Error estimations in the numerical solving of systems of equations in metric spaces

9 years ago

Abstract

Let $X_{1},X_{2}$ be two complete metric spaces, $X=X_{1}\times X_{2}$ and the nonlinear mappings $F_{1}:X\rightarrow X_{1},\ F_{2}:X\rightarrow X_{2}$. In order to solve the nonlinear system $x_{1}=F_{1}\left( x_{1},x_{2}\right),\ x_{2}=F_{2}\left( x_{1},x_{2}\right)$ we consider the Gauss-Seidel type method \[x_{n}=F_1 \left(x_{n-1},y_{n-1}\right), \\ y_{n}=F_2 \left( x_{n},y_{n-1}\right) .\] We obtain error estimations when the nonlinear mappings $F_{1}$ and $F_{2}$ are approximated by other mappings.

Authors

Ion Păvăloiu
(Tiberiu Popoviciu Institute of Numerical Analysis)

Title

Original title (in French)

Estimation des erreurs dans le résolution numérique des systèmes d’équations dans des espaces métriques

English translation of the title

Error estimations in the numerical solving of systems of equations in metric spaces

Keywords

system of equations in metric space; Gauss-Seidel type method; error estimations

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Cite this paper as:

I. Păvăloiu, Estimation des erreurs dans le résolution numérique des systèmes d’équations dans des espaces métriques, Seminar on functional analysis and numerical methods, Preprint no. 1 (1987), pp. 121-129 (in French).

About this paper

Journal

Seminar on functional analysis and numerical methods,
Preprint

Publisher Name

“Babes-Bolyai” University,
Faculty of Mathematics and Physics,
Research Seminars

DOI

Not available yet.

References

[1] Pavaloiu, I., La resolution des systemes operationnelles a l’aide des methodes iteratives, Mathematica, 11(34), 1969, 137–141.

[2] Urabe, M., Error estimation in numerical solution of equation by iteration processes, J. Sci. Hiroshima Univ., Ser. A-I, 26 (1962), 77–91.

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"Babeş-Bolyai" University

Faculty of Mathematics and Physics

Research Seminars

Seminar on Functional Analysis and Numerical Methods

Preprint Nr.1, 1987, pp.121-129

Error estimations in the numerical solving of systems of equations in metric spaces

Ion Păvăloiu

Let us designate by $(X_{i}, ρ_{i}), i = 1, 2$ two complete metric spaces and by $X = X_{1} \times X_{2}$ the Cartesian product of these spaces.

We designate by $F_{1} : X \to X_{1}$ And $F_{2} : X \to X_{2}$ two applications and we consider the following system of equations:

(1)		$x_{1}$	$= F_{1} (x_{1}, x_{2})$
	$x_{2}$	$= F_{2} (x_{1}, x_{2}), (x_{1}, x_{2}) \in X .$

For the resolution of the system of equations ( 1 ) we consider the following iterative process, of the Gauss-Seidel type:

(2)		$x_{1}^{(n + 1)}$	$= F_{1} (x_{1}^{(n)}, x_{2}^{(n)})$
	$x_{2}^{(n + 1)}$	$= F_{2} (x_{1}^{(n + 1)}, x_{2}^{(n)}), n = 0, 1, \dots, (x_{1}^{(0)}, x_{2}^{(0)}) \in X .$

Concerning the convergence of sequences we have the following Theorem [ 1 ] :

THEOREM 1 .

If the applications $F_{1}$ And $F_{2}$ check the conditions

	$ρ_{1} (F_{1} (x_{1}, y_{1}), F_{1} (x_{2}, y_{2}))$	$\leq α ρ_{1} (x_{1}, x_{2}) + β ρ_{2} (y_{1}, y_{2}),$
	$ρ_{2} (F_{2} (x_{1}, y_{1}), F_{2} (x_{2}, y_{2}))$	$\leq has ρ_{1} (x_{1}, x_{2}) + b ρ_{2} (y_{1}, y_{2})$

for all $(x_{1}, y_{1}), (x_{2}, y_{2}) \in X$ , Or $α, β, has$ And $b$ are nonnegative real numbers;

