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The continuum expression is the large-cavity leading term; using standing-wave boundary conditions gives the same bulk count. Both photon polarizations must be included. This is density per angular frequency, not per ordinary frequency.

Each mode is an oscillator with zero photon chemical potential in thermal equilibrium. Multiplying its mean thermal energy by the mode density per volume gives the angular-frequency Planck energy density

u (ω, T) = \frac{ℏ ω ^{3}}{π ^{2} c ^{3}} \frac{1}{e ^{ℏ ω / (k_{B} T)} - 1} .

(4)

At small

ℏ ω / (k_{B} T)

it reduces to the Rayleigh–Jeans result

k_{B} T ω^{2} / (π^{2} c^{3})

; exponential suppression at high frequency prevents a divergent thermal energy. With

x = ℏ ω / (k_{B} T)

u (T) = \int_{0}^{\infty} u (ω, T) d ω = \frac{k _{B}^{4} T ^{4}}{π ^{2} ℏ ^{3} c ^{3}} \int_{0}^{\infty} \frac{x ^{3}}{e ^{x} - 1} d x = \frac{π ^{2} k _{B}^{4}}{15 ℏ ^{3} c ^{3}} T^{4} .

(5)

The integral is the stipulated

π^{4} /15

. This is a volume energy density,

a T^{4}

; for isotropic cavity radiation the outward black-surface flux is

c u /4 = σ_{SB} T^{4}

, with

σ_{SB} = π^{2} k_{B}^{4} / (60 ℏ^{3} c^{2})

. Distinguishing flux, spectral density and total volume energy avoids extra factors of

c

4

2 π

Figure 1.
Oscillator and radiation energy
. Thermal excitation and dimensionless Planck spectra. **Details**.

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Past exam of the physics B course of the University of Cambridge / 2013 / Paper 1 / Section B / B7 / Solution 2026-10-20

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For distributed inductance

L

and capacitance

C

, a small segment obeys

V_{z} = - L I_{t}

and

I_{z} = - C V_{t}

. Differentiation gives

V_{zz} = L C V_{tt}

and

I_{zz} = L C I_{tt}

. For a forward wave,

v = \frac{1}{L C}, Z = \frac{V ^{+}}{I ^{+}} = L / C .

(1)

A backward voltage wave has current

I^{-} = - V^{-} / Z

. This sign matters at an impedance discontinuity.

The source figure is the stacked-strip parallel-plate transmission line geometry, despite its use of the word coplanar. Gauss's law gives

C = ϵ_{0} ϵ_{r} a / d

. Opposite surface currents give magnetic field

B = μ_{0} I / a

between the strips by Ampère's law; the magnetic flux per unit length is

B d

, hence

L = B d / I = μ_{0} d / a

. Therefore

v = \frac{c}{2.3} ≃ 1.98 \times 1 0^{8} m s^{- 1}, Z = \frac{d}{a} \frac{μ _{0}}{2.3 ϵ _{0}} ≃ 248.4 \frac{d}{a} Ω.

(2)

No numerical

a / d

is given; the impedance is appropriately left in terms of that ratio. The line is taken to be nonmagnetic and lossless.

Increasing the gap by

α

changes

L

αL

and

C

C / α

, so

Z_{2} = α Z_{1}

while

v

is unchanged. voltage and current continuity at the step give

V_{i} + V_{r} = V_{t}, \frac{V _{i} - V _{r}}{Z _{1}} = \frac{V _{t}}{Z _{2}} .

(3)

Consequently the voltage and energy-flux transmission coefficients are

t_{V} = \frac{2 α}{1 + α}, T = \frac{Z _{1}}{Z _{2}} ∣ t_{V} ∣^{2} = \frac{4 α}{( 1 + α ) ^{2}}, r_{V} = \frac{α - 1}{α + 1} .

(4)

Here amplitudes are peak voltage amplitudes, with mean power

∣ V ∣^{2} / (2 Z)

. The reciprocal gaps have

r_{V} (1/ α) = - r_{V} (α)

: equal reflected energy but opposite voltage phase. A voltage transmission coefficient above one does not imply energy gain.

For a finite impedance section of a transmission line, put

ℓ = l_{2} - l_{1}

and

δ = ω ℓ / v

. Multiple reflections inside the altered section interfere. Summing their geometric series, relative to the first boundary,

t_{whole} = \frac{( 1 - r _{V}^{2} ) e ^{i δ}}{1 - r _{V}^{2} e ^{2 i δ}}, T_{whole} = [1 + \frac{( α ^{2} - 1 ) ^{2}}{4 α ^{2}} sin^{2} δ]^{- 1} .

(5)

Transmission is unity at

δ = nπ

and minimal at

δ = (n + 1/2) π

, with minima

4 α^{2} / (1 + α^{2})^{2}

. Resonances are separated by

Δ f = v / (2 ℓ)

; the long-wavelength limit also transmits fully. Matching the far termination removes end echoes, not the two internal reflections.

Figure 1.
Original calculated power transmission through a finite changed-impedance section between matched lines
. **Details**.

