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## Search found 590 matches

- Fri Nov 06, 2020 11:59 am
- Forum: Linear Algebra
- Topic: Rank of product of matrices
- Replies:
**1** - Views:
**1238**

### 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:
**1357**

### 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:
**998**

### 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} \\ &=\left...

- Fri Nov 06, 2020 6:36 am
- Forum: Algebraic Structures
- Topic: Sum equals to zero
- Replies:
**1** - Views:
**638**

### Re: Sum equals to zero

Let us suppose that $|\mathcal{G}| = \kappa$ and $x= \frac{1}{\kappa} \sum \limits_{g \in \mathcal{G}} g $. We note that for every $h \in \mathcal{G}$ the depiction $\varphi: \mathcal{G} \rightarrow \mathcal{G}$ such that $\varphi(g)=h g $ is $1-1$ and onto. Thus: \begin{align*} x^2 &=\left ( \frac{...

- Fri Nov 06, 2020 5:28 am
- Forum: Algebraic Structures
- Topic: Isomorphic groups
- Replies:
**1** - Views:
**1034**

### Re: Isomorphic groups

Using $x^{-1}yx = y^{-1}$ or equivalently $yx = xy^{-1}$ we can write each element of $\mathcal{Q}_{2^n}$ in the form $x^ry^s$ where $r,s \in \mathbb{N} \cup \{0\}$. Using $x^2 = y^{2^{n-2}}$ we may assume that $r\in \{0,1\}$. Using $y^{2^{n-1}} = 1$ we may also assume that $s\in \{0,1,\ldots,2^{n...

- Tue Jun 09, 2020 11:33 am
- Forum: Functional Analysis
- Topic: Inner product space
- Replies:
**1** - Views:
**770**

### Re: Inner product space

**Hint:**Equality holds when vectors are parallel i.e, $u=kv$, $k \in \mathbb{R}^+$ because $u \cdot v= \|u \| \cdot \|v\| \cos \theta$ when $\cos \theta=1$, the equality of the Cauchy-Schwarz inequality holds.

- Sun Dec 15, 2019 10:51 pm
- Forum: Calculus
- Topic: Digamma and Trigamma series
- Replies:
**0** - Views:
**2445**

### Digamma and Trigamma series

Let $\psi^{(0)}$ and $\psi^{(1)}$ denote the digamma and trigamma functions respectively. Prove that:

\[\sum_{n=1}^{\infty} \left ( \psi^{(0)}(n) - \ln n + \frac{1}{2} \psi^{(1)}(n) \right ) = 1+ \frac{\gamma}{2} - \frac{\ln 2\pi}{2}\]

where $\gamma$ denotes the Euler – Mascheroni constant.

\[\sum_{n=1}^{\infty} \left ( \psi^{(0)}(n) - \ln n + \frac{1}{2} \psi^{(1)}(n) \right ) = 1+ \frac{\gamma}{2} - \frac{\ln 2\pi}{2}\]

where $\gamma$ denotes the Euler – Mascheroni constant.

- Sun Oct 13, 2019 1:06 pm
- Forum: Archives
- Topic: Mathematical newspaper
- Replies:
**1** - Views:
**1998**

### Re: Mathematical newspaper

The second issue of the JoM Journal is now out. You may download it from this web address. Hope you find something interesting within its $97$ pages.

- Sat Oct 12, 2019 12:26 pm
- Forum: Blog Discussion
- Topic: A logarithmic Poisson integral
- Replies:
**1** - Views:
**1684**

### A logarithmic Poisson integral

A logarithmic Poisson integral by Tolaso J Kos Let $a \geq 0$. We will prove that $$I(a)=\int_{0}^{\pi}\ln\left(1-2a\cos x+a^2\right) \, \mathrm{d}x = \left\{\begin{matrix} 0 & , & \left | a \right | \leq 1 \\ 2 \pi \ln \left | a \right | &, & \text{otherwise} \end{matrix}\right.$$ Background: This...

- Thu Oct 03, 2019 9:38 am
- Forum: Meta
- Topic: Welcome to the new and improved mathimatikoi.org
- Replies:
**7** - Views:
**6378**

### Re: Welcome to the new and improved mathimatikoi.org

As of today we have the ability to include xy.pic into our posts. Unfortunately, the rendering of all equations takes a little time to complete. We'll see if we can overcome this problem.