The convergence of the Regula Falsi method

1 Introduction

Regula Falsi method appears in many undergraduate numerical analysis documents and books (e.g., [1, 2, 3, 4, 6, 7, 8]). In [1, 7], this method is also called the chord method. The goal of this method is to generate a sequence ${x_{n}}_{n = 0}^{\infty}$ that converges to a zero $x^{*}$ of $f$ when $n$ tends to infinity. Some provided the proofs under some assumptions, such as both the first and second derivatives of $f$ do not change sign on [ $a, b$ ] (e.g., [1, 3, 4, 7]). In this paper, we remove those assumptions and prove the convergence of the method for all continuous functions.

2 Main result

In this paper, the symbol $N$ denotes the set of natural numbers ${0, 1, 2, \dots}$ and $N^{*} = N ∖ {0}$ . Let $f : [a, b] \to R$ be a continuous function on the interval $[a, b]$ and $f (a) f (b) < 0$ . We construct a sequence ${x_{n}}_{n = 0}^{\infty}$ as follows.

Set

u_{1} := a

v_{1} := b

if $f (x_{1}) f (a) > 0$ then

x_{0} := b

else

x_{0} := a

for $n = 1, 2, 3, \dots$ do

x_{n} := \frac{u_{n} f (v_{n}) - v_{n} f (u_{n})}{f (v_{n}) - f (u_{n})} .

(1)

if $f (x_{n}) f (u_{n}) > 0$ then

u_{n + 1} := x_{n},

v_{n + 1} := v_{n}

else

u_{n + 1} := u_{n}

v_{n + 1} := x_{n}

Algorithm 1 Regula Falsi

The main aim of this paper is to prove the following result.

Theorem 1.

For any continuous function $f$ on a closed, bounded interval $[a, b]$ that satisfies $f (a) f (b) < 0$ , the sequence ${x_{n}}_{n = 0}^{\infty}$ generated by Algorithm 1 converges to a zero $x^{*}$ of $f$ , i.e., $f (x^{*}) = 0$ .

The proof of this theorem is lengthy, and hence, it will be divided into several parts. Clearly, if $f (x_{n_{0}}) = 0$ for some $n_{0} \in N$ , then simple induction implies that $x_{n} = x_{n_{0}}$ for all $n \geq n_{0}$ and the desired convergence is obvious. Now, we consider the case where $f (x_{n}) \neq 0$ for all $n$ .

Lemma 1.

Considering the sequence ${r_{n}}_{n = 1}^{\infty}$ defined by

r_{n} := \frac{v_{n + 1} - u_{n + 1}}{v_{n} - u_{n}} .

(2)

Then for each $n$ , we have that:

r_{n} = ⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ \begin{matrix} \frac{1}{1 + | \frac{f (u_{n})}{f (v_{n})} |} if u_{n + 1} = x_{n} and v_{n + 1} = v_{n}, \frac{1}{1 + | \frac{f (v_{n})}{f (u_{n})} |} if u_{n + 1} = u_{n} and % v_{n + 1} = x_{n} . \end{matrix}

(3)

Proof.

Without loss of generality, we consider only the case where $u_{n + 1} = u_{n}$ and $v_{n + 1} = x_{n}$ . Thus, we have

v_{n + 1} - u_{n + 1} = x_{n} - u_{n} = \frac{u_{n} f (v_{n}) - v_{n} f (u_{n})}{f (v_{n}) - f (u_{n})} - u_{n} = \frac{(u_{n} - v_{n}) f (u_{n})}{f (v_{n}) - f (u_{n})} .

It follows by $(???)$ that

r_{n} = \frac{- f (u_{n})}{f (v_{n}) - f (u_{n})} = \frac{1}{1 - \frac{f (v_{n})}{f (u_{n})}} = \frac{1}{1 + | \frac{f (v_{n})}{f (u_{n})} |},

which is (3) exactly.
Analoguously, in the second case, where $u_{n + 1} = x_{n}$ and $v_{n + 1} = v_{n}$ , we have that

r_{n} = \frac{f (v_{n})}{f (v_{n}) - f (u_{n})} = \frac{1}{1 - \frac{f (u_{n})}{f (v_{n})}} = \frac{1}{1 + | \frac{f (u_{n})}{f (v_{n})} |} .

