Can i make 1/(x^2) into -1/(((1- ((x-1)^2)+2x)))
and use f(x) = 1/(1-u) where u = ((x-1)^2)+2x))
and get -sigma : ((x-1)^2 + 2x))^n
??
Or is that illegal
How about 1/(x^2) = d/dx [-1/x]
but what is the taylor series for 1/x
Can i make 1/(x^2) into -1/(((1- ((x-1)^2)+2x)))
and use f(x) = 1/(1-u) where u = ((x-1)^2)+2x))
and get -sigma : ((x-1)^2 + 2x))^n
??
Or is that illegal
How about 1/(x^2) = d/dx [-1/x]
but what is the taylor series for 1/x
Does (which is what your exprssion reduces to) have the form of a Taylor Series expanded around x = 1 (your post title suggests that you want it to .....)?
Last edited by mr fantastic; Oct 8th 2008 at 07:19 PM.
Reason: Fixed the sigma
Can i make 1/(x^2) into -1/(((1- ((x-1)^2)+2x)))
and use f(x) = 1/(1-u) where u = ((x-1)^2)+2x))
and get -sigma : ((x-1)^2 + 2x))^n
??
Or is that illegal
How about 1/(x^2) = d/dx [-1/x]
but what is the taylor series for 1/x
You can just apply the definition of Taylor series.