A hypergeometric series

January 25, 2019 / Leave a comment

Let $\alpha, \beta \in \mathbb{R}$ such that $0<\alpha<\beta$ . Evaluate the series:

$\mathcal{S} = \sum_{n=0}^{\infty} \frac{\alpha \left ( \alpha+1 \right ) \cdots \left ( \alpha+n \right )}{\left (\beta+1 \right )\left ( \beta+2 \right )\cdots\left ( \beta+n \right )}$

Solution

Lemma 1: For the $\mathrm{B}, \Gamma$ functions , it holds that:

$\mathrm{B}(x, y) = \int_{0}^{1} t^{x-1} \left ( 1-t \right )^{y-1}\, \mathrm{d}t = \frac{\Gamma(x) \Gamma(y)}{\Gamma(x+y)}$

Lemma 2: It holds that:

$\frac{\Gamma\left ( \alpha+n+1 \right )}{\Gamma\left ( \beta+n+2 \right )}= \frac{1}{\Gamma\left ( \beta-\alpha+1 \right )} \cdot \mathrm{B} \left ( n+\alpha+1, \beta-\alpha+1 \right )$

Proof: Simple calculations using Lemma $(1)$ reveal the identity.

Lemma 3: Using Lemma 2 it holds that

$\mathrm{B} \left ( n+\alpha+1, \beta-\alpha+1 \right ) = \frac{1}{\Gamma\left ( \beta-\alpha+1 \right )} \cdot \int_{0}^{1} t^{n+\alpha} \left ( 1-t \right )^{\beta-\alpha} \, \mathrm{d}t$

and as a consequence

$\frac{\Gamma\left ( \alpha+n+1 \right )}{\Gamma\left ( \beta+n+2 \right )} = \frac{1}{\Gamma\left ( \beta-\alpha+1 \right )} \cdot \int_{0}^{1} t^{n+\alpha} \left ( 1-t \right )^{\beta-\alpha} \, \mathrm{d}t$

Then, successively we have that:

$\begin{aligned} \sum_{n=0}^{\infty} \frac{\alpha \left ( \alpha+1 \right ) \cdots \left ( \alpha+n \right )}{\left ( \beta+1 \right )\left ( \beta+2 \right )\cdots \left ( \beta+n \right )} &= \sum_{n=0}^{\infty} \frac{\alpha \left ( \alpha+1 \right )_n}{\left ( \beta+1 \right )_n} \\ &=\sum_{n=0}^{\infty} \frac{\frac{\Gamma\left ( \alpha+n+1 \right )}{\Gamma\left ( \alpha \right )}}{\frac{\Gamma\left ( \beta+n+2 \right )}{\Gamma\left ( \beta \right )}} \\ &= \frac{\Gamma\left ( \beta \right )}{\Gamma\left ( \alpha \right )} \cdot \sum_{n=0}^{\infty} \frac{\Gamma\left ( \alpha+n+1 \right )}{\Gamma\left ( \beta+n+2 \right )}\\ &= \frac{\Gamma\left ( \beta \right )}{\Gamma\left ( \alpha \right ) \cdot \Gamma \left(\beta-\alpha+1 \right)} \cdot \sum_{n=0}^{\infty} \mathrm{B} \left ( n+\alpha+1, \beta-\alpha+1 \right )\\ &= \frac{\Gamma\left ( \beta \right )}{\Gamma\left (\alpha \right ) \cdot \Gamma \left(\beta-\alpha+1 \right)} \cdot \sum_{n=0}^{\infty} \int_{0}^{1} t^{n+\alpha} \left ( 1-t \right )^{\beta-\alpha} \, \mathrm{d}t \\ &= \frac{\Gamma\left ( \beta \right )}{\Gamma\left (\alpha \right ) \cdot \Gamma \left(\beta-\alpha+1 \right)}\int_{0}^{1} t^\alpha \left ( 1-t \right )^{\beta-\alpha} \sum_{n=0}^{\infty} t^n \, \mathrm{d}t \\ &=\frac{\Gamma\left ( \beta \right )}{\Gamma\left ( \alpha \right ) \cdot \Gamma \left(\beta-\alpha+1 \right)} \int_{0}^{1} t^\alpha \left ( 1-t \right )^{\beta-\alpha} \cdot \frac{\mathrm{d}t}{1-t} \\ &= \frac{\Gamma\left ( \beta \right )}{\Gamma\left ( \alpha \right ) \cdot \Gamma \left(\beta-\alpha+1 \right)} \cdot \int_{0}^{1} t^\alpha \left ( 1-t \right )^{\beta-\alpha-1} \, \mathrm{d}t \\ &= \frac{\Gamma\left ( \beta \right )}{\Gamma\left (\alpha \right ) \cdot \Gamma \left(\beta-\alpha+1 \right)} \cdot \mathrm{B} \left ( \alpha+1, \beta-\alpha \right ) \\ &=\frac{\Gamma\left ( \beta \right )}{\Gamma\left (\alpha \right ) \cdot \Gamma \left(\beta-\alpha+1 \right)} \cdot \frac{\Gamma\left ( \alpha+1 \right ) \Gamma\left ( \beta-\alpha \right )}{\Gamma\left ( \beta+1 \right )} \\ &= \frac{\Gamma\left ( \beta \right )}{\Gamma\left ( \alpha \right ) \cdot \left( \beta -\alpha \right) \cdot \Gamma \left(\beta-\alpha \right)} \cdot \frac{\alpha \Gamma \left (\alpha \right ) \Gamma\left ( \beta - \alpha \right )}{\beta \Gamma \left ( \beta \right )} \\ &= \frac{\alpha}{\beta} \cdot \frac{1}{\beta-\alpha} \end{aligned}$

