Re: Explicit solution to Root[]
- To: mathgroup at smc.vnet.net
- Subject: [mg58443] Re: Explicit solution to Root[]
- From: "Mukhtar Bekkali" <mbekkali at gmail.com>
- Date: Sat, 2 Jul 2005 04:07:03 -0400 (EDT)
- References: <da2mmv$932$1@smc.vnet.net>
- Sender: owner-wri-mathgroup at wolfram.com
I have Mathematica 5.0.1.0. Here is the exact code \!\(\* RowBox[{\(Clear["\<Global`*\>"]\), "\n", \(Remove["\<Global`*\>"]\), "\ \[IndentingNewLine]", \(Off[General::"\<spell1\>"\ ]\), \ "\[IndentingNewLine]", \(Needs["\<Miscellaneous`RealOnly`\>"]\), "\[IndentingNewLine]", RowBox[{ RowBox[{ RowBox[{ StyleBox["{", FontColor->RGBColor[1, 0, 1]], RowBox[{ StyleBox["λ", FontColor->RGBColor[1, 0, 1]], StyleBox[",", FontColor->RGBColor[1, 0, 1]], StyleBox["θ", FontColor->RGBColor[1, 0, 1]], StyleBox[",", FontColor->RGBColor[1, 0, 1]], StyleBox["α", FontColor->RGBColor[1, 0, 1]], StyleBox[",", FontColor->RGBColor[1, 0, 1]], StyleBox["r", FontColor->RGBColor[1, 0, 1]], StyleBox[",", FontColor->RGBColor[1, 0, 1]], StyleBox[\(γ\_0\), FontColor->RGBColor[1, 0, 1]], StyleBox[",", FontColor->RGBColor[1, 0, 1]], StyleBox["n", FontColor->RGBColor[1, 0, 1]], StyleBox[",", FontColor->RGBColor[1, 0, 1]], SubscriptBox[ StyleBox["I", FontColor->RGBColor[1, 0, 1]], "0"], ",", \(I\_n\)}], StyleBox["}", FontColor->RGBColor[1, 0, 1]]}], StyleBox["=", FontColor->RGBColor[1, 0, 1]], RowBox[{ StyleBox["{", FontColor->RGBColor[1, 0, 1]], RowBox[{ StyleBox["3", FontColor->RGBColor[1, 0, 1]], StyleBox[",", FontColor->RGBColor[1, 0, 1]], StyleBox[\(2\/3\), FontColor->RGBColor[1, 0, 1]], StyleBox[",", FontColor->RGBColor[1, 0, 1]], StyleBox[\(1\/2\), FontColor->RGBColor[1, 0, 1]], StyleBox[",", FontColor->RGBColor[1, 0, 1]], StyleBox["2", FontColor->RGBColor[1, 0, 1]], StyleBox[",", FontColor->RGBColor[1, 0, 1]], \(2\/3\), StyleBox[",", FontColor->RGBColor[1, 0, 1]], StyleBox["2", FontColor->RGBColor[1, 0, 1]], StyleBox[",", FontColor->RGBColor[1, 0, 1]], StyleBox["0", FontColor->RGBColor[1, 0, 1]], StyleBox[",", FontColor->RGBColor[1, 0, 1]], StyleBox["1", FontColor->RGBColor[1, 0, 1]]}], StyleBox["}", FontColor->RGBColor[1, 0, 1]]}]}], StyleBox[";", FontColor-> RGBColor[1, 0, 1]]}], "\[IndentingNewLine]", \(CS := â??\+\(i = 1\)\%n Integrate[u\_i, {β, I\_\(i - 1\), I\_i}]; PS := â??\+\(i = 1\)\%k R\_i; B := â??\+\(i = 1\)\%n\( m\_i\) b\_i;\), "\[IndentingNewLine]", StyleBox[\(k = 1;\), FontColor->RGBColor[1, 0, 0]], "\[IndentingNewLine]", \(Table[γ\_i = γ\_0, {i, 1, k}]; Table[γ\_i = 1 - γ\_0, {i, k + 1, n}];\), "\[IndentingNewLine]", \(Table[{p\_i, e\_i, s\_i} = {p\_i, e\_i, s\_i}, {i, 1, n}];\), "\[IndentingNewLine]", \(L\_p = Table[p\_i, {i, 1, n}]; L\_q = Join[Table[e\_i, {i, 1, n}], Table[s\_i, {i, 1, n}]];\), "\[IndentingNewLine]", \(L0\_q = Join[Table[Chop[N[\(I\_n\) i\/\(n + 1\)]], {i, 1, n}], Table[Chop[N[\(I\_n\) \((1 - i\/\(n + 1\))\)]], {i, 1, n}]];\), "\[IndentingNewLine]", RowBox[{ RowBox[{ StyleBox[\(T\_q\), FontColor->RGBColor[0, 0, 1]], StyleBox["=", FontColor->RGBColor[0, 0, 1]], RowBox[{ StyleBox["Table", FontColor->RGBColor[0, 0, 1]], StyleBox["[", FontColor->RGBColor[0, 0, 1]], RowBox[{ RowBox[{ StyleBox[\(q\_i\), FontColor->RGBColor[0, 0, 1]], StyleBox["=", FontColor->RGBColor[0, 0, 1]], \(e\_i + s\_i\)}], ",", StyleBox[\({i, 1, n}\), FontColor->RGBColor[0, 0, 1]]}], StyleBox["]", FontColor->RGBColor[0, 0, 1]]}]}], StyleBox[";", FontColor-> RGBColor[0, 0, 1]], \(T\_b = Table[b\_i = e\_i\/\(e\_i + s\_i\), {i, 1, n}]\), ";", \(T\_u = Table[u\_i = q\_i - \((b\_i - β)\)\^2 - p\_i, {i, 1, n}]\), ";"}], "\[IndentingNewLine]", \(T\_I = Table[I\_i = β /. \(Solve[ u\_i \[Equal] u\_\(i + 1\), β]\)[\([1]\)], {i, 1, n - 1}]\), "\[IndentingNewLine]", \(T\_m = Table[m\_i = I\_i - I\_\(i \ - 1\), {i, 1, n}]; T\_R = Table[R\_i = \(p\_i\) m\_i - \((\(γ\_i\) e\_i\^r\/r + \((1 - γ\_i)\) s\_i\^r\/r)\), {i, 1, n}];\), "\[IndentingNewLine]", \(FOCP = \ Table[D[R\_i, p\_i], {i, 1, n}]; L\_p = L\_p /. \(Solve[FOCP \[Equal] 0, L\_p]\)[\([ 1]\)]; Table[p\_i = L\_p[\([ i]\)], {i, 1, n}] // Simplify\), "\[IndentingNewLine]", \(FOCQ = \ Join[Table[D[R\_i, e\_i], {i, 1, n}], Table[D[R\_i, s\_i], { i, 1, n}]];\), "\[IndentingNewLine]", \(e\_1 = s\_2; e\_2 = s\_1;\), "\ \[IndentingNewLine]", \(N[ToRules[ Reduce[{0 â?¤ s\_2 â?¤ s\_1, 1\/2 â?¤ η < 1, FOCQ[\([3]\)] \[Equal] 0, FOCQ[\([ 4]\)] \[Equal] 0}, {s\_1, s\_2}, Reals, Backsubstitution \[Rule] True]]]\)}]\) After you run it you would see output among which there would be s2. Now, I just copied it and pasted as a new input expression, like this \!\(expr = Root[\(-2\)\ #1\^3 + 4\ #1\^4 + 5\ #1\ Root[\(-24\) + 6\ #1 + 51\ #1\^2 - 40\ #1\^3 - 54\ #1\^4 + 54\ #1\^5 &, 1] - 6\ #1\^2\ Root[\(-24\) + 6\ #1 + 51\ #1\^2 - 40\ \ #1\^3 - 54\ #1\^4 + 54\ #1\^5 &, 1] + 12\ #1\^3\ Root[\(-24\) + 6\ #1 + 51\ #1\^2 - 40\ #1\^3 - \ 54\ #1\^4 + 54\ #1\^5 &, 1] + Root[\(-24\) + 6\ #1 + 51\ #1\^2 - 40\ #1\^3 - 54\ \ #1\^4 + 54\ #1\^5 &, 1]\^2 - 6\ #1\ Root[\(-24\) + 6\ #1 + 51\ #1\^2 - 40\ #1\^3 - 54\ #1\^4 + 54\ #1\^5 &, 1]\^2 + 12\ #1\^2\ Root[\(-24\) + 6\ #1 + 51\ #1\^2 - 40\ #1\^3 - 54\ #1\^4 + 54\ #1\^5 &, 1]\^2 - 2\ Root[\(-24\) + 6\ #1 + 51\ #1\^2 - \ 40\ #1\^3 - 54\ #1\^4 + 54\ #1\^5 &, 1]\^3 + 4\ #1\ Root[\(-24\) + 6\ #1 + 51\ \ #1\^2 - 40\ #1\^3 - 54\ #1\^4 + 54\ #1\^5 &, 1]\^3 &, 2]\[IndentingNewLine] expr // N\) And now function N does not work. However, if you shutdown the kernel and run the second input N produces a number. I do not know what is going on. Must be something with Mathematica memory.