Page 300 - Schaum's Outline of Theory and Problems of Advanced Calculus

CHAP. 11] INFINITE SERIES 291

2
nx
11.37. Let S n ðxÞ¼ nxe ; n ¼ 1; 2; 3; ... ; 0 @ x @ 1.
1 1
ð ð
Determine whether lim lim S n ðxÞ dx:
ðaÞ S n ðxÞ dx ¼
n!1 0 0 n!1
Explain the result in (aÞ:
ðbÞ
ð 1 ð 1 2 2
1 nx 1
n
1
nxe nx dx ¼ e j 0 ¼ ð1 e Þ: Then
ðaÞ s n ðxÞ dx ¼ 2 2
0 0
1
ð
n
1
lim S n ðxÞ dx ¼ lim ð1 e Þ¼ 1
n!1 0 n!1 2 2
2
SðxÞ¼ lim S n ðxÞ¼ lim nxe nx ¼ 0; whether x ¼ 0or0 < x @ 1: Then,
n!1 n!1
1
ð
SðxÞ dx ¼ 0
0
ð 1 ð 1
It follows that lim S n ðxÞ dx 6¼ lim S n ðxÞ dx, i.e., the limit cannot be taken under the integral
sign. n!1 0 0 n!1
(b) The reason for the result in (a)isthat although the sequence S n ðxÞ converges to 0, it does not converge
ffiffiffiffiffi
p
uniformly to 0. To show this, observe that the function nxe nx 2 has a maximum at x ¼ 1= 2n (by the
q ffiffiffiffiffi
usual rules of elementary calculus), the value of this maximum being 1 n e 1=2 . Hence, as n !1,
2
S n ðxÞ cannot be made arbitrarily small for all x and so cannot converge uniformly to 0.
sin nx 1
1 ð 1
X X
: Prove that f ðxÞ dx ¼ 2 .
n
11.38. Let f ðxÞ¼ 3 4
n¼1 0 n¼1 ð2n 1Þ

sin nx 1
We have @ . Then by the Weierstrass M test the series is uniformly convergent for all x,in
n 3 n 3
particular 0 @ x @ , and can be integrated term by term. Thus
!
1 ð
ð ð X sin nx X sin nx
1
dx
f ðxÞ dx ¼ 3 dx ¼ 3
0 0 n¼1 n n¼1 0 n
1 cos n 1 1 1 1
1 1
X X
¼ 2 þ ¼ 2
n 1 3 5
¼ 4 4 þ 4 þ 4 4
n¼1 n¼1 ð2n 1Þ
POWER SERIES
1 1
X X
n
11.39. Prove that both the power series a n x and the corresponding series of derivatives na n x n 1
n¼0 n¼0
have the same radius of convergence.
n
Let R > 0be theradius of convergence of a n x .Let 0 < jx 0 j < R. Then, as in Problem 11.33, we can
1
choose N as that ja n j < for n > N.
n
jx 0 j
Thus, the terms of the series jna n x n 1 j¼ nja n jjxj n 1 can for n > N be made less than corresponding
n 1
terms of the series n jxj , which converges, by the ratio test, for jxj < jx 0 j < R.
n
jx 0 j
Hence, na n x n 1 converges absolutely for all points x 0 (no matter how close jx 0 j is to R).
n
If, however, jxj > R, lim a n x 6¼ 0 and thus lim na n x n 1 6¼ 0, so that na n x n 1 does not converge.
n!1 n!1