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PureMathematics
n
Ø
ê
Æ
,2021,11(1),41-46
PublishedOnlineJanuary2021inHans.http://www.hanspub.org/journal/pm
https://doi.org/10.12677/pm.2021.111007
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HigherOrderofSchwarzianDerivativeof
theHarmonicFunctions
YutongLiu,YiQi
SchoolofMathematicalScience,BeihangUniversity,Beijing
Received:Dec.12
th
,2020;accepted:Jan.11
th
,2021;published:Jan.18
th
,2021
Abstract
Inthispaper,wedefinethehigherorderofSchwarzianderivativeoftheharmonic
©
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n
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DOI:10.12677/pm.2021.111007
4
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functions.Wealsoprovethatit isstillM¨obiusinvariant.Finally, wegiveanequivalent
characterizationofthehigherorderofSchwarzianderivativeoftheharmonicfunctions.
Keywords
HarmonicFunction,HigherOrderofSchwarzianDerivative
Copyright
c
2021byauthor(s)andHansPublishersInc.
This work is licensed undertheCreative Commons Attribution International License (CCBY 4.0).
http://creativecommons.org/licenses/by/4.0/
1.
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DOI:10.12677/pm.2021.11100742
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DOI:10.12677/pm.2021.11100743
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§
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(
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(
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(
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+
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+
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(
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•
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(
w
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σ
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(
f
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)
·
(
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0
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−
1
y
²
:
n
= 3
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¤
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(
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1)
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(
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(
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σ
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−
(
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(
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(
f
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+(
k
−
1)(
σ
k
(
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(
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k
−
2
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(
k
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1)(
σ
k
(
f
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)(
P
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(
f
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)(
R
0
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−
(
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−
1)(
σ
k
(
f
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◦
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·
(
R
0
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k
−
1
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00
= (
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k
+1
(
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◦
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(
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k
.
·
‚
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ò
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8
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y
²
±
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2.2.
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f
=
h
+
g
´
ü
S
Û
Ü
ü
“
N
Ú¼
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b
¼
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f
•
3
‡
¼
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f
−
1
§
…
f
−
1
•
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Ú
¼
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K
∂
n
−
3
∂w
n
−
3
σ
3
(
f
−
1
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f
= (
h
0
J
f
)
n
−
1
·
σ
n
(
f
)(7)
y
²
µ
DOI:10.12677/pm.2021.11100744
n
Ø
ê
Æ
4
ˆ
Õ
§
Ö
À
Ä
k
§
w
=
f
(
z
)
§
·
‚
•
§
f
−
1
w
=
h
0
J
f
,f
−
1
w
=
−
g
0
J
f
,J
f
−
1
=
|
f
−
1
w
|
2
−|
f
−
1
w
|
2
=
1
J
f
.
¿
…
½
Â
∂z
∂w
=
f
−
1
w
"
n
= 3
ž
§
σ
3
(
f
) =
∂
2
∂z
2
logJ
f
−
1
2
(
∂
∂z
logJ
f
)
2
=
∂
2
∂z
2
J
f
J
f
−
3
2
(
∂
∂z
J
f
)
2
J
2
f
Ù
¥
§
∂
∂z
J
f
=
h
00
h
0
−
g
00
g
0
,
∂
2
∂z
2
J
f
=
h
000
h
0
−
g
000
g
0
.
a
q
/
σ
3
(
f
−
1
) =
∂
2
∂w
2
J
f
−
1
J
f
−
1
−
3
2
(
∂
∂w
J
f
−
1
)
2
J
2
f
−
1
K
σ
3
(
f
−
1
)
◦
f
=
J
f
·
(
∂
2
∂z
2
1
J
f
)
·
(
∂z
∂w
)
2
+
J
f
·
(
∂
∂z
1
J
f
)
·
(
∂z
∂w
)
·
∂
∂z
(
∂z
∂w
)
−
3
2
J
2
f
(
∂
∂z
1
J
f
)
2
·
(
∂z
∂w
)
2
=
(
h
0
)
2
J
f
·
(
∂
2
∂z
2
J
f
)
J
2
f
−
2(
∂
∂z
J
f
)
2
J
f
J
4
f
−
(
h
0
∂
∂z
J
f
J
2
f
)
2
+
3
2
(
h
0
)
2
(
∂
∂z
J
f
J
2
f
)
2
=
(
h
0
)
2
J
2
f
·
(
∂
2
∂z
2
J
f
J
f
−
3
2
(
∂
∂z
J
f
)
2
J
2
f
)
.
¤
±
§
n
= 3
ž
§
(7)
ª
¤
á
"
n>
3
ž
§
¦
^
ê
Æ
8
B
{
y
²
"
b
n
=
k
ž
§
(
Ø
¤
á
"
n
=
k
+1
ž
§
∂
k
−
2
∂w
k
−
2
σ
3
(
f
−
1
)
◦
f
=
∂
∂z
(
h
0
J
f
)
k
−
1
·
σ
k
(
f
)
∂z
∂w
=
−
(
k
−
1)
σ
k
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