# On Approximating the Toader Mean by Other Bivariate Means

Authors: Jun-Li Wang; Wei-Mao Qian; Zai-Yin He; Yu-Ming Chu
Journal: Journal of Function Spaces
 (2019)

---

## Abstract

In the article, we provide several sharp bounds for the Toader mean by use of certain combinations of the arithmetic, quadratic, contraharmonic, and Gaussian arithmetic-geometric means.


---

## Body

## 1. Introduction

Letq be a real number, 0
<
u
<
1 and x
,
y
∈
R
+ with x
≠
y. Then the complete elliptic integrals K
(
u
) and E
(
u
) [1–22] of the first and second kinds, geometric mean G
(
x
,
y
), logarithmic mean L
(
x
,
y
), arithmetic mean A
(
x
,
y
), quadratic mean Q
(
x
,
y
), contraharmonic mean C
(
x
,
y
), second contraharmonic mean C
¯
(
x
,
y
), qth Hölder mean H
q
(
x
,
y
), and Toader mean T
(
x
,
y
) [23–27] of x and y are given by(1)
K
u
=
∫
0
π
/
2
1
-
u
2
sin
2
⁡
v
-
1
/
2
d
v
,
E
u
=
∫
0
π
/
2
1
-
u
2
sin
2
⁡
v
d
v
,
G
x
,
y
=
x
y
,
L
x
,
y
=
x
-
y
log
⁡
x
-
log
⁡
y
,
A
x
,
y
=
x
+
y
2
,
Q
x
,
y
=
x
2
+
y
2
2
,
C
x
,
y
=
x
2
+
y
2
x
+
y
,
C
¯
x
,
y
=
x
3
+
y
3
x
2
+
y
2
,
H
q
x
,
y
=
x
q
+
y
q
2
1
/
q
q
≠
0
,
H
0
x
,
y
=
x
yand(2)
T
x
,
y
=
2
π
∫
0
π
/
2
x
2
cos
2
⁡
v
+
y
2
sin
2
⁡
v
d
v
=
2
x
π
E
1
-
y
x
2
,
x
>
y
,
2
y
π
E
1
-
x
y
2
,
x
<
y
,respectively.The classical Gaussian arithmetic-geometric meanA
G
M
(
u
,
v
) of two positive real numbers u and v is defined by the common limit of the sequences u
n and v
n, which are given by (3)
u
0
=
u
,
v
0
=
v
,
u
n
+
1
=
u
n
+
v
n
2
,
v
n
+
1
=
u
n
v
n
.The well-known Gaussian identity [10] and (2) show that(4)
T
1
,
u
=
2
E
u
′
π
,
A
G
M
1
,
u
=
π
2
K
u
′for all 0
<
u
<
1, where u
′
=
1
-
u
2
1
/
2.Ifx
≠
y, then it is well known that the function q
↦
H
q
(
x
,
y
) is strictly increasing on the interval (
-
∞
,
∞
) and the inequalities(5)
G
x
,
y
=
H
0
x
,
y
<
L
x
,
y
<
A
x
,
y
=
H
1
x
,
y
<
Q
x
,
y
=
H
2
x
,
y
<
C
x
,
y
<
C
¯
x
,
yare valid.Recently, the bounds for the Toader meanT
(
x
,
y
) and Gaussian arithmetic-geometric means A
G
M
(
x
,
y
) have attracted the interest of many mathematicians. The following inequalities(6)
L
x
,
y
<
A
G
M
x
,
y
<
