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Elliptic integral

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Special function defined by an integral

Inintegral calculus, anelliptic integral is one of a number of related functions defined as the value of certain integrals, which were first studied byGiulio Fagnano andLeonhard Euler (c. 1750). Their name originates from their connection with the problem of finding the arc length of anellipse.

Modern mathematics defines an "elliptic integral" as anyfunctionf which can be expressed in the form

$f(x)=\int _{c}^{x}R{\left({\textstyle t,{\sqrt {P(t)}}}\right)}\,dt,$

whereR is arational function of its two arguments,P is apolynomial of degree 3 or 4 with no repeated roots, andc is a constant.

In general, integrals in this form cannot be expressed in terms ofelementary functions. Exceptions to this general rule are whenP has repeated roots, whenR(x,y) contains no odd powers ofy, and when the integral is pseudo-elliptic. However, with the appropriatereduction formula, every elliptic integral can be brought into a form that involves integrals over rational functions and the threeLegendre canonical forms, also known as the elliptic integrals of the first, second and third kind.

Besides the Legendre form given below, the elliptic integrals may also be expressed inCarlson symmetric form. Additional insight into the theory of the elliptic integral may be gained through the study of theSchwarz–Christoffel mapping. Historically,elliptic functions were discovered as inverse functions of elliptic integrals.

Argument notation

Incomplete elliptic integrals are functions of two arguments;complete elliptic integrals are functions of a single argument. These arguments are expressed in a variety of different but equivalent ways as they give the same elliptic integral. Most texts adhere to a canonical naming scheme, using the following naming conventions.

For expressing one argument:

α, themodular angle
k = sinα, theelliptic modulus oreccentricity
m =k² = sin²α, theparameter

Each of the above three quantities is completely determined by any of the others (given that they are non-negative). Thus, they can be used interchangeably.

The other argument can likewise be expressed asφ, theamplitude, or asx oru, wherex = sinφ = snu andsn is one of theJacobian elliptic functions.

Specifying the value of any one of these quantities determines the others. Note thatu also depends onm. Some additional relationships involvingu include $\cos \varphi =\operatorname {cn} u,\quad {\textrm {and}}\quad {\sqrt {1-m\sin ^{2}\varphi }}=\operatorname {dn} u.$

The latter is sometimes called thedelta amplitude and written asΔ(φ) = dnu. Sometimes the literature also refers to thecomplementary parameter, thecomplementary modulus, or thecomplementary modular angle. These are further defined in the article onquarter periods.

In this notation, the use of a vertical bar as delimiter indicates that the argument following it is the "parameter" (as defined above), while the backslash indicates that it is the modular angle. The use of a semicolon implies that the argument preceding it is the sine of the amplitude: $F(\varphi ,\sin \alpha )=F\left(\varphi \mid \sin ^{2}\alpha \right)=F(\varphi \setminus \alpha )=F(\sin \varphi ;\sin \alpha ).$ This potentially confusing use of different argument delimiters is traditional in elliptic integrals and much of the notation is compatible with that used in the reference book byAbramowitz and Stegun and that used in the integral tables byGradshteyn and Ryzhik.

There are still other conventions for the notation of elliptic integrals employed in the literature. The notation with interchanged arguments,F(k,φ), is often encountered; and similarlyE(k,φ) for the integral of the second kind.Abramowitz and Stegun substitute the integral of the first kind,F(φ,k), for the argumentφ in their definition of the integrals of the second and third kinds, unless this argument is followed by a vertical bar: i.e.E(F(φ,k) |k²) forE(φ |k²). Moreover, their complete integrals employ theparameterk² as argument in place of the modulusk, i.e.K(k²) rather thanK(k). And the integral of the third kind defined byGradshteyn and Ryzhik,Π(φ,n,k), puts the amplitudeφ first and not the "characteristic"n.

Thus one must be careful with the notation when using these functions, because various reputable references and software packages use different conventions in the definitions of the elliptic functions. For example,Wolfram'sMathematica software andWolfram Alpha define the complete elliptic integral of the first kind in terms of the parameterm, instead of the elliptic modulusk.

Incomplete elliptic integral of the first kind

Theincomplete elliptic integral of the first kindF is defined as

$F(\varphi ,k)=F\left(\varphi \mid k^{2}\right)=F(\sin \varphi ;k)=\int _{0}^{\varphi }{\frac {d\theta }{\sqrt {1-k^{2}\sin ^{2}\theta }}}.$

This is Legendre's trigonometric form of the elliptic integral; substitutingt = sinθ andx = sinφ, one obtains Jacobi's algebraic form:

$F(x;k)=\int _{0}^{x}{\frac {dt}{\sqrt {\left(1-t^{2}\right)\left(1-k^{2}t^{2}\right)}}}.$

Equivalently, in terms of the amplitude and modular angle one has: $F(\varphi \setminus \alpha )=F(\varphi ,\sin \alpha )=\int _{0}^{\varphi }{\frac {d\theta }{\sqrt {1-\left(\sin \theta \sin \alpha \right)^{2}}}}.$

Withx = sn(u,k) one has: $F(x;k)=u;$ demonstrating that thisJacobian elliptic function is a simple inverse of the incomplete elliptic integral of the first kind.

