Applied Mathematics, 2014, 5, 2493-2517
Published Online September 2014 in SciRes. http://www.scirp.org/journal/am
http://dx.doi.org/10.4236/am.2014.516241
Compound Means and Fast Computation of
Radicals
Jan Šustek
Department of Mathematics, Faculty of Science, University of Ostrava, Ostrava, Czech Republic
Email: [email protected]
Received 28 June 2014; revised 2 August 2014; accepted 16 August 2014
Copyright © 2014 by author and Scientific Research Publishing Inc.
This work is licensed under the Creative Commons Attribution International License (CC BY).
http://creativecommons.org/licenses/by/4.0/
Abstract
In last decades, several algorithms were developed for fast evaluation of some elementary functions with very large arguments, for example for multiplication of million-digit integers. The
present paper introduces a new fast iterative method for computing values k x with high accuracy, for fixed k ∈  and x ∈  + . The method is based on compound means and Padé approximations.
Keywords
Compound Means, Padé Approximation, Computation of Radicals, Iteration
1. Introduction
In last decades, several algorithms were developed for fast evaluation of some elementary functions with very
large arguments, for example for multiplication of million-digit integers. The present paper introduces a new
iterative method for computing values k x with high accuracy, for fixed k ∈  and x ∈  + .
The best-known method used for computing radicals is Newton’s method used to solve the equation
f ( t ) = t k − x = 0.
Newton’s method is a general method for numerical solution of equations and for particular choice of the
equation it can lead to useful algorithms, for example to algorithm for division of long numbers. This method
converges quadratically. Householder [1] found a generalization of this method. Let d ∈  0 be a parameter of
the method. When solving the equation f ( t ) = 0 the iterations converging to the solution are
How to cite this paper: Šustek, J. (2014) Compound Means and Fast Computation of Radicals. Applied Mathematics, 5,
2493-2517. http://dx.doi.org/10.4236/am.2014.516241
J. Šustek
(d )
an +1
1
  ( an )
f
= an + ( d + 1)  ( d +1)
.
1
( an )
 
 f 
(1.1)
The convergence has order d + 2 . The order of convergence can be made arbitrarily large by the choice of
d . But for larger values of d it is necessary to perform too many operations in every step and the method gets
slower.
The method (3.4) presented in this paper involves compound means. It is proved that this method performs
less operations and is faster than the former methods.
Definition 1. A function P :  + ×  + →  + is called mean if for every t , u ∈  +
min ( t , u ) ≤ P ( t , u ) ≤ max ( t , u ) .
A mean P is called strict if
t ≠ u ⇒ min ( t , u ) < P ( t , u ) < max ( t , u ) .
A mean P is called continuous if the function P is continuous.
A known class of means is the power means defined for p ≠ 0 by
1
tp +up p
M p ( t , u ) := 
 ;
 2 
for p = 0 we define
M 0 ( t , u ) : lim
M p (t, u )
=
=
p →0
tu .
The most used power means are the arithmetic mean A = M 1 , the geometric mean G = M 0 and the harmonic mean H = M −1 . All power means are continuous and strict. There is a known inequality between power
means
α ≤ β ⇒ Mα ≤ M β ,
see e.g. [2]. From this one directly gets the inequality between arithmetic mean and geometric mean. For other
classes of means see e.g. [3].
Taking two means, one can obtain another mean by composing them by the following procedure.
Definition 2. Let P, Q be two means. Given two positive numbers t, u , put
=
a0 : t ,=
b0 : u ,
=
an +1 : P=
( an , bn ) , bn +1 : Q ( an , bn ) ,
n ≥ 0 . If these two sequences converge to a common limit then this limit is denoted by
P◊Q ( t , u ) :=lim an =lim bn .
n →∞
n →∞
The function P◊Q :  ×  →  is called compound mean.
The best known application of compound means is Gauss’ arithmetic-geometric mean [4]
+
+
+
π
A◊G ( t , u ) =
.
π 2
dθ
2 ∫0
t 2 cos 2 θ + u 2 sin 2 θ
(1.2)
Iterations of the compound mean then give a fast numerical algorithm for computation of the elliptic integral
(1.2).
Matkowski [5] proved the following theorem on existence of compound means.
Theorem 1. Let P, Q be continuous means such that at least one of them is strict. Then the compound mean
P◊Q exists and is continuous.
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J. Šustek
2. Properties
We call a mean P homogeneous if for every c, t , u ∈  +
P ( ct , cu ) = cP ( t , u ) .
All power means are homogeneous. If two means are homogeneous then their compound mean is also
homogeneous.
Homogeneous mean P can be represented by its trace
p ( t ) = P (1, t ) .
Conversely, every function p :  + →  + with property
min (1, t ) ≤ p ( t ) ≤ max (1, t )
(2.1)
represents homogeneous mean
u
P ( t , u ) = tp   .
t
Theorem 2. If the compound mean P◊Q exists then it satisfies the functional equation
P ◊Q ( P ( t , u ) , Q ( t , u ) ) =
P ◊Q ( t , u ) .
(2.2)
On the other hand, there is only one mean R satisfying the functional equation
R ( P (t, u ) , Q (t, u )) = R (t, u ).
Easy proofs of these facts can be found in [5].
2tu
t +u
and the harmonic mean H ( t , u ) =
. The aritht +u
2
metic-harmonic mean A◊H exists by Theorem 1 and Theorem 2 implies that A◊H =
G . Hence the iterations
of the arithmetic-harmonic mean A◊H (1, x ) can be used as a fast numerical method of computation of
G (1, x ) = x . This leads to a well known Babylonian method
Example 1. Take the arithmetical mean A ( t , u ) =
1
x 
 an + 
2
an 
a0 1, =
an +1
=
with a quadratical convergence to lim an = x .
n →∞
The Babylonian method is in fact Newton’s method used to solve the equation t 2 − x =
0 . Using Newton’s
method to solve the equation t k − x =
0 leads to iterations
a0 = 1,
an +1 =
1
x 
 ( k − 1) an + k −1 
k
an 
(2.3)
with a quadratical convergence to lim an = k x .
n →∞
3. Our Method
In the present paper we will proceed similarly as in Example 1. For a fixed integer k ≥ 2 and a positive real
number x we will find a sequence of approximations converging fast to k x .
We need the following lemma.
j
Lemma 1. Let function p satisfy p ( ) (1) =
q (t ) =
( )
k
( j)
(1)
for j = 0, , w and let p (
w +1)
be bounded. Let
t
. Assume that p and q satisfy (2.1) strictly if t ≠ 1 . Then the function q satisfies
p k −1 ( t )
j
j
q ( ) (1) = p ( ) (1) for j = 0, , w and q ( ) is bounded. Let P and Q be the homogeneous means corresponding to traces p and q , respectively. Then the compound mean P◊Q exists and its convergence has
w +1
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J. Šustek
order w + 1 .
Proof. The assumption implies that
p ( t ) =k t + O ( t − 1)
w +1
.
Then
t
=
k −1
p (t )
=
q (t )
= k −1
k
t
(
t
t + O ( t − 1)
k
t
+ O ( t − 1)
w +1
)
w +1 k −1
=k t + O ( t − 1)
w +1
,
w +1
hence q (1) = p (1) for j = 0, , w and q ( ) is bounded.
The compound mean P◊Q exists by Theorem 1. Let an and bn be the iterations of P◊Q . To find the
b − an
order of convergence put δ n := n
. Then
an
( j)
( j)
an +1 = p ( an , an (1 + δ n ) ) = an (1 + δ n ) = an ∑
w
p(
j)
j =0
bn +1 = p ( an , an (1 + δ n ) ) = an (1 + δ n ) = an ∑
w
q(
j)
(
)
w +1
(1)
j!
j =0
and bn +1 − an +1 = O δ nw +1 = O ( bn − an )
(1)
j!
δ nj + O (δ nw+1 ) ,
δ nj + O (δ nw+1 )
□
.
Take the mean Rk ( t , u ) := k t k −1u . This mean is strict, continuous and homogeneous and it has the property
Rk (1, x ) = k x . We will construct two means P, Q such that Rk = P◊Q .
Let s ∈  and let p ( t ) be the Padé approximation of the function k t of order [ s, s ] around t = 1 ,
s
∑e j t j
p (t ) =
j =0
s
∑ f jt
=k t + O ( t − 1)
2 s +1
.
(3.1)
j
j =0
The exact formula for e j , f j will be derived in Lemma 15. In Lemma 20 we will prove that p satisfies
(
)
(
min 1, k t < p ( t ) < max 1, k t
)
(3.2)
for every t ∈  + . Hence it is a trace of a strict homogeneous mean
u
j
u
∑ f j  t 
j =0
j
s
u
P=
( t , u ) tp=
  t
t
∑e j  t 
j =0
s
.
Relation Rk = P◊Q and Theorem 2 imply that
Q (t, u ) =
t k −1u
P k −1 ( t , u )
(3.3)
and its trace is
=
q ( t ) Q=
(1, t )
t
p
k −1
(t )
.
Inequalities (3.2) imply that Q is a strict homogeneous mean too.
As in Definition 2, denote the sequences given by the compound mean P◊Q by an , bn , starting with
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J. Šustek
a0 = 1 and b0 = x . From (3.3) we obtain that
ank −1bn
ank −1bn
=
,
P k −1 ( an , bn )
ank+−11
=
bn +1 Q=
( an , bn )
hence ank+−11bn +=

