Here is a funny exercise $$\sin(x  y) \sin(x + y) = (\sin x  \sin y)(\sin x + \sin y).$$ (If you prove it don't publish it here please). Do you have similar examples?

8[This](http://mathoverflow.net/questions/38856) is related. – J. M. ain't a mathematician Nov 03 '10 at 22:35

Also $\cos(x  y) \cos(x + y) = (\cos x  \sin y)(\cos x + \sin y) $. – Quixotic Nov 04 '10 at 11:05

3I noticed that your expression can be also written as $\sin(x  y) \sin(x + y) = (\cos y + \cos x) (\cos y  \cos x) $ – Quixotic Nov 04 '10 at 11:09

188I have tripped up many calculus students with this one: $log(1+2+3)=log1+log2+log3$. I am evil... – Dec 08 '12 at 01:23

11@SteveD If only we could find an odd example... – yearning4pi Jan 13 '13 at 00:31

11Almost an identity: $$\sqrt{123456790}\approx 11111.11111\,.$$ – Jul 12 '14 at 18:47

2[$$\int_{\infty}^{\infty} \frac{dx}{1 + (x + \tan x)^2}=\pi$$](http://math.stackexchange.com/q/1015462/146687) – Venus Nov 14 '14 at 22:05

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. – skan Nov 26 '16 at 17:08

$$\prod_{n\geq1}\frac1{4en}\bigg(\frac{(16n^29)^3}{16n^21}\bigg)^{1/4}\bigg(\frac{(4n+3)(4n1)}{(4n3)(4n+1)}\bigg)^n=\sqrt{\frac{2}{3\pi\sqrt{3}}}\exp\bigg(\frac{G}{\pi}+\frac12\bigg)$$ $G$ is Catalan's constant – clathratus Dec 31 '18 at 07:10

A few more examples : [the Senior's dream](https://twitter.com/InertialObservr/status/1166045556990332928) [(1)](https://i.stack.imgur.com/uBwub.jpg), or [here](https://twitter.com/InertialObservr/status/1159281436156805123) [(2)](https://i.stack.imgur.com/zMfgl.jpg). – Watson Dec 27 '19 at 21:40
63 Answers
$$\int_0^1\frac{\mathrm{d}x}{x^x}=\sum_{k=1}^\infty \frac1{k^k}$$
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17$$\int_0^1 {x^x}\mathrm{d}x=\sum_{k=1}^\infty\frac{(1)^{k1}}{k^k}$$ – user85798 Nov 02 '13 at 18:29
$$\left(\sum\limits_{k=1}^n k\right)^2=\sum\limits_{k=1}^nk^3 .$$
The two on the left is not a typo.
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18If you're physicsminded, the 2 and 3 are not a surprise: $n$ and $k$ must have the same dimension, so the right hand side has this dimension to the 4. So the only possible exponent on the left is 2. – Joce NoToPutinsWarInUkraine Nov 16 '16 at 15:01

3@Joce "so the right hand side has this dimension to the 4" I don't get that. Even though $n$ and $k$ have the same dimension, how can you *add* them? Adding the three dimensions of $k^3$ and one dimension of $n$ seems as a nonsense... The mathematics is fine but I don't understand your dimensional analysis analogy at all. – Tu1 Sep 21 '19 at 20:34

@Joce I mean, of course you can add $n$ and $k$ of the same dimension, but how can you do that when $n$ is just a limit? You don't really add it to $k$ here. $k$ is just a dummy variable, when you write it out you get $(1+2+3+\cdots +n)^2=1^3+2^3+3^3+\cdots +n^3$, no $k$ here. – Tu1 Sep 21 '19 at 20:44

I meant (three years ago) that $\sum_{k\leq n} k$ is behaving like $n^2$, while $\sum_{k\leq n} k^3$ is behaving like $n^4$, so of course if a power of the first is equal to the second the exponent has to be 2. – Joce NoToPutinsWarInUkraine Sep 25 '19 at 07:47

7But how do you know that $\sum_{k\le n}k$ is behaving like $n^2$ without knowing that $\sum_{k\le n}k=\frac{n^2+n}{2}$ beforehand? – Tu1 Sep 26 '19 at 15:45

1This is called [Nicomachus's theorem](https://en.wikipedia.org/wiki/Squared_triangular_number) – MCCCS Aug 13 '20 at 14:07
$$ \infty! = \sqrt{2 \pi} $$
It comes from the zeta function.

