Does the exponential of a function converge? What can we do with it?

Question

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?

Answer 1

I think $\exp(t)$ converges for every possible input of $t$, but I haven’t found a proof yet.

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.

Question: Does the exponential “converge” for function inputs?

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 $\exp(f(t))$.

We also define $\exp(t)$ for complex numbers, matrices, and basically anything you want to plug in.

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:

1. You need to know how to add elements of $X$. Typically $X$ would be an abelian group.
2. You need to know how to multiply elements of $X$ by rationals, or at least rationals of the form $\frac{1}{n!}$. So scalar multiplication. Typically this means a vector space over rationals, or even reals.
3. You need to know how to multiply elements of $X$ with each other, or at least form powers $x^n$. Typically this means an associative algebra.
4. You need some form of convergence, e.g. a topology.

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.

Answer 2

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.

The proof may depend on the type of the input. If $t$ is a real/complex number, then convergence criteria like the ratio test may be applied (or, equivalently, calculating the radius of convergence). If $t$ is an element of some complete normed $\mathbb{R}$- (or $\mathbb{C}$-) vector space, then it suffices to show that the series is absolutely convergent (see e.g. here).

I would also like to know if for some $\exp(t)$ where $t \in U$, $\exp(t)$ will also be $\in U$.

This depends on what $U$ is. For example, if $U = [1, 2]$, this is not the case, since $2 \in U$ but $\exp(2) \notin U$. Another counterexample would be $U = \mathbb{Q}$ and $t = 1$.

Does the exponential “converge” for function inputs?

The exponential series $\sum_{n = 0}^\infty \frac{x^n}{n!}$ converges for every $x \in \mathbb{C}$. In particular, If we have some function $f : D \to \mathbb{C}$ and some $t \in D$, $\sum_{n = 0}^\infty \frac{f(t)^n}{n!}$ converges.

I mean it certainly approaches one single infinite series (polynomial for polynomial inputs) but do we consider that convergence?

From what you wrote I think you are asking whether the resulting function will be an analytic function? This will certainly be the case if $f$ itself is an analytic function, since analytic functions are closed under composition.

Also, is there anything useful you can take away from taking the exponential of a function?

I cannot really answer this part, but there are definitly cases where this is done, e.g. when solving a linear ode of first order with variable coefficients.

Answer 3

The important thing to understand is the convergence of the series for $\exp(t)$.

We'd like to find out if the sum $$\sum_{n=0}^\infty \frac{t^n}{n!}$$ is defined for all real $t$ (I'm just going to focus on reals here).

Dealing with infinite sums can be tricky. If you're unsure about whether a particular series converges, and start manipulating it as if it does, you can get weird (and sometimes wrong) answers.

Finite sums, on the other hand, are much safer to deal with. So one approach for questions about convergence is to consider the sequence of "partial sums": in this case, define $$S_N(t)=\sum_{n=0}^N \frac{t^n}{n!}$$

This is defined for all real $t$ and for all non-negative whole numbers $N$ - that is, given a $t$ value and an $N$ value we could (in theory) work out this sum. (The motivation for doing this is that it's often a lot easier to work with sequences than series.)

Let's start with positive $t$. The sequence $S_N$ is certainly increasing (it's formed by adding up positive numbers). If we can show it's bounded above, then we have proved it converges.

Now, for any positive real $t$, let $T$ be the largest integer less than or equal to $t$. Then for any $n>T$, $$\frac{t^n}{n!}=\frac{t^T \cdot t^{n-T}}{T!\cdot (T+1)\cdots n} \le \frac{t^T}{T!}\cdot \left(\frac{t}{T+1}\right)^{n-T}$$

But the number $u=\frac{t}{T+1}$ is positive and less than $1$ (by definition). So for any $N>T$ we have $$S_N(t) \le S_T(t)+\frac{t^T}{T!}\sum_{n=T+1}^N p^{n-T}=S_T(t)+\frac{t^T}{T!}\sum_{k=1}^{N-T} p^k$$

But now we have a finite number ($S_T(t)$) and the sum of a geometric series with a positive ratio smaller than $1$. This proves that the sequence $S_N(t)$ is bounded above for any positive $t$, so it must converge.

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OK. This proof ran a bit long because of the commentary. The next step is to consider negative $t$; perhaps you can have a go at that? (A similar approach works: define the partial sums as before, and prove they converge. This time the sequence of partial sums is not strictly increasing, since every other term is negative, so you might need a subsequence).