Алгебраическая теорема Вейерштрасса

$\forall{f(x)}~-$ непрерывна на $[a,b]$ $\exists{\{P_{n}(x)\}}_{n=1}^{\infty}~~P_{m}(x) = \sum_{k=1}^{n} a_{k}x^{k}~~~~~P_{n}(x) \rightrightarrows_{n \to \infty}^{[a,b]} f(x)$ $\forall{\varepsilon>0}~~\exists{N}~~\forall{n>N}~~\forall{x \in [a,b]}\mathpunct{:}$ $$|P_{n}(x)-f(x)|<\varepsilon$$

$b_{n,k} = x^{k}(1-x)^{n-k},~~~~n \in N, k = 0,1,2,3\dots n$ $b^{'}_{n,k} = x^{k-1}(1-x)^{n-k-1}$ $(k(1-x) - (1-x))^{n-k-1}$ $x = 0,~x=1~-min$ $k-kx-xn-kx=k-xn=0,~~~x=\dfrac{k}{n}~- \max$ $ 1.$ Пусть $[a,b] = [0,1]$ $B_{n}(x) = \sum_{k=0}^{n} C_{n}^{k} x^{k}(1-x)^{n-k}$ $P_{n}(x) = \sum_{k=0}^{n} f\left( \dfrac{k}{n} \right) * b_{n,k} (x)$

1) $\sum_{k=1}^{n} b_{n,k}(x)=1$ 2) $\sum_{k=1}^{n} \dfrac{k}{n} b_{n,k} (x) = x$ 3) $\sum_{k=1}^{n} \left( \dfrac{k}{n} - x \right)^{2} - b_{n,k}(x) = \dfrac{x(1-x)}{n}$

${} 1.~ {}$$\sum_{k=-0}^{n} C_{n}^{k} x^{k} (1-x)^{n-k} = (x+(1-x))^{n} = 1^{n} = 1$ $ 2.~$$\sum_{k=-0}^{n} \dfrac{k}{n} C_{n}^{k} x^{k} (1-x)^{n-k}=$ $=\sum_{k=0}^{n} \dfrac{k}{n} * \dfrac{n!}{k!(n-k)!} x^{k}(1-x)^{n-k}$ $= x \sum_{k=0}^{n} \dfrac{(n-1)!}{(k-1)!(n-k)!} x^{k-1}(1-x)^{n-k}=$ $= x \sum_{k=0}^{n} \dfrac{(n-1)!}{k!(n-1-k)!} x^{k}(1-x)^{n-1-k} =$ $= x \sum_{k=0}^{n} C^{k}_{n} x^{k} (1-x)^{n-1-k}$ $ 3.~$$\left( \sum_{k=0}^{n} \left( \dfrac{k}{n} -x \right) b_{n,k}(x) \right) = 0$ $-\sum_{k=0}^{n} b_{n,k}(x) + \sum_{k=0}^{n} \left( \dfrac{k}{n} - x \right)^{2} b_{n,k}(x) \dfrac{n}{x(1-x)} = 0$ Что такое $b_{n,k}\mathpunct{:}~~$ $$b_{n,k} = C_{n}^{k} (k-nx)x^{k-1}(1-x)^{n-k-1} = C_{n}^{k} x^{k}(1-x)^{n-k} \left( \dfrac{k}{n} -x \right) \dfrac{n}{x(1-x)}$$ $\square$