Coefficient Restriction

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Nirjhor
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Coefficient Restriction

Unread post by Nirjhor » Thu Aug 28, 2014 1:26 am

Let \(P(x)\) be a polynomial whose coefficients are either \(1\) or \(-1\). For example \(P(x)\) can be \(x^2+x-1\). Given that the roots of \(P\) are all real, prove that \(\deg P\le 3\). FYI \(\deg P\) denotes the degree of \(P(x)\).

PC: Mursalin Habib. (memberlist.php?mode=viewprofile&u=2242)
- What is the value of the contour integral around Western Europe?

- Zero.

- Why?

- Because all the poles are in Eastern Europe.


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mutasimmim
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Re: Coefficient Restriction

Unread post by mutasimmim » Thu Aug 28, 2014 10:14 pm

Is the constant term considered as a coefficient?

Nirjhor
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Re: Coefficient Restriction

Unread post by Nirjhor » Thu Aug 28, 2014 11:17 pm

Yes of course.
- What is the value of the contour integral around Western Europe?

- Zero.

- Why?

- Because all the poles are in Eastern Europe.


Revive the IMO marathon.

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Phlembac Adib Hasan
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Re: Coefficient Restriction

Unread post by Phlembac Adib Hasan » Sun Aug 31, 2014 1:29 am

Nirjhor wrote:Let \(P(x)\) be a polynomial whose coefficients are either \(1\) or \(-1\). For example \(P(x)\) can be \(x^2+x-1\). Given that the roots of \(P\) are all real, prove that \(\deg P\le 3\). FYI \(\deg P\) denotes the degree of \(P(x)\).

PC: Mursalin Habib. (memberlist.php?mode=viewprofile&u=2242)
Misinterpreted. The roots should be distinct. Otherwise counter-examples like $P(x)=x^9-x^8$ can be constructed easily.
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Nirjhor
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Joined:Thu Aug 29, 2013 11:21 pm
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Re: Coefficient Restriction

Unread post by Nirjhor » Sun Aug 31, 2014 11:15 am

Phlembac Adib Hasan wrote:
Nirjhor wrote:Let \(P(x)\) be a polynomial whose coefficients are either \(1\) or \(-1\). For example \(P(x)\) can be \(x^2+x-1\). Given that the roots of \(P\) are all real, prove that \(\deg P\le 3\). FYI \(\deg P\) denotes the degree of \(P(x)\).

PC: Mursalin Habib. (memberlist.php?mode=viewprofile&u=2242)
Misinterpreted. The roots should be distinct. Otherwise counter-examples like $P(x)=x^9-x^8$ can be constructed easily.
That's not a valid construction. You are considering the coefficients of \(x^i\) for \(i\in\{0,1,...,7\}\) to be \(0\), which is not allowed.
- What is the value of the contour integral around Western Europe?

- Zero.

- Why?

- Because all the poles are in Eastern Europe.


Revive the IMO marathon.

Nirjhor
Posts:136
Joined:Thu Aug 29, 2013 11:21 pm
Location:Varies.

Re: Coefficient Restriction

Unread post by Nirjhor » Fri Sep 05, 2014 11:21 pm

Since this is going unsolved for too long.

Hint 1
Vieta's formulae.
Hint 2
AM-GM Inequality.
My Solution
Let \(P(x)=\displaystyle \sum_{k=0}^n a_k x^k\) so that \(\deg P=n\) and let it's roots be \(r_i\in\mathbb{R}\) for \(1\le i\le n\). WLOG let \(a_n=1\). Applying Vieta's we get

\[\sum_{cyc} r_i^2=\left(\sum_{cyc}r_i\right)^2-2\sum_{cyc} r_ir_j=3.\]

Notice that we must have \(\displaystyle \sum_{cyc} r_ir_j=-1\) because otherwise the sum of squares become negative. Finally applying Vieta's once again along with AM-GM inequality gives

\[n=n\sqrt[n]{\prod_{cyc} r_i^2}\le \sum_{cyc} r_i^2=3\]

which is \(\deg P\le 3\) as desired. \(\square\)
- What is the value of the contour integral around Western Europe?

- Zero.

- Why?

- Because all the poles are in Eastern Europe.


Revive the IMO marathon.

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