1PHNG PHP TNHGing vin: L TRUNG NGHAS tn ch: 2Chng 1SAISI. Sai s tuyt i v sai s tng i.Cc php oCc phng php tnh gn ngGi tr gn ngca i tngCn xc nh sai s.a. Sai s tuyt i.A - i lng ng;a gi tr gn ng ca A; (a xp x A hay aA) a A - sai s tuyt i ca A; c c lng bngmt s dng a no .a A a; (1. 1)a - sai s tuyt i gii hn; mi a > a - u l sai s tuyt i gii hn ca a.Chn a l s dng nh nht tho mn iu kin (1. 1)ngha l a aAa + a; (1. 3)Quy c vit: A=aa; (1. 2)b. Sai s tng i.Sai s tng i gii hn ca A: ;aaa= (1.4); .a aa = (1. 5)(1. 2) A = a(1 a); (1. 6)Thc t : av a sai s tuyt i v tng i.V d:A = e = 2,718281. . .2,71 < e < 2,73 = 2,71 + 0,01 c th chn 2a = 0,01;2,71 < e < 2,7183 = 2,71 + 0,0083 c th chn 2a = 0,0083;Sai s tng i cht lng php o. II. Cch vit s xp x. 1. Ch s c ngha: nhng ch so tnh t ch so khac 0 au tient ben trai tnh sang.2. Ch s ng tin. Mi s thp phn c th vit:; 10 .ssa = (1. 7)s- nhng s nguyn t 0 n 9; s = 1; 2; 3; . . .*/V d: 56, 708 = 5.101+ 6.100+7.10-1+ 0.10-2+ 8.10-3.125,018 = 1.102+ 2.101+ 5.100+ 0.10-1+ 1.10-2+ 8.10-3.vi 1= 5; 0= 6; -1 =7; -2= 0; -3= 8.2= 1;1= 2; 0= 5; -1 =0; -2= 1; -3= 8.*/Nu a - xp x ca A; a sai s tuyt i ca a:- a 0,5.10ss- ch s ng tin;- a> 0,5.10ss- ch s ng nghi;Mi ch s c ngha bn tri s ng tin; bn phi ng nghi*/ V d: s 56,7082- a= 0,0043 Cc s 5; 6; 7; 0: ng tin; 8; 2 : ng nghi;- a= 0,0067 5; 6; 7 ng tin;0; 8; 2 ng nghi.3. Cch vit s xp x.- Vit km sai s : a a- Vit theo quy c: mi ch s c ngha u ng tins gn ng c sai s tuyt i khng ln hn mt na n v hng cui cng.III. Quy trn s v sai s quy trn.Tnh ton thng quy trn s sai s quy trn.Sai s quy trn tuyt i = s quy trn - s cha quy trn.Nguyn tc : Sai s tuyt i quy trn khng ln hn nan v hng gi li cui cng bn phi hay 5 n v hngb i u tin bn tri.1. Quy trn s.2. Sai s ca s quy trn. A - s ng;- Quy trn a a ; sai s quy trn tuyt i a;; '' aa a (1. 8)a - s xp x ca A; a sai s tuyt i ca a.Sai s tuyt i a ca a:; ' '' a aA a a a A a + + ; '' ' a a aA a = + a>aquy trn lm tng sai s tuyt i3. nh hng ca sai s quy trn. V d tnh nh thc Niutn:Nu thay bng cc s quy trn: 2; 2 2378 3363 ) 10 2 (10 = ; 41421356 , 1 2 = vi2V tri V phi1,41,411,4141,414211,4142135630,00010485760,000134226590,000147912000,000148663990,0001486767833,810,020,5080,008620,0001472Cn quan tm n sai s quy trn khi tnh ton.2IV. Cc quy tc tnh sai s.1. Sai s ca tng: u = x + y;u= x+ y;Sai s tuyt i ca mt tng bng tng cc sai s tuyt ica cc s hng.