A Semilocal Convergence for a Uniparametric Family of Efficient Secant-Like Methods

Theorem 1. Under the (C) hypotheses, further suppose that

and for λ ∈ [1, λ₀], where Then, sequence {x_n} generated by secant-like method (4) is well defined, remains in $$ for each n = 0,1, 2, …, and converges to a solution $$ of equation F(x) = 0. Moreover, the following estimates hold Furthermore, the solution x^* is unique in 𝒟₀ = U(x₀, σ₀)∩D, where σ₀ = (1/H) − λc − R, provided that

Proof. The proof with the exception of the uniqueness part is given in Theorem 3 in [5] if we use $$ instead of F and set b = 1, where $$.

To prove the uniqueness of the solution, let us assume that y^* ∈ 𝒟₀ is a solution of F(x) = 0. Let L = [y^*, x^*; F]. Then, using (C₇) and the definition of σ₀, we get in turn that

It follows from (15) and the Banach lemma on invertible operators [1, 2, 4, 6–8, 14] that L⁻¹ ∈ ℒ(𝒴, 𝒳). Using the identity 0 = F(y^*) − F(x^*) = L(y^* − x^*), we deduce that x^* = y^*. That completes the proof of the theorem.

Remark 2. If K = H, Theorem 1 reduces to Theorem 3 in [5]. Otherwise, that is, if H < K, then our Theorem 1 constitutes an improvement over Theorem 3, since

where where given in [5] (for b = 1). Hence, (16) justify our claim for this section which was made in the Introduction of this study.

Lemma 3. Let c ≥ 0, η > 0, H > 0, K > 0 and λ ≥ 1. Set t₋₁ = 0, t₀ = c, and t₁ = c + η. Define scalar sequences {q_n}, {t_n}, {α_n} for each n = 0,1, … by

functions f_n on [0,1) for each n = 1,2, … by and polynomial p on [0,1) by Denote by α the only root of polynomial p in (0,1). Suppose that Then, sequence {t_n} is nondecreasing, bounded from above by t^** that is defined by and converges to its unique least upper bound t^* which satisfies Moreover, the following estimates are satisfied for each n = 0,1, 2, …:

Proof. We will first show that polynomial p has roots in (0,1). Indeed, we have p(0) = −λK < 0 and p(1) = 2Hλ > 0. Using the intermediate value theorem, we deduce that there exists at least one root of p in (0,1). Moreover p^′(t) > 0. Hence, p crosses the positive axis only once. Denote by α the only root of p in (0,1). It follows from (18) and (19) that estimate (25) is certainly satisfied if

Estimate (27) is true by (22) for n = 0. Then, we have by (18) that Suppose that Estimate (27) will be true for k + 1 replacing n if or where f_k is defined by (20). We need a relationship between two consecutive recurrent functions f_k for each k = 1,2…. Using (20) and (21), we deduce that since p(α) = 0. Define function f_∞ on (0,1) by Then, we get from (20) and (33) that Hence, by (32)–(34), (31) is satisfied if which is true by (22). The induction for (25) is complete. That is, sequence {t_n} is nondecreasing, bounded from above by t^** that is given by (23), and as such it converges to some t^* which satisfies (24). Estimate (26) follows from (25) by using standard majorization techniques [1, 2, 4, 6–8, 14]. The proof of Lemma 3 is complete.

Lemma 4. Let c ≥ 0, η > 0, H₀ > 0, H₁ > 0, H > 0, K > 0, and λ ≥ 1. Set s₋₁ = 0, s₀ = c, and s₁ = c + η. Define scalar sequences {s_n}, {b_n} for each n = 1,2, … by

and functions g_n on [0,1) by

Suppose that

where α is defined in Lemma 3. Then, sequence {s_n} is nondecreasing, bounded from above by s^** that is defined by and converges to its unique least upper bound s^* which satisfies Moreover, the following estimates are satisfied for each n = 1,2, …:

Proof. We will show using induction that

Estimate (43) is true for n = 0 by (39). Then, we have by (36) that Suppose that (43) holds for each n ≤ k. Then, using (36), we get that Estimate (43) will be satisfied if Using (38), we get the following relationship between two consecutive recurrent functions g_k: Define function g_∞ on [0,1) by Then, we get from (38) that Then, (46) is satisfied if which is true by the choice of α and the right hand side inequality in hypothesis (39). The induction for (43) (i.e., (42)) is complete. The rest of the proof as identical to Lemma 3 is omitted. The proof is complete.

