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By the chain rule, $y^{\prime \prime }=f_x+f_{y}f$ along the exact solution. Applying Taylor's theorem to the exact increment and the two arguments of $f$ givesThe assumed differentiability makes the remainders uniform on $0\le x\le 1-h$. Subtracting gives the consistency errorHere $O$ denotes Big O notation. The general order of a numerical method is therefore one for $\beta \ne 1/2$ and at least two for $\beta =1/2$. For example, on $y^{\prime }=y$ with unit initial value, $y_N=(1+h+\beta h^2{)}^{1/h}$. Expanding its logarithm gives $y_N=e[1+(\beta -1/2)h+O(h^2)]$, so the first-order term is actually nonzero when $\beta \ne 1/2$.

To control accumulation, the mean value theorem and the bound on $f_y$ givePut $K_h=L(1+\beta hL)$, $q_h=1+h{K}_h$ and $e_n=y(x_n)-y_n$. The exact and numerical updates yield $|e_{n+1}|\le q_h|e_n|+h|T_n|$, with $e_0=0$. Iterating this recurrence, as in the Lipschitz one-step error bound, givesBecause $K_h\le L(1+L)$ for $h\le 1$, the prefactor is bounded independently of the mesh. When $\beta =1/2$, the consistency error is $O(h^2)$, hence the maximum global error is $O(h^2)$. This proves at least second-order numerical convergence of the explicit midpoint method.

Choose each factor with its argument between zero and $\pi$ in the upper half-plane. On an interval between consecutive real prevertices, each factor has a constant boundary argument, so $f^{\prime }$ has constant argument and the image is a straight side. Crossing $X_j$ from left to right decreases $\arg (Z-X_j)$ by $\pi$, hence increases $\arg f^{\prime }$ by ${\beta }_j\pi$, giving the prescribed exterior turn. Locally,so an upper half-disc maps to angle $(1-{\beta }_j)\pi$, the required interior angle. The overall constant rotates and scales the polygon; integration adds translation. For a general prescribed polygon, the prevertex positions must also satisfy the side-length conditions: the derivative formula fixes angles but arbitrary choices do not fix arbitrary side lengths.

For the slit domain, the slit tip has interior angle $2\pi$, so its exterior parameter is $-1$. The two vertices at infinity give the factors at $\pm 1$, leading to $f^{\prime }(Z)$ proportional to $Z(1-Z^2{)}^{-3/2}$. Integrating and imposing the tip value givesThe square root is the principal one: for $\Im Z>0$, $1-Z^2$ never lies on the nonpositive real axis. Thus $\Re f>0$. Boundary points $-1<X<1$ map to the two banks of the slit, with $X=0$ at the tip; the outside intervals map to the imaginary axis. The derivative is nonzero in the open half-plane. Conversely each point of the slit domain determines exactly one root $Z$ of $Z^2=1-z^{-2}$ with positive imaginary part, proving bijectivity.

The Möbius map $(Z-1)/(Z+1)$ takes the upper half-plane to itself, so its principal argument lies in $(0,\pi )$. The imaginary part of its analytic logarithm is harmonic, equals $\pi$ on $(-1,1)$ and zero on the other real intervals. ThusIt lies between zero and one and has the required boundary limits away from the two discontinuities. Pulling this harmonic function back by the inverse conformal map yieldsThis specifies all branches and remains bounded even as $z$ tends to either end at infinity.

To exhibit nonuniqueness without the growth restriction, putThis is harmonic in the domain. On its boundary the corresponding $Z$ is real and the rational expression is real, except at $\pm 1$, which represent infinity. At $Z=\infty$, representing $z=0$, the expression tends to zero. Thus $H$ has zero finite boundary data. It is nonzero, since for $Z=iY$ its value is $Y/(1+Y^2)>0$, and it is unbounded as $Z$ approaches either pole $\pm 1$ non-tangentially. Therefore $u+cH$ gives distinct solutions with the same finite boundary data when boundedness at infinity is removed.

