INFINITE SERIES in .NET

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INFINITE SERIES
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Now Rn x Rn Rn 1 xn Rn 1 Rn 2 xn 1 Rn 2 Rn 3 xn 2 Rn xn Rn 1 xn 1 xn Rn 2 xn 2 xn 1 xn fRn 1 x Rn 1 Rn 2 x Rn 3 x2 g Hence, for 0 @ x < 1, jRn x j @ jRn j 1 x jRn 1 j jRn 2 jx jRn 3 jx2 1
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Since ak converges by hypothesis, it follows that given  > 0 we can choose N such that jRk j < =2 for all k A n. Then for n > N we have from (1),        2 jRn x j @ 1 x x x2  2 2 2 2 2 2 since 1 x 1 x x2 x3 1 (if 0 @ x < 1). Also, for x 1; jRn x j jRn j <  for n > N. Thus, jRn x j <  for all n > N, where N is independent of the value of x in 0 @ x @ 1, and the required result follows. Extensions to other power series are easily made.
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11.43. Prove Abel s limit theorem (see Page 272).
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As in Problem 11.42, assume the power series to be Then we must show that lim
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1 X k 0
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ak xk , convergent for 0 @ x @ 1.
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x!1
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1 X k 0
ak .
This follows at once from Problem 11.42, which shows that ak xk is uniformly convergent for 0 @ x @ 1, and from Problem 11.34, which shows that ak xk is continuous at x 1. Extensions to other power series are easily made.
11.44. (a) Prove that tan 1 x x 1 @ x @ 1. (b) Prove that
x3 x5 x7 where the series is uniformly convergent in 3 5 7
 1 1 1 1 . 4 3 5 7
1 1 x2 x4 x6 1 x2
(a) By Problem 2.25 of 2, with r x2 and a 1, we have 1<x<1 1
Integrating from 0 to x, where 1 < x < 1, yields x dx x3 x5 x7 tan 1 x x 2 3 5 7 0 1 x
using Problems 11.33 and 11.35. Since the series on the right of (2) converges for x 1, it follows by Problem 11.42 that the series is uniformly convergent in 1 @ x @ 1 and represents tan 1 x in this interval. (b) By Problem 11.43 and part (a), we have lim tan 1 x lim x x3 x5 x7 3 5 7 ! or  1 1 1 1 4 3 5 7
x!1
x!1
1 e x 11.45. Evaluate dx to 3 decimal place accuracy. x2 0
INFINITE SERIES
[CHAP. 11
We have eu 1 u
u2 u3 u4 u5 ; 2! 3! 4! 5!
1 < u < 1:
Then if u x2 ; e x 1 x2
x4 x6 x8 x10 ; 2! 3! 3! 5!
1 < x < 1:
Thus
1 e x x2 x4 x6 x8 1 : 2 2! 3! 4! 5! x
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
2 1  1 e x x3 x5 x7 x9 dx x   3 2! 5 3! 7 4! 9 5! x2 0 0 1 1 1 1 1 3 2! 5 3! 7 4! 9 5! 1 0:16666 0:03333 0:00595 0:00092 0:862
Note that the error made in adding the rst four terms of the alternating series is less than the fth term, i.e., less than 0.001 (see Problem 11.15).
MISCELLANEOUS PROBLEMS 11.46. Prove that y Jp x de ned by (16), Page 276, satis es Bessel s di erential equation x2 y 00 xy 0 x2 p2 y 0
The series for Jp x converges for all x [see Problem 11.110(a)]. Since a power series can be di erentiated term by term within its interval of convergence, we have for all x, y y0 y 00
1 X 1 n xp 2n 2p 2n n! n p ! n 0 1 X 1 n p 2n xp 2n 1 n 0
2p 2n n! n p ! 2p 2n n! n p !
1 X 1 n p 2n p 2n 1 xp 2n 2 n 0
Then, x2 p2 y xy 0 x2 y 00
1 1 X 1 n xp 2n 2 X 1 n p2 xp 2n 2p 2n n! n p ! n 0 2p 2n n! n p ! n 0 1 X 1 n p 2n xp 2n n 0 1 X
2p 2n n! n p !
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