3 2 n!( 2 )n 2 · 21 = n + 12 2 . = 2n + 1 =
n
1
Problem 127. Show that the polynomial yn (x) = (n!)−1 (1 − x2 ) 2 Pn ((1 − x2 )− 2 ) has the generating relation X ∞ 1 2 2 e 0 F1 −; t; x t = yn (x)tn 4 n=0 t
and that Theorem 45, page 133, is applicable to this yn (x). Translate the result into a property of Pn (x), obtaining equation (7), page 159. Solution 127. Consider the polynomial n
(1 − x2 ) 2 yn (x) = Pn n! We find that
1 √ 1 − x2
.
SOLUTIONS TO RAINVILLE’S “SPECIAL FUNCTIONS” (1960)
∞ X
yn (x)tn
n=0
=
∞ P X n
91
√ (t 1 − x2 )n
√ 1 1−x2
n!
n=0
−;
1 t2 (1 − x2 ) −1 1 − x2 4
= et 0 F1
1; −;
t 2 x2 . 4
= et 0 F1 1;
Problem 128. Let the polynomials wn (x) be defined by xt
2
2
e ψ[t (x − 1)] =
∞ X
wn (x)tn ,
n=0
with ψ(u) =
∞ X
γn un .
n=0
Show that ∞ X
n
(c)n wn (x)t = (1 − xt)
−c
n=0
∞ X
(c)2kγk
k=0
t2 (x2 − 1) (1 − xt)2
k
and thus obtain a result parallel to that in Theorem 46, page 134. Apply your new theorem to Legendre polynomials to derive equation (2), page 164. Solution 128. 14. Chapter 11 Solutions Problem 129. Use the fact that exp(2xt − t2 ) = exp(2xt − x2 t2 ) exp[t2 (x2 + 1)] to obtain the expansion n
Hn (x) =
[2] X n!Hn−2k (1)xn−2k (x2 + 1)k k=0
k!(n − 2k)!
.
Solution 129. From exp(2xt − t2 ) = exp(2xt − x2 t2 )exp[t2 (x2 + 1)] we obtain ∞ X
=
n=0
∞ X Hn (1)(xt)n n! n=0 n
=
!
∞ X (x2 + 1)n t2n n! n=0
[2] ∞ X X Hn−2k (1)(x)n−2k (x2 + 1)k tn n=0 k=0
k!(n − 2k)!
.
!
92 SOLUTIONS BY SYLVESTER J. PAGANO AND LEON HALL; EDITED BY TOM CUCHTA
It follows that n
[2] X n!Hn−2k (1)xn−2k (x2 + 1)k
Hn (x) =
k!(n − 2k)!
k=0
.
Problem 130. Use the expansion of xn in a series of Hermite polynomials to show that √ Z ∞ π 2 n −2k exp(−x )x Hn−2k (x)dx = 2 n! . k! −∞ Note in particular the special case k = 0. Solution 130. We know that n
[2] X n!Hn−2s (x) . x = 2n s!(n − 2s)! s=0 n
Then Z
n
∞ 2
n
exp(−x )x Hn−2k (x)dx = −∞
[2] X s=0
n! n s!(n − 2k)!
Z
∞
2
e−x Hn−2k (x)Hn−2s (x)dx.
−∞
The integrals involved on the right are zero except for s = k. Hence Z ∞ 2 n! 2 e−x Hn−2k (x)dx n 2 k!(n − 2k)! ∞√ −∞ n!2n−2k (n − 2k)! π = n √2 k!(n − 2k)! n! π = 2k . 2 k! √ For k = 0 the right member becomes n! π. Z
∞
exp(−x2 )xn Hn−2k (x)dx
=
Problem 131. Use the integral evaluation in equation (4), page 192, to obtain the result Z
∞
exp(−x2 )H2k (x)H2n+1 (x)dx =
0
(−1)k+s 22k+2s ( 12 )k ( 32 )s . 2s + 1 − 2k
Solution 131. We know that Z ∞ 2 e−x Hn (x)Hm (x)dx = 0, m 6= n. −∞
If m and n are both odd or both even, the integrand above is an even function of x and we obtain Z ∞ 2 2 e−x Hn (x)Hm (x)dx = 0 0
for m ≡ n mod 2. Now consider
SOLUTIONS TO RAINVILLE’S “SPECIAL FUNCTIONS” (1960)
∞
Z
93
2
e−x H2n (x)H2s+1 (x)dx.
0
By equation (4), page 331, we get ∞
Z
h i∞ 2 0 = e−x H2n (x)H2s+1 (x) − 202n (x)H2s+1 (x)
2
e−x H2n (x)H2s+1 (x)dx
2(2n − 2s − 1)
0
0
0 0 = −H2n (0)H2s+1 (0) + H2n (0)H2s+1 (0) 0 = −H2n (0)H2s+1 (0).
0 Using the values of H2n (0) and H2s+1 (0) from page 323, we get ∞
Z
−(−1)n 22n ( 21 )n (−1)s 22s+1 ( 32 )s 2(2n − 2s − 1) (−1)n+s 22n+2s ( 12 )n ( 23 )s . = 2s + 1 − 2n
2
e−x H2n (x)H2s+1 (x)dx
=
0
Problem 132. By evaluating the integral on the right, using equation (2), page 187, ans term-by-term integration, show that (A)
Pn (x) =
Z
2 √
n! π
∞
exp(−t2 )tn Hn (xt)dt,
0
which is Curzon’s integral for Pn (x), equation (4), page 191. Solution 132. Consider the integral
2 √ n! π
Z
∞
−t2 n
e
t Hn (xt)dt
0
2 =√ π
n
∞
Z
e
−t2 n
t
0
k=0
[n 2]
=
X (−1)k (2k)n−2k k=0 [n 2]
=
[2] X (−1)k (2xt)n−2k dt
k!(n − 2k)!
k!(n − 2k)! Z ∞ 1 1 √ e−β β n−k− 2 dβ π 0
X (−1)k (2x)n−2k Γ(n − k + 1 ) 2 k!(n − 2k)! Γ( 12 )
k=0 [n 2]
=
X (−1)k ( 1 )n−k (2x)n−2k 2
k=0
k!(n − 2k)!
.
Hence 2 √ n! π
Z
∞
2
e−t tn Hn (xt)dt = Pn (x).
0
Problem 133. Let vn (x) denote the right member of equation (A) of Exercise 132. Prove (A) by showing that ∞ X n=0
1
vn (x)y n = (1 − 2xy + y 2 )− 2 .
94 SOLUTIONS BY SYLVESTER J. PAGANO AND LEON HALL; EDITED BY TOM CUCHTA
n! π
vn (x)y
n
n=0
2
e−t tn Hn (xt)dt.
0
Then ∞ X
∞
Z
2 √
Solution 133. Put vn (x) =
Z ∞ ∞ n n 2 X t Hn (xt)t 2 =√ e−t dt n! π 0 n=0 Z ∞ 2 2 2 2 e−t e2(xt)yt−y t dt =√ π Z0 ∞ 2 2 2 =√ e−t [1−2xy+y ] dt. π 0
p Use t 1 − 2xy + y 2 = β to get ∞ X
2 vn (x)y n = √ p π 1 − 2xy + y 2 n=0
∞
Z
2
1
e−β dβ = (1 − 2xy + y 2 )− 2 =
0
∞ X
Pn (x)y n .
n=0
Hence vn (x) = Pn (x). Problem 134. Evaluate the integral on the right in Hn (x) = 2n+1 exp(x2 )
(B)
Z
∞
exp(−t2 )tn+1 Pn
x
x t
dt
by using n
(2t)n Pn
x t
=
[2] X n!(x2 − t2 )k (2x)n−2k k=0
(k!)2 (n − 2k)!
derived from equation (1), page 164, and term-by-term integration to prove the validity of (B), which is equation (5), page 191. Solution 134. We know n
Pn (x) =
[2] X n!(x2 − 1)k xn−2k
22k (k!)2 (n − 2k)!
k=0
from (1), page 280. Then n
n
(2t) Pn
x t
=
[2] X n!(x2 − t2 )k (2x)n−2k k=0
(k!)2 (n − 2k)!
which leads to the result
2
n+1 x2
Z
n
∞
e
e
−t2 n+1
t
Pn
x
Put t − x2 = β. Then
x t
dt =
[2] X k=0
2
n!ex 2 (k!) (n − 2k)!
Z
∞
x
2
e−t (x2 −t2 )k (2x)n−2k (2tdt).
SOLUTIONS TO RAINVILLE’S “SPECIAL FUNCTIONS” (1960)
n+1 x2
2
Z
n
∞
e
95
e
−t2 n+1
t
Pn
[2] Z ∞ X n!(2x)n−2k dt = e−β (−1)k β k dβ t (k!)2 (n − 2k)! 0
x
x
k=0 [n 2]
=
X n!(−1)k (2x)n−2k k! (k!)2 (n − 2k)!
k=0 [n 2]
=
X n!(−1)k (2x)n−2k k!(n − 2k)!
k=0
Hence 2
2n+1 ex
Z
∞
2
e−t tn+1 Pn
x
x
t
dt = Hn (x).
Problem 135. Use the Rodrigues formula exp(−x2 )Hn (x) = (−1)n Dn exp(−x2 ); D =
d dx
an iteration integration by parts to show that Z
∞
exp(−x2 )Hn (x)Hm (x)dx =
0 √ 2n n! π
−∞ 2
; m 6= n ;m = n
2
Solution 135. We know Hn (x) = (−1)n ex D n e−x . Then Z
∞
2
e−x Hn (x)Hm (x)dx
∞
Z
2
[D n e−x ]Hm (x)dx ∞ Z hn o i∞ 2 = (−1)n D n−1 e−x Hm (x) + (−1)n+1 = (−1)n
−∞
∞
= ... = (−1)2n
Z
∞
∞
−∞
2
e−x [D n Hm (x)]dx.
−∞
Hence Z
∞
2
e−x Hn (x)Hm (x)dx = 0, m 6= n
−∞
and, since Hn (x) = 2n xn + πn−2 (x) (where πn−2 (x) denotes a polynomial of degree (n − 2)), Z
∞
−∞
2
e−x Hn2 (x)dx
Z
∞
2
e−x [D n Hn (x)]dx −∞ Z ∞ 2 n = 2 n! e−x dx √−∞ = 2n n! π. =
2
[D n−1 e−x ][DHm (x)]dx
96 SOLUTIONS BY SYLVESTER J. PAGANO AND LEON HALL; EDITED BY TOM CUCHTA
15. Chapter 12 Solutions Problem 136. Show that −1
H2n (x) = (−1)n 22n n!Ln 2 (x2 ), 1
H2n+1 (x) = (−1)n 22n+1 n!xLn2 (x2 ). n
Solution 136. From Hn (x) =
[2] X (−1)k n!(2x)n−2k k=0
H2n (x)
= =
we get
k!(n − 2k)!
n X (−1)k (2n)!(2x)2n−2k k=0 n X k=0 n X
k!(2n − 2k)! (−1)n−k (2n)!(2x)2k (2k)!(n − k)!
(−1)n (2n)!(−n)k x2k k!( 12 )k n! k=0 1 1 2 n 2k = (−1) 2 1 F1 −n; ; x 2 2 n 1 n! (− 21 ) 2 n 2n = (−1) 2 Ln (x ). 2 n ( 12 )n
=
Hence (− 12 )
H2n (x) = (−1)n 22n n!Ln
(x2 ).
In the same way
H2n+1 (x)
= =
n X (−1)k (2n + 1)!(2x)2n+1−2k k=0 n X k=0 n X
k!(2n + 1 − 2k)! n−k
(−1) (2n + 1)!(2x)2k+1 (n − k)!(2k + 1)!
(−1)n (−n)k 22n n!( 23 )n 22k+1 x2k+1 n!22k k!( 32 )k k=0 3 2 n 2n 3 = (−1) 2 ( 2 )n (2x)1 F1 −n; ; x 2 (1) = (−1)n 22n n!(2x)Ln 2 (x2 ). =
Hence (1)
H2n+1 (x) = (−1)n 22n+1 n!xLn 2 (x2 ). Problem 137. Use Theorem 65, page 181, and the method of Section 118 above to derive the result
SOLUTIONS TO RAINVILLE’S “SPECIAL FUNCTIONS” (1960)
1 1 − (n − k), − (n − k − 1); 2 2
Pn(α) (x) =
n X
F 2 3 3 k=0 1 1 + k, (1 + α + k), (2 + α + k); 2 2 2
97
1 4
(−1)k (1 + α)n (2k + 1)Pk (x) . 2k (n − k)!( 3 ) (1 + α) k 2 k
Solution 137. We know that [n]
2 X (2x)n (2n − 4k + 1)Pn−2k (x) . = n! k!( 32 )n−k
k=0
Then ∞ X
Ln(α) (x)tn
=
n=0
=
∞ X n X (−1)s (1 + α)n xs tn n=0 s=0 ∞ X
s!(n − s)!(1 + α)s
(−1)s (1 + α)n+s xs tn+s s!n!(1 + α)s n,s=0 s
=
[2] ∞ X X (−1)s (1 + α)n+s (2s − 4k + 1)Ps−2k (x)tn+s
2s n!(1 + α)s k!( 23 )s−k
n,s=0 k=0 ∞ X
(−1)s (1 + α)n+s+2k (2s + 1)Ps (x)tn+s+2k 2s+2k k!n!(1 + α)s+2k ( 23 )s+k n,k,s=0 ∞ k n+k+2s change letters X (−1) (1 + α)n+k+2s (2k + 1)Pk (x)t = 3 k+2s 2 s!n!(1 + α)k+2s ( 2 )k+s n,k,s=0
=
n
[2] ∞ X X (−1)k (1 + α)n+k (2k + 1)Pk (x)tn+k = s!(n − 2s)!(1 + α)k+2s ( 32 )k+s 2k+2s n,k=0 s=0 n
[2] ∞ X X
(−n)2s 2−2s (−1)k (1 + α)n+k (2k + 1)Pk (x)tn+k 3 s!(1 + α + k)2s ( 2 + k)s (1 + α)k ( 32 )k 2k n! n,k=0 s=0 n n−1 ; − ,− 2 2 ∞ k n+k X 1 (−1) (1 + α)n+k (2k + 1)Pk (x)t = F . 2 3 3 k 4 2 (1 + α)k ( 2 )k n! 3 n,k=0 1+α+k 2+α+k + k, , ; 2 2 2 =
We thus get ∞ X
Ln(α) (x)tn
n=0
Hence
∞ X n X
−
n−k n−k−1 ,− ; 2 2
= 2 F3 3 n=0 k=0 1+α+k 2+α+k + k, , ; 2 2 2
1 4
(−1)k (1 + α)n (2k + 1)Pk (x)tn . 2k (1 + α)k ( 32 )k (n − k)!
98 SOLUTIONS BY SYLVESTER J. PAGANO AND LEON HALL; EDITED BY TOM CUCHTA
Ln(α) (x)
−
n X
n−k n−k−1 ,− ; 2 2
= 2 F3 3 k=0 1+α+k 2+α+k + k, , ; 2 2 2
1 4
(−1)k (1 + α)n (2k + 1)Pk (x) 2k (1 + α) ( 3 ) (n − k)! . k 2 k
Problem 138. Use formula (4), page 194, and the method of Section 118 to derive the result
1 1 − (n − k), − (n − k − 1); 2 2 n X L(α) (x) = F 2 2 n 1 k=0 1 (1 + α + k), (2 + α + k); 2 2
1 4
(−1)k (1 + α)n Hk (x) k!(n − k)!2k (1 + α)k .
Solution 138. From
[n]
2 X Hn−2s (x) (2x)n = n! s!(n − 2s)! s=0
we obtain
∞ X
Ln(α) (x)tn
=
n=0
=
∞ X n X (−1)k (1 + α)n xk tn n=0 k=0 ∞ X n,k=0
k!(n − k)!(1 + α)k
(−1)k (1 + α)n+k xk tn+k k!n!(1 + α)k k
[2] ∞ X X (−1)k (1 + α)n+k Hk−2s (x)tn+k = n!(1 + α)k 2k s!(k − 2s)! n,k=0 s=0 ∞ X (−1)k+2s (1 + α)n+k+2s Hk (x)tn+k+2s = n!(1 + α)k+2s 2k+2s s!k!
=
n,k,s=0 [n ∞ X 2] X
n,k=0 s=0
Thus we obtain
(−1)k (1 + α)n+k Hk (x)tn+k . s!(n − 2s)!(1 + α)k+2s 2k+2s k!
SOLUTIONS TO RAINVILLE’S “SPECIAL FUNCTIONS” (1960)
∞ X
99
n
[2] ∞ X X
(−n)2s 2−2s (−1)k (1 + α)n+k Hk (x)tn+k s!(1 + α + k)2s n!(1 + α)k 2k k! n=0 n,k=0 s=0 n n−1 − ,− ; 2 2 ∞ k n+k X 1 (−1) (1 + α)n+k Hk (x)t = F 2 2 k k!n!2 (1 + α)k 4 1+α+k 2+α+k n,k=0 , ; 2 2 n−k n−k−1 − , − ; 2 2 ∞ X n k n X 1 (−1) (1 + α)n Hk (x)t . = 2 F2 k 4 2 k!(n − k)!(1 + α)k 1+α+k 2+α+k n=0 k=0 , ; 2 2 Problem 139. Use the results in Section 56, page 102, to show that Ln(α) (x)tn
=
Z
t
Ln [x(t − x)]dx = 0
(−1)n H2n+1 ( 2t ) . 22n ( 32 )n
Solution 139. We use Theorem 37 to evaluate t
Z
Ln [x(t − x)]dx =
t
Z
0
1 F1 (−n; 1; x(t
− x))dx.
0
In Theorem 37 we use α = 1, β = 1, p = 1, q = 1, a1 = −n, b1 1, c = 1 to get −n, 1, 1; Z t t2 Ln [x(t − x)]dx = B(1, 1)t3 F3 4 0 2 3 1, , ; 2 2 3 t2 = t1 F1 −n; ; 2 4 n! 1 2 ( 12 ) = 3 Ln t . 4 ( 2 )n
= 1, k = 1, s =
By Exercise 136 above we have (1) Ln 2
1 2 t 4
=
(−1)n H2n+1 ( 12 t) . 22n tn!
Hence Z 0
t
Ln [x(t − x)]dx =
(−1)n H2n+1 22n ( 32 )n
1 t . 2
Problem 140. Use the results in Section 56, page 102, to show that Z 0
t
p H2n ( x(t − x))dx 1 1 2 p = (−1)n π22n Ln t . 2 n 4 x(t − x)
100SOLUTIONS BY SYLVESTER J. PAGANO AND LEON HALL; EDITED BY TOM CUCHTA
Solution 140. Since H2n (x) = (−1)n 22n
1 1 2 F −n; ; x , 1 1 2 n 2
by Exercise 136, we may write t
Z
Z t p p 1 n 2n 1 − 21 − 21 H2n ( x(t − x)) x(t − x)dx = (−1) 2 ( )n x (t−x) 1 F1 −n; ; x(t − x) dx. 2 2 0
0
We then apply Theorem 37 with α = 1 , c = 1, k = 1, s = 1, to get 2
1 1 , β = , p = 1, q = 1, a1 = −n, b1 = 2 2
1 1 −n, , ; 2 2 1 1 0 = (−1)n 22n ( 12 )n B , t 3 F3 2 2 1 1 2 , , ; 2 2 2 −n; Γ( 12 )Γ( 12 ) t2 n 2n 1 = (−1) 2 ( 2 )n 1 F1 Γ(1) 4 1; 1 = (−1)n 22n ( 12 )n πLn t2 . 4
Z 0
t
p H2n ( x(t − x)) p dx x(t − x)
t2 4
Problem 141. Show that if m is a non-negative integer and α is not a negative integer, L(α) n (x)
(α) (α+2m) n X (−m)k Lk (−x)Ln−k (x) (1 + α)n = . (1 + α)k (1 + 12 α + 12 m)n k=0
Solution 141. On page 366 we had
(1)
Ln(α) (x) =
(α) (2c−α−2) n (x) (1 + α)n X (1 + α − c)k Lk (−x)Ln−k . (c)n (1 + α)k k=0
1 1 1 1 In (1) choose c = 1 + α + m. Then 2c − α − 2 = m and 1 + α − c = α − m. 2 2 2 2 Hence Ln(α) (x)
(α) (m n X ( α−m (1 + α)n 2 )k Lk (−x)Ln−k (x) = . (1 + α)k (1 + 12 α + 21 m)n k=0
Problem 142. Show that if m is a non-negative integer and α is not a negative integer, L(α) n (x)
(α) (α+2m) ∞ (1 + α)n (1 + α)m X (−m)k Lk (−x)Ln−k (x) = . (1 + α)m+n (1 + α)k k=0
SOLUTIONS TO RAINVILLE’S “SPECIAL FUNCTIONS” (1960)
101
Solution 142. In equation (1) of Exercise 141 above choose c = 1 + α + m. Thus we get (α)
Ln (x)
(α) (α+2m) n (x) (1 + α)n X (−m)k Lk (−x)Ln−k (1 + α + m)n (1 + α)k k=0 (α) (α+2m) n (x) (1 + α)n (1 + α)m X (−m)k Lk (−x)Ln−k
= =
(1 + α)m+n
(1 + α)k
k=0
.
Problem 143. Use integration by parts and equation (2), page 202, to show that Z
∞
(α)
e−y Ln(α) (y)dy = e−x [Ln(α) (x) − Ln−1 (x)].
x
Solution 143. From (2), page 350, we get (α)
Ln(α) (x) = DLn(α) (x) − DLn+1 (x). Now put Z
∞
An =
e−y Ln(α) (y)dy
x (α)
and integrate by parts with u = e−y and dv = Ln (y)dy −→ du = −e−y dy and (α) (α) v = Ln (x) − Ln+1 (x) to get h
An = e
−y
n oi∞ Z (α) (α) Ln (y) − Ln+1 (y) + x
∞
h i (α) e−y Ln(α) (y) − Ln+1 (y) dy.
x
h
i (α) An = e−x Ln+1 (x) − Ln(α) (x) + An − An+1 . Hence h i (α) An+1 = e−x Ln+1 (x) − Ln(α) (x) , so that, by a shift of index, we get Z
∞
An =
h i (α) e−y Ln(α) (y)dy = e−x Ln(α) (x) − Ln−1 (x) .
x
Problem 144. Show that Z
t
xα (t − x)β−1 Ln(α) (x)dx =
0
Γ(1 + α)Γ(β) (1 + α)n tα+β (α+β) L (t). Γ(1 + α + β) (1 + α + β)n n
Solution 144. Let us use Theorem 37 to evaluate Z 0
t
xα (t − x)β−1 Ln(α) (x)dx =
(1 + α)n n!
Z
t
xα (t − x)β−1 1 F1 (−n; 1 + α; x)dx.
0
In Theorem 37 use α = α + 1, β = β, p = 1, q = 1, a1 = −n, b1 = 1 + α, c = 1, k = 1, s = 0 to arrive at
102SOLUTIONS BY SYLVESTER J. PAGANO AND LEON HALL; EDITED BY TOM CUCHTA
t
Z
xα (t − x)β−1 Ln(α) (x)dx
0
=
(1 + α)n B(α + 1, β)tα+β 2 F2 n!
