Is there a proof $e^ae^b=e^{a+b}$ using its power series expansion?

Question

We know $e^ae^b=e^{a+b}$. It iss a standard property of exponentiation; that it provides an isomorphism from $(\mathbb{R},+)$ to $\mathbb{R}^+,\times)$.

However, I don't see any natural reason for this based on the definition of the exponential function $e^x=\sum^\infty_{k=0}\frac{x^k}{k!}$, or from $e^0=1, \frac{\mathrm{d}e^x}{\mathrm{d}x}=e^x$.

Is there a way to prove $e^ae^b=e^{a+b}$ from these properties? Or is it just a mathematical "coincidence"?

I would guess something along the lines of $\frac{\mathrm{d}e^ae^b}{\mathrm{d}x}=e^ae^b\frac{\mathrm{d}a}{\mathrm{d}x}+e^ae^b\frac{\mathrm{d}b}{\mathrm{d}x}=e^ae^b\frac{\mathrm{d}(a+b)}{\mathrm{d}x}$ while $\frac{\mathrm{d}e^{a+b}}{\mathrm{d}x}=e^{a+b}\frac{\mathrm{d}(a+b)}{\mathrm{d}x}$ which both satisfy $f'=(a+b)'f$.

From J.G.'s comment, I also see that $e^{a+b}=\sum^\infty_{k=0}\frac{(a+b)^k}{k!}=\sum^\infty_{k=0}\frac{\sum^\infty_{i=0}{k\choose i}a^ib^{k-i}}{k!}$ which will be equal to $\sum^\infty_{k=0}\frac{(a)^k}{k!}\sum^\infty_{k=0}\frac{(b)^k}{k!}$. Expanding both:

$\sum^\infty_{k=0}\sum^{i=k}_{i=0}\frac{a^ib^{k-i}}{i!(k-i)!}=\sum^\infty_{k=0}\frac{a^k}{k!}\sum^\infty_{k=0}\frac{b^k}{k!}$

$e^{a+b}=\sum^\infty_{k=0}\frac{(a+b)^k}{k!}=\sum^\infty_{k=0}\frac{\sum^\infty_{i=0}{k\choose i}a^ib^a{k-i}}{k!}$ which will be equal to $\sum^\infty_{k=0}\frac{(a)^k}{k!}\sum^\infty_{k=0}\frac{(b)^k}{k!}$ — Numeral, Feb 16 '23 at 22:15
@J.G. I understand I can expand these more fully, but I am wary of changing the order of terms in an infinite series. — Numeral, Feb 16 '23 at 22:16
Since you mentioned differential equations, the following thread does in fact answer your question: Philosophical question about Pi and connections in maths. Also look at the post it links to for a completely elementary proof of that characterizing property of exp. — user21820, Mar 11 '24 at 07:04

score 3 · Answer 1 · edited Feb 16 '23 at 22:23

$$\begin{align*} e^{a+b} &= \sum_{n=0}^\infty \frac{(a+b)^n}{n!} \tag{1}\\ &= \sum_{n=0}^\infty \frac{1}{n!} \sum_{k=0}^n \binom n k a^k b^{n-k}\tag{2}\\ &= \sum_{k=0}^\infty \sum_{n=k}^\infty \frac{1}{n!} \binom n k a^k b^{n-k}\tag{3}\\ &= \sum_{k=0}^\infty \frac{a^k}{k!} \sum_{n=k}^\infty \frac{b^{n-k}}{(n-k)!} \tag{4}\\ &= e^a e^b \tag{5} \end{align*} $$

$(1)$ follows from the series definition of $e^x$
$(2)$ follows by the binomial theorem
In $(3)$, we bring the $1/n!$ inside, and swap the other of summation. The swapping of summations can be justified in various ways, e.g. Fubini's theorem.
In $(4)$, we note that $$ \binom n k = \frac{n!}{k!(n-k)!} $$ We "attach" the $1/k!$ factor to $a^k$, the $1/(n-k)!$ to $b^{n-k}$, and cancellation results among the $n!$ factors. We pull out the $k$-dependent terms from the inner sum.
In $(5)$, a simply reindexing (say, $m := n-k$) and reapplication of the definition shows that the inner and outer sums evaluate to $e^a$ and $e^b$ respectively, allowing us to conclude.

And in step (3), in justifying the use of Fubini's theorem, you can make a quite similar argument in conjunction with Tonelli's theorem to conclude that the double sum is absolutely convergent (and the sum of absolute values is equal to $e^{|a|+|b|}$). — Daniel Schepler, Feb 16 '23 at 22:41

score 2 · Accepted Answer · answered Feb 16 '23 at 23:01

Yes, Let $$e_n(a)=1+a+\frac{a^2}{2!}+...+\frac{a^n}{n!}$$ $$\text{Let }(a,b)_i^n=\text{binomial expansion of }(a+b)^n \text{with 1st & last $i$ terms omited}$$ Then $$e_n(a)e_n(b)=e_n(a+b)+\frac{(a,b)_1^{n+1}}{(n+1)!}+\frac{(a,b)_2^{n+2}}{(n+2)!}+...+\frac{(a,b)_n^{2n}}{(2n)!}$$ Take absolute values $$\vert e_n(a)e_n(b)-e_n(a+b)\vert \le e_{2n}(\vert a\vert +\vert b \vert)-e_{n}(\vert a\vert +\vert b \vert)$$ Take the limit as $n \text{ goes to } \infty$.Note that this proof also works for commuting $n \times n$ real or complex matrices.

Is there a proof $e^ae^b=e^{a+b}$ using its power series expansion?

2 Answers2