This completes the proof of Lemma 1. ∎

Lemma 2.

For each fixed $n_{0} \in N^{*}$ , the sequence ${x_{n}}_{n \geq n_{0}}$ is either upper or lower bounded by $x_{n_{0}}$ .

Proof.

Firstly, we notice the decreasing property of the sequence ${[u_{n}, v_{n}]}_{n = 1}^{\infty}$ , i.e.

[u_{1}, v_{1}] ⊋ [u_{2}, v_{2}] ⊋ \dots ⊋ [u_{n}, v_{n}] ⊋ \dots

In addition, since for any $n$ the point $x_{n}$ is either $u_{n + 1}$ or $v_{n + 1}$ , the desired claim follows directly. ∎

Due to Lemma 1, we have that $0 < r_{n} < 1 for all n \in N^{*} .$ Let $r = sup n r_{n}$ , we see that $0 < r \leq 1$ . Now we discuss two cases whether $r$ is strictly smaller than 1 or not in the following proposition.

Proposition 1.

Consider the sequences ${x_{n}}_{n = 0}^{\infty}$ generated by Algorithm 1 and ${r_{n}}_{n = 0}^{\infty}$ defined by (2), (3). Then the following claims hold true.
i) If $sup n r_{n} < 1$ then the sequence ${x_{n}}_{n = 0}^{\infty}$ converges to a zero of $f$ .
ii) If $sup n r_{n} = 1$ then there exists a subsequence ${x_{k_{n}}}_{n = 0}^{\infty}$ converges to a zero of $f$ .

Proof.

i) If $r < 1$ then $v_{n + 1} - u_{n + 1} \leq r (v_{n} - u_{n}) \leq \dots \leq r^{n} (v_{1} - u_{1}) \to 0$ when $n$ tends to infinity. This implies that ${[u_{n}, v_{n}]}_{n = 1}^{\infty}$ is a decreasing nested sequence of non-empty, closed and bounded intervals in $R$ and their lengths strictly decrease to $0$ . So, due to Cantor intersection theorem, see e.g. [5], there exists $x^{*} \in [a, b]$ such that ${x^{*}} = \cap_{n} [u_{n}, v_{n}]$ . Evidently, $lim n \to \infty x_{n} = lim n \to \infty u_{n} = lim n \to \infty v_{n} = x^{*}$ . Since $f (u_{n}) f (v_{n}) < 0$ for all $n$ , due to the continuity of $f$ , let $n \to \infty$ then $f (x^{*}) f (x^{*}) \leq 0$ , which only holds if $f (x^{*}) = 0$ . Consequently, the sequence ${x_{n}}_{n = 0}^{\infty}$ generated by Algorithm 1 converges to a zero $x^{*}$ of $f$ .

ii) Now we consider the case where $r = 1$ . Firstly, we observe that by the construction of the sequence ${x_{n}}_{n = 0}^{\infty}$ , for each $n \in N^{*}$ , there exist $p, q \in N$ depending on $n$ such that $u_{n} = x_{p}$ and $v_{n} = x_{q}$ . Since $sup n r_{n} = 1$ , there exists a subsequence ${r_{m_{n}}}_{n = 1}^{\infty}$ of ${r_{n}}_{n = 1}^{\infty}$ that converges to 1. Let $M = max x \in [a, b] | f (x) |$ then the continuity of $f$ implies that $M < \infty$ . Due to Lemma 1 we have that

r_{m_{n}} < \frac{1}{1 + \frac{| f (x_{m_{n}}) |}{M}} < 1 for all n \in N^{*} .

