# On the Convergence Ball and Error Analysis of the Modified Secant Method

Authors: Rongfei Lin; Qingbiao Wu; Minhong Chen; Xuemin Lei
Journal: Advances in Mathematical Physics
 (2018)

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## Abstract

We aim to study the convergence properties of a modification of secant iteration methods. We present a new local convergence theorem for the modified secant method, where the derivative of the nonlinear operator satisfies Lipchitz condition. We introduce the convergence ball and error estimate of the modified secant method, respectively. For that, we use a technique based on Fibonacci series. At last, some numerical examples are given.


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## Body

## 1. Introduction

A large number of nonlinear dynamic systems and scientific engineering problems can be concluded to the form of nonlinear equation(1)fx=0,where f is a nonlinear operator defined on a convex subset D of a complex dimension space C. Hence, finding the roots of the nonlinear (1) is widely required in both mathematical physics and nonlinear dynamic system. Iterative methods are considerable methods. There are many iterative methods for solving the nonlinear equation.Secant method [1, 2], which uses divided differences instead of the first derivative of the nonlinear operator, is one of the most famous iterative methods for solving the nonlinear equation. Secant method reads as follows:(2)xn+1=xn-xn-1,xn;f-1fxn,n≥0x0,x-1∈Dwhere the operator x,y;f is called a divided difference of first-order for the operator f on the points x and y(x≠y) if the following equality holds: (3)x,y;fx-y=fx-fy.Due to the well performance of the secant method, secant method and secant-like methods have been widely studied by many authors [3–11]. The authors [12] proposed a new method for solving the nonlinear equation.Convergence ball is a very important issue in the study of the iterative procedures. When nonlinear operatorf is first-order differentiable convex subset D can be open or closed, suppose x∗ is the root of the equation f(x)=0, an open area B(x∗,R) is called the convergence ball of the iterative algorithm. Authors [13–17] have discussed the convergence of the iterative methods using a convergence ball B(x∗,R) with center x∗ and radius R. For example, Ren and Wu [15] discussed the convergence of the secant method under Hölder continuous divided differences using a convergence ball.In this study, we consider the modified secant method with the below form based on [12](4)xn+1=xn-fxnxn-xn-13fxn+fxn-1-4fxn+xn-1/2n≥0,and we will establish the convergence ball and give the error analysis of the modified secant method for the nonlinear equation.


