Without using a priori knowledge of $e$ or the natural logarithm, I'm looking for a shorter proof to the statement $$\exists e>0:\forall x\in\mathbb{R}:\frac{d}{dx}e^x=e^x.\tag{1}$$

The shortest I have come up with is this, in outline form:

  1. Derive the Maclaurin series $f(x)=\sum_{n=0}^\infty\frac{x^n}{n!}$ from the 2 conditions $f(x)=f'(x)$ and $f(0)=1$.
  2. Prove absolute convergence for all $x\in\mathbb{R}$ using a standard convergence theorem.
  3. Derive $f(x+y)=f(x)f(y)$ from multiplying out $f(x+y)=\sum_{n=0}^\infty\frac{(x+y)^n}{n!}$, using the binomial theorem, and re-arranging the infinite series into the product of $\sum_{n=0}^\infty\frac{x^n}{n!}$ and $\sum_{n=0}^\infty\frac{y^n}{n!}$.
  4. Show $\forall x\in\mathbb{R}:0<f(x)$ from $f(x)=f(x/2)^2\ge0$, and $f(x)f(-x)=1$ so $f(x)\neq0$.
  5. $\forall x\in\mathbb{N}:\forall y\in\mathbb{R}:f(xy)=f(y)^x$ from induction on $x$.
  6. $\forall x\in\mathbb{Z}:\forall y\in\mathbb{R}:f(xy)=f(y)^x$ from $f(-xy)=1/f(xy)=1/f(y)^x=f(y)^{-x}$ for $x\in\mathbb{Z}_+$.
  7. $\forall x\in\mathbb{Q}:\forall y\in\mathbb{R}:f(xy)=f(y)^x$ from $f(ym/n)=f(y/n)^m=f(y/n)^{mn/n}=f(yn/n)^{m/n}=f(y)^{m/n}$ for $(m,n)\in\mathbb{Z}\times\mathbb{Z}_+$.
  8. $\forall x\in\mathbb{R}:\forall y\in\mathbb{R}:f(xy)=f(y)^x$ from $f(xy)=\lim_{i\rightarrow\infty}f(x_iy)=\lim_{i\rightarrow\infty}f(y)^{x_i}=f(y)^x$ for any rational sequence $x_i$ that converges to $x$.
  9. Finally, define $e\equiv f(1)$. $f(x)=f(1)^x=e^x$.

Is there a shorter proof?

  • 5
    $\begingroup$ Given $a>0$, how are you defining $a^x$? $\endgroup$ – Git Gud Sep 7 '13 at 18:57
  • $\begingroup$ @Git Gud $a^x$ is well-defined for $x\in\mathbb{Q}$. I.e. $a^{m/n}=\sqrt[n]{a^m}$ where $m$ carries the sign. From this, under the (reasonable) assumption that $a^x$ is continuous, it is uniquely defined. $\endgroup$ – Matt Sep 7 '13 at 19:13

I like to start at $f(x+y) = f(x)f(y) $, since that's why we use the exponential function.

Also, as Ellington didn't say, it don't mean a thing if it ain't got that derivative, so I assume that $f$ is differentiable.

Putting $y=0$, $f(x)f(0) = f(x)$, so $f(0) = 1$.


$\begin{array}\\ f(x+h)-f(x) &=f(x)f(h)-f(x)\\ &=f(x)(f(h)-1)\\ \text{so}\\ \dfrac{f(x+h)-f(x)}{h} &=f(x)\dfrac{f(h)-1}{h}\\ &=f(x)\dfrac{f(h)-f(0)}{h}\\ \text{Letting } h \to 0\\ f'(x) &=f(x)f'(0)\\ \end{array} $

From here on, other properties follow as above. The regular exponential function is characterized by having $f'(0) = 1$.

