260
$\begingroup$

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?

  • 7
    [This](http://mathoverflow.net/questions/38856) is related.2010-11-03
  • 0
    Also $\cos(x - y) \cos(x + y) = (\cos x - \sin y)(\cos x + \sin y) $.2010-11-04
  • 2
    I noticed that your expression can be also written as $\sin(x - y) \sin(x + y) = (\cos y + \cos x) (\cos y - \cos x) $2010-11-04
  • 171
    I have tripped up many calculus students with this one: $log(1+2+3)=log1+log2+log3$. I am evil...2012-12-08
  • 9
    @SteveD If only we could find an odd example...2013-01-13
  • 7
    Almost an identity: $$\sqrt{123456790}\approx 11111.11111\,.$$2014-07-12
  • 1
    [$$\int_{-\infty}^{\infty} \frac{dx}{1 + (x + \tan x)^2}=\pi$$](http://math.stackexchange.com/q/1015462/146687)2014-11-14
  • 0
    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.2016-11-26
  • 0
    $$\prod_{n\geq1}\frac1{4en}\bigg(\frac{(16n^2-9)^3}{16n^2-1}\bigg)^{1/4}\bigg(\frac{(4n+3)(4n-1)}{(4n-3)(4n+1)}\bigg)^n=\sqrt{\frac{2}{3\pi\sqrt{3}}}\exp\bigg(\frac{G}{\pi}+\frac12\bigg)$$ $G$ is Catalan's constant2018-12-31

63 Answers 63

220

$$\int_0^1\frac{\mathrm{d}x}{x^x}=\sum_{k=1}^\infty \frac1{k^k}$$

  • 49
    Sophomore's Dream?2012-06-19
  • 11
    $$\int_0^1 {x^x}\mathrm{d}x=\sum_{k=1}^\infty\frac{(-1)^{k-1}}{k^k}$$2013-11-02
154

$$\left(\sum\limits_{k=1}^n k\right)^2=\sum\limits_{k=1}^nk^3 .$$

The two on the left is not a typo.

  • 12
    If you're physics-minded, 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 2.2016-11-16
106

$$ \infty! = \sqrt{2 \pi} $$

It comes from the zeta function.

  • 2
    Can you help me understand $\infty!$? I don't know what to make of it.2010-11-04
  • 10
    @don: it's $\exp(-\zeta^{\prime}(0))$, where $\zeta^{\prime}(z)$ is *formally* $-\sum_{k=1}^\infty \frac{\ln\;k}{k^z}$2010-11-04
  • 0
    @a little don, You can read about it here http://katlas.math.toronto.edu/drorbn/MathBlog/2008-11/one/Gillet@FI-_What_is_infinity_factorial_(and_why_might_we_care)Q.pdf2010-11-04
  • 0
    Neat! I wonder whether "solving" this identity for $\infty$ [also yields $-\frac12$](http://math.stackexchange.com/questions/1074870/is-infty-frac12) *edit* Hm, since $(-\frac12)!=\sqrt\pi$ not :/2014-12-19
93

Ah, this is one identity which comes into use for proving the Euler's Partition Theorem. The identity is as follows: $$ (1+x)(1+x^{2})(1+x^{3}) \cdots = \frac{1}{(1-x)(1-x^{3})(1-x^{5}) \cdots}$$

78

Machin's Formula: \begin{eqnarray} \frac{\pi}{4} = 4 \arctan \frac{1}{5} - \arctan \frac{1}{239}. \end{eqnarray}

73

$$\frac{1}{\sin(2\pi/7)} + \frac{1}{\sin(3\pi/7)} = \frac{1}{\sin(\pi/7)}$$

  • 3
    I thought this was going to be hard to prove...It just took three lines!2014-02-01
69

The Frobenius automorphism

$$(x + y)^p = x^p + y^p$$

  • 56
    Only in a field of prime characteristic $p$ (much to the chagrin of my calculus students).2012-06-19
  • 25
    @AustinMohr: Not just _in a field_ of prime characteristic $p$, it holds in any commutative ring of characteristic $p$.2013-09-30
67

\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 number-of-unitary-divisors (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 $k$.

