Prove $\max_{|x|\leq \frac13}|f(x)-p_2(x)|\leq \frac{1}{2\cdot 3^{\frac{11}{2}}}\,\max_{|x|\leq \frac13}|f^{(4)}(x)|$ for Runge's function.

Question

Let $f(x)=\frac{1}{1+x^2}$ be Runge's function. If $p_2(x)$ is the interpolating polynomial of $f$ regarding the nodes $-\frac13,\,0,\frac13$, prove that:

$$\max_{|x|\leqslant \frac13}|f(x)-p_2(x)|\leqslant \frac{1}{2\cdot 3^{\frac{11}{2}}}\,\max_{|x|\leqslant \frac13}|f^{(4)}(x)|.$$

Attempt. Since $f$ is $C^3$ we use the error estimate:

$$\max_{|x|\leqslant \frac13}|f(x)-p_2(x)|\leqslant \frac{1}{3!} \max\left|x\left(x-\frac13\right)\left(x+\frac13\right)\right|\,\max_{|x|\leqslant \frac13}|f'''(x)|,$$

which involves $f'''$, instead of $f^{(4)}$. Even if we use the estimate $\left|\prod_{i=0}^n(x-x_i)\right|\leqslant \frac{n!}{4}\,h^{n+1}$ for $n=2$ and $h=\frac13$ (nodes are equidistant), we would get the estimate $\frac{1}{4\cdot 3^4}\,\max_{|x|\leqslant \frac13}|f'''(x)|,$ still involving $f'''$. At first, I thought that Runge's function seems irrelevant, but is there some property (even etc) that enables us to use $f^{(4)}$?

Thanks for the help.

Answer 1

You need to regard your interpolation scheme as $-\frac{1}{3}$, $0$, $0$, $\frac{1}{3}$. Then you will find the leading Newton form coefficient

$f[-\frac{1}{3}, 0, 0, \frac{1}{3}] = \frac{1}{h^3}\left( f(\frac{1}{3}) - 3f(0) + 3f(0) - f(-\frac{1}{3})\right) = 0$.

which reduces your polynomial to $p_2$.