i Hc Thi Nguyn
Trng i Hc Khoa Hc
Hong Vn Qu
Chui lu tha hnh thc v hm sinh
Chuyn ngnh : Phng Php Ton S Cp
M s: 60.46.40
Lun Vn Thc S Ton Hc
Ngi hng dn khoa hc: PGS.TS. m Vn Nh
Thi Nguyn - 2011
S ha bi Trung tm Hc liu i hc Thi Nguyn http://www.lrc-tnu.edu.vn
Cng trnh c hon thnh ti
Trng i Hc Khoa Hc - i Hc Thi Nguyn
Phn bin 1: . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
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Phn bin 2: . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
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Lun vn s c bo v trc hi ng chm lun vn hp ti:
Trng i Hc Khoa Hc - i Hc Thi Nguyn
Ngy.... thng.... nm 2011
C th tm hiu ti
Th Vin i Hc Thi Nguyn
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Mc lc
1 Kin thc chun b 4
1.1 Khi nim vnh v ng cu . . . . . . . . . . . . . . . . . . . 4
1.1.1 Vnh . . . . . . . . . . . . . . . . . . . . . . . . . . . 4
1.1.2 c ca khng. Min nguyn . . . . . . . . . . . . . . 4
1.1.3 ng cu . . . . . . . . . . . . . . . . . . . . . . . . . 5
1.1.4 Trng . . . . . . . . . . . . . . . . . . . . . . . . . . 5
1.2 Vnh a thc v nghim . . . . . . . . . . . . . . . . . . . . . 5
2 Vnh cc chui ly tha hnh thc 11
2.1 Vnh cc chui ly tha hnh thc . . . . . . . . . . . . . . . 11
2.2 Dy hiu ca mt dy . . . . . . . . . . . . . . . . . . . . . . 17
2.3 Hm sinh thng v dy Fibonacci, dy Catalan . . . . . . . . 20
2.4 Hm sinh m v dy s Stirling . . . . . . . . . . . . . . . . . 24
2.5 Hm sinh ca dy cc a thc Bernoulli . . . . . . . . . . . . 27
2.6 Hm sinh Dirichlet v hm Zeta-Riemann . . . . . . . . . . . 34
2.7 Tch v hn . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37
2.8 ng nht thc Newton . . . . . . . . . . . . . . . . . . . . . 41
2.9 Dy truy hi vi hm sinh . . . . . . . . . . . . . . . . . . . . 48
1
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M u
Trong ton hc vic s dng cc kin thc ton cao cp gii quyt cc bi
ton ph thng l iu rt quan trng. N khng ch gip ngi lm ton
c nhiu phng php la chn li gii, m rng tm hiu bit ton hc m
cn pht huy c s thng minh v sc sng to, tm bao qut bi ton, m
rng bi ton di nhiu hng khc nhau.
S dng cc kin thc v chui s gii quyt cc bi ton v dy s
l mt vn nh vy. Nh chng ta bit cc vn lin quan n dy
s l mt phn quan trng ca i s v gii tch ton hc. Khi tip cn vn
ny cc em hc sinh gii, sinh vin v kh nhiu thy c gio ph thng
thng rt phi i mt vi rt nhiu bi ton kh lin quan n chuyn
ny.
Trong cc k thi hc sinh gii quc gia, thi Olimpic ton quc t, thi
Olimpic ton sinh vin gia cc trng i hc, cao ng, cc bi ton lin
quan n dy s cng hay c cp v thng loi rt kh, i hi ngi
hc, ngi lm ton phi c mt tm hiu bit rng v rt su sc cc kin
thc v dy s v chui s mi a ra cc phng php gii ton hay v hon
thin c bi ton.
phc v cho vic bi dng hc sinh gii v vic trao i kinh nghim
vi cc thy c gio bi dng hc sinh gii quan tm v tm hiu thm v
phn ny, c s hng dn ca thy m Vn Nh tc gi hc tp thm
v vit ti " Chui lu tha hnh thc v hm sinh".
ti gii quyt cc vn trng tm :
Chng I : Kin thc chun b .Tc gi nhc li cc kin thc c bn nht
v :
1.1 Khi nim vnh v ng cu
1.1.1 Vnh.
1.1.2 c ca khng. Min nguyn.
2
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31.1.3 ng cu.
1.1.4 Trng.
1.2 Vnh a thc v nghim.
Chng II : Vnh cc chui lu tha hnh thc. Tc gi gii thiu cc kin
thc.
2.1 Vnh cc chui lu tha hnh thc.
2.2 Dy hiu ca mt dy .
2.3 Hm sinh thng v dy Fibonacci, dy Catalan.
2.4 Hm sinh m v dy s Stirling.
2.5 Hm sinh ca dy cc a thc Bernoulli.
2.6 Hm sinh Dirichlet v hm Zeta-Riemann.
2.7 Tch v hn.
2.8 ng nht thc Newton.
2.9 Dy truy hi vi hm sinh.
Lun vn ny c hon thnh di s hng dn v ch bo tn tnh ca
PGS.TS m Vn Nh - i hc S Phm H Ni. Thy dnh nhiu thi
gian hng dn v gii p cc thc mc ca tc gi trong sut qu trnh lm
lun vn. Tc gi xin by t lng bit n su sc n Thy.
Tc gi xin gi ti cc thy (c) khoa Ton, phng o to Trng i
Hc Khoa Hc - i Hc Thi Nguyn, cng cc thy c tham gia ging
dy kha Cao hc 2009-2011 li cm n su sc v cng lao dy d trong
thi gian qua. ng thi xin gi li cm n tp th lp Cao hc Ton K3B
Trng i Hc Khoa Hc ng vin gip tc gi trong qu trnh hc
tp v lm lun vn ny.
Tc gi xin cm n ti S Ni V, S Gio dc v o to Bc Ninh, Ban
gim hiu v t Ton trng THPT Lng Ti 2 to iu kin gip
tc gi hon thnh kha hc ny.
Tc gi
Hong Vn Qu
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Chng 1
Kin thc chun b
1.1 Khi nim vnh v ng cu
1.1.1 Vnh
nh ngha . Ta gi l vnh mt tp hp X cng vi hai php ton hai ngi
cho trong X k hiu theo th t bng cc du + v . (ngi ta thng k
hiu nh vy) v gi l php cng v php nhn sao cho cc iu kin sau
tha mn:
1) X cng vi php cng l mt nhm aben.
2) X cng vi php nhn l mt na nhm.
3) Php nhn phn phi vi php cng: Vi cc phn t ty x, y, z X tac:
x(y + z) = xy + xz(y + z)x = yx+ zx
Phn t trung lp ca php cng th k hiu l 0 v gi l phn t khng.
Phn t i xng (i vi php cng ) ca mt phn t x th k hiu l -x
v gi l i ca x . Nu php nhn l giao hon th ta bo vnh X l giao
hon. Nu php nhn c phn t trung lp th phn t gi l phn t n
v ca x v thng k hiu l e hay 1 .
1.1.2 c ca khng. Min nguyn
nh ngha1 : Ta gi l c ca 0 mi phn t a 6= 0 sao cho c b 6= 0 thamn quan h ab=0.
nh ngha2 : Ta gi min nguyn mt vnh c nhiu hn mt phn t, giao
4
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5hon, c n v, khng c c ca 0.
1.1.3 ng cu
nh ngha. Mt ng cu (vnh) l mt nh x t mt vnh X n mt
vnh Y sao cho:
f (a+ b) = f (a) + f (b)f (ab) = f (a) f (b)
vi mi a, b X. Nu X = Y th ng cu f gi l mt t ng cu ca X .Ta cng nh ngha n cu, ton cu, ng cu tng t nh nh ngha
trong nhm.
1.1.4 Trng
nh ngha: Ta gi l trng mt min nguyn X trong mi phn t khc
khng u c mt nghch o trong v nhm nhn X. Vy mt vnh X giao
hon, c n v, c nhiu hn mt phn t l mt trng nu v ch nu
X {0} l mt nhm i vi php nhn ca X.
1.2 Vnh a thc v nghim
Kt qu chnh
Cho vnh giao hon R v mt bin x trn R. Vi cc n N, xt tp hp:
R[x] = {a0 + a1x+ a2x2 + + anxn | ai R} ={ n
i=0
aixi | ai R
}.
Mi phn t f(x) R[x] c gi l mt a thc ca bin x vi cc h sai thuc vnh R. H s an c gi l h s cao nht, cn h s a0 c gil h s t do ca f(x). Khi an 6= 0 th n c gi l bc ca f(x) v c
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6k hiu deg f(x). Ring a thc 0 c quy nh c bc l hoc 1.Nu f(x) =
ni=0
aixi, g(x) =
mi=0
bixi R[x] th
f(x) = g(x) khi v ch khi m = n, ai = bi vi mi 0 6 i 6 n
f(x) + g(x) =i=0
(ai + bi)xi, f(x)g(x) =
i=0
(i
j=0
aijbj)xi.
nh l 1.2.1. Ta c R[x] l mt vnh giao hon. Hn na, nu R l mtmin nguyn th R[x] cng l mt min nguyn.
nh l 1.2.2. Gi s k l mt trng. Vi cc a thc f(x), g(x) k[x] vg(x) 6= 0 c hai a thc duy nht q(x), r(x) sao cho f(x) = q(x)g(x)+r(x)vi deg r(x) < deg g(x).
V d 1.2.3. Cho hai s t nhin n v p vi n > p > 1. Tm iu kin cnv xn an chia ht cho xp ap vi a R, a 6= 0.Bi gii: Biu din n = qp+ r trong Z vi 0 6 r < p. Khi c biu din
xn an = (xp ap)(xnp + apxn2p + + a(q1)pxnqp) + aqp(xr ar).Vy, iu kin cn v xn an chia ht cho xp ap l n : p.nh l 1.2.4. Gi s k l mt trng. Khi vnh k[x] l mt vnh chnhv n l vnh nhn t ha.
Gi s R v a thc f(x) =ni=0
aixi R[x]. Biu thc f() =
ni=0
aii R c gi l gi tr ca f(x) ti . Nu f() = 0 th cgi l mt nghim ca f(x) trong R. Gi s s nguyn m > 1 v k.f() = 0 c gi l mt nghim bi cp m ca f(x) trong k nu f(x) chiaht cho (x )m v f(x) khng chia ht cho (x )m+1.nh l 1.2.5. a thc f(x) k[x] bc n > 1. Khi ta c cc kt qu sau:(i) Nu k l nghim ca f(x) th f(x) = (x)g(x) vi g(x) k[x].(ii) f(x) c khng qu n nghim phn bit trong k.
