8
$\begingroup$

Suppose $f$ is entire and that in every power series $f(z)=\sum_{n=0}^\infty c_n(z-a)^n$ at least one coefficient is $0$. Prove that $f$ is a polynomial.

$\endgroup$
  • 2
    $\begingroup$ Welcome to Math.SE! Please, consider updating your question to include what you have tried / where you are getting stuck. You will find that people on this site will be significantly faster to help you if you do that; that way, we know exactly what help you need. $\endgroup$ – Did Aug 16 '13 at 0:22
7
$\begingroup$

Hint: For each complex number $a$, there exists an $n$ for which $c_n(a) = 0$, where $c_n(a)$ is the $n$th coefficient for the expansion at $a$. There are uncountably many complex numbers, but only countably many naturals, so....

$\endgroup$
  • 3
    $\begingroup$ Ah so there exists a $n$ with uncountable number of zeros. Such that $0=c_n(a)=\frac{f^{(n)}(a)}{n!}$ for uncountably many $a$. Since each derivative is entire, there are only countable number of zeros. Hence the $n$th derivative is 0, and $f$ is a polynomial. $\endgroup$ – anon Aug 16 '13 at 0:59
  • $\begingroup$ @user61527:I have one doubt regarding above question.Is it necessary that n is fixed?If n is not fixed then how can we say that $c_n$ is a function of $z$? $\endgroup$ – ASHWINI SANKHE May 19 '18 at 6:29

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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