GlAl TlCH SOl1rinh /nh 'gocI9ZOO9iMuc dich cua giai tich so:phat trien cac phuong phap hieu qua va chinhxac detinh xap xi cac dai luong ma khohoac khong thenhan duoc lang cacphuong tien giai tich.Moc dcbI. Cung cap kien thuc co sovephuong phap tinh.Z. Sinh vien liet ap dung, thuc hien (lang may tinh) va phan tich ket quatinh toan.J. Sinh vien cothetudoc cac sach, lai lao vephuong phap tinh.Noi dongI. Sai sovasohoc dau cham dongZ. Hephuong trnh dai sotuyen tinhJ. 'oi suy+. Cac phuong trnh phi tuyen5. Lao ham vatich phan soo. lhuong trnh vi phan thuongTai Iico doc tbcmDavid kincaid and Ward Cbcncy, 'umerical /nalysis Mathematics of Scien-tific Computing, BrooksCole lullishing Company I99I.Cap nhat: 9ZOIIiiCboong 1Sai sovasoboc dao cbam dongCac loai sai so:- sai sokhi thiet lap mohnh (toan hoc);- sai sodo phep do dulieu cua lai toan;- sai so do phuong phap tinh, goi la sai so roi rac hoa (discretization error)hay sai sochat cut (truncation error);- sai so do phep lieu dien so (lang mot so huu han cac lit) va tinh toantrong may tinh, goi lasai solam tron (roundoff error).1.1 Cac kbai nicm co banHai cach do dochinh xac cua mot dai luong xap xi: sa| s.uye. 1| (alsoluteerror) vasa| s.u+n 1| (relative error).|nb ngbia 1.1. Cho . lagiatri xap xi cua.. 1h sai sotuyet doi trong . la^. = . ..vaneu. ,= 0 sai sotuong doi la^..=. ...Sai sotuong doi thehien dochinh xac tot hon,nhung sai sotuyet doilai coich khi giatri chinh xac gan lang khong.IZ CHlO!C l. S^l SOV^SOHOC D^| CH^M DO!CBa| .an s(numerical prollem) lamoi quan heham giua dulieu nhap(input data) - lien doc lap trong lai toan - vadulieu xuat (output data) -ket quacan tm. Dulieu nhap vaxuat gom mot sohuu han cac dai luongthuc (hoac phuc) vanhu vay duoc lieu dien loi cac vecto cokich thuoc huuhan. Moi quan heham cothelieu dien duoi dang an hoac hien. 1huong tadoi hoi dulieu xuat phai duoc xac dinh duy nhat vaphuthuoc lien tuc vaodulieu nhap.T|ua. .an (algorithm) cho mot lai toan solasumotaday ducac pheptoan xac dinh tot qua domoi vecto dulieu nhap khadI duoc chuyen thanhmot vecto dulieu xuat. Cacphep toan oday duoc hieu lacac phep toansohoc valogic mamay tinh cothethuc hien duoc, hoac latham chieu dennhung thuat toan daliet.I|u+np|aps(numerical method) lamot thutuc dexap xi mot laitoan toan hoc lang mot lai toan sohay degiai mot lai toan so(hay it nualadan novemot lai toan don gian hon). 1huat nguphuong phap sotongquat va duoc dung rong rai hon thuat toan, noit nhan manh vao cac chi tiettinh toan.Tb do1.1. Xac dinh nghiem thuc lon nhat cua phuong trnh lac la(:) = a0:3a1:2a2:1a3 = 0voi cac hesothuca0. a1. a2. a3, lamot lai toan so. \ecto dulieu nhap la(a0. a1. a2. a3). Dulieu xuat langhiem.can tm. Mot thuat toan cho laitoan nay cotheduoc xay dung dua vao phuong phap 'ewton,cung quy tacchon giatri dau vadieu kien dung thuat toan. Cung cothexay dung thuattoan tren co socong thuc (cho nghiem chinh xac) cua Cardano. Cong thucnay chua can lac hai valac la v vay ta can chi dinh thuat toan tinh caccan thuc nay Mot lai toan toan hoc voi dulieu nhap.vadulieu xuat , =J(.)duoc goi la1|eu '|en .. (well-conditioned) neu nhung thay doi nho trong. dan den nhung thay doi nho trong,. 'eu suthay doi trong, lon,laitoan duoc goi la(otrong) 1|eu '|en xau (ill-condition). Lieu kien tot hay xaucothephuthuoc vao cach do suthay doi. \ephuong dien tinh toan,dieukien cua lai toan colien hevoi tinh on dinh (stalility) cua thuat toan. Motthuat toan lan 1(n| (stalle) neu nhung thay doi nho trong dulieu nhapdan den nhung thay doi nho trong dulieu xuat. 1ruong hop nguoc lai, tanoi thuat toan '|n n 1(n| (unstalle).Tb do1.2. Cho hamJ(.) khavi. Ciasudoi sonhap. cosai sotuong doilangc, khi dosai sotuyet doi trong giatri xuat cuaJ(.) laJ(.) J(. c.) ~ c.J0(.).l.l. C^C lH^l !lLM CO B^! JSai sotuong doi laJ(.) J(. c.)J(.)= c.J0(.)J(.).1ruong hopJ(.) =ex,sai sotuyet doi trong giatri cua ham mugayra do sai so c. trong doi so . duoc xap xi loi c.ex, vasai sotuong doi angchung c.. Khi . lon, dieu kien cua phep (lai toan) danh giaham nay doivoi sai sotuong doic nhophuthuoc rat nhieu vao viec chon cach do sai so.1ruong hopJ(.) = cos(.), ogan. = ,2, sai sotuyet doi do sunhieu. thanh. c. xap xi lang c. sin(.) ~ c,2.Sai so tuong doi tai ,2 khongxac dinh. 1uy nhien,cac giatri chinh xaccos(1.57079) = 0.63268 105. cos(1.57078) = 1.6327 105cho thay mot thay doi rat nho trong doi sogan,2 cothedan toi sai sotuong doi trong giatri ham rat lon (oI) Tb do1.3. 1ich phan tung phan thuong duoc dung dethiet lap cong thuctruy hoi. 1hi du, xet1n =_10.nex1J. voin = 1. 2. . . . (I.I)1u(I.I) ta congay11> 12> . . .> 0. (I.Z)/p dung cong thuc tich phan tung phan,sau mot solien doi,ta duoccong thuc truy hoi.1n = 1 n1n1. (I.J)1hanh phan dau11 = 1 _10ex1J. = 1,e.+ CHlO!C l. S^l SOV^SOHOC D^| CH^M DO!Cdung Matlal (ILLL-o+I, chinh xac don) ta tinh duoc:11= 0.367912= 0.2642. . .116= 0.055511T= 0.05721n khong giam'11S= 0.02951n khong duong'. . .120= 30.19241n khong nam giua O vaI'Lay la mot thi duve thuat toan khong on dinh. lhan tich:Ciasu ta lat daulang11 = 11, vacac tinh toan theo sau khong cosai so. 1h12= 1 2 11 = 1 2 11 2 = 12 2.13= 1 3 12 = 1 3 126 = 136.. . .1n= 1n(1)n1n.Mot thay doi nhotrong giatri dau11 lon len rat nhanh trong1n sau do./nh huong nay laxau v cac dai luong1n giam khi n tang.Mot phuong phap thuong dung de cai thien tinh on dinh la viet lai congthuc hoac thay doi thututinh toan. Ciasuta liet giatri xap xi11cua11voiNnao do, ta cothedanh giatich phan theo cong truy hoi nguoc:1n1 =1 1nnn = N. N 1. . . . . 2. (I.+)IILLL, viet tat cua Institute of Llectrical and Llectronics Lngineers,lamot hiep hoi thegioi cac chuyen gia kythuat.l.2. BlL| DlL! SOTnO!C M^Y Tl!H 5'ghien cuu su on dinh cua thuat toan giong nhu tren. 'eu11 = 11c th111=1 11N=1 11NcN= 111cN112= 112cN(N 1)...11= 11cN.Bang cong thuc (I.+) ta cothetinh xap xi 1n lat dau tu11voi Ndulon.1hat vay, tulat dang thuc0 < 1n0 vamoi Ji(i =2. 3. . . .) nhan mot trong cac giatri 0. . . . . 9.lhan.J1J2. . . duoc goi laphan dinh tri (mantissa),.J1J2. . . = J1 101J2 102. . .Cac tinh toan tren may tinh duoc thuc hien tren hethong so dau chamdong (floating point). Lay la hethong so dung mot so huu han cac con so dexap xi hethong sothuc (vohan). 1at cacac sothuoc hethong vois con sodung co soIO codang. = .J1J2. . . Jx 10e(I.o)o CHlO!C l. S^l SOV^SOHOC D^| CH^M DO!Ctrong dom _ e _ M.So khong la mot truong hop dac liet, noduoc viet nhula0.0 . . . 0 10n()Tb do 1.4. 'eus = 1, m = 1, M = 1, th tap hop cac sodau cham dong la0.1 101. 0.2 101. . . . . 0.9 101.0.1 100. 0.2 100. . . . . 0.9 100.0.1 101. 0.2 101. . . . . 0.9 101cung voi sokhong0.0 101. 'hu vay cotat ca55 con so Hnh I.I: lhan losodau cham dong voi = 4. s = 1. m = 1. M = 1.Do tap hop sodau cham dong, kyhieuJ,lahuu han nen mot sodaucham dong lieu dien (xap xi) nhieu sothuc. Khi somu e trong (I.5) lon honMth. khong theduoc lieu dien trong hethong dau cham dong nay. 'eutrong quatrnh tinh toan xuat hiensovoi e>Mth ta noi tinh toan daoverflow. Cac hedieu hanh khigap truong hop nay sedung lai viec tinhtoan. 1ruong hop khi e 12 ma tranHn cuc kyxau ngay cavoi sochinh xac kep'1heo mot ket quacua C. Szeg o ta couoc luong sauk(Hn) ~(_2 1)4(nC1)215{4_n- e3.5nTb do2.S. Cho A =_1 23 4_. 1m [A[, [A1[, cond(A).[A[ = max{[1[ [2[. [3[ [4[] = 7.2.+. SlCHl!H X^C 5I1heo cong thuc tinh ma tran nghich dao,A1=_ 2 13212_.nen[A1[ = max{[ 2[ [1[. [3,2[ [ 1,2[] = 3.\ay, cond(A) = [A[[A1[ = 21 Tb do2.6. Ma tran A =_1 11 1 105_ gan kydi vA1=_1 105105105105_.[A[ = 2, [A1[ = 2 105vacond(A) = 4 105.Linh lyZ.I khang dinh suton tai mot ma tran suy lien sai khac (tuongdoi) voi A khoang1,cond(A) = 2.5 106. Mac dukhong hoan toan gan A,ma tran don gian S =_1 11 1_ lakydi va [S A[[A[= 5 106Tb do 2.7. Cia su ta giai phuong trnh Ax = b tren mot may voi u = 51011vanhan duocz =_6.2341518.6243_. cond(A) = 1.0 104.Ciasu du lieu la chinh xac de cho [^A[,[A[ ~ 51010, tu (Z.ZJ), chan trencua sai sotuong doi la[^x[[x[~~ 104 5 1010= 5 106.5Z CHlO!C 2. HLIHlO!C TnI!H D^l SOT|YL! Tl!H'eu du lieu nhap co sai so, chang han, [^A[,[A[ ~ 106, [^b[,[b[ ~106, th chan tren cua sai sotuong doi la[^x[[x[~~ 104 2 106= 0.02.Lay [x[ ~ [z[ ~18.6 chan tren sai sotuyet doi la 0.37. \ vay phan tichnay cho.1 = 6.23 0.37. .2 = 18.62 0.37 2.S Cboong trnbMuc nay gioi thieu hai function viet lang ngon ngulap trnh Matlal. Sinhvien tunghien cuu vachay thucho cac lai tap.2.S.1 FactorMuc dich:lhan tich ma tran / lang cach dung phep khuCauss va danh giasodieu kien cua no. Factor duoc dung chung voiSolve degiai /*x=l.Loi sonhap:A - ma tranneq dong va cols cot can duoc phan tich.Loi soxuat:A - chua ma tran tam giac trenU trong phan tren cua novamot phienlan hoan vi cua ma tran tam giac duoi (I-L).'han tuhoa thoa (ma tranhoan vi)*A=L*U.flag - thong lao suthanh cong hay that lai. flag = 0 chi suthanhcong. 'euflag>0, mot phan tutrulang khong vadung tinh toan.pivots - lan ghi cac hoan vi dong. Lua vao pivots(neq)=(-1)^(so cuadong thay doi).Khiflag>0, dinh thuc cuaA lang O vakhi flag = 0, det(A)=pivots(neq)* A(1,1)** A(neq,neq).Loi soxuat tuy chon:Cond - khi flag>=0,mot danh giasodieu kien cuaA trong chuanvocung.Factor.m2.5. CHlO!C TnI!H 5Jfunction[A,flag,pivots,Cond]= Factor(A)[neq,cols]= size(A);flag = 0;pivots = zeros(neq,1);pivots(neq)= 1;if nargout == 4% InitializeCond for A thatis numericallysingular.Cond = realmax;% Computethe infinitynorm of A beforethe matrix is% overwrittenby its factorization.Anorm = norm(A,inf);endif neq == 1if A(1,1)== 0flag = 1;elseif nargout== 4Cond = 1;endreturnend% Gaussian eliminationwith partialpivoting.for k = 1:neq-1% Determinethe row m containingthe largestelement in% magnitudeto be used as a pivotand its magnitudebiggest.[biggest,occurred]= max(abs(A(k:neq,k)));m = occurred + k - 1;% If all possiblepivots are zero, A is numericallysingular.if biggest== 0flag = k;returnendpivots(k)= m;if m ~= k% Interchangethe currentrow k with the pivot row m.A([m k],k:neq)= A([k m],k:neq);pivots(neq)= - pivots(neq);end% Eliminatesubdiagonalentriesof column k.for i = k+1:neqt = A(i,k)/ A(k,k);A(i,k) = - t;5+ CHlO!C 2. HLIHlO!C TnI!H D^l SOT|YL! Tl!Hif t ~= 0A(i,k+1:neq)= A(i,k+1:neq)- t * A(k,k+1:neq);endendendif A(neq,neq)== 0flag = neq;returnendif nargout == 4% Estimatethe conditionnumber of A by computingthe infinity% norm of A directlyand a lower boundfor the norm of A^(-1).% A lowerbound for the normof A^(-1)is providedby the ratio% norm(y)/norm(d)for any vectorssuch that A*y = d and d ~= 0.% A "large"ratio is obtainedby computingy as one iterationof% inverseiterationfor the smallestsingularvalue of A, i.e.,% by solving for y such that(A*A)*y = e. This exploitsthe% fact that an LU decompositionof A can be used to solvethe% linear systemA*d = e as well as A*y = d. The entriesof e% are +1 or -1 with the signchosen during the computationof d% to increase the size of the entry of d and so make a "large"% lower bound for the norm of A^(-1) more likely.% Solve A*d = e using the decompositionof A.d = zeros(neq,1);d(1) = -1 / A(1,1);for k = 2:neqt = A(1:k-1,k)* d(1:k-1);if t < 0ek = -1;elseek = 1;endd(k) = -(ek + t) / A(k,k);endfor k = neq-1:-1:1d(k) = d(k) + A(k+1:neq,k)*d(k+1:neq);m = pivots(k);d([m k]) = d([k m]);end% Solve A*y = d.y = Solve(A,pivots,d);2.5. CHlO!C TnI!H 55% Computethe infinitynormsof the vectors.dnorm = norm(d,inf);ynorm = norm(y,inf);Cond = max(Anorm* ynorm / dnorm, 1);end2.S.2 SoIvcMuc dich:Ciai heneq phuong trnh tuyen tinh theoneq an lang cach dungphan tich LU nhan duoc lang cach goiFactor.Loi sonhap:A - output cuaFactor.pivots - output cuaFactor.b - vephai, vecto cododaineq.Loi soxuat:x - vecto nghiem cocung kich thuoc nhub.Solve.mfunctionx = Solve(A,pivots,b)neq = length(b);x = b;if neq == 1x(1) = x(1)/A(1,1);else% Forwardelimination.for k = 1:neq-1m = pivots(k);x([m k]) = x([k m]);x(k+1:neq)= x(k+1:neq)+ A(k+1:neq,k)*x(k);end% Back substitution.x(neq) = x(neq) / A(neq,neq);for i = neq-1:-1:1x(i) = (x(i)- A(i,i+1:neq)*x(i+1:neq))/ A(i,i);endend5o CHlO!C 2. HLIHlO!C TnI!H D^l SOT|YL! Tl!H2.6 Ma tran cocao troc dac bictHau het cac logiai (solver) hephuong trnh dai sotuyen tinh deu dua trenphep khuCauss voi phep xoay cuc lo. Khi ma tran A cotinh chat dac liet,ta cothegiam lot viec luu truvachi phi cho viec giai he.Khiquatrnhkhukhongcan den phepxoay cuc loth viec luu trugiam di cung nhuquatrnh tinh senhanh honrat nhieu. Cohai loai matran, noi chung, khong can den phep xoay cuc lo.Mot ma trannn A duocgoi latroi tren duong cheo (diagonally dominant), neu voi moi cot[}}[ _n