  • 1
    $\begingroup$ You'll also need the existence of such an $f(x)$ first, presumably using OP's assumption that $a^x$ is defined, from which it will follow, along with uniqueness. $\endgroup$ – Alex R. Mar 26 '16 at 22:15
  • $\begingroup$ Yep. However, a differentiable function that does not exist is unusual, and is probably paradoxical. $\endgroup$ – marty cohen Mar 26 '16 at 22:20
  • $\begingroup$ A silly example: declare $f(x)=1+f(x)$ and assume $f$ is differentiable. $\endgroup$ – Alex R. Mar 26 '16 at 22:23
  • $\begingroup$ Actually, once we have shown that $f'(x) = f'(0) f(x)$, we can then show that $f$ exists. This makes $f$ feel much better. $\endgroup$ – marty cohen Mar 26 '16 at 22:27
  • 1
    $\begingroup$ Thanks, I like this approach. You have basically replaced my first 3 steps with the assumption that there exists a differentiable group homomorphism $f:(\mathbb{R},+)\rightarrow(\mathbb{R}_{>0},\times)$ which is, in my opinion, a "deeper" characterization of $e$ than a function's derivative being equal to itself. $\endgroup$ – Matt Mar 27 '16 at 9:45

Here is seven(six) steps:

  1. Define $\log (x)$ by $\log(1) = 0$ and $ \log(x)' = \frac{1}{x}$.
  2. Prove $\log(xy) = \log(x) + \log(y)$ by considering the function $f(x) = \log(xy) - \log(x) - \log(y)$ which can be shown to have derivative zero and to be equal to zero at $x=1$.
  3. Define the exponential using a power series.
  4. Prove the exponential series converges.
  5. Use another differentiation argument to prove $\log(\exp(x)) = x$. Do this by defining $g(x) = \log(\exp(x)) - x$ then showing $g' = 0$.
  6. Define $x^y \triangleq \exp(y \log(x))$ as is standard.
  7. Notice $\exp (1)^x$ which is defined to be $\exp(x \log(\exp(1)))$ is equal to $\exp(x 1) = \exp(x)$.

Part of the reason this proof has to have so many steps is that in order to prove the number $e$ exists we must first define what the function $e: \mathbb R \rightarrow \mathbb R$ $ x \mapsto e^x$ should mean.

  • $\begingroup$ You mean "...equal to zero at $x=1$", right? $\endgroup$ – Pedro Tamaroff Sep 7 '13 at 19:51
  • $\begingroup$ We can't simply define $x^y$. We should check that our definition agrees with the "rational" definition. $\endgroup$ – savick01 Sep 7 '13 at 20:01
  • $\begingroup$ When $y$ is rational then we can say $x^y = \exp(y\log(x))$ by definition since we have already proves $\log$ and $\exp$ are each others' inverses. $\endgroup$ – Daron Sep 8 '13 at 12:19
  • $\begingroup$ It seems as if the extra steps taken to prove $x^y=\exp(y\log(x))$ as $y$ generalizes to larger sets $\mathbb{N}\subset\mathbb{Z}\subset\mathbb{Q}$ have been assumed obvious (which I have bothered to spell out), and that for $y\in\mathbb{R}\setminus\mathbb{Q}$ is where this definition applies. In any case I take your answer as an affirmation that I probably haven't missed something too obvious. $\endgroup$ – Matt Sep 8 '13 at 19:16

OP assumes that the function $a^{x}$ is defined for all $a, x \in \mathbb{R}$ and $a > 0$ such that usual properties of the exponents hold. And we are expected to show that there is a unique number $e$ such that $$\frac{d}{dx}e^{x} = e^{x}$$ for all $x \in \mathbb{R}$.

This is easy (because the biggest hurdle is to define $a^{x}$ without any use of $e$ and this has been assumed to be done somehow). Let $f(x) = a^{x}$ and then we have \begin{align} f'(x) &= \lim_{h \to 0}\frac{a^{x + h} - a^{x}}{h}\notag\\ &= \lim_{h \to 0}\frac{a^{x}a^{h} - a^{x}}{h} \text{ (via property of exponents)}\notag\\ &= a^{x}\lim_{h \to 0}\frac{a^{h} - 1}{h}\tag{1} \end{align} The challenge is to analyze the limit $$\lim_{h \to 0}\frac{a^{h} - 1}{h}$$ for all $a > 0$ and this we do now.