= a0k −1b=
ank −1b=
x and bn =
1
n
0
x
k −1
n
a
. So the iterations of the compound mean P◊Q are
j
 x 
ej  k 
∑

x 
j = 0  an 
an +1 P=
an
an
=
=
( an , bn ) P  a=
n , k −1 
j
s
 x 
 an 
∑ f j  ak 
j =0
 n
s
∑e j x j ( ank )
s
j =0
s
s− j
∑ f j x (a )
j
k
n
j =0
Note that we don’t have to compute the sequence bn .
According to (3.1) and Lemma 1 the convergence of the sequence (3.4) to its limit
k
s− j
.
(3.4)
x has order 2 s + 1 .
4. Complexity
Let M ( N ) denote the time complexity of multiplication of two N-digit numbers. The classical algorithm of
multiplication has asymptotic complexity M ( N ) = O N 2 . But there are also algorithms with asymptotic
complexity
( )
(
)
(
)
=
M ( N ) O=
N log 2 3 O N 1.585 ,
M ( N ) = O ( N log N log log N )
or
(
M ( N ) = O N log N 2log
∗
N
),
see Karatsuba [6], Schönhage and Strassen [7] or Fürer [8], respectively. The fastest algorithms have large
asymptotic constants, hence it is better to use the former algorithms if the number N is not very large.
The complexity of division of two N-digit numbers differs from the complexity of multiplication only by
some multiplicative constant D . Hence the complexity of division is DM ( N ) . Analysis in [9] shows that this
7
constant can be as small as D = .
2
We will denote by σ ( k ) the minimal number of multiplications necessary to compute the power t  t k .
See [10] for a survey on known results about the function σ .
Before the main computation of complexity we need this auxiliary lemma.
Lemma 2. Assume that r ∈  and that M ( N ) is a function such that for some ω ∈ [1, 2] the function
M (N)
is nondecreasing with
f (N) =
Nω
f (N) = N
For every δ > 0 put gδ ( N ) =
o (1)
.
N ω +δ
and assume that
M (N )
1) for every T the image set gδ ([1, T ]) is bounded and
2) there is N 0 (δ ) with gδ ( n ) ≥ gδ ( m ) for every n ≥ m ≥ N 0 (δ ) .
Then
log r N 
N
M j 
1
r 
j =1
.
= ω
M (N)
r −1
∑
lim
N →∞
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(4.1)
J. Šustek
Proof. From the monotonicity of f we have for every N
log r N 
log N 
=j
ω
N  r N
N
M j 
f  j  log N 
∑
ωj
r
r
1
 
r   r  1
1=
j 1
.
=
≤ ∑ ωj ≤ ω
ω
M (N)
N f (N)
r −1
j =1 r
∑
(4.2)
Let δ > 0 . Then (4.1) implies lim gδ ( N ) = ∞ . From this and properties 1 and 2 we deduce that there
N →∞
exists a number N1 (δ ) such that gδ ( n ) ≥ gδ ( m ) for every n ≥ N1 (δ ) and every m ≤ n . This implies for
every N ≥ N1 (δ )
log r N 
∑
log r N 
N
M j 
r 
j =1
=
M (N)
∑
j =1
N ω +δ
1
N
ω +δ j
r ( ) gδ  j  log r N 
1 − (ω +δ ) log N 
 r 
1
r
 ≥
r
.
=
∑
ω +δ
ω +δ
ω +δ ) j
(
−1
N
r
j =1 r
gδ ( N )
(4.3)
Inequalities (4.2) and (4.3) yield
log r N 
1
=j
ω +δ
N →∞
r
−1
≤ lim inf
log r N 
N
N
M j 
M j 
∑
r
1
 
r 
1=
j 1
≤ lim sup
≤ ω
.
M (N)
M (N)
r −1
N →∞
∑
□
Passing to the limit δ → 0 implies the result.
Note that all the above mentioned functions M ( N ) satisfy all assumptions of Lemma 2 with ω = 2 ,
ω = log 23 , ω = 1 and ω = 1 , respectively.
Now we compute the complexity V ( N ) of algorithm computing k x to within N digits. The functions
M ( N ) and V ( N ) have asymptotically the same order as N → ∞ , see for instance Theorem 6.3 in [3].
Hence all fast algorithms for computing k x differ only in the asymptotic constants.
Let the algorithm for computing k x
• performs Z multiplications of two N-digit numbers before the iterations,
• performs A multiplications and B divisions of two long numbers in every step,
• has order of convergence r .
The accuracy to within N digits is necessary only in the last step. In the previous step we need accuracy
N
only to within
digits and so on. Hence
r
N
V ( N=
) V   + ( A + BD ) M ( N ) + O ( N ) .
 r 
The error term O ( N ) corresponds to additions of N-digit numbers. This and Lemma 2 imply that1
V (N) =
( A + BD )
log r N 
∑
j =0
N
M j
r

 + ZM ( N ) + O ( N )



1

= 1 + ω
+ o (1)  ( A + BD ) + Z  M ( N ) + O ( N )

  r −1

1 


=  1 + ω
 ( A + BD ) + Z + o (1)  M ( N ) .
  r −1 

4.1. Complexity of Newton’s Method
Newton’s method (2.3) has order 2 and in every step it performs σ ( k − 1) multiplications (evaluation of ank −1 )
1
In the last line we assume the hypothesis that all multiplication algorithms satisfy
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N
= o (1) , see [12].
M (N)
J. Šustek
and 1 division. So the complexity is
V N=
(k, N )
( S ( k ) + o (1) ) M ( N )
N
where
1 

S N ( k ) = 1 + ω
 (σ ( k − 1) + D ) .
 2 −1 
For the choice of multiplication and division algorithms with ω = 1 and D =
V N ( k ,=
N)
7
we have
2
( 2σ ( k − 1) + 7 + o (1) ) M ( N ) .
4.2. Complexity of Householder’s Method
Consider Householder’s method (1.1) applied to the equation f ( t ) = t k − x = 0 . Let d ∈  0 . An easy
calculation (see [11] for instance) leads to iterations
an +1 = an
∑λ j x j ( ank )
d
d− j
j =0
d
d− j
∑µ j x j ( ank )
.
j =0
where λ j ( k , d ) and µ j ( k , d ) are suitable constants. The method has order d + 2 , before the iterations it
performs d − 1 multiplications of N-digit numbers (evaluation of x 2 , , x d ) and in every step it performs
σ ( k ) + 2d − 1 multiplications (evaluation of ank , then evaluations of numerator and denominator by Horner’s
method, and then the final multiplication) and 1 division. So the complexity of Householder’s algorithm is
VdH =
(k, N )
( S ( k ) + o (1) ) M ( N )
H
d
where


1
 (σ ( k ) + 2d − 1 + D ) + d − 1.
SdH ( k ) = 1 +
ω
 ( d + 2) −1 


For the choice of multiplication and division algorithms with ω = 1 and D =
7
we have
2


7 2σ ( k ) + 1
+ o (1)  M ( N ) .
VdH ( k , N )=  3d + σ ( k ) + +
2
2d + 2


The optimal value of d which minimizes the complexity is in this case
  2σ ( k ) + 1   2σ ( k ) + 1  
− 1 , 
− 1  .
d ∈ 
6
6
 
 
 
4.3. Complexity of Our Method
Given s ∈  , our method (3.4) has order 2 s + 1 , before the iterations it performs s − 1 multiplications of
N-digit numbers (evaluation of x 2 , , x s ) and in every step it performs σ ( k ) + 2 s − 1 multiplications
(evaluation of ank , then evaluations of numerator and denominator by Horner’s method2, and then the final
multiplication) and 1 division. So the complexity of our algorithm is
VsS (=
k, N )
( S ( k ) + o (1) ) M ( N )
S
s
where
2
Not always Horner’s method is optimal, see [13]. In those cases Householder’s and our algorithms are faster.
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J. Šustek


1
 (σ ( k ) + 2 s − 1 + D ) + s − 1.
S sS ( k ) = 1 +
ω
 ( 2 s + 1) − 1 


For the choice of multiplication and division algorithms with ω = 1 and D =
7
we have
2


5 2σ ( k ) + 5
VsS ( k , N )=  3s + σ ( k ) + +
+ o (1)  M ( N ) .
2
4s


The optimal value of s which minimizes the complexity is in this case
(4.4)
  2σ ( k ) + 5   2σ (k ) + 5  
,
s ∈ 
 .
12
12
 

