Elliptic integral

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 (c. 1750). Their name originates from their connection with the problem of finding the arc length of an ellipse.

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

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f(x) = \int_{c}^{x} R{\left({\textstyle t, \sqrt{ P(t)} }\right)} \, dt,}

where R is a rational function of its two arguments, P is a polynomial of degree 3 or 4 with no repeated roots, and c is a constant.

In general, integrals in this form cannot be expressed in terms of elementary functions. Exceptions to this general rule are when P has repeated roots, when R(x, y) contains no odd powers of y, and when 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, 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 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 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, the modular angle
• k = sin α, the elliptic modulus or eccentricity
• m = k² = sin² α, 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 φ, the amplitude, or as x or u, where x = sin φ = sn u and sn is one of the Jacobian elliptic functions.

Specifying the value of any one of these quantities determines the others. Note that u also depends on m. Some additional relationships involving u include Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \cos \varphi = \operatorname{cn} u, \quad \textrm{and} \quad \sqrt{1 - m \sin^2 \varphi} = \operatorname{dn} u.}

The latter is sometimes called the delta amplitude and written as Δ(φ) = dn u. 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: Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 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 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, F(k, φ), is often encountered; and similarly E(k, φ) for the integral of the second kind. Abramowitz and Stegun substitute the integral of the first kind, F(φ, k), for the argument φ 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) for E(φ | k²). Moreover, their complete integrals employ the parameter k² as argument in place of the modulus k, i.e. K(k²) rather than K(k). And the integral of the third kind defined by Gradshteyn and Ryzhik, Π(φ, n, k), puts the amplitude φ 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's Mathematica software and Wolfram Alpha define the complete elliptic integral of the first kind in terms of the parameter m, instead of the elliptic modulus k.

Incomplete elliptic integral of the first kind

The incomplete elliptic integral of the first kind F is defined as

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 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; substituting t = sin θ and x = sin φ, one obtains Jacobi's algebraic form:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 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: Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle F(\varphi \setminus \alpha) = F(\varphi, \sin \alpha) = \int_0^\varphi \frac{d\theta}{\sqrt{1-\left(\sin \theta \sin \alpha\right)^2}}.}

With x = sn(u, k) one has: Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle F(x;k) = u;} 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 theorem: Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle F\left[\arctan(x),k\right] + F\left[\arctan(y),k\right] = 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] }

The elliptic modulus can be transformed that way: Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle F\left[\arcsin(x),k\right] = \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]} }

Incomplete elliptic integral of the second kind

The incomplete elliptic integral of the second kind E in Legendre's trigonometric form is

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 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.}

Substituting t = sin θ and x = sin φ, 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: Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle E(\varphi \setminus \alpha) = E(\varphi, \sin \alpha) = \int_0^\varphi \sqrt{1-\left(\sin \theta \sin \alpha\right)^2} \, d\theta.}

Relations with the Jacobi elliptic functions include Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \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}}

The meridian arc length from the equator to latitude φ is written in terms of E: Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle m(\varphi) = a\left(E(\varphi,e)+\frac{d^2}{d\varphi^2}E(\varphi,e)\right),} where a is the semi-major axis, and e is the eccentricity.

The incomplete elliptic integral of the second kind has following addition theorem: Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{align} &E{\left[\arctan(x), k\right]} + E{\left[\arctan(y), k\right]}\\[1ex] &\quad = 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]}\\[1ex] &\qquad + \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) \end{align}}

The elliptic modulus can be transformed that way: Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{align} 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]}\\[.5ex] &\quad - \sqrt{1-k^2} F{\left[\arcsin(x),k\right]} + \frac{k^2x\sqrt{1-x^2}}{1+\sqrt{1-k^2x^2}} \end{align}}

Incomplete elliptic integral of the third kind

The incomplete elliptic integral of the third kind Π is Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \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

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \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) }}.}

The number n is called the characteristic and can take on any value, independently of the other arguments. Note though that the value Π(1; π/2 | m) is infinite, for any m.

