Power series of a strange function

[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.

 

[vela]

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.

 

[Mark44]

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

You're off to a bad start here. f(x) = e^-x2, not f'(x).
Still, I would follow vela's advice on this.

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.

 

[HallsofIvy]

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

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

Still, I would follow vela's advice on this.

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

 

[bonildo]

Following vela's advice look what I got:

http://www.wolframalpha.com/input/?i=integrate+(sum_(k=0)^infinity+(-x^2)^k/(k!))+from+0+to+x+#_=_

 

[Ray Vickson]

Following vela's advice look what I got:

http://www.wolframalpha.com/input/?i=integrate (sum_(k=0)^infinity (-x^2)^k/(k!)) from 0 to x #_=_

That is not what you were asked to get. You were asked for the infinite series, not for a formula for its sum. Also: what is the region of convergence?