Hermite integrators and Riordan arrays

Hermite integrators and Riordan arrays
When one feels tired in round-oﬀ error
Keigo Nitadori
keigo@riken.jp
Co-Design Team, Flagship 2020 Project
RIKEN Advanced Institute for Computational Science
January 7, 2016

Today I talk about
A general form of the correctors for the family of 2-step Hermite
integrators
Up to p-th order derivative of the force is calculated directly to
obtain 2(p + 1)-th order accuracy
A mathematical proof in elementary algebra is presented
In my last talk in MODEST 15-S, a simple and beautiful proof
made by Satoko Yamamoto were mostly omitted, but today I
would like you to see the full story.

References
I appreciate for many useful online documents:
1. R. Sprugnoli (2006). An Introduction to Mathematical Methods
in Combinatorics.
http://www.dsi.unifi.it/~resp/Handbook.pdf
2. R. Sprugnoli (2006). Riordan Array Proofs of Identities in
Gould’s Book.
http://www.dsi.unifi.it/~resp/GouldBK.pdf
3. D. Merlini (2011). A survey on Riordan arrays.
https://lipn.univ-paris13.fr/~banderier/
Seminaires/Slides/merlini.pdf
4. P. Barry (2005) A Catalan Transform and Related
Transformations on Integer Sequences. Journal of Integer
Sequences 8. https://cs.uwaterloo.ca/journals/JIS/
VOL8/Barry/barry84.html

Polynomial shift by Pascal matrix I
We consider to shift a ﬁnite order polynomial f (t), of which up to the
p-th derivatives are f (n) (t) = dn
dtn f (t) (0 ≤ n ≤ p). Adjusting the
dimension of the derivatives by a step size h, a vector
F(t) =
f (t)
h f (1) (t)
h2 f (2) (t)/2!
...
hn f (p) (t)/n!
. (1)
obeys a diﬀerential equation
d
dt
F(t) =
1
h
0 1 0 · · · 0
0 0 2 · · · 0
0 0 0
...
...
...
...
...
... p
0 0 0 · · · 0
F(t). (2)

Polynomial shift by Pascal matrix II
A formal solution at t + h is,
F(t + h) = exp

0 1 0 · · · 0
0 0 2 · · · 0
0 0 0
...
...
...
...
...
... p
0 0 0 · · · 0

F(t)
=
0
0
1
0
2
0 · · · p
0
0 1
1
2
1 · · · p
1
0 0 2
2 · · · p
2
...
...
...
...
...
0 0 0 · · · p
p
F(t). (3)
This matrix is referred to as upper triangle Pascal matrix.

Polynomial shift by Pascal matrix III
Example
For p = 9,
1 1 1 1 1 1 1 1 1 1
0 1 2 3 4 5 6 7 8 9
0 0 1 3 6 10 15 21 28 36
0 0 0 1 4 10 20 35 56 84
0 0 0 0 1 5 15 35 70 126
0 0 0 0 0 1 6 21 56 126
0 0 0 0 0 0 1 7 28 84
0 0 0 0 0 0 0 1 8 36
0 0 0 0 0 0 0 0 1 9
0 0 0 0 0 0 0 0 0 1

Construction of the Hermite integrators I
For simplicity, we integrate from t = −h to t = h, thus a timestep
becomes ∆t = 2h (not h).
The inputs are additions and subtractions
F+
n =
1
2
hn
n!
f (n)
(h) + f (n)
(−h) , (4)
F−
n =
1
2
hn
n!
f (n)
(h) − f (n)
(−h) , (5)
for 0 ≤ n ≤ p. The outputs are ﬁtting polynomial at t = 0,
Fn =
hn
n!
f (n)
(0), (6)
for 0 ≤ n ≤ 2p + 1.
The integral includes only the even order terms:
∆v =
h
−h
f (t)dt = F0 +
1
3
F2 +
1
5
F4 + · · · +
1
2p + 1
F2p, (7)
for a 2(p + 1)-th order integrator.