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Past exam of the physics course of the year 1 of the natural sciences course of the University of Cambridge / 2018 / Section A / 4 / Solution 2026-10-18

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First intensity minimum. In Fraunhofer diffraction, the path difference between adjacent slits is

d sin θ

, so the relative wave phase is

δ = 2 π d sin θ / λ_{1}

. Summing the

N

equal complex amplitudes as a geometric series gives the diffraction grating interference factor

I (θ) \propto \sum_{j = 0}^{N - 1} e^{ij δ}^{2} = \frac{s i n ^{2} ( N δ /2 )}{s i n ^{2} ( δ /2 )} .

(1)

The central maximum has

δ = 0

. Its nearest zero occurs at

N δ /2 = π

, where the denominator is nonzero. Thus

sin θ_{m i n} = \frac{λ _{1}}{N d} = \frac{640 \times 1 0 ^{- 9}}{10 ( 80 \times 1 0 ^{- 6} )} = 8.00 \times 1 0^{- 4}, ∣ θ_{m i n} ∣ = 8.00 \times 1 0^{- 4} rad = 0.045 8^{\circ} .

(2)

There is a symmetric first zero on each side of the incident direction. The narrow slits are treated as having a sufficiently broad single-slit diffraction envelope.

Wavelength resolution and the missing order specification. Adopt the Rayleigh criterion: the principal maximum of one wavelength lies at the adjacent minimum of the other in the same nonzero diffraction order

m

. For

λ_{2} > λ_{1}

and

m > 0

, those directions obey

d sin θ = m λ_{2} = (m + \frac{1}{N}) λ_{1} .

(3)

Consequently the ideal resolving power of a diffraction grating is

λ_{1} /Δ λ = m N

, giving

Δ λ_{m i n, m} = \frac{λ _{1}}{m N} = \frac{64}{m} nm .

(4)

The usual first-order reading therefore gives $64 nm$ . Zero diffraction order cannot separate the two wavelengths, because both central maxima coincide.

The paper does not specify which diffraction order is observed, so

64 nm

is not an order-independent minimum. In the ideal infinitely narrow-slit model, the largest usable order for this positive wavelength separation must satisfy

(m + 1/ N) λ_{1} \leq d

. Since

d / λ_{1} = 125

, this permits

m \leq 124

, giving the formal all-order result

Δ λ_{m i n, 124} = 0.516 nm .

(5)

That pair meets the criterion at

θ ≃ 83. 1^{\circ}

. Order

125

λ_{1}

is at grazing incidence and the corresponding longer-wavelength peak cannot propagate. In a real device, finite slit width, weak high-order intensity and screen acceptance may prevent such an extreme order from being used; those data are absent. Thus the specified-order expression, and

64 nm

when first order is intended, are the defensible resolution answers under the stated criterion.

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Past exam of the physics course of the year 1 of the natural sciences course of the University of Cambridge / 2024 / Paper 1 / Section B / 6 / g / Solution 2026-10-20

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For actual complete bounce distances

D_{1} = 4 m

and

D_{n + 1} = r D_{n}

, the geometric series gives

D_{total} = \frac{4}{1 - r} m, 0 \leq r < 1.

(1)

A natural intended estimate compares the observed first round trip with the loss-free round trip from the initially compressed position,

D_{ideal} = 2 (H_{s} + x) = 5.13764 m .

(2)

If the reduction factor is taken to be

r = D_{1} / D_{ideal} = 0.77857

, the answer is $D_{total} ≃ 18.1 m$ . The first term is the actual 4 m, not the hypothetical loss-free distance.

This numerical inference needs the stated comparison assumption. Strictly, knowing the first actual bounce distance and saying subsequent distances share a ratio does not independently specify that ratio. A change in spring compression also means that an energy-loss fraction is not automatically the same as a vertical-distance ratio.

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Past exam of the probability course of the University of Oxford / 2017 / 3 / b / i / Solution 2026-10-31

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The difference process $D_{n}$ is a Markov chain. If the columns are equal, the next block creates a height difference of one. If the difference is

k \geq 1

, landing in the taller column increases it to

k + 1

, while landing in the shorter column makes it zero. Therefore its only nonzero transition probabilities are

P (0, 1) = 1, P (k, k + 1) = P (k, 0) = \frac{1}{2} (k \geq 1) .

(1)

These probabilities depend only on the present difference, not on the absolute heights or their history. This is the reset-or-increment Markov chain with

p = 1/2

For its stationary distribution, the incoming-flow equations give

π_{1} = π_{0}

and

π_{k} = π_{k - 1} /2

for

k \geq 2

. Normalization uses a geometric series:

1 = π_{0} + \sum_{k = 1}^{\infty} π_{0} 2^{- (k - 1)} = 3 π_{0} .

(2)

Thus

π_{0} = \frac{1}{3}, π_{k} = \frac{1}{3 2 ^{k - 1}} (k \geq 1) .

(3)

The remaining balance equation is also satisfied: the flow into zero is

(1 - π_{0}) /2 = 1/3

. Every state can reach zero and every positive state can be reached from zero, so it is an irreducible Markov chain. Its invariant probability distribution makes it positive recurrent, and that distribution is unique.