Using the inequality (which can be proven algebraically) $$\frac{a^{r} - 1}{r} > \frac{a^{s} - 1}{s}\tag{2}$$ for $r > s > 0, a > 1$ it is seen that $g(h) = (a^{h} - 1)/h$ is strictly increasing function of $h$ for $a > 1, h > 0$ and clearly $g(h) > 0$ for $h > 0$ so that $\lim_{h \to 0^{+}}g(h)$ exists. Further putting $h = -k$ we can see that $$\lim_{h \to 0^{-}}g(h) = \lim_{k \to 0^{+}}\frac{a^{-k} - 1}{-k} = \lim_{k \to 0^{+}}\frac{a^{k} - 1}{k}\cdot\frac{1}{a^{k}} = \lim_{h \to 0^{+}}g(h)$$ and therefore $$f'(0) = \lim_{h \to 0}g(h) = \lim_{h \to 0}\frac{a^{h} - 1}{h}$$ exists. Clearly this limit is dependent on the value of $a$ and let's denote it by $L(a)$.

We then have $L(1) = 0$ obviously and $L(a)$ exists and is non-negative for $a > 1$. If $0 < a < 1$ then we can see easily prove (by setting $b = 1/a$) that $L(a) = -L(1/a)$ so that $L(a) = \lim_{h \to 0}g(h)$ exists for all $a > 0$. Using further inequality (which can be proven algebraically) $$a^{h - 1}(a - 1) \leq \frac{a^{h} - 1}{h} \leq a - 1\tag{3}$$ for $h > 0, a > 1$ we can see that that $$L(a) = \lim_{h \to 0}\frac{a^{h} - 1}{h} \geq \frac{a - 1}{a}$$ for $a > 1$. Thus $L(a) > 0$ for $a > 1$ and since $L(a) = -L(1/a)$ it follows that $L(a) < 0$ for $0 < a < 1$.

Using the definition of $L(a)$ we can easily show that $$L(ab) = L(a) + L(b), L(a/b) = L(a) - L(b)\tag{4}$$ for positive $a, b$. From the second relation above we see that $L(a)$ is a strictly increasing function of $a$. Again from inequality $(3)$ we can see that $$\frac{a - 1}{a}\leq L(a) \leq a - 1$$ for $a > 1$ and hence by squeeze theorem $$\lim_{a \to 1^{+}}\frac{L(a)}{a - 1} = 1\tag{5}$$ If $a \to 1^{-}$ we can put $b = 1/a$ and use $L(a) = -L(b)$ to see that $$\lim_{a \to 1^{-}}\frac{L(a)}{a - 1} = 1$$ and thus we have finally $$\lim_{x \to 1}\frac{L(x)}{x - 1} = 1 = \lim_{x \to 0}\frac{L(1 + x)}{x}\tag{6}$$ From the above equation and $L(a/b) = L(a) - L(b)$ it can be easily seen that $L(x)$ is continuous and differentiable on $(0, \infty)$ and $L'(x) = 1/x$.

Further we know that $L(2) > 0$ and using $(4)$ we can get $$L(2^{n}) = nL(2), L(2^{-n}) = -nL(2)$$ so that range of $L(x)$ is $(-\infty, \infty)$. Since $L(x)$ is strictly increasing there is a unique number $e > 1$ such that $L(e) = 1$. Our job is now done. In the beginning we had shown that if $f(x) = a^{x}$ then $f'(x) = f(x)L(a)$ and hence if $f(x) = e^{x}$ then $f'(x) = f(x)L(e) = f(x)$. The function $L(x)$ is traditionally denoted by $\log x$.

The content of this answer is largely based on my blog post.

  • $\begingroup$ Thanks for the alternate approach, but is this a shorter proof? $\endgroup$ – Matt Mar 28 '16 at 9:53
  • $\begingroup$ @Matt: I have tried to explain in detail my steps so it may look bit long, but the essential thing is to identify properties of function $L(x)$ namely that it is continuous and maps $(0, \infty)$ to $(-\infty, \infty)$. And this is based on inequality $(x - 1)/x \leq L(x) \leq (x - 1)$. The derivative $L'(x)=1/x$ I have provided extra and it is not needed to answer your question. Moreover it does not use any concepts beyond elementary calculus. $\endgroup$ – Paramanand Singh Mar 29 '16 at 3:34

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