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# Fourier Series watch

1. Hi, I have a problem with a Fourier series question and would like some assistance please:

The question is: find the Fourier series of x^2 in the range of -1<=x<=1.

Where f(x) = 0.5A0 + SUM OF(An cos(pi*n*x) + Bn sin(pi*n*x))

I think I have found the Fourier series correctly, with An = 4Cos(pi*n)/(pi^2*n^2) = 4(-1)^n/(pi^2*n^2), A0 = 2/3 and Bn = 0.

So the fourier series equation becomes: f(x) = 1/3 + 4/pi^2 * SUM of((-1)^n*cos(pi*n*x)/(n^2))

The next question is to hence show that the answer to the SUM of ( 1 - 1/2^2 + 1/3^2 - 1/4^2 +.....) = Pi^2/12

If my fourier series equation is correct, then that sum can only be obtained when x=1/n
(not sure if you can even do that). But then I get an answer of -(pi)^2/12 as n -> infinity.

Can anyone tell me if I have gone totally wrong or something I'm missing?

Thank you.
Hi, I have a problem with a Fourier series question and would like some assistance please:

The question is: find the Fourier series of x^2 in the range of -1<=x<=1.

Where f(x) = 0.5A0 + SUM OF(An cos(pi*n*x) + Bn sin(pi*n*x))

I think I have found the Fourier series correctly, with An = 4Cos(pi*n)/(pi^2*n^2) = 4(-1)^n/(pi^2*n^2), A0 = 2/3 and Bn = 0.

So the fourier series equation becomes: f(x) = 1/3 + 4/pi^2 * SUM of((-1)^n*cos(pi*n*x)/(n^2))

The next question is to hence show that the answer to the SUM of ( 1 - 1/2^2 + 1/3^2 - 1/4^2 +.....) = Pi^2/12

If my fourier series equation is correct, then that sum can only be obtained when x=1/n
(not sure if you can even do that). But then I get an answer of -(pi)^2/12 as n -> infinity.

Can anyone tell me if I have gone totally wrong or something I'm missing?

Thank you.
It has to be a silly sign error somewhere ...
I cannot possibly see it unless I do the question or post some workings
Hi, I have a problem with a Fourier series question and would like some assistance please:

The question is: find the Fourier series of x^2 in the range of -1<=x<=1.

Where f(x) = 0.5A0 + SUM OF(An cos(pi*n*x) + Bn sin(pi*n*x))

I think I have found the Fourier series correctly, with An = 4Cos(pi*n)/(pi^2*n^2) = 4(-1)^n/(pi^2*n^2), A0 = 2/3 and Bn = 0.

So the fourier series equation becomes: f(x) = 1/3 + 4/pi^2 * SUM of((-1)^n*cos(pi*n*x)/(n^2))

The next question is to hence show that the answer to the SUM of ( 1 - 1/2^2 + 1/3^2 - 1/4^2 +.....) = Pi^2/12

If my fourier series equation is correct, then that sum can only be obtained when x=1/n
(not sure if you can even do that). But then I get an answer of -(pi)^2/12 as n -> infinity.

Can anyone tell me if I have gone totally wrong or something I'm missing?

Thank you.
I'm a bit rusty on this so I hope I don't get this wrong.

You have

Rearranging you get

Now ask yourself what value of x makes this work?
4. (Original post by TeeEm)
It has to be a silly sign error somewhere ...
I cannot possibly see it unless I do the question or post some workings

I'm not sure, I just re did it and still get to the same point. Unless I am making the same mistake. I have attached the question and my working out.

(Original post by BuryMathsTutor)
I'm a bit rusty on this so I hope I don't get this wrong.

You have

Rearranging you get

Now ask yourself what value of x makes this work?
But if my equation up to that point is correct, then surely the only value of x that should work is if x=1/n so that the n=1 value is positive and then they continue to alternate signs. I did try something simmilar to what you did and got an answer of x = +- ROOT(6) over 3.

edit: for some reason my working out doesn't want to attach at the right rotation: http://tinypic.com/view.php?pic=205s...8#.VK_6a3uoP5k

Attached Images

I'm not sure, I just re did it and still get to the same point. Unless I am making the same mistake. I have attached the question and my working out.

But if my equation up to that point is correct, then surely the only value of x that should work is if x=1/n so that the n=1 value is positive and then they continue to alternate signs.

I did try something simmilar to what you did and got an answer of x = +- ROOT(6) over 3.

all correct so far

I can tell what you cannot see!!!

after you let x = 0 in both sides and you rearrange, the resulting sum is not the one you are after but the negative of that

write out a few terms to see it

your sum is pi squared over 12
6. (Original post by TeeEm)
all correct so far

I can tell what you cannot see!!!

after you let x = 0 in both sides and you rearrange, the resulting sum is not the one you are after but the negative of that

write out a few terms to see it

your sum is pi squared over 12
Ah yeah, I see that now. Thank you.

I guess that means that you cannot do the substitution x=1/n. I did think that you couldn't, but I was stuck in the idea that I had to get the signs of the SUM correct. In reality, -1* the answer for the same sum but with signs swapped is correct.
Ah yeah, I see that now. Thank you.

I guess that means that you cannot do the substitution x=1/n. I did think that you couldn't, but I was stuck in the idea that I had to get the signs of the SUM correct. In reality, -1* the answer for the same sum but with signs swapped is correct.
the substitutions in these types of questions will usually be 0, pi/2, pi and so on
it is just that the sum is -1+1/4-1/9+1/16 ... and so on
Ah yeah, I see that now. Thank you.

I guess that means that you cannot do the substitution x=1/n. I did think that you couldn't, but I was stuck in the idea that I had to get the signs of the SUM correct. In reality, -1* the answer for the same sum but with signs swapped is correct.
9. (Original post by BuryMathsTutor)
what program are you using?
10. (Original post by TeeEm)
what program are you using?
http://www.graphmatica.com/
11. thanks

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