5Can you help me understand $\infty!$? I don't know what to make of it. – futurebird Nov 04 '10 at 00:55

12@don: it's $\exp(\zeta^{\prime}(0))$, where $\zeta^{\prime}(z)$ is *formally* $\sum_{k=1}^\infty \frac{\ln\;k}{k^z}$ – J. M. ain't a mathematician Nov 04 '10 at 05:01

1@a little don, You can read about it here http://katlas.math.toronto.edu/drorbn/MathBlog/200811/one/Gillet@FI_What_is_infinity_factorial_(and_why_might_we_care)Q.pdf – Nov 04 '10 at 17:29

1Neat! I wonder whether "solving" this identity for $\infty$ [also yields $\frac12$](http://math.stackexchange.com/questions/1074870/isinftyfrac12) *edit* Hm, since $(\frac12)!=\sqrt\pi$ not :/ – Tobias Kienzler Dec 19 '14 at 19:41
Ah, this is one identity which comes into use for proving the Euler's Partition Theorem. The identity is as follows: $$ (1+x)(1+x^{2})(1+x^{3}) \cdots = \frac{1}{(1x)(1x^{3})(1x^{5}) \cdots}$$
Machin's Formula: \begin{eqnarray} \frac{\pi}{4} = 4 \arctan \frac{1}{5}  \arctan \frac{1}{239}. \end{eqnarray}
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$$\frac{1}{\sin(2\pi/7)} + \frac{1}{\sin(3\pi/7)} = \frac{1}{\sin(\pi/7)}$$
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5I thought this was going to be hard to prove...It just took three lines! – chubakueno Feb 01 '14 at 21:14
The Frobenius automorphism
$$(x + y)^p = x^p + y^p$$

58Only in a field of prime characteristic $p$ (much to the chagrin of my calculus students). – Austin Mohr Jun 19 '12 at 02:06

29@AustinMohr: Not just _in a field_ of prime characteristic $p$, it holds in any commutative ring of characteristic $p$. – Marc van Leeuwen Sep 30 '13 at 06:32
\begin{eqnarray} 1^{3} + 2^{3} + 2^{3} + 2^{3} + 4^{3} + 4^{3} + 4^{3} + 8^{3} = (1 + 2 + 2 + 2 + 4 + 4 + 4 + 8)^{2} \end{eqnarray} More generally, let $D_{k} = \{ d\}$ be the set of unitary divisors of a positive integer $k$, and let $\mathsf{d}^{*} \colon \mathbb{N} \to \mathbb{N}$ denote the numberofunitarydivisors (arithmetic) function. Then \begin{eqnarray} \sum_{d \in D} \mathsf{d}^{*}(d)^{3} = \left( \sum_{d \in D} \mathsf{d}^{*}(d) \right)^{2} \end{eqnarray}
Note that $\mathsf{d}^{*}(k) = 2^{\omega(k)}$, where $\omega(k)$ is the number distinct prime divisors of $k$.

There is no reason to restrict this to unitary divisors: Liouville's result still works if you replace "unitary divisor" with "divisor" throughout, which affords a much richer variety of sets $D_k$. The present formulation does not even generalize the standard sum $\sum n^3$ and describes a vanishingly small collection of subsets. – Erick Wong Mar 23 '17 at 23:25

Sure, but then you could still greatly simplify the description by replacing "unitary divisors" with "divisors" and restricting $k$ to be squarefree, there's no need to introduce two new notations. But at this point it feels like trying to dissect a proverbial joke :). – Erick Wong Apr 03 '18 at 19:15
$$\large{1,741,725 = 1^7 + 7^7 + 4^7 + 1^7 + 7^7 + 2^7 + 5^7}$$
and
$$\large{111,111,111 \times 111,111,111 = 12,345,678,987,654,321}$$
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19is there any way to generalise $$\large{1,741,725 = 1^7 + 7^7 + 4^7 + 1^7 + 7^7 + 2^7 + 5^7}$$? – pipi Nov 16 '12 at 07:32

6Do a search for Armstrong numbers and/or narcissistic numbers. Or type 1741725 into the Online Encyclopedia of Integer Sequences. – Gerry Myerson Sep 26 '13 at 13:22

@ThomasWeller The same trick works with $11\times11$, $111\times 111$, $1111\times 1111$, etc., if the given version is too large to fit. (The given example is the largest of its type, though, because otherwise the digits overflow into adjacent locations and it doesn't look as nice.) – Mario Carneiro Jul 12 '16 at 02:21
$$\sec^2(x)+\csc^2(x)=\sec^2(x)\csc^2(x)$$
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1Is the natural logarithm function, or the exponential function related to this? – Doug Spoonwood Feb 13 '12 at 03:24

8@Doug Spoonwood: If you multiply both sides by $\sin^2(x)\cos^2(x)$ you get the Pythagorean identity. Whether that's related to logarithm/exponential, I don't know. Just a test question I gave my students that I thought looked neat. – J126 Feb 13 '12 at 14:15

5And because of this identity we have $\frac{\,d}{\,dx} \left[e^{\tan{x}} \cdot e^{\cot{x}}\right] = \frac{\,d}{\,dx} \left[e^{\tan{x}}\right] \cdot \frac{\,d}{\,dx} \left[e^{\cot{x}}\right]$. – Ant Jun 10 '17 at 03:56
\[\sqrt{n^{\log n}}=n^{\log \sqrt{n}}\]
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Yeah, I'm sure there's a zillion related identities and generalisations (and I should probably be more careful about the domain of n). But this one in particular came up in my research and I thought it was funny  I couldn't decide whether or not to write $\sqrt{n}^{\log n}$ or $n^{\log \sqrt{n}}$. – Douglas S. Stones Nov 30 '10 at 10:22

