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Pythagorean theorem, written on the unit circle: sin²θ + cos²θ = 1.

The Math Flow

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63,394 просмотров • 3 дней назад •via X (Twitter)

Образование Наука и технологии

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Euler's formula states that e^(iθ) = cos θ + i sin θ. As the angle θ increases, e^(iθ) traces the unit circle in the complex plane. Plotted against θ, that rotation becomes a helix. The real part of the helix projects as the cosine wave Re(z) = cos θ. The imaginary part projects as the sine wave Im(z) = sin θ.

Euler's formula states that e^(iθ) = cos θ + i sin θ. As the angle θ increases, e^(iθ) traces the unit circle in the complex plane. Plotted against θ, that rotation becomes a helix. The real part of the helix projects as the cosine wave Re(z) = cos θ. The imaginary part projects as the sine wave Im(z) = sin θ.

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A radius turns at a steady rate inside a regular hexagon, a decagon and a circle, and the height y = r(θ) sin θ of its tip is traced against time. For an n-gon of circumradius R with a vertex at θ = 0, r(θ) = R cos(π/n) / cos((θ mod 2π/n) − π/n). Each edge is one piece of the wave; the flat top edge is a plateau at R cos(π/n), 0.866R for n = 6, 0.951R for n = 10. As n grows the plateaus narrow and rise to R, and the wave tends to R sin θ.

A radius turns at a steady rate inside a regular hexagon, a decagon and a circle, and the height y = r(θ) sin θ of its tip is traced against time. For an n-gon of circumradius R with a vertex at θ = 0, r(θ) = R cos(π/n) / cos((θ mod 2π/n) − π/n). Each edge is one piece of the wave; the flat top edge is a plateau at R cos(π/n), 0.866R for n = 6, 0.951R for n = 10. As n grows the plateaus narrow and rise to R, and the wave tends to R sin θ.

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A disk of radius a rolls on a table at tilt θ, its contact point circling at rate Ω. Gravity against the gyroscopic torque fixes Ω² = 4g/(a sin θ), with energy (3/2)Mga sin θ. Rolling resistance μ drains it as dθ/dt = −(2/3)μΩ, so near the end θ√θ falls linearly to zero: a finite-time stop. With a = 0.40 m and μ = 0.0086, θ drops from 70° to 1.4° in 15 s while Ω climbs from 1.63 to 10.08 Hz. The disk itself turns ever slower, at Ω sin θ.

A disk of radius a rolls on a table at tilt θ, its contact point circling at rate Ω. Gravity against the gyroscopic torque fixes Ω² = 4g/(a sin θ), with energy (3/2)Mga sin θ. Rolling resistance μ drains it as dθ/dt = −(2/3)μΩ, so near the end θ√θ falls linearly to zero: a finite-time stop. With a = 0.40 m and μ = 0.0086, θ drops from 70° to 1.4° in 15 s while Ω climbs from 1.63 to 10.08 Hz. The disk itself turns ever slower, at Ω sin θ.

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Graphical visualization of the unit circle, the functions sine and cosine and why cos²(x) + sin²(x) = 1

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A visual demonstration of the Pythagorean Theorem

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my parasite pen finally came in the mail ϵ( ♡Θ♡ )϶

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An electric dipole p(t) = p₀ sin(ωt) ẑ oscillating at a point radiates a field that, far away, falls off only as 1/r. Its radiation term is E = (μ₀/4π) p̈(t − r/c) sin θ / r along θ̂, with B = r̂ × E / c along φ̂, so E, B and the direction of travel r̂ are mutually perpendicular. The intensity goes as sin²θ / r²: zero along the axis, strongest at the equator. Every sphere around the source carries the same mean power, p₀²ω⁴ / (12πε₀c³).

An electric dipole p(t) = p₀ sin(ωt) ẑ oscillating at a point radiates a field that, far away, falls off only as 1/r. Its radiation term is E = (μ₀/4π) p̈(t − r/c) sin θ / r along θ̂, with B = r̂ × E / c along φ̂, so E, B and the direction of travel r̂ are mutually perpendicular. The intensity goes as sin²θ / r²: zero along the axis, strongest at the equator. Every sphere around the source carries the same mean power, p₀²ω⁴ / (12πε₀c³).

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$A mathematical construction in which one strip completes exactly ten turns. The polar angle of the generating curve is controlled by the exponential φ(θ) = (π/2) e^{-θ/8π}. Successive windings therefore nest inside one another, forming a compact layered rose with an exponentially tapering profile.$

A mathematical construction in which one strip completes exactly ten turns. The polar angle of the generating curve is controlled by the exponential φ(θ) = (π/2) e^{-θ/8π}. Successive windings therefore nest inside one another, forming a compact layered rose with an exponentially tapering profile.

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Ok, this is amazing. I asked Claude 3 to generate an animation of the Pythagorean Theorem and this is what it created:

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A rose curve is r = sin(kθ). Here k = b/a, with a = 1, 3, 7 down the rows and b = 2, 4, 8 across. With k = p/q in lowest terms and pq even, the rose has 2p petals and closes after θ = 2qπ. Each a is odd and each b a power of 2, so p = b and q = a: each column has 2b petals, 4, 8 and 16, and a only sets how far they wind: 2π, 6π, 14π. Swept over θ ∈ [0, 2aπ] in equal times, kθ = 2πb·t/T, so a column's three pens reach their petal tips together.

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The Hidden Geometry of Trigonometric Functions and the Unit Circle.

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You don't like math because you don't follow me. Check out the beautiful, intricate pattern generated by this equation: sin (sin (x) + cos (y)) = cos (sin (x*y) + cos (x))

You don't like math because you don't follow me. Check out the beautiful, intricate pattern generated by this equation: sin (sin (x) + cos (y)) = cos (sin (x*y) + cos (x))

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