The Dynamics of Even Odd Splitting by liaoqinmei

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```									            The Dynamics of Even-Odd Splitting
a
Emilia Huerta-S´nchez                            Aida Navarro-L´pezo
Mills College                              University of Puerto Rico
Oakland, CA                                     Humacao, PR
David Uminsky
Harvey Mudd College
Claremont
August 10, 2000

Abstract
The idea of splitting integrals of rational functions into its even and odd
parts is used to improve a new method of integration. It has been shown
to
that the even part of a function is easier √ deal √with. The integration of the
R( x)−R(− x)
odd part yields the map F(R(x)) =               √
2 x
on the space of rational
functions. The properties of F are examined in this paper. The case of the
quadratic rational function leads to a second order recurrence for an asssociated
family of polynomials. We also examine the class of functions R such that
F (n) (R(x)) = 0.

1    Introduction
The question of integration of rational functions is one of the basic problems in
Calculus. The classical method of partial fractions reduces the problem to that of
solving an algebraic equation: the evaluation of
b
P (x)
I=                  dx                            (1.1)
a       Q(x)
requires the factorization

Q(x) = (x − x1 )n1 × (x − x2 )n2 × · · · (x − xj )nj                 (1.2)

where xi : 1 ≤ i ≤ j are the roots of Q(x) = 0 counted according to multiplicity.
The factorization (1.2) can be reduced to a real form by combining the non-real roots
in conjugate pairs. The ﬁnal result has the form

Q(x) =            (x − x1 )n1 × (x − x2 )n2 × · · · (x − xk )nk ×

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(x2 + 2a1 x + a2 + b2 )m1 · · · (x2 + 2ap x + a2 + b2 )mp .
1    1                          p    p                   (1.3)

The diﬃculty associated with this method is that it is impossible to solve an
equation of degree 5 or more by radicals. This is a classical result of Algebra.
Therefore an interesting question is to classify the rational functions R for which the
integral (1.1) can be evaluated without factoring the polynomial Q.
The integration of even rational functions seems to be an easier problem. The
classical Wallis’s formula
∞
dx        π   2m
= 2m+1                                  (1.4)
0       (x2   + 1)m+1  2      m

and the evaluation in [1] of
∞
dx
N0,4 (a; m) :=                                                     (1.5)
0       (x4   + 2ax2 + 1)m+1

π
N0,4 (a; m) =                                     Pm (a)           (1.6)
2m+3/2 (a          + 1)m+1/2
where                                    m
−2m                     2m − 2k         m+k
Pm (a) = 2                   2k                           (a + 1)k         (1.7)
k=0
m−k             m
are indications of this fact. The special case
π
N0,4 (a; 0) =                                         (1.8)
2 2(a + 1)

will be employed later on.

The splitting of a rational function into its even and odd parts yields
∞                         ∞                        ∞
R(x)dx =                  Re (x)dx +               Ro (x)dx    (1.9)
0                         0                        0

and while the ﬁrst integral is easier to evaluate, the change of variables t = x2 yields
the identity
∞               ∞
1 ∞
R(x)dx =        Re (x)dx +        F(R(x))dx               (1.10)
0               0               2 0
where                                      √        √
R( x) − R(− x)
F(R(x)) =       √        .                                   (1.11)
2 x

In this paper we discuss properties of the map F(R).

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2    Some elementary properties of F
In this section we describe some elementary properties of the map F.
Proposition 2.1 Let R(x) be a rational function. Then F(R(x)) is also rational.
Proof. We write R(x) = P (x)/Q(x) and a direct calculation shows that
√       √        √      √
(P ( x)Q(− x) − P (− x)Q( x))
F(R(x)) =              √    √ √             .             (2.1)
Q( x)Q(− x)2 x
√
Now observe that Q(t)Q(−t) is an even polynomial in t = x, so it is a polynomial
in t2 = x. Similarly P (t)Q(−t) − P (−t)Q(−t) is an odd polynomial in t, so the
numerator in (2.1) is also a polynomial in x.

