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I have two polynomials:

$Q(z)=q_0 +q_1 z + \cdots q_mz^m$ and its reflection $ Q^'(z)=q_0 z^m +q_1z^{m-1}+ \cdots q_m$. I'd like to find a relation between them (i.e. $Q(z)= \phi(Q'(z))$, so far for I could only show that for $Q(z)=q_0+q_1 z$ and $Q'(z)=q_0z+q_1$ $Q^2(z)-Q'^2(z)=(1-z^2)(q_0^2-q_1^2)$.

There is probably some well-known solution to this problem. Please don't solve it for me, just point in the right direction.

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Consider $z^mQ(1/z)$ –  Bill Cook Oct 19 '11 at 17:34
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Or consider that if $a\neq 0$ is a root of $Q(z)$, then $\frac{1}{a}$ is a root of $Q'(z)$ and vice-versal; think about what that implies about the factorization of $Q'$ versus that of $Q$. –  Arturo Magidin Oct 19 '11 at 18:17

1 Answer 1

up vote 3 down vote accepted

To elaborate on my comment...don't read what follows if you want to work this out for yourself...

$Q(z^{-1})=a_mz^{-m}+\cdots+a_1z^{-1}+a_0$. So $z^mQ(z^{-1})=a_m+\cdots+a_1z^{m-1}+a_0z^m$. Thus the reflection of $Q(z)$ is just $z^mQ(z^{-1})$.

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yeah I sorted it out almost immediately haveing read your comment. –  sigma.z.1980 Oct 20 '11 at 9:38

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