## 2. Convergence Ball Study

Theorem 1.
Supposex∗ is the root of the equation f(x)=0 and f′(x∗)≠0. f is first-order differentiable, where the derivative of f satisfies the Lipschitz condition: |f′(x∗)-1(f′(x)-f′(y))|≤K|x-y| for all x,y∈D and K>0. Then, the sequence xn generated by the modified secant method (4), starting from any two initial points x0,x-1∈B(x∗,R), converges to the solution x∗. x∗ is the unique solution in B(x∗,2/K), where B(x∗,R)⊂B(x∗,2/K). Moreover, the following error estimate holds:(5)x∗-xn⩽Rx∗-x0RFnx∗-x-1RFn-1=25K5K2x∗-x0Fn5K2x∗-x-1Fn-1n≥1.Here, R=2/5K; Fn is a Fibonacci series, F0=F1=1,Fn+1=Fn+Fn-1,(n≥1).Proof.
From the condition of Theorem1, we know x0,x-1∈B(x∗,R). Assume x1,x2,…,xn are generated by the modified secant (4) and xk∈B(x∗,R)(0≤k≤n). Following, we will prove that xn+1∈B(x∗,R); we have(6)x∗-xn+1=x∗-xn+fxnxn-xn-13fxn+fxn-1-4fxn+xn-1/2=x∗-xn+fxn-fx∗xn-xn-13fxn,xn+xn-1/2xn-xn+xn-1/2+fxn-1,xn+xn-1/2xn-1-xn+xn-1/2=x∗-xn+fxn,x∗xn-x∗xn-xn-13fxn,xn+xn-1/2-fxn-1,xn+xn-1/2xn-xn-1/2=x∗-xn+2fxn,x∗xn-x∗3fxn,xn+xn-1/2-fxn-1,xn+xn-1/2=x∗-xn1-2fxn,x∗3fxn,xn+xn-1/2-fxn-1,xn+xn-1/2=x∗-xn3fxn,xn+xn-1/2-fxn-1,xn+xn-1/2-2fxn,x∗3fxn,xn+xn-1/2-fxn-1,xn+xn-1/2,f is first-order differentiable, so in convex domain D, first-order difference of f can be written in the following integral form:(7)fx,y=fx-fyx-y=∫01f′tx+1-tydtx,y∈D.Now, we give the estimate of (8)3fxn,xn+xn-1/2-fxn-1,xn+xn-1/2-2fxn,x∗3fxn,xn+xn-1/2-fxn-1,xn+xn-1/2.Obviously(9)f′x∗-13fxn,xn+xn-12-fxn-1,xn+xn-12-2fxn,x∗=∫01f′x∗-13f′txn+1-txn+xn-12-f′txn-1+1-txn+xn-12-2f′txn+1-tx∗dt,and(10)2-f′x∗-13fxn,xn+xn-12-fxn-1,xn+xn-12=f′x∗-12f′x∗-3fxn,xn+xn-12+fxn-1,xn+xn-12=∫01f′x∗-12f′x∗-3f′txn+1-txn+xn-12+f′txn-1+1-txn+xn-12dt.Using Lipschitz condition with the above (9) and (10), we have(11)f′x∗-13fxn,xn+xn-12-fxn-1,xn+xn-12-2fxn,x∗⩽K∫01txn+1-txn+xn-12-txn-1+1-txn+xn-12dt+2K∫01txn+1-txn+xn-12-txn+1-tx∗dt⩽Kxn-x∗+x∗-xn-12+2Kxn-x∗+xn-1-x∗4⩽Kx∗-xn+Kx∗-xn-1,and(12)2-f′x∗-13fxn,xn+xn-12-fxn-1,xn+xn-12⩽∫012Kx∗-txn+1-txn+xn-12+Ktxn-1+1-txn+xn-12-txn+1-txn+xn-12dt⩽2K∫011+t2x∗-xn+1-t2x∗-xn-1dt+K∫01txn-1-xndt≤2K34x∗-xn+14x∗-xn-1+K2x∗-xn+xn-1-x∗≤2Kx∗-xn+Kx∗-xn-1.We divide above inequality (12) number 2, so(13)1-f′x∗-123fxn,xn+xn-12-fxn-1,xn+xn-12≤Kx∗-xn+K2x∗-xn-1.According to the definition of R and xn,xn-1∈B(x∗,R), we get Kx∗-xn+K/2x∗-xn-1<3/5<1, by Banach Lemma, so 3f[xn,xn+xn-1/2]-f[xn-1,xn+xn-1/2] is reversible and also(14)2f′x∗3fxn,xn+xn-12-fxn-1,xn+xn-12-1⩽11-Kx∗-xn-K/2x∗-xn-1.Dividing (14) inequality number 2, we can get(15)f′x∗3fxn,xn+xn-12-fxn-1,xn+xn-12-1⩽12-2Kx∗-xn-Kx∗-xn-1,and with (11) and (15), we get following estimate formula:(16)3fxn,xn+xn-1/2-fxn-1,xn+xn-1/2-2fxn,x∗3fxn,xn+xn-1/2-fxn-1,xn+xn-1/2=f′x∗-13fxn,xn+xn-1/2-fxn-1,xn+xn-1/2-2fxn,x∗×f′x∗3fxn,xn+xn-12-fxn-1,xn+xn-12-1⩽Kx∗-xn+Kx∗-xn-12-2Kx∗-xn-Kx∗-xn-1,and with (6) and (16) and xn,xn-1∈B(x∗,R), we get following estimate formula:(17)x∗-xn+1≤x∗-xn3fxn,xn+xn-1/2-fxn-1,xn+xn-1/2-2fxn,x∗3fxn,xn+xn-1/2-fxn-1,xn+xn-1/2≤x∗-xnKx∗-xn+Kx∗-xn-12-2Kx∗-xn-Kx∗-xn-1≤x∗-xnKR+KR2-2KR-KR=x∗-xn<R.This means that xn+1∈B(x∗,R). So, from any x0,x-1∈B(x∗,R), the sequence {xn} of the modified secant method is convergent, the root x∗∈B(x∗,R), and by mathematical induction xn∈B(x∗,R)(n≥1).
In the following, we will derive the estimate of the modified secant method. Denoteen=x∗-xn,ρn=en/R,(n≥-1); from the above proof we can get 0≤ρn<1(n≥-1); from inequality (17), it is known that en+1≤en and(18)en+1≤enKen+Ken-12-2Ken-Ken-1=Ken2+Ken-1en2-2Ken-Ken-1<Ken2+Ken-1en2-2KR-KR.Hence, ρn+1≤ρn,(n≥-1); moreover, we have (19)ρn+1≤Kρn2R+Kρnρn-1R2-3KR≤Kρnρn-1R+Kρnρn-1R2-3KR=2KR2-3KRρnρn-1=ρnρn-1,so, we obtain the inequality (20)ρn≤ρn-1ρn-2,n≥1.
Now, we use mathematical induction to proof that the inequalityρk≤ρ0Fkρ-1Fk-1,(k≥1) is correct. (21)ρ1≤ρ0ρ-1=ρ0F1ρ-1F0,ρ2≤ρ1ρ0≤ρ0ρ-1ρ0=ρ0F2ρ-1F1,ρ3≤ρ2ρ1≤ρ0ρ-1ρ0ρ0ρ-1=ρ0F3ρ-1F2.Suppose the inequality ρk≤ρ0Fkρ-1Fk-1,(k≥1) is correct when 3≤k≤n-1; here {Fk} is Fibonacci sequence, F0=F1=1,Fk+1=Fk+Fk-1,(k≥1). So, when k=n, we have (22)ρn≤ρn-1ρn-2≤ρ0Fn-1ρ-1Fn-2ρ0Fn-2ρ-1Fn-3=ρ0Fnρ-1Fn-1.That means the inequality ρk≤ρ0Fkρ-1Fk-1,(k≥1) has been proved. So(23)ρn≤ρ0Fnρ-1Fn-1n≥1.From the definition of ρn and above formulation, we can get(24)x∗-xn⩽Rx∗-x0RFnx∗-x-1RFn-1=25K5K2x∗-x0Fn5K2x∗-x-1Fn-1n≥1.At last, we show the uniqueness of the solution in the area B(x∗,2/K). Assume that there exists another solution y∗∈B(x∗,2/K), y∗≠x∗. We consider the operator A=∫01‍f′(tx∗+(1-t)y∗)dt. Since A[y∗-x∗]=f(y∗)-f(x∗)=0, if operator A is invertible, then y∗=x∗. Indeed from (24), we have (25)1-f′x∗-1A=f′x∗-1∫01f′x∗-f′tx∗+1-ty∗dt⩽K∫011-tx∗-y∗dt=K2x∗-y∗<1.Then, by Banach lemma, we can tell that operator A is invertible. From the definition of radius R, it is easy to verify that the ball B(x∗,2/K) is bigger than B(x∗,R).
That completes the proof of Theorem1.