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7i khi tm mi lin h gia cc nghim hay mt tnh cht no ca
nghim a thc ta thng s dng kt qu sau y:
nh l 1.2.6. [Vit] Gi s x1, . . . , xn l n nghim ca a thc bc n sauy: f(x) = xn 1xn1 + 2xn2 + (1)nn. Khi c cc h thc
1 = x1 + x2 + + xn2 = x1x2 + x1x3 + + xn1xn...
n = x1x2 . . . xn.
nh l 1.2.7. Gi s f(x1, x2, . . . , xn) k[x1, x2, . . . , xn] l mt a thci xng khc 0. Khi tn ti mt v ch mt a thc s(x1, x2, . . . , xn) k[x1, x2, . . . , xn] sao cho f(x1, x2, . . . , xn) = s(1, 2, . . . , n).
Mt s v d
V d 1.2.8. Gi s f(x) = x4 5x3 + 9x2 10x+ 28. Tnh f(1 + 33).Bi gii: V 1 + 3
3 l nghim ca g(x) = x3 3x2 + 3x 4 = 0 v
f(x) = (x 2)g(x) + 20 nn f(1 + 33) = 20.V d 1.2.9. [VMO 1990] Gi s f(x) = a0x
n+a1xn1+ +an1x+an
R[x] vi a0 6= 0 v tha mn f(x)f(2x2) = f(2x3 + x) vi mi gi tr thcx. Chng minh rng f(x) khng th c nghim thc.
Bi gii: So snh h s ca x3n v x0 hai v, nn t f(x)f(2x2) = f(2x3+x) ta suy ra a20 = a0 v a
2n = an.V a0 6= 0 nn a0 = 1; cn an = 0 hoc an =
1. Nu an = 0 th f(x) = xrg(x) vi g(0) 6= 0. Vy xrg(x)2rx2rg(2x2) =
xr(2x2 + 1)rg(2x3 + x) hay g(x)2rx2rg(2x2) = (2x2 + 1)rg(2x3 + x). Vg(0) 6= 0 nn ta nhn c g(0) = 0 : mu thun. Vy an = 1. Gi sf(x) = 0 c nghim thc x0. Khi x0 6= 0 v an 6= 0. V f(2x30 + x0) =f(x0)f(2x
20) = 0 nn x1 = 2x
30 + x0 cng l nghim thc ca f(x). V hm
y = 2x3 + x l n iu tng nn dy (xr+1 = 2x3r + xr)r>0 v x0 6= 0 lmt dy v hn v mi s hng u l nghim ca f(x) hay f(x) c nhiuv hn nghim: mu thun theo nh l 1.2.5. Vy f(x) khng c nghimthc.
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8V d 1.2.10. [IMO 1991] Gi s s hu t a (0; 1) tha mn phngtrnh cos 3pia+ 2 cos 2pia = 0. Chng minh rng a =
2
3.
Bi gii: t x = cospia. Khi 4x3+4x23x2 = 0 hay (2x+1)(2x2+x2) = 0. Nu cos pia = x = 1
2th a =
2
3. Nu x 6= 1
2th 2x2+x2 =
0, v nh vy x l s v t. Do |x| 6 1 nn cospia = x = 1 +
17
4. Bng
quy np, c th ch ra cos 2npia =an + bn
17
4vi s nguyn l an, bn. V
an+1 + bn+1
17
4= cos 2n+1pia = 2 cos2 2npia 1 = 2[an + bn
17
4]2 1
nn an+1 =a2n + 17b
2n 8
2> an. Do dy (an) l mt dy tng nghim
ngt v nh vy tp cc gi tr ca cos 2npia vi n = 0, 1, 2, ... l tp vhn (*) v
17 l s v t. Nhng do a l s hu t nn tp cc gi trca cosmpia vi m = 0, 1, 2, ... phi l hu hn: mu thun vi (*). Do d
a =2
3.
V d 1.2.11. Gi thit a thc f(x) bc n c tt c cc nghim u thc.Khi tt c cc nghim ca af(x) + f (x) cng l nhng s thc.
Bi gii: Gi s f(x) c cc nghim thc x1, x2, . . . , xk vi bi tng ngr1, r2, . . . , rk v ta sp xp x1 < x2 < < xk. Hm s
g(x) =f (x)f(x)
=1
x x1 +1
x x2 + +1
x xkl hm lin tc trong cc khong (;x1), (x1;x2), . . . , (xk1;xk), (xk;).Da vo s bin thin ca cc hm
1
x xj , phng trnh g(x) = a c thmk nghim mi na khc x1, x2, . . . , xk khi a 6= 0. Vy f(x)[g(x) +a] = 0 ctt c (r1 1) + + (rk 1) + k = deg f(x) nghim thc. Vy tt c ccnghim ca af(x)+f (x) u thc. Khi a = 0 th g(x) = 0 c k1 nghimthc mi na. Vy f(x)[g(x)+0] = 0 c tt c (r11)+ +(rk1)+k1 =deg f (x). Tm li tt c cc nghim ca af(x)+f (x) l nhng s thc.
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9V d 1.2.12. Gi thit tt c cc nghim ca a thc f(x) v a thc g(x) =a0x
n + a1xn1 + + an u l nhng s thc. Khi tt c cc nghimca F (x) = a0f(x) + a1f
(x) + + anf (n)(x) cng u l nhng s thc.Bi gii: Biu din g(x) = a0(x+ 1)(x+ 2) . . . (x+ n) vi cc j thc.K hiu F0(x) = a0f(x), F1(x) = F0(x) + 1F
0(x) = a0[f(x) + 1f
(x)],F2(x) = F1(x)+2F
1(x) = a0[f(x)+(1+2)f
(x)]+12f (x)],v.v... cuicng Fn(x) = Fn1(x) + nF n1(x) = a0f(x) + a1f
(x) + + anf (n)(x).Theo V d 1.2.11 suy ra tt c cc nghim ca F0, F1, . . . , Fn u thc.
V d 1.2.13. Cho f = cosu + C1n cos(u + )x + + Cnn cos(u + n)xn.Gii phng trnh f(x) = 0.
Bi gii: t g = sinu+ C1n sin(u+)x+ + Cnn sin(u+n)xn. Khi f + ig = z + C1n ztx+ + Cnn ztnxn = z(1 + tx)nf ig = z + C1n ztx+ + Cnn ztnxn = z(1 + tx)n
z = cosu+ i sinu
t = cos + i sin.
Do 2f = z(1 + tx)n + z(1 + tx)n. Phng trnh f(x) = 0 tng
ng vi z(1 + tx)n + z(1 + tx)n = 0 hay(1 + tx
1 + tx
)n= z
z= z2.
Nh vy
(1 + tx1 + tx
)n= cos(2u + pi) + i sin(2u + pi) v c
1 + tx
1 + tx=
cos(2u+ pi + k2pi
n) + i sin(
2u+ pi + k2pi
n) vi k = 0, 1, . . . , n 1. T c x.
V d 1.2.14. Gi s a1, . . . , an, b R \ {0} v 1, . . . , n l nhng s thcphn bit. Khi f(x) = b+
nk=1
a2kx k ch c nghim thc.
Bi gii: Ta c f(c+ id) = b+nk=1
a2kc+ id k = b+
nk=1
a2k(c k id)(c k)2 + d2 .
Phn o Im(f(c + id)) = dnk=1
a2k(a k)2 + b2 6= 0 khi d 6= 0. Vy f(c +
id) 6= 0 khi d 6= 0. Khng hn ch c th coi 1 < 2 < < n1 < n.Hin nhin f(x) = 0 c n 1 nghim thc k tha mn
1 < 1 < 2 < 2 < < n1 < n1 < n
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10
v thm ng mt nghim tha mn hoc (, 1) hoc (n,+). T suy ra hm f(x) ch c cc nghim thc.V d 1.2.15. Cho a thc P (x) = 1 + x2 + x9 + xn1 + ...+ xns + x1992
vi n1, ..., ns l cc s t nhin cho trc tha mn 9 < n1 < ... < ns k(k + 1) nn|b0ak1 + b1ak2 + + bk2a1 + bk1| 6 2
(|ak1|+ |ak2|+ + |a1|+ 1)v nh th k(k + 1) < |kak| 6 2
(k + (k 1) + + 2 + 1) = k(k + 1) :mu thun. Nh vy |an| 6 n+ 1 vi mi n.V d 2.1.12. Chng minh rng 2
12 (22)
122 (2n) 12n < 4 vi mi s nguyndng n.
Bi gii: V 212 (22)
122 (2n) 12n = 2
nk=1
k
2knn ch cn chng minh
nk=1
k
2k 1 :an = 2an1 + (2n 1)2an2, bn = 2bn1 + (2n 1)2bn2.
Bi gii: Bng quy np theo n ta nhn c cc cng thc bn =nk=0
(2k+ 1)
v an = (2n + 1)an1 + (1)nn1k=0
(2k + 1). Vyanbn
=an1bn1
+(1)n2n+ 1vi
mi s nguyn n > 0. Nh vy anbn
= 1 13
+1
5+ + (1)
n
2n+ 1. Chuyn
qua gii hn ta c limn+
anbn
= arctan 1 =pi
4.
V d 2.1.14. Cho hai dy s nguyn (an) v (bn) tha mn:{a0 = 1, b0 = 1an = 2n 1, bn = n2, n > 1.Xy dng hai dy cc s nguyn (An) v (Bn) nh sau:{
A0 = 0, B0 = 1, A1 = 1, B1 = a1
An+1 = an+1An + bnAn1, Bn+1 = an+1Bn + bnBn1, n > 1.
(i) Tnh An, Bn theo n.
(ii) Chng minh
AnBn Q \ Z.
(iii) Tm limn
BnAn
.
(iv) Chng minh
nk=1
1
k=
1
1 12
3 22
5 32
.
.
. 2n 3 (n 1)
2
2n 1
.
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17
Bi gii: Bng quy np theo n ta nhn c cc cng thc Bn = n! v
An = (nk=1
1
k)n!.
(ii) Ta c
AnBn
=nk=1
1
k Q \ Z.
(iii) V
BnAn
=1nk=1
1
k
nn limn
BnAn
= limx1+
1
ln(1 + x)= 0.
(iii) Do
nk=1
1
k=AnBn
=1
1 12
3 22
5 32
.
.
. 2n 3 (n 1)
2
2n 1
.
2.2 Dy hiu ca mt dy
nh ngha 2.2.1. Cho dy s {an} = {an}nN. Dy {Dan}nN vi Dan =an+1 an, n > 0, c gi l dy hiu ca dy {an}.V dy hiu cng l mt dy s nn ta c th lp dy hiu ca n v k hiu
qua {D2an}. Hin nhinD2an = Dan+1 Dan = an+2 2an+1 + an.Tng qut Dk+1an = D
kan+1 Dkan v Dk(Dhan) = Dk+han.V d 2.2.2. Vi s nguyn dng r, dy (an), trong an =
(nr
), tha mnh thc Dan = an+1 an =
(nr1).
B 2.2.3. Vi hai dy s {an} v {bn} ta cD(ran+sbn) = rDan+sDbnv Dk(ran + sbn) = rD
kan + sDkbn vi mi s r, s v s t nhin k, n.
