The coefficient of $T^n$ is $\int \left(\genfrac{}{}{0ex}{}{x}{n}\right)d\mu (x)$. The Mahler expansion of continuous p-adic functions identifies $I$-valued p-adic measures on ${\mathbb{Z}}_p$ with $I[[T]]$. If $D=(1+T)d/dT$, then $D^k{\mathcal{A}}_{\mu }(0)=\int x^{k}d\mu (x)$. A measure supported on ${\mathbb{Z}}_p^{\times }$ has ${\sum }_{{\zeta }^p=1}{\mathcal{A}}_{\mu }(\zeta (1+T)-1)=0$, because the sum of ${\zeta }^x$ is zero for $p\nmid x$.

Let $I$ be a complete discrete valuation ring with a continuous ${\mathbb{Z}}_p$-algebra structure. Every continuous function on ${\mathbb{Z}}_p$ with values in $I$ has a unique uniformly convergent expansion in the binomial polynomials, with $a_n={\Delta }^{n}f(0)$. This is the coefficient-ring version of the Mahler theorem. To see that $a_n\to 0$, approximate the function modulo any power of the maximal ideal by a function of period $p^k$. In the expansion of $(E-1{)}^{p^k}$, all intermediate binomial coefficients are divisible by $p$, while $(E^{p^k}-1)$ kills the periodic function. Iterating makes its differences vanish modulo the prescribed ideal power. Finite differences do not increase the uniform norm, so the same conclusion holds for the original function. Binomial inversion gives agreement at the nonnegative integers, and their density in the p-adic integers gives the expansion everywhere. This gives the dual description of p-adic measures by arbitrary bounded sequences of binomial moments, and hence the Amice transform.

Write ${\mu }_r=\left(\genfrac{}{}{0ex}{}{n}{r}\right)p^{\left(\genfrac{}{}{0ex}{}{r}{2}\right)}$ and $T_s=\left(\genfrac{}{}{0ex}{}{r}{s}\right)\left(\genfrac{}{}{0ex}{}{n-r}{r-s}\right)p^{-\left(\genfrac{}{}{0ex}{}{s}{2}\right)}/\left(\genfrac{}{}{0ex}{}{n}{r}\right)$. The first mean condition implies $p^{-1}\le (en/r{)}^{2/(r-1)}$ eventually. For $2\le s\le r/2$, this yields $T_s\le (C/s{)}^s$, and each fixed-$s$ term tends to zero when $n\ge r^3$. The second mean condition implies $p\le ((r+1)/n{)}^{2/r}$ eventually. Writing $s=r-j$ gives $T_{r-j}={\mu }_r^{-1}\left(\genfrac{}{}{0ex}{}{r}{j}\right)\left(\genfrac{}{}{0ex}{}{n-r}{j}\right)p^{j(2r-j-1)/2}\le {\mu }_r^{-1}(C/j{)}^j$ for $1\le j\le r/2$, and $T_r={\mu }_r^{-1}$. Both bounding series are summable. The overlap formula for the variance of a clique count therefore gives relative variance tending to zero; the second moment method proves an $r$-clique exists with high probability.

Underfitting occurs when a prediction rule fails to capture useful structure in the target relation. A straight-line polynomial regression fitted to a genuinely quadratic noiseless target illustrates a representational limitation. Poor prediction alone does not identify the cause: compare candidate rules using cross-validation or a validation set, while keeping the final test set reserved for assessment.

The structure containing the angiosperm female gametophyte; a fertilized ovule can develop into a seed.

For a nontrivial Non-Archimedean absolute value on $\mathbb{Q}$, every integer has absolute value at most one. Some prime $p$ must have value less than one, or the value would be trivial on every rational number. Two distinct such primes are impossible by a Bezout identity. Prime factorization then gives the displayed formula. Distinct primes give inequivalent topologies, since powers of $p$ tend to zero only for the $p$-adic absolute value among these values.

