m0_58132032 2021-10-09 14:15 采纳率: 77.8%
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利用Matlab解决以下实验问题

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  • joel_1993 2021-10-09 14:33
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    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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问题事件

  • 系统已结题 10月20日
  • 已采纳回答 10月12日
  • 创建了问题 10月9日

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