# Administrative Information - **Class Meeting**: Monday, September 14, 2026, 1:00–1:50pm — **Coolbaugh 212**. *(No class Wednesday 9/16 — Career Day.)* - Previous meeting: [[MATH310F26(Day 8) - The spectrum of a pure tone]] - 📄 [[Teaching/MATH310/MATH310F26/Administration/MATH310_F26_Syllabus_1.0.pdf|MATH 310 F26 Syllabus, v1.0]] > [!info]- Admin. Notes > Expect an LR and WB for each lecture. WBs target 15-minute engagements (longer early on); LRs are short, ~5 minutes. > > - 🎯 **[LR] [Day 9 Reflection Questions](https://elearning.mines.edu/courses/81632/quizzes/121253)** — 4 questions drawn from today's material, unlimited attempts, highest kept — due **Friday 9/18, 11:59pm** (no class Wednesday, Career Day) > - 🎯 **[LR] [Day 9 Reflection Statement](https://elearning.mines.edu/courses/81632/quizzes/121254)** — one free response, no wrong answers — due **Friday 9/18, 11:59pm** (no class Wednesday, Career Day) > - 🎯 **[LR] [Day 9 Reflection Questions](https://elearning.mines.edu/courses/81632/quizzes/121253)** — 4 questions drawn from today's material, unlimited attempts, highest kept — due **Tuesday 9/16, 11:59pm** > - 🎯 **[LR] [Day 9 Reflection Statement](https://elearning.mines.edu/courses/81632/quizzes/121254)** — one free response, no wrong answers — due **Tuesday 9/16, 11:59pm** > - 🎯 **[LR] [Day 8 Reflection Questions](https://elearning.mines.edu/courses/81632/quizzes/121215)** — 4 questions drawn from last Friday's material, unlimited attempts, highest kept — due **Monday 9/14, 11:59pm** > - 🎯 **[LR] [Day 8 Reflection Statement](https://elearning.mines.edu/courses/81632/quizzes/121216)** — one free response, no wrong answers — due **Monday 9/14, 11:59pm** > - 🎯 **[WB] [Workbook 6](https://elearning.mines.edu/courses/81632/assignments/586618)** — conjugate pairs and the exponential representations of the sinusoids; upload one PDF > - 📄 [[MATH310F26_WB6.pdf|Workbook 6 (blank, 1 page)]] · [[MATH310F26(Workbook 6 Companion) - What a conjugate pair extracts|companion]] > - 🎯 **[WB] [Workbook 7](https://elearning.mines.edu/courses/81632/assignments/586619)** — Fourier transformation of sinusoids; upload one PDF > - 📄 [[MATH310F26_WB7.pdf|Workbook 7 (blank, 1 page)]] · [[MATH310F26(Workbook 7 Companion) - Two tones and a constant, transformed|companion]] > - A **dice-data report** was promised for Monday — watch Canvas > [!example]- 🗃️ COMAP Corner — three from the archive > Three from the archive — one data, one computational, one weird — for a day of images and their spectra: > > - **Data/stats — The Hydrographic Data Problem (1986 MCM A).** Sparse depth soundings scattered across a bay, a ship with a five-foot draft: reconstruct the whole seafloor surface from point data and say where it's safe to sail — and how much to trust the reconstruction. 📄 [[1986_MCM_Problem_A_The_Hydrographic_Data_Problem.pdf|problem PDF]] · [comap.com](https://www.contest.comap.com/undergraduate/contests/matrix/PDF/1986/1986A.pdf) > - **Computational — The Scanner Problem (1998 MCM A).** An MRI stores a 3-D grid of density pixels but only shows slices along the main axes. Design an algorithm that renders a sharp cross-section at *any* plane orientation — image reconstruction, the working mathematician's version of today's game. 📄 [[1998_MCM_Problem_A_The_Scanner_Problem_Problem_B_The_Grade_Inflation_Problem.html|problem page (archive)]] · [comap.com](https://www.contest.comap.com/undergraduate/contests/matrix/PDF/1998mcmProblems.htm) > - **Just weird — Drone Clusters as Sky Light Displays (2017 HiMCM A).** Design three synchronized aerial drone images — a Ferris wheel, a dragon, and an original — specifying every drone's position and flight path. Drawing pictures with point sources in the sky: the image game, run in reverse. 📄 [[2017_HiMCM_Problem_A_Drone_Clusters_as_Sky_Light_Displays_Problem_B_Ski_Slope.html|problem page (archive)]] · [comap.com](https://www.contest.comap.com/highschool/contests/himcm/2017problems.html) --- # Deliverables - 🎯 **[LR] [Day 9 Reflection Questions](https://elearning.mines.edu/courses/81632/quizzes/121253)** — 4 questions drawn from today's material, unlimited attempts, highest kept - 🎯 **[LR] [Day 9 Reflection Statement](https://elearning.mines.edu/courses/81632/quizzes/121254)** — one free response, no wrong answers Questions and statement are due Friday 9/18 at 11:59 pm (no class Wednesday, Career Day). --- # Lecture Boards + Transcript + GenAI **Ninth meeting, September 14 — seeing the Fourier transform.