# newton's interpolation error for non-differentiable function

I was given this function: $$f(x) = \begin{cases} x^3 & {\text{if}}\ x>0 \\ 0 & {\text{if}}\ x\leq0\ \end{cases}$$

and I was asked to give an upper bound on it's interpolation error expressed only by h when $x\in[-h,h]$ and using the known points $f(-x_0)=-h, f(x_1)=0, f(x_2)=h$.

So by newton's interpolation method the interpolation polynomial is: $$P_2(x) = f(x_0)+f[x_0,x_1,x_2](x-x_0)(x-x_1)$$

and the error is given by: $$E(x) = f[x_0,x_1,x_2,x](x-x_0)(x-x_1)(x-x_2)$$

But i have no idea of what is the way to get an upper bound on $|E(x)|$ in this case because clearly the third derivative doesn't exist here and therefore i cannot use the formula $f[x_0,x_1,x_2,x]=f'''(\xi)/3!$.

Any ideas?

• Are you sure that the third derivative doesn’t exist? $f’(x) = 3x^2$, $f^2(x)= 6x$, and $f^3(x) =6$ – Joe Goldiamond Aug 11 '18 at 12:15
• The third derivative at the point $x=0$ is not defined because of the different behavior of the function from the two sides of this point. Am i missing something? – Jakov Zingerman Aug 11 '18 at 12:25
• Correct, but I replied on your statement that the third derivative didn’t exist. So you have to adjust that sentence a little.... – Joe Goldiamond Aug 12 '18 at 9:53