Chng minh: VD(ran+sbn) = rDan+sDbn = r(an+1an)+s(bn+1bn)nn c ngay kt qu D(ran + sbn) = rDan + sDbn. Tng qut D
k(ran +sbn) = rD
kan + sDkbn c chng minh d dng bng qui np theo k.
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18
B 2.2.4. Cho dy s {an}. Nu Dr+1an = 0 vi mi n > 0 th r + 1 shng a0, Da0, . . . , D
ra0 xc nh hon ton tt c cc Dkan vi mi k, n.c bit, nu dy s {bn} tha mn Djb0 = Dja0 v Dr+1bn = 0 vi min > 0, 0 6 j 6 r, th an = bn vi mi n.Chng minh: Hin nhin.
nh l 2.2.5. Cho dy s {an}. Nu c a thc p(x) bc r tha mn an =p(n) vi mi n > 0 thDr+1an = 0 vi mi n > 0. Ngc li, nuDr+1an =0 vi mi n > 0 th
an =
(n
0
)a0 +
(n
1
)Da0 + +
(n
s
)Dsa0 + +
(n
r
)Dra0.
Chng minh: Gi s a thc p(x) bc r tha mn an = p(n) vi mi n > 0.Ta ch ra Dr+1an = 0 bng phng php qui np theo r. Khi r = 0 can = p(n) = a. Vy D
1an = a a = 0. Gi s kt lun ng cho r 1 vp(x) = crx
r+ +c0. V an = p(n) vi mi n > 0 nn Dan = an+1an =p(n+ 1) p(n). t q(x) = p(x+ 1) p(x) tha mn Dan = q(n). V q(x)l a thc bc r 1 nn Dr(Dan) = 0 theo gi thit qui np. Vy ta nhnc Dr+1an = 0.Gi thit dy {an} tha mn Dr+1an = 0 vi mi n > 0. nh ngha dymi {bn} xc nh bi:
bn =
(n
0
)a0 +D
(n
1
)a0 + +Ds
(n
s
)a0 + +Dr
(n
r
)a0, n > 0.
Theo B 2.2.3 ta c ngay
Dbn = D
(n
0
)a0 +D
2
(n
1
)a0 + +Dr+1
(n
r
)a0
=
(n
0
)Da0 +
(n
1
)D2a0 + +Dr+1
(n
r 1)Dra0
v Dr+1a0 = 0. Lp li, vi D2, . . . , Dj v ta nhn c
Djbn =
(n
0
)Dja0 +
(n
1
)Dj+1a0 + +
(n
r j)Dra0
v n Drbn = Dra0(nrr)
= Dra0. Do Dr+1bn = D
r+1a0 = 0 vi min > 0 v Djb0 = Dja0 vi mi 0 6 j 6 r. Vy theo B 2.2.4 c an = bnvi mi n > 0.
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19
V d 2.2.6. Cho dy (an) vi a0 = 3 v an+1 = an + 4n+ 1 vi mi n > 0.Chng minh rng vi mi s t nhin dngm u c n an3n1 = 2m2.Bi gii: V Dan = an+1 an = 4n + 1, D2an = Dan+1 Dan = 4(n +1) + 1 4n 1 = 4 v D3an = 0 nn theo nh l 2.2.5 ta c ngayan = 3
(n0
)+Da0
(n1
)+D2a0
(n2
)+ 0 = 3 +n+ 2n(n 1) = 2n2n+ 3 hay
an 3n 1 = 2(n 1)2 vi mi n > 0. Vy vi mi m c am+1 3(m+1) 1 = 2m2.V d 2.2.7. Cho dy (an) vi a0 = 3, a1 = 2 v an+2 = 3an+1 2an 6n2 + 14n 5 vi mi n > 0. Xc nh an theo n.Bi gii: V an+2 2an+1 = an+1 2an 6n2 + 14n 5 nn khi tbn = an+1 2an ta s c bn+1 = bn 6n2 + 14n 5 v dy (bn) vib0 = 4, bn+1 = bn 6n2 + 14n 5 vi mi n > 0. V Dbn = bn+1 bn =6n2 + 14n 5, D2bn = Dbn+1 Dbn = 6(n + 1)2 + 14(n + 1) 5 +6n2 14n + 5 = 12n + 8 v D3bn = 12, D4bn = 0 nn theo nh l2.2.5 ta c ngay bn = 4
(n0
)+Da0
(n1
)+D2a0
(n2
)+D3a0
(n3
)hay
bn = 4 5n+ 8(n
2
) 12
(n
3
)= 2n3 + 10n2 13n 4.
Vy a0 = 3, an+1 = 2an 2n3 + 10n2 13n 4 vi mi n > 0. Vi dykiu ny, ta xt an = u.2
n + an4 + bn3 + cn2 + dn+ e. T y d dng suyra an.
V d 2.2.8. Cho dy (an) vi a0 = 5, a1 = 1 v an+1 = an + 6an1 6n2 + 26n 25 vi mi n > 1. Chng minh rng vi mi t nhin n u can 2.3n + n2(mod 2n).Bi gii: Vi dy kiu ny, trc tin xt a thc c trng x2 x 6 =(x 3)(x+ 2). Tip theo an = u3n + v(2)n + +an3 + bn2 + cn+ d v xt
5 = a0 = u+ v + d
1 = a1 = 3u 2v + a+ b+ c+ d34 = a2 = 9u+ 4v + 8a+ 4b+ 2c+ d
39 = a3 = 27u 8v + 27a+ 9b+ 3c+ d226 = a4 = 81u+ 16v + 64a+ 16b+ 4c+ d
415 = a5 = 243u 32v + 125a+ 25b+ 5c+ d.
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20
Gii h c u = 2, v = 3, b = 1 v a = c = d = 0. Nh vy c cng thcan = 2.3
n + 3.(2)n + n2 v an 2.3n + n2(mod 2n) vi mi n > 0.Ch 2.2.9. Nu c a thc g(x) tha mn an = g(n) th an+1 an =g(n+ 1) g(n) l mt a thc ca n vi bc nh i 1.
2.3 Hm sinh thng v dy Fibonacci, dy Catalan
Mt trong nhng ngun gc dn n khi nim hm sinh chnh l nh l
khai trin nh thc Newton (1 + x)n =nk=0
(nk
)xk v khai trin thnh chui
ly tha ca hm phn thc
1
1 x = 1 + x + x2 + + xn + . y lmt k thut gii tch vi nhiu ng dng trong t hp v nghin cu dy s.
Hm sinh c phn ra lm hai loi: Hm sinh thng v Hm sinh m. Ta
bt u vi khi nim hm sinh thng di y:
nh ngha 2.3.1. Cho dy s {an}, hoc tng qut hn l dy hm {an =an(x)}. Chui lu tha hnh thc f(x) =
n=0
anxnc gi l hm sinh
thng ca dy {an}.
Kt qu chnh
nh l sau y c chng minh trong Gii tch.
nh l 2.3.2. Cho s thc dng . Nu chui lu tha hnh thci=0
aixi
hi t ti h(x) th cho mi x (, ) hm h(x) c h(x) =i=1
iaixi1v
h0
f(t)dt =i=0
aixi+1
i+ 1.
nh l 2.3.3. Dy s {an} c gi l dy xc nh kiu tuyn tnh nu dyc dng : a0 = 0, . . . , as1 = s1 v an+s = 1an+s1 + 2an+s2 + +san, n > 0. Khi hm sinh thng f(x) ca dy {an} l mt hm hut
b0 + b1x+ + brxr1 + 1x+ + sxs , r < s, khi v ch khi {an} l dy xc nh kiutuyn tnh.
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21
Chng minh: Gi thit a0 = 0, . . . , as1 = s1 v an+s = 1an+s1 +2an+s2 + + san, n 0. t f(x) = a0 + a1x+ a2x2 + a3x3 + . Sdng an+s = 1an+s1 + 2an+s2 + + san, n 0, ta c (1x+ 2x2 + +sxs)f(x) = p(x)+f(x), trong p(x) l a thc vi bc deg p(x) < s.t q(x) = 1x + 2x
2 + + sxs. Ta c (q(x) 1)f(x) = p(x). Vyf(x) =
p(x)
q(x) 1 hay hm sinh thng f(x) ca dy l mt hm hu t.
Ngc li, cho f(x) =b0 + b1x+ + brxr1 + 1x+ + sxs =
p(x)
q(x), r < s.V f(x)q(x) =
p(x) nn khi so snh h s ca cc xn ta c hai ha0 = b0a01 a1 = b1...
a0s1 + a1s2 + as1 = bs1 va0s + a1s1 + as = 0a1s + a2s1 + as+1 = 0a2s + a3s1 + as+2 = 0...
Gii h u c nghim a0 = 0, . . . , as1 = s1. T h sau ta suy raan+s = 1an+s1 + 2an+s2 + + san, n 0. Do {an} l dy xcnh kiu tuyn tnh.
nh l 2.3.4. Nu u(x) = xs + 1xs1 + + s = 0 c cc nghimr1, . . . , rt vi cc bi tng ng 1, . . . , t. Khi tn ti cc a thc p1(n),p2(n), . . . , pt(n) tha mn an = p1(n)r
n1 + p2(n)r
n2 + + pt(n)rnt v
0 6 deg pi(n) 6 i 1 vi i = 1, 2, . . . , t.
Mt vi v d
V d 2.3.5. Dy s a0 = 5, a1 = 13, a2 = 35 v an+3 = 6an+211an+1+6anvi n > 0. Chng minh rng an 2n+1(mod 3n+1) v an 3n+1(mod 2n+1).Bi gii: a thc p(x) = x3 6x2 + 11x 6 = (x 1)(x 2)(x 3). Vyan = a2
n + b3n + c. T iu kin ban u suy ra an = 2n+1 + 3n+1.
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22
V d 2.3.6. Xt dy s a0 = 11, a1 = 6, a2 = 18, a3 = 104, a4 = 346 v
an+5 = 6an+4 13an+3 + 14an+2 12an+1 + 8an, n > 0.Tm s nguyn dng ln nht m a2011 chia ht cho 2
m.
Bi gii: a thc p(x) = x5 6x4 + 13x3 14x2 + 12x 8 c vitthnh p(x) = (x 2)3(x i)(x + i). Vy an = p1(n)2n + ain + b(i)n.T iu kin ban u suy ra an = (n
2 + n + 1)2n + 5in + 5(i)n. Vya2011 = (2011
2 + 2011 + 1)22011. S m cn tm bng 2011.
S dng khi nim hm sinh v chui lu tha hnh thc xt mt s
dy s c bit.
V d 2.3.7. Xt dy s Fibonacci a0 = 0, a1 = 1, an+1 = an + an1, n > 1.Cng thc ng cho hm sinh thng ca dy l f(x) =
x
1 x x2 . Tm
an theo n v ch ran=0
an4n+1
=1
11.
