I am unsure how to find the error bounds for Hermite interpolation. I have some kind of idea but I have a feeling that I am going wrong somewhere.

$f(x)=3xe^x-e^{2x}$ with my x-values being 1 and 1.05

My hermite interpolating polynomial is: $H(x)=.7657893864+1.5313578773(x-1)-2.770468386(x-1)^2-4.83859508(x-1)^2(x-1.05)$

Error Bound: $\large{f^{n+1}(\xi)\over (n+1)!}*(x-x_0)(x-x_1)...(x-n)$

$$\large{f^3 (\xi) \over 3!}(x-1)^2(x-1.5)$$


We must find the maximum point of this cubic function which is at $(1.0333,1.8518463*10^{-5})$ $$\large{f^3 (\xi) \over 3!}*1.8518463*10^{-5}$$ Am I on the correct path and How would I continue from here?


1 Answer 1


I think that should be $(x-1)(x-1.05)$ instead of $(x-1)^2(x-1.5)$

  • $\begingroup$ If you want to, you might go on. The further question is "How would I continue from here?" $\endgroup$
    – 311411
    Aug 12, 2021 at 2:44
  • $\begingroup$ I am not sure how to continue, but I just find there maybe a little mistake in the middle step. that I mentioned. I feel it's worth to point out. $\endgroup$
    – Jiaxin Li
    Aug 13, 2021 at 3:05
  • 1
    $\begingroup$ Agreed. Especially since no one said anything for eight years :) $\endgroup$
    – 311411
    Aug 13, 2021 at 3:16

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .