Recent content by kottur

No, unfortunately I only know what it is supposed to give me. I know that it approximates a function but how exactly I don't know.

No, I don't think so, I only know what it's supposed to do unfortunately.

I think I know what you mean about the orthogonality. And about the Fourier series. I find the series to try to approximate a certain function. :smile:

But how is my answer in #5 a series? Shouldn't I have a sum there or something...? I'm sorry I'm confused now. And I don't know what the series expansion for sin(3x) is, I would probably try to use the formulas for it.

No I don't have to use them, that's just the only Idea I've got. The problem is exactly like this: Expand cos^{4}(x) into Fourier series. Nothing else given.

Yes but if I have to use these formulas: a_0=\frac{1}{2L}\int^{L}_{-L}f(x)dx a_n=\frac{1}{L}\int^{L}_{-L}f(x)cos(\frac{n\pi x}{L})dx (n\geq1) Then I need to know what L is... And it is half of the period of the function f. So it would be \frac{\pi}{2}? No?

The period is \pi so should I use the interval [0;\pi]?

Ahh shouldn't I use the period of the function as the interval?

Okay HallsofIvy, that would be: \frac{1}{8}(4cos(2x)+cos(4x)+3) But what should I do about the interval that I don't have?

Okay well it's an even function so b_{n}=0. And no I'm not given any interval so I'm not entirely sure how to find a_0 and a_n. I thought I was supposed to find them with: a_0=\frac{1}{2L}\int^{L}_{-L}f(x)dx a_n=\frac{1}{L}\int^{L}_{-L}f(x)cos(\frac{n\pi x}{L})dx (n\geq1) [-L;L]...

Homework Statement Put cos^{4}(x) into a Fourier series. Homework Equations cos^{4}(x)=(\frac{e^{ix}+e^{-ix}}{2})^{4} a_0+\sum^{\infty}_{n=1}(a_{n}cos(nx)+b_{n}sin(nx)) The Attempt at a Solution I don't get what I'm supposed to use as a_{0}, a_{n} and b_{n} so I'm stuck...

LCKurtz thank you for your help. I used a totally different method after all this, finding a, b and c with linear algebra.

Okay I thought about what you said with g(z) but I just used f(x) and L=1/2. I hope that was okay. So this is what I got: a_{0}=0 a_{n}=\frac{-cos(\pi-2\pi n)}{(\pi-2\pi n)}-\frac{cos(\pi+2\pi n}{(\pi+2\pi n)}=\frac{cos(2\pi n)}{(\pi-2\pi n)}+\frac{cos(2\pi n}{(\pi+2\pi n)}...

Hmm okay, well I'm supposed to approximate f(x) with the polynomial with a method called least sum of squares or something like that. It's hard to translate it. Okay I will try to do it with g(z) now. :smile:

The assignment is to Fourier approximate. I've got these formulas in my notes: For a function with period T and T=2L we have: f(x)=a_{0}+\sum^{\infty}_{n=1}(a_{n}cos(\frac{n\pi x}{L})+b_{n}sin(\frac{n\pi x}{L})) a_{0}=\frac{1}{2L}\int^{L}_{-L}f(x)dx...