Mathematical Equation of the Week (MEoW)

This number amazed me:

$1741725= 1^7+7^7+4^7+1^7+7^7+2^7+5^7$

For $n = 1,2,3,4,5,6$ is true that:

$14^n+16^n+45^n+54^n+73^n+83^n = 3^n+ 5^n + 28^n + 34^n+ 65^n + 66^n + 84^n$

A number which rearrange itself upto multiplication by 6

142857 $\times$ 1 = 142857

142857 $\times$ 2 = 285714

142857 $\times$ 3 = 428571

142857 $\times$ 4 = 571428

142857 $\times$ 5 = 714285

142857 $\times$ 6 = 857142

Here is the interesting part

142857 $\times$ 7 = 999999

$\displaystyle e^{\pi \sqrt(163)} \approx 262537412640768743.99999999999925007 \approx 640320^3+744-0.00000000000075$ Coincidence

What do you think?

Credits : Tumblr (& The Incredibles), Ramanujan, Wikipedia.

From http://math.stackexchange.com/questions/8814/funny-identities?rq=1

$\frac{\pi}{2} = \frac{2}{1}\cdot\frac{2}{3}\cdot\frac{4}{3}\cdot\frac{4}{5}\cdot\frac{6}{5}\cdot\frac{6}{7}\cdot\frac{8}{7}\cdot\ldots\\ \frac{\pi}{4} = 1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+\frac{1}{9}+\ldots\\ \frac{\pi^2}{6} = 1+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+\ldots\\ \frac{\pi^3}{32} = 1-\frac{1}{3^3}+\frac{1}{5^3}-\frac{1}{7^3}+\frac{1}{9^3}+\ldots\\ \frac{\pi^4}{90} = 1+\frac{1}{2^4}+\frac{1}{3^4}+\frac{1}{4^4}+\frac{1}{5^4}+\ldots\\ \frac{2}{\pi} = \frac{\sqrt{2}}{2}\cdot\frac{\sqrt{2+\sqrt{2}}}{2}\cdot\frac{\sqrt{2+\sqrt{2+\sqrt{2}}}}{2}\cdot\ldots\\ \pi = \cfrac{4}{1+\cfrac{1^2}{3+\cfrac{2^2}{5+\cfrac{3^2}{7+\cfrac{4^2}{9+\ldots}}}}}\\$

Fermat's last theorem.. $a^{n}+b^{n}=c^{n}$ for all a,b,c,n being integers gives real integral solutions only if n<=2 . The proof of the solution took 7 years of hard solitary work!!!

Honestly, the equation that comes into my mind was

$x^2-x-1=0$

which is related to the golden ratio and the Fibonacci Sequence.

I loved this one as well.

$\dfrac {d}{dx} (e^x) = e^x$

$a^2 + b^2 = c^2$ (Pythagoras Theorem)

Edit-These positive integers $a,b,c$ form the sides of a right triangle necessarily.

$\sin{(2°\times{10}^{-n})}-\sin{(2°\times{10}^{-(n+1)})}\approx\pi\times{10}^{-(n+2)}$

Try to derive it and you'll see the magic.

Inclusion-Exclusion: Either you like something, or you don't.

If you like this, then upvote it, or else, if you don't like it, then downvote it.

it is...........................

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1 = 1

$\zeta(-1) = 1+2+3+\dots = -\frac{1}{12}$

where $\zeta(s)$ is the Riemann zeta function.

Taylor Series -

$f(x) = \sum _{ n\quad =\quad }^{ \infty }{ \frac { { f }^{ (n) }(0) }{ n! } { x }^{ n } }$

I have one more(if it is allowed):

https://qph.is.quoracdn.net/main-qimg-7beca9cf6c835bdfeabac3541a778973?converttowebp=true

And one more:

http://en.wikipedia.org/wiki/Tupper%27sself-referentialformula

*The last 2 are just for sharing

1+(e^(i*pi))=0

five fundamental numbers in one equation... almost all you guys are already familiar with this

${ G }_{ \mu v }=8\pi G\left( { T }_{ \mu v }+{ \rho }_{ \Lambda }{ g}_{ \mu v } \right)$

Einstein field equation

$V-E+F=2$

This is not for competition. However this is what I remember since I was of teen age.$\color{#3D99F6}{152207 * 73= 11111111.(Eight~ '1's)\\~ It~ is ~ clear~ that ~if ~we~use~N*73, N=1,2,...9,~~\\we~ get~ EIGHT~ Ns.~\\152207~is ~divisible ~by ~11. ~(11|152207)}$

$\Im(i^{i})=0$

substitute x=pi. then eqution becomes e^(i pi)+1=0. this relation connects 5 most important numbers of mathematics, which makes this most beautiful relation. :)

Haha...The abbreviation is Meow :)

You all have really good entries. I don't know I only have one equation that may seem not as good as yours. Still have a look. e^{i(pi)}= -1 This equation is derived from De Moivre's Theorem(Sorry Sharky Kesa) But I loved it more than the original theorem since it is of the form An irrational number raised to the product of an imaginary and irrational number which gives us a real number. P S I don't know how to write 'pi' in Greek.

The sequence 1-1+1-1+1-1... an Oscillatory series equals 1/2 even though all are integers .