Convergence Radii for Eigenvalues of Tri--diagonal Matrices

0.19 MB·English

by  J. Adduci

CONVERGENCE RADII FOR EIGENVALUES OF
TRI–DIAGONAL MATRICES
9
J. ADDUCI,P. DJAKOV,AND B. MITYAGIN
0
0
2 Abstract. Consider a family of infinite tri–diagonal matrices of the
n formL+zB,wherethematrixLisdiagonalwithentriesLkk =k2,and
a the matrix B is off–diagonal, with nonzero entries Bk,k+1 = Bk+1,k =
J kα, 0 ≤ α < 2. The spectrum of L+zB is discrete. For small |z| the
6 n-th eigenvalue En(z), En(0)=n2, is a well–defined analytic function.
2 Let Rn be the convergence radius of its Taylor’s series about z = 0. It
is provedthat

P] Rn ≤C(α)n2−α if 0≤α<11/6.
S
.
h
t
a
m 1. Introduction
[
Since the famous 1969 paper of C. Bender and T. Wu [2], branching

1 points andthe crossings of energy levels have been studiedintensively in the
v mathematical and physical literature (e.g., [8, 1, 4, 3] and the bibliography
1
there). In this paper our goal is to analyze – mostly along the lines of J.
3
0 MeixnerandF.Scha¨fkeapproach[10]–atoymodeloftri–diagonalmatrices.
4 We consider the operator family L + zB, where L and B are infinite
.
1 matrices of the form
0
q 0 0 0 0 b 0 0
9 1 1
· ·
0 0 q2 0 0 c1 0 b2 0
 ·  ·
: (1.1) L = 0 0 q 0 , B = 0 c 0 b
v 3 2 3
· ·
i 0 0 0 q  0 0 c 0 
X  4 ·  3 ·
   
r  · · · · ·  · · · · ·
a    
with
(1.2) q = k2,
k
(1.3) b , c Mkα,
k k
| | | | ≤
(1.4) α < 2.
Sometimes we impose a symmetry condition:
(1.5) b = c¯.
k k
2000 Mathematics Subject Classification. 47B36 (primary), 47A10 (secondary).
Key words and phrases. tri–diagonal matrix, operator family, eigenvalues.
B. Mityagin acknowledges the support of the Scientific and Technological Research
Council of Turkeyand the hospitality of Sabanci University,April–June, 2008.
1
2 J.ADDUCI,P.DJAKOV,ANDB.MITYAGIN
Under the conditions (1.2)–(1.4) the spectrum of L+zB is discrete. If
α< 1 then a standard use of perturbation theory shows that there is r > 0
such that for z < r
| |
(1.6) Sp(L+zB)= E (z) ∞ , E (0) = n2,
{ n }n=1 n
where each E (z) is well–defined analytic function in the disc z : z < r .
n
{ | | }
If α [1,2), then in general there is no such r > 0. But the fact that n2
∈
is a simple eigenvalue of L guarantees (see [9], Chapter 7, Sections 1-3) that
for each n there exists r > 0 such that, on the disc z : z < r , there are
n n
{ | | }
an analytic function E (z) and an analytic eigenvector function ϕ (z) with
n n
(1.7) (L+zB)ϕ (z) = E (z)ϕ (z), z < r ,
n n n n
| |
(1.8) ϕ (0) = e , E (0) = n2.
n n n
Let
∞
(1.9) E (z) = a (n)zk
n k
k=0
X
bethe Taylor series of E (z) about0, and let R ,0 < R , beits radius
n n n
≤ ∞
of convergence. The asymptotic behavior of the sequence (R ) is one of the
n
main topics of the present paper.
It may happen that R > r . Then, by (1.9), E (z) is defined in the disc
n n n
z : z < R as anextension oftheanalytic function (1.7)in z : z < r .
n n
{ | | } { | | }
But are its values E (z) eigenvalues of L+zB if z is in the annulus r
n n
≤
z < R ?Theanswerispositiveasonecanseefromthenextconsiderations.
n
| |
In a more general context let us define Spectral Riemann Surface
(1.10) G = (z,E) : g Dom(L), g = 0 (L+zB)g = Eg .
{ ∃ ∈ 6 | }
This notion is justified by the following statement (coming from K. Weier-
strass, H. Poincare, T. Carlemann – see discussions on the related history
in [6, 11, 7]).
