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I'm reading Barbeau's polynomials, and he states:

For the polynomial $a_nt^n+a_{n-1}t^{n-1}+...+a_1t+a_0$, with $a_n \neq0 $, the numbers $a_i$ $(0 \leq i \leq n)$ are called coefficients.

Some pages later, there's a question:

Is $deg(p \circ q)$ related in any way to $deg(q \circ p)$?

Which I answered:

$deg(p\circ q)= deg(q\circ p)$ iff both polynomials have at least $a_1$, because we could have a constant polynomial in which I guess that the statement above would make no sense.

And then, when I went for the answer, I've found:

$deg$ $p\circ q$ $=$ $deg$ $q\circ p =(deg \, p)(deg \, q)$

So, considering two polynomials P and Q, both of $deg=0$, is it still possible to perform polynomial composition?

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What could prevent you to consider poq when p and q have degree zero? – Did Sep 15 '12 at 22:21
Not considering such a polynomial as a polynomial? – Voyska Sep 15 '12 at 22:46
I see no reason not to. – Did Sep 15 '12 at 22:48
Yep, it's because the definition of polynomial, it says: $a_i$ $(0 \leq i \leq n)$, it should be: $a_i$ $(1 \leq i \leq n)$, isn't it? – Voyska Sep 15 '12 at 22:58
You might want to write down in the question the definition of polynomial you found (so that one can see if indeed the trouble lies there). – Did Sep 15 '12 at 23:10
up vote 3 down vote accepted

If we think of polynomials as functions so that we associate the polynomial $a_nt^n+...+a_1t+a_0$ to the function $f(t)=a_nt^n+...+a_1t+a_0$, then we only have to understand function composition.

For example, let $f(t)=t^2-25$ and $g(t)=5$. Then $(f\circ g)(t)=5^2-25=0$ and $(g\circ f)(t)=5$ so that the degree formula may fail if we take the zero polynomial to have degree $-\infty$. But we can still perform composition.

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The zero polynomial should have degree $-\infty$. This is a little clearer if one thinks of the degree as a property of a Laurent polynomial $f(t) = a_n t^n + ... + a_{-m} t^{-m}$ rather than a polynomial. – Qiaochu Yuan Sep 16 '12 at 8:26

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