Ne1, Ne2 - Bisection, false position and secant methods | OCR A-Level Further Mathematics B (MEI)

Locate a root by halving an interval or by intersecting a chord with the horizontal axis. You will distinguish false position from the secant method, implement their different updates, and justify the decimal places claimed for a numerical root.

Bisection keeps a root inside a shrinking interval

Two methods start from $(1,-1)$ and $(2,2)$ on $y=x^2-2$ . Both draw the same chord and obtain $x=4/3$ . Must all their later answers agree too?

No. Choosing the next pair of points is part of the method, not a detail. First meet the method with the clearest guarantee: preserve opposite signs and halve the interval.

You already know

A continuous function with opposite signs at two endpoints has a zero between them. Cell references can retain a fixed parameter or follow the previous row. Keep full stored precision and round only the requested output.

Continuity is essential. A sign change across a discontinuity, such as $1/x$ across zero, need not enclose a root. Also, a sign change guarantees a root, not uniqueness; a monotonicity argument can establish uniqueness where needed.

Bisection

Suppose $f$ is continuous on $[a,b]$ and $f(a)f(b)<0$ . Form

c=\frac{a+b}{2}

If $f(c)=0$ , stop. Otherwise retain whichever of $[a,c]$ or $[c,b]$ has opposite endpoint signs.

After $n$ completed halvings, the bracket width is $(b_0-a_0)/2^n$ . Its midpoint has error magnitude at most half that width.

Not in the formula booklet: learn it.

The error bound is geometric: every point of the retained interval is within half its width of its midpoint. Count completed halvings from the original bracket, rather than count the printed rows indiscriminately.

[DIAGRAM: asset_name: The first bisection retains opposite signs; asset_slug: bisection_bracket; caption: The midpoint splits the interval and the sign change selects the retained half; description: The graph y=x squared minus 2 is drawn on positive x. Points at x=1 and x=2 have opposite signs. The first midpoint x=1.5 has positive function value, so a blue interval from 1 to 1.5 is retained and the interval from 1.5 to 2 is pale. The root is inside the retained half.]
The midpoint splits the interval and the sign change selects the retained half

Worked example 1: three halvings

Question. Apply three bisections to $f(x)=x^2-2$ , starting from $[1,2]$ . Give the final bracket and its midpoint error bound.

Plan. Calculate each midpoint and retain the sign-changing half.

Halving	Midpoint $c$	$f(c)$	Retained bracket
$1$	$1.5$	$0.25$	$[1,1.5]$
$2$	$1.25$	$-0.4375$	$[1.25,1.5]$
$3$	$1.375$	$-0.109375$	$[1.375,1.5]$

\begin{aligned} b_3-a_3&=1.5-1.375=\frac18 && \text{three halvings}\\ m_3&=\frac{1.375+1.5}{2}=1.4375 && \text{midpoint of final bracket}\\ |m_3-\alpha|&\le\frac1{16}=0.0625 && \text{half-width bound} \end{aligned}

Why this step? The last tested point is an endpoint of the final bracket; its midpoint gives the stated half-width guarantee.

Answer. $1.375<\alpha<1.5$ ; midpoint $1.4375$ , error magnitude at most $0.0625$ .

Check. The original width $1$ , divided by $2^3$ , is $1/8$ , matching the retained bracket.

Change only the starting bracket to $[1.4,1.5]$ , which is already narrower.

Worked example 2: a tighter initial bracket

Question. Repeat three bisections for the same function, starting from $[1.4,1.5]$ .

Plan. Use the same sign rule, starting from $f(1.4)=-0.04$ and $f(1.5)=0.25$ .

Halving	$c$	$f(c)$	Retained bracket
$1$	$1.45$	$0.1025$	$[1.4,1.45]$
$2$	$1.425$	$0.030625$	$[1.4,1.425]$
$3$	$1.4125$	$-0.00484375$	$[1.4125,1.425]$

\begin{aligned} b_3-a_3&=\frac{0.1}{8}=0.0125 && \text{smaller original width}\\ m_3&=1.41875 && \text{midpoint}\\ |m_3-\alpha|&\le0.00625 && \text{half the final width} \end{aligned}

Why this step? A better starting bracket improves the guarantee without changing the number of iterations.

Answer. Bracket $[1.4125,1.425]$ , midpoint $1.41875$ , error magnitude at most $0.00625$ .

