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Sharp feather tips can be approximated by adding smooth sine and cosine waves. This hummingbird’s outline is described by two Fourier series, one for each coordinate. Their rotating chains position the pen where the horizontal and vertical guides meet. Low frequencies capture the broad shape; higher frequencies recover smaller... show more

Philosophy Of Physics

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37,965 views • 5 days ago •via X (Twitter)

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Peter MOUEZA🇲🇫5 days ago

By Fourier

Klaudivan5 days ago

@threadreaderapp unroll please

Giovanni5 days ago

love that. wow

Master Steward of New London5 days ago

This is how they program drawing automatons, by the way

Ariz Ashraf5 days ago

Basically the first few waves just sketch the rough outline of the bird, then the higher ones fill in the sharp feathers and finer bits. Gets pretty accurate once you add enough of them.

TheRoloway5 days ago

Using Fourier series to redraw a hummingbird outline is a beautiful way to show how low frequencies hold the shape and the high ones chase the feathers.

Nícolas Abreu5 days ago

@grok me explique o que tá acontecendo de forma leiga e resumida.

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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.

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.

matharium

51,299 views • 11 days ago

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.

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.

matharium

92,000 views • 6 days ago

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.

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.

matharium

66,592 views • 5 days ago

I believe this is a projectile launched from a cloaked Pleiadian craft which is used to eliminate undesired frequencies in a specific location. ------- #NEIOH - "This Is A Laser Focused Frequency Sent From A Pleiadian Craft. The Energy Is Powerful And What Is Seen Is The Trail From A Small Pod Like Container That Will Dissolve Quickly When The Target Is Hit. This Is A Secret Operation Of The Galactic Federation. The Mothership Is Miles Away And The Fleet Of Crafts Has Been Dispersed In Many Directions With Each Projecting Frequencies Of Light. This Speed Can Only Be Attained By The Direction From Our Crafts. The Purpose Is To Eliminate Areas Of Frequencies Where Dark Forces Have Their Own Agenda." Malvern Hills, England - August 5, 2025

I believe this is a projectile launched from a cloaked Pleiadian craft which is used to eliminate undesired frequencies in a specific location. ------- #NEIOH - "This Is A Laser Focused Frequency Sent From A Pleiadian Craft. The Energy Is Powerful And What Is Seen Is The Trail From A Small Pod Like Container That Will Dissolve Quickly When The Target Is Hit. This Is A Secret Operation Of The Galactic Federation. The Mothership Is Miles Away And The Fleet Of Crafts Has Been Dispersed In Many Directions With Each Projecting Frequencies Of Light. This Speed Can Only Be Attained By The Direction From Our Crafts. The Purpose Is To Eliminate Areas Of Frequencies Where Dark Forces Have Their Own Agenda." Malvern Hills, England - August 5, 2025

Kab

339,033 views • 1 year ago

Discrete Fourier Transform by hand ✍️ ~ 12 steps walkthrough below Here is a little-known secret about the DFT and the inverse DFT: it is just matrix multiplication in both directions, one the transpose of the other, exactly like the forward pass and backpropagation I drew in other examples. Goal: recover which cosine waves a signal is made of, using nothing but multiplication and addition. = 1. Given = Three signals written as sums of cosines, and a fourth, X, that we do not know yet. = 2. Frequency matrix F = Let us write the coefficients as a matrix. Each signal is a row, each frequency a column, so A = cos(w) + 2cos(2w) becomes [1, 2, 0, 0]. = 3. Sample the waves = We read the four cosine waves at ten discrete time points. That word "discrete" is the whole difference between this and the continuous transform. = 4. Cosine matrix W = Let us write those samples as a matrix: each frequency a row, each time point a column. = 5. Frequency to time = We multiply F by W. That combines the four cosine waves in the proportions F specifies, and the result T is the three signals as they would look in time. = 6. Transpose = Let us stand each signal up as a column. = 7. Time to frequency = We multiply W by that transpose. Every cell is the dot product of one signal with one cosine wave, which measures how much of that wave the signal contains. Zero means none of it. = 8. Scale = Let us multiply by 2/n, with n = 10. The projections come out five times too large, and this is the correction. = 9. Transpose back = We turn it back around, and it is F again, exactly. That is the check: the transform recovered the coefficients we started from. = 10. Now solve for X = Let us run the same multiplication on the one signal whose recipe we never knew. = 11. Scale = We divide by 5 again. = 12. Transpose back = And X reads [0, 0, 3, 2], which says X = 3cos(3w) + 2cos(4w). Note: I originally drew this to show that the DFT is a special case of a convolution layer, its filters fixed to sine and cosine waves rather than learned. No wonder, then, that a convolution layer free to learn its own filters can be trained to process signals. 💾 Save this post!