- Trng hp u = x y; ( x, y cng du);y x uy xuu + == Khi x yrt b th urt ln trnh tr cc s gn nhau.2. Sai s ca tch: u = xy;;y x ux y + = ;y xyx uuy x u + =+==3. Sai s tng i ca mt thng:u = x/y; vi y 0.u= x + ySai s tng i ca mt tch hoc thng bng tng sai stng i ca cc s hng.V. Sai s tnh ton .Cc loi sai s thng gp phi:- Sai s cc s liu ( do o c hay tc nghim );- Sai s ca gi thit ( l tng ho, m hnh ho bi ton);-Sai s phng php (dng cc phng php gn ng gii cc bi ton phc tp;-Sai s ca php tnh (do thc hin cc php tnh i vi sgn ng, quy trn cc kt qu trung gian).Sai s tnh ton = sai s phng php + sai s php tnh.V d:a/ Tnh tng;6151413121113 3 3 3 3 3 + + = A- Tnh trc tip A qua cc phn s Khng c sai s ph/php;- Tnh n 3 s l v nh gi sai s quy trn tng ng000 , 111113= =vi 1= 0;125 , 081213= = 2= 0;037 , 0271313= = 3= 1.10-4; (0,037037)016 , 0641413= = 4= 4.10-4; (0,015625)008 , 01251513= = 5= 0;005 , 02161613= = 6= 4.10-4; (0,004629)A = 1,000 0,125 + 0,037 0,016 +0,008 0,005 = 0,899.; 10 . 946 5 4 3 2 1= + + + + + a Aa=9.10-4; A = 0,899 9.10-4.b/ Tnh tng ;1) 1 (413121111313 3 3 3 + + + =nBnVi sai s tuyt i khng qu 5.10-3.;1) 1 (413121111313 3 3 3nB Bnn + + = nB B- Sai s phng php.n cn : + sai s php tnh 1;~ = + +nn i n iq y ykhi nqu trnh tnh khng n nh.3Chng 2NI SUY V LY XP X HM SI. Ni suy. 1. t vn .- khng bit biu thc gii tch ca hm;y = f ( x );- bit gi tr ca hm ti mt s hu hn im trn on[ a, b] ( bng o c hoc thc nghim):xy=f( x )xo x1. . . xi. . . xn-1xnyoy1. . . yi. . . yn-1yn- Tm gi tr ca hm s ti mt s im trung gian khc.-Xy dng mt hm ( x ) c biu thc n gin, c gi trtrng vi gi tr ca f ( x ) ti cc im , cn ticcim khc trn on [a, b] th ( x ) kh gn f( x ) [phnnh gn ng quy lut f( x ) ]c th suy ra gi tr gn ngca f( x ) ti cc gi tr x bt k tho mn xo< x < xn.xo x1. . . xnBi ton ni suy:- ( x ) c gi l hm ni suy ca f( x ) trn on [a, b].- ngha hnh hc: xy dng ng cong y = ( x ) i qua ccim cho trc (xi, yi), I = 0, 1, . . . , n.2. a thc ni suy.Thng chn a thc lm hm ni suy v:- a thc l loi hm n gin;- Lun c o hm v nguyn hm;- Vic tnh gi tr ca chng n gin.- Trn on a x b cho mt li cc im chia (im nt)xi; i = 0, 1, 2, . . ., n; vi a xo, x1, x2, . . . .xn b.- Cho gi tr tng ng ca hm y = f( x ) ti cc nt:xy=f( x )xo x1. . . xi. . . xn-1xnyoy1. . . yi. . . yn-1yn- Cn xy dng mt a thc bc n:Pn( x ) = aoxn+ a1xn-1+ . . . + an-1x + an.sao cho Pn(x) trng vi f(x) ti cc nt xi,Pn(xi) = yi; i= 0, 1, 2, . . . , n.Bi ton:( 1 )S duy nht ca a thc ni suy: a thc ni suy Pn( x ) cahm f( x ) nh ngha trn nu c th ch c mt m thi.a thc ni suy c th xy dng theo nhiu cch nhng do tnh duy nht nn cc dng ca n u c th quy v nhau.3. a thc ni suy Lagrng.trong ;) )...( )( )...( () )...( )( )...( () (1 11 1n i i i i i o in i i oix x x x x x x xx x x x x x x xx l =+ + li(x) a thc bc nPn(x) a thc bc n= ) (j ix l1khi j = i;0 j i ;Pn(xi)= yi; i = 0, 1, 2, . . ., n.); ( ) (0x l y x Pinii n == ( 2 )( 2 ) a thc ni suy Lagrng.*/ Ni suy tuyn tnh. ( n = 1 )xyxox1yoy1( 2 ) ); ( ) ( ) (1 1 1x l y x l y x Po o+ = ( 3 )oooox xx xx lx xx xx l==1111) ( ; ) (; ) (1111oooo nx xx xyx xx xy x P+=C dng P1(x) = Ax + B - bc nht i vi x.