# Hermite interpolation

I need a help to my problem, I would be grateful if anyone could help.

Let $\epsilon \in [0,1]$ and for an integer $n$ we consider a set of nodes $T_n={t_0,t_1,....t_n}$. We define the function $f(x)=\frac{1}{x^2}$ in the interval $[\epsilon,1]$ and we consider the Hermite interpolation polynomial $p_{2n-1}(f;.)$ corresponding to the function $f$ and the set of nodes $T_n$. Applying Rolles theorem several times into the function $g_x(t)=f(t)-p_{2n-1}(f;t)-\frac{f(x)-p_{2n-1}(f;x)}{\prod_{i=1}^n(x-t_i)^2}\prod_{i=1}^n(t-t_i)^2$ we can prove that. for ever $x \in [\epsilon,1]$ there exist $\tau(x)$ sth: $f(x)=p_{2n-1}(f;x)+\frac{f^{(2n)}(\tau(x))}{(2n)!}\prod_{i=1}^n(x-t_i)^2$.

Now my question is to prove or disaprove that $\tau(x)>> \epsilon$ as $\epsilon$ goes to $0$ (for this case I think the answer will be no for $x\sim \epsilon$). and the same question if I consider that nodes set $T_n(\epsilon)$ depnds on $\epsilon$ (for example if we consider that $t_0(\epsilon)\sim\epsilon$ as $\epsilon$ goes to $0$ ).

Sorry that I don't have precise question, because it is an open problem and any suggestion will be helpfull.

Thank you.

• You mean that $f-p_{2n-1}$ has a double root in each node? – Fedor Petrov Oct 19 '15 at 8:30
• $p_{2n-1}$ is the Hermite interpolation polynomial of degree $2n-1$ statisfying: $p_{2n-1}(t_k)=f(t_k)$ and $p_{2n-1}^{'}(t_k)=f^{'}(t_k)$ for each $t_k$ – spray_user Oct 19 '15 at 9:03