** The roadmap on the way in: today we *see* the Fourier transform, Friday we *hear* it, modeling questions land this week, and after Fourier the course transitions into **stochastic processes**. The course is front-loaded — "grind fest on lecture" — but in-class collaboration days are coming, and presentation times get scheduled later in the semester. From the notes board: **753 random numbers from dice now exist** (the promised weekend look at the Yahtzee data happened), and another **modeling Q&A** is planned, with the goal of everyone having *direction* by the end of next week. Also, a public-service tangent: Crayola allegedly has the precise wax proportions of "red" locked down, so anyone modeling candles or crayons for their project has been warned. That doesn't seem to be useful. Moving on. ## The single finite pulse Adopt $x$ as the independent variable (it will help the properties read cleanly), and define, for $L\in(0,\infty)$ and $A\in\mathbb{R}$, $f(x)=\begin{cases}A, & x\in(-L,L),\\[2pt] 0, & x\notin(-L,L).\end{cases}$ Two behaviors: the height $A$ between $-L$ and $L$, the number $0$ outside — hence *single finite pulse*. (It looks a lot like a uniform distribution on $[-L,L]$; if $x$ were time it would be nothing, nothing, nothing, then a constant displacement, then nothing again.) In signal analysis this is the **rect function**, because rectangle. A good question from the room: what happens *at* the points $x=\pm L$? Answer: define $f(\pm L)$ however you like — the Fourier tool doesn't care, because **the integral doesn't see points; it sees accumulation of continuum**: changing a function on finitely many points — more generally, on any set of *measure zero* — leaves its transform untouched. (The in-class shorthand that the transform "averages the two sides" belongs, stated precisely, to the *return trip*: under the usual inversion hypotheses, the reconstruction at a jump comes back as the average $\tfrac{1}{2}\big(f(x^-)+f(x^+)\big)$ of the one-sided limits — that is where the $\tfrac{A}{2}$ lives.) A deeper version of the remark waits in Fourier-series land: a reconstruction can disagree with the original on such a negligible set and still deserve the equals sign — countably many points being the classroom-sized case of measure zero. That's troubling, but negligible, so it's OK. ## The computation $\hat f(\omega)=\int_{-\infty}^{\infty}f(x)\,e^{-i\omega x}\,dx=\int_{-L}^{L}A\,e^{-i\omega x}\,dx$ — the orange zero regions kill everything outside $[-L,L]$, and the constant $A$ doesn't care about being in the integral, so out it comes: $\hat f(\omega)=A\cdot\frac{1}{-i\omega}\,e^{-i\omega x}\Big|_{x=-L}^{x=L}=\frac{A}{-i\omega}\Big(e^{-i\omega L}-e^{+i\omega L}\Big).$ (A sign slip on the board — the lower bound's exponent "came at us with a plus sign" — was caught and repaired live.) Now the exponentials call out to Workbook 6: bust each into $\cos(\omega L)\mp i\sin(\omega L)$, add, and watch the cosines vanish while the sines double up: $\hat f(\omega)=\frac{A\,(-2i\sin(\omega L))}{-i\omega}=\frac{2A\sin(\omega L)}{\omega}.$ Negatives: toast. $is: also gone. One last cosmetic move — the sine's argument is $\omega L$, and the denominator would read better as $\omega L$ too, so multiply by one in the form $L/L$ (the yellow didn't affect anything): $\boxed{\ \hat f(\omega)=2AL\,\frac{\sin(\omega L)}{\omega L}\ }$ — *sine of stuff over that same stuff*, the shape called $\operatorname{sinc}$. ## Notes on $\hat f$ **The hole at $\omega=0$, spackled.** The denominator forbids $\omega=0$, but we want frequency zero back. As $\omega\to 0$ the form is indeterminate, so evoke L'Hôpital (little $h$ over the equals sign): differentiate top and bottom with respect to $\omega$, get $L\cos(\omega L)/L\to 1$. So the limit exists, the point was a *hole*, not a singularity, and we spackle it in: $\hat f(\omega)=\begin{cases}2AL\,\dfrac{\sin(\omega L)}{\omega L}, & \omega\neq 0,\\[4pt] 2AL, & \omega=0.