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# Recurrence relations for AQA A-level Maths

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.

The Newton-Raphson method uses tangent lines to find approximations of roots of equations in the form f(x) = 0. A value for xā 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.

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.

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

# 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 ā cos(3 Ć 1.6) = 0.931.
f'(1.6) = 3sin(3 Ć 1.6) = ā2.988.
Using the Newton-Raphson formula, xā = 1.6 ā 0.931/ā2.988 = 1.912.

# 2

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

f'(x) = 5cos(5x) ā 2.
f(0.3) = sin(5 Ć 0.3) ā 2(0.3) = 0.397.
f'(0.3) = 5cos(5 Ć 0.3) ā 2 = ā1.646.
Using the Newton-Raphson formula, xā = 0.3 ā 0.397/ā1.646 = 0.541.

# 3

The recurrence relationship of a sequence is given by u_(n + 1) = u_nĀ² ā 3, where uā = 1. Calculate the sum of the first 50 terms.

Generating the first few terms, 1, ā2, 1, ā2, ā¦
So, the first 50 terms have twenty-five 1 and twenty-five ā2.
So, sum = 25(1) + 25(ā2) = ā25.

# 4

The recurrence relationship of a sequence is given by u_(n+1) = 100 ā u_n, where uā = 18. Calculate the next three terms of the sequence.

uā = 100 ā 18 = 82.
uā = 100 ā 82 = 18.
uā = 100 ā 18 = 82.

# 5

Given the recurrence relationship u_(n + 1) = u_nĀ² + 10, fill in the blanks of the sequence.

_____ _____ 42446

42446 = uāĀ² + 10, so uā = ā42436 = 206.
206 = uāĀ² + 10, so uā = ā196 = 14.

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