3    An example of the dynamics
The purpose of this section is to describe the behavior of the rational function
1
R(a, x) =                                                 (3.1)
x2   + 2ax + 1

If we apply the even-odd splitting, the even part of the function results in
1 + x2
Re (a, x) = 4                                                  (3.2)
x + (2 − 4a2 )x2 + 1

We observe that the integral of the even part is then
∞                                               ∞
1 + x2                                              dx
dx = 2                                            dx
0       x4 + (2 − 4a2 )x2 + 1                   0       x4   + (2 − 4a2 )x2 + 1

where we have used the change of variables x → 1/x to reduce one of the integrals.
The resulting integral can be evaluated using (1.8) to produce
∞
π
Re (a, x)dx = 2N0,4 (1 − 2a2 , 0) = √       .
0                                          2 1 − a2

We now evaluate the rational function F(R(x)): direct calculation suggests that
the iterates of R under F have the form
Gm (a)
F (m) (R(a, x)) =                                               (3.3)
x2   + Hm (a)x + 1

where Gm (a) and Hm (a) are polynomials in a. This can be written as

F (m) (R(a, x)) = Gm (a) × R (Hm (a)/2, x)                         (3.4)

This is established in the next theorem.

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Theorem 3.1. The functions Hm (a) and Gm (a) satisfy the recursion

Hm+1 (a) = 2 − Hm (a)2                          (3.5)
Gm+1 (a) = −Gm (a)Hm (a)

with initial conditions G1 (a) = 1 and Hm (a) = 2a. It follows that Hm (a) and Gm (a)
are polynomials in a.
Proof. The proof is by induction so we assume that

Gm (a)
F (m) (R(x)) =
x2   + Hm (a)x + 1

and evaluate F (m+1) (R(x)) to obtain the indicated recursion.
Deﬁne Rm (x) = F (m) (R(x)) and compute

(odd)          1           Gm (a)          Gm (a)
Rm (x) =                               − 2                    .     (3.6)
2    x2   + Hm (a)x + 1 x − Hm (a)x + 1
√
We now substitute x →        x and combine the two fractions to produce
√
(m+1)              −2Gm (a)Hm (a) x
F       (R(x)) = 2           2
√
(x + (2 − Hm (a))x + 1)2 x

which simpliﬁes to

−Gm (a)Hm (a)
F (m+1) (R(x)) =                                 .
(x2           2
+ (2 − Hm (a))x + 1)

It follows that F (m+1) (R(x)) has the required form and that the function Hm (a) and
Gm (a) satisfy the recursion stated above.
The next proposition states the recursion satisﬁed by the polynomials Gm (a). We
present two slightly diﬀerent proofs.
Proposition 3.2. The polynomial Gm (a) satisfy the recursion

G3 (a)
m+1
Gm+2 (a) =               − 2Gm+1 (a).                   (3.7)
G2 (a)
m

Proof. Using the expression above

−Gm (a)Hm (a)             Gm (a)Hm (a)
F (m+2) (R(x)) =                                + 2
(x2             2 (a))x + 1)
+ (2 − Hm                        2
(x − (2 − Hm (a))x + 1)

we ﬁnd the common denominator to combine the two fractions
3
(4Gm (a)Hm (a) − 2Gm (a)Hm (a))x
F (m+2) (R(x))odd =                    2
x4 + Hm+2 (a)x2 + 1

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√
we substitute x →            x and it simpliﬁes to
3
2Gm (a)Hm (a) − Gm (a)Hm (a)
F (m+2) (R(x)) =                                  .
(x2 + Hm+2 (a)x + 1)

From Proposition 3.1, we know that Gm+1 (a) = −Gm (a)Hm (a) so it follows that

−Gm+1 (a)
Hm (a) =             .
Gm (a)

We conclude that
−Gm+1 (a)                      Gm+1 (a)
2Gm (a)     Gm (a)
+ Gm (a)          G3 (a)
(m+2)                                                          m
F           (R(x)) =
x2 + Hm+2 (a)x + 1

from where we obtain
G3 (a)
m+1
Gm+2 (a) =           − 2Gm+1 (a).
G2 (a)
m

Note. It is not trivial to assume that Gm+2 is a polynomial, but an elementary
inductive argument on Gm+1 results that G is a polynomial.

