गणितीय सम्मिश्र विश्लेषण में, थीटा फलन कई जटिल चर के कार्यों का एक विशेष वर्ग बनाते हैं। इन कार्यों की व्यवस्थित रूप से गणितज्ञ श्रीनिवास रामानुजन और कार्ल गुस्ताव जैकब जैकोबी द्वारा जांच की गई थी। थीटा फलन अण्डाकार कार्यों में से हैं। उनका उपयोग गणितीय विश्लेषण और उष्मागतिकी में किया जाता है। ग्रासमैन के बीजगणित के लिए सामान्यीकृत, थीटा फलन प्रमात्रा क्षेत्र सिद्धान्त में भी दिखाई होते हैं।
मूल जैकोबी थीटा फलन अर्ध-दोगुने आवधिक अण्डाकार कार्य हैं और उन्हें अनंत राशि के रूप में परिभाषित किया गया है:
ϑ
1
(
z
;
w
)
=
∑
k
=
−
∞
∞
(
−
1
)
k
−
1
/
2
exp
[
(
2
k
+
1
)
i
z
+
(
k
+
1
2
)
2
ln
(
w
)
]
{\displaystyle \vartheta _{1}(z;w)=\sum _{k=-\infty }^{\infty }(-1)^{k-1/2}\exp[(2k+1)iz+(k+{\tfrac {1}{2}})^{2}\ln(w)]}
ϑ
2
(
z
;
w
)
=
∑
k
=
−
∞
∞
exp
[
(
2
k
+
1
)
i
z
+
(
k
+
1
2
)
2
ln
(
w
)
]
{\displaystyle \vartheta _{2}(z;w)=\sum _{k=-\infty }^{\infty }\exp[(2k+1)iz+(k+{\tfrac {1}{2}})^{2}\ln(w)]}
ϑ
3
(
z
;
w
)
=
∑
k
=
−
∞
∞
exp
[
2
k
i
z
+
k
2
ln
(
w
)
]
{\displaystyle \vartheta _{3}(z;w)=\sum _{k=-\infty }^{\infty }\exp[2kiz+k^{2}\ln(w)]}
ϑ
4
(
z
;
w
)
=
∑
k
=
−
∞
∞
(
−
1
)
k
exp
[
2
k
i
z
+
k
2
ln
(
w
)
]
{\displaystyle \vartheta _{4}(z;w)=\sum _{k=-\infty }^{\infty }(-1)^{k}\exp[2kiz+k^{2}\ln(w)]}
प्रोटेस्टेंट जर्मन गणितज्ञ कार्ल गुस्ताव जैकोब जैकोबी ने 1829 में इन विश्लेषणात्मक कार्यों की शुरुआत की।
उन्होंने उन्हें "Fundamenta nova theoriae functionum ellipticarum" पुस्तक में नोट किया।
अतिरिक्त जैकोबी थीटा फलन[ 1]
ϑ
00
(
x
;
y
)
=
∏
n
=
1
∞
(
1
−
y
2
n
)
[
1
+
2
cos
(
2
x
)
y
2
n
−
1
+
y
4
n
−
2
]
{\displaystyle \vartheta _{00}(x;y)=\prod _{n=1}^{\infty }(1-y^{2n})[1+2\cos(2x)y^{2n-1}+y^{4n-2}]}
ϑ
01
(
x
;
y
)
=
∏
n
=
1
∞
(
1
−
y
2
n
)
[
1
−
2
cos
(
2
x
)
y
2
n
−
1
+
y
4
n
−
2
]
{\displaystyle \vartheta _{01}(x;y)=\prod _{n=1}^{\infty }(1-y^{2n})[1-2\cos(2x)y^{2n-1}+y^{4n-2}]}
ϑ
10
(
x
;
y
)
=
2
y
1
/
4
cos
(
x
)
∏
n
=
1
∞
(
1
−
y
2
n
)
[
1
+
2
cos
(
2
x
)
y
2
n
+
y
4
n
]
{\displaystyle \vartheta _{10}(x;y)=2y^{1/4}\cos(x)\prod _{n=1}^{\infty }(1-y^{2n})[1+2\cos(2x)y^{2n}+y^{4n}]}
ये तीन फलन गणित में नियमित रूप से उपयोग किए जाते हैं और उपरोक्त चार कार्यों से बीजगणितीय रूप से संबंधित हैं।
इन तीन थीटा फलनों का उपयोग करके "sn", "cn" और "dn" फलनों[ 2]
इन कार्यों के लिए तथाकथित थीटा शून्य मान (जर्मन भाषा में: Theta-Nullwerte) भी परिभाषित किए गए हैं:
ϑ
00
(
0
;
y
)
=
ϑ
00
(
y
)
=
∑
k
=
−
∞
∞
y
k
2
{\displaystyle \vartheta _{00}(0;y)=\vartheta _{00}(y)=\sum _{k=-\infty }^{\infty }y^{k^{2}}}
ϑ
01
(
0
;
y
)
=
ϑ
01
(
y
)
=
∑
k
=
−
∞
∞
(
−
1
)
k
y
k
2
{\displaystyle \vartheta _{01}(0;y)=\vartheta _{01}(y)=\sum _{k=-\infty }^{\infty }(-1)^{k}y^{k^{2}}}
ϑ
10
(
0
;
y
)
=
ϑ
10
(
y
)
=
∑
k
=
−
∞
∞
y
(
k
+
1
2
)
2
{\displaystyle \vartheta _{10}(0;y)=\vartheta _{10}(y)=\sum _{k=-\infty }^{\infty }y^{(k+{\frac {1}{2}})^{2}}}
यहां दिखाए गए अनंत योग बिल्कुल वही मान देते हैं जो x-मान शून्य के लिए उल्लिखित अनंत उत्पाद हैं।
इस प्रकार मुख्य थीटा फ़ंक्शन को तथाकथित अनुचित इंटीग्रल का उपयोग करके प्रदर्शित किया जा सकता है:
ϑ
00
(
v
)
=
1
+
4
v
ln
(
1
/
v
)
π
∫
0
∞
exp
[
−
ln
(
1
/
v
)
x
2
]
{
1
−
v
2
cos
[
2
ln
(
1
/
v
)
x
]
}
1
−
2
v
2
cos
[
2
ln
(
1
/
v
)
x
]
+
v
4
d
x
{\displaystyle \vartheta _{00}(v)=1+{\frac {4v{\sqrt {\ln(1/v)}}}{\sqrt {\pi }}}\int _{0}^{\infty }{\frac {\exp[-\ln(1/v)\,x^{2}]\{1-v^{2}\cos[2\ln(1/v)\,x]\}}{1-2v^{2}\cos[2\ln(1/v)\,x]+v^{4}}}\,\mathrm {d} x}
थीटा उदाहरण
θ
3
(
1
2
)
{\displaystyle \theta _{3}({\tfrac {1}{2}})}
θ
3
(
1
3
)
{\displaystyle \theta _{3}({\tfrac {1}{3}})}
ϑ
00
(
1
2
)
=
1
+
2
∑
n
=
1
∞
1
2
n
2
=
1
+
2
π
−
1
/
2
ln
(
2
)
∫
0
∞
exp
[
−
ln
(
2
)
x
2
]
{
16
−
4
cos
[
2
ln
(
2
)
x
]
}
17
−
8
cos
[
2
ln
(
2
)
x
]
d
x
{\displaystyle \vartheta _{00}{\bigl (}{\frac {1}{2}}{\bigr )}=1+2\sum _{n=1}^{\infty }{\frac {1}{2^{n^{2}}}}=1+2\pi ^{-1/2}{\sqrt {\ln(2)}}\,\int _{0}^{\infty }{\frac {\exp[-\ln(2)\,x^{2}]\{16-4\cos[2\ln(2)\,x]\}}{17-8\cos[2\ln(2)\,x]}}\,\mathrm {d} x}
ϑ
00
(
1
2
)
=
2.128936827211877158669
…
{\displaystyle \vartheta _{00}{\bigl (}{\frac {1}{2}}{\bigr )}=2.128936827211877158669\ldots }
ϑ
00
(
1
3
)
=
1
+
2
∑
n
=
1
∞
1
3
n
2
=
1
+
4
3
π
−
1
/
2
ln
(
3
)
∫
0
∞
exp
[
−
ln
(
3
)
x
2
]
{
81
−
9
cos
[
2
ln
(
3
)
x
]
}
82
−
18
cos
[
2
ln
(
3
)
x
]
d
x
{\displaystyle \vartheta _{00}{\bigl (}{\frac {1}{3}}{\bigr )}=1+2\sum _{n=1}^{\infty }{\frac {1}{3^{n^{2}}}}=1+{\frac {4}{3}}\pi ^{-1/2}{\sqrt {\ln(3)}}\,\int _{0}^{\infty }{\frac {\exp[-\ln(3)\,x^{2}]\{81-9\cos[2\ln(3)\,x]\}}{82-18\cos[2\ln(3)\,x]}}\,\mathrm {d} x}
ϑ
00
(
1
3
)
=
1.691459681681715341348
…
{\displaystyle \vartheta _{00}{\bigl (}{\frac {1}{3}}{\bigr )}=1.691459681681715341348\ldots }
ϑ
00
(
1
5
)
=
1
+
2
∑
n
=
1
∞
1
5
n
2
=
1
+
4
5
π
−
1
/
2
ln
(
5
)
∫
0
∞
exp
[
−
ln
(
5
)
x
2
]
{
625
−
25
cos
[
2
ln
(
5
)
x
]
}
626
−
50
cos
[
2
ln
(
5
)
x
]
d
x
{\displaystyle \vartheta _{00}{\bigl (}{\frac {1}{5}}{\bigr )}=1+2\sum _{n=1}^{\infty }{\frac {1}{5^{n^{2}}}}=1+{\frac {4}{5}}\pi ^{-1/2}{\sqrt {\ln(5)}}\,\int _{0}^{\infty }{\frac {\exp[-\ln(5)\,x^{2}]\{625-25\cos[2\ln(5)\,x]\}}{626-50\cos[2\ln(5)\,x]}}\,\mathrm {d} x}
ϑ
00
(
1
5
)
=
1.40320102401310720671
…
{\displaystyle \vartheta _{00}{\bigl (}{\frac {1}{5}}{\bigr )}=1.40320102401310720671\ldots }
थीटा कार्यों के लिए जोड़ प्रमेय इस प्रकार हैं:
ϑ
00
(
x
1
+
x
2
;
y
)
ϑ
00
(
x
1
−
x
2
;
y
)
ϑ
00
(
y
)
2
=
ϑ
01
(
x
1
;
y
)
2
ϑ
01
(
x
2
;
y
)
2
+
ϑ
10
(
x
1
;
y
)
2
ϑ
10
(
x
2
;
y
)
2
{\displaystyle \vartheta _{00}(x_{1}+x_{2};y)\vartheta _{00}(x_{1}-x_{2};y)\vartheta _{00}(y)^{2}=\vartheta _{01}(x_{1};y)^{2}\vartheta _{01}(x_{2};y)^{2}+\vartheta _{10}(x_{1};y)^{2}\vartheta _{10}(x_{2};y)^{2}}
ϑ
01
(
x
1
+
x
2
;
y
)
ϑ
01
(
x
1
−
x
2
;
y
)
ϑ
01
(
y
)
2
=
ϑ
00
(
x
1
;
y
)
2
ϑ
00
(
x
2
;
y
)
2
−
ϑ
10
(
x
1
;
y
)
2
ϑ
10
(
x
2
;
y
)
2
{\displaystyle \vartheta _{01}(x_{1}+x_{2};y)\vartheta _{01}(x_{1}-x_{2};y)\vartheta _{01}(y)^{2}=\vartheta _{00}(x_{1};y)^{2}\vartheta _{00}(x_{2};y)^{2}-\vartheta _{10}(x_{1};y)^{2}\vartheta _{10}(x_{2};y)^{2}}
गणितज्ञ श्रीनिवास रामानुजन ने इस पहचान[ 3] Modular Equations and Approximations to π " में लिखा:
∏
n
=
1
∞
(
1
−
x
2
n
−
1
)
=
(
x
;
x
2
)
∞
=
2
1
/
6
x
1
/
24
ϑ
10
(
x
)
−
1
/
6
ϑ
00
(
x
)
−
1
/
6
ϑ
01
(
x
)
1
/
3
{\displaystyle \prod _{n=1}^{\infty }(1-x^{2n-1})=(x;x^{2})_{\infty }=2^{1/6}x^{1/24}\vartheta _{10}(x)^{-1/6}\vartheta _{00}(x)^{-1/6}\vartheta _{01}(x)^{1/3}}
