matlab使用法.docx

1、matlab使用法二分法function x0,k=bisect1(fun1,a,b,ep)if nargin0 x0=fa,fb; k=0; return;endk=1;while abs(b-a)/2ep x=(a+b)/2; fx=feval(fun1,x); if fx*fa fun1=inline(x3-x-1); x0,k=bisect1(fun1,1.3,1.4,1e-4)x0 = 1.3247k = 7简单迭代法function x0,k=iterate1(fun1,x0,ep,N)if nargin4 N=500;endif narginep & k fun1=inline(

2、x+1)(1/3); x0,k=iterate1(fun1,1.5)x0 = 1.3247k = 7 fun1=inline(x3-1); x0,k=iterate1(fun1,1.5)x0 = Infk = 9Steffesen加速迭代（简单迭代法的加速）function x0,k=steffesen1(fun1,x0,ep,N)if nargin4 N=500;endif narginep & k fun1=inline(x+1)(1/3); x0,k=steffesen1(fun1,1.5)x0 = 1.3247k = 3 fun1=inline(x3-1); x0,k=steffese

3、n1(fun1,1.5)x0 = 1.3247k = 6Newton迭代function x0,k=Newton7(fname,dfname,x0,ep,N)if nargin5 N=500;endif narginep & k fname=inline(x-cos(x); dfname=inline(1+sin(x); x0,k=Newton7(fname,dfname,pi/4,1e-8)x0 = 0.7391k =4非线性方程求根的Matlab函数调用举例：1.求多项式的根： 求f(x)=x3-x-1=0的根： roots(1 0 -1 -1)ans = 1.3247 -0.6624 +

4、 0.5623i -0.6624 - 0.5623i2.求一般函数的根 fun=inline(x*sin(x2-x-1),x)fun = Inline function: fun(x) = x*sin(x2-x-1) fplot(fun,-2 0.1);grid on x=fzero(fun,-2,-1)x = -1.5956 x=fzero(fun,-1 -0.1)x = -0.6180x,f,h=fsolve(fun,-1.6)x = -1.5956f = 1.4909e-009h = 1（h0表示收敛，h A=2 3 4;3 5 2;4 3 30; b=6,5,32b = 6 5 32

5、A,x=gauss3(A,b)A = 2.0000 3.0000 4.0000 6.0000 0 0.5000 -4.0000 -4.0000 0 0 -2.0000 -4.0000x = -13 8 2列选主元的高斯消元法：function A,x=gauss5(A,b)%本算法用列选主元的高斯消元法求解线性方程组n=length(b);A=A,b;for k=1:n-1 %选主元 ap,p=max(abs(A(k:n,k); p=p+k-1; if pk t=A(k,:); A(k,:)=A(p,:); A(p,:)=t; end %消元 A(k+1):n,(k+1):(n+1)=A(k+

6、1):n,(k+1):(n+1)-A(k+1):n,k)/A(k,k)*A(k,(k+1):(n+1); A(k+1):n,k)=zeros(n-k,1); end%回代 x=zeros(n,1); x(n)=A(n,n+1)/A(n,n); for k=n-1:-1:1 x(k)=(A(k,n+1)-A(k,(k+1):n)*x(sk+1:n)/A(k,k);end A=2 3 4;3 5 2;4 3 30; b=6,5,32; A,x=gauss5(A,b)A = 4.0000 3.0000 30.0000 32.0000 0 2.7500 -20.5000 -19.0000 0 0 0.

7、1818 0.3636x = -13 8 2三角分解法：Doolittle 分解function L,U=doolittle1(A)n=length(A);U=zeros(n);L=eye(n);U(1,:)=A(1,:);L(2:n,1)=A(2:n,1)/U(1,1);for k=2:n U(k,k:n)=A(k,k:n)-L(k,1:k-1)*U(1:k-1,k:n); L(k+1:n,k)=A(k+1:n,k)-L(k+1:n,1:k-1)*U(1:k-1,n)/U(k,k);Endy=zeros(n,1);x=y;y(1)=b(1);for i=2:n y(i)=b(i)-L(i,1

8、:i-1)*y(1:i-1);endx(n)=y(n)/U(n,n);for i=n-1:-1:1 x(i)=(y(i)-U(i,i+1:n)*x(i+1:n)/U(i,i);end A=1 2 3;2 5 2 ;3 1 5;b=14 18 20; L,U,x=doolittle1(A,b)L = 1 0 0 2 1 0 3 -8 1U = 1 2 3 0 1 -4 0 0 -36x = 2.8333 1.33332.8333 平方根法：function L,x=choesky3(A,b)n=length(A);L=zeros(n);L(:,1)=A(:,1)/sqrt(A(1,1);for

9、k=2:n L(k,k)=A(k,k)-L(k,1:k-1)*L(k,1:k-1); L(k,k)=sqrt(L(k,k); for i=k+1:n L(i,k)=(A(i,k)-L(i,1:k-1)*L(k,1:k-1)/L(k,k); endend y=zeros(n,1);x=y;y(1)=b(1)/L(1,1);for i=2:n y(i)=(b(i)-L(i,1:i-1)*y(1:i-1)/L(i,i);endx(n)=y(n)/L(n,n);for i=n-1:-1:1 x(i)=(y(i)-L(i+1:n,i)*x(i+1:n)/L(i,i);end A=4 -1 1;-1 4.