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@ChantryCargill Not when you consider the fact that not many people know (surprisingly) that $\sqrt{x} \equiv x^{\frac{1}{2}}$ – Cole Tobin Jul 12 '14 at 18:31
$\displaystyle\big(a^2+b^2\big)\cdot\big(c^2+d^2\big)=\big(ac \mp bd\big)^2+\big(ad \pm bc\big)^2$
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There are also the related Lagrange's identity and the corresponding one for eight squares. – Yuval Filmus Nov 16 '11 at 13:28

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Facts about $\pi$ are always fun!
\begin{equation} \frac{\pi}{2} = \frac{2}{1}\cdot\frac{2}{3}\cdot\frac{4}{3}\cdot\frac{4}{5}\cdot\frac{6}{5}\cdot\frac{6}{7}\cdot\frac{8}{7}\cdot\ldots\\ \end{equation} \begin{equation} \frac{\pi}{4} = 1\frac{1}{3}+\frac{1}{5}\frac{1}{7}+\frac{1}{9}\ldots\\ \end{equation} \begin{equation} \frac{\pi^2}{6} = 1+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+\ldots\\ \end{equation} \begin{equation} \frac{\pi^3}{32} = 1\frac{1}{3^3}+\frac{1}{5^3}\frac{1}{7^3}+\frac{1}{9^3}\ldots\\ \end{equation} \begin{equation} \frac{\pi^4}{90} = 1+\frac{1}{2^4}+\frac{1}{3^4}+\frac{1}{4^4}+\frac{1}{5^4}+\ldots\\ \end{equation} \begin{equation} \frac{2}{\pi} = \frac{\sqrt{2}}{2}\cdot\frac{\sqrt{2+\sqrt{2}}}{2}\cdot\frac{\sqrt{2+\sqrt{2+\sqrt{2}}}}{2}\cdot\ldots\\ \end{equation} \begin{equation} \pi = \cfrac{4}{1+\cfrac{1^2}{3+\cfrac{2^2}{5+\cfrac{3^2}{7+\cfrac{4^2}{9+\ldots}}}}}\\ \end{equation}
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Well, i don't know whether to classify this as funny or surprising, but ok it's worth posting.
 Let $(X,\tau)$ be a topological space and let $A \subset X$ . By iteratively applying operations of closure and complemention, one can produce at most 14 distinct sets. It's called as the Kuratowski's Closure complement problem.

6An example achieving the is $[0,1] \cup (2,3) \cup \{(4,5) \cap \mathbb{Q}\} \cup \{(6,8)  \{7\}\} \cup \{9\}$. See section 9 of http://austinmohr.com/Work_files/730.pdf for details. – Austin Mohr Jun 19 '12 at 02:09

5I think you mean $[0, 1]\cup (2, 3)\cup((4, 5)\cap\mathbb{Q})\cup(6, 7)\cup(7, 8)\cup\{9\}$. The set you wrote isn't a subset of $\mathbb{R}$ as it contains $(4, 5)\cap\mathbb{Q}$ as an element. – Michael Albanese Jan 13 '13 at 15:08

1@MichaelAlbanese: Anyone able to understand the contents of this thread instantly recognizes those braces are there to give the intersection operator a higher precedence than the neighboring union operators. One must travel far out of one's mathematical way to arrive at your alternative interpretation... – mathematrucker Jun 03 '17 at 15:42
The following number is prime
$p = 785963102379428822376694789446897396207498568951$
and $p$ in base 16 is
$89ABCDEF012345672718281831415926141424F7$
which includes counting in hexadecimal, and digits of $e$, $\pi$, and $\sqrt{2}$.
Do you think this's surprising or not?
$$11 \times 11 = 121$$ $$111 \times 111 = 12321$$ $$1111 \times 1111 = 1234321$$ $$11111 \times 11111 = 123454321$$ $$\vdots$$
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24The prime is unsurprising  the final `F7` doesn't seem to mean anything, and about one in 111 numbers of that size is prime. So it's not very remarkable that there's a prime among the 256 40hexdigit numbers that start with those particular 38 chosen digits. – hmakholm left over Monica Nov 20 '13 at 18:04

1I remember that last from reading "The number devil"! And it works for other bases too; for a base $b$, until $\left(\sum_{n=0}^{b1}\left(b^n\right)\right)^2=123...\ \text{digit } b1\ ...321$. – JMCF125 Nov 24 '13 at 11:06

I find (1....1)^n interesting, it's also nearly impossible to caluclate by hand without messing it up. – HopefullyHelpful Mar 04 '16 at 23:19
\begin{align} \frac{\mathrm d}{\mathrm dx}(x^x) &= x\cdot x^{x1} &\text{Power Rule?}&\ \text{False}\\ \frac{\mathrm d}{\mathrm dx}(x^x) &= x^{x}\ln(x) &\text{Exponential Rule?}&\ \text{False}\\ \frac{\mathrm d}{\mathrm dx}(x^x) &= x\cdot x^{x1}+x^{x}\ln(x) &\text{Sum of these?}&\ \text{True}\\ \end{align}
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16This is a special case of $\frac{d}{dx} h(f(x),g(x)) = \partial_1 h f' + \partial_2h g'$ – ronno Dec 20 '13 at 13:56

