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.