लियोनार्ड ओइलर द्वारा निम्नलिखित उत्पाद का शोध किया गया था:
∏
n
=
1
∞
(
1
−
x
n
)
=
(
x
;
x
)
∞
=
2
−
1
/
6
x
−
1
/
24
ϑ
10
(
x
)
1
/
6
ϑ
00
(
x
)
1
/
6
ϑ
01
(
x
)
2
/
3
{\displaystyle \prod _{n=1}^{\infty }(1-x^{n})=(x;x)_{\infty }=2^{-1/6}x^{-1/24}\vartheta _{10}(x)^{1/6}\vartheta _{00}(x)^{1/6}\vartheta _{01}(x)^{2/3}}
यदि शर्त "0 < s < 1" लागू होती है, तो निम्न समीकरण मान्य है:
∑
k
=
−
∞
∞
2
s
k
s
2
k
+
1
=
ϑ
00
(
s
)
2
{\displaystyle \sum _{k=-\infty }^{\infty }{\frac {2s^{k}}{s^{2k}+1}}=\vartheta _{00}(s)^{2}}
अण्डाकार नोम फलन की यह परिभाषा है:
q
(
ε
)
=
exp
[
−
π
K
(
1
−
ε
2
)
K
(
ε
)
−
1
]
{\displaystyle q(\varepsilon )=\exp[-\pi K({\sqrt {1-\varepsilon ^{2}}})K(\varepsilon )^{-1}]}
फलन "के" निम्नलिखित अभिन्न द्वारा परिभाषित किया गया है:
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 }
K
(
r
)
=
∫
0
1
2
(
z
2
+
1
)
2
−
4
r
2
z
2
d
z
{\displaystyle K(r)=\int _{0}^{1}{\frac {2}{\sqrt {(z^{2}+1)^{2}-4r^{2}z^{2}}}}\,\mathrm {d} z}
कुछ फलन मानों की गणना निम्न सूत्र से की जा सकती है:
ϑ
00
[
q
(
ε
)
]
=
2
π
−
1
K
(
ε
)
{\displaystyle \vartheta _{00}[q(\varepsilon )]={\sqrt {2\pi ^{-1}K(\varepsilon )}}}
ϑ
01
[
q
(
ε
)
]
=
1
−
ε
2
4
2
π
−
1
K
(
ε
)
{\displaystyle \vartheta _{01}[q(\varepsilon )]={\sqrt[{4}]{1-\varepsilon ^{2}}}{\sqrt {2\pi ^{-1}K(\varepsilon )}}}
जैकोबी पहचान इन समीकरणों से उत्पन्न होती है:
ϑ
10
(
x
)
=
ϑ
00
(
x
)
4
−
ϑ
01
(
x
)
4
4
{\displaystyle \vartheta _{10}(x)={\sqrt[{4}]{\vartheta _{00}(x)^{4}-\vartheta _{01}(x)^{4}}}}
थीटा फलनों में निम्नलिखित मान[ 4]
ϑ
00
[
exp
(
−
π
)
]
=
π
1
/
4
Γ
(
3
4
)
−
1
{\displaystyle \vartheta _{00}[\exp(-\pi )]=\pi ^{1/4}\Gamma ({\tfrac {3}{4}})^{-1}}
ϑ
01
[
exp
(
−
π
)
]
=
2
−
1
/
4
π
1
/
4
Γ
(
3
4
)
−
1
{\displaystyle \vartheta _{01}[\exp(-\pi )]=2^{-1/4}\pi ^{1/4}\Gamma ({\tfrac {3}{4}})^{-1}}
ϑ
10
[
exp
(
−
π
)
]
=
2
−
1
/
4
π
1
/
4
Γ
(
3
4
)
−
1
{\displaystyle \vartheta _{10}[\exp(-\pi )]=2^{-1/4}\pi ^{1/4}\Gamma ({\tfrac {3}{4}})^{-1}}
ϑ
00
[
exp
(
−
2
π
)
]
=
2
−
1
/
8
π
1
/
2
Γ
(
3
4
)
−
1
/
2
Γ
(
5
8
)
−
1
{\displaystyle \vartheta _{00}[\exp(-{\sqrt {2}}\,\pi )]=2^{-1/8}\pi ^{1/2}\Gamma ({\tfrac {3}{4}})^{-1/2}\Gamma ({\tfrac {5}{8}})^{-1}}
ϑ
00
[
exp
(
−
3
π
)
]
=
2
5
/
6
3
−
5
/
8
π
1
/
2
Γ
(
2
3
)
−
3
/
2
{\displaystyle \vartheta _{00}[\exp(-{\sqrt {3}}\,\pi )]=2^{5/6}3^{-5/8}\pi ^{1/2}\Gamma ({\tfrac {2}{3}})^{-3/2}}
कई फलन मान निर्धारित करने के लिए निम्नलिखित पहचान सूत्रों का उपयोग किया जाता है:
ϑ
00
[
q
(
ε
)
2
]
=
cos
[
1
2
arcsin
(
ε
)
]
ϑ
00
[
q
(
ε
)
]
{\displaystyle \vartheta _{00}[q(\varepsilon )^{2}]=\cos[{\tfrac {1}{2}}\arcsin(\varepsilon )]\vartheta _{00}[q(\varepsilon )]}
ϑ
01
[
q
(
ε
)
2
]
=
(
1
−
ε
2
)
1
/
8
ϑ
00
[
q
(
ε
)
]
{\displaystyle \vartheta _{01}[q(\varepsilon )^{2}]=(1-\varepsilon ^{2})^{1/8}\vartheta _{00}[q(\varepsilon )]}
27
ϑ
00
[
q
(
ε
)
3
]
8
ϑ
00
[
q
(
ε
)
]
8
=
18
ϑ
00
[
q
(
ε
)
3
]
4
ϑ
00
[
q
(
ε
)
]
4
+
8
cos
[
2
arcsin
(
ε
)
]
ϑ
00
[
q
(
ε
)
3
]
2
ϑ
00
[
q
(
ε
)
]
2
+
1
{\displaystyle 27{\frac {\vartheta _{00}[q(\varepsilon )^{3}]^{8}}{\vartheta _{00}[q(\varepsilon )]^{8}}}=18{\frac {\vartheta _{00}[q(\varepsilon )^{3}]^{4}}{\vartheta _{00}[q(\varepsilon )]^{4}}}+8\cos[2\arcsin(\varepsilon )]{\frac {\vartheta _{00}[q(\varepsilon )^{3}]^{2}}{\vartheta _{00}[q(\varepsilon )]^{2}}}+1}
27
ϑ
01
[
q
(
ε
)
3
]
8
ϑ
01
[
q
(
ε
)
]
8
=
18
ϑ
01
[
q
(
ε
)
3
]
4
ϑ
01
[
q
(
ε
)
]
4
+
8
sec
[
2
arctan
(
ε
)
]
ϑ
01
[
q
(
ε
)
3
]
2
ϑ
01
[
q
(
ε
)
]
2
+
1
{\displaystyle 27{\frac {\vartheta _{01}[q(\varepsilon )^{3}]^{8}}{\vartheta _{01}[q(\varepsilon )]^{8}}}=18{\frac {\vartheta _{01}[q(\varepsilon )^{3}]^{4}}{\vartheta _{01}[q(\varepsilon )]^{4}}}+8\sec[2\arctan(\varepsilon )]{\frac {\vartheta _{01}[q(\varepsilon )^{3}]^{2}}{\vartheta _{01}[q(\varepsilon )]^{2}}}+1}
{
ϑ
01
[
q
(
ε
)
5
]
2
ϑ
01
[
q
(
ε
)
]
2
−
1
}
{
5
ϑ
01
[
q
(
ε
)
5
]
2
ϑ
01
[
q
(
ε
)
]
2
−
1
}
5
=
64
tan
[
2
arctan
(
ε
)
]
2
ϑ
01
[
q
(
ε
)
5
]
2
ϑ
01
[
q
(
ε
)
]
2
{\displaystyle {\biggl \{}{\frac {\vartheta _{01}[q(\varepsilon )^{5}]^{2}}{\vartheta _{01}[q(\varepsilon )]^{2}}}-1{\biggr \}}{\biggl \{}5{\frac {\vartheta _{01}[q(\varepsilon )^{5}]^{2}}{\vartheta _{01}[q(\varepsilon )]^{2}}}-1{\biggr \}}^{5}=64\tan[2\arctan(\varepsilon )]^{2}{\frac {\vartheta _{01}[q(\varepsilon )^{5}]^{2}}{\vartheta _{01}[q(\varepsilon )]^{2}}}}
गणना के उदाहरण:
exp
(
−
π
)
=
q
(
1
2
2
)
{\displaystyle \exp(-\pi )=q({\tfrac {1}{2}}{\sqrt {2}})}
exp
(
−
2
π
)
=
q
(
2
−
1
)
{\displaystyle \exp(-{\sqrt {2}}\,\pi )=q({\sqrt {2}}-1)}
exp
(
−
3
π
)
=
q
[
1
4
(
6
−
2
)
]
{\displaystyle \exp(-{\sqrt {3}}\,\pi )=q[{\tfrac {1}{4}}({\sqrt {6}}-{\sqrt {2}})]}
इन मानों को समीकरणों में डालने और फिर उपरोक्त समीकरणों को हल करने से निम्नलिखित मान उत्पन्न होते हैं:
ϑ
00
[
exp
(
−
3
π
)
]
ϑ
00
[
exp
(
−
π
)
]
=
108
−
1
/
8
3
+
1
{\displaystyle {\frac {\vartheta _{00}[\exp(-3\pi )]}{\vartheta _{00}[\exp(-\pi )]}}=108^{-1/8}{\sqrt {{\sqrt {3}}+1}}}
ϑ
00
[
exp
(
−
3
2
π
)
]
ϑ
00
[
exp
(
−
2
π
)
]
=
3
−
1
/
2
6
+
2
−
1
{\displaystyle {\frac {\vartheta _{00}[\exp(-3{\sqrt {2}}\,\pi )]}{\vartheta _{00}[\exp(-{\sqrt {2}}\,\pi )]}}=3^{-1/2}{\sqrt {{\sqrt {6}}+{\sqrt {2}}-1}}}
ϑ
00
[
exp
(
−
3
3
π
)
]
ϑ
00
[
exp
(
−
3
π
)
]
=
3
−
3
/
4
(
2
3
+
1
)
{\displaystyle {\frac {\vartheta _{00}[\exp(-3{\sqrt {3}}\,\pi )]}{\vartheta _{00}[\exp(-{\sqrt {3}}\,\pi )]}}=3^{-3/4}({\sqrt[{3}]{2}}+1)}
ϑ
01
[
exp
(
−
3
π
)
]
ϑ
01
[
exp
(
−
π
)
]
=
3
−
3
/
8
2
+
3
4
{\displaystyle {\frac {\vartheta _{01}[\exp(-3\pi )]}{\vartheta _{01}[\exp(-\pi )]}}=3^{-3/8}{\sqrt {{\sqrt {2}}+{\sqrt[{4}]{3}}}}}
ϑ
01
[
exp
(
−
3
2
π
)
]
ϑ
01
[
exp
(
−
2
π
)
]
=
3
−
1
/
2
3
+
2
{\displaystyle {\frac {\vartheta _{01}[\exp(-3{\sqrt {2}}\,\pi )]}{\vartheta _{01}[\exp(-{\sqrt {2}}\,\pi )]}}=3^{-1/2}{\sqrt {{\sqrt {3}}+{\sqrt {2}}}}}
ϑ
01
[
exp
(
−
3
3
π
)
]
ϑ
01
[
exp
(
−
3
π
)
]
=
108
−
1
/
4
(
2
2
3
+
3
−
1
)
{\displaystyle {\frac {\vartheta _{01}[\exp(-3{\sqrt {3}}\,\pi )]}{\vartheta _{01}[\exp(-{\sqrt {3}}\,\pi )]}}=108^{-1/4}(2{\sqrt[{3}]{2}}+{\sqrt {3}}-1)}
ϑ
01
[
exp
(
−
5
π
)
]
ϑ
01
[
exp
(
−
π
)
]
=
5
−
1
/
2
3
+
2
5
4
{\displaystyle {\frac {\vartheta _{01}[\exp(-5\pi )]}{\vartheta _{01}[\exp(-\pi )]}}=5^{-1/2}{\sqrt {3+2{\sqrt[{4}]{5}}}}}
ϑ
01
[
exp
(
−
5
2
π
)
]
ϑ
01
[
exp
(
−
2
π
)
]
=
4
15
10
cos
(
1
10
π
)
cosh
[
1
3
artanh
(
3
8
6
)
]
+
1
15
5
tan
(
1