(4)

Therefore, let $n$ tend to infinity we have $lim n \to \infty f (x_{m_{n}}) = 0$ . Again, Bolzano-Weierstrass theorem applied to the bounded sequence ${x_{m_{n}}}_{n = 1}^{\infty}$ implies the existence of a convergent subsequence ${x_{k_{n}}}_{n = 0}^{\infty}$ , whose limit is denoted by $x^{*}$ . The continuity of $f$ , therefore, implies that $f (x^{*}) = 0$ . ∎

Clearly, from Proposition 1, the remaining case is where $r = 1$ , and we need to prove the convergence of the whole sequence ${x_{n}}_{n = 0}^{\infty}$ instead of only the subsequence ${x_{k_{n}}}_{n = 0}^{\infty}$ . We will perform this task by disproof, i.e., the following assumption holds.

Assumption 1.

Suppose that the sequence ${x_{n}}_{n = 0}^{\infty}$ generated by Algorithm 1 diverges.

Under Assumption 1, there exists $ϵ > 0$ and a subsequence ${x_{i_{n}}}_{n = 0}^{\infty}$ of ${x_{n}}_{n = 0}^{\infty}$ such that

| x_{i_{n}} - x^{*} | > ϵ for all n \geq 0.

(5)

Since the sequence ${x_{k_{n}}}_{n = 0}^{\infty}$ converges to $x^{*}$ , there exists $n_{0} \in N$ such that

| x_{k_{n}} - x^{*} | < ϵ for all n \geq n_{0} .

(6)

By suitably removing a finite number of starting elements of two above sequences, one can assume that $n_{0} = 0$ and $i_{0} > k_{0}$ .

Lemma 3.

i) The point $x^{*}$ must lie between $x_{k_{0}}$ and $x_{i_{0}}$ .
ii) Two sequences ${x_{k_{n}}}_{n = 0}^{\infty}, {x_{i_{n}}}_{n = 0}^{\infty}$ are 2 monotone sequences in opposite directions.

Proof.

i) Due to (5) and (6), we first observe that $x_{i_{0}}$ could not lie between $x_{k_{0}}$ and $x^{*}$ . In case $x_{k_{0}}$ is between $x^{*}$ and $x_{i_{0}}$ , we suppose that $x^{*} < x_{k_{0}} < x_{i_{0}}$ , since similar arguments can be made for $x_{i_{0}} < x_{k_{0}} < x^{*}$ . Due to Lemma 2, $x_{n} > x_{k_{0}}$ for all $n > k_{0}$ . Thus, there does not exist any subsequence of ${x_{n}}_{n = 0}^{\infty}$ converging to $x^{*}$ , which contradicts the fact that $lim n \to \infty x_{k_{n}} = x^{*}$ . Therefore, $x^{*}$ must lie between $x_{k_{0}}$ and $x_{i_{0}}$ .
ii) Without loss of generality, we suppose that $x_{k_{0}} > x^{*} > x_{i_{0}}$ . The case where $x_{k_{0}} < x^{*} < x_{i_{0}}$ is on the analogy. Since $lim n \to \infty x_{k_{n}} = x^{*}$ , there exists a $k_{n_{1}}$ such that $k_{n_{1}} > i_{0} > k_{0}$ and $x_{k_{0}} > x_{k_{n}} > x_{i_{0}}$ . Due to Lemma 2 we obtain that ${x_{n}}_{n > i_{0}}$ is upper bounded by $x_{k_{0}}$ and lower bounded by $x_{i_{0}}$ . For each fixed $k_{n}$ , if there exists an $i_{m}$ such that $i_{m} > k_{n}$ and $x_{i_{m}} > x_{k_{n}}$ then $x_{k_{0}} > x_{i_{m}} > x_{k_{n}}$ , and hence, $| x^{*} - x_{i_{m}} | < max {| x^{*} - x_{k_{n}} |, | x^{*} - x_{k_{0}} |} < ϵ$ , which contradicts (5). Thus, $x_{i_{m}} < x_{k_{n}}$ for all $i_{m} > k_{n}$ . This, due to by Lemma (2), follows that $x_{j} < x_{k_{n}}$ for all $j > k_{n}$ , and hence, $x_{k_{n + 1}} < x_{k_{n}}$ . In other words, the sequence ${x_{k_{n}}}_{n = 0}^{\infty}$ is strictly decreasing. Similarly, the sequence ${x_{i_{n}}}_{n = 0}^{\infty}$ is strictly increasing. ∎