−n,
α+1 ; 1
t
1 + α, α + β + 1; (1 + α)n Γ(1 + |alpha)Γ(β) α+β = t 1 F1 (−n; α + β + 1; t) n! Γ(α + β + 1) α+β n! Γ(1 + α + n) Γ(1 + α)Γ(β)t (α+β) Ln = (t) n!Γ(1 + α) Γ(1 + α + β) (1 + α + β)n (1 + α)n (Γ(1 + α)Γ(β) α+β (α+β) t Ln (t). = (1 + α + β)n Γ(1 + α + β)
Problem 145. Show that the Laplace transform of Ln (t) is n Z ∞ 1 1 −st 1− e Ln (t)dt = . s 2 0 Solution 145. Z
∞
e
−st
Ln (t)dt
Z
∞
e−st
=
0
0 n X
n X (−1)k n!tk dt (k!)2 (n − k)!
k=0
(−1)k n! = k!(n − k)!sk+1 k=0 n 1 1 1− . = s s Problem 146. Show by the convolution theorem for Laplace transforms, or otherwise, that Z t Ln (t − x)Lm (x)dx = Lm+n (x)dx = Lm+n (t) − Lm+n+1 (t). 0
n 1 1 1− = Ln (t). (where L denotes Solution 146. We know that L s s Laplace transform and Ln denotes the Laguerre polynomial). Then by the convolution theorem, n m Z t 1 1 1 1 L Ln (t − x)Lm (x)dx = 1− 1− s" s s s 0 n+m # 1 1 1 = 1− . s s s −1
Hence Z
t
Ln (t − x)Lm (x)dx = L
0
−1
( " n+m #) 1 1 1 1− . s s s
But L Hence
−1
( n+m ) 1 1 1− = Ln+m (t). s s
SOLUTIONS TO RAINVILLE’S “SPECIAL FUNCTIONS” (1960)
Z
t
Ln (t − x)Lm (x)dx =
103
t
Z
Ln+m (β)dβ.
0
0
But, by (2), page 350, Ln (x) = DLn (x) − DLn+1 (x), so that Z
t
t
Ln (t − x)Lm (x)dx
= [Ln+m (β) − Ln+m+1 (β)]0
0
= Ln+m (t) − Ln+m+1 (t).
Problem 147. Evaluate the integral Z ∞ 2 xα e−x [L(α) n (x)] dx 0
of (7), page 206, by the following method. From (4), Section113, show that ∞ Z X
∞ α −x
x e
2 2n [L(α) n (x)] dxt
0
n=0
∞
−x(1 + t) = (1 − t) x exp dx 1−t 0 = (1 − t2 )−1−α Γ(1 + α) ∞ X Γ(1 + α + n)t2n = . n! n=0 −2−2α
Z
α
Solution 147. We know that (1 − t)−1−α exp
−xt 1−t
=
∞ X
Ln(α) (x)tn .
n=0
Then (1 − t)−2−2α exp
−2xt 1−t
=
n ∞ X X
(α)
(α)
Lk (x)Ln−k (x)tn .
n=0 k=0 (α)
Then, because of the orthogonality property of Ln (x) we get Z
∞
xα e−x
0
∞ X ∞ X
(α)
(α)
Lk (x)Ln−k (x)tn dx = (1−t)−2−2α
Z
n=0 k=0
0
∞
2xt xα exp −x − dt, 1−t
or ∞ Z X n=0
0
∞
Z h i2 xα e−x Ln(α) (x) dxt2n = (1 − t)−2−2α 0
∞
xα exp
−x(1 + t) dx. 1−t
1+t In the last integral put x = β and thus obtain 1−t α+1 Z ∞ α+1 Z ∞ −x(1 + t) 1−t 1−t α α −β x exp dx = β e dβ = Γ(1 + α) . 1−t 1+t 1+t 0 0 Therefore we have
104SOLUTIONS BY SYLVESTER J. PAGANO AND LEON HALL; EDITED BY TOM CUCHTA
∞ Z X n=0
∞
h i2 xα e−x Ln(α) (x) dxt2n
= (1 − t)−1−α (1 + t)−1−α Γ(1 + α)
0
= Γ(1 + α)(1 − t2 )−1−α ∞ X (1 + α)n Γ(1 + α)t2n = n! n=0 ∞ X Γ(1 + α + n)t2n = . n! n=0
Hence ∞
Z 0
h i2 Γ(1 + α + n) xα e−x Ln(α) (x) dx = . n! 16. Chapter 13 Solutions
Problem 148. Prove Theorem A (Sheffer): If φn (x) is of Sheffer A-type zero, gn (m, x) = Dm φn+m (x) is also of Sheffer A-type zero and belongs to the same operator as does φn (x). Solution 148. Let φn (x) be of Sheffer A-type zero. Put gn (m, x) = D m φn+m (x); D ≡
d . dx
We know there exists an A, ψ, and H such that (1)
A(t)ψ(xH(t))
= =
∞ X
φn (x)tn
n=0 ∞ X
φn+m (x)tn+m .
n=−m
Now the φn (x) form a simple set. Hence D m φk (x) = 0 for k < m. Therefore, using the operator D m on (1) we obtain A(t)[H(t)]m ψ (m) (xH(t)) =
∞ X
D m φn+m (x)tn+m .
n=0
Since H(0) = 0, we may write (2)
A(t)
H(t) t
m
ψ (m) (xH(t)) =
∞ X
D m φn+m (x)tn ,
n=0
from which it follows at once (Theorem 72) that D m φn+m (x) are polynomials of Sheffer A-type zero an that they belong to the same operator as do the φn (x), because the H(t) in (1) and (2) are the same function. Problem 149. Prove Theorem B: If φn (x) is of Sheffer A-type zero, ψn (x) = φn (x)
"m Y i=1
is of Sheffer A-type m.
#−1 (1 + ρi )n
SOLUTIONS TO RAINVILLE’S “SPECIAL FUNCTIONS” (1960)
105
Solution 149. We are given that φn (x) is of Sheffer A-type zero and wish to consider φn (x) =
φn (x) m Y
.
(1 + ρi )n
i=1
Let φn (x) belong to the operator J(D). By Theorem 74, there exists µk , vk such that n−1 X
(µk + xvk )D k+1 φn (x) = nφn (x).
k=0
Hence there exists operator B = B(x, D) =
∞ X
(µk + xvk )D k+1 = B1 (D) + xB2 (D)
k=0
such that Bφn (x) = nφn (x). Then (B = ρi )φn (x) = (n + ρi )φN (x), so that ( J(D)
m Y
) [B(x, D) + ρi ] φn (x) = J(D)
i=1
φn (x) m Y
=
(1 + ρi )n−1
i=1
φn−1 (x) m Y
= ψn−1 (x).
(1 + ρi )n−1
i=1
Since ψn (x) belongs to one operator whose coefficients have maximum degree m, ψn (x) is of Sheffer A-type m and ψn (x) belongs to the operator J1 (x, D) = J(D)
m Y
[ρi + B(x, D)].
i=1
Two operators arise here. Above implies commutativity of the operators [ρi + B(x, D)] and it implies that J1 is a proper operator, that is transforms every polynomial into one of degree one lower than the original. Let us now prove the desired commutativity property. Let εi = ρi +
∞ X
µk D k+1 + x
k=0
∞ X
vk D k+1 .
k=0
We shall show that ε1 ε2 = ε2 ε1 . Since D m+1 [ρ+µk D k+1 +xvk D k+1 ] = ρD m+1 +µk D m+k+2 +xvk D m+k+2 +(m+1)vk B m+k+1 , we obtain
106SOLUTIONS BY SYLVESTER J. PAGANO AND LEON HALL; EDITED BY TOM CUCHTA
∞ X
" ε1 ε2
= ρ1 +
µm D
m+1
+x
m=0
= ρ1 ρ2 + ρ1
∞ X
∞ X
#" vm D
m+1
ρ2 +
m=0
µk D
k+1
+ ρ1 x
k=0
∞ X
vk D
k+1
+ ρ2
∞ X k=0 ∞ X
µk D
k+1
+x
k=0
µm D
m+1
m=0
k=0
∞ X
+ xρ2
# vk D ∞ X
k+1
" vm D
m=0
which is symmetric in ρ1 and ρ2 . Hence ε1 ε2 = ε2 ε1 . We shall now prove a theorem of value in discussing what operators transform polynomials of degree n into polynomials of degree (n − 1). Theorem : If the operator J(x, D) is such that a simple set of polynomials ψn (x) belongs to it in the Sheffer sense, J(x, D)ψn (x) = ψn−1 (x), J(x, D)ψ0 (x) = 0,
(1)
then J(x, D) transforms every polynomial of degree precisely n into a polynomial of degree precisely n − 1. Proof : Let gn (x) be of degree exactly n. Then we know there exists the expansion gn (x) =
n X
A(k, n)ψn−k (x), A(0, n) 6= 0,
k=0
merely because the ψn (x) form a simple set. Then, because of (1), (2)
J(x, D)gn (x) =
n−1 X
A(k, n)ψn−1−k (x),
k=0
and the right member of (2) is of degree exactly (n − 1). Problem 150. Show that Hn (x) n! is of Sheffer A-type zero, obtain the associated functions J(t), H(t), A(t), and draw what conclusions you can from Theorems 73-76. Solution 150. Since exp(2xt + t2 ) = exp(−t2 ) exp(2xt) = it follows that
∞ X Hn (x)tn , n! n=0
Hn (x) is of Sheffer A-type zero. Furthermore, n! 2
A(t) = e−t , H(t) = 2t. THen K(t) =
1 t and we may proceed to use Theorems 73-76. 2 (α)
Problem 151. SHow that Ln (x) is of Sheffer A-type zero, and proceed as in Exercise 150.
m+1
+
∞ X m=0
(µm + xvm )D m
SOLUTIONS TO RAINVILLE’S “SPECIAL FUNCTIONS” (1960)
107
Solution 151. Since (1 − t)−1−α exp the polynomials
(α) Ln (x)
−xt 1−t
=
∞ X
Ln(α) (x)tn ,
n=0
are of Sheffer A-type zero and
A(t) = (1 − t)−1−α , H(t) =
−t . 1−t
−t . 1−t Problem 152. Show that the Newtonian polynomials Then J(t) =
(−1)n (−x)n n! are of Sheffer A-type zero, and proceed as in Exercise 150. Nn (x) =
Solution 152. Consider Nn (x) =
(−1)n (−x)n , n!
from which ∞ X
Nn (x)tn =
n=0
∞ X (−1)n (−x)n tn = (1 + t)x = exp[x log(1 + t)]. n! n=0
Then the polynomials Nn (x) are of Sheffer A-type zero with A(t) = 1, H(t) = log(1 + t). We obtain J(t) = et − 1. Problem 153. Show that (α)
Ln (x) (1 + α)n is of Sheffer A-type unity but that with σ chosen to be φn (x) =
σ = D(θ + α) the polynomials φn (x) are of σ-type zero. Solution 153. Consider (α)
φn (x) =
Ln (x) . (1 + α)n
(α)
Since Ln (x) is of Sheffer A-type zero (Exercise 151), it follows by Exercise 149 that φn (x) is of Sheffer A-type one. Now ∞ (α) X Ln (x)tn = et 0 F1 (−; 1 + α; −xt), (1 + α) − n n=0
so that, by Theorem 79, φn (x) is of σ-type zero with σ = D(θ + 1 + 1 − a) = D(θ + α).
108SOLUTIONS BY SYLVESTER J. PAGANO AND LEON HALL; EDITED BY TOM CUCHTA
Problem 154. Determine the Sheffer operator associated wit hthe set Ln (x) , (n!)2 and thus show that φn (x) is of Sheffer A-type 2. φn (x) =
Ln (x) . Since Ln (x) is of Sheffer A-type zero (n!)2 (by Exercise 151), we may use Exercise 149 to conclude that φn (x) is of Sheffer A-type 2. We now wish to find the operator to which φn (x) belongs. Let
Solution 154. Consdier φn (x) =
J(x, D) =
∞ X
Tk (x)D k+1 .
k=0
We first proceed by brute strength methods. We know that L0 (x) = 1, L1 (x) = 1 − x, 1 L2 (x) = 1 − 2x + x2 , 2 1 3 L3 (x) = 1 − 3x + x2 − x3 , 2 6 2 1 L4 (x) = 1 − 4x + 3x2 − x3 + x4 , 3 24 then J(x, D)φn (x) = φn−1 (x) requires that T0 (x)D
1−x 1 = ; T0 (x)(−1) = 1, T0 (x) = −1. (1!)2 (0!)2
Then also [T0 (x)D + T1 (x)D 2 ]
1 1 L2 (x) = L1 (x), 2 (2!) (1!)2
or 1 [D + T1 (x)D 2 ] 1 − 2x + x2 = 4(1 − x) 2 −(−2 + x) + T1 (x)(1) = 4 − 4x; T1 (x) = 2 − 3x. Next we have
3 1 [−D + (2 − 3x)D + T2 (x)D ] 1 − 3x + x2 − x3 2 6 or 2
3
1 2 = 9 1 − 2x + x . 2
1 2 − −3 + 3x − x + (2 − 3x)(3 − x) + T2 (x)(−1) = 9 − 18x + 9x2 , 2
SOLUTIONS TO RAINVILLE’S “SPECIAL FUNCTIONS” (1960)
109
or 1 9 3 − 3x + x2 + 6 − 11x + 3x2 − T2 (x) = 9 − 18x + x2 , 2 2 −T2 (x) = −4x + x2 ; T2 (x) = 4x − x2 Then 2 1 3 1 [−D+(2−3x)D 2 +(4x−x2 )D 3 +T4 (x)D 4 ] 1 − 4x + 3x2 − x3 + x4 = 16 1 − 3x + x2 − x3 , 3 24 2 6 or 1 8 1 − −4 + 6x − 2x2 + x3 +(2−3x) 6 − 4x + x2 +(4x−x2 )(−4+x)+T4 (x) = 16−48x+24x2 − x3 . 6 2 3 Thus we get x3 (1 + 9 + 6 − 16) = x2 ; T4 (x) = x2 . 6 Let us consider the effect of the operator T4 (x) = x2 +
J(x, D) = −D + (2 − 3x)D 2 + (4x − x2 )D 3 + x2 D 4 Ln (x) upon φn (x) = . (n!)2 Problem 155. Prove that if we know a generating function y(x, t) =
∞ X
φn (x)tn
n=0
or the simple set of polynomials φn (x) belonging to a Sheffer operator J(x, D), no matter what the A-type of φn (x), we can obtain a generating fuction (sum the series) ∞ X
φn (x)tn
n=0
for any other polynomial φn (x) belonging to the same operator J(x, D). Solution 155. We are given that φn (x) is a simple set of polynomials and (1)
y(xt) =
∞ X
φn (x)tn .
n=0
Let φn (x) belong to the operator J(x, D) in the Sheffer sense. Let ψn (x) belong to the same operator J(x, D). Then, by Theorem 71, there exist constants bk such that (2)
ψn (x) =
n X k=0
It follows from (1) and (2) that
bk φn−k (x).
110SOLUTIONS BY SYLVESTER J. PAGANO AND LEON HALL; EDITED BY TOM CUCHTA
∞ X
φn (x)tn
∞ X n X
bk φn−k (x)tn n=0 k=0 ! ∞ ! ∞ X X n n = bn t φn (x)t =
n=0
n=0
n=0
= B(t)y(x, t). Note that the bk ’s can be computed identically. Problem 156. Obtain a theorem analogous to that of Exercise 155 but with Sheffer A-type replaced by σ-type. Solution 156. Follow the proof in Exercise 155 except for two substitutions: (1): Replace J(x, D) by J(x, σ); (2): Replace Theorem 71 by Theorem 78. Problem 157. Show that if Pn (x) is the Legendre polynomial, 1 x (1 + x2 ) 2 n Pn √ φn (x) = n! 1 + x2 is a simple set of polynomials of Sheffer A-type zero. Solution 157. Consider n
(1 + x2 ) 2 φn (x) = Pn n!
x √ 1 + x2
.
We know that ∞ X Pn (u)v n v 2 (u2 − 1) uv = e 0 F1 −; 1; . n! 4 n=0 Therefore ∞ X n=0
φn (x)t
n
=
∞ P X n
√x 1+x
n
(1 + x2 ) 2 tn
n!
n=0
=e
√ t 1+x2 √
x 1+x2
−;
xt
= e 0 F1 1;
0 F1
−; t2 (1 + x2 )
1;
4
x2 1+x2
−1
−t2 . 4
t2 Hence φn (x) is of Sheffer A-type zero with H(t) = t, A(t) = 0 F1 −; 1; − . 4 Problem 158. Let Pn (x) be the Legendre polynomial. Choose σ = Dθ and show that the polynomials (x−)n x+1 φn (x) = Pn (n!)2 x−1
SOLUTIONS TO RAINVILLE’S “SPECIAL FUNCTIONS” (1960)
111
are of σ-type zero for that σ. Solution 158. Consider (x − 1)n Pn (n!)2 where Pn (x) is the Legendre polynomial. We know that
φn (x) =
(1)
0 F1
v(u − 1) −; 1; 2
In (1) put v = t(x − 1), u =
0 F1
x+1 x−1
,
v(u + 1) −; 1; 2
=
∞ X Pn (u)v n . (n!)2 n=0
x+1 u−1 1 u+1 x . Then = , = , and x1 2 x−1 2 x−1
(1) yields (2)
0 F1 (−; 1; t)0 F1 (−; 1; xt) =
∞ X
φn (x)tn .
n=0
Using Theorem 79 we may conclude from (2) that φn (x) is of σ-type zero with σ = D(θ + 1 − 1) = Dθ and A(t) = 0 F1 (−; 1; t), H(t) = t, φ(t) = 0 F1 (−; 1; t). Problem 159. Prove that if the operator J(x, D) is such that a simple set of polynomials φn (x) belongs to it in the Sheffer sense, then J(x, D) transforms every polynomial of degree precisely n into a polynomial of degree precisely (n − 1). Solution 159. 17. Chapter 14 Solutions Problem 160. Show that the Bateman’s polynomial (see Section 146, page 285) Zn (t) = 2 F2 (−n, n + 1; 1, 1; t) has the recurrence relation
n2 (2n−3)Zn (t)−(2n−1)[3n2 −6n+2−2(2n−3)t]Zn−1 (t)+(2n−3)[3n2 −6n+2+2(2n−1)t]Zn−2 (t)−(2n−1)(n−2)2 Z Solution 160. Let Zn (t) = 2 F2 (−n, N = 1; 1, 1; t) =
n X (−1)k (n + k)!tk k=0
or Zn (t) =
∞ X k=0
γ(k, n).
(k!)3 (n − k)!
112SOLUTIONS BY SYLVESTER J. PAGANO AND LEON HALL; EDITED BY TOM CUCHTA
Then Zn−1 (t) =
∞ X (−1)k (n − 1 + k)!tk k=0 ∞ X
Zn−2 (t) =
k=0
(k!)3 (n − 1 − k)!
=
∞ X n−k k=0
n+k
γ(k, n),
(n − k)(n − k − 1)(n − k − 2) γ(k, n). (n + k)(n + k − 1)(n + k − 2)
Also
tZn−1 (t) =
∞ X (−1)k (n − 1 + k)!xk+1
(k!)3 (n − 1 − k)!
k=0
=
∞ X (−1)k−1 (n − 2 + k)!xk k=1
[(k − 1)!]3 (n − k)!
or tZn−1 (t) =
∞ X k=0
tZn−2 (t) =
∞ X k=0
−k 3 γ(k, n) (n + k)(n + k − 1) ∞
X (−1)k−1 (n − 3 + k)!xk (−1)k (n − 2 + k)!xk+1 = 3 (k!) (n − 2 − k)! [(k − 1)!]3 (n − 1 − k)!
tZn−2 (t) =
k=1
∞ X k=0
−k 3 (n − k) γ(k, n). (n + k)(n + k − 1)(n + k − 2)
We may now proceed to a recurrence relation of the form Zn (t) + [A + Bt]Zn−1 (t) + [C + Dt]Zn−2 (t) + EZn−3 (t) = 0, in which A, B, C, D, E depend upon k. For the determination of those coefficients we need to satisfy the identity in k: (n + k)(n + k − 1)(n + k − 2) + A(n − k)(n + k − 1)(n + k − 2) − Bk 3 (n + k − 2) +C(n − k)(n − k − 1)(n + k − 2) − Dk 3 (n − k) + E(n − k)(n − k − 1)(n − k − 2) = 0. Use k=n:
2n(2n − 1)(2n − 2) − Bn3 (2n − 2) = 0 :
coef f k 4 :
−B + D = 0 :
k = 2 − n : −D(2 − n)3 (2n − 2) + E(2n − 2)(2n − 3)(2n − 4) = 0 D(n − 2)3 + 2E(n − 2)(2n − 3) = 0
k = 1 − n : −B(1 − n)3 (−1) + C(2n − 1)(2n − 2)(−1) − D(1 − n)3 (2n − 1) + E(2n − 1)(2n − 2)(2n − 3) = 0, or −B(n − 1)3 − 2C(n − 1)(2n − 1) + D(n − 1)3 (2n − 1) + E · 2(n − 1)(2n − 1)(2n − 3) = 0 Thus we have
2(2n − 1) n2 2(2n − 1) D= n2 B=
(n − 2)2 D 2(2n − 3) (2n − 1)(n − 2)2 =− n2 (2n − 3) E=−
SOLUTIONS TO RAINVILLE’S “SPECIAL FUNCTIONS” (1960)
113
= 2E(2n − 1)(2n − 3) + D(n − 1)2 (2n − 1) − B(n − 1)2 2(n − 1)2 (2n − 1) 2(2n − 1)2 (n − 2)2 (2n − 3) 2(n − 1)2 (2n − 1)2 − . =− + n2 (2n − 3) n2 n2
2C(2n − 1)
THen n2 C
= −(2n − 1)(n − 2)2 + (n − 1)2 (2n − 1) − (n − 1)2 = (2n − 1)[−n2 + 4n − 4 + n2 − 2n + 1] − (n − 1)2 = (2n − 1)(2n − 3) − (n − 1)2 = 4n2 − 8n + 3 − n2 + 2n − 1 = 3n2 − 6n + 2
Hence C=
3n2 − 6n + 2 . n2
(2n − 1)(2n − 2)(2n − 3) + A(1)(2n − 2)(2n − 3) − B(n − 1)3 (2n − 3) − D(n − 1)3 (1) = 0 or 2(2n − 1)(2n − 3) + 2A(2n − 3) − B(n − 1)2 (2n − 3) − D(n − 1)2 = 0, or 2(2n − 1)(n − 1)2 (2n − 3) 2(2n − 1)(n − 1)2 2A(2n − 3) = −2(2n − 1)(2n − 3) + + . n2 n2 Hence we get
k =n−1:
n2 (2n − 3)A
= −n2 (2n − 1)(2n − 3) + (2n − 1)(n − 1)2 (2n − 3) + (2n − 1)(n − 1)2 = (2n − 1)[(2n − 3)(−n2 + n2 − 2n + 1) + (n − 1)2 ] = (2n − 1)[−4n2 + 8n − 3 + n2 − 2n + 1] = −(2n − 1)(3n2 − 6n + 2).
We now have 2(2n − 1) (2n − 1)(3n2 − 6n + 2) 3n2 − 6n + 2 2(2n − 1 (2n − 1)(n − 2)2 ,B = . , C+ ,D = ,E = − 2 2 2 2 n (2n − 3) n n n n2 (2n − 3) The desired recurrence relation is therefore
A=−
n2 (2n − 3)Zn (t) − (2n − 1) 3n2 − 6n + 2 − 2(2n − 3)t Zn−1 (t) +(2n − 3)[3n2 − 6n + 2 + 2(2n − 1)t]Zn−2 (t) − (n − 2)2 (2n − 1)Zn−3 (t). Note the one sign change from that in Sister Celine’s paper. The sign was correct in her thesis. Problem 161. Show that Sister Celine’s polynomial fn (a; −; x), or 1 Fn (x) = 3 F2 −n, n + 1, a; 1, ; x 2 has the recurrence relation nfn (x)−[3n−2−4(n−1+a)x]fn−1 (x)+[3n−4−4(n−1−a)x]fn−2 (x)−(n−2)fn−3 (x) = 0.
