数值分析英文版课件 (2)
基于MATLAB的数值分析2PPT课件

【例】 clf;t=0.1:.1:3; y=exp(t.*t); semilogy(t,y) grid xlabel('t'); ylabel('exp(t.*t) ');
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若干特殊图形
x=[1:10]; y=[5 6 3 4 8 1 10 3 5 6]; subplot(2,3,1),bar(x,y),axis([1 10 1 11]) subplot(2,3,2),hist(y,x),axis([1 10 1 4]) subplot(2,3,3),stem(x,y,'k'),axis([1 10 1 11]) subplot(2,3,4),stairs(x,y,'k'), axis([1 10 1 11]) subplot(2,3,5), x = [1 3 0.5 5];explode = [0 0 0
【例】用图形表示离散函数 y(n6)1 。
n=(0:12)'; y=1./abs(n-6); plot(n,y,'r*',…
'MarkerSize',20) grid on
2021/3/9
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axis指令
axis([xmin xmax ymin ymax]): 设定二维图形的x和y坐标的范围;
t=(0:pi/50:2*pi)';
k=0.4:0.1:1;
Y=cos(t)*k;
数值分析英文版chapter 1

Errors
§1 .1 Errors and Significant Digits
1.1.1 Truncation error and round off error
Truncation error: made by numerical algorithms, arise from taking finite number of steps in computation
| x * x1 | 0.00173 0.5 102 4 , | x * x2 | 0.00073 0.5 102 4 , | x * x3 | 0.00027 0.5 1025 ,
According to Definition 1.4, x1, x2 and x3 have 4,4 and 5 significant digits respectively.
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Integration example (cont.)
Choosing a width of 3, we have
x 2 dx ( x 2 )
3 9 x 3
(6 3) ( x 2 )
x 6
(9 6)
(32 )3 (6 2 )3
27 108 135
Actual value is given by
x 3 93 33 2 x dx 3 3 234 3 3
9
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Truncation error is then
234 135 99
Can you find the truncation error with 4 rectangles? 2013-12-9
If only 3 terms are used,
数值分析(双语版)a12

1 2
n−1 −
…………
f [ x, x0 , ... , xn−1 ] = f [ x0 , ... , xn ] + ( x − xn ) f [ x, x0 , ... , xn ] n−
+ f [ x , x0 , ... , xn ]( x − x0 )...( x − xn−1 )( x − xn )
Nn(x)
ai = f [ x0, …, xi ]
Rn(x)
§2 Newton’s Interpolation
注:
由唯一性可知 Nn(x) ≡ Ln(x), 只是算法不同,故其 , 只是算法不同, 余项也相同, 余项也相同,即 f ( n +1 ) (ξ x ) f [ x , x 0 , ... , x n ]ω k +1 ( x ) = ω k +1 ( x ) ( n + 1) !
§2 牛顿插值
/* Newton’s Interpolation */
Lagrange 插值虽然易算,但若要增加一个节点时, 插值虽然易算,但若要增加一个节点时, 都需重新算过。 全部基函数 li(x) 都需重新算过。
? ? 将 Ln(x) 改写成 a0?+ a1( x − x0 ) + a2 ( x − x0 )(x − x1 ) + ... + a? ( x − x0 )...(x − xn−1 ) 的形式,希望每加一个节点, 的形式,希望每加一个节点, n
1 + (x − x0) × 2 + … … + (x − x0)…(x − xn−1) × −
数值分析英文课件

or relative forward error of about 1 percent. Since 1.96 = 1.4 , the absolute backward error is
ˆ ∆x = x − x = 1.96 − 2 = 0.04
Computational error = Truncation error + rounding error
• Propagated (传播) vs. computational error 传播)
– x = exact value, – f = exact function,
ˆ x = approx. value ˆ f = its approximation
Backward vs. forward errors
Suppose we want to compute y = f ( x ) , where f : ℜ → ℜ ˆ but obtain approximate value y
Forward Error:
ˆ ˆ ∆y = y − y = f ( x ) − f ( x )
Example of Ill-Posed Problem
x 1 x 1 x 11 1 + 2 + 3 = 2 3 6 1 1 1 13 x1 + x2 + x3 = 3 4 12 2 1 x1 + 1 x2 + 1 x3 = 47 3 4 5 60
2 significant digits rounding
• Problems that are not well-posed are ill-posed.
数 值 分 析Numerical Analysis

Time Limit Exceeded: Your program tried to run during too much time. This error does not allow you to know if your program would reach the correct solution to the problem.
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Laboratory Grade (30) = Lab ( i ) i 1
Numerical Analysis Laboratory Projects
1. Input and Output Your program must read from a file “in.txt” (if there is
Accepted: OK! Your program is correct! You will obtain 2 points for correctly solving one problem.
Presentation Error: Your program outputs are correct, but are not presented in the correct way. Check for spaces, justify line feeds...