  • 0
    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.2017-03-23
  • 0
    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 :).2018-04-03
64

$$\large{1,741,725 = 1^7 + 7^7 + 4^7 + 1^7 + 7^7 + 2^7 + 5^7}$$

and

$$\large{111,111,111 \times 111,111,111 = 12,345,678,987,654,321}$$

  • 17
    is there any way to generalise $$\large{1,741,725 = 1^7 + 7^7 + 4^7 + 1^7 + 7^7 + 2^7 + 5^7}$$?2012-11-16
  • 6
    Do a search for Armstrong numbers and/or narcissistic numbers. Or type 1741725 into the Online Encyclopedia of Integer Sequences.2013-09-26
  • 0
    @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.)2016-07-12
61

$$\sec^2(x)+\csc^2(x)=\sec^2(x)\csc^2(x)$$

  • 0
    Is the natural logarithm function, or the exponential function related to this?2012-02-13
  • 6
    @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.2012-02-13
  • 3
    And because of this identity we have $\frac{\,d}{\,dx} \left[e^{\tan{x}} \cdot e^{-\cot{x}}\right] = \frac{\,d}{\,dx} \left[e^{\tan{x}}\right] \cdot \frac{\,d}{\,dx} \left[e^{-\cot{x}}\right]$.2017-06-10
58

\[\sqrt{n^{\log n}}=n^{\log \sqrt{n}}\]

  • 9
    $a^{\log b} = b^{\log a}$ for a and b at least 1.2010-11-30
  • 0
    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 $n^{\log \sqrt{n}}$.2010-11-30
  • 12
    Isn't this kind of trivial?2014-02-26
  • 0
    @ChantryCargill Not when you consider the fact that not many people know (surprisingly) that $\sqrt{x} \equiv x^{\frac{1}{2}}$2014-07-12
57

$\displaystyle\big(a^2+b^2\big)\cdot\big(c^2+d^2\big)=\big(ac \mp bd\big)^2+\big(ad \pm bc\big)^2$

  • 41
    $(a^2+b^2)\cdot(c^2+d^2)$ is obviously $c^4+c^2d^2$2011-04-18
  • 6
    $c^2(c^2+d^2)$??... what do you mean?2011-04-18
  • 31
    $a^2+b^2=c^2$2011-04-18
  • 0
    There are also the related Lagrange's identity and the corresponding one for eight squares.2011-11-16
  • 7
    The Brahmagupta-Fibonacci identity.2012-05-01
53

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}

51

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.
  • 5
    An 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.2012-06-19
  • 4
    I 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.2013-01-13
  • 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...2017-06-03
49

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?

$$11 \times 11 = 121$$ $$111 \times 111 = 12321$$ $$1111 \times 1111 = 1234321$$ $$11111 \times 11111 = 123454321$$ $$\vdots$$

  • 21
    The 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 40-hex-digit numbers that start with those particular 38 chosen digits.2013-11-20
  • 1
    I remember that last from reading "The number devil"! And it works for other bases too; for a base $b$, until $\left(\sum_{n=0}^{b-1}\left(b^n\right)\right)^2=123...\ \text{digit } b-1\ ...321$.2013-11-24
  • 0
    I find (1....1)^n interesting, it's also nearly impossible to caluclate by hand without messing it up.2016-03-04
40

$$ \frac{e}{2} = \left(\frac{2}{1}\right)^{1/2}\left(\frac{2\cdot 4}{3\cdot 3}\right)^{1/4}\left(\frac{4\cdot 6\cdot 6\cdot 8}{5\cdot 5\cdot 7\cdot 7}\right)^{1/8}\left(\frac{8\cdot 10\cdot 10\cdot 12\cdot 12\cdot 14\cdot 14\cdot 16}{9\cdot 9\cdot 11\cdot 11\cdot 13\cdot 13\cdot 15\cdot 15}\right)^{1/16}\cdots $$ [Nick Pippenger, Amer. Math. Monthly, 87 (1980)]. Set all the exponents to 1 and you get the Wallis formula for $\pi/2$.

40

\begin{align} \frac{\mathrm d}{\mathrm dx}(x^x) &= x\cdot x^{x-1} &\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^{x-1}+x^{x}\ln(x) &\text{Sum of these?}&\ \text{True}\\ \end{align}

  • 11
    This is a special case of $\frac{d}{dx} h(f(x),g(x)) = \partial_1 h f' + \partial_2h g'$2013-12-20
  • 24
    "I just add those 2 together so that I can get partial credit."2014-08-11
  • 0
    @Derek朕會功夫 God the puns xD2017-03-21
37

\begin{eqnarray} \sum_{i_1 = 0}^{n-k} \, \sum_{i_2 = 0}^{n-k-i_1} \cdots \sum_{i_k = 0}^{n-k-i_1 - \cdots - i_{k-1}} 1 = \binom{n}{k} \end{eqnarray}

