पांचवीं घात वाले समीकरण (quintic equation) बहुपद समीकरणों के समूह से संबंधित हैं। इन समीकरणों में कम से कम एक अज्ञात मान निर्धारित किया जाना है और कम से कम दो स्थिर गुणांक हैं। आधार के रूप में इस अज्ञात मान की घातें और घातांक के रूप में 0 से 5 तक की प्राकृतिक संख्याएं इन समीकरणों में एक रैखिक संयोजन में जुड़ी हुई हैं।
पांचवीं घात वाले समीकरण का सर्वसामान्य रूप यह है:
a
x
5
+
b
x
4
+
c
x
3
+
d
x
2
+
e
x
+
f
=
0
{\displaystyle ax^{5}+bx^{4}+cx^{3}+dx^{2}+ex+f=0}
यहां दिखाया गया मान x अज्ञात है जिसे निर्धारित किया जाना है।
पांचवीं घात वाले समीकरण में, अज्ञात का उपयोग आधार के रूप में किया जाता है और 0 से 5 तक की संख्याओं को घातांक के रूप में उपयोग किया जाता है। परिणामी शक्तियों को निरंतर गुणांक से गुणा किया जाता है और फिर योग में जोड़ा जाता है। सामान्य तौर पर, अज्ञात मान x के निर्धारण के लिए गैर-प्राथमिक[ 1] मूलक अभिव्यक्तियों के उपयोग की आवश्यकता होती है।
दूसरे, तीसरे और चौथे घात वाले समीकरणों को हमेशा मूलों का उपयोग करके हल किया जा सकता है। 1545 में, गेरोलामो कार्डानो नामक गणितज्ञ ने "Ars magna de Regulis Algebraicis" नामक अपने काम में तीसरी घात के सामान्य समीकरणों का हल प्रकाशित किया:
c
x
3
+
d
x
2
+
e
x
+
f
=
0
{\displaystyle cx^{3}+dx^{2}+ex+f=0}
c, d, e, f के वास्तविक मान के लिये इसके मूल निम्नलिखित होंगे-
x
=
−
d
3
c
−
1
3
c
d
3
−
9
2
c
d
e
+
27
2
c
2
f
+
(
d
3
−
9
2
c
d
e
+
27
2
c
2
f
)
2
−
(
d
2
−
3
c
e
)
3
3
−
{\displaystyle x=-{\frac {d}{3c}}-{\frac {1}{3c}}{\sqrt[{3}]{d^{3}-{\frac {9}{2}}cde+{\frac {27}{2}}c^{2}f+{\sqrt {{\bigl (}d^{3}-{\frac {9}{2}}cde+{\frac {27}{2}}c^{2}f{\bigr )}^{2}-{\bigl (}d^{2}-3ce{\bigr )}^{3}}}}}-}
−
1
3
c
d
3
−
9
2
c
d
e
+
27
2
c
2
f
−
(
d
3
−
9
2
c
d
e
+
27
2
c
2
f
)
2
−
(
d
2
−
3
c
e
)
3
3
{\displaystyle -{\frac {1}{3c}}{\sqrt[{3}]{d^{3}-{\frac {9}{2}}cde+{\frac {27}{2}}c^{2}f-{\sqrt {{\bigl (}d^{3}-{\frac {9}{2}}cde+{\frac {27}{2}}c^{2}f{\bigr )}^{2}-{\bigl (}d^{2}-3ce{\bigr )}^{3}}}}}}
लोदोविको फेरारी के साथ, कार्डानो ने चौथी डिग्री के सामान्य समीकरणों के लिए एक समाधान भी विकसित किया।
चौथी डिग्री के समीकरणों के समाधान में हमेशा तीसरी डिग्री के संबंधित समीकरणों के समाधान के लिए द्विघात मूल सूत्र अभिव्यक्ति होती है:
x
4
−
3
(
2
m
x
2
+
4
n
x
+
m
2
+
1
)
=
0
{\displaystyle x^{4}-3(2mx^{2}+4nx+m^{2}+1)=0}
x
12
=
3
n
{
2
sinh
[
1
3
arsinh
(
4
m
3
+
3
m
−
9
n
2
)
]
+
m
}
2
+
3
m
2
+
3
±
{\displaystyle x_{12}={\frac {3n}{\sqrt {\{2\sinh[{\tfrac {1}{3}}{\text{arsinh}}(4m^{3}+3m-9n^{2})]+m\}^{2}+3m^{2}+3}}}\pm }
±
{
2
sinh
[
1
3
arsinh
(
4
m
3
+
3
m
−
9
n
2
)
]
+
m
}
2
+
3
m
2
+
3
+
sinh
[
1
3
arsinh
(
4
m
3
+
3
m
−
9
n
2
)
]
+
2
m
{\displaystyle \pm {\sqrt {{\sqrt {\{2\sinh[{\tfrac {1}{3}}{\text{arsinh}}(4m^{3}+3m-9n^{2})]+m\}^{2}+3m^{2}+3}}+\sinh[{\tfrac {1}{3}}{\text{arsinh}}(4m^{3}+3m-9n^{2})]+2m}}}
अभी दिखाया गया समाधान सूत्र सभी वास्तविक-मूल्यवान संख्यात्मक मानों m और n पर लागू होता है।
क्योंकि सभी चतुर्थ-डिग्री बहुपदों को "एक रैखिक बहुपद के द्विघात बहुपद ऋण वर्ग का वर्ग" के रूप के अंतर के रूप में दर्शाया जा सकता है:
x
4
−
x
−
1
=
{
x
2
+
2
3
3
sinh
[
1
3
arsinh
(
3
16
3
)
]
}
2
−
{
2
3
27
4
sinh
[
1
3
arsinh
(
3
16
3
)
]
x
+
1
4
3
4
csch
[
1
3
arsinh
(
3
16
3
)
]
}
2
{\displaystyle x^{4}-x-1={\bigl \{}x^{2}+{\tfrac {2}{3}}{\sqrt {3}}\sinh {\bigl [}{\tfrac {1}{3}}{\text{arsinh}}({\tfrac {3}{16}}{\sqrt {3}}){\bigr ]}{\bigr \}}^{2}-{\bigl \{}{\tfrac {2}{3}}{\sqrt[{4}]{27}}{\sqrt {\sinh {\bigl [}{\tfrac {1}{3}}{\text{arsinh}}({\tfrac {3}{16}}{\sqrt {3}}){\bigr ]}}}\,x+{\tfrac {1}{4}}{\sqrt[{4}]{3}}{\sqrt {{\text{csch}}{\bigl [}{\tfrac {1}{3}}{\text{arsinh}}({\tfrac {3}{16}}{\sqrt {3}}){\bigr ]}}}{\bigr \}}^{2}}
संक्षिप्त रूप sinh और arsinh अतिपरवलयिक फलन और उनके स्वयं के प्रतिलोम फलन को दर्शाते हैं:
sinh
[
1
3
arsinh
(
s
)
]
=
1
2
s
2
+
1
+
s
3
−
1
2
s
2
+
1
−
s
3
{\displaystyle \sinh[{\tfrac {1}{3}}{\text{arsinh}}(s)]={\tfrac {1}{2}}{\sqrt[{3}]{{\sqrt {s^{2}+1}}+s}}-{\tfrac {1}{2}}{\sqrt[{3}]{{\sqrt {s^{2}+1}}-s}}}
जियानफ्रांसेस्को मालफट्टी ने 1771 में पांचवीं डिग्री के समीकरणों को हल करने के लिए एक विधि की खोज की थी। हालाँकि, यह दृष्टिकोण केवल रूट एक्सप्रेशन द्वारा सॉल्वेबिलिटी के मामले में काम करता है। गणितज्ञ पाओलो रफिनी ने तब 1799 में 5वीं डिग्री सामान्य समीकरण की गैर-सॉल्वबिलिटी का थोड़ा त्रुटिपूर्ण प्रमाण प्रकाशित किया। अंत में, 1824 में, नील्स हेनरिक एबेल ने सफलतापूर्वक एक पूर्ण प्रमाण प्रस्तुत किया कि पांचवीं डिग्री के सामान्य समीकरण को प्राथमिक कट्टरपंथी मूल अभिव्यक्तियों द्वारा हल नहीं किया जा सकता है। यह "हाबिल-रफिनी प्रमेय" (Abel–Ruffini-Theorem) कहता है।
