Proof.
Let
{
x
n
}
be a sequence of approximations generated by an iterative method with memory (12). If this sequence converges to the zero
α
of
f
with the
R
-order (
≥
r
) of (12), then we write [11]
(13)
e
n
+
1
~
D
n
,
r
e
n
r
,
e
n
=
x
n
-
α
,
where
D
n
,
r
tends to the asymptotic error constant
D
r
of (12) when
n
→
∞
. Therefore,
(14)
e
n
+
1
~
D
n
,
r
(
D
n
-
1
,
r
e
n
-
1
r
)
r
=
D
n
,
r
D
n
-
1
,
r
r
e
n
-
1
r
2
.
Assume that the iterative sequences
{
w
n
}
,
{
y
n
}
, and
{
z
n
}
have the
R
-orders
q
,
p
, and
s
, respectively; then, bearing in mind (13), we obtain
(15)
e
¯
n
~
D
n
,
q
e
n
q
=
D
n
,
q
(
D
n
-
1
,
r
e
n
-
1
r
)
q
=
D
n
,
q
D
n
-
1
,
r
q
e
n
-
1
r
q
,
(16)
e
~
n
~
D
n
,
p
e
n
p
=
D
n
,
p
(
D
n
-
1
,
r
e
n
-
1
r
)
p
=
D
n
,
p
D
n
-
1
,
r
p
e
n
-
1
r
p
,
(17)
e
^
n
~
D
n
,
s
e
n
s
=
D
n
,
s
(
D
n
-
1
,
r
e
n
-
1
r
)
s
=
D
n
,
s
D
n
-
1
,
r
s
e
n
-
1
r
s
.
Let
ϕ
k
for
k
=
1
be defined by (2). We now obtain the order of convergence of the methods with memory (12), where
β
n
is calculated from (6). The error relations with the self-accelerating parameter
β
n
for (12) are in what follows (where
c
i
=
f
(
i
)
(
α
)
/
(
i
!
f
′
(
α
)
)
,
i
=
2,3
,
…
):
(18)
e
¯
n
=
w
n
-
α
~
(
1
+
β
n
f
′
(
α
)
)
e
n
,
(19)
e
~
n
=
y
n
-
α
~
c
2
(
1
+
β
n
f
′
(
α
)
)
e
n
2
,
(20)
e
^
n
=
z
n
-
α
~
c
n
,
4
(
1
+
β
n
f
′
(
α
)
)
2
e
n
4
,
(21)
e
n
+
1
=
x
n
+
1
-
α
~
c
n
,
8
(
1
+
β
n
f
′
(
α
)
)
3
e
n
8
.
In order to find the error relation for (12), we need to find the expression for
1
+
β
n
f
′
(
α
)
. Using a symbolic software such as
M
a
t
h
e
m
a
t
i
c
a
8 with the use of (6), we attain that
(22)
1
+
β
n
f
′
(
α
)
~
c
4
e
n
-
1
e
~
n
-
1
e
^
n
-
1
.
According to (13), (16), (17), (19), and (22), we obtain
(23)
e
~
n
~
c
2
c
4
e
n
-
1
e
~
n
-
1
e
^
n
-
1
e
n
2
~
c
2
c
4
e
n
-
1
D
n
-
1
,
p
e
n
-
1
p
D
n
-
1
,
s
e
n
-
1
s
(
D
n
-
1
,
r
e
n
-
1
r
)
2
~
c
2
c
4
D
n
-
1
,
p
D
n
-
1
,
s
D
n
-
1
,
r
2
e
n
-
1
1
+
p
+
s
+
2
r
.
Similarly, by (13), (16), (17), (20), and (22), we can write
(24)
e
^
n
~
c
n
,
4
(
c
4
e
n
-
1
e
~
n
-
1
e
^
n
-
1
)
2
e
n
4
~
c
n
,
4
(
c
4
e
n
-
1
D
n
-
1
,
p
e
n
-
1
p
D
n
-
1
,
s
e
n
-
1
s
)
2
(
D
n
-
1
,
r
e
n
-
1
r
)
4
~
c
n
,
4
c
4
2
e
n
-
1
2
D
n
-
1
,
p
2
e
n
-
1
2
p
D
n
-
1
,
s
2
e
n
-
1
2
s
D
n
-
1
,
r
4
e
n
-
1
4
r
~
c
n
,
4
c
4
2
D
n
-
1
,
p
2
D
n
-
1
,
s
2
D
n
-
1
,
r
4
e
n
-
1
2
+
2
p
+
2
s
+
4
r
.
Combining (13), (16), (17), (21), and (22) yields
(25)
e
n
+
1
~
c
n
,
8
(
c
4
e
n
-
1
e
~
n
-
1
e
^
n
-
1
)
3
e
n
8
~
c
n
,
8
(
c
4
e
n
-
1
D
n
-
1
,
p
e
n
-
1
p
D
n
-
1
,
s
e
n
-
1
s
)
3
(
D
n
-
1
,
r
e
n
-
1
r
)
8
~
c
n
,
8
c
4
3
e
n
-
1
3
D
n
-
1
,
p
3
e
n
-
1
3
p
D
n
-
1
,
s
3
e
n
-
1
3
s
D
n
-
1
,
r
8
e
n
-
1
8
r
~
c
n
,
8
c
4
3
D
n
-
1
,
p
3
D
n
-
1
,
s
3
D
n
-
1
,
r
8
e
n
-
1
3
+
3
p
+
3
s
+
8
r
.
