Discriminant of cubic equation in terms of coefficients

  • Thread starter Dowland
  • Start date
  • #1
Dowland
51
0
1. Background/theory

We know that if the equation x3+px2+qx+r=0 has solutions x1, x2, x3 then

x1 + x2 + x3 = -p
x1x2 + x2x3 + x3x1 = q
x1x2x3 = -r

2. Problem statement

Find (x1 - x2)2(x2 - x3)2(x3 - x1)2 as an expression containing p,q,r.

That is, I'm supposed to find the discriminant of the above cubic equation in terms of its coefficients.

The Attempt at a Solution


Now, I have approached the problem in two ways. First, I expanded the whole expression out and tried to manipulate the expression to get it in the desired terms. It is a very lenthy and laborious process though and I had trouble getting the right expression. So there must be some more elegant solution to this problem than just multiplying the expression out and manipulate it.

My best attempt at a "more elegant" solution was to reason as follows:

When we expand the expression (x1 - x2)2(x2 - x3)2(x3 - x1)2 we get a polynomial with terms of degree 6 in x1, x2, x3. Now, p has degree 1, q has degree 2 and r has degree 6 so the possible terms in the expression is:

p6, q3, r2, pqr, p3r, p4q, p2q2.

So we can write:

(x1 - x2)2(x2 - x3)2(x3 - x1)2 = Ap6 + Bq3 + Cr2 + Dpqr + Ep3r + Fp4q + Gp2q2 where A,...,G are constants.

Then I tried to determine A,...,G by finding cubic polynomials with given roots and thus given values of the discriminant and p, q, r. I easily determined the terms -27r2 and -4q3 by just letting p=0 but when I tried to determine the other constants I got weird numbers like A=770,4 and G=87,7 etc, whereas I expected those constants to be zero (since the discriminant does not contain p2q2, for instance). I can't explain why...

I am grateful for any guidance here! Also, I apologize for possible language errors, English is not my native.
 

Answers and Replies

  • #2
36,243
13,294
Then I tried to determine A,...,G by finding cubic polynomials with given roots and thus given values of the discriminant and p, q, r. I easily determined the terms -27r2 and -4q3 by just letting p=0 but when I tried to determine the other constants I got weird numbers like A=770,4 and G=87,7 etc, whereas I expected those constants to be zero (since the discriminant does not contain p2q2, for instance). I can't explain why...
I guess you made a calculation error somewhere, the approach looks possible (but not very elegant - you have to hope that the expression can be expressed like that, you do not show it).
 
  • #3
Dowland
51
0
I guess you made a calculation error somewhere, the approach looks possible (but not very elegant - you have to hope that the expression can be expressed like that, you do not show it).
Well, I gave the argument that in terms of p, q, r, the listed terms are the only possible terms in the expression. So if there is an expression in p, q, r, it must consist of some collection of one or more of those terms. Isn't that enough?
 
  • #4
36,243
13,294
So if there is an expression in p, q, r
That's the point - you just assume this (it is right here).
 
  • #5
Dowland
51
0
That's the point - you just assume this (it is right here).
Yes, but it's more or less assumed in the problem statement. The author of the textbook has written: "Find (x1 - x2)2(x2 - x3)2(x3 - x1)2 as an expression containing p, q, r." I think it implies that I may assume that there is such an expression.

Perhaps it's not very elegant then but I still think it's O.K. given the level of the problem. Thanks for pointing it out though, I see that it's definitely not something obvious to assume...
 
Last edited:
  • #6
epenguin
Homework Helper
Gold Member
3,959
995
First, I expanded the whole expression out and tried to manipulate the expression to get it in the desired terms. It is a very lenthy and laborious process though and I had trouble getting the right expression. So there must be some more elegant solution to this problem than just multiplying the expression out and manipulate it.

It doesn't have to be as difficult as all that. Essential feature of it is its symmetry. Look at both your discriminant expression and your equations for coefficients and notice they are unchanged by any exchange of the roots in the expressions, such as exchanging x1 and x3.