 
5. Examples
Example 2. Compare the algorithms for computation of 14 x . For k = 14 we have σ ( k ) = 14 and, according to (4.4), the optimal value of s for our algorithm is s = 1 . Padé approximation of the function 14 t
around t = 1 is
15t + 13
3
14
t
=
+ O ( t − 1) .
13t + 15
Hence the iterations of our algorithm are
15 x + 13a14
n
=
a0 : 1,=
an +1 : an
(5.1)
13x + 15a14
n
with convergence of order 3. For computation of N digits of
14
x we need to perform
 57

V1S (14, =
N )  + o (1)  M ( N )
 4

operations.
Newton’s method
a0 : 1, =
an +1 :
=
1
x 
13an + 13 
14 
an 
has order 2 and for computation of N digits it needs (σ (13) = 5 )
V N (14, N=
)
(17 + o (1) ) M ( N )
operations. Hence our method saves 16% of time compared to Newton’s method.
For Householder’s method the optimal value is d = 1 and it leads to the same iterations (5.1) as our method.
Example 3. Compare the algorithms for computation of 179 x . For k = 179 we have σ ( k ) = 10 and,
according to (4.4), the optimal value of s for our algorithm is s = 2 . Padé approximation of the function
179
t around t = 1 is
179
=
t
10770t 2 + 42721t + 10591
5
+ O ( t − 1) .
10591t 2 + 42721t + 10770
Hence the iterations of our algorithm are
=
a0 : 1,=
an +1 : an
358
10770 x 2 + 42721xa179
n + 10591an
358
10591x 2 + 42721xa179
n + 10770an
with convergence of order 5. For computation of N digits of
179
x we need to perform
 173

V2S (179,=
N) 
+ o (1)  M ( N )
 8

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J. Šustek
operations.
Newton’s method
1 
x 
178an + 178 
179 
an 
=
a0 : 1,=
an +1 :
has order 2 and for computation of N digits it needs
V N (179, N
=
)
( 27 + o (1) ) M ( N )
operations.
For Householder’s method the optimal value is d = 1 and it leads to iterations
=
a0 : 1,=
an +1 : an
180 x + 178a179
n
.
178 x + 180a179
n
This method has order 3 and for computation of N digits it needs
 87

V1H (179, =
N )  + o (1)  M ( N )
 4

operations.
Hence our method saves 20% of time compared to Newton’s method and saves 0.6% of time compared to
Householder’s method.
Example 4. Compare the algorithms for computation of 1234567890133 x . For k = 1234567890133 the exact
value of σ ( k ) is not known. We assume that σ ( k ) = 59 . According to (4.4), the optimal value of s for our
algorithm is s = 3 . The iterations of our algorithm are
a0 := 1,
 94083818620004407126666191157329760 x3



2 1234567890133
 +846754367579125169414786634836768160 x an


2469135780266 
 +846754367578439298370880283615327200 xan

 +94083818619724978182852492511557517 a 3703703670399

n

an +1 := an 
3
 94083818619724978182852492511557517 x


2 1234567890133 
 +846754367578439298370880283615327200 x an

 +846754367579125169414786634836768160 xa 2469135780266 
n


 +94083818620004407126666191157329760a 3703703670399

n


with convergence of order 7. For computation of N digits of 1234567890133 x we need to perform
 323

+ o (1)  M ( N )
V3S (1234567890133,=
N) 
 4

operations.
For Householder’s method the optimal value is d = 3 and it leads to iterations
a0 := 1,
 1881676372394245537344489473235577128 x3



2 1234567890133
 +20698440096361087436794958174946153000 x an


2469135780266 
+
xa
20698440096379377331299127540851245280
n


 +1881676372400342168845879261870607880a 3703703670399

n
.
an +1 := an 
3
 1881676372388148905843114499415227960 x


2 1234567890133 
 +20698440096342797542290773994226379160 x an

 +20698440096397667225803282091941655928 xa 2469135780266 
n


 +1881676372406438800347283865320320240a 3703703670399

n


2501
J. Šustek
This method has order 5 and for computation of N digits it needs
 691

V3H (1234567890133,=
N) 
+ o (1)  M ( N )
 8

operations.
Newton’s method
a0 := 1,
an +1 :
=


1
x
1234567890132an + 1234567890132 
1234567890133 
an

has order 2 and for computation of N digits it needs (assuming that σ ( k − 1) =
59 )
V N (1234567890133,=
N)
(125 + o (1) ) M ( N )
operations.
Hence our method saves 35% of time compared to Newton’s method and saves 7% of time compared to
Householder’s method.
6. Proofs
In this section we will prove that function p defined by (3.1) satisfies inequalities (3.2). For the sake of
brevity, will use the symbol  u for the set   [u , ∞ ) .
6.1. Combinatorial Identities
First we need to prove several combinatorial identities. Our notation will be changed in this subsection. Here,
n will be a variable used in mathematical induction, k will be a summation index, and A, B will be
additional parameters. The change of notation is made because of easy application of the following methods
based on [14]. For a function ϕ ( n, k ) we will denote its differences by
∆ nϕ ( n, k ) := ϕ ( n + 1, k ) − ϕ ( n, k ) ,
∆ k ϕ ( n, =
k ) : ϕ ( n, k + 1) − ϕ ( n, k ) .
Given a function F ( n, k ) , there is some function G ( n, k ) satisfying some relation between ∆ n F ( n, k )
and ∆ k G ( n, k ) . This new function is then used for easier evaluation of sums containing F ( n, k ) . Recall that
1
= 0 for negative integer n .
n!
For k ∈  0 and n, A, B ∈ 1 with n ≥ max ( A, B ) put
F1 ( n, k , A, B ) :=
( n − A)!( n + A − B )! B ! A!( n − B )!( 2n − B + k )!
,
( 2n − B )!( n + A)!k !( B − k )!( n − B + k )!( A − k )!( n − A − B + k )!
R1 ( n, k , A, B ) :=
(
),
k 2 B 2 − 6nB − 2kB − 4 B + kA − A + 4n 2 + 3kn + 5n + 3k + 1
( A + n + 1)( B − 2n − 2 )( B − 2n − 1)
G1 ( n, k , A, B ) := F1 ( n, k , A, B ) R1 ( n, k , A, B ) ,
n
S1 ( n, A, B ) := ∑F1 ( n, k , A, B ) .
k =0
Lemma 3. For every n, A, B ∈ 1 satisfying n ≥ max ( A, B ) we have
S1 ( n, A, B )
n
( n − A)!( n + A − B )! B ! A!( n − B )!( 2n − B + k )!
=
∑
( 2n − B )!( n + A)!k !( B − k )!( n − B + k )!( A − k )!( n − A − B + k )!
k =0
Proof. From the polynomial identity
2502
1.
( ( n + 1 − A)( n + 1 + A − B )( n + 1 − B )( 2n + 1 − B + k )( 2n + 2 − B + k )
J. Šustek
− ( 2n + 1 − B )( 2n + 2 − B )( n + 1 + A )( n + 1 − B + k )( n + 1 − A − B + k ) )
× ( ( k + 1)( n − B + k + 1)( n − A − B + k + 1)( A + n + 1)( B − 2n − 2 )( B − 2n − 1) )
=
( ( 2n + 1 − B )( 2n + 2 − B )( n + 1 + A)( n + 1 − B + k )( n + 1 − A − B + k ) )
(
(
× ( 2n − B + k + 1)( B − k )( A − k )( k + 1) 2 B 2 − 6nB − 2kB − 6 B + kA
) (
+ 3kn + 5n + 3k + 1) ( k + 1)( n − B + k + 1)( n − A − B + k + 1) )
+4n + 3kn + 8n + 3k + 4 − k 2 B − 6nB − 2kB − 4 B + kA − A
2
+4n 2
2
we immediately obtain
∆ n F1 ( n, k , A, B )
 ( n + 1 − A )( n + 1 + A − B )( n + 1 − B )( 2n + 1 − B + k )( 2n + 2 − B + k ) 
− 1 F1 ( n, k , A, B )

 ( 2n + 1 − B )( 2n + 2 − B )( n + 1 + A )( n + 1 − B + k )( n + 1 − A − B + k ) 
 ( n + 1 − A )( n + 1 + A − B )( n + 1 − B )( 2n + 1 − B + k )( 2n + 2 − B + k ) 


 − ( 2n + 1 − B )( 2n + 2 − B )( n + 1 + A )( n + 1 − B + k )( n + 1 − A − B + k ) 
F1 ( n, k , A, B )
=
2n + 1 − B )( 2n + 2 − B )( n + 1 + A )( n + 1 − B + k )( n + 1 − A − B + k ) 
 (





(

( 2n − B + k + 1)( B − k )( A − k )( k + 1) 2 B 2 − 6nB − 2kB − 6 B + kA 




+4n 2 + 3kn + 8n + 3k + 4 − k 2 B 2 − 6nB − 2kB − 4 B + kA − A




+4n 2 + 3kn + 5n + 3k + 1 ( k + 1)( n − B + k + 1)( n − A − B + k + 1)
 F ( n, k , A, B )
=
 ( k + 1)( n − B + k + 1)( n − A − B + k + 1)( A + n + 1)( B − 2n − 2 )( B − 2n − 1)  1










