Joy of Mathematics

“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!

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

$F(2) = \frac{5}{2}$

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

Hence by Parseval’s Theorem it follows that:

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

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

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

Determinant

February 22, 2020 / Leave a comment

Let $x>0$ and $A \in \mathbb{R}^{2 \times 2}$ such that $\det \left(A^2 + x \mathbb{I}_{2 \times 2} \right) = 0$ . Prove that

$\det \left( A^2 + A + x \mathbb{I}_{2 \times 2} \right) = x$

Solution

Note that

$\det \left(A^2 + x \mathbb{I}_{2 \times 2} \right) = \det \left(A + i \sqrt {x} \mathbb{I}_{2 \times 2} \right) \det \left(A - i\sqrt{x} \mathbb{I}_{2 \times 2} \right)$

Since $A$ is real, its complex eigenvalues come in conjugate pairs. Thus, in this case we conclude that $A$ has eigenvalues $\pm i \sqrt x$ .

Now, if $\lambda$ is an eigenvalue of $A$ , then $\lambda^2 + \lambda + x$ is an eigenvalue of $A^2 + A + x \mathbb{I}_{2 \times 2}$ . Thus, the matrix $A^2 + A + x \mathbb{I}_{2 \times 2}$ has eigenvalues $(i\sqrt {x} )^2 + i\sqrt {x} + x = i\sqrt {x}$ and $(-i\sqrt {x})^2 - i\sqrt {x} + x = -i\sqrt {x}$ .