114SOLUTIONS BY SYLVESTER J. PAGANO AND LEON HALL; EDITED BY TOM CUCHTA
Solution 161. Consider
1 fn (x) = 3 F2 −n, n + 1, a; 1, ; x , 2 for which we seek a pure recurrence relation. Now fn (x) =
n X (−1)k (n + k)!(a)k xk
k!(n − k)!k!(k)k
k=0
=
n X (−1)k (n + k)!(a)k (4x)k
k!(2k)!(n − k)!
k=0
.
Put (k, n) =
(−1)k (n + k)!(a)k (4x)k . k!(2k)!(n − k)!
Then fn (x) =
∞ X
(k, n),
k=0
fn−1 (x) =
∞ X (−1)k (n − 1 + k)!(a)k (4x)k
k!(2k)!(n − 1 − k)!
k=0
fn−2 (x) =
∞ X n−k k=0
∞ X (n − k)(n − k − 1) k=0
fn−3 (x) =
=
(n + k)(n + k − 1)
n+k
(k, n),
∞ X (n − k)(n − k − 1)(n − k − 2) k=0
(k, n),
(n + k)(n + k − 1)(n + k − 2)
(k, n).
Furthermore, (4x)fn−1 (x)
= = =
∞ X (−1)k (n − 1 + k)!(a)k (4x)k+1 k=0 ∞ X k=1 ∞ X k=0
k!(2k)!(n − 1 − k)! k−1
(−1) (n − 2 + k)!(a)k−1 (4x)k (k − 1)!(2k − 2)!(n − k)! −k · 2k(2k − 1) (k, n), (n + k)(n + k − 1)(a + k − 1)
and
(4x)fn−2 (x) =
∞ X (−1)k (n − 2 + k)!(a)k (4x)k+1 k=0
k!(2k)!(n − 2 − k)!
=
∞ X (−1)k−1 (n − 3 + k)!(a)n−1 (4x)k k=1
(k − 1)!(2k − 2)!(n − 1 − k)!
or (4x)fn−2 (x) =
∞ X k=0
−k(2k)(2k − 1)(n − k) (k, n). (n + k)(n + k − 1)(n + k − 2)(a + k − 1)
We therefore seek an identity of the form fn (x) + [A + B(4x)]fn−1 (x) + [C + D(4x)]fn−2 (x) + Efn−3 (x) = 0,
SOLUTIONS TO RAINVILLE’S “SPECIAL FUNCTIONS” (1960)
115
in which A, B, C, D, E may depend upon n. We are thus led to the identity (in k) (1)
1+A
n−k 2k 2 (2k − 1) (n − k)(n − k − 1) −B +C n+k (n + k)(n + k − 1)(a + k − 1) (n + k)(n + k − 1)
2k 2 (2k − 1)(n − k) (n − k)(n − k − 1)(n − k − 2) +E ≡ 0, (n + k)(n + k − 1)(n + k − 2)(a + k − 1) (n + k)(n + k − 1)(n + k − 2) or
−D
(2)
(n+k)(n+k−1)(n+k−2)(a+k−1)+A(n−k)(n+k−1)(n+k−2)(a+k−1) −2Bk 2 (2k − 1)(n + k − 2) + C(n − k)(n − k − 1)(n + k − 2)(a + k − 1) −2Dk 2 (2k − 1)(n − k) + E(n − k)(n − k − 1)(n − k − 2)(a + k − 1) ≡ 0.
We need five equations for the determination of A, B, C, D, E. Use k=n:
2n(2n − 1)(2n − 2)(a + n − 1) − 2Bn2 (2n − 1)(2n − 2) = 0, or a + n − 1 − Bn = 0
k =1−a:
B=
a+n−1 . n
−2B(1 − a)2 (1 − 2a)(n − 1 − a) − 2D(1 − a)2 (1 − 2a)(n − 1 + a) = 0, or B(n − 1 − a) + D(n − 1 + a) = 0
(n − 1 − a)B ;D = − a+n−1 n−1−a =− n
k = 2 − n : −2D(2 − n)2 (3 − 2n)(2n − 2) + E(2n − 2)(2n − 3)(2n − 4)(a − n + 1) = 0, or D(n − 2) + E(a − n + 1) = 0;
k=0: and coeff k 4 : k=0:
n(n − 1)(n − 2)(a − 1) + An(n − 1)(n − 2)(a − 1) + Cn(n − 1)(n − 2)(a − 1) +En(n − 1)(n − 2)(a − 1) = 0 1 − a − 4B + C + 4D − E = 0 1 + A + C + E = 0.
Hence 2 + 2C − 4B + 4D = 0. Therefore we have
n−2 E=− 1+a−n n−2 =− n
116SOLUTIONS BY SYLVESTER J. PAGANO AND LEON HALL; EDITED BY TOM CUCHTA
= −1 + 2B − 2D 2E + 2n − 2 2n − 2 − 2a = −1 + + n n
C
from which we get C=
3n − 4 . n
Finall, A = −1 − C − E = −1 − so A=−
3n − 4 n − 2 + n n
3n − 2 . n
Thus we obtain nfn (x)−[(3n−2)−(4x)(a+n−1)]fn−1 (x)+[3n−4−4x(n−1−a)]fn−2 (x)−(n−2)fn−3 (x) = 0. Problem 162. Show that Rice’s polynomial (see Section 147, page 287) Hn = Hn (ζ, p, v) = 3 F2 (−n, n + 1, ζ; 1, p; v) satisfies the relation
n(2n−3)(p+n−1)Hn −(2n−1)[(n−2)(p−n+1)+2(n−2)(2n−3)−2(2n−3)(ζ+n−1)v]Hn−1 +(2n−3)[2(n−1)2 −n(p− Solution 162. Now Hn (ζ, p, v)
= 3 F2 (−n, n + 1, ζ; 1, p; v) n X (−1)k (n + k)!(ζ)k v k = (k!)2 (n − k)!(p)k k=0 ∞ X (k, n). = k=0
Then Hn−1 =
∞ X (−1)k (ζ)k (n − 1 + k)!v k k=0
(k!)2 (p)k (n − 1 − k)!
Hn−2 =
∞ X n−k k=0
∞ X (n − k)(n − k − 1) k=0
Hn−3 =
=
(n + k)(n + k − 1)
n+k
(k, n),
(k, n),
∞ X (n − k)(n − k − 1)(n − k − 2) (k, n), ((n + k)(n + k − 1)(n + k − 2)
k=0
vHn−1 =
∞ X (−1)k (ζ)k (n − 1 + k)!v k+1 k=0
(k!)2 (p)k (n − 1 − k)!
vHn−1 =
∞ X k=0
=
∞ X (−1)k−1 (ζ)k−1 (n − 2 + k)!v k k=1
[(k − 1)!]2 (p)k−1 (n − k)!
−k 2 (p + k − 1) (k, n), (n + k)(n + k − 1)(ζ + k − 1)
SOLUTIONS TO RAINVILLE’S “SPECIAL FUNCTIONS” (1960)
117
and vHn−2 =
∞ X k=0
−k 2 (p + k − 1)(n − k) (k, n). (n + k)(n + k − 1)(n + k − 2)(ζ + k − 1)
Then there exists a relation Hn + (A + Bv)Hn−1 + (C + Dv)Hn−2 + EHn−3 = 0, in which A, B, C, D, E depend upon n alone. We are led to the identity in k: (n + k)(n + k − 1)(n + k − 2)(ζ + k − 1) + A(n − k)(n + k − 1)(n + k − 2)(ζ + k − 1) −Bk 2 (p + k − 1(n + k − 1) + C(n − k)(n − k − 1)(nk − 1)(n + k − 2)(ζ + k − 1) − Dk 2 (p + k − 1)(n − k) +E(n − k)(n − k − 1)(n − k − 2)(ζ + k − 1) = 0 We now solve for A, B, C, D, E : k =1−ζ :
−B(1 − ζ)2 (p − ζ)(n − ζ − 1) − D(1 − ζ)2 (p − ζ)(n − 1 + ζ);
k=n:
(2n)(2n − 1)(2n − 2)(ζ + n − 1) − Bn2 (p + n − 1)(2n − 2) = 0;
k = 2 − n : −D(2 − n)2 (p + 1 − n)(2n − 2) + E(2n − 2)(2n − 3)(2n − 4)(ζ + 1 − n) = 0;
−B(n − ζ − 1) . n+ζ −1 2(2n − 1)(n + ζ − 1) B= n(n + p − 1) −2(2n − 1)(n − ζ − 1) D= n(n + p − 1) (n − 2)(p + 1 − n) E= D 2(2n − 3)(ζ + 1 − n) (n − 2)(2n − 1)(p + 1 − n) = n(2n − 3)(n + p − 1) D=
k = 1 − n : −B(1 − n)2 (p − n)(−1) + C(2n − 1)(2n − 2)(−1)(ζ − n) − D(1 − n)2 (p − n)(2n − 1) +E(2n − 1)(2n − 2)(2n − 3)(p − n) = 0; we have B(n−1)(p−n)−2C(2n−1)(ζ−n)−D(n−1)(p−n)(2n−1)+2E(2n−1)(2n−3)(p−n) = 0, or 2(2n − 1)(n − ζ − 1)(n − 1)(p − n)(2n − 1) −(2n − 1)(n + p − 1)(n − 1)(p − n) −2C(2n−1)(ζ−n)+ n(n + p − 1) n(n + p − 1) +
2(n − 2)(2n − 1)(p + 1 − n)92n − 1)(2n − 3)(ζ − n) = 0, n(2n − 3)(n − p − 1)
or n(ζ − n)(n + p − 1)C
Hence
= (n + ζ − 1)(n − 1)(p − n) + (n − ζ − 1)(n − 1)(p − n)(2n − 1) +(n − 2)(p + 1 − n)(2n − 1)(p − n) = (n − 1)(p − n)[n + ζ − 1 + (n − ζ − 1)(2n − 1)] + (n − 2)(p + 1 − n)(2n − 1)(ζ − n) = −2(n − 1)2 (p − n)(ζ − n) + (n − 2)(p + 1 − n)(2n − 1)(ζ − n).
118SOLUTIONS BY SYLVESTER J. PAGANO AND LEON HALL; EDITED BY TOM CUCHTA
n(n + p − 1)C
= −2(n − 1)2 (p − n) + (n − 2)(2n − 1)(p + 1 − n) = −2(n − 1)2 (p − n + 1) + 2(n − 1)2 + (n − 2)(2n − 1)(p + 1 − n) = 2(n − 1)2 + (p − n + 1)[−2n2 + 4n − 2 + 2n2 − 5n + 2] = 2(n − 1)2 − n(p − n + 1),
so C=
2(n − 1)2 − n(p − n + 1) . n(p + n − 1)
Finally, coeff k 4 : 1 − A − B + C + D − E = 0. Then A=1−B+C +D−E so n(2n − 3)(p + n − 1)A
= n(2n − 3)(p + n − 1) − 2(2n − 3)(2n − 1)(n + ζ − 1) +(2n − 3)[2(n − 1)2 − n(p − n + 1)] − 2(2n − 3)(2n − 2)(n − ζ − 1)
−(n − 2)(2n − 1)(p + 1 − n), or n(2n − 3)(p + n − 1)A
= (2n − 3)[pn + n2 − n + 2n2 − 4n + 2 − pn + n2 − n] −(2n − 1)[2(2n − 3)(n + ζ − 1) + 22(2n − 3)(n − ζ − 1) + (n − 2)(p + 1 − n)] = 2(2n − 3)(n − 1)(2n − 1) − 2(2n − 1)(2n − 3)(n + ζ − 1) −(2n − 1)[2(2n − 3)(n − p − 1) + (1 − 2)(p + 1 − n)] = (2n − 1)[2(2n − 3)(−n − ζ + 1 − n + ζ + 1) + 2(n − 1)(2n − 3) − (n − 2)(p + 1 − n) = (2n − 1)[−4(n − 1)(2n − 3) + 2(n − 1)(2n − 3) − (n − 2)(p + 1 − n)] = −(2n − 1)[(n − 2)(p + 1 − n) + 2(n − 1)(2n − 3)].
Hence we arrive at the recurrence relation n(2n − 3)(p + n − 1)Hn −(2n − 1)[(n − 2)(p + 1 − n) + 2(n − 1)(2n − 3) − 2(2n − 3)(n + p − 1)v]Hn−1 +(2n − 3)[2(n − 1)2 − n(p + 1 − n) − 2(2n − 1)(n − p − 1)v]Hn−2 +(n − 3)(2n − 1)9p + 1 − n)Hn−3 = 0. Problem 163. Show that the polynomial fn (x) = 1 F2 (−n; 1 + α, 1 + β; x), which is intimately related to Bateman’s Jnu,v of Section 147, page 287, satisfies the relation
(α+n)(β+n)fn (x)−[3n2 −3n+1+(2n−1)(α+β)+αβ−x]fn−1 (x)+(n−1)(3n−3+α+β)fn−2 (x)−(n−1)(n−2)fn−3 (
SOLUTIONS TO RAINVILLE’S “SPECIAL FUNCTIONS” (1960)
Solution 163. Let fn (x) = 1 F2 (−m; 1+α, 1+β; x) =
n X k=0
119
(−1)k nxk . k!(1 + α)k (1 + β)k (n − k)!
fn (x) Put gn (x) = . Then n! n ∞ X X (−1)k xk gn (x) = = (k, n). k!(1 + α)k (1 + β)k (n − k)! k=0
k=0
Now gn−1 (x) =
∞ X k=0
∞
X (−1)k xk = (n − k)(k, n) k!(1 + α)k (1 + β)k (n − 1 − k)! k=0
and gn−2 (x) =
∞ X
(n − k)(n − k − 1)(k, n)
k=0
gn−3 (x) =
∞ X
(n − k)(n − k − 1)(n − k − 2)(k, n).
k=0
Also
xgn−1 (x) =
∞ X k=0
∞
X (−1)k−1 xk (−1)k xk+1 = k!(1 + α) − k(1 + β)k (n − 1 − k)! (k − 1)!(1 + α)k−1 (1 + β)k−1 (n − k)! k=1
or xgn−1 (x) =
∞ X
−k(α + k)(β + k)(k, n).
k=0
Then there exists the relation gn (x) + (A + Bx)gn−1 (x) + Cgn−2 (x) + Dgn−3 (x) = 0, with A, B, C, D dependent only on n. We are led to the identity in k 1+A(n−k)−Bk(α+k)(β+k)+C(n−k)(n−k−1)+D(n−k)(n−k−1)(n−k−2) = 0. Then use 1 . n(α + n)(β + n) −1 coeff k 3 : −B − D = 0; D= n(α + n)(β + n) (n − 1)(α + n − 1)(β + n − 1) − n(α + n)(β + n) k = n − 1 : 1 + A − B(n − 1)(α + n − 1)(β + n − 1) = 0; A = n(α + n)(β + n) 3n2 − 3n + 1 + (2n − 1)(α + β) + αβ =− n(α + n)(β + n) k=0: 1 + An + Cn(n − 1) + Dn(n − 1(n − 2) = 0 k=n:
Then
1 − Bn(a + n)(β + n) = 0;
B=
120SOLUTIONS BY SYLVESTER J. PAGANO AND LEON HALL; EDITED BY TOM CUCHTA
n(n − 1)(α + n)(β + n)C
= −(α + n)(β + n) + 3n − 3n + 1 + (α + β)(2n − 1) + αβ + (n − 1)(n − 2) = −n2 − (α + β)n − αβ + 3n2 − 3n + 1 + (α + β)(2n − 1) + αβ + n2 − 3n + 2 = 3n2 − 6n + 3 + (α + β)(−n + 2n − 1) = 3(n − 1)2 + (n − 1)(α + β) = (n − 1)[3n − 3 + α + β]
So C=
3n − 3 + α + β . n(α + n)(β + n)
We thus find that n(α + n)(β + n)gn (x) − [3n2 − 3n + 1 + βn − 1)(α + β) + αβ − x]gn−1 (x) +(3n − 3 + α + β)gn−2 (x) − gn−3 (x) = 0. fn (x) . Hence we get Now gn (x) = n! (α + n)(β + n) fn−1 (x) 3n − 3 + α + β fn−3 (x) fn (x)−[3n2 −3n+1+βn−1)(α+β)+αβ−x] + fn−2 (x)− = 0, (n − 1)! (n − 1)! (n − 2)! (n − 3)! or (α + n)(β + n)fn (x) − [3n2 − 3n + 1 + (2n − 1)(α + β) + αβ − x]fn−1 (x) +(n − 1)[3n − 3 + α + β]fn−2 (x) − (n − 1)(n − 2)fn−3 (x) = 0. Problem 164. Define the polynomial wn (x) by wn (x) =
n X (−1)k n!Lk (x) k=0
(k!)2 (n − k)!
in terms of the Laguerre polynomial Lk (x). Show that wn (x) possesses the pure recurrence relation n2 wn (x)−[(n−1)(4n−3)+x]wn−1 (x)+(6n2 −19n+16+x)wn−2 (x)−(n−2)(4n−9)wn−3 (x)+(n−2)(n−3)wn−4 (x) = 0. Solution 164. Let wn (x) =
n X (−1)k n!Lk (x) k=0
(k!)2 (n − k)!
in terms of the simple Laguerre polynomial. We know that (1) nLn (x) − (2n − 1 − x)Ln−1 (x) + (n − 1)Ln−2 (x) = 0 and we seek a recurrence relation for wn (x). wn (x) (−1)k Lk (x) Put = φn (x) and = gk (x). n! (k!)2 Then (−1)k (k!)2 kgk (x)−(−1)k−1 [(k−1)!]2 (2k−1−x)gk−1 (x)+(−1)k−2 [(k−2)!]2 (k−1)gk−2 (x) = 0,
SOLUTIONS TO RAINVILLE’S “SPECIAL FUNCTIONS” (1960)
121
or k 3 (k − 1)gk (x) + (k − 1)(2k − 1 − x)gk−1 (x) + gk−2 (x) = 0. Also φn (x) =
n ∞ X X gk (x) gk (x) = . (n − k)! (n − k)!
k=0
k=0
Then φn−1 (x) = φn−3 (x) =
∞ X k=0 ∞ X k=0
(n − k)
gk (x) , (n − k)!
φn−2 (x) =
gk (x) , (n − k)(n − k − 1)(n − k − 2) (n − k)!
φn−4 (x) =
∞ X
(n − k(N − k − 1)
k=0 ∞ X
gk (x) , (n − k)!
(n − k)(n − k − 1)(n − k − 2)(n − k − 3)
k=0
gk (x) . (n − k)!
We first wish to fins A, B, C, D, E so that Aφn (x) + Bφn−1 (x) + Cφn−2 (x) + Dφn−3 (x) + Eφn−4 (x) =
∞ X k 3 (k − 1)gk (x) k=0
(n − k)!
.
The above requires that A to E satisfy the identity in k, A+B(n−k)+C(n−k)(n−k−1)+D(n−k)(n−k−1)(n−k−2)+E(n−k)(n−k−1)(n−k−2)(n−k−3) = k 3 (k−1). We thus get A = n3 (n − 1) E=1 −D − E(n + n − 1 + n − 2 + n − 3) = −1 or D + 4n − 6 = 1; D = −(4n − 7) k = n − 1 : A + B(1) = (n − 1)3 (n − 2) B = (n − 1)3 (n − 2) − n3 (n − 1) = (n − 1)[(n − 1)2 (n − 3) − n3 ] B = −(n − 1)(4n2 − 5n + 2). k = n − 2 : A + 2B + 2 · 1 · C = (n − 2)3 (n − 3) 2C = (n − 2)3 (n − 3) − N 3 (n − 1) + 2(n − 1)(4n2 − 5n + 2) = 12n2 − 30n + 20
k=n: coeff k 4 : coeff k 3 :
So C = 2n2 − 15n + 10. We now have (1)
n3 (n − 1)φn (x) − (n − 1)(4n2 − 5n + 2)φn−1 (x) + (5n2 − 15n + 10)φn−2 (x) ∞ X k 3 (k − 1)gk (x) −(4n − 7)φn−3 (x) + φn−4 (x) = . (n − k)! k=0
From φn−1 (x) =
∞ X k=0
gk (x) , (n − 1 − k)!
we get, with our usual convention that gs (x) ≡ 0 for s < 0,
122SOLUTIONS BY SYLVESTER J. PAGANO AND LEON HALL; EDITED BY TOM CUCHTA
φn−1 (x) =
∞ X gk−1 (x) (n − k)!
k=0
and in the same way φn−2 (x) =
∞ X (n − k)gk−1 (x) k=0
φn−3 (x) =
∞ X
(n − k)!
(n − k)(n − k − 1)
k=0
,
gk−1 (x) , etc (n − k)!
We now wish to find F, G, H, so that F φn−1 (x) + Gφn−2 (x) + Hφn−3 (x) =
∞ X
(k − 1)(2k − 1)
k=0
gk−1 (x) . (n − k)!
We thus arrive at the identity F + G(n − k) + H(n − k)(n − k − 1) = (k − 1)(2k − 1). Hence k=n: F = (n − 1)(2n − 1) coeff k 2 : H = 2 coeff k : −G − H(n + n − 1) = −3 G + 2(2n − 1) = 3 G = −(4n − 5) Therefore we get (2)
(n−1)(2n−1)φn−1 (x)−(4n−5)φn−2 (x)+2φn−3 (x) =
∞ X (k − 1)(2k − 1)gk−1 (x)
(n − k)!
k=0
From A1 φn−1 (x) + B1 φn−2 (x) =
∞ X −(k − 1)gk−1 (x) k=0
(n − k)!
,
we determine A1 and B1 by A1 + B1 (n − k) ≡ −(k − 1). Then A1 = −(n − 1), B1 = 1. Hence − (n − 1)φn−1 (x) + φn−2 (x) =
(3)
∞ X −(k − 1)gk−1 (x)
(n − k)!
k=0
.
Also (4)
φn−2 (x) =
∞ X k=0
∞
X gk−2 (x) gk (x) = . (n − 2 − k)! (n − k)! k=0
Since ∞ X k 3 (k − 1)gk (x) + (k − 1)(2k − 1 − x)gk−1 (x) + gk−2 (x) k=0
(n − k)!
= 0,
.
SOLUTIONS TO RAINVILLE’S “SPECIAL FUNCTIONS” (1960)
123
we may combine (1), (2), (3), (4) to get n3 (n − 1)φn (x) − (n − 3)(4n2 − 5n + 2)φn−1 (x) + (6n2 − 15n + 10)φn−2 (x) −(4n − 7)φn−3 (x) + φn−4 (x) + (n − 1)(2n − 1)φn−1 (x) − (4n − 5)φn−3 (x) +2φn−3 (x) − (n − 1)xφn−1 (x) + xφn−2 (x) + φn−2 (x) = 0, or n3 (n − 1)φn (x) − (n − 1)[(4n2 − 7n + 3 + x]φn−1 (x) + (6n2 − 19n + 16 + x)φn−2 (x) −(4n − 9)φn−3 (x) + φn−4 (x) = 0 We recall that φn (x) =
wn (x) and use it to obtain the final result: n!
n2 wn (x)−(4n2 −7n+3+x)wn−1 (x)+(6n2 −19n+16+x)wn−2 (x)−(n−2)(4n−9)wn−3 (x)+(n−2)(n−3)wn−4 (x) = 0. Problem 165. Show that the polynomial wn (x) of Exercise 164 may be written
wn (x) =
n X
1 F1
−n + k;
1 + k;
k=0
k
(−n)k (−x) 1 . (k!)3
Solution 165. Consider
wn (x) =
n X (−1)k n!Lk (x)
(k!)2 (n − k)!
k=0
.