*/ Ni suy bc 2. ( n = 2 )xyxox1 x2yoy1y2); ( ) ( ) ( ) (2 2 1 1 2x l y x l y x l y x Po o+ + = ( 4 )2 1 211 2 1 1 2( )( ) ( )( )( ) ; ( )( )( ) ( )( )ooo o ox x x x x x x xl x l xx x x x x x x x = = ;) )( () )( () (1 2 212x x x xx x x xx loo =P2(x) c dng : P2(x) = Ax2+ Bx + C - bc 2 i vi x.*/ Sai s ni suy.nh l. Nu hm f(x) lin tc trn [a, b] v c trong [a, b] ohm lin tc n cp n+1 th sai s ni suy rn(x) =f(x) Pn(x) c biu thc:[ ] b a cnxc f x rnn, ;)! 1 () () ( ) () 1 (+=+( 5 )) )...( )( ( ) (1 n ox x x x x x x = ([a, b] - khong cha cc nt xi)Gi [ ] b a x x f Mn, ; ) ( max) 1 ( =+th; ) ()! 1 () ( xnMx rn+-u im ca a thc ni suy Lagrng : n gin;- Nhc im : thm mt nt th phi tnh li ton b.V DU----------------------------------------------------------------------------------------------------------------------Bang 4 moc 1, 2, 3, 4 ; 4 gia tr 5, 7, 8, 9. Viet ra bieu thc cac a thc noi suy c s. Tnh gia tr bang tai x = 3.5?2(3.5) l =3(3.5) l =( )( )( )( )( )( )( )02 3 41 2 1 3 1 4x x xl x = ( )03.5 0.0625 l =( )( )( )( )( )( )( )13.5 1 3.5 3 3.5 43.52 1 2 3 2 4l = = ( )3 0 1 2 3( ) 5 ( ) 7 ( ) 8 9 ( ) P x l x l x l x l x = + + + ( )33.5 8.4375 P =Viet bieu thc lk(x) (Khong tnh!) Thay x Gia tr44. a thc ni suy Niutn.a/ Sai phn hu hn.y = f(x) c gi tr yi= f(xi) ti cc nt xicch u nhau vixi+1 xi = h = const; i = 0, 1, 2, . . ., nx0; x1= xo+ h;x2= x0+ 2h; . . .xi= xo+ ih . . .nh ngha sai phn hu hn ca hm y = f(x):Sai phn cp 1(hng 1): yi= yi+1 yi;Sai phn cp hai: 2yi= yi+1 yi;nyi= (n-1yi) = n-1yi+1 n-1yi;Sai phn cp n l sai phn ca sai phn cp n-1:Sai phn cp 3:3yi= 2yi+1 2yi;;1 o oy y y = ; 2 ) ( ) (1 2 1 1 22o o oy y y y y y y y + = = ; 3 31 2 33o oy y y y y + = ; ) 1 ( ) 1 (! 3) 2 )( 1 (! 2) 1 (! 1113 2 1 on nn n n n ony y yn n nyn nyny y + + + = . . . . . . . . . . . . . . . . . . . . .b/a thc ni suy Niutn tin (ni suy v pha phi).Trng hp cc nt cch u, Niutn a thc c dng:) ( ) ( ) )( ( ) ( ) (1 1 2 1 + + + + =n o n o o o nx x x x a x x x x a x x a a x P- x = xoao= Pn(xo) = yo;- x = x1;) (1111hyhy yx xa x Pao ooo n===- x = x2;! 2 222) ( 2. 22) )( () ( ) (2221 221 221 2 22 1 22hyhy y yhy y y yh hhhyy yx x x xx x a a x Pao o o ooooo o n=+ = = = =;!ioiih iya=; ) (i i ny x P =n oa a a a , , , ,2 1 Xc nh t iu kin); ( ) (!) )( (! 2) (! 1) (1 122 + + + + =n ononooooo nx x x xh nyx x x xhyx xhyy x Pi bin, thx xto= x = xo+ t.hxi= xo+ i.h x - xi= x - xo- i.h = t.h - i.h = (t-i)h;( )21( 1) ( 1)...( 1)( ) ;2! !nn o o o ot t t t t nPx N t y t y y yn += = + + + + Thng dng tnh cc gi tr ca hm gn xo u bng.3/ a thc ni suy Niutn li (ni suy v pha tri)Vi cch lm tng t nhng bt u vi x = xn); ( ) (!) )( (! 2) (! 1) (1 12221x x x xh nyx x x xhyx xhyy x Pnnonn nnnnn n + + + + = ( )22 1 2( 1) ( 1)...( 1)( ) ;1! 