0
$\begingroup$

I'm studying power series / taylor series, but I'm having understanding how the different pieces relate. I have a few questions:

1) Both the power and Taylor series has a center, but what does c represent? (i.e. if at infinity, the polynomial approximation is very close to the original function, why not just set c=0, or remove the variable altogether)

2) How does c relate to interval of convergence if at all?

3) Why does adding n-th derivative makes the polynomial a more accurate representation of the original function? (i.e. why not subtract, or multiply?)

$\endgroup$
  • $\begingroup$ Important fact: the correct context for the power series is the complex analysis. $\endgroup$ – Martín-Blas Pérez Pinilla Mar 22 '17 at 15:39
0
$\begingroup$
  1. $c$ represents the point where you know the values $f(c)$ and the derivatives $f'(c), f''(c)$ etc. The more derivatives you know, the more terms you can add to your Taylor series.

  2. $c$ is the center of the interval of convergence.

  3. It does not always do that. You can have crazy functions for which the Taylor expansion of a smaller order is a more accurate approximation than some higher orders, although if you have sufficiently many derivatives, the Taylor polynomial does get closer to the function, only it need not do that at every single step.

$\endgroup$
  • $\begingroup$ the more terms you can add to your Taylor series. Is this because there are (n-1), where n is the exponent of x, inflection points in a graph, so more terms equal greater detail? c is the center of the interval of convergence. If ratio test for convergence is -1 < c < 1, does this mean the polynomial approximation can only be applied to that interval? $\endgroup$ – Daniel Reed Mar 22 '17 at 16:16
0
$\begingroup$

For (1) and (2): consider the power series of $f(z) = 1/z$ around $c\ne 0$. You can check easily that the radius of convergence is $|c| =$ distance from the center to the singularity.

(3) With the exception of the "crazy functions" of the other answer (i.e. if the function is regular enough) adding more terms means that the approximation will look more like the function locally. About "why not subtract, or multiply?": can be done. See Padé approximant (approximation by rational functions).

$\endgroup$

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.