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.