If the numbers $α, β, has$ And $b$ check the relationships

	$α + b + has β$	$< 2$
	$(1 - α) (1 - b)$	$> has β, b > 0, α > 0$

then the system ( 1 ) admits a single solution $({\bar{x}}_{1}, {\bar{x}}_{2}) \in X$ and the sequels ${(x_{1}^{(n)})}_{n = 0}^{\infty}, {(x_{2}^{(n)})}_{n = 0}^{\infty}$ are convergent:

lim_{n \to \infty} x_{1}^{(n)} = {\bar{x}}_{1}, lim_{n \to \infty} x_{2}^{(n)} = {\bar{x}}_{2} .

Demonstration.

We designate by ${(f_{n})}_{n = 0}^{\infty}$ And ${(g_{n})}_{n = 0}^{\infty}$ two sequences of non-negative numbers whose elements verify the relations

(3)		$f_{n}$	$\leq α f_{n - 1} + β g_{n - 1}$
	$g_{n}$	$\leq has f_{n} + b g_{n - 1}, n = 1, 2, \dots$

We associate with relations ( 3 ) the system of equations in the unknowns $h$ And $k$ following

(4)		$α + β h$	$= h k$
	$has k + b$	$= n k$

We will subsequently show that the system ( 4 ) admits a real solution $(h_{i}, k_{i})$ for which $0 \leq h_{i} k_{i} < 1$ if and only if the numbers $has, b, α$ And $β$ fulfill the conditions of Theorem 1 .

We assume that the system ( 4 ) admits the solutions $(h_{i}, k_{i})$ , $i = 1, 2$ for which the conditions $0 < h_{i} k_{i} < 1$ are fulfilled. We immediately check that the following equations result from the system ( 4 ):

(5)		$β h^{2} - (b + β has - α) h - α has$	$= 0$
	$has k^{2} - (α + β has - b) k - β b$	$= 0$

And

(6)

p^{2} - (b + β has + α) p - b α = 0

where we have designated by $p$ the product $h k .$ Equations ( 5 ) show us that the solutions $(h_{i}, k_{i}), i = 1, 2$ of the system ( 4 ) are real and of the equation ( 6 ) and the conditions $0 < p_{i} < 1,$ $i = 1, 2$ it results $f (0) > 0$ , $f (1) > 0$ And $b + β has + α < 2,$ Or $f (p) = p^{2} - (b + β has + α) p + b α$ . It is obvious that the condition $f (0) > 0$ is filled because $α > 0, b > 0$ , $f (1) > 0$ And $b + has β + α < 2$ represent the relations of the hypothesis of Theorem 1 .

If we now assume that the relations of the hypothesis of Theorem 1 are verified, then it is obvious that $f (1) > 0,$ $f (0) > 0$ And $b + has β + α < 2,$ which shows us that equation ( 6 ) has both positive roots and roots less than unity. Taking into account equations ( 5 ), we easily see that system ( 4 ) admits a solution $(h_{1}, k_{1})$ for which $h_{1} > 0, k_{1} > 0 .$

We now show that if the elements of the sequences ${(f_{n})}_{n = 0}^{\infty}$ And ${(g_{n})}_{n = 0}^{\infty}$ verify the relations ( 3 ) where the numbers $α, β, has$ And $b$ verify the hypothesis of Theorem 1 , then there exists a constant $C > 0$ , independent of $n$ such that for each $n = 1, 2, \dots$ relationships take place

(7)		$f_{n}$	$\leq C h_{1}^{n - 1} k_{1}^{n - 1}$
	$g_{n}$	$\leq C h_{1}^{n} k_{1}^{n - 1}$

and the series $\sum_{i = 0}^{\infty} f_{i}$ And $\sum_{i = 0}^{\infty} g_{i}$ are convergent.

Let us designate by $C$ a real number that satisfies the inequality

(8)

C \geq \max {α f_{0} + β g_{0}, \frac{has f_{1} + b g_{0}}{h_{1}}}

Or $(h_{1}, k_{1})$ is the positive solution of the system ( 4 ).