Let us form
ψ
= = = = =
∞ X wn (x)tn n! n=0 ∞ X n X (−1)k Lk (x)tn n=0 k=0 ∞ X
(−1)k Lk (x)tn+k (k!)2 n! n,k=0 ∞ X ∞ X (−1)k+s xs tn+k
n,k=0 s=0 ∞ X k,s,n=0
Then
(k!)2 (n − k)!
k!n!(s!)2 (k − s)!
(−1)k xs tn+k (s!)2 k!k!(k + s)!
124SOLUTIONS BY SYLVESTER J. PAGANO AND LEON HALL; EDITED BY TOM CUCHTA
ψ
∞ X n X
(−1)k xs tn+s (s!)2 k!(n − k)!(k + s)! n,s=0 k=0 ∞ −n; X xs tn+s 1 = 1 F1 n!(s!)3 n,s=0 1 + s; ∞ X n −n + s; X xs t n 1 = . 1 F1 3 (s!) (n − s)! n=0 s=0 1 + s;
=
Hence
wn (x)
=
n X
1 F1
=
1 F1
1 + s; −n + s;
s=0
1
s=0 n X
−n + s;
1 + s;
n!xs − s)!
(s!)3 (n
s
(−n)s (−x) 1 , (s!)3
as desired. Problem 166. Define the polynomial vn (x) by (see Section 131, page 251) vn (x) =
n X (−1)k n!Pk (x) k=0
(k!)2 (n − k)!
in terms of the Legendre polynomial Pk (x). Show that vn (x) satisfies the recurrence relation n2 vn (x)−[4n2 −5n+2−(2n−1)x]vn−1 (x)+[6n2 −15n+11−(4n−5)x]vn−2 (x)−(n−2)(4n−7−2x)vn−3 (x) +(n − 2)(n − 3)vn−4 (x) = 0. Solution 166. Consider vn (x) =
n X (−1)k n!Pk (x) k=0
(k!)2 (n − k)!
(−1)k Pk (x) vn (x) = σn (x) and = gk (x). n! (k!)2 We know that Pk (x) satisfies the relation Put
kPk (X) − (2k − 1)xPk−1 (x) + (k − 1)Pk−2 (x) = 0. Hence gk (x) satisfies (−1)k k(k!)2 gk (x)−(−1)k−1 [(k−1)!]2 (2k−1)xgk−1 (x)+(−1)k−2 (k−1)[(k−2)!]2 gk−2 (x) = 0, or (1) Also
k 3 (k − 1)gk (x) + (k − 1)(2k − 1)xgk−1 (x) + gk−2 (x) = 0.
SOLUTIONS TO RAINVILLE’S “SPECIAL FUNCTIONS” (1960)
(2)
σn (x) =
125
∞ X gk (x) . (n − k)!
k=0
Because of the identity of form of equation (2) with the corresponding relations of Exercise 164, we may write immediately (3) n3 (n − 1)σn (x) − (n − 1)(4n2 − 5n + 2)σn−1 (x) + (6n2 − 15n + 10)σn−2 (x) −(4n − 7)σn−3 (x) + σn−4 (x) =
∞ X k 3 (k − 1) k=0
(n − k)!
gk (x).
In the same way we use the work in Exercise 164 to conclude that
(4)
(n−1)(2n−1)σn−1 (x)−(4n−5)σn−2 (x)+2σn−3 (x) =
∞ X (k − 1)(2k − 1)gk−1 (x)
(n − k)!
k=0
and (5)
σn−2 (x) =
∞ X gk−2 (x) . (n − k)!
k=0
Since ∞ X k 3 (k − 1)gk (x) + (k − 1)(2k − 1)xgk−1 (x) + gk−2 9x)
(n − k)!
k=0
= 0,
we get from (3), (4), (5) that n3 (n − 1)σn (x) − (n − 1)[4n2 − 5n + 2 − (2n − 1)x]σn−1 (x) +[6n2 − 15n + 11 − (4n − 5)x]σn−2 (x) − [4n − 7 − 2x]σn−3 (x) + σn−4 (x) = 0. But σn (x) =
vn (x) . Hence we arrive at the result n!
n2 vn (x) − [4n2 − 5n + 2 − (2n − 1)x]vn−1 (x) + [6n2 − 15n + 11 − (4n − 5)x]vn−2 (x) −(n − 2)(4n − 7 − 2x)vn−3 (x) + (n − 2)(n − 3)vn−4 (x) = 0. Problem 167. Show that the vn (x) of Exercise 166 satisfies the relations (1 − x2 )vn00 (x) − 2xvn0 (x) + n(n = 1)vn (x) = 2n2 vn−1 9x) − n(n − 1)vn−2 (x) and 0 (1−x2 )vn0 (x)+nxvn (x) = [(2n−1)x−1]vn−1 (x)−(n−1)xvn−2 (x)+(1−x2 )vn−1 (x).
Solution 167.
,
126SOLUTIONS BY SYLVESTER J. PAGANO AND LEON HALL; EDITED BY TOM CUCHTA
Problem 168. Let (−1)k ( 12 )n−k (2x)n−2k , k!(n − 2k)! so that the Legendre polynomial of Chapter 10 may be written γ(k, n) =
Pn (x) =
∞ X
γ(k, n).
k=0
Show that
xPn−1 (x) =
∞ X (n − 2k)γ(k, n)
2n − 2k − 1
k=0
xPn0 (x) = 0 Pn−1 (x)
∞ X
,
(n − 2k)γ(k, n),
k=0 ∞ X
=
∞ X −2kγ(k, n) , 2n − 2k − 1 k=0 ∞ X 0 Pn+1 (1 + 2n − 2k)γ(k, n), (x) =
Pn−2 (x) =
k=0
−2kγ(k, n),
etc.
k=0
Use Sister Celine’s method to discover the various differential recurrence relations and the pure recurrence relation for Pn (x). Solution 168. Let γ(x, n) =
(−1)k ( 21 )n−k (2x)n−2k . k!(n − k)!
Then Pn (x) =
∞ X
γ(k, n)
k=0
with our usual conventions. Also xPn−1 (x) =
∞ X (−1)k ( 1 )n−1−k (2x)n−2k 2
k=0
2 · k!(n − 1 − 2k)!
=
∞ X
n − 2k γ(k, n) 2(n − k − 12 k=0
or xPn−1 (x) =
∞ X k=0
n − 2k γ(k, n). 2n − 2k − 1
Next Pn−2 (x)
= =
∞ X (−1)k ( 1 )n−2−k (2x)n−2−2k 2
k=0 ∞ X k=1
k!(n − 2 − 2k)! (−1)k−1 ( 21 )n−1−k (2x)n−2k , (k − 1)!(n − 2k)!
so that Pn−2 (x) = Now
∞ X
∞ X −k −2k γ(k, n) = γ(k, n). 1 2n − 2k − 1 n−k− 2 k=0 k=0
SOLUTIONS TO RAINVILLE’S “SPECIAL FUNCTIONS” (1960)
xPn0 (x) =
∞ X (−1)k ( 1 )n−k (2x)n−2k 2
k!(n − 1 − 2k)!
k=0
=
∞ X
127
(n − 2k)γ(k, n),
k=0
and 0 Pn+1 (x) =
∞ X (−1)k ( 1 )n+1−k · 2(2x)n−2k 2
k!(n − 2k)!
k=0
=
∞ X
(2n − 2k + 1)γ(k, n).
k=0
Finally, 0 Pn−1 (x)
= = =
∞ X 9 − 1)k ( 1 )n−1−k 2(2x)n−2−2k 2
k=0 ∞ X k=1 ∞ X
k!(n − 2 − 2k)! (−1)k−1 ( 21 )N −k 2(2x)n−2k (k − 1)!(n − 2k)! −2kγ(k, n).
k=0
Problem 169. Apply sister Celine’s method to discover relations satisfied by the Hermite polynomials of Chapter 11. Solution 169. Problem 170. Find the various relations of Section 114 on Laguerre polynomials by using Sister Celine’s technique. Solution 170. Problem 171. Consider the pseudo-Laguerre polynomials (Boas and Buck [2;16]) fn (x) defined for nonintegral λ by n
X (−λ)n−k xk (−λ)n fn (x) = . 1 F1 (−n; 1 + λ − n; x) = n! k!(n − k)! k=0
Show that the polynomials fn (x) are not orthogonal with respect to any weight function over any interval because no relation of the form fn (x) = (An + Bn x)fn−1 (x) + Cn fn−2 (x) is possible. Obtain the pure recurrence relation nfn (x) = (x + n − 1 − λ)fn−1 (x) − xfn−2 (x). Solution 171. Consider fn (x) defined by n
fn (x) =
X (−λ)n−k xk (−λ)n . 1 F1 (−n; 1 + λ − n; x) = n! k!(n − k)! k=0
Put (k, n) = Then
(−λ)n−k xk . k!(n − k)!
128SOLUTIONS BY SYLVESTER J. PAGANO AND LEON HALL; EDITED BY TOM CUCHTA
fn (x) =
∞ X
(k, n)
k=0
fn−1 (x) =
fn−2 (x) =
∞ ∞ X X (−λ)n−1−k xk n−k = (k, n) k!(n − 1 − k)! λ+n−1−k
k=0 ∞ X
k=0
k=0
(n − k)(n − k − 1) (k, n) (−λ + n − 1 − k)(−λ + n − 2 − k)
and
xfn−1 (x) =
∞ X (−λ)n−1−k xk+1 k=0
xfn−2 (x) =
k!(n − 1 − k)!
∞ X (−λ)n−2−k xk+1 k=0
k!(n − 2 − k)!
=
∞ X k=0
=
∞ X k=1
∞
X (−λ)n−k xk = k(k, n) (k − 1)!(n − k)! k=0
∞
X (−λ)n−1−k xk k(n − k) = (k, n). (k − 1)!(n − 1 − k)! −λ + n − 1 − k k=0
Now consider Tn ≡ fn (x) − An fn−1 9x) − Bn xfN −1 (x) − Cn fn−2 (x). At once ∞ X Tn ≡ 1 − An k=0
n−k (n − k)(n − k − 1) − Bn k − Cn (k, n). −λ + n − 1 − k (−λ + n − 1 − k)(−λ + n − 2 − k)
If Tn ≡ 0, then the following must be an identity in k with λ independent of n: (1)
(−λ+n−1−k)(−λ+n−2−k)+An (n−k)(−λ+n−2−k)−Bn k(−λ+n−1−k)(λ+n−2−k)−Cn (n−k)(n−k−1) ≡ 0. From the coefficient of k 3 we see that Bn = 0. Then Tn ≡ 0 is impossible because fn−1 (x) and fn−2 (x) are of lower degree than fn (x). Now consider the identity fn (x) − Dn xfn−1 (x) − En fn−1 (x) − Fn xfn−2 (x) = 0. We need to determine Dn , En , Fn to satisfy 1 − Dn k − En
n−k k(n − k) − Fn = 0, −λ + n − 1 − k −λ + n − 1 − k
or (2)
− λ + n − 1 − k − Dn k(−λ + n − 1 − k) − En (n − k) − Fn k(n − k) ≡ 0.
1 −λ − 1 − Dn n(−λ − 1) = 0; Dn = or λ = −1 n 1 coeff k 2 : Dn + Fn = 0; Fn = − n −λ + n − 1 k = −λ + n − 1 : (−λ − 1)En + (λ − 1)Fn (−λ + n − 1) = 0; En = . n k=n:
SOLUTIONS TO RAINVILLE’S “SPECIAL FUNCTIONS” (1960)
129
Hence, either λ = −1 or nfn (x) − xfn−1 (x) − (−λ + n − 1)fn−1 (x) + xfn−2 (x) = 0, or nfn (x) = [x + n − 1 − λ]fn−1 (x) − xfn−2 (x) = 0, as desired. For the polynomials in each of the following examples, use Sister Celine’s technique to discover the pure recurrence relation and whatever mixed relations exist. Problem 172. The Bessel polynomials of Section 150. Solution 172. Problem 173. Bendient’s polynomials Rn of Section 151. Solution 173. Problem 174. Bendient’s polynomials Gn of Section 151. Solution 174. Problem 175. Shively’s polynomials Rn of Section 152. Solution 175. Problem 176. Shively’s polynomials σn of Section 152. Solution 176. 18. Chapter 15 Solutions In Exercises 177-184, Hn (x), Pn (x), Ln (x) denote the Hermite, Legendre, and simple Laguerre polynomials, respectively. Derive each of the stated symbolic relations. Problem 177. Hn (x + y) + [H(x) + 2y]n . Solution 177. Consider ∞ X Hn (x + y)tn n! n=0
= exp[2(x + y)t − t2 ] 2 = exp(2yt) exp(2xt ! −∞t ) ! ∞ n n X X (2y) t Hn (x)tn = n! n! n=0 n=0 n n−k X X (2y) H (x) k = )∞ tn . n=0 k!(n − k)! k=0
It follows that Hn (x + y) =
n X n!(2y)n−k Hk (x) k=0
or its equivalent
k!(n − k)!
130SOLUTIONS BY SYLVESTER J. PAGANO AND LEON HALL; EDITED BY TOM CUCHTA
Hn (x + y) + {H(x) + 2y}n . Problem 178. 1
n
[H(x) + H(y)]n + 2 2 Hn (2− 2 (x + y)). Solution 178. Consider ∞ X [H(x) + H(y)]n tn n! n=0
+
∞ X n X Hk (x)Hn−k (y)tn
k!(n − k)! ! ∞ ! X Hn (y)tn Hn (x)tn = n! n! n=0 n=0 = exp(23xt − t2 ) exp(2yt − t2 ) 2 = exp[2(x + y)t − 2t ] √ 2 x+y √ √ = exp 2 ( 2t) − ( 2t) . 2 n=0 k=0 ∞ X
It follows that ∞ ∞ Hn X X [H(x) + H(y)]n tn + n! n=0 n=0
x+y √ 2
√ ( 2t)n ,
or n
{H(x) + H(y)} + 2 Problem 179.
1 2n
Hn
x+y √ 2
.
n 1−x 1+x Pn (x) + n! L −L 2 2
Solution 179. Consider ∞ X Pn (x)tn t(x + 1) t(x − 1) F = F −; 1; . −; 1; 0 1 0 1 (n!)2 2 2 n=0 Now et 0 F1 (−; 1; −yt) =
∞ X Ln (y)tn . n! n=0
Hence ∞ X Pn (x)tn (n!)2 n=0
t(x − 1) t(x + 1) = et 0 F1 −; 1; · e−t 0 F1 −; 1; 2 2 n! ∞ n! ∞ 1−x 1+x n X X Ln 2 t (−1) Ln 2 t = n! n! n=0 n=0 ∞ n 1+x 1−x k X X (−1) Lk ( 2 )Ln−k 2 = tn . k!(n − k)! n=0 k=0
Hence
SOLUTIONS TO RAINVILLE’S “SPECIAL FUNCTIONS” (1960)
Pn (x) = n!
n X n!(−1)k L ( 1+x )Ln−k ( 1−x ) 2
k!(n − k)!
k=)
2
,
or n 1−x 1+x Pn (x) + n! L −L . 2 2 Problem 180. Hn
1 P (x) + [H(x) − P (x)]n . 2
Solution 180. Consider ∞ X Hn n=0
1 2 P (x)
tn
n
+
[2] ∞ X X (−1)k Pn−2k (x)tn
k!(n − 2k)! ! ∞ ! X Pn (x)tn (−1) t = n! n! n=0 n=0 2 2 t (x − 1) −t2 xt = e e 0 F1 −; 1; ! ∞ 4 ! ∞ n X X (−1)n Pn (x)tn Hn (x)t = n! n! n=0 n=0 ∞ X n k X (−1) Pk (x)Hn−k (x)tn = . k!(n − k)! n=0
n!
n=0 k=0 ∞ X
n 2n
k=0
It follows that Hn
1 P (x) 2
+
n X (−1)k n!Pk (x)Hn−k (x)
k!(n − k)! n + {H(x) − P (x)} . k=0
Problem 181. [H(x) − 2P (x)]2n+1 + 0,
[H(x) − 2P (x)]2n + (−1)n 22n Solution 181. Consider
1 Ln (x2 − 1). 2 n
131
132SOLUTIONS BY SYLVESTER J. PAGANO AND LEON HALL; EDITED BY TOM CUCHTA
ψ
= +
∞ X [1 + (x) − 2P (x)]n tn n! n=0 ∞ X n k X (−2) Pk (x)Hn−k (x)tn
k!(n − k)! ! ∞ ! X Pn (x)(−2t)n Hn (x)tn = n! n! n=0 n=0 4t2 (x2 − 1) = exp(2xt − t2 ) exp(−2xt)0 F1 −; 1; 4 = exp(−t2 )0 F1 (−; 1; t2 (x2 − 1)) ∞ X (−1)n Ln (x2 − 1)t2n = . n! n=0 n=0 k=0 ∞ X
It follows that [H(x) − 2P (x)]2n+1 + 0 and 2n
[H(x) − 2P (x)]
(−1)n (2n)! Ln (x2 − 1) = (−1)n 22n + n!
1 Ln (x2 − 1). 2 n
Problem 182. [H(x) − P (2x)]2n+1 + 0, [H(x) − P (2x)]2n + (−1)n 22n
1 1 Ln x2 − . 2 n 4
Solution 182. Consider ∞ X [H(x) − P (2x)]n tn n! n=0
+
∞ X n X (−1)k Pk (2x)Hn−k (x)tn
k!(n − k)! ! ∞ ! X Pn (2x)(−t)n Hn (x)tn = n! n! n=0 n=0 t2 (4x2 − 1) = exp(2xt − t2 ) exp(−2xt)0 F1 −; 1; 4 1 2 2 2 = exp(−t )0 F1 −; 1; −(−t )(x − ) 4 ∞ X (−1)n Ln (x2 − 41 )t2n = . n! n=0 n=0 k=0 ∞ X
We may conclude that [H(x) − P (2x)]2n+1 + 0 and [H(x) − P (2x)]2n +
(−1)n (2n)!Ln (x2 − 41 ) = (−1)n 22n n!
1 1 L n x2 − . 2 n 4
SOLUTIONS TO RAINVILLE’S “SPECIAL FUNCTIONS” (1960)
Problem 183.
Hn
1 H(x) 2
n
133
1
+ 2 2 Hn (2− 2 x).
Solution 183. From n
∞ X Hn ( 21 H(x))tn n! n=0
+
[2] ∞ X X (−1)k Hn−2k (x)tn
k!(n − 2k)! ! ∞ ! X Hn (x)tn (−1) t = n! n! n=0 n=0 = exp(−t2 ) exp(2xt − t2 ) 2 = exp(2xt − 2t) √ 2 x √ 2t − ( 2t) = exp 2 √ 2 √ x ∞ H ( √ )( 2t)n X n 2 = n! n=0 n=0 k=0 ∞ X
n 2n
we may conclude that Hn
n 1 x H(x) + 2 2 Hn √ . 2 2
Problem 184. Hn (xy) + [H(x) + 2x(y − 1)]n . Solution 184. From ∞ X Hn (xy)tn n! n=0
= exp(2xyt − t2 ) = exp(2xt − t2 ) exp[2x(y ! ∞ − 1)t] ! ∞ X X [2x(y − 1)]tn Hn (x)tn = n! n! n=0 n=0 n ∞ X k X [2x(y − 1)] Hn−k (x)tn = , k!(n − k)! n=0 k=0
it follows that Hn (xy) + [H9x) + 2x(y − 1)]n . Problem 185. Use Laplace’s first integral for Pn (x) to derive equation (20), preceding these exercises. Solution 185. We know that Pn (x) =
1 Γ
Z
π
[x +
p
x2 − 1 cos φ]n dφ.
0
We have defined vn (x) by vn (x) + Ln (P (x)) =
n X (−1)k n!Pk (x) k=0
(k!)n (n − k)!
.
134SOLUTIONS BY SYLVESTER J. PAGANO AND LEON HALL; EDITED BY TOM CUCHTA
Hence √ Z n 1 π X (−1)k n![x + x2 − 1 cos φ vn (x) = dφ π 0 (k!)2 (n − k)! Z π k=0 p 1 = Ln x + x2 − 1 cos φ dφ. π 0 Problem 186. For the vn (x) of equation (15), page 251, evaluate Z
1
vm (x)Pk (x)dx, −1
and use your result to establish equation (21), preceding these exercises. Z
1
Solution 186. For the vn (x) of Exercise 185 we wish to evaluate
vm (x)Pk (x)dx. −1
Now 1
Z
1
Z vm (x)Pk (x)dx =
Pk (x)
−1
−1
m X (−1)s m!Ps (x)dx s=0
(s!)2 (m − s)!
.
Hence, if m < k, then s < k and we obtain Z
1
vm (x)Pk (x)dx = 0, m < k. −1
If m ≥ k, we get Z
1
vm (x)Pk (x)dx −1
Z
m X
Z 1 (−1)s m! = Pk (x)Ps (x)dx (s!)2 (m − s)! −1 s=0 Z 1 (−1)k m! = P 2 (x)dx (k!)2 (m − k)! −1 k (−1)k 2m! . = (k!)2 (m − k)!(2k + 1)
1
Now consider
vN (x)vm (x)dx. −1
Let m ≥ n. Then Z
1
Z
1
vn (x)vm (x)dx = −1
vn (x) −1
m X (−1)k m!Pk (x) k=0
(k!)2 (m − k)!
dx.
For k > n the integrals involved in the sum are zero. We may therefore write
SOLUTIONS TO RAINVILLE’S “SPECIAL FUNCTIONS” (1960)
1
Z
vn (x)vm (x)dx
=
−1
=
n X k=0 n X k=0 n X
(−1)k m! (k!)2 (m − k)!
Z
135
1
vn (x)Pk (x)dx −1
(−1)k m! (−1)k 2 · n! (k!)2 (m − k)! (k!)2 (n − k)!(2k + 1)
(−n)k (−m)k (k!)4 (k + 21 ) k=0 n X (−n)k (−m)k ( 1 )k 2 =2 . 4( 3 ) (k!) k 2 k=0 =
Hence, since m and n are interchangeable, 1 −n, −m, ; 2 1 vn (x)vm (x)dx = 23 F4 −1 3 1, 1, 1, ; 2 Problem 187. Show that y−x n n (y + x) Pn + n! {L(x) − L(y)} . y+x
Z
1
.
Solution 187. Consider ∞ n X (y + x)n Pn ( y−x y+x )t n=0
(n!)2
−; 1;
t(y + x)[ y−x y+x − 1]
!
0 F1 −; 1; 2 = 0 F1 (−; 1; −xt)0 F1 (−; 1; yt) −t = et 0 F1 (−; 1; −xt)e 0 F1 (−; 1; −y(−t))! ! ∞ ∞ 4 X X Ln (x)t (−1)n Ln (y)tn = n! n! n=0 n=0 n ∞ k X X (−1) Lk (y)Ln−k (x)tn = . k!(n − k)! n=0
= 0 F1
k=0
We may conclude that n
(y + x) Pn
y−x y+x
= n!
n X (−1)k n!Lk (y)Ln−k (x)
k!(n − k)! k=0 n + n! {L (x) − L (y)} ,
as desired. Problem 188. Define polynomials φn (x, y) by φn (x, y) + Hn (xL(y)) . Show that ∞ X φn (x, y)tn = exp(2xt − t2 )0 F1 (−; 1; −2xyt). n! n=0
t(y + x)[ y−x y+x + 1] 2
!
