A Complete Introduction to the Bisection Method with Worked Examples and Solutions

The Intermediate Value Theorem

The mathematical justification for the bisection method comes from the following theorem. Which states:

Suppose that \( f(x) \) is continuous on the closed interval \( [a,b] \). If \( N \) is any number between \( f(a) \) and \( f(b) \), then there exists a number \( c \in (a,b) \) such that

$$f(c) = N.$$

The bisection method uses the special case where

$$N = 0.$$

Therefore, if \( f(x) \) is continuous on \( [a,b] \) and

$$f(a)f(b) < 0,$$

then there exists at least one number \( c \in (a,b) \) such that

$$f(c) = 0.$$

In other words, if a continuous function has opposite signs at the endpoints of an interval, the function must cross the \( x \)-axis somewhere within that interval.

Why a Sign Change Indicates a Root

The condition

$$f(a)f(b) < 0$$

plays a central role in the bisection method. This inequality means that one endpoint has a positive function value, while the other has a negative function value.

For a continuous function, its graph cannot jump directly from negative values to positive values. Instead, it must pass through the value zero at some point in the interval.

Geometrically, this means that the graph of the function must intersect the \( x \)-axis at least once between \( a \) and \( b \).

Bracketing a Root

The process of identifying an interval containing a root is known as bracketing a root. An interval \( [a,b] \) is said to bracket a root if the function \( f(x) \) is continuous on \( [a,b] \), and the endpoints satisfy \( f(a)f(b) < 0 \).

The bisection method begins by finding a bracketed interval and then repeatedly replacing it with smaller bracketed intervals. Because each new interval still satisfies the conditions of the Intermediate Value Theorem, the existence of a root is preserved throughout the entire process.

This idea of maintaining an interval that contains a root at every iteration is what makes the bisection method one of the most reliable root-finding algorithms.

Why Does the Bisection Method Work?

One of the greatest strengths of the bisection method is that its convergence is guaranteed under relatively mild assumptions. The bisection method always converges to a root provided that the function \( f(x) \) is continuous on the interval \( [a,b] \), and \( f(a)f(b) < 0 \).

The reason the method works is that it repeatedly constructs smaller intervals that continue to contain a root. Because these intervals become arbitrarily small, their endpoints and midpoints eventually approach the exact root.

The Interval-Halving Strategy

Suppose that \( f(x) \) is continuous on the interval \( [a,b] \) and that

$$f(a)f(b) < 0.$$

By the Intermediate Value Theorem, there exists at least one root

$$c \in (a,b).$$

The bisection method begins by computing the midpoint

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

There are three possibilities:

• If \( f(c) = 0 \), then \( c \) is the exact root, and the algorithm terminates.
• If \( f(a)f(c) < 0 \), then the root lies in the interval \( [a,c] \).
• If \( f(c)f(b) < 0 \), then the root lies in the interval \( [c,b] \).

In either case, the original interval is replaced by a smaller interval that still contains a root.

Thus, the interval containing the root becomes smaller as the number of iterations increases.

Proof of Convergence

Let

$$[a_n, b_n]$$

denote the interval obtained after the \( n \)-th iteration of the bisection method.

Since the intervals are nested,

$$[a_0, b_0] \supset \cdots \supset [a_n, b_n],$$

and their lengths satisfy

$$b_n – a_n = \frac{b_0-a_0}{2^n},$$

the interval lengths approach zero as

$$n \to \infty.$$

By the Nested Interval Property, there exists a unique point

$$r$$

Such that

$$r \in [a_n, b_n] \forall n.$$

Since each interval contains a root and the interval lengths converge to zero, the endpoints and midpoints of the intervals converge to the same point \( r \).

Because \( f(x) \) is continuous and every interval contains a root, taking the limit yields

$$f(r) = 0.$$

Therefore, the sequence of approximations generated by the bisection method converges to a root of the equation

$$f(x) = 0.$$

Why the Bisection Method Is Guaranteed to Converge

The bisection method succeeds because it combines the Intermediate Value Theorem with the repeated halving of the interval, which guarantees that the approximations become arbitrarily accurate.

As long as the function is continuous and the initial interval contains a sign change, the method cannot diverge, oscillate, or miss the root contained within the interval. This guaranteed convergence is what makes the bisection method one of the most reliable and widely used algorithms in numerical analysis.

