Here is a funny exercise $\sin(x - y) \sin(x + y) = (\sin x - \sin y)(\sin x + \sin y).$ (If you prove it don't publish it here please). Do you have similar examples?
Funny identities
-
0$\prod_{n\geq1}\frac1{4en}\bigg(\frac{(16n^2-9)^3}{16n^2-1}\bigg)^{1/4}\bigg(\frac{(4n+3)(4n-1)}{(4n-3)(4n+1)}\bigg)^n=\sqrt{\frac{2}{3\pi\sqrt{3}}}\exp\bigg(\frac{G}{\pi}+\frac12\bigg)$ $G$ is Catalan's constant – 2018-12-31
63 Answers
$\int_0^1\frac{\mathrm{d}x}{x^x}=\sum_{k=1}^\infty \frac1{k^k}$
-
11$\int_0^1 {x^x}\mathrm{d}x=\sum_{k=1}^\infty\frac{(-1)^{k-1}}{k^k}$ – 2013-11-02
$\left(\sum\limits_{k=1}^n k\right)^2=\sum\limits_{k=1}^nk^3 .$
The two on the left is not a typo.
-
12If you're physics-minded, the 2 and 3 are not a surprise: $n$ and $k$ must have the same dimension, so the right hand side has this dimension to the 4. So the only possible exponent on the left is 2. – 2016-11-16
$ \infty! = \sqrt{2 \pi} $
It comes from the zeta function.
-
0Neat! I wonder whether "solving" this identity for $\infty$ [also yields $-\frac12$](http://math.stackexchange.com/questions/1074870/is-infty-frac12) *edit* Hm, since $(-\frac12)!=\sqrt\pi$ not :/ – 2014-12-19
Ah, this is one identity which comes into use for proving the Euler's Partition Theorem. The identity is as follows: $ (1+x)(1+x^{2})(1+x^{3}) \cdots = \frac{1}{(1-x)(1-x^{3})(1-x^{5}) \cdots}$
Machin's Formula: \begin{eqnarray} \frac{\pi}{4} = 4 \arctan \frac{1}{5} - \arctan \frac{1}{239}. \end{eqnarray}
$\frac{1}{\sin(2\pi/7)} + \frac{1}{\sin(3\pi/7)} = \frac{1}{\sin(\pi/7)}$
-
3I thought this was going to be hard to prove...It just took three lines! – 2014-02-01
The Frobenius automorphism
$(x + y)^p = x^p + y^p$
-
25@AustinMohr: Not just _in a field_ of prime characteristic $p$, it holds in any commutative ring of characteristic $p$. – 2013-09-30
\begin{eqnarray} 1^{3} + 2^{3} + 2^{3} + 2^{3} + 4^{3} + 4^{3} + 4^{3} + 8^{3} = (1 + 2 + 2 + 2 + 4 + 4 + 4 + 8)^{2} \end{eqnarray} More generally, let $D_{k} = ${ $d$ } be the set of unitary divisors of a positive integer $k$, and let $\mathsf{d}^{*} \colon \mathbb{N} \to \mathbb{N}$ denote the number-of-unitary-divisors (arithmetic) function. Then \begin{eqnarray} \sum_{d \in D} \mathsf{d}^{*}(d)^{3} = \left( \sum_{d \in D} \mathsf{d}^{*}(d) \right)^{2} \end{eqnarray}
Note that $\mathsf{d}^{*}(k) = 2^{\omega(k)}$, where $\omega(k)$ is the number distinct prime divisors of $k$.
