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How can I prove or disprove ?: Can there exist an analytic function $f(1/n)=(1/2)^n$[consider any suitable domain];$n\in N$ and $n\ge 2......(A)$
Now if I have to prove or disprove :$f(1/n)=(-1)^n/n $ exists or doesn't.I can first apply IVT to show there does exist a sequence of zeros of $f$.Then by, Idntity theorem(a.k.a. Uniqueness theorem),I can show that such $f$ is impossible to exist.
But in (A),I think there exists such function but don't have any idea how to prove it(or disprove it if my assumption is incorrect). Please help me through this problem.I'll appreciate any help towards this.
Thanks in advance!!

Koro
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  • Sorry but this is pretty badly stated. For example, where is $f$ defined and analytic? – zhw. Sep 22 '15 at 06:23
  • @zhw.,$f(1/n)=(1/2)^n$ is given and the question is whether such an analytic $f$ exists or not? – Koro Sep 22 '15 at 06:24
  • An analytic $f$ defined where? – zhw. Sep 22 '15 at 06:25
  • @zhw.,Conider any suitable domain.Can such an analytic function exist? – Koro Sep 22 '15 at 06:28
  • Well would a suitable domain be the open right half plane? – zhw. Sep 22 '15 at 06:38
  • @zhw.,you just have to prove such an analytic function exists or doesn't whatever domain you take.Does there exists any such analytic $f$ satisfying the conditions stated in OP? – Koro Sep 22 '15 at 06:41
  • Related (possibly a duplicate, depending on your assumptions): http://math.stackexchange.com/questions/392682 – mrf Sep 22 '15 at 07:50
  • @mrf,my doubt didn't clear . – Koro Sep 22 '15 at 08:45
  • (1)By analytic, without any conditions, means analytic on the whole complex plane – DanielWainfleet Sep 22 '15 at 14:48
  • @user254665,Please read OP. – Koro Sep 22 '15 at 14:50
  • For both, the existence is decided by whether $0$ belongs to the domain of the function or not. – Daniel Fischer Sep 22 '15 at 15:21
  • @DanielFischer,suppose $0$ belongs to the domain of $f$,then I define $h(z)=f(z)-(1/2)^{1/z}$,I want to use identity theorem which can't be used for $z=0$ because $h$ is not entire.So how to proceed. – Koro Sep 22 '15 at 15:47
  • If $0$ is in the domain, $f$ has a Taylor expansion with centre $0$. Show that that isn't compatible with $f(1/n) = 2^{-n}$. – Daniel Fischer Sep 22 '15 at 15:55
  • @DanielFischer,I found that $f^{(n)}=0$ for all $n$,that is $f(1/n)=0$ which is not compatible with $f(1/n)=2^{-n}$.Is it correct? – Koro Sep 22 '15 at 16:00
  • If your argument for $f^{(n)}(0) = 0$ for all $n$ is correct, that is correct. – Daniel Fischer Sep 22 '15 at 16:02
  • @DanielFischer,I proved it like this by cauchy 's ML inequality:$f^{(m)}\le \left(\frac{M_R\fact m}{R^m}\right)$[where $M_R$ is max. value of $f(1/n)$ on $|z|=R$.Am I right? – Koro Sep 22 '15 at 16:08
  • From the single point $1/n$ on the circle, you have not enough information to use that. $M_R$ could be anything. – Daniel Fischer Sep 22 '15 at 17:55
  • @DanielFischer,but I am assuming $f$ to be analytic.So $M_R $ will exist. – Koro Sep 22 '15 at 18:17
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    Sure, it will exist (for $R$ small enough that the circle $\lvert z\rvert = R$ is contained in the domain of $f$). But you don't know how large it will be, since you only know what would be $f(1/n)$ for positive integers $n$. – Daniel Fischer Sep 22 '15 at 18:28

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