Paper 1

words. Choose that fibre and, if necessary, discard surplus words. Because all retained words agree in the deleted coordinate, their mutual distances do not change. Hence one can always obtain

\overline{C} with guaranteed parameters [n - 1, ⌈ m /2 ⌉, d] .

(23)

Without discarding words, its size is that of the chosen fibre and its minimum distance is at least

d

For the final calculation, the given code is the whole space

F_{2}^{3}

, so its parity extension is the length-four even-weight binary code. In a binary symmetric channel, parity fails to notice a nonzero error exactly when an even number of bits flips. The possible error weights are therefore two and four. Their total probability is

(2 4) p^{2} (1 - p)^{2} + p^{4} = 6 p^{2} (1 - p)^{2} + p^{4} .

(24)

This excludes the weight-zero event, since it is not an undetected error.

Solved by gpt-5.6-sol high.

4F (Automata and Formal Languages)

Words: 209 Articles: 1

Solution

Words: 209

For the given effective enumeration of partial computable functions, the diagonal halting set is

K = {n \in N : f_{n, 1} (n) is defined} .

(25)

A many-one reduction

A \leq_{m} B

is a total computable function

g : N \to N

such that

x \in A ⟺ g (x) \in B .

(26)

The S-m-n theorem states that for every

m, k \geq 1

there is a total computable function

s_{k}^{m}

satisfying

f_{s_{k}^{m} (e, x_{1}, \dots, x_{m}), k} (y_{1}, \dots, y_{k}) = f_{e, m + k} (x_{1}, \dots, x_{m}, y_{1}, \dots, y_{k})

(27)

whenever either side is defined.

Suppose first that

X

is recursively enumerable. Choose a program that halts exactly on inputs in

X

, and let

e

index a two-variable program that, on

(x, y)

, ignores

y

and runs that program on

x

. By the S-m-n theorem,

g (x) = s_{1}^{1} (e, x)

(28)

is total computable and indexes the unary function

f_{g (x), 1} (y) = f_{e, 2} (x, y) .

(29)

Therefore

x \in X ⟺ f_{g (x), 1} (g (x)) is defined ⟺ g (x) \in K .

(30)

Thus

X \leq_{m} K

Conversely, suppose

X \leq_{m} K

through a total computable

g

. On input

x

, compute

g (x)

and simulate machine

g (x)

on its own code. This procedure halts exactly when

g (x) \in K

, equivalently exactly when

x \in X

. Hence

X

is recursively enumerable. We have proved the many-one completeness of the halting problem:

X is recursively enumerable ⟺ X \leq_{m} K .

(31)

Finally, use the stated fact that

0 \in / K

and define

S = {0} \cup {2 n + 1 : n \in K} .

(32)

Then

0 \in S ∖ K

, so

S \neq = K

. The total computable map

n \mapsto 2 n + 1

satisfies

n \in K ⟺ 2 n + 1 \in S,

(33)

and therefore

K \leq_{m} S

. This is a distinct representative of the halting many-one degree.

Solved by gpt-5.6-sol high.

5J (Statistical Modelling)

Words: 69 Articles: 1

Solution

Words: 69

Expand the exponent in the Inverse Gaussian distribution with unit shape parameter:

- \frac{( x - μ ) ^{2}}{2 μ ^{2} x} = - \frac{x}{2 μ ^{2}} + \frac{1}{μ} - \frac{1}{2 x} .

(34)

Consequently its probability density function can be written as

f (x; μ) = h (x) exp {η x - A (η)},

(35)

where

h (x) = \frac{1}{2 π x ^{3}} exp (- \frac{1}{2 x}), η = - \frac{1}{2 μ ^{2}} < 0, A (η) = - - 2 η .

(36)

Indeed,

- A (η) = - 2 η = 1/ μ

. This is the canonical form of a one-parameter exponential family, and its natural parameter is therefore

η = - \frac{1}{2 μ ^{2}} .

(37)

Here the natural statistic is

T (X) = X

. The exponential-family derivative identities give

E_{η} X = A^{'} (η) = (- 2 η)^{- 1/2} = μ

(38)

and

var_{η} X = A^{''} (η) = (- 2 η)^{- 3/2} = μ^{3} .

(39)

Thus

E X = μ, var X = μ^{3} .

(40)

Solved by gpt-5.6-sol high.

6E (Mathematical Biology)

Words: 327 Articles: 12

a

Words: 88 Articles: 4

i

Words: 30 Articles: 1

Solution

Words: 30

The per-capita net growth rate is the birth rate minus the death rate,

b e^{- a N} - d .

(41)

Multiplying by the population size gives the ordinary differential equation

\frac{d N}{d t} = N (b e^{- a N} - d) .

(42)

This is the exponential density-dependent birth model.

Solved by gpt-5.6-sol high.

ii

Words: 58 Articles: 1

Solution

Words: 58

At a positive equilibrium, division by

N

gives

b e^{- a N} = d .

(43)

Since

b > d

, this has the positive solution

N^{*} = \frac{1}{a} lo g \frac{b}{d} .

(44)

Write the vector field as

F (N) = N (b e^{- a N} - d)

. Its derivative is

F^{'} (N) = b e^{- a N} - d - ab N e^{- a N} .

(45)

N = N^{*}

, the first two terms cancel and

b e^{- a N^{*}} = d

, so

F^{'} (N^{*}) = - a d N^{*} < 0.

(46)

The one-dimensional linear stability analysis therefore shows that

N^{*}

is locally asymptotically stable.

Solved by gpt-5.6-sol high.

b

Words: 239 Articles: 6

i

Words: 85 Articles: 1

Solution

Words: 85

The susceptible equation may be written

\dot{S} = birth of susceptibles b e^{- a S} S - infection βS I - natural death d S .

(47)

The factor

b e^{- a S}

is a density-dependent per-capita birth rate. The law of mass action gives the infection incidence

βS I

, which transfers individuals out of the susceptible compartment.

Likewise,

\dot{I} = new infections βS I - natural death d I - disease-induced death δ I .

(48)

There is no birth term in the infective equation, so only susceptible individuals reproduce in this model. Infectives die at total per-capita rate

d + δ

. These terms define the susceptible-infective model with exponential density-dependent birth.

Solved by gpt-5.6-sol high.

ii

Words: 60 Articles: 1

Solution

Words: 60

The disease-free equilibrium is

(S, I) = (N^{*}, 0)

. For

F (S, I) = S (b e^{- a S} - β I - d), G (S, I) = I (βS - d - δ),

(49)

the Jacobian matrix there is

J (N^{*}, 0) = (- a d N^{*} 0 - β N^{*} β N^{*} - (d + δ)) .

(50)

Because this matrix is triangular, its eigenvalue are its diagonal entries:

- a d N^{*} < 0, β N^{*} - (d + δ) .

(51)

The second eigenvalue is positive exactly when

N^{*} > \frac{d + δ}{β} .

(52)

Under this condition the linear stability analysis gives an unstable disease-free equilibrium, agreeing with the epidemic invasion threshold.

Solved by gpt-5.6-sol high.

iii

Words: 94 Articles: 1

Solution

Words: 94

At an endemic equilibrium,

I^{*} > 0

. The equation

\dot{I} = 0

then gives

S^{*} = \frac{d + δ}{β} .

(53)

Since also

S^{*} > 0

, the equation

\dot{S} = 0

gives

β I^{*} + d = b e^{- a S^{*}}, I^{*} = \frac{b e ^{- a S^{*}} - d}{β} .

(54)

This infective population is positive precisely when

S^{*} < \frac{1}{a} lo g \frac{b}{d} = N^{*},

(55)

which is exactly the assumed instability condition for the disease-free equilibrium. Thus the endemic equilibrium of the susceptible-infective model with exponential birth exists.

At this equilibrium the relations above reduce the Jacobian matrix to

J (S^{*}, I^{*}) = (- a S^{*} b e^{- a S^{*}} β I^{*} - β S^{*} 0) .

(56)

Hence

tr J = - a S^{*} b e^{- a S^{*}} < 0, det J = β^{2} S^{*} I^{*} > 0.

(57)

By the trace-determinant stability criterion, both eigenvalues have negative real part. The linearization stability theorem therefore proves that the endemic equilibrium is locally asymptotically stable.

Solved by gpt-5.6-sol high.

7E (Further Complex Methods)

Words: 99 Articles: 1

Solution

Words: 99

Consider the meromorphic function

f (z) = \frac{e ^{i z}}{z ( z ^{2} - 1 )} .

(58)

Close the contour with a large semicircle in the upper half-plane, indenting above the three real pole

- 1, 0, 1

. The large-arc integral vanishes by Jordan's lemma, while the upper-half-plane indentation rule gives

PV \int_{- \infty}^{\infty} \frac{e ^{i x}}{x ( x ^{2} - 1 )} d x = iπ \sum_{a \in {- 1, 0, 1}} Res (f, a) .

(59)

The three residue are

Res (f, - 1) = \frac{e ^{- i}}{2}, Res (f, 0) = - 1, Res (f, 1) = \frac{e ^{i}}{2} .

(60)

Their sum is

cos 1 - 1

, so the residue theorem yields

PV \int_{- \infty}^{\infty} \frac{e ^{i x}}{x ( x ^{2} - 1 )} d x = iπ (cos 1 - 1) .

(61)

The real part of the integrand is an odd function, so its full-line principal value is zero. Its imaginary part,

\frac{s i n x}{x ( x ^{2} - 1 )},

(62)

is an even function. Taking imaginary parts and halving the full-line integral therefore gives

PV \int_{0}^{\infty} \frac{sin x}{x ( x ^{2} - 1 )} d x = \frac{π}{2} (cos 1 - 1) .

(63)

Solved by gpt-5.6-sol high.

8D (Classical Dynamics)

Words: 137 Articles: 4

a

Words: 59 Articles: 1

Solution

Words: 59

The kinetic energy of the two equal masses is

T = \frac{m}{2} (\overset{x}{˙}_{1}^{2} + \overset{x}{˙}_{2}^{2}) .

(64)

The extensions of the left, coupling, and right springs are respectively

x_{1}

x_{2} - x_{1}

, and

x_{2}

. Their potential energy is therefore

U = \frac{1}{2} k_{1} x_{1}^{2} + \frac{1}{2} k_{3} (x_{2} - x_{1})^{2} + \frac{1}{2} k_{2} x_{2}^{2} .

(65)

Thus the mechanical Lagrangian

L = T - U

has the requested quadratic form with the symmetric small-oscillation mass and stiffness matrices

(T_{ij}) = (m 0 0 m), (V_{ij}) = (k_{1} + k_{3} - k_{3} - k_{3} k_{2} + k_{3}) .

(66)

Solved by gpt-5.6-sol high.

b

Words: 78 Articles: 1

Solution

Words: 78

The Euler--Lagrange equations for the quadratic Lagrangian are

m \overset{x}{¨} + V x = 0.

(67)

For a normal mode

x (t) = a e^{iω t}

, they become the generalized eigenvalue problem for small oscillations

V a = m ω^{2} a .

(68)

Put

λ = m ω^{2} / k

. With the stated spring constants,

\frac{V}{k} = (1 + ε (1 + δ) - ε - ε 1 + ε (1 - δ)) = (1 + ε) I + ε (δ - 1 - 1 - δ) .

(69)

The final matrix has characteristic polynomial

ν^{2} - (1 + δ^{2}),

(70)

so its eigenvalue are

ν_{\pm} = \pm 1 + δ^{2}

. Hence

λ_{\pm} = 1 + ε \pm ε 1 + δ^{2} = 1 + ε (1 \pm 1 + δ^{2}) .

(71)

The two angular frequencies therefore satisfy

ω_{\pm}^{2} = λ_{\pm} \frac{k}{m}, λ_{\pm} = 1 + ε (1 \pm 1 + δ^{2}) .

(72)

This is the spectrum recorded by normal modes of two equal masses between three springs.

Solved by gpt-5.6-sol high.

9B (Cosmology)

Words: 258 Articles: 6

a

Words: 94 Articles: 1

Solution

Words: 94

In the homogeneous background,

v = 0

and the physical velocity is the Hubble flow

u = a H x = \overset{a}{˙} x .

(73)

The comoving coordinate

x

is held fixed by the time derivative in the fluid equation, so

\frac{\partial u}{\partial t} = \overset{a}{¨} x .

(74)

Moreover, spatial homogeneity gives

\nabla \overset{ˉ}{P} = 0

. The Euler equation therefore reduces to

\overset{ρ}{ˉ} a \overset{a}{¨} x = - \overset{ρ}{ˉ} \nabla \overset{ˉ}{Φ}, \nabla \overset{ˉ}{Φ} = - a \overset{a}{¨} x .

(75)

Taking the divergence in three spatial dimensions gives

\nabla^{2} \overset{ˉ}{Φ} = - 3 a \overset{a}{¨} .

(76)

Consistency with the Poisson equation,

\nabla^{2} \overset{ˉ}{Φ} = \frac{4 π G}{c ^{2}} \overset{ρ}{ˉ} a^{2},

(77)

then requires

\frac{a ¨}{a} = - \frac{4 π G}{3 c ^{2}} \overset{ρ}{ˉ} .

(78)

This is the pressureless, zero-cosmological-constant Raychaudhuri equation, obtained by the Newtonian fluid derivation of the Raychaudhuri equation.

Solved by gpt-5.6-sol high.

b

Words: 86 Articles: 1

Solution

Words: 86

The elementary Newtonian derivation of the Friedmann equation follows a test particle on the boundary of a finite expanding sphere. It therefore introduces a distinguished centre and can misleadingly suggest that all matter expands away from one origin.

The fluid derivation is local: the Euler equation and Poisson equation must be compatible in every comoving neighbourhood. It consequently describes a genuinely homogeneous infinite universe, in which every comoving observer sees the same Hubble flow, without selecting a physical centre. This is the improvement captured by the Newtonian fluid derivation of the Raychaudhuri equation.

Solved by gpt-5.6-sol high.

c

Words: 78 Articles: 1

Solution

Words: 78

Insert

ρ = \overset{ρ}{ˉ} + ϵ δ ρ, v = ϵ δ v

(79)

into the continuity equation. Its zeroth-order part is the homogeneous cosmological continuity equation

\dot{\overset{ρ}{ˉ}} + 3 H \overset{ρ}{ˉ} = 0.

(80)

Since

\overset{ρ}{ˉ}

is spatially homogeneous, the terms of first order in

ϵ

give

\frac{\partial δ ρ}{\partial t} + 3 Hδ ρ + \frac{ρ ˉ}{a} \nabla \cdot δ v = 0.

(81)

On the other hand, the quotient rule and the background equation imply

\frac{\partial}{\partial t} (\frac{δ ρ}{ρ ˉ}) = \frac{1}{ρ ˉ} \frac{\partial δ ρ}{\partial t} - \frac{δ ρ}{ρ ˉ ^{2}} \dot{\overset{ρ}{ˉ}} = \frac{1}{ρ ˉ} \frac{\partial δ ρ}{\partial t} + 3 H \frac{δ ρ}{ρ ˉ} .

(82)

Substitution of the first-order equation yields the linearized cosmological continuity equation

\frac{\partial}{\partial t} (\frac{δ ρ}{ρ ˉ}) = - \frac{1}{a} \nabla \cdot δ v .

(83)

The ratio

δ ρ / \overset{ρ}{ˉ}

is the density contrast.

Solved by gpt-5.6-sol high.

10D (Quantum Information and Computation)

Words: 219 Articles: 4

a

Words: 78 Articles: 1

Solution

Words: 78

With equal priors, the probability of a correct result from this quantum measurement is

P_{S} = \frac{1}{2} ∣ ⟨ b_{0} ∣ a_{0} ⟩ ∣^{2} + \frac{1}{2} ∣ ⟨ b_{1} ∣ a_{1} ⟩ ∣^{2} .

(84)

The two required amplitudes are

⟨ b_{0} ∣ a_{0} ⟩ = cos β, ⟨ b_{1} ∣ a_{1} ⟩ = \frac{c o s β - s i n β}{2} .

(85)

Therefore

P_{S} = \frac{1}{2} cos^{2} β + \frac{1}{4} (cos β - sin β)^{2} = \frac{1}{2} + \frac{1}{4} (cos 2 β - sin 2 β) = \frac{1}{2} + \frac{2}{4} cos (2 β + \frac{π}{4}) .

(86)

Its maximum occurs at

β = - π /8

and is

P_{S}^{m a x} = \frac{1}{2} + \frac{1}{2 2} = \frac{1}{2} (1 + \frac{1}{2}) .

(87)

Here

∣ ⟨ a_{0} ∣ a_{1} ⟩ ∣ = 1/ 2

, so

θ = π /4

and

sin θ = 1/ 2

. The stated Helstrom-Holevo bound is consequently exactly the same value, and the basis above is the Helstrom measurement for the zero and plus states.

Solved by gpt-5.6-sol high.

b

Words: 141 Articles: 1

Solution

Words: 141

The no-cloning theorem for two pure states states that, for distinct nonorthogonal states

∣ c_{0} ⟩

and

∣ c_{1} ⟩

, there is no unitary operator

U

and fixed blank state

∣0 ⟩

such that

U (∣ c_{j} ⟩ ∣0 ⟩) = ∣ c_{j} ⟩ ∣ c_{j} ⟩, j = 0, 1.

(88)

To derive this from state discrimination, let

s = ∣ ⟨ c_{0} ∣ c_{1} ⟩ ∣, 0 < s < 1,

(89)

and suppose such a cloner existed. Applying it repeatedly would produce

N

copies. The inner product of the two possible

N

-copy states has magnitude

s^{N}

, so their optimal success probability under the Helstrom-Holevo bound would be

P_{N} = \frac{1}{2} (1 + 1 - s^{2 N}) .

