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Discriminant of cubic equation in terms of coefficients

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  • #26
Dick
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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
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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
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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
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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
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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:
 
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  • #31
Dick
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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
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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.
 
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  • #33
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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
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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
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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.
 
  • #36
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Cool, and thanks for your kind words epenguin. :smile: And thank you so much for your help, Dick and epenguin, I am grateful to you for you took the time and effort to help me with this!
 

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