Computer-assisteed proof of a periodic solution in a non-linear feedback DDE

0.26 MB·English

by  Mikolaj Zalewski

Computer-assisteed proof of a periodi
solution in a non-linear feedba
k DDE
Mikoªaj Zalewski
Institute of Mathemati
s, Jagiellonian University
7 Reymonta 5, 30-059 Kraków, Poland
0
0 E-mail: Mikolaj.Zalewskiim.uj.edu.pl
2

n
a
J
9 Abstra
t

] In this paper we rigorously prove thxe′(etx)is=tanK
esionf(xa(tnon1-t)r)ivial
S
periodi
orbit for the non-linear DDE: − for
D K = 1.6
. We show that the equations on the Fourier equations have
.
h a solution by
omputing the lo
al Brower degree. This degree
an
t
a be
omputed by using a homotopy whi
h validity
an be
he
ked by
m

he
king a (cid:28)nite number of inequalities. Che
king these inequalities is
[ done by a
omputer program.

1
v
5
1 Introdu
tion
6
2
1 We will show that the equation:
0
7 x′(t) = Ksin(x(t 1))
0 − − (1)
/
h K = 1.6 1
t has a periodi
solution for . By a solution we mean a C fun
tion
a
m that satis(cid:28)es the equation. π < K < 5.1
v: Numeri
alsimulationssugges0ts that for 2 there isa attra
ting
periodi
orbitos
illatingaround . Aninformalargument fortheexisten
e of
i
X
su
h an orbit also
omes from the analysis of eigenvalues of the linearization
K = 1.6
r
a [Wis
hert et al 94℄. We will prove rigorously that for this orbit
exists.
In the remainder of the introdu
tion we brie(cid:29)y outline our method. We
will use the Fourier
oe(cid:30)
ients of the periodi
orbit. First we res
ale the
2π x˜(t) = x t τ
time to obtain a period. We make a substitution: τ where is
τ 2π
a parameter. If is equal to T (where T is the period of the s(cid:0)ol(cid:1)ution in the
2π
original equation) we will have a -periodi
solution of the new equation:
1
1 t K t
x˜′(t) = x′ = sin x 1 =
τ τ − τ τ −
(cid:18) (cid:19) (cid:18) (cid:18) (cid:19)(cid:19)
K t τ K
= sin x − = sin(x˜(t τ))
− τ τ − τ −
(cid:18) (cid:18) (cid:19)(cid:19)
x˜ x
By renaming to we obtain:
K
x′(t) = sin(x(t τ))
− τ − (2)
τ
We will prove for this equation that there exist a from a small interval
2π
su
h there exist a -periodi
orbit in a small neighborhood of a spe
i(cid:28)ed
fun
tion. Note that we don't obtain the exa
t value of the period (even if
4 τ = π τ
numeri
al solution suggests it is , i.e. 2) but treat as a variable.
Toprovetheexisten
eoftheorbitwewillusethemethodofself-
onsistent
bounds that was introdu
ed in the
ontext of Kuramoto-Shivashinsky PDEs
in [Zgli
zy«ski 01℄, [Zgli
zy«ski 04℄. When applied to the boundary value
problem for ODEs or DDEs this method is similar to the Cesari method
introdu
ed in [Cesari 64℄ but doesn't require one of the
onditions (cid:21) see
Se
tion 2.4 in [Zgli
zy«ski 01℄ for a
omparison. Below we brie(cid:29)y des
ribe
the method.
We will derive from (2) the equation on Fourier
oe(cid:30)
ients. Let's (cid:28)rst
write (2) with a Taylor expansion instead of sinus:
K ∞ ( 1)k
x′(t) = − [x(t τ)]2k+1
− τ (2k +1)! −
n=0
X
The equations on the Fourier
oe(cid:30)
ients will look similar:
K ∞ ( 1)k
n Z : inc = − c∗(2k+1) e−inτ
∀ ∈ n − τ (2k +1)! n
k=0
X (cid:0) (cid:1)
(c∗(2k+1)) c 2k +1
n
where means
onvoluting the sequen
e with itself times
n
and then taking the -th
oe(cid:30)
ient. The operation of
onvolution and this
notation is introdu
ed in details in se
tion 2.
We will show that su
h a fun
tion has a non-trivial zero:
∞ ( 1)k ∞
F(τ,c) = inτeinτc +K − c∗(2k+1)
n (2k +1)! n
( )
Xk=0 (cid:0) (cid:1) n=−∞
Where the domain is:
2
β
X := x : Z C n Z : x x = x
β → ∀ ∈ | n| ≤ ( n +1)2 and n −n
(cid:26) (cid:12) | | (cid:27)
(cid:12)
(cid:12)
β > 0 X
β (cid:12)
forsome . On we
onsidertheprodu
ttopology(whi
h isequivalent
to
omponent-wise
onvergen
e). Itturnsoutthatonsu
hadomainthesums
F F c
in the de(cid:28)nition of are
onvergent, is
ontinuous and ea
h sequen
e ∈
X F(c) = 0
β

orresponds to a real-valued
ontinuous fun
tion. Moreover if
1
then the fun
tion is C .
We willsear
h fora zero intheneighborhoodof an approximatedsolution
cˆ
obtained from numeri
al simulations. We will denote by this approxima-
n n 5
tion. For ≤ ≤ it is equal to:
n cˆ
n
0 0
1 0.1521000000 0.1163508047i
− −
2 0
3 0.0001123121 0.0002746107i
−
4 0
5 0.0000008173 0.0000001014i
− −
n > 5 cˆ n < 0 cˆ = cˆ τ
n n −n
For the is zero, for we have . The in the
τˆ = 1.570796
approximatioln isN: . P Q
l l
For ea
h ∈ let us de(cid:28)ne the Galerkin proje
tion and immersion :
P : CZ c (c ,...,c ) R Cl
l 0 l
∋ → ∈ ×
Q : R Cl (c ,...,c ) (...,0,0,c ,...,c ,...,c ,0,0,...) CZ
l 0 l −l 0 l
× ∋ → ∈
Let's note that we need only the non-negative terms in the (cid:28)nite spa
e
c = c
0 −0
as the necgativRe terms
an be obtained by X
onjugation. Also as so
0 β
we have ∈ . From the
ompa
tness of it will be easy to show that:
β > 0 l > 0 τ,τ R l > l
0 0
Lemma. 15 Let , , ∈ be (cid:28)xed. If for ea
h there
cl X τ [τ,τ] P F(τ ,cl) = 0
β l l l
is a ∈ and ∈ su
h that then there exist a
(c0,τ ) X [τ,τ] F(τ ,c0) = 0
0 β 0
∈ × su
h that .
cl D c0 D
Moreover if all the are in a
losed set then ∈ .
F
Thus it's enough that there is a zero for every Galerkin proje
tion of :
F˜ : R R Cl (τ,c) P F(τ,Q(c)) R Cl
l l l
× × ∋ → ∈ ×
(cid:0) (cid:1)
This will allow us to use a (cid:28)nite-dimensional topologi
al method to show
F 2l + 2
a zero of . Let's note that the real dimension of the domain is
3
2l + 1
while of the image is so if there is a zero then one
an expe
t a 1-
x( )
dimensionalmanifold of zeros. This manifold
an be ea˜sily identi(cid:28)ed (cid:21) if ·
x( +φ) F (τ,c ,c ,...,c ) = 0
l 0 1 l
is a soF˜lu(tτi,ocn,tehieφnc ,..·.,eilφcis)a=lso.0This means that if
l 0 1 l
then what
an be easy
he
ked. To use the
topologi
al method we wacnt tocˆh+avRe the zero isolated so we will limit the
1 1
domain (cid:21) we assume that ∈ . Of
ourFse (cid:28)nding a zero in aFl˜imited
l l
domain implies a zero of the full system. By we will denote the with
the smaller domain.
F
l
The before-mentioned method to prove the existen
e of a zero of is
the lo
al Brouwer degree (introdu
ed e.g. in [Smoller 83℄). Let's denote by
deg(F ,U,x) x U
l
the degree of on . It is known that if the degree is non-zero
y U : f(y) = x cˆ U x = 0
then ∃ ∈ . We will use a neighborhood of as and .
To
ompute the degree we will use the homotopy invarian
e of the lo
al
H F
l l
Brouwer degree. As our homotopy we will use a linear deformation of
G F
l l
into a fun
tion that
ontains the most important terms of (that is not
F
l
stri
tly a linearization of but it is
lose to):
H (h,τ,c) = hF (τ,c)+(1 h)G (τ,c)
l l l
−
G
l
The degree of will be easy to show that it is non-zero. What we will
0 / H ([0;1];∂U)
l
need toshow isthat ∈ . To show that it'senough toshow that
G F
l l
the terms in dominates the (mainly non-linear) terms that are in but
G G F G
l l l l
not (cid:21) that | | ≥ | − |. That part of the proof is
omputer-assisted (cid:21)
the proofs of the estimates are in this paper but
omputing the exa
t values
and
he
king that the inequalitiesholds isdone by a
omputer programusing
the CAPD pa
kage for rigorous interval arithmeti
s.
1.1 The program
This program is written in C++ and
an be downloaded from
http://www.im.uj.edu.pl/MikolajZalewski/dl/delay-sin.tgz . The rounding-
mode
hanging
ode required by the interval arithmeti
is system-dependent
and has been
he
ked to work on PCs (both 32-bit and 64-bit) on both
Windows (
ompiled with
ygwin) and Linux (
ompiled with g

