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MathGroup Archive 2007

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Re: Re: question 1

  • To: mathgroup at smc.vnet.net
  • Subject: [mg75919] Re: [mg75873] Re: [mg75810] question 1
  • From: Andrzej Kozlowski <akoz at mimuw.edu.pl>
  • Date: Sun, 13 May 2007 05:50:31 -0400 (EDT)
  • References: <200705110924.FAA05900@smc.vnet.net> <A52D6750-3017-4EF6-82F7-DFE4CB83812F@mimuw.edu.pl> <CB1CAE8E-3B30-4645-9279-14C936512446@mimuw.edu.pl> <3201069.1178958084224.JavaMail.root@m35> <op.tr8whmu1qu6oor@monster.ma.dl.cox.net>

This is of course the right approach when you know you have a "power  
sum" and want to find a suitbale representation. However, if you  
wanted to use this as a posssible transofrmation for Simplify it  
would not be much use since, while:

  powerSum[Expand[(x - 2)^4 + (y + 3)^5], {x, y}]
y + 3)^5 + (x - 2)^4

works nicely,

powerSum[Expand[(x - 2)^4 + (y + 3)^5] + 11, {x, y}]

is a disaster.

Note that with:

transf[f_, {e_, o_}] :=
  With[{a = Integrate[D[f, e], e], b = Integrate[D[f, o], o]},
   FullSimplify[Simplify[a] + Simplify[b] + Simplify[(f - (a + b))],
    ExcludedForms -> {(e + c_)^n_, (o + d_)^m_}]]

we get

  transf[Expand[(x - 2)^4 + (y + 3)^5] - 11, {x, y}]
(y + 3)^5 + (x - 2)^4 - 11

and even when we have cross terms, the answer returned is often  
better than that returend by FullSimplify:

  a = transf[Expand[(x - 2)^4 + (y + 3)^5] + x*y, {x, y}]

  (y + 3)^5 + x*(x*((x - 8)*x + 24) + y - 32) + 16

  b = FullSimplify[Expand[(x - 2)^4 + (y + 3)^5] +  x*y]

x^4 - 8*x^3 + 24*x^2 + (y - 32)*x + y*(y*(y*(y*(y + 15) + 90) + 270)  
+ 405) + 259

LeafCount /@ {a, b}
{21, 37}

My own intention was not so much to find a way of expressing a given  
expression as a sum of powers when we know that it cna be so  
expressed as to find a transformation that might be added to the  
transformations used by Simplify and that would help it to find such  
representations when they existed and would return something  
reasonably simple (better than FullSimplify does by default) in other  
cases.
Of course I don't really think this project is practical because of  
the high complexity any such transformation would probably have to have.

Andrzej Kozlowski



On 13 May 2007, at 12:11, DrMajorBob wrote:

> Clear[powerSum, test]
> powerSum[polynomial_, vars_List] /; PolynomialQ[polynomial, vars] :=
>   Module[{powers, coefficients, roots, soln},
>    powers = Exponent[polynomial, #] & /@ vars;
>    coefficients = Table[Unique[c], {Length@vars}];
>    roots = Table[Unique[r], {Length@vars}];
>    soln = SolveAlways[coefficients.(vars - roots)^powers ==  
> polynomial,
>        vars][[1]];
>    {roots, coefficients} = {roots, coefficients} /. soln;
>    coefficients.(vars - roots)^powers
>    ]



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