Ʃx1 = x1 + x2 + x3 = -p
Ʃx1x2 = x1x2 + x2x3 + x3x1 = q
Ʃx1x2x3 = x1x2x3 = -r

Here I'm using the convenient 'Ʃ notation' I hope is obvious.

When you expand your expression every term is going to be of form aijkx1ix2jx3j where aijk is a coefficient (integer) and i+j+k = 6 .

Now once you have got the coefficient of a term, say for example of x13x22x3 the symmetry tells you the coefficient of x12x23x3 is exactly the same, and that of x13x2x32 is the same too. You don't have to work them out separately. And there are 6 of them but you can write them down all together as just Ʃx13x22x3. Or rather if I am right that the coefficient is 2 in your expression as 2Ʃx13x22x3.

So the first stage is writing all your terms and overall expression down like that.

The second stage is putting that in terms of your p, q, r (which actually a theorem says you can always do for any degree polynomial). You may recognise one like a222x12x22x32 term that you'll know how to relate. Others, most, need a bit more work but are not all that difficult. If you get stuck at that stage come back and show how far you got.

Beware of superficial misunderstanding of the Ʃ notation and do not be misled into thinking say x1 or Ʃx1 is a factor Ʃx1x2, you may have to rewrite terms out explicitly till you get used to it.

More explanations are certainly in the next section or chapter of your textbook under some title like 'symmetric polynomials'.
 
Last edited:
  • #7
epenguin
Homework Helper
Gold Member
3,959
995
For the first step you have only five coefficients to work out!
 
Last edited:
  • #8
Dowland
51
0
This is really interersting epenguin. My textbook doesn't seem to cover the idea of symmetry - or it doesn't mention anything about it explicitly I should say because when I think about it, I have definitely encountered some symmetrical expressions earlier in the book. (Perhaps it wants the reader to figure it out by oneself or later as one learn more mathematics.) Where can I learn more about this?

And regarding the solution of the problem, I mentioned in the first post that I tried to expand the expression and write it in the desired terms, however unsuccessfully. What I didn't mention was that I was very close! I got the terms right in p, q, r but the coefficients was wrong. However, the coefficients of the expanded expression was not a problem I think because I noticed immediately when there where any "irregularities" (somehow I expected the expression to be "regular", in the way you describe it in your post), went backed and checked and coreccted my error. I must have screwed up when I tried to apply some algebraic identities in the manipulation process. I'll go back and see where I screwed up but it's two full pages of algebraic manipulations and I'm getting sick and tired by just looking at it right now. :tongue2: Perhaps I'll give the problem a new try later by applying what you just taught me and see if it goes smoother!

Thanks for your reply, I appreciate it!
 
  • #9
Dick
Science Advisor
Homework Helper
26,263
621
This is really interersting epenguin. My textbook doesn't seem to cover the idea of symmetry - or it doesn't mention anything about it explicitly I should say because when I think about it, I have definitely encountered some symmetrical expressions earlier in the book. (Perhaps it wants the reader to figure it out by oneself or later as one learn more mathematics.) Where can I learn more about this?

And regarding the solution of the problem, I mentioned in the first post that I tried to expand the expression and write it in the desired terms, however unsuccessfully. What I didn't mention was that I was very close! I got the terms right in p, q, r but the coefficients was wrong. However, the coefficients of the expanded expression was not a problem I think because I noticed immediately when there where any "irregularities" (somehow I expected the expression to be "regular", in the way you describe it in your post), went backed and checked and coreccted my error. I must have screwed up when I tried to apply some algebraic identities in the manipulation process. I'll go back and see where I screwed up but it's two full pages of algebraic manipulations and I'm getting sick and tired by just looking at it right now. :tongue2: Perhaps I'll give the problem a new try later by applying what you just taught me and see if it goes smoother!

Thanks for your reply, I appreciate it!

