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["Î»",
FontColor->RGBColor[1, 0, 1]],
StyleBox[",",
FontColor->RGBColor[1, 0, 1]],
StyleBox["Î¸",
FontColor->RGBColor[1, 0, 1]],
StyleBox[",",
FontColor->RGBColor[1, 0, 1]],
StyleBox["Î±",
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[\(Î³\_0\),
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, {Î², 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]",
\(Table[Î³\_i =
Î³\_0, {i,
1, k}]; Table[Î³\_i = 1 - Î³\_0, {i, k + 1, n}];\),
"\[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 -
\((b\_i -
Î²)\)\^2 - p\_i, {i,
1, n}]\), ";"}],
"\[IndentingNewLine]", \(T\_I = Table[I\_i = Î² /. \(Solve[
u\_i \[Equal] u\_\(i + 1\), Î²]\)[\([1]\)], {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 - \((\(Î³\_i\) e\_i\^r\/r + \((1 -
Î³\_i)\)
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?¤ Î· <
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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