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$$ \frac{e}{2} = \left(\frac{2}{1}\right)^{1/2}\left(\frac{2\cdot 4}{3\cdot 3}\right)^{1/4}\left(\frac{4\cdot 6\cdot 6\cdot 8}{5\cdot 5\cdot 7\cdot 7}\right)^{1/8}\left(\frac{8\cdot 10\cdot 10\cdot 12\cdot 12\cdot 14\cdot 14\cdot 16}{9\cdot 9\cdot 11\cdot 11\cdot 13\cdot 13\cdot 15\cdot 15}\right)^{1/16}\cdots $$ [Nick Pippenger, Amer. Math. Monthly, 87 (1980)]. Set all the exponents to 1 and you get the Wallis formula for $\pi/2$.
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\begin{eqnarray} \sum_{i_1 = 0}^{nk} \, \sum_{i_2 = 0}^{nki_1} \cdots \sum_{i_k = 0}^{nki_1  \cdots  i_{k1}} 1 = \binom{n}{k} \end{eqnarray}
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M.V Subbarao's identity: an integer $n>22$ is a prime number iff it satisfies,
$$n\sigma(n)\equiv 2 \pmod {\phi(n)}$$
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@PAD: See [M. V. Subbarao, *On two congruences for primality,* Pacific Journal of Mathematics, Volume 52, Number 1 (1974), 261268](http://projecteuclid.org/euclid.pjm/1102912230) ([Another PDF](http://www.math.ualberta.ca/~subbarao/documents/Subbarao1974.pdf)). The precise theorem is that $n\sigma(n) \equiv 2 \mod \phi(n)$ if and only if $n$ is prime or one of $1, 4, 6, 22$. – ShreevatsaR Oct 03 '13 at 23:28
$${\Large% \sqrt{\,\vphantom{\huge A}\color{#00f}{20}\color{#c00000}{25}\,}\, =\ \color{#00f}{20}\ +\ \color{#c00000}{25}\ =\ 45} $$
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$$\sum\limits_{n=1}^{\infty} n = 1 + 2 + 3 + \cdots \text{ad inf.} = \frac{1}{12}$$
You can also see many more here: The EulerMaclaurin formula, Bernoulli numbers, the zeta function, and realvariable analytic continuation
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1@fmartin: it's $\zeta(1)$, which can be shown to be expressible in terms of Bernoulli numbers. – J. M. ain't a mathematician Nov 08 '10 at 00:55

10@J.M.: I still fail to see how an infinite summation of positive numbers can result in a negative number. – Fernando Martin Nov 15 '10 at 23:26

5@fmartin: I agree it's counterintuitive; properly explaining this mathematical joke requires a foray into complex analysis (the magic words are "analytic continuation"), which I'll leave to more eloquent users to explain. – J. M. ain't a mathematician Nov 16 '10 at 07:05

9It's particularly a string theory joke, since this is the trick they use to regularize certain sums in their theories. That's how they arrive at 26 dimensions (in nonsupersymmetric theories), because the regularization only works for that many dimensions. I suppose the argument works in the same way in supersymmetric theories, but they then get 10 dimensions. – Raskolnikov Nov 27 '10 at 11:25

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Infinite sum is a function defined by it's properties, it is consistent with those properties to say $1+2+3...\infty =  \frac 1 {12}$. It isn't defined as "an infinite number of additions", which trips people up. – DanielV Nov 03 '13 at 05:34


Related: [Why does $1+2+3+\dots = \frac{1}{12}$?](http://math.stackexchange.com/questions/39802/whydoes123dotsfrac112) – Cole Tobin Jul 12 '14 at 18:35

There is a very nice video from 3Blue1Brown on Youtube for the visualizing of the Riemann zeta function and its analytic continuation: https://www.youtube.com/watch?v=sD0NjbwqlYw – Stefan Gruenwald Sep 11 '18 at 06:38
Two related integrals:
$$\int_0^\infty\sin\;x\quad\mathrm{d}x=1$$
$$\int_0^\infty\ln\;x\;\sin\;x\quad \mathrm{d}x=\gamma$$
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10They are Abelsummable integrals; e.g. the first one is properly interpreted as $\lim_{\epsilon\to 0}\int\exp(\epsilon x)\sin\;x\quad \mathrm{d}x$ – J. M. ain't a mathematician Nov 03 '10 at 22:53

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@J.M. I have not thought about this, but is that unique? I don't think so, but why look at the $\exp$ "kernel"? – AD  Stop Putin  Jul 04 '12 at 19:31

@AD. I'm just going from the definition of Abelsummable integrals. Maybe the use of a different kernel might yield different results, but it is noteworthy that the use of numerical methods yield results consistent with the Abel summation. – J. M. ain't a mathematician Jul 05 '12 at 01:44

$$ 10^2+11^2+12^2=13^2+14^2 $$
There's a funny Abstruse Goose comic about this, which I can't seem to find at the moment.
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By excluding the first two primes, Euler's Prime Product becomes a square:
$$\prod _{n=3}^{\infty } \frac{1}{1\frac{1}{(p_n)^{2}}}=\frac{\pi ^2}{9}$$
By using multiples of the product of the first two primes, we get the square root:
$$\prod _{n=1}^{\infty } \frac{1}{1\frac{1}{(n p_1 p_2)^{2}}}=\frac{\pi }{3}$$
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5It doesn't make sense to speak of "perfect squares" for positive real numbers... but this is a nice identity though. – Patrick Da Silva Jun 19 '12 at 19:53

