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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: Jun 4, 2010 12:39 PM
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On Jun 4, 12:47 am, stargene <> 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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