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This is in continuance to my earlier post : Deriving expression for general cubic equation solution. Also, the links are repeated for the first four pages of the book by Dickson, titled "Introduction to Theory of Algebraic Equations" are given as : page #1,$\,$page #2,$\,$ page #3, $\,$page #4

On page #4, the second question asks to show :
The cubic ($2$), i.e. the reduced cubic equation: $y^3+py+q=0$, with $$p = c_2-\frac{1}{3}c_1^2,\,\,\,\,\,\,\,\,\,\, q= -c_3+\frac{1}{3}c_1c_2 - \frac{2}{27}c_1^3$$ $$R = \frac{1}{4}q^2 + \frac{1}{27}p^3=\frac{1}{4}(-c_3+(\frac{1}{3}c_1c_2 - \frac{2}{27}c_1^3))^2 + \frac{1}{27}(c_2-\frac{1}{3}c_1^2)^3 = \frac{1}{4}(c_3^2+ (\frac{1}{9}c_1^2c_2^2 + \frac{4}{27^2}c_1^6 - \frac{4}{81}c_1^4c_2)-\frac{2}{3}c_1c_2c_3+\frac{4}{27}c_1^3c_3)+\frac{1}{27}(c_2^3+\frac{1}{27}c_1^6-c_2^2c_1^2+\frac{1}{3}c_2c_1^4) $$ $$=\frac{1}{4}c_3^2 + \frac{1}{36}c_1^2c_2^2+ \frac{1}{27^2}c_1^6-\frac{1}{81}c_1^4c_2 -\frac{1}{6}c_1c_2c_3 +\frac{1}{27}c_1^3c_3+\frac{1}{27}c_2^3+\frac{1}{27^2}c_1^6-\frac{1}{27}c_2^2c_1^2 +\frac{1}{81}c_2c_1^4$$ $$=\frac{1}{4}c_3^2+\frac{1}{36}c_1^2c_2^2+\frac{2}{27^2}c_1^6-\frac{1}{6}c_1c_2c_3+\frac{1}{27}c_1^3c_3+\frac{1}{27}c_2^3-\frac{1}{27}c_2^2c_1^3$$
has :
(i) one real root and two imaginary roots if $R\gt 0$,
(i) three real root, two of which are equal if $R= 0$,
(i) three real & distinct roots (irreducible case) if $R\lt 0$

jiten
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1 Answers1

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This is generally not true for $p,q \in \mathbb{C}$. For example, consider the polynomial $y^3+i=0$. Then $p=0, q=i$, and $R=\frac14q^2+\frac1{27}p^3=-\frac14 <0$ but it has no real root.

However, this is true for $p,q \in \mathbb{R}$. If $p,q \in \mathbb{R}$, we have at least one real root and any complex roots comes in complex conjugate pair.

\begin{align}(y_1-y_2)^2(y_2-y_3)^2(y_3-y_1)^2&=-27q^2-4p^3\\&=-(27q^2+4p^3)\\&=-4(27)\left( \frac14q^2+\frac1{27}p^3\right)\\&=-4(27)R \end{align}

If $R>0$, then $(y_1-y_2)^2(y_2-y_3)^2(y_3-y_1)^2<0$, hence we must have an imaginary root. Hence we have one real root and two imaginary roots.

If $R=0$, then we must have at least a pair of repeating root as $(y_1-y_2)^2(y_2-y_3)^2(y_3-y_1)^2=0$

If $R<0$, then $(y_1-y_2)^2(y_2-y_3)^2(y_3-y_1)^2>0$. Let's rule out the possibility that we have any complex root. Suppose $y_1 \in \mathbb{R}, y_2, y_3$ are complex conjugate. Then

\begin{align}(y_1-y_2)^2(y_2-y_3)^2(y_3-y_1)^2&=(y_1-y_2)^2(y_2-\bar{y_2})^2(\bar{y_2}-y_1)^2\\&=|y_1-y_2|^4(2\Im(y_2))i)^2\\&=-4 \Im(y_2)^2|y_1-y_2|^4 \le 0\end{align} which is a contradicition.

