The Ultimate Resource for Understanding the Secant Method
Understanding the secant method is an important step for anyone studying numerical analysis. By replacing the tangent line required in Newton’s method with a secant line through two nearby points, the secant method offers an elegant approach to finding roots without requiring derivative calculations. Because of this, it has become a standard topic in numerical analysis courses and is frequently used in applications.
In this guide, you’ll build an understanding of the secant method. We’ll begin by introducing and deriving the secant method. From there, we’ll explore why the method works, examine the conditions under which it succeeds or fails, and discuss its convergence. You’ll also learn how to compute approximation errors, choose appropriate stopping criteria, and follow a step-by-step algorithm for implementing the method. To reinforce these concepts, we’ll work through several detailed examples.
By the end of this article, you’ll not only know how to apply the secant method to solve equations but also understand the mathematical principles that make it possible.
What Is the Secant Method?
The secant method is an iterative numerical algorithm used to approximate the roots of an equation of the form
$$f(x) = 0.$$
The idea behind the secant method comes directly from the concept of a secant line in calculus. If you’ve read our article Secant and Tangent Lines in Calculus: A Complete Beginner’s Guide, you’ll recall that a secant line passes through two distinct points on the graph of a function.
As shown in that article, a secant line often provides a good approximation to the curve over a small interval. The secant method takes advantage of this observation by repeatedly drawing secant lines through two current approximations of a root. Instead of stopping after constructing the secant line, however, it computes the \( x \)-intercept of that line and uses the intercept as the next approximation. By repeating this process, the approximations often move closer to the actual root.
In contrast, Newton’s method uses a tangent line instead of a secant line. As discussed in our article Newton’s Method Explained for Beginners, Newton’s method constructs the tangent line at a single point and uses its \( x \)-intercept to generate the next approximation. Because the slope of the tangent line depends on the derivative \( f'(x)\ \), Newton’s method requires derivative calculations at every iteration.
The secant method avoids this requirement by approximating the derivative with the slope of a secant line through two nearby points. Although this approximation makes the secant method converge more slowly than Newton’s method, it often performs well while requiring less information about the function.
Over successive iterations, the secant lines generally become better approximations of the curve near the root, causing the estimated solution to improve. Under favorable conditions, this sequence converges rapidly to the desired root, making the secant method an effective algorithm for solving equations without derivatives.
In the next section, we’ll derive the secant method formula from the equation of a secant line.
The Secant Method Formula
The power of the secant method lies in its iteration formula. Starting with two initial approximations of a root, the formula computes a new approximation by finding the \( x \)-intercept of the secant line passing through the corresponding points on the graph of the function. Repeating this process produces a sequence of approximations that, under favorable conditions, converges to the desired root.
Like the iteration formula used in How to Solve Equations Using Fixed Point Iteration: A Step-by-Step Guide, the secant method repeatedly generates better approximations by applying the same function. Rather than evaluating a transformed function, the secant method constructs a new secant line at each iteration and uses its \( x \)-intercept as the next approximation.
Deriving the Formula
Suppose we have two current approximations of the root, \( x_{n – 1} \) and \( x_n \). These approximations correspond to the two points
$$(x_{n – 1}, f(x_{n – 1})),$$
and
$$(x_n, f(x_n))$$
on the graph of \( y = f(x) \).
The slope of the secant line through these points is given by
$$m = \frac{f(x_n) – f(x_{n – 1})}{x_n – x_{n – 1}}.$$
Using the point-slope form of a line, the equation of the secant line becomes
$$y – f(x_n) = \frac{f(x_n) – f(x_{n – 1})}{x_n – x_{n-1}}(x-x_n).$$
Since we want the point where the secant line crosses the \( x \)-axis, we let \( y = 0 \):
$$-f(x_n) = \frac{f(x_n) – f(x_{n – 1})}{x_n – x_{n – 1}}(x – x_n).$$
Multiplying both sides by
$$\frac{x_n – x_{n – 1}}{f(x_n) – f(x_{n – 1})}$$
gives
$$x – x_n = -\frac{f(x_n)(x_n – x_{n – 1})}{f(x_n) – f(x_{n – 1})}.$$
Finally, solving for \( x \) yields the next approximation, which we denote by \( x_{n + 1} \):
$$x_{n + 1} = x_n – \frac{f(x_n)(x_n – x_{n – 1})}{f(x_n) – f(x_{n – 1})}.$$
This equation is known as the secant method iteration formula.
