TITLE: Showing $f'(x) = f(x)$ implies an exponential function
QUESTION [12 upvotes]: Possible Duplicate:
Proof that $\exp(x)$ is the only function for which $f(x) = f'(x)$

How can I show the statement $f'(x) = f(x)$ implies the function is defined as $f: \mathbb{R} \rightarrow \mathbb{R} : x \rightarrow a\cdot \exp(x)$ without using integrals.
My attempt at a solution:
I tried to show that the Taylor series of $f$ has the same structure as $a\cdot \exp(x)$, but I fail at showing that the error on the series converges to zero.

REPLY [2 votes]: Just for fun (using the "Taylor approach"):
Suppose $f_1$ and $f_2$ are two functions both of which satisfy the conditions
$$\tag{1}
f'(x)=f(x),\quad \text{for all }\ x\in\Bbb R
$$
and
$$\tag{2}
f(0)=1.
$$
Let $F=f_1-f_2$. One easily verifies that $F$ satisfies $(1)$ and $F(0)=0$. One also easily verifies, by induction, that $F^{(n)}(x)=F(x)$ for all positive integers  $n$ and all $x\in\Bbb R$.
Now, fix $x\in\Bbb R$. Let $I_x$ be the closed interval with endpoints $0$ and $x$. By Taylor's Theorem we have for $n$ a fixed positive integer:
$$
F(x)=F(0)+{F'(0)\over 1!}x +\cdots+{F^{(n-1)}(0)\over (n-1)!}x^{n-1} +{F^{(n )}(c_n)\over  n  !}x^{n }
$$
for some $c_n$ between $0$ and $x$.
But $F^{(k)}(0)=F(0)=0$ for each $1\le k\le n-1$; so,
$$\tag{3}
F(x) = {F^{(n )}(c_n)\over  n  !}x^{n }={F(c_n)\over n!}x^n.
$$
Since $F$ is continuous on $I_x$, it follows from $(3)$ that
$$\tag{4}
|F(x) |\le {K\over  n  !}|x|^{n },
$$
for some constant $K$ independent of $n$.
As $n$ was arbitrary it follows from $(4)$ that $F(x)=0$. 
As $x$ was arbitrary, it follows that $f_1=f_2$.
From this it follows that the unique function satisfying $(1)$ and $(2)$ is $f(x)=e^x$;
whence the unique function satisfying $f'=f$ and $f(0)=a$ is $f(x)=ae^x$.