Phương pháp xác định phương trình dạng ẩn cho đường cong và mặt tham số hữu tỷ7_2

Phl;t ll;tc A
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THU~TTOANTINHB~CANCUAD~NGTHAMSO
Chung ta bier ding yoi m6i d~;lllgtham s6 hoa cua m~t huu ty se t6n
t(;limOt s6 nguyen n ~ 1 sao cho mOt diSm tren m~t tu'dng ling yoi n gia tri
tham s6. s6 n du'QcgQi 1fts6 tu'dng ling cua d(;lngtham s6 boa. Khi n = 1 ta
~

noi d(;lng tham s6 hoa 1ft chinh qui, ngu'Qc l(;li 1ftkhong chinh qui. Theo Dinh
19 Bezout [14], b~c cua da thlic f b~ng s6 giao diSm cua m~t d(;li s6 f

=0

yft hai m~t ph~ng. Nhu'ng giao cua hai m~t ph~ng 1ft mOt du'ong th~ng, y~y

su dvng hai du'ong th~ng khac nhau chung ta se Hnh du'Qcb~c cua da thlic f.
Thm}t toan A.I. Tinh b~c in cua m~t tham s6 huu ty (1.3) [4].
Nh4p:

Cac da thlic a,b,c,d E k[s,t].

Xudt:

I Ia b~c in cua m~t tham s6 huu ty (1.3).

Rude 1:

TIm biSu di~n in cua mOt du'ong th~ng toy 9 ca:t m~t tham
s6 huu ty alx + (31Y+ "YlZ+ 81 = a2x

Rude 2:

+ (32Y+ "f2Z + 82 = o.

Cac gia tri s tu'dng ling yoi cac giao diSm cua du'ong th~ng
trong Rude 1 yoi m~t tham s6 1ft nghi~m cua k~t thuc
h(s) = Res(ala + (31b+ "flC+ 81d,a2a + (32b+ "f2C+ 82d,t).

Rude 3:

Neu m~t tham s6 co chlia cac diSm co so thl h(s) co cac
nghi~m ngo(;li lai. DS lo(;li bo chung, chung ta Hm biSu
di~n in cua mOt du'ong th~ng toy
trong Rude 1 Ia alx

9 khac

+ !31Y+ ilZ + 81=

yoi du'ong th~ng

a2x + !32Y
+ i2Z

+ 82

= o.
Rude 4:

Tu'dng tv xac dinh da thlic co cling s6 nghi~m ngo(;lilai yoi

.


41
da thuc h(s) trong Budc 2 Ia fi(s) = Res(ala + ,BIb+ "YIC
+
81d, a2a + ,B2b + "Y2C+ 82d, i).

Budc 5:

Xua't l:= deg(h(s)) - deg(gcd(h(s),fi(s))).


Ph1;t l1;tc B

CACTHU!TTOANCOBANvtMATR!NDATHDc
Trang ph§n nay chung toi se trinh bay tom tilt mQt s6 khai nit?m va cac
thu~t roan Hnh roan cd ban vS ma tr~n da thuc lam cd sd cho vit?c xa~ dinh
modun syzygy [14, 19].

Ma tr~n F ca'p m x l voi cac ph§n td' thuQc k[xl'...,xIJ

duQc gQi Ia ma

tr~n da thuc, ky hit?u FE k1nXI[Xl"",Xn]. Ma tr~n F co h~ng Ia T', ky hi~u
rankF = T', ne'u tan t~i mQt dinh thuc con ca'p T' khac khong va ta't ca cac dinh
thuc con ca'p T'+ 1 dSu b~ng khong. Ne'u rank(F)

= min(m, l) ta noi F 1a h~ng

d§y duo B~c cua F, ky hit?u deg(F), 1a gia tri IOn nha't cua b~c cua cac ph§n td'

cua F.
Cho ME k1nX1n[Xl"",XIJ,khi do dinh thuc cua M 1a mQt da thuc thuQc
k[Xl"",Xn]' Ta noi M khong suy bie'n ne'u det(M) +=O. Ne'u det(M)