(90)

Already for

N = 2

P_{2} > P_{1} = \frac{1}{2} (1 + 1 - s^{2}),

(91)

because

0 < s < 1

implies

s^{4} < s^{2}

. But cloning followed by the two-copy Helstrom measurement for two pure states would itself be a quantum procedure applied to the original single state, contradicting the optimal one-copy bound

P_{1}

. Hence the assumed unitary cloner cannot exist.

Equivalently, clone-assisted asymptotic state discrimination would give

P_{N} \to 1

, although two nonorthogonal states are not perfectly distinguishable. Thus the Helstrom--Holevo theorem implies the no-cloning theorem.

Solved by gpt-5.6-sol high.

11K (Coding and Cryptography)

Words: 273 Articles: 1

Solution

Words: 273

Take every logarithm in base two. The information entropy of

X

H (X) = - i = 1 \sum N p_{i} lo g_{2} p_{i},

(92)

with

0 lo g_{2} 0 = 0

. If the codeword

c (μ_{i})

has length

s_{i}

, then its expected codeword length is

E (S) = i = 1 \sum N p_{i} s_{i} .

(93)

For any decipherable code, the Kraft inequality gives

K = \sum_{i = 1}^{N} 2^{- s_{i}} \leq 1.

(94)

Define the probability distribution

q_{i} = 2^{- s_{i}} / K

. By Gibbs inequality,

0 \leq \sum_{i} p_{i} lo g_{2} \frac{p _{i}}{q _{i}} = - H (X) + E (S) + lo g_{2} K .

(95)

Since

lo g_{2} K \leq 0

E (S) \geq H (X) - lo g_{2} K \geq H (X) .

(96)

Conversely, choose the integer lengths

s_{i} = ⌈ - lo g_{2} p_{i} ⌉ .

(97)

Then

2^{- s_{i}} \leq p_{i}

, so their Kraft sum is at most one and the converse part of Kraft's inequality supplies a binary prefix code. Moreover,

E (S) = \sum_{i} p_{i} ⌈ - lo g_{2} p_{i} ⌉ < \sum_{i} p_{i} (- lo g_{2} p_{i} + 1) = H (X) + 1.

(98)

Thus the minimum expected length

S^{*}

satisfies the Shannon noiseless coding theorem

H (X) \leq S^{*} < H (X) + 1 .

(99)

For a decipherable code with arbitrary lengths

s_{1}, \dots, s_{N}

, apply the lower bound to the uniform source

p_{i} = 1/ N

. Its entropy is

lo g_{2} N

and its expected length is

(s_{1} + \dots + s_{N}) / N

. Hence

N lo g_{2} N \leq s_{1} + \dots + s_{N},

(100)

the total length lower bound for a decipherable binary code.

Now consider the cumulative construction. Since

b_{j} - b_{i} = \sum_{k = i}^{j - 1} p_{k} \geq p_{i} \geq 2^{- s_{i}} (j > i),

(101)

b_{i}

and

b_{j}

cannot have the same first

s_{i}

binary digits: numbers sharing those digits lie in one half-open dyadic interval of length

2^{- s_{i}}

. Also,

p_{i} \geq p_{j}

implies

s_{i} \leq s_{j}

. Therefore no earlier codeword

b_{i}^{*}

is a prefix of a later one, and no later, weakly longer codeword can be a prefix of an earlier one. The Cumulative Shannon code is thus a prefix code, hence decipherable.

An optimal code minimizes expected codeword length over all decipherable codes for the specified source probabilities. The cumulative construction need not be optimal. For example, if

(p_{1}, p_{2}) = (0.6, 0.4),

(102)

then

(s_{1}, s_{2}) = (1, 2)

and the construction gives codewords

0

and

10

, with expected length

1.4

. The prefix code

0, 1

has expected length

1

, so the constructed code is not optimal. In general, Huffman coding produces an optimal prefix code.

Solved by gpt-5.6-sol high.

12F (Automata and Formal Languages)

Words: 352 Articles: 6

i

Words: 100 Articles: 1

Solution

Words: 100

A partial recursive function is obtained from the three kinds of initial function of recursion theory—
the zero function, successor function, and projection function—by finitely many applications of function composition in recursion theory, primitive recursion, and unbounded minimization. Explicitly, primitive recursion has the form

f (x, 0) = g (x), f (x, n + 1) = h (x, n, f (x, n)),

(103)

while minimization takes the least

y

for which

g (x, y) = 0

and is undefined if no such

y

is found.

For the function in part (i), define

s (0) = 1, s (n + 1) = 0.

(104)

The initial value

1 = S (Z (0))

and the identically zero recursion step are primitive recursive, so this is a primitive-recursive definition of the required zero test. It uses no minimization.

Solved by gpt-5.6-sol high.

ii

Words: 43 Articles: 1

Solution

Words: 43

Using the zero test

s

from part (i), define by primitive recursion

r (0) = 0, r (n + 1) = s (r (n)) .

(105)

Since

s (0) = 1

and

s (1) = 0

, the values alternate:

r (2 k) = 0, r (2 k + 1) = 1.

(106)

Thus

r

is the required parity function and, like

s

, is primitive recursive without minimization.

Solved by gpt-5.6-sol high.

iii

Words: 209 Articles: 1

Solution

Words: 209

Let

g (n, k) = r (n)

, where

r

is the total primitive-recursive parity function from part (ii), and apply unbounded minimization in

k

p (n) = μ k [g (n, k) = 0] .

(107)

n

is even, then

g (n, 0) = 0

and

p (n) = 0

. If

n

is odd, then

g (n, k) = 1

for every

k

, so the search never terminates and

p (n)

is undefined. Hence

p

is partial recursive. Because every function built without minimization is total,

p

cannot be defined without minimization; the functions

s

and

r

can.

It remains to characterize the functions built using only the initial functions and composition. They are exactly the functions

f (x_{1}, \dots, x_{k}) = c or f (x_{1}, \dots, x_{k}) = x_{i} + c

(108)

for some

c \in N

and some input coordinate

i

Indeed, the zero function is the first form, the successor function and each projection function have the second form, and composing functions of these forms preserves the classification: a constant outer function remains constant, while an outer function

y_{j} + c

selects one inner function and adds

c

. This proves necessity by structural induction. Conversely, applying the successor function

c

times to zero produces the constant

c

, and applying it

c

times to

π_{i}^{k}

produces

x_{i} + c

Every function in this class is directly computable: an algorithm either writes the fixed constant

c

, or copies the

i

th input and performs

c

successor steps.

Solved by gpt-5.6-sol high.

13J (Statistical Modelling)

Words: 596 Articles: 12

a

Words: 328 Articles: 6

i

Words: 105 Articles: 1

Solution

Words: 105

Let

Y_{g o}

be the count for treatment

g \in {C, L, S}

(109)

(Control, LD, or SD) and outcome

o \in {B, W}

(110)

(Better or Worse), and write

μ_{g o} = E Y_{g o}

. Both fits take the six cell counts to be independent Poisson variables.

The additive log-linear model fitted by fit1 is

Y_{g o} \sim Pois (μ_{g o}), lo g μ_{g o} = λ + α_{g} + β_{o},

(111)

with reference constraints

α_{C} = 0

and

β_{B} = 0

. Thus treatment changes the overall group count but not the relative frequencies of the two outcomes.

The interaction model fitted by fit2 is

Y_{g o} \sim Pois (μ_{g o}), lo g μ_{g o} = λ + α_{g} + β_{o} + γ_{g o},

(112)

where, under reference coding,

γ_{CB} = γ_{C W} = γ_{L B} = γ_{SB} = 0

(113)

and the two free interactions are

γ_{L W}

and

γ_{S W}

. This is the saturated log-linear model for the

3 \times 2

table.

Solved by gpt-5.6-sol high.

ii

Words: 116 Articles: 1

Solution

Words: 116

Let

N_{g} = Y_{g B} + Y_{g W}

(114)

and condition on

N_{g} = n_{g}

. The corresponding model for each treatment group is

(Y_{g B}, Y_{g W}) ∣ N_{g} = n_{g} \sim Mult (n_{g}; π_{g B}, π_{g W}) .

(115)

For fit1, the additive Poisson means imply

π_{g o} = \frac{e ^{β_{o}}}{e ^{β_{B}} + e ^{β_{W}}},

(116)

so the outcome probabilities are common to all three treatments. For fit2,

π_{g o} = \frac{e ^{β_{o} + γ_{g o}}}{\sum _{r \in {B, W}} e ^{β_{r} + γ_{g r}}},

(117)

so each treatment may have its own outcome probabilities.

The Poisson trick is Poisson-multinomial conditioning: independent Poisson cell counts, conditional on their total, are multinomial with probabilities proportional to their means. With free treatment main effects, profiling the Poisson likelihood over the group totals gives the multinomial likelihood up to a parameter-independent factor. The Poisson regressions therefore fit these multinomial models and give the same likelihood-ratio comparisons for the outcome parameters.

Solved by gpt-5.6-sol high.

iii

Words: 107 Articles: 1

Solution

Words: 107

The additive model has four free coefficients: an intercept, two treatment contrasts, and one outcome contrast. The interaction model adds

(3 - 1) (2 - 1) = 2

(118)

treatment-by-outcome interaction term, making six parameters for six cells. It is saturated, so the residual degrees of freedom fall from two to zero and the likelihood-ratio test has

6 - 4 = 2 degrees of freedom .

(119)

The deviance reduction is

44.48

, with p-value

2.194 \times 1 0^{- 10}

. The analysis of deviance for nested generalized linear models therefore gives overwhelming evidence against a common outcome distribution across the three groups. The observed proportions and the negative treatment--Worse interactions show that both doses reduce the chance of worsening relative to Control, with a larger reduction for LD.

Solved by gpt-5.6-sol high.

b

Words: 268 Articles: 4

i

Words: 101 Articles: 1

Solution

Words: 101

Because the second model is saturated, its fitted worsening probabilities are the observed proportions:

p_{C} = \frac{101}{5728 + 101} = 0.01733, p_{L} = \frac{3}{1364 + 3} = 0.002195, p_{S} = \frac{27}{4413 + 27} = 0.006081.

(120)

Thus the drug efficacy as a risk reduction is

E_{L} = 1 - \frac{p _{L}}{p _{C}} = 1 - 0.1266 = 87.3%,

(121)

and

E_{S} = 1 - \frac{p _{S}}{p _{C}} = 1 - 0.3509 = 64.9% .

(122)

These are reasonably similar to the published estimates of

90.0%

and

62.1%

The interaction coefficients provide the same quick comparison on the odds scale:

e^{- 2.08156} = 0.125 and e^{- 1.05847} = 0.347.

(123)

They are treatment-versus-control odds ratio. Since worsening is rare in every group, each odds ratio is close to its risk ratio, giving the approximate efficacies

87.5%

and

65.3%

Solved by gpt-5.6-sol high.

ii

Words: 167 Articles: 1

Solution

Words: 167

Equal LD and SD efficacy is equivalent to

p_{L} = p_{S},

(124)

because both efficacy definitions use the same control probability. In this two-outcome table that is also equivalent to equality of the LD and SD log-odds interactions,

H_{0} : γ_{L W} = γ_{S W} .

(125)

The summary gives the two estimates and their individual standard error, but a test of their difference needs

Var (γ_{L W} - γ_{S W}) = Var (γ_{L W}) + Var (γ_{S W}) - 2 Cov (γ_{L W}, γ_{S W}) .

(126)

The required covariance is absent from the displayed table, so the two separate coefficient p-values cannot test equality.

Instead, fit the equal-efficacy Poisson log-linear model. For example, create one indicator for a Worse outcome in either treated group:

r
data$treated_worse <- with(data, treatment != "Control" & outcome == "Worse")
fit_equal <- glm(count ~ treatment + outcome + treated_worse,
                 family = poisson, data = data)
anova(fit_equal, fit2, test = "LRT")

The reduced model has one common treated-versus-control outcome interaction, while retaining separate LD and SD main effects for their different group totals. The full model has two such interactions. Their analysis of deviance for nested generalized linear models is therefore a one-degree-of-freedom likelihood-ratio test of equal efficacy.

Solved by gpt-5.6-sol high.

14E (Further Complex Methods)

Words: 366 Articles: 8

a

Words: 45 Articles: 1

Solution

Words: 45

The identity theorem states that if two holomorphic functions on a connected open set

D

agree on a subset having an accumulation point in

D

, then they agree throughout

D

. In particular, agreement on any nonempty open subset is sufficient.

Solved by gpt-5.6-sol high.

b

Words: 80 Articles: 1

Solution

Words: 80

An analytic continuation of

f_{1}

from

D_{1}

D_{2}

is a holomorphic function

f_{2}

D_{2}

whose restriction to the nonempty overlap

D_{1} \cap D_{2}

agrees with

f_{1}

Suppose

f_{2}

and

f_{2}

are two such continuations. Both equal

f_{1}

D_{1} \cap D_{2}

, so

f_{2} = f_{2}

(127)

on that nonempty open set. Since

D_{2}

is connected, the identity theorem gives

f_{2} = f_{2} throughout D_{2} .

(128)

Thus analytic continuation to

D_{2}

is unique.

Solved by gpt-5.6-sol high.

c

Words: 123 Articles: 1

Solution

Words: 123

For

Im z > 0

, close the

t

-contour in the upper half-plane. Jordan lemma removes the large semicircle, and the only enclosed pole is the order-

n

pole at

t = z

. By the residue at a pole of order n,

Res_{t = z} \frac{e ^{i t}}{( t - z ) ^{n}} = \frac{1}{( n - 1 )!} \frac{d ^{n - 1}}{d t ^{n - 1}} e^{i t}_{t = z} = \frac{i ^{n - 1} e ^{i z}}{( n - 1 )!} .

(129)

The residue theorem therefore gives

F (z) = \frac{2 π i ^{n} e ^{i z}}{( n - 1 )!} .

(130)

To continue the integral itself, move the

t

-contour downward locally as

z

approaches and crosses the real axis, always keeping the pole above the contour. This analytic continuation by contour deformation defines

F (z) = \int_{C_{z}} \frac{e ^{i t}}{( t - z ) ^{n}} d t,

(131)

where

C_{z}

passes below

z

. Closing

C_{z}

upward continues to enclose the pole, even when

Im z \leq 0

, so

F (z) = \frac{2 π i ^{n} e ^{i z}}{( n - 1 )!}

(132)

there as well. The right-hand side is an entire function, so it is the unique analytic continuation of

F

to the whole complex plane.

Solved by gpt-5.6-sol high.

d

Words: 118 Articles: 1

Solution

Words: 118

Im z > 0

, the same upper-half-plane contour calculation gives

G (z) = \frac{2 π i ^{n} e ^{i z}}{( n - 1 )!} .

(133)

Im z < 0

, the pole lies below the real integration contour. Closing upward encloses no pole, so Jordan lemma and the Cauchy integral theorem give

G (z) = 0.

(134)

Hence at a real point

x

the upper and lower boundary values satisfy

G (x + i 0) - G (x - i 0) = \frac{2 π i ^{n} e ^{i x}}{( n - 1 )!} .

(135)

Thus crossing from the upper half-plane to the lower half-plane produces the negative of this jump.

An analytic continuation must be holomorphic, hence continuous, across every point where it is defined. The undeformed real-axis formula for

G

has the nonzero jump above, while the continuation constructed in part (c) retains

2 π i^{n} e^{i z} / (n - 1)!

below the axis. Therefore

G

cannot be the analytic continuation of

F

Solved by gpt-5.6-sol high.

15B (Cosmology)

Words: 345 Articles: 8

a

Words: 93 Articles: 1

Solution

Words: 93

Write the action density as

L = a^{3} [\frac{1}{2} \dot{ϕ}^{2} - \frac{c ^{2}}{2 a ^{2}} \nabla ϕ \cdot \nabla ϕ - V (ϕ)] .

(136)

Under a compactly supported variation

ϕ \mapsto ϕ + ϵη

δ S = \int d^{3} x d t [a^{3} \dot{ϕ} \overset{η}{˙} - a c^{2} \nabla ϕ \cdot \nabla η - a^{3} V^{'} (ϕ) η] .

(137)

Integrating the first term by parts in time and the second in space, and discarding the boundary terms, gives

δ S = \int d^{3} x d t [- \frac{\partial}{\partial t} (a^{3} \dot{ϕ}) + a c^{2} \nabla^{2} ϕ - a^{3} V^{'} (ϕ)] η .

(138)

The principle of stationary action requires this to vanish for every

η

. Since

\frac{\partial}{\partial t} (a^{3} \dot{ϕ}) = a^{3} (\ddot{ϕ} + 3 H \dot{ϕ}),

(139)

division by

- a^{3}

yields the field Euler-Lagrange equation

\ddot{ϕ} + 3 H \dot{ϕ} - \frac{c ^{2}}{a ^{2}} \nabla^{2} ϕ + \frac{d V}{d ϕ} = 0 .

(140)

This is the equation generated by the inflaton action in an expanding universe.

Solved by gpt-5.6-sol high.

b

Words: 49 Articles: 1

Solution

Words: 49

When

V

is approximately constant,

d V / d ϕ ≃ 0

, and exponential expansion gives the constant Hubble parameter

H = H_{inf}

. Take the spatial Fourier transform

ϕ (x, t) = \int \frac{d ^{3} k}{( 2 π ) ^{3}} ϕ_{k} (t) e^{i k \cdot x} .

