Show that there is no entire function such that $f(\frac
1n)=\frac{n}{3n-2},\forall n\in\mathbb{N}$
My attempt:
Suppose that there is such entire function $f$. Define $g:z\mapsto \frac{z^{-1}}{3z^{-1}-2}$. Then $f(\frac 1n)=g(\frac 1n),\forall n\in\mathbb{N}$. This implies that $f=g$ on $\mathbb{C}$. The function $g$, however, is not entire since it has a pole in $3/2$ – a contradiction! Therefore, such entire function does not exist.
Is this a good approach?
Best Answer
No, it is not correct. You cannot deduce that $f=g$ simply from the fact that$$(\forall n\in\mathbb N):f\left(\frac1n\right)=g\left(\frac1n\right).\tag1$$For instance, if $f$ is the null function and $g(z)=\sin\left(\frac\pi z\right)$, but in that case you don't have $f=g$, right?!
Instead, define$$\begin{array}{rccc}h\colon&D_{\frac32}\left(0\right)&\longrightarrow&\mathbb C\\&z&\mapsto&\frac1{3-2z}.\end{array}$$Then $(\forall n\in\mathbb N):f\left(\frac1n\right)=h\left(\frac1n\right)$ and so$$f(0)=\lim_{n\to\infty}f\left(\frac 1n\right)=\lim_{n\to\infty}h\left(\frac 1n\right)=h(0)$$and therefore$$\left\{z\in D_{\frac32}\left(0\right)\,\middle|\,f(z)=h(z)\right\}\supset\{0\}\cup\left\{\frac1n\,\middle|\,n\in\mathbb N\right\}.$$Since the set $\left\{z\in D_{\frac32}\left(0\right)\,\middle|\,f(z)=h(z)\right\}$ contains a non-isolated point ($0$), it follows from the identity theorem that$$\left(\forall z\in D_{\frac32}(0)\right):f(z)=h(z).$$But $\lim_{z\to\frac32}f(z)=f\left(\frac32\right)$, whreas the limit $\lim_{z\to\frac32}h(z)$ doesn't exist (in $\mathbb C$).