Find a approximate value of e^(-4) using Taylor series, so that error is less than 10^(-3)

530 Views Asked by At

So, the Taylor series for $e^x$ is $e^x = 1 + x + \frac{x^2}{2!}+ \ldots + \frac{x^n}{n!}$. In this instance I have $x = -4$, so the series looks as follows: $$1 - 4 + \frac{4^2}{2!} - \frac{4^3}{3!} + \ldots + (-1)^n\frac{4^n}{n!}.$$

So what is the way I can find a value of the function with a error smaller than $10^{-3}$?

1

There are 1 best solutions below

0
On

You have an alternating series.

As you know the error in alternating series is less than the absolute value of the next term.

Therefore you need to find $n$ such that $|4^n|/{n!}<0.001$ I found that for $n=15$ the error term is less than $0.001$ so you need to add up the first $14$ terms to satisfy the tolerance of $0.001$