I have something of the form

$$\frac{(A*G+C)(B*G)}{(G^2+junk4)(G + junk3)}$$

I know G is very large, so my largest term will look like:

$\frac{ABG^3}{G^4} = \frac{AB}{G}$

And my second largest term will look like $\frac{CB}{G^2}$

When expressions get long and complicated, it would be nice to have a function that can perform this calculation for me. Intuitively, I would think this type of simplification using G>>all other terms is pretty useful - so I figured there would be a built in function that can handle this. I haven't found anything so far. Should I try to write my own function to perform this behavior, or is there some built in tools that can do it for me?

  • $\begingroup$ How about the Limit as g-> Infinity? $\endgroup$ – bill s Feb 4 '19 at 23:55
  • $\begingroup$ Nah, in this case it would just return zero, since the denominator gets larger than the numerator. $\endgroup$ – Steven Sagona Feb 4 '19 at 23:58
expr = (A*G + C) (B*G)/((G^2 + junk4) (G + junk3));

For large G expand about Infinity

approx1 = Series[expr, {G, Infinity, 2}] // Normal

(* (A B)/G + (B C - A B junk3)/G^2 *)

For small junk3 expand about 0

approx2 = Series[approx1, {junk3, 0, 0}] // Normal // Expand

(* (B C)/G^2 + (A B)/G *)

Or in a single step

approx = Series[expr, {G, Infinity, 2}, {junk3, 0, 0}] // Normal

(* (B C)/G^2 + (A B)/G *)
| improve this answer | |
  • $\begingroup$ Expanding out at infinity is what I'm looking for. Thanks very much. $\endgroup$ – Steven Sagona Feb 5 '19 at 0:00

The accepted answer is what I'm looking for. But in case anyone ever might be interested, here's a more crude way that I did it if "expressionp" is the name of the expression to simplify.

  expression_] := (numerTerms = 
   CoefficientList[Expand[expression], G];
  polynum = Dimensions[numerTerms] - 1;
  final = numerTerms[[polynum + 1]]*G^polynum; 

SimplifyFraction[expression_] := 
   BiggestTerm[Denominator[Expressionp]] // Simplify

| improve this answer | |

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.