Proofs

Give it a tolerance ε — any width of the amber band — and it hands you a threshold N past which the sequence never leaves it (green points are inside the band; grey ones, before N, are still escaping). Shrink ε and N marches right; that march is the entire meaning of aₙ → L. “Chasing epsilons” is just this game, played in symbols. Here aₙ = 1 + sin(2n)/(n+2).

The gold polynomial hugs the curve near a and peels away past it — add a degree and the good region widens. The leftover between the two curves is exactly the remainder R_n; since every derivative of sine is at most 1, it can never exceed |x−a|ⁿ⁺¹/(n+1)! — the Lagrange bound shown above, made visible. That bound going to zero is why the series above stays faithful.

The Intermediate Value Theorem, live. Slide the level line y = k: whenever it sits inside the bracket [f(a), f(b)], a continuous graph cannot get from one end to the other without crossing it — so a solution x★ with f(x★) = k must exist. Here f(x) = sin x + 0.35x on [a, b]; continuity is the whole hypothesis (a jump would let the graph skip k).

The Mean Value Theorem, live. Drag the endpoints a and b: the green tangent at c = x★ is always parallel to the amber secant — somewhere inside the interval your instantaneous slope equals your average slope. Here f(x) = sin x + 0.3x; the proof below is why it can never fail.

Calculus finds the best point; the calculus of variations finds the best function. Fix the endpoints, add a bump A·η that vanishes at both ends, and watch the trial path pull away from the straight line. The functional J(A) — arc length here — bottoms out at A=0: any perturbation only lengthens it. That the first variation vanishes for every admissible bump is exactly the Euler–Lagrange equation. An original demonstration — the proof is in the gallery below.

The slope points uphill, so you step the other way — that's the whole algorithm. The only knob is the learning rate: nudge it up and the crawl speeds into a clean glide, up more and it starts zig-zagging across the minimum, and past a threshold each step overshoots farther than the last and it diverges. Every model you'll train lives somewhere on that dial.

Here is the surprise: a Gaussian is densest at its center — the single likeliest point is the origin. So why does the cloud avoid it? Because the chance of landing at length r is the density times how much room sits at that length — and the room at radius r is the surface of a sphere, $r^{n-1}$. The bell pulls inward, the shell pushes outward, and they balance on a thin skin at √n. In 1D the bell wins and you sit near 0; by n = 120 the shell wins so completely that every draw lands within a few percent of the same radius.

That balancing factor $r^{n-1}$ is exactly the surface area of the n-sphere, $S_{n-1}(r)=\frac{2{\pi }^{n/2}}{\Gamma (n/2)}{r}^{n-1}$ — the same change of variables that turns $\int e^{-x^2}dx$ into $\sqrt{\pi }$ at n = 2 (Izenman, Modern Multivariate Statistical Techniques, Prob. 2.4). So the Gamma normaliser, the Gaussian integral, and concentration of measure are one fact: sample that high-dimensional Gaussian and its persistent homology reads a sphere — the modular-forms-and-TDA crossover. See Γ built three ways in the distribution foundry.

• Probability space — the shaded area over an interval is a probability; slide k and read the 68 / 95 / 99.7 rule off ±1, 2, 3 σ.
• Likelihood space — fix an observation and read the height as the mean moves instead: that's the likelihood, area vs height.
• √π space — turn on density and the height is no longer free: A = 1/(σ√2π), forced so the total area is exactly 1 — that √2π is the Gaussian integral (√π, rescaled) doing its one job.

It is the same surface $f(x\mid \mu ,\sigma )$, sliced two ways. Probability fixes the parameters and varies the data — a normalized distribution whose area over a region is the probability. Likelihood fixes the observed data and varies the parameter — the height at $x^{\ast }$ read across hypotheses, peaking at the MLE but never enclosing area 1. Bayes is the machine that turns that height into a probability over $\mu$: multiply by a prior and divide by the evidence — the area that renormalizes it. $p(\mu \mid x)\propto f(x\mid \mu )p(\mu )$.