−Gm+1 (a)
Second Proof. Observe that Hm (a) =                   Gm (a)
,   the recursion for Gm now follows
from the recursion for Hm (a).
We summarize our discussion in a theorem:
Theorem 3.1: The family of functions

Gm (a)
R(a; x) :=                      a∈R
x2   + Hm (a)x + 1

is closed under the operation F. The integral formula
∞                                                        ∞
Gm (a)dx               πGm (a)                            Gm+1 (a)dx
2 + H (a)x + 1
=                        +                                     (3.8)
0       x      m                         2
1 − Hm (a)           0       x2   + Hm+1 (a)x + 1

holds.

4    A modiﬁed function
Let us modify our F such that, after we apply the mapping function F, we perform
a second substitution x → −x. We will deﬁne this modiﬁed mapping function to be
F . Using F we can now model a more complicated function deﬁned as

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1
R(x) =                                             (4.1)
+  x3 + bx3 + 1
ax2
The result is a similar, more complicated form of (3.3)

Qm (a, b) + Rm (a, b)x
F (m) (R(x)) =                                                 (4.2)
x3   + Sm (a, b)x2 + Tm (a, b)x + 1

Where Qm (a, b), Rm (a, b), Sm (a, b), and Tm (a, b) are polynomials on a and b.

Proposition 3.3 The PolynomialSm (a, b) satisfy the recursion
1 4            2                      1 2
Sm+2 (a, b) = − Sm (a, b) + 2Sm ((a, b)Sm+1 (a, b)) + Sm+1 (a, b) + 4Sm (a, b)   (4.3)
2                                     2

Proof. We assume the form
Qm (a, b) + Rm (a, b)x
F (m) (R(x)) =                                                 (4.4)
x3   + Sm (a, b)x2 + Tm (a, b)x + 1

and evaluate F (m+1) (R(x)) to obtain

Qm (a, b) + Rm (a, b)x
F (m) (R(x))odd =
x3
+        2                         2
− 2Tm (a, b))x2 + (Tm (a, b) − 2Sm (a, b))x + 1
(Sm (a, b)
Qm (a, b) − Rm (a, b)x
−    3 + (S 2 (a, b) − 2T (a, b))x2 − (T 2 (a, b) − 2S (a, b))x + 1
−x       m             m               m            m

Applying the substitution x → −x proves that F (m) (R(x)) has the desired form and
we obtain the recursions
2
Sm+1 (a, b) = Sm (a, b) − 2Tm (a, b)                  (4.5)
2
Tm+1 (a, b) = Tm (a, b) − 2Sm (a, b).                  (4.6)
Solving (4.5) for Tm (a, b) we get
1 2
Tm (a, b) = (Sm (a, b) − Sm+1 (a, b))
2
Taking this result and substituting into (4.6) produces
1 2
(S   (a, b) − Sm+2 (a, b)) =
2 m+1
1 4              2                      2
(Sm (a, b) − 2Sm (a, b)Sm+1 (a, b) + Sm+1 (a, b) − 2Sm (a, b)      (4.7)
4

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which simpliﬁes to
1 4            2                    1 2
Sm+2 (a, b) = − Sm (a, b) + 2Sm (a, b)Sm+1 (a, b) + Sm+1 (a, b) + 4Sm (a, b)
2                                   2

Proposition 3.4 The polynomial Tm (a, b) satisfy the recursion
1 4            2                    1 2
Tm+2 (a, b) = − Tm (a, b) + 2Tm (a, b)Tm+1 (a, b) + Tm+1 (a, b) + 4Tm (a, b)(4.8)
2                                   2

Proof. This is similar to the one for Sm (a, b) given above.
We now consider the family of integrals obtained by the iteration of F described
above. The previous process is completely formal, that is, the convergence of these
integrals has not been taken into account. This is presently considered. As shown
in [4], x3 +ax21
+2bx+1
converges on the interval 0 to ∞ if a and b satisfy the condition
3               3
4a − 18ab + 4b + 27 > 0. It is an open question whether one can ﬁnd conditions on a
1
for F (m) x2 +2ax+1 . Similarly, it is an open question whether one can ﬁnd conditions
on a and b for F (m) x3 +ax21+2bx+1
.