136SOLUTIONS BY SYLVESTER J. PAGANO AND LEON HALL; EDITED BY TOM CUCHTA
Solution 188. We define φn (x, y) by φn (x, y) + Hn (xL (y)) [n 2] X (−1)k n!2n−2k xn−2k Ln−2k (y) = . k!(n − 2k)! k=0
Then ∞ X φn (x, y)tn n! n=0
∞ X (−1)k (2x)n Ln (y)tn+2k k!n! n,k=0 ∞ X Ln (y)(2xt)n = exp(−t2 ) n! n=0 2 = exp(−t ) exp(2xt)0 F1 (−; 1; −2xgt) = exp(2xt − t2 )0 F1 (−; 1; −2xyt),
=
as desired. 19. Chapter 16 Solutions Problem 189. Let
1−x 1+x gn (x) = . (1 + α)n (1 + β)n Use Bateman’s generating function, page 256, to see that (α,β)
(1 + x)n Pn
∞ X
gn (x)tn = 0 F1 (−; 1 + α; −xt)0 F1 (−; 1 + β; t)
n=0
and thus show that, in the sense of Section 126, gn (x) is of σ-type zero with σ = D(θ + α). Show also that gn (x) is of Sheffer A-type unity. Solution 189. Let gn (x) =
(α,β) 1−x ( 1+x )
(1 + x)n Pn
(1 + α)n (1 + β)n
.
Then ∞ X n=0
gn (x)tn
=
(α,β) 1−x ∞ X Pn ( 1+x )[t(1 + x)]n
(1 + α)n (1 + β)n h i 1−x t(1 + x) + 1 1+x t(1 + x) 1 − x −; 1 + β; −1 = 0 F1 −; 1 + α; 0 F1 2 1+x 2 n=0
= 0 F1 (−; t + α; −xt)0 F1 (−; 1 + β; t). We now see that with σ = D(θ+α), gn (x) is of σ-type zero with A(t) = 0 F1 (−; 1+ β; t), H(t) = −t, J(t) = −t. Since σgn (x) = gn−1 (x), it follows that in the Sheffer classification notation
SOLUTIONS TO RAINVILLE’S “SPECIAL FUNCTIONS” (1960)
J(x, D)
137
= D(θ + α) = D(xD + α) = (1 + α)D + xD 2 .
Thus T0 (x) = 1 + α, T1 (x) = x, Tk (x) ≡ 0 for k ≥ 2. So gn (x) is of Sheffer A-type unity. (α,β) It might be of interest to see what we get on Pn (x) because of the fact that gn (x) is of σ-type zero. Problem 190. Show that (α,β)
(α,β)
2x(α+β+n)DPn(α,β) (x)+[x(α−β)−(α+β+2n)]DPn−1 (x) = (α+β+n)[2nPn(α,β) (x)−(α−β)Pn−1 (x)], which reduces to equation (2), page 159, for α = β = 0. Solution 190. From equation (3), page 457 and (6) page 450 we get, with D ≡
d , dx
(α,β)
(α,β)
(1)
(x−1)(α+β+n)DPn(α,β) (x)+(x−1)(α+n)DPn−1 (x) = n(α+β+n)Pn(α,β) (x)−(α+n)(α+β+n)Pn−1 (x),
(2)
(x+1)(α+β+n)DPn(α,β) (x)−(x+1)(β+n)DPn−1
(α,beta)
(α,β)
(x) = n(α+β+n)Pn(α,β) (x)+(β+n)(α+β+n)Pn−1 (x).
We add (1) and (2) to get (3)
(α,β)
(α,β)
2x(α+β+n)DPn(α,β) (x)+[x(α−β)−(α+β+2n)]DPn−1 (x) = (α+β+n)[2nPn(α,β) (x)−(α−β)Pn−1 (x)]. In (3) put α = β = 0 to obtain 0 nxPn0 (x) − nPn−1 (x) = n2 Pn (x),
or 0 nPn (x) = xPn0 (x) − Pn−1 (x)
which is (2), page 270. Problem 191. Show that (α,β)
(α,β)
2(α+β+n)DPn(α,β) (x)+[α−β−x(α+β+2n)]DPn−1 (x) = (α+β+n)(α+β+2n)Pn−1 (x), which reduces to equation (6), page 159, for α = β = 0. Solution 191. In Exercise 190 substract (1) from (2) to get (α,β)
(α,β)
2(α+β+n)DPn(α,β) (x)+[α−β−x(α+β+2n)]DPn−1 (x) = (α+β+n)(α+β+2n)Pn−1 (x). In the above put α = β = 0 and divide by 2n to get 0 Pn0 (x) − xPn−1 (x) = nPn−1 (x),
which is (6), page 271, with a shift of index.
138SOLUTIONS BY SYLVESTER J. PAGANO AND LEON HALL; EDITED BY TOM CUCHTA
Problem 192. Show that (α,β)
(α,β)
2n(α+β+n+1)DPn+1 (x)+[(α+β)(n+2)x+n(α−β)]DPn(α,β) (x)+(n+1)[(α−β)x−(α+β+2n)]DPn−1 (x) (α,β)
= n[2(n+1)(α+β+n)+(α+β+n+1)(α+β+2n+2)]Pn(α,β) (x)−(α−β)(n+1)(α+β+n)Pn−1 (x), which reduces to equation (5), page 159, for α = β = 0. Solution 192. From Exercise 191 with a shift of index we g et (α,β)
2(α+β+n+1)DPn+1 (x)+[(α−β)−x(α+β+2n+2)]DPn(α,β) (x) = (α+β+n+1)(α+β+2n+2)Pn(α,β) (x). From Exercise 190 we get (α,β)
2x(n + 1)(α + β + n)DPn(α,β) (x) + (n + 1)[x(α − β) − (α + β + 2n)]DPn−1 (x) (α,β)
= 2n(n + 1)(α + β + n)Pn(α,β) (x) − (α − β)(n = 1)(α + β + n)Pn−1 (x). Let us form n times the first equation plus the second equation, we thus get (α,β)
2n(α + β + n + 1)DPn+1 (x) +[n(α − β) + x(−nα − nβ − 2n2 − 2n + 2nα + 2nβ + 2n2 + 2α + 2β + 2n)]DPn(α,β) (x) (α,β)
+(n + 1)[x(α − β) − (α + β + 2n)]DPn−1 (x) (α,β)
= n[(α+β+n+1)(α+β+2n+2)+2(n+1)(α+β+n)]Pn(α,β) (x)−(α−β)(n+1)(α+β+n)Pn−1 (x), or
(α,β)
(α,β)
2n(α+β+n+1)DPn+1 (x)+[n(α−β)+x(α+β)(n+2)]DPn(α,β) (x)+(n+1)[x(α−β)−(α+β+2n)]DPn−1 (x) (α,β)
= n[(α+β+n+1)(α+β+2n+2)+2(n+1)(α+β+n)]Pn(α,β) (x)−(α−β)(n+1)(α+β+n)Pn−1 (x), which is the desired result. On the above equation put α = β = 0 and divide by 2n(n + 1) to get 0 0 Pn+1 (x) − Pn−1 (x) = (2n + 1)Pn (x), which is equation 5, page 271.
Problem 193. Use the method of Section 142 to show that Pn(α,β) (x) =
n (α,0) (x) (1 + α)n X (−1)n−k (β)n−k (1 + α + β)n+k (1 + α + 2k)Pk . (1 + α + β)n (n − k)!(1 + α)n+k+1 k=0
(α,β)
(α,β)
Solution 193. We wish to expand Pn (x) in a series involving Pk Now we have from (2), page 457, with p = 0, (1)
1−x s
s = (1 + α)s
and from (2), page 476,
(x).
s (α,0) X (1 + α + 2k)(1 + α)k P (x) (−1)k s! k
k=0
(1 + α)s+1+k (1 + α)k
(s − k)!
SOLUTIONS TO RAINVILLE’S “SPECIAL FUNCTIONS” (1960)
(α,β)
(1 + α + β)n Pn (1 + α)n
(2)
(x)
=
139
n X (−1)s (1 + α + β)n+s ( 1−x )s s
s!(n − s)!(1 + α)s
s=0
.
Consider the series ∞ (α,β) X (1 + α + β)n Pn (x)tn (1 + α)n n=0 ∞ s X (−1) (1 + α + β)n+2s ( 1−x )s tn+s 2 = s!n!(1 + α) s n,s=0
ψ(x, t)
=
=
∞ X s (α,0) X (−1)k s!(1 + α + 2k)(−1)s (1 + α + β)n+2s Pk (x)tn+s (s − k)!(1 + α)s+1+k s!n! n,s=0 k=0
∞ (α,0) X (−1)s (1 + α + 2k)(1 + α + β)n+2k+2s Pk (x)tn+s+k = s!(1 + α)s+1+2k n! n,k,s=0 ∞ n X X (−1)s (1 + α + β)n+2k+s (1 + α + 2k)Pk (x)(α,0) tn+k = s!(1 + α)2k+1+s (n − s)! s=0 n,k=0
Then
ψ(x, t)
= = = =
∞ X n,k=0 ∞ X n,k=0 ∞ X
2 F1
−n, 1 + α + β + n + 2k;
(α,0)
(1 + α + β)n+2k (1 + α + 2k)Pk 1 n!(1 + α)2k+1
(x)tn+k
2 + α + 2k; (α,0) Γ(2 + α + 2k)Γ(1 − β)(1 + α + β)n+2k (1 + α + 2k)Pk (x)tn+k Γ(2 + α + 2k + n)Γ(1 − β − n)n!(1 − α)2k+1 (α,0)
(1 + α)2k+1 (−1)n (β)n (1 + α + β)n+2k (1 + α + 2k)Pk (1 + α)n+2k+1 n!(1 + α)2k+1
n,k=0 n ∞ X X
n=0 k=0
(α,0)
(−1)n−k (β)n−k (1 + α + β)n+k (1 + α + 2k)Pk (n − k)!(1 + α)n+k+1
(x)tn
(x)tn+k
.
We may now conclude that Pn(α,β) (x) =
n (α,0) (x) (1 + α)n X (−1)n−k (β)n−k (1 + α + β)n+k (1 + α + 2k)Pk . (1 + α + β)n (n − k)!(1 + α)n+k+1 k=0
Problem 194. Use the result obtained in Section 142 to evaluate Z
1
−1
(α,α)
(1 − x2 )α Pn(α,β) (x)Pk
(x)dx.
Solution 194. We know from Section 142 that Pn(α,β) (x)
n (α,α) (x) (1 + α)n X (−1)n−k (β − α)n−k (1 + |alpha + β)n+k (1 + 2α)k (1 + 2α + 2k)Pk = . (1 + α + β)n (n − k)!(1 + 2α)n+k+1 (1 + α)k k=0
(α,β)
By using β = α in the orthogonality property of Pn
(x) we get
140SOLUTIONS BY SYLVESTER J. PAGANO AND LEON HALL; EDITED BY TOM CUCHTA
Z
1 (α,α) (1 − x2)α Pn(α,α) (x)Pm (x)dx = 0, m 6= n
−1
and from (11), page 454, Z
1
(1 − x2 )α Pn(α,α) (x)]2 dx =
−1
21+2α Γ(1 + α + n)Γ(1 + α + n) . n!(1 + 2α + 2n)Γ(1 + 2α + n)
We may now write 1
Z A(k, n)
(α,α)
= −1
(1 − x2 )α Pn(α,β) (x)Pk
(x)dx
Z n (1 + α)n X (−1)n−s (β − α)n−s (1 + α + β)n+s (1 + 2α)s (1 + 2α + 2s) 1 (α,α) (1 − x2 )α Ps(α,α) (x)Pk = (x)dx. (1 + α + β)n s=0 (n − s)!(1 + 2α)n+s+1 (1 + α)s −1 If k > n, then k > s and we get A(k, n) = 0 for k > n. If 0 ≤ k ≤ n, then the integrals in the sum are zero except for the one in which s = k. We thus arrive at A(k, n) =
(1 + α)n (−1)n−k (β − α)n−k (1 + α + β)n+k (1 + 2α)k (1 + 2α + 2k)21+2α Γ(1 + α + k)Γ(1 + α + k) (n − k)!k!Γ(1 + α + β + n)Γ(1 + 2α + n − k + 1)Γ(1 + α + k)Γ(1 + 2α + k)
or A(k, n)
22+2α (−1)n−k (β − α)n−k Γ(1 + α + n)Γ(1 + α + β + n + k)Γ(1 + 2α + k)Γ(1 + α + k)Γ(1 + α + k) (n − k)!k!Γ(1 + α + β + n)Γ(1 + 2α + n + k + 1)Γ(1 + α + k)Γ(1 + 2α + k) 21+2α (−1)n−k (β − α)n−k Γ(1 + α + k)Γ(1 + α + n)Γ(1 + α + β + n + k) = k!(n − k)!Γ(1 + α + β + n)Γ(2 + 2α + n + k =
for 0 ≤ k ≤ n. Problem 195. Use the result in Exercise 193 above to evaluate Z
1
−1
(α,0)
(1 − x)α Pn(α,β) (x)Pk Z
1
Solution 195. We wish to evaluate −1
(x)dx. (α,0)
(1 − x)α Pn(α,β) (x)Pk
(x)dx.
We know that Z
1
−1
(α,0)
(1−x)α Pk
(x)Ps(α,0) (x)dx
= 0; k 6= s 21+α 21+α Γ(1 + α + k)Γ(1 + k) = ;s = k = k!(1 + α + 2k)Γ(1 + α + k) 1 + α + 2k
Now, by Exercise 193 we get Z
1 α
(1−x) −1
(α,0) Pn(α,β) (x)Pk (x)dx
Z n (1 + α)n X (−1)n−s (β)n−s (1 + α + β)n+s (1 + α + 2s) 1 (α,0) = (1−x)α Ps(α,0) (x)Pk (x)dx. (1 + α + β)n s=0 (n − s)!(1 + α)n+s+1 −1
SOLUTIONS TO RAINVILLE’S “SPECIAL FUNCTIONS” (1960)
141
If k > n, then k > s, so we get Z
1
−1
(α,0)
(1 − x)α Pn(α,β) (x)Pk
(x)dx = 0, k > n.
If 0 ≤ k ≤ n, each integral in the sum is zero except for the one in which s = k. Hence 1
(1 + α)n (−1)n−k (β)n−k (1 + α + β)n+k (1 + α + 2k)21+α . (1 + α + β)n (n − k)!(1 + α)n+k+1 (1 + α + 2k) −1 v Problem 196. Use Theorem 84, page 269, with y = x = − to conclude that 1−v 1 1 a, b; a, b; a, b, (a + b), (a + b + 1); 2 2 v 2 F1 v = 4 F3 2 F1 4v(1 − v) . c; 1 + a + b − c; a + b, c, 1 + a + b − c;
Z
(α,0)
(1−x)α Pn(α,β) (x)Pk
(x)dx =
Solution 196. Theorem 84 is
a, b;
2 F1
−x 2 F1 1−x
a, b;
−y −x −y , . = F4 a, b; c, 1 − c + a + b; 1−y (1 − x)(1 − y) (1 − x)(1 − y)
1 − c + a + b; −x Now put x = y = . Then the above becomes 1−x a, b; a, b; v 2 F1 v = F4 (a, b; c, 1 − c + a + b; v(1 − v), v(1 − v)) 2 F1 c; 1 − c + a + b; ∞ X (a)n+k (b)n+k v n+k (1 − v)n+k = k!n!(c)k (1 − c + a + b)n n,k=0 n ∞ X X (a)n (b)n v n (1 − v)n = k!(n − k)!(c)k (1 − c + a + b)n−k n=0 k=0 ∞ −n, c − a − b − n; X (a)n (b)n v n (1 − v)n 1 = 2 F1 n!(1 − c + a + b)n n=0 c; ∞ n X Γ(c)Γ(a + b + 2n)(a)n (b)n v (1 − v)n = Γ(c + n)Γ(a + b + n)n!(1 − c + a + b)n n=0 ∞ X (a + b)2n (a)n (b)n v n (1 − v)n = (c) (a + b)n n!(1 − c + a + b)n n=0 n a+b a+b+1 a, b, , ; 2 2 = 4 F3 4v(1 − v) , c;
c, a + b, 1 − c + a + b; as desired. Problem 197. Use the result in Exercise 196 above, and Theorem 25, page 67, to show that
142SOLUTIONS BY SYLVESTER J. PAGANO AND LEON HALL; EDITED BY TOM CUCHTA
2
2 2a, 2b, a + b; y = 3 F2 1 2a + 2b, a + b + ; 2
a, b;
F1
1 a+b+ ; 2
y .
Solution 197. In Exercise 196 above replace a by (2a), b by (2b), and then put 1 1 1 c = a + b + to get (since 1 + 2a + 2b − a − b − = a + b + ) 2 2 2 2 1 2a, 2b; 2a, 2b, a + b, a + b + ; 2 v 4v(1 − v) = 4 F3 2 F1 1 1 1 a+b+ ; 2a + 2b, a + b + , a + b + ; 2 2 2 2a, 2b, a + b; 4v(1 − v) . = 3 F2 1 2a + 2b, a + b + ; 2 By Theorem 25, page 114, we may now write
2 F1
a, b;
1 a+b+ ; 2
2a, 2b, a + b; 4v(1 − v) = F 3 2 1 2a + 2b, a + b + ; 2
4v(1 − v) ,
or 2
F1
2a, 2b, a + b; y = F 3 2 1 2a + 2b, a + b + ; 2
a, b; 1 a+b+ ; 2
y .
20. Chapter 17 Solutions Problem 198. Show that the Gegenbauer polynomial Cnv (x) and the Hermite polynomial Hn (x) are related by n
Cnv (x)
=
[2] X
2 F0 (−k, v
+ n − k; −; 1)
k=0
(−1)k (v)n−k Hn−2k (x) . k!(n − 2k)!
Solution 198. From n
Cnv (x)
=
[2] X (−1)s (v)n−s (2x)n−2s s=0
s!(n − 2s)!
we get ∞ X n=0
We know that
Cnv (x)tn
∞ X (−1)s (v)n+s (2x)n tn+2s = . s!n! n,s=0
SOLUTIONS TO RAINVILLE’S “SPECIAL FUNCTIONS” (1960)
[n]
2 X (2x)n Hn−2k (x) = . n! k!(n − 2k)!
k=0
Then ∞ X
n
[2] ∞ X X (−1)s (v)n+s Hn−2k (x)tn+2s = s!k!(n − 2k)! n,s=0 k=0 ∞ X (−1)s (v)n+s+2k Hn (x)tn+2k+2s = s!k!n!
Cnv (x)tn
n=0
=
n,k,s=0 ∞ X k X
n,k=0 s=0 ∞ X k X
(−1)s (v)n+2k−s Hn (x)tn+2k s!(k − s)! n!
(1 )k+s (v)n+k+s Hn (x)tn+2k s!(k − s)! n! n,k=0 s=0 ∞ −k, v + n + k; X (−1)k (v)n+k Hn (x)tn+2k 1 = 2 F0 k!n! −; n,k=0 n [2] ∞ X X (−1)k (v)n−k Hn−2k (x)tn = . 2 F0 (−k, v + n − k; −; 1) k!(n − 2k)! n=0 =
k=0
Hence n
Cnv (x)
=
[2] X
2 F0 (−k, v
+ n − k; −; 1)
k=0
(−1)k (v)n−k Hn−2k (x) . k!(n − 2k)!
Problem 199. Show that [n]
2 v (v + n − 2k)Cn−2k (x) Hn (x) X = (−1)k 1 F1 (−k; 1 + v + n − 2k; 1) . n! k!(v)n+1−2k
k=0
Solution 199. Consider ∞ ∞ X X Hn 9x)tn (−1)s (2x)n tn+2s = . n! s!n! n,s=0 n=0
Now by equation (36), [n]
2 X (v + n − 2k)Cn−2k (x) (2x)n = . n! v · k!(v)n+1−k
k=0
Hence
143
144SOLUTIONS BY SYLVESTER J. PAGANO AND LEON HALL; EDITED BY TOM CUCHTA
n
∞ X Hn (x)tn n! n=0
[2] ∞ X v X (−1)s (v + n − 2k)Cn−2k (x)tn+2s
= = = = =
s!k!(v)n+1−k
n,s=0 k=0 ∞ X
(−1) (v + n)Cnv (x)tn+2k+2s s!k!(v)n+1+k
n,k,s=0 ∞ X k X n,k=0 s=0 ∞ X k X n,k=0 s=0 ∞ X
s
(−1)s (v + n)Cnv (x)tn+2k s!(k − s)!(v)n+1+k−s (−1)k−s (v + n)Cnv (x)tn+2k s!(k − s)!(v)n+1+s
1 F1 (−k; 1
n,k=0
+ v + n; 1)
(−1)k (v + n)Cnv (x)tn+2k k!(v)n+1
Then
[n]
2 v (1 )k (v + n − 2k)Cn−2k (x) Hn (x) X = . 1 F1 (−k; 1 + v + n − 2k; 1) n! k!(v)n+1−2k
k=0
Problem 200. Show, using the modified Bessel function of Section 65, that
e
xt
=
−v ∞ X 1 t Γ(v) (v + n)Iv+n (t)Cnv (x). 2 n=0
Solution 200. We know that
( 2t )v+n t2 Iv+n (t) = 0 F1 −; 1 + v + n; Γ(1 + v + n) 4 ∞ X ( 2t )v+n+2k = k!Γ(1 + v + n + k) k=0
Now
[n]
2 v X (v + n − 2k)Cn−2k (x) xn = . n n! 2 k!(v)n+1−k
k=0
Hence
SOLUTIONS TO RAINVILLE’S “SPECIAL FUNCTIONS” (1960)
C xt
=
∞ X xn t n n! n=0 n
= =
[2] ∞ X v X (v + n − 2k)Cn−2k (x)tn
2n k!(v)n+1−k
n=0 k=0 ∞ X
(v + n)Cnv (x)tn+2k 2n+2k k!(v)n+1+k n,k=0 ∞ ∞ XX ( t )n+2k Γ(v)
2 (v + n)Cnv (x) k!Γ(v + n + 1 + k) −v X ∞ ( 2t )n+v+2k t (v + n)Cnv (x). = Γ(v) 2 k!Γ(v + n + 1 + k) n=0
=
n=0 k=0
Therefore
e
xt
−v X ∞ t = |Gamma(v) (v + n)Iv+n (t)Cnv (x). 2 n=0
Problem 201. Show that
n
[2] X (v − 12 )k (v)n−k (1 + 2n − 4k)Pn−2k (x) v Cn (x) = . k!( 23 )n−k k=0
Solution 201. From
n
Cnv (x)
=
[2] X (−1)k (v)n−k (2k)n−2k k=0
k!(n − 2k)!
and
[n]
2 X (2x)n (2n − 4x + 1)Pn−2k (x) = n! k!( 32 )n−k
k=0
we obtain
145
146SOLUTIONS BY SYLVESTER J. PAGANO AND LEON HALL; EDITED BY TOM CUCHTA
∞ X
n
Cnv (x)tn
=
n=0
=
[2] ∞ X X (−1)s (v)n−s (2x)n−2s tn
s!(n − 2s)!
n=0 s=0 ∞ X
s
(−1) (v)n+s (2x)n tn+2s s!n! n,s=0 n
[2] ∞ X X (−1)s (v)n+s (2n + 1)Pn (x)tn+2k+2s = s!k!( 32 )n+k n,s=0 k=0 ∞ X (−1)s (v)(2n + 1)Pn (x)tn+2k+2s = s!k!( 32 )n+k n,k,s=0
=
∞ X k X (−1)s (v)(2n + 1)Pn (x)tn+2k n,k=0 s=0 ∞ X k X
s!(k − s)!( 32 )n+k−s
(−1)k−s (v)n+k+s (2n + 1)Pn (x)tn+2k s!(k − s)!( 23 )n+s n,k=0 s=0 −k, v + n + k; ∞ k n+2k X 1 (−1) (v)n+k (2n + 1)Pn (x)t = . 2 F1 3 k!( 32 )n n,k=0 + n; 2 Therefore we get, using Example 5, page 69, =
∞ X
Cnv (x)tn
=
n=0
=
=
∞ X (−1)k ( 23 )n (v − 12 )k (−1)k (v)n+k (2n + 1)Pn (x)tn+2k ( 32 )n+k k!( 32 )n n,k=0 ∞ X (v − 1 )k (v)n+k (2n + 1)Pn (x)tn+2k 2
k!( 32 )n+k
n,k=0 [n ∞ X 2] X
(v − 12 )k (v)n−k (2n − 4k + 1)Pn−2k (x)tn . k!( 23 )n−k n=0 k=0
Hence n
Cnv (x)
[2] X (v − 12 )k (v)n−k (2n − 4k + 1)Pn−2k (x) = . k!( 23 )n−k k=0
21. Chapter 18 Solutions Problem 202. For the Bernoulli polynomial of Section 153 show that xn =
n X k=0
n!Bk (x) . k!(n − k + 1)!
