What are some surprising equations/identities that you have seen, which you would not have expected?
This could be complex numbers, trigonometric identities, combinatorial results, algebraic results, etc.
I'd request to avoid 'standard' / well-known results like $ e^{i \pi} + 1 = 0$.
Please write a single identity (or group of identities) in each answer.
I found this list of Funny identities, in which there is some overlap.
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$$\sum_{n=1}^\infty \frac{1}{n^2}=\frac{\pi^2}{6}$$
was surprising to me when I saw it for the first time.
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$$ \begin{align}\frac{\pi}{4} = 4 \arctan \frac{1}{5} - \arctan \frac{1}{239} \\\,\\\,\\ \frac{\pi}{4} = 5 \arctan \frac{1}{7} + 2 \arctan \frac{3}{79}\end{align}$$
Both can be shown easily using polar form, complex multiplication.
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$$10^2+11^2+12^2=13^2+14^2$$ I found that one stunning.
P.S. In general, for $n>0$, the sum of $n+1$ consecutive squares starting with $x_1 = 2n^2+n$ is equal to $n$ consecutive squares starting with $y_1 = x_1+(n+1)$. Hence,
$$3^2+4^2 = 5^2$$
$$10^2+11^2+12^2=13^2+14^2$$
$$21^2+22^2+23^2+24^2 = 25^2+26^2+27^2$$
and so on.
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$3^3 + 4^3 + 5^3 = 6^3$.
Also,
$1/89 = 0.01 + 0.001 + 0.0002 + 0.00003 + 0.000005 + 0.0000008 + 0.00000013 + \cdots$.
Let $S = \sum \frac{F_n} {k^n}$. Then $S + kS = 1 + \sum \frac{ F_{n} + F_{n-1} } {k^n} = 1 + \sum \frac {F_{n+1}}{k^n} = 1 + k^2S -1 - k$
In particular, for $k=10$, we get $ S = \frac{10}{89}$. Divide by 10 to get the second equation.
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Where $\varphi = \frac{1 + \sqrt{5}}{2}$ a golden ratio, $$\int_0^\infty\frac{1}{(1+x^\varphi)^\varphi}\mathrm dx = 1.$$
This follows immediately from the substitution $t=[x^{\varphi}(1+x^{\varphi})^{-1}]^{\varphi}$.
Proof (below) by filmor
$$\begin{align} \int_0^\infty\frac{1}{(1+x^\varphi)^\varphi}\mathrm dx &= \varphi^{-1}\int_0^\infty\frac{y^{\varphi^{-1} - 1}}{(1+y)^\varphi}\mathrm dy \\ &= \varphi^{-1}\int_0^\infty\frac{y^{\varphi - 2}}{(1+y)^\varphi}\mathrm dy \\ &= \varphi^{-1} B\bigl(\varphi - 1, 1\bigr) \\ &= \varphi^{-1}\frac{\Gamma(\varphi-1)\ \Gamma(1)}{\Gamma(\varphi)} \\ &= \varphi^{-1} \frac{1}{\varphi - 1} = 1 \end{align}$$
One more thing with golden ratio: by Ramanujan, $$r=\dfrac{e^{-2\pi/5}}{1 + \dfrac{e^{-2\pi}}{ 1 + \dfrac{e^{-4\pi}}{1 + \cdots}}} = \sqrt{ \sqrt{5}\varphi} - \varphi$$
and even more bizarrely (found based on the work of Vidunas), the hypergeometric function $N=\,_2F_1\big(\tfrac{19}{60},\tfrac{-1}{60},\tfrac{4}{5},1\big)$ is a deg-80 algebraic number given by,
$$N=\frac{1}{(r^{20}-228r^{15}+494r^{10}+228r^5+1)^{1/20}}$$
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$$(x_1^2+x_2^2+x_3^2+\dots+x_{16}^2)(y_1^2+y_2^2+y_3^2+\dots+y_{16}^2) = z_1^2+z_2^2+z_3^2+\dots+z_{16}^2$$
where the $z_i$ are rational functions of the $x_i, y_i$. One would have thought that $n$ square identities are only for $n = 1,2,4,8$, but non-bilinear ones in fact are for all $n = 2^m$.
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Easy geometric series but I found this one charming when I found out:
1/7 = 0,142857...
= 0,14 +
0,0028 +
0,000056 +
0,00000112 +
0,0000000224 + ... (double the value and shift it by two spaces)
A rather simple one $$2^4 = 4^2$$
You can use this one to "proof" (as a prank) that $x^y = y^x$
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This is slightly contrived, but consider a situation where you have two balls, of mass $M$ and $m$, where $M=16\times100^N\times m$ for some integer $N$. The balls are placed against a wall as shown:
We push the heavy ball towards the lighter one and the wall. The balls are assumed to collide elastically with the wall and with each other. The smaller ball bounces off the larger ball, hits the wall and bounces back. At this point there are two possible solutions: the balls collide with each other infinitely many times until the larger ball reaches the wall (assume they have no size), or the collisions from the smaller ball eventually cause the larger ball to turn around and start heading in the other direction - away from the wall.
