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. Thus its second derivative is continuous and piecewise of polynomial degree at most one: $S^{''}$ is a linear spline on the same spline knots. Its third derivative is constant on each open interval and can jump across the spline knots; its fourth derivative vanishes inside each open interval.

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

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Put

E = T - S

. Matching interpolation data and endpoint derivatives gives

E (x_{i}) = 0 (0 \leq i \leq n), E^{'} (x_{0}) = E^{'} (x_{n}) = 0.

(1)

The endpoint slope constraints make

S

a clamped cubic spline. The decisive step in its minimum roughness property of a clamped cubic spline is the orthogonality identity

\int_{x_{0}}^{x_{n}} S^{''} (x) E^{''} (x) d x = 0.

(2)

Prove it separately on each interval. Two applications of integration by parts, using the vanishing fourth derivative of the cubic piece, give

\int_{x_{i - 1}}^{x_{i}} S^{''} E^{''} d x = [S^{''} E^{'}]_{x_{i - 1}}^{x_{i}} - [S^{'''} E]_{x_{i - 1}}^{x_{i}} = [S^{''} E^{'}]_{x_{i - 1}}^{x_{i}} .

(3)

The second boundary term is zero because

E

vanishes at both endpoints of this interval. Sum over

i

. At each interior spline knot, the two values of

S^{''} E^{'}

agree because both factors are continuous, so the contributions cancel. The remaining outer boundary terms vanish because

E^{'}

is zero at both outer endpoints. This proves the identity.

Expanding the square now gives the stronger exact equality

\int_{x_{0}}^{x_{n}} (T^{''})^{2} d x = \int_{x_{0}}^{x_{n}} (S^{''})^{2} d x + \int_{x_{0}}^{x_{n}} (T^{''} - S^{''})^{2} d x .

(4)

The last integral is nonnegative, proving the required inequality. It measures the excess second derivative roughness penalty of

T

over

S

The argument applies directly to twice continuously differentiable competitors. More generally it holds when

T^{'}

is an absolutely continuous function and

T^{''}

is square-integrable, the usual finite-energy class for this second derivative roughness penalty. Mere existence of pointwise second derivatives does not by itself guarantee that all displayed integrals exist; the variational statement is understood for admissible competitors with this regularity.

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Past exam of the numerical analysis course of the University of Oxford / 2021 / 3 / a / i / Solution 2026-10-25

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Here

y^{'} = x / (1 + y^{2})

and

y (0) = 1

. Since

x \geq 0

, the solution is nondecreasing, so

y (x) \geq 1

. In fact

0 \leq y^{'} \leq 1/2

[0, 1]

, which also bounds the solution and ensures it extends across this entire interval. The smooth right-hand side makes

y

twice continuously differentiable, with second derivative

y^{''} = \frac{1}{1 + y ^{2}} - \frac{2 x ^{2} y}{( 1 + y ^{2} ) ^{3}} = \frac{( 1 + y ^{2} ) ^{2} - 2 x ^{2} y}{( 1 + y ^{2} ) ^{3}} .

(1)

For

y \geq 1

and

0 \leq x \leq 1

(1 + y^{2})^{2} \geq 4 y^{2} \geq 2 y \geq 2 x^{2} y,

(2)

y^{''} \geq 0

. The subtracted term in the first expression is nonnegative, giving the useful bound

0 \leq y^{''} (x) \leq \frac{1}{1 + y ( x ) ^{2}} \leq \frac{1}{2} (0 \leq x \leq 1) .

(3)

The integral form of the Taylor remainder now gives

τ_{n} = \frac{y ( x _{n + 1} ) - y ( x _{n} )}{h} - f (x_{n}, y (x_{n})) = \frac{1}{h} \int_{x_{n}}^{x_{n + 1}} (x_{n + 1} - t) y^{''} (t) d t .

(4)

Therefore the normalized local truncation error satisfies the stronger estimate

0 \leq τ_{n} \leq \frac{h}{4} \leq \frac{h}{2} .

(5)

This proves the requested bound for

n = 1, \dots, N - 1

, and also for

n = 0

, which will be needed to start the global error estimate. For

N = 1

the stated range is empty, but the estimate at

n = 0

still applies.

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Past exam of the numerical solutions of partial differential equations course of the University of Oxford / 2023 / 1 / c / Solution 2026-10-25

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Use

u_{i} = u (x_{i})

for the sampled exact solution, and let

e_{i} = U_{i} - u_{i}

. The Dirichlet boundary conditions imply

e_{0} = e_{N} = 0

. The definition of the truncation error

φ

gives the exact discrete equation

- D_{x}^{+} D_{x}^{-} u_{i} + u_{i} + u_{i}^{3} = f_{i} - φ_{i} .

