# Prove $\exp(x+y) = \exp(x) \exp(y)$ for $\exp(x) = \sum_{n=0}^\infty \frac {x^n}{n!}$

I am trying to prove $\exp(x+y) = \exp(x) \exp(y)$.

I may use that $$\exp(x) = \sum_{n=0}^\infty \frac {x^n}{n!}$$ I further know how to multiply two power series in one point, i.e. if $f(x) = \sum_{n=0}^\infty c_n(x-a)^n$ and $g(x) = \sum_{k=0}^\infty d_n(x-a)^n$ then $$f(x)g(x) = \sum_{n=0}^\infty e_n(x-a)^n$$ with $$e_n = \sum_{m=0}^n c_md_{n-m}$$

• What's the question? Commented Jan 23, 2013 at 18:54
• Presumably you tried to multiply the power series for $\exp(x)$ and $\exp(y)$. What did you get stuck on? Commented Jan 23, 2013 at 18:54
• The title is the question @nbubis. This question is from Tao Analysis II and he gives the hint with the multiplication of power series. I dont see how to apply that hint.
– user42761
Commented Jan 23, 2013 at 18:57
• Ok I got it :) Can I delete this question ?
– user42761
Commented Jan 23, 2013 at 19:00
• This must be a duplicate... Commented Jan 23, 2013 at 19:02

\begin{align} \exp(x+y)&=\sum_n\frac{(x+y)^n}{n!} \\\\ &=\sum_{n}\frac{1}{n!}\sum_{a+b=n} {n \choose a} x^ay^b \\\\ &= \sum_{n}\frac{n!}{n!}\sum_{a+b=n}\frac{x^a}{a!}\frac{y^b}{b!} \\\\ &= \sum_{a,b} \frac{x^a}{a!}\frac{y^b}{b!} \\\\ &= \exp(x)\cdot\exp(y) \end{align}

• You can write faster than me :D
– user42761
Commented Jan 23, 2013 at 19:16
• You overcomplicated things in your solution. There is no need to work with a generating function as you have done. Also, you left out some steps in your first equality when going from a product of two generating functions into a single generating function. Commented Jan 23, 2013 at 19:17
• I gave the legitimation for that in my question.
– user42761
Commented Jan 23, 2013 at 19:19
• Oh, I see now. Thanks for pointing that out :) Commented Jan 23, 2013 at 19:20
• you have used this equality: $$\bigsqcup _n\{(a,b) \mid a+b=n\}=\{(a,b)\mid \}$$ and absolute convergence.
– user59671
Commented May 9, 2013 at 11:05

This can actually be done without writing a single sum. Consider the function $$f(x, y) = \frac{e^x e^y}{e^{x+y}}.$$ Observe that $$\frac{\partial f}{\partial x} = \frac{e^x e^y e^{x+y} - e^x e^y e^{x+y}}{(e^{x+y})^2} = 0.$$ Similarly, $$\frac{\partial f}{\partial y} = \frac{e^x e^y e^{x+y} - e^x e^y e^{x+y}}{(e^{x+y})^2} = 0.$$ This shows that $f$ is a constant function. Now, we need only to use the series definition to show $f(0, 0) = 1$. Then, by rearrangement, we have the desired result: $$e^{x+y} = e^x e^y.$$

• Don't we need to assume that $e^{x+y} \neq 0$ to define $f$? But the usual proofs of this fact use the fact the contradiction that $1 = e^0 = e^{c-c} = e^ce^{-c} = 0$. Isn't the reasoning circular? Commented Oct 22, 2023 at 0:16

My solution

Let $x,y \in \mathbb R$ and

$f(z) := \sum_{n=0}^\infty \left(\frac {x^n}{n!} \right )z^n$ and $g(z) := \sum_{n=0}^\infty \left(\frac {y^n}{n!} \right )z^n$. Then $\exp(x) \exp(y) = f(1)g(1)$. That is $$f(z)g(z) = \sum_{n=0}^\infty \left( \sum_{k=0}^m \frac {x^m y^{n-m}}{m! (n-m)!} \right)z^n$$ $$= \sum_{n=0}^\infty \frac 1 {n!} (x+y)^n z^n$$ thus $f(1)g(1) = \exp(x+y)$.

