I have a question: How can we prove that the Bernstein polynomial $$p_{n}(x)=\sum_{l=0}^{n} e^{l\over n}\begin{pmatrix} n\\ l \end{pmatrix}x^l(1-x)^{n-l}$$ uniformly converges $e^x$ in the interval [0,1]?
My thoughts: given any $\alpha>0$,then we can find a $\beta$>0 such that |$e^x-e^y$|<$\alpha\over 2$, as long as |x-y|<$\beta$, then $$|p_{n}(x)-e^x|=\sum_{|{l\over n}-x|>=\beta} |e^{l\over n}-e^x|\begin{pmatrix} n\\ l \end{pmatrix}x^l(1-x)^{n-l}+\sum_{|{l\over n}-x|<\beta} |e^{l\over n}-e^x|\begin{pmatrix} n\\ l \end{pmatrix}x^l(1-x)^{n-l}$$,where the second sum is smaller than $\alpha \over 2$, but I don't know how to continue? can someone tell me how to prove this question without using stone weierstrass theorem directly. Also you don't need to use my methods, can someone tell me how to prove it ?
Since we have known that $$p_{n}(x)=\sum_{l=0}^{n} e^{l\over n}\begin{pmatrix} n\\ l \end{pmatrix}x^l(1-x)^{n-l}=(e^{1\over n}x+1-x)^{n}$$.
If we don't need to use the stone weierstrass theorem, then we can change our mind let us to prove whether $(e^{1\over n}x+1-x)^{n}$->$e^x$ uniformly.
So we can just need to prove that $n log( e^{1\over n}x+1-x)$ ->x uniformly, so we can just use taylor series to prove it, the next proof just some simple computation.
If we use taylor series, we can easily prove $n log( e^{1\over n}x+1-x)$ ->x uniformly. I think it is a simple method