2! !nn n n n ot t t t t t nPx N t y y y yn + + + = = + + + + u im ca cng thc ni suy Niutn: thm nt ch cn thms hng, khng cn phi tnh li. thun tin tnh ton thng lp bng sai phn ng cho.x y y 2y 3y 4y 5y 6yxox1x2x3x4x5x6yoy1y2y3y4y5y6yoy1y2y3y4y52yo2y12y22y32y43yo3y13y23y34yo4y14y25yo5y16yoBng sai phn ng cho ca cng thc ni suy tinx y y 2y 3y 4y 5y 6yxn-6xn-5xn-3xn-2xn-1xnyn-6yn-5yn-4yn-3yn-2yn-12yn-62yn-52yn-42yn-32yn-23yn-63yn-53yn-43yn-34yn-64yn-54yn-45yn-65yn-56yn-6Bng sai phn ng cho ca cng thc ni suy lixn-4yn-6yn-5yn-3yn-2yn-1ynyn-44/ Sai s ca php ni suy Niutn.Vn dng cng thc sai s bit trong phn ni suy Lagrngnhng thay o hm hng n+1 bng sai phn hng n+1- Vi cng thc ni suy tin:;)! 1 () )...( 1 () ( ) ( ) (1onnynn t t tx P x f x r++ =- Vi cng thc ni suy li:;)! 1 () )...( 1 () ( ) ( ) (1onnynn t t tx P x f x r+++ + =5V DU NOI SUY NEWTON------------------------------------------------------------------------------------------------------------------Chobanggia trsinxt 15 55 .Xaydnga thc noi suy tien (lui) cap 3 & tnh sin16 (sin54 )0.8192 550.7660 500.7071 450.6428 400.5736 350.5 300.4226 250.3420 200.2588 153y 2y y y xV DU NOI SUY NEWTON---------------------------------------------------------------------------------------------------------------------------Tat ca sai phan: Noi suy Newton Lagrange!a thc noi suy tien:x 15 ( ) ( )( )! 32 10006 . 0210026 . 0 0832 . 0 2588 . 0 ) (1 + =t t t t tt t Nt xxt 5 15515+ = = x = 16 t = 0.2 N1(0.2) = 0. 2756 sin16 = 0. 2756a thc noi suy lui:x 55 t xxt 5 55555+ = = ( ) ( )( )! 32 10003 . 0210057 . 0 0532 . 0 8192 . 0 ) (2+ ++ + =t t t t tt t Nx = 54 t = 0.2 N2(0.2) = 0.80903 sin54 = 0. 8090Cau hoi: Tnh tai x = 54 vi Noi suy tien. Nhan xet?II. Ly xp x hm s. Phng php bnh phng nh nht.1. t vn .*/Ly xp x bng cc a thc ni suy c nhng nhc im:- Nu nhiu nt bc ca a thc ni suy s rt lntnh ton khng thun tin-Vic buc Pn(xi) = yikhng hp l ( cc s liuo c, thcnghim c th khng chnh xc sai s ln khi ni suy)- Khng tht ph hp nu f(x) l hm tun hon.*/ Thng c nhu cu lm trn cc ng cong thc nghim, hoc biu din cc quan h thc nghim di dng mt hms bit no , v d:y = a + bx;y = a + bx + cx2;y = a + bcosx + csinx;y = aebx; y = axb; . . .Ly xp x mt hm s l tm mt hm s khc hoc mt thp cc hm s khc m sai khc vi hm s cho btheo mt ngha no .2. Ly xp x bng phng php bnh phng nh nht.- Cho hm s f(x) v a thc:); ( ) ( ) ( ) ( ) (2 2 1 1 0x C x C x C x C x Qm m o + + + + =( a )m i xi,..., 2 , 1 , 0 ); ( = - h cc hm s no ca x;m oC C C C ,..., , ,2 1- cc h s.-Cn xc nh sao cho Q(x) trn tp choX xsai khc vi f(x) nh nht c th.m oC C C C ,..., , ,2 1Q(x) a thc xp x ca f(x).- Khi , ) ( ;... ) ( ; ) ( ; 1 ) (22 1mm ox x x x x x x = = = = ; ... ) (22 1mm o mx C x C x C C x Q + + + + =( b )l mt a thc i s thng thng.Phng php thng dng khi ly xp x[ xc nh cc h sca ( a ) hoc ( b ) ] l phng php bnh phng nh nht.