Properties of Roots in Univariate Polynomial of Degree n

[knowLittle]

Homework Statement

1. Let $p(x) = a_{0} x^{n} + a_{1} x^{n−1} + ... + a_{n} , a_{0} \neq 0$be an univariate polynomial of degree n.
Let r be its root, i.e. p(r) = 0. Prove that
$|r| \leq max(1, \Sigma_{1 \leq i \leq n} | \dfrac{a_{i} }{ a_{0} } | )$
Is it always true that?
$|r| \leq \Sigma_{1 \leq i \leq n} | \dfrac{a_{i} }{ a_{0} } | )$

Homework Equations

$ar + br^{2} =0$
$r(a+br)=0$
$r=0 & r = -a/b$

The Attempt at a Solution

The last equation satisfies what they state but I don't know how to proceed?
Any help?

 

[epenguin]

Homework Statement

1. Let $p(x) = a_{0} x^{n} + a_{1} x^{n−1} + ... + a_{n} , a_{0} \neq 0$be an univariate polynomial of degree n.
Let r be its root, i.e. p(r) = 0. Prove that
$|r| \leq max(1, \Sigma_{1 \leq i \leq n} | \dfrac{a_{i} }{ a_{0} } | )$
Is it always true that?
$|r| \leq \Sigma_{1 \leq i \leq n} | \dfrac{a_{i} }{ a_{0} } | )$

Homework Equations

$ar + br^{2} =0$
$r(a+br)=0$
$r=0 & r = -a/b$

The Attempt at a Solution

The last equation satisfies what they state but I don't know how to proceed?
Any help?

For almost all such polynomial probs there is no loss of generality in making a₀ = 1 and then you don't have that denominator in the expression.

Not sure I know now to proceed either but since they bring in 1 as a case of maxima, it is suggestive when you put r = 1 perhaps.

I don't follow what you are saying with your relevant equations but when you can't see how to do it in general it is certainly good to try with n=1 and 2.

 

[knowLittle]

I sort of proved it by contradiction.
Let's say that
$p(x)= ax^{2}+bx=0$
$0=x(ax+b), \text{ then the roots are }r_{1}=0 \text{, } r_{2}=\dfrac{-b}{a}$
$|r| \geq \dfrac{a+b}{a} \\$
$| r_{2}|=| \dfrac{-b}{a}| \geq | \dfrac{ a+b}{a}| \text{, will never be true.}$

 

[epenguin]

I sort of proved it by contradiction.
Let's say that
$p(x)= ax^{2}+bx=0$
$0=x(ax+b), \text{ then the roots are }r_{1}=0 \text{, } r_{2}=\dfrac{-b}{a}$
$|r| \geq \dfrac{a+b}{a} \\$
$| r_{2}|=| \dfrac{-b}{a}| \geq | \dfrac{ a+b}{a}| \text{, will never be true.}$

That I think is not right, you have misunderstood the formula in the original problem. The Ʃ subscript contains a 1≤ not a 0 ≤. The theorem then is rather trivially true for the first degree case with the ≤ it was required to prove reducing to = . Also your factor x seems an irrelevancy. But there is maybe the germ of an argument.

As I said you avoid something unnecessary if you make, without loss of generality, a₀ = 1. So then the question becomes:
Let $p(x) = x^{n} + a_{1} x^{n−1} + ... + a_{n} ,$

then prove

|r| ≤ max( 1, | a₁ + a₂ +... + a_n| ).

I think my previous idea leads there but maybe extending what you were trying does too.

 

[Dick]

Here's another hint. Since you have the 1 in the max, it's trivial if |r|<=1. So you only have to do some work for the case |r|>=1. You'll want to use the triangle inequality.