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This is similar to the how-to-replace-variable-with-power question, but here it involves two variables with different power combinations.

This quesion arises from paper forward kinematics of the 6-6 Stewart platform. I’m trying to solve the forward kinematics problem with Mathematica. Mathematica really helped—I’ve nearly nailed it. However I’m stuck on the final key step.

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I cannot eliminate $c1$ $c2$ with power {{8,0},{7,1},{6,2},...}.

ClearAll[c1, c2] phi3A29= U1 + {U2j0, U2j1, U2j2}.Table[c1^(2 - j) c2^j, {j, 0, 2}] + {U3j0, U3j1, U3j2, U3j3, U3j4}.Table[c1^(4 - j) c2^j, {j, 0, 4}] + {U4j0, U4j1, U4j2, U4j3, U4j4, U4j5, U4j6}. Table[c1^(6 - j) c2^j, {j, 0, 6}] + {U5j0, U5j1, U5j2, U5j3, U5j4, U5j5, U5j6, U5j7, U5j8}. Table[c1^(8 - j) c2^j, {j, 0, 8}] ruleA32= Thread[Table[c1^(4 - j) c2^j, {j, 0, 4}] -> Table[Symbol["H" <> ToString[j] <> ToString[jj]], {j, 0, 4}, {jj, 1, 3}] . {c1^2, c1 c2, c2^2}/E0]; ruleA32// MatrixForm nu = phi3A29 //. ruleA32// Together // Numerator co = CoefficientRules[nu, {c1, c2}]; co // Keys

I supposed that $co$ should be $\{\{2,0\}, \{1,1\}, \{0,2\}, \{0,0\}\}$, which matches equation (34) in the paper, the order does not matter in $\{ \}$. I have a rough idea of what’s causing the issue, but I’d like to make better use of Mathematica to solve it. Could anyone help me out? Thanks in advance!

a rather simple (and maybe a bit clumsy) idea : manually write out a substitution rule for $c1^{8-j}c2^j$ using combinations of ${c1}^{4-j}{c2}^j$, $c1^2$ and $c2^2$.

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    $\begingroup$ Change Rule to a subtraction in ruleA32. Then do red = PolynomialReduce[phi3A29, ruleA32, c3, {c1, c2}, MonomialOrder -> EliminationOrder, CoefficientDomain -> RationalFunctions][[2]] $\endgroup$ Commented Jun 3 at 18:32

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PolynomialReduce[] seems made to order:

nu = Last[PolynomialReduce[phi3A29, Subtract @@@ ruleA32, {c1, c2}, MonomialOrder -> DegreeLexicographic]]; co = CoefficientRules[nu, {c1, c2}]; co // Keys (* {{2, 0}, {1, 1}, {0, 2}, {0, 0}} *)

An alternative is to use MonomialOrder -> DegreeReverseLexicographic, and it produces the same coefficient rules.

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As far as I understand your question, you want to replace all c1^i c2^j.. terms with i+j==4 by your rules.

For a start, replace works purely structural, it does not make any evaluation. E.g. consider:

c (c+1) /. c^2->a c (1 + c)

There is no c^2 to match. Now consider e.g. (For readability I replaced E[EmptyCircle] by E0):

phi3A29 //. ruleA32 ... (c2^2 (c1^2 H01 + c1 c2 H02 + c2^2 H03) U4j2)/E0 ...

Here we have the same problem as above. However, you may use "Expand":

phi3A29 //. ruleA32 // Expand ... (c1^2 c2^2 H01 U4j2)/E0 + (c1 c2^3 H02 U4j2)/E0 + (c2^4 H03 U4j2)/E0 ...

Now a second application of the rules will work:

... (phi3A29 //. ruleA32 // Expand ) //. ruleA32 (H01 (c1^2 H21 + c1 c2 H22 + c2^2 H23) U4j2)/E0^2 + (H02 (c1^2 H31 + c1 c2 H32 + c2^2 H33) U4j2)/E0^2 + (H03 (c1^2 H41 + c1 c2 H42 + c2^2 H43) U4j2)/E0^2 ...

Note the braces. We need this because /. has a higher precedence as //.... Finally we would then have:

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  • $\begingroup$ Thank you for your help — it might not quite solve the problem, but it reminded me of a rather simple (and maybe a bit clumsy) idea: I think I can manually write out a substitution rule for $c1^{8-j}c2^j$ using combinations of ${c1}^{4-j}{c2}^j$, $c1^2$ and $c2^2$. $\endgroup$ Commented Jun 3 at 13:13

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