Question

The first four Hermite polynomials are f(x) = 1,g(x) = 2t, h(x) = 2−4t +t², and p(x) =

6−18t +9t²−t³

. Show that these polynomials form a basis for P3.

Answer 1

$\alpha _1f+\alpha _2g+\alpha _3h+\alpha _4p=0\Rightarrow \\\Rightarrow \alpha _1+2\alpha _2t+\alpha _3\left( 2-4t+t^2 \right) +\alpha _4\left( 6-18t+9t^2-t^3 \right) =0\Rightarrow \\\Rightarrow \left\{ \begin{array}{c} \alpha _1+2\alpha _3+6\alpha _4=0\\ 2\alpha _2-4\alpha _3-18\alpha _4=0\\ a_3+9\alpha _4=0\\ -\alpha _4=0\\\end{array} \right. \Rightarrow \left\{ \begin{array}{c} \alpha _4=0\\ \alpha _3=0\\ 2\alpha _2=0\\ \alpha _1=0\\\end{array} \right. \Rightarrow \alpha _1=\alpha _2=\alpha _3=\alpha _4=0\\The\,\,system\,\,is\,\,linearly\,\,independent. Since\,\,the\,\,dimension\,\,of\,\,P_3\,\,is\,\,4, \\the\,\,system\,\,forms\,\,a\,\,basis$