For a fixed positive integer $m$, if $a_n\to 0$ and the series of the displayed block sums converges, then the original series converges to the same limit of a sequence. Its partial sum at a block endpoint is exactly a partial sum of $\sum b_n$; the difference at an intermediate endpoint has at most $m-1$ terms, all tending to zero. This justification is essential when block cancellation proves convergence without absolute convergence. It does not apply without further control to blocks whose lengths grow.

A matrix summability method takes the ordinary limit of a sequence of the row sums $A_i(x)$, whenever every row series converges and the resulting sequence has a finite limit. For rows in the absolutely summable sequence space with uniformly bounded norms, the matrix defines a bounded linear operator from the l-infinity sequence space to itself.

The sequence is bounded and $|a_{n+1}-a_n|\le \sqrt{n+1}-\sqrt{n}\to 0$ by the mean value theorem. Nevertheless, along the integer parts of $(\pi /2+2\pi j{)}^2$ its terms tend to one, and along the integer parts of $(3\pi /2+2\pi j{)}^2$ they tend to minus one. The square root rounding error is $O(1/j)$. Thus a bounded sequence with vanishing successive increments need not be a convergent sequence.

If all moments of real random variables converge to those of a moment-determinate limit law, then the variables converge to that law in the sense of convergence in distribution. Moment determinacy is essential; the standard normal distribution meets this condition.

Let $Z$ take values $1,-1$ with equal probabilities. Set $X_n=Z$ for every $n$, and $Y_n=Z$ for even $n$, $Y_n=-Z$ for odd $n$. Both marginal laws are constant, but $X_n+Y_n$ alternates between the law of $2Z$ and a point mass at zero. Thus separate weak convergence of random variables does not determine a joint limit or a sum limit. The obstruction disappears when one marginal limit is constant, as in the Slutsky theorem.

If $Z_n$ has probability at least p at $z_n$, with $p>0$ and $z_n\to z$, every weak limit has probability at least p at z. Apply the closed-set bound of the Portmanteau theorem to arbitrarily small closed intervals containing z and eventually $z_n$, then decrease their radii. Thus a persistent atom prevents convergence to a continuous law, including a chi-squared law. This provides a direct check when bounded-information estimation fails standard asymptotic-normal theory.

A Markov decision process specifies states, admissible actions, action-dependent transition probabilities and rewards. A policy chooses actions from observed information; a time-homogeneous problem uses transition and reward rules independent of the date. A Bellman equation optimizes current reward plus the conditional value of subsequent decisions. For nonnegative rewards, finite-horizon values increase as the horizon grows, while identification of their limit with the infinite-horizon value requires an approximation argument for admissible policies.

For $Y$ uniform on $(0,2)$, put $g(y)=1/(\pi (1+y^2))$ and $H(y)=y^2-4/3$. Then $\mathbb{E}H=0$, and $1/2-2\overline{g-\beta H}$ estimates the standard Cauchy distribution probability above two. Its variance is $4\mathrm{Var}(g-\beta H)/n$, minimized by the displayed coefficient. Here $\mathrm{Var}(H)=64/45$ and $\mathrm{Cov}(g,H)=1/\pi -7\arctan (2)/(6\pi )$, so the optimal coefficient is negative. A coefficient estimated using a separate pilot can be fixed for a fresh unbiased production sample.

If $Y$ is an unbiased estimator and a vector $C$ has known mean, then $Y-{\beta }^T(C-\mathbb{E}C)$ remains unbiased for a deterministic coefficient vector. Its variance is $\mathrm{Var}Y-2{\beta }^{T}c+{\beta }^T\Sigma \beta$, where $c=\mathrm{Cov}(C,Y)$ and $\Sigma =\mathrm{Cov}(C)$. For invertible $\Sigma$, completing the square gives the displayed optimum and minimum variance $\mathrm{Var}Y-c^T{\Sigma }^{-1}c$. With singular $\Sigma$, use a generalized inverse on its range. Estimating coefficients from the same simulation can introduce finite-sample bias; fixed pilot coefficients or independent training preserve the elementary unbiasedness assertion.

A control polygon joins successive control points by line segments. It gives an editable geometric skeleton, not usually the exact curve. An open control polygon has two ends; a closed control polygon also joins the last control point to the first. Store the first control point only once when using cyclic indices, so the closing edge is counted exactly once. Positive partition of unity basis functions place the curve inside the convex hull of its active control points; a convergent subdivision curve rule refines the control polygon toward its limit, with a certified error when an appropriate bound is available.

A mesh connecting a surface's control points according to the representation's indexing topology. A tensor-product surface basis gives a rectangular index grid and quadrilateral cells, even when its points in space are neither planar nor rectangular. The drawn mesh is a geometric skeleton, not generally the exact limit surface or an enclosing solid. Positive partition of unity coefficients give a convex hull bound on the relevant controls.