A relation with the Jacobian elliptic functions is Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \Pi\left(n; \,\operatorname{am}(u;k); \,k\right) = \int_0^u \frac{dw} {1 - n \,\operatorname{sn}^2 (w;k)}.}

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

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle m(\varphi)=a\left(1-e^2\right)\Pi\left(e^2 ; \varphi \,|\,e^2\right).}

Complete elliptic integral of the first kind

File:Mplwp complete ellipticKk.svg

Plot of the complete elliptic integral of the first kind K(k)

Elliptic Integrals are said to be 'complete' when the amplitude φ = π/2 and therefore x = 1. The complete elliptic integral of the first kind K may thus be defined as Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 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 a power series Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 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 \left(P_{2 n}(0)\right)^2 k^{2n},}

where P_n is the Legendre polynomials, which is equivalent to

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 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),}

where n!! denotes the double factorial. In terms of the Gauss hypergeometric function, the complete elliptic integral of the first kind can be expressed as

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle K(k) = \tfrac{\pi}{2} \,{}_2F_1 \left(\tfrac{1}{2}, \tfrac{1}{2}; 1; k^2\right).}

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: Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle K(k) = \frac{\pi}{2\operatorname{agm}\left(1,\sqrt{1-k^2}\right)}.}

Therefore, the modulus can be transformed as:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{align} 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}}{1+\sqrt{1-k^2}}\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}}

This expression is valid for all Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle n \isin \mathbb{N}} and 0 ≤ k ≤ 1:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 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

If k² = λ(i√r) and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle r \isin \mathbb{Q}^+} (where λ is the modular lambda function), then K(k) is expressible in closed form in terms of the gamma function. For example, r = 2, r = 3 and r = 7 give, respectively,

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 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

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle K\left(\frac{\sqrt{3}-1}{2\sqrt{2}}\right)=\frac{1}{8\pi}\sqrt[4]{3}\,\sqrt[3]{4}\,\Gamma\left(\frac{1}{3}\right)^3}

and

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 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 Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{iK'}{K}=\frac{iK\left(\sqrt{1-k^2}\right)}{K(k)}} be in an imaginary quadratic field is sufficient. For instance, if k = e^5πi/6, then iKTemplate:Prime/K = e^2πi/3 and

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 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 Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \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 Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{\Gamma\left(\frac{1}{k}\right)^{k/2}}{\sqrt{\pi}} } is a period for all even divisors Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle k} of Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 24} :

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{align} \frac{\Gamma\left(\frac{1}{4}\right)^2}{\sqrt{\pi}} &= 4 \, K\left( \tfrac{1}{\sqrt{2}} \right) \\[1ex] \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 \\[1ex] \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 \\[1ex] \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 \\[1ex] \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\!\left( (2 - \sqrt{3})(\sqrt{3} - \sqrt{2}) \right)^6 \end{align}}

Asymptotic expressions

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 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 than 3×10⁻⁴ for k < 1/2. Keeping only the first two terms is correct to 0.01 precision for k < 1/2.

Differential equation

The differential equation for the elliptic integral of the first kind is Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \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 Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle K\left(\sqrt{1-k^2}\right)} . This solution satisfies the relation Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{d}{dk}K(k) = \frac{E(k)}{k\left(1-k^2\right)}-\frac{K(k)}{k}.}

Continued fraction

A continued fraction expansion is: Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{align} \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}}}}}, \end{align}} where the nome is Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 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 the parameter Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle m} instead, because the squaring function introduces problems when inverting in the complex plane. So let Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle K[m]=\int_0^{\pi/2}\dfrac{d\theta}{\sqrt{1-m\sin^2\theta}}} and let Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \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,} Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \theta_3(\tau)=1+2\sum_{n=1}^\infty q^{n^2},\quad q=e^{\pi i\tau},\,\operatorname{Im}\tau >0} be the theta functions.