Ia mQt

h~ng khac khong thuQc k ta noi M Ia ddn modun.
Djnh nghia B.l. Cha FE k1nXI[Xl'...,xnJ,m < l. Khi do F dll(1CgQi [a:
(1)

Nghi~m nguyen ((5 trai (ZLP) ntu khong t6n tCfimQt n-bQ (zf, ..., z~) E kn [a

nghi~m chung cila tat cd cac dinh thac can cap m cila F.
(2) Dinh thac nguyen t6 trai (MLP) ntu tat cd cac dinh thac can cap m cila F
nguyen to' cung nhau.

(3) Thaa s6 nguyen t6 trai (FLP) ntu F

= ~F;, trangdo ~

dan modun.

Cac khai ni~m nghi~m nguyen t6 phdi (ZRP), dinh thac nguyen t6 phdi (MRP) va
thaa s6 nguyen tlf phdi (FRP) dU:(lcdinh nghia tllang fT!.


43

Vdi n = 1,2 thl MLP

va MRP

$.

FLP. Vdi n

2::

FLP va MRP

FRP. Vdi n

2::

3 thl MLP $. FLP

1 thl MLP ~ FLP va MRP ~ FRP [18].

B.I Phan tich nhan tii'ma tr~n da thuc
~

Thu~t tmin B.l. Phan tich nhan tii' tnii cho ma tr~n da thuc hai bie'n h~ng d~y du
[12, 16].

< l.

Nhljp:

Ma tr~n FE kmXI[XpX2Jco h~ng d~y du, m

Xudt:

Hai ma tr~n L E kmxm[xpx2J, R E kmXI[XpX2JsaD cho F = LR

va det L = 9 vdi 9 E k[XIJIa dung cua tide chung ldn nha't cua
cae dinh thuc ca'p m cua F.
BucJe 1:

TIm da thuc 9 E k[XIJIa dung cua tide chung ldn nha't cua cac
dinh thuc ca'p m cua F. Phan tich 9 thanh tich cua cae da thuc
ba't kha guy trang k[XIJ.
L = 1m;R = F.

Bude 2:

Ne'u da: xet he't cae thanh ph~n ba't kha guy thl de'n Rude 6.
NgtiQc I~i, vdi thanh ph~n ba't kha guy p E k[XIJ
i

Rude 3:

=

1; j

Trang ma tr~n

= 1; R = R

(mod p ).

R, ne'u t6n t~i mQt hang io, vdi i < io < m, ma

ta't ca cae ph~n tii' deu b~ng kh6ng thl Do = diag(l,...,p,...,l)
Ia mQt tidc trai cua F
L = LDo; R = D;;lR.

Quay I~i Bude 2 vdi thanh ph~n ba't kha guy ke'tie'p.


44

BUdC 4:

R, tlm cQt jo d~u tien, vdi j < jo < l, co it nha't

Trang ma tr~n

mQt ph~n ta khac kh6ng trong cac hang tu i de'n m

= )0'

)

BUdC 5:

Trang cQt j cua ma tr~n R, tu hang i de'n m Hmph~n ta co
~

b~c theo x2 nho nha't, gQi la Rj~' D~t D1 la ma tr~n co du<;1ctu
1mbiing cach hoan vi hai dong i va
-

R

-

~

-1

= D1R; L = LD1;R = D1 R.

Giii sa CQtj co d~ng (...,ai,...,am)T. Hc$s6 chu d~o cua cac da
thuc ai' ..., am E k(X1)[X2] la cac da thuc bi, ..., bm E k[xd. Do
bi va p nguyen t6 clIng nhau nen theo thu~t toan chia Euclide
tan t~i cac da thuc g, h E k[X1]sao cho
gbi = 1- hp.

