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Properties of Roots in Univariate Polynomial of Degree n

  • Thread starter knowLittle
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  • #1
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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 dont know how to proceed?
Any help?
 

Answers and Replies

  • #2
epenguin
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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 dont 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.
 
  • #3
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I sorta 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.}##
 
Last edited:
  • #4
epenguin
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I sorta 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 +... + an| ).

I think my previous idea leads there but maybe extending what you were trying does too.
 
Last edited:
  • #5
Dick
Science Advisor
Homework Helper
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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.
 

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