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