Paper 1

The Huffman coding theorem proves optimality, so an optimal coding with all but one word of the same length does exist. This is an instance of a nonunique optimal prefix code: a different resolution of the ties also gives the optimal profile

(2, 2, 2, 3, 3)

An optimal code cannot have five distinct lengths. By the deepest sibling property of an optimal prefix code, at least two maximal-length codewords have the same length. Thus the answers are respectively

(i) yes, (ii) no .

(31)

Solved by gpt-5.6-sol high.

4I (Automata & Formal Languages)

Words: 300 Articles: 11

a

Words: 53 Articles: 1

Solution

Words: 53

A context-free grammar is in Chomsky normal form when every production is of one of the forms

A \to BC, A \to a,

(32)

where

A, B, C

are nonterminal symbols and

a

is a terminal symbol. The conventional exceptional production

S \to ϵ

may be allowed when the language contains the empty word, with

S

excluded from right-hand sides.

Solved by gpt-5.6-sol high.

b

Words: 30 Articles: 1

Solution

Words: 30

The length is

2 n - 1 .

(33)

Indeed, the parse tree has

n

terminal-production nodes and, as a full binary tree with

n

leaves,

n - 1

binary-production nodes. This is the Chomsky-normal-form derivation length.

Solved by gpt-5.6-sol high.

c

Words: 69 Articles: 1

Solution

Words: 69

One parse tree can be recorded unambiguously by the nested expression

S (A (A (a), B (b)), C (B (b), A (A (A (a), B (b)), B (B (b), A (a))))) .

(34)

Every internal node uses one of the given productions, and the leaves read from left to right are

a, b, b, a, b, b, a .

(35)

Equivalently, the corresponding leftmost grammar derivation is

S \Rightarrow A C \Rightarrow A BC \Rightarrow a BC \Rightarrow ab C \Rightarrow ab B A \Rightarrow abb A \Rightarrow abb A B \Rightarrow abb A BB \Rightarrow abba BB \Rightarrow abbab B \Rightarrow abbab B A \Rightarrow abbabb A \Rightarrow abbabba .

(36)

Hence the required word is generated by the grammar.

Solved by gpt-5.6-sol high.

d

Words: 148 Articles: 4

i

Words: 86 Articles: 1

Solution

Words: 86

Take

G^{'} = ({a, b}, {S, A, B}, P^{'}, S),

(37)

where

P^{'} = {S \to A B A, A \to A B, B \to B A, A \to a, B \to b} .

(38)

This grammar is in Weak Chomsky normal form. In the original grammar,

C

occurs only in

S \to A C

and has the sole production

C \to B A

. Thus every use of

S \to A C \to A B A

(39)

is replaced in

G^{'}

S \to A B A

, while every later production is unchanged. Conversely, the ternary production can always be expanded into those two original steps. Therefore

G^{'}

is equivalent to

G

in the sense of an equivalent grammar.

Solved by gpt-5.6-sol high.

ii

Words: 62 Articles: 1

Solution

Words: 62

The compressed initial production gives the leftmost derivation

S \Rightarrow A B A \Rightarrow A BB A \Rightarrow a BB A \Rightarrow ab B A \Rightarrow abb A \Rightarrow abb A B \Rightarrow abb A BB \Rightarrow abba BB \Rightarrow abbab B \Rightarrow abbab B A \Rightarrow abbabb A \Rightarrow abbabba .

(40)

It has

12

steps, whereas part (b) gives

2 \cdot 7 - 1 = 13

for every derivation in Chomsky normal form. The Weak Chomsky normal form ternary rule has compressed exactly one pair of binary productions, so the derivation is strictly shorter as required.

Solved by gpt-5.6-sol high.

5J (Statistical Modelling)

Words: 118 Articles: 1

Solution

Words: 118

For

X = x

, the last flip must be the

r

th head. Among the preceding

x + r - 1

flips, exactly

x

are tails and

r - 1

are heads. There are

(x x + r - 1)

choices for their positions, and every resulting sequence has probability

(1 - p)^{x} p^{r}

. The negative binomial stopping argument therefore gives

P (X = x) = (x x + r - 1) (1 - p)^{x} p^{r}, x = 0, 1, \dots,

(41)

X

has the failures-before-the-

r

th-success negative binomial distribution.

To put the mass function into exponential family form, set

θ = lo g (1 - p), - \infty < θ < 0.

(42)

Since

p = 1 - e^{θ}

P_{θ} (X = x) = (x x + r - 1) exp {θ x + r lo g (1 - e^{θ})} = h (x) exp {θT (x) - A (θ)},

(43)

where

T (x) = x, A (θ) = - r lo g (1 - e^{θ}), h (x) = (x x + r - 1) .

(44)

Thus the natural parameter of an exponential family is

θ

, and the Fisher-Neyman factorization theorem shows that

T (X) = X

is a sufficient statistic. This is the negative binomial exponential family.

The exponential-family derivative identities now give

E [X] = A^{'} (θ) = \frac{r e ^{θ}}{1 - e ^{θ}} = \frac{r ( 1 - p )}{p}

(45)

and

var (X) = A^{''} (θ) = \frac{r e ^{θ}}{( 1 - e ^{θ} ) ^{2}} = \frac{r ( 1 - p )}{p ^{2}} .

(46)

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6C (Mathematical Biology)

Words: 145 Articles: 1

Solution

Words: 145

With population

n

, the transition diagram has the two outgoing arrows

n λ n + 3, n β n n - 1.

(47)

Thus this is a batch-birth linear-death process. If

P_{n} (t) = P (N_{t} = n)

and

P_{j} = 0

for

j < 0

, probability enters state

n

from

n - 3

by a birth and from

n + 1

by a death. Hence the master equation is

\dot{P}_{n} = λ P_{n - 3} + β (n + 1) P_{n + 1} - (λ + β n) P_{n} .

(48)

The Markov jump-process generator is

(L f) (n) = λ (f (n + 3) - f (n)) + β n (f (n - 1) - f (n)) .

(49)

For

f (n) = n

, this gives

L n = 3 λ - β n

. Therefore the expected value

m (t) = E [N_{t}]

satisfies

m^{'} = 3 λ - β m, m (0) = n_{0},

(50)

m (t) = \frac{3 λ}{β} (1 - e^{- βt}) + n_{0} e^{- βt} .

(51)

For

f (n) = n^{2}

L n^{2} = λ (6 n + 9) + β n (- 2 n + 1) .

(52)

Writing

s = E [N_{t}^{2}]

and using the generator identity gives

s^{'} = 6 λm + 9 λ - 2 β s + β m .

(53)

The variance

v = s - m^{2}

consequently obeys

v^{'} = s^{'} - 2 m m^{'} = 9 λ + β m - 2 β v .

(54)

v (0) = v_{0} < \infty

, substitution of the formula for

m

yields

v (t) = v_{0} e^{- 2 βt} + \frac{6 λ}{β} (1 - e^{- 2 βt}) + (n_{0} - \frac{3 λ}{β}) (e^{- βt} - e^{- 2 βt}) .

(55)

All transient terms vanish, and the moments of a batch-birth linear-death process therefore give

var (N_{t}) ⟶ \frac{6 λ}{β} .

(56)

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7E (Further Complex Methods)

Words: 120 Articles: 1

Solution

Words: 120

Taking reciprocals in the finite product defining the gamma function gives

\frac{1}{Γ ( z )} = z lim_{n \to \infty} n^{- z} \prod_{k = 1}^{n} (1 + \frac{z}{k}) .

(57)

Let

H_{n} = \sum_{k = 1}^{n} k^{- 1}

. Since

n^{- z} = e^{z (H_{n} - l o g n)} e^{- z H_{n}} = e^{z (H_{n} - l o g n)} \prod_{k = 1}^{n} e^{- z / k}

(58)

and

H_{n} - lo g n \to γ

, where

γ

is the Euler--Mascheroni constant, passage to the limit gives the Weierstrass product for the reciprocal gamma function

\frac{1}{Γ ( z )} = z e^{γ z} k = 1 \prod \infty (1 + \frac{z}{k}) e^{- z / k} .

(59)

The logarithmic derivative of this identity is

- ψ (z) = \frac{1}{z} + γ + \sum_{k = 1}^{\infty} (\frac{1}{z + k} - \frac{1}{k}) .

(60)

Since

k^{- 1} - (z + k)^{- 1} = z / [k (z + k)]

, the digamma function therefore satisfies

ψ (z) = - γ - \frac{1}{z} + z k = 1 \sum \infty \frac{1}{k ( z + k )} .

(61)

For real

z > 0

, termwise differentiation gives the trigamma function

ψ^{'} (z) = \frac{1}{z ^{2}} + \sum_{k = 1}^{\infty} \frac{1}{( z + k ) ^{2}} > 0.

(62)

Thus

ψ

is strictly increasing on the positive real axis. At the two positive integers needed here, telescoping gives

ψ (1) = - γ < 0, ψ (2) = 1 - γ > 0.

(63)

The intermediate value theorem supplies a zero in

(1, 2)

, and strict increase makes it unique. This is the positive zero of the digamma function.

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8D (Classical Dynamics)

Words: 166 Articles: 6

a

Words: 42 Articles: 1

Solution

Words: 42

Under the stated gauge transformation, the charged-particle electromagnetic Lagrangian becomes

L^{'} = \frac{1}{2} m \overset{r}{˙}^{2} - q (ϕ - \partial_{t} f - \overset{r}{˙} \cdot (A + \nabla f)) = L + q (\partial_{t} f + \overset{r}{˙} \cdot \nabla f) = L + q \frac{df ( r ( t ) , t )}{d t} .

(64)

Thus

L^{'} - L = q df / d t .

(65)

The action changes only by

q [f (t_{2}) - f (t_{1})]

, whose variation vanishes for fixed endpoints. By total-time-derivative invariance of a Lagrangian, the Euler-Lagrange equation is unchanged.

Solved by gpt-5.6-sol high.

b

Words: 51 Articles: 1

Solution

Words: 51

In Cartesian components,

\frac{\partial L}{\partial r ˙ _{i}} = m \overset{r}{˙}_{i} + q A_{i}

(66)

and

\frac{d}{d t} \frac{\partial L}{\partial r ˙ _{i}} = m \overset{r}{¨}_{i} + q \partial_{t} A_{i} + q \overset{r}{˙}_{j} \partial_{j} A_{i}, \frac{\partial L}{\partial r _{i}} = - q \partial_{i} ϕ + q \overset{r}{˙}_{j} \partial_{i} A_{j} .

(67)

The Euler-Lagrange equation therefore gives

m \overset{r}{¨}_{i} = q [- \partial_{i} ϕ - \partial_{t} A_{i} + \overset{r}{˙}_{j} (\partial_{i} A_{j} - \partial_{j} A_{i})] .

(68)

Using

E = - \nabla ϕ - \partial_{t} A, B = \nabla \times A,

(69)

and

[\overset{r}{˙} \times B]_{i} = \overset{r}{˙}_{j} (\partial_{i} A_{j} - \partial_{j} A_{i})

, this becomes the Lorentz force

m \overset{r}{¨} = q (E + \overset{r}{˙} \times B) .

(70)

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c

Words: 73 Articles: 1

Solution

Words: 73

Let

b = B /∣ B ∣

. For a constant uniform magnetic field, choose the symmetric gauge

A = \frac{1}{2} B \times r,

(71)

which has

\nabla \times A = B

and

A \cdot b = 0

. With

E = 0

, one may take

ϕ = 0

. The coordinate

r_{∥} = r \cdot b

is then absent from the Lagrangian, so it is an ignorable coordinate. Its generalized momentum

p_{∥} = \frac{\partial L}{\partial r ˙ _{∥}} = (m \overset{r}{˙} + q A) \cdot b = m \overset{r}{˙} \cdot b

(72)

is conserved. Hence

\frac{d}{d t} (p \cdot b) = 0,

(73)

which is the canonical momentum parallel to a uniform magnetic field in this gauge.

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9B (Cosmology)

Words: 147 Articles: 6

a

Words: 40 Articles: 1

Solution

Words: 40

For a comoving region,

E = ρ V

and

V = a^{3} V_{0}

. The work law gives

d (ρ V) = - P d V,

(74)

so differentiation with respect to time yields

V \overset{ρ}{˙} + (ρ + P) \dot{V} = 0.

(75)

Since

\dot{V} / V = 3 \overset{a}{˙} / a

, division by

V

gives the cosmological perfect-fluid continuity equation

\overset{ρ}{˙} + 3 \frac{a ˙}{a} (ρ + P) = 0 .

(76)

Solved by gpt-5.6-sol high.

b

Words: 46 Articles: 1

Solution

Words: 46

Put

K = 8 π G / (3 c^{2})

. Differentiating

Q = K ρ a^{2} - \overset{a}{˙}^{2}

gives

\dot{Q} = K (a^{2} \overset{ρ}{˙} + 2 a \overset{a}{˙} ρ) - 2 \overset{a}{˙} \overset{a}{¨} .

(77)

The continuity equation and the Friedmann acceleration equation imply

a^{2} \overset{ρ}{˙} + 2 a \overset{a}{˙} ρ = - a \overset{a}{˙} (ρ + 3 P)

(78)

and

- 2 \overset{a}{˙} \overset{a}{¨} = K a \overset{a}{˙} (ρ + 3 P) .

(79)

These terms cancel, so

\dot{Q} = 0 .

(80)

This is the First integral of the Friedmann acceleration equation.

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c

Words: 61 Articles: 1

Solution

Words: 61

For the barotropic equation of state

P = wρ

, the continuity equation separates as

\frac{d ρ}{ρ} = - 3 (1 + w) \frac{d a}{a} .

(81)

Since

a (t_{0}) = 1

ρ = ρ_{0} a^{- 3 (1 + w)}, ρ_{0} = ρ (t_{0}),

(82)

which is the constant-equation-of-state density scaling.

When

Q = 0

, the expanding branch satisfies

\overset{a}{˙} = \frac{8 π G ρ _{0}}{3 c ^{2}} a^{- (1 + 3 w) /2} = H_{0} a^{- (1 + 3 w) /2},

(83)

where

H_{0} = \overset{a}{˙} (t_{0}) > 0

. Because

w > - 1

, integration gives

a (t)^{3 (1 + w) /2} = 1 + \frac{3}{2} (1 + w) H_{0} (t - t_{0}),

(84)

and hence the flat constant-equation-of-state scale factor

a (t) = [1 + \frac{3}{2} (1 + w) H_{0} (t - t_{0})]^{2/ [3 (1 + w)]} .

(85)

It vanishes at

t_{*} = t_{0} - \frac{2}{3 ( 1 + w ) H _{0}} < t_{0},

(86)

∣ w ∣ < 1

ensures

1 + w > 0

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10D (Quantum Information and Computation)

Words: 314 Articles: 6

a

Words: 93 Articles: 1

Solution

Words: 93

Apply the hypothetical quantum cloning device repeatedly to obtain

N

copies of the unknown state. If

c = ∣ ⟨ ϕ ∣ ψ ⟩ ∣ < 1

, the two possibilities are

∣ ϕ ⟩^{\otimes N}, ∣ ψ ⟩^{\otimes N},

(87)

whose overlap has modulus

c^{N}

. For equal prior probabilities, the Helstrom-Holevo bound gives optimal success probability

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

(88)

Thus, for any

ε > 0

, sufficiently many clones permit success probability above

1 - ε

. Orthogonal states already have

c = 0

and can be distinguished perfectly. This is clone-assisted asymptotic state discrimination.

Here distinct states means distinct rays. Vectors differing only by a global phase represent the same quantum state and cannot be distinguished.

Solved by gpt-5.6-sol high.

b

Words: 88 Articles: 1

Solution

Words: 88

Run the perfect discriminator on the supplied state and record whether the result is

i = 0

i = 1

. For each

i

, choose a unitary

V_{i}

satisfying

V_{i} ∣0 ⟩ = ∣ ϕ_{i} ⟩,

(89)

which is possible by extending

∣ ϕ_{i} ⟩

to an orthonormal basis. Prepare two fresh systems in

∣0 ⟩ ∣0 ⟩

and, controlled by the classical outcome

i

, apply

V_{i} \otimes V_{i}

. The output is

∣ ϕ_{i} ⟩ ∣ ϕ_{i} ⟩ .

(90)

The discriminator may destroy its input; the two freshly prepared outputs still implement cloning on this known pair. This is the construction that perfect discrimination implies cloning for a known state family.

Solved by gpt-5.6-sol high.

c

Words: 133 Articles: 1

Solution

Words: 133

Suppose first that the stated unitary and environment states exist. Preservation of the inner product gives

⟨ ϕ_{0} ∣ ϕ_{1} ⟩ = ⟨ ψ_{0} ∣ ψ_{1} ⟩ ⟨ e_{0} ∣ e_{1} ⟩ .

