数值ys银计算方法上机实习

x00whileabs(b-a)>5*1e-ifexp(c)+10*c-2>0elsea=c;结果：cx00xk1 abs(x-a)>5*1e-x=(2-

2ex；结果：x=whileabs(b-a)>5*1e-4z=exp(y)+10*y-结果：x=x00whileabs(b-a)>5*1e-4

x=可得x23456789yb解：设y=f(x)具有指数形式y

（a>0,b<0。对此式两边取对数，得xx23456789t=[0.50000.33330.25000.20000.16670.14290.12500.11110.09090.08330.07690.07140.0667z=[1.85942.10412.25972.25132.27212.30262.29562.30162.35992.36092.37952.36092.3888ans=- 由此可得A=2.4578，B=-1.1107，aeA11.6791，b=B=-y

1xx=[2:16];y=[6.428.29.589.59.7109.939.9910.4910.5910.6010.610.910.76];f(x)=11.6791*exp(-1.1107./x);forc=c+(a-结果：averge

0 0

0 5

2

，b

，Ax 0

5

2 426 426 JACOBIGAUSS-SEIDEL

xk1

SOR（

functiony=jacobi(a,b,x0)U=-whilenorm(y-x0)>1e-4a=[4-10-100;-14-10-10;0-14-10-1;-10-14-1-10-14-1;00-10-1b=[05-25-2x0=[00000y=nGAUSS-SEIDELfunctiony=seidel(a,b,x0)U=-whilenorm(y-x0)>10^(-4)na=[4-10-100;-14-10-10;0-14-10-1;-10-14-10;0-0-14-1;00-10-1b=[05-25-2x0=[00000ynSOR（functiony=sor(a,b,w,x0)U=-L=-tril(a,-whilenorm(y-x0)>10^(-4)

na=[4-10-100;-14-10-10;0-14-10-1;-10-14-10;0-0-14-1;00-10-1b=[05-25-2x0=[00000c=[1.3341.95forynyn=yn A11

1 1解：算法为结合原点平移的反幂法，编写程序如下functiont=maxnorm(A)forifabs(A(i))-abs(t)>=0A=[631;321;11v=[11forfprintf('n=%d,x=[%f%f%f],lamda=%f\n',i-ifabs(1/lam(i+1)-1/lam(i))<1e-3;故所求特征值为7.873473，特征向量为[1.0000000.5493000.225573用经典R-K

y12y1y22sin

y1(0)2yy2y2cosx2sin

y(0)

y12y1y22sin

y10)2y998y999y999cosx999sin

y(0) y(x)2exsin和精确解

2(x)2excos2解：编写程序如下Y2=2*exp(-ydot1=@(x,y)[-2*y(1)+y(2)+2*sin(x);y(1)-2*y(2)+2*cos(x)-gridy(1x2)y用有限差分法求解边值问题

y(1y(11解：通过差 近导数，将微分方程离散为差分方 [2(1x)2h2]y

,(iy0y20编写程序如下fori=1:19;a(i)=1;b(i)=- d(1)=d(1)-d(19)=d(19)--------------------程序如下：a=[11111111111111111b=[-2.0181-2.0164-2.0149-2.0136-2.0125--2.0109-2.0104--2.0101-2.0104-2.0109-2.0116-2.0125-2.0136--c=[1111111111111111d=[-100000000000000000-x=n=length(x);forj=1:n-

mu=b(j+1)=b(j+1)-x(j+1)=x(j+1)-x(n)=x(n)/b(n);forj=n-1:-1:1x(j)=(x(j)-forh=-gridon;y=asinbxxy解：使用非线性拟合函数nlinfit，编写程序如下ify=[0.61.11.61.82.01.91.7beta0=[11]';fprintf('Thenonlinearleastsquarefittingy=a*sin(b*x)forfprintf('\n\nis\n\ty=%7.4f*sin(%7.4f*x)\n\n',beta);gridfunctionyb=mymodel(beta,xb)Thenonlinearleastsquarefittingy=a*sin(b*x)fory=1.9750*sin(f（x）≈（a+bx）/(1+cx)的函数的一种快速方法是将最小二乘法用于下列问题：f（x）(1+cx)≈（a+bx，试用这一方法拟合表4-4给出的4-次序年份第一次第二次第三次第四次第五次f（xx=[1953196419821900y=[5.826.9510.0811.34A=[ones(5,1)x-plot(x,y,'b-+',v,(a+b*v)./(1+c*v),'k-a=2.9456,b=-0.0014,c=-已知20世纪人口的统计数字如表（单位：百万000052773054试拟合人口20世纪的增长率x=[76.092.0106.5 131.7150.7 204.0226.5%2-pointv=p./x(1:end-fprintf('%4d%6.3f %3-p=[pv=[vfprintf('%4d 二点结果年份19001910192019301940195019601970年增