-
0Sure, but then you could still greatly simplify the description by replacing "unitary divisors" with "divisors" and restricting $k$ to be squarefree, there's no need to introduce two new notations. But at this point it feels like trying to dissect a proverbial joke :). – 2018-04-03
$\large{1,741,725 = 1^7 + 7^7 + 4^7 + 1^7 + 7^7 + 2^7 + 5^7}$
and
$\large{111,111,111 \times 111,111,111 = 12,345,678,987,654,321}$
-
0@ThomasWeller The same trick works with $11\times11$, $111\times 111$, $1111\times 1111$, etc., if the given version is too large to fit. (The given example is the largest of its type, though, because otherwise the digits overflow into adjacent locations and it doesn't look as nice.) – 2016-07-12
$\sec^2(x)+\csc^2(x)=\sec^2(x)\csc^2(x)$
-
3And because of this identity we have $\frac{\,d}{\,dx} \left[e^{\tan{x}} \cdot e^{-\cot{x}}\right] = \frac{\,d}{\,dx} \left[e^{\tan{x}}\right] \cdot \frac{\,d}{\,dx} \left[e^{-\cot{x}}\right]$. – 2017-06-10
\[\sqrt{n^{\log n}}=n^{\log \sqrt{n}}\]
-
0@ChantryCargill Not when you consider the fact that not many people know (surprisingly) that $\sqrt{x} \equiv x^{\frac{1}{2}}$ – 2014-07-12
$\displaystyle\big(a^2+b^2\big)\cdot\big(c^2+d^2\big)=\big(ac \mp bd\big)^2+\big(ad \pm bc\big)^2$
-
7The Brahmagupta-Fibonacci identity. – 2012-05-01
Facts about $\pi$ are always fun!
\begin{equation} \frac{\pi}{2} = \frac{2}{1}\cdot\frac{2}{3}\cdot\frac{4}{3}\cdot\frac{4}{5}\cdot\frac{6}{5}\cdot\frac{6}{7}\cdot\frac{8}{7}\cdot\ldots\\ \end{equation} \begin{equation} \frac{\pi}{4} = 1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+\frac{1}{9}-\ldots\\ \end{equation} \begin{equation} \frac{\pi^2}{6} = 1+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+\ldots\\ \end{equation} \begin{equation} \frac{\pi^3}{32} = 1-\frac{1}{3^3}+\frac{1}{5^3}-\frac{1}{7^3}+\frac{1}{9^3}-\ldots\\ \end{equation} \begin{equation} \frac{\pi^4}{90} = 1+\frac{1}{2^4}+\frac{1}{3^4}+\frac{1}{4^4}+\frac{1}{5^4}+\ldots\\ \end{equation} \begin{equation} \frac{2}{\pi} = \frac{\sqrt{2}}{2}\cdot\frac{\sqrt{2+\sqrt{2}}}{2}\cdot\frac{\sqrt{2+\sqrt{2+\sqrt{2}}}}{2}\cdot\ldots\\ \end{equation} \begin{equation} \pi = \cfrac{4}{1+\cfrac{1^2}{3+\cfrac{2^2}{5+\cfrac{3^2}{7+\cfrac{4^2}{9+\ldots}}}}}\\ \end{equation}
Well, i don't know whether to classify this as funny or surprising, but ok it's worth posting.
- Let $(X,\tau)$ be a topological space and let $A \subset X$ . By iteratively applying operations of closure and complemention, one can produce at most 14 distinct sets. It's called as the Kuratowski's Closure complement problem.
-
1@MichaelAlbanese: Anyone able to understand the contents of this thread instantly recognizes those braces are there to give the intersection operator a higher precedence than the neighboring union operators. One must travel far out of one's mathematical way to arrive at your alternative interpretation... – 2017-06-03
The following number is prime
$p = 785963102379428822376694789446897396207498568951$
and $p$ in base 16 is
$89ABCDEF012345672718281831415926141424F7$
which includes counting in hexadecimal, and digits of $e$, $\pi$, and $\sqrt{2}$.
Do you think this's surprising or not?
