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[?] Solve powerless polynomial equations?

永康杨

永康杨, Wuhan Institude of Technology

Posted 6 years ago

Hello everyone. Not much to say, paste the Wolfram Mathematica code Solve[FullSimplify@ TrigExpand[({{2, 0}, {1, 1}, {1, 0}, {0, 2}, {0, 1}} /. CoefficientRules[((x^2/a^2 + y^2/b^2 - 1) /. Solve[Dot[{{Cos[\[Theta]], -Sin[\[Theta]]}, {Sin[\[Theta]], Cos[\[Theta]]}}, {x + x0, y + y0}] == {x1, y1}, {x, y}])[[1]], {x1, y1}]) == (x0^2/a^2 + y0^2/b^2 - 1) {a1, b1, c1, d1, e1} /. {\[Theta] -> 2 ArcTan[u]}], {a, b, x0, y0, u}] Unfortunately, Wolfram took a long time and didn't solve it (even I used the meta-form) I have to use the maple to solve the problem, the maple is very good. Maple code solve([-(4a^2u^2+b^2(u^2-1)^2)/((u^2+1)^2(-b^2x0^2+a^2(b+y0)(b-y0))) = a1, -(4(a-b))(a+b)u(u^2-1)/((u^2+1)^2(-b^2x0^2+a^2(b+y0)(b-y0))) = b1, c1+(2b^2(u^2-1)x0+4a^2uy0)/((u^2+1)(-b^2x0^2+a^2(b+y0)(b-y0))) = 0, -(4b^2u^2+a^2(u^2-1)^2)/((u^2+1)^2(-b^2x0^2+a^2(b+y0)(b-y0))) = d1, (4b^2ux0-2a^2(u^2-1)y0)/((u^2+1)(-b^2x0^2+a^2(b+y0)(b-y0))) = e1], [a, b, x0, y0, u])

Hello everyone. Not much to say, paste the Wolfram Mathematica code

Solve[FullSimplify@
  TrigExpand[({{2, 0}, {1, 1}, {1, 0}, {0, 2}, {0, 1}} /. 
       CoefficientRules[((x^2/a^2 + y^2/b^2 - 1) /. 
           Solve[Dot[{{Cos[\[Theta]], -Sin[\[Theta]]}, {Sin[\[Theta]],
                 Cos[\[Theta]]}}, {x + x0, y + y0}] == {x1, y1}, {x, 
             y}])[[1]], {x1, y1}]) == (x0^2/a^2 + y0^2/b^2 - 1) {a1, 
       b1, c1, d1, e1} /. {\[Theta] -> 2 ArcTan[u]}], {a, b, x0, y0, 
  u}]

Unfortunately, Wolfram took a long time and didn't solve it (even I used the meta-form) I have to use the maple to solve the problem, the maple is very good. Maple code

solve([-(4*a^2*u^2+b^2*(u^2-1)^2)/((u^2+1)^2*(-b^2*x0^2+a^2*(b+y0)*(b-y0))) = a1, -(4*(a-b))*(a+b)*u*(u^2-1)/((u^2+1)^2*(-b^2*x0^2+a^2*(b+y0)*(b-y0))) = b1, c1+(2*b^2*(u^2-1)*x0+4*a^2*u*y0)/((u^2+1)*(-b^2*x0^2+a^2*(b+y0)*(b-y0))) = 0, -(4*b^2*u^2+a^2*(u^2-1)^2)/((u^2+1)^2*(-b^2*x0^2+a^2*(b+y0)*(b-y0))) = d1, (4*b^2*u*x0-2*a^2*(u^2-1)*y0)/((u^2+1)*(-b^2*x0^2+a^2*(b+y0)*(b-y0))) = e1], [a, b, x0, y0, u])

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POSTED BY: 永康杨

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Hans Dolhaine

Hans Dolhaine, retired

Posted 6 years ago

Are you sure that Maple gives you a correct answer? I tried to figure out what you are doing and with eq = Simplify /@ Thread[TrigExpand[({{2, 0}, {1, 1}, {1, 0}, {0, 2}, {0, 1}} /. CoefficientRules[((x^2/a^2 + y^2/b^2 - 1) /. Solve[Dot[{{Cos[\[Theta]], -Sin[\[Theta]]}, {Sin[\[Theta]], Cos[\[Theta]]}}, {x + x0, y + y0}] == {x1, y1}, {x, y}])[[1]], {x1, y1}]) == (x0^2/a^2 + y0^2/b^2 - 1) {a1, b1, c1, d1, e1} /. {\[Theta] -> 2 ArcTan[u]}]] I arrive at five equations for your five unknowns. But note that u is of 4th order and with Solve[eq[[1, 1]]/eq[[4, 1]] == eq[[1, 2]]/eq[[4, 2]], u] // FullSimplify ( or Solve[ List @@ eq[[1]]/List @@ eq[[4]] /. List -> Equal , u] // FullSimplify ) you get four more or less complicated expressions for u. Each of these has to be inserted in the other equations to solve for a, b, x0, y0, which are quadratic and therefore give in general two solutions. So things are for sure not simple and I guess there is a set of different solutions and not a single one. Did you check it numerically?

