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.
$\displaystyle f(x)=\frac{1}{x^2}\implies f(1)=1$
$\displaystyle f'(x)=-\frac{2}{x^3}\implies f'(1)=-2$
$\displaystyle f''(x)=\frac{6}{x^4}\implies f''(1)=6$
$\displaystyle f'''(x)=-\frac{24}{x^5}\implies f'''(1)=-24$
$\displaystyle f^{(4)}(x)=\frac{120}{x^6}\implies f^{(4)}(1)=120$
$\displaystyle \therefore \frac{1}{x^2}=1-2(x-1)+\frac{6(x-1)^2}{2!}-\frac{24(x-1)^3}{3!}+\frac{120(x-1)^4}{4!}+\dots$ $\displaystyle =1-2(x-1)+3(x-1)^2-4(x-1)^3+5(x-1)^4+\dots+(-1)^n(n+1)(x-1)^n$
$\displaystyle \therefore \sum_{k=0}^{\infty}(-1)^k(k+1)(x-1)^{k}=\frac{1}{x^2}$ at the point $\displaystyle x=1$.
Does this make sense?