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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.

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    $\begingroup$ I really can't believe no one has posted this yet: xkcd.com/687 $\endgroup$ – MikeTheLiar Sep 26 '13 at 14:48
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    $\begingroup$ 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
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    $\begingroup$ 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
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    $\begingroup$ 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

100 Answers 100

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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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  • $\begingroup$ What are the patterns of the coefficients? Clearly all $1$ for the first line and natural numbers for the second, but with so few terms I can't decide on the next three lines $\endgroup$ – Brevan Ellefsen Mar 4 '17 at 21:23
  • $\begingroup$ @Brevan - Binomial-coefficients. I've found them when playing with the Faulhaber-polynomials using matrices ("P" = "Pascal-matrix" = Binomial-coefficients, "Gp"-matrix (own creature)) and the inverses of that matrices $\endgroup$ – Gottfried Helms Mar 5 '17 at 1:52
  • $\begingroup$ so is it a finite summation? Each line of Pascal's triangle is finite... $\endgroup$ – Brevan Ellefsen Mar 5 '17 at 1:53
  • $\begingroup$ @Brevan: well, the rows are finite, but not the columns... $\endgroup$ – Gottfried Helms Mar 5 '17 at 1:54
  • $\begingroup$ ooooh, the diagonals. OK, that makes sense. I wasn't thinking. Thanks. $\endgroup$ – Brevan Ellefsen Mar 5 '17 at 1:56
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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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  • $\begingroup$ Do you mean 9 instead? $\endgroup$ – Calvin Lin Sep 17 '14 at 12:29
  • $\begingroup$ No… e.g. $1,2,3,4$ are in arithmetical progression, and $(4^2-1^2)/(3^2-2^2)= 15/5 = 3$. $\endgroup$ – Kieren MacMillan Sep 17 '14 at 13:32
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    $\begingroup$ Write $(a,b,c,d) = (a,a+\delta,a+2\delta,a+3\delta)$, and by substitution \begin{align} \frac{(d^2 - a^2)}{(c^2 - b^2)} = \frac{(d-a)(d+a)}{(c-b)(c+b)} &= \frac{(3\delta)(2a+3\delta)}{(\delta)(2a+3\delta)} = 3. \end{align} $\endgroup$ – Kieren MacMillan Sep 17 '14 at 13:37
  • $\begingroup$ @CalvinLin: Surprising, yes? =) $\endgroup$ – Kieren MacMillan Sep 17 '14 at 20:56
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    $\begingroup$ Oh sorry, I completely misread it as (d-a)^2 / (c-b)^2. Not sure why. Yes, this is initially surprising :) $\endgroup$ – Calvin Lin Sep 17 '14 at 21:35
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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):

