Proof that the Function $f(x) = \sum_{n=0}^{\infty} \frac{x^n}{n!} $ Satisfies $ f(x+y) = f(x)f(y) $

Question

I am attempting to prove that the function $ f(x) = \sum_{n=0}^{\infty} \frac{x^n}{n!} $ satisfies the functional equation $f(x+y) = f(x)f(y) $ for all real $ x $ and $y$, without relying on the knowledge that $ f(x) $ is the exponential function or using Taylor series expansions. I have been unable to establish the relationship rigorously.It may not be correct what i tried and i may have an error in all equations, but i have tried using the binomial theorem and limit theorems. We know that $ f(x+y) = \sum_{n=0}^{\infty} \frac{(x+y)^n}{n!} $, so by the binomial theorem, it becomes:

$ \sum_{i=0}^{n} \left( \sum_{k=0}^{i} \frac{x^{k-i} y^i}{i!(k-i)!} \right) $

Similarly, I represented $ f(x)f(y) $ as:

$ \sum_{i=0}^{n} \left( \sum_{k=0}^{n} \frac{y^k x^i}{k! i!} \right) $

Then, I considered their difference in module and tried to find ways to show that they can become arbitrarily close.

Another approach I tried was induction. I wanted to prove that

$ \sum_{i=0}^{n} \left( \sum_{k=0}^{n} \frac{y^k x^i}{k! i!} \right) $

can be represented as

$ \sum_{i=0}^{n} \left( \sum_{k=0}^{i} \frac{x^{k-i} y^i}{i!(k-i)!} \right) + S_n $

where $ S_n \rightarrow 0 $ as $ n \rightarrow \infty $, but I couldn't achieve that.

I thought that the proof of inductive step would work for n+1 like this: Suppose we assume $(1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+...+ \frac{x^n}{n!})(1+y+\frac{y^2}{2!} +...+ \frac{y^n}{n!})$, this can be represented as $1+(y+x)+(\frac{x+y}{2!})^2+...+(\frac{x+y}{n!})^n + S_n$, where I think $S_n$ is a term that may approach 0, which is what I want to show.

For the $n+1$ step we want to show that:

$(1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+...+ \frac{x^n}{n!} + \frac{x^{n+1}}{{n+1}!})(1+y+\frac{y^2}{2!} +...+ \frac{y^n}{n!} + \frac{y^{n+1}}{{n+1}!}) = 1+(y+x)+(\frac{x+y}{2!})^2+...+(\frac{x+y}{n!})^n +\frac{(x+y)^{n+1}}{(n+1)!} + S_{n+1}$,

we can write as follows:

$((1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+...+ \frac{x^n}{n!}) + \frac{x^{n+1}}{(n+1)!})((1+y+\frac{y^2}{2!} +...+ \frac{y^n}{n!}) + \frac{y^{n+1}}{(n+1)!}) = (1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+...+ \frac{x^n}{n!})(1+y+\frac{y^2}{2!} +...+ \frac{y^n}{n!}) + \frac{y^{n+1}}{(n+1)!}(1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+...+ \frac{x^n}{n!})+ \frac{x^{n+1}}{(n+1)!}(1+y+\frac{y^2}{2!} +...+ \frac{y^n}{n!}) + \frac{x^{n+1}y^{n+1}}{(n+1)!(n+1)!}$

$= 1+(y+x)+(\frac{x+y}{2!})^2+...+(\frac{x+y}{n!})^n + S_n + \frac{y^{n+1}}{(n+1)!}(1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+...+ \frac{x^n}{n!}) + \frac{x^{n+1}}{(n+1)!}(1+y+\frac{y^2}{2!} +...+ \frac{y^n}{n!}) + \frac{x^{n+1}y^{n+1}}{(n+1)!(n+1)!}$

But I haven't tried how to construct $\frac{(x+y)^{n+1}}{(n+1)!} + S_{n+1}$, $S_{n+1}$ approach to 0, from $S_n + \frac{y^{n+1}}{(n+1)!}(1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+...+ \frac{x^n}{n!}) + \frac{x^{n+1}}{(n+1)!}(1+y+\frac{y^2}{2!} +...+ \frac{y^n}{n!}) + \frac{x^{n+1}y^{n+1}}{(n+1)!(n+1)!}$. what is a good proof? thanks!