When Does the Bisection Method Succeed?

One of the most attractive features of the bisection method is its reliability. Unlike many other numerical root-finding methods, the bisection method is guaranteed to converge whenever a few simple conditions are satisfied. Understanding these conditions is important because they determine when the method can be applied successfully and when another numerical technique may be more appropriate.

The bisection method succeeds when two requirements are met:

1. The function must be continuous on the interval of interest.
2. The chosen interval must contain a sign change.

When both conditions hold, the Intermediate Value Theorem guarantees the existence of at least one root, and the interval-halving process ensures convergence to that root.

Condition 1: The Function Must Be Continuous

The first requirement is that the function \( f(x) \) be continuous on the interval

$$[a,b].$$

Continuity is necessary because the bisection method relies on the Intermediate Value Theorem. If the function contains a discontinuity, then a sign change does not necessarily imply the existence of a root.

Condition 2: The Initial Interval Must Contain a Sign Change

The second requirement is that the function values at the endpoints of the interval have opposite signs. That is

$$f(a)f(b) < 0.$$

This condition guarantees that at least one root exists within the interval \( (a,b) \).

When Does the Bisection Method Fail?

Although the bisection method is one of the most reliable root-finding algorithms, it is not universally applicable. The method succeeds only when its underlying assumptions are satisfied. When these assumptions fail, the bisection method may be unable to locate a root or may incorrectly conclude that no root exists.

Understanding the limitations of the bisection method is important because it helps determine when to use the method and when an alternative numerical technique is more appropriate.

The bisection method can fall for several reasons:

• The function is not continuous.
• The initial interval does not contain a sign change.
• The root has even multiplicity.
• The chosen interval brackets multiple roots.

Failure Due to Lack of a Sign Change

The most common reason for failure is choosing an interval whose endpoints have the same sign. Recall that the bisection method requires

$$f(a)f(b) < 0.$$

If this condition is not satisfied, the Intermediate Value Theorem cannot guarantee the existence of a root within the interval.

Failure Due to Discontinuity

The bisection method also fails when the function is not continuous on the interval of interest.

The Intermediate Value Theorem requires continuity, since a discontinuous function may exhibit a sign change without containing a root.

Failure for Roots of Even Multiplicity

Another limitation of the bisection method occurs when a root has even multiplicity.

This occurs because roots of even multiplicity do not result in a sign change. As a result, the bisection method cannot detect such roots.

Failure to Find Multiple Roots

The bisection method can guarantee the existence of at least one root in a bracketed interval, but it cannot determine how many roots are present.

Error Analysis for the Bisection Method

One of the major advantages of the bisection method is that its error can be estimated precisely. Unlike many numerical algorithms, whose errors can be difficult to analyze, the bisection method provides a guaranteed upper bound on the approximation error after each iteration.

This error analysis allows us to answer important practical questions, such as:

• How accurate is the current approximation?
• How many iterations are required to achieve a desired accuracy?
• How quickly does the method converge?

The answers to these questions follow directly from the interval-halving property of the bisection method.

Approximation Error

Suppose that the exact root of the equation

$$f(x) = 0$$

is \( r \), and let \( x_n \) denote the approximation obtained after the \( n \)-th iteration of the bisection method.

The absolute error is defined as

$$\epsilon_n = |r – x_n|.$$

Unfortunately, because the exact root is usually unknown, the absolute error cannot be computed directly. Instead, we estimate the error by using the length of the interval that contains the root.

Derivation of the Error Bound

Suppose the initial interval is

$$[a,b].$$

After one iteration, the interval length becomes

$$\frac{b – a}{2}.$$

After two iterations, the interval length is

$$\frac{b – a}{2^2}.$$

Continuing this process, after \( n \) iterations the interval containing the root has length

$$\frac{b – a}{2^n}.$$

Since the exact root \( r \) lies somewhere within this interval, the maximum possible error in the approximation cannot exceed the interval length. Therefore,

$$|r – x_n| \leq \frac{b – a}{2^n}.$$

This inequality is the error bound for the bisection method.

Convergence Rate of the Bisection Method

The error formula shows that each iteration reduces the error by approximately one-half.

Thus, the bisection method exhibits geometric convergence, meaning that the error decreases by a constant factor during each iteration.

