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What are some surprising equations/identities that you have seen, which you would not have expected?

This could be complex numbers, trigonometric identities, combinatorial results, algebraic results, etc.

I'd request to avoid 'standard' / well-known results like $ e^{i \pi} + 1 = 0$.

Please write a single identity (or group of identities) in each answer.

I found this list of Funny identities, in which there is some overlap.

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    $\begingroup$ I really can't believe no one has posted this yet: xkcd.com/687 $\endgroup$ Sep 26, 2013 at 14:48
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    $\begingroup$ This is not in line with what you are looking for, but as a child I discovered that 10million pi is the number of seconds in a year to 1/2% accuracy. This is useful for quick back of envelope calculations, where seconds are involved. $\endgroup$ Sep 26, 2013 at 19:48
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    $\begingroup$ Pi seconds is a nanocentury! $\endgroup$ Oct 1, 2013 at 21:59
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    $\begingroup$ The three trigonometric identities in the following exercises of my Wikibook: en.wikibooks.org/wiki/On_2D_Inverse_Problems/… $\endgroup$
    – DVD
    Oct 12, 2015 at 2:51
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    $\begingroup$ This shouldn't be closed. There are still so many nice equations.In spherical trigonometry $$\frac{sin(A)}{sin(a)}=\frac{sin(B)}{sin(b)}=\frac{sin(B)}{sin(b)}$$ where the capital letters are the angles and lowercase are the opposite sides. $\endgroup$
    – skan
    Nov 26, 2016 at 17:12

118 Answers 118

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A fun one from probability: If $U, V, W$ are independent random variables, each uniform on $(0,1)$, then $(UV)^W$ is also uniform on $(0,1)$.

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An impressive relationship connecting three important functions in number theory $$\sigma(n)+ \phi(n) = n \cdot d(n) \qquad \text{for } n \in \mathbb P $$ $\sigma(n)$ - sum of positive divisors of n

$\phi(n)$ - Euler's totient function

$d(n)$ - number of positive divisors of n

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    $\begingroup$ I don't understand. $\sigma(6) + \phi(6) = 6 + 2 = 8 \neq 18 = 6\cdot3 = 6\cdot d(6)$ $\endgroup$ Dec 28, 2019 at 0:09
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    $\begingroup$ @Oscar, you are absolutely right. I totally forgot to mention, that the equation is true iff n is a prime number. $\endgroup$
    – ThomasL
    Dec 28, 2019 at 19:03
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A surprising family of series for $e$ can be derived by algebraically combining the terms in Newton's series expansion:

\begin{equation} e=\sum_{k=0}^{\infty } \dfrac{1}{k!}=\dfrac{1}{0!}+\dfrac{1}{1!}+\dfrac{1}{2!}+\dfrac{1}{3!}+\dfrac{1}{4!}+\dfrac{1}{5!}+\ldots. \end{equation}

Here are a few examples:

\begin{equation} e=\sum _{k=0}^{\infty } \frac{2k+1}{(2k)!}=\frac{1}{0!}+\frac{3}{2!}+\frac{5}{4!}+\frac{7}{6!}+\frac{9}{8!}+\frac{11}{10!}+\ldots \end{equation}

\begin{equation} 2e=\sum _{k=0}^{\infty } \frac{k+1}{k!}=\frac{1}{0!}+\frac{2}{1!}+\frac{3}{2!}+\frac{4}{3!}+\frac{5}{4!}+\frac{6}{5!}+\ldots \end{equation}

\begin{equation} 1/e=\sum _{k=0}^{\infty } \frac{1-2k}{(2k)!}=\frac{1}{0!}-\frac{1}{2!}-\frac{3}{4!}-\frac{5}{6!}-\frac{7}{8!}-\frac{9}{10!}-\ldots~. \end{equation}

Beyond being pretty, these series converge substantially faster than Newton's series. For more formulas and details on derivation see: http://www.brotherstechnology.com/math/cmj-supp.html