Due to Lemma 3, without loss of generality, we suppose that $x_{k_{0}} > x^{*} > x_{i_{0}}$ . In this case, the sequence ${x_{k_{n}}}_{n = 0}^{\infty}$ is strictly decreasing and the sequence ${x_{i_{n}}}_{n = 0}^{\infty}$ is strictly increasing. Furthermore, their boundedness implies that they converge to some limits. As we know, $x^{*} = lim n \to \infty x_{k_{n}}$ . Let us denote by $y$ the limit of ${x_{i_{n}}}_{n = 0}^{\infty}$ . Following from (5), for all $n$ , we have $x_{i_{n}} < x^{*} - ϵ$ , and hence, $y \leq x^{*} - ϵ < x^{*}$ .

Lemma 4.

i) There does not exist any $x_{n}$ such that $y \leq x_{n} \leq x^{*}$ .
ii) The subsequences of ${x_{n}}_{n = 0}^{\infty}$ belonging to the intervals $[x_{i_{0}}, y)$ and $(x^{*}, x_{k_{0}}]$ are the strictly monotone. Moreover, they converge to $y$ and $x^{*}$ respectively.

Proof.

i) Suppose the contrary, i.e., there exists $x_{n}$ such that $y \leq x_{n} \leq x^{*}$ . Then, there exists $m \in N$ such that $i_{m} > n, k_{m} > n$ and $x_{i_{m}} < x_{n} < x_{k_{m}}$ , which contradicts Lemma 2.

ii) Firstly, we consider the subsequence of ${x_{n}}_{n = 0}^{\infty}$ lying within $[x_{i_{0}}, y)$ . Because $lim m \to \infty x_{i_{m}} = y$ , for each such $x_{n}$ , there exists $m \in N$ such that $x_{n} < x_{i_{m}} < y$ . Lemma 2 implies that $x_{k} > x_{n}$ for all $k > n$ . This leads to the monotonicity of the subsequence being considered. Consequently, this subsequence converges to $y$ . Similarly, the subsequence lying within $(x^{*}, x_{k_{0}}]$ is strictly decreasing and converges to $x^{*}$ . ∎

Making use of Lemma 4, for notational convernience, from now we use the notations ${x_{i_{n}}}_{n = 0}^{\infty}$ and ${x_{k_{n}}}_{n = 0}^{\infty}$ to denote the subsequences of ${x_{n}}_{n = 0}^{\infty}$ lying within $[x_{i_{0}}, y)$ and $(x^{*}, x_{k_{0}}]$ , respectively. Notice that, due to Lemma 4, we have

{x_{n}}_{n = i_{0}}^{\infty} = {x_{i_{n}}}_{n = 0}^{\infty} \cup {x_{k_{n}}}_{n = p}^{\infty}

where $p$ satisfies $k_{p} > i_{0} > k_{p - 1} .$

Lemma 5.

Consider the two sequences ${u_{n}}_{n = 0}^{\infty}$ and ${v_{n}}_{n = 0}^{\infty}$ generated by Algorithm 1. Under Assumption 1, the following claims hold true.
i) The sequence ${u_{n}}_{n = 0}^{\infty}$ (resp. ${v_{n}}_{n = 0}^{\infty}$ ) lies within the interval $[x_{i_{0}}, y)$ (resp. $(x^{*}, x_{k_{0}}]$ ). Moreover, $lim n \to \infty u_{n} = y$ and $lim n \to \infty v_{n} = x^{*}$ .
ii) For any large enough $n$ , $| f (v_{i_{n}}) | > | f (u_{i_{n}}) |$ and $| f (v_{k_{n}}) | < | f (u_{k_{n}}) |$ .
iii) The point $y$ is also a zero of $f$ , i.e., $f (y) = 0$ .