From ( 8 ) and ( 3 ) we deduce for $n = 1$

f_{1} \leq C and g_{1} \leq C h_{1}

so relations ( 7 ) are verified for $n = 1 .$

We assume that the following inequalities hold:

f_{k - 1} \leq C h_{1}^{k - 2} \cdot k_{1}^{k - 2}, g_{k - 1} \leq C h_{1}^{k - 1} \cdot k_{1}^{k - 2}, k = 2, \dots, n

From ( 3 ) and ( 4 ) we deduce

	$f_{n}$	$\leq α f_{n - 1} + β g_{n - 1} \leq C h_{1}^{n - 2} k_{1}^{n - 2} (α + β h_{1}) = C h_{1}^{n - 1} k_{1}^{n - 1}$
	$g_{n}$	$\leq has f_{n} + b g_{n - 1} \leq C h_{1}^{n - 1} k_{1}^{n - 2} (has k_{1} + b) = C h_{1}^{n} k_{1}^{n - 1}$

which shows us that relations ( 7 ) take place for all $n \in ℕ$ .

It is obvious that the series $\sum_{i = 0}^{\infty} f_{i}$ And $\sum_{i = 0}^{\infty} g_{i}$ are convergent, because from ( 7 ) it follows that they are bounded above by two geometric series with a ratio less than unity.

From ( 2 ) and the hypothesis of Theorem 1 we deduce the relations:

(9)		$ρ_{1} (x_{1}^{(n + 1)}, x_{1}^{(n)})$	$\leq α ρ_{1} (x_{1}^{(n)}, x_{1}^{(n - 1)}) + β ρ_{2} (x_{2}^{(n)}, x_{2}^{(n - 1)})$
	$ρ_{2} (x_{2}^{(n + 1)}, x_{2}^{(n)})$	$\leq has ρ_{1} (x_{1}^{(n + 1)}, x_{1}^{(n)}) + b ρ_{2} (x_{2}^{(n)}, x_{2}^{(n - 1)}), n = 0, 1, \dots$

We now set in ( 9 ) $f_{i} = ρ_{1} (x_{1}^{(i + 1)}, x_{1}^{(i)}), g_{i} = ρ_{2} (x_{2}^{(i + 1)}, x_{2}^{(i)}),$ $i = 0, 1, 2, \dots$ and we obtain the relations ( 3 ). Taking into account the hypotheses of Theorem 1 we deduce the relations

(10)		$ρ_{1} (x_{1}^{(i + 1)}, x_{1}^{(i)})$	$\leq C h_{1}^{i - 1} \cdot k_{1}^{i - 1}$
	$ρ_{2} (x_{2}^{(i + 1)}, x_{2}^{(i)})$	$\leq C h_{1}^{i} \cdot k_{1}^{i - 1}, i = 1, 2, \dots$

We will now show that suites ${(x_{1}^{(n)})}_{n = 0}^{\infty}$ And ${(x_{2}^{(n)})}_{n = 0}^{\infty}$ are convergent.

We have indeed

(11)		$ρ_{1} (x_{1}^{(n + s)}, x_{1}^{(n)})$	$\leq f_{n + s - 1} + f_{n + s - 2} + \dots + f_{n}$
		$\leq p_{1}^{n - 1} C (1 + p_{1} + p_{1}^{2} + \dots + p_{1}^{s - 1}) \leq \frac{C p_{1}^{n - 1}}{1 - p_{1}}$

Or $p_{1} = h_{1} k_{1}$ .

We deduce in the same way

(12)

ρ_{2} (x_{2}^{(n + s)}, x_{2}^{(n)}) \leq g_{n + s - 1} + g_{n + s - 2} + \dots + g_{n} \leq \frac{C h_{1} p_{1}^{n - 1}}{1 - p_{1}} .

Taking into account that $0 < p_{1} < 1$ and the fact that spaces $X_{1}$ And $X_{2}$ are complete, it follows that the sequences ${(x_{1}^{(n)})}_{n = 0}^{\infty}$ And ${(x_{2}^{(n)})}_{n = 0}^{\infty}$ are convergent.