In fact, it is the second scenario which occurs: the larger ball eventually heads away from the wall. Denote by $p(N)$ the number of collisions between the two balls before the larger one changes direction, and gaze in astonishment at the values of $p(N)$ for various $N$:
\begin{align} p(0)&=3\\ p(1)&=31\\ p(2)&=314\\ p(3)&=3141\\ p(4)&=31415\\ p(5)&=314159\\ \end{align}
and so on. $p(N)$ is the first $N+1$ digits of $\pi$!
This can be made to work in other bases in the obvious way.
See 'Playing Pool with $\pi$' by Gregory Galperin.
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Do logic answers count? I like the Drinker Paradox, which isn't really a paradox but actually a theorem of logic:
$\exists x.\ [D(x) \rightarrow \forall y.\ D(y)]$
For every bar there is a person for whom, if that person is drinking, then everyone is drinking.
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Some zeta-identies have been much surprising to me.
Let's denote the value $\zeta(s)-1$ as $\zeta_1(s)$ then $$ \small \begin{array} {} 1 \zeta_1(2) &+&1 \zeta_1(3)&+&1 \zeta_1(4)&+&1 \zeta_1(5)&+& ... &=&1\\ 1 \zeta_1(2) &+&2 \zeta_1(3)&+&3 \zeta_1(4)&+&4 \zeta_1(5)&+& ... &=&\zeta(2)\\ & &1 \zeta_1(3)&+&3 \zeta_1(4)&+&6 \zeta_1(5)&+& ... &=&\zeta(3)\\ & & & &1 \zeta_1(4)&+&4 \zeta_1(5)&+& ... &=&\zeta(4)\\ & & & & & &1 \zeta_1(5)&+& ... &=&\zeta(5)\\ ... & & & & & & & &... &= & ... \end{array} $$ There are very similar stunning alternating-series relations:
$$ \small \begin{array} {} 1 \zeta_1(2) &-&1 \zeta_1(3)&+&1 \zeta_1(4)&-&1 \zeta_1(5)&+& ... &=&1/2\\ & &2 \zeta_1(3)&-&3 \zeta_1(4)&+&4 \zeta_1(5)&-& ... &=&1/4\\ & & & &3 \zeta_1(4)&-&6 \zeta_1(5)&+& ... &=&1/8\\ & & & & & &4 \zeta_1(5)&-& ... &=&1/16\\ ... & & & & & & & &... &= & ... \end{array} $$
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This one by Ramanujan gives me the goosebumps:
$$ \frac{2\sqrt{2}}{9801} \sum_{k=0}^\infty \frac{ (4k)! (1103+26390k) }{ (k!)^4 396^{4k} } = \frac1{\pi}. $$
P.S. Just to make this more intriguing, define the fundamental unit $U_{29} = \frac{5+\sqrt{29}}{2}$ and fundamental solutions to Pell equations,
$$\big(U_{29}\big)^3=70+13\sqrt{29},\quad \text{thus}\;\;\color{blue}{70}^2-29\cdot\color{blue}{13}^2=-1$$
$$\big(U_{29}\big)^6=9801+1820\sqrt{29},\quad \text{thus}\;\;\color{blue}{9801}^2-29\cdot1820^2=1$$
$$2^6\left(\big(U_{29}\big)^6+\big(U_{29}\big)^{-6}\right)^2 =\color{blue}{396^4}$$
then we can see those integers all over the formula as,
$$\frac{2 \sqrt 2}{\color{blue}{9801}} \sum_{k=0}^\infty \frac{(4k)!}{k!^4} \frac{29\cdot\color{blue}{70\cdot13}\,k+1103}{\color{blue}{(396^4)}^k} = \frac{1}{\pi} $$
Nice, eh?
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Another one, which occured to me when I began to learn about double-sums in the context of divergent summation. I really had to chew on this, that the sum of the vertical sums can be different from the sum of the horizontal sums... And just different by the exact value of 1. So this had some appeal as another example of Where is the missing 1 in the equation? (From an older essay of mine):
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Taken from the first question I posed upon joining M.SE:
Define a function $f(\alpha, \beta)$, $\alpha \in (-1,1)$, $\beta \in (-1,1)$ as
$$ f(\alpha, \beta) = \int_0^{\infty} dx \: \frac{x^{\alpha}}{1+2 x \cos{(\pi \beta)} + x^2}$$
You can use, for example, the Residue Theorem to show that
$$ f(\alpha, \beta) = \frac{\pi \sin{\pi \alpha \beta}}{ \sin{\pi \alpha} \sin{\pi \beta}} $$
Clearly, from this latter expression, $f(\alpha, \beta) = f(\beta, \alpha)$. But from where does such a symmetric result come? The integral itself does not lend itself to predicting any such symmetry so far as I (and many others so far) can see.
