I have a question

[Anonymous No.1386827]

How did they calculate that Pi=3.14159.....
Is it some kind of integral or limit?

[Anonymous No.1386834]

this

[Anonymous No.1386836]

>>1386827

use google

[Anonymous No.1386843]

a series

[Anonymous No.1386841]

it took me less then 15 seconds to go to Wikipedia and save that then post here

[Anonymous No.1386840]

they divided 22 by 7 then just switched around some of the thousandths, hundred thousandths etc

[Anonymous No.1386837]

[Anonymous No.1386861]

>>1386834
this for early calculations

I think modern calculations use some kind of taylor series expansion

[Anonymous No.1386872]

That's a big question.
Read up on the history of pi and you'll see that over the past few thousand years there have been many different ways to approximate and calculate pi.

Some of the best original techniques were to place polygons in a unit circle, and the unit cirlce and work out the are of the two polygons, then get an approximate value for pi.

after the invention/discovery of the calculus, people were able to come up with many elegant expression for pi using series and integrals.

For example if you know the series expansion for artan, then you can plug 1 into it and get a series expression for pi.

See http://mathworld.wolfram.com/PiFormulas.html for a list of nice formulas for pi, some of them with their origin explained.

[Josef @home !!cvJKiVKmWz2 No.1386876]

You'll love this one.
http://en.wikipedia.org/wiki/Bailey%E2%80%93Borwein%E2%80%93Plouffe_formula

[Anonymous No.1386881]

>>1386872

> and the unit circle inside another polygon, then..**

[Anonymous No.1386889]

I'm moving, and I just packed away "e The story of a number". It contained a pretty elegant limit type explanation of pi, having to do with dividing the circle into progressively smaller triangular wedges. Damned if I can remember specifics.

[Anonymous No.1386903]

The first way was a method of exhaustion. You approximated the circumference of a unit circle by using regular polygons. Nowadays you can use a ton of different methods, integrals, continued fractions, sequences, series.

[Anonymous No.1386941]

To calculate pi to any precision you want, Taylor expansion is a common way to do that.

[Anonymous No.1386955]

Curvature of the earth

then an integral

[Anonymous No.1386967]

>>1386941

That isn't a common way to calculate pi. You would be hard pushed to find software or calculators that use Taylor series to calculate pi, they're very inefficient.

The Brent-Salamin Algorithm is much more common.

[Josef @home !!cvJKiVKmWz2 No.1386970]

>>1386941
Taylor expanding <span class="math">\pi[/spoiler]? The resulting Taylor polynom is <span class="math">\pi[/spoiler]. Now what?

[Anonymous No.1386978]

>>1386970
Now calculate it! It's very common.

[Anonymous No.1386995]

There are a ton of ways to calculate pi. Check out wikipedia's entry on "continued fractions." There's a very rapidly converging continued fraction for pi. As I recall, the best of the three they show is the one with all the sixes. My continued fraction programs are all at home, so I can't remember which I chose.

[Anonymous No.1387001]

C=Pi x R so Pi = C/R
C = circumference
R = Radius
Pi = Pie

[Anonymous No.1387008]

>>1386970
I lol'd. Usually people push the taylor expansion of arctan 1 to calculate pi/4. I think that's where one of the continued fraction expansions comes from (euler's cf applied to taylor expansion of arctan).

[Josef @home !!cvJKiVKmWz2 No.1387004]

>>1386978
<div class="math">\mathcal T(P(x)=\pi)_{x_0=0} = \sum_{k=0}^\infty \frac{\partial^k\pi}{\partial x^k}\frac{x^k}{k!} = \frac{\partial^0\pi}{\partial x^0} = \pi</div>
;)

[Josef @home !!cvJKiVKmWz2 No.1387022]

>>1387008
Ah, Taylor <span class="math">\mathrm{arctan}(x)[/spoiler] around x=1? Sounds reasonable.

[Anonymous No.1387054]

>>1387022
I'm sure that's what that anon meant.

[Anonymous No.1387055]

>>1386970
for example:

<span class="math"> \tan^{-1}(x) = x - x^3/3 + x^5/5 - x^7/7 + ... [/spoiler]

let <span class="math"> x = 1 [/spoiler] and the series gives <span class="math"> \pi/4 [/spoiler]

some math guy pointed out that this converges more slowly than some sophisticated algorithm out there, but this works

[Anonymous No.1387060]

>>1387055
well fuck, they beat me to it

[Anonymous No.1387073]

Arctan comes from trigonometric circular function and the value of pi is assumed in it already, I think you are using pi to define pi, which is not legit

just saying

[Anonymous No.1387076]

>>1387073

No you're not. It's legit.

[Josef @home !!cvJKiVKmWz2 No.1387083]

>>1387073
There are many ways of defining arctan, and the trigonometric approach is the least used one (read: not at all) in maths.

[Anonymous No.1387085]

>>1387073
Not really assumed, no.

[Anonymous No.1387994]

Has anyone here heard of infinite powers and fractions???

Also,
Pi = 2x((2x2x4x4x6x6...)/(1x1x3x3x5x5x7...))

Using arctan gives a decent approximation, but less percent change of error occurs using infinite fractions and powers.

[Anonymous No.1388030]

<span class="math">\pi = 4 \sum_{k = 0}^{\infty} \frac{1}{(16^k)(8k+1)} - 2 \sum_{k = 0}^{\infty} \frac{1}{(16^k)(8k+4)} - \sum_{k = 0}^{\infty} \frac{1}{(16^k)(8k+5)}
- \sum_{k = 0}^{\infty} \frac{1}{(16^k)(8k+6)}. [/spoiler]

[Anonymous No.1388040]

>>1388030

= (>>1386876)

Well done for reading the thread first.

[ego !TuVtA3NpiI No.1388071]

(P(x)=)x0=0=k=0xkkk!xk=x00=

bullshit brainwash code cause you're dumb as fuck

[Anonymous No.1388094]

If you want to calculate pi, just type in "3.141" into your calculator and hit random numbers afterwards.

The government pays big money to people for doing that day after day, and they leave it up to the best mathematicians to figure out what they typed.

[Anonymous No.1388723]

bump

[Anonymous No.1388920]

The spigot algorithms are pretty interesting but the nth hexadecimal digit is... kind of a specialized need.