Bi gii: t f(x) = a0+a1x+a2x2+a3x
3+ .Khi (1xx2)f(x) =x hay ta c f(x) =
x
1 x x2 . Vi a =1 +
5
2v b =
152
, biu din
f(x) qua chui ly tha f(x) =15
( 11 ax
1
1 bx). Vy c
f(x) =15
((1 + ax+ a2x2 + ) (1 + bx+ b2x2 + )).So snh h s ca xn hai v c an =
an bn5v cng thc ng f(x) =
x
1 x x2 . Vi x =1
4ta nhn c
n=0
an4n+1
=1
11.
T kt qu an =an bn
5ta suy ra cc ng nht thc sau y:
V d 2.3.8. Xt dy s Fibonacci a0 = 0, a1 = 1, an+1 = an + an1, n > 1.Khi ta c
(i) [Phng trnh Biner] an =an bn
5, Ln = a
n + bn [S Lucas]. Do
limn
an+1an
= a.
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23
(ii) an =1
2n1[n12 ]k=0
(n
2k+1
)5k =
[n12 ]k=0
(nk1
k
).
(iii) a2n+1 = a2n + a
2n+1, a3n = a
3n + a
3n+1 a3n1.(iv) an = aan + an1, bn = ban + an1.
(v) a3n + (1)nan : a2n.(vi) Nu p > 5 l s nguyn t th ap : p v a2 + 1 = 2 : 2, a3 + 1 =
3 : 3, a5 = 5 : 5.
Bi gii: (i),(ii),(iii),(iv) v (v) u c suy ra t cng thc an =an bn
5.
Vi p > 5 l s nguyn t th t (ii) c ap =1
2p1[p12 ]k=0
(p
2k+1
)5k : p. Vi
V d 2.3.9. Xt dy Catalan a0 = 1, an+1 = a0an+a1an1 + +an1a1 +ana0, n > 0. Cng thc ng cho hm sinh ca dy l f(x) =
11 4x2x
.
Tm cng thc tnh an theo n.
Bi gii: t f(x) = a0 + a1x+ a2x2 + a3x
3 + . Khi f(x)f(x) = a20 + (a0a1 + a1a0)x+ (a0a2 + a1a1 + a2a0)x
2 + = a1 + a2x+ a3x
2 + a4x3 + = f(x) 1
x.
Vy x[f(x)]2 f(x) + 1 = 0. Gii phng trnh ny v do f(0) = 0 nn
f(x) =11 4x
2x=
1
2x 1
2x(1 4x) 12 .
Cng thc ng cho hm sinh f(x) l f(x) =11 4x
2x. Biu din
hm ny qua chui lu tha f(x) =1
2x
[2x +
22
2!x2 +
1.3
3!23x3 + +
1.3.5...(2n 3)n!
2nxx + ]. So snh h s c an = 1n+ 1
(2n
n
)vi mi
s nguyn n > 1.
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24
V d 2.3.10. Dy s (an) c xc nh nh sau: a1 = 1 v an =1
n!+
a1(n 1)! +
a2(n 2)! + +
an11!vi n > 1. Xc nh cng thc ng ca
f(x) v ch ra an =F (n)(0)
n!, trong F (x) =
1ex 2 vi mi n.
Bi gii: t a0 = 1. Xt hm sinh f(x) =n=0
anxn. Khi f(x)(ex1) =(
a0+a1x+a2x2+ +anxn+
)( 11!x+
1
2!x2+ + 1
n!xn+ ) = f(x)1.Vy f(x) =
1ex 2 . Da vo Cng thc khai trin Taylor-Maclaurin ta c
an =F (n)(0)
n!, trong F (x) =
1ex 2 vi mi n.
2.4 Hm sinh m v dy s Stirling
Kt qu chnh
Nh mt s tip tc, khi nim hm sinh m s c nh ngha di y.
nh ngha 2.4.1. Cho dy s {an}. Chui lu tha hnh thc biu din trongdng f(x) =
n=0
ann!xn c gi l hm sinh m ca dy {an}.
V d 2.4.2. Vi dy s ((mn
)) hm sinh thng f(x) =
n=0
(mn
)xn c vit
thnh f(x) =n=0
m!
(m n)!n!xn =
n=0
Anmn!xn l hm sinh m ca dy (Anm).
(m l s c nh cho trc )
nh ngha 2.4.3. Cho mt tp hu hn S khc rng. Mt phn hoch caS thnh k phn, vi 1 6 k 6 n, l mt h cc tp con S1, . . . , Sk tha mnba iu kin sau y :
ki=1
Si = S
Si 6= vi mi iSi Sj = vi mi i, j, i 6= j.
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25
nh ngha 2.4.4. Cho tp hu hn S khc rng. S cc phn hoch ca tpS thnh k phn c k hiu l S(n, k) v c gi l s Stirling (loi 2).
nh l 2.4.5. Vi hai s nguyn dng n, k tha mn 1 6 k 6 n ta lun c
k!S(n, k) =ki=0
(1)ki(k
i
)in.
Chng minh: Xt tp S = {a1, a2, . . . , an} v R = {1, 2, . . . , k}. TheoBi tp 1.2.22, s cc ton nh t S ln R bng k!S(n, k). Mt khc, biu
din nh x f : S R qua(
a1 a2 . . . anf(a1) f(a2) . . . f(an)
)ta c ngay
{f(a1), f(a2), . . . , f(an)} = {1, 2, . . . , k}. Vit dy s f(a1) . . . f(an) nhmt chnh hp lp chp n ca k s. Nh vy, tng ng mi nh x f
vi ng mt chnh hp lp chp n ca k s. S chnh hp lp ny ngbng kn. K hiu A l tp tt c cc nh x t S vo R v cho mi i k
hiu Ai l tp con ca A gm tt c cc nh x t S vo R \ {i}. Ta c|A| = kn, |Ai| = (k 1)n v
sj=1
Aij = (k s)n. Tp tt c cc ton nh t
S ln R ng bng A \ki=1
Ai. Theo nh l 2.4.5, ta nhn c
k!S(n, k) = |A| |ki=1
Ai| = kn (k
1
)(k 1)n
+
(k
2
)(k 2)n + (1)k
(k
k
)(k k)n.
Vy k!S(n, k) =ki=0
(1)i(ki)(k i)n = ki=0
(1)ki(ki)in.H qu 2.4.6. Vi hai s nguyn dng n, k tha mn 1 6 k 6 n ta lun c
kn =ki=0
(k
i
)i!S(n, i).
Chng minh: t ak = k!S(n, k) v bi = in. Theo nh l 2.4.5, ta c
ak =ki=0
(1)ki(ki)bi. t ck = (1)kak. Khi ck = ki=0
(1)i(ki)bi. K
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26
hiu ng thc ny nh sau ck = (1 b)k v hiu l sau khi khai trin thay bibi bi.Vi k hiu hiu hnh thc, ng vi mi gi tr ca x, c th vit ngnht thc nh sau: (c+ x)k = (b+ 1 + x)k. Cho x = 1 ta c (1)kbk =(c 1)k hay bk = (1 c)k =
ki=0
(1)i(ki)ci. Vy kn = ki=0
(ki
)i!S(n, i).
nh l 2.4.7. Hm sinh m ca dy s Stirling l f(x) =n=0
S(n, k)xn
n!c
cng thc ng bng
(ex 1)kk!
.
Chng minh: Theo nh l 2.4.5, f(x) =n=0
1
n!
( 1k!
ki=0
(1)ki(ki)in)xn.Do k!f(x) =
ki=0
(1)ki(ki) n=0
(ix)n
n!=
ki=0
(1)ki(ki)eix = (ex 1)khay f(x) =
(ex 1)kk!
.
Mt vi v d
V d 2.4.8. K hiu D(n) l s cc hon v ca n phn t khng c phn tc nh, chng hn: S cc php hon v pi Sn sao cho pi(k) 6= k vi mi k.D dng ch raD(n) = n!
( 10! 1
1!+
1
2! + (1)
n
n!
). Hm sinh m ca dy
(D(n)) l f(x) =n=0
D(n)
n!xn. Xc nh cng thc ng ca f(x) v chng
minh D(n) = nD(n 1) + (1)n, D(n) = (n 1)(D(n 1) +D(n 2)).Bi gii: Do bi f(x) =
n=0
D(n)
n!xn =
n=0
( 10! 1
1!+
1
2! + (1)
n
n!
)xn
nn f(x) =( k=0
(1)kk!
xk)(
k=0
xk)
= ex.1
1 x =1
ex(1 x) .
T f(x)(1x) = ex suy ra D(n)n!D(n 1)
(n 1)! =(1)nn!v nh vyD(n) =
nD(n 1) + (1)n. T{D(n) = nD(n 1) + (1)nD(n 1) = (n 1)D(n 2) + (1)n1 suyra D(n) + D(n 1) = nD(n 1) + (n 1)D(n 2) hay D(n) = (n 1)(D(n 1) +D(n 2)).
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27
V d 2.4.9. K hiu D(n) l s cc hon v ca n phn t khng c phn
t c nh. Chng minh rng
mn=0
(mn
)D(n) = m! vi mi s nguyn m > n.
Bi gii: Do bi f(x) =n=0
D(n)
n!xn =
1
ex(1 x) nn f(x)ex =
1
1 x.
Vy
( n=0
D(n)
n!xn)(
n=0
1
n!xn)
=n=0
xn. So snh h s ca xm hai v ta
c
mn=0
(mn
)D(n) = m!.
V d 2.4.10. Dy s Bell (Bn) c xc nh nh sau: Bn =nk=1
S(n, k)
vi n > 1 v B0 = 1. Xc nh cng thc ng ca f(x) =n=0
Bnn!xn.
Bi gii: D dng ch ra Bn+1 =nk=0
(nk
)Bk. t f(x) =
n=0
Bnn!xn. Vy
f(x)ex =n=0
1
n!Bn+1x
n =n=0
(n + 1)Bn+1
(n+ 1)!xn = f (x). T y suy ra f (x)
f(x)dx =
ex dx hay ln(f(x)) = ex + C. V f(0) = B0 = 1 nn
C = 1 v nh th ln(f(x)) = ex 1. Do f(x) = eex1.
2.5 Hm sinh ca dy cc a thc Bernoulli
nh ngha 2.5.1. Cc a thc Bernoulli {Bn(x)} l nhng a thc tha mnba iu kin sau y:
(i) B0(x) = 1
(ii) Bn(x) = nBn1(x) vi n > 1
(iii)
10
Bn(x)dx = 0 vi n > 1.
nh ngha 2.5.2. S Bernoulli th n l Bn(0) v c k hiu qua Bn vin = 0, 1, 2, . . . .
B 2.5.3. o hm cp s ca Bn(x) l B(s)n (x) = n(n 1) . . . (n s +
1)Bns(x) v Bn(x) l a thc bc n.
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28
Chng minh: B(s)n (x) = n(n 1) . . . (n s + 1)Bns(x) c suy ra tiu kin (ii). V B
(n)n (x) = n!B0(x) = n! nn degBn(x) = n.
nh l 2.5.4. Ta c ngay cc h thc sau y: Bn(x) =ns=0
(ns
)Bsx
nsv
B0 = 1,ns=0
(n+1s
)Bs = 0, Bn Q.
