Page 303 - Schaum's Outline of Theory and Problems of Advanced Calculus

294 INFINITE SERIES [CHAP. 11

u 2 u 3 u 4 u 5
u
2! 3! 4! 5!
We have e ¼ 1 þ u þ þ þ þ þ ; 1 < u < 1:

2 2 x 4 x 6 x 8 x 10
2
x
Then if u ¼ x ; e þ ; 1 < x < 1:
2! 3! þ 3! ¼ 5!
¼ 1 x þ
2 2 4 6 8
x
1 e x x x x
Thus ¼ 1 þ þ :
x 2 2! 3! 4! 5!
Since the series converges for all x and so, in particular, converges uniformly for 0 @ x @ 1, we can
integrate term by term to obtain
1 1 e x x x x
ð x 2 3 5 7 9 1

þ

dx ¼ x
þ

0 x 2 3 2! 5 3! 7 4! 9 5! 0
1 1 1 1
3 2! 5 3! 7 4! 9 5!
¼ 1 þ þ
¼ 1 0:16666 þ 0:03333 0:00595 þ 0:00092 ¼ 0:862
Note that the error made in adding the first four terms of the alternating series is less than the fifth term,
i.e., less than 0.001 (see Problem 11.15).
MISCELLANEOUS PROBLEMS
11.46. Prove that y ¼ J p ðxÞ defined by (16), Page 276, satisfies Bessel’s differential equation

2
2
x y þ xy þðx p Þy ¼ 0
2 00
0
The series for J p ðxÞ converges for all x [see Problem 11.110(a)]. Since a power series can be differ-
entiated term by term within its interval of convergence, we have for all x,
1 n pþ2n
X ð 1Þ x
y ¼ pþ2n
2 n!ðn þ pÞ!
n¼0
ð 1Þ ð p þ 2nÞx
1 n pþ2n 1
X
0
y ¼ pþ2n
2 n!ðn þ pÞ!
n¼0
ð 1Þ ð p þ 2nÞð p þ 2n 1Þ x
1 n pþ2n 2
X
00
y ¼ pþ2n
2 n!ðn þ pÞ!
n¼0
Then,
1 n pþ2nþ2 1 n 2 pþ2n
X ð 1Þ x X ð 1Þ p x
2 2
ðx p Þy ¼ pþ2n pþ2n
2 n!ðn þ pÞ! 2 n!ðn þ pÞ!
n¼0 n¼0
1 n pþ2n
X ð 1Þ ðp þ 2nÞx
0
xy ¼ pþ2n
2 n!ðn þ pÞ!
n¼0
1 n pþ2n
X ð 1Þ ð p þ 2nÞð p þ 2n 1Þx
2 00
x y ¼ pþ2n
2 n!ðn þ pÞ!
n¼0