5      Mapping F(R) to Even Functions
Finding a class of functions that map to even rational function using F(R) is an
important question in context of integration. Any such rational function can be
integrated in two steps, provided one has an eﬃcient algorithm for the integration of
even functions. The beginnings of such an algorithm is described in [3].
First observe that if R is an even function, then F(R(x)) = 0. We now describe
an example of a family of rational functions that becomes even after one application
of F, this family satisﬁes F (2) (R(x)) = 0.
Theorem 5.1 Let P, Q, and V be polynomials in x and consider the rational function

xP (x4 ) + x2 Q(x2 )
R(x) =                        .                  (5.1)
R(x4 )

Then F(R(x)) is an even function.
Proof. The odd part of R(x) is computed as

xP (x4 ) + x2 Q(x2 )             xP (x4 ) + x2 Q(x2 ) (−x)P ((−x)4 ) + (−x)2 Q((−x)2 )
=                       −
V (x4 )          odd             V (x4 )                  V ((−x)4 )

which simpliﬁes to
2xP (x4 )
V (x4 )

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√
substituting x →          x results in
√     √
2 xP (( x)4 ) P (x2 )
√     √ 4 =
2 xQ(( x) )   V (x2 )
and this function is even.
Note: Observe that the resulting even function is independent of Q.
Example. Consider the case P (x) = bx + c, Q(x) = dx + e and V (x) = x2 + 2ax + 1.
Then we are trying to evaluate the integral
∞                                                           ∞
x(bx4 + c) + x2 (dx2 + e)                                   bx5 + dx4 + ex2 + cx
dx =                                                   dx   (5.2)
0            x8 + 2ax4 + 1                                  0           x8 + 2ax4 + 1
The even part reduces to the integral
∞                  ∞
dx4 + ex2
Re (x) =                         dx                         (5.3)
0                  0       x8 + 2ax4 + 1
and the new integral coming from F is
∞                              ∞
bx2 + c
F(R(x))dx =                                    dx.                  (5.4)
0                              0         x4 + 2ax2 + 1
This part of the problem can be reduced to a sum of two integrals of the type N0,4
describe above.

6    The power series representation
It is a classical result that the power series expansion of a rational function about
x = 0 given by
∞
R(x) =               aj x j                             (6.1)
j=0

has only a ﬁnite number of negative terms and the coeﬃcients satisfy a periodicity
condition: for j suﬃciently large there exist an m ∈ N such that aj = aj+m . This
property of a rational function is key to ﬁnding a suﬃcient condition on R(x) such
that
F (n) (R(x)) = E(x)                          (6.2)
where E(x) is an even function.

Theorem 6.1. Let R(x) be deﬁned by (6.1). Then

F (n) (R(x)) = E(x) ⇐⇒ a2n+1 j+2n+1 −1 = 0                            ∀j ∈ N.         (6.3)

where E(x) is an even function.

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First we need to state a simple Lemma.
Lemma 1.                                                                      ∞
(n)
F         (R(x)) =                  a2n j+2n −1 xj                                            (6.4)
j=0

Proof. The proof is elementary.

Now we can prove Theorem 6.1,
Proof. By Induction.

Base Case: n = 1. (⇒) F (1) (R(x)) is even
∞                             ∞
(1)                       (1)                       j
F         (R(x)) = F                             aj x             =          a2j+1 x2j+1
j=0                           j=0

√
Let x →       x,
∞                      1              ∞
j=0   a2j+1 xj+ 2
√          ⇒                         a2j+1 xj
2 x                             j=0

Now splitting F (1) (R(x)) into its odd and even part yields,
∞                         ∞                                 ∞
a2j+1 xj =                 a4j+3 x2j+1                        a4j+1 x2j
j=0                          j=0                            j=0

