Change of Sign, Cobweb/Staircase Iteration, Newton-Raphson & Trapezium Rule
Most equations encountered in real-world engineering and science, such as e^x = 4 - x² or sin(x) = ln(x) - cannot be solved algebraically, and many definite integrals such as ∫ e^(-x²) dx have no elementary antiderivative. Numerical methods provide systematic algorithms to approximate roots and integrals to any desired degree of accuracy. At A-Level, you must write watertight change-of-sign proofs, classify fixed-point iterations x_(n+1) = g(x_n) using staircase and cobweb diagrams, apply and explain failure cases of the Newton-Raphson method, and determine whether the Trapezium Rule overestimates or underestimates an integral.
- Locating Roots by Change of Sign
- To show that an equation has a root α in the interval [a, b] (or to verify α = 2.34 to 2 decimal places by testing the bounds a = 2.335 and b = 2.345), first rearrange the equation into the form f(x) = 0. Evaluate f(a) and f(b) to at least 3 significant figures, showing that one is positive and one is negative. To earn the final mark, your concluding statement must include three things: 1. There is a change of sign; 2. f(x) is continuous on [a, b]; 3. Therefore a root α lies in the interval (a, b).
- When Change-of-Sign Methods Fail
- A change-of-sign test can fail in three situations: 1. Discontinuity (Vertical Asymptote): f(x) changes sign across a vertical asymptote (such as f(x) = 1 / x on [-1, 1] or tan(x) across π/2) where there is no root at all; 2. Even Number of Roots: if there are two (or any even number of) roots inside [a, b], f(a) and f(b) have the same sign so the roots are missed; 3. Repeated Root (Touching the x-axis): if y = f(x) touches the x-axis at a turning point without crossing it, f(x) does not change sign.
- Fixed-Point Iteration: Staircase and Cobweb Diagrams
- Rearranging f(x) = 0 into x = g(x) defines the iterative formula x_(n+1) = g(x_n), whose roots correspond to the intersections of the curve y = g(x) and the line y = x. To draw the diagram starting at x₀ on the x-axis: move vertically to the curve y = g(x), then horizontally to the line y = x, then vertically to the curve y = g(x), and so on. The iteration converges to the root α if |g'(α)| < 1: it forms a staircase diagram if 0 < g'(α) < 1 (positive gradient) and a cobweb diagram if -1 < g'(α) < 0 (negative gradient). If |g'(α)| > 1, the iteration diverges.
Change of Sign, Cobweb/Staircase Iteration, Newton-Raphson & Trapezium Rule: Wave Equation (v = fλ) & Force Vector (F = ma) Simulator
Adjust physical variables on the sliders below to see real-time changes in wave geometry, resultant forces, and acceleration.
F_net = 80 N − 25 N = 55 NAcceleration: a = F_net ÷ m = 55 ÷ 10 = 5.50 m/s²Key Concepts & Exam Definitions (6)
- Locating Roots by Change of Sign
- To show that an equation has a root α in the interval [a, b] (or to verify α = 2.34 to 2 decimal places by testing the bounds a = 2.335 and b = 2.345), first rearrange the equation into the form f(x) = 0. Evaluate f(a) and f(b) to at least 3 significant figures, showing that one is positive and one is negative. To earn the final mark, your concluding statement must include three things: 1. There is a change of sign; 2. f(x) is continuous on [a, b]; 3. Therefore a root α lies in the interval (a, b).
- When Change-of-Sign Methods Fail
- A change-of-sign test can fail in three situations: 1. Discontinuity (Vertical Asymptote): f(x) changes sign across a vertical asymptote (such as f(x) = 1 / x on [-1, 1] or tan(x) across π/2) where there is no root at all; 2. Even Number of Roots: if there are two (or any even number of) roots inside [a, b], f(a) and f(b) have the same sign so the roots are missed; 3. Repeated Root (Touching the x-axis): if y = f(x) touches the x-axis at a turning point without crossing it, f(x) does not change sign.
- Fixed-Point Iteration: Staircase and Cobweb Diagrams
- Rearranging f(x) = 0 into x = g(x) defines the iterative formula x_(n+1) = g(x_n), whose roots correspond to the intersections of the curve y = g(x) and the line y = x. To draw the diagram starting at x₀ on the x-axis: move vertically to the curve y = g(x), then horizontally to the line y = x, then vertically to the curve y = g(x), and so on. The iteration converges to the root α if |g'(α)| < 1: it forms a staircase diagram if 0 < g'(α) < 1 (positive gradient) and a cobweb diagram if -1 < g'(α) < 0 (negative gradient). If |g'(α)| > 1, the iteration diverges.
