Re: What is the error of the truncated exponential series?

[rusty]

heinrich_neumaier@xxxxxxxxx wrote:

> 
> Summing from 0 to n the terms 1/n!, one gets an approximation for
> e=2.71828...
> 
> Is there a formula for the difference between the result  (as a
> function of n) and e?

You can use the formulas for the remainder term in Taylor's theorem.
For example

f(x) -\sum_{k=0}^n f^(k)(0) x^k/k! = \int_0^x (x-t)^n/n! f^{(n+1)}(t) dt

This gives

e -\sum_{k=0}^n 1/k! = e \int_0^1 e^{-s} s^n/n! ds.   
-- 
rusty