courses › Beyond College

Numerical methods

level 54 course

How computers find roots and areas step by step.

Pen and paper is fine · no calculator needed why?

Opens at level 54.

Builds on: Using derivatives (not open yet) · Integrals (not open yet)

the lesson

The idea, the techniques and a tip for each skill, right here. The Learn page adds worked examples for every skill and untimed practice.

Read the lesson · about 3 minutes

The idea

Many equations and integrals have no tidy formula, so computers approximate them with fixed recipes.

Newton’s method moves a guess to where the tangent line meets the x-axis, usually closer to the root. Bisection halves an interval that holds a root. The trapezoid rule and Simpson’s rule estimate the area under a curve from its heights at equally spaced points.

Techniques

One Newton step

Improving a guess x₀ for a root of f(x) = 0.

Work out f(x₀) and the slope f′(x₀).
x₁ = x₀ − f(x₀)/f′(x₀): where the tangent line crosses the x-axis.

worked example

Example: Apply one step of Newton’s method to f(x) = x² − 7, starting from x₀ = 3. What is x₁? Give an exact fraction.

f(3) is 9 − 7, which is 2. f′(x) is 2x, so f′(3) is 6.
x₁ = 3 − 2/6 = 3 − 1/3 = 8/3.

Answer: 8/3

Halve and keep the sign change

f(a) and f(b) have opposite signs, so a root lies between a and b.

Find the sign of f at the midpoint.
Keep the half whose ends give opposite signs. Repeat as many times as asked.
Answer with the midpoint of the last interval.

worked example

Example: Start with [1, 2] for f(x) = x² − 2. Test the midpoint and keep the half where f changes sign. Do this twice. What is the midpoint of the interval you end with? Give the answer as a decimal.

f(1) is negative. f(1.5) is 0.25, positive: keep [1, 1.5].
f(1.25) is −0.4375, negative: keep [1.25, 1.5].
Its midpoint is 1.375.

Answer: 1.375

Weight the heights

Estimating an integral from heights at equally spaced points, h apart.

Work out f at each point.
Trapezoid rule: weights 1, 2, 2, …, 2, 1, then multiply the sum by h/2.
Simpson’s rule, 2 strips: weights 1, 4, 1, then multiply by h/3. It is exact for polynomials up to x³.

worked example

Example: Use the trapezoid rule with 2 strips of width 1 to estimate ∫₁³ x² dx. Give the estimate as a decimal.

At x = 1, 2 and 3, x² is 1, 4 and 9.
1 + 2 × 4 + 9 = 18.
Times h/2: 18 ÷ 2 = 9.
The exact value is 26/3 ≈ 8.67, so the estimate is a little high.

Answer: 9

Tips by skill

TipNewton’s method: x₁ = x₀ − f(x₀)/f′(x₀). Find f and its slope at x₀, and watch the sign of f(x₀).
TipBisection: Each time, test the midpoint and keep the half whose ends give opposite signs. Answer with the last interval’s midpoint.
TipTrapezoid rule: Weights 1, 2, 2, …, 2, 1 on the heights, then multiply the sum by half the strip width.
TipSimpson’s rule: Weights 1, 4, 1 on the three heights, then multiply the sum by a third of the strip width.

Watch out for

Getting the sign of the Newton step wrong. When f(x₀)/f′(x₀) is negative, subtracting it makes x₁ bigger than x₀.
Giving the last point you tested instead of the midpoint of the last interval.
Forgetting to double the inside heights, or using the trapezoid weights for Simpson’s rule.

skills · practice stats

From rounds of this course only: box, review and test-out answers are left out. Once a skill has 40 tries, it compares your first 20 tries with your last 20.

Newton’s method par 30 s not tried yet
worked example
Apply one step of Newton’s method to f(x) = x³ − 58, starting from x₀ = 3. What is x₁? Give an exact fraction.

Answer: 112/27
1. x₁ = x₀ − f(x₀)/f′(x₀), with f′(x) = 3x².
2. f(3) = 27 − 58 = −31 and f′(3) = 27.
3. x₁ = 3 − (−31)/27 = 3 + 31/27 = 112/27 ≈ 4.148.
Bisection par 30 s not tried yet
worked example
Start with [2, 3] for f(x) = x³ − 16. Test the midpoint and keep the half where f changes sign. Do this twice. What is the midpoint of the interval you end with? Give the answer as a decimal.

Answer: 2.625
1. 2.5³ = 15.625 < 16, so f(2.5) < 0: keep [2.5, 3].
2. 2.75³ = 20.796875 > 16, so f(2.75) > 0: keep [2.5, 2.75].
3. The midpoint of [2.5, 2.75] is 2.625.
Trapezoid rule par 30 s not tried yet
worked example
Use the trapezoid rule with 4 strips of width 1 to estimate ∫₂⁶ (x² − 2x) dx. Give the estimate as a decimal.

Answer: 38
1. The points are x = 2, 3, 4, 5, 6, where f(x) = x² − 2x takes the values 0, 3, 8, 15, 24.
2. (h/2)(f₀ + 2f₁ + 2f₂ + 2f₃ + f₄) = ½ × (0 + 2 × 3 + 2 × 8 + 2 × 15 + 24) = ½ × 76 = 38.
3. The exact value is 112/3, so the estimate is 2/3 too high.
Simpson’s rule par 30 s not tried yet
worked example
Use Simpson’s rule with 2 strips of width 1 to estimate ∫₀² (x³ − 3x² + x − 4) dx.

Answer: -10
1. The points are x = 0, 1, 2, where f(x) = x³ − 3x² + x − 4 takes the values −4, −5, −6.
2. (h/3)(f₀ + 4f₁ + f₂) = ⅓ × (−4 + 4 × (−5) + (−6)) = ⅓ × (−30) = −10.
3. Simpson’s rule is exact for polynomials up to degree 3, so −10 is also the exact value of the integral.

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