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# The convergence of the Regula Falsi method

Regula Falsi, or the method of false position, is a numerical method for finding an approximate solution to f(x) = 0 on a finite interval [a, b], where f is a real-valued continuous function on [a, b] and satisfies f(a)f(b) < 0. Previous studies proved the convergence of this method under certain assumptions about the function f, such as both the first and second derivatives of f do not change the sign on the interval [a, b]. In this paper, we remove those assumptions and prove the convergence of the method for all continuous functions.

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## 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 that converges to a zero of when tends to infinity. Some provided the proofs under some assumptions, such as both the first and second derivatives of do not change sign on [] (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 denotes the set of natural numbers and . Let be a continuous function on the interval and . We construct a sequence as follows.

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

###### Theorem 1.

For any continuous function on a closed, bounded interval that satisfies , the sequence generated by Algorithm 1 converges to a zero of , i.e., .

The proof of this theorem is lengthy, and hence, it will be divided into several parts. Clearly, if for some , then simple induction implies that for all and the desired convergence is obvious. Now, we consider the case where for all .

###### Lemma 1.

Considering the sequence defined by

 rn:=vn+1−un+1vn−un . (2)

Then for each , we have that:

 rn=⎧⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎨⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎩11+|f(un)f(vn)|  if un+1=xn and vn+1=vn,11+|f(vn)f(un)|  if un+1=un and % vn+1=xn. (3)
###### Proof.

Without loss of generality, we consider only the case where and . Thus, we have

 vn+1−un+1=xn−un=unf(vn)−vnf(un)f(vn)−f(un)−un=(un−vn)f(un)f(vn)−f(un).

It follows by that

 rn=−f(un)f(vn)−f(un)=11−f(vn)f(un)=11+|f(vn)f(un)| ,

which is (3) exactly.
Analoguously, in the second case, where and , we have that

 rn=f(vn)f(vn)−f(un)=11−f(un)f(vn)=11+|f(un)f(vn)| .

This completes the proof of Lemma 1. ∎

###### Lemma 2.

For each fixed , the sequence is either upper or lower bounded by .

###### Proof.

Firstly, we notice the decreasing property of the sequence , i.e.

 [u1,v1]⊋[u2,v2]⊋…⊋[un,vn]⊋…

In addition, since for any the point is either or , the desired claim follows directly. ∎

Due to Lemma 1, we have that Let , we see that . Now we discuss two cases whether is strictly smaller than 1 or not in the following proposition.

###### Proposition 1.

Consider the sequences generated by Algorithm 1 and defined by (2), (3). Then the following claims hold true.
i) If then the sequence converges to a zero of .
ii) If then there exists a subsequence converges to a zero of .

###### Proof.

i) If then when tends to infinity. This implies that is a decreasing nested sequence of non-empty, closed and bounded intervals in and their lengths strictly decrease to . So, due to Cantor intersection theorem, see e.g. [5], there exists such that . Evidently, . Since for all , due to the continuity of , let then , which only holds if . Consequently, the sequence generated by Algorithm 1 converges to a zero of .

ii) Now we consider the case where . Firstly, we observe that by the construction of the sequence , for each , there exist depending on such that and . Since , there exists a subsequence of that converges to 1. Let then the continuity of implies that . Due to Lemma 1 we have that

 rmn<11+|f(xmn)|M<1 for all n∈N∗. (4)

Therefore, let tend to infinity we have . Again, Bolzano-Weierstrass theorem applied to the bounded sequence implies the existence of a convergent subsequence , whose limit is denoted by . The continuity of , therefore, implies that . ∎

Clearly, from Proposition 1, the remaining case is where , and we need to prove the convergence of the whole sequence instead of only the subsequence . We will perform this task by disproof, i.e., the following assumption holds.

###### Assumption 1.

Suppose that the sequence generated by Algorithm 1 diverges.

Under Assumption 1, there exists and a subsequence of such that

 |xin−x∗|>ϵ for all n≥0. (5)

Since the sequence converges to , there exists such that

 |xkn−x∗|<ϵ for all n≥n0. (6)

By suitably removing a finite number of starting elements of two above sequences, one can assume that and .