The incomplete elliptic integral of the first kind has following addition theorem^{[citation needed]}: $F{\bigl [}\arctan(x),k{\bigr ]}+F{\bigl [}\arctan(y),k{\bigr ]}=F\left[\arctan \left({\frac {x{\sqrt {k'^{2}y^{2}+1}}}{\sqrt {y^{2}+1}}}\right)+\arctan \left({\frac {y{\sqrt {k'^{2}x^{2}+1}}}{\sqrt {x^{2}+1}}}\right),k\right]$

The elliptic modulus can be transformed that way: $F{\bigl [}\arcsin(x),k{\bigr ]}={\frac {2}{1+{\sqrt {1-k^{2}}}}}F\left[\arcsin \left({\frac {\left(1+{\sqrt {1-k^{2}}}\right)x}{1+{\sqrt {1-k^{2}x^{2}}}}}\right),{\frac {1-{\sqrt {1-k^{2}}}}{1+{\sqrt {1-k^{2}}}}}\right]$

Incomplete elliptic integral of the second kind

Theincomplete elliptic integral of the second kindE in Legendre's trigonometric form is

$E(\varphi ,k)=E\left(\varphi \,|\,k^{2}\right)=E(\sin \varphi ;k)=\int _{0}^{\varphi }{\sqrt {1-k^{2}\sin ^{2}\theta }}\,d\theta .$

Substitutingt = sinθ andx = sinφ, one obtains Jacobi's algebraic form:

$E(x;k)=\int _{0}^{x}{\frac {\sqrt {1-k^{2}t^{2}}}{\sqrt {1-t^{2}}}}\,dt.$

Equivalently, in terms of the amplitude and modular angle: $E(\varphi \setminus \alpha )=E(\varphi ,\sin \alpha )=\int _{0}^{\varphi }{\sqrt {1-\left(\sin \theta \sin \alpha \right)^{2}}}\,d\theta .$

Relations with theJacobi elliptic functions include ${\begin{aligned}E{\left(\operatorname {sn} (u;k);k\right)}=\int _{0}^{u}\operatorname {dn} ^{2}(w;k)\,dw&=u-k^{2}\int _{0}^{u}\operatorname {sn} ^{2}(w;k)\,dw\\[1ex]&=\left(1-k^{2}\right)u+k^{2}\int _{0}^{u}\operatorname {cn} ^{2}(w;k)\,dw.\end{aligned}}$

Themeridian arc length from theequator tolatitudeφ is written in terms ofE: $m(\varphi )=a\left(E(\varphi ,e)+{\frac {d^{2}}{d\varphi ^{2}}}E(\varphi ,e)\right),$ wherea is thesemi-major axis, ande is theeccentricity.

The incomplete elliptic integral of the second kind has following addition theorem^{[citation needed]}: ${\begin{aligned}&E{\left[\arctan(x),k\right]}+E{\left[\arctan(y),k\right]}\\[.5ex]&\quad =E{\left[\arctan \left({\frac {x{\sqrt {k'^{2}y^{2}+1}}}{\sqrt {y^{2}+1}}}\right)+\arctan \left({\frac {y{\sqrt {k'^{2}x^{2}+1}}}{\sqrt {x^{2}+1}}}\right),k\right]}\\[1ex]&\qquad +{\frac {k^{2}xy}{k'^{2}x^{2}y^{2}+x^{2}+y^{2}+1}}\left({\frac {x{\sqrt {k'^{2}y^{2}+1}}}{\sqrt {y^{2}+1}}}+{\frac {y{\sqrt {k'^{2}x^{2}+1}}}{\sqrt {x^{2}+1}}}\right)\end{aligned}}$

The elliptic modulus can be transformed that way: ${\begin{aligned}E{\left[\arcsin(x),k\right]}&=\left(1+{\sqrt {1-k^{2}}}\right)E{\left[\arcsin \left({\frac {\left(1+{\sqrt {1-k^{2}}}\right)x}{1+{\sqrt {1-k^{2}x^{2}}}}}\right),{\frac {1-{\sqrt {1-k^{2}}}}{1+{\sqrt {1-k^{2}}}}}\right]}\\[.5ex]&\quad -{\sqrt {1-k^{2}}}F{\left[\arcsin(x),k\right]}+{\frac {k^{2}x{\sqrt {1-x^{2}}}}{1+{\sqrt {1-k^{2}x^{2}}}}}\end{aligned}}$

Incomplete elliptic integral of the third kind

Theincomplete elliptic integral of the third kindΠ is $\Pi (n;\varphi \setminus \alpha )=\int _{0}^{\varphi }{\frac {1}{1-n\sin ^{2}\theta }}{\frac {d\theta }{\sqrt {1-\left(\sin \theta \sin \alpha \right)^{2}}}}$

or

$\Pi (n;\varphi \,|\,m)=\int _{0}^{\sin \varphi }{\frac {1}{1-nt^{2}}}{\frac {dt}{\sqrt {\left(1-mt^{2}\right)\left(1-t^{2}\right)}}}.$

The numbern is called thecharacteristic and can take on any value, independently of the other arguments. Note though that the valueΠ(1;⁠π/2⁠ |m) is infinite, for anym.

A relation with the Jacobian elliptic functions is $\Pi {\bigl (}n;\,\operatorname {am} (u;k);\,k{\bigr )}=\int _{0}^{u}{\frac {dw}{1-n\,\operatorname {sn} ^{2}(w;k)}}.$

The meridian arc length from the equator to latitudeφ is also related to a special case ofΠ:

$m(\varphi )=a\left(1-e^{2}\right)\Pi \left(e^{2};\varphi \,|\,e^{2}\right).$

Complete elliptic integral of the first kind

Plot of the complete elliptic integral of the first kindK(k)

Elliptic Integrals are said to be 'complete' when the amplitudeφ =⁠π/2⁠ and thereforex = 1. Thecomplete elliptic integral of the first kindK may thus be defined as $K(k)=\int _{0}^{\tfrac {\pi }{2}}{\frac {d\theta }{\sqrt {1-k^{2}\sin ^{2}\theta }}}=\int _{0}^{1}{\frac {dt}{\sqrt {\left(1-t^{2}\right)\left(1-k^{2}t^{2}\right)}}},$ or more compactly in terms of the incomplete integral of the first kind as $K(k)=F\left({\tfrac {\pi }{2}},k\right)=F\left({\tfrac {\pi }{2}}\,|\,k^{2}\right)=F(1;k).$