\end{cases}$ **Even.** Throw in $-\omega$: the sine tosses its negative out front, the denominator supplies another, they cancel — $\hat f(-\omega)=\hat f(\omega)$. Practical payoff: graph half, copy the other half over. **Decay.** As $|\omega|\to\infty$, the sine stays trapped in $[-1,1]$ while the denominator grows without bound: $\hat f\to 0$. **Zeros.** $\hat f(\omega)=0$ exactly when $\omega L=n\pi$, $n\in\mathbb{Z}\setminus\{0\}$ — the roots march along at $\omega=\pi/L,\ 2\pi/L,\ 3\pi/L,\dots$ ## Squiddy Put it together: peak $2AL$ at the center (the spackled point), oscillating arms decaying to either side, roots at $n\pi/L$. In keeping with a course tradition, the graph receives little eyes, is recognized as looking a lot like an octopus or a squid, and is henceforth **Squiddy**: ![[day9_squiddy.png]] Read Squiddy as instructions to the inverse transform: the rectangle is *almost* a constant, so take a lot of the $\omega\approx 0$ non-wave (that's $e^{i\cdot 0\cdot t}$, a constant); but it isn't *completely* constant, so add corrective waves in decreasing amounts, dipping negative in the first side lobes, then positive, and so on — and at the roots $\omega=n\pi/L$, take **none at all**. Those are the frequencies the rectangle simply doesn't need. > [!tip]- The uncertainty principle, by squid > Stretch the rectangle ($L$ bigger) and watch Squiddy: the roots $n\pi/L$ crowd inward, and the whole creature **compresses**. Squeeze the rectangle and Squiddy sprawls. Widening in $x$ shrinks in $\omega$, always — stretching the pulse compresses the Fourier *scale*, wholesale. (The in-class variance phrasing deserves one refinement: $\hat f$ itself is not a probability density — the rigorous Heisenberg inequality is stated for the densities $|\psi|^2$ and $|\hat\psi|^2$ of $L^2$ wavefunctions, and for the hard-edged rectangle the frequency-side variance is in fact *infinite* — but this scaling trade-off is exactly what that inequality makes precise.) The chemistry majors have met this: electron *clouds* — shouldn't the electron be someplace? That is the **Heisenberg uncertainty principle**, and it is not an add-on: it is *baked into the Fourier transform*. The math quantum mechanics uses to describe position and momentum says this trade-off is simply what has to be going on. > > One more reading, for what's coming below: if you don't care about the sign of the amplitude, reflect the negative arms up ($|\hat f|$) and read the plot as **brightness** — blazing at the center, dimming outward, with **dark points** at the roots. Hold that thought. ## The applet — play with all of this yourself Something realized over the weekend: **apps can be embedded into this website.** (Inspiration: the excellent [Math Insight](https://mathinsight.org/) applets from Calc 2/3.) The rectangular pulse from today is the first one below — drag $L$ and watch the uncertainty principle happen; tick *unit area* and drag $L\to 0$ to watch the pulse sharpen into a delta while Squiddy flattens toward the constant $1$: **deltas come at us in many forms, even the oscillatory types.** An assignment built around these is coming; for now, play. <div class="applet" data-applet="rect-transform"></div> The same story in two dimensions — the 2-D rect transforms to a product of sincs (a Squiddy horizontally times a Squiddy vertically, decaying in both directions off-axis), and **rotating the rectangle rotates the cross**: <div class="applet" data-applet="rect-transform-2d"></div> And the discrete version on actual images — presets, your own uploaded file, or **draw a shape** (in-class finding: clearly S is the best letter): <div class="applet" data-applet="image-dft"></div> Two full pages of these live on the site now, with the audio: [[Fourier Series]] (build a wave from harmonics, and *hear* each one — Friday's preview) and [[Fourier Transforms]] (the rect ↔ sinc pair, the pulse train, duality, the 2-D bar, and the image DFT). ## The picture game The class then played a guessing game from the projector: **see an image, commit to what its Fourier transform must look like, then reveal.