# Q7_130 - As the function is odd we know that a k = 0 for k...

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Math 415 Quiz 7A Find the Fourier series for f ( x ) = x if - 1 x < 0 , 0 if 0 x < 1 , f ( x + 2) = f ( x ) . What function does it actually converge to? This is not an even or odd function, so none of our coeﬃcients are automatically zero. Also, since f ( x ) is zero on the interval [ - 1 , 0), our integrals will be simpliﬁed somewhat. a 0 = Z 1 0 xdx = 1 2 , a n = Z 1 0 x cos( nπx ) dx = ( - 1) n - 1 n 2 π 2 , b n = Z 1 0 x sin( nπx ) dx = ( - 1) n +1 . So the Fourier series is 1 4 + X n =1 ( - 1) n - 1 n 2 π 2 cos( nπx ) + ( - 1) n +1 sin( nπx ) . By the Fourier convergence theorem this converges to ˜ f ( x ) = x if - 1 < x < 0 , 0 if 0 < x < 1 , - 1 2 if x = 1 f ( x + 2) = f ( x ) . Prove that the derivative of an even function is odd. If f is an even function then f ( - x ) = f ( x ). Taking the derivative of both sides gives us (using the chain rule) - f 0 ( - x ) = f 0 ( x ). In other words, f 0 ( - x ) = - f 0 ( x ), and f 0 is thus odd. 1

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Math 415 Quiz 7B Find the Fourier series for f ( x ) = - x, - π x < π ; f ( x + 2) = f ( x ) . What function does it actually converge to?
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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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Q7_130 - As the function is odd we know that a k = 0 for k...

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