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Personally I find this very interesting: $$ \lim_{n\to\infty} e^{-n}\sum_{k=0}^n \frac{n^k}{k!}=\frac{1}{2}. $$
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$$ \int_0^1 \frac{\ln(1+t^{4+\sqrt{15}})}{1+t}dt= -\frac{\pi^2}{12}(\sqrt{15}-2)+\ln 2\cdot \ln(\sqrt{3}+\sqrt{5})+\ln\frac{1+\sqrt{5}}{2}\cdot \ln(2+\sqrt{3}) $$
For references, see http://ega-math.narod.ru/Chowla/index.htm (there is a scan of a paper of Herglotz where it is proved).
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There are many involving infinite sums of number theoretic functions: $$\sum_{i = 1}^\infty \frac{\phi(i)}{i^k} = \frac{\zeta(k - 1)}{\zeta(k)}$$ $$\sum_{i = 1}^\infty \frac{\tau(i)}{i^k} = \zeta(k)^2$$ $$\sum_{i = 1}^\infty \frac{\sigma(i)}{i^k} = \zeta(k - 1)\zeta(k)$$ $$\sum_{i = 1}^\infty \frac{\mu(i)}{i^k} = \frac{1}{\zeta(k)}$$
And for some reason, the Riemann zeta function pops up in each. (I have no idea if these are well-known or not, I just thought they were very surprising, because I learned about these functions in a very discrete, number theory context, and the zeta function in, well, not that.)
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$$\cos \left(20\right) \cos \left(40\right) \cos \left(80\right) = \frac{1}{8}$$ for angles in degrees. This identity is interesting for its historical association with the teenage Richard Feynman. From Genius by James Gleick:
"He and his friends traded mathematical tidbits like baseball cards. If a boy named Morrie Jacobs told him that the cosine of 20 degrees multiplied by the cosine of 40 degrees multiplied by the cosine of 80 degrees equaled exactly one-eighth, he would remember that curiosity for the rest of his life, and he would remember that he was standing in Morrie's father's leather shop when he heard it."
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I know it's incredibly simple, but I'm always awed by $$ 2+2 = 2 \cdot 2 = 2^2 = \;^2 2. $$ Two is where addition, multiplication and exponentiation meet. And: tetration.
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The square root of 2 is also the only real number other than 1 whose infinite tetrate is equal to its square...
$$\sqrt{2}^{\sqrt{2}^{\sqrt{2}^{.^{.^.}}}}=2.$$
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This one is no less-
Let $d$ be the distance between Incenter($r$) and Circumcenter ($R$) Then-
$$R^2-d^2=2Rr$$ and this one $$\frac{1}{R-d}+ \frac{1}{R+d}=\frac{1}{r}$$
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$y(x)=\left \{ \frac{x- \frac {\left \lceil \frac {\sqrt {1+8x}-1} {2} \right \rceil \left ( 1-\left \lceil \frac {\sqrt {1+8x}-1} {2} \right \rceil \mod2 \right)} {2}} {\left \lceil \frac {\sqrt {1+8x}-1} {2} \right \rceil}\right \}$
This function has the following slopes: 1 at [0,1), 2 at [1,3) and so on
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By far my favorite identity: $\displaystyle\int_{-\infty}^{\infty} \frac{\sin \left( x\right )}{x} \mathrm{d}x = \int_{-\infty}^{\infty} \frac{\sin ^ 2\left( x\right )}{x^2} \mathrm{d}x$
The fun part about this one (for me) is that it looks absolutely false at first glance. They both evaluate to $\pi$.
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Something very exotic... and I do not know, whether this example fits the bill for this question here. But let's see.
In some sense it seems to be possible to assign equality $$ e = \tfrac 1{e^1} \tfrac 1{e^2} \tfrac 1{e^4}\tfrac 1{e^8}...$$
Initially I thought that this were an example where the rule of the closed form of the geometric series might break. The equality seems impossible because the product is even of only decreasing factors and should so be convergent and moreover converge to zero:
$$e^{-(1+2+4+8+16+...)} = \tfrac 1{e^1} \tfrac 1{e^2} \tfrac 1{e^4}... $$
so
$$ e \overset{???}{=} \tfrac 1{e^1} \tfrac 1{e^2} \tfrac 1{e^4}\tfrac 1{e^8}...$$
Well, in some circumstances in the context of divergent series we observe strange things - but here the factors are all nicely decreasing and no unexpected effect should occur.
But - that this actually holds in some sense was mentioned by Robert Israel here in MSE
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$$\sum_{k=-\infty}^\infty 2^k = 0$$ as can be shown from the regularization $\sum\limits_{k=1}^\infty 2^k=-1$. I'm wondering whether this is not actually the case for all ("sensible") two-sided regularizations, see here
Too explain this sum, note how for $|q|<1$ the geometric series $1+q+q^2+...$ converges to $\frac1{1-q}$. This works fine for the negative powers of two, i.e. $q=\frac12$ such that $\sum\limits_{k=-\infty}^02^k=\sum\limits_{k=0}^\infty\left(\frac12\right)^k=2$. Regularization now basically consists of stating "Ok, outside the convergence region (the positive powers of two in this case) just claim $1+q+q^2+...$ is still "equal" to $\frac1{1-q}$", i.e. $\sum\limits_{k=0}^\infty 2^k "=" \frac1{1-2}=-1$. Subtract the $1=2^0$ counted twice from these two sums to get above result.
