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A point traces each closed mathematical curve at constant speed. Its vertical coordinate versus elapsed time produces a unique periodic waveform. 4-petal rose r = a cos(2θ). Astroid hypocycloid x = a cos³θ, y = a sin³θ. Squircle |x|⁴ + |y|⁴ = a⁴. 4-cusped epicycloid. Heart curve (x² +...

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Matheorems

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10,607 просмотров • 15 дней назад •via X (Twitter)

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What Curve Beats a Straight Line? Take two fixed points A and B. A bead slides without friction under gravity. Among all possible curves connecting A to B, which one gives the smallest travel time? This is the Brachistochrone problem. The time functional is T[y] = ∫ √((1 + y′²)/(2gy)) dx where y is the vertical drop from the starting point. The winning curve is not a straight line. It is a cycloid x(θ) = a(θ - sin θ) y(θ) = a(1 - cos θ) The bead first falls steeply to gain speed, then uses that speed across the remaining distance. The path is longer than the straight segment, yet the travel time is smaller. The shortest path solves geometry. The fastest path solves variation. #CalculusOfVariations #Brachistochrone #Cycloid #ClassicalMechanics #MathematicalPhysics #Optimization #DifferentialEquations #PhysicsAnimation #Mathematics
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What Curve Beats a Straight Line? Take two fixed points A and B. A bead slides without friction under gravity. Among all possible curves connecting A to B, which one gives the smallest travel time? This is the Brachistochrone problem. The time functional is T[y] = ∫ √((1 + y′²)/(2gy)) dx where y is the vertical drop from the starting point. The winning curve is not a straight line. It is a cycloid x(θ) = a(θ - sin θ) y(θ) = a(1 - cos θ) The bead first falls steeply to gain speed, then uses that speed across the remaining distance. The path is longer than the straight segment, yet the travel time is smaller. The shortest path solves geometry. The fastest path solves variation. #CalculusOfVariations #Brachistochrone #Cycloid #ClassicalMechanics #MathematicalPhysics #Optimization #DifferentialEquations #PhysicsAnimation #Mathematics

Mathelirium

17,600 просмотров • 3 месяцев назад

a=(x,y,d=5*cos(o=mag(k=x/8-25,e=y/8-25)/3))=>[(q=x/2+k/atan(9*cos(e))*sin(d*4-t))*sin(c=d/3-t/8)+200,(y/3+d+q)/2*cos(c)+200] t=0,draw=$=>{t||createCanvas(w=400,w);background(6,19);for(t+=PI/60,y=99;++y<300;)for(x=99;++x<300;stroke(x,y,y+x,36))point(...a(x,y))} #つぶやきProcessing
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a=(x,y,d=5*cos(o=mag(k=x/8-25,e=y/8-25)/3))=>[(q=x/2+k/atan(9*cos(e))*sin(d*4-t))*sin(c=d/3-t/8)+200,(y/3+d+q)/2*cos(c)+200] t=0,draw=$=>{t||createCanvas(w=400,w);background(6,19);for(t+=PI/60,y=99;++y<300;)for(x=99;++x<300;stroke(x,y,y+x,36))point(...a(x,y))} #つぶやきProcessing

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a=(x,y,d=5*cos(o=mag(k=x/8-25,e=y/8-25)/3))=>[(q=x/2+k/atan(9*cos(e))*sin(d*4-t))*sin(c=d/3-t/8)+200,(y/4+5*o*o+q)/2*cos(c)+200] t=0,draw=$=>{t||createCanvas(w=400,w);background(6).stroke(255,96);for(t+=PI/60,y=99;++y<300;)for(x=99;++x<300;)point(...a(x,y))} //#つぶやきProcessing
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a=(x,y,d=5*cos(o=mag(k=x/8-25,e=y/8-25)/3))=>[(q=x/2+k/atan(9*cos(e))*sin(d*4-t))*sin(c=d/3-t/8)+200,(y/4+5*o*o+q)/2*cos(c)+200] t=0,draw=$=>{t||createCanvas(w=400,w);background(6).stroke(255,96);for(t+=PI/60,y=99;++y<300;)for(x=99;++x<300;)point(...a(x,y))} //#つぶやきProcessing