The equation Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \tau=i\frac{K[1-m]}{K[m]}} can then be solved (provided that a solution Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle m} exists) by Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle m=\frac{\theta_2(\tau)^4}{\theta_3(\tau)^4}} which is in fact the modular lambda function.

For the purposes of computation, the error analysis is given by Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \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}\;} Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \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}\;} where Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle N\in\mathbb{Z}_{\ge 1}} and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \operatorname{Im}\tau >0} .

Also Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle K[m]=\frac{\pi}{2}\theta_3(\tau )^2,\quad \tau=i\frac{K[1-m]}{K[m]}} where Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle m\in\mathbb{C}\setminus\{0,1\}} .

Complete elliptic integral of the second kind

File:Mplwp complete ellipticEk.svg

Plot of the complete elliptic integral of the second kind E(k)

The complete elliptic integral of the second kind E is defined as

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 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 kind E(φ,k) as

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle E(k) = E\left(\tfrac{\pi}{2},k\right) = E(1;k).}

For an ellipse with semi-major axis a and semi-minor axis b and eccentricity e = √1 − b²/a², the complete elliptic integral of the second kind E(e) is equal to one quarter of the circumference C of the ellipse measured in units of the semi-major axis a. In other words:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle C = 4 a E(e).}

The complete elliptic integral of the second kind can be expressed as a power series

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 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

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 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 the Gauss hypergeometric function, the complete elliptic integral of the second kind can be expressed as

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle E(k) = \tfrac{\pi}{2} \,{}_2F_1 \left(\tfrac12, -\tfrac12; 1; k^2 \right).}

The modulus can be transformed that way: Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 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 the arithmetic–geometric mean.

Define sequences a_n and g_n, where a₀ = 1, g₀ = √1 − k² = kTemplate:Prime and the recurrence relations a_{n + 1} = a_n + g_n/2, g_{n + 1} = √a_n g_n hold. Furthermore, define Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle c_n=\sqrt{\left|a_n^2-g_n^2\right|}.}

By definition,

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle a_\infty = \lim_{n\to\infty} a_n = \lim_{n\to\infty} g_n = \operatorname{agm}\left(1, \sqrt{1-k^2}\right).}

Also

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \lim_{n\to\infty} c_n=0.}

Then

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 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 relation c_{n + 1} = c_n²/4a_{n + 1} can be used.

Furthermore, if k² = λ(i√r) and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle r \isin \mathbb{Q}^+} (where λ is the modular lambda function), then E(k) is expressible in closed form in terms of Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 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 and r = 7 give, respectively,

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle E{\left(\frac{1}{\sqrt{2}}\right)} = \frac{1}{2} K{\left(\frac{1}{\sqrt{2}}\right)} + \frac{\pi}{4 K{\left(\frac{1}{\sqrt{2}}\right)}},}

and

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 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}}{12 K{\left(\frac{\sqrt{3}-1}{2\sqrt{2}}\right)}},}

and

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 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

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{dE(k)}{dk} = \frac{E(k)-K(k)}{k}} Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \left(k^2-1\right) \frac{d}{dk} \left( k \;\frac{dE(k)}{dk} \right) = k E(k)}

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

Complete elliptic integral of the third kind

File:Mplwp complete ellipticPi nfixed k.svg

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

The complete elliptic integral of the third kind Π can be defined as

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \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 the characteristic n, Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \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.

Partial derivatives

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \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}}

Jacobi zeta function

In 1829, Jacobi defined the Jacobi zeta function: Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle Z(\varphi,k)=E(\varphi,k)-\frac{E(k)}{K(k)}F(\varphi,k).} It is periodic in Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \varphi} with minimal period Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \pi} . It is related to the Jacobi zn function by Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle Z(\varphi,k)=\operatorname{zn}(F(\varphi,k),k)} . In the literature (e.g. Whittaker and Watson (1927)), sometimes Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle Z} means Wikipedia's Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \operatorname{zn}} . Some authors (e.g. King (1924)) use Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle Z} for both Wikipedia's Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle Z} and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \operatorname{zn}} .