Answer 1

Perhaps the cleanest approach is to use a combination of the binomial theorem and Cauchy Products. It also seems easier to manipulate $f(x)f(y)$ into $f(x+y)$ than to do the reverse direction.

Here's one approach, done in three steps.

Step 1. Cauchy Products

We can write $f(x)f(y)$ in the following way $$ f(x)f(y) = \left ( \sum_{n=0}^\infty \frac{x^n}{n!} \right )\left ( \sum_{m=0}^\infty \frac{y^m}{m!} \right ) = \sum_{k=0}^\infty \sum_{n=0}^k \frac{x^ny^{k-n}}{n!(k-n)!} $$ This last formula is sometimes called a Cauchy product formula. Please look at the comments on your question for more information about it.

Step 2. Binomial Theorem

Focus on a single fixed $k$ term and recall that $${k \choose n} = \frac{k!}{n!(k-n)!} \iff \frac{1}{k!}{k \choose n} = \frac{1}{n!(k-n)!}.$$ From this it follows $$\sum_{n=0}^k \frac{x^ny^{k-n}}{n!(k-n)!} = \sum_{n=0}^k \frac{1}{k!}{k \choose n}x^ny^{k-n} = \frac{1}{k!} \sum_{n=0}^k {k \choose n}x^ny^{k-n} = \frac{1}{k!}(x+y)^k$$ where the last equality is the binomial theorem.

Step 3. Concluding

Putting the identity from step 2 for each $k$ into the formula in step 1 gices $$f(x)f(y) = \sum_{k=0}^\infty \sum_{n=0}^k \frac{x^ny^{k-n}}{n!(k-n)!} = \sum_{k=0}^\infty \frac{(x+y)^k}{k!} = f(x+y).$$

Answer 2

Math Werenski already gave an answer using Cauchy product so I'll try proving that using differentiation

First of all it's possible to prove that $f'(x) = f(x)$ using the standard definition of derivative

$$ \begin{align} & & f'(x) &=\lim_{\epsilon\rightarrow 0}\frac{f(x+\epsilon)-f(x)}{\epsilon} = \lim_{\epsilon \rightarrow 0} \sum_{n=1}^{\infty}\frac{((x+\epsilon)^n-x^n}{n!\epsilon}\\ && &= \lim_{\epsilon \rightarrow 0} \sum_{n=1}^{\infty}\frac{\sum_{k=0}^{n-1}x^{k}(x+\epsilon)^{n-1-k}}{n!} = \lim_{\epsilon \rightarrow 0} \sum_{n=0}^{\infty}\frac{\sum_{k=0}^{n}x^{k}(x+\epsilon)^{n-k}}{(n+1)!}\\ & &\Rightarrow f(x)'-f(x) &= \lim_{\epsilon \rightarrow 0} \sum_{n=0}^{\infty}\frac{\sum_{k=0}^{n-1}(x^{k}(x+\epsilon)^{n-k}-x^n)}{(n+1)!}\\ & & &= \lim_{\epsilon \rightarrow 0} \;\epsilon \,\sum_{n=0}^{\infty}\frac{\sum_{k=0}^{n-1}\sum_{l=0}^{n-k}x^{k+l}(x+\epsilon)^{n-k-l-1}}{(n+1)!} = 0\\ \end{align} $$

Then we have that $\frac{d}{dx}(f(x)f(-x)) = f(x)f(-x) - f(x)f(-x) = 0$ hence $f(x)f(-x) = f^2(0) = 1$ therefore $\forall x, f(x) \neq 0$.

Then for a given y we have: $$\frac{d}{dx}\frac{f(x+y)}{f(x)} = \frac{f(x+y)f(x)-f(x+y)f(x)}{f^2(x)} = 0$$ Therefore $f(x+y) = cf(x)$ where $c$ is independent of $x$

Let $x = 0$ then $c = f(y)$

Therefore $f(x+y) = f(x)f(y)$