Let $E:\mathbb{R} \to \mathbb{R}$ be an infinitely continuously differentiable function and $E$ is not zero function 
such that $$E(u+v)=E(u)E(v).$$
Show that $E(x)=e^{ax}$ for some $a\in \mathbb{R}$.

My partial answer:

The function $x\mapsto e^{ax}$ satisfies the properties immediately.
For $y=0$, we have $E(x)=E(x)E(0)$. Thus, $E(0)=1$. Let $y\in \mathbb{R}$ is fixed, then 
$$E'(x+y)=E(y)E'(x)$$.

I don't know how to continue the answer. 
Could you give me some hint? 
Thanks.
['https://math.stackexchange.com/questions/612405', 'https://math.stackexchange.com', 'https://math.stackexchange.com/users/27594/']
This is a classic problem, first solved by the Dutch mathematician J.C. Kluyver in 1905. He derives the result given in the [answer](https://math.stackexchange.com/a/612395/87355) of Felix Marin, that the probability to return to the unit disc after $n$ steps is $1/(n+1)$:

![](https://ilorentz.org/beenakker/MO/Kluyver1.png)

![](https://ilorentz.org/beenakker/MO/Kluyver2.png)

For a purely geometric derivation, see [arXiv:1007.4870](https://arxiv.org/abs/1007.4870). (There the result is attributed to Rayleigh, but it appears Rayleigh only considered the large-$n$ limit.) See also this [MO posting.](https://mathoverflow.net/a/347336/11260)
This addresses the t=3 case:

Assume you start at origin ($P\_0$), and without loss of generality, assume first move is to $P\_1=(1, 0)$. Your second move will be to $P\_2=(1 + cos(\theta), sin(\theta))$, where theta is uniform [0, $\pi$] (can ignore the [$\pi,2\pi$] range due to symmetry). Density function of $\theta$ is just $f(\theta)=\frac{1}{\pi}$

Draw unit-circles around $P\_0$ and $P\_2$, and let X be the second intersection point of these two circles (aside from $P\_1$).

Now draw the quadrilateral $\overline{P\_0P\_1P\_2X}$. Turns out this is actually a rhombus, with each side of length 1. From geometry of parallel-lines, one can see that the angle $\angle P\_1P\_2X$ equals theta. Also, this angle $\theta$ subtends the arc of the $P\_2$ circle that is inside the $P\_0$ circle.

Therefore, for any given $\theta$, the probability that a chosen point on the $P\_2$ circle will be inside the $P\_0$ circle is simply $\frac{\theta}{2\pi}$.

Putting this all together, the probability that we return to the unit-circle after 3 moves is:
$\int^\pi\_0(1/\pi) \* (\theta/2\pi) d\theta = 1/4$