0
$\begingroup$

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?

Thanks!

$\endgroup$
  • $\begingroup$ 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
1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.