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    $\begingroup$ $$ 1=\sum_{k=0}^{\infty } \frac{1}{k!(k+2)}=\frac{1}{0!2}+\frac{1}{1!3}+\frac{1}{2!4}+\frac{1}{3!5}+\frac{1}{4!6}+\frac{1}{5!7}+\ldots $$ $\endgroup$ Jan 30, 2014 at 21:56
  • $\begingroup$ Nice, @FredKline. The above link leads to a list of formulas at: brotherstechnology.com/math/e-formulas.html. This is a variation of Formula (26) with n=1: \begin{equation} 1=\sum _{k=1}^{\infty } \frac{k}{(k+1)!}=\frac{1}{2!}+\frac{2}{3!}+\frac{3}{4!}+\frac{4}{5!}+\frac{5}{6!}+\frac{6}{7!}+\ldots~. \end{equation} $\endgroup$
    – Harlan
    Feb 1, 2014 at 0:23
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It has surprised me that

$$F_n^\star=\frac{\phi^n-(-\phi)^{-n}}{\sqrt5}$$

where $\phi$ is the golden ratio and $F_n^\star$ is the nth Fibonacci number. Then, one stumbles upon characteristic equations:

$$a_kF_{n+k}=a_{k-1}F_{n+k-1}+a_{k-2}F_{n+k-2}+\dots+a_0F_n$$

$$\implies F_n=b_kr_k^n+b_{k-1}r_{k-1}^n+\dots+b_1r_1^n$$

where $r$ is a solution of the equation

$$a_kr^k=a_{k-1}r^{k-1}+a_{k-2}r^{k-2}+\dots+a_0$$

assuming the roots do not repeat. The coefficients are determined based on initial values.

It then surprises me further that this extends to differential equations:

$$a_ky^{(k)}+a_{k-1}y^{(k-1)}+\dots+a_0y=0$$

$$\implies y=b_ke^{r_kx}+b_{k-1}e^{r_{k-1}x}+\dots+b_1e^{r_1x}$$

where $r$ is a root of the equation

$$a_kr^k+a_{k-1}r^{k-1}+\dots+a_0=0$$

again assuming roots do not repeat.

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Prime solutions for the Prouhet-Tarry-Escott problem of size ten ($10$ primes on the left side, other $10$ primes on the right side) .

$$ 2589701^k + 2972741^k + 6579701^k + 9388661^k + 9420581^k + 15740741^k + 15772661^k + 18581621^k + 22188581^k + 22571621^k $$ $$ = 2749301^k + 2781221^k + 6835061^k + 8399141^k + 10314341^k + 14846981^k + 16762181^k + 18326261^k + 22380101^k + 22412021^k $$

( Prime solution, $ k = 1, 2, 3, 4, 5, 6, 7, 8, 9 $)

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$$\prod_{n=1}^{\infty}\frac1{4en}\bigg(\frac{(16n^2-9)^3}{16n^2-1}\bigg)^{1/4}\bigg(\frac{(4n+3)(4n-1)}{(4n-3)(4n+1)}\bigg)^n=\sqrt{\frac{2}{3\pi\sqrt{3}}}\exp\bigg(\frac{G}{\pi}+\frac12\bigg)$$ Where $G$ is Catalan's constant.

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•I will love to add this nested radical from Ramanujan: $$\sqrt{1+2\sqrt{1+3\sqrt{1+4\sqrt{\cdots}}}}=3$$ •Also like to add this too:$$1+2+3+4+\cdots=-\frac{1}{12}$$ •This famous Stirling'sapproximation$$n!\approx\sqrt{2\pi n}{\left(\frac{n}{e}\right)}^n$$

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Möbius inversion formula may be an example. Also Ramanujan's Partition Congruences were surprising to me when I first saw them.

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Simple and surprising $$7! = 7\cdot{8}\cdot{9}\cdot{10}$$ $$\frac{1}{2}! = \sqrt{\frac{\pi}{4}}$$ $$\frac{123456789}{987654321} = 0.1249999988$$

EDIT: here's one interesting and surprising relationship I came up with myself, some years ago $$\color{purple}{\sin{\frac{\pi}{36}} \approx \frac{1}{6+\sqrt{3}+\sqrt{14}}}$$

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  • $\begingroup$ And $$ 6!\cdot 7!=10!$$ how many other pairs is this true I wonder.... $\endgroup$ Nov 2, 2023 at 11:20
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Something very exotic... and I do not know, whether this example fits the bill for this question here. But let's see.

