I have the taylor series $f(z)=f(x_0)+(x-x_0)f'(z)+1/2(x-x_0)^2f''(z) ...$

and I am told that "As a first order approximation," $x-x_0$ ~ $\frac{f(x)-f(x_0)}{f'(x_0)}$ assuming $f'(x_0) \neq 0$

I see how to solve for $x-x_0$ in the original by ignoring all but the first two terms of Taylor series, but can someone explain why this is a "first order approximation" and why this is how you arrive at it?


  • $\begingroup$ Are you just wondering why the approximation is termed "first order", or why it is an approximation at all? $\endgroup$ – mb7744 Jun 24 '14 at 1:49

It is a first order approximation because the polynomial used to approximate $ f(z) $ is first order (i.e. of degree 1). This is simply a name for the approximation, so when we say we want the second order approximation, we are looking for the Taylor series written to more terms.


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