Landen's transformation

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Landen's transformation is a mapping of the parameters of an elliptic integral, which shows how the value of the integral, changes when its parameters:amplitude and modular angle changes following some dependency. As a special case, we can see when the transformation does not change the value of the integral.

It was originally due to John Landen, although independently rediscovered by Carl Friedrich Gauss.

For example,the incomplete elliptic integral of the first kind $F$ is

$F(\varphi ,k)=F(\varphi \,|\,k^{2})=F(\sin \varphi ;k)=\int _{0}^{\varphi }{\frac {d\theta }{{\sqrt {1-k^{2}\sin ^{2}\theta }}}}.$

If parameters will be φ1 and k1 then $F(\varphi 1,k1)=\int _{0}^{{\varphi {1}}}{\frac {d\theta }{{\sqrt {1-k1^{2}\sin ^{2}\theta }}}}.$

Landen's transformation shows how one can calculate second integral through first and formula connecting k and k1.

A comprehensive list of transformations included in the tables 21.6-2 and 21.6-3 "Mathematical Handbook for scientists and engineers" by G.A. Korn and T.M. Korn. Accordingly 21.6-3

$K(k)={\frac {2}{1+k1}}K({\frac {1-k1}{1+k1}})$

(aa)

Where

$k1={{\sqrt {1-k^{2}}}}$

$K(k)=\int _{0}^{{{\frac {\pi }{2}}}}{\frac {d\theta }{{\sqrt {1-k^{2}\sin ^{2}\theta }}}}.$

Consider an example when the transformation does not change the value of the integral.

Let

$I=\int _{0}^{{{\frac {\pi }{2}}}}{\frac {1}{{\sqrt {a^{2}\cos ^{2}(\theta )+b^{2}\sin ^{2}(\theta )}}}}\,d\theta$

and $\scriptstyle {a}$ and $\scriptstyle {b}$ are replaced by their arithmetic and geometric means respectively, that is

$a_{1}={\frac {a+b}{2}},\qquad b_{1}={\sqrt {ab}}.\,$

$I_{1}=\int _{0}^{{{\frac {\pi }{2}}}}{\frac {1}{{\sqrt {a_{1}^{2}\cos ^{2}(\theta )+b_{1}^{2}\sin ^{2}(\theta )}}}}\,d\theta$

Obviously

$I={\frac {1}{a}}K({\frac {{\sqrt {(a^{2}-b^{2})}}}{a}})$

$I_{1}={\frac {2}{a+b}}K({\frac {a-b}{a+b}})$

Accordingly formula (aa)

$K({\frac {{\sqrt {(a^{2}-b^{2})}}}{a}})={\frac {2a}{a+b}}K({\frac {a-b}{a+b}})$

(aaa)

As follows from the formula (aaa)

$I_{1}=I$

The same equation can be proved using a simple mathematical analysis.

The transformation, may be achieved purely by integration by substitution. It is convenient to first cast the integral in an algebraic form by a substitution of $\scriptstyle {\theta =\arctan \left({\frac {x}{b}}\right)}$ , $\scriptstyle {d\theta =\left({\frac {1}{b}}\cos ^{{2}}(\theta )\right)dx}$ giving

$I=\int _{0}^{{{\frac {\pi }{2}}}}{\frac {1}{{\sqrt {a^{2}\cos ^{2}(\theta )+b^{2}\sin ^{2}(\theta )}}}}\,d\theta =\int _{0}^{\infty }{\frac {1}{{\sqrt {(x^{2}+a^{2})(x^{2}+b^{2})}}}}\,dx$

A further substitution of $\scriptstyle {x=t+{\sqrt {t^{{2}}+ab}}}$ gives the desired result (in the algebraic form)

${\begin{aligned}I&=\int _{0}^{\infty }{\frac {1}{{\sqrt {(x^{2}+a^{2})(x^{2}+b^{2})}}}}\,dx\\&=\int _{{-\infty }}^{\infty }{\frac {1}{2{\sqrt {\left(t^{2}+\left({\frac {a+b}{2}}\right)^{2}\right)(t^{2}+ab)}}}}\,dt\\&=\int _{0}^{\infty }{\frac {1}{{\sqrt {\left(t^{2}+\left({\frac {a+b}{2}}\right)^{2}\right)\left(t^{2}+\left({\sqrt {ab}}\right)^{2}\right)}}}}\,dt\end{aligned}}$

This latter step is facilitated by writing the radical as

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

and the infinitesimal as

$dx={\frac {x}{{\sqrt {t^{2}+ab}}}}\,dt$

so that the factor of $\scriptstyle {x}$ is easily recognized and cancelled between the two factors.

Arithmetic-geometric mean and Legendre's first integral

If the transformation is iterated a number of times, then the parameters $\scriptstyle {a}$ and $\scriptstyle {b}$ 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 $\scriptstyle {a}$ and $\scriptstyle {b}$ , $\scriptstyle {\operatorname {AGM}(a,b)}$ . In the limit, the integrand becomes a constant, so that integration is trivial

$I=\int _{0}^{{{\frac {\pi }{2}}}}{\frac {1}{{\sqrt {a^{2}\cos ^{2}(\theta )+b^{2}\sin ^{2}(\theta )}}}}\,d\theta =\int _{0}^{{{\frac {\pi }{2}}}}{\frac {1}{\operatorname {AGM}(a,b)}}\,d\theta ={\frac {\pi }{2\,\operatorname {AGM}(a,b)}}$

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

$I={\frac {1}{a}}\int _{0}^{{{\frac {\pi }{2}}}}{\frac {1}{{\sqrt {1-k^{2}\sin ^{2}(\theta )}}}}\,d\theta ={\frac {1}{a}}F\left({\frac {\pi }{2}},k\right)={\frac {1}{a}}K(k)$

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

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

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

$a_{{-1}}=a+{\sqrt {a^{2}-b^{2}}}\,$

$b_{{-1}}=a-{\sqrt {a^{2}-b^{2}}}\,$

$\operatorname {AGM}(a,b)=\operatorname {AGM}(a+{\sqrt {a^{2}-b^{2}}},a-{\sqrt {a^{2}-b^{2}}})\,$

the relationship may be written as

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

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

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

The definition adopted here for $\scriptstyle {K(k)}$ , differs from that used in the arithmetic-geometric mean article, such that $\scriptstyle {K(k)}$ here is $\scriptstyle {K(m^{{2}})}$ in that article.

References

Louis V. King On The Direct Numerical Calculation Of Elliptic Functions And Integrals (Cambridge University Press, 1924)

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