Ramanujan in his first paper states that the Bernoulli numbers satisfy the following equations
1 0 1
0 0 1
1
3
0 0 1
0
520 0 1
0 0 11 0 0 1
1
5
0 0
14340 0 1
0 4 0 0
28630 0 1
0 0
20450 0 221 0 0 1
1
7
0 0
193870 0
323070 0 1
0
1120 0
710650 0
355340 0 1
· · · · · · · · · · · ·
B
2(0) B
4(0) B
6(0) B
8(0) B
10(0) B
12(0) B
14(0) B
16(0) B
18(0) B
20(0) B
22(0)
·
=
1/6
−1/30 1/42 1/45
−1/132 4/455 1/120
−1/306 3/665 1/231
−1/552
·
(actually, since Ramanujan Bernoulli numbers are the moduli of ours, his equa- tions are a bit different).
Call
R
B
2(0) B
4(0) B
6(0)
.. .
=
f
1f
2f
3.. .
(RamanSyst)
such semi-infinite Ramanujan linear system. We observe that such system is equivalent to the following linear systems:
R
2!
x 4!
x2
. . .
x 2!
x2 4!
. . .
B
2(0) B
4(0) B
6(0)
.. .
=
f
1f
2f
3.. .
2!
x 4!
x2
. . .
R ˜
x 2!
x2 4!
. . .
B
2(0) B
4(0) B
6(0)
.. .
=
f
1f
2f
3.. .
R ˜
x 2!
x2 4!
. . .
B
2(0) B
4(0) B
6(0)
.. .
=
x 2!
x2 4!
. . .
f
1f
2f
3.. .
, ˜ R =
+∞
X
j=0
α
jZ
j=
+∞
X
s=0
2 x
3s(6s + 2)!(2s + 1) Z
3s,
α 0 = 1, α 1 = α 2 = 0, α 3 = 8!3 2 x 3 , α 4 = α 5 = 0, α 6 = 14!5 2 x 6 , α 7 = α 8 = 0, α 9 = 20!7 2 x 9 , . . . . The symbol Z denotes the semi-infinite lower shift matrix:
Z =
0 1 0
1 0
· ·
So, the Ramanujan system is equivalent to a lower triangular Toeplitz (sparse)
linear system. Can the latter system be obtained from the lower triangular
Toeplitz (dense) even and odd linear systems introduced in toe 1n ? First let
us prepare things.
Theorem.
Let Z n−1 and Z n be the upper-left n − 1 × n − 1 and n × n submatrices of Z.
Then
n−1
X
j=0
α
jZ
jn
B
0(0)
x 2!
B
2(0)
x2 4!
B
4(0)
·
xn−1
(2(n−1))!
B
2(n−1)(0)
=
x 2!
f
1 x2 4!f
2·
xn−1 (2(n−1))!
f
n−1
+B
0(0)
ARB
α
1α
2· α
n−1
=
w
0 x 2!w
1 x24!
w
2·
xn−1 (2(n−1))!
w
n−1
if and only if
n−2
X
j=0
α
jZ
jn−1
x 2!
B
2(0)
x2 4!
B
4(0)
·
xn−1
(2(n−1))!
B
2(n−1)(0)
=
x 2!
f
1 x2 4!f
2·
xn−1 (2(n−1))!
f
n−1
=
x 2!
w
1 x24!
w
2·
xn−1 (2(n−1))!
w
n−1
−B
0(0)
α
1α
2· α
n−1
The implication “⇐” holds if α 0 B 0 (0) = + B 0 (0) · ARB (α 0 B 0 (0) = w 0 ).
Application of the Theorem:
We have shown that the Ramanujan system is equivalent to the system
(∗) P
n−2j=0
α
jZ
n−1j
x 2!
B
2(0)
x2 4!
B
4(0)
·
xn−1
(2(n−1))!
B
2(n−1)(0)
=
x 2!
f
1 x24!
f
2·
xn−1 (2(n−1))!
f
n−1
, α
j= δ
j=0mod
3 2xj (2j+2)!(23j+1),
f
1=
16, f
2= −
301, f
3=
421, f
4=
451, f
5= −
1321, f
6=
4554, f
7=
1201, f
8=, −
3061f
9=
6653, f
10=
2311, f
11= −
5521, . . .
Then, by the Theorem,
n−1
X
j=0
α
jZ
nj
B
0(0)
x 2!