This is a very interesting sort of problem. There is a theorem that any symmetric polynomial like yours is a polynomial expression in the elementary symmetric polynomials (x1+x2+x3), (x1*x2+x2*x3+x1*x3) and x1*x2*x3 that define your p, q and r. So you don't have to assume that there is one. There is a proof that there is. And one of the ways of proving that is an algorithm to construct a solution. http://en.wikipedia.org/wiki/Elementary_symmetric_polynomial

Using the algorithm, I managed to get the answer. It took a while (both to understand the algorithm and to lay out the solution). I might not have attempted it if I didn't have a CAS (computer algebra system) to do the actual work. I used maxima. It is tedious and error prone. But it is fun to see it finally work. And maybe there is a clever way to bypass the tedium but I'm not seeing it yet. Maybe have a look back... And maybe epenguin has something simpler in mind.
 
Last edited:
  • #10
Dick
Science Advisor
Homework Helper
26,263
621
For the first step you have only five coefficients to work out!

Ah, right. I'm looking back now. So p^6 and p^4*q can't be in the solution. So picking five different (x1,x2,x3) combinations in the right way should give you five linearly independent equations to solve for the coefficients. Still pretty complex. But probably easier than hacking through the polynomial expansions.

And ok, now that I'm done amusing myself I note the OP already correctly has the terms -27r^2 and -4q^3. This is starting to seem easier and easier. Now there's only three variables to determine. I think Dowland was already on the right track and just made a mistake. The coefficents aren't weird, they are all integers. And p^2*q^2 is in the answer. Why does Dowland think it isn't? Maybe that's the mistake.
 
Last edited:
  • #11
Dowland
51
0
Ah, right. I'm looking back now. So p^6 and p^4*q can't be in the solution. So picking five different (x1,x2,x3) combinations in the right way should give you five linearly independent equations to solve for the coefficients. Still pretty complex. But probably easier than hacking through the polynomial expansions.

And ok, now that I'm done amusing myself I note the OP already correctly has the terms -27r^2 and -4q^3. This is starting to seem easier and easier. Now there's only three variables to determine. I think Dowland was already on the right track and just made a mistake. The coefficents aren't weird, they are all integers. And p^2*q^2 is in the answer. Why does Dowland think it isn't? Maybe that's the mistake.
I have tried it twice and have deduced the five equations further down by the following method:
1. Construct a polynomial of degree 3 and compute its discriminant Δ. (I have constructed polynomials where P=-1 in every polynomial.)
2. Set Δ=Ap6 - 4q3 - 27r2 + Dpqr + Ep3r + Fp4q + Gp2q2.
3. Compue the terms -4q3, -27r2 and move them over to the left hand side of the equation.

320=A+16D-4E-4F+16G
1575=A+81D-9E-9F+81G
4208=A+256D-16E-16F+256G
11975=A+625D-25E-25F+625G
24768=A+1296D-36E-36F+1296G

This gives weird non-integer values for some of the unknowns.
 
  • #12
Dowland
51
0
This is a very interesting sort of problem. There is a theorem that any symmetric polynomial like yours is a polynomial expression in the elementary symmetric polynomials (x1+x2+x3), (x1*x2+x2*x3+x1*x3) and x1*x2*x3 that define your p, q and r. So you don't have to assume that there is one. There is a proof that there is. And one of the ways of proving that is an algorithm to construct a solution. http://en.wikipedia.org/wiki/Elementary_symmetric_polynomial

Using the algorithm, I managed to get the answer. It took a while (both to understand the algorithm and to lay out the solution). I might not have attempted it if I didn't have a CAS (computer algebra system) to do the actual work. I used maxima. It is tedious and error prone. But it is fun to see it finally work. And maybe there is a clever way to bypass the tedium but I'm not seeing it yet. Maybe have a look back... And maybe epenguin has something simpler in mind.
Very interesting, thank you for the link! This looks like quite profound mathematical theory above my level and my brain is going berserk trying to read it haha... But it's nice to know that there is a theorem that supports my assumption although I don't understand it yet.
 
  • #13
Dick
Science Advisor
Homework Helper
26,263
621
I have tried it twice and have deduced the five equations further down by the following method:
1. Construct a polynomial of degree 3 and compute its discriminant Δ. (I have constructed polynomials where P=-1 in every polynomial.)
2. Set Δ=Ap6 - 4q3 - 27r2 + Dpqr + Ep3r + Fp4q + Gp2q2.
3. Compue the terms -4q3, -27r2 and move them over to the left hand side of the equation.