The maximum-height process $H_{n}$ is not Markov. Starting empty, condition first on the positive-probability history

(H_{1}, H_{2}, H_{3}) = (1, 1, 2)

. After two blocks the columns were equal, so after three their difference is one. Hence

P (H_{4} = 3 ∣ H_{1} = 1, H_{2} = 1, H_{3} = 2) = 1/2

For the history

(1, 2, 2)

, the first two blocks built one column and the third caught the other column up, leaving equal heights. Hence

P (H_{4} = 3 ∣ H_{1} = 1, H_{2} = 2, H_{3} = 2) = 1

. Both histories have the same time and present height

H_{3} = 2

, but different predictions. This violates the Markov property and illustrates why a function of a Markov chain need not be Markov.

Figure 1.
The height difference has reset-or-increment transitions and a geometric stationary tail
. Original transition diagram and stationary probabilities. Independent simulated height averages illustrate the proved limiting growth rate; the finite diagram continues indefinitely to the right. **Details**.

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Past exam of the Probability, statistics and options II course of the University of Oxford / 2007 / M / 5 / b / Solution 2026-10-31

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The state-zero stationary distribution equation is

π_{0} = q π_{0} + q π_{1}

, so

π_{1} = (p / q) π_{0}

. The state-one equation is

π_{1} = p π_{0} + q π_{2}

, giving

π_{2} = (p / q) π_{1}

. More generally, the stationary probability flow across the cut

{0, \dots, i}

is balanced, so

p π_{i} = q π_{i + 1}

. This also follows inductively from

π_{i} = p π_{i - 1} + q π_{i + 1}

Hence

π_{i} = (p / q)^{i} π_{0}

. Since

p < q

, the geometric series is summable, and normalization gives

π_{i} = (1 - \frac{p}{q}) (\frac{p}{q})^{i}, i \geq 0.

(1)

These probabilities satisfy detailed balance and therefore all the stationarity equations. The Markov chain is an irreducible Markov chain, so this normalized stationary distribution is unique.

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Past exam of the Probability, statistics and options II course of the University of Oxford / 2008 / M / 4 / Solution 2026-10-31

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The moment generating function is

M_{X} (t) = E (e^{tX})

wherever that expectation is finite. A useful continuity theorem says that if the moment generating functions

M_{n}

are finite on a common open interval containing zero and converge pointwise there to the moment generating function of a probability distribution, then the corresponding variables converge in distribution to that law. Equivalently one may assume a finite limiting function on such a neighborhood, with the usual continuity at zero, and obtain the uniquely determined limiting law.

For the gamma density, integrating the exponential against the density and changing variable to

(λ - t) x

gives

M_{X} (t) = (\frac{λ}{λ - t})^{α}, t < λ .

(1)

For the waiting time to one head, independence of successive tosses gives

Pr (Y = j) = p (1 - p)^{j - 1}

for

j \geq 1

. Summing the geometric series gives

M_{Y} (t) = \frac{p e ^{t}}{1 - ( 1 - p ) e ^{t}}, (1 - p) e^{t} < 1.

(2)

The waiting times between consecutive heads are independent copies of

Y

, so their sum

N

has moment generating function

M_{Y} (t)^{k}

. Consequently

M_{2 pN} (t) = [\frac{p e ^{2 pt}}{1 - ( 1 - p ) e ^{2 pt}}]^{k} ⟶ (1 - 2 t)^{- k} .

(3)

For fixed

t < 1/2

, numerator and denominator have leading terms

p

and

p (1 - 2 t)

p ↓ 0

. A fixed neighborhood of zero is inside the convergence domain for all sufficiently small

p

, so the continuity theorem applies. The limiting distribution is

Gamma (shape k, rate 1/2),

(4)

equivalently a chi-squared distribution with

2 k

degrees of freedom.

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Past exam of the Probability, statistics and options II course of the University of Oxford / 2009 / S / S2 / b / Solution 2026-10-31

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For one proposal,

P (Y = x, accept) = q (x) \tilde{p} (x) / (M \tilde{q} (x)) = \tilde{p} (x) / (M Z_{q})

. Summing gives total acceptance probability

a = Z_{p} / (M Z_{q}) > 0

. Conditioning on acceptance therefore gives

P (Y = x ∣ accept) = \tilde{p} (x) / Z_{p} = p (x)

The first accepted result has this law too: the probability of

k - 1

failures followed by acceptance at

x

(1 - a)^{k - 1} \tilde{p} (x) / (M Z_{q})

. Sum the geometric series over

k \geq 1

to obtain

P (X = x) = p (x)

. Positive acceptance probability also proves almost-sure termination.

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Past exam of the Probability, statistics and options II course of the University of Oxford / 2011 / M / M4 / c / Solution 2026-10-31

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At each index the two independent uniform symbols match with probability

p = k / k^{2} = 1/ k

. The match indicators are independent across indices. Consequently the waiting times between successive matches are independent geometric random variables on

1, 2, \dots

with success probability

p

, and

S_{m}

is their sum.

