10

In my calculus class, $e$ was defined to be the number such that $\frac{d}{dx}e^x = e^x$.

From the definition of the derivative, we have

\begin{align*} \frac{d}{dx}a^x &= \lim_{h \to 0} \frac{a^{x+h} - a^x}{h}\\ &= a^x \lim_{h \to 0} \frac{a^h - 1}{h} \end{align*}

Thus $e$ is the number such that

$$ \lim_{h \to 0} \frac{e^h - 1}{h} = 1 $$

But how is it proven that there exists such number?

WeakestTopology
  • 1,167
  • 2
    Possible duplicate of Is $e$ a coincidence? – lioness99a Jun 06 '17 at 13:41
  • 1
    Here's a hella long proof of it existing: https://www2.bc.edu/robert-c-haraway/exist.pdf – Brenton Jun 06 '17 at 13:42
  • Rather then treating it as a limit, pretend $h$ is "very small" and solve the equation for $e$. It's not rigorous, but this should give you the common definition of $e$. Now, all you would have to do is prove it is a limit that exists. – Kaynex Jun 06 '17 at 13:46
  • 2
    You can show that $\lim_{n\to\infty} \left(1+1/n\right)^n$ exists and call this limit $e$. Then $\lim_{h \to 0} \frac{e^h - 1}{h} = 1$ follows from that and $\lim_{n\to\infty} \left(1+x/n\right)^n=e^x.$ – sharding4 Jun 06 '17 at 13:49
  • 1
    How do you define $a^h$ and $a^x$ if you don't have $e^x$? Usually you start with $e^x$, define $\log x$ from that (or vice versa), then define $a^x=e^{x \log a}$. There are a number of ways to define $e^x$. You pick one, then prove the rest as theorems. If you are going to follow your approach, you have to prove the function $e^x$ exists and is unique from the definition, then $e=e^1$ is easy. You need $e^0=1$ as part of your definition. – Ross Millikan Jun 06 '17 at 14:56
  • @Brenton: that paper assumes you already have $a^x$, though I have usually seen $a^x$ defined in terms of $e^x$. I don't know how it is defined. – Ross Millikan Jun 06 '17 at 15:02
  • @RossMillikan: You can define $a^x$ for rational $x$ in the usual algebraic fashion, and extend by continuity. Although, I suppose, this presumes we've shown that $n$-th roots of positive numbers exist. –  Jun 06 '17 at 16:16
  • If it can be shown that $f(a) := \left. \frac{d}{dx}(a^x) \right|_{x=0}$ is continuous and that $f(2) < 1$ and $f(3) > 1$ then by the Intermediate value theorem there must be an $e$ with $2 < e < 3$ such that $f(e) = 1$. – md2perpe Jun 06 '17 at 19:08
  • See this answer https://math.stackexchange.com/a/678361/72031 – Paramanand Singh Jun 08 '17 at 05:04
  • This is yet another question whose proper solution lies in a rigorous theory of exponential and logarithmic functions. Many calculus textbooks (especially the introductory ones) don't provide this and instead give some incomplete definitions like the one used in this question. For the benefit of those who are perplexed by such presentation in textbooks I have written some blog posts starting with http://paramanands.blogspot.com/2014/05/theories-of-exponential-and-logarithmic-functions-part-1.html where these questions receive full attention. – Paramanand Singh Jun 08 '17 at 05:15

3 Answers3

7

Let $a>0$ and $a\ne1$. First we have to prove the existence of $\displaystyle \lim_{h\to0}\frac{a^h-1}{h}$.

Assume that $r>1$ and let $f(x)=x^r-rx+r-1$ for $x>0$. Then

$$f'(x)=r(x^{r-1}-1)\begin{cases}<0 &\text{if }0<x<1\\ =0 &\text{if }x=1\\ >0 &\text{if }x>1 \end{cases}$$

Therefore, $f$ attains its absolute minimum at $x=1$. So for all $x>0$, we have

$$f(x)\ge f(1)=0$$

$$x^r\ge rx+1-r$$

So when $r>1$ and $h>0$, $\displaystyle\frac{a^{rh}-1}{rh}\ge\frac{ra^h+1-r-1}{rh}$ and hence

\begin{align} \frac{a^{rh}-1}{rh}-\frac{a^h-1}{h}\ge0 \end{align}

When $r>1$ and $h<0$, $\displaystyle\frac{a^{rh}-1}{rh}\le\frac{ra^h+1-r-1}{rh}$ and hence

\begin{align} \frac{a^{rh}-1}{rh}-\frac{a^h-1}{h}\le0 \end{align}

Therefore, $\displaystyle \frac{a^h-1}{h}$ is an increasing function in $h$. As it is bounded below by $0$ on $(0,\infty)$, $\displaystyle \lim_{h\to0^+}\frac{a^h-1}{h}$ exists.