Determining the Required Number of Iterations

A common practical question is determining how many iterations are needed to achieve a specified error tolerance.

Suppose we want the approximation error to satisfy

$$|r – x_n| < \epsilon,$$

where \( \epsilon \) is a prescribed tolerance. Using the error bound formula,

$$\frac{b – a}{2^n} < \epsilon.$$

Multiplying both sides by \( 2^n \) gives

$$b – a < \epsilon 2^n.$$

Dividing by \( \epsilon \) yields

$$\frac{b-a}{\epsilon} < 2^n.$$

Taking the logarithm of both sides gives

$$\log(\frac{b – a}{\epsilon}) < n\log(2).$$

Therefore,

$$n > \frac{\log(\frac{b-a}{\epsilon})}{\log(2)}.$$
Equivalently,

$$n > \log_2(\frac{b – a}{\epsilon}).$$

This formula allows us to determine the minimum number of iterations required before beginning the computation.

Relative Error Approximation

In practice, the exact error is usually unavailable. Consequently, many implementations estimate the error using the difference between successive approximations:

$$\epsilon \approx x_n – x_{n-1}.$$

This is called the relative error.

Step-by-Step Procedure for the Bisection Method

Now that we understand the mathematical foundation and error analysis of the bisection method, we can describe the algorithm itself. The idea is to repeatedly divide an interval containing a root into two smaller intervals, selecting the subinterval that still contains the root.

The following procedure describes the bisection method.

Step 1: Choose an Initial Interval

Begin by selecting an interval

$$[a,b]$$

such that \( f(x) \) is continuous on \( [a,b] \), and the endpoints satisfy

$$f(a)f(b) < 0.$$

This condition guarantees that the interval contains at least one root.

Step 2: Compute the Midpoint

Compute the midpoint of the interval using the formula

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

The midpoint serves as the next approximation to the root.

Step 3: Evaluate the Function at the Midpoint

Evaluate the function at the midpoint.

If \( f(c) = 0 \), then \( c \) is the exact root and the algorithm terminates. Otherwise, proceed to the next step.

Step 4: Determine the New Interval

Determine which half of the interval contains the root by checking the sign of the function.

If \( f(a)f(c) < 0 \), then the root lies in the interval \( [a,c] \), so let \( b = c \). Otherwise, \( f(c)f(b) < 0 \), and the root lies in the interval \( [c,b] \), so let \( a = c \).

In either case, the new interval is half the length of the previous interval.

Step 5: Check the Stopping Criterion

After each iteration, determine whether the desired accuracy has been achieved.

Worked Out Examples

The first example shows how to use the bisection method.

Example 1: Approximate the root of \( x^3 – x – 2 = 0 \), rounded to 2 decimal places.

