Re: Can someone explain this proof, please?

Timothy Murphy wrote:

I take it that your proof shows that if the solution is written
f(x) g(x) = C(x),
then we get the same solution on replacing x by 1/x
ie (x^n f(1/x)) (x^m g(1/x)) = x^{m+n} C(1/x) = C(x).

An easy way to see that f and g are reciprocal polynomials
is to note that each zero z of f is a root of unity
and so has modulus 1. As f has real coefficients the
conjugate of z which is z^{-1} is also a root. Hence
f is reciprocal.

Victor Meldrew

.

Relevant Pages

  • Re: Can someone explain this proof, please?
    ... then we get the same solution on replacing x by 1/x ... An easy way to see that f and g are reciprocal polynomials ... is to note that each zero z of f is a root of unity ... could have been proved first, ...
    (sci.math)
  • Re: root in FC 10
    ... Timothy Murphy wrote: ... I must say I haven't noticed that, as I haven't logged in as root for years. ... But I do recall one time when it was useful, ... for thou art crunchy and taste good with Ketchup! ...
    (Fedora)
  • Re: Where can I download SuSE on campus?
    ... root wrote: ... > news server retention is at least a week long, ... Timothy Murphy ... e-mail: tim /at/ birdsnest.maths.tcd.ie ...
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  • Re: reciprocal polynomial
    ... of a regular n-gon with conjugate symmetry. ... of any root is also a root. ... Now apply this insight to my previous ... = product of reciprocal polynomials of gand h. ...
    (comp.dsp)
  • Re: Cyclotomic field
    ... Timothy Murphy wrote: ... z_k is the primitive k-th root of unity. ... Gerry Myerson ...
    (sci.math)