(91)

The environment states are normalized, so the Cauchy-Schwarz inequality gives

∣ ⟨ e_{0} ∣ e_{1} ⟩ ∣ \leq 1

. Therefore

∣ ⟨ ϕ_{0} ∣ ϕ_{1} ⟩ ∣ \leq ∣ ⟨ ψ_{0} ∣ ψ_{1} ⟩ ∣.

(92)

Conversely, write

a = ⟨ ϕ_{0} ∣ ϕ_{1} ⟩, b = ⟨ ψ_{0} ∣ ψ_{1} ⟩

(93)

and assume

∣ a ∣ \leq ∣ b ∣

. If

b \neq = 0

, put

c = a / b

, so

∣ c ∣ \leq 1

, and choose

∣ e_{0} ⟩ = ∣0 ⟩, ∣ e_{1} ⟩ = c ∣0 ⟩ + 1 - ∣ c ∣^{2} ∣1 ⟩ .

(94)

Then

⟨ e_{0} ∣ e_{1} ⟩ = c

and the desired output vectors have inner product

b c = a

, exactly matching the input vectors. If

b = 0

, the inequality forces

a = 0

, and one may take

∣ e_{0} ⟩ = ∣ e_{1} ⟩ = ∣0 ⟩

In either case the input pair and output pair have the same Gram matrix. The isometry taking one pair to the other extends to a unitary matrix, so the required

U

exists. Hence

U exists ⟺ ∣ ⟨ ϕ_{0} ∣ ϕ_{1} ⟩ ∣ \leq ∣ ⟨ ψ_{0} ∣ ψ_{1} ⟩ ∣,

(95)

which is the environment-assisted two-state pure-state transformation criterion.

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11I (Coding and Cryptography)

Words: 331 Articles: 6

a

Words: 137 Articles: 1

Solution

Words: 137

A binary symmetric channel independently flips each input bit with probability

p

, so, with rows indexed by the input and columns by the output, its channel matrix is

(1 - p p p 1 - p) .

(96)

p > 1/2

, complementing every received bit converts the channel into one with crossover probability

1 - p < 1/2

without changing its information-carrying ability. The case

p = 1/2

has identical rows and capacity zero, so it suffices to study

p < 1/2

The Shannon second coding theorem states that for every rate below the channel capacity there are arbitrarily long codes whose decoding error tends to zero, whereas no sequence of codes with rate above capacity can have vanishing error. It also identifies

C = max_{P_{X}} I (X; Y) .

(97)

For the present channel,

I (X; Y) = H (Y) - H (Y ∣ X) = H (Y) - h_{2} (p) .

(98)

The output entropy is at most one bit, with equality for a uniform input. Hence the binary symmetric channel capacity is

C = 1 - h_{2} (p)

(99)

bits per channel use.

Solved by gpt-5.6-sol high.

b

Words: 114 Articles: 1

Solution

Words: 114

Let

X \in {0, 1}

denote whether

00

11

was sent, and write

Y_{j} = X \oplus E_{j},

(100)

where

E_{1}, E_{2}

are independent Bernoulli variables of parameter

p

. Because

X

is uniform,

Y_{1}

is uniform, while

H (Y_{1} ∣ X) = h_{2} (p)

. Thus the mutual information in the first received digit is

I (X; Y_{1}) = 1 - h_{2} (p) .

(101)

By the chain rule for mutual information, the extra information in the second digit is

I (X; Y_{2} ∣ Y_{1})

. Conditional on

X

, the second channel error is independent of the first output, so

H (Y_{2} ∣ X, Y_{1}) = H (E_{2}) = h_{2} (p) .

(102)

Moreover,

P (Y_{2} \neq = Y_{1}) = P (E_{2} \neq = E_{1}) = 2 p (1 - p) .

(103)

Given either value of

Y_{1}

, the second output therefore differs from it with probability

2 p (1 - p)

, and

H (Y_{2} ∣ Y_{1}) = h_{2} (2 p (1 - p)) .

(104)

Consequently the incremental information in a twofold binary repetition code is

I (X; Y_{2} ∣ Y_{1}) = h_{2} (2 p (1 - p)) - h_{2} (p)

(105)

bits.

Solved by gpt-5.6-sol high.

c

Words: 80 Articles: 1

Solution

Words: 80

Every row of the channel matrix is a permutation of

(1 - 2 α, α, α)

, and every column sum is one. It is therefore a weakly symmetric channel. The uniform input makes the output uniform, with entropy

lo g_{2} 3

, while the conditional output entropy is the entropy of one row:

H (Y ∣ X) = - (1 - 2 α) lo g_{2} (1 - 2 α) - 2 α lo g_{2} α .

(106)

The weakly symmetric channel capacity formula now gives the ternary symmetric channel capacity

C = lo g_{2} 3 + (1 - 2 α) lo g_{2} (1 - 2 α) + 2 α lo g_{2} α

(107)

bits per channel use, for

0 \leq α \leq 1/2

, with the convention

0 lo g_{2} 0 = 0

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12I (Automata & Formal Languages)

Words: 429 Articles: 12

i

Words: 35 Articles: 1

Solution

Words: 35

The relation

A \leq_{m} B

means that there is a total computable function

g : W \to W

such that

w \in A ⟺ g (w) \in B

(108)

for every word

w

. Such a

g

is a many-one reduction.

Solved by gpt-5.6-sol high.

ii

Words: 40 Articles: 1

Solution

Words: 40

A set

A

C

-complete under many-one reductions when

A \in C and (\forall B \in C) B \leq_{m} A .

(109)

The first condition is membership in the class; the second is

C

-hardness. This is a many-one complete set.

Solved by gpt-5.6-sol high.

iii

Words: 55 Articles: 1

Solution

Words: 55

A set

A \subseteq W

is a Sigma-1 set when it is the domain of a partial computable function: there is a partial computable

f

such that

w \in A ⟺ f (w) ↓ .

(110)

Equivalently,

w \in A

iff

\exists t R (w, t)

for some computable predicate

R

; one may take

R (w, t)

to say that the truncated computation function observes a halt within

t

steps.

Solved by gpt-5.6-sol high.

iv

Words: 45 Articles: 1

Solution

Words: 45

For an effective enumeration

(W_{w})_{w \in W}

of the computably enumerable languages, the diagonal halting problem is

K = {w \in W : w \in W_{w}} .

(111)

Equivalently, if

f_{w, 1}

is the unary partial function coded by

w

, then

w \in K ⟺ f_{w, 1} (w) ↓ .

(112)

Thus

K

is the diagonal halting set.

Solved by gpt-5.6-sol high.

v

Words: 130 Articles: 1

Solution

Words: 130

First,

K \in Σ_{1}

: the partial procedure on input

w

simulates

f_{w, 1} (w)

and halts exactly when that computation halts.

For hardness, let

A \in Σ_{1}

. Choose a partial computable function

g

with

w \in A ⟺ g (w) ↓ .

(113)

For each fixed

w

, define a unary program

P_{w}

as follows: on any input

x

, simulate

g (w)

; if that simulation halts, halt and output

x

, and otherwise run forever. The S-m-n theorem gives a total computable function

h

that maps

w

to a code for

P_{w}

. Its accepted language is

W_{h (w)} = {W, \emptyset, g (w) ↓, g (w) ↑ .

(114)

In particular,

w \in A ⟺ h (w) \in W_{h (w)} ⟺ h (w) \in K .

(115)

Thus

h

is a many-one reduction

A \leq_{m} K

. Since

A

was arbitrary,

K

Σ_{1}

-hard; together with membership this proves the many-one completeness of the halting problem.

Solved by gpt-5.6-sol high.

vi

Words: 124 Articles: 1

Solution

Words: 124

Membership is immediate from the supplied characterization of

Π_{2}

. Define the partial computable function

F (v, x) = f_{v, 1} (x) .

(116)

Then

v \in Tot ⟺ W_{v} = W ⟺ (\forall x \in W) F (v, x) ↓,

(117)

Tot \in Π_{2}

Now let

P \in Π_{2}

. There is a partial computable

f

such that

w \in P ⟺ (\forall x \in W) f (w, x) ↓ .

(118)

For each

w

, define a unary partial function

g_{w} (x) = {x, undefined, f (w, x) ↓, f (w, x) ↑ .

(119)

Operationally, its program simulates

f (w, x)

and returns

x

if that computation halts. By the S-m-n theorem, a total computable map

h

produces an index

h (w)

for

g_{w}

. Therefore

w \in P ⟺ (\forall x) g_{w} (x) ↓ ⟺ W_{h (w)} = W ⟺ h (w) \in Tot .

(120)

Thus every Pi-2 set many-one reduces to

Tot

. Combined with membership, this proves that the totality problem is

Π_{2}

-complete.

Solved by gpt-5.6-sol high.

13J (Statistical Modelling)

Words: 339 Articles: 10

i

Words: 110 Articles: 1

Solution

Words: 110

Write

Y = μ + ε, Y^{*} = μ + ε^{*},

(121)

where

ε

and

ε^{*}

are independent, centered, and have covariance matrix

σ^{2} I_{n}

. Then

H Y - Y^{*} = - (I - H) μ + H ε - ε^{*} .

(122)

The deterministic term is orthogonal in expectation to the two centered random terms, and the random terms are independent. Since the hat matrix is a symmetric idempotent projection of rank

p

E ∥ H ε ∥^{2} = σ^{2} tr (H^{T} H) = σ^{2} tr H = p σ^{2},

(123)

while

E ∥ ε^{*} ∥^{2} = n σ^{2}

. Therefore

E ∥ H Y - Y^{*} ∥^{2} = ∥ (I - H) μ ∥^{2} + (n + p) σ^{2} .

(124)

In this bias-variance decomposition for linear prediction,

∥ (I - H) μ ∥^{2}

is squared model bias,

p σ^{2}

is variance from fitting

p

coefficients, and

n σ^{2}

is irreducible noise in the future response. Enlarging the model tends to reduce the first term while increasing the fitted-model variance, which is the bias-variance tradeoff.

Solved by gpt-5.6-sol high.

ii

Words: 57 Articles: 1

Solution

Words: 57

Because

I - H

is an orthogonal projection of rank

n - p

E ∥ Y - H Y ∥^{2} = E ∥ (I - H) (μ + ε) ∥^{2} = ∥ (I - H) μ ∥^{2} + σ^{2} tr (I - H) = ∥ (I - H) μ ∥^{2} + (n - p) σ^{2} .

(125)

It follows that Mallows Cp satisfies

E [C_{p}] = E ∥ Y - H Y ∥^{2} + 2 p σ^{2} = ∥ (I - H) μ ∥^{2} + (n + p) σ^{2} = E ∥ H Y - Y^{*} ∥^{2},

(126)

where the last equality is part (i). Thus

C_{p}

is an unbiased estimator of the independent-copy prediction error, as summarized by the unbiased prediction-error identity for ordinary least squares.

Solved by gpt-5.6-sol high.

iii

Words: 41 Articles: 1

Solution

Words: 41

Let

β = (X^{T} X)^{- 1} X^{T} Y, σ^{2} = \frac{∥ Y - X β ∥ ^{2}}{n - p} .

(127)

The pivotal quadratic form has an

F_{p, n - p}

distribution, so the

(1 - α)

normal linear-model confidence ellipsoid is

{b \in R^{p} : \frac{( b - β ) ^{T} X ^{T} X ( b - β )}{p σ ^{2}} \leq F_{p, n - p} (1 - α)} .

(128)

Here

F_{p, n - p} (1 - α)

is the

(1 - α)

quantile of the indicated

F

distribution.

Solved by gpt-5.6-sol high.

iv

Words: 72 Articles: 1

Solution

Words: 72

Let

β_{(i)}

be the least-squares estimate after deleting observation

i

, and put

Y = X β

and

Y_{(i)} = X β_{(i)}

. Cook's distance is

D_{i} = \frac{∥ Y _{(i)} - Y ∥ ^{2}}{p σ ^{2}} = \frac{( β _{(i)} - β ) ^{T} X ^{T} X ( β _{(i)} - β )}{p σ ^{2}} .

(129)

Thus

D_{i}

is the squared displacement caused by deleting observation

i

, measured in the same

X^{T} X

metric and scale as the confidence ellipsoid. In particular,

β_{(i)}

lies outside the

(1 - α)

ellipsoid centered at

β

exactly when

D_{i} > F_{p, n - p} (1 - α) .

(130)

Solved by gpt-5.6-sol high.

v

Words: 59 Articles: 1

Solution

Words: 59

Here

p = 4

and

n - p = 96

. The supplied quantile gives

F_{4, 96} (0.9) = 2.00.

(131)

Since

D_{i} = 3.1 > 2.00,

(132)

the coefficient estimate after deleting this observation lies outside the

90%

confidence ellipsoid centered at the full-data estimate. It also greatly exceeds the common rough reference value

D_{i} = 1

. The observation is therefore highly influential and merits investigation:

yes, its influence is concerning .

(133)

Solved by gpt-5.6-sol high.

14E (Further Complex Methods)

Words: 591 Articles: 8

i

Words: 256 Articles: 1

Solution

Words: 256

After division by

z

, the coefficient functions are

P (z) = \frac{1}{z} - 1, Q (z) = \frac{λ}{z} .

(134)

Both

z P (z) = 1 - z

and

z^{2} Q (z) = λ z

are analytic at zero, so the regular singular point criterion for a second-order equation shows that

z = 0

is regular singular. Substitution of

y = z^{r}

into the leading terms gives the indicial equation

r (r - 1) + r = r^{2} = 0.

(135)

The exponent is repeated. Directly substituting

y = \sum_{n \geq 0} a_{n} z^{n}

gives

(n + 1)^{2} a_{n + 1} + (λ - n) a_{n} = 0, a_{n + 1} = \frac{n - λ}{( n + 1 ) ^{2}} a_{n} .

(136)

Thus

a_{0}

determines the entire power series, so there is only one such solution

y_{1}

up to scale. The logarithmic solution from a repeated Frobenius exponent predicts

y_{2} (z) = y_{1} (z) lo g z + a power series,

(137)

so its leading nonanalytic term is proportional to

lo g z

For the contour ansatz, differentiation under the integral and one integration by parts give

0 = = \int_{γ} e^{z t} {z (t^{2} - t) + (t + λ)} f (t) d t [e^{z t} t (t - 1) f (t)]_{\partial γ} + \int_{γ} e^{z t} {(t + λ) f - [t (t - 1) f]^{'}} d t .

(138)

The integral vanishes when

t (t - 1) f^{'} + (t - 1 - λ) f = 0,

(139)

whose solution is the Laguerre contour-integral amplitude

f (t) = t^{- λ - 1} (t - 1)^{λ} .

(140)

The contour and branches must make

f

single-valued along the traversed path and must kill the endpoint term

[e^{z t} t^{- λ} (t - 1)^{λ + 1}]_{\partial γ} .

(141)

Now suppose

λ < 0

is nonintegral. Near

t = 0

f (t) = O (t^{- λ - 1})

is integrable and the endpoint factor is

O (t^{- λ}) \to 0

. We may therefore choose

γ_{1} : q u a d 0 ⟶ one counterclockwise loop around 1 ⟶ 0,

(142)

a finite loop based at the branch point

0

and avoiding

1

except by encirclement. A second choice is

γ_{2} : q u a d 0 ⟶ - \infty

(143)

along one bank of the negative real axis, with a consistent branch. At

- \infty

e^{z t}

kills the algebraic endpoint factor because

Re z > 0

The

γ_{1}

integral is analytic in

z

because its contour is finite, and it is a nonzero solution. By uniqueness of the analytic local solution, it is a constant multiple of

y_{1}

. This is the Finite Laguerre contour solution construction.

Solved by gpt-5.6-sol high.

ii

Words: 99 Articles: 1

Solution

Words: 99

Let

λ > 0

be nonintegral. This time

f (t) = O ((t - 1)^{λ})

is integrable at

t = 1

, and the endpoint factor behaves as

(t - 1)^{λ + 1} \to 0

. Suitable contours are therefore

γ_{1} : q u a d 1 ⟶ one counterclockwise loop around 0 ⟶ 1

(144)

and

γ_{2} : q u a d 1 ⟶ - \infty,

(145)

where the second path makes a fixed detour above or below the branch point

0

and the branch of

t^{- λ - 1} (t - 1)^{λ}

is continued along it. The exponential again controls the endpoint at

- \infty

for

Re z > 0

As in part (i), the finite

γ_{1}

integral is analytic near zero and hence is a constant multiple of the unique power-series solution

y_{1}

Solved by gpt-5.6-sol high.

iii

Words: 118 Articles: 1

Solution

Words: 118

Write

λ = - m

with

m \in N

. The amplitude is now the rational function

f (t) = \frac{t ^{m - 1}}{( t - 1 ) ^{m}} .

