Polynomial of the nth degree and its elements.












2












$begingroup$


I have to prove, that polynomial:



$ x^n + a_{n-1}x^{n-1}+...+a_2x^2+a_1x+a_0=0$
, where $ a_i= [-1,0,1] $ and $ i=0,1,.....,n $

doesn't have any solutions in the set
$$
begin{aligned}
( -infty,-2) cup (2, infty)
end{aligned}
$$
We know, that $ a_n=1$, so we know, that $ x_1+x_2+....+x_n=-2$.
It's all I have though, any
suggestions, what can I do in next step?






proof-verification polynomials







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    elcome to MSE. Your question is phrased as an isolated problem, without any further information or context. This does not match many users' quality standards, so it may attract downvotes, or be put on hold. To prevent that, please edit the question. This will help you recognise and resolve the issues. Concretely: please provide context, and include your work and thoughts on the problem. These changes can help in formulating more appropriate answers.
    $endgroup$
    – José Carlos Santos
    Dec 2 '18 at 8:31










  • $begingroup$
    "elcome?" 😁😁😁
    $endgroup$
    – Robert Lewis
    Dec 2 '18 at 22:44
















2












$begingroup$


I have to prove, that polynomial:



$ x^n + a_{n-1}x^{n-1}+...+a_2x^2+a_1x+a_0=0$
, where $ a_i= [-1,0,1] $ and $ i=0,1,.....,n $

doesn't have any solutions in the set
$$
begin{aligned}
( -infty,-2) cup (2, infty)
end{aligned}
$$
We know, that $ a_n=1$, so we know, that $ x_1+x_2+....+x_n=-2$.
It's all I have though, any
suggestions, what can I do in next step?






proof-verification polynomials







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    elcome to MSE. Your question is phrased as an isolated problem, without any further information or context. This does not match many users' quality standards, so it may attract downvotes, or be put on hold. To prevent that, please edit the question. This will help you recognise and resolve the issues. Concretely: please provide context, and include your work and thoughts on the problem. These changes can help in formulating more appropriate answers.
    $endgroup$
    – José Carlos Santos
    Dec 2 '18 at 8:31










  • $begingroup$
    "elcome?" 😁😁😁
    $endgroup$
    – Robert Lewis
    Dec 2 '18 at 22:44














2












2








2


2



$begingroup$


I have to prove, that polynomial:



$ x^n + a_{n-1}x^{n-1}+...+a_2x^2+a_1x+a_0=0$
, where $ a_i= [-1,0,1] $ and $ i=0,1,.....,n $

doesn't have any solutions in the set
$$
begin{aligned}
( -infty,-2) cup (2, infty)
end{aligned}
$$
We know, that $ a_n=1$, so we know, that $ x_1+x_2+....+x_n=-2$.
It's all I have though, any
suggestions, what can I do in next step?






proof-verification polynomials







share|cite|improve this question











$endgroup$




I have to prove, that polynomial:



$ x^n + a_{n-1}x^{n-1}+...+a_2x^2+a_1x+a_0=0$
, where $ a_i= [-1,0,1] $ and $ i=0,1,.....,n $

doesn't have any solutions in the set
$$
begin{aligned}
( -infty,-2) cup (2, infty)
end{aligned}
$$
We know, that $ a_n=1$, so we know, that $ x_1+x_2+....+x_n=-2$.
It's all I have though, any
suggestions, what can I do in next step?






proof-verification polynomials




proof-verification polynomials






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 2 '18 at 8:48







Kukoz

















asked Dec 2 '18 at 8:29









KukozKukoz

278




278








  • 1




    $begingroup$
    elcome to MSE. Your question is phrased as an isolated problem, without any further information or context. This does not match many users' quality standards, so it may attract downvotes, or be put on hold. To prevent that, please edit the question. This will help you recognise and resolve the issues. Concretely: please provide context, and include your work and thoughts on the problem. These changes can help in formulating more appropriate answers.
    $endgroup$
    – José Carlos Santos
    Dec 2 '18 at 8:31