We Theoretical Physicists tend to do things like this regularly, which mostly means we were too eager on swapping $\lim$s at some point before ;)
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Further working on TobiMcNamobi 's identity:
$$ \begin{array}{rcl} \dfrac{1}{7} & = & \sum\limits_{k=1}^{\infty} 2^k\times7\times10^{-2k} \\ \dfrac{1}{49} & = & \sum\limits_{k=1}^{\infty} 2^k\times10^{-2k} \\ & = & \sum\limits_{k=1}^{\infty} 2^k\times100^{-k} \\ & = & \sum\limits_{k=1}^{\infty} \left(\dfrac{2}{100}\right)^k \\ & = & \sum\limits_{k=1}^{\infty} \left(\dfrac{1}{50}\right)^k \end{array} \\ \\ \boxed{\dfrac{1}{49} = \dfrac{1}{50} + \dfrac{1}{2500} + \dfrac{1}{625000} + \cdots} $$
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$$\frac{\Gamma\left(\frac15\right)\Gamma\left(\frac4{15}\right)}{\Gamma\left(\frac13\right)\Gamma\left(\frac2{15}\right)}=\frac{\sqrt2\,\,\sqrt[20]3}{\sqrt[6]5\,\sqrt[4]{5-\frac{7}{\sqrt{5}}+\sqrt{6-\frac{6}{\sqrt{5}}}}}=\frac{\phi \,\, \sqrt[20]3 \,\, \sqrt{\!\sqrt 3 \cdot \sqrt[4] 5-\phi^{3/2}}}{\sqrt 2 \,\, \sqrt[24] 5}$$
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$$\sum_{n=1}^\infty(n\,\operatorname{arccot}n-1)=\frac12+\frac{17\pi}{24}-\ln\sqrt{e^{2\pi}-1}+\frac1{4\pi}\operatorname{Li}_2\left(e^{-2\pi}\right),$$ where $\operatorname{Li}_2$ is the dilogarithm.
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When I began my serious encounter with number theory and looked at properties of prominent combinatorical matrices I found this identity. This impressed me so much (even a bit philosophically) that I wanted to printed it on a t-shirt (but the white-on-black printing was then too expensive). The german phrase means "the exponential of the counting is the binomial"
Here is, how it looked asymptotically:
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This formula thrills me and stirs my mind as nothing could for years... $$\int^\infty_{0}\!\!e^{-3\pi x^2}\frac{\sinh(\pi x)}{\sinh(3 \pi x)}\,dx=\frac{1}{e^{2\pi/3}\cdot \sqrt3} \sum^\infty_{n=0}\frac{e^{-2n(n+1)\pi}}{(1+e^{-\pi})^{2}(1+e^{-3\pi})^{2}\cdots(1+e^{-(2n+1)\pi})^{2}}$$
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It is still strange for me $$ i^i = e^{\pi(2k-\frac{1}{2})}. $$ And so, one could say $i^i\in\mathbb R$.
Note that $i^i$ is a sequence of real numbers and actually $i^i\not\in\mathbb R$, but still $i^i\subset\mathbb R$.
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I find this identity due to Euler particularly striking (and not obvious at all): $$\prod_{n=1}^\infty (1-x^n) = \sum_{k=-\infty}^\infty (-1)^k\,x^{p(k)}$$
where the $p(k) = \dfrac{k(3k-1)}{2}$ are the generalized pentagonal numbers. This is what these numbers look like us for $1 \leq k \leq 5$,
[image created by Aldoaldoz]
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There are many fantastic equations that I've seen, but the one that definitely sticks out as top in my mind is the Atiyah-Singer Index theorem (it can be written in many ways; this is the way that I first learned it).
$$\operatorname{Ind}(D) = (-1)^n \int_M \frac{\operatorname{ch}(E)-\operatorname{ch}(F)}{\operatorname{e}(TM)} \operatorname{Td}^{hol}(TM \otimes \mathbb C)$$
Here $M$ is a $2n$-dimensional smooth compact manifold, with $E$, $F$ vector bundles over $M$, $D:\Omega^0(E) \rightarrow \Omega^0(F)$ is an elliptic differential operator, $\operatorname{Ind}(D) = \dim \ker D - \dim \operatorname{coker} D$, $\operatorname{ch}$ denotes the Chern class, $\operatorname{e}$ is the Euler class, and $\operatorname{Td}^{hol}$ is the holomorphic Todd class. Note that in this formulation the choice of the divisor depends naturally on the choice of $D$, a fact which is obscured in the notation.
It's probably the only equation I've ever seen in math which forced me to think about two entire fields in a different way. The mere fact that something like this connecting analysis and topology at such a deep level could be true is really incredible.