B
2(0)
x2 4!
B
4(0)
·
xn−1
(2(n−1))!
B
2(n−1)(0)
=
x 2!
f
1x2 4!
f
2·
xn−1 (2(n−1))!
f
n−1
+B
0(0)
ARB
α
1α
2· α
n−1
, +B
0(0)·ARB = α
0B
0(0),
(I+ 2
8!3 x
3Z
3+ 2
14!5 x
6Z
6+ 2
20!7 x
9Z
9+. . .)
B
0(0)
x 2!
B
2(0)
x2 4!
B
4(0)
x3 6!
B
6(0)
x4 8!
B
8(0)
x5 10!
B
10(0)
x6 12!
B
12(0)
.. .
=
1
x 2!
1 6 x2
4!
(−
301)
x3 6!
1 42
+
2x8!33x4 8! 1
45 x5 10!
(−
1321)
x6 12! 4
455
+
14!52x6x7 14!
1 120 x8 16!
(−
3061)
x9 18!
3 665
+
20!72x9x10 20!
1 231 x11
22!
(−
5521) .. .
=
1 0!
(
1·11)
x 2!
(
3·21)
x2
4!
(
5·31−
5·21)
x3 6!
(
7·41)
x4 8!
(
9·51)
x5
10!
(
11·61−
11·41)
x6 12!
(
13·71)
x7 14!
(
15·81)
x8
16!
(
17·91−
17·61)
x9 18!
(
19·101)
x10 20!
(
21·111)
x11
22!
(
23·121−
23·81) .. .
,
+∞
X
j=0
α j Z j D x b = D x q R ,
α j = δ j=0 mod 3
2x j
(2j + 2)!( 2 3 j + 1) , q R = ( 1
(2i + 1)(i + 1) (1−δ i=2 mod 3 3
2 )), i = 0, 1, 2, 3, . . . .
Note that from the explicit expression of q R it follows an explicit expression for the f i of the Ramanujan system:
f i = 1
(2i + 1)(i + 1) (1 − δ i=2 mod 3 3
2 − δ i=0 mod 3 1
2
3 i + 1 ), i = 1, 2, 3, . . . . Note also that (*) can be rewritten as
L(α n−1 )I n 2 D x b = diag (z i , i = 1, 2, . . . , n − 1)I n 2 D x q R for suitable z i . Let us look for such z i :
(1 − δ i=2 mod 3 3
2 − δ i=0 mod 3 1
2
3
i+1 ) = z i (1 − δ i=2 mod 3 3 2 ), z i = 1 −
δ
i=0mod
323i+111−δ
i=2
mod
332= 1 − δ i=0 mod 323i+1 1 .
Now let us consider the two even and odd systems introduced in toe 1n whose solution is again the vector of Bernoulli numbers:
+∞
X
j=0
x j
(2j + 2)! Z j D x b = D x q e , q e =
1 2 1 6 1 10
·
,
+∞
X
j=0
x j
(2j + 1)! Z j D x b = D x q o , q o =
1
1 2 1 2
·
;
n−1
X
j=0
x j
(2j + 2)! Z n j D x b = D x q e ,
n−1
X
j=0
x j
(2j + 1)! Z n j D x b = D x q o . And apply to them the Theorem:
n−2
X
j=0
x
j(2j + 2)! Z
n−1j
x 2!
B
2(0)
x2 4!
B
4(0)
·
xn−1
(2(n−1))!
B
2(n−1)(0)
=
x 2!
1 6 x2
4!
1 10
·
xn−1 (2(n−1))! 1
4n−2
−B
0(0)
x 4!
x2 6!
·
xn−1 (2n)!
=
x
4!1x
2 26!x
3 38!· x
n−1 n−1(2n)!
n−2
X
j=0
x
j(2j + 1)! Z
n−1j
x 2!
B
2(0)
x2 4!
B
4(0)
·
xn−1
(2(n−1))!
B
2(n−1)(0)
=
x 2!1
2 x2
4!1 2
·
xn−1 (2(n−1))!
1 2
−B
0(0)
x 3!
x2 5!
·
xn−1 (2n−1)!
=
x
3!21x
2 35!2x
3 57!2· x
n−1(2n−1)!22n−3
from which it follows
n−2
X
j=0
x
j(2j + 2)! Z
n−1j
x 2!
