The Point Zoro Symmetric Single-Step Procedure for Simultaneous Estimation of Polynomial Zeros

Definition 2.1. If there exists a p ≥ 1 such that for any null sequence {w^(k)}  generated from {x^(k)},   then the R-factor of the sequence {w^(k)}  is defined to be

where R_p  is independent of the norm ∥·∥.

Definition 2.2. We next define the R-order of the procedure I in terms of the R-factor as

Suppose that R_p(w^(k)) < 1,   then it follows from Ortega and Rheinboldt [13] that the R-order of I satisfies the inequality O_R(I, x^*) ≥ p.

Theorem 2.3. Let I be an iterative procedure and let Ω(I, x^*) be the set of all sequences {x^(k)}  generated by I which converges to the limit x^*. Suppose that there exists a p ≥ 1 and a constant γ such that for any {x^(k)} ∈ Ω(I, x^*),

Then, it follows that R-order of I satisfies the inequality O_R(I, x^*) ≥ p.

Lemma 3.1. If

• (i)

  p : C → C is defined by (2.1);

• (ii)

  p_i : C → C is defined by

  $$ ()

• (iii)

  q_i : C → C is defined by

  $$ ()
  where $$ and $$ (j, m = 1, …, n; j ≠ m);

• (iv)

  ϕ_i : C → C is defined by

  $$ ()
  then
  $$ ()

Lemma 3.2. If (i)–(iv) of Lemma 3.1 are valid; (v) $$ (i = 1, …, n) are such that $$ (i = 1, …, n), $$ (m = 1, …, i − 1), $$ (m = i + 1, …, n), and

  $$, $$, and $$ (i = 1, …, n), then where

Lemma 3.3. If (i)–(v) of Lemma 3.2 are valid; (vi) $$ and $$ (i = 1, …, n), where $$ and 0 < θ < 1, then $$ (i = 1, …, n).

Theorem 3.4. If (i) p : C → C defined by (2.1) has n distinct zeros $$ (i = 1, …, n); (ii) $$ (i = 1, …, n), where 0 < θ < 1 and $$, and the sequences $$ (i = 1, …, n) are generated from PZSS1 (i.e., from (3.1a)–(3.1e)), then $$ (k → ∞) (i = 1, …, n) and O_R(PZSS1, x^*) ≥ 4.

Proof. For  i = 1, …, n, let

Then, by (3.5) and (3.6),

where p_i(x) is defined by (3.4).

By Lemmas 3.1 and 3.2 with q_i = q_1,i, $$, $$, $$, ϕ_i = ϕ_1,i (i = 1, …, n), it follows that, for i = 1, …, n, k ≥ 0,

where Similarly, by Lemmas 3.1 and 3.2, with q_i = q_2,i, $$, $$, $$, ϕ_i = ϕ_2,i (i = 1, …, n), it follows that, for i = 1, …, n, k ≥ 0, where Similarly, by Lemma 3.1 and Lemma 3.2, with q_i = q_3,i, $$, $$, $$, ϕ_i = ϕ_3,i (i = 1, …, n), it follows that, for i = 1, …, n, k ≥ 0, where It follows from (3.13)-(3.14) and Lemma 3.3 that $$ (i = 1, …, n), and it follows from (3.15)-(3.16) and Lemma 3.3 that $$ (i = 1, …, n) follows from (3.1e). It follows from (3.17)-(3.18) and Lemma 3.3 that whence $$ (i = 1, …, n). It then follows by induction on k that, for all k ≥ 0, whence $$, (i = 1, …, n). Let Then, by (3.13), (3.15), (3.17), and (3.21), for i = 1, …, n, (recall (3.1b), (3.1c), (3.1d)) for i = n, …, 1, and for i = 1, …, n, Let For r = 1,2, 3,    let Then, by (3.25)–(3.26), for all  k ≥ 1, Suppose, without loss of generality, that Then, by a lengthy inductive argument, it follows from (3.21)–(3.29) that for i = 1, …, n, for all k ≥ 1, whence, by (3.28) and (3.1e), for all k ≥ 1, By (3.21) for m = 3, then by (3.1e), So, Let Then, by (3.22)–(3.35) So, Therefore (Ortega and Rheindboldt [13]),