D~t a;

= gai (modp) E k[XpX2]'Khi do tan t~i cac c~p da thuc

qk' ..., rk E k[X1,X2]sao cho
*

ak = qkak + rk'

vdi k = i + 1, ..., m va degt rk < degt a; ho~c rk
1

1
D3=

1
-xqi+1

-xqm

1
1

= O. D~t


45

-

-

-1

R = D3R;L = LD3 ;R = D3R.
Khi do cQt j co d(;lng(...,ai,1i+1,...,r;n)T.

R = R (modp).
Ne'u cac da thuc 1i+1= ... = r;n= 0 thl

i = i + 1;j = j + 1;
va quay l(;liRUdC 3. NguQc l(;li, quay l(;liRUdC 5.
RUdC 6:

Xua't L, R.

Ne'u m > l, ta cling co phan rich nhan tii' phai cho P, nghla la tlm duQc
hai ma tr~n L E k1nXl[XpX2] va R E k1Xl[X1'
X2] sao cho P = LR. Ap d\lllg ma tr~n

chuyS'n vi ta co thS' xay dvng thu~t roan phan rich nhan tii'phai nhu sau.
Thm}t tmin B.2. Phan rich nhan tii' phai cho ma tr~n da thuc hai bie'n h(;lngdfiy
duo

> l.

Nhqp:

Ma tr~n P E k1nXl[Xp
X2]co h(;lng dfiy du, m

Xudt:

Hai ma tr~n L E k1nXl[X1'
X2]' R E k1Xl[X1'
X2] sao cho P

= LR va

det R = g voi g E k[X1]la dung cua uoc chung IOn nha't cua
cac dinh thuc ca'p l cua P.
RUdC 1:

Di;it p' = pT E k1X1n[X1'
X2]' dung Thu~t roan B.1 tlm hai ma
tr~n L' E k1Xl[X1'
X2] va R' E k1X1n[xp
X2]sao cho p'

RUdC 2:

Ta co (pT)T = pIT = (L'R')T hay P = R'TLff = LR. Di;it
L = R'T;R = Lff.

RUdC 3:

= L'R'.

Xua't L, R.

B.2 Tim d~ng Hermite


46

Cho ma tr;%n F E k1nXI[Xl'X2]' d,.lllg Hermite

cua F la ma tr;%n H = UF vdi

hij = 0 nSu i > j va degx2 hii > degx2hij nSu i < j, trong do U E k1nX1n[xl'
X2] va

ddn modun. Sa dl;lllgky thu;%tkha Gauss de tim d~ng Hermite cho tru'ong hQp
mQt biSn [12], ta co thu;%ttoan sau.
Thu~t toan B.3. Tim d~ng Hermite cho ma tr;%nda thti'c hai biSn h~ng d~y du
[12, 14, 19].
Nhqp:

Ma tr;%nF E k1nXI[Xl'
X2]co h~ng d~y duo

Xudt:

Ma tr;%nHermite HE k1nXI[Xl'X2]cua F.

Bu(]c 1:

Xem F E k1nX\Xl)[X2]'sa d\lng ky thu;%tkha Gauss ta tim du'Qc
ma tr;%nHermite II E k1nXI(Xl)[X2]cua F va ma tr;%n 0 E

k1nX1n(

Xl)[X2] sao cho II = OF.
Bu(]c 2:

GQi hi vdi i = 1,..., m la bQi chung nho nha't cua cae m§:uthti'c
hang thti' i. D~t D = diag(~,...,h1n)'

H=DH;U=DU.
Bu(]c 3:

Xua't H.

B.3 Rut trich tide chung IOn nha't
A, B la hai ma tr;%nda thti'c cung sO'hang (cQt). A, B du'QcgQi la nguyen
to' cung nhau trai (phai) nSu t6n t~i hai ma tr;%n A, B va ma tr;%nda thti'c ddn

modun C sao cho A = CA,B = CB (A = AC,B = BC).


47

Ma tr;%nda thuc vuong D duQc gQi la uoc chung tnii (phiii) IOn nha't cua

A va B ne'u tan t~i hai ma tr;%nda thuc 1,
sao cho A

13nguyen

to' cung nhau trai (phiii)

= DA,B = DB (A = AD,B = BD).