B
2(0)
x2 4!
B
4(0)
·
xn−1
(2(n−1))!
B
2(n−1)(0)
=
x 2!
1 4·3 x2 4!
2 6·5 x3 6! 3
8·7
·
xn−1 (2(n−1))!
n−1 2n(2n−1)
= diag ( i
i + 1 , i = 1 . . . n−1)I
n−11(Z
nTD
xq
e)
n−2
X
j=0
x
j(2j + 1)! Z
n−1j
x 2!
B
2(0)
x2 4!
B
4(0)
·
xn−1
(2(n−1))!
B
2(n−1)(0)
=
x 2!
1 2·3 x2
4! 3 2·5 x3
6! 5 2·7
·
xn−1 (2(n−1))!
2n−3 2(2n−1)
= diag ( 2i − 1
2i + 1 , i = 1 . . . n−1)I
n−11(Z
nTD
xq
o)
Set
b =
B 0 (0) B 2 (0) B 4 (0)
.. .
, D x = diag ( x i
(2i)! , i = 0, 1, 2, . . .).
Then
L(α)D x b = D x q, L(α)Z T D x b = d(z)Z T D x q,
where the vectors α = (α i ) +∞ i=0 , q = (q i ) +∞ i=0 , and z = (z i ) +∞ i=1 , can assume respectively the values:
α R i = δ i=0 mod 3 2x
i
(2i+2)!(
23i+1) , q R i = (2i+1)(i+1) 1 (1 − δ i=2 mod 3 3 2 ), i = 0, 1, 2, 3, . . . z i R = 1 − δ i=0 mod 3
1
2
3
i+1 , i = 1, 2, 3, . . . , α e i = (2i+2)! x
i, q e i = 2(2i+1) 1 , i = 0, 1, 2, 3, . . .
z i e = i+1 i , i = 1, 2, 3, . . . ,
α o i = (2i+1)! xi , i = 0, 1, 2, 3, . . . , q 0 o = 1, q i o = 1 2 , i = 1, 2, 3, . . . z i o = 2i−1 2i+1 , i = 1, 2, 3, . . . .
( Z is the semi-infinite lower shift matrix
Z =
0 1 0
1 0
· ·
,
L(α) is the semi-infinite lower triangular Toeplitz matrix with first column α, i.e.
L(α) =
+∞
X
i=0
α i Z i =
α 0
α 1 α 0
α 2 α 1 α 0
α 3 α 2 α 1 α 0
· · · · ·
. )
Exercise.
Try to prove that the system L(α R )D x b = D x q R (the lower triangular Toeplitz sparse system equivalent to the Ramanujan lower triangular sparse system) can be obtained as a consequence of the system L(α o )D x b = D x q o (or L(α e )D x b = D x q e ).
For example, try to find β e , β o such that L(β e )L(α e ) = L(α R ) = L(β o )L(α o ) [?? L(β e )D x q e = D x q R = L(β o )D x q o ??], i.e. L(α e )β e = α R = L(α o )β o . Find γ such that L(γ)L(α e ) = L(α) (or L(γ)L(α o ) = L(α)) with α more sparse than α R .
Exercise.
Prove that d(z)Z T L(α)(D x b) = L(α)Z T (D x b).
Note that
L(a) =
1 a
11 a
2a
11 a
3a
2a
11 a
4a
3a
2a
11 a
5a
4a
3a
2a
11
· · · · · · ·
1
−a
1a
2−a
3a
4−a
5·
=
1 0 2a
2− a
210 2a
4− 2a
1a
3+ a
220
2a
6− 2a
1a
5+ 2a
2a
4− a
230
2a
8− 2a
1a
7+ 2a
2a
6− 2a
3a
5+ a
240
·
,
L(a) e
1+
+∞
X
i=1
(−1)
ia
ie
i+1= e
1+
+∞
X
i=1
δ
i=0mod
22a
i+
i−1
X
j=1
(−1)
ja
ja
i−je
i+1. If we introduce the following two semi-infinite matrices,
I ± = diag ((−1) i , i = 0, 1, 2, 3, . . .), E =
1 0 0 0 0 1 0 0 0 0 0 0 1 0 0 0 0 0
· · · · ·
,
then the previous equality can be rewritten as
L(a)I ± a = Ea (1) , a (1) 0 = 1, a (1) s = 2a 2s +
2s−1
X
j=1
(−1) j a j a 2s−j , s = 1, 2, 3, . . . .