(146)

Choose

γ_{1}

to be a small positively oriented circle around the pole

t = 1

, and choose

γ_{2}

from

t = 0

along the negative real axis to

- \infty

. At zero the endpoint factor behaves as

t^{m}

and vanishes; at negative infinity the exponential decays.

By the residue theorem, the finite-contour solution is

y_{γ_{1}} (z) = \frac{2 πi}{( m - 1 )!} \frac{d ^{m - 1}}{d t ^{m - 1}} (e^{z t} t^{m - 1})_{t = 1} .

(147)

Every derivative term contains

e^{z}

times a power of

z

of degree at most

m - 1

, and the highest-degree term is nonzero. Hence

y_{1} (z) = e^{z} P_{m - 1} (z)

(148)

up to a constant, where

P_{m - 1}

is a polynomial of degree

m - 1

. This is the negative-integer case of the Integer-parameter Laguerre contour residues.

Solved by gpt-5.6-sol high.

iv

Words: 118 Articles: 1

Solution

Words: 118

Write

λ = N

with

N \in Z_{\geq 0}

. Then

f (t) = \frac{( t - 1 ) ^{N}}{t ^{N + 1}} .

(149)

Take

γ_{1}

to be a small positively oriented circle around

t = 0

. For

γ_{2}

, start at

t = 1

, travel to

- \infty

, and avoid the pole at zero by a fixed upper or lower detour. The endpoint factor vanishes at

t = 1

, and the exponential controls

- \infty

The finite contour extracts the residue at zero:

y_{γ_{1}} (z) = 2 πi [t^{N}] (e^{z t} (t - 1)^{N}) .

(150)

Only powers

z^{0}, \dots, z^{N}

occur, and the coefficient of

z^{N}

is nonzero. Thus this solution is a polynomial of degree

N

and, up to normalization, is the Laguerre polynomial

L_{N} (z)

y_{1} (z) \propto L_{N} (z) .

(151)

This also follows from the recurrence in part (i), because

a_{N + 1} = 0

and every later coefficient vanishes.

Solved by gpt-5.6-sol high.

15B (Cosmology)

Words: 291 Articles: 8

a

Words: 57 Articles: 1

Solution

Words: 57

Write

A = 8 π G / (3 c^{2})

and

H = \overset{a}{˙} / a

. Differentiating the Friedmann equation gives

2 H \dot{H} - 2 \frac{K c ^{2}}{a ^{2}} H = A \overset{ρ}{˙} .

(152)

The cosmological perfect-fluid continuity equation says

\overset{ρ}{˙} = - 3 H (ρ + P)

, so, first where

H \neq = 0

and then everywhere by continuity,

\dot{H} - \frac{K c ^{2}}{a ^{2}} = - \frac{4 π G}{c ^{2}} (ρ + P) .

(153)

Using

K c^{2} / a^{2} = A ρ - H^{2}

from the original equation gives

\dot{H} + H^{2} = - \frac{4 π G}{3 c ^{2}} (ρ + 3 P) .

(154)

Since

\dot{H} + H^{2} = \overset{a}{¨} / a

for the Hubble parameter, this is the Friedmann acceleration equation.

Solved by gpt-5.6-sol high.

b

Words: 76 Articles: 1

Solution

Words: 76

Separate continuity gives

ρ_{r} = ρ_{r 0} a^{- 4} = β \frac{3 c ^{2} H _{0}^{2}}{8 π G} a^{- 4}, ρ_{Λ} = ρ_{Λ0} = \frac{3 c ^{2} H _{0}^{2}}{8 π G},

(155)

using the scalings for radiation in cosmology and cosmological constant energy. At

t = t_{0}

, where

a = 1

, the Friedmann equation becomes

H_{0}^{2} + K c^{2} = (β + 1) H_{0}^{2},

(156)

K c^{2} = β H_{0}^{2} .

(157)

At general

a

H^{2} + \frac{β H _{0}^{2}}{a ^{2}} = H_{0}^{2} (\frac{β}{a ^{4}} + 1),

(158)

which rearranges to

H^{2} = \frac{H _{0}^{2}}{a ^{4}} (β - β a^{2} + a^{4}) .

(159)

Part (a), applied to

ρ = ρ_{r} + ρ_{Λ}

and

P = ρ_{r} /3 - ρ_{Λ}

, gives

\dot{H} + H^{2} = - H_{0}^{2} (\frac{β}{a ^{4}} - 1) .

(160)

Subtracting the displayed formula for

H^{2}

yields

\dot{H} = - β \frac{H _{0}^{2}}{a ^{4}} (2 - a^{2}) .

(161)

These relations define the closed radiation--cosmological-constant turning polynomial.

Solved by gpt-5.6-sol high.

c

Words: 107 Articles: 1

Solution

Words: 107

Put

x = a^{2}

. Apart from the positive factor

H_{0}^{2} / a^{4}

H^{2} has the sign of F (x) = x^{2} - β x + β .

(162)

For

β > 4

, its discriminant

β (β - 4)

is positive and its two roots are

x_{\pm} = \frac{β \pm β ^{2} - 4 β}{2} .

(163)

Both are positive. Moreover

F (0) > 0

and

F (2) = 4 - β < 0

, so

0 < x_{-} < 2 < x_{+} .

(164)

A universe expanding from

a = 0

follows the allowed region

0 \leq a^{2} \leq x_{-}

and reaches

a^{2} = x_{-}

with

H = 0

. There,

\dot{H} = - β \frac{H _{0}^{2}}{a ^{4}} (2 - a^{2}) < 0,

(165)

H

crosses from positive to negative: expansion reverses into contraction.

a \to 0

on the contracting branch,

H \sim - \frac{H _{0} β}{a ^{2}}, \overset{a}{˙} = a H \sim - \frac{H _{0} β}{a} .

(166)

Hence

a d a \sim - H_{0} β d t

, so zero scale factor is reached in finite time. The universe therefore undergoes the supercritical closed radiation--cosmological-constant recollapse to a Big Crunch.

Solved by gpt-5.6-sol high.

d

Words: 51 Articles: 1

Solution

Words: 51

For

β = 4

H^{2} = \frac{H _{0}^{2}}{a ^{4}} (a^{2} - 2)^{2} .

(167)

The branch expanding from

a = 0

has

H = \frac{H _{0} ( 2 - a ^{2} )}{a ^{2}}

(168)

until it approaches

a^{2} = 2

. With

x = a^{2}

\overset{x}{˙} = 2 a \overset{a}{˙} = 2 a^{2} H = 2 H_{0} (2 - x), x (0) = 0.

(169)

Solving this linear equation gives the critical closed radiation--cosmological-constant solution

a (t) = 2 (1 - e^{- 2 H_{0} t}) .

(170)

For

H_{0} t ≪ 1

1 - e^{- 2 H_{0} t} \sim 2 H_{0} t, a (t) \sim 2 H_{0} t .

(171)

For

H_{0} t ≫ 1

a (t) = 2 (1 - \frac{1}{2} e^{- 2 H_{0} t} + O (e^{- 4 H_{0} t})),

(172)

a (t) ⟶ 2

(173)

exponentially from below.

Solved by gpt-5.6-sol high.

16H (Logic and Set Theory)

Words: 550 Articles: 7

a

Words: 293 Articles: 1

Solution

Words: 293

The compactness theorem says that a set

T

of first-order sentences has a model if and only if every finite subset of

T

has a model. One direction is immediate: a model of

T

models every subset. Conversely, suppose every finite subset is satisfiable. If

T

had no model, then

T ⊨ ⊥

. The Godel completeness theorem would give a formal proof

T ⊢ ⊥

. Every proof is finite, so it would use only a finite subset

T_{0} \subseteq T

, making

T_{0}

inconsistent and hence unsatisfiable. This contradiction proves compactness.

The Upward Lowenheim-Skolem theorem says that a first-order theory with an infinite model has models of arbitrarily large cardinality. Fix a cardinal

κ

and expand the language by constants

c_{α}

for

α < κ

. Add the sentences

c_{α} \neq = c_{β} (α \neq = β)

(174)

T

. Any finite subset mentions only finitely many new constants, which can be interpreted as distinct elements of the given infinite model. Compactness therefore gives a model of the expanded theory in which the

c_{α}

are all distinct. Its reduct is a model of

T

of cardinality at least

κ

. Together with the downward theorem, this gives a model of exactly

κ

whenever

κ \geq ∣ L ∣ + ℵ_{0}

The Downward Lowenheim-Skolem theorem says that if

A

is an infinite

L

-structure and

∣ L ∣ + ℵ_{0} \leq κ \leq ∣ A ∣,

(175)

then

A

has an elementary substructure of cardinality

κ

. To see why, first choose

κ

elements. For every existential formula, add a Skolem function selecting a witness whenever one exists, and repeatedly close the chosen set under all language operations and these witness functions. There are at most

κ

functions of finite arity, so the closure still has size

κ

. Every existential statement true in

A

with parameters from the closure has a witness in the closure; the Tarski--Vaught test therefore says that the resulting substructure is elementary.

Solved by gpt-5.6-sol high.

b

Words: 257 Articles: 4

i

Words: 174 Articles: 1

Solution

Words: 174

Expand

L

by a unary function symbol

f

. Let

T^{*}

contain:

every sentence of $T$ ;
the first-order axiomatization of an automorphism saying that $f$ is bijective and preserves every function and relation of $L$ ;
the sentence $\exists x [f (x) \neq = x]$ ;
for each $n \geq 1$ , a sentence $θ_{n}$ asserting the existence of at least $n$ distinct elements.

Every finite subset

S \subseteq T^{*}

contains only finitely many size requirements, say no requirement beyond

θ_{N}

. By hypothesis there is a finite nonrigid model

A ⊨ T

with at least

N

elements. Interpret

f

as a nonidentity automorphism of

A

. This expanded structure satisfies

S

; indeed it satisfies every automorphism axiom, including any finitely many that occur in

S

Thus every finite subset of

T^{*}

is satisfiable. By the compactness theorem,

T^{*}

has a model

M^{*}

. The sentences

θ_{n}

make its domain infinite, while

\exists x [f (x) \neq = x]

and the automorphism axioms make

f^{M^{*}}

a nonidentity automorphism. The

L

-reduct

M

is therefore an infinite nonrigid model of

T

. This is the compactness transfer of finite nonrigidity.

Solved by gpt-5.6-sol high.

ii

Words: 83 Articles: 1

Solution

Words: 83

No. Take

L = {<}

and let

T

be the theory of linear orders. It has finite models of every positive size. Every finite linear order is a rigid first-order structure: an order automorphism fixes the least element, then the next element, and inductively every element.

However,

(Z, <)

is an infinite model of

T

and is not rigid. For example,

n ⟼ n + 1

(176)

is a nonidentity order automorphism. Hence arbitrarily large rigid finite models do not force all infinite models to be rigid.

Solved by gpt-5.6-sol high.

17H (Graph Theory)

Words: 327 Articles: 7

a

Words: 134 Articles: 1

Solution

Words: 134

Fix

k \geq 1

and sample

G

from the Erdős-Rényi model with

p = \frac{1}{2} n^{- 2/3}

. Let

X

count copies of the complete bipartite graph

K_{3, 3}

. By the expected subgraph count in the Erdős-Rényi model,

E X = \frac{1}{2} (3 n) (3 n - 3) p^{9} ⟶ \frac{1}{72 2 ^{9}} < 1.

(177)

Put

m = ⌈ n / k ⌉

and let

Z

count independent set of size

m

. If the chromatic number satisfies

χ (G) \leq k

, some colour class has at least

m

vertices, so

Z \geq 1

. Moreover,

E Z = (m n) (1 - p)^{(2 m)} \leq (\frac{e n}{m})^{m} exp (- p (2 m)) ⟶ 0,

(178)

because the positive part of the exponent is

O (n)

whereas

p (2 m)

has order

n^{4/3}

. The first moment method therefore gives

P (χ (G) \leq k) \leq E Z = o (1)

Using Markov inequality for

X

and the union bound,

P (X = 0 and χ (G) > k) \geq 1 - P (X \geq 1) - P (χ (G) \leq k) \geq 1 - E X - o (1) > 0

(179)

for all sufficiently large

n

. Hence at least one such

G

contains no

K_{3, 3}

and has

χ (G) > k

Solved by gpt-5.6-sol high.

b

Words: 193 Articles: 4

i

Words: 94 Articles: 1

Solution

Words: 94

Let

X

count copies of the complete graph

K_{4}

G

. Its expected value is

μ = E X = (4 n) p^{6} \sim \frac{( l o g n ) ^{6}}{24} ⟶ \infty.

(180)

Write

X

as a sum of indicator random variable. Indicators for two distinct copies are independent when the copies share at most one vertex, since their edge sets are then disjoint. Pairs sharing two vertices have eleven edges in their union, while pairs sharing three have nine. It follows that

var (X) \leq μ + O (n^{6} p^{11}) + O (n^{5} p^{9}) = O ((lo g n)^{6}) + o (1) .

(181)

Thus

var (X) / μ^{2} \to 0

. By the second moment method,

X / μ \to 1

in probability, so

P (X \geq 100) \to 1

. This is the sparse clique-count concentration estimate.

Solved by gpt-5.6-sol high.

ii

Words: 99 Articles: 1

Solution

Words: 99

Let

Y

count unordered pairs of distinct

K_{4}

copies that have a common vertex. Classifying such a pair by whether it shares one, two, or three vertices gives

E Y = O (n^{7} p^{12}) + O (n^{6} p^{11}) + O (n^{5} p^{9}) = o (1) .

(182)

Indeed, after substituting

p = n^{- 2/3} lo g n

, these terms are respectively

O (n^{- 1} (lo g n)^{12}), O (n^{- 4/3} (lo g n)^{11}), O (n^{- 1} (lo g n)^{9}) .

(183)

The first moment method gives

P (Y > 0) \leq E Y \to 0

. Part (i) says that

X \geq 100

with probability tending to one, so the union bound shows that, with probability tending to one, both

X \geq 100

and

Y = 0

. On that event any one hundred of the

K_{4}

copies are pairwise vertex-disjoint, proving vertex-disjoint sparse clique copies in this case.

Solved by gpt-5.6-sol high.

18I (Galois Theory)

Words: 487 Articles: 11

a

Words: 118 Articles: 1

Solution

Words: 118

A finite extension

L / K

is normal when every irreducible polynomial in

K [T]

having one root in

L

splits completely over

L

Suppose first that

L / K

is normal. Choose

α_{1}, \dots, α_{m}

with

L = K (α_{1}, \dots, α_{m})

, and let

f_{j}

be the minimal polynomial of

α_{j}

over

K

. Every

f_{j}

splits in

L

by normality, and

L

is generated by their roots because it is already generated by the particular roots

α_{j}

. Thus

L

is the splitting field of

f_{1} \dots f_{m}

Conversely, suppose that

L

is the splitting field of

f \in K [T]

. Every

K

-embedding

σ : L \to \overline{K}

permutes the roots of

f

, so

σ (L) = L

. By the embedding characterization of normality,

L / K

is normal. This proves the finite normal extension as a splitting field criterion.

Solved by gpt-5.6-sol high.

b

Words: 153 Articles: 6

i

Words: 54 Articles: 1

Solution

Words: 54

This is true. If

M / K

is normal, the finite normal extension as a splitting field criterion writes

M

as the splitting field over

K

of some

f \in K [T]

. Since

K \subseteq L \subseteq M

, the same field

M

is the splitting field of

f

, now regarded as a polynomial over

L

. Hence

M / L

is normal.

Solved by gpt-5.6-sol high.

ii

Words: 54 Articles: 1

Solution

Words: 54

This is false. Let

K = Q, qq u a d L = Q (32), qq u a d M = Q (32, ω),

(184)

where

ω

is a primitive cube root of unity. Then

M

is the splitting field of

T^{3} - 2

over

Q

, so

M / K

is normal. The irreducible polynomial

T^{3} - 2

has a root in

L

but does not split there, so

L / K

is not normal.

Solved by gpt-5.6-sol high.

iii

Words: 45 Articles: 1

Solution

Words: 45

This is false. Take

K = Q, qq u a d L = Q (2), qq u a d M = Q (42) .

(185)

Each of

L / K

and

M / L

is a quadratic extension and hence normal. However,

T^{4} - 2

has the root

42

M

while its roots

i 42

and

- i 42

are absent. Thus

M / K

is not normal.

