> > 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? :-(