But we know that F (1) (R(x)) which can only occur if the odd part of F (1) (R(x)),
∞           2j+1
j=0 a4j+3 x      , which results in a22 j+3 = 0 ∀j ∈ N.
(⇐) a4j+3 = 0 ∀j ∈ N. Well R(x) can be expressed in the form
∞                   ∞                       ∞                                       ∞                          ∞
j                         4j                             4j+1                             4j+2
R(x) =            aj x =                 a4j x +                a4j+1 x                  +             a4j+2 x          +         a4j+3 x4j+3
j=0                  j=0                     j=0                                     j=0                        j=0

But we know that,
∞
a4j+3 x4j+3 = 0
j=0

and
∞                                                       ∞
(1)                          4j+2                           (1)
F                        a4j+2 x                = 0, F                             a4j x4j       =0
j=0                                                        j=0

because they are even functions so
∞                                                ∞                                            ∞
(1)                          4j+1           1                          √                                        √
F                  a4j+1 x                    ⇒√                    a4j+1 ( x)4j+1 ⇒                         a4j+1 ( x)2j
j=0
x         j=0                                       j=0

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which is even.
Induction Hypothesis.(IHOP) Assume F (n) (R(x)) is even ⇐⇒ a2n+1 j+2n+1 −1 = 0  ∀j ∈
(n+1)
N, is it true for n + 1? i.e. F       (R(x)) is even ⇐⇒ a2n+2 j+2n+2 −1 = 0 ∀j ∈ N.
(⇒) F (n+1) (R(x)) by (6.4)
∞
(n)
F         (R(x)) =            a2n j+2n −1 xj
j=0

so by IHOP,
∞                                ∞
F (n+1) (R(x)) = F                       a2n j+2n −1 xj      ⇒           a2n+1 j+2n+1 −1 xj
j=0                             j=0

Splitting F (n+1) (R(x)) into it’s odd and even part results in,
∞                                    ∞
(n+1)                                               2j
F           (R(x)) =               a2n+2 j+2n+1 −1 x +               a2n+2 j+2n+2 −1 x2j+1
j=0                                   j=0

but F (n+1) (R(x)) is even so a2n+2 j+2n+2 −1 must be 0.
(⇐) a2n+2 j+2n+2 −1 = 0. By our lemma,
∞
F (n) (R(x)) =                a2n j+2n −1 xj
j=0

which can be written as
F (n) (R(x)) =
∞                            ∞                                     ∞                                 ∞
a2n+2 j+2n −1 x4j +         a2n+2 j+2n+1 −1 x4j+1 +               a2n+2 j+3(2n )−1 x4j+2 +           a2n+2 j+2n+2 −1 x4j+3
j=0                         j=0                                   j=0                                j=0
so
∞                                                ∞
F (n+1) (R(x)) ⇒ F                 a2n+2 j+2n −1 x4j            = 0, F             a2n+2 j+3(2n )−1 x4j+2    =0
j=0                                             j=0

because they are even and we know that (⇐) a2n+2 j+2n+2 −1 = 0, so
∞
F (n+1) (R(x)) = F                    a2n+2 j+2n+1 −1 x4j+1         ⇒
j=0

∞
a2n+2 j+2n+1 −1 x2j
j=0

which is even.
This is a very important result because now we can take any rational function and
express it as a power series and determine if it will every result in an even function.

66
7      A Special Property of Mapping Polynomials
The deﬁnition of F shows that every even function is mapped to 0. The converse
is also true: if F(R(x)) = 0, then R(x) is even. An interesting open problem is to
classify all functions R for which there exists an integer n such that F (n) (R(x)) = 0.
This is the kernel of the iterate F (n) : for each n ∈ N we deﬁne the sets
Kn := {R rational : F (n) (R(x)) = 0}
We say that a rational function R has order n if n is the samllest positive integer
such that R ∈ Kn . We say that R has inﬁnite order if R ∈ Kn for every n ∈ N.
An interesting property of the map F is that some iteration of it all polynomials
that are mapped by F(R(x)) will eventually result in zero. In addition the number
of mappings of the polynomials is deﬁnable.

