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1. Hermite integrators and 2-parameter subgroup of
Riordan group
Keigo Nitadori
keigo@riken.jp
RIKEN Advanced Institute for Computational Science
July 13, 2016

2. Abstract
A general form 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
Now, a simple and general form for the predictor is available
The mathematics was slightly simplified and generalized from the
last talk (99% of the proof was invented by Satoko Yamamoto)

3. Upper triangular Pascal matrix
This following upper triangular Pascal matrix appearers frequently in
this story:
(1) Upas ij
=
j
i
, U−1
pas ij
= (−1)i+j j
i
.
A (p + 1) × (p + 1) sized one looks like,
(2) Upas =
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
.

4. Polynomial shift
A vector of scaled force
(3) F(t) =
f (t)
h f (1) (t)
h2 f (2) (t)/2!
...
hn f (p) (t)/n!
,
where f (n) (t) = dn
dtn f (t) (0 ≤ n ≤ p), obeys a transformation
F(t + h) =
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) = UpasF(t).(4)

5. Linear equation to solve
For even order polynomial coefficients:
(5)
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
...
.
For odd order polynomial coefficients:
(6)
F−
0
F+
1
F−
2
F+
3
...
=
1
0
3
0
5
0
7
0 . . .
1
1
3
1
5
1
7
1 . . .
0 3
2
5
2
7
2 . . .
0 3
3
5
3
7
3 . . .
...
...
...
...
...
F1
F3
F5
F7
...
.
(definitions: F+
i = 1
2 (FR
i + FL
i ), F−
i = 1
2 (FR
i − FL
i ).)

6. Generalization
We modify the upper triangular Pascal matrix with 2-parameter (a, b),
as in
(7) [M(a,b)]ij =
aj + b
i
.
This is no longer an upper triangular matrix.
Our interest is to find the solutions
M−1
(2,0) for the even order terms, and
M−1
(2,1) for the odd order terms.
Unfortunately, explicit forms of their inverses seem to be hardly
available.

7. Matrix inverse (conjecture)
It turned out that, by multiplying (inverse) Pascal matrices, we can
write
(8) M−1
(a,b) = U−1
pas M(1/a,−b/a) U−1
pas,
or equivalently,
(9) M(a,b)U−1
pas
−1
= M(1/a,−b/a)U−1
pas.
Thus, if we define
(10) L(a,b) = M(a,b)U−1
pas,
then
L−1
(a,b) =L(1/a,−b/a),(11)
M−1
(a,b) =U−1
pasL(1/a,−b/a).(12)

8. Interpretation
A 2-parameter dense matrix M(a,b) can be factorized into
(13) M(a,b) = L(a,b)Upas
and we will soon see that L(a,b) is upper triangular.
Though explicit matrix elements of L(a,b) in general are hard to obtain
(see (3.150) GouldBK.pdf, Sprugnoli, 2006), all we need are
M−1
(2,0) =U−1
pasL(1/2,0),(14)
M−1
(2,1) =U−1
pasL(1/2,−1/2).(15)
Fortunately, matrix elements of U−1
pas, L(1/2,0), and L(1/2,−1/2) are
available explicitly.

9. Example (for p = 7, 16th-order) I
(16) [L1/2,0]nk =
(−2)k
(−4)n
k
n
2n − k − 1
n − k
1 0 0 0 0 0 0 0
0 1
2 0 0 0 0 0 0
0 −1
8
1
4 0 0 0 0 0
0 1
16 −1
8
1
8 0 0 0 0
0 − 5
128
5
64 − 3
32
1
16 0 0 0
0 7
256 − 7
128
9
128 − 1
16
1
32 0 0
0 − 21
1024
21
512 − 7
128
7
128 − 5
128
1
64 0
0 33
2048 − 33
1024
45
1024 − 3
64
5
128 − 3
128
1
128

10. Example (for p = 7, 16th-order) II
(17) [L1/2,−1/2]nk =
(−2)k
(−4)n
2n − k
n − k
1 0 0 0 0 0 0 0
−1
2
1
2 0 0 0 0 0 0
3
8 −3
8
1
4 0 0 0 0 0
− 5
16
5
16 −1
4
1
8 0 0 0 0
35
128 − 35
128
15
64 − 5
32
1
16 0 0 0
− 63
256
63
256 − 7
32
21
128 − 3
32
1
32 0 0
231
1024 − 231
1024
105
512 − 21
128
7
64 − 7
128
1
64 0
− 429
2048
429
2048 − 99
512
165
1024 − 15
128
9
128 − 1
32
1
128

11. L(a,b): 2-parameter lower-triangular matrix
We compute the matrix element from its definition:
[L(a,b)]ij = M(a,b)U−1
pas ij
(18)
=
∞
k=0
ak + b
i
(−1)k+j j
k
=
∞
k=0
[yi
](1 + y)ak+b
· [tk
](−1 + t)j
=
∞
k=0
[tk
](−1 + t)j
· [yi
](1 + y)b
(1 + y)a k
=[ti
](1 + t)b
−1 + (1 + t)a j
= R (1 + t)b
, −1 + (1 + t)a
ij
.
R(d(t), h(t)) is a Riordan array of which element is [ti]d(t)h(t)j.

12. Composition and convolution
To remove the summation on k, we used a relation
∞
k=0
[tk
] f (t) · [yn
]h(y)g(y)k
=
∞
m=0
∞
k=0
[tk
] f (t) · [yn−m
]hmg(y)k
(19)
=
∞
m=0
hm[tn−m
] f (g(t))
=
∞
m=0
[ym
]h(y) · [tn−m
] f (g(t))
=[tn
]h(t) f (g(t))
with f (t) = (−1 + t)j, g(y) = (1 + y)a, and h(t) = (1 + t)b.

