For $r=-\lambda >0$ and vanishing solid-to-liquid diffusivity ratio, $H(r)=\sqrt{\pi }r{e}^{r^2}\mathrm{erfc}(-r)$ gives the interfacial liquid concentration. It increases from zero to $C_s$ as $r$ increases. Its derivative is $C_s{H}^{\prime }/(1+H{)}^2$, which decreases from $\sqrt{\pi }{C}_s$ to zero. For a direct proof, write $A=H/r=\sqrt{\pi }{e}^{r^2}\mathrm{erfc}(-r)$. Then $A^{\prime }=2rA+2$, so $A\ge \sqrt{\pi }+2r$. Differentiation gives $2H^{\prime 2}-(1+H)H^{\prime \prime }=2[(1+r^2+2r^4)A^2+r(4r^2-1)A+2r^2-2]>0$: the bracket increases with $A$, and substitution of its lower bound gives only positive polynomial coefficients, including constant $\pi -2$. Thus the concentration is strictly concave as a function of melting speed parameter $r$. Combining this with a linearized thermal balance produces one fold on the melting branch and, with the freezing branch, an interval of three possible interface speeds.

For incident and reflected harmonic arterial pressure waves with real $0<r<1$, the local pressure and flow phasors are proportional to $1+r{e}^{-i\vartheta }$ and $1-r{e}^{-i\vartheta }$. Their maximum-time difference is the displayed phase lag, on the branch $0<\vartheta <\pi$. Here $\vartheta =2\omega d/c$ is the round-trip phase to the reflector. Its derivative has the sign of $\cos \vartheta$. Raising the wave speed reduces the lag when $0<\vartheta <\pi /2$, but increases it when $\pi /2<\vartheta <\pi$. This conditional harmonic model does not establish an age trend for arbitrary vascular geometry or measurement stations beyond the sole reflector.

The graph kernel of a bicyclic graph satisfies ${\sum }_v(\mathrm{deg}v-2)=2$, with all degrees at least three. It therefore has either one degree-four vertex carrying two loops, a bouquet graph, or two degree-three vertices. In the latter case connectivity and degree counting give either three parallel edges, the three-edge Theta graph, or one edge between the vertices and one loop at each, a dumbbell graph. All bicyclic cores are subdivisions of these three kernels, subject to simplicity restrictions on loop lengths and parallel paths.

If $b_k$ counts labelled bicyclic cores on $k$ vertices, the bicyclic kernel classification gives $b_k\le Ak!k^2$. Three internally nonempty paths between two branching vertices give $b_k\ge k!\left(\genfrac{}{}{0ex}{}{k-3}{2}\right)/12$ for $k\ge 5$. Attach a forest with those core vertices as roots to obtain $C(n,n+1)={\sum }_k\left(\genfrac{}{}{0ex}{}{n}{k}\right)b_{k}k{n}^{n-k-1}$. After division by $n^{n+1}$ the relevant sum is $n^{-2}{\sum }_k{k}^3(n{)}_k/n^k$. Its upper bound follows from $(n{)}_k/n^k\le e^{-k(k-1)/(2n)}$; its lower bound uses $\sqrt{n}\le k\le 2\sqrt{n}$, where this product stays bounded below. The sum is bounded above and below by positive constants. The restriction $n\ge 4$ is necessary: $C(3,4)=0$.

Put $b_i={\lambda }_i+N-i$. In the bialternant formula, substitute $x_j=e^{t{y}_j}$ with distinct $y_j$ and compare the first nonzero determinant coefficients as $t\to 0$. They arise from the distinct degrees $0,\dots ,N-1$ in the exponential expansions. The common factor in the $y_j$ and the factorials cancel, leaving the product of differences of the $b_i$ divided by the differences of $N-i$. This proves the displayed integral specialization without dividing by a vanishing Vandermonde value.

A bialgebra equipped with a coquasitriangular structure, namely a convolution-invertible scalar form $r:H\otimes H\to k$ satisfying $r(ab,c)=\sum r(a,c_{(1)})r(b,c_{(2)})$, $r(a,bc)=\sum r(a_{(1)},c)r(a_{(2)},b)$ and $\sum r(a_{(1)},b_{(1)})a_{(2)}{b}_{(2)}=\sum b_{(1)}{a}_{(1)}r(a_{(2)},b_{(2)})$, with unit normalization. The induced right-comodule braiding is $v\otimes w\mapsto \sum w_{(0)}\otimes v_{(0)}r(v_{(1)},w_{(1)})$.

For an injective function that does not increase distances, one stretch-factor convention measures the largest compression that must be undone:If the sharp Lipschitz constant of $f$ is one, this equals bi-Lipschitz distortion. Fixing the scale matters: uniform compression $f(x)=x/2$ has stretch factor two in this convention, but bi-Lipschitz distortion one. In a geometric spanner, the corresponding stretch factor is the largest graph-route distance divided by direct Euclidean distance; the graph-to-Euclidean identification does not increase distances. ^[ref].

