拉格朗日插值方法

Question

我使用卷积和for循环(太多的循环)来计算插值使用 Lagrange's method,这里是主要代码:

function[p] = lagrange_interpolation(X,Y)
L = zeros(n);
p = zeros(1,n);


% computing L matrice, so that each row i holds the polynom L_i
% Now we compute li(x) for i=0....n  ,and we build the polynomial 

for k=1:n
    multiplier = 1;
    outputConv = ones(1,1);
    for index = 1:n
        if(index ~= k && X(index) ~= X(k))
            outputConv = conv(outputConv,[1,-X(index)]);
            multiplier = multiplier * ((X(k) - X(index))^-1);
        end
    end
    polynimialSize = length(outputConv);
    for index = 1:polynimialSize
        L(k,n - index + 1) = outputConv(polynimialSize - index + 1);
    end
    L(k,:) = multiplier .* L(k,:);
end

% continues 


end

这些对于计算循环来说太多了l_i(x)(这是在最后一次计算之前完成的P_n(x) = Sigma of y_i * l_i(x)).

有什么建议让它更正式matlab吗？

谢谢

Answer 1

是的,一些建议(在下面的版本1中实现):if循环可以与for上面的组合(只需通过下面的内容index跳过); 总是相等的,总是相等的(因为你有数据点,带系数的多项式),所以最后一个循环和下一行也可以用简单替换kjr(jr~=j)polynomialSizelength(outputConv)nn(n-1)thnforL(k,:) = multiplier * outputConv;

所以我复制的例子http://en.wikipedia.org/wiki/Lagrange_polynomial(并通过他们的j-m符号,但对我来说j去1:n和m是1:n和m~=j),因此我的初始化看起来像

clear; clc;
X=[-9 -4 -1 7]; %example taken from http://en.wikipedia.org/wiki/Lagrange_polynomial
Y=[ 5  2 -2 9];
n=length(X); %Lagrange basis polinomials are (n-1)th order, have n coefficients
lj = zeros(1,n); %storage for numerator of Lagrange basis polyns - each w/ n coeff
Lj = zeros(n); %matrix of Lagrange basis polyns coeffs (lj(x))
L  = zeros(1,n); %the Lagrange polynomial coefficients (L(x))

然后v 1.0看起来像

jr=1:n; %j-range: 1<=j<=n
for j=jr %my j is your k
    multiplier = 1;
    outputConv = 1; %numerator of lj(x)
    mr=jr(jr~=j); %m-range: 1<=m<=n, m~=j
    for m = mr %my m is your index
        outputConv = conv(outputConv,[1 -X(m)]);
        multiplier = multiplier * ((X(j) - X(m))^-1);
    end
    Lj(j,:) = multiplier * outputConv; %jth Lagrange basis polinomial lj(x)
end
L = Y*Lj; %coefficients of Lagrange polinomial L(x)

如果你意识到l_j(x)的分子只是一个具有特定根的多项式,那么可以进一步简化 - 因为在matlab中有一个很好的命令 - poly.类似地,分母只是在X(j)处评估的多数 - 因为有polyval.因此,v 1.9:

jr=1:n; %j-range: 1<=j<=n
for j=jr
    mr=jr(jr~=j); %m-range: 1<=m<=n, m~=j
    lj=poly(X(mr)); %numerator of lj(x)
    mult=1/polyval(lj,X(j)); %denominator of lj(x)
    Lj(j,:) = mult * lj; %jth Lagrange basis polinomial lj(x)
end
L = Y*Lj; %coefficients of Lagrange polinomial L(x)

为什么版本1.9而不是2.0？好吧,有可能有办法摆脱这最后的循环,并将其全部写入1行,但我现在想不到它 - 它是v 2.0的todo :)

而且,对于甜点,如果你想获得与维基百科相同的图片:

figure(1);clf
x=-10:.1:10;
hold on
plot(x,polyval(Y(1)*Lj(1,:),x),'r','linewidth',2)
plot(x,polyval(Y(2)*Lj(2,:),x),'b','linewidth',2)
plot(x,polyval(Y(3)*Lj(3,:),x),'g','linewidth',2)
plot(x,polyval(Y(4)*Lj(4,:),x),'y','linewidth',2)
plot(x,polyval(L,x),'k','linewidth',2)
plot(X,Y,'ro','linewidth',2,'markersize',10)
hold off
xlim([-10 10])
ylim([-10 10])
set(gca,'XTick',-10:10)
set(gca,'YTick',-10:10)
grid on

产生

享受并随意重用/改进