13. Riordan array
is an infinitesimal lower triangle matrix, taking 2 formal power series.
The element at the n-th row and the k-th column is
(20) R d(t), h(t)
nk
= [tn
]d(t)h(t)k
.
Matrix multiplication is given by
R d(t), h(t) ◦ R f (t), g(t) = R d(t) · f h(t) , g h(t) .(21)
For our 2-parameter matrices
L(a,b) =R (1 + t)b
, −1 + (1 + t)a
,
L(c,d) =R (1 + t)d
, −1 + (1 + t)c
,
we have a multiplication
L(a,b) L(c,d) = R (1 + t)b
(1 + t)ad
, −1 + (1 + t)ac
= L(ac,ad+b).
(22)

14. Group structure
L(a,b) forms a subgroup of the Riordan group when a 0.
Multiplication:
L(a,b) L(c,d) = L(ac,ad+b).
Identity:
L(1,0) L(a,b) = L(a,b) L(1,0) = L(a,b)
Inverse:
L−1
(a,b) = L(1/a,−b/a).
The following 2 × 2 matrix preserves the same structure.
(23)
1 0
b a
1 0
d c
=
1 0
ad + b ac
.

15. Application to the Hermite integrators
Remember:
M−1
(2,0) =U−1
pasL−1
(2,0) = U−1
pasL(1/2,0),(24)
M−1
(2,1) =U−1
pasL−1
(2,1) = U−1
pasL(1/2,−1/2).(25)
All we need is the matrices elements of U−1
pas, L(1/2,0), and L(1/2,−1/2).
The first one is very simple: [U−1
pas]ij = (−1)i+j j
i .

16. Implementation (p = 7, 16th-order interpolation) I
Even order coefficients are computed in
F8
F10
F12
F14
=
1 −5 15 −35
0 1 −6 21
0 0 1 −7
0 0 0 1
×
(26)
5
64 − 3
32
1
16 0 0 0
− 7
128
9
128 − 1
16
1
32 0 0
21
512 − 7
128
7
128 − 5
128
1
64 0
− 33
1024
45
1024 − 3
64
5
128 − 3
128
1
128
F+
2 − 1
2 F−
1
F−
3
F+
4
F−
5
F+
6
F−
7
.

17. Implementation (p = 7, 16th-order interpolation) II
And the odd order ones
F9
F11
F13
F15
=
1 −5 15 −35
0 1 −6 21
0 0 1 −7
0 0 0 1
×
(27)
− 35
128
15
64 − 5
32
1
16 0 0 0
63
256 − 7
32
21
128 − 3
32
1
32 0 0
− 231
1024
105
512 − 21
128
7
64 − 7
128
1
64 0
429
2048 − 99
512
165
1024 − 15
128
9
128 − 1
32
1
128
F+
1 − F−
0
F−
2
F+
3
F−
4
F+
5
F−
6
F+
7
.

18. Implementation (p = 7, 16th-order interpolation) III
Right shift for the next predictor by right-bottom submatrix of Upas:
FR
8
FR
9
FR
10
FR
11
FR
12
FR
13
FR
14
FR
15
=
1 9 45 165 495 1287 3003 6435
0 1 10 55 220 715 2002 5005
0 0 1 11 66 286 1001 3003
0 0 0 1 12 78 364 1365
0 0 0 0 1 13 91 455
0 0 0 0 0 1 14 105
0 0 0 0 0 0 1 15
0 0 0 0 0 0 0 1
F8
F9
F10
F11
F12
F13
F14
F15
(28)

19. Matrix element of L(1/2, 0)
We use Lagrange inverse formula with w(t) = −1 +
√
1 + t. For
t = w(w + 2), φ(t) = 1/(2 + t) satisfies w = tφ(w).
[L1/2, 0]nk =[tn
] −1 + (1 + t)1/2 k
(29)
= [tn
]wk
=
k
n
[tn−1
]tk−1
φ(t)n
=
k
n
[tn−k
](2 + t)−n
= 2k−2n k
n
−n
n − k
= (−1)n+k
2k−2n k
n
2n − k − 1
n − k
.
In case n = 0, it is δk0.

20. Matrix element of L(1/2, −1/2)
Similarly,
[L1/2, −1/2]nk =[tn
](1 + t)−1/2
−1 + (1 + t)1/2 k
(30)
= [tn
]
wk
1 + w
= [tn
]
tk
1 + t
φ(t)n−1
φ(t) − tφ (t)
= [tn
]
tk
1 + t
1
(2 + t)n−1
2(1 + t)
(2 + t)2
= 2[tn−k
](2 + t)−n−1
= 2k−2n −n − 1
n − k
= (−1)n+k
2k−2n 2n − k
n − k
.

21. Review
We have concerned with a 2-parameter matrix
(31) [M(a,b)]ij =
aj + b
i
,
which has an LU factorized form
(32) M(a,b) = L(a,b)Upas
where Upas = M(1,0) is an upper triangle Pascal matrix and L(a,b) is a
lower triangle matrix expressed in a Riordan array form
(33) L(a,b) = R (1 + t)b
, −1 + (1 + t)a
.
Matrix inverse M−1
(a,b) is available in
(34) M−1
(a,b) = U−1
pasL−1
(a,b) = U−1
pasL(1/a,−b/a) = U−1
pas M(1/a,−b/a) U−1
pas,

22. Summary
We have generalized the 2-step Hermite integrators
Predictor and corrector
General and explicit matrix elements
Simple and economical formulation, easy to implement
Mathematical combinatorics have performed an essential role for
the derivation and proof
Formal power series, Lagrange inverse formula, Riordan array
Implementers only need to know about matrix vector
multiplications and binomial coefficients, not the full mathematics