Construction of the Hermite integrators II
A linear equation we want to solve is
F+
0
F−
1
F+
2
F−
3
...
=
0
0
2
0
4
0
6
0 . . .
0 2
1
4
1
6
1 . . .
0 2
2
4
2
6
2 . . .
0 0 4
3
6
3 . . .
...
...
...
...
...
F0
F2
F4
F6
...
, (8)
of size (p + 1). The element of matrix is 2j
i (counted from (0, 0)),
which means we have picked up the even number columns from the
Pascal matrix.

Construction of the Hermite integrators III
Example
By solving
F+
0
F−
1
F+
2
=
1 1 1
0 2 4
0 1 6
F0
F2
F4
, (9)
we have
F0
F2
F4
=
1
8
8 −5 2
0 6 −4
0 −1 2
F+
0
F−
1
F+
2
. (10)
The sixth-order integrator is thus
∆v =2 F0 +
1
3
F2 +
1
5
F4 h
= F+
0 −
2
5
F−
1 +
2
15
F+
2 ∆t (11)

Coeﬃcients table
F+
0 F−
1 F+
2 F−
3 F+
4 F−
5 F+
6 F−
7
A2 1
H4 1 −
1
3
H6 1 −
2
5
2
15
H8 1 −
3
7
4
21
−
2
35
H10 1 −
4
9
6
27
−
2
21
8
315
H12 1 −
5
11
8
33
−
4
33
8
165
−
8
693
H14 1 −
6
13
10
39
−
20
143
48
715
−
32
1287
16
3003
H16 1 −
7
15
12
45
−
2
13
16
195
−
16
429
64
5005
−
16
6435
Table: Coeﬃcients for up to the 16th-order Hermite integrator

General form
For the 2(p + 1)-th order integrator (p ≥ 0), the k-th term
(0 ≤ k ≤ p) is,
c(p)
k =
1
(−2)k
(2k)!!
(2k + 1)!!
p − k + m
p − k
(2k + 1)!!
(2k + 1 − 2m)!!
(2p + 1 − 2m)!!
(2p + 1)!!
,
(12)
with m = (k + 1)/2 . A recurrent form starting from the diagonal
element to the bottom is
c(p)
k =



1
(−2)p
(2p)!!
(2p + 1)!!
(p = k)
p − k + m
p − k
2p + 1 − 2m
2p + 1
c(p−1)
k (p > k)
. (13)
It is a rational recurrent form, however, we still seek for a diﬀerential
recurrent form in c(p)
k − c(p−1)
k for our proof.

Coﬀee break (double factorial)
Deﬁnition:
n!! = n · (n − 2)!! (for n ≥ 2), 1!! = 0!! = 1. (14)
(There are products of odd numbers, and even numbers)
Properties:
(2n)!! =2n
n!, (15)
(2n + 1)!!(2n)!! =(2n + 1)!, (16)
(2n + 1)!! =
(2n + 1)!
2nn!
=
(n + 1)!
2n
2n + 1
n
, (17)
(2n)!!(2n − 1)!! =(2n)!, (18)
(2n − 1)!! =
(2n)!
2nn!
=
n!
2n
2n
n
=
n!
2n
2n
n
2n − 1
n − 1
, (19)
etc.

Outline of the proof
1. We wrote an expected form of the coefficients, c(p)
k
2. Then calculate a differential recurrence, c(p)
k − c(p−1)
k
3. We set a linear equation Ax = b, of which solution x should
correspond to the expected coefficients
4. LU decomposition, A = LU
5. The proof for the form of inverse matrices L−1 and U−1 will be
shown later, by using of a modern tool Riordan arrays
6. Calculate L−1b
7. Finally we see that the solution x = U−1L−1b obeys the same
recursion to c(p)
k , given a matrix size (p + 1)

