Polynomial of the nth degree and its elements.
$begingroup$
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
asked Dec 2 '18 at 8:29
KukozKukoz
278
$endgroup$
add a comment |
$begingroup$
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
asked Dec 2 '18 at 8:29
KukozKukoz
278
$endgroup$
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.
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– José Carlos Santos
Dec 2 '18 at 8:31
$begingroup$
"elcome?" 😁😁😁
$endgroup$
– Robert Lewis
Dec 2 '18 at 22:44
add a comment |
$begingroup$
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
asked Dec 2 '18 at 8:29
KukozKukoz
278
$endgroup$
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
asked Dec 2 '18 at 8:29
KukozKukoz
278
asked Dec 2 '18 at 8:29
KukozKukoz
278
edited Dec 2 '18 at 8:48
Kukoz
asked Dec 2 '18 at 8:29
KukozKukoz
278
asked Dec 2 '18 at 8:29
KukozKukoz
278
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
add a comment |
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
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
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.
answered Dec 2 '18 at 8:55
yahooyahoo
606412
$endgroup$
$begingroup$
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
add a comment |
Your Answer
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1 Answer
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active
oldest
votes
1 Answer
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active
oldest
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$begingroup$
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.
answered Dec 2 '18 at 8:55
yahooyahoo
606412
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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.
answered Dec 2 '18 at 8:55
yahooyahoo
606412
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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.
answered Dec 2 '18 at 8:55
yahooyahoo
606412
$endgroup$
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.
answered Dec 2 '18 at 8:55
yahooyahoo
606412
answered Dec 2 '18 at 8:55
yahooyahoo
606412
answered Dec 2 '18 at 8:55
yahooyahoo
606412
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
add a comment |
$begingroup$
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
add a comment |
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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