# Thread: Maclaurin series error of approximation!?!

1. ## 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?

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

3. Originally Posted by uottawakid
You're in Steve Desjardins' class? Aren't you?
Yup...

4. 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.