Prove that $C e^x$ is the only set of functions for which $f(x) = f'(x)$ I was wondering on the following and I probably know the answer already: NO.
Is there another number with similar properties as $e$? So that the derivative of $ e^x$ is the same as the function itself.
I can guess that it's probably not, because otherwise $e$ wouldn't be that special, but is there any proof of it?
 A: Let $f(x)$ be a differentiable function such that $f'(x)=f(x)$. This implies that the $k$-th derivative, $f^{(k)}(x)$, is also equal to $f(x)$. In particular, $f(x)$ is $C^\infty$ and we can write a Taylor expansion for $f$:
$$T_f(x) = \sum_{k=0}^\infty c_k x^k.$$
Notice that the fact that $f(x)=f^{(k)}(x)$, for all $k\geq 0$, implies that the Taylor series $T_f(x_0)$ converges to $f(x_0)$ for every $x_0\in \mathbb{R}$ (more on this later), so we may write $f(x)=T_f(x)$. Since $f'(x) = \sum_{k=0} (k+1)c_{k+1}x^k = f(x)$, we conclude that $c_{k+1} = c_k/(k+1)$. The value of $c_0 = f(0)$, and therefore, $c_k = f(0)/k!$ for all $k\geq 0$. Hence:
$$f(x) = f(0) \sum_{k=0}^\infty \frac{x^k}{k!} = f(0) e^x,$$
as desired.
Addendum: About the convergence of the Taylor series. Let us use Taylor's remainder theorem to show that the Taylor series for $f(x)$ centered at $x=0$, denoted by $T_f(x)$, converges to $f(x)$ for all $x\in\mathbb{R}$. Let $T_{f,n}(x)$ be the $n$th Taylor polynomial for $f(x)$, also centered at $x=0$. By Taylor's theorem, we know that 
$$|R_n(x_0)|\leq |f^{(n+1)}(\xi)|\frac{ |x_0 - 0|^{n+1}}{(n+1)!},$$ 
where $R_n(x_0)=f(x) - T_{f,n}(x)$ and $\xi$ is a number between $0$ and $x_0$. Let $M=M(x_0)$ be the maximum value of $|f(x)|$ in the interval $I=[-|x_0|,|x_0|]$, which exists because $f$ is differentiable (therefore, continuous) in $I$. Since $f(x)=f^{(n+1)}(x)$, for all $n\geq 0$, we have:
$$|R_n(x_0)|\leq |f^{(n+1)}(\xi)|\frac{ |x_0|^{n+1}}{(n+1)!}\leq |f(\xi)|\frac{ |x_0|^{n+1}}{(n+1)!}\leq M \frac{|x_0|^{n+1}}{(n+1)!} \longrightarrow 0 \ \text{ as } \ n\to \infty.$$
The limit goes to $0$ because $M$ is a constant (once $x_0$ is fixed) and $A^n/n! \to 0$ for all $A\geq 0$. Therefore, $T_{f,n}(x_0) \to f(x_0)$ as $n\to \infty$ and, by definition, this means that $T_f(x_0)$ converges to $f(x_0)$. 
A: Yet another way: By the chain rule, ${\displaystyle {d \over dx} \ln|f(x)| = {f'(x) \over f(x)} = 1}$. Integrating, you get $\ln |f(x)| = x + C$. Taking $e$ to both sides, you obtain $|f(x)| = e^{x + C} = C'e^x$, where $C' > 0$. 
As a result, $f(x) = C''e^x$, where $C''$ is an arbitrary constant.  
If you are worried about $f(x)$ being zero, the above shows $f(x)$ is of the form $C''e^x$ on any interval for which $f(x)$ is nonzero. Since $f(x)$ is continuous, this implies $f(x)$ is always of that form, unless $f(x)$ is identically zero (in which case we can just take $C'' = 0$ anyhow). 
A: 
Let $x \in C^1$ on the whole line be a solution to $\dot{x}(t) = x(t)$, $x(0) = 1$. Using the Taylor expansion with remainder, show that necessarily $x(t) = e^t$.

We have that $\dot{x} = x$ implies $x^{(n)} = x^{(n-1)}$ for all $n \ge 1$, and by induction on $n$, we have that $x(t)$ is $C^\infty$ with $x^{(n)} = x$ for all $n$. Thus, if $x(0) = 1$ and $\dot{x} = x$, Taylor's Theorem gives$$x(t) = \left( \sum_{k=0}^{N-1} {{t^k}\over{k!}}\right) + {{x^{(N)}(t_1)}\over{N!}}t^N,$$for $t_1$ between $0$ and $t$. But $x^{(N)} = x$, so if$$M = \max_{|t_1| \le |t|} |x(t)|,$$which we know exist by compactness of $[-|t|, |t|]$, then$$\left| x(t) - \sum_{k=0}^{N-1} {{t^k}\over{k!}}\right| < {{Mt^N}\over{N!}}.$$The right-hand side heads to $0$ as $N \to \infty$, so the series for $e^t$ converges to $x(t)$.