(1)

Subtracting it from the numerical equation therefore gives

- D_{x}^{+} D_{x}^{-} e_{i} + e_{i} + (U_{i}^{3} - u_{i}^{3}) = φ_{i} .

(2)

The sign on the right is positive because

φ_{i}

was defined as the difference approximation minus the exact second derivative. Multiply by

h e_{i}

, sum over the interior nodes and apply discrete summation by parts:

∥ D_{x}^{-} e] ∥_{h}^{2} + ∥ e ∥_{h}^{2} + h \sum_{i = 1}^{N - 1} (U_{i}^{3} - u_{i}^{3}) e_{i} = (φ, e)_{h} .

(3)

Again

(U_{i}^{3} - u_{i}^{3}) e_{i} = e_{i}^{2} (U_{i}^{2} + U_{i} u_{i} + u_{i}^{2}) \geq 0

. By Young's inequality for products,

(φ, e)_{h} \leq \frac{1}{2} ∥ φ ∥_{h}^{2} + \frac{1}{2} ∥ e ∥_{h}^{2}

. Hence

∥ D_{x}^{-} U - D_{x}^{-} u] ∥_{h}^{2} + \frac{1}{2} ∥ U - u ∥_{h}^{2} \leq \frac{1}{2} ∥ φ ∥_{h}^{2} .

(4)

No fourth derivative is needed for this step: the assumed twice continuous differentiability suffices to define

φ

and obtain the discrete energy estimate.

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Past exam of the optimal transport and partial differential equations course of the University of Oxford / 2025 / 2 / b / Solution 2026-10-25

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Set

q = y - x

and

z_{t} = x + tq

. The definition of pushforward measure gives the confinement energy

V [μ_{t}] = \int V (z_{t}) d Π

. The chain rule and differentiation under the integral sign, whenever justified, give

\frac{d}{d t} V [μ_{t}] = \int \nabla V (z_{t}) \cdot q d Π, \frac{d ^{2}}{d t ^{2}} V [μ_{t}] = \int q^{T} D^{2} V (z_{t}) q d Π.

(1)

For example, bounded Hessian matrix is sufficient: it gives

∣\nabla V (z) ∣ \leq C (1 + ∣ z ∣)

, and the first two differentiated integrands are bounded by

C^{'} (1 + ∣ x ∣^{2} + ∣ y ∣^{2})

. The dominated convergence theorem applies because

Π

has finite second absolute moments. Compactly supported endpoints also suffice because all interpolating points lie in one compact set.

The printed growth assumption alone does not ensure a classical second derivative. Here is a counterexample within the stated assumptions. On the line take the bounded nonnegative smooth function

V (x) = 1 - cos (2 π x^{3}), p_{n} = c n^{- 4} (n \geq 1), c^{- 1} = \sum_{n \geq 1} n^{- 4},

(2)

and the probability measures

μ = \sum_{n} p_{n} δ_{n - 1/2}

ν = \sum_{n} p_{n} δ_{n + 1/2}

. Both have finite second absolute moments. Their translation transport plan is optimal: its cost is one, while every coupling has cost at least the square of its mean displacement, also one. At

t_{0} = 1/2

z_{t_{0}} = n

, so

F (t) = V [μ_{t}]

satisfies

F (t_{0}) = 0

. Each summand is nonnegative, and the Fatou lemma gives

lim inf_{h \to 0} \frac{F ( t _{0} + h ) + F ( t _{0} - h ) - 2 F ( t _{0} )}{h ^{2}} \geq \sum_{n} p_{n} V^{''} (n) = 36 π^{2} c \sum_{n} 1 = \infty.

(3)

Thus the requested classical identity requires an additional integrability assumption on the Hessian matrix; quadratic growth controls the energy but not its second derivative.

For a confinement potential

V

that is a

λ

-strongly convex function, the derivative formula, where valid, yields

\frac{d ^{2}}{d t ^{2}} V [μ_{t}] \geq λ \int ∣ q ∣^{2} d Π = λ D^{2} .

(4)

The underlying displacement convexity conclusion is valid under the printed assumptions without that extra differentiation hypothesis. Apply convexity of

V (x) - λ ∣ x ∣^{2} /2

pointwise and integrate:

V [μ_{t}] \leq (1 - t) V [μ] + t V [ν] - \frac{λ}{2} t (1 - t) D^{2} .

(5)

The same argument on every subinterval makes

t \mapsto V [μ_{t}] - λ D^{2} t^{2} /2

convex, so the second derivative lower bound also holds in the sense of distributions.

