Editing Elliptic integral (section)

=== 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.