If we pose ${\bar{x}}_{1} = lim_{n \to \infty} x_{1}^{(n)}$ and ${\bar{x}}_{2} = lim_{n \to \infty} x_{2}^{(n)}$ then taking into account the continuity of applications $F_{1}$ And $F_{2}$ and passing to the limit in the equalities ( 2 ) when $n \to \infty$ it follows that $({\bar{x}}_{1}, {\bar{x}}_{2})$ represents a solution of the system ( 1 ).

Regarding the uniqueness of the solution, we will assume by contradiction that the system ( 1 ) does not have a unique solution.

Let us designate by $({\bar{x}}_{1}, {\bar{x}}_{2})$ And $({\bar{y}}_{1}, {\bar{y}}_{2})$ two solutions of the system ( 1 ). It follows from the first condition of Theorem 1 :

	$ρ_{1} ({\bar{x}}_{1}, {\bar{y}}_{1})$	$\leq α ρ_{1} ({\bar{x}}_{1}, {\bar{y}}_{1}) + β ρ_{2} ({\bar{x}}_{2}, {\bar{y}}_{2})$
	$ρ_{2} ({\bar{x}}_{2}, {\bar{y}}_{2})$	$\leq has ρ_{1} ({\bar{x}}_{1}, {\bar{y}}_{1}) + b ρ_{2} ({\bar{x}}_{2}, {\bar{y}}_{2})$

	$(1 - α) ρ_{1} ({\bar{x}}_{1}, {\bar{y}}_{1})$	$\leq β ρ_{2} ({\bar{x}}_{2}, {\bar{y}}_{2})$
	$(1 - b) ρ_{2} ({\bar{x}}_{1}, {\bar{y}}_{2})$	$\leq has ρ_{1} ({\bar{x}}_{1}, {\bar{y}}_{1}) .$

We deduce from this

ρ_{1} ({\bar{x}}_{1}, {\bar{y}}_{1}) \leq \frac{β has}{(1 - α) (1 - b)} ρ_{1} ({\bar{x}}_{1}, {\bar{y}}_{1})

And

ρ_{2} ({\bar{x}}_{2}, {\bar{y}}_{2}) \leq \frac{β has}{(1 - α) (1 - b)} ρ_{2} ({\bar{x}}_{2}, {\bar{y}}_{2}),

but $\frac{β has}{(1 - α) (1 - b)} < 1$ therefore the last inequalities take place if and only if ${\bar{x}}_{1} = {\bar{y}}_{1}$ And ${\bar{x}}_{2} = {\bar{y}}_{2} .$

It follows from ( 11 ) and ( 12 ) that $s \to \infty$

ρ_{1} ({\bar{x}}_{1}, x_{1}^{(n)}) \leq \frac{C \cdot p_{1}^{n - 1}}{1 - p_{1}} e t ρ_{2} ({\bar{x}}_{2}, x_{2}^{(n)}) \leq \frac{C h_{1} p_{1}^{n}}{1 - p_{1}}, n = 1, 2, \dots

∎

Let us now designate by $F_{1}^{*} : X \to X_{1}, F_{2}^{*} : X \to X_{2}$ two other apps that check with $F_{1}$ And $F_{2}$ the conditions

(13)		$ρ_{1} (F_{1}^{*} (u, v), F_{1} (u, v))$	$\leq δ_{1}$
	$ρ_{2} (F_{2}^{*} (u, v), F_{2} (u, v))$	$\leq δ_{2}$

for everything $(u, v) \in X,$ Or $δ_{1} > 0, δ_{2} > 0$ are two given real numbers.

Besides the iterative process ( 2 ) we consider the following iterative process:

(14)		$ξ_{1}^{(n + 1)}$	$= F_{1}^{*} (ξ_{1}^{(n)}, ξ_{2}^{(n)})$
	$ξ_{2}^{(n + 1)}$	$= F_{2}^{*} (ξ_{1}^{(n + 1)}, ξ_{2}^{(n)}), n = 0, 1, \dots, (ξ_{1}^{(0)}, ξ_{2}^{(0)}) \in X .$