गणितज्ञ गैलोइस ने बाद में 1830 में यह निर्धारित करने के लिए तरीके विकसित किए कि गणितीय जड़ों का उपयोग करके दिया गया समीकरण हल करने योग्य है या नहीं। इन मूलभूत परिणामों पर निर्माण करते हुए, जॉर्ज पैक्सटन यंग और कार्ल रनगे ने 1885 में एक स्पष्ट मानदंड साबित किया कि क्या जड़ों के साथ दी गई पांचवीं डिग्री समीकरण हल करने योग्य है या नहीं। उन्होंने दिखाया कि ब्रिंग-जेरार्ड रूप[ 2]
x
5
+
5
μ
4
(
4
ν
+
3
)
ν
2
+
1
x
=
4
μ
5
(
2
ν
+
1
)
(
4
ν
+
3
)
ν
2
+
1
{\displaystyle x^{5}+{\frac {5\mu ^{4}(4\nu +3)}{\nu ^{2}+1}}x={\frac {4\mu ^{5}(2\nu +1)(4\nu +3)}{\nu ^{2}+1}}}
अभी दिखाए गए समीकरण का वास्तविक-मूल्यवान समाधान निम्नलिखित तरीके से स्थापित किया गया है:
x
=
2
μ
20
ν
+
15
5
ν
2
+
1
4
cosh
{
1
5
arcosh
[
125
ν
2
+
1
4
(
2
ν
2
+
1
+
2
ν
−
1
)
(
20
ν
+
15
)
3
/
2
]
}
−
{\displaystyle x={\frac {2\mu {\sqrt {20\nu +15}}}{5{\sqrt[{4}]{\nu ^{2}+1}}}}\cosh {\biggl \{}{\frac {1}{5}}{\text{arcosh}}{\biggl [}{\frac {125{\sqrt[{4}]{\nu ^{2}+1}}(2{\sqrt {\nu ^{2}+1}}+2\nu -1)}{(20\nu +15)^{3/2}}}{\biggr ]}{\biggr \}}-}
−
2
μ
20
ν
+
15
5
ν
2
+
1
4
sinh
{
1
5
arsinh
[
125
ν
2
+
1
4
(
2
ν
2
+
1
−
2
ν
+
1
)
(
20
ν
+
15
)
3
/
2
]
}
{\displaystyle -{\frac {2\mu {\sqrt {20\nu +15}}}{5{\sqrt[{4}]{\nu ^{2}+1}}}}\sinh {\biggl \{}{\frac {1}{5}}{\text{arsinh}}{\biggl [}{\frac {125{\sqrt[{4}]{\nu ^{2}+1}}(2{\sqrt {\nu ^{2}+1}}-2\nu +1)}{(20\nu +15)^{3/2}}}{\biggr ]}{\biggr \}}}
अब मान युग्म μ = 1 और ν = 0 के उदाहरण का वर्णन किया गया है:
x
5
+
15
x
=
12
{\displaystyle x^{5}+15x=12}
x
=
2
5
15
cosh
[
1
5
arcosh
(
5
9
15
)
]
−
2
5
15
sinh
[
1
5
arsinh
(
5
3
15
)
]
{\displaystyle x={\tfrac {2}{5}}{\sqrt {15}}\cosh[{\tfrac {1}{5}}{\text{arcosh}}({\tfrac {5}{9}}{\sqrt {15}})]-{\tfrac {2}{5}}{\sqrt {15}}\sinh[{\tfrac {1}{5}}{\text{arsinh}}({\tfrac {5}{3}}{\sqrt {15}})]}
इसलिए निम्नलिखित समीकरण को नीचे दिखाए गए तरीके से हल किया जा सकता है:
x
5
+
x
=
2
(
1
+
y
−
y
2
)
2
+
2
y
2
5
5
(
2
y
5
−
y
6
)
(
1
+
2
y
)
4
{\displaystyle x^{5}+x={\frac {2(1+y-y^{2}){\sqrt {2+2y^{2}}}}{5{\sqrt[{4}]{5(2y^{5}-y^{6})(1+2y)}}}}}
x
=
2
5
y
−
1
/
4
10
+
15
y
−
10
y
2
4
cosh
{
1
5
arcosh
[
5
5
+
5
y
2
(
1
+
2
y
)
4
+
6
y
−
4
y
2
]
}
−
{\displaystyle x={\frac {2}{5}}y^{-1/4}{\sqrt[{4}]{10+15y-10y^{2}}}\cosh {\biggl \{}{\frac {1}{5}}{\text{arcosh}}{\biggl [}{\frac {5{\sqrt {5+5y^{2}}}}{(1+2y){\sqrt {4+6y-4y^{2}}}}}{\biggr ]}{\biggr \}}-}
−
2
5
y
−
1
/
4
10
+
15
y
−
10
y
2
4
sinh
{
1
5
arsinh
[
5
y
5
+
5
y
2
(
2
−
y
)
4
+
6
y
−
4
y
2
]
}
{\displaystyle -{\frac {2}{5}}y^{-1/4}{\sqrt[{4}]{10+15y-10y^{2}}}\sinh {\biggl \{}{\frac {1}{5}}{\text{arsinh}}{\biggl [}{\frac {5y{\sqrt {5+5y^{2}}}}{(2-y){\sqrt {4+6y-4y^{2}}}}}{\biggr ]}{\biggr \}}}
सूत्रों का यह युग्म 0 से 2 तक की सभी वास्तविक संख्याओं y के लिए मान्य है।
अण्डाकार समाधान[ 3]
x
5
+
x
=
w
{\displaystyle x^{5}+x=w}
x
=
2
5
y
−
1
/
4
10
+
15
y
−
10
y
2
4
cosh
{
1
5
arcosh
[
5
5
+
5
y
2
(
1
+
2
y
)
4
+
6
y
−
4
y
2
]
}
−
{\displaystyle x={\frac {2}{5}}y^{-1/4}{\sqrt[{4}]{10+15y-10y^{2}}}\cosh {\biggl \{}{\frac {1}{5}}{\text{arcosh}}{\biggl [}{\frac {5{\sqrt {5+5y^{2}}}}{(1+2y){\sqrt {4+6y-4y^{2}}}}}{\biggr ]}{\biggr \}}-}
−
2
5
y
−
1
/
4
10
+
15
y
−
10
y
2
4
sinh
{
1
5
arsinh
[
5
y
5
+
5
y
2
(
2
−
y
)
4
+
6
y
−
4
y
2
]
}
{\displaystyle -{\frac {2}{5}}y^{-1/4}{\sqrt[{4}]{10+15y-10y^{2}}}\sinh {\biggl \{}{\frac {1}{5}}{\text{arsinh}}{\biggl [}{\frac {5y{\sqrt {5+5y^{2}}}}{(2-y){\sqrt {4+6y-4y^{2}}}}}{\biggr ]}{\biggr \}}}
y
=
5
ϑ
00
⟨
q
{
ctlh
[
1
2
aclh
(
5
4
5
4
w
)
]
2
}
5
⟩
2
2
ϑ
00
⟨
q
{
ctlh
[
1
2
aclh
(
5
4
5
4
w
)
]
2
}
⟩
2
−
1
2
{\displaystyle y={\frac {5\vartheta _{00}{\bigl \langle }q\{{\text{ctlh}}[{\tfrac {1}{2}}{\text{aclh}}({\tfrac {5}{4}}{\sqrt[{4}]{5}}\,w)]^{2}\}^{5}{\bigr \rangle }^{2}}{2\vartheta _{00}{\bigl \langle }q\{{\text{ctlh}}[{\tfrac {1}{2}}{\text{aclh}}({\tfrac {5}{4}}{\sqrt[{4}]{5}}\,w)]^{2}\}{\bigr \rangle }^{2}}}-{\frac {1}{2}}}
दिखाए गए फलनों को परिभाषित करने के लिए निम्नलिखित सूत्रों का उपयोग किया जाता है:
ϑ
00
(
z
)
=
1
+
2
∑
k
=
1
∞
z
k
2
{\displaystyle \vartheta _{00}(z)=1+2\sum _{k=1}^{\infty }z^{k^{2}}}
ϑ
00
(
z
)
=
∏
k
=
1
∞
(
1
−
z
2
k
)
(
1
+
z
2
k
−
1
)
2
{\displaystyle \vartheta _{00}(z)=\prod _{k=1}^{\infty }(1-z^{2k})(1+z^{2k-1})^{2}}