Solution 202. By definition of Bn (x) we have ∞ X text Bn (x)tn = . t e − 1 n=0 n!
Then
SOLUTIONS TO RAINVILLE’S “SPECIAL FUNCTIONS” (1960)
ext
∞ X 1 Bn (x)tn = (et − 1) = t n! n=0
∞ n−1 X t n! n=1
!
∞ X Bn (x)tn n! n=0
147
! ,
from which ∞ X xn tn n! n=0
∞ X
! ∞ ! X Bn (x)tn tn = (n + 1)! n! n=0 n=0 ∞ X n n X Bk (x)t = k!(n + 1 − k)! n=0 k=0
Hence xn =
n X k=0
n!Bk (x) . k!(n + 1 − k)!
Problem 203. Let Bn (x) and Bn = Bn (0) denote the Bernoulli polynomials and numbers as t reated in Section 153. Define the differential operator A(c, D) by A(x, D) =
∞ X 1 B2k D2k . x− D− 2 (2k)! k=1
Prove that A(x, D)Bn (x) = nBn (x). Solution 203. From ∞ X text Bn (x)tn = et − 1 n=0 n!
Bn (x) is of Sheffer A-type zero. We have, in the Shefer notation, n! t H(t) = t, J(t) = t, A(t) = t . e −1 We wish to apply Theorem 74, page 391. Now we see that
log A(t) = log t − log(et − 1) 1 et 1 1 A0 (t) = − t = −1− t A(t) t e −1 t e −1 ∞ X tA0 (t) t Bn t n =1−t− t =1−t− . A(t) e −1 n! n=0
1 We know that B0 = 1, B1 = − , B2n+1 = 0 for n ≥ 1. Hence 2 ∞ 0 tA (t) t X B2k t2k =− − . A(t) 2 (2k)! k=1
In the terminology of Theorem 74, since u = J(t) = t, ∞ X k=0
∞
t X B2k t2k µk tk=1 = − − , 2 (2k)! k=1
148SOLUTIONS BY SYLVESTER J. PAGANO AND LEON HALL; EDITED BY TOM CUCHTA
∞ X
vk tk+1 = tH 0 (t) = t.
k=0
1 Therefore v0 = 1, vk = 0 for k ≥ 1, and µ0 = − , µ2k = 0 for k ≥ 1, µ2k−1 = 2 B2k for k ≥ 1. − (2k)! The identity ∞ X
(µk + vk )D k+1 φn (x) = nφn (x)
k=0
of Theorem 74 becomes "
1 x− 2
D−
∞ X B2k D 2k k=1
(2k)!
#
nBn (x) Bn (x) = . n! n!
Hence, with ∞ X 1 B2k D 2k A(x, D) = x − D− , 2 (2k)! k=1
we may conclude that A(x, D)Bn (x) = nBn (x). Note that in Theorem 74, since D k+1 φn (x) = 0 f or k ≥ n, n−1 X
(µk + xvk )D k+1
k=0
may be replaced by ∞ X
(µk + xvk )D k+1 .
k=0
Problem 204. Consider the polynomials ψn (c, x, y) =
(−1)n ( 21 + 21 x)n 3 F2 (c)n
−n,
1 1 − x, 1 − c − n; 2 2 c,
1 1 − x − n; 2 2
y .
Show that 1 F1
X ∞ 1 1 1 1 ψn (c, x, y)tn − x; c; yt 1 F1 = x; c; −t = , 2 2 2 2 n! n=0
SOLUTIONS TO RAINVILLE’S “SPECIAL FUNCTIONS” (1960)
1 1 1 1 − x, + x; 2 2 2 2 − 12 + 12 x − 12 − 21 x (1−yt) (t+t) 2 F1
149
∞ X (c)n ψn (c, x, y)tn −yt2 , = n! (1 − yt)(1 + t) n=0
c; and that ψn (1, x, 1) = Fn (x), where Fn is Bateman’s polynomial of Section 148. Solution 204. We define ψn (c, x, y) by ψn (c, x, y) =
(−1)n ( 1+x 2 )n 3 F2 (c)n
−n,
1−x , 1 − c − n; 2 c,
y .
1−x − n; 2
Then
ψn (c, x, y) =
n X (−1)n+k n!( 1+x )n ( 1−x )(c)n ( 1+x )n−k y k 2
2
2
k!(n − k)!(c)n (c)k (c)n−k ( 1+x 2 )n
k=0
or
ψn (c, x, y) =
n X (−1)n+k n!( 1−x )k ( 1+x )n−k y k 2
2
k!(n − k)!(c)k (c)n−k
k=0
.
First consider the series ∞ X ψn (c, x, y)tn n! n=0
=
∞ X n X (−1)n−k ( 1−x )k ( 1+x )n−k y k tn 2
2
k!(n − k)!(c)k (c)n−k n=0 k=0 1+x 1−x = 1 F1 ; 0; yt 1 F1 ; c; −t , 2 2
as desired. The use of c = 1, y = 1 in the above generating relation yields 1 F1
1−x ; 1; t 2
1 F1
1+x ; 1; −t 2
=
∞ X ψn (1, x, 1)tn . n! n=0
Brafman had 1 F1
X ∞ 1−x 1+x Fn (x)tn ; 1; t 1 F1 ; 1; −t = . 2 2 n! n=0
Hence ψn (1, x, 1) = Fn (x).
150SOLUTIONS BY SYLVESTER J. PAGANO AND LEON HALL; EDITED BY TOM CUCHTA
Next consider the series ∞ ∞ X n 1+x k n X X (−1)n+k (c)n ( 1−x (c)n ψn (c, x, y)tn 2 )k ( 2 )n−k y t = n! k!(n − k)!(c)k (c)n−k n=0 n=0 k=0 ∞ n X (−1) (c)n+k ( 1−x )k ( 1+x )n y k tn+k 2 2 = k!n!(c)k (c)n n,k=0 ∞ X ∞ k n n 1+x X (c + n)k ( 1−x 2 )k (yt) (−1) ( 2 )n t = k!(c)k n! n=0 k=0 1−x ∞ c + n, ; n 1+x n X (−1) ( 2 )n t 2 = F 2 1 yt n! n=0 c; 1−x −n, ; ∞ (−1)n ( 1+x ) tn X 2 1−x 2 n −yt (1 − yt)− 2 2 F1 = n! 1 − yt n=0 c; n ∞ X k k k n 1+x n X (−1)k n!( 1−x 1−x 2 )k (−1) y t (−1) ( 2 )n t = (1 − yt)− 2 k!(n − k)!(c)k (1 − yt)k n! n=0 k=0 ∞ 1−x 1+x n+k k n+2k X ( y t 1−x 2 )k ( 2 )n+k (−1) = (1 − yt)− 2 k k!n!(c)k (1 − yt) n,k=0 ∞ X ∞ 1+x 1+x k 2k X ( 2 + k)n (−t)n (−1)K ( 1−x 1−x 2 )k ( 2 )k y t = (1 − yt)− 2 n! k!(c)k (1 − yt)k k=0 n=0 ∞ 1−x 1+x k k 2k X( 1+x 1−x 2 )k ( 2 )k (−1) y t = (1 − yt)− 2 (1 + t)− 2 k!(c)k (1 + t)k (1 − yt)k k=0
We thus arrive at 1−x 1+x , ; ∞ 2 2 X 1+x (c)n ψn (c, x, y)tn − 1−x − = (1−yt) 2 (1+t) 2 2 F1 n! n=0
−yt2 . (1 − yt)(1 + t)
c; Now put c = 1, y = 1 in the above. We get 1−x 1+x , ; 2 2 1+x − 1−x − −t2 = (1 − t) 2 (1 + t) 2 1 F1 1 − t2 1; 1+x 1+x , ; 1+x 1+x 2 2 − 2 − 1−x = (1 − t) 2 (1 + t) 2 (1 − t ) 2 2 F1
∞ X n=0
ψn (1, x, 1)tn
1; 1+x 1+x , ; 2 2 x = (1 − t) 2 F1
1;
t2 .
t2
SOLUTIONS TO RAINVILLE’S “SPECIAL FUNCTIONS” (1960)
In Theorem 23, page 110, put a =
1+x 1 , b = to get 2 2
1+x 1 , ; 2 2 −1−x F 2 = (1 − t) 2 1 (−t) 1;
1+x 1+x , ; 2 2 F
1;
151
−4t . (1 − t)2
We may therefore write 1+x 1 , ; ∞ 2 2 X n −1 ψn (1, x, 1)t = (1 − t) 2 F1 n=0 1;
∞ X −4t = Fn (x)tn , 2 (1 − t) n=0
by (2), page 505. Hence ψn (1, x, 1) = Fn (x), (again). Problem 205. Sylvester (1879) studied polynomials, xn 1 . ψn (x) = 2 F0 −n, x; −; − n! x Show that (1 − t)−x ext =
∞ X
φn (x)tn ,
n=0
(1 − xt)−c 2 F0 c, x; −;
t 1 − xt
∼ =
∞ X
(c)n φn (x)tn .
n=0
Solution 205. Consider φn (x) =
xn 1 F . −n, x; −; − 2 0 n! x
THen φn (x) =
n X (−1)k n!xn (x)k (−1)k x−k k=0
k!(n − k)!n!
n X (x)k xn−k = . k!(n − k)! k=0
Hence ∞ X n=0
Next consider
φn (x)tn
=
∞ X n X (c)k xn−k tn
k!(n − k)! ! ∞ ! X xn t n (x)n tn = n! n! n=0 n=0 −x xt = (1 − t) e . n=0 k=0 ∞ X
152SOLUTIONS BY SYLVESTER J. PAGANO AND LEON HALL; EDITED BY TOM CUCHTA
∞ X
(c)n φn (x)tn
∞ X n X (c)n (x)k xn−k tn
=
n=0
k!(n − k)!
n=0 k=0 ∞ X
(c)n+k (x)k xn tn+k k!n! n,k=0 ∞ X ∞ X (c + k)n (xt)n (c)k (x)k tk
= =
n!
k=0 n=0 ∞ X
=
k=0
k!
k
(c)k (x)k t . k!(1 − xt)c+k
Therefore ∞ X
(c)n φn (x)tn = (1 − xt)−c 2 F0 c, x; −;
n=0
t 1 − xt
.
Problem 206. For Sylvester’s polynomials of Exercise 205 find what properties you can from the fact that φn (x) is of Sheffer A-type zero. Solution 206. For the φn (x) of Exercise 205 recall that (1 − t)−x ext =
∞ X
φn (x)tn .
n=0
Now (1 − t)−x = exp[−x log(1 − t)]. Hence ∞ X
φn (x)tn = exp[x{t − log(1 − t)}].
n=0
Therefore φn (x) are of Sheffer A-type zero with A(t) = 1, H(t) = t − log(1 − t). Problem 207. Show that Bateman’s Zn (x), the Legendre plynomial Pn (x), and the Laguerre polynomial Ln (x) are related symbolically by Zn (x) + Pn (2L(x) − 1). Solution 207. We know that Bateman’s Zn (x) has the generating relation (1)
∞ X
n
−1
Zn (x)t = (1 − t)
n=0
1 F1
1 −4xt ; 1; 2 (1 − t)2
and that the Laguerre polynomial satisfies X ∞ (c)n Ln (x)tn −xt (1 − t)−c 1 F1 c; 1; = , 1−t n! n=0 including the special case X ∞ ( 21 )n Ln (x)v n 1 −xv ; 1; = . 2 1−v n! n=0 2 4t 1−t −v −4t In (2) put v = . Then 1 − v = (1 + t)2 1+t 1 − v (1 − t)2 (2)
1
(1 − v)− 2 1 F1
SOLUTIONS TO RAINVILLE’S “SPECIAL FUNCTIONS” (1960)
153
and (2) becomes −1
(1 − t)
(3)
(1 + t)1 F1
1 −4xt ; 1; 2 (1 − t)2
=
∞ X ( 1 )k Lk (x)22k tk 2
k=0
k!(1 + t)2k
,
or −1
(1 − t)
1 F1
1 −4xt ; 1; 2 (1 − t)2
=
∞ X ( 1 )k 22k Lk (x)tk 2
k!(1 + t)1+2k
k=0
.
Therefore we have ∞ X
Zn (x)tn
=
n=0
= =
∞ X (−1)n 22k ( 21 )k (1 + 2k)n Lk (x)tn+k k!n! n.k=0 ∞ n X (−1) (n + 2k)!Lk (x)tn+k n,k=0 n ∞ X X n=0 k=0
(k!)2 n! (−1)n+k (n + k)!Lk (x) n t . (k!)2 (n − k)!
We know from (3), page 285 that
−n.n + 1;
1+x . 2
Pn (x) = (−1)n 2 F1 1; Hence Pn (2y − 1)
−n.n + 1;
= (−1)n 2 F1 =
y
1; n X (−1)n+k (n + k)!y k k=0
(k!)2 (n − k)!
.
We therefore have, from
Zn (x) =
n X (−1)n+k (n + k)!Lk (x) k=0
(k!)2 (n − k)!
= (−1)n
n X (−n)k (n + 1)k Lk (x) k=0
the symbolic result Zn (x) + Pn (2L (x) − 1). Problem 208. Show that Sister Celine’s polynomial 1 fn (x) = 2 F2 −n, n + 1; 1, ; x 2 of equation (16), page 292, is such that Z ∞ e−x fn (x)dx = (−1)n (2n + 1). 0
k!k!
154SOLUTIONS BY SYLVESTER J. PAGANO AND LEON HALL; EDITED BY TOM CUCHTA
Solution 208. We know that fn (x) = 2 F2
1 −n, n + 1; 1, ; x . 2
Then ∞ X
fn (x)tn =
n=0
∞ X n X (−1)s (n + s)!xs tn n=0 s=0
s!s!( 21 )s (n − s)!
.
We also know from (19), page 373, that s
X (−1)s Lk (x) xs = . 2 (s!) k!(s − k)! k=0
Hence, ∞ X
fn (x)tn
=
n=0
= =
s n X ∞ X X (−1)k+s (n + s)!Lk (x)tn n=0 s=0 k=0 ∞ X s X
( 12 )s k!(s − k)!(n − s)!
(−1)k+s (n + 2s)!Lk (x)tn+s ( 12 )s k!(s − k)!n! n,s=0 k=0 ∞ X (−1)s (n + 2k + 2s)!Lk (x)tn+k+s
n,k,s=0 ∞ X n X
s!k!n!( 12 )s+k
(−1)s (n + 2k + s)!Lk (x)tn+k s!( 12 )k+s (n − s)!k! n,k=0 s=0 −n, 1 + n + 2k; ∞ n+k X 1 (n + 2k)!Lk (x)t . = F 1 k!( 12 )k n! n,k=0 + k; 2 By Example 5, page 119, we get =
2 F1
−n, 1 + n + 2k;
1 + k; 2
n 1 = (−1) (1 + 1 + 2k − ( 12 + k)n
1 2
− k)n
=
(−1)n ( 23 + k)n (−1)n ( 23 )n+k ( 12 )k = . ( 12 + k)n ( 32 )k ( 12 )n+k
Then ∞ X
fn (x)tn
n=0
∞ X (−1)n ( 23 )n+k ( 12 )k (n + 2k)!Lk (x)tn+k ( 23 )k ( 12 )n+k k!( 12 )k n! n,k=0 ∞ X n X (−1)n−k ( 32 )n (n + k)!Lk (x)tn = . ( 23 )k ( 12 )n k!(n − k)! n=0 k=0
=
Therefore, since
(1)
( 32 )n (n + 21 ) = = 2n + 1, ( 12 )n ( 12 ) n X (−n)k (n + 1)k Lk (x) fn (x) = (−1)n (2n + 1) . k!( 23 )k k=0
SOLUTIONS TO RAINVILLE’S “SPECIAL FUNCTIONS” (1960)
155
Equation (1) will also be of use to us in later work. We now note that Z
∞
e−x fn (x)dx = (−1)n (2n + 1)
0
Z n X (−n)k (n + 1)k k=0
k!( 32 )k
∞
e−x Lk (x)dx.
0
Z
∞
The integral in the sum is zero except for k = 0 and
e−x L0 (x)dx = 1.
0
Hence ∞
Z
e−x fn (x)dx = (−1)n (2n + 1).
0
Problem 209. For Sister Celine’s fn (x) of Exercise 208 show that ∞
Z
−n, n + 1, −m, m + 1;
e−x fn (x)fm (x)dx = (−1)n+m (2n+1)(2m+1)4 F3
0
∞
Z 0
n (−1) (2n + 1)(−n)k (n + 1)k −x e Lk (x)fn (x)dx = k!( 32 )k 0
Z 0
3 3 1, , ; 2 2
∞
1 ,
; 0 ≤ k ≤ n, ; k > n,
−n, n + 1, −k, k + 1;
e−x fn (x)Zk (x)dx = (−1)n+k (2n + 1)4 F3
3 1, 1, ; 2
1 .
1 Solution 209. Again we use fn (x) = 2 F2 −n, n + 1; 1, ; x as in Exercise 208. 2 Recall that fn (x) = (−1)n (2n + 1)
(1)
n X (−n)k (n + 1)k Lk (x) k=0
k!( 32 )k
and, from Exercise 207, that (2)
Zn (x) = (−1)n
n X (−n)k (n + 1)k Lk (x) k=0
k!k!
.
At once Z 0
∞
e−x fn (x)fm (x)dx = (−1)n+m (2n+1)(2m+1)
Z n X m X (−n)k (n + 1)k (−m)s (m + 1)s k=0 s=0
k!( 32 )k s!( 32 )s
Now, by the orthogonality property of Laguerre polynomials, and the fact that Z ∞ e−x Ln2 (x)dx = 1, we get 0
0
∞
e−x Lk (x)Ls (x)dx.
156SOLUTIONS BY SYLVESTER J. PAGANO AND LEON HALL; EDITED BY TOM CUCHTA
min(n,m)
∞
Z
(−n)k (n + 1)k (−m)k (m + 1)k k!( 32 )k k!( 32 )k k=0 −n, n + 1, −m, m + 1; 1 . = (−1)n+m (2n + 1)(2m + 1)4 F3 3 3 1, , ; 2 2
e−x fn (x)fm (x)dx
= (−1)n+m (m + 1)(2m + 1)
0
X
Next consider Z
∞
e−x Lk (x)fn (x)dx.
0
At once ∞
Z
e−x Lk (x)fn (x)dx = 0; k > n,
0
from the orthogonality property of Lk (x). If 0 ≤ k ≤ n, ∞
Z
e−x Lk (x)fn (x)dx
0
= (−1)n (2n + 1)
Z n X (−n)s (n + 1)s
s!( 32 )s s=0 n (−1) (2n + 1)(−n)k (n + 1)k = . k!( 32 )k
∞
e−x Lk (x)Ls (x)dx
0
Finally, using (1) and (2) above we get
Z
∞
e−x fN (x)Zk (x)dx
= (−1)n+k (2n + 1)
0
Z k n X X (−n)s (n + 1)s (−k)i (k + 1)i s=0 i=0 min(n,k)
s!( 23 )s i!i!
0
(−n)s (n + 1)s (−k)s (k + 1)s s!( 32 )s s!s! s=0 −n, n + 1, −k, k + 1; 1 . = (−1)n+k (2n + 1)4 F3 3 1, 1, ; 2
= (−1)n+k (2n + 1)
X
Problem 210. Show that
Z 0
∞
−n, n + 1, −k, k + 1;
e−x Zn (x)Zk (x)dx = (−1)n+k 4 F3
1 .
1, 1, 1, ;
Solution 210. Using equation (2) of Exercise 209 above, we get
∞
e−x Ls (x)Li (x)dx
SOLUTIONS TO RAINVILLE’S “SPECIAL FUNCTIONS” (1960)
∞
Z
e
−x
Zn (x)Zk (x)dx
= (−1)
n+k
0
157
Z n X k X (−n)s (n + 1)s (−k)i (k + 1)i s=0 i=0 min(n,k)
s!s!i!i!
∞
e−x Ls (x)Li (x)dx
0
(−n)s (n + 1)s (−k)s (k + 1)s s!s!s!s! s=0 −n, n + 1, −k, k + 1; 1 . = (−1)n+k 4 F3 1, 1, 1; = (−1)n+k
X
Problem 211. Gottlieb introduced the polynomials φn (x; λ) = e−nλ 2 F1 (−n, −x; 1; 1 − eλ ). Show that ∞ X φn (x; λ)tn = et 1 F1 (1 + x; 1; −t(1 − e−λ )), n! n=0
φn (x, λ + µ) =
n (eµ − 1)n X n!(1 − eλ+µ )k φk (x; λ) , eµn (1 − eλ )n k!(n − k)!(eµ − 1)k k=0
∞
Z
e−st φn (x; −t)dt =
0
Γ(s − n)Γ(s + x + 1) , Γ(s + 1)Γ(s + x − n + 1)
(x − n)φn (x; λ) = xφn (x − 1; λ) − ne−λ φn−1 (x; λ), n[φn (x; λ) − φn−1 (x; λ)] = (x + 1)[φn (x + 1; λ) − φn (x; λ)], (x + n + 1)φn (x; λ) = xeλ φn (x − 1; λ) + (n + 1)eλ φn+1 (x; λ), and see Gottlieb for many other results on φn (x; λ). Solution 211. Problem 212. With fn denoting Sister Celine’s polynomials of Section 149, show that Z
t
1
1
x 2 (t − x) 2 fn (−; −; x)dx = πZn (t), 0 t
Z
1
1
x 2 (t − x) 2 fn (−; −; x(t − x))dx = πZn 0
1 2 t , 4
and Z 0
t
1 x (t − x) fn −; ; x(t − x) dx = πLn (t)Ln (−t). 2 1 2
1 2
158SOLUTIONS BY SYLVESTER J. PAGANO AND LEON HALL; EDITED BY TOM CUCHTA
Solution 212. We shall use Theorem 37, page (??), to evaluate the integrals in question. 1 Recall that fn (−; −; x) = 2 F2 −n, n + 1; 1, ; x . 2 1 1 1 Then use α = , β = , p = 2, q = 2, a1 = −n, a2 = n + 1, b1 = 1, b2 = , c = 2 2 2 1, k = 1, s = 0 in Theorem 37 to get 1 −n, n + 1, ; 2 1 1 =B , F 2 2 1 1, , 1; 2 −n, n + 1; 1 1 Γ( 2 )Γ( 2 ) = F 2 2 Γ(1) 1, 1; = πZn (t).