Take two reservoirs separated by a semipermeable membrane that passes water and retains one neutral solute. Stipulate $T=300\mathrm{K}$, $c_A=10\mathrm{mol}{\mathrm{m}}^{-3}$ and $c_B=30\mathrm{mol}{\mathrm{m}}^{-3}$, equivalent to $10$ and $30\mathrm{mmol}{\mathrm{L}}^{-1}$. Treat both chemical solutions as dilute and ideal. These are invented model conditions, not measurements or recommended biological conditions.

For pressures measured at the same height beside the membrane, define $\Delta P=P_B-P_A$. The model predictsAt $\Delta P=0$, net water transfer favors A to B. At $\Delta P\simeq 49.89\mathrm{kPa}$, the opposing pressure balances the osmotic pressure difference, giving zero net transfer with unequal concentrations. At $\Delta P=75\mathrm{kPa}$, the sign reverses and net water transfer favors B to A, toward the more dilute reservoir. Raising pressure changes the total water potential, rather than changing the membrane into a one-way gate.

Fixed concentrations require large maintained reservoirs or an instantaneous comparison. In a closed membrane osmometer, entering water causes dilution of the receiving chemical solution and changes its volume; the final balance must use final concentrations. A supported cell wall provides another way to develop the opposing pressure, as explained under turgor pressure.

A circular distant barycentric Kepler orbit around a circular star–planet binary need not have constant star-centred osculating orbital elements. Let $\mu$ be the planet's fraction of the binary mass, $\alpha =a_{\mathrm{pl}}/a\ll 1$, and $\epsilon =\mu /\sqrt{\alpha }\ll 1$. Subtracting the stellar reflex velocity produces radial and tangential perturbations $(-v_0\epsilon \sin \Delta ,v_0\epsilon \cos \Delta )$, where $\Delta$ is the planet's longitude minus the particle's longitude. The vis-viva equation gives the displayed semi-major axis variation. The eccentricity vector components follow from $e_r=r{v}_{\theta }^2/(G{M}_{\star })-1$ and $e_{\theta }=-r{v}_r{v}_{\theta }/(G{M}_{\star })$. The errors are $O(\mu +{\epsilon }^2)$; the spatial translation itself is only $O(\mu \alpha )$. These are changes of instantaneous orbital description, even when the assumed barycentric trajectory is fixed.

Tensor products of the assigned vectors represent strong graph powers, and tensor products of the handle have squared inner products equal to products of the original ones. The vectors assigned to an independent set are orthonormal. Bessel inequality then bounds its size by the largest inverse squared handle inner product. Taking roots and minimizing proves $c(G)\le \vartheta (G)$.

Represent unit-radius de Sitter spacetime by the spacelike unit hyperboloid in five-dimensional Minkowski spacetime. Appending its position vector as the last column of an oriented pseudo-orthonormal frame produces a matrix in $SO(4,1)$. Projection to that last column identifies $SO(4,1)$ with the oriented orthonormal frame bundle, a principal bundle with group $SO(3,1)$. The full bundle allowing both orientations is $O(4,1)$ with group $O(3,1)$. Requiring a time orientation as well selects the corresponding identity components.

An oriented pseudo-orthonormal frame in Minkowski spacetime consists of its base point $a$ and a Lorentz matrix $L$ acting on a fixed reference frame. The pairs $(a,L)$ form the Poincaré group, and right multiplication by $(0,h)$ sends the frame to $(a,Lh)$ while fixing its base point. Thus the displayed projection is a principal bundle with group $SO(3,1)$. The full frame bundle uses $O(3,1)$; a specified time orientation restricts to the corresponding connected component.

Represent the Anti-de Sitter spacetime quadric by $x\cdot x=-{\ell }^2$ in ${\mathbb{R}}^{3,2}$. Appending its timelike unit normal $x/\ell$ to an oriented tangent pseudo-orthonormal frame identifies the frame bundle with $SO(3,2)$. The stabilizer of that normal is $SO(3,1)$, so projection realizes $SO(3,2)/SO(3,1)$ as the base of a principal bundle. For the commonly used universal cover of the anti-de Sitter quadric, pull this bundle back along the covering map; the quadric itself has a periodic timelike coordinate.

