some property of bijective polynomial on the unit disc
in [Math]
|
Prev: about understanding conjugate roots in Q[x] - correction andsupplement
Next: LG Cookie Pep GD510 Price
From: quasi on 3 Aug 2010 13:11 On Tue, 3 Aug 2010 09:06:42 -0700 (PDT), achille <achille_hui(a)yahoo.com.hk> wrote: >On Aug 4, 1:04 am, quasi <qu...(a)null.set> wrote: >> On Tue, 03 Aug 2010 11:53:43 -0500, quasi <qu...(a)null.set> wrote: >> >On Tue, 3 Aug 2010 02:48:32 -0700 (PDT), John <to1m...(a)yahoo.com> >> >wrote: >> >> >>Let P(z) = a_n z^n + ... + a_1 z + a_0 be polynomial over C, such that >> >>it is bijective in the unit disc. >> >>Prove that |a_n| <= |a_1| / n. >> >> >>What useful we can learn from the bijectivity ? >> >> >I can see 3 possibly useful things ... >> >> >(1) For all c in the unit disc, |f(c)| <= 1. In particular, >> >> > |f(0)| <= 1. >> >> >Actually, based on (3) below, we get >> >> > |f(0)| < 1. >> >> >(2) For any c in the unit disk, the polynomial equation f(z) = c has >> >exactly one root in the unit disk. Hence all the other roots are >> >outside the unit disk. Hence if the roots (including repetitions, if >> >any) are z_1, ..., z_n with z_1 in the unit disk, then >> >> > |f(z_k)| > 1 for k > 1. >> >> >(3) By topological considerations, f must take the unit circle >> >bijectively onto itself. Thus, >> >> > |f(z)| = 1 for all z with |z| = 1 >> >> >and >> >> > |f(z)| < 1 for all z with |z| < 1. >> >> >As to how to use the above, I see some immediate moves, but I'll let >> >you play. >> >> Achille's hint is key. >> >> Consider the roots of f'(z). Where do they live? >> >> Now look at the coefficients of f'(z). >> >> quasi > >Welcome back, quasi ;-) Thanks.
From: master1729 on 3 Aug 2010 14:29 indeed wb quasi. i left sci.math for a while to work on prime twins. my polynomials skills got rusty and need practice ... this seems the perfect thread for it. no bieberbach :) im not sure exactly what way you have in mind for proof , i assume multiple proofs exists ... oh its been a while. i assume you use newton formula's for the sum of zero's of a polynomial in terms of the polynomial coef. the zero's of f' must lie within the polygon containing the original zero's ... although perhaps not my best idea , i assume the converse holds as well , not ? thus Let P(z) = a_n z^n + ... + a_1 z + a_0 be ENTIRE over C, such that it is bijective in OUTSIDE the unit disc. Prove that |a_n| => |a_1| / n. the following idea might be in your mind , or not. exchanging the coefficients of a univariate polynomial end-to-end produces a polynomial whose roots are reciprocals of the original roots. im tired .. greetz
From: Ken Barrett on 3 Aug 2010 14:34 Been awhile since I dealt with something like that, heh. I remember having to prove that back in college.
From: achille on 3 Aug 2010 20:38 On Aug 4, 6:29 am, master1729 <tommy1...(a)gmail.com> wrote: > indeed wb quasi. > > i left sci.math for a while to work on prime twins. > > my polynomials skills got rusty and need practice ... > > this seems the perfect thread for it. > > no bieberbach :) > > im not sure exactly what way you have in mind for proof , i assume multiple proofs exists ... > > oh its been a while. > > i assume you use newton formula's for the sum of zero's of a polynomial in terms of the polynomial coef. > > the zero's of f' must lie within the polygon containing the original zero's ... > > although perhaps not my best idea , i assume the converse holds as well , not ? > > thus > > Let P(z) = a_n z^n + ... + a_1 z + a_0 be > ENTIRE over C, such that > it is bijective in OUTSIDE the unit disc. > Prove that |a_n| => |a_1| / n. > > the following idea might be in your mind , or not. > > exchanging the coefficients of a univariate polynomial end-to-end produces a polynomial whose roots are reciprocals of the original roots. > > im tired .. > > greetz There are no polynomial of degree > 1 that is injective OUTSIDE the unit disc. Let n = deg(P) > 1 and M = max { P(z) : |z| <= 1 }. Let a be any complex number such that |a| > M. f(z) = a has n root (counting multiplicity) outside the unit disk. Let z_a be one of these roots. If f(z) is injective outside the unit disk, then remaining n-1 root of f(z) must coincides with z_a, ie. f(z) has the form a_n (z-z_a)^n. Apply the same argument to another complex number b <> a with |b| > M, we get f(z) has the form a_n (z-z_b)^n with z_a <> z_b. Impossible! Now pick
From: quasi on 4 Aug 2010 02:29
On Tue, 03 Aug 2010 18:29:04 EDT, master1729 <tommy1729(a)gmail.com> wrote: >indeed wb quasi. Thanks. >i left sci.math for a while to work on prime twins. > >my polynomials skills got rusty and need practice ... > >this seems the perfect thread for it. > >no bieberbach :) > >im not sure exactly what way you have in mind for proof , i assume multiple proofs exists ... > >oh its been a while. > >i assume you use newton formula's for the sum of zero's of a polynomial in terms of the polynomial coef. In this case, the product of the roots, but the roots of f', not of f. >the zero's of f' must lie within the polygon containing the original zero's ... More important is to determine whether or not f' has roots in the interior of the unit disk. See achille's hint. >although perhaps not my best idea , i assume the converse holds as well , not ? > >thus > >Let P(z) = a_n z^n + ... + a_1 z + a_0 be >ENTIRE over C, such that >it is bijective in OUTSIDE the unit disc. >Prove that |a_n| => |a_1| / n. As achille notes, if n > 1, bijectivity outside the unit disk is impossible. >the following idea might be in your mind , or not. > >exchanging the coefficients of a univariate polynomial end-to-end produces a polynomial whose roots are reciprocals of the original roots. Such a transformation might help -- I'm not sure. But the problem resolves directly without. >im tired .. > >greetz Greetings to you as well. quasi |