Proposition 1. If (1.1)–(1.4) hold, then there exists a nonzero entire func-
tion Φ(z,w) such that
(1.11) G= (z,w) C2 : Φ(z,w) = 0 .
{ ∈ }
Proof. The identity
(1.12) (L+zB)g = wg, g = 0, g Dom(L)
6 ∈
is equivalent to
(1.13) (1 A(z,w))h = 0 with h = L1/2g Dom(L1/2), h= 0,
− ∈ 6
where
(1.14) A(z,w) = zL−1/2BL−1/2+wL−1.
−
Therefore, w is an eigenvalue of the operator L+zB if and only if 1 is an
eigenvalue of the operator A(z,w).
CONVERGENCE RADII FOR EIGENVALUES 3
On the space S of trace class operators T the determinant
1
(1.15) d(T) = det(1 T)
−
is well defined (see [6], Chapter 4, Section 1 or [12], Chapter 3, Theorem
3.4), and 1 Sp(T) if and only if d(T) = 0 (see [12], Theorem 3.5 (b)).
∈
Of course, the second term L−1 in (1.14) is an operator of trace class
(even in S ,p > 1/2) by (1.2). But (1.3)–(1.4) imply that L−1/2BL−1/2 is
p
in the Schatten class S ,p > 1/(2 α); only α < 1 would guarantee that it
p
−
is of trace class.
However, (1.15) could be adjusted (see [6] Chapter 4, Section 2 or [12],
Chapter 9, Lemma 9.1 and Theorem 9.2). Namely, for any positive integer
p 2 we set
≥
(1.16) d (T) = det(1 Q (T))
p p
−
where
T2 Tp−1
Q (T) = 1 (1 T)exp T + + + .
p
− − 2 ··· p 1
(cid:18) − (cid:19)
Then Q (T) S if T S , so d is a well-defined function of T S and
p 1 p p p
∈ ∈ ∈
1 Sp(T) if and only if d (T)= 0.
p
∈
In our context we define, with A(z,w) (1.14) and p > 1/(2 α),
∈ −
(1.17) Φ(z,w) = det[(1 Q (A(z,w))].
p
−
Now, from Claim 8, Section 1.3, Chapter 4 in [6] it follows that Φ(z,w) is
an entire function on C2.
The function Φ vanishes at (z,w) if and only if 1 is an eigenvalue of the
operatorA(z,w),i.e.,ifandonlyif(z,w) G. Thiscompletestheproof. (cid:3)
∈
In particular, the above Proposition implies that Φ(z,E (z)) = 0 if z <
n
| |
r , so by analyticity and uniqueness Φ(z,E (z)) = 0 if r z < R .
n n n n
≤ | |
Equivalenceofthetwodefinitions(1.10)and(1.11)fortheSpectralRiemann
Surface G explains now that E (z) is an eigenvalue function in the disc
n
z : z < R .
n
{ | | }
Our main focus in the search for an understanding of the behavior of R
n
will be on the special case where
(1.18) 0 α < 2,
≤
(1.19) b = c¯ = kα.
k k
If α = 0 in (1.19), we have the Mathieu matrices. They arise if Fourier’s
method is used to analyze the Hill–Mathieu operator on I = [0,π]
Ly = y′′+2a(cos2x)y, y(π) = y(0), y′(π) = y′(0).
−
In this case J. Meixner and F. W. Scha¨fke proved ([10], Thm 8, Section 1.5;
[11], p. 87) the inequality R Cn2 and conjectured that the asymptotic
n
≤
R n2 holds. This has been proved 40 years later by H. Volkmer [13].
n
≍
4 J.ADDUCI,P.DJAKOV,ANDB.MITYAGIN
But what can be said if 0 < α < 2? Proposition 4 in [5] shows that if
(1.1)–(1.3) and (1.18) hold, then
(1.20) R cn1−α.
n
≥
This estimate from below cannot be improved in the class (1.1)–(1.3),
(1.18) as examples in Section 4 show. But in the special case (1.18)–(1.19)
one could expect the asymptotic
(1.21) R n2−α.
n
≍
We show that
R Cn2−α,
n
≤
at least for 0 < α < 11/6.
Notice that in the Hill–Mathieu case we have α = 0, b = 1 k, so the
k
∀
operator B is bounded, while it could be unbounded in the case α > 0. We
use the approach of Meixner and Scha¨fke [10], but complement it with an
additional argument to help us deal with the cases where the operator B is
unbounded (but relatively compact with respect to L). The main result is
the following.
