From an equation to an iteration | CAIE 9709 Pure Mathematics 2 Notes

Chapters

Pure Mathematics 2

Mathematics9709

Read the subscripts as step numbers

One iteration produces one new value

current input

xₙ

↓

apply the rule

↓

next output

xₙ₊₁

Feed this output back as the next input ↻

xₙ₊₁ = F(xₙ)

The output becomes the next input.

x_{n+1}=F(x_n)

$x_0$ is the stated or chosen starting value.
$x_n$ is the current approximation. The subscript $n$ labels its position in the sequence.
$x_{n+1}$ is the next approximation.
Applying $F$ once is one iteration. Its output is fed back as the next input.

Common mistake

A subscript is not multiplication.

x_{n+1}

means the next term, not

x(n+1)

Rearrange first, then add subscripts

An iterative formula starts from an equivalent equation $x=F(x)$ , meaning an equation with the same solutions as the original one. This is not yet a solved value because $x$ still appears on both sides. It is a rule for making a new approximation from the current one.

Worked example

Build a logarithmic iteration

1+e^{2x}=|x-4|

The modulus $|x-4|$ is the non-negative distance from $x$ to 4. The graph places the root below 4, so $x-4<0$ and $|x-4|=4-x$ on the relevant interval.

1+e^{2x}=4-x

e^{2x}=3-x

Taking $\ln$ undoes the exponential:

2x=\ln(3-x)

x=\frac12\ln(3-x)

Now replace the old input and new output with subscripts:

x_{n+1}=\frac12\ln(3-x_n)

(9709/22/O/N/24 Q4(b)–(c))

Use the exact rearrangement supplied or requested in the question.
Preserve the domain, meaning the allowed inputs. Here $3-x_n>0$ is needed because a logarithm takes a positive input.
Different correct rearrangements can produce different sequences; some converge and some do not.

Show what a converged value must satisfy

Suppose the sequence converges to $\alpha$ . Then the current and next approximations approach the same number, so:

\alpha=F(\alpha)

This is a fixed point: putting $\alpha$ into the rule returns $\alpha$ . Rearrange this equation back to the original equation to show that the limit is the required root.

Worked example

Check the limit of the logarithmic iteration

\alpha=\frac12\ln(3-\alpha)

2\alpha=\ln(3-\alpha)

e^{2\alpha}=3-\alpha

1+e^{2\alpha}=4-\alpha=|\alpha-4|

The last equality uses the known interval $\alpha<4$ . The limit therefore satisfies the original equation.

Examiner note

This argument begins with “if the sequence converges”. It identifies the value a convergent sequence must approach; it does not prove that the sequence will converge.

Each output becomes the next input

grey: y = x

blue: y = F(x)

xₙ₊₁ = ½ ln(3 − xₙ)

Each vertical-and-horizontal pair turns xₙ into xₙ₊₁; the path closes in on the fixed point.

n = 0·x₀ = 0.45000·still changing

Recognise convergence and failure

Check whether the values get closer

one-sided

0.40 → 0.46 → 0.465

getting closer

alternating

0.45 ↗ 0.468 ↘ 0.464

getting closer

fails

1 → 3 → 9 → 81

moving away

Alternating values can still converge.

A sequence may approach from one side, alternate around the root, or fail.

Values may approach the root from one side.
They may alternate above and below the root while their distance from it becomes smaller.
They may repeat the same short list of values in a cycle, move farther away, or leave the formula's valid domain. These are signs that the iteration is failing.

The 9709 syllabus does not require a derivative condition for convergence. You do need to understand that convergence is not guaranteed and recognise the behaviour from the generated values.

Your turnIndependent check [4 marks]

Rearrange $e^x-1=2x$ into $x=F(x)$ , write the iterative formula, and show that any converged value is a root of the original equation.