Thu~t tmin B.4. TIm uoc chung phiii Ion nha't cua hai ma tr;%nda thuc [12, 16].
~

Nhqp:

Hai ma tr;%n da thuc

A E kmXI[Xl'X2] va B E knXI[XpX2] sao cho

(AT,BT)T E k(m+n)xl[Xl,X2]la h~ng d~y duo
Xu{{t:

Ma tr;%nda thuc DE kIXI[Xl,X2]la uoc chung Ion nha't phiii cua

A va B.
RUdC1:

Dung Thu;%ttoan B.2 phan tfch nhan tu phiii ma tr;%n

= [~]R.

[~]
RUdC 2:

Dung Thu;%ttoan B.3 tlm d~ng Hermite cua ma tr;%n (iF, IF)T

u[~J = [~J.

RUdC3:

Dung Thu;%ttoan B.l phan tich nhan tu trai ma tr;%n
Rv

RUdc 4:

Xua't D

= LR.

= RR.

B.4 Dc) phuc t~p Hnh toaD
BQ phuc t~p cua Thu;%ttoan BA phl;1thuQc vao dQ phuc t~p cua Thu;%t
toan B.l d RUdC1 va 3 va dQ phuc t~p cua Thu;%ttoan B.3 d RUdC2. Cac Thu;%t
toan B.l va B.3 d€u duQc xay d1.,1'ng
d1.,1'a
tren cach tie'p c;%nc6 dien la phuong
phap khii' Gauss. Banh gia dQ phuc t~p cua Thu;%ttoan B.l tuy khong don giiin


48

nhrtng chUng ta co th€ thiy cd ban no se g6rn di) ph",c taP d€ Hnh

(7]

dinh thuc

ca'p m d Budc 1 va dQ phuc t""p cua cac bu'oc khii' Gauss con I""i. Nhu' v~y dQ
phuc t""p cua Thu~t roan B.I cling Ia mQt ham da thuc theo m, Z. DQ phuc t~p
,

cua Thu~t roan B.3 Ia O(mZ2d2)[20], trong do d Ia b~c cua ma tr~n da thuc. V~y
dQ phuc t""pcua Thu~t roan BA Ia mQt ham da thuc.


Phl;t ll;tc C
A

""

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THU~T TOAN XAC DJNH JL-COSO
Trang ph~n nay chung Wi trlnh bay thu~t toan tim f.L-cdsd cua duong cQng

ph~ng va m~t thalli s6 hull ty duqc sa d1;lllgtrong Thu~t toan 3.3 va 3.4 trong
Chuang 3.
c.t Thn~t toaD Om p,-cd sO'cua du'ong cong tham s(f hun ty phdng
Cd sd de xay dl;1'ngthu~t toan xac dinh f.L-cd sd cua duongcong

thalli s6

hull ty ph~ng la cac ke't qua sau.

Dinh Iy C.t. Cho p, q E Syz(a,b,c) la hai dudng thdng di d(}ng sinh dudng Gong
tham sf;' (1.2) bfjc n. Gid sii deg(p)

< deg( q),

khi do p va q tCJothanh f.L-casO

cua dudng Gong(1.2) niu va chi niu m(}t trong cac ddu ki~n sau thoa
(1) M(}t dudng thdng di d(}ng L sinh dudng Gongtham s{f (1.2) co thi duf/c biiu
diln boi (2.3) wii deg(~p) :::;deg(L) va deg(h2q) < deg(L).
(2) M(}t dudng thdng di d(}ng L sinh dudng Gong tham s{f (1.2) co thi duf/c bilu
diln boi (2.3) va hai vectd LV(p) va LV(q) d(}c lfjp tuyin tinh tren JR.
(3) M(}t dudng thdng di d(}ng L sinh dudng Gong tham s{f (1.2) co thi duf/c bilu

diln boi (2.3) va deg(p) + deg(q) = n.
Chung minh: xem [2, tr. 374].
M~nh d~ C.2.
(1) Modun syzygy Syz(a,b,c) cua dudng Gong tham s{f hilu ty phdng (1.2) duf/c


50

sinh biJi ba duilng thling di dQng VI = (-b, a,0), V2= (-c, 0,a) va V3= (D,c,

-b).
(2)

rank(vI,v2,V3)=2.

