OT2.2, OT2.5, OT3.2, OT3.3 - Applying and extending mathematics from an unfamiliar passage | OCR A-Level Mathematics B (MEI)

Build a connected solution from a supplied update rule, decide when its target is met and judge how much confidence the calculation deserves. You will use a model with suitable inputs, interpret its outputs and separate numerical accuracy from the limitations of the procedure being modelled.

Reading an update rule

Can a setting move closer to its target at every update, yet never reach it exactly? Suppose a brightness setting starts at $120$ , aims for $180$ , and closes one quarter of the remaining gap each time. The first update adds $15$ ; the next adds only $11.25$ . This is no ordinary arithmetic sequence.

You already know

A geometric sequence multiplies by the same ratio each time. Subscripts count positions or updates: starting at $r_0$ , one update produces $r_1$ . Logarithms turn powers into products, but dividing an inequality by a negative number reverses its direction.

Read this passage as a model with declared assumptions. You do not need prior knowledge of display electronics; every rule needed here is supplied.

A gradual display adjustment

[D1] A display has a numerical brightness setting between $0$ and $255$ . To move gradually towards a fixed target $L$ , a designer uses
$r_{n+1}=r_n+\lambda(L-r_n),\qquad n=0,1,2,\ldots$

> > Here $r_0$ is the initial setting and $0<\lambda<1$ is a fixed adjustment fraction. The model uses real-number settings, without rounding, clipping or response delay. > > **[D2]** In the example, $L=180$, $r_0=120$ and $\lambda=\frac14$. The design target is a setting within $0.5$ of $180$, including the boundary. The designer asks for the least number of updates. > > **[D3]** A practical implementation might round every new setting. That would change the procedure, so its accuracy must be checked separately. The adjustment is a fraction of the **remaining gap**, not a fixed fraction of the target. D2 begins as follows. | Updates $n$ | Setting $r_n$ | Signed gap $180-r_n$ | |---|---|---| | $0$ | $120$ | $60$ | | $1$ | $135$ | $45$ | | $2$ | $146.25$ | $33.75$ | | $3$ | $154.6875$ | $25.3125$ | The settings are not geometric, but the gaps are. Naming the right quantity reveals the familiar structure hidden inside the passage.

Signed deviation

Define $e_n=L-r_n$ . It is positive below the target, zero at the target and negative above it. The distance from the target is $|e_n|$ , not necessarily $e_n$ .

Subtract D1 from $L$ to obtain a simpler update rule:

\begin{aligned} e_{n+1}&=L-\bigl[r_n+\lambda(L-r_n)\bigr] && \text{use D1}\\ &=(1-\lambda)(L-r_n) && \text{factor the gap}\\ &=(1-\lambda)e_n && \text{use the definition} \end{aligned}

Repeated multiplication, starting at index zero, therefore gives

e_n=(L-r_0)(1-\lambda)^n,\qquad r_n=L-(L-r_0)(1-\lambda)^n

These are passage-specific results to derive from D1; they are not in the formula sheet. Check the initial condition and the update relation, rather than relying on a few matching numerical terms.

Worked example 1: approaching from below

Question. For D1 with $L=180$ , $\lambda=\frac14$ and $r_0=120$ , find $r_1$ , $r_2$ and an expression for $r_n$ . Explain whether the sequence overshoots the target.

Plan. Calculate two updates, then use the signed geometric deviation to describe every update.

Apply D1 twice:

r_1=120+\frac14(60)=135,\qquad r_2=135+\frac14(45)=146.25

Use the initial gap and its multiplier:

e_0=60,\qquad e_{n+1}=\frac34e_n\quad\Longrightarrow\quad e_n=60\left(\frac34\right)^n

Convert the deviation back into a setting:

r_n=180-60\left(\frac34\right)^n<180

Why this step? The positive deviation stays positive while shrinking; it cannot change sign and produce an overshoot.

Answer. $r_1=135$ , $r_2=146.25$ and $r_n=180-60(3/4)^n$ . The settings approach the target from below without overshooting.

Check. At $n=0$ , the formula gives $120$ . Multiplying its gap by $3/4$ gives the next required gap, so the formula satisfies D1 at every step.

Worked example 2 changes only the initial setting to one above the target. Keep the sign of the initial deviation; the rule itself is unchanged.

Worked example 2: approaching from above

Question. For D1 with $L=180$ , $\lambda=\frac14$ and $r_0=240$ , find $r_1$ , $r_2$ and an expression for $r_n$ . Explain whether the sequence overshoots the target.

Plan. Calculate the first updates, then retain the negative signed deviation in the geometric expression.