Differential recurrence I
This is a hand exercise (remember m = (k + 1)/2 ):
c(p)
k − c(p−1)
k =c(p)
k 1 −
p − k + m
p − k
2p + 1 − 2m
2p + 1
=c(p)
k
m(2k + 1 − 2m)
(p − k + m)(2p + 1 − 2m)
(20)
Now, the difference is simplified as in
1
(−2)k
(2k)!!
(2k + 1)!!
p − k + m
m
(2k + 1)!!
(2k + 1 − 2m)!!
(2p + 1 − 2m)!!
(2p + 1)!!
×
m(2k + 1 − 2m)
(p − k + m)(2p + 1 − 2m)
=
(−1)k k!
(2p + 1)!!
p − k + m − 1
m − 1
(2p − 1 − 2m)!!
(2k − 1 − 2m)!!
=
(−1)k k!
(2p + 1)!!
(p − k + m − 1)!
(p − k)!(m − 1)!
2k−m
2p−m
(k − m)!
(p − m)!
(2p − 2m)!
(2k − 2m)!
(21)

Diﬀerential recurrence II
Let us now introduce b = k mod 2 ∈ {0, 1}, and ¯b = 1 − b, hence
k = 2m − b. Some useful properties are:
(n + b)!
n!
= (n + 1)b
,
(n − b)!
(n − 1)!
= n
¯b
, (n − 1)b
a
¯b
= n − b,
(−1)k
2p−k
(2m − b)!
(2p + 1)!!
(p − m − 1 + b)!
(p − 2m + b)!(m − 1)!
(m − b)!
(p − m)!
(2p − 2m)!
(2m − 2b)!
=
(−1)k
2p−k
(2m − 1)b
(2p + 1)!!
(2p − 2m)!
(p − 2m + b)!
m
¯b
(p − m)¯b
×
2
¯b
2¯b
=
(−1)k
2p−k (2p + 1)!!
(2m − 1)b (2m)
¯b
(p − 2m + b)!
(2p − 2m)!
(2p − 2m)¯b
=
(−1)k
2p−k (2p + 1)!!
2m − b
(p − 2m + b)!
(2p − 2m + b − 1)! ×
(2p)!!
2p p(p − 1)!
=
(−1)k
22p−k
(2p)!!
(2p + 1)!!
k
p
(2p − k − 1)!
(p − k)!(p − 1)!
(22)

Diﬀerential recurrence III
Finally, we have
c(p)
k =



1
(−2)p
(2p)!!
(2p + 1)!!
(p = k)
c(p−1)
k +
(−1)k
22p−k
(2p)!!
(2p + 1)!!
k
p
2p − k − 1
p − k
(p > k)
.
(23)
for which we are going to make the proof. Note that this form does not
include m = (k + 1)/2 .

Diﬀerential recurrence (old version) I
When k is even (k = 2m),
(2m)!
(2p + 1)!!
p − m − 1
m − 1
(2p − 2m − 1)!!
(2m − 1)!!
=
(2m)!!
(2p + 1)!!
(p − m − 1)!
(p − 2m)!(m − 1)!
(2p − 2m − 1)!
(2p − 2m − 2)!!
=
1
2p−m−1
2mm!
(2p + 1)!!
(2p − 2m − 1)!
(p − 2m)!(m − 1)!
×
m!
m!
=
k
2p−k
1
(2p + 1)!!
(2p − 2m − 1)!
(p − 2m)!
×
(p − 1)!
(p − 1)!
=
k
2p−k
(p − 1)!
(2p + 1)!!
2p − k − 1
p − k
×
p
p
(2p)!!
(2p)!!
=
k
2p−k
(2p)!!
(2p + 1)!!
(p − 1)!
2p p!
2p − k − 1
p − k
=
1
22p−k
(2p)!!
(2p + 1)!!
k
p
2p − k − 1
p − k
. (24)

Diﬀerential recurrence (old version) II
When k is odd (k = 2m − 1),
−
(2m − 1)!
(2p + 1)!!
p − m
m − 1
(2p − 2m − 1)!!
(2m − 3)!!
= −
(2m − 1)(2m − 2)!!
(2p + 1)!!
(p − m)(p − m − 1)!
(p − 2m + 1)!(m − 1)!
(2p − 2m − 1)!
(2p − 2m − 2)!!
= −
2m−1
2p−m−1
(2m − 1)(p − m)
(2p + 1)!!
(2p − 2m − 1)!
(p − 2m + 1)!
×
2p − 2m
2p − 2m
= −
1
2p−k−1
k
(2p + 1)!!
p − m
2p − 2m
(2p − 2m)!
(p − 2m + 1)!
×
(p − 1)!
(p − 1)!
= −
k
2p−k
(p − 1)!
(2p + 1)!!
2p − k − 1
p − k
×
p
p
(2p)!!
(2p)!!
= −
1
22p−k
(2p)!!
(2p + 1)!!
k
p
2p − k − 1
p − k
. (25)

