View Single Post
quietrain
#6
Apr2-12, 05:29 AM
P: 651
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) + ...


Quote Quote by I like Serena View Post
Hi quietrain! :)

Suppose you expand f(a-kx).

You would get: f(a-kx) = f(a) - kx f'(a) + (1/2) k2 x2 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)(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?




Quote Quote by Ray Vickson View Post
Fix x, and let g(k) = f(x-kx). Expand as g(k) = g(0) + k*g'(0) + (k^2/2)*g''(0) + ...
g(0) = f(x), g'(k) = -x*f'(x-kx) --> g'(0) = -x*f'(x), g''(k) = x^2*f''(x-kx) --> 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!