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a=(x,y,d=abs(e=cos(k=x/8-12.5)+sin(y/24)+cos(k/2)))=>point((q=x/4+90+d*k*(1+cos(d*4-t*2+y/72)))*cos(c=y*e/594-t/8+d/6)+200,(q/2+99*cos(c/2))*sin(c)+e*6+200) t=0,draw=$=>{t||createCanvas(w=400,w);background(6).stroke(w,36);for(t+=PI/60,i=4e4;i--;)a(i%200,i/200)}#つぶやきProcessing
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a=(x,y,d=abs(e=cos(k=x/8-12.5)+sin(y/24)+cos(k/2)))=>point((q=x/4+90+d*k*(1+cos(d*4-t*2+y/72)))*cos(c=y*e/594-t/8+d/6)+200,(q/2+99*cos(c/2))*sin(c)+e*6+200) t=0,draw=$=>{t||createCanvas(w=400,w);background(6).stroke(w,36);for(t+=PI/60,i=4e4;i--;)a(i%200,i/200)}#つぶやきProcessing

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a=(x,y,o=mag(k=x/4-12.5,e=y/9+5)/9)=>point((q=x+99+tan(1/k)+o*k*(cos(e*9)/4+cos(y/2))*sin(o*4-t))*.7*sin(c=o*e/30-t/8)+9*cos(y/19+t)+200,200+q/2*cos(c)) t=0,draw=$=>{t||createCanvas(w=400,w);background(6).stroke(w,46);for(t+=PI/90,i=2e4;i--;)a(i%100,i/100)} //#つぶやきProcessing
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a=(x,y,o=mag(k=x/4-12.5,e=y/9+5)/9)=>point((q=x+99+tan(1/k)+o*k*(cos(e*9)/4+cos(y/2))*sin(o*4-t))*.7*sin(c=o*e/30-t/8)+9*cos(y/19+t)+200,200+q/2*cos(c)) t=0,draw=$=>{t||createCanvas(w=400,w);background(6).stroke(w,46);for(t+=PI/90,i=2e4;i--;)a(i%100,i/100)} //#つぶやきProcessing

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a=(x,y,o=mag(k=x/4-12.5,e=y/9)/9)=>point((q=x/3+99+3/k*sin(y)+k*(1+cos(y)/3+sin(e+o*4-t*2)))*cos(c=o/5+e/4-t/8)+200,(q+49)*sin(c)*cos(c)-q/3+30*o+220) t=0,draw=$=>{t||createCanvas(w=400,w);background(6).stroke(w,46);for(t+=PI/90,i=2e4;i--;)a(i%100,i/350)}// #つぶやきProcessing
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a=(x,y,o=mag(k=x/4-12.5,e=y/9)/9)=>point((q=x/3+99+3/k*sin(y)+k*(1+cos(y)/3+sin(e+o*4-t*2)))*cos(c=o/5+e/4-t/8)+200,(q+49)*sin(c)*cos(c)-q/3+30*o+220) t=0,draw=$=>{t||createCanvas(w=400,w);background(6).stroke(w,46);for(t+=PI/90,i=2e4;i--;)a(i%100,i/350)}// #つぶやきProcessing

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a=(x,y,o=mag(k=x/8-12.5,e=y/8-9)/4)=>stroke(255,36).point((q=x+y+e/k+o*k*cos(y/3)*sin(o*2-t)*o/2)/3*atan(3*sin(c=o*e/40-t/8))+200,o*sin(x/9)-y*cos(c)+q/4*cos(c)+200) t=0,draw=$=>{t||createCanvas(w=400,w);background(6);for(t+=PI/120,i=2e4;i--;)a(i%200,i/200<<1)}#つぶやきProcessing
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a=(x,y,o=mag(k=x/8-12.5,e=y/8-9)/4)=>stroke(255,36).point((q=x+y+e/k+o*k*cos(y/3)*sin(o*2-t)*o/2)/3*atan(3*sin(c=o*e/40-t/8))+200,o*sin(x/9)-y*cos(c)+q/4*cos(c)+200) t=0,draw=$=>{t||createCanvas(w=400,w);background(6);for(t+=PI/120,i=2e4;i--;)a(i%200,i/200<<1)}#つぶやきProcessing

