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The previous question was:

Find the fourier coefficients of $f(x)=x^2+1$

To which I found $a_0=\frac{\pi^2}{3}+1$ and $a_n=\frac{2}{n^2}(-1)^n$ (I am unsure on $b_n$ I get $b_n=(\frac{-2}{n\pi}-\frac{2\pi}{n}+\frac{4}{n^3\pi})(-1)^n$ but this is more method than correcting that)

I want to complete the next part:

By considering the derivative of $f$, write down the fourier series of $g(x)=x $ without explicitly calculating the coefficients

So as $\frac{1}{2}f'(x)=g(x)$ I get (again ignoring $b_n$ for now):

\begin{align} g(x) & =\frac{1}{2}\int_{-\pi}^\pi \left(\frac{\pi^2}{3} + 1 + \sum_{n=1}^\infty \frac{2}{n^2} (-1)^n\cos(nx)\,dx\right) \\[10pt] & =\frac{1}{2} \left[\frac{\pi^2x}{3} + x + \sum_{n=1}^\infty \frac{2}{n^3}(-1)^n\cos(nx)\right]_{-\pi}^\pi =\frac{2\pi^3}{6} + \frac{\pi}{2} + \sum_{n=1}^\infty \frac{1}{n^3} \end{align}

Is this correct?

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  • $\begingroup$ You have $g(x) = \text{an expression with no }x\text{ in it.}$ That can't be right. $\qquad$ $\endgroup$ – Michael Hardy Aug 16 '16 at 11:51
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Firstly: $f(x)$ is an even function, so you expect all $b_n = 0$.

Now for $g(x)$. You are right to note that $g(x) = \frac{1}{2} f'(x)$, but why are you integrating? Futhermore: the last equation: LHS is a function of $x$ and RHS is just a constant.

To solve the problem write $f(x)$ as a sum of cosines (so in in it's Fourier form), take the derivative and then divide by $2$ to get $g(x)$. From there you can get the coefficients without much hustle.

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  • $\begingroup$ Oh yeah obviously I should have differentiated not integrated! So $$g(x)=\frac{1}{2} \frac{dn}{dx} \frac{\pi^2}{3} + 1 + \sum_{n=1}^\infty \frac{2}{n^2} (-1)^n\cos(nx)= \sum_{n=1}^{\infty} -n\frac{2}{n^2} (-1)^n\sin(nx)$$ $\endgroup$ – Lauren Aug 16 '16 at 12:03
  • $\begingroup$ Note how your expression now has $\frac{1}{n}$ in it. This corresponds to power of x in your function. $x^2$ will give you $\frac{1}{n^2}$ etc etc. Next thing: when asking for Fourier series you should always state the interval. Right now you are working on $[-\pi, \pi]$. $\endgroup$ – Piotr Benedysiuk Aug 16 '16 at 12:36
  • $\begingroup$ From $\sum_{n=1}^{\infty}\frac{-2}{n}(-1)^n \sin (nx)$ how do I then put this into the standard $g(x)=a_0+\sum_{n=1}^{\infty}a_n \cos (nx) + b_n \sin (nx)$ form? $\endgroup$ – Lauren Aug 16 '16 at 13:25
  • $\begingroup$ Realize that $g(x) = \sum\limits_{n =1}^{\infty} \frac{-2}{n} (-1)^n \sin(nx)$ is already in standard form. What then must $a_0$, $a_n$ and $b_n$ be? $\endgroup$ – Piotr Benedysiuk Aug 16 '16 at 16:39

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