Mathematics 9709 · Pure Mathematics 2

Rearranging f(x)=0 to x=F(x), and Proving It Converges to the Root

$x=F(x)$ works like a machine: feed it a guess, it gives back a better guess.
The answer is the number the machine can no longer improve. That one idea earns three different exam marks.

A root $\alpha$ is a fixed point: feed it in and you get it straight back, $F(\alpha)=\alpha$ . That is just what the rearranged equation says. So the value that fits $x=F(x)$ is the same as the root of $f(x)=0$ . One idea, three marks: it justifies the rearrangement, it gives the convergence proof, and it tells you when to stop.

Rearranging f(x) = 0 into x = F(x)

To iterate you first need the equation as $x=F(x)$ (x equals some function of x). Put a single x on the left. Move everything else into $F(x)$ on the right.

Three moves cover almost every paper:

The awkward term	The move	Example
plain $x$ terms	make $x$ the subject	$x^2+x-3=0\Rightarrow x=\tfrac{3}{x}-1$
an $e^{(\dots)}$	isolate it, take $\ln$	$e^x=2x+1\Rightarrow x=\ln(2x+1)$
a trig term	isolate it, apply the inverse	$\tfrac12 x=\tan(\pi-x)\Rightarrow x=\pi-\tan^{-1}(\tfrac12 x)$

Worked example

Real question: show $x=e^x-3$ rearranges to $x_{n+1}=\ln(3+x_n)$

9709/31 O/N 2023 Q8: the rearrangement the iterative formula is built on. The $e^x$ is the awkward term. So isolate it and take $\ln$ .

Isolate the $e^x$ : from $x=e^x-3$ , add 3 to both sides → $e^x=x+3$ .
Take $\ln$ of both sides — that undoes the $e$ : $x=\ln(x+3)=\ln(3+x)$ .
Hang the subscripts on it for the iterative formula.

Answer

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

Show every line, skip nothing. That is what an “answer given” step needs. (9709/31/O/N/23 Q8)

Common mistake

When the question says “show that it can be written as

x=F(x)

”, the answer is already printed for you. You must show every algebra line down to that exact form. No calculator shortcut, or you lose the method mark. (9709/31/O/N/23 Q8)

The rearrangement can come out of an integral (9709/31/M/J/23 Q9)

The “show that $a=\dots$ ” step need not start from a tidy polynomial. In one real paper you are told $\displaystyle\int_0^{a}xe^{-2x}\,dx=\tfrac18$ . Doing the integration (by parts) and tidying up gives $a=\tfrac12\ln(4a+2)$ . Only then do you iterate. The rearranging skill is the same: isolate the awkward term, take $\ln$ , make the variable the subject. So this chapter often turns up inside integration questions.(9709/31/M/J/23 Q9)

The iterative formula

Once you have $x=F(x)$ , iterating just means feed the answer back in as the next input:

$x_{n+1}=F(x_n)$

Start from a first guess $x_1$ inside the range you found in §6.1. Then keep running the machine: $x_2=F(x_1)$ , $x_3=F(x_2)$ , and so on.
From $e^x-3=x$ the easy rearrangement $e^x=3+x$ gives the iterative formula $x_{n+1}=\ln(3+x_n)$ .

y = x and y = F(x) = ln(3 + x) cross at the fixed point. The iteration steps the guess in toward it.

The picture is just for understanding. You never have to draw or analyse it for marks.

Press Step below to run $x_{n+1}=\ln(3+x_n)$ one iteration at a time:

n = 1·x₁ = 0.4000·still climbing

Each step goes up to the curve (working out $F(x_n)$ ). Then it goes across to the line $y=x$ (feeding the answer back in as the next $x$ ). It closes in on the root.
From $x_1=0.4$ the values settle to $1.5052$ in just a few steps. (9709/31/O/N/23 Q8)

Proving it converges to the root

The whole proof in three lines. If it converges, the values stop moving, so $x_{n+1}=x_n=x$ . Put that into the formula to get $x=F(x)$ . Un-rearrange and you get back the original equation, so the limit is the root.

This is worth a real $1$ – $2$ marks.
Here it is on the real example, where $x_{n+1}=\ln(3+x_n)$ came from $x=e^x-3$ .

Worked example

Show that if $x_{n+1}=\ln(3+x_n)$ converges, it converges to the root of $e^x-3=x$

If it converges, consecutive values settle to one number $x$ , so set $x_{n+1}=x_n=x$ : $x=\ln(3+x)$ .
Un-rearrange — undo the $\ln$ : $e^x=3+x$ .
Tidy to the original equation: $x=e^x-3$ .

Answer

x=\ln(3+x)\Rightarrow e^x=3+x\Rightarrow x=e^x-3:\ \text{the limit is the root}

The mark scheme wants exactly this: write $x=F(x)$ , then rearrange back to the given equation. (9709/31/O/N/23 Q8)

Common mistake

Don't “prove” convergence by running the iteration and showing the numbers settle. That earns nothing here. The mark is for the un-rearrange argument: write

x=F(x)

, then show it goes back to the original equation.

Your turn— tap to reveal the worked answer (9709/31/M/J/24 Q6(c))

Show that if $x_{n+1}=\ln\!\big(\operatorname{cosec}(\tfrac12 x_n)+3\big)$ converges, it converges to the root of $\operatorname{cosec}(\tfrac12 x)=e^x-3$ . Use the same method.

