Mixing
How do the roots behave asymoptotically?
$begingroup$
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?
complex-analysis polynomials roots
asked Jan 8 at 0:40
MyCindy2012MyCindy2012
627
$endgroup$
add a comment |
$begingroup$
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?
complex-analysis polynomials roots
asked Jan 8 at 0:40
MyCindy2012MyCindy2012
627
$endgroup$
$begingroup$
I would write $ω_m^j=(ω_m)^j$ for the unit roots, it seems more natural.
$endgroup$
– LutzL
Jan 11 at 12:01
add a comment |
$begingroup$
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?
complex-analysis polynomials roots
asked Jan 8 at 0:40
MyCindy2012MyCindy2012
627
$endgroup$
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?
complex-analysis polynomials roots
complex-analysis polynomials roots
asked Jan 8 at 0:40
MyCindy2012MyCindy2012
627
asked Jan 8 at 0:40
MyCindy2012MyCindy2012
627
edited Jan 8 at 1:02
MyCindy2012
asked Jan 8 at 0:40
MyCindy2012MyCindy2012
627
asked Jan 8 at 0:40
MyCindy2012MyCindy2012
627
asked Jan 8 at 0:40
MyCindy2012MyCindy2012
627
627
$begingroup$
I would write $ω_m^j=(ω_m)^j$ for the unit roots, it seems more natural.
$endgroup$
– LutzL
Jan 11 at 12:01
add a comment |
$begingroup$
I would write $ω_m^j=(ω_m)^j$ for the unit roots, it seems more natural.
$endgroup$
– LutzL
Jan 11 at 12:01
$begingroup$
I would write $ω_m^j=(ω_m)^j$ for the unit roots, it seems more natural.
$endgroup$
– LutzL
Jan 11 at 12:01
$begingroup$
I would write $ω_m^j=(ω_m)^j$ for the unit roots, it seems more natural.
$endgroup$
– LutzL
Jan 11 at 12:01
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Let $o(|u|^r)$ ($rge0$) denote the class (and also an element of it) of functions $q(u)inmathbb{C}$ 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 $omega = e^{frac{2pi i}{m}}$ 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 $cge 0$ (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 $-ato a$, the given equation becomes
$$
x^m g(x) = acdot h(x).tag{*}
$$ 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 $lim_{ato 0^+}F_a(z) = F_0 (z)=z^m -h(0)$ 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 $epsilonin (0,frac{|h(0)|^{frac{1}{m}}}{100})$, there exists $a^*>0$ such that for all $ain [0,a^*)$, $F_a(z)=0$ has exactly one root in each $B(zeta_j,epsilon)$.
Proof: Let $epsilonin (0,frac{|h(0)|^{frac{1}{m}}}{100})$ 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 $zinpartial B_j$, 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 $ain [0,a_j^*)$ for some $a_j^*>0$. Define $$
N(a) = frac{1}{2pi i}int_{partial B_j}frac{F_a'(z)}{F_a(z)}dz
$$ for $ain [0,a_j^*)$. 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 $ain [0,a^*_j)$. 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 $ainmathbb{R}$. 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.
answered Jan 10 at 15:28
SongSong
11k628
$endgroup$
$begingroup$
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
|
show 3 more comments
$begingroup$
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 $aapprox 0$.
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);
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()
answered Jan 15 at 9:48
LutzLLutzL
57.5k42054
$endgroup$
add a comment |
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2 Answers
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2 Answers
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$begingroup$
Let $o(|u|^r)$ ($rge0$) denote the class (and also an element of it) of functions $q(u)inmathbb{C}$ 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 $omega = e^{frac{2pi i}{m}}$ 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 $cge 0$ (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 $-ato a$, the given equation becomes
$$
x^m g(x) = acdot h(x).tag{*}
$$ 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 $lim_{ato 0^+}F_a(z) = F_0 (z)=z^m -h(0)$ 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 $epsilonin (0,frac{|h(0)|^{frac{1}{m}}}{100})$, there exists $a^*>0$ such that for all $ain [0,a^*)$, $F_a(z)=0$ has exactly one root in each $B(zeta_j,epsilon)$.
Proof: Let $epsilonin (0,frac{|h(0)|^{frac{1}{m}}}{100})$ 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 $zinpartial B_j$, 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 $ain [0,a_j^*)$ for some $a_j^*>0$. Define $$
N(a) = frac{1}{2pi i}int_{partial B_j}frac{F_a'(z)}{F_a(z)}dz
$$ for $ain [0,a_j^*)$. 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 $ain [0,a^*_j)$. 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 $ainmathbb{R}$. 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.
answered Jan 10 at 15:28
SongSong
11k628
$endgroup$
$begingroup$
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
|
show 3 more comments
$begingroup$
Let $o(|u|^r)$ ($rge0$) denote the class (and also an element of it) of functions $q(u)inmathbb{C}$ 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 $omega = e^{frac{2pi i}{m}}$ 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 $cge 0$ (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 $-ato a$, the given equation becomes
$$
x^m g(x) = acdot h(x).tag{*}
$$ 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 $lim_{ato 0^+}F_a(z) = F_0 (z)=z^m -h(0)$ 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 $epsilonin (0,frac{|h(0)|^{frac{1}{m}}}{100})$, there exists $a^*>0$ such that for all $ain [0,a^*)$, $F_a(z)=0$ has exactly one root in each $B(zeta_j,epsilon)$.