320=A+16D-4E-4F+16G
1575=A+81D-9E-9F+81G
4208=A+256D-16E-16F+256G
11975=A+625D-25E-25F+625G
24768=A+1296D-36E-36F+1296G

This gives weird non-integer values for some of the unknowns.

If you think about it, you'll never get a power of any single variable that's greater than four out of your polynomial. That means you don't need Ap^6 and Fp^4q in there, and there's really only three variables.What values of x1,x2 and x3 are you using?
 
  • #14
epenguin
Homework Helper
Gold Member
3,959
995
I don't say there is only one way to do it.

The way I suggested is you are asked about

(x1 - x2)2(x2 - x3)2(x3 - x1)2

I mentioned that the terms when expanding this are all of

aijkx1ix2jx3j where aijk is a coefficient (integer) and i+j+k = 6

and there are just 5 such terms.

It is easy to wrtie down the five ijk combinations. Not too hard to find the coefficients. No one has tried this yet, so if you get stuck you can.
 
Last edited:
  • #15
Dick
Science Advisor
Homework Helper
26,263
621
I don't say there is only one way to do it.

The way I suggested is you are asked about

(x1 - x2)2(x2 - x3)2(x3 - x1)2

I mentioned that the terms when expanding this are all of

aijkx1ix2jx3j where aijk is a coefficient (integer) and i+j+k = 6

and there are just 5 such terms.

It is easy to wrtie down the five ijk combinations. Not too hard to find the coefficients. No one has tried this yet, so if you get stuck you can.

Why do you think there are 5 terms in the expansion? I count 19.
 
  • #16
epenguin
Homework Helper
Gold Member
3,959
995
Why do you think there are 5 terms in the expansion? I count 19.

I'm sorry, I missed out the Ʃ in a formula, the post had been tricky typographically to write and I think I deleted the symbol by mistake.

In the sense of the Ʃ notation of #6.

A phrase in #6 should read:

aijkƩx1ix2jx3j where aijk is a coefficient (integer) and i+j+k = 6

I say there are just five such terms.
 
  • #17
Dick
Science Advisor
Homework Helper
26,263
621
I'm sorry, I missed out the Ʃ in a formula, the post had been tricky typographically to write and I think I deleted the symbol by mistake.

In the sense of the Ʃ notation of #6.

A phrase in #6 should read:

aijkƩx1ix2jx3j where aijk is a coefficient (integer) and i+j+k = 6

I say there are just five such terms.

Ok, I see what you are saying. The 'sum' here means sum over all permutations of i,j,k. So then, yes, there are 5. ijk=420,411,330,321,222. So you get those coefficients, then go and figure out what the contributions are to each from 5 possible terms r^2, pqr, p^3r, q^3, p^2q^2 (this is where it might get complicated) and match them up in a system of 5 linear equations in 5 unknowns. Sounds complicated, but certainly doable.

But practically I think Dowland should just stick to what I assume is the original program and pick enough numerical values of x1,x2,x3 and plugging them into both sides of the equation to get a system of linear equations. Dowland already has two coefficients and counting correctly there are only 3 left. It worked out pretty easily for me. I just think there may be numerical mistakes in the the equations derived.
 
Last edited:
  • #18
epenguin
Homework Helper
Gold Member
3,959
995
Ok, I see what you are saying. The 'sum' here means sum over all permutations of i,j,k. So then, yes, there are 5. ijk=420,411,330,321,222. So you get those coefficients, then go and figure out what the contributions are to each from 5 possible terms r^2, pqr, p^3r, q^3, p^2q^2 (this is where it might get complicated) and match them up in a system of 5 linear equations in 5 unknowns. Sounds complicated, but certainly doable.

But practically I think Dowland should just stick to what I assume is the original program and pick enough numerical values of x1,x2,x3 and plugging them into both sides of the equation to get a system of linear equations. Dowland already has two coefficients and counting correctly there are only 3 left. It worked out pretty easily for me. I just think there may be numerical mistakes in the the equations derived.

Yes he would do well to do that, since you have found it can be done, as it is an original approach.

