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Re: 0,1,2,0,1,2,3,0,1,2,3,4,...

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  • richard042@yahoo.com
    ... number). ... Take any two univariate polynomials, P(x) and K(x), such that their orders sum to p(n), then yes, you can multiply them together to create a
    Message 1 of 4 , Sep 4, 2004
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      --- In primenumbers@yahoogroups.com, REZA RANJBAR
      <saeedgeometr22@y...> wrote:
      > Let P(x) is none-zero polynomial on R (the set of all real
      number).
      > Show that.
      > There is none –zero polynomial K(x) which
      >
      > P(x)*K(x)=a(0) + a(1)x^2 + a(2)x^3 +a(3)x^5 +a(4)x^7+ … a(n)x^p(n).
      >
      > for some p(n).
      > Such that p(n) is nst. Prime number.
      > Remark: a(i) is not zero for all i such that 0<=i<=n.

      Take any two univariate polynomials, P(x) and K(x), such that their
      orders sum to p(n), then yes, you can multiply them together to
      create a new polynomial with order p(n).

      >
      > Send me.thank you.
      >
      > Saeed ranjbar.

      I think it's as simple as that, but perhaps I am missing something?

      Regards,

      -Dick
    • Edwin Clark
      ... It seems that Reza is not asking for just a polynomial of degree p(n), but a polynomial of the form a(0) + sum(a(i)x^p(i),i=1..n) where each a(i) is not
      Message 2 of 4 , Sep 5, 2004
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        On Sun, 5 Sep 2004 richard042@... wrote:

        > --- In primenumbers@yahoogroups.com, REZA RANJBAR
        > <saeedgeometr22@y...> wrote:
        > > Let  P(x) is none-zero polynomial on  R (the set of all real
        > number).
        > > Show that.
        > > There is none –zero  polynomial    K(x)  which
        > >
        > > P(x)*K(x)=a(0) + a(1)x^2 + a(2)x^3 +a(3)x^5 +a(4)x^7+ … a(n)x^p(n).
        > >
        > > for some p(n).
        > > Such that  p(n) is nst. Prime number.
        > > Remark: a(i) is not zero for all i such that  0<=i<=n.
        >
        > Take any two univariate polynomials, P(x) and K(x), such that their
        > orders sum to p(n), then yes, you can multiply them together to
        > create a new polynomial with order p(n).

        It seems that Reza is not asking for just a polynomial of degree p(n), but
        a polynomial of the form a(0) + sum(a(i)x^p(i),i=1..n) where each a(i)
        is not zero and p(i) is the ith prime.

        Clearly you must rule out polynomials of the form P(x) = x^j*f(x)
        where j > 0 since that would force a(i) = 0 for i < j.

        But even if you make the assumption that x does not divide P(x) the
        result is not true. Take P(x) = 1 + x^2 + x^4. I claim
        that if P(x)K(x) has 0 coefficients for x^j when j > 0 and not prime
        then it will also have 0 coefficients for x^0 and x^2.


        Edwin
      • Cletus Emmanuel
        No homework requests please!!! REZA RANJBAR wrote: Let P(x) is none-zero polynomial on R (the set of all real number). Show that.
        Message 3 of 4 , Sep 7, 2004
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          No homework requests please!!!

          REZA RANJBAR <saeedgeometr22@...> wrote:
          Let P(x) is none-zero polynomial on R (the set of all real number).



          Show that.



          There is none �zero polynomial K(x) which



          P(x)*K(x)=a(0) + a(1)x^2 + a(2)x^3 +a(3)x^5 +a(4)x^7+ � a(n)x^p(n).

          for some p(n).

          Such that p(n) is nst. Prime number.

          Remark: a(i) is not zero for all i such that 0<=i<=n.



          Send me.thank you.



          Saeed ranjbar.



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