Series of Bessel function

Let J_n denote the Bessel function of the first kind. Prove that

    \[\sum_{n=-\infty}^{\infty}\left|J_n(z)\right|^2 =1\]


The Jacobi – Anger expansion tells us that

(1)   \begin{equation*} e^{iz\sin\theta} = \sum_{n=-\infty}^{\infty}J_n(z) e^{in\theta}  \end{equation*}

Hence by Parseval’s Theorem it follows that

    \begin{align*} \sum_{n=-\infty}^{\infty}\left|J_n(z)\right|^2 &= \frac{1}{2\pi}\int_{0}^{2\pi}e^{iz\sin \theta} e^{iz\sin(-\theta)}\,\mathrm{d} \theta \\ &= \frac{1}{2\pi}\int_{0}^{2\pi}\mathrm{d} \theta \\ &=1 \end{align*}

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Proof of “Fermat’s last theorem”

Let x,y,z,n\in \mathbb{N}^* and n \geq z. Prove that the equation


has no solution.


Without loss of generality , assume that x<y. If x^n+y^n=z^n held , then it would be z^n > y^n thus z^n \geq (y+1)^n. It follows from Bernoulli’s inequality that,

    \begin{align*} z^n &=x^n +y^n\\ &< 2y^n \\ &\leq \left(1 + \frac {n}{z} \right) y^n \\ &\leq \left(1 + \frac {1}{z} \right)^ny^n\\ &< \left(1 + \frac {1}{y} \right)^ny^n \\ &= (y+1)^n \\ &\leq z^n \end{align*}

which is an obscurity. The result follows.

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Limit of a sequence

Let f:[0, 1] \rightarrow (0, +\infty) be a continuous function and A be the set of all positive integers n such that there exists x_n such that

    \[\int_{x_n}^{1} f(t) \, \mathrm{d}t = \frac{1}{n}\]

Prove that \{x_n\}_{n \in A} is infinite and evaluate the limit

    \[\ell = \lim_{n \rightarrow +\infty} n \left( x_n -1 \right)\]

A factorial limit

Let \cdot! denote the factorial of a real number; that is x!=\Gamma(x+1). Evaluate the limit:

    \[\ell = \lim_{x \rightarrow n} \frac{x!-n!}{x-n}\]


It holds that

    \begin{align*} \lim_{x\rightarrow n} \frac{x!-n!}{x-n} &= \lim_{x\rightarrow n} \frac{\Gamma(x+1)- \Gamma(n+1)}{x-n} \\ &=\Gamma'(n+1) \\ &=\Gamma(n+1) \psi^{(0)}(n+1) \\ &=n! \left ( \mathcal{H}_n - \gamma \right ) \end{align*}

where \mathcal{H}_n denotes the n-th harmonic number and \gamma the Euler – Mascheroni constant.

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Linear isometry

Let f:\mathbb{R}^2 \rightarrow \mathbb{R}^2. If:

  • f(\mathbf{0})=\mathbf{0}
  • \left| {f\left( {\bf{u}} \right) - f\left( {\bf{v}} \right)} \right| = \left| {{\bf{u}} - {\bf{v}}} \right| for all {{\bf{u}},{\bf{v}}}

then prove that f is linear.


For convenience, identify \mathbb{R}^2 with \mathbb{C} here. Then note that for any such function f:\mathbb{C} \to \mathbb{C}, also z_1 \cdot f(z) a solution for any point z_1 on the unit circle. Also \overline{f(z)} is a solution. Note that \vert f(1)\vert=1 and hence we can wlog assume that f(1)=1. So f(i) is a point on the unit circle with distance \sqrt{2} to 1. Hence f(i) =\pm i, so w.l.o.g. assume that f(i)=i. But then for any z \in \mathbb{C}, both z and f(z) have the same distance to 0,1 and i. So supposing z \ne f(z), all 0,1,i lie on the perpendicular bisector between these points and in particular 0,1 and i are collinear which clearly is absurd. Hence f(z)=z for all z which proves the claim.

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