mot_phuong_phap_giai_pt_ham_tren_n.pdf

MT K THUT GII PHNG TRNH HM TRN N

Nguyn Trng Tun PTNK - i hc quc gia thnh ph H Ch Minh

M U: Cc bi ton v phng trnh hm i hi phi bit vn dng tng hp cc kin thc v hm s. i vi lp cc phng trnh hm trn N, chng ta cn phi bit vn dng cc kin thc ca l thuyt s. Ta bit cc kt qu rt c bn v quen thuc sau y : 1.Mi s t nhin u c th phn tch c mt cch duy nht ( khng k th t) thnh tch ca cc tha s nguyn t: 1 21 2 ... kkn p p paa a= y pi l cc s nguyn t phn bit v i Na 2. Mi s t nhin n u c biu din duy nht di dng 2 .mn k= , y m l s t nhin v k l s l. 3. Vi hai s t nhin p khc 0 cho trc, mi s n u c th vit dc duy nht di dang n = kp + r vi 0 r p -1 . Cc kt qu trn rt hu ch khi gii mt s phng trnh hm trn N. Ni dung ca phng php ny c th tm tt nh sau : - Tm cng thc xc nh f(n) vi n thuc v mt tp hp T N no . - Biu din n theo cc phn t ca tp hp T. Ni tm lai, xc nh hm f(n) ta ch cn xc nh hm f(n) trn T. Tp hp T c th v hn hoc hu hn, thm ch T c th gm 1 hoc 2 phn t. V d 1. Tm tt c cc hm f: N* N* tha mn iu kin f(f(m)+ f(n)) = m +n vi mi n N* ? Gii. D`dng chng minh c f l hm c tnh cht n nh. Gi s m,n,p,q l 4 s t nhin tha mn m +n = p + q. Khi f(f(m)+f(n))= f(f(p) + f(q)) Do f l n nh nn ta c f(m) + f(n) = f(p) + f(q) T ng thc trn suy ra: f(m) - f(m-1) = f(m-1) - f(m-2) = = f(3) - f(2) = f(2) - f(1) Nh th dy (f(m)) l mt cp s cng vi s hng u l f(1) v cng sai l d = f(2) - f(1). Nh th xc nh c f(m) ta ch cn xc nh f(2) v f(1).

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Ta c f(m) = f(1) +(f(2)-f(1))(m-1) vi mi m N*. Thay biu thc ca f(m) vo h thc u bi ta c f(2f(1) + (m-1)d + (n-1)d) = m + n Hay f(1) + [2f(1)+(m-1)d + (n-1)d -1 ] d = m + n Suy ra f(1) = d =1 . T hm s duy nht cn tm l f(n) = n vi mi n N*. V d 2. C bao nhiu hm s f: N N tha mn iu kin f(f(n)) = n +4 vi mi n N ? Gii Gi s f(m) = n. Khi f(n) = f(f(m)) = m+4 . Suy ra f(m+4) = f(f(n)) = n + 4 hay f(m+4) = f(m) + 4. Bng qui np theo k, d dng chng minh c f(m+4k) = f(m) + 4k. Vit m di dng m = 4k + r , y r {0,1,2,3}. Khi f(m) = f(4k+r) = f(r)+4k. Nh vy, xc nh hm s ta ch cn xc nh cc gi tr f(0),f(1),f(2),f(3). Vi mi r {0,1,2,3}, ta vit f(r) di dng duy nht f(r) = 4p + s vi s {0,1,2,3} Khi f(4p+s) =f(f(r)) , suy ra 4p + f(s) = r + 4. Nh th r = f(s) + 4(p-1), t dn ti p = 0 hoc p = 1. Xt p = 0. Ta c f(r) = s v f(s) = r + 4. Xt p = 1. Ta c f(r) = s + 4 v f(s) = r Ch rng ta khng th c f(r) = r. T ta c th xy dng tt c cc hm f tha mn yu cu bi ton nh sau: - Chia tp hp {0,1,2,3} thnh 2 tp ri nhau {a,b} v {c,d} - Nu f(a) = c ,f(b) = d th f(c)=a+4,f(d)=b+4 Nu f(a) = d ,f(b) = c th f(d)=a+4,f(c)=b+4 Chng hn, ta c th ch ra mt hm nh sau: f(1) = 3, f(4) = 2 , f(3) = 5 , f(2) = 8 v f(4k +r ) = 4k +f(r) , r {0,1,2,3} D thy c tt c 12 hm s tha mn yu cu bi ton. Nhn xt: Bng phng php trn ta c th gii bi ton tng qut: Cho b l s t nhin , tm tt c cc hm f: N N tha mn iu kin f(f(n)) = n + b vi mi n N. V d 3. Cho hm f: N N tha mn iu kin