** The stills in the slide deck came from a lecture video, so the images below have been *recomputed from scratch* for this page (zero-mean 2-D FFT, $\log(1+|F|)$, computed in one sandbox run); the photograph is the field's standard *cameraman* test image (copyright owned by MIT). The reveals are folded: **guess first, then click.** **Round 1 — one ripple.** A pure horizontal-banded cosine, rippling down the page. A slice down any column reads down-up-down-up: one cosine, one frequency. How did we build a cosine today? Out of *two* conjugate tones. ![[day9_grating_h.png]] > [!question]- Reveal — its Fourier transform > ![[day9_grating_h_ft.png]] > > Two dots on the **vertical** frequency axis — the 2-D version of $\pi\delta(\omega-\omega_{0})+\pi\delta(\omega+\omega_{0})$: ideal localization at the ripple's one frequency, in the ripple's one direction (plus its mirror, because the image is real). **Round 2 — three variations.** Vertical stripes; the same stripes crammed finer; stripes at $45°$. ![[day9_gratings3.png]] > [!question]- Reveal — their Fourier transforms > ![[day9_gratings3_ft.png]] > > Rotate the ripples and **the dots rotate**. Cram the ripples together and the dots **move farther from the origin** — that is a higher carrier frequency, not uncertainty at work. (The *uncertainty* version of this experiment: keep the spacing but confine the ripples to a small patch of the image, and each dot smears into a blob — localize in $\mathbf{x}$, spread in $\mathbf{k}$. The wedge caveat at the end of this page is the same fact from the other side.) Nothing else changes: still one frequency each, still ideally localized. **Round 3 — a square.** No more ripples; now a blockage with an amount let through, in both directions. ![[day9_square.png]] > [!question]- Reveal — its Fourier transform > ![[day9_square_ft.png]] > > A **cross**: this direction gives Squiddy vibes and so does that one — a sinc horizontally times a sinc vertically, with the off-axis regions doubly decayed into darkness. The high-frequency ripples along each arm are the squid arms, rendered as brightness and dark roots. **Round 4 — a rotated bar.** Not a square anymore: elongated one way, thin the other, and tilted. ![[day9_bar.png]] > [!question]- Reveal — its Fourier transform > ![[day9_bar_ft.png]] > > The cross **rotates with the bar**, and the two arms trade widths: the elongated direction of the bar is the *compressed* direction in frequency, and the thin direction sprawls. (Rendered hot, as the class saw it.) **Round 5 — a dot.** "Just for funsies." ![[day9_disk.png]] > [!question]- Reveal — its Fourier transform > ![[day9_disk_ft.png]] > > Concentric **rings** — a circular object has no preferred direction, so neither does its transform; the radial profile is the circular cousin of the sinc. We don't *need* this one, but it's a cool picture. **Round 6 — an actual photograph.** The classic *cameraman* test image, loaded live into the applet above: jacket against sky, tripod legs against grass, a field of not-much-changing and then boom, sky. ![[day9_photo.png]] > [!question]- Reveal — its Fourier transform > ![[day9_photo_ft.png]] > > **Streaks.** Streaks in Fourier space feel like they came from Squiddy, and Squiddy comes from rectangles — but forget the precision: rectangles are really **edges**. Every abrupt straight transition in the photo — the jacket against the sky, the black legs on the white-gray ground, field-then-boom-sky — throws a bright line across frequency space, perpendicular to the edge. Line-like structure in $\hat f$ ⇒ abrupt straight change in $f$. That is a diagnostic you can use. ### The \$10 question, settled A hand went up on exactly the right thing: *what are those two lines near the top of the transform?* The live answer — "I have no idea, but I bet you \$10 they're the jacket" — turned into a follow-up computation after class, and the bet pays. The trick is that **the transform is invertible, so any piece of the spectrum can be turned back into a piece of the picture.