-Ni dung ca phng php bnh phng nh nht l tmcc tiu ca hm[ ]21( ) ( ) ;ni iiM f x Qx== ( c )f(xi), i = 0, 1, . . ., n l nhng gi tr bit ca f(x) trn tpim xo, x1, . . ., xn.Coi cc h s Ck l cc n phi tm, M cc tiu phi c:) ,..., 1 , 0 ( ; 0 m kCMk= =( d )T ( d ) tm c cc h s Ck.f(xi) Q(xi) - Sai s ti xikhi thay f(x) bng Q(x).3. Mt s trng hp thng gp.a/ a thc xp x c chn di dng cc hm tuyn tnh.xy=f(x) x1. . . xi. . . xn-1xny1. . . yi. . . yn-1yn*/ Q(x) = a + bx( c ):[ ] ( ); ) ( ) (0202 = = = =nii inii ib ax y x Q x f M M b nht:; 0 ; 0 ==bMaM2;;i ii i i ib x an yb x a x x y + = + = thun tin tnh ton thng lp bng:xiyixi2xiyi. . .. . .. . .. . .. . .. . .. . .. . .Thay cc s liu bng vo v gii ph/trnh a, b.6*/ Q(x) = a + bx + cx2.( )221;ni i iiM y cx bx a== ; 0 ; 0 ; 0 ===cMbMaM4 3 2 23 22;;;i i i i ii i i i ii i ic x b x a x y xc x b x a x y xc x b x an y + + = + + = + + = xiyixi2xi2yi. . .. . .. . .. . .. . .. . .. . .. . .yi. . .. . .. . .. . .. . .. . .. . .. . .xi3xi4xiyi15.6 13. 12.5 10.1 8.8 7.0 5.0 4.2 3.5 1.3 yk10 9 8 7 6 5 4 3 2 1 xk15.6 13. 12.5 10.1 8.8 7.0 5.0 4.2 3.5 1.3 yk10 9 8 7 6 5 4 3 2 1 xk1/ VD: Tm ham bac 1 xap x bang sau theo ngha BPCTxiyixi2 xiyi 11.311.3 23.547 34.2912.6 45.01620 57.02535 68.83652.8 710.14970.7 812.564100 913.081117 n=10 1015.6100156 5581385572.4 2385 55 572.455 10 811.53820.36i i i ii ib x a x x yb x an ya ba bba + =+ =+ = + = Vy Q(x) = 1.5382x 0.36Dang 1: = = + y f(x) a bx Cach x ly may tnh bo tui MS: Bc 1 : Mod Mod Reg Lin(linear) Bc 2 : Nhap tng cap :1 1, x y M + 2 2,,n nx y Mx y M + +Bc 3 : Shift 2 Bm phm qua phi 2 ln tnh A, B ES: Bc 1 : Mod Stat A+BX Bc 2 : Nhap bang d lie u :1 1( )n nX FXx yx y AC Bc 3 : Mod Reg A, B 2/ Cho quan h thc nghim:xi12345yi9.817,2 25,8 37,1 49,7Biu din mi quan h bng hm bc 2y = a + bx + cx2Xc nh cc h s a, b, c theo p/p bnh phng nh nht.12345159,817,225,837,149,7139,6149162555116812566259799,868,8232,2593,61242,52146,9xiyixi2xi3xi4yixiyixi2n = 5182764125225H phng trnh:( a )9,834,477,4148,4248,5518,52;i i ic x b x an y + + = 4 3 2 2;i i i i ic x b x a x y x + + = 3 2;i i i i ic x b x a x y x + + = Xt h23 24 3 2 255 15 5 139.6 0.9357225 55 15 518.5 4.3557979 225 55 2146.9 4.56i i ii i i i ii i i i ic x b x an yc x b x a x x yc x b x a x x ya b c ca b c ba b c a + + =+ + =+ + =+ + = + + = + + = = Vy y=0.9357 + 4.3557x + 4.56x27Dang 2:2y f(x) a bx cx = = + + Cach x ly may tnh bo tui MS: Bc 1 : Mod Mod Reg Bm phm qua phi Quad (Quadratic) Bc 2 : Nhap tng cap :1 1, x y M + 2 2,,n nx y Mx y M + +Bc 3 : Shift 2 Bm phm qua phi 2 ln tnh A, B, C b/ a thc xp x l nhng hm phi tuyn.Trong nhiu trng hp c th tuyn tnh ho vic tnhton thun li:*/ 1; ya bx=+t Y = 1/y Y = a + bx;*/;b axxy+= t Y = 1/yv X = 1/x Y = a + bX;*/ ;bxae y = vi a > 0;Ly logarit thp phn hoc npec 2 v:logy = loga + xbloge hay lny = lna + bx;t lny = Y; X = x; A = lna; B = b Y = A + BXChuyn bng s liu thc nghim sang quan h gia Xiv Yi tm A, B. Sau a = 10A; b = B/loge.