Page 364 - Schaum's Outline of Theory and Problems of Advanced Calculus

CHAP. 13] FOURIER SERIES 355

M
a 0 X
Then
2 þ ða n cos nx þ b n sin nxÞ
S M ðxÞ¼
n¼1

M ð
1 1 X
ð
¼ f ðuÞ du þ f ðuÞ cos nðu xÞ du
2
n¼1
M
1 ð ( 1 X )
cos nðu xÞ du
¼ f ðuÞ þ
2
n¼1
1 ð 1
2 du
sinðM þ Þðu xÞ
1
¼ f ðuÞ
2 sin ðu xÞ
2
using Problem 13.18. Letting u x ¼ t,we have
1
1 ð x sinðM þ Þt
2
S M ðxÞ¼ f ðt þ xÞ 1 dt
2 sin t
x 2
Since the integrand has period 2 ,wecan replace the interval x; x by any other interval of
length 2 ,in particular, ; . Thus, we obtain the required result.
13.22. Prove that
!
1
ð 0
sinðM þ Þtdt
f ðx þ 0Þþ f ðx 0Þ f ðt þ xÞ f ðx 0Þ 1
S M ðxÞ ¼ 1 2
2 2 sin t
2
!
1 ð f ðt þ xÞ f ðx þ 0Þ
1
sinðM þ Þtdt
þ 1 2
0 2 sin t
2
From Problem 13.21,
1
1
1 ð 0 sinðM þ Þt 1 ð sinðM þ Þt
2
2
S M ðxÞ¼ f ðt þ xÞ 1 dt þ f ðt þ xÞ 1 dt ð1Þ
2 sin t 0 2 sin t
2
2
Multiplying the integrals of Problem 13.18(b)by f ðx 0Þ and f ðx þ 0Þ, respectively,
1
1
1 ð 0 sinðM þ Þt 1 ð sinðM þ Þt
2 2 dt
f ðx þ 0Þþ f ðx 0Þ
1
1
2 ¼ f ðx 0Þ 2 sin t dt þ 0 f ðx þ 0Þ 2 sin t ð2Þ
2 2
Subtracting (2) from (1)yields the required result.
13.23. If f ðxÞ and f ðxÞ are piecewise continuous in ð ; Þ, prove that
0
f ðx þ 0Þþ f ðx 0Þ
lim S M ðxÞ¼
2
M!1
The function f ðt þ xÞ f ðx þ 0Þ is piecewise continuous in 0 < t @ because f ðxÞ is piecewise con-
1
2
tinous. 2 sin t
t
Also, lim f ðt þ xÞ f ðx þ 0Þ ¼ lim f ðt þ xÞ f ðx þ 0Þ ¼ lim f ðt þ xÞ f ðx þ 0Þ exists,
1
1
2 sin t t 2 sin t t
2 2
t!0þ t!0þ t!0þ
since by hypothesis f ðxÞ is piecewise continuous so that the right-hand derivative of f ðxÞ at each x exists.
0
Thus, f ðt þ xÞ f ðx 0Þ is piecewise continous in 0 @ t @ .
1
2 sin t
2
Similarly, f ðt þ xÞ f ðx 0Þ is piecewise continous in @ t @ 0.
1
2 sin t
2