5
π
)
{\displaystyle {\frac {\vartheta _{01}[\exp(-5{\sqrt {2}}\,\pi )]}{\vartheta _{01}[\exp(-{\sqrt {2}}\,\pi )]}}={\tfrac {4}{15}}{\sqrt {10}}\cos({\tfrac {1}{10}}\pi )\cosh[{\tfrac {1}{3}}{\text{artanh}}({\tfrac {3}{8}}{\sqrt {6}})]+{\tfrac {1}{15}}{\sqrt {5}}\tan({\tfrac {1}{5}}\pi )}
ϑ
01
[
exp
(
−
5
3
π
)
]
ϑ
01
[
exp
(
−
3
π
)
]
=
2
15
10
3
(
3
−
3
)
sin
(
1
5
π
)
+
1
15
15
cot
(
1
10
π
)
{\displaystyle {\frac {\vartheta _{01}[\exp(-5{\sqrt {3}}\,\pi )]}{\vartheta _{01}[\exp(-{\sqrt {3}}\,\pi )]}}={\tfrac {2}{15}}{\sqrt[{3}]{10}}(3-{\sqrt {3}})\sin({\tfrac {1}{5}}\pi )+{\tfrac {1}{15}}{\sqrt {15}}\cot({\tfrac {1}{10}}\pi )}
अण्डाकार नोम और अण्डाकार अभिन्न के बीच संबंध के बारे में पहले बताए गए सूत्र के आधार पर, कुछ अतिरिक्त मान भी कुशलतापूर्वक स्थापित किए जा सकते हैं:
ϑ
00
[
q
(
ε
)
]
=
[
2
π
K
(
ε
)
]
1
/
2
{\displaystyle \vartheta _{00}[q(\varepsilon )]={\biggl [}{\frac {2}{\pi }}K(\varepsilon ){\biggr ]}^{1/2}}
अब कुछ उदाहरण दिये जायेंगे:
ϑ
00
[
exp
(
−
5
π
)
]
=
⟨
2
π
K
{
sin
[
1
2
arcsin
(
5
−
2
)
]
}
⟩
1
/
2
{\displaystyle \vartheta _{00}{\bigl [}\exp(-{\sqrt {5}}\,\pi ){\bigr ]}={\biggl \langle }{\frac {2}{\pi }}K{\bigl \{}\sin {\bigl [}{\tfrac {1}{2}}\arcsin({\sqrt {5}}-2){\bigr ]}{\bigr \}}{\biggr \rangle }^{1/2}}
ϑ
00
[
exp
(
−
6
π
)
]
=
{
2
π
K
[
(
2
−
3
)
(
3
−
2
)
]
}
1
/
2
{\displaystyle \vartheta _{00}{\bigl [}\exp(-{\sqrt {6}}\,\pi ){\bigr ]}={\biggl \{}{\frac {2}{\pi }}K{\bigl [}(2-{\sqrt {3}}\,)({\sqrt {3}}-{\sqrt {2}}\,){\bigr ]}{\biggr \}}^{1/2}}
ϑ
00
[
exp
(
−
7
π
)
]
=
{
2
π
K
[
1
8
(
3
2
−
14
)
]
}
1
/
2
{\displaystyle \vartheta _{00}{\bigl [}\exp(-{\sqrt {7}}\,\pi ){\bigr ]}={\biggl \{}{\frac {2}{\pi }}K{\bigl [}{\frac {1}{8}}{\bigl (}3{\sqrt {2}}-{\sqrt {14}}\,{\bigr )}{\bigr ]}{\biggr \}}^{1/2}}
ϑ
00
[
exp
(
−
10
π
)
]
=
{
2
π
K
[
(
10
−
3
)
(
2
−
1
)
2
]
}
1
/
2
{\displaystyle \vartheta _{00}{\bigl [}\exp(-{\sqrt {10}}\,\pi ){\bigr ]}={\biggl \{}{\frac {2}{\pi }}K{\bigl [}({\sqrt {10}}-3)({\sqrt {2}}-1)^{2}{\bigr ]}{\biggr \}}^{1/2}}
ϑ
00
[
exp
(
−
11
π
)
]
=
{
2
π
K
[
1
16
(
22
+
3
2
)
(
1
3
6
3
+
2
11
3
−
1
3
6
3
−
2
11
3
+
1
3
11
−
1
)
4
]
}
1
/
2
{\displaystyle \vartheta _{00}{\bigl [}\exp(-{\sqrt {11}}\,\pi ){\bigr ]}={\biggl \{}{\frac {2}{\pi }}K{\bigl [}{\frac {1}{16}}{\bigl (}{\sqrt {22}}+3{\sqrt {2}}\,{\bigr )}{\bigl (}{\frac {1}{3}}{\sqrt[{3}]{6{\sqrt {3}}+2{\sqrt {11}}}}-{\frac {1}{3}}{\sqrt[{3}]{6{\sqrt {3}}-2{\sqrt {11}}}}+{\frac {1}{3}}{\sqrt {11}}-1{\bigr )}^{4}{\bigr ]}{\biggr \}}^{1/2}}
ϑ
00
[
exp
(
−
13
π
)
]
=
⟨
2
π
K
{
sin
[
1
2
arcsin
(
5
13
−
18
)
]
}
⟩
1
/
2
{\displaystyle \vartheta _{00}{\bigl [}\exp(-{\sqrt {13}}\,\pi ){\bigr ]}={\biggl \langle }{\frac {2}{\pi }}K{\bigl \{}\sin {\bigl [}{\tfrac {1}{2}}\arcsin(5{\sqrt {13}}-18){\bigr ]}{\bigr \}}{\biggr \rangle }^{1/2}}
ϑ
00
[
exp
(
−
14
π
)
]
=
⟨
2
π
K
{
tan
[
1
4
arccsc
(
8
2
+
11
)
]
}
⟩
1
/
2
{\displaystyle \vartheta _{00}{\bigl [}\exp(-{\sqrt {14}}\,\pi ){\bigr ]}={\biggl \langle }{\frac {2}{\pi }}K{\bigl \{}\tan {\bigl [}{\tfrac {1}{4}}\operatorname {arccsc}(8{\sqrt {2}}+11){\bigr ]}{\bigr \}}{\biggr \rangle }^{1/2}}
संज्ञा के परिवर्तनों पर[ 5]
ϑ
10
(
x
2
)
=
1
2
2
[
ϑ
00
(
x
)
2
−
ϑ
01
(
x
)
2
]
{\displaystyle \vartheta _{10}(x^{2})={\tfrac {1}{2}}{\sqrt {2[\vartheta _{00}(x)^{2}-\vartheta _{01}(x)^{2}]}}}
ϑ
00
(
x
2
)
=
1
2
2
[
ϑ
00
(
x
)
2
+
ϑ
01
(
x
)
2
]
{\displaystyle \vartheta _{00}(x^{2})={\tfrac {1}{2}}{\sqrt {2[\vartheta _{00}(x)^{2}+\vartheta _{01}(x)^{2}]}}}
ϑ
01
(
x
2
)
=
ϑ
01
(
x
)
ϑ
00
(
x
)
{\displaystyle \vartheta _{01}(x^{2})={\sqrt {\vartheta _{01}(x)\vartheta _{00}(x)}}}
जैकोबी पहचान के अनुसार, तीन थीटा शून्य-मूल्य फ़ंक्शन के वर्ग भी वर्ग फ़ंक्शन द्वारा एक आंतरिक फ़ंक्शन पाइथागोरस ट्रिपल के रूप में बनते हैं। इसके अतिरिक्त, निम्नलिखित परिवर्तन लागू होते हैं:
ϑ
00
(
x
4
)
=
1
2
ϑ
00
(
x
)
+
1
2
ϑ
01
(
x
)
{\displaystyle \vartheta _{00}(x^{4})={\tfrac {1}{2}}\vartheta _{00}(x)+{\tfrac {1}{2}}\vartheta _{01}(x)}
27
ϑ
00
(
x
3
)
8
−
18
ϑ
00
(
x
3
)
4
ϑ
00
(
x
)
4
−
ϑ
00
(
x
)
8
=
8
ϑ
00
(
x
3
)
2
ϑ
00
(
x
)
2
[
2
ϑ
01
(
x
)
4
−
ϑ
00
(
x
)
4
]
{\displaystyle 27\,\vartheta _{00}(x^{3})^{8}-18\,\vartheta _{00}(x^{3})^{4}\vartheta _{00}(x)^{4}-\,\vartheta _{00}(x)^{8}=8\,\vartheta _{00}(x^{3})^{2}\vartheta _{00}(x)^{2}[2\,\vartheta _{01}(x)^{4}-\vartheta _{00}(x)^{4}]}
27
ϑ
01
(
x
3
)
8
−
18
ϑ
01
(
x
3
)
4
ϑ
01
(
x
)
4
−
ϑ
01
(
x
)
8
=
8
ϑ
01
(
x
3
)
2
ϑ
01
(
x
)
2
[
2
ϑ
00
(
x
)
4
−
ϑ
01
(
x
)
4
]
{\displaystyle 27\,\vartheta _{01}(x^{3})^{8}-18\,\vartheta _{01}(x^{3})^{4}\vartheta _{01}(x)^{4}-\,\vartheta _{01}(x)^{8}=8\,\vartheta _{01}(x^{3})^{2}\vartheta _{01}(x)^{2}[2\,\vartheta _{00}(x)^{4}-\vartheta _{01}(x)^{4}]}
[
ϑ
00
(
x
)
2
−
ϑ
00
(
x
5
)
2
]
[
5
ϑ
00
(
x
5
)
2
−
ϑ
00
(
x
)
2
]
5
=
256
ϑ
00
(
x
5
)
2
ϑ
00
(
x
)
2
ϑ
01
(
x
)
4
[
ϑ
00
(
x
)
4
−
ϑ
01
(
x
)
4
]
{\displaystyle [\vartheta _{00}(x)^{2}-\vartheta _{00}(x^{5})^{2}][5\,\vartheta _{00}(x^{5})^{2}-\vartheta _{00}(x)^{2}]^{5}=256\,\vartheta _{00}(x^{5})^{2}\vartheta _{00}(x)^{2}\vartheta _{01}(x)^{4}[\vartheta _{00}(x)^{4}-\vartheta _{01}(x)^{4}]}
[
ϑ
01
(
x
5
)
2
−
ϑ
01
(
x
)
2
]
[
5
ϑ
01
(
x
5
)
2
−
ϑ
01
(
x
)
2
]
5
=
256
ϑ
01
(
x
5
)
2
ϑ
01
(
x
)
2
ϑ
00
(
x
)
4
[
ϑ
00
(
x
)
4
−
ϑ
01
(
x
)
4
]
{\displaystyle [\vartheta _{01}(x^{5})^{2}-\vartheta _{01}(x)^{2}][5\,\vartheta _{01}(x^{5})^{2}-\vartheta _{01}(x)^{2}]^{5}=256\,\vartheta _{01}(x^{5})^{2}\vartheta _{01}(x)^{2}\vartheta _{00}(x)^{4}[\vartheta _{00}(x)^{4}-\vartheta _{01}(x)^{4}]}
ऐसी सर्वसमिकाएँ हैं, जो दीर्घवृत्तीय नोम की तीसरी घात और तीसरे मूल के थीटा मानों को एक दूसरे के संबंध में निर्धारित करती हैं।
[
ϑ
00
(
x
1
/
3
)
2
ϑ
00
(
x
)
2
−
3
ϑ
00
(
x
3
)
2
ϑ
00
(
x
)
2
]
2
=
4
−
4
[
2
ϑ
10
(
x
)
2
ϑ
01
(
x
)
2
ϑ
00
(
x
)
4
]
2
/
3
{\displaystyle {\biggl [}{\frac {\vartheta _{00}(x^{1/3})^{2}}{\vartheta _{00}(x)^{2}}}-{\frac {3\,\vartheta _{00}(x^{3})^{2}}{\vartheta _{00}(x)^{2}}}{\biggr ]}^{2}=4-4{\biggl [}{\frac {2\,\vartheta _{10}(x)^{2}\vartheta _{01}(x)^{2}}{\vartheta _{00}(x)^{4}}}{\biggr ]}^{2/3}}
[
3
ϑ
01
(
x
3
)
2
ϑ
01
(
x
)
2
−
ϑ
01
(
x
1
/
3
)
2
ϑ
01
(
x
)
2
]
2
=
4
+
4
[
2
ϑ
10
(
x
)
2
ϑ
00
(
x
)
2
ϑ
01
(
x
)
4
]
2
/
3
{\displaystyle {\biggl [}{\frac {3\,\vartheta _{01}(x^{3})^{2}}{\vartheta _{01}(x)^{2}}}-{\frac {\vartheta _{01}(x^{1/3})^{2}}{\vartheta _{01}(x)^{2}}}{\biggr ]}^{2}=4+4{\biggl [}{\frac {2\,\vartheta _{10}(x)^{2}\vartheta _{00}(x)^{2}}{\vartheta _{01}(x)^{4}}}{\biggr ]}^{2/3}}
