# Taylor approximation.

• Oct 4th 2010, 05:30 PM
Pinkk
Taylor approximation.
Use Taylor approximation on $\displaystyle e^{-x^{2}}$ to compute $\displaystyle \int_{0}^{1} e^{-x^{2}}\,dx$ to three decimal places and prove the accuracy of your answer using the theorem that if $\displaystyle f$ is of class $\displaystyle C^{k+1}$ on an interval $\displaystyle I$ and $\displaystyle |f^{(k+1)}(x)|le M$ for $\displaystyle x\in I$, then $\displaystyle |R_{a, k}(h)|\le \frac{M}{(k+1)!}|h|^{k+1}$.

So I know how to find the taylor expansion of $\displaystyle e^{-x^{2}}$ about zero and then integrate each term, but I do not know how to approximate to ensure that the approximation is good to three decimal places nor do I know how to apply that theorem to prove it. Any help would be appreciated.
• Oct 5th 2010, 05:05 AM
HallsofIvy
You have that the error in the polynomial value is less than $\displaystyle \frac{M}{(k+1)!}|h|^{k+1}$ (M is an upper bound on $\displaystyle e^{-x^2}$) so the error in the integral is less than the integral of that from 0 to 1. And since that is a constant, it is just that number times the length of the interval, which is 1.
• Oct 5th 2010, 12:24 PM
Pinkk
Ah, okay. And I guess to ensure that my approximation is to three decimal places, I show that that remainder is less than $\displaystyle 0.5 \times 10^{-4}$, right?
• Oct 6th 2010, 04:02 AM
HallsofIvy
Yes.