*/ y = axb( a > 0) logy = loga + b.logx hay lny = lna + blnxt Y = lny; A = lna; X = lnx; B = b; Y = A + BXChuyn bng s liu sang quan h ca Xiv Yi tm A, B.Dang 3:bxy f(x) ae,a 0 = = > . Cach x ly ma y tnh bo tui MS: Bc 1 : Mod Mod Reg Exp (Exponent) Bc 2 : Nhap tng cap :1 1, x y M + 2 2,,n nx y Mx y M + +Bc 3 : Shift 2 Bm phm qua phi 2 ln tnhA, B V du : Cho ba ng no i suy : x 0, 23 1,15 2, 04 2, 57 3, 01 y 2, 64 3,14 3, 71 4, 08 4, 43 Xa p x a th c no i suy da ng bxy f(x) ae,( a 0) = = >Giai La y logarit c so e hai ve , ta co : lny = lna + bx (*) a t : Y = lny ; A = lna ; B = b ; X = x Khi o , (*) tr tha nh : Y = A + BXBa ng no i suy theo X, Y nh sau : X 0, 23 1,15 2, 04 2, 57 3, 01 Y 0, 970779 1,144223 1, 311032 1, 406097 1, 488400 V i A, B tho a hephng trnh : n ni ii 1 i 1n n n2i i i ii 1 i 1 i 1nA B X YA X B X X Y= == = =+ =+ = 5A9B 6,3205319A 21, 202B 12, 307394 + = + = A 0, 929282B 0,186013 = = Aa e 2, 532690b B 0,186013 = = = = Va y : 0,186013xy 2, 532690.e =Dang 4:by f(x) ax,a 0 = = > . Cach x ly ma y tnh bo tui MS: Bc1:Mod Mod Reg Bmphmquaphi Pwr (Power) Bc 2 : Nhap tng cap :1 1, x y M + 2 2,,n nx y Mx y M + +Bc 3 : Shift 2 Bm phm qua phi 2 ln tnh A, B 8V du : Cho ba ng no i suy : x 10 20 30 40 50 60 70 80y 1, 06 1, 33 1, 52 1, 68 1,81 1, 91 2, 01 2,11 Xa p x a th c no i suy da ng by f(x) ax,( a 0) = = >vatnh y(2,5)? Gia i La y logarit c so e hai ve , ta co : lny = lna + blnx (*) a t : Y = lny ; A = lna ; B = b ; X = lnx Khi o , (*) tr tha nh : Y = A + BXBa ng no i suy theo X, Y nh sau :

X 2,302585 2, 995732 3, 401197 3, 688879 3, 912023 4, 094345 4,248495 4,382027Y 0, 058269 0, 285179 0, 418710 0, 518794 0, 593327 0, 647103 0,698135 0, 746688V i A, B tho a hephng trnh : n ni ii 1 i 1n n n2i i i ii 1 i 1 i 1nA B X YA X B X X Y= == = =+ =+ = 8A29,025248B 3, 96620529,025248A 108, 771736B 15, 534974 + = + = A 0, 703656B 0,330589 = = Aa e 0, 494773b B 0, 330589 = = = = Va y : 0,330589y 0, 494773.x =Chng 3GII GN NG PHNG TRNH I S V SIU VITI.Tm nghim thc ca mt phng trnh.a. Nghim thc ca phng trnh mt n ngha hnh hc.f(x) = 0; ( 1 )f hm cho trc ca i s x - nghim thc ca ( 1 )f() = 0; ( 2 )- V th y = f(x)Honh im Mnghim .Oyx Mf(x)OyxMg(x)h(x)~ g(x) = h(x) th y1 = g(x) v y2= h(x)- hoc (1)b. S tn ti ca nghim thc.nh l. Nu c hai s thc a, b (a < b)sao cho f(a) v f(b) tridu, tc lf(a).f(b) < 0 ( 3 )ng thi f(x) lin tc trn [a, b] th trong khong [a, b] t nht cmt nghim thc ca phngtrnh f(x) = 0.OyxABabc. Khong phn ly nghim (tch nghim)nh ngha.Khong [a, b]no