Chng minh: Ta lun cBn(x) =ns=0
B(s)n (0)xs
s!. Theo B 2.5.3,Bn(x) =
ns=0
n!Bns(0)xs
s!(n s)! =ns=0
(ns
)Bsx
ns. Cho n 1, t iu kin10
Bn(x)dx = 0
ta suy ra h thc
10
( ns=0
(ns
)Bsx
ns)dx = 0 hay 1n+ 1
ns=0
(n+1s
)Bs = 0. V
B0 = 1,ns=0
(n+1s
)Bs = 0 nn Bn Q vi mi n (bng quy np).
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29
T kt qu ny, d dng nhn c mt s ng nht thc v a thc sau:
B0 = 1
2B1 +B0 = 0
3B2 + 3B1 +B0 = 0
4B3 + 6B2 + 4B1 +B0 = 0
5B4 + 10B3 + 10B2 + 5B1 +B0 = 0
6B5 + 15B4 + 20B3 + 15B2 + 6B1 +B0 = 0
. . .ns=0
(n+ 1
s
)Bs = 0.
B0(x) = 1
B1(x) = x 12
B2(x) = x2 x+ 1
6
B3(x) = x3 3
2x2 +
1
2x
B4(x) = x4 2x3 + x2 1
30
B5(x) = x5 5
2x4 +
5
3x3 1
6x.
H qu 2.5.5. Ta c Bn(x) = Bn(x+ 1)Bn(x) = nxn1.Chng minh: Ta lun c h thc di y:
Bn(x+ 1) =ns=0
B(s)n (1)xs
s!=
ns=0
(n
s
)Bs(1)x
ns.
Hin nhin B0(1) = B0(0) = 1. Cho n > 2, t iu kin10
Bm(x)dx = 0
khi m > 1 suy ra 0 =10
nBn1(x)dx =10
Bn(x)dx = Bn(1) Bn(0). Dovy, khi xt b(x) = Bn(x+ 1)Bn(x), t nh l 2.5.4 ta c
b(x) =ns=0
(n
s
)[Bs(1)Bs(0)
]xns =
(n
1
)[B1(1)B1(0)
]xn1.
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30
V B1(1)B1(0) = 1 nn Bn(x) = Bn(x+ 1)Bn(x) = nxn1.
V d 2.5.6. Vi s nguyn dng n, s cn1k=0
ks =1
s+ 1
sk=0
(s+1k
)Bkn
s+1k.
Bi gii: Theo H qu 2.5.5 c
n1k=0
ks =1
s+ 1
n1k=0
(Bs+1(k+1)Bs+1(k)
).
Nh vy
n1k=0
ks =1
s+ 1
(Bs+1(n) Bs+1(0)
)=
1
s+ 1
sk=0
(s+1k
)Bkn
s+1k
theo nh l 2.5.4.
V d 2.5.7. Tnh tng T =nk=0
k4 theo n.
Bi gii: Theo v d trn c
nk=0
k4 =1
5
(B5(n + 1) B5
)v nh vy nhn
c T =n5
5+n4
2+n3
3 n
30.
H qu 2.5.8. Ta c Bn =(1)nn!
det
(20
) (21
)0 ... 0(
30
) (31
) (32
)... 0
... ... ... ... ...(n0
) (n1
)... ...
(nn1)(
n+10
) (n+11
)... ...
(n+1n1)
.
Chng minh: T B0 = 1,ni=0
(n+1i
)Bi = 0, ta c h phng trnh tuyn tnh
vi cc n (B0, B1, . . . , Bn) :
(10
)B0 = 1(
20
)B0 +
(21
)B1 = 0(
30
)B0 +
(31
)B1 +
(32
)B2 = 0
...(n+10
)B0 +
(n+11
)B1 + +
(n+1n
)Bn = 0.
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31
H ny c nghim (B0, B1, . . . , Bn). Tnh Bn qua nh thc ta c
Bn det
(10
)0 0 ... 0 0(
20
) (21
)0 ... 0 0(
30
) (31
) (32
)... 0 0
... ... ... ... ... ...(n0
) (n1
)... ...
(nn1)
0(n+10
) (n+11
)... ...
(n+1n1) (
n+1n
)
= det
(10
)0 0 ... 0 1(
20
) (21
)0 ... 0 0(
30
) (31
) (32
)... 0 0
... ... ... ... ... ...(n0
) (n1
)... ...
(nn1)
0(n+10
) (n+11
)... ...
(n+1n1)
0
hay Bn =(1)nn!
det
(20
) (21
)0 ... 0(
30
) (31
) (32
)... 0
... ... ... ... ...(n0
) (n1
)... ...
(nn1)(
n+10
) (n+11
)... ...
(n+1n1)
.
Xt dy s Bernoulli {Bn} vi B0 = 1,ni=0
(n+1i
)Bi = 0, n > 1. Hm sinh
m ca dy cc s Bernoulli {Bn} l B(x) =n=0
Bnxn
n!.
nh l 2.5.9. Cng thc ng ca hm sinh m B(x) ca dy cc s
Bernoulli l
x
ex 1 .
Chng minh: V
ns=0
(n+1s
)Bs = 0 nnBn+1 =
n+1s=0
(n+1s
)Bs vi n = 1, 2, . . . .
Th n + 1 qua n c Bn =ns=0
(ns
)Bs vi n = 2, 3, . . . . V B0 =
(00
)B0
v B1 =(10
)B0 +
(11
)B1 1 nn B(x) = x +
n=0
( ns=0
(ns
)Bs)xnn!
=
x+n=0
ns=0
(Bsxss!
)( xns(n s)!
)= x+B(x)ex. Vy B(x) = x
ex 1 .
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32
H qu 2.5.10. Cho n 6= 1 v n l c Bn = 0.
Chng minh: V 1 +n=2
Bnxn
n!= B(x) B1x = x
ex 1 +x
2nn 1 +
n=2
Bnxn
n!=x(ex + 1)
2(ex 1) . Vx(ex + 1)
2(ex 1) l hm chn nn Bn = 0 khi n 6= 1 vn l s l.
nh l 2.5.11. [Euler] Ta c
n=0
Bn(x)
n!zn =
zexz
ez 1 .
Chng minh: V Bn(x) =ns=0
(ns
)Bsx
nstheo nh l 2.5.4 nn ta nhn
c biu din
n=0
Bn(x)
n!zn =
n=0
ns=0
(ns
)Bsx
ns
n!zn
=n=0
ns=0
Bsxns
s!(n s)!zn =
s=0
Bszs
s!
r=0
xrzr
r!.
Theo nh l 2.5.9 ta c
n=0
Bn(x)
n!zn =
z
ez 1exz =
zexz
ez 1 .
H qu 2.5.12. Ta lun c
(i)
n1s=0
(ns
)Bs(x) = nx
n1vi mi n > 2.
(ii) Bn(x+ y) =ns=0
(ns
)Bs(x)y
ns.
Chng minh: (i) Theo nh l 2.5.11, ta c biu din
n=0
Bn(x+ 1)
n!zn =
zexz
ez 1ez =
( r=0
Br(x)
r!zr)(
s=0
zs
s!
). So snh h s ca zn hai v, ta c
Bn(x+1) =ns=0
(ns
)Bs(x) hayBn(x+1) = Bn(x)+
n1s=0
(ns
)Bs(x).VBn(x+
1)Bn(x) = nxn1 theo H qu 2.5.5 nnn1s=0
(ns
)Bs(x) = nx
n1, n > 2.
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33
(ii) T
ze(x+y)z
ez 1 =zexz
ez 1eyzta suy ra s bng nhau gia hai chui
n=0
Bn(x+ y)
n!zn =
( r=0
Br(x)
r!zr)(
s=0
yszs
s!
)v nh vy
n=0
Bn(x+ y)
n!zn =
r,s=0
Bs(x)yr
s!r!zs+r. So snh h t ca zn c
Bn(x+ y) =ns=0
(ns
)Bs(x)y
ns.
V d 2.5.13.
n=0
Bn(mx)
n!zn =
1
m
m1k=0
n=0
mnzn
n!Bn(x +
k
m) vi mi s
nguyn m > 1.
Bi gii: V
1
ez 1 =1 + ez + e2z + + e(m1)z
emz 1 nn theo nh l 2.5.11,
ta c biu din
n=0
Bn(mx)
n!zn =
zemxz
ez 1 =1
m
mzemxz
ez 1 hayn=0
Bn(mx)
n!zn =
1
m
emxzmz(1 + ez + + e(m1)z)emz 1
=1
m
m1k=0
mzemz(x+
k
m)
emx 1
=1
m
m1k=0
n=0
mnzn
n!Bn(x+
k
m).
Vy
n=0
Bn(mx)
n!zn =
1
m
m1k=0
n=0
mnzn
n!Bn(x+
k
m) vi mi m > 1.
V d 2.5.14. Vi hai s nguyn m,n > 1 hy tm tt c cc a thc f(x)
vi h t cao nht bng 1 tha mn mnf(mx) =1
m
m1k=0
f(x+k
m).
Bi gii: T
n=0
Bn(mx)
n!zn =
1
m
m1k=0
n=0
mnzn
n!Bn(x+
k
m) theo V d 2.5.13
ta suy ra mnBn(mx) =1
m
m1k=0
Bn(x+k
m). Vy Bn(x) tha mn u bi.
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34
Tnh duy nht: Gi s c hai a thc phn bit p(x) = xn + axn1 + vq(x) = xn+bxn1+ tha mn u bi. Khi hiu (x) = p(x)q(x) =cxd + vi c 6= 0 v d < n cng tha mn u bi. V mn(mx) =1
m
m1k=0
(x +k
m) nn ta c c = mdnc hay c = 0 : v l, do m > 1 v
d < n.
V d 2.5.15. Nu c
x
ex 1 = 1 +n=1
unn!xn th cc un Q vi mi n.
Bi gii: Gi s
x
ex 1 = 1 +n=1
unn!xn. Khi , qua quy ng v gin c
x, nhn c ng nht thc 1 =(1 +
n=1
1
(n+ 1)!xn)(
1 +n=1
unn!xn). So
snh h s ca xn hai v, cunn!1!
+un1
(n 1)!2! + +u1
1!n!+
1
(n+ 1)!= 0
vi mi n > 1. Biu din dng t hp qua vic nhn hai v vi (n+ 1)! :(n+ 1
1
)un +
(n+ 1
2
)un1 + +
(n+ 1
n
)u1 + 1 = 0, n > 1,
hay vit theo kiu hnh thc (u+ 1)n+1 un+1 = 0 vi ch : Sau khi khaitrin xong phi vit uk thnh uk. T ng nht thc ny ta c h phng
trnh tuyn tnh v hn di y:
2u1 + 1 = 0
3u2 + 3u1 + 1 = 0
4u3 + 6u2 + 4u1 + 1 = 0
5u4 + 10u3 + 10u2 + 5u1 + 1 = 0
6u5 + 15u4 + 20u3 + 15u2 + 6u1 + 1 = 0
. . . .