Remark 5. (a) Let us consider an interesting choice for λ. Let λ = 1 (secant method). Then, using (21) and (22), we have that

The corresponding condition for the secant method is given by [2, 4, 9, 23] as follows:

Condition (52) can be weaker than (53) (see also the numerical examples at the end of the study). Moreover, the majorizing sequence {u_n} for the secant method related to (53) is given by

A simple inductive argument shows that if H < K, then for each n = 2,3, …:

(b) The majorizing sequence {v_n} used in [5] is essentially given by

Then, again we have

Moreover, our sufficient convergence conditions can be weaker than [5].

(c) Clearly, iteration {s_n} is tighter than {t_n} and as we have in (57) than for H₀ < K or H₁ < H as follows:

Lemma 6. Let N = 0,1, 2, … be fixed. Suppose that

Then, sequence {t_n} generated by (19) is nondecreasing, bounded from above by t^**, and converges to t^* which satisfies t^* ∈ [t_N+1, t^*]. Moreover, the following estimates are satisfied for each n = 0,1, …:

Lemma 7. Let N = 1,2, … be fixed. Suppose that

Then, sequence {s_n} generated by (36) is nondecreasing, bounded from above by s^** and converges to s^* which satisfies s^* ∈ [s_N+1, s^*]. Moreover, the following estimates are satisfied for each n = 0,1, …

Theorem 8. Suppose that the (C), Lemma 3 (or Lemma 6) conditions and

hold. Then, sequence {x_n} generated by secant-like method is well defined, remains in U for each n = −1,0, 1,2, …, and converges to a solution $$ of equation F(x) = 0. Moreover, the following estimates are satisfied for each n = 0,1, …: Furthermore, if there exists T ≥ t^* − c such that then, the solution x^* is unique in $$.

Proof. We use mathematical induction to prove that

for each k = −1,0, 1, …. Let $$. Then, we obtain that which implies that $$. Let also $$. We get that hence, $$. Note that ∥x₋₁ − x₀∥ ≤ c = t₀ − t₋₁ and $$. That is, $$. Hence, estimates (67) and (68) hold for k = −1 and k = 0. Suppose that (67) and (68) hold for all n ≤ k. Then, we obtain that Hence, x_k+1, $$.

Using (C₇), Lemma 3, and the introduction hypotheses, we get that

It follows from (72) and the Banach lemma on invertible operators [1, 2, 4, 6–8, 14] that $$ exists and In view of (4), we obtain the following identity: Using (4), (34), and the induction hypotheses, we get in turn that It now follows from (4), (18), (74), and (75) that which completes the induction for (67). Moreover, let $$. Then, we get that which implies that $$. The induction for (68) is complete.

Lemma 3 implies that {t_k} is a complete sequence. It follows from (67) and (68) that {x_k} is a complete sequence in a Banach space 𝒳 and as such it converges to some $$ (since $$ is a closed set). By letting k → +∞ in (75), we obtain F(x^*) = 0. Furthermore, estimate (65) follows from (64) by using standard majorization techniques [1–4, 6–8]. To show the uniqueness part, let $$ be such that F(y^*) = 0. We have that

It follows from (78) and the Banach lemma on invertible operators that [y^*, x^*; F] ⁻¹ exists. Then, using the identity 0 = F(y^*) − F(x^*) = [y^*, x^*; F](y^* − x^*), we deduce that x^* = y^*. The proof of Theorem 8 is complete.

Remark 9. (a) The limit point t^* can be replaced in Theorem 8 by t^** which is given in closed form by (23).

(b) It follows from the proof of Theorem 8 that {s_n} is also a majorizing sequence for {x_n}. Hence, Lemma 4 (or Lemma 7), {s_n}, s^* can replace Lemma 3 (or Lemma 6) {t_n}, t^* in Theorem 8.

