In order to cleanse this board of brainlets and pop sci faggots, everyone post your favorite non-trivial formula for pi

All 10/10 answers

Pi == ((2^n/n) Integrate[Sin[t]^n/t^n, {t, 0, Infinity}])/ Sum[((-1)^k (n - 2 k)^(n - 1))/(k! (n - k)!), {k, 0, Floor[(n - 1)/2]}] /; Element[n, Integers] && n > 0

The formulas are derived using calculus and trigonometry. It's not a case of evaluating the formula and checking the digits against pi

fug :DD my formula didn't work

[eqn]1 + 1/4 + ... = pi^2/6[/eqn]

Forgot a \ there

yes

learn [math]\LaTeX[/math] for fucks sake

>What's the proof that each of these actually amount to Pi? Who's to say that every digit is identical and spot on except for the 500^2'th digit?

This is done using calculus, or if you want to be fancy then you can call it real anal.

When you do anal for real, you define a sequence and you "think of" a number that it converges to. Then, through some weird mathematical machinery, you can prove that your given sequence does converge to that number.

In this case, these people constructed this sequences probably by using some method they found for approximating Pi, and then naturally you think that if you were to continue this approximation you would, at infinity, reach Pi.

Then, after you think that, you find the definition of convergence and prove rigorously that indeed that is the case by doing a lot of gay inequalities.

Someone found a formula similar to OP's which yields the first billion digits but then begins to differ from pi. The trick is to start with a well known identity such as the gaussian integral [math]\left(\int e^{-x^2}\right)^2 = 2\pi[/math] and approximate it with a discrete series, e.g. writing the integral as a Riemann sum.

[math]\pi = \text{ }^{\color{#571da2}{\displaystyle\text{W}}}\text{ }^{^{^{^{\color{#462eb9}{\displaystyle\text{h}}}}}}\text{ }^{^{^{^{^{^{^{\color{#3f47c8}{\displaystyle\text{y}}}}}}}}}\text{ }^{^{^{^{^{^{^{^{^{^{\color{#3f62cf}{\displaystyle\text{ }}}}}}}}}}}}\text{ }^{^{^{^{^{^{^{^{^{^{^{^{^{\color{#437ccc}{\displaystyle\text{i}}}}}}}}}}}}}}}\text{ }^{^{^{^{^{^{^{^{^{^{^{^{^{^{^{\color{#4b90bf}{\displaystyle\text{s}}}}}}}}}}}}}}}}}\text{ }^{^{^{^{^{^{^{^{^{^{^{^{^{^{^{^{^{\color{#56a0ae}{\displaystyle\text{ }}}}}}}}}}}}}}}}}}}\text{ }^{^{^{^{^{^{^{^{^{^{^{^{^{^{^{^{^{^{^{\color{#62ab99}{\displaystyle\text{t}}}}}}}}}}}}}}}}}}}}}\text{ }^{^{^{^{^{^{^{^{^{^{^{^{^{^{^{^{^{^{^{^{\color{#71b484}{\displaystyle\text{h}}}}}}}}}}}}}}}}}}}}}}\text{ }^{^{^{^{^{^{^{^{^{^{^{^{^{^{^{^{^{^{^{^{\color{#82ba70}{\displaystyle\text{i}}}}}}}}}}}}}}}}}}}}}}\text{ }^{^{^{^{^{^{^{^{^{^{^{^{^{^{^{^{^{^{^{^{\color{#96bc5f}{\displaystyle\text{s}}}}}}}}}}}}}}}}}}}}}}\text{ }^{^{^{^{^{^{^{^{^{^{^{^{^{^{^{^{^{^{^{^{\color{#a9bd52}{\displaystyle\text{ }}}}}}}}}}}}}}}}}}}}}}\text{ }^{^{^{^{^{^{^{^{^{^{^{^{^{^{^{^{^{^{^{\color{#bcbb48}{\displaystyle\text{o}}}}}}}}}}}}}}}}}}}}}\text{ }^{^{^{^{^{^{^{^{^{^{^{^{^{^{^{^{^{\color{#ceb541}{\displaystyle\text{n}}}}}}}}}}}}}}}}}}}\text{ }^{^{^{^{^{^{^{^{^{^{^{^{^{^{^{\color{#dcab3c}{\displaystyle\text{ }}}}}}}}}}}}}}}}}\text{ }^{^{^{^{^{^{^{^{^{^{^{^{^{\color{#e39938}{\displaystyle\text{/}}}}}}}}}}}}}}}\text{ }^{^{^{^{^{^{^{^{^{^{\color{#e68033}{\displaystyle\text{s}}}}}}}}}}}}\text{ }^{^{^{^{^{^{^{\color{#e3632d}{\displaystyle\text{c}}}}}}}}}\text{ }^{^{^{^{\color{#de4227}{\displaystyle\text{i}}}}}}\text{ }^{\color{#da2121}{\displaystyle\text{/}}} [/math]

[math]2\cdot \int_{-1}^{1}\sqrt{1-x^2}dx[/math]
or how about [math]\frac{ln(-1)}{i}[/math]
or one for engineers [math]\pi=3[/math]

Screw Pi.
youtube.com/watch?v=jG7vhMMXagQ

355/113

22/7

...

0+e+2pi+i=-1

*sigh*...upvoted

Baddest Wallis since Marcellus...