I only said if you expand the discriminant (x1 - x2)2(x2 - x3)2(x3 - x1)2 you get an expression made up of the five Ʃs and integers. The expressions for p, q, r are also Ʃs. They are different Ʃs than the 6-degree ones of the discriminant, but the latter can be expressed in terms of the former without too much difficulty.
 
Last edited:
  • #19
Dick
Science Advisor
Homework Helper
26,263
621
Yes he would do well to do that, since you have found it can be done, as it is an original approach.

I only said if you expand the discriminant (x1 - x2)2(x2 - x3)2(x3 - x1)2 you get an expression made up of the five Ʃs and integers. The equations for p, q, r are also Ʃs. They are different Ʃs than the 6-degree ones of the discriminant, but the latter can be expressed in terms of the former without too much difficulty.

I agree. I didn't actually do it your way, so I'm really not sure how complicated it would get. But Dowland is already pretty close. Putting x1=1, x2=1, x3=0, gives you one of the 3 unknown coefficients directly. It looks downhill from there.
 
  • #20
epenguin
Homework Helper
Gold Member
3,959
995
So we can write:

(x1 - x2)2(x2 - x3)2(x3 - x1)2 = Ap6 + Bq3 + Cr2 + Dpqr + Ep3r + Fp4q + Gp2q2 where A,...,G are constants.

Then later on you write "there cannot be any p6 term".

I hope you are not concluding from the fact that there is no x16 term that A = 0 ?

It does give you some relation between the constants though, so I think you could reduce the number of them.

I trust you are using special cases so as to reduce the number of unknowns to be solved for at a time? :wink:

Ah, I see that is what Dick is saying.
 
  • #21
Dick
Science Advisor
Homework Helper
26,263
621
Then later on you write "there cannot be any p6 term".

I hope you are not concluding from the fact that there is no x16 term that A = 0 ?

It does give you some relation between the constants though, so I think you could reduce the number of them.

I trust you are using special cases so as to reduce the number of unknowns to be solved for at a time? :wink:

Ah, I see that is what Dick is saying.

But I think you can conclude that. p^6 is the only term that can give you an x1^6 contribution and there isn't one. So its coefficient must be 0. Likewise p^4q isn't going to work.
 
  • #22
epenguin
Homework Helper
Gold Member
3,959
995
But I think you can conclude that. p^6 is the only term that can give you an x1^6 contribution and there isn't one. So its coefficient must be 0. Likewise p^4q isn't going to work.

Oops yes, I was thinking without examining that there would be another term that could give x6 that might for instance have coefficient equal to -A, but examining the terms you are right that there isn't one. So A = 0

And there is another term that is 0 for similar reasons if I am not mistaken.
 
  • #23
Office_Shredder
Staff Emeritus
Science Advisor
Gold Member
5,461
1,391
I noticed something that saves a bit of algebra in the problem by doing some of the factorization for you if you have calculus at your disposal. If
[tex] f(x) = (x-x_1)(x-x_2)(x-x_3) [/tex]
then
[tex] f'(x) = (x-x_1)(x-x_2) + (x-x_1)(x-x_3) + (x-x_2)(x-x_3) [/tex]
and
[tex] f'(x_1) = (x_1-x_2)(x_1-x_3),\ f'(x_2) = (x_2-x_3)(x_2-x_1),\ f'(x_3) = (x_3-x_2)(x_3-x_1) [/tex]
which means
[tex] (x_1-x_2)^2 (x_1-x_3)^2 (x_2-x_3)^2 = -f'(x_1)f'(x_2)f'(x_3) [/tex]
You still can't evaluate the derivatives without having some xis involved, but if you group up things in the multiplication in the obvious way then this has basically taken care of the hard half of the work for you (since many p's and q's have been factored out for you already). From this I was able to do the remaining algebra in about a minute by just examining the remaining expressions and guessing what they were in terms of p,q and r.
 
  • #24
Dowland
51
0
If you think about it, you'll never get a power of any single variable that's greater than four out of your polynomial. That means you don't need Ap^6 and Fp^4q in there, and there's really only three variables. What values of x1,x2 and x3 are you using?
Ah, this makes sense! I honestly think that I reasoned like this for a minute buth dismissed the argument because I argued something like this: sometimes when you factor an expression you rewrite it by adding and subtracting terms - what says you won't add any terms where the variables have powers of 6? But when factoring an expression it doesn't make sense adding and subtracting variables with higher powers than the already given variables, right?