The function $e^{-x^2}/x$ is strictly decreasing from infinity to zero on $x>0$, so there is a unique positive root, which tends to infinity. Let $L=\log (1/ϵ)$. Taking logarithms givesFirst $x^2\sim L$. Put $x=\sqrt{L}+d$ with $d=o(\sqrt{L})$, expand, and cancel $L$:The leading balance gives $d=-(\log L)/(4\sqrt{L})$. Substitution leaves a residual $O((\log L{)}^2/L)$; the derivative of $x^2+\log x$ is of order $\sqrt{L}$, so the root error is $O((\log L{)}^2/L^{3/2})$. ThusThe second term is logarithmic rather than a constant multiple of $L^{-1/2}$. This is logarithmic inversion of a Gaussian divided by its argument.

To supply nonzero channel flux while letting the speed decay in the large upper region, take a logarithmic velocity potential:Use the logarithm analytic on the upper half-plane, with argument between zero and $\pi$. Its complex velocity isAt upper infinity it tends to zero. At the channel end $Z\to 0$, it tends to $1/i=-i$, which means $(u,v)=(0,1)$, as required. The streamfunction is constant on each real boundary interval, and differs by two between the two channel walls. That flux equals the channel width two times the unit upward speed, providing an independent normalization check.

On the fluid free boundary write $q=e^{-i\theta }$, with $-\pi <\theta <0$: the lip has $q=-1$ and the downstream limit has $q=1$. ThenSubstitution into the relation from part (ii) givesThe surface lies on the top of the velocity potential strip, so its logarithm has imaginary part $\pi$, givingFinally $dz/dw=1/q=e^{i\theta }$, and thereforeThe source's typeset $\cot \theta /2$ is interpreted as $\cot (\theta /2)$, as the derivation requires.

First state the normalization needed for the displayed logarithmic factorization: choose a continuous holomorphic function branch $L=\log F$ in the strip that tends to zero at both horizontal ends and has sufficient decay for its Cauchy integrals. In particular the index of the kernel along a horizontal line must be zero.

Choose $\alpha <c_{+}<c_{-}<\beta$ and let ${\Gamma }_{\pm }=\{\xi +i{c}_{\pm }:\xi \in \mathbb{R}\}$, both oriented from left to right. PutThe first is analytic above the lower line and the second below the upper line. On the region between the lines, closing a rectangle gives $C_{+}L-C_{-}L=L$: the lower horizontal side is traversed forward and the upper side backward. HenceThe equation $F{w}_{+}=w_{-}$ becomes $F_{+}{w}_{+}=F_{-}{w}_{-}$ on the overlap region. With the canonical bounded factors normalized at infinity, these two products glue to a bounded entire function. Part (a) givesThis is Wiener-Hopf factorization with the quotient convention used in this question.

The source hypotheses require a qualification: nonvanishing in the strip and $F\to 1$ do not by themselves force index zero or a logarithm tending to zero at both ends. For exampleis analytic and nonzero for $|\mathrm{Im}z|<1/2$, tends to one at both ends, but winds once around zero along the real line. A logarithm continued along that line changes by $2\pi i$, so it cannot have the stated normalization at both ends. One must first separate the index factor, or allow appropriately subtracted factors and a correspondingly different common-function growth class. The displayed constant conclusion follows under the canonical zero-index assumptions above; it is not an unconditional consequence of the two printed properties of $F$. This is the issue measured by the Index of a Wiener-Hopf kernel.

Over the plotted range, approximate the surface tension by $\gamma (T)={\gamma }_0-\beta T$ with constant positive $\beta =-{\gamma }^{\prime }$, and take $C_A=70\mathrm{mJ}{\mathrm{K}}^{-1}=0.070\mathrm{J}{\mathrm{K}}^{-1}$. A reversible adiabat hasIntegrating givesThe arbitrary fixed reference temperature $T_{\ast }$ makes the logarithm dimensionless; changing it only shifts the integration constant. This is the dimensionally precise version of writing $C_A\ln T+\beta A$ as constant. Each term has units of entropy. Between two states,If the graph had substantial curvature, the entropy differential would still hold, but this particular constant-$\beta$ exponential would no longer follow.

