 Solution

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

R \geq S

about zero. The Cauchy estimate gives, for each integer

k > n

∣ f^{(k)} (0) ∣ \leq k! C R^{n - k} ⟶ 0 (R \to \infty) .

Therefore all these derivatives vanish. The Taylor series of an entire function converges to it everywhere, and hence

f (z) = k = 0 \sum n \frac{f ^{(k)} ( 0 )}{k !} z^{k} .

As usual

n

is a nonnegative integer; if a real exponent is allowed, the degree is at most its integer part. A negative growth exponent forces

f = 0

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