Proof: Let $epsilonin (0,frac{|h(0)|^{frac{1}{m}}}{100})$ 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 $zinpartial B_j$, 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 $ain [0,a_j^*)$ for some $a_j^*>0$. Define $$
N(a) = frac{1}{2pi i}int_{partial B_j}frac{F_a'(z)}{F_a(z)}dz
$$ for $ain [0,a_j^*)$. 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 $ain [0,a^*_j)$. 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 $ainmathbb{R}$. 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.
answered Jan 10 at 15:28
SongSong
11k628
$endgroup$
$begingroup$
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
|
show 3 more comments
$begingroup$
Let $o(|u|^r)$ ($rge0$) denote the class (and also an element of it) of functions $q(u)inmathbb{C}$ 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 $omega = e^{frac{2pi i}{m}}$ 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 $cge 0$ (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 $-ato a$, the given equation becomes
$$
x^m g(x) = acdot h(x).tag{*}
$$ 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 $lim_{ato 0^+}F_a(z) = F_0 (z)=z^m -h(0)$ 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 $epsilonin (0,frac{|h(0)|^{frac{1}{m}}}{100})$, there exists $a^*>0$ such that for all $ain [0,a^*)$, $F_a(z)=0$ has exactly one root in each $B(zeta_j,epsilon)$.
Proof: Let $epsilonin (0,frac{|h(0)|^{frac{1}{m}}}{100})$ 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 $zinpartial B_j$, 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 $ain [0,a_j^*)$ for some $a_j^*>0$. Define $$
N(a) = frac{1}{2pi i}int_{partial B_j}frac{F_a'(z)}{F_a(z)}dz
$$ for $ain [0,a_j^*)$. 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 $ain [0,a^*_j)$. 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 $ainmathbb{R}$. 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.
answered Jan 10 at 15:28
SongSong
11k628
$endgroup$
Let $o(|u|^r)$ ($rge0$) denote the class (and also an element of it) of functions $q(u)inmathbb{C}$ 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 $omega = e^{frac{2pi i}{m}}$ 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 $cge 0$ (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 $-ato a$, the given equation becomes
$$
x^m g(x) = acdot h(x).tag{*}
$$ 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 $lim_{ato 0^+}F_a(z) = F_0 (z)=z^m -h(0)$ 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 $epsilonin (0,frac{|h(0)|^{frac{1}{m}}}{100})$, there exists $a^*>0$ such that for all $ain [0,a^*)$, $F_a(z)=0$ has exactly one root in each $B(zeta_j,epsilon)$.
Proof: Let $epsilonin (0,frac{|h(0)|^{frac{1}{m}}}{100})$ 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 $zinpartial B_j$, 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 $ain [0,a_j^*)$ for some $a_j^*>0$. Define $$
N(a) = frac{1}{2pi i}int_{partial B_j}frac{F_a'(z)}{F_a(z)}dz
$$ for $ain [0,a_j^*)$. 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 $ain [0,a^*_j)$. 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 $ainmathbb{R}$. 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.
answered Jan 10 at 15:28
SongSong
11k628
edited Jan 11 at 19:35
answered Jan 10 at 15:28
SongSong
11k628
answered Jan 10 at 15:28
SongSong
11k628
answered Jan 10 at 15:28
SongSong
11k628
11k628
$begingroup$
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
|
show 3 more comments
$begingroup$
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
$begingroup$
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
$begingroup$
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
$begingroup$
@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
$begingroup$
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
$begingroup$
@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
$begingroup$
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
|
show 3 more comments
$begingroup$
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 $aapprox 0$.
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);
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()
answered Jan 15 at 9:48
LutzLLutzL
57.5k42054
$endgroup$
add a comment |
$begingroup$
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 $aapprox 0$.
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);
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()
answered Jan 15 at 9:48
LutzLLutzL
57.5k42054
$endgroup$
add a comment |
$begingroup$
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 $aapprox 0$.
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);
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()
answered Jan 15 at 9:48
LutzLLutzL
57.5k42054
$endgroup$
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 $aapprox 0$.
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);
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()
answered Jan 15 at 9:48
LutzLLutzL
57.5k42054
answered Jan 15 at 9:48
LutzLLutzL
57.5k42054
answered Jan 15 at 9:48
LutzLLutzL
57.5k42054
answered Jan 15 at 9:48
LutzLLutzL
57.5k42054
57.5k42054
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$begingroup$
I would write $ω_m^j=(ω_m)^j$ for the unit roots, it seems more natural.
$endgroup$
– LutzL
Jan 11 at 12:01