数四数值分析 期中考试题与参考解答

1、數四數值分析 期中考試題與參考解答科目代碼：BMA42401 11/07/20081. 20% For , (a) Compute the values of f(x) at x=10-2, 10-4, 10-6, 10-8, 10-10(b) Analytically find the limit . (c) Do you think the values in (a) all are correct ? If your answer is 'no' explain what causes the problem and how can you do an improveme

2、nt ? 解：(a) x= 10-2 10-4 10-6 10-8 10-10 ; y= (x-sin(x)./(x.3 ) y = 0.1667 0.1667 0.1667 0 0 (b) (c) No. The loss of accuracy is due to round-off error. The subtraction of nearly equal numbers ( as ) causes the loss of significant digits. By Taylor formula therefore, as x0, we define the function gx=

3、1/6-(x.2)/120+(x.4)/5040 gx = 0.1667 0.1667 0.1667 0.1667 0.1667 2. 25% Consider the equation ex = 3x on 1, 2 . (a) Show that there exists a unique solution in the interval 1,2 . (b) To find a bound for the number of iterations needed to achieve an approximation solution with accuracy 10-6 by the bi

4、section method . And generate the values of the iterations. (c) Using Aitkensmethod to improve the convergence in part (b). 解：(a) Let f(x)= ex - 3x , we obtain f(1)=-0.2817<0 and f(2)=1.3891>0 Hence, there exists a solution in1,2 . Since f '(x)= ex 3 , f(x) is increasing onln3, 2 and decre

5、asing on1,ln3 ln3 1.0986123, and f(1)= -0.2817. Therefore, the solution is unique. (b) a= 1; b=2; tor=10(-6) ; t=6*log2(10) t = 19.9316 Take n = 20 s='3*x-exp(x)' ; n=20; format longp=bisection1(s,a,b, n,tor) n c f(c) error 0 1.500000 0.01831093 0.5 1 1.750000 -0.5046027 0.25 2 1.625000 -0.2

6、03419 0.125 3 1.562500 -0.08323318 0.0625 4 1.531250 -0.03020315 0.03125 5 1.515625 -0.005390404 0.015625 6 1.507813 0.006598107 0.0078125 7 1.511719 0.0006384471 0.00390625 8 1.513672 -0.002367313 0.001953125 9 1.512695 -0.0008622684 0.0009765625 10 1.512207 -0.0001113698 0.0004882813 11 1.511963 0

7、.0002636738 0.0002441406 12 1.512085 7.618579e-005 0.0001220703 13 1.512146 -1.758357e-005 6.103516e-005 14 1.512115 2.930322e-005 3.051758e-005 15 1.512131 5.860353e-006 1.525879e-005 16 1.512138 -5.861477e-006 7.629395e-006 17 1.512135 -5.286926e-010 3.814697e-006 18 1.512133 2.92992e-006 1.907349

8、e-006 19 1.512134 1.464698e-006 9.536743e-007convergent or f(x) is almost 0p = Columns 1 through 4 1.500000000000000 1.750000000000000 1.625000000000000 1.562500000000000 Columns 5 through 8 Columns 9 through 12 Columns 13 through 16 Columns 17 through 20 (d) generate the new sequence : type deltasq

9、r.m function rs=deltasqr(p)%this function is generate the Aitken's delta squaren=length(p) ;for j=1:n-2 h=p(j+1)-p(j); k=p(j+2)-p(j+1); rs(j)=p(j)-(h2)/(k-h) ;end q=deltasqr(p) q = Columns 1 through 4 1.666666666666667 1.500000000000000 1.500000000000000 1.500000000000000 Columns 5 through 8 Col

10、umns 9 through 12 Columns 13 through 16 Columns 17 through 18 compare Matlab fzero('exp(x)-3*x',1 2)format short ans = 3. 25% For the equation x2 = 1+x on 0, 2 . (a) Show that there exists a unique solution in the interval 0,2 . (b) Find two functions g1(x) and g2(x) then use fixed-point ite

