Bound on function in disk using Cauchy Estimate

Solution 1:

You are right, the bound

$$\frac{n! M}{R^n}$$

is a bound for $f^{(n)}(0)$ only, not for $\lvert z\rvert \leqslant \rho$. For $\lvert z\rvert \leqslant \rho$, the standard estimate yields

$$\left\lvert f^{(n)}(z)\right\rvert \leqslant \frac{n! MR}{(R-\rho)^{n+1}}$$

(which for $\rho > 0$ is not a sharp bound, since all of the circle except at most one point is at a distance $> R-\rho$ from $z$ then). A better, but hard to evaluate bound would be

$$\left\lvert f^{(n)}(z)\right\rvert \leqslant \frac{n!MR}{2\pi} \int_0^{2\pi} \frac{dt}{\lvert Re^{it} - z\rvert^{n+1}} \leqslant \frac{n! MR}{2\pi} \int_0^{2\pi} \frac{dt}{\lvert Re^{it} - \rho\rvert^{n+1}}.$$