Solving Fourier Series for Periodic Functions

[NutriGrainKiller]

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.

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

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

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

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

This is of course only the case with periodic functions. Depending on how the graph looks it is possible to derive f(x), 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 f(x), all we do is find a_{0}, a_{n} and B_{n} 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:

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

 

[Office_Shredder]

The very important, first latex image isn't working. Neither is the image labelled (2)

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

and

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

Did I get those right? I'm not a latex expert, I just fixed up some underscores and hoped for the best :D

 

[NutriGrainKiller]

Thanks OS..formatting is working now, still need help though.

 

[Swapnil]

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

You are missing a plus sign in the middle. It should be
f(x) = \frac{a_0}{2} + \sum_{n=1}^{\infty}(a_n}\cos{nx} + b_{n}\sin{nx})

if we are given f(x), all we do is find a_{0}, a_{n} and B_{n} then plug into the first equation. Is this right?

Yes, that is the way you do it.

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:

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

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

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

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

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

where {\omega}_0 = \frac{2\pi}{T} and \int_{t_0}^{t_0 + T} means that the integral is over any particular period. In your example, the period is 2\lambda so {\omega}_0 = \frac{2\pi}{2\lambda} = \frac{\pi}{\lambda} and your integral should look something like this: \int_{0}^{2\lambda}.

 

[Swapnil]

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.