De Casteljau's algorithm evaluates a Bézier curve by repeated linear interpolation between neighboring control points. Fix $t\in [0,1]$, start with $P_i^{(0)}(t)=P_i$, and computeEach new row has one fewer point. The final point is $P_0^{(n)}(t)=C(t)$. Use the same $t$ at every stage: it describes the fraction of each line segment traversed from its first endpoint toward its second. A cubic Bézier curve needs three levels containing $3$, $2$, and $1$ linear interpolations, respectively.^[ref]

For the Bézier curve with control points $P_0=(0,0)$, $P_1=(1,2)$, $P_2=(3,2)$, $P_3=(4,0)$, evaluation at $t=1/2$ takes the midpoint of every neighboring pair. The first level givesThe second level givesand the third level givesThe intermediate control polygons shrink toward the evaluated point. Each step is a convex combination, giving a geometric explanation of the convex hull property; every step also commutes with an affine map.

The same construction splits the Bézier curve exactly, with no approximation. The outer edges of the triangular array give the new control points: for a cubic, $(P_0,Q_0,R_0,S)$ specify the left piece and $(S,R_1,Q_2,P_3)$ specify the right piece.^[ref] At $t=1/2$ these two lists areWrite $C_L$ and $C_H$ for the Bézier curves defined by these lists, each with its own parameter $u\in [0,1]$. ThenTheir images reproduce the original curve, while their derivatives account for the shorter parameter intervals. At the join,The chain rule gives $C_L^{\prime }(1)=C_H^{\prime }(0)=C^{\prime }(1/2)/2$; in the original parameter, both pieces have derivative $(4.5,0)$. This is the interval scaling in the parametric first-derivative join of Bézier curves.

Repeated midpoint splitting is useful for drawing vector graphics paths: each piece has a smaller control polygon and its own convex hull enclosure. A renderer can refine pieces until they are sufficiently close to line segments at the chosen display scale. Evaluating $C(t)$ and splitting it are thus two results of the same short algorithm.

For a proof outline, expanding the recursion givesPascal's identity proves this by induction; at $r=n$, $i=0$ it is precisely the Bernstein basis formula for the Bézier curve. The geometric construction remains useful even when expanding the polynomial by hand would be cumbersome.

// Target: analysis.bigb

For a degree-$n$ Bézier curve, the $r$th derivative has degree at most $n-r$ and control points $n!{\Delta }^r{P}_j/(n-r)!$. The first derivative follows by differentiating the Bernstein basis and shifting one summation index; iteration proves the general formula. A parameter interval of length $h$ multiplies these derivative controls by $h^{-r}$.

For real $p,q>0$ and $0\le z\le 1$, the displayed integral is the incomplete beta function. The regularized version is $I_z(p,q)=B_z(p,q)/B_1(p,q)$, the cumulative distribution function of the beta distribution. It has endpoint values zero and one and derivative $z^{p-1}(1-z{)}^{q-1}/B_1(p,q)$. This is a useful explicit scale function of a one-dimensional diffusion when the scale density has two power-law endpoint singularities.

independent gamma distributions $S\sim \Gamma (\alpha ,r)$ and $T\sim \Gamma (\beta ,r)$ with a common rate yield a Beta distribution $S/(S+T)\sim \mathrm{Beta}(\alpha ,\beta )$. The change of variables $S=wu$, $T=w(1-u)$ has Jacobian determinant $w$, and factors the joint probability density function into a Beta distribution density in $u$ and a gamma distribution density in $w$. In particular, independent chi-squared distributions with $2\alpha$ and $2\beta$ degrees of freedom give the required ratio.

For positive integers $\alpha ,\beta$, the $\alpha$th order statistic from $\alpha +\beta -1$ independent uniform distributions on $(0,1)$ has Beta distribution $\mathrm{Beta}(\alpha ,\beta )$. Selecting the observations below and above a value $u$ gives its probability density function proportional to $u^{\alpha -1}(1-u{)}^{\beta -1}$, with normalizing factor $(\alpha +\beta -1)!/((\alpha -1)!(\beta -1)!)$.

Moment matching for a beta prior turns a desired mean and variance for an unknown probability into the parameters of a beta distribution prior. Choose $0<m<1$ andSolving the beta distribution mean and variance identities givesThe strict upper bound on the variance makes $\kappa >0$, so both parameters are positive. Substitution gives mean $\mathbb{E}[P]=m$ and variance $\mathrm{Var}(P)=v$ exactly. A smaller specified variance at fixed mean produces a more concentrated prior distribution.

For example, a desired prior mean $m=0.3$ and variance $v=0.01$ give $\kappa =0.21/0.01-1=20$, hence the beta distribution prior $\mathrm{Beta}(6,14)$. With the same mean and variance $v=0.035$, the calculation instead gives $\kappa =5$ and the broader beta distribution prior $\mathrm{Beta}(1.5,3.5)$. Both appear in the concentration diagram above.

