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Math Help - Roots of equation?

  1. #1
    Super Member fardeen_gen's Avatar
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    Roots of equation?

    If x_1,x_2,x_3,\mbox{...},x_n are the roots of the equation x^n + p_1x^{n - 1} + p_2x^{n - 2} + ... + p_{n - 1}x + p_n = 0, where coefficient p_1,p_2,p_3,\mbox{...},p_n are real, show that (1 + x_1^2)(1 + x_2^2)(1 + x_3^2)\mbox{...}(1 + x_n^2) = (1 - p_1 + p_4 + \mbox{...})^2 + (p_1 - p_3 + p_5 + \mbox{...})^2
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  2. #2
    Lord of certain Rings
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    Quote Originally Posted by fardeen_gen View Post
    If x_1,x_2,x_3,\mbox{...},x_n are the roots of the equation x^n + p_1x^{n - 1} + p_2x^{n - 2} + ... + p_{n - 1}x + p_n = 0, where coefficient p_1,p_2,p_3,\mbox{...},p_n are real, show that (1 + x_1^2)(1 + x_2^2)(1 + x_3^2)\mbox{...}(1 + x_n^2) = (1 - p_1 + p_4 + \mbox{...})^2 + (p_1 - p_3 + p_5 + \mbox{...})^2
    I think it should be (1 + x_1^2)(1 + x_2^2)(1 + x_3^2)\mbox{...}(1 + x_n^2) = (1 - p_2 + p_4 + \mbox{...})^2 + (p_1 - p_3 + p_5 + \mbox{...})^2
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  3. #3
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    Induction

    Base case (n=1): x+p_1=0~~\rightarrow~~1+x_1^2=1+(-p_1)^2=(1)^2+(p_1)^2

    Define Q(x)=\sum_{i=0}^{n-1}q_ix^i=\prod_{i=1}^{n-1}(x-x_i)=0 and P(x)=\sum_{i=0}^np_ix^i=\prod_{i=1}^n(x-x_i)= (x-x_n)Q(x)=xQ(x)-x_nQ(x)=0 . So p_i=q_i-x_nq_{i-1} for i\in\mathbb{N} and p_0=q_0=1

    Assume (1+x_1^2)(1+x_2^2)...(1+x_{n-1}^2)=(1-q_2+q_4-q_6+...)^2+(q_1-q_3+q_5-q_7+...)^2=Q_{even}^2+Q_{odd}^2

    Show: (1+x_1^2)(1+x_2^2)...(1+x_n^2)=(1-p_2+p_4-p_6+...)^2+(p_1-p_3+p_5-p_7+...)^2=P_{even}^2+P_{odd}^2

    Proof: P_{even}^2+P_{odd}^2=(1-p_2+p_4-p_6+...)^2+(p_1-p_3+p_5-p_7+...)^2

    = [1-(q_2-x_nq_1)+(q_4-x_nq_3)-(q_6-x_nq_5)+...]^2+ [(q_1-x_n)-(q_3-x_nq_2)+(q_5-x_nq_4)-(q_7-x_nq_6)+...]^2

    = [(1-q_2+q_4-q_6+...)+x_n(q_1-q_3+q_5-q_7+...)]^2+ [(q_1-q_3+q_5-q_7+...)-x_n(1-q_2+q_4-q_6+...)]^2

    = [Q_{even}+x_nQ_{odd}]^2+[Q_{odd}-x_nQ_{even}]^2

    = [Q_{even}^2+Q_{odd}^2](1+x_n^2)

    = (1+x_1^2)(1+x_2^2)...(1+x_{n-1}^2)(1+x_n^2)

    QED
    Last edited by Media_Man; June 6th 2009 at 02:09 PM. Reason: clearer definition
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  4. #4
    Senior Member pankaj's Avatar
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    (x-x_{1})(x-x_{2}).....(x-x_{n})=x^n+p_{1}x^{n-1}+p_{2}x^{n-2}+...+p_{n}
    is an identity in x

    Put x=i
    (i-x_{1})(i-x_{2}).....(i-x_{n}) =(p_{n}-p_{n-2}+p_{n-4}-p_{n-6}+...)+i(p_{n-1}-p_{n-3}+p_{n-5}-.....)

    Put x=-i
    (-i-x_{1})(-i-x_{2}).....(-i-x_{n}) =(p_{n}-p_{n-2}+p_{n-4}-p_{n-6}+...)-i(p_{n-1}-p_{n-3}+p_{n-5}-.....)

    Multiplying 1. and 2.
    (1+x_{1}^2)(1+x_{2}^2)...(1+x_{n}^2)=(p_{n}-p_{n-2}+p_{n-4}-p_{n-6}+...)^2+(p_{n-1}-p_{n-3}+p_{n-5}-.....)^2

    You will get the required result whatever n may be either even or odd.
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