UPSC 2024 Maths Optional Paper 1 Q1d — Step-by-Step Solution

10 marks · Section A

Taylor's theorem with remainders · Calculus · asked 3× in 14 yrs · Read the full method →

Question

Expand $\ln(x)$ in powers of $(x-1)$ by Taylor’s theorem and hence find the value of $\ln(1.1)$ correct up to four decimal places.

Technique

Compute derivatives of $\ln x$ at $x=1$ ; write the Taylor series; evaluate at $x=1.1$ and truncate using the alternating-series bound.

Solution

Step 1 — Derivatives at $x=1$ .

Let $f(x)=\ln x$ . Then $f^{(n)}(x)=(-1)^{n-1}(n-1)!/x^n$ for $n\ge 1$ , so $f^{(n)}(1)=(-1)^{n-1}(n-1)!$ and $f(1)=0$ .

Step 2 — Taylor series.

\ln x=\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n}(x-1)^n=(x-1)-\frac{(x-1)^2}{2}+\frac{(x-1)^3}{3}-\cdots,

valid for $0<x\le 2$ .

Step 3 — Evaluate $\ln(1.1)$ .

Set $x=1.1$ , so $x-1=0.1$ . Successive terms:

$n$	Term	Decimal
1	$0.1$	$0.1000000$
2	$-0.01/2$	$-0.0050000$
3	$0.001/3$	$0.0003333$
4	$-0.0001/4$	$-0.0000250$
5	$0.00001/5$	$0.0000020$

The 6th term ( $\approx 1.7\times10^{-7}$ ) is below $5\times10^{-7}$ , so terms 1–5 suffice. By the alternating-series error bound, the truncation error is less than the first omitted term.

Sum: $0.1000000-0.0050000+0.0003333-0.0000250+0.0000020=0.0953103$ .

Answer

\boxed{\;\ln(1.1)\approx 0.0953.\;}