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Right now I'm on a journey to clear up my confusion about exponential functions. Thanks to your help in this stack, I was able to derive the basic properties of the ln function from the "position of defining ln in terms of integrals".

In this stack, the definitions of exp and e are as follows;

  • $$\exp(x):= 1+\frac{x}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!}+\ldots,\ $$
  • $$e:=\lim_{n\to \infty} \left( 1+\frac{1}{n} \right)^{n} , n\in \mathbb{N}.$$

When defined in this way, the relationship between $\exp(x)$ and $e$ seems to be non-trivial . We further assume that;

  • The theory on limit operations such as the scissors-out theorem and the theory on convergence conditions for power series are assumed to exist separately and independently.
  • For any integer $a>0$ where $n$ is an integer, $a^n$ is assumed to be defined in an inductive way.

My question.

Under these assumptions, how can we mathematically prove the following? $$\exp(1)=e$$

There is a similar stack, but it does not seem to go into the definition of Napier-number. If possible, we focus my questions on the following "How to evaluate from the lower side.

As far as I tried, I was probably able to suppress it from the upper side. In other words, I believe I was able to prove the following.

$$e\le \exp(1)$$

Therefore, perhaps the question is how to find the series to suppress from below.

This means that the most desirable answer is to show $$e\ge \exp(1)$$

Proof of $e\le exp(1)$
From the binomial theorem,

$$\left( 1+\frac{1}{n} \right)^{n} =\sum_{k=0}^{n} {}_{n} \mathrm{C}_{k} \frac{1}{n^k}\\ =\sum_{k=0}^{n} \frac{1}{k!} \left( 1- \frac{1}{n} \right)\cdots \left( 1- \frac{1-k}{n} \right)\\ \le \sum_{k=0}^{n} \frac{1}{k!} \le \exp(1) $$

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    $\begingroup$ I think you would try to prove $$\exp(x) = \lim_{n\to \infty} \left( 1+\frac{x}{n} \right)^{n} , n\in \mathbb{N}.$$ $\endgroup$
    – GEdgar
    Feb 14, 2021 at 16:29
  • $\begingroup$ Thanks for the comment. I think $(1+x/n)^n \le \exp(x)$ can be shown similarly to my proof. However, appropriate way to evaluate the lower side. $\endgroup$ Feb 14, 2021 at 16:33
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    $\begingroup$ Does this answer your question? Bernoulli's representation of Euler's number, i.e $e=\lim \limits_{x\to \infty} \left(1+\frac{1}{x}\right)^x $ $\endgroup$ Feb 14, 2021 at 16:37
  • $\begingroup$ @ Kyk The big picture is the same. But the detailed positions and focus points are very different. As you know, very basic topics are completely different if the assumptions are different. I take the position that "exponents" are only allowed for natural numbers. (That is, something like $e^\pi$ does not yet exist in this world! ) Also, the concept of exp is defined analytically. $\endgroup$ Feb 14, 2021 at 16:52
  • $\begingroup$ See also math.stackexchange.com/questions/4777547/… $\endgroup$ Sep 29, 2023 at 10:49

3 Answers 3

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Let $N\in\Bbb N$. Then, for $n\geqslant N$,\begin{align}\left(1+\frac1n\right)^n&=\sum_{k=0}^n\frac1{k!}\left(1-\frac1n\right)\cdots\left(1-\frac{k-1}n\right)\\&\geqslant\sum_{k=0}^N\frac1{k!}\left(1-\frac1n\right)\cdots\left(1-\frac{k-1}n\right).\end{align}Therefore,$$e\geqslant\sum_{k=0}^N\frac1{k!}\left(1-\frac1n\right)\cdots\left(1-\frac{k-1}n\right)$$and so$$e\geqslant\lim_{n\to\infty}\sum_{k=0}^N\frac1{k!}\left(1-\frac1n\right)\cdots\left(1-\frac{k-1}n\right)=\sum_{k=0}^N\frac1{k!}.$$Since this takes place for each $N\in\Bbb N$,$$e\geqslant\sum_{k=0}^\infty\frac1{k!}=\exp(1).$$

