Recursive least square

给定一组样本点：

求拟合函数 $G(x)$ 使得 $\sum_{i=1}^k (G(x_i)-y_i)^2$ 取最小值。

设 $\phi_0, \phi_1, ... \phi_n$ 是n+1个线性无关的连续函数，$G_n$ 是由 $\phi_i(i=0,1...n)$ 的所有线性组合构成的集合，记 $G_n = span \{ \phi_0, \phi_1,...,\phi_n \}$ ，$G(x)\in G_n$ 。

将 $G(x)$ 写成矩阵的形式：

$$\begin{bmatrix} G(x_1)\\ G(x_2) \\ \vdots\\ G(x_k) \end{bmatrix}=\begin{bmatrix} \phi_0(x_1) & \phi_1(x_1) & \cdots & \phi_n(x_1) \\ \phi_0(x_2) & \phi_1(x_2) & \cdots & \phi_n(x_2) \\ \vdots & \vdots & \vdots & \vdots \\ \phi_0(x_k) & \phi_1(x_k) & \cdots & \phi_n(x_k) \\ \end{bmatrix}\begin{bmatrix} a_0\\ a_1 \\ \vdots\\ a_n \end{bmatrix}$$

所求的 $G(x)$ 满足如下条件 ：

$$\sum_{i=0}^k(\sum_{j=0}^na_j\phi_j(x_i)-y_i)\phi_h(x_i)=0,h=0,1,...,n.$$

以矩阵形式表示如下：

为$\Phi_k^T(\Phi_kA_k-Y_k)=0$ ($A_k$表示有 $k$ 组样本点时对应的 $n+1$ 个参数)

其中，$\Phi_k=\begin{bmatrix} \phi_0(x_1) & \phi_1(x_1) & \cdots & \phi_n(x_1) \\ \phi_0(x_2) & \phi_1(x_2) & \cdots & \phi_n(x_2) \\ \vdots & \vdots & \vdots & \vdots \\ \phi_0(x_k) & \phi_1(x_k) & \cdots & \phi_n(x_k) \\ \end{bmatrix}$， $A_k=\begin{bmatrix} a_0\\ a_1 \\ \vdots\\ a_n \end{bmatrix}$，$Y_k=\begin{bmatrix} y_1\\ y_2 \\ \vdots\\ y_k \end{bmatrix}$，

$$A_k=(\Phi_k^T\Phi_k)^{-1}\Phi_k^TY_k$$

目标：$A_{k+1}= H(A_k)$

对 $A_k=(\Phi_k^T\Phi_k)^{-1}\Phi_k^TY_k$ 进行推导变换：

令 $P_k=\Phi_k^T\Phi_k$ ，$Q_k=\Phi_k^TY_k$ ，则 $A_k=P_k^{-1}Q_k$ .

$$P_k=\Phi_k^T\Phi_k=\begin{bmatrix} \phi_0(x_1) & \phi_0(x_2) & \cdots & \phi_0(x_k) \\ \phi_1(x_1) & \phi_1(x_2) & \cdots & \phi_1(x_k) \\ \vdots & \vdots & \vdots & \vdots \\ \phi_n(x_1) & \phi_n(x_2) & \cdots & \phi_n(x_k) \\ \end{bmatrix}\begin{bmatrix} \phi_0(x_1) & \phi_1(x_1) & \cdots & \phi_n(x_1) \\ \phi_0(x_2) & \phi_1(x_2) & \cdots & \phi_n(x_2) \\ \vdots & \vdots & \vdots & \vdots \\ \phi_0(x_k) & \phi_1(x_k) & \cdots & \phi_n(x_k) \\ \end{bmatrix}$$

令 $\overrightarrow{\phi_i}^T=\begin{bmatrix} \phi_0(x_i)\\ \phi_1(x_i) \\ \vdots\\ \phi_n(x_i) \end{bmatrix},i=1,...k$.

$$P_k=\Phi_k^T\Phi_k=\begin{bmatrix} \overrightarrow{\phi_1}^T & \overrightarrow{\phi_2}^T &\cdots & \overrightarrow{\phi_k}^T \\ \end{bmatrix}\begin{bmatrix} \overrightarrow{\phi_1} \\ \overrightarrow{\phi_2} \\ \vdots\\ \overrightarrow{\phi_k} \end{bmatrix}=\sum_{i=1}^k\overrightarrow{\phi_i}^T\overrightarrow{\phi_i}=\sum_{i=1}^{k-1}\overrightarrow{\phi_i}^T\overrightarrow{\phi_i}+\overrightarrow{\phi_k}^T\overrightarrow{\phi_k}=P_{k-1}+\overrightarrow{\phi_k}^T\overrightarrow{\phi_k}$$ $$Q_k=\Phi_k^TY_k=\begin{bmatrix} \overrightarrow{\phi_1}^T & \overrightarrow{\phi_2}^T &\cdots & \overrightarrow{\phi_k}^T \\ \end{bmatrix}\begin{bmatrix} y_1\\ y_2 \\ \vdots\\ y_k \end{bmatrix}=\sum_{i=1}^k\overrightarrow{\phi_i}^Ty_i=\sum_{i=1}^{k-1}\overrightarrow{\phi_i}^Ty_i+\overrightarrow{\phi_k}^Ty_k=Q_{k-1}+\overrightarrow{\phi_k}^Ty_k$$ $$A_{k-1}=P_{k-1}^{-1}Q_{k-1} => Q_{k-1}=P_{k-1}A_{k-1}$$ $$A_k=P_k^{-1}Q_k$$ $$=P_k^{-1}(Q_{k-1}+\overrightarrow{\phi_k}^Ty_k)$$ $$=P_k^{-1}(P_{k-1}A_{k-1}+\overrightarrow{\phi_k}^Ty_k)$$ $$=P_k^{-1}[(P_k-\overrightarrow{\phi_k}^T\overrightarrow{\phi_k})A_{k-1}+\overrightarrow{\phi_k}^Ty_k]$$ $$=A_{k-1}-P_k^{-1}\overrightarrow{\phi_k}^T\overrightarrow{\phi_k}A_{k-1}+P_k^{-1}\overrightarrow{\phi_k}^Ty_k$$

function Ak = nafit1( A, x, xk, yk, m)
Pni=inv(P(x,m));
phik2phik=phi(xk,m);
phi2y=phiy(xk, yk, m);
Ak=A'-Pni*phik2phik*A'+Pni*phi2y;
Ak=fliplr(Ak');
end

function phi2y = phiy(xk, yk, m)
phi2y = zeros(m+1, 1);
for i=0:m
    phi2y(i+1)=xk^i*yk;
end
end

function phik2phik = phi(xk, m)
phik2phik = zeros(m+1, m+1);
for i=0:m
    for j=0:m
        phik2phik(i+1,j+1)=xk^(i+j);
    end
end
end

function Pres = P(x, m)
len = length(x);
Pres = zeros(m+1, m+1);
for k=1:len
    Pres = Pres + phi(x(k), m);
end
end

代码的使用方法