Nearest-neighbor (faint) greedily hops to the closest unvisited city and leaves crossings; 2-opt (solid) uncrosses edge pairs until no swap helps — trimming the tour by ~6% here, with no optimality guarantee.

The same five are ahead, but there is no spur — only a large stranger standing beside you on a footbridge over the track. Shoving them off would stop the trolley and save the five, but now their death is the means, not a side effect.

Score each option under each theory, weight by your credence, compare expected values. The switch and the footbridge share the same arithmetic (1 vs 5) yet pull apart the instant a death becomes the means rather than a side effect — slide a single credence to 100% and watch the recommendation flip.

The relative frequency wanders at first, then is dragged toward the amber p as flips pile up. The shaded funnel is ±2·$\sqrt{p(1-p)/n}$ around the true p — the estimate's sampling spread. It closes like $1/\sqrt{n}$: quadruple the flips to halve the error, and the frequency stays inside it about 95% of the time.

At n = 1 you're just looking at the source — skewed, split, whatever it is. Push n up and the distribution of the average forgets that shape and becomes the same bell every time, narrowing like σ/√n. And that rate is not a guess: the measured spread of the 1500 sample means stays locked to the predicted σ/√n at every n — the match holds near 1.00, the standard error confirmed by counting. That inevitability is why the Gaussian is everywhere, and why an average is trustworthy long before a single draw is.

A monic polynomial is just its roots, $p(z)={\prod }_{k=1}^n(z-r_k)$. This is its phase portrait over a patch of the complex plane: each point z is coloured by the angle $\arg p(z)$ and shaded by its size $|p(z)|$, so every root becomes a dark pinwheel where all the hues meet. Drag a root and the whole field reorganises around it.

The winding number of a closed curve about a point is the net number of full counter-clockwise turns it makes around that point — a signed integer you read off by following the angle and counting revolutions. Here the curve is $p(\partial D)$, the image of the dashed contour $\partial D$ (a circle) under the polynomial, shown in the corner inset.

The argument principle says that winding number about the origin equals the number of roots the contour encloses:

We trace that angle the long way round and it lands on the honest count every time. Press enclose alland the winding equals the degree: a degree-$n$ polynomial has exactly $n$ roots — the Fundamental Theorem of Algebra, measured rather than asserted. The $z^n-1$ preset drops the roots onto the unit circle as the $n$th roots of unity, the cyclotomic gateway that, run forward, becomes the modular world next door.

Start with the base cases. $p(z)=z^2-1$ has roots $\pm 1$: Newton cleaves the plane into two half-planes, each point sliding to its nearer root, and the border between them is a straight line. Add one more root — $z^3-1$ — and that border can no longer stay straight: three basins cannot meet along a clean curve without every meeting point touching all three at once, so it shreds into a fractal. Watch it happen with the two presets below.

Same roots, a different question: not where is p but where does Newton go. From each starting point we run $z\leftarrow z-ap/p^{\prime }$ until it lands on a root, then colour that pixel by which root caught it and brighten it by how fast. Near a root the basin is calm; on the shared border the colours shred into a Julia set, where an arbitrarily small move changes the destination, and (famously) every boundary point touches all the basins at once. Because $p^{\prime }/p={\sum }_{k}1/(z-r_k)$, the step needs no coefficients — the roots are the whole story. Drag a root and the basins heave; turn the relaxation $a$ and the fractal re-grows. The $z^n-1$ preset is Newton on the roots of unity from next door — the picture that started the whole subject of complex dynamics.

The Kauffman bracket splits every crossing two ways, sums the 2ⁿ resulting pictures in powers of A, corrects for writhe, and substitutes A = t^−1/4 — the Jones polynomial V(t), computed live from the diagram alone. The three unknot drawings all get V = 1 (try one), while the left and right trefoils get mirror polynomials — the Jones polynomial proves the trefoil is chiral.