q
(
ε
)
=
exp
[
−
π
K
(
1
−
ε
2
)
K
(
ε
)
−
1
]
{\displaystyle q(\varepsilon )=\exp[-\pi K({\sqrt {1-\varepsilon ^{2}}})K(\varepsilon )^{-1}]}
K
(
r
)
=
∫
0
π
/
2
1
1
−
r
2
sin
(
φ
)
2
d
φ
{\displaystyle K(r)=\int _{0}^{\pi /2}{\frac {1}{\sqrt {1-r^{2}\sin(\varphi )^{2}}}}\mathrm {d} \varphi }
निम्नलिखित अतिपरवलयिक लेमनिसकट फलन (hyperbolic lemniscate function)[ 4]
ctlh
[
1
2
a
c
l
h
(
s
)
]
2
=
(
2
s
2
+
2
+
2
s
4
+
1
)
−
1
/
2
(
s
4
+
1
+
1
+
s
)
{\displaystyle {\text{ctlh}}{\bigl [}{\tfrac {1}{2}}\mathrm {aclh} (s){\bigr ]}^{2}=(2s^{2}+2+2{\sqrt {s^{4}+1}})^{-1/2}({\sqrt {{\sqrt {s^{4}+1}}+1}}+s)}
लेमनिसकट फलनों के की परिभाषाएँ:
s
l
(
φ
)
=
tan
⟨
2
arctan
{
4
G
sin
(
φ
G
)
∑
k
=
1
∞
cosh
[
(
2
k
−
1
)
π
]
cosh
[
(
2
k
−
1
)
π
]
2
−
cos
(
φ
/
G
)
2
}
⟩
{\displaystyle \mathrm {sl} (\varphi )=\tan {\biggl \langle }2\arctan {\biggl \{}{\frac {4}{G}}\sin {\bigl (}{\frac {\varphi }{G}}{\bigr )}\sum _{k=1}^{\infty }{\frac {\cosh[(2k-1)\pi ]}{\cosh[(2k-1)\pi ]^{2}-\cos(\varphi /G)^{2}}}{\biggr \}}{\biggr \rangle }}
c
l
(
φ
)
=
tan
⟨
2
arctan
{
4
G
cos
(
φ
G
)
∑
k
=
1
∞
cosh
[
(
2
k
−
1
)
π
]
cosh
[
(
2
k
−
1
)
π
]
2
−
sin
(
φ
/
G
)
2
}
⟩
{\displaystyle \mathrm {cl} (\varphi )=\tan {\biggl \langle }2\arctan {\biggl \{}{\frac {4}{G}}\cos {\bigl (}{\frac {\varphi }{G}}{\bigr )}\sum _{k=1}^{\infty }{\frac {\cosh[(2k-1)\pi ]}{\cosh[(2k-1)\pi ]^{2}-\sin(\varphi /G)^{2}}}{\biggr \}}{\biggr \rangle }}
[
sl
(
φ
)
2
+
1
]
[
cl
(
φ
)
2
+
1
]
=
2
{\displaystyle [{\text{sl}}(\varphi )^{2}+1][{\text{cl}}(\varphi )^{2}+1]=2}
ctlh
(
ϱ
)
=
cl
(
1
2
2
ϱ
)
[
sl
(
1
2
2
ϱ
)
2
+
1
sl
(
1
2
2
ϱ
)
2
+
cl
(
1
2
2
ϱ
)
2
]
1
/
2
{\displaystyle {\text{ctlh}}(\varrho )=\operatorname {cl} ({\tfrac {1}{2}}{\sqrt {2}}\varrho ){\biggl [}{\frac {\operatorname {sl} ({\tfrac {1}{2}}{\sqrt {2}}\varrho )^{2}+1}{\operatorname {sl} ({\tfrac {1}{2}}{\sqrt {2}}\varrho )^{2}+\operatorname {cl} ({\tfrac {1}{2}}{\sqrt {2}}\varrho )^{2}}}{\biggr ]}^{1/2}}
aclh
(
s
)
=
1
2
2
π
G
−
∫
0
1
s
s
4
t
4
+
1
d
t
{\displaystyle {\text{aclh}}(s)={\frac {1}{2}}{\sqrt {2}}\,\pi \,G-\int _{0}^{1}{\frac {s}{\sqrt {s^{4}t^{4}+1}}}\,\mathrm {d} t}
कार्ल फ्रेडरिक गाउस के अनुसार गाउस स्थिरांक को G अक्षर से निरूपित किया जाता है।
G
=
1
2
2
π
Γ
(
3
4
)
−
2
{\displaystyle G={\tfrac {1}{2}}{\sqrt {2\pi }}\,\Gamma ({\tfrac {3}{4}})^{-2}}
चार्ल्स हर्मिट[ 5] [ 6] थीटा फलन का उपयोग करके पांचवीं डिग्री सामान्य समीकरण को हल करने में सफल रहे।
गणितज्ञ श्रीनिवास रामानुजन ने लियोनार्ड जेम्स रोजर्स के साथ मिलकर निरंतर अंश फलनों का आविष्कार किया:
R
(
z
)
=
z
1
/
5
∏
n
=
1
∞
(
1
−
z
5
n
−
1
)
(
1
−
z
5
n
−
4
)
(
1
−
z
5
n
−
2
)
(
1
−
z
5
n
−
3
)
{\displaystyle R(z)=z^{1/5}\prod _{n=1}^{\infty }{\frac {(1-z^{5n-1})(1-z^{5n-4})}{(1-z^{5n-2})(1-z^{5n-3})}}}
R
(
z
)
=
tan
{
1
2
arctan
[
ϑ
00
(
z
1
/
2
)
2
2
ϑ
00
(
z
5
/
2
)
2
−
1
2
]
}
2
/
5
tan
{
1
2
arccot
[
ϑ
00
(
z
1
/
2
)
2
2
ϑ
00
(
z
5
/
2
)
2
−
1
2
]
}
1
/
5
{\displaystyle R(z)=\tan {\biggl \{}{\frac {1}{2}}\arctan {\biggl [}{\frac {\vartheta _{00}(z^{1/2})^{2}}{2\vartheta _{00}(z^{5/2})^{2}}}-{\frac {1}{2}}{\biggr ]}{\biggr \}}^{2/5}\tan {\biggl \{}{\frac {1}{2}}\operatorname {arccot} {\biggl [}{\frac {\vartheta _{00}(z^{1/2})^{2}}{2\vartheta _{00}(z^{5/2})^{2}}}-{\frac {1}{2}}{\biggr ]}{\biggr \}}^{1/5}}
R
(
z
2
)
=
tan
{
1
2
arctan
[
ϑ
00
(
z
)
2
2
ϑ
00
(
z
5
)
2
−
1
2
]
}
2
/
5
tan
{
1
2
arccot
[
ϑ
00
(
z
)
2
2
ϑ
00
(
z
5
)
2
−
1
2
]
}
1
/
5
{\displaystyle R(z^{2})=\tan {\biggl \{}{\frac {1}{2}}\arctan {\biggl [}{\frac {\vartheta _{00}(z)^{2}}{2\vartheta _{00}(z^{5})^{2}}}-{\frac {1}{2}}{\biggr ]}{\biggr \}}^{2/5}\tan {\biggl \{}{\frac {1}{2}}\operatorname {arccot} {\biggl [}{\frac {\vartheta _{00}(z)^{2}}{2\vartheta _{00}(z^{5})^{2}}}-{\frac {1}{2}}{\biggr ]}{\biggr \}}^{1/5}}
S
(
z
)
=
tan
{
1
2
arctan
[
ϑ
00
(
z
)
2
2
ϑ
00
(
z
5
)
2
−
1
2
]
}
1
/
5
cot
{
1
2
arccot
[
ϑ
00
(
z
)
2
2
ϑ
00
(
z
5
)
2
−
1
2
]
}
2
/
5
{\displaystyle S(z)=\tan {\biggl \{}{\frac {1}{2}}\arctan {\biggl [}{\frac {\vartheta _{00}(z)^{2}}{2\vartheta _{00}(z^{5})^{2}}}-{\frac {1}{2}}{\biggr ]}{\biggr \}}^{1/5}\cot {\biggl \{}{\frac {1}{2}}\operatorname {arccot} {\biggl [}{\frac {\vartheta _{00}(z)^{2}}{2\vartheta _{00}(z^{5})^{2}}}-{\frac {1}{2}}{\biggr ]}{\biggr \}}^{2/5}}
S
(
z
)
=
R
