Question

Let p0, p1, and p2 be the orthogonal polynomials described below, where the inner product on P4 is given by evaluation at -2, -1, 0, 1, 2. Find the othogonal projection of 3t^(3) onto Span{p0,p1,p2}.
p0(t) = 4
p1(t) = 3t
p2(t) = t^(2) -2

Answer 1

Answer:

$\frac{51}{5}t$

Step-by-step explanation:

Let W = $(p_0, p_1, p_2)$  be orthogonal polynomials which is equal to $(4, 3t, t^2 -2)$, which defines the inner products as

$(f,g)=f(-2)g(-2)+f(-1)g(-1)+f(0)g(0)+f(1)g(1)+f(2)g(2)$

Now, we find the orthogonal projection of $p=3t^3$ on W.

So the projection is

$Proj_W p = \frac{(p_0,p)}{(p_0,p_0)}p_0+\frac{(p_1,p)}{(p_1,p_1)}p_1+\frac{(p_2,p)}{(p_2,p_2)}p_2$

$(p_0,p)=p_0(-2)p(-2)+p_0(-1)p(-1)+p_0(0)p(0)+p_0(1)p(1)+p_0(2)p(2)$

          $=4(-24)+4(-3)+4(0)+4(3)+4(24)=0$

$(p_0,p_0)=p_0(-2)p_0(-2)+p_0(-1)p_0(-1)+p_0(0)p_0(0)+p_0(1)p_0(1)+p_0(2)p_0(2)$

            $=4(4)+4(4)+4(4)+4(4)+4(4)=80$

$(p_1,p)=p_1(-2)p(-2)+p_1(-1)p(-1)+p_1(0)p(0)+p_1(1)p(1)+p_1(2)p(2)$

          $=(-6)(-24)+(-3)(-3)+0(0)+3(3)+6(24)=306$

$(p_1,p_1)=p_1(-2)p_1(-2)+p_1(-1)p_1(-1)+p_1(0)p_1(0)+p_1(1)p_1(1)+p_1(2)p_1(2)$

            $=(-6)(-6)+(-3)(-3)+0(0)+3(3)+6(6)=90$

$(p_2,p)=p_2(-2)p(-2)+p_2(-1)p(-1)+p_2(0)p(0)+p_2(1)p(1)+p_2(2)p(2)$

         $=2(-24)+(-1)(-3)+(-2)(0)+(-1)(3)+2(24)=0$

$(p_2,p_2)=p_2(-2)p_2(-2)+p_2(-1)p_2(-1)+p_2(0)p_2(0)+p_2(1)p_2(1)+p_2(2)p_2(2)$

            $=(2)(2)+(-1)(-1)+(-2)(-2)+(-1)(-1)+2(2)=14$

Therefore,

$Proj_W p = \frac{(p_0,p)}{(p_0,p_0)}p_0+\frac{(p_1,p)}{(p_1,p_1)}p_1+\frac{(p_2,p)}{(p_2,p_2)}p_2$

              $=\frac{0}{80}(4)+\frac{306}{90}(3t)+\frac{0}{14}(t^2-2)$

              $=\frac{51}{5}t$