It can be expressed as apower series $K(k)={\frac {\pi }{2}}\sum _{n=0}^{\infty }\left({\frac {(2n)!}{2^{2n}(n!)^{2}}}\right)^{2}k^{2n}={\frac {\pi }{2}}\sum _{n=0}^{\infty }{\bigl (}P_{2n}(0){\bigr )}^{2}k^{2n},$

whereP_n is theLegendre polynomials, which is equivalent to

$K(k)={\frac {\pi }{2}}\left(1+\left({\frac {1}{2}}\right)^{2}k^{2}+\left({\frac {1\cdot 3}{2\cdot 4}}\right)^{2}k^{4}+\cdots +\left({\frac {\left(2n-1\right)!!}{\left(2n\right)!!}}\right)^{2}k^{2n}+\cdots \right),$

wheren!! denotes thedouble factorial. In terms of the Gausshypergeometric function, the complete elliptic integral of the first kind can be expressed as

$K(k)={\tfrac {\pi }{2}}\,{}_{2}F_{1}\left({\tfrac {1}{2}},{\tfrac {1}{2}};1;k^{2}\right).$

The complete elliptic integral of the first kind is sometimes called thequarter period. It can be computed very efficiently in terms of thearithmetic–geometric mean:^[1] $K(k)={\frac {\pi }{2\operatorname {agm} \left(1,{\sqrt {1-k^{2}}}\right)}}.$

Therefore, the modulus can be transformed as:

${\begin{aligned}K(k)&={\frac {\pi }{2\operatorname {agm} \left(1,{\sqrt {1-k^{2}}}\right)}}\\[4pt]&={\frac {\pi }{2\operatorname {agm} \left({\frac {1}{2}}+{\frac {\sqrt {1-k^{2}}}{2}},{\sqrt[{4}]{1-k^{2}}}\right)}}\\[4pt]&={\frac {\pi }{\left(1+{\sqrt {1-k^{2}}}\right)\operatorname {agm} \left(1,{\frac {2{\sqrt[{4}]{1-k^{2}}}}{\left(1+{\sqrt {1-k^{2}}}\right)}}\right)}}\\[4pt]&={\frac {2}{1+{\sqrt {1-k^{2}}}}}K\left({\frac {1-{\sqrt {1-k^{2}}}}{1+{\sqrt {1-k^{2}}}}}\right)\end{aligned}}$

This expression is valid for all $n\in \mathbb {N}$ and0 ≤k ≤ 1:

$K(k)=n\left[\sum _{a=1}^{n}\operatorname {dn} \left({\frac {2a}{n}}K(k);k\right)\right]^{-1}K\left[k^{n}\prod _{a=1}^{n}\operatorname {sn} \left({\frac {2a-1}{n}}K(k);k\right)^{2}\right]$

Relation to the gamma function

Ifk² =λ(i√r) and $r\in \mathbb {Q} ^{+}$ (whereλ is themodular lambda function), thenK(k) is expressible in closed form in terms of thegamma function.^[2] For example,r = 2,r = 3 andr = 7 give, respectively,^[3]

$K\left({\sqrt {2}}-1\right)={\frac {\Gamma \left({\frac {1}{8}}\right)\Gamma \left({\frac {3}{8}}\right){\sqrt {{\sqrt {2}}+1}}}{8{\sqrt[{4}]{2}}{\sqrt {\pi }}}},$

and

$K\left({\frac {{\sqrt {3}}-1}{2{\sqrt {2}}}}\right)={\frac {1}{8\pi }}{\sqrt[{4}]{3}}\,{\sqrt[{3}]{4}}\,\Gamma {\biggl (}{\frac {1}{3}}{\biggr )}^{3}$

and

$K\left({\frac {3-{\sqrt {7}}}{4{\sqrt {2}}}}\right)={\frac {\Gamma \left({\frac {1}{7}}\right)\Gamma \left({\frac {2}{7}}\right)\Gamma \left({\frac {4}{7}}\right)}{4{\sqrt[{4}]{7}}\pi }}.$

More generally, the condition that ${\frac {iK'}{K}}={\frac {iK\left({\sqrt {1-k^{2}}}\right)}{K(k)}}$ be in animaginary quadratic field^{[note 1]} is sufficient.^[4]^[5] For instance, ifk =e^5πi/6, then⁠iK′/K⁠ =e^2πi/3 and^[6]

$K\left(e^{5\pi i/6}\right)={\frac {e^{-\pi i/12}\Gamma ^{3}\left({\frac {1}{3}}\right){\sqrt[{4}]{3}}}{4{\sqrt[{3}]{2}}\pi }}.$

The second formula above, written as ${\frac {\Gamma \left({\frac {1}{3}}\right)^{3}}{\pi }}=2^{7/3}\,3^{-1/4}\,K\left({\tfrac {{\sqrt {3}}-1}{2{\sqrt {2}}}}\right)$ , can be completed by 5 equations showing that ${\frac {\Gamma \left({\frac {1}{k}}\right)^{k/2}}{\sqrt {\pi }}}$ is aperiod for all even divisors $k {\displaystyle k}$ of $24 {\displaystyle 24}$ :

${\frac {\Gamma \left({\frac {1}{4}}\right)^{2}}{\sqrt {\pi }}}=4\,K\left({\tfrac {1}{\sqrt {2}}}\right)$

${\frac {\Gamma \left({\frac {1}{6}}\right)^{3}}{\sqrt {\pi }}}=2^{11/3}\cdot 3\cdot K\left({\tfrac {{\sqrt {3}}-1}{2{\sqrt {2}}}}\right)^{2}$

${\frac {\Gamma \left({\frac {1}{8}}\right)^{4}}{\sqrt {\pi }}}=2^{17/2}\,K\left({\tfrac {1}{\sqrt {2}}}\right)\,K\left({\sqrt {2}}-1\right)^{2}$

${\frac {\Gamma \left({\frac {1}{12}}\right)^{6}}{\sqrt {\pi }}}=2^{55/6}\,3^{7/4}\,({\sqrt {3}}+1)^{3}\,K\left({\tfrac {{\sqrt {3}}-1}{2{\sqrt {2}}}}\right)^{2}\,K\left({\tfrac {1}{\sqrt {2}}}\right)^{3}$