i6D}[i}[.Cothethay, voi ma tran nay ta khong can den phep xoay cuc lotrong quatrnh khuCauss. Loai ma tran con lai lama tran doi xung, A = AT.2.6.1 Ma tran bang'hac lai rang trong thuat toan khuCauss,vong lap trong cung cotheduoclodi khi nhan tu t = 0. Lieu nay phan anh su kien lien (tuong ung) khonghien dien trong phuong trnh vav vay khong can den phep khu. Khi matran A gan voi ma tran tam giac tren,viec kiem tra nhan tulang khongcothetiet kiem duoc luong tinh toan. Mot loai ma tran cuc kyquan trongtrong nhieu lanh vuc tinh toan lama tran lang. Ma tran A = ai}| duoc goilama tran lang khi moi phan tukhac khong nam trong mot dai doc theoduong cheo chinh. Cu the, khiai} = 0 neui > mI va i > mu, ma tranduoc goi lacochieu rong lang duoi mI,chieu rong lang trenmu,vachieurong langmImu1. Mot thi ducua ma tran voimI = 2, mu = 1 la______+ + 0 0 0+ + + 0 0+ + + + 00 + + + +0 0 + + +__day + chi phan tucothekhac khong. Khi tien hanh phep khutren matran nhu vay, toi damIphan tuphai duoc khuomoi luoc. Xemxet cac2.o. M^ Tn^! COC^| Tn|C D^C BlLT 57phan tutren duong cheo chung torang nhieu phan tukhong van giulangkhong. 1hat vay, phepxoay cuc losedelai cac sokhongtronga(k)i}voi i > mImu. \ voi cac nhan tu khong, ta cothetang toc qua trnh tinhtoan lang cach nhan dien cac phan tukhong van con lang khong. Quansat quan trong khac la, lang cach dung so doluu trudac liet, khong can luutrucac phan tu a(k)i}voii >mI, hay i >mImu. Macai dat phepkhuCauss dac liet cho cac ma tran lang cothetm thay trong LI'l/CKcung nhu L/l/CK. Mac ducokhokhan trong viec cai dat A theo cach luutrudac liet, nhung sutiet kiem cotherat lon. Cac ket qua solagiong nhau,nhung viec luu truodang lang xap xi n(2mImu) thay vn2. Khoi luongtinh toan vao khoangnmI(mImu) thay vn3,3,vacosuthuan loi tuongtutrong phep thetien valui.Bay giota xet mot dang khac cua phep khuCauss thuan tien cho cachcai dat ma tran lang. Ciasuphep phan tich A = t ton tai. 1ruoc het chuyranga11 =n

nD11nun1 = 11u11v cac ma tran latam giac. Chon11 ,= 0 thu11 = a11,11.\oii > 1,ai1 =n

nD1inun1 = i1u11.v vayi1 = ai1,u11voii = 2. . . . . n.Cung vay, voi> 1,a1} =n

nD11nun} = 11u1}.5o CHlO!C 2. HLIHlO!C TnI!H D^l SOT|YL! Tl!Hvay,u1} = a1},11voi = 2. . . . . n.'oi chung, moi lan ta lap mot cot cua t vamot dong cua . Ciasuta datinh duoc cac cot1. . . . . k 1 cua t vacac dong1. . . . . k 1 cua . 1hakk =n

nD1knunk = kkukkk1

nD1knunk.Cac sohang trong tong cuoi cung ladaliet. Chonkk, thukk =_akkk1

nD1knunk_,kk.\oii > k,aik =n

nD1inunk = ikukkk1

nD1inunk.v vayik =_aikk1

nD1inunk_,ukkvoii = k 1. . . . . n.\oi> k,ak} =n

nD1knun} = kkuk} k1

nD1knun}.vay,uk} =_ak} k1

nD1knun}_,kkvoi = k 1. . . . . n.2.o. M^ Tn^! COC^| Tn|C D^C BlLT 59'eu tat cacac phan tutren duong cheo chinh cua t lay lang I,thuat toannay laphep khuCauss khong dung phep xoay cuc lo. 1rong lan luan cuachung ta vephep khuap dung cho ma tran lang A, ta thay nhieu cong viecva vung luu tru cotheduoc tiet kiem. 'eu A lama tran lang voi chieu ronglang duoi mIvachieu rong lang trenmu,th t cung lama tran lang voichieu rong lang duoimI va lama tran lang voi chieu rong lang trenmu.'eu ta chon cac phan tucheo cua t lang I, th khong can luu truchung vagiong nhu truong hop ma tran day du, nhan tu t va co the duoc viet chonglen A khi chung duoc tinh toan.2.6.2 Ma tran ba doong cbcoKhi mI =mu =1 ma tran cac he soduoc goilama tran la duong cheo(tridiagonal matrix). \iec giai sophuong trnh dao ham rieng thuong dan veviec giai cac hephuong trnh gom mot so rat lon cac an, co thelen den hangngan. Khi ay viec dung lactorSolve lakhong thich hop, nhung voi thuattoan comuc dich dac liet th gap khokhan nay. Ciasuhela duong cheoduoc viet duoi dang_______a1c1b2a2c20.........0 bn1an1cn1ancn_________.1.2....n1.n__=_______J1J2...Jn1Jn__.Khi khong dung phep xoay cuc lo, khucacbita duoc_______1c12c20.........0 n1cn1ncn_________.1.2....n1.n__=_______e1e2...en1en__.'hu ta thay caccikhong thay doi. Bay giothiet lap cong thuc choi, ei,truoc het ta thay1 =a1, e1 =J1. Lekhu b2,khong dung phep xoay cucoO CHlO!C 2. HLIHlO!C TnI!H D^l SOT|YL! Tl!Hlo, nhan tu m2 = b2,1, suy ra:2= a2 m2c1.c2= c2 m2 0 = c2.e2= J2 m2J1.Lehoan tat viec thiet lap ta dung quy nap. Ciasurang ta dathiet lap duociva ei den dongk. 1h ta co0 kck0 [ ek0 bkC1akC1ckC1[ JkC1tren dongk va k 1. Rorang nhan tula mkC1 =bkC1,k. lhep khudongk 1 cho:kC1= akC1 mkC1ck.ckC1= ckC1 mkC1 0 = ckC1.ekC1= JkC1 mkC1Jk.\iec luu trucotheduoc tochuc cuc kyhieu qua. Mot ma tran tongquat capn can luu tru n2phan tu, nhung mot ma tran tam giac tren chi canluu tru 3n 2 phan tukhac khong. Mot so dotunhien laluu trula dai ak,bkva cknhu lala vecto chieu da n. 1a cothevietmk lenbkva k lenakkhi chung duoc tinh; theo cac luoc thetien va luiek va.k co theviet chonglenJkdecho chi can them mot vecto chieu dai n. 1huat toan tren khongdung phep xoay cuc lonen ket qua socotherat xau.\oi hela duong cheocomot dieu kien don gian,thuong thoa man trong thuc hanh,lao dam ketquathu duoc latot. 1ruoc het chuyrang neu lat ky ck haybk triet tieu, hecothele thanh cac henhohon macung lala duong cheo. Suy ra, ta cothegiasu ck va bk khac khong voi moik. Ciathiet then chot la[a1[ > [b2[.[ak[ _ [bkC1[ [ckC1[. k = 2. . . . . n 1.[an[ > [cn1[.Lieu kien nay manh hon dieu kien troi tren duong cheo, cho phep chung toma tran khong suy lien.2.6.3 Ma tran doi xong'eu ma tran A cothephan tich thanh T voi lama tran tam giac tren,th A lama tran doi xung, xac dinh duong. 'guoc lai, mot ma tran doi xung,2.. C^C IHlO!C IH^I l^I oIxac dinh duong A cothe phan tich thanh tich T voi lama tran tam giactren khong suy lien. Bang thutuc trnh lay otren ta cothexac dinh . 1aphai cotT= nena11 = 11u11 = u211.suy rau11 =_a11, vanhu truocu1} = ai},u11 = 2. . . . . n.Bay gioakk =n