Linear equation to solve
What we want to solve is
Ax = b, (A ∈ N
(p+1)×(p+1)
0 , x, b ∈ Qp+1
) (26)
where
Ai j =
2i
j
and bj =
1
2j + 1
, (0 ≤ i, j ≤ p) (27)
For example when p = 5,
1 0 0 0 0 0
1 2 1 0 0 0
1 4 6 4 1 0
1 6 15 20 15 6
1 8 28 56 70 56
1 10 45 120 210 252
1
−5/11
8/33
−4/33
8/165
−8/693
=
1
1/3
1/5
1/7
1/9
1/11
. (28)
gives the coeﬃcients of the 12th-order integrator

LU decomposition
An LU decomposition A = LU is available in,
Ai j =
2i
j
, Li j =
i
j
, Ui j = 22i−j i
j − i
, (29)
irrespective to the matrix size (p + 1). Thus, LUx = b can be solved
as x = U−1L−1b.
Example
1 0 0 0 0 0
1 2 1 0 0 0
1 4 6 4 1 0
1 6 15 20 15 6
1 8 28 56 70 56
1 10 45 120 210 252
=
1 0 0 0 0 0
1 1 0 0 0 0
1 2 1 0 0 0
1 3 3 1 0 0
1 4 6 4 1 0
1 5 10 10 5 1
1 0 0 0 0 0
0 2 1 0 0 0
0 0 4 4 1 0
0 0 0 8 12 6
0 0 0 0 16 32
0 0 0 0 0 32
(30)
For the proof, we prepare the tools in mathematical combinatorics.

Formal power series, and coeﬃcient extraction
Formal power series
f (t) =
∞
k=0
tk
fk
f (t) is referred to as a generating function of a sequence
( f0, f1, f2, . . .).
Coeﬃcient extraction
[tn
] f (t) = fn
This is an operator but with a weak associativity.
Shifting
[tn
]tk
f (t) = [tn−k
] f (t)
Newton’s binomial theorem
[tk
](1 + t)n
=
n
k

Proof for A = LU
2i
j
=[tj
](1 + t)2i
=[tj
] 1 + t(t + 2)
i
=[tj
]
∞
k=0
i
k
tk
(t + 2)k
=[tj−k
]
∞
k=0
i
k
∞
=0
k
t 2k−
=
∞
k=0
i
k
k
j − k
22k−j
. (31)
This discussion is valid for ﬁnite matrices, for k iterates from 0 to
min(i, j), touching only the non-zero elements of the upper and lower
triangular matrices.

Inverse triangle matrices, L−1 and U−1
For the lower triangle Li j =
i
j
,
L−1
i j
= (−1)i+j i
j
, (32)
and for the upper triangle Ui j = 22i−j i
j − i
,
[U−1
]i j =



(−1)i+j
22j−i
i
j
2j − i − 1
j − i
(1 ≤ i ≤ j)
1 (i = j = 0)
0 (otherwise)
. (33)
The both do not depend on the matrix size 0 ≤ i, j ≤ p. The proof is a
little bit technical and we put them later.

Ux = L−1b
L−1
b
i
=
∞
j=0
(−1)i+j i
j
1
2j + 1
=
∞
j=0
(−1)i+j i
j
1
0
x2j
dx
=(−1)i
1
0
1 − x2 i
dx
=(−1)i (2i)!!
(2i + 1)!!
. (34)
This vector does not depend on the vector length p. A proof of the
fourth identity (an integral to double factorials) is in the next slide.

