# How do the roots behave asymoptotically?

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?

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

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.

• 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. – MyCindy2012 Jan 10 at 18:42
• @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. – Song Jan 10 at 18:47
• 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. – MyCindy2012 Jan 10 at 21:10
• @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. – Song Jan 10 at 21:21
• Is this inevitable? I mean even if we treat this separately, how do we cope the situation when $a < 0$? – MyCindy2012 Jan 10 at 21:26

## 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 $$a\approx 0$$.

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);