nD1knunk =n

nD1u2nk. (Z.Z+)tuday suy raukk =_akkk1

nD1u2nk_1{2.Roi, cung nhu truoc,uk} =_ak} k1

nD1unkun}_,ukk. = k 1. . . . . n.1u(Z.Z+) ta thayakk _ u2nk:suy ra[unk[ __akkvoi moim _ k, voi moik. Lieu nay noi rang cac nhan tukhong thelon doivoi A. lhep phan tich nay goi laphuong phap Cholesky hay phep phan tichcan lac hai. 'olao vetot hon cau truc lang cua ma tran.2.7 Cac pboong pbap Iap1rong muc nay ta xet hai phuong phap lap |acoli va Causs-Seidel cho phuongtrnh Ax = b. \iet lai phuong trnh duoi dangMx = b (M A)x.oZ CHlO!C 2. HLIHlO!C TnI!H D^l SOT|YL! Tl!Htrong doM lama tran gan A; tinh day nghiem xap xi x(k)loiMx(kC1)= b (M A)x(k).lhep lap |acoli codang nay voi M lama tran cheo voi duong cheo chinh laduong cheo chinh cua ma tran A. 1uong tu,phep lap Causs-Seidel ung voitruong hop M lama tran tam giac duoi gom duong cheo chinh cua M vacacphan tulen duoi duong cheo chinh cua no. Rorang rat degiai cac phuongtrnh odang nay.Qua gioi han ca hai ve cua phuong trnh xac dinh phep lap, ta thay neuxap xi hoi tuth chung phai hoi tuvex. Sai soc(k)= x x(k)thoac(kC1)= M1(M A)c(k).suy ra[c(kC1)[ _ [M1(M A)[[c(k)[.'euj = [M1(M A)[0,ton tai da thuc1(.) sao cho [ 1[> options= optimset(Display,iter);>> x=fzero(f,2,options)Ket quatraveIJO CHlO!C +. !CHlLM IHlO!C TnI!H IHl T|YL!Search for an intervalaround 2 containinga sign change:Func-count a f(a) b f(b) Procedure1 2 2 2 2 initial interval3 1.94343 1.77693 2.05657 2.22947 search5 1.92 1.6864 2.08 2.3264 search7 1.88686 1.56025 2.11314 2.46535 search9 1.84 1.3856 2.16 2.6656 search11 1.77373 1.1461 2.22627 2.9563 search13 1.68 0.8224 2.32 3.3824 search15 1.54745 0.394607 2.45255 4.01499 search16 1.36 -0.1504 2.45255 4.01499 searchSearch for a zero in the interval[1.36, 2.45255]:Func-count x f(x) Procedure16 1.36 -0.1504 initial17 1.39945 -0.0415434 interpolation18 1.41435 0.000384399 interpolation19 1.41421 -2.01697e-006 interpolation20 1.41421 -9.69058e-011 interpolation21 1.41421 4.44089e-016 interpolation22 1.41421 4.44089e-016 interpolationZero found in the interval [1.36,2.45255]x =1.41424.6 Hcpboong trnb pbi toycnMot lai toan thuong xuat hien trong toan hoc tinh toan latm mot vai hoactat ca cac nghiem cua mot he gom n phuong trnh phi tuyen voi n an. 'hunglai toan nhu vay tong quat vakhohon nhieu so voi lai toan mot phuongtrnh mot an so.Co the thay ngay phuong phap chia doi khong ap dung duoc(morong duoc) cho truong hop nay. lhuong phap cat tuyen cothetong quathoa cho truong hop nay nhung cach lam khong hien nhienv suphuc taphnh hoc khi sochieu lon. \oi phuong phap 'ewton th khac, sutong quathoa ra truong hopn phuong trnhn an sorat tunhien varat . . . dep. Ledongian viec trnh lay, xet hegom hai phuong trnh theo hai an:(.. ,) = 0g(.. ,) = 0(+.Io)+.o. HLIHlO!C TnI!H IHl T|YL! IJI\iet duoi dang vectob(w) = 0.trong dow =_.,_. b =_g_.1uongtu nhutrongtruonghopI-chieu(phuongphap'ewton), khai trienham. g nhocong thuc 1aylor chi giulai cac sohang lac nhat(.0. ,0) dd.(.0. ,0)(. .0) dd,(.0. ,0)(, ,0) = 0g(.0. ,0) dgd.(.0. ,0)(. .0) dgd,(.0. ,0)(, ,0) = 0Xap xi ketiep(.1. ,1) = (.1^.1. ,1^,1) cothetm lang cach giai hephuong trnhdd.(.0. ,0)^.1dd,(.0. ,0)^,1=(.0. ,0)dgd.(.0. ,0)^.1dgd,(.0. ,0)^,1=g(.0. ,0)xac dinh(^.1. ^,1).'eu kyhieu)(wk) =_J(Jx(.k. ,k)J(J,(.k. ,k)JJx(.k. ,k)JJ,(.k. ,k)_lama .ran ja.'| cua hephuong trnh (+.Io). 1h phuong phap 'ewton cho hehai phuong trnh hai an la)(wk)^wk = b(wk). (+.I9)IJZ CHlO!C +. !CHlLM IHlO!C TnI!H IHl T|YL!(.0. ,0) 'ghiem(.. ,) Solan lap(1.2. 2.5) (1.3364. 1.7542) 4(2.0. 2.5) (0.9013. 2.0866) 9(1.2. 2.5) (0.9013. 2.0866) 4(2.0. 2.5) (3.0016. 0.1481) 19Bang +.I: Ket quasothi du+.+Tb do4.4. Ciai hephuong trnh.2.,3 9 = 03.2, ,3 4 = 0(+.ZO)\dd.= 2. ,3.dd,= 3.,2.dgd.= 6.,.dgd,= 3.2 3,2.heduoc giai tai moi luoc lap la_2.k,3k3.k,2k6.k,k3.2k 3,2k_ _^.k^,k_= _.2k .k,3k 93.2k,k ,3k 4_.Bang +.+ cho ket qua so voi cac diem khoi dau (.0. ,0) khac nhau.1rongtat cacac truong hop phep lap dung khimax_[b[.[^w[[w[__ 106.Ket quatinh toan chung torang hecoit nhat la nghiem, moi nghiem duoctm thay phuthuoc vao diem khoi dau(.0. ,0).+.o. HLIHlO!C TnI!H IHl T|YL! IJJCungnhu phuongphap 'ewtoncho ham mot lien, cothe chung torang neu b hai lan khavi gan nghiem a cua b(w) = 0,neu ma tran |acolitai a, )(a), khong suy lien,vaneu w0 dugan a, th phuong phap 'ewton sehoi tuvea vasuhoi tulacap hai.Mot khokhan trong thuc hanh latm diem khoi dau dugan dequatrnh lap dan toi nghiem can tm. Suhieu liet velai toan vasumorongnghiem cotherat huu ich o day. Mot cach tiep can tong quat la lien heviectm nghiem w cua b(w) = 0 voi sucuc tieu hoa thang du1(w) = [b(w)[2=(w)|2 g(w)|2. Rorang ham naycocuc tieula 0 tai moi nghiemcuab(w) = 0. Ytuong lachuyden sulien thien^wktinh tuphuong phap'ewton nhu viec cho huong theo dota tm mot giatri z sao cho phep qualapwkC1 = wkz^wkcho giatri nhocua thang du:1(wkC1) < 1(wk).Lieu nay luon luon thuc hien duoc v cho den khi nhan duoc nghiem,_ddz1(wkz^wk)_2D0= 21(wk) < 0.Co nhieu chi tiet thuc hanh phai duoc vach ra.Chang han, khong nhat thiet,hay ngay ca mong muon, tm giatri cuaz cuc tieu hoa thang du. Cac phuongphapthuoc loainayduoc goilacac phuongphap'ewtonchung(damped'ewton methods). Mot cai dat can than phuong phap nay sephat sinh daylap hoi tukhi maphuong phap 'ewton that lai. 1rong truong hop cahaiphuong phap deu hoi tu th hieu quacua chung lanhu nhau.Cao boi vabai tap4.1. 1hang du cua mot nghiem dua rarcuaJ(.) la J(r). 1a thuong thayphat lieu: thang du lanho nen nghiem phai tot. Lieu nay coxac thuckhong! \ai trocua viec. \iec dinh ti ledong vai trog!4.2. lhan liet nghiem don vanghiem loi lang dothi nhu thenao! Ciaithich lang dothi cach xac dinh tot cac nghiem. So sanh voi lai tap +.I.IJ+ CHlO!C +. !CHlLM IHlO!C TnI!H IHl T|YL!Hnh +.o: Bai tap +.J.4.3. Lanh gialang hnh hoc nghiem cua hamJ(.) ma do thi duoc cho lenduoi.(a) \oi khoang chua nghiem lan dau0.0. 1.0| la khoang chua nghiemketiep lanhung khoang nao!(l) 'eu.1 = 0.0 va .2 = 1.0, danh dau tren dothi vi tri xap xi cua.3lang cach dung mot luoc cua phuong phap cat tuyen.(c) 'eu.1 =0.5,danh dau tren dothi vi tri xap xi cua.2 va .3 langcach dung hai luoc cua phuong phap 'ewton.4.4. La thuc(.) = .3 2. 5 comot nghiem trong2. 3|.(a) Chung torang2. 3| lakhoang chua nghiem cua(.).(l) /p dung lon luoc cua phuong phap chia doi degiam khoang chuanghiem xuong con1,16.(c) 1inh.3 va .4lang phuong phap cat tuyen lat dau voi .1 =3 va.2 = 2.(d) 1inh.2, .3, va .4 dung phuong phap 'ewton voi.1 = 2.4.S. Letm noi sin . = .,2 voi.> 0,(a) 1m mot khoang thich hop chua nghiem cua mot ham phu hop (.).(l) /p dung lon luoc cua phuong phap chia doi degiam khoang chuanghiem xuong con1,16.(c) 1inh.3 va.4 lang phuong phap cat tuyen lat dau voi.1 va.2 laylang cac giatri cua khoang.(d) 1inh.2, .3,va .4 dung phuong phap 'ewton voi .1ladiem giuacua khoang.4.6. Colao nhieu phep danh giaham trong phuong phap chia doi!4.7. 1rong chung minh hoi tucua phuong phap cat tuyen, phat lieu rang,neuc = max{c0. c1] < 1 th lat dang thucciC1 _ cici1+.o. HLIHlO!C TnI!H IHl T|YL! IJ5am chici _ ci.vai =1_5___1 _52_iC1_1 _52_iC1__.Hay thiet lap dieu nay.4.8. \iet maMatlal cho phuong phap 'ewton.4.9. Ham dac lieterf(.) =2__x0et2Jt.goi la|am sa| (error function),duoc dung trong lythuyet xac suat vanhieulanh vuc cua khoa hoc vakythuat.\ ham duoi dau tich phan laduong voimoi t ,ham tang ngat vav vay coham nguoc. = erf1(,). Ham sai nguoclamot ham quan trong do khanang ap dung cua novacothedanh giavoi,cho truoc lang cach giai phuong trnh erf(.) =,. lunctionerfinv cuaMatlal cho ham sai nguoc. Hay dung phuong phap 'ewton detm ham sainguoc, so sanh voi ket quatinh nho erfinv.4.10. 1m cap hoi tucua cac day sau:a).n =_1,n; l).n =n_n; c).n =_1 1,n; .nC1 = arctg.n.4.11. Chung tohamJ(.) =4.(1 .) anh xadoan0. 1| vao chinh novakhong co. Chung torang nocodiem lat dong. 1ai sao dieu nay khong mauthuan voi dinh lyanh xaco!4.12. Chung tohamJ(.) =2 . arctg. cotinh chat [J0(.)[ < 1 nhungnokhong codiem lat dong. Ciai thich lydo dieu nay khong mau thuan voidinh lyanh xaco.4.13. Chung tophuong phap tinh_1 duoi day hoi tucap la:.nC1 =.n(.2n31)3.2n1.IJo CHlO!C +. !CHlLM IHlO!C TnI!H IHl T|YL!4.14. Dung phuong phap lap 'ewton tm mot nghiem (voi sai sotuyet doitoi da la 104) gan diem(0.5. 1.0. 0.0) cua hephi tuyen2.2 . ,2 : = 0.32.2 ,2 20: = 0.,2 14.: = 0.4.1S. 1m la tham so . va ;trong mohnh(.) = ex;.lang phep noi suy la diem(1. 10). (2. 12) va (3. 18).Dung phep lap 'ewtontm la tham sovoi la chusoconghIa.Cboong STcb pban soXap xi _bo(.)J. lang soduoc goi latich phan sohay cau phuong. Hau hetchuong nay lien quan den khoang huu hana. b|, nhung comot vai lan luanvetich phan voi a vahayblavohan. 1hinh thoang dua vao ham trongluongn(.) > 0 lahuu ich vanhu vay xap xi tich phan dang _bo(.)n(.)J..Conhieu lydo nghien cuu tich phan so. 'guyen ham (antiderivative) cuaco thekhong liet hay khong la ham so cap.1ich phan co the khong co hieuluc v ham duoc xac dinh loi cac giatri trong lang hay loi mot chuongtrnh con. Hay, cac tich phan xac dinh phai duoc xap xi nhu thanh phan cuaso dotinh toan phuc tap hon, chang han nhu giai cac phuong trnh vi phanlang phan tuhuu han nhocac phuong phap lien phan hay Calerkin.Mot nguyen lyco lan trong giai tich so la neu ta khong thelam dieu tamuon voi mot ham(.) cho truoc, ta xap xi nolang mot ham mavoi notaco the thuc hien duoc.1huong ham xap xi la mot da thuc noi suy.Bang cachdung nguyen lynay ta sethiet lap mot vai quy tac cau phuong va nghien cuusai socua chung. Khi xap xi ham ta thay rang da thuc noi suy tung manhtien loi hon da thuc noi suy, o day dieu nay cung dung. Cach noi suy da thuctung manh latunhien cho cau phuong v dung ham nhu vay chung qui lalekhoang lay tich phan thanh cac manh vaxap xi lang da thuc tren moimanh ay. Ytuong then chot trong cau phuong laphai tinh den dang dieucua(.) khi chia tach khoang. lhep cau phuong thich ung nay duoc motatrong muc 5.Z va maduoc lan luan trong muc tiep theo. lhep cau phuongthich ung lavan dechinh cua chuong,nhung vai chuyduoc cho cho tichphan cua lang dulieu vacho tich phan cua cac ham hai lien. Chuydacliet duoc danh cho cac lai toan chuan li cho loi giai cohieu quacua chunglang cac macua loai phat trien oday.IJ7IJo CHlO!C 5. TlCH IH^! SOS.1 Cac qoy tac cao pboong co banLexap xi_bo(.)n(.)J. (5.I)giasudaliet giatri cua taiNdiem phan liet.1. .2. . . . . .1. Coi 11(.)lada thuc noi suy tai cac diem nay. Dang Lagrange cua11(.) dedangdan den xap xi_bo(.)n(.)J. ~_bo11(.)n(.)J. =_bo1

iD1(.i)1i(.)n(.)J. =1

iD1(.i)_bo1i(.)n(.)J. =1

iD1i(.i). (5.Z)day giathiet cac trong luongiton tai. Lieu nay tuong duong voi sutontai cac tich phan_bo.}n(.)J. voi = 0. 1. .... N 1.1rong truong hopn(.) =1, a va b huu han, th giathiet nay ladung. 1uynhien, neukhoangla vo han(e.g., _10(.)J.), tiepcantrenthatlaivkhong co .}nao cotich phan tren khoang nay.Khokhan co lan cua viec tiep can, trong truong hop _10(.)J., lanoduoc dat co sotren suxap xi(.) loi da thuc, macac da thuc khong cotichphan huu han tren khoang vohan. \ tich phan(.) ton tai, nophai dantoi khong that nhanh khi . o. Mot loi khuyen huu ich laphai colapdang dieu khac da thuc vao ham trong luong. Chang han, neu ta dua vaoham trong luongn(.) = exva dinh nghIaJ(.) = (.)ex, tich phan co theviet lai nhu la _10J(.)exJ.. Khong phuc tap lam denhan duoc cong thuccho cac tich phan dang _10J(.)exJ. v cac tich phan _10.}exJ. ton taivoi moi . Loi khuyen nay cocho mot xap xi tot _10(.)J. hay khong lavan de J(.) coung xugiong mot da thuc hon(.) hay khong.5.l. C^C _|Y T^C C^| IHlO!C CO B^! IJ91ich phan tren khoang vohan laloai lai toan chua nhieu khokhan.1ich phan voi ham duoi dau tich phan cokydi cung chua dung khokhanv chung ung xukhong giong da thuc. 1hong thuong viec dung ham trongluong lacach tot dedoi xuvoi nhung lai toan nhu vay. Chang han,trongloi giai cac lai toan thevi phang lang phuong phap phan tulien,can xapxi cac tich phan thuoc dang_10J(.) ln .J.(vasau dogiaihecac phuong trnhtuyen tinh decoduoc nghiemsochophuong trnh tich phan cua lythuyet thevi). Ham ln . cotheduoc xem nhuham trong luong v nokhong duong tren khoang(0. 1) vacac tich phan_10.}ln(.)J.tontai voimoi (hamtrongluongn(.)trong(5.I) co the laylaln .).1uong tucach lam trong thi dulay tich phan tren khoang vohan, neu tamuon tinh _10(.)J.va (.) ung xugiong ln.khi . 0,ta cotheduavao ln . nhu ham trong luong vavietJ(.) =(.), ln(.). Lng xugiongkhi . 0 conghIa lalimx!0(.)ln(.)= c.1uday vesau dieu nay seduoc viet la (.) -c ln(.). \J(.) comot gioihan huu han tai . =0,noduoc xap xi loi da thuc tot hon(.),malavohan odo.Mot cong thuc dang1

iD1i(.i) (5.J)dexap xi (5.I) duoc goi lacong thuc hay quy tac cau phuong. So dodephatsinh cac quy tac vua mota dan den cac quy tac cau phuong noi suy. Mot quyI+O CHlO!C 5. TlCH IH^! SOtac nhu vay setich phan chinh xac da thuc lat kycolac nhohonN. Lolav neu(.) lada thuc lac nhohonN, th loi tinh duy nhat cua da thuc noisuy, 11(.) (.),vaquy tac duoc xay dung detich phan11(.) lachinhxac.|nb ngbia S.1. Sai so tuyet doi cua cong thuc cau phuong dang (5.J) la luong1( ) =_bo(.)n(.)J. 1