Theorem 2. If the conditions (1.2) and (1.19) hold, then for each α
[0, 11) there exist constants C > 0 and N N such that ∈
6 α α ∈
(1.22) R C n2−α, n N .
n α α
≤ ≥
Proof is given in Section 3. It has two parts. In Section 2, we prove
an upper bound for Taylor coefficients a (n) in terms of k, n, R and α
k n
| |
(see Theorem 3). In Section 3 we show how a certain lower bound on
a (n) , in terms of k,n, and α, can be used to prove the desired inequality
k
| |
on particular subsets of [0,2). In the same section we provide such lower
bounds for a (n), a (n),..., a (n). This general scheme could be used
2 4 12
| | | | | |
in an attempt to prove (1.22) for larger subsets of [0,2). One would then
need to compute (and manipulate) a (n) for values of k > 12. See Section
k
3 for details.
2. An upper bound for a (n)
k
| |
In what follows in this section, suppose that n is a fixed positive integer.
Theorem 3. In the above notations, and under the conditions (1.2) and
(1.3), if
(a) α [0,2) and (1.5) holds, or (b) α [0,1),
∈ ∈
then
(2.1) ak(n) Cρ−(k−1) nα+ρ2−αα , 0 <ρ < Rn,
| | ≤
(cid:16) (cid:17)
where C = C(α,M).
CONVERGENCE RADII FOR EIGENVALUES 5
Proof. For r > 0, let
∆ = z C : z < r , C = z C : z = r .
r r
{ ∈ | | } { ∈ | | }
Let us choose, for every z ∆ , an eigenvector g(z) = (g (z))∞ such
∈ Rn n n=1
that g(z) = 1 (this is possible by Proposition 1). Then
ℓ2
k k
(2.2) (L+zB)g(z) = E (z)g(z), g(z) = 1,
n ℓ2
k k
which implies (after multiplication from the right by g(z))
(2.3) ℓ(z)+zb(z) = E (z), z ∆ ,
n ∈ Rn
where
∞
(2.4) ℓ(z) := Lg(z),g(z) = k2 g (z)2,
k
h i | |
k=1
X
and
∞
(2.5) b(z) := Bg(z),g(z) = c g (z)g (z)+b g (z)g (z) .
k k k+1 k k+1 k
h i
Xk=1(cid:16) (cid:17)
The functions ℓ(z) and b(z) are bounded if z ρ < R . Indeed, by (2.4)
n
| | ≤
we have ℓ(z) > 0. By (2.5) and (1.3)
∞ ∞
(2.6) b(z) Mkα g (z)2 + g 2 2M kα g (z)2,
k k+1 k
| | ≤ | | | | ≤ | |
k=1 k=1
X (cid:0) (cid:1) X
so, estimating the latter sum by H¨older’s inequality, we get
(2.7) b(z) 2M(ℓ(z))α/2.
| | ≤
Therefore, in view of (2.3).
ℓ(z) E (z) + zb(z) E (z) +2Mρ(ℓ(z))α/2, z ρ.
n n
≤| | | | ≤ | | | | ≤
Now, Young’s inequality implies
α 2
ℓ(z) En(z) +(1 α/2)22−α(2Mρ)2−α +(α/4) ℓ(z),
≤ | | − ·
so, in view of (1.18), ℓ(z) is bounded by
α 2
ℓ(z) 2En(z) +2(1 α/2)22−α(2Mρ)2−α, z ρ.
≤ | | − | | ≤
By (2.7), the function b(z) is also bounded if z ρ.
| | ≤
Since in (2.2) the vectors g(z), z ∆ , are chosen in an arbitrary way,
∈ Rn
we cannot expect the function z g(z) to be continuous, or even measur-
→
able. But the functions ℓ(z) and b(z) are measurable. The explanation of
this fact is the only difference in the proof of (2.1) in the cases (a) and (b).
(a) The functions ℓ(z) and b(z) are continuous on ∆ ( R ,R ).
Rn \ − n n
Indeed, in view of (2.5) the symmetry assumption (1.5) implies that
the function b(z) is real–valued. Therefore, from (2.3) it follows yb(z) =
6 J.ADDUCI,P.DJAKOV,ANDB.MITYAGIN
ImE (z)withz = x+iy,soℓ(z)andb(z)arecontinuouson∆ ( R ,R )
n Rn\ − n n
because
1 x
(2.8) b(z) = Im(E (z)), ℓ(z) = Re(E (z)) Im(E (z)), y = 0.
n n n
y − y 6
(b) For every z such that E (z) is a simple eigenvalue of L + wB the
n
values ℓ(z) and b(z) are uniquely determined by (2.4) and (2.5) and do not
depend on the choice of the vector g(z) in (2.2). Therefore, the functions
ℓ(z) and b(z) are uniquely determined on the set
U = z ∆ : E (z) is a simple eigenvalue of L+zB .
