# Functional equation related to $\sin$: $f(x+y)=f(x)f'(y)+f'(x)f(y)$

Find all differentiable $f:\mathbb R \to \mathbb R$ such that $$\forall (x,y)\in \mathbb R^2, f(x+y)=f(x)f'(y)+f'(x)f(y)$$

It's easy to check that the only constant solution is $0$ and the only polynomial solution is $x\mapsto x$. Besides, it's also easy to check that $\sin$ is a solution, as well as $\sinh$. More generally, $x\mapsto \frac {\sin(ax)}{a}$ and $x\mapsto \frac {\sinh(ax)}{a}$ are solutions.

Setting $x=y=0$ yields $f(0)(1-2f'(0))=0$.
By letting $y=0$, one gets the ODE $f(0)f'(x)+(f'(0)-1)f(x)=0$.

If $f(0)\neq 0$, then $f'(0)=\frac 12$ and this is easily solved as $x\mapsto \frac {\exp(ax)}{2a}$
If $f(0)=0$, either $f'(0)\neq 1$ and then $f=0$ , or $f'(0)=1$ and the ODE is now useless.

How should I continue ? Are there other solutions ?

• Differentiating w.r.t. $x$ gives $f'(x+y)=f'(x)f'(y)+f''(x)f(y)$. W.r.t. $y$ gives $f'(x+y)=f(x)f''(y)+f'(x)f'(y)$. Therefore $f(x)f''(y)=f(y)f''(x)$ for all $x,y$. Implying that $f(x)/f''(x)$ must be a constant. Probably we can make progress from here on. Also, probably there's a cleaner way. Gotta go, sorry. – Jyrki Lahtonen Sep 3 '18 at 15:10
• Looks like a special case of Solution to functional equation. – Martin R Sep 3 '18 at 15:11
• $f$ must be twice differentiable since $f'(x+y)=f'(x)f'(y)+f''(x)f(y)$ proves it. – Marco Sep 3 '18 at 15:21
• @LorenzoQuarisa: Fix any $y$ with $f(y) \ne 0$ and solve the functional equation for $f'(x)$. It follows that $f'$ is differentiable. – Martin R Sep 3 '18 at 15:22
• Possible duplicate of Solution to functional equation – Jyrki Lahtonen Sep 3 '18 at 17:59

With $y=0$ you get $f(x)=f(x)f'(0)+f'(x)f(0)$.

If $f(0)\ne0$, we obtain $f'(x)=\frac{1}{f(0)}f(x)(1-f'(0))$. The case $f'(0)=1$ yields $f'(x)=0$, so $f$ is constant $c$, which implies $c=0$: a contradiction.

If $f'(0)\ne1$, we can write $f'(x)=kf(x)$, with $k\ne0$, so $f(x)=ae^{kx}$ (with $a\ne0$). The main relation now is $$ae^{k(x+y)}=ae^{kx}ake^{ky}+ake^{kx}ae^{ky}$$ that implies $1=2ak$. This is a solution.

If $f(0)=0$, the equation becomes $f(x)=f(x)f'(0)$ or $(1-f'(0))f(x)=0$. If $f'(0)\ne 1$, we get the constant $0$ function.

If $f(0)=0$ and $f'(0)=1$, the business becomes interesting.

Since $f'(0)\ne1$, the function is not constant, so there is $y_0$ with $f(y_0)\ne0$. In particular, for all $x$, $$f'(x)=\frac{1}{f(y_0)}(f(x+y_0)-f(x)f'(y_0))$$ which shows $f'$ is differentiable (and also that $f$ is infinitely differentiable). Thus we can differentiate the main relation with respect to $x$ and $y$: \begin{align} f'(x+y)&=f'(x)f'(y)+f''(x)f(y) \\ f'(x+y)&=f(x)f''(y)+f'(x)f'(y) \end{align} that implies $f(x)f''(y)=f''(x)f(y)$. Therefore, for all $x$, $$f''(x)=rf(x)$$ where $r=f''(y_0)/f(y_0)$.

This is an easy differential equation.

• @GabrielRomon That's not a contradiction: $0=f'(0)\ne1$. – egreg Sep 3 '18 at 17:27

Just suggesting an approach. Its clear that its infinitely differentiable.Let $f$ be analytic. $f(x) = \sum_{i=0}^{\infty} a_i (x-a)^i$.

Then $f(x+y) = f(x) f'(y) + f'(x) f(y)$ implies

$\sum_{i=0}^{\infty} a_i (x+y-a)^i = (\sum_{i=0}^{\infty} a_i (x-a)^i)(\sum_{i=1}^{\infty} i a_i (y-a)^{i-1}) + (\sum_{i=0}^{\infty} a_i (y-a)^i)(\sum_{i=1}^{\infty} i a_i (x-a)^{i-1})$.

Now solve for $a_i$.

For example: Let $a=0$. setting $x=0,y=0$

$a_0 = a_0 a_1 + a_0 a_1$.

Hence $a_1 = \frac{1}{2}$ or $a_0=0$.

Now set $a=0,x=0$: We get $\sum_{i=0}^{\infty} a_i y^i = a_0 \sum_{i=1}^{\infty} i a_i y^{i-1} + a_1 \sum_{i=0}^{\infty} a_i y^{i}$

Hence $a_i = (i+1) a_0 a_{i+1} + a_1 a_i$ Hence $a_{i+1} = \frac{a_i(1-a_1)}{(i+1)a_0}$. (Assuming $a_0 \neq 0$)

• Infinitely differentiable does not imply analytic though. – Gabriel Romon Sep 3 '18 at 16:21
• Atleast $a_1$ is solved for... If you take $a=0$, this might help. – Balaji sb Sep 3 '18 at 16:28
• I just edited it ..its either $a_1=1/2$ or $a_0=0$ – Balaji sb Sep 3 '18 at 16:29
• I updated with calcuation for $a_i$ assuming $f$ is analytic near 0. – Balaji sb Sep 3 '18 at 16:47