284
$\begingroup$

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.

$\endgroup$
  • 61
    $\begingroup$ I really can't believe no one has posted this yet: xkcd.com/687 $\endgroup$ – MikeTheLiar Sep 26 '13 at 14:48
  • 23
    $\begingroup$ 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. $\endgroup$ – JoeTaxpayer Sep 26 '13 at 19:48
  • 12
    $\begingroup$ Pi seconds is a nanocentury! $\endgroup$ – Oscar Cunningham Oct 1 '13 at 21:59
  • $\begingroup$ @CalvinLin Is that until you get the most rare badge? I got the 81st favorite too! $\endgroup$ – zerosofthezeta Oct 2 '13 at 4:32
  • 1
    $\begingroup$ The three trigonometric identities in the following exercises of my Wikibook: en.wikibooks.org/wiki/On_2D_Inverse_Problems/… $\endgroup$ – DVD Oct 12 '15 at 2:51

101 Answers 101

0
$\begingroup$

Further working on TobiMcNamobi 's identity:

$$ \begin{array}{rcl} \dfrac{1}{7} & = & \sum\limits_{k=1}^{\infty} 2^k\times7\times10^{-2k} \\ \dfrac{1}{49} & = & \sum\limits_{k=1}^{\infty} 2^k\times10^{-2k} \\ & = & \sum\limits_{k=1}^{\infty} 2^k\times100^{-k} \\ & = & \sum\limits_{k=1}^{\infty} \left(\dfrac{2}{100}\right)^k \\ & = & \sum\limits_{k=1}^{\infty} \left(\dfrac{1}{50}\right)^k \end{array} \\ \\ \boxed{\dfrac{1}{49} = \dfrac{1}{50} + \dfrac{1}{2500} + \dfrac{1}{625000} + \cdots} $$

$\endgroup$
  • 8
    $\begingroup$ Well, that's just the GP formula. $\endgroup$ – Calvin Lin Sep 28 '13 at 4:45
0
$\begingroup$

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.$54,63,72,81,90$

$\endgroup$
  • 2
    $\begingroup$ I used this to help me learn my $9$ times tables back in Grade 2 $\endgroup$ – TrueDefault Feb 28 '14 at 2:28
0
$\begingroup$

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:
$\pi(x)\sim \frac{x}{logx}$

$\endgroup$
0
$\begingroup$

Wheatstone's identity, which shows how powers can be constructed via arithmetic progressions: $$n^a = \sum_{k=0}^{t-1}\biggl(\frac{n^a}{t}-\frac{\delta(t-1)}{2}+k\delta\biggr).$$ Put into words: An arithmetic progression of $t$ terms with constant difference $\delta$ and first term $\tfrac{1}{t}n^a-\tfrac{1}{2}\delta(t-1)$ will sum to $n^a$.

$\endgroup$
0
$\begingroup$

Möbius inversion formula may be an example. Also Ramanujan's Partition Congruences were surprising to me when I first saw them.

$\endgroup$
0
$\begingroup$

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 $)

$\endgroup$
0
$\begingroup$

The formulas obtained by Robert Scheider and involving The Golden Ratio $\phi$, the Euler Totient $\varphi (n)$ and the Moebius Function $\mu(n)$, are rather surprising $$ \begin{array}{*{20}c} {\phi = - \sum\limits_{1\, \le \,k} {\frac{{\varphi (k)}}{k}\ln \left( {1 - \frac{1}{{\phi ^{\,k} }}} \right)} } & {e^{\,\phi } = \prod\limits_{1\, \le \,k} {\left( {1 - \frac{1}{{\phi ^{\,k} }}} \right)^{\, - \;\frac{{\varphi (k)}}{k}} } } \\ {\frac{1}{\phi } = - \sum\limits_{1\, \le \,k} {\frac{{\mu (k)}}{k}\ln \left( {1 - \frac{1}{{\phi ^{\,k} }}} \right)} } & \begin{array}{l} e^{\,{{1\,} \mathord{\left/ {\vphantom {{1\,} {\,\phi }}} \right. } {\,\phi }}} = \prod\limits_{1\, \le \,k} {\left( {1 - \frac{1}{{\phi ^{\,k} }}} \right)^{\, - \;\frac{{\mu (k)}}{k}} } = \\ = \frac{1}{e}\prod\limits_{1\, \le \,k} {\left( {1 - \frac{1}{{\phi ^{\,k} }}} \right)^{\, - \;\frac{{\varphi (k)}}{k}} } \\ \end{array} \\ {1 = \sum\limits_{1\, \le \,k} {\frac{{\mu (k) - \varphi (k)}}{k}\ln \left( {1 - \frac{1}{{\phi ^{\,k} }}} \right)} } & {e = \prod\limits_{1\, \le \,k} {\left( {1 - \frac{1}{{\phi ^{\,k} }}} \right)^{\,\frac{{\mu (k) - \varphi (k)}}{k}} } } \\ \end{array} $$

Also refer to this Wikipedia article.

$\endgroup$
0
$\begingroup$

$$\prod_{n=1}^{\infty}\frac1{4en}\bigg(\frac{(16n^2-9)^3}{16n^2-1}\bigg)^{1/4}\bigg(\frac{(4n+3)(4n-1)}{(4n-3)(4n+1)}\bigg)^n=\sqrt{\frac{2}{3\pi\sqrt{3}}}\exp\bigg(\frac{G}{\pi}+\frac12\bigg)$$ Where $G$ is Catalan's constant.