Integral and double factorial
We apply the integration by parts:
In =
1
0
1 + x2 n
dx
=
1
0
x 1 + x2 n
dx
= x 1 − x2 n 1
0
−
1
0
x 1 − x2 n
dx
= − 2n
1
0
−x2
1 + x2 n−1
dx
= − 2n
1
0
1 − x2
1 + x2 n−1
− 1 + x2 n−1
dx
= − 2n (In − In−1) . (35)
For In =
2n
2n + 1
In−1 and I0 = 1, we have In =
(2n)!!
(2n + 1)!!
.

The solution, x = U−1L−1b
The last element of unknown vector x(p) is now available in
x(p)
p = L−1
b
p
/ [U]pp =
1
(−2)p
(2p)!!
(2p + 1)!!
. (36)
In a general case,
x(p)
k =
p
j=k
U−1
k j
L−1
b
j
. (37)
From the p-dependency of the solution vector x(p), we have a
recursion
x(p)
k − x(p−1)
k = U−1
k p
L−1
b
p
=
(−1)k
22p−k
k
p
(2p)!!
(2p + 1)!!
2p − k − 1
p − k
.
(38)
This is (23), exactly what we wanted!!

Riordan array
For the proof of our matrix form L−1 and U−1, we make a brief review
of Riordan arrays.
A Riordan array is an infinite lower triangular matrix taking a pair of
formal power series d(t) (d0 0) and h(t) (h0 = 0, h1 0) 1, of
which element in the n-th row k-th column (counted from (0, 0)) is
R d(t), h(t)
n,k
= [tn
]d(t) (h(t))k
(39)
They make a group, providing explicit forms of
matrix multiplication, identity, and inverse.
In this study, we exploit the Riordan arrays to find matrix inverses of
which coefficients are expressed in binomials.
1Notice that Handbook.pdf by Sprugnoli takes a slightly different notation
R(d(t), th(t)) with d0, h0 0, which does not make a practical difference.

Matrix multiplication
By deﬁnition, the product of two Riordan arrays is
R d(t), h(t) R a(t), b(t)
n,k
=
∞
j=0
[tn
]d(t) (h(t))j
· [yj
]a(y) (b(y))k
= [tn
]d(t)
∞
j=0
(h(t))j
· [yj
]a(y) (b(y))k
= [tn
]d(t) · a (h(t)) (b (h(t)))k
= R d(t) · a(h(t)), b(h(t))
n,k
. (40)
In the third equality, we used a relation
∞
j=0
xj
· [yj
] f (y) = f (x), (41)
with x = (h(t))j and f (y) = a(y)(b(y))k .

Riordan group
Identity:
R 1, t
n,k
= [tn
]tk
= δnk (42)
Inverse:
R d(t), h(t)
−1
= R
1
d(¯h(t))
, ¯h(t) (43)
where ¯h(t) is an inverse function of h(t) such that
¯h(h(t)) = h(¯h(t)) = t.
Proof.
Just substitute a(t) = 1/d(¯h(t)) and b(t) = ¯h(t) for the matrix
multiplication to obtain an identity R(1, t) as a product.

Maxima script
Since I am not a mathematician but rather a programmer as you know,
I need concrete examples and numeric dumps to study the arrays.
Maxima is a nice free software for such purposes.
RAelem(d, h, n, k)
:= coeff(taylor(d*h^k, t, 0, n), t^n);
genRA(d, h, nmax)
:= genmatrix(
lambda([i,j], RAelem(d, h, i-1, j-1)),
nmax,
nmax);
Today, these software are not clever enough to ﬁnd a general form or a
proof automatically, but they help us greatly.

Pascal matrix (the matrix L) I
We consider d(t) = 1/(1 − t) and h(t) = t/(1 − t),
L = R
1
1 − t
,
t
1 − t
(44)
The elements are binomial coeﬃcients:
Ln,k = [tn
]
1
1 − t
t
1 − t
k
=[tn−k
](1 − t)−k−1
=[tn−k
]
∞
=0
−k − 1
1−k−1−
(−t)
=(−1)n−k −k − 1
n − k
=
n
k
. (45)
Here, we used −n
k = (−1)k n+k−1
k .

