# Maclaurin Series for Natural Log Function

1. Aug 4, 2013

### Justabeginner

1. The problem statement, all variables and given/known data
Use x=-1/2 in the MacLaurin series for e^x to approximate 1/sqrt(e) to four decimal places.

2. Relevant equations

3. The attempt at a solution
$\sum_{n=0}^\infty \frac{x^n}{n!} = 1 + x + x^2/2 + x^3/6 + ...$

For this particular power series, I have:

$\sum_{n=0}^\infty \frac{(x+0.5)^n}{sqrt(e)(n!)} = 1/sqrt(e) + (x+0.5)/(sqrt(e)) + (x+0.5)^2/sqrt(e) + (x+0.5)^3/6*sqrt(e)...$

(-1/2)^(n+1)/(n+1)! <= 0.00004
n= 2

(1/sqrt(e) + (x+0.5)/sqrt(e) + (x+0.5)^(2)/2*sqrt(e)) = (1/sqrt(e) + 0 + 0)= 1/sqrt(e)= 0.6065

But this doesn't seem right because I should end up with the actual number instead of the 1/sqrt(e) itself. I thought I was doing the method correctly, but I guess not. I would appreciate any insight on this at all, as I don't know of any other method to apply here. Thanks!

2. Aug 4, 2013

### lurflurf

I don't know why you did that. From
$$e^x \sim \sum_{k=0}^\infty \frac{x^k}{k!}$$
you should simply have
$$e^{-1/2} \sim \sum_{k=0}^\infty \frac{(-1/2)^k}{k!}$$.
A partial sum will be a good estimate if enough terms are used. Do you know how to determine how many terms are needed?

3. Aug 4, 2013

### Justabeginner

I think that I have to set it to be less than or equal to 0.00004 as I did above? So n=2?