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?

  • $\begingroup$ Are you sure that the third derivative doesn’t exist? $f’(x) = 3x^2$, $f^2(x)= 6x$, and $f^3(x) =6$ $\endgroup$ – Joe Goldiamond Aug 11 '18 at 12:15
  • $\begingroup$ 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? $\endgroup$ – Jakov Zingerman Aug 11 '18 at 12:25
  • $\begingroup$ Correct, but I replied on your statement that the third derivative didn’t exist. So you have to adjust that sentence a little.... $\endgroup$ – Joe Goldiamond Aug 12 '18 at 9:53

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