In the following we will proceed to the delimitation of errors in case the root $({\bar{x}}_{1}, {\bar{x}}_{2})$ of the system ( 1 ) is approximated by elements of the sequences ${(ξ_{1}^{(n)})}_{n = 0}^{\infty} {(ξ_{2}^{(n)})}_{n = 0}^{\infty}$ generated using the method ( 14 ). We obviously have no information concerning the applications $F_{1}^{*}$ And $F_{2}^{*}$ if not that they verify the relations ( 13 ), this is why we cannot affirm anything relative to the convergence of the sequences ${(ξ_{1}^{(n)})}_{n = 0}^{\infty}$ And ${(ξ_{2}^{(n)})}_{n = 0}^{\infty} .$

We will show later that the process of calculating the elements of sequences ${(ξ_{1}^{(n)})}_{n = 0}^{\infty}$ And ${(ξ_{2}^{(n)})}_{n = 0}^{\infty}$ can certainly be stopped when $ρ_{1} (ξ_{1}^{(n + 1)}, ξ_{1}^{(n)}) < ε_{1}$ And $ρ_{2} (ξ_{2}^{(n + 1)}, ξ_{2}^{(n)}) < ε_{2},$ if $ε_{1}$ And $ε_{2}$ are suitably chosen with respect to the numbers $δ_{1}, δ_{2}, has, b, α$ And $β .$

We have indeed

	$ρ_{1} (ξ_{1}^{(n + 1)}, ξ_{1}^{(n)}) =$	$ρ_{1} (F_{1}^{} (ξ_{1}^{(n)}, ξ_{2}^{(n)}), F_{1}^{} (ξ_{1}^{(n - 1)}, ξ_{2}^{(n - 1)}))$
	$\leq$	$ρ_{1} (F_{1}^{*} (ξ_{1}^{(n)}, ξ_{2}^{(n)}), F_{1} (ξ_{1}^{(n)}, ξ_{2}^{(n)}))$
		$+ ρ_{1} (F_{1} (ξ_{1}^{(n)}, ξ_{2}^{(n)}), F_{1} (ξ_{1}^{(n - 1)}, ξ_{2}^{(n - 1)}))$
		$+ ρ_{1} (F_{1} (ξ_{1}^{(n - 1)}, ξ_{2}^{(n - 1)}), F_{1}^{*} (ξ_{1}^{(n - 1)}, ξ_{2}^{(n - 1)}))$
	$\leq$	$α ρ_{1} (ξ_{1}^{(n)}, ξ_{1}^{(n - 1)}) + β ρ_{2} (ξ_{2}^{(n)}, ξ_{2}^{(n - 1)}) + 2 δ_{1} .$

For $ρ_{2} (ξ_{2}^{(n + 1)}, ξ_{2}^{(n)})$ we have the same way

ρ_{2} (ξ_{2}^{(n + 1)}, ξ_{2}^{(n)}) \leq has ρ_{1} (ξ_{1}^{(n + 1)}, ξ_{1}^{(n)}) + b ρ_{2} (ξ_{2}^{(n)}, ξ_{2}^{(n - 1)}) + 2 δ_{2} .

If we write now $f_{n}^{*} = ρ_{1} (ξ_{1}^{(n + 1)}, ξ_{1}^{(n)}), g_{n}^{*} = ρ_{2} (ξ_{2}^{(n + 1)}, ξ_{2}^{(n)}),$ $n = 0, 1, \dots$ then the above inequalities are written

(15)		$f_{n}^{*}$	$\leq α f_{n - 1}^{} + β g_{n - 1}^{} + 2 δ_{1}$
	$g_{n}^{*}$	$\leq has f_{n}^{} + b g_{n - 1}^{} + 2 δ_{2} .$

If we place ourselves in the hypotheses of Theorem 1 , then there exists a constant $C_{1}$ independent of $n$ , such that we have the following relations:

(16)		$f_{n}^{*}$	$\leq C_{1} h_{1}^{n - 1} k_{1}^{n - 1} + \frac{2 [β δ_{2} + (1 - b) δ_{1}]}{(1 - α) (1 - b) - has β}$
	$g_{n}^{*}$	$\leq C_{1} h_{1}^{n} k_{1}^{n - 1} + \frac{2 [(1 - α) δ_{2} + has δ_{1}]}{(1 - α) (1 - b) - has β}, n = 1, 2, \dots$

Or $(h_{1}, k_{1})$ is the solution of the system ( 4 ) which verifies the conditions: $0 < h_{1} k_{1} < 1, h_{1} > 0, k_{1} > 0$ .