$11 \times 11 = 121$ $111 \times 111 = 12321$ $1111 \times 1111 = 1234321$ $11111 \times 11111 = 123454321$ $\vdots$
-
0I find (1....1)^n interesting, it's also nearly impossible to caluclate by hand without messing it up. – 2016-03-04
\begin{align} \frac{\mathrm d}{\mathrm dx}(x^x) &= x\cdot x^{x-1} &\text{Power Rule?}&\ \text{False}\\ \frac{\mathrm d}{\mathrm dx}(x^x) &= x^{x}\ln(x) &\text{Exponential Rule?}&\ \text{False}\\ \frac{\mathrm d}{\mathrm dx}(x^x) &= x\cdot x^{x-1}+x^{x}\ln(x) &\text{Sum of these?}&\ \text{True}\\ \end{align}
-
0@Derek朕會功夫 God the puns xD – 2017-03-21
$ \frac{e}{2} = \left(\frac{2}{1}\right)^{1/2}\left(\frac{2\cdot 4}{3\cdot 3}\right)^{1/4}\left(\frac{4\cdot 6\cdot 6\cdot 8}{5\cdot 5\cdot 7\cdot 7}\right)^{1/8}\left(\frac{8\cdot 10\cdot 10\cdot 12\cdot 12\cdot 14\cdot 14\cdot 16}{9\cdot 9\cdot 11\cdot 11\cdot 13\cdot 13\cdot 15\cdot 15}\right)^{1/16}\cdots $ [Nick Pippenger, Amer. Math. Monthly, 87 (1980)]. Set all the exponents to 1 and you get the Wallis formula for $\pi/2$.
\begin{eqnarray} \sum_{i_1 = 0}^{n-k} \, \sum_{i_2 = 0}^{n-k-i_1} \cdots \sum_{i_k = 0}^{n-k-i_1 - \cdots - i_{k-1}} 1 = \binom{n}{k} \end{eqnarray}
$\sum\limits_{n=1}^{\infty} n = 1 + 2 + 3 + \cdots \text{ad inf.} = -\frac{1}{12}$
You can also see many more here: The Euler-Maclaurin formula, Bernoulli numbers, the zeta function, and real-variable analytic continuation
-
0There is a very nice video from 3Blue1Brown on Youtube for the visualizing of the Riemann zeta function and its analytic continuation: https://www.youtube.com/watch?v=sD0NjbwqlYw – 2018-09-11
Two related integrals:
$\int_0^\infty\sin\;x\quad\mathrm{d}x=1$
$\int_0^\infty\ln\;x\;\sin\;x\quad \mathrm{d}x=-\gamma$
-
0@J.M. Ok Thanks. – 2012-07-05
M.V Subbarao's identity: an integer $n>22$ is a prime number iff it satisfies,
$n\sigma(n)\equiv 2 \pmod {\phi(n)}$
-
0@PAD: See [M. V. Subbarao, *On two congruences for primality,* Pacific Journal of Mathematics, Volume 52, Number 1 (1974), 261-268](http://projecteuclid.org/euclid.pjm/1102912230) ([Another PDF](http://www.math.ualberta.ca/~subbarao/documents/Subbarao1974.pdf)). The precise theorem is that $n\sigma(n) \equiv 2 \mod \phi(n)$ if and only if $n$ is prime or one of $1, 4, 6, 22$. – 2013-10-03
$32768=(3-2+7)^6 / 8$
Just a funny coincidence.
${\Large% \sqrt{\,\vphantom{\huge A}\color{#00f}{20}\color{#c00000}{25}\,}\, =\ \color{#00f}{20}\ +\ \color{#c00000}{25}\ =\ 45} $
By excluding the first two primes, Euler's Prime Product becomes a square:
$\prod _{n=3}^{\infty } \frac{1}{1-\frac{1}{(p_n)^{2}}}=\frac{\pi ^2}{9}$
By using multiples of the product of the first two primes, we get the square root:
$\prod _{n=1}^{\infty } \frac{1}{1-\frac{1}{(n p_1 p_2)^{2}}}=\frac{\pi }{3}$
-
2@BrunoJoyal If I'm not mistaken, periods include $\zeta(3)$ and many more - they are basically anything you can get with integration. If I recall correctly, it is not known whether or not $\dfrac1\pi$ is a period. – 2014-08-28
$ 10^2+11^2+12^2=13^2+14^2 $
There's a funny Abstruse Goose comic about this, which I can't seem to find at the moment.
-
11http://abstrusegoose.com/63 – 2013-09-27
$\left|z+z'\right|^{2}+\left|z-z'\right|^{2}=2\times\left(\left|z\right|^{2}+\left|z'\right|^{2}\right)$
The sum of the squares of the sides equals the sum of the squares of the diagonals.