For one waiting time

G

, summing the geometric series gives

E e^{tG} = p e^{t} / [1 - (1 - p) e^{t}]

. Independence yields

M_{S_{m}} (t) = (\frac{e ^{t} / k}{1 - ( 1 - 1/ k ) e ^{t}})^{m}, E S_{m} = mk .

(1)

For

k > 1

the moment-generating function is finite for

t < lo g (k / (k - 1))

. When

k = 1

, every index matches,

S_{m} = m

deterministically and the formula reduces to

e^{m t}

for every real

t

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Past exam of the real and harmonic analysis course of the University of Oxford / 2007 / 3 / c / Solution 2026-10-31

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Write

A_{N} (f) = (N + 1)^{- 1} \sum_{n = 0}^{N} f (e^{in ξ})

. For a monomial

z^{k}

k \in Z

, the geometric series gives

A_{N} (z^{k}) = ⎩ ⎨ ⎧ 1, \frac{1 - e ^{ik (N + 1) ξ}}{( N + 1 ) ( 1 - e ^{ik ξ} )}, e^{ik ξ} = 1, e^{ik ξ} \neq = 1.

(1)

Thus every such average converges, and by linearity so does the average of every trigonometric polynomial. Also

∣ A_{N} (f) ∣ \leq ∥ f ∥_{\infty}

uniformly in

N

Given

ε > 0

, choose a trigonometric polynomial

P

with

∥ f - P ∥_{\infty} < ε

. Then

∣ A_{N} (f) - A_{M} (f) ∣ \leq 2 ε + ∣ A_{N} (P) - A_{M} (P) ∣.

(2)

The last term tends to zero as

N, M \to \infty

. The scalar sequence is Cauchy and therefore the limit exists for every continuous $f$ and every real $ξ$ .

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Past exam of the real and harmonic analysis course of the University of Oxford / 2007 / 4 / b / Solution 2026-10-31

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A summability kernel on the circle is a sequence of continuous periodic functions with

\int k_{n} d m = 1, sup_{n} ∥ k_{n} ∥_{1} = C < \infty, \int_{δ \leq ∣ t ∣ \leq π} ∣ k_{n} (t) ∣ d m (t) ⟶ 0 (0 < δ < π) .

(1)

The frequently used stronger condition of uniform convergence to zero off each neighborhood of zero implies the last condition. For

f

in a homogeneous Banach space,

∥ f * k_{n} - f ∥_{B} \leq \int ∣ k_{n} (s) ∣∥ τ_{s} f - f ∥_{B} d m (s) .

(2)

Given

ε > 0

, translation continuity bounds the difference inside a sufficiently small neighborhood by

ε

; its integral is at most

Cε

. Outside, use

2∥ f ∥_{B}

and the vanishing tail condition. Hence

f * k_{n} \to f

B

With the indexing needed in part (c), the Fejér kernel is

K_{n} (t) = \frac{1}{n + 1} \sum_{r = 0}^{n} e^{i r t}^{2} = \sum_{∣ j ∣ \leq n} (1 - \frac{∣ j ∣}{n + 1}) e^{ij t}, n \geq 0.

(3)

In particular

K_{0} = 1

. The squared-modulus expression proves

K_{n} \geq 0

, and its constant Fourier coefficient proves

\int K_{n} d m = ∥ K_{n} ∥_{1} = 1

. The geometric series yields

K_{n} (t) = \frac{1}{n + 1} \frac{s i n ^{2} (( n + 1 ) t /2 )}{s i n ^{2} ( t /2 )} \leq \frac{1}{( n + 1 ) s i n ^{2} ( δ /2 )} (δ \leq ∣ t ∣ \leq π) .

(4)

Thus these are summability kernels. For

T_{f} : g \mapsto f * g

, part (a) gives

∥ T_{f} ∥ \leq ∥ f ∥_{B}

. Since

∥ K_{n} ∥_{1} = 1

and

T_{f} K_{n} \to f

B

, the reverse bound follows:

∥ T_{f} : L^{1} (T) \to B ∥ = ∥ f ∥_{B} .

(5)

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Past exam of the short options paper of the physics Prelims of the University of Oxford / 2021 / S01 / 1 / b / Solution 2026-10-25

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The geometric series for

(2 + z)^{- 1}

converges for

∣ z ∣ < 2

, while the cancelled sine terms have an everywhere convergent Taylor series. Hence the requested Laurent series has no negative powers:

f (z) = n = 0 \sum \infty \frac{( - 1 ) ^{n} z ^{n}}{2 ^{n + 1}} + k = 1 \sum \infty \frac{( - 1 ) ^{k} z ^{2 k - 1}}{( 2 k + 1 )!}, 0 < ∣ z ∣ < 2.

(1)

For example,

f (z) = \frac{1}{2} - \frac{5 z}{12} + \frac{z ^{2}}{8} - \frac{13 z ^{3}}{240} + \frac{z ^{4}}{32} + O (z^{5}) .

(2)

The extension through zero also makes this a Taylor series. The pole at

- 2

fixes its radius of convergence at two.

