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MathGroup Archive 2005

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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 � s\_2 � s\_1, 1\/2 � 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.


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