Check. The bound is one tenth of the previous one, matching the one-tenth starting width.

To obtain midpoint error magnitude below a specified tolerance $\varepsilon$ , choose $n$ so that $(b_0-a_0)/2^{n+1}<\varepsilon$ . This is an error bound, not automatically a guarantee about every decimal-rounding boundary.

Watch out

Wrong: replace the left endpoint whenever the midpoint is negative, regardless of the endpoint signs. Right: retain the sign-changing half; the sign at the original left endpoint need not always be negative.

Use a new function and keep the same three-halving procedure.

False position uses a chord and keeps the bracket

A midpoint ignores the actual heights of the two endpoints. False position uses the straight line joining their graph points, then retains opposite signs as before.

The chord through $(a,f(a))$ and $(b,f(b))$ has equation

y=f(a)+\frac{f(b)-f(a)}{b-a}(x-a)

Setting $y=0$ gives its horizontal-axis intercept.

False-position step

For a continuous function with $f(a)f(b)<0$ ,

c=\frac{a f(b)-b f(a)}{f(b)-f(a)}

Stop if $f(c)=0$ . Otherwise replace the endpoint whose function value has the same sign as $f(c)$ , preserving the sign-changing bracket.

Not in the formula booklet: learn it.

The intercept lies between opposite-sign endpoints. It need not halve the interval, and one endpoint can remain fixed for several steps.

Worked example 3: one endpoint stays fixed

Question. Apply three false-position steps to $f(x)=x^2-2$ , starting from $[1,2]$ . Record the new intercepts to eight decimal places.

Plan. Find each chord intercept and replace only the endpoint with the same sign.

\begin{aligned} c_1&=\frac{1(2)-2(-1)}{2-(-1)}=\frac43 && \text{first chord}\\ f(c_1)&=-\frac29<0 && \text{replace the left endpoint}\\ c_2&=\frac{(4/3)2-2(-2/9)}{2+2/9}=\frac75 && \text{inputs }4/3,2\\f(c_2)&=-\frac1{25}<0 && \text{left endpoint moves}\\ c_3&=\frac{(7/5)2-2(-1/25)}{2+1/25}=\frac{24}{17} && \text{inputs }7/5,2\\f(c_3)&=-\frac2{289}<0 && \text{right endpoint still }2 \end{aligned}

Why this step? The right endpoint $2$ is retained because its function value remains the opposite sign.

Answer. $c_1=1.33333333$ , $c_2=1.40000000$ , $c_3=1.41176471$ , with final bracket $[24/17,2]$ .

Check. All three intercepts lie below the positive root, and $f(2)=2>0$ still supplies the other bracket endpoint.

Change only the initial right endpoint from $2$ to $1.5$ .

Worked example 4: a closer opposite-sign endpoint

Question. Apply three false-position steps to the same function on $[1,1.5]$ .

Plan. Retain $1.5$ while the successive intercepts have negative function values.

\begin{aligned} c_1&=\frac{1(0.25)-1.5(-1)}{0.25-(-1)}=\frac75 && \text{new first chord}\\ f(c_1)&=-\frac1{25}<0 && \text{left endpoint moves}\\ c_2&=\frac{(7/5)(1/4)-(3/2)(-1/25)}{1/4+1/25}=\frac{41}{29} && \text{right endpoint still }1.5\\ c_3&=\frac{(41/29)(1/4)-(3/2)(-1/841)}{1/4+1/841}=\frac{239}{169} && \text{same update rule} \end{aligned}

Why this step? Replacing the right endpoint merely because it is older would turn this into a different method.

Answer. The last two function values are $-1/841$ and $-1/28561$ , so both new intercepts replace the left endpoint. The recorded intercepts are $1.40000000,1.41379310,1.41420118$ , with final bracket $[239/169,1.5]$ .

Check. The closer initial endpoint gives better intercepts here, but the final bracket remains much wider than the last two intercepts' difference.

A sequence of close intercepts is not the same object as the maintained sign bracket. Do not silently use the last two negative intercepts as opposite-sign bounds.

Watch out

Wrong: keep the newest two points after every false-position step. Right: keep the pair with opposite signs, even if one endpoint is old.

Use a new equation with the same positive-side endpoint pattern.

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Ne1, Ne2 - Bisection, false position and secant methods

Bisection keeps a root inside a shrinking interval

False position uses a chord and keeps the bracket

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