Hence if $R<0$, every root is distinct.

Siong Thye Goh
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  • Thanks a lot. The third case is a special case, as dealt here at mse: https://math.stackexchange.com/q/145590/424260. The link to this was got by me from another post at: https://math.stackexchange.com/q/257388/424260. Although, unable to understand it, expect would give more insight. – jiten Apr 30 '18 at 00:21
  • Would request some corrective comments for my post concerning 'logic for substitutions for cubic' at : https://math.stackexchange.com/q/2758275/424260. I feel that the equation of circle & hyperbola derived by me is doubtful, for second (b) substitution. The doubt in my little explanation is hindering my further thought process. In particular, I am confused as to how to interpret $p_1 = z^2$, there is only one way, i.e. to take $z$ as radius, but then the usual equation of circle is of the form : $(x-h)^2+(y-k)^2 =r^2$, but here have $radius = z = \sqrt{p_1}$. – jiten Apr 30 '18 at 01:12
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    no comment for now. – Siong Thye Goh Apr 30 '18 at 03:26
  • i don't have any geometric intuition so I am not the right person. unlikely i can answer u using any geometric approach. – Siong Thye Goh Apr 30 '18 at 05:53
  • Sorry for being too late. I had a simple circle's equation that can be at best stated for representation as a circle as: $\sqrt{p_1}^2=z^2$, i.e. square on both sides. I just want to know how this form can be used to interpret a circle, given $z$ is derived from $y$, which is a horizontally translated root derived from the root $x$. Or simply, how it can be fitted to the form $(x'-h)^2 + (y'-k)^2 = \sqrt{p_1}^2$, where the notation $x', y'$ is chosen to look different from $x,y$. – jiten Apr 30 '18 at 21:48
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    Fixing $p_1$, $z^2=p_1$ is not a circle, there are only two complex numbers that satisfies that. Locus of circle about the origin with radius $r$ can be written as $|z|=r$. – Siong Thye Goh Apr 30 '18 at 21:55
  • Thanks a lot for the enlightening answer. Will $z = \sqrt{p_1}$ work then. If yes, then how. By your earlier comment, it should be the locus approach. So, the usual circle equation is not applicable here, and locus approach works. – jiten Apr 30 '18 at 21:59
  • Let us continue this discussion in chat. – Siong Thye Goh Apr 30 '18 at 22:00
  • The stated post for the page $4$ statement about the sextic equation ($5$) having factor = $\frac{1}{3}$ is stated at : https://math.stackexchange.com/q/2761061/424260 – jiten Apr 30 '18 at 23:51
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    i don't think it means $\frac13$ is a root. but yup, will visit it when i have more time. – Siong Thye Goh Apr 30 '18 at 23:55
  • Two issues: (i) In second case above, have shown at least a pair of roots is repeating as at least one term involving any of the $3$ subtractions is $0$. But, not excluded the case that all three are equal. (ii) Concerns with the starting line of answer, regarding the scope of theorems for polynomials:- Not sure, but hope that all theorems (as conjugate root theorem) must be applicable for real coefficients' polynomials only. Say, for $x^2 +(2+i)x +2i=0$ have two roots $-i, -2$. Here, roots are not conjugates. Request references for helping me have firm ground with complex coefficients. – jiten May 09 '18 at 22:52
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    For the second case, if all three roots are equal, say $y_0$ then the coefficient of $y^2$ must be $-3y_0$, but we know the equation is depressed, hence $y_0=0$. For real coefficient, we have conjugate root theorem, it is not true in general. I have no recommendation for references. – Siong Thye Goh May 10 '18 at 02:01