A: Hint $\rm\displaystyle\:\ \begin{align} f{\:'}\!\! &=\ \rm a\ f \\ \rm \:\ g'\!\! &=\ \rm a\ g \end{align} \iff \dfrac{f{\:'}}f = \dfrac{g'}g \iff \bigg(\!\!\dfrac{f}g\bigg)' =\ 0\ \iff W(f,g) = 0\:,\ \ W = $ Wronskian
This is a very special case of the uniqueness theorem for linear differential equations, esp. how the Wronskian serves to measure linear independence of solutions. See here for a proof of the less trivial second-order case (that generalizes to n'th order). See also the classical result below on Wronskians and linear dependence from one of my sci.math posts on May 12, 2003.
Theorem $\ \ $  Suppose  $\rm\:f_1,\ldots,f_n\:$   are $\rm\:n-1\:$ times differentiable on interval $\rm\:I\subset \mathbb R\:$
and suppose they have Wronskian $\rm\: W(f_1,\ldots,f_n)\:$ vanishing at all points in $\rm\:I\:.\:$ Then  $\rm\:f_1,\ldots,f_n\:$  are linearly dependent on some subinterval of $\rm\:I\:.$
Proof $\ $  We employ the following easily proved Wronskian identity:
$\rm\qquad\ W(g\ f_1,\ldots,\:g\ f_n)\ =\ g^n\ W(f_1,\ldots,f_n)\:.\ $  This immediately implies
$\rm\qquad\quad\ \ \ W(f_1,\ldots,\: f_n)\ =\ f_1^{\:n}\ W((f_2/f_1)',\ldots,\:(f_n/f_1)'\:)\quad $ if  $\rm\:\ f_1 \ne 0 $
Proceed by induction on  $\rm\:n\:.\:$ The Theorem is clearly true if  $\rm\:n = 1\:.\:$ Suppose that $\rm\: n > 1\:$  and  $\rm\:W(f_1,\ldots,f_n) = 0\:$  for all $\rm\:x\in I.\:$
If  $\rm\:f_1 = 0\:$  throughout $\rm\:I\:$ then $\rm\: f_1,\ldots,f_n\:$  are dependent on $\rm\:I.\:$ Else $\rm\:f_1\:$ is nonzero at some point of $\rm\:I\:$ so also throughout some subinterval $\rm\:J \subset I\:,\:$ since $\rm\:f_1\:$ is continuous (being differentiable by hypothesis). By above   $\rm\:W((f_2/f_1)',\ldots,(f_n/f_1)'\:)\: =\: 0\:$ throughout $\rm\:J,\:$ so by induction there exists a subinterval $\rm\:K \subset J\:$
where the arguments of the Wronskian are linearly dependent, i.e.
on $\rm\ K:\quad\ \ \ c_2\ (f_2/f_1)' +\:\cdots\:+ c_n\ (f_n/f_1)'\:  =\ 0,\ \ $  all $\rm\:c_i'\:=\ 0\:,\ $ some $\rm\:c_j\ne 0 $
$\rm\qquad\qquad\: \Rightarrow\:\ \ ((c_2\ f_2 +\:\cdots\: + c_n\ f_n)/f_1)'\: =\ 0\ \ $   via $({\phantom m})'\:$ linear
$\rm\qquad\qquad\: \Rightarrow\quad\ \  c_2\ f_2 +\:\cdots\: + c_n\ f_n\  =\  c_1 f_1\ \ $   for some $\rm\:c_1,\ c_1'\: =\: 0 $
Therefore  $\rm\ f_1,\ldots,f_n\:$  are linearly dependent on  $\rm\:K \subset I\:.\qquad$ QED
This theorem has as immediate corollaries the well-known results
that the vanishing of the Wronskian on an interval $\rm\: I\:$ is
a necessary and sufficient condition for linear dependence of
$\rm\quad (1)\ $ functions analytic on $\rm\: I\:$
$\rm\quad (2)\ $ functions satisfying a monic homogeneous linear differential
equation
$\rm\quad\phantom{(2)}\ $ whose coefficients are continuous throughout $\rm\: I\:.\:$
A: Of course $C e^x$ has the same property for any $C$ (including $C = 0$). But these are the only ones.