और ऐसी भी पहचानें हैं, जो दीर्घवृत्तीय नोम की पांचवीं शक्ति और पांचवें मूल के थीटा मानों को एक दूसरे के संबंध में निर्धारित करती हैं।
2
arctan
[
ϑ
00
(
x
5
)
ϑ
00
(
x
1
/
5
)
−
ϑ
00
(
x
)
2
ϑ
00
(
x
1
/
5
)
2
+
5
ϑ
00
(
x
5
)
2
−
ϑ
00
(
x
)
2
−
3
ϑ
00
(
x
5
)
ϑ
00
(
x
1
/
5
)
]
+
arctan
(
2
)
=
{\displaystyle 2\arctan {\biggl [}{\frac {\vartheta _{00}(x^{5})\,\vartheta _{00}(x^{1/5})-\vartheta _{00}(x)^{2}}{\vartheta _{00}(x^{1/5})^{2}+5\,\vartheta _{00}(x^{5})^{2}-\vartheta _{00}(x)^{2}-3\,\vartheta _{00}(x^{5})\,\vartheta _{00}(x^{1/5})}}{\biggr ]}+\arctan(2)=}
arctan
[
ϑ
00
(
x
1
/
5
)
2
2
ϑ
00
(
x
)
2
−
1
2
]
+
arctan
[
5
ϑ
00
(
x
5
)
2
2
ϑ
00
(
x
)
2
−
1
2
]
{\displaystyle \arctan {\biggl [}{\frac {\vartheta _{00}(x^{1/5})^{2}}{2\,\vartheta _{00}(x)^{2}}}-{\frac {1}{2}}{\biggr ]}+\arctan {\biggl [}{\frac {5\,\vartheta _{00}(x^{5})^{2}}{2\,\vartheta _{00}(x)^{2}}}-{\frac {1}{2}}{\biggr ]}}
पहले इस्तेमाल की गई जैकोबी फलन पहचान[ 6] [ 7]
ϑ
00
[
q
(
k
)
]
=
ϑ
00
{
exp
[
−
π
K
′
(
k
)
K
(
k
)
]
}
=
∑
n
=
−
∞
∞
exp
[
−
n
2
π
K
′
(
k
)
K
(
k
)
]
=
[
2
π
K
(
k
)
]
1
/
2
{\displaystyle \vartheta _{00}{\bigl [}q(k){\bigr ]}=\vartheta _{00}{\biggl \{}\exp {\biggl [}-\pi \,{\frac {K'(k)}{K(k)}}{\biggr ]}{\biggr \}}=\sum _{n=-\infty }^{\infty }\exp {\biggl [}-n^{2}\pi \,{\frac {K'(k)}{K(k)}}{\biggr ]}={\biggl [}{\frac {2}{\pi }}K(k){\biggr ]}^{1/2}}
और दीर्घवृत्तीय मापांक
k
{\displaystyle k}
k
′
=
1
−
k
2
{\displaystyle k'={\sqrt {1-k^{2}}}}
ϑ
00
[
q
′
(
k
)
]
=
ϑ
00
{
exp
[
−
π
K
(
k
)
K
′
(
k
)
]
}
=
∑
n
=
−
∞
∞
exp
[
−
n
2
π
K
(
k
)
K
′
(
k
)
]
=
[
2
π
K
′
(
k
)
]
1
/
2
{\displaystyle \vartheta _{00}{\bigl [}q'(k){\bigr ]}=\vartheta _{00}{\biggl \{}\exp {\biggl [}-\pi \,{\frac {K(k)}{K'(k)}}{\biggr ]}{\biggr \}}=\sum _{n=-\infty }^{\infty }\exp {\biggl [}-n^{2}\pi \,{\frac {K(k)}{K'(k)}}{\biggr ]}={\biggl [}{\frac {2}{\pi }}K'(k){\biggr ]}^{1/2}}
संख्या सिद्धांत पर आधारित "पॉइसन योग सूत्र" इस प्रकार उभरता है:
ϑ
00
[
q
′
(
k
)
]
ϑ
00
[
q
(
k
)
]
=
[
K
′
(
k
)
K
(
k
)
]
1
/
2
{\displaystyle {\frac {\vartheta _{00}{\bigl [}q'(k){\bigr ]}}{\vartheta _{00}{\bigl [}q(k){\bigr ]}}}={\biggl [}{\frac {K'(k)}{K(k)}}{\biggr ]}^{1/2}}
उल्लिखित अंतिम दो सूत्रों का भागफल सीधे पॉइसन योग सूत्र में परिणत होता है।
परिणाम प्रदर्शित करने का निम्नलिखित तरीका प्रत्येक स्थान पर ''वास्तविक अवधि अनुपात'' दर्शाता है:
ϑ
00
{
exp
[
−
π
K
(
k
)
÷
K
′
(
k
)
]
}
ϑ
00
{
exp
[
−
π
K
′
(
k
)
÷
K
(
k
)
]
}
=
[
K
′
(
k
)
K
(
k
)
]
1
/
2
{\displaystyle {\frac {\vartheta _{00}{\bigl \{}\exp {\bigl [}-\pi \,K(k)\div K'(k){\bigr ]}{\bigr \}}}{\vartheta _{00}{\bigl \{}\exp {\bigl [}-\pi \,K'(k)\div K(k){\bigr ]}{\bigr \}}}}={\biggl [}{\frac {K'(k)}{K(k)}}{\biggr ]}^{1/2}}
अब, अवधि अनुपात को पैरामीटर
p
=
K
′
(
k
)
÷
K
(
k
)
{\displaystyle p=K'(k)\div K(k)}
ϑ
00
[
exp
(
−
π
÷
p
)
]
ϑ
00
[
exp
(
−
π
×
p
)
]
=
p
{\displaystyle {\frac {\vartheta _{00}{\bigl [}\exp(-\pi \div p){\bigr ]}}{\vartheta _{00}{\bigl [}\exp(-\pi \times p){\bigr ]}}}={\sqrt {p}}}
इस प्रकार सूत्र को दीर्घवृत्तीय समाकलनों की सहायता से सिद्ध किया जाता है।
निम्नलिखित गणना उदाहरण सटीक रूप से तैयार किए जाएंगे:
ϑ
00
[
exp
(
−
1
2
4
3
π
)
]
=
2
6
ϑ
00
[
exp
(
−
2
3
π
)
]
{\displaystyle \vartheta _{00}{\bigl [}\exp {\bigl (}-{\frac {1}{2}}{\sqrt[{3}]{4}}\,\pi {\bigr )}{\bigr ]}={\sqrt[{6}]{2}}\,\vartheta _{00}{\bigl [}\exp {\bigl (}-{\sqrt[{3}]{2}}\,\pi {\bigr )}{\bigr ]}}
ϑ
00
[
exp
(
−
1
3
81
5
π
)
]
=
3
10
ϑ
00
[
exp
(
−
3
5
π
)
]
{\displaystyle \vartheta _{00}{\bigl [}\exp {\bigl (}-{\frac {1}{3}}{\sqrt[{5}]{81}}\,\pi {\bigr )}{\bigr ]}={\sqrt[{10}]{3}}\,\vartheta _{00}{\bigl [}\exp {\bigl (}-{\sqrt[{5}]{3}}\,\pi {\bigr )}{\bigr ]}}
ϑ
00
[
exp
(
−
1
5
15625
7
π
)
]
=
5
14
ϑ
00
[
exp
(
−
5
7
π
)
]
{\displaystyle \vartheta _{00}{\bigl [}\exp {\bigl (}-{\frac {1}{5}}{\sqrt[{7}]{15625}}\,\pi {\bigr )}{\bigr ]}={\sqrt[{14}]{5}}\,\vartheta _{00}{\bigl [}\exp {\bigl (}-{\sqrt[{7}]{5}}\,\pi {\bigr )}{\bigr ]}}
विषम अंकीय[ 8] [ 9]
∑
n
=
1
∞
1
F
2
n
−
1
=
5
2
∑
n
=
1
∞
2
(
Φ
−
2
)
n
−
1
/
2
1
+
(
Φ
−
2
)
2
n
−
1
=
5
4
∑
a
=
−
∞
∞
2
(
Φ
−
2
)
a
−
1
/
2
1
+
(
Φ
−
2
)
2
a
−
1
=
{\displaystyle \sum _{n=1}^{\infty }{\frac {1}{F_{2n-1}}}={\frac {\sqrt {5}}{2}}\,\sum _{n=1}^{\infty }{\frac {2(\Phi ^{-2})^{n-1/2}}{1+(\Phi ^{-2})^{2n-1}}}={\frac {\sqrt {5}}{4}}\sum _{a=-\infty }^{\infty }{\frac {2(\Phi ^{-2})^{a-1/2}}{1+(\Phi ^{-2})^{2a-1}}}=}
=
5
4
ϑ
10
(
Φ
−
2
)
2
=
5
8
[
ϑ
00
(
Φ
−
1
)
2
−
ϑ
01
(
Φ
−
1
)
2
]
{\displaystyle ={\frac {\sqrt {5}}{4}}\,\vartheta _{10}(\Phi ^{-2})^{2}={\frac {\sqrt {5}}{8}}{\bigl [}\vartheta _{00}(\Phi ^{-1})^{2}-\vartheta _{01}(\Phi ^{-1})^{2}{\bigr ]}}
इन सूत्रों के लिए
Φ
=
5
+
1
2
{\displaystyle \Phi ={\tfrac {{\sqrt {5}}+1}{2}}}
फाइबोनैचि संख्याओं के वर्गों के व्युत्क्रमों का अनंत योग:
∑
n
=
1
∞
1
F
n
2
=
5
24
[
2
ϑ
10
(
Φ
−
2
)
4
−
ϑ
00
(
Φ
−
2
)
4
+
1
]
=
5
24
[
ϑ
00
(
Φ
−
2
)
4
−
2
ϑ
01
(
Φ
−
2
)
4
+
1
]
{\displaystyle \sum _{n=1}^{\infty }{\frac {1}{F_{n}^{2}}}={\frac {5}{24}}{\bigl [}2\vartheta _{10}(\Phi ^{-2})^{4}-\vartheta _{00}(\Phi ^{-2})^{4}+1{\bigr ]}={\frac {5}{24}}{\bigl [}\vartheta _{00}(\Phi ^{-2})^{4}-2\vartheta _{01}(\Phi ^{-2})^{4}+1{\bigr ]}}
विषम अंकीय पेल संख्याओं के व्युत्क्रमों का अनंत योग:
∑
n
=
1
∞
1
P
2
n
−
1
=
1
2
ϑ
10
[
(
2
−
1
)
2
]
2
=
1
2
2
[
ϑ
00
(
2
−
1
)
2
−
ϑ
01
(
2
−
1
)
2
]
{\displaystyle \sum _{n=1}^{\infty }{\frac {1}{P_{2n-1}}}={\frac {1}{\sqrt {2}}}\,\vartheta _{10}{\bigl [}({\sqrt {2}}-1)^{2}{\bigr ]}^{2}={\frac {1}{2{\sqrt {2}}}}{\bigl [}\vartheta _{00}({\sqrt {2}}-1)^{2}-\vartheta _{01}({\sqrt {2}}-1)^{2}{\bigr ]}}
योग श्रृंखला जिसका आधार योग सूचकांक के संबंध में स्थिर है और एक घातांक जो योग सूचकांक के संबंध में वर्गाकार है, को हमेशा फ़ंक्शन ϑ₀₀ के प्राथमिक रैखिक संयोजनों के रूप में व्यक्त किया जा सकता है:
∑
k
=
−
∞
∞
x
a
k
2
+
b
k
+
c
=
exp
[
4
a
c
−
b
2
4
a
ln
(
x
)
]
[
−
π
a
ln
(
x
)
]
1
/
2
ϑ
00
{
π
b
2
a
;
exp
[
π
2
a
ln
(
x
)
−
1
]
}
{\displaystyle \sum _{k=-\infty }^{\infty }x^{ak^{2}+bk+c}=\exp {\bigl [}{\frac {4ac-b^{2}}{4a}}\ln(x){\bigr ]}{\bigl [}{\frac {-\pi }{a\ln(x)}}{\bigr ]}^{1/2}\vartheta _{00}{\bigl \{}{\frac {\pi b}{2a}};\exp {\bigl [}{\frac {\pi ^{2}}{a}}\ln(x)^{-1}{\bigr ]}{\bigr \}}}
संख्या x का मान धनात्मक होना चाहिए.
