
One question I have though.. How is the computer calculating the trig functions? I like the sum one because I know what it is doing.

The x86 math coprocessors (x87 if x <= 3) have instructions that can be used to calculate trig functions.
I don't know how accurate (or precise) they are, though... probably only to the machine floatingpoint precision (either 32 or 64 bits, or 80 if your compiler supports "long double").

the computer isn't...
with the atan suggestions, the taylor series expansion is used...
atan(x) = x  x^3/3 + x^5/5  x^7/7 ...
or
atan(x) = summation from n=1 to infinity of (1)^(n+1)*x^(2n1) / (2n1)
so each iteration you just add another term and it gets closer to the exact value of atan(x), which can be used to find pi in a number of ways.

Are you sure the computer uses Taylor series? They're very slow; I think x86s have some binary method which involves mathematical rotations.
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sorry, misunderstanding. no, i highly doubt that the computer uses the taylor series. i'm saying the programmer could use that method to get perfect accuracy for atan, instead of using the machine's approximation of atan.

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Originally posted by o0zi
Why calculate when you know it to several places from memory?
3.141592635358979323846264338327950288419716939937
(yes, that was from memory)
Damn! That is three digits more than I have memorized!
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just use the pi button on a calculator.
ha ha.
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You might as well ask the same of most of mathematics  it's an intellectual pursuit which many people enjoy merely for the fun of it, and not the practical value.
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Originally posted by o0zi
Why calculate when you know it to several places from memory?
3.141592635358979323846264338327950288419716939937
(yes, that was from memory)
Don't know if this was a typo or not, but a published value is:
3.141592653589793238462643383279502884197169399375 11
(you've got an extra 3 in the 8th digit to the right of the decimal point).

Yup, sorry, typo
3.141592653589793238462643383279502884197169399375 10582
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You just put that there to make us think you took it from memory.
Impostor.

Originally posted by drummerboy195
we are working on sums of sequences in clac right now, and if it wasnt for the pesky change in signs every other number, it could be expressed as the sum of 4(1/(2i1)) as i goes from 1 to infinity. except, the real sum should be 4(1/(2i1)1/(2i/1)+1/(2i1)1/(2i/1)+1/(2i1)........) anybody know how to set that up?
4*sum((1)^(i+1)/(2*i1),i,1,infinity)=4(11/3+1/51/7...)

Re: Million Dollar que$tion
Originally posted by knute
Why?
That's how nerds entertain themselves.
Andrew Wiles locked himself up in an attic for 7 years working up a proof to Fermat's Last Theorem. Nobody knows what to do with it now, but at least Wiles had fun.
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