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Math Help - polynomial height

  1. #1
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    polynomial height

    let' s define 'height' of a polynomial the max of the absolute value of its coefficients.
    Do exist P(x) and Q(x) TWO polynomials at integer coefficient with heights equal or greater than 2011, whose product is a polynomial with height =1?

    I have tried with cyclotomic polynomials, that we know can have any coefficient among the integers, and whose product of convenient cyclotomic gives x^n-1 (whoe height is 1), but I cannot demonstrate that I can find exactlky TWO polynomials whith height >=2011 whose product is a polyn. with height=1.

    has anyone a clue onhow to solve with cyclotomic or with any other method?
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  2. #2
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    Re: polynomial height

    Quote Originally Posted by pincopallino View Post
    let' s define 'height' of a polynomial the max of the absolute value of its coefficients.
    Do exist P(x) and Q(x) TWO polynomials at integer coefficient with heights equal or greater than 2011, whose product is a polynomial with height =1?

    I have tried with cyclotomic polynomials, that we know can have any coefficient among the integers, and whose product of convenient cyclotomic gives x^n-1 (whoe height is 1), but I cannot demonstrate that I can find exactlky TWO polynomials whith height >=2011 whose product is a polyn. with height=1.

    has anyone a clue onhow to solve with cyclotomic or with any other method?
    I am suspicious of questions that involve the number of the current year, because these often feature in olympiad-type competitions. If you can convince me that this problem does not come from a current competition then I can give some hints on how to solve it.
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    Re: polynomial height

    Quote Originally Posted by Opalg View Post
    I am suspicious of questions that involve the number of the current year, because these often feature in olympiad-type competitions. If you can convince me that this problem does not come from a current competition then I can give some hints on how to solve it.
    No competition related. Actually I could have written any integer (prime as 2011 or not)!
    I have succeeded to find a polynomial with any height that muliplied by a polyn with height=1 gives a result with height 1, but two polynomials not yet!
    some hint?
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  4. #4
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    Re: polynomial height

    Quote Originally Posted by pincopallino View Post
    let' s define 'height' of a polynomial the max of the absolute value of its coefficients.
    Do exist P(x) and Q(x) TWO polynomials at integer coefficient with heights equal or greater than 2011, whose product is a polynomial with height =1?
    Okay, here is the idea. Given a positive integer N, we want to find polynomials P(x), Q(x), with integer coefficients, each having height at least N, such that P(x)Q(x) has height 1.

    Even for N = 2 this takes a bit of thought. Start with the polynomial

    (1-x^2)(1-x^4) = 1-x^2-x^4+x^6,

    which clearly has height 1. Now by repeatedly factorising the difference of two squares, write that polynomial as

    (1-x^2)(1-x^4) = (1-x^2)^2(1+x^2) = (1-x)^2(1+x)^2(1+x^2).

    Take P(x) = (1-x)^2 and Q(x) = (1+x)^2(1+x^2). Then each of P(x), Q(x) has height at least 2, and their product has height 1.

    Now do the same sort of thing for N = 3, looking at the polynomial (1-x^2)(1-x^4)(1-x^8). Once you have sorted that one out, you will see how to do it for any N. (It's really a very simple-minded construction, nothing as sophisticated as cyclotomic polynomials!)
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    Re: polynomial height

    Quote Originally Posted by Opalg View Post
    Okay, here is the idea. Given a positive integer N, we want to find polynomials P(x), Q(x), with integer coefficients, each having height at least N, such that P(x)Q(x) has height 1.

    Even for N = 2 this takes a bit of thought. Start with the polynomial

    (1-x^2)(1-x^4) = 1-x^2-x^4+x^6,

    which clearly has height 1. Now by repeatedly factorising the difference of two squares, write that polynomial as

    (1-x^2)(1-x^4) = (1-x^2)^2(1+x^2) = (1-x)^2(1+x)^2(1+x^2).

    Take P(x) = (1-x)^2 and Q(x) = (1+x)^2(1+x^2). Then each of P(x), Q(x) has height at least 2, and their product has height 1.

    Now do the same sort of thing for N = 3, looking at the polynomial (1-x^2)(1-x^4)(1-x^8). Once you have sorted that one out, you will see how to do it for any N. (It's really a very simple-minded construction, nothing as sophisticated as cyclotomic polynomials!)
    brilliant! definetly very simple!!!!! thanks a lot!!!!!! the trick is to multiply with such polyn with degree such that the product gives degrees that whose terms do not sum and such that the different facorization gives the desired coefficients!!!!

    the question I would like to ask is: how one can have this 'enlightment' to try this factorization? experience? chance? trials on basic polynomials?...
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  6. #6
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    Re: polynomial height

    Quote Originally Posted by pincopallino View Post
    brilliant! definetly very simple!!!!! thanks a lot!!!!!! the trick is to multiply with such polyn with degree such that the product gives degrees that whose terms do not sum and such that the different facorization gives the desired coefficients!!!!

    the question I would like to ask is: how one can have this 'enlightment' to try this factorization? experience? chance? trials on basic polynomials?...
    As I said, it takes a bit of thought, even for N=2. See the Newton quote below! In fact, I spent a long evening struggling with the N=2 case, doing fruitless computations, until the factorisation idea suddenly jumped out at me.
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    Re: polynomial height

    I see! <wait till the first dawnings open little by little into the full light>

    (1-x^2)(1-x^4)(1-x^8)...(1-x^2^^(^n)) . = (1-x)^n (1+x)^n (1+x^2)^n^-^1 (1+x^4)^n^-^2 (1+x^8)^n^-^3))... (1+x^2^^(^n^-^1^)) .
    Last edited by pincopallino; August 3rd 2011 at 01:26 PM.
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