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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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    $\begingroup$ This is related. $\endgroup$ – J. M. isn't a mathematician Nov 3 '10 at 22:35
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    $\begingroup$ I noticed that your expression can be also written as $\sin(x - y) \sin(x + y) = (\cos y + \cos x) (\cos y - \cos x) $ $\endgroup$ – Quixotic Nov 4 '10 at 11:09
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    $\begingroup$ I have tripped up many calculus students with this one: $log(1+2+3)=log1+log2+log3$. I am evil... $\endgroup$ – user641 Dec 8 '12 at 1:23
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    $\begingroup$ @SteveD If only we could find an odd example... $\endgroup$ – peoplepower Jan 13 '13 at 0:31
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    $\begingroup$ Almost an identity: $$\sqrt{123456790}\approx 11111.11111\,.$$ $\endgroup$ – Jakob Werner Jul 12 '14 at 18:47

63 Answers 63

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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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  • $\begingroup$ Is this true for any exponent, or just $\pi$ and $e$? $\endgroup$ – Akiva Weinberger Aug 17 '15 at 19:43
  • $\begingroup$ It's a simple exercise, try it. $\endgroup$ – Vladimir Reshetnikov Aug 17 '15 at 21:41
  • $\begingroup$ @Vladimir: Would you mind giving a hint? $\endgroup$ – Jason DeVito Nov 28 '15 at 19:51
  • $\begingroup$ The identity also holds if you replace $\pi$ with Euler's $\gamma$ :) $\endgroup$ – Vladimir Reshetnikov Nov 29 '15 at 1:19
  • $\begingroup$ Obviously... :) $\endgroup$ – ParaH2 Feb 21 '16 at 14:01
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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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    $\begingroup$ Hooray for Umbral calculus! $\endgroup$ – Steven Stadnicki Jun 18 '12 at 22:41
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    $\begingroup$ 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?) $\endgroup$ – Akiva Weinberger Aug 28 '14 at 4:00
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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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$$ 71 = \sqrt{7! + 1}. $$

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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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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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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We have by block partition rule for determinant $$ \det \left[ \begin{array}{cc} U & R \\ L & D \end{array} \right] = \det U\cdot \det ( D-LU^{-1}R) $$ But if $U,R,L$ and $D$ commute we have that $$ \det \left[ \begin{array}{cc} U & R \\ L & D \end{array} \right] = \det (UD-LR) $$

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  • $\begingroup$ Actually, only U and L need to commute. $\endgroup$ – Dirk Aug 22 '18 at 14:08
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Heres a interesting one again
$3435=3^3+4^4+3^3+5^5%$

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

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  • $\begingroup$ Similar results for $1/((9...9)^2)$, and any further analysis will explain why :-) $\endgroup$ – Mark Hurd Jul 12 '12 at 4:03
  • $\begingroup$ Indeed, I saw it explained here initially :-) $\endgroup$ – preferred_anon Jul 14 '12 at 20:19
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    $\begingroup$ 1/98.99 = 0.010102030508132134... $\endgroup$ – Empy2 Sep 26 '13 at 14:41
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$$ \dfrac{1}{2}=\dfrac{\dfrac{1}{2}}{\dfrac{1}{2}+\dfrac{\dfrac{1}{2}}{\dfrac{1}{2}+\dfrac{\dfrac{1}{2}}{\dfrac{1}{2}+\dfrac{\dfrac{1}{2}}{\dfrac{1}{2}+\dfrac{\dfrac{1}{2}}{\dfrac{1}{2}+\dfrac{\dfrac{1}{2}}{\dfrac{1}{2}+\cdots}}}}}} $$

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

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

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    $\begingroup$ 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 $\mathrm{Re}(z)>-1$. $\endgroup$ – robjohn Jun 21 '12 at 0:52
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$(x-a)(x-b)(x-c)\ldots(x-z) = 0$

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  • $\begingroup$ I don't get it? $\endgroup$ – Akiva Weinberger Aug 17 '15 at 19:52
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    $\begingroup$ @columbus8myhw $(x-a)(x-b)(x-c)\ldots (x-w){\color{red}{(x-x)}}(x-y)(x-z)=0$ $\endgroup$ – Surb Aug 23 '15 at 16:00
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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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\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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    $\begingroup$ I don't see the 'punch' here. Isn't that just reversing the order of summation and truncating some zeros? $\endgroup$ – Ofir Jan 13 '13 at 0:10
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$\lnot$(A$\land$B)=($\lnot$A$\lor$$\lnot$B) and $\lnot$(A$\lor$B)=($\lnot$A$\land$$\lnot$B), because they mean that negation is an "equal form".