-
0The so-called parallelogram identity. – 2013-08-19
What is 42?
$ 6 \times 9 = 42 \text{ base } 13 $ I always knew that there is something wrong with this universe.
The product of any four consecutive integers is one less than a perfect square.
To phrase it more like an identity:
For every integer $n$, there exists an integer $k$ such that $n(n+1)(n+2)(n+3) = k^2 - 1.$
-
11You can also write that as $n(n+1)(n+2)(n+3)=((n+1)^2+1)^2 - 1$ – 2012-06-19
I have one: In a $\Delta ABC$, $\tan A+\tan B+\tan C=\tan A\tan B\tan C.$
Considering the main branches
$i^i = \exp\left(-\frac{\pi}{2}\right)$
$\root i \of i = \exp\left(\frac{\pi}{2}\right) $
And $ \frac{4}{\pi } = \displaystyle 1 + \frac{1}{{3 +\displaystyle \frac{{{2^2}}}{{5 + \displaystyle\frac{{{3^2}}}{{7 +\displaystyle \frac{{{4^2}}}{{9 +\displaystyle \frac{{{n^2}}}{{\left( {2n + 1} \right) + \cdots }}}}}}}}}} $
$\int_0^\infty\frac1{1+x^2}\cdot\frac1{1+x^\pi}dx=\int_0^\infty\frac1{1+x^2}\cdot\frac1{1+x^e}dx$
-
0Obviously... :) – 2016-02-21
Let $f$ be a symbol with the property that $f^n = n!$. Consider $d_n$, the number of ways of putting $n$ letters in $n$ envelopes so that no letter gets to the right person (aka derangements). Many people initially think that $d_n = (n-1)! = f^{n-1}$ (the first object has $n-1$ legal locations, the second $n-2$, ...). The correct answer isn't that different actually:
$d_n = (f-1)^n$.
-
1To make this more rigorous, in a sense: We can define a linear operator $L$ acting on $\mathbb{C}[f]$ such that $L(f^n)=n!$ and $L(1)=1$. Thus, we can write $d_n=L\left((f-1)^n\right)$. (Am I doing this right?) – 2014-08-28
Best near miss
$\int_{0}^{\infty }\cos\left ( 2x \right )\prod_{n=0}^{\infty}\cos\left ( \frac{x}{n} \right )~\mathrm dx\approx \frac{\pi}{8}-7.41\times 10^{-43}$
One can easily be fooled into thinking that it is exactly $\dfrac{\pi}{8}$.
References:
- Wikipedia
- Future Prospects for Computer-Assisted Mathematics, by D.H. Bailey and J.M. Borwein
I actually think currying is really cool:
$(A \times B) \to C \; \simeq \; A \to (B \to C)$
Though not strictly an identity, but an isomorphism.
When I met it for the first time it seemed to be a bit odd but it is so convenient and neat. At least in programming.
$ 71 = \sqrt{7! + 1}. $
Besides the amusement of reusing the decimal digits $7$ and $1$, this is conjectured to be the last solution of $n!+1 = x^2$ in integers. ($n=4$ and $n=5$ also work.) Even finiteness of the set of solutions is not known except using the ABC conjecture.
