Re: Re: question 1
- To: mathgroup at smc.vnet.net
- Subject: [mg75902] Re: [mg75873] Re: [mg75810] question 1
- From: DrMajorBob <drmajorbob at bigfoot.com>
- Date: Sun, 13 May 2007 05:41:45 -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.tr8w3vygqu6oor@monster.ma.dl.cox.net>
- Reply-to: drmajorbob at bigfoot.com
OK, this is better:
Clear[powerSum, ruleReverse, test]
ruleReverse[var_List][Rule[a_, b_]] /; MemberQ[var, a] := Rule[b, a]
ruleReverse[var_List][Rule[a_, b_]] := Rule[a, b]
powerSum[polynomial_, vars_List] /; PolynomialQ[polynomial, vars] :=
Module[{powers = Exponent[polynomial, #] & /@ vars, coefficients,
roots, soln},
roots = Table[Unique[a], {Length@vars}];
coefficients = Coefficient[polynomial, #] & /@ vars^powers;
soln = ruleReverse[Variables@polynomial] /@
SolveAlways[coefficients.(vars - roots)^powers == polynomial,
vars][[1]];
roots = roots /. soln;
coefficients.(vars - roots)^powers
]
test[p_] := powerSum[Expand@p, Variables@p]
test /@ {(-2 + x)^3 + (1 +
y)^3, (x - 5)^4 + (y - 3)^3, (x - 2)^4 + (y + 1)^2, (a +
3)^5 + (x - 7)^6, (o - 8)^4 - (e + 4)^8}
{(-2 + x)^3 + (1 + y)^3, (-5 + x)^4 + (-3 + y)^3, (-2 + x)^4 + (1 +
y)^2, (3 + a)^5 + (-7 + x)^6, -(4 + e)^8 + (-8 + o)^4}
powerSum[Expand[(a - x)^3 - (b - y)^4], {a, b}]
(a - x)^3 - (b - y)^4
powerSum[Expand[(x - a)^3 - (y - b)^4], {x, y}]
(-a + x)^3 - (-b + y)^4
Bobby
> On Sat, 12 May 2007 02:11:17 -0500, Andrzej Kozlowski <akoz at mimuw.edu.pl>
> wrote:
>
>>
>> On 12 May 2007, at 10:37, Andrzej Kozlowski wrote:
>>
>>>
>>> On 12 May 2007, at 10:23, Andrzej Kozlowski wrote:
>>>
>>>> On 11 May 2007, at 18:24, dimitris wrote:
>>>>
>>>>> I have
>>>>>
>>>>> In[5]:=
>>>>> f = (o - 8)^4 - (e + 4)^8
>>>>>
>>>>> Out[5]=
>>>>> -(4 + e)^8 + (-8 + o)^4
>>>>>
>>>>> In[6]:=
>>>>> ff = Expand[f]
>>>>>
>>>>> Out[6]=
>>>>> -61440 - 131072*e - 114688*e^2 - 57344*e^3 - 17920*e^4 - 3584*e^5 =
-
>>>>> 448*e^6 - 32*e^7 - e^8 - 2048*o + 384*o^2 - 32*o^3 + o^4
>>>>>
>>>>> Is it possible to simplify ff to f again?
>>>>>
>>>>> Thanks!
>>>>>
>>>>>
>>>>
>>>> I don't think Mathematica can do it automatically, because the
>>>> most obvious way of carrying out this simplification relies on
>>>> transformations of the kind that that Simplify or FullSimplify
>>>> never use (such as adding and simultaneously subtracting some
>>>> number, then rearranging the whole expression and factoring parts
>>>> of it). These kind of transformations could be implemented but
>>>> they would work in only a few cases, and would considerably
>>>> increase the time complexity of simplifying. Here is an example of
>>>> another transformation that will work in this and some similar
>>>> cases:
>>>>
>>>> transf[f_, {e_, o_}] :=
>>>> With[{a = Integrate[D[f, e], e], b = Integrate[D[f, o], o]},
>>>> Simplify[a] + Simplify[b] + Simplify[(f - (a + b))]]
>>>>
>>>> (one can easily implment a version with more than two variables).
>>>>
>>>> This works in your case:
>>>>
>>>> ff = Expand[(o - 8)^4 - (e + 4)^8];
>>>>
>>>> transf[ff, {e, o}]
>>>> (o - 8)^4 - (e + 4)^8
>>>>
>>>> and in quite many cases like yours :
>>>>
>>>> gg = Expand[(a + 3)^5 + (x - 7)^6];
>>>>
>>>> transf[gg, {a, x}]
>>>> (x - 7)^6 + (a + 3)^5
>>>>
>>>> but not in all
>>>>
>>>> hh = Expand[(x - 5)^4 + (y - 3)^3];
>>>>
>>>> transf[hh, {x, y}]
>>>> (x - 5)^4 + y*((y - 9)*y + 27) - 27
>>>>
>>>> Even this, however, is better than the answer FullSimplify gives:
>>>>
>>>> FullSimplify[hh]
>>>>
>>>> (x - 10)*x*((x - 10)*x + 50) + y*((y - 9)*y + 27) + 598
>>>>
>>>> Unfortunately transf also has very high complexity (it uses
>>>> Integrate) and is unlikely to be useful in cases other than sums
>>>> of polynomials in different variables (without "cross terms") so I
>>>> doubt that it would be worth implementing some version of it in
>>>> FullSimplify.
>>>>
>>>>
>>>> Andrzej Kozlowski
>>>>
>>>
>>> I forgot that Integrate already maks use of Simplify, so (probably)
>>> the folowing version of transf will work just as well and faster:
>>>
>>> transf[f_, {e_, o_}] :=
>>> With[{a = Integrate[D[f, e], e], b = Integrate[D[f, o], o]},
>>> a + b + Simplify[(f - (a + b))]]
>>>
>>> Andrzej Kozlowski
>>
>> Sorry, that last remark just not true. Cleary Integrate uses Simplify=
>> before integration but does not Simplify its output (obviously to
>> save time) so including Simplify does make a difference. Note also
>> the following works much better, but is,, of course, much more tiem
>> consuming:
>>
>>
>> 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 -> {(x + c_)^n_, (y + d_)^m_}]]
>>
>>
>> This will deal with many cases that would not work before:
>>
>> hh = Expand[(x - 5)^4 + (y - 3)^3];
>>
>> transf[hh, {x, y}]
>> (x - 5)^4 + (y - 3)^3
>>
>> p = Expand[(x - 2)^4 + (y + 1)^2];
>>
>> transf[p, {x, y}]
>> (x - 2)^4 + (y + 1)^2
>>
>> but at least one power has to be higher than 3. This this will work
>>
>> q = Expand[(x - 2)^4 + (y + 1)^3];
>> transf[q, {x, y}]
>> (x - 2)^4 + (y + 1)^3
>>
>> but this will not (all powers are <=3)
>>
>> q = Expand[(x - 2)^3 + (y + 1)^3];
>>
>> transf[q, {x, y}]
>>
>> x*((x - 6)*x + 12) + y*(y*(y + 3) + 3) - 7
>>
>> Andrzej Kozlowski
>>
>>
>>
>>
>>
>
>
>
-- =
DrMajorBob at bigfoot.com
- References:
- question 1
- From: dimitris <dimmechan@yahoo.com>
- question 1