  • $begingroup$
    "elcome?" 😁😁😁
    $endgroup$
    – Robert Lewis
    Dec 2 '18 at 22:44














  • 1




    $begingroup$
    elcome to MSE. Your question is phrased as an isolated problem, without any further information or context. This does not match many users' quality standards, so it may attract downvotes, or be put on hold. To prevent that, please edit the question. This will help you recognise and resolve the issues. Concretely: please provide context, and include your work and thoughts on the problem. These changes can help in formulating more appropriate answers.
    $endgroup$
    – José Carlos Santos
    Dec 2 '18 at 8:31










  • $begingroup$
    "elcome?" 😁😁😁
    $endgroup$
    – Robert Lewis
    Dec 2 '18 at 22:44








1




1




$begingroup$
elcome to MSE. Your question is phrased as an isolated problem, without any further information or context. This does not match many users' quality standards, so it may attract downvotes, or be put on hold. To prevent that, please edit the question. This will help you recognise and resolve the issues. Concretely: please provide context, and include your work and thoughts on the problem. These changes can help in formulating more appropriate answers.
$endgroup$
– José Carlos Santos
Dec 2 '18 at 8:31




$begingroup$
elcome to MSE. Your question is phrased as an isolated problem, without any further information or context. This does not match many users' quality standards, so it may attract downvotes, or be put on hold. To prevent that, please edit the question. This will help you recognise and resolve the issues. Concretely: please provide context, and include your work and thoughts on the problem. These changes can help in formulating more appropriate answers.
$endgroup$
– José Carlos Santos
Dec 2 '18 at 8:31












$begingroup$
"elcome?" 😁😁😁
$endgroup$
– Robert Lewis
Dec 2 '18 at 22:44




$begingroup$
"elcome?" 😁😁😁
$endgroup$
– Robert Lewis
Dec 2 '18 at 22:44










1 Answer
1






active

oldest

votes


















1












$begingroup$

Let $f(x)=0$, then
begin{align}
0=&|x^n+a_{n-1}x^{n-1}+cdots+a_1x+a_0|\
geq&|x|^n-|a_{n-1}||x|^{n-1}-cdots-|a_1||x^1|-|a_0|
end{align}
Write it as
begin{align}
|x|^nleq & |a_{n-1}||x|^{n-1}+cdots+|a_1||x^1|+|a_0|\
leq & |x|^{n-1}+|x|^{n-2}+cdots+|x|+1\
= & frac{|x|^{n}-1}{|x|-1}
end{align}
If $|x|>2$, we get $|x|-1>1$ and then
$$
|x|^nleq |x|^n-1.
$$
It is impossible.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Could you tell me, why did you change $|a_0|$ to $1$?
    $endgroup$
    – Kukoz
    Dec 2 '18 at 13:24






  • 1




    $begingroup$
    @Kukoz Because $a_0in{-1,0,1}$, hence $|a_0|leq 1$.
    $endgroup$
    – yahoo
    Dec 2 '18 at 13:53













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1 Answer
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1 Answer
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1












$begingroup$

Let $f(x)=0$, then
begin{align}
0=&|x^n+a_{n-1}x^{n-1}+cdots+a_1x+a_0|\
geq&|x|^n-|a_{n-1}||x|^{n-1}-cdots-|a_1||x^1|-|a_0|
end{align}
Write it as
begin{align}
|x|^nleq & |a_{n-1}||x|^{n-1}+cdots+|a_1||x^1|+|a_0|\
leq & |x|^{n-1}+|x|^{n-2}+cdots+|x|+1\
= & frac{|x|^{n}-1}{|x|-1}
end{align}
If $|x|>2$, we get $|x|-1>1$ and then
$$
|x|^nleq |x|^n-1.
$$
It is impossible.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Could you tell me, why did you change $|a_0|$ to $1$?
    $endgroup$
    – Kukoz
    Dec 2 '18 at 13:24






  • 1




    $begingroup$
    @Kukoz Because $a_0in{-1,0,1}$, hence $|a_0|leq 1$.
    $endgroup$
    – yahoo
    Dec 2 '18 at 13:53


















