# 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