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## Every expansion has a 0

Suppose is entire such that for each at least one of the coefficients of the expansion is equal to 0. Prove that is a polynomial.

*Solution:* The statement of the problem is equivalent to such that . Therefore, if we denote we have that is closed for any (because is entire and is continuous ) and . Since is a complete metric space, by Baire’s Category Theorem, one of the sets has non-void interior. This means that there exists such that for in an open disk. This means that is a polynomial in that disk, and furthermore, by analytic continuation, must be a polynomial everywhere.

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Categories: Complex analysis
complex, entire, series, taylor

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