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#12
Apr2-12, 08:20 AM
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Quote Quote by quietrain View Post
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)(x-a)+1/2! f''(a)(x-a)2+...
f(x)=f(0)+f'(0)(x-0)+1/2! f''(0)(x-0)2+...
f(x)=1+k(x-0)+1/2! 0.(x-0)2+ 0 + ...
f(x)=1+kx

Hey! But you already had that! :)