iD1i(.i). (5.+)1anoi congthuccauphuongdang(5.J) co 'a..|/n|xa.(degreeofprecision)J _ 0 neu:(i)1(.}) = 0, = 0. 1. . . . . J, va(ii) 1(.dC1) ,= 0.Sau nay ta sethay, mot suchon lua dung dan cac diem noi suy.ikhixay dung (5.Z) dan den cong thuc voi lac chinh xac lon honN 1. 1huongth cac.ithuoca. b|, nhung thuc ra khong nhat thiet phai nhu vay. Changhan, cong thuc /dams cho loi giai phuong trnh vi phan dua tren quy tac cauphuong dung cac diem nut,ngoai truhai diem cuoi a va b,nam len ngoaikhoang nay. Lieu nay cung dung voi phuong phap tich phan theo lang dulieu seduoc decap den sau nay.Linhly duoi day phattrienmot vaichantrensai so cuacongthucvoi lac chinh xacJ. 'oduoc phat lieu lang cach dung kyhieu [ [ chomaximum tren a. b| cua [(.)[.Cung vay, nhu trong chuong J, Mq duoc dungdechi [(q)[.|nb Iy S.1. !eu .n .|u. .au p|u+n (5.2) .'a. .|/n| xa.J, .|| t+| 'a. 'y1a.|u.(.) 'a.q _ J,[1( )[ _ [ [__bon(.)J. 1

iD1[i[_. (5.5)!eu m| i> 0, .||[1( )[ _ 2[ [_bon(.)J.. (5.o)5.l. C^C _|Y T^C C^| IHlO!C CO B^! I+IC|un m|n|. \oi(.) lada thuc lat kylacq _ J,[1( )[ __bo(.)n(.)J. _bo((.) (.))n(.)J.1

iD1i(.i) 1

iD1i((.i) (.i))_ [1()[ _bo[(.) (.)[n(.)J. 1

iD1[i[[(.i) (.i)[_ [ [__bon(.)J. 1

iD1[i[_.trong dota dadung1() = 0. Lay la(5.5). Khi moii> 0 th dau tri tuyetdoi trong (5.5) cothelo. \ cong thuc cau phuong chinh xac khi (.) 1nen1

iD1i1 =_bon(.)1J..vata co(5.o).Hcqoa 1. !eu(.) . J 1 1a |am '|en .u. .rena. b|, .||[1( )[ __b a2_dC1MdC1(J 1)__bon(.)J. 1

iD1[i[_. (5.7)!eu m| i> 0, .||[1( )[ __b a2_dC1MdC1(J 1)2_bon(.)J.. (5.o)I+Z CHlO!C 5. TlCH IH^! SOC|un m|n|. \ chan cua dinh ly 5.I dung voi moi da thuc (.) co lac q _ J,ta cothedung(.) lada thuc 1aylor khai trien(.) tai .0 =(a b),2,n = q:(.) = (.0) . .010(.0) (. .0)qq(q)(.0)va1qC1(.) =(. .0)qC1(q 1)(qC1)(:)voi: nam giua.0 va .. Lieu nay am chi[ [ =maxoxb(. .0)qC1(q 1)(qC1)(:)__b a2_qC1MqC1(q 1). (5.9)1hay dieu nay voiq = Jvao (5.5) hay (5.o) ta nhan duoc (5.7) hay (5.o).Nban xct S.1. Khi nghien cuu noi suy da thuc, ta da liet cac noi suy lac caocokhanang dao dong vagay ra nhung thay doi khong phuhop. 1nh hnhlay gioth khac v nolaxap xi dien tich len duoi duong cong va duong nhucac dao dong seli trung lnh hoa. Lieu nay quan trong doi voi truong hopdac liet cua cong thuc khi machan sai socua dinh ly5.I ladung, vatat cai>0. Lang tiec, cac cong thuc cau phuong noi suy dua tren {.i] cach deutronga. b|, goi la cac cong thuc cau phuong !eu.n - C.es, co mot vai i laygiatri am ngay ca voi cac lac chinh xac vua phai. Ket qua cua cac cong thucnay cothekhong hoi tutoi giatri cua tich phan khi lac gia tang.Nban xct S.2. 1rong cac chan (5.5),(5.o) ta cothelay da thuc(.) lat kyvoi lacq _J. \oi a. b huu han,ton tai da thuc

(.) voi lac _Jgannhat theo nghIa[

[ =mindeg ]q[ [.Khi do,[1( )[ _ [

[__bon(.)J. 1

iD1[i[_. (5.IO)5.l. C^C _|Y T^C C^| IHlO!C CO B^! I+J'eu moii> 0, th[1( )[ _ 2[

[_bon(.)J.. (5.II)Cac lat dang thuc (5.IO),(5.II) cho chan sai sodua tren xap xi tot nhat cua(.); chung huu ich khi ham lay tich phan khong dutron.Mot phan tich sai sochi tiet hon chung to rang sai so 1( ) cotheduoclieu dien nhu la1( ) = c_b a2_dC2(dC1)() (5.IZ)voic R va (a. b). 'eu mot cong thuc cau phuong colac chinh xacJ,th1(.}) = 0. = 0. 1. . . . . J (5.IJ)1(.dC1) ,= 0. (5.I+)'eu ta giasurang sai socodang (5.IZ), dedang tmc tu1(.dC1) = c_b a2_dC2(J 1). (5.I5)Cac phuong trnh (5.IJ), (5.I+) cung cap cach khac desinh ra cac quy tac cauphuong. Cach tiep can duoc liet nhu lap|u+np|ap|es'a. 1(n|. 1rongcach tiep can nay cac heso iduoc xem nhu cac an duoc tm lang suthoaman hephuong trnh tuyen tinh (5.IJ voi Jlon cothe. 1ruoc khi cho cacthi du,ta chuyrang, nen ap dung phuong phap hesolat dinh cho khoangchuan1. 1| varoi lien doi thanh khoang tong quata. b| lang mot phepdoi lien don gian. 'eu ta co_11(.)J. =1

iD1i(.i) c(dC1)().I++ CHlO!C 5. TlCH IH^! SOdatt =b a2. a b2.1hJt = (b a)J.,2 va_bo(t )Jt =b a2_11_b a2(. 1) a_J.=b a21

iD1i_b a2.i a b2_b a21( ).\JJ.=JtJ.JJt=b a2JJt=JdC1J.dC1 =_b a2_dC1JdC1JtdC1.nen phep doi lien cho_bo(t )Jt =1

iD1_b a2i__b a2.i a b2__b a2_dC2(dC1)().Tb doS.1. 1m cong thuc cau phuong dang_11(.)J. = 1(1) 2(1) 1( ).1heo phuong phap hesolat dinh(.) = 1 =2 = 12.(.) = . =0 = 12.Suy ra: 1 =2 =1. 1a cung thay rang,lang cach xay dung, J _1. 1h(.) = .2nhan duoc23= 121(.2) =1(.2) = 43.5.l. C^C _|Y T^C C^| IHlO!C CO B^! I+5Hnh 5.I: Quy tac hnh thang.\1(.2) ,= 0 dieu nay noi rangJ = 1 va c = 1(.2),2 = 2,3, nghIa la_11(.)J. = (1) (1) 2300()voi (1. 1) .\oi khoanga. b| tong quat,ap dungcong thuc doi lienta co(congthuc trong dau |)_uy .a. ||n| .|an (trapezoid rule)_bo(.)J. =b a2((a) (b)) (b a)31200(). (5.Io)trong do (a. b).Tb doS.2. 1m cong thuc chinh xac nhat dang_11(.)J. = 1(1) 2(0) 3(1) 1( ).1heo phuong phap hesolat dinh(.) = 1 =2 = 123.(.) = . =0 = 13.(.) = .2=2,3 = 13.I+o CHlO!C 5. TlCH IH^! SOSuy ra: 1 = 3 = 1,3, 2 = 4,3.Letm lac chinh xac ta kiem1( ) ,=0 voi (.) lada thuc lac caohon Z. 'eu(.) = .3ta duoc0 = 131(.3) =1(.3) = 0:nghIa la, quy tac colac chinh xac lon hon Z. 'eu lay(.) = .4ta duoc25= 131(.4) =1(.4) = 415=c = 190.'hu vay, lac chinh xacJ = 3, va_11(.)J. =13(1) 43(0) 13(1) 29(4)()voi (1. 1) . Cung dung cong thuc doi lien ta suy ra cong thuc tong quat._uy .a. S|mpsn_bo(.)J. =b a6_(a) 4_a b2_(b)_(b a)52880(4)(). (5.I7)trong do (a. b).Cahai cong thuc thuoc lop .n .|u. !eu.n-C.es v cac nut cach deutronga. b|. 1hutuc thiet lap cong thuc lao gom viec chon truoc cac nut.ivaroi giai hephuong trnh tuyen tinh xac dinh cac trong luongi. 'hungneu cac.i chua liet (duoc phep chon)!Soan can tm segap doi, 2N, ocachsap xep cua ta, cothehy vong tm cac cong thuc voi lac chinh xac cao honnhieu, nhu sethay duoi day, co cong thuc voi lac chinh xac 2N 1 madungchi Ngiatri cua . Lang tiec,hephuong trnh choiva .ilaphi tuyen.Khong hien nhien hedoconghiem thuc, maneu co, lam thenao denhanduoc chung. Causs dagiai quyet van demot cach thanh lich voi Ntongquat,ngay cavoi cac ham trong luong tong quat hon vacac khoang lavohan. Ket quaduoc liet nhu la .n .|u. .au p|u+n Causs.Mot sotruong hopdac liet cothechi ra theo cach so cap.5.l. C^C _|Y T^C C^| IHlO!C CO B^! I+7Tb doS.3. ChoN = 1 cong thuc Causs codang_11(.)J. = 1(.1) 1( ).Bang phuong phap hesolat dinh(.) = 1 =2 = 1.(.) = 0 =0 = 1.1.suy ra1 = 2 va .1 = 0. Lexac dinh sai so, ta thu(.) = .2=23= 2 0 1(.2).vathayJ = 1, c = 1,3, va_11(.)J. = 2(0) 1300().1rena. b| cong thuc nay trothanh_bo(.)J. =_(b a)_a b2__(b a)32400(). (5.Io)Cong thuc nay duoc liet nhu laquy tac diem giua Tb doS.4. ChoN = 3 cong thuc Causs codang_11(.)J. = 1(.1) 2(.2) 3(.3) 1( ).Do tinh doi xung cua khoang1. 1|,cothecho rang1 =3, .2 =0,va.1 = .3, nghIa la_11(.)J. = 1(.1) 2(0) 1(.1) 1( ).I+o CHlO!C 5. TlCH IH^! SOBang phuong phap hesolat dinh,(.) = 1 =2 = 212.(.) = . =0 = 1.11(.1) (tudong thoa).(.) = .2=2,3 = 21.21.(.) = .3=0 = 1.31 1(.31) (tudong thoa).(.) = .4=2,5 = 21.51.Ciai henay, ta duoc: 1 = 5,9, 2 = 8,9, .1 = _3,5 = .3.Letm sai so, thu(.) = .5=0 = 1.51 1(.51) 1(.5) =1(.5) = 0'hu vay, lac chinh xac lon hon4. 1iep tuc voi(.) = .6, ta co:2,7 = 21.61 1(.6) =6251(.6).suy ra: J = 5, c = 1,15750, va_11(.)J. =19_5__35_8(0) 5__35__115750(6)().1rena. b|, kyhieu.0 = (a b),2, h = (b a),2, ket quala.n .|u..au p|u+n Causs -1|em_bo(.)J. =h9_5_.0 h_35_8(.0) 5_.0h_35__hT15750(6)(). (5.I9)\oiNlon phuong phap hesolat dinh dethiet lap quy tac cau phuongCauss lakhong thuc te. Ben canh do, van deton tai cong thuc valac chinhxac tot nhat cothelacau hoi con demotrong cach tiep can nay. Causs dadung lythuyet cac da thuc truc giao detra loi. Loi giai cua Causs khong duoctrnh lay oday, nhung ta cothexem lac chinh xac cua cong thuc cao nhu5.l. C^C _|Y T^C C^| IHlO!C CO B^! I+9thenao. \oi cac dieu kien phuhop trenn(.) va a. b|,ta liet rang ton taimot day cac da thuc01C1(.), N = 0. 1. . . . sao cho01C1(.) lalacNva_bo.}01C1(.)n(.)J. = 0 khi < N. (5.ZO)Khi n(.) =1, a = 1, b =1,cac da thuc nay la1a .|u. leenJre. 1a cungliet rangNnghiemphanliet cua01C1(.) lathuc vanam trong khoang(a. b). Ciasurang cong thuc cau phuong noi suy (5.Z) dua tren co sonoi suytai cac nghiem cua 01C1(.). 'eu(.) lada thuc lac 2N 1, nocotheduocviet(.) = q(.)01C1(.) r(.).trong doda thuc thuongq(.) vada thuc dur(.) colac toi daN 1. 1h_bo(.)n(.)J. =_boq(.)01C1(.)n(.)J._bor(.)n(.)J. =_bor(.)n(.)J..trong dosohang dau triet tieu do (5.ZO). \oi cach chon lat kycac nut.i,cong thuc (5.Z) tich phan da thuc lacNcach chinh xac, vay_bor(.)n(.)J. =1

iD1ir(.i). (5.ZI)Cong thuc ap dung cho(.) codang1

iD1i(.i) =1

iD1iq(.i)01C1(.i) 1

iD1ir(.i).Bay giota dung sukien.ilacac nghiem cua01C1(.) dethay_bo(.)n(.)J. =1

iD1i(.i) =1

iD1ir(.i) =_bor(.)n(.)J..I5O CHlO!C 5. TlCH IH^! SODau lang cuoi cung lado (5.ZI). \ da thuc lat ky (.) colac2N 1 duoctich phan chinh xac, nen cong thuc nay colac chinh xac it nhat la 2N 1.Conhieu cach thuan tien vephuong dientinh toan dethiet lap caccong thuc cau phuong Causs, vacac cong thuc cothetm thay trong cac sachchuyen khao. Cac cong thuc Causs cogiatri v chung cung cap lac chinhxac cao nhat voi socac giatri (.). Mot sukien quan trong vecong thucCauss latat cacacideu duong. 'hu dalan trong phan chan sai so,dieunay conghIa lata cothedung cong thuc voi lac chinh xac cao, ngay cakhiham duoi dau tich phan khong tron. Cong thuc Causs ket hop chat chevoicac ham trong luong lacong cudac liet quan trong khi doi xuvoi cac tichphan maham duoi dau tich phan cokydi hoac cac khoang lay tich phan vohan. Ducoham trong luong hay khong, tat cacac nut cua cong thuc Caussdeu nam trong khoang mo (a. b) (khong dung den(a) va (b)). Lieu nayrat huu ich khi doi xuvoi tich phan maham duoi dau tich phan kydi.S.2 Qoy tac cao pboong da bopCho den nay ta chi xet cac thutuc dua tren xap xi ham(.) tren toan lokhoanga. b|. Cung nhu noi suy da thuc,sai sophuthuoc manh vao chieudaicuakhoang. Lieunaydenghi ta phanhoachkhoangvaxap xi hamlang ham da thuc tung manh. Cach tiep can don gian nhat lachia khoangthanhnhungkhoang conchi dinh truoc. 'euta phan hoacha. b| thanha = .1< .2< . . .< .nC1 = b, th_bo(.)J. =n

