Inner Product Examples, what is the points?

Example: For $-\pi<x<\pi$,

$$x =-2 \sum_{n=1}^{\infty} \frac{(-1)^n}{n} \sin(nx)$$

and

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

by using inner products of these two functions, the value of

$$\sum_{n=1}^{\infty} \frac{1}{n^4}$$

is equal to $\frac {\pi^4}{90}$ .

My question is what is the point in this example that the author gives the solution without any detail? any users can tips me how this value is reached?

• It was long an open problem what the value of $\sum_{n\geq 1}\frac{1}{n^2}$ was. It was Euler who found a remarkable reasoning to show that $\sum_{n\geq 1} \frac{1}{n^2}=\frac{\pi^2}{6}$. However, his reasoning used some facts that were not proved in his time. The exercise above asks you to do more or less the same, you have to find the value of $\sum_{n\geq 1}\frac{1}{n^4}$. You can do this by considering the inner product of those two functions and using Parseval's theorem. – Mathematician 42 Jul 26 '16 at 11:44
• Also see this math.stackexchange.com/questions/650966/… – Mathematician 42 Jul 26 '16 at 11:44
• @Mathematician42 I think it's $\pi^2/6$ – GeorgSaliba Jul 26 '16 at 11:46
• @GeorgSaliba: It is, thank you, corrected it. :) – Mathematician 42 Jul 26 '16 at 11:48
• @Mathematician42 also, I think the second series in the question is for $x^3$, not $x$... – GeorgSaliba Jul 26 '16 at 12:04

HINT: $$\frac 1{\pi}\int_{-\pi}^\pi x^4dx=4\sum_{n=1}^\infty \frac{\pi^2}{n^2}-\frac 6{n^4}$$ FURTHER EXPLANATION:

You have two functions $f(x)=x$ and $g(x)=x^3$. The coefficients of the expansion of $f$ in the basis $\sin(nx)$ are of the form $$a_n=-2\times \frac{(-1)^n}{n}$$

and the coefficients for $g(x)$ are: $$b_n=-2\times\left(\frac{\pi^2}{n^2}-\frac 6{n^3}\right)\times (-1)^n$$

Then the inner product is given by: $$\frac 1\pi\int_{-\pi}^{\pi}f(x)g(x)=\sum_{n=1}^\infty a_n b_n$$

Replacing $f,g,a_n$ and $b_n$, using the fact that $\sum 1/n^2=\pi^2/6$ and manipulating, gives you the answer.

SOLUTION:

$$\frac 1{\pi}\int_{-\pi}^\pi x^4dx=4\sum_{n=1}^\infty \left(-2\times \frac{(-1)^n}{n}\right)\times(-2)\left(\left(\frac{\pi^2}{n^2}-\frac 6{n^3}\right)\times (-1)^n\right)$$ $$\frac 1{\pi}\int_{-\pi}^\pi x^4dx=4\sum_{n=1}^\infty \frac{\pi^2}{n^2}-\frac 6{n^4}$$ $$\frac{2\pi^5}{5\pi}=4\pi^2\times\frac{\pi^2}6-24\sum_{n=1}^\infty\frac 1{n^4}$$ $$24\sum_{n=1}^\infty\frac 1{n^4}=\pi^4\left( \frac 23 - \frac 25\right)$$ $$\sum_{n=1}^\infty\frac 1{n^4}=\frac{\pi^4}{90}$$

• @user355834 I have added more details. The link contains another method using only one function $x^2$. You're asking for the inner product of $x^3$ and $x$ not the inner product of $x^2$ and itself (which uses the Parseval identity). – GeorgSaliba Jul 26 '16 at 12:19
• Very creative answer, very nice, Please wait until again and again I read it... thanks so much – user355834 Jul 26 '16 at 12:20
• maybe has some unclear points for me. would you please say how I can continue from $\frac 1\pi\int_{-\pi}^{\pi}f(x)g(x)=\sum_{n=1}^\infty a_n b_n$? I couldn't reach to last answer :) – user355834 Jul 26 '16 at 12:40
• @user355834 I have added a full solution. Just hover your mouse over the yellow box. – GeorgSaliba Jul 26 '16 at 12:48