Re: question 1 (Integrate and simplification)

*To*: mathgroup at smc.vnet.net*Subject*: [mg75965] Re: question 1 (Integrate and simplification)*From*: dimitris <dimmechan at yahoo.com>*Date*: Mon, 14 May 2007 03:37:51 -0400 (EDT)*References*: <200705110924.FAA05900@smc.vnet.net>

>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... Hmmm! I am a bit confused! Until now I think Integrate use Simplify not in order to simplify the input but also to simplify its output. In fact there was a recent post about a complicated integral with log where two successive executions succeded in different outputs! The reason was that Integrate use even (very little!) FullSimplify to its output. See here... http://groups.google.gr/group/comp.soft-sys.math.mathematica/browse_thread/= thread/51a2b235eebfd92d/004350a53d3f071e?lnk=gst&q=a+log+integral&rnum= =3&hl=el#004350a53d3f071e To this end, add the following rules In[24]:= Unprotect[FullSimplify]; FullSimplify[a___] := Null /; (Print[InputForm[fullsimplify[a]]]; False) Unprotect[Simplify]; Simplify[a___] := Null /; (Print[InputForm[simplify[a]]]; False) Then In[29]:= Integrate[1/Sqrt[Sin[x]], {x, 0, Pi/2}] >From In[29]:= simplify[1/Sqrt[Sin[x]], TimeConstraint -> 1/2, Assumptions -> True] >From In[29]:= simplify[True, Assumptions -> True] >From In[29]:= simplify[1/Sqrt[Sin[x]], Assumptions -> 0 < x < Pi/2] >From In[29]:= simplify[Re[Sin[x]], Assumptions -> Im[x] == 0 && Re[x] >= 0 && Re[x] <= Pi/2] >From In[29]:= simplify[Im[Sin[x]], Assumptions -> Im[x] == 0 && Re[x] >= 0 && Re[x] <= Pi/2] >From In[29]:= simplify[False, Assumptions -> True] >From In[29]:= simplify[False, Assumptions -> True] >From In[29]:= simplify[{1, 0, 0, 0, 1/12}, True] >From In[29]:= simplify[Infinity, Assumptions -> x > 0 && x < 1/12, TimeConstraint -> 1] >From In[29]:= simplify[1/(x*(1/Sqrt[x] + x^(3/2)/12)), Assumptions -> True] >From In[29]:= simplify[Infinity, Assumptions -> x > 0 && x < 1/12, TimeConstraint -> 1] >From In[29]:= simplify[{1, 0, 1/4}, True] >From In[29]:= simplify[False, Assumptions -> True] >From In[29]:= simplify[False, Assumptions -> True] >From In[29]:= simplify[False, Assumptions -> True] >From In[29]:= simplify[False, Assumptions -> True] >From In[29]:= simplify[1/Sqrt[Sin[x]], TimeConstraint -> 1/2, Assumptions -> True] >From In[29]:= simplify[Re[Sin[x]], Assumptions -> Im[x] == 0 && Re[x] >= 0 && Re[x] <= Pi/2] >From In[29]:= simplify[Im[Sin[x]], Assumptions -> Im[x] == 0 && Re[x] >= 0 && Re[x] <= Pi/2] >From In[29]:= simplify[False, Assumptions -> True] >From In[29]:= simplify[False, Assumptions -> True] >From In[29]:= simplify[-2*EllipticF[(Pi/2 - x)/2, 2], Assumptions -> True] >From In[29]:= simplify[{}, Assumptions -> True] >From In[29]:= simplify[1, Assumptions -> True] >From In[29]:= simplify[0, Assumptions -> True] >From In[29]:= simplify[False, Assumptions -> True] >From In[29]:= simplify[True, Assumptions -> True] >From In[29]:= simplify[Pi/2, Assumptions -> True] >From In[29]:= simplify[-2*EllipticF[(Pi/2 - x)/2, 2], Assumptions -> True] >From In[29]:= simplify[0, Assumptions -> True] >From In[29]:= simplify[-2*EllipticF[(Pi/2 - x)/2, 2], Assumptions -> True] >From In[29]:= simplify[2*EllipticF[Pi/4, 2], Assumptions -> True] >From In[29]:= simplify[2*EllipticF[Pi/4, 2], Assumptions -> True] >From In[29]:= simplify[True, Assumptions -> True] >From In[29]:= simplify[2*EllipticF[Pi/4, 2], Assumptions -> True] >From In[29]:= fullsimplify[2*EllipticF[Pi/4, 2], True] >From In[29]:= simplify[False, Assumptions -> True] >From In[29]:= simplify[False, Assumptions -> True] >From In[29]:= simplify[2*EllipticF[Pi/4, 2], TimeConstraint -> 1/2, Assumptions -> True] >From In[29]:= fullsimplify[2*EllipticF[Pi/4, 2], TimeConstraint -> 1/2, Assumptions - > True] Out[29]= \!\(2\ EllipticF[=F0\/4, 2]\) It is clear the attempt to (Full)Simplify the output BUT in very little timing! Dimitris =CF/=C7 Andrzej Kozlowski =DD=E3=F1=E1=F8=E5: > 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

**References**:**question 1***From:*dimitris <dimmechan@yahoo.com>

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