Derive the coefficients bn
bn=(1/pi)intg(0-2pi)[f(x)*sin(nx)dx] n =1,2,...

Question

Derive the coefficients bn 
bn=(1/pi)intg(0->2pi)[f(x)*sin(nx)dx] n =1,2,...

jamesj · Answer

Fourier coefficients, of what function f(x)?

anonymous · Answer

$$b_{n}=(1/\pi)\int\limits\limits_{0}^{2\pi}f(x)*\sin(nx)dx$$

anonymous · Answer

sorry, yes of f(x). I'm assuming so. The question simply states find b_(n) for this equation.

anonymous · Answer

Actually, it says derive. So I am wondering if he is wanting me to derive this equation from a generalized equation.

jamesj · Answer

As it stands this is a definition of the numbers $ b_ n $. So in this form, this question is incomplete. But these are Fourier coefficients. You might find this helpful: http://ocw.mit.edu/courses/mathematics/18-03-differential-equations-spring-2010/video-lectures/lecture-15-introduction-to-fourier-series/

anonymous · Answer

$$f (x) ~ a_{0}/2 + \sum_{n=1}^{\infty}[a_{n}\cos nx + b_{n} \sin nx]$$

anonymous · Answer

Sorry, the only place it states anything about what f(x) would be is the line above,
f(x) ~ to the equation above

jamesj · Answer

Right.  The video lecture I just linked to will give you the derivation of the formula for b_n.

anonymous · Answer

Alright, thanks.