The Secant Method Formula
The iteration formula is
$$x_{n + 1} = x_n – \frac{f(x_n)(x_n – x_{n – 1})}{f(x_n) – f(x_{n – 1})}.$$
where \( x_n \) is the current approximation of the root, \( x_{n – 1} \) is the previous approximation, \( f(x_n) \) is the value of the function at the current approximation, \( f(x_{n -1}) \) is the value of the function at the previous approximation, and \( x_{n + 1} \) is the new approximation.
Beginning with two initial guesses, \( x_0 \) and \( x_1 \), this formula is applied repeatedly to generate the sequence
$$\{x_0, x_1, x_2, \ldots\}$$
until the approximations satisfy a chosen stopping criterion.
Now that we have derived the secant method formula, the next step is to understand why this iterative process converges toward a root and under what conditions it produces accurate approximations.
Why the Secant Method Works
To understand why the secant method works, it helps to examine both the geometric intuition and the mathematical properties of the algorithm.
The Geometric Intuition
Suppose we want to solve the equation
$$f(x) = 0.$$
Finding a root of this equation is equivalent to finding the point where the graph of \( y = f(x) \) crosses the \( x \)-axis.
Assume we have two approximations of the root, \( x_{n – 1} \) and \( x_n \). These correspond to the points
$$(x_{n-1}, f(x_{n-1}))$$
and
$$(x_n,f(x_n))$$
on the graph of the function.
Rather than trying to determine the exact shape of the curve between these points, the secant method approximates that portion of the graph with the straight line joining them. Because a straight line is much simpler to analyze than a curved function, we can compute where the secant line crosses the \( x \)-axis. This intersection becomes the next approximation of the root.
If the function is smooth and the two approximations lie close to the true solution, the secant line closely resembles the curve over that small interval. As a result, its \( x \)-intercept is usually a better approximation than either of the previous estimates.
Each new approximation produces another secant line that more accurately reflects the behavior of the function near the root. As the iterations continue, the secant lines generally become better approximations of the curve, causing the estimated root to move closer to the true solution.
Why the Approximations Improve
Each iteration uses the two most recent approximations to construct a new secant line.
If the previous approximations are close to the root, then the secant line becomes a better approximation of the curve, its \( x \)-intercept moves closer to the actual root, the next secant line becomes even more accurate, and the process repeats.
This means improved approximations generate increasingly accurate secant lines, which in turn produce even better approximations.
Under suitable conditions, the error decreases rapidly from one iteration to the next.
Superlinear Convergence
One of the major reasons the secant method is so popular is its convergence rate.
The bisection method reduces the error at a linear rate because each iteration simply halves the search interval.
Newton’s method converges quadratically, meaning that once the approximation is sufficiently close to the root, the number of correct digits roughly doubles with each iteration.
The secant method falls between these two algorithms. It converges superlinearly, with an order of convergence approximately equal to
$$\frac{1 + \sqrt{5}}{2} \approx 1.618,$$
which is the golden ratio.
Although this convergence rate is slower than quadratic convergence, it is considerably faster than the linear convergence of the bisection method. Since the secant method also avoids computing derivatives, it often provides a balance between computational cost and convergence speed.
Why Smooth Functions Help
The secant method performs best when the function is smooth and well-behaved near the root.
For smooth functions, nearby points have similar slopes; the secant line closely approximates the curve, successive approximations remain stable, and convergence is rapid.
Conversely, if the function has discontinuities, sharp corners, multiple roots, or rapidly changing slopes, the secant line may provide a poor approximation of the curve. In these situations, the algorithm may converge slowly, oscillate, or fail to converge altogether.
We’ll examine these situations in detail in the next couple of sections.
When the Secant Method Succeeds
Although the secant method is not guaranteed to converge for every equation, it performs exceptionally well when certain conditions are satisfied. Under these circumstances, the algorithm often converges rapidly, producing highly accurate approximations with few iterations. Understanding these conditions can help you decide when the secant method is an appropriate choice and when another root-finding algorithm may be more reliable.
Smooth, Continuous Functions
The secant method works best when the function is continuous and differentiable in a neighborhood containing the root. For a review of continuous functions, please refer to the article A Complete Introduction to the Bisection Method with Worked Examples and Solutions.
For smooth functions, the graph changes gradually, nearby points have similar slopes, the secant line closely approximates the curve, and successive approximations tend to move steadily toward the root.
Because the method relies on straight-line approximations, abrupt changes in the shape of the function reduce its effectiveness.
