 Solution

ID: past-exam-of-the-algebra-analysis-and-differential-equations-1-course-of-the-university-of-oxford/2010/6/ii/solution

M / r^{k + 1}

, and the circle has length

2 π r

. The ML inequality therefore gives

∣ f^{(k)} (z_{0}) ∣ \leq \frac{k !}{2 π} (2 π r) \frac{M}{r ^{k + 1}} = \frac{k ! M}{r ^{k}} .

The maximum

M

exists because the modulus of the continuous function

f

attains a maximum on the compact circle. These are the Cauchy estimate.

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