Periodic Functions & Orthogonal Functions — Question 4

Question 4 – Solution

Strategy. Compute the product integral before deciding whether a pair is orthogonal or merely normalized.

Step 1: Determine the window dependence. Using $\sin x\cos x=\tfrac 12\sin 2x$ gives $$\langle u,v\rangle_{a,L}=\frac{\cos 2a-\cos(2a+2L)}4 =\boxed{\frac 12\sin L\sin(2a+L).}$$ It vanishes for all $a$ exactly when $\sin L=0$. The positive lengths are $\boxed{L=m\pi,\ m=1,2,\ldots}$. If $\sin L\ne 0$, one can choose $a$ with $\sin(2a+L)=1$, proving necessity.

Step 2: Locate the shorter orthogonal windows. For $L=\pi/2$, orthogonality requires $\sin(2a+\pi/2)=0$. Hence $$\boxed{a=-\pi/4+j\pi/2,\qquad j\in\mathbb Z.}$$ The product changes sign; shifting the window can balance its signed areas. At $a=0$, however, the product is positive inside the interval.

Step 3: Test individual normalization. On $[0,\pi/2]$, $\|u\|^2=\|v\|^2=\pi/4$ and $\langle u,v\rangle=1/2$. The unit-norm functions are $\widehat u=2\sin x/\sqrt\pi$ and $\widehat v=2\cos x/\sqrt\pi$, with $$\boxed{\langle\widehat u,\widehat v\rangle=2/\pi\ne 0.}$$ Nonzero scaling cannot turn a nonzero inner product into zero.

Step 4: Normalize on the correct windows. Over any length $m\pi$, the oscillatory term in either squared integral cancels, so both squared norms equal $m\pi/2$. Thus $\boxed{\sqrt{2/(m\pi)}\sin x,\ \sqrt{2/(m\pi)}\cos x}$ are orthonormal on every such window. The graph shows equal positive and negative product areas on $[0,\pi]$; the integral calculation establishes their exact cancellation.