A
x
,
y
<
π
2
L
x
,
y
,
(7)
A
1
/
2
x
,
y
G
1
/
2
x
,
y
<
A
G
M
x
,
y
<
A
x
,
y
+
G
x
,
y
2
2for all x
,
y
∈
R
+ with x
≠
y were established in [28–32].Alzer and Qiu [12] and Barnard, Pearce, and Richards [33] proved that the double inequalities (8)
2
π
1
L
x
,
y
+
1
-
2
π
1
A
x
,
y
<
1
A
G
M
x
,
y
<
3
4
1
L
x
,
y
+
1
4
1
A
x
,
y
,
H
3
/
2
x
,
y
<
T
x
,
y
<
H
log
⁡
2
/
log
⁡
π
-
log
⁡
2
x
,
yhold for all x
,
y
∈
R
+ with x
≠
y.In [23, 34], the authors proved that α
=
1
/
2, β
=
(
4
-
π
)
/
[
(
2
-
1
)
π
]
=
0.6597
⋯, λ
=
1
/
4 and μ
=
4
/
π
-
1
=
0.2732
⋯ are the best possible parameters such that the double inequalities(9)
α
Q
x
,
y
+
1
-
α
A
x
,
y
<
T
x
,
y
<
β
Q
x
,
y
+
1
-
β
A
x
,
y
,
(10)
λ
C
x
,
y
+
1
-
λ
A
x
,
y
<
T
x
,
y
<
μ
C
x
,
y
+
1
-
μ
A
x
,
yhold for all x
,
y
∈
R
+ with x
≠
y.Qian, Song, Zhang, and Chu [35] proved that the two-sided inequalities (11)
λ
1
C
¯
x
,
y
+
1
-
λ
1
A
x
,
y
<
T
x
,
y
<
μ
1
C
¯
x
,
y
+
1
-
μ
1
A
x
,
y
,
C
¯
λ
2
x
+
1
-
λ
2
y
,
λ
2
y
+
1
-
λ
2
x
<
T
x
,
y
<
C
¯
μ
2
x
+
1
-
μ
2
y
,
μ
2
y
+
1
-
μ
2
xare valid for all x
,
y
∈
R
+ with x
≠
y if and only if λ
1
≤
1
/
8, μ
1
≥
4
/
π
-
1
=
0.2732
⋯, λ
2
≤
1
/
2
+
2
/
8
=
0.6767
⋯ and μ
2
≥
1
/
2
+
(
4
-
π
)
/
(
3
π
-
4
)
/
2
=
0.6988
⋯ if λ
1
,
μ
1
∈
(
0,1
/
2
) and λ
2
,
μ
2
∈
(
1
/
2,1
).Inequalities (5), (6), and (9) and the identity Q
(
x
,
y
)
=
A
(
x
,
y
)
C
(
x
,
y
) lead to(12)
A
G
M
x
,
y
<
A
x
,
y
<
T
x
,
y
<
Q
x
,
y
<
C
x
,
y
,
(13)
T
x
,
y
<
Q
x
,
y
=
A
x
,
y
C
x
,
y
,
(14)
T
x
,
y
>
Q
x
,
y
+
A
x
,
y
2
>
Q
x
,
y
A
x
,
y
>
Q
x
,
y
A
G
M
x
,
yfor all x
,
y
∈
R
+ with x
≠
y.Motivated by inequalities (12)–(14), in this article we deal with the optimality of the parameters α
i and β
i
(
i
=
1,2
,
3,4
) on the interval (
0,1
) such that (15)
α
1
C
x
,
y
+
1
-
α
1
A
G
M
x
,
y
<
T
x
,
y
<
β
1
C
x
,
y
+
1
-
β
1
A
G
M
x
,
y
,
α
2
Q
x
,
y
+
1
-
α
2
A
G
M
x
,
y
<
T
x
,
y
<
β
2
Q
x
,
y
+
1
-
β
2
A
G
M
x
,
y
,
α
3
C
x
,
y
+
1
-
α
3
A
G
M
x
,
y
<
T
2
x
,
y
A
x
,
y
<
β
3
C
x
,
y
+
1
-
β
3
A
G
M
x
,
y
,
α
4
Q
x
,
y
+
1
-
α
4
A
G
M
x
,
y
<
T
2
x
,
y
Q
x
,
y
<
β
4
Q
x
,
y
+
1
-
β
4
A
G
M
x
,
yfor all x
,
y
∈
R
+ with x
≠
y.


