91Ó°ÊÓ

Let's denote the given set of functions as \(\{u_1(x) = 1, u_2(x) = x, u_3(x) = x^2\}\). We will apply the Gram-Schmidt process to find a new set \(\{v_1(x), v_2(x), v_3(x)\}\) that is orthogonal with respect to the inner product. First, let \(v_1(x) = u_1(x) = 1\). Next, let \(v_2(x) = u_2(x) - \frac{\langle u_2, v_1 \rangle}{\langle v_1, v_1 \rangle} v_1(x)\): Calculate \(\langle u_2, v_1 \rangle\) and \(\langle v_1, v_1 \rangle\): \(\langle u_2, v_1 \rangle = \int_{-1}^{1} x \cdot 1 dx = \left[ \frac{1}{2}x^2 \right]_{-1}^{1} = 0\) \(\langle v_1, v_1 \rangle = \int_{-1}^{1} 1 \cdot 1 dx = \left[ x \right]_{-1}^{1} = 2\) So, \(v_2(x) = x - \frac{0}{2} \cdot 1 = x\). Finally, let \(v_3(x) = u_3(x) - \frac{\langle u_3, v_1 \rangle}{\langle v_1, v_1 \rangle} v_1(x) - \frac{\langle u_3, v_2 \rangle}{\langle v_2, v_2 \rangle} v_2(x)\): Calculate \(\langle u_3, v_1 \rangle\), \(\langle u_3, v_2 \rangle\), and \(\langle v_2, v_2 \rangle\): \(\langle u_3, v_1 \rangle = \int_{-1}^{1} x^2 \cdot 1 dx = \left[ \frac{1}{3}x^3 \right]_{-1}^{1} = \frac{2}{3}\) \(\langle u_3, v_2 \rangle = \int_{-1}^{1} x^2 \cdot x dx = \left[ \frac{1}{4}x^4 \right]_{-1}^{1} = 0\) \(\langle v_2, v_2 \rangle = \int_{-1}^{1} x^1 \cdot x^1 dx = \left[ \frac{1}{3}x^3 \right]_{-1}^{1} = \frac{2}{3}\) So, \(v_3(x) = x^2 - \frac{\frac{2}{3}}{2} \cdot 1 - \frac{0}{\frac{2}{3}} \cdot x = x^2 - \frac{1}{3}\).

Now, we have an orthogonal set \(\{v_1(x) = 1, v_2(x) = x, v_3(x) = x^2 - \frac{1}{3}\}\). We will normalize each of these functions to obtain an orthonormal basis: Normalize \(v_1(x) = 1\): \(\|v_1\|^2 = \int_{-1}^{1} 1^2 dx = [\,x\,]_{-1}^{1} = 2\) \(e_1(x) = \frac{v_1(x)}{\sqrt{\|v_1\|^2}} = \frac{1}{\sqrt{2}}\) Normalize \(v_2(x) = x\): \(\|v_2\|^2 = \int_{-1}^{1} x^2 dx = [\frac{1}{3}x^3\,]_{-1}^{1} = \frac{2}{3}\) \(e_2(x) = \frac{v_2(x)}{\sqrt{\|v_2\|^2}} = \sqrt{3}x\) Normalize \(v_3(x) = x^2 - \frac{1}{3}\): \(\|v_3\|^2 = \int_{-1}^{1} \left(x^2 - \frac{1}{3}\right)^2 dx = \int_{-1}^{1} \left(x^4 - \frac{2}{3}x^2 + \frac{1}{9}\right)dx = \frac{8}{45}\) \(e_3(x) = \frac{v_3(x)}{\sqrt{\|v_3\|^2}} = \frac{3\sqrt{5}}{2}\left(x^2 - \frac{1}{3}\right)\) Now, we have an orthonormal basis for the subspace spanned by \(1\), \(x\), and \(x^2\): \(\left\{ e_1(x) = \frac{1}{\sqrt{2}}, \, e_2(x) = \sqrt{3}x, \, e_3(x) = \frac{3\sqrt{5}}{2}\left(x^2 - \frac{1}{3}\right) \right\}\).