(141)

Since

\nabla^{2} e^{i k \cdot x} = - k^{2} e^{i k \cdot x},

(142)

each independent inflationary scalar Fourier mode satisfies

\ddot{ϕ}_{k} + 3 H_{inf} \dot{ϕ}_{k} + \frac{c ^{2} k ^{2}}{a ^{2}} ϕ_{k} = 0 .

(143)

Solved by gpt-5.6-sol high.

c

Words: 86 Articles: 1

Solution

Words: 86

Conformal time is defined by

d τ = \frac{d t}{a ( t )} .

(144)

For

a = a_{0} e^{H_{inf} t}

, choose

τ \to 0

t \to + \infty

. Integration gives

τ = - \frac{e ^{- H_{inf} t}}{a _{0} H _{inf}} = - \frac{1}{a H _{inf}}, - \infty < τ < 0,

(145)

as recorded by Conformal time during de Sitter expansion.

Writing a prime for

d / d τ

, the relations

\frac{d}{d t} = \frac{1}{a} \frac{d}{d τ}, \frac{a ^{'}}{a} = - \frac{1}{τ}

(146)

turn the Fourier-mode equation into

ϕ_{k}^{''} + 2 \frac{a ^{'}}{a} ϕ_{k}^{'} + c^{2} k^{2} ϕ_{k} = 0.

(147)

Set

ϕ_{k} = a ϕ_{k} = - \frac{1}{H _{inf} τ} ϕ_{k} .

(148)

A direct substitution gives

ϕ_{k}^{''} + (c^{2} k^{2} - \frac{a ^{''}}{a}) ϕ_{k} = 0.

(149)

Here

a^{''} / a = 2/ τ^{2}

, so the Canonically rescaled de Sitter scalar mode obeys

\frac{d ^{2} ϕ _{k}}{d τ ^{2}} + (c^{2} k^{2} - \frac{2}{τ ^{2}}) ϕ_{k} = 0 .

(150)

Solved by gpt-5.6-sol high.

d

Words: 117 Articles: 1

Solution

Words: 117

In the far past,

∣ c k τ ∣ ≫ 1

, the term

2/ τ^{2}

is negligible and the rescaled mode is a quantum harmonic oscillator of angular frequency

c k

. The stated choice of its ground state is the Bunch-Davies vacuum.

At late times, after the mode is far outside the Hubble scale,

∣ c k τ ∣ ≪ 1

. Its variance becomes

P_{k} = \frac{ℏ H _{inf}^{2}}{2 c ^{3} k ^{3}} (1 + c^{2} k^{2} τ^{2}) ⟶ \frac{ℏ H _{inf}^{2}}{2 c ^{3} k ^{3}} .

(151)

Since

P_{k}

has dimensions of length cubed, the dimensionless power is proportional to

k^{3} P_{k}

. More precisely, with the standard isotropic Fourier convention,

Δ^{2} (k) = \frac{k ^{3} P _{k}}{2 π ^{2}} ⟶ \frac{ℏ H _{inf}^{2}}{4 π ^{2} c ^{3}},

(152)

which is independent of the wavenumber

k

. Thus equal logarithmic intervals in

k

carry equal late-time power. This is the sense in which inflation naturally generates a scale-invariant inflationary power spectrum.

Solved by gpt-5.6-sol high.

16G (Logic and Set Theory)

Words: 321 Articles: 6

a

Words: 71 Articles: 1

Solution

Words: 71

A set

S

is a deductively closed set of formulae when every consequence of

S

already belongs to it:

S ⊢ φ ⟹ φ \in S .

(153)

It is a consistent set of formulae when it does not prove a contradiction,

S ⊬ ⊥.

(154)

S

is inconsistent, there is a formal derivation of

⊥

from assumptions in

S

. By the finite character of formal proofs, that derivation mentions only finitely many assumptions

s_{1}, \dots, s_{m} \in S

. Hence

{s_{1}, \dots, s_{m}} ⊢ ⊥,

(155)

so this finite subset of

S

is already inconsistent.

Solved by gpt-5.6-sol high.

b

Words: 70 Articles: 1

Solution

Words: 70

Suppose the finitary set of formulae

S

is equivalent to the finite set

F = {f_{1}, \dots, f_{m}} .

(156)

For each

f_{j}

, the relation

S ⊢ f_{j}

has a finite proof, so it uses a finite set

R_{j} \subseteq S

of assumptions. Put

R = ⋃_{j = 1}^{m} R_{j} .

(157)

This is a finite subset of

S

, and

R ⊢ f_{j}

for every

j

, hence

R ⊢ F

. Since equivalence also gives

F ⊢ S

, transitivity of deduction yields

R ⊢ S .

(158)

Solved by gpt-5.6-sol high.

c

Words: 180 Articles: 1

Solution

Words: 180

If some

T_{i}

were inconsistent, it would prove every formula by the classical principle of explosion. Because

T_{i}

is a deductively closed set of formulae, it would then contain every formula and could have no proper extension. This contradicts

T_{i} ⊊ T_{i + 1}

. Thus every

T_{i}

is consistent.

Let

T = ⋃_{i = 0}^{\infty} T_{i} .

(159)

T

were inconsistent, the finite character of formal proofs would give a finite inconsistent subset

R \subseteq T

. The chain is increasing, so all finitely many members of

R

lie in some common

T_{N}

. Then

T_{N}

would be inconsistent, a contradiction. Hence

T

is consistent.

T ⊢ φ

, a proof again uses only finitely many assumptions from

T

, all lying in some

T_{N}

. Thus

T_{N} ⊢ φ .

(160)

Since

T_{N}

is deductively closed,

φ \in T_{N} \subseteq T

. Therefore

T

is deductively closed.

Finally, suppose that

T

were finitary. Part (b) would provide a finite

R \subseteq T

such that

R ⊢ T

. Choose

N

with

R \subseteq T_{N}

. Then

T_{N} ⊢ T

; since

T_{N}

is deductively closed, this implies

T \subseteq T_{N}

. But

T_{N} ⊊ T_{N + 1} \subseteq T,

(161)

a contradiction. Consequently

T

is not finitary, proving the increasing union of deductively closed sets.

Solved by gpt-5.6-sol high.

17G (Graph Theory)

Words: 313 Articles: 6

a

Words: 78 Articles: 1

Solution

Words: 78

The binomial random graph

G (n, p)

has vertex set

[n] = {1, \dots, n}

, and each unordered pair of distinct vertices is included as an edge independently with probability

p

Let

X_{t}

count copies of the complete graph

K_{t}

. Each

t

-element vertex set forms a copy with probability

p^{(2 t)}

, so linearity of expectation gives

E X_{t} = (t n) p^{(2 t)} .

(162)

The event

E_{t}

{X_{t} \geq 1}

. By the first moment method,

P (E_{t}) \leq E X_{t} \leq \frac{n ^{t}}{t !} p^{t (t - 1) /2} = \frac{1}{t !} (p n^{2/ (t - 1)})^{t (t - 1) /2} ⟶ 0.

(163)

This is the clique count in a binomial random graph.

Solved by gpt-5.6-sol high.

b

Words: 106 Articles: 1

Solution

Words: 106

Chebyshev inequality states that, for a random variable

Y

of finite variance and any

a > 0

P (∣ Y - E Y ∣ \geq a) \leq \frac{var Y}{a ^{2}} .

(164)

Let

X = \sum_{T} I_{T}

count triangles, where

T

ranges over the three-element vertex sets and

I_{T}

is the corresponding indicator random variable. Then

μ = E X = (3 n) p^{3} .

(165)

Two distinct triangle indicators are independent unless their triangles share an edge. There are

(2 n) (2 n - 2)

(166)

unordered pairs sharing an edge, and each covariance is

p^{5} - p^{6} \leq p^{5} .

(167)

Therefore

var X \leq (3 n) p^{3} + 2 (2 n) (2 n - 2) p^{5} .

(168)

It follows that

\frac{var X}{μ ^{2}} = O (\frac{1}{n ^{3} p ^{3}} + \frac{1}{n ^{2} p}) ⟶ 0

(169)

when

n p \to \infty

. Chebyshev's inequality now gives

P (X = 0) \leq P (∣ X - μ ∣ \geq μ) \leq \frac{var X}{μ ^{2}} ⟶ 0.

(170)

Hence

P (E_{3}) ⟶ 1,

(171)

as summarized by the triangle count in a binomial random graph.

Solved by gpt-5.6-sol high.

c

Words: 129 Articles: 1

Solution

Words: 129

Partition the vertex set into disjoint sets

A

and

B

of sizes

⌊ n /2 ⌋

and

⌈ n /2 ⌉

. The induced subgraph on

A

has distribution

G (⌊ n /2 ⌋, p) .

(172)

For

p = n^{- 0.9}

p ∣ A ∣ \sim \frac{1}{2} n^{0.1} ⟶ \infty,

(173)

so part (b) shows that

A

contains a triangle with probability tending to one.

On that event, choose a triangle using only the edges internal to

A

. The edges from its three vertices to

B

remain independent of this choice. The probability that all

3∣ B ∣

such edges are absent is

(1 - p)^{3∣ B ∣} \leq e^{- 3 p ∣ B ∣} ⟶ 0,

(174)

because

p ∣ B ∣ \sim n^{0.1} /2

. Thus, with probability tending to one, some vertex of

B

is adjacent to a vertex of the triangle. Those four vertices and the four required edges form the given graph

H

, a triangle with an attached leaf in a binomial random graph. Therefore

P (F) ⟶ 1 .

(175)

Solved by gpt-5.6-sol high.

18I (Galois Theory)

Words: 479 Articles: 4

a

Words: 254 Articles: 1

Solution

Words: 254

A field extension

L / K

is a splitting field for

f \in K [X]

when

f

splits into linear factors over

L

and

L

is generated over

K

by the roots of

f

. Equivalently, it is the smallest extension inside an algebraic closure over which

f

splits.

To prove existence, choose an irreducible factor of

f

, adjoin one of its roots, and factor

f

over the resulting finite extension. If a nonlinear factor remains, adjoin one of its roots and repeat. Each step lowers the total degree of the unsplit factors, so after finitely many finite extensions the polynomial splits. The field generated by the roots obtained is therefore a splitting field.

The precise uniqueness theorem says that if

L

and

L^{'}

are splitting fields of the same polynomial

f

over

K

, then there is a field embedding isomorphism

φ : L ⟶ L^{'}

(176)

whose restriction to

K

is the identity. More generally, an isomorphism of base fields carrying one polynomial to another extends to an isomorphism of their splitting fields. Thus splitting fields are unique up to base-field isomorphism; inside a fixed algebraic closure, the field generated by all the roots is literally unique. These facts are summarized by existence and uniqueness of splitting fields.

Let

α = 32 > 0, ω = e^{2 πi /3} .

(177)

The roots of

X^{3} - 2

are

α, ω α, ω^{2} α

, so

L = Q (α, ω)

(178)

is the splitting field of x cubed minus two. As a subfield of

C

it is unique, because any splitting field in

C

must contain all three roots and hence must contain both

α

and

ω = \frac{ω α}{α};

(179)

minimality then forces it to equal

Q (α, ω)

Solved by gpt-5.6-sol high.

b

Words: 225 Articles: 1

Solution

Words: 225

The seventh cyclotomic polynomial is

Φ_{7} (X) = X^{6} + X^{5} + X^{4} + X^{3} + X^{2} + X + 1.

(180)

It is irreducible over

Q

, and all its roots

ζ_{7}^{a}

a = 1, \dots, 6

, lie in

L = Q (ζ_{7})

. Hence

L / Q

is a degree-six Finite Galois extension, with

G = Gal (L / Q) = {σ_{a} : a \in (Z /7 Z)^{\times}}, σ_{a} (ζ_{7}) = ζ_{7}^{a} .

(181)

The group

(Z /7 Z)^{\times}

is cyclic of order six; for example,

G = ⟨ σ_{3} ⟩ ≅ C_{6} .

(182)

By the Galois correspondence, the subfields are the fixed fields of the four subgroups of

C_{6}

Gal (L / M) G {σ_{1}, σ_{2}, σ_{4}} {σ_{1}, σ_{6}} {σ_{1}} [M : Q] 1236 M Q Q (η), η = ζ_{7} + ζ_{7}^{2} + ζ_{7}^{4} Q (θ), θ = ζ_{7} + ζ_{7}^{- 1} L .

(183)

Thus these are all the subfields.

For the Quadratic Gaussian period in the seventh cyclotomic field, put

\overset{η}{ˉ} = ζ_{7}^{3} + ζ_{7}^{5} + ζ_{7}^{6} .

(184)

The cyclotomic relation gives

η + \overset{η}{ˉ} = - 1

, while direct multiplication gives

η \overset{η}{ˉ} = 2

. Hence

η^{2} + η + 2 = 0.

(185)

Its discriminant is

- 7

, so

η = \frac{- 1 + - 7}{2}, m_{η, Q} (X) = X^{2} + X + 2.

(186)

For the real cubic subfield of the seventh cyclotomic field, divide

1 + ζ_{7} + \dots + ζ_{7}^{6} = 0

(187)

ζ_{7}^{3}

. With

θ = ζ_{7} + ζ_{7}^{- 1}

ζ_{7}^{2} + ζ_{7}^{- 2} = θ^{2} - 2, ζ_{7}^{3} + ζ_{7}^{- 3} = θ^{3} - 3 θ,

(188)

θ^{3} + θ^{2} - 2 θ - 1 = 0.

(189)

This cubic has no rational root, and therefore

m_{θ, Q} (X) = X^{3} + X^{2} - 2 X - 1.

(190)

The requested primitive elements, minimal polynomials, and automorphism groups are consequently

M Q Q (η) Q (θ) L primitive element 0 η θ ζ_{7} m (X) X X^{2} + X + 2 X^{3} + X^{2} - 2 X - 1 Φ_{7} (X) Aut (M / Q) 1 C_{2} C_{3} C_{6} Aut (L / M) C_{6} {σ_{1}, σ_{2}, σ_{4}} ≅ C_{3} {σ_{1}, σ_{6}} ≅ C_{2} 1.

(191)

More explicitly, the nontrivial automorphism of

Q (η)

sends

η ⟼ \overset{η}{ˉ} = - 1 - η,

(192)

and the three automorphisms of

Q (θ)

cyclically permute

ζ_{7} + ζ_{7}^{- 1}, ζ_{7}^{2} + ζ_{7}^{- 2}, ζ_{7}^{4} + ζ_{7}^{- 4} .

(193)

Finally,

G

is abelian, so every subgroup is normal. The subextensions of an abelian Galois extension theorem shows that

all four subfields M are Galois over Q .

(194)

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19I (Representation Theory)

Words: 466 Articles: 11

a

Words: 210 Articles: 1

Solution

Words: 210

A representation is completely reducible, or semisimple, when it is a direct sum of irreducible representation. Equivalently, every invariant subspace has an invariant complement.

Maschke's theorem says that every finite-dimensional representation of a finite group

G

over a field of characteristic zero is completely reducible. More generally, it is enough that the field characteristic does not divide

∣ G ∣

Schur lemma states the following.

If $V$ and $W$ are irreducible $G$ -representations, every intertwining operator $T : V \to W$ is either zero or an isomorphism.
If $V$ is finite-dimensional over an algebraically closed field, every $G$ -endomorphism of $V$ is a scalar multiple of the identity.

Indeed,

ker T \subseteq V

and

im T \subseteq W

are invariant subspaces. If

T \neq = 0

, irreducibility forces

ker T = 0, im T = W,

(195)

T

is an isomorphism. For an endomorphism of a finite-dimensional complex representation, choose an eigenvalue

λ

T

. The intertwiner

T - λ I

has nonzero kernel, so the first part forces

T - λ I = 0.

(196)

This proves the Proof of Schur lemma.

Now let

ρ : G \to GL (V)

be faithful and irreducible over

C

. For every

z \in Z (G)

, the operator

ρ (z)

commutes with every

ρ (g)

. Schur's lemma therefore gives

ρ (z) = λ_{z} I

(197)

for some

λ_{z} \in C^{\times}

. Faithfulness makes

z \mapsto λ_{z}

an embedding

Z (G) ↪ C^{\times} .

(198)

Since every finite subgroup of the multiplicative complex numbers is cyclic,

Z (G) is cyclic .

(199)

This is the cyclic-center obstruction to a faithful irreducible representation.

Solved by gpt-5.6-sol high.

b

Words: 112 Articles: 1

Solution

Words: 112

The complex regular representation has basis

(e_{h})_{h \in G}

and left action

ρ (g) e_{h} = e_{g h} .

(200)

ρ (g)

is the identity, then

e_{g} = ρ (g) e_{1} = e_{1},

(201)

g = 1

. Thus

C G

is a faithful representation.

By Maschke's theorem, the regular representation decomposes as a direct sum of irreducibles,

C G = ⨁_{j} V_{j} .

(202)

Its kernel is the intersection of the kernels of the constituent representations:

ker ρ = ⋂_{j} ker (ρ ∣_{V_{j}}) = {1} .

(203)

Each constituent kernel is a normal subgroup of

G

. If

G

is simple, every such kernel is either

{1}

G

. They cannot all be

G

, because then their intersection would be

G

, contradicting faithfulness of the regular representation. Hence some

V_{j}

has trivial kernel. Therefore every finite simple group has a faithful irreducible representation of a finite simple group.

Solved by gpt-5.6-sol high.

c

Words: 144 Articles: 6

i

Words: 51 Articles: 1

Solution

Words: 51

Yes. If

g

generates the cyclic group

C_{n}

, define the one-dimensional complex representation

ρ (g) = e^{2 πi / n} .

(204)

It is irreducible because it is one-dimensional, and it is faithful because the displayed scalar has multiplicative order

n

. Thus every

C_{n}

, including the trivial case

n = 1

, has a faithful irreducible representation.