- The Newton-raphson method
- Starting from an initial approximation x₀ to a root of f(x) = 0, the Newton-Raphson formula draws the tangent to y = f(x) at (x_n, f(x_n)) and finds where that tangent crosses the x-axis to get the next approximation: x_(n+1) = x_n - f(x_n) / f'(x_n). Newton-Raphson converges much faster (quadratically) than simple fixed-point iteration, but it fails if x₀ is chosen at a stationary point where f'(x₀) = 0 (since the tangent is horizontal and never crosses the x-axis, causing division by zero), or if f'(x₀) is very close to 0 (sending x₁ far away to a different root or into an undefined region).
- The Trapezium Rule for Numerical Integration
- To approximate ∫_a^b y dx using n strips (which requires n + 1 ordinates y₀, y₁, ., y_n), the strip width is h = (b - a) / n. The Trapezium Rule formula is: ∫_a^b y dx ≈ (h / 2)[y₀ + y_n + 2(y₁ + y₂ + . + y_(n-1))]. Remember the mnemonic: 'half the width, times (first + last + twice the rest)'.
- Overestimates vs Underestimates in the Trapezium Rule
- Whether the Trapezium Rule gives an overestimate or an underestimate depends on the concavity of the curve (the sign of d²y/dx²): 1. If the curve is concave upward (convex, d²y/dx² > 0, like y = e^x or y = x²), the straight-line chords of the trapezia lie above the curve, giving an overestimate. 2. If the curve is concave downward (concave, d²y/dx² < 0, like y = ln x or y = sin x on [0, π]), the chords lie below the curve, giving an underestimate.
Quick-Fire Flashcards (6)
Click any card to check the answer, then mark it as Known to track your progress.
To prove that a root α equals 1.46 correct to 2 decimal places, what two x-values must you substitute into f(x)?
What three points must be written in your conclusion for a change-of-sign proof?
When drawing a cobweb or staircase diagram for x_(n+1) = g(x_n), do you move from x₀ on the x-axis to the line y = x or the curve y = g(x) first?
State the Newton-Raphson iterative formula and explain geometrically why it fails at a turning point.
In the Trapezium Rule on [1, 3] with 4 strips (5 ordinates), what is the strip width h?
How do you prove whether the Trapezium Rule overestimates or underestimates an integral?
Exam-Style Questions & Mark Schemes (4)
Write or think through your answer first, then click Check Mark Scheme to see the exact points examiners award marks for.
- Q1
Let f(x) = x³ - 4x - 2. (a) Show that the equation f(x) = 0 has a root α in the interval [2, 3]. (b) Taking x₀ = 2.2 as a first approximation to α, apply the Newton-Raphson method once to find a second approximation x₁ to 3 decimal places. (c) By considering a change of sign in a suitable interval, verify that α = 2.214 correct to 3 decimal places. (7 marks)
Click “Check Mark Scheme” to compare your answer, 1. (a) Evaluate f(2) and f(3): f(2) = 2³ - 4(2) - 2 = -2 < 0, and f(3) = 3³ - 4(3) - 2 = +13 > 0 [1 mark]. 2. Since there is a change of sign and f(x) is continuous on [2, 3], a root α lies in the interval (2, 3) [1 mark]. 3. (b) Differentiate f(x): f'(x) = 3x² - 4 [1 mark]. 4. Evaluate f(2.2) and f'(2.2): f(2.2) = (2.2)³ - 4(2.2) - 2 = 10.648 - 10.8 = -0.152, and f'(2.2) = 3(2.2)² - 4 = 14.52 - 4 = 10.52 [1 mark]. 5. Apply Newton-Raphson x₁ = x₀ - f(x₀) / f'(x₀): x₁ = 2.2 - (-0.152 / 10.52) = 2.214448. = 2.214 (3 d.p) [1 mark]. 6. (c) Test the bounds 2.2135 and 2.2145: f(2.2135) = (2.2135)³ - 4(2.2135) - 2 = -0.00825 < 0, and f(2.2145) = (2.2145)³ - 4(2.2145) - 2 = +0.00235 > 0 [1 mark]. 7. Since there is a change of sign and f(x) is continuous, 2.2135 < α < 2.2145, so α = 2.214 correct to 3 decimal places [1 mark].