iD1_xiC1xi(.)J..ta cotheap dung cac quy tac cau phuong chuan chon tich phan ovephai.Ketqua duoc lietnhula quy.a.1a|+p(compositerule) hayquy.a.|ep(compoundrule). 'guoitathuongdungphanhoachdeukhoang a. b| vadung cung cong thuc cau phuong tren moi khoang con,nhungdieu nay lakhong nhat thiet.Tb doS.S (Quytac hnhthang da hop). Quy tac hnhthang da hop xapxi 1 =_bo(.)J.lang cach phan hoacha. b| thanhn khoang con dodaih = (b a),n va ap dung cong thuc cau phuong hnh thang cho moi khoang.5.2. _|Y T^C C^| IHlO!C D^ HOI I5I\oi dinh nghIa.i = a ih.1 ~ Tn=h2(.0) (.1)| h2(.1) (.2)| . . . h2(.n) (.nC1)|.thu gon, ta duoc:Tn = h_12(.0) (.1) (.2) . . . (.n) 12(.nC1)_ 1heo .n .|u. .n Lu'er-Ma.'aur|n, neu(2)(.) lien tuc trena. b|, thton tai (a. b) sao cho1 = Tn1

kD1h2k(2k)T2k(2k1)(b) (2k1)(a)| nh2C1(2k)T2(2)() sai so.Cac heso T2k xuat hien oday duoc liet nhu lacac sBernu''|. \ai sohangdau cua khai trien sai sola1 = Tnh2120(b) 0(a)| h4720(3)(b) (3)(a)| Quy tac hnh thang ap dung cho mot khoang do dai h co sai so dan ve khongnhuh3. Khi n =(b a),h tuduoc tohop,sai socua xap xi tich phan danvekhong nhuh2. 1uy nhien,neu xay ra0(b) =0(a), cong thuc sechinhxac hon lnh thuong. 'eu them vao cac dao ham khac tai cac diem cuoi cuakhoang tich phan lang nhau, th cong thuc con chinh xac hon nua. Khi tichphan ham tuan hoan tren mot loi cua chu ky,tat cadao ham tai cac diemcuoi cua khoang lay tich phan lalang nhau vacong thuc nay cuc kychinhxac. 1huc ra,neu ham tuan hoan lagiai tich,nocodao ham moi cap,thTn 1nhanh hon lat kyluy thua nao cuah' Mac dukhadac liet, nhungdieu nay lacuc kyquan trong trong giai tich lourier.Sai socuaTn cotheduoc danh gialang cach so sanh novoi ket quachinh xac hon, T2n, nhan duoc lang cach chia doi moi khoang con. Mot cachI5Z CHlO!C 5. TlCH IH^! SOthuan tien dedanh giacong thuc laT2nDh2_12f.x0/ Cf.x1{2Cf.x1/ C: : : Cf.xn1/ Cf.xn1{2/ C12f.xn/_D12.TnCMn/;trong doMn = hn

kD1(a (k 1,2)h).Chuyrang tat cacac danh giacuathuc hien trongTn deu duoc dung laitrongT2n.Comot cach khai thac khai trien sai socua quy tac hnh thang da hopdo Romlerg tm ra rat pholien cho ham duoi dau tich phan tong quat. Ytuong latohopTn va T2n denhan duoc ket quacolac chinh xac cao hon.1ucac khai trien sai so1 = Tnh2120(b) 0(a)| h4720(3)(b) (3)(a)| = T2n(h,2)2120(b) 0(a)| (h,2)4720(3)(b) (3)(a)| ta suy ra1 =22T2n Tn22 1_22 2422 1_(h,2)4720(3)(b) (3)(a)| Cong thuc1 ~22T2n Tn22 1chinh xac cao hon moi cong thuc thanh phan. 1hutuc to hop trnh lay trenduoc goi lap|ep na| suy nm'er.1ich phan Romlerg thuong rat hieu qua.'othich ung lac cua phuongphap voi lai toan. 1uy nhien, ket qua phuthuoc vao tinh tron cua ham duoi5.. C^| IHlO!C THlCH l!C I5Jdau tich phan. Cung vay,nodanh gia (.) tai cac diem cuoi cua khoang,madieu nay doi luc gay lat tien. 'eu cokydi tai diem cuoi cua khoanghoac quatrnhkhonghoituth nendungquy tac diem giua cho khoangchua diem cuoi nay vanen chia nhokhoang nay thanh Z hay J khoang con.S.3 Cao pboong tbcb ongMuc nay trnh lay ytuong co lan - cau phuong thich ung - cho cac chuongtrnh tinh xap xi tich phan dat dochinh xac theo yeu cau nguoi dung. Lieunay duoc thuc hien lang cach chia nhokhoanga. b| thanh cac khoang conva ap dung cong thuc cau phuong co lan cho moi khoang. Khoang duoc phanhoach theo cach thich ung voi dang dieu cua ham (.), dua tren ket qua viecdanh giasai so.'eu sai sokhong chap nhan duoc, viec chia nhokhoang conseduoc tiep tuc. 'hu ta dathay,ngay cavoi cong thuc cau phuong colacchinh xac vua phai, sugiam thieu dodai cua khoang, veco lan, lam tang suchinh xac cua xap xi.1ien hanh theo cach nay, cong thuc duoc ap dung trentoan locac khoang con trong do (.) duoc xap xi tot hon. Ytuong duoctrnh lay oday laco socac chuong trnh con cua mot sothu vien chuongtrnh nhu QU/Dl/CK, '/C vaIMSL; tham chi trong cac moi truong tinhtoan nhu Matlal.1rong cac chuong trnh cau phuong thich ung, khi manhan duoc cacdung saiI(tolerance) cua sai sotuyet doiabserr vasai sotuong doirelerr,nosecogang tinh giatrians sao cho[1 ans[ _ max {abserr. relerr [1[]. (5.ZZ)Khi cai dat (5.ZZ) ta dung danh giasai soerr ~ 1 ansvathay1langans trong (5.ZZ)[err[ _ max {abserr. relerr [ans[]. (5.ZJ)Luu y, ta khong liet giatri dung cua1.Khong thecoduoc xap xi tich phan chinh xac hon giatri dung cua noduoc lam tron,v vay layrelerr 10u. Cung vay, abserr> 0 dedoi xuvoi truong hop1 = 0.lhuongphapthichungdungtrongma chiakhoang a. b| thanhcackhoangcon. |, trendo quytaccauphuongcolanduocdung, dat dochinh xac yeu cau. Lequyet dinh xem ket quatinh toan daduchinh xacchua ta phai danh giasai so cua cong thuc. Lieu nay duoc thuc hien dua trennguyen lyco lan cua giai tich so, ay ladanh gia sai so cua ket qua lang cachso sanh novoi ket quachinh xac hon. Coi Q laket quatinh toan vaQ laket quachinh xac hon, ta dung danh giasai so:err =Q Q.nghIala, thay1 langQ. 1hi du, Qduoctinhlangquytachnhthang(theo() va ()),Q duoc tinh lang quy tac Simpson (theo(), () va(( ),2)),duoc liet lachinh xac hon. Cung cothelayQ nhu laketqua tinh lang cung phuong phap (hnh thang) nhung khoang. | duoc chiadoi. Mot luu ykhi thiet kechuong trnh layeu cau vetinh kinh te. 'enchon phuong phap tinhQ sao cho nothua huong cac danh gia (.) dacokhi tinhQ. 'hu trong thi duvua neu, detinhQ ta chi can danh giathem(( ),2), cac danh gia (), () duoc thua huong tuqua trnh tinhQ.S.4 Cac cboong trnb con1rong muc nay ta lam quen voi mot sochuong trnh con dang function tinhtich phan so.function smpsns la chuong trnh con tinh tich phan lang phuongphap Simpson (simpson's method) voi Ndoan. Ham(.) cothecokyditai avahaybnhungtich phanhoitu. 1hi du[(.)[ o khi . a,can duoi seduoc doilena(1 EPS) hoaca EPS (detranh truong hopa EPS underflow). Cung detranh truong hop(.i) overflow, ta dat(.i)lang realmax neu(.i) lang inf.functionINTf=smpsns(f,a,b,N,varargin)% tich phan cua f(x) tren [a,b]bang quy tac Simpsonvoi N doanEPS=1e-12;if nargin> f = @(x) x.*sin(10.*x);>> smpsns(f,a,b,20)Ket quatraveans =0.0785'eu dung phuong phap Simpson thich ung voitol= 106:>> [INTf,points,err]=asmpsn(f,0,1,10^-6);>> INTfINTf =0.0785>> length(points)ans =65S.S Mot sovan dctboc banbCac cong thuc cau phuong xap xi tich phan dung mot sohuu han cac diem.'eu giatri ham tai cac diem nay khong lieu dien tot ham th ket quatinh5.5. MOT SOV^! DLTHlC H^!H I57toanco the khongchinhxacmacdu saiso danhgia la chapnhanduoc.'guyen nhan latich phan xap xi vasai sodanh giadua tren giathiet hamduoi dau tich phan latron giua cac diem.Chuong trnh cau phuong thich ung cho ket qua tot neu nhan dang duocdang dieu cua ham duoi dau tich phan. 'hung neu(.) codiem nhon hoacdao dong nhieu trong khoang lay tich phan th ket quakhong luong truocduoc. Loi khi ta phai cat khoang lay tinh phan thanh nhung khoang phuhop decac diem danh gia(phat sinh trong macau phuong thich ung) namtrong nhung vung covan de. Xem thi dusau.Hocac tich phan1n =_t0sin2n.J..cothetinh lang cong thuc truy hoi1n =2n 12n1n1. 10 = .Khi nlonhamduoi dautichphanco motdinhnhontaodiemgiuacuakhoang.Dung functionasmpsn voin = 200, tol= 106, ket quatraveINTf = 0.12525310615320509044501307016617err = 7.8051e-011voi129 diem danh gia.'eu so voi gia tri dung: 0.1252531061532049786372206411372, ta thayket quacho rat chinh xac.Ma asmpsn khong gap khokhan v dinh nhon ladiem danh gia. 'hung neu khoang lay tich phan duoc tach thanh0. 2.6| va2.6. |. Khi dodinh nhon sekhong roi vao cac diem danh gia. 1ich phantren hai doan nay, dungasmpsn, roi cong lai, ta duoc ket qua la1.5417e-007,tong cong chi co 6 diem danh gia. Ket quasai liet rat nhieu so voi giatridung''hu vay, voi mot chut khao sat toan hoc ham duoi dau tich phan, ta cothetranh duoc sai sot khi ap dung cac macau phuong. Lieu nay chi cothelam duoc neu ta liet thuat toan cua chuong trnh.Ham dooi dao tcb pban dao dong'eu ham duoi dau tich phan (.) la tuan hoan chu ky, i.e., (.) =I5o CHlO!C 5. TlCH IH^! SO(.) voi moi., va b a lamot loi cua chu ky, a b = n, th_bo(.)J. = n_]0(.)J..Khi do, chi can ap dung quy tac cau phuong cho tich phan trong mot chu ky.Loi voi ham dao dong khong tuan hoan van decokhokhan hon. 'oichung, khoang lay tich phan nen duoc phan thanh nhieu khoang con sao chomoi khoang con chi chua vai dao dong. 1hi du, tich phan_t0sin(20.)1 .2J..cotheviet lai:2

} D10_}t{20(} 1)t{20sin(20.)1 .2J..khi doviec ap dung macau phuong thich ung cho ket quatot.Tcb pban voi can vobanMa cau phuong thich ung khong the ap dung truc tiep de tinh tich phanvoi van vohan. Mot cach deap dung noladung dinh nghIa_1o(.)J. =limb!1_bo(.)J..Ytuong laxac dinh mot chan giai tich cho phan du [_1b(.)J.[. 'honocanb duoc chon dulon decho _bo(.)J.xap xi _1o(.)J.voi dochinhxac yeu cau. Khong thanh van deneub lon hon giatri can thiet, v vay motchan tho cho phan du ladu.Mot cach khac ladoi lien thich hop decotich phan voi can huu han.Chang han,dedanh giatich phan _11exsin .J., dung lien moi s =1,.,tich phan thanh_10e1{xsin(1,s),sJs5.o. TlCH IH^! C|^ B^!C DlllL| I59tren khoang huu han0. 1|. 'oi chung, dieu nay sinh ra khokhan khac, tichphan co ky di o diem cuoi.1rong truong hop dac liet o day, limx!0C e1{xsin(1,s),s =0, v vay ham duoi dau tich phan lien tuc tais = 0, vamacau phuong thichung cotheap dung duoc.S.6 Tcb pban coa bang doIicoBai toan duoc lan den oday laxap xi _bo(.)J.machi duoc cho(.n. ,n)voi1 < n < N, trong do ,n= (.n). Cac chuong trnh con cau phuong thichung khong thedung duoc v chung tudong chon cac diem odo (.) duocdanh giama cac diem nay co thekhong nam trong dulieu {.i] duoc cho cuaham. Cach tiep can co lan: xap xi(.) lang da thuc tung manhJ(.), roitich phan ham nay cach chinh xac.\ spline lac la cho xap xi tot nen cach chon tunhien hamJ(.) laspline lac la. Ledon gian, giasu a =.1va b =.1. Dung kyhieu cuachuong J cho spline,_boS(.)J. =11

nD1_xnC1xnS(.)J.=11

nD1_anhnbnh2n2cnh3n3Jnh4n4_.1hay cac lieu thuc cuaan, bn va Jn theo dulieu (n) va cn, ta duoc_boS(.)J. =11

