Bracketing and Bisection

The Central Idea

Suppose \(f\) is continuous on

and the endpoint values have opposite signs:

Then at least one root lies between \(a\) and \(b\).

The bisection method repeatedly cuts this interval in half and retains the half that still contains a sign change.

What Is Bracketing?

A root is bracketed when two values \(a\) and \(b\) surround a root in a way that can be verified numerically.

For a continuous function, the standard bracketing condition is

This means one endpoint gives a negative function value and the other gives a positive function value.

The Intermediate Value Theorem

The mathematical foundation for sign-change bracketing is the Intermediate Value Theorem.

If \(f\) is continuous on \([a,b]\) and

Example: Bracketing \(\sqrt{2}\)

Consider

Evaluate the endpoints \(1\) and \(2\):

Since

the positive root is bracketed:

What Is Bisection?

Bisection means dividing something into two parts.

In the bisection method, the current search interval

is divided at its midpoint

The sign of \(f(m)\) determines which half should be retained.

The Three Possible Midpoint Outcomes

After computing

evaluate \(f(m)\).

Three cases are possible:

1. If \(f(m)=0\), then \(m\) is an exact root.
2. If \(f(a)f(m)<0\), keep \([a,m]\).
3. If \(f(m)f(b)<0\), keep \([m,b]\).

One Bisection Step

Begin with

Compute

Then replace the original interval with either

or

whichever still brackets the root.

Example: First Bisection for \(\sqrt{2}\)

Start with

The midpoint is

Evaluate:

Since

and

the sign change occurs on

Second Bisection

Now bisect

The midpoint is

Evaluate:

Since

and

keep

Third Bisection

Bisect

The midpoint is

Then

Since

and

the new interval is

Fourth Bisection

Bisect

The midpoint is

Evaluate:

Therefore, the sign change lies in

Bisection Produces Nested Intervals

The intervals generated above are

Each interval is contained inside the previous interval.

The root remains trapped while the interval becomes smaller.

Interval Width Is Halved Each Time

Suppose the original interval has width

After one bisection, the retained interval has width

After two bisections, the width is

After \(n\) bisections,

Predictable Error Control

The predictable shrinking of the interval is one of the greatest strengths of bisection.

If the final root lies somewhere in

then the midpoint

is at most half the interval width from any point in the interval.

Therefore,

where \(r\) is the bracketed root.

Error Bound from the Original Interval

Since

the midpoint error satisfies

This lets us determine in advance how many bisections are sufficient for a desired accuracy.

How Many Bisections Are Needed?

Suppose we want the interval width to be at most \(\varepsilon\).

We require

Rearranging gives

Taking base-\(2\) logarithms,

The required iteration count is the next integer at or above this value.

Example: Required Number of Steps

Suppose the initial interval is

and we want the final interval width to be at most

We need

Since

we have

Therefore,

Bracketing versus Guessing

A bracket is not merely a rough guess.

For a continuous function, the condition

provides a mathematical guarantee that at least one root lies between the endpoints.

This guarantee is what makes bisection unusually reliable.

Example: Solving \(x^3-x-2=0\)

Define

Evaluate:

Therefore,

brackets at least one root.

First Bisection for \(x^3-x-2=0\)

Midpoint:

Evaluate:

Since

and

retain

Second Bisection

Midpoint:

Evaluate:

Since

and

retain

Third Bisection

Midpoint:

Evaluate:

Therefore, retain

Fourth Bisection

Midpoint:

Evaluate:

Since the value is positive while

the root lies in

Continuing the Process

Repeating the method eventually produces intervals such as

and then still narrower intervals.

The real root is approximately

Bisection approaches this value from a sequence of guaranteed brackets.

A Bisection Table

Repeated bisection is often organized in a table.

The Bisection Algorithm

A standard bisection procedure is:

1. Choose \(a\) and \(b\) so that \(f(a)f(b)<0\).
2. Check that \(f\) is continuous on \([a,b]\).
3. Compute \(m=(a+b)/2\).
4. If \(f(m)=0\), stop.
5. If \(f(a)f(m)<0\), replace \(b\) with \(m\).
6. Otherwise, replace \(a\) with \(m\).
7. Repeat until the stopping condition is satisfied.

