1. ## Series Approximation Help

Do i need a calculator to find the approximation with the error 10^-8?

The series i got from f(x) is the series from n=0 to inf of (-1)^n(x)^(5n) / n!

Then i evaluated the series' integral from 0 to 0.1.

I wasnt sure how i could go about finding the value of this integral within 10^-8.

I am using the alternating series estimation error to do this but i find it difficult to do it without checking my calculator.

2. ## Re: Series Approximation Help

You missed an $x$ in the series.

$\displaystyle xe^{-x^5} = x\sum_{n\ge 0} \dfrac{(-x^5)^n}{n!} = \sum_{n\ge 0} (-1)^n \dfrac{x^{5n+1}}{n!}$

Integrating term-by-term, we get:

$\displaystyle \sum_{n\ge 0}(-1)^n \dfrac{(0.1)^{5n+2}}{n!(5n+2)}$

When $n=2$, you have a term that is less than $10^{-11}$. Since this is a telescoping sum, that means that you only need to add the first two terms to find the value within $10^{-8}$.

$\displaystyle \dfrac{10^{-2}}{2}-\dfrac{10^{-7}}{7} \approx 0.00499999$

3. ## Re: Series Approximation Help

Oops, i gave you the incorrect answer. Forgot to multiply the series with an x. Thanks. I understand it now