# Surprising identities / equations

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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I really can't believe no one has posted this yet: xkcd.com/687 –  mikeTheLiar Sep 26 '13 at 14:48
This is not in line with what you are looking for, but as a child I discovered that 10million pi is the number of seconds in a year to 1/2% accuracy. This is useful for quick back of envelope calculations, where seconds are involved. –  JoeTaxpayer Sep 26 '13 at 19:48
Pi seconds is a nanocentury! –  Oscar Cunningham Oct 1 '13 at 21:59

$$\sum_{n=1}^\infty(n\,\operatorname{arccot}n-1)=\frac12+\frac{17\,\pi}{24}-\ln\sqrt{e^{2\pi}-1}+\frac1{4\pi}\operatorname{Li}_2\frac1{e^{2\pi}},$$ where $\operatorname{Li}_2$ is the dilogarithm.

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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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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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There are not other integer solutions (except for $x=y$). –  pts Sep 26 '13 at 11:02
@pts There are infinitely many rational solutions, though, with $(e,e)$ as a limit point. –  A Walker Sep 27 '13 at 5:25

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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$$\cos \left(20\right) \cos \left(40\right) \cos \left(80\right) = \frac{1}{8}$$ 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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Hint: Generalisation: $\cos(60-x)\cos(60+x)\cos(x)=\cos(3x)/4$. It gets even more interesting with tangents: $\tan(60-x)\tan(60+x)\tan(x)=\tan(3x)$ –  chubakueno Sep 29 '13 at 18:51

$$\ \ \ \ \ 2592=2^59^2\ \ \ \ \$$

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It is still strange for me $$i^i = e^{\pi(2k-\frac{1}{2})}.$$ And so, one could say $i^i\in\mathbb R$.

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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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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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That extends beyond exponentiation, as well, in the sense that 2 op 2 has the same value for every binary operation op in Goodstein's infinite sequence of hyperoperations (+, *, ↑, ↑↑, ↑↑↑, ...). –  r.e.s. Sep 27 '13 at 4:23
@r.e.s. Can you explain that? Isn't 2 ↑↑ 2 = 16 ? –  MrZander Sep 28 '13 at 0:04
@MrZander For any op beyond addition, in the expression x op y the y specifies how many x's are to be "combined" using the hyperoperator at the next lower level. E.g., 2↑↑4 = 2↑2↑2↑2 = 2↑2↑4 = 2↑16 = 65536, 2↑↑3 = 2↑2↑2 = 2↑4 = 16, 2↑↑2 = 2↑2 = 4. –  r.e.s. Sep 28 '13 at 3:57

1. $$e^{\pi i} + 1 = 0$$

This simple equation links five fundamental mathematical constants:

• The number 0, the additive identity.
• The number 1, the multiplicative identity.
• The irrational number π (pi), pivotal in trigonometry and geometry.
• The transcendental constant e, the base of the natural logarithm, widely used in scientific analysis.
• The number i (iota), the imaginary unit of complex numbers, and the square root of -1.

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: $$e^{ix}=cosx+ i sinx$$ 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: $$\lim_{n\to\infty}\frac{F(n+1)}{F(n)}=\phi$$

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

$$\frac{1}{\pi} = \frac{2\sqrt{2}}{9801} \sum^\infty_{k=0} \frac{(4k)!(1103+26390k)}{(k!)^4 396^{4k}}$$

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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Thanks! This is a good list :) –  Calvin Lin Oct 2 '13 at 4:00

$$(1+2+3+\cdots+n)!=1!3!5!\cdots(2n-1)!$$for $n=0,1,2,3,4$.

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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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This one is one of my favorite:

# $$\log(1+2+3) = \log(1)+\log(2)+\log(3)$$

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$\log(1+1+2+4)=\log(1)+\log(1)+\log(2)+\log(4)$ The general case is analyzed in www-users.mat.umk.pl/~anow/ps-dvi/si-krl-a.pdf –  Mark S. Dec 13 '13 at 3:23

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

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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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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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Each of the Dirichlet series at left has an Euler product. Since the zeta function does as well, we can verify these identities by looking at the factors associated to each prime. Since the factors coming from the zeta function are so simple, it's not too surprising that they show up other places. Or, you prove each of these identities by using Dirichlet convolution: $\phi * 1 = N$, $\tau = 1 * 1$, $\sigma = 1* N$, and $\mu *1 = \delta_1$ (the indicator function of $1$). –  A Walker Sep 27 '13 at 5:38

$$\sum_{i=0}^N {{N}\choose{i}}=2^N$$

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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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How can the sum from 1 to infinity of positive integers lead to a negative number? –  Mew Sep 28 '13 at 2:27
@Chris Sorry for the brevity, I expanded a bit on this –  Tobias Kienzler Sep 28 '13 at 7:22
@Chris You need to review the "allowable" operations on summations. SumToInfinity(F(X)) is not the same as F(0) + F(1) + F(2) + ..., otherwise you'd have paradoxes even in convergent sums due to the fact that you can arbitrary choose the order of which terms to sum. –  DanielV Sep 28 '13 at 14:11
The Casimir effect needs a trick like that. –  Felix Marin Sep 29 '13 at 9:52

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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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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+1 Interesting. BTW, didn't you mean to write "the Erdos-Selfridge result"? –  r.e.s. Oct 3 '13 at 3:23

from $1$ years ago

$$\tan x=\cfrac{x}{1-\cfrac{x^2}{3-\cfrac{x^2}{5-\cfrac{x^2}{7-...}}}}$$

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If $x^n + y^n + z^n=0$ and $xyz \ne 0$, then $$\frac{(x^n-y^n)^2}{(xy)^n} + \frac{(y^n-z^n)^2}{(yz)^n} + \frac{(z^n-x^n)^2}{(zx)^n} = -9.$$

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A related — and similarly surprising — identity is $$\biggl(\frac{x^2}{yz}\biggr)^{\!n} + \biggl(\frac{y^2}{xz}\biggr)^{\!n} + \biggl(\frac{z^2}{xy}\biggr)^{\!n} = 3.$$ –  Kieren MacMillan Jun 26 at 22:56

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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Thanks. Those do seem strange, till you see how it is set up. –  Calvin Lin Sep 27 '13 at 13:11

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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@rbm: Simplify and solve the polynomial equation: $$(x-1)^2+x^2+(x+1)^2=(x+2)^2+(x+3)^2$$ –  Asaf Karagila Sep 29 '13 at 9:04

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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$$2^{67}-1 = 193,707,721 × 761,838,257,287$$ 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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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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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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If $a,b,c,d$ are in arithmetical progression, then $$\frac{d^2-a^2}{c^2-b^2}=3.$$

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## protected by Zev ChonolesSep 27 '13 at 7:46

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