$\endgroup$
-2
$\begingroup$

Not sure if it's an equation but.. I was very surprised that among all two-dimensional compact orientable surfaces, a 2-sphere $S^2$ is the only one that has nontrivial higher homotopy groups. So, from the point of view of homotopy, sphere is the "most complicated" surface in some sense.

$\endgroup$
  • $\begingroup$ Isn't this just because you measure homotopy by using the embeddings of spheres into your space? $\endgroup$ – Oscar Cunningham Mar 1 '16 at 9:33
  • $\begingroup$ @OscarCunningham Unfortunately, I don't understand what you mean by "measuring homotopies by embeddings...". $\endgroup$ – Peter Franek Sep 15 '16 at 8:45
  • $\begingroup$ I mean that the second homotopy group of a space $X$ is defined as the set of injections $S^2\to X$, along with some multiplication operation. So it's not so surprising that $S^2$ has an interesting second homotopy group, since $S^2$ is itself involved in the definition. $\endgroup$ – Oscar Cunningham Sep 15 '16 at 8:49
  • $\begingroup$ @OscarCunningham No, homotopy groups are not defined by injections and/or embeddings, but rather by continuous maps (up to homotopy). Also $\pi_2$ is nontrivial for the projective space but I don't think you can find an embedding. And of course, I talked about higher homotopy groups as well, not just the second! Don't you find it surprising that it's so enormly complex (and still open) for the sphere but they are all completely trivial for a sphere with handles? $\endgroup$ – Peter Franek Sep 15 '16 at 8:54
  • $\begingroup$ Oh you're right (I've been thinking about knot theory recently, where you do only care about injections). It's not so surprising to me that adding handles to a sphere stops there from being any non-trivial function from a sphere. What is surprising to me is that this doesn't happen in higher dimensions! $\endgroup$ – Oscar Cunningham Sep 15 '16 at 9:19
-6
$\begingroup$

$e=2.71\ldots$
$\pi=3.141592\dots$
$i^2=-1$
then
It is amazing that
$e^{i\pi}+1=0$

$\endgroup$
-7
$\begingroup$

Jean Nicod's first single axiom DDpDqrDDtDttDDsqDDpsDps, where D stands for the Sheffer stroke ("NAND"), and we write in Polish notation. Under the rule {D$\alpha$D$\beta$$\gamma$, $\alpha$} $\vdash$ $\gamma$, all tautologies (containing just "D" and variables) in classical logic can get derived.

CCCpqrCCrpCsp. This is Lukasiewicz's 13 letter axiom which suffices as the only axiom, under detachment, to prove all formulas of the classical pure implicational calculus of propositions.

Meredith's 21 letter axiom CCCCCpqCNrNsrtCCtpCsp which suffices as the only axiom for C-N classical logic under detachment (it's not known if this is the shortest such axiom).

(x(y(((zz')(uy)')x))') = u. This is a product and inverse single axiom for group theory. There are no shorter product and inverse single axioms for group theory.

((x|z)|y)|((x|(x|y))|x)=y, (x|((y|x)|x))|(y|(z|x))=y, ((x|y)|z)|(x|((x|z)|x))=z, and ((x|(z|x))|x)|(z|(y|x))=z are among the shortest possible single axiom for Boolean algebra in the Sheffer stroke.

XCB: EpEEEpqErqr was the last proved shortest single axiom and completes the set of shortest single axioms for equivalential calculus under the rule {E$\alpha$$\beta$\, $\alpha$ $\vdash$ $\beta$}. The proof involved a computer program.

C$\delta$pC$\delta$Np$\delta$q, where $\delta$ comes as a functional variable of one argument, consists of a single axiom which under uniform substitution and detachment comes as sufficient to deduce all C-N-$\delta$ tautologies.

C$\delta$CpqCp$\delta$q suffices as a single axiom to prove a deduction theorem for a propositional calculus.

C$\delta$CpqC$\delta$p$\delta$q, from what I have read though I have not seen a proof, suffices as a single axiom for all C-$\delta$ propositions.

C$\delta$$\delta$0$\delta$p. This 6 letter single axiom under detachment, substitution, and the rules for quantifiers, I have read, suffices for all theorems or axioms of classical propositional logic with functional variables, quantifiers of propositions, and quantifiers of functional variables.

$\endgroup$
  • 1
    $\begingroup$ In what sense these are identities? Why do you find them surprising? $\endgroup$ – Moishe Kohan May 24 '14 at 2:24
  • $\begingroup$ @studiosus The single axioms for various propositional calculi here all evaluate to the designated value for any valuation of their variables. In that sense, they refer to identities (though when working with a propositional calculus, you don't usually use the identity CCCpqrCCrpCsp=1, where "1" indicates the designated value, because you would then make a reference to meaning). I find them surprising in that, in a sense, we can compress what is needed for various propositional calculi into that small of a space (with the rules implicit). Also, the group/Boolean axioms are identities. $\endgroup$ – Doug Spoonwood May 24 '14 at 3:04

protected by Zev Chonoles Sep 27 '13 at 7:46

Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).

Would you like to answer one of these unanswered questions instead?

Not the answer you're looking for? Browse other questions tagged or ask your own question.