数值分析答案

1、习题三上机1.解：雅可比迭代法：代码：function S=GS(A,b,xO)D=diag(diag(A);L=-tril(A,-l);U=-triu(A,1);G=(D-L)U;f=(D-L)b;S=G*xO+f；while norm(S-xO)=1.Oe-4xO=S;S=G*xO+f;End运行结果：Command WindowNew to MATLAB? Watch this Viduo, see Dwmo, or read Getting Started. A =5 2 1;-1 4 2; 2 -3 10:b= -12;20;3;x0 二0:0:0: j ac (A? b? xO)a

2、rts =-4. 00003. 00002. 0000高斯一赛德尔迭代法：代码：function J=jac(A,b,x0)D=diag(diag(A);D=inv(D);L=tril(A,T);U=triu(A,1);B=-D* (L+U)；f=D*b；J=B*x0+f；while norm(J-xO)=1.Oe-4xO=J;J=B*x0+f;end运行结果：Command Window New to MATLAB? Watch this Video, see Demoj or read Getting Started. A=5 2 1;-1 4 2; 2-3 10;b=-12:20：3;x

3、0= 0:0:0:GSCA.b.xO)ans =-4. 00003.00002. 000015 .解：代码：function x=Gauss(A,b)m,n=size(A);if m=nerror(!0O62EQ -061)returnendB=A,b;n=length(A);for j=l:n-1q= zeros (j-lf 1)j) ; $C 为列主元,t 为所在行crr=max(abs(q);if s = jtemp=B (j, :) ;&交换两行B(j,：)=B(，:)；B(:)=temp;endfor i=j+1:nendendx(n)=B(n,n+l)/B(n,n);for i=n

4、-l:-l:1for j=i:n-lB(i,n+l)=B(i,n+l)-B(i,j+1)*x(j+1);endx(i)=B(irn+l)/B(iri);end运行结果：A=0.4096,0.1234,0.3678,0.2943;0.2246,0.3872,0.40150.1129;0.3645,0.1920,0.3781,0.0643;0.1784,0.4002,0.2786,0.3927;b=0.4043;0.15S0;0.4240;0.2557;Command Windowi- Nev/ to MATLAB? V/atch this Video, see Demos, or read Ge

5、tting Startec. A= 0.4096,0.1234,0. 367&0. 2943:0.2246,0.3872,0.4015,0. 1129:0.3645,0. 1920, 0.3781, 0.0643;0. 1784, 0.4002, 0.2786, 0.3927: b= 0.4043:0.155。； 0.4240; 0.2557; x=Gauss(AJb)3.46061.5610 -2.9342-0.4301习题四.9.设f二土,用等距节点分卜5, 5成10个子区间。试用 l+xz分段线性插值求各子区间中点的值，并且估计误差。解：(1)先用Mat lab模拟一下函数图象，.m源文

6、件代码如下：xl=-5:1:5;yl=l./(1+xl.A2);plot(xl,yl, T rT);hold on;x2=-5:1/3000:5;y2=l./(l+x2.A2);plot(x2,y2);title ( !y=l/(l+xA2 ) !);xlabel(T xT);ylabel(T yT);gtext (Ty=l/ (l+xA2 ) -5,5分成十等分，红色，);gtext ( !y=l/ (l+xA2 ) -5,5原函数图象，蓝色 D;画出的函数图象如下:a%竖管耍化，a住 File Edit View Insert Tools Desktop V/indow Help10.90

7、.8070.6a 0.50.40.30.20.10-5-4-3-2-1012345(2)分段线性插值求各子区间中点的值由于对称性，就选择左边区间0, 5上取等距插值01234510.50.20.10.058820.03846x-6 + 1)xkP =.c 1、凡 + . 1A h +lk-(k + l)(k + l)-k=_爪很_1)+月满_幻一 (x-l)+0.5x,-0.5(x-2)+0.2(x-lX p(x) = -o.2(x - 3)+ 0.1(x - 2),-0.1(x-4)+0.05882(x-3),*0,1 xg1,2 W n3,4-0.05882(* -5)+0.03846(x 一 4), x 4,5在每个分段区间k.k+l于是， p(4.5)=-0.05882X(4.5-5)+0.03846X(4.5-4)=0.04864实际值：f(4.5)=0.047058