# 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.

## 1 Answer

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 $a_1 = \ln(a)$.

• So it was that easy! This means that all those conditions on the coefficients that scared me are automatically satisfied, because we already know that the answer is correct. Otherwise we need to prove it, for example veryfing that this series expansion is indeed the solution of the functional equation, but this is not interesting because we know how to do this, or prove directly that all other conditions are satified. Do you think a direct proof is that easy as well? – Massimo Mar 9 '17 at 11:52
• 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