matharium
@matharium • 10,067 subscribers
Math was never the problem. The explanation was.
Shorts
![A turnstile antenna is two crossed dipoles driven 90° apart, one dipole p turning at ω about its mast. Its far field E ∝ r̂ × (r̂ × p̈(t − r/c)) has phase ωt − kr − φ at every latitude, so each wavefront is the spiral r = (ωt − φ − 2πn)/k. Along the mast E ∝ (1/r)[cos(kr − ωt) ŷ + sin(kr − ωt) x̂] keeps its length and turns once per period, and B = (1/c) r̂ × E turns with it. In the plane of the cross the same wave is linearly polarized.](https://image.24vids.com/tw-2106418848027299848/amplify_video_thumb/2106418431650418688/img/AxlwU3tXg4lXbw-v.jpg)
![A double integral is the limit of a sum of prisms: cut [0, π]² into n × n squares of area ΔA = (π/n)² and raise each to the height of z = sin³x sin³y at its centre. The volume is (∫₀^π sin³x dx)² = (4/3)² = 16/9 ≈ 1.7777778. The prisms give 1.7602152 at n = 4, 1.7768141 at n = 8 and 1.7777776 at n = 64, always from below. Since sin³x = (3 sin x − sin 3x)/4, the h² errors of the two waves cancel, and the gap shrinks like 7π⁴/(180n⁴).](https://image.24vids.com/tw-2106381302031393274/amplify_video_thumb/2106381195407937536/img/-FsmOkG9RkHlxZ_Q.jpg)
Videos
Each coordinate of a closed outline has its own real Fourier series: x(t) = ∑ (aₖ cos kt + bₖ sin kt), and the same for y. A cosine and sine of frequency k make one arm of length √(aₖ² + bₖ²) turning k times per lap. One chain sets x, one sets y, and the pen sits where their guides cross. With N harmonics per axis the knight is off by 30.2 % of its height at N = 1, 0.70 % at N = 50. At N = 1 both coordinates are pure sinusoids: an ellipse.
matharium1,370,332 次观看 • 9 天前
Joseph Fourier’s 1823 engraving, traced as 57 ink shapes joined by thin bridges into one closed outline, is a sum of circles: z(t) = ∑ cₖ eⁱᵏᵗ, the cₖ from its discrete Fourier transform. With the N largest circles the widest gap to the outline is 42.0 % of the portrait’s height at N = 1, 3.16 % at 128 and 0.92 % at 512, and the circles shrink like k^−1.47. The tip’s distance |z(t)| plays as pitch.
matharium105,487 次观看 • 4 天前
The treble clef 𝄞, its four contours joined by thin bridges and sampled at 4096 points, is a sum of circles: z(t) = ∑ cₖ eⁱᵏᵗ. With the N largest circles the widest gap to the outline falls from 35.7 % of its height at N = 1 to 0.37 % at 256. The tip's distance |z(t)| is played as pitch, floored to C major. One circle keeps |z| fixed: one note. With 64 the tip plays the outline's own note 93 % of the lap; with 256, 98 %.
matharium189,597 次观看 • 9 天前
The curves y = x² and y = 1.1ˣ cross exactly three times: at x = −0.955, at x = 1.051 and at x = 95.717. Between the second and third crossings the square leads, by a lot: 50² = 2500 while 1.1⁵⁰ ≈ 117. The ratio 1.1ˣ/x² bottoms out at x = 2/ln 1.1 ≈ 21 and grows without bound after it, so the third crossing is final. By x = 200, 1.1ˣ ≈ 4 700 · x².
matharium47,724 次观看 • 2 天前
Each coordinate of the Koch snowflake's outline has its own real Fourier series, x(t) = ∑ (aₖ cos kt + bₖ sin kt), and one chain of arms per axis draws it where their guides cross. The snowflake's six-fold symmetry leaves only k ≡ ±1 (mod 6), so five arms in six are zero, and its mirror leaves x only sines and y only cosines. The error falls from 11.5 % of the height at N = 1 to 0.60 % at N = 193. At N = 1 the two arms are equal: a circle.
matharium92,000 次观看 • 6 天前
The Moore curve of order 4 is a loop of 256 unit steps through every cell of a 16 × 16 grid. Each coordinate has its own real Fourier series, and one chain of arms per axis draws it. The second half of the loop is the first half turned 180°, so every even harmonic is zero, and its mirror leaves x only sines and y only cosines. A curve that wiggles everywhere converges slowly: 32.3 % of the height at N = 1, still 0.47 % at N = 255.
