Runge's phenomenon-question

Question

I am looking at the Runge's phenomenon and I have a question. We have the interval $[a,b]=[-1,1]$ and $f\in C^{n+1}[-1,1]$. We know that $\forall x \in [-1,1] $ $\exists$ $\xi\in[-1,1]$ so that: $$f(x)-p_{n}(x)=\frac{1}{(n+1)!}f^{(n+1)}(\xi)\varPhi_{n+1}(x)$$ where $\varPhi_{n+1}(x)=\prod_{i=0}^{n}(x-x_{i})$ So,$$||f-p_{n}||_{\infty}=max|f(x)-p_{n}(x)|\leq\frac{1}{(n+1)!}||\varPhi_{n+1}||_{\infty}||f^{n+1}||_{\infty} $$ Then,we suppose that $x_{i}=-1+ih , h=\frac{b-a}{n}$ Then we say that $||\varPhi_{n+1}||_{\infty}\leq \frac{n!}{4}h^{n+1}$ ,but I haven't understood why..Could you explain it to me?

Answer 1

Because

\begin{align} |Φ_{n+1}(x)|&=(x-0h)\,(x-1h)\dots(x-jh)\;((j+1)h-x)\dots(nh-x)\\ &\le((j+1)h)\,(jh)\dots(2h)\;\left(\tfrac{(x-jh)+((j+1)h-x)}2\right)^2\;(2h)\dots((n-j)h)\\ &=\tfrac14(j+1)!(n-j)!\;h^{n+1} \end{align}

for all $x\in[jh,(j+1)h]$, $j=0,1,\dots,n-1$.