Exercise 6.C.12

Proof. Define $U$ by

Note that $U$ is a subspace of $P_3(ℝ)$. Let $u\in U$. Since $p\in P_3(ℝ)$, we have

for some $a,b,c,d\in ℝ$. $p(0)=0$ and $p^{\prime }(0)=0$ now imply that $c=d=0$ and so $U\subset \mathrm{span}<!--\; FUNCTION\; APPLICATION\; -->(x^2,x^3)$. Because $x^2,x^3\in U$, it follows that $x^2,x^3$ is a basis of $U$. Applying the Gram_Schmidt Procedure to this basis, we get

Now, using the formula from part(i) in 6.55 to get $P_U(2+3x)$, we get

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