# tex-sandbox (from)

\begin{aligned} 1&=sF_n+tF_{n+1}\\ &=s(F_{n+2}-F_{n+1})+tF_{n+1}\\ &=sF_{n+2}+(t-s)F_{n+1}\;, \end{aligned}

$\left(\begin{array}{cc}1&2\\3&4\end{array}\right)$

MATH-LATEX

$\Sigma\Pi\Gamma$

$\bullet\bullet\mbox{coset maps}\bullet\bullet$

$^{a}f(H_x)=f(Hxa)$

$f_a(H_x)=\overline{x}a{\overline{xa}}^{-1}$

$f_{ab}(H_x)=f_a (Hx)\ ^{a}f_b(Hx)$

$f_{ab}(H_x)=f_a (Hx)\ f_b(Hxa)$

$f_{ab}(H_x)=\overline{x}a{\overline{xa}}^{-1}\overline{xa}b{\overline{xab}}^{-1}$

$f_{ab}(H_x)=\overline{x}ab{\overline{\bar{x}ab}}^{-1}$

$f_{abc}(H_x)=f_a (Hx)\ ^{a}f_b(Hx)\ ^{ab}f_c(Hx)$

$f_{abc}(H_x)=f_a (Hx)\ f_b(Hxa)\ f_c(Hxab)$

$f_{abcd}(H_x)=f_a (Hx)\ ^{a}f_b(Hx)\ ^{ab}f_c(Hx)\ ^{abc}f_d(Hx)$

$f_{abcd}(H_x)=f_a (Hx)\ f_b(Hxa)\ f_c(Hxab)\ f_d(Hxabc)$

$\Box$

$\Xi$

$\int_a^bf(x)dx=F(b)-F(a)$

$b_t={C^s}_te_s$

$b_t={C^s}_te_s$

$\begin{pmatrix}x\y\end{pmatrix}\stackrel{T}\mapsto\begin{pmatrix}1 & 2\3 & 4\-6 & 5 \end{pmatrix}\begin{pmatrix}x\y\end{pmatrix}=\begin{pmatrix}x+2y\3x+4y\-6x+5y\end{pmatrix} &bg=ccccff&fg=000066&s=2$

$\cdots\left>\sqsubset\Box\overbrace{\underbrace{\Omega\qquad\Lambda\Phi\Box\Box\Box\Phi\Lambda\qquad\Omega}_{\sqcap}}^{\sqcup} \Box\sqsupset\right>$

A

$D_TT=\left(\begin{array}{ccc}\frac{\partial}{\partial x}(\frac{dx}{dt})&\frac{\partial}{\partial y}(\frac{dx}{dt})&\frac{\partial}{\partial z}(\frac{dx}{dt})\\\frac{\partial}{\partial x}(\frac{dy}{dt})&\frac{\partial}{\partial y}(\frac{dy}{dt})&\frac{\partial}{\partial z}(\frac{dy}{dt})\\\frac{\partial}{\partial x}(\frac{dz}{dt})&\frac{\partial}{\partial y}(\frac{dz}{dt})&\frac{\partial}{\partial z}(\frac{dz}{dt})\end{array}\right) \left(\begin{array}{c}\frac{dx}{dt}\\\frac{dy}{dt}\\\frac{dz}{dt}\end{array}\right)$

$=\left(\begin{array}{c}\frac{\partial}{\partial x}(\frac{dx}{dt})\frac{dx}{dt}+\frac{\partial}{\partial y}(\frac{dx}{dt})\frac{dy}{dt}+\frac{\partial}{\partial z}(\frac{dx}{dt})\frac{dz}{dt}\\\frac{\partial}{\partial x}(\frac{dy}{dt})\frac{dx}{dt}+\frac{\partial}{\partial y}(\frac{dy}{dt})\frac{dy}{dt}+\frac{\partial}{\partial z}(\frac{dy}{dt})\frac{dz}{dt}\\\frac{\partial}{\partial x}(\frac{dz}{dt})\frac{dx}{dt}+\frac{\partial}{\partial y}(\frac{dz}{dt})\frac{dy}{dt}+\frac{\partial}{\partial z}(\frac{dz}{dt})\frac{dz}{dt}\end{array}\right)$

$=T'$

$\frac{d}{dx}\left({2n\choose n}^{-1}x^{n+1}\right)=(n+1){2n\choose n}^{-1}x^n$

$\frac{d}{dx}\left(\frac{x^{n+1}}{{2n\choose n}}\right)=\frac{(n+1)x^n}{{2n\choose n}}$

SUM OF RECIPROCALS: from a given strictly increasing sequence $n\to a_n\in\mathbb{Z}$, one can ask for a generating function such that its expansion at the origin throws  a series $\sum_{n=0}^{\infty}\frac{x^n}{a_n}$ with a non zero convergence radius.

Are there names associated to this game? Euler, Gauss, Riemann, Dirichlet, Sprugnoli, Iñiguez,…

An example $\sum_{n=0}^{\infty}\frac{x^n}{(n+1)^n}$

$\Sigma\Pi\Gamma$

$\int_a^bf(x)dx=F(b)-F(a)$

$b_t={C^s}_te_s$

$\left\langle\left(\begin{array}{c}v^1\\ v^2\\\vdots\\v^n\end{array}\right),\left(\begin{array}{c}w^1\\ w^2\\\vdots\\w^n\end{array}\right)\right\rangle=v^1w^1+v^2w^2+\cdots+v^nw^n$

$\left(\begin{array}{c}v^1\\v^2\\\vdots\\v^n\end{array}\right)\mapsto[a_1,a_2,...,a_n]\left(\begin{array}{c}v^1\\v^2\\\vdots\\v^n\end{array}\right) =a_1v^1+a_2v^2+\cdots+a_nv^n=a_sv^s$

$[g_{ij}]=\left(\begin{array}{cccc}g_{11}&g_{12}&\cdots&g_{1n}\\g_{21}&g_{22}&\cdots&g_{2n}\\\vdots&&&\vdots\\g_{n1}&g_{n2}&\cdots&g_{nn}\end{array}\right)$

 $\left(\begin{array}{cc}1&0\\1&1\end{array}\right)$ $\left(\begin{array}{cc}1&0\\-1&1\end{array}\right)$ $\left(\begin{array}{cc}1&-1\\1&1\end{array}\right)$ $\left(\begin{array}{cc}0&1\\-1&0\end{array}\right)$ $\left(\begin{array}{cc}1&0\\3&1\end{array}\right)$ $\left(\begin{array}{cc}1&0\\-1&1\end{array}\right)$ $\left(\begin{array}{cc}1&0\\23&1\end{array}\right)$ (1,1,1) (1,1,2) (1,1,3) (1,1,3) (1,1,3) (1,1,3) (1,1,3)

### 2 responses to “tex-sandbox (from)”

1. $(\mathbb{H}\setminus\{\left[\begin{matrix}0 & 0\\ 0 & 0\end{matrix}\right]\},\cdot)$

2. Manuel

Faltaba checar 3 axiomas de espacios vectoriales:

$(\alpha+\gamma)B(v,w)=(\alpha+\gamma)v^tBw$
$=(\alpha v^tBw)+(\gamma v^tBw)$
$=\alpha B(v,w)+\gamma B(v,w)$

$\alpha(\gamma B(v,w))=\alpha(\gamma v^tBw)$
$=\alpha\gamma v^tBw$
$=\alpha\gamma B(v,w)$

$1B(v,w)=1(v^{t}Bw)$
$=v^{t}Bw$
$=B(v,w)$