# Funny identities [closed]

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?

• This is related. Nov 3, 2010 at 22:35
• I noticed that your expression can be also written as $\sin(x - y) \sin(x + y) = (\cos y + \cos x) (\cos y - \cos x)$ Nov 4, 2010 at 11:09
• I have tripped up many calculus students with this one: $log(1+2+3)=log1+log2+log3$. I am evil...
– user641
Dec 8, 2012 at 1:23
• @SteveD If only we could find an odd example... Jan 13, 2013 at 0:31
• Almost an identity: $$\sqrt{123456790}\approx 11111.11111\,.$$
– user158047
Jul 12, 2014 at 18:47

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

• Sophomore's Dream? Jun 19, 2012 at 20:50
• $$\int_0^1 {x^x}\mathrm{d}x=\sum_{k=1}^\infty\frac{(-1)^{k-1}}{k^k}$$ Nov 2, 2013 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.

• 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.
– Joce
Nov 16, 2016 at 15:01
• @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, 2019 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, 2019 at 20:44
• But 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, 2019 at 15:45
• This is called Nicomachus's theorem Aug 13, 2020 at 14:07

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

It comes from the zeta function.

• Can you help me understand $\infty!$? I don't know what to make of it. Nov 4, 2010 at 0:55
• @don: it's $\exp(-\zeta^{\prime}(0))$, where $\zeta^{\prime}(z)$ is formally $-\sum_{k=1}^\infty \frac{\ln\;k}{k^z}$ Nov 4, 2010 at 5:01
• @a little don, You can read about it here katlas.math.toronto.edu/drorbn/MathBlog/2008-11/one/…
– anon
Nov 4, 2010 at 17:29
• Neat! I wonder whether "solving" this identity for $\infty$ also yields $-\frac12$ edit Hm, since $(-\frac12)!=\sqrt\pi$ not :/ Dec 19, 2014 at 19:41

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

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

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

• I thought this was going to be hard to prove...It just took three lines! Feb 1, 2014 at 21:14

The Frobenius automorphism

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

• Only in a field of prime characteristic $p$ (much to the chagrin of my calculus students). Jun 19, 2012 at 2:06
• @AustinMohr: Not just in a field of prime characteristic $p$, it holds in any commutative ring of characteristic $p$. Sep 30, 2013 at 6:32

$$\begin{eqnarray} 1^{3} + 2^{3} + 2^{3} + 2^{3} + 4^{3} + 4^{3} + 4^{3} + 8^{3} = (1 + 2 + 2 + 2 + 4 + 4 + 4 + 8)^{2} \end{eqnarray}$$ 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 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 $$k$$.

• 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. Mar 23, 2017 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 :). Apr 3, 2018 at 19:15

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

and

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

• is 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, 2012 at 7:32
• Do a search for Armstrong numbers and/or narcissistic numbers. Or type 1741725 into the Online Encyclopedia of Integer Sequences. Sep 26, 2013 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.) Jul 12, 2016 at 2:21

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

• Is the natural logarithm function, or the exponential function related to this? Feb 13, 2012 at 3:24
• @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, 2012 at 14:15
• And because of this identity we have $\frac{\,d}{\,dx} \left[e^{\tan{x}} \cdot e^{-\cot{x}}\right] = \frac{\,d}{\,dx} \left[e^{\tan{x}}\right] \cdot \frac{\,d}{\,dx} \left[e^{-\cot{x}}\right]$.
– Ant
Jun 10, 2017 at 3:56

$\sqrt{n^{\log n}}=n^{\log \sqrt{n}}$

• $a^{\log b} = b^{\log a}$ for a and b at least 1.
– Wok
Nov 30, 2010 at 10:03
• 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 $n^{\log \sqrt{n}}$. Nov 30, 2010 at 10:22
• Isn't this kind of trivial?
– user109879
Feb 26, 2014 at 4:28
• @ChantryCargill Not when you consider the fact that not many people know (surprisingly) that $\sqrt{x} \equiv x^{\frac{1}{2}}$ Jul 12, 2014 at 18:31

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

• $(a^2+b^2)\cdot(c^2+d^2)$ is obviously $c^4+c^2d^2$ Apr 18, 2011 at 2:30
• $c^2(c^2+d^2)$??... what do you mean? Apr 18, 2011 at 13:43
• $a^2+b^2=c^2$ Apr 18, 2011 at 16:06
• There are also the related Lagrange's identity and the corresponding one for eight squares. Nov 16, 2011 at 13:28
• The Brahmagupta-Fibonacci identity. May 1, 2012 at 5:17

Facts about $\pi$ are always fun!

