Search found 177 matches

by Riemann
Fri Mar 22, 2024 7:16 pm
Forum: High-school maths
Topic: Fun integral problem
Replies: 2
Views: 152

Re: Fun integral problem

First of all we note that $\mathrm{P}(x) + \mathrm{P} \left ( 12 - x \right ) = 12$ since \begin{align*} \mathrm{P}(x) - 12 &= \left ( x -4 \right ) \left ( x - 5 \right ) \left ( x -9 \right ) \\ &= -\left ( 4 - x \right ) \left ( 5- x \right ) \left ( 9 - x \right ) \\ &= - \left [ \le...
by Riemann
Thu Mar 16, 2023 3:14 pm
Forum: General Mathematics
Topic: An inequality
Replies: 0
Views: 2140

An inequality

Let $x, y, z$ be positive real numbers. Prove that:

$$ \sqrt{\frac{x}{x+y}} + \sqrt{\frac{y}{y+z}} + \sqrt{\frac{z}{z+x}} \leq \frac{3}{\sqrt{2}}$$
by Riemann
Tue Feb 08, 2022 9:31 pm
Forum: Calculus
Topic: Integral
Replies: 0
Views: 3227

Integral

Let $a\geq 1$. Prove that:

$$\int_{0}^{\pi} \left ( 1 + \cos x \right ) \ln \left ( a + \cos x \right )\, \mathrm{d}x = \pi \left ( a - \sqrt{a^2-1} \right ) + \pi \ln \left ( \frac{a + \sqrt{a^2-1}}{2} \right )$$
by Riemann
Sat Nov 27, 2021 7:07 pm
Forum: Calculus
Topic: Trigonometric logarithmic integral
Replies: 2
Views: 3270

Re: Trigonometric logarithmic integral

We are using the following results: 1. $\displaystyle \sum_{n=1}^{\infty}\frac{x^n \cos (na)}{n}=-\frac{1}{2}\ln \left ( x^2-2x \cos a+1 \right ) \; , \; \left | x \right |<1$ 2. $\displaystyle \sum_{n=1}^{\infty}(-1)^n \frac{\sin n x}{n}=-\frac{x}{2}\; , \; \left | x \right |<\pi$ 3. $\displaystyle...
by Riemann
Sat Nov 27, 2021 7:02 pm
Forum: Calculus
Topic: Trigonometric logarithmic integral
Replies: 2
Views: 3270

Re: Trigonometric logarithmic integral

We note that $$\ln \left ( \frac{x^2+2kx \cos b+k^2}{x^2+2kx\cos a+k^2} \right )=\ln \left ( x^2+2kx \cos \alpha +k^2 \right ) \bigg|_{\alpha =a}^{\alpha =b}$$ Hence, \begin{align*} \int^{\infty}_{0} \frac{1}{x}\ln\left(\frac{x^2+2kx\cdot \cos b+k^2}{x^2+2k x\cdot \cos a+k^2}\right)\,dx &= \int_...
by Riemann
Sat Nov 27, 2021 6:55 pm
Forum: Calculus
Topic: A series involving alternating Harmonic numbers
Replies: 1
Views: 3456

Re: A series involving alternating Harmonic numbers

We are proving that $\displaystyle \sum_{n=1}^{\infty} \frac{(-1)^{n+1} \mathcal{H}_n}{n} = \frac{\pi^2}{12} - \frac{\ln^2 2}{2}$. Indeed, note that: \begin{align*} \sum_{n=1}^N\frac{(-1)^{n-1}}{n}H_n &=\sum_{n=1}^N\frac{(-1)^{n-1}}{n^2}+\sum_{n=2}^N\frac{(-1)^{n-1}}{n}H_{n-1}\\ &=\sum_{n=1}...
by Riemann
Fri Nov 06, 2020 5:35 am
Forum: Algebraic Structures
Topic: Not a Hopfian group
Replies: 1
Views: 3071

Re: Not a Hopfian group

Well, we define $f:\mathcal{G} \rightarrow \mathcal{G}$ by $f(x)=x^2$ and $f(y)=y$ and extend it to $\mathcal{G}$ homomorphically. Since $\mathcal{G}$ is well defined then $f$ is a surjective because $$f\left ( y^{-1} xy x^{-1} \right ) = x$$ but not an isomorphism because if we take $z=y^{-1} x y$ ...
by Riemann
Sat Dec 14, 2019 3:16 pm
Forum: Calculus
Topic: \(\sum_{n=2}^{\infty} \frac{(-1)^n}{n\log{n}}\)
Replies: 1
Views: 3986

Re: \(\sum_{n=2}^{\infty} \frac{(-1)^n}{n\log{n}}\)

Basically it equals to

$$\int_1^\infty \left( 1+\left(2^{1-s}-1\right)\zeta(s) \right) \, \mathrm{d}s$$

However, the $\zeta$ function does not behave well under integrals. So, I would not expect a closed form to exist ... !
by Riemann
Wed Oct 23, 2019 7:58 pm
Forum: General Mathematics
Topic: An inequality
Replies: 1
Views: 6314

Re: An inequality

The Engels form of the Cauchy – Schwartz inequality gives us: \begin{align*} \sum \frac{\log_{x_1}^4 x_2}{x_1+x_2} & \geq \frac{\left (\sum \log_{x_1}^2 x_2 \right )^2}{\sum (x_1+x_2)} \\ &= \frac{\left ( \sum \log_{x_1}^2 x_2 \right )^2}{2\sum x_1} \\ &\!\!\!\!\!\!\overset{\text{AM-GM}}...
by Riemann
Sun Oct 20, 2019 6:15 pm
Forum: General Mathematics
Topic: Arithmotheoretic limit
Replies: 0
Views: 6171

Arithmotheoretic limit

Evaluate the limit:

$$\ell= \lim_{n \rightarrow +\infty} \frac{1}{n^2} \sum_{m=1}^{n} n \pmod m$$