Editing Elliptic integral (section)

==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<ref>{{cite web|access-date=2022-11-29|language=de|title=Legendre-Relation|url=https://www.spektrum.de/lexikon/mathematik/legendre-relation/5873}}<!-- auto-translated by Module:CS1 translator --></ref><ref>{{cite web|access-date=2022-11-29|title=Legendre Relation|url=https://archive.lib.msu.edu/crcmath/math/math/l/l179.htm}}<!-- auto-translated by Module:CS1 translator --></ref> 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's transformation|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<ref>{{cite web|access-date=2023-02-10|language=en|title=integration - Proving Legendres Relation for elliptic curves|url=https://math.stackexchange.com/questions/701515/proving-legendres-relation-for-elliptic-curves}}<!-- auto-translated by Module:CS1 translator --></ref><ref>{{citation|access-date=2023-02-10|author=Internet Archive|date=1991|isbn=0-387-97509-8|publisher=New York : Springer-Verlag|title=Paul Halmos celebrating 50 years of mathematics|url=https://archive.org/details/paulhalmoscelebr0000unse}}<!-- auto-translated by Module:CS1 translator --></ref>  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.