Theorem 10. Suppose that the (C) conditions, Lemma 4 (or Lemma 7), and

hold. Then, sequence {x_n} generated by secant-like method is well defined, remains in U for each n = −1,0, 1,2, …, and converges to a solution x^* ∈ U(x₀, s^* − c) of equation F(x) = 0. Moreover, the following estimates are satisfied for each n = 0,1, …: Furthermore, if there exists T ≥ s^* − c such that then, the solution x^* is unique in $$.

Lemma 11. Let c ≥ 0, η > 0, H₀ > 0, H₁ > 0, H > 0, M₀ > 0, M > 0, K > 0, and λ ≥ 1. Set γ₋₁ = 0, γ₀ = c, and γ₁ = c + η. Define scalar sequences {γ_n}, {δ_n} by

and functions h_n on [0,1) by Suppose that function φ given by has a minimal zero a in [0,1) and where α was defined in Lemma 3. Then, sequence {γ_n} is nondecreasing, bounded from above by γ^** that is defined by and converges to its unique least upper bound γ^* which satisfies Moreover, the following estimates are satisfied for each n = 1,2, …:

Proof. Simply use {γ_n}, {δ_n}, {h_n}, φ, a instead of {s_n}, {b_n}, {g_n}, p, α in the proof of Lemma 4.

Theorem 12. Suppose that the (C^*), Lemma 11 conditions,

hold, where U was defined in Theorem 8 and $$. Then, sequence {x_n} generated by the secant-like method (83) in well defined, remains in U for each n = −1,0, 1,2, …, and converges to a solution $$ of equation F(x) + G(x) = 0. Moreover, the following estimates are satisfied for each n = 0,1, …: Furthermore, if there exists $$ such that then, the solution x^* is unique in $$.

Proof. The proof until the uniqueness part follows as in Theorem 8 but using the following identity:

instead of (74). Finally, for the uniqueness part, let $$ be such that F(y^*) + G(y^*) = 0. Then, we get from (83) the identity This identity leads to Hence, we deduce lim _n→+∞x_n = y^*. But we know that lim _n→+∞x_n = x^*. That is, we conclude that x^* = y^*. That completes the proof of the theorem.

Example 1. Let 𝒳 = 𝒴 = 𝒞[0,1], equipped with the max-norm. Consider the following nonlinear boundary value problem:

It is well known that this problem can be formulated as the integral equation where 𝒬 is the Green function as follows: We observe that Then, problem (101) is in the form (1), where F : 𝒟 → 𝒴 is defined as The Fréchet derivative of the operator F is given by Then, we have that Hence, if 2γ < 5, then It follows that F^′(x₀) ⁻¹ exists and We also have that ∥F(x₀)∥ ≤ 1 + γ. Define the divided difference defined by Choosing x₋₁(s) such that ∥x₋₁ − x₀∥ ≤ c and l₀c < 1. Then, we have for λ = 1, where l₀ is such that Set u₀(s) = s and 𝒟 = U(u₀, R₀). It is easy to verify that U(u₀, R₀) ⊂ U(0, R₀ + 1) since ∥u₀∥ = 1. If 2γ < 5 and l₀c < 1, the operator F^′ satisfies conditions of Theorem 8, with

Choosing R₀ = 1, γ = 0.5, and c = 1, we obtain that

Moreover, we obtain that a₋₁ = 0.317477 and b₋₁ = 0.251336, but conditions of Theorem 1 are not satisfied since

Notice also that the popular condition (53) is also not satisfied, since $$. Hence, there is no guarantee under the old conditions that the secant-type method converges to x^*. However, conditions of Lemma 3 are satisfied since The convergence of the secant-type method is also ensured by Theorem 8.

Example 2. Let 𝒳 = 𝒴 = ℝ, and consider the real function

and we are going to apply secant-type method with λ = 2.5. We take the starting points x₀ = 1, x₋₁ = 0.25 and we consider the domain Ω = B(x₀, 3/4). In this case, we obtain Notice that the conditions of Theorem 1 and Lemma 3 are satisfied, but since H < K, Remark 2 ensures that our uniqueness ball is larger. It is clear as R₁ = 1.83333 ⋯ >0.193452 ⋯ = R₀.