Removing factorial growth from the coefficients of a formal inverse-power series produces its Borel transform. When continuation and exponential growth bounds allow it, ${\int }_{\mathcal{C}}{e}^{-zt}\mathcal{B}(t)dt$ reconstructs the series by Laplace transform. A singularity on the integration direction requires a lateral contour prescription. A simple pole contributes a residue when that prescription is changed, explaining how Stokes phenomenon arises from otherwise identical algebraic expansions.

If the events $A_n$ are mutually independent events and ${\sum }_{n=1}^{\infty }\mathbb{P}(A_n)=\infty$, then $\mathbb{P}({\mathrm{limsup}}_n{A}_n)=1$. For an infinite family, mutual independence means every finite subfamily satisfies the factorization defining independent events.^[ref]

Proof outline: for $M\ge m$, independent events and $1-x\le e^{-x}$ giveContinuity of probability makes each infinite tail union have probability one; their countable intersection has probability one as well.

For independent standard normal random variables $X_n$, the displayed limit superior equals two almost surely. The Mills ratio asymptotic gives, for $a>0$,The series converges for $a>2$ and diverges for $0<a\le 2$. The two Borel-Cantelli lemmas, applied at countably many thresholds approaching two, give the result. In particular, the value two is also a threshold exceeded infinitely often, despite being the almost-sure limit superior.

Take independent random variables $U_n$, each with the continuous uniform distribution on $[0,1]$, and set $A_n=\{U_n\le p_n\}$. Then $\mathbb{P}(A_n)=p_n$, and the events are mutually independent events. The product probability space supplies a model for the entire infinite sequence; a finite list of simulated values is only an illustration.

For $p_n=1/(n+1{)}^2$, the p-series is summable:The Basel problem supplies the total sum, and the integral test for convergence gives the displayed tail bound. The First Borel-Cantelli lemma makes the limsup of events have probability zero. In fact, the tail bound implies that the probability of even one alarm at or after $m$ is at most $1/m$; for $m=100$, it is at most $0.01$. This bound remains valid for dependent events with the same individual probabilities.

For $p_n=1/(n+1)$, the harmonic series diverges, so the Second Borel-Cantelli lemma makes the limsup of events have probability one. There is also an exact finite-horizon calculation using independent events:For any fixed $m$, this finite product of numbers tends to zero as $M\to \infty$. Every infinite tail therefore contains an alarm almost surely, even though the individual probabilities tend to zero.

Circuit depth is the largest number of counted gates on any input-to-output path in the Boolean circuit. For an AND/OR circuit with negations on input literals, only the AND/OR gates are counted. This convention makes the restriction argument for the Linial-Mansour-Nisan theorem precise.

The size of a Boolean circuit counts its gates, or in another common convention its wires. The convention must be fixed when stating quantitative bounds. In switching arguments, new clauses created from a decision tree are not separate original gates to be charged in the failure union bound.

For the number $N$ of occurring events, write $S_j=\mathbb{E}\left(\genfrac{}{}{0ex}{}{N}{j}\right)$ and $T_m={\sum }_{r=0}^m(-1{)}^r(\genfrac{}{}{0ex}{}{k+r}{k})S_{k+r}$. Then odd truncations bound $\mathrm{Pr}(N=k)$ below and even truncations above. For each integer $N>k$, the truncated binomial sum equals $(-1{)}^m(\genfrac{}{}{0ex}{}{N}{k}\left)\right(\genfrac{}{}{0ex}{}{N-k-1}{m})$ when $m<N-k$, and is zero otherwise. Its sign proves the bounds on taking expectations. This is the exact-count version of the Bonferroni inequalities.

For the normalized isotropic Maxwell molecule collision operator, put $a=f^{\prime }{f}_{\ast }^{\prime }$ and $b=f{f}_{\ast }$. The integral is over position, both velocities and collision direction. The symmetrized weak identity yields the displayed derivative of the Boltzmann H functional. It is nonpositive because $(a-b)\log (a/b)\ge 0$ for positive $a,b$. Spatial transport contributes only a boundary flux, which must vanish for the total-space inequality.

Keep the gap $\Delta$ fixed and vary positive temperature. With $x=k_{B}T/\Delta$, the Boltzmann factor is $q=e^{-1/x}$. In the four-microstate example, as $T\to 0^{+}$, $q\to 0$: the unique ground state has probability tending to one, and the other microstates have probabilities tending to zero. If several microstates instead shared the lowest energy, their state degeneracy would split that limiting population equally.

As $T\to \infty$ in this fixed finite model, all four individual Boltzmann factors tend to one, so all four individual microstates have probability tending to $1/4$. The three aggregate probabilities instead approachThe equal-population limit applies to this finite set of microstates. An infinite collection of microstates cannot all have the same positive normalized probability, so the statement cannot be extended to an unbounded energy spectrum. Real systems may also acquire additional accessible microstates as temperature rises.