Solved by gpt-5.6-sol high.

ii

Words: 56 Articles: 1

Solution

Words: 56

Yes. Write

D_{8} = ⟨ r, s : r^{4} = s^{2} = 1, srs = r^{- 1} ⟩ .

(205)

The Faithful irreducible representation of D8 is

ρ (r) = (i 0 0 - i), ρ (s) = (0110) .

(206)

These matrices satisfy the defining relations. The two eigenspaces of

ρ (r)

are exchanged by

ρ (s)

, so there is no common one-dimensional invariant subspace; the representation is irreducible. Its eight matrices

ρ (r)^{j}

and

ρ (r)^{j} ρ (s)

are distinct, so it is faithful.

Solved by gpt-5.6-sol high.

iii

Words: 37 Articles: 1

Solution

Words: 37

No. The center of a group satisfies

Z (C_{2} \times D_{8}) = C_{2} \times Z (D_{8}) ≅ C_{2} \times C_{2},

(207)

because

Z (D_{8}) = {1, r^{2}} ≅ C_{2}

. This centre is not cyclic. The cyclic-center obstruction to a faithful irreducible representation therefore rules out a faithful irreducible complex representation of

C_{2} \times D_{8}

Solved by gpt-5.6-sol high.

20G (Number Fields)

Words: 318 Articles: 6

a

Words: 66 Articles: 1

Solution

Words: 66

The element

α

is a root of

f (T) = T^{3} - 5 T + 8.

(208)

By the rational root theorem, any rational root of this monic polynomial must be one of

\pm 1, \pm 2, \pm 4, \pm 8.

(209)

Direct substitution shows that none is a root. A reducible cubic over a field has a linear factor, so

f

is an irreducible polynomial over

Q

. It is therefore the minimal polynomial of

α

, and

[K : Q] = de g f = 3 .

(210)

Solved by gpt-5.6-sol high.

b

Words: 152 Articles: 1

Solution

Words: 152

Put

β = \frac{α + α ^{2}}{2} .

(211)

Using

α^{3} = 5 α - 8

, multiplication by

β

on the ordered basis

(1, α, β)

is determined by

β \cdot 1 = β, β α = - 4 + 2 α + β, β^{2} = - 4 - α + 3 β .

(212)

Its matrix, with these coordinate vectors as columns, is

M_{β} = 001 - 4 21 - 4 - 1 3 .

(213)

The characteristic polynomial is

det (T I - M_{β}) = T^{3} - 5 T^{2} + 11 T - 12.

(214)

By the Cayley-Hamilton theorem, this monic integer polynomial annihilates

β

, so

β

is an algebraic integer.

Let

A = Z ⟨ 1, α, β ⟩ .

(215)

The element

α

is also integral because it satisfies the monic polynomial

f

. Hence

A \subseteq O_{K}

. Since

α^{2} = 2 β - α,

(216)

the lattice

Z [α]

has index two in

A

. The given power-basis discriminant is

Δ (1, α, α^{2}) = - 4 \cdot 307,

(217)

so the discriminant-index formula for an integral lattice gives

Δ (1, α, β) = \frac{- 4 \cdot 307}{2 ^{2}} = - 307.

(218)

m = [O_{K} : A]

, then

- 307 = m^{2} d_{K} .

(219)

Because

307

is prime, the square

m^{2}

can divide

307

only when

m = 1

. Thus

O_{K} = A = Z ⟨ 1, α, β ⟩,

(220)

(1, α, β)

is an integral basis of the cubic field of discriminant minus 307.

Finally, the relation

β^{2} = - 4 - α + 3 β

gives

α = - 4 + 3 β - β^{2} .

(221)

Consequently

O_{K} = Z [β]

, which will allow direct use of Dedekind's theorem.

Solved by gpt-5.6-sol high.

c

Words: 100 Articles: 1

Solution

Words: 100

The minimal polynomial of

β

g (T) = T^{3} - 5 T^{2} + 11 T - 12,

(222)

and part (b) showed that

O_{K} = Z [β]

. Modulo

3

\overline{g} (T) = T (T^{2} + T - 1) .

(223)

The quadratic has no root in

F_{3}

, so it is irreducible. The Dedekind factorization theorem therefore gives

(3) = (3, β) (3, β^{2} + β - 1) .

(224)

The two factors are distinct prime ideal, with norms

3

and

3^{2} = 9

, respectively.

Modulo

2

\overline{g} (T) = T (T^{2} + T + 1),

(225)

and the quadratic is irreducible over

F_{2}

. Hence

(2) = (2, β) (2, β^{2} + β + 1) .

(226)

This is a product of two distinct proper prime ideals, so

(2)

is not prime. These decompositions are collected in prime ideals above two and three in the cubic field of discriminant minus 307.

Solved by gpt-5.6-sol high.

21F (Algebraic Topology)

Words: 295 Articles: 6

a

Words: 37 Articles: 1

Solution

Words: 37

Spaces

X

and

Y

are homotopy equivalent if there are continuous maps

f : X \to Y, g : Y \to X

(227)

such that

g \circ f ≃ id_{X}, f \circ g ≃ id_{Y} .

(228)

The maps

f

and

g

are called homotopy inverses.

Solved by gpt-5.6-sol high.

b

Words: 80 Articles: 1

Solution

Words: 80

A subspace

Y \subseteq X

is a retract of

X

if there is a continuous retraction

r : X \to Y

(229)

such that

r (y) = y

for every

y \in Y

. A space is a contractible space if its identity map is homotopic to a constant map.

Let

i : Y ↪ X

be the inclusion and let

H : X \times I \to X

(230)

be a contraction, with

H (x, 0) = x

and

H (x, 1) = x_{0}

. Then

K (y, t) = r (H (i (y), t))

(231)

is continuous and satisfies

K (y, 0) = r (i (y)) = y, K (y, 1) = r (x_{0}) .

(232)

Thus

K

contracts

Y

, proving that every retract of a contractible space is contractible.

Solved by gpt-5.6-sol high.

c

Words: 178 Articles: 1

Solution

Words: 178

Choose a contraction

h : A \times I \to A, h (a, 0) = a, h (a, 1) = a_{0} .

(233)

Apply the homotopy extension property to the initial map

id_{X} : X \to X

and to

h

, regarded as a homotopy into

X

. It supplies

H : X \times I \to X

(234)

such that

H (x, 0) = x, H (a, t) = h (a, t) .

(235)

t = 1

, the map

H_{1}

sends every point of

A

a_{0}

. It is therefore constant on the equivalence classes of the quotient topology

X / A

. By the universal property of the quotient topology, there is a continuous map

g : X / A \to X

(236)

such that

g \circ q = H_{1},

(237)

where

q : X \to X / A

is the quotient map. The homotopy

H

immediately gives

g \circ q = H_{1} ≃ H_{0} = id_{X} .

(238)

For every

t

, the map

q \circ H_{t}

is constant on

A

, because

H (a, t) = h (a, t) \in A

. It therefore descends to

\overline{H} : (X / A) \times I \to X / A, \overline{H} ([x], t) = q (H (x, t)) .

(239)

This descended map is continuous: the product

q \times id_{I}

is a quotient map because

I

is compact Hausdorff, and

q \circ H

is constant on its fibres.
At the endpoints,

\overline{H} ([x], 0) = [x], \overline{H} ([x], 1) = q (g ([x])) .

(240)

Thus

q \circ g ≃ id_{X / A} .

(241)

The maps

q

and

g

are homotopy inverses, so

X ≃ X / A .

(242)

This proves the theorem on collapsing a contractible cofibration.

Solved by gpt-5.6-sol high.

22H (Linear Analysis)

Words: 489 Articles: 14

a

Words: 162 Articles: 4

i

Words: 62 Articles: 1

Solution

Words: 62

Assume

x_{n} ⇀ x_{\infty}

. For each

n

, define the bounded functional

T_{n} (h) = ⟨ h, x_{n} ⟩ .

(243)

Weak convergence makes

(T_{n} (h))_{n}

convergent, hence bounded, for every

h \in H

. The Uniform boundedness principle gives

sup_{n} ∥ T_{n} ∥ < \infty.

(244)

By the Riesz representation theorem,

∥ T_{n} ∥ = ∥ x_{n} ∥

, so

(x_{n})

is bounded.

Taking

h = e_{i}

in the definition of weak convergence in a Hilbert space gives

⟨ x_{n}, e_{i} ⟩ ⟶ ⟨ x_{\infty}, e_{i} ⟩

(245)

for every

i

. Thus (i) implies (ii).

Solved by gpt-5.6-sol high.

ii

Words: 100 Articles: 1

Solution

Words: 100

Conversely, suppose

∥ x_{n} ∥ \leq C

(246)

and define

a_{i} = lim_{n \to \infty} ⟨ x_{n}, e_{i} ⟩ .

(247)

For every

N

, Bessel inequality and passage to the limit give

\sum_{i = 1}^{N} ∣ a_{i} ∣^{2} = lim_{n \to \infty} \sum_{i = 1}^{N} ∣ ⟨ x_{n}, e_{i} ⟩ ∣^{2} \leq C^{2} .

(248)

Hence

\sum_{i} ∣ a_{i} ∣^{2} \leq C^{2}

. The partial sums

\sum_{i = 1}^{N} a_{i} e_{i}

are therefore Cauchy; completeness of the Hilbert space gives an element

x_{\infty} = \sum_{i = 1}^{\infty} a_{i} e_{i}, ∥ x_{\infty} ∥ \leq C .

(249)

Let

P_{N} h = \sum_{i = 1}^{N} ⟨ h, e_{i} ⟩ e_{i}

. For fixed

N

, coordinate convergence gives

⟨ x_{n} - x_{\infty}, P_{N} h ⟩ ⟶ 0.

(250)

The Hilbertian basis property gives

P_{N} h \to h

, and

∣ ⟨ x_{n} - x_{\infty}, h - P_{N} h ⟩ ∣ \leq (∥ x_{n} ∥ + ∥ x_{\infty} ∥) ∥ h - P_{N} h ∥ \leq 2 C ∥ h - P_{N} h ∥.

(251)

First choose

N

large and then

n

large. This proves

⟨ x_{n}, h ⟩ ⟶ ⟨ x_{\infty}, h ⟩

(252)

for every

h \in H

, so

x_{n} ⇀ x_{\infty}

. This establishes the coordinate criterion for weak convergence in a separable Hilbert space.

Solved by gpt-5.6-sol high.

b

Words: 89 Articles: 1

Solution

Words: 89

Let

∥ x_{n} ∥ \leq C

. For each

i

, Bessel inequality gives

∣ ⟨ x_{n}, e_{i} ⟩ ∣ \leq C,

(253)

so the

i

th coordinate sequence is bounded. By the Bolzano-Weierstrass theorem, choose a subsequence on which the first coordinate converges. From it choose a further subsequence on which the second coordinate converges, and continue.

Take the diagonal subsequence

(x_{ϕ (n)})

. For every fixed

i

, its

i

th coordinate converges, and it remains bounded by

C

. The coordinate criterion for weak convergence in a separable Hilbert space supplies an

x_{\infty} \in H

such that

x_{ϕ (n)} ⇀ x_{\infty} .

(254)

This proves the theorem on a weak subsequence of a bounded Hilbert-space sequence.

Solved by gpt-5.6-sol high.

c

Words: 238 Articles: 6

i

Words: 96 Articles: 1

Solution

Words: 96

Suppose first that

∥ x_{n} - x_{\infty} ∥ \to 0

. The reverse triangle inequality gives

∥ x_{n} ∥ - ∥ x_{\infty} ∥ \leq ∥ x_{n} - x_{\infty} ∥ ⟶ 0,

(255)

so (i) implies (ii).

It also implies (iii). Given

ϵ > 0

, choose

N

so that

∥ x_{n} - x_{\infty} ∥^{2} < \frac{ϵ}{8} (n \geq N) .

(256)

By the Parseval identity for a Hilbertian basis, choose

I_{0}

so that the basis tail of

x_{\infty}

beyond

I_{0}

is less than

ϵ /8

. For

n \geq N

and

I \geq I_{0}

\sum_{i \geq I} ∣ ⟨ x_{n}, e_{i} ⟩ ∣^{2} \leq 2∥ x_{n} - x_{\infty} ∥^{2} + 2 \sum_{i \geq I} ∣ ⟨ x_{\infty}, e_{i} ⟩ ∣^{2} < \frac{ϵ}{2} .

(257)

There are only finitely many

n < N

, so increasing

I

makes each of their tails less than

ϵ

. Thus one

I

works for every

n

, proving (iii).

Solved by gpt-5.6-sol high.

ii

Words: 37 Articles: 1

Solution

Words: 37

Assume (ii). Weak convergence gives

⟨ x_{n}, x_{\infty} ⟩ ⟶ ⟨ x_{\infty}, x_{\infty} ⟩ = ∥ x_{\infty} ∥^{2} .

(258)

Using the inner product identity,

∥ x_{n} - x_{\infty} ∥^{2} = ∥ x_{n} ∥^{2} + ∥ x_{\infty} ∥^{2} - 2 Re ⟨ x_{n}, x_{\infty} ⟩ ⟶ ∥ x_{\infty} ∥^{2} + ∥ x_{\infty} ∥^{2} - 2∥ x_{\infty} ∥^{2} = 0.

(259)

Hence (ii) implies (i); this is the Radon-Riesz theorem in a Hilbert space.

Solved by gpt-5.6-sol high.

iii

Words: 105 Articles: 1

Solution

Words: 105

Assume (iii), and let

δ > 0

. Choose

I

such that

\sum_{i \geq I} ∣ ⟨ x_{n}, e_{i} ⟩ ∣^{2} < δ

(260)

for every

n

. Coordinate convergence implies, for every finite

J \geq I

\sum_{i = I}^{J} ∣ ⟨ x_{\infty}, e_{i} ⟩ ∣^{2} = lim_{n \to \infty} \sum_{i = I}^{J} ∣ ⟨ x_{n}, e_{i} ⟩ ∣^{2} \leq δ .

(261)

Letting

J \to \infty

shows that the corresponding tail of

x_{\infty}

is also at most

δ

By the Parseval identity for a Hilbertian basis,

∥ x_{n} - x_{\infty} ∥^{2} = i < I \sum ∣ ⟨ x_{n} - x_{\infty}, e_{i} ⟩ ∣^{2} + i \geq I \sum ∣ ⟨ x_{n} - x_{\infty}, e_{i} ⟩ ∣^{2} .

(262)

The finite first sum tends to zero by weak coordinate convergence, while the second is at most

2 \sum_{i \geq I} ∣ ⟨ x_{n}, e_{i} ⟩ ∣^{2} + 2 \sum_{i \geq I} ∣ ⟨ x_{\infty}, e_{i} ⟩ ∣^{2} \leq 4 δ .

(263)

Since

δ

is arbitrary,

∥ x_{n} - x_{\infty} ∥ \to 0

. Thus (iii) implies (i), completing the equivalence and proving the uniform basis-tail criterion for strong convergence.

Solved by gpt-5.6-sol high.

23H (Analysis of Functions)

Words: 300 Articles: 4

a

Words: 83 Articles: 1

Solution

Words: 83

For an integrable function

f : R^{n} \to C

, the Lebesgue differentiation theorem states that

lim_{r ↓ 0} \frac{1}{λ ( B ( x , r ))} \int_{B (x, r)} ∣ f (y) - f (x) ∣ d λ (y) = 0

(264)

for almost everywhere

x \in R^{n}

. In particular,

lim_{r ↓ 0} \frac{1}{λ ( B ( x , r ))} \int_{B (x, r)} f (y) d λ (y) = f (x)

(265)

at every Lebesgue point of

f

Almost every

x \in R

is a Lebesgue point of

g

. At such an

x

, for

t > 0

\frac{G ( x + t ) - G ( x )}{t} - g (x) \leq \frac{1}{t} \int_{x}^{x + t} ∣ g (y) - g (x) ∣ d λ (y) ⟶ 0.

(266)

For

t < 0

, the same estimate over

[x + t, x]

gives the identical limit. Therefore

G^{'} (x) = g (x)

(267)

at every Lebesgue point of

g

, so

G

is differentiable

λ

-almost everywhere. This is the differentiation of an indefinite Lebesgue integral.

Solved by gpt-5.6-sol high.

b

Words: 217 Articles: 1

Solution

Words: 217

Every Lebesgue measurable set

A

differs from a Borel set

B

by a Lebesgue-null set

N

. The given function

h

is a homeomorphism from

R

onto the interval

h (R)

. Hence

h (B)

is Borel relative to

h (R)

and therefore Lebesgue measurable in

R

. Moreover,

h (A) △ h (B) \subseteq h (A △ B) \subseteq h (N),

(268)

and

h (N)

is null by Lusin condition N. The Completeness of Lebesgue measure now shows that

h (A)

is measurable, so

ν (A) := λ (h (A))

(269)

is defined for every

A \in M

Because

h

is injective, it maps pairwise disjoint sets to pairwise disjoint sets and commutes with arbitrary unions. The countable additivity of

λ

therefore gives

ν (⋃_{j = 1}^{\infty} A_{j}) = λ (⋃_{j = 1}^{\infty} h (A_{j})) = \sum_{j = 1}^{\infty} λ (h (A_{j})) = \sum_{j = 1}^{\infty} ν (A_{j}) .

(270)

Thus

ν

is the image-length measure of a strictly increasing continuous function. If

λ (A) = 0

, the null-set hypothesis gives

ν (A) = λ (h (A)) = 0

, so

ν ≪ λ .