(3)

rank(LV(vI)' LV(V2)'LV(vJ) = 2.

Chung minh: xem [2, tr. 376

- 377].

Thu~t toaD C.l. Tim p,-ed sa eua dttong eong tham s6 huu ty phing [2].

Nh(jp:

Cae da thlie a,b,c E k[t].

Xudt:

Hai da thlie p, q E k[x, y, t] 13.p,-ed sa eua dttong eong tham s6

huu ty phing.
Rude 1:

Df;it
VI =

(-b,a,D), V2= (-c,D,a), V3= (D,c,-b);

mI = LV(VI)' m2=LV(v2)' m3=LV(v3)'
Rude 2:

Df;it ni = deg(vi)' vdi i = 1,2,3. Khong ma't tinh tang quat, ta
siip xe'p 1~i cae Vi theo thli t1! giam d~n eua cae ni.

Rude 3:

Tim cae s6 th1!e aI' a2' a3 (co it nha't 2 s6 khae thong) sao eho
aImI + a2m2 + a3m3

Rude 4:

Ne'u al

-:;r:.

V1 --

Ne'u

al

=

D.

D thi
a 1V1 + a 2tnl-n2v

~

= LV(VI);

nI

= deg(

2

+ a 3 tnl-n3v 3'.

VI)'

= D khi do a2,a3 -:;r:.
0 thi
V2 --

r" V
"'2 2

+

a3

t n2-n3V3''


51

m2
n2

Buac 5:

=

LV(V2);

= deg( V2).

Ne'u mQt trong cac Vi = 0, gia SITla VI= 0, thl d~t p = V2va
q = V3 Ia hai da thuc t<:;lO
thanh f-J,-Cdsa cua du'ong cong t~ham

s6 huu ty ph£ng. Ngu'qc l<:;li,
quay v6 Buac 2.
C.2 Thn~ t toaD tim p,-cd sd cua m~t tham sf{ hun ty
Cd sa quan trQng de giup Hm f-J,-cdsa cua m~t tham s6 hUll ty la hai ke't
qua sau.
Dinh Iy C.3. Cho mift tham sd hiiu ty (1.3), khi d6 fuon tan ((;iiba phdng di dqng
p, q,r saG cho (2.4) thoa. Ban niia, mqt ca sa p, q,r tfly

y caa modun

Syz(a, b,c,

d) Gang thoa (2.4).
Chung minh: xem [1, tr. 694].

Dinh Iy C.4. Cho p, q,r fa mQt f-J,-casa caa mift tham so' hiiu ty (1.3). Khi do
p, q,r fa mQt ca sa caa modun Syz(a, b,c,d), ngh'ia fa vai ba'tky phdng di dQng P
sinh mift tham so' co bilu ddn duy nha't
P = ~p + ~q + h3r
vai ~,~,

h3 E IR.[s,t]. Ban niia degt(~p), degt(~q), degt(h3r)

+ degt(q) + degt(r)

-

n; degs(~p), degs(~q), degs(h3r)

:::;

< degt(P) + degt(p)

degs(P) + degs(p) + degs(

q)+ degs(r) - m ntu mift fa song btJ-c(m, n) va deg(~p), deg(~q), deg(h3r) < deg
(P) + deg(p) + deg(q) + deg(r)
Chung minh: xem [1, tr. 695].

-

n ntu mift fa phan tam gidc btJ-cn.


