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UPSC 2017 Maths Optional Paper 2 Q4a — Step-by-Step Solution

15 marks · Section A

Taylor's Series for Analytic Functions · Complex Analysis · Read the full method →

Question

For a function $f:\mathbb C\to\mathbb C$ and $n\ge1$, let $f^{(n)}$ denote the $n$th derivative of $f$ and $f^{(0)}=f$. Let $f$ be an entire function such that for some $n\ge1$, $f^{(n)}\!\left(\dfrac{1}{k}\right)=0$ for all $k=1,2,3,\ldots$. Show that $f$ is a polynomial.

Technique

Derivatives of entire functions are entire; Identity Theorem on $f^{(n)}$ using the accumulation point $0$ of $\{1/k\}$; then $f^{(n)}\equiv0\Rightarrow f$ polynomial of degree $<n$.

Solution

The strategy: derivatives of an entire function are entire, so $f^{(n)}$ is entire; it vanishes on a set with an accumulation point, hence is identically zero by the Identity Theorem; an entire function whose $n$th derivative is $0$ is a polynomial of degree $<n$.

Step 1 — $f^{(n)}$ is entire

Since $f$ is entire (analytic on all of $\mathbb C$), it is infinitely complex-differentiable, and every derivative of an analytic function is again analytic. In particular $g:=f^{(n)}$ is entire.

Step 2 — The zero set has an accumulation point in $\mathbb C$

By hypothesis $g\!\left(\tfrac1k\right)=f^{(n)}\!\left(\tfrac1k\right)=0$ for all $k=1,2,3,\ldots$. The points $z_k=\tfrac1k$ are distinct, lie in $\mathbb C$, and satisfy

So the zero set $\{z_k\}$ has the accumulation point $0$, and crucially $0\in\mathbb C$ — it is in the domain $\mathbb C$ on which $g$ is analytic. (By continuity $g(0)=\lim_k g(1/k)=0$ too, so $0$ is itself a zero, but the accumulation is the key point.)

Step 3 — Identity Theorem $\Rightarrow g\equiv0$

Identity Theorem. If a function analytic on a connected open set has zeros accumulating at a point inside that set, then it is identically zero there.

$\mathbb C$ is connected and open; $g=f^{(n)}$ is analytic on it; its zeros $\{1/k\}$ accumulate at $0\in\mathbb C$. Therefore

Step 4 — $f^{(n)}\equiv0$ forces $f$ to be a polynomial of degree $<n$

Integrating $n$ times: $f^{(n)}\equiv0$ gives $f^{(n-1)}\equiv c_{n-1}$ (constant), then $f^{(n-2)}(z)=c_{n-1}z+c_{n-2}$, and inductively $f$ is a polynomial of degree at most $n-1$. (Equivalently, in the Taylor expansion $f(z)=\sum_{m\ge0}a_m z^m$, we have $f^{(n)}(z)=\sum_{m\ge n}\frac{m!}{(m-n)!}a_m z^{m-n}\equiv0$, which forces $a_m=0$ for all $m\ge n$.)

Hence $f$ is a polynomial:

Answer