The excluded boundary cases have clear meanings. Zero variance would require $P=m$ with certainty, not a beta distribution with finite positive parameters. At the other boundary, any random variable $P\in [0,1]$ satisfies $P^2\le P$, hence variance at most $m(1-m)$; equality requires $P$ to take only the endpoint values zero and one. This is a Bernoulli distribution, not a beta distribution. Thus $v\ge m(1-m)$ cannot be matched by positive finite parameters, and $m=0$ or $m=1$ is also excluded.

After observing $n=h+t>0$ Bernoulli trials, the Beta-binomial conjugacy makes the posterior mean a weighted average of the prior mean $m$ and observed binomial proportion $h/n$:This formula gives $\kappa$ a precise interpretation as the prior's weight relative to the sample size in this posterior mean. It does not require fictitious observed counts: the beta distribution parameters can be any positive real numbers. For the eight-head, two-tail example above, $\kappa =4$ and $m=1/2$, so the posterior mean is $(4\cdot 1/2+10\cdot 0.8)/14=5/7$.

// Target: statistical-inference.bigb

For real functions, let $r=f-p_{\ast }$ and put $\mathrm{sign}(0)=0$. If $\int v\mathrm{sign}(r)=0$ for every $v\in U$, then $p_{\ast }$ is a best approximation in a normed space using the $L^1$ norm. The proof is the pointwise inequality $|r-v|\ge |r|-v\mathrm{sign}(r)$, followed by integration. This is a sufficient criterion even if the residual vanishes on a positive-measure set; a general necessary criterion may choose any subgradient in $[-1,1]$ on that zero set.

For $1<p<\infty$ and $n\ge 1$, existence follows from finite-dimensional compactness and uniqueness from strictly convex normed space geometry. Translation by $2\pi /n$ and negative translation by $\pi /n$ preserve the target and the error norm. Uniqueness forces the minimizing trigonometric polynomial to be invariant under the first operation and the negative of itself under the second. Invariance removes every nonzero frequency below $n$; negative invariance removes the remaining constant.

Put $\delta =t^{2/3}h(t)$ in the Bessel density modes of a polytropic expanding universe equation. It becomes $h^{\prime \prime }+8h^{\prime }/(3t)+{\Lambda }^2{t}^{-2\nu -2}h=0$. Substituting $h=1+a{\Lambda }^2{t}^{-2\nu }+\cdots$ gives $a=[2\nu (5/3-2\nu ){]}^{-1}$. Since $ck/R=\Lambda t^{-\nu -1}$, the fractional pressure correction is small when $(ck/R{)}^2{t}^2\ll 2\nu (5/3-2\nu )$. For $0<\nu <1/3$, this is the order-of-magnitude condition $ck/R\ll \sqrt{6\pi \nu G\rho }$. It describes approach to the dust-like growing branch, not an exact universal time at which every initial perturbation starts increasing. The instantaneous Jeans instability balance instead compares $(ck/R{)}^2$ with $4\pi G\rho$.

A Perron tree is a finite triangular splitting-and-sliding arrangement retaining a prescribed angular range while overlapping its pieces strongly. The compression parameter and the number of generations can make union area an arbitrarily small fraction of total piece area. Suitable longitudinal extensions separate the pieces into disjoint rectangles. This stronger finite geometry is used by the ball-multiplier counterexample; a compact measure-zero Besicovitch set alone does not assert that arbitrary segment translates are disjoint.

Choose a compact $K\subset [0,1]\times \mathbb{R}$ with full horizontal projection and zero-length projections $y+tx$ for $t\in [0,1]$. Then the displayed set is compact, has zero area by the Fubini theorem, and contains a segment of every slope in $[0,1]$. Such parameter sets exist by the Baire category theorem in the Hausdorff metric: fine parallel segments approximate any parameter set while making a fixed band of projections uniformly small. Finite grids and a countable intersection make every required projection null. Four rotated copies supply every planar direction.

The typical sequence theorem turns a probability concentration into a coding construction. Assign one binary number index to each weakly typical sequence, plus a reserved index for every word outside the set. The decoder recovers every typical word exactly; the reserved index has one predetermined output. The block code usesbits and has reconstruction-error probability at most $1-P(T_{\epsilon }^{(n)})$.

For this example and $\epsilon =0.1$, $n=40$ needs 34 bits but permits error up to about $0.3611$. At $n=640$, 570 bits suffice and the error bound is about $0.0002215$. These are existence calculations for an explicitly enumerated collection, rather than claims about an efficient practical encoder. The reduction comes from protecting the whole high-probability collection, rather than just its single most likely member.

The count $K_n$ has binomial distribution with mean $n/4$ and variance $3n/16$. The information content per symbol is an affine function of $K_n/n$, so its variance isChebyshev's inequality therefore gives the finite-sample lower boundThe bound is often loose, but it proves that every fixed positive tolerance eventually captures nearly all the probability. Direct summation of the binomial distribution's probability masses gives $P(T_{0.1}^{(640)})\approx 0.999779$. A finite sample does not need to have a nearly full typical set: the 40-symbol example above captures only about $0.639$.