《数值分析》黄仿伦改编英文版课后习题答案

Answers for Exercises —Numerical methods using MatlabChapter 1P10 2. Solution (a) )(x g x = produces an equation 0862=+-x x . Solving it gives the roots 2=x and 4=x .Since 2)2(=g and 4)4(=g , thus, both 2=P and 4=P are fixed points of )(x g . (b) –(d) The iterative rule using )(x g is 22144n n n p p p ---=. The results for part (b)-(d) with starting value 9.10=p and 8.30=p are listed in Table 1.(e) Calculate values of x x g -='4)( at 2=x and 4=x .12)2(>='g , and 10)4(<='g .Since )(x g ' is continuous, there exists a number 0>δ such that1)(<'x g for all ]4,4[δ+δ-∈x .There also exists a number 0>λ such that1)(>'x g for all ]2,2[λ+λ-∈x .Therefore, 4=p is an attractive fixed point. The sequence generated by22144n n n p p p ---=with starting value 8.30=p converges to 4=p . 2=p is a repelling fixed point. The sequence generated by 22144n n n p p p ---=with starting value 9.10=p does not converge to 2=p .P11 4. Find the fixed point for )(x g : )(x g x = gives 2±=p . Find the derivative: 12)(+='x x g .Evaluate )2(-'g and )2(g ': 3)2(-=-'g , 5)2(='g .Both 2-=p and 2=p gives 1)(>'p g . There is no reason to find the solution(s)using the fixed-point iteration.P11 6. Proof ))(()()(010112p p g p g p g p p -ξ'=-=-)()()( 0101p p K p p g -<-ξ'≤P214. False position method: Assume that ],[n n b a contains the root. The equation of the secand line through ))(,(n n a f a and ))(,(n n b f b is )()()()(n nn n n n b x a b a f b f b f y ---=-. Itintersects x -axise at)()())((n n n n n n n a f b f a b b f b c ---= (Eq. 1.36, p18)1981.0)6.1()(,4907.0)4.2()(00-=-==-=f b f f a f ,8301.1)()())((0000000-=---=a f b f a b b f b c ;Since 0095.0)(0-=c f , then ]8301.1,4.2[],[11--=b a . Similarly, we have1.84093- 1=c , ]1.84093- ,4.2[],[22-=b a 1.84139- 2=c , ]1.84139- ,4.2[],[33-=b a -1.841403=c10. Bisection method: Assume that ],[n n b a contains the root. Then 2nn n b a c +=. (a) 1587.1)4(,4;1425.0)3(,300==-==f b f a , then 5.30=c .Since 03746.0)5.3()(0>==f c f , then ]5.3,3[],[11=b a .Similarly, we can obtain ,,,321c c c . The results are listed in Table 3.The values of tan(x) at midpoints are going to zero while the sequence converges(b) Since 0)3tan(<=, there exist a root in )3,1(..0-=, 055741425.1tan(>)1The results using Bisection method are listed in Table 4.Although the sequence converges, the values of tan (x) at midpoints are not going to zero.P36 2. 3)(2--=x x x f has two zeros 2131±=x . (3028.2,3028.121≈-≈x x ) The first derivative of 3)(2--=x x x f is 12)(-='x x f .The Newton-Raphson iterative function is 123)()()(2-+='-=x x x f x f x x g . The Newton-Raphson formula is 12321-+=+n nn p p p , ,2,1,0=n . The results are listed in Table 5 with starting value p 0=1.6 and p 0=0.0 respectively.Obviously, the sequence generated by the starting value p 0=0.0 does not converge.11. Use Newton-raphson method to solve 0)(3=-=A x x f .The derivative of )(x f is 23)(x x f ='.3232)()()(223x Ax x A x x f x f x x g +=+='-=.Newton-Raphoson formula is 32211--+=n n n p Ap p , ,2,1=n .Since 3A p = is a zero of A x x f -=3)( and 10332)(33<=⎥⎦⎤⎢⎣⎡-='=Ap x A p g ,The sequence generated by the recursive formula 32211--+=n n n p Ap p will converge to3A p = for any starting value ],[330δδ+-∈A A p , where 0>δ.·Answers for Exercises —Numerical methods using MatlabChapter 2P44 2. Solution The 4th equation yields 24=x .Substituting 24=x to the 3rd equation gives 53=x .Substituting both 24=x and 53=x to the 2nd equation produces 32-=x . 21=x is obtained by sustituting all 32-=x , 53=x and 24=x to the 1st equation. The value of the determinant of the coefficient matrix is 115573115=⨯⨯⨯=D .4. Proof (a) Calculating the product of the two given upper-triangular matrices gives⎥⎥⎥⎦⎤⎢⎢⎢⎣⎡++++=⎥⎥⎥⎦⎤⎢⎢⎢⎣⎡⎥⎥⎥⎦⎤⎢⎢⎢⎣⎡=33333323232222223313231213112212121111113323221312113323221312110000b a b a b a b a b a b a b a b a b a b a b b b b b b a a a a a a B A . It is also an upper-triangular matrix.