Periodic Functions & Orthogonal Functions — Question 5

Question 5 – Solution

Strategy. Use parity to eliminate unnecessary projections, then compute the remaining moments exactly.

Step 1: Construct the monic family. An odd integrand integrates to zero on $[-1,1]$, so every even polynomial is orthogonal to every odd polynomial. For $p_2=x^2-c$, the equation $\langle p_2,1\rangle=2/3-2c=0$ gives $c=1/3$. For $p_3=x^3-dx$, $\langle p_3,x\rangle=2/5-2d/3=0$ gives $d=3/5$. Thus $$\boxed{p_0=1,\quad p_1=x,\quad p_2=x^2-1/3,\quad p_3=x^3-3x/5.}$$ Parity and these two moment equations verify every distinct pair.

Step 2: Compute and apply the norms. Integrating the squares term by term yields $$\boxed{\|p_0\|^2=2,\quad\|p_1\|^2=2/3,\quad \|p_2\|^2=8/45,\quad\|p_3\|^2=8/175.}$$ An orthonormal set is therefore $\boxed{p_0/\sqrt 2,\ \sqrt{3/2}\,p_1,\ \sqrt{45/8}\,p_2, \ \sqrt{175/8}\,p_3}$. Monic normalization fixes the leading coefficient; unit normalization fixes the squared integral and generally changes that coefficient.

Step 3: Find the best quadratic approximation. The decomposition $x^3=3x/5+p_3$ has its residual orthogonal to every quadratic. For any other quadratic $q$, the cross term vanishes, giving $$\|x^3-q\|^2=\|p_3\|^2+\|3x/5-q\|^2.$$ Hence $\boxed{q_*=3x/5,\quad\min\|x^3-q\|^2=8/175}$. Equality forces the continuous polynomial $q-3x/5$ to have zero norm and therefore vanish identically, proving uniqueness.

Step 4: Reject normalization without orthogonalization. The normalized versions of $1$ and $x^2$ are $1/\sqrt 2$ and $\sqrt{5/2}\,x^2$. Their inner product is $\boxed{\sqrt 5/3\ne 0}$. Normalizing lengths alone does not remove projections onto the preceding functions.