\begin{align*} & \exp(x+y)=\sum_{n=0}^{\infty}\frac{1}{n!}(x+y)^n = \sum_{n=0}^{\infty}\frac{1}{n!}\sum_{k=0}^{n}\frac{n!}{(n-k)!k!} x^ky^{n-k} \\ =& \sum_{n=0}^{\infty}\sum_{k=0}^{n}\frac{x^k}{k!}\frac{y^{n-k}}{(n-k)!} = \lim_{N \to \infty}\sum_{n=0}^{N}\sum_{k=0}^{n} \frac{x^k}{k!}\frac{y^{n-k}}{(n-k)!}\\ & \mathrm{\quad make \; a \; bijective \;map\; between} \left\{(n,k):0 \le n \le N,0 \le k\le n\right\} \\&\mathrm{\; and } \left\{(n,k):0 \le k \le N,k \le n\le N\right\} \\ =& \lim_{N \to \infty}\sum_{n=0}^{N}\sum_{k=0}^{n} \frac{x^k}{k!}\frac{y^{n-k}}{(n-k)!}= \lim_{N \to \infty}\sum_{k=0}^{N}\sum_{n=k}^{N}\frac{x^k}{k!}\frac{y^{n-k}}{(n-k)!} \\ = &\lim_{N \to \infty} \sum_{k=0}^{N}\frac{x^k}{k!}\sum_{n=k}^{N}\frac{y^{n-k}}{(n-k)!} = \lim_{N \to \infty} \sum_{k=0}^{N}\frac{x^k}{k!}\sum_{n=0}^{N-k}\frac{y^{n}}{n!} = \sum_{k=0}^{\infty}\frac{x^k}{k!}\sum_{n=0}^{\infty}\frac{y^{n}}{n!} \\ =&\sum_{k=0}^{\infty}\frac{x^k}{k!}\sum_{n=0}^{\infty}\frac{y^{n}}{n!}=\sum_{k=0}^{\infty}\sum_{n=0}^{\infty}\frac{x^k}{k!}\frac{y^{n}}{n!} = \exp(x)\exp(y) \end{align*}

$A(t)=\exp(x) = \sum_{n=0}^\infty \frac {x^n}{n!}t^n$
$B(t)=\exp(y) = \sum_{n=0}^\infty \frac {y^n}{n!}t^n$
$C(t) = A(t)*B(t)=\sum_{n=0}^\infty (\sum_{k+z=n}^\ \frac {x^k}{k!}*\frac {y^z}{z!})t^n=\sum_{n=0}^\infty \frac {(x+y)^n}{n!}t^n=exp(x+y)$

and use $t=1$

sry i was too late^^

• Duplicate of the two answers ^^
– user42761
Commented Jan 23, 2013 at 19:23
• Actually, both you and @nordmann use an extraneous generating function [in the variables $t$,$z$ respectively]. The reason why I say "extraneous" is because if you look at my solution above, no new variable need be introduced. Commented Jan 23, 2013 at 19:43
• sry im new to this latex stuff^^ Commented Jan 23, 2013 at 23:16

Euler formula says that (exponential form of a complex number) $$e^{i\theta}=\cos\theta+i\sin\theta.$$ Therefor $$e^{x+y}=\cos(-i(x+y))+i\sin(-i(x+y))\\=\cos(xi+yi)-i\sin(xi+yi)\\=(\cos ix\cos iy-\sin ix\sin iy)+i(\sin ix\cos iy+\cos ix\sin iy)\\=(\cos ix+i\sin iy)(\cos iy+i\sin iy)\\=e^xe^y.$$

Assume the function $\frac{exp(x+y)}{exp(x)}$, where $y$ is some constant. Differentiate with respect to $x$ and find that the derivative is zero. Therefore $\frac{exp(x+y)}{exp(x)}=c\in\mathbb{R}$. From the fact that $exp(0)=1$ we get $exp(y)=c$ which proves what was asked.