# 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, Live Forever, apnorton, user2345215Jan 31 '15 at 18:16

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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
$$\int_0^1 {x^x}\mathrm{d}x=\sum_{k=1}^\infty\frac{(-1)^{k-1}}{k^k}$$ – Sekots Nov 2 '13 at 18:29

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

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

The two on the left is not a typo.

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@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

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

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

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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 $\mathrm{Re}(z)>-1$. – robjohn Jun 21 '12 at 0:52

\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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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

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

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

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

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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 $\sin(\theta+120^{\circ})$.) – Hans Lundmark Nov 4 '10 at 9:56
considering your first two identities the thirth should be $$\tan \theta \cdot \tan \bigl(60 - \theta \bigr) \cdot \tan \bigl(60 + \theta \bigr) = \tan 3\theta$$ – Neves Mar 6 '11 at 16:08

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

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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

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

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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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I don't see the 'punch' here. Isn't that just reversing the order of summation and truncating some zeros? – Ofir Jan 13 '13 at 0:10

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

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

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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
The Brahmagupta-Fibonacci identity. – The Chaz 2.0 May 1 '12 at 5:17

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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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

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

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$a^{\log b} = b^{\log a}$ for a and b at least 1. – Wok Nov 30 '10 at 10:03
Isn't this kind of trivial? – Chantry Cargill Feb 26 '14 at 4:28

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

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

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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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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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$\lnot$(A$\land$B)=($\lnot$A$\lor$$\lnotB) and \lnot(A\lorB)=(\lnotA\land$$\lnot$B), because they mean that negation is an "equal form".

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

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@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

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

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

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.

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$$\binom{n}{0}+\binom{n}{1}+\binom{n}{2}+\cdots+\binom{n}{n}=2^n$$

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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 $$p(A) = \det(AI_n-A) = \det(0-0) = 0$$

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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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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