Maclaurin series error of approximation!?!

So it says find the Maclaurin series of
$f(x)=e^(2x)$
Then the part I don't get: How many terms of the series are required so that the error of the approximation is at most 0.01 for all x in the interval [−2, 2] ?
I know the Mac. Series is (2x)^n/n! but how to find the error?

You're in Steve Desjardins' class? Aren't you?

Quote:

Originally Posted by uottawakid

You're in Steve Desjardins' class? Aren't you?

Yup...

Quote:

Originally Posted by calculuskid1

So it says find the Maclaurin series of
$f(x)=e^(2x)$
Then the part I don't get: How many terms of the series are required so that the error of the approximation is at most 0.01 for all x in the interval [−2, 2] ?
I know the Mac. Series is (2x)^n/n! but how to find the error?

If you truncate the MacLaurin series at n - 1 (including the n - 1 term in your approximation), then the error is:

$\Delta f(x) = \frac{x^n}{n!}\frac{d^n}{dx^n}f(\xi),~\xi \in [0, x]$

$\xi$ is here some unknown value between 0 and x.

$\frac{d^n}{dx^n}e^{2x} = 2^n e^{2x}~\Rightarrow~\Delta f(x) = \frac{(2x)^n}{n!}e^{2\xi}$

Demanding a maximum error of $\epsilon = 0.01$ means that we want:

$\epsilon > \left|\frac{(2x)^n}{n!}e^{2\xi}\right|$

for ALL values of x and $\xi$ in the area $x \in [-2, 2],~\xi \in [-2, 2]$ . The expression has a maximum value at $x = \xi = 2$ . So then we should find the smallest value of n possible for which:

$\epsilon > \frac{4^n}{n!}e^4$

I don't have my calculator for the moment, but I guess you can check for yourself for what n this is true. :)