The plotted crossing occurs at $x=1/\ln 2$, where $P(0)=P(\Delta )=4/9$. For larger positive temperatures, the middle macrostate is more probable than the lowest-energy macrostate, although the lowest-energy individual microstate remains more probable than either individual middle microstate. The expected value of energy rises from zero toward $\Delta$ in this finite model.

Consider two distinguishable, noninteracting two-level systems, each with energies $0$ and $\Delta >0$. Their four microstates $(0,0),(0,1),(1,0),(1,1)$ have total energies $0,\Delta ,\Delta ,2\Delta$. Thus the three energy macrostates have state degeneracies $1,2,1$. This is a stipulated finite model, not a measured material.

Set $q=e^{-\beta \Delta }$. The four individual Boltzmann factors are $1,q,q,q^2$, so the partition function and aggregate probabilities areEach of the two middle microstates has probability $q/Z$, while their shared energy macrostate has probability $2q/Z$.

At the temperature satisfying $\Delta =k_{B}T\ln 2$, the Boltzmann factor is $q=1/2$ and the partition function is $Z=9/4$. ThereforeThe middle energy is as likely as the ground state energy because its state degeneracy is twice as large. Yet each individual middle microstate is only half as likely as the ground state microstate. The diagram places the four energies beside their individual probabilities, then shows the sums within each macrostate.

The expected value of energy follows directly by weighting the three possible energies:At $q=1/2$ this is $2\Delta /3$. An individual energy measurement still returns $0$, $\Delta$ or $2\Delta$; the expected value describes the ensemble average, rather than a fourth possible measured energy.

Group microstates by their energy $E$. The state degeneracy $g(E)$ counts how many microstates have that energy; their aggregate probability isHere $P(E)$ is the probability of the macrostate retaining only energy, while $p_i$ is the probability of one individual microstate. For two such macrostates,Counting more microstates can compensate for a smaller Boltzmann factor. Omitting the state degeneracy confuses a sum over microstates with a sum over distinct energies. ^[ref]

For two individual microstates, the partition function cancels:At positive temperature, a higher-energy individual microstate is less probable, and equal-energy individual microstates are equally probable. For example, a gap $E_i-E_j=k_{B}T\ln 2$, using the natural logarithm, gives $p_i/p_j=1/2$. This comparison concerns individual microstates; a whole macrostate can contain several of them.

Changing the reference for energy leaves every probability unchanged. If all energies become $E_i^{\prime }=E_i+C$ for the same constant $C$, thenThe partition function itself changes, but its common factor cancels in the Boltzmann distribution. Thus choosing a ground state energy of zero simplifies calculations without changing populations. The expected value of energy shifts by $C$, as it must when its reference changes.

For a cyclic group $G$, a rank-$d$ phase-radius-$\rho$ Bohr set contains a symmetric proper generalized arithmetic progression of rank at most $d+1$ and cardinality at least $c(d,\rho )|G|$. The constant can be taken to depend only on $d$ and $\rho$. This geometry-of-numbers consequence follows by applying successive minima to the lattice recording the simultaneous character congruences; any cyclic subgroup factor can be represented by one additional progression coordinate. It is stronger than finding just one long ordinary progression.

Choose a Hilbert-Schmidt inner product orthonormal Hermitian operator basis with final element $I/\sqrt{N}$. A density operator has real coefficients $r_k=\mathrm{Tr}({\sigma }_k\rho )$ and fixed final coefficient $1/\sqrt{N}$. The remaining coefficients form its generalized Bloch vector. Positivity restricts these real vectors to a compact convex body; for $N>2$ this body is not the entire Euclidean ball allowed by a purity bound.

The qubit density operators $\rho =(I+\mathbf{s}\cdot \mathbf{\sigma })/2$ correspond exactly to the closed unit ball: their eigenvalues are $(1\pm |\mathbf{s}|)/2$, so positivity is equivalent to $|\mathbf{s}|\le 1$. The boundary is the Bloch sphere of pure states, and the centre is $I/2$. A depolarizing channel scales this ball isotropically, possibly with a sign reversal in the larger completely positive parameter interval.

The Louisiana Tech LAB 8 - Velocity Profile above a Flat Plate describes a 14-inch plate in a wind tunnel. A Pitot-static tube of 0.10-inch outside diameter traverses normally about 12 inches downstream; its screw scale has 0.001-inch increments. A water-column manometer reads the difference between stagnation pressure and static pressure. The procedure establishes a wall position, zeros the tunnel indicator, sets fan speed, and records at least ten positions inside the boundary layer, together with room temperature and barometric pressure. The instrument list includes a flow-stream thermometer implemented with a temperature-sensitive resistor. The manual requests comparison of outer velocity with the tunnel indicator and of the recovered velocity profile with laminar or turbulent predictions; it provides no observed dataset. These are apparatus specifications and instructions, not a reported confirmation of laminar flow.