enter image description here

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    $\begingroup$ It's nice to see a pair of natural computations with divergent sums that do not miraculously coincide. Reading Euler's work, as one can in Lagarias's article in the Bulletin right now, you get the impression that there is a kind of mystic unity to the spectrum of cleverly done sums so that when done "right" they reveal some consistency in our aesthetic choices in extending math. Fortunately not. $\endgroup$ – Ryan Reich Sep 26 '13 at 21:55
  • $\begingroup$ Is there a simple proof explaining why the difference is always one? Or a name of the result that one can google? $\endgroup$ – Aaron Sep 27 '13 at 16:50
  • $\begingroup$ I don't have it at hand, but it can be reconstructed when one does Ramanujan-summation or uses the Euler/MacLaurin-formula and replaces the occuring Bernoulli-numbers by zetas. In the Ram.-summation we have one additional integral for such sums, and in the Euler/MacLaurin occurs the Bernoulli-number $B_0$ which can be understood as renormalized ratio of $\zeta(1)/\Gamma(0)$ equalling 1 (or -1). In the above formula would the latter idea occur with an additional row above the main-matrix with the series of $1+1/2+1/3+...=\zeta(1)$ and denominator of $(-1)!$ , whose ratio is normalized $-1$ $\endgroup$ – Gottfried Helms Sep 27 '13 at 17:29
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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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  • $\begingroup$ Can you generalize it? $\endgroup$ – Abramo Dec 12 '13 at 19:58
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    $\begingroup$ $\log(1+1+2+4)=\log(1)+\log(1)+\log(2)+\log(4)$ The general case is analyzed in www-users.mat.umk.pl/~anow/ps-dvi/si-krl-a.pdf $\endgroup$ – Mark S. Dec 13 '13 at 3:23
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    $\begingroup$ Your equation is true because $1+2+3=6$, and $1\times 2\times 3=6$. So, $\log(1+2+3)=\log(1\times 2\times 3)=\log(1)+\log(2)+\log(3)$ (because $\log(x+y+z)=\log(x)+\log(y)+\log(z)$) $\endgroup$ – TrueDefault Feb 28 '14 at 2:29
  • $\begingroup$ Ok JChau, I knew about this! ;-) $\endgroup$ – Abramo Mar 1 '14 at 14:55
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$$\int_0^1\frac{x^4(1-x)^4}{1+x^2}\mathrm{d}x=\frac{22}{7}-\pi$$

It's interesting how something so bizarre on the left hand side yields the tiniest of errors in one of the most famous approximations of $\pi$.

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  • $\begingroup$ This was a Putnam Competition problem. $\endgroup$ – DanielWainfleet Nov 18 '17 at 8:09
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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: $$\lim_{n\to\infty}\frac{F(n+1)}{F(n)}=\phi$$


3. Tupper's Self Referential Formula

enter image description here

When plotted with k=960939379918958884971672962127852754715004339660129306651505519271702802395266424689642842174350718121267153782770623355993237280874144307891325963941337723487857735749823926629715517173716995165232890538221612403238855866184013235585136048828693337902491454229288667081096184496091705183454067827731551705405381627380967602565625016981482083418783163849115590225610003652351370343874461848378737238198224849863465033159410054974700593138339226497249461751545728366702369745461014655997933798537483143786841806593422227898388722980000748404719

$0 \le x \le 106$ and $k \le y \le k + 17$, the resulting graph looks like this:

enter image description here


4. Ramanujan's golden ratio equation:enter image description here


5. Gaussian integral:

$$\int_{-\infty}^\infty \! e^{-x^2}dx = \sqrt{\pi}$$


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: enter image description here


9. The Schrodinger Equation:

$$H\Psi(x,t) = i\hbar\frac{\partial}{\partial t}\Psi(x,t)$$

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    $\begingroup$ Thanks! This is a good list :) $\endgroup$ – Calvin Lin Oct 2 '13 at 4:00
  • $\begingroup$ I would rewrite the 2. formula to make it even more symmetric/astonishing/mystical as $$ \phi^{-1}=\phi-1$$ $\endgroup$ – Gottfried Helms Nov 19 '13 at 10:14
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    $\begingroup$ @GottfriedHelms Even better as $$\phi^1-\phi^0=\phi^{-1}$$ $\endgroup$ – Alice Ryhl Sep 20 '14 at 9:58
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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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  • $\begingroup$ Seen this? mathworld.wolfram.com/NestedRadical.html $\endgroup$ – Bennett Gardiner Oct 2 '13 at 23:24
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    $\begingroup$ @BennettGardiner there is nothing given on the above radical?? $\endgroup$ – Shobhit Oct 4 '13 at 10:21
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    $\begingroup$ True, I missed the negatives, what is the pattern for the minus signs? $\endgroup$ – Bennett Gardiner Oct 4 '13 at 12:07
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    $\begingroup$ @BennettGardiner ++-+++-++++-+++++-.... $\endgroup$ – Shobhit Oct 4 '13 at 12:12
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    $\begingroup$ @BennettGardiner The pattern is actually ++-+ infinitely repeating, and the proof is actually fairly simple. We have $x = \sqrt{5 + \sqrt{5 + \sqrt{5 - \sqrt{5 + x}}}}$, solving this equation gives the above value by solving $(((x^2 - 5)^2 - 5)^2 - 5)^2 - 5 - x = 0$. How Ramanujan did this before computer algebra systems I don't immediately know, but I assume there's a shortcut to solving the polynomial. $\endgroup$ – orlp Dec 27 '16 at 21:06
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Pfister's 16-Square Identity:

$$(x_1^2+x_2^2+x_3^2+\dots+x_{16}^2)(y_1^2+y_2^2+y_3^2+\dots+y_{16}^2) = z_1^2+z_2^2+z_3^2+\dots+z_{16}^2$$

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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Another surprising equation $$\bbox[7pt,border:3px #FF69B4 solid]{\color{red}{\large 213 \times 122 = 25986}}$$ Now read above expression in reverse order $$\bbox[7pt,border:3px #FF69B4 solid]{\color{red}{\large 68952 = 221 \times 312}}$$

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    $\begingroup$ Here is another one $$\bbox[7pt,border:3px #FF69B4 solid]{\color{red}{\large 221 \times 113 = 24973}}$$ Now read above expression in reverse order $$\bbox[7pt,border:3px #FF69B4 solid]{\color{red}{\large 37942 = 311 \times 122}}$$ $\endgroup$ – Venus Jan 7 '15 at 19:01
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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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    $\begingroup$ It's not so surprising that $i^i$ should be real since we'd like $\overline{i^i} = (-i)^{-i} = i^i$ $\endgroup$ – user399601 Jan 1 '17 at 18:24
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Using the mystical ennead to calculate the decimal expressions for fractions of 7. Start with this figure:

numbered ennead

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

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

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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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  • $\begingroup$ Very cool. Is there a short way to prove this? $\endgroup$ – DanielV Jun 25 '14 at 3:02
  • $\begingroup$ @DanielV: I found this in my sketchbook from February 2008, but unfortunately there was no derivation included. However, it's fairly easy to derive in any number of ways. Here's one: \begin{align} 0^3 &= (x^n+y^n+z^n)^3 \\ &= x^{3n}+y^{3n}+z^{3n} + 3(x^n+y^n)(y^n+z^n)(z^n+x^n) \\ &= x^{3n}+y^{3n}+z^{3n} - 3(xyz)^n \\ 3(xyz)^n &= x^n(-x^n)^2 + y^n(-y^n)^2 + z^n(-z^n)^2 \\ &= x^n(y^n+z^n)^2 + y^n(z^n+x^n)^2 + z^n(x^n+y^n)^2 \\ &= x^n(y^n-z^n)^2 + y^n(z^n-x^n)^2 + z^n(x^n-y^n)^2 + 12(xyz)^n, \end{align} and since $xyz\ne0$ (by standard FLT hypothesis), the identity quickly follows. $\endgroup$ – Kieren MacMillan Jun 26 '14 at 12:22
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    $\begingroup$ A related — and similarly surprising — identity is $$\biggl(\frac{x^2}{yz}\biggr)^{\!n} + \biggl(\frac{y^2}{xz}\biggr)^{\!n} + \biggl(\frac{z^2}{xy}\biggr)^{\!n} = 3.$$ $\endgroup$ – Kieren MacMillan Jun 26 '14 at 22:56
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I've found this to be rather surprising:

  • $\displaystyle\sum\limits_{n=1}^{\infty}\frac{1}{2^n}=1$

  • $\displaystyle\sum\limits_{n=1}^{\infty}\frac{1}{2^n\ln(2^n)}=1$

As it essentially yields the identity:

$$\sum\limits_{n=1}^{\infty}\frac{1}{2^n}=\sum\limits_{n=1}^{\infty}\frac{1}{2^n\ln(2^n)}$$

It is surprising because obviously:

$$\forall{n\in\mathbb{N}}:\frac{1}{2^n}\neq\frac{1}{2^n\ln(2^n)}$$

In fact, the above inequity holds for every value of $n$, except for $n=\log_2e$.