nD1_nhn_nC1 nhn23cnhn13cnC1hn_h2n2cnh3n3cnC1 cn3hnh4n4_=11

nD1_h2(nnC1) h3n12(cncnC1)_. (5.Z+)IoO CHlO!C 5. TlCH IH^! SOMot so do duoc dung rong rai dat co sotren noi suy lac hai dia phuong.Lexap xi(.) lang da thuc lac hai tren.n. .nC1| can phai noi suy ham taila diem. 1a cothenoi suy tai.n1. .n. .nC1 hoac.n. .nC1. .nC2. Khong colydo g dekhang dinh cach nao cho ket quatot hon,v vay mot cach phuhop latinh lang cahai cach roi lay trung lnh. Ket qualamot cong thucdoi xung lam tron hoa sai socotrong lang dulieu. 1at nhien,ocac khoangchua diem dau, cuoi (n = 1, n = N 1), chi dung mot trong hai phep noi suy.S.7 Tcb pban boi1ich phan xac dinh theo hai hay nhieu lien hon, noi chung, khoxap xi honnhieu, chuyeu do hnh hoc cua mien lay tich phan. Muc nay chi dua ra motsolan luan cho truong hop hai lien,dac liet,nhung van decolien quanden phuong phap phan tuhuu han.1ich phan tren hnh chunhat,1( ) =_b1o1_b2o2(.. ,)J.J,.cothethuc hien dedang nhocong thuc cho truong hop mot lien, lang tinhtich phan lap. Lau tien xap xi1( ) ~11

iD1i_b2o2(.i. ,)J,voi quy tac cau phuong dungN1 diem {.i], varoi1( ) ~11

iD1i__12

} D1T}(.i. ,})__.dung quy tacN2 diem {,}]. Cach lam nay cothetong quat hoa cho truonghop1( ) =_b1o1_x2(,)x1(,)(.. ,)J,J..5.. TlCH IH^! BOl IoIBac chinh xac lay giotham chieu den cac da thuc theo hai lien, v vaymot cong thuc,chang han,colac chinh xac laZ th phai tich phan chinhxac tat cacac da thuc codanga0,0aI,0. a0,I, a2,0.22 aI,I., a0,2,2tren mien dang xet. Ciong nhu trong truong hop mot lien, ta co the thiet lapcong thuc cau phuong lang cach noi suy(.. ,) vatich phan ham noi suy.Lieu nay hoan toan thuc hien duoc tren hnh vuong hay tam giac nhu chi raotren. So dotinh cho hnh chunhat dua tren tich phan lap cohieu quakhicong thuc cho truong hop mot lien laquy tac Causs. 'hung chung khongnhat thiet lacai tot nhat. Cung nhu truong hop mot lien,cothethiet lapcong thuc cac danh gia (.. ,) latoi thieu voi lac chinh xac cho truoc.1uynhien, cong thuc hieu quanhat cothekhong hap dan trong thuc hanh.Cach tiep can dua tren noi suy rat tien loi khi phep noi suy duoc thuc hientai cac diem duoc quan tam v lydo khac,nhu trong phuong phap phan tuhuu han. Cach tiep can lang tich phan lap cotherat thuan tien v tinh dongian vatong quat cua no.1rong mot chieu lien doi khoang huu han lat ky a. b| thanh khoangchuan1. 1| latam thuong. 1rong truong hop hai chieu th van detronenquan trong va khohon nhieu. Ciasu ta can tinh tich phan tren mien1 tongquat.Khi do, 1 phai duoc phan nhothanh cac manh cothelien doi thanhhnh vuong hay tam giac chuan. \iec roi rac hoa mien1 theo cach nay laphan quan trong trong maphan tuhuu han. 'eu mien1 duoc phan hoachthanh cac tam giac (voi canh thang) th phep lien doi laaffine rat don gian.Mot tich phan tren tam giac tong quatTtrothanh tich phan tren tam giacchuanT