## 3. Numerical Examples

In this section, the convergence ball results were applied to numerical examples.Example 1.
Let us consider(26)fx=x2-1,x∈0,2.It is obviously that f′(x)=2x.f(x)=0 has a root x∗=1 and f′(x∗)=2. It is easy to know (27)f′x∗-1f′x-f′y⩽x-y.
According to Theorem1, we can obtain the fact that the radius of the convergence ball of the modified secant method is R=2/5K=2/5=0.4 at least.Example 2.
Let us consider the following numerical problem which has been studied in [4, 11, 13]: (28)fx=ex-1,D=-1,1.f′(x)=ex,x∗=0, and f′(x∗)=1.
We know|ex-ey|⩽e|x-y|; hence, (29)f′x∗-1f′x-f′y⩽ex-y.So K=e in this problem.
By Theorem1, we can obtain the fact that the radius of the convergence ball of the modified secant method is R=2/5K=2/5e≈0.1472 at least.Example 3.
Let us consider the nonlinear equation(30)f′x=25sin⁡x+x,x∈0,2.Here, f′(x)=2/5cos⁡x+1,x∗=0, and f′x∗=7/5.
We know thatcos⁡x-cos⁡y≤|x-y|; then it is obvious that(31)f′x∗-1f′x-f′y≤27x-y.In this case, the radius of the convergence ball of the modified secant method is R=2/5K=7/5=1.4 at least, according to Theorem 1.



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