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 5 Aug 2010 12:59 On Thu, 5 Aug 2010 04:59:47 -0700 (PDT), achille <achille_hui(a)yahoo.com.hk> wrote: >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. For the original problem, I guess the interpretation of "bijective" was intended to be (1) above. In fact, are there any polynomials f(z) other than f(z) = c z, where |c| = 1 which are literally bijective (i.e., using interpretation 3) on the unit disk? quasi
From: achille on 5 Aug 2010 23:30 On Aug 6, 12:59 am, quasi <qu...(a)null.set> wrote: > On Thu, 5 Aug 2010 04:59:47 -0700 (PDT), achille > > > > <achille_...(a)yahoo.com.hk> wrote: > >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. > > For the original problem, I guess the interpretation of "bijective" > was intended to be (1) above. > > In fact, are there any polynomials f(z) other than > > f(z) = c z, where |c| = 1 > > which are literally bijective (i.e., using interpretation 3) on the > unit disk? > > quasi There are no other polynomials which are literally bijective. The group of biholomorphic maps of the unit disk onto itself is a subgroup of the Mobius transforms and has elements of the form: c ( z - a )/(1 + \bar{a} z ) where |a| < 1 and |c| = 1. Other than the trivial c z, the others are not polynomials.
From: quasi on 6 Aug 2010 00:38 On Thu, 5 Aug 2010 20:30:29 -0700 (PDT), achille <achille_hui(a)yahoo.com.hk> wrote: >On Aug 6, 12:59 am, quasi <qu...(a)null.set> wrote: >> On Thu, 5 Aug 2010 04:59:47 -0700 (PDT), achille >> >> >> >> <achille_...(a)yahoo.com.hk> wrote: >> >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. >> >> For the original problem, I guess the interpretation of "bijective" >> was intended to be (1) above. >> >> In fact, are there any polynomials f(z) other than >> >> f(z) = c z, where |c| = 1 >> >> which are literally bijective (i.e., using interpretation 3) on the >> unit disk? >> >> quasi > >There are no other polynomials which are literally bijective. >The group of biholomorphic maps of the unit disk onto itself >is a subgroup of the Mobius transforms and has elements of >the form: > > c ( z - a )/(1 + \bar{a} z ) where |a| < 1 and |c| = 1. > >Other than the trivial c z, the others are not polynomials. That agrees with what I expected. Thanks. quasi
From: master1729 on 5 Aug 2010 23:39 > On Thu, 5 Aug 2010 04:59:47 -0700 (PDT), achille > <achille_hui(a)yahoo.com.hk> wrote: > > >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. > > For the original problem, I guess the interpretation > of "bijective" > was intended to be (1) above. > > In fact, are there any polynomials f(z) other than > > f(z) = c z, where |c| = 1 > > which are literally bijective (i.e., using > interpretation 3) on the > unit disk? > > quasi i was assuming 3 all the time. does that make proof different ? now im confused ...
From: achille on 6 Aug 2010 03:59
On Aug 6, 3:39 pm, master1729 <tommy1...(a)gmail.com> wrote: > > i was assuming 3 all the time. > > does that make proof different ? > > now im confused ... When I first see the question, my first response is to interpret it as (3). But then I remember the only biholomorphic functions that map unit disk onto itself are the Mobius transform, interpretation (3) doesn't make sense. The proof is the same, relaxing the constraint simply allow you ignore irrelevant misleading hints ;-) |