). It should
also work on SPARC and Ma
OS X although that has not been tested.
Using other CPUs or
ompilers might require modi(cid:28)
ations to the rounding

ode.
4
2 Fourier
oe(cid:30)
ients
x ∞ y ∞
We willuse the followingnotation: if { n}n=−∞ and { n}n=−∞ are sequen
es
x y
with
omplex values then by ∗ we will denote the
onvolution of the
sequen
es:
∞ ∞
x y := x y
k n−k
∗
( )
kX=−∞ n=−∞
It is well known that if the sum
onverges the operation of
onvolution is
(...) n
n
asso
iative. We will also use(xtheyn)ot=ation∞ xfoyr the -th
oe(cid:30)
ient of the
sequen
e in bra
kets (cid:21) e.g. ∗ n k=−∞ k n−k. Also:
P
x∗k := x x k 1
∗···∗ (for ≥ )
k
times
| {z 1} n = 0
(x∗0) := if
n 0 n = 0
(cid:26) if 6
Of
ourse the sum in the de(cid:28)nition of the
onvolution may be not
on-
verging so we will limit our attention to a domain where the sum will be
always
onvergent. The following lemma holds:
x y α 2,β ,β > 0 : n : x
1 2 n
Lemma 1. If and is su
h that ∃ ≥ ∀ | | ≤
β1 , y β2 (x y) n (x y)
(|n|+1)α | n| ≤ (|n|+1)α, then ∗ n is
onvergent for ea
h and | ∗ n| ≤
C β1β2 C = 2(2α+1)
(|n|+1)α where α−1
n 0 (x y)
n
Proof: Let's assume ≥ and let's estimate | ∗ |:
∞ ∞ n−1 ∞
x y x y + x y + x y
k n−k −k n+k k n−k n+k −k
≤
(cid:12) (cid:12) (cid:12) (cid:12) (cid:12) (cid:12) (cid:12) (cid:12)
(cid:12)kX=−∞ (cid:12) (cid:12)Xk=0 (cid:12) (cid:12)Xk=1 (cid:12) (cid:12)Xk=0 (cid:12)
(cid:12) (cid:12) (cid:12) (cid:12) (cid:12) (cid:12) (cid:12) (cid:12)
(cid:12) (cid:12) (cid:12) (cid:12) (cid:12) (cid:12) (cid:12) (cid:12)
(cid:12) n−1x(cid:12) y (cid:12) β β(cid:12) (cid:12)n−1 1 (cid:12) 1 (cid:12) (cid:12)
k=1 k n−k ≤ 1 2 k=1 (k+1)α(n−k+1)α
2β β 1 1
(cid:12)(cid:12)P (cid:12)(cid:12) ≤ 2β1β2P 1≤k≤n2 (k+11)α(n−k1+1)α
≤ 2β11β222αP1≤k≤n2 (k+1)1α((n/2)+1)α
≤ 2(βn1+β222)ααP 1≤k≤n2 1(k+1)α
≤ (n+2)α Pk≥1 (k+1)α
∞ x y β β P∞ 1 1
| k=0 −k n+k| ≤ 1 2 k=0 (k+1)α(n+k)α+1
β1β2 1
P ≤ (n+1)αP k≥0 (k+1)α
P
5
Analogi
ally:
∞
β β 1