Good Initial Guesses
Perhaps the most important factor affecting convergence is the choice of the initial approximations.
If the starting values \( x_0 \) and \( x_1 \) are reasonably close to the desired root, the secant line is more likely to provide an accurate approximation of the curve. Consequently, the next approximation is often even closer to the solution, allowing the method to converge.
Poor initial guesses, on the other hand, can produce secant lines that point away from the root or lead the iteration into regions where the function behaves poorly.
Simple Roots
The secant method performs particularly well when the equation has a simple root, that is, a root \( r \) satisfying
$$f(r) = 0$$
and
$$f'(r) \neq 0.$$
Near a simple root, the graph crosses the \( x \)-axis with a nonzero slope, making the secant line a good approximation of the local behavior of the function.
A Nonzero Derivative Near the Root
Even though the secant method does not explicitly compute derivatives, it still benefits when the derivative near the root is nonzero.
If the graph is nearly horizontal near the solution, the secant line may also become nearly horizontal. In this situation, small changes in the function values can produce very large changes in the computed approximation, making convergence slower or less stable.
When the derivative remains nonzero, the secant lines generally intersect the \( x \)-axis in a predictable manner, leading to steady improvement in the approximations.
For a review of derivatives, please refer to the article How to Differentiate a Function Step by Step: A Beginner’s Guide.
When these conditions are satisfied, the secant method often converges rapidly, making it one of the most used algorithms for solving equations. In the next section, we’ll examine situations in which these assumptions break down and explore why the secant method may converge slowly, oscillate, or fail altogether.
When the Secant Method Fails
Although the secant method is an efficient root-finding algorithm, it is not guaranteed to converge. The secant method can behave unpredictably when the function or the initial approximations are unfavorable. In some situations, it converges slowly; in others, it may oscillate indefinitely or diverge completely.
Understanding these reasons for failure is just as important as understanding why the method succeeds. Recognizing them allows you to choose better initial approximations, identify when the algorithm is struggling, and determine when a different numerical method would be more appropriate.
Poor Initial Guesses
The most common reason the secant method fails is that the initial approximations are poorly chosen.
Because the secant line determines each new approximation through the previous two approximations, the direction of the secant line depends entirely on the starting values. If \( x_0 \) and \( x_1 \) are far from the desired root, the secant line may point in the wrong direction, producing approximations that move farther away instead of closer.
Whenever possible, choosing initial approximations near the expected solution greatly improves the likelihood of convergence.
Division by Zero
The secant method iteration formula is
$$x_{n + 1} = x_n – \frac{f(x_n)(x_n – x_{n – 1})}{f(x_n) – f(x_{n – 1})}.$$
Notice that the denominator is
$$f(x_n) – f(x_{n – 1}).$$
If
$$f(x_n) = f(x_{n – 1}),$$
the denominator becomes zero, making the next approximation undefined.
Even when the denominator is not exactly zero, it may become small. In this case, the computed approximation can be very large, causing the algorithm to jump far from the desired root and potentially fail to converge.
Nearly Horizontal Secant Lines
Another difficulty occurs when the secant line is nearly horizontal.
If the function values satisfy
$$f(x_n) \approx f(x_{n – 1}),$$
the slope of the secant line is very small.
A nearly horizontal line intersects the \( x \)-axis far from the current approximations, producing a large correction.
This behavior is especially common near turning points of the function where the slope is close to zero.
Multiple Roots
The secant method generally performs best when the equation has a simple root satisfying
$$f(r) = 0$$
and
$$f'(r) \neq 0.$$
When the equation has a multiple root, the graph can touch the \( x \)-axis without crossing it.
Oscillation
Sometimes the approximations alternate between different regions of the graph without showing consistent improvement. This often results in a lack of convergence.
Divergence
In some cases, the approximations simply move farther away from the desired solution. This is known as divergence.
Rather than decreasing, the error increases with each iteration.
Discontinuous Functions
If the function contains discontinuities, jump discontinuities, removable discontinuities, or vertical asymptotes, the secant line may bear little resemblance to the actual function.
In these situations, the computed \( x \)-intercept can be misleading.
Consequently, the secant method should generally be applied only to continuous functions.
The secant method can fail for several reasons, including poor initial approximations, division by zero, nearly horizontal secant lines, multiple roots, oscillation, divergence, and discontinuous functions. Most of these difficulties arise because the method relies on local linear approximations and does not guarantee that the approximations remain close to the desired root.