52

Do d6 de rim f-L-cosd cua mi,it tham s6 huu ty (1.3) chung ta chi cffn xac
dinh co sd cua m6dun Syz(a, b,c,d). Chung ta bie't ding vi~c xac dinh co sd cho
mQt m6dun tl,l' do la kh6ng d~ dang. Trong khi d6 ta c6 the sa d\lng

y tudng cua

thu~t roan Buchberger de xay dl,l'ngt~p hqp cac phffn ta sinh cho m6dun sy-zygy

[10] va sa d\lng ke't qua cua M~nh d€ 2.20 de rim co sd cho m6dun Syz(a,b,c,d).
Nhung phuong phap nay ding chua hi~u qua.
Sa d\lng cac ke't qua nghien CUll v€ m6dun syzygy cua h~ cac phuong

trinh tuye'n tinh thuffn nha't voi cac h~ s6 la da thuc nhi€u bie'n dl,l'atren ly

thuye'tcac ma tr~n da thuc [18], chung ta c6 the xay dl,l'ngthu~t roan xac dinh f-Lco sd cho mi,it tham s6 huu ty [12]. Hon nua phuong phap nay cling c6 the duqc

ap d\lng de rim f-L-cosd cho duong cong tham s6 huu ty hi~u qua hon Thu~t roan

D.l.
Cho ma tr~n da thuc F =

(A,...,ft) E k7nXI[Xl"",XrJ Khi

d6 (~,...,hl)T

E kl[

Xl,...,X,J duqc gQi Ia mQt syzygy cua F ne'u
I

Lhd:
= o.
i=l

(D. 1)

T~p hqp ta't ca cac syzygy cua F duqc gQila m6dun syzygy cua F, ky hi~u la

Syz(F). T~p hqp cac ~,...,hs E kl[xl1'",xnJ duqc gQi Ia t~p cac phffn ta sinh cua
Syz(F) ne'u
Fhi

=0

voi i = 1,...,8

(D.2)

va voi mQi syzygy t E Syz(F) t6n t~i duy nha't cac da thuc fl1""fs E k[xl1""x,J
saD cho


53

t

= h~ +...

(D.3)

Ishs'

Ne'u ta d~t h = (hi,...,fmJT,

hj

= (~j,...,hlj)Tva t = (~,...,tl)T voi i = 1,...,[ va

j = 1,...,s thl cac bi€u thuc (C.2) va (C.3) trd thanh

hI

hi )(~I

... ~s
=0

Imi ... ImLj
l hn ...
..

his

Ii

~

F

va

;. _f;I
-

...

;sH~

Ihn

. ..

his Ills

tl

Khi do H = (~,...,hs) E kIXS[xu...,x,J duQc gQi la ma tr~n sinh cua Syz(F). Cau
hoi d~t ra la khi nao thl ma tr~n H co s6 cQt nho nha't va co tan t~i hay kh6ng
mQt ma tr~n sinh co s6 cQt nho nha't voi F ba't ky? Cac ke't qua sau se tra loi
cho va'n d€ nay.
Mc$nh d@ C.S. Cho F

= (-fib) E kmXI[Xl1'..,X,J co h{mg fa m, wii 1> m va b

kmxm[xu ..., x,J khong suy bitn. Diit T = [-

E

m, khi do modun Syz(F) co ma trlJ.n

sinh dip [x T ntu va chi ntu tan t{li m(}t ma trtJ.nMRP HE kIXr[XI,...,x,J SaD cho
FH = O. Hefn nila H chinh fa ma trtJ.nsinh.
Chung minh: xem [18, tr. 80].

Voi n ::;2 ta lu6n Hm duQc mQt ma tr~n sinh co s6 cQt nho nha't cho F tuy y.


54

co hc;mgla m, veli l> m va DE
Mc%nhd~ C.6. Cho F = (-Njj) E klnXI[Xl,X2]
klnXIn[xl'X2] khong suy bien. Dijt r
klXr[xl' X2] cho modun

= l-

m, khi do tan tc;limQt ma tr4n sinh H E

Syz(F).

