Show that there does not exist any analytic function for which $f\left(\dfrac{1}{n}\right)=\dfrac{1}{2^n}$

Question

Show that there does not exist any analytic function $f$ on the open disc for which $f\left(\dfrac{1}{n}\right)=\dfrac{1}{2^n}$.$\forall n\in \Bbb N$

Suppose such a function say $f$ exists then $f\left(\dfrac{1}{n}\right)=\dfrac{1}{2^n}$. By continuity of $f$ we have $\lim _{n\to \infty}f\left(\dfrac{1}{n}\right)=\lim_{n\to \infty}\dfrac{1}{2^n}\implies f(0)=0$

Also $f$ has a power series representation about $0$ i.e $f(z)=\sum_{n=0}^\infty \dfrac{f^n(0)}{n!}z^n;\forall z\in D$ where $D$ is open disc.

But I am failing to arrive at a contradiction from here.Please give some hints.

Answer 1

Suppose $f(z) = \sum_n a_n z^n$. You know that $a_0 = 0$.

Suppose $a_0,...,a_{k-1} $ are all zero. Then $\lim_{z \to 0} {f(z) \over z^k} = a_k$, and using the above formula, we can show that $a_k = 0$. Hence $f=0$ which is a contradiction.

Note that ${f({1 \over n}) \over {1 \over n^k}} = {n^k \over 2^n}$. This converges to zero as $n \to \infty$.

Answer 2

U can also use the the identity theorem. 1st u gotta write $\frac{1}{2^n}=\frac{1}{2^{1/\frac{1}{n}}}$ so we start with $f(z)\ ,\ g(z)=2^{-\frac{1}{z}}$ we can see that $f(z) = g(z)$ when $z\in \{1,\frac{1}{2},\frac{1}{3}...\}$ se we can use identity theorem since $\ 0$ $\ $is an accumulation point so $f(z)\equiv g(z)$ $ \ $ but $ \ $$\lim_{z\rightarrow0}f(z)=f(0)$ and $\lim_{z\rightarrow0_{Re(z)>0}}2^{-\frac{1}{z}} \neq \lim_{z\rightarrow0_{Im(z)>0}}2^{-\frac{1}{z}}$ so we contradicted that $f(z)\equiv g(z)$ so there does not exist any analytic function ff on the open disc for which $f(\frac{1}{n})=\frac{1}{2^n} \ ,\ ∀n∈ℕ$