f(f(n)) = 2n vi mi n N. Xc nh gi tr nh nht c th c ca f(2010). Gii. Gi s f(m) = n , khi f(n) = f(f(m)) =2m v nh th f(2m) = f(f(n)) = 2n hay f(2m) =2f(m) vi mi m N. Bng qui np, ta chng minh c (2 ) 2 ( ) ,k kf m f m m N= " Mi s t nhin n lun vit c duy nht di dng 2 .mn a= , trong m l s t nhin , cn a l mt s l. Ta c ( ) (2 ) 2 ( )n nf n f a f a= = Nh vy xc nh hm s, ta ch cn xc nh f(a) vi a l. By gi, gi s ( ) 2 pf a b= vi p l s t nhin v b l. Khi (2 ) ( ( )pf b f f a= hay 12 ( ) 2 2 ( )p pf b a f b a-= = Do a l nn p = 1 v t f(a)= 2b v f(b) = a. Hin nhin rng a b v nu f(a) = 2b=2a th f(2a) = f(f(a))= 2a hay 4b = 2a , mu thun. Nh vy, c th xy dng hm f nh sau : - Chia tp hp v hn cc s l thnh hai tp con v hn ri nhau A v B. Vi mi a A t ty f(a) = b B v f(b) = 2a. - Vi mi 2 .mn a= vi a l , t ( ) 2 ( )mf n f a= . Ta c 2010 = 2. 1005 T f(2010) =2f(1005). Suy ra f(2010) nh nht khi f(1005)=1 ng vi hm f m f(1005)=1 , f(1) = 2010 v cc gi tr k l khc ta t ty theo qui tc trn. V d 4. Cho hm s f: N* N* tha mn iu kin: 2( ( )) ( ), , *f mf n n f m m n N= " Xc nh gi tr nh nht c th c ca f(2010). Gii. D thy f l hm n nh. Cho m = n = 1 ta c f(f(1)) = f(1) , suy ra f(1) = 1. Cho m = 1 ta c 2( ( )) , *f f n n n N= " Thay m bi f(m) ta c: 2 2 2( ( ) ( )) ( ( )) ( ( ))f f m f n n f f m n m f f mn= = = Do f n nh nn f(mn) = f(m)f(n) vi mi m,n N*. Vy f c tnh cht nhn . T xc nh hm f ta ch cn xc nh gi tr ca n ti cc im p nguyn t. Gi p l s nguyn t v gi s f(p) = ab. Khi f(ab) = f(f(p)) hay 2( ) ( )f a f b p=


Do p nguyn t nn ch xy ra kh nng f(a) = f(b) = p , suy ra a = b. Vy f(p) = a2 v f(a2) = f(f(p) = p2 . Nh th f(a) = p . Ta s chng minh a l mt s nguyn t. Tht vy, nu a = mn vi m,n > 1 th p= f(a) = f(m)f(n) , v l v f(m) >1 v f(n) > 1. Tm li, nu p l s nguyn t th f(p) hoc l s nguyn t, hoc l bnh phng ca mt s nguyn t. Ch rng khng th c kh nng f(p)= p hoc f(p) =p2 vi p nguyn t. T ta c th xy dng hm f tha mn yu cu bi ton nh sau : - Chia tp hp cc s nguyn t thnh v hn cp ri nhau (p,q) v t f(p) = q , f(q) = p2. - Vi mi s 1 21 2 ... kkn p p paa a= ta t 1 21 2( ) ( ) ( ) ... ( ) kkf n f p f p f p aa a= Ta c 2010 = 22.3.5.67 nn gi tr nh nht c th c ca f(2010) ng vi hm f m f(2) = 3 , f(3) = 4 , f(5) = 7 , f(67) =11 v f(2012) = 32.4.7.11 = 2772. Hy vng rng qua 4 v d trn , chng ta s phn no nm c k thut gii mt lp cc phng trnh hm trn N. Sau y l mt s bi tp rn luyn. BI TP 1. Chng minh rng khng tn ti hm s f: N* N* tha mn cc iu kin f(mf(n)) = n + f(2m) vi mi m,n N*. 2.Tm tt c cc hm f: N N tha mn iu kin f(f(n)) = n + 2 vi mi n N.

3. Cho hm f: N N tha mn iu kin f(f(n)) = 3n vi mi n N.

Xc nh gi tr nh nht c th c ca f(2012).

4. Tm tt c cc hm f: N N tha mn iu kin f( n+f(n)) = 2n + 2 vi mi n N.

5. Tm tt c cc hm f: N* N* tha mn cc iu kin i) f(mf(n)) = n2f(mn) vi mi n N*. ii) Nu p nguyn t th f(p) l hp s nhng khng phi l s chnh phng.


6. Tm tt c cc hm f: N N tha mn iu kin f(mf(n)) = n3f(m) vi mi n N*. 7. (IMO 1998) Xt cc hm s f: N* N* tha mn cc iu kin f(m2f(n)) = n(f(m))2 vi mi m,n N*.


mot_phuong_phap_giai_pt_ham_tren_n.pdf

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