A square root limit

January 24, 2019 / Leave a comment

Let $c>0$ . Prove that

$\lim _{n\rightarrow ++\infty} \left ( \frac {1+ \sqrt [n] {c}}{2} \right ) ^n= \sqrt {c}$

Solution

It holds that

$x^{1/n} = 1 + \frac{\log x}{n} + \mathcal{O} \left ( \frac{1}{n^2} \right )$

Thus,

$\begin{align*} \lim_{n \rightarrow +\infty} \left ( \frac{1+\sqrt[n]{c}}{2} \right )^n &= \lim_{n \rightarrow +\infty} \left ( \frac{1+c^{1/n}}{2} \right )^n \\ &=\lim_{n \rightarrow +\infty} \left ( \frac{1}{2} +\frac{1}{2}\left ( 1+ \frac{\log c}{n} + \mathcal{O}\left ( \frac{1}{n^2} \right ) \right ) \right )^n \\ &= \lim_{n \rightarrow +\infty} \left ( 1 + \frac{\log c}{2n} + \frac{1}{2} \mathcal{O}\left ( \frac{1}{n^2} \right ) \right )^n \\ &= \lim_{n \rightarrow +\infty} \left ( 1+ \frac{\log \sqrt{c}}{n} + \frac{1}{2} \mathcal{O}\left ( \frac{1}{n^2} \right ) \right )^n \\ &= \exp \left ( \log \sqrt{c} \right ) \\ &= \sqrt{c} \end{align*}$

Note: Similarly, it holds that $\lim \limits_{n \rightarrow +\infty} \left ( 2\sqrt[n]{x}-1 \right )^n =x^2$ .

Searching for the …function

January 23, 2019 / Leave a comment

Find all $\mathcal{C}^1$ functions $f:[0, 1] \rightarrow (0, +\infty)$ such that $\frac{f(1)}{f(0)} = e$ and

$\int_{0}^{1} \frac{\mathrm{d}x}{f^2(x)} + \int_{0}^{1} \left( f'(x) \right)^2 \, \mathrm{d}x \leq 2$

Solution

First of all we note that

$\int_{0}^{1} \frac{f'(x)}{f(x)} \, \mathrm{d}x = \left [ \ln f(x) \right ]_0^1 = 1$

Thus,

$\begin{align*} \int_{0}^{1} \frac{\mathrm{d}x}{f^2(x)} + \int_{0}^{1} \left( f'(x) \right)^2 \, \mathrm{d}x \leq 2 &\Leftrightarrow \int_{0}^{1} \frac{\mathrm{d}x}{f^2(x)} + \int_{0}^{1} \left( f'(x) \right)^2 \, \mathrm{d}x \\ & \quad \quad \quad \leq 2 \int_{0}^{1} \frac{f'(x)}{f(x)} \, \mathrm{d}x\\ &\Leftrightarrow \int_{0}^{1} \frac{\mathrm{d}x}{f^2(x)} - 2 \int_{0}^{1} \frac{f'(x)}{f(x)} \, \mathrm{d}x + \\ & \quad \quad \quad + \int_{0}^{1} \left( f'(x) \right)^2 \, \mathrm{d}x \leq 2 \\ &\Leftrightarrow \int_{0}^{1} \left ( \frac{1}{f(x)} - f'(x) \right )^2 \, \mathrm{d}x \leq 0 \end{align*}$

Since the integrand is positive it only remains that

$\begin{align*} \frac{1}{f(x)}- f'(x) =0 &\Leftrightarrow \frac{1}{f(x)} = f'(x) \\ &\Leftrightarrow f'(x) f(x) = 1 \\ &\Leftrightarrow f^2(x) = 2x+ c \end{align*}$

Setting $x=0$ and $x=1$ at the last equation we have that:

$\begin{matrix} f^2(0) & = & c \\ f^2(1) &= &2+c \end{matrix}\Rightarrow \left ( \frac{f(0)}{f(e)} \right )^2 = \frac{c}{2+c} \Rightarrow \frac{1}{e^2} = \frac{c}{2+c} \Rightarrow c = \frac{2}{e^2-1}$

Since $f$ is positive we conclude that

$\begin{align*} f^2(x) = 2x+ \frac{2}{e^2-1} &\Leftrightarrow f(x) = \sqrt{2x + \frac{2}{e^2-1}} \; , \; x \in [0, 1] \end{align*}$

which satisfies the given conditions.

Inequality of a concave function

January 23, 2019 / Leave a comment

Let $f:[0, 1] \rightarrow [0, +\infty)$ be a concave function. Prove that

$\int_0^1 x^2 f(x) \, \mathrm{d}x \leq \frac{1}{2} \int_0^1 f(x) \,\mathrm{d}x$

Solution

Since $f$ is concave , it holds that

$f\left ( ax + \left ( 1-a \right )y \right ) \geq a f(x) + \left ( 1-a \right ) f(y) \geq a f(x) \quad \text{forall} \;\; a, x, y \in [0, 1]$

By setting $y=0$ and $x=a$ we get that $x f(x) \leq f \left ( x^2 \right )$ . Thus,

$\begin{align*} \int_{0}^{1} x^2 f(x) \, \mathrm{d}x &= \frac{1}{2} \int_{0}^{1} x f(x) 2x \, \mathrm{d}x \\ &\leq \frac{1}{2}\int_{0}^{1} f \left ( x^2 \right ) 2x \, \mathrm{d}x \\ &=\frac{1}{2} \int_{0}^{1} f \left ( x^2 \right ) \, \mathrm{d} \left ( x^2 \right ) \\ &= \frac{1}{2} \int_{0}^{1} f(x) \, \mathrm{d}x \end{align*}$