Pascal matrix (the matrix L) II
Now we seek for its inverse.
h =
t
1 − t
⇐⇒ t =
h
1 + h
⇐⇒ ¯h(t) =
t
1 + t
(46)
Thus,
L−1
= R
1
1 + t
,
t
1 + t
(47)
L−1
n,k
[tn
]
1
1 + t
t
1 + t
k
=[tn−k
](1 + t)−k−1
=
−k − 1
n − k
=(−1)n−k n
k
(48)

Pascal matrix (the matrix L) III
The inverse relation to the infinite triangle matrices is valid for the
finite ones, thus, for example:
1 0 0 0 0 0
1 1 0 0 0 0
1 2 1 0 0 0
1 3 3 1 0 0
1 4 6 4 1 0
1 5 10 10 5 1
−1
=
1 0 0 0 0 0
−1 1 0 0 0 0
1 −2 1 0 0 0
−1 3 −3 1 0 0
1 −4 6 −4 1 0
−1 5 −10 10 −5 1
This confirms the relation Li j =
i
j
and [L−1]i j = (−1)i+j i
j
in the
main story.

Catalan’s triangle (the matrix U) I
Let us consider special case that d(t) = 1, and set h(t) = t(1 − t).
B =R 1, t(1 − t) (49)
Bn,k =[tn
](t(1 − t))k
=[tn−k
](1 − t)k
=(−1)n−k k
n − k
(50)
1 0 0 0 0 0 0 0
0 1 0 0 0 0 0 0
0 −1 1 0 0 0 0 0
0 0 −2 1 0 0 0 0
0 0 1 −3 1 0 0 0
0 0 0 3 −4 1 0 0
0 0 0 −1 6 −5 1 0
0 0 0 0 −4 10 −6 1

Catalan’s triangle (the matrix U) II
Now, let us ﬁnd the inverse.
h = t(1 − t) ⇐⇒ t =
1 ±
√
1 − 4h
2
, but ¯h(t) =
1 −
√
1 − 4t
2
B−1
=R 1,
1 −
√
1 − 4t
2
(51)
B−1
n,k
= [tn
]
1 −
√
1 − 4t
2
k
=
k
n
2n − k − 1
n − k
(52)
for 1 ≤ k ≤ n, otherwise δnk .
1 0 0 0 0 0 0 0
0 1 0 0 0 0 0 0
0 1 1 0 0 0 0 0
0 2 2 1 0 0 0 0
0 5 5 3 1 0 0 0
0 14 14 9 4 1 0 0
0 42 42 28 14 5 1 0
0 132 132 90 48 20 6 1

Catalan’s triangle (the matrix U) III
Proof. If we put
w(t) =
1 −
√
1 − 4t
2
(53)
Then, t =
w
1 − w
or w =
t
1 − w
. What we want is
[B−1]n,k = [tn](w(t))k and we can apply the following Lagrange
inverse theorem:
[tn
] F(w) w = tφ(w) =
1
n
[tn−1
]F (t) φ(t)
n
(54)
In our case, F(w) = wk and φ(t) = (1 − t)−1.

Catalan’s triangle (the matrix U) IV
[tn
] wk
w =
t
1 − w
=
1
n
[tn−1
]ktk−1
(1 − t)−n
=
k
n
[tn−k
](1 − t)−n
=
k
n
(−1)n−k −n
n − k
=
k
n
2n − k − 1
n − k
(55)
for 1 ≤ k ≤ n. For other cases n = 0, k = 0, or k > n, we have trivial
elements [B−1]n,k = δnk .

Catalan’s triangle (the matrix U) V
Just transposing and multiplying diagonal matrices from our result
({ai bj Mi j }−1 = {a−j b−i[M−1]i j })
Bi j =(−1)i+j j
i − j
,
B−1
i j
=
j
i
2i − j − 1
i − j
for 1 ≤ j ≤ i otherwise δi j,
we have the last piece needed in our study:
Ui j =(−1)i+j
22i−j
Bji = 22i−j i
j − i
[U−1
]i j =
(−1)i+j
22j−i
B−1
ji
=
(−1)i+j
22j−i
i
j
2j − i − 1
j − i
for 1 ≤ i ≤ j otherwise δi j,
except that Lagrange inverse theorem is used without proof.

