数值分析作业

问题：

分别使用Lagrange和Newton插值公式计算函数的值。

对于Newton插值要求先输出差商表，在计算x=7.5时4次插值多项式的值，并与四次Lagrange插值结果进行比较

x	4	5	6	7	8
y	2.0000	2.23607	2.44949	2.64575	2.82843

Lagrange

function y = Lagrange(a,b,x,n)
%Lagrange插值法 a,b分别为x以及对应的f(x),x为插入数据，n为插值多项式次数
% a=4:8
% b=[2 2.23607 2.44949 2.64575 2.82843]
if length(a)~=length(b)
    error('输入数据不同维')
end
if n>=length(a)
    error('n值太大')
end
d=abs(x-a);
[~,idx]=sort(d);
S=0;
for i=1:n+1
    t1=1;
    t2=1;
    for j=1:n+1
        if j~=i
            t1=t1*(x-a(idx(j)));
            t2=t2*(a(idx(i))-a(idx(j)));
        end
    end
    S=S+t1/t2*b(idx(i));
end
y=S;
end

Newton

•  Chashang_List

function A = Chashang_List(a,b)
%Newton插商计算函数
if length(a)~=length(b)
    error('输入数据不同维')
end
N=length(a);
A=zeros(N,N+1);
A(:,1)=a;
A(:,2)=b;
for j=3:N+1
    for i=j-1:N
        A(i,j)=(A(i,j-1)-A(i-1,j-1))/(A(i,1)-A(i-(j-2),1));
    end
end
ChaShang_List=A
end

---

function y = Newton(a,b,x,n)
% Newton插值法 a,b分别为x以及对应的f(x),x为插入数据，n为插值多项式次数
% a=4:8
% b=[2 2.23607 2.44949 2.64575 2.82843]
if n>=length(a)
    error('n值太大')
end
A=Chashang_List(a,b);
S=A(1,2);
X=cumprod(x-a);
for i=1:n
    S=S+A(i+1,i+2)*X(i);
end
y=S;
end

欧拉 

function y = Euler (fun, x0, y0, h, b)
n = (b - x0)/h;
A = zeros (n+1, 4);
A(1,1) = x0;
A(1,2) = y0;
A(1,1) = y0;
for i = 1 : n 
A(i+1,1) = A (i,1) + h;
A(i+1,2) = A (i,2) + h*fun (A(i,1), A(i,2));
A(i+1,3) = A (i,2) + h*fun (A(i+1,1), A(i+1,2));
yp = A(i,4) + h*fun (A (i,1), A(i,4));
yc = A(i,4) + h*fun (A (i+1,1), yp);
A (i+1,4) = 1/2*(yp + yc);
end 
y = A;


fun=@(x,y)-y+x+1
y = Euler(fun, 0, 1, 0.5, 1)

y =

    1.0000    1.0000         0         0
    1.5000    1.5000    1.5000    0.8750
    2.0000    2.0000    2.0000    1.6094

龙贝格

function y = Runge_Kutta (fun, x0, y0, h, b)
n = (b-x0)/h;
A = zeros (n+1, 6);
A(1,1) = x0;
A(1,6) = y0;

for i = 1 : n 
A(i+1,1) = A (i,1) + h;  
A(i+1,2) = fun (A(i,1), A(i,2));  %k1
A(i+1,3)=fun (A(i,1)+h/2, A(i, 6)+h/2* A(i+1,2));  %k2
A(i+1,4)=fun (A(i,1)+h/2, A(i, 6)+h/2* A(i+1,3));  %k3
A(i+1,5)=fm(A(i,1)+h, A(i,6)+h* A(i+1,4));  %k4
A(i+1,6)=A(i,6)+h/6*(A(i+1,2)+2*A(i+1,3)+2*A(i+1,4)+A(i+1,5));
end 
y = A;


fun=@(x,y)-3*y+2*x+2
y = Euler(fun, 0, 1, 0.1, 1)

y =

    1.0000    1.0000         0         0
    1.1000    1.1000    1.0900    0.3500
    1.2000    1.1900    1.1830    0.6278
    1.3000    1.2730    1.2681    0.8517
    1.4000    1.3511    1.3477    1.0355
    1.5000    1.4258    1.4234    1.1894
    1.6000    1.4980    1.4964    1.3211
    1.7000    1.5686    1.5675    1.4362
    1.8000    1.6380    1.6372    1.5390
    1.9000    1.7066    1.7061    1.6326
    2.0000    1.7746    1.7742    1.7193

 用迭代法求方程 e^x +10x - 2 = 0 的近似根， 取初值 x = 0 ， 迭代公式，要求精度为0.5 *10^-3 ;

X0=0;
errf=1;
i=0;
while errf>0.5*10^(-3)
i=i+1;
x=(2-exp(X0))/10;
errf=abs((x-X0)/X0);
x0=x;
end