# Power series solution of $f(x+y) = f(x)f(y)$ functional equation

Here on StackExchange I read a lot of interesting questions and answers about functional equations, for example a list of properties and links to questions is Overview of basic facts about Cauchy functional equation.

I'm interested in the following problem: if $f:\mathbb{R} \rightarrow \mathbb{R}$ is a continuous function verifying the functional equation $f(x+y)=f(x)f(y), \ \forall x,y\in \mathbb{R}$, find its non identically zero solution using power series.

My attempt so far using power series:
let $$f(x) = \sum_{n=0}^{\infty} a_{n} \, x^{n}$$ so $$f(y) = \sum_{n=0}^{\infty} a_{n} \, y^{n}$$ and $$f(x+y) = \sum_{n=0}^{\infty} a_{n} \, (x+y)^{n}$$

The functional equation $f(x+y)=f(x)f(y)$ leads to $$\sum_{n=0}^{\infty} a_{n} \, (x+y)^{n}=\sum_{n=0}^{\infty} a_{n} \, x^{n}\sum_{n=0}^{\infty} a_{n} \, y^{n}$$

Using the binomial theorem $$(x+y)^{n} = \sum_{k=0}^{n}\binom{n}{k}x^ky^{n-k}$$ and the Cauchy product of series $$\sum_{n=0}^{\infty} a_{n} \, x^{n}\sum_{n=0}^{\infty} a_{n} \, y^{n} = \sum_{n=0}^{\infty}(\sum_{k=0}^n a_k a_{n-k}x^k y^{n-k})$$
it follows $$\sum_{n=0}^{\infty} a_{n} (\sum_{k=0}^{n}\binom{n}{k}x^ky^{n-k})=\sum_{n=0}^{\infty}(\sum_{k=0}^n a_k a_{n-k}x^k y^{n-k})$$ $$\sum_{n=0}^{\infty}(\sum_{k=0}^{n} a_{n} \binom{n}{k}x^ky^{n-k})=\sum_{n=0}^{\infty}(\sum_{k=0}^n a_k a_{n-k}x^k y^{n-k})$$

Now I need to equate the coefficients: $$\forall n\in\mathbb N, \;\;\;\; \; a_{n} \binom{n}{k} = a_k a_{n-k} \;\; \textrm{for } k= 0,1,...,n$$

The first equation, for $n=0$, is $a_0=a_0a_0$, that is $a_0(a_0-1)=0$ with solutions $a_0=0$ and $a_0=1$. If $a_0=0$ every coefficient would be zero, so we have found the first term of the power series: $a_0=1$.

Now the problem is to determine the remaining coefficients. I tried, but it's too difficult to me.

From $a_n{n\choose n-1} = a_{n-1}a_1$ we have $a_n = a_{n-1}\dfrac{a_1}{n}$. So $a_n = \dfrac{a_1^n}{n!}$.
We know that the functional equation has as solutions the expnential functions $f(x) = a^x$ for some positive real number $a$. We are insterested to know if there is a relation between $a$ and the coefficient $a_1$.
Let us call $f_{a_1}(x)$ the solution of the functional equation where the coefficients are $(a_1)^n/n!$ and let $e$ be the real number defined by $f_1(x)$, i.e. $f_1(x) = e^x$. Then the series expansion tells us that $f_1(a_1x) = f_{a_1}(x)$, i.e. $e^{a_1x} = a^x$. For $x = 1$ we have that $e^{a_1} = a$.
From the series expansion, one sees that $e^x$ is a strictly increasing function and it's continuous by definition. Thus it has a continuous inverse. Let us call $\ln(x) = f^{-1}_{1}(x)$. Then $a_1 = \ln(a)$.
• This proves that if the functional equation has a series expansion, then it must be this one. To prove that it does indeed, one should prove that it's differentiable, i.e. the following limit exists: $$\lim_{h\to0} \dfrac{f(x+h)-f(x)}{h} = f(x)\lim_{h\to 0}\dfrac{f(h)-1}{h}.$$ Note that if it does, then $f'(x)$ is proportional to $f(x)$, so the function would be $\mathcal{C}^\infty$ and $f'(0) = a_1$. – Darth Geek Mar 9 '17 at 12:29