I’m in middle school (6th grade) and I have a question related to the exponential function (the teacher couldn’t help me):
We define the exponential as follows:
$$ \exp(t) = \sum_{n = 0}^{\infty} \frac{t^n}{n!}$$
For whole number inputs of $t$, this corresponds to raising $e$ to that power.
I think $\exp(t)$ converges for every possible input of $t$, but I haven’t found a proof yet. If anyone can provide a proof or disproof then that would be helpful. I would also like to know if for some $\exp(t)$ where $t \in U$, $\exp(t)$ will also be $\in U$. However, these are secondary questions and you can ignore them (but do answer if you want).
We also define $\exp(t)$ for complex numbers, and matrices.
Today I thought about the exponential of a function:
$\exp(f(t))$.
For example, let’s take $\exp(t^2)$. This would result in something like:
$\exp(t^2) = \sum_{n = 0}^{\infty} \frac{(t^2)^n}{n!} = 1 + t^2 + \frac{t^4}{2!} + \frac{t^6}{3!} + \cdots$
Which we could write as a regular polynomial as
$1 + t^2 + \frac{1}{2}t^4 + \frac{1}{6}t^6 + \frac{1}{24}t^8 + \cdots$
Another example: $\exp\left(\sin(t)\right)$:
$1 + \sin(t) + \frac{1}{2}\sin^2 (t) + \frac{1}{6}\sin^3 (t) + \cdots$
Actually, this is the same thing as raising $e$ to the power of a function. If we took $\exp(e^x)$ it would be the same as $e^{(e^x)}$:
$1 + \frac{1}{2}e^x + \frac{1}{6}e^{2x} + \cdots$
Question: Does the exponential “converge” for function inputs? Also, is there anything useful you can take away from taking the exponential of a function?
Best Answer
For all real and complex inputs of $t$, yes. This is not something trivial to show. You need proper background in set theory, how naturals, integers, reals and complex numbers work. You need to understand not only what it means for a series to converge, but what a series even is. Convergence tests are important as well. And how all of that works in complex numbers is even more complicated. You can start by reading some book on mathematical analysis, e.g. Robert Magnus "Fundamental Mathematical Analysis".
Depending on how much you already know this question might have an easy answer. For example we can apply d'Alembert's criterion to $a_n=\frac{t^n}{n!}$ to get
$$\lim_{n\to\infty}\bigg|\frac{a_{n+1}}{a_n}\bigg|=\lim_{n\to\infty}\frac{t}{n+1}=0$$ and so the series $\sum_{n=0}^\infty a_n$ converges absolutely for each complex $t$. But still, my advice is to read some book.
Yes. And this is actually quite simple. It doesn't matter that the input is of the form $f(x)$, and that the particular series is the exponential function. As long as a series $S(t)=\sum_{n=0}^\infty \alpha(t, n)$ is well defined/convergent at some $y$, and $f(x)=y$ then $S(f(x))$ is well defined, in fact equal to $S(y)$.
In particular since $\exp(y)$ is defined for all real $y$, then $\exp(t^2)$ is also well defined, because $t^2$ is a real. So is $\exp(sin(t))$. And so is any $\exp(f(t))$ as long as $f(t)$ is a real/complex.
So for example you already know that $\exp(1)$ converges. Does it make a difference if I write it as $\exp(sin(\pi/2))$? Because that's literally what putting a function inside does: it rewrites arguments. The same reasoning works backwards, if we start with $\exp(f(t))$.
Not anything. That's a serious exaggeration. In order for this definition to make sense on a set $X$ you need the following structures on it:
So this is an absolute minimum. But in order to prove things you will need more, e.g. those structures have to be related with each other. Additionally having a complete metric would be useful. In fact a proper setup for considering exponential functions is Banach algebra. Which covers all those examples you've mentioned. But it is a quite restrictive structure, and not used commonly.
So in my opinion its the opposite: there aren't many mathematical structures which you can use as inputs for the exponential function. For most of the time it is enough to consider reals and complex numbers only.