Nh vy un = Bn Q vi mi n.
2.6 Hm sinh Dirichlet v hm Zeta-Riemann
nh ngha 2.6.1. Vi hm s hc f : N C, v s > 1, chui lu tha hnh
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35
thc F (s) =n=1
f(n)
nsc gi l chui Dirichlet tng ng f. Cho dy s
{an}. Chui g(s) =n=1
annscn c gi l hm sinh Dirichlet.
Ta cng nhn hai nh l sau v tnh duy nht, tnh nhn ca chui Dirichlet:
nh l 2.6.2. Cho hai chui Dirichlet F (s) =n=1
f(n)
nsv G(s) =
n=1
g(n)
ns
tng ng cc hm s hc f, g : N C. Nu c s R sao cho F (s) =G(s) vi mi s > th f(n) = g(n) cho mi n.
nh l 2.6.3. Cho ba chui Dirichlet Fi(s) =n=1
fi(n)
nstng ng ba hm
s hc fi : N C, i = 1, 2, 3. Nu f3(n) =
u,v,uv=nf1(u)f2(v) vi mi
n N th F3(s) = F1(s)F2(s).nh l 2.6.4. Cho s s > 1 v hai dy s {an}, {bn}. Xt hai hm sinhDirichlet g(s) =
n=1
annsv h(s) =
n=1
bnns. Gi thit hai chui
n=1
|an|nsv
n=1
|bn|nshi t trong khong (s0,). Khi g(s)h(s) cng l mt hm sinhDirichlet sinh ra bi dy {cn} vi cn =
uv=n
aubv trong khong (s0,).
Chng minh: Theo php nhn cc chui ta c
g(s)h(s) = (u=1
auus
)(v=1
bvvs
) =u=1
v=1
aubv(uv)s
=n=1
uv=n
aubv
ns.
Vy g(s)h(s) l mt hm sinh Dirichlet sinh ra bi dy {cn} vi cn =uv=n
aubv.
nh l 2.6.5. Nu f l mt hm nhn thn=1
f(n)
ns=p
(i=0
f(pi)
pis), trong
tch ly theo tt c cc s nguyn t p. Chuin=1
f(n)
nsc gi l hm
sinh Dirichlet ca hm s hc f.
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36
Chng minh: Khai trin tch v phi ca h thc trn ta cp
(i=0
f(pi)
pis) =
j=1
(1 +
i=1
f(pij)
pisj
)= f(pj1j1 )f(pj2j2 ) . . . f(pjrjr )
psj1j1
psj2j2
. . . psjrjr
=
f(p11 p22 . . . p
rr ), v f l hm s nhn,
=n=1
f(n)
ns, n = p
j1j1pj2j2. . . p
jrjr,
trong pt l s nguyn t th t. Nh vy h thc trn l ng.
Ch 2.6.6. Ta s dng tch hu hn Tk =ki=1
( j=1
f(pji )
pjsi
). Sau cho
k .nh ngha 2.6.7. Cho s s > 1 v dy s {an}. Chui lu tha hnh thc(s) =
n=1
1
nsc gi l hm zeta Riemann.
nh l 2.6.8. Vi s > 1 ta c
(i) 2(s) =n=1
d(n)
ns.
(ii) (s) =j=1
1
1 psj.
(iii)
1
(s)=n=1
(n)
ns.
Chng minh: (i) Theo nh l 2.6.4, 2(s) =n=1
u|n
1.1
ns=n=1
d(n)
ns.
(ii) Theo nh l 2.6.5 vi f(n) = 1 c (s) =j=1
(i=0
1
pisj) =
j=1
1
1 psj.
(iii) K hiu G(s) l hm sinh Dirichlet ca hm s hc . Khi
G(s) =n=1
(n)
ns=j=1
(i=0
(pij)
pisj) =
j=1
(1 psj ) =1
(s).
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37
Nh vy
1
(s)=n=1
(n)
ns.
nh l 2.6.9. Ta c (2) =pi2
6.
Chng minh: Cng thc khai trin Fourier ca hm f(x) trn [pi, pi] :
f(x) =a02
+n=1
(an cosnx+ bn sinnx),
an =
1
pi
pipif(x) cosnx dx
bn =1
pi
pipif(x) sinnx dx
n > 1.
Vi hm s chn f(x) = x2 c x2 =a02
+n=1
an cosnx, an =2
pi
pi0
x2 cosnx dx
cho mi n 0. Vy x2 = pi2
3+ 4
n=1
(1)n cosnxn2
. Cho x = pi ta c
(2) =pi2
6.
2.7 Tch v hn
nh ngha 2.7.1. Vi dy s (an) ta t tch a1a2 . . . an . . . =n=1
an. Tch
ny c gi l mt tch v hn. K hiuAk =k
n=1an.Nu tn ti lim
n+ An =
A th A c gi l gi tr ca tch v hnn=1
an v vit A =n=1
an.
Vn t ra: Khi no tn ti gi tr ca tch v hn
n=1
an i vi mt dy
s cho trc (an).Ta c kt qu sau y ca G.M. Fichtenholz v tnh hi t ca mt tch qua
hi t ca mt tng tng ng:
nh l 2.7.2. [ Fichtenholz] Cho dy cc s dng (an) v dy cc s m(bn). Khi
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38
(i) Tch v hn
i=1
(1 + an) hi t khi v ch khi tng v hni=1
an hi t.
(ii) Tch v hn
i=1
(1 + bn) hi t khi v ch khi tng v hni=1
bn hi t.
Chng minh: (i) Hin nhin, nu tch v hn
i=1
(1 + an) hoc tng v hn
i=1
an hi t th limn+ an = 0. Khi limn+
ln(1 + an)
an= 1. Do bi tch v
hn P =i=1
(1 + an) hi t khi v ch khi tng v hn lnP =i=1
ln(1 + an)
hi t. Vy vic hi t hay phn k ca
i=1
ln(1 + an) vi=1
an l tng
ng. T y suy ra tch v hn
i=1
(1 + an) hi t khi v ch khi tng v
hn
i=1
an hi t.
V d 2.7.3. Gi s dy (an) c xc nh nh sau: a1 = 1 v an+1 =(n+ 1)3 1(n+ 1)3 + 1
an vi mi n > 1. Tm limn+ an v tch v hn
n=2
(n3 1n3 + 1
).
Bi gii: Hin nhin ak =k
n=2
(n3 1n3 + 1
). Do bi n + 1 = (n + 2) 1 v
n2 + n + 1 = (n + 1)2 (n + 1) + 1 nn ak = 2(k2 + k + 1)
3k(k + 1). Do
limn+ an =
2
3v tch v hn
n=2
(n3 1n3 + 1
)=
2
3.
V d 2.7.4. t In =pi/20
sinn x dx . Khi In+1 =n
n+ 1In1 vi mi n > 1.T suy ra
(i)
pi/20
sin2n x dx =(2n 1)(2n 3) . . . 3.1
2n(2n 2) . . . 4.2pi
2pi/20
sin2n+1 x dx =2n(2n 2) . . . 4.2
(2n+ 1)(2n 1) . . . 3.1 .
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39
(ii)
[ (2n)!!(2n 1)!!
]2 12n+ 1
2
3.
Bi gii: Bin i c T = 43 = 4xyz. V 1 = xy+yz+zx > 3 3
(xyz)2
nn T 6 4
3
9. Ta cn c P = 21 + 73 > 21. V (x + y + z)2 >
3(x+ yz + zx) = 3 nn P > 2
3.
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47
V d 2.8.10. Chng minh rng nu a, b, c l ba s thc phn bit th c
a3(b2 c2) + b3(c2 a2) + c3(a2 b2)a2(b c) + b2(c a) + c2(a b) < a
2 + b2 + c2.
Bi gii: D thy c t v mu u chia ht cho (a b)(b c)(c a). VyV T = ab+ bc+ ca < a2 + b2 + c2 v a, b, c phn bit.
V d 2.8.11. Chng minh rng nu a, b, c l ba s thc phn bit th
a2(a+ b)(a+ c)
(a b)(a c) +b2(b+ c)(b+ a)
(b c)(b a) +c2(c+ a)(c+ b)
(c a)(c b) > 3(ab+ bc+ ca).
Bi gii: Sau khi quy ng, c t v mu u chia ht cho (ab)(bc)(ca).Vy V T = (a+ b+ c)2 > 3(ab+ bc+ ca) v a, b, c phn bit.
V d 2.8.12. Chng minh rng nu a, b, c, d R tha mn ab+ ac+ ad+bc+ bd+ cd = 0 th
a3 + b3 + c3 + d3 3(bcd+ cda+ dab+ abc) = (a+ b+ c+ d)3.
Bi gii: t
1 = a+ b+ c+ d
2 = ab+ ac+ ad+ bc+ bd+ cd
3 = abc+ abd+ acd+ bcd
4 = abcd
Nt = at + bt + ct + dt, t = 1, 2, . . . , N0 = 4.
Theo nh
l 2.8.1 c N3 N21 + N12 33 = 0. Vy N3 33 = N21 N12 =31 312. V 2 = 0 nn a3 + b3 + c3 + d3 3
(bcd+ cda+ dab+ abc
)=
(a+ b+ c+ d)3.
V d 2.8.13. a thc T = 2(x7 + y7 + z7) 7xyz(x4 + y4 + z4) c nhnt l x+ y + z.
Bi gii: t A = x + y + z, B = xy + yz + zx, C = xyz. t an =xn + yn + zn. Khi x, y, z l ba nghim ca t3 At2 + Bt C = 0v
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48
an+3 = Aan+2 Ban+1 + Can vi s nguyn n > 0 v a0 = 3. Ch a0 = 3
a1 = A
a2 = A2 2B
a3 = Aa2 Ba1 + Ca0 = A3 3AB + 3C = Ak3 + 3Ca4 = Aa3 Ba2 + Ca1 = Ak4 + 2B2a5 = Ak5 5BCa6 = Ak6 B3 + 3C2a7 = Ak7 + 7B
2C.
Vy T = 2a7 7Ca4 = A(2k7 7k4C) c nhn t A = x+ y + z.
2.9 Dy truy hi vi hm sinh
V d 2.9.1. Dy s (an) xc nh bi:
{a0 = 2, a1 = 4, a2 = 31
an+3 = 4an+2 + 3an+1 18anvi mi n > 0. Chng minh rng a2010 1(mod 2011).Bi gii: t f(x) = a0 + a1x+ a2x
2 + a3x3 + . Khi c quan h
f(x)(4x+ 3x2 18x3) = f(x) 9x2 + 4x 2
hay f(x) =9x2 4x+ 2
18x3 3x2 4x+ 1 =1
1 + 2x+
1
(1 3x)2 . T y suy ra
f(x) =n=0
((2)n + (n + 1)3n)xn v c an = (2)n + (n + 1)3n vi mi
n > 0. Nh vy a2010 1(mod 2011).V d 2.9.2. Dy (an) xc nh qua a1 = 1 v an = 1.2.an1 + 2.3.an2 + + (n 1).n.a1 vi mi s nguyn n > 2. Chng minh ng nht thcan+3 4an+2 an+1 = 2
nk=1
ak khi n > 2.
