Chapter 22 — Quiz
Twelve conceptual checks. Answer before expanding. If you can explain each in a sentence, you have the chapter's spine: a Fourier coefficient is an orthogonal projection.
Q1. What is the inner product of two functions $f$ and $g$ on $[-\pi,\pi]$, and what role does it play?
Answer
$\langle f,g\rangle = \int_{-\pi}^{\pi} f(x)g(x)\,dx$. It is the function-space analogue of the dot product: it defines length ($\lVert f\rVert = \sqrt{\langle f,f\rangle}$), angle, and orthogonality, turning the space of functions into an inner product space where all of Part IV's machinery applies.Q2. State the orthogonality relation for two sines $\sin kx$ and $\sin mx$ on $[-\pi,\pi]$, including both cases.
Answer
$\int_{-\pi}^{\pi}\sin kx\,\sin mx\,dx = \pi$ when $k = m$, and $0$ when $k \neq m$ (for positive integers $k,m$). So distinct sine frequencies are orthogonal, and each $\sin kx$ has squared norm $\pi$.Q3. A Fourier coefficient is "a projection." Onto what, and using what formula?
Answer
It is the orthogonal projection of the signal $f$ onto a single basis frequency. For example $b_k = \tfrac1\pi\langle f, \sin kx\rangle$ — the inner product of $f$ with $\sin kx$, divided by the squared norm $\langle \sin kx,\sin kx\rangle = \pi$. This is exactly the projection-onto-an-orthogonal-direction formula from Chapter 19.Q4. Why is the factor $1/\pi$ present in $a_k$ and $b_k$, but $1/(2\pi)$ in $a_0$?
Answer
Because the projection formula divides by the *squared norm* of the direction. The sines and cosines have squared norm $\pi$, giving $1/\pi$; the constant function $1$ has squared norm $\langle 1,1\rangle = 2\pi$, giving $1/(2\pi)$. The factors normalize an orthogonal-but-not-unit basis.Q5. What single property of the basis makes each Fourier coefficient computable independently of all the others?
Answer
Orthogonality. When you integrate $f(x)\sin kx$ to find $b_k$, every other basis function contributes zero because it is orthogonal to $\sin kx$. So no other frequency leaks in, and each coefficient is its own independent integral — the central gift of orthogonality from Part IV.Q6. The square wave's Fourier series contains only odd-harmonic sines. Give the coefficient formula and the reason the even harmonics and all cosines vanish.
Answer
$b_k = 4/(\pi k)$ for odd $k$, and $0$ otherwise; all $a_k = 0$. The square wave is *odd*, so it is orthogonal to every (even) cosine and to the constant — those projections are zero. Its half-wave symmetry then also makes it orthogonal to the even sines, leaving only odd harmonics.Q7. What is the Gibbs phenomenon, and does it go away as you add more terms?
Answer
Near a jump discontinuity, every partial sum overshoots the target by about $9\%$ of the jump height. It does **not** go away with more terms: the overshoot's *height* stays near $1.179$ (for a jump to $1$) while its *width* shrinks toward the jump. It is the gap between mean-square convergence (which holds) and pointwise convergence (which fails *at* the jump).Q8. In what sense, then, does the Fourier series of the square wave "converge" to it?
Answer
In the mean-square (energy) sense: $\lVert f - S_N\rVert \to 0$ as $N \to \infty$. Equivalently, the integrated squared error vanishes. It also converges *pointwise* wherever $f$ is continuous; at the jump it converges to the midpoint of the jump. The Gibbs spike contributes ever less *energy* even though its height persists.Q9. Why is the truncated Fourier series the best approximation of $f$ using those frequencies?
Answer
Because truncation is orthogonal projection onto the subspace spanned by those frequencies, and orthogonal projection gives the closest point in a subspace (Chapter 19). No other combination of the same sines and cosines comes closer to $f$ in the norm $\lVert\cdot\rVert$. Truncating a Fourier series is least-squares approximation.Q10. What does Parseval's identity say, and why does it justify "keep the big coefficients" as a compression strategy?
Answer
Parseval says $\lVert f\rVert^2 = 2\pi a_0^2 + \pi\sum(a_k^2 + b_k^2)$ — the total energy equals the sum of the energies in each frequency (the Pythagorean theorem in an orthonormal basis). Because energy is additive across orthogonal coordinates, discarding a small coefficient loses exactly its small energy and nothing else, so keeping the large coefficients preserves almost all the signal.Q11. How does the continuous Fourier series relate to the discrete Fourier transform (DFT) that a computer runs?
Answer
They are the same projection at different resolutions. The series projects a continuous function onto an infinite orthogonal basis using integrals; the DFT projects a *sampled* vector onto a *finite* orthogonal basis of sampled exponentials using dot products. Replace the integral by a Riemann sum and the series becomes the DFT. The FFT is just a fast $N\log N$ algorithm for the DFT.Q12. Sines and cosines are the eigenfunctions of which operator, and why does that foreshadow Part V?