hi everyone thanks for helping
but the taylor expansion was for xkx
or perhaps only kx was expanded?
but anyway it was given as x  kx f'(x) + ...
Quote by I like Serena
Hi quietrain! :)
Suppose you expand f(akx).
You would get: f(akx) = f(a)  kx f'(a) + (1/2) k^{2} x^{2} f''(a) + ...
Now replace "a" with "x"...
Apparently you've got a typo in the first term.

with regards to the above, how did you get that formula?
because i only know this f(x) = f(a) + f'(a)(xa) + 1/2! f''(a)(xa)
^{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?
Quote by Ray Vickson
Fix x, and let g(k) = f(xkx). Expand as g(k) = g(0) + k*g'(0) + (k^2/2)*g''(0) + ...
g(0) = f(x), g'(k) = x*f'(xkx) > g'(0) = x*f'(x), g''(k) = x^2*f''(xkx) > g''(0) = x^2*f''(x), etc.
RGV

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?
thanks!