) (
)


( 2n − B + k + 1)( B − k )( A − k )
= 
R1 ( n, k + 1, A, B ) − R1 ( n, k , A, B )  F1 ( n, k , A, B )
 ( k + 1)( n − B + k + 1)( n − A − B + k + 1)

= F1 ( n, k + 1, A, B ) R1 ( n, k + 1, A, B ) − F1 ( n, k , A, B ) R1 ( n, k , A, B )
= ∆ k G1 ( n, k , A, B ) .
(6.1)
1
For fixed k > n we have max ( A, B ) < k and hence =
( A − k )!
F1 ( n + 1, k =n + 1, A, B ) =0.
1
= 0 . Thus
( B − k )!
(6.2)
Similarly for k = 0 and k= n + 1 we have
G=
0,=
A, B ) 0
1 ( n, k
(6.3)
and
G1 ( n, k =
n + 1, A, B ) =
0.
Then (6.1), (6.2), (6.3) and (6.4) imply
2503
(6.4)
J. Šustek
∆ n S1 ( n, A, B )
n
= F1 ( n + 1, k =n + 1, A, B ) + ∑∆ n F1 ( n, k , A, B )
k =0
n
= F1 ( n + 1, k =n + 1, A, B ) + ∑∆ k G1 ( n, k , A, B )
(6.5)
k =0
=
F1 ( n + 1, k =
n + 1, A, B ) + G1 ( n, k =
n + 1, A, B ) − G1 ( n, k =
0, A, B )
= 0.
1
= 0 for k < B and
( k − B )!
If A = n and B ≤ n then
1
= 0 for k > B . Hence the only nonzero
( B − k )!
summand in S1 ( n, A = n, B ) is the one for k = B and
0!( 2n − B ) ! B !n !( n − B ) !( 2n − B + k ) !
S1 (=
n, A n=
, B)
∑ ( 2n − B )!( 2n )!k !( B − k )!( n − B + k )!( n − k )!( k −=
B )!
1.
(6.6)
k =B
Similarly for A < n and B = n we obtain
S1 ( n, A < n, B =n ) =
1.
From this and (6.6) we obtain that
=
S1 ( n max
=
( A, B ) , A, B ) 1.
Equation (6.5) implies that S1 will not change for greater n .
For n, k , A ∈  0 with 1 ≤ A ≤ n and k ≤ n − A put
□
( −1)
( n + k )!( n + 1 − A)!( n + 1)!
F2 ( n, k , A ) :=
,
k!( A + k )!( n + 1 − A − k )!( 2n + 1 − A ) !
k +n− A
T2 ( n, k , A ) :=
S2 ( n, A ) :=
k(A+ k)
( n + 1)( A − n − 1)
,
n− A
∑ F2 ( n, k , A) .
k =0
Lemma 4. For every n, A ∈ 1 with A ≤ n
=
S2 ( n, A )
( −1)
( n + k )!( n + 1 − A)!( n + 1)!
= 1.
∑
k = 0 k !( A + k ) !( n + 1 − A − k ) !( 2n + 1 − A ) !
n− A
k +n− A
Proof. From the polynomial identity
( n + k + 1)( n + 1 − A − k ) + k ( A + k ) =− ( n + 1)( A − n − 1) .
we obtain for k ≤ n − A that
( n + k + 1)( n + 1 − A − k ) + k ( A + k )
∆ k ( F2 ( n, k , A ) T2 ( n, k , A ) ) =
−
F2 ( n, k , A ) =
F2 ( n, k , A ) .
( n + 1)( A − n − 1)
This and the fact that T2 (=
n, k 0,=
A ) 0 imply
S2 ( n, A=
)
n− A
n− A
∑ F2 ( n, k , A=) ∑∆ k ( F2 ( n, k , A) T2 ( n, k , A) )
=
k 0=
k 0
= F2 ( n, k = n − A + 1, A ) T2 ( n, k = n − A + 1, A ) − F2 ( n, k = 0, A ) T2 ( n, k = 0, A )
( 2n − A + 1)!( n + 1 − A)!( n + 1)!( n − A + 1)( n + 1)
= 1.
( n − A + 1)!( n + 1)!0!( 2n + 1 − A)!( n + 1)( A − n − 1)
( −1)
2 n − 2 A +1
2504
□
J. Šustek
For n, k , A, B ∈  0 with 1 ≤ A ≤ B < n and k ≤ n − A put
F3 ( n, k , A, B ) :=
( −1) ( B + k )!( n − k − 1)!
,
B
( − k )!k !( A + k )!( n − A − k )!
k
( −1) ( B + k )!( n − k )!
G3 ( n, k , A, B ) :=
,
( B − k )!( k − 1)!( A + k − 1)!( n − A − k + 1)!
k +1
S3 ( n, A, B ) :=
n− A
∑ F3 ( n, k , A, B ) .
k =0
Lemma 5. For every n, A, B ∈ 1 with A ≤ B < n
=
S3 ( n, A, B )
( −1) ( B + k )!( n − k − 1)!
= 0.
∑
k = 0 ( B − k ) ! k !( A + k ) !( n − A − k ) !
k
n− A
Proof. From the polynomial identity
( B − n )( B + n + 1)( n − A − k + 1) − ( n + 1)( A − n − 1)( n − k )
= ( B + k + 1)( B − k )( n − A − k + 1) + ( n − k ) k ( A + k )
we obtain for A ≤ B < n and k ≤ n − A that
( −1) ( B + k )!( n − k )!
( −1) ( B + k + 1)!( n − k − 1)!
=
∆ k G3 ( n, k , A, B )
−
( B − k − 1)!k !( A + k )!( n − A − k )! ( B − k )!( k − 1)!( A + k − 1)!( n − A − k + 1)!

(n − k ) k ( A + k ) 
=  ( B + k + 1)( B − k ) +
 F3 ( n, k , A, B )
k +1
k
n − A − k +1 

( n + 1)( A − n − 1)( n − k ) 
=  ( B − n )( B + n + 1) −
 F3 ( n, k , A, B )
n − A − k +1


= ( B − n )( B + n + 1) F3 ( n, k , A, B ) − ( n + 1)( A − n − 1) F3 ( n + 1, k , A, B ) .

(6.7)
For k = 0 we have G=
0,=
A, B ) 0 . This and (6.7) imply
3 ( n, k
( B − n )( B + n + 1) S3 ( n, A, B ) − ( n + 1)( A − n − 1) S3 ( n + 1, A, B )
n− A
∑ ( ( B − n )( B + n + 1) F3 ( n, k , A, B ) − ( n + 1)( A − n − 1) F3 ( n + 1, k , A, B ) )
=
k =0
− ( n + 1)( A − n − 1) F3 ( n + 1, k = n − A + 1, A, B )
=
n− A
∑∆ k G3 ( n, k , A, B ) − ( n + 1)( A − n − 1) F3 ( n + 1, k = n − A + 1, A, B )
(6.8)
k =0
= G3 ( n, k = n − A + 1, A, B ) − G3 ( n, k = 0, A, B ) − ( n + 1)( A − n − 1) F3 ( n + 1, k = n − A + 1, A, B )
( n + 1)( A − n − 1)( n − A + B + 1)!( A − 1)!
( −1) ( n − A + B + 1)!( A − 1)! ( −1)
0.
=
−
=
( A + B − n − 1)!( n − A + 1)!( n + 1)!1!
( A + B − n − 1)!( n − 1)!n !0!
n− A
n − A +1
Lemma 4 for n = B implies
( −1)
( B + k )!( B + 1 − A)!( B + 1)!
∑ k !( A + k )!( B + 1 − A − k )!( 2 B + 1 − A)! = 1,
k =0
B− A
k + B− A
hence
( −1) ( B + k )!
( −1) ( 2 B + 1 − A)!
∑ k !( A + k )!( B + 1 − A − k )! = ( B + 1 − A)!( B + 1)!
k =0
B− A
B− A
k
and
2505
J. Šustek
( −1) ( B + k )!
S3 ( n =
B + 1, A, B ) =
0.
∑ k !( A + k )!( B + 1 − A − k )! =
k =0
B − A +1
k
(6.9)
For n ≥ B + 1 we have A − n − 1 > 0 and (6.8) yields
( B − n )( B + n + 1)
S3 ( n + 1, A, B ) =
S ( n, A, B ) ,
( n + 1)( A − n − 1) 3
□
This with (6.9) implies the result for every n > B .
For n, k , A ∈  0 with A ≤ n put
F4 ( n, k , A ) :=
( 2n − k )!k !( n − A)!( A + n + 1)!
,
( k − A)!( n − k )!n !( 2n + 1)! A!
G4 ( n, k , A ) := −
S4 ( n, A ) :=
( 2n − k + 1)!k !( n − A)!( A + n + 1)!
,
( k − A − 1)!( n − k + 1)!( n + 1)!( 2n + 3)! A!
n
∑ F4 ( n, k , A) .
k=A
Lemma 6. For every A, n ∈  0 with A ≤ n
S 4 ( n, A )
=
n
( 2n − k )!k !( n − A)!( A + n + 1)!
=
∑
( k − A)!( n − k )!n !( 2n + 1)! A!
1.
k=A
Proof. From the polynomial identity
( 2n − k + 1)( 2n − k + 2 )( n − A + 1)( A + n + 2 ) − ( n + 1 − k )( n + 1)( 2n + 2 )( 2n + 3)
= ( 2n − k + 1)( 2nA − kA + 2 A + kn + n + k + 1)( k − A )
− ( k + 1)( 2nA − kA + A + kn + 2n + k + 2 )( n − k + 1)
we obtain for 0 ≤ A ≤ n and k ∈  0 that
∆ n F4 ( n, k , A )
( 2n − k + 1)( 2n − k + 2 )( n − A + 1)( A + n + 2 ) − ( n + 1 − k )( n + 1)( 2n + 2 )( 2n + 3)
F4 ( n, k , A )
( n + 1 − k )( n + 1)( 2n + 2 )( 2n + 3)
 − ( k + 1)( 2nA − kA + A + kn + 2n + k + 2 )( n − k + 1) 


