Funny identities

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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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
Also this $\cos(x - y) \cos(x + y) = (\cos x - \cos y)(\cos x + \cos y)$ holds ;) – Quixotic Nov 4 '10 at 8:38
I have tripped up many calculus students with this one: $log(1+2+3)=log1+log2+log3$. I am evil... – Steve D Dec 8 '12 at 1:23
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40 Answers

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

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Sophomore's Dream? – rotskoff Jun 19 '12 at 20:50
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$$\left(\sum\limits_{k=1}^n k\right)^2=\sum\limits_{k=1}^nk^3 .$$

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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

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

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

It comes from the zeta function.

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@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
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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} 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 $c^4+c^2d^2$ – muntoo Apr 18 '11 at 2:30
$c^2(c^2+d^2)$??... what do you mean? – Neves Apr 18 '11 at 13:43
$a^2+b^2=c^2$ – muntoo Apr 18 '11 at 16:06
The Brahmagupta-Fibonacci identity. – The Chaz 2.0 May 1 '12 at 5:17
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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.
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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
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$\sqrt{n}^{\log n}=n^{\log \sqrt{n}}$

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$a^{\log b} = b^{\log a}$ for a and b at least 1. – wok Nov 30 '10 at 10:03
You should have written $\sqrt{n^{\log n}}$ – ypercube Feb 25 '11 at 22:13
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$$\sum\limits_{n=1}^{\infty} n = 1 + 2 +3 + \cdots \text{ad inf.} = - \frac{1}{12}$$

Now doesn't this sound funny. You can also see many more on http://terrytao.wordpress.com/2010/04/10/the-euler-maclaurin-formula-bernoulli-numbers-the-zeta-function-and-real-variable-analytic-continuation/

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What am I missing here? – Fernando Martin 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. – Fernando Martin 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 $\zeta(-1)$. – pbs Aug 14 '12 at 11:43
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$$\large{1,741,725 = 1^7 + 7^7 + 4^7 + 1^7 + 7^7 + 2^7 + 5^7}$$

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

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

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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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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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$$\frac{e}{2} = \left(\frac{2}{1}\right)^{1/2}\left(\frac{2\cdot 4}{3\cdot 3}\right)^{1/4}\left(\frac{4\cdot 6\cdot 6\cdot 8}{5\cdot 5\cdot 7\cdot 7}\right)^{1/8}\left(\frac{8\cdot 10\cdot 10\cdot 12\cdot 12\cdot 14\cdot 14\cdot 16}{9\cdot 9\cdot 11\cdot 11\cdot 13\cdot 13\cdot 15\cdot 15}\right)^{1/16}\cdots$$ [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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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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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

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

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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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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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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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Shouldn't that be $6\times 9=42\mod 12$? – Joel Reyes Noche Feb 6 '12 at 11:48
@JoelReyesNoche $6\times 9 =54 = 4\times 13 +2$ from here – draks ... Feb 6 '12 at 12:02
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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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You can also write that as $$n(n+1)(n+2)(n+3)=((n+1)^2+1)^2 - 1$$ – AD. Jun 19 '12 at 6:53
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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 at 0:10