Is e^x the unique function of which derivative is itself? Can you prove it?

Is ${e}^{x}$ the uniq function of which derivative is itself? Can you prove it? $f ' \left(x\right) = f \left(x\right) \implies f \left(x\right) = {e}^{x}$ , is this true everytime? Is there any second function like this?

Nov 14, 2017

Almost. If $k$ is any constant then $f \left(x\right) = k {e}^{x}$ satisfies $f ' \left(x\right) = f \left(x\right)$.

Uniqueness of ${e}^{x}$ is given by requiring $f \left(x\right)$ to be defined everywhere, and $f \left(0\right) = 1$.

Explanation:

Any function of the form $f \left(x\right) = k {e}^{x}$ satisfies $f ' \left(x\right) = f \left(x\right)$

In order for ${e}^{x}$ to be the only solution we need an extra condition such as $f \left(0\right) = 1$

Suppose $f \left(x\right)$ is a well behaved function from $\mathbb{R}$ to $\mathbb{R}$ with $f \left(0\right) = 1$.

We can write the Maclaurin series for $f \left(x\right)$ as:

$f \left(x\right) = {\sum}_{n = 0}^{\infty} {a}_{n} {x}^{n} \text{ }$ for some constants ${a}_{0} , {a}_{1} , \ldots$

Then:

$1 = f \left(0\right) = {a}_{0}$

and:

$f ' \left(x\right) = {\sum}_{n = 0}^{\infty} \left(\frac{d}{\mathrm{dx}} {a}_{n} {x}^{n}\right)$

$\textcolor{w h i t e}{f ' \left(x\right)} = {\sum}_{n = 0}^{\infty} \left(n {a}_{n} {x}^{n - 1}\right)$

$\textcolor{w h i t e}{f ' \left(x\right)} = {\sum}_{n = 0}^{\infty} \left(n + 1\right) {a}_{n + 1} {x}^{n}$

Since we want $f ' \left(x\right) = f \left(x\right)$, we have:

${\sum}_{n = 0}^{\infty} \left(n + 1\right) {a}_{n + 1} {x}^{n} = {\sum}_{n = 0}^{\infty} {a}_{n} {x}^{n}$

So equating the coefficients, we find:

$\left\{\begin{matrix}{a}_{0} = 1 \\ {a}_{n + 1} = {a}_{n} / \left(n + 1\right) \text{ for "n >= 0}\end{matrix}\right.$

So:

1/a_0 = 1/1 = 0!

1/a_1 = 1/a_0 = 1!

1/a_2 = 2/a_1 = 2!

1/a_3 = 3/a_2 = 3!

and:

1/a_(n+1) = (n+1) (1/a_n) = (n+1)!

Hence:

f(x) = sum_(n=0)^oo x^n/(n!)

which is one of the definitions of ${e}^{x}$

Then if $k$ is any constant, we find:

$\frac{d}{\mathrm{dx}} k {e}^{x} = k \frac{d}{\mathrm{dx}} {e}^{x} = k {e}^{x}$

So:

$f \left(x\right) = k {e}^{x}$

is also a solution of $f \left(x\right) = f ' \left(x\right)$

Footnote

There are some non well behaved functions that also satisfy $f ' \left(x\right) = f \left(x\right)$

For example:

$f \left(x\right) = \left\{\begin{matrix}\text{undefined" " if " x=0 \\ abs(e^x)" if } x \ne 0\end{matrix}\right.$