Fourier series question, either i'm wrong or the answers are wrong

[imsleepy]

Homework Statement

Homework Equations

The Attempt at a Solution

i'm pretty much learning Fourier series from scratch today after only looking at it in lectures (and my exam is tomorrow lol @ me) and I'm sort of stuck.
The solutions have something different to what i have, i think it might be an error in their integration by parts (which i also learned today, despite how simple it seems).

this is my working out: http://i.imgur.com/6OXlr.jpg
and this is the worked solutions: http://i.imgur.com/1RAJL.png

doesnt the integral of (1/n)*sin(nx) equal to (-1/n^2)*cos(nx)? like i have in my working?
because the solutions are saying it integrates into (-1/n)*cos(nx).
or is mine set out incorrectly after I've integrated by parts?

edit: I've only shown both workings from where i had to find a subscript n, before that everything was fine and correct.

 

[lanedance]

little hard to follow but i think maybe you're just missing the limits in you int by parts...
$$u = x$$
$$du = 1$$
$$dv = cos(nx)$$
$$v = \frac{1}{n}.sin(nx)$$

$$\int_{a}^{b} u.dv = (u.v)|_{a}^{b} - \int_{a}^{b}du.v$$
$$\int_0^{\pi} x.cos(nx).dx$$
$$= (x.\frac{1}{n}.sin(nx))|_{0}^{\pi} - \int_{0}^{\pi} \frac{1}{n}.sin(nx).dx$$
$$= 0 - \frac{1}{n} \int_{0}^{\pi} sin(nx).dx$$
$$= \frac{1}{n^2} (cos(nx).dx)|_{0}^{\pi}$$

 

[lanedance]

it is true that
$$\frac{1}{n}\int sin(nx).dx = -\frac{1}{n^2}cos(nx)$$

just differentiate to check

 

[imsleepy]

little hard to follow but i think maybe you're just missing the limits in you int by parts...
$$\int_{a}^{b} u.dv = (u.v)|_{a}^{b} - \int_{a}^{b}du.v$$

$$\int_0^{/pi} x.cos(nx).dx = (x.n.sin(nx))|_{0}^{\pi} - \int_{0}^{\pi} cos(nx).dx = 0 - \int_{0}^{\pi} cos(nx).dx = -(\frac{1}{n}.sin(nx))|_{0}^{\pi}$$
[/tex]

i don't understand what you're doing?

what have u used as u and v?
shouldnt (x.n.sin(nx) be (x/n).sin(nx) after the first = sign?

 

[lanedance]

sorry, did that a too quick, corrected and lines up with yours now

notice the sine terms cancel as $sin(0) = sin(\pi) = 0$

 

[imsleepy]

ok thanks.

also where do the cos's disappear to?
i think something about if that part of the function is odd then you make it -1 or something? and/or if it's even make it 1, i don't know :S

 

[lanedance]

which cos's? in the first step the sin appear instead of cos's due to the integration by parts

now you should be considering the whole integral from -pi to pi, braking it up into two pieces note that
cos(-x) = cos(x)
and
sin(-x) = -sin(x)

you should be able to show for any function
- if the function is odd, it only has sin components (sin is an odd function)
- if the function is even, it only has cos components (cos is an even function)

 

[imsleepy]

yeah i underrstand, thanks a lot!

also regarding the values for a0, an, and bn (http://i.imgur.com/KIjJQ.png) they are all integrals from -L -> L, but in all the examples I've done or seen (about 3), they've taken the integral from 0 -> L and then multiplied the entire thing by 2.

Now i understand they're halving the region of integration and then doubling the overall answer, but can this always be done when a question asks for a Fourier series of a piecemeal function, whether the fn is even or odd? (with a question of the same style as the one in the OP)
And if so, any reason why my formula sheet gives us a0 etc as integrals from -L to L when we can just make it twice the integral from 0-L?

 

[lanedance]

write out the whole integral, break it into 2 pieces at 0, then use the properties of sin or cos as described last time and it should become obvious. this will only work for odd or even functions