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The derivative of a degree $5$ polynomial $p\in\mathbb{C}[z]$ is a degree four polynomial $p'\in\mathbb{C}[z]$, and as such, the zeros of $p'$ may be found explicitly using the quartic formulae.

One may think of a finite Blaschke product as playing the same role on the unit disk that a polynomial plays on the plane (in some ways!). In particular, a degree $5$ finite Blaschke product has $5$ zeros and $4$ critical points in the disk. However in general, a finite Blaschke product $B$ will have additional critical points outside the unit disk.

Suppose $B$ is a finite Blaschke product of degree $n=5$. Let $w_1,w_2,w_3,w_4\in\mathbb{D}$ be the critical points of $B$ which lie in the disk. Assume that no $w_i=0$. Due to the conjugate symmetry of $B$, the critical points of $B$ outside the unit disk are exactly $1/\bar{w_1},1/\bar{w_2},1/\bar{w_3},1/\bar{w_4}\in\mathbb{D}\setminus\overline{\mathbb{C}}$. Thus if $B'=P/Q$ for relatively prime polynomials $P$ and $Q$, then $\deg(P)=8$.

Can the known conjugate symmetry of the degree $8$ polynomial $P$ across the unit circle allow us to find its zeros using the quartic formulae?

NOTE 1: The above is a narrowing of this question on MSE, wherein I also asked about whether a degree $8$ polynomial $p(z)$ whose set of zeros is known to consist of four conjugate pairs could be factored somehow using the quartic formula.

NOTE 2: The same question above goes for degree $4$ finite Blaschke products. Of course for a degree $3$ or less Blaschke product, there are at most $4$ critical points anyway, so one may just use the quartic formula straightforwardly.

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Even for degree $4$ Blaschke products, the critical points need not be contained in a solvable Galois extension of the field of coefficients of $P$. For instance, let $t$ be another variable which is supposed to parametrize the purely imaginary numbers. Take e.g. $w_k=1/(k+1)+t$. Then the Blaschke product is, up to a constant factor, $B(z)=\prod\frac{z-1/(k+1)-t}{(1/(k+1)-t)z-1}$. The numerator $P(z)$ of the derivative of $B(z)$ is a degree $6$ polynomial in $z$ over the field $\mathbb Q(t)$. It is not hard to see that the Galois group of $P(z)$ over $\mathbb C(t)$ is the symmetric group $S_6$. So there is no quartic formula expressing the $w_k$'s in terms of the coefficients of $P(z)$.

In order to get an arithmetic example, note that the Galois group of $P(z)$ over the smaller field $\mathbb Q(\sqrt{-1},t)$ is $S_6$ even more. So generically (for instance use Hilbert's irreducibility theorem) upon specializing $t$ to $\sqrt{-1}a$ for most rationals $a$ we get that the Galois group of $P(z)$ is still $S_6$. Instead of using HIT, one may use explicit examples. For instance $a=1/2$ works.

The same kind of construction works for degree $5$ too.

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For example, consider the Blaschke product $$ B(z) = {\frac { \left( z-1/2 \right) \left( z-1/3 \right) \left( z-1/4 \right) \left( z-1/5 \right) \left( z-1/6 \right) }{ \left( 1-z/2 \right) \left( 1-z/3 \right) \left( 1-z/4 \right) \left( 1-z/5 \right) \left( 1-z/6 \right) }} $$ The numerator of $B'(z)$ is $$13356\,{z}^{8}-222040\,{z}^{7}+1402555\,{z}^{6}-4211480\,{z}^{5}+ 6172978\,{z}^{4}-4211480\,{z}^{3}+1402555\,{z}^{2}-222040\,z+13356 $$ which is an irreducible polynomial with galois group "8T44" (according to Maple). This is a solvable group, so the polynomial should be solvable by radicals.

Indeed, if $w = z - 1/z$ for a root $z$, then $w$ is a root of $$ {w}^{8}-{\frac {26555231\,{w}^{6}}{455058}}+{\frac {1322220803\,{w}^{4 }}{1213488}}-{\frac {61878800\,{w}^{2}}{8427}}+{\frac{38769600}{2809}} $$ This is a quartic in $w^2$, so can be solved using the quartic formula. And then $$ z = \frac{w \pm \sqrt{w^2 +4}}{2}$$

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  • $\begingroup$ I believe that the point of the question is whether there are cases where the critical points can not be computed by a quartic formula. $\endgroup$ – Peter Mueller May 9 '17 at 21:22

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