 + ( 2n − k + 1)( 2nA − kA + 2 A + kn + n + k + 1)( k − A ) 
=
 F4 ( n, k , A )
( n − k + 1)( n + 1)( 2n + 2 )( 2n + 3)


=




 ( k + 1)( 2nA − kA + A + kn + 2n + k + 2 ) ( 2n − k + 1)( 2nA − kA + 2 A + kn + n + k + 1)( k − A ) 
=
+
 −
 F4 ( n, k , A )
( n − k + 1)( n + 1)( 2n + 2 )( 2n + 3)
( n + 1)( 2n + 2 )( 2n + 3)


( 2n − k )!( k + 1)!( n − A)!( A + n + 1)!( 2nA − kA + A + kn + 2n + k + 2 )
= −
( k − A)!( n − k )!( n + 1)!( 2n + 3)!A!
( 2n − k + 1)!k!( n − A)!( A + n + 1)!( 2nA − kA + 2 A + kn + n + k + 1)
( k − A − 1)!( n − k + 1)!( n + 1)!( 2n + 3)!A!
∆ k G4 ( n, k , A ) .
+
=
(6.10)
For k = A we have G4 (=
n, k A=
, A ) 0 . From this, (6.10) and the polynomial identity
2506
2nA − ( n + 1) A + 2 A + ( n + 1) n + 2n + 2 =
J. Šustek
( n + 1)( A + n + 2 )
we obtain
n
∆ n S4 ( n, A ) =F4 ( n + 1, k =n + 1, A ) + ∑∆ n F4 ( n, k , A )
k=A
n
=F4 ( n + 1, k =n + 1, A ) + ∑∆ k G4 ( n, k , A )
k=A
=
F4 ( n + 1, k =
n + 1, A ) + G4 ( n, k + 1, A ) − G4 ( n, k =
A, A )
( n + 1)!( n + 1)!( n + 1 − A)!( A + n + 2 )!
( n + 1 − A)!0!( n + 1)!( 2n + 3)! A!
n !( n + 1)!( n − A )!( A + n + 1)!( 2nA − ( n + 1) A + 2 A + ( n + 1) n + 2n + 2 )
−
( n − A)!0!( n + 1)!( 2n + 3)! A!
( n + 1)!( A + n + 2 )! n !( A + n + 1)!( n + 1)( A + n + 2 )
=
−
( 2n + 3 ) ! A!
( 2n + 3 ) ! A!
=
(6.11)
= 0.
For n = A the sum S4 ( n, A ) contains only one nonzero summand for k = A and we have S=
A=
, A) 1 .
4 (n
□
Equation (6.11) implies that S4 will not change for greater n .
For n, k , A, B ∈  0 with A ≤ B < n put
F5 ( n, k , A, B ) :=
( 2 B − k )!( n − B + k − 1)!( B − A)!( n + A)!
,
( k − A)!( B − k )! B !( n + B )!( n − B + A − 1)!
G5 ( n, k , A, B ) := −
S5 ( n, A, B ) :=
( 2 B − k + 1)!( n − B + k − 1)!( B − A)!( n + A)!
,
( k − A − 1)!( B − k )! B !( n + B + 1)!( n − B + A)!
B
∑ F5 ( n, k , A, B ) .
k=A
Lemma 7. For every A, B, n ∈  0 with A ≤ B < n
S5 ( n, A, B )
=
B
( 2 B − k )!( n − B + k − 1)!( B − A)!( n + A)!
=
∑
( k − A)!( B − k )! B !( n + B )!( n − B + A − 1)!
1.
k=A
Proof. From the polynomial identity
( n − B + k )( n + A + 1) − ( n + B + 1)( n − B + A)= ( 2 B − k + 1)( k − A) − ( n − B + k )( B − k )
we obtain for 0 ≤ A ≤ B < n and k ∈  0
∆ n F5 ( n, k , A, B )
( n − B + k )( n + A + 1) − ( n + B + 1)( n − B + A)
F5 ( n, k , A, B )
( n + B + 1)( n − B + A)

( 2 B − k + 1)( k − A) 
( n − B + k )( B − k )
=
+
 −
 F5 ( n, k , A, B )
 ( n + B + 1)( n − B + A ) ( n + B + 1)( n − B + A ) 
( 2 B − k )!( n − B + k )!( B − A)!( n + A)!
= −
−
k
( A)!( B − k − 1)! B !( n + B + 1)!( n − B + A)!
( 2 B − k + 1)!( n − B + k − 1)!( B − A)!( n + A)!
+
( k − A − 1)!( B − k )! B !( n + B + 1)!( n − B + A)!
= ∆ k G5 ( n, k , A, B ) .
=
2507
(6.12)
J. Šustek
For k = A and k= B + 1 we have
G
=
A=
, A, B ) 0
5 ( n, k
and
G5 ( n, k =
B + 1, A, B ) =
0.
From this and (6.12) we obtain
∆ n S5 ( n, A, B )
B
B
=
∑∆ n F5 ( n, A, B ) =
∑∆ n G5 ( n, A, B )
(6.13)
=
k A=
k A
=
G5 ( n, k =
B + 1, A, B ) − G5 ( n, k =
A, A, B ) =
0.
Lemma 6 for n = B yields
S5 ( n= B + 1, A, B )
=
( 2 B − k )!k !( B − A)!( A + B + 1)!
B
=
∑
k = A ( k − A ) !( B − k ) ! B !( 2 B + 1) ! A!
S=
B=
, A ) 1.
4 (n
Equation (6.13) implies that S5 ( n, A, B ) has the same value for greater n .
For n, B ∈ 1 and k ∈  0 put
F6 ( n, k , B ) :=
□
( −1) ( n + k )!( 2n + B )!2 ( B − 1)!( 2n + B − k − 1)!n !
,
( n − k )!2 k !( n + B + k )!( n + B − 1)!2 ( 3n + B )!
k
T6 ( n, k , B ) := 3nB 4 − kB 4 + 3B 4 + 24n 2 B 3 − 9knB 3 + 39nB 3 − k 2 B 3 − 6kB 3
+ 15 B 3 + 69n3 B 2 − 26kn 2 B 2 + 151n 2 B 2 − 7 k 2 nB 2 − 30knB 2
+ 104nB 2 − 5k 2 B 2 − 7 kB 2 + 22 B 2 + 84n 4 B − 28kn3 B + 225n3 B
−16k 2 n 2 B − 38kn 2 B + 211n 2 B − 23k 2 nB − 9knB + 80nB − 8k 2 B
+ 2kB + 10 B + 37 n5 − 10kn 4 + 116n 4 − 11k 2 n3 − 14kn3 + 136n3
− 22k 2 n 2 − 2kn 2 + 74n 2 − 13k 2 n + 2kn + 19n − 2k 2 + 2,
G6 ( n, k , B ) :=
( −1) ( n + k )!( 2n + B )!2 ( B − 1)!( 2n + B − k )!n!T6 ( n, k , B )
,
( n − k + 1)!2 ( k − 1)!( n + B + k )!( n + B )!2 ( 3n + B + 3)!
k +1
n
S6 ( n, B ) := ∑F6 ( n, k , B ) .
k =0
Lemma 8. For every n, B ∈ 1
( −1) ( n + k )!( 2n + B )!2 ( B − 1)!( 2n + B − k − 1)!n !
=
S 6 ( n, B ) ∑
=
( n − k )!2 k !( n + B + k )!( n + B − 1)!2 ( 3n + B )!
k =0
k
n
1.
Proof. From the polynomial identity
( n + k + 1)( 2n + B + 1) ( 2n + B + 2 ) ( 2n + B − k )( 2n + B − k + 1)( n + 1)
2
2
− ( n − k + 1) ( n + B + k + 1)( n + B ) ( 3n + B + 1)( 3n + B + 2 )( 3n + B + 3)
2
= ( n + k + 1) T6 ( n, k + 1, B )( n − k + 1) + ( 2n + B − k ) T6 ( n, k , B ) k ( n + B + k + 1)
2
2
we obtain
2508
J. Šustek
∆ n F6 ( n, k , B )
 ( n + k + 1)( 2n + B + 1)2 ( 2n + B + 2 )2 ( 2n + B − k )( 2n + B − k + 1)( n + 1) 


