METH@usc-isi.arpa (12/26/85)
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)
I am FASCINATED by the fact that this equality holds. What
is special about these two angles to give such an EXACT result.
Other examples of these unexpected rational results of
combinations of functions with irrational solutions abound, such
as
cos(20degrees)*cos(40degrees)*cos(80degrees) = 1/8
Does anyone have mathematical insight as to why these things
are true?
-Sheldon Meth (METH@ISIA.ARPA)
-------ags@pucc-h (Dave Seaman) (12/27/85)
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
steve@anasazi.UUCP (Steve Villee) (12/30/85)
> pi/4 = 4*arctan(1/5) - arctan(1/239) > > I am FASCINATED by the fact that this equality holds. What > is special about these two angles to give such an EXACT result. > > cos(20degrees)*cos(40degrees)*cos(80degrees) = 1/8 > > Does anyone have mathematical insight as to why these things > are true? From the definitions, cos(x) = (e**(x*i) + e**(-x*i)) / 2 sin(x) = (e**(x*i) - e**(-x*i)) / (2*i) tan(x) = sin(x)/cos(x) we can derive cos(-x) = cos(x), sin(-x) = - sin(x), tan(-x) = - tan(x) cos(x+y) = cos(x)*cos(y) - sin(x)*sin(y) sin(x+y) = cos(x)*sin(y) - cos(y)*sin(x) tan(x+y) = (tan(x) + tan(y)) / (1 - tan(x)*tan(y)) Now let a = arctan(1/5), b = arctan(1/239). tan(a) = 1/5 tan(2*a) = 5/12 tan(4*a) = 120/119 tan(b) = 1/239 tan(4*a - b) = 1 4*a - b = pi/4, since a and b are between 0 and pi/4. For the other one, let r = e**(2*pi*i/9). Then r is a primitive ninth root of 1 (that is, r**9 = 1, but this is not true for any smaller positive exponent). cos(20degrees) = - cos(160degrees) = - (r**4 + r**5) / 2 cos(40degrees) = (r + r**8) / 2 cos(80degrees) = (r**2 + r**7) / 2 Expanded out, the product of the three is r**7 + r**8 + r**5 + r**6 + r**3 + r**4 + r + r**2 - -------------------------------------------------- 8 The numerator is just the sum of all the ninth roots of 1 except 1 itself. The sum of all nine roots is 0 (the r**8 term in "r**9 - 1" is 0), so the numerator must be -1. So the product is 1/8. There is probably an easier way to show this, but I have a particular fascination for roots of 1. --- Steve Villee (ihnp4!terak!anasazi!steve) International Anasazi, Inc. 7500 North Dreamy Draw Drive, Suite 120 Phoenix, Arizona 85020 (602) 870-3330