Proof.

i) Observe that at the n-th step, $u_{n}, v_{n}$ are some elements of the sequence ${x_{n}}_{n = 0}^{\infty}$ and either $x_{n} = u_{n + 1}$ or $x_{n} = v_{n + 1}$ . If at some $n_{0}$ -th step, both $u_{n_{0}}$ and $v_{n_{0}}$ belong to the same one of intervals $[x_{i_{0}}, y)$ and $(x^{*}, x_{k_{0}}]$ such as $(x^{*}, x_{k_{0}}]$ then $x_{n} \in (x^{*}, x_{k_{0}}]$ for all $n > n_{0}$ , which is a contradiction. So, $u_{n} \in [x_{i_{0}}, y)$ and $v_{n} \in (x^{*}, x_{k_{0}}]$ for all $n \geq i_{0}$ . Moreover, we see that ${u_{n}}_{n = 0}^{\infty}$ is not decreasing and has ${x_{i_{n}}}_{n = 0}^{\infty}$ as a subsequence. Thus, $lim n \to \infty u_{n} = y$ . Similarly, $lim n \to \infty v_{n} = x^{*}$ .

ii) Firstly, due to the fact that $lim n \to \infty x_{i_{n}} = lim n \to \infty u_{i_{n}} = y$ , $lim n \to \infty x_{k_{n}} = lim n \to \infty v_{k_{n}} = x^{*}$ , we see that for large enough $n$ , the following inequality holds.

| x_{i_{n}} - u_{i_{n}} | < x^{*} - y < | x_{i_{n}} - v_{i_{n}} | .

This inequality, together with the identity $∣ ∣ ∣ \frac{f (v_{i_{n}})}{f (u_{i_{n}})} ∣ ∣ ∣ = \frac{| x_{i_{n}} - v_{i_{n}} |}{| x_{i_{n}} - u_{i_{n}} |}$ , imply that $| f (v_{i_{n}}) | > | f (u_{i_{n}}) |$ . Similarly, we have $| f (v_{k_{n}}) | < | f (u_{k_{n}}) |$ .

iii) Since ii) taking the limits as $n \to \infty$ , we obtain that $| f (x^{*}) | \geq | f (y) |$ and $| f (x^{*}) | \leq | f (y) |$ . This implies $| f (y) | = | f (x^{*}) | = 0$ , i.e., $f (y) = 0$ . ∎

From Lemma 5, since both sequences ${f (x_{i_{n}})}_{n = 0}^{\infty}$ and ${f (x_{k_{n}})}_{n = 0}^{\infty}$ converge to zero, we have

lim n \to \infty | f (x_{n}) | = 0.

(7)

Now, we point out a sub-sequence of the sequence ${x_{n}}_{n = 0}^{\infty}$ such that its image under the function $| f |$ is strictly increasing. Firstly, denote by $χ := χ_{(x^{*}, x_{k_{0}}]}$

the characteristic function of the interval

(x^{*}, x_{k_{0}}]

, i.e.,

χ (x) = {\begin{matrix} 1 & for all x \in (x^{*}, x_{k_{0}}] %, 0 & otherwise. \end{matrix}

Because there are infinitely many $x_{n}$ in each of two intervals $[x_{i_{0}}, y)$ and $(x^{*}, x_{k_{0}}]$ , there are infinitely many $x_{n}$ such that each of the following two cases occurs.

$χ (x_{n}) - χ (x_{n + 1}) = 1$ , i.e., $x_{n} \in (x^{*}, x_{k_{0}}]$ and $x_{n + 1} \in [x_{i_{0}}, y)$ .
$χ (x_{n}) - χ (x_{n + 1}) = - 1$ , i.e., $x_{n} \in [x_{i_{0}}, y)$ and $x_{n + 1} \in (x^{*}, x_{k_{0}}]$ .

Now, we call $n$ a jump if $χ (x_{n}) - χ (x_{n + 1}) \neq 0$ . Let $p$ , $q$ be two consecutive jumps such that $χ (x_{q}) - χ (x_{q + 1}) = 1$ and $χ (x_{p}) - χ (x_{p + 1}) = - 1$ . For large enough $p, q$ , we consider two cases where $p > q$ and $p < q$ .