gi l khong phn ly nghimca phng trnh f(x) = 0 nu ncha mt v ch mt nghimca phng trnh . Mun th:trong [a, b] :- hm f(x) n iuOyxABabf(x) khng i duII. Cc phng php xc nh gn ng nghim thc camt phng trnh.1. Phng php th2. Phng php th.V d : Tm nghim ca phng trnh:f(x) = xlogx 1,2 = 0;xf(x)1 2 3 4- 1,2 - 0,5980 0,2313 1,2084- [2, 3] - khong phn ly nghim;- tip tc chia nh khong [a, b];- Tnh th gi tr cc nt; khong cha nghim mi;- Lp li cc bc trn cho n khi t chnh xc cn thit.nh l. Nu hm s f(x) lin tc v n iu trn khong [a, b], ng thi f(a) v f(b) tri du th [a, b] l khong phn ly nghimca phng trnh f(x) = 0.3. Phng php chia i. Cho phng trnh f(x) = 0;- Gi s [a, b] l khong phn ly nghim ca phng trnh.- Chia i khong [a, b];2b ac+=- Tnh f(c)f(c) = 0 c = (nghim);f(c) 0 Xt du f(c).f(a) v f(c).f(b);Khong phn ly nghim mi [a1, b1];); (211 1a b a b = - Tip tc qu trnh chia i [a2, b2]); (21) (2121 1 2 2a b a b a b = = . . . . . . . . .. . . . . . . . . . . . . . . .); (21a b a bnn n = Vi an bn.9- Ly anhoc bnlm gi tr gn ng ca nghim;- Sai s:;2nn n na ba b a= ( 4 ) ;2nn n na ba b b= ( 5 )V d: Tm nghim ca phng trnh f(x) = x4+ 2x3x 1 = 0;f(0) = -1; f(1) = 1 [ ]; 1 , 0 f(0,5) = -1,9[ ]; 1 , 5 , 0 f(0,75) = -0,59[ ]; 1 , 75 , 0 f(0,875) = +0,05[ ]; 875 , 0 , 75 , 0 f(0,8125) = -0,304[ ]; 875 , 0 , 8125 , 0 f(0,8438) = -0,135[ ]; 875 , 0 , 8348 , 0 f(0,8594) = -0,043[ ]; 875 , 0 , 8594 , 0 Ly[ ]; 867 , 02875 , 0 8594 , 0=+ Sai s mc phi:7 71 0 1;2 2 2n nb aa = =Cc bc tnh: Cho phng trnh f(x) = 0;- n nh sai s cho php;- Xc nh khong phn ly nghim (p2 th, p2th . . .);- Gii theo s : c = (a+b)/2; f(c)f(c).f(a) 0;- xntheo (14) khi n* Sai s. Ly xnnghim gn ng sai s:;) (mx fxnn ; ) ( ' 0 x f m < ; b x a viTrong thc t, thng dng qu trnh tnh khi:xn xn-1 < (sai s cho php)( 15 )( 16 )* Ch .-Phng trnh (13) thay cho (1) l tuyn tnh i vi x nnphng php Niutn cng gi l phg php tuyn tnh ho;- (14) p/php Niutn cng l p/php lp vi hm lp:;) ( ') () (x fx fx x = S tm tt cc bc gii:1/ Cho phng trnh f(x) = 0;2/ n nh sai s cho php ;3/ Tm khong phn ly nghim;;) ( ') (1ooox fx fx x =;1 ox x e =e < xo= x1x = 5/ Tnh sai s mc phi theo (15)4/ Chn u tnh xo f(xo).f(xo) > 0;V DU LAP NEWTON TIEP TUYEN--------------------------------------------------------------------------------------------------------------------------Giai xap x f(x) = x cosx = 0tren [0, 1], sai so 108 6. Phng php dy cung.