In some sense it seems to be possible to assign equality $$ e = \tfrac 1{e^1} \tfrac 1{e^2} \tfrac 1{e^4}\tfrac 1{e^8}...$$


Originally I thought I had a mathematical contradiction when I wrote: $$ \text{ How can } \qquad e^{-1-2-4-8-16-...} = e^{-1/(1-2)}= e^{+1} = e \qquad \text{?}$$

Initially I thought that this were an example where the rule of the closed form of the geometric series might break. The equality seems impossible because the product is even of only decreasing factors and should so be convergent and moreover converge to zero: $$e^{-(1+2+4+8+16+...)} = \tfrac 1{e^1} \tfrac 1{e^2} \tfrac 1{e^4}... $$ so $$ e \overset{???}{=} \tfrac 1{e^1} \tfrac 1{e^2} \tfrac 1{e^4}\tfrac 1{e^8}...$$ Well, in some circumstances in the context of divergent series we observe strange things - but here the factors are all nicely decreasing and no unexpected effect should occur.
But - that this actually holds in some sense was mentioned by Robert Israel here in MSE

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    $\begingroup$ This follows from the fallacious argument that $1 +2 + 4 + 8 + \ldots = \frac{1}{1-2} $, which doesn't hold because to apply the GP sum to infinity, we need $|x|<1$. $\endgroup$
    – Calvin Lin
    Sep 27, 2013 at 13:00
  • $\begingroup$ After looking over Robert's answer, he said that "I would avoid writing it as $e$". And in fact, you have not shown that it is $e$, because you do not know what the analytic continuation is. I do not think you can extend it beyond the unit circle. $\endgroup$
    – Calvin Lin
    Sep 27, 2013 at 13:10
  • $\begingroup$ @Calvin: Here is another link where I stumbled into this problem earlier, discussed it with some arguments and halfbrewn counterarguments - maybe somehow instructive, too (not only for historical reasons) : math.eretrandre.org/tetrationforum/showthread.php?tid=420 $\endgroup$ Sep 27, 2013 at 13:13
  • $\begingroup$ @Calvin: Well, he wrote "... it is indeed true that ... has an analytical continuation of.. with value $e$ at $z=2$ ..." . But true, he also said he would avoid to write the equality in the product-notation. That is why I added the phrase "in some sense" to the equation. All in all - maybe the given example is too complicated here for that list of curious examples... on the other hand, the OP didn't want to get the standard ones... Hmm, I don't know whether I should delete it? (Upps- the OP: that were you ;-) - sorry) $\endgroup$ Sep 27, 2013 at 13:33
  • $\begingroup$ I think you should justify the "in some sense". I'm fine with leaving it up and letting the community vote, and you can always delete it later. E.g. if you simply wrote $1+2 + 4 + \ldots = -\frac{1}{2}$, I think that will result in a lot of down votes immediately. My main concern is that the GP doesn't converge on the unit circle, and so I don't understand the analytic continuation argument that allows you to push through this boundary. $\endgroup$
    – Calvin Lin
    Sep 27, 2013 at 13:38
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Further working on TobiMcNamobi 's identity:

$$ \begin{array}{rcl} \dfrac{1}{7} & = & \sum\limits_{k=1}^{\infty} 2^k\times7\times10^{-2k} \\ \dfrac{1}{49} & = & \sum\limits_{k=1}^{\infty} 2^k\times10^{-2k} \\ & = & \sum\limits_{k=1}^{\infty} 2^k\times100^{-k} \\ & = & \sum\limits_{k=1}^{\infty} \left(\dfrac{2}{100}\right)^k \\ & = & \sum\limits_{k=1}^{\infty} \left(\dfrac{1}{50}\right)^k \end{array} \\ \\ \boxed{\dfrac{1}{49} = \dfrac{1}{50} + \dfrac{1}{2500} + \dfrac{1}{625000} + \cdots} $$

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    $\begingroup$ Well, that's just the GP formula. $\endgroup$
    – Calvin Lin
    Sep 28, 2013 at 4:45
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I believe that one that should be mentioned is the prime number theorem:
Let $\pi(x)$ be the number of primes not exceeding $x$. Then:
$\pi(x)\sim \frac{x}{logx}$

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Wheatstone's identity, which shows how powers can be constructed via arithmetic progressions: $$n^a = \sum_{k=0}^{t-1}\biggl(\frac{n^a}{t}-\frac{\delta(t-1)}{2}+k\delta\biggr).$$ Put into words: An arithmetic progression of $t$ terms with constant difference $\delta$ and first term $\tfrac{1}{t}n^a-\tfrac{1}{2}\delta(t-1)$ will sum to $n^a$.