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

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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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    $\begingroup$ See also the comments here: math.stackexchange.com/q/8385/1242 $\endgroup$ – Hans Lundmark Nov 4 '10 at 8:53
  • $\begingroup$ Not much idea what you want me to understand ? :) $\endgroup$ – Quixotic Nov 4 '10 at 9:38
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    $\begingroup$ 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 $\sin(\theta+120^{\circ})$.) $\endgroup$ – Hans Lundmark Nov 4 '10 at 9:56
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    $\begingroup$ considering your first two identities the thirth should be $$ \tan \theta \cdot \tan \bigl(60 - \theta \bigr) \cdot \tan \bigl(60 + \theta \bigr) = \tan 3\theta $$ $\endgroup$ – Neves Mar 6 '11 at 16:08
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$$ \sum_{n=1}^{+\infty}\frac{\mu(n)}{n}=1-\frac12-\frac13-\frac15+\frac16-\frac17+\frac1{10}-\frac1{11}-\frac1{13}+\frac1{14}+\frac1{15}-\cdots=0 $$ 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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$\tan^{-1}(1)+\tan^{-1}(2)+\tan^{-1}(3) = \pi$ (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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$$27\cdot56=2\cdot756,$$ $$277\cdot756=27\cdot7756,$$ $$2777\cdot7756=277\cdot77756,$$ and so on.

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    $\begingroup$ How does it work? $\endgroup$ – Maxwell Oct 2 '15 at 10:45
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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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$$ \begin{array}{rcrcl} \vdots & \vdots & \vdots & \vdots & \vdots \\[1mm] \int{1 \over x^{3}}\,{\rm d}x & = & -\,{1 \over 2}\,{1 \over x^{2}} & \sim & x^{\color{#ff0000}{\large\bf -2}} \\[1mm] \int{1 \over x^{2}}\,{\rm d}x & = & -\,{1 \over x} & \sim & x^{\color{#ff0000}{\large\bf -1}} \\[1mm] \int{1 \over x}\,{\rm d}x & = & \ln\left(x\right) & \sim & x^{\color{#0000ff}{\LARGE\bf 0}} \color{#0000ff}{\LARGE\quad ?} \\[1mm] \int x^{0}\,{\rm d}x & = & x^{1} & \sim & x^{\color{#ff0000}{\large\bf 1}} \\[1mm] \int x\,{\rm d}x & = & {1 \over 2}\,x^{2} & \sim & x^{\color{#ff0000}{\large\bf 2}} \\[1mm] \vdots & \vdots & \vdots & \vdots & \vdots \end{array} $$

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  • $\begingroup$ $\ln(x)$ converges slowly enough, so indeed $\ln(x) \sim 1$ might not be completely stupid :p $\endgroup$ – Thomas Sep 28 '13 at 8:31
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    $\begingroup$ 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. $\endgroup$ – Travis Bemrose Sep 28 '13 at 10:11
  • $\begingroup$ @Travis Wow, nice one! But I think you took my comment too seriously. $\endgroup$ – Thomas Sep 28 '13 at 13:13
  • $\begingroup$ The mystery, perhaps, lies in the constant of integration. All of the other integrals are evaluated from 0 to x, while the $\dfrac1x$ one is evaluated from 1 to x. (Note, by the way, that $\displaystyle\ln x=\lim_{t\to0}\frac{x^t-1}t$, which perhaps fits the pattern better.) $\endgroup$ – Akiva Weinberger Aug 28 '14 at 4:08
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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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    $\begingroup$ Remind me of this: http://xkcd.com/184/ $\endgroup$ – alex.jordan Nov 3 '13 at 17:31
  • $\begingroup$ Or maybe the "proof" is through taking the closure? $\endgroup$ – alex.jordan Dec 15 '13 at 5:57
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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.

Link to proof: Prove that $\prod_{k=1}^{n-1}\sin\frac{k \pi}{n} = \frac{n}{2^{n-1}}$

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\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}

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

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Voronoi summation formula:

$\sum \limits_{n=1}^{\infty}d(n)(\frac{x}{n})^{1/2}\{Y_1(4\pi \sqrt{nx})+\frac{2}{\pi}K_1(4\pi \sqrt{nx})\}+x \log x +(2 \gamma-1)x +\frac{1}{4}=\sum \limits _{n\leq x}'d(n)$

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$\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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  • $\begingroup$ how do you obtain $ \dfrac{six}{1} $ ? $\endgroup$ – ASB Jan 29 '15 at 13:49
  • $\begingroup$ @Tgymasb Denominator: $1/n\cdot n=1$. Numerator: $1/\textbf{n}\cdot\operatorname{si}\!\textbf{n}\, x=\operatorname{si}x$. $\endgroup$ – triple_sec Jan 29 '15 at 14:00
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I have another one, but I'm quite unwilling to post this here because it's MINE, I haven't found it anywhere, so don't steal this.

Let us take the four most important mathematical constants: The Euler number $e$, the Aurea Golden Ratio $\phi$, the Euler-Mascheroni constant $\gamma$ and finally $\pi$. Well we can see easily that

$$e\cdot\gamma\cdot\pi\cdot\phi \approx e + \gamma + \pi + \phi$$

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    $\begingroup$ I wouldn't take credit for this dude $\endgroup$ – user285523 Nov 11 '15 at 2:48

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