Howto understand this periodic fourier series

[Susanne217]

Homework Statement

I am given this function

f(x) = \left\{\begin{array}{cccc} x^2 \ \mathrm{where} \ \frac{-\pi}{2} < x < \frac{\pi}{2} \\ \ \frac{1}{4}\pi^2 \ \mathrm{where} \ \frac{\pi}{2} < x < \frac{3\pi}{2} \end{array}


Doesn't this mean that the function is periodic Fourier which is defined on

[-\frac{L}{2}, \frac{3L}{2}]?

Anyway I have formula to find the corresponding Fourier series had been defined on [-L,L] but do I still use this formula eventhough the interval is different?

is then true that if p = pi/2

then

a_0: = \frac{1}{2\pi} \int_{\frac{-\pi}{2}}^{\frac{\pi}{2}} x^2 dx + \frac{1}{2\pi} \int_{\frac{\pi}{2}}^{\frac{3\pi}{2}} \frac{1}{4} \cdot \pi^2 dx = \frac{\pi^2}{6}

/Susanne

 

[LCKurtz]

The short answer to your question is yes. Your function is defined on [-\pi/2,3\pi/2] which is an interval of length 2\pi. If you extend it periodically it will have period 2\pi. For a periodic function of period P, it is true that integrating it over any period will give the same result as integrating over any other period:

\int_a^{a+P}f(x)\,dx = \int_b^{b+P}f(x)\,dx

So you don't have to integrate over (-\pi,\pi) and in fact you really want to integrate over [-\pi/2,3\pi/2], because that is where you have the formulas given. If, for some reason, you decided to integrate from (2\pi,4\pi) you couldn't use x^2 because it would have to be translated.

 

[Susanne217]

Hi again,

Is my a_n then

a_n: = \frac{1}{2\pi} \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} x^2 \cdot cos(\frac{nx\pi}{2\pi}) dx + \frac{1}{2\pi} \int_\frac{\pi}{2}^{\frac{3\pi}{2}} \frac{1}{4}\pi^2 cos(\frac{nx\pi}{2\pi}) dx?

and my b_n is

b_n: = \frac{1}{2\pi} \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} x^2 \cdot sin(\frac{nx\pi}{2\pi}) dx + \frac{1}{2\pi} \int_\frac{\pi}{2}^{\frac{3\pi}{2}} \frac{1}{4}\pi^2 sin(\frac{nx\pi}{2\pi}) dx?

what makes this problem complicated is that most of the formulas I has deals with nice even interval :(

 

[LCKurtz]

Hi again,

Is my a_n then

a_n: = \frac{1}{2\pi} \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} x^2 \cdot cos(\frac{nx\pi}{2\pi}) dx + \frac{1}{2\pi} \int_\frac{\pi}{2}^{\frac{3\pi}{2}} \frac{1}{4}\pi^2 cos(\frac{nx\pi}{2\pi}) dx?

and my b_n is

b_n: = \frac{1}{2\pi} \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} x^2 \cdot sin(\frac{nx\pi}{2\pi}) dx + \frac{1}{2\pi} \int_\frac{\pi}{2}^{\frac{3\pi}{2}} \frac{1}{4}\pi^2 sin(\frac{nx\pi}{2\pi}) dx?

what makes this problem complicated is that most of the formulas I has deals with nice even interval :(

Why don't you cancel the \pi's?

The period is 2\pi so 2p=2\pi, so p=\pi so you should just have nx inside the integrals.

 

[Susanne217]

Why don't you cancel the \pi's?

The period is 2\pi so 2p=2\pi, so p=\pi so you should just have nx inside the integrals.

Okay, but besides from that I hopefully used the formel definition for a_n and b_n correctly ?

a_n: = \frac{1}{\pi} \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} x^2 \cdot cos(\frac{nx\pi}{\pi}) dx + \frac{1}{\pi} \int_\frac{\pi}{2}^{\frac{3\pi}{2}} \frac{1}{4}\pi^2 cos(\frac{nx\pi}{\pi}) dx?

and my b_n is

b_n: = \frac{1}{\pi} \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} x^2 \cdot sin(\frac{nx\pi}{\pi}) dx + \frac{1}{\pi} \int_\frac{\pi}{2}^{\frac{3\pi}{2}} \frac{1}{4}\pi^2 sin(\frac{nx\pi}{\pi}) dx?

 

[LCKurtz]

Yes. And of course a₀ is different.

 

[Susanne217]

Yes. And of course a₀ is different.

thanks