Theorem 7.1 Let P be a polynomial then there exists some n such that
F n (P (x)) = 0,              n ∈ Z+                         (7.1)

PROOF. Let P (x) be a polynomial of the form P (x) = a0 + a1 x + a2 x2 + · · · + am xm
It follows that x with an even degree will be equal to zero after one iteration of as
shown:
x2k − x2k
F(x2k ) =            =0                            (7.2)
2
m
So we need to look at power of x of the form x2 p−1 , where m is a positive integer
m
and p is odd, and there exists a positive integer n such that F n (x2 p−1 ) = 0. Then
m p−1            m−1 p−1
F 1 (x2           ) = x2
and
m p−1            m−2 p−1
F 2 (x2           ) = x2
Iterating this procedure yields
m p−1
F m (x2           ) = xp−1
Notice that xp−1 is an even power of x. So by 7.2,
m p−1
F m+1 (x2           ) = F(xp−1 ) = 0

k                  k              kn p −1
COROLLARY.Let P (x) = a1 x2 1 p1 −1 + a2 x2 2 p2 −1 + ... + an x2          n
where pi is odd,
and 1 ≤ i ≤ n. Deﬁne
k = max(ki ), 1 ≤ i ≤ n
Then
F k+1 (P (x)) = 0
Proof: This is the speciﬁc case for m = k.

67
8    Conclusion
We explored many aspects of this mapping function F and have found some important
properties. In the context of integrating, we have been able to deﬁne and prove (6.3)
when this mapping function F is optimal to use in order to reduce a seemingly
diﬃcult integral into easier even parts. This was done by looking at the power series
expansion of the rational function. In addition, we found a particular class that maps
to even functions after one interation to an even function. This particular class (5.1)
of functions is an example of the result of the more general, theorem 6.1. We were
also able to prove in (7.1) that polonomials will always map to even functions after n
interations. In addition we showed exactly how to determine n exactly for any given
polynomial. In the future, ﬁnding other particular classes of functions such as (5.1)
for the second, third and nth interation of F would be very useful for recognizing
when a rational function would eventually map to an even function. As of now, our
general theorem (6.3) can tell us this by looking at the power series expansion but
speciﬁc cases would facilitate this process. Finally it would very useful to continue
looking at theorem 6.1, and use it to ﬁnd an explicit formula for R(x) which will
determine wether a function will become even after n iterations of F.
Unfortunately, many rational functions will never map to even functions. For
these rational functions, describing their behavior under this mapping function (6.3)
becomes very complicated. In the case of the quadratic, applying our mapping func-
tion results in a standard formula 3.3 that reﬂects its behavior. After obtaining
these formulas, and placing conditions on their coeﬃcients to conﬁrm convergence,
we are able to establish wether integrability over certain limits is possible. Modeling
the quadratic recursively was straightforward, but looking at the cubic resulted in a
much more complicated model. And, as of now, we have an incomplete model that
illustrates the behavior of the cubic under this mapping function. We were able to
ﬁnd recursion formulas for the coeﬃcients in the denominator, but for the numerator,
this has not been accomplished yet. This area of our research has the most unﬁn-
ished work. Modeling these functions allow us to ﬁnd recursive functions for them.
An area that ties directly to convergence is to use these recursive functions to classify
the behavior of the roots of a rational function. A large undertaking would include
writing algorithms to study the behavior of functions that map back to themselves,
to classify ﬁxed points, and to measure the length of their orbits. This approach to
the map F is very similar to the research done in [2].

9    Acknowledgements
We would like to thank Herbert Medina, Ivelisse Rubio, Olgamary, Monica and the
entire SIMU staﬀ for organizing SIMU and giving us the opportunity to participate in
an undergraduate research project. We would also like to thank Jean Carlos Cortissoz
and Victor Moll for their guidance and contributions to our work.

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References
[1]   BOROS, G. - MOLL, V.: An integral hidden in Grashteyn and Rhyzik. Jour.
Comp. Appl. Math. 106, 1999, 361-368.

[2]   BOROS, G. - JOYCE, M. - MOLL. V.: A map of rational functions. Preprint.

[3]   BOROS, G. - MOLL, V.: Landen transformations and the integration of ra-
tional functions. Preprint.

[4]   GUTIERREZ, E. - ROSARIO-GARCIA, M. - TORRES, M. - Cortissoz, J. -
MOLL, V. : Convergence of a Landen transformation of degree 6. Preprint.

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