 − ( n − k + 1)2 ( n + B + k + 1)( n + B )2 ( 3n + B + 1)( 3n + B + 2 )( 3n + B + 3) 
 F6 ( n, k , B )
=
2
2
 ( n − k + 1) ( n + B + k + 1)( n + B ) ( 3n + B + 1)( 3n + B + 2 )( 3n + B + 3) 






( n + k + 1) T6 ( n, k + 1, B )( n − k + 1) + ( 2n + B − k ) T6 ( n, k , B ) k ( n + B + k + 1)
=
F6 ( n, k , B )
2
2
( n + B + k + 1)( n − k + 1) ( n + B ) ( 3n + B + 1)( 3n + B + 2 )( 3n + B + 3)
2

( n + k + 1) T6 ( n, k + 1, B )
=
 ( n + B + k + 1)( n + B )2 ( 3n + B + 1)( 3n + B + 2 )( 3n + B + 3)

+
(6.14)

( 2n + B − k ) T6 ( n, k , B ) k
 F6 ( n, k , B )
2
( n − k + 1) ( n + B ) ( 3n + B + 1) × ( 3n + B + 2 )( 3n + B + 3) 
2
( −1) ( n + k + 1)!( 2n + B )!2 ( B − 1)!( 2n + B − k − 1)!n!T6 ( n, k + 1, B )
( n − k )!2 k!2 ( n + B + k + 1)!( n + B )!2 ( 3n + B + 3)!
k +1
( −1) ( n + k )!( 2n + B )!2 ( B − 1)!( 2n + B − k )!n!T6 ( n, k , B )
−
( n − k + 1)!2 ( k − 1)!( n + B + k )!( n + B )!2 ( 3n + B + 3)!
∆ k G6 ( n, k , B ) .
k +2
=
=
For k = 0 we have G6 (=
n, k 0,=
B ) 0 . From this, (6.14) and the polynomial identity
T6 ( n, k = n + 1, B ) = ( 2n + 2 )( 2n + B + 1) ( 2n + B + 2 )( n + B )
2
we obtain
n
∆ n S6 ( n, B ) = F6 ( n + 1, k =n + 1, B ) + ∑∆ n F6 ( n, k , B )
k =0
n
= F6 ( n + 1, k =n + 1, B ) + ∑∆ k G6 ( n, k , B )
k =0
0, B )
=
F6 ( n + 1, k =
n + 1, B ) + G6 ( n, k =
n + 1, B ) − G6 ( n, k =
( −1) ( 2n + 2 )!( 2n + B + 2 )!2 ( B − 1)!( n + B )!( n + 1)!
=
0!2 ( n + 1) !( 2n + B + 2 ) !( n + B ) !2 ( 3n + B + 3) !
n+2
( −1) ( 2n + 1)!( 2n + B )!2 ( B − 1)!( n + B − 1)!n!T6 ( n, k = n + 1, B )
+
0!2 n!( 2n + B + 1)!( n + B ) !2 ( 3n + B + 3) !
n +1
( −1) ( 2n + 1)!( 2n + B )!2 ( B − 1)!( n + B − 1)!
=
( 2n + B + 1)!( n + B )!2 ( 3n + B + 3)!
n +1
(
× ( 2n + 2 )( 2n + B + 1) ( 2n + B + 2 )( n + B ) − T6 ( n, k = n + 1, B )
2
(6.15)
)
= 0.
For n = 1 we have
( B − 1)!( B + 2 )!2 2 ( B − 1)!( B + 2 )! ( B + 1)( B + 2 ) − 2
S6 ( n =
1, B ) =
−
=
=
1.
B !( B + 3)!
B ( B + 3)
B !2 ( B + 3)!
Equation (6.15) implies that S6 ( n, B ) has the same value for greater n .
2509
□
J. Šustek
For n, A, B ∈ 1 and k ∈  0 with n ≥ 2 A + B put
F7 ( n, k , A, B ) :=
( −1) ( A + k )!( 2 A + B )!( B − 1)!( n − k − 1)!( n − A − B )!n !
,
( A − k )!k !( A + B + k )!( A + B − 1)!( n − A − 1)!( n − A − B − k )!( n + A)!
G7 ( n, k , A, B ) :=
( −1) ( A + k )!( 2 A + B )!( B − 1)!( n − k )!( n − A − B )!n !
,
( A − k )!( k − 1)!( A + B + k − 1)!( A + B + 1)!( n − A)!( n − A − B − k + 1)!( n + A + 1)!
k
k +1
A
S7 ( n, A, B ) := ∑F7 ( n, k , A, B ) .
k =0
Lemma 9. For every A, B, n ∈ 1 with n ≥ 2 A + B
S7 ( n, A, B )
A
( −1) ( A + k )!( 2 A + B )!( B − 1)!( n − k − 1)!( n − A − B )!n!
= 1.
∑
−
A
k
)!k!( A + B + k )!( A + B − 1)!( n − A − 1)!( n − A − B − k )!( n + A)!
k =0 (
k
Proof. From the polynomial identity
( n − k )( n − A − B + 1)( n + 1) − ( n − A)( n − A − B − k + 1)( n + A + 1)
= ( A + k + 1)( A − k )( n − A − B − k + 1) + ( n − k ) k ( A + B + k )
we obtain for n, A, B ∈ 1 and k ∈  0 with n ≥ 2 A + B
∆ n F7 (n, k , A, B)
( n − k )( n − A − B + 1)( n + 1) − ( n − A)( n − A − B − k + 1)( n + A + 1)
F7 ( n, k , A, B )
( n − A)( n − A − B − k + 1)( n + A + 1)
( A + k + 1)( A − k )( n − A − B − k + 1) + ( n − k ) k ( A + B + k )
=
F7 ( n, k , A, B )
( n − A)( n − A − B − k + 1)( n + A + 1)
 ( A + k + 1)( A − k )

(n − k ) k ( A + B + k )
= 
+
 F7 ( n, k , A, B )
 ( n − A )( n + A + 1) ( n − A )( n − A − B − k + 1)( n + A + 1) 
=
(6.16)
( −1) ( A + k + 1)!( 2 A + B )!( B − 1)!( n − k − 1)!( n − A − B )!n!
=
( A − k − 1)!k!( A + B + k )!( A + B + 1)!( n − A)!( n − A − B − k )!( n + A + 1)!
k +1
( −1) ( A + k )!( 2 A + B )!( B − 1)!( n − k )!( n − A − B )!n!
−
( A − k )!( k − 1)!( A + B + k − 1)!( A + B + 1)!( n − A)!( n − A − B − k + 1)!( n + A + 1)!
= ∆ k G7 ( n, k , A, B ) .
k +2
For k = 0 and k= A + 1 we have
G=
0,=
A, B ) 0
7 ( n, k
and
G7 ( n, k =
A + 1, A, B ) =
0.
From this and (6.16) we obtain
A
A
∆ n S7 ( n, A, B ) =
∑∆ n F7 ( n, k , A, B ) =
∑∆ k G7 ( n, k , A, B )
=
k 0=
k 0
=
G7 ( n, k =
A + 1, A, B ) − G7 ( n, k =
0, A, B ) =
0.
For =
n 2 A + B Lemma 8 implies
2510
(6.17)
J. Šustek
S7 (=
n 2 A + B, A, B )
( −1) ( A + k )!( 2 A + B )!( B − 1)!( 2 A + B − k − 1)!A!( 2 A + B )!
k = 0 ( A − k ) !k!( A + B + k ) !( A + B − 1) !( A + B − 1) !( A − k ) !( 3 A + B ) !
= S=
A=
, B ) 1.
6 (n
Equation (6.17) implies that S7 ( n, B ) has the same value for greater n .
k
A
=∑
□
6.2. Formulas
Now the symbol k again has its original meaning.
For k ∈  2 and i ∈ 1 put
g 0, k : 1=
=
and gi , k :
( −1)
i −1
k ii!
i −1
∏ ( jk − 1) .
j =1
The numbers gi , k are the coefficients of Taylor’s polynomial of the function
Lemma 10. For k ∈  2 and t ∈ ( 0, 2 )
∞
k
t.
∑gi ,k ( t − 1) .
k
=
t
i
i =0
Proof. See for instance Theorem 159 in [15].
Now we prove a technical lemma that we need later.
Lemma 11. For k ∈  , s ∈ 1 and j = 1, , s
□
( 2s − j )! s
( 2s − j )! s
=
( hk + 1)
∏
∏ ( hk − 1)
j !( s − j ) ! h =s − j +1
j !( s − j ) ! h =s − j +1
 ( −1)i
−∑ 
i!
i =1 

j
s

( 2s − j + i )! i −1
( hk − 1) ∏ ( hk − 1)  .
∏
( j − i )!( s − j + i )! h = 1
h = s − j + i +1