This is a phase portrait of ζ(s) over the critical strip: hue is arg ζ, brightness is |ζ|. The nontrivial zeros are the dark points where all the hues spiral together — and they sit, one above the next, on the single dashed line Re(s) = ½. Drag the coloured slice: at σ = ½ its |ζ| profile (right) pinches to zero exactly level with each pinwheel; leave the line and every dip lifts off zero. The Riemann Hypothesis is the claim that this is no accident — that every nontrivial zero lies on this line. Verified for the first ten trillion, unproven for all.

What you’re looking at. Ordinary numbers sit on a line; p-adic numbers sit on a tree. Every leaf below is a residue $\mathrm{mod}{p}^n$, and two leaves are close when they share many low digits — so “distance” is how far up you climb to their common branch. Drag hops to send a random walker across the tree: each jump climbs to the common ancestor of where it was and where it lands (that height is the distance), then drops to the new leaf.

The middle inequality is the whole story: distance on a tree is ultrametric — every triangle is isosceles, so “how far apart” only ever means “which branch do we share.” Jumps follow the Vladimirov kernel (right); the slider $\alpha$ tunes how often the walker takes a long, root-reaching hop.

Two facts the tree forces, checked live in the readout: (1) a jump escapes a level-$m$ ball only when its separation $s>m$, so escape times are geometric and scale by $p^{\alpha }$ — the discrete scale invariance that gives p-adic diffusion its log-periodic clock. (2) Sampling the tree uniformly (Haar measure) and averaging $|x{|}_p^s$ lands on a clean closed form,${\int }_{{\mathbb{Z}}_p}|x{|}_p^{s}dx=\frac{1-p^{-1}}{1-p^{-(s+1)}}$ — the local zeta integral.

The same math, verified in dependency-free C behind an adversarial-review + sanitizer gate, lives at github.com/tbrown25/p-adic-mc — ${\mathbb{Z}}_p$ arithmetic → the ultrametric walk → ensemble diffusion → Haar Monte-Carlo, and it connects to the number-theory road of the Eigenbook.

Ramanujan’s three differential equations, starting from $E_2=E_4=E_6=1$ at $q=0$, regenerate the Eisenstein series. The RK4 solution (lines) lands on their q-series (dots) — and from $E_4,E_6$ alone you get the discriminant $\Delta$ and the $j$-invariant.

With $(P,Q,R)=(E_2,E_4,E_6)$, these are a closed system — three equations that generate all of modular forms. The discriminant $\Delta =\frac{1}{1728}(E_4^3-E_6^2)$ is the first cusp form; $j=1728E_4^3/(E_4^3-E_6^2)$ is the modular invariant. It connects to the modular-forms road of the Eigenbook.

Kepler's equation M = E − e·sin E ties the clock (mean anomaly M) to position on the ellipse (eccentric anomaly E) — and there is no closed form for E. The natural fix is a power series in the eccentricity: E = M + c₁e + c₂e² + ⋯. Laplace found it has a hard wall. For e < λ ≈ 0.6627 the series locks onto the true orbit (the error plunges); for e > λ it diverges no matter how many terms you keep — the eccentric anomaly is still perfectly well-defined (Newton finds it instantly), the series just can't reach it. That wall is a radius of convergence — a Cauchy–Hadamard number, fixed by the nearest singularity of E(e) in the complex plane.

Seven lengths, four equations: a²+b²=d², a²+c²=e², b²+c²=f², a²+b²+c²=g². Over ℝ they are no constraint at all — every box solves them. The entire difficulty is the demand that all seven be integers (and since the equations are homogeneous, rational ⇒ integer by clearing denominators, so ℚ and ℤ ask the same thing). So the only meaningful gap is ℝ versus ℚ = ℤ — climb the ladder ℕ → ℤ → ℚ → ℝ and the very same Pythagoras flips from one of the hardest open questions to trivially solvable, without touching the equations. That is what makes “does an Euler brick exist?” a question about what a number is: a perfect cuboid would be a rational point on a K3-type surface, forcing a congruent-number elliptic curve of rank ≥ 2 — tying this brick to Birch–Swinnerton-Dyer. The Euler brick is built from Pythagorean triples; the perfect cuboid is the one box we have never been able to build, or to forbid.