求连续函数最佳平方逼近的步骤:

求连续函数最佳平方逼近的步骤:

  1. 给定[a,b]上的连续函数f(x),及子空间

Φ = s p a n ( ϕ 0 ( x ) , ϕ 1 ( x ) , ⋯   , ϕ n ( x ) ) \Phi = span(\phi_{0}(x),\phi_{1}(x),\cdots,\phi_{n}(x)) Φ=span(ϕ0(x),ϕ1(x),,ϕn(x))

  1. 利用内积

( ϕ i , ϕ k ) = ∫ a b ϕ i ( x ) ϕ k ( x ) d x (\phi_i,\phi_k)=\int_{a}^{b}\phi_{i}(x)\phi_{k}(x)dx (ϕi,ϕk)=abϕi(x)ϕk(x)dx

( f , ϕ k ) = ∫ a b f ( x ) ϕ k ( x ) d x (f,\phi_k)=\int_{a}^{b}f(x)\phi_{k}(x)dx (f,ϕk)=abf(x)ϕk(x)dx

给出法方程组:
[ ( ϕ 0 , ϕ 0 ) ( ϕ 1 , ϕ 0 ) ⋯ ( ϕ n , ϕ 0 ) ( ϕ 0 , ϕ 1 ) ( ϕ 1 , ϕ 1 ) ⋯ ( ϕ n , ϕ 1 ) ⋯ ⋯ ⋯ ( ϕ 0 , ϕ n ) ( ϕ 1 , ϕ n ) ⋯ ( ϕ n , ϕ n ) ] [ c 0 c 1 ⋮ c n ] = [ ( f , ϕ 0 ) ( f , ϕ 1 ) ⋮ ( f , ϕ n ) ] \begin{bmatrix} (\phi_{0},\phi_{0}) & (\phi_{1},\phi_{0}) & \cdots & (\phi_{n},\phi_{0}) \\(\phi_{0},\phi_{1}) & (\phi_{1},\phi_{1}) & \cdots & (\phi_{n},\phi_{1}) \\ \cdots & \cdots & & \cdots \\ (\phi_{0},\phi_{n}) & (\phi_{1},\phi_{n}) & \cdots & (\phi_{n},\phi_{n})\\ \end{bmatrix} \begin{bmatrix} c_{0} \\ c_{1} \\ \vdots \\ c_{n} \\ \end{bmatrix}= \begin{bmatrix} (f,\phi_{0}) \\ (f,\phi_{1}) \\ \vdots \\ (f,\phi_{n}) \\ \end{bmatrix} (ϕ0,ϕ0)(ϕ0,ϕ1)(ϕ0,ϕn)(ϕ1,ϕ0)(ϕ1,ϕ1)(ϕ1,ϕn)(ϕn,ϕ0)(ϕn,ϕ1)(ϕn,ϕn)c0c1cn=(f,ϕ0)(f,ϕ1)(f,ϕn)

  1. 求出法方程组的解 c 0 ∗ , c 1 ∗ , ⋯   , c n ∗ c_0^*,c_1^*,\cdots,c_n^* c0,c1,,cn,得到最佳平方逼近

p n ∗ ( x ) = c 0 ∗ ϕ 0 ( x ) + c 0 ∗ ϕ 1 ( x ) + ⋯ + c n ∗ ϕ n ( x ) p_{n}^*(x)=c_{0}^*\phi_0(x)+c_{0}^*\phi_1(x)+\cdots+c_n^*\phi_n(x) pn(x)=c0ϕ0(x)+c0ϕ1(x)++cnϕn(x)

  1. 求出误差

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