Pressure and its physical terms. Solving the molar Van der Waals equation for pressure givesThe excluded-volume parameter $b$ reduces the volume available for translational motion, raising the kinetic pressure relative to $RT/V$. The attraction parameter $a$ lowers the pressure because attractive interactions reduce the momentum transferred to the boundary. This mean-field model treats $a,b$ as temperature-independent material constants. For one mole, $V$ is the sample volume with the molar amount already fixed; $b$ has units of that volume and $a$ of pressure times volume squared.

volume dependence of entropy. The Helmholtz free energy has differential $dF=-SdT-PdV$. Equality of mixed derivatives yields the Maxwell relationFor $a=0$ this derivative is $R/(V-b)$, givingEquivalently, $S=R\ln (V-b)+f(T)$ with a common reference volume understood inside the logarithm. The equation of state fixes this volume dependence, not the otherwise unspecified function of temperature.

Excess entropy at fixed volume and temperature. The ideal gas has $(\partial S_{\mathrm{id}}/\partial V{)}_T=R/V$. ThereforeIntegrate from infinite volume, where the entropy difference is specified to vanish:It is negative because excluding configurations reduces the accessible state count. It tends to zero as $V\to \infty$ and has leading dilute limit $-Rb/V$. This is the entropy excess of a van der Waals gas.

Effect of attraction. Restoring a temperature-independent $a$ changes the pressure by $-a/V^2$, but not its derivative with respect to temperature. With the same infinite-volume entropy reference, it therefore does not change $\Delta S^{\mathrm{ex}}$ at fixed $V,T$. This can also be seen fromThe attraction affects internal energy and the equilibrium volume, so it can affect entropy indirectly when different state points are compared. If $a$ were temperature dependent, its derivative would contribute directly to the entropy; that is outside the stipulated constant-parameter model.

Coexistence isotherm. Below the Van der Waals critical point, the formal homogeneous isotherm has a local minimum and maximum. Its middle branch has $\partial P/\partial V>0$, hence negative isothermal compressibility, and is mechanically unstable. Portions of the negative-slope branches can be metastable rather than globally stable. The equilibrium isotherm replaces the loop between liquid volume $V_l$ and vapour volume $V_g$ by a constant saturation pressure $P_s$. The two volume are fixed byThe Maxwell equal-area construction expresses equality of the two phases' chemical potentials. Intermediate total volume are mixtures: changing the vapour fraction changes total volume at the same pressure, rather than following the unstable homogeneous curve. The spinodals are the turning points of the formal isotherm; they are distinct from the coexistence endpoints.

Dense-phase volume at low temperature. At a general subcritical temperature, the two coexistence equations above must be solved together. For this equation of state the area condition is explicitlyThis supplies an exact implicit determination of the liquid volume even when a simple closed form is unavailable.

For the asymptotically low-temperature limit, put $\tau =RTb/a\ll 1$. The liquid is dense, $V_l\simeq b$, while the vapour is dilute, $P_s{V}_g\simeq RT$ and $V_g\gg V_l$. The area condition then givesThus the vapour volume grows exponentially in $1/\tau$, and the saturation pressure is exponentially small. It is consequently consistent to neglect $P_s$ in the liquid equation to algebraic order in $\tau$:Choose the root near $b$, not the large-volume zero-pressure root:Expanding gives the leading requested dense-phase behavior,This is the low-temperature liquid volume of a van der Waals gas. Setting $P_s=0$ is an asymptotic approximation, not an exact coexistence condition at every temperature below $T_c$. The figure illustrates generic subcritical coexistence; its finite-temperature liquid volume is determined by the full equal-area equations. The van der Waals liquid also ceases to be a realistic description at sufficiently low physical temperature, where other phases or quantum effects can matter.

Configuration count. Choose the end-to-end direction as positive and let $N_R,N_L$ count rightward and leftward links. ThenA configuration is determined by choosing the positions of its leftward links, soThe counts must be nonnegative integers; in particular, allowed signed extensions occur in steps of $2a$. If distance means $|L|$ with neither direction selected, there are two corresponding sets for $L>0$, adding a factor of two. That constant does not affect the tension. There is no excluded-volume penalty in this one-dimensional idealized count.

Entropy from the multiplicity. Put $x=L/(Na)$, so $|x|<1$. Applying the leading Stirling's approximation to all three large factorials givesTerms subleading in $N$, including logarithmic factorial prefactors, have been omitted consistently. Expanding near zero extension givesStretching reduces the number of available configurations and hence the entropy.