Despite these limitations, the secant method remains one of the most effective root-finding algorithms when applied under appropriate conditions. By understanding the situations in which it can fail, you can recognize convergence problems early and decide whether to modify the algorithm or choose a different method.
Error Computation
In numerical analysis, obtaining an approximation to a root is only part of the problem. Equally important is determining how accurate that approximation is. Since the exact root is often unknown, we need practical ways to measure the quality of our approximations and decide when to stop iterating. Error computation provides this information.
Absolute Error
The absolute error measures the distance between the approximate root and the exact root. If the exact root is denoted by \( r \) and the approximation is \( x_n \), then the absolute error is
$$\epsilon = |r – x_n|.$$
An absolute error of zero means the approximation is exact.
Although absolute error is one of the most natural measures of accuracy, it is rarely available in practice because the exact root is usually unknown. Consequently, it is primarily used to analyze equations whose exact solutions are already known.
Relative Error
Sometimes the magnitude of the root itself should be considered. The relative error is defined by
$$\epsilon = \frac{|r – x_n|}{|r|},$$
Relative error measures the size of the error compared to the size of the exact solution, making it particularly useful when comparing approximations of quantities with different magnitudes.
Like absolute error, relative error requires knowledge of the exact root and is therefore rarely used in practice.
Approximate Relative Error
Since the exact root is generally unknown, the secant method usually estimates the error by comparing consecutive approximations.
The approximate relative error is given by
$$\epsilon = |\frac{x_n – x_{n – 1}}{x_n}|.$$
When expressed as a percentage, the formula becomes
$$\epsilon = |\frac{x_n – x_{n-1}}{x_n}| \times 100\%.$$
This is one of the most commonly used stopping criteria because it requires only the approximations generated during the iteration process.
Choosing a Tolerance
Iterations generally stop once the error becomes smaller than a predetermined tolerance. The appropriate tolerance depends on the problem being solved. For example, an implementation may terminate when the approximate relative error is sufficiently small or the maximum number of iterations is reached.
Step-by-Step Procedure
Now that we’ve explored the theory behind the secant method, it’s time to put the algorithm into practice. The secant method follows a well-defined sequence of steps.
Step 1: Choose Two Initial Approximations
Begin by selecting two initial approximations,
$$x_0$$
and
$$x_1.$$
Choosing values that are reasonably close to the desired solution generally improves the likelihood of rapid convergence.
Step 2: Evaluate the Function
Compute the function values corresponding to the two initial approximations
$$f(x_0)$$
and
$$f(x_1).$$
Step 3: Compute the Next Approximation
Use the secant method iteration formula
$$x_{n + 1} = x_n – \frac{f(x_n)(x_n – x_{n – 1})}{f(x_n) – f(x_{n – 1})}$$
to calculate the next approximation.
Step 4: Check the Stopping Criterion
If the stopping criterion is satisfied, the algorithm terminates, and \( x_{n + 1} \) is accepted as the approximate root.
Step 5: Update the Approximations
If the stopping criterion has not been satisfied, replace the previous approximations according to
$$x_n \rightarrow x_{n – 1},$$
$$x_{n + 1} \rightarrow x_n.$$
Step 6: Repeat the Process
Evaluate the function at the new approximation and repeat Steps 3 through 6 until one of the stopping criteria is met or the maximum number of iterations has been reached.
Worked Out Examples
The first example shows how to use the secant method.
Example 1: Using initial approximations of \( x_0 = 1 \) and \( x_1 = 2 \), approximate the root of \( x^3 – x – 2 = 0 \). Use 6 iterations of the secant method.