(6.18)
Proof. On both sides of (6.18) there are polynomials of degree j in variable k . Therefore it suffices to
1
prove the equality for k → ∞ and for j values k =
with m = s − j + 1, , s .
m
For k → ∞ we immediately have
LHS ∼
For k =
( 2s − j )! j s
k ∏ h ∼ RHS.
j !( s − j ) ! h =s − j +1
1
with m = s − j + 1, , s we obtain on the left-hand side of (6.18)
m
=
LHS
( 2s − j )! s  h  ( 2s − j )! 1 ( m + s )!
=
.
∏  + 1
j !( s − j ) ! h =s − j +1  m  j !( s − j ) ! m j ( m + s − j ) !
For such numbers m we have
h

0.
 − 1 =
m


h =s − j +1
s
∏
Therefore we obtain on the right-hand side of (6.18)
( 2s − j )! s  h  j ( −1)  ( 2s − j + i )! i −1  h  s  h  

∏  − 1 − ∑
∏  − 1 ∏  − 1 
j !( s − j ) ! h =s − j +1  m  i =1 i !  ( j − i )!( s − j + i )! h =1  m  h =s − j + i +1  m  
j 
s

1 ( 2s − j + i )!
1 i −1
∑  i ! ( j − i )!( s − j + i )! m j −1 ∏ ( m − h ) ∏ ( h − m )  .
i =1 
h= 1
h = s − j + i +1

i
=
RHS
=
2511
J. Šustek
The first product on the last line is equal to zero for i ≥ m + 1 , the second product on the last line is equal to
zero for i ≤ m + j − s − 1 . For other values of i both products contain only positive terms. Hence
s
 1 ( 2s − j + i )!

1 i −1
m − h ) ∏ ( h − m ) 

j −1 ∏ (
i = m − s + j  i ! ( j − i ) !( s − j + i ) ! m
h= 1
h = s − j + i +1

min ( j , m )
∑
=
LHS
=
( 2s − j + i )!m !( s − m )!
.
j
i = m − s + j i !( j − i ) !( s − j + i ) ! m ( m − i ) !( s − m − j + i ) !
min ( j , m )
∑
This implies that
( 2s − j + i )!m!( s − m )!j!( s − j )!( m + s − j )!
RHS min ( j , m )
= ∑
.
LHS i = m − s + j i!( j − i ) !( s − j + i ) !( m − i ) !( s − m − j + i ) !( 2 s − j ) !( m + s ) !
For i < m − s + j we have
1
= 0 , for i > j we have
( s − m − j + i )!
1
= 0 and for i > m we have
( j − i )!
1
= 0 . Then Lemma 3 implies
s
−
( i )!
=
1 S1=
, A m=
, B j)
( n s=
( 2s − j + i )!m !( s − m )!j !( s − j )!( m + s − j )!
i = 0 i !( j − i ) !( s − j + i ) !( m − i ) !( s − m − j + i ) !( 2 s − j ) !( m + s ) !
min ( j , m )
( 2s − j + i )!m !( s − m )!j !( s − j )!( m + s − j )!
= ∑
i = m − s + j i !( j − i ) !( s − j + i ) !( m − i ) !( s − m − j + i ) !( 2 s − j ) !( m + s ) !
s
=∑
=
RHS
.
LHS
1
with m = s − j + 1, , s .
m
For k ∈  2 , s ∈ 1 and j = 0, , s put
Hence equality (6.18) follows for k =
c j ,k ,s :
=
( 2s − j )! s − j s
k
∏ ( hk + 1),
j !( s − j )!
h =s − j +1
=
d j ,k ,s :
( 2s − j )! s − j s
k
∏ ( hk − 1) .
j !( s − j )!
h =s − j +1
□
For k ∈  2 and s ∈ 1 put
s
pk , s ( t ) :=
∑c j ,k , s ( t − 1)
j
∑d j ,k , s ( t − 1)
j
j =0
s
.
j =0
We will prove that (6.19) is Padé approximation of k t .
Lemma 12. For k ∈  2 , s ∈ 1 the numbers c j , k , s and d j , k , s satisfy the system of equations
j
∑d j=
−i , k , s gi , k
i =0
c=
j 0, , s.
j ,k ,s ,
Proof. Lemma 11 implies
2512
(6.19)
J. Šustek
( 2s − j )! s − j s
k
∏ ( hk + 1)
j !( s − j ) !
h =s − j +1
=
c j ,k ,s
j  −
( 1) ( 2s − j + i )! s − j
( 2s − j )! s − j s

−
−
k
hk
k
1
)
(
∑
∏
j !( s − j ) !
i =1 
h =s − j +1
 i ! ( j − i ) !( s − j + i ) !
i
=
i −1
×∏ ( hk − 1)
h= 1

s
∏ ( hk − 1) 

h = s − j + i +1
□
 ( 2s − j + i )!
=
k
k s − j +i
( hk − 1) + ∑ 
∏
−
−
+
j !( s − j ) !
j
i
s
j
i
!
!
(
)
(
)
i =1 
h =s − j +1
i −1
s

( −1) i −1
× ∏ ( hk − 1) i ∏ ( hk − 1) 

k i ! h=
1
h =s − j + i +1

( 2s − j )!
j
s
s− j
j
= d j , k , s g 0, k + ∑d j −i , k , s gi , k
i =1
Lemma 13. For k ∈  2 , s ∈ 1 the numbers c j , k , s and d j , k , s satisfy the system of equations
s
∑d s −i ,k=
,s g j +i,k
0,=
j 1, , s.
i =0
Proof. For j = 1, , s we obtain on the left-hand side
i + j −1
 ( s + i )! i ( −1)i + j −1 s

∑  ( s − i )!i ! k k i + j ( i + j )! ∏ ( hk − 1) ∏ ( hk − 1)  .
=i 0=i 0
h=
i +1
h=
1


j
This expression, multiplied by k , is a polynomial of degree s + j − 1 in variable k . Therefore it suffices
s
1
1

=
m 1, , s + j . For m ≤ s we have ∏  h −  =
to prove the equality for s + j values k =
with
0 and
m
k

h =1 
hence the whole expression is equal to zero. For m ≥ s + 1 we obtain
=
LHS
s
∑d s −i,k , s g j +=i
s
( s + i )! i + j −1
 s −1 s
 s ( −1)
LHS =
 k ∏ ( h − m )  ∑ s − i !i ! i + j ! ∏ ( h − m ) .
) ( ) h = i +1
h= 1

 i =0 (
i + j −1
i m − j . Then Lemma
The second product is equal to zero for i > m − j , therefore the summations ends for =
5 implies
 s −1 s
 m − j ( −1) ( s + i ) ! i + j −1
LHS =
k
h
−
m
(
)
∏

 ∑ s − i !i ! i + j ! ∏ ( m − h )
) ( ) h = i +1
h= 1

 i =0 (
i
=
( −1) ( s + i )!( m − i − 1)!
∑ ( s − i )!i !( i + j )!( m − i − j )=!
i =0
i
m− j
□
S3 ( =
n m, A
= j , B= s=
) 0.
Lemma 14. Function pk , s ( t ) is the Padé approximation of the function
Proof. For t → 1 Lemma 10, Lemma 12 and Lemma 13 imply
k
t of order
[ s, s ]
around t = 1 .
∞
 s
 s
i k
i 
i 
t  ∑di , k , s ( t − 1)  ∑gi , k ( t − 1) 
 ∑di , k , s ( t − 1)  =
=
=
i 0

 i 0=
 i 0

=
s
j
s
j
=j 0=i 0
=
s
∑ ( t − 1) ∑d j −i ,k , s gi ,k + ∑ ( t − 1) ∑d s −i ,k , s g j +i ,k + O ( t − 1)
s
∑c j ,k , s ( t − 1)
j =0
s+ j
=j 1
j
+ O ( t − 1)
2 s +1
2 s +1
=i 0
.
□
The result follows.
2513
J. Šustek
Now we find the coefficients ei , k , s and fi , k , s of the Padé approximation
s
pk , s ( t ) =:
∑ei ,k , s t i
i =0
s
∑ fi , k , s t
.
i
i =0
Lemma 15. For every s ∈ 1 , i = 0, , s and k ∈ 
s
s s
ei , k , s =   ∏ ( hk − 1) ∏ ( hk + 1) ,
h = s − i +1
 i  h =i +1
(6.20)
s
s s
fi , k , s =   ∏ ( hk + 1) ∏ ( hk − 1) .
h = s − i +1
 i  h =i +1
(6.21)
Proof. First we prove (6.21). From (6.19) we obtain
s
s
∑d j ,k , s ( t − 1)
=
∑ fi , k , s t i
j
.
=i 0=j 0
Binomial theorem then implies that
fi , k , s
=
s
 j
j =i
 