Discrete Fourier Transform by hand ✍️ ~ 12 steps walkthrough below Here is a little-known secret about the DFT and the inverse DFT: it is just matrix multiplication in both directions, one the transpose of the other, exactly like the forward pass and backpropagation I drew in other examples. Goal: recover which cosine waves a signal is made of, using nothing but multiplication and addition. = 1. Given = Three signals written as sums of cosines, and a fourth, X, that we do not know yet. = 2. Frequency matrix F = Let us write the coefficients as a matrix. Each signal is a row, each frequency a column, so A = cos(w) + 2cos(2w) becomes [1, 2, 0, 0]. = 3. Sample the waves = We read the four cosine waves at ten discrete time points. That word "discrete" is the whole difference between this and the continuous transform. = 4. Cosine matrix W = Let us write those samples as a matrix: each frequency a row, each time point a column. = 5. Frequency to time = We multiply F by W. That combines the four cosine waves in the proportions F specifies, and the result T is the three signals as they would look in time. = 6. Transpose = Let us stand each signal up as a column. = 7. Time to frequency = We multiply W by that transpose. Every cell is the dot product of one signal with one cosine wave, which measures how much of that wave the signal contains. Zero means none of it. = 8. Scale = Let us multiply by 2/n, with n = 10. The projections come out five times too large, and this is the correction. = 9. Transpose back = We turn it back around, and it is F again, exactly. That is the check: the transform recovered the coefficients we started from. = 10. Now solve for X = Let us run the same multiplication on the one signal whose recipe we never knew. = 11. Scale = We divide by 5 again. = 12. Transpose back = And X reads [0, 0, 3, 2], which says X = 3cos(3w) + 2cos(4w). Note: I originally drew this to show that the DFT is a special case of a convolution layer, its filters fixed to sine and cosine waves rather than learned. No wonder, then, that a convolution layer free to learn its own filters can be trained to process signals. 💾 Save this post!

Tom Yeh

13,602 views • 1 month ago

This shows ~1,200 years of seawater erosion of the Khafre Pyramid, compressed into one minute. For the first ~500 years a high ocean stand holds about 300 ft above the base; its waves lap the soft Tura limestone casing and dissolve it (carbonic acid) at roughly 0.08 ft/year — about one inch of perpendicular recession annually, and ~14× faster than the slow dissolution on submerged surfaces below. As the sea recedes, it then pauses at successively lower stands, each held ~75–100 years, and each carves a distinct horizontal band where the casing is stripped to the bare core. Loosened blocks slough off at their own pace and fall down each face, piling into talus that mounds toward the center of all four sides. The smooth white cap above the highest waterline is never reached and survives intact.

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Sterling Cooley

642,387 views • 4 months ago

Discrete Fourier Transform by hand ✍️ ~ 12 steps walkthrough below Here is a little-known secret about the DFT and the inverse DFT: it is just matrix multiplication in both directions, one the transpose of the other, exactly like the forward pass and backpropagation I drew in other examples. Goal: recover which cosine waves a signal is made of, using nothing but multiplication and addition. = 1. Given = Three signals written as sums of cosines, and a fourth, X, that we do not know yet. = 2. Frequency matrix F = Let us write the coefficients as a matrix. Each signal is a row, each frequency a column, so A = cos(w) + 2cos(2w) becomes [1, 2, 0, 0]. = 3. Sample the waves = We read the four cosine waves at ten discrete time points. That word "discrete" is the whole difference between this and the continuous transform. = 4. Cosine matrix W = Let us write those samples as a matrix: each frequency a row, each time point a column. = 5. Frequency to time = We multiply F by W. That combines the four cosine waves in the proportions F specifies, and the result T is the three signals as they would look in time. = 6. Transpose = Let us stand each signal up as a column. = 7. Time to frequency = We multiply W by that transpose. Every cell is the dot product of one signal with one cosine wave, which measures how much of that wave the signal contains. Zero means none of it. = 8. Scale = Let us multiply by 2/n, with n = 10. The projections come out five times too large, and this is the correction. = 9. Transpose back = We turn it back around, and it is F again, exactly. That is the check: the transform recovered the coefficients we started from. = 10. Now solve for X = Let us run the same multiplication on the one signal whose recipe we never knew. = 11. Scale = We divide by 5 again. = 12. Transpose back = And X reads [0, 0, 3, 2], which says X = 3cos(3w) + 2cos(4w). Note: I originally drew this to show that the DFT is a special case of a convolution layer, its filters fixed to sine and cosine waves rather than learned. No wonder, then, that a convolution layer free to learn its own filters can be trained to process signals. 💾 Save this post!

Discrete Fourier Transform by hand ✍️ ~ 12 steps walkthrough below Here is a little-known secret about the DFT and the inverse DFT: it is just matrix multiplication in both directions, one the transpose of the other, exactly like the forward pass and backpropagation I drew in other examples. Goal: recover which cosine waves a signal is made of, using nothing but multiplication and addition. = 1. Given = Three signals written as sums of cosines, and a fourth, X, that we do not know yet. = 2. Frequency matrix F = Let us write the coefficients as a matrix. Each signal is a row, each frequency a column, so A = cos(w) + 2cos(2w) becomes [1, 2, 0, 0]. = 3. Sample the waves = We read the four cosine waves at ten discrete time points. That word "discrete" is the whole difference between this and the continuous transform. = 4. Cosine matrix W = Let us write those samples as a matrix: each frequency a row, each time point a column. = 5. Frequency to time = We multiply F by W. That combines the four cosine waves in the proportions F specifies, and the result T is the three signals as they would look in time. = 6. Transpose = Let us stand each signal up as a column. = 7. Time to frequency = We multiply W by that transpose. Every cell is the dot product of one signal with one cosine wave, which measures how much of that wave the signal contains. Zero means none of it. = 8. Scale = Let us multiply by 2/n, with n = 10. The projections come out five times too large, and this is the correction. = 9. Transpose back = We turn it back around, and it is F again, exactly. That is the check: the transform recovered the coefficients we started from. = 10. Now solve for X = Let us run the same multiplication on the one signal whose recipe we never knew. = 11. Scale = We divide by 5 again. = 12. Transpose back = And X reads [0, 0, 3, 2], which says X = 3cos(3w) + 2cos(4w). Note: I originally drew this to show that the DFT is a special case of a convolution layer, its filters fixed to sine and cosine waves rather than learned. No wonder, then, that a convolution layer free to learn its own filters can be trained to process signals. 💾 Save this post!

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