Solution: Consider the function
$$f(x) = x^3 – x – 2.$$
Evaluating the function at \( x_0 = 1 \) we obtain
$$f(1) = 1^3 – 1 – 2.$$
Cubing we find
$$f(1) = 1 – 1 – 2.$$
Subtracting gives
$$f(1) = -2.$$
Evaluating the function at \( x_1 = 2 \) we obtain
$$f(2) = 2^3 – 2 – 2.$$
Cubing we find
$$f(2) = 8 – 2 – 2.$$
Subtracting gives
$$f(2) = 4.$$
The next approximation is
$$x_2 = 2 – 4\frac{2 – 1}{4 – (-2)}.$$
Subtracting we find
$$x_2 = 2 – 4\frac{1}{6}.$$
Dividing we get
$$x_2 = 2 – 4(0.166666).$$
Multiplying, this is
$$x_2 = 2 – 0.666666.$$
Subtracting gives
$$x_2 = 1.333333.$$
Evaluating the function at \( x_2 = 1.333333 \) we obtain
$$f(1.333333) = 1.333333^3 – 1.333333 – 2.$$
Cubing we find
$$f(1.333333) = 2.370370 – 1.333333 – 2.$$
Subtracting gives
$$f(1.333333) = -0.962963.$$
The next approximation is
$$x_3 = 1.333333 – (-0.962963)\frac{1.333333 – 2}{-0.962963 – 4}.$$
Subtracting we find
$$x_3 = 1.333333 – (-0.962963)\frac{-0.666666}{-4.962963}.$$
Dividing we get
$$x_3 = 1.333333 – (-0.962963)(0.134328).$$
Multiplying this is
$$x_3 = 1.333333 – (-0.129353).$$
Subtracting gives
$$x_3 = 1.462686.$$
Evaluating the function at \( x_3 = 1.462686 \) we obtain
$$f(1.462686) = 1.462686^3 – 1.462686 – 2.$$
Cubing we find
$$f(1.462686) = 3.129347 – 1.462686 – 2.$$
Subtracting gives
$$f(1.462686) = -0.333338.$$
The next approximation is
$$x_4 = 1.462686 – (-0.333338)\frac{1.462686 – 1.333333}{-0.333338 – (-0.962963)}.$$
Subtracting we find
$$x_4 = 1.462686 – (-0.333338)\frac{0.129353}{0.629624}.$$
Dividing we get
$$x_4 = 1.462686 – (-0.333338)(0.205445).$$
Multiplying this is
$$x_4 = 1.462686 – (-0.068482).$$
Subtracting gives
$$x_4 = 1.531169.$$
Evaluating the function at \( x_4 = 1.531169 \) we obtain
$$f(1.531169) = 1.531169^3 – 1.531169 – 2.$$
Cubing we find
$$f(1.531169) = 3.589795 – 1.531169 – 2.$$
Subtracting gives
$$f(1.531169) = 0.058626.$$
The next approximation is
$$x_5 = 1.531169 – 0.058626\frac{1.531169 – 1.462686}{0.058626 – (-0.333338)}.$$
Subtracting we find
$$x_5 = 1.531169 – 0.058626\frac{0.068482}{0.391965}.$$
Dividing we get
$$x_5 = 1.531169 – 0.058626(0.174716).$$
Multiplying this is
$$x_5 = 1.531169 – 0.010243.$$
Subtracting gives
$$x_5 = 1.520926.$$
Evaluating the function at \( x_5 = 1.520926 \) we obtain
$$f(1.520926) = 1.520926^3 – 1.520926 – 2.$$
Cubing we find
$$f(1.520926) = 3.518233 – 1.520926 – 2.$$
Subtracting gives
$$f(1.520926) = -0.002693.$$
The next approximation is
$$x_6 = 1.520926 – (-0.002693)\frac{1.520926 – 1.531169}{-0.002693 – 0.058626}.$$
Subtracting we find
$$x_6 = 1.520926 – (-0.002693)\frac{-0.010243}{-0.061319}.$$
Dividing we get
$$x_6 = 1.520926 – (-0.002693)(0.167042).$$
Multiplying this is
$$x_6 = 1.520926 – (-0.000449).$$
Subtracting gives
$$x_6 = 1.521376.$$
Evaluating the function at \( x_6 = 1.521376 \) we obtain
$$f(1.521376) = 1.521376^3 – 1.521376 – 2.$$
Cubing we find
$$f(1.521376) = 3.521356 – 1.521376 – 2.$$
Subtracting gives
$$f(1.521376) = -0.000020.$$
The next approximation is
$$x_7 = 1.521376 – (-0.000020)\frac{1.521376 – 1.520926}{-0.000020 – (-0.002693)}.$$
Subtracting we find
$$x_7 = 1.521376 – (-0.000020)\frac{0.000449}{0.002673}.$$
Dividing we get
$$x_7 = 1.521376 – (-0.000020)(0.168301).$$
Multiplying this is
$$x_7 = 1.521376 – (-0.000003).$$
Subtracting gives a final answer of
$$x_7 = 1.521379.$$
The next example illustrates one of the primary applications of numerical analysis, namely solving equations that cannot be solved analytically.
Example 2: Using initial approximations of \( x_0 = 0 \) and \( x_1 = 1 \), approximate the root of \( \cos x = x \), Use 4 iterations of the secant method.