1 2
x y
(cid:12) n+k −k(cid:12) ≤ (n+1)α (k +1)α
(cid:12)Xk=0 (cid:12) Xk≥0
(cid:12) (cid:12)
(cid:12) (cid:12)
∞ (cid:12) 1 (cid:12)
The sums k=k0 kα+1
an be estimated by integrals:
P∞ 1 ∞ 1 α
1+ dx =
(k +1)α ≤ (x+1)α α 1
k=0 Z0 −
X
∞ 1 1 ∞ 1 α+1
+ dx =
(k +1)α ≤ 2α (x+1)α 2α(α 1)
k=1 Z1 −
X
From that we obtain:
∞
β β α+1 α β β 2(2α+1)
x y 1 2 2α+1 +2 1 2
k n−k
≤ (n+1)α 2α(α 1) α 1 ≤ (n+1)α α 1
(cid:12) (cid:12)
(cid:12)kX=−∞ (cid:12) (cid:20) − − (cid:21) −
(cid:12) (cid:12)
(cid:12) (cid:12)
n < 0
(cid:12) (cid:12)
The result for
an be obtained by analogous estimations or by
x˜ y˜ x˜ := x y˜ := y
n −n n −n
taking a sequen
es , : , and applying for them the
n > 0
result for .
α = 2 α = 2
With one ex
eption we will use this lemma for . For we have
C = 10
.
x α 2,β > 0 : n : x β
Observation 1. If is su
h that ∃ ≥ ∀ | n| ≤ (|n|+1)α then
(x∗k) Ck−1 βk C = 2(2α+1)
n ≤ (|n|+1)α where α−1
(cid:12) (cid:12)
x y x = x y = y (x y) =
(cid:12) (cid:12) −n n −n n −n
Lemma 2. If and are su
h that , then ∗
(x y)
n
∗
x X y X x y X
Thus we have that if ∈ β1 and ∈ β2 then ∗ ∈ Cβ1β2 (where
X C
β
was de(cid:28)ned in the introdu
tion and is from Lemma 1). It will be also
X
β
useful to de(cid:28)ne a set of sequen
es from any :
X := X
β
β>0
[
X
We will limit ourselves to the sequen
es in . For them we have:
c X ∞ c eint
Lemma 3. Let ∈ . Then n=−∞ n
onverges to a real-valued
on-
tinuous fun
tion P
6
N c eint
cPro=of:c Thecfuenin
tti+oncs e−n=in−tN nR are
ontinuous and real-valued
n −n n −n
as∞ c eint ⇒ ∞ Pβ < ∈ . They uniformly
onvergent as
n=−∞| n | ≤ n=−∞ (|n|+1)2 ∞. Thus the limit is also real-valued
aPnd
ontinuous. P
c X 1
Let's note that ∈ doesn't guarantee that the fun
tion is C :
We use the
onvolution be
ause of this well known fa
t:
c X x(t) d X
Lemma 4. If ∈ are the Fourier
oe(cid:30)
ients of , ∈ are the
y(t) x(t) y(t) c d

oe(cid:30)
ients of then the Fourier
oe(cid:30)
ients of · are ∗ . As a
xn(t) c∗n