(b) Let N N ij a A ⨯=)( and N N ij b B ⨯=)( where 0=ij a and 0=ij b when j i >.Let N N ij c B A C ⨯==)(. According to the definition of product of the two matrices, we have ∑==Nk kjik ij b ac 1for all N j i ,,2,1, =.0=ij c when j i > because 0=ij a and 0=ij b when j i >.That means that the product of the two upper-triangular matrices is also upper triangular.5. Solution From the first equation we have 31=x .Substituting 31=x to the second equation gives 22=x .13=x is obtained from the third equation and 14-=x is attained from the last equation.The value of the determinant of the coefficient is 243)1(42)det(-=⨯-⨯⨯=A7. Proof The formula of the back substitution for an N N ⨯upper-triangular system is N NN a b x =and kkNk j jkj k k a x a b x ∑+=-=1 for 1,,2,1 --=N N k .The process requiresN N=+++111 divisions, 22)1()1(212NN N N N -=-=-+++ multiplications, and2)1(212NN N -=-+++ additions or subtractions.P53 1. Solution Using elementary transformations for the augmented matrix gives330012630464275101263046425232103514642],[3231213121⎥⎥⎥⎦⎤⎢⎢⎢⎣⎡--−−→−⎥⎥⎥⎦⎤⎢⎢⎢⎣⎡---−−−→−⎥⎥⎥⎦⎤⎢⎢⎢⎣⎡--=++-+-r r r r r r B AThat means that ⎪⎩⎪⎨⎧=++=++-=-+523 1035 4642321321321x x x x x x x x x is equivalent to⎪⎩⎪⎨⎧==+-=-+33 1263 4642332321x x x x x x The set of solutions is .3,2,1123-===x x x11. Solution Using the algorithm of Gaussian Elimination gives12420010324050110700211242001032409013270021],[212⎥⎥⎥⎥⎦⎤⎢⎢⎢⎢⎣⎡----−−−→−⎥⎥⎥⎥⎦⎤⎢⎢⎢⎢⎣⎡--=+-r r B A ⎥⎥⎥⎥⎦⎤⎢⎢⎢⎢⎣⎡------−−→−+1242001032005011070021324r r ⎥⎥⎥⎥⎦⎤⎢⎢⎢⎢⎣⎡------−−→−+21000103200501107002143r r The set of solutions of the system is obtained by the back substitutions,3,2,2234==-=x x x and .11=x(Chasing method for solving tridiagonal linear systems)14. (a) (i) Solution Applying Gaussian elimination with partial pivoting to the augment matrix results in⎥⎥⎥⎦⎤⎢⎢⎢⎣⎡---−−→−⎥⎥⎥⎦⎤⎢⎢⎢⎣⎡---=↔1100320001.0101001.01003001.010030001.010*******],[31r r B A ⎥⎥⎥⎦⎤⎢⎢⎢⎣⎡--−−→−⎥⎥⎥⎦⎤⎢⎢⎢⎣⎡--−−→−↔+-+-00043.03333.43019933.996667.630001.0100319933.996667.63000043.03333.430001.01003 3231213231r r r r r r ⎥⎥⎥⎦⎤⎢⎢⎢⎣⎡---−−−−→−+-6806.00625.680019933.996667.630001.01003326667.633333.43r rThe set of solutions is,101.0524,0100.0-623⨯==x x and .105.2400 -61⨯=x15. Solution The N N ⨯Hilbert matrix is defined byN N ij H H ⨯=)( where 11-+=j i H ij for N j i ≤≤,1.(a) The inverse of the 44⨯ Hilbert matrix is⎥⎥⎥⎥⎦⎤⎢⎢⎢⎢⎣⎡--------=-280042001680140420064802700240168027001200120140240120161H The exact solution is T X )140,240,120,16(--=.(b) The solution is T X )0881.185,0628.310,6053.149,7308.18(--=.>>1 H is ill-conditioned. A miss is as good as a mile. (失之毫厘,谬以千里)P62 5 (a) Solving B LY = gives TY )2,12,6,8(-=. From Y UX = we have TX )2,1,1,3(-=. The product of A and X is TAX )4,10,4,8(--=.That means B AX =(b) Similarly to the part (a), we haveTY )1,12,6,28(=, TX )1,2,1,3(=, and B AX T==)4,23,13,28(.6. ⎥⎥⎥⎥⎦⎤⎢⎢⎢⎢⎣⎡---=175.113011*********L , ⎥⎥⎥⎥⎦⎤⎢⎢⎢⎢⎣⎡-----=5.70001040085304011UP72 7. (a) Jacobi Iterative formula is ()⎪⎪⎩⎪⎪⎨⎧-+=+-=++-=+++)()()1()()()1()()()1(226141358k k k k k k k k k y x z z x y z y x for ,2,1,0=kResults for ),,()()()(k k k k z y x P =’, ,3,2,1=k are listed in Table 2.1 with starting value )0,0,0(0=P .The numerical results show that Jacobi iteration does not converge.(b) Gauss-Seidel Iterative formula is()⎪⎪⎩⎪⎪⎨⎧-+=+-=++-=++++++)1()1()1()()1()1()()()1(226141358k k k k k k k k k y x z z x y z y x for ,2,1,0=kResults ),,()()()(k k k k z y x P =’, ,3,2,1=k are listed in Table 2.2 with starting value )0,0,0(0=P ’Reasons:Conside the eigenvalues of iterative matricesSplit the coefficient matrix ⎥⎥⎥⎦⎤⎢⎢⎢⎣⎡----=612114151A into three matrices⎥⎥⎥⎦⎤⎢⎢⎢⎣⎡-⎥⎥⎥⎦⎤⎢⎢⎢⎣⎡---⎥⎥⎥⎦⎤⎢⎢⎢⎣⎡-=--=000100150012004000600010001U L D A .The iterative matrix of Jacobi iteration is⎥⎥⎥⎥⎦⎤⎢⎢⎢⎢⎣⎡--=⎥⎥⎥⎦⎤⎢⎢⎢⎣⎡----⎥⎥⎥⎥⎦⎤⎢⎢⎢⎢⎣⎡=+=-061311041500121041506100010001)(1U L D T JThe spectral raduis of J T is 16800.5)(>=ρJ T . )1176.0,4546405880(i . .-±=λ’ So Jacobi method doesnot converge.Similarly, the iterative matrix of Gauss-Seidel iteration is⎥⎥⎥⎥⎦⎤⎢⎢⎢⎢⎣⎡--=-=-65503200150)(1U L D T G .The spectral radius of G T is 2532.19)(=ρG T >1. )0866.0,2532.19,0(-=λ’ So Gauss-Seidel method does not converge.8. (a) Jacobi Iterative formula is()⎪⎩⎪⎨⎧-+=-+=+-=+++6/225/)8(4/)13()()()1()()()1()()()1(k k k k k k k k k y x z z x y z y x for ,2,1,0=k ),,()()()(k k k k z y x P =’ for 10,,2,1 =k are listed in Table 2.3 with starting value )0,0,0(0=P .Jacobi iteration converges to the solution (3, 2, 1)’(b) Gauss-Seidel iterative formula is()⎪⎪⎩⎪⎪⎨⎧+---=+---=+-=++++++)1()1()1()()1()1()()()1(22614/)8(4/)13(k k k k k k k k k y x z z x y z y x for ,2,1,0=k ),,()()()(k k k k z y x P =’ for 10,,2,1 =k are listed in Table 2.4 with starting value )0,0,0(0=PGauss-Seidel iteration converges to the solution (3, 2, 1)’Answers for Exercises —Numerical methods using MatlabChapter 3P99 1. Solution (a) The nth order derivative of )sin()(x x f = is )2sin()()(π+=n x x f n .Therefore, !5!3)(535x x x x P +-=, !7!5!3)(7537x x x x x P -+-= and !9!7!5!3)(97539x x x x x x P +-+-=.