matharium66,592 次观看 • 5 天前
A square wave is a sum of odd sines, f(x) = (4/π) ∑ sin((2n−1)x)/(2n−1), and each term is a circle of radius 4/(π(2n−1)) turning 2n−1 times per period. Chained end to end, the height of the n-th joint is the partial sum Sₙ(x). The overshoot at each jump tends to 0.179, about 9 % of the jump, however many terms are added. At x = π/2 every arm points straight up or down and f = 1, which is Leibniz's series: π/4 = 1 − 1/3 + 1/5 − 1/7 + ⋯.
matharium47,203 次观看 • 8 天前
The lemniscates of the Mandelbrot set are the curves |pₙ(c)| = 2, with p₁(c) = c and pₙ₊₁ = pₙ² + c. At n = 1 the curve is the circle |c| = 2. Each lies inside the last, and all pass through c = −2, where pₙ = 2 for n ≥ 2. The area inside falls from 4π ≈ 12.57 to 1.92 at n = 12, a curve of degree 4096. They close in on the set: c lies in M exactly when |pₙ(c)| ≤ 2 for every n, since an orbit that leaves the disc of radius 2 escapes.
matharium38,435 次观看 • 7 天前
The outline of π, sampled at 2048 points, is a sum of circles: z(t) = ∑ cₖ eⁱᵏᵗ, the cₖ from its discrete Fourier transform. Circle k has radius |cₖ| and turns k times per lap. With the N largest, the widest gap to the outline is 41.6 % of the letter's height at N = 1, 1.37 % at 64, 0.36 % at 256. Finitely many circles make a smooth curve, and π has four sharp corners, so no chain is exact. The error stays below the sum of the radii left out.
matharium51,299 次观看 • 11 天前
A string fixed at both ends vibrates in standing waves sin(nπx/L), with frequencies fₙ = n·f₁ = (n/2L)√(T/μ). On the A string, f₁ = 110 Hz, the first eight are A2, A3, E4, A4, C#5, E5, G5, A5. Six sit within 2 cents of a piano key; the 5th is 14 cents flat and the 7th, 770 Hz, is 31 cents flat of G5. A pluck at p = 1/5 is the sum of all modes with bₙ = 2h sin(nπp)/(n²π²p(1−p)). Since sin(π) = 0, it has no 5th harmonic.
matharium26,780 次观看 • 8 天前
Ten functions y(x) are played as pitch, f = 130.8 · 2ʸ Hz, with x advancing 2.5 units per second, so every graph is four bars of music. A kick is y = −1.6 + 5e^(−kx): a five octave jump that falls back toward 43 Hz. ⌊·⌋ snaps two sines onto steps of 7/12 of an octave, a fifth. arctan(s sin 4πx)/arctan s squares a vibrato as s climbs from 0.5 to 40, and sin(πp²/4) quickens its own notes.
matharium33,662 次观看 • 11 天前
The first phrase of Beethoven's Ode to Joy is a single formula: d = ⌊2.27 + 1.2 sin(π(s + 1)/8) − cos(πs/4)⌋ with s = ⌊x⌋ + ½, played at f = 261.63 · 2ʸ Hz. The tune moves only by scale steps, so two harmonics carry its shape and the floor snaps each beat onto a degree: E E F G G F E D C C D E E D D. Then ⌊(12d + 5)/7⌋ = 0, 2, 4, 5, 7, 9, 11 turns degrees into semitones. The major scale itself is a floor: seven notes spread evenly over twelve.
matharium13,817 次观看 • 9 天前
A square wave flipping between ±1 at multiples of π is the limit of f_N(t) = (4/π) ∑ sin((2k+1)t)/(2k+1) for k < N. Only odd harmonics appear, since f(t + π) = −f(t). Term k is a circle of radius (4/π)/(2k+1) turning 2k+1 times as fast as the first; the chain ends at height f_N(t). The sum overshoots each jump. Its first peak, at t = π/(2N), is 1.182 for N = 5 and 1.179 for N = 100, about 9 % of the jump. More terms narrow it, never remove it.
matharium23,614 次观看 • 23 天前
A polyrhythm is two even pulses sharing one span. One hand turns every 3 s and sounds each corner it crosses: the square's four, every 0.75 s, against a regular n-gon's n, every 3/n s. The two pulses meet gcd(n, 4) times a turn: once for 3:4, 5:4, 7:4, 9:4 and 15:4, twice for 6:4 and 10:4, and on every square beat for 8:4 and 20:4.
matharium17,238 次观看 • 23 天前
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