Mathematics 9709 · Pure Mathematics 2

Running the Iteration to a Required Accuracy (and Confirming the Root)

This is the safe way to finish the question, and the easiest $3$ marks in the chapter.
A fixed calculator routine, no thinking needed. Just keep going until the answer stops changing.

Set radians. Type $x_1$ then $=$ . Then type $F(\mathrm{Ans})$ and press $=$ again and again, writing each value to $4$ dp. Stop when two in a row match to the asked accuracy.

The calculator drill (Ans key)

The Ans key holds your last result, so pressing $=$ again runs the machine again.

Put the calculator in radians if there is any trig.
Type the starting value $x_1$ and press $=$ (now $\mathrm{Ans}=x_1$ ).
Type the formula with Ans in place of $x$ (for example $\ln(3+\mathrm{Ans})$ ), then press $=$ . That is $x_2$ .
Keep pressing $=$ . Each press is the next iteration. Copy each value to $4$ dp into a little table.
Stop when two consecutive values are the same to the required accuracy, then quote the root to that accuracy.

Worked example

Iterate $x_{n+1}=\ln(3+x_n)$ from $x_1=1.5$ to $2$ dp

Write each value to $4$ dp, then give the final root to $2$ dp.

n	xₙ (to 4 dp)
$x_1$	$1.5000$
$x_2$	$1.5041$
$x_3$	$1.5050$
$x_4$	$1.5052$
$x_5$	$1.5052$

By $x_4$ and $x_5$ the values have stopped moving: both are $1.5052$ , which rounds to $1.51$ .

Answer

\alpha=1.51\ \text{(2 dp)}

Show enough iterations (to at least $4$ dp) so the $2$ dp answer is backed up. (9709/31/O/N/23 Q8)

Worked example

Real question: iterate the cosec formula from $x_1=1.4$ (radians!) to $2$ dp

9709/31 M/J 2024 Q6(d), iterating $x_{n+1}=\ln(\operatorname{cosec}(\tfrac12 x_n)+3)$ . The value jumps above and below before it settles (a cobweb, not a staircase), but it still settles.

n	xₙ (to 4 dp)
$x_1$	$1.4000$
$x_2$	$1.5156$
$x_3$	$1.4940$
$x_4$	$1.4978$
$x_5$	$1.4971$
$x_6$	$1.4972$

By $x_5$ and $x_6$ the value has settled to $1.497\ldots$ , which rounds to $1.50$ .

Answer

\alpha=1.50\ \text{(2 dp)}

In degrees the very first step gives nonsense. The mark scheme gives zero for that. (9709/31/M/J/24 Q6)

Three ways students lose these marks

Common mistake

1. Calculator in degrees. If the formula has any trig and you are not in radians, every value is wrong and you lose all the marks. Check the mode before you start. This is the biggest mark-loser in the topic.

Common mistake

2. Only the final number, no working. The mark scheme gives no marks for a bare calculator answer. You must show the iteration values. A lone

\alpha=1.51

with nothing above it scores zero, even when it is right.

Common mistake

3. Rounding too early, or giving the wrong precision. Keep full precision in the calculator. Use Ans, never retype a rounded value. Round only the final answer. Give it to the precision asked: a

2

dp answer, even though you wrote the iterations to

4

dp.

Confirming the root (the half-interval check)

To prove a rounded root like $\alpha=1.51$ , you can instead do a change-of-sign test on the half-interval $(1.505,\,1.515)$ . The mark scheme accepts either way.

The root rounds to $1.51$ (2 dp) exactly when it lies between $1.505$ and $1.515$ .
So put those two endpoints into $f(x)=e^x-3-x$ and show a sign change ( $f(1.505)<0$ , $f(1.515)>0$ ). This is the §6.1 change-of-sign idea used again to finish.

The root is often not called a root (9709/32/F/M/24 Q7)

Examiners often hide an iteration inside a calculus problem. In one real paper the minimum point $M$ of $y=xe^{2x}-5x$ sits at $x=a$ . Differentiate and set $\tfrac{dy}{dx}=0$ to get $a=\tfrac12\ln\!\big(\tfrac{5}{1+2a}\big)$ , then iterate to $a=0.47$ . The word “root” never shows up. But the $x$ -coordinate of a stationary point is the root of $\tfrac{dy}{dx}=0$ , so the same three steps run. Spotting that is the whole skill. (9709/32/F/M/24 Q7)

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

You have iterated $x_{n+1}=\ln(\operatorname{cosec}(\tfrac12 x_n)+3)$ (radians) to $\alpha\approx1.4972$ . Check the 2 dp answer with a half-interval sign change. The root rounds to $1.50$ , so test $(1.495,\,1.505)$ in $f(x)=\operatorname{cosec}(\tfrac12 x)-e^x+3$ .

