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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.
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?
${1\over 2} < \left\lfloor \mathrm{mod}\left(\left\lfloor {y \over 17} \right\rfloor 2^{-17 \lfloor x \rfloor - \mathrm{mod}(\lfloor y\rfloor, 17)},2\right)\right\rfloor$
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.
$\mathrm{GCD}(F_{n},F_{m}) = F_{\mathrm{GCD}(n,m)}$ where $F_n$ is the $n$th Fibonacci number.
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.
$$\sum_{n=1}^\infty \frac{1}{n^2}=\frac{\pi^2}{6}$$
was surprising to me when I saw it for the first time.
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 $\pi$.
$$10^2+11^2+12^2=13^2+14^2$$ 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.
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.
$3^3 + 4^3 + 5^3 = 6^3$.
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.
$$\frac{\Gamma\left(\frac15\right)\Gamma\left(\frac4{15}\right)}{\Gamma\left(\frac13\right)\Gamma\left(\frac2{15}\right)}=\frac{\sqrt2\,\,\sqrt[20]3}{\sqrt[6]5\,\sqrt[4]{5-\frac{7}{\sqrt{5}}+\sqrt{6-\frac{6}{\sqrt{5}}}}}=\frac{\phi \,\, \sqrt[20]3 \,\, \sqrt{\!\sqrt 3 \cdot \sqrt[4] 5-\phi^{3/2}}}{\sqrt 2 \,\, \sqrt[24] 5}$$
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,
$$N=\frac{1}{(r^{20}-228r^{15}+494r^{10}+228r^5+1)^{1/20}}$$
If $A+B+C=180^\circ$ then $$\tan(A)+\tan(B)+\tan(C)=\tan(A)\tan(B)\tan(C)$$
Here is a mathematical scherzo.
$$\left(\sum_{k=1}^n k\right)^2 = \sum_{k=1}^n k^3.$$
$\displaystyle\sum_{k=1}^{24} k^2=70^2$ is novel.
$$ {a \over b} = {c \over d} \quad\Longrightarrow\quad {a + b\over a - b} = {c + d \over c - d} $$
$$\int_0^\infty\frac1{1+x^2}\cdot\frac1{1+x^\pi}dx=\int_0^\infty\frac1{1+x^2}\cdot\frac1{1+x^e}dx$$
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:
This bit of notational juggling may cause one do double take...
$$\huge \sqrt[\sqrt{2}]{2} = \sqrt{2}^\sqrt{2}$$
By definition, the LHS is the number $x$ such that $x^\sqrt{2} = 2$. It is simple to check that the RHS also has this property.
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.
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)
7 * 14 = 100 - 2, therefore you start with 14, divide by 100 and multiply by 2. You could also use 7 * 143 = 1000 - -1 for an even simpler series.
$\endgroup$
Oct 2, 2013 at 11:35
$$\int_{0}^{1}\sin{(\pi x)}x^x(1-x)^{1-x}dx=\dfrac{\pi e}{4!}$$
$ \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 $.
$$ \int_0^1 \frac{\ln(1+t^{4+\sqrt{15}})}{1+t}dt= -\frac{\pi^2}{12}(\sqrt{15}-2)+\ln 2\cdot \ln(\sqrt{3}+\sqrt{5})+\ln\frac{1+\sqrt{5}}{2}\cdot \ln(2+\sqrt{3}) $$
For references, see http://ega-math.narod.ru/Chowla/index.htm (there is a scan of a paper of Herglotz where it is proved).
The square root of 2 is also the only real number other than 1 whose infinite tetrate is equal to its square...
$$\sqrt{2}^{\sqrt{2}^{\sqrt{2}^{.^{.^.}}}}=2.$$
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]
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} ] $
If $a,b,c,d$ are in arithmetical progression, then $$\frac{d^2-a^2}{c^2-b^2}=3.$$
$3^3 + 4^4 + 3^3 + 5^5 = 3435$
$1^1=1$ is the only other such number.