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Past exam of the optimal transport and partial differential equations course of the University of Oxford / 2025 / 2 / c / Solution 2026-10-25

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Use an independent copy

(x^{'}, y^{'})

(x, y)

under the product measure

Π \otimes Π

. Write

q = y - x

q^{'} = y^{'} - x^{'}

, and

r_{t} = z_{t} - z_{t}^{'} = (1 - t) (x - x^{'}) + t (y - y^{'}), a = q - q^{'} .

(1)

Since the pushforward measure of

Π \otimes Π

under

(z_{t}, z_{t}^{'})

μ_{t} \otimes μ_{t}

, the interaction energy is

W [μ_{t}] = \frac{1}{2} \iint W (r_{t}) d Π (x, y) d Π (x^{'}, y^{'}) .

(2)

Two applications of the chain rule give

\frac{d ^{2}}{d t ^{2}} W [μ_{t}] = \frac{1}{2} \iint (q - q^{'})^{T} D^{2} W (z_{t} - z_{t}^{'}) (q - q^{'}) d Π d Π^{'} .

(3)

The factor

1/2

is retained. This classical identity requires differentiation under the integral sign to be justified, just as in part (b). Bounded Hessian matrix of

W

is sufficient, since

∣ q - q^{'} ∣^{2} \leq 2∣ q ∣^{2} + 2∣ q^{'} ∣^{2}

is integrable; compactly supported endpoints are another sufficient case. The printed quadratic-growth condition alone does not provide this domination.

The failure can occur for the interaction energy too. Take the even interaction potential

W (x) = 1 - cos (2 π x^{3})

and replace the endpoint probability measures in part (b) by

\frac{1}{2} δ_{0} + \frac{1}{2} μ

and

\frac{1}{2} δ_{0} + \frac{1}{2} ν

. Keep the origin fixed and translate every other atom by one. This coupling is increasing on the line, hence is optimal: its assigned values extend to a nondecreasing transport map, which is the gradient of a convex function. Differences between translated atoms remain integers, so their mutual interaction vanishes. Only interactions with the origin contribute, giving

W [μ_{t}] = F (t) /4

, with

F

from part (b). It therefore also lacks a finite classical second derivative at

t = 1/2

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Past exam of the options II course of the University of Oxford / 2005 / J / 2 / Solution 2026-10-31

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Set

u = r ψ

, choose

α > 0

for a negative bound-state energy, and write

κ = 2 m K / ℏ^{2}

. Multiplying the radial equation by

r

gives

u^{''} + [- α^{2} + \frac{κ}{r} - \frac{ℓ ( ℓ + 1 )}{r ^{2}}] u = 0.

(1)

For

u = e^{- α r} f

, its second derivative is

e^{- α r} (f^{''} - 2 α f^{'} + α^{2} f)

. Substitution cancels the two

α^{2}

terms and gives

f^{''} - 2 α f^{'} - \frac{ℓ ( ℓ + 1 )}{r ^{2}} f + \frac{κ}{r} f = 0.

(2)

For the physical hydrogen problem, the angular eigenvalue has

ℓ

a nonnegative integer. Regularity at the origin selects the Frobenius exponent

u \sim r^{ℓ + 1}

, rather than the singular branch. Put

f = r^{ℓ + 1} \sum_{j = 0}^{\infty} c_{j} r^{j} .

(3)

The equation for the series factor

v = f / r^{ℓ + 1}

r v^{''} + (2 ℓ + 2 - 2 α r) v^{'} + (κ - 2 α (ℓ + 1)) v = 0

, and coefficient comparison yields

c_{j + 1} = \frac{2 α ( j + ℓ + 1 ) - κ}{( j + 1 ) ( j + 2 ℓ + 2 )} c_{j} .

(4)

The regular confluent-hypergeometric solution generically grows like

e^{2 α r}

in this series factor; after the extracted

e^{- α r}

it is not normalizable. A bound state requires termination at some degree

q \geq 0

, namely

2 α (q + ℓ + 1) = κ

. Thus

n = q + ℓ + 1

is an integer greater than

ℓ

, and

α = \frac{m K}{ℏ ^{2} n}, E_{n} = - \frac{m K ^{2}}{2 ℏ ^{2} n ^{2}} .

(5)

Regularity, normalizability and the integer angular index are the physical conditions selecting this spectrum; the differential equation without those conditions would admit many other solutions.

For

n = 1

, necessarily

ℓ = 0

and

q = 0

. With

a_{0} = ℏ^{2} / (m K)

f_{1} (r) = C r, ψ_{1} (r) = C e^{- r / a_{0}} .