32

$$\sum\limits_{n=1}^{\infty} n = 1 + 2 + 3 + \cdots \text{ad inf.} = -\frac{1}{12}$$

You can also see many more here: The Euler-Maclaurin formula, Bernoulli numbers, the zeta function, and real-variable analytic continuation

  • 1
    @fmartin: it's $\zeta(-1)$, which can be shown to be expressible in terms of Bernoulli numbers.2010-11-08
  • 9
    @J.M.: I still fail to see how an infinite summation of positive numbers can result in a negative number.2010-11-15
  • 4
    @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.2010-11-16
  • 9
    It'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 non-supersymmetric 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.2010-11-27
  • 4
    Isn't this Ramanujan's interpretation of $\zeta(-1)$.2012-08-14
  • 0
    This is an example of zeta regularization.2013-09-27
  • 0
    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.2013-11-03
  • 0
    http://www.youtube.com/watch?v=w-I6XTVZXww2014-03-13
  • 0
    Related: [Why does $1+2+3+\dots = -\frac{1}{12}$?](http://math.stackexchange.com/questions/39802/why-does-123-dots-frac112)2014-07-12
  • 0
    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=sD0NjbwqlYw2018-09-11
29

Two related integrals:

$$\int_0^\infty\sin\;x\quad\mathrm{d}x=1$$

$$\int_0^\infty\ln\;x\;\sin\;x\quad \mathrm{d}x=-\gamma$$

  • 5
    can you give a hint for those of us who don't see it? :)2010-11-03
  • 8
    They are Abel-summable integrals; e.g. the first one is properly interpreted as $\lim_{\epsilon\to 0}\int\exp(-\epsilon x)\sin\;x\quad \mathrm{d}x$2010-11-03
  • 5
    Doesn't the first integral diverge?2012-06-19
  • 0
    @jak, see my previous comment.2012-06-19
  • 0
    @J.M. I have not thought about this, but is that unique? I don't think so, but why look at the $\exp$ "kernel"?2012-07-04
  • 0
    @AD. I'm just going from the definition of Abel-summable 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.2012-07-05
  • 0
    @J.M. Ok Thanks.2012-07-05
28

M.V Subbarao's identity: an integer $n>22$ is a prime number iff it satisfies,

$$n\sigma(n)\equiv 2 \pmod {\phi(n)}$$

  • 0
    do you have a reference for this result? Very interesting!2012-07-13
  • 0
    @PAD: See [M. V. Subbarao, *On two congruences for primality,* Pacific Journal of Mathematics, Volume 52, Number 1 (1974), 261-268](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 $1, 4, 6, 22$.2013-10-03
26

$32768=(3-2+7)^6 / 8$

Just a funny coincidence.

24

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:

$$\prod _{n=1}^{\infty } \frac{1}{1-\frac{1}{(n p_1 p_2)^{2}}}=\frac{\pi }{3}$$

  • 5
    It doesn't make sense to speak of "perfect squares" for positive real numbers... but this is a nice identity though.2012-06-19
  • 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.2013-09-26
  • 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.2014-08-28
24

$$ 10^2+11^2+12^2=13^2+14^2 $$

There's a funny Abstruse Goose comic about this, which I can't seem to find at the moment.

  • 11
    http://abstrusegoose.com/632013-09-27
24

$${\Large% \sqrt{\,\vphantom{\huge A}\color{#00f}{20}\color{#c00000}{25}\,}\, =\ \color{#00f}{20}\ +\ \color{#c00000}{25}\ =\ 45} $$

23

Parallelogram

$$\left|z+z'\right|^{2}+\left|z-z'\right|^{2}=2\times\left(\left|z\right|^{2}+\left|z'\right|^{2}\right)$$

The sum of the squares of the sides equals the sum of the squares of the diagonals.

  • 0
    The so-called parallelogram identity.2013-08-19
23

What is 42?

$$ 6 \times 9 = 42 \text{ base } 13 $$ I always knew that there is something wrong with this universe.