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Not sure if it's an equation but.. I was very surprised that among all two-dimensional compact orientable surfaces, a 2-sphere $S^2$ is the only one that has nontrivial higher homotopy groups. So, from the point of view of homotopy, sphere is the "most complicated" surface in some sense.

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  • $\begingroup$ Isn't this just because you measure homotopy by using the embeddings of spheres into your space? $\endgroup$ Mar 1, 2016 at 9:33
  • $\begingroup$ @OscarCunningham Unfortunately, I don't understand what you mean by "measuring homotopies by embeddings...". $\endgroup$ Sep 15, 2016 at 8:45
  • $\begingroup$ I mean that the second homotopy group of a space $X$ is defined as the set of injections $S^2\to X$, along with some multiplication operation. So it's not so surprising that $S^2$ has an interesting second homotopy group, since $S^2$ is itself involved in the definition. $\endgroup$ Sep 15, 2016 at 8:49
  • $\begingroup$ @OscarCunningham No, homotopy groups are not defined by injections and/or embeddings, but rather by continuous maps (up to homotopy). Also $\pi_2$ is nontrivial for the projective space but I don't think you can find an embedding. And of course, I talked about higher homotopy groups as well, not just the second! Don't you find it surprising that it's so enormly complex (and still open) for the sphere but they are all completely trivial for a sphere with handles? $\endgroup$ Sep 15, 2016 at 8:54
  • $\begingroup$ Oh you're right (I've been thinking about knot theory recently, where you do only care about injections). It's not so surprising to me that adding handles to a sphere stops there from being any non-trivial function from a sphere. What is surprising to me is that this doesn't happen in higher dimensions! $\endgroup$ Sep 15, 2016 at 9:19
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I have found the identity below from which can be deduced infinitely many others of increasing degree by the elementary theory of elliptic curves.

The 6-tuples indicate the coefficients of, respectively $n^5$, $n^4$, $n^3$, $n^2$, $n$ and $1$
$$ A = (1, 10, -8, 16, 64, -32) \\ B = (1, -10, -8, -16, 64, 32) \\ C = (-1, 8, 8, -16, 80, 32) \\ D = (-1, -8, 8, 16, 80, -32)$$

Then it is verified the identity $$n(6n^4 +24n^2 + 96)^3 = A^3 + B^3 + C^3 + D^3$$

The factor of $n$ at the left is never nul for any integer $n$.

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  • $\begingroup$ I wanted to write 6$n^4$ + 24$n^2$ + 96 at the left but I could not do $\endgroup$
    – Piquito
    Apr 22, 2015 at 0:14
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For the orthogonal ($\vec{v}_\perp$) and parallel component ($\vec{v}_\parallel$) of a $\Bbb R^3$ vector $\vec{v}$ along another vector $\vec{k}$ with lenght $k$ we have $$ k\,\vec{v}\, k= \underbrace{\vec{k}\,(\vec{v}\cdot\vec{k})}_{k \vec{v}_\parallel k} + \underbrace{\vec{k}\times(\vec{v}\times\vec{k})}_{k \vec{v}_\perp k} $$

or a bit less showy

$$\vec{v}= \frac{\vec{k}\,(\vec{v}\cdot\vec{k}) + \vec{k}\times(\vec{v}\times\vec{k})}{k^2},$$

which for unit length vectors $\vec{k}$ gives obvioulsy

$$ \vec{v}= \underbrace{\vec{k}\,(\vec{v}\cdot\vec{k})}_{\vec{v}_\parallel} + \underbrace{\vec{k}\times(\vec{v}\times\vec{k})}_{\vec{v}_\perp}. $$

To me this looked a first quite surprising since the scalar, the cross and the product of scalar and vector, are combined in a rather symmetric and harmonic fashion here. This is kind of unexpected, since as we know the one is a vector and the other one is a scalar. Though its of course very easy to show that this is correct and also the one contains the $\sin$ and the other the $\cos$ of the angle in between $\vec{k}$ and $\vec{v}$. But its more the vector algebra which kind of amazed me.