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:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 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: Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 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:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (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: Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \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 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:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{\mathrm{d}}{\mathrm{d}y} \,K{\left(\frac{1}{\sqrt{2}}\right)} - F\left[\arccos (xy);\frac{1}{\sqrt{2}}\right] = \frac{\sqrt{2}\,x}{\sqrt{1 - x^4 y^4}}} Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{\mathrm{d}}{\mathrm{d}y} \,2E{\left(\frac{1}{\sqrt{2}}\right)} - K{\left(\frac {1}{\sqrt{2}}\right)} - 2E\left[\arccos(xy);\frac{1}{\sqrt{2}}\right] + F\left[\arccos(xy );\frac{1}{\sqrt{2}}\right] = \frac{\sqrt{2}\,x^3 y^2}{\sqrt{1 - x^4 y^4}} }

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

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle K{\left(\frac{1}{\sqrt{2}}\right)} - F{\left(\arccos (x);\frac{1}{\sqrt{2}}\right)} = \int_0^1 \frac{\sqrt{2}\,x}{\sqrt{1 - x^4 y^4}} \,\mathrm{d}y} Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 2E{\left(\frac{1}{\sqrt{2}}\right)} - K{\left(\frac {1}{\sqrt{2}}\right)} - 2E{\left(\arccos(x); \frac{1}{\sqrt{2}}\right)} + F{\left(\arccos(x);\frac{1}{\sqrt{2}}\right)} = \int_0^1 \frac{\sqrt{2}\,x^3 y^2}{\sqrt{1 - x^4 y^4}} \,\mathrm{d}y }

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

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{\sqrt{2}}{\sqrt{1 - x^4}} \left\{2E\left(\frac{1}{\sqrt{2}}\right) - K\left(\frac{1}{\sqrt{2}}\right) - 2E\left[\arccos(x);\frac{1}{\sqrt{2}}\right] + F\left[\arccos( x);\frac{1}{\sqrt{2}}\right]\right\} \,+ } Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle + \,\frac{\sqrt{2} \,x^2}{\sqrt{1 - x^4}} \left\{K\left(\frac{1}{\sqrt{2}}\right) - F\left[\arccos(x);\frac{1}{\sqrt{2}}\right]\right\} = \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 the Product rule this formula results:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{align} &\left[K{\left(\frac{1}{\sqrt{2}}\right)} - F{\left(\arccos(x);\frac{1}{\sqrt{2}}\right)}\right] \left[2E{\left(\frac{1}{\sqrt{2}}\right)} - K{\left(\frac{1}{\sqrt{2}}\right)} - 2E{\left(\arccos(x);\frac{1}{\sqrt{2}}\right)} + F{\left(\arccos(x);\frac{1}{\sqrt{2}}\right)}\right] \\[1ex] &=\int_0^1 \frac{1}{y^2}(y^2 + 1)\left[\text{artanh}(y^2) - \text{artanh} \left(\frac{\sqrt{1 - x^4}\,y^2}{\sqrt{1 - x^4 y^4}}\right)\right] \mathrm{d}y \end{align} }

If the value Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x = 1} is inserted in this integral identity, then the following identity emerges:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{align} K{\left(\frac{1}{\sqrt{2}}\right)} \left[2\,E\left(\frac{1}{\sqrt{2}}\right) - K\left (\frac{1}{\sqrt{2}}\right)\right] &= \int_0^1 \frac{1}{y^2}(y^2 + 1) \,\text{artanh}(y^2) \,\mathrm{d}y \\ &= \left[2\arctan(y) - \frac{1}{y}(1 - y^2)\,\text{artanh}(y^2)\right]_{y = 0}^{y = 1} \\ &= 2\arctan(1) = \frac{\pi}{2} \end{align} }

