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函数的最佳平方逼近

岁杪二四...大约 1 分钟计算方法/数值分析学习笔记

研究在区间[a,b][a,b]上一般的最佳平方逼近问题,对f(x)C[a,b]f(x)\in C[a,b]C[a,b]C[a,b]中的一个子集φ=span{φ0(x),φ1(x),,φn(x)}\varphi=span\{\varphi_0(x),\varphi_1(x),\cdots,\varphi_n(x)\},若存在S(x)φS^*(x)\in\varphi,使

f(x)S(x)22=minS(x)φf(x)S(x)22=minS(x)φabφ(x)[f(x)S(x)]2dx \begin{equation} \begin{aligned} \parallel f(x)-S^*(x)\parallel_2^2&=\mathop{\min}_{S(x)\in\varphi}\parallel f(x)-S(x)\parallel_2^2 \\&=\mathop{\min}_{S(x)\in\varphi}\int_a^b\varphi(x)[f(x)-S(x)]^2dx \end{aligned} \end{equation}

则称S(x)S^*(x)使f(x)f(x)在子集φC[a,b]\varphi\subset C[a,b]中的最佳平方逼近函数

为了求S(x)S^*(x),可等价于求多元函数

I(a0,a1,,an)=abρ(x)[j=0najφj(x)f(x)]2dx I(a_0,a_1,\cdots,a_n)=\int_a^b\rho(x)\left[\sum_{j=0}^{n}a_j\varphi_j(x)-f(x)\right]^2dx

最小值问题

为了确定参数 ak(k=0,1,,n)a_k\quad(k=0,1,\cdots,n) ,由多元函数极值存在的必要条件,有

Iak=0(k=0,1,,n)Iuk=2abρ(x)[j=0najφj(x)f(x)]φk(x)dx=0(k=0,1,,n)即有j=0n(φj,φk)aj=(f,φk),(k=0,1,,n) \begin{aligned} &\frac{\partial I}{\partial a_k}=0\quad(k=0,1,\cdots,n) &\\&\frac{\partial I}{\partial u_k}=2\int_a^b\rho(x)\biggl[\sum_{j=0}^na_j\varphi_j(x)-f(x)\biggr]\varphi_k(x)dx=0\\&&(k=0,1,\cdots,n)\\&\text{即有}\sum_{j=0}^n(\varphi_j,\varphi_k)a_j=(f,\varphi_k),\quad(k=0,1,\cdots,n) \end{aligned}

而对应j=0n(φj,φk)aj=(f,φk),(k=0,1,,n)\sum_{j=0}^n(\varphi_j,\varphi_k)a_j=(f,\varphi_k),\quad(k=0,1,\cdots,n)即为:

[(φ0,φ0)(φ0,φ1)(φ0,φn)(φ1,φ0)(φ1,φ1)(φ1,φn)(φn,φ0)(φn,φ1)(φn,φn)][a0a1an]=[(φ0,f)(φ1,f)(φn,f)] \begin{bmatrix} (\varphi_0,\varphi_0) & (\varphi_0,\varphi_1) & \cdots & (\varphi_0,\varphi_n) \\ (\varphi_1,\varphi_0) & (\varphi_1,\varphi_1) & \cdots & (\varphi_1,\varphi_n) \\ \vdots & \vdots & \vdots & \vdots \\ (\varphi_n,\varphi_0) & (\varphi_n,\varphi_1) & \cdots & (\varphi_n,\varphi_n) \end{bmatrix} {\color{red}{\begin{bmatrix} \textcolor{red}{a_0} \\ \textcolor{red}{a_1} \\ \vdots \\ \textcolor{red}{a_n} \end{bmatrix}}} = \begin{bmatrix} (\varphi_0,f) \\ (\varphi_1,f) \\ \vdots \\ (\varphi_n,f) \end{bmatrix}

红色部分即为所求。这是关于未知数a0,a1,,ana_0,a_1,\cdots,a_n的线性代数方程组,称为法方程组,由于φ0,φ1,,φn\varphi_0,\varphi_1,\cdots,\varphi_n线性无关,故系数行列式G(φ0,φ1,,φn)0G(\varphi_0,\varphi_1,\cdots,\varphi_n)\neq0,于是方程组有唯一解ak=ak(k=0,1,,n)a_k=a_k^*(k=0,1,\cdots,n),从而得到

S(x)=a0φ0(x)++anφn(x) S^*(x)=a_0^*\varphi_0(x)+\cdots+a_n^*\varphi_n(x)

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