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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 π Z π-π-x sin kxdx =-2 π Z π 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-π < x < π for x = π ; f ( x + 2 π ) = f ( x ) . Prove that the derivative of an odd function is even. An odd function satisﬁes 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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