Solved by gpt-5.6-sol high.

c

Words: 216 Articles: 1

Solution

Words: 216

Put

α = 47

. The roots of

T^{4} - 7

are

α, i α, - α, - i α

, so its splitting field is

L = Q (α, i) .

(186)

The polynomial is irreducible over

Q

by the Eisenstein criterion at

7

, and

Q (α) \subset R

does not contain

i

. The tower law for field extensions therefore gives

[L : Q] = 8

Define automorphisms

r (α) = i α, q u a d r (i) = i, qq u a d s (α) = α, q u a d s (i) = - i .

(187)

They satisfy

r^{4} = s^{2} = 1

and

srs = r^{- 1}

. The eight maps

r^{j}

and

r^{j} s

are distinct, so they exhaust the Galois group. Hence

Gal (L / Q) ≅ D_{8},

(188)

the dihedral group of order eight.

The Galois correspondence gives all intermediate fields. The full list, with a single generator for each field, is

subgroup D_{8} ⟨ r ⟩ ⟨ r^{2}, s ⟩ ⟨ r^{2}, rs ⟩ ⟨ r^{2} ⟩ ⟨ s ⟩ ⟨ r^{2} s ⟩ ⟨ rs ⟩ ⟨ r^{3} s ⟩ {1} fixed field Q = Q (0) Q (i) Q (7) Q (- 7) Q (7 + i) Q (α) Q (i α) Q ((1 + i) α) Q ((1 - i) α) L = Q (α + i) normal over Q yes yes yes yes yes no no no no yes .

(189)

For example,

7 + i

has stabilizer

⟨ r^{2} ⟩

, and

(1 + i) α

has stabilizer

⟨ rs ⟩

; the other entries follow similarly. Also

α + i

has trivial stabilizer, so its orbit has eight elements and the primitive element theorem conclusion

L = Q (α + i)

holds explicitly.

Finally, the normal subextension criterion says that normal intermediate fields correspond exactly to normal subgroups. In

D_{8}

these are

D_{8}

⟨ r ⟩

, the two index-two Klein four-groups,

⟨ r^{2} ⟩

, and

{1}

, which gives exactly the “yes” rows above. This is the subfields of the splitting field of x to the fourth minus seven lattice.

Solved by gpt-5.6-sol high.

19H (Representation Theory)

Words: 285 Articles: 1

Solution

Words: 285

A complex representation of a finite group

G

is a homomorphism

ρ : G ⟶ GL (V)

(190)

for a finite-dimensional complex vector space

V

. It is an irreducible representation when its only invariant subspaces are

0

and

V

, and its degree is

dim V

. Two representations are isomorphic when there is an invertible linear map intertwining their

G

-actions.

Write the dihedral group as

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

(191)

In an irreducible representation choose an eigenvector

0 \neq = v \in V

ρ (r)

, which is possible because

r^{n} = 1

. If its eigenvalue is

λ

, then

ρ (r) ρ (s) v = ρ (s) ρ (r)^{- 1} v = λ^{- 1} ρ (s) v .

(192)

Thus

W = span {v, ρ (s) v}

is invariant under both

r

and

s

. Irreducibility forces

W = V

, proving

dim V \leq 2

as in irreducible complex representations of a finite dihedral group.

Let

ω = e^{2 πi / n}

. Besides the one-dimensional representations described below, define for each indicated

j

ρ_{j} (r) = (ω^{j} 0 0 ω^{- j}), ρ_{j} (s) = (0110) .

(193)

These matrices satisfy the defining relations. Since

ω^{j} \neq = ω^{- j}

, the only one-dimensional

r

-invariant subspaces are the two coordinate axes, and

s

interchanges them. Hence

ρ_{j}

is irreducible. Their characters satisfy

χ_{j} (r) = 2 cos \frac{2 πj}{n},

(194)

whose values are distinct in the ranges below, so the

ρ_{j}

are pairwise nonisomorphic. This is the family of two-dimensional representations of a finite dihedral group.

n

is odd, a one-dimensional character must send

r

to a scalar

ε

satisfying

ε^{n} = 1

and

ε = ε^{- 1}

. Thus

ε = 1

, while

s

may independently map to

1

- 1

. These give two one-dimensional irreducibles. Taking

j = 1, \dots, \frac{n - 1}{2}

(195)

gives

(n - 1) /2

two-dimensional irreducibles, for a total of

2 + \frac{n - 1}{2} = \frac{n + 3}{2} .

(196)

n

is even,

r

may map to either

1

- 1

, and again

s

may map independently to either sign. This gives four pairwise nonisomorphic one-dimensional irreducibles. Taking

j = 1, \dots, \frac{n}{2} - 1

(197)

gives

n /2 - 1

two-dimensional irreducibles, for a total of

4 + (\frac{n}{2} - 1) = \frac{n + 6}{2} .

(198)

Different dimensions distinguish the one- and two-dimensional families, completing the required construction and justification.

Solved by gpt-5.6-sol high.

20H (Number Fields)

Words: 369 Articles: 7

a

Words: 184 Articles: 4

i

Words: 71 Articles: 1

Solution

Words: 71

By unique factorization of ideals in a number field, each nonzero integral ideal has a unique expression

a = \prod_{p} p^{e_{p}},

(199)

with only finitely many nonzero exponents. Since the ideal norm is multiplicative, formally expanding one geometric series for each prime ideal gives

p \prod (1 - N (p)^{- s})^{- 1} = p \prod e \geq 0 \sum N (p)^{- es} = 0 \neq = a \subseteq O_{K} \sum N (a)^{- s} = ζ_{K} (s) .

(200)

Uniqueness of ideal factorization says that every term occurs exactly once. This proves the Euler product for the Dedekind zeta function as a formal identity.

Solved by gpt-5.6-sol high.

ii

Words: 113 Articles: 1

Solution

Words: 113

For each rational prime

p

, define

χ (p) = ⎩ ⎨ ⎧ 1, - 1, 0, (p) = p_{1} p_{2} with p_{1} \neq = p_{2}, (p) remains prime in O_{K}, (p) = p^{2} is ramified .

(201)

These are the three cases in splitting of rational primes in a quadratic field.

Set

x = p^{- s}

. In the split case the local factor of the Dedekind zeta function is

(1 - x)^{- 2}

; in the inert case it is

(1 - x^{2})^{- 1}

because the prime ideal has norm

p^{2}

; and in the ramified case it is

(1 - x)^{- 1}

. On the other hand, the product of the local factors of

ζ_{Q} (s)

and

L (χ, s)

(1 - x)^{- 1} (1 - χ (p) x)^{- 1},

(202)

which gives exactly those three expressions when

χ (p) = 1, - 1, 0

. Multiplying over all rational primes proves the quadratic Dedekind zeta factorization

ζ_{K} (s) = ζ_{Q} (s) L (χ, s)

(203)

formally.

Solved by gpt-5.6-sol high.

b

Words: 185 Articles: 1

Solution

Words: 185

Because

p \equiv 11 (mod 12)

, we have

p \equiv 3 (mod 4)

and

O_{K} = Z [\frac{1 + - p}{2}] .

(204)

For

θ = (1 + - p) /2

, the minimal polynomial is

f (T) = T^{2} - T + \frac{p + 1}{4} .

(205)

Since

p \equiv 2 (mod 3)

, reduction modulo

3

gives

f (T) \equiv T (T - 1)

. Its two roots are distinct, so

(3) = p_{1} p_{2}

(206)

with distinct prime ideals of norm

3

. Thus

3

splits completely, as recorded in splitting of three in Q of square root minus p.

The ideal class group

Cl_{K}

is the group of nonzero fractional ideals of

O_{K}

modulo the subgroup of principal fractional ideals. Saying that

p_{1}

has order

n

means that

n

is the least positive integer for which

p_{1}^{n}

is principal.

Suppose for contradiction that

n

is odd and

p_{1}

has order

n

. Then

p_{1}^{n} = (α)

for some

α \in O_{K}

, and taking norms gives

∣ N_{K / Q} (α) ∣ = N (p_{1})^{n} = 3^{n} .

(207)

Write

α = (a + b - p) /2

, where

a, b \in Z

have the same parity. Then

4 \cdot 3^{n} = a^{2} + p b^{2} .

(208)

b \neq = 0

, the right-hand side is at least

p

, contradicting

p > 3^{n + 2} > 4 \cdot 3^{n}

. Hence

b = 0

, so

α

is a rational algebraic integer and therefore an integer. But then

3^{n} = N (α) = α^{2}

, impossible when

n

is odd. This proves the odd-order obstruction for a split prime in an imaginary quadratic field and shows that the order cannot equal

n

Solved by gpt-5.6-sol high.

21G (Algebraic Topology)

Words: 316 Articles: 1

Solution

Words: 316

Given group homomorphisms

i_{A} : C \to A

and

i_{B} : C \to B

, their amalgamated free product is a group

A *_{C} B

with maps

j_{A} : A \to A *_{C} B

and

j_{B} : B \to A *_{C} B

satisfying

j_{A} i_{A} = j_{B} i_{B}

, such that every pair

f_{A} : A \to H

f_{B} : B \to H

satisfying

f_{A} i_{A} = f_{B} i_{B}

factors through a unique homomorphism

A *_{C} B \to H

The Seifert-van Kampen theorem says that if

X = U_{1} \cup U_{2}

, where

U_{1}, U_{2}

, and

U_{1} \cap U_{2}

are open and path-connected and

x_{0} \in U_{1} \cap U_{2}

, then the inclusion maps induce

π_{1} (X, x_{0}) ≅ π_{1} (U_{1}, x_{0}) *_{π_{1} (U_{1} \cap U_{2}, x_{0})} π_{1} (U_{2}, x_{0}) .

(209)

Here is a direct proof of the requested generation statement. For a based loop

γ

X

, the Lebesgue number lemma supplies a subdivision

0 = t_{0} < t_{1} < \dots < t_{m} = 1

(210)

such that every arc

γ ∣_{[t_{j - 1}, t_{j}]}

lies in one

U_{ε_{j}}

. Combine adjacent arcs assigned to the same set, so each transition point

z_{j} = γ (t_{j})

lies in

U_{1} \cap U_{2}

. Since the intersection is path-connected, choose a path

q_{j}

in it from

x_{0}

z_{j}

, taking

q_{0} = q_{m}

constant. Then

q_{j - 1} * γ ∣_{[t_{j - 1}, t_{j}]} * \overline{q_{j}}

(211)

is a loop in

U_{ε_{j}}

. In the product of their classes, each

\overline{q_{j}} * q_{j}

cancels by path reversal, leaving

[γ]

. Thus the two inclusion images generate

π_{1} (X, x_{0})

, as summarized by generation of a fundamental group by two open sets.

Now let

a, b

be the standard generators in the fundamental group of the torus. A Möbius band retracts to a core circle, while its boundary traverses that core twice. If

x

and

y

denote the core classes of the two attached bands, the Seifert-van Kampen theorem gives

π_{1} (Y) ≅ ⟨ a, b, x, y ∣ [a, b] = 1, x^{2} = ab, y^{2} = a^{2} b^{3} ⟩ .

(212)

The two attaching classes

A = ab

and

B = a^{2} b^{3}

commute. Their exponent matrix has determinant

det (1213) = 1,

(213)

so they form a basis of

⟨ a, b ⟩ ≅ Z^{2}

, with

a = A^{3} B^{- 1}, b = A^{- 2} B .

(214)

Substituting

A = x^{2}

and

B = y^{2}

eliminates

a, b

and leaves the two-generator one-relator presentation

π_{1} (Y) ≅ ⟨ x, y ∣ [x^{2}, y^{2}] = 1 ⟩ .

(215)

This is the calculation in two Möbius bands attached to a torus.

Finally send

x

(12)

and

y

(23)

in the symmetric group

S_{3}

. Both squares are the identity, so the relator is satisfied, and these transpositions generate

S_{3}

. We obtain a surjective group homomorphism from

π_{1} (Y)

onto the nonabelian group

S_{3}

. Therefore

π_{1} (Y)

is nonabelian.

Solved by gpt-5.6-sol high.

22F (Linear Analysis)

Words: 381 Articles: 6

a

Words: 124 Articles: 1

Solution

Words: 124

The open mapping theorem states that every surjective bounded linear map

S : X \to Y

between Banach spaces maps open sets to open sets. Equivalently, there is

c > 0

such that

B_{Y} (0, c) \subseteq S (B_{X} (0, 1)) .

(216)

The closed graph theorem states that a linear map

T : X \to Y

between Banach spaces is continuous if and only if its graph

Γ (T) = {(x, T x) : x \in X}

(217)

is closed in

X \times Y

Assume the open mapping theorem and suppose

Γ (T)

is closed. As a closed subspace of the Banach space

X \times Y

, the graph is itself Banach. The coordinate projection

P : Γ (T) ⟶ X, (x, T x) ⟼ x

(218)

is a bounded linear bijection. By the open mapping theorem,

P^{- 1}

is bounded. The other coordinate projection

Q : Γ (T) \to Y

is bounded, and

T = Q \circ P^{- 1} .

(219)

Thus

T

is bounded and hence continuous. This is the closed graph theorem from the bounded inverse theorem argument.

Solved by gpt-5.6-sol high.

b

Words: 119 Articles: 1

Solution

Words: 119

The continuous dual space of a normed vector space

V

V^{*} = {f : V \to F : f is linear and continuous},

(220)

with the operator norm

∥ f ∥ = sup_{∥ v ∥ \leq 1} ∣ f (v) ∣.

(221)

Let

(f_{n})

be a Cauchy sequence in this norm. For each

v \in V

∣ f_{n} (v) - f_{m} (v) ∣ \leq ∥ f_{n} - f_{m} ∥ ∥ v ∥,

(222)

(f_{n} (v))

is Cauchy in the scalar field. Define

f (v) = lim_{n} f_{n} (v)

. Passing to limits in the linearity identities shows that

f

is linear.

A norm-Cauchy sequence is bounded, say

∥ f_{n} ∥ \leq M

. Hence

∣ f (v) ∣ \leq M ∥ v ∥

, so

f \in V^{*}

. Given

ε > 0

, choose

N

such that

∥ f_{n} - f_{m} ∥ < ε

for

m, n \geq N

. Letting

m \to \infty

gives

∣ f_{n} (v) - f (v) ∣ \leq ε ∥ v ∥

(223)

for every

v

, and therefore

∥ f_{n} - f ∥ \leq ε

. Thus

f_{n} \to f

in operator norm, proving completeness of the dual space:

V^{*}

is Banach.

Solved by gpt-5.6-sol high.

c

Words: 138 Articles: 1

Solution

Words: 138

Write

T_{v} = T (v)

. We prove that

T

has a closed graph. Suppose

v_{n} ⟶ v in V, T_{v_{n}} ⟶ f in V^{*} .

(224)

Put

u_{n} = v_{n} - v

and

g = f - T_{v}

. Then

u_{n} \to 0

and

T_{u_{n}} \to g

V^{*}

Fix

w \in V

and

t \in R

. Positivity and linearity give

0 \leq T_{u_{n} + tw} (u_{n} + tw) = T_{u_{n}} (u_{n}) + t (T_{u_{n}} (w) + T_{w} (u_{n})) + t^{2} T_{w} (w) .

(225)

Because

(T_{u_{n}})

converges in the continuous dual space, its norms are bounded, so

∣ T_{u_{n}} (u_{n}) ∣ \leq ∥ T_{u_{n}} ∥ ∥ u_{n} ∥ ⟶ 0.

(226)

Also

T_{u_{n}} (w) \to g (w)

, while

T_{w} (u_{n}) \to 0

because the fixed functional

T_{w}

is continuous. Passing to the limit yields

0 \leq t g (w) + t^{2} T_{w} (w)

(227)

for every real

t

. If

g (w) \neq = 0

, a sufficiently small

t

of the opposite sign makes the right-hand side negative. Hence

g (w) = 0

for every

w

, so

g = 0

and

f = T_{v}

The graph is therefore closed. The closed graph theorem now proves that

T : V \to V^{*}

is continuous. This is precisely continuity of a positive linear map into a dual space.

Solved by gpt-5.6-sol high.

23F (Analysis of Functions)

Words: 278 Articles: 6

a

Words: 122 Articles: 1

Solution

Words: 122

For

s > \frac{1}{2}

, the Sobolev trace theorem states that the restriction map initially defined on smooth functions extends uniquely to a bounded linear operator

γ : H^{s} (R^{n}) ⟶ H^{s - 1/2} (R^{n - 1}), (γ u) (x^{'}) = u (x^{'}, 0) .

(228)

Take first

u

in the Schwartz space. Up to the harmless constant determined by the Fourier transform convention,

γ u (ξ^{'}) = \int_{R} u (ξ^{'}, ξ_{n}) d ξ_{n} .