The Cayley-Hamilton theorem:
If $A \in \mathbb{R}^{n \times n}$ and $I_{n} \in \mathbb{R}^{n \times n}$ is the identity matrix, then the characteristic polynomial of $A$ is $p(\lambda) = \det(\lambda I_n - A)$. Then the Cayley Hamilton theorem can be obtained by "substituting" $\lambda = A$, since $p(A) = \det(AI_n-A) = \det(0-0) = 0$
We have by block partition rule for determinant $ \det \left[ \begin{array}{cc} U & R \\ L & D \end{array} \right] = \det U\cdot \det ( D-LU^{-1}R) $ But if $U,R,L$ and $D$ commute we have that $ \det \left[ \begin{array}{cc} U & R \\ L & D \end{array} \right] = \det (UD-LR) $
-
0Actually, only U and L need to commute. – 2018-08-22
$\frac{1}{998901}=0.000001002003004005006...997999000001...$
-
41/98.99 = 0.010102030508132134... – 2013-09-26
Heres a interesting one again
$3435=3^3+4^4+3^3+5^5%$
$ \frac{1}{2}=\frac{\frac{1}{2}}{\frac{1}{2}+\frac{\frac{1}{2}}{\frac{1}{2}+\frac{\frac{1}{2}}{\frac{1}{2}+\frac{\frac{1}{2}}{\frac{1}{2}+\frac{\frac{1}{2}}{\frac{1}{2}+\frac{\frac{1}{2}}{\frac{1}{2}+\cdots}}}}}} $
and more generally we have $ \frac{1}{n+1}=\frac{\frac{1}{n(n+1)}}{\frac{1}{n(n+1)}+\frac{\frac{1}{n(n+1)}}{\frac{1}{n(n+1)}+\frac{\frac{1}{n(n+1)}}{\frac{1}{n(n+1)}+\frac{\frac{1}{n(n+1)}}{\frac{1}{n(n+1)}+\frac{\frac{1}{n(n+1)}}{\frac{1}{n(n+1)}+\frac{\frac{1}{n(n+1)}}{\frac{1}{n(n+1)}+\ddots}}}}}} $
\begin{eqnarray} \zeta(0) = \sum_{n \geq 1} 1 = -\frac{1}{2} \end{eqnarray}
-
1Actually, this can be made rigorous by noting that $ \zeta(z)=\lim_{n\to\infty}\left(\sum_{k=1}^nk^{-z}-\frac{1}{1-z}n^{1-z}-\frac12n^{-z}\right) $ for \mathrm{Re}(z)>-1. – 2012-06-21
$(x-a)(x-b)(x-c)\ldots(x-z) = 0$
-
4@columbus8myhw $(x-a)(x-b)(x-c)\ldots (x-w){\color{red}{(x-x)}}(x-y)(x-z)=0$ – 2015-08-23
\begin{eqnarray} \sum_{k = 0}^{\lfloor q - q/p) \rfloor} \left \lfloor \frac{p(q - k)}{q} \right \rfloor = \sum_{k = 1}^{q} \left \lfloor \frac{kp}{q} \right \rfloor \end{eqnarray}
-
4I don't see the 'punch' here. Isn't that just reversing the order of summation and truncating some zeros? – 2013-01-13
\begin{align}\frac{64}{16}&=\frac{6\!\!/\,4}{16\!\!/}\\&=\frac41\\&=4\end{align}
For more examples of these weird fractions, see "How Weird Are Weird Fractions?", Ryan Stuffelbeam, The College Mathematics Journal, Vol. 44, No. 3 (May 2013), pp. 202-209.
-
0Also [Project Euler problem number 33](https://projecteuler.net/problem=33). – 2016-10-24
$\lnot$(A$\land$B)=($\lnot$A$\lor$\lnot$B) and $\lnot$(A$\lor$B)=($\lnot$A$\land$\lnot$B), because they mean that negation is an "equal form".
$2592=2^59^2$ Found this in one of Dudeney's puzzle books
$ \sin \theta \cdot \sin \bigl(60^\circ - \theta \bigr) \cdot \sin \bigl(60^\circ + \theta \bigr) = \frac{1}{4} \sin 3\theta$
$ \cos \theta \cdot \cos \bigl(60^\circ - \theta \bigr) \cdot \cos \bigl(60^\circ + \theta \bigr) = \frac{1}{4} \cos 3\theta$
$ \tan \theta \cdot \tan \bigl(60^\circ - \theta \bigr) \cdot \tan \bigl(60^\circ + \theta \bigr) = \tan 3\theta $
-
1considering your first two identities the thirth should be $ \tan \theta \cdot \tan \bigl(60 - \theta \bigr) \cdot \tan \bigl(60 + \theta \bigr) = \tan 3\theta $ – 2011-03-06
$ \sum_{n=1}^{+\infty}\frac{\mu(n)}{n}=1-\frac12-\frac13-\frac15+\frac16-\frac17+\frac1{10}-\frac1{11}-\frac1{13}+\frac1{14}+\frac1{15}-\cdots=0 $ This relation was discovered by Euler in 1748 (before Riemann's studies on the $\zeta$ function as a complex variable function, from which this relation becomes much more easier!).
