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{\bf Question}

Show that $(f_1^2)"(p)=0$ for a period-doubling point of $f_a$.
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{\bf Answer}

$f_a^2(p)-f_a(f_a(p))$ so $(f_a^2)'(p)={f_a}'(f_a(p)) \cdot
{f_a}'(p).$

Then $(f_a^2)''(p)={f_a}''(f_a(p)) \cdot
({f_a}'(p))^2+{f_a}'(f_a(p)) \cdot {f_a}''(p).$

But $f_a(p)=p$ and ${f_a}'(p)=-1$, giving $(f_a^2)''(p)=0.$

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