TRNG I HC KHOA HC T NHINKHOA TON - C - TIN HCBi tp kim tra gia kMn: Gii tch liGing vin: PGS. TS. Nguyn Hu inHc vin: Phm Th ThoLp: Cao hc ton 09-11Chuyn ngnh: Ton hc tnh tonH Ni - 2010Cc bi tp c giao: Bi 9/88, bi 30/ 92, bi 41/94, bi 10/209, bi 23/210.Bi 9/88. Cho Ci Rnili ng khc rng v xi Ci(i= 1, k). Gi C := C1 C1 Ck. Chng minh rngNC(x1, x2, . . . , xk) = NC1(x1)NC2(x2) NCk(xk), (1)TC(x1, x2, . . . , xk) = TC1(x1) TC2(x2)TCk(xk). (2)Chng minh. Ta c: TC(x) = FC(x) trong FC(x) = d Rn[ t0 : x + td C t [0, t0]l nn cc hng chp nhn c.+) Ta chng minh ng thc: FC(x1, x2, . . . , xk) = FC1(x1)FC2(x2) FCk(xk).Ly a FC(x) suy ra: t0 : x + ta C t [0, t0] .x + ta C _x1, x2, . . . , xk_+ t_a1, a2, . . . , ak_ C xit + ai Cii = 1, k ai FCi(xi) i = 1, k hay a FC1(x1)FC2(x2) FCk(xk).Ngc li, ly b FC1(x1)FC2(x2) FCk(xk).Ta c bi FCi(xi)nn ti0 : xi+ tbi Ci t _0, ti0.Ly t/0= minti0 suy ra xit + bi Cit _0, t/0_, i = 1, k.Vy b FC(x).+) Ta chng minh ng thc X = X1X2 Xktrong X = X1X2 Xk.Ly y X yk X : yk y (k ).Do yn=_y1n, y2n, . . . , ykn_ y yi Xii v y X1X2 Xk.Ngc li,y = (y1, y2, . . . , yk) X1X2 Xkth vi mi i = 1, k: _yin_ Xi : yin yi. Suy ra tn ti dy_yn= (y1n, y2n, . . . , ykn)_ X : yn y.Suy ra iu phi chng minh.1p dng hai kt qu va chng minh c ta suy raTC(x1, x2, . . . , xk) = FC(x1, x2, . . . , xk)= FC1(x1)FC2(x2) FCk(xk)= TC1(x1) TC2(x2)TCk(xk).Vy (2) c chng minh.S dng kt qu nn php tuyn v nn tip xc l i cc ca nhau ta suy raNC(x1, x2, . . . , xk) = TC(x1, x2, . . . , xk)= TC1(x1) TC2(x2)TCk(xk)= NC1(x1)NC2(x2) NCk(xk).Vy (1) c chng minh.Bi 41/94. Cho A l ma trn cp mn, c Rn. Khi trong hai h sau c ng mth c nghimAx 0, x 0, cTx > 0, (1)ATy c, y 0. (2)Chng minh.+) Gi s (2) c nghim ta chng minh (1) khng c nghim.Gi s Ax 0, x 0, nhn hai v ca (2) vi xTta cxTATy xTc. (2/)(2/) yTAx cTxDoAx 0, y 0_ cTx yTAx 0.Vy (1) khng c nghim.+) Gi s (2) khng c nghim, ta chng minh (1) c nghim.t X =_x[ y 0 : ATy = x_, suy ra X li ng.Nhn thy 0 X, c/ X. Nh vy, tn ti mt siu phng tch mnh c v X, tcl:p ,=0 v s R pTc >> pTx vi mi x thuc X.V 0 X nn > 0 suy ra_pTc > 0 ()> pTx x X ()T (*) cTp > 0.T (**) > pTATy hay T> yT(Ap).2Mt khc, nu x X th vi 0 th x X nn ta ca y c th ln ty .TT> yT(Ap) ta suy ra Ap 0.Vy p l nghim ca phng trnh (1).Bi 30/92. Cho C l mt tp li ng. Chng minh rngi) x C, tp NC(x) l mt nn li ng.ii)pC(x) = (I + NC)1(x) xChng minh.i) Ta c NC(x) = w[ w, y x 0 y C.