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a=(x,y,k=x/8-12.5,d=abs(e=cos(k)+sin(y/3)+cos(k/2)))=>point((q=x/4+90+d*k*(.5+cos(d*4-t*2+y/12)))*.7*(cos(c=y*e/198-t/8+d/5))+200,(q+70)*.7*sin(c)+200) t=0,draw=$=>{t||createCanvas(w=400,w);background(0,96).stroke(w,26);for(t+=PI/90,i=4e4;i--;)a(i%200,i/500)}//#つぶやきProcessing

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$=[],t=0,draw=_=>t++?background(0,14)+($=[...$.slice(-5e3),...[...'#つぶやきProcessing'].map(p5.Vector.random3D)]).map(_=>stroke(255,96).point((_.x+=sin(r=((_.x*2*(k=_.x*4+2^_.y*4|2+_.z*4)^_.y*k+t/299)&1)*2-1)/99*_.z)*109+270,(_.y+=cos(r)/99*_.z)*109+270)):createCanvas(540,540)

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Quantum mechanics has a reputation for being mystical mainly because people skip the rules and jump to interpretations. In this lecture series, we’re doing the opposite. We start from the rules, follow the algebra, and let the picture be the calculation. Classical Probability Theory combines alternatives by adding their probabilities. Quantum Theory combines them one step earlier…add complex amplitudes first, then square at the end. That swap in order is everything. Expand |a₁ + a₂|² and you don’t just get |a₁|² + |a₂|²…you get a cross-term, 2 Re(a₁ a₂*). Its sign is set by phase, so the same two contributions can reinforce or cancel. Interference is just the algebra of squaring a sum. In the 3D render, the surface height is proportional to |a(x)| (so peaks become bright bands after squaring), while the surface skin is colored by the local phase arg(a(x)). As the phase knob φ(t) is swept on path 2, the cross-term oscillates, and you literally watch the interference ridges slide across the screen. We model a detector screen with coordinates x in R² (think x = (x,y)). A quantum state assigns a complex amplitude a(x). The rule for outcomes is p(x) = |a(x)|² Now the key situation: two coherent alternatives contribute to the same outcome x. Let their amplitudes be a₁(x) and a₂(x). Quantum says a(x) = a₁(x) + a₂(x) So the probability density becomes p(x) = |a₁(x) + a₂(x)|² Expand it (this is the whole episode): p(x) = (a₁ + a₂)(a₁* + a₂*) = |a₁|² + |a₂|² + a₁ a₂* + a₁* a₂ = |a₁|² + |a₂|² + 2 Re(a₁ a₂*) That last term is the interference term. It can be positive or negative. To see phase explicitly, write each contribution in polar form: a₁(x) = r₁(x) exp(i θ₁(x)) a₂(x) = r₂(x) exp(i θ₂(x)) Then a₁ a₂* = r₁ r₂ exp(i(θ₁ − θ₂)) So the cross-term is 2 Re(a₁ a₂*) = 2 r₁ r₂ cos(θ₁(x) − θ₂(x)) That’s the fringe engine: p(x) = r₁² + r₂² + 2 r₁ r₂ cos(Δθ(x)) Now the phase knob we animate: Add a controllable phase shift φ to path 2: a₂(x) → a₂(x) exp(i φ) Then Δθ(x) → Δθ(x) − φ, so p(x; φ) = r₁² + r₂² + 2 r₁ r₂ cos(Δθ(x) − φ) As φ changes smoothly, the bright/dark pattern slides continuously. Same setup, same geometry, same magnitudes r₁,r₂, only phase changed. #QuantumMechanics #WaveInterference #ComplexAmplitudes #DoubleSlit #Physics #Mathematics
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Quantum mechanics has a reputation for being mystical mainly because people skip the rules and jump to interpretations. In this lecture series, we’re doing the opposite. We start from the rules, follow the algebra, and let the picture be the calculation. Classical Probability Theory combines alternatives by adding their probabilities. Quantum Theory combines them one step earlier…add complex amplitudes first, then square at the end. That swap in order is everything. Expand |a₁ + a₂|² and you don’t just get |a₁|² + |a₂|²…you get a cross-term, 2 Re(a₁ a₂*). Its sign is set by phase, so the same two contributions can reinforce or cancel. Interference is just the algebra of squaring a sum. In the 3D render, the surface height is proportional to |a(x)| (so peaks become bright bands after squaring), while the surface skin is colored by the local phase arg(a(x)). As the phase knob φ(t) is swept on path 2, the cross-term oscillates, and you literally watch the interference ridges slide across the screen. We model a detector screen with coordinates x in R² (think x = (x,y)). A quantum state assigns a complex amplitude a(x). The rule for outcomes is p(x) = |a(x)|² Now the key situation: two coherent alternatives contribute to the same outcome x. Let their amplitudes be a₁(x) and a₂(x). Quantum says a(x) = a₁(x) + a₂(x) So the probability density becomes p(x) = |a₁(x) + a₂(x)|² Expand it (this is the whole episode): p(x) = (a₁ + a₂)(a₁* + a₂*) = |a₁|² + |a₂|² + a₁ a₂* + a₁* a₂ = |a₁|² + |a₂|² + 2 Re(a₁ a₂*) That last term is the interference term. It can be positive or negative. To see phase explicitly, write each contribution in polar form: a₁(x) = r₁(x) exp(i θ₁(x)) a₂(x) = r₂(x) exp(i θ₂(x)) Then a₁ a₂* = r₁ r₂ exp(i(θ₁ − θ₂)) So the cross-term is 2 Re(a₁ a₂*) = 2 r₁ r₂ cos(θ₁(x) − θ₂(x)) That’s the fringe engine: p(x) = r₁² + r₂² + 2 r₁ r₂ cos(Δθ(x)) Now the phase knob we animate: Add a controllable phase shift φ to path 2: a₂(x) → a₂(x) exp(i φ) Then Δθ(x) → Δθ(x) − φ, so p(x; φ) = r₁² + r₂² + 2 r₁ r₂ cos(Δθ(x) − φ) As φ changes smoothly, the bright/dark pattern slides continuously. Same setup, same geometry, same magnitudes r₁,r₂, only phase changed. #QuantumMechanics #WaveInterference #ComplexAmplitudes #DoubleSlit #Physics #Mathematics

