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AN INTERESTING RELATION

  I just copy here an answer of mine of this post, just in case OP decided to delete it.  Link: https://math.stackexchange.com/q/5005043/1231540 First we state two results:  - **Proposition 1:** Let $n \ge 3$. The minimal polynomial of $\cos(2\pi/n)$ - denoted by $\psi_n$ - has degree $\phi(n)/2.$ Here $\phi$ is Euler totient function.  - **Proposition 2:** The roots of $\psi_n$ are $\cos(2k\pi/n)$ where $\gcd(k,n)=1$ and $0 \le k \le \lfloor \frac{n}{2}\rfloor$. The proofs of the above two propositions can be found in this paper: https://www.jstor.org/stable/2324301 ***Proof of the first identity:*** It can be observed that $4\sin^2(\pi/n) = 2-4\cos(2\pi/n)$, so the polynomial $\Phi_n$ and $\psi_n$ have the same degree, which is $\phi(n)/2$. In particular, $\Phi_{2m}$ has degree $\phi(2m)/2$.  If $m$ is odd, we clearly have $\phi(2m) = \phi(m)$, which means $\Phi_{2m}$ and $\Phi_m$ have the same degrees. By definition, $\Phi_{2m}(x)$ is irreducible and has...

ON THE EULER'S SUM

 In this short post, I want to illustrate how to use Fourier series to attain the following remarkable identity of Euler:

\[\sum_{n=1}^\infty \dfrac{1}{n^2} = \dfrac{\pi^2}{6}\]

First we recall some related results that will be used below.

Theorem 1: The trigonometric system 

\[\dfrac{1}{\sqrt{2\pi}}, \dfrac{\cos(nx)}{\sqrt{\pi}}, \dfrac{\sin(mx)}{\sqrt{\pi}}, \quad n,m =1,2,\ldots\]

forms an orthonormal basis of the Hilbert space $L^2[0,2\pi]$.

Theorem 2: Suppose that $\left\lbrace x_k\right\rbrace_{k=1}^\infty$ is an orthogonal sequence in a Hilbert space $H$. Then the following are equivalent

  •   $\sum_{k=1}^\infty x_k $ converges.
  •  $\sum_{k=1}^\infty ||x_k||^2 $ converges.

Theorem 3: (Parseval's equality) Using the same notation in the previous theorem and denoting $x = \sum_{k=1}^\infty x_k$, then we have:

\[||x||^2 = \sum_{k=1}^\infty ||x_k||^2\]

These three results are quite standard in any Functional analysis text, so I don't give any proof here. For further details please check the reference.

Now we turn to some definition. Recall that over $L^2[0,2\pi]$, the norm is defined as 

\[ \left\langle f, g \right\rangle = \int_0 ^{2\pi} f\overline{g}dm,\]

where $m$ denotes the Lebesgue measure. Then theorem 1 says that for every $f \in L^2[0,2\pi]$ and $a_n = \left\langle f,\dfrac{\cos(nx)}{\sqrt{\pi}}\right\rangle $ and $b_n =\left\langle f,\dfrac{\sin(nx)}{\sqrt{\pi}}\right\rangle $ then 

\[f = \dfrac{a_0}{2} +\sum_{k=1}^{\infty} a_k\cos(kx) + b_k\sin(kx). \]

The RHS of the above equation is usually called Fourier series of $f$.


Now consider the Fourier series of $f(x)=x$. It can be verified that 

\[f(x) = 2 \sum_{n=1}^\infty \dfrac{(-1)^{n+1}}{n}\sin(nx)\]

Then the Fourier coefficient $a_n =\dfrac{2(-1)^{n+1}\sqrt{\pi}}{n}$. Then by theorem 2 and 3, the sum $\sum_{n=1}^\infty |c_n|^2$ converges to the norm of $f(x)=x$. Thus 

\begin{align*}\sum_{n=1}^\infty \dfrac{4\pi}{n^2} = \sum_{n=1}^\infty |c_n|^2=||x||^2 =\int_0^{2\pi} x^2 dx=\dfrac{2\pi^3}{3}\end{align*}

Dividing both sides by $4\pi$ gives the desired result.

Reference: 

Real Analysis: Measure Theory, Integration, and Hilbert Spaces


Comments

  1. Euler sum is called Basel problem. Euler found many sums of an infinite series. And most famous one is Basel problem

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