${\frac {\Gamma \left({\frac {1}{24}}\right)^{12}}{\sqrt {\pi }}}=2^{89/3}3^{25/4}({\sqrt {2}}+1)^{6}({\sqrt {3}}-1)^{3}K\!\left({\tfrac {1}{\sqrt {2}}}\right)^{3}K\!\left({\tfrac {{\sqrt {3}}-1}{2{\sqrt {2}}}}\right)^{4}K\!{\Bigl (}(2-{\sqrt {3}})({\sqrt {3}}-{\sqrt {2}}){\Bigr )}^{6}$

Asymptotic expressions

$K\left(k\right)\approx {\frac {\pi }{2}}+{\frac {\pi }{8}}{\frac {k^{2}}{1-k^{2}}}-{\frac {\pi }{16}}{\frac {k^{4}}{1-k^{2}}}$ This approximation has a relative precision better than3×10⁻⁴ fork <⁠1/2⁠. Keeping only the first two terms is correct to 0.01 precision fork <⁠1/2⁠.^{[citation needed]}

Differential equation

The differential equation for the elliptic integral of the first kind is ${\frac {d}{dk}}\left(k\left(1-k^{2}\right){\frac {dK(k)}{dk}}\right)=k\,K(k)$

A second solution to this equation is $K\left({\sqrt {1-k^{2}}}\right)$ . This solution satisfies the relation ${\frac {d}{dk}}K(k)={\frac {E(k)}{k\left(1-k^{2}\right)}}-{\frac {K(k)}{k}}.$

Continued fraction

Acontinued fraction expansion is:^[7] ${\frac {K(k)}{2\pi }}=-{\frac {1}{4}}+\sum _{n=0}^{\infty }{\frac {q^{n}}{1+q^{2n}}}=-{\frac {1}{4}}+{\cfrac {1}{1-q+{\cfrac {\left(1-q\right)^{2}}{1-q^{3}+{\cfrac {q\left(1-q^{2}\right)^{2}}{1-q^{5}+{\cfrac {q^{2}\left(1-q^{3}\right)^{2}}{1-q^{7}+{\cfrac {q^{3}\left(1-q^{4}\right)^{2}}{1-q^{9}+\cdots }}}}}}}}}},$ where thenome is $q=q(k)=\exp[-\pi K'(k)/K(k)]$ in its definition.

Inverting the period ratio

Here, we use the complete elliptic integral of the first kind with theparameter $m {\displaystyle m}$ instead, because the squaring function introduces problems when inverting in the complex plane. So let

K[m]=\int _{0}^{\pi /2}{\dfrac {d\theta }{\sqrt {1-m\sin ^{2}\theta }}}

and let

\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,

\theta _{3}(\tau )=1+2\sum _{n=1}^{\infty }q^{n^{2}},\quad q=e^{\pi i\tau },\,\operatorname {Im} \tau >0

be thetheta functions.

The equation

\tau =i{\frac {K[1-m]}{K[m]}}

can then be solved (provided that a solution $m {\displaystyle m}$ exists) by

m={\frac {\theta _{2}(\tau )^{4}}{\theta _{3}(\tau )^{4}}}

which is in fact themodular lambda function.

For the purposes of computation, the error analysis is given by^[8]

\left|{e}^{-\pi i\tau /4}\theta _{2}\!\left(\tau \right)-2\sum _{n=0}^{N-1}{q}^{n\left(n+1\right)}\right|\leq {\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}}\;

\left|\theta _{3}\!\left(\tau \right)-\left(1+2\sum _{n=1}^{N-1}{q}^{n^{2}}\right)\right|\leq {\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}}\;

where $N\in \mathbb {Z} _{\geq 1}$ and $\operatorname {Im} \tau >0$ .

Also

K[m]={\frac {\pi }{2}}\theta _{3}(\tau )^{2},\quad \tau =i{\frac {K[1-m]}{K[m]}}

where $m\in \mathbb {C} \setminus \{0,1\}$ .

Complete elliptic integral of the second kind

Plot of the complete elliptic integral of the second kindE(k)

Thecomplete elliptic integral of the second kindE is defined as

$E(k)=\int _{0}^{\tfrac {\pi }{2}}{\sqrt {1-k^{2}\sin ^{2}\theta }}\,d\theta =\int _{0}^{1}{\frac {\sqrt {1-k^{2}t^{2}}}{\sqrt {1-t^{2}}}}\,dt,$

or more compactly in terms of the incomplete integral of the second kindE(φ,k) as

$E(k)=E\left({\tfrac {\pi }{2}},k\right)=E(1;k).$

For an ellipse with semi-major axisa and semi-minor axisb and eccentricitye =√1 −b²/a², the complete elliptic integral of the second kindE(e) is equal to one quarter of thecircumferenceC of the ellipse measured in units of the semi-major axisa. In other words:

The complete elliptic integral of the second kind can be expressed as apower series^[9]

$E(k)={\frac {\pi }{2}}\sum _{n=0}^{\infty }\left({\frac {(2n)!}{2^{2n}\left(n!\right)^{2}}}\right)^{2}{\frac {k^{2n}}{1-2n}},$

which is equivalent to

$E(k)={\frac {\pi }{2}}\left(1-\left({\frac {1}{2}}\right)^{2}{\frac {k^{2}}{1}}-\left({\frac {1\cdot 3}{2\cdot 4}}\right)^{2}{\frac {k^{4}}{3}}-\cdots -\left({\frac {(2n-1)!!}{(2n)!!}}\right)^{2}{\frac {k^{2n}}{2n-1}}-\cdots \right).$

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

$E(k)={\tfrac {\pi }{2}}\,{}_{2}F_{1}\left({\tfrac {1}{2}},-{\tfrac {1}{2}};1;k^{2}\right).$