10、25 2.75;1 2.75 3.5A = 4.0000 -1.0000 1.0000 -1.0000 4.2500 2.7500 1.0000 2.7500 3.5000 b=4 6 7.25b = 4.0000 6.0000 7.2500L,x=choesky3(A,b)L = 2.0000 0 0 -0.5000 2.0000 0 0.5000 1.5000 1.0000x = 1 1 1迭代法求方程组的解Jacobi迭代法：function x,k=jacobi2(a,b,x0,ep,N)%本算法用Jacobi迭代求解ax=b,用分量形式n=length(b);k=0;if nargi

11、n5 N=500;endif nargin4 ep=1e-5;endif narginep & kN k=k+1; x0=x; for i=1:n y(i)=b(i); for j=1:n if j=i y(i)=y(i)-a(i,j)*x0(j); end end if abs(a(i,i)1e-10|k=N warning(a(i,i) is too small); return end y(i)=y(i)/a(i,i); end x=y; enda=4 3 0;3 4 -1; 0 -1 4;b=24 30 -24；x,k=jacobi2(a,b)x = 3.0000 4.0000 -5.

12、0000k =59Gauss-seidel迭代法：function x,k=gaussseide2(a,b,x0,ep,N)%本算法用Gauss-seidel迭代求解ax=b,用分量形式n=length(b);k=0;if nargin5 N=500;endif nargin4 ep=1e-5;endif narginep & kN k=k+1; x0=x; y=x; for i=1:n z(i)=b(i); for j=1:n if j=i z(i)=z(i)-a(i,j)*x(j); end end if abs(a(i,i) u,v=sort(-abs(diag(a)u = -1.190

13、9 -0.4326 -0.1867 -0.1139v = 2 1 3 4A=a(v,v)A = 1.1909 -1.6656 0.1746 2.1832 -1.1465 -0.4326 0.3273 -0.5883 1.1892 0.1253 -0.1867 -0.1364 -0.0376 0.2877 0.7258 0.1139求二次多项式的根，避免上下溢，避免两个数相加或者相减接近0function x flag=root2qe(a,b,c)if nargin=1 c=a(3);b=a(2);a=a(1);endif a=0 M=max(abs(a,b,c); a=a/M;b=b/M;c=

14、c/M; s=sign(b); if s=0 s=1; end x1=(-b-s*sqrt(b2-4*a*c)/(2*a); x2=c/(a*x1); x=x1 x2; flag=two solutions;elseif b=0 x=-c/b; flag=one solution;elseif c=0 x= ; flag=no solutionelse x=1; flag=all x are solutions;end x flag=root2qe(6,10,-4)x = -2.0000 0.3333flag =two solutions x flag=root2qe(6e+154,10e+1

15、54,-4e+154)x = -2.0000 0.3333flag =two solutions x flag=root2qe(0,1,1)x = -1flag =one solution x flag=root2qe(1,-1e+5,1)x = 1.0e+004 * 10.0000 0.0000flag =two solutions计算如下的递推数列：x1=1/3; x2=1/12; x(k+1)=2.25*x(k)-0.5*x(k-1); 可以证明， xk=4(1-k)/3; 是一个单调递减数列，但是计算数据如下：function ex2x(1)=1/3;x(2)=1/12;for k=3

16、:60 x(k)=2.25*x(k-1)-0.5*x(k-2);endplot(x(1:60)想想是什么原因导致这种现象？找出函数y=arctan(x)在区间【1,6】的11个等分点，做10次插值多项式。打印出Newton形式的系数，在同一个图上作出在区间【0,8】的33个等分点上插值多项式的值与y=arctan(x)的值，由此得出什么结论？function ex3n=11;x=linspace(1,6,n);h=(6-1)/(n-1);y=atan(x);for j=2:n y(1:n+1-j,j)=diff(y(1:n+2-j,j-1)/(j-1)*h);endy=y(1,:);forma

17、t long efprintf(n the coefficients of Newton interpolation is :nn);fprintf(%12.9f,y);pz=;v=linspace(0,8,33);for t=v, z=y(n); for j=n-1:-1:1 z=z*(t-x(j)+y(j); end pz=pz z;endplot(v,pz,g*-,v,atan(v),b+:);fprintf(n xi, p(xi), atan(xi), errorn);for j=1:length(v) fprintf(%12.6f%12.6f%12.6f%12.6f,. v(j),pz(j),atan(v(j),abs(pz(j)-atan(v(j);end （注：可编辑下载，若有不当之处，请指正，谢谢!）