(b) Estimating the remainder term gives71091075574.2!101!10)5sin()(-⨯≤≤π+=x c x E for 1≤x .(c) Substituting 4π=x to )2sin()()(π+=n x x f n gives ,22)4()4(,22)4()4()3(-=π=π''=π'=πf f f f and 22)4()4()5()4(-=π=πf f .By using Taylor polynomial we have!5)4(22!4)4(22!3)4(22!2)4(22)4(2222)(54325π-+π-+π--π--π-+=x x x x x x P P108 1. (a) Using th e Horner ’s method to find )4(P givesSo )4(P =1.18.(b) From part (a) we have 12.002.002.0)(2-+-=x x x Q . )4()4(Q P =' can be also obtained byusing Horner ’s method.So )4(P '=-0.36 Another method:Hence, P(4)=-0.36.(c) Find )4(I and )1(I firstly.Then=-=⎰)1()4()(41I I dx x P 4.3029.(d) Use Horner ’s method to evaluate P (5.5)Hence, P (5.5)=0.2575.(d) Let 012233)(a x a x a x a x P +++=. There are 4 coefficients needed to found.Substituting four known point ),(i i y x , i =1, 2, 3, 4, into )(x P gives four linear equations with unknowni a , i =1, 2, 3, 4.54.10123=+++a a a a 5.12480123=+++a a a a 42.139270123=+++a a a a 66.05251250123=+++a a a aThe coefficients can be found by solving this linear system: .66.1,2.0,1.0,02.00123=-==-=a a a aP120 1. The values of f (x ) at the given points are listed in Table 3.1:(a) Find the Lagrange coefficient polynomials and 010)(0,1x x x L -=---=.1101)(1,1+=++=x x x LThe interpolating polynomial is x x L f x L f x P =+-=)()0()()1()(1,10,11. (b) ),(21)11()1()(20,2x x x x x L -=----=,110)1)(1()(21,2x x x x L -=--+=),(212)1()(22,2x x x x x L +=+=x x L f x L f x L f x P =++-=)()1()()0()()1()(2,21,20,22. (c) ),2)(1(61)21)(11()2)(1()(0,3---=-------=x x x x x x x L),2)(1)(1(21)20)(10)(10()2)(1)(1()(1,3--+=--+--+=x x x x x x x L),2)(1(21)21(1)11()2()1()(2,3-+-=-+-+=x x x x x x x L),1)(1(61)12(2)12()1()1()(3,3-+=-+-+=x x x x x x x L33,32,31,30,33)()2()()1()()0()()1()(x x L f x L f x L f x L f x P =+++-=(d) ,2212)(0,1x x x L -=--=,1121)(0,1-=--=x x x L 67)()2()()1()(1,10,11-=+=x x L f x L f x P . (e) ),23(21)20)(10()2)(1()(20,2+-=----=x x x x x L ),2()21(1)2()(21,2x x x x x L --=--=),(21)12(2)1()(20,2x x x x x L -=--=.23)()2()()1()()0()(22,21,20,22x x x L f x L f x L f x P -=++=7. (a) Note that each Lagrange polynomial )(,2x L k is of degree at most 2 and )(x g is a combination of)(,2x L k . Hence )(x g is also a polynomial of degree at most 2.(b) For each k x , 2,1,0=k , the Lagrange coefficient polynomial 1)(,2=k k x L , and 0)(,2=k j x L for k j ≠, 2,1,0=j . Therefore, 01)()()()(2,21,20,2=-++=k k k k x L x L x L x g .(c) )(x g is a polynomial of degree 2≤n and has n ≥ 3 zeroes. According to the fundamental theorem of algebra, 0)(=x g for all x .9. Let )()()(x P x f x E N N -=. )(x E N is a polynomial of degree N ≤.)(x f is degree with )(x P N at N +1 points N x x x ,,,10 implies that )(x E N has N +1 zeroes. Therefore, 0)(=x E N for all x , that is, )()(x P x f N = for all x .P131 6. (a) Find the divided-difference table:(b) Find the Newton polynomials with order 1, 2, 3 and 4.)0.1(80.16.3)(1--=x x P , )0.2)(0.1(6.0)0.1(80.160.3)(2--+--=x x x x P ,)0.3)(0.2)(0.1(15.0)0.2)(0.1(6.0)0.1(80.16.3)(3------+--=x x x x x x x P , )0.4)(0.3)(0.2)(0.1(03.0 )0.3)(0.2)(0.1(15.0)0.2)(0.1(6.0)0.1(80.16.3)(4----+------+--=x x x x x x x x x x x P .(c)–(d) The results are listed in Table 3.2P143 6. x x x T 32)(323-=, ]1,1[-∈x .The derivative of )(3x T is 323)(223-⋅='x x T . 