$b_n$ Fourier series $f(x)=x^3$

Question

I'm having trouble to solve the $b_n$ coefficient of Fourier series of $x^3$ function on the $-\pi<x<\pi$. The result is almost right, but the powers of $n$ are one less degree I mean

The result is with $(-1)^n[12/n^2]$ (should be $/n^3$) less $2\pi^2$ (should be $/n$). There must be something I'm doing wrong but I simply can't find my error... Can you please help me out on this?

Thanks Ams

There are directions in the FAQ section to write properly mathematics in this wite with LaTeX. Use that, please. — DonAntonio, Apr 24 '13 at 16:09
And after you learn how to write mathematics here (it takes about 5-6 minutes TOPS), describe to us what is it you did that "must be wrong". — DonAntonio, Apr 24 '13 at 16:10
See this. http://math.stackexchange.com/questions/360705/calculating-the-fourier-series-of-x3/362374#362374 — Ron Gordon, Apr 24 '13 at 16:58
Thanks but I need to show the results like I did in the other exercises. But there is something wrong on my calculus ... — Deiota, Apr 24 '13 at 17:30
This is the result I'm searching for:(-1)^{n}(12/n^3-2pi^2/n) I must be doing some calculus wrong but I don't know what!!!! — Deiota, Apr 24 '13 at 17:56
You mean $b_n={\frac {2( \left( -1 \right) ^{n+1}{n}^{2}{\pi }^{2}+6, \left( -1 \right) ^{n})}{{n}^{3}}}.$ — Mhenni Benghorbal, Apr 25 '13 at 06:46
You need to post your work in order for people to spot your mistake. — Mhenni Benghorbal, Apr 25 '13 at 07:20
No. What I mean the result I need is (((-'1)^n)((12/(n^3))-((2pi^2)/n))) and what I have is ((-'1)^n)*((12/(n^2))-(2pi^2)) — Deiota, Apr 25 '13 at 08:22
I can't post my calculus yeti because e can't post pictures yet. As soon as I can i post my calculus :) — Deiota, Apr 25 '13 at 08:23
This is my calculation for bn... Can you please see what is wrong???
http://www.image-share.com/ijpg-2169-123.html — Deiota, Apr 25 '13 at 11:34
The answer for $b_n$ in my above comment is correct. I am sure of it. — Mhenni Benghorbal, Apr 25 '13 at 11:39

score 2 · Accepted Answer · answered Apr 25 '13 at 11:47

2

Just take your time to do the calculations. Here is the Fourier series

$$ x^3 = 2\sum_{n=1}^{\infty} {\frac{ \left( -1 \right) ^{n+1}{n}^{2}{\pi }^{2}+6\, \left( -1 \right)^{n}}{{n}^{3}}} \sin{nx}. $$

answered Apr 25 '13 at 11:47

Mhenni Benghorbal

47,431

This is the right result for the x3 function on the Fourier series ? – Deiota Apr 25 '13 at 12:35
@Amccds: Yes, it is. – Mhenni Benghorbal Apr 25 '13 at 12:43
What am I do doing wrong??? – Deiota Apr 25 '13 at 13:08
Can someone please check my image I posted to see my resolution of the integer to help me see what I did wrong??? Please???? Thanks – Deiota Apr 25 '13 at 20:07
Just got the right answer!!!! Can close this topic!!! – Deiota Apr 25 '13 at 21:06
@Amccds: good. It is a matter of calculations. I recommand you not to rush when you do any calculations. Just go step by step. – Mhenni Benghorbal Apr 25 '13 at 21:24

$b_n$ Fourier series $f(x)=x^3$

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