Figure 1: In case

p > q

In case $p > q$ , due to $x_{q + 1}, \dots, x_{p} \in [x_{i_{0}}, y)$ , it follows by i) of Lemma 5 that $x_{m} = u_{m + 1}$ for all $m = q + 1, \dots, p$ . So, $v_{m} = v_{m + 1}$ for all $m = q + 1, \dots, p$ , i.e., $v_{q + 1} = v_{q + 2} = \dots = v_{p + 1}$ . On the other hand, $x_{q} \in (x^{*}, x_{k_{0}}]$ , so $x_{q} = v_{q + 1}$ . Since ii) of Lemma 5, for large enough $p$ we have $| f (u_{p + 1}) | > | f (v_{p + 1}) |$ . Therefore,

| f (x_{p}) | = | f (u_{p + 1}) | > | f (v_{p + 1}) | = | f (v_{p}) | = \dots = | f (v_{q + 1}) | = | f (x_{q}) | .

Figure 2: In case

p < q

Analoguously, if $q > p$ then

| f (x_{q}) | = | f (v_{q + 1}) | > | f (u_{q + 1}) | = | f (u_{q}) | = \dots = | f (u_{p + 1}) | = | f (x_{p}) | .

To sum up, if $q < p$ are two consecutive jumps then $| f (x_{q}) | < | f (x_{p}) |$ . Let us denote by ${j_{n}}_{n = 0}^{\infty}$ the sequence of jumps. Since some large enough $n_{0}$ , the sequence ${| f (x_{j_{n}}) |}_{n = n_{0}}^{\infty}$ is strictly increasing. This contradicts to (7), and therefore, implies that Assumption 1 is wrong. This completes the proof of Theorem 1.

3 Conclusion

In this paper, we prove the convergence of Regula Falsi method for all continuous functions. In the future, it is of great interest to provide theoretical proofs about the convergence rate of the method.

References

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[3] Gautschi, W. (2012). Numerical Analysis, 2sd ed. Basel: Birkhauser.
[4] Stoer, J., Bulirsch, R. (2002). Introduction to Nummerical Analysis, 3rd ed. (Bartels, R., Gautschi, W., Witzgall, C., trans.) New York: Springer, pp. 339-340.
[5] Korner, T. W. (2003). A Companion to Analysis: A Second First and First Second Course in Analysis, 2sd ed. American Mathematical Society.
[6] Sureshkumar, R., Rutledge, G.C., The Method of False Position, http://web.mit.edu/10.001/Web/Course_Notes/NLAE/node5.html
[7] Vinh, L.T. (2007). Giao trinh giai tich so (in Vietnamese), Ha Noi: Vietnam Science and Technics Publishing House, pp. 26-28.
[8] Weisstein, E., Method of False Position —From MathWorld, A Wolfram Web Resource, http://mathworld.wolfram.com/MethodofFalsePosition.html

The convergence of the Regula Falsi method

An iterative method for estimation the roots of real-valued functions

Constructing subgradients from directional derivatives for functions of two variables

Interpolation of Set-Valued Functions

About some aspects of function interpolation by trigonometric splines

A Bisection Method Like Algorithm for Approximating Extrema of a Continuous Function

Parameter Identification with Finite-Convergence Time Alertness Preservation

Cutoff profile of the Metropolis biased card shuffling

1 Introduction

2 Main result

Theorem 1.

Lemma 1.

Proof.

Lemma 2.

Proof.

Proposition 1.

Proof.

Assumption 1.

Lemma 3.

Proof.

Lemma 4.

Proof.

Lemma 5.

Proof.

3 Conclusion

References

The convergence of the Regula Falsi method

Related Research

An iterative method for estimation the roots of real-valued functions

Constructing subgradients from directional derivatives for functions of two variables

Interpolation of Set-Valued Functions

About some aspects of function interpolation by trigonometric splines

A Bisection Method Like Algorithm for Approximating Extrema of a Continuous Function

Parameter Identification with Finite-Convergence Time Alertness Preservation

Cutoff profile of the Metropolis biased card shuffling

1 Introduction

2 Main result

Theorem 1.

Lemma 1.

Proof.

Lemma 2.

Proof.

Proposition 1.

Proof.

Assumption 1.

Lemma 3.

Proof.

Lemma 4.

Proof.

Lemma 5.

Proof.

3 Conclusion

References