(6)

A normalized spatial ground state is

Ψ_{1} = (π a_{0}^{3})^{- 1/2} e^{- r / a_{0}}

. In spherical coordinates the radial probability has weight

r^{2} d r

. Constants and the angular integral cancel in the tail ratio. The supplied integral with

λ = 2/ a_{0}

gives

Pr (r > a_{0}) = \frac{\int _{a_{0}}^{\infty} r ^{2} e ^{- 2 r / a_{0}} d r}{\int _{0}^{\infty} r ^{2} e ^{- 2 r / a_{0}} d r} = e^{- 2} (1 + 2 + 2) = 5 e^{- 2} .

(7)

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Past exam of the physics A course of the University of Cambridge / 2015 / Paper 2 / Section B / B7 / Solution 2026-10-20

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For a monochromatic wave, a constant phase

k x - ω t

moves at phase velocity

v_{p} = ω / k

. For a narrow wave packet around

k

, expansion

ω (k + Δ k) ≃ ω (k) + (d ω / d k) Δ k

shows that its envelope moves at group velocity

v_{g} = d ω / d k

. The envelope interpretation requires a sufficiently narrow spectrum and weak distortion; it is not a general identification with a causal signal front. Since

ω = k v_{p}

and

λ = 2 π / k

v_{g} = v_{p} + k \frac{d v _{p}}{d k} = v_{p} - λ \frac{d v _{p}}{d λ} .

(1)

For the gravity-capillary wave, put

s = σ / ρ

and take

k > 0

. Direct differentiation gives

v_{p} = g / k + s k, v_{g} = \frac{g + 3 s k ^{2}}{2 g k + s k ^{3}}, \frac{v _{g}}{v _{p}} = \frac{g + 3 s k ^{2}}{2 ( g + s k ^{2} )} = 1 - \frac{g / k ^{2} - s}{2 ( g / k ^{2} + s )} .

(2)

Thus

v_{g} = v_{p}

when

s k_{*}^{2} = g

, or

λ^{*} = 2 π \frac{σ}{ρ g} = 2.89 \times 1 0^{- 2} m .

(3)

At this wavelength the gravity and surface-tension restoring terms contribute equally. Also

d (v_{p}^{2}) / d k = - g / k^{2} + s = 0

with positive second derivative: this is the minimum phase speed of a gravity-capillary wave,

v_{p, m i n} = (4 g σ / ρ)^{1/4} = 0.214 m s^{- 1}

. Locally the carrier crests and narrow envelope travel together. The medium has not become nondispersive:

d^{2} ω / d k^{2} = k_{*} d^{2} v_{p} / d k^{2} \neq = 0

there, so a finite wave packet can still spread. In the gravity-dominated limit

v_{g} / v_{p} \to 1/2

; in the capillary-dominated limit it tends to

3/2

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Past exam of the Probability, statistics and options II course of the University of Oxford / 2009 / H / H2 / Solution 2026-10-31

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The lower mass has coordinates

(x + l sin θ, - l cos θ)

. Its squared speed is

\overset{x}{˙}^{2} + 2 l cos θ \overset{x}{˙} \dot{θ} + l^{2} \dot{θ}^{2}

. Adding the bead's kinetic energy gives the kinetic part of the supplied Lagrangian. Equal spring prestresses give spring energy

k x^{2}

up to a constant: the two squared extensions are proportional to

(a + x - ℓ_{0})^{2} + (a - x - ℓ_{0})^{2}

. The gravitational potential is

- m_{2} g l cos θ

. Thus

L = T - V

has exactly the stated form, with constant energy offsets discarded.

Since

L

has no explicit time dependence, the conserved energy is

E = \frac{1}{2} (m_{1} + m_{2}) \overset{x}{˙}^{2} + m_{2} l cos θ \overset{x}{˙} \dot{θ} + \frac{1}{2} m_{2} l^{2} \dot{θ}^{2} + k x^{2} - m_{2} g l cos θ .

(1)

The two Euler–Lagrange equations are

(m_{1} + m_{2}) \overset{x}{¨} + m_{2} l cos θ \ddot{θ} - m_{2} l sin θ \dot{θ}^{2} + 2 k x m_{2} l cos θ \overset{x}{¨} + m_{2} l^{2} \ddot{θ} + m_{2} g l sin θ = 0, = 0.

(2)

At equilibrium,

x = 0

and

sin θ = 0

. There are two geometrically distinct positions,

(x, θ) = (0, 0), (0, π)

modulo

2 π

. The downward one has positive potential Hessian

diag (2 k, m_{2} g l)

and is stable; the inverted one has a negative angular curvature and is unstable.