(271)

The measure is sigma-finite, since

ν ([- m, m]) = h (m) - h (- m) < \infty.

(272)

The Radon-Nikodym theorem consequently supplies a nonnegative locally integrable function

f

such that

ν (A) = \int_{A} f d λ .

(273)

For

a < x

, strict increase and continuity give

h ([a, x]) = [h (a), h (x)]

(274)

and hence

h (x) - h (a) = ν ([a, x]) = \int_{a}^{x} f (t) d t .

(275)

The analogous oriented identity holds for

x < a

. By differentiation of an indefinite Lebesgue integral,

h^{'} (x) = f (x) for almost every x .

(276)

Yes, the differentiability conclusion still holds when

h

is merely non-decreasing. In fact, the Lebesgue theorem on differentiability of monotone functions says that every monotone function is differentiable almost everywhere; neither continuity nor the null-set hypothesis is needed for that conclusion. Flat intervals mean that the preceding image-set argument no longer gives disjoint images, so its natural replacement is the Lebesgue-Stieltjes measure determined by

μ_{h} ((a, b]) = h (b) - h (a) .

(277)

Solved by gpt-5.6-sol high.

24F (Riemann Surfaces)

Words: 449 Articles: 8

a

Words: 96 Articles: 1

Solution

Words: 96

Suppose first that

u

is harmonic. Since

\frac{\partial}{\partial y} u_{x} = \frac{\partial}{\partial x} u_{y} and \frac{\partial}{\partial x} u_{x} = - \frac{\partial}{\partial y} u_{y},

(278)

the Cauchy-Riemann equations show that

g := u_{x} - i u_{y}

(279)

is holomorphic on

D

. An open disc is a simply connected domain, so

g

has a holomorphic primitive

f

. If

f = p + i q

, then

p_{x} = Re f^{'} = u_{x}, p_{y} = - Im f^{'} = u_{y} .

(280)

Thus

p - u

has zero gradient and is constant on the connected disc. Subtracting that real constant from

f

gives

u = Re f .

(281)

Conversely, if

u = Re f

for a holomorphic

f = u + i v

, differentiating the Cauchy-Riemann equations gives

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

(282)

Hence

u

is harmonic. This proves the harmonic function as the real part of a holomorphic function criterion on

D

Solved by gpt-5.6-sol high.

b

Words: 73 Articles: 1

Solution

Words: 73

Take the punctured complex plane

D = C^{*} = C ∖ {0}

(283)

and

u (z) = lo g ∣ z ∣.

(284)

In polar coordinates,

u (r, θ) = lo g r

, so

Δ u = u_{rr} + \frac{1}{r} u_{r} + \frac{1}{r ^{2}} u_{θθ} = - \frac{1}{r ^{2}} + \frac{1}{r ^{2}} = 0.

(285)

Thus

u

is harmonic on

D

u = Re f

for a holomorphic function

f

, the calculation from part (a) would give

f^{'} (z) = u_{x} - i u_{y} = \frac{1}{z} .

(286)

The integral of a derivative around a closed curve is zero, whereas the residue theorem gives

\int_{∣ z ∣ = 1} \frac{d z}{z} = 2 πi .

(287)

This contradiction proves that no such

f

exists. It is precisely the log modulus has no global harmonic conjugate on the punctured plane obstruction.

Solved by gpt-5.6-sol high.

c

Words: 169 Articles: 1

Solution

Words: 169

Lift

u

through the complex exponential by setting

v (w) = u (e^{w}), w \in C .

(288)

The conformal invariance of harmonicity makes

v

harmonic. Since

e^{w + 2 πi} = e^{w}

and

u (2 z) = u (z)

v (w + 2 πi) = v (w), v (w + lo g 2) = v (w) .

(289)

Thus

v

is bounded on the compact rectangle

{s lo g 2 + 2 πi t : 0 \leq s, t \leq 1},

(290)

and its two periods make it bounded on the whole plane. The Harmonic Liouville theorem makes

v

constant. Since the exponential map is surjective onto

C^{*}

u

is constant.

For the requested counterexample after deleting a countable set, take

S = {2^{m} : m \in Z}

(291)

and let

℘_{Λ}

be the Weierstrass elliptic function for the lattice

Λ = (lo g 2) Z + 2 πi Z .

(292)

Define

u (z) = Re ℘_{Λ} (lo g z), z \in C^{*} ∖ S .

(293)

Changing a branch of the logarithm adds

2 πi

, a period of

℘_{Λ}

, so

u

is well defined. Its poles project precisely to

S

, and the conformal invariance of harmonicity makes

u

harmonic away from them. The other period gives

u (2 z) = Re ℘_{Λ} (lo g z + lo g 2) = u (z) .

(294)

Moreover,

z \in / S

implies

2 z \in / S

. Finally,

u

is nonconstant because as positive real

z \to 1

℘_{Λ} (lo g z) \sim \frac{1}{( l o g z ) ^{2}},

(295)

which is real and unbounded. This is a scale-periodic harmonic function from an elliptic function.

Solved by gpt-5.6-sol high.

d

Words: 111 Articles: 1

Solution

Words: 111

Since

C

is simply connected, write

u = Re f

(296)

for an entire function

f

. If

u

were bounded above, then

∣ e^{f} ∣ = e^{u}

(297)

would be bounded, so the Liouville theorem would make

e^{f}

constant. Its derivative

f^{'} e^{f}

would then vanish, whence

f

and

u

would be constant. Applying the same argument to

e^{- f}

rules out a lower bound. Thus a nonconstant

u

is unbounded both above and below.

Given any

c \in R

, choose

z_{-}, z_{+} \in C

with

u (z_{-}) < c < u (z_{+}) .

(298)

The restriction of

u

to the line segment from

z_{-}

z_{+}

is continuous, so the intermediate value theorem supplies a point on that segment where

u = c

. Therefore

u (C) = R .

(299)

This proves the surjectivity of a nonconstant entire harmonic function.

Solved by gpt-5.6-sol high.

25I (Algebraic Geometry)

Words: 406 Articles: 9

Solution

Words: 110

The Hilbert basis theorem makes

k [x_{1}, \dots, x_{n}]

a Noetherian ring. Reversing the inclusion of ideals shows that its Zariski-closed subsets satisfy the descending-chain condition, so every affine variety is a Noetherian topological space.

Suppose some nonempty closed subset had no finite decomposition into irreducible closed subsets. It would be reducible, so it could be written

V = V_{1} \cup V_{2}

(300)

with

V_{1}, V_{2}

proper closed subsets. At least one of these would again admit no finite irreducible decomposition. Repeating that choice would produce an infinite strictly descending chain of closed subsets, contradicting Noetherianity. Therefore

V is a finite union of irreducible subvarieties .

(301)

Removing any member contained in another leaves exactly the irreducible components.

Solved by gpt-5.6-sol high.

i

Words: 55 Articles: 1

Solution

Words: 55

A point

p

belongs to

V (I_{1} + I_{2})

exactly when every polynomial

f_{1} + f_{2}

, with

f_{j} \in I_{j}

, vanishes at

p

. This implies that every member of each ideal vanishes by taking the other summand to be zero, and the converse follows by addition. Hence

V (I_{1} + I_{2}) = V (I_{1}) \cap V (I_{2}) .

(302)

This is the first vanishing loci of ideal sums and intersections identity.

Solved by gpt-5.6-sol high.

ii

Words: 47 Articles: 1

Solution

Words: 47

Since

I_{1} \cap I_{2} \subseteq I_{j}

, every point of either

V (I_{j})

lies in

V (I_{1} \cap I_{2})

. Conversely, suppose

p \in / V (I_{1}) \cup V (I_{2}) .

(303)

Choose

f_{j} \in I_{j}

with

f_{j} (p) \neq = 0

. Then

f_{1} f_{2} \in I_{1} I_{2} \subseteq I_{1} \cap I_{2}

(304)

but

(f_{1} f_{2}) (p) \neq = 0

, so

p \in / V (I_{1} \cap I_{2})

. Therefore

V (I_{1} \cap I_{2}) = V (I_{1}) \cup V (I_{2}) .

(305)

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Decomposition of the given vanishing locus

Words: 65 Articles: 1

Solution

Words: 65

Subtracting the two defining equations gives

y^{2} + z^{2} = (y - i z) (y + i z) = 0.

(306)

Consequently the variety is

V (y - i z, x^{2} - z^{2} - 1) \cup V (y + i z, x^{2} - z^{2} - 1) .

(307)

Each member is irreducible. Indeed, after imposing its linear equation, its coordinate ring is

C [x, z] / (x^{2} - z^{2} - 1) .

(308)

The invertible linear substitution

u = x - z, v = x + z

(309)

turns this into

C [u, v] / (uv - 1) ≅ C [u, u^{- 1}],

(310)

which is an integral domain. The two curves are distinct and neither contains the other, so they are precisely the irreducible components of two intersecting complex hyperbolas.

Solved by gpt-5.6-sol high.

Non-isomorphism of the two unions of lines

Words: 129 Articles: 1

Solution

Words: 129

The three coordinate axes are the three irreducible components of

V = V (x y, x z, yz) \subseteq A_{k}^{3} .

(311)

Their common intersection is the origin. Likewise, if

ℓ_{1}, ℓ_{2}, ℓ_{3}

are distinct linear forms defining the three lines in

A_{k}^{2}

, then

W = V (ℓ_{1} ℓ_{2} ℓ_{3}),

(312)

and its three irreducible components have only the origin in common. Any isomorphism of algebraic varieties would permute irreducible components and therefore map the origin of

V

to the origin of

W

Every defining polynomial

x y, x z, yz

has zero differential at the origin, so

T_{0} V = k^{3} and dim T_{0} V = 3.

(313)

The defining polynomial of

W

is homogeneous of degree three, so its differential also vanishes at the origin, giving

T_{0} W = k^{2} and dim T_{0} W = 2.

(314)

An isomorphism induces an isomorphism of Zariski tangent spaces, which is impossible here. Hence

V

and

W

are not isomorphic. This is the tangent-space obstruction between two unions of three lines.

Solved by gpt-5.6-sol high.

26F (Differential Geometry)

Words: 339 Articles: 10

a

Words: 53 Articles: 1

Solution

Words: 53

For each

p \in S

, the first fundamental form is defined without coordinates by

I_{p} (v, w) = v \cdot w, v, w \in T_{p} S,

(315)

the restriction of the Euclidean inner product to the tangent space of

S

ϕ (u, v)

is a local parametrization and

ξ = ξ^{1} \partial_{u} + ξ^{2} \partial_{v}, η = η^{1} \partial_{u} + η^{2} \partial_{v},

(316)

then

d ϕ (ξ) = ξ^{1} ϕ_{u} + ξ^{2} ϕ_{v}, d ϕ (η) = η^{1} ϕ_{u} + η^{2} ϕ_{v} .

(317)

Taking their dot product gives

I = E d u^{2} + 2 F d u d v + G d v^{2},

(318)

where

E = ϕ_{u} \cdot ϕ_{u}, F = ϕ_{u} \cdot ϕ_{v}, G = ϕ_{v} \cdot ϕ_{v} .

(319)

Solved by gpt-5.6-sol high.

b

Words: 75 Articles: 1

Solution

Words: 75

Since

z > 0

, the cone has the parametrization

ϕ (r, θ) = (r cos θ, r sin θ, a r), r > 0.

(320)

Its coordinate derivatives satisfy

ϕ_{r} \cdot ϕ_{r} = 1 + a, ϕ_{r} \cdot ϕ_{θ} = 0, ϕ_{θ} \cdot ϕ_{θ} = r^{2},

(321)

so its first fundamental form is

I = (1 + a) d r^{2} + r^{2} d θ^{2} .

(322)

On a sufficiently narrow angular patch, define

R = 1 + a r, Θ = \frac{θ}{1 + a},

(323)

and map the cone to the plane point with polar coordinates

(R, Θ)

. The Euclidean metric pulls back as

d R^{2} + R^{2} d Θ^{2} = (1 + a) d r^{2} + r^{2} d θ^{2} = I .

(324)

The map is therefore a local isometry from a circular cone to the plane, proving that

Σ

is locally isometric to the Euclidean plane.

Solved by gpt-5.6-sol high.

c

Words: 211 Articles: 5

Solution

Words: 64

For a chosen unit normal field

N

, the second fundamental form is

I I_{p} (v, w) = ⟨ - d N_{p} (v), w ⟩ .

(325)

In a local parametrization

ϕ

, its coefficients, up to the equivalent simultaneous sign convention, are

e = N \cdot ϕ_{uu}, f = N \cdot ϕ_{uv}, g = N \cdot ϕ_{vv} .

(326)

The Gaussian curvature is the determinant of the shape operator, hence

K = det (- d N) = \frac{e g - f ^{2}}{EG - F ^{2}} .

(327)

Theorema Egregium states that

K

is determined entirely by the first fundamental form. In particular, it is intrinsic and is preserved by every local isometry.

Solved by gpt-5.6-sol high.

i

Words: 64 Articles: 1

Solution

Words: 64

Let

F (x, y, z) = x^{2} + y^{2} + z^{2} - 2 x y - 2 yz .

(328)

Its gradient is

\nabla F = 2 (x - y, y - x - z, z - y) .

(329)

If this vanishes, the first and third coordinates give

x = y = z

, and the middle coordinate then gives

y = 0

. Thus the origin is the only critical point of

F

. It has been removed from

V

, so

0

is a regular value of the restriction under consideration. The regular level set theorem shows that

V = F^{- 1} (0) ∖ {0}

(330)

is a smooth surface.

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ii

Words: 83 Articles: 1

Solution

Words: 83

Completing the square gives

F = (x - y)^{2} + (z - y)^{2} - y^{2} .

(331)

Thus

V

is a two-napped cone. On the nappe where

ε y > 0

, with

ε \in {1, - 1}

, use

ϕ_{ε} (r, θ) = (ε r + r cos θ, ε r, ε r + r sin θ), r > 0.

(332)

This has the conical form

ϕ_{ε} (r, θ) = r c_{ε} (θ) .

(333)

Its second derivatives obey

ϕ_{rr} = 0, ϕ_{r θ} = c_{ε}^{'} (θ) = \frac{1}{r} ϕ_{θ} .

(334)

The latter vector is tangent, so for either choice of unit normal,

e = N \cdot ϕ_{rr} = 0, f = N \cdot ϕ_{r θ} = 0.

(335)

Since the parametrization is regular,

EG - F^{2} > 0

, and therefore

K = \frac{e g - f ^{2}}{EG - F ^{2}} = 0

(336)

at every point of

V

. This is the Gaussian curvature of a cone away from its vertex.

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27H (Probability and Measure)

Words: 152 Articles: 6

a

Words: 71 Articles: 1

Solution

Words: 71

Fatou lemma states that for any sequence

(f_{n})

of nonnegative measurable functions on a measure space,

\int n \to \infty lim inf f_{n} d μ \leq n \to \infty lim inf \int f_{n} d μ .

(337)

For the proof, define

g_{n} = in f_{k \geq n} f_{k} .

(338)

Then

(g_{n})

is a nonnegative increasing sequence and

g_{n} ↑ lim inf_{k \to \infty} f_{k} .

(339)

The monotone convergence theorem therefore gives

\int lim inf_{k} f_{k} d μ = lim_{n \to \infty} \int g_{n} d μ .

(340)

For each

k \geq n

, we have

g_{n} \leq f_{k}

, so monotonicity of the Lebesgue integral yields

\int g_{n} d μ \leq in f_{k \geq n} \int f_{k} d μ .

(341)

Taking the limit in

n

proves the asserted inequality and the proof of Fatou lemma.

Solved by gpt-5.6-sol high.

b

Words: 34 Articles: 1

Solution

Words: 34

On the probability space

([0, 1], B, λ)

, let

f_{n} = n 1_{(0, 1/ n)} .

(342)

For every

x \in [0, 1]

, eventually

x \in / (0, 1/ n)

, so

f_{n} (x) \to 0

. Nevertheless,

\int_{0}^{1} f_{n} d λ = 1

(343)

for every

n

. Consequently

\int lim inf_{n} f_{n} d λ = 0 < 1 = lim inf_{n} \int f_{n} d λ .

(344)

This gives strict inequality in Fatou lemma.

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c

Words: 47 Articles: 1

Solution

Words: 47

Yes. Almost-sure convergence gives

X = lim inf_{n \to \infty} X_{n}

(345)

almost surely. Since the random variables are nonnegative, Fatou lemma applies and gives

E X \leq lim inf_{n \to \infty} E X_{n} \leq sup_{n \in N} E X_{n} .

(346)

The inequalities remain valid in the extended nonnegative reals, so no integrability assumption beyond nonnegativity is required.

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28K (Applied Probability)

Words: 256 Articles: 6

a

Words: 113 Articles: 1

Solution

Words: 113

Let

N_{j} (A)

be the number of particles of gas

j

in a bounded Lebesgue measurable set

A

. Because Lebesgue measure is diffuse, two independent spatial Poisson point processes have no common particle almost surely. Hence the mixture count is

N (A) = N_{1} (A) + N_{2} (A) .

(347)

Writing

λ_{j} = z_{j} ∣ A ∣

, independence gives the probability generating function

E [s^{N (A)}] = E [s^{N_{1} (A)}] E [s^{N_{2} (A)}] = exp (λ_{1} (s - 1)) exp (λ_{2} (s - 1)) = exp ((z_{1} + z_{2}) ∣ A ∣ (s - 1)) .

(348)

Thus

N (A) \sim Pois ((z_{1} + z_{2}) ∣ A ∣) .