Z
t
1
1
x 2 (t − x)− 2 fn (−; −; x)dx
0
t t
Next, make the same substitution except that s = 1 instead of s = 0. We thus get 1 1 −n, n + 1, , ; 2 2 = πF 1 1 2 1, , , ; 2 2 2 −n, n + 1; t2 = π 2 F2 4 1, 1; 1 2 = πZn t . 4
Z
t
1
1
x− 2 (t − x)− 2 fn (−; −; −x(t − x))dx
0
t2 4
Finally, consider
1 fn −; ; z 2
−n, n + 1;
= 2 F3
1 In Theorem 37 we now use α = , β = 2 1 1, b2 = 1, b − 3 = , c = 1, k = 1, s = 1, to 2 Z t 1 1 1 x 2 (t − x) 2 fn −; ; x(t − x) dx 2 0
1 1 1, , ; 2 2
z .
1 , p = 2, q = 2, a1 = −n, a2 = n + 1, b1 = 2 get
=
Γ( 21 )Γ( 12 ) F Γ(1)
−n, n + 1;
= πLn (t)Ln (−t),
1 1, 1, ; 2
t2 4
SOLUTIONS TO RAINVILLE’S “SPECIAL FUNCTIONS” (1960)
by equation (9), page 509. Recall Ramanujan’s theorem, Example 5, page 180, β β 1 x2 − 1F1 (α; β; x)1 F1 (α; β; −x) = 2 F3 α, β − α; β, , + ; . 2 2 2 4 Use α = −n, β = 1, x = t to get 1 t2 , 1 F1 (−n; 1; t)1 F1 (−n; 1; −t) = 2 F3 −n, 1 + n; 1, , 1; 2 4 which is the result we used from page 509. Problem 213. Define polynomials φn (x) by φn (x) =
(c)n 2 F2 (−n, c + n; 1, 1; x). n!
Show that φn (x) =
n X (−1)n−k (c − 1)n−k (c)n+k (2k + 1)Zk (x)
(n − k)!(n + k + 1)!
k=0
.
Solution 213. Let us define φn (x) by φn (x) =
(c)n 2 F2 (−n, c + n; 1, 1; x). n!
Then φn (x) =
n X (−1)k (c)n+k xk k=0
k!(n − k)!(k!)2
,
so ∞ X
φn (x)t
n
=
n=0
=
∞ X n X (−1)s (c)n+s xs tn n=0 s=0 ∞ X
(s!)3 (n − s)!
(−1)s (c)n+2s xs tn+s . (s!)3 n! n,s=0
We know (page 199) that xs = (s!)2
s X (−s)k (2k + 1)Zk (x) k=0
(s + k + 1)!
.
Then ∞ X
φn (x)t
n=0
n
∞ X s X (−1)s+k (c)n+2s (2k + 1)Zk (x)tn+s = n!(s − k)!(s + k + 1)! n,s=0 k=0 ∞ X (−1)s (c)n+2k+2s (2k + 1)Zk (x)tn+k+s = n!s!(s + 2k + 1)! n,k,s=0 ∞ X n s X (−1) (c)n+2k+s (2k + 1)Zk (x)tn+k = . s!(n − s)!(s + 2k + 1)! s=0 n,k=0
We now have
159
160SOLUTIONS BY SYLVESTER J. PAGANO AND LEON HALL; EDITED BY TOM CUCHTA
∞ X
φn (x)tn
∞ X
=
n=0
= = =
n,k=0 ∞ X n,k=0 ∞ X
2 F1
−n, c + n + 2k;
n+k
(c)n+2k (2k + 1)Zk (x)t 1 n!(2k + 1)!
2 + 2k; (c)n+2k (2k + 1)Zk 9x)tn+k Γ(2 + 2k)Γ(2 − c) Γ(2 + 2k + n)Γ(2 − c − n) n!(2k + 1)!
(−1)n (c − 1)n (c)n+2k (2k + 1)!Zk (x)tn+k (n + 2k + 1)!n! n,k=0 ∞ n X X (−1)n−k (c)n+k (c − 1)n−k (2k + 1)Zk (x) (n − k)!(n + k + 1)!
n=0 k=0
tn .
Hence φn (x) =
n X (−1)n−k (c − 1)n−k (c)n+k (2k + 1)Zk (x) k=0
(n − k)!(n + k + 1)!
.
Problem 214. Show that the F (t) of equation (10), page 286, can be put in the form
v;
−4zt . (1 − z)2
F (t) = tb e−t (1 − z)−2v 1 F1 b + 1; Solution 214. From equation (10), page 500, we get
F (t)
−n, 2v + n; ∞ n X (2v)n z t = tb e−t 2 F2 1 n! n=0 v + , b + 1; 2 n ∞ X X (−1)k n!(2v)n (2v + n)k tk z n b −t =t e k!(n − k)!n!(v + 12 )k (b + 1)k n=0 k=0 ∞ X (−1)k (2v)n+2k tk z n+k = tb e−t k!n!(v + 12 )k (b + 1)k n,k=0 ∞ X ∞ X (2v + 2k)n z n (−1)k (2v)2k tk z k = tb e−t n! k!(v + 12 )k (b + 1)k k=0 n=0 ∞ k k k 2k X (−1) t z 2 (v)k (v + 1 )k 2 = tb e−t 1 2v+2k k!(v + ) (b + 1) (1 − z) k k 2 k=0 ∞ X (−1)k 22k (v)k tk z k = tb (1 − z)−2v e−t k!(b + 1)k (1 − z)2k k=0 −4tz = tb (1 − z)−2v e−t 1 F1 v; b + 1; . (1 − z)2 22. Chapter 19 Solutions
No problems in Chapter 19.
SOLUTIONS TO RAINVILLE’S “SPECIAL FUNCTIONS” (1960)
161
23. Chapter 20 Solutions 1
Problem 215. Show that θ10 = 2q 4 G3 . Solution 215. We know that θ10
1 4
= 2p G
∞ Y
(1 − q 2n )2
n=1
and that G=
∞ Y
(1 − q 2n ).
n=1
Hence 1
θ10 = 2q 4 G3 . Problem 216. The following formulas are drawn from Section 168. Derive (9) and (10): (8)
θ42 θ12 (z) + θ32 θ22 (z) = θ22 θ32 (z),
(9)
θ32 θ12 (z) + θ42 θ22 (z) = θ22 θ42 (z),
(10)
θ22 θ12 (z) + θ42 θ32 (z) = θ32 θ42 (z),
(11)
θ22 θ22 (z) + θ42 θ42 (z) = θ32 θ32 (z).
Solution 216. We are given θ42 θ12 (z) + θ32 θ22 (z) = θ22 θ32 (z). π Use the basic table with z||z + to obtain 2 (8)
θ42 θ22 (z) + θ32 θ12 (z) = θ22 θ42 (z) or θ32 θ12 (z) + θ42 θ22 (z) = θ22 θ42 (z). 1 In (9) use the basic table with z||z + πτ to get 2 (9)
1
1
1
−θ32 q − 2 e−2iz θ42 (z) + θ42 q − 2 e−2it θ32 (z) = −θ22 q − 2 e−2iz θ12 (z) or θ32 θ42 (z) − θ42 θ32 (z) = θ22 θ12 (z) from which we get (10)
θ22 θ12 (z) + θ42 θ32 (z) = θ32 θ42 (z).
Problem 217. Use (8) and (10) of Exercise 216 and the equation θ24 + θ44 = θ34 to show that θ14 (z) + θ34 (z) = θ24 (z) + θ44 (z).
162SOLUTIONS BY SYLVESTER J. PAGANO AND LEON HALL; EDITED BY TOM CUCHTA
Solution 217. We may write (8) and (10) of Exercise 216 in the form (8) (10)
θ42 θ12 (z) − θ22 θ32 (z) = −θ32 θ22 (z), θ22 θ12 (z) + θ42 θ32 (z) = θ32 θ42 (z).
We square each member and add to obtain (θ44 + θ24 )[θ14 (z) + θ34 (z)] = θ34 [θ24 (z) + θ44 (z)]. But θ24 + θ34 = θ34 , so we get θ14 (z) + θ34 (z) = θ24 (z) + θ44 (z). Problem 218. The first of the following relations was derived in Section 169. Obtain the other three by using appropriate changes of variable andthe basic table, 1 1 page 319. For example, change x to x + π , or x to x + πτ , etc. 2 2 θ22 θ1 (x + y)θ1 (x − y) = θ12 (x)θ22 (y) − θ22 (x)θ12 (y), θ22 θ2 (x + y)θ2 (x − y) = θ22 (x)θ22 (y) − θ12 (x)θ12 (y), θ22 θ3 (x + y)θ3 (x − y) = θ32 (x)θ22 (y) + θ42 (x)θ12 (y), θ22 θ4 (x + y)θ4 (x − y) = θ42 (x)θ22 (y) + θ32 (x)θ12 (y). Solution 218. We are given from Section 169 that θ22 θ1 (x + y)θ1 (x − y) = θ12 (x)θ22 (y) − θ22 (x)θ12 (y). π In (A) change x to x + and use the basic table to get 2 (A)
θ22 θ2 (x + y)θ2 (x − y) = θ22 (x)θ22 (y) − θ12 (x)θ12 (y). πτ Now in last equation above change x to x + to obtain 2 1
1
1
θ22 q − 2 e−2ix θ2 (x + y)θ3 (x − y) = q − 2 e−2ix θ32 (x)θ22 (y) + q − 2 e−2ix θ42 (x)θ12 (y), or θ22 θ3 (x + y)θ3 (x − y) = θ32 (x)θ22 (y) + θ42 (x)θ12 (y). π In last equation above change x to x + to arrive at 2 θ22 θ4 (x + y)θ4 (x − y) = θ42 (x)θ22 (y) + θ32 (x)θ12 (y).
SOLUTIONS TO RAINVILLE’S “SPECIAL FUNCTIONS” (1960)
163
Problem 219. Use the identities in Exercise 216 and the relation θ24 + θ44 = θ34 to transform the first relation of Exercise 218 into the first relation below. Then obtain the remaining three relations with the aid of the basic table, page 319. θ22 θ1 (x + y)θ1 (x − y) = θ42 (x)θ32 (y) − θ32 (x)θ42 (y), θ22 θ2 (x + y)θ2 (x − y) = θ32 (x)θ32 (y) − θ42 (x)θ42 (y), θ22 θ3 (x + y)θ3 (x − y) = θ22 (x)θ32 (y) + θ12 (x)θ42 (y), θ22 θ4 (x + y)θ4 (x − y) = θ12 (x)θ32 (y) + θ22 (x)θ42 (y). Solution 219. From Exercise 218 we obtain θ22 θ1 (x + y)θ1 (x − y) = θ12 (x)θ22 (y) − θ22 (x)θ12 (y). From (10) and (11) of Exercise 216 we get θ12 (x) = θ2−2 θ32 θ42 (x) − θ42 θ32 (x) θ22 (y) = θ2−2 θ32 θ32 (y) − θ42 θ42 (y) . Hence we obtain θ22 θ1 (x + y)θ1 (x − y)
= θ2−4 θ34 θ42 (x)θ32 (y) + θ44 θ32 (x)θ42 (y) − θ32 θ44 θ32 (x)θ32 (y) + θ42 (x)θ42 (y) −θ2−2 θ34 θ42 (y)θ32 (x) + θ44 θ32 (y)θ42 (x) − θ22 θ42 θ32 (y)θ32 (x) + θ42 (y)θ42 (x) = θ2−4 (θ34 − θ44 ){θ42 (x)θ32 (y) − θ32 (x)θ42 (y)} .
Now θ34 − θ44 = θ24 so that we may conclude that θ22 θ1 (x + y)θ1 (x − y) = θ42 (x)θ32 (y) − θ32 (y) − θ32 (x)θ42 (y).
(A)
We shall now transform (A) by means of the basic table. Note that (A) is the first of the required results 219. inπExercise In (A) replace x by x + to get 2 θ22 θ2 (x + y)θ2 (x − y) = θ32 (x)θ32 (y) − θ42 (x)θ42 (y). πτ In (B) replace x by x + to obtain 2 (B)
1
1
1
q − 2 e−2ix θ22 θ3 (x + y)θ3 (x − y) = q − 2 e−2ix θ22 (x)θ32 (y) + q − 2 e−2ix θ12 (x)θ42 (y), or θ22 θ3 (x + y)θ3 (x − y) = θ22 (x)θ32 (y) + θ12 (x)θ42 (y). π In (C) replace x by x + to arrive at 2 (C)
(D)
θ22 θ4 (x + y)θ4 (x − y) = θ12 (x)θ32 (y) + θ22 (x)θ42 (y).
Equations (A), (B), (C), (D) are those required.
164SOLUTIONS BY SYLVESTER J. PAGANO AND LEON HALL; EDITED BY TOM CUCHTA
Problem 220. Use equation (10) of Exercise 216 and the first relation of Exercise 219 to obtain the first relation below; then use the basic table to get the remaining three results. θ32 θ1 (x + y)θ1 (x − y) = θ12 (x)θ32 (y) − θ32 (x)θ12 (y), θ32 θ2 (x + y)θ2 (x − y)θ22 (x)θ32 (y) − θ42 (x)θ12 (y), θ32 θ3 (x + y)θ3 (x − y) = θ32 (x)θ32 (y) + θ12 (x)θ12 (y), θ32 θ4 (x + y)θ4 (x − y) = θ42 (x)θ32 (y) + θ22 (x)θ12 (y). Solution 220. From Exercise 219 we use θ22 θ1 (x + y)θ1 (x − y) = θ42 (x)θ32 (y) − θ32 (x)θ42 (y). From Exercise 216 we use equation (10) in the form (10)
θ42 (x) = θ3−2 [θ22 θ12 (x) + θ42 θ32 (x)].
Thus we obtain θ22 θ1 (x+y)θ1 (x−y) = θ3−2 [θ22 θ12 (x)θ32 (y)+θ42 θ32 (x)θ32 (y)−θ22 θ32 (x)θ12 (y)−θ42 θ32 (x)θ32 (x)]. Two terms drop out and we are left with (A) θ32 θ1 (x + y)θ1 (x − y) = θ12 (x)θ32 (y) − θ32 (x)θ12 (y), which was desired. π to get In (A) replace x by x + 2 θ32 θ2 (x + y)θ2 (x − y) = θ22 (x)θ32 (y) − θ42 (x)θ12 (y). πτ In (B) replace x by x + to get 2 (B)
(C)θ32 θ3 (x + y)θ3 (x − y) = θ32 (x)θ32 (y) + θ12 (x)θ12 (y), 1
from which we have removed the common factor q − 2 e−2ix . π In (C) replace x by x + to get 2 (D) θ32 θ4 (x + y)θ4 (x − y) = θ42 (x)θ32 (y) + θ22 (x)θ12 (y). Equations (A), (B), (C), (D) are the required ones. Problem 221. Use the first relation of Exercise 220 and the identities from Exercise 216 to obtain the frist relation below. Derive the other three relations from the first one. θ32 θ1 (x + y)θ1 (x − y) = θ42 (x)θ22 (y) − θ22 (x)θ42 (y), θ32 θ2 (x + y)θ2 (x − y) = θ32 (x)θ22 (y) − θ12 (x)θ42 (y), θ32 θ3 (x + y)θ3 (x − y) = θ22 (x)θ22 (y) + θ42 (x)θ42 (y),
SOLUTIONS TO RAINVILLE’S “SPECIAL FUNCTIONS” (1960)
165
θ32 θ4 (x + y)θ4 (x − y) = θ12 (x)θ22 (y) + θ32 (x)θ42 (y). Solution 221. From (9) and (11) of Exercise 216 we may write θ12 (x) = θ3−2 [θ22 θ42 (x) − θ42 θ22 (x)] and θ32 (y) = θ3−2 [θ22 θ22 (y) + θ42 θ42 (y)], which we substitute into equation (A) of Exercise 219 above. The result is θ32 θ1 (x + y)θ1 (x − y)
= θ3−4 [θ24 θ42 (x)θ22 (y) − θ44 θ22 (x)θ42 (y) + θ22 θ42 {θ42 (x)θ42 (y) − θ22 (x)θ22 (y)}] −θ3−4 [θ24 θ42 (y)θ22 (x) − θ44 θ22 (y)θ42 (x) + θ22 θ42 {θ42 (y)θ42 (x) − θ22 (y)θ22 (x)}].
We thus obtain θ32 θ1 (x + y)θ1 (x − y) = θ3−4 [(θ24 + θ44 ){θ42 (x)θ22 (y) − θ22 (x)θ42 (x)θ42 (y)}]. Since θ24 + θ44 = θ34 , we arrive at the desired result θ32 θ1 (x + y)θ1 (x − y) = θ42 (x)θ22 (y) − θ22 (x)θ42 (y). π to obtain In (A) replace x by x + 2 (A)
θ32 θ2 (x + y)θ2 (x − y) = θ32 (x)θ22 (y) − θ12 (x)θ42 (y). πτ 1 and remove the factor q − 2 e−2ix to get In (B) replace x by x + 2 (B)
θ32 θ3 (x + y)θ3 (x − y) = θ22 (x)θ22 (y) + θ42 (x)θ42 (y). π In (C) replace x by x + to obtain the formula of the requires results: 2 (C)
(C)
θ32 θ4 (x + y)θ4 (x − y) = θ12 (x)θ22 (y) + θ32 (x)θ42 (y).
Problem 222. Use equation (10) of Exercise 216 and the first relation of Exercise 220 to obtain the first relation below. Derive the others with the aid of the basic table, page 319. θ42 θ1 (x + y)θ1 (x − y) = θ12 (x)θ42 (y) − θ42 (x)θ12 (y), θ42 θ2 (x + y)θ2 (x − y) = θ22 (x)θ42 (y) − θ32 (x)θ12 (y), θ42 θ3 (x + y)θ3 (x − y) = θ32 (x)θ42 (y) − θ22 (x)θ12 (y), θ42 θ4 (x + y)θ4 (x − y) = θ42 (x)θ42 (y) − θ12 (x)θ12 (y).
166SOLUTIONS BY SYLVESTER J. PAGANO AND LEON HALL; EDITED BY TOM CUCHTA
Solution 222. From (10) of Exercise 216 we g et θ32 (y) = θ4−2 [θ32 θ42 (y) − θ22 θ12 (y)] and a similar result in x. The first equation in Exercise 220 is θ32 θ1 (x + y)θ1 (x − y) = θ12 (x)θ32 (y) − θ32 (x)θ12 (y). Then two applications of (10) yield θ32 θ1 (x+y)θ1 (x−y) = θ4−2 [θ32 θ12 (x)θ42 (y)−θ22 θ12 (x)θ12 (y)]−θ4−2 [θ32 θ42 (x)θ12 (y)−θ22 θ12 (x)θ12 (y)] In the above, two terms drop out and θ32 cancels out to yield θ42 θ1 (x + y)θ1 (x − y) = θ12 (x)θ42 (y) − θ42 (x)θ12 (y). π and see the basic table to get In (A) replace x by x + 2 (A)
θ2 θ (x + y)θ (x − y) = θ22 (x)θ42 (y) − θ32 (x)θ12 (y). 4 2 πτ 2 1 In (B) replace x by x + and remove the common factor q − 2 e−2ix to obtain 2 (B)
θ42 θ3 (x + y)θ3 (x − y) = θ32 (x)θ42 (y) − θ22 (x)θ12 (y). π and thus arrive at In (C) replace x by x + 2 (C)
(D)
θ42 θ4 (x + y)θ4 (x − y) = θ42 (x)θ42 (y) − θ12 (x)θ12 (y).
Problem 223. The first relation below was derived in Section 169. Obtain the other three. θ42 θ1 (x + y)θ1 (x − y) = θ32 (x)θ22 (y) − θ22 (x)θ32 (y), θ42 θ2 (x + y)θ2 (x − y) = θ42 (x)θ22 (y) − θ12 (x)θ32 (y), θ42 θ3 (x + y)θ3 (x − y) = θ42 (x)θ32 (y) − θ12 (x)θ22 (y), θ42 θ4 (x + y)θ4 (x − y) = θ32 (x)θ32 (y) − θ22 (x)θ22 (y). Solution 223. From Section 169, equation (7), page 371, we obtain θ42 θ1 (x + y)θ1 (x − y) = θ32 (x)θ22 (y) − θ22 (x)θ32 (y). π In (A) replace x by x + to get 2 (A)
θ42 θ2 (x + y)θ2 (x − y) = θ42 (x)θ22 (y) − θ12 (x)θ32 (y). πτ 1 In (B) replace x by x + and remove the factor q − 2 e−2ix to arrive at 2 (B)
θ42 θ3 (x + y)θ2 (x − y) = θ42 (x)θ32 (y) − θ12 (x)θ22 (y). π In (C) replace x by x + to obtain 2 (C)
(D)
θ42 θ4 (x + y)θ4 (x − y) = θ32 (x)θ32 (y) − θ22 (x)θ22 (y).
SOLUTIONS TO RAINVILLE’S “SPECIAL FUNCTIONS” (1960)
167
Problem 224. Use the method of Section 169 together with the basic table, page 319, to derive the following results, one of which was obtain in Section 169. θ3 θ4 θ1 (x + y)θ2 (x − y) = θ1 (x)θ2 (x)θ3 (y)θ4 (y) + θ1 (y)θ2 (y)θ3 (x)θ4 (x), θ2 θ4 θ1 (x + y)θ3 (x − y) = θ1 (x)θ2 (y)θ3 (x)θ4 (y) + θ1 (y)θ2 (x)θ3 (y)θ4 (x), θ2 θ3 θ3 (x + y)θ4 (x − y) = θ1 (x)θ2 (y)θ3 (y)θ4 (x) + θ1 (y)θ2 (x)θ3 (x)θ4 (y), θ2 θ3 θ2 (x + y)θ3 (x − y) = θ2 (x)θ3 (x)θ2 (y)θ3 (y) − θ1 (x)θ4 (x)θ1 (y)θ4 (y), θ2 θ4 θ2 (x + y)θ4 (x − y) = θ2 (x)θ4 (x)θ2 (y)θ4 (x) − θ1 (x)θ3 (x)θ1 (y)θ3 (y), θ3 θ4 θ3 (x + y)θ4 (x − y) = θ3 (x)θ4 (x)θ3 (y)θ4 (y) − θ1 (x)θ2 (x)θ1 (y)θ2 (y). Solution 224. For brevity, since we know how to discover such formulas, let us consider the function θ1 (x)θ2 (x)θ3 (y)θ4 (y) + θ1 (y)θ2 (y)θ3 (x)θ4 (y) . θ1 (x + y)θ2 (x − y) we shall show that φ1 (x) is an elliptic function of order less than two, and then obtain its constant value. Let N1 (x) = the numerator of φ1 (t) and D1 (x) be the denominator of φ1 (x). Then at once the basic table yields φ1 (x) =
N1 (x + πτ ) = θ2 (x)θ1 (x)θ3 (y)θ4 (y) + θ1 (y)θ2 (y)θ4 (x)θ3 (x) = N1 (x) and D1 (x + πτ ) = −θ1 (x + y)[−θ2 (x − y)] = D1 (x). Hence φ1 (x + πτ ) = φ1 (x). The periods π and πτ are also those of the zeros of the four theta functions. π Hence the only poles of φ1 (x) in a cell are simple ones at x = y = 0 and x − y = . 2 But x = −y is also a zero of the numerator because N1 (−y) = −θ1 (y)θ2 (x)θ3 (y)θ4 (y) + θ1 (y)θ2 (y)θ3 (y)θ4 (y) = 0. Hence φ1 (x) is an elliptic function order less than two and is constant. Now 0 + θ1 (y)θ2 (y)θ3 θ4 = θ3 θ4 , θ1 (y)θ2 (−y) since θ2 (y) is an even function. Thus we obtain the first of the derived relations, φ1 (0) =
(A)
θ3 θ4 θ1 (x + y)θ2 (x − y) = θ1 (x)θ2 (x)θ3 (y)θ4 (y) + θ1 (y)θ2 (y)θ3 (x)θ4 (x).