Mathelirium

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Trifolium Rose Curve via Parametric Gear Coupling Gear ratio 3:1. A fixed point on the linkage traces a perfect periodic path in the rotating frame of the larger gear. Orthogonal projections (blue) physically realize: r = a cos(3θ)
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Trifolium Rose Curve via Parametric Gear Coupling Gear ratio 3:1. A fixed point on the linkage traces a perfect periodic path in the rotating frame of the larger gear. Orthogonal projections (blue) physically realize: r = a cos(3θ)

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a=(x,y,o=mag(k=x/4-12.5,e=y/9)/9)=>point((q=99+3*(tan(y/2)/2+cos(y))/k+k*(3+cos(y)/3+sin(e+o*4-t*2)))*cos(c=o/4+e/4-t/8)*cos(c/2-e/3+t/8)+200,q*sin(c)+200) t=0,draw=$=>{t||createCanvas(w=400,w);background(6).stroke(w,46);for(t+=PI/90,i=3e4;i--;)a(i%100,i/350)}#つぶやきProcessing
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a=(x,y,o=mag(k=x/4-12.5,e=y/9)/9)=>point((q=99+3*(tan(y/2)/2+cos(y))/k+k*(3+cos(y)/3+sin(e+o*4-t*2)))*cos(c=o/4+e/4-t/8)*cos(c/2-e/3+t/8)+200,q*sin(c)+200) t=0,draw=$=>{t||createCanvas(w=400,w);background(6).stroke(w,46);for(t+=PI/90,i=3e4;i--;)a(i%100,i/350)}#つぶやきProcessing

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a=(x,y,d=mag(k=x/8-25,e=y/8-25)**2/99)=>[(q=x/3+k*.5/cos(y*5)*sin(d*d-t))*sin(c=d/2-t/8)+e*sin(d+k-t)+200,(q+y/8+d*9)*cos(c)+200] t=0,draw=$=>{t||createCanvas(w=400,w);background(6).stroke(255,96);for(t+=PI/60,y=99;y<300;y+=5)for(x=99;++x<300;)point(...a(x,y))}#つぶやきProcessing
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a=(x,y,d=mag(k=x/8-25,e=y/8-25)**2/99)=>[(q=x/3+k*.5/cos(y*5)*sin(d*d-t))*sin(c=d/2-t/8)+e*sin(d+k-t)+200,(q+y/8+d*9)*cos(c)+200] t=0,draw=$=>{t||createCanvas(w=400,w);background(6).stroke(255,96);for(t+=PI/60,y=99;y<300;y+=5)for(x=99;++x<300;)point(...a(x,y))}#つぶやきProcessing

