Taylor approximation.

**Pinkk** · October 4th 2010, 05:30 PM

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

So I know how to find the taylor expansion of $e^{-x^{2}}$ 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.

**HallsofIvy** · October 5th 2010, 05:05 AM

You have that the error in the polynomial value is less than $\frac{M}{(k+1)!}|h|^{k+1}$ (M is an upper bound on $e^{-x^2}$ ) 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.

**Pinkk** · October 5th 2010, 12:24 PM

Ah, okay. And I guess to ensure that my approximation is to three decimal places, I show that that remainder is less than $0.5 \times 10^{-4}$ , right?

**HallsofIvy** · October 6th 2010, 04:02 AM

Yes.

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