** Take $\hat f(\mathbf{k})$, choose a region $R$ of the $k$-plane — say a thin wedge through the origin along one streak — and build a mask $M(\mathbf{k})$ that is $1$ inside $R$ and $0$ everywhere else. Multiply and invert: $g(\mathbf{x})=\mathcal{F}^{-1}\big[\,M(\mathbf{k})\,\hat f(\mathbf{k})\,\big].$ The image $g$ contains only the sinusoids whose frequencies live in $R$ — since the full image is the sum of all its sinusoids, $g$ is *literally the part of the picture that lives in $R$*. Where $g$ is large, the picture has structure of that kind; where it is nearly zero, it doesn't. (Linearity is doing the work: cut the spectrum into regions, invert each, and the pieces add back to the original.) To paint that as a color map, don't use the oscillating values of $g$ themselves — they swing positive and negative across every edge — but their strength: the **local energy** $e(\mathbf{x})=(g^{2}*\text{Gaussian})(\mathbf{x})$. Each spectral region gets its own $e$, and each pixel wears the color of whichever region carries the most energy there: ![[day9_spoke_attribution.png]] Read the verdicts. The vertical spoke is the **horizon and skyline** (horizontal structure lives on the vertical frequency axis — perpendicular, always). The horizontal spoke is the **tripod's vertical post**. The two oblique spokes are the **two tripod legs**, one each. The wider band — the two lines the question asked about — is the **coat's outline and folds**: ten dollars, please. And the high-$|k|$ halo, far from the origin in every direction, returns the **fine structure**. > [!note]- The caveat, which is the uncertainty principle again > A thin wedge keeps almost no frequencies *along* the edge's direction, so $g$ can barely vary in that direction — a horizontal edge comes back as a stripe across the full width of the frame. The wedge tells you which **rows** the horizon is in, not which columns. Widen the wedge and $g$ localizes along the edge too, at the cost of admitting nearby orientations. Narrow in $\mathbf{k}$ means spread out in $\mathbf{x}$: it is Squiddy's trade-off wearing image-processing clothes, and it is why the colors in the figure are ribbons rather than dots. **Final round — the million-dollar question.** (Well, not anymore — pretend it's ten years ago, before the neural networks ate it.) Three different people write the letter **A**: ![[day9_letters_A.png]] > [!question]- Reveal — their Fourier transforms > ![[day9_letters_A_ft.png]] > > Don't those A's feel like **tripods**? Then their transforms should share a distinctive tripod signature — and they do: the same three-way star, robust across handwriting that varies enough to fool an undertrained eye in image space. And then **B**, **C**, **D**: ![[day9_letters_BCD.png]] > [!question]- Reveal — their Fourier transforms > ![[day9_letters_BCD_ft.png]] > > Each letter carries its own spectral fingerprint — D's straight back, for instance, plants a clean stripe its curvy neighbors lack. So a classifying machine that is *not* a neural network can look at data twice: the letters, **and** their transforms — a cheap **dual perspective** in which every A looks like every other A. That is the modeling point of the whole game: the transform is extra data you already own. > [!success]- Numbers checked in one sandbox run > The boxed pair was verified numerically at $A=1.3$, $L=0.8$: $\hat f(0)=2.08=2AL$; $\hat f(\pi/L)$, $\hat f(2\pi/L)$, $\hat f(3\pi/L)$ all $\sim 10^{-17}$; $\hat f(2)-\hat f(-2)=0$ exactly (even); and $\sin(\omega L)/\omega L\to 1$ as $\omega\to 0$, confirming the L'Hôpital spackle. Every image pair above is a fresh zero-mean 2-D FFT ($\log(1+|F|)$, spikes dilated for visibility where the exact transform is a pair of single-pixel dots); the letter images are synthetic type in varied faces and slants, standing in for handwriting. The spoke-attribution figure was computed in a companion session: wedge/band/halo masks on $\hat f$, inverse transform, local energy $g^{2}*\text{Gaussian}$, winner-take-most coloring. > [!example]- Board scans — September 14, 2026 > ![[MATH310F26-Day9-Board-1.jpg]] > ![[MATH310F26-Day9-Board-2.jpg]] > ![[MATH310F26-Day9-Board-3.jpg]] > ![[MATH310F26-Day9-Board-4.jpg]] > > 📄 [[MATH310F26-Day9-Boards-2026-09-14.pdf|Full board capture (4 pages, PDF)]]