On 9/28/2013 4:24 PM, Richard Tobin wrote: > In article <9cSdnYGhH7CMpNrPnZ2dnUVZ_sCdnZ2d@megapath.net>, > Hetware <email@example.com> wrote: > >> So the answer is consensus among mathematicians holds that F(t) = (t^2 - >> 9)/(t - 3) is undefined at t=3? > > Yes. > >> Perhaps what I should have said at the >> outset is something along the lines of: on any given day, if I'm setting >> up an equation in physics, and produce an expression such as F(t) = (t^2 >> - 9)/(t - 3), I treat it as t+3, and do not expect any adverse >> consequence from doing so. > > Your simplification is not valid for t=3. If there is a real > physical interpretation, perhaps you can derive the formula t+3 > without going through the intermediate form (t^2-9)/(t-3). Or > consider the special case t=3 to show that the result is indeed > t+3 in that case too. > > In fact, I would be interested to see a physical problem where you > can't do that. > > -- Richard >
I believe most mathematicians solving for x as a function of t given
t^2 - 9 = x (t - 3)
would not hesitate to factor the left hand side and divide both sides by t - 3 without treating t = 3 as a special case. Doing so repeats the sin of dividing by zero twice. We can certainly solve
t^2 - 9 = 6 (t - 3)
without dividing by zero which seems to justify our implied sin.