Re: Legendre,Chebychev polynomials

From: G. A. Edgar (edgar_at_math.ohio-state.edu.invalid)
Date: 03/26/05

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    Date: Sat, 26 Mar 2005 15:41:27 -0500

    Maybe you need to explain your notation better...

    In article <117ac0e5.0503252040.8d0bf2f@posting.google.com>, Alex.Lupas
    <alexandru.lupas@ulbsibiu.ro> wrote:

    > In the following, let us denote :
    > SUM{A_k}:=A_1+A_2+...+A_n ; (n >= 1 )

    so SUM{A_k} no longer depends on k?

    > P_n(x)=k_n*[(x^2-1)^n]^{(n)} , P_n(1)=1 , be the Legendre polynomial
    > having the roots a_1,a_2,...,a_n ;
    > U_n(x)=sin((n+1)*arccos(x))/((n+1)*sqrt(1-x^2)) , U_n(1)=1 , be the Chebychev
    > polynomial of second kind with the roots
    > b_k=cos(k*pi/(n+1)) , k in {1,2,...,n} .
    >
    > [A.] I need elementary proofs of following identities /if true/:
    >
    >
    > (1) P_n(b_k)=SUM{(b_k)^n *U_n((b_k)^2)}

    How can this be if SUM no longer depends on k ??

    >
    > (2) U_n(a_k)=-SUM{(b_k)^n *U_n(a_k*b_k)}
    >
    > It's true that
    > =======================================
    > (1-2) P_n(x)-U_n(x)=SUM{(b_k)^n*U_n(x*b_k)} ??
    > =======================================
    >
    > [B.] Let c_k=cos((2k-1)pi/(2n)) , k in {1,2,...,n}, and denote
    >
    > Q_{2n}(x):= SUM{ P_n(x^2+(1-x^2)*c_k) } .
    >
    > It's true that Q_{2n}(x) has only double roots ?? 

    -- 
G. A. Edgar                               http://www.math.ohio-state.edu/~edgar/
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