Answer

x=\ln(\operatorname{cosec}(\tfrac12 x)+3)\Rightarrow e^x=\operatorname{cosec}(\tfrac12 x)+3\Rightarrow \operatorname{cosec}(\tfrac12 x)=e^x-3

The big idea of the whole chapter, in one line

Everywhere else in P3 you solve for the answer in one go. Here you can't: $x=e^x-3$ has no algebra answer at all. So the chapter uses a loop instead. A wrong guess gives a less-wrong one. Repeat, and you land on the one number the loop can't improve. That number is the root. That swap, solve once for repeat many times, is the whole idea. Everything else is just the steps.

PreviousLocating a Root: Sign Change and Sketching a Pair of Graphs NextRunning the Iteration to a Required Accuracy (and Confirming the Root)

Rearranging f(x)=0 to x=F(x), and Proving It Converges to the Root

$x=F(x)$ works like a machine: feed it a guess, it gives back a better guess.
The answer is the number the machine can no longer improve. That one idea earns three different exam marks.

Rearranging f(x) = 0 into x = F(x)

To iterate you first need the equation as $x=F(x)$ (x equals some function of x). Put a single x on the left. Move everything else into $F(x)$ on the right.

Three moves cover almost every paper:

The awkward term	The move	Example
plain $x$ terms	make $x$ the subject	$x^2+x-3=0\Rightarrow x=\tfrac{3}{x}-1$
an $e^{(\dots)}$	isolate it, take $\ln$	$e^x=2x+1\Rightarrow x=\ln(2x+1)$
a trig term	isolate it, apply the inverse	$\tfrac12 x=\tan(\pi-x)\Rightarrow x=\pi-\tan^{-1}(\tfrac12 x)$

Worked example

Real question: show $x=e^x-3$ rearranges to $x_{n+1}=\ln(3+x_n)$

9709/31 O/N 2023 Q8: the rearrangement the iterative formula is built on. The $e^x$ is the awkward term. So isolate it and take $\ln$ .

Isolate the $e^x$ : from $x=e^x-3$ , add 3 to both sides → $e^x=x+3$ .
Take $\ln$ of both sides — that undoes the $e$ : $x=\ln(x+3)=\ln(3+x)$ .
Hang the subscripts on it for the iterative formula.

Answer

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

Show every line, skip nothing. That is what an “answer given” step needs. (9709/31/O/N/23 Q8)

Common mistake

When the question says “show that it can be written as

x=F(x)

”, the answer is already printed for you. You must show every algebra line down to that exact form. No calculator shortcut, or you lose the method mark. (9709/31/O/N/23 Q8)

The rearrangement can come out of an integral (9709/31/M/J/23 Q9)

The iterative formula

Once you have $x=F(x)$ , iterating just means feed the answer back in as the next input:

$x_{n+1}=F(x_n)$

Start from a first guess $x_1$ inside the range you found in §6.1. Then keep running the machine: $x_2=F(x_1)$ , $x_3=F(x_2)$ , and so on.
From $e^x-3=x$ the easy rearrangement $e^x=3+x$ gives the iterative formula $x_{n+1}=\ln(3+x_n)$ .

y = x and y = F(x) = ln(3 + x) cross at the fixed point. The iteration steps the guess in toward it.

The picture is just for understanding. You never have to draw or analyse it for marks.

Press Step below to run $x_{n+1}=\ln(3+x_n)$ one iteration at a time:

n = 1·x₁ = 0.4000·still climbing

Each step goes up to the curve (working out $F(x_n)$ ). Then it goes across to the line $y=x$ (feeding the answer back in as the next $x$ ). It closes in on the root.
From $x_1=0.4$ the values settle to $1.5052$ in just a few steps. (9709/31/O/N/23 Q8)

Proving it converges to the root

This is worth a real $1$ – $2$ marks.
Here it is on the real example, where $x_{n+1}=\ln(3+x_n)$ came from $x=e^x-3$ .

Worked example

Show that if $x_{n+1}=\ln(3+x_n)$ converges, it converges to the root of $e^x-3=x$

If it converges, consecutive values settle to one number $x$ , so set $x_{n+1}=x_n=x$ : $x=\ln(3+x)$ .
Un-rearrange — undo the $\ln$ : $e^x=3+x$ .
Tidy to the original equation: $x=e^x-3$ .

Answer

x=\ln(3+x)\Rightarrow e^x=3+x\Rightarrow x=e^x-3:\ \text{the limit is the root}

The mark scheme wants exactly this: write $x=F(x)$ , then rearrange back to the given equation. (9709/31/O/N/23 Q8)

Common mistake

Don't “prove” convergence by running the iteration and showing the numbers settle. That earns nothing here. The mark is for the un-rearrange argument: write

x=F(x)

, then show it goes back to the original equation.

Your turn— tap to reveal the worked answer (9709/31/M/J/24 Q6(c))

Show that if $x_{n+1}=\ln\!\big(\operatorname{cosec}(\tfrac12 x_n)+3\big)$ converges, it converges to the root of $\operatorname{cosec}(\tfrac12 x)=e^x-3$ . Use the same method.

Answer

x=\ln(\operatorname{cosec}(\tfrac12 x)+3)\Rightarrow e^x=\operatorname{cosec}(\tfrac12 x)+3\Rightarrow \operatorname{cosec}(\tfrac12 x)=e^x-3

The big idea of the whole chapter, in one line