onsequen
e the
oe(cid:30)
ients of are .
Now we
an write the equation (2) on the Fourier
oe(cid:30)
ients.
τ x(t) : R R 2π
Theorem 1. Let be (cid:28)xed and → be a -periodi
fun
tion with
c X
the Fourier
oe(cid:30)
ients ∈ . Then:
n Z
(i) for ea
h ∈ the sum on the right-hand side of the following equation

onverges:
K ∞ ( 1)k
inc = − c∗(2k+1) e−inτ
n − τ (2k +1)! n (3)
k=0
X (cid:0) (cid:1)
x(t) 2π c
(ii) is a -periodi
solution of (2) ⇔ satis(cid:28)es the equations (3).
To prove the theorem we will need some lemmas. As it has been already
c X x(t) 1 c
mentioned that ∈ doesn't imply that is C . However if is the
solution of equation (3) then we have the following lemma that will allow us
x(t) 1 x(t) ∞
to show that is C ( is even C ):
c β > 0,α 2 n :
Lcemma β5. If satiβs(cid:28)′e:s enqu:action (3) aβn′ d ≥ are su
h that ∀
| n| ≤ (|n|+1)α then ∃ ∀ | n| ≤ (|n|+1)α+1
C := 2(2α+1)
Proof: Let α−1 . We have that:
K ∞ (−1)k c∗(2k+1) e−inτ K ∞ 1 c∗(2k+1)
−τ k=0 (2k+1)! n ≤ τ k=0 (2k+1)!| n|
(cid:12)(cid:12) P (cid:0) (cid:1) (cid:12)(cid:12) ≤ Kτ P∞k=0 (2k+11)!C(cid:0)2k(|βn2|+k+11)(cid:1)α (4)
(cid:12) (cid:12) = K 1 sinh(Cβ)
τ CP(|n|+1)α
inc = n c
n n
We also have | | | || |. The two sides of the equation must be
equal hen
e we obtain:
K 1
sinh(Cβ) n c
τ C( n +1)α ≥ | || n|
| |
7
n = 0 β′ = 2K sinh(Cβ)
Thus for 6 the Cτ satis(cid:28)es the assertion. If it is not
n = 0 β′
satis(cid:28)ed for we
an in
rease the .
α
In
reasing the is important as we have:
c c β
Lemma 6. Let be a sequen
e of
omplex values satisfying | n| ≤ (|n|+1)3
β c = c c
n −n
for some and . Then the sequen
e is a sequen
e of Fourier
1 x(t)

