← 2016 Paper 2

UPSC 2016 Maths Optional Paper 2 Q4c — Step-by-Step Solution

20 marks · Section A

Power Series of Analytic Functions; Radius of Convergence · Complex Analysis · Read the full method →

Question

Prove that every power series represents an analytic function inside its circle of convergence.

Technique

Cauchy–Hadamard ($n^{1/n}\to1$ keeps $R$ fixed) + Weierstrass $M$-test (geometric majorant on compact subdiscs) + uniform-limit-of-holomorphic theorem (Morera/Cauchy) to justify term-by-term differentiation.

Solution

Let $f(z)=\sum_{n=0}^\infty a_n(z-z_0)^n$ with radius of convergence $R>0$. We must show $f$ is analytic (complex-differentiable) at every point of the open disc $|z-z_0|<R$. The plan: (i) the formally differentiated series has the same radius $R$; (ii) the series converges uniformly on compact subdiscs (Weierstrass $M$-test); (iii) term-by-term differentiation is valid, so $f'$ exists and equals the differentiated series. WLOG $z_0=0$.

Step 1 — The differentiated series has the same radius of convergence

The radius is given by Cauchy–Hadamard: $\dfrac1R=\limsup_{n\to\infty}|a_n|^{1/n}$. The termwise derivative is $g(z)=\sum_{n=1}^\infty n a_n z^{n-1}$. Its radius $R'$ satisfies

using $n^{1/n}\to1$. Hence $R'=R$: the differentiated series converges on the same open disc $|z|<R$.

Step 2 — Absolute & uniform convergence on compact subdiscs (Weierstrass $M$-test)

Fix $r$ with $0<r<R$ and consider the closed disc $\overline D_r=\{|z|\le r\}$. Pick $\rho$ with $r<\rho<R$. Since $\sum a_n\rho^n$ converges, $a_n\rho^n\to0$, so $|a_n\rho^n|\le C$ for some constant $C$ and all $n$. Then for $|z|\le r$,

The bound $M_n=C\theta^n$ is summable ($\sum\theta^n=\frac1{1-\theta}<\infty$). By the Weierstrass $M$-test, $\sum a_nz^n$ converges uniformly (and absolutely) on $\overline D_r$. The same estimate (with $n|a_n|r^{n-1}\le \frac Cr\, n\theta^n$ and $\sum n\theta^n<\infty$) shows $\sum n a_n z^{n-1}$ converges uniformly on $\overline D_r$.

Step 3 — Term-by-term differentiation gives $f'=g$

Each partial sum $S_N(z)=\sum_{n=0}^N a_nz^n$ is a polynomial, hence holomorphic, with $S_N'(z)=\sum_{n=1}^N na_nz^{n-1}$. On $\overline D_r$ we have shown:

• $S_N\to f$ uniformly;
• $S_N'\to g=\sum_{n\ge1}na_nz^{n-1}$ uniformly.

Theorem (uniform limit of holomorphic functions; Weierstrass). If holomorphic $S_N$ converge uniformly on compact subsets of an open set to $f$, then $f$ is holomorphic and $S_N'\to f'$.

Proof via Morera + Cauchy. For any triangle (or closed contour) $\Gamma\subset\overline D_r$, $\oint_\Gamma S_N\,dz=0$ (each $S_N$ holomorphic, by Cauchy/Goursat). Uniform convergence lets us pass the limit through the integral: $\oint_\Gamma f\,dz=\lim_N\oint_\Gamma S_N\,dz=0$. Since $f$ is continuous (uniform limit of continuous functions) and integrates to $0$ around every triangle, Morera’s theorem gives $f$ holomorphic on $|z|<R$. Moreover, for $z$ in the interior, Cauchy’s integral formula for the derivative,

with $C$ a small circle about $z$ inside $\overline D_r$, and uniform convergence $S_N\to f$ on $C$ gives $S_N'(z)\to f'(z)$. Hence $f'(z)=g(z)=\sum_{n\ge1}na_nz^{n-1}$. $\square$

Step 4 — Conclusion

$r<R$ was arbitrary, so $f$ is complex-differentiable at every point of the open disc $|z|<R$; i.e. $f$ is analytic (holomorphic) there, with

Answer