Pure: Numerical Methods
Numerical methods are techniques for finding an approximate solution to an equation that cannot be solved exactly by algebra, using a sequence of closer and closer estimates. In A Level Maths this covers locating a root with a change of sign, refining it by iteration or the Newton-Raphson method, and estimating areas under curves with the trapezium rule.
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Method
- To locate a root in an interval [a,b], evaluate f(a) and f(b); if f is continuous and the values have opposite signs, a root lies between a and b (the change of sign test).
- To use iteration, rearrange f(x) = 0 into the form x = g(x), then apply x_(n+1) = g(x_n) repeatedly, starting from the given x0, rounding each answer to the number of decimal places asked.
- To use the Newton-Raphson method, differentiate f(x) to find f'(x), then apply x_(n+1) = x_n - f(x_n)/f'(x_n) repeatedly from the given starting value.
- To confirm a root correct to a stated number of decimal places, evaluate f at the value rounded up and rounded down (e.g. 1.3275 and 1.3285 for 1.328 to 3dp) and check for a change of sign.
- To use the trapezium rule, split the interval into equal strips of width h, list the y-values at each x-value, and apply (h/2)[y0 + yn + 2(sum of the middle y-values)].
- To decide whether the trapezium rule over- or under-estimates the true area, consider whether the curve is convex or concave on that interval: convex curves give an overestimate, concave curves give an underestimate.
Worked example
Show that f(x) = x^3 - 4x + 1 = 0 has a root between x = 0 and x = 1. Using the iteration x_(n+1) = (x_n^3 + 1)/4 with x0 = 0, find x1, x2 and x3, each to 4 decimal places, and hence show the root is x = 0.254 correct to 3 decimal places.
- Evaluate f(0) = 1 and f(1) = 1 - 4 + 1 = -2; opposite signs, so a root lies in (0,1).
- Rearrange 4x = x^3 + 1 to get x = (x^3 + 1)/4, so x_(n+1) = (x_n^3 + 1)/4.
- Substitute x0 = 0: x1 = (0^3 + 1)/4 = 0.2500.
- Substitute x1 = 0.2500: x2 = (0.2500^3 + 1)/4 = 0.2539 (4dp).
- Substitute x2 = 0.2539: x3 = (0.2539^3 + 1)/4 = 0.2541 (4dp).
- Check f(0.2535) = 0.00229 (>0) and f(0.2545) = -0.00152 (<0); change of sign confirms the root is x = 0.254 correct to 3dp.
Practice questions
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Q1State the condition on f(x) required for the change of sign test to guarantee a root of f(x) = 0 lies in the interval [a,b].Show answer
Answer: f must be continuous on [a,b] (and f(a), f(b) must have opposite signs).
Q2Show that f(x) = x^3 + x - 3 has a root between x = 1 and x = 2.Show answer
Answer: f(1) = -1, f(2) = 7; opposite signs, so a root lies in (1,2).
Q3The equation x^3 + x - 3 = 0 can be rearranged to x = (3 - x)^(1/3). Using x0 = 1.5, find x1, giving your answer to 4 decimal places.Show answer
Answer: x1 = 1.1447 (substitute x0 = 1.5 into (3 - x)^(1/3)).
Q4The curve y = x^2 + 1 is defined for 0 <= x <= 3, with y = 1, 2, 5, 10 at x = 0, 1, 2, 3. Use the trapezium rule with all 3 strips to estimate the integral of x^2 + 1 from 0 to 3.Show answer
Answer: 12.5 (h = 1: (1/2)[1 + 10 + 2(2+5)] = 12.5).
Q5Let f(x) = x^2 - 7. Using the Newton-Raphson method with x0 = 2, find x1.Show answer
Answer: x1 = 2.75 (x1 = 2 - f(2)/f'(2) = 2 - (-3)/4).
Q6Let f(x) = x^3 - 2x - 5. Using the Newton-Raphson method with x0 = 2, find x1 and x2, each to 4 decimal places.Show answer
Answer: x1 = 2.1000, x2 = 2.0946 (using f'(x) = 3x^2 - 2).
Exam-style questions
Written in the style of a A Level Maths exam paper, with a full mark scheme.
The curve y = 1/(x+1) is defined for 0 <= x <= 4. The table gives values of y at x = 0, 1, 2, 3 and 4: y = 1, 0.5, 0.3333, 0.25, 0.2. Use the trapezium rule with all 4 strips to find an estimate for the integral of 1/(x+1) from 0 to 4, giving your answer to 4 decimal places.
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Let f(x) = x^3 - 6x + 2. (a) Show that f(x) = 0 has a root between x = 0 and x = 1. (2) (b) Using the iteration x_(n+1) = (x_n^3 + 2)/6 with x0 = 0.5, find x1 and x2, each to 4 decimal places. (3)
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The equation x^2 - 10 = 0 has a positive root, r. The iteration x_(n+1) = (x_n + 10/x_n)/2 is used to estimate r, starting from x0 = 3. (a) Find x1 and x2, each to 4 decimal places. (3) (b) State what happens to the sequence x0, x1, x2, ... as n increases. (1)
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