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ladinh thuc |acoli cua phep lien doi.1ren day chi lanhung lan luan veytuong co lan, ap dung cong thucmot lien cho truong hop nhieu lien. Conhieu van denay sinh lien quanden phep phan hoach mien, cac phep lien doi moi manh tong quat vemienchuan. Lanhvuctichphanhamnhieulienchodennayvancondangnghien cuu.Cao boi vabai tapIoZ CHlO!C 5. TlCH IH^! SOS.1. Dung phuong phap hesolat dinh dethiet lap quy tac 'ewton3,8_11(.)J. = 1(1) 2_13_3_13_4(1) c(dC1)().1inh1. 2. 3. 4. Jva c.S.2. Dung phuong phap hesolat dinh detm cong thuc cau phuong CaussZ-diem voi sai solien ket. Bat dau lang_11(.)J. = 1(.1) 2(.1) 1( )vatinh1 va .1. Ciasu 1( ) =c(dC1)(),tmJva c. Cong thuc trongtruong hop tong quat, khoanga. b|.S.3. Cai dat quy tac cau phuong hnh thang da hop vaap dung nodetinh_t0J.4 sin(20.).1at nhien lan phai chonh dunhovaduoc lay trong moi chu ky. Xap xitich phan voi mot socach chonh dan ve 0. 1heo lythuyet thTn hoi turatnhanh. day lan thay g!Cboong 6Fboong trnb vi pban tboong6.1 Co soIytboyctCho ham(.. ,) lien tuc (theo lien.) trong doana. b| voi moi,. lhuongtrnh vi phan cap mot tong quat codang,0(.) = (.. ,(.)) (o.I)voi moi . (a. b). 1rong chuong nay ta xet lai toan tm nghiem,(.),laham cua. codao ham lien tuc khi . (a. b), thoa phuong trnh (o.I) vagiatri cua notai diem dau cua khoang:,(a) = (o.Z)lhuong trnh (o.Z) duoc goi ladieu kien dau, vatohop (o.I) va(o.Z) duoc goila'a| .an |a.r( 1au hay 'a| .an Cau.|y cho phuong trnh vi phan.Mot dieu kien don gian lao dam suton tai vaduy nhat nghiem cotheduoc thiet lap nhocach(.. ,) phuthuoc,.Ham (.. ,) thoa dieu kien Lipschitz theo , neu voi moi . trong khoanga. b| vavoi moiu. ,[(.. u) (.. )[ _ 1[u [ (o.J)voi1 la hang so, sau nay duoc goi la hang soLipschitz. 1ruong hopcodaoham rieng lien tuc theo lien thuhai,[(.. u) (.. )[ =dd,(.. n)[u [IoJIo+ CHlO!C o. IHlO!C TnI!H Vl IH^! THlO!Cvoi n ogiuau va ,vaneud,d, li chan voi moi doi so,th thoa dieukien Lipschitz vahang so 1 lat kysao chodd,(.. n)_ 1voi moi .tronga. b| vavoi moi nlamot hangLipschitz. 'eu dao hamrieng khong li chan, cothechi ra rang lat dang thuc (o.J) khong thedungvoi moiu. vavoi moi. tronga. b|, vaykhong thoa dieu kien Lipschitz.Tb do6.1. Ham(.. ,) = .2cos2, , sin2.,xac dinh voi [.[ _ 1 vamoi,, laLipschitz voi hang so 1 = 3. Lethay dieu nay, dao ham doi voi, chodd,= 2.2cos , sin , sin2..vanhu vay voi moi., [.[ _ 1, ta codd,_ 2 1 1 1 = 3 Tb do6.2. Ham(.. ,) =_[,[ khong thoa dieu kien Lipschitz v nocodao ham rieng lien tuc voi,> 0, khong li chan khi, 0:dd,=12_,Mot truong hop quan trong cua (o.I) la phuong trnh vi phan tuyen tinh,(.. ,) =g(.), h(.). Ham(.. ,) lien tuc theo(.. ,) tuong duong voig(.) va h(.) lien tuc theo.. \dd,= g(.)va vhamlientucg(.)li chantrenkhoanghuuhana. b| latky, nenphuong trnh tuyen tinh thoa dieu kien Lipschitz trong hau het cac truonghop thuc hanh.o.l. CO SOlYTH|YLT Io5Tb do6.3. 1ich phan Dawson laham,(.) = ex2_x0et2Jt.Cothekiem tra rang tich phan tren langhiem cua lai toan giatri dau chophuong trnh vi phan tuyen tinh,0= 1 2.,.,(0) = 0.1renkhoang0. b|voi lat ky b ,=0, ham(.. ,) =1 2.,lientuc vaLipschitz voi hang soLipschitz1 = 2[b[ Cac dieu kien dudephuong trnh vi phan ton tai vaduy nhat nghiemcotheduoc phat lieu mot cach hnh thuc:|nb Iy 6.1. C|(.. ,) '|en .u. t+| m|. .rn '|an a. b| tam|,, ta.|a(o.). T|| t+| 'a. 'ys ,'a| .an |a.r( 1au,0=(.. ,), ,(a) = .n||emJuy n|a.,(.) xa. 1(n| t+| m| . .|u. '|ana. b|.Cho den nay ta danoi vemot phuong trnh vi phan voi mot an,(.).Mot hephuong trnh vi phan cap mot voim an laY01= J1(.. Y1. Y2. . . . . Yn)Y02= J2(.. Y1. Y2. . . . . Yn)...Y0n= Jn(.. Y1. Y2. . . . . Yn)(o.+)Cung voi cac phuong trnh (o.+) cocac dieu kien dauY1(a) = 1Y2(a) = 2...Yn(a) = n(o.5)Ioo CHlO!C o. IHlO!C TnI!H Vl IH^! THlO!C'eu datY(.) =_____Y1Y2...Yn__. A =_____12...n__. F(.. Y) =_____J1(.. Y)J2(.. Y)...Jn(.. Y)__(o.o)th (o.+) va(o.5) trothanhY0= F(.. Y). (o.7)Y(a) = A. (o.o)Mot lan nua ta xem to hop cua (o.+) va(o.5) nhu lalai toan giatri dau.Bangcach dung kyhieu vecto lam cho truong hopm an trong giong nhu truonghop mot an. Mot trong cac khia canh may man cua lythuyet lai toan giatri dau lalythuyet cho hephuong trnh vi phan cap mot cot yeu giong nhutruong hop mot an. Cac chung minh cho hechinh ladua vao cac vecto vachuan cua chung odau cocac vohuong vacac giatri tuyet doi trong chungminh cho mot an. \oi ham vecto F(.. Y) thoa dieu kien Lipschitz, dieu kiendu la moi Ji(.. Y1. Y2. . . . . Yn) thoa dieu kien Lipschitz doi voi moiY}; nghIala, ton tai cac hang so 1i}sao cho[Ji(.. Y1. . . . . Y} 1. u. Y} C1. . . . . Yn)Ji(.. Y1. . . . . Y} 1. . Y} C1. . . . . Yn)[ _ 1i}[u[voi moii. . \oi dieu nay, dinh lytuong tudinh lyo.I cho truong hopm andung. \ lythuyet cac phuong phap socho hecac phuong trnh vecot yeucung giong nhu voi mot phuong trnh, nen ta tuhan chechi doi xuchi tietvoi truong hop mot phuong trnh vaphat lieu ket quatuong tucho he.Hau het chuongtrnhmay tinh doi hoilai toan phaiduoc cho duoidang chuan (o.+) va (o.5), nhung cac phuong trnh xuat hien trong nhieu dangkhac nhau. Chang han, phuong trnh cap hai, nghIa la cac phuong trnh dang,00= g(.. ,. ,0).thuonggaptrongcactai lieuve he dongluc. LinhnghIave nghiemlamorong hien nhien cua truong hop cap mot vadieu kien dau thich hop la,(a) =1. ,0(a) =2. Lay laphuong trnh vi phan cap hai cho mot dailuong chua liet, ,(.). Mot lai toan tuong duong odang chuan (o.+) cotheduoc tm lang cach dua vao hai dai luong chua liet vatm hai phuong trnhvi phan cap mot duoc thoa loi chung. Mot trong hai an moi phai cho chungo.l. CO SOlYTH|YLT Io7ta an goc, vay ta layY1(.) = ,(.). 1a lay an con lai ladao ham cua an goc,Y2(.) = ,0(.). Lao ham cac an moi, ta thu duocY01= ,0(.) = Y2(.).Y02= ,00(.) = g(.. ,(.). ,0(.)) = g(.. Y1(.). Y2(.)).Bang cach nay ta di den hehai phuong trnh vi phan cap mot theo hai an:Y01= Y2.Y02= g(.. Y1. Y2).Lay la dang chuan va lythuyet co theap dung cho nodeket luan su ton tainghiem duy nhatY1(.) va Y2(.) thoa dieu kien dauY1(a) = 1.Y2(a) = 2.'ghiem cua lai toan goc nhan duoc tu,(.) = Y1(.).Le kiem dieu nay, truochet chuyrang mot phuong trnh phat lieu rang,0(.) =Y01(.) =Y2(.),vaphuong trnh con lai phat lieu rang,00(.) = Y02(.) = g(.. Y1(.). Y2(.)) = g(.. ,(.). ,0(.)).1uong tu, cothethay rang cac dieu kien dau duoc thoa.lhuong trnh vi phan capm tong quat mot an,,(n)= g(.. ,. ,0. . . . . ,(n1)).,(a) = 1. ,0(a) = 2. . . . . ,(n1)(a) = nco the duoc dat thanh dang chuan theo m an Y1(.) = ,(.). Y2(.) = ,0(.). . . . . Yn(.) =,(n1)(.) vaJ1(.. Y1. Y2. . . . . Yn) = Y2J2(.. Y1. Y2. . . . . Yn) = Y3...Jn1(.. Y1. Y2. . . . . Yn) = YnJn(.. Y1. Y2. . . . . Yn) = g(.. Y1. Y2. . . . . Yn).Tb do6.4. Lechuyen lai toan giatri dau,00(,2 1),0, = 0. ,(0) = 1. ,0(0) = 4Ioo CHlO!C o. IHlO!C TnI!H Vl IH^! THlO!Cthanh hephuong trnh vi phan cap mot, datY1(.) = ,(.). Y2(.) = ,0(.).1hY01= ,0= Y2Y02= ,00= (Y21 1)Y2 Y1vaY1(0) = 1. Y2(0) = 4.Bai toan nay cothedat thanh dang (o.+) lang cach dinh nghIaY =_Y1Y2_. A =_14_. F(.. Y) =_Y2(Y21 1)Y2 Y1_ 6.2 Mot so dosodon gianXet lai toan giatri dau (o.I) va(o.Z),,0= (.. ,),(a) = .tren khoanga. b|. Cac phuong phap sota xet sinh ra mot lang cac giatrixap xi cho,(.). 1am thoi ta giasurang cac diem nhap vao cach deu theolien khong gian.. 'ghIa la, ta chon mot so nguyenNva voi h = (b a),N,ta xay dung xap xi tai cac diem.n =a nh voi n =0. 1. . . . . N. Kyhieu,(.n) duoc dung cho nghiem cua (o.I) va(o.Z) duoc danh giatai. = .n, con,n duoc dung cho mot xap xi cua,(.n).lhuong trnh vi phan khong co kyuc. 'eu ta liet gia tri ,(.n), Linhlyo.I ap dung cho lai toanu0= (.. u)u(.n) = ,(.n)noi rang nghiemcua lai toan giatri dau nay tren khoang.n. b| chinh la,(.). jSauhet, ,(.)la nghiemva dinhly noi rangchi co motnghiem.|'ghIa la, cac gia tri cua ,(.) voi . o truoc .n, khong anh huong truc tiep dennghiem cua phuong trnh vi phan voi .osau.n. Mot vai phuong phap soo.2. MOT SO DOSODO! Cl^! Io9cho phuong trnh vi phan cokyuc vamot vai phuong phap th khong. Lopcac phuong phap duoc lietnhu la p|u+np|apm. 'u+. (one-step method)khong cokyuc - cho truoc,n, comot cong thuc cho giatri,nC1 phuthuocvao.n. ,n. va h. Bat dau voi giatri lan dau hien nhien,0 =,phuongphap mot luoc sinh ra mot lang giatri,(.) lang cach thuc hien lap lai motluoc theo. voi dodaih desinh ra day lien tiep,1. ,2. . . .1hi dudon gian nhat cua phuong phap mot luoc lap|u+n p|ap Lu'er.1a nghiencuu nov cac chi tiet khong lam modi ytuong vatruong hoptong quat larat giong. Khai trien 1aylor,(.) quanh. = .n, cho,(.nC1) = ,(.n) h,0(.n) h22,00(n)voi .n 0vaH =_2tthhn ~_[,00(.n)[H = (.n)HI7+ CHlO!C o. IHlO!C TnI!H Vl IH^! THlO!Cxac dinh(.). ChuyrangH =0(t1{2) demax[,(.n) ,n[ la 0(t1{2) chophuong phap Luler voi suchon lua tudong kich thuoc luoc.6.3 Cac pboong pbap mot boocBay giota xet cac phuong phap mot luoc vadat cac giathiet cua chung duatheo phuong phap Luler. Cong thuc tong quat codang,0= .,nC1= ,nh(.n. ,n. . h). n = 0. 1. . . . (o.I7)lhuong phap khong co kyuc, nen chi phuthuoc vao cac doi so.n. ,n. . h.1hong thuong va h duoc lodi trong kyhieu. Ciasu lien tuc theo.va ,. lhuong phap Luler lay(.. ,) =(.. ,) vadieu kien Lipschitz duocdung lacot yeu. \ay, voi cong thuc tong quat ta giasurang[(.. u) (.. )[ _ 1[u [ (o.Io)khi a _. _b,voi moi 0 1 .|,(.), .|| t+| 'a. 'y .n = a nh a. b|[,(.n) ,n[ _Ch]1(e1(xno) 1). (o.ZI)o.. C^C IHlO!C IH^I MOT BlOC I75C|un m|n|. 'hu truoc, dat1n = ,(.n) ,n vatru(o.I7) cho (o.I9) ta duoc1nC1 = 1nh(.n. ,(.n)) (.n. ,n)| hjnDungdieu kienLipschitz(o.Io)vagiathiet phuong phap lacap, ta thayrang[1nC1[ = (1 h1)[1n[ Ch].Bay giodinh lylaket quacua Bodeo.Z vasukien10 = 0.Cung nhu lan luan cua phuong phap Luler, ket quacua dinh lycho suhoi tu 0(h]). Lieu nay giai thich viec ta goi phuong phap lacap cho,(.).1huat ngu phuong phap thuoc cap duoc dung de mo ta mot phuong phapma thuoc cap neula du tron.Cap cua su hoi tu la thap hon khikhongtron nhu vay.'hu dagiai thich trong moi lien hevoi phuong phap Luler,doan machon tudong kich thuoc luoc degiucho sai soluon nhohon mot dung sait. Long thoi chung cogang dung mot luoc dulon. Mot mohnh hop lycuathuat toan tm kich thuoc luoc nhu vay dan den mot kich thuoc luochn tai.n cho loihn= (.n)Hvoi mot ham lien tuc tung khuc(.) voi0< 0 _(.) _ 1 trena. b|. \oikich thuoc luoc duoc chi dinh theo cach nay,chung minh suhoi tucothethay doi dedang deket luan rang sai sola 0(H]) = 0(t1{]).Cong viec quan trong nhat con lai lay gio la phai tm cac ham khongdat tien khi danh giavathuoc cap voi tron. 1ucong thuc (o.I9) tacanjn = 0(h]). Khai trien 1aylor cua,(.) chung torang,(.nC1) = ,(.n) h_,0(.n) . . . h]1(),(])(.n)_h]C1( 1),(]C1)(n)neu,(.) C]C1a. b|. \ay, ta tm xem, neu phuong phap lacap, th nophai co(.. ,(.)) = ,0(.) h2,00(.) . . . h]1(),(])(.) (.).voi (.) =0(h]). \ ,(.)la nghiemcuaphuongtrnhvi phan,0(.) =(.. ,(.) cac dao ham cua, cotheduoc lieu dien nhodao ham toan phanI7o CHlO!C o. IHlO!C TnI!H Vl IH^! THlO!Ccua . Dung kyhieu(n)(.. ,(.)) dechi dao ham toan phan capm cuavachi soduoi dechi dao ham rieng, hethuc la,(n)= (n)(.. ,(.)).trong do(1)== x(.. ,(.)) ,(.. ,(.))(.. ,(.)).(n)== (n1)x(.. ,(.)) (n1),(.. ,(.))(.. ,(.)). m = 2. 3. . . .Bieu thuc cho(.. ,) trothanh(.. ,) = (.. ,) h2(1)(.. ,) . . . h]1(]1)(.. ,) 0(h]). (o.ZZ)Mot chon lua hien nhien cho lahamT(.. ,)T(.. ,) = (.. ,) h2(1)(.. ,) . . . h]1(]1)(.. ,).cung cap mot ho cac phuong phap mot luoc, goi la cac p|u+n p|ap .|u| .ay'r(1aylor series methods). lhuong phap Luler latruong hop = 1. Mot khi cothetinh duoc cac dao ham th cac phuong phap nay rat cohieu qua.Cacp|u+np|apnune-lu..adungtohop tuyentinh nhieudanh giacua(.. ,) dexap xi,(.). 1ruong hop don gian nhat laphuong phap Lulerchi dung mot danh gia. Bay giota thiet lap mot thutuc dung hai danh gia(.n. ,n) va (.n1h. ,n2h(.n. ,n)), trong do 1 va 2 lacac thamso. 1h voi ta dung tohop tuyen tinh1(.. ,):1(.n. ,n) = a1(.n. ,n) a2(.n1h. ,n2h(.n. ,n)).1rong lieu thuc nay ta tudo chon cac giatri huu dung cho1. 2. a1, va a2.Muc dich lachon cac tham sodecho lieu dien (o.ZO) dung voi giatri cuacang lon cang tot. Lethuc hien dieu nay ta khai trien tat cacac luong trongchuoi 1aylor theoh vadong nhat cac hesocua luy thua. Ledon gian cachkyhieu,cac doi soduoc viet ra neu chung khac(.n. ,n). 1a tien hanh nhuo.. C^C IHlO!C IH^I MOT BlOC I77sau.1 = a1 a1(.n1h. ,n2h )= a1 a2_(.n. ,n2h ) 1hx(.n. ,n2h )21h22xx(.n. ,n2h ) 0(h3)_= a1 a2_ 2h,22h222,,0(h3)1hx12h2x, 0(h3) 21h22xx0(h3)_= (a1a2) a2h(2,1x)a2h22(222,, 212x,21xx) 0(h3).Bay giota muon chon cac tham sodecho1 = h2(1)h26(2)0(h3).hay viet tuong minh la1 = h2(,x) h26(2,,2x,xxx, 2, ) 0(h3).Can lang cac hesoluy thua cuah cung lac, ta duoca1a2= 1.a22= 1,2.a21= 1,2.Laya2 = th voi giatri lat kycua tham so ,a2 = . a1 = 1 I7o CHlO!C o. IHlO!C TnI!H Vl IH^! THlO!Ccho cong thuc phuhop voi dang thuc dau.Hon nua, neu doi hoi ,= 0, chon1 = 2 =12.cho cong thuc phuhop voi hai dang thuc cuoi. 1om lai,1(.. ,) = (1 )(.. ,) _. h2. , h2(.. ,)_cho mot hocac phuong phap mot luoc cap Z khi ,= 0 va dutron.Mot vai thanh vien cua hocong thuc nay coten. lhuong phap Lulerco =0 vacap =1. lhuong phap Heun (con goi laphuong phap Lulercai tien) latruong hop =1,2,vaphuong phap Luler diem giua (midpointLuler method) hay phuong phap Luler hieu chinh (modified Luler method) latruong hop = 1. Lethay khanang ap dung cac cong thuc nay ta can lietdieu kien can dedinh lyhoi tucohieu luc.1inh lien tuc cua1 hien nhienduoc suy ra tutinh lien tuc cua . Lieu kien Lipschitz tren1 cung rut ra tu .