However, I solved the problem now by assuming that p6 and p4q can't be terms in the expression. It went fairly simple and smooth by using special cases, like letting p=0, r=0, q=0 etc. Thanks so much for your help epenguin and Dick!

One thing I still don't understand is why my first method (as shown in post #1) didn't work, when I had all seven terms in the expression? I really don't think I messed up the numbers anywhere.
 
  • #25
Dowland
51
0
Then later on you write "there cannot be any p6 term".

I hope you are not concluding from the fact that there is no x16 term that A = 0 ?
I don't think I understand. On what basis do you exclude the possibility of a p6 term? You say that when expanding the expression we get aijkƩx1ix2jx3k where aijk is a coefficient (integer) and i+j+k = 6. But that doesn't exclude the possibility of the case where i=6, j=0, k=0?
 
  • #26
Dick
Science Advisor
Homework Helper
26,263
621
I don't think I understand. On what basis do you exclude the possibility of a p6 term? You say that when expanding the expression we get aijkƩx1ix2jx3k where aijk is a coefficient (integer) and i+j+k = 6. But that doesn't exclude the possibility of the case where i=6, j=0, k=0?

p^6 contains a term x1^6. None of your other forms do, and neither does the original polynomial. Hence it's coefficient must be 0. Same reasoning with p^4q. It contains an x1^5*x2 term.
 
  • #27
Dick
Science Advisor
Homework Helper
26,263
621
Ah, this makes sense! I honestly think that I reasoned like this for a minute buth dismissed the argument because I argued something like this: sometimes when you factor an expression you rewrite it by adding and subtracting terms - what says you won't add any terms where the variables have powers of 6? But when factoring an expression it doesn't make sense adding and subtracting variables with higher powers than the already given variables, right?

However, I solved the problem now by assuming that p6 and p4q can't be terms in the expression. It went fairly simple and smooth by using special cases, like letting p=0, r=0, q=0 etc. Thanks so much for your help epenguin and Dick!

One thing I still don't understand is why my first method (as shown in post #1) didn't work, when I had all seven terms in the expression? I really don't think I messed up the numbers anywhere.

You're welcome! Your first attempt should have worked if you had gotten all of the numbers right and you managed to get 5 independent equations. That's why I was asking how you picked x1,x2,x3 for each one.
 
  • #28
Dick
Science Advisor
Homework Helper
26,263
621
One thing I still don't understand is why my first method (as shown in post #1) didn't work, when I had all seven terms in the expression? I really don't think I messed up the numbers anywhere.

Now that you know the correct values for A,B,C,D,E,F,G you should be able to substitute them into your equations from post 11 and see if the numbers check. I tried that and they all check out except for the third one. What happened there?
 
  • #29
epenguin
Homework Helper
Gold Member
3,959
995
I don't think I understand. On what basis do you exclude the possibility of a p6 term? You say that when expanding the expression we get aijkƩx1ix2jx3k where aijk is a coefficient (integer) and i+j+k = 6. But that doesn't exclude the possibility of the case where i=6, j=0, k=0?

That was a mistake. I thought you had picked it up too.

re your last sentence, starting from your original equation for the disciminant in #1; (x1 - x2)2(x2 - x3)2(x3 - x1)2 you have any root (and x1 means any root, have to get used to the idea that nothing in the general algebra ever distinguishes one root from another) to the fourth power at most. So there was never any reason for a p6 term and the p4q term. My #6 should better have specified i,j,k≤4 as well as i + j + k = 6.
 
  • #30
epenguin
Homework Helper
Gold Member
3,959
995
It now seems to be taking a long time. It seems worth completing since I think you will not find it done this way in any Algebra textbook when treating discriminants. And it is possibly a quicker calculation than what you will find. (They don't do it that way because it is not the lesson they are trying to impart.)

