7
$\begingroup$

Let $g, h \in \mathbb C[x]$ and $$ f(x, a) = (x-x_0)^m g(x) + h(x) (a-a_0),$$ where $m \ge 2$, $a \in \mathbb R$ and $a_0$ is a fixed real number. Suppose $g(x_0) \neq 0$ and $h(x_0) \neq 0$. By this setup, we should be able to have $n$ continuous functions $\alpha_1, \dots, \alpha_n: \mathbb R \to \mathbb C$ such that for each $t \in \mathbb R$, $\alpha_1(t), \dots, \alpha_n(t)$ constituents the zeros of $f(x, t)$. If $a \to a_0$, then we should have $m$ functions converges to $x_0$. I am wondering how these functions behave. More specifically, it seems to me: we can set $$ (x-x_0)^m g(x) + h(x)(a-a_0) = 0.$$ As $a \to a_0$, $g(x) \to g(x_0)$ and $h(x) \to h(x_0)$. If I am allowed to hand wave a little bit, then in a neighborhood of $a_0$ $$ \alpha_j(a) \approx x_0 + \left( \frac{-h(x_0)}{g(x_0)} (a-a_0) \right)^{1/m} \omega_j^m, \text{ for } j=1, \dots, m,$$ where we assume $\alpha_1, \dots, \alpha_m$ are functions converging to $x_0$ and $\omega_j^m$ are solutions to $x^m =1 $. Is there a way to a rigorous statement on the asymptotic behavior of $\alpha_j$'s?

$\endgroup$
  • $\begingroup$ I would write $ω_m^j=(ω_m)^j$ for the unit roots, it seems more natural. $\endgroup$ – LutzL Jan 11 at 12:01
1
+50
$\begingroup$

Let $o(|u|^r)$ ($r\ge0$) denote the class (and also an element of it) of functions $q(u)\in\mathbb{C}$ such that $\lim_{|u|\to 0^+} \frac{|q(u)|}{|u|^r} = 0$, that is, for all $\epsilon>0$ there exists $u^*>0$ such that $|q(a)|<\epsilon |u|^r$ for all $|u|\in (0,u^*)$. We will show that $$ \alpha_j(a) = x_0 +\left(-(a-a_0)\frac{h(x_0)}{g(x_0)}\right)^{\frac{1}{m}}\omega^j + o(|a-a_0|^{\frac{1}{m}}) $$ where $\omega = e^{\frac{2\pi i}{m}}$ is the primitive $m$-th root of unity. We denote by $z^{\frac{1}{m}}$ a fixed root $w$ of $w^m=z$, which is arbitrarily chosen. Since the roots differ by $\omega^r$ multiplicatively, the choice of a particular $(-(a-a_0)\frac{h(x_0)}{g(x_0)})^{\frac{1}{m}}$ does not affect validity of the statement. In what follows, $c^{\frac{1}{m}}$ is also understood in the same way unless $c\ge 0$ (as long as validity is not affected.)

Without loss of generality, we may assume that $x_0 = a_0 = 0$ and $g(0)=1$. By changing $-a\to a$, the given equation becomes $$ x^m g(x) = a\cdot h(x).\tag{*} $$ Assume $a>0$. By the change of variable $z =\frac{x}{a^{\frac{1}{m}}}$ we get modified equation: $$ z^m g(a^{\frac{1}{m}}z)=h(a^{\frac{1}{m}}z). $$ Let $F_a(z) = z^m g(a^{\frac{1}{m}}z)-h(a^{\frac{1}{m}}z).$ We can see that $\lim_{a\to 0^+}F_a(z) = F_0 (z)=z^m -h(0)$ and that $F_0(z)$ has $$\zeta_j = [h(0)]^{\frac{1}{m}}\omega^j,\quad j=1,2,\ldots,m$$ as its roots.

Claim: For all $\epsilon\in (0,\frac{|h(0)|^{\frac{1}{m}}}{100})$, there exists $a^*>0$ such that for all $a\in [0,a^*)$, $F_a(z)=0$ has exactly one root in each $B(\zeta_j,\epsilon)$.

Proof: Let $\epsilon\in (0,\frac{|h(0)|^{\frac{1}{m}}}{100})$ be given. Fix $j$ and let us consider an open ball $B_j =B(\zeta_j,\epsilon)$ centered at $\zeta_j$. Note that $B_j$ are disjoint. If $z\in\partial B_j$, then there exists $\eta>0$ such that $|z^m - h(0)|\ge \eta$ by the compactness of $\partial B_j$. Since $\frac{h(a^{\frac{1}{m}}z)}{g(a^{\frac{1}{m}}z)}\to h(0)$ uniformly on $\partial B_j$, it says that $F_a(z)$ does not vanish on $\partial B_j$ for all $a\in [0,a_j^*)$ for some $a_j^*>0$. Define $$ N(a) = \frac{1}{2\pi i}\int_{\partial B_j}\frac{F_a'(z)}{F_a(z)}dz $$ for $a\in [0,a_j^*)$. By Cauchy's argument principle, $N(a)$ gives the number of zeros of $F_a$ in $B_j$. By the construction, $N(a)$ is an integer-valued continuous function with $N(0)=1$. This gives $N(a) \equiv 1$. This means $F_a(z)=0$ has exactly one root in $B_j$ for all $a\in [0,a^*_j)$. Now, let $a^* = \min_j a^*_j>0$, then the claim follows.$\blacksquare$

