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“Upper” bound

February 22, 2020 / Leave a comment

Let f:\mathbb{R}^+ \rightarrow \mathbb{R} be a function satisfying

    \[\left | f\left ( a + b \right ) - f(a)\right | \leq \frac{b}{a}\]

for all positive real numbers a and b. Prove that

    \[\left | f(1) - f(x) \right | \leq \left |\ln x \right | \quad \text{for all} \;\; x>0\]

Solution

For starters, let us assume that x>1. Dividing the interval (1, x) into n subintervals each of length h so that h=\frac{x-1}{n}. Thus,

    \begin{align*} \left | f(1) - f(x) \right | &= \left | \sum_{k=0}^{n-1} f \left ( 1 + h(k+1) \right ) - f \left ( 1+kh \right ) \right | \\ &\leq \sum_{k=0}^{n-1} \left | f \left ( 1 + h(k+1) \right ) - f \left ( 1+kh \right ) \right | \\ &=\sum_{k=0}^{n-1} \left | f\left ( \left ( 1+kh \right ) + h\right ) - f \left ( 1+kh \right ) \right | \end{align*}

The inequality \left | f\left ( a + b \right ) - f(a)\right | \leq \frac{b}{a} implies that

    \[\left | f\left ( \left ( 1+kh \right ) + h\right ) - f \left ( 1+kh \right ) \right | \leq \frac{h}{1+kh}\]

Hence,

    \[\left | f(1) - f(x) \right | \leq \sum_{k=0}^{n-1} \frac{h}{1+kh}\]

The limit \displaystyle \sum_{k=0}^{n-1} \frac{h}{1+kh} exists and equals to \ln x. Hence , the inequality is proved for x>1.

Now, assume that x<1. Dividing the interval (x, 1) into n subintervals each of length h so that h=\frac{1-x}{n}. Thus,

    \begin{align*} \left | f(1) - f(x) \right | &= \left | \sum_{k=0}^{n-1} f \left ( 1 - h(k+1) \right ) - f \left ( 1 - kh \right ) \right | \\ &\leq \sum_{k=0}^{n-1} \left | f \left ( 1 - h(k+1) \right ) - f \left ( 1 - kh \right ) \right | \\ &=\sum_{k=0}^{n-1} \left | f\left ( \left ( 1 - kh \right ) - h\right ) - f \left ( 1 - kh \right ) \right | \end{align*}

The inequality \left | f\left ( a + b \right ) - f(a)\right | \leq \frac{b}{a} implies that

    \[\left | f\left ( \left ( 1 - kh \right ) - h\right ) - f \left ( 1 - kh \right ) \right | \leq \frac{h}{1-kh}\]

Hence,

    \[\left | f(1) - f(x) \right | \leq \sum_{k=0}^{n-1} \frac{h}{1+kh}\]

The limit \displaystyle \sum_{k=0}^{n-1} \frac{h}{1-kh} exists and equals to -\ln x. Hence , the inequality is also proved for x<1. This completes the proof!

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Sum over all positive rationals

February 22, 2020 / Leave a comment

For a rational number x that equals \frac{a}{b} in lowest terms , let f(x)=ab. Prove that:

    \[\sum_{x \in \mathbb{Q}^+} \frac{1}{f^2(x)} = \frac{5}{2}\]

Solution

First of all we note that

    \[F(s) = \sum_{x \in \mathbb{Q}^+} \frac{1}{f^s(x)} = \sum_{\substack{a,b=1 \\ \gcd(a, b)=1}}^{\infty} \frac{1}{\left ( ab \right )^s}\]

Moreover for s>1 we have that

    \begin{align*} \zeta^2(s) &= \left ( \sum_{a=1}^{\infty} \frac{1}{a^s} \right )^2 \\ &=\sum_{a, b=1}^{\infty} \frac{1}{(ab)^s} \\ &=\sum_{d=1}^{\infty} \sum_{\substack{a, b=1 \\\gcd(a, b)=d}}^{\infty} \frac{1}{\left ( ab \right )^s} \\ &= \sum_{d=1}^{\infty} \frac{1}{d^{2s}} \sum_{\substack{a, b=1 \\\gcd(a, b)=1}}^{\infty} \frac{1}{\left ( ab \right )^s} \\ &= \zeta(2s) F(s) \end{align*}

Hence for s=2 we have that

    \[F(2) = \frac{5}{2}\]

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Bessel series

February 22, 2020 / Leave a comment

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

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

Solution

The Jacobi – Anger expansion tells us that

    \[e^{iz\sin\theta} = \sum_{n=-\infty}^{\infty}J_n(z) e^{in\theta}\]

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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Pell – Lucas series

February 22, 2020 / Leave a comment

The Pell – Lucas numbers \mathcal{Q}_n are defined as follows \mathcal{Q}_0 = \mathcal{Q}_1 =2 and for every n \geq 2 it holds that

    \[\mathcal{Q}_n = 2\mathcal{Q}_{n-1} +\mathcal{Q}_{n-2}\]

Prove that

    \[\sum_{n=1}^{\infty} \arctan \frac{2}{\mathcal{Q}_n} \arctan \frac{2}{\mathcal{Q}_{n+1}} = \frac{\pi^2}{32}\]

 

Definite logarithmic integral

January 25, 2020 / Leave a comment

Let 0<\alpha<\beta. Evaluate the integral:

    \[\mathcal{J}=\int_\alpha^\beta \frac{\ln x}{(x+\alpha)(x+\beta)}\, \mathrm{d}x\]

Solution

We have successively:

    \begin{align*} \mathcal{J} &= \int_{\alpha}^{\beta} \frac{\ln x}{\left ( x+\alpha \right )\left ( x+\beta \right )}\, \mathrm{d}x \\ &\!\!\!\!\!\!\!\overset{x \mapsto \alpha \beta/x}{=\! =\! =\! =\! =\!=\!} \int_{\alpha}^{\beta} \frac{\ln \alpha \beta- \ln x}{\left ( x + \alpha \right ) \left ( x + \beta \right )} \, \mathrm{d}x\\ &=\frac{1}{2} \int_{\alpha}^{\beta} \frac{\ln \alpha \beta}{\left ( x+\alpha \right )\left ( x + \beta \right )} \, \mathrm{d}x \\ &=\frac{\ln \alpha \beta}{\beta-\alpha} \ln \left ( \frac{(\alpha+\beta)^2}{4\alpha \beta} \right ) \end{align*}

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