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Suppose that we got $f\in\mathcal{C}^{n+1}[-1,1]$ and $x_{0},...,x_{n}\in [-1,1]$ , uniformly separated.With $x_{i}=-1+ih, \ \ i=0,..,n$ , $h=2/n$ and $p_{n}\in \mathcal{P}_{n}$(polynomial).

We also know that the approximation of error can be expressed as

$f(x)-p_{n}(x)=\frac{1}{(n+1)!}\Phi_{n+1}(x)f^{(n+1)}({\xi})$, $\xi \in [-1,1] , x\in[-1,1]$

where $\Phi_{n+1}(x)=\prod_{i=0}^{n}(x-x_{i})$

My problem is that in my notes I have that $\left \| \Phi_{n+1} \right \|_{\infty}\leq \frac{n!}{4}h^{n+1}$, but there is no proof and I really don't know how to prove it.

Jonathan1234
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1 Answers1

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Similar to here in Theorem 2, we can estimate $\|\Phi_{n+1}\|_{\infty}$ the following way:

First set $h := 2/n$. Assume $x_j \leq x < x_{j+1}$. We see that

$$|\Phi_{n+1}(x)| = |x-x_0|\dots|x-x_n|$$

Now, each $|x-x_j|$ is smaller or equal to $jh$ respectively. This is because $x_{j+1}-x_j = h$ by construction. So we get

\begin{align} |\Phi_{n+1}(x)| &\leq jh \cdot (j-1)h \cdot \dots 2h \cdot (x-x_j) (x_{j+1} - x) \cdot 2h \dots \cdot(n+1-j)h \\ &= (x-x_j)(x_{j+1}-x)h^{n-1}j!(n+1-j)!. \end{align}

Now since we had $x_j \leq x < x_{j+1}$ by assumption we can further estimate the term:

\begin{align} (x-x_j)(x_{j+1}-x)h^{n-1}j!(n+1-j)! \leq \frac{(x_{j+1} - x_j)^2}{4} h^{n-1} j!(n+1-j)!\leq \frac{n!}{4}h^{n+1}. \end{align}

TheWaveLad
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