{ ∈ Rn n }
On the other hand, the set ∆ U is at most countable and has no finite
Rn \
accumulation points (see Section 5.1 in [5]).
If w U, then it is known ([9], Ch.VII, Sect. 1-3, in particular, Theorem
∈
1.7) that there is a disc D(w,τ) with center w and radius τ such that E (z)
n
is a simple eigenvalue of the operator L + zB for z D(w,τ) and there
∈
exists an analytic eigenvector function ψ(z) defined in D(w,τ), i.e.,
(L+zB)ψ(z) = E (z)ψ(z), ψ(z) = 0, z D(w,τ).
n
6 ∈
Let g(z) = ψ(z)/ ψ(z) for z D(w,τ). Then the coordinate functions
ℓ2
k k ∈
g (z) are continuous, and by (2.4) the function ℓ(z), z D(w,τ), is a sum
k
∈
of a series of positive continuous terms. Therefore, the function ℓ(z) is lower
semi–continuous in D(w,τ), so it is lower semi–continuous in U. Thus, ℓ(z)
is measurable on ∆ . By (2.3) we have b(z) = (E (z) ℓ(z))/z for z = 0.
Rn n − 6
Thus, b(z) is measurable in ∆ as well.
Rn
Foreachρ (0,R ),considerthespaceL2(C )withthenorm defined
n ρ ρ
∈ k·k
by f 2 = 1 2π f(ρeiθ)2dθ. The functions ℓ(z) and b(z) are integrable on
k kρ 2π 0 | |
each circle C , ρ < R because they are bounded and measurable on C .
ρ n ρ
R
From (2.7) and H¨older’s inequality it follows that
(2.9) b(z) 2M ℓ(z) α/2.
k kρ ≤ k kρ
Since ℓ(z) > 0, by (2.3) and (2.7) we have
Im(E (z) n2) = Im(zb(z)) ρb(z).
n
| − | | | ≤ | |
Therefore,
(2.10) Im(E (z) n2) ρ b(z) .
n ρ ρ
k − k ≤ ·k k
If f is an analytic function defined on ∆ with f(0) = 0, then Re(f) =
Rn k kρ
Im(f) . In particular, we have
ρ
k k
Re(E (z) n2) = Im(E (z) n2) ,
n ρ n ρ
k − k k − k
which implies, by (2.10),
(2.11) E (z) n2 √2ρ b(z) .
n ρ ρ
k − k ≤ ·k k
In view of (2.3) and (2.11), the triangle inequality implies
ℓ n2+ E (z) n2 + b(z) n2+(1+√2)ρ b(z) .
ρ n ρ ρ ρ
k k ≤ k − k k k ≤ ·k k
CONVERGENCE RADII FOR EIGENVALUES 7
Therefore, from (2.9) it follows that
(2.12) ℓ n2+5Mρ ℓ α/2.
k kρ ≤ k kρ
Now, Young’s inequality yields
5Mρkℓkαρ/2 ≤ 1−α/2)(5M2α/2ρ 2−2α + α4kℓkρ ≤ C1ρ2−2α + 21kℓkρ,
(cid:16) (cid:17)
2
with C = 1 α/2)(5M2α/2 2−α . Thus, by (2.12), we have
1
−
2
(cid:0) ℓ(cid:1) 2n2+2C1ρ2−α.
k k ≤
In view of (2.11) and (2.9), this implies
(2.13) En(z) n2 ρ 3Mρ 2α/2nα+(2C1)α/2ρ2−αα .
k − k ≤
(cid:16) (cid:17)
By Cauchy’s formula, we have
1 E (ζ) n2
n
a (n) = − dζ.
k 2πi ζk+1
Z∂∆ρ
From (2.13) it follows that
ak(n) ρ−k En(z) n2 ρ 3Mρ−k+1 2α/2nα+(2C1)α/2ρ2−αα ,
| | ≤ k − k ≤
(cid:16) (cid:17)
which implies (2.1) with C = 3M(2+2C )α/2. This completes the proof of
1
Theorem 3. (cid:3)
Remark. In fact, to carry out the proof of Theorem 3 we need only to
know that there exists a pair of functions ℓ(z) and b(z) which satisfy (2.3)
and (2.7), and are integrable on each circle C , ρ < R . We explained that
ρ n
the pair defined by (2.2), (2.4) and (2.5) has these properties. In the case
(a) of Theorem 3 the same argument could be used to define a pair of real
analytic functions functions ℓ(z) and b(z) which satisfy (2.3) and (2.7).
