Could you please show us your n+m factor here: Take this function without any abs or if-then-else:
f(x) = sqrt(x^2)
Or as a parse tree:
f(x) = "sqrt" | "^2" | x
Whats the derivative at x=0? Is it dividable by (n+m)? What result do you get?
Am Samstag, 11. November 2017 14:43:48 UTC+1 schrieb John Gabriel: > This moron claims to be studying for a PhD!!! BWaaaa haaaa haaaaa > > > If you are his teacher, you need to be ashamed of yourself for producing such a clueless idiot!