[1(.. u) 1(.. )[ =(1 )(.. u) (.. )| __. h2. u h2(.. u)__. h2. h2(.. )___ (1 )1[u [ [[1u h2(.. u) (.. )|_ (1 )1[u [ [[1[u [ h212[u [__(1 ) [[ h2_1[u [voi moi0 < h < h0, vata cothelay hang soLipschitz cho1 la_(1 ) [[ h2_1.o.. C^C IHlO!C IH^I MOT BlOC I79\ vay, neu phuong trnh vi phan thoa cac dieu kien cua dinh lyo.I, vaneu hamcodao ham den cap Z lien tuc jnhu vay nghiem,(.) C3a. b||,thanh vien lat kycua hovoi ,= 0 hoi tucap Z.Cac thutuc cap cao lao gom nhieu thay thehon cotheduoc thiet laptheo cung mot cach, mac dumot cach tunhiencac khai trien tronen ratdai dong vatenhat. 'hu xay ra, thutuc thuoc cap can danh giaomoiluoc khi =1. 2. 3. 4 nhung khong nhu vay khi =5. \ lydo nay,caccong thuc cap lon voi kich thuoc luoc hang thuong duoc dung detich phansophuong trnh vi phan. Ciong nhu trong truong hop cap hai, comot hocacthutuc cap lon phuthuoc nhieu tham so. Cach chon codien cac tham sodan den thuat toan,0 = .vakhin = 0. 1. . . .k0= (.n. ,n).k1= _.nh2. ,nh2k0_.k2= _.nh2. ,nh2k1_.k3= (.nh. ,nhk2).,nC1= ,nh6(k02k12k2k3).Loi voi hephuong trnh vi phan cap I,Y0= F(.. Y).Y(a) = A.mot cach tunhienY0 = A.vakhin = 0. 1. . . .IoO CHlO!C o. IHlO!C TnI!H Vl IH^! THlO!Cthuat toan Runge-Kutta codien lak0= F(.n. Yn).k1= F_.nh2. Ynh2k0_.k2= F_.nh2. Ynh2k1_.k3= F(.nh. Ynhk2).YnC1= Ynh6(k02k12k2k3).Mot thutuc cap lon khac, hoan toan tuong tuk0= F(.n. Yn). (o.ZJ)k1= F_.nh2. Ynh2k0_.k2= F_.nh2. Ynh4k0h4k1_.k3= F(.nh. Yn hk12hk2). (o.Z+)YnC1= Ynh6(k04k2k3).6.4 Sai sod|a pboong vatoan cocCac mahiennay cho lai toan giatri dau khongdung kich thuoc luoc codinh. Sai soomoi luoc duoc danh giava h duoc dieu chinhlai denhanduoc xap xi duchinh xac. Comot nham lan dang tiec tunhieu nguoi dungmavoi danh giasai sovecai duoc do valien hecua novoi sai sothuc.Ham,(.) kyhieu nghiem duy nhat cua lai toan,0= (.. ,).,(a) = .o.+. S^l SODI^ IHlO!C V^TO^! C|C IoISai sothuc hay toan tuc tai.xC1 la,(.nC1) ,nC1.'hung tiec lacokhokhan vaton kem dedanh giadai luong nay,v trongluoc tinh.nC1thutuc sochi cung cap.n. ,ndedanh gia . 'ghiem diaphuong tai.n langhiemu(.)u0= (.. u).u(.n) = ,n.Sai sodia phuong lau(.nC1) ,nC1.Lay lasai sodo xap xi nghiem phuong trnh vi phan goc tai (.n. ,n) langmot luoc. Sai sonay duoc minh hoa tren hnh o.I. \iec doi hoi thutuc sogiucho sai sonay nholahop ly. Sai sonay anh huong len sai sotoan cucphuthuoc vao lan than phuong trnh vi phan. Sau het,,(.nC1) ,nC1 = ,(.nC1) u(.nC1)| u(.nC1) ,nC1|. (o.Z5)Lai luong,(.nC1) u(.nC1)lasodo suon dinh cua phuong trnh vi phan v nolahau qua(tai.nC1) cuasusai liet lan dau,(.n) ,n tai.n. 'eu dai luong nay gia tang ngay canglon, th lai toan duoc dat xau hay dieu kien xau hay khong on dinh.Tb do6.6. Xet,0= ,voi lahang so. 1a cosau mot sotinh toan:,(.) = ,(.n)e(xxn).u(.) = ,ne(xxn):hon nua,,(.nC1) u(.nC1) = ,(.n) ,n|eh. (o.Zo)'eu> 0, cac duong cong nghiem trai rong ra (Hnh o.Za), cang nhieu khilon. 1u lieu thuc (o.Zo) ro rang sai so dia phuong nho tai moi luoc khong chosai sotoan cuc nho.Mat khac, neu< 0, cac duong cong tuvao nhau (Hnho.Zl) va (o.Zo) chung to rang su dieu khien sai so dia phuong se dieu khien saisotoan cuc. \oi cac ham(.. ,) tong quat dieu kien Lipschitz khong thetien doan dang dieu nay,v voi thi dunay hang soLipschitz la[[ trong cahai truong hop IoZ CHlO!C o. IHlO!C TnI!H Vl IH^! THlO!CHnh o.I: Sai sodia phuong vasai sotoan cuc.Hnh o.Z: Cac duong cong nghiem voi: (a),0= 2,; (l),0= 2,.o.+. S^l SODI^ IHlO!C V^TO^! C|C IoJSai sodia phuong lien hevoi sai sochat cut dia phuong. 1hat vay, nodung langh lan sai sochat cut dia phuong, j, voi nghiem dia phuongu(.):sai sodia phuong = u(.nC1) ,nC1= (,n) h(.n. ,n) hjn) ,nC1= hjn.Chang han, khi,n la nghiem cua ,0= (.. ,), ta da thay phuong phap Lulerco,(.nC1) = ,(.n)h(.n. ,(.n))h22(.n. ,(.n)),(.n. ,(.n))x(.n. ,(.n))| = 0(h3)./p dung chou(.), ta cosai sodia phuong =h22(,x) 0(h3).1uong tuvoi cong thuc Rung-Kutta cap2 ( ,= 0), ta cou(.nC1) = ,nh_ h2(1)h26(2)_0(h4)th xap xi sothoa ,nC1 = ,nh_ h2(,x) h28(2,,2x,xx)_0(h4).Lieu nay dan densai sodia phuong = h jn = h3_1618_(2,,2x,xx)h36(x,2, )0(h4).Cac lieu thuc nay denghi mot cach danh giasai sodia phuong. Ciasutatinh,nC1 lang phuong phap Luler vata cung tinh mot xap xi nghiem ,nC1Io+ CHlO!C o. IHlO!C TnI!H Vl IH^! THlO!Clang mot trong cac cong thuc Runge-Kutta cap Z. Bieu thuc tren chung torang ,nC1 ,nC1 =h22(,x) 0(h3) = hjn0(h3).'ghIa la,sukhac nhau giua hai giatri cho danh giasai solang cong thuccap thap hon. Lieu nay giong nguyen lydung trong chuong 5 dedanh giacac sai socau phuong. 'oi chung, giasurang them vao giatri,nC1 = ,nh(.n. ,n)voi sai sochat cutjn = 0(h]), ta tinh xap xi khac ,nC1 = ,nh(.n. ,n)voi sai sochat cut jn = 0(hq) cocap cao hon, q> . 1h loi dinh nghIau(.nC1) = ,nh(.n. ,n) hjn = ,nC1hjnva, tuong tu,u(.nC1) = ,nC1h jn.ma, lang cach trunhau, chung torang ,nC1 ,nC1 = hjn h jn = hjn0(hqC1).\h jn dan vekhong nhanh honhjn, ta cothedanh giasai sodia phuongloisai sodia phuong = hjn ~ ,nC1 ,nC1.1a muon xap xi nghiem dia phuongu(.nC1). \ su kien ta co mot danhgiasai sotrong,nC1 tot,tai sao khong cogang cai thien nolang cach loailosai so! Quatrnh nay,goi lana| suy 1(a p|u+n (local extrapolation),oday tuong duong cach hnh thuc voi viec dexuat phep tich phan lang xapxi cap cao hon ,n loi vu(.nC1 = ,nC1hjn ~ ,nC1( ,nC1 ,nC1) = ,nC1.Lieu nay lao cho chung ta rang ngoai suy dia phuong senang cap hieu quacua cap tu lenq. 'hu vay ta cothenghI vedieu dang xay ra trong haicach. Cong thuc capdang duoc dung voi ket quacua noduoc cai thiennhongoai suy dia phuong. Cong thuc con lai, capq dang duoc dung voi kichthuoc luoc duoc chon cach dedat nhodoi hoi rang luoc duoc lay voi congo.+. S^l SODI^ IHlO!C V^TO^! C|C Io5thuc cap thap hon. Boi v ngoai suy dia phuong gia tang suchinh xac makhonggia tang suton kem, tat cacac masan xuat hiennay dua tren cacphuong phap Runge-Kutta hien deu dung no.CongthucRunge-Kuttacap+doi hoi (it nhat) londanhgia cuaFo moi luoc va motcongthuccungloai cap 5doihoi itnhatsau. Lungnhu cau phuong Causs-Kronrod, thuthuat cohieu qualaphai thiet lap congthuc nhu mot cap trong docac danh giaham duoc dung trong cahai congthuc. R. Lngland dacong lomot cap cong thuc nhu vay trong jJ|. Letientu .nden.n h, onglay luoc dodai h,2 voi (o.Z+) decoket quacap +YnC1{2~Y(.n h,2) va roi luockhacdo dai h,2de co ketqua cap+YnC1 ~ Y(.nh). Bang cach thuc hien hai luoc mot nua, ong ta coducacdanh giaham cohieu luc mavoi chi them mot danh gia,ong ta cothelapmot xap xi cap 5YnC1cho YnC1. Bang cach nay,them mot danh giahamduoc thuc hien omoi hai luoc mot nua decodanh giasai so. Mot danh giasai soduoc dung dedieu khien sai sodia phuong vanhu vay cho sutin caynao dovao nghiem tinh toan. 'ocung cho phep machon lua kich thuocluoc lon nhat maket quavan qua duoc sukiem tra sai so. 'goai trucactruong hopkhongthongthuong, su thichung kihthuocluoc chonghiemtheo cach nay gia tang tinh hieu qua cua phep tich phan rat nhieu. 'otuongung voi cac so docau phuong thich ung cua chuong 5.Ioo CHlO!C o. IHlO!C TnI!H Vl IH^! THlO!CCong thuc cua Lngland lanhu sau.k0= F(.n. Yn).k1= F_.nh4. Ynh4k0_.k2= F_.nh4. Ynh8(k0k1)_.k3= F_.nh2. Ynh2k1hk2_.YnC1{2= Ynh12(k04k2k3):k4= F_.nh2. YnC1{2_.k5= F_.n3h4. YnC1{2h4k4_.k6= F_.n3h4. YnC1{2h8(k4k5)_.kT= F_.nh. YnC1{2h2k5hk6_.YnC1= YnC1{2h12(k44k6kT):kS= F_.nh. Ynh12(k0 96k192k2 121k3144k4k5 12k6_.YnC1= Ynh180(14k064k2k3 8k464k615kT kS).Matkhongthuanloi cuacacthuattoangiai laitoangia tri daulachung sinh ra mot lang cac giatri xap xi trong khi nghiem toan hoc,(.) lao.+. S^l SODI^ IHlO!C V^TO^! C|C Io7mot ham lien tuc. Cothexap xi nghiem cho moi. lang noi suy.Cao boi vabai tap6.1. 'hu mot thi dukhong duy nhat nghiem, kiem tra rang voi hang so clat ky, 0 _ c _ b, ham,(.) xac dinh loi,(.) =_0. neu0 _ . _ c14(. c)2. neuc< . _ blamot nghiem cua lai toan giatri dau,0=_[,[,(0) = 0.6.2. Xet lai toan,0=_[1 ,2[,(0) = 1.Kiem tra rang(a),(.) = 1 langhiem tren khoang lat kychua. = 0,(l),(.) = cosh. langhiem tren0. b| voi lat ky b> 0, va(c),(.) = cos . langhiem tren khoang thich hop.Cai g lakhoang lon nhat chua. = 0 tren docos . langhiem!6.3. Dung phuong phap Luler cho cac lai toan sau lang cach dung kich thuocluoc codinhh =1.0,varoi h =0.5. 1rong moi truong hop tinh sai sotai. = 1.0.(a),0= ,,(. 1) voi,(0) = 1, vay,(.) = 1,(. 1).(l),0= ,3,2 voi,(0) = 1, vay,(.) = l,_1 ..6.4. /p dung phuong phap Luler dedanh gianghiem cua lai toan giatridau trong lai tap o.Jl. Dungh = 1,40 va h = 1,80. 1inh sai sotai. = 0.5va . =1.0 dethay neu chung duoc chia doi mot cach thonhuh la. Lanhgiaxemh can nholao nhieu desai sotuyet doi nhohon106vedolon.Ioo CHlO!C o. IHlO!C TnI!H Vl IH^! THlO!CHoong dan & ap sobai tapCa. 'a|.ap .rn .a| '|eu nay n|am |up s|n| t|en .u'|em .ra '|en .|u.,|a. 'sun .a. 1|em '|n 1u+. .r|n| 'ay .rn 'a| |an. S|n| t|en nen .an .u|a|.a. 'a| .ap. C|/ nen .|am '|a '+| |a| .+1ay sau '|| 1a|a| 1u+. (s san| .|m.a.| |a| .. |+n), |a. sau '|| 1a.an n||eu 'an n|un '|n .|an| .n.Cu| .un 'a m. s .|un m|n| .a. 'e. qua ('|n 1u+. .|un m|n| 1ay 1u).rn .a| '|eu.ai tap cboong 1I.I CoiNlachi solat dau cua thuat toan dung cong thuc truy hoi lui (I.+),c lasai sotuyet doi cua11,11 11 = c. 1a co:^11i =11i 11i = (1)icN(N 1)(N i 1).Mat khac, tulat dang thuc0 < 11> A=[1 1 1; 1 1 0; 0 1 1]A =1 1 11 1 00 1 1>> b=[110;78.33;58.33]b =110.000078.330058.3300>> [A,flag,pivots,Cond]= Factor(A)I9o Hu+n Jan c Dap s'a| .apA =1 1 1-1 1 10 0 -1flag =0pivots =13-1Cond =7>> x = Solve(A,pivots,b)x =51.670026.660031.6700'ghiem chinh xacx =5167/1001333/503167/100Ket quagiai lang lactorSolve cho ket quachinh xac'Z.9 Do AA1= l nen ta cothetm cac cot cua ma tran A1lang cach giaiAx = li, trong dolilacot thu icua ma tran don vi.>> clear all>> A=[1 2 3; 4 5 6; 7 8 9.01];>> I1=[1;0;0];>> I2=[0;1;0];>> I3=[0;0;1];>> [A,flag,pivots,Cond]= Factor(A)A =7.0000 8.0000 9.0100-0.5714 0.8571 1.7129Hu+n Jan c Dap s'a| .ap I99-0.1429 -0.5000 -0.0050flag =0pivots =331Cond =1.4370e+004>> B(:,1)=Solve(A,pivots,I1);>> B(:,2)=Solve(A,pivots,I2);>> B(:,3)=Solve(A,pivots,I3)B =98.3333-199.3333 100.0000-198.6667 399.6667-200.0000100.0000-200.0000 100.0000Sodieu kienCond=1.4370e+004 qualon. Ma tran ladieu kien xau. Kiem>> A*Bans =0 0 1.0000-55.1905 113.9048 -57.285784.7857-170.3571 85.2143ans khong gan voi ma tran don vi'Z.IO a)>> clear all>> A=[0.2170.732 0.414;0.508 0.809 0.376;0.795 0.886 0.338];>> b=[0.741;0.613; 0.485];>> [A,flag,pivots,Cond]= Factor(A)A =0.7950 0.8860 0.3380-0.6390 0.4902 0.3217-0.2730 -0.4955 0.0006flag =ZOO Hu+n Jan c Dap s'a| .ap0pivots =331Cond =4.9409e+003>> x = Solve(A,pivots,b)x =0.0000-0.41602.5254l) Chuan cua A vab:>> normA=norm(A,inf)norm =2.0190>> normb=norm(b,inf)normb =0.7410'hu vay,[^A[[A[=3 0.00052.0190= 7.4294 104.[^b[[b[=0.00050.7410= 6.7476 104.1heo lat dang thuc (Z.ZI), ta co[^x[[x[~~ Cond(7.4294 1046.7476 104) = 7.0047 ~ 700 .Ket quakhong dang tin cay.c) danh cho sv.ai tap cboong 3J.J Dung MatlalHu+n Jan c Dap s'a| .ap ZOI>> clear all>> xn=[1 2];>> fn=[2 4];>> syms x>> L1=(x-xn(2))/(xn(1)-xn(2))L1 =2 - x>> L2=(x-xn(1))/(xn(2)-xn(1))L2 =x - 1>> P2=fn(1)*L1+fn(2)*L2P2 =2*xCothechonQ(.) = 12(.) (. 1)(. 2). Lieu nay khong mau thuan voitinh duy nhat cua da thuc noi suy. \ theo chung minh dinh lyJ.I suduynhat hieu theo nghIa cac da thuc lac _ N 1.>> Q=simplify(P2+(x-xn(1))*(x-xn(2)))Q =x^2 - x + 2>> ezplot(P2,[12])>> hold on>> ezplot(Q,[12])>> hold offJ.+ lhuong trnh11(.) = c1c2. . . . c1.11cotheviet duoi dang ma tran:11(.) = 1..2. . . .11|_____c1c2...c11__.ZOZ Hu+n Jan c Dap s'a| .apHnh o.+: Lothi ham12(.), Q(.), lai tap J.J.'hu vay,11(.}) = 1.}.2}. . . .11}|_____c1c2...c11__= (.}) phuong trnh thu }. = 1. . . . . N.Hephuong trnh xac dinh cac heso c1. . . . . c11:_____1 .1.21. . . .1111 .2.22. . . .112............1 .1.21. . . .111__ M_____c1c2...c11__ c=_____(.1)(.2)...(.1)__ f.1huat toanI. 'hap du lieu: .n = .(1). .(2). . . . . .(N)|; n = .(1). .(2). . . . . .(N)|T.Hu+n Jan c Dap s'a| .ap ZOJZ. Lap ma tran Mfor i=1:NM(i,1)=1;for j=2:NM(i,j)=M(i,j-1)*xn(i);endendJ. Ciai phuong trnh Mc = f.+. Xuat ket qua.Ket quacua thuat toan lacac heso c1. . . . . c11. Cac hesonay hoan toanxac dinh mot da thuc.J.o a) La thuc noi suy ham(.):11(.) =1

kD1k1

} D1,} 6Dk(. .})(.k .})=1

kD1k.11

1} D1,} 6Dk(.k .})=_1

kD1k

1} D1,} 6Dk(.k .})_.111uday suy ra dieu phai chung minh.J.I5 1huat toanI. 'hap day.1. .2. . . . . .n va ..Z. k = 1.J. 'eu (.k _ . ck< n) thk = k 1, trolai J.+. Xuatk (khoang.k1. .k| chua.).1hehien thuat toan lang Matlalfunction[k] = tim_khoang(d,x)ZO+ Hu+n Jan c Dap s'a| .ap% functiontim_khoangtra ve chi so cua phan tu lon hon% va gan x nhat% d .....day so thuc duoc sap theo thu tu tang dan% x .....so thuc nam giua x1,xnn=length(d);k=1;while d(k)