Fourier series

1. Jul 1, 2007

prasanaharani

expand the function y=sinx in a series of cosines in the interval (0 to 180)
i want to know only the value of f(x) for solving this.what is the value of
f(x).

2. Jul 1, 2007

prasanaharani

expand the function y=sinx in a series of cosines in the interval (0 to 180)
i want to know only the value of f(x) for solving this.what is the value of
f(x).

3. Jul 1, 2007

chroot

Staff Emeritus
I suggest you post the entire problem, exactly as it was given to you. Next, show us your attempts at a solution, and where you've gotten stuck.

- Warren

4. Jul 2, 2007

prasanaharani

to solve this problem first we need to know the value of An and A0
ie the series is A0/2+SUMATION n=1 to infinity (An cosnx)where An is given by An=2/pie integral of -pie to +pie f(x)cosnx dx
now what is the value of f(x) to substitute in that place to solve it.pls tell me.

5. Jul 2, 2007

prasanaharani

To solve this problem first we need to know the value of An and A0
ie the series is A0/2+SUMATION n=1 to infinity (An cosnx)where An is given by An=2/pie integral of -pie to +pie f(x)cosnx dx
now what is the value of f(x) to substitute in that place to solve it.pls tell me.

6. Jul 2, 2007

CompuChip

$$f(x) = \sin(x)$$?
Why do you have two names (y and f(x) for the same thing?)
At least, that's what you said in your first (two) post(s).

Last edited: Jul 2, 2007
7. Aug 13, 2007

matematikawan

cos(nx) is an even function. So you have to even extend the function$$f(x) = \sin(x)$$ first.
It will be a periodic function of period $$2L = \pi$$.
Then the coefficient
$$a_n = \frac{2}{\pi/2} \int{\sin(x)\cos(2nx) dx}$$
integrate from 0 to $$\frac{\pi}{2}$$
which simplify to
$$a_n = -\frac{2}{\pi (4n^2-1)}$$

The Fourier series is then
$$\sin(x) = \frac{2}{\pi} -\frac{4}{\pi}(\frac{\cos(2x)}{1.3} + \frac{\cos(4x)}{3.5} + . . . )$$

Last edited: Aug 13, 2007