# 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?

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## closed as primarily opinion-based by Najib Idrissi, Mike Miller, ᴡᴏʀᴅs, apnorton, user2345215Jan 31 '15 at 18:16

Many good questions generate some degree of opinion based on expert experience, but answers to this question will tend to be almost entirely based on opinions, rather than facts, references, or specific expertise.If this question can be reworded to fit the rules in the help center, please edit the question.

Maybe a moderator should put the zeta ones together since there are three already? – anon Nov 3 '10 at 22:29
Perhaps this should be a community wiki question. – Nuno Nov 3 '10 at 22:31
This is related. – J. M. Nov 3 '10 at 22:35
I have tripped up many calculus students with this one: $log(1+2+3)=log1+log2+log3$. I am evil... – user641 Dec 8 '12 at 1:23
@SteveD If only we could find an odd example... – peoplepower Jan 13 '13 at 0:31

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

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I had to do something about my accept range :) – AD. May 17 '12 at 4:47
Sophomore's Dream? – rotskoff Jun 19 '12 at 20:50
$$\int_0^1 {x^x}\mathrm{d}x=\sum_{k=1}^\infty\frac{(-1)^{k-1}}{k^k}$$ – bwv869 Nov 2 '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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I think the first exponent should be a 3. – Daron Nov 3 '13 at 17:19
@Daron nope -- it is in fact a $2$. – oldrinb Nov 3 '13 at 17:36

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

It comes from the zeta function.

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Can you help me understand $\infty!$? I don't know what to make of it. – futurebird Nov 4 '10 at 0:55
@don: it's $\exp(-\zeta^{\prime}(0))$, where $\zeta^{\prime}(z)$ is formally $-\sum_{k=1}^\infty \frac{\ln\;k}{k^z}$ – J. M. Nov 4 '10 at 5:01
@a little don, You can read about it here katlas.math.toronto.edu/drorbn/MathBlog/2008-11/one/… – anon Nov 4 '10 at 17:29
Oh wow, this is the best. – Newb Oct 30 '13 at 8:29
Neat! I wonder whether "solving" this identity for $\infty$ also yields $-\frac12$ 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: $$(1+x)(1+x^{2})(1+x^{3}) \cdots = \frac{1}{(1-x)(1-x^{3})(1-x^{5}) \cdots}$$

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This identity was my favourite class of undergrad combinatorics! – Johanna Dec 24 '14 at 4:26

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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I thought this was going to be hard to prove...It just took three lines! – chubakueno Feb 1 '14 at 21:14

The Frobenius automorphism

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

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Only in a field of prime characteristic $p$ (much to the chagrin of my calculus students). – Austin Mohr Jun 19 '12 at 2:06
@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 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 $k$.

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

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$(a^2+b^2)\cdot(c^2+d^2)$ is obviously $c^4+c^2d^2$ – Mateen Ulhaq Apr 18 '11 at 2:30
$c^2(c^2+d^2)$??... what do you mean? – Neves Apr 18 '11 at 13:43
$a^2+b^2=c^2$ – Mateen Ulhaq Apr 18 '11 at 16:06
There are also the related Lagrange's identity and the corresponding one for eight squares. – Yuval Filmus Nov 16 '11 at 13:28
The Brahmagupta-Fibonacci identity. – The Chaz 2.0 May 1 '12 at 5:17

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

and

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

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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 '12 at 7:32
Do 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
+1 The latter one is really hard to prove, because most calculators don't have so many digits. – Thomas Weller Oct 27 '14 at 20:59
@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 at 2:21

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

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$a^{\log b} = b^{\log a}$ for a and b at least 1. – Wok Nov 30 '10 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 $n^{\log \sqrt{n}}$. – Douglas S. Stones Nov 30 '10 at 10:22
Isn't this kind of trivial? – Chantry Cargill Feb 26 '14 at 4:28
@ChantryCargill Not when you consider the fact that not many people know (surprisingly) that $\sqrt{x} \equiv x^{\frac{1}{2}}$ – Cole Johnson Jul 12 '14 at 18:31

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

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Is the natural logarithm function, or the exponential function related to this? – Doug Spoonwood Feb 13 '12 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. – Joe Johnson 126 Feb 13 '12 at 14:15

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.
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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. – Austin Mohr Jun 19 '12 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. – Michael Albanese Jan 13 '13 at 15:08

Facts about $\pi$ are always fun!

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

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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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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. – Henning Makholm Nov 20 '13 at 18:04
I remember that last from reading "The number devil"! And it works for other bases too; for a base $b$, until $\left(\sum_{n=0}^{b-1}\left(b^n\right)\right)^2=123...\ \text{digit } b-1\ ...321$. – 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 4 at 23:19

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

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

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What am I missing here? – F M Nov 6 '10 at 5:42
@J.M.: I still fail to see how an infinite summation of positive numbers can result in a negative number. – F M Nov 15 '10 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. – J. M. Nov 16 '10 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. – Raskolnikov Nov 27 '10 at 11:25
Isn't this Ramanujan's interpretation of $\zeta(-1)$. – poirot Aug 14 '12 at 11:43

\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}

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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{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}

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This 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
"I just add those 2 together so that I can get partial credit." – Derek 朕會功夫 Aug 11 '14 at 5:33

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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can you give a hint for those of us who don't see it? :) – anon Nov 3 '10 at 22:48
They are Abel-summable integrals; e.g. the first one is properly interpreted as $\lim_{\epsilon\to 0}\int\exp(-\epsilon x)\sin\;x\quad \mathrm{d}x$ – J. M. Nov 3 '10 at 22:53
Doesn't the first integral diverge? – Hawk Jun 19 '12 at 1:52
@jak, see my previous comment. – J. M. Jun 19 '12 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"? – AD. Jul 4 '12 at 19:31

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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do you have a reference for this result? Very interesting! – PAD Jul 13 '12 at 11:55
@PAD: See M. V. Subbarao, On two congruences for primality, Pacific Journal of Mathematics, Volume 52, Number 1 (1974), 261-268 (Another 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 3 '13 at 23:28

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

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

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abstrusegoose.com/63 – Memming Sep 27 '13 at 22:59

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

Just a funny coincidence.

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

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

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The so-called parallelogram identity. – nayrb Aug 19 '13 at 19:38

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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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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It 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
@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
@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 3:56

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

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Also $\cot(A/2)+\cot(B/2)+ \cot(C/2)=\cot(A/2)\cot(B/2)\cot(C/2)$. – N. S. Apr 11 '13 at 22:39

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

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

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

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Hooray for Umbral calculus! – Steven Stadnicki Jun 18 '12 at 22:41
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?) – Akiva Weinberger Aug 28 '14 at 4:00