I want some help here. I want to define a function called Lg[z] that has the property that Lg[a/b] is always converted automatically to the form Lg[a] - Lg[b] , when a and b are polynomials of some variable say t.

I tried many things nothing works. I don't want to replace one by one by hand as it takes too long.


1 Answer 1


This could be a start, but it performs the transformation irrespective of whether a and b are polynomials in t or not.

Lg[HoldPattern[Times[a__]]] := Plus @@ Lg /@ {a};
Lg[HoldPattern[Power[a_, p_]]] := p  Lg[a];

Here a few usage examples:

Lg[a b]
Lg[a/ b]
Lg[Sum[t^k, {k, 0, 3}]/ Sum[k t^k, {k, 0, 3}]]

Lg[a] + Lg[b]

Lg[a] - Lg[b]

p Lg[a]

Lg[1 + t + t^2 + t^3] - Lg[t + 2 t^2 + 3 t^3]

One could make the patterns more restrictive with PolynomialQ, but maybe this already suffices for your task?

  • $\begingroup$ what's the advantage of using a HoldPattern? $\endgroup$
    – Roman
    Dec 26, 2019 at 12:11
  • 6
    $\begingroup$ @Roman: the pattern Times[a__] evaluates to a__ during the assignment in SetDelayed, so if you don't use HoldPattern, you won't get the correct downvalues. $\endgroup$ Dec 26, 2019 at 12:15
  • 1
    $\begingroup$ I see, thanks! A bit simpler would be Lg[a_*b_] = Lg[a] + Lg[b] and Lg[a_^p_] = p*Lg[a]. $\endgroup$
    – Roman
    Dec 26, 2019 at 12:34
  • 6
    $\begingroup$ @Roman - Henrik's approach is more efficient for more complicated arguments. Compare timings for Lg[a b c d e f g h i j k l m n o p q r s t u v w x y z] $\endgroup$
    – Bob Hanlon
    Dec 26, 2019 at 13:13
  • $\begingroup$ @BobHanlon yes Henrik's is much faster because mine relies on recursion. Cheers! $\endgroup$
    – Roman
    Dec 26, 2019 at 13:39

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