(
z
4
)
R
(
z
2
)
R
(
z
)
{\displaystyle S(z)={\frac {R(z^{4})}{R(z^{2})R(z)}}}
और लेमनिसकट ज्या फलन के लिए यह सूत्र लागू होता है:
sl
[
1
2
2
a
c
l
h
(
s
)
]
=
s
4
+
1
−
s
2
{\displaystyle {\text{sl}}{\bigl [}{\tfrac {1}{2}}{\sqrt {2}}\,\mathrm {aclh} (s){\bigr ]}={\sqrt {{\sqrt {s^{4}+1}}-s^{2}}}}
सूत्रों का यह युग्म सभी वास्तविक संख्याओं w के लिए मान्य है:
x
5
+
x
=
w
{\displaystyle x^{5}+x=w}
x
=
S
⟨
q
{
ctlh
[
1
2
aclh
(
5
4
5
4
w
)
]
2
}
⟩
2
−
R
⟨
q
{
ctlh
[
1
2
aclh
(
5
4
5
4
w
)
]
2
}
2
⟩
S
⟨
q
{
ctlh
[
1
2
aclh
(
5
4
5
4
w
)
]
2
}
⟩
2
×
{\displaystyle x={\frac {S{\bigl \langle }q\{{\text{ctlh}}[{\tfrac {1}{2}}{\text{aclh}}({\tfrac {5}{4}}{\sqrt[{4}]{5}}\,w)]^{2}\}{\bigr \rangle }^{2}-R{\bigl \langle }q\{{\text{ctlh}}[{\tfrac {1}{2}}{\text{aclh}}({\tfrac {5}{4}}{\sqrt[{4}]{5}}\,w)]^{2}\}^{2}{\bigr \rangle }}{S{\bigl \langle }q\{{\text{ctlh}}[{\tfrac {1}{2}}{\text{aclh}}({\tfrac {5}{4}}{\sqrt[{4}]{5}}\,w)]^{2}\}{\bigr \rangle }^{2}}}\times }
×
1
−
R
⟨
q
{
ctlh
[
1
2
aclh
(
5
4
5
4
w
)
]
2
}
2
⟩
S
⟨
q
{
ctlh
[
1
2
aclh
(
5
4
5
4
w
)
]
2
}
⟩
R
⟨
q
{
ctlh
[
1
2
aclh
(
5
4
5
4
w
)
]
2
}
2
⟩
2
×
{\displaystyle \times {\frac {1-R{\bigl \langle }q\{{\text{ctlh}}[{\tfrac {1}{2}}{\text{aclh}}({\tfrac {5}{4}}{\sqrt[{4}]{5}}\,w)]^{2}\}^{2}{\bigr \rangle }\,S{\bigl \langle }q\{{\text{ctlh}}[{\tfrac {1}{2}}{\text{aclh}}({\tfrac {5}{4}}{\sqrt[{4}]{5}}\,w)]^{2}\}{\bigr \rangle }}{R{\bigl \langle }q\{{\text{ctlh}}[{\tfrac {1}{2}}{\text{aclh}}({\tfrac {5}{4}}{\sqrt[{4}]{5}}\,w)]^{2}\}^{2}{\bigr \rangle }^{2}}}\times }
×
ϑ
00
⟨
q
{
ctlh
[
1
2
aclh
(
5
4
5
4
w
)
]
2
}
5
⟩
ϑ
00
⟨
q
{
ctlh
[
1
2
aclh
(
5
4
5
4
w
)
]
2
}
1
/
5
⟩
2
−
5
ϑ
00
⟨
q
{
ctlh
[
1
2
aclh
(
5
4
5
4
w
)
]
2
}
5
⟩
3
2
20
4
sl
[
1
2
2
aclh
(
5
4
5
4
w
)
]
ϑ
00
⟨
q
{
ctlh
[
1
2
aclh
(
5
4
5
4
w
)
]
2
}
⟩
3
{\displaystyle \times {\frac {\vartheta _{00}{\bigl \langle }q\{{\text{ctlh}}[{\tfrac {1}{2}}{\text{aclh}}({\tfrac {5}{4}}{\sqrt[{4}]{5}}\,w)]^{2}\}^{5}{\bigr \rangle }\,\vartheta _{00}{\bigl \langle }q\{{\text{ctlh}}[{\tfrac {1}{2}}{\text{aclh}}({\tfrac {5}{4}}{\sqrt[{4}]{5}}\,w)]^{2}\}^{1/5}{\bigr \rangle }^{2}-5\,\vartheta _{00}{\bigl \langle }q\{{\text{ctlh}}[{\tfrac {1}{2}}{\text{aclh}}({\tfrac {5}{4}}{\sqrt[{4}]{5}}\,w)]^{2}\}^{5}{\bigr \rangle }^{3}}{2{\sqrt[{4}]{20}}\,{\text{sl}}[{\tfrac {1}{2}}{\sqrt {2}}\,{\text{aclh}}({\tfrac {5}{4}}{\sqrt[{4}]{5}}\,w)]\,\vartheta _{00}{\bigl \langle }q\{{\text{ctlh}}[{\tfrac {1}{2}}{\text{aclh}}({\tfrac {5}{4}}{\sqrt[{4}]{5}}\,w)]^{2}\}{\bigr \rangle }^{3}}}}
ये दो उदाहरण गणनाएं हैं:
x
5
+
x
=
3
{\displaystyle x^{5}+x=3}
x
=
S
⟨
q
{
ctlh
[
1
2
aclh
(
15
4
5
4
)
]
2
}
⟩
2
−
R
⟨
q
{
ctlh
[
1
2
aclh
(
15
4
5
4
)
]
2
}
2
⟩
S
⟨
q
{
ctlh
[
1
2
aclh
(
15
4
5
4
)
]
2
}
⟩
2
×
{\displaystyle x={\frac {S{\bigl \langle }q\{{\text{ctlh}}[{\tfrac {1}{2}}{\text{aclh}}({\tfrac {15}{4}}{\sqrt[{4}]{5}})]^{2}\}{\bigr \rangle }^{2}-R{\bigl \langle }q\{{\text{ctlh}}[{\tfrac {1}{2}}{\text{aclh}}({\tfrac {15}{4}}{\sqrt[{4}]{5}})]^{2}\}^{2}{\bigr \rangle }}{S{\bigl \langle }q\{{\text{ctlh}}[{\tfrac {1}{2}}{\text{aclh}}({\tfrac {15}{4}}{\sqrt[{4}]{5}})]^{2}\}{\bigr \rangle }^{2}}}\times }
×
1
−
R
⟨
q
{
ctlh
[
1
2
aclh
(
15
4
5
4
)
]
2
}
2
⟩
S
⟨
q
{
ctlh
[
1
2
aclh
(
15
4
5
4
)
]
2
}
⟩
R
⟨
q
{
ctlh
[
1
2
aclh
(
15
4
5
4
)
]
2
}
2
⟩
2
×
{\displaystyle \times {\frac {1-R{\bigl \langle }q\{{\text{ctlh}}[{\tfrac {1}{2}}{\text{aclh}}({\tfrac {15}{4}}{\sqrt[{4}]{5}})]^{2}\}^{2}{\bigr \rangle }\,S{\bigl \langle }q\{{\text{ctlh}}[{\tfrac {1}{2}}{\text{aclh}}({\tfrac {15}{4}}{\sqrt[{4}]{5}})]^{2}\}{\bigr \rangle }}{R{\bigl \langle }q\{{\text{ctlh}}[{\tfrac {1}{2}}{\text{aclh}}({\tfrac {15}{4}}{\sqrt[{4}]{5}})]^{2}\}^{2}{\bigr \rangle }^{2}}}\times }
×
ϑ
00
⟨
q
{
ctlh
[
1
2
aclh
(
15
4
5
4
)
]
2
}
5
⟩
ϑ
00
⟨
q
{
ctlh
[
1
2
aclh
(
15
4
5
4
)
]
2
}
1
/
5
⟩
2
−
5
ϑ
00
⟨
q
{
ctlh
[
1
2
aclh
(
15
4
5
4
)
]
2
}
5
⟩
3
2
20
4
sl
[
1
2
2
aclh
(
15
4
5
4
)
]
ϑ
00
⟨
q
{
ctlh
[
1
2
aclh
(
15
4
5
4
)
]
2
}
⟩
3