Then one of the most impressive formulas is the functional equation for the $\zeta$ function, in its asimmetric form: it highlights a very very deep and smart connection between the $\Gamma$ and the $\zeta$: $ \pi^{\frac s2}\Gamma\left(\frac s2\right)\zeta(s)= \pi^{\frac{1-s}2}\Gamma\left(\frac{1-s}2\right)\zeta(1-s)\;\;\;\forall s\in\mathbb C\;. $
Moreover no one seems to have wrote the Basel problem (Euler, 1735): $ \sum_{n=1}^{+\infty}\frac1{n^2}=\frac{\pi^2}{6}\;\;. $
$\tan^{-1}(1)+\tan^{-1}(2)+\tan^{-1}(3) = \pi$ (using the principal value), but if you blindly use the addition formula $\tan^{-1}(x) + \tan^{-1}(y) = \tan^{-1}\dfrac{x+y}{1-x y}$ twice, you get zero:
$\tan^{-1}(1) + \tan^{-1}(2) = \tan^{-1}\dfrac{1+2}{1-1*2} =\tan^{-1}(-3)$; $\tan^{-1}(1) + \tan^{-1}(2) + \tan^{-1}(3) =\tan^{-1}(-3) + \tan^{-1}(3) =\tan^{-1}\dfrac{-3+3}{1-(-3)(3)} = 0$.
$27\cdot56=2\cdot756,$ $277\cdot756=27\cdot7756,$ $2777\cdot7756=277\cdot77756,$ and so on.
-
1How does it work? – 2015-10-02
$\frac{\pi}{4}=\sum_{n=1}^{\infty}\arctan\frac{1}{f_{2n+1}}, $ where $f_{2n+1}$ there are fibonacci numbers, $n=1,2,...$
$\lim_{\omega\to\infty}3=8$ The "proof" is by rotation through $\pi/2$. More of a joke than an identity, I suppose.
-
0Or maybe the "proof" is through taking the closure? – 2013-12-15
For all $n\in\mathbb{N}$ and $n\neq1$ $\prod_{k=1}^{n-1}2\sin\frac{k \pi}{n} = n$
For some reason, the proof involves complex numbers and polynomials.
Link to proof: Prove that $\prod_{k=1}^{n-1}\sin\frac{k \pi}{n} = \frac{n}{2^{n-1}}$
Here's one clever trigonometric identity that impressed me in high-school days. Add $\sin \alpha$, to both the numerator and the denominator of $\sqrt{\frac{1-\cos \alpha}{1 + \cos \alpha}}$ and get rid of the square root and nothing changes. In other words:
$\frac{1 - \cos \alpha + \sin \alpha}{1 + \cos \alpha + \sin \alpha} = \sqrt{\frac{1-\cos \alpha}{1 + \cos \alpha}}$
If you take a closer look you'll notice that the RHS is the formula for tangent of a half-angle. Actually if you want to prove those, nothing but the addition formulas are required.