We have got an equation in 5 unknowns:

Bq3 + Cr2 + Dpqr + Ep3r + Gp2q2 = 0

Making x1 = 0 , r = 0 you get B immediately. Making e.g. x1 = x2 = 1, x3 = -2 you have p = 0, and get C immediately. From what's left you can make 3 equations in the remaining 3 unknowns with simple number coefficients etc.

For the conventional method of getting the discriminant I tried different notations and variations of approach. The best* seemed to be the same form, calling the roots α, β, γ but writing it:

2 - 2αβ + β2)(α2 - 2αγ + γ2)(β2 - 2βγ + γ2)

As mentioned, the symmetry reduces the calculation considerably - you have only to calculate the coefficients of the terms

α4β2, α4βγ, α3β3, α3β2γ, α2β2γ2

Then the coefficients of
Ʃα4β2, Ʃα4βγ, Ʃα3β3, Ʃα3β2γ, Ʃα2β2γ2
are just the same.

Whilst calculating some individuals of the five expressions above you may find it convenient to write out the expression without γ2 and/or γ if you know they are not going to appear in the final result, e.g. for calculating

α3β2γ

it may help to write out

2 - 2αβ + β2)(α2 - 2αγ)(β2 - 2βγ)

Calculating these Ʃ's is considered a preliminary to calculating your other Ʃ's that appear in the determinant expression of #1.

*this one took me maybe 25 min., two other approaches a bit longer plus bumbling. This is not a race. But explains why helpers might like to see students come back with their results and the problem finished. :wink:
 
Last edited:
  • #31
Dick
Science Advisor
Homework Helper
26,263
621
Well, Dowland said the answer was obtained in post #24. I don't have any idea whether it's correct or not since the solution wasn't presented. I was also kind of curious as to why Dowland was convinced the linear equations were correct when one (I think) wasn't. Guess we'll never know.
 
  • #32
Dowland
51
0
Sorry guys! I have been too lazy to write down the solution here. :redface: Anyways, here is the full solution I wrote down in my papers (I also remind you that English is not my native so have forgiveness for any language errors):

When we expand the expression Δ = (x1 - x2)2(x2 - x3)2(x3 - x1)2 we get a polynomial with terms of degree 6 in x1, x2, x3. Now, p has degree 1, q has degree 2 and r has degree 3 so no other terms than the following are possible in the expression:

p6, q3, r2, pqr, p3r, p4q, p2q2

Furthermore, when we expand the above expression we never get a term where a single variable x1, x2, x3 has a degree higher than 4. Thus, we may exclude the terms p6 and p4q as possible terms in the expression since they require the variables to have a degree higher than 4. We can now write:

Δ = Aq3 + Br2 + Cp2q2 + Dp3r + Epqr where A,...,E are constants.

Consider the expression x3 - x = x(x+1)(x-1). If we compare it to the general expression x3 + px2 + qx + r we see that p=0, r=0 so the discriminant only contains the term Aq3 = -A since q = -1. We compute the discriminant to 4, so -A = 4, or A = 4.

Now consider x3 - 7x - 6 = (x-1)(x-2)(x+3). p = 0 so Δ = -4q3 + Br2 = -4(-7)3 + 36B, since q = -7, r = -6. We compute the discriminant to 400, so 36B = 400 + Aq3, hence B = -27.

Consider x3 - 3x2 + 2x = x(x-1)(x-2). r = 0 so Δ = -4q3 + Cp2q2 = -4*23 + C(-3)2*22. We compute that Δ = 4, so 36C = 36, hence C = 1.

Consider x3 - 7x2 + 36 = (x+2)(x-3)(x-6). q = 0 so Δ = -27r2 + Dp3r = -27*362 + D(-7)3*36. We compute that Δ = 14400, so -12348D = 49392D, hence D = -4.

Until now we know that Δ = -4q3 - 27r2 + p2q2 - 4p3r + Epqr where E is some constant which remains to be determined.

Consider x3 - 2x2 - x + 2 = (x-1)(x+1)(x-2). We see that Δ = -4(-1)3 - 27*22 + (-2)2(-1)2 - 4(-2)3*2 + E(-2)(-1)*2 = 4E - 36. Furthermore, we compute that Δ = 36, so 4E = 72, hence E = 18.