On its maximal domain in ${\ell }^2$, put $g_n=2^n{f}_n$. Then $g_n-g_{n-1}=-2^{-n}(Qf{)}_n$, so $g_n$ has a finite limit ${\ell }_f$. Summation gives $\mathrm{Re}⟨Qf,f⟩=-\frac{1}{2}(|g_0{|}^2+|{\ell }_f{|}^2+{\sum }_{n\ge 1}|g_n-g_{n-1}{|}^2)$. For $\alpha >0$, the resolvent equation is the recursion $g_n=(g_{n-1}+2^{-n}{h}_n)/(1+\alpha 4^{-n})$, which produces a bounded $g_n$ and hence $f\in {\ell }^2$. The Lumer-Phillips theorem proves contraction-semigroup generation.

// Target: analysis.bigb

A semigroup is hypercontractive when evolution improves integrability through norm bounds $\Vert P_{t}f{\Vert }_{q(t)}\le \Vert f{\Vert }_p$ with $q(t)>p$ for positive times. For a symmetric Markov semigroup satisfying a logarithmic Sobolev inequality with the joint-energy normalization, one may take $q(t)=1+(p-1)e^{4t/c_{\mathrm{LS}}}$.

// Target: analysis.bigb

The same self-map can be a contraction mapping for one metric but not for a bilipschitz equivalent metric. On three points with distances $d^{\prime }(x,y)=d^{\prime }(x,z)=2$, $d^{\prime }(y,z)=1$, the map $f(x)=y$, $f(y)=f(z)=z$ has Lipschitz constant $1/2$. For the discrete metric, its Lipschitz constant is one. The metrics satisfy $d\le d^{\prime }\le 2d$, so they preserve topology and completeness but not the strict numerical contraction condition. A contraction mapping theorem fixed-point proof therefore needs the specified metric and its contraction constant.

For the Levi-Civita connection, the second Bianchi identity gives ${\nabla }_i{R}^{ik}{}_{mn}={\nabla }_m{R}^k{}_n-{\nabla }_n{R}^k{}_m$. Applying another covariant derivative and the contracted Bianchi identity reduces its divergence to a commuting scalar Hessian plus curvature commutators on the Ricci tensor. The latter are the difference of symmetric tensors: a square of the Ricci tensor and $R^{ab}{R}_{anbm}$, which is symmetric in $m,n$ by curvature pair exchange and Ricci symmetry. Both differences vanish. This identity does not require vacuum field equations and is independent of the overall curvature sign convention.

The coordinate function has spectrum $[0,1]$: outside the interval its difference from a scalar has a continuously differentiable reciprocal, and on the interval it has a zero. Every character of an algebra therefore takes the coordinate to some $t\in [0,1]$. Polynomial density in the norm containing both function and derivative, together with automatic continuity of characters, forces evaluation at $t$ on every function. The maximal ideals are exactly $\{f:f(t)=0\}$.

For finitely many independent uniform arrival times on $[0,1]$, and fixed positive weights, normalize the unarrived mass by the remaining time. Conditional on not having arrived by $s$, each arrival remains uniform on $(s,1)$, giving the martingale identity. The process is càdlàg, nonnegative, and has constant mean equal to the total weight. It vanishes eventually before time one on each typical path, so it is not uniformly integrable when the total weight is positive.

A continuous uniform distribution on a finite interval with endpoints $a<b$ has probability density functionChoosing open or closed endpoints does not change this probability distribution, because single points have zero probability. Its cumulative distribution function is zero below $a$, $(x-a)/(b-a)$ between the endpoints, and one above $b$. Consequently a subinterval $[c,d]\subseteq [a,b]$ has probability $(d-c)/(b-a)$.

Integration gives expected value $(a+b)/2$ and variance $(b-a{)}^2/12$. If $U$ is uniform on $(0,1)$, then $a+(b-a)U$ is uniform on $(a,b)$, a basic input transformation for Monte Carlo method and inverse transform sampling. For example, on $(2,5)$ the expected value is $7/2$, the variance is $3/4$, and $\mathbb{P}(3<X<4)=1/3$. A uniform distribution is relative to its coordinate or reference measure: taking its logarithm generally does not preserve uniformity, as the log-uniform distribution illustrates.

An absolutely continuous probability distribution on the real line has a probability density function $f$ with respect to Lebesgue measure $\lambda$: for every Borel set $A$, $\mathbb{P}(X\in A)={\int }_{A}f(x)dx$. Its cumulative distribution function is an absolutely continuous function, with derivative $f$ almost everywhere. The probability density function and cumulative distribution function example under probability distribution illustrates this class.

The standard arcsine distribution is a continuous probability distribution on $[0,1]$ with probability density functionIt is the beta distribution with both parameters equal to $1/2$. The density grows without bound toward both endpoints, yet neither endpoint has any probability mass. Its name comes from the inverse sine in its cumulative distribution function.^[ref]

A useful way to see the shape is to choose a point uniformly around a circle and project it onto a diameter. Equal angular intervals do not give equal horizontal distances: the projection moves slowly near either end. An ideal harmonic oscillator sampled at a uniformly chosen time in one complete period has the same endpoint concentration after rescaling its position. The distribution translates a simple geometric slowdown into a precise probability calculation.

For an original comparison, take equal-length regions: the middle half $[1/4,3/4]$, and the two outside quarters together. ThenA uniform distribution on $[0,1]$ instead assigns each region probability $1/2$. Both distributions have mean $1/2$; the mean alone hides their different concentration.