# Fourier series of complex function

1. Nov 12, 2013

### gl0ck

1. The problem statement, all variables and given/known data
Hello guys,

I have problem with the fourier series, since we had only one lecture about it and I cannot find anything similar to my problem in internet.
should we consider for the first f(x+1) integrated from -1 to 0 ?
http://img819.imageshack.us/img819/3508/wbve.jpg [Broken]
when I use that i can find Ao= 3/2?
and about the extra info the lecturer told us if it is even it converges to 1/2 of the value, so may I consider if the series is even it converges to 1.5?
I think 2nd one is similar to the 1st one

2. Relevant equations

3. The attempt at a solution

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• ###### fourier.jpg
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Last edited by a moderator: May 6, 2017
2. Nov 12, 2013

### HallsofIvy

Staff Emeritus
Where did that "f(x+1)" come from? You are given that f(x)= x+ 1 for $-1\le x\le 0$ and f(x)= x for $0< x\le 1$.

3. Nov 12, 2013

### vela

Staff Emeritus
Say you want to calculate
$$a_n = \frac{2}{L}\int_{-L/2}^{L/2} f(x)\cos\left(\frac{2\pi n}{L} x\right)\,dx.$$ For the function you're given, L=2. Because it's piecewise continuous, you need to break up the interval of integration to correspond to each piece.
$$a_n = \int_{-1}^{1} f(x)\cos(2\pi n x)\,dx = \int_{-1}^0 f(x)\cos(2\pi n x)\,dx + \int_0^1 f(x)\cos(2\pi n x)\,dx.$$ In the first integral on the righthand side, $x$ is between -1 and 0, so in that integral, you replace f(x) by $x+1$. In the second integral, $x$ is between 0 and 1, so you replace f(x) by 1. You need to understand this part because what you wrote down was gibberish.

You now have
$$a_n = \int_{-1}^0 (x+1)\cos(2\pi n x)\,dx + \int_0^1 \cos(2\pi n x)\,dx,$$ which is left for you to evaluate. You calculate $b_n$ similarly.

Last edited by a moderator: May 6, 2017
4. Nov 13, 2013

### gl0ck

Hey guys, I've made some progress, I think I've solved the problems

http://imageshack.us/a/img59/9217/s0du.jpg [Broken]

For the 1st question I got for An and Bn these things:
http://imageshack.us/a/img543/6600/6kl1.jpg [Broken]

and the final answer should be multiplied by 1/2 and 3/2 ( Ao ) which forgot to include.
http://imageshack.us/a/img689/6442/f6yz.png [Broken]
2nd question I got only cosine terms since the function is even with period 2pi
so for An I got this : http://imageshack.us/a/img35/7937/06ha.jpg [Broken]
and for the final answer when I subsitute the d^2y/dx^2 with $\sum-Cnn^2cos(nx)$
and y with $\sum-Cncos(nx)$ to be equal to An

and got this:
http://imageshack.us/a/img5/4756/89rb.jpg [Broken]

Hope I am right and thanks

Last edited by a moderator: May 6, 2017
5. Nov 13, 2013

### vela

Staff Emeritus
I know that using images may be convenient for you, but it's annoying for the helpers. Perhaps someone else will be kind enough to go to the trouble of opening up each image to check your work.

6. Nov 13, 2013

### gl0ck

I thought it pops up in the whole threat just like the formulas used in the forum , or it just happens to me? Don't know what you mean