intermediate steps of simplification of algebraic expression without evaluation

Question

I am using MMA do do some algebra and I need to do some simplification. For example, I have the following expression $$ x^2 \left( \frac{6 \left(2 c_1 x^6+c_2\right)}{x^4} \right)-x \left( 4 c_1 x^3-\frac{2 c_2}{x^3}\right)-8\left(c_1 x^4+\frac{c_2}{x^2}\right ) $$ Can I get intermediate steps to reach the final answer? For example, I want to factor out $c_1$ and $c_2$ and write the expression as $c_1 (x\dots) + c_2 (x\dots)$ but the terms $x\dots $ need not be evaluated. I can to the calculation as it is simple, but if I could do this from MMA, it would greatly save my time.

I always have some terms that could be factored out. It could be constant $c_k$ or polynomial $x^n$ or trig functions and there could be multiple terms but they are always unique and do not conflict with each other.

Answer 1

You have your expression

expression = 
 x^2 (6 (2 c1 x^6 + c2)/x^4) - x (4 c1 x^3 - 2 c2/x^3) - 
  8 (c1 x^4 + c2/x^2)
(* -x (-((2 c2)/x^3) + 4 c1 x^3) - 8 (c2/x^2 + c1 x^4) + (
 6 (c2 + 2 c1 x^6))/x^2 *)

for now it hasn't been evaluated to zero yet, but if you try most anything it will do so:

Expand@expression
Series[expression, {c1, 0, 2}]
Coefficient[expression, c2]
(* 0 *)
(* O[c1]^3 *)
(* 0 *)

One way to get what you are looking for is to change the Head of your expression from Plus to List,

List @@ expression
(* {-x (-((2 c2)/x^3) + 4 c1 x^3), -8 (c2/x^2 + c1 x^4), (
 6 (c2 + 2 c1 x^6))/x^2} *)

Now you can get the coefficients for c1 and c2 for each of the terms in the sum,

Coefficient[#, c1] & /@ (List @@ expression)
(* {-4 x^4, -8 x^4, 12 x^4} *)

Coefficient[#, c2] & /@ (List @@ expression)
(* {2/x^2, -(8/x^2), 6/x^2} *)

I don't know of a general way to do what you want, but this may work with some input from you.

Answer 2

I generalize Jason B's answer to group by like terms in any number of coefficients, as well as in cases where you have polynomials multiplying one another:

It sounds like you mostly want Plus to keep from evaluating. You can do this by replacing Plus with plus, where plus is a Flat function (since we might as well keep associativity of addition)

SetAttributes[plus, Flat];

expr = (a + b c) (a - b c);
expr /. Plus->plus

plus[a, -b c] plus[a, b c]

Now make plus distrbutive (including over itself -- i.e., when it appears in an integer power) :

sep = % //. {  a_ b_plus :> (a*# &) /@ b,  Power[a_plus, n_ /; n > 0 && IntegerQ@n] :> (Power[a, n - 1]*# &) /@ a  }

plus[a^2, -a b c, a b c, -b^2 c^2]

Now say you want to sort the output based on powers of a, b, and c. To this end, I define a getPower function :

(* operator form *)
getPower[var_][expr_] := getPower[expr, var]

(* Thread over a list of variables *)
getPower[expr_, varList_List] := getPower[expr, #] & /@ varList

getPower[expr_, var_] /; FreeQ[expr, var] := 0
getPower[expr_Times, var_] := Plus @@ getPower[var] /@ List @@ expr
getPower[(expr_)^n_, var_] := n getPower[expr, var]
getPower[var_, var_] := 1

(Note that getPower requires expr to be a monomial in the supplied variables; otherwise the output will contain getPower. Furthermore, note that getPower[ 1/(a+x), x] returns 0, but getPower[ 1/x, x] returns -1. )

Then you can use GatherBy to sort the output by coefficient:

GatherBy[List @@ sep, getPower@{a, b, c}]

{{a^2}, {-a b c, a b c}, {-b^2 c^2}}

It might be desirable to convert the sublists back to plus :

plus@@@%

{plus[a^2], plus[-a b c, a b c], plus[-b^2 c^2]}

You can imagine generalizing the above to pull the coefficients out front.

Note that it may or may not be useful to have given plus attribute OneIdentity from the start, depending on how you'd like to generalize.