Equating the exponents of the error
e
n
-
1
in pairs of relations (19) and (23), (20) and (24), and then (21) and (25), we arrive at the following system of equations:
(26)
r
p
-
1
-
p
-
s
-
2
r
=
0
,
r
s
-
2
-
2
p
-
2
s
-
4
r
=
0
,
r
2
-
3
-
3
p
-
3
s
-
8
r
=
0
.
Positive solution of this system is
p
=
(
1
/
6
)
(
7
+
109
)
,
s
=
(
1
/
3
)
(
7
+
109
)
, and
r
=
(
1
/
2
)
(
11
+
109
)
=
10.7202
. Therefore, the
R
-order of the methods with memory (12), when
β
n
is calculated by (6), is at least 10.7202.
Now, using a symbolic software such as
M
a
t
h
e
m
a
t
i
c
a
8 with the use of (7), we attain that
(27)
1
+
β
n
f
′
(
α
)
~
c
4
e
¯
n
-
1
e
~
n
-
1
e
^
n
-
1
.
Combining (13), (15), (16), (17), (18), and (27), we obtain
(28)
e
¯
n
~
c
4
e
¯
n
-
1
e
~
n
-
1
e
^
n
-
1
e
n
~
c
4
D
n
-
1
,
q
e
n
-
1
q
D
n
-
1
,
p
e
n
-
1
p
D
n
-
1
,
s
e
n
-
1
s
D
n
-
1
,
r
e
n
-
1
r
~
c
4
D
n
-
1
,
q
D
n
-
1
,
p
D
n
-
1
,
s
D
n
-
1
,
r
e
n
-
1
q
+
p
+
s
+
r
.
In the similar way, we find the following error relations:
(29)
e
~
n
~
c
2
c
4
e
¯
n
-
1
e
~
n
-
1
e
^
n
-
1
e
n
2
~
c
2
c
4
D
n
-
1
,
q
e
n
-
1
q
D
n
-
1
,
p
e
n
-
1
p
D
n
-
1
,
s
e
n
-
1
s
(
D
n
-
1
,
r
e
n
-
1
r
)
2
~
c
2
c
4
D
n
-
1
,
q
D
n
-
1
,
p
D
n
-
1
,
s
D
n
-
1
,
r
2
e
n
-
1
q
+
p
+
s
+
2
r
,
(30)
e
^
n
~
c
n
,
4
(
c
4
e
¯
n
-
1
e
~
n
-
1
e
^
n
-
1
)
2
e
n
4
~
c
n
,
4
(
c
4
D
n
-
1
,
q
e
n
-
1
q
D
n
-
1
,
p
e
n
-
1
p
D
n
-
1
,
s
e
n
-
1
s
)
2
(
D
n
-
1
,
r
e
n
-
1
r
)
4
~
c
n
,
4
c
4
2
D
n
-
1
,
q
2
e
n
-
1
2
q
D
n
-
1
,
p
2
e
n
-
1
2
p
D
n
-
1
,
s
2
e
n
-
1
2
s
D
n
-
1
,
r
4
e
n
-
1
4
r
~
c
n
,
4
c
4
2
D
n
-
1
,
q
2
D
n
-
1
,
p
2
D
n
-
1
,
s
2
D
n
-
1
,
r
4
e
n
-
1
2
q
+
2
p
+
2
s
+
4
r
,
(31)
e
n
+
1
~
c
n
,
8
(
c
4
e
¯
n
-
1
e
~
n
-
1
e
^
n
-
1
)
3
e
n
8
~
c
n
,
8
(
c
4
D
n
-
1
,
q
e
n
-
1
q
D
n
-
1
,
p
e
n
-
1
p
D
n
-
1
,
s
e
n
-
1
s
)
3
(
D
n
-
1
,
r
e
n
-
1
r
)
8
~
c
n
,
8
c
4
3
D
n
-
1
,
q
3
e
n
-
1
3
q
D
n
-
1
,
p
3
e
n
-
1
3
p
D
n
-
1
,
s
3
e
n
-
1
3
s
D
n
-
1
,
r
8
e
n
-
1
8
r
~
c
n
,
8
c
4
3
D
n
-
1
,
q
3
D
n
-
1
,
p
3
D
n
-
1
,
s
3
D
n
-
1
,
r
8
e
n
-
1
3
q
+
3
p
+
3
s
+
8
r
.
Comparing the exponents of
e
n
-
1
on the right hand sides of (18) and (28), (19) and (29), (20) and (30), and then (21) and (31), we arrive at the following system of equations:
(32)
r
q
-
q
-
p
-
s
-
r
=
0
,
r
p
-
q
-
p
-
s
-
2
r
=
0
,
r
s
-
2
q
-
2
p
-
2
s
-
4
r
=
0
,
r
2
-
3
q
-
3
p
-
3
s
-
8
r
=
0
.