Bi gii: Xt f(x) = a1x+ a2x2 + + anxn + . Khi ta c h thc
f(x)(1.2.x+ 2.3.x2 + + n.(n+ 1).xn + ) = f(x) x.
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49
T
1
1 x = 1 + x + x2 + x3 + ta suy ra 1x2 + 2x3 + = x
2
(x 1)2 .
Ly o hm hai v c 1.2.x + 2.3.x2 + = 2x(x 1)3 . Vy f(x) = x +
2x2x3 3x2 + 5x 1 . T f(x)(x
33x2+5x1) = x43x3+3x2x ta nhn rav so snh h s ca xn, n > 4 hai v, nhn c an+3 = 5an+23an+1+anvi mi s nguyn n > 2. Biu din cc mi quan h trong bng h thc sau
y:
a1 = 1a2 + 5a1 = 3a3 + 5a2 3a1 = 3a4 + 5a3 3a2 + a1 = 1a5 + 5a4 3a3 + a2 = 0a6 + 5a5 3a4 + a3 = 0 = an + 5an1 3an2 + an3 = 0an+1 + 5an 3an1 + an2 = 0an+2 + 5an+1 3an + an1 = 0an+3 + 5an+2 3an+1 + an = 0.
Cng v vi v c an+3 +
4an+2 + an+1 + 2nk=1
ak = 0. Nh vy an+3 4an+2 an+1 = 2nk=1
ak.
B 2.9.3. Gi s dy (an) c hm sinh thng f(x) =n=0
anxntha mn
f(s) c xc nh vi s = 0, 1, . . . , k v l cn nguyn thy bc k ca
n v. Khi c
n=0
ank =f(1) + f(2) + + f(k1)
k.
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50
Bi gii: Vi hm sinh thng f(x) = a0 + a1x+ + anxn + c
f(1) =n=0
(ank + ank+1 + + a(n+1)k1
)f() =
n=0
(ank + ank+1 + + a(n+1)k1k1
)f(2) =
n=0
(ank + ank+1
2 + + a(n+1)k12(k1))
= f(k1) =
n=0
(ank + ank+1
k1 + + a(n+1)k1(k1)(k1)).
Bi v 1+s+2s+ +s(k1) = 0 vi s = 1, 2, . . . , k1, nn khi cng kng nht thc trn ta nhn c f(1)+f(2)+ +f(k1) = k
n=0
ank.
V d 2.9.4. Vi s nguyn dng n, hy tnh tng an =nk=0
(1)k( n3k). Xttnh tun hon ca dy (an) v ch ra an : 3
[n/2]1.
Bi gii: Xt hm sinh thng f(x) ca dy (bk = (1)k(nk
)). Khi ta c
h thc f(x) =nk=0
(1)k(nk)xk = (1 x)n. Nh vy, vi = 12 + i
3
2c
an =nk=0
(1)k(n
3k
)=
nk=0
(1)3k(n
3k
)=
1
3
(f(1) + f() + f(2)
)
hay an =(1 )n + (1 2)n
3=
(32 i
3
2
)n+(3
2+ i
3
2
)n3
. Tm li
tng an =2(
3)n
3cos
npi
6=
2.33k1(1)k khi n = 6k33k(1)k khi n = 6k + 133k(1)k khi n = 6k + 20 khi n = 6k + 3
33k+1(1)k+1 khi n = 6k + 433k+2(1)k+1 khi n = 6k + 5.
D dng
suy ra (an) khng tun hon v an : 3[n/2]1.
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51
V d 2.9.5. Vi mi s nguyn dng n, hy tnh tng an =nk=0
(1)k(5n5k).Bi gii: Xt hm sinh thng f(x) ca dy (bk = (1)k
(5nk
)). Ta c h thc
f(x) =nk=0
(1)k(5nk )xk = (1x)5n. Nh vy, vi = cos 2pi5 + i sin 2pi5 can =
5nk=0
(1)k(
5n
5k
)=
5nk=0
(1)5k(
5n
5k
)=
1
5
4k=0
f(k)
hay
(1 )5n + (1 2)5n + (1 3)5n + (1 4)5n5
.Vy nhn c tng
a2m =210m+1(1)m
5
[sin10m
pi
5+ sin10m
2pi
5
]v a2m+1 = 0.
V d 2.9.6. Dy (an) tha mn a1 = 12v an+1 =
(n+11
)an +
(n+12
)an1 +
+ (n+1n )a1 + 1 vi mi s nguyn n > 1. Chng minh rng a2011 nguynv chia ht cho 2011.
Bi gii: D dng kim tra
( k=0
xk
(k + 1)!
)(1 +
n=1
anxn
n!
)= 1. Nh vy
1 +n=1
anxn
n!=
x
ex 1 =n=0
Bnxn
n!theo nh l 2.5.9. Do an = Bn
vi mi n. c bit a2011 = B2011 = 0 theo H qu 2.5.10 v suy ra a2011nguyn v chia ht cho 2011.
V d 2.9.7. Dy (an) xc nh qua a1 = 1 v an = 1.2.an1 2.3.an2 + + (1)n(n 1).n.a1 vi mi s nguyn n > 2. Chng minh rng, khin > 2 lun c an+3 + 2an+2 + 5an+1 + 6
nk=1
ak = 8.
Bi gii: Xt f(x) = a1x+ a2x2 + + anxn + . Khi ta c h thc
f(x)(1.2.x 2.3.x2 + + (1)n+1n.(n+ 1).xn + ) = f(x) x.
T
1
1 + x= 1x+x2x3+ ta suy ra1+2x3x2+ = 1
(x+ 1)2.
Ly o hm hai v c 1.2.x2.3.x2+ = 2x(x+ 1)3v nh vy nhn c
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52
f(x) =x4 + 3x3 + 3x2 + x
x3 + 3x2 + x+ 1. T f(x)(x3+3x2+x+1) = x4+3x3+3x2+x
ta nhn ra v so snh h s ca xn, n > 4 hai v, c a1 = 1, a2 = 2,v a3 = 2, a4 + a3 + 3a2 = 0, an+3 + an+2 + 3an+1 + an = 0 vi mi snguyn n > 2. Biu din cc mi quan h trong bng h thc sau y:
a1 = 1
a2 + a1 = 3
a3 + a2 + 3a1 = 3
a4 + a3 + 3a2 + a1 = 1
a5 + a4 + 3a3 + a2 = 0
a6 + a5 + 3a4 + a3 = 0
= an + an1 + 3an2 + an3 = 0an+1 + an + 3an1 + an2 = 0an+2 + an+1 + 3an + an1 = 0an+3 + an+2 + 3an+1 + an = 0.
Cng v vi v c an+3 + 2an+2 + 5an+1 + 6nk=1
ak = 8.
V d 2.9.8. Dy (an) xc nh bi an =1.3.5 . . . (2n+ 1)
2011n.n!vi mi s nguyn
n > 0. Tnhn=0
an.
Bi gii: Hin nhin an+1 =2n+ 3
2011(n+ 1)an vi mi n > 1. Khi n +
th
an+1an 2
2011. Nh vy chui ly tha f(x) = a0 + a1x+ a2x
2 + +anx
n + lun lun hi t. T h thc 2011(n + 1)an+1 = 2nan + 3ansuy ra 2011(n + 1)an+1x
n = 2x(nanxn1) + 3anxn. Cho n = 0, 1, 2, . . . v
ly tng tt c c 2011( k=0
ak+1xk+1)
= 2x( k=0
akxk)
+ 3( k=0
akxk).
Khi ta c h thc 2011(f(x) 1) = 2xff (x) + 3f(x) hay (2011
2x)f (x) = 3f(x). T y suy ra
f (x)f(x)
=3
2011 2x. Ly tch phn hai
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53
v c ln f(x) = 32
ln(2011 2x) + a hay f(x) = (2011 2x)3/2.ea.V f(0) = 1 nn 1 = 20113/2.ea hay ea = 20113/2. Tm li ta c f(x) =
20113/2.(2011 2x)3/2. Vi x = 1 c
n=0an =
(20112009
)32 .
V d 2.9.9. Dy (an) xc nh qua a1 = 1 v an = 1an1+2an2+ +(n1)a1 vi mi s nguyn n > 2.Chng minh a3 = 3a2 v an+23an+1+an = 0vi mi s nguyn n > 2 v xc nh an theo n. T suy ra a2k+1 chia htcho 3 khi k > 1.Bi gii: Xt f(x) = a1x+ a2x
2 + + anxn + . Khi ta cf(x)(1x+ 2x2 + + nxn + ) = f(x) x.T
1
1 x = 1 + x + x2 + x3 + ta suy ra 1x + 2x2 + = x
(x 1)2 .
Vy f(x) = x +x2
x2 3x+ 1 v c f(x)(x2 3x + 1) = x3 2x2 + x.Nhn ra v so snh h s ca xn, n > 1 hai v, nhn c a1 = 1, a2 = 1,a3 = 3a2, v an+2 3an+1 + an = 0 vi mi s nguyn n > 2. T d ccng thc xc nh an.
V d 2.9.10. Dy (an) xc nh qua a1 = 1 v an+1 =2n+ 3
4(n+ 1)an vi mi
s nguyn n > 0. Tnh tng T =n=0
an.
Bi gii: Do 4(n+1)an+1 = 2nan+3an nn 4(n+1)an+1xn = 2xnanx
n1+3anx
nvi mi s nguyn n > 0. Cng tt c cc h thc ny ta nhn c
4( n=0
an+1xn+1)
= 2x( n=0
anxn)
+ 3n=0
anxn.
t f(x) =n=0
anxn. Khi 4(f 1) = 2xf + 3f hay (4 2x)f = 3f v
suy ra
f
f=
3
2
1
2 x. Nh vy (ln f) = 3
2(ln(2x)). dng c f(x) =
(2 x)3
2ea. V f(0) = a0 = 1 nn ea = 2
3
2 . Tm li f(x) =(
1 x2
)32 .
Vi x = 1 c T = f(1) = 2
2.
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54
V d 2.9.11. Dy (an) xc nh qua a1 = 1 v an+1 =2n+ 5
3(n+ 2)an vi mi
s nguyn n > 0. Chng minh rngk
n=0an 0. Cng tt c cc h thcny ta nhn c
3( n=0
an+1xn+2)
= 2x( n=0
anxn+1)
+ 3n=0
anxn+1.
t f(x) =n=0
anxn+1. Khi 3(f x) = 2xf + 3f hay (3 2x)f =
3f + 3. t g(x) = f(x) + 1. Khi g
g=
3
3 2x. Nh vy (ln g) =
32
(ln(3 2x)). dng c g(x) = (3 2x)3
2ea. V g(0) = a0 = 1 nn
ea = 3
3
2 . Tm li g(x) = 1 +n=0
anxn+1 =
(1 2x
3
)32 . Vi x = 1 c
T = g(1) 1 = 33 1. Do vyk
n=0an 0. Xc nh an theo n v chng minh
n=0
an+1n!