The modulus can be transformed that way: $E(k)=\left(1+{\sqrt {1-k^{2}}}\right)\,E\left({\frac {1-{\sqrt {1-k^{2}}}}{1+{\sqrt {1-k^{2}}}}}\right)-{\sqrt {1-k^{2}}}\,K(k)$

Computation

Like the integral of the first kind, the complete elliptic integral of the second kind can be computed very efficiently using thearithmetic–geometric mean.^[1]

Define sequencesa_n andg_n, wherea₀ = 1,g₀ =√1 −k² =k′ and the recurrence relationsa_{n + 1} =⁠a_n +g_n/2⁠,g_{n + 1} =√a_n g_n hold. Furthermore, define $c_{n}={\sqrt {\left|a_{n}^{2}-g_{n}^{2}\right|}}.$

By definition,

$a_{\infty }=\lim _{n\to \infty }a_{n}=\lim _{n\to \infty }g_{n}=\operatorname {agm} \left(1,{\sqrt {1-k^{2}}}\right).$

Also

$\lim _{n\to \infty }c_{n}=0.$

Then

$E(k)={\frac {\pi }{2a_{\infty }}}\left(1-\sum _{n=0}^{\infty }2^{n-1}c_{n}^{2}\right).$

In practice, the arithmetic-geometric mean would simply be computed up to some limit. This formula converges quadratically for all|k| ≤ 1. To speed up computation further, the relationc_{n + 1} =⁠c_n²/4a_{n + 1}⁠ can be used.

Furthermore, ifk² =λ(i√r) and $r\in \mathbb {Q} ^{+}$ (whereλ is themodular lambda function), thenE(k) is expressible in closed form in terms of $K(k)={\frac {\pi }{2\operatorname {agm} \left(1,{\sqrt {1-k^{2}}}\right)}}$ and hence can be computed without the need for the infinite summation term. For example,r = 1,r = 3 andr = 7 give, respectively,^[10]

$E\left({\frac {1}{\sqrt {2}}}\right)={\frac {1}{2}}K\left({\frac {1}{\sqrt {2}}}\right)+{\frac {\pi }{4K\left({\frac {1}{\sqrt {2}}}\right)}},$

and

$E\left({\frac {{\sqrt {3}}-1}{2{\sqrt {2}}}}\right)={\frac {3+{\sqrt {3}}}{6}}K\left({\frac {{\sqrt {3}}-1}{2{\sqrt {2}}}}\right)+{\frac {\pi {\sqrt {3}}}{12K\left({\frac {{\sqrt {3}}-1}{2{\sqrt {2}}}}\right)}},$

and

$E\left({\frac {3-{\sqrt {7}}}{4{\sqrt {2}}}}\right)={\frac {7+2{\sqrt {7}}}{14}}K\left({\frac {3-{\sqrt {7}}}{4{\sqrt {2}}}}\right)+{\frac {\pi {\sqrt {7}}}{28K\left({\frac {3-{\sqrt {7}}}{4{\sqrt {2}}}}\right)}}.$

Derivative and differential equation

${\frac {dE(k)}{dk}}={\frac {E(k)-K(k)}{k}}$ $\left(k^{2}-1\right){\frac {d}{dk}}\left(k\;{\frac {dE(k)}{dk}}\right)=kE(k)$

A second solution to this equation isE(√1 −k²) −K(√1 −k²).

Complete elliptic integral of the third kind

Plot of the complete elliptic integral of the third kindΠ(n,k) with several fixed values ofn

Thecomplete elliptic integral of the third kindΠ can be defined as

$\Pi (n,k)=\int _{0}^{\frac {\pi }{2}}{\frac {d\theta }{\left(1-n\sin ^{2}\theta \right){\sqrt {1-k^{2}\sin ^{2}\theta }}}}.$

Note that sometimes the elliptic integral of the third kind is defined with an inverse sign for thecharacteristicn, $\Pi '(n,k)=\int _{0}^{\frac {\pi }{2}}{\frac {d\theta }{\left(1+n\sin ^{2}\theta \right){\sqrt {1-k^{2}\sin ^{2}\theta }}}}.$

Just like the complete elliptic integrals of the first and second kind, the complete elliptic integral of the third kind can be computed very efficiently using the arithmetic-geometric mean.^[1]

Partial derivatives

${\begin{aligned}{\frac {\partial \Pi (n,k)}{\partial n}}&={\frac {1}{2\left(k^{2}-n\right)(n-1)}}\left(E(k)+{\frac {1}{n}}\left(k^{2}-n\right)K(k)+{\frac {1}{n}}\left(n^{2}-k^{2}\right)\Pi (n,k)\right)\\[8pt]{\frac {\partial \Pi (n,k)}{\partial k}}&={\frac {k}{n-k^{2}}}\left({\frac {E(k)}{k^{2}-1}}+\Pi (n,k)\right)\end{aligned}}$

Jacobi zeta function

In 1829, Jacobi defined theJacobi zeta function: $Z(\varphi ,k)=E(\varphi ,k)-{\frac {E(k)}{K(k)}}F(\varphi ,k).$ It is periodic in $\varphi$ with minimal period $\pi$ . It is related to theJacobi zn function by $Z(\varphi ,k)=\operatorname {zn} (F(\varphi ,k),k)$ . In the literature (e.g. Whittaker and Watson (1927)), sometimes $Z {\displaystyle Z}$ means Wikipedia's $\operatorname {zn}$ . Some authors (e.g. King (1924)) use $Z {\displaystyle Z}$ for both Wikipedia's $Z {\displaystyle Z}$ and $\operatorname {zn}$ .

