# Power series of a strange function

1. Dec 5, 2014

### bonildo

1. Write ∫e^(-t^2)dt with 0<=t<=x , as power series around 0. For what values of x this series converge ?

attempt at a solution:

f' = e^(-x^2) => f'(0) = 1
f''= -2x*e^(-x^2) => f''(0)= 0
f'''= -2e^(−x2) +4*x^2*e^(−x^2) => f'''(0)=-2

I tried to find a general rule for the derivatives but with no sucess.

2. Dec 5, 2014

### vela

Staff Emeritus
You'd have to take more derivatives to hope to find a pattern. I would try an alternative approach, however. Expand the integrand as a series and then integrate term by term.

3. Dec 6, 2014

### Staff: Mentor

You're off to a bad start here. f(x) = e-x2, not f'(x).

4. Dec 6, 2014

### HallsofIvy

Staff Emeritus
No, he is taking f to be the integral itself so f' is the integrand.

Agreed. bonildo, write the Taylor's series for e^x, replace x with -x^2, then integrate term by term.

5. Dec 6, 2014

6. Dec 6, 2014

### Ray Vickson

Last edited: Dec 6, 2014