उदाहरण के लिए, वह अनंत योग निम्नलिखित मान देता है:
∑
k
=
−
∞
∞
(
7
11
)
5
k
2
+
3
k
+
2
=
(
7
11
)
31
/
20
(
π
5
)
1
/
2
ln
(
11
7
)
−
1
/
2
ϑ
00
{
3
π
10
;
exp
[
1
5
π
2
ln
(
7
11
)
−
1
]
}
{\displaystyle \sum _{k=-\infty }^{\infty }{\bigl (}{\frac {7}{11}}{\bigr )}^{5k^{2}+3k+2}={\bigl (}{\frac {7}{11}}{\bigr )}^{31/20}{\bigl (}{\frac {\pi }{5}}{\bigr )}^{1/2}\ln {\bigl (}{\frac {11}{7}}{\bigr )}^{-1/2}\vartheta _{00}{\bigl \{}{\frac {3\pi }{10}};\exp {\bigl [}{\frac {1}{5}}\pi ^{2}\ln {\bigl (}{\frac {7}{11}}{\bigr )}^{-1}{\bigr ]}{\bigr \}}}
∑
k
=
−
∞
∞
(
11
13
)
7
k
2
+
5
k
+
3
=
(
11
13
)
59
/
28
(
π
7
)
1
/
2
ln
(
13
11
)
−
1
/
2
ϑ
00
{
5
π
14
;
exp
[
1
7
π
2
ln
(
11
13
)
−
1
]
}
{\displaystyle \sum _{k=-\infty }^{\infty }{\bigl (}{\frac {11}{13}}{\bigr )}^{7k^{2}+5k+3}={\bigl (}{\frac {11}{13}}{\bigr )}^{59/28}{\bigl (}{\frac {\pi }{7}}{\bigr )}^{1/2}\ln {\bigl (}{\frac {13}{11}}{\bigr )}^{-1/2}\vartheta _{00}{\bigl \{}{\frac {5\pi }{14}};\exp {\bigl [}{\frac {1}{7}}\pi ^{2}\ln {\bigl (}{\frac {11}{13}}{\bigr )}^{-1}{\bigr ]}{\bigr \}}}
∑
k
=
−
∞
∞
(
13
17
)
11
k
2
+
7
k
+
5
=
(
13
17
)
171
/
44
(
π
11
)
1
/
2
ln
(
17
13
)
−
1
/
2
ϑ
00
{
7
π
22
;
exp
[
1
11
π
2
ln
(
13
17
)
−
1
]
}
{\displaystyle \sum _{k=-\infty }^{\infty }{\bigl (}{\frac {13}{17}}{\bigr )}^{11k^{2}+7k+5}={\bigl (}{\frac {13}{17}}{\bigr )}^{171/44}{\bigl (}{\frac {\pi }{11}}{\bigr )}^{1/2}\ln {\bigl (}{\frac {17}{13}}{\bigr )}^{-1/2}\vartheta _{00}{\bigl \{}{\frac {7\pi }{22}};\exp {\bigl [}{\frac {1}{11}}\pi ^{2}\ln {\bigl (}{\frac {13}{17}}{\bigr )}^{-1}{\bigr ]}{\bigr \}}}
थीटा शून्य मान फलनों[ 10]
d
d
x
ϑ
10
(
x
)
=
1
2
π
x
ϑ
10
(
x
)
ϑ
00
(
x
)
2
E
[
ϑ
10
(
x
)
2
ϑ
00
(
x
)
2
]
{\displaystyle {\frac {\mathrm {d} }{\mathrm {d} x}}\,\vartheta _{10}(x)={\frac {1}{2\pi x}}\vartheta _{10}(x)\vartheta _{00}(x)^{2}E{\biggl [}{\frac {\vartheta _{10}(x)^{2}}{\vartheta _{00}(x)^{2}}}{\biggr ]}}
d
d
x
ϑ
00
(
x
)
=
ϑ
00
(
x
)
[
ϑ
00
(
x
)
2
+
ϑ
01
(
x
)
2
]
{
1
2
π
x
E
[
ϑ
00
(
x
)
2
−
ϑ
01
(
x
)
2
ϑ
00
(
x
)
2
+
ϑ
01
(
x
)
2
]
−
ϑ
01
(
x
)
2
4
x
}
{\displaystyle {\frac {\mathrm {d} }{\mathrm {d} x}}\,\vartheta _{00}(x)=\vartheta _{00}(x){\bigl [}\vartheta _{00}(x)^{2}+\vartheta _{01}(x)^{2}{\bigr ]}{\biggl \{}{\frac {1}{2\pi x}}E{\biggl [}{\frac {\vartheta _{00}(x)^{2}-\vartheta _{01}(x)^{2}}{\vartheta _{00}(x)^{2}+\vartheta _{01}(x)^{2}}}{\biggr ]}-{\frac {\vartheta _{01}(x)^{2}}{4x}}{\biggr \}}}
d
d
x
ϑ
01
(
x
)
=
ϑ
01
(
x
)
[
ϑ
00
(
x
)
2
+
ϑ
01
(
x
)
2
]
{
1
2
π
x
E
[
ϑ
00
(
x
)
2
−
ϑ
01
(
x
)
2
ϑ
00
(
x
)
2
+
ϑ
01
(
x
)
2
]
−
ϑ
00
(
x
)
2
4
x
}
{\displaystyle {\frac {\mathrm {d} }{\mathrm {d} x}}\,\vartheta _{01}(x)=\vartheta _{01}(x){\bigl [}\vartheta _{00}(x)^{2}+\vartheta _{01}(x)^{2}{\bigr ]}{\biggl \{}{\frac {1}{2\pi x}}E{\biggl [}{\frac {\vartheta _{00}(x)^{2}-\vartheta _{01}(x)^{2}}{\vartheta _{00}(x)^{2}+\vartheta _{01}(x)^{2}}}{\biggr ]}-{\frac {\vartheta _{00}(x)^{2}}{4x}}{\biggr \}}}
दूसरे प्रकार के पूर्ण अण्डाकार अभिन्न अंग की यह परिभाषा है:
E
(
r
)
=
∫
0
π
/
2
1
−
r
2
sin
(
φ
)
2
d
φ
{\displaystyle E(r)=\int _{0}^{\pi /2}{\sqrt {1-r^{2}\sin(\varphi )^{2}}}\,\mathrm {d} \varphi }
E
(
r
)
=
∫
0
1
2
(
z
2
+
1
)
2
−
4
r
2
z
2
(
z
2
+
1
)
2
d
z
{\displaystyle E(r)=\int _{0}^{1}{\frac {2{\sqrt {(z^{2}+1)^{2}-4r^{2}z^{2}}}}{(z^{2}+1)^{2}}}\,\mathrm {d} z}
यहां उल्लिखित तीन थीटा फ़ंक्शंस में से दो से भागफल के व्युत्पन्न हमेशा उन तीन फ़ंक्शंस के साथ तर्कसंगत संबंध रखते हैं:
d
d
x
ϑ
10
(
x
)
ϑ
00
(
x
)
=
ϑ
10
(
x
)
ϑ
01
(
x
)
4
4
x
ϑ
00
(
x
)
{\displaystyle {\frac {\mathrm {d} }{\mathrm {d} x}}\,{\frac {\vartheta _{10}(x)}{\vartheta _{00}(x)}}={\frac {\vartheta _{10}(x)\vartheta _{01}(x)^{4}}{4x\,\vartheta _{00}(x)}}}
d
d
x
ϑ
10
(
x
)
ϑ
01
(
x
)
=
ϑ
10
(
x
)
ϑ
00
(
x
)
4
4
x
ϑ
01
(
x
)
{\displaystyle {\frac {\mathrm {d} }{\mathrm {d} x}}\,{\frac {\vartheta _{10}(x)}{\vartheta _{01}(x)}}={\frac {\vartheta _{10}(x)\vartheta _{00}(x)^{4}}{4x\,\vartheta _{01}(x)}}}
d
d
x
ϑ
00
(
x
)
ϑ
01
(
x
)
=
ϑ
00
(
x
)
5
−
ϑ
00
(
x
)
ϑ
01
(
x
)
4
4
x
ϑ
01
(
x
)
{\displaystyle {\frac {\mathrm {d} }{\mathrm {d} x}}\,{\frac {\vartheta _{00}(x)}{\vartheta _{01}(x)}}={\frac {\vartheta _{00}(x)^{5}-\vartheta _{00}(x)\vartheta _{01}(x)^{4}}{4x\,\vartheta _{01}(x)}}}
ये समाकलन[ 11]
∫
0
1
ϑ
00
(
x
)
d
x
=
∑
k
=
−
∞
∞
1
k
2
+
1
=
π
coth
(
π
)
≈
3,153
348
{\displaystyle \int _{0}^{1}\vartheta _{00}(x)\,\mathrm {d} x=\sum _{k=-\infty }^{\infty }{\frac {1}{k^{2}+1}}=\pi \coth(\pi )\approx 3{,}153348}
∫
0
1
ϑ
01
(
x
)
d
x
=
∑
k
=
−
∞
∞
(
−
1
)
k
k
2
+
1
=
π
csch
(
π
)
≈
0,272
029
{\displaystyle \int _{0}^{1}\vartheta _{01}(x)\,\mathrm {d} x=\sum _{k=-\infty }^{\infty }{\frac {(-1)^{k}}{k^{2}+1}}=\pi \,\operatorname {csch} (\pi )\approx 0{,}272029}
∫
0
1
ϑ
10
(
x
)
d
x
=
∑
k
=
−
∞
∞
4
(
2
k
+
1
)
2
+
4
=
π
tanh
(
π
)
≈
3,129
881
{\displaystyle \int _{0}^{1}\vartheta _{10}(x)\,\mathrm {d} x=\sum _{k=-\infty }^{\infty }{\frac {4}{(2k+1)^{2}+4}}=\pi \tanh(\pi )\approx 3{,}129881}
दिखाए गए अंतिम परिणाम कॉची के सामान्य सूत्रों पर आधारित हैं।
निम्नलिखित रूप में, पांचवीं घात वाले समीकरणों [ 12] [ 13] [ 14] [ 15] [ 16]
c
∈
R
{\displaystyle c\in \mathbb {R} }
ब्रिंग-जेरार्ड रूप में पांचवीं घात वाले समीकरण
x
5
+
5
x
=
4
c
{\displaystyle x^{5}+5\,x=4\,c}
Q
=
q
[
(
2
c
2
+
2
+
2
c
4
+
1
)
−
1
/
2
(
c
4
+
1
+
1
+
c
)
]
{\displaystyle Q=q{\bigl [}(2\,c^{2}+2+2{\sqrt {c^{4}+1}})^{-1/2}({\sqrt {{\sqrt {c^{4}+1}}+1}}+c){\bigr ]}}
u
=
2
ϑ
00
(
Q
5
)
ϑ
00
(
Q
1
/
5
)
−
2
ϑ
00
(
Q
)
2
ϑ
00
(
Q
1
/
5
)
2
+
5
ϑ
00
(
Q
5
)
2
−
4
ϑ
00
(
Q
)
2
{\displaystyle u={\frac {2\,\vartheta _{00}(Q^{5})\,\vartheta _{00}(Q^{1/5})-2\,\vartheta _{00}(Q)^{2}}{\vartheta _{00}(Q^{1/5})^{2}+5\,\vartheta _{00}(Q^{5})^{2}-4\,\vartheta _{00}(Q)^{2}}}}
x
=
c
u
(
5
u
2
−
10
u
+
4
)
5
u
2
−
6
u
+
2
{\displaystyle x={\frac {c\,u\,(5\,u^{2}-10\,u+4)}{5\,u^{2}-6\,u+2}}}
यह मान c = 1 के लिए पहला सटीक गणना उदाहरण है:
x
5
+
5
x
=
4
{\displaystyle x^{5}+5\,x=4}
Q
=
q
[
(
4
+
2
2
)
−
1
/
2
(
2
+
1
+
1
)
]
≈
0.1852028700803001414251518230736124606036037762504611138834393086
…
{\displaystyle Q=q{\bigl [}(4+2{\sqrt {2}})^{-1/2}({\sqrt {{\sqrt {2}}+1}}+1){\bigr ]}\approx 0.1852028700803001414251518230736124606036037762504611138834393086\ldots }
u
=
2
ϑ
00
(
Q
5
)
ϑ
00
(
Q
1
/
5
)
−
2
ϑ
00
(
Q
)
2
ϑ
00
(
Q
1
/
5
)
2
+
5
ϑ
00
(
Q
5
)
2
−
4
ϑ
00
(
Q
)
2
≈
0.3447357019439680107642844248203058218212701865144922007257031
…
{\displaystyle u={\frac {2\,\vartheta _{00}(Q^{5})\,\vartheta _{00}(Q^{1/5})-2\,\vartheta _{00}(Q)^{2}}{\vartheta _{00}(Q^{1/5})^{2}+5\,\vartheta _{00}(Q^{5})^{2}-4\,\vartheta _{00}(Q)^{2}}}\approx 0.3447357019439680107642844248203058218212701865144922007257031\ldots }
x
=
u
(
5
u
2
−
10
u
+
4
)
5
u
2
−
6
u
+
2
≈
0.75192639869405948026865366345020738740978383913037835
…
{\displaystyle x={\frac {u\,(5\,u^{2}-10\,u+4)}{5\,u^{2}-6\,u+2}}\approx 0.75192639869405948026865366345020738740978383913037835\ldots }
यह मान c = 2 के लिए दूसरा सटीक गणना उदाहरण है:
x
5
+
5
x
=
8
{\displaystyle x^{5}+5\,x=8}
Q
=
q
[
(
10
+
2
17
)
−
1
/
2
(
17
+
1
+
2
)
]
≈
0.3063466544466074265361088194021326272090461143559097382981847
…
{\displaystyle Q=q{\bigl [}(10+2{\sqrt {17}})^{-1/2}({\sqrt {{\sqrt {17}}+1}}+2){\bigr ]}\approx 0.3063466544466074265361088194021326272090461143559097382981847\ldots }
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0.26117326232214439979677125632439205703620052387333832254673599
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{\displaystyle u={\frac {2\,\vartheta _{00}(Q^{5})\,\vartheta _{00}(Q^{1/5})-2\,\vartheta _{00}(Q)^{2}}{\vartheta _{00}(Q^{1/5})^{2}+5\,\vartheta _{00}(Q^{5})^{2}-4\,\vartheta _{00}(Q)^{2}}}\approx 0.26117326232214439979677125632439205703620052387333832254673599\ldots }
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1.16703618370164304731101943199639610129755211048801991
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{\displaystyle x={\frac {2\,u\,(5\,u^{2}-10\,u+4)}{5\,u^{2}-6\,u+2}}\approx 1.16703618370164304731101943199639610129755211048801991\ldots }
यह मान c = 3 के लिए तीसरा सटीक गणना उदाहरण है:
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{\displaystyle x^{5}+5\,x=12}
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0.370664951152024075624432522177568657151868089959747395750974
…
{\displaystyle Q=q{\bigl [}(20+2{\sqrt {82}})^{-1/2}({\sqrt {{\sqrt {82}}+1}}+3){\bigr ]}\approx 0.370664951152024075624432522177568657151868089959747395750974\ldots }