Solution: Subtracting x from both sides, we get
$$\cos x – x = 0.$$
Consider the function
$$f(x) = \cos x – x.$$
Evaluating the function at \( x_0 = 0 \) we obtain
$$f(0) = \cos 0 – 0.$$
Evaluating cosine, we find
$$f(0) = 1 – 0.$$
Subtracting gives
$$f(0) = 1.$$
Evaluating the function at \( x_1 = 1 \) we obtain
$$f(1) = \cos 1 – 1.$$
Evaluating cosine, we find
$$f(1) = 0.540302 – 1.$$
Subtracting gives
$$f(1) = -0.459697.$$
The next approximation is
$$x_2 = 1 – (-0.459697)\frac{1 – 0}{-0.459697 – 1}.$$
Subtracting we find
$$x_2 = 1 – (-0.459697)\frac{1}{-1.459697}.$$
Dividing we get
$$x_2 = 1 – (-0.459697)(-0.685073).$$
Multiplying this is
$$x_2 = 1 – 0.314926.$$
Subtracting gives
$$x_2 = 0.685073.$$
Evaluating the function at \( x_2 = 0.685073 \) we obtain
$$f(0.685073) = \cos 0.685073 – 0.685073.$$
Evaluating cosine, we find
$$f(0.685073) = 0.774372 – 0.685073.$$
Subtracting gives
$$f(0.685073) = 0.089299.$$
The next approximation is
$$x_3 = 0.685073 – 0.089299\frac{0.685073 – 1}{0.089299 – (-0.459697)}.$$
Subtracting we find
$$x_3 = 0.685073 – 0.089299\frac{-0.314926}{0.548997}.$$
Dividing we get
$$x_3 = 0.685073 – 0.089299(-0.573640).$$
Multiplying this is
$$x_3 = 0.685073 – (-0.051225).$$
Subtracting gives
$$x_3 = 0.736299.$$
Evaluating the function at \( x_3 = 0.736299 \) we obtain
$$f(0.736299) = \cos 0.736299 – 0.736299.$$
Evaluating cosine, we find
$$f(0.736299) = 0.740959 – 0.736299.$$
Subtracting gives
$$f(0.736299) = 0.004660.$$
The next approximation is
$$x_4 = 0.736299 – 0.004660\frac{0.736299 – 0.685073}{0.004660 – 0.089299}.$$
Subtracting we find
$$x_4 = 0.736299 – 0.004660\frac{0.051225}{-0.084639}.$$
Dividing we get
$$x_4 = 0.736299 – 0.004660(-0.605223).$$
Multiplying this is
$$x_4 = 0.736299 – (-0.002820).$$
Subtracting gives
$$x_4 = 0.739119.$$
Evaluating the function at \( x_4 = 0.739119 \) we obtain
$$f(0.739119) = \cos 0.739119 – 0.739119.$$
Evaluating cosine, we find
$$f(0.739119) = 0.739062 – 0.739119.$$
Subtracting gives
$$f(0.739119) = -0.000057.$$
The next approximation is
$$x_5 = 0.739119 – (-0.000057)\frac{0.739119 – 0.736299}{-0.000057 – 0.004660}.$$
Subtracting we find
$$x_5 = 0.739119 – (-0.000057)\frac{0.002820}{-0.004717}.$$
Dividing we get
$$x_5 = 0.739119 – (-0.000057)(-0.597873).$$
Multiplying this is
$$x_5 = 0.739119 – 0.000034.$$
Subtracting gives a final answer of
$$x_5 = 0.739085.$$
Conclusion
Understanding the secant method is an important step in learning numerical analysis. By repeatedly replacing a nonlinear function with a secant line and using its x-intercept as the next approximation, the secant method provides an efficient way to approximate roots.
In this guide, we’ve explored the secant method from both theoretical and practical perspectives. We introduced the motivation behind the algorithm, derived the secant method formula, explained why the method converges under favorable conditions, and examined the situations in which it may fail. We also discussed common error measures and a step-by-step procedure for implementing the algorithm. Finally, through worked examples involving both polynomial and transcendental equations, we saw how the secant method produces increasingly accurate approximations with only a handful of iterations.
Further Reading
How to Find Intervals of Increase and Decrease: A Step-by-Step Guide – Earlier, we mentioned that the secant method fails when the derivative is zero at or near the root. This guide explains how functions behave when the derivative is zero.
