RootSum

*To*: mathgroup at smc.vnet.net*Subject*: [mg75231] RootSum*From*: dimitris <dimmechan at yahoo.com>*Date*: Sat, 21 Apr 2007 23:09:45 -0400 (EDT)

Although I understand the presence of RootSum in cases of rational functions P(x)/Q(x) where Q(x) is a polynomial of degree 5 or higher (recall the Abel's impisibility theorem) I believe that for degree less than 5 (in particular for 3 and 4) Integrate could not generate RootSum. In[307]; Quit[] In[1]:= $Version Out[1]= "5.2 for Microsoft Windows (June 20, 2005)" Consider the function In[2]:= g[x_] := (2*x)/((x + 1)*(x^3 + 3*x^2 + 2*x + 1)) The following analysis show that the function is integrable in the range [0,Infinity). In[3]:= Plot[g[x], {x, 0, 10}] Out[3]= Graphics[] In[4]:= (g[x] + O[x, #1]^2 & ) /@ {0, Infinity} Out[4]= {SeriesData[x, 0, {2}, 1, 2, 1], SeriesData[x, Infinity, {}, 2, 2, 1]} In[5]:= Reduce[Denominator[g[x]] == 0 && x > 0, x, Reals] Out[5]= False Mathematica tries really hard but it returns the definite integral back. In[6]:= Timing[Integrate[g[x], {x, 0, Infinity}]] Out[6]= {142.078*Second, Integrate[(2*x)/((1 + x)*(1 + 2*x + 3*x^2 + x^3)), {x, 0, Infinity}]} Let's obtain the (Mathematica's) antiderivative of g(x) In[6]:= G[x_] = Integrate[g[x], x] Out[6]= 2*(-Log[1 + x] + RootSum[1 + 2*#1 + 3*#1^2 + #1^3 & , (Log[x - #1] + 2*Log[x - #1]*#1 + Log[x - #1]*#1^2)/ (2 + 6*#1 + 3*#1^2) & ]) Note at this point how Integrate works. First Apart is called in order to write the integrand as follows In[8]:= Apart[g[x]] Out[8]= -(2/(1 + x)) + (2*(1 + 2*x + x^2))/(1 + 2*x + 3*x^2 + x^3) and then Integrate...integrates each term of the sum. In[9]:= (Integrate[#1, x] & ) /@ % Out[9]= -2*Log[1 + x] + 2*RootSum[1 - #1 + #1^3 & , (Log[1 + x - #1]*#1^2)/(-1 + 3*#1^2) & ] Note also that version 5.2 is more careful and doesn't return Infinity for the definite integral as version 4.0 does (I think because it considers the integrals of the parts in the last output, which both are divergent at Infinity). So, the generation of the RootSum object in the antiderivative is due to the failure if the Apart function to do the partial fraction decomposition of (2*(1 + 2*x + x^2))/(1 + 2*x + 3*x^2 + x^3). Note that I lack serious knowledge on symbolics algebra and so, I may miss something fundamental but (although a few months ago I have a query about this behavior of Apart) I still don't understand why there are cases that Apart returns its argument back. Here are two examples In[12]:= Apart /@ {x^2/(x^2 + 3*x + 2), x/(x^3 + 2*x^2 + x + 1)} Out[12]= {1 + 1/(1 + x) - 4/(2 + x), x/(1 + x + 2*x^2 + x^3)} It seems that Apart does the PFD provided it can evaluate the roots of the denominator of the rational function. In the second case it can't evaluate and that's why it fails to do the PFD. Could/Should Apart have an option to make it do something like? In[13]:= Simplify[x /. Solve[1 + x + 2*x^2 + x^3 == 0, x]] Simplify /@ Factor[1 + x + 2*x^2 + x^3, Extension -> %] Simplify /@ Apart[x/%] (*output is ommited*) I think if Apart could do this, then the generation of RootSum objects could be avoided (especially for definite integrals where one or both limits of integration is infinity, and undoubtfully Integrate faces problems. Of course may be what I ask is too difficult and in view of the output of In[13] it seems quite complicated, and in the very next version RootSum at infinity does not face problems; If I knew that I would strop right now the message! If the other CAS I use can already do it properly despite the generation of similar to RootSum objects I think (and I hope!) our lovely Mathematica will be able to do better!). Returning back to the antiderivative In[14]:= Timing[(Limit[G[x], x -> #1] & ) /@ {0, Infinity}] Out[14]= {191.422*Second, {2*RootSum[1 + 2*#1 + 3*#1^2 + #1^3 & , (Log[-#1] + 2*Log[-#1]*#1 + Log[-#1]*#1^2)/(2 + 6*#1 + 3*#1^2) & ], 0}} In[15]:= (Plus[#2 - #1] & ) @@ %[[2]] N[%] Out[15]= -2*RootSum[1 + 2*#1 + 3*#1^2 + #1^3 & , (Log[-#1] + 2*Log[-#1]*#1 + Log[-#1]*#1^2)/(2 + 6*#1 + 3*#1^2) & ] Out[114]= 0.37121697526024766 + 0.*I Of course the result is correct but 190 seconds of timing!?!? Now you understand probably why I have sticked somehow with RootSum! Unecessarly very long time for evaluating something that at infinity is zero. (The value at zero is returned almost immediately). Dimitris