0)(3='x T yields 21±=x . Evaluating )(3x T at 21±=x and 1±=x gives 1)1(3-=-T , 1)21(3=-T , 1)21(3-=T and 1)1(3=T .Therefore, 1))(max(3=x T , 1))(min(3-=x T .10. When 2=N , the Chebyshev nodes are ,23)6/5cos(0-=π=x ,01=x and 23)6/cos(2=π=x .Calculating the Lagrange coefficient polynomials based on 210,,x x x can produce the following results:,323)232(23)23()(20,2x x x x x L +-=⨯-⨯--=,341)23(23)23()23()(21,2x x x x L -=-⨯-+=.32323232)23()(22,2x x x x x L +=⨯⨯+=The proof is finished.Answers for Exercises —Numerical methods using MatlabChapter 4P157 1(a). Solution The sums for obtaining Normal equations are listed in Table 4.1The normal equations are ,710=A 135=B . Then ,7.0=A 6.2=B .The least-squares line is 6.27.0+=x y .2449.0)((51)(215122=⎪⎪⎭⎫ ⎝⎛-=∑=k k k x f y f EP158 4. Proof Suppose the linear-squares line is B Ax y += where A and B satisfiesthe Normal equations ∑∑===+N k k Nk ky xAB N 11and ∑∑∑====+Nk k k N k k N k k y x x A x B 1121.y y N x A B N N B x N A B x A N k k Nk k N k k ==⎪⎪⎭⎫ ⎝⎛+=+⎪⎪⎭⎫ ⎝⎛=+∑∑∑===111111 meas thatthe point ),(y x lies on the linear-squares line B Ax y +=.5. First eliminating B on the Normal equations∑∑===+Nk k Nk k y x A B N 11and ∑∑∑====+Nk k k Nk k Nk k y x x A x B 1121gives⎪⎪⎭⎫ ⎝⎛-=∑∑∑===Nk k N k k N k k k y x y x N D A 1111 where 2112⎪⎪⎭⎫ ⎝⎛-=∑∑==N k k N k k x x N D . Substituting A into the first equation gets⎪⎪⎭⎫⎝⎛⎪⎪⎭⎫ ⎝⎛+-=∑∑∑∑∑=====Nk k Nk k N k k k N k k N k k y x N y x x y N D D B 12111111. Note that ∑∑∑∑∑∑∑∑========⎪⎪⎭⎫ ⎝⎛-=⎪⎪⎭⎫ ⎝⎛⎪⎪⎭⎫ ⎝⎛-=N k k N k k N k k N k k N k k N k k N k k Nk k y x N y x y x x N N y N D 12111212112111. Simplifying B gives⎪⎪⎭⎫ ⎝⎛-=∑∑∑∑====Nk k k N k k Nk k N k k y x x y x D B 111121.8(b). The sums needed in the Normal equations are listed in Table 4.26177.142==∑∑k kk x y x A )2(=M5606.063==∑∑kkkxy x B )3(=MHence, 26177.1x y = and 35606.0x y =.0.3594 )(51)(21512222=⎪⎪⎭⎫ ⎝⎛-=∑=k k k Ax y Ax E , 1.1649 )(51)(21512332=⎪⎪⎭⎫ ⎝⎛-=∑=k k k Bx y Bx E .26175.1x y =fits the given data better.P171 2(c). The sums for normal equations are listed in Table 4.3.Using the formula∑∑∑∑∑∑∑========++=++52545352515135125151512515k kk k k k k k k k kk k k ky x x A x B x C y x A xB Cproduces the system with unkowns A , B , and CSolving the obove system gives .6.0,1.0,5.2-=-==C B A The fitting curve is .6.01.05.22--=x x yP172 4. (a) Translate points in x-y plane into X-Y plane using y Y x X ln ,==. The results arelisted in Table 4.4.The Normal equationsgive the system .793410,110,22105=+-==+A C B A C ∑∑∑∑∑======+=+515125151515k kk k k k k k kk k Y X X A X B Y X A B .8648.0155,2196.455-=+=+A B A BThen -0.50844=A , 1.3524=B . Thus 866731.3524.e e C B ===.The fitting curve is xe .y 50844.086673-=, and 1190.0)((51)(215122=⎪⎪⎭⎫ ⎝⎛-=∑=k k k x f y f E .(b) Translate points in x-y plane into X-Y plane using yY x X 1,==. The results are listed in Table 4.5.The Normal equationsgive the system Then 2432.0=A , 30280.0=B .The fitting curve is 30280.02432.01+=x y and 5548.4)((51)(215122=⎪⎪⎭⎫ ⎝⎛-=∑=k k k x f y f E .(c) It is easy to see that the exponential function is better comparing with errors in part (a) and part (b)..1620.5155,7300.255=+=+A B A B ∑∑∑∑∑======+=+515125151515k kk k k k k k kk k Y X X A X B Y X A BP188 1. (a) Derivativing )(x S gives 232132)(x a x a a x S ++='. Substituting the conditions intothe derivative pruduces the system of equations . 012428420321 32132103213210⎪⎪⎩⎪⎪⎨⎧=++=+++=++=+++a a a a a a a a a a a a a a (b) Solving the linear system of equations in (a) gives 29,,12,63210-==-==a a a a . The cubic polynomial is 3229126)(x x x x S -+-=.Figure: Graph of the cubic polynomial4. Step 1 Find the quantities: 3,1210===h h h , 21/)20(/)(0010-=-=-=h y y d13/)03(/)(1121=-=-=h y y d , 6667.03/)31(/)(2232-=-=-=h y y d18)(6011=-=d d u , 10)(6122-=-=d d uStep 2 Use ⎪⎪⎩⎪⎪⎨⎧-'-=⎪⎭⎫ ⎝⎛++'--=+⎪⎭⎫⎝⎛+))((3232))((32232322211100121110d x S u m h h m h x S d u m h m h h to obtain the linear system⎩⎨⎧-=+=+0001.155.1032135.72121m m m m .The solutions are 5161.2,8065.321-==m m . Step 3 Compute 0m and 3m using clamaped boundary.4.90322))((310000-=-'-=m x S d h m , 2.92482))((322323=--'=md x S h m Step 4 Find the spline coefficients16)2(,210001,000,0-=+-===m m h d s y s , 1.45166,-2.45162013,002,0=-===h m m s m s ;-1.54856)2(,021111,110,1=+-===m m h d s y s , -0.35136,1.903321123,112,1=-===h m m s m s ;0.38716)2(,332221,220,2=+-===m m h d s y s , 0.30236,-1.258122233,222,2=-===h m m s m s ;Step 5 The cubic spline is320)3(4516.1)3(4516.2)3(2)()(+++-+-==x x x x S x S for 23-≤≤-x ,321)2(3513.0)2(903.1)2(5484.1)()(+-+++-==x x x x S x S for 12≤≤-x , and322)1(3023.0)1(2581.1)1(3871.03)()(-+---+==x x x x S x S for 41≤≤x .5. Calculate the quantities: 3,1210===h h h , 20-=d ,11=d , 6667.02-=d ,181=u , 102-=u . ( Same values as Ex. 4)Substituting }{j h , }{j d and }{j u into ()()⎩⎨⎧=++=++22211112111022u m h h m h u m h m h h gives ⎩⎨⎧-=+=+1012318382121m m m mSolve the linear equation to obtain .5402.1,8276.221-==m m In addition, .030==m m Use formula (4. 