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Past exam of the simulation and statistical programming course of the University of Oxford / 2021 / 2 / a / ii / Solution 2026-10-25

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Integrating the conditional distribution against the exponential distribution probability density function gives, for integer

k \geq 0

P (Y = k) = \int_{0}^{\infty} \frac{e ^{- x} x ^{k}}{k !} λ e^{- λ x} d x = \frac{λ}{k !} \frac{Γ ( k + 1 )}{( 1 + λ ) ^{k + 1}} = \frac{λ}{1 + λ} (\frac{1}{1 + λ})^{k} .

(1)

Here the gamma function identity

Γ (k + 1) = k!

evaluates the integral. Hence

Y \sim Geom (\frac{λ}{1 + λ})

(2)

{0, 1, 2, \dots}

. Put

r = 1/ (1 + λ)

and

ρ = 1 - r

. The geometric series gives

F (k) = 1 - r^{k + 1}

. Its generalized inverse, the quantile function, is

F^{- 1} (u) = ⌈ \frac{lo g ( 1 - u )}{lo g r} ⌉ - 1, 0 < u < 1.

(3)

This is the smallest nonnegative integer

k

for which

1 - r^{k + 1} \geq u

; dividing by

lo g r < 0

reverses the inequality. Draw

U

from the uniform distribution on

(0, 1)

and use this formula. The commonly used expression

⌊ lo g (1 - U) / lo g r ⌋

has the same probability distribution, but differs from the exact generalized inverse when

U

is precisely a jump value of the cumulative distribution function. Such values have probability zero for an ideal continuous uniform draw.

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Past exam of the simulation and statistical programming course of the University of Oxford / 2021 / 3 / b / ii / Solution 2026-10-25

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Let

m = E X

and

v = Var (X)

, evaluated from the given polylogarithm formulas. Generate a large sample of independent and identically distributed random variables with r_sample_f, and compute its sample mean

\overline{X}_{n}

and sample variance

S_{n}^{2}

using mean and var. The central limit theorem gives the first standardized check:

Z_{m} = \frac{n ( X _{n} - m )}{v} \Rightarrow N (0, 1) .

(1)

For the second check, the correct asymptotic variance of the sample variance involves the fourth central moment

μ_{4} = E [(X - m)^{4}]

Z_{v} = \frac{n ( S _{n}^{2} - v )}{μ _{4} - v ^{2}} \Rightarrow N (0, 1) .

(2)

The denominator is not

2 v^{2}

unless the probability distribution has the requisite fourth central moment, as does the normal distribution. Here the probability density function decays exponentially, so all required moments are finite.

To compute

μ_{4}

, expand the denominator of the probability density function as a geometric series and integrate each nonnegative term. The gamma function integral gives the raw moments

m_{j} := E [X^{j}] = - \frac{j ! Li _{j + 1} ( 1 - p )}{β ^{j} l o g p}, j \geq 1.

(3)

Thus

v = m_{2} - m_{1}^{2}

and

μ_{4} = m_{4} - 4 m_{1} m_{3} + 6 m_{1}^{2} m_{2} - 3 m_{1}^{4}

. The following R code implements both checks for chosen interior parameters and large n:

p <- 0.4
beta <- 1.3
n <- 100000
raw_m <- vapply(1:4, function(j) {
  -factorial(j) * polylog(j + 1, 1 - p) / (beta^j * log(p))
}, numeric(1))
m <- raw_m[1]
v <- raw_m[2] - m^2
mu4 <- raw_m[4] - 4 * m * raw_m[3] +
  6 * m^2 * raw_m[2] - 3 * m^4
x <- r_sample_f(n, p, beta)
c(mean_check = sqrt(n) * (mean(x) - m) / sqrt(v),
  variance_check = sqrt(n) * (var(x) - v) / sqrt(mu4 - v^2))

For each check separately, an absolute value below

1.96

is consistent with its approximate

95%

confidence interval. Repeat with many independent samples: the two standardized statistics should each resemble a standard normal distribution, with about

5%

outside that range. Also vary

p

β

and n; a systematic discrepancy that grows with

n

suggests a bias. The two checks from the same sample are generally dependent, so passing both is not itself a joint

95%

statement, nor can a finite test prove that a generator is correct.

The second central limit theorem follows from the central limit theorem for the sample variance:

n (S_{n}^{2} - v)

differs by a quantity tending to zero in probability from

n^{- 1/2} \sum_{i} [(X_{i} - m)^{2} - v]

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Past exam of the simulation and statistical programming course of the University of Oxford / 2021 / 3 / b / i / Solution 2026-10-25

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A numerical power series evaluation needs a stopping rule. For

r = ∣ z ∣ < 1

and integer

a \geq 0

, the absolute remainder after

N

terms satisfies the polylogarithm series remainder bound

Li_{a} (z) - k = 1 \sum N \frac{z ^{k}}{k ^{a}} \leq \frac{r ^{N + 1}}{( N + 1 ) ^{a} ( 1 - r )} .