1












$begingroup$

Let $f(x)=0$, then
begin{align}
0=&|x^n+a_{n-1}x^{n-1}+cdots+a_1x+a_0|\
geq&|x|^n-|a_{n-1}||x|^{n-1}-cdots-|a_1||x^1|-|a_0|
end{align}
Write it as
begin{align}
|x|^nleq & |a_{n-1}||x|^{n-1}+cdots+|a_1||x^1|+|a_0|\
leq & |x|^{n-1}+|x|^{n-2}+cdots+|x|+1\
= & frac{|x|^{n}-1}{|x|-1}
end{align}
If $|x|>2$, we get $|x|-1>1$ and then
$$
|x|^nleq |x|^n-1.
$$
It is impossible.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Could you tell me, why did you change $|a_0|$ to $1$?
    $endgroup$
    – Kukoz
    Dec 2 '18 at 13:24






  • 1




    $begingroup$
    @Kukoz Because $a_0in{-1,0,1}$, hence $|a_0|leq 1$.
    $endgroup$
    – yahoo
    Dec 2 '18 at 13:53
















1












1








1





$begingroup$

Let $f(x)=0$, then
begin{align}
0=&|x^n+a_{n-1}x^{n-1}+cdots+a_1x+a_0|\
geq&|x|^n-|a_{n-1}||x|^{n-1}-cdots-|a_1||x^1|-|a_0|
end{align}
Write it as
begin{align}
|x|^nleq & |a_{n-1}||x|^{n-1}+cdots+|a_1||x^1|+|a_0|\
leq & |x|^{n-1}+|x|^{n-2}+cdots+|x|+1\
= & frac{|x|^{n}-1}{|x|-1}
end{align}
If $|x|>2$, we get $|x|-1>1$ and then
$$
|x|^nleq |x|^n-1.
$$
It is impossible.






share|cite|improve this answer









$endgroup$



Let $f(x)=0$, then
begin{align}
0=&|x^n+a_{n-1}x^{n-1}+cdots+a_1x+a_0|\
geq&|x|^n-|a_{n-1}||x|^{n-1}-cdots-|a_1||x^1|-|a_0|
end{align}
Write it as
begin{align}
|x|^nleq & |a_{n-1}||x|^{n-1}+cdots+|a_1||x^1|+|a_0|\
leq & |x|^{n-1}+|x|^{n-2}+cdots+|x|+1\
= & frac{|x|^{n}-1}{|x|-1}
end{align}
If $|x|>2$, we get $|x|-1>1$ and then
$$
|x|^nleq |x|^n-1.
$$
It is impossible.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 2 '18 at 8:55









yahooyahoo

606412




606412












  • $begingroup$
    Could you tell me, why did you change $|a_0|$ to $1$?
    $endgroup$
    – Kukoz
    Dec 2 '18 at 13:24






  • 1




    $begingroup$
    @Kukoz Because $a_0in{-1,0,1}$, hence $|a_0|leq 1$.
    $endgroup$
    – yahoo
    Dec 2 '18 at 13:53




















  • $begingroup$
    Could you tell me, why did you change $|a_0|$ to $1$?
    $endgroup$
    – Kukoz
    Dec 2 '18 at 13:24






  • 1




    $begingroup$
    @Kukoz Because $a_0in{-1,0,1}$, hence $|a_0|leq 1$.
    $endgroup$
    – yahoo
    Dec 2 '18 at 13:53


















$begingroup$
Could you tell me, why did you change $|a_0|$ to $1$?
$endgroup$
– Kukoz
Dec 2 '18 at 13:24




$begingroup$
Could you tell me, why did you change $|a_0|$ to $1$?
$endgroup$
– Kukoz
Dec 2 '18 at 13:24




1




1




$begingroup$
@Kukoz Because $a_0in{-1,0,1}$, hence $|a_0|leq 1$.
$endgroup$
– yahoo
Dec 2 '18 at 13:53






$begingroup$
@Kukoz Because $a_0in{-1,0,1}$, hence $|a_0|leq 1$.
$endgroup$
– yahoo
Dec 2 '18 at 13:53




















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