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Update Math4202_L23.md
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Zheyuan Wu
2026-03-24 15:47:02 -05:00
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@@ -62,7 +62,6 @@ Therefore $f$ must have a root in $B^2$.
If $\|a_{n-1}\|+\|a_{n-2}\|+\cdots+\|a_0\|< R$ has a root in the disk $B^2_R$. (and $R\geq 1$, otherwise follows part 1) If $\|a_{n-1}\|+\|a_{n-2}\|+\cdots+\|a_0\|< R$ has a root in the disk $B^2_R$. (and $R\geq 1$, otherwise follows part 1)
Consider $\tilde{f}(x)=f(Rx)$. Consider $\tilde{f}(x)=f(Rx)$.
$$ $$
\begin{aligned} \begin{aligned}
\tilde{f}(x) \tilde{f}(x)
@@ -73,7 +72,7 @@ $$
$$ $$
\begin{aligned} \begin{aligned}
\|\frac{a_{n-1}}{R}\|+\|\frac{a_{n-2}}{R^2}\|+\cdots+\|\frac{a_0}{R^n}\|&=\frac{1}{R}\|a_{n-1}\|+\frac{1}{R^2}\|a_{n-2}\|+\cdots+\frac{1}{R^n}\|a_0\|\\ \left\|\frac{a_{n-1}}{R}\right\|+\left\|\frac{a_{n-2}}{R^2}\right\|+\cdots+\left\|\frac{a_0}{R^n}\right\|&=\frac{1}{R}\|a_{n-1}\|+\frac{1}{R^2}\|a_{n-2}\|+\cdots+\frac{1}{R^n}\|a_0\|\\
&<\frac{1}{R}\left(\|a_{n-1}\|+\|a_{n-2}\|+\cdots+\|a_0\|\right)\\ &<\frac{1}{R}\left(\|a_{n-1}\|+\|a_{n-2}\|+\cdots+\|a_0\|\right)\\
&<\frac{1}{R}<1 &<\frac{1}{R}<1
\end{aligned} \end{aligned}