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L23 - Fourier Series For the motivation of the problem see...

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Unformatted text preview: Fourier Series For the motivation of the problem see the previous lecture. We’ll actually extend the scope some more and address the fol- lowing Problem: Given a function f : [- L,L ] → R , can we find coefficients a ,a 1 ,a 2 ,... and b 1 ,b 2 ,.. such that f ( x ) = a 2 + ∞ X n =1 a n cos nπx L + ∞ X n =1 b n sin nπx L ? If such coefficients exist, the right hand side above is termed the Fourier Expan- sion of f . On the way to the (very) positive answer, we need some information: Z L- L cos nπx L cos mπx L dx = if m 6 = n L if m = n Z L- L sin nπx L sin mπx L dx = if m 6 = n L if m = n Z L- L cos nπx L sin mπx L dx = 0 . If f : [- L,L ] → R has a representation as a superposition of sines and cosines as in ? , then 2 f ( x ) = a 2 + ∞ X n =1 a n cos nπx L + ∞ X n =1 b n sin nπx L 3 f ( x ) cos mπx L = a 2 cos mπx L + ∞ X n =1 a n cos nπx L cos mπx L + ∞ X n =1 b n sin nπx L cos mπx L 4 Z L- L f ( x ) cos mπx L dx = a 2 Z L- L cos mπx L dx + ∞ X n =1 a n Z L- L cos nπx L cos mπx L dx + ∞ X n =1 b n Z L- L sin nπx L cos mπx L dx 5 Z L- L f ( x ) cos mπx L dx = a 2 Z L- L cos mπx L dx | {z } =0 if m 6 =0 , = La if m =0 + ∞ X n =1 a n Z L- L cos nπx L cos mπx L dx | {z } =0 if n 6 = m , = L if n = m | {z } = La m + ∞ X n =1 b n Z L- L sin nπx L cos mπx L dx | {z } =0 Hence a m = 1 L Z L- L f ( x ) cos mπx L dx . m = 0 , 1 ,......
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L23 - Fourier Series For the motivation of the problem see...

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