Still, when summing up each of these infinite sequences, the result is $1$ in both cases.

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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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    $\begingroup$ There are not other integer solutions (except for $x=y$). $\endgroup$ – pts Sep 26 '13 at 11:02
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    $\begingroup$ @pts There are infinitely many rational solutions, though, with $(e,e)$ as a limit point. $\endgroup$ – awwalker Sep 27 '13 at 5:25
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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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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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    $\begingroup$ Thanks. Those do seem strange, till you see how it is set up. $\endgroup$ – Calvin Lin Sep 27 '13 at 13:11
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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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    $\begingroup$ Hint: Generalisation: $\cos(60-x)\cos(60+x)\cos(x)=\cos(3x)/4$. It gets even more interesting with tangents: $\tan(60-x)\tan(60+x)\tan(x)=\tan(3x)$ $\endgroup$ – chubakueno Sep 29 '13 at 18:51
  • $\begingroup$ Also $\sin(60-x)\sin(60+x)\sin(x)=\sin(3x)/4$ (just multiple both your $\cos$ and $\tan$ identities). $\endgroup$ – user26486 Apr 3 '15 at 13:09
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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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    $\begingroup$ +1 Interesting. BTW, didn't you mean to write "the Erdos-Selfridge result"? $\endgroup$ – r.e.s. Oct 3 '13 at 3:23
  • $\begingroup$ You are right, of course. Thanks. I will fix. $\endgroup$ – marty cohen Oct 5 '13 at 22:38
  • $\begingroup$ Oh nice, didn't expect it to be finite. $\endgroup$ – Calvin Lin Oct 10 '13 at 0:34
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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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    $\begingroup$ I would classify this as a "'standard' / well-known result," but it's cool nonetheless $\endgroup$ – MCT Feb 28 '14 at 2:29
  • $\begingroup$ Unfortunately, this formula converges pathetically slowly to be of any use =( $\endgroup$ – Trogdor Sep 7 '15 at 17:06
  • $\begingroup$ @Trogdor That's why we have things like the Euler sum, and you can find more such formulas here. $\endgroup$ – Simply Beautiful Art Mar 20 '17 at 23:27
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One of the most worth-mentioning identities may probably be Carl Friedrich Gauss's compution of $\cos(\frac{2\pi}{17})$:

\begin{align} \cos(\frac{2\pi}{17})&= \frac1{16}[-1+\sqrt{17} + \sqrt{34-2\sqrt{17}}+2\sqrt{17+3\sqrt{17}-\sqrt{34-2\sqrt{17}}-2\sqrt{34+2\sqrt{17}}}]\\ \end{align}

which has a significant role in regular heptadecagon construction.

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  • $\begingroup$ As Gauss himself noted, no one before him had ever found a regular polygon with an odd number, $n$, of sides, with $n>5$, that could be constructed with only drawing compass and unmarked straight-edge. $\endgroup$ – DanielWainfleet Nov 18 '17 at 8:33
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$$\sum_{i=0}^N {{N}\choose{i}}=2^N$$

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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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    $\begingroup$ How can the sum from 1 to infinity of positive integers lead to a negative number? $\endgroup$ – Kenshin Sep 28 '13 at 2:27
  • $\begingroup$ @Chris Because it isn't a sum in the normal sense. It's a regularized divergence. $\endgroup$ – Potato Sep 28 '13 at 5:48
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    $\begingroup$ @Chris Sorry for the brevity, I expanded a bit on this $\endgroup$ – Tobias Kienzler Sep 28 '13 at 7:22
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    $\begingroup$ @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. $\endgroup$ – DanielV Sep 28 '13 at 14:11
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    $\begingroup$ The Casimir effect needs a trick like that. $\endgroup$ – Felix Marin Sep 29 '13 at 9:52
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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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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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$$(1+i)(1+2i)(1+3i) = (1-i)(1-2i)(1-3i)$$