Proposition: Let $f : \mathbb{R} \to \mathbb{R}$ be a differentiable function such that $f(0) = 1$ and $f'(x) = f(x)$. Then it must be the case that $f = e^x$.
Proof. Let $g(x) = f(x) e^{-x}$. Then 
$$g'(x) = -f(x) e^{-x} + f'(x) e^{-x} = (f'(x) - f(x)) e^{-x} = 0$$
by assumption, so $g$ is constant. But $g(0) = 1$, so $g(x) = 1$ identically. 
N.B. Note that it is also true that $e^{x+c}$ has the same property for any $c$. Thus there exists a function $g(c)$ such that $e^{x+c} = g(c) e^x = e^c g(x)$, and setting $c = 0$, then $x = 0$, we conclude that $g(c) = e^c$, hence $e^{x+c} = e^x e^c$. 
This observation generalizes to any differential equation with translation symmetry. Apply it to the differential equation $f''(x) + f(x) = 0$ and you get the angle addition formulas for sine and cosine. 
A: Here is a different take on the question. There is a whole spectrum of different discrete "calculi" which converge to the continuous case, each of which has it's special "$e$".
Pick some $t>0$. Consider the equation $$f(x)=\frac{f(x+t)-f(x)}{t}$$
It is not hard to show by induction that there is a function $C_t:[0,t)\to \mathbb{R}$ so that $$f(x)=C_t(\{\frac{x}{t}\})(1+t)^{\lfloor\frac{x}{t}\rfloor}$$
where $\{\cdot\}$ and $\lfloor\cdot\rfloor$ denote fractional and integer part, respectively. If I take Qiaochu's answer for comparison, then $C_t$ plays the role of the constant $C$ and $(1+t)^{\lfloor\frac{x}{t}\rfloor}$ the role of $e^x$. Therefore for such a discrete calculus the right value of "$e$" is $(1+t)^{1/t}$. Now it is clear that as $t\to 0$ the equation becomes $f(x)=f'(x)$, and $(1+t)^{1/t}\to e$.
A: The solutions of $f(x) = f'(x)$ are exactly $f(x) = f(0) e^x$. But you can also write it as $b a^x$, if you want a different basis. Then $f'(x) = b \log(a) a^x$, and so if you want $f'=f$ you need $\log(a)=1$ and $a=e$ (except for the trivial case $b=0$).
A: The proof they use at High School, so not as deep or instructive, but it doesn't require as much knowledge.
$$\begin{eqnarray*}
\frac{dy}{dx} &=& y\\
\frac{dx}{dy} &=& \frac 1 y\\
x &=& \log |y| + C\\
y &=& A\exp(x) \end{eqnarray*} $$
A: Ok, so all of the above answers either assume that the derivative of $e^x$ is itself, the derivative of $\log(x)$ is $1/x$, or some other theorem regarding Taylor series. Here is an argument which is, in some sense, from first principles.
Suppose you want to find a function which is it's own derivative. Suppose we have a real valued function $f(x)$ which is suitably well behaved such that every higher order "antiderivative" of $f$ to exist.
Let us denote the antiderivative of $f_0=f$ to be $f_1$, the antiderivative of $f_1$ to be $f_2$ and so on.
Suppose we consider the sum
\begin{equation}
E(x) = \sum_{i=0}^\infty f_i(x).
\end{equation}
Suppose we assume that the series is absolutely and uniformly convergent so that it can be differentiated term by term (this will be justified later on).
We end up with the equation
$$
\frac{d}{dx} E(x) = E(x) - f(x).
$$
Since $E$ is our guess answer for the question, we take our $f$ to be identically zero. This gives us that $f_1(x) = c_1$, $f_2(x) = c_1x+c_2$ and so on. But note here that the constants $c_1,c_2,\ldots$ should be chosen such that the series defining $E$ should converge absolutely and uniformly. Equivalently we can choose $c_i$ such that $C=\sum c_i < \infty$. Therefore we end up with the expression,
$$
E(x) = C + C\sum_{i=1}^\infty \frac{x^i}{i!}.
$$
This seems to answer the question but there is a slight caveat here.

Suppose $E$ is a function which is its own derivative, why should it be the sum of some $f_i$'s as given above?

Well, this question is answered if we can characterize the starting function $f$ in terms of $E$, which can be seen from the equation $f(x) = E(x) - E'(x)$. In fact this forces the choice of $f$ to be zero! Now, the claim is proved.
