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The series $\sum\limits_{n=0}^\infty {a_{n}}(x-c)^n $ is a polynomial in $x$.

(This is from this question.)

For $c=0$ this clearly means that, for some $n>k, \space a_n=0$, almost by definition.

Is it also true for all other values of $c$ that there exists a $k$ such that $n>k, \space a_n=0$?

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Mark Hurd
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  • Of course it should be true "by translation of $c$; the polynomial is still a polynomial", but the actual algebra doesn't seem to be as easy. – Mark Hurd Dec 08 '12 at 09:32
  • I'm not too clear on what the actual question here is. Are you asking that if the series is a polynomial for some value of $c$ then it is a polynomial for all values of $c$? – EuYu Dec 08 '12 at 09:35
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    @EuYu: The question is: if $c\neq0$ is given and $\sum\limits_{n=0}^\infty {a_{n}}(x-c)^n$ is a polynomial, does it follow that for some $k$ one has $a_n=0$ for all $n>k$. The answer is yes. – Marc van Leeuwen Dec 08 '12 at 09:37
  • @EuYu What Marc said; I have updated the question. – Mark Hurd Dec 08 '12 at 09:39

1 Answers1

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The substitution $x:=x-a$ maps polynomials to polynomials, and since its inverse is the substitution $x:=x+a$, the result of the substitution into a series $S$ is a polynomial if and only if $S$ is itself a polynomial. Therefore $\sum_{n=0}^\infty {a_{n}}(x-c)^n$ is a polynomial if and only if $\sum_{n=0}^\infty {a_{n}}x^n$ is a polynomial.

Marc van Leeuwen
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  • Yeah, so as I said, a polynomial translated is still a polynomial; ignore the complex algebra that would otherwise occur. – Mark Hurd Dec 08 '12 at 09:55