Legendre's relation

TheLegendre's relation orLegendre Identity shows the relation of the integrals K and E of an elliptic modulus and its anti-related counterpart^[11]^[12] in an integral equation of second degree:

For two modules that are Pythagorean counterparts to each other, this relation is valid:

$K(\varepsilon )E\left({\sqrt {1-\varepsilon ^{2}}}\right)+E(\varepsilon )K\left({\sqrt {1-\varepsilon ^{2}}}\right)-K(\varepsilon )K\left({\sqrt {1-\varepsilon ^{2}}}\right)={\frac {\pi }{2}}$

For example:

K({\color {blueviolet}{\tfrac {3}{5}}})E({\color {blue}{\tfrac {4}{5}}})+E({\color {blueviolet}{\tfrac {3}{5}}})K({\color {blue}{\tfrac {4}{5}}})-K({\color {blueviolet}{\tfrac {3}{5}}})K({\color {blue}{\tfrac {4}{5}}})={\tfrac {1}{2}}\pi

And for two modules that are tangential counterparts to each other, the following relationship is valid:

(1+\varepsilon )K(\varepsilon )E({\tfrac {1-\varepsilon }{1+\varepsilon }})+{\tfrac {2}{1+\varepsilon }}E(\varepsilon )K({\tfrac {1-\varepsilon }{1+\varepsilon }})-2K(\varepsilon )K({\tfrac {1-\varepsilon }{1+\varepsilon }})={\tfrac {1}{2}}\pi

For example:

{\tfrac {4}{3}}K({\color {blue}{\tfrac {1}{3}}})E({\color {green}{\tfrac {1}{2}}})+{\tfrac {3}{2}}E({\color {blue}{\tfrac {1}{3}}})K({\color {green}{\tfrac {1}{2}}})-2K({\color {blue}{\tfrac {1}{3}}})K({\color {green}{\tfrac {1}{2}}})={\tfrac {1}{2}}\pi

The Legendre's relation for tangential modular counterparts results directly from the Legendre's identity for Pythagorean modular counterparts by using theLanden modular transformation on the Pythagorean counter modulus.

Special identity for the lemniscatic case

For the lemniscatic case, the elliptic modulus or specific eccentricity ε is equal to half the square root of two. Legendre's identity for the lemniscatic case can be proved as follows:

According to theChain rule these derivatives hold:

{\frac {\mathrm {d} }{\mathrm {d} y}}\,K{\bigl (}{\frac {1}{2}}{\sqrt {2}}{\bigr )}-F{\biggl [}\arccos(xy);{\frac {1}{2}}{\sqrt {2}}{\biggr ]}={\frac {{\sqrt {2}}\,x}{\sqrt {1-x^{4}y^{4}}}}

{\frac {\mathrm {d} }{\mathrm {d} y}}\,2E{\bigl (}{\frac {1}{2}}{\sqrt {2}}{\bigr )}-K{\bigl (}{\frac {1}{2}}{\sqrt {2}}{\bigr )}-2E{\biggl [}\arccos(xy);{\frac {1}{2}}{\sqrt {2}}{\biggr ]}+F{\biggl [}\arccos(xy);{\frac {1}{2}}{\sqrt {2}}{\biggr ]}={\frac {{\sqrt {2}}\,x^{3}y^{2}}{\sqrt {1-x^{4}y^{4}}}}

By using theFundamental theorem of calculus these formulas can be generated:

K{\bigl (}{\frac {1}{2}}{\sqrt {2}}{\bigr )}-F{\biggl [}\arccos(x);{\frac {1}{2}}{\sqrt {2}}{\biggr ]}=\int _{0}^{1}{\frac {{\sqrt {2}}\,x}{\sqrt {1-x^{4}y^{4}}}}\,\mathrm {d} y

2E{\bigl (}{\frac {1}{2}}{\sqrt {2}}{\bigr )}-K{\bigl (}{\frac {1}{2}}{\sqrt {2}}{\bigr )}-2E{\biggl [}\arccos(x);{\frac {1}{2}}{\sqrt {2}}{\biggr ]}+F{\biggl [}\arccos(x);{\frac {1}{2}}{\sqrt {2}}{\biggr ]}=\int _{0}^{1}{\frac {{\sqrt {2}}\,x^{3}y^{2}}{\sqrt {1-x^{4}y^{4}}}}\,\mathrm {d} y

TheLinear combination of the two now mentioned integrals leads to the following formula:

{\frac {\sqrt {2}}{\sqrt {1-x^{4}}}}{\biggl \{}2E{\bigl (}{\frac {1}{2}}{\sqrt {2}}{\bigr )}-K{\bigl (}{\frac {1}{2}}{\sqrt {2}}{\bigr )}-2E{\biggl [}\arccos(x);{\frac {1}{2}}{\sqrt {2}}{\biggr ]}+F{\biggl [}\arccos(x);{\frac {1}{2}}{\sqrt {2}}{\biggr ]}{\biggr \}}\,+

+\,{\frac {{\sqrt {2}}\,x^{2}}{\sqrt {1-x^{4}}}}{\biggl \{}K{\bigl (}{\frac {1}{2}}{\sqrt {2}}{\bigr )}-F{\biggl [}\arccos(x);{\frac {1}{2}}{\sqrt {2}}{\biggr ]}{\biggr \}}=\int _{0}^{1}{\frac {2\,x^{3}(y^{2}+1)}{\sqrt {(1-x^{4})(1-x^{4}\,y^{4})}}}\,\mathrm {d} y

By forming the original antiderivative related to x from the function now shown using theProduct rule this formula results:

{\biggl \{}K{\bigl (}{\frac {1}{2}}{\sqrt {2}}{\bigr )}-F{\biggl [}\arccos(x);{\frac {1}{2}}{\sqrt {2}}{\biggr ]}{\biggr \}}{\biggl \{}2E{\bigl (}{\frac {1}{2}}{\sqrt {2}}{\bigr )}-K{\bigl (}{\frac {1}{2}}{\sqrt {2}}{\bigr )}-2E{\biggl [}\arccos(x);{\frac {1}{2}}{\sqrt {2}}{\biggr ]}+F{\biggl [}\arccos(x);{\frac {1}{2}}{\sqrt {2}}{\biggr ]}{\biggr \}}=

