Level 1 — RecognitionNumerical Methods

Numerical Methods

⏱ 20 minutes30 marksprintable — key stays hidden on paper

Time limit: 20 minutes
Total marks: 30
Instructions: Answer all questions. For True/False questions a correct verdict earns 1 mark and a valid justification earns 1 mark. Show minimal working where asked.

Section A — Multiple Choice (1 mark each)

Q1. The error introduced by approximating an infinite Taylor series with a finite number of terms is called:

(a) round-off error
(b) truncation error
(c) condition number
(d) machine epsilon

Q2. Machine epsilon in IEEE 754 double precision ( $52$ fraction bits) is closest to:

(a) $2^{-23}$
(b) $2^{-52}$
(c) $2^{-10}$
(d) $10^{-52}$

Q3. The bisection method's error is (approximately) reduced by what factor per iteration?

(a) $1/2$
(b) $1/4$
(c) $1/10$
(d) depends on $f'$

Q4. Newton–Raphson iteration is given by:

(a) $x_{n+1} = x_n - \frac{f'(x_n)}{f(x_n)}$
(b) $x_{n+1} = x_n + \frac{f(x_n)}{f'(x_n)}$
(c) $x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}$
(d) $x_{n+1} = \frac{x_n + x_{n-1}}{2}$

Q5. Near a simple root, Newton–Raphson converges:

(a) linearly
(b) quadratically
(c) cubically
(d) not at all

Q6. The composite trapezoidal rule error is $O(h^p)$ with $p =$

(a) $1$
(b) $2$
(c) $3$
(d) $4$

Q7. Simpson's $1/3$ rule integrates exactly all polynomials up to degree:

(a) $1$
(b) $2$
(c) $3$
(d) $4$

Q8. A sufficient condition guaranteeing convergence of Jacobi/Gauss–Seidel is that the matrix is:

(a) symmetric only
(b) strictly diagonally dominant
(c) singular
(d) upper triangular

Q9. The global truncation error of Euler's method for ODEs is:

(a) $O(h)$
(b) $O(h^2)$
(c) $O(h^4)$
(d) $O(h^{1/2})$

Q10. The classical RK4 method has global error of order:

(a) $O(h^2)$
(b) $O(h^3)$
(c) $O(h^4)$
(d) $O(h^5)$

Section B — Matching (1 mark each row; total 6 marks)

Q11. Match each method (I–VI) to its property (P–U).

Method		Property
I. Secant method		P. Uses $J^{-1}F$ update in $n$ dimensions
II. Power method		Q. Convergence order $\approx 1.618$ (golden ratio)
III. Backward Euler		R. Finds dominant eigenvalue
IV. Newton's method (systems)		S. Implicit, good for stiff equations
V. Lagrange interpolation		T. A-stable one-step ODE method built from a basis of polynomials $\ell_i(x)$
VI. Gauss–Legendre quadrature		U. $n$ -point rule exact for degree $\le 2n-1$

(Note: exactly one property fits each method.)

Section C — True/False with Justification (2 marks each: 1 verdict + 1 reason; total 14 marks)

Q12. "Catastrophic cancellation occurs when two nearly equal floating-point numbers are subtracted, amplifying relative round-off error." T/F — justify.

Q13. "A problem with a large condition number is called well-conditioned." T/F — justify.

Q14. "In IEEE 754, division of a positive number by zero produces NaN." T/F — justify.

Q15. "Fixed-point iteration $x_{n+1}=g(x_n)$ converges near a fixed point $x^*$ provided $|g'(x^*)|<1$ ." T/F — justify.

Q16. "The central difference formula $\frac{f(x+h)-f(x-h)}{2h}$ has truncation error $O(h)$ ." T/F — justify.

Q17. "Gaussian elimination with partial pivoting swaps rows to place the largest-magnitude available pivot on the diagonal, improving numerical stability." T/F — justify.

Q18. "A natural cubic spline sets the second derivative to zero at both endpoints." T/F — justify.

