(a) This follows at once from the results of Problems 13.2 and 13.3. (b) By Problem 13.3, L in VS .NET

Generate QR Code JIS X 0510 in VS .NET (a) This follows at once from the results of Problems 13.2 and 13.3. (b) By Problem 13.3, L

(a) This follows at once from the results of Problems 13.2 and 13.3. (b) By Problem 13.3, L
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FOURIER SERIES !2 r 1 mx dx 1; sin L L L Also,
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[CHAP. 13 !2 r 1 mx dx 1 cos L L
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Thus the required orthonormal set is given by 1 1 x 1 x 1 2x 1 2x p ; p sin ; p cos ; p sin ; p cos ;... L L L L L L L 2L L
MISCELLANEOUS PROBLEMS 13.26. Find a Fourier series for f x cos x;  @ x @ , where 6 0; 1; 2; 3; . . . .
We shall take the period as 2 so that 2L 2; L . Since the function is even, bn 0 and 2 L 2  f x cos nx dx cos x cos nx dx an L 0  0  1 fcos n x cos n xg dx  0 & ' 1 sin n  sin n  2 sin  cos n  n n  2 n2 2 sin  0  Then cos x
1 sin  2 sin  X cos n cos nx   2 n2 n 1   sin  1 2 2 2 2 cos x 2 cos 2x 2 cos 3x  12 22 32
x2 13.27. Prove that sin x x 1 2 
x2 1 2 2
! x2 1 . 3 2
Let x  in the Fourier series obtained in Problem 13.26. Then   sin  1 2 2 2 2 2 2 cos  12 22 32 or  cot  1 2 2 2 2 12 2 22 2 32 1
This result is of interest since it represents an expansion of the contangent into partial fractions. By the Weierstrass M test, the series on the right of (1) converges uniformly for 0 @ j j @ jxj < 1 and the left-hand side of (1) approaches zero as ! 0, as is seen by using L Hospital s rule. Thus, we can integrate both sides of (1) from 0 to x to obtain  x x x 1 2 2  cot  d d d 2 1 2 2 0 0 0 2 or ! !   2 2 sin  x  ln 1 x ln 1 x ln  0 12 22
CHAP. 13] 
FOURIER SERIES ! ! !  sin x x2 x2 x2 lim ln 1 2 ln 1 2 ln 1 2 ln n!1 x 1 2 n ( ! ! !) 2 2 2 x x x lim ln 1 2 1 2 1 2 n!1 1 2 n ( ! ! !) 2 2 x x x2 ln lim 1 2 1 2 1 2 n!1 1 2 n ! ! ! ! ! x2 x2 x2 x2 1 2 1 2 1 2 1 2 2 n 1 2 ! ! x2 1 2 2
i.e.,
so that sin x x2 lim 1 2 n!1 x 1 Replacing x by x=, we obtain
x2 sin x x 1 2 
called the in nite product for sin x, which can be shown valid for all x. The result is of interest since it corresponds to a factorization of sin x in a manner analogous to factorization of a polynomial.
13.28. Prove that
 2 2 4 4 6 6 8 8... . 2 1 3 3 5 5 7 7 9...
Let x 1=2 in equation (2) of Problem 13.27. Then,         2 1 1 1 1 3 3 5 5 7 1 2 1 2 1 2  2 2 4 4 6 6 2 4 6 Taking reciprocals of both sides, we obtain the required result, which is often called Wallis product.
Supplementary Problems
FOURIER SERIES 13.29. Graph each of the following functions and nd their corresponding Fourier series using properties of even and odd functions wherever applicable. & & 8 0<x<2 x 4 @ x @ 0 a f x Period 4 b f x Period 8 8 2 < x < 4 x 0@x@4 & c f x 4x; 0 < x < 10; Period 10 d f x 2x 0 0@x<3 3 < x < 0 Period 6
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