The discussion focuses on calculating the Fourier series for the function f(x) = sin(x) over the interval 0 < x < π. The Fourier series is expressed as f(x) = ∑(A_n cos(nx) + B_n sin(nx)), where A_n and B_n are coefficients derived from integrals. For f(x) = sin(x), it is established that A_n = 0 for all n, B_1 = 1, and B_n = 0 for n > 1. This indicates that the Fourier series representation accurately reflects the periodic nature of the sine function.
PREREQUISITESStudents and educators in mathematics, particularly those studying Fourier analysis, as well as professionals in engineering and physics who apply Fourier series in practical scenarios.
KAS90 said:Homework Statement
find Fourier series for:
sinnπ=0
(n=0,+1,-1,+2,-2...)
I really can't understand how Fourier series works.. Like I tried solving problems..but till now, it didn't sink in my brain..
I just want 2 know the basic way of solving a problem regarding Fourier series..then it will be much easier to understand I guess..
Thanx in advance..
hi markMark44 said:What's the function that you want to find the Fourier series for? sin (n*pi) = 0 for all integers, as you point out, but it's otherwise not very interesting. Is the function f(x) = 0?
If so, all of the coefficients in the Fourier series would be 0.
Is there more to this problem that you haven't shown us?
HallsofIvy said:A Fourier series for a function f(x) is an infinite sum such that
f(x)= \sum_{m=0}^\infty A_n cos(nx)+ B_n sin(nx) (or other (x/L) or whatever inside the trig functions).
Of course, the right side of that is periodic with period 2\pi so if f is not itself periodic, that can only be true on some interval (which is one reason why you might need that "/L" to alter the interval).
But if f(x)= sin(x) is not only periodic with period 2\pi, it is already of that form and it is obvious that A_n= 0 for all n, while B_1= 1 and B_n= 0 for all n greater than 1.
More generally, if f(x) is integrable on the interval [0, L], then on that interval f(x)= A_n cos(2n x\pi/L)+ B_n sin(2n x\pi/L) where
A_0= \frac{1}{L}\int_0^L f(x) dx
A_n= \frac{1}{2L}\int_0^L f(x)cos(2n x/L)
for n> 0 and
B_n= \frac{1}{2L}\int_0^L f(x)sin(2n x/L).