1. Introduction
In this work we deal with finding a fixed point 
 of the equation
      
      where 
F is a Fréchet-differentiable operator defined on a convex subset 
D of a Banach space 
X with values into itself. By 
I we denote the identity linear operator in 
. The symbol 
 stands for the space of bounded linear operators from X into X.
Many applications from different areas, including education, reduce to dealing with Equation (
1) utilizing mathematical modelling [
1,
2,
3,
4,
5,
6,
7,
8,
9,
10,
11,
12,
13,
14,
15,
16,
17,
18,
19,
20,
21,
22,
23,
24]. However, the solution 
 is found in closed form only in rare cases. This problem leads to the usage of methods that are iterative in nature.
We study Stirling’s method given for all 
 by
      
      where 
. Further we will introduce an operator 
 such that 
 with 
, and denote 
 for use in later Sections.
This method converges quadratically as Newton’s method does, and also requires the same computational effort (see details in [
1,
22]). It is considered to be a useful alternative in cases where Newton’s method fails to converge (see such examples in [
22]). However, the usage of Stirling’s method has a drawback, since the convergence criteria require contractions. We have detected some other problems listed in Remarks 3 and 4. These drawbacks limit the applicability of Stirling’s method. In order to extend its applicability, we do not use contractive conditions in our semi-local as well as the local convergence results.
The rest of the work is structured as follows. 
Section 2 includes the semi-local convergence analysis. 
Section 3 contains the local analysis. The numerical results are given in 
Section 4.
  2. Semi-Local Convergence Analysis
Let 
 and 
. Consider a real sequence 
 as
      
Next, we study the convergence of sequence  by developing relevant lemmas and theorems.
Lemma 1. Then, sequence  generated for  by (4) is increasing, converges to its unique least upper bound , so thatwhereand  Proof.  It is convenient to first simplify sequence 
. Define sequence 
 by 
. Then, by (
4) we can write 
. Moreover, define sequence 
 by 
. Then, we can write 
. We have by (
4) that 
 and 
. Suppose that 
 and 
. Then, we get in turn that
        
        and
        
Hence, 
 is a decreasing sequence, so 
 and 
 are also decreasing sequences. In particular,
        
From , we get , so . That is sequence  is increasing, bounded from above by , so it converges to .
Next, we show (
4). We can write
        
        or
        
Using the estimate
        
        we get first an upper bound for 
 by (
5) and (
6) and the inequality 
 for 
:
        
        which together with 
 imply 
. The lower bound in (
4) is obtained similarly from the estimate:
        
 □
 Lemma 2. Then, sequence  is increasingly converging to .
 Proof.  We have 
 and 
. Then, by (
8), we get 
.
 In what follows the set denoted by  is a ball with center  and of radius .
To simplify, the notation, by  in this work, we denote the operator norm or the norm on the Banach space. The semi-local convergence analysis is based on the conditions ():
- ()
-  is a Fréchet differentiable operator and there exist  -  such that  -  with
           - 
          and
           
- ()
- There exist  -  such that for each  - 
          and
           
- ()
- Let  -  and  - . There exist  -  such that for each  - 
          and
           
- ()
- Hypotheses of Lemmas 1 and 2 hold with
           - 
          and
           
- ()
- ()
- The ball  -  is constructed such that
           
We suppose from now on that the conditions () hold.
Next, the semi-local convergence result is given for Stirling’s method (
2).
Theorem 1. Under conditions (), sequence  generated by Stirling’s method (2) is well defined, remains in  for each  and converges to  which satisfies  with Q-order of convergence 2. Moreover, the following estimates holdand  is the only fixed point of F in , with  Proof.  Let 
. We get by (
) and (
) that
        
        so 
. Using (
) and the Lemmas 1 and 2, we have in turn that
        
By the Lemma of Banach on invertible operators [
21] (Perturbation Lemma 2.3.2, p. 45) 
, and
        
Using Stirling’s method (
2):
        
Then, in view of (
), (
) and Equation (
11), we obtain in turn that
        
Next, we can connect the preceding estimates on sequence 
 with 
. Indeed, we get by (
) and Equation (
3) that
        
By induction, Equations (
3), (
4), (
10) and (
12), we have in turn that
        
Hence, 
 defined by Equation (
3) is a majorizing sequence for 
. By Lemmas 1 and 2, sequence 
 is complete as convergent to 
. It then follows by Equation (
13) that sequence 
 is also complete so it converges to some 
. By the estimate (see (
12))
        
        we deduce that 
 by letting 
. Estimate 
 follows from Equation (
13) and for 
, we get that
        
        which implies that the 
Q-order convergence of Stirling’s method (
2) is two. Furthermore, to show the uniqueness part, let 
 with 
. Define the operator 
Q by 
 In view of (
) and (
), we obtain in turn that
        
