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Re: Is there a compact form for n-tuple relativistic additions of
velocities?
Posted:
Jun 4, 2010 12:39 PM
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On Jun 4, 12:47 am, stargene <starg...@sbcglobal.net> wrote:
> Hi, and several googols of thanks for your help. When I tried your
>
> vsum(n copies of x) = 1-2/(2*AT(x))^n
>
> it did not work for me, probably due to a misunderstanding on my
> part. But your
>
> vsum(n copies of x) = T(n*AT(x))
>
> worked perfectly on my Haxial calculator, reproducing results
> identical to my own tedious calculations, eg: with n = 5, 10 and 50,
> using recursive versions of SR's original relation. Using your
> relation and pushing n to 10^7 and then 10^9, it also shows that
> vsum(n copies of x) converges quickly to
>
> v = .761594155... Co,
>
> instead of Co itself. This is unexpected, though I already knew
> that for n = 2, 3, 5 and 10 (with v = Co/2 , Co/3 , Co/5 and Co/10 ),
> the resultant velocities <decreased>, ie:
>
> 0.8 , 0.777 , 0..7672 and 0.76299 times Co ,
>
> respectively (where Co = unity). This bothered me, especially since
> initially it seemed conceivable that the sum might even converge to
> 0.0 as n --> infinity and v --> 0.0 Co! Nevertheless, the actual con-
> vergence is still counter-intuitive, having expected the sum to rise
> eventually to Co, as I'm guessing you did too.
>
> Interesting...though what it might mean physically is anybody's guess,
> without a ouiji board and Prof. Einstein.
Take a look at my previous correction.
The correct formula is vsum(n copies of x) = 1-2/E(2*AT(x))^n.
This DOES approach 1 for large n.
If x is small, this is about 1-2/(1+2*x)^n.
Sorry about the mistake.
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