65) to find the spline coefficients:4713.26)2(,210001,000,0-=+-===m m h d s y s , 4713.06,020013,002,0=-===h m m s m s ;-1.05756)2(,021111,110,1=+-===m m h d s y s , -0.24276,4138.121123,112,1=-===h m m s m s ;8735.06)2(,332221,220,2=+-===m m h d s y s , 0856.06,7701.0-22233,222,2=-===h m m s m s .Therefore, 30)3(4713.0)3(4713.22)(+++-=x x x S , for 23-≤≤-x ;321)2(2427.0)2(4138.1)2(0575.1)(+-+++-=x x x x S , for 12≤≤-x322)1(0856.0)1(7701.0)1(8735.03)(-+---+=x x x x S for 41≤≤x .Answers for Exercises —Numerical methods using MatlabChapter 5P209 1(b). Solution LetThe result of using the trapezoidal rule with h =1 isUsing Simpson’s rule with h=1/2, we haveFor Simpson’s 3/8 rule with h=1/3, we obtainThe result of using the Boole’s rule with h=1/4 is4. Proof Integrate )(1x P over ],[10x x .110102012101)(2)(2)(xx x x x x x x h f x x h f dx x P -+--=⎰=)(210f f h +. The Quadrature formula )(2)(1010f f hdx x f x x +≈⎰is called the trapezoidal rule.6. Solution The Simpson ’s rule is)4(3)(21010f f f hdx x f x x ++≈⎰. It will suffice to apply Simpson ’s rule over the interval [0, 2] with the test functions32,,,1)(x x x x f = and 4,x . For the first four functions, since)1141(31212+⨯+==⎰dx , )2140(31220+⨯+==⎰xdx , )4140(3138202+⨯+==⎰dx x , )8140(314203+⨯+==⎰dx x , the Simpson ’s rule is exact. But for 4)(x x f =,)16140(3153224+⨯+≠=⎰dx x . .Therefore, the degree of precision of Simpson ’s rule is n =3.T he Simpson’s rule and the Simpson ’s 3/8 rule have the same degree of precision n =3.).4cos(1)(x e x f x -+=..f f f f h dx x f 3797691))1()0((21)(2)(1010=+=+≈⎰.9583190))1()5.0(4)0((61)4(3)(21010. f f f f f f h dx x f =++=++≈⎰.9869270 ))1()3/2(3)3/1(3)0(( 8/1 )33(83)(321010.f f f f f f f f hdx x f =+++=+++≈⎰.008761 ))1(7)4/3(32)2/1(12)4/1(32)0(7( 90/1 )73212327(452)(432101.f f f f f f f f f f hdx x f =++++=++++≈⎰P220 3(a) Solution When 3)(x x f =for 10≤≤x , ⎰+π=123912dx x x area .The values of 2391)(x x x g +=at 11 sample points (M =10) are listed in the Table 5.1:(i) Using the composite Trapezoidal rule ∑-=++=110)()()((2),(M k k M x g h x g x g hh g T , the computation is)9156.11084.16098.03719.01563.00710.00280.00081.00010.0(101)1623.30(201)101,(++++++++++=g T=)2160.4(101)1623.3(201+=0.1576+0.4216=0.5792.(ii) Using the composite Simposon ’s rule ∑∑-=--=+++=11121120)(34)(32)()((3),(M k k M k k M x g h x g h x g x g h h g S , the computation is)9156.16098.01563.00280.00010.0(304)1084.13719.00710.00081.0(302 )1623.30(301)101,(++++++++++=g S=)7106.2(304)5054.1(302 )1623.3(301++=0.5672.7. (a) Because the formula)2()1()0()(2102g w g w g w dt t g ++=⎰is exact for the three functions 1)(=t g ,x t g =)(, and 2)(x t g =, we obtain three equations with unkowns 0w , 1w , and 2w :2210=++w w w , 2221=+w w ,38421=+w w . Solving this linear system gives 310=w , 341=w and 312=w .Thus, ())2()1(4)0(31)(20g g g dt t g ++=⎰(b) Let ht x x +=0 and denote ,01h x x +=.202h x x +=Then the change of variable ht x x +=0 translates ],[20x x into [0, 2] and converts the integral expresion dx x f )( into dt ht x hf )(0+. Hence,()())()(4)(3)2()1(4)0(3)()()(21022002x f x f x f h g g g hdt t g h dt ht x f h dx x f x x ++=++==+=⎰⎰⎰. The formula ())()(4)(3)(21020x f x f x f hdx x f x x ++=⎰ is known as the Simpson ’s rule over ],[20x x .8(a).9(a).P234 1(a) Let 212sin )(x xx f +=. The Romberg table with three rows for ⎰+3212sin dx x xis given as follows:Where04191.0)02794.0(23)106sin 0(23))3()0((23)0()0,0(-=-=+=+==f f T R , 04418.0)5.113sin (5.1204191.0)5.1(5.12)0()1()0,1(2=++-=+==f T T R ,3800.0)25.215.4sin 75.015.1sin (75.0204418.0))25.2()75.0((75.02)1()2()0,2(22=++++=++==f f T T R , 07288.03)04191.0(04418.043)0,0()0,1(4)1()1,1(=--⨯=-==R R S R ,].6/,6/[],[ and cos )(Let ππ-==b a x x f have we , and cos )( ,sin )( Since Mab h x x f x x f -=-=''-='.10513/123/ )(12),(922-⨯<⨯⎪⎭⎫ ⎝⎛ππ≤''--=M c f h a b h f E T .1039.2/)( and )4375( 9.4374 So,4-⨯≈-==>M a b h M M ].6/,6/[],[ andcos )(Let ππ-==b a x x f ,cos )( ,sin )( , cos )( ,sin )( Since )4(x x f x x f x x f x x f =='''-=''-='.105123/1803/ )(180),(92)4(4-⨯<⨯⎪⎭⎫ ⎝⎛ππ≤--=M c f h a b h f E S havewe ,2 and M a b h -=.1027.92/)( and )565( 8.564 So,4-⨯≈-==>M a b h M M4919.0304418.03800.043)0,1()0,2(4)2()1,2(=-⨯=-==R R S R ,5198.0307288.04919.01615)1,1()1,2(16)2()2,2(=-⨯=-==R R B R ,2. Proof If L J T J =∞→)(lim , thenL LL J T J T J S J J =-=--=∞→∞→343)1()(4lim)(lim andL LL J S J S J B J J =-=--=∞→∞→151615)1()(16lim )(lim .9. (a) Let 78)(x x f =. 