Search found 179 matches
- Sat Apr 06, 2024 8:50 am
- Forum: Analytic Geometry
- Topic: Equality of vectors
- Replies: 0
- Views: 1271
Equality of vectors
In the following figure $\mathrm{AE}$ is the exterior angle bisector of the angle $\mathrm{A}$ of the triangle $\mathrm{AB} \Gamma$. quicklatex.com-4a9d5cf71853d480c0d967a5686094ba_l3.png Prove that $\displaystyle \overrightarrow{\mathrm{AE}} = \frac{1}{\beta-\gamma} \left ( \beta \overrightarrow{\m...
- Sat Apr 06, 2024 8:37 am
- Forum: Calculus
- Topic: Nice integral
- Replies: 1
- Views: 4659
Re: Nice integral
Let $\displaystyle f(z) = \frac{e^{iz}}{\left ( 1 + z + z^2 \right )^2+1}$ and consider a semicircle $\gamma$ of radius $R$ at the upper half plane. The poles of $f$ are $z_1=i$ and $z_2=-1+i$. Then, $$\oint \limits_{\gamma} f(z) \, \mathrm{d}z = \int_{-R}^{R} f(x) \, \mathrm{d}x + \int \limits_{\te...
- Fri Mar 22, 2024 7:16 pm
- Forum: High-school maths
- Topic: Fun integral problem
- Replies: 2
- Views: 2351
Re: Fun integral problem
First of all we note that $\mathrm{P}(x) + \mathrm{P} \left ( 12 - x \right ) = 12$ since \begin{align*} \mathrm{P}(x) - 12 &= \left ( x -4 \right ) \left ( x - 5 \right ) \left ( x -9 \right ) \\ &= -\left ( 4 - x \right ) \left ( 5- x \right ) \left ( 9 - x \right ) \\ &= - \left [ \le...
- Thu Mar 16, 2023 3:14 pm
- Forum: General Mathematics
- Topic: An inequality
- Replies: 0
- Views: 4756
An inequality
Let $x, y, z$ be positive real numbers. Prove that:
$$ \sqrt{\frac{x}{x+y}} + \sqrt{\frac{y}{y+z}} + \sqrt{\frac{z}{z+x}} \leq \frac{3}{\sqrt{2}}$$
$$ \sqrt{\frac{x}{x+y}} + \sqrt{\frac{y}{y+z}} + \sqrt{\frac{z}{z+x}} \leq \frac{3}{\sqrt{2}}$$
Integral
Let $a\geq 1$. Prove that:
$$\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 )$$
$$\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 )$$
- Sat Nov 27, 2021 7:07 pm
- Forum: Calculus
- Topic: Trigonometric logarithmic integral
- Replies: 2
- Views: 5055
Re: Trigonometric logarithmic integral
We are using the following results: 1. $\displaystyle \sum_{n=1}^{\infty}\frac{x^n \cos (na)}{n}=-\frac{1}{2}\ln \left ( x^2-2x \cos a+1 \right ) \; , \; \left | x \right |<1$ 2. $\displaystyle \sum_{n=1}^{\infty}(-1)^n \frac{\sin n x}{n}=-\frac{x}{2}\; , \; \left | x \right |<\pi$ 3. $\displaystyle...
- Sat Nov 27, 2021 7:02 pm
- Forum: Calculus
- Topic: Trigonometric logarithmic integral
- Replies: 2
- Views: 5055
Re: Trigonometric logarithmic integral
We note that $$\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}$$ Hence, \begin{align*} \int^{\infty}_{0} \frac{1}{x}\ln\left(\frac{x^2+2kx\cdot \cos b+k^2}{x^2+2k x\cdot \cos a+k^2}\right)\,dx &= \int_...
- Sat Nov 27, 2021 6:55 pm
- Forum: Calculus
- Topic: A series involving alternating Harmonic numbers
- Replies: 1
- Views: 5285
Re: A series involving alternating Harmonic numbers
We are proving that $\displaystyle \sum_{n=1}^{\infty} \frac{(-1)^{n+1} \mathcal{H}_n}{n} = \frac{\pi^2}{12} - \frac{\ln^2 2}{2}$. Indeed, note that: \begin{align*} \sum_{n=1}^N\frac{(-1)^{n-1}}{n}H_n &=\sum_{n=1}^N\frac{(-1)^{n-1}}{n^2}+\sum_{n=2}^N\frac{(-1)^{n-1}}{n}H_{n-1}\\ &=\sum_{n=1}...
- Fri Nov 06, 2020 5:35 am
- Forum: Algebraic Structures
- Topic: Not a Hopfian group
- Replies: 1
- Views: 4622
Re: Not a Hopfian group
Well, we define $f:\mathcal{G} \rightarrow \mathcal{G}$ by $f(x)=x^2$ and $f(y)=y$ and extend it to $\mathcal{G}$ homomorphically. Since $\mathcal{G}$ is well defined then $f$ is a surjective because $$f\left ( y^{-1} xy x^{-1} \right ) = x$$ but not an isomorphism because if we take $z=y^{-1} x y$ ...
- Sat Dec 14, 2019 3:16 pm
- Forum: Calculus
- Topic: \(\sum_{n=2}^{\infty} \frac{(-1)^n}{n\log{n}}\)
- Replies: 1
- Views: 4660
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 ... !
$$\int_1^\infty \left( 1+\left(2^{1-s}-1\right)\zeta(s) \right) \, \mathrm{d}s$$
However, the $\zeta$ function does not behave well under integrals. So, I would not expect a closed form to exist ... !