\begin{align} E &= \sqrt{\left(pc\right)^{2} + \left(mc^{2}\right)^{2}} = mc^{2} + \left[\sqrt{\left(pc\right)^{2} + \left(mc^{2}\right)^{2}} - mc^{2}\right] \\[3mm]&= mc^{2} + {\left(pc\right)^{2} \over \sqrt{\left(pc\right)^{2} + \left(mc^{2}\right)^{2}} + mc^{2}} = mc^{2} + {p^{2}/2m \over 1 + {\sqrt{\left(pc\right)^{2} + \left(mc^{2}\right)^{2}} - mc^{2} \over 2mc^{2}}} \\[3mm]&= mc^{2} + {p^{2}/2m \over 1 + {p^{2}/2m \over \sqrt{\left(pc\right)^{2} + \left(mc^{2}\right)^{2}} + mc^{2}}} = mc^{2} + {p^{2}/2m \over 1 + {p^{2}/2m \over 1 + {p^{2}/2m \over \sqrt{\left(pc\right)^{2} + \left(mc^{2}\right)^{2}} - mc^{2}}}} \end{align}
$ \begin{array}{rcrcl} \vdots & \vdots & \vdots & \vdots & \vdots \\[1mm] \int{1 \over x^{3}}\,{\rm d}x & = & -\,{1 \over 2}\,{1 \over x^{2}} & \sim & x^{\color{#ff0000}{\large\bf -2}} \\[1mm] \int{1 \over x^{2}}\,{\rm d}x & = & -\,{1 \over x} & \sim & x^{\color{#ff0000}{\large\bf -1}} \\[1mm] \int{1 \over x}\,{\rm d}x & = & \ln\left(x\right) & \sim & x^{\color{#0000ff}{\LARGE\bf 0}} \color{#0000ff}{\LARGE\quad ?} \\[1mm] \int x^{0}\,{\rm d}x & = & x^{1} & \sim & x^{\color{#ff0000}{\large\bf 1}} \\[1mm] \int x\,{\rm d}x & = & {1 \over 2}\,x^{2} & \sim & x^{\color{#ff0000}{\large\bf 2}} \\[1mm] \vdots & \vdots & \vdots & \vdots & \vdots \end{array} $
-
0The mystery, perhaps, lies in the constant of integration. All of the other integrals are evaluated from 0 to x, while the $\dfrac1x$ one is evaluated from 1 to x. (Note, by the way, that $\displaystyle\ln x=\lim_{t\to0}\frac{x^t-1}t$, which perhaps fits the pattern better.) – 2014-08-28
$ \frac{\pi}{2}=1+2\sum_{k=1}^{\infty}\frac{\eta(2k)}{2^{2k}} $ $ \frac{\pi}{3}=1+2\sum_{k=1}^{\infty}\frac{\eta(2k)}{6^{2k}} $ where $ \eta(n)=\sum_{k=1}^{\infty}\frac{(-1)^{k+1}}{k^{n}} $
Voronoi summation formula:
$\sum \limits_{n=1}^{\infty}d(n)(\frac{x}{n})^{1/2}\{Y_1(4\pi \sqrt{nx})+\frac{2}{\pi}K_1(4\pi \sqrt{nx})\}+x \log x +(2 \gamma-1)x +\frac{1}{4}=\sum \limits _{n\leq x}'d(n)$
$\textbf{Claim:}\quad$\frac{\sin x}{n}=6 for all $n,x$ ($n\neq 0).
\textit{Proof:}\quad$\frac{\sin x}{n}=\frac{\dfrac{1}{n}\cdot\sin x}{\dfrac{1}{n}\cdot n}=\frac{\operatorname{si}x}{1}=\text{six}.\quad\blacksquare
-
0@Tgymasb Denominator: $1/n\cdot n=1$. Numerator: $1/\textbf{n}\cdot\operatorname{si}\!\textbf{n}\, x=\operatorname{si}x$. – 2015-01-29
Let $\sigma(n)$ denote the sum of the divisors of $n$.
If $p=1+\sigma(k),$ then $p^a=1+\sigma(kp^{a-1})$ where $a,k$ are positive integers and $p$ is a prime such that $p\not\mid k$.
If we define $P$ as the infinite lower triangular matrix where $P_{i,j} = \binom{i}{j}$ (we can call it the Pascal Matrix), then $P^k_{i,j} = \binom{i}{j}k^{i-j}$
where $P^k_{i,j}$ is the element of $P^k$ in the position $i,j.$
I have another one, but I'm quite unwilling to post this here because it's MINE, I haven't found it anywhere, so don't steal this.
Let us take the four most important mathematical constants: The Euler number $e$, the Aurea Golden Ratio $\phi$, the Euler-Mascheroni constant $\gamma$ and finally $\pi$. Well we can see easily that
$e\cdot\gamma\cdot\pi\cdot\phi \approx e + \gamma + \pi + \phi$
-
1I wouldn't take credit for this dude – 2015-11-11
$ \int_{-\infty}^{\infty}{\sin\left(x\right) \over x}\,{\rm d}x = \pi\int_{-1}^{1}\delta\left(k\right)\,{\rm d}k $