Keeping only quadratic terms about the downward equilibrium gives

L_{2} = \frac{1}{2} (m_{1} + m_{2}) \overset{x}{˙}^{2} + m_{2} l \overset{x}{˙} \dot{θ} + \frac{1}{2} m_{2} l^{2} \dot{θ}^{2} - k x^{2} - \frac{1}{2} m_{2} g l θ^{2} .

(3)

For unit parameters, the mass and stiffness matrices are

M = (2111)

and

K = (2001)

. The equation

det (K - ω^{2} M) = 0

ω^{4} - 4 ω^{2} + 2 = 0

. Therefore the normal angular frequencies are

ω_{-} = 2 - 2, ω_{+} = 2 + 2 .

(4)

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Past exam of the Probability, statistics and options II course of the University of Oxford / 2010 / U / U2 / b / iii / Solution 2026-10-31

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Here

f^{'} (x) = 4 x^{3} - 12 x^{2} + 4 x + 12

and

f^{''} (x) = 12 x^{2} - 24 x + 4

. This implementation uses the undamped Newton step, checks a stationary point's curvature, and reports failure on a two-cycle or exhausted iteration limit:

R
newton_min <- function(x0, tol = 1e-10, maxit = 100L) {
  fp <- function(x) 4*x^3 - 12*x^2 + 4*x + 12
  fpp <- function(x) 12*x^2 - 24*x + 4
  x <- x0
  trace <- x
  for (k in seq_len(maxit)) {
    g <- fp(x)
    h <- fpp(x)
    if (abs(g) <= tol) {
      return(list(x = x, converged = h > 0,
                  reason = if (h > 0) "minimum" else "not a strict minimum",
                  trace = trace))
    }
    if (!is.finite(h) || abs(h) <= .Machine$double.eps)
      stop("Newton step has zero or invalid curvature")
    next_x <- x - g/h
    if (!is.finite(next_x)) stop("Nonfinite Newton iterate")
    trace <- c(trace, next_x)
    n <- length(trace)
    if (n >= 3L && abs(trace[n] - trace[n - 2L]) <= tol*(1 + abs(next_x)))
      return(list(x = next_x, converged = FALSE,
                  reason = "two-cycle", trace = trace))
    x <- next_x
  }
  list(x = x, converged = FALSE, reason = "iteration limit", trace = trace)
}

A sufficiently close start near

- 0.7692923542

converges to the minimum. The derivative cubic has exactly one real root: after

x = 1 + t

it is proportional to

t^{3} - 2 t + 2

, whose local maximum and minimum are both positive, so it crosses zero only once, on the left. Since the quartic tends to

+ \infty

in both directions and its curvature at that stationary point is positive, this is the unique global minimum. The function reports convergence rather than assuming that every starting point must reach it.

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Past exam of the Probability, statistics and options II course of the University of Oxford / 2010 / U / U2 / b / iv / Solution 2026-10-31

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x = 1

, the derivative and curvature are

f^{'} (1) = 8

f^{''} (1) = - 8

, so the next iterate is

2

. At

x = 2

, they are

f^{'} (2) = 4

f^{''} (2) = 4

, so the next iterate is

1

. Hence

1 ⟼ 2 ⟼ 1 ⟼ 2 ⟼ \dots .

(1)

The undamped algorithm does not converge to a minimum; it has an exact two-cycle. The implementation in (iii) detects this after returning to one and reports failure. This example illustrates why the local convergence result in (ii) does not justify global success.

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Past exam of the quantum condensed matter course of the University of Cambridge / 2012 / 1 / c / Solution Created 2026-10-18 Updated 2026-10-20

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The band effective mass is determined by the second derivative of the dispersion relation:

\frac{1}{m ^{*} ( k )} = \frac{1}{ℏ ^{2}} \frac{d ^{2} E}{d k ^{2}} .

(1)

With the stated sign convention and

t > 0

, the band minimum is at

k = 0

. Expanding the cosine, or differentiating twice, gives

E (k) = E_{0} - 2 t + t a^{2} k^{2} + O (k^{4}), \frac{d ^{2} E}{d k ^{2}}_{0} = 2 t a^{2} .

(2)

Comparison with a free electron's kinetic energy

ℏ^{2} k^{2} / (2 m_{e})

therefore requires

t = \frac{ℏ ^{2}}{2 m _{e} a ^{2}} = 6.78 \times 1 0^{- 20} J = 0.423 eV .

(3)

The hopping integral has units of energy. A negative

t

would move the minimum to the edge of the Brillouin zone and give the same positive bottom-of-band effective mass with

∣ t ∣

in this expression.

