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From: ags@pucc-h (Dave Seaman)
Newsgroups: net.micro
Subject: Re: Series for Pi and Other Trig Oddities
Message-ID: <2537@pucc-h>
Date: Fri, 27-Dec-85 12:34:09 EST
Article-I.D.: pucc-h.2537
Posted: Fri Dec 27 12:34:09 1985
Date-Received: Sat, 28-Dec-85 01:51:18 EST
References: <1043@brl-tgr.ARPA>
Reply-To: ags@pucc-h.UUCP (Dave Seaman)
Organization: Purdue University Computing Center
Lines: 46

In article <1043@brl-tgr.ARPA> METH@usc-isi.arpa writes:
>
>     My  formula  for  pi,  while  correct,  was  pointed  out  by 
>LINDSAY@TL-20B.ARPA to be very, very slowly converging.
>
>     I have seen on the net other,  more quickly converging series 
>such as:
>
>     pi/4 = 4*arctan(1/5) - arctan(1/239)
>
>     Does  anyone have mathematical insight as to why these things 
>are true?

Let x = arctan(1/5).  Then tan x = 1/5.

	tan 2x = (2 * tan x) / (1 - (tan x)^2)
	       = (2 * 1/5) / (1 - (1/5)^2)
	       = (2/5) / (24/25)
	       = 5/12

	tan 4x = (2 * tan 2x) / (1 - (tan 2x)^2)
	       = (2 * 5/12) / (1 - (5/12)^2)
	       = (5/6) / (119/144)
	       = 120/119.

Since tan (pi/4) = 1, we have

	tan (4x - pi/4) = (tan 4x - tan(pi/4)) / (1 + (tan 4x) * (tan (pi/4)))
			= (120/119 - 1) / (1 + (120/119) * 1)
			= (1/119) / (239/119)
			= 1/239

Since everything is in the first quadrant:

	4x - pi/4       = arctan(1/239)

	pi/4		= 4x - arctan(1/239)
			= 4*arctan(1/5) - arctan(1/239)

You can easily derive other series for pi by playing with the tangent formula.
The tangent series converges very rapidly for small arguments and very slowly
for arguments close to 1.  In this case the arctan(1/5) series is slow to
converge, compared to arctan(1/239), so it would be advantageous to look
for identities with smaller arguments to the arctan series.
-- 
Dave Seaman	  					pur-ee!pucc-h!ags