(349)

A_{1}, \dots, A_{m}

are pairwise disjoint, the family

{N_{j} (A_{r}) : j = 1, 2, 1 \leq r \leq m}

(350)

is independent: counts in disjoint sets are independent within each gas, and the two gases are independent. Therefore the sums

N (A_{r})

are independent. These two facts prove directly that the mixture is a Poisson point process. By the Superposition theorem for Poisson point processes, its activity is

z_{1} + z_{2} .

(351)

Solved by gpt-5.6-sol high.

b

Words: 55 Articles: 1

Solution

Words: 55

Put

λ_{i} = z ∣ V_{i} ∣

. Then

N (V_{i}) \sim Pois (λ_{i}), E N (V_{i}) = λ_{i},

(352)

and

λ_{i} \to \infty

. The Poisson central limit theorem gives

\frac{N ( V _{i} ) - λ _{i}}{λ _{i}} d N (0, 1) .

(353)

Since

λ_{i} = z ∣ V_{i} ∣

\frac{N ( V _{i} ) - E N ( V _{i} )}{∣ V _{i} ∣} = z \frac{N ( V _{i} ) - λ _{i}}{λ _{i}} d N (0, z) .

(354)

Equivalently, the characteristic function of the left-hand side is

exp {λ_{i} (e^{i t / ∣ V_{i} ∣} - 1 - \frac{i t}{∣ V _{i} ∣})} ⟶ e^{- z t^{2} /2},

(355)

which is the characteristic function of that centered normal distribution.

Solved by gpt-5.6-sol high.

c

Words: 88 Articles: 1

Solution

Words: 88

Let

S_{g} = \sum_{x \in Π} g (x) .

(356)

The compact support of

g

ensures that this sum has only finitely many nonzero terms almost surely. The intensity measure of the ideal gas is

μ (d x) = z d x .

(357)

Applying the first-moment identity in Campbell theorem gives

E S_{g} = z \int_{R^{3}} g (x) d x .

(358)

For the second moment, separate equal and distinct particles:

E S_{g}^{2} = E x \in Π \sum g (x)^{2} + E x, y \in Π x \neq = y \sum g (x) g (y) = z \int_{R^{3}} g (x)^{2} d x + z^{2} (\int_{R^{3}} g (x) d x)^{2} .

(359)

Subtracting

(E S_{g})^{2}

yields

Var (S_{g}) = z \int_{R^{3}} g (x)^{2} d x .

(360)

These are the mean and variance formulas in Campbell's theorem for a homogeneous Poisson point process.

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29J (Principles of Statistics)

Words: 254 Articles: 6

a

Words: 73 Articles: 1

Solution

Words: 73

Let

ℓ_{n} (θ) = lo g f_{θ} (X_{1}, \dots, X_{n})

(361)

be the log-likelihood. The score function and scalar Fisher information are

S_{n} (θ) = \frac{\partial}{\partial θ} ℓ_{n} (θ), I_{n} (θ) = E_{θ} [S_{n} (θ)^{2}] .

(362)

Under the usual regularity conditions the score has mean zero, so this is also its variance.

The information tensorizes when it is additive over observations. For identically distributed observations,

I_{n} (θ) = n I_{1} (θ) .

(363)

For an independent sample the joint score is the sum of the individual scores, and the mean-zero score identity makes all cross covariances vanish. This is the Tensorization of Fisher information.

Solved by gpt-5.6-sol high.

b

Words: 63 Articles: 1

Solution

Words: 63

Write

f_{θ} (x)

for the joint density and assume regularity permits differentiation under the integral. Since

\int f_{θ} (x) d x = 1,

(364)

one differentiation gives

E_{θ} S_{n} (θ) = \int \frac{\partial _{θ} f _{θ}}{f _{θ}} f_{θ} d x = \partial_{θ} \int f_{θ} d x = 0.

(365)

Differentiating the score itself,

\partial_{θ} S_{n} = \frac{\partial _{θ}^{2} f _{θ}}{f _{θ}} - (\frac{\partial _{θ} f _{θ}}{f _{θ}})^{2} .

(366)

Taking expectation gives

E_{θ} [\partial_{θ} S_{n} (θ)] = \int \partial_{θ}^{2} f_{θ} (x) d x - E_{θ} [S_{n} (θ)^{2}] = - I_{n} (θ) .

(367)

Thus the information identity is

I_{n} (θ) = - E_{θ} [\frac{\partial}{\partial θ} S_{n} (θ)] = - E_{θ} [ℓ_{n}^{''} (θ)] .

(368)

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c

Words: 118 Articles: 1

Solution

Words: 118

The model factors into the normal distribution densities

X_{1} \sim N (θ, 1)

(369)

and, conditionally on

X_{i - 1}

X_{i} ∣ X_{i - 1} \sim N (θ (1 - γ) + γ X_{i - 1}, 1 - γ) .

(370)

Ignoring constants independent of

θ

, the log likelihood is

ℓ_{n} (θ) = - \frac{1}{2} (X_{1} - θ)^{2} - \frac{1}{2 ( 1 - γ )} \sum_{i = 2}^{n} [X_{i} - γ X_{i - 1} - θ (1 - γ)]^{2} .

(371)

Its score is

S_{n} (θ) = (X_{1} - θ) + \frac{1 - γ}{1 - γ} \sum_{i = 2}^{n} [X_{i} - γ X_{i - 1} - θ (1 - γ)] .

(372)

In terms of the independent innovations,

S_{n} (θ) = ε_{1} + \frac{1 - γ}{1 - γ} \sum_{i = 2}^{n} ε_{i} .

(373)

Variance additivity for independent random variables therefore yields

I_{n} (θ) = 1 + (n - 1) \frac{( 1 - γ ) ^{2}}{1 - γ} = 1 + (n - 1) \frac{1 - γ}{1 + γ} .

(374)

Since

I_{1} (θ) = 1

, the equality

I_{n} = n I_{1}

holds for all

n

exactly when

\frac{1 - γ}{1 + γ} = 1,

(375)

namely when

γ = 0 .

(376)

Thus the information tensorizes only in the independent case.

Finally, the Cramer-Rao bound gives every unbiased estimator

θ

the lower bound

Var_{θ} (θ) \geq \frac{1}{I _{n} ( θ )} = [1 + (n - 1) \frac{1 - γ}{1 + γ}]^{- 1} .

(377)

This is the Fisher information in a stationary Gaussian autoregressive location model.

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30K (Stochastic Financial Models)

Words: 413 Articles: 10

a

Words: 55 Articles: 1

Solution

Words: 55

A process

(X_{n})_{n \geq 0}

is a martingale with respect to a filtration

(F_{n})_{n \geq 0}

when:

$X_{n}$ is $F_{n}$ -measurable for every $n$ , so the process is adapted;
$E ∣ X_{n} ∣ < \infty$ for every $n$ ;
for every $n \geq 0$ ,
$E [X_{n + 1} ∣ F_{n}] = X_{n} almost surely .$
(378)

Equivalently, every increment has conditional mean zero given the past.

Solved by gpt-5.6-sol high.

b

Words: 67 Articles: 1

Solution

Words: 67

Because

X

is a martingale,

E [ξ_{n} ∣ F_{n - 1}] = 0.

(379)

Writing

p_{n} = P (ξ_{n} = 1 ∣ F_{n - 1}),

(380)

and using

ξ_{n} \in {- 1, 1}

gives

0 = p_{n} - (1 - p_{n}) = 2 p_{n} - 1.

(381)

Thus

P (ξ_{n} = 1 ∣ F_{n - 1}) = P (ξ_{n} = - 1 ∣ F_{n - 1}) = \frac{1}{2} .

(382)

For signs

e_{1}, \dots, e_{m} \in {- 1, 1}

, repeated use of the tower property of conditional expectation gives

P (ξ_{1} = e_{1}, \dots, ξ_{m} = e_{m}) = E [1_{{ξ_{1} = e_{1}, \dots, ξ_{m - 1} = e_{m - 1}}} P (ξ_{m} = e_{m} ∣ F_{m - 1})] = \frac{1}{2} P (ξ_{1} = e_{1}, \dots, ξ_{m - 1} = e_{m - 1}) = 2^{- m} .

(383)

Hence the increments are IID symmetric signs, proving the symmetric signs forced by the martingale property result. Thus

X_{n} - X_{0}

is a simple symmetric random walk.

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c

Words: 73 Articles: 1

Solution

Words: 73

Because

H_{k}

is previsible,

H_{k}

F_{k - 1}

-measurable. The finite sum

X_{n}

is therefore

F_{n}

-measurable. Boundedness of

H

and integrability of the martingale increments imply

E ∣ X_{n} ∣ < \infty.

(384)

Finally,

E [X_{n} - X_{n - 1} ∣ F_{n - 1}] = E [H_{n} (X_{n} - X_{n - 1}) ∣ F_{n - 1}] = H_{n} E [X_{n} - X_{n - 1} ∣ F_{n - 1}] = 0.

(385)

Thus

E [X_{n} ∣ F_{n - 1}] = X_{n - 1},

(386)

X

is a martingale. It is the martingale transform of

X

H

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d

Words: 100 Articles: 1

Solution

Words: 100

Define

H_{n} = {1, - 1, n \leq T_{a}, n > T_{a} .

(387)

The event

{n \leq T_{a}}

means that level

a

has not been reached by time

n - 1

, so it belongs to

F_{n - 1}

. Hence

H

is a bounded predictable process. A direct check across the hitting time gives

X_{n} - X_{n - 1} = H_{n} (X_{n} - X_{n - 1}),

(388)

and therefore

X_{n} = \sum_{k = 1}^{n} H_{k} (X_{k} - X_{k - 1}) .

(389)

Part (c) shows that

X

is a martingale. Its increments still take values in

{- 1, 1}

, so part (b) shows that they are IID symmetric signs. Consequently

X

is a simple symmetric random walk. This is the path transformation behind the reflection principle for simple symmetric random walk.

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e

Words: 118 Articles: 1

Solution

Words: 118

The reflection map from part (d) bijects paths that hit

a

and finish below

a

with paths whose reflected endpoint lies strictly above

a

. Since the reflected walk has the same distribution,

P (M_{n} \geq a) = P (X_{n} \geq a) + P (X_{n} > a) = P (X_{n} \geq a) + P (X_{n} \geq a + 1) .

(390)

Subtracting the corresponding identity at level

a + 1

gives

P (M_{n} = a) = P (X_{n} = a) + P (X_{n} = a + 1) .

(391)

For a simple symmetric random walk,

P (X_{n} = k) = 2^{- n} (( n + k ) /2 n)

(392)

when

∣ k ∣ \leq n

and

n + k

is even, and it is zero otherwise. Therefore the explicit answer is

P (M_{n} = a) = 2^{- n} [(( n + a ) /2 n) + (( n + a + 1 ) /2 n)],

(393)

where a binomial coefficient is interpreted as zero when its lower argument is not an integer in

{0, \dots, n}

. This is the point probability for the maximum of simple symmetric random walk.

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31J (Mathematics of Machine Learning)

Words: 254 Articles: 1

Solution

Words: 254

For points

x_{1}, \dots, x_{n} \in X

, the restriction of

H

to those points is the set of binary vectors

H ∣_{x_{1 : n}} = {(h (x_{1}), \dots, h (x_{n})) : h \in H} .

(394)

The shattering coefficient is

s (H, n) = x_{1}, \dots, x_{n} \in X sup ∣ H ∣_{x_{1 : n}} ∣ .

(395)

The VC dimension is

VC (H) = sup {n : s (H, n) = 2^{n}},

(396)

with value infinity when arbitrarily large finite sets are shattered.

H^{'} \subseteq H

, then for every choice of points,

H^{'} ∣_{x_{1 : n}} \subseteq H ∣_{x_{1 : n}} .

(397)

Thus every set shattered by

H^{'}

is also shattered by

H

, and

VC (H^{'}) \leq VC (H) .

(398)

Now put

d = dim F

and suppose that

x_{1}, \dots, x_{n}

are shattered by

H = {1_{{u : f (u) \leq 0}} : f \in F^{'}} .

(399)

Consider the linear map

T : F ⟶ R^{n}, T f = (f (x_{1}), \dots, f (x_{n})) .

(400)

n > d

, the rank-nullity theorem shows that

im T

is a proper subspace of

R^{n}

. Choose a nonzero

α \in (im T)^{⊥}

(401)

and replace

α

- α

if necessary so that at least one coordinate is positive.

Ask for the labeling that assigns label zero when

α_{i} > 0

and label one when

α_{i} < 0

; coordinates with

α_{i} = 0

may be labeled arbitrarily. Shattering would supply

f \in F^{'}

such that

α_{i} > 0 ⟹ f (x_{i}) > 0, α_{i} < 0 ⟹ f (x_{i}) \leq 0.

(402)

Every product

α_{i} f (x_{i})

is then nonnegative, and at least one is strictly positive. Hence

α \cdot T f > 0,

(403)

contradicting

α ⊥ im T

. Therefore

n \leq d

, proving

VC (H) \leq dim F .

(404)

This is the VC dimension of a vector space argument.

Finally, a closed Euclidean ball with center

c

and radius

r

{x : ∥ x - c ∥_{2}^{2} \leq r^{2}} = {x : f_{c, r} (x) \leq 0},

(405)

where

f_{c, r} (x) = ∥ x ∥_{2}^{2} - 2 c^{T} x + ∥ c ∥_{2}^{2} - r^{2} .

(406)

Every such function lies in

F = span {∥ x ∥_{2}^{2}, x_{1}, \dots, x_{d}, 1},

(407)

whose dimension is at most

d + 2

. Applying the result just proved gives the VC dimension upper bound for Euclidean balls

VC (H) \leq d + 2 .

(408)

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32A (Dynamical Systems)

Words: 611 Articles: 11

a

Words: 109 Articles: 1

Solution

Words: 109

Let

x_{*}

be an equilibrium point of a dynamical system. A Lyapunov function on a neighbourhood of

x_{*}

is a continuously differentiable function

V

such that

V (x_{*}) = 0, V (x) > 0 (x \neq = x_{*}),

(409)

and whose derivative along trajectories satisfies

\dot{V} (x) = \nabla V (x) \cdot f (x) \leq 0.

(410)

The First Lyapunov theorem says that the existence of such a positive-definite

V

proves Lyapunov stability of

x_{*}

. If

\dot{V} < 0

away from

x_{*}

, the Second Lyapunov theorem strengthens this to asymptotic stability.

LaSalle invariance principle says that if a trajectory remains in a compact positively invariant set

Ω

on which

\dot{V} \leq 0

, then it approaches the largest invariant subset of

{x \in Ω : \dot{V} (x) = 0} .

(411)

In particular, if that largest invariant subset consists only of

x_{*}

, every trajectory in

Ω

approaches

x_{*}

Solved by gpt-5.6-sol high.

b

Words: 112 Articles: 1

Solution

Words: 112

Set

U (x) = \frac{x ^{2}}{( 1 + x ^{2} ) ^{2}}, E (x, y) = \frac{1}{2} y^{2} + U (x) .

(412)

The stated identity gives

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

(413)

Hence, along the system,

\dot{E} = y \overset{y}{˙} + U^{'} (x) \overset{x}{˙} = y [- U^{'} (x) - k y] + U^{'} (x) y = - k y^{2} \leq 0.

(414)

Near the origin,

E

is positive definite because

U (x) > 0

for

x \neq = 0

. The First Lyapunov theorem therefore gives stability.

Choose a sufficiently small compact invariant sublevel set

Ω_{c} = {(x, y) : E (x, y) \leq c}, 0 < c < \frac{1}{4} .

(415)

{\dot{E} = 0}

we have

y = 0

. A trajectory can remain there only if

\overset{y}{˙} = - U^{'} (x) = 0,

(416)

which occurs at

x = 0, \pm 1

. The choice

c < 1/4 = U (\pm 1)

excludes the two latter points, so the largest invariant subset of

{\dot{E} = 0} \cap Ω_{c}

is the origin. LaSalle invariance principle now gives convergence to the origin. Thus, for every

k > 0

(0, 0) is asymptotically stable .

(417)

This is an instance of damped mechanical energy as a Lyapunov function.

Solved by gpt-5.6-sol high.

c

Words: 268 Articles: 4

i

Words: 106 Articles: 1

Solution

Words: 106

When

k = 0

, the system is conservative and

E (x, y) = \frac{1}{2} y^{2} + \frac{x ^{2}}{( 1 + x ^{2} ) ^{2}}

(418)

is constant. The fixed points are

(0, 0), (1, 0), (- 1, 0) .

(419)

The potential

U

has a strict minimum

U (0) = 0

, strict maxima

U (\pm 1) = \frac{1}{4},

(420)

and tends to zero as

∣ x ∣ \to \infty

. The conservative planar phase portrait therefore has:

a center at the origin, surrounded by closed periodic energy curves for $0 < E < 1/4$ ;
saddles at $(\pm 1, 0)$ ;
the level $E = 1/4$ as the separatrix, including the two heteroclinic branches joining the saddles around the central periodic region;
unbounded outer trajectories on either side of the barriers, and crossing trajectories when $E > 1/4$ .

The portrait is symmetric under

(x, y) \mapsto (- x, - y)

and under time reversal

(x, y, t) \mapsto (x, - y, - t)

Solved by gpt-5.6-sol high.

ii

Words: 162 Articles: 1

Solution

Words: 162

For

0 < k ≪ 1

, the same three fixed points remain. At the origin the Jacobian is

(0 - 2 1 - k) .

(421)

Its eigenvalues have negative real part; for small

k

they are complex, so the origin is a stable spiral. At

x = \pm 1

U^{''} (\pm 1) < 0

, making the Jacobian determinant negative, so both points remain saddles.