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0.2090460949758925319033273467025369651597551071050541509568731
…
{\displaystyle u={\frac {2\,\vartheta _{00}(Q^{5})\,\vartheta _{00}(Q^{1/5})-2\,\vartheta _{00}(Q)^{2}}{\vartheta _{00}(Q^{1/5})^{2}+5\,\vartheta _{00}(Q^{5})^{2}-4\,\vartheta _{00}(Q)^{2}}}\approx 0.2090460949758925319033273467025369651597551071050541509568731\ldots }
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1.38409179582314635924775512626713547488593506018067645
…
{\displaystyle x={\frac {3\,u\,(5\,u^{2}-10\,u+4)}{5\,u^{2}-6\,u+2}}\approx 1.38409179582314635924775512626713547488593506018067645\ldots }
महत्वपूर्ण अतिरिक्त जानकारी:
q
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{\displaystyle q{\bigl [}(2\,c^{2}+2+2{\sqrt {c^{4}+1}})^{-1/2}({\sqrt {{\sqrt {c^{4}+1}}+1}}+c){\bigr ]}=}
=
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{\displaystyle =\exp {\biggl \{}-\pi {\biggl [}\int _{0}^{1}2{\bigl (}z^{4}+2c{\sqrt {2{\sqrt {c^{4}+1}}-2c^{2}}}\,z^{2}+1{\bigr )}^{-1/2}\,\mathrm {d} z{\biggr ]}\div {\biggl [}\int _{0}^{1}2{\bigl (}z^{4}-2c{\sqrt {2{\sqrt {c^{4}+1}}-2c^{2}}}\,z^{2}+1{\bigr )}^{-1/2}\,\mathrm {d} z{\biggr ]}{\biggr \}}}
↑ "Derivative of the Jacobi theta function: Introduction to the Jacobi theta functions" . functions.wolfram.com . अभिगमन तिथि 2021-07-28 .↑ "What is a Theta Function?" . Mathematics Stack Exchange . अभिगमन तिथि 2021-07-28 .↑ Weisstein, Eric W. "Ramanujan g- and G-Functions" . mathworld.wolfram.com (अंग्रेज़ी में). अभिगमन तिथि 2021-07-28 . ↑ "Theta-function identities and the explicit formulas for theta-function and their applications" . Journal of Mathematical Analysis and Applications (अंग्रेज़ी में). 292 (2): 381–400. 2004-04-15. आइ॰एस॰एस॰एन॰ 0022-247X . डीओआइ :10.1016/j.jmaa.2003.12.009 .↑ एंड्रियास डाइकमैन: अनंत उत्पादों की तालिका अनंत रकम अनंत श्रृंखला, एलिप्टिक थीटा।
↑ एरिक डब्ल्यू वेइसटीन , मैथवर्ल्ड पर Jacobi Theta Functions ↑ http://wayback.cecm.sfu.ca/~pborwein/TEMP_PROTECTED/pi-agm.pdf ↑ Landau (1899) zitiert nach Borwein , Page 94, Exercise 3.
↑ साँचा:Internetquelle ↑ एरिक डब्ल्यू वेइसटीन , मैथवर्ल्ड पर Elliptic Alpha Function ↑ साँचा:Internetquelle ↑ Young, George Paxton (1885-01-01). Solution of Solvable Irreducible Quintic Equations, without the Aid of a Resolvent Sextic ↑ Runge, C. (1900-01). "Über die auflösbaren Gleichungen von der Form x5+ux+v=o" . Acta Mathematica . 7 (none): 173–186. आइ॰एस॰एस॰एन॰ 0001-5962 . डीओआइ :10.1007/BF02402200 . ↑ Emil Jann Brahmeyer (2024-06-06), Wiederholung der Gleichungen fünften Grades in BJ-Form , अभिगमन तिथि 2024-06-16 ↑ Emil Jann Brahmeyer (2024-05-24), Gleichungen fünften Grades Gleichungsrennen Zwei , अभिगमन तिथि 2024-06-16 ↑ 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 .
![{\displaystyle \vartheta _{1}(z;w)=\sum _{k=-\infty }^{\infty }(-1)^{k-1/2}\exp[(2k+1)iz+(k+{\tfrac {1}{2}})^{2}\ln(w)]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/10c0333dea5cba2907dbe8ea54e20d87fe60210c)
![{\displaystyle \vartheta _{2}(z;w)=\sum _{k=-\infty }^{\infty }\exp[(2k+1)iz+(k+{\tfrac {1}{2}})^{2}\ln(w)]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0064ae5ac867024074af029a747568538389f12e)
![{\displaystyle \vartheta _{3}(z;w)=\sum _{k=-\infty }^{\infty }\exp[2kiz+k^{2}\ln(w)]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/15e72dc038a3b081b9ed137b0b93e85fe1231f1c)
![{\displaystyle \vartheta _{4}(z;w)=\sum _{k=-\infty }^{\infty }(-1)^{k}\exp[2kiz+k^{2}\ln(w)]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1883f4d67d248437cd2b7f2b5de9271a5bc8e1e7)
![{\displaystyle \vartheta _{00}(x;y)=\prod _{n=1}^{\infty }(1-y^{2n})[1+2\cos(2x)y^{2n-1}+y^{4n-2}]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/31517b9d6574cf66118c97bdae43a14d730ad026)
![{\displaystyle \vartheta _{01}(x;y)=\prod _{n=1}^{\infty }(1-y^{2n})[1-2\cos(2x)y^{2n-1}+y^{4n-2}]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/4248e41592bc7ef203c97bcc6dd7f6bbef3eb1ca)
![{\displaystyle \vartheta _{10}(x;y)=2y^{1/4}\cos(x)\prod _{n=1}^{\infty }(1-y^{2n})[1+2\cos(2x)y^{2n}+y^{4n}]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6be72557fd8c5a2b56d1c999900683e005d27150)
![{\displaystyle \vartheta _{00}(v)=1+{\frac {4v{\sqrt {\ln(1/v)}}}{\sqrt {\pi }}}\int _{0}^{\infty }{\frac {\exp[-\ln(1/v)\,x^{2}]\{1-v^{2}\cos[2\ln(1/v)\,x]\}}{1-2v^{2}\cos[2\ln(1/v)\,x]+v^{4}}}\,\mathrm {d} x}](https://wikimedia.org/api/rest_v1/media/math/render/svg/521d054f6c58690a1b026e8a250dc18cd71ccd39)
![{\displaystyle \vartheta _{00}{\bigl (}{\frac {1}{2}}{\bigr )}=1+2\sum _{n=1}^{\infty }{\frac {1}{2^{n^{2}}}}=1+2\pi ^{-1/2}{\sqrt {\ln(2)}}\,\int _{0}^{\infty }{\frac {\exp[-\ln(2)\,x^{2}]\{16-4\cos[2\ln(2)\,x]\}}{17-8\cos[2\ln(2)\,x]}}\,\mathrm {d} x}](https://wikimedia.org/api/rest_v1/media/math/render/svg/eab6b3d48c8ae4405ca75db86f40267d60873d46)
![{\displaystyle \vartheta _{00}{\bigl (}{\frac {1}{3}}{\bigr )}=1+2\sum _{n=1}^{\infty }{\frac {1}{3^{n^{2}}}}=1+{\frac {4}{3}}\pi ^{-1/2}{\sqrt {\ln(3)}}\,\int _{0}^{\infty }{\frac {\exp[-\ln(3)\,x^{2}]\{81-9\cos[2\ln(3)\,x]\}}{82-18\cos[2\ln(3)\,x]}}\,\mathrm {d} x}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e0c534a207e59dc4513d7518b645c0b63e2d939c)
![{\displaystyle \vartheta _{00}{\bigl (}{\frac {1}{5}}{\bigr )}=1+2\sum _{n=1}^{\infty }{\frac {1}{5^{n^{2}}}}=1+{\frac {4}{5}}\pi ^{-1/2}{\sqrt {\ln(5)}}\,\int _{0}^{\infty }{\frac {\exp[-\ln(5)\,x^{2}]\{625-25\cos[2\ln(5)\,x]\}}{626-50\cos[2\ln(5)\,x]}}\,\mathrm {d} x}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6e44ac49667d6429020931cca2bbe8dc7f990282)
![{\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 \vartheta _{00}[q(\varepsilon )]={\sqrt {2\pi ^{-1}K(\varepsilon )}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f62a25adecf660aa0d24611b65a4fee95a7e95e8)
![{\displaystyle \vartheta _{01}[q(\varepsilon )]={\sqrt[{4}]{1-\varepsilon ^{2}}}{\sqrt {2\pi ^{-1}K(\varepsilon )}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0f10b2062dae698c5286d188e05a86107c76d4e0)
![{\displaystyle \vartheta _{10}(x)={\sqrt[{4}]{\vartheta _{00}(x)^{4}-\vartheta _{01}(x)^{4}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/68c5ae92bc640738ba5a1e90fff0886f9124d337)
![{\displaystyle \vartheta _{00}[\exp(-\pi )]=\pi ^{1/4}\Gamma ({\tfrac {3}{4}})^{-1}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/2769cb1e29c33be4806f1077155eb1baa62beac5)
![{\displaystyle \vartheta _{01}[\exp(-\pi )]=2^{-1/4}\pi ^{1/4}\Gamma ({\tfrac {3}{4}})^{-1}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0151b294d38ae0b2180e49357c763efda6522232)
![{\displaystyle \vartheta _{10}[\exp(-\pi )]=2^{-1/4}\pi ^{1/4}\Gamma ({\tfrac {3}{4}})^{-1}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8abda280e2cfb61e8008b8eda62c653b92c2edbf)
![{\displaystyle \vartheta _{00}[\exp(-{\sqrt {2}}\,\pi )]=2^{-1/8}\pi ^{1/2}\Gamma ({\tfrac {3}{4}})^{-1/2}\Gamma ({\tfrac {5}{8}})^{-1}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5406357a490bd21d276c89e3b04f6d4c42ffc886)
![{\displaystyle \vartheta _{00}[\exp(-{\sqrt {3}}\,\pi )]=2^{5/6}3^{-5/8}\pi ^{1/2}\Gamma ({\tfrac {2}{3}})^{-3/2}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a4cca4d1e9e491686d5cbb9c02e8acca096faa9e)
![{\displaystyle \vartheta _{00}[q(\varepsilon )^{2}]=\cos[{\tfrac {1}{2}}\arcsin(\varepsilon )]\vartheta _{00}[q(\varepsilon )]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0633b78ed20d4a51fc22b5691025508dfedf81ec)
![{\displaystyle \vartheta _{01}[q(\varepsilon )^{2}]=(1-\varepsilon ^{2})^{1/8}\vartheta _{00}[q(\varepsilon )]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/934ad5207f7906a55180385764587bd5bc857a24)
![{\displaystyle 27{\frac {\vartheta _{00}[q(\varepsilon )^{3}]^{8}}{\vartheta _{00}[q(\varepsilon )]^{8}}}=18{\frac {\vartheta _{00}[q(\varepsilon )^{3}]^{4}}{\vartheta _{00}[q(\varepsilon )]^{4}}}+8\cos[2\arcsin(\varepsilon )]{\frac {\vartheta _{00}[q(\varepsilon )^{3}]^{2}}{\vartheta _{00}[q(\varepsilon )]^{2}}}+1}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c31a935cbb4cfcafda5c2dff3d2dc9099b9460c5)
![{\displaystyle 27{\frac {\vartheta _{01}[q(\varepsilon )^{3}]^{8}}{\vartheta _{01}[q(\varepsilon )]^{8}}}=18{\frac {\vartheta _{01}[q(\varepsilon )^{3}]^{4}}{\vartheta _{01}[q(\varepsilon )]^{4}}}+8\sec[2\arctan(\varepsilon )]{\frac {\vartheta _{01}[q(\varepsilon )^{3}]^{2}}{\vartheta _{01}[q(\varepsilon )]^{2}}}+1}](https://wikimedia.org/api/rest_v1/media/math/render/svg/cb249fba89a15424e685c0bee08a4d871931682a)
