# Can I get a better bound on this function?

Question: if f is analytical and $|f(z)| < M$ for $|z| =< R$ find an upper bound for $|f('n)(z)|$ in $|z| =< \rho < R$ (where $f('n)$ means the nth derivative of $f$).

So I cited Cauchy's Inequality, to find the bound $\frac{n!M_R} {R^n}$ , but since we have the additional limit of $\rho$ I think we should be able to find a better bound. The geometric interpretation is that the the circle of radius $\rho$ limits the radius of $|z|$. Is there further geometric intuition that I should see to solve this problem?

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No, there is no better bound. Consider the function $f(z)=M \frac{z^n}{R^n}$ which fulfills the requirement (ok, that $|f(z)|$ is really smaller than $M$ one should replace $M$ by $M-\epsilon$ and in the end let $\epsilon\to0$). It is easy to check that $$f^{(n)}(z) = \frac{n! M}{R^n},$$ i.e., it exhausts the Cauchy bound. Note however that the $n$-th derivative is independent on $z$. So there is no way you get a better bound by restricting $|z|<\rho$.
This is a question of Lars V. Ahlfors, Complex Analysis, Third Edition, McGraw-Hill International Editions, 1979, at the end of the subsection "Higher Derivatives" of the section "Cauchy's Integral Formula". In this context, the answer should take this into account. My suggestion is to apply the formula $f^{\left(n\right)}\left(z\right)=\frac{n!}{2\pi i}\int_{C}\frac{f\left(\zeta\right)d\zeta}{\left(\zeta-z\right)^{n+1}}$. Thus we have
$\begin{split}\left|f^{\left(n\right)}\left(z\right)\right| & =\left|\frac{n!}{2\pi i}\int_{C}\frac{f\left(\zeta\right)d\zeta}{\left(\zeta-z\right)^{n+1}}\right|\\ & =\frac{n!}{2\pi}\left|\int_{C}\frac{f\left(\zeta\right)d\zeta}{\left(\zeta-z\right)^{n+1}}\right|\\ & \leq\frac{n!}{2\pi}\int_{C}\frac{\left|f\left(\zeta\right)\right|d\zeta}{\left|\zeta-z\right|^{n+1}}\\ & \leq\frac{n!}{2\pi}M\int_{C}\frac{d\zeta}{R^{n+1}}\\ & =\frac{n!M}{2\pi R^{n+1}}\int_{C}d\zeta\\ & \leq\frac{n!M}{2\pi R^{n+1}}2\pi R\\ & =\frac{n!M}{R^{n}}.\end{split}$