(1)

Indeed

k^{- a} \leq (N + 1)^{- a}

for

k > N

, and the remaining powers sum as a geometric series. This also covers negative

z

. The code sums terms in vectors of at most 4096 entries, checking the bound after each block; there is no loop over individual terms.

polylog <- function(a, z, tol = 1e-10, max_terms = 1000000) {
  scalar <- function(v) {
    is.numeric(v) && !is.complex(v) && length(v) == 1L &&
      is.finite(v)
  }
  if (!scalar(a) || a < 0 || a != floor(a))
    stop("a must be a nonnegative integer")
  if (!scalar(z) || abs(z) >= 1)
    stop("z must be real with abs(z) < 1")
  if (!scalar(tol) || tol <= 0 || tol >= 1)
    stop("tol must lie strictly between 0 and 1")
  if (!scalar(max_terms) || max_terms < 1 ||
      max_terms != floor(max_terms))
    stop("max_terms must be a positive integer")
  if (z == 0) return(0)
  r <- abs(z)
  N <- 0
  ans <- 0
  repeat {
    if (N >= max_terms)
      stop("requested truncation tolerance not reached")
    B <- min(4096, max_terms - N)
    k <- seq.int(N + 1, N + B)
    ans <- ans + sum(z^k * exp(-a * log(k)))
    N <- N + B
    log_bound <- (N + 1) * log(r) - a * log(N + 1) - log1p(-r)
    if (log_bound <= log(tol)) return(ans)
  }
}

This accepts every positive integer order in the question, and also order zero if

0 \in N

is the convention adopted. The bound controls absolute truncation error; floating-point rounding is a separate source of error. Close to

∣ z ∣ = 1

, convergence can be slow. If the resource limit is reached, the function reports failure instead of silently claiming the requested accuracy. Useful exact checks are

Li_{1} (z) = - lo g (1 - z)

and

Li_{0} (z) = z / (1 - z)

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Past exam of the simulation and statistical programming course of the University of Oxford / 2023 / 2 / b / iii / Solution 2026-10-25

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For a measurable set

C

, (i) gives the joint acceptance probability

P (Y \in C, accept) = \int_{C} \frac{q ( y )}{Z _{q}} \frac{p ( y )}{M q ( y )} d y = \frac{1}{M Z _{q}} \int_{C} p (y) d y .

(1)

Divide by

Z_{p} / (M Z_{q})

from (ii). The resulting conditional probability is

P (Y \in C ∣ accept) = \frac{1}{Z _{p}} \int_{C} p (y) d y = \int_{C} p (y) d y .

(2)

Thus every accepted value has target probability density function

p

. Repeating rejected trials with fresh independent random variables does not change this law: the contribution from acceptance on trial

j

is multiplied by

(1 - Z_{p} / (M Z_{q}))^{j - 1}

, and summing that geometric series gives the same conditional law. Independent repetitions produce independent target draws.

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Past exam of the theoretical physics II course of the University of Cambridge / 2014 / Paper 1 / 4 / c / Solution 2026-10-20

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Set

q = e^{- β ℏ ω}

. The oscillator populations are

(1 - q) q^{n}

, and differentiating the geometric series gives

\overset{n}{ˉ} = q / (1 - q)

. The canonical oscillator coordinate variance follows from

2 \overset{n}{ˉ} + 1 = (1 + q) / (1 - q) = coth (β ℏ ω /2)

, and

⟨ x^{2} ⟩_{T} = \frac{ℏ}{2 mω} coth (\frac{ℏ ω}{2 k _{B} T}) .

(1)

It approaches

ℏ/ (2 mω)

at zero temperature and

k_{B} T / (m ω^{2})

at high temperature, agreeing respectively with ground-state fluctuations and classical equipartition.

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Past exam of the theoretical physics II course of the University of Cambridge / 2015 / Paper 1 / 4 / c / Solution 2026-10-20

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The stated annihilation conditions can be solved algebraically, even though (b) shows they are not ground-state conditions for the full printed Hamiltonian. Invert the canonical transformation:

A = a cosh θ + b^{†} sinh θ, B = b cosh θ + a^{†} sinh θ .

(1)

For

∣ ψ_{λ} ⟩ = N e^{λ a^{†} b^{†}} ∣0, 0 ⟩

, the commutators give

a ∣ ψ_{λ} ⟩ = λ b^{†} ∣ ψ_{λ} ⟩, b ∣ ψ_{λ} ⟩ = λ a^{†} ∣ ψ_{λ} ⟩ .

(2)

Thus both

A

and

B

annihilate it precisely when

λ = - tanh θ .

(3)

The normalized two-mode squeezed vacuum is

∣ ψ_{λ} ⟩ = 1 - ∣ λ ∣^{2} \sum_{n = 0}^{\infty} λ^{n} ∣ n ⟩_{x} ∣ n ⟩_{y}, ∣ λ ∣ < 1.

(4)

The factorial from the exponential cancels those generated by the two creation operators. Normalization follows from a geometric series; the problem does not require its value, but it makes the subsequent partial trace explicit.