![{\displaystyle {\biggl \{}{\frac {\vartheta _{01}[q(\varepsilon )^{5}]^{2}}{\vartheta _{01}[q(\varepsilon )]^{2}}}-1{\biggr \}}{\biggl \{}5{\frac {\vartheta _{01}[q(\varepsilon )^{5}]^{2}}{\vartheta _{01}[q(\varepsilon )]^{2}}}-1{\biggr \}}^{5}=64\tan[2\arctan(\varepsilon )]^{2}{\frac {\vartheta _{01}[q(\varepsilon )^{5}]^{2}}{\vartheta _{01}[q(\varepsilon )]^{2}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c27cf20ed1406a755cbb49ba3c7a507791ab67f0)
![{\displaystyle \exp(-{\sqrt {3}}\,\pi )=q[{\tfrac {1}{4}}({\sqrt {6}}-{\sqrt {2}})]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ebc39dc73ec8303afd8b04823d87a91c5e986e39)
![{\displaystyle {\frac {\vartheta _{00}[\exp(-3\pi )]}{\vartheta _{00}[\exp(-\pi )]}}=108^{-1/8}{\sqrt {{\sqrt {3}}+1}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1912ef1637ef157a9d8f07287cef05654d6542ca)
![{\displaystyle {\frac {\vartheta _{00}[\exp(-3{\sqrt {2}}\,\pi )]}{\vartheta _{00}[\exp(-{\sqrt {2}}\,\pi )]}}=3^{-1/2}{\sqrt {{\sqrt {6}}+{\sqrt {2}}-1}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/2181af31a17a9ee56a43d65225c0edb54bb08653)
![{\displaystyle {\frac {\vartheta _{00}[\exp(-3{\sqrt {3}}\,\pi )]}{\vartheta _{00}[\exp(-{\sqrt {3}}\,\pi )]}}=3^{-3/4}({\sqrt[{3}]{2}}+1)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/03eb54f6c6cf0d015381a479a63cd19ecea6b9f2)
![{\displaystyle {\frac {\vartheta _{01}[\exp(-3\pi )]}{\vartheta _{01}[\exp(-\pi )]}}=3^{-3/8}{\sqrt {{\sqrt {2}}+{\sqrt[{4}]{3}}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/201e1e8ce6eeba27e46d08ff27ca483346c81be1)
![{\displaystyle {\frac {\vartheta _{01}[\exp(-3{\sqrt {2}}\,\pi )]}{\vartheta _{01}[\exp(-{\sqrt {2}}\,\pi )]}}=3^{-1/2}{\sqrt {{\sqrt {3}}+{\sqrt {2}}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/95087e5acb126d7839e81c3b40384f2af09852c3)
![{\displaystyle {\frac {\vartheta _{01}[\exp(-3{\sqrt {3}}\,\pi )]}{\vartheta _{01}[\exp(-{\sqrt {3}}\,\pi )]}}=108^{-1/4}(2{\sqrt[{3}]{2}}+{\sqrt {3}}-1)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7edff75a9fbea2b38c1d0cd318367e75556d66ef)
![{\displaystyle {\frac {\vartheta _{01}[\exp(-5\pi )]}{\vartheta _{01}[\exp(-\pi )]}}=5^{-1/2}{\sqrt {3+2{\sqrt[{4}]{5}}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8dfe2849cd21be45bf3031beebbcef401205190f)
![{\displaystyle {\frac {\vartheta _{01}[\exp(-5{\sqrt {2}}\,\pi )]}{\vartheta _{01}[\exp(-{\sqrt {2}}\,\pi )]}}={\tfrac {4}{15}}{\sqrt {10}}\cos({\tfrac {1}{10}}\pi )\cosh[{\tfrac {1}{3}}{\text{artanh}}({\tfrac {3}{8}}{\sqrt {6}})]+{\tfrac {1}{15}}{\sqrt {5}}\tan({\tfrac {1}{5}}\pi )}](https://wikimedia.org/api/rest_v1/media/math/render/svg/17e38a69f3323f1c34ad77d8624eba123dfeb95e)
![{\displaystyle {\frac {\vartheta _{01}[\exp(-5{\sqrt {3}}\,\pi )]}{\vartheta _{01}[\exp(-{\sqrt {3}}\,\pi )]}}={\tfrac {2}{15}}{\sqrt[{3}]{10}}(3-{\sqrt {3}})\sin({\tfrac {1}{5}}\pi )+{\tfrac {1}{15}}{\sqrt {15}}\cot({\tfrac {1}{10}}\pi )}](https://wikimedia.org/api/rest_v1/media/math/render/svg/9eb97d0f3d2b68f594ed49a4f85fd4a09b71d6a8)
![{\displaystyle \vartheta _{00}[q(\varepsilon )]={\biggl [}{\frac {2}{\pi }}K(\varepsilon ){\biggr ]}^{1/2}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/cc65a15f3eab9f308495d0ee51d1957f466a4a09)
![{\displaystyle \vartheta _{00}{\bigl [}\exp(-{\sqrt {5}}\,\pi ){\bigr ]}={\biggl \langle }{\frac {2}{\pi }}K{\bigl \{}\sin {\bigl [}{\tfrac {1}{2}}\arcsin({\sqrt {5}}-2){\bigr ]}{\bigr \}}{\biggr \rangle }^{1/2}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f8531c3f3cbb37059a68d552ae79486f638c139e)
![{\displaystyle \vartheta _{00}{\bigl [}\exp(-{\sqrt {6}}\,\pi ){\bigr ]}={\biggl \{}{\frac {2}{\pi }}K{\bigl [}(2-{\sqrt {3}}\,)({\sqrt {3}}-{\sqrt {2}}\,){\bigr ]}{\biggr \}}^{1/2}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f86e63b0a369b2bc0c126c1e19cd80ab94eec7d9)
![{\displaystyle \vartheta _{00}{\bigl [}\exp(-{\sqrt {7}}\,\pi ){\bigr ]}={\biggl \{}{\frac {2}{\pi }}K{\bigl [}{\frac {1}{8}}{\bigl (}3{\sqrt {2}}-{\sqrt {14}}\,{\bigr )}{\bigr ]}{\biggr \}}^{1/2}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/22722fbb10b2426f8658c9806770bef7e358aba3)
![{\displaystyle \vartheta _{00}{\bigl [}\exp(-{\sqrt {10}}\,\pi ){\bigr ]}={\biggl \{}{\frac {2}{\pi }}K{\bigl [}({\sqrt {10}}-3)({\sqrt {2}}-1)^{2}{\bigr ]}{\biggr \}}^{1/2}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/dae3fca860ebac2549c1cd76a3d622dea687de0b)
![{\displaystyle \vartheta _{00}{\bigl [}\exp(-{\sqrt {11}}\,\pi ){\bigr ]}={\biggl \{}{\frac {2}{\pi }}K{\bigl [}{\frac {1}{16}}{\bigl (}{\sqrt {22}}+3{\sqrt {2}}\,{\bigr )}{\bigl (}{\frac {1}{3}}{\sqrt[{3}]{6{\sqrt {3}}+2{\sqrt {11}}}}-{\frac {1}{3}}{\sqrt[{3}]{6{\sqrt {3}}-2{\sqrt {11}}}}+{\frac {1}{3}}{\sqrt {11}}-1{\bigr )}^{4}{\bigr ]}{\biggr \}}^{1/2}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d3726b8b838b686dfcdbed4dcabb8d3f0c945009)
![{\displaystyle \vartheta _{00}{\bigl [}\exp(-{\sqrt {13}}\,\pi ){\bigr ]}={\biggl \langle }{\frac {2}{\pi }}K{\bigl \{}\sin {\bigl [}{\tfrac {1}{2}}\arcsin(5{\sqrt {13}}-18){\bigr ]}{\bigr \}}{\biggr \rangle }^{1/2}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/4144d4f024327d1552dbecbb63d7ad6a3a871588)
![{\displaystyle \vartheta _{00}{\bigl [}\exp(-{\sqrt {14}}\,\pi ){\bigr ]}={\biggl \langle }{\frac {2}{\pi }}K{\bigl \{}\tan {\bigl [}{\tfrac {1}{4}}\operatorname {arccsc}(8{\sqrt {2}}+11){\bigr ]}{\bigr \}}{\biggr \rangle }^{1/2}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/38a6c5c7f7e664093294fa5072777d3a080385c2)
![{\displaystyle \vartheta _{10}(x^{2})={\tfrac {1}{2}}{\sqrt {2[\vartheta _{00}(x)^{2}-\vartheta _{01}(x)^{2}]}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/dc9bd97eafef2da346b986c44278e04f423d95bd)
![{\displaystyle \vartheta _{00}(x^{2})={\tfrac {1}{2}}{\sqrt {2[\vartheta _{00}(x)^{2}+\vartheta _{01}(x)^{2}]}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a5a54d7761b6db328cc78be2dee8bc20a83184e8)
![{\displaystyle 27\,\vartheta _{00}(x^{3})^{8}-18\,\vartheta _{00}(x^{3})^{4}\vartheta _{00}(x)^{4}-\,\vartheta _{00}(x)^{8}=8\,\vartheta _{00}(x^{3})^{2}\vartheta _{00}(x)^{2}[2\,\vartheta _{01}(x)^{4}-\vartheta _{00}(x)^{4}]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/4fe6c0b4ed1a7010a802637bbbc211ffb24e3526)
![{\displaystyle 27\,\vartheta _{01}(x^{3})^{8}-18\,\vartheta _{01}(x^{3})^{4}\vartheta _{01}(x)^{4}-\,\vartheta _{01}(x)^{8}=8\,\vartheta _{01}(x^{3})^{2}\vartheta _{01}(x)^{2}[2\,\vartheta _{00}(x)^{4}-\vartheta _{01}(x)^{4}]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/2d981666970f7bc4250a2030d31e4ffac5b96608)
![{\displaystyle [\vartheta _{00}(x)^{2}-\vartheta _{00}(x^{5})^{2}][5\,\vartheta _{00}(x^{5})^{2}-\vartheta _{00}(x)^{2}]^{5}=256\,\vartheta _{00}(x^{5})^{2}\vartheta _{00}(x)^{2}\vartheta _{01}(x)^{4}[\vartheta _{00}(x)^{4}-\vartheta _{01}(x)^{4}]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/34d31f83d29921403d740556f198dc1e222c079a)
![{\displaystyle [\vartheta _{01}(x^{5})^{2}-\vartheta _{01}(x)^{2}][5\,\vartheta _{01}(x^{5})^{2}-\vartheta _{01}(x)^{2}]^{5}=256\,\vartheta _{01}(x^{5})^{2}\vartheta _{01}(x)^{2}\vartheta _{00}(x)^{4}[\vartheta _{00}(x)^{4}-\vartheta _{01}(x)^{4}]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8fe7af92061688ce1c235885e4fc5c57ca2625b4)
![{\displaystyle {\biggl [}{\frac {\vartheta _{00}(x^{1/3})^{2}}{\vartheta _{00}(x)^{2}}}-{\frac {3\,\vartheta _{00}(x^{3})^{2}}{\vartheta _{00}(x)^{2}}}{\biggr ]}^{2}=4-4{\biggl [}{\frac {2\,\vartheta _{10}(x)^{2}\vartheta _{01}(x)^{2}}{\vartheta _{00}(x)^{4}}}{\biggr ]}^{2/3}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/51e3e9e97311736c32ad24533b1730a6556e4d20)
![{\displaystyle {\biggl [}{\frac {3\,\vartheta _{01}(x^{3})^{2}}{\vartheta _{01}(x)^{2}}}-{\frac {\vartheta _{01}(x^{1/3})^{2}}{\vartheta _{01}(x)^{2}}}{\biggr ]}^{2}=4+4{\biggl [}{\frac {2\,\vartheta _{10}(x)^{2}\vartheta _{00}(x)^{2}}{\vartheta _{01}(x)^{4}}}{\biggr ]}^{2/3}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/42623c1f5ba6b7c0b4bd68e80a41d51487a7489f)
![{\displaystyle 2\arctan {\biggl [}{\frac {\vartheta _{00}(x^{5})\,\vartheta _{00}(x^{1/5})-\vartheta _{00}(x)^{2}}{\vartheta _{00}(x^{1/5})^{2}+5\,\vartheta _{00}(x^{5})^{2}-\vartheta _{00}(x)^{2}-3\,\vartheta _{00}(x^{5})\,\vartheta _{00}(x^{1/5})}}{\biggr ]}+\arctan(2)=}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e6806acdc71122a3b8906a42ec3ae226654080e7)
