Re: cubic polynomial

*To*: mathgroup at smc.vnet.net*Subject*: [mg22683] Re: [mg22646] cubic polynomial*From*: Andrzej Kozlowski <andrzej at tuins.ac.jp>*Date*: Sat, 18 Mar 2000 01:27:57 -0500 (EST)*Sender*: owner-wri-mathgroup at wolfram.com

on 3/17/00 3:10 PM, Andrzej Kozlowski at andrzej at tuins.ac.jp wrote: > on 3/17/00 5:48 AM, Andrzej Kozlowski at andrzej at tuins.ac.jp wrote: > >> on 3/16/00 3:11 PM, E.Carletti1 at E.Carletti1 at lse.ac.uk wrote: >> >>> >>> I have a problem: I want to solve this equation: >>> >>> x^3+(3c-1)(x^2)-4cx-4(c^2)=0 with respect to x. >>> >>> It has to have a real solution because it is continuous >>> in x. c is a positive parameter. >>> If i solve it numerically, plugging numbers for c, then it is fine. But >>> I would like an analytical solution: in this case i get only one >>> solution which should give me a real value for x (the other two are >>> imaginary) but >>> it has a square root with all negative members (all terms with c, which >>> is positive, with a sign - in front of it, so immaginary). How is that >>> possible? What procedure does mathematica use to solve cubic expression? >>> >>> How can I express the expression in a nicer way to get rid of these >>> negative terms? My feeling is that the programm is not able to simplify >>> the expression for the solution. >>> Could you please help me? if I write the expression I find in the paper >>> I am writing, noone will believe it is real! >>> And even more funny, if I plug number into the solution I find for x, it >>> comes the same number as I plug numbers directly into the function I >>> want to solve, except for the last part which is an imaginary number >>> which shoult tend to zero. >>> What is going on? >>> >>> Thank your for your help >>> >>> Elena Carletti >>> Financial Markets Group >>> LSE >>> Houghton Street >>> London WC2A 2AE >>> >>> Tel. 0044 (0)20 7955 7896 >>> Fax 0044 (0) 20 7242 1006 >>> >>> >>> >> >> I am afraid this your problem is as unsolvable as it is typical. Mathematica >> solves cubic equations using the famous formula attributed to the 16 century >> Italian mathematician Cardano (though actually it was discovered by another >> 16th century Italian Tartaglia). This formula (and another one for fourth >> degree equations discovered by Cardano) inevitably involves complex numbers >> even when the roots are real. >> >> For example consider the following equation: >> x^3 + x^2 - 12 x - 6 == 0 >> >> This has three real roots. You can see this just by looking at the graph >> Plot[x^3 + x^2 - 12 x - 6, {x, -4, 4}] >> >> (I personally prefer a different proof. Just evaluate >> In[3]:= >> Experimental`CylindricalAlgebraicDecomposition[x^3 + x^2 - 12 x - 6 == 0, x] >> Out[3]= >> 3 2 >> x == Root[#1 + #1 - 12 #1 - 6 & , 1] || >> >> 3 2 >> x == Root[#1 + #1 - 12 #1 - 6 & , 2] || >> >> 3 2 >> x == Root[#1 + #1 - 12 #1 - 6 & , 3] >> >> This shows that all three roots must be real!) >> >> >> Now look at the answer Solve gives >> >> In[4]:= >> Solve[x^3 + x^2 - 12 x - 6 == 0, x] >> >> Out[4]= >> 1 37 >> {{x -> -(-) + ------------------------- + >> 3 1/3 >> 3 (26 + 9 I Sqrt[617]) >> >> 1 1/3 >> - (26 + 9 I Sqrt[617]) }, >> 3 >> >> 1 37 (1 + I Sqrt[3]) >> {x -> -(-) - ------------------------- - >> 3 1/3 >> 6 (26 + 9 I Sqrt[617]) >> >> 1 1/3 >> - (1 - I Sqrt[3]) (26 + 9 I Sqrt[617]) }, >> 6 >> >> 1 37 (1 - I Sqrt[3]) >> {x -> -(-) - ------------------------- - >> 