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Recurrence relations for Edexcel A-level Maths

Recurrence relations

This page covers the following topics:

1. Recurrence relationships
2. Graphical representation of Netwon-Raphson method
3. Newton-Raphson method
4. Newton-Raphson method problems in context

A recurrence relationship can be used to generate all the terms of a sequence. It describes each term as a function of the previous term.

Recurrence relationships

The Newton-Raphson method uses tangent lines to find approximations of roots of equations in the form f(x) = 0. A value for xโ‚€ is chosen and a tangent at that point is drawn. The next x value is taken to be the point at which the tangent intersects the x-axis. This process is continued to find increasingly accurate approximations of the root.

Graphical representation of Netwon-Raphson method

The Newton-Raphson formula can be used to calculate increasingly accurate approximations of a root, given a starting value. If the starting value is chosen to be a turning point, the formula cannot be used, since its derivative will be 0 and division by 0 in the formula will not be possible. If the starting value is chosen to be near a turning point, the gradient will be small, therefore the tangent will intersect the x-axis a long way away from the starting value, and therefore the Newton-Raphson method may fail.

Newton-Raphson method

The Newton-Raphson method can be used to model situations and find their solutions.

Newton-Raphson method problems in context

1

Using x = 1.6 as the first approximation, calculate a second approximation for the given root of the graph.

f'(x) = 3sin(3x).
f(1.6) = 1 โˆ’ cos(3 ร— 1.6) = 0.931.
f'(1.6) = 3sin(3 ร— 1.6) = โˆ’2.988.
Using the Newton-Raphson formula, xโ‚ = 1.6 โˆ’ 0.931/โˆ’2.988 = 1.912.

Using x = 1.6 as the first approximation, calculate a second approximation for the given root of the graph.

2

Using x = 0.3 as a first approximation, calculate a second approximation for the root of the given function.

f'(x) = 5cos(5x) โˆ’ 2.
f(0.3) = sin(5 ร— 0.3) โˆ’ 2(0.3) = 0.397.
f'(0.3) = 5cos(5 ร— 0.3) โˆ’ 2 = โˆ’1.646.
Using the Newton-Raphson formula, xโ‚ = 0.3 โˆ’ 0.397/โˆ’1.646 = 0.541.

Using x = 0.3 as a first approximation, calculate a second approximation for the root of the given function.

3

The recurrence relationship of a sequence is given by u_(n + 1) = u_nยฒ โˆ’ 3, where uโ‚ = 1. Calculate the sum of the first 50 terms.

Generating the first few terms, 1, โˆ’2, 1, โˆ’2, โ€ฆ
So, the first 50 terms have twenty-five 1 and twenty-five โˆ’2.
So, sum = 25(1) + 25(โˆ’2) = โˆ’25.

The recurrence relationship of a sequence is given by u_(n + 1) = u_nยฒ โˆ’ 3, where uโ‚ = 1. Calculate the sum of the first 50 terms.

4

The recurrence relationship of a sequence is given by u_(n+1) = 100 โˆ’ u_n, where uโ‚ = 18. Calculate the next three terms of the sequence.

uโ‚‚ = 100 โˆ’ 18 = 82.
uโ‚ƒ = 100 โˆ’ 82 = 18.
uโ‚„ = 100 โˆ’ 18 = 82.

The recurrence relationship of a sequence is given by u_(n+1) = 100 โˆ’ u_n, where uโ‚ = 18. Calculate the next three terms of the sequence.

5

Given the recurrence relationship u_(n + 1) = u_nยฒ + 10, fill in the blanks of the sequence.

_____ _____ 42446

42446 = uโ‚‚ยฒ + 10, so uโ‚‚ = โˆš42436 = 206.
206 = uโ‚ยฒ + 10, so uโ‚ = โˆš196 = 14.

Given the recurrence relationship u_(n + 1) = u_nยฒ + 10, fill in the blanks of the sequence. 

_____ _____ 42446

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