Numerical Methods in Context
Choosing, applying, and interpreting numerical methods for real-world problems
SPEC 9.5 • NUMERICAL METHODS
Why Numerical Methods?
Many equations and integrals cannot be solved exactly using algebraic methods.
Key Insight
Numerical methods give us approximate solutions to problems that have no closed-form answer.
Unsolvable Equations
x = cos x
eˣ = 3x
x⁵ + 3x − 7 = 0
Unsolvable Integrals
∫ e^(x²) dx
∫ sin(x²) dx
∫ √(1 + x³) dx
Remember
Numerical methods are essential tools, not just alternatives — they solve problems that algebra cannot.
Your Numerical Toolkit
You have learned four main techniques:
Problem TypeMethods Available
Finding roots of f(x) = 0Change of sign (interval bisection) Fixed point iteration Newton-Raphson
Approximating definite integralsTrapezium rule
Exam Tip
Questions often ask you to choose an appropriate method and justify your choice. Know the strengths and weaknesses of each!
Comparing Root-Finding Methods
Change of Sign
Simple and reliable
Always converges
✗ Slow convergence✗ Needs interval with sign change
Best for: Initial root location, guaranteed convergence
Fixed Point Iteration
Visual (cobweb/staircase)
Can be fast
✗ May diverge if |g'(x)| > 1✗ Different rearrangements → different roots
Best for: When suitable rearrangement exists with |g'(x)| < 1
Newton-Raphson
Very fast (quadratic convergence)
Only needs one starting value
✗ Needs derivative f'(x)✗ May fail near stationary points
Best for: Rapid accurate solutions when derivative is easy to find
Which Method Should I Use?
Start: Need to solve f(x) = 0
Can you find f'(x) easily?
↓ YES
Any risk of f'(x) = 0 near root?
↓ NO
Use Newton-Raphson
If derivative is hard
Try iteration xₙ₊₁ = g(xₙ) with |g'(x)| < 1 near the root
If convergence is uncertain
Use change of sign to locate root first, then refine
Important
In exams, if a specific method is stated, use it! Only choose when the question asks you to select a method.
01Worked Example
The equation x³ + 2x − 7 = 0 has a root between x = 1 and x = 2.
(a) Show there is a root in [1, 2] using change of sign.
(b) Use Newton-Raphson with x₀ = 1.5 to find the root to 4 decimal places.
(a) Change of sign:
•
f(x) = x³ + 2x − 7
•
f(1) = 1 + 2 − 7 = −4 < 0
•
f(2) = 8 + 4 − 7 = 5 > 0
•
Sign change ⟹ root in [1, 2] ✓
(b) Newton-Raphson:
•
f(x) = x³ + 2x − 7, f'(x) = 3x² + 2
•
x₁ = 1.5 − (1.5³ + 2(1.5) − 7)/(3(1.5)² + 2) = 1.5 − (−0.625)/8.75 = 1.5714...
•
x₂ = 1.5714 − f(1.5714)/f'(1.5714) = 1.5689...
•
x₃ = 1.5689 (converged)
Answer: x = 1.5689 (4 d.p.)
02Worked Example
The curves y = eˣ and y = 4 − x² intersect at a point P.
(a) Show that the x-coordinate of P satisfies eˣ + x² − 4 = 0.
(b) Show there is a root in [1, 1.5].
(c) Using the iteration xₙ₊₁ = ln(4 − xₙ²) with x₀ = 1, find x₃.
(a) At intersection: eˣ = 4 − x² ⟹ eˣ + x² − 4 = 0 ✓
(b) f(x) = eˣ + x² − 4
•
f(1) = e¹ + 1 − 4 = 2.718 − 3 = −0.282 < 0
•
f(1.5) = e^1.5 + 2.25 − 4 = 4.482 − 1.75 = 2.732 > 0
•
Sign change ⟹ root in [1, 1.5] ✓
(c) Iteration: xₙ₊₁ = ln(4 − xₙ²)
•
x₁ = ln(4 − 1²) = ln 3 = 1.0986
•
x₂ = ln(4 − 1.0986²) = ln(2.793) = 1.0272
•
x₃ = ln(4 − 1.0272²) = ln(2.945) = 1.0804
Answer: x₃ = 1.0804
03Worked Example
A population P at time t years is modelled by P = 500e^(0.03t).
The population is sustainable when P < 800 + 20t.
Find the time when the population first becomes unsustainable.
Step 1: Set up the equation
•
Unsustainable when: 500e^(0.03t) = 800 + 20t
•
Rearrange: 500e^(0.03t) − 20t − 800 = 0
Step 2: Locate the root
•
f(t) = 500e^(0.03t) − 20t − 800
•
f(20) = 500e^0.6 − 400 − 800 = 911.8 − 1200 = −288 < 0
•
f(30) = 500e^0.9 − 600 − 800 = 1229.7 − 1400 = −170 < 0
•
f(40) = 500e^1.2 − 800 − 800 = 1660 − 1600 = 60 > 0
•
Root in [30, 40]
Step 3: Use Newton-Raphson with t₀ = 35
•
f'(t) = 15e^(0.03t) − 20
•
After iterations: t ≈ 37.2 years
Answer: The population becomes unsustainable after approximately 37 years.
04Worked Example
A company's revenue is R(x) = 50x − 0.1x² and cost is C(x) = 200 + 12x.
Find the break-even point (where R = C) using a numerical method.
