背景知識(shí)

The method of least squares is a standard approach in regression analysis to the approximate solution of overdetermined systems, i.e., sets of equations in which there are more equations than unknowns. "Least squares" means that the overall solution minimizes the sum of the squares of the errors made in the results of every single equation.

多項(xiàng)式擬合

The first clear and concise exposition of the method of least squares was published by Legendre in 1805.[5] The technique is described as an algebraic procedure for fitting linear equations to data and Legendre demonstrates the new method by analyzing the same data as Laplace for the shape of the earth. The value of Legendre's method of least squares was immediately recognized by leading astronomers and geodesists of the time.

實(shí)驗(yàn)一

%%給定數(shù)據(jù)對(duì)，(x0,y0)

x0=0:0.1:1;

y0=[-.447,1.978,3.11,5.25,5.02,4.66,4.01,4.58,3.45,5.35,9.22];

%%求擬合多項(xiàng)式

n=3;%假設(shè)數(shù)據(jù)對(duì)(x0,y0)一共有m對(duì)，則通常情況下，應(yīng)當(dāng)保證n

%%求得x0,y0數(shù)組所給數(shù)據(jù)的n階擬合多項(xiàng)式系數(shù)向量p,為了保證較好的擬合效果，多項(xiàng)式的階數(shù)要取得適當(dāng)，過低，殘差較大；過高,擬合模型將包含噪聲影響。

P=polyfit(x0,y0,n)

%圖示擬合情況

xx=0:0.01:1;

yy=polyval(P,xx);

%?y?=?polyval(p,x)?returns?the?value?of?a?polynomial?of?degree?n?evaluated

%?at?x.?The?input?argument?p?is?a?vector?of?length?n+1?whose?elements?are

%?the?coefficients?in?descending?powers?of?the?polynomial?to?be?evaluated.

%

%?y?=?p1xn?+?p2xn–1?+?…?+?pnx?+?pn+1

plot(xx,yy,'-b',x0,y0,'.r','MarkerSize',20)

legend('擬合曲線','原始數(shù)據(jù)','Location','SouthEast')

xlabel('x')

最小二乘

Least squares problems fall into two categories: linear or ordinary least squares and non-linear least squares, depending on whether or not the residuals are linear in all unknowns. The linear least-squares problem occurs in statistical regression analysis; it has a closed-form solution. The non-linear problem is usually solved by iterative refinement; at each iteration the system is approximated by a linear one, and thus the core calculation is similar in both cases.

實(shí)驗(yàn)二

%%給定自變量x數(shù)據(jù)組

x0=(0:0.1:1);

%%給定自變量y數(shù)據(jù)組

y0=[-.447,1.978,3.11,5.25,5.02,4.66,4.01,4.68,4.45,5.35,9.22]';

m=length(x0);

n=3;

%多項(xiàng)式階數(shù)

%%初始化數(shù)據(jù)陣X

X=zeros(m,n+1);

%%將X設(shè)置成各列等比的形式

for?k=1:n

X(:,n-k+1)=(x0.^k);

end

X(:,n+1)=ones(m,1);

%%利用最小二乘原理求解多項(xiàng)式系數(shù)

aT=(X\y0)'

aT =

51.9713 ?-80.2234 ? 37.7844 ? -0.8078

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