{\displaystyle \times {\frac {\vartheta _{00}{\bigl \langle }q\{{\text{ctlh}}[{\tfrac {1}{2}}{\text{aclh}}({\tfrac {15}{4}}{\sqrt[{4}]{5}})]^{2}\}^{5}{\bigr \rangle }\,\vartheta _{00}{\bigl \langle }q\{{\text{ctlh}}[{\tfrac {1}{2}}{\text{aclh}}({\tfrac {15}{4}}{\sqrt[{4}]{5}})]^{2}\}^{1/5}{\bigr \rangle }^{2}-5\,\vartheta _{00}{\bigl \langle }q\{{\text{ctlh}}[{\tfrac {1}{2}}{\text{aclh}}({\tfrac {15}{4}}{\sqrt[{4}]{5}})]^{2}\}^{5}{\bigr \rangle }^{3}}{2{\sqrt[{4}]{20}}\,{\text{sl}}[{\tfrac {1}{2}}{\sqrt {2}}\,{\text{aclh}}({\tfrac {15}{4}}{\sqrt[{4}]{5}})]\,\vartheta _{00}{\bigl \langle }q\{{\text{ctlh}}[{\tfrac {1}{2}}{\text{aclh}}({\tfrac {15}{4}}{\sqrt[{4}]{5}})]^{2}\}{\bigr \rangle }^{3}}}}
अनुमानित संख्यात्मक मान:
q
{
ctlh
[
1
2
aclh
(
15
4
5
4
)
]
2
}
≈
0.452374059450344348576600264284387826377845763909
{\displaystyle q\{{\text{ctlh}}[{\tfrac {1}{2}}{\text{aclh}}({\tfrac {15}{4}}{\sqrt[{4}]{5}})]^{2}\}\approx 0.452374059450344348576600264284387826377845763909}
x
≈
1.132997565885065266721141634288532379816526027727
{\displaystyle x\approx 1.132997565885065266721141634288532379816526027727}
समीकरणों की निम्नलिखित प्रणाली को समान रूप से स्थापित किया जा सकता है:
x
5
+
x
=
7
{\displaystyle x^{5}+x=7}
x
=
S
⟨
q
{
ctlh
[
1
2
aclh
(
35
4
5
4
)
]
2
}
⟩
2
−
R
⟨
q
{
ctlh
[
1
2
aclh
(
35
4
5
4
)
]
2
}
2
⟩
S
⟨
q
{
ctlh
[
1
2
aclh
(
35
4
5
4
)
]
2
}
⟩
2
×
{\displaystyle x={\frac {S{\bigl \langle }q\{{\text{ctlh}}[{\tfrac {1}{2}}{\text{aclh}}({\tfrac {35}{4}}{\sqrt[{4}]{5}})]^{2}\}{\bigr \rangle }^{2}-R{\bigl \langle }q\{{\text{ctlh}}[{\tfrac {1}{2}}{\text{aclh}}({\tfrac {35}{4}}{\sqrt[{4}]{5}})]^{2}\}^{2}{\bigr \rangle }}{S{\bigl \langle }q\{{\text{ctlh}}[{\tfrac {1}{2}}{\text{aclh}}({\tfrac {35}{4}}{\sqrt[{4}]{5}})]^{2}\}{\bigr \rangle }^{2}}}\times }
×
1
−
R
⟨
q
{
ctlh
[
1
2
aclh
(
35
4
5
4
)
]
2
}
2
⟩
S
⟨
q
{
ctlh
[
1
2
aclh
(
35
4
5
4
)
]
2
}
⟩
R
⟨
q
{
ctlh
[
1
2
aclh
(
35
4
5
4
)
]
2
}
2
⟩
2
×
{\displaystyle \times {\frac {1-R{\bigl \langle }q\{{\text{ctlh}}[{\tfrac {1}{2}}{\text{aclh}}({\tfrac {35}{4}}{\sqrt[{4}]{5}})]^{2}\}^{2}{\bigr \rangle }\,S{\bigl \langle }q\{{\text{ctlh}}[{\tfrac {1}{2}}{\text{aclh}}({\tfrac {35}{4}}{\sqrt[{4}]{5}})]^{2}\}{\bigr \rangle }}{R{\bigl \langle }q\{{\text{ctlh}}[{\tfrac {1}{2}}{\text{aclh}}({\tfrac {35}{4}}{\sqrt[{4}]{5}})]^{2}\}^{2}{\bigr \rangle }^{2}}}\times }
×
ϑ
00
⟨
q
{
ctlh
[
1
2
aclh
(
35
4
5
4
)
]
2
}
5
⟩
ϑ
00
⟨
q
{
ctlh
[
1
2
aclh
(
35
4
5
4
)
]
2
}
1
/
5
⟩
2
−
5
ϑ
00
⟨
q
{
ctlh
[
1
2
aclh
(
35
4
5
4
)
]
2
}
5
⟩
3
2
20
4
sl
[
1
2
2
aclh
(
35
4
5
4
)
]
ϑ
00
⟨
q
{
ctlh
[
1
2
aclh
(
35
4
5
4
)
]
2
}
⟩
3
{\displaystyle \times {\frac {\vartheta _{00}{\bigl \langle }q\{{\text{ctlh}}[{\tfrac {1}{2}}{\text{aclh}}({\tfrac {35}{4}}{\sqrt[{4}]{5}})]^{2}\}^{5}{\bigr \rangle }\,\vartheta _{00}{\bigl \langle }q\{{\text{ctlh}}[{\tfrac {1}{2}}{\text{aclh}}({\tfrac {35}{4}}{\sqrt[{4}]{5}})]^{2}\}^{1/5}{\bigr \rangle }^{2}-5\,\vartheta _{00}{\bigl \langle }q\{{\text{ctlh}}[{\tfrac {1}{2}}{\text{aclh}}({\tfrac {35}{4}}{\sqrt[{4}]{5}})]^{2}\}^{5}{\bigr \rangle }^{3}}{2{\sqrt[{4}]{20}}\,{\text{sl}}[{\tfrac {1}{2}}{\sqrt {2}}\,{\text{aclh}}({\tfrac {35}{4}}{\sqrt[{4}]{5}})]\,\vartheta _{00}{\bigl \langle }q\{{\text{ctlh}}[{\tfrac {1}{2}}{\text{aclh}}({\tfrac {35}{4}}{\sqrt[{4}]{5}})]^{2}\}{\bigr \rangle }^{3}}}}
अनुमानित संख्यात्मक मान:
q
{
ctlh
[
1
2
aclh
(
35
4
5
4
)
]
2
}
≈
0.53609630892200161460073096549143569900990236
{\displaystyle q\{{\text{ctlh}}[{\tfrac {1}{2}}{\text{aclh}}({\tfrac {35}{4}}{\sqrt[{4}]{5}})]^{2}\}\approx 0.53609630892200161460073096549143569900990236}
x
≈
1.4108138510595771319852918753499397839215989
{\displaystyle x\approx 1.4108138510595771319852918753499397839215989}
↑ Weisstein, Eric W. "Quintic Equation" . mathworld.wolfram.com (अंग्रेज़ी में). अभिगमन तिथि 2022-01-16 . ↑ Weisstein, Eric W. "Bring-Jerrard Quintic Form" . mathworld.wolfram.com (अंग्रेज़ी में). अभिगमन तिथि 2022-01-16 . ↑ Brioschi, F. (1858-12-01). "Sulla risoluzione delle equazioni del quinto grado: Hermite — Sur la résolution de l'Équation du cinquiéme degré Comptes rendus —. N. 11. Mars. 1858" . डीओआइ :10.1007/bf03197334 . ↑ Deng, Ji-En; Chen, Chao-Ping (2014-01-24). "Sharp Shafer-Fink type inequalities for Gauss lemniscate functions" . Journal of Inequalities and Applications . 2014 (1): 35. आइ॰एस॰एस॰एन॰ 1029-242X . डीओआइ :10.1186/1029-242X-2014-35 . ↑ "Charles Hermite - Biography" . Maths History (अंग्रेज़ी में). अभिगमन तिथि 2022-01-16 .↑ Weisstein, Eric W. "Jacobi Theta Functions" . mathworld.wolfram.com (अंग्रेज़ी में). अभिगमन तिथि 2022-01-16 .