For a thermal inertial Langevin equation conditioned on fixed initial position and velocity, the centered displacement is $R_i={\zeta }^{-1}{\int }_0^t[1-e^{-\zeta (t-s)}]A_i(s)ds$. With the fluctuation-dissipation relation for a Langevin particle, $D=k_{B}T/(m\zeta )$ and integrating the squared kernel gives the displayed variance. At short times it is $2D{\zeta }^2{t}^3/3$ per component; at long times it is $2Dt+O(1)$. Averaging initial velocity over thermal equilibrium adds a ballistic contribution and gives $2D[t-(1-e^{-\zeta t})/\zeta ]$ per component.

Let $d\sigma =-\sigma dt/\tau +\sqrt{\kappa }dW$, ${\sigma }_0=0$, and $\dot{B}=\sigma B$, $B_0>0$. The logarithm $\log (B_t/B_0)={\int }_0^t{\sigma }_{s}ds$ is a centered Gaussian random variable of variance $V(t)=\kappa {\tau }^2[t-2\tau (1-e^{-t/\tau })+\tau (1-e^{-2t/\tau })/2]$. Squaring the stochastic kernel $\sqrt{\kappa }\tau [1-e^{-(t-u)/\tau }]$ and integrating proves this formula. The exponential moment of a Gaussian linear functional gives $\mathbb{E}{B}_t^n=B_0^n{e}^{n^{2}V(t)/2}$ and the displayed long-time growth exponent. Positive moment growth with zero mean strain reflects the breadth of the log-normal distribution; it does not imply positive growth of the mean logarithm.

The linear stochastic differential equation has the explicit solution $Y_t=e^{-\lambda t}{Y}_0+\sigma {\int }_0^t{e}^{-\lambda (t-s)}d{W}_s$. Its integrating factor is $e^{\lambda t}$. For $\lambda >0$, an independent initial normal variable of variance ${\sigma }^2/(2\lambda )$ makes its zero-mean Gaussian process stationary.

For $\lambda >0$, set ${\mathcal{G}}_t={\mathcal{F}}_{e^{2\lambda t}}$ and $W_t=(2\lambda {)}^{-1/2}{\int }_1^{e^{2\lambda t}}{u}^{-1/2}d{B}_u$. This is a Brownian motion in the new filtration; its independent Gaussian increments have variance equal to elapsed time. The Itô product rule gives $dY=\sqrt{2\lambda }dW-\lambda Ydt$. Its initial value is $B_1$, independent of future W increments, and its covariance function is $e^{-\lambda |t-s|}$. Consequently it is a stationary Ornstein-Uhlenbeck process. The quadratic variation of the unscaled time change is $e^{2\lambda t}-1$, not $e^{2\lambda t}$.

For an adapted stationary process with the displayed bounds, the Itô formula and stationarity give ${\int }_0^t(\mathbb{E}[Z_s{g}_s]+\lambda )ds=0$. The Cauchy-Schwarz inequality makes each integrand nonnegative, so equality holds almost everywhere. Then $\mathbb{E}(g_s+\lambda Z_s{)}^2\le 0$, proving $g=-\lambda Z$ for almost every time and outcome. Thus the process solves the Ornstein-Uhlenbeck stochastic differential equation and its initial value, rather than stationarity alone, determines its coupling with W.

For any step $\Delta >0$, set $a=e^{-\theta \Delta }$. The Ornstein-Uhlenbeck transition distribution gives the exact updateusing a fresh independent standard normal random variable for each interval. The update reproduces the joint probability distribution on the selected time grid. It can start from a fixed value or use Ornstein-Uhlenbeck stationary initialization. On an equally spaced grid it is an autoregressive process of order one after subtracting $\mu$. Its stationary variance follows directly from $v=a^{2}v+v_{\infty }(1-a^2)$.

For the stipulated example with $\Delta =0.1$, $a=e^{-0.1}\approx 0.90484$ and the innovation variance is $1-e^{-0.2}\approx 0.18127$. Given $X_t=6$, the next conditional expected value is about $5.61935$ and the conditional standard deviation is about $0.42576$.

