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Topic: Is there a compact form for n-tuple relativistic additions of

Replies: 7   Last Post: Jun 4, 2010 6:08 PM

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Martin Cohen

Posts: 350
Registered: 12/4/04
Re: Is there a compact form for n-tuple relativistic additions of

Posted: May 31, 2010 10:00 AM
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On May 29, 12:00 pm, stargene <> 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.

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