Indeed,by(1.5)theoperatorB isaself–adjoint, soL+xB, x R,isself–
∈
adjoint as well. Thus, the function E (z) takes real values on the real line
n
and its Taylor’s coefficients are real. Since the quotients 1 Im(x+iy)k, k
y ∈
N, are polynomials of y, it is easy to see by the Taylor series of E (z) that
n
1Im(E (z)) (defined properly for y = 0) is a real analytic function in ∆ .
y n Rn
Therefore, if one defines a pair of functions ℓ˜(z) and ˜b(z) by (2.8), then
(2.3) holds immediately, and (2.7) follows becauseon ∆ ( R ,R ) these
Rn\ − n n
functions coincide with ℓ(z) and b(z).
3. An upper bound for R
n
In this section we use (2.1) in the case of (1.19) to prove Theorem 2.
Roughly speaking, the bound (1.22) will be achieved for α [0, 11) by
∈ 6
inserting the known (from [5]) formulas for a (α,n),...,a (α,n) into in-
2 12
equality (2.1). With our approach, using only a , k 6, it is possible to
2k
≤
get good lower bounds only if 0 α< 11/6.
≤
8 J.ADDUCI,P.DJAKOV,ANDB.MITYAGIN
We begin with the following observation.
Lemma 4. Suppose the conditions (1.2),(1.3) and (1.18) hold.
(a) If for some fixed k,n N and α [0,2 2) we have a (n) = 0, then
∈ ∈ − k k 6
R < .
n
∞
(b) If R = , then E (z) is a polynomial such that degE (z) α .
n ∞ n n ≤ 2−α
Proof. Let a = a (n) > 0. Then, by Theorem 3,
k
| |
(3.1) aρk−1 C nα+ρ2−αα , ρ < Rn.
≤ ∀
(cid:16) (cid:17)
The condition α [0,2 2) implies k 1 > α ; therefore, (3.1) fails for
∈ − k − 2−α
sufficiently large ρ. Thus, R sup ρ : ρ (3.1) < , which proves (a).
n
≤ { ∈ } ∞
If R = , then (a) shows that a (n) = 0 for all k such that k > α .
n ∞ k 2−α
This proves (b).
(cid:3)
Lemma 5. Suppose that conditions (1.2) and (1.3) hold. If for some fixed
k,n N, A > 0 and α [0,2 2) we have
∈ ∈ − k
(3.2) Ankα−2(k−1) a (n),
k
≤ | |
then
(3.3) R C˜n2−α,
n
≤
where C˜ = C˜(α,M,A,k).
Proof. It is enough to prove that
(3.4) ρ C˜n2−α, ρ (0,R ).
n
≤ ∀ ∈
Then (3.3) follows if we let ρ R .
n
→
By (2.1) we have
Ankα−2(k−1) ak(n) 2C(α,M)ρ−(k−1)max(nα,ρ2−αα).
≤ | |≤
If nα ρ2−αα, then we get (3.4) with C˜ = 1.
≥ α α α
Suppose that nα < ρ2−α. Then max(nα,ρ2−α) = ρ2−α, so
Aρk−2−2α 2C(α,M)(n2−α)k−2−2α.
≤
Thus,wheneverα < 2 2/k,thisinequality implies(3.3)withC˜ = (2C/A)γ,
where γ = (2 α)/(k(−2 α) 2). (cid:3)
− − −
According to the preceding lemma, all one needs in order to get an upper
boundon R of theform (3.3)(or even to explain thatR is finite)is to find
n n
alowerboundon a (n) oftheform(3.2)(oratleasttoexplainthata (n) =
k k
| | 6
0). We now describe a technique to provide such lower bounds. Theorem 2
will follow when we get such lower bounds for a (n),..., a (n).