![{\displaystyle \arctan {\biggl [}{\frac {\vartheta _{00}(x^{1/5})^{2}}{2\,\vartheta _{00}(x)^{2}}}-{\frac {1}{2}}{\biggr ]}+\arctan {\biggl [}{\frac {5\,\vartheta _{00}(x^{5})^{2}}{2\,\vartheta _{00}(x)^{2}}}-{\frac {1}{2}}{\biggr ]}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0583863b785318fc5a66cc97b19f6058efc14486)
![{\displaystyle \vartheta _{00}{\bigl [}q(k){\bigr ]}=\vartheta _{00}{\biggl \{}\exp {\biggl [}-\pi \,{\frac {K'(k)}{K(k)}}{\biggr ]}{\biggr \}}=\sum _{n=-\infty }^{\infty }\exp {\biggl [}-n^{2}\pi \,{\frac {K'(k)}{K(k)}}{\biggr ]}={\biggl [}{\frac {2}{\pi }}K(k){\biggr ]}^{1/2}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/4e4e517a60b8a1d52ad7878de4276ab9e98e8a1e)
![{\displaystyle \vartheta _{00}{\bigl [}q'(k){\bigr ]}=\vartheta _{00}{\biggl \{}\exp {\biggl [}-\pi \,{\frac {K(k)}{K'(k)}}{\biggr ]}{\biggr \}}=\sum _{n=-\infty }^{\infty }\exp {\biggl [}-n^{2}\pi \,{\frac {K(k)}{K'(k)}}{\biggr ]}={\biggl [}{\frac {2}{\pi }}K'(k){\biggr ]}^{1/2}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a5a0098026855b83ff163afd9bca2e30b0afa190)
![{\displaystyle {\frac {\vartheta _{00}{\bigl [}q'(k){\bigr ]}}{\vartheta _{00}{\bigl [}q(k){\bigr ]}}}={\biggl [}{\frac {K'(k)}{K(k)}}{\biggr ]}^{1/2}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/827177642ed5d6d9e1feba09ff9898a94b207292)
![{\displaystyle {\frac {\vartheta _{00}{\bigl \{}\exp {\bigl [}-\pi \,K(k)\div K'(k){\bigr ]}{\bigr \}}}{\vartheta _{00}{\bigl \{}\exp {\bigl [}-\pi \,K'(k)\div K(k){\bigr ]}{\bigr \}}}}={\biggl [}{\frac {K'(k)}{K(k)}}{\biggr ]}^{1/2}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/514290161b9676362ba02b4f61f60c28a02fd2e5)
![{\displaystyle {\frac {\vartheta _{00}{\bigl [}\exp(-\pi \div p){\bigr ]}}{\vartheta _{00}{\bigl [}\exp(-\pi \times p){\bigr ]}}}={\sqrt {p}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1513bc2c0a668a1dfa4a430e9d493bc946eb8cc8)
![{\displaystyle \vartheta _{00}{\bigl [}\exp {\bigl (}-{\frac {1}{2}}{\sqrt[{3}]{4}}\,\pi {\bigr )}{\bigr ]}={\sqrt[{6}]{2}}\,\vartheta _{00}{\bigl [}\exp {\bigl (}-{\sqrt[{3}]{2}}\,\pi {\bigr )}{\bigr ]}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/534853d3f82566d88122934866709078645ac6e9)
![{\displaystyle \vartheta _{00}{\bigl [}\exp {\bigl (}-{\frac {1}{3}}{\sqrt[{5}]{81}}\,\pi {\bigr )}{\bigr ]}={\sqrt[{10}]{3}}\,\vartheta _{00}{\bigl [}\exp {\bigl (}-{\sqrt[{5}]{3}}\,\pi {\bigr )}{\bigr ]}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/abe063bef25404ad28cce925e71073b0bf1662e1)
![{\displaystyle \vartheta _{00}{\bigl [}\exp {\bigl (}-{\frac {1}{5}}{\sqrt[{7}]{15625}}\,\pi {\bigr )}{\bigr ]}={\sqrt[{14}]{5}}\,\vartheta _{00}{\bigl [}\exp {\bigl (}-{\sqrt[{7}]{5}}\,\pi {\bigr )}{\bigr ]}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/15bd82ca3015263e65e0dd5467d67b37fd85ee83)
![{\displaystyle ={\frac {\sqrt {5}}{4}}\,\vartheta _{10}(\Phi ^{-2})^{2}={\frac {\sqrt {5}}{8}}{\bigl [}\vartheta _{00}(\Phi ^{-1})^{2}-\vartheta _{01}(\Phi ^{-1})^{2}{\bigr ]}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8c906dd29e77af17ce6f59a1fc787537c3343d4e)
![{\displaystyle \sum _{n=1}^{\infty }{\frac {1}{F_{n}^{2}}}={\frac {5}{24}}{\bigl [}2\vartheta _{10}(\Phi ^{-2})^{4}-\vartheta _{00}(\Phi ^{-2})^{4}+1{\bigr ]}={\frac {5}{24}}{\bigl [}\vartheta _{00}(\Phi ^{-2})^{4}-2\vartheta _{01}(\Phi ^{-2})^{4}+1{\bigr ]}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/081afe6f5a8e85c66841ed1269b92c192b709528)
![{\displaystyle \sum _{n=1}^{\infty }{\frac {1}{P_{2n-1}}}={\frac {1}{\sqrt {2}}}\,\vartheta _{10}{\bigl [}({\sqrt {2}}-1)^{2}{\bigr ]}^{2}={\frac {1}{2{\sqrt {2}}}}{\bigl [}\vartheta _{00}({\sqrt {2}}-1)^{2}-\vartheta _{01}({\sqrt {2}}-1)^{2}{\bigr ]}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8b1d871a2d6fdca3df549fc1d00d4a4070c595d5)
![{\displaystyle \sum _{k=-\infty }^{\infty }x^{ak^{2}+bk+c}=\exp {\bigl [}{\frac {4ac-b^{2}}{4a}}\ln(x){\bigr ]}{\bigl [}{\frac {-\pi }{a\ln(x)}}{\bigr ]}^{1/2}\vartheta _{00}{\bigl \{}{\frac {\pi b}{2a}};\exp {\bigl [}{\frac {\pi ^{2}}{a}}\ln(x)^{-1}{\bigr ]}{\bigr \}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0e559b587188b55187ca9b42d13b4ff6813a3956)
![{\displaystyle \sum _{k=-\infty }^{\infty }{\bigl (}{\frac {7}{11}}{\bigr )}^{5k^{2}+3k+2}={\bigl (}{\frac {7}{11}}{\bigr )}^{31/20}{\bigl (}{\frac {\pi }{5}}{\bigr )}^{1/2}\ln {\bigl (}{\frac {11}{7}}{\bigr )}^{-1/2}\vartheta _{00}{\bigl \{}{\frac {3\pi }{10}};\exp {\bigl [}{\frac {1}{5}}\pi ^{2}\ln {\bigl (}{\frac {7}{11}}{\bigr )}^{-1}{\bigr ]}{\bigr \}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/52dde98bebae1dcaf7583e1b2ceea687ae65733c)
![{\displaystyle \sum _{k=-\infty }^{\infty }{\bigl (}{\frac {11}{13}}{\bigr )}^{7k^{2}+5k+3}={\bigl (}{\frac {11}{13}}{\bigr )}^{59/28}{\bigl (}{\frac {\pi }{7}}{\bigr )}^{1/2}\ln {\bigl (}{\frac {13}{11}}{\bigr )}^{-1/2}\vartheta _{00}{\bigl \{}{\frac {5\pi }{14}};\exp {\bigl [}{\frac {1}{7}}\pi ^{2}\ln {\bigl (}{\frac {11}{13}}{\bigr )}^{-1}{\bigr ]}{\bigr \}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/282e155fee6e9f25fb2a344dc77c886fe0ab8b98)
![{\displaystyle \sum _{k=-\infty }^{\infty }{\bigl (}{\frac {13}{17}}{\bigr )}^{11k^{2}+7k+5}={\bigl (}{\frac {13}{17}}{\bigr )}^{171/44}{\bigl (}{\frac {\pi }{11}}{\bigr )}^{1/2}\ln {\bigl (}{\frac {17}{13}}{\bigr )}^{-1/2}\vartheta _{00}{\bigl \{}{\frac {7\pi }{22}};\exp {\bigl [}{\frac {1}{11}}\pi ^{2}\ln {\bigl (}{\frac {13}{17}}{\bigr )}^{-1}{\bigr ]}{\bigr \}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/4b6c8ab8569e20a0d7d5634bd343882ff804c18b)
![{\displaystyle {\frac {\mathrm {d} }{\mathrm {d} x}}\,\vartheta _{10}(x)={\frac {1}{2\pi x}}\vartheta _{10}(x)\vartheta _{00}(x)^{2}E{\biggl [}{\frac {\vartheta _{10}(x)^{2}}{\vartheta _{00}(x)^{2}}}{\biggr ]}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e84cd43077be6186141b44bfcaa9c1575bafada2)
![{\displaystyle {\frac {\mathrm {d} }{\mathrm {d} x}}\,\vartheta _{00}(x)=\vartheta _{00}(x){\bigl [}\vartheta _{00}(x)^{2}+\vartheta _{01}(x)^{2}{\bigr ]}{\biggl \{}{\frac {1}{2\pi x}}E{\biggl [}{\frac {\vartheta _{00}(x)^{2}-\vartheta _{01}(x)^{2}}{\vartheta _{00}(x)^{2}+\vartheta _{01}(x)^{2}}}{\biggr ]}-{\frac {\vartheta _{01}(x)^{2}}{4x}}{\biggr \}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/bf281c3f5561e384cc0481d485fde75ff3293f67)
![{\displaystyle {\frac {\mathrm {d} }{\mathrm {d} x}}\,\vartheta _{01}(x)=\vartheta _{01}(x){\bigl [}\vartheta _{00}(x)^{2}+\vartheta _{01}(x)^{2}{\bigr ]}{\biggl \{}{\frac {1}{2\pi x}}E{\biggl [}{\frac {\vartheta _{00}(x)^{2}-\vartheta _{01}(x)^{2}}{\vartheta _{00}(x)^{2}+\vartheta _{01}(x)^{2}}}{\biggr ]}-{\frac {\vartheta _{00}(x)^{2}}{4x}}{\biggr \}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6ebd20be5b0d91ff51a243c43e3ef95e8153e2aa)
![{\displaystyle Q=q{\bigl [}(2\,c^{2}+2+2{\sqrt {c^{4}+1}})^{-1/2}({\sqrt {{\sqrt {c^{4}+1}}+1}}+c){\bigr ]}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/cc53425461de700284c45d892dcd95413b1e1ecf)
![{\displaystyle Q=q{\bigl [}(4+2{\sqrt {2}})^{-1/2}({\sqrt {{\sqrt {2}}+1}}+1){\bigr ]}\approx 0.1852028700803001414251518230736124606036037762504611138834393086\ldots }](https://wikimedia.org/api/rest_v1/media/math/render/svg/6ece3fc33b4cdbeb2df0575c200c3b889ae8d7f3)
![{\displaystyle Q=q{\bigl [}(10+2{\sqrt {17}})^{-1/2}({\sqrt {{\sqrt {17}}+1}}+2){\bigr ]}\approx 0.3063466544466074265361088194021326272090461143559097382981847\ldots }](https://wikimedia.org/api/rest_v1/media/math/render/svg/e162c22bfbdebe15705da6128b37aa69363a8275)
![{\displaystyle Q=q{\bigl [}(20+2{\sqrt {82}})^{-1/2}({\sqrt {{\sqrt {82}}+1}}+3){\bigr ]}\approx 0.370664951152024075624432522177568657151868089959747395750974\ldots }](https://wikimedia.org/api/rest_v1/media/math/render/svg/ec41e00c4c6492d95e6382bfaa1e9be6fa9a3f1b)
![{\displaystyle q{\bigl [}(2\,c^{2}+2+2{\sqrt {c^{4}+1}})^{-1/2}({\sqrt {{\sqrt {c^{4}+1}}+1}}+c){\bigr ]}=}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d31f8c23fd292956b3857995e0ebd91e1da3b832)
![{\displaystyle =\exp {\biggl \{}-\pi {\biggl [}\int _{0}^{1}2{\bigl (}z^{4}+2c{\sqrt {2{\sqrt {c^{4}+1}}-2c^{2}}}\,z^{2}+1{\bigr )}^{-1/2}\,\mathrm {d} z{\biggr ]}\div {\biggl [}\int _{0}^{1}2{\bigl (}z^{4}-2c{\sqrt {2{\sqrt {c^{4}+1}}-2c^{2}}}\,z^{2}+1{\bigr )}^{-1/2}\,\mathrm {d} z{\biggr ]}{\biggr \}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b6a67caf01791b94991f61cb5fd9b9f068eeeba2)