Solution: Let

$$f(x) = x^3 – x – 2.$$

Consider

$$f(1)f(2) = (1^3 – 1 – 2)(2^3 – 2 – 2).$$

Exponentiating we find

$$f(1)f(2) = (1 – 1 – 2)(8 – 2 – 2).$$

Subtracting gives

$$f(1)f(2) = (–2)(4).$$

Multiplying we obtain

$$f(1)f(2) = –8.$$

Since \( f(1)f(2) < 0 \), a root exists in

$$[1, 2].$$

The midpoint is given by

$$c = \frac{1 + 2}{2}.$$

Adding we get

$$c = \frac{3}{2}.$$

Dividing we obtain

$$c = 1.5.$$

Next, we compute

$$f(1.5)f(2) = (1.5^3 – 1.5 – 2)(2^3 – 2 – 2).$$

Exponentiating we find

$$f(1.5)f(2) = (3.375 – 1.5 – 2)(8 – 2 – 2).$$

Subtracting gives

$$f(1.5)f(2) = (–0.125)(4).$$

Multiplying we obtain

$$f(1.5)f(2) = –0.5.$$

Since \( f(1.5)f(2) < 0 \), the new interval is

$$[1.5, 2].$$

The midpoint is given by

$$c = \frac{1.5 + 2}{2}.$$

Adding we get

$$c = \frac{3.5}{2}.$$

Dividing we obtain

$$c = 1.75.$$

Next, we compute

$$f(1.5)f(1.75) = (1.5^3 – 1.5 – 2)(1.75^3 – 1.75 – 2).$$

Exponentiating we find

$$f(1.5)f(1.75) = (3.375 – 1.5 – 2)(5.359375 – 1.75 – 2).$$

Subtracting gives

$$f(1.5)f(1.75) = (–0.125)(1.609375).$$

Multiplying we obtain

$$f(1.5)f(1.75) = –0.201171.$$

Since \( f(1.5)f(1.75) < 0 \), the new interval is

$$[1.5, 1.75].$$

The midpoint is given by

$$c = \frac{1.5 + 1.75}{2}.$$

Adding we get

$$c = \frac{3.25}{2}.$$

Dividing we obtain

$$c = 1.625.$$

Next, we compute

$$f(1.5)f(1.625) = (1.5^3 – 1.5 – 2)(1.625^3 – 1.625 – 2).$$

Exponentiating we find

$$f(1.5)f(1.625) = (3.375 – 1.5 – 2)(4.291015 – 1.625 – 2).$$

Subtracting gives

$$f(1.5)f(1.625) = (–0.125)(0.666015).$$

Multiplying we obtain

$$f(1.5)f(1.625) = –0.083251.$$

Since \( f(1.5)f(1.625) < 0 \), the new interval is

$$[1.5, 1.625].$$

The midpoint is given by

$$c = \frac{1.5 + 1.625}{2}.$$

Adding we get

$$c = \frac{3.125}{2}.$$

Dividing we obtain

$$c = 1.5625.$$

Next, we compute

$$f(1.5)f(1.5625) = (1.5^3 – 1.5 – 2)(1.5625^3 – 1.5625 – 2).$$

Exponentiating we find

$$f(1.5)f(1.5625) = (3.375 – 1.5 – 2)(3.814697 – 1.5625 – 2).$$

Subtracting gives

$$f(1.5)f(1.5625) = (–0.125)(0.252197).$$

Multiplying we obtain

$$f(1.5)f(1.5625) = –0.031524.$$

Since \( f(1.5)f(1.5625) < 0 \), the new interval is

$$[1.5, 1.5625].$$

The midpoint is given by

$$c = \frac{1.5 + 1.5625}{2}.$$

Adding we get

$$c = \frac{3.0625}{2}.$$

Dividing we obtain

$$c = 1.53125.$$

Next, we compute

$$f(1.5)f(1.53125) = (1.5^3 – 1.5 – 2)(1.53125^3 – 1.53125 – 2).$$

Exponentiating we find

$$f(1.5)f(1.53125) = (3.375 – 1.5 – 2)(3.590362 – 1.53125 – 2).$$

Subtracting gives

$$f(1.5)f(1.53125) = (–0.125)(0.059112).$$

Multiplying we obtain

$$f(1.5)f(1.53125) = –0.007389.$$

Since \( f(1.5)f(1.53125) < 0 \), the new interval is

$$[1.5, 1.53125].$$

The midpoint is given by

$$c = \frac{1.5 + 1.53125}{2}.$$

Adding we get

$$c = \frac{3.03125}{2}.$$

Dividing we obtain

$$c = 1.515625.$$

Next, we compute

$$f(1.515625)f(1.53125) = (1.515625^3 – 1.515625 – 2)(1.53125^3 – 1.53125 – 2).$$

Exponentiating we find

$$f(1.515625)f(1.53125) = (3.481571 – 1.515625 – 2)(3.590362 – 1.53125 – 2).$$

Subtracting gives

$$f(1.515625)f(1.53125) = (–0.034053)(0.059112).$$

Multiplying we obtain

$$f(1.515625)f(1.53125) = –0.002013.$$

Since \( f(1.515625)f(1.53125) < 0 \), the new interval is

$$[1.515625, 1.53125].$$

The midpoint is given by

$$c = \frac{1.515625 + 1.53125}{2}.$$

Adding we get

$$c = \frac{3.046875}{2}.$$

Dividing we obtain

$$c = 1.523437.$$

Next, we compute

$$f(1.515625)f(1.523437) = (1.515625^3 – 1.515625 – 2)(1.523437^3 – 1.523437 – 2).$$

Exponentiating we find

$$f(1.515625)f(1.523437) = (3.481571 – 1.515625 – 2)(3.535687 – 1.523437 – 2).$$

Subtracting gives

$$f(1.515625)f(1.523437) = (–0.034053)(0.012250).$$

Multiplying we obtain

$$f(1.515625)f(1.523437) = –0.000417.$$

Since $$f(1.515625)f(1.523437) < 0,$$ the new interval is

$$[1.515625, 1.523437].$$

Since both endpoints, rounded to 2 decimal places, are 1.52, all points within the interval round to 1.52. Hence, the final approximated root is

$$x = 1.52.$$

The next example illustrates one of the primary applications of numerical analysis, namely solving equations that cannot be solved analytically.