When $h<0$,

$$\frac{a^h-1}{h}=a^h\left(\frac{a^{-h}-1}{-h}\right)$$

As $\displaystyle \lim_{h\to0^-}a^h$ exists and equals $1$, $\displaystyle \lim_{h\to0^-}\frac{a^h-1}{h}$ exists and $\displaystyle \lim_{h\to0^-}\frac{a^h-1}{h}= \lim_{h\to0^+}\frac{a^h-1}{h}$.

Therefore, $\displaystyle \lim_{h\to0}\frac{a^h-1}{h}$ exists.

Now we are ready to prove that there exists an $e$ such that $\displaystyle \lim_{h\to0}\frac{e^h-1}{h}=1$.

Define $e=a^\frac{1}{k}$, where $\displaystyle k=\lim_{h\to0}\frac{a^h-1}{h}$. Then

\begin{align} \lim_{h\to0}\frac{e^h-1}{h}&=\lim_{h\to0}\left(\frac{a^\frac{h}{k}-1}{\frac{h}{k}}\cdot \frac{1}{k}\right)\\ &=k\cdot\frac{1}{k}\\ &=1 \end{align}

This number $e$ is unique. Indeed, if $b>0$ and $\displaystyle \lim_{h\to0}\frac{b^h-1}{h}=1$, then we can prove that $b=e$.

Let $p=\log_eb$. Then $b=e^p$.

\begin{align} \lim_{h\to 0}\frac{b^h-1}{h}-\lim_{h\to 0}\frac{e^h-1}{h}&=1-1\\ \lim_{h\to 0}\frac{e^{ph}-e^h}{h}&=0\\ \lim_{h\to 0}\left[(p-1)e^h\cdot\frac{e^{(p-1)h}-1}{(p-1)h}\right]&=0\\ (p-1)(1)(1)&=0\\ p&=1 \end{align}

Hence $b=e$.

CY Aries
  • 23,393
  • Nice proof about existence of $e$ so +1 but you need to show that the $e$ defined in this manner is unique (ie independent of the number $a$). – Paramanand Singh Jun 08 '17 at 05:08
  • Also your starting line does it seem to be related to rest of your answer. Your answer does not in anyway depend on log function. – Paramanand Singh Jun 08 '17 at 05:11
  • @ParamanandSingh I originally planned to define $e$ by logarithm. That's why I have put the first line. It is deleted now. Thanks for pointing out the uniqueness problem. I have added the proof in my answer. – CY Aries Jun 08 '17 at 06:21
  • The part about uniqueness has some issue. Since $e$ is not unique you are not allowed to take logs to the base $e$. Uniqueness follows by the following result $$\lim_{h\to 0}\frac{a^{h}-1}{h}=0\Rightarrow a=1$$ you should try to prove this. One way is to show that the limit defines a function of $a$ which is strictly increasing. – Paramanand Singh Jun 08 '17 at 07:03
  • @ParamanandSingh I starts with an arbitrary $a$ and define $e$ by $e=a^{1/k}$. This $e$ is not guaranteed to be unique. But then I suppose that there is a $b$ such that $\lim_{h\to0}\frac{b^h-1}{h}=1$ and let $p$ be the logarithm of this $b$ to this particular $e$ (I am not assuming the uniqueness of $e$ here). And I finally proved that $p=1$ and conclude that $b=e$. So I essentially proved that when $b$ has the property that $\lim_{h\to0}\frac{b^h-1}{h}=1$, this $b$ must be equal $e$. Is this approach correct? – CY Aries Jun 08 '17 at 07:50
  • I am not sure how to pinpoint the issue here. Since $e$ is not itself any specific value, the equation $b=e$ also implies that $b$ does not have any specific value. The right approach is to use my hint in previous comments. Also you don't need to use the logarithm function. – Paramanand Singh Jun 08 '17 at 08:03
4

It depends on how you define $e$. In some senses, you could define $$e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!} $$ and then from here set $x=1$ and you can prove that $e$ has all the properties we know and love. You can also show that this is equivalent to the limit definition of $e$.

It exists because its definition is based on concepts that are already well-defined (and the power series is convergent for all $x \in \mathbb{R}$ and so defines a function on $\mathbb{R}$).

Trevor Gunn
  • 27,041
PhysicsMathsLove
  • 3,142
  • 1
  • 21
  • 39
0

Here's a proof for the alternative form $e := \lim_n (1+1/n)^n$.

Note first that the binomial theorem yields

$$e = \lim_n \sum_{k=0}^n \binom{n}{k} \left ( \frac{1}{n} \right )^k$$

and the elementary bounds $\frac{n^k}{k^k} \le \binom{n}{k} \le \frac{n^k}{k!}$ yield respective bounds

$$\lim_n \sum_{k=0}^n \frac{1}{k^k} \le e \le \lim_n \sum_{k=0}^n \frac{1}{k!}$$

Fixing any $n > 1$ and avoiding the limit for the lower bound gives the weaker bound $2 < e$, while noting that $k! \ge 2^{k-1}$ for $k \ge 1$ gives the weaker bound $e \le 1 + \lim_n \sum_{k=0}^n 2^{-k} = 3$, i.e., $2 < e \le 3$.

(Note that the sophomore's dream has made an appearance here.)

S Huntsman
  • 465