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Past exam of the quantum condensed matter course of the University of Cambridge / 2016 / 3 / Solution Created 2026-10-18 Updated 2026-10-20

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mass as band curvature. In a periodic potential, a weakly forced electron is usefully described by a wave packet confined to one energy band. Its group velocity is

v_{i} = \frac{1}{ℏ} \frac{\partial ε _{n}}{\partial k _{i}}, ℏ \dot{k}_{j} = F_{j} .

(1)

Differentiating the velocity gives

\overset{v}{˙}_{i} = \sum_{j} (m^{*- 1})_{ij} F_{j}, (m^{*- 1})_{ij} = \frac{1}{ℏ ^{2}} \frac{\partial ^{2} ε _{n}}{\partial k _{i} \partial k _{j}} .

(2)

This effective-mass tensor describes the band response, not a change of the electron's bare rest mass. Along a principal direction, a steep upward curvature gives a small positive mass and a flatter curvature a larger one. Near a band maximum the electron curvature and effective mass are negative; the corresponding missing-electron description uses positive electron hole masses. Anisotropy produces distinct principal masses, and strong nonparabolicity makes them momentum dependent. The single-band semiclassical description assumes fields weak and slowly varying enough to avoid substantial interband transitions.

Lattice symmetry and Bloch's theorem. Use the active spatial translation operator

T_{a} ψ (r) = ψ (r - a)

. A lattice-periodic scalar potential satisfies

V (r + a) = V (r)

, while the kinetic term is translation invariant. Therefore

[H, T_{a}] = 0

. The lattice translations also commute with one another, so they can be diagonalized simultaneously with

H

, including within a degenerate energy subspace.

For a common eigenstate, write

T_{a} ψ = t (a) ψ

. Unitarity gives

∣ t (a) ∣ = 1

, and the group law gives

t (a + b) = t (a) t (b)

. On primitive Bravais vectors

a_{j}

, choose

t (a_{j}) = e^{- i k \cdot a_{j}}

. Every lattice vector is an integer combination of these generators, hence

T_{a} ψ_{n k} = e^{- i k \cdot a} ψ_{n k}, ψ_{n k} (r + a) = e^{i k \cdot a} ψ_{n k} (r) .

(3)

The vector

k

is defined modulo a reciprocal lattice vector. The band index labels the different energy eigenstates with the same translation character. Define

u_{n k} (r) = e^{- i k \cdot r} ψ_{n k} (r)

. The phase factors cancel under a lattice translation, proving Bloch's theorem in its usual form:

ψ_{n k} (r) = e^{i k \cdot r} u_{n k} (r), u_{n k} (r + a) = u_{n k} (r) .

(4)

This is the Translation-character proof of Bloch theorem; the commutator guarantees a basis of such states, even when an arbitrary superposition of different translation characters is not itself a translation eigenstate.

Periodic Hamiltonian. Acting once with momentum gives

p (e^{i k \cdot r} u) = e^{i k \cdot r} (p + ℏ k) u

. Applying it again and cancelling the common plane-wave factor yields

H (k) u_{n k} = [\frac{( p + ℏ k ) ^{2}}{2 m} + V (r)] u_{n k} = ε_{n} (k) u_{n k} .

(5)

The Bloch Hamiltonian acts on lattice-periodic functions in a unit cell. Expanding its square separates a zero-momentum Hamiltonian, a term linear in

k

and an identity term quadratic in

k

Expansion about the zone centre. Let

∣ n ⟩ = u_{n 0}

E_{n} = ε_{n} (0)

, and normalize all these states using a unit-cell inner product. For the real scalar Hamiltonian in the question, a nondegenerate zone-centre eigenfunction can be chosen real. Then

⟨ n ∣ p ∣ n ⟩ = - \frac{i ℏ}{2} \int_{cell} \nabla (u_{n 0}^{2}) d^{3} r = 0,

(6)

because opposite unit-cell faces cancel. Equivalently, time-reversal symmetry forbids a nonzero electric current in this nondegenerate spinless state. This explains why no linear energy term appears; it should not simply be omitted without a symmetry or extremum argument.

The first-order energy shift is therefore

ℏ^{2} k^{2} / (2 m)

. In the second-order sum the quadratic term is proportional to the identity and has no off-diagonal matrix elements. Only the linear momentum vertex contributes. Second-order nondegenerate perturbation theory gives the momentum matrix-element expansion of band energy

ε_{n} (k) = E_{n} + \frac{ℏ ^{2} k ^{2}}{2 m} + \frac{ℏ ^{2}}{m ^{2}} l \neq = n \sum \frac{∣ ⟨ n ∣ k \cdot p ∣ l ⟩ ∣ ^{2}}{E _{n} - E _{l}} + O (∣ k ∣^{4}) .

