On May 29, 12:00 pm, stargene <starg...@sbcglobal.net> wrote: > I seek advice on whether a certain very long function can be made > significantly more compact and therefore easier to compute. > > I am interested in calculating the value of a function which resembles > both a continued fraction and an iterated function, where the basic > unit of the iteration is the right hand side of the formula for > relativistic > addition of two (collinear) velocities: > > (1) W_1 = (u + v) / (1+uv) . > > u and v are the summed velocities, given as decimal fractions of the > velocity of light C. Ie: u,v range from 0.0 to 1.0 . Their resultant > velocity is W_n. > > The essential idea is to divide C into n <equal> velocities (v) and > then > add them sequentially. My procedure is to take the first sum, W, and > then add to it a third identical velocity v, which gives a second > resultant > velocity > > (2) W_2 = (W_1 + v) / (1 + vW_1) . > > This is an iteration of the basic form in (1). > > Similarly, a fourth identical velocity v is added to W_2, giving > > (3) W_3 = (W_2 + v) / (1+ vW_2) , > > and so on and on... > > This is to be repeated n times, where n becomes very large and > v commensurately smaller. I wish to know the values of this function > as n becomes extremely large. I especially want to know what the > function converges to as n ---> infinity and C/n ---> zero velocity. > > A partial glimpse of this procedure as a single large relation > resembles > an unusual variety of continued fraction. I omit rendering it in ASC > II > form here since it rapidly becomes very confusing and opaque. > > Any feedback will be very appreciated. > > Thanks, > Gene
iirc, the addition formula for the inverse hyperbolic tangent is the same as the addition formula for velocities in SR. i.e., atanh(x)+atanh(y) = atanh((x+y)/(1+xy)). Take tanh of both sides and have fun.