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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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closed as primarily opinion-based by Najib Idrissi, Mike Miller, 1999, apnorton, user2345215 Jan 31 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.

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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
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Perhaps this should be a community wiki question. –  Nuno Nov 3 '10 at 22:31
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This is related. –  Guess who it is. Nov 3 '10 at 22:35
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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
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@SteveD If only we could find an odd example... –  peoplepower Jan 13 '13 at 0:31

66 Answers 66

up vote 159 down vote accepted

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

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I had to do something about my accept range :) –  AD. May 17 '12 at 4:47
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Sophomore's Dream? –  rotskoff Jun 19 '12 at 20:50
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$$\int_0^1 {x^x}\mathrm{d}x=\sum_{k=1}^\infty\frac{(-1)^{k-1}}{k^k}$$ –  LTS Nov 2 '13 at 18:29

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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An awesome pattern.

    1 x 1 = 1
   11 x 11 = 121
  111 x 111 = 12321
 1111 x 1111 = 1234321
11111 x 11111 = 123454321
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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
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"I just add those 2 together so that I can get partial credit." –  Derek 朕會功夫 Aug 11 '14 at 5:33

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

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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? –  F M Nov 6 '10 at 5:42
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@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
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@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. –  Guess who it is. Nov 16 '10 at 7:05
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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
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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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$$ \frac{\pi}{2}=1+2\sum_{k=1}^{\infty}\frac{\eta(2k)}{2^{2k}} $$ $$ \frac{\pi}{3}=1+2\sum_{k=1}^{\infty}\frac{\eta(2k)}{6^{2k}} $$ 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.

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

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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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See also the comments here: math.stackexchange.com/q/8385/1242 –  Hans Lundmark Nov 4 '10 at 8:53
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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
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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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$$ \begin{array}{rcrcl} \vdots & \vdots & \vdots & \vdots & \vdots \\[1mm] \int{1 \over x^{3}}\,{\rm d}x & = & -\,{1 \over 2}\,{1 \over x^{2}} & \sim & x^{\color{#ff0000}{\large\bf -2}} \\[1mm] \int{1 \over x^{2}}\,{\rm d}x & = & -\,{1 \over x} & \sim & x^{\color{#ff0000}{\large\bf -1}} \\[1mm] \int{1 \over x}\,{\rm d}x & = & \ln\left(x\right) & \sim & x^{\color{#0000ff}{\LARGE\bf 0}} \color{#0000ff}{\LARGE\quad ?} \\[1mm] \int x^{0}\,{\rm d}x & = & x^{1} & \sim & x^{\color{#ff0000}{\large\bf 1}} \\[1mm] \int x\,{\rm d}x & = & {1 \over 2}\,x^{2} & \sim & x^{\color{#ff0000}{\large\bf 2}} \\[1mm] \vdots & \vdots & \vdots & \vdots & \vdots \end{array} $$

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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

$$ \int_{-\infty}^{\infty}{\sin\left(x\right) \over x}\,{\rm d}x = \pi\int_{-1}^{1}\delta\left(k\right)\,{\rm d}k $$

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