Proving f is a Polynomial with Degree at Most n in Complex Analysis

[Poopsilon]

Let D ⊂ ℂ be a domain and let f be analytic on D. Show that if there is an a ∈ D such that the kth derivative of f at a is zero for k=n, n+1, n+2,..., then f is a polynomial with degree at most n.

So I believe I have a proof, but the theorems are so powerful I feel like I might be overlooking something, here it is:

Basically f being analytic at a means it can be expanded as a convergent power-series in some open neighborhood around a. Now since the kth derivative and beyond is zero at a, we know this power-series is just at most an nth degree polynomial, which coincides with the values of f for all z in our neighborhood.

Now there is a theorem in my book which says that if g:D→ℂ is an analytic function on a domain D which is not identically zero, then the set of zeros of g is discrete in D. By discrete we mean that the set of zeros does not contain an accumulation point.

Thus since the kth derivative and beyond of f is analytic on D and are zero on an open neighborhood of D, and since this neighborhood is open in ℂ, then it clearly contains accumulation points, and thus the kth derivative and beyond of f must be zero on all of D in order not to contradict the theorem given above.

Therefore f is a polynomial with degree at most n.

Thanks =].

 

[mighty2000]

Your proof looks correct to me! You have used the fact that the kth derivative and beyond being zero at a implies that the power-series expansion of f at a is just an nth degree polynomial, and then used the theorem about the set of zeros of an analytic function being discrete to show that f must be a polynomial on all of D. Great job!