Every nonstationary trajectory moves across the former energy contours in the direction of decreasing

E = \frac{1}{2} y^{2} + U (x), \dot{E} = - k y^{2} < 0.

(422)

The unstable branches of the two saddles directed into the central region spiral into the origin, while their outward branches escape toward

x = \pm \infty

. The stable manifolds of

(\pm 1, 0)

are the separatrices that form the boundary of the basin of attraction of the origin. Initial points between these stable separatrices lose enough energy to remain between the barriers and spiral to the origin; points outside escape to one of the two infinities.

This gives the weakly damped damped rational double-barrier phase portrait and identifies the requested domain of stability as the open region bounded by those stable manifolds.

Solved by gpt-5.6-sol high.

d

Words: 122 Articles: 1

Solution

Words: 122

(1, y_{0})

the trajectory initially moves into

x > 1

. Suppose

y

had a first zero at time

t_{*} > 0

. Before that time

y > 0

, hence

x (t_{*}) > 1

. Since

U^{'} (x) < 0

for

x > 1

\overset{y}{˙} (t_{*}) = - U^{'} (x (t_{*})) > 0,

(423)

which is incompatible with crossing from positive values to zero. Thus

y (t) > 0, x (t) > 1

(424)

for all

t > 0

The energy from part (b) decreases strictly because

\dot{E} = - k y^{2} < 0.

(425)

Since

U \geq 0

\frac{1}{2} y (t)^{2} \leq E (t) < E (0) = \frac{1}{2} y_{0}^{2} + \frac{1}{4} .

(426)

Therefore

0 < y (t) < y_{0}^{2} + \frac{1}{2} (t > 0) .

(427)

Finally, fix

ε > 0

. If the trajectory never entered

0 < y < ε

, positivity would force

y (t) \geq ε

for all

t > 0

. Then

\dot{E} (t) = - k y (t)^{2} \leq - k ε^{2},

(428)

E (t) \leq E (0) - k ε^{2} t,

(429)

which becomes negative for large

t

, contradicting

E \geq 0

. Hence the trajectory must enter every such strip. This proves outward escape from the rational potential barrier.

Solved by gpt-5.6-sol high.

33D (Integrable Systems)

Words: 242 Articles: 6

a

Words: 73 Articles: 1

Solution

Words: 73

The system prescribes both first derivatives of the same vector

ψ

, so it is overdetermined: the two mixed derivatives must agree. Differentiating gives

\partial_{\overset{z}{ˉ}} \partial_{z} ψ \partial_{z} \partial_{\overset{z}{ˉ}} ψ = (\partial_{\overset{z}{ˉ}} U) ψ + U \partial_{\overset{z}{ˉ}} ψ = (\partial_{\overset{z}{ˉ}} U + U V) ψ, = (\partial_{z} V) ψ + V \partial_{z} ψ = (\partial_{z} V + V U) ψ .

(430)

Thus every solution satisfies

(\partial_{\overset{z}{ˉ}} U - \partial_{z} V + [U, V]) ψ = 0.

(431)

For the system to possess arbitrary nontrivial local solution data, the coefficient matrix must vanish:

\partial_{\overset{z}{ˉ}} U - \partial_{z} V + [U, V] = 0 .

(432)

This is the zero-curvature condition, or the compatibility condition for an overdetermined linear system.

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b

Words: 60 Articles: 1

Solution

Words: 60

Substitute

U = \frac{1}{2} (u_{z} λ^{- 1} e^{u} λ^{- 1} e^{- u} - u_{z}), V = \frac{1}{2} (- u_{\overset{z}{ˉ}} λ e^{- u} λ e^{u} u_{\overset{z}{ˉ}}) .

(433)

The off-diagonal entries of

U_{\overset{z}{ˉ}} - V_{z} + [U, V]

(434)

cancel identically. Its upper-left entry is

u_{z \overset{z}{ˉ}} + \frac{1}{4} (e^{- 2 u} - e^{2 u}) = u_{z \overset{z}{ˉ}} - \frac{1}{2} sinh (2 u),

(435)

and its lower-right entry is the negative of this. Hence compatibility is equivalent to

u_{z \overset{z}{ˉ}} = \frac{1}{2} sinh (2 u) .

(436)

This is the Sinh-Gordon equation in complex light-cone coordinates.

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c

Words: 109 Articles: 1

Solution

Words: 109

When

u

is independent of

y

and

z = x + i y

\partial_{z} = \partial_{\overset{z}{ˉ}} = \frac{1}{2} \frac{d}{d x} .

(437)

Put

L = U - V, M = U + V .

(438)

The zero-curvature equation becomes

\frac{1}{2} L_{x} + [U, V] = 0.

(439)

Since

[M, L] = [U + V, U - V] = - 2 [U, V],

(440)

we obtain the isospectral Lax equation

L_{x} = [M, L] .

(441)

Therefore, for every positive integer

n

, the cyclic property of the matrix trace gives

\frac{d}{d x} tr (L^{n}) = n tr (L^{n - 1} [M, L]) = n tr (L^{n - 1} M L - M L^{n}) = 0.

(442)

These are the trace invariants of a Lax equation.

The PDE from part (b) reduces to

\frac{1}{4} u_{xx} = \frac{1}{2} sinh (2 u) .

(443)

With

ϕ = 2 u

, this is

ϕ^{''} = 4 sinh ϕ .

(444)

To extract a nontrivial invariant, use

n = 2

. Writing

p = u_{x}

, direct multiplication gives

tr (L^{2}) = \frac{1}{2} [p^{2} + λ^{2} + λ^{- 2} - 2 cosh (2 u)] .

(445)

The terms involving

λ

are constant, and

p = ϕ^{'} /2

. Hence

\frac{1}{2} (ϕ^{'})^{2} - 4 cosh ϕ = C

(446)

is a first integral. Direct differentiation verifies it:

\frac{d}{d x} [\frac{1}{2} (ϕ^{'})^{2} - 4 cosh ϕ] = ϕ^{'} (ϕ^{''} - 4 sinh ϕ) = 0.

(447)

Solved by gpt-5.6-sol high.

34B (Principles of Quantum Mechanics)

Words: 309 Articles: 4

a

Words: 148 Articles: 1

Solution

Words: 148

The Born rule depends only on transition probabilities

∣ ⟨ ϕ ∣ ψ ⟩ ∣^{2}

(448)

between rays. A physical transformation must preserve these probabilities. By Wigner theorem, such a ray transformation is induced by a unitary or antiunitary operator. For ordinary continuously connected transformations one chooses the unitary branch, so

U (g)^{†} U (g) = I_{H} .

(449)

The states

U (g_{1}) U (g_{2}) ∣ ψ ⟩

and

U (g_{1} g_{2}) ∣ ψ ⟩

represent the same ray. They may therefore differ by a state-independent quantum phase:

U (g_{1}) U (g_{2}) = e^{i ϕ (g_{1}, g_{2})} U (g_{1} g_{2}) .

(450)

Thus quantum transformations form a projective unitary representation.

g

is a symmetry of the Hamiltonian operator

H

, it also obeys

U (g)^{†} H U (g) = H,

(451)

equivalently

[U (g), H] = 0

. For a differentiable one-parameter subgroup, write

U (s) = e^{- i s Q /ℏ}

(452)

with a self-adjoint generator

Q

. Differentiating the symmetry equation at

s = 0

yields

[Q, H] = 0.

(453)

For a time-independent

Q

, the Heisenberg picture equation is

\frac{d Q}{d t} = \frac{i}{ℏ} [H, Q] = 0.

(454)

Hence the generator is a conserved observable. Conversely, a self-adjoint conserved

Q

commutes with

H

and its unitary group generates symmetries. This is the conserved generator of a continuous quantum symmetry.

Solved by gpt-5.6-sol high.

b

Words: 161 Articles: 1

Solution

Words: 161

The unitary time evolution is

U (t) = e^{- i H t /ℏ} .

(455)

First suppose that time reversal

T

is linear and unitary. Differentiating

U (t) T = T U (- t)

(456)

t = 0

gives

- \frac{i}{ℏ} H T = T \frac{i}{ℏ} H,

(457)

and linearity permits the scalar

i

to pass through

T

. Hence

H T = - T H .

(458)

H ∣ E ⟩ = E ∣ E ⟩

, then

H (T ∣ E ⟩) = - E (T ∣ E ⟩) .

(459)

Thus every energy has its negative as an energy. The Coulomb Hamiltonian has arbitrarily large positive energies in its continuum, so this symmetry would force energies arbitrarily far below zero. It would have no lowest energy eigenvalue and therefore no stable ground state. This is the unitary time reversal reverses the energy spectrum obstruction.

Now let

T

be antiunitary. It is conjugate linear, so

T (i ∣ ψ ⟩) = - i T ∣ ψ ⟩ .

(460)

Differentiating the same intertwining equation gives

- \frac{i}{ℏ} H T ∣ ψ ⟩ = T (\frac{i}{ℏ} H ∣ ψ ⟩) = - \frac{i}{ℏ} T H ∣ ψ ⟩ .

(461)

Therefore

H T = T H .

(462)

Time reversal now maps an energy eigenstate to another state with the same energy:

H (T ∣ E ⟩) = E (T ∣ E ⟩) .

(463)

The spectrum is preserved rather than reflected through zero, so the lower-bounded Coulomb spectrum and its stable ground state are compatible with time-reversal symmetry. This is why antiunitary time reversal preserves the energy spectrum.

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35B (Application of Quantum Mechanics)

Words: 383 Articles: 11

a

Words: 114 Articles: 1

Solution

Words: 114

Let

H

be the one-dimensional Hamiltonian operator and let

E_{0} \leq E_{1} \leq E_{2} \leq \dots

(464)

be its discrete energy eigenvalues. Expanding a normalized trial wavefunction in an orthonormal energy basis,

∣ ψ ⟩ = \sum_{n} c_{n} ∣ n ⟩,

(465)

gives

⟨ ψ ∣ H ∣ ψ ⟩ = \sum_{n} ∣ c_{n} ∣^{2} E_{n} \geq E_{0} \sum_{n} ∣ c_{n} ∣^{2} = E_{0} .

(466)

Thus the Rayleigh-Ritz variational principle gives

E_{0} \leq \frac{\int _{R} ψ ^{*} ( x ) H ψ ( x ) d x}{\int _{R} ∣ ψ ( x ) ∣ ^{2} d x}

(467)

for every nonzero admissible trial wavefunction. One chooses a parameterized family and minimizes the quotient.

When

V (x) = V (- x)

, parity commutes with

H

. In one dimension the nondegenerate ground state is even and the first excited state is odd. Restricting the trial family to odd wavefunctions makes every trial state orthogonal to the ground state, so

E_{1} \leq \frac{⟨ ψ _{odd} ∣ H ∣ ψ _{odd} ⟩}{⟨ ψ _{odd} ∣ ψ _{odd} ⟩} .

(468)

This is the odd-state variational principle for an even potential.

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b

Words: 269 Articles: 8

i

Words: 65 Articles: 1

Solution

Words: 65

The condition

2 m = ℏ^{2}

makes

H = - \frac{d ^{2}}{d x ^{2}} - V_{0} e^{- x^{2}} .

(469)

For

ψ_{a} (x) = e^{- a x^{2} /2}, a > 0,

(470)

the squared norm is the Gaussian integral

⟨ ψ_{a} ∣ ψ_{a} ⟩ = \int_{R} e^{- a x^{2}} d x = \frac{π}{a} .

(471)

Since

ψ_{a}^{'} = - a x ψ_{a}

, integration by parts gives the kinetic quotient

\frac{\int ψ _{a} ( - ψ _{a}^{''} ) d x}{\int ψ _{a}^{2} d x} = \frac{\int ∣ ψ _{a}^{'} ∣ ^{2} d x}{\int ψ _{a}^{2} d x} = a^{2} \frac{\int x ^{2} e ^{- a x^{2}} d x}{\int e ^{- a x^{2}} d x} = \frac{a}{2} .

(472)

The potential quotient is

- V_{0} \frac{\int e ^{- (a + 1) x^{2}} d x}{\int e ^{- a x^{2}} d x} = - V_{0} \frac{a}{1 + a} .

(473)

The variational principle therefore yields the Gaussian variational bound for an attractive Gaussian well

E_{0} \leq E (a) = \frac{a}{2} - V_{0} \frac{a}{1 + a} .

(474)

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ii

Words: 41 Articles: 1

Solution

Words: 41

The positive zero of

E (a)

satisfies

\frac{a}{2} = V_{0} \frac{a}{1 + a} .

(475)

Squaring and using

a > 0

gives

a (1 + a) = 4 V_{0}^{2},

(476)

a_{0} = \frac{1 + 16 V _{0}^{2} - 1}{2} .

(477)

Furthermore,

E^{'} (a) = \frac{1}{2} - \frac{V _{0}}{2 a ( 1 + a ) ^{3/2}} .

(478)

At an extremum this vanishes, and hence

V_{0} = a (1 + a)^{3/2} .

(479)

Squaring gives the required equation

(1 + a)^{3} = \frac{V _{0}^{2}}{a} .

(480)

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iii

Words: 104 Articles: 1

Solution

Words: 104

a ↓ 0

E (a) = - V_{0} a + O (a),

(481)

E (a) < 0

for all sufficiently small positive

a

. On the other hand,

E (a) ⟶ + \infty (a \to \infty) .

(482)

The derivative satisfies

E^{'} (a) \to - \infty (a ↓ 0), E^{'} (a) \to \frac{1}{2} (a \to \infty) .

(483)

The extremum equation

a (1 + a)^{3} = V_{0}^{2}

(484)

has exactly one positive solution because its left side is strictly increasing. Thus

E

has one global minimum, and that minimum is negative.

The potential tends to zero at spatial infinity, so zero is the threshold of the continuous spectrum. A trial state with negative energy expectation forces the spectrum below zero; the corresponding lowest spectral value is a normalizable bound state. Hence every

V_{0} > 0

produces at least one bound state.

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iv

Words: 59 Articles: 1

Solution

Words: 59

Since

- V_{0} < V (x) \leq 0

(485)

except at the single point

x = 0

, and the kinetic-energy expectation is nonnegative, every normalized wavefunction satisfies

⟨ H ⟩ > - V_{0} .

(486)

Therefore

E_{0} > - V_{0} .

(487)

For the upper bound, use the admissible choice

a = V_{0}^{2} .

(488)

Then

E (V_{0}^{2}) = \frac{1}{2} V_{0}^{2} - \frac{V _{0}^{2}}{1 + V _{0}^{2}} = - \frac{1}{2} V_{0}^{2} + \frac{1}{2} V_{0}^{4} + O (V_{0}^{6}) .

(489)

The Rayleigh-Ritz variational principle now gives

- V_{0} < E_{0} \leq - \frac{1}{2} V_{0}^{2} + O (V_{0}^{3}) .

(490)

Indeed, the displayed calculation provides the sharper remainder

O (V_{0}^{4})

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36C (Statistical Physics)

Words: 305 Articles: 8

a

Words: 38 Articles: 1

Solution

Words: 38

For a fixed number

N

of particles, the ideal gas equation of state is

p V = N k_{B} T .

(491)

Its internal energy depends only on temperature, and

d E = C_{V} d T .

(492)

When the heat capacity at constant volume is constant, a choice of the energy zero gives

E = C_{V} T .

(493)

Solved by gpt-5.6-sol high.

b

Words: 35 Articles: 1

Solution

Words: 35

At constant volume, the First law of thermodynamics gives

δ Q = d E = C_{V} d T .

(494)

At constant pressure,

δ Q = C_{p} d T = d E + p d V .

(495)

Differentiating

p V = N k_{B} T

at fixed

p

gives

p d V = N k_{B} d T .

(496)

Consequently,

C_{p} d T = (C_{V} + N k_{B}) d T,

(497)

and the Mayer relation is

C_{p} - C_{V} = N k_{B} .

(498)

The specific-heat ratio is therefore

γ = \frac{C _{p}}{C _{V}} = 1 + \frac{N k _{B}}{C _{V}} .

(499)

Solved by gpt-5.6-sol high.

c

Words: 47 Articles: 1

Solution

Words: 47

An adiabatic process exchanges no heat with its surroundings, so

δ Q = 0

. For a reversible process it is also isentropic. The first law then gives

C_{V} d T = d E = - p d V .

(500)

Using

p = N k_{B} T / V

and the Mayer relation,

\frac{d T}{T} = - \frac{N k _{B}}{C _{V}} \frac{d V}{V} = - (γ - 1) \frac{d V}{V} .

(501)

Integration yields

T V^{γ - 1} = constant .

(502)

Substituting

T = p V / (N k_{B})

gives the reversible ideal-gas adiabat

p V^{γ} = constant .

(503)

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d

Words: 185 Articles: 1

Solution

Words: 185

In the

(p, V)

plane, the Otto cycle consists of:

$A \to B$ : an adiabatic compression along $p V^{γ} = constant$ from $V_{1}$ to $V_{2}$ ;
$B \to C$ : a vertical constant-volume rise at $V_{2}$ ;
$C \to D$ : an adiabatic expansion back to $V_{1}$ ;
$D \to A$ : a vertical constant-volume fall at $V_{1}$ .

It is a clockwise loop, so its enclosed area is the positive net work output. In the

(T, S)

plane, the two reversible adiabats

A \to B

and

C \to D

are vertical constant-entropy segments. Constant-volume heating

B \to C

moves upward and to the right, while constant-volume cooling

D \to A

moves downward and to the left.

Let

r = \frac{V _{1}}{V _{2}} > 1.

(504)

The two adiabatic relations give

T_{B} = T_{A} r^{γ - 1}, T_{C} = T_{D} r^{γ - 1} .

