# Roots of equation?

• May 25th 2009, 10:05 AM
fardeen_gen
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 $(1 + x_1^2)(1 + x_2^2)(1 + x_3^2)\mbox{...}(1 + x_n^2) = (1 - p_1 + p_4 + \mbox{...})^2 + (p_1 - p_3 + p_5 + \mbox{...})^2$
• May 31st 2009, 02:36 AM
Isomorphism
Quote:

Originally Posted by fardeen_gen
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 $(1 + x_1^2)(1 + x_2^2)(1 + x_3^2)\mbox{...}(1 + x_n^2) = (1 - p_2 + p_4 + \mbox{...})^2 + (p_1 - p_3 + p_5 + \mbox{...})^2$
• Jun 6th 2009, 01:55 PM
Media_Man
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 (Rock)
• Jun 8th 2009, 08:34 AM
pankaj
$(x-x_{1})(x-x_{2}).....(x-x_{n})=x^n+p_{1}x^{n-1}+p_{2}x^{n-2}+...+p_{n}$
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.