Quote by quietrain
does it then mean i cannot taylor expand things like 1+kx? i have to expand only curves?

If you expand 1+kx around x=0, you are effectively defining f(x)=1+kx, and a=0.
So using your formula to expand it, you get:
f(x)=1+kx
f'(x)=k
f''(x)=0
f'''(x)=0
So:
f(x)=f(a)+f'(a)(xa)+1/2! f''(a)(xa)
^{2}+...
f(x)=f(0)+f'(0)(x0)+1/2! f''(0)(x0)
^{2}+...
f(x)=1+k(x0)+1/2! 0.(x0)
^{2}+ 0 + ...
f(x)=1+kx
Hey! But you already had that! :)