Fourier series of complex function

gl0ck
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Homework Statement


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
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

Homework Equations


The Attempt at a Solution

 

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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.
 
gl0ck said:

Homework Statement


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
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

Homework Equations





The Attempt at a Solution

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


http://imageshack.us/a/img59/9217/s0du.jpg

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

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
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
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

Hope I am right and thanks
 
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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.
 
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
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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