bracketleftbig f x f x bracketrightbig 200 In particular at x T we have a 2

# Bracketleftbig f x f x bracketrightbig 200 in

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bracketleftbig f ( x + ) + f ( x )bracketrightbig . (200) In particular, at x = ± T , we have a 0 2 + summationdisplay n =1 braceleftBig a n cos n π x T + b n sin n π x T bracerightBig = 1 2 bracketleftbig f ( T + ) + f ( T )bracketrightbig . (201) Remark 19. Recall that a function is “piecewise continuous” on [ a, b ] if it is continuous at every point in [ a, b ] except possibly for a finite number of points at which f has a “jump discontinuity”, that is f ( x + ) , f ( x ) both exist but are not equal. Recall that f ( x + ) = lim h> 0 ,h 0 f ( x + h ) , f ( x ) = lim h< 0 ,h 0 f ( x + h )= lim h> 0 ,h 0 f ( x h ) . (202) Remark 20. In particular, for functions specified in the theorem, we have f ( x )= a 0 2 + summationdisplay n =1 braceleftBig a n cos n π x T + b n sin n π x T bracerightBig (203) whenever f is continuous at x . Example 21. (10.3.17) Consider the Fourier series for f ( x ) with period 2 π and f ( x ) = x for π < x < π . f is continuous for π < x < π but f ( π ) = π π = f ( π +) . As a consequence its Fourier series will converge to f ˜ with period 2 π and f ˜ ( x )= braceleftbigg x π < x < π 0 x = ± π (204) Example 22. (10.3.19) Consider the Fourier series for f ( x ) with period 4 and f ( x )= braceleftbigg 1 2 < x < 0 x 0 < x < 2 . (205) We see that f is continuous for 2 < x < 0 and 0 < x < 2 , while has jump discontinuity at 2 , 0 , 2 . As a consequence its Fourier series converges to f ˜ with period 4 and f ˜ ( x )= 1 2 < x < 0 1/2 x =0 x 0 < x < 2 3/2 x = ± 2 . (206) Remark 23. (Gibbs phenomenon, 10.3.39) When jump discontinuities are present, the convergence around them suffers from the so-called “Gibbs phenomenon”, that is, the partial sum a 0 2 + summationdisplay 1 N braceleftBig a n cos n π x T + b n sin n π x T bracerightBig (207) always overshoot f by approximately 9% of the size of the jump. As a consequence the convergence is not uniform – which means it may be misleading to use the graph of the partial sum as an indicator of what the actually function looks like. Gibbs phenomenon disappears as soon as the function is continuous everywhere. In fact, in this case the convergence is uniform: Theorem 24. (Uniform convergence of Fourier series) Let f be a continuous functions on ( − ∞ , ) and periodic of period 2 T . If f is piecewise continuous on [ T , T ] , then the Fourier series for f con- verges uniformly to f on [ T , T ] and hence on any interval. That is, for each ε > 0 , there exists an integer N 0 (that depends on ε ) such that vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle f ( x ) bracketleftBigg a 0 2 + summationdisplay 1 N braceleftBig a n cos n π x T + b n sin n π x T bracerightBig bracketrightBiggvextendsingle vextendsingle vextendsingle vextendsingle vextendsingle < ε (208) for all N greaterorequalslant N 0 and all x ( − ∞ , ) . For f with appropriate continuity/smoothness, we can differentiate and integrate term by term.  #### You've reached the end of your free preview.

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