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:
$\large{f^{n+1}(\xi)\over (n+1)!}*(x-x_0)(x-x_1)…(x-n)$
$$\large{f^3 (\xi) \over 3!}(x-1)^2(x-1.5)$$
$(x-1)^2(x-1.5)=x^3-3.05x^2+3.1x-1.05$
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?
Best Answer
I think that should be $(x-1)(x-1.05)$ instead of $(x-1)^2(x-1.5)$