(7)

Hermitian operator makes the two reciprocal matrix elements complex conjugates. All other periodic eigenstates at

k = 0

enter the sum. The remainder is even because time-reversal symmetry gives

ε_{n} (k) = ε_{n} (- k)

near this isolated state. At a degeneracy the denominators signal the need for degenerate perturbation theory instead.

mass ratio and the printed formula. Write

p_{n l}^{i} = ⟨ n ∣ p_{i} ∣ l ⟩

. Differentiating the quadratic energy yields

(m^{*- 1})_{ij} = \frac{δ _{ij}}{m} + \frac{2}{m ^{2}} l \neq = n \sum \frac{Re ( p _{n l}^{i} p _{l n}^{j} )}{E _{n} - E _{l}} .

(8)

For a specified unit direction

e

, the resulting scalar mass ratio is

\frac{m}{m _{e}^{*}} = 1 + \frac{2}{m} l \neq = n \sum \frac{∣ ⟨ n ∣ e \cdot p ∣ l ⟩ ∣ ^{2}}{E _{n} - E _{l}} .

(9)

In an isotropic three-dimensional band, averaging equivalent directions gives

\frac{m}{m ^{*}} = 1 + \frac{2}{3 m} l \neq = n \sum \frac{∣ ⟨ n ∣ p ∣ l ⟩ ∣ ^{2}}{E _{n} - E _{l}} .

(10)

Coupling to lower bands contributes positively to inverse mass; coupling to higher bands contributes negatively. Their signs and matrix elements determine the actual curvature.

The original PDF prints

m^{*} = ℏ^{2} [\nabla_{k}^{2} ε_{n}]^{- 1}

. Taken literally as a three-dimensional Laplacian, this misses a factor of three for an isotropic scalar mass. The correct identities are

\frac{\nabla _{k}^{2} ε _{n}}{ℏ ^{2}} = tr (m^{*- 1}), m^{*} = \frac{3 ℏ ^{2}}{\nabla _{k}^{2} ε _{n}} in three isotropic dimensions .

(11)

The printed expression would return

m /3

even for the free dispersion

ℏ^{2} k^{2} / (2 m)

. It is correct in one dimension, or if its second derivative is interpreted directionally. For completeness, its literal three-dimensional trace ratio would instead be

3 + (2/ m) \sum_{l \neq = n} ∣ ⟨ n ∣ p ∣ l ⟩ ∣^{2} / (E_{n} - E_{l})

; that is the trace of the dimensionless inverse-mass tensor, not the usual scalar

m / m^{*}

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Past exam of the short options paper shared by Prelims and Parts A and B of the physics course of the University of Oxford / 2017 / S1 / 1 / 1 / Solution 2026-10-25

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Complex differentiability gives

u_{x} = v_{y}

u_{y} = - v_{x}

. Differentiating once more, for continuously differentiable first derivatives with the needed second derivatives, yields

u_{xx} + u_{yy} = v_{y x} - v_{x y} = 0

and

v_{xx} + v_{yy} = - u_{y x} + u_{x y} = 0

. Thus both components satisfy Laplace's equation. The result is local on the domain of holomorphy; it makes no assertion across a branch cut.

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Past exam of the year 1 of the computer science course of the University of Oxford / 2021 / Discrete mathematics, continuous mathematics, and probability / 4 / e / Solution 2026-10-25

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For negative

x

, the mean value theorem gives

f^{'} (x) = 3 + x f^{''} (ξ_{x})

. In particular,

f^{'} (- 2) \leq - 1 < 0, f^{'} (- 1) > 0.

(1)

Since

f^{'}

is continuous and strictly increasing, it vanishes at exactly one point

x_{m} \in (- 2, - 1)

. The positive second derivative makes

f

a strictly convex function on the entire real line. Its derivative changes from negative to positive at

x_{m}

, so $x_{m}$ is the unique global minimum. One may also use

f (x) \geq x^{2} + 3 x - 5 \to \infty

to verify coercivity.

In the endpoint-bracketing form of golden-section search, begin with $[- 2, - 1]$ , a width-one interval known to contain this unique minimum, and place the two standard interior probes. The attainable position accuracy is different from that for a simple root. Around

x_{m}

, Taylor's theorem gives

f (x_{m} + h) - f (x_{m}) = \frac{1}{2} f^{''} (ξ) h^{2} .

