Re: Trouble with a system of equations

*To*: mathgroup at smc.vnet.net*Subject*: [mg77519] Re: [mg77488] Trouble with a system of equations*From*: DrMajorBob <drmajorbob at bigfoot.com>*Date*: Mon, 11 Jun 2007 04:27:33 -0400 (EDT)*References*: <13301479.1181524668855.JavaMail.root@m35>*Reply-to*: drmajorbob at bigfoot.com

> 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)} Here's another way of stating the same definition: eq = {a + b + c + d == 4*m0, b + d == 4*m1, c + d == 4*m2, d == 4*m3}; rules = {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)}; eqns = eq /. rules {t0/(1 + t0) + (t0 t1)/(1 + t0 t1) + (t0 t2)/(1 + t0 t2) + ( t0 t1 t2 t3)/(1 + t0 t1 t2 t3) == 4 m0, (t0 t1)/(1 + t0 t1) + (t0 t1 t2 t3)/(1 + t0 t1 t2 t3) == 4 m1, (t0 t2)/(1 + t0 t2) + (t0 t1 t2 t3)/(1 + t0 t1 t2 t3) == 4 m2, (t0 t1 t2 t3)/(1 + t0 t1 t2 t3) == 4 m3} The result is something Solve can't solve for the t variables (for whatever reason). But you didn't have to complicate things that way. Instead, you can solve the problems separately: s1 = Solve[eq, {a, b, c, d}] {{a -> 4 (m0 - m1 - m2 + m3), b -> 4 (m1 - m3), c -> 4 (m2 - m3), d -> 4 m3}} s2 = Solve[rules /. Rule -> Equal, {t0, t1, t2, t3}] {{t3 -> (a (1 - b - c + b c) d)/((-1 + a) b c (-1 + d)), t1 -> (-b + a b)/(a (-1 + b)), t2 -> ((-1 + a) c)/(a (-1 + c)), t0 -> -a/(-1 + a)}} and combine the two solutions: s2[[1]] /. s1 {{t3 -> ((1 - 4 (m1 - m3) - 4 (m2 - m3) + 16 (m1 - m3) (m2 - m3)) m3 (m0 - m1 - m2 + m3))/((m1 - m3) (m2 - m3) (-1 + 4 m3) (-1 + 4 (m0 - m1 - m2 + m3))), t1 -> (-4 (m1 - m3) + 16 (m1 - m3) (m0 - m1 - m2 + m3))/( 4 (-1 + 4 (m1 - m3)) (m0 - m1 - m2 + m3)), t2 -> ((m2 - m3) (-1 + 4 (m0 - m1 - m2 + m3)))/((-1 + 4 (m2 - m3)) (m0 - m1 - m2 + m3)), t0 -> -(4 (m0 - m1 - m2 + m3))/(-1 + 4 (m0 - m1 - m2 + m3))}} eqns /. % // Simplify {{True, True, True, True}} Always take advantage of what you know about the problem structure. Bobby On Sun, 10 Jun 2007 06:20:40 -0500, Yaroslav Bulatov <yaroslavvb at gmail.com> 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}]]] > > > -- DrMajorBob at bigfoot.com