12.3 Alternating series and Fourier series 617Then tell why (12.37) must be valid and substitute in it to obtain
ax 4 ax 1 3- I 5ax 1 lax
SinL =7r[sinL +3sin L +5sin-I-+7sin
L +Observe that this implies that
a ax (^1) Sax (^1) Sax 1 lax
sin L + 3 sin L + sin L + sin L +
when 0 < x < L and that putting x = L/2 gives the formula
a=1
4 -sf5 v v
which we have seen before.
(^10) Engineers who are interested in fully rectified (or full-wave rectified)
alternating currents would want x replaced by wt in the formula
4 r1 cos 2x cos 4x cos 6x cos 8x 11
sin x L2 - 1 3 -
3.5 - 5.7 - 7.9 -
...J'
but this shift is easily made. Work out the formula and observe that it is correct
when x = 0 because
2
= (^113315) + 517 + 719 + ...
(^11) Sketch a graph of the even function f of period 2a for which f(x) = x when
0 =< x < a. Show that
f(x) =a 2 - a4 (cosx +cos 3z 32 +cos 5x 52 +cos 7x 72 +
12 Use the formulas following (12.384) to show that, when n is a positive
integer,
B2^(0) = (-1)"+1
(27r)2
2n (2n)
and hence that B2,(0) is a small number like 1/621 when n is large. Use the
formula Bk = k! Bk(0) and the Stirling formula of Section 12.2, Problem 19, to
show that
2n
Ben = (-1)n+14 1/n7r(n7re S(2n)02^124n
Remark: Even crude estimates show that JB2nj is very large when n is large.
Since re < 9, we have n/ire > 10 and JB2n1 > 102n when n > 90. Similarly,
JB2nI > 10002n when n? 900.
13 Our work with Fourier series has involved Fourier analysis. We started
with given functions and found their Fourier series. While proofs of results lie
beyond the scope of this course, we take brief cognizance of a problem in Fourier