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  • $\begingroup$ Interesting, it's actually correct! $\endgroup$ – DanielV Nov 22 '14 at 17:53
  • $\begingroup$ Ah , I get it, this is a consequence of the $$\begin{align}\arctan(1) + \arctan(2) + \arctan(3) = \pi \text{ radians } \\ = -\pi \text{ radians } = \arctan(-1) + \arctan(-2) + \arctan(-3)\end{align}$$ $\endgroup$ – DanielV Nov 22 '14 at 17:59
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The Fibonacci sequence occurs in the decimal expansion of some "special" fractions.

$\begin{array}{ccccccccc} \frac{1}{89} &= &0.\color{blue}{0} \\ &+ & &\color{blue}{1} \\ &+ & & &\color{blue}{1} \\ &+ & & & &\color{blue}{2} \\ &+ & & & & &\color{blue}{3} \\ &+ & & & & & &\color{blue}{5} \\ &+ & & & & & & &\color{blue}{8} \\ &+ & & & & & & & &\color{blue}{13} \\ &+ & & & & & & & & &\ddots \\ \end{array}\\ \begin{array}{cc}\frac{1}{89} &=0.\color{blue}{011235}\color{red}{9}\ldots \end{array}$

Eventually the pattern is destroyed by carries in the decimal digits of the sum. If you want to go further, simply take a different fraction:

$\begin{array}{lllllllll} \frac{1}{9899} &= &0. &\underbrace{00}_{F_0} &\underbrace{01}_{F_1} &\underbrace{01}_{F_2} &\underbrace{02}_{F_3} &\underbrace{03}_{F_4} &\underbrace{05}_{F_5} &\underbrace{08}_{F_6} &\underbrace{13}_{F_7} &\underbrace{21}_{F_8} &\underbrace{34}_{F_9} &\underbrace{55}_{F_{10}} &\ldots \\ \end{array}$

Again the pattern from this point forward is obscured by carries in the sum (the next two digits are $90$ not $89$).

These patterns are a consequence of the fact that the generating function for the Fibonacci sequence (with $F_0=0, F_1=1$) is

$$f(x)=\frac{x}{1-x-x^2}$$

and taking $x=0.1,0.01,0.001,\ldots$ ensures that the terms in the power series expansion are synonymous with places in the decimal expansion. These values of $x$ are identified with the "special" denominators $89,9899,998999,\ldots$ which are really just entries in the sequence $a_n=10^{2n}-10^n-1$ for $n=1,2,3,\ldots$.

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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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    $\begingroup$ 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 $1$). $\endgroup$ – awwalker Sep 27 '13 at 5:38
  • $\begingroup$ I guess it depends on if you know about Euler products already (I didn't before the counselor showed me the proof). $\endgroup$ – Henry Swanson Sep 27 '13 at 6:05
  • $\begingroup$ (2) and (3) are special cases of $\sum_{i = 1}^\infty \dfrac{\sigma_a(i)}{i^k} = \zeta(k)\zeta(k-a)$. Ramanujan gave a similar identity; $\sum_{i = 1}^\infty \dfrac{\sigma_a(i)\sigma_b(i)}{i^k} = \dfrac{\zeta(k)\zeta(k-a)\zeta(k-b)\zeta(k-a-b)}{\zeta(2k-a-b)}$. $\endgroup$ – Jaycob Coleman Sep 28 '13 at 7:42
  • $\begingroup$ Now that I am older and (at least a smidgen) wiser, this is much less surprising to me than it used to be. Really, this should have emphasized to me exactly why the zeta function is important in number theory! $\endgroup$ – Henry Swanson Nov 10 '17 at 7:49

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

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