5@PatrickDaSilva It might, if you know that the values of $L$functions sometimes land in a special ring which is strictly between algebraic numbers and transcendental numbers. This is the ring of 'periods'. I don't believe that it is closed under taking square roots, so to say that something is the square of a period might not be completely silly. – Bruno Joyal Sep 26 '13 at 21:29

2@BrunoJoyal If I'm not mistaken, periods include $\zeta(3)$ and many more  they are basically anything you can get with integration. If I recall correctly, it is not known whether or not $\dfrac1\pi$ is a period. – Akiva Weinberger Aug 28 '14 at 03:56
The product of any four consecutive integers is one less than a perfect square.
To phrase it more like an identity:
For every integer $n$, there exists an integer $k$ such that $$n(n+1)(n+2)(n+3) = k^2  1.$$
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I wonder now if this isn't a proper identity because of the existential quantifier. Any thoughts? – Austin Mohr Jun 19 '12 at 03:15

12You can also write that as $$n(n+1)(n+2)(n+3)=((n+1)^2+1)^2  1$$ – AD  Stop Putin  Jun 19 '12 at 06:53
$$\leftz+z'\right^{2}+\leftzz'\right^{2}=2\times\left(\leftz\right^{2}+\leftz'\right^{2}\right)$$
The sum of the squares of the sides equals the sum of the squares of the diagonals.
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What is 42?
$$ 6 \times 9 = 42 \text{ base } 13 $$ I always knew that there is something wrong with this universe.
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Considering the main branches
$$i^i = \exp\left(\frac{\pi}{2}\right)$$
$$\root i \of i = \exp\left(\frac{\pi}{2}\right) $$
And $$ \frac{4}{\pi } = \displaystyle 1 + \frac{1}{{3 +\displaystyle \frac{{{2^2}}}{{5 + \displaystyle\frac{{{3^2}}}{{7 +\displaystyle \frac{{{4^2}}}{{9 +\displaystyle \frac{{{n^2}}}{{\left( {2n + 1} \right) + \cdots }}}}}}}}}} $$
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$$\int_0^\infty\frac1{1+x^2}\cdot\frac1{1+x^\pi}dx=\int_0^\infty\frac1{1+x^2}\cdot\frac1{1+x^e}dx$$
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1The identity also holds if you replace $\pi$ with Euler's $\gamma$ :) – Vladimir Reshetnikov Nov 29 '15 at 01:19


@JasonDeVito Substitue $\tan x = t$ and $t' = \frac{\pi}{2}  t$. Then it is easy to show $I = \int_{0}^{\frac{\pi}{2}} \frac{1}{2} dt$. The $\pi$ or $e$ is just a trick  it holds for all real numbers. – Vue Apr 11 '22 at 12:44

@Vue: I certainly don't practice integrals in my spare time or anything, but I'm still not able to get it. With $a$ being used to denote $\pi$, $e$, $\gamma$, or whatever, making the first substitution you suggest gives $\int_0^{\pi/2} \frac{1}{1+\arctan(t)^2} \frac{1}{1+\arctan(t)^a} \frac{1}{1+t^2}\; dt$, and then the second substitution just seems to make it worse. Then I thought maybe you meant $x = \tan t$, which leads to the much nicer $\int_0^{\pi/2} \frac{1}{1+\tan^a(x)}\; dx$. The second substitution then gives $\int_0^{\pi/2} \frac{1}{1+\cot^a(x)}\; dx$. What now? – Jason DeVito Apr 11 '22 at 14:31

@JasonDeVito My bad, it was $x = \tan t$. Multiply $\tan^a x$ on both the denominator and the numerator  then the sum of the two integrals, which is $2I$ equals $\int_{0}^{\frac{\pi}{2}} 1 dt$. It's a nice trick that is only possible on definite integrals. – Vue Apr 11 '22 at 14:33

