Recreational Multiplication
Proving that exp(x) exp(y) = exp(x+y)
Define the exponential function
\[\exp(x) := 1 + x + \frac{x^2}{2!} + \cdots = \sum_{k=0}^{\infty} \frac{x^k}{k!}\]From this definition as a power series, it isn’t so clear that our function really does behave like an exponential. Any exponent satisfies \(a^{m+n} = a^m a^n\). So our goal here is to show that \(\exp(x+y) = \exp(x) \exp(y)\).
As a warmup, first we will show the easier \(\exp(2x) = \exp(x)\exp(x)\).
\[\exp(x) \exp(x) = \sum_{p=0}^{\infty} \frac{x^p}{p!} \sum_{q=0}^{\infty} \frac{x^q}{q!} = \sum_{p, \ q \geq 0} \frac{x^p x^q}{p! \ q!}\]Denote by \([x^k]\) the coefficient of \(x^k\). To find it, set $p + q = k$. Then
\[\begin{align*} [x^k] &= \sum_{p + q = k} \frac{1}{p! \ q!} \\ &= \frac{1}{0! \, k!} + \frac{1}{1! \, (k-1)!} + \dots + \frac{1}{k! \, 0!} \\ &= \frac{1}{k!} \sum_{p=0}^{k} \binom{k}{p} \\ &= \frac{2^k}{k!} \end{align*}\]Aha! The coefficient of \(x^k\) is \(\frac{2^k}{k!}\), so the $k$-th term is given by \(\frac{(2x)^k}{k!}\), as desired.
Now that we have a hold on a specific case, let us move on to the general one.
\[\exp(x) \exp(y) = \sum_{p=0}^{\infty} \frac{x^p}{p!} \sum_{q=0}^{\infty} \frac{y^q}{q!} = \sum_{p, \ q \geq 0} \frac{x^p y^q}{p! \ q!}\]Now letting $p + q = k$ and summing over all $k$ is the same as the sum above1
\[\begin{align*} \exp(x) \exp(y) &= \sum_{k=0}^{\infty} \sum_{p + q = k} \frac{x^p y^q}{p! \ q!} \\ &= \sum_{k=0}^{\infty} \sum_{p=0}^k \frac{x^p y^{k-p}}{p! \ (k-p)!} \\ &= \sum_{k=0}^{\infty} \frac{1}{k!} \sum_{p=0}^k \binom{k}{p} x^p y^{k-p} \\ &= \sum_{k=0}^{\infty} \frac{1}{k!} (x+y)^k \quad \href{https://en.wikipedia.org/wiki/Binomial_theorem}{\text{(binomial theorem)}} \\ &= \sum_{k=0}^{\infty} \frac{(x+y)^k}{k!} \\ &= \exp(x+y) \quad \blacksquare \end{align*}\]And so we are done! What we are doing here is called the Cauchy product.
Exercise left to the reader: now prove the formula for \(\sin(x+y)\)… 😜
Let us now work with another definition of \(\exp\). Define \(\exp(x)\) to be the solution to the equation \(f'(x) = f(x)\) with \(f(0) = 1\). Now our goal is to show that \(\exp(x+y) = \exp(x) \exp(y)\). Let \(g(x) = \exp(x+y)\) and \(h(x) = \exp x \exp y\). Then it is trivial to check that these satisfy the differential equation:
\[g'(x) = \frac{d}{dx} \exp(x+y) = \exp(x+y) = g(x)\]by the chain rule and
\[h'(x) = \frac{d}{dx} \exp x \exp y = \exp x \exp y = h(x).\]And also that \(g(0) = h(0) = \exp y\), using \(\exp 0 =1\).
Now by Picard–Lindelöf, we have two functions satisfying the same ODE with the same initial values, and so they are the same function by uniqueness. \(\blacksquare\)
The reader is encouraged to also show this for the limit definition,
\[\exp x := \lim_{n \to \infty} \left(1 + \frac{x}{n} \right)^n\]Heuristic:
\[\left(1+\frac{x}{n}\right)^n\left(1+\frac{y}{n}\right)^n = \left(1+\frac{x+y}{n}+\frac{xy}{n^2}\right)^n = \cdots\]This rearrangement holds by Fubini’s theorem, as the exponential series converges absolutely. ↩︎