Now denote each root in $B_j$ of $F_a(z)$ by $\gamma_j(a)$. Then by the above claim we can write $$ \gamma_j(a) = \zeta_j+o(1). $$ Since the roots $\beta_j(a)$ of $(*)$ can be expressed as $a^{\frac{1}{m}}\gamma_j(a)$, we get $$ \beta_j(a) = a^{\frac{1}{m}}\zeta_j + o(|a|^{\frac{1}{m}})=\left(ah(0)\right)^{\frac{1}{m}}\omega^j + o(|a|^{\frac{1}{m}}). $$ Now, we deal with the case where $a<0$. We can modify $(*)$ as $$ x^m g(x) = (-a)\cdot(-h(x)). $$ By letting $b=-a>0$ and $k(x)=-h(x)$, as a corollary of the above argument we have that $$ \tilde{\beta}_j(b) = \left(bk(0)\right)^{\frac{1}{m}}\omega^j + o(|b|^{\frac{1}{m}})=\left(ah(0)\right)^{\frac{1}{m}}\omega^j + o(|a|^{\frac{1}{m}}). $$ Relabeling $\tilde{\beta}_j(b)$ as $\beta_j(a)$, we get $$ \beta_j(a) =\left(ah(0)\right)^{\frac{1}{m}}\omega^j + o(|a|^{\frac{1}{m}}). $$ for all $a\in\mathbb{R}$. Now, turning back to the original equation, we finally get for all $a$, $$ \alpha_j(a) = x_0 +\left(-(a-a_0)\frac{h(x_0)}{g(x_0)}\right)^{\frac{1}{m}}\omega^j + o(|a-a_0|^{\frac{1}{m}}). $$ This gives the desired result.

$\endgroup$
  • $\begingroup$ Could you give a few more explanations on why we can take $x_0 = 0$? I could not understand why the zeros of $x^m g(x) + a h(x) =0$ can tell us the zeros of $(x-x_0)^m g(x) + ah(x) = 0$? Thanks. $\endgroup$ – MyCindy2012 Jan 10 at 18:42
  • $\begingroup$ @MyCindy2012 Actually, it is obtained by change of variable $x-x_0 = z$. Then the equation becomes $z^mg(z+x_0)+ah(z+x_0)=0$. Accordingly, we should also substitute $g'(z) = g(z+x_0)$ and $h'(z) = h(z+x_0)$. Perhaps this confusion is because I labeled $g,g'$ and $h,h'$ using the same notation. $\endgroup$ – Song Jan 10 at 18:47
  • $\begingroup$ Thanks so much for your clarification. I do have another question coming to mind: why do you consider one-sided limit $a \to 0+$? I cannot see why the limiting could not be treated simultaneously. $\endgroup$ – MyCindy2012 Jan 10 at 21:10
  • $\begingroup$ @MyCindy2012 Well, my concern was about $a^{1/m}$ when $a<0$. Of course we can treat it as one of the roots of $w^m = a$, but it looks ugly (let $\omega=e^{\pi i/m}$ and $a^{1/m}:=(-a)^{1/m}\omega$, etc)... However, as you pointed out, we may be able to treat both cases simultaneously. $\endgroup$ – Song Jan 10 at 21:21
  • $\begingroup$ Is this inevitable? I mean even if we treat this separately, how do we cope the situation when $a < 0$? $\endgroup$ – MyCindy2012 Jan 10 at 21:26
1
$\begingroup$

Illustration to the answer of @Song

Set $g(x)=1+x^3$, $h(x)=1-x^2$, $m=5$ and $a_0=0=x_0$. Then the polynomial is $$ f(x)=x^5(1-x^3)+a(1+x^2) $$ Plot the roots for $a=\pm b^5$ for $b$ in some arithmetic sequence spanning $[0,1]$. Plot the roots (left) and the roots divided by $a^{1/5}=\pm b$ (right). The bold points left are the roots for $a=\pm 1$, blue for positive, red for negative $a$, while the bold points on the right are the locations of the scaled roots for $a\approx 0$.

root plot

fig, ax = plt.subplots(1,2,figsize = (2*8, 8))

def F(a): return [1, 0, 0, -1, 0, 0, a, 0, a]
z = np.roots(F(1)); ax[0].plot(z.real, z.imag, 'ob', ms=6);
z = np.roots(F(-1)); ax[0].plot(z.real, z.imag, 'or', ms=6);
ax[1].add_artist(plt.Circle((0,0),1, color='k', fill=False))
for b in np.linspace(0.01,1,11)[::-1]:
    z = np.roots(F(b**5)); ax[0].plot(z.real, z.imag, 'ob', ms=2);
    z = z/b; ax[1].plot(z.real, z.imag, 'ob', ms=2);
    z = np.roots(F(-b**5)); ax[0].plot(z.real, z.imag, 'or', ms=2);
    z = -z/b; ax[1].plot(z.real, z.imag, 'or', ms=2);
z = np.roots(F(b**5))/b; ax[1].plot(z.real, z.imag, 'ob', ms=6);
z = -np.roots(F(-b**5))/b; ax[1].plot(z.real, z.imag, 'or', ms=6);
r=1.5; ax[1].set_ylim([-r,r]); ax[1].set_xlim([-r,r])
plt.show()
$\endgroup$

Your Answer

By clicking "Post Your Answer", you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.