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UPSC 2024 Maths Optional Paper 2 Q3b — Step-by-Step Solution

20 marks · Section A

Uniform Convergence: Term-by-Term Differentiation · Real Analysis · Read the full method →

Question

The sum of the first $n$ terms of $\sum U_n(x)$ on $[0,1]$ is $S_n(x)=\dfrac{1}{2n^2}\log(1+n^4 x^2)$. Show that the series can be differentiated term-by-term, though $\sum U_n'(x)$ does not converge uniformly on $[0,1]$.

Technique

Compute pointwise limits of $S_n$ and $S_n'$; locate the maximum of $S_n'$ to show non-uniformity.

Solution

Step 1 — Pointwise limit $S(x)$

For $x=0$: $S_n(0)=0$ for all $n$. For $x\in(0,1]$: $\log(1+n^4 x^2)\sim 4\log n$, so $S_n(x)\sim 2\log n/n^2\to 0$.

Therefore $S(x)=0$ on $[0,1]$, and $S'(x)=0$.

Step 2 — Pointwise limit of $S_n'$

For $x\in(0,1]$: $S_n'(x)\sim 1/(n^2 x)\to 0$ as $n\to\infty$. Also $S_n'(0)=0$.

So $\lim_n S_n'(x)=0=S'(x)$ pointwise on $[0,1]$.

Therefore $\sum U_n$ can be differentiated term-by-term: both the sum and the derivative of the sum equal $0$ on $[0,1]$.

Step 3 — $\sum U_n'$ does not converge uniformly

If convergence were uniform, then $\sup_{x\in[0,1]}|S_n'(x)|\to 0$. Maximising $S_n'(x)=n^2 x/(1+n^4 x^2)$: setting the derivative to zero gives $x^\star=1/n^2$, at which

So $\sup|S_n'(x)|\ge 1/2$ for every $n$ — the supremum never tends to $0$. Hence $S_n'$ does not converge uniformly to $S'=0$ on $[0,1]$.

Answer