← 2019 Paper 2

UPSC 2019 Maths Optional Paper 2 Q5b — Step-by-Step Solution

10 marks · Section B

Newton-Raphson method (convergence, geometric meaning) · Numerical Analysis · asked 7× in 14 yrs · Read the full method →

Question

Apply Newton-Raphson method, to find a real root of transcendental equation $x\log_{10}x=1.2$, correct to three decimal places.

Technique

Newton–Raphson on $f(x)=x\log_{10}x-1.2$, using $f'(x)=\log_{10}x+M$ with $M=1/\ln10=0.434294$.

Solution

Setup. Let $f(x)=x\log_{10}x-1.2$. Recall $\log_{10}x=\dfrac{\ln x}{\ln 10}$, $\ln 10=2.302585$, and $\dfrac{1}{\ln 10}=0.434294$ ($=M$, the modulus).

Newton iteration: $\displaystyle x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}$.

Step 1 — Locate the root

$f(2)=2\log_{10}2-1.2=2(0.30103)-1.2=-0.59794<0$. $f(3)=3\log_{10}3-1.2=3(0.47712)-1.2=+0.23136>0$. Root in $(2,3)$. Take $x_0=2.5$.

Step 2 — Iterations

$x_0=2.5$: $\log_{10}2.5=0.39794$. $f=2.5(0.39794)-1.2=-0.20515$, $f'=0.39794+0.43429=0.83223$.

$x_1=2.74650$: $\log_{10}2.74650=0.438756$. $f=2.74650(0.438756)-1.2=1.205112-1.2=0.005112$, $f'=0.438756+0.434294=0.873050$.

$x_2=2.740645$: $f=2.740645\,(0.437809)-1.2\approx 0.000003\approx 0$.

Converged. To three decimals:

Answer