oe(cid:30)
ients of a real-valued C fun
tion .
n c eikt n ikc eikt
Proof: The sequen
es k=−n k and k=−n k are real-valued,
the se
ond is the derivativePof the (cid:28)rst one anPd are uniformly
onvergent as
n
→ ∞. Hen
e both
onverge tox(
to)n=tinuo∞us func
etiioktn and1 the se
ond is the
derivative of the (cid:28)rst one. Thus k=−∞ k is C .
Proof of the Theorem 1: P
x X
Ad (i): The
onvolutions are
onvergent be
ause ∈ . From the
x X
β
equation (4) from the proof of Lemma 5 we have that if ∈ then
∞ (−1)k c∗(2k+1) e−inτ 1 sinh(Cβ) <
k=0 (2k+1)! n ≤ C(|n|+1)2 ∞
(cid:12) (cid:12) inc ∞
P Ad(cid:12)(ii): (cid:0)Impli
at(cid:1)ion ⇒:(cid:12)It's enough to show that { n}n=−∞ are the
(cid:12) x′(t) (cid:12) K ∞ (−1)k c∗(2k+1) e−inτ
Fourier
oe(cid:30)
ients of while −τ k=0 (2k+1)! n are the
K sinx(t τ)
Fourier
oe(cid:30)
ients of −τ − . P (cid:0) (cid:1)
2πx′(t)eiktdt
The (cid:28)rst part
an be obtained by integrating 0 by parts.
K N (−1)k (x(t τ))2k+1 K sinx(t
Asforthese
ondwehavethat−τ k=0 (2k+1)!R − → −τ −
τ) N x(t τ) d :=
c ea−sinτ ∞→ ∞. The Fourier
oe(cid:30)
Pientsx(otf τ)2−k+1 ared∗e(2qku+a1l) to
{ n }n=−∞. Hen
e the
oe(cid:30)
ients of − are whi
h is
c∗(2k+1)e−inτ
equal to . K N (−1)k (x(t τ))2k+1
Hen
ewehavethattheFourier
oe(cid:30)
ientsof−τ k=0 (2k+1)! −
∞
K N (−1)k c∗(2k+1) e−inτ P
are equal to −τ k=0 (2k+1)! n n=−∞. It has been shown
n N o n
P (cid:0) (cid:1)
thKatsitnhxis(tsequτe)n
e inK
on∞verg(e−n1t)kasc∗(2k+→1) ∞e−tihnτus the -th
oe(cid:30)
ient of
−τ − is −τ k=0 (2k+1)! n . This ends the proof of
this
ase. P (cid:0) (cid:1)
x(t) 1
Impli
ation⇐: fromLemmas5and6we obtainthat isaC fun
tion.
K ∞ (−1)k c∗(2k+1) e−inτ
We know that −τ k=0 (2k+1)! n are the Fourier
oe(cid:30)
ients of
K sinx(t τ) inc x′(t)
−τ − . TPhey are equ(cid:0)al to (cid:1)n (cid:21) the Fourier
oe(cid:30)
ients of .
K sinx(t τ) x′(t) 2π
So both −τ − and are
ontinuous and -periodi
fun
tions
with equal Fourier
oe(cid:30)
ients. Hen
e the fun
tions themselves are equal and
equation (2) is satis(cid:28)ed. F : R X CZ
CZ Let's de(cid:28)ne a fun
tion × → (with the produ
t topology on
)
orresponding to the equation (3):
8
∞ ( 1)k ∞
F(τ,c) := inτeinτc +K − c∗(2k+1)
n
(2k +1)!
( )
Xk=0 (cid:0) (cid:1) n=−∞
F(τ,c) = 0 τ,c
Of
ourse having is equivalent to the fa
t that satis(cid:28)es
F
equation (3). In the rest of the paper we will show that has a nontrivial
F(τ,c) = F(τ,c)
n −n
zero. Let's note that .
F
As we will use topologi
al tools we will want to be
ontinuous. First
X
let's note that the
onvolution is not
ontinuous on the whole (as written
X
above we use the produ
t topology on ) but we have:
X
β
Lemma 7. The operation of
onvolution is
ontinuous on ea
h
x0,y0 X δ
β
Proof: Let's (cid:28)x some ∈ . Let's (cid:28)x some and take some
x,y X (x0,y0) x,y n N :
β δ
∈ from a neighborhood of : are su
h that ∀| | ≤
x (x0) , y (y0) < δ N n > N :
n n n n δ δ
| − | | − | where is large enough that for | |
x,y,x0,y0 X x (x0) , y (y0) < δ
β n n n n
∈ ⇒ | − | | − | . We have that:
x0 y0 x y x0 (y0 y) + (x0 x) y
∗ − ∗ n ≤ ∗ − n | − ∗ n|
(cid:12)(cid:0) (cid:1) (cid:12) (cid:12)(cid:0) (cid:1) (cid:12) (cid:0) (cid:1)
(cid:12) (cid:12) (cid:12) (cid:12)
∞ ∞
1
x0 (y0 y) x0 y0 y βδ
∗ − n ≤ j · n−j − n−j ≤ ( j +1)2
(cid:12)(cid:0) (cid:1) (cid:12) j=X−∞(cid:12)(cid:12)(cid:0) (cid:1) (cid:12)(cid:12) (cid:12)(cid:12)(cid:0) (cid:1) (cid:12)(cid:12) j=X−∞ | |
(cid:12) (cid:12)
(cid:12) (cid:12) (cid:12) (cid:12)
δ 0 ((x0 x) y)
Thistendstozeroas → . Afterananalogousestimationfor| − ∗ n|
we have that the
onvolution is
ontinuous.
F X
β
Lemma 8. The fun
tion is
ontinuous on ea
h .
X
β
Proof: The
onvolutions are
ontinuous on ea
h and the result of a