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${1\over 2} < \left\lfloor \mathrm{mod}\left(\left\lfloor {y \over 17} \right\rfloor 2^{-17 \lfloor x \rfloor - \mathrm{mod}(\lfloor y\rfloor, 17)},2\right)\right\rfloor$
The above is the most interesting inequality in mathematics. If you plot it so that areas satisfying the inequality are shaded, this is what you get:
This is known as Tupper's self referential formula.
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Given a polynomial $p(x)$ of degree $n$, let $a$ be the leading coefficient. Then:
$$\sum_{k=0}^n (-1)^k{n\choose k}p(x-k)=an!$$
This happens to be equivalent to:
$$p^{(n)}(x)=an!$$
where $p^{(k)}$ is the $k$th derivative of $p(x)$.
The surprising part is that the sum can actually obtain the leading coefficient without any remaining reference to the polynomial aside from the factorial of the degree.
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Something I recently saw on Abstruse Goose (although I don't recall the exact link).
$$10^2+11^2+12^2=13^2+14^2$$
Moreover, one can easily prove that this is the only sequence of five consecutive positive numbers which have this property!
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One more with continued fractions. In 2003 there was a discussion in sci.math about the continued fractions of powers of $e$ - if I recall correctly, then that of even powers are somehow folklore.
But examining the pattern to the depth we came to the following infinite continued fraction with a variable parameter:
$$ \operatorname{cfe}(x)= [1,\tfrac1x-1,1, \quad 1,\tfrac3x-1,1, \quad 1,\tfrac5x-1,1, \quad \ldots ]$$
where the pattern is easily recognizable.
Then "generalize" the continued fraction and allow irrational values for $x$. Then
$$ x = \operatorname{cfe}( \ln(x) ) \qquad \qquad x \ne 1$$ or $$ x= 1+\cfrac{1} {(\tfrac1{\ln x}-1) + \cfrac{1} {1+\cfrac{1} {1+\cfrac{1} {(\tfrac3{\ln x}-1) + \cfrac{1} {1+\cfrac{1} {1+\cfrac{1} {(\tfrac5{\ln x}-1) + \cfrac{1} {1+\cfrac{1} {...}}}} }}}}}$$
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1. $$ e^{\pi i} + 1 = 0$$
This simple equation links five fundamental mathematical constants:
Moreover, the three basic arithmetic operations occur exactly once each: addition, multiplication and exponentiation; and these are magically wound into one single relation$(=).$
The beauty lies in the fact that an irrational number, raised to the power of an imaginary number multiplied with another irrational number, exactly becomes zero when added to $1.$
As quoted by Benjamin Peirce, a noted American 19th-century philosopher,mathematician, and professor at Harvard University, "it is absolutely paradoxical; we cannot understand it, and we don't know what it means, but we have proved it, and therefore we know it must be the truth."
This identity is a special case of Euler's Formula: $$e^{ix}=\cos(x)+ i\sin(x)$$ It's almost mystical that these values are even related to one another.
2. The solution to this equation: $$1+\frac{1}{\phi}=\phi$$ Which is The golden ratio:$$\phi=\frac{1+\sqrt5}{2}=1.6180339887 . . .$$Which can turn into recurrence equation: $$\phi^{n+1}=\phi^n+\phi^{n-1}$$ Beautiful how it is also related to Fibonacci numbers: $$1 , 1 , 2 , 3 , 5 , 8 , 13 , 21 , ...$$ Where if you divide any consecutive Fibonacci numbers, in the infinite horizon will converge to, again, the golden ratio: $$\lim_{n\to\infty}\frac{F(n+1)}{F(n)}=\phi$$
3. Tupper's Self Referential Formula
When plotted with k=960939379918958884971672962127852754715004339660129306651505519271702802395266424689642842174350718121267153782770623355993237280874144307891325963941337723487857735749823926629715517173716995165232890538221612403238855866184013235585136048828693337902491454229288667081096184496091705183454067827731551705405381627380967602565625016981482083418783163849115590225610003652351370343874461848378737238198224849863465033159410054974700593138339226497249461751545728366702369745461014655997933798537483143786841806593422227898388722980000748404719
$0 \le x \le 106$ and $k \le y \le k + 17$, the resulting graph looks like this:
4. Ramanujan's golden ratio equation:
$$\int_{-\infty}^\infty \! e^{-x^2}dx = \sqrt{\pi}$$
6. Cauchy's Integral Formula: $${f^{\left( n \right)}}\left( a \right) = \frac{{n!}}{{2\pi i}}\oint_\gamma {\frac{{f\left( z \right)}}{{{{\left( {z - a} \right)}^{n + 1}}}}dz}$$ The derivative of a analytic function given as a closed path integral in the complex plane.