Of course, L(a)L(I ± a) = L(I ± a)L(a) = L(Ea (1) ), so, for example, a lower triangular Toeplitz (l.t.T.) system of the type L(a)z = c is equivalent to the l.t.T. system L(Ea (1) )z = L(I ± a)c whose coefficient matrix has null diagonals alternating with non-null ones. We shall introduce, as a consequence of this result, an algorithm of complexity O(n(log 2 n) 2 ) for solving a lower triangular Toeplitz system (of order n = 2 k ).
Exercise. Apply the above result to the odd (even) system (the one solved by Bernoulli numbers) in order to obtain a system where null diagonals alternate with the non-null ones. Is it possible to write an explicit formula for the entries of the non-null diagonals?
Exercise. Use twice the above result in order to show that a l.t.T. system of the
type L(a)z = c is equivalent to a l.t.T. system where the coefficient matrix has
three-null diagonals alternating with non-null ones.
Exercise. Answer to the quotation marks in the following equality:
L(a) =
1 a
11 a
2a
11 a
3a
2a
11 a
4a
3a
2a
11 a
5a
4a
3a
2a
11
· · · · · · ·
1
?
?
?
?
?
·
=
1 0 0
? 0 0
? 0 0
?
·
.
As a consequence of this result,
(i) introduce an algorithm of complexity O(n(log 3 n) 2 ) for solving a lower triangular Toeplitz system (for example, in case the system has order n = 27 such algorithm is more efficient rather than embed the system in one of order n = 32 and use the algorithm ok for n = 2 k );
(ii) obtain the Ramanujan system solved by Bernoulli numbers as a conse- quence of the odd (even) system.
A lower triangular Toeplitz system solver of complexity O(n(log 2 n) 2 ) Set
n = 2
k, a = a
(0)=
1 a
1· a
n−1
=
1 a
(0)1· a
(0)n−1
, L(a) =
1 a
11 a
2a
11
· · · ·
a
n−1· a
2a
11
,
and let I ± and E denote the n × n upper-left submatrices of the previously defined semi-infinite matrices I ± and E. In the following we multiply L(a) on the left by suitable l.t.T matrices so to reduce the number of its non-null diagonals; we do this in k = log 2 n steps:
step 1
L(a)I ± a =: Ea (1) ,
a (1) 0 = 1, a (1) s = 2a (0) 2s + P 2s−1
j=1 (−1) j a (0) j a (0) 2s−j , s = 1, . . . , n 2 − 1, a (1) s = 0, s = n 2 , . . . , n − 1, L(a)L(I ± a) = L(I ± a)L(a) = L(Ea (1) )
Note: One here needs to compute a (1) , i.e. ( n 2 − 1 odd entries of) the vector L(a) · I ± a . Denote by ϕ n the cost (the number of multiplications) of such operation.
step 2
L(Ea (1) )EI ± a (1) =: E 2 a (2) , a (2) 0 = 1, a (2) s = 2a (1) 2s + P 2s−1
j=1 (−1) j a (1) j a (1) 2s−j , s = 1, . . . , n 4 − 1, a (2) s = 0, s = n 4 , . . . , n − 1, L(Ea (1) )L(EI ± a (1) ) = L(EI ± a (1) )L(Ea (1) ) = L(E 2 a (2) )
Note: One here needs to compute a (2) , i.e. ( n 4 − 1 odd entries of) the order n 2
non null first part of the vector L(a (1) ) · I ± a (1) . Denote by ϕ
n2 the cost of such
operation.
step 3
L(E 2 a (2) )E 2 I ± a (2) =: E 3 a (3) , a (3) 0 = 1, a (3) s = 2a (2) 2s + P 2s−1
j=1 (−1) j a (2) j a (2) 2s−j , s = 1, . . . , n 8 − 1, a (3) s = 0, s = n 8 , . . . , n − 1, L(E 2 a (2) )L(E 2 I ± a (2) ) = L(E 2 I ± a (2) )L(E 2 a (2) ) = L(E 3 a (3) )
Note: One here needs to compute a (3) , i.e. ( n 8 − 1 odd entries of) the order n 4 non null first part of the vector L(a (2) ) · I ± a (2) . Denote by ϕ
n4
the cost of such operation.