Step 1: Set up the equation
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R(x) = C(x): 50x − 0.1x² = 200 + 12x
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Rearrange: −0.1x² + 38x − 200 = 0
•
Or: 0.1x² − 38x + 200 = 0
Step 2: This is actually quadratic! We can solve exactly:
•
x = (38 ± √(1444 − 80))/0.2 = (38 ± √1364)/0.2
•
x = (38 ± 36.93)/0.2
•
x = 5.35 or x = 374.65
Step 3: Interpret in context
•
Two break-even points exist
•
First at x ≈ 5 units, second at x ≈ 375 units
Answer: Break-even at x = 5.35 units or x = 374.65 units
Trapezium Rule in Context
The trapezium rule approximates definite integrals when:
The antiderivative cannot be found (e.g., ∫ e^(−x²) dx)
You only have data points (no formula for f(x))
The function is too complex to integrate analytically
Trapezium Rule
where h = (b − a)/n
Real-World Applications
Physics: Work done = ∫ F dx (when force varies)
Statistics: Probability = ∫ f(x) dx (for normal distribution)
Economics: Consumer surplus, total revenue
Accuracy and Limitations
Sources of Error
Rounding errors: Calculator limitations, significant figures
Truncation errors: Stopping iteration too early
Method errors: Trapezium rule approximates curves with lines
Model limitations: Equation may not perfectly represent reality
Improving Accuracy
More iterations (root-finding)
More strips (trapezium rule)
Better starting value
Check with alternative method
Always Check
Does the answer make sense?
Is it in the expected range?
Substitute back to verify
05Worked Example
The equation x = 2 + ln x has a root near x = 3.1.
Compare the convergence of fixed point iteration and Newton-Raphson.
Fixed Point Iteration: xₙ₊₁ = 2 + ln xₙ, x₀ = 3.1
•
x₁ = 2 + ln 3.1 = 3.1314
•
x₂ = 2 + ln 3.1314 = 3.1415
•
x₃ = 2 + ln 3.1415 = 3.1447
•
x₄ = 3.1457, x₅ = 3.1460
Newton-Raphson: f(x) = x − 2 − ln x, f'(x) = 1 − 1/x
•
x₁ = 3.1 − (3.1 − 2 − ln 3.1)/(1 − 1/3.1) = 3.1459
•
x₂ = 3.1462 (converged to 4 d.p.)
Comparison:
•
Iteration: 5 steps to reach 4 d.p. accuracy
•
Newton-Raphson: 2 steps to reach 4 d.p. accuracy
Conclusion: Newton-Raphson converges much faster
Exam Strategy
Key Phrases to Look For
Question SaysYou Should"Show that..."Use change of sign to verify root location"Use the iteration..."Apply given formula repeatedly"Using Newton-Raphson..."Find f'(x) and apply xₙ₊₁ = xₙ − f(xₙ)/f'(xₙ)"Estimate using trapezium rule..."Calculate h, find y-values, apply formula"Explain why this method fails..."Discuss divergence, stationary points, or |g'(x)| > 1
Common Mistakes
Forgetting to show sign change explicitly
Not giving enough iterations
Rounding too early (keep extra d.p. during working)
Not interpreting the answer in context
Practice
Question 1: The equation x³ − 3x + 1 = 0 has three roots.
(a) Show there is a root in the interval [−2, −1].
(b) Using the iteration xₙ₊₁ = (xₙ³ + 1)/3 with x₀ = 0, find x₃.
(c) State which root this iteration converges to.
(a) f(x) = x³ − 3x + 1
f(−2) = −8 + 6 + 1 = −1 < 0
f(−1) = −1 + 3 + 1 = 3 > 0
Sign change ⟹ root in [−2, −1] ✓
(b) x₁ = (0 + 1)/3 = 0.333...
x₂ = (0.333³ + 1)/3 = 0.346
x₃ = (0.346³ + 1)/3 = 0.347
(c) This converges to the root near x = 0.347, which is the root in [0, 1], NOT the one in [−2, −1].
Practice
Question 2: A tank contains water at temperature T°C after t minutes, modelled by T = 80e^(−0.05t) + 20.
Room temperature is 20°C.
(a) Find the initial temperature of the water.
(b) Find the time when T = 50°C using Newton-Raphson with t₀ = 10.
(c) State one limitation of this model.
(a) When t = 0: T = 80e⁰ + 20 = 80 + 20 = 100°C
(b) Solve: 80e^(−0.05t) + 20 = 50 → 80e^(−0.05t) = 30
f(t) = 80e^(−0.05t) − 30, f'(t) = −4e^(−0.05t)
t₁ = 10 − (80e^(−0.5) − 30)/(−4e^(−0.5)) = 10 − (48.52 − 30)/(−2.43)
t₁ = 10 + 7.62 = 17.62
t₂ = 17.62 − ... ≈ 19.6 minutes
(c) Limitations: Model assumes constant room temperature; doesn't account for heat loss through evaporation; assumes perfect insulation except at surface.
Summary
Choosing the Right Method
If you need...Use...
To show a root existsChange of sign
Fast convergence with easy derivativeNewton-Raphson
Iteration when given x = g(x)Fixed point iteration
To approximate an integralTrapezium rule
Key Exam Skills
Set up equations correctly from contextual problems
Justify your method choice when asked
Interpret solutions in context — what does the answer mean?
Discuss limitations of both the method and the model
Check your answer makes sense
Next: Why Numerical Methods?