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jverzani
2024-10-15 17:17:25 -04:00
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@@ -481,7 +481,7 @@ Now consider the related polynomial, $q$, where we multiply $p$ by $x^n$ and sub
#| hold: true
p = a*x^2 + b*x + c
n = 2 # the degree of p
q = expand(x^n * p(x => 1/x))
expand(x^n * p(x => 1/x))
```
In particular, from the reversal, the behavior of $q$ for large $x$ depends on the sign of $a_0$. As well, due to the $1/x$, the behaviour of $q$ for large $x>0$ is the same as the behaviour of $p$ for small *positive* $x$. In particular if $a_n > 0$ but $a_0 < 0$, then `p` is eventually positive and `q` is eventually negative.