Somemathhere

          Let $Z$ be a random variable which takes values in the interval [0,1] with continuous density function $f_Z$. Now we have a coin with probability for heads equal to $Z$. Let $X_n$ be the number of heads when flipping the coin $n$ times. Now consider the function                     $$f_n(x)=\begin{cases}                       0, & x <0   \\                       0, & x\ge 1 \\                         (n+1)\operatorname{P}(X_n=d),& x\in \Big[\frac{d}{n+1}, \frac{d+1}{n+1}\Big)\ \text{for some }d\in \{0,1,\ldots,n\}              \end{cases}$$           It turns out that $f_n$ converges to $f_Z$ in some sense. We will prove now pointwise convergence.                 Proof. Fix $x\in(0,1)$ and we should prove that $f_n(x)\to f_Z(x)$ as $n$ tends to infinity (from now on all our considerations will be for the intervarl $[0,1]$). For a fixed $n$ $x \in \Big[\frac{\left \lfloor{(n+1)x}\right \rfloor }{n+1}, \frac{\left \lfloor{(n+1)x}\right \rfloor+1}{n+1}\B

Given $n$ complex numbers $z_1,z_2,\ldots,z_n$ such that $$z_1^k+z_2^k+\cdots +z_n^k=0$$ for all $k\in\mathbb{N}$. Prove that $z_1=z_2=\cdots=z_n=0$. Solution: Lets call the sum of $k$-th powers $S_k$. Since all $S_k$-s are homogeneous polynomials we know that for any $\lambda\in\mathbb{C}$ the numbers $\lambda z_1,\lambda z_2,\ldots ,\lambda z_n$ will also fulfill these equations and call the corresponding polynomials $S_k(\lambda)$. Choose $\lambda'$, such that $|\lambda' z_i|<1$ for all $i=1,2,\ldots,n$. Now for $\lambda: \ |\lambda|\le |\lambda'|$ examine $$f(\lambda;z_i)=\sum_{k=1}^{\infty}\frac{(\lambda z_i)^k}{k}$$ This is a series for logarithm, so $$S:=\sum_{i=1}^{n}f(\lambda;z_i)=-\sum_{i=1}^{n}\log(1-\lambda z_i)=-\log((1- \lambda z_1)\cdots(1- \lambda z_n))$$ On the other hand $$S=\sum_{i=1}^{n}\sum_{k=1}^{\infty}\frac{(\lambda z_i)^k}{k}=\sum_{k=1}^{\infty}\frac{S_k(\lambda)}{k}=0$$ Thus we have $\log((1-\lambda z_1)\cdots(1-\lambda z_n))=0$ hence $$g(\l

Дадена е редица, на която първият член е $1$, а всеки следващ се получава като към предишния се добави най-малката му ненулева цифра. Да се определи колко цифри има числото на позиция $9\cdot 1000^{1000}$

Let $$x*y=\frac{x + y}{1 + xy}$$ Evaluate $$(\cdots(2*3)*4)*\cdots*2018)$$ A solution with polynomials is not difficult to find. Insightful solution with hyperbolic functions also available, thans to Zamkovoy, Bozhilov.

Нека $X$ и $Y$ са независими стандартно нормално разпределени случайни величини. Да се пресметне $\text{E}(X|X>Y)$ Решение. Да означим с $F$ и $f$ функциите съответно на разпределение и на плътност на стандартното нормално разпределение (съответно на величините $X$ и $Y$). Нека $Z=X|_{X>Y}$. Така търсим $\text{E}(Z)$. За целта да пресметнем функцията на разпределение $F_Z$ и функцията на плътност $f_Z$. $F_Z(z)=\text{P}(Z<z)=\operatorname{P}(X<z|X>Y)=\frac{\text{P}(X<z \bigcap X>Y)}{P(X>Y)}$. Заради симетрия $\text{P}(X>Y)=\text{P}(Y>X)\Rightarrow \text{P}(X>Y)=\frac{1}{2}$. Ще пресметнем $\text{P}(X<z \bigcap X>Y)$ чрез формулата за пълна вероятност. Зaбелязваме, че $$\text{P}(X<z \bigcap X>Y|X=s)=\begin{cases} 0, & s>z \\ \text{P}(Y<s), & s\le z \end{cases}$$ Тогава $$\text{P}(X<z \bigcap X>Y)=\int_{-\infty}^{\infty}\text{P}(X<z \bigcap X>Y|X=s)f(s)\mathrm{d}s=\int_{-\infty}^{z}F(s)f(s)\mathrm{d}s$$ Така $$F_Z(z)