@Vue: That's very slick! I'm quite confident I would have never figured that out on my own. Thanks for the help! – Jason DeVito Apr 11 '22 at 15:47
Best near miss
$$\int_{0}^{\infty }\cos\left ( 2x \right )\prod_{n=0}^{\infty}\cos\left ( \frac{x}{n} \right )~\mathrm dx\approx \frac{\pi}{8}7.41\times 10^{43}$$
One can easily be fooled into thinking that it is exactly $\dfrac{\pi}{8}$.
References:
 Wikipedia
 Future Prospects for ComputerAssisted Mathematics, by D.H. Bailey and J.M. Borwein
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Let $f$ be a symbol with the property that $f^n = n!$. Consider $d_n$, the number of ways of putting $n$ letters in $n$ envelopes so that no letter gets to the right person (aka derangements). Many people initially think that $d_n = (n1)! = f^{n1}$ (the first object has $n1$ legal locations, the second $n2$, ...). The correct answer isn't that different actually:
$d_n = (f1)^n$.
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2To make this more rigorous, in a sense: We can define a linear operator $L$ acting on $\mathbb{C}[f]$ such that $L(f^n)=n!$ and $L(1)=1$. Thus, we can write $d_n=L\left((f1)^n\right)$. (Am I doing this right?) – Akiva Weinberger Aug 28 '14 at 04:00
$$ 71 = \sqrt{7! + 1}. $$
Besides the amusement of reusing the decimal digits $7$ and $1$, this is conjectured to be the last solution of $n!+1 = x^2$ in integers. ($n=4$ and $n=5$ also work.) Even finiteness of the set of solutions is not known except using the ABC conjecture.
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We have by block partition rule for determinant $$ \det \left[ \begin{array}{cc} U & R \\ L & D \end{array} \right] = \det U\cdot \det ( DLU^{1}R) $$ But if $U,R,L$ and $D$ commute we have that $$ \det \left[ \begin{array}{cc} U & R \\ L & D \end{array} \right] = \det (UDLR) $$
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The CayleyHamilton theorem:
If $A \in \mathbb{R}^{n \times n}$ and $I_{n} \in \mathbb{R}^{n \times n}$ is the identity matrix, then the characteristic polynomial of $A$ is $p(\lambda) = \det(\lambda I_n  A)$. Then the Cayley Hamilton theorem can be obtained by "substituting" $\lambda = A$, since $$p(A) = \det(AI_nA) = \det(00) = 0$$
$(xa)(xb)(xc)\ldots(xz) = 0$
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9@columbus8myhw $(xa)(xb)(xc)\ldots (xw){\color{red}{(xx)}}(xy)(xz)=0$ – Surb Aug 23 '15 at 16:00
$$\frac{1}{998901}=0.000001002003004005006...997999000001...$$
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Similar results for $1/((9...9)^2)$, and any further analysis will explain why :) – Mark Hurd Jul 12 '12 at 04:03

Indeed, I saw it explained [here](http://www.youtube.com/watch?v=daro6K6mym8) initially :) – preferred_anon Jul 14 '12 at 20:19

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$$ \dfrac{1}{2}=\dfrac{\dfrac{1}{2}}{\dfrac{1}{2}+\dfrac{\dfrac{1}{2}}{\dfrac{1}{2}+\dfrac{\dfrac{1}{2}}{\dfrac{1}{2}+\dfrac{\dfrac{1}{2}}{\dfrac{1}{2}+\dfrac{\dfrac{1}{2}}{\dfrac{1}{2}+\dfrac{\dfrac{1}{2}}{\dfrac{1}{2}+\cdots}}}}}} $$
and more generally we have $$ \dfrac{1}{n+1}=\frac{\dfrac{1}{n(n+1)}}{\dfrac{1}{n(n+1)}+\dfrac{\dfrac{1}{n(n+1)}}{\dfrac{1}{n(n+1)}+\dfrac{\dfrac{1}{n(n+1)}}{\dfrac{1}{n(n+1)}+\dfrac{\dfrac{1}{n(n+1)}}{\dfrac{1}{n(n+1)}+\dfrac{\dfrac{1}{n(n+1)}}{\dfrac{1}{n(n+1)}+\dfrac{\frac{1}{n(n+1)}}{\dfrac{1}{n(n+1)}+\ddots}}}}}} $$
\begin{eqnarray} \zeta(0) = \sum_{n \geq 1} 1 = \frac{1}{2} \end{eqnarray}
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1Actually, this can be made rigorous by noting that $$ \zeta(z)=\lim_{n\to\infty}\left(\sum_{k=1}^nk^{z}\frac{1}{1z}n^{1z}\frac12n^{z}\right) $$ for $\mathrm{Re}(z)>1$. – robjohn Jun 21 '12 at 00:52
\begin{align}\frac{64}{16}&=\frac{6\!\!/\,4}{16\!\!/}\\&=\frac41\\&=4\end{align}
For more examples of these weird fractions, see "How Weird Are Weird Fractions?", Ryan Stuffelbeam, The College Mathematics Journal, Vol. 44, No. 3 (May 2013), pp. 202209.
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See also http://math.stackexchange.com/questions/49657/badfractionreductionthatactuallyworks – Nate Eldredge Nov 15 '13 at 16:59