∑  i  ( −1)
j −i
d j ,k ,s .
Thus equality (6.21) is equivalent to
s
s s
hk
1
+
− 1)
(
)
( hk=
∏
∏
 
h = s − i +1
 i  h = i +1
s
 j 
j =i
 
∑   i  ( −1)
j −i

( 2s − j )! s − j s
k
( hk − 1)  .
∏
j !( s − j ) !
h = s − j +1

(6.22)
On both sides of (6.22) there are polynomials of degree s in variable k . Therefore it suffices to prove the
1
equality for every k =
with r ∈ 1 , r ≥ s + 1 .
r
Lemma 7 implies
1 S5 =
A i,=
B s)
=
( n r ,=
( 2s − j )!( r − s + j − 1)!( s − i )!( r + i )!
j = i ( j − i ) !( s − j ) ! s !( r + s ) !( r − s + i − 1) !
s
 j  ( 2 s − j )! ( r − s + j − 1)!
∑  i  j !( s − j )! ( r − s − 1)!
j =i  
=
 s  ( r + s )! ( r − s + i − 1)!
 
 i  ( r + i )! ( r − s − 1)!
s
s
 j
j −i ( 2s − j )!
∑  i  ( −1) j !( s − j )! ∏ ( h − r )
j =i  
h =s − j +1
s
=∑
=
s
s s
  ∏ (h + r ) ∏ (h − r )
h = s − i +1
 i  h =i +1
s
s
 j
j −i ( 2s − j )!
∑  i  ( −1) j !( s − j )! k s − j ∏ ( hk − 1)
j =i  
h =s − j +1
=
,
s
s s
  ∏ ( hk + 1) ∏ ( hk − 1)
h = s − i +1
 i  h =i +1
hence (6.22) and (6.21) follow.
Putting k := −k into (6.21) and applying binomial theorem in the same way we obtain (6.20).
2514
□
J. Šustek
6.3. Bounds
By Lemma 14 the function pk , s is Padé approximation, hence we know its properties in the neighbourhood of
1. Here we find global bounds for pk , s that are necessary for functionality of our algorithm.
We need another two technical lemmas.
Lemma 16. For k ∈  and s, m ∈ 1
s + m + i −1
( −1) ( s + i )! s
( s + m − 1)! s
=
− 1) ∏ ( hk − 1)
( hk + 1) .
∑ ( s − i )!i !( s + m + i )! ∏ ( hk
( 2s + m )!( m − 1)! ∏
i =0
h=
i +1
h=
s+m
h=
1
s −i
s
Proof. On both sides there are polynomials of degree s in variable k . Therefore it suffices to prove the
1
equality for every k =
with r ∈ 1 , r ≥ 2 s + m .
r
Lemma 9 implies
1 S7=
, A s=
, B m)
=
( n r=
( −1) ( s + i )!( 2s + m )!( m − 1)!( r − i − 1)!( r − s − m )!r !
=∑
s
−
i
)!i !( s + m + i )!( s + m − 1)!( r − s − 1)!( r − s − m − i )!( r + s )!
i =0 (
i
s
s + m + i −1
s
 ( −1)i ( s + i ) !( 2 s + m ) !( m − 1) ! s

−1
∑  ( s − i )!i !( s + m + i )!( s + m − 1)! ∏ ( r − h ) ∏ ( r − h ) ∏ ( h + r ) 
i =0
h=
i +1
h=
s+m
h=
1


s
=
s + m + i −1
s
 ( −1)s −i ( s + i )!( 2 s + m )!( m − 1)! s

−1
= ∑
hk − 1) ∏ ( hk − 1) ∏ ( hk + 1)  .
(
∏

i = 0  ( s − i ) !i !( s + m + i ) !( s + m − 1) ! h =
i +1
h=
s+m
h=
1


s
□
This implies the result.
Lemma 17. For k ∈  2 and j ∈ 1 with j ≥ 2 s + 1
s
∑ ( −1)
i =0
s −i
d s −i , k , s g j − s + i , k > 0.
Proof. Put m :=j − 2 s ≥ 1 . Then
s
∑ ( −1)
=
LHS
s −i
i =0
d s −i , k , s g j − s + i , k
j − s + i −1
s
s 

1
s −i ( s + i )!
=
∑  ( −1) ( s − i )!i ! k i ∏ ( hk − 1) k j − s +i ( j − s + i )! ∏ ( hk − 1) 
i =0 
h=
i +1
h=
1

s −i
j − s + i −1
s 

( −1) ( s + i )! s

−
hk
1
)∑
( hk − 1) ∏ ( hk − 1) 
∏
j−s ∏ (
k
i = 0 ( s − i ) !i !( j − s + i ) ! h =
h=
i +1
h=
j−s
1


s −i
+
+
−
s
m
s
i
1
m + s −1
s 

( −1) ( s + i )!
1
hk − 1) ∏ ( hk − 1)  .
= m + s ∏ ( hk − 1) ∑ 
(
∏

k
i = 0  ( s − i ) !i !( m + s + i ) ! h =
i +1
h=
m+ s
h =1


1
=
j − s −1
Lemma 16 implies
LHS
=
m + s −1
( s + m − 1)! s
1
+
hk
(
)
∏
∏ ( hk − 1) > 0.
!( m − 1)! h 1 =h 1
k m + s ( 2 s + m )=
1
□
Now we find lower and upper bounds for the function pk , s .
Lemma 18. For k ∈  2 , s ∈ 1 and t ∈ ( 0,1) we have pk , s ( t ) > k t .
Proof. Put
∞

s


 ∞
i 
  ∑gi , k ( t − 1)  .
=
 i 0=
 i 0


s
k
1) :  ∑d j , k , s ( t − 1) =
 t  ∑di , k , s ( t − 1)
∑α j ,k , s ( t −=
=j 0
j

=j 0
j
2515
i
(6.23)
J. Šustek
Lemma 12 and Lemma 13 imply
2s
s
1)
∑α j ,k , s ( t −=
∑c j ,k , s ( t − 1)
j
j
.
(6.24)
=j 0=j 0
From (6.23) we obtain for j ≥ 2 s + 1
s
α j , k , s = ∑d s −i , k , s g j − s + i , k .
i =0
We have g j − s + i , k =
( −1)
j − s + i −1
g j − s + i , k . Lemma 17 implies
∞
∑ α j ,k , s ( t − 1)
j
2 s +1
j=
∞
 s

s −i
j
=
− ∑  ∑ ( −1) d s −i , k , s g j − s + i , k  (1 − t ) < 0.
2 s +1  i =
0
j=

(6.25)
From (6.23), (6.24) and (6.25) we obtain
 s
j k
 ∑d j , k , s ( t − 1) =
 t
0
 j=

2s
∑α j ,k , s ( t − 1)
j
+
∞
∑ α j ,k , s ( t − 1)
j
j=
j=
0
2 s +1
s
< ∑c j , k , s ( t − 1) .
j
j=
0
Lemma 15 implies that
s
∑d j ,k , s ( t − 1)
j
> 0,
j =0
hence
s
=
pk , s ( t )
∑c j ,k , s ( t − 1)
j
∑d j ,k , s ( t − 1)
s
j =0
s
□
> k t.
j =0
Lemma 19. For every k ∈  2 , s ∈ 1 and x > 1 we have pk , s ( t ) > 1 .
Proof. Directly from the definition of c j , k , s and d j , k , s we obtain
s
=
pk , s ( t )
∑c j ,k , s ( t − 1)
s
 ( 2s − j )!

s
j
∑  j !( s − j )! k s − j ∏ ( hk + 1)  ( t − 1)
j
h =s − j +1


> 1.
 ( 2s − j )! s − j s

j
∑  j !( s − j )! k ∏ ( hk − 1)  ( t − 1)
j =0 
h =s − j +1

j =0
j =0
s
=
j
∑d j ,k , s ( t − 1)
s
j =0
(6.26)
The inequality is strict, since for j ≥ 1 the main bracket in the numerator is greater than the main bracket in
the denominator.
□
Now we prove that function pk , s ( t ) satisfies inequalities (3.2).
Lemma 20. For every k ∈  2 , s ∈ 1 and t ∈  +  {1}
(
)
(
)
min 1, k t < pk , s ( t ) < max 1, k t .
Proof. The proof splits into four cases.
1) For 0 < t < 1 Lemma 18 implies that k t < pk , s ( t ) .
2) Lemma 15 implies that e j , k , s = f s − j , k , s for every j , k , s . Hence
1
1
pk , s   =
.
 t  pk , s ( t )
Using this and the first case we obtain that for t > 1 we have pk , s ( t ) < k t .
3) For t > 1 Lemma 19 implies that pk , s ( t ) > 1 .
4) Using this and (6.27) we obtain for 0 < t < 1 that pk , s ( t ) < 1 .
(6.27)
□
Acknowledgements
The author would like to thank to professor Andrzej Schinzel for recommendation of the paper [14] and also to
2516
J. Šustek
author’s colleagues Kamil Brezina, Lukáš Novotný and Jan Štĕpnička for checking the results. Publication of
this paper was supported by grant P201/12/2351 of the Czech Science Foundation, by grant 01798/2011/RRC of
the Moravian-Silesian region and by grant SGS08/PrF/2014 of the University of Ostrava.
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