Tension and the range of Hooke's law. The model assigns no extension-dependent internal energy. At fixed temperature, the externally applied tension is $\mathcal{F}=(\partial F/\partial L{)}_T=-T(\partial S/\partial L{)}_T$. Differentiating the full leading entropy givesIt is the external force needed to oppose the chain's restoring entropic force. Only when $|L|\ll Na$ can one put $\mathrm{artanh}x\simeq x$, yieldingThe paper's displayed linear law therefore requires this small-extension approximation, which is not explicitly stated in its premise $L<L_0$. It is not the exact force law up to full extension. The entropic spring constant grows with temperature because stronger thermal disorder makes a given reduction in configurational entropy cost more free energy. This is the one-dimensional entropic chain force-extension relation.

A sparse number of backward links. Write $q=N_L\ll N$, so $N_R=N-q$ and $L=a(N-2q)$. Exactly,For fixed $q$ and large $N$, each numerator factor approaches $N$, givingMore precisely, the logarithm of the product is $-q(q-1)/(2N)+O(q^3/N^2)$. A small relative error in the multiplicity requires $q^2/N\ll 1$; merely $q/N\ll 1$ need not make the omitted product close to one. The weaker sparse-link condition does suffice for the leading entropy density and leading high-extension tension. These are sparse backward links near full chain extension.

Tension in that limit. In the overlap regime $1\ll q\ll N$, use Stirling also for $q!$:ThusThe full leading entropy would give $(k_{B}T/(2a))\ln [(N-N_L)/N_L]$, which has the same dominant logarithm when $N_L/N\to 0$.

Approaching full extension. The continuum force law diverges logarithmically as $N_L\to 0$, or $L\to Na$. One must not literally apply Stirling to a zero factorial: at the discrete endpoint there is exactly one directed configuration, $\Omega =1$, and its entropy is zero. A continuous fixed-extension derivative is not defined at that endpoint.

An independent force-ensemble calculation makes the limiting physical statement precise. For applied tension $\mathcal{F}$, a link has partition function $z=2\cosh (\beta \mathcal{F}a)$, giving the exact mean extensionIt approaches $Na$ only as $\mathcal{F}\to \infty$ at positive temperature, and inversion reproduces the logarithmic force law for mean extension. In a real chain, bond extension or failure eventually replaces the ideal inextensible-link model.

Temperature as an equilibrium relation. Classical thermodynamics begins with the zeroth law of thermodynamics: thermal equilibrium is transitive. A thermometer can therefore assign a common empirical temperature to systems in mutual thermal equilibrium. Different thermometer properties initially give different monotone empirical scales; a resistance reading by itself does not define the absolute scale.

The Second law of thermodynamics provides a substance-independent thermodynamic scale. A reversible engine operating between reservoirs has a heat ratio depending only on their temperatures. One chooses absolute temperature so that $Q_h/Q_c=T_h/T_c$, with $Q_h,Q_c$ the positive heat magnitudes. Its reversible efficiency is $1-T_c/T_h$. Equivalently, reversible heat defines entropy through $dS=\delta Q_{\mathrm{rev}}/T$. For an equilibrium simple compressible system,Classical thermodynamics uses this relation without needing a microscopic account of entropy. A dilute-gas thermometer realizes an empirical scale approaching the thermodynamic scale after corrections for nonideal behavior.

Statistical meaning of the same derivative. In the microcanonical description, an equilibrium energy shell containing $\Omega (U,V,N)$ accessible states has $S=k_B\ln \Omega$. Its statistical temperature is consequentlyTo see why this is an equilibrium temperature, let two weakly coupled systems exchange a small energy while their total energy stays fixed. The combined multiplicity is ${\Omega }_A(U_A){\Omega }_B(U_{\mathrm{tot}}-U_A)$. Maximizing its logarithm giveswhich is equality of the two inverse temperatures. Stability requires the total entropy maximum, not only stationarity. The overwhelmingly dominant macrostate in a large system therefore realizes the thermodynamic equilibrium condition.

