Numerical Methods

The equation x³ − 2x − 5 = 0 has a root in which of the following intervals, found by a change of sign?

• Abetween x = 0 and x = 1
• Bbetween x = 1 and x = 2
• Cbetween x = 2 and x = 3✓
• Dbetween x = 3 and x = 4

Answer: Let f(x) = x³ − 2x − 5. Then f(2) = 8 − 4 − 5 = −1 (negative) and f(3) = 27 − 6 − 5 = 16 (positive). The sign change shows a root lies between x = 2 and x = 3.

The iteration xₙ₊₁ = √(2xₙ + 3) is used with x₀ = 2. Find x₁ correct to 3 decimal places.

• A2.000
• B3.500
• C1.871
• D2.646✓

Answer: x₁ = √(2 × 2 + 3) = √7 = 2.6457…, which is 2.646 to 3 decimal places.

The Newton-Raphson method is applied to f(x) = x² − 5 with x₀ = 2. Find x₁.

• A2.25✓
• B2.5
• C1.75
• D2.2

Answer: f'(x) = 2x. Then x₁ = x₀ − f(x₀)/f'(x₀) = 2 − (2² − 5)/(2 × 2) = 2 − (−1)/4 = 2 + 0.25 = 2.25.

Which formula correctly states the Newton-Raphson iteration for solving f(x) = 0?

• Axₙ₊₁ = xₙ + f(xₙ)/f'(xₙ)
• Bxₙ₊₁ = xₙ − f(xₙ)/f'(xₙ)✓
• Cxₙ₊₁ = xₙ − f'(xₙ)/f(xₙ)
• Dxₙ₊₁ = f(xₙ)/f'(xₙ)

Answer: The Newton-Raphson formula is xₙ₊₁ = xₙ − f(xₙ)/f'(xₙ), subtracting the ratio of the function to its derivative at the current estimate.

Use the trapezium rule with 2 strips of width 1 to estimate the area under y = x² from x = 0 to x = 2. The ordinates at x = 0, 1, 2 are 0, 1, 4.

• A5
• B4
• C3✓
• D2.5

Answer: Trapezium rule: area ≈ (h/2)[y₀ + 2y₁ + y₂] = (1/2)[0 + 2(1) + 4] = (1/2)(6) = 3.

The trapezium rule is used to estimate the area under the curve y = x², which is convex (it bends upward). The estimate will be:

• Aan underestimate, because the tops of the strips lie below the curve
• Bexactly the true area, because the curve is a polynomial
• Can underestimate, because the curve is increasing
• Dan overestimate, because the chords lie above the convex curve✓

Answer: For a curve that is convex (concave up), each chord joining the tops of a strip lies above the curve, so the trapezia include extra area. The trapezium rule therefore gives an overestimate.

A continuous function f has f(2) = −3 and f(3) = 4. What can be concluded?

• AThere is exactly one root between x = 2 and x = 3
• BThere is a root of f(x) = 0 somewhere between x = 2 and x = 3✓
• CThere is a stationary point between x = 2 and x = 3
• Df is increasing throughout the interval

Answer: A sign change across an interval, for a function continuous on it, guarantees at least one root inside — that is the change-of-sign rule. It does not promise exactly one: the curve could cross three times. Nor does it say anything about stationary points or overall behaviour.

Apply one Newton-Raphson iteration to f(x) = x² − 5 starting from x₀ = 2. Find x₁ as a fraction.

• A2
• B7/4
• C−1/4
• D9/4✓

Answer: Newton-Raphson is x₁ = x₀ − f(x₀)/f′(x₀). Here f(2) = −1 and f′(2) = 4, so x₁ = 2 − (−1)/(4) = 9/4. Adding instead of subtracting is the error that makes the iteration diverge.

Apply one Newton-Raphson iteration to f(x) = x³ − 2 starting from x₀ = 1. Find x₁ as a fraction.

• A−1/3
• B2/3
• C4/3✓
• D1

Answer: Newton-Raphson is x₁ = x₀ − f(x₀)/f′(x₀). Here f(1) = −1 and f′(1) = 3, so x₁ = 1 − (−1)/(3) = 4/3. Adding instead of subtracting is the error that makes the iteration diverge.

f(x) is continuous on [2, 3], with f(2) = −3 and f(3) = 4. What can be concluded?

• AA root of f(x) = 0 lies somewhere between x = 2 and x = 3✓
• BA root of f(x) = 0 lies somewhere between x = 3 and x = 4
• CThere is exactly one root, and it sits at the midpoint x = 2.5
• DNo root can exist in the interval because the signs differ here

Answer: A change of sign across an interval on which the function is continuous guarantees at least one root inside it. It does not tell you how many roots there are, nor whereabouts in the interval they sit. Continuity is the condition that makes the argument valid at all.

An iteration xₙ₊₁ = g(xₙ) produces 2, 2.5, 2.7, 2.75, 2.76, 2.76. What does this suggest?

• AThe iteration is converging to a root near 2.76✓
• BThe iteration is diverging
• CThe iteration is oscillating without settling
• DThe root is exactly 2.76

Answer: Successive terms are changing by less and less and settling on one value, which is convergence. It does NOT prove the root is exactly 2.76 — that is only the limit to the accuracy shown, and a change-of-sign check would be needed to confirm the root to a given number of decimal places.

What does the change-of-sign method locate?

• AA stationary point where the gradient of the curve is momentarily zero
• BA root of the equation✓
• CA point of inflection at which the concavity of the curve reverses
• DThe maximum value taken by the function across the interval

Answer: If a continuous function changes sign across an interval it must cross zero somewhere inside, so the method brackets a ROOT. It says nothing about gradients, so stationary points and inflections are out of scope.