Wikipedia states that Bernstein's constant is defined as:
Let $E_n(f)$ be the error of the best uniform approximation to a real function $f$ on the interval $[-1,1]$ by real polynomials of no more than degree $n$. In the case of $f(x)=|x|$, we have that $$\beta=\lim_{n\rightarrow\infty}2nE_{2n}(f)$$I am having trouble visualizing this number. A more mathematical definition for $E_n(f)=\sup\left(|f(x)-f_p(x)|\bigg|x\in[-1,1]\right)$where $f_p(x)$ is the polynomial the approximates $f(x)$. This polynomial could be a Bernstein polynomial as they approximate functions well, so $$f_p(x)=\sum_{k=0}^n f\left(\frac kn\right)\binom{n}{k}x^k(1-x)^{n-k}$$ For $f(x)=|x|$, we get that $$E_n(f)=\sup\left(\left||x|-\sum_{k=1}^n \frac kn\binom{n}{k}x^k(1-x)^{n-k}\right|\text{ where }x\in[-1,1]\right)$$I graphed this on Desmos. The error is supposed to approach $0$ so that $\beta$ could exist. But for some reason, when $x\in[-1,1]$, $E_n(f)$ approaches $1$. What is going wrong here?
Edit: I realized that $E(x)$ actually approximates $|x|$ pretty well, so I wrote $2a||x|-E(x)|$ instead, but it still explodes.
The error is in fact 1. This becomes obvious if you note that $$ \sum_{k=0}^n \frac kn \binom{n}{k}x^k (1-x)^{n-k} = x. $$