. . .
step k − 1 = log 2 n − 1
L(E
k−2a
(k−2))E
k−2I
±a
(k−2)=: E
k−1a
(k−1), a
(k−1)0= 1,
a
(k−1)s= 2a
(k−2)2s+ P
2s−1j=1
(−1)
ja
(k−2)ja
(k−2)2s−j, s = 1, . . . ,
2k−1n− 1, a
(k−1)s= 0, s =
2k−1n, . . . , n − 1,
L(E
k−2a
(k−2))L(E
k−2I
±a
(k−2)) = L(E
k−2I
±a
(k−2))L(E
k−2a
(k−2)) = L(E
k−1a
(k−1)) Note: One here needs to compute a (k−1) , i.e. ( 2k−1n − 1 odd entries of) the order
n
2
k−2non null first part of the vector L(a (k−2) ) · I ± a (k−2) . Denote by ϕ
n2k−2
the cost of such operation.
step k = log 2 n
L(E
k−1a
(k−1))E
k−1I
±a
(k−1)=: E
ka
(k), a
(k)0= 1,
a
(k)s= 0, s =
2nk, . . . , n − 1,
L(E
k−1a
(k−1))L(E
k−1I
±a
(k−1)) = L(E
k−1I
±a
(k−1))L(E
k−1a
(k−1)) = L(E
ka
(k)) (E k = e 1 e T 1 , a (k) = e 1 , L(E k a (k) ) = I)
Note: One here needs to compute a (k) , i.e. ( 2 nk − 1 odd entries of) the order
n
2
k−1non null first part of the vector L(a (k−1) ) · I ± a (k−1) . Denote by ϕ
n2k−1
the cost of such operation (ϕ
n2k−1
= 0).
Finally note that the following equality holds
L(E
k−1I
±a
(k−1))L(E
k−2I
±a
(k−2)) · · · L(E
2I
±a
(2))L(EI
±a
(1))L(I
±a ) h L(a) i
= I, in other words we have transformed L(a) into the identity matrix.
Example: n = 8
n = 2
3, a = a
(0)=
1 a
1a
2a
3a
4a
5a
6a
7
=
1 a
(0)1a
(0)2a
(0)3a
(0)4a
(0)5a
(0)6a
(0)7
, L(a) =
1 a
11 a
2a
11 a
3a
2a
11 a
4a
3a
2a
11 a
5a
4a
3a
2a
11 a
6a
5a
4a
3a
2a
11 a
7a
6a
5a
4a
3a
2a
11
,
and let I ± and E denote the 8 × 8 upper-left submatrices of the previously
defined semi-infinite matrices I ± and E.
step 1
L(a)I
±a =
1 a
11 a
2a
11 a
3a
2a
11 a
4a
3a
2a
11 a
5a
4a
3a
2a
11 a
6a
5a
4a
3a
2a
11 a
7a
6a
5a
4a
3a
2a
11
1
−a
1a
2−a
3a
4−a
5a
6−a
7
=: Ea
(1)=
1 0 a
(1)10 a
(1)20 a
(1)30
,
a
(1)0= 1, a
(1)s= 2a
(0)2s+ P
2s−1j=1
(−1)
ja
(0)ja
(0)2s−j, s = 1, . . . ,
82− 1, a
(1)s= 0, s =
82, . . . , 8 − 1, a
(1)1= 2a
2− a
21, a
(1)2= 2a
4− 2a
1a
3+ a
22, a
(1)3= 2a
6− 2a
1a
5+ 2a
2a
4− a
23,
L(a)L(I
±a ) = L(I
±a )L(a) = L(Ea
(1))
Note: One here needs to compute a (1) , i.e. ( 8 2 − 1 odd entries of) the vector L(a) · I ± a . Denote by ϕ 8 the cost of such operation (ϕ 8 ≤ 6 if multiplications by 2 are not counted).