## 2. Lemmas

It is well known thatK
(
u
) and E
(
u
) satisfy the following formulas (see [10]): (16)
d
K
u
d
u
=
E
u
-
u
′
2
K
u
u
u
′
2
,
d
E
u
d
u
=
E
u
-
K
u
u
,
d
E
u
-
u
′
2
K
u
d
u
=
u
K
u
,
E
2
u
1
+
u
=
2
E
u
-
u
′
2
K
u
1
+
u
,
K
2
u
1
+
u
=
1
+
u
K
u
,
lim
u
→
0
+
⁡
K
u
=
lim
u
→
0
+
⁡
E
u
=
π
2
,
lim
u
→
1
-
⁡
K
u
=
∞
,
lim
u
→
1
-
⁡
E
u
=
1
.Lemma 1 (See [10, Theorem 1.25]).
Leta
,
b
∈
R with a
<
b, f
,
g
:
[
a
,
b
]
↦
R be continuous real-valued functions and differentiable on (
a
,
b
) with g
′
(
x
)
≠
0. Then both the functions (
f
(
x
)
-
f
(
a
)
)
/
(
g
(
x
)
-
g
(
a
)
) and (
f
(
x
)
-
f
(
b
)
)
/
(
g
(
x
)
-
g
(
b
)
) are (strictly) increasing (decreasing) on (
a
,
b
) if the function f
′
(
x
)
/
g
′
(
x
) is (strictly) increasing (decreasing) on (
a
,
b
).Lemma 2.
For the complete elliptic integralsK
(
u
) and E
(
u
), we have the following monotonicity results:
(1) The functionu
↦
E
(
u
)
-
u
′
2
K
(
u
)
/
u
2 is strictly increasing from (
0,1
) onto (
π
/
4,1
).
(2) The functionu
↦
K
(
u
) is strictly increasing from (
0,1
) onto (
π
/
2
,
∞
) and the function u
↦
E
(
u
) is strictly decreasing from (
0,1
) onto (
1
,
π
/
2
).
(3) The functionu
↦
r
′
λ
K
(
u
) is strictly decreasing from (
0,1
) onto (
0
,
π
/
2
) if λ
≥
1
/
2.
(4) The functionu
↦
(
1
+
u
′
2
)
K
(
u
)
-
2
E
(
u
) is strictly increasing from (
0,1
) onto (
0
,
∞
).
(5) The functionu
↦
K
(
u
)
-
E
(
u
)
/
u
2 is strictly increasing from (
0,1
) onto (
π
/
4
,
∞
).
(6) The functionu
↦
2
E
(
u
)
-
u
′
2
K
(
u
) is strictly increasing from (
0,1
) onto (
π
/
2,2
).
(7) The functionu
↦
φ
1
(
u
)
=
u
′
2
/
E
(
u
) is strictly decreasing from (
0,1
) onto (
0,2
/
π
).
(8) The functionu
↦
φ
2
(
u
)
=
1
+
u
2
E
(
u
)
-
u
′
2
K
(
u
)
/
u
2 is strictly increasing from (
0,1
) onto (
π
/
4
,
2
).
(9) The functionu
↦
φ
3
(
u
)
=
2
E
(
u
)
-
u
′
2
K
(
u
)
/
1
+
u
2 is strictly decreasing from (
0,1
) onto (
2
,
π
/
2
).
(10) The functionu
↦
φ
4
(
