4

Question is prove or disprove: There exists an analytic function such that $f(1/n)=(-1)^n/n, n=1,2\dots$, with $0$ in the domain of $f.$

My attempt: If it exists, then clearly $f(z)=z^kg(z)$, where $k\ge0$ is the order of zero [or possible zero,I have included this by $k\gt0$] $z=0$ of $f(z).$Also,$g(0)\ne0$
So, $g(1/n)=f(1/n)n^k\implies g(1/n)=n^k\{\frac{(-1)^n}n\}$
Now the $\lim_{n\to\infty}\{\frac{(-1)^nn^k}n\}$ does not exist[of course,when $k\ge1$]$\implies g(0)$ doesn't exist.
But this is a contradiction to my assumption that $f$ is analytic.So such an $f$ doesn't exist.
And in case of $k=0$,above limit $\lim_{n\to\infty}g(1/n)=0$,which is again a contradiction.
Is there something wrong in above procedure?Please guide me through if there is some.I'll appreciate any help towards this.Thanks in advance!
PS:I stated that $g(0)\implies$ "contradiction to my assumption that $f$ is analytic".Well,I stated so because since by assumption $f$ is analytic and $z^k$ is entire[$k\ge0$],so $g$ must be analytic in the domain of $f$.

zhw.
  • 105,693
Koro
  • 11,402
  • You have the right idea. But the limit you say does not exist actually does exist - limit as n goes to infinity of (-1)^n/n is 0. So you need to fix your proof. – ykm Sep 23 '15 at 22:28
  • @ykm,thanks .I fixed it (actually I edited it).So will you please have a look at it now? – Koro Sep 23 '15 at 23:47
  • oh, i see, that was a typo. yeah, it looks okay to me. – ykm Sep 24 '15 at 18:50

1 Answers1

9

Another proof: Assume $f$ is entire for convenience and that $f(1/n) = (-1)^n/n, n = 1,2,\dots$ Then $f(z) = -z$ on the set $\{1,1/3,1/5,\dots \},$ which has the limit point $0.$ By the identity principle, $f(z) = -z$ everywhere. But $f(1/2) = 1/2,$ contradiction. Hence there is no such $f.$

zhw.
  • 105,693
  • Just a quick question: suppose $n=2,3,\dots$, so $n\neq 1$. Could this still apply? I suppose $f(z)=-z$ then on the set ${1/3,1/5, \dots}$, so I could still say it has a limit point of $0$ and apply identity principle... I was just curious if $1$ needed to be in the set where $f(z)=-z$ for this to hold. – User7238 Aug 03 '21 at 19:24
  • 1
    No, $1$ does not matter. Neither do $2,3,\dots 10,000.$ Tail wags the dog here. – zhw. Aug 03 '21 at 19:58
  • Thank you. I didn't think it did, I was just second guessing myself. For instance, if our domain was $\mathbb{D}$, I was trying to think if a point on our boundary had to be included, but I know that isn't part of the hypothesis of the identity principle. Anyway, I really appreciate you replying to a message from an answer from 6 years ago! Thank you so much! – User7238 Aug 03 '21 at 20:01
  • 1
    About the boundary: Any crazy thing can happen out there. Inside is where things are nice (at least locally). – zhw. Aug 03 '21 at 20:55
  • Thank you so much! – User7238 Aug 03 '21 at 21:02