7. Ramanujan's Infinite series for calculation of $\pi$. It converges faster
$$ \frac{1}{\pi} = \frac{2\sqrt{2}}{9801} \sum^\infty_{k=0} \frac{(4k)!(1103+26390k)}{(k!)^4 396^{4k}}$$
8. Batman Curve
The batman curve is a piecewise curve in the shape of the logo of the Batman superhero originally posted on reddit.com on Jul. 28, 2011. It can written as two functions, one for the upper part and the other for the lower part, as:
9. The Schrodinger Equation:
$$H\Psi(x,t) = i\hbar\frac{\partial}{\partial t}\Psi(x,t)$$
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Using the mystical ennead to calculate the decimal expressions for fractions of 7. Start with this figure:
Then follow the connected path, giving the sequence 1 4 2 8 5 7. Then you write this sequence starting on each digit, in order, giving
. 1 4 2 8 5 7 = 1/7
. 2 8 5 7 1 4 = 2/7
. 4 2 8 5 7 1 = 3/7
. 5 7 1 4 2 8 = 4/7
. 7 1 4 2 8 5 = 5/7
. 8 5 7 1 4 2 = 6/7
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$$ {a \over b} = {c \over d} \quad\Longrightarrow\quad {a + b\over a - b} = {c + d \over c - d} $$
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Ramanujan stated this radical in his lost notebook:
$$\sqrt{5+\sqrt{5+\sqrt{5-\sqrt{5+\sqrt{5+\sqrt{5+\sqrt{5-\dots}}}}}}} = \frac{2+\sqrt 5 +\sqrt{15-6\sqrt 5}}{2}$$
I still don't have any idea on this one.
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$$
2^{67}-1 = 193,707,721 × 761,838,257,287
$$
This identity was found by Cole in the early 20th century. He later said that the result had taken him "three years of Sundays" to find.
There's also the fact that:
$2^{127} -1 $ is indeed prime, as Mersenne claimed. This was the largest known prime number for 75 years, and the largest ever calculated by hand. Édouard Lucas proved its primality in 1876.
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Let $p_n$ be the probability that a random permutation in the symmetric group $S_n$ doesn't have fixed points. Then $\lim_{n\to\infty}p_n=\frac{1}{e}$.
I was amazed the first time I saw this exercise!
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The infinite Fibonacci sequence $F = (0, 1, 1, 2, 3, 5, 8, \dots)$ and the infinite Fibonacci string $S = 1011010110110 \dots$ are related by the following remarkable identity for all real or complex $\beta$ such that $\ |\beta| > 1$:
$$[0 \ ; \ \beta^{F_0}, \beta^{F_1}, \beta^{F_2}, \dots] = (\beta - 1) \cdot (0.S)_\beta$$
where $[0 \ ; \ \beta^{F_0}, \beta^{F_1}, \beta^{F_2}, \dots]$ denotes the continued fraction
$$\frac{1}{\beta^{F_0} + \frac{1}{\beta^{F_1} + \frac{1}{\beta^{F_2} + \cdots}}} $$
and $(0.S)_\beta = (0.1011010110110 \dots)_\beta$ denotes the number obtained by reading $0.S$ as a "base-$\beta$ numeral"; that is, $(0.S)_\beta$ denotes the sum of the infinite series
$$S[1] \beta^{-1} + S[2]\beta^{-2} + S[3]\beta^{-3} + \cdots$$
where $S[n]$ is the $n$th element of string $S$.
E.g., the so-called rabbit constant $(0.S)_2$ = 0.709803... in decimal , $(0.S)_\pi$ = 0.362011... in decimal, etc.
($F$ and $S$ are also related by the fact that both are generated by recursions of the form $x_{n+1} = x_{n} + x_{n-1} ;\ x_0 = 0, \ x_1 = 1$, where the $+$ is interpreted in one case as arithmetic addition, and in the other case as string concatenation.)
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I think this is simple but I want to post it:
$$1 \times 9=9\implies(0+9=9)$$ $$2\times 9=18\implies(1+8=9)$$ $$3\times 9=27\implies(2+7=9)$$ $$4\times 9=36\implies(3+6=9)$$ $$5\times 9=45\implies(4+5=9)$$ $$6\times 9=54\implies(5+4=9)$$ $$7\times 9=63\implies(6+3=9)$$ $$8\times 9=72\implies(7+2=9)$$ $$9\times 9=81\implies(8+1=9)$$ $$10\times 9=90\implies(9+0=9)$$ and also no. are in a pattern $09,18,27,36,45\;$then reverse the no.$54,63,72,81,90$
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This is the most surprising result that I am the discoverer of.
Consider the diophantine equation $$x(x+1)...(x+n-1) -y^n = k$$
where $x, y, n,$ and $k$ are integers, $x \ge 1$, $y \ge 1$, and $n \ge 3$.
I was led to consider considering this by trying to generalize the Erdos-Selfridge result that the product of consecutive integers could never be a power.
I phrased this as "How close and how often can the product of $n$ consecutive integers be to an $n$-th power?"
Looking at this equation, it seemed reasonable to think that, for fixed $k$ and $n$, there were only a finite number of $x$ and $y$ that satisfied it. This was not too hard to prove.
What greatly surprised me was that I was able to prove that for any fixed $k$, there were only a finite number of $n$, $x$, and $y$ that satisfied it.
The proof went like this:
I first showed that any solution must have $y \le |k|$. This was moderately straightforward, and involved considering the three cases $y < x$, $x \le y \le x+n-1$, and $y \ge x+n$.
