Discriminant of cubic equation in terms of coefficients

[Dick]

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.

 

[Dowland]

Sorry guys! I have been too lazy to write down the solution here. 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.

 

[Dowland]

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...

 

[Dick]

Sorry guys! I have been too lazy to write down the solution here. 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!

 

[epenguin]

That seems exactly right. Congratulations! 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.

 

[Dowland]

Cool, and thanks for your kind words epenguin. 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!