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t=0,d=5e-4,draw=_=>{t++||createCanvas(w=400,w);background(9).stroke(w,96);for(x=y=z=9,i=3e4;i--;point((q=x*(e=sin(t*PI/20-x*x/99+i%9)+1)+89)*cos(k=z/59-e/29+t*PI/480+i%9*8)+200,200-(q+60*cos(k/2))*sin(k)))[x,y,z]=[x+9*(y-x)*d,y+(x*(28-z)-y)*d,z+(x*y-z-z)*d]}//#つぶやきProcessing
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t=0,d=5e-4,draw=_=>{t++||createCanvas(w=400,w);background(9).stroke(w,96);for(x=y=z=9,i=3e4;i--;point((q=x*(e=sin(t*PI/20-x*x/99+i%9)+1)+89)*cos(k=z/59-e/29+t*PI/480+i%9*8)+200,200-(q+60*cos(k/2))*sin(k)))[x,y,z]=[x+9*(y-x)*d,y+(x*(28-z)-y)*d,z+(x*y-z-z)*d]}//#つぶやきProcessing

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59,919 просмотров • 5 месяцев назад

a=(x,y,d=.5+5*cos(o=mag(k=x/8-12.5,e=y/8-12.5)**2/169))=>stroke((d*sin(k)*sin(t*4+e))**2*w,36).point(x+d*k*sin(d*2+o+t)+e*cos(e+t)+100,o*135-y/4-d*6*cos(d*3+o*9+t)+125) t=0,draw=$=>{t||createCanvas(w=400,w);background(6);for(t+=PI/120,i=4e4;i--;)a(i%200,i/200)}#つぶやきProcessing
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a=(x,y,d=.5+5*cos(o=mag(k=x/8-12.5,e=y/8-12.5)**2/169))=>stroke((d*sin(k)*sin(t*4+e))**2*w,36).point(x+d*k*sin(d*2+o+t)+e*cos(e+t)+100,o*135-y/4-d*6*cos(d*3+o*9+t)+125) t=0,draw=$=>{t||createCanvas(w=400,w);background(6);for(t+=PI/120,i=4e4;i--;)a(i%200,i/200)}#つぶやきProcessing

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48,268 просмотров • 1 год назад

a=(x,y,d=5*cos(o=mag(k=x/8-12,e=y/13-14)/2))=> point((q=x/2+10+1/k+k*cos(e)*sin(d*8-t))*sin(c=d/3+t/8)+sin(d*2+t)*k+200,(y/4+5*o*o+q*cos(c*3))/2*cos(c)+200) t=0,draw=$=>{t||createCanvas(w=400,w);background(6).stroke(w,66);for(t+=PI/60,i=2e4;i--;)a(i%200,i/200)}#つぶやきProcessing
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a=(x,y,d=5*cos(o=mag(k=x/8-12,e=y/13-14)/2))=> point((q=x/2+10+1/k+k*cos(e)*sin(d*8-t))*sin(c=d/3+t/8)+sin(d*2+t)*k+200,(y/4+5*o*o+q*cos(c*3))/2*cos(c)+200) t=0,draw=$=>{t||createCanvas(w=400,w);background(6).stroke(w,66);for(t+=PI/60,i=2e4;i--;)a(i%200,i/200)}#つぶやきProcessing

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for(f=0,a=[],W=400,y=0;y<W;y+=10)for(x=0;x<W;x+=10)a.push([x,y]);draw=o=>{for(f||createCanvas(W,W),background(0,9),stroke(W),C=a=>constrain(a,100,300),i=W;i--;)d=PI/noise(i+sin(f)/13),point(a[i][0]=C(sin(d+f)+a[i][0]),a[i][1]=C(cos(d+f)+a[i][1]));f+=.01};//#つぶやきProcessing
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for(f=0,a=[],W=400,y=0;y<W;y+=10)for(x=0;x<W;x+=10)a.push([x,y]);draw=o=>{for(f||createCanvas(W,W),background(0,9),stroke(W),C=a=>constrain(a,100,300),i=W;i--;)d=PI/noise(i+sin(f)/13),point(a[i][0]=C(sin(d+f)+a[i][0]),a[i][1]=C(cos(d+f)+a[i][1]));f+=.01};//#つぶやきProcessing

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