  • Request elaboration for the statement that if all three roots are equal to some value, say $y_0$, then the coefficients of $y^2$ must be $-3y_0$. I have taken eqn. (2) as given on page $1$ of the book, and got the coefficient of $y^2=0$, due to the depressed form not having the second order term. But, how you got the coefficient of $y^2$ in the discriminant $R= \frac 14 q^2 + \frac{p^3}{27}= \frac{(y_1-y_2)^2(y_2-y_3)^2(y_3-y_1)^2}{-4\cdot(27)}=0$, as $-3y_0$ is not clear. Note I am taking the value of $R=0$ for the case of $y_1=y_2=y_3=y_0$. – jiten May 10 '18 at 04:15
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    try to expand $(y-y_0)^3$. – Siong Thye Goh May 10 '18 at 04:19
  • Although, am unable to find the significance of the third order term : $(y-y_0)^3$, but its expansion is : $y^3 -y_0^3 -3y^2y_0+3yy_0^2$. My guess is that for the depressed cubic equation (2): $y^3 +py +q = 0$, have taken root as $y=y_0$ leading to $(y-y_0)^3 +p(y-y_0) +q=0$. This can be further expanded to : $y^3 -y_0^3 -3y^2y_0+3yy_0^2 +p(y-y_0) +q =0$. – jiten May 10 '18 at 04:22
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    i mean we have proven for the cubic equation $y^3+py+q$ which we have derived from the original equation, if $R=0$, then it has repeating equation. I claimed that if it has three identical roots, $y_0$, then the root must be zero. as we can compare $y^3-3y_0y^2+3yy_0^2-y_0^3=y^3+py+q$ and by comparing coefficient of $y^2$, we get $-3y_0=0$ and hence $y_0=0$. – Siong Thye Goh May 10 '18 at 04:27
  • Thanks for that, but my confusion comes from being unable to do analysis for two repeating roots' case. Say, if two roots are equal (let, two roots are $y_0$, and the third is $y_1$), then can not use both roots at the same time. Will use either $y_0$ or $y_1$ at a time, as shown :>: (i) root is $y_0: (y-?)(y-y_0)^2 +p(y-y_0)+q = 0$; (ii) root is $y_1: (y-y_1)(y-?)^2+p(y-y_1)+q=0$. Sorry, but for two repeated roots' case, am unable to use your strategy to show the possibility in a similar manner; as you showed failure for all repeated roots' case. – jiten May 10 '18 at 04:36
  • Let us continue this discussion in chat. – Siong Thye Goh May 10 '18 at 04:59
  • Regarding formal proof for $y_0=0$ for $3$ equal roots($=y0$), $c1=3x_0,y=x−\frac{c1}{3}\implies y=x−x_0=0$, as $x=x0$. Have addtl. reason as book: 'Unattainable Attempt Avoid Casus Irreducibilis for Cubic Equations', by 'Sara Confalonieri', with image https://i.stack.imgur.com/UgzXg.png, for pg. #33, 34 takes instead a non-monic form of cubic as $\alpha_3x^3 + \alpha_2x^2 + \alpha_1x + \alpha_0$ with slight erratum (no errata page on springer or elsewhere) on top of pg. #34 as correct expression's for $p$'s denominator is $3\alpha_3$. Request chat as too much & not worthy of a seperate post. – jiten May 11 '18 at 08:25
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    hmm... I do not know what types of response are you expecting. – Siong Thye Goh May 11 '18 at 08:30
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    I am on the chat now. – Siong Thye Goh May 11 '18 at 08:33
  • Please see modified post of mine at : https://math.stackexchange.com/q/2758275/424260. Have the last section with 'update' from earlier post with inputs solely from the book as in my last comment earlier. The modification is aimed to clarify understanding, and request your critique and (if possible) discovery of the book's contents. – jiten May 13 '18 at 01:20
  • For the post at : https://math.stackexchange.com/q/2758275/424260, have got some material with first 2 pages of the chapter 3 of the book 'Polynomial Resolution' by 'William Hardy' at: https://i.stack.imgur.com/bjhTo.png, https://i.stack.imgur.com/Zf6PJ.png. – jiten May 26 '18 at 07:13
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    It is beyond my ability. perhaps reading up on Galois theory and Kummer theory might help you. – Siong Thye Goh May 27 '18 at 22:50