![{\displaystyle x=-{\frac {d}{3c}}-{\frac {1}{3c}}{\sqrt[{3}]{d^{3}-{\frac {9}{2}}cde+{\frac {27}{2}}c^{2}f+{\sqrt {{\bigl (}d^{3}-{\frac {9}{2}}cde+{\frac {27}{2}}c^{2}f{\bigr )}^{2}-{\bigl (}d^{2}-3ce{\bigr )}^{3}}}}}-}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f81b8a92fc669f4584c8d6c51de5cec1f66739e0)
![{\displaystyle -{\frac {1}{3c}}{\sqrt[{3}]{d^{3}-{\frac {9}{2}}cde+{\frac {27}{2}}c^{2}f-{\sqrt {{\bigl (}d^{3}-{\frac {9}{2}}cde+{\frac {27}{2}}c^{2}f{\bigr )}^{2}-{\bigl (}d^{2}-3ce{\bigr )}^{3}}}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6f9d99c1c657b4be6ba488d75f96d39bb83a4b66)
![{\displaystyle x_{12}={\frac {3n}{\sqrt {\{2\sinh[{\tfrac {1}{3}}{\text{arsinh}}(4m^{3}+3m-9n^{2})]+m\}^{2}+3m^{2}+3}}}\pm }](https://wikimedia.org/api/rest_v1/media/math/render/svg/fd79bdec6ee3f040ba5a3f4e9f19b30f265eb40e)
![{\displaystyle \pm {\sqrt {{\sqrt {\{2\sinh[{\tfrac {1}{3}}{\text{arsinh}}(4m^{3}+3m-9n^{2})]+m\}^{2}+3m^{2}+3}}+\sinh[{\tfrac {1}{3}}{\text{arsinh}}(4m^{3}+3m-9n^{2})]+2m}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a3021075946ca5915cb752a2e2677f6ff39b436a)
![{\displaystyle x^{4}-x-1={\bigl \{}x^{2}+{\tfrac {2}{3}}{\sqrt {3}}\sinh {\bigl [}{\tfrac {1}{3}}{\text{arsinh}}({\tfrac {3}{16}}{\sqrt {3}}){\bigr ]}{\bigr \}}^{2}-{\bigl \{}{\tfrac {2}{3}}{\sqrt[{4}]{27}}{\sqrt {\sinh {\bigl [}{\tfrac {1}{3}}{\text{arsinh}}({\tfrac {3}{16}}{\sqrt {3}}){\bigr ]}}}\,x+{\tfrac {1}{4}}{\sqrt[{4}]{3}}{\sqrt {{\text{csch}}{\bigl [}{\tfrac {1}{3}}{\text{arsinh}}({\tfrac {3}{16}}{\sqrt {3}}){\bigr ]}}}{\bigr \}}^{2}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6a053b4ddabebe55bd61bb338b8d7ce96038fd67)
![{\displaystyle \sinh[{\tfrac {1}{3}}{\text{arsinh}}(s)]={\tfrac {1}{2}}{\sqrt[{3}]{{\sqrt {s^{2}+1}}+s}}-{\tfrac {1}{2}}{\sqrt[{3}]{{\sqrt {s^{2}+1}}-s}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/45fa6c4662f61b67832e9d7a54d83563ad2211b5)
![{\displaystyle x={\frac {2\mu {\sqrt {20\nu +15}}}{5{\sqrt[{4}]{\nu ^{2}+1}}}}\cosh {\biggl \{}{\frac {1}{5}}{\text{arcosh}}{\biggl [}{\frac {125{\sqrt[{4}]{\nu ^{2}+1}}(2{\sqrt {\nu ^{2}+1}}+2\nu -1)}{(20\nu +15)^{3/2}}}{\biggr ]}{\biggr \}}-}](https://wikimedia.org/api/rest_v1/media/math/render/svg/4943c8736c95f3862b73b8ede4c00d38de1c4a71)
![{\displaystyle -{\frac {2\mu {\sqrt {20\nu +15}}}{5{\sqrt[{4}]{\nu ^{2}+1}}}}\sinh {\biggl \{}{\frac {1}{5}}{\text{arsinh}}{\biggl [}{\frac {125{\sqrt[{4}]{\nu ^{2}+1}}(2{\sqrt {\nu ^{2}+1}}-2\nu +1)}{(20\nu +15)^{3/2}}}{\biggr ]}{\biggr \}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f91c58be36a5c7838366db935249a7ba25067f58)
![{\displaystyle x={\tfrac {2}{5}}{\sqrt {15}}\cosh[{\tfrac {1}{5}}{\text{arcosh}}({\tfrac {5}{9}}{\sqrt {15}})]-{\tfrac {2}{5}}{\sqrt {15}}\sinh[{\tfrac {1}{5}}{\text{arsinh}}({\tfrac {5}{3}}{\sqrt {15}})]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d603eeb2e2c4d1b45289dae39bdc0147e0495975)
![{\displaystyle x^{5}+x={\frac {2(1+y-y^{2}){\sqrt {2+2y^{2}}}}{5{\sqrt[{4}]{5(2y^{5}-y^{6})(1+2y)}}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8554dc68d3fca04f91a7dce125ab1c3273e4043d)
![{\displaystyle x={\frac {2}{5}}y^{-1/4}{\sqrt[{4}]{10+15y-10y^{2}}}\cosh {\biggl \{}{\frac {1}{5}}{\text{arcosh}}{\biggl [}{\frac {5{\sqrt {5+5y^{2}}}}{(1+2y){\sqrt {4+6y-4y^{2}}}}}{\biggr ]}{\biggr \}}-}](https://wikimedia.org/api/rest_v1/media/math/render/svg/9c520010591fbdcf2f5f61bdb7454116b63c2072)
![{\displaystyle -{\frac {2}{5}}y^{-1/4}{\sqrt[{4}]{10+15y-10y^{2}}}\sinh {\biggl \{}{\frac {1}{5}}{\text{arsinh}}{\biggl [}{\frac {5y{\sqrt {5+5y^{2}}}}{(2-y){\sqrt {4+6y-4y^{2}}}}}{\biggr ]}{\biggr \}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/01f503a8d0944a1d12408b892d7c2cc400abab0d)
![{\displaystyle y={\frac {5\vartheta _{00}{\bigl \langle }q\{{\text{ctlh}}[{\tfrac {1}{2}}{\text{aclh}}({\tfrac {5}{4}}{\sqrt[{4}]{5}}\,w)]^{2}\}^{5}{\bigr \rangle }^{2}}{2\vartheta _{00}{\bigl \langle }q\{{\text{ctlh}}[{\tfrac {1}{2}}{\text{aclh}}({\tfrac {5}{4}}{\sqrt[{4}]{5}}\,w)]^{2}\}{\bigr \rangle }^{2}}}-{\frac {1}{2}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f34ae73ec4c35b4c1bff9dbb8a439d0f796f979d)
![{\displaystyle q(\varepsilon )=\exp[-\pi K({\sqrt {1-\varepsilon ^{2}}})K(\varepsilon )^{-1}]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6c64c06dc297c68502bb06e6c07e36bbccd609e2)
![{\displaystyle {\text{ctlh}}{\bigl [}{\tfrac {1}{2}}\mathrm {aclh} (s){\bigr ]}^{2}=(2s^{2}+2+2{\sqrt {s^{4}+1}})^{-1/2}({\sqrt {{\sqrt {s^{4}+1}}+1}}+s)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/fb0e85a53ae3dee4d6260bf34574bdda2677ab5e)
