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In this arxiv paper (p. 11, eq. (3.2)) the authors claim that equation (3.2) is

... a quartic equation [...] which can be solved explicitly.

The equation in question is \begin{equation} %\tag{3.2} \frac{1}{m} = -z+\frac{\vartheta y}{\vartheta m-\varphi} - \frac{\left(\vartheta+\varphi\right)\left(1+\vartheta\varphi\right)y}{\left(\vartheta m-\varphi\right)\sqrt{\left[\left(1-\varphi\right)^{2}+m\left(1+\vartheta\right)^{2}\right]\left[\left(1+\varphi\right)^{2}+m\left(1-\vartheta\right)^{2}\right]}} \end{equation} with $\vartheta,\varphi\in\mathbb{R}$, $\left|\varphi\right|>0$, $y>0$ and $z,m\in\mathbb{C}^{+}$.

After some algebra, I arrive at \begin{align} \tag{2} \left(\left(\vartheta+\varphi\right)\left(1+\vartheta\varphi\right)y\right)m ={}& \overbrace{\left(\vartheta y m-\left(1+z m\right)\left(\vartheta m-\varphi\right)\right)}^{A} \\ &\underbrace{\cdot\sqrt{\left[\left(1-\varphi\right)^{2}+m\left(1+\vartheta\right)^{2}\right]\left[\left(1+\varphi\right)^{2}+m\left(1-\vartheta\right)^{2}\right]}}_{B} %&\underbrace{\cdot\sqrt{\left(1-\varphi\right)^{2}+m\left(1+\vartheta\right)^{2}} \sqrt{\left(1+\varphi\right)^{2}+m\left(1-\vartheta\right)^{2}}}_{B} \text{.} \end{align}

The term on the LHS is linear in $m$. Heuristically, the term on the RHS should behave as follows: The first factor $A$ is quadratic in $m$, the discriminant of the second factor is also quadratic under the root. By taking the root one gets orders $m$ and $m^{\frac{1}{2}}$. Multiplying this with $m^{2}$ and $m$ from factor $A$, one should get terms of orders $m^{3},m^{\frac{5}{2}},m^{2},m^{\frac{3}{2}},m,m^{\frac{1}{2}}$ and a constant. If one substitutes $x^{2}=m$, one has accordingly terms of order $x^{6},x^{5},x^{4},x^{3},x^{2},x$ and a constant.

Mathematica solves (2) as a sextic equation and gives 6 roots (very long terms). Mathematica also gives 6 roots if I assume $\vartheta,\varphi\in\mathbb{R}$, $\left|\varphi\right|>0$, $y>0$ and $z,m\in\mathbb{C}^{+}$ as in the main paper.

Maybe the sextic equation collapses to quartic one by the assumption that $m\in\mathbb{C}^{+}$. If one substitutes $x^{2}=m$, the discriminant of $B$ is given by \begin{equation} %\tag{3} \left(\left(1-\varphi\right)^{2}+x^{2}\left(1+\vartheta\right)^{2}\right)\left(\left(1+\varphi\right)^{2}+x^{2}\left(1-\vartheta\right)^{2}\right) = \left(x\pm\frac{\left(\varphi+1\right)\mathsf{i}}{\vartheta-1}\right)\left(x\pm\frac{\left(\vartheta-1\right)\mathsf{i}}{\vartheta+1}\right) \text{,} \end{equation} where $\pm$ means that each factor occurs ones with positive and ones with negative sign. So in total we have on the RHS a product of four factors, each in $x$.

Because $\left|\varphi\right|<1$, the nominator of every root never changes in sign regardless of the value of $\varphi$. By the requirement that $m\in\mathbb{C}^{+}$, the possible values of $\vartheta$ get restricted. My argument goes like this: Since the roots of the discriminant are somehow related to the solution of the sextic equation, the sextic equation collapses to a quartic one?!

Is this reasoning right and can somebody please help to make it rigorous? And, most important, what are the four (assuming the authors in the linked paper are right) solutions to the first equation?

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(Too long for a comment.)

Hm, there may be a typo in the paper or an unqualified statement. When you have an equation in $x$ of form,

$$P_1(x) = P_2(x)\sqrt{P_3(x)}$$

like your $(2)$ above, the straightforward way to get its degree is to square both sides, then equate it to zero,

$$\big(P_1(x)\big)^2-\Big(P_2(x)\sqrt{P_3(x)}\Big)^2 = 0\tag3$$

As you surmised, the degree "apparently" is $6$th deg. However, your particular equation factors, into a linear and quintic factor in $m$. (Put your $(2)$ in the form of $(3)$ and factor it with Mathematica and see what happens.)

But for the special cases $\vartheta = 0$ and $\vartheta = \pm 1$, the leading coefficient of the quintic vanishes, and you end up with quartic.

Thus, either there is a typo in the paper, or I'm missing an assumption, or the comment by the authors re the quartic is only for the special cases of $\vartheta$ (which they do mention in the paragraph immediately after $(3.2)$).

But if indeed you just need a quartic, then an easy way to get the roots is to use the simplified solution in this post.

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  • $\begingroup$ Thank you for your help. The linear term in your mentioned factorization is $m\vartheta-\varphi$, so it vanishes only for $\vartheta=0$ or if $m=\frac{\varphi}{\vartheta}$. For both, $\vartheta=0$ and $\vartheta=\pm 1$ the quintic factor becomes quartic. But since the linear factor only vanishes for $\vartheta=0$, only then I get the quintic polynomial. The leading coefficient of the 6th degree polynomial is $-z^2\vartheta^2\left(1+\vartheta\right)^2\left(1-\vartheta\right)^2$, therefore it vanishes at $\left\{0,\pm 1\right\}$, as you said. But for $\vartheta=\pm 1$ I still have a quintic. $\endgroup$ – Marco Breitig Aug 3 '15 at 9:13
  • $\begingroup$ @MarcoBreitig: By focusing on the $6$th deg, you are making things overly complicated. If a polynomial is known to factor, then you just inspect the factors _individually_. Thus, when you have a "sextic" of form $P_6(m) = P_1(m) P_5(m) = 0$, then the 6 roots are given by $P_1(m) = 0$ and $P_5(m) = 0$. In general, the quintic is not solvable in radicals. Bu as you've seen, for the particular cases $\vartheta = 0$ and $\vartheta = \pm1$, your quintic reduces to a quartic. Thus, for those cases, the original sextic is solvable in radicals as well. I hope things are clear now. $\endgroup$ – Tito Piezas III Aug 3 '15 at 12:54
  • $\begingroup$ Piezas: For solving the polynomial I agree, factor the polynomial as much as possible and identify the roots of each factor. I was merely commenting on the overall degree, which only changed to a quintic. But I very much appreciate your help, because I was stuck and needed some clarification. $\endgroup$ – Marco Breitig Aug 3 '15 at 13:20

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