We conclude that Δ = -4q3 - 27r2 + p2q2 - 4p3r + 18pqr.
 
Last edited:
  • #33
Dowland
51
0
Now that you know the correct values for A,B,C,D,E,F,G you should be able to substitute them into your equations from post 11 and see if the numbers check. I tried that and they all check out except for the third one. What happened there?
Ah! I see now the mistake I made (I accidently wrote 6192 instead of 6912...). The third equation in #11 should be 4928 = A + 256D -16E - 16F + 256G (I used the values 1, 4, -4 for the roots). Can't believe I messed up with the numbers twice - I thought I was sooo cautious...
 
  • #34
Dick
Science Advisor
Homework Helper
26,263
621
Sorry guys! I have been too lazy to write down the solution here. :redface: Anyways, here is the full solution I wrote down in my papers (I also remind you that English is not my native so have forgiveness for any language errors):

When we expand the expression Δ = (x1 - x2)2(x2 - x3)2(x3 - x1)2 we get a polynomial with terms of degree 6 in x1, x2, x3. Now, p has degree 1, q has degree 2 and r has degree 3 so no other terms than the following are possible in the expression:

p6, q3, r2, pqr, p3r, p4q, p2q2

Furthermore, when we expand the above expression we never get a term where a single variable x1, x2, x3 has a degree higher than 4. Thus, we may exclude the terms p6 and p4q as possible terms in the expression since they require the variables to have a degree higher than 4. We can now write:

Δ = Aq3 + Br2 + Cp2q2 + Dp3r + Epqr where A,...,E are constants.

Consider the expression x3 - x = x(x+1)(x-1). If we compare it to the general expression x3 + px2 + qx + r we see that p=0, r=0 so the discriminant only contains the term Aq3 = -A since q = -1. We compute the discriminant to 4, so -A = 4, or A = 4.

Now consider x3 - 7x - 6 = (x-1)(x-2)(x+3). p = 0 so Δ = -4q3 + Br2 = -4(-7)3 + 36B, since q = -7, r = -6. We compute the discriminant to 400, so 36B = 400 + Aq3, hence B = -27.

Consider x3 - 3x2 + 2x = x(x-1)(x-2). r = 0 so Δ = -4q3 + Cp2q2 = -4*23 + C(-3)2*22. We compute that Δ = 4, so 36C = 36, hence C = 1.

Consider x3 - 7x2 + 36 = (x+2)(x-3)(x-6). q = 0 so Δ = -27r2 + Dp3r = -27*362 + D(-7)3*36. We compute that Δ = 14400, so -12348D = 49392D, hence D = -4.

Until now we know that Δ = -4q3 - 27r2 + p2q2 - 4p3r + Epqr where E is some constant which remains to be determined.

Consider x3 - 2x2 - x + 2 = (x-1)(x+1)(x-2). We see that Δ = -4(-1)3 - 27*22 + (-2)2(-1)2 - 4(-2)3*2 + E(-2)(-1)*2 = 4E - 36. Furthermore, we compute that Δ = 36, so 4E = 72, hence E = 18.

We conclude that Δ = -4q3 - 27r2 + p2q2 - 4p3r + 18pqr.

Yep, that's it. Well done!
 
  • #35
epenguin
Homework Helper
Gold Member
3,959
995
That seems exactly right. Congratulations! :approve: You have done it by a method which seems not found in a book for this problem.

I think the mathematicians should have a look at this.

I venture that it could be used for discriminants in general and wider polynomial algebra etc., whether the mathematicians would consider that of interest, novel, convenient or insightful is another matter, but IMO you can be pleased with yourself. :thumbs:

You may be taken through conventional symmetric polynomial theory etc. now, but you already have a few insights into it so I hope are well motivated by this success.
 

Suggested for: Discriminant of cubic equation in terms of coefficients

Replies
4
Views
1K
Replies
4
Views
482
Replies
6
Views
503
  • Last Post
Replies
3
Views
1K
Replies
2
Views
1K
Replies
21
Views
1K
Replies
10
Views
624
Replies
4
Views
666
Top