Does the Taylor series of $e^{f(x)}$ converge everywhere?

Question

In STAT 110, the professor says "the Taylor series of $e^x$ converges everywhere, and then proceeds to convert:

$${e}^{t^2/2} = \sum_{i=0}^\infty \frac{{(t^2/2)}^n}{n!}$$

I understand that the Taylor series for

$$e^x = \sum_{i=0}^\infty \frac{x^n}{n!}$$

but I'm having trouble understanding how $e^{f(x)}$ would have the general form of $\sum_{i=0}^\infty \frac{{(f(x))}^n}{n!}$

Any help would be greatly appreciated, this is all very new to me.

Thanks!

Answer 1

You can read $e^x=\sum_{n=0}^\infty \frac{x^n}{n!}$ as saying that you can compute the exponential of anything by raising it to powers, dividing by $n!$, and summing. Whether the argument of the exponential comes to you as $x$ or as some function $f(x)$, it does not matter.

One subtlety that you may not have noticed is that the series that you get might not be a Taylor series. When you replace $x=t^2/2$, you actually do get the Taylor series of $e^{t^2/2}$ at $t=0$. (This is not completely trivial, there is some "uniqueness of power series" going on here.) On the other hand, if you replace, say, $x=\sin(t)$ then it is not a Taylor series anymore.