On Jun 2, 8:50 pm, mjc <mjco...@acm.org> wrote: > On Jun 1, 10:30 am, stargene <starg...@sbcglobal.net> wrote: > > > > > > > > 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. > > > Thanks for your response. > > > Indeed, I looked it up and you are right. However, as I am > > often preternaturally dense, I don't understand how knowing > > that makes my task of adding, say, n = 10^6 identical little > > velocities, where each one is v = Co / 10^6, (identically one > > millionth the velocity of light) any easier without steam > > shooting out of my ears. > > > Eg: How would taking the tanh of both sides of your relation > > help me, and how does that take into account the totality of > > all my n= 10^6 additions? And when n = 10^9 or more? :-( > > > Can you please enlighten me? > > Gene > > Let T(a) = tanh(a) and AT(a) = atanh(a), for brevity. > > If we write vsum(x, y) = (x+y)/(1+xy), > then, if x = T(u) and y = T(v), > vsum(x, y) = (T(u)+T(v))/(a+T(u)T(v)) > so AT(vsum(x, y)) = AT((T(u)+T(v))/(a+T(u)T(v)) = AT(T(u)) + AT(T(v)) > = u + v = AT(x) + AT(y). > To get the vsum of 3 variables, > AT(vsum(vsum(x, y), z)) = AT(vsum(x, y)) + AT(z) = AT(x)+AT(y)+AT(z). > Note that this shows that we can write vsum(x, y, z) since the result > holds for and permutation of x, y, and z. > Continuing, by induction, AT(vsum(x, y, z, a, b, ...)) = AT(x)+AT(y) > + .... > Setting all of the variables equal, if there are n of then, > AT(vsum(n copies of x)) = n*AT(x) > or vsum(n copies of x) = T(n*AT(x)). > > We now need to look at T(x) for large values of x (= n*AT(y) for large > n). > Writing E(x) = exp(x), > T(x) = (E(x)-E(-x))/(E(x)+E(-x)) = (E(2x)-1)/(E(2x)+1) > = (E(2x)+1-2)/(E(2x)+1) = 1 - 2/(E(2x)+1) > ~ 1 - 2/E(2x) for large x. ("~" means "is close to") > > Putting this in the above, > vsum(n copies of x) = T(n*AT(x)) ~ 1 - 2/E(2*n*AT(x)). > > You can rewrite this in various ways. One is > vsum(n copies of x) = 1-2/(2*AT(x))^n, showing that > (1-vsum(n copies of x))/(1-vsum(n-1 copies of x)) ~ 1/(2*AT(x)). > This shows how fast the SR sum of n copies of x approaches 1.- Hide quoted text - > > - Show quoted text -
vsum(n copies of x) = 1-2/E(2*AT(x))^n (I lost the E)
A small addition:
If x is small, both T(x) and AT(x) are about x. so vsum(n copies of x) ~ 1-2/E(2*x)^n ~ ~ 1-2/(1+2*x)^n Since E(x) + 1+x for small x.