POSTED BY: Hans Dolhaine

Updating Name

Posted 6 years ago

In fact I think Maple did NOT arrive at a solution of your problem. As before with eq = Simplify /@ Thread[TrigExpand[({{2, 0}, {1, 1}, {1, 0}, {0, 2}, {0, 1}} /. CoefficientRules[((x^2/a^2 + y^2/b^2 - 1) /. Solve[Dot[{{Cos[\[Theta]], -Sin[\[Theta]]}, \ {Sin[\[Theta]], Cos[\[Theta]]}}, {x + x0, y + y0}] == {x1, y1}, {x, y}])[[1]], {x1, y1}]) == (x0^2/a^2 + y0^2/b^2 - 1) {a1, b1, c1, d1, e1} /. {\[Theta] -> 2 ArcTan[u]}]]; eq // ColumnForm you have 5 equations for your five unknowns. By forming some quotients you get another 4 equations eq[[1, 1]]/eq[[2, 1]] == eq[[1, 2]]/eq[[2, 2]] eq[[1, 1]]/eq[[3, 1]] == eq[[1, 2]]/eq[[3, 2]] eq[[1, 1]]/eq[[4, 1]] == eq[[1, 2]]/eq[[4, 2]] eq[[1, 1]]/eq[[5, 1]] == eq[[1, 2]]/eq[[5, 2]] and by dividing eq1 by eq4 you get a complicated solution (in fact 4 solutions) for u Solve[List @@ eq[[1]]/List @@ eq[[4]] /. List -> Equal,u] // FullSimplify which in fact is not very helpful at this stage. Proceeding further you can get rid of some inconveniernt numerators eq1 = Map[# a^2 b^2 (1 + u^2)^2 &, eq, {2}]; eq1 // ColumnForm Using again the quotient-trick you get q12 = eq1[[1, 1]]/eq1[[2, 1]] == eq1[[1, 2]]/eq1[[2, 2]] q13 = eq1[[1, 1]]/eq[[3, 1]] == eq1[[1, 2]]/eq1[[3, 2]] q14 = eq1[[1, 1]]/eq1[[4, 1]] == eq1[[1, 2]]/eq1[[4, 2]] q15 = eq1[[1, 1]]/eq[[5, 1]] == eq1[[1, 2]]/eq1[[5, 2]] By inspection you see that q13 and q15 contain x0 and y0, so these may be extracted / solved for lsgxy = Solve[{q13, q15}, {x0, y0}] // Flatten and q12 and q14 contain only a, b and u. So lsga = Solve[q12, a] gives two solutions for a. But inserted in q14 this equation is free of b. ????? Maybe b is a free parameter? Or the equations are chosen wrongly? Ignoring this one could decide to insert the meanwhile known parameters into e.g. the first of the equations eq1a = eq[[1]] /. lsgxy /. lsga[[1]] // FullSimplify // Together giving an expression which contains indeed only b and u as unknowns, Variables[List @@ eq1a] but u in a really high order. Impossible to solve for analytically. So I really doubt that Maple could come up with a solution of your problem.

In fact I think Maple did NOT arrive at a solution of your problem.

As before with

eq = Simplify /@ 
   Thread[TrigExpand[({{2, 0}, {1, 1}, {1, 0}, {0, 2}, {0, 1}} /. 
         CoefficientRules[((x^2/a^2 + y^2/b^2 - 1) /. 
             Solve[Dot[{{Cos[\[Theta]], -Sin[\[Theta]]}, \
{Sin[\[Theta]], Cos[\[Theta]]}}, {x + x0, y + y0}] == {x1, y1}, {x, 
               y}])[[1]],
          {x1, y1}]) == (x0^2/a^2 + y0^2/b^2 - 1) {a1, b1, c1, d1, 
         e1} /. {\[Theta] -> 2 ArcTan[u]}]];
eq // ColumnForm

you have 5 equations for your five unknowns. By forming some quotients you get another 4 equations

eq[[1, 1]]/eq[[2, 1]] == eq[[1, 2]]/eq[[2, 2]]
eq[[1, 1]]/eq[[3, 1]] == eq[[1, 2]]/eq[[3, 2]]
eq[[1, 1]]/eq[[4, 1]] == eq[[1, 2]]/eq[[4, 2]]
eq[[1, 1]]/eq[[5, 1]] == eq[[1, 2]]/eq[[5, 2]]

and by dividing eq1 by eq4 you get a complicated solution (in fact 4 solutions) for u

Solve[List @@ eq[[1]]/List @@ eq[[4]] /. List -> Equal,u] // FullSimplify

which in fact is not very helpful at this stage. Proceeding further you can get rid of some inconveniernt numerators

eq1 = Map[# a^2 b^2 (1 + u^2)^2 &, eq, {2}];
eq1 // ColumnForm

Using again the quotient-trick you get

q12 = eq1[[1, 1]]/eq1[[2, 1]] == eq1[[1, 2]]/eq1[[2, 2]]
q13 = eq1[[1, 1]]/eq[[3, 1]] == eq1[[1, 2]]/eq1[[3, 2]]
q14 = eq1[[1, 1]]/eq1[[4, 1]] == eq1[[1, 2]]/eq1[[4, 2]]
q15 = eq1[[1, 1]]/eq[[5, 1]] == eq1[[1, 2]]/eq1[[5, 2]]

By inspection you see that q13 and q15 contain x0 and y0, so these may be extracted / solved for

lsgxy = Solve[{q13, q15}, {x0, y0}] // Flatten

and q12 and q14 contain only a, b and u. So

lsga = Solve[q12, a]

gives two solutions for a. But inserted in q14 this equation is free of b. ?????

Maybe b is a free parameter? Or the equations are chosen wrongly?

Ignoring this one could decide to insert the meanwhile known parameters into e.g. the first of the equations

eq1a = eq[[1]] /. lsgxy /. lsga[[1]] // FullSimplify // Together

giving an expression which contains indeed only b and u as unknowns,

Variables[List @@ eq1a]

but u in a really high order. Impossible to solve for analytically. So I really doubt that Maple could come up with a solution of your problem.

POSTED BY: Updating Name

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