Bisection in Pseudocode

choose a and b with f(a)f(b) < 0

repeat
    m = (a + b) / 2

    if f(m) = 0
        stop
    else if f(a)f(m) < 0
        b = m
    else
        a = m
until stopping condition is satisfied

Stopping by Interval Width

One common stopping rule is

Once the interval is narrow enough, its midpoint can be reported as the approximate root.

Stopping by Midpoint Error Bound

If we want the midpoint error to be less than \(\varepsilon\), it is enough to require

Equivalently,

Stopping by Residual

We may also monitor

If

the midpoint produces a small residual.

However, a residual criterion and an interval-width criterion measure different things.

Why Interval Width Is Especially Reliable

A small residual does not always imply a small error in \(x\).

By contrast, if a bracket is valid and has width

then the root must lie somewhere inside that interval.

The interval itself gives direct location information.

Bisection Is Guaranteed under the Right Conditions

If

• \(f\) is continuous on the interval;
• the endpoints have opposite signs;
• the method always keeps the sign-changing half;

then bisection continues to trap at least one root in smaller and smaller intervals.

This reliability is a major advantage over many faster methods.

Bisection Is Usually Not the Fastest Method

Bisection reduces the interval width by a factor of

at each step.

This is predictable but relatively gradual.

Methods such as Newton's method can converge much faster when they work well, but they require additional assumptions and can fail from poor starting values.

Reliability versus Speed

Numerical methods often involve a tradeoff.

Bisection emphasizes

• simplicity;
• guaranteed interval reduction;
• clear error bounds;
• robustness.

Other methods may emphasize faster convergence.

When Bisection Works Well

Bisection is especially useful when:

• a continuous function is available;
• a sign-changing interval can be found;
• reliability is important;
• derivatives are unavailable or inconvenient;
• a guaranteed error bound is useful.

When Bisection Is Not Directly Applicable

Standard sign-change bisection is not directly applicable when:

• no sign-changing bracket is known;
• the function is discontinuous across the interval;
• the root has even multiplicity and the sign does not change;
• the target problem is not naturally expressed as \(f(x)=0\).

Even-Multiplicity Roots

Consider

The root is

But

and

There is no sign change.

A standard sign-change bracketing search can therefore miss this root.

Discontinuities Can Create False Sign Changes

Consider

Then

and

The signs differ, but the function has no zero in \([-1,1]\).

The discontinuity at

invalidates the continuity assumption required by the Intermediate Value Theorem.

A Wide Bracket Can Contain Several Roots

A sign change guarantees at least one root, not necessarily exactly one.

A broad interval may contain multiple roots.

If the goal is to isolate a particular root, a narrower initial bracket may be preferable.

Finding an Initial Bracket

Initial brackets can come from:

• algebraic estimates;
• numerical tables;
• graphs;
• physical constraints;
• known bounds;
• systematic interval searches.

Example: Bracketing from Algebra

To solve

note that

and

Therefore,

The interval \([3,4]\) provides a natural starting bracket.

Example: Bracketing from a Table

Suppose a numerical table gives

and

If \(f\) is continuous on \([2.1,2.2]\), then

brackets at least one root.

Example: Bracketing from a Graph

A graph may suggest that a function crosses the \(x\)-axis between

and

Numerical evaluation should then confirm

before the interval is treated as a formal bisection bracket.

Bisection and Decimal Accuracy

If a root has been trapped inside

then its location is known within an interval of width

This information can justify a decimal approximation more rigorously than simply reading digits from a calculator display.

Bisection and Computational Efficiency

Bisection requires one new function evaluation per iteration after the endpoint values have been established.

One endpoint from the previous interval is retained, so its function value can also be retained.

Good implementations avoid recomputing values that are already known.

Reuse Known Endpoint Values

Suppose the current interval is

and \(f(a)\) and \(f(b)\) are already known.

After computing

only

needs to be newly evaluated.

The retained endpoint value can be carried forward to the next iteration.

Common Error: Forgetting Continuity

The condition

is not enough by itself.

Continuity on the interval is essential to the usual root guarantee.

Common Error: Keeping the Wrong Half

After testing the midpoint, retain the half whose endpoints still have opposite signs.

If

then keep

not \([m,b]\) unless a separate sign change has been established there.

Common Error: Averaging Function Values Instead of Inputs

The bisection midpoint is

It is not

Bisection halves the input interval, not the output values.

