# Sequence of complex polynomials

1. Mar 6, 2009

### redrzewski

This is problem 13.3 from Rudin's Real and Complex analysis. It is not homework.

Is there a sequence of polynomials {Pn} such that Pn(0) = 1 for n = 1,2,3,... but Pn(z) -> 0 for all z != 0 as n -> infinity?

My guess here is no. Sketch of proof: Assume such a sequence existed. Then we should be able to contradict the maximum modulus theorem for any disk around 0 since all Pn(z) for |z| = r will be approaching 0 for large enough n, but Pn(0) = 1.

Is this correct?

thanks

2. Mar 8, 2009

### lurflurf

It depends what you mean by converges.
clearly if you fix n lim x->infinity pn(x)=infinity
but it may be that if you fix x lim n->infinity pn(x)=0
consider
f=(1-x^2/n^n)^n

3. Mar 8, 2009

### redrzewski

Can you clarify that? Say we fix an "r", and take the circle |z| = r. Then for any finite n, and z on this circle, with your f, f(z) will either be approching 0, or approaching infinity. It doesn't seem like all the values on the circle can be driven to 0 at the same time for a given n.

Your function will work fine for real x. But I can't seem to make it work for the complex case. What am I missing?

Also, I could be blowing the limit computation, but doesn't that f converge to 1 for all finite x as n goes to infinity?

And if we go with:
f = (1-x^2/n)^n, that'll converge to exp(-x^2), which has the same problem that I can't drive all the values on the circle to 0.

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