Also [Project Euler problem number 33](https://projecteuler.net/problem=33). – Mike Pierce Oct 24 '16 at 21:12
$$\frac{\pi}{4}=\sum_{n=1}^{\infty}\arctan\frac{1}{f_{2n+1}}, $$ where $f_{2n+1}$ there are fibonacci numbers, $n=1,2,...$
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\begin{eqnarray} \sum_{k = 0}^{\lfloor q  q/p) \rfloor} \left \lfloor \frac{p(q  k)}{q} \right \rfloor = \sum_{k = 1}^{q} \left \lfloor \frac{kp}{q} \right \rfloor \end{eqnarray}
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4I don't see the 'punch' here. Isn't that just reversing the order of summation and truncating some zeros? – Ofir Jan 13 '13 at 00:10
$$27\cdot56=2\cdot756,$$ $$277\cdot756=27\cdot7756,$$ $$2777\cdot7756=277\cdot77756,$$ and so on.
$$ \sum_{n=1}^{+\infty}\frac{\mu(n)}{n}=1\frac12\frac13\frac15+\frac16\frac17+\frac1{10}\frac1{11}\frac1{13}+\frac1{14}+\frac1{15}\cdots=0 $$ This relation was discovered by Euler in 1748 (before Riemann's studies on the $\zeta$ function as a complex variable function, from which this relation becomes much more easier!).
Then one of the most impressive formulas is the functional equation for the $\zeta$ function, in its asimmetric form: it highlights a very very deep and smart connection between the $\Gamma$ and the $\zeta$: $$ \pi^{\frac s2}\Gamma\left(\frac s2\right)\zeta(s)= \pi^{\frac{1s}2}\Gamma\left(\frac{1s}2\right)\zeta(1s)\;\;\;\forall s\in\mathbb C\;. $$
Moreover no one seems to have wrote the Basel problem (Euler, 1735): $$ \sum_{n=1}^{+\infty}\frac1{n^2}=\frac{\pi^2}{6}\;\;. $$
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$$ \sin \theta \cdot \sin \bigl(60^\circ  \theta \bigr) \cdot \sin \bigl(60^\circ + \theta \bigr) = \frac{1}{4} \sin 3\theta$$
$$ \cos \theta \cdot \cos \bigl(60^\circ  \theta \bigr) \cdot \cos \bigl(60^\circ + \theta \bigr) = \frac{1}{4} \cos 3\theta$$
$$ \tan \theta \cdot \tan \bigl(60^\circ  \theta \bigr) \cdot \tan \bigl(60^\circ + \theta \bigr) = \tan 3\theta $$
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1See also the comments here: http://math.stackexchange.com/q/8385/1242 – Hans Lundmark Nov 04 '10 at 08:53


1I just wanted to mention that your first identity is equivalent to the case $n=3$ of the formula for $\sin nx$ given there. (Just replace $\sin(60^{\circ}\theta)$ by $\sin(\theta+120^{\circ})$.) – Hans Lundmark Nov 04 '10 at 09:56

1considering your first two identities the thirth should be $$ \tan \theta \cdot \tan \bigl(60  \theta \bigr) \cdot \tan \bigl(60 + \theta \bigr) = \tan 3\theta $$ – Neves Mar 06 '11 at 16:08
$\lnot$(A$\land$B)=($\lnot$A$\lor$$\lnot$B) and $\lnot$(A$\lor$B)=($\lnot$A$\land$$\lnot$B), because they mean that negation is an "equal form".
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Here's one clever trigonometric identity that impressed me in highschool days. Add $\sin \alpha$, to both the numerator and the denominator of $\sqrt{\frac{1\cos \alpha}{1 + \cos \alpha}}$ and get rid of the square root and nothing changes. In other words:
$$\frac{1  \cos \alpha + \sin \alpha}{1 + \cos \alpha + \sin \alpha} = \sqrt{\frac{1\cos \alpha}{1 + \cos \alpha}}$$
If you take a closer look you'll notice that the RHS is the formula for tangent of a halfangle. Actually if you want to prove those, nothing but the addition formulas are required.
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$\tan^{1}(1)+\tan^{1}(2)+\tan^{1}(3) = \pi$ (using the principal value), but if you blindly use the addition formula $\tan^{1}(x) + \tan^{1}(y) = \tan^{1}\dfrac{x+y}{1x y}$ twice, you get zero:
$\tan^{1}(1) + \tan^{1}(2) = \tan^{1}\dfrac{1+2}{11*2} =\tan^{1}(3)$; $\tan^{1}(1) + \tan^{1}(2) + \tan^{1}(3) =\tan^{1}(3) + \tan^{1}(3) =\tan^{1}\dfrac{3+3}{1(3)(3)} = 0$.
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$$ \begin{array}{rcrcl} \vdots & \vdots & \vdots & \vdots & \vdots \\[1mm] \int{1 \over x^{3}}\,{\rm d}x & = & \,{1 \over 2}\,{1 \over x^{2}} & \sim & x^{\color{#ff0000}{\large\bf 2}} \\[1mm] \int{1 \over x^{2}}\,{\rm d}x & = & \,{1 \over x} & \sim & x^{\color{#ff0000}{\large\bf 1}} \\[1mm] \int{1 \over x}\,{\rm d}x & = & \ln\left(x\right) & \sim & x^{\color{#0000ff}{\LARGE\bf 0}} \color{#0000ff}{\LARGE\quad ?} \\[1mm] \int x^{0}\,{\rm d}x & = & x^{1} & \sim & x^{\color{#ff0000}{\large\bf 1}} \\[1mm] \int x\,{\rm d}x & = & {1 \over 2}\,x^{2} & \sim & x^{\color{#ff0000}{\large\bf 2}} \\[1mm] \vdots & \vdots & \vdots & \vdots & \vdots \end{array} $$
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$\ln(x)$ converges slowly enough, so indeed $\ln(x) \sim 1$ might not be completely stupid :p – Thomas Sep 28 '13 at 08:31