![{\displaystyle \mathrm {sl} (\varphi )=\tan {\biggl \langle }2\arctan {\biggl \{}{\frac {4}{G}}\sin {\bigl (}{\frac {\varphi }{G}}{\bigr )}\sum _{k=1}^{\infty }{\frac {\cosh[(2k-1)\pi ]}{\cosh[(2k-1)\pi ]^{2}-\cos(\varphi /G)^{2}}}{\biggr \}}{\biggr \rangle }}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1791550a2f621d4e3af70d6d487ed8d762a4a462)
![{\displaystyle \mathrm {cl} (\varphi )=\tan {\biggl \langle }2\arctan {\biggl \{}{\frac {4}{G}}\cos {\bigl (}{\frac {\varphi }{G}}{\bigr )}\sum _{k=1}^{\infty }{\frac {\cosh[(2k-1)\pi ]}{\cosh[(2k-1)\pi ]^{2}-\sin(\varphi /G)^{2}}}{\biggr \}}{\biggr \rangle }}](https://wikimedia.org/api/rest_v1/media/math/render/svg/814a459471e82cb54c3bc0988069b83be88ce2c4)
![{\displaystyle [{\text{sl}}(\varphi )^{2}+1][{\text{cl}}(\varphi )^{2}+1]=2}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5010c59ea047a03787ec123d3cddf7707106be7c)
![{\displaystyle {\text{ctlh}}(\varrho )=\operatorname {cl} ({\tfrac {1}{2}}{\sqrt {2}}\varrho ){\biggl [}{\frac {\operatorname {sl} ({\tfrac {1}{2}}{\sqrt {2}}\varrho )^{2}+1}{\operatorname {sl} ({\tfrac {1}{2}}{\sqrt {2}}\varrho )^{2}+\operatorname {cl} ({\tfrac {1}{2}}{\sqrt {2}}\varrho )^{2}}}{\biggr ]}^{1/2}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/835ac16e842bf820bd2afd699320c4bb8483be82)
![{\displaystyle R(z)=\tan {\biggl \{}{\frac {1}{2}}\arctan {\biggl [}{\frac {\vartheta _{00}(z^{1/2})^{2}}{2\vartheta _{00}(z^{5/2})^{2}}}-{\frac {1}{2}}{\biggr ]}{\biggr \}}^{2/5}\tan {\biggl \{}{\frac {1}{2}}\operatorname {arccot} {\biggl [}{\frac {\vartheta _{00}(z^{1/2})^{2}}{2\vartheta _{00}(z^{5/2})^{2}}}-{\frac {1}{2}}{\biggr ]}{\biggr \}}^{1/5}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e05fd6a136f16b01c04ca730bf084d9004b4581b)
![{\displaystyle R(z^{2})=\tan {\biggl \{}{\frac {1}{2}}\arctan {\biggl [}{\frac {\vartheta _{00}(z)^{2}}{2\vartheta _{00}(z^{5})^{2}}}-{\frac {1}{2}}{\biggr ]}{\biggr \}}^{2/5}\tan {\biggl \{}{\frac {1}{2}}\operatorname {arccot} {\biggl [}{\frac {\vartheta _{00}(z)^{2}}{2\vartheta _{00}(z^{5})^{2}}}-{\frac {1}{2}}{\biggr ]}{\biggr \}}^{1/5}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/fa7ecda83ca3a83d770dca4b783ba22086abd63a)
![{\displaystyle S(z)=\tan {\biggl \{}{\frac {1}{2}}\arctan {\biggl [}{\frac {\vartheta _{00}(z)^{2}}{2\vartheta _{00}(z^{5})^{2}}}-{\frac {1}{2}}{\biggr ]}{\biggr \}}^{1/5}\cot {\biggl \{}{\frac {1}{2}}\operatorname {arccot} {\biggl [}{\frac {\vartheta _{00}(z)^{2}}{2\vartheta _{00}(z^{5})^{2}}}-{\frac {1}{2}}{\biggr ]}{\biggr \}}^{2/5}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c0283e6b959ed0567349c3d9a92d9f5b73238462)
![{\displaystyle {\text{sl}}{\bigl [}{\tfrac {1}{2}}{\sqrt {2}}\,\mathrm {aclh} (s){\bigr ]}={\sqrt {{\sqrt {s^{4}+1}}-s^{2}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6417f3c2a1d5e63bededbc392ece42de7e2cf02d)
![{\displaystyle x={\frac {S{\bigl \langle }q\{{\text{ctlh}}[{\tfrac {1}{2}}{\text{aclh}}({\tfrac {5}{4}}{\sqrt[{4}]{5}}\,w)]^{2}\}{\bigr \rangle }^{2}-R{\bigl \langle }q\{{\text{ctlh}}[{\tfrac {1}{2}}{\text{aclh}}({\tfrac {5}{4}}{\sqrt[{4}]{5}}\,w)]^{2}\}^{2}{\bigr \rangle }}{S{\bigl \langle }q\{{\text{ctlh}}[{\tfrac {1}{2}}{\text{aclh}}({\tfrac {5}{4}}{\sqrt[{4}]{5}}\,w)]^{2}\}{\bigr \rangle }^{2}}}\times }](https://wikimedia.org/api/rest_v1/media/math/render/svg/8c7e6cbec3ecdeb892514072f67a3c67e970d6fb)
![{\displaystyle \times {\frac {1-R{\bigl \langle }q\{{\text{ctlh}}[{\tfrac {1}{2}}{\text{aclh}}({\tfrac {5}{4}}{\sqrt[{4}]{5}}\,w)]^{2}\}^{2}{\bigr \rangle }\,S{\bigl \langle }q\{{\text{ctlh}}[{\tfrac {1}{2}}{\text{aclh}}({\tfrac {5}{4}}{\sqrt[{4}]{5}}\,w)]^{2}\}{\bigr \rangle }}{R{\bigl \langle }q\{{\text{ctlh}}[{\tfrac {1}{2}}{\text{aclh}}({\tfrac {5}{4}}{\sqrt[{4}]{5}}\,w)]^{2}\}^{2}{\bigr \rangle }^{2}}}\times }](https://wikimedia.org/api/rest_v1/media/math/render/svg/9e2665ac8c95edccf1f8be194a49db90e21553ac)
![{\displaystyle \times {\frac {\vartheta _{00}{\bigl \langle }q\{{\text{ctlh}}[{\tfrac {1}{2}}{\text{aclh}}({\tfrac {5}{4}}{\sqrt[{4}]{5}}\,w)]^{2}\}^{5}{\bigr \rangle }\,\vartheta _{00}{\bigl \langle }q\{{\text{ctlh}}[{\tfrac {1}{2}}{\text{aclh}}({\tfrac {5}{4}}{\sqrt[{4}]{5}}\,w)]^{2}\}^{1/5}{\bigr \rangle }^{2}-5\,\vartheta _{00}{\bigl \langle }q\{{\text{ctlh}}[{\tfrac {1}{2}}{\text{aclh}}({\tfrac {5}{4}}{\sqrt[{4}]{5}}\,w)]^{2}\}^{5}{\bigr \rangle }^{3}}{2{\sqrt[{4}]{20}}\,{\text{sl}}[{\tfrac {1}{2}}{\sqrt {2}}\,{\text{aclh}}({\tfrac {5}{4}}{\sqrt[{4}]{5}}\,w)]\,\vartheta _{00}{\bigl \langle }q\{{\text{ctlh}}[{\tfrac {1}{2}}{\text{aclh}}({\tfrac {5}{4}}{\sqrt[{4}]{5}}\,w)]^{2}\}{\bigr \rangle }^{3}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/be4e06136ee2d75943c483914dcdf36f1489d0e3)