Apply D1 with a negative gap:

r_1=240+\frac14(-60)=225,\qquad r_2=225+\frac14(-45)=213.75

Track the signed deviation:

e_0=-60,\qquad e_n=-60\left(\frac34\right)^n

Recover the setting by subtracting that negative deviation:

r_n=180+60\left(\frac34\right)^n>180

Why this step? Replacing the initial deviation by its magnitude would put the setting on the wrong side of the target.

Answer. $r_1=225$ , $r_2=213.75$ and $r_n=180+60(3/4)^n$ . The settings approach from above without crossing below the target.

Check. At $n=0$ , the formula gives $240$ ; at $n=1$ , it gives $225$ , agreeing with the update.

There is also a direct range check. D1 can be written as $r_{n+1}=(1-\lambda)r_n+\lambda L$ , a weighted average with positive weights summing to one. It lies between the current value and the target. If both are in the device's range, the next value stays in that range too. Starting exactly at the target leaves the setting fixed.

Watch out

Wrong: add $\lambda L$ on every update. Right: add $\lambda(L-r_n)$ . The shrinking gap is the reason the increments shrink; a constant increment would describe a different procedure.

The next initial value is still above the target, but has a different signed gap.

A stopping condition

Turn a target into an exact inequality before deciding which integer number of updates answers the question.

D2 uses distance from the target. This works for starts above or below it, whereas a signed gap alone could be negative and appear to satisfy a tolerance immediately.

Tolerance

A tolerance $\delta>0$ specifies an allowed distance from a target. “Within $\delta$ , including the boundary” means $|L-r_n|\le\delta$ . “Strictly less than $\delta$ away” means $|L-r_n|<\delta$ .

For D2 the distance is $60(3/4)^n$ . It decreases at every update, so once a tolerance is met it stays met. Logarithms locate the first possible integer; neighbouring values provide a useful check.

Worked example 3: solving the passage's stopping request

Question. For D2, find the least number of updates needed to be within $0.5$ of the target, including the boundary. Justify minimality. Give neighbouring distances to $4$ significant figures.

Plan. Solve the distance inequality with logarithms, take the least allowed integer and check its predecessor.

Write the inclusive distance condition:

60\left(\frac34\right)^n\le0.5\quad\Longleftrightarrow\quad \left(\frac34\right)^n\le\frac1{120}

Take logarithms and divide by a negative number:

n\ln(3/4)\le\ln(1/120)\quad\Longrightarrow\quad n\ge\frac{\ln(1/120)}{\ln(3/4)}\approx16.6416

Test the neighbouring integer counts using the unrounded model:

|e_{16}|\approx0.6014>0.5,\qquad |e_{17}|\approx0.4510<0.5

Why this step? $\ln(3/4)<0$ , so the inequality reverses. The answer is an update count above the real threshold, not its ordinary rounded value.

Answer. The least count is $17$ updates. The distance decreases throughout, $16$ fails and $17$ succeeds.

Check. $r_{17}\approx179.5490$ , so its distance from $180$ is about $0.4510$ , consistent with the inequality.

Worked example 4 changes only the tolerance from $0.5$ to $1$ . A wider tolerance should require fewer updates; the same outputs and precision are requested.

Worked example 4: a wider permitted gap

Question. For D2's update model, find the least number of updates needed to be within $1$ of the target, including the boundary. Justify minimality. Give neighbouring distances to $4$ significant figures.

Plan. Replace the tolerance in the same inequality and retain full precision until choosing the integer.

Form the distance inequality:

60\left(\frac34\right)^n\le1\quad\Longleftrightarrow\quad \left(\frac34\right)^n\le\frac1{60}

Use the negative logarithm correctly:

n\ge\frac{\ln(1/60)}{\ln(3/4)}\approx14.2322

Check the two boundary counts:

|e_{14}|\approx1.069>1,\qquad |e_{15}|\approx0.8018<1

Why this step? The real threshold is just above $14$ ; rounding it to $14$ would select an update count that fails.

Answer. The least count is $15$ updates. Decreasing distance, together with failure at $14$ and success at $15$ , establishes minimality.

Check. The wider tolerance reduces the count from $17$ to $15$ , as expected. The unrounded setting at $15$ is about $179.1982$ .

Boundary wording can matter even when a logarithmic threshold is an exact integer. In a separate model with distance $32(1/2)^n$ , the distance at $n=5$ is exactly $1$ . An inclusive tolerance of $1$ first works at $5$ ; a strict distance less than $1$ first works at $6$ , when the distance is $1/2$ .

Watch out

Wrong: $n\le16.6416$ , after dividing by $\ln(3/4)$ . Right: $n\ge16.6416$ , because the divisor is negative. Check a larger count: it must produce a smaller distance.

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OT2.2, OT2.5, OT3.2, OT3.3 - Applying and extending mathematics from an unfamiliar passage

Reading an update rule

A stopping condition

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