0
$\begingroup$

I'd like to find the dominant terms of an expression given the conditions $x,y >> 1$:

$c_0 y+\frac{y^2 \left(4 c_0^2 y - 12 c_1 c_2^2+y^2\right)}{24 x^2 \left(\text{c1}^2+\text{c2}^2+y\right)}$

Is there an easy way to do this? I need the dominant terms for a given order O(x,y). It would be helpful to get the "small" contribution as a separate output. Thanks.

| improve this question | |
$\endgroup$
  • 1
    $\begingroup$ You can try Series[(*expression here*), {x,Infinity,0},{y,Infinity,0}] $\endgroup$ – QuantumDot Oct 5 '15 at 19:05
  • $\begingroup$ thanks! made an answer for future reference. $\endgroup$ – anon01 Oct 5 '15 at 19:21
  • $\begingroup$ You really require more information regarding relative sizes of x vs. y. $\endgroup$ – Daniel Lichtblau Oct 5 '15 at 20:30
  • $\begingroup$ ? I'm interested in your thoughts if you wish to clarify. I consider them to increase parametrically on the same scale. $\endgroup$ – anon01 Oct 5 '15 at 21:10
  • $\begingroup$ If they vary jointly e.g. y=k*x (or approximately equals) for some unspecified constant k!=0 then you can just replace y with k*x and take the Series. If they vary in an unspecified way e.g. y could be on the order of x^2 or x^(1/2) or Exp[x] then your asymptotic behavior will behave quite differently in different regions of the parametrized space. $\endgroup$ – Daniel Lichtblau Oct 6 '15 at 14:25
1
$\begingroup$

Convert both variables to the same, then evaluate as a series from infinity should work in most cases:

A straightforward example for some function f = f(x,y):

Series[f/.{x->q,y->q},{q,Infinity,2}]

Of course this requires that both variables grow parametrically at the same rate. This also allows you to take different limits: x->1/q for example if one of your variables limits toward zero. A thanks to QuantumDot.

| improve this answer | |
$\endgroup$

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.