![{\displaystyle x={\frac {S{\bigl \langle }q\{{\text{ctlh}}[{\tfrac {1}{2}}{\text{aclh}}({\tfrac {15}{4}}{\sqrt[{4}]{5}})]^{2}\}{\bigr \rangle }^{2}-R{\bigl \langle }q\{{\text{ctlh}}[{\tfrac {1}{2}}{\text{aclh}}({\tfrac {15}{4}}{\sqrt[{4}]{5}})]^{2}\}^{2}{\bigr \rangle }}{S{\bigl \langle }q\{{\text{ctlh}}[{\tfrac {1}{2}}{\text{aclh}}({\tfrac {15}{4}}{\sqrt[{4}]{5}})]^{2}\}{\bigr \rangle }^{2}}}\times }](https://wikimedia.org/api/rest_v1/media/math/render/svg/f0c89500c3bb1767fb3430d839d84e3ae480af8f)
![{\displaystyle \times {\frac {1-R{\bigl \langle }q\{{\text{ctlh}}[{\tfrac {1}{2}}{\text{aclh}}({\tfrac {15}{4}}{\sqrt[{4}]{5}})]^{2}\}^{2}{\bigr \rangle }\,S{\bigl \langle }q\{{\text{ctlh}}[{\tfrac {1}{2}}{\text{aclh}}({\tfrac {15}{4}}{\sqrt[{4}]{5}})]^{2}\}{\bigr \rangle }}{R{\bigl \langle }q\{{\text{ctlh}}[{\tfrac {1}{2}}{\text{aclh}}({\tfrac {15}{4}}{\sqrt[{4}]{5}})]^{2}\}^{2}{\bigr \rangle }^{2}}}\times }](https://wikimedia.org/api/rest_v1/media/math/render/svg/781028992714c0946b51d0f4beb4404be08a6957)
![{\displaystyle \times {\frac {\vartheta _{00}{\bigl \langle }q\{{\text{ctlh}}[{\tfrac {1}{2}}{\text{aclh}}({\tfrac {15}{4}}{\sqrt[{4}]{5}})]^{2}\}^{5}{\bigr \rangle }\,\vartheta _{00}{\bigl \langle }q\{{\text{ctlh}}[{\tfrac {1}{2}}{\text{aclh}}({\tfrac {15}{4}}{\sqrt[{4}]{5}})]^{2}\}^{1/5}{\bigr \rangle }^{2}-5\,\vartheta _{00}{\bigl \langle }q\{{\text{ctlh}}[{\tfrac {1}{2}}{\text{aclh}}({\tfrac {15}{4}}{\sqrt[{4}]{5}})]^{2}\}^{5}{\bigr \rangle }^{3}}{2{\sqrt[{4}]{20}}\,{\text{sl}}[{\tfrac {1}{2}}{\sqrt {2}}\,{\text{aclh}}({\tfrac {15}{4}}{\sqrt[{4}]{5}})]\,\vartheta _{00}{\bigl \langle }q\{{\text{ctlh}}[{\tfrac {1}{2}}{\text{aclh}}({\tfrac {15}{4}}{\sqrt[{4}]{5}})]^{2}\}{\bigr \rangle }^{3}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8d8c01516b64dd0311e47b51ba3d0d855ef50d55)
![{\displaystyle q\{{\text{ctlh}}[{\tfrac {1}{2}}{\text{aclh}}({\tfrac {15}{4}}{\sqrt[{4}]{5}})]^{2}\}\approx 0.452374059450344348576600264284387826377845763909}](https://wikimedia.org/api/rest_v1/media/math/render/svg/177ba0691acd01a59eda4ab442a6dd6b3995110c)
![{\displaystyle x={\frac {S{\bigl \langle }q\{{\text{ctlh}}[{\tfrac {1}{2}}{\text{aclh}}({\tfrac {35}{4}}{\sqrt[{4}]{5}})]^{2}\}{\bigr \rangle }^{2}-R{\bigl \langle }q\{{\text{ctlh}}[{\tfrac {1}{2}}{\text{aclh}}({\tfrac {35}{4}}{\sqrt[{4}]{5}})]^{2}\}^{2}{\bigr \rangle }}{S{\bigl \langle }q\{{\text{ctlh}}[{\tfrac {1}{2}}{\text{aclh}}({\tfrac {35}{4}}{\sqrt[{4}]{5}})]^{2}\}{\bigr \rangle }^{2}}}\times }](https://wikimedia.org/api/rest_v1/media/math/render/svg/07fff627922c2d8c366dc8825dc5126cecc8a01b)
![{\displaystyle \times {\frac {1-R{\bigl \langle }q\{{\text{ctlh}}[{\tfrac {1}{2}}{\text{aclh}}({\tfrac {35}{4}}{\sqrt[{4}]{5}})]^{2}\}^{2}{\bigr \rangle }\,S{\bigl \langle }q\{{\text{ctlh}}[{\tfrac {1}{2}}{\text{aclh}}({\tfrac {35}{4}}{\sqrt[{4}]{5}})]^{2}\}{\bigr \rangle }}{R{\bigl \langle }q\{{\text{ctlh}}[{\tfrac {1}{2}}{\text{aclh}}({\tfrac {35}{4}}{\sqrt[{4}]{5}})]^{2}\}^{2}{\bigr \rangle }^{2}}}\times }](https://wikimedia.org/api/rest_v1/media/math/render/svg/3e95f9ebcc86c1af3e2a91f7c05b13d3249dfeba)
![{\displaystyle \times {\frac {\vartheta _{00}{\bigl \langle }q\{{\text{ctlh}}[{\tfrac {1}{2}}{\text{aclh}}({\tfrac {35}{4}}{\sqrt[{4}]{5}})]^{2}\}^{5}{\bigr \rangle }\,\vartheta _{00}{\bigl \langle }q\{{\text{ctlh}}[{\tfrac {1}{2}}{\text{aclh}}({\tfrac {35}{4}}{\sqrt[{4}]{5}})]^{2}\}^{1/5}{\bigr \rangle }^{2}-5\,\vartheta _{00}{\bigl \langle }q\{{\text{ctlh}}[{\tfrac {1}{2}}{\text{aclh}}({\tfrac {35}{4}}{\sqrt[{4}]{5}})]^{2}\}^{5}{\bigr \rangle }^{3}}{2{\sqrt[{4}]{20}}\,{\text{sl}}[{\tfrac {1}{2}}{\sqrt {2}}\,{\text{aclh}}({\tfrac {35}{4}}{\sqrt[{4}]{5}})]\,\vartheta _{00}{\bigl \langle }q\{{\text{ctlh}}[{\tfrac {1}{2}}{\text{aclh}}({\tfrac {35}{4}}{\sqrt[{4}]{5}})]^{2}\}{\bigr \rangle }^{3}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7bb028d4ca204cfbcf87201b4b37c676ac96e017)
![{\displaystyle q\{{\text{ctlh}}[{\tfrac {1}{2}}{\text{aclh}}({\tfrac {35}{4}}{\sqrt[{4}]{5}})]^{2}\}\approx 0.53609630892200161460073096549143569900990236}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0eddcb1393869a86603d07d9ba98c79ccd0fd20a)