23

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 $$n(n+1)(n+2)(n+3) = k^2 - 1.$$

  • 0
    I wonder now if this isn't a proper identity because of the existential quantifier. Any thoughts?2012-06-19
  • 11
    You can also write that as $$n(n+1)(n+2)(n+3)=((n+1)^2+1)^2 - 1$$2012-06-19
23

I have one: In a $\Delta ABC$, $$\tan A+\tan B+\tan C=\tan A\tan B\tan C.$$

  • 7
    Also $\cot(A/2)+\cot(B/2)+ \cot(C/2)=\cot(A/2)\cot(B/2)\cot(C/2)$.2013-04-11
21

Considering the main branches

$$i^i = \exp\left(-\frac{\pi}{2}\right)$$

$$\root i \of i = \exp\left(\frac{\pi}{2}\right) $$

And $$ \frac{4}{\pi } = \displaystyle 1 + \frac{1}{{3 +\displaystyle \frac{{{2^2}}}{{5 + \displaystyle\frac{{{3^2}}}{{7 +\displaystyle \frac{{{4^2}}}{{9 +\displaystyle \frac{{{n^2}}}{{\left( {2n + 1} \right) + \cdots }}}}}}}}}} $$

17

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 $d_n = (n-1)! = f^{n-1}$ (the first object has $n-1$ legal locations, the second $n-2$, ...). The correct answer isn't that different actually:

$d_n = (f-1)^n$.

  • 1
    Hooray for Umbral calculus!2012-06-18
  • 1
    To 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 $d_n=L\left((f-1)^n\right)$. (Am I doing this right?)2014-08-28
17

$$\int_0^\infty\frac1{1+x^2}\cdot\frac1{1+x^\pi}dx=\int_0^\infty\frac1{1+x^2}\cdot\frac1{1+x^e}dx$$

  • 0
    Is this true for any exponent, or just $\pi$ and $e$?2015-08-17
  • 0
    It's a simple exercise, try it.2015-08-17
  • 0
    @Vladimir: Would you mind giving a hint?2015-11-28
  • 0
    The identity also holds if you replace $\pi$ with Euler's $\gamma$ :)2015-11-29
  • 0
    Obviously... :)2016-02-21
16

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:

15

I actually think currying is really cool:

$$(A \times B) \to C \; \simeq \; A \to (B \to C)$$

Though not strictly an identity, but an isomorphism.

When I met it for the first time it seemed to be a bit odd but it is so convenient and neat. At least in programming.

15

$$ 71 = \sqrt{7! + 1}. $$

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.

14

The Cayley-Hamilton 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 $$p(A) = \det(AI_n-A) = \det(0-0) = 0$$

14

We have by block partition rule for determinant $$ \det \left[ \begin{array}{cc} U & R \\ L & D \end{array} \right] = \det U\cdot \det ( D-LU^{-1}R) $$ But if $U,R,L$ and $D$ commute we have that $$ \det \left[ \begin{array}{cc} U & R \\ L & D \end{array} \right] = \det (UD-LR) $$

  • 0
    Actually, only U and L need to commute.2018-08-22
13

$$\frac{1}{998901}=0.000001002003004005006...997999000001...$$

  • 0
    Similar results for $1/((9...9)^2)$, and any further analysis will explain why :-)2012-07-12
  • 0
    Indeed, I saw it explained [here](http://www.youtube.com/watch?v=daro6K6mym8) initially :-)2012-07-14
  • 4
    1/98.99 = 0.010102030508132134...2013-09-26
12

\begin{eqnarray} \zeta(0) = \sum_{n \geq 1} 1 = -\frac{1}{2} \end{eqnarray}

  • 1
    Actually, this can be made rigorous by noting that $$ \zeta(z)=\lim_{n\to\infty}\left(\sum_{k=1}^nk^{-z}-\frac{1}{1-z}n^{1-z}-\frac12n^{-z}\right) $$ for $\mathrm{Re}(z)>-1$.2012-06-21
12

$$ \frac{1}{2}=\frac{\frac{1}{2}}{\frac{1}{2}+\frac{\frac{1}{2}}{\frac{1}{2}+\frac{\frac{1}{2}}{\frac{1}{2}+\frac{\frac{1}{2}}{\frac{1}{2}+\frac{\frac{1}{2}}{\frac{1}{2}+\frac{\frac{1}{2}}{\frac{1}{2}+\cdots}}}}}} $$

and more generally we have $$ \frac{1}{n+1}=\frac{\frac{1}{n(n+1)}}{\frac{1}{n(n+1)}+\frac{\frac{1}{n(n+1)}}{\frac{1}{n(n+1)}+\frac{\frac{1}{n(n+1)}}{\frac{1}{n(n+1)}+\frac{\frac{1}{n(n+1)}}{\frac{1}{n(n+1)}+\frac{\frac{1}{n(n+1)}}{\frac{1}{n(n+1)}+\frac{\frac{1}{n(n+1)}}{\frac{1}{n(n+1)}+\ddots}}}}}} $$

