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I am asked to devolop the function $f(x)=x^2$ in a series of sine and cosine (Fourier series) in the interval $[\frac{-1}{2},\frac{1}{2}]$. And use one of these series to calculate $$\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n^2}\hspace{0.5cm} \text{ and }\hspace{0.5cm} \sum_{n=1}^{\infty} \frac{1}{n^2}$$

Now, the Fourier series is

$S(f)(x)=a_0+2\sum_{n=1}^{\infty}a_n \cos(2\pi nx)+2\sum_{n=1}^{\infty}b_n \sin(2\pi nx)$

Since $f(x)=x^2$ is even, $b_n=0$. I understand this means that I have to use the cosine series the infinite series.

Using $a_0=\int_{0}^{1} f(x) \,\text{d}x$ and $a_n=\int_{0}^{1} f(x) \cos(2\pi nx) \,\text{d}x$

I got that $S(f)(x)=\frac{1}{12}+2\sum_{n=1}^{\infty}\frac{(-1)^n}{\pi^2n^2}\cos(2\pi nx)$

Now, how can I calculate the initial series?

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    $\begingroup$ plugging in a convenient value for $x$ is a good start $\endgroup$ Jan 30, 2018 at 19:53
  • $\begingroup$ I think the Fourier series converges to the initial function, since $x^2$ is continuous in all of $\mathbb{R}$, therefore $x=S(f)(x)$. Then, if I plug in $x=0$, I will have $0=\frac{1}{12}+2\sum_{n=1}^{\infty}\frac{(-1)^n}{n^2}$. This would mean that $\frac{2}{12}=\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n^2}$. Is this correct? I am not sure since I know that $\sum_{n=1}^{\infty}\frac{1}{n^2}=\frac{\pi^2}{6}$ (I know it from another problem) $\endgroup$ Jan 30, 2018 at 20:04
  • $\begingroup$ that computation doesn't look right, you dropped a sign and a $\pi^2$. Also I don't believe there should be a 2 $\endgroup$ Jan 30, 2018 at 20:10
  • $\begingroup$ Also why are you integrating on the interval $[0,1]$? $\endgroup$ Jan 30, 2018 at 20:16
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    $\begingroup$ If you shift the integral like this, you will phase shift all of your trig functions along with it. $\endgroup$
    – Doug M
    Jan 30, 2018 at 20:27

1 Answer 1

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The function $x^2$ is an even function in the given interval, $[\frac{-1}{2},\frac{1}{2}]$, thus the sine terms will vanish.

I calculated the Fourier series to be: $$f(x)=x^2=\frac{1}{12}+\sum_{n=1}^\infty {\frac{(-1)^n}{\pi^2n^2}cos({2\pi}nx)}$$ NOTE: There is a mistake in your equation above where you multiplied the summation by 2.

  1. In solving for $\sum_{n=1}^\infty {\frac{(-1)^{n+1}}{n^2}}$: Let $x=0$. The Fourier series becomes: $$f(0)=(0)^2=\frac{1}{12}+\sum_{n=1}^\infty {\frac{(-1)^n}{\pi^2n^2}cos({2\pi}n{0})}$$ Simplifying: $$0=\frac{1}{12}+\sum_{n=1}^\infty {\frac{(-1)^n}{\pi^2n^2}cos(0)} = \frac{1}{12}+\sum_{n=1}^\infty {\frac{(-1)^n}{\pi^2n^2}}$$ Multiplying by $-1$ gives: $$0=-\frac{1}{12}-\sum_{n=1}^\infty {\frac{(-1)^n}{\pi^2n^2}} = -\frac{1}{12}+\sum_{n=1}^\infty {\frac{(-1)^{n+1}}{\pi^2n^2}}$$ Simplifying further: $$\sum_{n=1}^\infty {\frac{(-1)^{n+1}}{\pi^2n^2}}=\frac{1}{12}$$ Finally, multiplying throughout by $\pi^2$ gives: $$\sum_{n=1}^\infty {\frac{(-1)^{n+1}}{n^2}}=\frac{\pi^2}{12}$$

  2. In solving for $\sum_{n=1}^\infty {\frac{1}{n^2}}$: If we substitute for $x=\frac{1}{2}$, the Fourier series gives:

$$f(\frac{1}{2})=(\frac{1}{2})^2=\frac{1}{12}+\sum_{n=1}^\infty {\frac{(-1)^n}{\pi^2n^2}cos({2\pi}n\frac{1}{2})}$$ Simplifying: $$\frac{1}{4}=\frac{1}{12}+\sum_{n=1}^\infty {\frac{(-1)^n}{\pi^2n^2}cos(n\pi)} = \frac{1}{12}+\sum_{n=1}^\infty {\frac{(-1)^n}{\pi^2n^2}(-1)^n}$$ But $$\frac{1}{12}+\sum_{n=1}^\infty {\frac{(-1)^n}{\pi^2n^2}(-1)^n}=\frac{1}{12}+\sum_{n=1}^\infty {\frac{(-1)^{2n}}{\pi^2n^2}}=\frac{1}{12}+\sum_{n=1}^\infty {\frac{1}{\pi^2n^2}}$$ Therefore $$\frac{1}{4}=\frac{1}{12}+\sum_{n=1}^\infty {\frac{1}{\pi^2n^2}}$$ Again simplifying gives: $$\frac{1}{6}=\sum_{n=1}^\infty {\frac{1}{\pi^2n^2}}$$ And finally, multiplying both sides by $\pi^2$ gives: $$\frac{\pi^2}{6}=\sum_{n=1}^\infty {\frac{1}{n^2}}$$

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