2
$\begingroup$

Given that $x$ is small (closer to zero), what should be the expression $(a x^{-2} + b x^{-4})^{-1}$ approximated with? Is it $a^{-1} x^2$ or $b^{-1} x^4$ or something else? Here, $a$ and $b$ are positive constants.

$\endgroup$
3
  • 3
    $\begingroup$ For $x$ near $0,$ the term $bx^{-4}$ is much larger in magnitude (i.e. in absolute value) than $ax^{-2},$ so when adding the two, one ignores $ax^{-2}.$ So to a first approximation we have $(bx^{-4})^{-1} = b^{-1}x^4.$ $\endgroup$ Jan 15, 2022 at 19:58
  • $\begingroup$ But for small $x$, doesn't it seem that $a^{-1}x^2$ will be dominant? $\endgroup$
    – User101
    Jan 15, 2022 at 21:18
  • 1
    $\begingroup$ Note that small + large is large, so we have $\frac{1}{\text{small + large}} \approx \frac{1}{\text{large}},$ which is small (what we know is the case), whereas what you suggest would have us using $\frac{1}{\text{small + large}} \approx \frac{1}{\text{small}},$ which is large (what we know is NOT the case). This is meant to be descriptive, so not all the small's are necessarily equal and not all the large's are necessarily equal. Also, $(U + V)^{-1} \neq U^{-1} + V^{-1},$ which you'll be using at some point if you tried to write everything out explicitly to obtain what you're thinking. $\endgroup$ Jan 15, 2022 at 21:57

1 Answer 1

2
$\begingroup$

So you have $$(ax^{-2}+bx^{-4})^{-1} = \dfrac{x^4}{ax^2+b}\approx \dfrac{x^4}{b}\left(1-\dfrac{a}{b}x^2+O(x^4)\right)=O(x^4)$$ by Taylor's approximation.

$\endgroup$
1
  • $\begingroup$ But for small $x$, doesn't it seem that $a^{-1}x^2$ will be dominant? $\endgroup$
    – User101
    Jan 15, 2022 at 21:18

Not the answer you're looking for? Browse other questions tagged .