Then, by (
15) 
. Finally, we obtain 
 using the identity
        
 □
 Remark 1. -  (a)
- The Stirling’s method usual conditions corresponding to () (first condition) are given by [22]: - ()’
-  for each  and . 
 - That is, operator F must be a contraction on D. Moreover, the convergence of Stirling’s method was shown in [22] under (),  and . However, in the present study no such assumption is made. Hence, the applicability of Stirling’s method (2) is extended. Notice also that we can have  and c can be chosen as . 
-  (b)
- Estimate (4) is similar to the sufficient convergence Kantorovich-type criteria for the semi-local convergence of Newton’s method given by us in [4]. However, the constants  and  are the center-Lipschitz and Lipschitz constants for operator F (see also part (e)). 
-  (c)
- If set  is switched by , since  and the iterates remain in  the results can be improved even further. The corresponding constants to b and  will be at least as small. 
-  (d)
- In view of the proof of Theorem 1, scalar sequence  defined byis also a majorizing sequence for Stirling’s method (2), whereand 
-  (e)
- Newton’s method for Equation (1) is given for all  by - Consider, items , , , , , , , , , ,  and , corresponding to c, γ, , L, , , , b, , ,  and h respectively as - The scalar sequence  is defined as - Then, Stirling’s method sufficient convergence criteria, error bounds and information on the uniqueness of the solution are better than Newton’s method when the "bar" constants and sets are smaller than the non bar constants. Similar favorable comparison can be made in the local convergence case that follows. 
   3. Local Convergence
The conditions (
) are used in the local convergence analysis of Stirling’s method (
2):
- ()
-  is a Fréchet differentiable operator, and there exists  such that  and . 
- ()
- There exist  -  such that for each  - 
          and
           
- ()
- Let  - . There exists  -  such that for each  
- ()
- The ball  -  is constructed such that  - ,
			where
           
Theorem 2. Suppose that conditions  hold. Then, sequence  generated for  by Stirling’s method (2) is well defined in , remains in  for each  and converges to . Moreover, the following inequality holds Furthermore, if ,  is the only fixed point of F on .
 Proof.  We shall show using mathematical induction that sequence 
 is well defined, remains in 
 and converges to 
 so that (
16) is satisfied. We have by (
) and (
) for 
 that
        
        so 
. Then by (
)
        
Hence, 
 and
        
In particular, (
18) holds for 
, which shows that 
 is well defined by Stirling’s method for 
. We can write by (
) that
        
We get in turn by (
) and (
)
        
Then, by (
18)–(
20), we get that also
        
        so (
16) holds for 
 and 
. Switch 
 by 
 in the preceding estimates, we arrive at (
16). In view of the estimate 
, we conclude that 
 and 
. Let 
 in (
15) to show the uniqueness part. □
 Remark 2. The local results in the literature use ()’ and . But () is weaker than ()’. Hence, we extend the applicability of Stirling’s method (2) in the local case too.    4. Numerical Example with Concluding Remarks
In the next example, we compare Stirling’s method with Newton’s method.
Example 1. Let . Consider function F on D as  Clearly, the quadratic polynomial joins smoothly with the linear parts.
      
- (I)
- Semilocal case (i). If we choose , we see that . Moreover, the semi-local convergence criteria of Theorem 1 are satisfied (with ,  and ). 
- (II)
- Local convergence criteria of Theorem 2 (with , since the derivative of the quadratic polynomial satisfies  for all ). 
- (III)
- In  Table 1-  and  Table 2-  we present some cases in which Stirling’s method stands better than Newton’s one. 
In the current study, we have successfully demonstrated our claims on Stirling’s method by focusing on very classic problems, but in the future we will consider studying other complex problems such us solving symmetric ordinary differential equations with a more favorable theory.
   
  
    Author Contributions
All authors have equally contributed to this work. All authors have read and agreed to the published version of the manuscript.
Funding
Research supported in part by Séneca 20928/PI/18 and by MINECO PGC2018-095896-B-C21.
Conflicts of Interest
The authors have no conflict of interest. The funders had no role in the design of the study; in the collection, analyses, or interpretation of data; in the writing of the manuscript, or in the decision to publish the results.
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    Table 1.
    Iteration of Newton’s and Stirling’s method with different starting points.
  
 
  
      Table 1.
    Iteration of Newton’s and Stirling’s method with different starting points.
      
        | Iteration | Newton’s Method | Stirling’s Method | 
|---|
| 0 |  |  | 
| 1 |  |  | 
| 2 |  | 0 | 
      
 
  
    
  
  
    Table 2.
    Iteration of Newton’s and Stirling’s method with different starting points.
  
 
  
      Table 2.
    Iteration of Newton’s and Stirling’s method with different starting points.
      
        | Iteration | Newton’s Method | Stirling’s Method | 
|---|
| 0 |  |  | 
| 1 | ∞ |  | 
| 2 | − | 0 | 
      
 
      
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