A large reservoir gives a small subsystem the canonical probabilities $p_i\propto e^{-\beta E_i}$, with $\beta =1/(k_{B}T)$. In that ensemble, $F=-k_{B}T\ln Z$ and $S=-k_B{\sum }_i{p}_i\ln p_i$ agree with macroscopic thermodynamic relations in their usual equilibrium regime. The equipartition identity for a classical dilute monatomic gas, $⟨p^2/(2m)⟩=3k_{B}T/2$, is a useful operational connection between microscopic motion and temperature, but is not a universal definition: quantum modes need not obey equipartition, and a system away from equilibrium need not have one temperature at all.

For finite systems one must specify the entropy convention and energy resolution; surface and fluctuation corrections can distinguish definitions that agree in the ordinary thermodynamic limit. Bounded-energy population-inverted systems can also have a negative entropy derivative and hence a negative statistical temperature. That possibility lies outside an ordinary positive-temperature ideal-gas thermometer; it does not mean temperature is merely an arbitrary average kinetic energy.

Obtaining the energy–temperature conversion. The Boltzmann constant connects entropy measured in $\mathrm{J}{\mathrm{K}}^{-1}$ with the dimensionless logarithm of a state count. Historically it could be determined from the ideal gas law $pV=N{k}_{B}T$: measure the molar gas constant $R$ and independently determine the Avogadro constant $N_A$, then obtain $k_B=R/N_A$. The independent particle-number determination matters; using $R$ alone would not fix energy per particle.

An electrical method uses Johnson-Nyquist noise. A resistor of resistance $R_e$ at known temperature has, in the classical range $hf\ll k_{B}T$, a one-sided open-circuit voltage-noise spectrum $S_V(f)=4k_{B}T{R}_e$. Over a calibrated flat bandwidth $\Delta f$,A practical measurement uses an equilibrium resistor, a calibrated low-noise amplifier and a known transfer function, subtracts or independently characterizes amplifier noise, and replaces a nominal bandwidth by the actual effective noise bandwidth. Varying temperature and fitting the mean-square voltage is more informative than treating one amplified voltage sample as the thermal variance. Other primary methods relate temperature to sound speed in a dilute gas or to calibrated particle diffusion and mobility.

There is a historically relevant unit-definition qualification: by the date of this exam, the revised SI had already taken effect on 20 May 2019. The kelvin is now defined by fixingThe same experiments now realize or compare thermodynamic temperatures using this fixed constant, whereas historically they measured the conversion constant on an independently defined temperature scale. The physical distinction between equilibrium temperature and microscopic state counting is unchanged.^[ref]

For $x>0$, the event $M_n/n\le x$ says that each of the independent random variables is at most $\lfloor nx\rfloor$. Hence, for all sufficiently large $n$,Since $n/(\lfloor nx\rfloor +1)\to 1/x$ and $\log (1-u)=-u+o(u)$ as $u\to 0$, taking logarithms gives the limit $e^{-1/x}$. For $x\le 0$ the probability is zero. ThereforeThis is the Fréchet limit for a reciprocal integer tail, with a Fréchet distribution of shape one. Its cumulative distribution function is continuous even at zero, so the pointwise calculation establishes convergence in distribution.

Taking logarithms of the product givesBy part (d)(i), $V_j=E_j-\log b$ in probability distribution, where $E_j$ has the exponential distribution of rate one. Thus $\mathbb{E}|V_j|\le 1+|\log b|<\infty$ and $\mathbb{E}{V}_j=1-\log b$. The strong law of large numbers givesIf $b<e$, choose $\delta =(1-\log b)/2>0$. Eventually $\log W_n\le -\delta n$, so $W_n\le e^{-\delta n}\to 0$. If $b>e$, choose $\delta =(\log b-1)/2>0$; eventually $W_n\ge e^{\delta n}\to \infty$. ThereforeIn particular, the second range cannot have almost sure convergence to zero. This is the almost-sure growth rate of independent products. At $b=e$ the logarithmic limit is zero, which by itself decides neither behavior; that boundary case is not requested.

Evaluate the source's printed logarithm using the numerical magnitude value 21.0, with base-ten logarithms:This is the literal printed-fit convention identified above. Convert the mean visual surface brightness to a linear luminosity per projected area. At 10 parsecs, one square arcsecond subtends $100/(206264.806{)}^2\mathrm{p}{\mathrm{c}}^2$. The definition of absolute magnitude consequently givesThe constant $21.572=5{\log }_{10}(206264.806)-5$ is the conversion between the angular and physical projected area. Since the effective radius encloses half the total luminosity,For the stipulated stellar population of Solar-type stars, approximate the stellar mass-to-light ratio by $M_{\odot }/L_{V,\odot }$. Summing 100 such galaxies then givesThis assumes negligible interstellar extinction and that the supplied surface brightness is an intrinsic or already corrected value; an observed cosmological surface brightness requires additional dimming and passband corrections before this conversion.