Indeed, if we choose $C_{1}$ such as

	$C_{1} \geq \max$	${\| α f_{0}^{} + β g_{0}^{} + 2 δ_{1} - \frac{2 [β δ_{2} + (1 - b) δ_{1}]}{(1 - α) (1 - b) - has β} \|,$
		$\frac{1}{h_{1}} \| has f_{1}^{} + b g_{0}^{} + 2 δ_{2} - \frac{2 [(1 - α) δ_{2} + has δ_{1}]}{(1 - α) (1 - b) - has β} \|}$

so it is obvious that

	$f_{1}^{*}$	$\leq α f_{0}^{} + β g_{0}^{} + 2 δ_{1} \leq C_{1} + \frac{2 [β δ_{2} + (1 - b) δ_{1}]}{(1 - α) (1 - b) - has β}$
	$g_{1}^{*}$	$\leq has f_{1}^{} + b g_{0}^{} + 2 δ_{2} \leq C_{1} h_{1} + \frac{2 [(1 - α) δ_{2} + has δ_{1}]}{(1 - α) (1 - b) - has β}$

and inequalities ( 16 ) are verified for $n = 1$ .

We assume that relations ( 16 ) hold for each $n = 1, 2, \dots s$ and we will show that they also take place for $n = s + 1$ . We deduce from ( 15 )

	$f_{s + 1}^{*}$	$\leq α C_{1} h_{1}^{s - 1} k_{1}^{s - 1} + β C_{1} h_{1}^{s} k_{1}^{s - 1} + \frac{2 α [β δ_{2} + (1 - b) δ_{1}]}{(1 - α) (1 - b) - has β} + \frac{2 β [(1 - α) δ_{2} + has δ_{1}]}{(1 - α) (1 - b) - has β} + 2 δ_{1}$
		$= C_{1} h_{1}^{s - 1} k_{1}^{s - 1} (α + β h_{1}) + \frac{2 [β δ_{2} + (1 - b) δ_{1}]}{(1 - b) (1 - α) - has β}$
		$= C_{1} h_{1}^{s} k_{1}^{s} + \frac{2 [β δ_{2} + (1 - b) δ_{1}]}{(1 - α) (1 - b) - has β} .$

We show in the same way that the second inequality ( 16 ) also holds for $n = s + 1$ .

If now we assume that

(17)		$ε_{1}$	$> \frac{2 [β δ_{2} + (1 - b) δ_{1}]}{(1 - α) (1 - b) - has β}$
	$ε_{2}$	$> \frac{2 [(1 - α) δ_{2} + has δ_{1}]}{(1 - α) (1 - b) - has β}$

then relations ( 16 ) assure us that there exists a $n^{'} \in N$ such as for all $n > n^{'}$ we have inequalities $ρ_{1} (ξ_{1}^{(n + 1)}, ξ_{1}^{(n)}) < ε_{1}$ And $ρ_{2} (ξ_{2}^{(n + 1)}, ξ_{2}^{(n)}) < ε_{2} .$ Let us designate by $ε_{1}$ And $ε_{2}$ two positive numbers which verify the relations ( 17 ) and $n > n^{'} + 1$ . We propose to evaluate under these conditions the distances between ${\bar{x}}_{1}$ And $ξ_{1}^{(n + 1)}$ And ${\bar{x}}_{2}$ And $ξ_{2}^{(n + 1)},$ that is, the delimitation of errors in the case of solving the system ( 1 ) using an approximation process of the form ( 14 ) where the applications $F_{1}^{*}$ And $F_{2}^{*}$ depend on $F_{1}$ And $F_{2}$ by the relations ( 13 ). Taking into account the hypotheses in which we have placed ourselves, we will have