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

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 2E\left(\frac{1}{\sqrt{2}}\right)K\left(\frac{1}{\sqrt{2}}\right) - K\left(\frac{1}{\sqrt{2}}\right)^2 = \frac{\pi}{2}}

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 Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \varepsilon } 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:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \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:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{\mathrm{d}}{\mathrm{d}\varepsilon} K(\varepsilon) = \frac{1}{\varepsilon(1-\varepsilon^2)} \left[E( \varepsilon) - (1-\varepsilon^2)K(\varepsilon)\right]} Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{\mathrm{d}}{\mathrm{d}\varepsilon} E(\varepsilon) = - \,\frac{1}{\varepsilon}\left[K(\varepsilon) - E (\varepsilon)\right]}

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

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{\mathrm{d}}{\mathrm{d}\varepsilon}K(\sqrt{1 - \varepsilon^2}) = \frac{1}{\varepsilon(1-\varepsilon^ 2)} \left[\varepsilon^2 K(\sqrt{1 - \varepsilon^2}) - E(\sqrt{1 - \varepsilon^2})\right]} Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{\mathrm{d}}{\mathrm{d}\varepsilon }E(\sqrt{1 - \varepsilon ^2}) = \frac{\varepsilon }{1 - \varepsilon ^2} \left[K(\sqrt{1 - \varepsilon^2}) - E(\sqrt{1 - \varepsilon^2})\right]}

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:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{\mathrm{d}}{\mathrm{d}\varepsilon}K(\varepsilon)E(\sqrt{1 - \varepsilon^2}) = \frac{1}{\varepsilon( 1-\varepsilon^2)} \left[E(\varepsilon)E(\sqrt{1 - \varepsilon^2}) - K(\varepsilon)E(\sqrt{1 - \varepsilon^2}) + \varepsilon^2 K(\varepsilon)K(\sqrt{1 - \varepsilon^2})\right]} Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{\mathrm{d}}{\mathrm{d}\varepsilon}E(\varepsilon)K(\sqrt{1 - \varepsilon^2}) = \frac{1}{\varepsilon( 1-\varepsilon^2)} \left[- E(\varepsilon)E(\sqrt{1 - \varepsilon^2}) + E(\varepsilon)K(\sqrt{1 - \varepsilon^2}) - (1 - \varepsilon^2) K(\varepsilon)K(\sqrt{1 - \varepsilon^2})\right]} Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{\mathrm{d}}{\mathrm{d}\varepsilon}K(\varepsilon)K(\sqrt{1 - \varepsilon^2}) = \frac{1}{\varepsilon( 1-\varepsilon^2)} \left[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})\right]}

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

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{\mathrm{d}}{\mathrm{d}\varepsilon} \left[K(\varepsilon)E(\sqrt{1 - \varepsilon^2}) + E(\varepsilon)K (\sqrt{1 - \varepsilon^2}) - K(\varepsilon)K(\sqrt{1 - \varepsilon^2})\right] = 0}

In relation to the Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \varepsilon } the equation balance constantly gives the value zero.

The previously determined result shall be combined with the Legendre equation to the modulus Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \varepsilon = 1/\sqrt{2}} that is worked out in the section before:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 2E\left(\frac{1}{\sqrt{2}}\right)K\left(\frac{1}{\sqrt{2}}\right) - K\left(\frac{1}{\sqrt{2}}\right)^2 = \frac{\pi}{2}}

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

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 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 Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \varepsilon } and the associated function graph is therefore a horizontal straight line.

See also

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

References

Sources

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• 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.
• Template:Dlmf
• 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 from the original (PDF) on 2011-07-14. Retrieved 2016-07-24.
• 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 from the original on 2011-08-11, retrieved 2011-08-09

External links

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• 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