= e.
Bi gii: Xt hm sinh m ca dy (an) l f(x) =n=0
ann!xn. Khi ta c
f =n=0
an+1n!
xn =n=0
nan 2n2 + 5n 3n!
xn
=n=0
nann!
xn n=0
2n2 5n+ 3n!
xn = xf (2x2 5x+ 3)ex.
Nh vy f = (2x3)ex v suy ran=0
an+1n!
xn =n=0
2n 3n!
xn. Vy an+1 =
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55
2n 3 hay an = 2n 5 vi n > 1. Tn=0
an+1n!
xn = (2x 3)ex ta nhn
c
n=0
an+1n!
= e khi x = 1.
V d 2.9.13. Chng minh rng
nk=0
(2kk
)k + 1
(2(nk)nk
)n k + 1 =
(2(n+1)n+1
)n+ 2
.
Bi gii: Xt f(x) =k=0
(2nn
)n+ 1
xn vi 0 < |x| < 14. Bi v 1+xf(x)2 f(x)
theo V d 2.3.9 nn
nk=0
(2kk
)k + 1
(2(nk)nk
)n k + 1 =
(2(n+1)n+1
)n+ 2qua vic so snh h s
ca xn+1 hai v.
V d 2.9.14. Xt dy s hu t a1 = 1, an = (an1
1!+an2
2!+ + a1
(n 1)!)vi mi s nguyn n > 2. Tm tt c cc s nguyn dng n n!an+1 = 1.
Bi gii: Ta c an = 2an +(an1
1!+an2
2!+ + a1
(n 1)!)vi mi s
nguyn n > 2. t f(x) = a1x+ a2x2 + a3x3 + . Khi
f(x)(
2 +x
1!+x2
2!+x3
3!+
)= 2a1x+ (2a2 +
a11!
)x2 + (2a3 +a21!
+a12!
)x3 + = 2a1x+ a2x
2 + a3x3 + a4x
4 + = f(x) + x.Vy f(x)(1 + ex) = f(x) + x hay f(x) = xex. T y suy ra ng nht
a1x+ a2x2 + a3x
3 + = x(
1 x1!
+x2
2! x
3
3!+
).
Do an+1 =(1)nn!vi mi s nguyn n > 1. n!an+1 = 1 cn v nl s chn.
V d 2.9.15. Xt n > 3 s nguyn dng a1 6 a2 6 a3 6 6 an1 6 anvi tnh cht: Khng c ba s no l di ba cnh mt tam gic khng suy
bin. Xc nh gi tr nh nht ca
ana1m n c th t c.
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56
Bi gii: Ta bit ba s 0 < a 6 b 6 c l di ba cnh tam gic khi vch khi c < a + b. Vy khng c ba s hng bt k ca dy ca dy l di ba cnh mt tam gic khng suy bin th ap > aq + ar vi mi p, q, r vp > q, r. V dy l dy khng gim nn ta ch cn xt ai + ai+1 6 ai+2 vii = 1, 2, . . . , 2009. Vi nh ngha
(ab
)>(cd
)khi v ch khi a > c v
b > d ta c bt ng thc(an+1an
)>(
1 11 0
)(anan1
)>(
1 11 0
)n1(a2a1
), n > 2.
Gi s
(1 11 0
)n1=
(a bc d
). Khi ta nhn c an > ca2 + da1 v
suy ra an > (c+ d)a1. Do ana1> c+ d. Do an
a1nh nht l bng c+ d
khi dy l n s hng u ca dy Fibonacci v gi tr nh nht ca t s
bng an =an bn
5vi a =
1 +
5
2v b =
152
.
V d 2.9.16. Xt dy a1 = 1, an = 12an1 + 22an2 + + (n 1)2a1 vimi s nguyn n > 2. Khi
{a2 = 4a1 3, a3 = 4a2 2a1 + 3an+3 = 4an+2 2an+1 + an, n > 2.Bi gii: t f(x) = a1x+ a2x
2 + a3x3 + . Khi tch hai chui
f(x)(
12x+ 22x2 + 32x3 + )
= 12a1x2 + (12a2 + 2
2a1)x3 + (12a3 + 2
2a2 + 32a1)x
4 + = a2x
2 + a3x3 + a4x
4 + a5x5 + = f(x) x.
T
1
1 x = 1 + x+ x2 + x3 + x4 + x5 + ta suy ra chui ly tha sau:
1
(1 x)2 = 1 + 2x+ 3x2 + 4x3 + 5x4 + 6x5 + . Do nhn c
x
(1 x)2 = 1x+ 2x2 + 3x3 + 4x4 + 5x5 + 6x6 + v c biu din
x(1 + x)
(1 x)3 = 12x+ 22x2 + 32x3 + 42x4 + 52x5 + 62x6 + . Nh th
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57
f(x)(x(1 + x)
(1 x)3)
= f(x)x hay f(x)[x32x2+4x1
]= x43x3+3x2x.T ng nht
[a1x+a2x
2+a3x3+
][x32x2+4x1
]= x43x3+3x2xsuy ra a3 = 4a2 2a1 + 3, an+3 = 4an+2 2an+1 + an vi mi n > 2.V d 2.9.17. Xt dy a1 = 1, an = 1an1+2an2 +(1)n1(n1)a1vi mi s nguyn n > 2. Khi ta c
(i) a2 = 1, a3 + 3a2 = 0, an+2 + 3an+1 + an = 0, n > 2.(ii) Tm d ca php chia an cho 3.
Bi gii: (i) t f(x) = a1x+ a2x2 + a3x
3 + . Tch hai chui ly tha
F (x) = f(x)( 1x+ 2x2 3x3 +
)= 1a1x2 + (1a2 + 2a1)x3 + (1a3 + 2a2 3a1)x4 + = a2x
2 + a3x3 + a4x
4 + a5x5 + = f(x) x.
T
1
1 + x= 1 x+ x2 x3 + x4 x5 + suy ra chui ly tha sau y:
1(1 + x)2
= 1 + 2x 3x2 + 4x3 5x4 + 6x5 . Do nhn cx
(1 + x)2= 1x+ 2x2 3x3 + 4x4 5x5 + 6x6 . Th vo F (x) c
f(x)( x
(1 + x)2
)= f(x) x hay f(x)
[x2 + 3x + 1
]= x3 + 2x2 + x. T
ng nht
[a1x+ a2x
2 + a3x3 +
][x2 + 3x+ 1
]= x3 + 2x2 + x s suy
ra ngay a1 = 1, a2 + 3a1 = 2, a3 + 3a2 = 0, an+2 + 3an+1 + an = 0, n > 2.(ii) Ta c a3 0(mod 3).V an+2+3an+1+an = 0 nn an+2+an 0(mod 3)
khi n > 2. Do , khi s nguyn k > 1 c
a2k+1 0(mod 3)a4k+2 a2 2(mod 3)a4k 1(mod 3).
V d 2.9.18. Xt dy (an), trong an =1.3.5.7 . . . (2n+ 1)
4n.n!vi mi s
nguyn n > 0. t f(x) =k=0
akxk. Tm cng thc ng v tnh f(1).
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58
Bi gii: V an+1 =2n+ 3
4(n+ 1)an vi mi n nn
an+1an 1
2< 1. Vy f(x)
hi t. D dng ch ra
(f(x)a0
)= 2xf (x)+3f(x). Vy
f (x)f(x)
=3
2
1
2 x
hay f(x) =(2 x)32ec. Bi v f(0) = a0 = 1 nn ec = 23/2. Tm li
f(x) =(2 x
2
)32v f(1) = 23/2.
Mt s v d tham kho
V d 2.9.19. [VMO-1997] Cho dy s nguyn (an) ,n N c xc nhnh sau: a0 = 1, a1 = 45 v an+2 = 45an+1 7an vi mi n = 0, 1, 2, . . . .Khi hy
(i) Tnh s c dng ca a2n+1 anan+2 theo n.(ii) Chng minh rng 1997a2n + 7
n+1.4 l s chnh phng vi mi n.
V d 2.9.20. [VMO-1998-A] Cho dy s nguyn (an) ,n N c xcnh nh sau: a0 = 20, a1 = 100 v an+2 = 4an+1 + 5an + 20 vi min = 0, 1, 2, . . . . Khi hy
(i) Tm s nguyn dng h nh nht c tnh cht an+h an chia ht cho1998.
(ii) Tm s hng tng qut ca dy.
V d 2.9.21. [VMO-2011] Cho dy s nguyn (an) ,n N c xc nhnh sau: a0 = 1, a1 = 1 v an = 6an1 + 5an2 vi mi n = 2, 3, 4 . . . .Khi hy
(i) Chng minh rng a2012 2010 chia ht cho 2011.(ii) Tm s hng tng qut ca dy .
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Kt lun ca lun vn
Trong lun vn tc gi trnh by c cc ni dung chnh sau y:
(1) Vnh, c ca khng, min nguyn, ng cu, trng, vnh a thc v
nghim.
(2) Vnh cc chui ly tha hnh thc, khi nim hm sinh m v hm sinh
thng cng mt vi dy s lin quan.
(3) Nghin cu mt s dy s Fibonacci, dy Catalan, dy Stirling v dy
cc a thc Bernoulli, Hm sinh Dirichlet v hm Zeta-Riemann, tch
v hn.
(4) Tnh c mt s cng thc ng ca mt s dy v chng minh ng
nht thc Newton.
Do thi gian v dung lng nn lun vn mi ch dng li mc tm hiu v
gii thiu v "Vnh cc chui lu tha hnh thc" v mt s "Hm sinh" c
bn v dy s. Trong thi gian ti, nu iu kin cho php, tc gi s nghin
cu, tm hiu k hn c th a ra mt s kt qu c tnh ng dng thc
tin hn phc v qu trnh hc tp v ging dy.
Trong qu trnh thc hin lun vn chc chn khng trnh khi thiu st.
Tc gi rt mong nhn c nhng kin ng gp ca thy c v bn b
hon thin lun vn tt hn.
Tc gi xin chn thnh cm n.
59
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Ti liu tham kho
[1] H.X. Snh, i s i cng, NXB Gio dc, 2001.
[2] N.V. Hi, N.K. Minh v H.Q. Vinh, Cc bi thi Olympic Ton THPT
Vit Nam (1990-2006), NXB Gio dc, 2007.
[3] R. Merris, Combinatorics, PWS publishing company 20 Park Plaza,
Boston, MA 02116-4324.
[4] K.H. Wehrahn, Combinatorics-An Introduction, Carslaw Publications
1992.
60
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