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Topic: holomorphic polynomials (Read 5274 times) |
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Mary I
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Hello!
If we know that f is holomorphic on D(0,1) and assume that f^2 is a holomorphic polynomial on D(0,1), does it follow that f is also a polynomial on D(0,1)?
I am almost sure that f is a polynomial but how can I show it?
I have tried to use the Cauchy product but it didn't work out.
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Eigenray
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Re: holomorphic polynomials
« Reply #1 on: Dec 2nd, 2005, 5:23am » |
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No. Note that any holomorphic function which has no zeroes on a simply-connected domain like D(0,1) has a holomorphic square root there.
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Icarus
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Boldly going where even angels fear to tread.
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Re: holomorphic polynomials
« Reply #2 on: Dec 2nd, 2005, 3:48pm » |
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For example, f(x) = sqrt(x2 + 4) satisfies all your conditions, but is not a polynomial.
(By the way, "holomorphic polynomial" is redundant. All polynomials are holomorphic.)
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"Pi goes on and on and on ...
And e is just as cursed.
I wonder: Which is larger
When their digits are reversed? " - Anonymous
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