Positive solution of this system is
q
=
2
,
p
=
3
,
s
=
6
, and
r
=
11
. Therefore, the
R
-order of the methods with memory (12), when
β
n
is calculated by (7), is at least 11.
Using a symbolic software such as
M
a
t
h
e
m
a
t
i
c
a
8 with the use of (8), we attain that
(33)
1
+
β
n
f
′
(
α
)
~
c
5
e
n
-
1
e
¯
n
-
1
e
~
n
-
1
e
^
n
-
1
.
Using (33) and previously derived relations, we obtain the following error relations for the intermediate approximations:
(34)
e
¯
n
~
c
5
e
n
-
1
e
¯
n
-
1
e
~
n
-
1
e
^
n
-
1
e
n
~
c
5
e
n
-
1
D
n
-
1
,
q
e
n
-
1
q
D
n
-
1
,
p
e
n
-
1
p
D
n
-
1
,
s
e
n
-
1
s
D
n
-
1
,
r
e
n
-
1
r
~
c
5
D
n
-
1
,
q
D
n
-
1
,
p
D
n
-
1
,
s
D
n
-
1
,
r
e
n
-
1
1
+
q
+
p
+
s
+
r
,
(35)
e
~
n
~
c
2
c
5
e
n
-
1
e
¯
n
-
1
e
~
n
-
1
e
^
n
-
1
e
n
2
~
c
2
c
5
e
n
-
1
D
n
-
1
,
q
e
n
-
1
q
D
n
-
1
,
p
e
n
-
1
p
D
n
-
1
,
s
e
n
-
1
s
(
D
n
-
1
,
r
e
n
-
1
r
)
2
~
c
2
c
5
D
n
-
1
,
q
D
n
-
1
,
p
D
n
-
1
,
s
D
n
-
1
,
r
2
e
n
-
1
1
+
q
+
p
+
s
+
2
r
,
(36)
e
^
n
~
c
n
,
4
(
c
5
e
n
-
1
e
¯
n
-
1
e
~
n
-
1
e
^
n
-
1
)
2
e
n
4
~
c
n
,
4
(
c
5
e
n
-
1
D
n
-
1
,
q
e
n
-
1
q
D
n
-
1
,
p
e
n
-
1
p
D
n
-
1
,
s
e
n
-
1
s
)
2
(
D
n
-
1
,
r
e
n
-
1
r
)
4
~
c
n
,
4
c
5
2
e
n
-
1
2
D
n
-
1
,
q
2
e
n
-
1
2
q
D
n
-
1
,
p
2
e
n
-
1
2
p
D
n
-
1
,
s
2
e
n
-
1
2
s
D
n
-
1
,
r
4
e
n
-
1
4
r
~
c
n
,
4
c
5
2
D
n
-
1
,
q
2
D
n
-
1
,
p
2
D
n
-
1
,
s
2
D
n
-
1
,
r
4
e
n
-
1
2
+
2
q
+
2
p
+
2
s
+
4
r
.
In the similar fashion, we find the final error relation (21) which is given by
(37)
e
n
+
1
~
c
n
,
8
(
c
5
e
n
-
1
e
¯
n
-
1
e
~
n
-
1
e
^
n
-
1
)
3
e
n
8
~
c
n
,
8
(
c
5
e
n
-
1
D
n
-
1
,
q
e
n
-
1
q
D
n
-
1
,
p
e
n
-
1
p
D
n
-
1
,
s
e
n
-
1
s
)
3
×
(
D
n
-
1
,
r
e
n
-
1
r
)
8
~
c
n
,
8
c
5
3
e
n
-
1
3
D
n
-
1
,
q
3
e
n
-
1
3
q
D
n
-
1
,
p
3
e
n
-
1
3
p
D
n
-
1
,
s
3
e
n
-
1
3
s
D
n
-
1
,
r
8
e
n
-
1
8
r
~
c
n
,
8
c
5
3
D
n
-
1
,
q
3
D
n
-
1
,
p
3
D
n
-
1
,
s
3
D
n
-
1
,
r
8
e
n
-
1
3
+
3
q
+
3
p
+
3
s
+
8
r
.
Comparing the exponents of
e
n
-
1
on the right hand sides of (18) and (34), (19) and (35), (20) and (36), and then (21) and (37), we arrive at the following system of equations:
(38)
r
q
-
q
-
p
-
s
-
r
-
1
=
0
,
r
p
-
q
-
p
-
s
-
2
r
-
1
=
0
,
r
s
-
2
q
-
2
p
-
2
s
-
4
r
-
2
=
0
,
r
2
-
3
q
-
3
p
-
3
s
-
8
r
-
3
=
0
.
Positive solution of this system is
q
=
(
1
/
3
)
(
1
+
2
7
)
,
p
=
(
2
/
3
)
(
2
+
7
)
,
s
=
(
4
/
3
)
(
2
+
7
)
, and
r
=
2
(
3
+
7
)
=
11.2915
. Therefore, the
R
-order of the methods with memory (12), when
β
n
is calculated by (8), is at least 11.2915.