168SOLUTIONS BY SYLVESTER J. PAGANO AND LEON HALL; EDITED BY TOM CUCHTA
Next let us consider θ1 (x)θ2 (y)θ3 (x)θ4 (y) + θ1 (y)θ2 (x)θ3 (y)θ4 (x) . θ1 (x + y)θ3 (x − y) Let as usual the numerator be N2 (x), the denominator D2 (x). Then φ2 (x) =
N2 (x + π) = θ1 (x)θ2 (y)θ3 (x)θ4 (y) − θ1 (y)θ2 (x)θ3 (x)θ4 (x), N2 (x + π) = −N − 2(x). Also D2 (x + π) = −θ1 (x + y)θ3 (x − y) = −D2 (x + π). Hence θ2 (x + π) = θ2 (x). Next N2 (x + πτ ) = −q −2 e−4ix N2 (x) and D2 (x + πτ ) = −q −2 e−2i(x+y) e−2i(x−y) D2 (x) = −q −2 e−4ix D2 (x). Hence also φ2 (x + πτ ) = φ2 (x). In a cell, with periods π, πτ, D2 (x) has simple zeros at x + y = 0 and x − y = π πτ + . But, for x = −y, 2 2 N2 (−y) = −θ1 (y)θ2 (y)θ3 (y)θ4 (y) + θ1 (y)θ2 (y)θ3 (y)θ4 (y) = 0. Therefore φ2 (x) is an elliptic function of order less than two. The constant value of φ2 (x) is 0 + θ1 (y)θ3 (y)θ4 = θ2 θ4 . θ1 (y)θ3 (−y) We may now conclude the validity of the equation φ2 (0) =
(B)
θ2 θ4 θ1 (x + y)θ3 (x − y) = θ1 (y)θ2 (y)θ3 (x)θ4 (y) + θ1 (y)θ2 (x)θ3 (y)θ4 (x).
From Section 169 we quote (C)
θ2 θ3 θ1 (x + y)θ4 (x − y) = θ1 (x)θ2 (y)θ3 (y)θ4 (x) + θ1 (y)θ2 (x)θ3 (x)θ4 (y).
Equations (A), (B), and (C) are the first three of our desired results. In (C) π to get replace x by x + 2 (D)
θ2 θ3 θ2 (x + y)θ3 (x − y) = θ2 (x)θ2 (y)θ3 (y)θ3 (x) − θ1 (y)θ1 (x)θ4 (x)θ4 (y),
SOLUTIONS TO RAINVILLE’S “SPECIAL FUNCTIONS” (1960)
169
which is the equation, slightly rewritten, of the desired results. In (B) fourth π replace x by x + to find that 2 θ2 θ4 θ2 (x + y)θ4 (x − y) = θ2 (x)θ2 (y)θ4 (x)θ4 (y) − θ1 (y)θ1 (x)θ3 (y)θ3 (x),
(E)
which is a rewritten form of a desired result. Finally, turn to (A) and replace x πτ by x + to get 2 1
1
1
θ3 θ4 iq − 2 e−i(x+y) e−i(x−y) θ4 (x+y)θ3 (x−y) = iq − 2 e−2ix θ4 (x)θ3 (x)θ3 (y)θ4 (y)+iq − 2 e−2ix θ1 (y)θ2 (y)θ2 (x)θ1 (x), or θ3 θ4 θ4 (x + y)θ3 (x − y) = θ3 (x)θ4 (x)θ3 (y)θ4 (y) + θ1 (x)θ2 (x)θ1 (y)θ2 (y). Now change y to (−y) to arrive at the desired result (F )
θ3 θ4 θ3 (x + y)θ4 (x − y) = θ4 (x)θ4 (x)θ3 (y)θ4 (y) − θ1 (x)θ2 (x)θ1 (y)θ − 2(y).
Problem 225. Use the results in Exercises 218-224 above to show that θ23 θ2 (2x) = θ24 (x) − θ14 (x) = θ34 (x) − θ44 (x), θ33 θ3 (2x) = θ14 (x) + θ34 (x) = θ24 (x) + θ44 (x), θ43 θ4 (2x) = θ44 (x) − θ14 (x) = θ34 (x) − θ24 (x), θ32 θ2 θ2 (2x) = θ22 (x)θ32 (x) − θ12 (x)θ42 (x), θ42 θ2 θ2 (2x) = θ22 (x)θ42 (x) − θ12 (x)θ32 (x), θ22 θ3 θ3 (2x) = θ22 (x)θ32 (x) + θ12 (x)θ42 (x), θ42 θ3 θ3 (2x) = θ32 (x)θ42 (x) − θ12 (x)θ22 (x), θ22 θ4 θ4 (2x) = θ12 (x)θ32 (x) + θ22 (x)θ42 (x). θ32 θ4 θ4 (2x) = θ12 (x)θ22 (x) + θ32 (x)θ42 (x). Solution 225. We shall use y = x in various equations of Exercises 218-224. From Exercise 218, second equation, we get θ22 θ2 (2x) = θ24 (x) − θ14 (x). From Exercise 218 third equation, we get θ22 θ3 θ3 (2x) = θ22 (x)θ32 (x) + θ12 (x)θ42 (x). From Exercise 218, fourth equation, we get θ22 θ4 θ4 (2x) = θ22 (x)θ42 (x) + θ12 (x)θ32 (x).
170SOLUTIONS BY SYLVESTER J. PAGANO AND LEON HALL; EDITED BY TOM CUCHTA
From Exercise 219, equations 1,2,3 in order, we obtain θ23 θ2 (2x) = θ34 (x) − θ44 (x), θ22 θ3 θ3 (2x) = θ22 (x)θ32 (x) − θ12 (x)θ42 (x), θ22 θ4 θ4 (2x) = θ12 (x)θ32 (x) + θ22 (x)θ42 (x). Next from Exercise 220, equations 2,3,4, we get θ2 θ32 θ2 (2x) = θ22 (x)θ32 (x) − θ12 (x)θ42 (x), θ33 θ3 (2x) = θ34 (x) + θ14 (x), θ32 θ4 θ4 (2x) = θ32 (x)θ42 (x) + θ12 (x)θ22 (x). From equations 2,3,4 of Exercise 221 we obtain θ2 θ32 θ2 (2x) = θ22 (x)θ32 (x) − θ12 (x)θ42 (x), θ33 θ3 (2x) = θ24 (x) + θ44 (x), θ32 θ4 θ4 (2x) = θ12 (x)θ22 (x) + θ32 (x)θ42 (x). From equations 2,3,4 of Exercise 222 we get (
θ2 θ42 θ2 (2x) = θ22 (x)θ42 (x) − θ1 x)θ32 (x), θ3 θ42 θ3 (2x) = θ32 (x)θ42 (x) − θ12 (x)θ22 (x), θ43 θ4 (2x) = θ44 (x) − θ14 (x). From equations 2,3,4 of Exercise 223 we get (
θ2 θ42 θ2 (2x) = θ22 (x)θ4 x) − θ12 (x)θ32 (x), θ3 θ42 θ3 (2x) = θ32 (x)θ42 (x) − θ12 (x)θ22 (x), θ43 θ4 (2x) = θ34 (x) − θ24 (x). We have now derived all 12 equations of Exercise 225, but we try Exercise 224 to see whether from each of the first three equations θ2 θ3 θ4 θ1 (2x) = 2θ1 (x)θ2 (x)θ3 (x)θ4 (x), (which we had in text), and θ22 θ3 θ2 (2x) = θ22 (x)θ32 (x) − θ12 (x)θ42 (x), θ2 θ42 θ2 (2x) = θ22 (x)θ42 (x) − θ12 (x)θ32 (x), and
SOLUTIONS TO RAINVILLE’S “SPECIAL FUNCTIONS” (1960)
171
θ3 θ42 θ3 (2x) = θ32 (x)θ42 (x) − θ12 (x)θ22 (x). The last two appear as answers to Exercise 225, parts 5 and 7. Problem 226. Use the method of Section 168 to show that 1
θ3 (z|τ ) = (−iτ )− 2 exp
z2 πiτ
θ3
z −t . τ τ
From the above identity, obtain corresponding ones for the other theta functions with the aid of the basic table, page 319. Solution 226. Consider the function ψ(z) = exp
θ3 (z|τ ) . θ3 ( τz | −1 τ )
z2 π|τ
First the theta functions are analytic for all finite z since Im(τ ) > 0 implies 1 Im − > 0. τ To get the zeros of the denominator, recall that θ3 (y|t) has its zeros all simple ones located at π πt + + nπ + mπt 2 2 z 1 for integral n and m. Then θ3 − has its zeros all simple ones located at τ τ z π π mπ = − + nπ − τ 2 2τ τ or at y=
z=
πτ π − + nπτ − mπ, 2 2
which is equivalent to z=
π πτ + + n1 π + m1 πτ, 2 2
z 1 for integral n1 and m1 . Therefore the zeros of θ3 (z|τ ) and those of θ3 − τ τ coincide. Hence the function ψ(z0 has no singular points in the finite plane. To show that ψ(z) has the desired two periods, consider ψ(z + π) and ψ(z + πτ ). ψ(z + π) = exp
θ3 (z + π|τ ) . 1 θ3 z+π τ |− τ
(z+π)2 πiτ
Recall that (basic table) θ3 (y + π, t) = θ3 (y|t) and that θ3 (y + πt|t) = e−iπt e−2iy θ3 (y|t). Since θ3 (z) is an even function of z, we may now write
172SOLUTIONS BY SYLVESTER J. PAGANO AND LEON HALL; EDITED BY TOM CUCHTA
ψ(z + π)
θ3 (z|τ ) π exp exp exp iτ θ3 − τz − πτ | − τ1 θ3 (z|τ ) = z2 πi iπ exp πiτ exp − 2iz exp − τ τ exp τ exp θ3 (z|τ ) = z2 exp πiτ θ3 τz | − τ1 = ψ(z). =
z2 πiτ
2z iτ
2iz τ
θ3 − τz | −
1 τ
Thus ψ(z) has a period π in z. Next consider ψ(z + πτ ) :
ψ(z + πτ ) =
θ3 (z + πτ |τ ) = (z+πτ )2 z+πτ 1 exp exp θ | − 3 πiτ τ τ
z2 πiτ
e−iπτ e−2iz θ3 (z|τ ) . θ3 τz + π| − τ1 exp πτ i
We thus obtain |psi(z + πτ ) =
exp
θ3 (z|τ ) = ψ(z). θ3 τz | − τ1
z2 πiτ
Therefore ψ(τ ) also has the period πτ in z. We can now conclude that ψ(z) is an elliptic function of order less than two. It is a constant with the value c = ψ(z) = ψ(0) =
θ3 (0|τ ) θ3 (0|τ ) = . e0 θ3 (0| − τ1 ) θ3 (0| − τ1 )
We shall evaluate c later. At present we have (A)
θ3 (z|τ ) = c exp
z2 πiτ
z 1 θ3 , − τ τ
upon which we shalluse our basic table. πτ In (A) replace z by z + to get 2 2 (z + πτ z π 1 πτ 2 ) θ3 + − . θ3 z + τ = c exp 2 πiτ τ 2 τ We conclude that (B)
θ2 (z|τ ) = c exp
z2 πiτ
θ4
z 1 . − τ τ
π Return to (A) and replace z by z + to get 2 (z + π2 )2 π 1 π z θ3 + θ3 z + τ = c exp − 2 πiτ τ 2τ τ or
SOLUTIONS TO RAINVILLE’S “SPECIAL FUNCTIONS” (1960)
173
π z z z2 π 1 1 exp θ3 − + − exp − iτ 4iτ τ 2 τ τ πiτ 2 π πi z z iz z 1 exp exp = c exp exp exp θ2 − − 4τ τ τ τ πiτ iτ 4iτ 2 z z 1 = c exp θ2 . − πiτ τ τ
θ4 (z|τ )
= c exp
Hence we arrive at (C)
z2 πiτ
z 1 θ2 . − τ τ
θ4 (z|τ ) = c exp πτ Now in (C) replace z by z + . We thus obtain 2 2 (z + πτ πτ π 1 z 2 ) θ4 z + θ2 + − , τ = c exp 2 πiτ τ 2 τ or
πiτ i exp − 4
exp(−iz)θ1 (z|τ ) = c exp
z2 πiτ
exp
z c
exp
πτ 4i
z 1 . (−1)θ1 − τ τ
we arrive in this way at z 1 (D) θ1 (z|τ ) = icθ1 . − τ τ We still need to find the value of c. From (D) we get ic 0 z 1 . (E) = θ1 − τ τ τ We wish to put z = 0 in (E) and use the known result that θ10 = θ2 θ3 θ4 . From (E) we thus get θ10 (z|τ )
θ2 (0|τ )θ3 (0|τ )θ4 (0|τ ) =
ic θ2 τ
0| −
1 τ
θ3
0| −
1 τ
θ4
or θ2 (0|τ ) θ3 (0|τ ) θ4 (0|τ ) ic 1 · 1 · 1 = τ . θ4 (0| − τ ) θ3 (0| − τ ) θ2 (0| − τ ) With the aid of (A), (B), (C), we may conclude that ccc =
ic , τ
or c2 = 1
1
Hence c = (−iτ )− 2 or c = −(−iτ )− 2 . Now θ3 (z) was defined by
−1 . iτ
0| −
1 τ
,
174SOLUTIONS BY SYLVESTER J. PAGANO AND LEON HALL; EDITED BY TOM CUCHTA
θ3 (z|τ ) = 1 + 2
∞ X
2
e−iπn
τ
cos(2nz).
n=1
It follows that, if τ is pure imaginary, then θ3 (0|τ ) = 1 + 2
∞ X
2
eπn
(−iτ )
>0
n=0
and, since −1|τ is also then pure imaginary, 1 θ3 0| − > 0. τ 1
We may therefore conclude that c = (−iτ )− 2 . Final result of our work is that the set of identities 2 z z 1 − 12 θ3 (z|τ ) = (−iτ ) exp θ3 , − πiτ τ τ 2 1 z z 1 θ2 (z|τ ) = (−iτ )− 2 exp θ4 − , πiτ τ τ 2 1 z z 1 θ4 (z|τ ) = i(−iτ )− 2 exp θ2 , − πiτ τ τ 2 z z 1 − 12 θ1 (z|τ ) = i(−iτ ) exp θ1 . − πiτ τ τ With regard to the usefulness of these formulas, note that q = exp(πiτ ) so that |q| = exp[−Im(τ )]. For Im(τ ) > 0, |q| < 1 and we have convergence of the series definitions of the theta functions but that convergence is slow if Im(τ ) is small so that |q| is near unity. Since 1 Im(τ ) Im − = , τ |τ |2 the parameter (−τ −1 ) will be useful in computations with Im(t) small and positive. 24. Chapter 21 Solutions Problem 227. Derive the preceding addition theorem (7) by the method of Section 179. You may use results from the exercises at the end of Chapter 20. Solution 227. We seek an addition theorem for cn(u). Now θ4 θ2 (uθ3−2 ) . θ2 θ4 (uθ3−2 ) At the end of Exercise 5, Chapter 20 we obtained cn(u) =
θ42 θ4 (x + 4)θ4 (x − y) = θ42 (x)θ42 (y) − θ12 (x)θ12 (y). The next to last equation in Exercise 10, Chapter 20, is
SOLUTIONS TO RAINVILLE’S “SPECIAL FUNCTIONS” (1960)
175
θ2 θ4 θ2 (x + y)θ4 (x − y) = θ2 (x)θ4 (x)θ2 (y)θ4 (y) − θ1 (x)θ3 (x)θ1 (y)θ3 (y). We divide the last equation above by the one preceding it to get θ2 (x + y) θ4 (x + y)
θ2 (x)θ4 (x)θ2 (y)θ4 (y) − θ1 (x)θ3 (x)θ1 (y)θ3 (y) θ42 (x)θ42 (y) − θ12 (x)θ12 (y) θ12 (x) θ22 (y) θ2 (x) θ2 (y) θ3 (x) θ1 (y) θ3 (y) = df rac θ4 (x) θ4 (y) − θθ14 (x) (x) θ4 (x) θ4 (y) θ4 (y) 1 − θ 2 (x) θ 2 (y) .
=
4
Now use x =
uθ3−2
and y =
θ2 θ2 cn(u + v) = θ4 θ4
vθ3−2
4
to obtain
θ2 θ2 θ4 cn(u) θ4 cn(v)
θ2 θ3 θ2 θ3 θ3 sn(u) θ4 dn(u) θ3 sn(v) θ4 dn(v) θ2 f racθ22 θ32 sn2 (u) θ22 sn2 (v) 3
1−
−
or cn(u + v) =
cn(u)cn(v) − sn(u)dn(u)sn(v)dn(v) , 1 − k 2 sn2 (u)sn2 (v)
as desired. Problem 228. Derive preceding (8) by the method of Section 179, with the aid of results from the exercises at the end of Chapter 20. Solution 228. We know that dn(u) =
θ2 θ4 θ3 (uθ3−2 ) ,k = 2. −2 θ3 θ4 (uθ3 ) θ3
From the last equation of Exercise 10, Chapter 20 we get θ3 θ4 θ3 (x + y)θ4 (x − y) = θ3 (x)θ4 (x)θ3 (y)θ4 (y) − θ1 (x)θ2 (x)θ1 (y)θ2 (y) and from the last equation of Exercise 8 Chapter 20 we get θ42 θ4 (x + y)θ4 (x − y) = θ42 (x)θ42 (y) − θ12 (x)θ12 (y). The two equations about at once yield θ3 θ3 dn(u + v) θ4 θ4
=
θ3 θ3 θ4 dn(u) θ4 dn(v)
−
1−
θ2 θ2 θ2 θ2 θ3 sn(u)) θ4 cn(u) θ3 sn(v) θ4 sn(v) . θ22 θ2 sn2 (u) θ22 sn2 (v) θ32 3
Hence we arrive at the desired result, dn(u)dn(v) − k 2 sn(u)cn(u)sn(v)cn(v) . 1 − k 2 sn2 (u)sn2 (v) Z 1 Problem 229. Show that cn3 (x)dn(x)dx = sn(x) − sn3 (x) + c. 3 dn(u + v) =
176SOLUTIONS BY SYLVESTER J. PAGANO AND LEON HALL; EDITED BY TOM CUCHTA
Solution 229. We wish to evaluate Z cn3 (x)dn(x)dx. We know that d dn(x) = cn(x)dn(x) dx and that cn2 (x) = 1 − sn2 (x). Hence Z
Z
3
cn (x)dn(x)dx
=
[1 − sn2 (x)]cn(x)dn(x)dx
1 = sn(x) − sn2 (x) + c. 3 Problem 230. Show that if g(x) and h(x) are any two different ones of the three functions sn(x), cn(x), dn(x), and if m is a non-negative integer, you can perform the integration Z g 2m+1 (x)h(x)dx. Solution 230. Consider Z A=
g 2m+1 (x)h(x)dx.
We know that d sn(x) = cn(x)dn(x); dx d cn(x) = −sn(x)dn(x); dx d dn(x) = −k 2 sn(x)cn(x). dx Now let g(x) and h(x) be any two different ones of sn(x), cn(x), dn(x). Let ψ(x) be the other of the three functions. THen g(x)h(x)dx = c1 d[ψ(x)], in which c1 = 1, −1 or −K −2 according to whether ψ(x) is sn(x), cn(x), or dn(x). Now we also know that sn2 (x) + cn2 (x) = 1, k 2 sn2 (x) + dn2 (x) = 1, dn2 (x) − k 2 cn2 (x) = x2 .
SOLUTIONS TO RAINVILLE’S “SPECIAL FUNCTIONS” (1960)
177
Therefore the square of any one of sn(x), cn(x), dn(x) is a linear function of the square of any other one of them. Hence we may write g 2 (x) = a1 + b1 ψ 2 (x). We know have Z
g 2m+1 (x)h(x)dx
Z
[g 2 (x)]m g(x)h(x)dx Z = c1 [a1 + b1 ψ 2 (x)]m dψ(x), =
which is integrable as a sum of powers. Problem 231. Obtain the result Z 1 sn(x)dx = log[dn(x) − kcn(x)] + c. k Z Solution 231. Consider sn(x)dx. We may write Z Z sn(x)[dn(x) − kcn(x)] sn(x)dx = dx dn(x) − kcn(x) Z sn(x)dn(x) − ksn(x)cn(x) = dx dn(x) − kcn(x) Z −d[dn(x)] + k1 d[dn(x)] = Z dn(x) − kcn(x) 1 d[dn(x)] − k[cn(x)] = k dn(x) − kcn(x) 1 = log[dn(x) − kcn(x)] + c. k Problem 232. Show that dn(2x) − kcn(2x) =
(1 − k)[1 + ksn2 (x)] . 1 − ksn2 (x)
Solution 232. We wish to show that (1 − k)[1 + ksn2 (x)] . 1 − ksn2 (x) Now from the addition formulas of Exercises 227 and 228 we obtain dn(2x) − kcn(2x) =
dn(2x) =
dn2 (x) − k 2 sn2 (x)dn2 (x) 1 − k 2 sn4 (x)
and cn2 (x) − sn2 (x)dn2 (x) . 1 − k 2 sn4 (x) Let us put each of the above in terms of sn(x). We get cn(2x) =
dn(2x) =
1 − k 2 sn2 (x) − k 2 sn2 (x)[1 − sn2 (x)] 1 − 2k 2 sn2 (x) + k 2 sn4 (x) = . 1 − k 2 sn4 (x) 1 − k 2 sn4 (x)
178SOLUTIONS BY SYLVESTER J. PAGANO AND LEON HALL; EDITED BY TOM CUCHTA
cn(2x) =
1 − sn2 (x) − sn2 (x)[1 − x2 sn2 (x)] 1 − 2k 2 sn2 (x) + k 2 sn4 (x) = 1 − k 2 sn4 (x) 1 − k 2 sn4 (x)
THen dn(2x) − kcn(2x)
1 − k 2 sn2 (x) − k 2 sn2 (x) + k 2 sn4 (x) − k + ksn2 (x) + ksn2 (k) − k 3 sn4 (x) 1 − k 2 sn4 (x) 2 2 1 + ksn (x) + ksn (x)[1 + ksn2 (x)] − k[1 + ksn2 (x)] − k 2 sn2 (x)[1 + ksn2 (x)] = [1 − ksn2 (x)][1 + ksn2 (x)] 2 2 2 1 + ksn (x) − k − k sn (x) = 1 − ksn2 (x) 1 − k + k(1 − k)sn2 (x) = . 1 − ksn2 (x) =
Hence dn(2x) − kcn(2x) =
(1 − k)[1 + ksn2 (x)] . 1 − ksn2 (x)