1Hmm, considering that logarithms get at the exponent, and $x$ has a constant exponent ... Since $\ln\left(x^a\right)=a\ln\left(x\right)$ (the log of an expression equals the exponent times the log of the base), then $\ln\left(x^1\right)=1\ln\left(x\right)=x^0\ln\left(x\right)$ might be saying something to the effect that it's more important that your exponent is a constant, than the fact that the log of your base $\ln\left(x\right)$ is growing slowly. – Travis Bemrose Sep 28 '13 at 10:11


The mystery, perhaps, lies in the constant of integration. All of the other integrals are evaluated from 0 to x, while the $\dfrac1x$ one is evaluated from 1 to x. (Note, by the way, that $\displaystyle\ln x=\lim_{t\to0}\frac{x^t1}t$, which perhaps fits the pattern better.) – Akiva Weinberger Aug 28 '14 at 04:08
$$\lim_{\omega\to\infty}3=8$$ The "proof" is by rotation through $\pi/2$. More of a joke than an identity, I suppose.
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For all $n\in\mathbb{N}$ and $n\neq1$ $$\prod_{k=1}^{n1}2\sin\frac{k \pi}{n} = n$$
For some reason, the proof involves complex numbers and polynomials.
Link to proof: Prove that $\prod_{k=1}^{n1}\sin\frac{k \pi}{n} = \frac{n}{2^{n1}}$
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\begin{align} E &= \sqrt{\left(pc\right)^{2} + \left(mc^{2}\right)^{2}} = mc^{2} + \left[\sqrt{\left(pc\right)^{2} + \left(mc^{2}\right)^{2}}  mc^{2}\right] \\[3mm]&= mc^{2} + {\left(pc\right)^{2} \over \sqrt{\left(pc\right)^{2} + \left(mc^{2}\right)^{2}} + mc^{2}} = mc^{2} + {p^{2}/2m \over 1 + {\sqrt{\left(pc\right)^{2} + \left(mc^{2}\right)^{2}}  mc^{2} \over 2mc^{2}}} \\[3mm]&= mc^{2} + {p^{2}/2m \over 1 + {p^{2}/2m \over \sqrt{\left(pc\right)^{2} + \left(mc^{2}\right)^{2}} + mc^{2}}} = mc^{2} + {p^{2}/2m \over 1 + {p^{2}/2m \over 1 + {p^{2}/2m \over \sqrt{\left(pc\right)^{2} + \left(mc^{2}\right)^{2}}  mc^{2}}}} \end{align}
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Voronoi summation formula:
$\sum \limits_{n=1}^{\infty}d(n)(\frac{x}{n})^{1/2}\{Y_1(4\pi \sqrt{nx})+\frac{2}{\pi}K_1(4\pi \sqrt{nx})\}+x \log x +(2 \gamma1)x +\frac{1}{4}=\sum \limits _{n\leq x}'d(n)$
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$\textbf{Claim:}\quad$$$\frac{\sin x}{n}=6$$ for all $n,x$ ($n\neq 0$).
$\textit{Proof:}\quad$$$\frac{\sin x}{n}=\frac{\dfrac{1}{n}\cdot\sin x}{\dfrac{1}{n}\cdot n}=\frac{\operatorname{si}x}{1}=\text{six}.\quad\blacksquare$$
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@Tgymasb Denominator: $1/n\cdot n=1$. Numerator: $1/\textbf{n}\cdot\operatorname{si}\!\textbf{n}\, x=\operatorname{si}x$. – triple_sec Jan 29 '15 at 14:00
$$ \frac{\pi}{2}=1+2\sum_{k=1}^{\infty}\frac{\eta(2k)}{2^{2k}} $$ $$ \frac{\pi}{3}=1+2\sum_{k=1}^{\infty}\frac{\eta(2k)}{6^{2k}} $$ where $ \eta(n)=\sum_{k=1}^{\infty}\frac{(1)^{k+1}}{k^{n}} $
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I have another one, but I'm quite unwilling to post this here because it's MINE, I haven't found it anywhere, so don't steal this.
Let us take the four most important mathematical constants: The Euler number $e$, the Aurea Golden Ratio $\phi$, the EulerMascheroni constant $\gamma$ and finally $\pi$. Well we can see easily that
$$e\cdot\gamma\cdot\pi\cdot\phi \approx e + \gamma + \pi + \phi$$
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Let $\sigma(n)$ denote the sum of the divisors of $n$.
If $$p=1+\sigma(k),$$ then $$p^a=1+\sigma(kp^{a1})$$ where $a,k$ are positive integers and $p$ is a prime such that $p\not\mid k$.
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If we define $P$ as the infinite lower triangular matrix where $P_{i,j} = \binom{i}{j}$ (we can call it the Pascal Matrix), then $$P^k_{i,j} = \binom{i}{j}k^{ij}$$
where $P^k_{i,j}$ is the element of $P^k$ in the position $i,j.$
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$$ \int_{\infty}^{\infty}{\sin\left(x\right) \over x}\,{\rm d}x = \pi\int_{1}^{1}\delta\left(k\right)\,{\rm d}k $$
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