some property of bijective polynomial on the unit disc
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From: achille on 4 Aug 2010 03:16 On Aug 4, 8:38 am, achille <achille_...(a)yahoo.com.hk> wrote: > 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 Oops, I make mistakes in my argument, should be: f(z) - a has the form a_n ( z - z_a)^n AND f(z) - b has the form a_n ( z - z_b)^n => a - b = a_n (( z - z_a)^n - (z - z_b)^n ) again impossible because the LHS is a constant while RHS is a polynomial of degree n-1. Sorry for the confusion ;-p
From: master1729 on 4 Aug 2010 04:46 quasi : > On Tue, 03 Aug 2010 18:29:04 EDT, master1729 > <tommy1729(a)gmail.com> > wrote: > > >indeed wb quasi. > > Thanks. Your welcome. > > >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. ah. lets try. i assume not because of the bijective property. a zero of a derivate means a local max or min , but since f is bijective within the disk , no local max or min occurs within the unit disk. what about on the unit disk btw ? and do roots of f' on the unit disk matter ? > > >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. ok bad idea. :) > > >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 and hello to achille as well. finally some more good people again :)
From: achille on 4 Aug 2010 10:28 On Aug 4, 8:46 pm, master1729 <tommy1...(a)gmail.com> wrote: > a zero of a derivate means a local max or min , but since f is bijective within the disk , no local max or min occurs within the unit disk. > > what about on the unit disk btw ? > It is possible for f to be bijective on the closed unit disc but f' vanishes on some points on the boundary. eg. (z+1)^2. However, I believe a higher order zero (ie. f' and f'' vanishes at the same time) is impossible (at least for those f that can be extended to an entire function over an open neighbourhood of the closed unit disc). > and do roots of f' on the unit disk matter ? > > >
From: master1729 on 5 Aug 2010 03:05 achille wrote : > On Aug 4, 8:46 pm, master1729 <tommy1...(a)gmail.com> > wrote: > > a zero of a derivate means a local max or min , but > since f is bijective within the disk , no local max > or min occurs within the unit disk. > > > > what about on the unit disk btw ? > > > > It is possible for f to be bijective on the closed > unit disc > but f' vanishes on some points on the boundary. eg. > (z+1)^2. but z = 1 gives (z+1)^2 = 4 ?? 4 is not within the unit disk ? > > However, I believe a higher order zero (ie. f' and > f'' vanishes > at the same time) is impossible (at least for those f > that can > be extended to an entire function over an open > neighbourhood of > the closed unit disc). > why do you believe that ? so you dont believe in a bending point on the edge of the unit radius ? i think that within is excluded by conformal mapping , but why excluding a bending point on the edge is less clear to me. > > > > and do roots of f' on the unit disk matter ? > > > > > > > > regards tommy1729
From: achille on 5 Aug 2010 07:59
On Aug 5, 7:05 pm, master1729 <tommy1...(a)gmail.com> wrote: > > but z = 1 gives (z+1)^2 = 4 ?? > > 4 is not within the unit disk ? > You can always scale and add a constant to (z+1)^2 to make the range of P(unit disk) falls within the unit disk. In fact, when one say a function f(z) is bijective on the closed unit disk D, there are several possible interpretations: f is bijective between D and the range f(D), and 1) the range can be anything. OR 2) the range f(D) is a subset of D. OR 3) the range f(D) = D. I'm using the interpretation (1) as whatever I say doesn't care whether the range P(D) is a subset of D or not. > > > > However, I believe a higher order zero (ie. f' and > > f'' vanishes > > at the same time) is impossible (at least for those f > > that can > > be extended to an entire function over an open > > neighbourhood of > > the closed unit disc). > > why do you believe that ? so you dont believe in a bending point on the edge of the unit radius ? > i think that within is excluded by conformal mapping , but why excluding a bending point on the edge is less clear to me. If f' and f'' = 0 on a point z_0 on the unit circle, and k >= 3 is the smallest integer such that f^(k)(z_0) <> 0, then for z sufficiently close to z_0, we have f(z) = f(z_0) + f^(k)(z_0)/k! (z-z_0)^k + O((z-z_0)^(k+1)) This means for any number s sufficiently close (but not equal ) to f(z_0), the equations f(z) = s has k roots r_0, ... r_{k-1} near z_0. Furthermore, they can be labeled such that r_j - z_0 = ( r_0 - z_0 ) exp( 2 pi i ( j / k ) ) + O(|r_0-z_0|^2) Since the angles between r_j - z_0 and r_(j-1) - z_0 ~ 2pi/k, we can move s around to make two of these roots falls within the unit cirlce. This will contradict with the requirement that f is injective on the unit disk. |