Английская Википедия:Elliptic integral

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Шаблон:Short description Шаблон:Use American English In integral calculus, an elliptic integral is one of a number of related functions defined as the value of certain integrals, which were first studied by Giulio Fagnano and Leonhard Euler (Шаблон:Circa). Their name originates from their originally arising in connection with the problem of finding the arc length of an ellipse.

Modern mathematics defines an "elliptic integral" as any function Шаблон:Math which can be expressed in the form

<math display="block"> f(x) = \int_{c}^{x} R{\left({\textstyle t, \sqrt{ P(t)} }\right)} \, dt,</math>

where Шаблон:Math is a rational function of its two arguments, Шаблон:Math is a polynomial of degree 3 or 4 with no repeated roots, and Шаблон:Math is a constant.

In general, integrals in this form cannot be expressed in terms of elementary functions. Exceptions to this general rule are when Шаблон:Math has repeated roots, or when Шаблон:Math contains no odd powers of Шаблон:Math or if the integral is pseudo-elliptic. However, with the appropriate reduction formula, every elliptic integral can be brought into a form that involves integrals over rational functions and the three Legendre canonical forms (i.e. the elliptic integrals of the first, second and third kind).

Besides the Legendre form given below, the elliptic integrals may also be expressed in Carlson symmetric form. Additional insight into the theory of the elliptic integral may be gained through the study of the Schwarz–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 (they give the same elliptic integral). Most texts adhere to a canonical naming scheme, using the following naming conventions.

For expressing one argument:

• Шаблон:Math, the modular angle
• Шаблон:Math, the elliptic modulus or eccentricity
• Шаблон:Math, the parameter

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 as Шаблон:Math, the amplitude, or as Шаблон:Math or Шаблон:Math, where Шаблон:Math and Шаблон:Math is one of the Jacobian elliptic functions.

Specifying the value of any one of these quantities determines the others. Note that Шаблон:Math also depends on Шаблон:Math. Some additional relationships involving Шаблон:Math include <math display="block">\cos \varphi = \operatorname{cn} u, \quad \textrm{and} \quad \sqrt{1 - m \sin^2 \varphi} = \operatorname{dn} u.</math>

The latter is sometimes called the delta amplitude and written as Шаблон:Math. Sometimes the literature also refers to the complementary parameter, the complementary modulus, or the complementary modular angle. These are further defined in the article on quarter 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: <math display="block"> F(\varphi, \sin \alpha) = F\left(\varphi \mid \sin^2 \alpha\right) = F(\varphi \setminus \alpha) = F(\sin \varphi ; \sin \alpha).</math> 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 by Abramowitz and Stegun and that used in the integral tables by Gradshteyn and Ryzhik.

There are still other conventions for the notation of elliptic integrals employed in the literature. The notation with interchanged arguments, Шаблон:Math, is often encountered; and similarly Шаблон:Math for the integral of the second kind. Abramowitz and Stegun substitute the integral of the first kind, Шаблон:Math, for the argument Шаблон:Mvar in their definition of the integrals of the second and third kinds, unless this argument is followed by a vertical bar: i.e. Шаблон:Math for Шаблон:Math. Moreover, their complete integrals employ the parameter Шаблон:Math as argument in place of the modulus Шаблон:Math, i.e. Шаблон:Math rather than Шаблон:Math. And the integral of the third kind defined by Gradshteyn and Ryzhik, Шаблон:Math, puts the amplitude Шаблон:Mvar first and not the "characteristic" Шаблон:Mvar.

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's Mathematica software and Wolfram Alpha define the complete elliptic integral of the first kind in terms of the parameter Шаблон:Math, instead of the elliptic modulus Шаблон:Math.

Incomplete elliptic integral of the first kind

The incomplete elliptic integral of the first kind Шаблон:Mvar is defined as

<math display="block"> 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}}.</math>

This is the trigonometric form of the integral; substituting Шаблон:Math and Шаблон:Math, one obtains the Legendre normal form:

<math display="block"> F(x ; k) = \int_{0}^{x} \frac{dt}{\sqrt{\left(1 - t^2\right)\left(1 - k^2 t^2\right)}}.</math>

Equivalently, in terms of the amplitude and modular angle one has: <math display="block"> F(\varphi \setminus \alpha) = F(\varphi, \sin \alpha) = \int_0^\varphi \frac{d\theta}{\sqrt{1-\left(\sin \theta \sin \alpha\right)^2}}.</math>

With Шаблон:Math one has: <math display="block">F(x;k) = u;</math> demonstrating that this Jacobian 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 theoremШаблон:Citation needed: <math display="block">F\bigl[\arctan(x),k\bigr] + F\bigl[\arctan(y),k\bigr] = F\left[\arctan\left(\frac{x\sqrt{k'^2y^2+1}}{\sqrt{y^2+1}}\right) + \arctan\left(\frac{y\sqrt{k'^2x^2+1}}{\sqrt{x^2+1}}\right),k\right] </math>

