Re: Partial fraction expansion

*To*: mathgroup at smc.vnet.net*Subject*: [mg30469] Re: Partial fraction expansion*From*: "Allan Hayes" <hay at haystack.demon.co.uk>*Date*: Thu, 23 Aug 2001 02:15:33 -0400 (EDT)*References*: <9lvh6b$4oc$1@smc.vnet.net>*Sender*: owner-wri-mathgroup at wolfram.com

Lee, Andrzej Here are some functions for factorizing and putting into partial fractions. The methods are different from Andrzej's and sometimes give different results. FactorC[p_, x_] := (*over complex numbers*) Times @@ Cases[Roots[p == 0, x, Cubics -> False], u_ == v_ -> x - v] FactorR[p_, x_] := (*over real numbers*) (Times @@ Join[Cases[#1, u_ == v_ /; Im[v] == 0 :> x - v], Cases[#1, u_ == v_ /; Im[v] > 0 :> x^2 - x*2*Re[v] + Abs[v]^2]] & )[ Roots[p == 0, x, Cubics -> False]] PartialFractionsC[p_, x_] := (*over complex numbers*) (Apart[Numerator[#1]/FactorC[Denominator[#1], x], x] & )[Together[p]] PartialFractionsR[p_, x_] := (*over real numbers*) (Apart[Numerator[#1]/FactorR[Denominator[#1], x], x] & )[Together[p]] PROGRAMMING NOTES The option Cubics->False is used to keep roots of cubics in Root[....] form. This more effIcient for computation. Re[v] and Abs[v]^2 are used rather than v+Conjugate[v] and v*Conjugate[v] to prevent Apart from factorising x^2 - x*2*Re[v] + Abs[v]^2] back to complex form. EXAMPLES pol = Expand[(x - 1)*(x + 1)^2*(x^2 + x + 1)^2*(x^2 + 4)]; f1 = FactorC[pol, x] (-1 + x)*(-2*I + x)*(2*I + x)*(1 + x)^2*((-1)^(1/3) + x)^2* (-(-1)^(2/3) + x)^2 f2 = FactorR[pol, x] (-1 + x)*(1 + x)^2*(4 + x^2)*(1 + x + x^2)^2 f3 = FactorR[x^3 + x + 1, x] (x - Root[1 + #1 + #1^3 & , 1])* (x^2 - 2*x*Root[-1 + 2*#1 + 8*#1^3 & , 1] + Root[-1 - #1^4 + #1^6 & , 2]^2) Root objects appear because of the option Cubics->False in Roots. We can sometimes get radical forms, but notice the complication. ToRadicals[f3] ((2/(3*(-9 + Sqrt[93])))^(1/3) - ((1/2)*(-9 + Sqrt[93]))^(1/3)/3^(2/3) + x)* (1/3 + (1/3)*(29/2 - (3*Sqrt[93])/2)^(1/3) + (1/3)*((1/2)*(29 + 3*Sqrt[93]))^(1/3) - 2*(((1/2)*(9 + Sqrt[93]))^(1/3)/(2*3^(2/3)) - 1/(2^(2/3)*(3*(9 + Sqrt[93]))^(1/3)))*x + x^2) Inexact forms can be found, from f3 : N[f3] (0.6823278038280193 + x)*(1.4655712318767682 - 0.6823278038280193*x + x^2) or directly f3 = FactorR[x^3 + x + 1//N, x] (0.6823278038280193 + x)*(1.4655712318767682 - 0.6823278038280193*x + x^2) Partial fractions pf1 = PartialFractionsR[(2 + x)/pol, x] 1/(60*(-1 + x)) - 1/(10*(1 + x)^2) - 39/(100*(1 + x)) + (-54 - 31*x)/(4225*(4 + x^2)) + (-1 + 3*x)/(13*(1 + x + x^2)^2) + (44 + 193*x)/(507*(1 + x + x^2)) pf2 = PartialFractionsR[(1 + x)/(1 - 3*x + x^2), x] (-5 - Sqrt[5])/(Sqrt[5]*(3 + Sqrt[5] - 2*x)) + (-5 + Sqrt[5])/(Sqrt[5]*(-3 + Sqrt[5] + 2*x)) Partial fractions will often involve Root objects (I delete the output for brevity) pf3 = PartialFractionsR[(1 + x)/(x^3 - x + 1), x] This ressult can in fact be put into radical form ToRadicals[pf3] We could have found the inexact form directly. PartialFractionsR[(1+x)/(x^3-x+1)//N,x] -(0.07614206365252976/(1.324717957244746 + 1.*x)) + (0.7982664819556426 + 0.07614206365252976*x)/ (0.754877666246693 - 1.324717957244746*x + 1.*x^2) -- Allan --------------------- Allan Hayes Mathematica Training and Consulting Leicester UK www.haystack.demon.co.uk hay at haystack.demon.co.uk Voice: +44 (0)116 271 4198 Fax: +44 (0)870 164 0565 "Lee Lin Yap" <kendallyap at hotmail.com> wrote in message news:9ki2fp$kcu$1 at smc.vnet.net... > Hi, > > I'm currently doing a software programming for my final year project. I need > to solve a partial fraction expansion in my program. I have a numerator and > I got the roots for it. But I can't solve the problem to get the answer that > I need. > > _________________________________________________________________ > Get your FREE download of MSN Explorer at http://explorer.msn.com/intl.asp > >