hi everyone thanks for helping
but the taylor expansion was for x-kx
or perhaps only -kx was expanded?
but anyway it was given as x - kx f'(x) + ...
with regards to the above, how did you get that formula?
because i only know this f(x) = f(a) + f'(a)(x-a) + 1/2! f''(a)(x-a)2
as the general formula?
in any case, does it mean if i want to expand say 1 + kx
then it would be f(1+kx) = f(1) + kx f'(1) + ...
so f(1) is 1 + k(1) = 1+k?
then f'(1) is k? then where do i put my 1 since i don't have x
also does it mean f'' onwards are all 0?
YES! this is the one. so it was an expansion about the small value k ? and not x? i see.
but in particular, how did you get g'(k) and g'(k)?
is it through the chain rule differentiation?