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

Incomplete elliptic integral of the second kind

The incomplete elliptic integral of the second kind Шаблон:Math in trigonometric form is

<math display="block"> 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.</math>

Substituting Шаблон:Math and Шаблон:Math, one obtains the Legendre normal form:

<math display="block"> E(x;k) = \int_0^x \frac{\sqrt{1-k^2 t^2} }{\sqrt{1-t^2}}\,dt.</math>

Equivalently, in terms of the amplitude and modular angle: <math display="block"> E(\varphi \setminus \alpha) = E(\varphi, \sin \alpha) = \int_0^\varphi \sqrt{1-\left(\sin \theta \sin \alpha\right)^2} \, d\theta.</math>

Relations with the Jacobi elliptic functions include <math display="block">\begin{align} 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{align}</math>

The meridian arc length from the equator to latitude Шаблон:Math is written in terms of Шаблон:Math: <math display="block">m(\varphi) = a\left(E(\varphi,e)+\frac{d^2}{d\varphi^2}E(\varphi,e)\right),</math> where Шаблон:Math is the semi-major axis, and Шаблон:Math is the eccentricity.

The incomplete elliptic integral of the second kind has following addition theoremШаблон:Citation needed: <math display="block">E{\left[\arctan(x), k\right]}

+ E{\left[\arctan(y), k\right]}

= E{\left[\arctan\left(\frac{x\sqrt{k'^2y^2+1}}{\sqrt{y^2+1}}\right) + \arctan\left(\frac{y\sqrt{k'^2x^2+1}}{\sqrt{x^2+1}}\right),k\right]}

+ \frac{k^2xy}{k'^2x^2y^2+x^2+y^2+1}\left(\frac{x\sqrt{k'^2y^2+1}}{\sqrt{y^2+1}}+\frac{y\sqrt{k'^2x^2+1}}{\sqrt{x^2+1}}\right) </math>

The elliptic modulus can be transformed that way: <math display="block">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^2x^2}}\right),\frac{1-\sqrt{1-k^2}}{1+\sqrt{1-k^2}}\right]}

- \sqrt{1-k^2} F{\left[\arcsin(x),k\right]}
 + \frac{k^2x\sqrt{1-x^2}}{1+\sqrt{1-k^2x^2}} </math>

Incomplete elliptic integral of the third kind

The incomplete elliptic integral of the third kind Шаблон:Math is <math display="block"> \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}}</math>

or

<math display="block"> \Pi(n ; \varphi \,|\,m) = \int_{0}^{\sin \varphi} \frac{1}{1-nt^2} \frac{dt}{\sqrt{\left(1-m t^2\right)\left(1-t^2\right) }}.</math>

The number Шаблон:Math is called the characteristic and can take on any value, independently of the other arguments. Note though that the value Шаблон:Math is infinite, for any Шаблон:Math.

A relation with the Jacobian elliptic functions is <math display="block"> \Pi\bigl(n; \,\operatorname{am}(u;k); \,k\bigr) = \int_0^u \frac{dw} {1 - n \,\operatorname{sn}^2 (w;k)}.</math>

The meridian arc length from the equator to latitude Шаблон:Math is also related to a special case of Шаблон:Math:

<math display="block">m(\varphi)=a\left(1-e^2\right)\Pi\left(e^2 ; \varphi \,|\,e^2\right).</math>

Complete elliptic integral of the first kind

Файл:Mplwp complete ellipticKk.svg

Plot of the complete elliptic integral of the first kind Шаблон:Math

Elliptic Integrals are said to be 'complete' when the amplitude Шаблон:Math and therefore Шаблон:Math. The complete elliptic integral of the first kind Шаблон:Math 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 Шаблон:Math 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 Шаблон:Math 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:Шаблон:Sfn <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 Шаблон:Math:

<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 Шаблон:Math and <math>r \isin \mathbb{Q}^+</math> (where Шаблон:Mvar is the modular lambda function), then Шаблон:Math is expressible in closed form in terms of the gamma function. For example, Шаблон:Math, Шаблон:Math and Шаблон:Math give, respectively,

<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 imaginary quadratic field is sufficient. For instance, if Шаблон:Math, then Шаблон:Math and

<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>

Relation to Jacobi theta function

The relation to Jacobi's theta function is given by <math display="block">K(k)=\frac{\pi}{2}\theta_3^2(q),</math> where the nome Шаблон:Math is <math display="block">q(k) = \exp\left(-\pi \frac{K\left(\sqrt{1-k^2}\right)}{K(k)}\right).</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 Шаблон:Val for Шаблон:Math. Keeping only the first two terms is correct to 0.01 precision for Шаблон:Math.Шаблон:Citation needed

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: <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 is <math> q = q(k) = \exp[-\pi K'(k)/K(k)] </math> in its definition.

