Tarski's high school algebra problem

2012-05-03T07:37:18Z

213.122.35.163: /* Generalisations */

'''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 {{math|F}} is

: <math> F(\varphi,k) = F(\varphi \,|\, k^2) = F(\sin \varphi ; k) = \int_0^\varphi \frac {d\theta}{\sqrt{1 - k^2 \sin^2 \theta}}.</math>
If parameters will be φ1 and k1 then
<math> F(\varphi1,k1) = \int_0^{\varphi{1}} \frac {d\theta}{\sqrt{1 - k1^2 \sin^2 \theta}}.</math>
: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
:<math>K(k)=\frac{2}{1+k1} K(\frac{1-k1}{1+k1})</math>
:(aa)
Where
:<math> k1={\sqrt{1 - k^2}} </math>
:<math> K(k) = \int_0^{\frac{\pi}{2}} \frac {d\theta}{\sqrt{1 - k^2 \sin^2 \theta}}.</math>

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

:<math>I = \int _0^{\frac{\pi}{2}}\frac{1}{\sqrt{a^2 \cos^2(\theta) + b^2 \sin^2(\theta)}} \, d \theta</math>

and <math>\scriptstyle{a}</math> and <math>\scriptstyle{b}</math> are replaced by their [[arithmetic mean|arithmetic]] and [[geometric mean]]s respectively, that is

:<math>a_1 = \frac{a + b}{2},\qquad b_1 = \sqrt{a b}.\,</math>
<math>I_1 = \int _0^{\frac{\pi}{2}}\frac{1}{\sqrt{a_1^2 \cos^2(\theta) + b_1^2 \sin^2(\theta)}} \, d \theta</math>
:Obviously
:<math>I=\frac{1}{a}K(\frac{\sqrt{(a^2 - b^2)}}{a}) </math>
:<math>I_1=\frac{2}{a+b}K(\frac{a-b}{a+b}) </math>
:Accordingly formula (aa)
:<math>K(\frac{\sqrt{(a^2 - b^2)}}{a})=\frac{2a}{a+b}K(\frac{a-b}{a+b}) </math>
:(aaa)
: As follows from the formula (aaa)
:<math>I_1=I</math>
: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 function|algebraic]] form by a substitution of <math>\scriptstyle{\theta = \arctan\left( \frac{x}{b}\right)}</math>, <math>\scriptstyle{d \theta = \left( \frac{1}{b}\cos^{2}(\theta)\right) d x}</math> giving

:<math>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</math>

A further substitution of <math>\scriptstyle{x = t + \sqrt{t^{2} + a b}}</math> gives the desired result (in the algebraic form)

:<math>\begin{align}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 + 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}</math>

This latter step is facilitated by writing the radical as

:<math>\sqrt{(x^2 + a^2) (x^2 + b^2)} = 2x \sqrt{t^2 + \left( \frac{a + b}{2}\right)^2}</math>

and the infinitesimal as

:<math> dx = \frac{x}{\sqrt{t^2 + a b}} \, dt</math>

so that the factor of <math>\scriptstyle{x}</math> 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 <math>\scriptstyle{a}</math> and <math>\scriptstyle{b}</math> 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 <math>\scriptstyle{a}</math> and <math>\scriptstyle{b}</math>, <math>\scriptstyle{\operatorname{AGM}(a,b)}</math>. In the limit, the integrand becomes a constant, so that integration is trivial

:<math>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)}</math>

The integral may also be recognized as a multiple of [[Legendre form|Legendre's complete elliptic integral of the first kind]]. Putting <math>\scriptstyle{b^2 = a^2 (1 - k^2)}</math>

:<math>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)</math>

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

:<math>K(k) = \frac{\pi a}{2 \, \operatorname{AGM}(a,a \sqrt{1 - k^2})}</math>

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

:<math>a_{-1} = a + \sqrt{a^2 - b^2} \, </math>

:<math>b_{-1} = a - \sqrt{a^2 - b^2} \, </math>

:<math>\operatorname{AGM}(a,b) = \operatorname{AGM}(a + \sqrt{a^2 - b^2},a - \sqrt{a^2 - b^2}) \, </math>

the relationship may be written as

:<math>K(k) = \frac{\pi a}{2 \, \operatorname{AGM}(a (1 + k),a (1 - k))} \, </math>

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

:<math> \operatorname{AGM}(u,v) = \frac{\pi (u + v)}{4 K\left( \frac{u - v}{v + u}\right)}.</math>

:''The definition adopted here for ''<math>\scriptstyle{K(k)}</math>'', differs from that used in the [[arithmetic-geometric mean]] article, such that ''<math>\scriptstyle{K(k)}</math>'' here is ''<math>\scriptstyle{K(m^{2})}</math>'' in that article.''

==References==
* Louis V. King ''[http://www.archive.org/details/onthenumerical032686mbp On The Direct Numerical Calculation Of Elliptic Functions And Integrals]'' (Cambridge University Press, 1924)

[[Category:Elliptic functions]]

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Tarski's high school algebra problem