=\int _{0}^{1}{\frac {1}{y^{2}}}(y^{2}+1){\biggl [}{\text{artanh}}(y^{2})-{\text{artanh}}{\bigl (}{\frac {{\sqrt {1-x^{4}}}\,y^{2}}{\sqrt {1-x^{4}y^{4}}}}{\bigr )}{\biggr ]}\mathrm {d} y

If the value $x=1$ is inserted in this integral identity, then the following identity emerges:

K{\bigl (}{\frac {1}{2}}{\sqrt {2}}{\bigr )}{\biggl [}2\,E{\bigl (}{\frac {1}{2}}{\sqrt {2}}{\bigr )}-K{\bigl (}{\frac {1}{2}}{\sqrt {2}}{\bigr )}{\biggr ]}=\int _{0}^{1}{\frac {1}{y^{2}}}(y^{2}+1)\,{\text{artanh}}(y^{2})\,\mathrm {d} y=

={\biggl [}2\arctan(y)-{\frac {1}{y}}(1-y^{2})\,{\text{artanh}}(y^{2}){\biggr ]}_{y=0}^{y=1}=2\arctan(1)={\frac {\pi }{2}}

This is how this lemniscatic excerpt from Legendre's identity appears:

2E{\bigl (}{\frac {1}{2}}{\sqrt {2}}{\bigr )}K{\bigl (}{\frac {1}{2}}{\sqrt {2}}{\bigr )}-K{\bigl (}{\frac {1}{2}}{\sqrt {2}}{\bigr )}^{2}={\frac {\pi }{2}}

Generalization for the overall case

Now the modular general case^[13]^[14] is worked out. For this purpose, the derivatives of the complete elliptic integrals are derived after the modulus $\varepsilon$ and then they are combined. And then the Legendre's identity balance is determined.

Because the derivative of thecircle function is the negative product of theidentical mapping function and the reciprocal of the circle function:

{\frac {\mathrm {d} }{\mathrm {d} \varepsilon }}{\sqrt {1-\varepsilon ^{2}}}=-\,{\frac {\varepsilon }{\sqrt {1-\varepsilon ^{2}}}}

These are the derivatives of K and E shown in this article in the sections above:

{\frac {\mathrm {d} }{\mathrm {d} \varepsilon }}K(\varepsilon )={\frac {1}{\varepsilon (1-\varepsilon ^{2})}}{\bigl [}E(\varepsilon )-(1-\varepsilon ^{2})K(\varepsilon ){\bigr ]}

{\frac {\mathrm {d} }{\mathrm {d} \varepsilon }}E(\varepsilon )=-\,{\frac {1}{\varepsilon }}{\bigl [}K(\varepsilon )-E(\varepsilon ){\bigr ]}

In combination with the derivative of the circle function these derivatives are valid then:

{\frac {\mathrm {d} }{\mathrm {d} \varepsilon }}K({\sqrt {1-\varepsilon ^{2}}})={\frac {1}{\varepsilon (1-\varepsilon ^{2})}}{\bigl [}\varepsilon ^{2}K({\sqrt {1-\varepsilon ^{2}}})-E({\sqrt {1-\varepsilon ^{2}}}){\bigr ]}

{\frac {\mathrm {d} }{\mathrm {d} \varepsilon }}E({\sqrt {1-\varepsilon ^{2}}})={\frac {\varepsilon }{1-\varepsilon ^{2}}}{\bigl [}K({\sqrt {1-\varepsilon ^{2}}})-E({\sqrt {1-\varepsilon ^{2}}}){\bigr ]}

Legendre's identity includes products of any two complete elliptic integrals. For the derivation of the function side from the equation scale of Legendre's identity, theProduct rule is now applied in the following:

{\frac {\mathrm {d} }{\mathrm {d} \varepsilon }}K(\varepsilon )E({\sqrt {1-\varepsilon ^{2}}})={\frac {1}{\varepsilon (1-\varepsilon ^{2})}}{\bigl [}E(\varepsilon )E({\sqrt {1-\varepsilon ^{2}}})-K(\varepsilon )E({\sqrt {1-\varepsilon ^{2}}})+\varepsilon ^{2}K(\varepsilon )K({\sqrt {1-\varepsilon ^{2}}}){\bigr ]}

{\frac {\mathrm {d} }{\mathrm {d} \varepsilon }}E(\varepsilon )K({\sqrt {1-\varepsilon ^{2}}})={\frac {1}{\varepsilon (1-\varepsilon ^{2})}}{\bigl [}-E(\varepsilon )E({\sqrt {1-\varepsilon ^{2}}})+E(\varepsilon )K({\sqrt {1-\varepsilon ^{2}}})-(1-\varepsilon ^{2})K(\varepsilon )K({\sqrt {1-\varepsilon ^{2}}}){\bigr ]}

{\frac {\mathrm {d} }{\mathrm {d} \varepsilon }}K(\varepsilon )K({\sqrt {1-\varepsilon ^{2}}})={\frac {1}{\varepsilon (1-\varepsilon ^{2})}}{\bigl [}E(\varepsilon )K({\sqrt {1-\varepsilon ^{2}}})-K(\varepsilon )E({\sqrt {1-\varepsilon ^{2}}})-(1-2\varepsilon ^{2})K(\varepsilon )K({\sqrt {1-\varepsilon ^{2}}}){\bigr ]}

Of these three equations, adding the top two equations and subtracting the bottom equation gives this result:

{\frac {\mathrm {d} }{\mathrm {d} \varepsilon }}{\bigl [}K(\varepsilon )E({\sqrt {1-\varepsilon ^{2}}})+E(\varepsilon )K({\sqrt {1-\varepsilon ^{2}}})-K(\varepsilon )K({\sqrt {1-\varepsilon ^{2}}}){\bigr ]}=0

In relation to the $\varepsilon$ the equation balance constantly gives the value zero.

