每週問題 December 17, 2012

本週問題是利用 Gram-Schmidt 正交化求正交基底函數。

Let C[0,1] be the space of continuous functions on the interval [0,1] with the inner product defined by

\displaystyle\left\langle f,g\right\rangle=\int_0^1f(x)g(x)dx.

Let \mathcal{P}_2=\mathrm{span}\{1,x,x^2\}. Apply the Gram-Schmidt algorithm to the basis 1, x, x^2 to obtain an orthogonal basis for \mathcal{P}_2.

 
參考解答:

\mathbf{u}_1=1\mathbf{u}_2=x\mathbf{u}_3=x^2。Gram-Schmidt 正交化程序如下:

\begin{aligned}  \mathbf{v}_1&=\mathbf{u}_1=1\\  \mathbf{v}_2&=\mathbf{u}_2-  \frac{\left\langle\mathbf{v}_1,\mathbf{u}_2\right\rangle}  {\left\langle\mathbf{v}_1,\mathbf{v}_1\right\rangle}\mathbf{v}_1=x-\frac{1/2}{1}1=  x-\frac{1}{2}\\  \mathbf{v}_3&=\mathbf{u}_3-  \frac{\left\langle\mathbf{v}_1,\mathbf{u}_3\right\rangle}  {\left\langle\mathbf{v}_1,\mathbf{v}_1\right\rangle}\mathbf{v}_1-  \frac{\left\langle\mathbf{v}_2,\mathbf{u}_3\right\rangle}  {\left\langle\mathbf{v}_2,\mathbf{v}_2\right\rangle}\mathbf{v}_2\\  &=x^2-\frac{1/3}{1}1-\frac{1/12}{1/12}\left(x-\frac{1}{2}\right)=x^2-x+\frac{1}{6}.\end{aligned}

再將 \mathbf{v}_1\mathbf{v}_2\mathbf{v}_3 予以標準化:

\begin{aligned}  \mathbf{q}_1&=\frac{1}{\sqrt{\left\langle \mathbf{v}_1,\mathbf{v}_1\right\rangle}}\mathbf{v}_1=\frac{1}{1}\mathbf{v}_1=1\\  \mathbf{q}_2&=\frac{1}{\sqrt{\left\langle \mathbf{v}_2,\mathbf{v}_2\right\rangle}}  \mathbf{v}_2=\frac{1}{\sqrt{1/12}}\left(x-\frac{1}{2}\right)=2\sqrt{3}  \left(x-\frac{1}{2}\right)\\  \mathbf{q}_3&=\frac{1}{\sqrt{\left\langle \mathbf{v}_3,\mathbf{v}_3\right\rangle}}  \mathbf{v}_3=\frac{1}{\sqrt{1/180}}\left(x^2-x+\frac{1}{6}\right)=6\sqrt{5}  \left(x^2-x+\frac{1}{6}\right).\end{aligned}

PowSol-Dec-17-12

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