利用Matlab解决以下实验问题

你好同学，代码仅供参考：
拉格朗日插值

function [y,L] = Lagrange_Interp(X,Y,x)
% X is a vector of independent variable values
% Y is a vector of measurements of dependent variable values
% x is the value of the independent variable to be interpolated
% y is the interpolated value, evaluated at x
% L is a vector of Lagrange polynomial values at x
%all row vectors

if size(X,2) == 1
    X = X';
end
l_X = length(X);
X = -X;
vector_product = zeros(l_X);
for k = 1:l_X
    X_copy = X;
    X_copy(:,k) = [];
    den = prod(-X(k) + X_copy);
    vector_product(k,:) =  (Y(k)/den) * my_prod(X_copy);
end
L = sum(vector_product,1);
a = x.^ (length(L)-1:-1:0);
y = sum(a.*L);

end

function [vector_product] = my_prod(X_copy)
prod1 = [1 X_copy(1)];
for k = 2:length(X_copy)
    prod2 = [1 X_copy(k)];
    prod_result = prod2;
    for m = 1:length(prod1)-1
    prod_result = cat(2,prod_result,(prod1(m+1)*prod2));
    end
    prod1 = my_sum(prod_result);
end
vector_product = prod1;
end

function [total] = my_sum(prod1)
m = 1;
total = zeros(1,(length(prod1)-2)/2);
for k = 2:2:length(prod1)-1
    total(m) = prod1(k) + prod1(k+1);
    m = m+1;
end
total = [prod1(1) total prod1(end)];
end

牛顿插值

function yp = Newtoninterp(x,y,xp)
% Inputs
% example data
%     x = [1 4 6 5]; % x vector
%     y = log(x); % y = f(x) vector
%     xp = 2 or xp = [2 3]; % x point(s) (scalar or vector)
% Calculations 
n = numel(x); deg = n-1; % number of elements and degree polynomial
np = numel(xp); % number of elements of xp
%     Example for degree = 3: convert to a for loop
%         y1 = diff(y)./(x(1+1:end)-x(1:end-1))
%         y2 = diff(y1)./(x(1+2:end)-x(1:end-2))
%         y3 = diff(y2)./(x(1+3:end)-x(1:end-3))
% interpolate
yp = zeros(size(xp)); % temporary yp
for j = 1:np
    dy = diff(y); % initial y difference
    ypj = y(1); % initial y point
    for i = 1:deg
        yi = dy./(x(1+i:end)- x(1:end-i)); % ith divided differences
        dy = diff(yi); % get and update the yi difference
        ypj = ypj+yi(1)*prod(ones(1,i)*xp(j)-x(1:i)); % y = f(x point) point per j
    end
    yp(j) = ypj; % y = f(x point) point(s) (scalar or vector)
end
end

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