step 2
L(Ea
(1))EI
±a
(1)=
1
1 a
(1)11
a
(1)11 a
(1)2a
(1)11
a
(1)2a
(1)11 a
(1)3a
(1)2a
(1)11
a
(1)3a
(1)2a
(1)11
1 0
−a
(1)10 a
(1)20
−a
(1)30
=: E
2a
(2)=
1 0 0 0 a
(2)10 0 0
,
a
(2)0= 1, a
(2)s= 2a
(1)2s+ P
2s−1j=1
(−1)
ja
(1)ja
(1)2s−j, s = 1, . . . ,
84− 1, a
(2)s= 0, s =
84, . . . , 8 − 1, a
(2)1= 2a
(1)2− (a
(1)1)
2,
L(Ea
(1))L(EI
±a
(1)) = L(EI
±a
(1))L(Ea
(1)) = L(E
2a
(2))
Note: One here needs to compute a (2) , i.e. ( 8 4 − 1 odd entries of) the order 8 2 non null first part of the vector L(a (1) ) · I ± a (1) . Denote by ϕ
82
the cost of such operation (ϕ
82
≤ 1 if multiplications by 2 are not counted).
step 3 = log 2 8
L(E 2 a (2) )E 2 I ± a (2) =
1
1 1
1
a (2) 1 1
a (2) 1 1
a (2) 1 1
a (2) 1 1
1 0 0 0
−a (2) 1 0 0 0
=: E 3 a (3) =
1 0 0 0 0 0 0 0
,
a (3) 0 = 1, a (3) s = 2a (2) 2s + P 2s−1
j=1 (−1) j a (2) j a (2) 2s−j , s = 1, . . . , 8 8 − 1, a (3) s = 0, s = 8 8 , . . . , 8 − 1, L(E 2 a (2) )L(E 2 I ± a (2) ) = L(E 2 I ± a (2) )L(E 2 a (2) ) = L(E 3 a (3) )
(E 3 = e 1 e T 1 , a (3) = e 1 , L(E 3 a (3) ) = I) Note: One here needs to compute a (3) , i.e. ( 8 8 − 1 odd entries of) the order 8 4 non null first part of the vector L(a (2) ) · I ± a (2) . Denote by ϕ
84
the cost of such operation (ϕ
84
= 0).
By using the obtained identities, one realizes that
L(E 2 I ± a (2) )L(EI ± a (1) )L(I ± a) L(a) = I, i.e. we have performed a kind of Gaussian elimination . . . . Upper bounds for the cost P 1
j=k ϕ 2
jof the computation of a (1) , a (2) , . . . , a (k−1) , a (k) Since the cost of a matrix-vector product involving a generic vector and a 2 j × 2 j lower triangular Toeplitz matrix with ones on the diagonal is obviously bounded by 1 + 2 + 3 + . . . + (2 j − 1) = 1 2 2 j (2 j − 1), we can say that
k
X
j=1
ϕ
2j≤ 1 2
k
X
j=1
(2
2j− 2
j) = 2
3 (2
k)
2− 2
k+ 1 3 .
However, by exploiting the particularity of our matrix-vector products, we observe that ϕ 2 = 0, ϕ 4 ≤ 1, ϕ 8 ≤ 3 + 1 + 2, ϕ 16 ≤ 7 + 1 + 2 + 3 + 4 + 5 + 6, ϕ 2j ≤ 1 2 2 j−1 (2 j−1 − 1). So,
k
X
j=1
ϕ
2j≤ 1 2
k
X
j=2
(2
2j−2− 2
j−1) = 1
6 (2
k)
2− 1 2 2
k+ 1
3 .
Finally, since a matrix-vector product involving a generic vector and a 2 j × 2 j lower triangular Toeplitz matrix can be computed by means of the three FFT of order 2 j+1 , i.e. with a cost O((j + 1)2 j+1 ), we have
k
X
j=1
ϕ
2j= O(k
22
k) = O(n(log
2n)
2) prove this!
Computing the first column of L(a) −1 Now observe that
L(E
k−1I
±a
(k−1))L(E
k−2I
±a
(k−2)) · · · L(E
2I
±a
(2))L(EI
±a
(1))L(I
±a ) h L(a) i
= I, L(a)z = c,
z = L(E
k−1I
±a
(k−1))L(E
k−2I
±a
(k−2)) · · · L(E
2I
±a
(2))L(EI
±a
(1))L(I
±a ) c
= L(I
±a )L(EI
±a
(1))L(E
2I
±a
(2)) · · · L(E
k−2I
±a
(k−2))L(E
k−1I
±a
(k−1)) c, L(a)z = ue
1+ ve
n2+1
,
z = L(I
±a )L(EI
±a
(1))L(E
2I
±a
(2)) · · · L(E
k−2I
±a
(k−2))L(E
k−1I
±a
(k−1)) (ue
1+ ve
n2+1
