Editing Elliptic integral (section)

==Complete elliptic integral of the second kind==
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[[Image:Mplwp complete ellipticEk.svg|thumb|300px|Plot of the complete elliptic integral of the second kind {{math|''E''(''k'')}}]]

The '''complete elliptic integral of the second kind''' {{math|''E''}} is defined as

<math display="block">E(k) = \int_0^\tfrac{\pi}{2} \sqrt{1-k^2 \sin^2\theta} \, d\theta = \int_0^1 \frac{\sqrt{1-k^2 t^2}}{\sqrt{1-t^2}} \, dt,</math>

or more compactly in terms of the incomplete integral of the second kind {{math|''E''(''φ'',''k'')}} as

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

For an ellipse with semi-major axis {{math|''a''}} and semi-minor axis {{math|''b''}} and eccentricity {{math|1=''e'' = {{sqrt|1 − ''b''<sup>2</sup>/''a''<sup>2</sup>}}}}, the complete elliptic integral of the second kind {{math|''E''(''e'')}} is equal to one quarter of the [[Ellipse#Circumference|circumference]] {{math|''C''}} of the ellipse measured in units of the semi-major axis {{math|''a''}}. In other words:

<math display="block">C = 4 a E(e).</math>

The complete elliptic integral of the second kind can be expressed as a [[power series]]<ref>{{Cite web|url=https://functions.wolfram.com/EllipticIntegrals/EllipticE/06/01/03/01/0003/|title=Complete elliptic integral of the second kind: Series representations (Formula 08.01.06.0002)}}</ref>

<math display="block">E(k) = \frac{\pi}{2}\sum_{n=0}^\infty \left(\frac{(2n)!}{2^{2n} \left(n!\right)^2}\right)^2 \frac{k^{2n}}{1-2n},</math>

which is equivalent to

<math display="block">E(k) = \frac{\pi}{2}\left(1-\left(\frac12\right)^2 \frac{k^2}{1}-\left(\frac{1\cdot 3}{2\cdot 4}\right)^2 \frac{k^4}{3}-\cdots-\left(\frac{(2n-1)!!}{(2n)!!}\right)^2 \frac{k^{2n}}{2n-1}-\cdots\right).</math>

In terms of the [[Gauss hypergeometric function]], the complete elliptic integral of the second kind can be expressed as

<math display="block">E(k) = \tfrac{\pi}{2} \,{}_2F_1 \left(\tfrac12, -\tfrac12; 1; k^2 \right).</math>

The modulus can be transformed that way:
<math display="block">E(k) = \left(1+\sqrt{1-k^2}\right)\,E\left(\frac{1-\sqrt{1-k^2}}{1+\sqrt{1-k^2}}\right) - \sqrt{1-k^2}\,K(k) </math>

===Computation===

Like the integral of the first kind, the complete elliptic integral of the second kind can be computed very efficiently using the [[arithmetic-geometric mean|arithmetic–geometric mean]].{{sfn|Carlson|2010|loc=19.8}}

Define sequences {{mvar|a<sub>n</sub>}} and {{mvar|g<sub>n</sub>}}, where {{math|1=''a''<sub>0</sub> = 1}}, {{math|1=''g''<sub>0</sub> = {{sqrt|1 − ''k''<sup>2</sup>}} = ''k''{{prime}}}} and the recurrence relations {{math|1=''a''<sub>''n'' + 1</sub> = {{sfrac|''a<sub>n</sub>'' + ''g<sub>n</sub>''|2}}}}, {{math|1=''g''<sub>''n'' + 1</sub> = {{sqrt|''a<sub>n</sub> g<sub>n</sub>''}}}} hold. Furthermore, define
<math display="block">c_n=\sqrt{\left|a_n^2-g_n^2\right|}.</math>

By definition,

<math display="block">a_\infty = \lim_{n\to\infty} a_n = \lim_{n\to\infty} g_n = \operatorname{agm}\left(1, \sqrt{1-k^2}\right).</math>

Also

<math display="block">\lim_{n\to\infty} c_n=0.</math>

Then

<math display="block">E(k) = \frac{\pi}{2a_\infty}\left(1-\sum_{n=0}^{\infty} 2^{n-1} c_n^2\right).</math>

In practice, the arithmetic-geometric mean would simply be computed up to some limit. This formula converges quadratically for all {{math|{{abs|''k''}} ≤ 1}}. To speed up computation further, the relation {{math|1=''c''<sub>''n'' + 1</sub> = {{sfrac|''c<sub>n</sub>''<sup>2</sup>|4''a''<sub>''n'' + 1</sub>}}}} can be used.

Furthermore, 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|''E''(''k'')}} is expressible in closed form in terms of
<math display="block">K(k)=\frac{\pi}{2\operatorname{agm}\left(1,\sqrt{1-k^2}\right)}</math>
and hence can be computed without the need for the infinite summation term. For example, {{math|1=''r'' = 1}}, {{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. 26, 161</ref>

<math display="block">E\left(\frac{1}{\sqrt{2}}\right)=\frac{1}{2}K\left(\frac{1}{\sqrt{2}}\right)+\frac{\pi}{4K\left(\frac{1}{\sqrt{2}}\right)},</math>

and

<math display="block">E\left(\frac{\sqrt{3}-1}{2\sqrt{2}}\right)=\frac{3+\sqrt{3}}{6}K\left(\frac{\sqrt{3}-1}{2\sqrt{2}}\right)+\frac{\pi\sqrt{3}}{12K\left(\frac{\sqrt{3}-1}{2\sqrt{2}}\right)},</math>

and

<math display="block">E\left(\frac{3-\sqrt{7}}{4\sqrt{2}}\right)=\frac{7+2\sqrt{7}}{14}K\left(\frac{3-\sqrt{7}}{4\sqrt{2}}\right)+\frac{\pi\sqrt{7}}{28K\left(\frac{3-\sqrt{7}}{4\sqrt{2}}\right)}.</math>

===Derivative and differential equation===
<math display="block">\frac{dE(k)}{dk} = \frac{E(k)-K(k)}{k}</math>
<math display="block">\left(k^2-1\right) \frac{d}{dk} \left( k \;\frac{dE(k)}{dk} \right) = k E(k)</math>

A second solution to this equation is {{math|''E''({{sqrt|1 − ''k''<sup>2</sup>}}) − ''K''({{sqrt|1 − ''k''<sup>2</sup>}})}}.