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

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

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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 $i,j.$

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

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.

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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? –  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

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

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

$$\sum_{n=1}^{+\infty}\frac{\mu(n)}{n}=1-\frac12-\frac13-\frac15+\frac16-\frac17+\frac1{10}-\frac1{11}-\frac1{13}+\frac1{14}+\frac1{15}-\cdots=0$$ 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): $$\sum_{n=1}^{+\infty}\frac1{n^2}=\frac{\pi^2}{6}\;\;.$$

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

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

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

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

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

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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.

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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 at 5:33

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

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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.

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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:

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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 $p\not\mid k$.

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

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\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.

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

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

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

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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

$\textbf{Claim:}\quad$$\frac{\sin x}{n}=6$$ for all$n,x$($n\neq 0$).$\textit{Proof:}\quad$$\frac{\sin x}{n}=\frac{\dfrac{1}{n}\cdot\sin x}{\dfrac{1}{n}\cdot n}=\frac{\operatorname{si}x}{1}=\text{six}.\quad\blacksquare$$

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$\tan^{-1}(1)+\tan^{-1}(2)+\tan^{-1}(3) = \pi$ (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:

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

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

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Remind me of this: http://xkcd.com/184/ –  alex.jordan Nov 3 '13 at 17:31

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

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

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\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} - 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. – Travis Bemrose Sep 28 '13 at 10:11 \int_{-\infty}^{\infty}{\sin\left(x\right) \over x}\,{\rm d}x = \pi\int_{-1}^{1}\delta\left(k\right)\,{\rm d}k - \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} - \[\sqrt{n^{\log n}}=n^{\log \sqrt{n}} - a^{\log b} = b^{\log a} for a and b at least 1. – Wok Nov 30 '10 at 10:03 Here is an Asian kid paradox - perhaps they will understand (apologies for not being strictly mathematical). IfStudy=No \;Fail$$and$$No \; Study=Fail$$then$$Study+No\;Study=Fail+No\;Fail\implies (1+No)Study=(1+No)Fail$$Cancelling gives$$Study=Fail Isn't that weird???

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$1+No$ is 0 in both sides. You can't cancel zero... –  CODE Jun 4 '13 at 16:47