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Consider the following example expression dependent on parameter t:

expr= t a ((t b+t c+ d )(t e+ f)+t g (t k+ l)+m (t n+ p));

This is a very short expression, which stands as an example for more complicated ones that are much larger, more nested and of higher overall degree in t.

Now I would like to take the limit:

Limit[expr/t^3,t->Infinity]

However, instead of doing it using this function, I would like to manipulate expr/t^3 such as to cancel most t parameters without opening any brackets, so that we get:

 expr2=myManipulate[expr/t^3]

a (( b+ c+ d/t )( e+ f/t)+ g ( k+ l/t)+m (n+ p/t)/t)

Now we can see that taking the limit amounts to just setting t->Infinity:

expr2/.t->Infinity

a ( e ( b+ c )+ g k)

We did not have to open any brackets and did not have to use Limit function. Is there a way to write an efficient myManipulate function that does this for arbitrarily more complicated (larger and more nested) input? Thanks for any suggestion!

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You should recursively reshape the expression.

Rules below work with your example, though I expect they wouldn't with a more complicated one. You'd probably want to divide innermost polynomial by its order, not just linear power y.

divide[x_ /; Depth[x] <= 3] := Expand[x]

divide[Times[x_Plus, y : _^_?Negative]] := Expand[x y]

divide[Times[x__Plus, y_^p_?Negative]] :=
 Module[{n, k},
  n = Length[{x}];
  k = Min[n, -p];
  (Times @@ Join[
      Table[divide[z/y], {z, Take[{x}, k]}],
      Drop[{x}, k]])/y^Max[-p - n, 0]]

divide[Times[x_, y_Plus, z : _^_?Negative]] := 
 Times[x, divide[z #] & /@ y]

Example

e1 = t a ((t b + t c + d) (t e + f) + t g (t k + l) + m (t n + p));

e2 = divide[e1/t^3]

a ((b + c + d/t) (e + f/t) + g (k + l/t) + m (p/t^2 + n/t))

e2 /. t -> Infinity

a ((b + c) e + g k)

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  • $\begingroup$ This looks useful! I am not sure I understand all the syntax. Does it specialize on cases when two things (like your x and y) are multiplied and would require a further rule for three things or more? $\endgroup$ – Kagaratsch Aug 12 '16 at 13:24
  • $\begingroup$ @Kagaratsch Check ?divide, rules are tried in that order. First rule matched, first fired. You should think about expressions in their FullForm. $\endgroup$ – BoLe Aug 12 '16 at 13:33

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