Chung minh: xem [18, tr. 81].
D~t F = (a,b,c,d) E eX4[s,t], theo M~nh de C.6 m6dun Syz(F) hay Syz(a,

b,c,d) co ma tr~n sinh H E k4X3[S,t]. Theo M~nh de 2.20, suy ra cac cQtcua H
t(;lOthanh mQt cd sa cua m6dun Syz(a, b,c,d). V~y ta co thu~t toan sau xac dinh
p,-cd sa cho m~t tham sO'hilu ty.
Thu~t toaD C.2. Tim p,-cd sa cho m~t tham sO'hilu ty [12].
Nh4p:

Cac da thuc a,b,c,d E k[s,t].

Xw1't:

Ba da thuc p, q,r E k[x,y, z, s,t] la p,-cd sa cua m~t tham sO'hilu
?

ty.

Buelc 1:

D~t

N = (-a,-b,-c)

va jj = (d). Xay dllng ma tr~n

p = jj-lN = (-ajd,-bjd,-cjd)
Trong do N = (-a,-b,-c)
BuelC 2:

= ND-1.

va D = diag(d,d,d).

Dung Thu~t toan B.4 till u'oc chung IOn nhtt C cua N va D.
Khi do t6n t(;lihai ma tr~n N, jj E k[s, t] sao cho
N = NC; D = DC;

trong do N va jj nguyen to' cung nhau phai. D~t
H

-T

-T T

= (N ,D ) = (hr,~,h3);

P = hrT(x,y,z,l);q= h;(x,y,z,l);r = hJ(x,y,z,l).

55

Blide 3:

Xua't p, q, r.

C.3 Dc} phuc t~ p Hnh tmin
DQ phuc t(;lp cua Thu~t roan C.! trong trliong hQp trung binh la O(n2) [2].

Trang khi do vi~c danh gia dQ phuc t(;lpcua Thu~t roan C.2 la khong don gian
va no ph\l thuQc chinh vao dQ phuc t(;lpcua cac thu~t roan v€ Hnh roan ma tr~n
da thuc trong Ph\ll\lc B. C\l the, neu bo qua dQ phuc t(;lpcua vi~c xay d1;fngcac
ma tr~n N va D t(;liBlide 1 thi dQ phuc t(;lpcua Thu~t roan C.2 ph\l thuQc chinh
vao dQ phuc t(;lpcua Thu~t roan B.4 d Blide 2. V~y ta co the xem dQ phuc t(;lp
cua Thu~t roan C.2 xa'p Xl dQ phuc t(;lpcua Thu~t roan B.4 va cling la mQt ham
da thuc.


Phl;t ll;tc D

cA C PHUONG TRINH M4.T HUU TY
Trang phgn nay chung Wi lit%tke phu'dng trinh cac m~t huu ty du'Qcdung
d€ daub gia cac thu~t roan gn hoa trong Chu'dng 3.
Sl

= -3s3t - 3,-4i

+ 6st,9s - 6,t;

s2' = it + 2st5 - st s3e - s2t3 - 2it2 + S2 - st - 2s", s - t - 2 s2t - se - 2st,

+ t, s2t+ st4 + st2 - 3s + t5 - 3t3, st +

S3

=

S4

= -18s4t + 27s3t2 - s2t3 - 8st4 + 2t5,

S2

4s3t3 + 6s3t2+ 2s3t -

2s2t4 - 3it3

e-

3, s2t + 2se - 3s + t3 - 3t;

- s2t2,

-3s3t2 - 3s3t - 8it3 - 8s2t2 - 6s2t - 6i + 3st4 + 3st3 + 2st2 + 2st,

6s3t2+ 6s3t - 5s2t3- 5it2 + st4 + st3;
S5

S6

=

S3

+ 5st2+ 2st+ t3 - 1,2S2+ 3st + 2t3 - 2,5s3+ 2it + 3se + 5st+ 5t2- 5,

S4

+ t4 - 1;

= 16s3+ 32it -120i - 56st + 128s+ 24t - 24,
48it + 8i + 16se - 456st + 384s + 16e -16t2 + 392t - 392,
16s3 + 32it + 192i + 48se -1328st + 1040s + 64e - 48e + 1232t -1248,
16s3 - 240st + 224s + 16e + 224t - 240;