11

Heres a interesting one again
$3435=3^3+4^4+3^3+5^5%$

10

$(x-a)(x-b)(x-c)\ldots(x-z) = 0$

  • 0
    I don't get it?2015-08-17
  • 4
    @columbus8myhw $(x-a)(x-b)(x-c)\ldots (x-w){\color{red}{(x-x)}}(x-y)(x-z)=0$2015-08-23
8

\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}

  • 4
    I don't see the 'punch' here. Isn't that just reversing the order of summation and truncating some zeros?2013-01-13
8

\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. 202-209.

  • 0
    See also http://math.stackexchange.com/questions/49657/bad-fraction-reduction-that-actually-works2013-11-15
  • 0
    Also [Project Euler problem number 33](https://projecteuler.net/problem=33).2016-10-24
7

$\lnot$(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".

6

$$ \sin \theta \cdot \sin \bigl(60^\circ - \theta \bigr) \cdot \sin \bigl(60^\circ + \theta \bigr) = \frac{1}{4} \sin 3\theta$$

$$ \cos \theta \cdot \cos \bigl(60^\circ - \theta \bigr) \cdot \cos \bigl(60^\circ + \theta \bigr) = \frac{1}{4} \cos 3\theta$$

$$ \tan \theta \cdot \tan \bigl(60^\circ - \theta \bigr) \cdot \tan \bigl(60^\circ + \theta \bigr) = \tan 3\theta $$

  • 1
    See also the comments here: http://math.stackexchange.com/q/8385/12422010-11-04
  • 0
    Not much idea what you want me to understand ? :)2010-11-04
  • 1
    I 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 $\sin(\theta+120^{\circ})$.)2010-11-04
  • 1
    considering your first two identities the thirth should be $$ \tan \theta \cdot \tan \bigl(60 - \theta \bigr) \cdot \tan \bigl(60 + \theta \bigr) = \tan 3\theta $$2011-03-06
6

$$2592=2^59^2$$ Found this in one of Dudeney's puzzle books

6

$$ \sum_{n=1}^{+\infty}\frac{\mu(n)}{n}=1-\frac12-\frac13-\frac15+\frac16-\frac17+\frac1{10}-\frac1{11}-\frac1{13}+\frac1{14}+\frac1{15}-\cdots=0 $$ 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 $\zeta$: $$ \pi^{\frac s2}\Gamma\left(\frac s2\right)\zeta(s)= \pi^{\frac{1-s}2}\Gamma\left(\frac{1-s}2\right)\zeta(1-s)\;\;\;\forall s\in\mathbb C\;. $$

Moreover no one seems to have wrote the Basel problem (Euler, 1735): $$ \sum_{n=1}^{+\infty}\frac1{n^2}=\frac{\pi^2}{6}\;\;. $$

5

$$27\cdot56=2\cdot756,$$ $$277\cdot756=27\cdot7756,$$ $$2777\cdot7756=277\cdot77756,$$ and so on.

  • 1
    How does it work?2015-10-02
5

$$\frac{\pi}{4}=\sum_{n=1}^{\infty}\arctan\frac{1}{f_{2n+1}}, $$ where $f_{2n+1}$ there are fibonacci numbers, $n=1,2,...$

4

$\tan^{-1}(1)+\tan^{-1}(2)+\tan^{-1}(3) = \pi$ (using the principal value), but if you blindly use the addition formula $\tan^{-1}(x) + \tan^{-1}(y) = \tan^{-1}\dfrac{x+y}{1-x y}$ twice, you get zero:

$\tan^{-1}(1) + \tan^{-1}(2) = \tan^{-1}\dfrac{1+2}{1-1*2} =\tan^{-1}(-3)$; $\tan^{-1}(1) + \tan^{-1}(2) + \tan^{-1}(3) =\tan^{-1}(-3) + \tan^{-1}(3) =\tan^{-1}\dfrac{-3+3}{1-(-3)(3)} = 0$.

4

$$\lim_{\omega\to\infty}3=8$$ The "proof" is by rotation through $\pi/2$. More of a joke than an identity, I suppose.

  • 3
    Remind me of this: [http://xkcd.com/184/](http://xkcd.com/184/)2013-11-03
  • 0
    Or maybe the "proof" is through taking the closure?2013-12-15
4

For all $n\in\mathbb{N}$ and $n\neq1$ $$\prod_{k=1}^{n-1}2\sin\frac{k \pi}{n} = n$$

For some reason, the proof involves complex numbers and polynomials.