The next step really surprised me. I showed that $n < e|k|$, where $e$ is the good old base of natural logarithms.
The proof was amazingly (to me) simple. Since $y \le |k|$ and $2(n/e)^n < n!$,
$\begin{align} 2(n/e)^n &< n!\\ &\le x(x+1)...(x+n-1)\\ &= y^n+k\\ &\le |k|^n+|k|\\ &\le |k|^n+|k|^n\\ &= 2|k|^n\\ \end{align} $
so $n < e |k|$.
I still remember staring at this in disbelief, over forty years later.
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Tetration :
consider the tower of taking infinite powers : $x^{x^{x^{x^{x^{x^{x^{.{^{.^{.}}}}}}}}}}$ .
At first its seems big mystery and undefined one for lots of real numbers.
Surprising fact is its indeed converges in an closed interval which is bounded by the fancy real numbers $e^{-e}$, $e^\frac{1}{e}$
So $x^{x^{x^{x^{x^{x^{x^{.{^{.^{.}}}}}}}}}}$ converges for $ x \in [e^{-e}, e^\frac{1}{e} ] $
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The continued fraction of The Golden Ratio:
$\frac{1+\sqrt5}{2}=[1;,1,1,1,\dots]=[1,\bar1]$
Also: $\frac{1+\sqrt5}{2}=\sqrt{1+\sqrt{1+\sqrt{1+\dots}}}$.
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$$ {\large\sqrt{\vphantom{\Large A}\,\color{#ff0000}{20}\color{#0000ff}{25}\,}\, = 45 = \color{#ff0000}{20} + \color{#0000ff}{25}} $$
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I believe that one that should be mentioned is the prime number theorem:
Let $\pi(x)$ be the number of primes not exceeding $x$. Then:
$\pi(x)\sim \frac{x}{logx}$
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A surprising family of series for $e$ can be derived by algebraically combining the terms in Newton's series expansion:
\begin{equation} e=\sum_{k=0}^{\infty } \dfrac{1}{k!}=\dfrac{1}{0!}+\dfrac{1}{1!}+\dfrac{1}{2!}+\dfrac{1}{3!}+\dfrac{1}{4!}+\dfrac{1}{5!}+\ldots. \end{equation}
Here are a few examples:
\begin{equation} e=\sum _{k=0}^{\infty } \frac{2k+1}{(2k)!}=\frac{1}{0!}+\frac{3}{2!}+\frac{5}{4!}+\frac{7}{6!}+\frac{9}{8!}+\frac{11}{10!}+\ldots \end{equation}
\begin{equation} 2e=\sum _{k=0}^{\infty } \frac{k+1}{k!}=\frac{1}{0!}+\frac{2}{1!}+\frac{3}{2!}+\frac{4}{3!}+\frac{5}{4!}+\frac{6}{5!}+\ldots \end{equation}
\begin{equation} 1/e=\sum _{k=0}^{\infty } \frac{1-2k}{(2k)!}=\frac{1}{0!}-\frac{1}{2!}-\frac{3}{4!}-\frac{5}{6!}-\frac{7}{8!}-\frac{9}{10!}-\ldots~. \end{equation}
Beyond being pretty, these series converge substantially faster than Newton's series. For more formulas and details on derivation see: http://www.brotherstechnology.com/math/cmj-supp.html
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It's funny noone mentioned the hockey-stick identity, partial sum of columns in a Pascal triangle:
$$ \sum_{k=0}^{m}\binom{n+k}{k}=\binom{n+m+1}{n} $$
http://www.artofproblemsolving.com/Wiki/index.php/Combinatorial_identity
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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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I have always been fascinated by Leibniz's formula for $\pi$: $$\dfrac{\pi}{4}=\sum\limits_{n=0}^\infty \dfrac{1}{2n+1}\times(-1)^{n}=1-\dfrac{1}{3}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{9}-\dfrac{1}{11}\dots$$ This can be used to determine the exact value of $\pi$, which is what makes it interesting. $$\displaystyle \boxed{\pi=4\sum\limits_{n=0}^\infty \dfrac{1}{2n+1}\times(-1)^{n}=4-\dfrac{4}{3}+\dfrac{4}{5}-\dfrac{4}{7}+\dfrac{4}{9}-\dfrac{4}{11}\dots}$$
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I apologize if this is already here. I thought it was but I can't find it, so I must have seen it somewhere else on the site.
$$\begin{matrix} f(x) & \displaystyle\int f(x)dx \\[6pt] \hline x^2 & \dfrac{x^3}{3} \\[6pt] x & \dfrac{x^2}{2} \\[6pt] 1 & x \\[6pt] \dfrac{1}{x} & \color{red}{\log(x)} \\[6pt] \dfrac{1}{x^2} & -\dfrac{1}{x} \\[6pt] \dfrac{1}{x^3} & -\dfrac{1}{2x^2} \\[6pt] \dfrac{1}{x^4} & -\dfrac{1}{3x^3} \end{matrix}$$
$\displaystyle{\int x^n dx} = \frac{x^{n+1}}{n+1}$
$\displaystyle{\lim_{n \rightarrow -1} \left(\frac{x^{n+1}}{n+1}\right)} = \log(x)$?
