approximating π

>>8578359
Do your own homework.

File: Skärmavbild 2017-01-03 kl. 14.20.14.png (46KB, 668x490px) Image search: [Google]

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>>8578359
>and it appears to be within one-half part per trillion of the actual value.
You sure about that, m8?

>>8578359
>>8578530

OP is a faggot and can't type 66317 properly.

>>8578535
probably why he can't do his homework

hard to decompose into partial fractions when you start with the wrong factors

>>8578535
oops

You'd think that with modern super computers, the last digit of pie would be found by now.

>>8578568
Pi actually ends with 51413, it's a palindromic number.

File: IMG_20170103_160613.jpg (4MB, 3120x4160px) Image search: [Google]

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>>8578359
How am I suppose to solve this?
>A B C

>>8578594
>66371 = 17 * 47 *83
try harder baguette

>>8578594
To fulfill this, you'd need, for example, this equality to be true as well:
6 = 8a (mod 17)

Up to you to figure out why.

>>8578594
you doing chemistry mate ?
what uni you at ?

>>8578607
The actual number es 66317, not 66371...

>>8578416
>>8578538
>homework
No, when I wrote that I found this written in the back of an old book (Mathematical Handbook by Murray R.Spiegel) that was true, unlike what you wrote.

>>8578851
Correct, I transposed the 1 and 7 because fatigue.

>>8580633
Stop talking like a stuck-up faggot. Just talk like a normal person and say "I fucked up".

>>8580634
stop talking like an inbred faggot and talk like a normal person

>>8578594
Find out how to solve linear Diophantine equations in three variables.
I found this solution:
A = -64169028
B = 176464827
C = 1666728

>>8578359
>it appears to be within one-half part per trillion of the actual value
it's only ~39 ppb by my calculation.

I found a other approximation which is slightly more accurate:
[math]\frac{312689}{99532} = \frac{A}{4} + \frac{B}{149} + \frac{C}{167}[/math]

>>8580717
positive integers A, B, C are all
smaller than your value for C

>>8578594
lol Americans...

>>8580634
your inferiority is showing

>>8581718
too lazy to find the particular solution :^)

>>8578359
[math]
\frac{5}{17} + \frac{37}/{47} + \frac{171}/{83}
[/math]

>>8581926
baka

>>8578359
[math]
\frac{5}{17} + \frac{37}{47} + \frac{171}{83}
[/math]

>>8581927
How did you arrive?

>>8581946
it's pretty trivial after re-expressing it as
[math]
799\,C+1411\,B+3901\,A=208341
[/math]
so trivial, in fact, that i used a CAS to give the general solution

>>8581953
I plugged it into Mathematica, but I don't know how to specify that I want only natural number solutions.

Does anybody know how to do that?

>>8582027
Reduce[3901 A + 1411 B + 799 C == 208341, {A, B, C}, Integers]
:^)

>>8582032
Cool, thx
(I see ", Integers" also works with Solve)

>>8581927
>>8581926
>>8581856
found this solution as well

>>8580800
[eqn]\frac{312689}{99532} = \frac{3}{4} + \frac{144}{149} + \frac{238}{167}[/eqn]

File: pi-approx.png (8KB, 526x235px) Image search: [Google]

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>>8580762
>within one-half part per trillion
>it's only ~39 ppb by my calculation
We were both wrong, it's 39 per trillion.

>>8581926
>>8581927
I'm a rebel
[eqn]\frac{22}{17} + \frac{37}{47} + \frac{87}{83}[/eqn]

>>8578594
>french
>page in the background literally says croissante
Really makes you think.

here's the solution I found with some modular algebra:
[math]208341=(47)(83)A+(17)(83)B+(17)(47)C[/math]
remainders on division by 17:
[math]6 \equiv 8A(mod~17)[/math] so A = 5,22,39...
remainders on division by 47:
[math]37 \equiv B(mod~47)[/math] so B = 37,84,131...
remainders on division by 83:
[math]11 \equiv 52C(mod~83)[/math] so C = 5,88,171...
picking B = 37 (for example) gives the choices
(A, C) = (5, 171), (22, 88), (39, 5)

>>8583465
>3.1295444607
Pi as fuck.

>>8584020
>Joke's on you I was only pretending to be retarded