 Solution

ID: past-exam-of-the-algebra-analysis-and-differential-equations-2-course-of-the-university-of-oxford/2005/analysis/4/i/solution

f

is holomorphic on a neighborhood of the closed disc, the differentiated Cauchy integral formula is

f^{(k)} (a) = \frac{k !}{2 πi} \int_{∣ z - a ∣ = r} \frac{f ( z )}{( z - a ) ^{k + 1}} d z,

with positive orientation. On the circle the integrand has modulus at most

M / r^{k + 1}

2 π r

. Thus

∣ f^{(k)} (a) ∣ \leq \frac{k ! M}{r ^{k}} .

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