Example 2: Approximate the root of \( \cos(x) – x = 0 \), rounded to 2 decimal places.

Solution: Let

$$f(x) = \cos(x) – x.$$

Consider

$$f(0)f(1) = (\cos(0) – 0)(\cos(1) – 1).$$

Evaluating cosine, we find

$$f(0)f(1) = (1 – 0)(0.540302 – 1).$$

Subtracting gives

$$f(0)f(1) = (1)(-0.459697).$$

Multiplying we obtain

$$f(0)f(1) = –0.459697.$$

Since \( f(0)f(1) < 0 \), a root exists in

$$[0, 1].$$

The midpoint is given by

$$c = \frac{0 + 1}{2}.$$

Adding we get

$$c = \frac{1}{2}.$$

Dividing we obtain

$$c = 0.5.$$

Next, we compute

$$f(0.5)f(1) = (\cos(0.5) – 0.5)(\cos(1) – 1).$$

Evaluating cosine, we find

$$f(0.5)f(1) = (0.877582 – 0.5)(0.540302 – 1).$$

Subtracting gives

$$f(0.5)f(1) = (0.377582)(-0.459697).$$

Multiplying we obtain

$$f(0.5)f(1) = –0.173573.$$

Since \( f(0.5)f(1) < 0 \), the new interval is

$$[0.5, 1].$$

The midpoint is given by

$$c = \frac{0.5 + 1}{2}.$$

Adding we get

$$c = \frac{1.5}{2}.$$

Dividing we obtain

$$c = 0.75.$$

Next, we compute

$$f(0.5)f(0.75) = (\cos(0.5) – 0.5)(\cos(0.75) – 0.75).$$

Evaluating cosine, we find

$$f(0.5)f(0.75) = (0.877582 – 0.5)(0.731688 – 0.75).$$

Subtracting gives

$$f(0.5)f(0.75) = (0.377582)(-0.018311).$$

Multiplying we obtain

$$f(0.5)f(0.75) = –0.006913.$$

Since \( f(0.5)f(0.75) < 0 \), the new interval is

$$[0.5, 0.75].$$

The midpoint is given by

$$c = \frac{0.5 + 0.75}{2}.$$

Adding we get

$$c = \frac{1.25}{2}.$$

Dividing we obtain

$$c = 0.625.$$

Next, we compute

$$f(0.625)f(0.75) = (\cos(0.625) – 0.625)(\cos(0.75) – 0.75).$$

Evaluating cosine, we find

$$f(0.625)f(0.75) = (0.810963 – 0.625)(0.731688 – 0.75).$$

Subtracting gives

$$f(0.625)f(0.75) = (0.185963)(-0.018311).$$

Multiplying we obtain

$$f(0.625)f(0.75) = –0.003405.$$

Since \( f(0.625)f(0.75) < 0 \), the new interval is

$$[0.625, 0.75].$$

The midpoint is given by

$$c = \frac{0.625 + 0.75}{2}.$$

Adding we get

$$c = \frac{1.375}{2}.$$

Dividing we obtain

$$c = 0.6875.$$

Next, we compute

$$f(0.6875)f(0.75) = (\cos(0.6875) – 0.6875)(\cos(0.75) – 0.75).$$

Evaluating cosine, we find

$$f(0.6875)f(0.75) = (0.772834 – 0.6875)(0.731688 – 0.75).$$

Subtracting gives

$$f(0.6875)f(0.75) = (0.085334)(-0.018311).$$

Multiplying we obtain

$$f(0.6875)f(0.75) = –0.001562.$$

Since \( f(0.6875)f(0.75) < 0 \), the new interval is

$$[0.6875, 0.75].$$

The midpoint is given by

$$c = \frac{0.6875 + 0.75}{2}.$$

Adding we get

$$c = \frac{1.4375}{2}.$$

Dividing we obtain

$$c = 0.71875.$$

Next, we compute

$$f(0.71875)f(0.75) = (\cos(0.71875) – 0.71875)(\cos(0.75) – 0.75).$$

Evaluating cosine, we find

$$f(0.71875)f(0.75) = (0.752629 – 0.71875)(0.731688 – 0.75).$$

Subtracting gives

$$f(0.71875)f(0.75) = (0.033879)(-0.018311).$$

Multiplying we obtain

$$f(0.71875)f(0.75) = –0.000620.$$

Since \( f(0.71875)f(0.75) < 0 \), the new interval is

$$[0.71875, 0.75].$$

The midpoint is given by