Lagrange inverse theorem I
From “Handbook of Mathematical Functions”
If y = f (x), y0 = f (x0), f (x0) 0, then
3.6.6
x = x0 +
∞
k=1
(y − y0)k
k!

dk−1
dxk−1
x − x0
f (x) − y0
k x=x0
3.6.7
g(x) = g(x0) +
∞
k=1
(y − y0)k
k!

dk−1
dxk−1
g (x)
x − x0
f (x) − y0
k x=x0
where g(x) is any function indeﬁnitely diﬀerentiable.
In formal power series,
[tn
] w(t) w = tφ(w) =
1
n
[tn−1
] φ(t)
n
, (56)
[tn
] F(w) w = tφ(w) =
1
n
[tn−1
]F (t) φ(t)
n
. (57)

Lagrange inverse theorem II
This is the last one piece for which we need to see a proof.
As an inverse of a Riordan array R(1, h(t)), that is R(1, ¯h(t)) for
d(t) = 1, we assume the following form:
dn,k = R 1, ¯h(t)
n,k
=
k
n
[tn−k
]
t
h(t)
n
. (58)
If the following relation
vn,k =
∞
j=0
dn, j[yj
] (h(y))k
= δnk, (59)
if satisﬁed, it is equivalent to
[tn
] ¯h(t)
k
= R (1, h(t))−1
n,k
= R 1, ¯h(t)
n,k
=
k
n
[tn−k
]
t
h(t)
n
.
(60)

Lagrange inverse theorem III
We start the multiplication
vn,k =
∞
j=0
j
n
[tn−j
]
t
h(t)
n
[yj
] (h(y))k
, (61)
and by applying a diﬀerentiation rule
j[yj
] (h(y))k
= [yj−1
] (h(y))k
= [yj
]y h (y) k (h(y))k−1
, (62)
we continue the calculation, as in
vn,k =
k
n
∞
j=0
[tn−j
]
t
h(t)
n
[yj
]yh (y) (h(y))k−1
=
k
n
[tn
]
t
h(t)
n
th (t) (h(t))k−1
. (63)
Here, we used a convolution rule:
[tn
] f (t)g(t) =
∞
j=0
[tj
] f (t) · [yn−j
]g(y). (64)

Lagrange inverse theorem IV
When k = n,
vn,n =[tn
]tn
t
h (t)
h(t)
=[t0
]
h (t)
h(t)/t
=[t0
]
h1 + 2h2t + 3h3t2 + · · ·
h1 + h2t + h3t2 + · · ·
= 1 (65)
When k n,
vn,k =
k
n
[tn
]tn+1
(h(t))k−n−1
h (t)
=
k
n
[t−1
]
1
k − n
(h(t))k−n
= 0 (66)
Here, (h(t))k−n is a formal Laurent series of which derivative cannot
contain a non-zero coeﬃcient for t−1.

Lagrange inverse theorem V
Let us set φ(t) = t/h(t) and w = ¯h(t). Then w(t) is an implicit
function of t such that w = tφ(w).
For a special case k = 1, we have
[tn
] w(t) w = tφ(w) = dn,1 =
1
n
[tn−1
] φ(t)
n
.
In more general cases, for a formal power series F(w), we have
[tn
] F(w) w = tφ(w) =[tn
]
∞
k=0
Fk wk
=
∞
k=0
Fk dn,k
=
∞
k=0
Fk
k
n
[tn−k
] φ(t)
n
=
1
n
[tn−1
]
∞
k=0
kFktk−1
φ(t)
n
=
1
n
[tn−1
]F (t) φ(t)
n
.

Summary
Combinatorics, a branch of mathematics, turned out to be a
powerful tool to study numerical integrators for the N-body
problem. Especially the formal power series, Lagrange inverse
theorem and modern Riordan arrays are practical tools for even
non-mathematicians.
LU decomposition is not only for Top500/Green500, but a
general tool for studying integers.
Mathematics is useful.
Satoko Yamamoto did a great work.

Hermite integrators and Riordan arrays

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