Landen's transformation

The incomplete elliptic integral of the first kind {{math|F}} is

:$F(\backslash varphi\; \backslash setminus\; \backslash alpha)\; =\; F(\backslash varphi,\; \backslash sin\; \backslash alpha)\; =\; \backslash int\_0^\backslash varphi\; \backslash frac\{d\; \backslash theta\}\{\backslash sqrt\{1-(\backslash sin\; \backslash theta\; \backslash sin\; \backslash alpha)^2\}\},$

where $\backslash alpha$ is the modular angle. Landen's transformation states that if $\backslash alpha\_0$, $\backslash alpha\_1$, $\backslash varphi\_0$, $\backslash varphi\_1$ are such that $(1\; +\; \backslash sin\backslash alpha\_1)(1\; +\; \backslash cos\backslash alpha\_0)\; =\; 2$ and $\backslash tan(\backslash varphi\_1\; -\; \backslash varphi\_0)\; =\; \backslash cos\backslash alpha\_0\; \backslash tan\; \backslash varphi\_0$, then{{AS ref}}

:$\backslash begin\{align\}$

F(\varphi_0 \setminus \alpha_0) &= (1 + \cos\alpha_0)^{-1} F(\varphi_1 \setminus \alpha_1) \\

&= \tfrac{1}{2}(1 + \sin\alpha_1) F(\varphi_1 \setminus \alpha_1).

\end{align}

Landen's transformation can similarly be expressed in terms of the elliptic modulus $k\; =\; \backslash sin\backslash alpha$ and its complement $k\text{'}\; =\; \backslash cos\backslash alpha$.

In Gauss's formulation, the value of the integral

:$I\; =\; \backslash int\_0^\{\backslash frac\{\backslash pi\}\{2\}\}\backslash frac\{1\}\{\backslash sqrt\{a^2\; \backslash cos^2(\backslash theta)\; +\; b^2\; \backslash sin^2(\backslash theta)\}\}\; \backslash ,\; d\; \backslash theta$

is unchanged if $a$ and $b$ are replaced by their arithmetic and geometric means respectively, that is

:$a\_1\; =\; \backslash frac\{a\; +\; b\}\{2\},\backslash qquad\; b\_1\; =\; \backslash sqrt\{a\; b\},$

:$I\_1\; =\; \backslash int\; \_0^\{\backslash frac\{\backslash pi\}\{2\}\}\backslash frac\{1\}\{\backslash sqrt\{a\_1^2\; \backslash cos^2(\backslash theta)\; +\; b\_1^2\; \backslash sin^2(\backslash theta)\}\}\; \backslash ,\; d\; \backslash theta.$

Therefore,

:$I=\backslash frac\{1\}\{a\}K\backslash left(\backslash frac\{\backslash sqrt\{a^2\; -\; b^2\}\}\{a\}\backslash right),$

:$I\_1=\backslash frac\{2\}\{a+b\}K\backslash left(\backslash frac\{a-b\}\{a+b\}\backslash right).$

From Landen's transformation we conclude

:$K\backslash left(\backslash frac\{\backslash sqrt\{a^2\; -\; b^2\}\}\{a\}\backslash right)=\backslash frac\{2a\}\{a+b\}K\backslash left(\backslash frac\{a-b\}\{a+b\}\backslash right)$

and $I\_1=I$.

The transformation may be effected by integration by substitution. It is convenient to first cast the integral in an algebraic form by a substitution of $\backslash theta\; =\; \backslash arctan\; (x/b)$, $d\; \backslash theta\; =\; (\backslash cos^\{2\}(\backslash theta)/b)\; d\; x$ giving

:$I\; =\; \backslash int\; \_0^\{\backslash frac\{\backslash pi\}\{2\}\}\backslash frac\{1\}\{\backslash sqrt\{a^2\; \backslash cos^2(\backslash theta)\; +\; b^2\; \backslash sin^2(\backslash theta)\}\}\; \backslash ,\; d\; \backslash theta\; =\; \backslash int\; \_0^\backslash infty\; \backslash frac\{1\}\{\backslash sqrt\{(x^2\; +\; a^2)\; (x^2\; +\; b^2)\}\}\; \backslash ,\; dx$

A further substitution of $x\; =\; t\; +\; \backslash sqrt\{t^\{2\}\; +\; a\; b\}$ gives the desired result

:

& = \int _{- \infty}^\infty \frac{1}{2 \sqrt{\left( t^2 + \left( \frac{a + b}{2}\right)^2 \right) (t^2 + a b)}} \, dt \\

& = \int _0^\infty\frac{1}{\sqrt{\left( t^2 + \left( \frac{a + b}{2}\right)^2\right) \left(t^2 + \left(\sqrt{a b}\right)^2\right)}} \, dt \end{align}

This latter step is facilitated by writing the radical as

:$\backslash sqrt\{(x^2\; +\; a^2)\; (x^2\; +\; b^2)\}\; =\; 2x\; \backslash sqrt\{t^2\; +\; \backslash left(\; \backslash frac\{a\; +\; b\}\{2\}\backslash right)^2\}$

and the infinitesimal as

:$dx\; =\; \backslash frac\{x\}\{\backslash sqrt\{t^2\; +\; a\; b\}\}\; \backslash ,\; dt$

so that the factor of $x$ is recognized and cancelled between the two factors.

If the transformation is iterated a number of times, then the parameters $a$ and $b$ converge very rapidly to a common value, even if they are initially of different orders of magnitude. The limiting value is called the arithmetic-geometric mean of $a$ and $b$, $\backslash operatorname\{AGM\}(a,b)$. In the limit, the integrand becomes a constant, so that integration is trivial

:$I\; =\; \backslash int\; \_0^\{\backslash frac\{\backslash pi\}\{2\}\}\; \backslash frac\{1\}\{\backslash sqrt\{a^2\; \backslash cos^2(\backslash theta)\; +\; b^2\; \backslash sin^2(\backslash theta)\}\}\; \backslash ,\; d\backslash theta\; =\; \backslash int\; \_0^\{\backslash frac\{\backslash pi\}\{2\}\}\backslash frac\{1\}\{\backslash operatorname\{AGM\}(a,b)\}\; \backslash ,\; d\backslash theta\; =\; \backslash frac\{\backslash pi\}\{2\; \backslash operatorname\{AGM\}(a,b)\}$

The integral may also be recognized as a multiple of Legendre's complete elliptic integral of the first kind. Putting $b^2\; =\; a^2\; (1\; -\; k^2)$

:$I\; =\; \backslash frac\{1\}\{a\}\; \backslash int\; \_0^\{\backslash frac\{\backslash pi\}\{2\}\}\; \backslash frac\{1\}\{\backslash sqrt\{1\; -\; k^2\; \backslash sin^2(\backslash theta)\}\}\; \backslash ,\; d\backslash theta\; =\; \backslash frac\{1\}\{a\}\; F\backslash left(\; \backslash frac\{\backslash pi\}\{2\},k\backslash right)\; =\; \backslash frac\{1\}\{a\}\; K(k)$

Hence, for any $a$, the arithmetic-geometric mean and the complete elliptic integral of the first kind are related by

:$K(k)\; =\; \backslash frac\{\backslash pi\; \}\{2\; \backslash operatorname\{AGM\}(1,\; \backslash sqrt\{1\; -\; k^2\})\}$

By performing an inverse transformation (reverse arithmetic-geometric mean iteration), that is

:$a\_\{-1\}\; =\; a\; +\; \backslash sqrt\{a^2\; -\; b^2\}\; \backslash ,$

:$b\_\{-1\}\; =\; a\; -\; \backslash sqrt\{a^2\; -\; b^2\}\; \backslash ,$

:$\backslash operatorname\{AGM\}(a,b)\; =\; \backslash operatorname\{AGM\}\backslash left(a\; +\; \backslash sqrt\{a^2\; -\; b^2\},a\; -\; \backslash sqrt\{a^2\; -\; b^2\}\backslash right)\; \backslash ,$

the relationship may be written as

:$K(k)\; =\; \backslash frac\{\backslash pi\}\{2\; \backslash operatorname\{AGM\}(1\; +\; k,\; 1\; -\; k)\}\; \backslash ,$

which may be solved for the AGM of a pair of arbitrary arguments;

:$\backslash operatorname\{AGM\}(u,v)\; =\; \backslash frac\{\backslash pi\; (u\; +\; v)\}\{4\; K\backslash left(\; \backslash frac\{u\; -\; v\}\{v\; +\; u\}\backslash right)\}.$

{{Reflist}}

• Louis V. King [https://archive.org/details/onthenumerical032686mbp On The Direct Numerical Calculation Of Elliptic Functions And Integrals] (Cambridge University Press, 1924)

Category:Elliptic functions