Show worked answer

e^x=2x+1

x=\ln(2x+1)

x_{n+1}=\ln(2x_n+1)

If $x_n\to\alpha$ , then $\alpha=\ln(2\alpha+1)$ . Exponentiating gives $e^\alpha=2\alpha+1$ , so:

Answer

e^\alpha-1=2\alpha

Read the subscripts as step numbers

One iteration produces one new value

current input

xₙ

↓

apply the rule

↓

next output

xₙ₊₁

Feed this output back as the next input ↻

xₙ₊₁ = F(xₙ)

The output becomes the next input.

x_{n+1}=F(x_n)

$x_0$ is the stated or chosen starting value.
$x_n$ is the current approximation. The subscript $n$ labels its position in the sequence.
$x_{n+1}$ is the next approximation.
Applying $F$ once is one iteration. Its output is fed back as the next input.

Common mistake

A subscript is not multiplication.

x_{n+1}

means the next term, not

x(n+1)

Rearrange first, then add subscripts

Worked example

Build a logarithmic iteration

1+e^{2x}=|x-4|

The modulus $|x-4|$ is the non-negative distance from $x$ to 4. The graph places the root below 4, so $x-4<0$ and $|x-4|=4-x$ on the relevant interval.

1+e^{2x}=4-x

e^{2x}=3-x

Taking $\ln$ undoes the exponential:

2x=\ln(3-x)

x=\frac12\ln(3-x)

Now replace the old input and new output with subscripts:

x_{n+1}=\frac12\ln(3-x_n)

(9709/22/O/N/24 Q4(b)–(c))

Use the exact rearrangement supplied or requested in the question.
Preserve the domain, meaning the allowed inputs. Here $3-x_n>0$ is needed because a logarithm takes a positive input.
Different correct rearrangements can produce different sequences; some converge and some do not.

Show what a converged value must satisfy

Suppose the sequence converges to $\alpha$ . Then the current and next approximations approach the same number, so:

\alpha=F(\alpha)

This is a fixed point: putting $\alpha$ into the rule returns $\alpha$ . Rearrange this equation back to the original equation to show that the limit is the required root.

Worked example

Check the limit of the logarithmic iteration

\alpha=\frac12\ln(3-\alpha)

2\alpha=\ln(3-\alpha)

e^{2\alpha}=3-\alpha

1+e^{2\alpha}=4-\alpha=|\alpha-4|

The last equality uses the known interval $\alpha<4$ . The limit therefore satisfies the original equation.

Examiner note

This argument begins with “if the sequence converges”. It identifies the value a convergent sequence must approach; it does not prove that the sequence will converge.

Each output becomes the next input

grey: y = x

blue: y = F(x)

xₙ₊₁ = ½ ln(3 − xₙ)

Each vertical-and-horizontal pair turns xₙ into xₙ₊₁; the path closes in on the fixed point.

n = 0·x₀ = 0.45000·still changing

Recognise convergence and failure

Check whether the values get closer

one-sided

0.40 → 0.46 → 0.465

getting closer

alternating

0.45 ↗ 0.468 ↘ 0.464

getting closer

fails

1 → 3 → 9 → 81

moving away

Alternating values can still converge.

A sequence may approach from one side, alternate around the root, or fail.

Values may approach the root from one side.
They may alternate above and below the root while their distance from it becomes smaller.
They may repeat the same short list of values in a cycle, move farther away, or leave the formula's valid domain. These are signs that the iteration is failing.

The 9709 syllabus does not require a derivative condition for convergence. You do need to understand that convergence is not guaranteed and recognise the behaviour from the generated values.

Your turnIndependent check [4 marks]

Rearrange $e^x-1=2x$ into $x=F(x)$ , write the iterative formula, and show that any converged value is a root of the original equation.

Show worked answer

e^x=2x+1

x=\ln(2x+1)

x_{n+1}=\ln(2x_n+1)

If $x_n\to\alpha$ , then $\alpha=\ln(2\alpha+1)$ . Exponentiating gives $e^\alpha=2\alpha+1$ , so:

Answer

e^\alpha-1=2\alpha