u
)
=
u
′
2
K
2
(
u
)
(
K
(
u
)
-
E
(
u
)
)
/
u
2 is strictly decreasing from (
0,1
) onto (
0
,
π
3
/
16
).
(11) The functionu
↦
φ
5
(
u
)
=
(
1
+
u
2
)
E
(
u
)
-
u
′
2
K
(
u
)
/
(
u
′
2
u
2
) is strictly increasing from (
0,1
) onto (
3
π
/
4
,
∞
).Proof.
Parts (1)–(6) can be found in [10, Theorem 3.21(1), (2), (7) and (8), and Exercise 3.43(4), (11) and (13)]. For part (7), it is not difficult to verify that(17)
φ
1
0
+
=
2
π
,
φ
1
1
-
=
0
,
(18)
φ
1
′
u
=
-
u
2
E
u
+
E
u
-
u
′
2
K
u
u
E
2
u
=
-
u
E
2
u
E
u
+
E
u
-
u
′
2
K
u
u
2
.
It follows from parts (1) and (2) together with (18) that(19)
φ
1
′
u
<
0for 0
<
u
<
1.
Therefore, part (7) follows from (17) and (19).
For part (8), simple computations lead to(20)
φ
2
0
+
=
π
4
,
φ
2
1
-
=
2
,
(21)
φ
2
′
u
=
2
K
u
-
2
E
u
-
u
2
E
u
u
3
1
+
u
2
=
1
+
u
′
2
K
u
-
2
E
u
u
3
1
+
u
2
+
u
1
+
u
2
K
u
-
E
u
u
2
.
From parts (4) and (5) together with (21) we clearly see that(22)
φ
2
′
u
>
0for 0
<
u
<
1.
Therefore, part (8) follows from (20) and (22).
For part (9), we clearly see that(23)
φ
3
0
+
=
π
2
,
φ
3
1
-
=
2
.
Differentiatingφ
3
(
u
) and making use of part (5) we get(24)
φ
3
′
u
=
-
u
u
′
2
1
+
u
2
3
/
2
K
u
-
E
u
u
2
<
0for 0
<
u
<
1.
Therefore, part (9) follows from (23) and (24).
For part (10), elaborated computation gives(25)
φ
4
0
+
=
π
3
16
,
φ
4
1
-
=
0
,
(26)
φ
4
′
u
=
-
K
u
u
K
u
u
2
+
K
u
-
E
u
u
2
1
+
u
′
2
K
u
-
2
E
u
.
It follows from parts (2), (4), and (5) together with (26) that(27)
φ
4
′
u
<
0for 0
<
u
<
1.
Therefore, part (10) follows from (25) and (27).
For part (11), it is not difficult to verify that(28)
φ
5
0
+
=
3
π
4
,
φ
5
1
-
=
∞
,
(29)
φ
5
′
u
=
u
4
+
5
u
2
-
2
E
u
+
2
2
u
4
-
3
u
2
+
1
K
u
u
3
u
′
4
.
Let(30)
φ
6
u
=
u
4
+
5
u
2
-
2
E
u
+
2
2
u
4
-
3
u
2
+
1
K
u
.
Then we clearly see that that(31)
φ
6
0
+
=
0
,
(32)
φ
6
′
u
=
11
u
3
5
11
E
u
+
E
u
-
u
′
2
K
u
u
2
.
Therefore, part (11) follows easily from parts (1) and (2) together with (28)–(32).


