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UPSC 2026 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

By using the Newton-Raphson iterative formula, establish the formula

to find the cube root of $N$. Use this formula, if established, to find the cube root of 63 correct to four decimal places with the initial approximation $3\cdot9$.

Technique

“Establish” means derive, not quote: choose $f(x)=x^{3}-N$ (whose positive root is $N^{1/3}$), write Newton–Raphson $x_{i+1}=x_i-f(x_i)/f'(x_i)$, and simplify. The numerical part is then three iterations from $x_0=3.9$; the marks in the second half are for displaying every iterate and for justifying the four-decimal claim — and the justification that actually holds water is a bracket $\alpha<N^{1/3}<\beta$, not “two iterates agreed”.

Solution

Step 1 — Choose the function and state Newton–Raphson.

To compute $N^{1/3}$ ($N>0$) we need the real root of

$f$ is continuous, $f(0)=-N<0$, $f(x)\to+\infty$, and $f'(x)=3x^{2}>0$ for $x>0$: so $f$ has exactly one positive zero, namely $x=N^{1/3}$, and it is a simple root ($f'(N^{1/3})=3N^{2/3}\neq0$) — which is what entitles us to expect second-order convergence.

The Newton–Raphson iteration (the tangent at $(x_i,f(x_i))$ met with the $x$-axis) is

Step 2 — Substitute and simplify.

This is the required formula. Two remarks that cost one line each and earn marks:

Fixed point. With $g(x)=\tfrac13\bigl(2x+Nx^{-2}\bigr)$, $\ g(N^{1/3})=\tfrac13\bigl(2N^{1/3}+N\cdot N^{-2/3}\bigr)=\tfrac13\cdot 3N^{1/3}=N^{1/3}$ — the exact cube root is the fixed point, as it must be.

Order. $g'(x)=\tfrac13\bigl(2-2Nx^{-3}\bigr)=\tfrac23\bigl(1-N/x^{3}\bigr)$, so $g'(N^{1/3})=0$: the iteration is at least second order, i.e. quadratically convergent.

Step 3 — An exact error relation (this replaces hand-waving later).

Let $r=N^{1/3}$, so $N=r^{3}$, and let $e_i=x_i-r$. Then

The numerator has $x_i=r$ as a double root; dividing twice by $(x_i-r)$,

Two consequences we will use:

1. Sign. For $x_i>0$ we have $2x_i+r>0$, so $e_{i+1}>0$ whenever $e_i\neq0$: every iterate from $x_1$ onwards overshoots above $r$. (Geometrically: $f$ is convex on $x>0$, so the tangent line lies below the curve and meets the axis to the right of the root.)
2. Monotonicity. For $x_i>r$, $x_i-x_{i+1}=\dfrac{x_i^{3}-N}{3x_i^{2}}>0$, so $x_1>x_2>x_3>\cdots>r$: the iterates decrease monotonically to $r$. This gives a rigorous upper bound for $r$ at every stage — and it is the honest half of the “four decimal places” claim.

Step 4 — The numerical work: $N=63$, $x_0=3\cdot9$.

Iteration 1. $x_0=3.9$, $x_0^{2}=15.21$, $\dfrac{63}{15.21}=\dfrac{6300}{1521}=\dfrac{700}{169}=4.1420118343$,

Iteration 2. $x_1^{2}=15.8457385168$, $\dfrac{63}{x_1^{2}}=3.9758323623$,

Iteration 3. $x_2^{2}=15.8329014670$, $\dfrac{63}{x_2^{2}}=3.9790559002$,

| $i$ | $x_i$ (10 d.p.) | $|x_i-x_{i-1}|$ | |---|---|---| | 0 | $3.9000000000$ | — | | 1 | $3.9806706114$ | $8.067\times10^{-2}$ | | 2 | $3.9790578617$ | $1.613\times10^{-3}$ | | 3 | $3.9790572079$ | $6.538\times10^{-7}$ |

The successive differences fall as $8\times10^{-2}\to1.6\times10^{-3}\to6.5\times10^{-7}$ — the number of correct digits roughly doubles each step, exactly the quadratic behaviour predicted by $(\ast)$ with asymptotic constant $1/r\approx0.2513$. A fourth iteration would change $x_3$ only in about the thirteenth decimal (by $(\ast)$, $e_4\approx e_3^{2}/r$ with $e_3\approx1.1\times10^{-13}$), so the iteration has converged to the precision carried.

Step 5 — Why the answer is correct to four decimal places (the part worth doing properly).

Rounding $x_3$ blindly is not an argument, and neither is ”$x_2$ and $x_3$ agree to six places” — agreement of successive iterates is evidence, not proof (it fails, for instance, for a slowly convergent linear iteration). Here is a proof, and it takes two lines.

Upper bound. By Step 3(2), the iterates decrease monotonically to $r$, so

Lower bound. $x\mapsto x^{3}$ is strictly increasing on $x>0$, so it suffices to test one cube:

Hence

so $r=3.9790\ldots$ with the fifth decimal digit $5$ followed by more, and

(For completeness the upper end of the four-decimal window also checks directly: $(3.97915)^{3}=63.004408>63$, so $r<3.97915$.)

Step 6 — A trap worth naming. A candidate who stops at $x_1=3.98067$ and rounds gets $3.9807$ — wrong in the third decimal. One iterate is never enough here because $e_1\approx1.6\times10^{-3}$; it is $e_2\approx6.5\times10^{-7}$ that first clears the $5\times10^{-5}$ needed for four decimals. Two iterations suffice for the stated accuracy; the third is written down to demonstrate convergence.

Answer