The Euler-Maruyama method instead approximates the same stochastic differential equation byFor $0<\theta \Delta <2$ its stationary variance solves $v=(1-\theta \Delta {)}^{2}v+{\sigma }^2\Delta$, givingAt $\Delta =0.1$ in the example, this is $1/0.95\approx 1.05263$, about $5.26\%$ above the continuous model's stationary variance. If $\theta \Delta \ge 2$, this Euler recurrence has no finite stationary variance despite the stable continuous model. Exact transitions avoid this discretization error at sampled times; neither update alone determines the unresolved motion between samples.

Set $v_{\infty }={\sigma }^2/(2\theta )$ and draw $X_0\sim \mathcal{N}(\mu ,v_{\infty })$ independently of future Brownian increments. The Ornstein-Uhlenbeck process is then a strictly stationary process, withEvery time has the same normal distribution, and the covariance function depends only on the time lag. For a Gaussian process these properties also make all its finite-dimensional distributions invariant under a common time shift. The stationary distribution balances the continual random forcing against the contracting drift.^[ref]

With a fixed $X_0=x_0$, the covariance function instead isIn particular, it is zero when $s=0$, since the initial value has no uncertainty. The second term records the fixed start; it is the general scaling of the zero-start Ornstein-Uhlenbeck covariance. The pure exponential covariance function displayed above assumes stationary initialization. For a fixed start, that function emerges as the late-time limit at fixed lag, while the one-time probability distributions approach $\mathcal{N}(\mu ,v_{\infty })$ by convergence in distribution.

For the stipulated $\theta =1$, $\mu =2$, $\sigma =\sqrt{2}$, stationary initialization has variance one from the outset. Its autocorrelation is $e^{-|h|}$, so the correlation time is one: a lag of one reduces the autocorrelation to $e^{-1}\approx 0.3679$. Its raw second moment $\mathbb{E}[X_s{X}_t]=4+e^{-|t-s|}$ includes the squared expected value; it must not be confused with the covariance.

The upper panel shows three simulated sequences with independent stationary initial values and exact transitions at spacing $0.05$. Joining their points only guides the eye; it does not simulate a continuous sample path between those points. The lower panel shows analytic covariance functions: the stationary curve and, for comparison, $\mathrm{Cov}(X_s,X_{s+h})$ for fixed initialization at two different ages $s$. These are simulated samples and model curves, not measured trajectories or empirical covariance estimates.

For a fixed initial value $X_0=x_0$, the Ornstein-Uhlenbeck process has the normal distributionThe two parameters of $\mathcal{N}(m,v)$ are the expected value and variance, not the standard deviation. At $t=0$ the variance is zero and the value is fixed. More generally, its transition kernel over any interval of length $h>0$ has conditional expected value $\mu +(x-\mu )e^{-\theta h}$ and conditional variance ${\sigma }^2(1-e^{-2\theta h})/(2\theta )$ when the current value is $x$. The coefficients depend on elapsed time; the initial probability distribution determines whether the resulting stochastic process is stationary.^[ref]

In the stipulated example $\theta =1$, $\mu =2$, $\sigma =\sqrt{2}$, $x_0=6$, the expected value at time two is $2+4e^{-2}\approx 2.5413$ and the variance is $1-e^{-4}\approx 0.9817$. The initial distance from two has fallen by a factor $e^{-2}$, but the standard deviation has grown to about $0.9908$. Approaching the reference level in expected value is compatible with growing uncertainty from a perfectly known initial value.

The solid probability density functions use times $0.25$, $1$ and $2$; the dashed curve is the limiting normal distribution with expected value two and variance one. The curves are exact model predictions. There is no ordinary density for the fixed value at time zero.

A short derivation uses the integrating factor $e^{\theta t}$. The Itô product rule gives $d[e^{\theta t}(X_t-\mu )]=\sigma e^{\theta t}d{W}_t$, henceThe stochastic integral has zero expected value. The Itô isometry gives its variance as ${\sigma }^2{\int }_0^t{e}^{-2\theta (t-u)}du$. Its deterministic integrand makes it a Gaussian random variable, yielding the displayed normal distribution. Shifting the integration interval gives the same transition kernel from any later time.