$$c = \frac{0.71875 + 0.75}{2}.$$

Adding we get

$$c = \frac{1.46875}{2}.$$

Dividing we obtain

$$c = 0.734375.$$

Next, we compute

$$f(0.734375)f(0.75) = (\cos(0.734375) – 0.734375)(\cos(0.75) – 0.75).$$

Evaluating cosine, we find

$$f(0.734375)f(0.75) = (0.742249 – 0.734375)(0.731688 – 0.75).$$

Subtracting gives

$$f(0.734375)f(0.75) = (0.007874)(-0.018311).$$

Multiplying we obtain

$$f(0.734375)f(0.75) = –0.000144.$$

Since \( f(0.734375)f(0.75) < 0 \), the new interval is

$$[0.734375, 0.75].$$

The midpoint is given by

$$c = \frac{0.734375 + 0.75}{2}.$$

Adding we get

$$c = \frac{1.484375}{2}.$$

Dividing we obtain

$$c = 0.742187.$$

Next, we compute

$$f(0.734375)f(0. 0.742187) = (\cos(0.734375) – 0.734375)(\cos(0.742187) – 0.742187).$$

Evaluating cosine, we find

$$f(0.734375)f(0.742187) = (0.742249 – 0.734375)(0.736991 – 0.742187).$$

Subtracting gives

$$f(0.734375)f(0.742187) = (0.007874)(-0.005195).$$

Multiplying we obtain

$$f(0.734375)f(0.742187) = –0.000040.$$

Since $$f(0.734375)f(0.742187) < 0,$$ the new interval is

$$[0.734375, 0.742187].$$

The midpoint is given by

$$c = \frac{0.734375 + 0.742187}{2}.$$

Adding we get

$$c = \frac{1.4765625}{2}.$$

Dividing we obtain

$$c = 0.738281.$$

Next, we compute

$$f(0.738281)f(0. 0.742187) = (\cos(0.738281) – 0.738281)(\cos(0.742187) – 0.742187).$$

Evaluating cosine, we find

$$f(0.738281)f(0.742187) = (0.739626 – 0.738281)(0.736991 – 0.742187).$$

Subtracting gives

$$f(0.738281)f(0.742187) = (0.001345)(-0.005195).$$

Multiplying we obtain

$$f(0.738281)f(0.742187) = –0.000006.$$

Since $$f(0.738281)f(0.742187) < 0,$$ the new interval is

$$[0.738281, 0.742187].$$

Since both endpoints, rounded to 2 decimal places, are 0.74, all points within the interval round to 0.74. Hence, the final approximated root is

$$x = 0.74.$$

Conclusion

In this guide, we have explored the theory and implementation of the bisection method. We began by introducing the basic idea of interval halving, examined the mathematical foundation provided by the Intermediate Value Theorem, and explained why the method is guaranteed to converge under appropriate conditions. We also discussed the situations in which the bisection method succeeds and fails, derived its error bounds, and presented a procedure for implementing the algorithm.

Throughout this guide on the bisection method with worked examples, we have seen how the method can be used to approximate the roots of both polynomial and transcendental equations. However, the bisection method converges more slowly than other algorithms; its simplicity, reliability, and guaranteed convergence make it a valuable tool.

Further Reading

Secant and Tangent Lines in Calculus: A Complete Beginner’s Guide – Other commonly used root-finding methods include the Secant Method and Newton’s Method. Both of these methods make use of secant and tangent lines.

Frequently Asked Questions