## 3. Main Results

Theorem 3.
The double inequality(33)
α
1
C
x
,
y
+
1
-
α
1
A
G
M
x
,
y
<
T
x
,
y
<
β
1
C
x
,
y
+
1
-
β
1
A
G
M
x
,
yholds for all x
,
y
∈
R
+ with x
≠
y if and only if α
1
≤
2
/
5 and β
1
≥
2
/
π
=
0.6366
⋯.Proof.
We clearly see thatC
(
x
,
y
), A
G
M
(
x
,
y
) and T
(
x
,
y
) are symmetric and homogenous of degree 1. Without loss of generality, we assume that x
>
y
>
0. Let u
=
(
x
-
y
)
/
(
x
+
y
). Then 0
<
u
<
1 and (2) and (4) lead to(34)
A
G
M
x
,
y
=
π
2
A
x
,
y
K
u
,
(35)
T
x
,
y
=
2
π
A
x
,
y
2
E
u
-
u
′
2
K
u
.
It follows from (34) and (35) together with C
(
x
,
y
)
=
A
(
x
,
y
)
(
1
+
u
2
) that(36)
T
x
,
y
-
A
G
M
x
,
y
C
x
,
y
-
A
G
M
x
,
y
=
2
/
π
2
E
u
-
u
′
2
K
u
-
π
/
2
K
u
1
+
u
2
-
π
/
2
K
u
=
2
/
π
K
u
2
E
u
-
u
′
2
K
u
-
π
/
2
1
+
u
2
K
u
-
π
/
2
≔
F
u
.
Letf
1
(
u
)
=
2
K
(
u
)
2
E
(
u
)
-
u
′
2
K
(
u
)
/
π
-
π
/
2 and g
1
(
u
)
=
(
1
+
u
2
)
K
(
u
)
-
π
/
2. Then elaborated computations lead to(37)
f
1
0
+
=
g
1
0
+
=
0
,
F
u
=
f
1
u
g
1
u
,
(38)
f
1
′
u
g
1
′
u
=
4
π
E
u
E
u
-
u
′
2
K
u
2
u
2
E
u
+
u
′
2
E
u
-
u
′
2
K
u
=
4
π
1
2
u
2
/
E
u
-
u
′
2
K
u
+
u
′
2
/
E
u
.
It follows from Lemma2(1) and (7) together with (38) that f
1
′
(
u
)
/
g
1
′
(
u
) is strictly increasing on (
0,1
). Then from Lemma 1 and (37) we know that F
(
u
) is strictly increasing on (
0,1
). Moreover,(39)
F
0
+
=
2
5
,
F
1
-
=
2
π
.
Therefore, Theorem3 follows from (36) and (39) together with the monotonicity of F
(
u
).Theorem 4.
The double inequality(40)
α
2
Q
x
,
y
+
1
-
α
2
A
G
M
x
,
y
<
T
x
,
y
<
β
2
Q
x
,
y
+
1
-
β
2
A
G
M
x
,
yholds for all x
,
y
∈
R
+ with x
≠
y if and only if α
2
≤
2
/
3 and β
2
≥
2
2
/
π
=
0.9003
⋯.Proof.
SinceQ
(
x
,
y
), A
G
M
(
x
,
y
), and T
(
x
,
y
) are symmetric and homogenous of degree 1, without loss of generality, we assume that x
>
y
>
0. Let u
=
(
x
-
y
)
/
(
x
+
y
). Then u
∈
(
0,1
), and (34) and (35) together with Q
(
x
,
y
)
=
A
(
x
,
y
)
1
+
u
2 lead to(41)
T
x
,
y
-
A
G
M
x
,
y
Q
x
,
y
-
A
G
M
x
,
y
=
2
/
π
2
E
u
-
u
′
2
K
u
-
π
/
2
K
u
1
+
u
2
-
π
/
2
K
u
≔
G
u
.
Letf
2
(
u
)
=
2
2
E
(
u
)
-
u
′
2
K
(
u
)
/
π
-
π
/
2
K
(
u
) and g
2
(
u
)
=
1
+
u
2
-
π
/
2
K
(
u
). Then simple computations lead to(42)
f
2
0
+
=
g
2
0
+
=
0
,
G
u
=
f
2
u
g
2
u
,
(43)
f
2
′
u
g
2
′
u
=
1
-
1
-
2
/
π
1
+
u
2
E
u
-
u
′
2
K
u
/
u
2
1
+
π
/
2
1
+
u
2
E
u
-
u
′
2
K
u
/
u
2
1
/
u
′
K
u
2
It follows from Lemma2(3) and (8) together with (43) that f
2
′
(
u
)
/
g
2
′
(
u
) is strictly increasing on (
0,1
). Then from Lemma 1 and (42) we know that G
(
u
) is strictly increasing on (
0,1
). Moreover,(44)
G
0
+
=
2
3
,
G
1
-
=
2
2
π
.
Therefore, Theorem4 follows from (41) and (44) together with the monotonicity of G
(
u
).Theorem 5.
The double inequality(45)
α
3
C
x
,
y
+
1
-
α
3
A
G
M
x
,
y
<
T
2
x
,
y
A
x
,
y
<
β
3
C
x
,
y
+
1
-
β
3
A
G
M
x
,
yholds for all x
,
y
∈
R
+ with x
≠
y if and only if α
3
≤
3
/
5 and β
