Mathematical method in elliptic functions
Landen's transformation is a mapping of the parameters of an elliptic integral , useful for the efficient numerical evaluation of elliptic functions. It was originally due to John Landen and independently rediscovered by Carl Friedrich Gauss .[1]
Statement [ edit ] The incomplete elliptic integral of the first kind F is
F
(
φ
∖
α
)
=
F
(
φ
,
sin
α
)
=
∫
0
φ
d
θ
1
−
(
sin
θ
sin
α
)
2
,
{\displaystyle F(\varphi \setminus \alpha )=F(\varphi ,\sin \alpha )=\int _{0}^{\varphi }{\frac {d\theta }{\sqrt {1-(\sin \theta \sin \alpha )^{2}}}},}
where
α
{\displaystyle \alpha }
modular angle . Landen's transformation states that if
α
0
{\displaystyle \alpha _{0}}
α
1
{\displaystyle \alpha _{1}}
φ
0
{\displaystyle \varphi _{0}}
φ
1
{\displaystyle \varphi _{1}}
(
1
+
sin
α
1
)
(
1
+
cos
α
0
)
=
2
{\displaystyle (1+\sin \alpha _{1})(1+\cos \alpha _{0})=2}
tan
(
φ
1
−
φ
0
)
=
cos
α
0
tan
φ
0
{\displaystyle \tan(\varphi _{1}-\varphi _{0})=\cos \alpha _{0}\tan \varphi _{0}}
[2]
F
(
φ
0
∖
α
0
)
=
(
1
+
cos
α
0
)
−
1
F
(
φ
1
∖
α
1
)
=
1
2
(
1
+
sin
α
1
)
F
(
φ
1
∖
α
1
)
.
{\displaystyle {\begin{aligned}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{aligned}}}
Landen's transformation can similarly be expressed in terms of the elliptic modulus
k
=
sin
α
{\displaystyle k=\sin \alpha }
k
′
=
cos
α
{\displaystyle k'=\cos \alpha }
Complete elliptic integral [ edit ] In Gauss's formulation, the value of the integral
I
=
∫
0
π
2
1
a
2
cos
2
(
θ
)
+
b
2
sin
2
(
θ
)
d
θ
{\displaystyle I=\int _{0}^{\frac {\pi }{2}}{\frac {1}{\sqrt {a^{2}\cos ^{2}(\theta )+b^{2}\sin ^{2}(\theta )}}}\,d\theta }
is unchanged if
a
{\displaystyle a}
b
{\displaystyle b}
arithmetic and geometric means respectively, that is
a
1
=
a
+
b
2
,
b
1
=
a
b
,
{\displaystyle a_{1}={\frac {a+b}{2}},\qquad b_{1}={\sqrt {ab}},}
I
1
=
∫
0
π
2
1
a
1
2
cos
2
(
θ
)
+
b
1
2
sin
2
(
θ
)
d
θ
.
{\displaystyle I_{1}=\int _{0}^{\frac {\pi }{2}}{\frac {1}{\sqrt {a_{1}^{2}\cos ^{2}(\theta )+b_{1}^{2}\sin ^{2}(\theta )}}}\,d\theta .}
Therefore,
I
=
1
a
K
(
a
2
−
b
2
a
)
,
{\displaystyle I={\frac {1}{a}}K\left({\frac {\sqrt {a^{2}-b^{2}}}{a}}\right),}
I
1
=
2
a
+
b
K
(
a
−
b
a
+
b
)
.