11、ration to find an approximation solution with accuracy 10-6 by g1(x) and g2(x) respectively. (c) Plot the graphs of g1(x) and y=x together and locate the points of iterations. 解：(a) Let f(x)= x2 - x- 1 , we obtain f(0)=-1<0 and f(2)=1>0 Hence, there exists a solution in0,2 . Since f '(x)=

12、2x- 1 , f(x) is increasing on0.5, 2 and decreasing on0,0.5 f(0.5)=-1.25 and f(0)=-1 . So the solution is unique. (b) Let g1(x)= , g2(x)= type fixpt.m function fixpt( s, a, n, tor)%fixed-point method%g=inline(s,'x') ;ga = feval(g,a); %evaluate g(a)fprintf('initial a= %10.6f g(a)= %0.7gnn&

13、#39;,a ,ga)%start iterationfprintf(' n x g(x)n')for j = 1: n fprintf('%3d %10.6f %0.7gn',j ,a ,ga) if abs(ga-a) < tor fprintf('nconvergencen') return end a=ga ;%generate next point ga = feval(g,a);end %for s='sqrt(x+1) ' ; a=0 ; % initial n=20; tor= 10(-6) ; fixpt(

14、s, a,n,tor) initial a= 0.000000 g(a)= 1 n x g(x) 1 0.000000 1 2 1.000000 1.414214 3 1.414214 1.553774 4 1.553774 1.598053 5 1.598053 1.611848 6 1.611848 1.616121 7 1.616121 1.617443 8 1.617443 1.617851 9 1.617851 1.617978 10 1.617978 1.618017 11 1.618017 1.618029 12 1.618029 1.618032 13 1.618032 1.6

15、18033 14 1.618033 1.618034convergence s='x2-x-1' newton(s,a,n,tor) initial a= 0.000000 f(a)= -1 n x f(x) 1 -1.000000 1 2 -0.666667 0.1111111 3 -0.619048 0.002267574 4 -0.618034 1.026516e-006 5 -0.618034 2.109424e-013convergence s='x2-x-1' a=1.2; newton(s,a,n,tor) initial a= 1.200000

16、f(a)= -0.76 n x f(x) 1 1.742857 0.2946939 2 1.624302 0.01405529 3 1.618051 3.907091e-005 4 1.618034 3.052978e-010 5 1.618034 0convergence ( c) t=0:01:2 ; gt=sqrt(t+1) ; plot(t,t,t,gt) 4. 25% Nevilles method is used to approximate f(0.4), giving the following table x0 = 0 P0 =1 x1 = 0.25 P1 =2 P01 =2

17、.6 x2 = 0.5 P2 =? P12 =? P012 =? x3 = 0.75 P3 =8 P23 =2.4 P123 =2.96 P0123 =3.016 (a) Determine P2 , P12 and P012 . (b) Use the data in the table to construct the Newtons divided-difference coefficients and find the approximation of f(0.4) by forward-difference formula . 解：(a) Similarly, 或執行m-file n

18、evl.m得下列結果path('d:numerical',path) n=4; t=0.4;x=0 0.25 0.5 0.75 ; y=1 2 4 8 ; rs= nevl(x,y,n,t) rs = 1.0000 0 0 0 2.0000 2.6000 0 0 4.0000 3.2000 3.0800 0 8.0000 2.4000 2.9600 3.0160 (b) 執行m-file coeff.m得Newton's divided-difference coefficients C=coeff(x,y,n) ;t = 0.4; for j =1:nc (j)= C(j, j) ; %take the main diagonal entries end val = evalnt(c, x, t, n) ;fprintf('The value of p(x) at t=%f is :%f',t,val) The value of p(x) at t=0.400000 is :3.016000 與上面Neville's P0123=3.016是一致的。5. 15% (a) A natural cubic spline S on 0,2 is defined by Find b, c, and d . (b) Plot th