For comparison only, if the empirical coefficients and intercept were instead meant for ${\log }_{10}({\mathcal{I}}_e/[L_{V,\odot }\mathrm{p}{\mathrm{c}}^{-2}])$, substituting $140.9$ in that fit would give $R_e=1.09\mathrm{kpc}$, $L_V=1.06\times 10^9{L}_{V,\odot }$, and $M_{\ast ,\mathrm{cluster}}=1.06\times 10^{11}{M}_{\odot }$. Those are conditional alternative results, not an unannounced correction of the printed units.

A branch point obstructs a single-valued analytic continuation around a closed circuit. Continuing $\log z=\log |z|+i\theta$ once around zero increases its value by $2\pi i$. A branch cut removes a curve joining branch points or extending to infinity so that a continuous argument can be chosen on the remainder. Zero and infinity are logarithmic branch points; the positive-axis cut used here selects $0<\arg z<2\pi$. The cut is a choice of representation, not a line of poles of the multivalued logarithm.

The large circle contributes $O((\log R{)}^n/R)$ and the small circle contributes $O(ϵ|\log ϵ{|}^n)$, so both disappear. The enclosed simple poles are $i$ and $-i$, with logarithms $i\pi /2$ and $3i\pi /2$. Their residues are $(i\pi /2{)}^n/(2i)$ and $-(3i\pi /2{)}^n/(2i)$. The positive orientation of the keyhole therefore givesThe upper bank runs outward and the lower bank inward. Their logarithms differ by $2\pi i$; keeping that orientation is essential for the sign.

At half ionization, electron and neutral densities are each $n_0/2$, so the left side of Saha's equation is $n_0/2$. Its logarithm changes by $d\log n_0=(3/2+R/(k_B{T}^{\prime }))d\log T^{\prime }$. To first order,This holds for small fractional changes and the supplied ground-state partition approximation, not an exact finite-difference identity.

At a regular point, $f(z_0+\zeta )-f(z_0)=f^{\prime }(z_0)\zeta +O({\zeta }^2)$. Multiplication by a nonzero complex number rotates every tangent by the same angle and multiplies every length by the same positive factor. Oriented angles are preserved. If the first derivative vanishes but the second does not, the leading term is $f^{\prime \prime }(z_0){\zeta }^2/2$: tangent arguments are doubled, so angles are doubled, modulo the identification of rays. This is a local statement; holomorphy alone does not imply global injectivity.

Write $z=x+iy$. ThenFor $-1<x<1$ and $y>0$, the imaginary part is positive. The bottom edge maps monotonically to $[-1,1]$; the left and right vertical edges map to $(-\infty ,-1]$ and $[1,\infty )$ respectively. The inverse below establishes that every point of the open upper half-plane occurs exactly once. At fixed $y=a$, putting $u=\mathrm{Re}w$, $v=\mathrm{Im}w$, one obtainsThis is the upper half of an ellipse, traversed from its left endpoint to its right endpoint.

Choose the branch $s(w)=\sqrt{1-w^2}$ analytic in the upper half-plane and positive on the boundary interval $(-1,1)$. Then the inverse branch into the strip isThe logarithm is chosen continuously so that $-1<\mathrm{Re}z<1$ and $\mathrm{Im}z>0$; for instance $w=iv$, $v>0$, gives $z=(2i/\pi )\mathrm{arsinh}v$. Differentiation gives $dz/dw=2/(\pi \sqrt{1-w^2})$. The only zeros of the original derivative in the closed strip are $z=\pm 1$, with images $w=\pm 1$. The inverse has square-root branch points there, and cannot extend holomorphically across them. At either corner the original map doubles the right angle to a straight angle; a holomorphic inverse with nonzero derivative could not halve it. The open strip interior has no such failure.

