Блог пользователя gatrmani

Need someone who can teach me everyday master
Автор gatrmani, 8 часов назад, По-английски

need someone who can teach me every day complex code and math equations and level up in contests fast Greatly Appreciated ... Thank You master

$$$\widehat F(z, t) = \sum_{n\ge 1}\sum_{k=1}^n A_{n,k}\frac{z^nt^k}{n!}$$$
$$$\sum_{j=1}^m \sum_{S \subseteq \{1, 2, \cdots m-1\}} (-1)^{|S|} Q(S, z, t, j)$$$
$$$\left([z^n]\frac{t(\mathrm e^{z(1-t)}-1)}{(1-z) (1-t \mathrm e^{z(1-t)})}\right) + 1 = (1-t)[z^n] \frac 1{(1-z)(1-t\mathrm e^{z(1-t)})}$$$
$$$[z^n]\frac 1{(1-z)(1-t\mathrm e^{z(1-t)})} = (1-t)^n[u^n] \frac1{(1-\frac u{1-t})(1-t\mathrm{e}^u)}$$$
$$$\begin{aligned} (1-t)[z^n] \frac 1{(1-z)(1-t\mathrm e^{z(1-t)})} &= [u^n]\frac{(1-t)^{n+2}}{(1-\frac{t}{1-u})(1-t\mathrm e^u)(1-u)}\\&= (1-t)^{n+2} [u^n] \left(\frac{-\mathrm e^u}{\left(\mathrm e^u u-\mathrm e^u+1\right) \left(1-t \mathrm e^u\right)}+\frac{\frac{1}{1-u}}{\left(\mathrm e^u u-\mathrm e^u+1\right) (1-\frac{t}{1-u})}\right)\end{aligned}$$$
$$$\Delta_{X_{d-1},X_d}\colon K[ [ X_1,\dots,X_d ] ] \to K[ [ X_1,\dots,X_{d-2},U ] ]$$$
$$$\left(\sum_{i_1,\dots,i_d\geq 0} f_{i_1,\dots,i_d} X_1^{i_1}\cdots X_d^{i_d}\right) \mapsto \sum_{i_1,\dots,i_{d-2},j\geq 0} f_{i_1,\dots,i_{d-2},j,j} X_1^{i_1}\cdots X_{d-2}^{i_{d-2}} U^j.$$$
$$$\begin{align*} &\quad \langle g_m(x)(1-x)^m, 1 + \cdots + x^m \rangle\\ &= \sum g_{m,k} \langle x^k(1-x)^m, 1+\cdots+x^m \rangle\\ &= \sum g_{m,k} [x^k] x(x-1)^{m-1}\\ &= \langle g_m(x), x(x-1)^{m-1} \rangle\\ &= \langle h_m(u), x(x-1)^{m-1} \rangle\\ &= \langle h_m(te^t), x(x-1)^{m-1} \rangle. \end{align*}$$$
$$$\begin{align*} &\quad \langle t^k, x(x-1)^{m-1} \rangle\\ &= \left\langle \left(\frac{x}{1-x}\right)^k, x(x-1)^{m-1} \right\rangle\\ &= [x^m]\left(\frac{x}{1-x}\right)^k \cdot (1-x)^{m-1}\\ &= [x^m] x^k (1-x)^{m-k-1}\\ &= [k=m]. \end{align*}$$$
$$$\exp\left( \sum_{k\geq 3} \frac{(n-1)!x^n/2}{n!} \right) = \exp\left(-\frac x 2 - \frac{x^2}4\right)\frac 1{\sqrt{1-x}},$$$
$$$a_n = \frac1{2^{2n}}\sum_{i,j,k} 2^i (-1)^j (-1)^k \binom{n}{i,j,k,n-i-j-k}2^i2^{2j}\binom{n-i-j}{k} k! 2^k (2n-2i-j-2k)!$$$
$$$f(x) = \sum_{i=1}^{n-1} f(q^i) \frac{\prod_{j=0}^{n-1} (x-q^j)}{(x-q^i) \prod_{j\in \{0,\dots,\overline i,\dots,n-1\}} (q^i-q^j)},$$$
$$$\sum_i q_i(t-1) \partial_t^i F_1 = \sum_{1\leq j\leq \ell} \epsilon_j (t-1)^{k+j}.$$$
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