Editing Elliptic integral (section)

==Complete elliptic integral of the first kind==
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[[Image:Mplwp complete ellipticKk.svg|thumb|300px|Plot of the complete elliptic integral of the first kind {{math|''K''(''k'')}}]]
Elliptic Integrals are said to be 'complete' when the amplitude {{math|1=''φ'' = {{sfrac|π|2}}}} and therefore {{math|1=''x'' = 1}}. The '''complete elliptic integral of the first kind''' {{math|''K''}} may thus be defined as
<math display="block">K(k) = \int_0^\tfrac{\pi}{2} \frac{d\theta}{\sqrt{1-k^2 \sin^2\theta}} = \int_0^1 \frac{dt}{\sqrt{\left(1-t^2\right)\left(1-k^2 t^2\right)}},</math>
or more compactly in terms of the incomplete integral of the first kind as
<math display="block">K(k) = F\left(\tfrac{\pi}{2},k\right) = F\left(\tfrac{\pi}{2} \,|\, k^2\right) = F(1;k).</math>

It can be expressed as a [[power series]]
<math display="block">K(k) = \frac{\pi}{2}\sum_{n=0}^\infty \left(\frac{(2n)!}{2^{2 n} (n!)^2}\right)^2 k^{2n} = \frac{\pi}{2} \sum_{n=0}^\infty \bigl(P_{2 n}(0)\bigr)^2 k^{2n},</math>

where {{math|''P''<sub>''n''</sub>}} is the [[Legendre polynomials]], which is equivalent to

<math display="block">K(k) = \frac{\pi}{2}\left(1+\left(\frac{1}{2}\right)^2 k^2+\left(\frac{1\cdot 3}{2\cdot 4}\right)^2 k^4+\cdots+\left(\frac{\left(2n-1\right)!!}{\left(2n\right)!!}\right)^2 k^{2n}+\cdots\right),</math>

where {{math|''n''!!}} denotes the [[double factorial]]. In terms of the Gauss [[hypergeometric function]], the complete elliptic integral of the first kind can be expressed as

<math display="block">K(k) = \tfrac{\pi}{2} \,{}_2F_1 \left(\tfrac{1}{2}, \tfrac{1}{2}; 1; k^2\right).</math>

The complete elliptic integral of the first kind is sometimes called the [[quarter period]]. It can be computed very efficiently in terms of the [[arithmetic–geometric mean]]:{{sfn|Carlson|2010|loc=19.8}}
<math display="block">K(k) = \frac{\pi}{2\operatorname{agm}\left(1,\sqrt{1-k^2}\right)}.</math>

Therefore, the modulus can be transformed as:

<math display="block">\begin{align}
K(k) &= \frac{\pi}{2\operatorname{agm}\left(1,\sqrt{1-k^2}\right)} \\[4pt]
& = \frac{\pi}{2\operatorname{agm}\left(\frac12+\frac\sqrt{1-k^2}{2},\sqrt[4]{1-k^2}\right)} \\[4pt]
&= \frac{\pi}{\left(1+\sqrt{1-k^2}\right)\operatorname{agm}\left(1,\frac{2\sqrt[4]{1-k^2}}{\left(1+\sqrt{1-k^2}\right)}\right)} \\[4pt]
& = \frac{2}{1+\sqrt{1-k^2}}K\left(\frac{1-\sqrt{1-k^2}}{1+\sqrt{1-k^2}}\right)
\end{align}</math>

This expression is valid for all <math>n \isin \mathbb{N}</math> and {{math|0 ≤ ''k'' ≤ 1}}:

<math display="block">K(k) = n\left[\sum_{a = 1}^{n} \operatorname{dn}\left(\frac{2a}{n}K(k);k\right)\right]^{-1}K\left[k^n\prod_{a=1}^{n}\operatorname{sn}\left(\frac{2a-1}{n}K(k);k\right)^2\right] </math>