	$ρ_{1} ({\bar{x}}_{1}, ξ_{1}^{(n + 1)}) =$	$ρ_{1} (F_{1} ({\bar{x}}_{1}, {\bar{x}}_{2}), F_{1}^{*} (ξ_{1}^{(n)}, ξ_{2}^{(n)}))$
	$\leq$	$ρ_{1} (F_{1} ({\bar{x}}_{1}, {\bar{x}}_{2}), F_{1} (ξ_{1}^{(n)}, ξ_{2}^{(n)}))$
		$+ ρ_{1} (F_{1} (ξ_{1}^{(n)}, ξ_{2}^{(n)}), F_{1}^{*} (ξ_{1}^{(n)} L ξ_{2}^{(n)}))$
	$\leq$	$α ρ_{1} ({\bar{x}}_{1}, ξ_{1}^{(n)}) + β ρ_{2} ({\bar{x}}_{2}, ξ_{2}^{(n)}) + δ_{1} .$

We have similarly for $ρ_{2} ({\bar{x}}_{2}, ξ_{2}^{(n + 1)}) :$

ρ_{2} ({\bar{x}}_{2}, ξ_{2}^{(n + 1)}) \leq has ρ_{1} ({\bar{x}}_{1}, ξ_{1}^{(n + 1)}) + b ρ_{2} ({\bar{x}}_{2}, ξ_{2}^{(n)}) + δ_{2}

We deduce from the above inequalities

	$(1 - α) ρ_{1} ({\bar{x}}_{1}, ξ_{1}^{(n + 1)})$	$\leq α ε_{1} + β ρ_{2} ({\bar{x}}_{2}, ξ_{2}^{(n)}) + δ_{1}$
	$(1 - b) ρ_{2} ({\bar{x}}_{2}, ξ_{2}^{(n + 1)})$	$\leq has ρ_{1} ({\bar{x}}_{1}, ξ_{1}^{(n + 1)}) + b ε_{2} + δ_{2}$

that's to say

(18)		$ρ_{1} ({\bar{x}}_{1}, ξ_{1}^{(n + 1)})$	$\leq (\frac{α}{1 - α} ε_{1} + \frac{β}{1 - α} ρ_{2} ({\bar{x}}_{1}, ξ_{2}^{(n)}) + \frac{δ_{1}}{1 - α})$
	$ρ_{2} ({\bar{x}}_{2}, ξ_{2}^{(n + 1)})$	$\leq \frac{has}{1 - b} ρ_{1} ({\bar{x}}_{1}, ξ_{1}^{(n + 1)}) + \frac{b}{1 - b} ε_{2} + \frac{δ_{2}}{1 - b}$

from which it results

(19)

ρ_{2} ({\bar{x}}_{2}, ξ_{2}^{(n + 1)}) \leq \frac{has (α ε_{1} + β ε_{2}) + b (1 - α) ε_{2}}{(1 - α) (1 - b) - has β} + \frac{ε_{2}}{2}

Because $n > n^{'} + 1$ , it follows that the second inequality of ( 18 ) also holds in the case where we put $n$ instead of $n + 1,$ that's to say

ρ_{2} ({\bar{x}}_{2}, ξ_{2}^{(n)}) \leq \frac{has}{1 - b} ρ_{1} ({\bar{x}}_{1}, ξ_{1}^{(n)}) + \frac{b ε_{2}}{1 - b} + \frac{δ_{2}}{1 - b}

This inequality and the first inequality ( 18 ) give us

(20)

ρ_{1} ({\bar{x}}_{1}, ξ_{1}^{(n + 1)}) \leq \frac{β (has ε_{1} + b ε_{2}) + α ε_{1} (1 - b)}{(1 - α) (1 - b) - has β} + \frac{ε_{1}}{2}

Bibliography

[1] ^†^†margin: $\leftarrow$ clickable Păvăloiu, I., The resolution of operational systems using iterative methods, Mathematica, 11(34), 1969, 137–141.
[2] Urabe, M., Error estimation in numerical solution of equation by iteration processes , J. Sci. Hiroshima Univ., Ser. AI, 26 (1962), 77–91.

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Error estimations in the numerical solving of systems of equations in metric spaces

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