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**Unformatted text preview: **As the function is odd, we know that a k = 0 for k = 0, 1, 2, . . . So we just have to calculate b k : b k = 1 Z --x sin kxdx =-2 Z x sin kxdx = 2(-1) k k . And so the Fourier series for f is X k =1 2(-1) k k sin kx. By the Fourier convergence theorem it converges to f = -x if- < x < for x = ; f ( x + 2 ) = f ( x ) . Prove that the derivative of an odd function is even. An odd function satises the condition f (-x ) =-f ( x ). If we take the derivative of this (using the chain rule) we get-f (-x ) =-f ( x ). Or in other terms: f (-x ) = f ( x ), i.e. f is even. 1...

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