It's been awhile since calc II and I don't feel like finding the problem/ formula... but the volume of Gabriel's Horn was pretty cool.

square root of 10

Pi = ln (-1)/ i

π=π

I feel like the left side of the equation should have one more part before the ellipses

sin(1/5555555555555555555555555555)

...

2*arcsin(1)

those are the primes, not the odds right? Also, if you multiply that by 2 to get tau, the 6's match up, and you gain an additional "prime" sqrt(2).

>know Ramanujan's formula
>haven't gone through all of A Synopsis of Elementary Results in Pure and Applied Mathematics

[math]
\pi = \sqrt{6}\sqrt{1 + \frac{1}{4} + \frac{1}{9} + \frac{1}{16} + ... + \frac{1}{n^2}}
[/math]

neat

1. Let [math]F_n[/math] be the nth Fibonacci number. Then [eqn]\sum_{n \ge 1} \arctan \frac{1}{F_{2n-1}}= \frac{\pi}{4}[/eqn]
2. [eqn]\prod_{n \ge 1} \left(1 + \frac{1}{n^3} \right) = \frac{\cosh \frac{\pi \sqrt{3}}{2}}{\pi}[/eqn]
3. [eqn]\lim_{x \rightarrow \infty}\frac{\sqrt{x}}{1+\frac{2x}{2 + \frac{3x}{3 + \frac{4x}{4 + \cdots}}}}=\sqrt{\frac{2}{\pi}} [/eqn]
4. For n greater than 0 [eqn]\int_0^\infty \frac{\sin nx}{x+\frac{1}{x+\frac{2}{x+\frac{3}{x+\cdots}}}}\, \mathrm{d}x = \frac{\sqrt{\frac{\pi}{2}}}{n+\frac{1}{n+\frac{2}{n+\frac{3}{n+\cdots}}}}[/eqn]
5. [eqn]\int_{-\infty}^\infty \frac{1+ \left(\frac{x}{2}\right)^2}{1+x^2} \, \frac{1+ \left(\frac{x}{4}\right)^2}{1+ \left(\frac{x}{3} \right)^2} \, \frac{1+ \left(\frac{x}{6}\right)^2}{1+ \left(\frac{x}{5} \right)^2} \cdots \, \mathrm{d}x = \frac{8}{\pi}[/eqn]
6. [eqn]\int_{-\infty}^\infty \frac{1}{1+x^2}\,\frac{1 - x^2/e^{2\pi}}{1+x^2/e^{2\pi}}\,\frac{1 - x^2/e^{4\pi}}{1+x^2/e^{4\pi}}\,\frac{1 - x^2/e^{6\pi}}{1+x^2/e^{6\pi}} \cdots \mathrm{d} x = \frac{\pi^{5/4}}{\Gamma \left(\frac{3}{4} \right)}[/eqn]
7. [eqn]\lim_{n \rightarrow \infty} \prod_{r = n}^{2n} \frac{\pi}{2 \arctan{r}} = 4^{1/\pi}[/eqn]

Those are the odd squares.

how the fuck Ramanujan even got that?

en.wikipedia.org/wiki/Ramanujan–Sato_series

Pi == 4

22/7

ln (-1)/i = π
>tfw brainlet

Pi doesn't exist.

Proof: give me the exact value of pi.

That is neat and all, but why would you use more than 16 digits to get 16 digits?
An explanation of why that produces near Pi and why it breaks at the 17th digit would be nice.
If it has nothing to do with the properties of a circle, and is merely an overly complicated way to calculate 16 digits of a known constant, then where is the value?

[math]1_{\pi}[/math]
:^)

>1 is prime

>brainlet
I can tell. Because that doesn't properly work. You could also write it as [math]\mathrm {ln} \frac { -1} {\mathrm i} = 180 °[/math]. The [math]\pi[/math] comes only from choosing radians.

wat
if e^iπ = -1
then ln(-1) = iπ
ln(-1)/i = π
why would I need radians or whatever?

...

it's 10
holy fuck you suck

the engineer's pi

sqrt(2)*((17 e!)/5 + 9/2 - 69/(5 e) - (43 e)/10)

Heh, nice try, kiddos. Watch this

[math] \pi = \big( \int_0^\infty \! x^{\frac{1}{2}}e^{-x}\, \mathrm{d}x \big)^{2} [/math]

how autistic are you to seek 7 pi definitions on the internet and then bother to tex em all

fucking loser lol

> [math] \ln (-1) [/math]

HAHAHAHAHAHAHAHAHAH

shit, you are so edgy you have infinite edges and are convex! u is so kewl!

>retard doesn't know a definition that's not alredy in thread and tries to get away from his true brainlet nature by using babby tier set properties.

lmao

le epic trolle

how many more mistakes are you going to make in your attempt at explaining the expansion of log domain into complex numbers?

[math] p(t) := {\mathrm e}^{it} [/math]

[math] {\mathrm d}\log(p) = \frac{p'(t)}{p(t)} {\mathrm d}t = i \, {\mathrm d}t [/math]

[math] \log(p(T)) = \log(p(0)) + \int_{p(0)}^{p(T)} {\mathrm d}\log(p) = 0 + i \int_0^T {\mathrm d}t = i T [/math]

Let's find out. I'm just trying to provide value

just pullin' your leg mate, you're alright

Pi=Pi