3
≥
8
/
π
2
=
0.8105
⋯.Proof.
Without loss of generality, we assume thatx
>
y
>
0. Let u
=
(
x
-
y
)
/
(
x
+
y
)
∈
(
0,1
). Then it follows from (34), (35) and C
(
x
,
y
)
=
A
(
x
,
y
)
(
1
+
u
2
) that(46)
T
2
x
,
y
/
A
x
,
y
-
A
G
M
x
,
y
C
x
,
y
-
A
G
M
x
,
y
=
4
/
π
2
2
E
u
-
u
′
2
K
u
2
-
π
/
2
K
u
1
+
u
2
-
π
/
2
K
u
≔
H
u
.
Letf
3
(
u
)
=
4
2
E
(
u
)
-
u
′
2
K
(
u
)
2
/
π
2
-
π
/
2
K
(
u
) and g
3
(
u
)
=
(
1
+
u
2
)
-
π
/
2
K
(
u
). Then elaborated computations lead to(47)
f
3
0
+
=
g
3
0
+
=
0
,
H
u
=
f
3
u
g
3
u
,
(48)
f
3
′
u
g
3
′
u
=
1
-
1
-
4
/
π
2
2
E
u
-
u
′
2
K
u
E
u
-
u
′
2
K
u
/
u
2
1
+
π
/
4
E
u
-
u
′
2
K
u
/
u
2
1
/
u
′
K
u
2
.It follows from Lemma2(1), (3), and (6) together with (48) that f
3
′
(
u
)
/
g
3
′
(
u
) is strictly increasing on (
0,1
). Then from Lemma 1 and (47) we know that H
(
u
) is strictly increasing on (
0,1
). Moreover,(49)
H
0
+
=
3
5
,
H
1
-
=
8
π
2
.Therefore, Theorem5 follows from (46) and (49) together with the monotonicity of H
(
u
).Theorem 6.
The double inequality(50)
α
4
Q
x
,
y
+
1
-
α
4
A
G
M
x
,
y
<
T
2
x
,
y
Q
x
,
y
<
β
4
Q
x
,
y
+
1
-
β
4
A
G
M
x
,
yholds for all x
,
y
∈
R
+ with x
≠
y if and only if α
4
≤
1
/
3 and β
4
≥
8
/
π
2
=
0.8105
⋯.Proof.
Without loss of generality, we assume thatx
>
y
>
0. Let u
=
(
x
-
y
)
/
(
x
+
y
)
∈
(
0,1
). Then it follows from (34), (35) and Q
(
x
,
y
)
=
A
(
x
,
y
)
1
+
u
2 that(51)
T
2
x
,
y
/
Q
x
,
y
-
A
G
M
x
,
y
Q
x
,
y
-
A
G
M
x
,
y
=
4
/
π
2
2
E
u
-
u
′
2
K
u
2
-
π
1
+
u
2
/
2
K
u
1
+
u
2
-
π
1
+
u
2
/
2
K
u
=
1
-
1
-
4
/
π
2
2
E
u
-
u
′
2
K
u
/
1
+
u
2
2
1
-
π
/
2
1
+
u
2
K
u
≔
1
-
J
u
.
Letf
4
(
u
)
=
1
-
4
(
2
E
(
u
)
-
u
′
2
K
(
u
)
)
/
1
+
u
2
2
/
π
2 and g
4
(
u
)
=
1
-
π
/
2
1
+
u
2
K
(
u
). Then simple computations lead to(52)
f
4
0
+
=
g
4
0
+
=
0
,
J
u
=
f
4
u
g
4
u
,
(53)
f
4
′
u
g
4
′
u
=
16
π
3
2
E
u
-
u
′
2
K
u
1
+
u
2
u
′
2
K
u
2
K
u
-
E
u
1
+
u
2
E
u
-
u
′
2
K
u
=
16
π
3
2
E
u
-
u
′
2
K
u
1
+
u
2
u
′
2
K
2
u
K
u
-
E
u
/
u
2
1
+
u
2
E
u
-
u
′
2
K
u
/
u
′
2
u
2
.
It follows from Lemma2(9)–(11) and (53) that f
4
′
(
u
)
/
g
4
′
(
u
) is strictly increasing on (
0,1
). Then from Lemma 1 and (52) we know that J
(
u
) is strictly increasing on (
0,1
). Moreover,(54)
J
0
+
=
2
3
,
J
1
-
=
1
-
8
π
2
.
Therefore, Theorem6 follows from (51) and (54) together with the monotonicity of J
(
u
).Letx
=
1 and y
=
u
′. Then (2), (4) and Theorems 3–6 lead to Corollary 7 immediately.Corollary 7.
The double inequalities(55)
π
4
1
+
u
′
2
K
u
+
3
π
1
+
u
′
20
1
+
u
′
K
u
<
E
u
<
4
1
+
u
′
2
K
u
+
π
π
-
2
1
+
u
′
4
1
+
u
′
K
u
,
π
2
2
1
+
u
′
2
K
u
+
π
12
K
u
<
E
u
<
4
1
+
u
′
2
K
u
+
π
π
-
2
2
4
K
u
,
π
2
3
1
+
u
′
2
K
u
+
π
1
+
u
′
10
K
u
<
E
u
<
1
4
16
1
+
u
′
2
K
u
+
π
π
2
-
8
1
+
u
′
K
u
,
π
2
1
+
u
′
2
K
u
+
π
2
1
+
u
′
2
6
K
u
<
E
u
<
1
4
16
1
+
u
′
2
K
u
+
π
π
2
-
8
2
1
+
u
′
2
K
uhold for all 0
<
u
<
1.



---

*Source: 6082413-2019-03-03.xml*