Question

Let f(x)=4 g(x)=-4x+6 h(x)=2x^2+8x -3 inner product <p,q>=p(-1) q(-1) + P(0)q(0). +p(1)q(1) .usw gramschmidt to determine orthonormal basis for subspace p2 spanned by polynomials f(x) g(x) h(x)

Answer 1

$f\left( x \right) =4\\g\left( x \right) =-4x+6\\h\left( x \right) =2x^2+8x-3\\\\s_1\left( x \right) =f\left( x \right) =4,\left\| s_1 \right\| ^2=4\cdot 4+4\cdot 4+4\cdot 4=48\\s_2\left( x \right) =g\left( x \right) -\frac{<g,s_1>}{\left\| s_1 \right\| ^2}s_1\left( x \right) =-4x+6-\frac{10\cdot 4+6\cdot 4+2\cdot 4}{48}\cdot 4=-4x\\\left\| s_2 \right\| ^2=4\cdot 4+0\cdot 0+\left( -4 \right) \cdot \left( -4 \right) =32\\s_3\left( x \right) =h\left( x \right) -\frac{<h,s_1>}{\left\| s_1 \right\| ^2}s_1\left( s \right) -\frac{<h,s_2>}{\left\| s_2 \right\| ^2}s_2\left( x \right) =\\=2x^2+8x-3-\frac{\left( -9 \right) \cdot 4+\left( -3 \right) \cdot 4+7\cdot 4}{48}\cdot 4-\frac{-9\cdot 10+\left( -3 \right) \cdot 6+7\cdot 2}{32}\cdot \left( -4x \right) =\\=2x^2+\frac{175}{16}x-\frac{4}{3}\\\left\| s_3 \right\| ^2=\left( 2-\frac{175}{16}+\frac{4}{3} \right) ^2+\left( -\frac{4}{3} \right) ^2+\left( 2+\frac{175}{16}-\frac{4}{3} \right) ^2=\frac{111893}{576}\\Orthonormal\,\,basis:\\e_1=\frac{4}{\sqrt{48}}=0.57735\\e_2=\frac{-4}{\sqrt{32}}x=-0.707107x\\e_3=\frac{2x^2+\frac{175}{16}x-\frac{4}{3}}{\sqrt{\frac{111893}{576}}}=0.143496x^2+0.784744x-0.095664$