Search found 597 matches
- Wed Nov 01, 2023 8:33 pm
- Forum: Meta
- Topic: Sortables captcha upgrade
- Replies: 0
- Views: 10098
Sortables captcha upgrade
Dear guests, we would like to inform you that we have updated the Sortables Captcha extension due to increase volume of spams. We noticed that in the last days alone $3$ spambots managed to bypass our security measures and create an account here on mathimatikoi.org. As if that weren't enough, they w...
A series
Evaluate the series
$$\mathcal{S} = \sum_{n=1}^{\infty} (-1)^{n-1} \ln \frac{n+1}{n}$$
$$\mathcal{S} = \sum_{n=1}^{\infty} (-1)^{n-1} \ln \frac{n+1}{n}$$
- Wed Apr 12, 2023 8:01 pm
- Forum: Analytic Geometry
- Topic: Vector algebra
- Replies: 0
- Views: 2929
Vector algebra
Let $\mathbf{a} , \mathbf{b}, \mathbf{c}$ be three non coplanar vector. If $\displaystyle{\mathbf{a}' = \frac{\mathbf{b} \times \mathbf{c}}{\left [ \mathbf{a,b, c} \right ]} \; , \; \mathbf{b}' = \frac{\mathbf{c} \times \mathbf{a}}{\left [ \mathbf{a,b, c} \right ]} \; , \; \mathbf{c}' = \frac{\mathb...
- Sun Mar 12, 2023 3:03 pm
- Forum: Calculus
- Topic: An infinite product
- Replies: 0
- Views: 3449
An infinite product
Let $\mathcal{F}_n$ denote the $n$ -th Fibonacci number and $\mathcal{L}_n$ the $n$ – th Lucas. Prove that
$$\prod_{n=1}^{\infty} \left ( 1 + \frac{1}{\mathcal{F}_{2^n +1} \mathcal{L}_{2^n+1}} \right ) = \frac{3}{\varphi^2}$$
$$\prod_{n=1}^{\infty} \left ( 1 + \frac{1}{\mathcal{F}_{2^n +1} \mathcal{L}_{2^n+1}} \right ) = \frac{3}{\varphi^2}$$
- Sun Mar 12, 2023 2:52 pm
- Forum: Meta
- Topic: Forum upgrade to latest version
- Replies: 1
- Views: 12399
Re: Forum upgrade to latest version
Greetings everyone, we are pleased to announce that the forum software has been upgraded to the latest version hardening the security of our website. You will notice that many cosmetic things have been restored to normal. This new version is compatible with php 8.2 that our server is currently runni...
- Wed Nov 16, 2022 7:30 am
- Forum: Meta
- Topic: Forum upgrade to latest version
- Replies: 1
- Views: 12399
Forum upgrade to latest version
Greetings, we have updated the forum to its latest version phpbb 3.3.x. You will find that many bugs have been fixed in this latest version. We would like to also inform you that the ability to add tags has been restored and now it's working flawlessly. You can select among many different tags to ca...
- Sun Apr 10, 2022 6:24 am
- Forum: Complex Analysis
- Topic: Contour integral
- Replies: 1
- Views: 8272
Re: Contour integral
It follows from Taylor's theorem that $f(z)=\sum \limits_{n=0}^{\infty} c_n z^n$ and that the convergence is uniform. Thus, \begin{align*} \frac{1}{2\pi i }\oint \limits_{|z|=1} \frac{\overline{f(z)}}{z-\alpha} \,\mathrm{d}z &=\frac{1}{2\pi i }\oint \limits_{|z|=1} \sum_{n=0}^{\infty} \frac{\ove...
- Fri Nov 06, 2020 11:59 am
- Forum: Linear Algebra
- Topic: Rank of product of matrices
- Replies: 1
- Views: 8732
Re: Rank of product of matrices
It holds that $${\rm nul} (T_1 T_2) \leq {\rm nul} (T_1) + {\rm nul} (T_2)$$ where $T_1, \; T_2$ are the corresponding linear transformations. Proof: The proof of the lemma is based on the rank - nullity theorem. Based upon the above lemma we have that \begin{align*} {\rm rank} \left ( T_1 T_2 \rig...
- Fri Nov 06, 2020 11:57 am
- Forum: Linear Algebra
- Topic: On permutation
- Replies: 1
- Views: 4905
Re: On permutation
The sum of $D(\sigma)$ over the even permutations minus the one over the odd permutations is the determinant of the matrix $A$ with entries $a_{i,j}=\vert i-j\vert$ and this determinant is known to be
$$\det A = (-1)^{n-1} (n-1) 2^{n-2}$$
$$\det A = (-1)^{n-1} (n-1) 2^{n-2}$$
- Fri Nov 06, 2020 6:50 am
- Forum: Competitions
- Topic: An equality with matrices
- Replies: 1
- Views: 6973
Re: An equality with matrices
Let $A, B$ be elements of an arbitrary associative algebra with unit. Then: \begin{align*} \left ( A^{-1} +\left ( B^{-1} - A \right )^{-1} \right )^{-1} &= \left ( A^{-1} \left ( B^{-1} - A \right )\left ( B^{-1} - A \right )^{-1} + A^{-1} A \left ( B^{-1} - A \right )^{-1} \right )^{-1} \\ &am...