No. Let $g(x,n)=\frac{x^{n+1}}{n+1}$.
For $x\in(0,\infty)$, you have:
$g(x,n)>0$ as $n\rightarrow -1$ from above, and
$g(x,n)<0$ as $n\rightarrow -1$ from below, but
$\log(x)<0$ on $x\in(0,1)$ and $\log(x)>0$ on $x\in(1,\infty)$.
I wish I understood this better.
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I was really amazed by discovering that the squared arcsine function has a pretty nice Taylor series at the origin: $$\arcsin^2(z)=\frac{1}{2}\sum_{n\geq 0}\frac{(2z)^{2n}}{n^2\binom{2n}{n}}$$ Even more amazed by the variety of techniques one may employ to prove such identity: combinatorial convolutions, hypergeometric transformations, (poly)logarithmic integrals, the Lagrange inversion theorem, the residue theorem, Legendre polynomials, Euler's Beta function, creative telescoping... They're all pretty interesting.
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$$\sin \pi x=\pi x\prod_{n=1}^{\infty}\left(1-\frac {x^2}{n^2}\right).\quad \text {(L. Euler).}$$ Obviously the LHS and RHS have the same set of zeroes but that alone does not imply equality. And putting $x=1/2$ into it, we derive the Wallis product for $\pi, $ which itself is remarkable, especially as Wallis was Newton's immediate predecessor in the "Lucas chair" and obtained his product without the full generality of the methods of calculus developed by Newton..
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Prime solutions for the Prouhet-Tarry-Escott problem of size ten ($10$ primes on the left side, other $10$ primes on the right side) .
$$ 2589701^k + 2972741^k + 6579701^k + 9388661^k + 9420581^k + 15740741^k + 15772661^k + 18581621^k + 22188581^k + 22571621^k $$ $$ = 2749301^k + 2781221^k + 6835061^k + 8399141^k + 10314341^k + 14846981^k + 16762181^k + 18326261^k + 22380101^k + 22412021^k $$
( Prime solution, $ k = 1, 2, 3, 4, 5, 6, 7, 8, 9 $)
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I believe McShane's identity is very surprising, and is pretty different from the rest of these answers.
Let $\Sigma_{1,1}$ be a once punctured torus with a complete, finite volume, hyperbolic metric $g$. Let $C$ be the set of closed geodesic curves which do not intersect themselves (simple closed curves). Then we have the following identity:
$$ \sum_{c \in C} \frac{1}{1+e^{\mathrm{length}(c)}}=\frac{1}{2}. $$
This is extremely interesting because there are many such metrics up to isometry. In fact the space of such metrics is naturally isomorphic to the modular curve $\mathbb H^2/ \mathrm{PSL}(2,\mathbb{Z})$. In general the set of lengths of simple closed curves will be very different for different metrics. For example, for any $\epsilon >0$ you can find a metric with a $c \in C$ of that length. When $\epsilon$ is small the corresponding term is very close to $1/2$. There are also metrics where every curve in $C$ has length greater than 1.
This identity, and generalizations of this, end up being related to important work by Maryam Mirzakhani, who won the Fields medal.
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This expression is from Francois Viète: $$\pi = \frac{2}{1} \times \frac{2}{\sqrt2} \times \frac{2}{\sqrt{2+\sqrt2}}\times \frac {2}{\sqrt{2+\sqrt{2+\sqrt2}}} \times \cdots$$
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•I will love to add this nested radical from Ramanujan: $$\sqrt{1+2\sqrt{1+3\sqrt{1+4\sqrt{\cdots}}}}=3$$ •Also like to add this too:$$1+2+3+4+\cdots=-\frac{1}{12}$$ •This famous Stirling'sapproximation$$n!\approx\sqrt{2\pi n}{\left(\frac{n}{e}\right)}^n$$
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I was amazed at one of the results that Ramanujan sent to Hardy in his first letter.
$$ (1.10) \text{ If } u = \frac{x}{1+} \frac{x^5}{1+} \frac{x^{10}}{1+} \frac{x^{15}}{1+\dots},\, v = \frac{x^{1/5}}{1+} \frac{x}{1+}\frac{x^{2}}{1+}\frac{x^{3}}{1+\dots},\\ \text{then}\qquad\qquad v^5 = u\frac{1-2u+4u^2-3u^3+u^4} {1+3u+4u^2+2u^3+u^4} $$
This is just one of the modular equations satisfied by the Rogers-Ramanujan continued fraction. The Wikipedia article mentions other amazing facts about the continued fraction but this result stands out for me.
$ \tan 10^\circ = \tan 20^\circ \times \tan 30^\circ \times \tan 40^\circ $.
$\tan 80^\circ = \tan 70^\circ \times \tan 60^\circ \times \tan 50^\circ $.