Does the Taylor series expansion for e^x converge quickly?

[sandy.bridge]

Hello all,
My question is in regards to the Taylor series expansion of
f(x)=e^x=1+x+x^2/(2!)+x^3/(3!)...
I calculated the value of
e^(-2)
using the first 4 terms, 6 terms, and then the first 8 terms. I then calculated the relative error to compare it to the true value, depcited by my calculator to 6 significant figures. Using the first four terms, I found an error of 2609%. Using the first 6 terms I found an error of 2905%, and lastly using the first 8 terms I found an error of 2952%.

What can I conclude from this? Does the error increase (at a decreasing rate) until it begins decreasing (at an increasing rate)?

 

[CompuChip]

How did you get to 2609%?
If I take 4 terms I get an error of around 46%.

Are you sure you entered the even powers correctly? One of the most common mistakes is to enter (-2)² as -2² = -(2²).

 

[sandy.bridge]

Oops, I had a mistake.n However, upon doing it again, I get an error of 346.3.
relative error=100((e^(-2)-(sum of first four terms))/e^(-2))=100((e^(-2)-(-0.333333))/e^(-2))=346%

 

[pbuk]

Yes the error after 4 terms is 346%. But in the next 4 terms the errors are 146%, -51%, 15% and 4%; after another 4 terms the error is only 5 parts in a thousand and another four, -2 in 1 million. The error decreases in magnitude with each successive term (except for the second term, x) and converges fast enough for me.