For comparison, the actual ground wave function of the printed stable Hamiltonian is

ψ_{true} (x, y) = (\frac{m ^{2} ω _{+} ω _{-}}{π ^{2} ℏ ^{2}})^{1/4} exp [- \frac{m}{4ℏ} ((ω_{+} + ω_{-}) (x^{2} + y^{2}) + 2 (ω_{+} - ω_{-}) x y)] .

(5)

It follows directly from the two normal-mode Gaussian vacua. The proposed

∣ ψ_{λ} ⟩

is instead the exact ground state of the distinct pair-coupling Hamiltonian

ℏ ω (a^{†} a + b^{†} b + 1) + g (ab + a^{†} b^{†})

under its stability condition. Omitting the remaining terms of

k x y

silently would change the problem.

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Past exam of the theoretical physics major option course of the University of Oxford / 2017 / 3 / e / Solution 2026-10-25

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The rates in part d turn the generating-function equation into

\partial_{t} F = λ [(z - 1) + (z^{- 1} - 1)] z \partial_{z} F = λ (z - 1)^{2} \partial_{z} F, F (z, 0) = z .

(1)

For the distribution of a critical linear birth-death population, the method of characteristics gives

\overset{z}{˙} = - λ (z - 1)^{2}

and

\dot{F} = 0

. Setting

w = (1 - z)^{- 1}

gives

\overset{w}{˙} = - λ

, so the initial characteristic coordinate is

w_{0} = w + λ t

. Hence

F (z, t) = 1 - \frac{1 - z}{1 + λ t ( 1 - z )} .

(2)

It has the correct initial value and

F (1, t) = 1

. With

a = λ t

, rewrite the denominator as

(1 + a) [1 - a z / (1 + a)]

and expand its geometric series. The probabilities are

P_{0} (t) = \frac{a}{1 + a}, P_{n} (t) = \frac{a ^{n - 1}}{( 1 + a ) ^{n + 1}} (n \geq 1) .

(3)

t = 0

this is understood by continuity:

P_{1} = 1

and all other probabilities vanish. Summing the positive-state probabilities gives

1/ (1 + a)

, which together with

P_{0}

verifies normalization directly.

There is eventual extinction with probability one, because

P_{0} (t) \to 1

. Nevertheless

⟨ n ⟩ = F_{z} (1, t) = 1

at every finite time, while

Var (n) = 2 λ t

. There is no contradiction: the increasingly rare surviving populations have conditional mean

⟨ n ∣ n > 0 ⟩ = 1 + λ t

. This is the distinction between a constant ensemble mean and survival of a typical realization.

Figure 1.
Extinction probability and surviving-population distribution of the critical branching process
. **Details**.

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Past exam of the thermal and statistical physics course of the University of Cambridge / 2018 / Section B / B3 / Solution 2026-10-18

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Classical translational statistics. Write the number density as

n = N / V

and the thermal de Broglie wavelength as

λ = 2 π ℏ^{2} / (m k_{B} T)

. Quantum exchange is negligible when wave packets rarely overlap:

n λ^{3} ≪ 1 or n^{- 1/3} ≫ λ .

(1)

The equivalent state-counting statement is that the mean occupation of each thermally relevant one-particle state is much smaller than one. The equations in this question count one internal species; with

g

equally populated spin states the relevant density per species is

n / g

Let

z = e^{β μ}

be the dimensionless quantum gas fugacity, with the lowest one-particle energy chosen as zero. Both quantum distributions can be written

n_{η} (E) = \frac{1}{z ^{- 1} e ^{βE} - η} = \frac{z e ^{- βE}}{1 - ηz e ^{- βE}}, η = {+ 1 - 1 bosons, fermions .

(2)

z ≪ 1

, then

n_{η} (E) ≃ z e^{- βE} ≪ 1

for all

E \geq 0

, recovering Maxwell–Boltzmann statistics. Conversely, low occupation even at the spectrum's bottom requires

z / (1 - z) ≪ 1

for bosons or

z / (1 + z) ≪ 1

for fermions; either condition implies

z = e^{β μ} ≪ 1.

(3)

A single high-energy state can be sparsely occupied even in a degenerate gas, so the condition must hold across the relevant spectrum, not just for one arbitrarily chosen energy. Also, classical translational statistics does not imply that all internal excitation energies or other quantum effects can be ignored.

Grand partition function, density and pressure. Noninteracting particles have independent occupations of each one-particle state. Put

x_{k} = z e^{- β ε_{k}}

. Summing the allowed occupations gives

Ξ_{k} = {\sum_{r = 0}^{\infty} x_{k}^{r} = (1 - x_{k})^{- 1} 1 + x_{k} bosons, fermions, Ξ = \prod_{k} (1 - η x_{k})^{- η} .

(4)

Therefore

ln Ξ = - η \sum_{k} ln (1 - ηz e^{- β ε_{k}}) .

(5)

The grand potential is

Ω_{G} = - k_{B} T ln Ξ = - p V

for this homogeneous gas. Replacing the sum by

V \int d^{3} k / (2 π)^{3}

and using

ε_{k} = ℏ^{2} k^{2} / (2 m)

gives

βp = - η \int \frac{d ^{3} k}{( 2 π ) ^{3}} ln [1 - ηz e^{- β ℏ^{2} k^{2} / (2 m)}] .

