{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

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

This preview shows pages 1–7. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

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 ,......
View Full Document

{[ snackBarMessage ]}

### Page1 / 24

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

This preview shows document pages 1 - 7. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online