0)()8(=x f implies 4=K . Thus 256)4,4(=R .(b) Let 1011)(x x f =.0)()11(=x f implies 5=K . Thus 2048)5,5(=R .10. (a) Do variable translation t x =. Thendt t dt t t dx x tx ⎰⎰⎰=⋅===121122.That means the two integrals dx x ⎰1anddt t ⎰122have the same numerical value.(b) Let 22)(t t f = and x x g =)(.Use dt t R ⎰≈122)1,1( means that the truncation error is )()4(n f k ξ approximately.Note that 0)()3(=t f . It means )1,1(212R dt t =⎰.But for x x g =)(, 0)()(=x g n is not true for all ]1,0[∈x and any integer 0>n .Thus the Romberg sequence is faster for dt t ⎰122 than fordx x ⎰1even though they have the samenumerical value.P242 1 (a) Applying the change of variable 22ab x a b t ++-=to dt t ⎰256 givesdx x dt t x t ⎰⎰-+=+⋅==115125)1(66.Thus the two integrals are dt t ⎰256 anddx x ⎰-+⋅115)1(6equivalent.(b)315315311525)1(6)1(6)()1(66=-=-+++=≈+⋅=⎰⎰x x x x f G dx x dt t =0.0809 +58.5857=58.6667If using )(3f G to approximate the integral, The result is535055353115205)1(695)1(698)1(695)()1(66==-=-+++++=≈+⋅=⎰⎰x x x x x x f G dx x dt t64105.5965956.0000 98 0.0035 95=⨯+⨯+⨯=6. Analysis: The fact that the degree of precision of N -point Gauss-Legendra integration is 2N -1 impliesthat the error term can be represented in the form )()()2(c kf f E N N =.(a) Since dt t dx x tx ⎰⎰-+=+==117127)1(88, and ()0)1()8(7=+t implies 82=K . Thus =256)(4=f G .(b),)1(111111101210dt t dx x tx ⎰⎰-+=+==and ()0)1()11(10=+t implies 122=K .Thusdx x⎰21011=2048)(6=f G .7. The n th Legendre polynomial is defined by The first five polynomials areThe roots of them are same as ones in Table 5.8.11. The conditions that the relation is exact for the functions means the three equations:326.0 6.0 0)6.0( )6.0(2 3132111321=+=+-=++w w w w w w w Sloving the system gives 98 ,95 231===w w w . ))6.0((95)0(98))6.0((95)(212111f f f dx x f ++-≈⎰- is called three-point Gauss-Legendre rule.Answers for Exercises —Numerical methods using MatlabChapter 6P249 1. (a) Proof Differentiate 22)(2+-+=-t t Ce t y t .22)(-+-='-t Ce t y tSubstitute )(t y and )(t y ' into the right-hand side of the equation y t y -='2.side left )(22)22(side right 222='=-+-=+-+-=-=--t y t Ce t t Ce t y t t t(b) Solution Let y t y t f -=2),(. Then 1),(-=y t f y for any ),(y t .So, the Lipschitz constant is 1=L .()[],2,11!21)(1)(20=-⋅==n x dx d n x P x P nnn n n ()()()3303581)(3521)(1321)()(1)(244332210+-=-=-===x x x P x x P x x P x x P x P 2,,1)(x x x f =12 . Integrate both side of )(t f y =' over [a , b ]: ⎰⎰='=-babadt t f dt y a y b y )()()(.Then,)()()(a y b y dt t f ba-=⎰, where )(t y is the solution of the I. V . P)(t f y =', for b t a ≤≤ with 0)(=a y . That means that the definite integral⎰badt t f )( can becomputed using the two values )(a y and )(b y of the solution )(t y of the given I. V . P.. 14. Solution Separate the two variables of the equation 211t y +=' into the form dt tdy 211+=. Integrate dt tdy 211+=and yeild the general solution C t y +=arctan . The initial-value condition 0)0(=y means that 0=C . The solution for the I. V . P. is t y arctan =.P257 3. (a)-(c) The formula using Euler ’s method to solve the I. V . P. ty y -=', 1)0(=y canbe represented in the form k k k y ht y )1(1-=+. When 2.0=h and 1.0=h , the results are listed in Table 6.1.(d) The F. G . E. does decrease half approximatelly as expacted when h is halved.6. When 02.0=a , 00004.0=b and 10=h , the Euler ’s formula for 2bP aP P -=' is in the form210004.02.1k k k P P P -=+. With 1.760=P , the missing entries can be filled in the table.。
数值分析英文版课件 (6)

12
4.1 Matrix Algebra (III)
The matrix A is denoted as:
1.1 −0.12 3.0 6.2 0.0 0.15 0.6 −4.0 1.3 2.1 8.2 9.3 then a12 = 1.1, or A( 3,2 ) = −4.0 , or ( A )43 = 8.2 .
10
4.1 Matrix Algebra (I)
A matrix is a rectangular array of numbers such as
1.1 −0.12 3.0 6.2 0.0 0.15 , 0.6 −4.0 1.3 9.3 2.1 8.2 3.2 −17 , −4.7 0.11
18
4.1 Matrix Algebra (IX)
Theorem 1 (on equivalent systems):
If one system of equations is obtained from another by a finite sequence elementary operations, then the two systems are equivalent .