S7

= 16s3 + 32it -120i

- 56st + 128s + 24t - 24,

48s2t + 8i + 16se - 456st + 384s + 16t3 -16e + 392t - 392,
16s3 + 32it + 192i + 48st2 -1328st + 1040s + 64t3 - 48e + 1232t -1248,
- 2i + 8s - 6;
S8

= 684288s3-1419264it

- 608256i + 684288st2 + 1495296st - 836352s-

684288t2 - 76032t + 760320,983808s2t+ 1043712i - 8711424st +


57

7803648s + 684288t3 + 1119744t2+ 7043328t - 8847360, 4790016s3
6983424s2t -1126656s2

-15667200st

+ 17556480s + 1430784t2 +

20597760t - 21219840,6158592s3-19660032s2t -12780288i

74437632st- 62152704s - 4790016e

-

+

13996800e- 44987584t+

-

68774400;
S9

= -84ie

+ 7716st -14004s - 7632t + 14004,- 84se + 1056st -1692s-

972t + 1692, - 252s2t - 84st2 + 3360st - 4284s - 3024t + 4284, 6i 57s + 6t + 51;
SlO= - 2209it3 + 7359ie

- 4800it - 350i + 2225st3 - 7425se + 4800st +

400s, - 2209it3 - 732s2t2+ 3291s2t+ 2225st3 + 750se

-

3375st,

168ie -168it - i -168se + 168st+ 7,
24it2 - 24it - 24st2+ 24st + 1;
S11

= -2209it3 - 732ie + 3291s2t+ 2225st3+ 750se - 3375st,
2209s2t3- 7359ie + 4800it+ 350s2- 2225st3+ 7425se- 4800st - 400s,
168ie -168it

+ S2-168se + 168st + 7,

24s2e - 24s2t - 24st2 + 24st + 1;
S12

= 2209it3

-

7359it2 + 4800it + 350i

2209ie + 732ie
168ie

-168it

-

2225st3+ 7425st2- 4800st

- 3291s2t- 2225st3 - 750se

- S2 -168se

-

400s,

+ 3375st,

+ 168st + 7,

24it2 - 24s2t - 24st2 + 24st + 1;
S13

= 2209s2t3+ 732ie
-

-

3291it - 2225se - 750st2+ 3375st,

2209s2t3+ 7359ie - 4800s2t- 350s2+ 2225st3- 7425st2+ 4800st+ 400s,

-2209it3

+ 7359s2e - 4800it - 350i + 2225st3 - 7425se + 4800st + 400s,

24s2t2- 24it

-

24st2 + 24st + 1;


58

814

= 73682e -11182t2 - 600it - 25i - 7448t3+ 1448e + 6008t + 32t3 -132e +
100,- 2072it3 + 171982e + 378it + 20648t3-17288t2 - 3368t + 32t3 +
36t2 -168t, 54ie - 3682t2-1882t + i - 548e + 368t2+ 188t - 68 + 6,
18it3 -12ie

815

=

-

- 682t-188e + 128e + 68t + 1;

207282t3 + 171982t2+ 37882t + 20648t3 - 17288e - 3368t + 32t3 +

36t2 -168t, -100 - 32e + 25i - 73682t3 - 6008t + 132e -1448t2 +
7448t3 + 60082t + 11182t2,5482t3 - 1882t - 3682t2 +

188t816

=

-

68

i

- 548t3

+ 368e +

+ 6,1882t3-1282t2 - 6s2t -188e + 128e + 68t + 1;

736it3 + 111it2 + 60082t + 25i + 7448e -1448e

- 6008t - 32t3 +

132t2 -100, 207282t3 -171982t2 - 37882t - 20648t3 + 17288t2 + 3368t - 32t3 36t2 + 168t, 54ie

- 36it2 -18it

18it3 -1282t2 - 682t -188e

+ i - 548t3 + 368t2 + 168t - 68 + 6,

+ 128t2 + 68t + 1;