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Since $\pi$ is a zero of the Taylor series of $\sin{x}$, I wonder if all transcendental numbers (computable ones) are zeros of polynomials if we allow for infinite polynomials. Does anyone know about this?

If this is true, does this tell us anything important about the nature of Transcendental numbers? I mean, algebraic numbers are zeros of finite polynomials. So are Transcendental numbers limits of algebraic numbers as the polynomial length tends to infinity?

Ryder Rude
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    Transcendental numbers (computable or not) are even limits of rational numbers - think about their decimal expansions. – Noah Schweber Mar 31 '20 at 02:44
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    All numbers are roots of an entire function defined by a series with rational coefficients, see https://math.stackexchange.com/questions/895449/ring-of-rational-coefficient-power-series-defining-entire-functions – orangeskid Mar 31 '20 at 06:03

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