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    $\begingroup$ Your inequality evaluation is like watching a magic show. I was impressed. I immediately thought of the series to be suppressed from above, but I had no idea about the series to be suppressed from below. That was my main concern, so your answer is the best answer. Thank you!! $\endgroup$ Feb 14, 2021 at 17:01
  • $\begingroup$ If there is any concern, it is whether $(1-1/n)*\cdots (1-(N-1)/n)$ will converge. $\endgroup$ Feb 14, 2021 at 17:22
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    $\begingroup$ I do not mention $\left(1-\frac1n\right)\cdots\left(1-\frac{N-1}n\right)$ in my answer. What I mention is$$\sum_{k=0}^N\frac1{k!}\left(1-\frac1n\right)\cdots\left(1-\frac{k-1}n\right).$$And the limit of this expression, when $n$ goes to infinity, is $\sum_{k=0}^N\frac1{k!}$ because it is a finite sum and because, for each $k\in\{0,1,\ldots,N\}$,$$\lim_{n\to\infty}\frac1{k!}\left(1-\frac1n\right)\cdots\left(1-\frac{k-1}n\right)=\frac1{k!}.$$ $\endgroup$ Feb 14, 2021 at 17:30
  • $\begingroup$ Thanks for the comment; is the most important point that N is fixed, am I right? $\endgroup$ Feb 14, 2021 at 17:38
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    $\begingroup$ It sure is!${}$ $\endgroup$ Feb 14, 2021 at 17:40
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This appears as Theorem 8.1.6 in Strichartz's The Way of Analysis. The idea is the following:

  • First use the series definition to develop some basic properties of $\exp$: it is strictly positive on $\mathbb{R}$, it is its own derivative, it is strictly increasing, it satisfies $(\exp x)^n = \exp(nx)$.

  • Use the inverse function theorem to show that $\exp$ has an inverse function, call it $\ln$, which is differentiable. Use the chain rule to show $\frac{d}{dx} \ln x = \frac{1}{x}$. Also verify that $\ln(1)=0$ and $\ln (x^n) = n \ln x$.

  • Write $$ \lim_{n \to \infty} \ln \left(1 + \frac{1}{n}\right)^n = \lim_{n \to \infty} \frac{\ln\left(1 + \frac{1}{n}\right) - \ln(1)}{\frac{1}{n}} = \left.\frac{d}{dx}\right|_{x=1} \ln x = 1.$$

  • Take $\exp$ of both sides and use the continuity of $\exp$ to conclude.

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  • $\begingroup$ Thanks for the answer. That's a very smart way of doing things. However, I am wondering if it is possible to exchange ln and the limit. $\endgroup$ Feb 14, 2021 at 16:58
  • $\begingroup$ @BlueVarious: Yes, that would also work. $\endgroup$ Feb 14, 2021 at 16:59
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So you want to show that $$ \lim_{n\to\infty}\left(1+\frac1n\right)^n=\sum_{k=0}^\infty\frac{1}{k!}=:\exp(1) $$

Let $s_n$ be the partial sum of the series and $t_n=\left(1+\frac1n\right)^n$.

Then the binomial theorem implies that \begin{aligned} t_{n}=1+1+\frac{1}{2 !}\left(1-\frac{1}{n}\right)+\frac{1}{3 !}\left(1-\frac{1}{n}\right) &\left(1-\frac{2}{n}\right)+\cdots \\ &+\frac{1}{n !}\left(1-\frac{1}{n}\right)\left(1-\frac{2}{n}\right) \cdots\left(1-\frac{n-1}{n}\right) \end{aligned}

Thus $t_n\le s_n$, so that $$ \limsup t_n\le \exp(1)\tag{1} $$

On the other hand, if $n\ge m$, then $$ t_{n} \geq 1+1+\frac{1}{2 !}\left(1-\frac{1}{n}\right)+\cdots+\frac{1}{m !}\left(1-\frac{1}{n}\right) \cdots\left(1-\frac{m-1}{n}\right) $$ Let $n\to\infty$, keeping $m$ fixed you get $$ \liminf _{n \rightarrow \infty} t_{n} \geq 1+1+\frac{1}{2 !}+\cdots+\frac{1}{m !}=s_m\tag{2} $$

Combining (1) and (2) you are done.


Notes. This is Theorem 3.31 in Rudin's Principles of Mathematical Analysis.

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  • $\begingroup$ Thank you for your answer. I don't know the concept of limsup or liminf, but am I correct in understanding that your answer is essentially equivalent to @José Carlos Santos 's answer? $\endgroup$ Feb 14, 2021 at 17:10
  • $\begingroup$ @BlueVarious: you are welcome. they are slightly different. In my answer, it is not assumed that one knows a priori that the limit of $t_n$ exists. $\endgroup$
    – user9464
    Feb 14, 2021 at 18:01

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