- Q2
The equation e^(-x) - x + 2 = 0 can be rearranged into the iterative form x_(n+1) = e^(-x_n) + 2. Starting with x₀ = 2, calculate x₁, x₂, and x₃ to 4 decimal places, and state whether the iteration produces a staircase or a cobweb diagram, justifying your answer using calculus. (5 marks)
Click “Check Mark Scheme” to compare your answer, 1. Substitute x₀ = 2 into x_(n+1) = e^(-x_n) + 2: x₁ = e^(-2) + 2 = 2.135335. = 2.1353 (4 d.p) [1 mark]. 2. Substitute unrounded x₁ (using the ANS key): x₂ = e^(-2.135335.) + 2 = 2.1182 (4 d.p) [1 mark]. 3. Calculate x₃: x₃ = e^(-2.118204.) + 2 = 2.1202 (4 d.p) [1 mark]. 4. Let g(x) = e^(-x) + 2 and differentiate: g'(x) = -e^(-x) [1 mark]. 5. At the root α ≈ 2.12, g'(α) = -e^(-2.12) ≈ -0.120; since -1 < g'(α) < 0 (the gradient is negative and less than 1 in magnitude), the terms oscillate alternately above and below the root, producing a convergent cobweb diagram [1 mark].
- Q3
Use the Trapezium Rule with 4 strips to estimate the value of I = ∫₀^2 √(1 + x³) dx to 3 decimal places. State with a reason whether your estimate is an overestimate or an underestimate, given that y = √(1 + x³) is convex for x > 0. (5 marks)
Click “Check Mark Scheme” to compare your answer, 1. Calculate the strip width h = (2 - 0) ÷ 4 = 0.5 [1 mark]. 2. Construct the table of x and y = √(1 + x³) values: x₀ = 0 → y₀ = 1; x₁ = 0.5 → y₁ = √1.125 = 1.06066; x₂ = 1.0 → y₂ = √2 = 1.41421; x₃ = 1.5 → y₃ = √4.375 = 2.09165; x₄ = 2.0 → y₄ = √9 = 3 [1 mark]. 3. Substitute into the Trapezium Rule I ≈ (h / 2)[y₀ + y₄ + 2(y₁ + y₂ + y₃)]: I ≈ (0.5 / 2)[1 + 3 + 2(1.06066 + 1.41421 + 2.09165)] [1 mark]. 4. Evaluate: I ≈ 0.25 × [4 + 2(4.56652)] = 0.25 × 13.13305 = 3.283 (to 3 d.p) [1 mark]. 5. Because the curve is convex (concave upward) on (0, 2), the straight top edges of the trapezia lie above the curve, so 3.283 is an overestimate [1 mark].
- Q4
A student attempts to find a root of f(x) = x³ - 3x² + 5 = 0 using the Newton-Raphson method with initial approximation x₀ = 2. Explain with mathematical working why this attempt fails, and show what happens if x₀ = 1.9 is used instead. (4 marks)
Click “Check Mark Scheme” to compare your answer, 1. Differentiate f(x): f'(x) = 3x² - 6x = 3x(x - 2) [1 mark]. 2. Evaluate f'(2) = 3(2)(2 - 2) = 0; since (2, 1) is a stationary point (local minimum), the tangent at x₀ = 2 is horizontal and never intersects the x-axis, causing division by zero in x₁ = x₀ - f(x₀)/f'(x₀) [2 marks]. 3. With x₀ = 1.9, f(1.9) = (1.9)³ - 3(1.9)² + 5 = 1.029 and f'(1.9) = 3(1.9)(-0.1) = -0.57, so x₁ = 1.9 - (1.029 / -0.57) = 3.705 (because x₀ = 1.9 is to the left of the local minimum at x = 2, the negative gradient shoots x₁ to the right where there is no root) [1 mark].
Common Exam Questions (4)
How do I use the ANS button on my calculator to compute x₁, x₂, x₃ in 5 seconds?
First type the starting value x₀ (e.g., 2) and press '=' so it is stored in ANS. Then type the iterative formula using ANS in place of x_n (e.g., e^(-ANS) + 2). Every time you press '=', the calculator instantly outputs the next iterate x₁, x₂, x₃ without any rounding errors.
Why did I lose marks in Newton-Raphson even though my value of x₁ was right?
Mark schemes explicitly require you to write down the derivative f'(x) and show the numerical substitution into x₁ = x₀ - f(x₀) / f'(x₀) (e.g., x₁ = 2.2 - (-0.152 / 10.52)). Just writing down the final answer from your calculator earns zero method marks if a 'Show detailed reasoning' command is used.
How can I make the Trapezium Rule approximation more accurate?
Increase the number of strips n (which reduces the strip width h = (b - a) / n), so the straight chords of the trapezia follow the curvature of y = f(x) more closely.
If I know the Trapezium Rule gives an underestimate I_trap for ∫_a^b f(x) dx, how does that help me bound ∫_a^b [k - f(x)] dx?
Since ∫_a^b [k - f(x)] dx = k(b - a) - ∫_a^b f(x) dx, subtracting an underestimate from the exact rectangle area k(b - a) produces an overestimate.
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