How do we find the Fourier coefficient for a cosine term in a Fourier series?

[member 428835]

hey pf!

okay, so if you've studied PDEs you know the value of a Fourier series, and the difficulty of determining a Fourier coefficient. my question relates to finding this coefficient. briefly, i'll define a Fourier series as f(x)=\sum_{n=0}^{\infty} A_n\cos\frac{n\pi x}{L}+B_n\sin\frac{n\pi x}{L} when solving for A_n one way (the only way i know) is to multiply both sides by \cos\frac{m\pi x}{L}. now, when we integrate over [-L,L] the sine term vanishes by its orthogonality with cosine and we are left with \int_{-L}^{L}f(x)\cos\frac{m\pi x}{L}dx=\sum_{n=0}^{\infty} A_n \int_{-L}^{L} \cos\frac{m\pi x}{L}\cos\frac{n\pi x}{L}dx, where the r.h.s=0 if m \neq n thus we know m = n (for now let's assume m=n \neq 0). m = n \implies \int_{-L}^{L} \cos\frac{m\pi x}{L}\cos\frac{n\pi x}{L}dx=L thus \sum_{n=1}^{\infty}A_n=\frac{1}{L}\int_{-L}^{L}f(x)\cos\frac{n\pi x}{L}dx but every book i see states simply A_n=\frac{1}{L}\int_{-L}^{L}f(x)\cos\frac{n\pi x}{L}dx where did the sum go?

 

[jasonRF]

...
now, when we integrate over [-L,L] the sine term vanishes by its orthogonality with cosine and we are left with \int_{-L}^{L}f(x)\cos\frac{m\pi x}{L}dx=\sum_{n=0}^{\infty} A_n \int_{-L}^{L} \cos\frac{m\pi x}{L}\cos\frac{n\pi x}{L}dx, where the r.h.s=0 if m \neq n

Exactly. Which means only one term of the sum is left, the n=m term, so this equation is identical to

\int_{-L}^{L}f(x)\cos\frac{m\pi x}{L}dx= A_m \int_{-L}^{L} \cos\frac{m\pi x}{L}\cos\frac{m\pi x}{L}dx

Does that make sense?

jason

 

[member 428835]

Exactly. Which means only one term of the sum is left, the n=m term, so this equation is identical to

\int_{-L}^{L}f(x)\cos\frac{m\pi x}{L}dx= A_m \int_{-L}^{L} \cos\frac{m\pi x}{L}\cos\frac{m\pi x}{L}dx

wait, does this mean \sum\int_{-L}^{L}f(x)\cos\frac{m\pi x}{L}dx= A_m \int_{-L}^{L} \cos\frac{m\pi x}{L}\cos\frac{m\pi x}{L}dx is equivalent to \int_{-L}^{L}f(x)\cos\frac{m\pi x}{L}dx= \sum A_m \int_{-L}^{L} \cos\frac{m\pi x}{L}\cos\frac{m\pi x}{L}dx
could you explain?

 

[member 428835]

never mind, i see what you mean now! thanks!

 

[AlephZero]

"Where did the sum go?" is the wrong question.

The right question is "where do the parentheses go?"

Answer:
$$\int_{-L}^{L}f (x) \cos\frac{m\pi x}{L} dx = \sum_{n=0}^{\infty} \left( A_n \int_{-L}^{L} \cos \frac{m\pi x}{L} \cos \frac{n\pi x}{L} dx \right)$$

and all the integrals on the right hand side are zero, except when $m = n$.