Common Error: Assuming the Midpoint Is the Root

The midpoint is a test point.

It becomes an exact root only when

Otherwise, its role is to help select the next interval.

Common Error: Rounding Midpoints Too Aggressively

Excessive rounding can distort function values and may even create an incorrect sign decision near a root.

Retain sufficient precision during the iteration and round the final result according to the required tolerance.

Common Error: Confusing Width with Midpoint Error

If the final interval has width

then the midpoint is at most

from the root.

The interval width and midpoint error bound differ by a factor of \(2\).

A Reliable Bisection Workflow

1. Rewrite the problem as \(f(x)=0\).
2. Find endpoints \(a\) and \(b\) with opposite function signs.
3. Verify continuity on \([a,b]\).
4. Compute the midpoint \(m=(a+b)/2\).
5. Evaluate \(f(m)\).
6. If \(f(m)=0\), stop.
7. Otherwise keep the half that preserves the sign change.
8. Measure the new interval width.
9. Repeat until the stopping condition is met.
10. Report the midpoint or bracket with an appropriate error statement.

Example: Full Workflow for \(\sqrt{5}\)

Define

First,

So begin with

Midpoint:

Since

keep

Continue the \(\sqrt{5}\) Workflow

Midpoint of

is

Evaluate:

Therefore, keep

Refine Again

Midpoint:

Then

Therefore, keep

Repeating the process eventually produces

Practice 1: Verify a Bracket

For

show that \([1,2]\) brackets the positive root.

Practice 2: First Bisection

Using

and the bracket \([1,2]\), compute the midpoint and determine the next interval.

Practice 3: Second Bisection

Continue Practice 2 for one more bisection step.

Practice 4: Interval Width

Find the width of

Practice 5: Width after Bisections

An interval begins with width

What is its width after five bisections?

Practice 6: Number of Steps

An interval has initial width \(1\).

How many bisections are needed to make the width at most

Practice 7: Midpoint Error Bound

A root is known to lie in

If the midpoint is reported as the approximation, what is the largest possible error?

Practice 8: Continuity Warning

Explain why

cannot use the interval \([-1,1]\) as a valid bisection bracket for a root, even though the endpoint values have opposite signs.

Practice 9: Repeated Root

Explain why standard sign-change bisection does not naturally locate the root of

Practice 10: Interpret the Method

In one sentence, describe what bisection does at every iteration.

Practice Answers

Practice 1

The function is continuous and the signs differ, so

brackets the positive root.

Practice 2

The midpoint is

Evaluate:

Since

and

the next interval is

Practice 3

The midpoint of \([1.5,2]\) is

Evaluate:

Since

and

keep

Practice 4

Therefore, the width is

Practice 5

After five bisections,

Therefore,

Practice 6

We need

Since

the answer is

Practice 7

The interval width is

The midpoint error is at most half the width:

Practice 8

The function is discontinuous at

Therefore, the Intermediate Value Theorem does not apply on \([-1,1]\), and the sign change does not guarantee a zero.

Practice 9

Since

on both sides of \(x=4\), the function does not change sign across the root.

Standard sign-change bracketing therefore does not naturally detect it.

Practice 10

Bisection tests the midpoint of a valid bracket and keeps the half that still brackets the root.

What to Remember

• Bracketing traps a root inside a verified search interval.
• For continuous functions, \(f(a)f(b)<0\) guarantees at least one root in \((a,b)\).
• Bisection repeatedly tests the midpoint \(m=(a+b)/2\).
• At each step, keep the half whose endpoints still have opposite signs.
• The interval width is halved at every iteration.
• After \(n\) steps, the width is \((b-a)/2^n\).
• The midpoint error is at most half the current interval width.
• Bisection is reliable but generally slower than higher-order methods.
• Continuity must be checked before using a sign change as a root guarantee.
• Even-multiplicity roots may not be detected by sign-change bracketing.

Where We Go Next

Bisection refines a root by repeatedly shrinking a guaranteed interval. Another major approach is to update a single estimate by rewriting the equation into a form whose repeated evaluation may converge to the solution.

The next lesson, Fixed-Point Iteration, develops iteration rules of the form \(x_{n+1}=g(x_n)\), examines when they converge, and shows how fixed points can be used to solve equations numerically.