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- Thread starter Epic Sandwich
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phyzguy

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http://en.wikipedia.org/wiki/Pi

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We usually use an infinite series that converges to pi or some other iterative method.

For example arctan(x) = x - x^3/3 + x^5/5 - x^7/7 + ... up to infinity. If x=1 then arctan(1) = pi/4 so you can see that you could calculate pi in such a way. The problem is that this infinite series converges (approaches) pi very slowly. What this means is that you need to add a lot of terms before the number starts to look like pi.

Currently, there are much better infinite series to use for example: http://planetmath.org/encyclopedia/RamanujansFormulaForPi.html [Broken] is a much better infinite series to use to calculate pi.

Typically the computations are done, well, by computers that are able of doing calculations very fast. I think that currently pi has been calculated to like 10 billion digits or something ridiculous like that.

For example arctan(x) = x - x^3/3 + x^5/5 - x^7/7 + ... up to infinity. If x=1 then arctan(1) = pi/4 so you can see that you could calculate pi in such a way. The problem is that this infinite series converges (approaches) pi very slowly. What this means is that you need to add a lot of terms before the number starts to look like pi.

Currently, there are much better infinite series to use for example: http://planetmath.org/encyclopedia/RamanujansFormulaForPi.html [Broken] is a much better infinite series to use to calculate pi.

Typically the computations are done, well, by computers that are able of doing calculations very fast. I think that currently pi has been calculated to like 10 billion digits or something ridiculous like that.

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Is there a way to calculate the number of the nth digit without expanding out pi?

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What do you mean Anonymous217?

Oh! You mean like take pi and you want to find a number in pi at a particular digit?

http://www.math.hmc.edu/funfacts/ffiles/20010.5.shtml

Oh! You mean like take pi and you want to find a number in pi at a particular digit?

http://www.math.hmc.edu/funfacts/ffiles/20010.5.shtml

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^ Yes! That's what I was looking for. Thank you.

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