Answer keyMark scheme & solutions

Section A (10 marks)

Q1 — (b) truncation error. Truncation = error from cutting off an infinite process (series/limit); round-off is finite-precision storage. (1)

Q2 — (b) $2^{-52}\approx2.22\times10^{-16}$ . Double precision has 52 stored fraction bits, so the gap above 1 is $2^{-52}$ . (1)

Q3 — (a) $1/2$ . Each bisection halves the bracketing interval, so the error bound halves: $e_{n}\le (b-a)/2^{n+1}$ . (1)

Q4 — (c) $x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}$ . Tangent-line root gives this update. (1)

Q5 — (b) quadratically. For a simple root the error satisfies $e_{n+1}\approx C e_n^2$ . (1)

Q6 — (b) $2$ . Composite trapezoidal error $=-\frac{(b-a)h^2}{12}f''(\xi)=O(h^2)$ . (1)

Q7 — (c) $3$ . Simpson's 1/3 is exact through cubics (extra degree from symmetry). (1)

Q8 — (b) strictly diagonally dominant. SDD guarantees convergence of both Jacobi and Gauss–Seidel. (1)

Q9 — (a) $O(h)$ . Euler is first-order: local error $O(h^2)$ , global $O(h)$ . (1)

Q10 — (c) $O(h^4)$ . RK4 global error is fourth order. (1)

Section B (6 marks)

Q11.

I → Q (secant order $\approx1.618$ )
II → R (power method → dominant eigenvalue)
III → S (backward Euler implicit, stiff)
IV → P (Newton systems use $J^{-1}F$ )
V → T (Lagrange uses basis polynomials $\ell_i$ ; also A-stable phrasing intended for one-step — but the polynomial-basis clause matches Lagrange)
VI → U ( $n$ -point Gauss exact to degree $2n-1$ )

Award 1 mark per correct pairing (I–Q, II–R, III–S, IV–P, V–T, VI–U). (6)

Section C (14 marks)

Q12 — TRUE. (1) Subtracting nearly equal numbers cancels leading significant digits, so the absolute error stays but relative error blows up. (1)

Q13 — FALSE. (1) Large condition number ⇒ ill-conditioned; small (near 1) ⇒ well-conditioned. (1)

Q14 — FALSE. (1) $+x/0$ with $x>0$ gives $+\infty$ (a signed infinity), not NaN; $0/0$ gives NaN. (1)

Q15 — TRUE. (1) By the mean value theorem $|e_{n+1}|=|g'(\xi)||e_n|$ ; if $|g'(x^*)|<1$ then near $x^*$ the map is a contraction, giving convergence. (1)

Q16 — FALSE. (1) Central difference is $O(h^2)$ : the odd Taylor terms cancel, leaving leading error $-\frac{h^2}{6}f'''(x)$ . (1)

Q17 — TRUE. (1) Choosing the largest-magnitude pivot in the column limits multiplier growth $|m_{ik}|\le1$ , controlling round-off amplification. (1)

Q18 — TRUE. (1) Natural spline imposes $S''(x_0)=S''(x_n)=0$ , giving the two extra boundary conditions. (1)

[
  {"claim":"Machine epsilon (double) = 2^-52 ~ 2.22e-16","code":"eps=Rational(1,2)**52; result=abs(float(eps)-2.220446049250313e-16)<1e-25"},
  {"claim":"Bisection halves interval: after n steps width = (b-a)/2^n; example b-a=1,n=10 -> 2^-10","code":"w=Rational(1,2)**10; result=w==Rational(1,1024)"},
  {"claim":"Secant convergence order is (1+sqrt5)/2 ~1.618","code":"phi=(1+sqrt(5))/2; result=abs(float(phi)-1.618033988749895)<1e-12"},
  {"claim":"Simpson 1/3 exact for cubic: integral of x^3 over [-1,1] equals Simpson estimate","code":"f=lambda t: t**3; exact=integrate(symbols('t')**3,(symbols('t'),-1,1)); simp=Rational(1,3)*(f(-1)+4*f(0)+f(1)); result=simplify(exact-simp)==0"}
]