* Gi s: - ( 1 ) c nghim duy nht trn [a, b];- f(x) v f(x) khng i du trn [a, b];* Thay ng cong f(x) bng dy cung ni A[a,f(a)], B[b,f(b)];* Honh giao im ca dy cung AB vi trc honhnghim gn ng. O xyABa x1x2 b* Phng trnh dy cung AB:;) ( ) () (a bx Xa f b fa f Ya=Ti giao im:Y = 0; X = x1;;) ( ) () ( ) (1a f b fa f a ba x = ( 17 );) ( ) () ( ) (1a f b fa bf b afx= ( 18 )* Xt du f(a).f(x1)khong phn ly nghim mi, tip tc tnhn khi: a b < sai s cho php. * Sai s tnh theo ( 15 ).Cc bc tnh:1/ Cho f(x) = 0;2/ n nh sai s cho php;3/ Tm khong phn ly nghim;4/ Tnh theo s : ;) ( ) () ( ) (1a f b fa bf b afx=f(x1).f(a) 01,2552,0 -0,2796 5 1,2945-0,12531,29452,0 -0,1253 5> 01,3117-0,0548> 01,31172,0 -0,0548 5 1,3192-0,023> 01,31922,0 -0,023 5 1,3223-0,0103 > 01,32232,0 -0,0103 5 1,3237-0,004> 01,32372,0 -0,004 5 1,3242-0,002 > 01,32422,0 -0,002 5 1,3245-0,00098 > 01,32452,0 -0,00098 5 1,3246 -0,00037 > 01,32462,0 -0,00037 5 1,3247 -0,00011 > 01,32472,0 -0,00011 5 1,324715 -0,000013a b f(a) f(b) x f(x) f(x).f(a)C th ly = 1,324715 vi sai s < = 10-3.Chng 4TM NGHIM CA H PHNG TRNH (Tham kho)A. Nghim ca h phng trnh i s tuyn tnh.I. Khi nim chung.Xt h n phng trnh c n n s:a11x1+ a12x2+ . . . +a1nxn=f1;a21x1+ a22x2+ . . . +a2nxn=f2;an1x1+ an2x2+ . . . +annxn=fn;. . . . . . . . . . . . . . . . . . . . . . . . .( 20 )Ma trn h s ca phng trnh:a11a12. . .a1na21a22. . .a2nan1an2. . .ann. . . . . . . . . . . . . .A =( 21 )Vect v phi v vct n:f1f2fn...f = ;x1x2xn...x =( 22 )( 20 )Ax=f ( 23 )Hay f = [ f1, f2, , fn ]T; x = [ x1, x2, , xn]T;II. Tm nghim theo quy tc Cramer v phng php Gauss.S tn ti v duy nht ca nghim ca h. = det (A)- i suy ra t : thay ct th i bng v phi. a/ nh l Cramer: Nu 0 th h ( 20 )khng suy bin vc nghim duy nht c tnh bi cng thc:n i xii,..., 2 , 1 ; ==( 24 )Nhn xt: - Cng thc p, gn;- Khi n ln phi thc hin mt s lng rt ln cc php tnh.NC(n) - s lng php tnh cn lm khi h c n ph/trnh.NC(n) = (n+1)!nVi n = 15: NC(15) = 3.1014.- = 0ma trn A suy bin h ( 20 ) suy bin.b/ Phng php Gauss.Ni dung: Kh dn cc n v h tng ng c dngtam gic trn ri gii t di ln khng phi tnh nh thc.V d: a11x1+ a12x2+ a13x3= a14;a21x1+ a22x2+ a23x3= a24;a31x1+ a32x2+ a33x3= a34;Kh dn cc nx1+ b12x2+ b13x3= b14;x2+ b23x3= b24;x3= b34;Gii ngc t di ln tm c cc n.( 25 )( 26 )Qu trnh a (25) (26):qu trnh thun.Qu trnh gii (26): qu trnh ngc.Khi lng php tnh:;67 9 4) (2 3n n nn NG +=Vi n = 15; NG(15) = 2570; nh hn nhiu so vi ph/php trn.14c/ Tm ng dn nghim ca h phng trnh i s tuyntnh bng phng php lp n.Ni dung phng php:- Cho h phng trnh Ax =f ;- Bin i v dng tng ng: x =Bx + g ; ( 27 )b11b12. . .b1nb21b22. . .b2nbn1bn2. . .bnn.. . . . . . . . . . . .B = suy ra t A; g suy ra t f.-Chn x(0)no lm nghim gn ng u tin v tnh ccgn ng tip theo: x(1), x(2). ., x(m)theo cng thc lp:) 1 ( ;) 1 ( ) ( + =m g Bx xm m( 29 ) xo cho trc;( 30 )trong ; . ) (1jnjij ix b Bx ==Phng php tnh x(m)theo ( 28 ) phng php lp n;B ma trn lp.( 28 )S hi t v sai s ca phng php.nh ngha v s hi t. Gi s [ 1, 2, . . ., n]T- nghim ca(20) tc (23), nuimix ) (khi n i m ,..., 3 , 2 , 1 ; = phng php lp hi t.nh l v s hi t (ca phng php lp n).i vi ma trn B nu:; 1 ; ; ; max1 1 12 1 0