(6)

The mean number is

N = z \partial_{z} ln Ξ

, giving

\frac{N}{V} = \int \frac{d ^{3} k}{( 2 π ) ^{3}} \frac{1}{z ^{- 1} e ^{β ℏ^{2} k^{2} / (2 m)} - η} .

(7)

For bosons, the grand sum requires

z < 1

when the ground energy is zero. A macroscopic condensate at the limiting

z \to 1

must be separated from the continuum integral in the condensed phase. The following small-

z

calculation is in the normal, nondegenerate regime, so that complication does not enter.

Density expansion and inversion. Expand the occupation as a geometric series:

n_{η} (E) = z e^{- βE} + η z^{2} e^{- 2 βE} + z^{3} e^{- 3 βE} + \dots .

(8)

The Gaussian wavevector integral for its

ℓ

th term is

\int \frac{d ^{3} k}{( 2 π ) ^{3}} e^{- ℓ β ℏ^{2} k^{2} / (2 m)} = (\frac{m}{2 π ℓ β ℏ ^{2}})^{3/2} = \frac{1}{λ ^{3} ℓ ^{3/2}} .

(9)

It follows that

n λ^{3} = z + \frac{η z ^{2}}{2 ^{3/2}} + O (z^{3}) .

(10)

For the inverse series, put

y = n λ^{3}

and try

z = y + c y^{2} + O (y^{3})

. Substitution gives

0 = (c + η / 2^{3/2}) y^{2} + O (y^{3})

, hence

z = n λ^{3} - \frac{η}{2 ^{3/2}} (n λ^{3})^{2} + O ((n λ^{3})^{3}) .

(11)

The leading equality

n λ^{3} ≃ z

directly connects the density and occupancy criteria for classical behavior.

Pressure and its first quantum correction. Expanding the logarithm in the pressure expression gives one extra factor

1/ ℓ

compared with the density series:

βp λ^{3} = z + \frac{η z ^{2}}{2 ^{5/2}} + O (z^{3}) .

(12)

Insert the inverse series. To second order in

y = n λ^{3}

βp λ^{3} = y - \frac{η y ^{2}}{2 ^{3/2}} + \frac{η y ^{2}}{2 ^{5/2}} + O (y^{3}) = y - \frac{η y ^{2}}{2 ^{5/2}} + O (y^{3}) .

(13)

Thus

p V = N k_{B} T [1 - \frac{η}{2 ^{5/2}} \frac{N λ ^{3}}{V} + O ((N λ^{3} / V)^{2})] .

(14)

The first quantum correction at fixed density and temperature is

Δ p = - \frac{η}{2 ^{5/2}} k_{B} T n^{2} λ^{3} .

(15)

It is negative for bosons and positive for fermions, and vanishes in the classical density limit.

Virial coefficient. Define the number-density virial expansion by

\frac{p}{n k _{B} T} = 1 + B_{2} (T) n + B_{3} (T) n^{2} + \dots = \sum_{j = 1}^{\infty} B_{j} (T) n^{j - 1}, B_{1} = 1.

(16)

The coefficient

B_{j}

has units

volume^{j - 1}

in this convention. Comparing terms gives

B_{2}^{ex} (T) = - \frac{η}{2 ^{5/2}} λ^{3} = - \frac{η}{2 ^{5/2}} (\frac{2 π ℏ ^{2}}{m k _{B} T})^{3/2} .

(17)

This is the exchange contribution to the second virial coefficient. It arises without a pair-interaction potential; the underlying gas is noninteracting. A molar-density convention would multiply this

B_{2}

by the Avogadro constant. For

g

equally populated internal species, exchange occurs within each species and the coefficient in terms of total density is smaller by a factor

g

Why the two signs differ. Bosonic exchange symmetry enhances the probability of nearby identical bosons, whereas fermionic exchange symmetry suppresses nearby identical fermions in the same internal state. In the dilute regime, the normalized equal-time pair detection correlation is

g^{(2)} (r) = 1 + η e^{- 2 π r^{2} / λ^{2}} .

(18)

The exchange part follows by squaring the Gaussian one-particle coherence factor

e^{- π r^{2} / λ^{2}}

. At zero separation this gives

g^{(2)} (0) = 2

for bosons and

0

for one-component fermions. Integrating the exchange pair weight provides a direct coefficient check:

B_{2}^{ex} = - \frac{1}{2} \int [g^{(2)} (r) - 1] d^{3} r = - \frac{η}{2} (\frac{λ ^{2}}{2})^{3/2} = - \frac{η λ ^{3}}{2 ^{5/2}} .

(19)

Bunching therefore acts like an effective statistical attraction in the dilute equation of state, lowering the pressure needed at fixed density. Exclusion acts like an effective statistical repulsion, raising it. Neither description asserts an actual interparticle force. Different internal fermion species do not exclude each other from a common spatial location in the same way, explaining the importance of the species convention.

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