===Relation to the gamma function===
If {{math|1=''k''<sup>2</sup> = ''λ''(''i''{{sqrt|''r''}})}} and <math>r \isin \mathbb{Q}^+</math> (where {{mvar|λ}} is the [[modular lambda function]]), then {{math|''K''(''k'')}} is expressible in closed form in terms of the [[gamma function]].<ref>{{Cite book |last1=Borwein |first1=Jonathan M. |last2=Borwein| first2=Peter B. |title=Pi and the AGM: A Study in Analytic Number Theory and Computational Complexity |publisher=Wiley-Interscience |year=1987 |edition=First |isbn=0-471-83138-7}} p. 296</ref> For example, {{math|1=''r'' = 2}}, {{math|1=''r'' = 3}} and {{math|1=''r'' = 7}} give, respectively,<ref>{{Cite book |last1=Borwein |first1=Jonathan M. |last2=Borwein| first2=Peter B. |title=Pi and the AGM: A Study in Analytic Number Theory and Computational Complexity |publisher=Wiley-Interscience |year=1987 |edition=First |isbn=0-471-83138-7}} p. 298</ref>

<math display="block">K\left(\sqrt{2}-1\right)=\frac{\Gamma \left(\frac18\right)\Gamma \left(\frac38\right)\sqrt{\sqrt{2}+1}}{8\sqrt[4]{2}\sqrt{\pi}},</math>

and

<math display="block">K\left(\frac{\sqrt{3}-1}{2\sqrt{2}}\right)=\frac{1}{8\pi}\sqrt[4]{3}\,\sqrt[3]{4}\,\Gamma\biggl(\frac{1}{3}\biggr)^3</math>

and

<math display="block">K\left(\frac{3-\sqrt{7}}{4\sqrt{2}}\right)=\frac{\Gamma \left(\frac17\right)\Gamma \left(\frac27\right)\Gamma \left(\frac47\right)}{4\sqrt[4]{7}\pi}.</math>

More generally, the condition that
<math display="block">\frac{iK'}{K}=\frac{iK\left(\sqrt{1-k^2}\right)}{K(k)}</math>
be in an [[Quadratic field|imaginary quadratic field]]<ref group="note">{{mvar|K}} can be [[Analytic continuation|analytically extended]] to the [[complex plane]].</ref> is sufficient.<ref>{{Cite journal|title=On Epstein's Zeta Function (I).|last1=Chowla|first1=S.|last2=Selberg|first2=A.|journal=Proceedings of the National Academy of Sciences|year=1949|volume=35|issue=7|page=373|doi=10.1073/PNAS.35.7.371|pmid=16588908|pmc=1063041|bibcode=1949PNAS...35..371C|s2cid=45071481}}</ref><ref>{{Cite journal|url=https://eudml.org/doc/150803|title=On Epstein's Zeta-Function|last1=Chowla|first1=S.|last2=Selberg|first2=A.|journal=Journal für die Reine und Angewandte Mathematik|year=1967|volume=227|pages=86–110}}</ref> For instance, if {{math|1=''k'' = ''e''<sup>5''πi''/6</sup>}}, then {{math|1={{sfrac|''iK''{{prime}}|''K''}} = ''e''<sup>2''πi''/3</sup>}} and<ref>{{Cite web|url=https://fungrim.org/topic/Legendre_elliptic_integrals/|title = Legendre elliptic integrals (Entry 175b7a)}}</ref>

<math display="block">K\left(e^{5\pi i/6}\right)=\frac{e^{-\pi i/12}\Gamma ^3\left(\frac13\right)\sqrt[4]{3}}{4\sqrt[3]{2}\pi}.</math>

===Asymptotic expressions===
<math display="block">K\left(k\right)\approx\frac{\pi}{2}+\frac{\pi}{8}\frac{k^2}{1-k^2}-\frac{\pi}{16}\frac{k^4}{1-k^2}</math>
This approximation has a relative precision better than {{val|3|e=−4}} for {{math|''k'' < {{sfrac|1|2}}}}. Keeping only the first two terms is correct to 0.01 precision for {{math|''k'' < {{sfrac|1|2}}}}.{{citation needed|date=January 2017}}