Write $t=1-\rho \in (0,1)$. The target Poisson distribution and proposal geometric distribution have probability mass functions $p_k=e^{-\lambda }{\lambda }^k/k!$ and $q_k=(1-t)t^k$. Their ratio isThe successive ratio $R_{k+1}/R_k=\lambda /[t(k+1)]$ shows that a maximizing integer is $m=\lfloor \lambda /t\rfloor$, with a tie at $m-1$ when $\lambda /t$ is an integer. Thus the tight envelope isFor any fixed $t$, draw $Y\sim \mathrm{Geom}(1-t)$ using (ii)'s inversion formula with this new parameter, and draw an independent $U$ from the uniform distribution on $(0,1)$. Accept whenotherwise restart. The acceptance probability is $1/M(t)$ and the expected number of proposals is $M(t)$.

To minimize that expected count, set $K=\lceil \lambda \rceil$. The exact answer isHere is a proof that includes the changes of maximizing integer. Each $\log R_k(t)$ is a convex function of $t$, since its second derivative is $(1-t{)}^{-2}+k/t^2>0$. Their pointwise maximum $\log M(t)$ is also a convex function. When $m$ is fixed its derivative isIf $K-1<\lambda \le K^2/(K+1)$, take $t_{\ast }=\lambda /K$, where $m$ changes from $K$ on the left to $K-1$ on the right. The left derivative is nonpositive because $\lambda \le K^2/(K+1)$; the right derivative is nonnegative because $\lambda \ge K-1$. This is a minimum. If $K^2/(K+1)<\lambda \le K$, take $t_{\ast }=K/(K+1)$. There $K<\lambda /t_{\ast }\le K+1$ and the active branch has $m=K$ except for the endpoint tie; its derivative vanishes. Convexity again proves the global minimum. The formulas agree at the boundary.

For a fully specified optimal algorithm, compute $K,t_{\ast },{\rho }_{\ast }$ and $m_{\ast }=\lfloor \lambda /t_{\ast }\rfloor$ once, then use the preceding acceptance test with $t=t_{\ast }$ and $m=m_{\ast }$. At an integer tie either maximizing integer gives the same test. In code, factorials can be evaluated through the logarithm of the gamma function:and acceptance is $\log U\le \min (0,\log a_Y)$. This avoids overflowing factorials.

The optimum minimizes expected trials, the usual idealization of time when each proposal and test has constant cost. Actual runtime also depends on the implementation of discrete inversion and factorial evaluation. Simply minimizing an upper bound for $M(t)$ would not necessarily give the exact optimum above; see optimal geometric envelope for a Poisson distribution.

Solving $u=F(x)$ gives $1-(1-p)e^{-\beta x}=p^{1-u}$. Hence the quantile function isFor such $u$, $p<p^{1-u}<1$, making the logarithm's argument lie in $(0,1)$ and the output positive. Its endpoint limits are $F^{-1}(0)=0$ and $F^{-1}(1)=+\infty$. Substitution into (i)'s cumulative distribution function recovers $u$.

Taking the logarithm of the normalized probability density function, then summing over the independent observations, gives the log-likelihoodBoth $p$ and $\beta$ are parameters being inferred, so their contributions to the normalizing constant must be retained. The following R implementation validates the arguments and returns one number:

ll_f <- function(p, beta, x) {
  scalar <- function(v) {
    is.numeric(v) && !is.complex(v) && length(v) == 1L &&
      is.finite(v)
  }
  if (!scalar(p) || p <= 0 || p >= 1)
    stop("p must lie strictly between 0 and 1")
  if (!scalar(beta) || beta <= 0)
    stop("beta must be positive")
  if (!is.numeric(x) || is.complex(x) || !is.null(dim(x)) ||
      any(!is.finite(x)) || any(x <= 0))
    stop("x must be a vector of finite positive numbers")
  if (length(x) == 0L) return(0)
  t <- log1p(-p) - beta * x
  length(x) * (log(beta) - log(-log(p))) +
    sum(t - log(-expm1(t)))
}

Here $t_i=\log (1-p)-\beta x_i<0$ and -expm1(t) computes $1-e^t$ accurately near zero. The empty-data likelihood function is one, so its log-likelihood is zero. Invalid parameter values produce errors in this evaluator; the next algorithm rejects candidates outside the parameter support before calling it.