2 12
| | | |
CONVERGENCE RADII FOR EIGENVALUES 9
Lemma 6. Under conditions (1.4) and (1.19), for each fixed α < 2, the
coefficient a (n,α) can be written in the form
k
(3.5) a (n,α) = nkα−(k−1)f (1/n)
k α
where
∞
f (w) = P (j,α)wj
α k
j=0
X
is analytic on the disk w < 1/k, and P (j,α) are polynomials of α.
k
| |
Proof. We begin this proof by stating the equation (3.7) from [5]
1
(3.6) a (n) = (λ n2) R0(BR0)ke ,e dλ,
k 2πi  − h λ λ j ji
Z∂Π |j−n|≤k
X
 
whereR0 = (λ L)−1, e is the jth unitvector, and Πis thesquarecentered
λ − j
at n2 of width 2n. This formula appears in [5] only in the case of α [0,1),
∈
but its proof therein holds for α < 2 as well. It follows from (1.1) that for
each j N,
∈
(j−1)αe + jα e if j > 1
λ−j2 j−1 λ−j2 j+1
BR0e =
λ j 
 1 e if j = 1.
λ−1 2
So, (λ−n2)hRλ0(BRλ0)kej,eji can be written as a finite sum each of whose
terms is of the form
λ n2 k (n d′)α
− − i
λ (n j′)2 λ (n j′)2
− − 0 i=1 − − i
Y
with j′ and d′ integers satisfying j′ , d′ < k for each i. So, from a residue
i i | i| | i|
calculation on (3.6), a (n) can be written as a linear combination of terms
k
of the form
k−1 (n d )α
(3.7) (n d )α − i
− k n2 (n j )2
i
i=1 − −
Y
αk−1 α −1
d d j
= Cnkα−(k−1) 1 k 1 i 1 i
− n − n − 2n
" #
(cid:18) (cid:19) i=1 (cid:18) (cid:19) (cid:18) (cid:19)
Y
with C = k−1(2j )−1 and j , d < k for each i.
i=1 i | i| | i|
For n > k, we have d /n < 1 and j /(2n) < 1. Thus,
i i
Q | | | |
α 2
d d α(α 1) d
i i i
(3.8) 1 = 1 α + − +...
− n − n 2 n
(cid:18) (cid:19) (cid:18) (cid:19) (cid:18) (cid:19)
−1 2
j j j
i i i
(3.9) 1 = 1+ + +...
− 2n 2n 2n
(cid:18) (cid:19) (cid:18) (cid:19) (cid:18) (cid:19)
10 J.ADDUCI,P.DJAKOV,ANDB.MITYAGIN
are analytic functions of z = 1/n whenever n > k. Combining (3.7) with
(3.8)–(3.9), we deduce that a (n) can be written as in (3.5) with f (z)
k α
analytic for z < 1/k. (cid:3)
| |
The preceding lemma guarantees that whenever α < 2,
a (n,α) = P (0,α)nkα−(k−1) +O(nkα−k) as n .
k k
→ ∞
Whena (n),...,a (n)werecomputed(following theapproach of[5, p.305–
2 12
306]), an interesting phenomenon was observed. If 2 k 12, then
≤ ≤
(3.10) P (j,α) = 0 for each 0 j k 2.
k
≤ ≤ −
In particular, if (1.18) and (1.19) hold, then
(3.11) a (n) = P (k 1,α)nkα−2(k−1) +O(nkα−2k+1), n ;
k k
− → ∞
the polynomials P (k 1,α), k = 2,4,...,12, are given in the following
k
−
table.
k P (k 1,α)
k
−
2 α+ 1
− 2
4 α3+ 9α2 11α+ 5
− 4 − 8 32
6 9α5+ 73α4 27α3+ 281α2 147α+ 9
−4 8 − 2 32 − 64 64
8 61α7+ 2881α6 6875α5+ 33937α4 11437α3+ 64649α2 4507α+ 1469
− 9 72 − 72 288 − 144 2304 − 1024 8192
10 1525α9 + 23705α8 353023α7+ 648539α6 5774039α5+ 7955297α4
− 64 128 − 576 576 − 4608 9216
6626165α3+ 6173425α2 148881α+ 4471
− 18432 73728 − 16384 16384
12 221321α11+ 8544347α10 1207947α9 + 71029219α8 92577243α7+ 385333821α6
− 2400 9600 − 320 7680 − 6400 25600
16162765α5+ 9344339α4 583689039α3+ 296768801α2 12877899α+ 121191
− 1536 1920 − 409600 1228800 − 655360 262144