Solving Fourier Series for Periodic Functions

In summary: So, the Fourier series is just f(x) = \cos({\frac{\pi x}{\lambda}}).In summary, the Fourier Theorem helps us find the Fourier series of a periodic function by using the formulas for the coefficients a_{0}, a_{n}, and b_{n}. However, these formulas only work for functions with a period of 2\pi, so for more general functions, we need to adjust the formulas accordingly. Additionally, some functions may not have a Fourier series because they cannot be further decomposed.
  • #1
NutriGrainKiller
62
0
I understand what the Fourier Theorem means, as well as how it behaves, I just don't understand how the math actually pans out or in what order to do what.

I'm going to start off with what I know.

[tex]f(x) = \frac{a_0}{2} \sum_{n=1}^{\infty}(a_n}\cos{nx} + b_{n}\sin{nx})[/tex]

while,
[tex] a_{0} = \frac{1}{\pi} \int_{-\pi}^{\pi} f(x) dx[/tex]

[tex]a_n = \frac{1}{\pi} \int_{-\pi}^{\pi} f(x) \cos({nx})\ dx[/tex]

[tex]b_n = \frac{1}{\pi} \int_{-\pi}^{\pi} f(x) \sin({nx})\ dx[/tex]

This is of course only the case with periodic functions. Depending on how the graph looks it is possible to derive [tex]f(x)[/tex], the spatial period, and maybe even its tendency to be even or odd.

(Even/odd meaning whether or not the beginning of the wavelength is at the origin. If it is/does, it's odd and only contains cosine terms, if not and it behaves more like a sine wave (highest amplitude at origin) than it is even, thus not containing any cosine terms.)

if we are given [tex]f(x)[/tex], all we do is find [tex]a_{0}[/tex], [tex]a_{n}[/tex] and [tex] B_{n}[/tex] then plug into the first equation. Is this right? I am getting absurdly long answers doing this, and as far as I can tell I can't find any way of finding out whether I'm headed in the right direction or not.

Here is one of the problems I'm having trouble with:

[tex]f(x) = A\cos({\frac{\pi x}{\lambda}})[/tex], find the Fourier series (it is assumed the function is periodic on the interval [tex][0,2\lambda][/tex])
 
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  • #2
The very important, first latex image isn't working. Neither is the image labelled (2)

Try
[tex]f(x) = \frac{A_0}{2} \sum_{n=1}^{\infty}(A_n}\cos{k_nx} + B_{n}\sin{k_nx})[/tex]

and

[tex] = [\frac{\sin(n k x)}{(n k)^2} - \frac{x\cos(n k x)}{n k}]_{-\lambda/2}^{lambda/2}[/tex]

Did I get those right? I'm not a latex expert, I just fixed up some underscores and hoped for the best :D
 
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  • #3
Thanks OS..formatting is working now, still need help though.
 
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  • #4
NutriGrainKiller said:
[tex]f(x) = \frac{a_0}{2} \sum_{n=1}^{\infty}(a_n}\cos{nx} + b_{n}\sin{nx})[/tex]
You are missing a plus sign in the middle. It should be
[tex]f(x) = \frac{a_0}{2} + \sum_{n=1}^{\infty}(a_n}\cos{nx} + b_{n}\sin{nx})[/tex]

NutriGrainKiller said:
if we are given [tex]f(x)[/tex], all we do is find [tex]a_{0}[/tex], [tex]a_{n}[/tex] and [tex] B_{n}[/tex] then plug into the first equation. Is this right?
Yes, that is the way you do it.

NutriGrainKiller said:
I am getting absurdly long answers doing this, and as far as I can tell I can't find any way of finding out whether I'm headed in the right direction or not.

Here is one of the problems I'm having trouble with:

[tex]f(x) = A\cos({\frac{\pi x}{\lambda}})[/tex], find the Fourier series (it is assumed the function is periodic on the interval [tex][0,2\lambda][/tex])

This is because the formulas for the coefficients you have stated only hold true for functions whose period is [tex]2\pi[/tex]. The more general formulas are:

[tex] a_{0} = \frac{2}{T} \int_{t_0}^{t_0 + T} f(x) dx[/tex]

[tex]a_n = \frac{2}{T} \int_{t_0}^{t_0 + T} f(x) \cos({n{\omega}_0 x})\ dx[/tex]

[tex]b_n = \frac{2}{T} \int_{t_0}^{t_0 + T} f(x) \sin({n{\omega}_0 x})\ dx[/tex]

where [tex]{\omega}_0 = \frac{2\pi}{T}[/tex] and [tex]\int_{t_0}^{t_0 + T} [/tex] means that the integral is over any particular period. In your example, the period is [tex]2\lambda[/tex] so [tex]{\omega}_0 = \frac{2\pi}{2\lambda} = \frac{\pi}{\lambda}[/tex] and your integral should look something like this: [tex]\int_{0}^{2\lambda}[/tex].
 
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  • #5
Also, the function you have given in your example is very weird because the function and its Fourier series is the same. You can't decompose your function any further.
 

Related to Solving Fourier Series for Periodic Functions

1. What is a Fourier series?

A Fourier series is a mathematical representation of a periodic function as a sum of sine and cosine waves. It is used to decompose a complex function into simpler components, making it easier to analyze and manipulate.

2. How are Fourier series used in science?

Fourier series are used in many fields of science, including physics, engineering, and signal processing. They are particularly useful in analyzing and understanding the behavior of periodic phenomena such as sound waves, electromagnetic waves, and oscillations in physical systems.

3. What is the process for solving Fourier series for a periodic function?

The process for solving Fourier series involves finding the coefficients of the sine and cosine terms in the series using integration and the orthogonality properties of trigonometric functions. These coefficients can then be used to reconstruct the original periodic function.

4. Can Fourier series be used for non-periodic functions?

No, Fourier series can only be used for periodic functions. However, the concept of Fourier transform can be used to analyze and decompose non-periodic functions.

5. Are there any limitations to using Fourier series?

One limitation of using Fourier series is that it only works for functions that are periodic and have a well-defined period. It also assumes that the function is continuous and can be represented as an infinite sum of sine and cosine waves, which may not always be the case.

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