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Thus nbn is bounded as n ^ ^. Likewise, for an, we
1 ´ü . nnx
obtain nan = -— f (x)sin----------dx and hence nan is also
Ï J-L L
bounded as n ^ ^.
18b. Note that f and f' are continuous at all points where f" is continuous. Let x1, ..., xm be the points in (-L,L)
where f" is not continuous. By splitting up the interval of integration at these points, and integrating Eq.(3) by parts twice, we obtain
0 n X-1 nnxi n
n2bn = — V [f(xi + )-f(xi-)]cos--------— [f(L-)-f(-L+)]cosnn
Ï ^ L Ï
i = 1
L V-1 nnxi L ´ü nnx
-— V [f (xi+)-f (x1-)]sin -------- — f (x)sin---------dx, where
n2^ L n2J-L L
we have used the fact that cosine is continuous. We want the first two terms on the right side to be zero, for otherwise they grow in magnitude with n. Hence f must be continuous throughout the closed interval [-L,L]. The
last two terms are bounded, by the hypotheses on f and
22 f . Hence n bn is bounded; similarly n an is bounded.
Convergence of the Fourier series then follows by comparison with V n-2.
Section 10.4, Page 570
3. Let f(x) = tan2x, then
f(-x) = tan(-2x) = ---------------------------- = - = -tan2x = -f(x)
and thus tan2x is an odd function.
6. Let f(x) = e x, then f(-x) = ex so that f(-x) Ô f(x) and f(-x) Ô -f(x) and thus e-x is neither even nor odd.
13. By the hint f(-x) = g(-x) + h(-x) = g(x) - h(x), since g is an even function and f is an odd function. Thus f(x) + f(-x) = 2g(x) and hence g(x) = [f(x) + f(-x)]/2 defines g(x). Likewise
f(x)-f(-x) = g(x)-g(-x) + h(x)-h(-x) = 2h(x) and thus
h(x) = [f(x) - f(-x)]/2.
All functions and their derivatives in Problems 14 through 30 are piecewise continuous on the given intervals and their extensions. Thus the Fourier Theorem applies in all cases.
14. For the cosine series we use the even extension of the function given in Eq.(13) and hence
[ 0 -2 < x <-1
f(x) = < on the interval -2 < x < 0.
[ 1+x -1 < x < 0
However, we don't really need this, as the coefficients in this case are given by Eqs.(7), which just use the original values for f(x) on 0 < x < 2. Applying Eqs.(7) we have L = 2 and thus
a0 = (2/2) J1(1-x)dx + (2/2)J20dx = 1/2. Similarly,
an = (2/2)11(1-x)cos(nnx/2)dx = 4[1-cos(nn/2)]/n2n2 and
bn = 0. Substituting these values in the Fourier series yields the desired results.
For the sine series, we use Eqs.(8) with L = 2. Thus an = 0 and
bn = (2/2)11(1-x)sin(nnx/2)dx = 4[nn/2 - sin(nn/2)]/n2n2.
The graph of the function to which the series converges is shown in the figure. Using Eqs.(7) with L = 2 we have
a0 = 11dx = 1 and an = 11cos(nïx/2)dx = 2sin(nï/2)/nï.
Thus an = 0 for n even, an = 2/ïï for n = 1,5,9,...and an = -2/ïï for n = 3,7,11,... . Hence we may write
a2n = 0 and a2n-1 = 2(-1) ï+1/(2ï-1)ï, which when substituted into the series gives the desired answer.
The graph of the function to which the series converges is indicated in the figure.
1 A A . . 1 , 1 A .
•3n -2¿¿ -¿¿ ¿¿ 2¿¿ Ç¿¿
Since we want a sine series, we use Eqs.(8) to find, with
L = ï, that bn = (2/ï) Psinnxdx = 2[1-(-1) ï]/ïï and thus
bn = 0 for n even and bn = 4/ïï for n odd.
The graph of the function to which the series converges is shown in the figure.
We note that f(x) is specified over its entire fundamental period (T = 1) and hence we cannot extend f to make it either an odd or an even function. Using Eqs.(2) and (3) from Section 10.3 we have (L = 1/2)
a0 = 2 11xdx = 1, an = 2 I ^cos^^x^x = 0 and
bn = 2 I ^s^^^x^x = -1/ïï. [Note: We have used the
results of Problem 27 of Section 10.2 in writing these integrals. That is, if f(x) is periodic of period T, then every integral of f over an interval of length T has the same value. Thus we integrate from 0 to 1 here, rather than -1/2 to 1/2.] Substituting the above values into Eq.(1) of Section 10.3 yields the desired solution. It can also be observed from the above graph that g(x) = f(x) - 1/2 is an odd function. If Eqs.(8) are used with g(x), then it is found that
g(x) = (-1/n) sin(2nnx)/n and thus we obtain the same
series for f(x) as found above.
2 ¥2 2 nnx
bn = — (2-x2)sin--------dx