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/sci/ - Science & Math

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10952208 No.10952208 [Reply] [Original]

would you mind writing down what this number you call "pi" is anon?

>> No.10952210
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10952210

A number for virgins

>> No.10952218

2*arcsin(1)

>> No.10952221

>>10952210
how can tau be twice as transcendental as pi?

>> No.10952249

>>10952208
"10"
in base pi

>> No.10952253

>>10952221
By definition.
Tau=2pi

>> No.10952262

>>10952249
What

>> No.10952289

>>10952208
[eqn] \pi = 4 \sum_{k=0}^{\infty} \frac{(-1)^k}{2k+1} [/eqn]

>> No.10952298

>>10952262
protip
[math]
123 = 1 \cdot 10^2 + 2 \cdot 10^1 + 3\cdot 10^0
[/math]

>> No.10952336

>>10952210
Archmedes was concerned with circle areas, a place where pi is more natural than tau/2.
It's just that areas stopped being inreresting since Calculus was developed, while rotations were discovered to be very important in lots of aplications.

>> No.10952352

log(-1)/i

>> No.10952353

>>10952298
Ok, I know it
But how do you do it with a unlimited non periodic number such as pi?
What does it mean "10 in base pi"?

>> No.10952366

>>10952353
Most numbers become infinite series

>> No.10952376

>>10952366
>infinite series
Oh ok, that makes sense
lol sorry but I study medicine and my math knowledge finishes at derivatives and integrals
never studied Taylor series for example

>> No.10952377

>>10952353
In base pi, the everyday ten is a unlimited non periodic number.

>> No.10952382

>>10952249
>base pi
No such thing

>> No.10952383

>>10952376 me
>>10952366
Can you compute (is this the correct term) how the sum for pi would be?
Are there other than >>10952289 one?
I don't visualized how 1-1/2+1/3-1/4... is pi/4
How do I learn if a serie converge/diverge? And if it converge, how do I calculate exactly at which number?

>> No.10952397

>>10952382
kek, there's nothing mysterious about it
[math]
123_{10} = 1 \cdot 10^2 + 2 \cdot 10^1 + 3\cdot 10^0
\\

123_{ \pi} = 1 \cdot { \pi}^2 + 2 \cdot { \pi}^1 + 3\cdot { \pi}^0
[/math]

>> No.10952400

>>10952397
The concept of a base only makes sense for natural numbers. But you already know that since you are an obvious troll.

>> No.10952404

>>10952383
>Are there other
https://www.youtube.com/watch?v=HEfHFsfGXjs

>> No.10952407

>>10952400
bs, the only difference is convenience

>> No.10952414

>>10952382
https://en.wikipedia.org/wiki/Non-integer_representation#Base_%CF%80

>> No.10952423

>>10952404
fantastic
thanks anon

>> No.10952424

>>10952383
That series os for 4*arctan(1).
Arctan(1) = pi/4 (look at a trig circle).
If uv != 1 then arctan((u+v)/(1-uv)) = arctan(u) + arctan(v) and you can use this identity to create lots of other formulas for pi based on the arctan series.
This was how it was done in the 18th century, Euler computed pi to 300+ digits using one such formula and someone would compute pi up to some more digits every once in a while simply by reapplying the identity again and again.
There's also several convergence acceleration methods you could use to turn that series into another that also goes to pi.
But that's transforming the series you already have.
Thanks to a result due to Euler, we know that [math]\sum_i^\infty{1/i^2}[/math] = pi^2/6. So just multiply that by 6 and take a square root.
There areany other identities.
Archimedes method was aproximating a circle with other polygons, you can get a sequence aproximating pi pretty fast.
Lots of modern methods get pretty crazy too.

>> No.10952431

>>10952424
Fuck, the lower limit should've been i=1

>> No.10952437

>>10952424
>arctan((u+v)/(1-uv)) = arctan(u) + arctan(v)
Which rule is this? Demonstration?

>> No.10952442

>>10952437
Take the tangent of the sum of angles identity and make a change of variables to turn the tans into atans

>> No.10952447

>>10952442
Anon, I'm too much rusty in math right now
I've studied till integrals and derivatives and basic differentials in med school but never studied/done this

>> No.10952451

>>10952447 me
I'm a retard
I was thinking
>sum of angles identity
was another thing, got it now kek

>> No.10952453

>>10952210
>a quarter turn along the unit circle is pi/2 because
my sides are in orbit

>> No.10952466

>>10952424
[math] \displaystyle
\sum_{i=1}^{ \infty } \dfrac{1}{i^{2}} = \dfrac{ \pi ^{2}}{6}
[/math]
Optimized.

>> No.10952481

>>10952447
Ok, put x = tan(u), y = tan(u) and so u = arctan(x) and v = arctan(y).

Now consider the refered tan addition formula: tan(x+y) = (tan(x)+tan(y)/(1-tan(x)tan(y)). By the way we definied x, y it becomes:
tan(x+y) = (u+v)/(1-uv).
Apply arctan in both sides and it becomes:
x+y = arctan((u+v)/(1-uv)) (we are handwaving alway questiona about branches and variable ranges)
but x+y = arctan(u)+arctan(v) and so we have the desired identity:
arctan(u)+arctan(v) = arctan((u+v)/(1-uv)).

>> No.10952707

>>10952481
Thanks anon, thought it was something more difficult

>> No.10953037

>>10952210
>number of radians in 1 turn
But I just rotated "1 tau" about an axis in SU(2) and I've only gone halfway around. Sounds like 4pi is the more natural rotation constant instead.

>> No.10954785

3.14