Fourier series of complex function

In summary, the first function you are given is a function that is continuous at the boundaries of the interval of integration, but it is not piecewise continuous. For the first question, you get the answers for An and Bn using the formulas for simple harmonic functions. For the second question, you only get the cosine terms, so you use the same formula as for the first question, but you substitute d^2y/dx^2 for y. For the final answer, you use the same formula as for the first question, but you substitute An for y.
  • #1
gl0ck
85
0

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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  • #2
Where did that "f(x+1)" come from? You are given that f(x)= x+ 1 for [itex]-1\le x\le 0[/itex] and f(x)= x for [itex]0< x\le 1[/itex].
 
  • #3
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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  • #4
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 [itex]\sum-Cnn^2cos(nx)[/itex]
and y with [itex]\sum-Cncos(nx)[/itex] 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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  • #5
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
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
 

1. What is a Fourier series of a complex function?

A Fourier series of a complex function is a way of representing a periodic complex-valued function as a sum of simpler trigonometric functions. It is similar to a Fourier series of a real-valued function, but includes complex coefficients and complex trigonometric functions.

2. How is a Fourier series of a complex function calculated?

A Fourier series of a complex function is calculated by finding the complex coefficients using the complex form of the Fourier series formula. This involves integrating the function over one period and using complex integration techniques to find the coefficients.

3. What is a complex form of the Fourier series formula?

The complex form of the Fourier series formula is given by:
f(x) = a0 + ∑(n=1 to ∞) (ancos(nx) + bnsin(nx))
where an and bn are the complex coefficients.

4. What are the applications of Fourier series of complex functions?

Fourier series of complex functions have various applications in mathematics, physics, and engineering. They are used to solve differential equations, analyze periodic signals, and study the behavior of complex systems. They are also used in image and signal processing, as well as in the field of quantum mechanics.

5. Can a non-periodic complex function have a Fourier series?

No, a non-periodic complex function cannot have a Fourier series. The Fourier series representation is only valid for periodic functions, and non-periodic functions do not have a well-defined period. However, a non-periodic complex function can be approximated by a Fourier series over a finite interval, known as a truncated Fourier series.

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