Question

What is the interval of convergence for the Taylor series of `f(x)=e^(x^2)`?

Answer 1

The interval of convergence is `(-oo, oo)`

Explanation:

Recall that the Taylor series of `e^t` is

`e^t = sum_{n=0}^{oo}t^n/(n!)`

For this function, substitute `color(red)(t = x^2)`.

`e^color(red)(x^2) = sum_{n=0}^{oo}(color(red)(x^2))^n/(n!)`

Apply the ratio test.

`r = lim_{n->oo}|a_{n+1}/a_n|`

If `r < 0`, then the series is absolutely convergent.

`r = lim_{n->oo}|{(x^2)^{n+1}}/{(n+1)!} * {n!}/(x^2)^n|`

`r = lim_{n->oo}|((x^2)^n*x^2)/{(n+1) * n!} * {n!}/(x^2)^n|`

`r = lim_{n->oo}|(x^2)/(n + 1)|`

And since `x^2` is independent of the limit:

`r = x^2 * lim_{n->oo}|1/(n + 1)|`

`r = x^2 * 0`

`r = 0`

Regardless of the value of `x`, the Taylor series absolutely converges. The interval of convergence then must be `(-oo, oo)`.