Re: reciprocal polynomial

On 5月20日, 上午4时08分, "dvsarw...@xxxxxxxxx" <dvsarw...@xxxxxxxxx> wrote:
On May 19, 11:52 am, Martin Eisenberg <martin.eisenb...@xxxxxxx>
wrote:



then reciprocal polynomial is a factor of p^n + 1.

The roots of p^n+1 lie on the complex unit circle and are the corners
of a regular n-gon with conjugate symmetry. Therefore the reciprocal
of any root is also a root. Now apply this insight to my previous
paragraph's last statement.

Even more easily, and without having to know anything about the
roots of p^n + 1, note that (1/p)^n + 1 = g(1/p)h(1/p), and so

p^n((1/p)^n + 1) = p^n[g((1/p)h(1/p)] = [p^{n-k}g((1/p)][p^{k}h(1/p)]
= product of reciprocal polynomials of g(p) and h(p).

Since the left-hand side is just p^n + 1 (or 1 + p^n for the purists),
we deduce that the reciprocal polynomial of h(p) is a factor of p^n +
1.

Hope this helps

--Dilip Sarwate

Hi everyone,

thanks for your help. I'm studying Proakis's book "Digital
Communications" and confused by this question. Now I understand it.
Thank you very much.
.

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