(2)

Here the minimum value is nonzero and of order one: the lower quadratic bound gives

f (x_{m}) \geq - 29/4

, while

f (x_{m}) \leq f (- 1) < - 13/2

. Ordinary relative round-off error in these function values therefore hides changes of order

ϵ

, once

h^{2}

has that size. The function-value resolution near a nondegenerate minimum is consequently of order

∣ h ∣ \sim ϵ = 2^{- 20} \approx 9.54 \times 1 0^{- 7} .

(3)

Each golden-section search contraction multiplies the bracket width by

ρ = (5 - 1) /2

. Using

2^{- 20}

as the nominal target width gives

N \geq \frac{20 l o g 2}{- l o g ρ} = 28.81 \dots, N = 29 contractions .

(4)

Unknown evaluation-error constants can change the actual stopping point;

29

is the standard scale-based count. Requiring

2^{- 40}

position accuracy from these function-value comparisons would disregard the quadratic flatness of the minimum. The same precision effect is documented for one-dimensional numerical minimization.^[ref]

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Past exam of the year 1 of the computer science course of the University of Oxford / 2021 / Discrete mathematics, continuous mathematics, and probability / 7 / c / ii / Solution 2026-10-25

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Insert the Poisson distribution probability generating function into the preceding bound:

P (X \geq x) \leq exp (λ (s - 1) - x lo g s), s > 1.

(1)

For fixed

x > λ

, differentiate the exponent with respect to

s

. Its derivative is

λ - x / s

, and its second derivative is

x / s^{2} > 0

, so its unique minimum occurs at

s = x / λ > 1

. At that value,

λ (s - 1) - x lo g s = x - λ - x lo g (x / λ) .

(2)

Consequently

P (X \geq x) \leq e^{- λ} (\frac{e λ}{x})^{x} .

(3)

This Chernoff bound remains valid for noninteger positive thresholds

x > λ

; the event is then the same as exceeding or equalling its integer ceiling.

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Past exam of the year 1 of the computer science course of the University of Oxford / 2022 / Discrete mathematics, continuous mathematics, and probability / Continuous mathematics / 4 / a / ii / Solution 2026-10-25

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Expand before taking the second derivative:

f (x) = e^{3 x} + 3 e^{x} + 3 e^{- x} + e^{- 3 x}, f^{''} (x) = 9 e^{3 x} + 3 e^{x} + 3 e^{- x} + 9 e^{- 3 x} > 0.

(1)

The second-derivative criterion for convexity proves that $f$ is strictly convex, hence convex.

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The Hessian criterion for convexity, in the usual twice-differentiable multivariate setting, is

g is convex on R^{d} ⟺ v^{T} H (g) (x) v \geq 0 for all x, v \in R^{d} .

(1)

Thus the Hessian matrix must be a positive semidefinite matrix everywhere. The criterion follows by applying the second-derivative criterion for convexity to each line restriction

t \mapsto g (x + t v)

, whose second derivative is

v^{T} H (g) (x + t v) v

. Continuous second partials are a sufficient regularity hypothesis for this line-restriction argument.

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Past exam of the year 1 of the computer science course of the University of Oxford / 2022 / Discrete mathematics, continuous mathematics and probability resit / 6 / b / Solution 2026-10-25

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Substitution gives

f (1) = - 6, f^{'} (1) = f^{''} (1) = f^{(3)} (1) = 0, f^{(4)} (1) = 24.

(1)

In particular,

1

is a stationary point. For

x

sufficiently close to

1

, Taylor's theorem with fifth-order remainder gives

f (x) = - 6 + (x - 1)^{4} + \frac{f ^{(5)} ( ξ )}{120} (x - 1)^{5},

(2)

where

ξ

lies between

1

and

x

. The fifth derivative is bounded near

1

, so the remainder is

O ((x - 1)^{5})

and is smaller in magnitude than half the positive quartic term for sufficiently small nonzero

x - 1

. Therefore $x = 1$ is a strict local minimum, despite the vanishing second derivative.

An exact algebraic check gives the stronger global conclusion:

f (x) + 6 = \frac{( x - 1 ) ^{4}}{x ^{2}} \geq 0 (x \neq = 0) .

(3)

Equality holds only at

x = 1

, so it is the unique global minimum on the actual domain.

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Pinsker's inequality Created 2026-09-24 Updated 2026-11-05

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With Total variation distance defined by

d_{TV} (P, Q) = sup_{A} ∣ P (A) - Q (A) ∣

and relative entropy computed using natural logarithms, Pinsker's inequality states

d_{TV} (P, Q) \leq \frac{1}{2} D (P ∥ Q) .

(1)

P

is not absolutely continuous with respect to

Q

, the relative entropy is infinite and the bound is vacuous. A small relative entropy therefore forces small discrepancies in probabilities of all events. For example,