===Differential equation===
The differential equation for the elliptic integral of the first kind is
<math display="block">\frac{d}{dk}\left(k\left(1-k^2\right)\frac{dK(k)}{dk}\right) = k \, K(k)</math>

A second solution to this equation is <math>K\left(\sqrt{1-k^2}\right)</math>. This solution satisfies the relation
<math display="block">\frac{d}{dk}K(k) = \frac{E(k)}{k\left(1-k^2\right)}-\frac{K(k)}{k}.</math>

===Continued fraction===
A [[continued fraction]] expansion is:<ref>N.Bagis,L.Glasser.(2015)"Evaluations of a Continued fraction of Ramanujan". Rend.Sem.Mat.Univ.Padova, Vol.133 pp 1-10</ref> 
<math display="block">\frac{K(k)}{2\pi} = -\frac{1}{4} + \sum^{\infty}_{n=0} \frac{q^n}{1+q^{2n}} = -\frac{1}{4} + \cfrac{1}{1-q+ \cfrac{\left(1-q\right)^2}{1-q^3+ \cfrac{q\left(1-q^2\right)^2}{1-q^5+ \cfrac{q^2\left(1-q^3\right)^2}{1-q^7+\cfrac{q^3\left(1-q^4\right)^2}{1-q^9+\cdots}}}}},</math>
where the [[Nome (mathematics)|nome]] is <math> q = q(k) = \exp[-\pi K'(k)/K(k)] </math> in its definition.

===Inverting the period ratio===
Here, we use the complete elliptic integral of the first kind with the ''parameter'' <math>m</math> instead, because the squaring function introduces problems when inverting in the complex plane. So let
:<math>K[m]=\int_0^{\pi/2}\dfrac{d\theta}{\sqrt{1-m\sin^2\theta}}</math>
and let
:<math>\theta_2(\tau)=2e^{\pi i\tau/4}\sum_{n=0}^\infty q^{n(n+1)},\quad q=e^{\pi i\tau},\, \operatorname{Im}\tau >0,</math>
:<math>\theta_3(\tau)=1+2\sum_{n=1}^\infty q^{n^2},\quad q=e^{\pi i\tau},\,\operatorname{Im}\tau >0</math>
be the [[theta function]]s.

The equation
:<math>\tau=i\frac{K[1-m]}{K[m]}</math>
can then be solved (provided that a solution <math>m</math> exists) by
:<math>m=\frac{\theta_2(\tau)^4}{\theta_3(\tau)^4}</math>
which is in fact the [[modular lambda function]].

For the purposes of computation, the error analysis is given by<ref>{{cite web |url=https://fungrim.org/topic/Approximations_of_Jacobi_theta_functions/ |title=Approximations of Jacobi theta functions |last= |first= |date= |website=The Mathematical Functions Grimoire |publisher=Fredrik Johansson |access-date=August 29, 2024}}</ref>
:<math>\left|{e}^{-\pi i \tau / 4} \theta_{2}\!\left(\tau\right) - 2\sum_{n=0}^{N - 1} {q}^{n \left(n + 1\right)}\right| \le \begin{cases} \frac{2 {\left|q\right|}^{N \left(N + 1\right)}}{1 - \left|q\right|^{2N+1}}, & \left|q\right|^{2N+1} < 1\\\infty, & \text{otherwise}\\ \end{cases}\;</math>
:<math>\left|\theta_{3}\!\left(\tau\right) - \left(1+2\sum_{n=1}^{N - 1} {q}^{n^2}\right)\right| \le \begin{cases} \frac{2 {\left|q\right|}^{N^2}}{1 - \left|q\right|^{2N+1}}, & \left|q\right|^{2N+1} < 1\\\infty, & \text{otherwise}\\ \end{cases}\;</math>
where <math>N\in\mathbb{Z}_{\ge 1}</math> and <math>\operatorname{Im}\tau >0</math>.

Also
:<math>K[m]=\frac{\pi}{2}\theta_3(\tau )^2,\quad \tau=i\frac{K[1-m]}{K[m]}</math>
where <math>m\in\mathbb{C}\setminus\{0,1\}</math>.