Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 $n>k, \space a_n=0$?

share|improve this question
    
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 8 '12 at 9:32
    
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 8 '12 at 9:35
1  
@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 8 '12 at 9:37
    
@EuYu What Marc said; I have updated the question. –  Mark Hurd Dec 8 '12 at 9:39
add comment

1 Answer

up vote 5 down vote accepted

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.

share|improve this answer
    
Yeah, so as I said, a polynomial translated is still a polynomial; ignore the complex algebra that would otherwise occur. –  Mark Hurd Dec 8 '12 at 9:55
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.