Re: Trouble with a system of equations

*To*: mathgroup at smc.vnet.net*Subject*: [mg77500] Re: [mg77488] Trouble with a system of equations*From*: Andrzej Kozlowski <akoz at mimuw.edu.pl>*Date*: Mon, 11 Jun 2007 04:17:34 -0400 (EDT)*References*: <200706101120.HAA18117@smc.vnet.net>

On 10 Jun 2007, at 20:20, Yaroslav Bulatov wrote: > Hi, I'm trying to solve a certain kind of system of equations, and > while they are solvable by hand, Mathematica 6.0 has problems solving > it > > Here's an example > > eqns = {a + b + c + d == 4*m0, b + d == 4*m1, c + d == 4*m2, d == > 4*m3} /. {a -> t0/(1 + t0), b -> (t0*t1)/(1 + t0*t1), c -> (t0*t2)/(1 > + t0*t2), d -> (t0*t1*t2*t3)/(1 + t0*t1*t2*t3)} > Solve[eqns, {t0, t1, t2, t3}] > > The solution can be found by hand and verified below > > sol = {t0 -> a/(1/4 - a), t1 -> (b/(1/4 - b))*((1/4 - a)/a), t2 -> (c/ > (1/4 - c))*((1/4 - a)/a), t3 -> (m3/(1/4 - m3))*(a/(1/4 - a))*((1/4 - > b)/b)*((1/4 - c)/c)} /. {a -> m0 - m1 - m2 + m3, b -> m1 - m3, c -> m2 > - m3} > eqns /. sol // Simplify > > This is an example of estimating equations for a saturated logistic > regression model with 2 independent variables. I'd like to see if > formulas also exist for more variables, but they are too cumbersome to > solve by hand. Are there any Mathematica tricks I can use to answer > this question? > > Here's the procedure that generates the system of equations for d > variables (d=2 produces the system above) > > logeq[d_] := Module[{bounds, monomials, params, > partition,derivs,sums}, > xs = (Subscript[x, #1] & ) /@ Table[i, {i, 1, d}]; > monomials = Subsets[xs]; monomials = (Prepend[#1, 1] & ) /@ > monomials; > monomials = (Times @@ #1 & ) /@ monomials; > params = (Subscript[th, #1] & ) /@ Table[i, {i, 0, 2^d - 1}]; > monomials = (Times @@ #1 & ) /@ Thread[{params, monomials}]; > partition = Log[1 + Exp[Plus @@ monomials]]; > derivs = (D[partition, Subscript[th, #1]] & ) /@ > Table[i, {i, 0, 2^d - 1}]; bounds = ({#1, 0, 1} & ) /@ xs; > sums = (Table[#1, Evaluate[Sequence @@ bounds]] & ) /@ derivs; > sums = (Plus @@ #1 & ) /@ (Flatten[#1] & ) /@ sums; > Thread[sums == Table[Subscript[m, i], {i, 0, 2^d - 1}]]] > > Here is a simple way to solve your original system. I have not tried any of the more general ones. eqns = {a + b + c + d == 4*m0, b + d == 4*m1, c + d == 4*m2, d == 4*m3} /. {a -> t0/(1 + t0), b -> (t0*t1)/(1 + t0*t1), c -> (t0*t2)/(1 + t0*t2), d -> (t0*t1*t2*t3)/(1 + t0*t1*t2*t3)}; g = {t0/(t0 + 1) + (t1*t0)/(t0*t1 + 1) + (t2*t0)/(t0*t2 + 1) + (t1*t2*t3*t0)/(t0*t1*t2*t3 + 1) == 4*m0, (t0*t1)/(t0*t1 + 1) + (t0*t2*t3*t1)/(t0*t1*t2*t3 + 1) == 4*m1, (t0*t2)/(t0*t2 + 1) + (t0*t1*t3*t2)/(t0*t1*t2*t3 + 1) == 4*m2, (t0*t1*t2*t3)/(t0*t1*t2*t3 + 1) == 4*m3} /. Equal -> Subtract; g1 = Map[Numerator[Together[#]] &, g]; gr = GroebnerBasis[g1, {t0, t1, t2, t3}]; Simplify[Solve[gr == 0, {t0, t1, t2, t3}]] {{t0 -> -((4*(-m0 + m1 + m2 - m3))/(-4*m0 + 4*m1 + 4*m2 - 4*m3 + 1)), t1 -> ((-4*m0 + 4*m1 + 4*m2 - 4*m3 + 1)*(m3 - m1))/((-m0 + m1 + m2 - m3)* (-4*m1 + 4*m3 + 1)), t2 -> ((-4*m0 + 4*m1 + 4*m2 - 4*m3 + 1)* (m2 - m3))/ ((4*m2 - 4*m3 - 1)*(-m0 + m1 + m2 - m3)), t3 -> ((4*m1 - 4*m3 - 1)*(-m0 + m1 + m2 - m3)*m3*(-4*m2 + 4*m3 + 1))/ ((-4*m0 + 4*m1 + 4*m2 - 4*m3 + 1)*(m1 - m3)*(m3 - m2)*(4*m3 - 1))}} Andrzej Kozlowski

**References**:**Trouble with a system of equations***From:*Yaroslav Bulatov <yaroslavvb@gmail.com>