Q7A_130 - f ( x ) = x if-1 < x < , if...

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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
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Unformatted text preview: f ( x ) = x if-1 &lt; x &lt; , if 0 &lt; x &lt; 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 (-x ) = f ( x ). In other words, f (-x ) =-f ( x ), and f is thus odd. 1...
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