Under incompressible flow conditions, an ideal aligned Pitot tube gives $q(y)=p_0(y)-p(y)=\rho u(y{)}^2/2$. Convert the manometer head $h$ to $q=({\rho }_{\mathrm{water}}-\rho )gh$, then infer $u=\sqrt{2q/\rho }$. Here $g$ is gravitational acceleration; mass density $\rho$ can be estimated from temperature and absolute pressure using the ideal gas law. A normalized velocity profile rises from the unresolved near-wall region to a plateau, whose value defines $U$; downstream profiles should broaden and collapse when plotted against $\eta$ if the Blasius boundary layer assumptions hold. Integrating a recovered velocity profile gives displacement thickness and momentum thickness.

Additional controls for interpreting this experiment follow from the measurement model: check differential-pressure calibration with equal port pressures and known liquid heads; leak-check tubing; align the Pitot tube with the incoming velocity; convert wall contact to the inlet-centre coordinate rather than treating outside contact as $y=0$; repeat traverses and outer velocity readings to detect drift. Keep the plate smooth and aligned, and check outer static pressure at multiple downstream stations before assuming zero pressure gradient. Repeating at another station or outer velocity tests the predicted boundary-layer scaling more directly than fitting one curve. These controls are proposed checks, not actions documented as completed in the manual.

Finite probe size, wall interference, flow fluctuations, and weak near-wall dynamic pressure limit inference. Even the manual's probe-based Reynolds number criterion, ${\mathrm{Re}}_d=ud/\nu >1000$ for its ideal Bernoulli equation conversion, must be checked with local $u$, not just $U$. In the illustrative model at $x=0.3048\mathrm{m}$, ${\delta }_{99}\simeq 3.32\mathrm{mm}$; the quoted outside diameter is $2.54\mathrm{mm}$, a substantial fraction of that thickness. A real inlet's sensing footprint and correction cannot be deduced from outside diameter alone. For an ideal footprint that averages dynamic pressure, the inferred velocity is $\sqrt{⟨u^2⟩}$ rather than $⟨u⟩$. Differentiating such sparse readings to obtain wall shear stress magnifies measurement error; the momentum thickness route is less sensitive to this differentiation but still needs the unresolved near-wall contribution. The no-slip boundary condition is a model constraint here, not a direct reading from a probe centred on the wall.

The vega of an European option is the derivative of its price with respect to spot volatility, holding its other parameters fixed. A dividend-free European call option under the Black-Scholes formula has $\nu =S_0\sqrt{T}\varphi (d_1)>0$. Differentiation proves this because $S_0\varphi (d_1)=K{e}^{-rT}\varphi (d_2)$ and ${\partial }_{\sigma }(d_1-d_2)=\sqrt{T}$. Positivity proves uniqueness of Black-Scholes implied volatility for prices strictly between the limiting call-price bounds.

The claim $\max (K,S_T)$ for $K>0$ combines a guaranteed cash floor with a European call option, equivalently a stock with a put. In the Black-Scholes model, for $\tau =T-t>0$, its price is $V=S_t\Phi (d_1)+K{e}^{-r\tau }\Phi (-d_2)$, where $d_1=[\log (S_t/K)+(r+{\sigma }^2/2)\tau ]/(\sigma \sqrt{\tau })$ and $d_2=d_1-\sigma \sqrt{\tau }$. Under the risk-neutral lognormal law, the truncated stock moment is $S_t{e}^{r\tau }\Phi (d_1)$ and the probability of ending below the floor is $\Phi (-d_2)$, proving the price. Its option delta is $\Phi (d_1)$, since $S_t\varphi (d_1)=K{e}^{-r\tau }\varphi (d_2)$ cancels the derivative terms. The replicating strategy holds that many stock units and $K{e}^{-r\tau }\Phi (-d_2)/B_t$ bank-account units.

Under the Black-Scholes model with interest rate $\rho$, volatility $\sigma$ and remaining time $\Delta$, risk-neutral pricing gives $V(S)=S[\log S+(\rho +{\sigma }^2/2)\Delta ]$. This follows by differentiating the lognormal exponential moment. Its option delta is $\log S+1+(\rho +{\sigma }^2/2)\Delta$ and its second stock derivative is $1/S$.

Average constant-rate Black-Scholes model prices of a fixed terminal payoff over $\rho \sim N({\rho }_0,{\tau }^2)$. Tilting this Gaussian by the discount factor shifts its mean by $-{\tau }^{2}T$; combining the resulting log-price variance with stock variance gives the displayed effective parameters. Thus the average equals the Black-Scholes price with those parameters, whenever the expectation is well defined. This is an average of constant-parameter prices, rather than a specification of an adapted stochastic short-rate market.