Complete elliptic integral of the second kind

Файл:Mplwp complete ellipticEk.svg

Plot of the complete elliptic integral of the second kind Шаблон:Math

The complete elliptic integral of the second kind Шаблон:Math 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 Шаблон:Math as

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

For an ellipse with semi-major axis Шаблон:Math and semi-minor axis Шаблон:Math and eccentricity Шаблон:Math, the complete elliptic integral of the second kind Шаблон:Math is equal to one quarter of the circumference Шаблон:Math of the ellipse measured in units of the semi-major axis Шаблон:Math. 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

<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.Шаблон:Sfn

Define sequences Шаблон:Mvar and Шаблон:Mvar, where Шаблон:Math, Шаблон:Math and the recurrence relations Шаблон:Math, Шаблон:Math 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 Шаблон:Math. To speed up computation further, the relation Шаблон:Math can be used.

Furthermore, if Шаблон:Math and <math>r \isin \mathbb{Q}^+</math> (where Шаблон:Mvar is the modular lambda function), then Шаблон:Math 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, Шаблон:Math, Шаблон:Math and Шаблон:Math give, respectively,

<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 Шаблон:Math.

Complete elliptic integral of the third kind

Файл:Mplwp complete ellipticPi nfixed k.svg

Plot of the complete elliptic integral of the third kind Шаблон:Math with several fixed values of Шаблон:Mvar

The complete elliptic integral of the third kind Шаблон:Math can be defined as

<math display="block">\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}}.</math>

Note that sometimes the elliptic integral of the third kind is defined with an inverse sign for the characteristic Шаблон:Math, <math display="block">\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}}.</math>

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.Шаблон:Sfn

Partial derivatives

<math display="block">\begin{align} \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{align}</math>

Jacobi zeta function

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

Legendre's relation

The Legendre's relation or Legendre Identity shows the relation of the integrals K and E of an elliptic modulus and its anti-related counterpart in an integral equation of second degree:

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

<math display="block">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}</math>

For example:

<math>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</math>

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

<math>(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 </math>

For example:

<math>\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</math>

The Legendre's relation for tangential modular counterparts results directly from the Legendre's identity for Pythagorean modular counterparts by using the Landen 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 the Chain rule these derivatives hold:

<math>\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}}</math>

<math>\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}} </math>

By using the Fundamental theorem of calculus these formulas can be generated:

<math>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 </math>

<math>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 </math>

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

<math>

\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\} \,+ </math>

<math>

+ \,\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 </math>

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

<math> \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\} =

</math>

<math> = \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 </math>

If the value <math>x = 1</math> is inserted in this integral identity, then the following identity emerges:

<math> 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 = </math>

<math> = \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} </math>

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

<math>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}</math>

Generalization for the overall case

Now the modular general case is worked out. For this purpose, the derivatives of the complete elliptic integrals are derived after the modulus <math> \varepsilon </math> and then they are combined. And then the Legendre's identity balance is determined.

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

<math>\frac{\mathrm{d}}{\mathrm{d}\varepsilon}\sqrt{1 - \varepsilon^2} = -\,\frac{\varepsilon}{\sqrt{1 - \varepsilon^2}}</math>

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

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

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

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

<math>\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]</math>

<math>\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]</math>

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, the Product rule is now applied in the following:

<math>\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]</math>

<math>\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]</math>

<math>\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]</math>

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

<math>\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</math>

In relation to the <math> \varepsilon </math> the equation balance constantly gives the value zero.

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

<math>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}</math>

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

<math>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</math>

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 <math> \varepsilon </math> and the associated function graph is therefore a horizontal straight line.

See also

• Elliptic curve
• Schwarz–Christoffel mapping
• Carlson symmetric form
• Jacobi's elliptic functions
• Weierstrass's elliptic functions
• Jacobi theta function
• Ramanujan theta function
• Arithmetic–geometric mean
• Pendulum period
• Meridian arc

References

Sources

• Шаблон:AS ref
• Шаблон:Cite book
• Шаблон:Cite journal
• Шаблон:Dlmf
• Шаблон:Cite book
• Шаблон:Cite book
• Шаблон:Cite book
• Шаблон:Cite book
• Шаблон:Cite book
• Шаблон:Citation

External links

• Шаблон:Springer
• Eric W. Weisstein, "Elliptic Integral" (Mathworld)
• Matlab code for elliptic integrals evaluation by elliptic project
• Rational Approximations for Complete Elliptic Integrals (Exstrom Laboratories)
• A Brief History of Elliptic Integral Addition Theorems