onvolution lays in a Xβ′ for some β′ so K Nk=0 (2(−k+1)1k)! c∗(2k+1) n is
ontin-
N
uous for ea
h . From estimations as in eqPuation (4) w(cid:0)e have t(cid:1)hat for ea
h
N

oe(cid:30)
ientKthis∞seri(e−s1)
konvce∗r(2gke+s1)uniformly as → ∞ hen
e we obtain that
the limit k=0 (2k+1)! n is
ontinuous.
inτeinτc F
The termP n is(cid:0)also
o(cid:1)ntinuous so is
ontinuous.
3 Some estimates
As mentioned in the introdu
tion we will need some estimates to show that
cˆ
the inequality holds in the neighborhood of . We will use two kinds of sets:
9
cˆ+ X X β
β2 that will be used to obtain (cid:28)ner estimates and β1 (where 1 will
cˆ+X
be large enough to
ontain the whole set β2) for some more rough but
simpler ones.
cˆ
In this se
tion we will only assume of that almost all
oe(cid:30)
ients are
Y cˆ
l
equal to zero. By we will denote the spa
e of the possible values of (cid:21) the
l,...,l
set of sequen
es su
h that at most the elements − are non-zero:
Y := c X : n : n > l c = 0
l n
{ ∈ ∀ | | ⇒ }
First let's note two simple properties:
c Y c∗k Y
l kl
Lemma 9. If ∈ then ∈
x,y X
Lemma 10. If ∈ then
k=n
n
(x+y)∗n = x∗k y∗(n−k)
k ∗
k=0(cid:18) (cid:19)
X
K (x∗p)
To estimate p! n (cid:21) an element of the sum in the equation (3) (cid:21) for
x X c+X
β β
∈ it's enough to use Lemma 1. However for the sets of the form
we will use a more sophisti
ated estimation:
c Y p > 1 β > 0 x X
l β
Lemma 11. Let ∈ , , . Then for any ∈ :
K K p−1 p lk βp−k
(c+x)∗p c∗p + c∗k Cp−k−1
p! n ≤ p! | n | k j ( n j +1)2
(cid:12) (cid:12) k=0(cid:18) (cid:19)j=−lk | − | !
(cid:12) (cid:12) X X (cid:12) (cid:12)
(cid:12) (cid:12) (cid:12) (cid:12)
(cid:12) C = 10(cid:12)
Where
Proof: From Lemmas 1 and 10 we have:
p
K(x+c)∗p K p c∗k x∗(p−k)
p! n ≤ p! k=0 k ∗ n
(cid:18) (cid:19)
(cid:12)(cid:12) (cid:12)(cid:12) K Pp p (cid:12)(cid:12)(cid:0) ∞ c∗k (cid:1)x∗(cid:12)(cid:12)(p−k)
(cid:12) (cid:12) ≤ p! k=0 k j=−∞ j n−j
(cid:18) (cid:19)
(cid:12) (cid:12)
K Pc∗p + p−P1 p (cid:12)(cid:12) ∞(cid:12)(cid:12)(cid:12) c∗k(cid:12)Cp−k−1 βp−k
≤ p! | n | k=0 k j=(cid:12)−∞ j (cid:12) (|n−j|+1)2
(cid:18) (cid:18) (cid:19) (cid:19)
P P (cid:12) (cid:12)
c Y c∗k = 0 j > kl (cid:12) (cid:12)
We have ∈ l so j for what ends the proof.
Of
oursewe
anusethepreviouslemmaonlyfora(cid:28)nitenumberofterms.
To estimate the tail we will use a weaker estimationby using a neighborhood
of the se
ond type, applying Lemma 1 and summing the geometri
sequen
e:
10