Proof: Zeros Localisation Theorem I have looked at the suggested related questions before asking.
From a First Course In Mathematical Analysis, David Brannan, page 148.Is the proof wrong?
zeros Localisation Theorem
Let $p(x) = {x^n} + {a_{n - 1}}{x^{n - 1}} + ... + {a_1}x + {a_0}$,$x \in R$,be a polynomial.
Then all zeros of p line in $( - M,M)$, where $M = 1 + \max \{ \left| {{a_{n - 1}}} \right|,...,\left| {{a_1}} \right|\left| {{a_0}} \right|\} $
$r(x) = \frac{{p(x)}}{{{x^n}}} - 1 = \frac{{{a_{n - 1}}}}{x} + ... + \frac{{{a_1}}}{{{x^{n - 1}}}} + \frac{{{a_0}}}{{{x^n}}},x \in R - \{ 0\} $ then, using the triangle inequality for $\left| x \right| > 1$
$\left| {r(x)} \right| = \left| {\frac{{{a_{n - 1}}}}{x} + ... + \frac{{{a_1}}}{{{x^{n - 1}}}} + \frac{{{a_0}}}{{{x^n}}}} \right| \le \left| {\frac{{{a_{n - 1}}}}{x}} \right| + ... + \left| {\frac{{{a_0}}}{{{x^n}}}} \right|$
$ \le \max \{ \left| {{a_{n - 1}}} \right|,\left| {{a_1}} \right|,\left| {{a_0}} \right|\}\left( {\frac{1}{{\left| x \right|}} + ... + \frac{1}{{{{\left| x \right|}^{n - 1}}}} + \frac{1}{{{{\left| x \right|}^n}}}} \right)$
$ < M\left( {\frac{1}{{\left| x \right|}} + ... + \frac{1}{{{{\left| x \right|}^{n - 1}}}} + \frac{1}{{{{\left| x \right|}^n}}} + ...} \right)$
$ = M$$\frac{{{\textstyle{1 \over {\left| x \right|}}}}}{{1 - {\textstyle{1 \over {\left| x \right|}}}}} = \frac{M}{{\left| x \right| - 1}}$it follows that if $\ {x \ge M = 1 + \max \{ \left| {{a_{n - 1}}} \right|,...,\left| {{a_0}} \right|\} } $ then $\left| {r(x)} \right| < 1$.
But surely, if $\left| x \right| = M$ then $\left| {r(x)} \right| < \frac{M}{{\left| x \right| - 1}} = \frac{M}{{M - 1}}$ is greater than 1!
 A: I understand now. Because $\max \{ \left| {{a_{n - 1}}} \right|,...,\left| {{a_1}} \right|,\left| {{a_0}} \right|\}  \le M - 1 < M$. So that
$\max \{ \left| {{a_{n - 1}}} \right|,...,\left| {{a_1}} \right|,\left| {{a_0}} \right|\}  < M$ is true but now the tighter inequality is obliterated.
So looking back one line with $M$ in the sum causes a issue. $M - 1$ should have been carried through rather than $M$ to $\frac{M}{{\left| x \right| - 1}}$. So a better bound on ${r(x)}$ is $\frac{{M - 1}}{{\left| x \right| - 1}}$. I.e. $\left| {r(x)} \right| \le \frac{{M - 1}}{{\left| x \right| - 1}}$ then the argument from there is correct.
A: Your error is that you wrote that the max is less than $M$ but you really should have just denoted it by $M-1$ to keep things tighter. In the end the numerator here should be $M-1$ if $M$ is defined the way you defined it. This is why I don't like the style of injecting this $1+$ into the definition of symbols, I would prefer to define $M$ to be the maximum and use functions of that if need be in the proof (and even in the statement, in this situation).