3 1/3 >> 6 (26 + 9 I Sqrt[617]) >> >> 1 1/3 >> - (1 + I Sqrt[3]) (26 + 9 I Sqrt[617]) }} >> 6 >> >> All roots look complex, though we know they are real. In general it is >> impossible to give an expression for arbitrary real roots of polynomial >> equations in terms of radicals without using complex numbers, although one >> can >> give such expressions if one does not insist on using radicals. In the above >> case one can give real expressions for the roots using trigonometric >> functions: for example the first root above is actually equal to >> >> >> 1 2 1 9 Sqrt[617] >> -(-) + - Sqrt[37] Cos[- ArcTan[-----------]] >> 3 3 3 26 >> and one can find similar real expressions for all of the others. >> >> If your equation involves parameters the situation is even more confusing. >> You >> may get roots which seem to involve complex numbers and others which appear >> to >> be real. But actually which one is real and which is not will depend on the >> values of the parameters, and in fact for certain values the roots that look >> complex may turn out to be real while those which look real may result in >> complex values. >> >> The moral is that if you want to use algebraic solutions of equations you >> simply can't avoid complex numbers. This is exactly what Tartaglia and >> Cardano >> discovered and this is the reason why they invented what they called >> "imaginary" numbers. > > -- > > When I wrote my first reply to this message I forgot to consider the equation > that caused the original posting. To determine the real roots of an algebraic > equation the best way is, in my opinion, to use on eof the most powerful > functions in Mathematica 4.0 as follows: > > In[1]:= > Experimental`CylindricalAlgebraicDecomposition[ > x^3 + 3*c*x^2 - x^2 - 4*c*x - 4*c^2 == 0, {c, x}] > Out[1]= > 3 2 2 2 > c < 0 && x == Root[#1 + 3 c #1 - #1 - 4 c #1 - 4 c & , > > 1] || c == 0 && (x == 0 || x == 1) || > > 3 2 2 > c > 0 && (x == Root[#1 + 3 c #1 - #1 - 4 c #1 - > > 2 > 4 c & , 1] || x == > > 3 2 2 2 > Root[#1 + 3 c #1 - #1 - 4 c #1 - 4 c & , 2] || > > 3 2 2 2 > x == Root[#1 + 3 c #1 - #1 - 4 c #1 - 4 c & , 3]) > > What this means is this. If c<0 then there is only one real root, which, as a > Root object can be described as Root[#1^3 + 3*c*#1^2 - #1^2 - 4*c*#1 - 4*c^2 & > , 1]. If you want to express it in terms of radicals you can do so: > > In[3]:= > ToRadicals[Root[#1^3 + 3*c*#1^2 - #1^2 - 4*c*#1 - 4*c^2 & , > 1]] > Out[3]= > 1 1 3 2 > - (1 - 3 c) + - Power[-27 c + 27 c + 9 c + > 3 3 > > 5 4 3 > 6 Sqrt[3] Sqrt[-27 c - 9 c - c ] + 1, 1/3] - > > 2 > (-9 c - 6 c - 1) / > > 3 2 > (3 Power[-27 c + 27 c + 9 c + > > 5 4 3 > 6 Sqrt[3] Sqrt[-27 c - 9 c - c ] + 1, 1/3]) > > If c==0 then there are just two possible real roots, x==0 and x==1. And if c>0 > then the real root is just one of the roots of the original equation! Which > one? That, of course, depends, on the value of c. > > > Andrzej Kozlowski > Toyama International University > JAPAN > http://sigma.tuins.ac.jp > I have to send a correction to the last part of the above message. Writing in a hurry I misinterpreted the meaning of Mathematica's answer: which is that when c>0 all three roots are real. So when c<0 there is just one real root, for c=0 two and c<0 all three roots are real. -- Andrzej Kozlowski Toyama International University JAPAN http://sigma.tuins.ac.jp

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