3
INDEX
4.0 Introduction 4.1Matrix Algebra 4.2 LU and Cholesky Factorizations
4
INDEX
4.0 Introduction 4.1Matrix Algebra 4.2 LU and Cholesky Factorizations
p
(1 ≤ i ≤ix Algebra (VII)
数值分析(浙江大学)全套课件

数值分析 (第七版 影印版)
Richard L. Burden & J. Douglas Faires (高等教育出版社)
ห้องสมุดไป่ตู้ 学习方法
1.注意掌握各种方法的基本原理 2.注意各种方法的构造手法 3.重视各种方法的误差分析 4.做一定量的习题 5.注意与实际问题相联系
教材 (Text Book) 数值计算方法 郑慧娆等 编著 (武汉大学出版社)
参考书目 (Reference)
➢ Numerical Analysis:Mathematics of Scientific Computing (Third Edition)
数值分析 (英文版 第3版 )
David Kincaid & Ward Cheney(机械工业出版社)
10
n
0
1
102
0
10 1 101 0
2。与计算机不能分离:上机实习(掌握一 门语言:C语言,会用Matlab)
1.2 误差 ( Error )
§1 误差的背景介绍 ( Introduction ) 1. 来源与分类 ( Source & Classification ) 模型误差 ( Modeling Error ): 从实际问题中抽象出数 学模型
1 e x2 dx 0
(第七章的内容:数值积分)
数值分析的特点
1。近似: 由此产生“误差”
在计算数学和应用数学中一个有趣的问题: 什么是零?
1 10 1 10
原点附近
1
在纯数学中,认为此矩阵为满秩矩阵
10 1
但在计算数学中,它却是降秩矩阵 ?
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9
3.1.0 Introduction (2)
This is a consequence of the Intermediate-Value
Theorem (中值定理).
The bisection method exploits this idea in the
following way:
4
3.0 Introduction(2)
Examples of nonlinear equations can be found in
many applications.
In the theory of diffraction of light, we need the
roots of the equation
and if f (a) f (b) < 0, then f must have a zero in (a, b).
Since f (a) f (b) < 0, the function f changes sign
on the interval [a, b] and, therefore, it has at least one zero in the interval.
Next, e is the computation of the error bound that is established in the theorem to follow.
19
3.1.1Bisection Algorithm(3)
Notice the stopping criteria are present in the
21
3.1.2Error Analysis(1)
To analyze the bisection method, let us denote
the successive intervals that arise the process by [a0, b0], [a1,b1], and so on.
15
3.1.0 Introduction (8)
k 1 2 3 4 13 14 15 16 c 3.5000 3.2500 3.1250 3.1875 3.1829 3.1830 3.1831 3.1831 f c 0.321 0.694 10 1 0.605 10 1 0.625 10 1 0 .122 10 3 0 .193 10 4 0.124 1 0 4 0.345 10 5
14
3.1.0 Introduction (7)
When the bisection algorithm was run on a machine similar to the Marc-32.
The following output was produced, starting with the interval [-4,-3]
20
Topics of Today
3.0 Introduction 3.1 Bisection( Interval Halving) Method
3.1.0 Introduction 3.1.1 Bisection Algorithm 3.1.2 Error Analysis
3.2 Newton’s Method 3.3 Secant Method
16
Topics of Today
3.0 Introduction 3.1 Bisection( Interval Halving) Method
3.1.0 Introduction 3.1.1 Bisection Algorithm 3.1.2 Error Analysis
3.2 Newton’s Method 3.3 Secant Method
Here are some observations about the numbers:
a0 a1 a2 ... b0 b0 b1 b2 ... a0 1 bn 1 an 1 bn an 2
n 0
(1)
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3.1.2Error Analysis(2)
algorithm.
First, M gives the maximum number of steps that the user permits. Next, the calculation can be stopped when either the error is small enough or the value of f (c) is small enough.
If f (a) f (b) < 0, then we compute c = ½(a+b) and test whether f (a) f (c) < 0.
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3.1.0 Introduction (3)
If this is true, then f has a zero in [a, c]. So we rename c as b and start again with the new interval [a, b], which is half as large as the original interval. If f (a) f (c) < 0, then f (c) f (b) < 0, and in this case we rename c as a. In either case, a new interval containing a zero of f has been produced, and the process can be repeated.
Also, we discuss special methods for computing
the zeros of polynomials.
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Topics of Today
3.0 Introduction 3.1 Bisection( Interval Halving) Method
Chapter 3 Solution of Nonlinear Equations
Lecturer: GUO Tongtong Time: 15th December, 2010 The 11th Lecture
1
Topics of Today
3.0 Introduction 3.1 Bisection( Interval Halving) Method 3.2 Newton’s Method
in numerical calculations it is best to compute a quantity by adding a small correction term to a previous approximation.
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3.1.1Bisection Algorithm(2)
x tan x 0
5
3.0 Introduction(3)
In the calculation of planetary orbits, we need the
roots of Kepler’s equation
x a sin x b
for various values of a and b.
Second, it is better to determine whether the function changes sign over the interval using sign (w) ≠ sign (u) rather than wv < 0,
since the latter requires an unnecessary multiplication and could cause an underflow or overflow.
The objective of this chapter is :
To solve the roots of equations (or zeros of functions). To solve a system of nonlinear equations: to find x such that f (x) = 0 or finding X = (x1,x2,…,x n) T so that F (X) = 0.
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3.0 Introduction(4)
In this chapter, we begin with three simple
methods that are quite useful:
The bisection method, Newton’s method, and the secant method.
3.3 Secant Method
2
Topics of Today
3.0 Inon( Interval Halving) Method 3.2 Newton’s Method
3.3 Secant Method
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3.0 Introduction(1)
|f (c)| < 10 - 5.
The bisection method is also known as the
method of interval halving (区间减半法).
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3.1.0 Introduction (6)
EXAMPLE 1
Use the bisection method to find the root of the equation ex = sin x closest to 0.