Re: Help proving lemma about the roots of a polynomial

On Feb 4, 7:20 pm, quasi <qu...@xxxxxxxx> wrote:
On Wed, 4 Feb 2009 19:09:49 -0800 (PST), Mariano Suárez-Alvarez





<mariano.suarezalva...@xxxxxxxxx> wrote:
On Feb 5, 12:31 am, quasi <qu...@xxxxxxxx> wrote:
On Wed, 4 Feb 2009 17:55:26 -0800 (PST), "Achava Nakhash, the Loving

Snake" <ach...@xxxxxxxxxxx> wrote:
On Feb 4, 5:53 pm, "Achava Nakhash, the Loving Snake"
<ach...@xxxxxxxxxxx> wrote:
Hi group,

In Stolarsky's book on Diophantine Equations, he states a lemma
attributed to Cauchy that if x is a root of

a_n*x^n + ... + a_1*x + a_0 = 0    and if

M = max(i not equal n) |a_i| / |a_0)| then

1/(M+1) < |x| < M + 1

The proof of the right-hand inequality is fine and very clever.
However his suggested proof of the left-hand inequality yields me only
that 1/(N+1) < |x|  where n is the max of the |a_i| /|a_0| for i not
equal to 0.  There is only a problem here if |x| < 1 and also |a_0|  <
|a_n|   In all the examples I can come up with, the original form of
the inequality is valid.  Perhaps this can be shown not to happen?

The proof of the right inequality is as follows:

First note that it is obvious if |x| is not greater than 1.  So assume
|x| > 1,

Then 0 >= a_n*x^n*(1 - M(|x|^-1 - x^-2 - ... - |x|^-n) > a_n*x^n*(|x|
- 1 - M)/(|x| - 1)

from which the conclusion is evident.

The proof of the left-hand inequality is supposed to relate to the
fact that |x|^-1 satisfies the same polynomial with the order of the
coefficients reversed.  Then we apply the right-hand inequality to |x|
^-1.  But this doesn't work!  Neither do my efforts to fix it.

Can anybody help me with this?  The result is supposed to be on page
39 of a book by Marden called "Geometry of Polynomials" to which I
have no axis for at least a week.

Thanks in advance,
Achava

Little correction!  In above, M is the max over |a_i| / |a_n| and N is
the max over |a_i| / |a_0|.

Sorry about that,
Achava

Did you consider the possibility that the author made an error
(presumably a typo)?

Surely, the correct inequality is, in fact,

   1/(N+1) < |x| < M +1

As an example, try a the linear equation 2x - 1 = 0.

Then M = 1/2, N = 2, and the only root is x = 1/2.

The inequality 1/(N+1) < |x| < M+1 succeeds,

however the inequality 1/(M+1) < |x| fails.

quasi

Doesn't the inequality 1/(N+1) < |x|  simply
follow from considering the polynomial that
results from substituting x by x^{-1} and
multiplying by x^n ?

Yes.

The issue was an apparent error (probably a typo) on the part of the
author.

quasi-

Actually, whatever the the problem is, it is not a typo. The N that I
use is my own notation. Stolarsky simply mentions M and then has a
throwaway line about using the same techinique involving the
polynomial with 1/x as a root as was used in proving a weaker
inequality. I am interested in looking up the book reference that he
gives, but I have to get away from work for too long a time most of
the time to get to a library that might or might not have it. Of
course I could buy the book, which does look worth looking at, but it
is too expensive.

Regards,
Achava
.

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