2
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We might take:

s = {(x-y)^2, x*y+z};
e = (x+y)^2 + 4*z;

and linearDecompose[e,s] could give {1,4} for example.

I know Solve with universal quantifiers over all free variables would do it:

Solve[ForAll[{x,y,z},s.{q1,q2} == e], {q1,q2}] 

But is this the "right" way to do it? Does it get inefficient in time and memory as the number of variables increases?

Is there a more specialized function for this purpose or will Solve work with high efficiency?

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  • 1
    $\begingroup$ In[19]:= SolveAlways[e == {c1, c2}.s, {x, y}] Out[19]= {{c1 -> 1, c2 -> 4}} So that's one way. Now SolveAlways is a bit dated and not the most efficient function in general, but it should do quite well when the underlying problem is linear, as this is. $\endgroup$ – Daniel Lichtblau May 9 at 15:18
2
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s = {(x - y)^2, x*y + z};
e = (x + y)^2 + 4*z;

You can also use PolynomialReduce:

{q, r} = PolynomialReduce[e, s, {x, y, z}]
 {{1, 4}, 0}
e == q.s + r // Expand
 True
ClearAll[linDecomp]
linDecomp = If[VectorQ[#, NumericQ] && #2 === 0, #, {}] & @@ 
    PolynomialReduce[##, Variables@#] &;

linDecomp[e, s]
 {1, 4}
linDecomp[e, s + (x + y)^3]
 {}
| improve this answer | |
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    $\begingroup$ This is good but...in a "bad" case, one might add e.g. (x+y)^3 to both polynomials in s. That would make for trouble. $\endgroup$ – Daniel Lichtblau May 9 at 15:20
  • $\begingroup$ why would that be? $\endgroup$ – user3257842 May 9 at 17:51
  • $\begingroup$ It would fail to give the desired result because no division would take place. $\endgroup$ – Daniel Lichtblau May 9 at 21:48

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