[psi]z being (a/z)([phi](z+1)/[phi]z); of which observe that it diminishes without limit as z increases without limit. Accordingly, we have [psi]z = a/z+ [psi](z+1) = a/z+ a/(z+1)+ [psi](z+2) = a/z+ a/(z+1)+ a/(z+2)+ [psi](z+3), etc. And, [psi](z + n) diminishing without limit, we have a/z . [phi](z+1)/[phi]z = (a/z+) (a/(z+1)+) (a/(z+2)+) (a/((z+3)+ ...)) Let z = 1/2; and let 4a = -x^2. Then (a/z)[phi](z+1) is -(x^2/2) ( 1 - x^2/(2.3) + x^4/(2.3.4.5...)) or -(x/2) sin x. Again [phi]z is 1 - x^2/2 + x^4/(2.3.4) or cos x: and the continued fraction is (1/4)x^2/(1/2)+ (1/4)x^2/(3/2)+ (1/4)x^2/(5/2)+ ... or -x/2 x/1+ -x^2/3+ -x^2/5+ ... {371} whence tan x = x/1+ -x^2/3+ -x^2/5+ -x^2/7+ ... Or, as written in the usual way, tan x = x ------- 1 - x^2 ------- 3 - x^2 ------- 5 - x^2 ------- 7 - ... This result may be proved in various ways: it may also be verified by calculation. To do this, remember that if a_1/b_1+ a_2/b_2+ a_3/b_3+ ... a_n/b_n = P_n/Q_n; then P_1=a_1, P_2=b_2 P_1, P_3=b_3 P_2+a_3 P_1, P_4=b_4 P_3+a_4 P_2, etc. Q_1=b_1, Q_2=b_2 Q_1+a_2, Q_3=b_3 Q_2+a_3 Q_1, Q_4=b_4 Q_3+a_4 Q_2, etc. in the case before us we have a_1=x, a_2=-x^2, a_3=-x^2, a_4=-x^2, a_5=-x^2, etc. b_1=1, b_2=3, b_3=5, b_4=7, b_5=9, etc. P_1=x Q_1=1 P_2=3x Q_2=3-x^2 P_3=15x-x^3 Q_3=15-6x^2 P_4=105x-10x^3 Q_4=105-45x^2+x^4 P_5=945x-105x^3+x^5 Q_5=945-420x^2+15x^4 P_6=10395x-1260x^3+21x^5 Q_6=10395-4725x^2+210x^4-x^6 We can use this algebraically, or arithmetically. If we divide P_n by Q_n, we shall find a series agreeing with the known series for tan x, _as far as_ n _terms_. That series is x + x^3/3 + 2x^5/15 + 17x^7/315 + 62x^9/2835 + ... {372} Take P_5, and divide it by Q_5 in the common way, and the first five terms will be as here written. Now take _x_ = .1, which means that the angle is to be one tenth of the actual unit, or, in degrees 5 deg..729578. We find that when x = .1, P_6 = 1038.24021, Q_6 = 10347.770999; whence P_6 divided by Q_6 gives .1003346711. Now 5 deg..729578 is 5 deg.43'46-1/2"; and from the old tables of Rheticus[675]--no modern tables carry the tangents so far--the tangent of this angle is .1003347670. Now let x = (1/4)[pi]; in which case tan x = 1. If (1/4)[pi] be commensurable with the unit, let it be (m/n), m and n being integers: we know that (1/4)[pi] < 1. We have then 1=(m/n)/1- (m^2/n^2)/3- (m^2/n^2)/5- ... = m/n- m^2/3n- m^2/5n- m^2/7n- ...