1
$\begingroup$

If I have an equation given by

$$\displaystyle Y = \frac{a^2}{d^2}\frac{(1-c^2\frac{c}{a})}{(1-b^2)}$$

and I expand $a,b,c,d$ in a Taylor series, where $a$ is truncated at the $A^{th}$ order, $b$ is truncated at the $B^{th}$ order, $c$ is truncated at the $C^{th}$ order, and $d$ is truncated at the $D^{th}$ order, how do I go about determining the order of the truncation error for the full equation $ Y$?

$\endgroup$
1
$\begingroup$

Given two truncated Laurent series, $f = p(x)+O(x^p)$, $g = q(x)+O(x^q)$, the orders will be $$ f+g = p(x) + q(x) + O(x^{\min(p,q)}), $$ $$ fg = p(x) q(x) + O(x^{\min(p+q_0, q+p_0)}), $$ $$ 1/g = 1/q(x) + O(x^{q-2q_0}). $$ where $p_0$ and $q_0$ are the degrees (possibly negative) of the first nonzero monomials in $p$ and $q$. So for each subexpression it is necessary to determine the leading term (which is possibly non-constant, and might be singular), then it is straightforward algebra.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ I follow you with the addition and multiplication, but I'm a bit confused on 1/g. Where does the 2 come from? Also...how does raising a term to a power influence the truncation error? Following your example, if it's $f^2$ does the truncation error become $O(x^{2p})$ or $O(x^{2+p})$? Could you possibly cater your answer to directly reflect the equation I presented? Given 4 truncated Laurent series... $\endgroup$ – ThatsRightJack Jul 7 '16 at 23:21
  • $\begingroup$ You've left me hanging...just thought I'd try to get your attention to see if you're still available to offer further insight. $\endgroup$ – ThatsRightJack Jul 19 '16 at 23:00
  • $\begingroup$ could you possibly elaborate more based on my comment above? $\endgroup$ – ThatsRightJack Aug 4 '16 at 20:05

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.