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I work as a freelance math mentor, and have found that many high schoolers, like myself back then, are somewhat mystified with the exponential function. So I set myself the task of giving a rigorous listing & explanation of its properties, and just why it is such a singular and singularly useful function.

Starting from the wishful definition, that $\exp(x)$ is the unique function which satisfies $$\left.\begin{align} \exp(0)&=1 \\ \frac{\mathrm d}{\mathrm dx}\exp(x)&=\exp(x) \end{align}\right\}\tag{*}$$

I have a proof for uniqueness, and existence as an absolutely convergent sum (the Taylor/Maclaurin series), satisfying (*). And, starting with an example of interest rates, and with some Binomial Series Expansion shown that $$\lim_{n\rightarrow\infty}\left(1+\frac{1}{n}\right)^n = \sum_{n=0}^{\infty} \frac{1}{n!} = \exp(1) = e.$$

When you combine this with the equation $\exp(x+y)=\exp(x)\cdot \exp(y)$, which I have also shown to work, it is very tempting to assume that $\exp(x)$ is an exponent of Euler’s number $e$, as the former two equations, and the first part of (*) are compatible with this.

That is to say, $\exp(x)\equiv e^x$ is congruent with everything I’ve shown thus far, but I don’t have a strict proof that this is truly the right call. I’d like a rigorous proof that $\sum_{n=0}^{\infty} \frac{x^n}{n!}$ and $e^x$, that is raising eulers number $e$ to the power of $x$, are truly the same function.

Woodenplank
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  • See https://fr.wikipedia.org/wiki/Exponentielle_de_base_a#Par_la_propriété_algébrique – Anne Bauval Mar 24 '25 at 06:18
  • See the middle of my answer here, where I discuss the existence and uniqueness of $F_a$. In particular, it is the existence of the exponential function which is the tough part. We have uniqueness for $x\in\Bbb{Q}$ quite easily, and with the very mild extra assumption of continuity, we get uniqueness over all $x\in\Bbb{R}$ (simply because $\Bbb{Q}$ is dense in $\Bbb{R}$). – peek-a-boo Mar 24 '25 at 06:52
  • G.M. Fikhtengol'ts, A Course of Differential and Integral Calculus, I volume. – zkutch Mar 24 '25 at 07:13
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    Your question is not clear. Do you want 1) "a rigorous proof that there isn’t simply some other function satisfying $f(0)=1$, $f(1)=e$ and $f(x+y)=f(x)f(y)$" (which requires an additional continuity hypothesis), or 2) a proof that "$\exp(x)$ and $e^x$ are one and the same" (which requires a definition of $e^x$)? – Anne Bauval Mar 24 '25 at 11:14
  • The latter would be preferable. My apologies for being unclear. – Woodenplank Mar 24 '25 at 12:06
  • Thank you for your edit. Please, again: what is your definition of "raising eulers number $e$ to the power of $x$"? (I mean: I know how to do it but you should include some words about it in your post.) – Anne Bauval Mar 24 '25 at 22:34
  • @AnneBauval is right to tell you to articulate a definition of $e^x$. If you define $e^x$ for non-negative integers $x$, and extend this in obvious ways to negative integers and rational numbers, the functions will match on $\Bbb Q$. If you define $e^x$ to be continuous on $\Bbb R$, its values on $\Bbb Q$ specify the function, which completes the proof it's $\exp x$ for all $x\in\Bbb R$. – J.G. Mar 25 '25 at 08:22
  • I understand the first parts well enough, $e^n,n\in\mathbb{N}$ is of course just $e\cdot e\cdot e\cdot... n$ times. For $n\in\mathbb{z}$ you take $1/(e^n}$ for negatives. For rational numbers, $e^{n/m} = \sqrt[m]{e^n}$with $n,m = \mathbb{z}$. However, I don't understand what you mean by your last sentence? – Woodenplank Mar 26 '25 at 18:21
  • You've only defined $e^x$ for rational values of $x$. But $\exp(x)$ is defined for all real numbers $x$. You can use induction to show that those two definitions agree for all rational $x$. But before you can even ask whether it holds for irrational $x$, you need to define $e^x$ when $x$ is irrational. Because the rationals are dense in the reals (between every two real numbers there is a rational number), you can show that there is at most one continuous function $f(x)$ on $\Bbb R$ that agrees with $e^x$ on the rationals. Since $\exp(x)$ is such a function, it is the only one. – Paul Sinclair Mar 26 '25 at 20:19

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