Prev: about understanding conjugate roots in Q[x] - correction andsupplement
Next: LG Cookie Pep GD510 Price
From: John on 3 Aug 2010 05:48
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 ?

Thanks
From: achille on 3 Aug 2010 11:52
On Aug 3, 5:48 pm, 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 ?
>
> Thanks

Hint: If P(z) is injective on an open neighbour of
a point z_0, what can you say about P'(z_0)?
From: quasi on 3 Aug 2010 12:53
On Tue, 3 Aug 2010 02:48:32 -0700 (PDT), John <to1mmy2(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.

quasi
From: quasi on 3 Aug 2010 13:04
On Tue, 03 Aug 2010 11:53:43 -0500, quasi <quasi(a)null.set> wrote:

>On Tue, 3 Aug 2010 02:48:32 -0700 (PDT), John <to1mmy2(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
From: achille on 3 Aug 2010 12:06
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 ;-)
 |  Next  |  Last
Pages: 1 2 3 4 5 6
Prev: about understanding conjugate roots in Q[x] - correction andsupplement
Next: LG Cookie Pep GD510 Price