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One of students asked me about "some beautiful properties (or relation) of $e$". Then I list like below \begin{align} & e \equiv \lim_{x \to \infty} \left(1+\frac{1}{x} \right)^x\\[10pt] & e = \sum_{k=0}^\infty \frac{1}{k!}\\[10pt] & \frac{d}{dx} (e^x) = e^x\\[10pt] & e^{ix} = \cos x + i \sin x \quad \text{(Euler)}\\[10pt] & e^{i \pi} + 1 = 0 \end{align} After this, he asked me for more relation or properties. I said I'll think and answer ...

Now I want help to add some relation, properties, or visual things (like proof without words)

Please help me to add something more. Thanks in advance.

***The class was math. 1. engineering

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    $\begingroup$ Related question. Also, don't forget $\int_1^e 1/x\,dx=1$. $\endgroup$ – Arthur Jan 23 '17 at 22:02
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    $\begingroup$ $$ \frac{1}{\pi}\int_{\mathbb{R}}\frac{\cos(x)}{1+x^2}=\frac{1}e $$ $\endgroup$ – tired Jan 23 '17 at 22:05
  • $\begingroup$ $e^{a\frac{d}{dx}}f(x)=f(x+a)$ $\endgroup$ – tired Jan 23 '17 at 22:09
  • $\begingroup$ $\exp(\text{tr}(\log(A)))=\det(A)$ $\endgroup$ – tired Jan 23 '17 at 22:11
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    $\begingroup$ The number of derángements is $!n=\left[\frac{n!}{e}\right]$ for $n\ge 1$. $\endgroup$ – Masacroso Jan 23 '17 at 22:13
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One of my favourites is the following property of $e$.

Write down a random number between $0$ and $1$. Write down another one and add it to the previous one. If the total exceeds $1$, stop. Otherwise keep adding such random numbers until the total exceeds $1$ and stop.

The expected number of such random numbers is $e$.

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    $\begingroup$ +1. A high school student I tutor, asked me today "why do we need $e$"? I offered an answer on the grounds of "making our life easier while producing beautiful results". Next lesson your answer will be the first thing he hears about. $\endgroup$ – MathematicianByMistake Jan 23 '17 at 22:52
  • $\begingroup$ This is remarkable. Do you have a reference for the property? I carried out some numerical tests just for fun, and found that if you replace the threshold with $2$, then the expected number is $e^2-e$. But if you replace the threshold with $3$, then the expected number looks like $20/3$, which seems odd... $\endgroup$ – Théophile Jan 26 '17 at 21:57
  • $\begingroup$ @Théophile...there is a very elegant proof of this result I have seen somewhere on MSE involving the volume of a simplex being $\frac{1}{n!}$. If I can find it I will post a link. $\endgroup$ – David Quinn Jan 26 '17 at 22:20
  • $\begingroup$ I worked out the math; it comes down to solving a straightforward differential equation. If $f(x)$ is the expected number of steps to cross the finish line $x$ units away, then that function must satisfy $f(x) = 1 + \int_x^{x+1}f(t)\ \textrm dt$. It turns out that $f(3) = e^3-2e^2+e/2 = 6.66656\ldots$, a happy coincidence. $\endgroup$ – Théophile Feb 7 '17 at 17:13
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My favourite is about derangements.

$n$ people write their name on different envelopes and put the envelopes in a box. If they pick a random envelope back from the box, the probability that no one picks an envelope with its name is $\approx\frac{1}{e}$.


A number-theoretic one: if $\{p_k\}_{k\geq 1}$ is the sequence of primes, $$ \prod_{k=1}^{n} p_k = e^{n+o(n)}.$$

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  • $\begingroup$ I think the asymptotic approximation is rather $e^{(1+o(1))n \log n}$ (according to the Wikipedia article on primorials). $\endgroup$ – Théophile Jan 23 '17 at 23:14
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Here are some more relations which might be pleasing.

From section 1.3 of Mathematical Constants by S.R. Finch:

  • A Wallis-like infinite product is \begin{align*} e=\frac{2}{1}\cdot\left(\frac{4}{3}\right)^{\frac{1}{2}} \cdot\left(\frac{6\cdot 8}{5\cdot 7}\right)^{\frac{1}{4}} \cdot\left(\frac{10\cdot 12\cdot 14\cdot 16}{9\cdot 11\cdot 13\cdot 15}\right)^{\frac{1}{8}}\cdots \end{align*}

  • From Stirling's formula we derive \begin{align*} e=\lim_{n\rightarrow \infty}\frac{n}{(n!)^{\frac{1}{n}}} \end{align*}

  • Another continued fraction representation is \begin{align*} e&=2+\frac{\left.2\right|}{\left|2\right.} +\frac{\left.3\right|}{\left|3\right.} +\frac{\left.4\right|}{\left|4\right.} +\frac{\left.5\right|}{\left|5\right.}+\cdots\\ &=2+\frac{2}{2+\frac{3}{3+\frac{4}{4+\frac{5}{5+\cdots}}}} \end{align*}

$$ $$

In section Intriguing Results in Real Infinite Series by D.D. Bonar and M.J. Khoury we find

  • Gem 89 (American Math Monthly 42:2 pp. 111-112)

\begin{align*} e=\frac{1}{5}\left(\frac{1^2}{0!}+\frac{2^2}{1!}+\frac{3^2}{2!}+\frac{4^2}{3!}+\cdots\right) \end{align*}

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Here is the (simple) continued fraction for $e-1.$ The pattern continues forever, 1,1,2,1,1,4,1,1,6,1,1,8,1,1,10,1,1,12,...

To get $e$ itself add one, this is the same as changing the first $1$ to a $2$

$$ \begin{array}{ccccccccccccccccccccccc} & & 1 & & 1 & & 2 & & 1 & & 1 & & 4 & & 1 & & 1 & & 6 & \\ \frac{0}{1} & \frac{1}{0} & & \frac{1}{1} & & \frac{2}{1} & & \frac{5}{3} & & \frac{7}{4} & & \frac{12}{7} & & \frac{55}{32} & & \frac{67}{39} & & \frac{122}{71} & & \frac{799}{465} \\ \end{array} $$

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Let's say that a number is cut into equal parts and then multiply those parts together, for example take number $20$ and divide it by $4$ ($20/4=5$), as result we have:

$5\times5\times5\times5=625$

The question here is to find the number $n\in \mathrm{N}$ such that is maximizes

$r = \arg \max_{n,m} \prod_{i=1}^{n}m$

where

$m=(\mathrm{your\:choice\:of\:number})/n$

We can show that the number $r$ is maximized when $m\approx e$. In the case of choosing $20$ we have

$20/7\approx e$

so

$r=(2.8)^7$

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