(229)

The Cauchy-Schwarz inequality, with weights

1 + ∣ ξ^{'} ∣^{2} + ξ_{n}^{2}

, gives

∣ γ u (ξ^{'}) ∣^{2} \leq (\int_{R} (1 + ∣ ξ^{'} ∣^{2} + ξ_{n}^{2})^{s} ∣ u (ξ^{'}, ξ_{n}) ∣^{2} d ξ_{n}) \times (\int_{R} (1 + ∣ ξ^{'} ∣^{2} + ξ_{n}^{2})^{- s} d ξ_{n}) .

(230)

After the substitution

ξ_{n} = (1 + ∣ ξ^{'} ∣^{2})^{1/2} t

, the second factor is

C_{s} (1 + ∣ ξ^{'} ∣^{2})^{1/2 - s}, C_{s} = \int_{R} (1 + t^{2})^{- s} d t < \infty,

(231)

where finiteness is exactly the condition

s > 1/2

. Multiplying by

(1 + ∣ ξ^{'} ∣^{2})^{s - 1/2}

and integrating in

ξ^{'}

yields

∥ γ u ∥_{H^{s - 1/2} (R^{n - 1})} \leq C_{s}^{1/2} ∥ u ∥_{H^{s} (R^{n})} .

(232)

Density of the Schwartz space in the Sobolev space completes the unique bounded extension.

Solved by gpt-5.6-sol high.

b

Words: 81 Articles: 1

Solution

Words: 81

For

1 \leq p < \infty

, choose nonzero

ϕ \in C_{c} (R^{n - 1})

and

ψ \in C_{c} (R)

with

ψ (0) = 1

. Define

u_{k} (x^{'}, x_{n}) = ϕ (x^{'}) ψ (k x_{n}) .

(233)

These functions are continuous and belong to the Lebesgue space

L^{p} (R^{n})

, while their restrictions to the hyperplane are all the same:

u_{k} (x^{'}, 0) = ϕ (x^{'}) .

(234)

Changing variables in the normal coordinate gives

∥ u_{k} ∥_{L^{p} (R^{n})}^{p} = \frac{1}{k} ∥ ϕ ∥_{L^{p} (R^{n - 1})}^{p} ∥ ψ ∥_{L^{p} (R)}^{p} ⟶ 0.

(235)

If the claimed bounded

T

existed, then

0 < ∥ ϕ ∥_{p} = ∥ T u_{k} ∥_{p} \leq ∥ T ∥ ∥ u_{k} ∥_{p} ⟶ 0,

(236)

a contradiction. This scaling argument is the failure of an Lp hyperplane trace.

Solved by gpt-5.6-sol high.

c

Words: 75 Articles: 1

Solution

Words: 75

Compact support and the fundamental theorem of calculus give, for every

(x, y)

∣ u (x, y) ∣^{2} = \int_{- \infty}^{y} \partial_{t} ∣ u (x, t) ∣^{2} d t \leq 2 \int_{R} ∣ u (x, t) ∣ ∣ \partial_{2} u (x, t) ∣ d t \leq 2 A (x),

(237)

where

A (x) = \int_{R} ∣ u (x, t) ∣ ∣\nabla u (x, t) ∣ d t .

(238)

Applying the same argument in the other coordinate gives

∣ u (x, y) ∣^{2} \leq 2 B (y), B (y) = \int_{R} ∣ u (t, y) ∣ ∣\nabla u (t, y) ∣ d t .

(239)

Multiplication and Fubini's theorem now yield

\int_{R^{2}} ∣ u (x, y) ∣^{4} d x d y \leq 4 \int_{R^{2}} A (x) B (y) d x d y = 4 (\int_{R^{2}} ∣ u ∣ ∣\nabla u ∣)^{2} .

(240)

Finally, the Cauchy-Schwarz inequality gives

(\int_{R^{2}} ∣ u ∣ ∣\nabla u ∣)^{2} \leq (\int_{R^{2}} ∣ u ∣^{2}) (\int_{R^{2}} ∣\nabla u ∣^{2}) .

(241)

Thus the required estimate holds with

C = 4

; it is the Ladyzhenskaya inequality in two dimensions.

Solved by gpt-5.6-sol high.

24F (Riemann Surfaces)

Words: 405 Articles: 1

Solution

Words: 405

A germ of a holomorphic function at

z \in D

is an equivalence class

[f]_{z}

of pairs consisting of a neighbourhood

U

z

and a holomorphic function

f : U \to C

. Two pairs represent the same germ if their functions agree on some neighbourhood of

z

The space of germs of holomorphic functions is

G = {[f]_{z} : z \in D} .

(242)

For every holomorphic

f

on an open

U \subseteq D

, the set

U_{f} = {[f]_{z} : z \in U}

(243)

is declared open; these sets form a basis. The forgetful map

π : G \to D, π ([f]_{z}) = z,

(244)

maps each

U_{f}

homeomorphically onto

U

. The inverses of these restrictions give the charts defining the complex structure, so

π

is a local biholomorphism.

The evaluation map is

E : G \to C, E ([f]_{z}) = f (z) .

(245)

It is well-defined by the germ equivalence relation. On the chart

U_{f}

, its coordinate expression is

E \circ (π ∣_{U_{f}})^{- 1} (z) = f (z),

(246)

which is holomorphic. Hence

E

is analytic, as stated in evaluation map on a space of germs.

Now put

D = C ∖ {ζ : ζ^{8} = 1}, R = {(z, w) \in D \times C : w^{2} = z^{8} - 1} .

(247)

This gives the germ surface of the square root of z to the eighth minus one. An explicit gluing description is obtained by pairing the eight roots into four adjacent pairs and cutting the plane along four disjoint arcs joining the members of each pair. On the complement of the cuts choose one branch

q (z)

z^{8} - 1

. Take two copies, labelled by

q

and

- q

, and glue the upper bank of each cut in one copy to the lower bank in the other, and conversely. The cut interiors are restored by the gluing, while the eight endpoints remain absent because they are not in

D

On this surface define

π (z, w) = z, E (z, w) = w .

(248)

For each

(z, w) \in R

, the holomorphic function

z^{8} - 1

is nonzero near

z

, so it has a unique local square root

q

with

q (z) = w

. Define

Φ : R ⟶ G, Φ (z, w) = [q]_{z} .

(249)

This is well-defined, injective, and analytic in the displayed local sheets; its inverse on its image is

[q]_{z} \mapsto (z, q (z))

. Thus

Φ

is the required analytic embedding, and it intertwines both the forgetful and evaluation maps.

To compactify, first add one point above each of the eight roots of unity. These are simple branch points of the double cover. Since the polynomial has even degree eight, the even-degree hyperelliptic model has two distinct, unbranched points above infinity. Consequently

∣ \overline{R} ∖ R ∣ = 8 + 2 = 10.

(250)

Finally apply the Riemann-Hurwitz formula to the degree-two meromorphic map

\overset{π}{ˉ} : \overline{R} \to C_{\infty}

. Its only ramification consists of the eight simple finite branch points, so

2 g (\overline{R}) - 2 = 2 (2 g (C_{\infty}) - 2) + 8 = 2 (- 2) + 8 = 4.

(251)

Therefore

g (\overline{R}) = 3.

(252)

This is the compactification of y squared equals x to the eighth minus one.

Solved by gpt-5.6-sol high.

25G (Algebraic Geometry)

Words: 406 Articles: 1

Solution

Words: 406

Let

X \subseteq A^{n}

have ideal

I (X)

, and let

p \in X

. Its Zariski tangent space is

T_{p} X = {v \in k^{n} : \sum_{j = 1}^{n} \frac{\partial f}{\partial x _{j}} (p) v_{j} = 0 for every f \in I (X)} .

(253)

X

is irreducible,

p

is smooth when

dim T_{p} X = dim X;

(254)

otherwise it is singular. Equivalently, for generators of

I (X)

, the Jacobian has the maximum rank

n - dim X

p

Now let the irreducible affine plane cubic be

X = V (f) \subseteq A^{2}

. Suppose distinct points

p, q \in X

were both singular. Parametrize their joining line by

ℓ (t) = p + t (q - p)

(255)

and put

h (t) = f (ℓ (t))

. Then

de g h \leq 3

. At

t = 0

h (0) = 0, h^{'} (0) = \nabla f (p) \cdot (q - p) = 0,

(256)

so zero is a root of multiplicity at least two. The same argument at

t = 1

gives another root of multiplicity at least two. Therefore

h

has at least four roots counted with multiplicity and must vanish identically. The line through

p, q

is then contained in

X

, so its linear equation divides

f

, contradicting irreducibility. This proves singular points of an irreducible affine plane cubic are unique when they exist.

For the density statement, embed an irreducible affine variety

X

of dimension

d

A^{n}

and choose generators

f_{1}, \dots, f_{r}

of its ideal. By dimension from minimum tangent dimension, some point has tangent dimension

d

, so the Jacobian has rank

n - d

there. Some

(n - d) \times (n - d)

minor

Δ

is consequently nonzero on

X

. The distinguished open set

X_{Δ} = {p \in X : Δ (p) \neq = 0}

(257)

is nonempty, and at each of its points the Jacobian rank is at least

n - d

. Since every tangent space has dimension at least

d

, the rank is also at most

n - d

. Thus every point of

X_{Δ}

is smooth. A nonempty open subset of an irreducible topological space is dense, proving density of the smooth locus.

Finally write the smooth irreducible projective hypersurface as

X = V_{+} (F) \subseteq P^{n}

(258)

for an irreducible homogeneous polynomial

F

. The closure of

π^{- 1} (X)

is its affine cone over a projective hypersurface

Y = V (F) \subseteq A^{n + 1} .

(259)

At a nonzero point

y \in Y

, simultaneous vanishing of all partial derivatives of

F

would make the projective point

[y] \in X

singular. Since

X

is smooth, this cannot happen. Hence every nonzero point of

Y

is smooth, and the origin is its only possible singular point.

Both possibilities occur. If

X

is a projective hyperplane, then

F

is linear and

Y

is an affine linear subspace, hence smooth even at the origin. For a singular cone, take the smooth conic

X = V_{+} (x_{0} x_{2} - x_{1}^{2}) \subseteq P^{2} .

(260)

Its affine cone is

Y = V (x_{0} x_{2} - x_{1}^{2}) \subseteq A^{3} .

(261)

All first derivatives vanish at the origin, so the vertex is singular, while the preceding argument shows that it is the only singular point.

Solved by gpt-5.6-sol high.

26G (Differential Geometry)

Words: 519 Articles: 8

a

Words: 104 Articles: 1

Solution

Words: 104

A point

y_{0} \in Y

is a regular value of the smooth map

f : X \to Y

if, for every

x \in f^{- 1} (y_{0})

, the derivative

D f_{x} : T_{x} X ⟶ T_{y_{0}} Y

(262)

is surjective. A value outside the image is regular vacuously.

For a nonvacuous example, take

f : R^{2} \to R, f (x, y) = x^{2} + y^{2} .

(263)

The value

1

is regular: its fibre is the unit circle, and at every

(x, y)

on that circle,

D f_{(x, y)} (u, v) = 2 xu + 2 y v

(264)

is a nonzero linear functional and hence surjective.

Sard theorem states that the set of critical values of a smooth map between smooth manifolds has measure zero in the target. Equivalently, almost every target point is a regular value.

Solved by gpt-5.6-sol high.

b

Words: 148 Articles: 1

Solution

Words: 148

Assume first that

Y

is connected; otherwise the definition is made separately on each connected component. Choose a regular value

y \in Y

and define the degree modulo two by

de g_{2} (f) = ∣ f^{- 1} (y) ∣ (mod 2) .

(265)

The following course results make this well-defined.

By Sard theorem, regular values are dense, so one can choose $y$ .
By the preimage theorem, $f^{- 1} (y)$ is a zero-dimensional submanifold because $dim X = dim Y$ . It is discrete and closed in compact $X$ , hence finite.
The parity is independent of the regular value. Indeed, after choosing a path between two regular values transverse to $f$ , its inverse image is a compact one-dimensional manifold whose boundary is the disjoint union of the two fibres. Every compact one-manifold has an even number of boundary points.

The same transverse-preimage argument applied to a smooth homotopy proves homotopy invariance. Consequently the definition depends only on

f

, rather than on the chosen regular value.

Solved by gpt-5.6-sol high.

c

Words: 166 Articles: 1

Solution

Words: 166

For

v \in T_{p} S

, let

γ_{v}

be the maximal geodesic satisfying

γ_{v} (0) = p, \overset{γ}{˙}_{v} (0) = v .

(266)

The exponential map is

exp_{p} (v) = γ_{v} (1)

(267)

on the domain of the exponential map

D_{p} = {v \in T_{p} S : γ_{v} is defined throughout [0, 1]} .

(268)

This is an open, star-shaped neighbourhood of zero. In local coordinates the geodesic equation is a smooth ordinary differential equation, so smooth dependence on its initial position and velocity proves that

(p, v) \mapsto exp_{p} (v)

is smooth on its domain.

For both requested phenomena consider the embedded surface

S = {(x, y, 0) : (x, y) \neq = (0, 0)}

(269)

with

p = (1, 0, 0)

. Its geodesics are Euclidean straight lines for as long as they remain in

S

. The initial vector

v = (- 2, 0, 0)

would give

γ_{v} (t) = (1 - 2 t, 0, 0),

(270)

which reaches the missing origin at

t = 1/2

. Thus

v \in / D_{p}

, and the domain is not all of

T_{p} S

The point

q = (- 1, 0, 0)

is not in the image of

exp_{p}

. Any geodesic from

p

q

would have to be their unique Euclidean straight line, which passes through the omitted origin. Hence this exponential map is also not surjective, as summarized by exponential map of the punctured plane.

Solved by gpt-5.6-sol high.

d

Words: 101 Articles: 1

Solution

Words: 101

The induced metric makes the compact surface

S

a compact Riemannian manifold. By the Hopf-Rinow theorem it is geodesically complete, so every

exp_{p}

is defined on all of

T_{p} S

. In particular,

exp_{p} (V (p))

is defined for every

p

The global exponential map

Exp : TS \to S, (p, v) \mapsto exp_{p} (v),

(271)

is smooth by smooth dependence of geodesics on initial data. Since

p \mapsto (p, V (p))

is a smooth section of

TS

, their composition

ϕ (p) = Exp (p, V (p))

(272)

is smooth.

For

0 \leq t \leq 1

, define

ϕ_{t} (p) = exp_{p} (t V (p)) .

(273)

Completeness makes this a well-defined smooth homotopy, with

ϕ_{0} = id_{S}

and

ϕ_{1} = ϕ

. By homotopy invariance of degree modulo two,

de g_{2} (ϕ) = de g_{2} (id_{S}) = 1.

(274)

This is the exponential displacement map on a compact surface.

Solved by gpt-5.6-sol high.

27K (Probability and Measure)

Words: 378 Articles: 4

a

Words: 220 Articles: 1

Solution

Words: 220

Dynkin lemma, also called the pi-lambda theorem, states that if a Dynkin system

D

contains a pi-system

P

, then

σ (P) \subseteq D .

(275)

Let

λ (P)

be the smallest Dynkin system containing

P

. It is enough to prove that

λ (P)

is a sigma-algebra, because then

σ (P) \subseteq λ (P) \subseteq D .

(276)

First fix

A \in P

and define

G_{A} = {B \in λ (P) : A \cap B \in λ (P)} .

(277)

This is a Dynkin system. It contains

Ω

because

A \in λ (P)

; closure under relative complements follows from

A \cap B^{c} = A ∖ (A \cap B),

(278)

and closure under disjoint unions follows by distributing

A \cap -

over the union. Since

P

is closed under intersections,

P \subseteq G_{A}

. Minimality therefore gives

λ (P) \subseteq G_{A} .

(279)

Thus

A \cap B \in λ (P)

whenever

A \in P

and

B \in λ (P)

Now fix

B \in λ (P)

and define

H_{B} = {A \in λ (P) : A \cap B \in λ (P)} .

(280)

The same argument shows that

H_{B}

is a Dynkin system, and the preceding paragraph shows that it contains

P

. Hence

H_{B} = λ (P)

. We have proved that

λ (P)

is itself a pi-system.

A Dynkin system that is also a pi-system is a sigma-algebra: it is closed under arbitrary finite intersections, hence finite unions by complements, and any countable union can be disjointified before using closure under disjoint unions. Therefore

λ (P) = σ (P)

, proving the lemma.

Solved by gpt-5.6-sol high.

b

Words: 158 Articles: 1

Solution

Words: 158

Fix

A_{1} \in A_{1}

and let

D_{A_{1}} = {B \in F : P (A_{1} \cap B) = P (A_{1}) P (B)} .

