Page 363 - Schaum's Outline of Theory and Problems of Advanced Calculus

354 FOURIER SERIES [CHAP. 13

CONVERGENCE OF FOURIER SERIES
1
sinðM þ Þt
2
13.18. Prove that (a) 1
2 þ cos t þ cos 2t þ þ cos Mt ¼ 1
2 sin t
2
1 ð 0 1
1 sinðM þ Þt 1 1 sinðM þ Þt 1
ð
2
2
(b) 1 dt ¼ ; 1 dt ¼ :
0 2 sin t 2 2 sin t 2
2 2
1
1
1
1
(a)We have cos nt sin t ¼ fsinðn þ Þt sinðn Þtg.
2 2 2 2
Then summing from n ¼ 1to M,
1 3 1 5 3
2 2 2 2 2
sin tfcos t þ cos 2t þ þ cos Mtg¼ðsin t sin tÞþðsin t sin tÞ
1

1
þ þ sinðM þ Þt sinðM Þt
2 2
1 1 1
2 2 2
¼ fsinðM þ Þt sin tg
1
1
On dividing by sin t and adding ,the required result follows.
2 2
(b)Integrating the result in (a) from to 0 and 0 to , respectively. This gives the required results, since
the integrals of all the cosine terms are zero.
ð ð
13.19. Prove that lim f ðxÞ sin nx dx ¼ lim f ðxÞ cos nx dx ¼ 0if f ðxÞ is piecewise continuous.
n!1 n!1
2 1
a 0 X 2 2
This follows at once from Problem 13.15, since if the series þ ða n þ b n Þ is convergent, lim a n ¼
lim b n ¼ 0. 2 n¼1 n!1
n!1
The result is sometimes called Riemann’s theorem.

ð
1
13.20. Prove that lim f ðxÞ sinðM þ Þxdx ¼ 0if f ðxÞ is piecewise continuous.
2
M!1
We have
ð ð ð
1
1 1 f f ðxÞ cos xg sin Mx dx
2 2 2
f ðxÞ sinðM þ Þxdx ¼ f f ðxÞ sin xg cos Mx dx þ

Then the required result follows at once by using the result of Problem 13.19, with f ðxÞ replaced by
1
1
f ðxÞ sin x and f ðxÞ cos x respectively, which are piecewise continuous if f ðxÞ is.
2 2
The result can also be proved when the integration limits are a and b instead of and .
13.21. Assuming that L ¼ , i.e., that the Fourier series corresponding to f ðxÞ has period 2L ¼ 2 , show
that
M ð 1
X 1 sinðM þ Þt
a 0
2
2 2 sin t
S M ðxÞ¼ þ ða n cos nx þ b n sin nxÞ¼ f ðt þ xÞ 1 dt
n¼1 2
Using the formulas for the Fourier coefficients with L ¼ ,we have
1 1
ð ð
f ðuÞ sin nu du sin nx
a n cos nx þ b n sin nx ¼ f ðuÞ cos nu du cos nx þ

1 ð
f ðuÞ cos nu cos nx þ sin nu sin nxÞ du
¼ ð

1 ð
f ðuÞ cos nðu xÞ du
¼

1 ð
a 0
Also, f ðuÞ du
2 ¼ 2