Link to proof: Prove that $\prod_{k=1}^{n-1}\sin\frac{k \pi}{n} = \frac{n}{2^{n-1}}$

4

Here's one clever trigonometric identity that impressed me in high-school 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 half-angle. Actually if you want to prove those, nothing but the addition formulas are required.

3

\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}

3

$$ \begin{array}{rcrcl} \vdots & \vdots & \vdots & \vdots & \vdots \\[1mm] \int{1 \over x^{3}}\,{\rm d}x & = & -\,{1 \over 2}\,{1 \over x^{2}} & \sim & x^{\color{#ff0000}{\large\bf -2}} \\[1mm] \int{1 \over x^{2}}\,{\rm d}x & = & -\,{1 \over x} & \sim & x^{\color{#ff0000}{\large\bf -1}} \\[1mm] \int{1 \over x}\,{\rm d}x & = & \ln\left(x\right) & \sim & x^{\color{#0000ff}{\LARGE\bf 0}} \color{#0000ff}{\LARGE\quad ?} \\[1mm] \int x^{0}\,{\rm d}x & = & x^{1} & \sim & x^{\color{#ff0000}{\large\bf 1}} \\[1mm] \int x\,{\rm d}x & = & {1 \over 2}\,x^{2} & \sim & x^{\color{#ff0000}{\large\bf 2}} \\[1mm] \vdots & \vdots & \vdots & \vdots & \vdots \end{array} $$

  • 0
    $\ln(x)$ converges slowly enough, so indeed $\ln(x) \sim 1$ might not be completely stupid :p2013-09-28
  • 1
    Hmm, 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.2013-09-28
  • 0
    @Travis Wow, nice one! But I think you took my comment too seriously.2013-09-28
  • 0
    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 $\displaystyle\ln x=\lim_{t\to0}\frac{x^t-1}t$, which perhaps fits the pattern better.)2014-08-28
1

$$ \frac{\pi}{2}=1+2\sum_{k=1}^{\infty}\frac{\eta(2k)}{2^{2k}} $$ $$ \frac{\pi}{3}=1+2\sum_{k=1}^{\infty}\frac{\eta(2k)}{6^{2k}} $$ where $ \eta(n)=\sum_{k=1}^{\infty}\frac{(-1)^{k+1}}{k^{n}} $

1

Voronoi summation formula:

$\sum \limits_{n=1}^{\infty}d(n)(\frac{x}{n})^{1/2}\{Y_1(4\pi \sqrt{nx})+\frac{2}{\pi}K_1(4\pi \sqrt{nx})\}+x \log x +(2 \gamma-1)x +\frac{1}{4}=\sum \limits _{n\leq x}'d(n)$

0

$\textbf{Claim:}\quad$$$\frac{\sin x}{n}=6$$ for all $n,x$ ($n\neq 0$).

$\textit{Proof:}\quad$$$\frac{\sin x}{n}=\frac{\dfrac{1}{n}\cdot\sin x}{\dfrac{1}{n}\cdot n}=\frac{\operatorname{si}x}{1}=\text{six}.\quad\blacksquare$$

  • 0
    how do you obtain $ \dfrac{six}{1} $ ?2015-01-29
  • 0
    @Tgymasb Denominator: $1/n\cdot n=1$. Numerator: $1/\textbf{n}\cdot\operatorname{si}\!\textbf{n}\, x=\operatorname{si}x$.2015-01-29
0

Let $\sigma(n)$ denote the sum of the divisors of $n$.

If $$p=1+\sigma(k),$$ then $$p^a=1+\sigma(kp^{a-1})$$ where $a,k$ are positive integers and $p$ is a prime such that $p\not\mid k$.

0

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 $$P^k_{i,j} = \binom{i}{j}k^{i-j}$$

where $P^k_{i,j}$ is the element of $P^k$ in the position $i,j.$

0

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 Euler-Mascheroni constant $\gamma$ and finally $\pi$. Well we can see easily that

$$e\cdot\gamma\cdot\pi\cdot\phi \approx e + \gamma + \pi + \phi$$

  • 1
    I wouldn't take credit for this dude2015-11-11
-2

$$ \int_{-\infty}^{\infty}{\sin\left(x\right) \over x}\,{\rm d}x = \pi\int_{-1}^{1}\delta\left(k\right)\,{\rm d}k $$