(281)

This is a Dynkin system. It contains

Ω

. If

B \in D_{A_{1}}

, then

P (A_{1} \cap B^{c}) = P (A_{1}) - P (A_{1} \cap B) = P (A_{1}) (1 - P (B)) = P (A_{1}) P (B^{c}),

(282)

so it is closed under complements. Countable additivity gives closure under countable disjoint unions.

The hypothesis says

A_{2} \subseteq D_{A_{1}}

. Since

A_{2}

is a pi-system, Dynkin lemma yields

σ (A_{2}) \subseteq D_{A_{1}} .

(283)

Thus the factorization holds for every

A_{1} \in A_{1}

and every

B \in σ (A_{2})

Now fix such a

B

and define

E_{B} = {A \in F : P (A \cap B) = P (A) P (B)} .

(284)

The identical calculation makes

E_{B}

a Dynkin system. The first step gives

A_{1} \subseteq E_{B}

, so a second application of Dynkin lemma gives

σ (A_{1}) \subseteq E_{B} .

(285)

Therefore the probability factorization holds for every

A \in σ (A_{1}), B \in σ (A_{2}),

(286)

which is precisely independence of the two sigma-algebras. This is independence extended from generating pi-systems.

Solved by gpt-5.6-sol high.

28J (Applied Probability)

Words: 322 Articles: 8

a

Words: 71 Articles: 1

Solution

Words: 71

Superpose the two independent Poisson processes. The combined process has rate

9

, and each arrival is independently a Number 1 bus with probability

1/9

and a Number 8 bus with probability

8/9

. Therefore the number of Number 8 buses before the first Number 1 has the geometric distribution on

{0, 1, 2, \dots}

, and

P (exactly three Number 8 buses) = (\frac{8}{9})^{3} \frac{1}{9} = \frac{512}{6561} .

(287)

This is number of one Poisson type before another with rates

1

and

8

Solved by gpt-5.6-sol high.

b

Words: 93 Articles: 1

Solution

Words: 93

For

0 < j_{1} < \dots < j_{n} < t

, the probability of one jump in each infinitesimal interval around

j_{1}, \dots, j_{n}

and no other jump by time

t

gives the joint density

λ^{n} e^{- λ t}

(288)

for

(J_{1}, l d o t s, J_{n})

together with the event

N_{t} = n

. Since

P (N_{t} = n) = e^{- λ t} \frac{( λ t ) ^{n}}{n !},

(289)

the conditional density is

\frac{λ ^{n} e ^{- λ t}}{e ^{- λ t} ( λ t ) ^{n} / n !} = \frac{n !}{t ^{n}}

(290)

on that ordered simplex.

Now

n

independent uniform points on

[0, t]

have joint density

t^{- n}

on the cube. Each ordered vector has

n!

permutations, so their order statistics have density

n! / t^{n}

on the same simplex. The densities agree, proving Poisson process conditional arrival times.

Solved by gpt-5.6-sol high.

c

Words: 86 Articles: 1

Solution

Words: 86

Condition on

N_{t} = n

. By Poisson process conditional arrival times, the jump times are the ordered version of independent

U_{1}, \dots, U_{n}

uniformly distributed on

[0, t]

. The sum is symmetric in the marked pairs, and the marks are independent, so

E [exp {θ i = 1 \sum N_{t} g (J_{i}, X_{i})} N_{t} = n] = [\frac{1}{t} \int_{0}^{t} E (e^{θ g (s, X_{1})}) d s]^{n} .

(291)

Write the bracket as

a

. Averaging over the Poisson distribution of

N_{t}

gives

E exp {θ i = 1 \sum N_{t} g (J_{i}, X_{i})} = e^{- λ t} n = 0 \sum \infty \frac{( λ t a ) ^{n}}{n !} = exp {λ t (a - 1)} = exp {λ \int_{0}^{t} (E (e^{θ g (s, X_{1})}) - 1) d s} .

(292)

This is the exponential formula for a marked Poisson sum.

Solved by gpt-5.6-sol high.

d

Words: 72 Articles: 1

Solution

Words: 72

Use the convention

J_{0} = 0

, so

A (t) = t

when

N_{t} = 0

. Conditional on

N_{t} = n \geq 1

, the last jump

J_{n}

is the largest of

n

independent uniform points on

[0, t]

. The stated order statistic formula gives

E (J_{n} ∣ N_{t} = n) = \frac{n t}{n + 1},

(293)

and hence, for every

n \geq 0

E (A (t) ∣ N_{t} = n) = \frac{t}{n + 1} .

(294)

Putting

μ = λ t

and averaging over

N_{t}

E A (t) = t e^{- μ} n = 0 \sum \infty \frac{μ ^{n}}{n ! ( n + 1 )} = t e^{- μ} \frac{e ^{μ} - 1}{μ} = \frac{1 - e ^{- λ t}}{λ} .

(295)

This agrees with age of a Poisson process.

Solved by gpt-5.6-sol high.

29K (Principles of Statistics)

Words: 239 Articles: 6

a

Words: 33 Articles: 1

Solution

Words: 33

Under the usual regularity conditions, consistency and asymptotic normality of a maximum likelihood estimator give

n (θ_{n} - θ_{0}) d N_{p} (0, I (θ_{0})^{- 1}),

(296)

where

I (θ_{0})

is the one-observation Fisher information matrix. In a scalar model this reads

N (0, I (θ_{0})^{- 1})

Solved by gpt-5.6-sol high.

b

Words: 86 Articles: 1

Solution

Words: 86

The multivariate Wald statistic for a candidate parameter

θ \in R^{p}

W_{n} (θ) = n (θ_{n} - θ)^{T} I (θ_{n}) (θ_{n} - θ) .

(297)

Replacing

I (θ_{n})

by any consistent estimator of

I (θ_{0})

gives the same limit.

Under

P_{θ_{0}}

, part (a) and consistency of the information matrix imply, by Slutsky theorem, that

I (θ_{n})^{1/2} n (θ_{n} - θ_{0}) d N_{p} (0, I_{p}) .

(298)

Consequently

W_{n} (θ_{0}) d χ_{p}^{2},

(299)

using the characterization of the chi-squared distribution as the squared norm of a standard normal vector.

q_{p, 1 - α}

is the

(1 - α)

quantile of

χ_{p}^{2}

, an asymptotic

(1 - α)

confidence region is

C_{n} = {θ \in R^{p} : W_{n} (θ) \leq q_{p, 1 - α}} .

(300)

Solved by gpt-5.6-sol high.

c

Words: 120 Articles: 1

Solution

Words: 120

Let

R = (I_{k} 0) \in R^{k \times p}, r = R θ^{*} .

(301)

Then the null hypothesis is

R θ_{0} = r

. Define the Wald statistic for linear restrictions

W_{n, R} = n (R θ_{n} - r)^{T} [R I (θ_{n})^{- 1} R^{T}]^{- 1} (R θ_{n} - r) .

(302)

Under

H_{0}

, part (a) and the continuous mapping theorem give

n (R θ_{n} - r) = n R (θ_{n} - θ_{0}) d N_{k} (0, V_{0}),

(303)

where

V_{0} = R I (θ_{0})^{- 1} R^{T} .

(304)

The matrix

V_{0}

is positive definite because

I (θ_{0})

is positive definite and

R

has full row rank. Consistency gives

R I (θ_{n})^{- 1} R^{T} p V_{0} .

(305)

Therefore Slutsky theorem yields

V_{0}^{- 1/2} n (R θ_{n} - r) d N_{k} (0, I_{k}),

(306)

and applying the continuous squared-norm map proves rigorously that

W_{n, R} d χ_{k}^{2} .

(307)

Let

q_{k, 1 - α}

be the

(1 - α)

quantile of

χ_{k}^{2}

. Rejecting

H_{0}

when

W_{n, R} > q_{k, 1 - α}

(308)

has rejection probability tending to

α

under every fixed parameter satisfying

H_{0}

, and is therefore an asymptotically valid level-

α

test.

Solved by gpt-5.6-sol high.

30K (Stochastic Financial Models)

Words: 498 Articles: 8

a

Words: 76 Articles: 1

Solution

Words: 76

y_{1} \leq y_{2}

, then the utility function is increasing, so for every

X \in X

E [U (X + y_{1})] \leq E [U (X + y_{2})] .

(309)

Taking suprema gives

F (y_{1}) \leq F (y_{2})

For

0 \leq t \leq 1

, let

X_{1}

and

X_{2}

attain the suprema at

y_{1}

and

y_{2}

. Because

X

is a vector space,

X_{t} = t X_{1} + (1 - t) X_{2} \in X .

(310)

The concavity of

U

now gives

F (t y_{1} + (1 - t) y_{2}) \geq E [U (X_{t} + t y_{1} + (1 - t) y_{2})] \geq t E [U (X_{1} + y_{1})] + (1 - t) E [U (X_{2} + y_{2})] = tF (y_{1}) + (1 - t) F (y_{2}) .

(311)

Thus

F

is increasing and concave, as stated by optimized affine shift of concave utility.

Solved by gpt-5.6-sol high.

b

Words: 55 Articles: 1

Solution

Words: 55

For

0 \leq t \leq 1

, concavity of

U

implies

G (t s_{1} + (1 - t) s_{2}) = E [U (t (m + s_{1} Z) + (1 - t) (m + s_{2} Z))] \geq tG (s_{1}) + (1 - t) G (s_{2}),

(312)

G

is concave.

Since

E Z = 0

, Jensen inequality gives

G (s) = E [U (m + s Z)] \leq U (m + s E Z) = U (m) = G (0) .

(313)

0 \leq s < t

, concavity and

s = (1 - s / t) 0 + (s / t) t

yield

G (s) \geq (1 - \frac{s}{t}) G (0) + \frac{s}{t} G (t) \geq G (t) .

(314)

Hence

G

is decreasing on

[0, \infty)

; this is scaled centered risk under concave utility.

Solved by gpt-5.6-sol high.

c

Words: 161 Articles: 1

Solution

Words: 161

Let the value function at time

n

be the greatest conditional expected terminal utility attainable from wealth

x

. Independence of the innovations gives the Bellman equation

V (N, x) = U (x), V (n, x) = θ \in R^{d} sup E [V (n + 1, (1 + r) x + θ^{T} ξ_{n + 1})]

(315)

for

0 \leq n < N

. A policy obtained from the maximizing choices is admissible because each time-

n

holding is a predictable process, hence is known before

ξ_{n}

is observed.

We prove the claimed properties by backward induction. They hold at

n = N

because

V (N, \cdot) = U

is increasing and concave. Suppose they hold at time

n + 1

. For fixed

θ

, the map

x ⟼ E [V (n + 1, (1 + r) x + θ^{T} ξ_{n + 1})]

(316)

is increasing because

1 + r > 0

, and taking a supremum preserves this inequality.

For concavity, let

θ_{x}

and

θ_{y}

be optimal at wealths

x

and

y

. For

0 \leq t \leq 1

, use the admissible portfolio

θ_{t} = t θ_{x} + (1 - t) θ_{y} .

(317)

The next wealth is the same convex combination of the next wealths generated from

(x, θ_{x})

and

(y, θ_{y})

. Therefore

V (n, t x + (1 - t) y) \geq E [V (n + 1, (1 + r) (t x + (1 - t) y) + θ_{t}^{T} ξ_{n + 1})] \geq t V (n, x) + (1 - t) V (n, y) .

(318)

This completes the induction and proves monotonicity and concavity of a portfolio value function.

Solved by gpt-5.6-sol high.

d

Words: 206 Articles: 1

Solution

Words: 206

Condition on the information available before time

n

. The current wealth is then fixed, while

ξ_{n} \sim N (b, Σ)

is independent of that information. Put

B = b^{T} Σ^{- 1} b > 0.

(319)

For any portfolio

θ

, use the inner product induced by the positive-definite matrix

Σ

to write

θ = λ Σ^{- 1} b + η, λ = \frac{θ ^{T} b}{B}, η^{T} b = 0.

(320)

The linear image of a multivariate normal vector shows that the two portfolio returns are jointly normal, and

cov ((λ Σ^{- 1} b)^{T} ξ_{n}, η^{T} ξ_{n}) = λ η^{T} Σ Σ^{- 1} b = λ η^{T} b = 0.

(321)

Thus independence of uncorrelated jointly normal variables makes the residual return

η^{T} ξ_{n}

independent of the aligned return; it also has expected value zero.

The continuation value

V (n, \cdot)

is concave by part (c). Conditional Jensen inequality therefore shows that adding the independent centered residual cannot improve the objective:

E [V (n, m + (λ Σ^{- 1} b)^{T} ξ_{n} + η^{T} ξ_{n})] \leq E [V (n, m + (λ Σ^{- 1} b)^{T} ξ_{n})],

(322)

where

m = (1 + r) X_{n - 1}

. Since the optimal portfolio is unique, its residual must be zero.

It remains to determine the sign. If

λ < 0

, the portfolios

λ Σ^{- 1} b

and

- λ Σ^{- 1} b

have the same return variance, namely

λ^{2} B

, while their means are

λ B

and

- λ B

. The latter return has the distribution of the former plus the positive constant

- 2 λ B

. Since

V (n, \cdot)

is increasing, replacing

λ

- λ

cannot reduce the objective, contradicting uniqueness. Hence

λ \geq 0

Applying this conditional argument at every time gives nonnegative, past-measurable random variables

λ_{n}

such that

θ_{n}^{*} = λ_{n} Σ^{- 1} b, 1 \leq n \leq N .

(323)

This is the Gaussian one-fund theorem.

Solved by gpt-5.6-sol high.

31J (Mathematics of Machine Learning)

Words: 404 Articles: 12

a

Words: 30 Articles: 1

Solution

Words: 30

A set

C \subseteq R^{d}

is convex when every line segment between two of its points remains in the set: for every

x, y \in C

and

0 \leq t \leq 1

t x + (1 - t) y \in C .

(324)

Solved by gpt-5.6-sol high.

b

Words: 60 Articles: 1

Solution

Words: 60

A function

f : C \to R

is strictly convex when, for distinct

x, y \in C

and

0 < t < 1

f (t x + (1 - t) y) < t f (x) + (1 - t) f (y) .

(325)

Suppose distinct points

x

and

y

were both minimizers, with minimum value

m

. Since

C

is a convex set, their midpoint belongs to

C

, while strict convexity gives

f (\frac{x + y}{2}) < \frac{f ( x ) + f ( y )}{2} = m,

(326)

a contradiction. Thus any minimizer is unique, which is the uniqueness of a minimizer of a strictly convex function.

Solved by gpt-5.6-sol high.

c

Words: 52 Articles: 1

Solution

Words: 52

For

x \in R^{d}

, the euclidean projection onto a convex set

C

π_{C} (x) = z \in C argmin ∥ x - z ∥_{2} .

(327)

The squared distance

z \mapsto ∥ x - z ∥_{2}^{2}

is a strictly convex function. Its restriction to the convex set

C

can therefore have at most one minimizer by part (b). Closedness, together with finite dimensionality, supplies existence, although existence was not required to be proved.

Solved by gpt-5.6-sol high.

d

Words: 46 Articles: 1

Solution

Words: 46

Assume

π \in C

and

(x - π)^{T} (z - π) \leq 0

(328)

for every

z \in C

. Expanding the squared euclidean distance gives

∥ x - z ∥_{2}^{2} = ∥ (x - π) - (z - π) ∥_{2}^{2} = ∥ x - π ∥_{2}^{2} + ∥ z - π ∥_{2}^{2} - 2 (x - π)^{T} (z - π) \geq ∥ x - π ∥_{2}^{2} .

(329)

Hence

π

minimizes the distance from

x

over

C

, so

π = π_{C} (x)

. This proves the sufficient direction of the variational characterization of convex projection.

Solved by gpt-5.6-sol high.

e

Words: 131 Articles: 1

Solution

Words: 131

The set

C

is the second-order cone. Put

ρ = ∥ u ∥_{2}, α = \frac{1}{2} (1 + \frac{t}{ρ}) .

(330)

When

ρ \geq ∣ t ∣

and

ρ > 0

, one has

0 \leq α \leq 1

, and the proposed point

π = α (u, ρ)

(331)

lies on the boundary of

C

because

∥ αu ∥_{2} = α ρ

Set

δ = (ρ - t) /2 \geq 0

. Since

(u, t) - π = δ (\frac{u}{ρ}, - 1),

(332)

for any

(v, s) \in C

we obtain

((u, t) - π)^{T} ((v, s) - π) = δ (\frac{u ^{T} v}{ρ} - s) \leq δ (∥ v ∥_{2} - s) \leq 0,

(333)

where the first inequality is the Cauchy-Schwarz inequality and the second uses

∥ v ∥_{2} \leq s

. Part (d) therefore gives

π_{C} (u, t) = \frac{1}{2} (1 + \frac{t}{∥ u ∥ _{2}}) (u, ∥ u ∥_{2})

(334)

whenever

∥ u ∥_{2} \geq ∣ t ∣

and

u \neq = 0

. At

(u, t) = (0, 0)

the projection is plainly

(0, 0)

, consistent with the limiting formula.

