Pi by recursive decimal places

glenvb · May 12, 2004, 02:21 PM

Nice! Your basic idea is roughly similar to that of Archimedes, one of the first to calculate pi. He took the calculation to a 96-sided polygon, I believe. In the 15th century the Persian astronomer Jamshid al-Kashi computed pi to about 16 decimal places by taking this method out to a polygon of over 805 million sides. Yikes.

matte · Aug 23, 2007, 06:07 AM

2nrcos180((n-2)/(2n)) would be a more proper sollution actually for the distance around the polygon. Divide by 2 to get the value of pi.

ebaines · Aug 23, 2007, 12:50 PM

This is a fine technique, although there's a crucial step missing in how to calculate the cosine or sine of the angles (assuming that like Archimedes you don't have a calculator). One could use an infinite series, for example:

Sin(x) = x - x^3/3! + x^5/5! - x^7/7! + ...

But this is cumbersome, and requires a knowledge of calculus that I doubt the ancients had. So I wonder if they didn't use another technique, such as this: given the knowledge that cos(pi/4) = sqrt(2)/2, and the relationship cos(x) = sqrt[(cos(2x)-1)/2], you can play around using basic trig identities and you find that you get a neat recursive formula:

$ \sin $\frac {\pi} 4 $ = \frac {\sqrt 2} 2 \\ \sin $\frac {\pi} 8 $ = \frac 12 \sqrt {2 - \sqrt 2} \\ \sin $\frac {\pi} {16}$ = \frac 12 \sqrt {2 - \sqrt {2+ \sqrt 2}}\\ \sin $\frac {\pi} {32}$ = \frac 12 \sqrt {2 - \sqrt {2+ \sqrt {2 + \sqrt 2}}} $

Etc.

The length of the side of an n-sided polygon is 2* sin(pi/(n)), so if we choose polygons where each has twice the number of sides as its predecessor, the accuracy in determining pi is limited only by one's accuracy in calculating square roots. The formula for estimating the length of half the the perimeter (that is, pi) is:

$ \pi (est.) = N \times \sin(\pi/N). $

We choose the value for N to be 1, 2, 4, 8, 16, etc., and what we find is that the approximation for pi converges pretty quickly to a reasonable level of accuracy:

$ \begin{matrix} N & \Pi(est) \\ 4 & 2.828 \\ 8 & 3.061 \\ 16 & 3.121 \\ 32 & 3.136 \\ 64 & 3.140 \\ 128 & 3.1413 \\ 256 & 3.14151 \\ 512 & 3.14157 \end{matrix} $

Etc.

eodnohj · Oct 27, 2011, 05:40 PM

your answers are wrong for one crutial fact that is the sine function in pure math is in radians and thus nsin(180/n) becomes nsin(Pi/n). as n->oo sin nsinPi/n aproaches n*Pi/n (by small angle theorem) so you get the trivial solution of Pi=Pi. However if you want a mathematically correcct expression for pi then you can employ the use of a taylor series fot arctangent and evaluate at 1, knowing that tan(pi/4)=1. This becomes the infinte series of sum(4*(-1)^(i+1)*factorial(2*i-2)/factorial(2*i-1), i = 1 .. infinity), which is an expression of pi, mathematically correct, however it does not converge that well. Your answer allthouge logically correct is mathematically insound due to the fact that degrees are not pure numbers.

ebaines · Oct 28, 2011, 06:10 AM

Quote:

Originally Posted by eodnohj

your answers are wrong for one crutial fact that is the sine function in pure math is in radians and thus nsin(180/n) becomes nsin(Pi/n). .... Your answer allthouge logically correct is mathematically insound due to the fact that degrees are not pure numbers.

There is nothing wrong with using degrees as the measure of an angle. Degrees and radians are both dimensionless numbers, so I don't understand your point. When someone writes "sin(180/n)" it's pretty clear they're talking about degrees, not radians, so there should be no confusion.

Pi by recursive decimal places

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