+) NC(x) l nn: Ly u NC(x) u, y x 0 y C. > 0 : u, y x = u, y x 0 y C.u NC(x)+) NC(x) l nn li: Ly u, v bt k thuc NC(x) th [0, 1] : u + (1 ) v , y x= u, y x + (1 ) v, y x 0 y C.+)NC(x) ng: Ly dy wk NC(x) v limkwk=w. Ta chng minh w NC(x).C wk, y x 0 y C. Do tnh lin tc ca tch v hng nnlimkwk, y x =_limkwk, y x_ = w, y x y C.Li c wk, y x 0 y C nn suy raw, y x =limkwk, y x 0.Vy w NC(x).ii) t pC(x) = y ta suy ra x y NC(y) hay x y + NC(y).T y ta suy ra y = (I + NC)1(x).Vy pC(x) = (I + NC)1(x).Bi 23/210. Cho x l cc tiu ca hm li kh vi f trn tp C= x[ Ax = b trong A l ma trn cp mn, b Rm. Chng minh rng tn ti =(1, . . . m) 0sao chof (x) +TA = 0.3Chng minh.Gi s ma trn A gm cc vc t hng A1, . . . , Am. Ta cC = x[ Ax = b =_x[ Aix = bi, i = 1, m_t hi(x) = Aix bi, i = 1, m hi(x) l cc hm aphin hu hn trn X.D =_= (0, 1, . . . , m) Rm+1 x X : f (x) f (x) 0 v i= 0 vi mi i = 1, m th c th ly x = x.+) D l tp li trong Rm+1+)0/ D v nu 0 D th tn ti x mf (x) 0 ly 0=f (x) f (x) v i= hi(x) vi mi i = 1, m th D. ()T 0 hay0 ( f (x) f (x) ) +mi=1i hi(x) 0. 0f (x) +mi=1i hi(x) 0f (x) +mi=1i hi(x) 0.Cho 0 ta cL(x, ) L(x, ) x X.trong L(x, ) = 0 f (x) +mi=1ihi(x).Vfl hm li kh vi trn C nn iu kin trn tng ng viL(x, ) = 0.hay 0 f (x) +mi=1i hi(x) = 0. f (x) +mi=1i0hi(x) = 0.t i=i0th ta vit li phng trnh trn di dngf (x) +mi=1iAi= 0 f (x) +TA = 0.4Vy ta suy ra iu phi chng minh.Bi 10/208. Cho hm tch binf(x1, x2, . . . , xn) =ni=1fi(xi) xc nh trn siu hpC:= x Rn[l x u. Chng minh rng hm bao li ca hm tng bng tnghm bao li ca cc hm thnh phn trn mi cnh ca siu hp. Tc lcof (x) =ni=1cofi(xi),trong cofi(xi) l hm bao li cafi trn cnh th i ca siu hp.Nu thay C bng n hnh chun trong Rnth iu khng nh trn cn nghay khng?Chng minh.C l tp li nn theo nh ngha ta ccof (x) = inf___jJ(x)jf_xj_[x =jJ(x)jxj, j 0,jJ(x)j= 1___cofi(xi) = inf___jJ(xi)ji f_xji_[xi=jJ(x)jixj, xji 0,jJ(xi)xji= 1___ i = 1, n+) Chng minh cof (x) ni=1cofi(xi)Ta cjJ(x)jf _xj_ = jJ(x)jni=1fi_xij_ =ni=1jJ(x)fi_xij_ jJ(x)jf _xj_ ni=1cofi(xi) .Suy ra cof (x) ni=1cofi(xi).+) Chng minhni=1cofi(xi) cof (x)Ta c cc on [li, ui] R l cc tp li ng c th nguyn 1 nn bao li ca cchmfi, i = 1, n cn c th nh ngha di dngcofi(xi) = min_2j=1ji fi_xij_[xi=2j=1jixij, j 0,2j=1ji= 1_Gi s rng cofi(xi) =ifi_x1i_+ (1 i) fi_x2i_. Suy racofi(xi) =ifi_x1i_+ (1 i) fi_x2i_ =i_fi_x1i_ fi_x2i__+ fi_x2i_.5ni=1cofi(xi) =ni=1i_fi_x1i_ fi_x2i__+ f (x2).Ta lun chn c mt s (0, 1) tha mnni=1i_fi_x1i_ fi_x2i__ ni=1_fi_x1i_ fi_x2i__ni=1cofi(xi) f (x1) + (1 ) f (x2)ni=1cofi(xi) cof (x).Vy ng thc c chng minh.6