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Taken from Apostol Analysis, it says, find a continuous function that generates the fourier series:

$$ \sum_{n} \frac{\left(-1\right)^n}{n^3} \sin(nx) $$

I really have no idea how to solve this, instinctively I tried solving $\langle f,\sin(nx)\rangle = \frac{(-1)^n}{n^3}$ and $\langle f,\cos(nx)\rangle = 0$ and got nowhere, any help would be much appreciated.

garondal
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sr chunchurria
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    I would start by calculating the Fourier series of $ax^3 + bx$ on $(-\pi,\pi)$ with $2\pi$ periodic boundary conditions. But that is a guess based upon prior experience. – Stephen Montgomery-Smith Dec 16 '13 at 04:32
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    Differentiating the series formally (that is, without proper justification), we get $$f''(x) =- \sum_{n=1}^\infty \frac{(-1)^n}{n}\sin (nx)$$ which is a series you might recognize (it's a piecewise linear function, and you may have a similar example in the book). If you can identify this function, then $f$ can be found by integration. The lack of justification is not an issue then, because you can check directly that $f$ has the right Fourier coefficients. – Post No Bulls Dec 16 '13 at 05:33
  • thank you both for your answers, they really helped. If anyone is interested the solution is indeed a cubic polynomial of the form $\frac{ x^3 - x{\pi}^2}{12}$. – sr chunchurria Dec 19 '13 at 11:34

1 Answers1

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Observations:

  1. $f$ is odd
  2. $f$ is continuous on $\mathbb R$, because the series converges uniformly. Together with 1, this implies $f(\pi)=0$.
  3. $f$ is a cubic polynomial on $(-\pi,\pi)$, because the Fourier coefficients of $x^k$ involve $1/n^k$ (integration by parts happens $k$ times).

Odd cubic polynomials vanishing at $\pi$ are of the form $A(x^3-\pi^2 x)$. There are various ways to find $A$, including boring integration. A less boring way is to observe that
$$ f'(x)= \sum_{n=1}^\infty \frac{(-1)^{n}}{n^2}\cos nx$$ is continuous on $\mathbb R$ and $$f'(\pi)= \sum_{n=1}^\infty \frac{1}{n^2} = \frac{\pi ^2}{6}$$ Since $A(x^3-\pi^2 x)'=A(3x^2-\pi^2)$ evaluates to $2\pi^2A$ at $x=\pi$, we have $A=1/12$.

Post No Bulls
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  • Great answer! As a newbie I have two further questions. Can you suggest a source for item 3 that I can read and understand? Also, you state "odd cubic polynomials vanishing at $\pi$ are of the form...". Is there such a list to refer to? – ck1987pd Jun 16 '20 at 14:43
  • To find the Fourier coefficients of $x^kf(x)$ we can do $k$ integrations by parts ($u=f$, $v’=\cos/\sin$) to reduce it to the Fourier coefficients of $f$, but meanwhile, we gained a factor of $1/n^k$ (can be proved by induction on $k$). Take a general expression of a 3rd degree polynomial, check that the parity requirement throws out the even coefficients, then that the root at pi gives the expression above. – Bcpicao Jan 12 '23 at 19:21