Joy of Mathematics

Exponential fractional part

June 5, 2020 Tolaso

Let $\{ \cdot \}$ denote the fractional part of $x$ . Evaluate the integral

$\mathcal{J} = \int_0^1 \int_0^1 \left\{ \frac{e^x}{e^y} \right\} \, \mathrm{d}(x, y)$

Solution

By definition it holds that

$\left \{ x \right \} = x - \left \lfloor x \right \rfloor$

hence

$\iint \limits_{\left [ 0, 1 \right ]^2} \left \{ e^x e^{-y} \right \} \, \mathrm{d}(x, y)= \iint \limits_{\left [ 0, 1 \right ]^2} e^x e^{-y} \, \mathrm{d}(x, y)-\iint \limits_{\left [ 0, 1 \right ]^2} \left \lfloor e^x e^{-y} \right \rfloor \, \mathrm{d}(x, y)$

We note however that

$\left \lfloor e^x e^{-y} \right \rfloor = \left\{\begin{matrix} 0 & , & x< y \\ 2 & , & y< x - \ln 2 \\ 1 & , & \text{otherwise} \end{matrix}\right.$

Hence,

$\begin{align*} \iint \limits_{[0, 1]^2} \left \lfloor e^x e^{-y} \right \rfloor \, \mathrm{d} (x, y) &= \int_{0}^{1} \int_{x}^{1} 0 \, \mathrm{d}(y, x) + \int_{0}^{1} \int_{0}^{x} \, \mathrm{d} (y, x) + \\ & \quad \quad + \int_{\ln 2}^{1} \int_{0}^{x-\ln 2} 1 \, \mathrm{d} \left ( y, x \right ) \\ &= 1 - \ln 2 + \frac{\ln^2 2}{2} \end{align*}$

Summing up,

$\iint \limits_{[0, 1]^2} \left \{ e^x e^{-y} \right \} \, \mathrm{d}(x, y) = \frac{1}{e} +e + \ln 2 - \frac{\ln^2 2}{2} - 3$

Rational function and polynomial

May 9, 2020 Tolaso

Prove that there does not exist a rational function $f$ with real coefficients such that

$f \left ( \frac{x^2}{x+1} \right ) = \mathrm{P}(x)$

where $\mathrm{P}(x) \in \mathbb{R}[x]$ is a non constant polynomial.

Solution

Since polynomials are defined on $x=-1$ we have that

$\begin{align*} \mathbb{R} \ni \mathrm{P}(-1) & =\lim_{x \rightarrow -1^+} \mathrm{P}(x) \\ &=\lim_{x \rightarrow -1^+}f \left(\frac{x^2}{x+1} \right) \\ &=\lim_{x \rightarrow \infty} f\left(\frac{x^2}{x+1}\right) \\ &=\lim_{x \rightarrow \infty} \mathrm{P}(x) \end{align*}$

Since $\mathrm{P}$ tends to a finite value as $x \rightarrow \infty$ it must be a constant polynomial. In particular, $f$ must be constant in the range of $\frac{x^{2}}{x + 1}$

Sum over all positive rationals

February 22, 2020 Tolaso

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

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

Infinite Product

July 31, 2019July 31, 2019 Tolaso

Evaluate the infinite product:

$\Pi = \prod_{n=3}^{\infty} \frac{\left ( n^3+3n \right )^2}{n^6-64}$

(IMC 2019 / Day 1 / Problem 1)

Solution

$\begin{align*} \prod_{n=3}^{\infty} \frac{\left ( n^3+3n \right )^2}{n^6-64} &= \prod_{n=3}^{\infty} \frac{n^2 \left ( n^2+3 \right )^2}{n^6-69} \\ &=\prod_{n=3}^{\infty} \frac{n^2\left ( n^2+3 \right )^2}{\left ( n^3-8 \right )\left ( n^3+8 \right )} \\ &= \prod_{n=3}^{\infty} \frac{n^2\left ( n^2+3 \right )^2}{\left ( n-2 \right )\left ( n^2+2n+4 \right )\left ( n+2 \right )\left ( n^2-2n+4 \right )} \\ &=\prod_{n=3}^{\infty} \frac{n}{n-2} \cdot \frac{n}{n+2} \cdot \frac{n^2+3}{\left ( n-1 \right )^2+3} \cdot \frac{n^2+3}{\left ( n+1 \right )^2+3} \\ &= \lim_{N \rightarrow +\infty} \frac{N\left ( N-1 \right )}{1 \cdot 2} \cdot \frac{3 \cdot 4}{\left ( N+1 \right ) \left ( N+2 \right )} \cdot \frac{N^2+3}{2^2+3} \cdot \\ & \quad \quad \quad \quad \quad \cdot\frac{3^2+3}{\left ( N+1 \right )^2+3} \\ &= \frac{72}{7} \end{align*}$

since the product telescopes.

A divergent series

May 20, 2017 Tolaso

Let $a_n$ be a positive and strictly decreasing sequence such that $\lim a_n =0$ . Prove that the series

$\mathcal{S} = \sum_{n=1}^{\infty} \frac{a_n-a_{n+1}}{a_n}$

diverges.

Solution

We shall begin with a lemma.

Lemma: Let $x_1, \dots, x_n \in (0, 1)$ . It holds that

$\sum_{i=1}^n (1-x_i) \geqslant 1 - \prod_{i=1}^n x_i$

Proof: We are using induction on $n$ . For $n=1$ it is trivial. Suppose that it holds for $n=k$ . Then

$\begin{align*} \sum_{i=1}^{k+1} (1-x_i) &= \left( \sum_{i=1}^{k} (1-x_i)\right) + (1-x_{k+1}) \\ &\geqslant \left(1 - \prod_{i=1}^k x_i\right) + (1-x_{k+1}) \\ &= 1 - \prod_{i=1}^{k+1} x_i + \\ & \quad \quad +\left(1 - \prod_{i=1}^k x_i\right)\left(1-x_{k+1}\right) \\ &\geqslant 1 - \prod_{i=1}^{k+1} x_i \end{align*}$

Thus it holds for $n=k+1$ and the lemma is proved. Since $a_n>0$ and $\lim a_n =0$ I can find $1 = i_1 < i_2 < \cdots$ such that $\frac{a_{i_{r+1}}}{a_{i_r}} < \frac{1}{2}$ for all $r$ . But then

$\begin{align*} \sum_{n=1}^{i_k-1} \frac{a_n - a_{n+1}}{a_n} &= \sum_{r=1}^{k-1} \sum_{n=i_r}^{i_{r+1}-1}\left(1 - \frac{a_{n+1}}{a_n} \right) \\ &\geqslant \sum_{r=1}^{k-1} \left(1 - \frac{a_{i_{r+1}}}{a_{i_r}} \right) \\ &> \sum_{r=1}^{k-1} \frac{1}{2} \\ &> \frac{k-1}{2} \end{align*}$

where in the first inequality the lemma was used.

The exercise can also be found at mathematica.gr . It can also be found in Problems in Mathematical Analysis v1 W.J.Kaczor M.T.Nowak as exercise 3.2.43 page 80 .