Proving the uniqueness of $e$ (defined as the number such that $(e^x)'=e^x$)

Question

If we define $e$ as such a number, that $(e^x)' = e^x$, then how do we prove its uniqueness?

Answer 1

$\def\RR{\mathbb R}$One intepretation I can make of what you wrote — that has the advantage of being true — is:

For each positive real number $a$ let $f_a:\RR\to\RR$ be the function such that $f_a(t)=a^t$ for all $t\in\RR$. Then $$f_a'=f_a\iff a=e.$$

The implication $\Longrightarrow$ is possibly the uniqueness you are after. Let me make three observations:

• It is important to notice that for this interpretation to make sense, we need (i) to have a good definition of what we mean by $a^t$ for all $a>0$ and all $t\in\RR$, so as to actually be able to define all those functions $f_a$, and (ii) to know that the functions $f_a$ that we obtain in this way have derivatives. I will not enter into this.

• I will not consider the implication $\Longleftarrow$, either, but I will assume that we already know it holds: this is the claim that the function $t\mapsto e^t$ is its own derivative. You seem to know this already, and hopefully you also know how to prove it!

• Finally, I will need the fact that the exponential function $t\mapsto e^r$ does not vanish and that its value at $1$ is $e$

These facts about exponentials can be proved in various ways — depending on how exactly you define the exponential functions — and are needed for what follows.

With that out of our way, let us check the implication $\Longrightarrow$. We will proceed in two steps. First, we show that

if $f,~g:\RR\to\RR$ are two smooth functions such that $f'=f$ and $g'=g$, then either $f$ has a zero somewere or there is a number $\lambda$ such that $g=\lambda f$.

To do this, we consider two smooth functions $f,~g:\RR\to\RR$ such that $f'=f$ and $g'=g$, suppose that $f$ is nowhere zero, and show that then there exists a scalar $\lambda$ such that $g=\lambda f$. Let us consider the function $h=g/f$: since we are supposing that $f$ is nowhere zero, this is well-defined on the whole of $\RR$ and moreover smooth. We know that its derivative is $h'=(g'f-gf')/f^2$, and the hypothesis on $f$ and $g$ implies that this is in fact $$h'=\frac{g'f-gf'}{f^2}=\frac{gf-gf}{f^2}=0.$$ We thus see that $h$ is constant: there is a real number $\lambda$ such that $h(t)=\lambda$ for all $t\in\RR$, and this means, of course, that $g(t)=\lambda f(t)$ for all $t\in\RR$. This is what we wanted.

Now we can take the second step:

If $a$ is positive number such that $f_a'=f_a$, then $a=e$.

To see this, suppose that we have a positive number $a$ such that $f_a'=f_a$. Since we also have that $f_e'=f_e$ and we know that $f_e$ is nowhere zero, what we have just done tells us that there is a number $\lambda$ such that $f_a(t)=\lambda f_e(t)$ for all $t\in\RR$. In particular, we can take $t=0$ in this equality, and find that $1=a^0=f_1(0)=\lambda f_e(0)=\lambda e^0=\lambda$. We can thus conclude that in fact we have $f_a(t)=f_e(t)$ for all $t\in\RR$. Now we take $t=1$ in this equality, and find that $a=a^1=f_a(1)=f_e(1)=e^1=e$.

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It may be of interest to you to notice what we have done above is enough to prove the following better statement:

if $f:\RR\to\RR$ is any smooth function such that $f'=f$ and $f(0)=1$, then $f(t)=e^t$ for all $t\in\RR$.

This is better because: our original statement claimed that $t\mapsto e^t$ is the unique among the functions of the form $t\mapsto a^t$ which is equal to its out derivative, while the second statement claims that the function $t\mapsto e^t$ is the unique among all smooth functions that are equal to their derivative and take the value $1$ at $0$ — a set of functions that appears at first to be much, much larger.

Answer 2

$$\left(\frac{a^x}{e^x}\right)'=\frac{(a^x)'e^x-a^x(e^x)'}{e^{2x}}=\frac{a^xe^x-a^xe^x}{e^{2x}}=0\implies \frac{a^x}{e^x}=c.$$ $$x=0\implies c=1.$$ $$x=1\implies a=e.$$

Answer 3

Let $b>0$ be positive real number, and define the function $f(x) = b^x$.

Then, by the definition of the derivative, $$ f'(x) = \lim_{y\to 0} \frac{f(x+y) - f(x)}{y} = \lim_{y\to 0} \frac{b^xb^y - b^x}{y} = L(b) b^x$$ where $L(b)$ is the following limit, $$ L(b) = \lim_{y\to 0} \frac{b^y - 1}{y} $$

It is not difficult to check that, $$ L(b^p) = p L(b) $$ Indeed, $$ L(b^p) = \lim_{y\to 0} \frac{b^{py} - 1}{y} = p \lim_{y\to 0} \frac{b^{(py)} - 1}{(py)} = p L(b) $$

If you are willing to accept (based on numerical approximations) that $L(2) < 1$ and $L(3) > 1$ then you can choose a number $2<e<3$ such that $L(e) = 1$. Now we claim that $L(b) = \log_e(b)$. To see why, $$ L(b) = L(e^{\log_e b}) = (\log _e b)L(e) = \log_e b $$ Hence, we are forced to conclude there is one such number $e$ which has this property.

Answer 4

Effectively, you are asking whether the exponential function is the only function that is its own derivative.

This amounts to solving the differential equation $y' = y$ , the solution to which is $y=Ce^x$ where $C$ is a constant (this is solved relatively simply by using separation of variables).

From this solution, you can see that there is one other function that isn't exponential that differentiates into itself, when $C=0$ , $y=0$. Clearly the derivative of $y=0$ is itself.