Need Help with Fourier Coefficients for 1-t Function?

[TheBigDig]

Homework Statement

[/B]
I am looking for help with part (d) of this question

2. Homework Equations

The Attempt at a Solution

I have attempted going through the integral taking L = 4 and t₀ = -2. I was able to solve for a₀ but I keep having the integrate by parts on this one. I've tried it out twice and messed up. Any help would be appreciated.

 

[BvU]

Hi TBD,

Well, it's nice to tell us you have attempted, but the idea is that you post your work and then we can comment, hint, etc.

 

[LCKurtz]

You haven't told us what $L$ represents. Half period or full period? And your half range formula needs to be consistent with your choice. Also, in half range expansions, you would normally integrate from $0$ to half a period.

 

[BvU]

I keep having the integrate by parts on this one

Can't be right. Show us.

 

[TheBigDig]

Can't be right. Show us.

And you'd be right.

You haven't told us what $L$ represents. Half period or full period? And your half range formula needs to be consistent with your choice. Also, in half range expansions, you would normally integrate from $0$ to half a period.

Apologies for not being more clear. L represents the fundamental period of the function. In this case L = 4 because f(t+4) = f(t).

Sorry for not posting my workings but as it happens they were pretty much useless (I will do this in future though). I ended up splitting the integral up into 4 parts: $$ \frac{1}{2} \int_{-2}^{-1}f(t) cos(\frac{\pi n t}{2}) dt + \frac{1}{2} \int_{-1}^{0}f(t) cos(\frac{\pi n t}{2}) dt + \frac{1}{2} \int_{0}^{1}f(t) cos(\frac{\pi n t}{2}) dt + \frac{1}{2} \int_{1}^{2}f(t) cos(\frac{\pi n t}{2}) dt $$
Since everything else goes to #0# we only need to consider the integral between -1 and 1. The integral from -1 to 1 is equivalent to twice the integral from 0 to 1:
$$ \frac{1}{2} \int_{-1}^{1}f(t) cos(\frac{\pi n t}{2})dt = 2*\frac{1}{2} \int_{0}^{1}f(t) cos(\frac{\pi n t}{2}) dt $$and because the values of x are positive, the function $1-|x| = 1-x$ meaning we only have to integrate that and not a nasty ass modulus. In the end I got an answer of
$a_n = \frac{4}{\pi^2 n^2}$

 

[LCKurtz]

And you'd be right.
Apologies for not being more clear. L represents the fundamental period of the function. In this case L = 4 because f(t+4) = f(t).

Sorry for not posting my workings but as it happens they were pretty much useless (I will do this in future though). I ended up splitting the integral up into 4 parts: $$ \frac{1}{2} \int_{-2}^{-1}f(t) cos(\frac{\pi n t}{2}) dt + \frac{1}{2} \int_{-1}^{0}f(t) cos(\frac{\pi n t}{2}) dt + \frac{1}{2} \int_{0}^{1}f(t) cos(\frac{\pi n t}{2}) dt + \frac{1}{2} \int_{1}^{2}f(t) cos(\frac{\pi n t}{2}) dt $$
Since everything else goes to #0# we only need to consider the integral between -1 and 1. The integral from -1 to 1 is equivalent to twice the integral from 0 to 1:
$$ \frac{1}{2} \int_{-1}^{1}f(t) cos(\frac{\pi n t}{2})dt = 2*\frac{1}{2} \int_{0}^{1}f(t) cos(\frac{\pi n t}{2}) dt $$and because the values of x are positive, the function $1-|x| = 1-x$ meaning we only have to integrate that and not a nasty ass modulus. In the end I got an answer of
$a_n = \frac{4}{\pi^2 n^2}$

Your last integral is correct. But you have made a mistake in your integration because it doesn't give you $a_n = \frac{4}{\pi^2 n^2}$. All I can suggest is for you to recheck your integration since you haven't shown your steps. Also, you will need to calculate $a_0$ separately.

 

[TheBigDig]

Your last integral is correct. But you have made a mistake in your integration because it doesn't give you $a_n = \frac{4}{\pi^2 n^2}$. All I can suggest is for you to recheck your integration since you haven't shown your steps. Also, you will need to calculate $a_0$ separately.

Okay, no problem. I am right in saying $1-|x| = 1-x$ though amn't I? It's similar to an example our professor gave us which is how I made the leap. Probably just made some small error somewhere in the integration. I did calculate $a_0$ separately and got $\frac{1}{2}$.

 

[LCKurtz]

Okay, no problem. I am right in saying $1-|x| = 1-x$ though amn't I?

Yes. As I said earlier, your last integral is correct with $f(t) = 1-t$.

It's similar to an example our professor gave us which is how I made the leap. Probably just made some small error somewhere in the integration. I did calculate $a_0$ separately and got $\frac{1}{2}$.

You should get $a_0 =\frac 1 4$. And your answer for $a_n$ should have $\cos(\frac{n\pi} 2)$ involved in its formula.

 

[TheBigDig]

Yes. As I said earlier, your last integral is correct with $f(t) = 1-t$.
You should get $a_0 =\frac 1 4$. And your answer for $a_n$ should have $\cos(\frac{n\pi} 2)$ involved in its formula.

Okay must have made some massive integration errors. At least I know I have the integral right and that's half the battle :D! Thanks for your comments