Euler integration formula is Taylor expansion using only 2 terms, good ? how is done?: taylor: x = a f(x) = f(a) + [f'(a)/1! ](x - a) + [f''(a)/2! ](x - a)^2 + ... so: if I call f like x x(t0) = x0 x(t0 + h) = x0 + x'(t0)(t0 + h - t0) = x0 + x'(t0)*h I don't know if I have done it good, if no, anyone can tell me how to deduce euler from taylor? thnx