Definition of e, and proof of existence.

Question

So, I was wondering what is the proof of the existence of $e$. I want to define it as the number such that for which the derivative of $(b^x)'=b^x$. Now, I have to show it exists. So, I have to show that there is a number $b$ such that $\lim{\frac{b^h-1}{h}}=1$ but how would I prove that. Any nice and easy proofs? Maybe it is easier to use another definition of $e$ and show that that $e$ also fulfils this definition?

Answer 1

To do this the "logarithm first" way, write

$$ \frac{b^h-1}{h} = \int_1^b x^{h-1} \, dx. $$ We now have to use a result about interchanging limits and integrals: a simple one is that since $x^{h-1}$ are continuous functions, converging uniformly to the continuous function $x^{-1}$ on $[0,b]$ as $h \to 0$, the limit and integral can be interchanged, so $$ \lim_{h \to 0} \int_1^b x^{h-1} \, dx = \int_1^b \frac{dx}{x}. $$ The latter, normally called the logarithm of $b$, $\log{b}$, is an increasing function of $b$ that starts from $0$ and is unbounded above, so the intermediate value theorem implies there is a value of $b$ for which it has the value $1$.

Answer 2

As suggested by Jack d'Aurizio via comment, you have to proceed step by step:

• Define $b^{x} =\lim_{n\to\infty} b^{x_{n}} $ where $x_{n} $ is a sequence of rationals converging to $x$. You need to show that this definition makes sense: the limit exists and is independent of the specific sequence $x_{n} $.
• Establish that the limit $\lim_{x\to 0}(b^{x}-1)/x$ exists for all $b>0$ and defines a function of $b$ say $L(b) $.
• Show that $L(b) $ is a strictly increasing function of $b$ and maps interval $(0,\infty)$ to whole of $\mathbb{R} $. And hence $L(b) =1$ for a unique number $b>0$ which is traditionally denoted by $e$. Your problem is solved now but do continue further.
• Show that $L(ab) =L(a) +L(b) $ and $\dfrac{d}{db} L(b) =\dfrac{1}{b}$.

All the above steps are carried out in almost the same order in my blog post. The function $L$ is traditionally denoted by $\log$. By the way the approach outlined above is intuitive but difficult. There are easier approaches available to develop the theory of exponential and logarithmic functions which are described here and here.

Answer 3

Expanding on the comment of Francesco, let's start out with your hypothetical number $b$, such that the function $b^x$ differentiates to itself. In addition, let's blindly hope that this function is given by a (absolutely convergent) power series:

$$ b^x = \sum_{n = 0}^\infty a_n x^n $$

For now we just hope that this works, and see what the coefficients would have to be. Since $b^0 = 1$, we have that $a_0 = 1$. Taking derivatives on both sides gives that $(b^x)' = b^x = a_1 + 2 a_2 x + 3 a_3 x^2 + \cdots$, and so $b^0 = 1$ gives $a_1 = 1$. Keep differentiating and evaluating at $x = 0$ to find that in general, $a_n = \frac{1}{n!}$. So we have a candidate function:

$$ b^x = \sum_{n = 0}^\infty \frac{1}{n!}x^n = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots$$

You can apply some standard analysis tests (root test, ratio test, etc) to show that this function is convergent on the whole of $\mathbb{R}$. (This is essentially the "hard" step in the existence here). Furthermore, we have that

$$ \frac{b^h - 1}{h} = \frac{h + \frac{h^2}{2!} + \frac{h^3}{3!} + \cdots}{h} = 1 + \frac{h}{2!} + \frac{h}{3!} + \cdots$$

Which has a limit of $1$ as $h \to 0$. Finally, you can recover the number $e$ as a value of this function: $e = b^1$.

As an interesting fact, notice here that the only properties of the function $b^x$ actually used were that it was its own derivative, and $b^0 = 1$. This was enough to completely determine the exponential function.