The previously determined result shall be combined with the Legendre equation to the modulus $\varepsilon =1/{\sqrt {2}}$ that is worked out in the section before:

2E{\bigl (}{\frac {1}{2}}{\sqrt {2}}{\bigr )}K{\bigl (}{\frac {1}{2}}{\sqrt {2}}{\bigr )}-K{\bigl (}{\frac {1}{2}}{\sqrt {2}}{\bigr )}^{2}={\frac {\pi }{2}}

The combination of the last two formulas gives the following result:

K(\varepsilon )E({\sqrt {1-\varepsilon ^{2}}})+E(\varepsilon )K({\sqrt {1-\varepsilon ^{2}}})-K(\varepsilon )K({\sqrt {1-\varepsilon ^{2}}})={\tfrac {1}{2}}\pi

Because if the derivative of a continuous function constantly takes the value zero, then the concerned function is a constant function. This means that this function results in the same function value for each abscissa value $\varepsilon$ and the associated function graph is therefore a horizontal straight line.

See also

References

Notes

^K can beanalytically extended to thecomplex plane.

References

^^a ^b ^cCarlson 2010, 19.8.
^Borwein, Jonathan M.; Borwein, Peter B. (1987).Pi and the AGM: A Study in Analytic Number Theory and Computational Complexity (First ed.). Wiley-Interscience.ISBN 0-471-83138-7. p. 296
^Borwein, Jonathan M.; Borwein, Peter B. (1987).Pi and the AGM: A Study in Analytic Number Theory and Computational Complexity (First ed.). Wiley-Interscience.ISBN 0-471-83138-7. p. 298
^Chowla, S.; Selberg, A. (1949)."On Epstein's Zeta Function (I)".Proceedings of the National Academy of Sciences.35 (7): 373.Bibcode:1949PNAS...35..371C.doi:10.1073/PNAS.35.7.371.PMC 1063041.PMID 16588908.S2CID 45071481.
^Chowla, S.; Selberg, A. (1967)."On Epstein's Zeta-Function".Journal für die Reine und Angewandte Mathematik.227:86–110.
^"Legendre elliptic integrals (Entry 175b7a)".
^N.Bagis,L.Glasser.(2015)"Evaluations of a Continued fraction of Ramanujan". Rend.Sem.Mat.Univ.Padova, Vol.133 pp 1-10
^"Approximations of Jacobi theta functions".The Mathematical Functions Grimoire. Fredrik Johansson. RetrievedAugust 29, 2024.
^"Complete elliptic integral of the second kind: Series representations (Formula 08.01.06.0002)".
^Borwein, Jonathan M.; Borwein, Peter B. (1987).Pi and the AGM: A Study in Analytic Number Theory and Computational Complexity (First ed.). Wiley-Interscience.ISBN 0-471-83138-7. p. 26, 161
^"Legendre-Relation" (in German). Retrieved2022-11-29.
^"Legendre Relation". Retrieved2022-11-29.
^"integration - Proving Legendres Relation for elliptic curves". Retrieved2023-02-10.
^Internet Archive (1991),Paul Halmos celebrating 50 years of mathematics, New York : Springer-Verlag,ISBN 0-387-97509-8, retrieved2023-02-10

Sources

Abramowitz, Milton;Stegun, Irene Ann, eds. (1983) [June 1964]."Chapter 17".Handbook of Mathematical Functions with Formulas, Graphs, and Mathematical Tables. Applied Mathematics Series. Vol. 55 (Ninth reprint with additional corrections of tenth original printing with corrections (December 1972); first ed.). Washington D.C.; New York: United States Department of Commerce, National Bureau of Standards; Dover Publications. p. 587.ISBN 978-0-486-61272-0.LCCN 64-60036.MR 0167642.LCCN 65-12253.
Byrd, P. F.; Friedman, M.D. (1971).Handbook of Elliptic Integrals for Engineers and Scientists (2nd ed.). New York: Springer-Verlag.ISBN 0-387-05318-2.
Carlson, B. C. (1995). "Numerical Computation of Real or Complex Elliptic Integrals".Numerical Algorithms.10 (1):13–26.arXiv:math/9409227.Bibcode:1995NuAlg..10...13C.doi:10.1007/BF02198293.S2CID 11580137.
Carlson, B. C. (2010),"Elliptic integral", inOlver, Frank W. J.; Lozier, Daniel M.; Boisvert, Ronald F.; Clark, Charles W. (eds.),NIST Handbook of Mathematical Functions, Cambridge University Press,ISBN 978-0-521-19225-5,MR 2723248.
Erdélyi, Arthur;Magnus, Wilhelm; Oberhettinger, Fritz; Tricomi, Francesco G. (1953).Higher transcendental functions. Vol II(PDF). McGraw-Hill Book Company, Inc., New York-Toronto-London.MR 0058756. Archived fromthe original(PDF) on 2011-07-14. Retrieved2016-07-24.
Gradshteyn, Izrail Solomonovich;Ryzhik, Iosif Moiseevich;Geronimus, Yuri Veniaminovich;Tseytlin, Michail Yulyevich; Jeffrey, Alan (2015) [October 2014]. "8.1.". In Zwillinger, Daniel;Moll, Victor Hugo (eds.).Table of Integrals, Series, and Products. Translated by Scripta Technica, Inc. (8 ed.).Academic Press, Inc.ISBN 978-0-12-384933-5.LCCN 2014010276.
Greenhill, Alfred George (1892).The applications of elliptic functions. New York: Macmillan.
Hancock, Harris (1910).Lectures on the Theory of Elliptic Functions. New York: J. Wiley & sons.
King, Louis V. (1924).On The Direct Numerical Calculation Of Elliptic Functions And Integrals. Cambridge University Press.
Press, W. H.; Teukolsky, S. A.; Vetterling, W. T.; Flannery, B. P. (2007),"Section 6.12. Elliptic Integrals and Jacobian Elliptic Functions",Numerical Recipes: The Art of Scientific Computing (3rd ed.), New York: Cambridge University Press,ISBN 978-0-521-88068-8, archived fromthe original on 2011-08-11, retrieved2011-08-09

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