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The formulas obtained by Robert Scheider and involving The Golden Ratio $\phi$, the Euler Totient $\varphi (n)$ and the Moebius Function $\mu(n)$, are rather surprising $$ \begin{array}{*{20}c} {\phi = - \sum\limits_{1\, \le \,k} {\frac{{\varphi (k)}}{k}\ln \left( {1 - \frac{1}{{\phi ^{\,k} }}} \right)} } & {e^{\,\phi } = \prod\limits_{1\, \le \,k} {\left( {1 - \frac{1}{{\phi ^{\,k} }}} \right)^{\, - \;\frac{{\varphi (k)}}{k}} } } \\ {\frac{1}{\phi } = - \sum\limits_{1\, \le \,k} {\frac{{\mu (k)}}{k}\ln \left( {1 - \frac{1}{{\phi ^{\,k} }}} \right)} } & \begin{array}{l} e^{\,{{1\,} \mathord{\left/ {\vphantom {{1\,} {\,\phi }}} \right. } {\,\phi }}} = \prod\limits_{1\, \le \,k} {\left( {1 - \frac{1}{{\phi ^{\,k} }}} \right)^{\, - \;\frac{{\mu (k)}}{k}} } = \\ = \frac{1}{e}\prod\limits_{1\, \le \,k} {\left( {1 - \frac{1}{{\phi ^{\,k} }}} \right)^{\, - \;\frac{{\varphi (k)}}{k}} } \\ \end{array} \\ {1 = \sum\limits_{1\, \le \,k} {\frac{{\mu (k) - \varphi (k)}}{k}\ln \left( {1 - \frac{1}{{\phi ^{\,k} }}} \right)} } & {e = \prod\limits_{1\, \le \,k} {\left( {1 - \frac{1}{{\phi ^{\,k} }}} \right)^{\,\frac{{\mu (k) - \varphi (k)}}{k}} } } \\ \end{array} $$

Also refer to this Wikipedia article.

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I was amazed at one of the results that Ramanujan sent to Hardy in his first letter.

$$ (1.10) \text{ If } u = \frac{x}{1+} \frac{x^5}{1+} \frac{x^{10}}{1+} \frac{x^{15}}{1+\dots},\, v = \frac{x^{1/5}}{1+} \frac{x}{1+}\frac{x^{2}}{1+}\frac{x^{3}}{1+\dots},\\ \text{then}\qquad\qquad v^5 = u\frac{1-2u+4u^2-3u^3+u^4} {1+3u+4u^2+2u^3+u^4} $$

This is just one of the modular equations satisfied by the Rogers-Ramanujan continued fraction. The Wikipedia article mentions other amazing facts about the continued fraction but this result stands out for me.

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Perhaps this is mainly for the programmers among us, a few fun 'cyclical' identities between bases $10$ and $16$:

$$8200_{10}=2008_{16}$$ $$4100_{10}=1004_{16}$$ $$1041_{10}=0411_{16}$$

In case you're confused, the rule for the above is:

$$\mathrm{CycleDigitsLeft}(N_{10},1)=N_{16}$$

I thought it was cute :)

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Sophomore's $\color{red}{\text{Eternal}}$ Dream:

$$\lim_{n\to\infty}\int_0^1\cdot\cdot\cdot\int_0^1\int_0^1 (x_1x_2\cdot\cdot\cdot x_n)^{x_1x_2\cdot\cdot\cdot x_n} dx_1 dx_2\cdot\cdot\cdot dx_n=1$$

Here is the solution link.

