例
设 f ( x ) = 1 + x 2 f(x)=\sqrt{1+x^2} f(x)=1+x2，求 [ 0 , 1 ] [0,1] [0,1]上的一个一次最佳平方逼近多项式。
解 ：
d 0 = ∫ 0 1 1 + x 2 d x = 1 2 ln ( 1 + 2 ) + 2 2 ≈ 1.147 d_0=\int_{0}^{1}\sqrt{1+x^2}dx=\frac{1}{2}\ln(1+\sqrt{2})+\frac{\sqrt{2}}{2}\approx 1.147 d0=∫011+x2dx=21ln(1+2)+22≈1.147
d 1 = ∫ 0 1 x 1 + x 2 d x = 1 3 ( 1 + x 2 ) 3 2 ∣ 0 1 = 2 2 − 1 3 ≈ 0.609 d_1=\int_{0}^{1}x\sqrt{1+x^2}dx=\left.\frac{1}{3}(1+x^2)^{\frac{3}{2}}\right|_{0}^{1}=\frac{2\sqrt{2}-1}{3}\approx 0.609 d1=∫01x1+x2dx=31(1+x2)23 01=322−1≈0.609
由方程组
( 1 1 2 1 2 1 3 ) ( a 0 a 1 ) = ( 1.147 0.609 ) , \begin{pmatrix} 1 & \frac{1}{2}\\ \frac{1}{2} & \frac{1}{3} \end{pmatrix}\begin{pmatrix} a_0\\ a_1 \end{pmatrix}=\begin{pmatrix} 1.147\\ 0.609 \end{pmatrix}, (1212131)(a0a1)=(1.1470.609),
解出 a 0 = 0.934 , a 1 = 0.426 , S 1 ∗ ( x ) = 0.934 + 0.426 x a_0=0.934,\quad a_1=0.426,\quad S_1^*(x)=0.934+0.426x a0=0.934,a1=0.426,S1∗(x)=0.934+0.426x
平方误差 ∥ δ ∥ 2 2 = ( f , f ) − ( S 1 ∗ , f ) = ∫ 0 1 ( 1 + x 2 ) d x − 0.426 d 1 − 0.934 d 0 = 0.0026 \|\delta\|^2_2=(f,f)-(S_1^*,f)=\int_{0}^{1}(1+x^2)dx-0.426d_1-0.934d_0=0.0026 ∥δ∥22=(f,f)−(S1∗,f)=∫01(1+x2)dx−0.426d1−0.934d0=0.0026
最大误差 ∥ δ ∥ ∞ = max 0 ≤ x ≤ 1 ∣ 1 + x 2 − S 1 ∗ ( x ) ∣ ≈ 0.066 \|\delta\|_\infty=\max_{0\leq x\leq1}|\sqrt{1+x^2}-S_1^*(x)|\approx 0.066 ∥δ∥∞=0≤x≤1max∣1+x2−S1∗(x)∣≈0.066