ρ \leq - t

, then for every

(v, s) \in C

(u, t)^{T} (v, s) \leq ρ ∥ v ∥_{2} + t s \leq (ρ + t) s \leq 0.

(335)

Applying part (d) with

π = 0

gives

π_{C} (u, t) = (0, 0) .

(336)

Together with the trivial case

ρ \leq t

, this is the full projection onto the second-order cone.

Solved by gpt-5.6-sol high.

f

Words: 85 Articles: 1

Solution

Words: 85

Using the harmless normalization

1/ (2 n)

, the squared-loss empirical risk for linear prediction is

R_{n} (β) = \frac{1}{2 n} \sum_{i = 1}^{n} (β^{T} X_{i} - Y_{i})^{2},

(337)

with explicit gradient

\nabla R_{n} (β) = \frac{1}{n} i = 1 \sum n (β^{T} X_{i} - Y_{i}) X_{i} .

(338)

Starting from any

β_{0} \in C

, projected gradient descent with step sizes

η_{k} > 0

β_{k + 1} = β_{k} - η_{k} \nabla R_{n} (β_{k}), β_{k + 1} = π_{C} (β_{k + 1}) .

(339)

For a fully explicit update, write

β_{k + 1} = (u_{k}, t_{k})

and

ρ_{k} = ∥ u_{k} ∥_{2}

. The projection from part (e) is

π_{C} (u_{k}, t_{k}) = ⎩ ⎨ ⎧ (u_{k}, t_{k}), (0, 0), \frac{1}{2} (1 + \frac{t _{k}}{ρ _{k}}) (u_{k}, ρ_{k}), ρ_{k} \leq t_{k}, ρ_{k} \leq - t_{k}, ρ_{k} > ∣ t_{k} ∣.

(340)

Every iterate therefore lies in the prescribed hypothesis class, and any convergent run under the standard convex-optimization step-size conditions targets its empirical-risk minimizer.

Solved by gpt-5.6-sol high.

32A (Dynamical Systems)

Words: 399 Articles: 4

a

Words: 151 Articles: 1

Solution

Words: 151

Dulac theorem states the following. Let

D

be a simply connected region for the planar system

\overset{z}{˙} = f (z)

, and suppose there is a continuously differentiable function

B : D \to R

for which

\nabla \cdot (B f)

(341)

has one strict sign throughout

D

. Then there is no periodic orbit contained in

D

To prove it, suppose a periodic orbit

Γ

existed and let

A

be its interior. Simple connectedness ensures

A \subset D

. By the planar divergence theorem,

\iint_{A} \nabla \cdot (B f) d A = \int_{Γ} B f \cdot n d s .

(342)

The vector field

f

is tangent to its trajectory

Γ

, so its scalar product with the outward normal

n

vanishes. The right-hand side is therefore zero. The left-hand side cannot be zero because its integrand has one strict sign, a contradiction.

The Poincare-Bendixson theorem states that if a forward trajectory of a smooth planar system remains in a compact set and its nonempty omega-limit set contains no equilibrium point, then that omega-limit set is a periodic orbit.

Solved by gpt-5.6-sol high.

b

Words: 248 Articles: 1

Solution

Words: 248

At an equilibrium point of a dynamical system, the second equation gives

x (s - x y) = 0.

(343)

The choice

x = 0

is impossible because then

\overset{x}{˙} = r > 0

. Hence

y = s / x

, and substitution into the first equation gives

0 = r - (1 + s) x + s x = r - x .

(344)

Thus the unique fixed point is

(x_{*}, y_{*}) = (r, \frac{s}{r}) .

(345)

Choose

α = \frac{r}{1 + s},

(346)

which satisfies

0 < α < r

. Let

R = {(x, y) : x \geq α, 0 \leq y \leq \frac{s}{α}, x + y \leq r + \frac{s}{α}} .

(347)

We check the direction of the vector field on its four edges.

x = α

\overset{x}{˙} = r - (1 + s) α + α^{2} y = α^{2} y \geq 0.

(348)

y = 0

\overset{y}{˙} = s x > 0.

(349)

y = s / α

, where

x \geq α

\overset{y}{˙} = s x (1 - \frac{x}{α}) \leq 0.

(350)

Finally, throughout the system,

\overset{x}{˙} + \overset{y}{˙} = r - x .

(351)

On the sloping edge

x + y = r + s / α

, the bound

y \leq s / α

implies

x \geq r

, and hence

\overset{x}{˙} + \overset{y}{˙} \leq 0

. Every boundary edge therefore points inward. Violations of the corresponding four inequalities are driven toward the boundary in the same order, so positive trajectories enter this compact trapping region and then remain there. This is the Brusselator trapping region.

The Jacobian matrix at the unique fixed point is

J_{*} = (- (1 + s) + 2 x_{*} y_{*} s - 2 x_{*} y_{*} x_{*}^{2} - x_{*}^{2}) = (s - 1 - s r^{2} - r^{2}) .

(352)

Its determinant and trace are

det J_{*} = r^{2} > 0, tr J_{*} = s - 1 - r^{2} .

(353)

When

s - 1 > r^{2}

, the trace is positive, so linear stability of a planar equilibrium shows that the fixed point is a repeller. It lies in the interior of

R

Choose a trajectory in

R

other than the equilibrium. Compactness of

R

gives a nonempty compact omega-limit set, and the repelling equilibrium cannot belong to that set. Since there are no other equilibria, the Poincare-Bendixson theorem gives a periodic orbit. Hence

s - 1 > r^{2} ⟹ the system has a periodic orbit .

(354)

This is the Brusselator periodic-orbit criterion.

Solved by gpt-5.6-sol high.

33E (Integrable Systems)

Words: 306 Articles: 1

Solution

Words: 306

Equality of the mixed derivatives of the auxiliary vector

Ψ

gives

Ψ_{x t} = U_{t} Ψ + U V Ψ, Ψ_{t x} = V_{x} Ψ + V U Ψ.

(355)

Thus the zero-curvature condition for this convention is

U_{t} - V_{x} + [U, V] = 0 .

(356)

Substituting the displayed matrices and collecting powers of the spectral parameter

λ

makes every diagonal entry and every

λ

-dependent term cancel. The remaining matrix is

U_{t} - V_{x} + [U, V] = (0 i r_{t} + r_{xx} + 2 q r^{2} i q_{t} - q_{xx} - 2 r q^{2} 0) .

(357)

Consequently compatibility is equivalent to

i r_{t} + r_{xx} + 2 q r^{2} = 0, i q_{t} - q_{xx} - 2 r q^{2} = 0,

(358)

so the requested constant is

a = 2 .

(359)

This is the AKNS Lax pair for the nonlinear Schrodinger equation.

The reduction

q = \overline{r}

(360)

is preserved by the two equations, which become complex conjugates and reduce to the Focusing nonlinear Schrodinger equation

i r_{t} + r_{xx} + 2∣ r ∣^{2} r = 0 .

(361)

The analogous reduction

q = - \overline{r}

(362)

gives the Defocusing nonlinear Schrödinger equation

i r_{t} + r_{xx} - 2∣ r ∣^{2} r = 0 .

(363)

For a complex field

r

and rapidly decreasing boundary conditions, define

H_{foc} [r, \overline{r}] = \int_{R} (∣ r_{x} ∣^{2} - ∣ r ∣^{4}) d x, H_{def} [r, \overline{r}] = \int_{R} (∣ r_{x} ∣^{2} + ∣ r ∣^{4}) d x .

(364)

Their variational derivatives, after integration by parts, are

\frac{δ H _{foc}}{δ r} = - r_{xx} - 2∣ r ∣^{2} r, \frac{δ H _{def}}{δ r} = - r_{xx} + 2∣ r ∣^{2} r .

(365)

Both equations therefore have the Hamiltonian field equation

i r_{t} = \frac{δH}{δ r},

(366)

or, including the conjugate equation,

(r_{t} \overline{r}_{t}) = (0 i - i 0) (δH / δr δH / δ \overline{r}) .

(367)

This is the Hamiltonian form of the cubic nonlinear Schrodinger equations.

Now put

r (x, t) = e^{- i Et} f (x),

(368)

where

E

and

f

are real and

f

is smooth and rapidly decreasing. For the focusing equation,

f^{''} + E f + 2 f^{3} = 0.

(369)

Multiplication by

f^{'}

and use of the decay at infinity gives the first integral

(f^{'})^{2} = - E f^{2} - f^{4} .

(370)

A nonzero solution requires

E < 0

. Writing

E = - κ^{2}

with

κ > 0

, separation of variables, or direct substitution, gives

f (x) = \pm κ sech (κ (x - x_{0})) .

(371)

Hence

r (x, t) = \pm κ e^{i κ^{2} t} sech (κ (x - x_{0}))

(372)

is the Bright standing soliton of the focusing nonlinear Schrodinger equation.

For the defocusing equation the profile instead satisfies

f^{''} + E f - 2 f^{3} = 0,

(373)

with first integral

(f^{'})^{2} = f^{4} - E f^{2} = f^{2} (f^{2} - E) .

(374)

If a nonzero rapidly decreasing profile existed,

∣ f ∣

would attain a positive maximum

A

. At that point

f^{'} = 0

, so the identity forces

E = A^{2} > 0

. But along either tail, where

0 < ∣ f ∣ < A

, the right-hand side

f^{2} (f^{2} - E)

is negative, which is impossible. Thus there is no nonzero solution of the prescribed form, as recorded by No rapidly decaying standing wave for the defocusing cubic nonlinear Schrodinger equation.

Solved by gpt-5.6-sol high.

34B (Principles of Quantum Mechanics)

Words: 270 Articles: 7

a

Words: 128 Articles: 1

Solution

Words: 128

In units with

ℏ = 1

, the quantum harmonic oscillator Hamiltonian and number operator are

H = ω (N + \frac{1}{2}), N = A^{†} A .

(375)

The nontrivial commutators among the creation and annihilation operators and

N

are

[A, A^{†}] = 1, [N, A] = - A, [N, A^{†}] = A^{†} .

(376)

Their reversed commutators have the opposite signs, while the commutator of any operator with itself is zero.

The operator

N

is self-adjoint because

(A^{†} A)^{†} = A^{†} A .

(377)

Its eigenvalues are therefore real. Moreover, for every state

∣ ψ ⟩

⟨ ψ ∣ N ∣ ψ ⟩ = ⟨ A ψ ∣ A ψ ⟩ = ∥ A ∣ ψ ⟩ ∥^{2} \geq 0,

(378)

so every eigenvalue is nonnegative.

N ∣ ν ⟩ = ν ∣ ν ⟩

, the commutator with

A

gives

N (A ∣ ν ⟩) = (ν - 1) A ∣ ν ⟩,

(379)

and

∥ A ∣ ν ⟩ ∥^{2} = ν ∥ ∣ ν ⟩ ∥^{2} .

(380)

Repeated application of

A

lowers the eigenvalue by one. It must terminate before producing a negative eigenvalue. If

m

is the last occupied rung, then

A ∣ ν - m ⟩ = 0

, and the norm identity forces

ν - m = 0

. Hence

ν \in {0, 1, 2, \dots} .

(381)

This is the integer spectrum of the number operator.

Solved by gpt-5.6-sol high.

b

Words: 142 Articles: 4

i

Words: 42 Articles: 1

Solution

Words: 42

For occupation numbers

n, m \geq 0

, the Hamiltonian of the two commensurate quantum harmonic oscillators is

H_{0} = N_{A} + 2 N_{B} + \frac{3}{2}, E_{n, m} = n + 2 m + \frac{3}{2} .

(382)

The five lowest distinct levels and their product-state bases are

E \frac{3}{2} \frac{5}{2} \frac{7}{2} \frac{9}{2} \frac{11}{2} states ∣ n, m ⟩ ∣0, 0 ⟩ ∣1, 0 ⟩ ∣2, 0 ⟩, ∣0, 1 ⟩ ∣3, 0 ⟩, ∣1, 1 ⟩ ∣4, 0 ⟩, ∣2, 1 ⟩, ∣0, 2 ⟩ degeneracy 11223

(383)

The repeated levels exhibit quantum degeneracy.

Solved by gpt-5.6-sol high.

ii

Words: 100 Articles: 1

Solution

Words: 100

The degenerate

E_{0} = 9/2

subspace has ordered basis

∣3, 0 ⟩, ∣1, 1 ⟩ .

(384)

Using the ladder actions,

H^{'} ∣3, 0 ⟩ H^{'} ∣1, 1 ⟩ = A^{2} B^{†} ∣3, 0 ⟩ = 6 ∣1, 1 ⟩, = A^{† 2} B ∣1, 1 ⟩ = 6 ∣3, 0 ⟩ .

(385)

Thus degenerate perturbation theory asks us to diagonalize

H^{'}_{E_{0}} = (0660) .

(386)

Its normalized eigenvectors and first-order energies are

∣ ψ_{+} ⟩ = \frac{∣3 , 0 ⟩ + ∣1 , 1 ⟩}{2}, E_{+} = \frac{9}{2} + 6 λ,

(387)

∣ ψ_{-} ⟩ = \frac{∣3 , 0 ⟩ - ∣1 , 1 ⟩}{2}, E_{-} = \frac{9}{2} - 6 λ .

(388)

There is no order-

λ

admixture from other unperturbed levels.

In fact,

H^{'}

preserves

n + 2 m

, so this two-dimensional eigenspace of

H_{0}

is invariant. Directly,

H^{'} ∣ ψ_{\pm} ⟩ = \pm 6 ∣ ψ_{\pm} ⟩, H_{0} ∣ ψ_{\pm} ⟩ = \frac{9}{2} ∣ ψ_{\pm} ⟩ .

(389)

Therefore

H ∣ ψ_{\pm} ⟩ = (\frac{9}{2} \pm 6 λ) ∣ ψ_{\pm} ⟩

(390)

for every real

λ

, with no omitted higher-order terms. This is exact diagonalization within a degenerate subspace, specifically the resonant two-to-one oscillator coupling.

Solved by gpt-5.6-sol high.

35D (Applications of Quantum Mechanics)

Words: 349 Articles: 4

a

Words: 117 Articles: 1

Solution

Words: 117

For a wave incident from the left, define the reflection and transmission amplitudes by

ψ (x) \sim {e^{ik x} + r e^{- ik x}, t e^{ik x}, x \to - \infty, x \to + \infty, E = \frac{ℏ ^{2} k ^{2}}{2 m} .

(391)

Reflection invariance gives the same amplitudes for incidence from the right. If the incoming amplitudes from the left and right are ordered first, and the outgoing amplitudes toward the left and right second, the One-dimensional S-matrix is

S = (r t t r) .

(392)

For

V (x) = V_{0} δ (x)

, the wavefunction is continuous and its derivative has the jump

ψ^{'} (0^{+}) - ψ^{'} (0^{-}) = \frac{2 m V _{0}}{ℏ ^{2}} ψ (0) .

(393)

The scattering ansatz gives

1 + r = t, 2 ik r = \frac{2 m V _{0}}{ℏ ^{2}} t .

(394)

With

γ = \frac{m V _{0}}{ℏ ^{2} k},

(395)

we obtain the scattering by a delta potential amplitudes

t = \frac{1}{1 + iγ}, r = \frac{- iγ}{1 + iγ} .

(396)

They satisfy

∣ r ∣^{2} + ∣ t ∣^{2} = 1, r^{*} t + t^{*} r = 0.

(397)

Therefore

S^{†} S = (∣ r ∣^{2} + ∣ t ∣^{2} t^{*} r + r^{*} t r^{*} t + t^{*} r ∣ r ∣^{2} + ∣ t ∣^{2}) = I,

(398)

so the scattering matrix is unitary.

Solved by gpt-5.6-sol high.

b

Words: 232 Articles: 1

Solution

Words: 232

For a potential of period

a

, define the Floquet matrix for a one-dimensional periodic potential by

(ψ (x + a) ψ^{'} (x + a)) = M (E) (ψ (x) ψ^{'} (x)) .

(399)

The Schrodinger equation has no first-derivative term, so the Wronskian is constant and

det M = 1

. Its multipliers are therefore reciprocal. They lie on the unit circle precisely when

∣ tr M (E) ∣ \leq 2 .