$$\frac{\pi}{2} = \frac{2}{1}\cdot\frac{2}{3}\cdot\frac{4}{3}\cdot\frac{4}{5}\cdot\frac{6}{5}\cdot\frac{6}{7}\cdot\frac{8}{7}\cdot\ldots\\$$ $$\frac{\pi}{4} = 1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+\frac{1}{9}-\ldots\\$$ $$\frac{\pi^2}{6} = 1+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+\ldots\\$$ $$\frac{\pi^3}{32} = 1-\frac{1}{3^3}+\frac{1}{5^3}-\frac{1}{7^3}+\frac{1}{9^3}-\ldots\\$$ $$\frac{\pi^4}{90} = 1+\frac{1}{2^4}+\frac{1}{3^4}+\frac{1}{4^4}+\frac{1}{5^4}+\ldots\\$$ $$\frac{2}{\pi} = \frac{\sqrt{2}}{2}\cdot\frac{\sqrt{2+\sqrt{2}}}{2}\cdot\frac{\sqrt{2+\sqrt{2+\sqrt{2}}}}{2}\cdot\ldots\\$$ $$\pi = \cfrac{4}{1+\cfrac{1^2}{3+\cfrac{2^2}{5+\cfrac{3^2}{7+\cfrac{4^2}{9+\ldots}}}}}\\$$

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.
• 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 austinmohr.com/Work_files/730.pdf for details. Jun 19, 2012 at 2:09
• 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. Jan 13, 2013 at 15:08
• @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... Jun 3, 2017 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?

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

• 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. Nov 20, 2013 at 18:04
• I remember that last from reading "The number devil"! And it works for other bases too; for a base $b$, until $\left(\sum_{n=0}^{b-1}\left(b^n\right)\right)^2=123...\ \text{digit } b-1\ ...321$. Nov 24, 2013 at 11:06
• I find (1....1)^n interesting, it's also nearly impossible to caluclate by hand without messing it up. Mar 4, 2016 at 23:19
• 12343, 12347, 123457, 1234567891 are prime. Jun 14, 2023 at 22:25

\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}

• This is a special case of $\frac{d}{dx} h(f(x),g(x)) = \partial_1 h f' + \partial_2h g'$ Dec 20, 2013 at 13:56
• "I just add those 2 together so that I can get partial credit." Aug 11, 2014 at 5:33
• @Derek朕會功夫 God the puns xD Mar 21, 2017 at 1:16

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

\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}

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

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

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

• Kaprekar numbers Dec 25, 2022 at 8:17
• @ManasDogra Thanks. That's new for me. Dec 26, 2022 at 20:06

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

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

• @fmartin: it's $\zeta(-1)$, which can be shown to be expressible in terms of Bernoulli numbers. Nov 8, 2010 at 0:55
• @J.M.: I still fail to see how an infinite summation of positive numbers can result in a negative number. Nov 15, 2010 at 23:26
• @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. Nov 16, 2010 at 7:05
• 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. Nov 27, 2010 at 11:25
• Isn't this Ramanujan's interpretation of $\zeta(-1)$. Aug 14, 2012 at 11:43

Two related integrals:

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

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

• can you give a hint for those of us who don't see it? :)
– anon
Nov 3, 2010 at 22:48
• They are Abel-summable integrals; e.g. the first one is properly interpreted as $\lim_{\epsilon\to 0}\int\exp(-\epsilon x)\sin\;x\quad \mathrm{d}x$ Nov 3, 2010 at 22:53
• Doesn't the first integral diverge? Jun 19, 2012 at 1:52
• @jak, see my previous comment. Jun 19, 2012 at 2:05
• @J.M. I have not thought about this, but is that unique? I don't think so, but why look at the $\exp$ "kernel"? Jul 4, 2012 at 19:31

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:

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

• It doesn't make sense to speak of "perfect squares" for positive real numbers... but this is a nice identity though. Jun 19, 2012 at 19:53
• @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. Sep 26, 2013 at 21:29
• @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. Aug 28, 2014 at 3:56

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

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

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

Just a funny coincidence.

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

• Also $\cot(A/2)+\cot(B/2)+ \cot(C/2)=\cot(A/2)\cot(B/2)\cot(C/2)$. Apr 11, 2013 at 22:39
• @N.S. there's no need for them to be divided by 2 right? May 26, 2019 at 22:36
• @InertialObserver The tan identity applies to angles summing to two right angles, like angles in a triangle. The cot identity applies to angles summing to one right angle, so you'd need to halve angles in a triangle. Here's a result closely related to Heron's formula: $r^2s=(s-a)(s-b)(s-c)$ where $r$ is the radius of the incircle and $s$ the semi-perimeter $(a+b+c)/2$. Writing $x=s-a=r\cot(A/2)$, $y=s-b=r\cot(B/2)$, $z=s-c=r\cot(C/2)$, this becomes $xyz=r^2(x+y+z)$ and is equivalent to $\cot(A/2)+\cot(B/2)+\cot(C/2)=\cot(A/2)\cot(B/2)\cot(C/2)$ Aug 1, 2023 at 0:44

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

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

• The so-called parallelogram identity. Aug 19, 2013 at 19:38
• I remember similar identity in vector products. Aug 19, 2022 at 5:55

What is 42?

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

The product of any four consecutive integers is one less than a perfect square.

To phrase it more like an identity:

For every integer $$n$$, we have $$n(n+1)(n+2)(n+3) = ((n^2 + 1)^2 + n)^2 - 1.$$

• You can also write that as $$n(n+1)(n+2)(n+3)=((n+1)^2+1)^2 - 1$$ Jun 19, 2012 at 6:53
• @AD-StopPutin- Actually $$n(n+1)(n+2)(n+3) = ((n+1)^2+n)^2-1$$ Jun 28, 2023 at 8:35

Considering the main branches

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

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

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