# Show that no entire function $f\big(\frac{1}{n}\big)= \frac{n}{2n-1}$ exists using the identity theorem.

Show that no entire function $$f: \mathbb{C} \to \mathbb{C}$$ exists with $$f\bigg(\frac{1}{n}\bigg)= \frac{n}{2n-1}$$ for $$n \in \mathbb{N}$$. What is the domain of a holomorphic function with such values?

I was thinking of putting $$g(n) = \frac{1}{2-n}$$ with $$n \in \mathbb{N}$$ which is holomorphic on $$\mathbb{C}\setminus \{2\}$$ and then arguing that because both functions "meet" at $$0$$, according to identity theorem, $$g(n) = f\big(\frac{1}{n}\big)$$ and because $$g(n)$$ is not an entire function per definition, there is no entire function $$f\big(\frac{1}{n}\big)$$ on $$\mathbb{C}$$. The domain for which a function with such values is defined would be defined for values in $$(0, 1]$$, right?

Let $$g(z)=\frac1{2-z}$$, for each $$z\in\Bbb C\setminus\{2\}$$. Suppose that such an entire function $$f$$ exists. Then$$(\forall n\in\Bbb N):f\left(\frac1n\right)=\frac n{2n-1}=g\left(\frac1n\right).$$On the other hand, since $$f$$ is continuous,\begin{align}f(0)&=\lim_{n\to\infty}f\left(\frac1n\right)\\&=\lim_{n\to\infty}\frac n{2n-1}\\&=\frac12\\&=g(0).\end{align}So,$$\{z\in\Bbb C\setminus\{2\}\mid f(z)=g(z)\}\supset\{0\}\cup\left\{\frac1n\,\middle|\,n\in\Bbb N\right\}.$$In particular, the set $$\{z\in\Bbb C\setminus\{2\}\mid f(z)=g(z)\}$$ contains an accumulation point ($$0$$) and therefore, by the identity theorem and because $$\Bbb C\setminus\{2\}$$ is connected,$$(\forall z\in\Bbb C\setminus\{2\}):f(z)=g(z).$$But this is impossible, since the limit $$\lim_{z\to 2}f(z)$$ exists in $$\Bbb C$$ (it is $$f(2)$$), whereas the limit $$\lim_{z\to 2}g(z)$$ does not exist in $$\Bbb C$$.

• Thanks. And what would be the domain of definition of a function $f$ with such values?
– MJP
Jun 8 '20 at 21:29
• Since $f$ is an entire function, its domain would be $\Bbb C$. Jun 8 '20 at 21:31
• Wait, but I thought I was showing that there is no entire function $f$?
– MJP
Jun 8 '20 at 21:32
• Yes, what I did was to prove that there is no entire function $f$ such that $(\forall n\in\Bbb N):f\left(\frac1n\right)=\frac n{2n-1}$. Jun 8 '20 at 21:33
• Yes, that was my intent as well, but I was also trying to answer the part of the question where it is asking what the maximum domain of definition of such a holomorphic function is. Maybe I'm not understanding that second part of the question right...
– MJP
Jun 8 '20 at 21:37