Answer

f(1.495)>0,\ f(1.505)<0:\ \text{change of sign}\Rightarrow\alpha=1.50

Tying the three parts together

When a part says “use an iterative formula based on the equation in (a)”, you pick the rearrangement from §6.2. The formula is not given to you.

The chapter is one run of three steps:

§6.1 locate: sketch a pair of graphs (one root) and a change of sign (between $p$ and $q$ ).
§6.2 rearrange: get $x=F(x)$ and, if asked, prove convergence by un-rearranging.
§6.3 iterate: run the iteration on the calculator to the asked accuracy, showing the values, then confirm.

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

Running the Iteration to a Required Accuracy (and Confirming the Root)

This is the safe way to finish the question, and the easiest $3$ marks in the chapter.
A fixed calculator routine, no thinking needed. Just keep going until the answer stops changing.

Set radians. Type $x_1$ then $=$ . Then type $F(\mathrm{Ans})$ and press $=$ again and again, writing each value to $4$ dp. Stop when two in a row match to the asked accuracy.

The calculator drill (Ans key)

The Ans key holds your last result, so pressing $=$ again runs the machine again.

Put the calculator in radians if there is any trig.
Type the starting value $x_1$ and press $=$ (now $\mathrm{Ans}=x_1$ ).
Type the formula with Ans in place of $x$ (for example $\ln(3+\mathrm{Ans})$ ), then press $=$ . That is $x_2$ .
Keep pressing $=$ . Each press is the next iteration. Copy each value to $4$ dp into a little table.
Stop when two consecutive values are the same to the required accuracy, then quote the root to that accuracy.

Worked example

Iterate $x_{n+1}=\ln(3+x_n)$ from $x_1=1.5$ to $2$ dp

Write each value to $4$ dp, then give the final root to $2$ dp.

n	xₙ (to 4 dp)
$x_1$	$1.5000$
$x_2$	$1.5041$
$x_3$	$1.5050$
$x_4$	$1.5052$
$x_5$	$1.5052$

By $x_4$ and $x_5$ the values have stopped moving: both are $1.5052$ , which rounds to $1.51$ .

Answer

\alpha=1.51\ \text{(2 dp)}

Show enough iterations (to at least $4$ dp) so the $2$ dp answer is backed up. (9709/31/O/N/23 Q8)

Worked example

Real question: iterate the cosec formula from $x_1=1.4$ (radians!) to $2$ dp

9709/31 M/J 2024 Q6(d), iterating $x_{n+1}=\ln(\operatorname{cosec}(\tfrac12 x_n)+3)$ . The value jumps above and below before it settles (a cobweb, not a staircase), but it still settles.

n	xₙ (to 4 dp)
$x_1$	$1.4000$
$x_2$	$1.5156$
$x_3$	$1.4940$
$x_4$	$1.4978$
$x_5$	$1.4971$
$x_6$	$1.4972$

By $x_5$ and $x_6$ the value has settled to $1.497\ldots$ , which rounds to $1.50$ .

Answer

\alpha=1.50\ \text{(2 dp)}

In degrees the very first step gives nonsense. The mark scheme gives zero for that. (9709/31/M/J/24 Q6)

Three ways students lose these marks

Common mistake

2. Only the final number, no working. The mark scheme gives no marks for a bare calculator answer. You must show the iteration values. A lone

\alpha=1.51

with nothing above it scores zero, even when it is right.

Common mistake

2

dp answer, even though you wrote the iterations to

4

dp.

Confirming the root (the half-interval check)

To prove a rounded root like $\alpha=1.51$ , you can instead do a change-of-sign test on the half-interval $(1.505,\,1.515)$ . The mark scheme accepts either way.

The root rounds to $1.51$ (2 dp) exactly when it lies between $1.505$ and $1.515$ .
So put those two endpoints into $f(x)=e^x-3-x$ and show a sign change ( $f(1.505)<0$ , $f(1.515)>0$ ). This is the §6.1 change-of-sign idea used again to finish.

The root is often not called a root (9709/32/F/M/24 Q7)

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

Answer

f(1.495)>0,\ f(1.505)<0:\ \text{change of sign}\Rightarrow\alpha=1.50

Tying the three parts together

When a part says “use an iterative formula based on the equation in (a)”, you pick the rearrangement from §6.2. The formula is not given to you.

The chapter is one run of three steps:

§6.1 locate: sketch a pair of graphs (one root) and a change of sign (between $p$ and $q$ ).
§6.2 rearrange: get $x=F(x)$ and, if asked, prove convergence by un-rearranging.
§6.3 iterate: run the iteration on the calculator to the asked accuracy, showing the values, then confirm.