###### Lemma 3.

i) The point must lie between and .
ii) Two sequences are 2 monotone sequences in opposite directions.

###### Proof.

i) Due to (5) and (6), we first observe that could not lie between and . In case is between and , we suppose that , since similar arguments can be made for . Due to Lemma 2, for all . Thus, there does not exist any subsequence of converging to , which contradicts the fact that . Therefore, must lie between and .
ii) Without loss of generality, we suppose that . The case where is on the analogy. Since , there exists a such that and . Due to Lemma 2 we obtain that is upper bounded by and lower bounded by . For each fixed , if there exists an such that and then , and hence, , which contradicts (5). Thus, for all . This, due to by Lemma (2), follows that for all , and hence, . In other words, the sequence is strictly decreasing. Similarly, the sequence is strictly increasing. ∎

Due to Lemma 3, without loss of generality, we suppose that . In this case, the sequence is strictly decreasing and the sequence is strictly increasing. Furthermore, their boundedness implies that they converge to some limits. As we know, . Let us denote by the limit of . Following from (5), for all , we have , and hence, .

###### Lemma 4.

i) There does not exist any such that .
ii) The subsequences of belonging to the intervals and are the strictly monotone. Moreover, they converge to and respectively.

###### Proof.

i) Suppose the contrary, i.e., there exists such that . Then, there exists such that and , which contradicts Lemma 2.

ii) Firstly, we consider the subsequence of lying within . Because , for each such , there exists such that . Lemma 2 implies that for all . This leads to the monotonicity of the subsequence being considered. Consequently, this subsequence converges to . Similarly, the subsequence lying within is strictly decreasing and converges to . ∎

Making use of Lemma 4, for notational convernience, from now we use the notations and to denote the subsequences of lying within and , respectively. Notice that, due to Lemma 4, we have

 {xn}∞n=i0={xin}∞n=0 ∪ {xkn}∞n=p

where satisfies

###### Lemma 5.

Consider the two sequences and generated by Algorithm 1. Under Assumption 1, the following claims hold true.
i) The sequence (resp. ) lies within the interval (resp. ). Moreover, and .
ii) For any large enough , and .
iii) The point is also a zero of , i.e., .

###### Proof.

i) Observe that at the n-th step, are some elements of the sequence and either or . If at some -th step, both and belong to the same one of intervals and such as then for all , which is a contradiction. So, and for all . Moreover, we see that is not decreasing and has as a subsequence. Thus, . Similarly, .

ii) Firstly, due to the fact that , , we see that for large enough , the following inequality holds.

 |xin−uin|

This inequality, together with the identity  , imply that . Similarly, we have .

iii) Since ii) taking the limits as  , we obtain that and . This implies , i.e., . ∎

From Lemma 5, since both sequences and converge to zero, we have

 limn→∞|f(xn)|=0. (7)

Now, we point out a sub-sequence of the sequence such that its image under the function is strictly increasing. Firstly, denote by

the characteristic function of the interval

, i.e.,

 χ(x)={1 for all x∈(x∗,xk0]%,0 otherwise.

Because there are infinitely many in each of two intervals and , there are infinitely many such that each of the following two cases occurs.

• , i.e., and .

• , i.e., and .

Now, we call a jump if . Let , be two consecutive jumps such that and . For large enough , we consider two cases where and .

In case , due to , it follows by i) of Lemma 5 that for all . So, for all , i.e., . On the other hand, , so . Since ii) of Lemma 5, for large enough we have . Therefore,

 |f(xp)|=|f(up+1)|>|f(vp+1)|=|f(vp)|=…=|f(vq+1)|=|f(xq)|.

Analoguously, if then

 |f(xq)|=|f(vq+1)|>|f(uq+1)|=|f(uq)|=…=|f(up+1)|=|f(xp)|.

To sum up, if are two consecutive jumps then . Let us denote by the sequence of jumps. Since some large enough , the sequence 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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• [5] Korner, T. W. (2003). A Companion to Analysis: A Second First and First Second Course in Analysis, 2sd ed. American Mathematical Society.
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