Solving Pi

[John Gabriel]

I am entering this conversation not having seen previous correspondence,
but here is a trick from my high school days (in England during WWII).

arctan(x+h) = arctan(x) + arctan(h/(1 + h*x + x**2))

Therefore arctan(1) = arctan(1/2) + arctan(1/3)
	= arctan(1/4) + arctan(2/9) + arctan(1/6) + arctan(3/19)

and so on etc. If we stop at the above 4 way split, the error after N
terms is roughly 1 bit in 2*N. Thus for a 64 bit floating point number
with a 56 bit fractional part we need only 28 terms. A further split
reduces the problem to 8 series with the worst being a power series
in 1/8. Algebraic ingeneuity can combine terms so that your may not be
doing quite as much work as summing all the series.

After 8 terms you lose as much as you gain from further splits.

Better solutions for use in subroutine libraries exist and are given in

"A Software Manual for the Elementary Functions" by W.J. Cody and W.M.
Waite, Prentice Hall.

However if you want to compute Pi to a few thousand decimal places other
considerations arise like Do you want to use rational fraction
arithmetic?, should you be using methods like continued fraction
expansions?, and on and on.

I seem to recollect results sort of like

	pi/10 = arctan(r)

where r was some reasonable rational fraction, but I can't for the life
of me remember how it was obtained. There are of course the 
results from the fact that a polygon with 2**n + 1 sides can be solved
using nothing worse than square roots, and so once you have that
solution you also have a rapidly convergent series, e.g. for
arctan(2*pi/17). But you still must compute the square roots. On the
whole methods using repeated fractions are probably best. For details
about that, get in touch with Henry Thacher c/o Dept. of Computer Science,
U. of Kentucky, Lexington KY. by snail mail or telephone. Henry has
retired and lives somewhere in CA now, but the Dept should still know
where to find him.
				John Gabriel