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Is it possible to have a power series $f(x)$ like that: $f(x)=\left \lceil{x-\left \lfloor{x}\right \rfloor}\right \rceil$?

user26857
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pgp1
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    But $f(x)$ is discontinuous whereas any finite polynomial is continuous. – Peter Foreman Dec 24 '19 at 13:43
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    But there are discontinuous infinite polynomials – pgp1 Dec 24 '19 at 13:47
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    There's no such thing as an infinite polynomial, all polynomials have finite degree. – Peter Foreman Dec 24 '19 at 13:51
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    What about the Taylor expension of $\sin x$? – pgp1 Dec 25 '19 at 13:55
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    The Taylor series expansion for $\sin{(x)}$ is a representation for the function which uses a convergent power series. A power series is not the same as a polynomial. – Peter Foreman Dec 25 '19 at 14:17
  • If your power series converges at some non-zero $x_0$ then it is analytic for $|x|< x_0$ and the only analytic function which is constant $=1$ on $(0,|x_0|)$ is the trivial power series $1$. The first theorem to know is that a non-constant analytic function has isolated zeros (here you'll substract $1$ to find it has non-isolated zeros thus it is constant) – reuns Dec 26 '19 at 19:21

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Your $f(x)$ is $=0$ for all $x\in{\mathbb Z}$ and $=1$ for all $x\notin{\mathbb Z}$. It is sufficient to check the $x\in [0,1[\>$ to check this. There is no polynomial $f$ that can deliver such values, e.g., because polynomials are continuous. By the way, there are no "infinite polynomials".

Christian Blatter
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    Infinite polynomial is a polynom that there is no degree you can say it's the "highest" degree – pgp1 Dec 25 '19 at 13:54