{\displaystyle I_{1}={\frac {2}{a+b}}K\left({\frac {a-b}{a+b}}\right).}
From Landen's transformation we conclude
K
(
a
2
−
b
2
a
)
=
2
a
a
+
b
K
(
a
−
b
a
+
b
)
{\displaystyle K\left({\frac {\sqrt {a^{2}-b^{2}}}{a}}\right)={\frac {2a}{a+b}}K\left({\frac {a-b}{a+b}}\right)}
and
I
1
=
I
{\displaystyle 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
θ
=
arctan
(
x
/
b
)
{\displaystyle \theta =\arctan(x/b)}
d
θ
=
(
cos
2
(
θ
)
/
b
)
d
x
{\displaystyle d\theta =(\cos ^{2}(\theta )/b)dx}
I
=
∫
0
π
2
1
a
2
cos
2
(
θ
)
+
b
2
sin
2
(
θ
)
d
θ
=
∫
0
∞
1
(
x
2
+
a
2
)
(
x
2
+
b
2
)
d
x
{\displaystyle 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
x
=
t
+
t
2
+
a
b
{\displaystyle x=t+{\sqrt {t^{2}+ab}}}
I
=
∫
0
∞
1
(
x
2
+
a
2
)
(
x
2
+
b
2
)
d
x
=
∫
−
∞
∞
1
2
(
t
2
+
(
a
+
b
2
)
2
)
(
t
2
+
a
b
)
d
t
=
∫
0
∞
1
(
t
2
+
(
a
+
b
2
)
2
)
(
t
2
+
(
a
b
)
2
)
d
t
{\displaystyle {\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
(
x
2
+
a
2
)
(
x
2
+
b
2
)
=
2
x
t
2
+
(
a
+
b
2
)
2
{\displaystyle {\sqrt {(x^{2}+a^{2})(x^{2}+b^{2})}}=2x{\sqrt {t^{2}+\left({\frac {a+b}{2}}\right)^{2}}}}
and the infinitesimal as
d
x
=
x
t
2
+
a
b
d
t
{\displaystyle dx={\frac {x}{\sqrt {t^{2}+ab}}}\,dt}
so that the factor of
x
{\displaystyle x}
Arithmetic-geometric mean and Legendre's first integral [ edit ] If the transformation is iterated a number of times, then the parameters
a
{\displaystyle a}
b
{\displaystyle b}
arithmetic-geometric mean of
a
{\displaystyle a}
b
{\displaystyle b}
AGM
(
a
,
b
)
{\displaystyle \operatorname {AGM} (a,b)}
I
=
∫
0
π
2
1
a
2
cos
2
(
θ
)
+
b
2
sin
2
(
θ
)
d
θ
=
∫
0
π
2
1
AGM
(
a
,
b
)
d
θ
=
π
2
AGM
(
a
,
b
)
{\displaystyle 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
b
2
=
a
2
(
1
−
k
2
)
{\displaystyle b^{2}=a^{2}(1-k^{2})}
I
=
1
a
∫
0
π
2
1
1
−
k
2
sin
2
(
θ
)
d
θ
=
1
a
F
(
π
2
,
k
)
=
1
a
K
(
k
)
{\displaystyle 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
a
{\displaystyle a}
K
(
k
)
=
π
2
AGM
(
1
,
1
−
k
2
)
{\displaystyle K(k)={\frac {\pi }{2\operatorname {AGM} (1,{\sqrt {1-k^{2}}})}}}
By performing an inverse transformation (reverse arithmetic-geometric mean iteration), that is
a
−
1
=
a
+
a
2
−
b
2
{\displaystyle a_{-1}=a+{\sqrt {a^{2}-b^{2}}}\,}
b
−
1
=
a
−
a
2
−
b
2
{\displaystyle b_{-1}=a-{\sqrt {a^{2}-b^{2}}}\,}
AGM
(
a
,
b
)
=
AGM
(
a
+
a
2
−
b
2
,
a
−
a
2
−
b
2
)
{\displaystyle \operatorname {AGM} (a,b)=\operatorname {AGM} \left(a+{\sqrt {a^{2}-b^{2}}},a-{\sqrt {a^{2}-b^{2}}}\right)\,}
the relationship may be written as
K
(
k
)
=
π
2
AGM
(
1
+
k
,
1
−
k
)
{\displaystyle K(k)={\frac {\pi }{2\operatorname {AGM} (1+k,1-k)}}\,}
which may be solved for the AGM of a pair of arbitrary arguments;
AGM
(
u
,
v
)
=
π
(
u
+
v
)
4
K
(
u
−
v
v
+
u
)
.
{\displaystyle \operatorname {AGM} (u,v)={\frac {\pi (u+v)}{4K\left({\frac {u-v}{v+u}}\right)}}.}
References [ edit ]
^ Gauss, C. F.; Nachlass (1876). "Arithmetisch geometrisches Mittel, Werke, Bd. 3". Königlichen Gesell. Wiss., Göttingen : 361–403. ^ Abramowitz, Milton ; Stegun, Irene Ann , eds. (1983) [June 1964]. Handbook of Mathematical Functions with Formulas, Graphs, and Mathematical Tables ISBN 978-0-486-61272-0 LCCN 64-60036 . MR 0167642 . LCCN 65-12253 .