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I was calculating a Galois group and came across this weird looking identity that seemingly came out of nowhere:

$$ \left( \sqrt{3} \left(\frac{1+\sqrt{3}}{2}\right) (1+i) \sqrt{\frac{2}{\sqrt{3}}-1} \right)^4 = -3.$$

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Les $(e_n)_{n \ge 0}$ be the sequence defined by $e_n = (-1)^{s_2(n)}$ where $s_2(n)$ is the sum of the digits in the decomposition of $n$ in base 2. Then $$\prod_{k=0}^{\infty}\Big(\frac{2k+1}{2k+2}\Big)^{e_n} = \frac{\sqrt{2}}{2}.$$ Moreover, for every $n \ge 0$, $$e_n = 1 \text{ or } -1 \text{ according that } \prod_{k=0}^{n-1}\Big(\frac{2k+1}{2k+2}\Big)^{e_k} \text{ is grater or smaller than } \frac{\sqrt{2}}{2}.$$ https://cs.uwaterloo.ca/~shallit/Papers/ubiq15.pdf

Here is another amazing result (to my opinion). For every integer $n \ge 1$, call $t_n$ the unique solution of the equation $\tan t = t$ in the interval $]n\pi,n\pi+\pi/2[$. Then $$\sum_{n=1}^{+\infty} t_n^{-2} = \frac{1}{10}, \quad \sum_{n=1}^{+\infty} t_n^{-4} = \frac{1}{350}.$$ There exists a formula like (I am not completely sure)$$\frac{3}{t}+\sum_{n=1}^{+\infty} \frac{2t}{t^2-t_n^2} = \frac{\tan^2t}{\tan t - t} = \frac{\sin^2 t}{\cos t (\sin t - t\cos t)}.$$

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Some identities, mostly by Ramanujan, from this paper by S.G. Gindikin (in Russian).

$$\sum_{n=0}^\infty (-1)^n(4n+1)\left(\frac{(2n-1)!!}{(2n)!!}\right)^3=\frac 2\pi,$$

$$\sqrt{6+2\sqrt{7+3\sqrt{8+4\sqrt{9+\dots}}}}=4,$$

$$1+\frac 1{1\cdot 3}+\frac 1{1\cdot 3\cdot 5}+\frac 1{1\cdot 3\cdot 5\cdot 7} +\frac 1{1\cdot 3\cdot 5\cdot 7\cdot 9}+$$ $$\frac{1}{1+\frac 1{1+\frac 2{\frac{1+ 3}{1+\frac 4{1+\dots}}}}}=\sqrt{\frac{\pi e}2},$$

$$1+\frac x{1-x}+\frac{x^4}{(1-x)(1-x^2)}+\frac{x^9}{(1-x)(1-x^2)(1-x^3)}= \frac 1{(1-x)(1-x^6)(1-x^{11})\cdot \dots \cdot (1-x^4)(1-x^9)(1-x^{16})}.$$

PS. Because of bad image quality, I am not sure in some powers in the last identity.

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  • $\begingroup$ Feel free to delete this answer if the above identities were already posted. $\endgroup$ Dec 16, 2023 at 10:56
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Let $G$ be a finite group. Then $G$ is abelian if and only if

$$\frac{ \{g, h \in G: g^{-1}h^{-1}gh = e \}}{|G|^2} > \frac{5}{8}.$$

Essentially this is saying that if the proportion of a group that commutes exceeds a certain amount, then all of it must commute.

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The following identities are very simple, but maybe a long ago, when I was a child and played with a calculator, I was surprised.

$$\begin{align} 1^2 & = 1\\ 11^2 & = 121\\ 111^2 & = 12321\\ 1111^2 & = 1234321\\ 11111^2 & =123454321\\ 111111^2 & =12345654321\\ 1111111^2 & =1234567654321\\ 11111111^2 & =123456787654321\\ 111111111^2 & =12345678987654321 \end{align}$$

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  • $\begingroup$ Feel free to delete this answer if these identities were already posted. $\endgroup$ Dec 16, 2023 at 1:47
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The fact that taking Pascal's triangle mod 2 gives the Sierpiński triangle still blows my mind.

(See : this link for an explanation)

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Let $x\ge -1$ :

$$F(x)=(1/x!)!^{(1/x!)!^{\cdots}}$$

Then the two minima are equal one around $-0.59$ and the other around $2.085$

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Let define $x>-1$ :

$$F(x)=(1/x!)!^{(1/x!)!^{\cdots}}$$

Then the two minima are equal around $-0.59$ and $2.085$

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