(505)

Because heat is exchanged at constant volume,

Q_{1} = C_{V} (T_{C} - T_{B}), Q_{2} = C_{V} (T_{D} - T_{A}),

(506)

where

Q_{2}

denotes the positive amount rejected. Hence

\frac{Q _{2}}{Q _{1}} = \frac{T _{C} / r ^{γ - 1} - T _{A}}{T _{C} - T _{A} r ^{γ - 1}} = \frac{1}{r ^{γ - 1}} .

(507)

The thermal efficiency is therefore

η = \frac{W}{Q _{1}} = 1 - \frac{Q _{2}}{Q _{1}} = 1 - \frac{1}{r ^{γ - 1}} .

(508)

For a fixed gas,

γ

is fixed and the idealized efficiency increases monotonically with the compression ratio

r

, so it is maximized by making

r

as large as the physical constraints allow.

Solved by gpt-5.6-sol high.

37C (Electrodynamics)

Words: 357 Articles: 8

a

Words: 46 Articles: 1

Solution

Words: 46

With

\partial_{μ} = (\frac{1}{c} \frac{\partial}{\partial t}, \nabla),

(509)

the electromagnetic field tensor is the antisymmetric tensor

F^{μν} = \partial^{μ} A^{ν} - \partial^{ν} A^{μ} .

(510)

Under a Lorentz transformation

x^{' μ} = Λ^{μ}_{ν} x^{ν}

, the four-potential transforms as a four-vector and therefore

F^{' μν} (x^{'}) = Λ^{μ}_{ρ} Λ^{ν}_{σ} F^{ρ σ} (x) .

(511)

Thus

F

is an antisymmetric rank-two tensor under Lorentz transformations.

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b

Words: 85 Articles: 1

Solution

Words: 85

Define the dual tensor by

F^{μν} = \frac{1}{2} ϵ^{μν ρ σ} F_{ρ σ} .

(512)

For metric signature

(-, +, +, +)

and

ϵ^{0123} = + 1

, two independent quadratic electromagnetic field invariants are

F_{μν} F^{μν} = 2 (B^{2} - \frac{E ^{2}}{c ^{2}})

(513)

and

F_{μν} F^{μν} = - \frac{4}{c} E \cdot B .

(514)

Changing orientation or the sign convention for

F

can change the displayed overall signs but not their vanishing.

For a vacuum plane electromagnetic wave with wavevector

k

k \cdot E = 0, B = \frac{k \times E}{ω}, ω = c ∣ k ∣.

(515)

Hence

E \cdot B = 0, B^{2} = \frac{E ^{2}}{c ^{2}},

(516)

for every transverse polarization. Both Lorentz scalars therefore vanish.

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c

Words: 115 Articles: 1

Solution

Words: 115

Choose, for example,

B_{0} = (0, 0, B_{0}), E_{0} = (c B_{0}, 0, 0), B_{0} \neq = 0.

(517)

The fields are perpendicular and satisfy

E_{0}^{2} = c^{2} B_{0}^{2}

, so both invariants vanish. Every nonzero constant background with vanishing invariants can be brought to this form by a spatial rotation, up to signs.

In this frame, and with a convention consistent with the relativistic Lorentz force, the mixed tensor is

F^{μ}_{ν} = B_{0} 0100 10 - 1 0 01000000 .

(518)

Calling the displayed dimensionless matrix

N

, direct multiplication gives

N^{2} = 10 - 1 0 0000 10 - 1 0 0000, N^{3} = 0.

(519)

Consequently

F^{μ}_{ρ} F^{ρ}_{σ} F^{σ}_{ν} = 0 .

(520)

This is a tensor equation. Since it holds in the canonical frame, it holds in every Lorentz frame for any null electromagnetic field.

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d

Words: 111 Articles: 1

Solution

Words: 111

Let

X^{μ} (τ)

be the worldline,

u^{μ} = d X^{μ} / d τ

its four-velocity, and

Ω = \frac{q B _{0}}{m} .

(521)

The relativistic Lorentz force in the constant background is

\frac{d u ^{μ}}{d τ} = \frac{q}{m} F^{μ}_{ν} u^{ν} .

(522)

Because

F^{3} = 0

, its exponential terminates:

u (τ) = exp (\frac{q}{m} F τ) u (0) = [I + \frac{q}{m} F τ + \frac{1}{2} (\frac{q}{m} F τ)^{2}] u (0) .

(523)

The particle starts from rest, so

u (0) = (c, 0, 0, 0) .

(524)

Using the matrix from part (c) gives

u^{0} = c (1 + \frac{1}{2} Ω^{2} τ^{2}), u^{1} = c Ω τ, u^{2} = - \frac{c}{2} Ω^{2} τ^{2}, u^{3} = 0.

(525)

Integrating from the origin and using

X^{0} = c t

yields the parametric trajectory

t (τ) x (τ) y (τ) z (τ) = τ + \frac{Ω ^{2} τ ^{3}}{6}, = \frac{c Ω τ ^{2}}{2}, = - \frac{c Ω ^{2} τ ^{3}}{6}, = 0.

(526)

This is the relativistic trajectory in a constant null crossed field. The real cubic relation for

t (τ)

is strictly increasing, so it determines

τ

uniquely as a function of coordinate time if an explicit

x (t)

is desired.

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38C (General Relativity)

Words: 348 Articles: 9

a

Words: 98 Articles: 1

Solution

Words: 98

Contract the defining formula in its first and third indices. In four dimensions,

C^{α}_{β α δ} = = R_{β δ} - \frac{1}{2} (4 R_{β δ} + g_{β δ} R - R_{β δ} - R_{β δ}) + \frac{1}{6} (4 g_{β δ} - g_{β δ}) R 0.

(527)

Thus

C^{α}_{β α δ} = 0 .

(528)

The Weyl tensor has the same algebraic symmetries as the Riemann curvature tensor: it is antisymmetric within each index pair, symmetric under exchanging the pairs, and obeys the first Bianchi identity. A contraction within an antisymmetric pair vanishes against the symmetric metric. Every contraction between the two pairs can be converted by the pair symmetries to the one just calculated, possibly with a sign. Hence every contraction vanishes, proving the Trace-free property of the Weyl tensor.

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b

Words: 68 Articles: 1

Solution

Words: 68

Write

H_{α β} = \partial_{α} \partial_{β} ω_{p}, H = η^{α β} H_{α β} .

(529)

The Levi-Civita connection of

g_{α β} = e^{2 ω} η_{α β}

(530)

Γ_{α β}^{ρ} = δ_{α}^{ρ} \partial_{β} ω + δ_{β}^{ρ} \partial_{α} ω - η_{α β} \partial^{ρ} ω .

(531)

The assumed first-derivative condition makes

Γ_{α β}^{ρ} (p) = 0

. Differentiating once more gives, at

p

R_{α β γ δ} = e^{2 ω} (η_{α δ} H_{β γ} - η_{α γ} H_{β δ} - η_{β δ} H_{α γ} + η_{β γ} H_{α δ}) .

(532)

Its contractions are

R_{β δ} = - 2 H_{β δ} - η_{β δ} H, R = - 6 e^{- 2 ω} H .

(533)

Substitution into the defining expression for

C_{α β γ δ}

cancels every Hessian term, leaving

C_{α β γ δ} (p) = 0 .

(534)

Since

p

was arbitrary, this is the Weyl tensor of a conformally flat metric.

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c

Words: 182 Articles: 4

i

Words: 64 Articles: 1

Solution

Words: 64

The Schwarzschild exterior is vacuum, so

R_{μν} = 0, R = 0,

(535)

and consequently

C_{α β γ δ} = R_{α β γ δ}

. Put

f (r) = 1 - \frac{2 M}{r} .

(536)

The relevant Christoffel symbols are

Γ^{t}_{t r} = \frac{f ^{'}}{2 f}, Γ^{r}_{rr} = - \frac{f ^{'}}{2 f}, Γ^{r}_{tt} = \frac{1}{2} f f^{'} .

(537)

With the convention

R^{α}_{β γ δ} = \partial_{γ} Γ_{β δ}^{α} - \partial_{δ} Γ_{β γ}^{α} + \dots,

(538)

one finds

R_{t r t r} = \frac{1}{2} f^{''} (r) .

(539)

Since

f^{''} = - 4 M / r^{3}

C_{t r t r} = - \frac{2 M}{r ^{3}} (r ≫ 2 M) .

(540)

This is the leading Radial Weyl curvature of Schwarzschild spacetime. The opposite convention for the Riemann tensor reverses the sign but not the physical tidal magnitude.

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ii

Words: 118 Articles: 1

Solution

Words: 118

This curvature component produces radial tidal acceleration: nearby freely falling particles separated radially are stretched relative to one another, while transverse separations are compressed. At the Earth, after removing the Earth's own field, the Sun and especially the Moon produce the familiar ocean tides known since antiquity.

There is no conflict with the equivalence principle. A freely falling local frame can remove the uniform gravitational acceleration at one event by setting the connection to zero there. It cannot remove the Riemann curvature tensor, which controls relative acceleration through geodesic deviation and is detectable across an extended body such as the Earth. Tides measure this spatial variation of gravity rather than gravity at a single point.

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39A (Fluid Dynamics II)

Words: 458 Articles: 11

a

Words: 76 Articles: 1

Solution

Words: 76

For an incompressible Newtonian fluid with negligible inertia and no body force, the Stokes equations are

- \nabla p + μ \nabla^{2} u = 0, \nabla \cdot u = 0 .

(541)

They are linear because any linear combination of solutions corresponds to the same linear combination of boundary data and applied forces. They are reversible because replacing all imposed velocities and forces by their negatives sends

(u, p) ⟼ (- u, - p) .

(542)

Fluid particles therefore retrace their paths under reversal. This is kinematic reversibility of Stokes flow.

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b

Words: 382 Articles: 8

i

Words: 92 Articles: 1

Solution

Words: 92

The geometry and boundary velocities are axisymmetric and linear in the axial pseudovector

Ω

. The pressure is a true scalar. No nonzero parity-even scalar can be formed linearly from

Ω

and the radial vector

x

Ω \cdot x

is a pseudoscalar. Hence the flow-induced pressure is spatially constant.

Equivalently, taking the divergence of the Stokes momentum equation gives the Harmonic pressure and vorticity in Stokes flow result

\nabla^{2} p = 0,

(543)

and symmetry plus the homogeneous boundary data excludes every nonconstant harmonic pressure mode. Since pressure is defined up to an additive constant, we choose

p = 0 .

(544)

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ii

Words: 90 Articles: 1

Solution

Words: 90

Both candidate fields are divergence-free. For solid-body rotation,

u = Ω \times x,

(545)

each component is linear in

x

, so

\nabla^{2} u = 0

. Also,

u = Ω \times \nabla (\frac{1}{r})

(546)

is harmonic for

r > 0

, because

\nabla^{2} (1/ r) = 0

there and constant-coefficient derivatives commute with the Laplacian. Thus both solve the Stokes equations with

p = 0

The rotationally symmetric solution is consequently

u = (A + \frac{B}{r ^{3}}) Ω \times x .

(547)

The no-slip boundary condition gives

A + \frac{B}{a ^{3}} = 1, A + \frac{B}{b ^{3}} = 0.

(548)

Therefore

A = - \frac{a ^{3}}{b ^{3} - a ^{3}}, B = \frac{a ^{3} b ^{3}}{b ^{3} - a ^{3}},

(549)

and

u (x) = \frac{a ^{3}}{b ^{3} - a ^{3}} (\frac{b ^{3}}{r ^{3}} - 1) Ω \times x .

(550)

This is the Rotational Stokes flow between concentric spheres.

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iii

Words: 65 Articles: 1

Solution

Words: 65

Write

f (r) = A + \frac{B}{r ^{3}}, w = Ω \times x, u = f w .

(551)

The velocity gradient of the solid-rotation part is antisymmetric, so it drops out of the Newtonian fluid stress tensor. Since

\partial_{i} f = - \frac{3 B}{r ^{5}} x_{i},

(552)

and

p = 0

, the stress is

σ_{ij} = - \frac{3 μ B}{r ^{5}} [x_{i} (Ω \times x)_{j} + x_{j} (Ω \times x)_{i}] .

(553)

Equivalently, with

n = x / r

σ_{ij} = - \frac{3 μ B}{r ^{3}} [n_{i} (Ω \times n)_{j} + n_{j} (Ω \times n)_{i}] .

(554)

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iv

Words: 135 Articles: 1

Solution

Words: 135

r = c

, the traction

σ n

exerted by the outer fluid on the inner fluid is

σ n = - \frac{3 μ B}{c ^{3}} (Ω \times n) .

(555)

The inner fluid exerts the opposite traction on the outer fluid. Its couple is therefore

G = - \int_{r = c} x \times (σ n) d S = 3 μ B c^{- 2} \int_{r = c} n \times (Ω \times n) d S .

(556)

Using

n \times (Ω \times n) = Ω - n (n \cdot Ω)

(557)

and the supplied spherical integral,

\int_{r = c} n_{i} n_{j} d S = \frac{4 π c ^{2}}{3} δ_{ij},

(558)

we obtain

\int_{r = c} n \times (Ω \times n) d S = \frac{8 π c ^{2}}{3} Ω .

(559)

Thus the radius cancels:

G = 8 π μ B Ω = \frac{8 π μ a ^{3} b ^{3}}{b ^{3} - a ^{3}} Ω .

(560)

This is the Torque in rotational Stokes flow between concentric spheres.

When

a ≪ b

G \sim 8 π μ a^{3} Ω,

(561)

the rotational drag torque for a sphere in an effectively unbounded fluid. If

h = b - a ≪ a

, then

b^{3} - a^{3} \sim 3 a^{2} h, a^{3} b^{3} \sim a^{6},

(562)

G \sim \frac{8 π μ a ^{4}}{3 h} Ω,

(563)

the inverse-gap growth expected from a thin Couette flow.

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40A (Waves)

Words: 344 Articles: 6

a

Words: 118 Articles: 1

Solution

Words: 118

Use the acoustic velocity potential convention

u = \nabla ϕ, p = - ρ_{0} \frac{\partial ϕ}{\partial t} .

(564)

Put

k = \frac{ω}{c _{0}}, α = k R .

(565)

For a spherically symmetric harmonic solution, write

ϕ (r, t) = F (r) sin ω t .

(566)

The wave equation and

\nabla^{2} F = \frac{1}{r} \frac{d ^{2}}{d r ^{2}} (r F)

(567)

give the radial Helmholtz equation

\frac{d ^{2}}{d r ^{2}} (r F) + k^{2} (r F) = 0.

(568)

The rigid inner sphere requires

u_{r} (R, t) = F^{'} (R) sin ω t = 0.

(569)

A form satisfying this condition automatically is

F (r) = \frac{C}{r} [cos k (r - R) + \frac{1}{k R} sin k (r - R)] .

(570)

Indeed, if the bracket is

h (r)

, then

h (R) = 1

and

h^{'} (R) = 1/ R

, so

(h / r)^{'} = 0

R

At the outer sphere,

ε p_{0} cos ω t = - ρ_{0} ω F (2 R) cos ω t,

(571)

F (2 R) = - \frac{ε p _{0}}{ρ _{0} ω} .

(572)

Since

F (2 R) = \frac{C}{2 R} (cos α + \frac{s i n α}{α}),

(573)

we obtain

C = - \frac{2 Rε p _{0}}{ρ _{0} ω ( c o s α + α ^{- 1} s i n α )} .

(574)

Therefore

ϕ (r, t) = - \frac{2 Rε p _{0}}{ρ _{0} ω r} \frac{cos k ( r - R ) + ( k R ) ^{- 1} sin k ( r - R )}{cos α + α ^{- 1} sin α} sin ω t .

(575)

The assumption

α \leq π /2

keeps the displayed denominator positive.

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b

Words: 103 Articles: 1

Solution

Words: 103

At every radius, the pressure perturbation has the form

p (r, t) = - ρ_{0} ω F (r) cos ω t,

(576)

whereas the radial velocity is

u_{r} (r, t) = F^{'} (r) sin ω t .

(577)

Define the average over one forcing period

T = 2 π / ω

⟨ I_{r} ⟩ = \frac{1}{T} \int_{0}^{T} p (r, t) u_{r} (r, t) d t .

(578)

Then

⟨ I_{r} ⟩ = - ρ_{0} ω F (r) F^{'} (r) \frac{1}{T} \int_{0}^{T} cos ω t sin ω t d t = 0 .

(579)

The same is true for the net flux through every concentric sphere.

Pressure and velocity are in temporal quadrature, so this is a reactive acoustic energy flux. The forcing establishes a standing wave in the closed, lossless annulus: compressional and kinetic energy exchange during each cycle, but no energy is transported outward on average. This is the pressure-forced standing acoustic wave in a spherical annulus.

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c

Words: 123 Articles: 1

Solution

Words: 123

At larger frequencies the denominator

D (α) = cos α + \frac{s i n α}{α}

(580)

can vanish. Its zeros satisfy

tan α = - α

(581)

and are the normal-mode frequencies for a pressure node at the outer sphere and zero radial velocity at the inner sphere. Near one of these radial acoustic resonances, the ideal undamped harmonic response becomes very large; exactly at resonance the linear lossless initial-value problem develops secular growth rather than a bounded periodic solution.

In a physical system, fluid viscosity, thermal losses, acoustic radiation, and damping in the elastic shell keep the amplitude finite and shift its phase. A nonzero pressure-velocity phase component then supplies positive mean power. At sufficiently large amplitude the linear acoustic approximation can also fail, leading to nonlinear frequency shifts or shock formation.

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