> [!NOTE] Theorem (The characteristic property of the exponential) > For all $x,y\in\mathbb{R}$ $\exp(x+y)=\exp(x)\exp(y)$ where $\exp$ denotes the [[Real Exponential Function|real exponential function]]. ***Proof using power series definition***: This proof assumes the definition of $\exp$ as [[Real Exponential Function as Power Series|power series]]: $\exp(x)=\sum_{n=0}^{\infty} \frac{x^{n}}{n!}.$ STS $ \sum_0^{2 m} \frac{(x+y)^k}{k !}-\left(\sum_0^m \frac{x^i}{i !}\right)\left(\sum_0^m \frac{y^j}{j !}\right) \rightarrow 0 $because the first term converges to $\exp(x+y)$ and the second to $\exp(x)\exp (y)$. By [[Binomial Theorem|binomial theorem]], for each $k$, $ (x+y)^k=\sum_{i=0}^k \frac{k !}{i !(k-i) !} x^i y^{k-i}=\sum_{i+j=k} k ! \frac{x^i}{i !} \frac{y^j}{j !} $ So $\begin{aligned} \sum_0^{2 m} \frac{(x+y)^k}{k !}-\left(\sum_0^m \frac{x^i}{i !}\right)\left(\sum_0^m \frac{y^j}{j !}\right) \\ &=\sum_{i+j \leq 2 m} \frac{x^i}{i !} \frac{y^j}{j !}-\sum_{i \leq m, j \leq m} \frac{x^i}{i !} \frac{y^j}{j !} \\&=\sum_{\substack{i \geq m+1\\i+j\leq 2m}} \frac{x^i}{i !} \frac{y^j}{j !}+\sum_{\substack{j \geq m+1\\i+j\leq 2m}} \frac{x^i}{i !} \frac{y^j}{j !}\\&\leq\sum_{\substack{i \geq m+1\\i+j\leq 2m}} \frac{|x|^i}{i !} \frac{|y|^j}{j !}+\sum_{\substack{j \geq m+1\\i+j\leq 2m}} \frac{|x|^i}{i !} \frac{|y|^j}{j !}\\&\leq\sum_{\substack{i \geq m+1\\j\geq 0}} \frac{|x|^i}{i !} \frac{|y|^j}{j !}+\sum_{\substack{j \geq m+1\\i\geq 0}} \frac{|x|^i}{i !} \frac{|y|^j}{j !}\\&=\left( \sum_{i\geq m+1}\frac{|x|^{i}}{i!} \right)\left( \sum_{j= 0}^{\infty}\frac{|y|^{j}}{j!} \right)+\left( \sum_{i=0}^{\infty}\frac{|x|^{i}}{i!} \right)\left( \sum_{j\geq m+1}\frac{|y|^{j}}{j!} \right) \end{aligned}$which $\to 0$ since $\left( \sum_{i\geq m+1}\frac{|x|^{i}}{i!} \right)\to{0}$ and $\left( \sum_{j= 0}^{\infty}\frac{|y|^{j}}{j!} \right)\to e^{|y|}$ as $m\to \infty.$ ###### Proof using power series definition and Merten's theorem By [[Binomial Theorem|binomial theorem]] $\begin{align} \exp(x+y) &= \sum_{n=0}^{\infty} \frac{(x+y)^{n}}{n!} = \sum_{n=0}^{\infty} \sum_{m=0}^{n} \frac{x^{m}}{m!} \frac{y^{n-m}}{(n-m)!} \end{align}$By [[Cauchy Product of Univariate Real Power Series About Zero Converges to Product]] (or [[Merten's Convergence Theorem]]), the Cauchy product converges to the product: that is, $\sum_{n=0}^{\infty} \sum_{i+j=n} \frac{x^{i}}{i!} \frac{y^{j}}{j!} =\left( \sum_{n=0}^{\infty} \frac{x^{n}}{n!} \right) \left( \sum_{m=0}^{\infty} \frac{y^{m}}{m!} \right)=\exp(x)\exp(y).$ ***Proof by differentiation***: Fix $z\in \mathbb{R}$ and consider the function $x\mapsto \exp(x)\exp(z-x).$Using [[Derivative of Sum of Differentiable Real Functions|product rule]] and [[Chain Rule for Differentiability|chain rule]] its derivative w.r.t. $x$ is given by $\exp(x)\exp(z-x)-\exp(x)\exp(z-x)=0.$By [[Real Function with Zero Derivative is Constant|MVT]] the function is constant. At $x=0$, the function is $\exp(z)$ so we know that for all $x$, $\exp(x)\exp(z-x)=\exp(z).$Setting $z= x+y$ gives the desired result. $\square$