(400)

Those energies form continuous allowed bands of bounded Bloch theorem solutions. Values with

∣ tr M ∣ > 2

have a growing and a decaying multiplier and form forbidden gaps. The band edges have

tr M = \pm 2

, as in Floquet discriminant and energy bands.

For the delta-comb Kronig-Penney model, start immediately to the right of one delta function. Free propagation to the next delta and the derivative jump there are represented by

P = (cos ka - k sin ka \frac{sin ka}{k} cos ka), D = (1 \frac{2 m V _{0}}{ℏ ^{2}} 01) .

(401)

Thus

M (E) = D P = cos ka \frac{2 m V _{0}}{ℏ ^{2}} cos ka - k sin ka \frac{sin ka}{k} cos ka + \frac{2 m V _{0}}{ℏ ^{2} k} sin ka .

(402)

Consequently the Floquet discriminant of the delta-comb Kronig-Penney model is

\frac{1}{2} tr M (E) = cos (ka) + γ sin (ka), γ = \frac{m V _{0}}{ℏ ^{2} k} .

(403)

All band edges are therefore determined by

cos (ka) + γ sin (ka) = σ, σ = \pm 1.

(404)

In factorized form, the periodic edges

σ = 1

obey

sin \frac{ka}{2} = 0 or tan \frac{ka}{2} = γ,

(405)

while the antiperiodic edges

σ = - 1

obey

cos \frac{ka}{2} = 0 or tan \frac{ka}{2} = - \frac{1}{γ} .

(406)

To express the same edges through the single-barrier scattering data, the One-dimensional transfer matrix from scattering amplitudes gives

tr M = \frac{( t ^{2} - r ^{2} ) e ^{ika} + e ^{- ika}}{t} .

(407)

At an edge put

z = e^{- ika}

and

tr M = 2 σ

. Then

z^{2} - 2 σ t z + t^{2} - r^{2} = 0,

(408)

so the scattering-amplitude equations for one-dimensional band edges are

e^{- ika} = σ t \pm r .

(409)

Explicitly,

tr M = 2 tr M = - 2 e^{- ika} = t + r or t - r e^{- ika} = - t + r or - t - r .

(410)

For the delta barrier,

t - r = 1

and

t + r = \frac{1 - iγ}{1 + iγ},

(411)

which reproduces the four factorized edge equations above.

Solved by gpt-5.6-sol high.

36A (Statistical Physics)

Words: 379 Articles: 11

a

Words: 139 Articles: 1

Solution

Words: 139

The microcanonical ensemble describes an isolated system at fixed energy

E

, volume

V

, and particle number

N

, assigning equal weight to accessible states in a narrow energy shell.

The canonical ensemble describes a system that can exchange energy with a heat reservoir at fixed temperature

T

, while

V

and

N

remain fixed. A microstate

s

has probability

p_{s} = \frac{e ^{- β E_{s}}}{Z}, Z = \sum_{s} e^{- β E_{s}}, β = \frac{1}{k _{B} T} .

(412)

The grand canonical ensemble allows exchange of both energy and particles at fixed

T

V

, and chemical potential

μ

. Its state weights are

p_{s} = \frac{e ^{- β (E_{s} - μ N_{s})}}{Ξ} .

(413)

The choice is irrelevant for bulk thermodynamic observables in the thermodynamic limit when the system is additive, has sufficiently short-range interactions, and lies away from singular phase-coexistence regimes. Then relative energy and particle-number fluctuations vanish as the system size grows. These are the standard conditions for equivalence of statistical ensembles.

Solved by gpt-5.6-sol high.

b

Words: 240 Articles: 8

i

Words: 47 Articles: 1

Solution

Words: 47

The classical canonical partition function uses one phase-space cell

h^{2}

per state. In polar coordinates,

Z_{1} = \frac{1}{h ^{2}} \int_{R^{2}} e^{- β p^{2} / (2 m)} d^{2} p \int_{r < R} e^{- β κ r^{2} /2} d^{2} x = \frac{1}{h ^{2}} (\frac{2 πm}{β}) (2 π \int_{0}^{R} r e^{- β κ r^{2} /2} d r) = \frac{4 π ^{2} m}{h ^{2} β ^{2} κ} (1 - e^{- β κ R^{2} /2}) .

(414)

Equivalently, using

h = 2 π ℏ

Z_{1} = \frac{m}{ℏ ^{2} β ^{2} κ} (1 - e^{- β κ R^{2} /2}) .

(415)

This is the particle in a finite two-dimensional harmonic trap partition function.

Solved by gpt-5.6-sol high.

ii

Words: 60 Articles: 1

Solution

Words: 60

Put

a = \frac{1}{2} κ R^{2} .

(416)

The thermodynamic derivatives of the canonical partition function give

⟨ E ⟩ = - \frac{\partial}{\partial β} lo g Z_{1} = - \frac{\partial}{\partial β} [- 2 lo g β + lo g (1 - e^{- β a}) + constant] = \frac{2}{β} - \frac{a}{e ^{β a} - 1} .

(417)

The two momentum coordinates are unrestricted quadratic degrees of freedom, so the equipartition theorem gives mean kinetic energy

1/ β

. Hence

⟨ V ⟩ = ⟨ E ⟩ - \frac{1}{β} = \frac{1}{β} - \frac{a}{e ^{β a} - 1} .

(418)

The same result follows from

⟨ V ⟩ = - (κ / β) \partial_{κ} lo g Z_{1}

Solved by gpt-5.6-sol high.

iii

Words: 99 Articles: 1

Solution

Words: 99

a = \frac{1}{2} κ R^{2} ≫ k_{B} T,

(419)

then

e^{β a}

is large and the wall correction is exponentially small:

⟨ E ⟩ \sim 2 k_{B} T .

(420)

This is exactly the equipartition theorem result from the two kinetic and two harmonic-potential quadratic terms.

a ≪ k_{B} T

, use

\frac{a}{e ^{β a} - 1} = \frac{1}{β} - \frac{a}{2} + O (β a^{2}) .

(421)

Then

⟨ E ⟩ = k_{B} T + \frac{κ R ^{2}}{4} + O (\frac{κ ^{2} R ^{4}}{k _{B} T}) \sim k_{B} T .

(422)

The leading term is the kinetic energy. The particle is nearly uniform in the disk, so

⟨ r^{2} ⟩ \to R^{2} /2

and

⟨ V ⟩ \to κ R^{2} /4

, much smaller than

k_{B} T

. The hard wall therefore suppresses the positional equipartition contribution, as in the hard-wall correction to harmonic equipartition.

Solved by gpt-5.6-sol high.

iv

Words: 34 Articles: 1

Solution

Words: 34

For

N

identical, noninteracting classical particles, the one-particle phase-space integrals factorize and indistinguishability supplies the Gibbs factor. Thus the classical ideal-gas partition function rule gives

Z_{N} = \frac{Z _{1}^{N}}{N !} = \frac{1}{N !} [\frac{4 π ^{2} m}{h ^{2} β ^{2} κ} (1 - e^{- β κ R^{2} /2})]^{N} .

(423)

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37A (Electrodynamics)

Words: 366 Articles: 8

a

Words: 106 Articles: 1

Solution

Words: 106

For a timelike worldline,

- η_{μν} \overset{x}{˙}^{μ} \overset{x}{˙}^{ν} > 0

. Its square root is a Lorentz scalar, and

- η_{μν} \overset{x}{˙}^{μ} \overset{x}{˙}^{ν} d λ = - η_{μν} d x^{μ} d x^{ν} = c d τ .

(424)

Thus the first term is the Lorentz-invariant free-particle action

- m c^{2} \int d τ

. Written as a line element, it is also invariant under a change of the worldline parameter

λ

The interaction is the scalar line integral

q \int A_{μ} d x^{μ} .

(425)

Under an electromagnetic gauge transformation

A_{μ} ⟼ A_{μ} + \partial_{μ} χ,

(426)

it changes only by

q \int \partial_{μ} χ d x^{μ} = q [χ]_{endpoints} .

(427)

This endpoint term does not affect a fixed-endpoint variational problem. Hence both Lorentz invariance and gauge invariance select the displayed relativistic charged-particle action, with the expected free-particle and minimal-coupling terms.

Solved by gpt-5.6-sol high.

b

Words: 75 Articles: 1

Solution

Words: 75

Write

L = - m c - \overset{x}{˙}^{2} + q A_{ν} (x) \overset{x}{˙}^{ν} .

(428)

Since

- \overset{x}{˙}^{2} = c \frac{d τ}{d λ}, \overset{x}{˙}^{μ} = \frac{d τ}{d λ} u^{μ},

(429)

its derivatives are

\frac{\partial L}{\partial x ˙ ^{μ}} = m u_{μ} + q A_{μ}, \frac{\partial L}{\partial x ^{μ}} = q (\partial_{μ} A_{ν}) \overset{x}{˙}^{ν} .

(430)

The Euler-Lagrange equation therefore gives

0 = \frac{d}{d λ} (m u_{μ} + q A_{μ}) - q (\partial_{μ} A_{ν}) \overset{x}{˙}^{ν} = m \frac{d u _{μ}}{d λ} + q (\partial_{ν} A_{μ} - \partial_{μ} A_{ν}) \overset{x}{˙}^{ν} .

(431)

Using the electromagnetic field tensor

F_{μν} = \partial_{μ} A_{ν} - \partial_{ν} A_{μ}

(432)

and dividing by

d τ / d λ

gives

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

(433)

Raising the first index yields the equivalent requested form

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

(434)

This is the Lorentz-force equation from the worldline action.

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c

Words: 75 Articles: 1

Solution

Words: 75

The four-velocity and particle energy are

u^{μ} = γ (c, v), E = γm c^{2}, p^{0} = m u^{0} = \frac{E}{c} .

(435)

With the conventions for the given metric,

F^{0 i} = E_{i} / c

. The temporal component of the covariant Lorentz force is therefore

\frac{d ( E / c )}{d τ} = q F^{0 i} u_{i} = \frac{q}{c} γ E \cdot v .

(436)

Since

d t / d τ = γ

, cancellation of

γ / c

gives the relativistic electromagnetic work-energy theorem

\frac{d E}{d t} = q E \cdot v .

(437)

The absence of

B

reflects that magnetic force is perpendicular to the velocity and does no work.

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d

Words: 110 Articles: 1

Solution

Words: 110

The spatial part of the Lorentz force is

\frac{d}{d t} (γm v) = q (E + v \times B) .

(438)

Part (c), together with

E = γm c^{2}

, gives

\frac{d γ}{d t} = \frac{q}{m c ^{2}} E \cdot v .

(439)

Writing

a = d v / d t

and expanding the momentum derivative,

mγ a + m v \frac{d γ}{d t} = q (E + v \times B) .

(440)

Substitution of

d γ / d t

gives the coordinate acceleration of a relativistic charged particle:

\frac{d v}{d t} = \frac{q}{mγ} [E + v \times B - \frac{1}{c ^{2}} v (v \cdot E)] .

(441)

In the nonrelativistic limit

∣ v ∣/ c \to 0

, one has

γ \to 1

and the final term is smaller by order

v^{2} / c^{2}

. Hence

m \frac{d v}{d t} = q (E + v \times B),

(442)

which is precisely Newton's second law with the ordinary Lorentz force.

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38B (General Relativity)

Words: 294 Articles: 6

a

Words: 60 Articles: 1

Solution

Words: 60

Put

Y = \nabla_{ρ} ϕ \nabla^{ρ} ϕ .

(443)

Using metric compatibility and the symmetric Hessian of a scalar field, the divergence of the proposed tensor is

2 \nabla^{μ} T_{μν} = (\nabla^{μ} \nabla_{μ} ϕ) \nabla_{ν} ϕ + \nabla^{μ} ϕ \nabla_{μ} \nabla_{ν} ϕ + A \nabla_{ν} Y + 2 Bϕ \nabla_{ν} ϕ = m^{2} ϕ \nabla_{ν} ϕ + \frac{1}{2} \nabla_{ν} Y + A \nabla_{ν} Y + 2 Bϕ \nabla_{ν} ϕ = (A + \frac{1}{2}) \nabla_{ν} Y + (m^{2} + 2 B) ϕ \nabla_{ν} ϕ .

(444)

For this to vanish for arbitrary solutions and local field data, the two coefficients must vanish. Hence

A = - \frac{1}{2}, B = - \frac{m ^{2}}{2} .

(445)

The resulting tensor is the Conserved Klein-Gordon scalar stress-energy ansatz.

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b

Words: 72 Articles: 1

Solution

Words: 72

Write the Einstein field equations in

D

dimensions as

R_{μν} - \frac{1}{2} g_{μν} R = κ T_{μν} .

(446)

Taking the trace gives

(1 - \frac{D}{2}) R = κ T, R = - \frac{2 κ}{D - 2} T .

(447)

Therefore the Trace-reversed Einstein field equations in D dimensions are

R_{μν} = κ (T_{μν} - \frac{1}{D - 2} g_{μν} T) .

(448)

For the values from part (a), let

Y = \nabla_{ρ} ϕ \nabla^{ρ} ϕ

. Then

T_{μν} = \frac{1}{2} \nabla_{μ} ϕ \nabla_{ν} ϕ - \frac{1}{4} g_{μν} (Y + m^{2} ϕ^{2}),

(449)

and

T = \frac{2 - D}{4} Y - \frac{D}{4} m^{2} ϕ^{2} .

(450)

Substitution cancels the

Y g_{μν}

terms and yields

R_{μν} = \frac{κ}{2} [\nabla_{μ} ϕ \nabla_{ν} ϕ + \frac{m ^{2}}{D - 2} g_{μν} ϕ^{2}] .

(451)

For the convention

κ = 8 π G

, the prefactor is

4 π G

. This is the Ricci curvature sourced by a Klein-Gordon scalar field.

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c

Words: 162 Articles: 1

Solution

Words: 162

Because the condition is local and covariant, choose normal coordinates at the point and a Lorentz frame in which the timelike observer is

X^{μ} = δ_{0}^{μ}

. Rescaling a timelike vector by a positive factor does not affect the sign, so it suffices to use this unit vector. With signature

(-, +, \dots, +)

\nabla_{ρ} ϕ \nabla^{ρ} ϕ = - (\partial_{0} ϕ)^{2} + \sum_{i = 1}^{D - 1} (\partial_{i} ϕ)^{2} .

(452)

The measured energy density is therefore

2 T_{μν} X^{μ} X^{ν} = 2 T_{00} = (\partial_{0} ϕ)^{2} - A [- (\partial_{0} ϕ)^{2} + i \sum (\partial_{i} ϕ)^{2}] - B ϕ^{2} = (1 + A) (\partial_{0} ϕ)^{2} - A i \sum (\partial_{i} ϕ)^{2} - B ϕ^{2} .

(453)

The value of

ϕ

, its time derivative, and its spatial first derivatives can be varied independently at a point. Nonnegativity for all such data is therefore equivalent to nonnegativity of all three coefficients:

1 + A \geq 0, - A \geq 0, - B \geq 0.

(454)

Thus the most general constraints are

- 1 \leq A \leq 0, B \leq 0 .

(455)

These conditions are also sufficient in every timelike frame, because any timelike vector can be brought to the chosen rest frame. This is the weak energy condition for a quadratic scalar stress-energy ansatz.

The conserved values from part (a),

A = - \frac{1}{2}, B = - \frac{m ^{2}}{2},

(456)

satisfy these inequalities for real

m

, as required.

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39C (Fluid Dynamics II)

Words: 326 Articles: 9

a

Words: 61 Articles: 1

Solution

Words: 61

The steady incompressible Stokes flow equations are

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

(457)

Taking the curl eliminates the pressure gradient and commutes with the Laplacian, so for the vorticity

ω = \nabla \times u

\nabla^{2} ω = 0 .

(458)

This is the vorticity part of Harmonic pressure and vorticity in Stokes flow.

For a two-dimensional incompressible flow, choose the stream function convention

u_{x} = ψ_{y}, u_{y} = - ψ_{x} .

(459)

Then

ω_{z} = \partial_{x} u_{y} - \partial_{y} u_{x} = - \nabla^{2} ψ .

(460)

Applying the harmonic-vorticity equation gives

\nabla^{4} ψ = 0,

(461)

the Biharmonic stream function for planar Stokes flow equation.

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b

Words: 265 Articles: 6

i

Words: 68 Articles: 1

Solution

Words: 68

In the exterior domain

r > a

, the streamfunction obeys

\nabla^{4} ψ = 0 .

(462)

Using the Streamfunction in polar coordinates convention

u_{r} = \frac{1}{r} ψ_{θ}, u_{θ} = - ψ_{r},

(463)

the no-slip condition on the stationary disk is

u_{r} (a, θ) = u_{θ} (a, θ) = 0.

(464)

Thus

ψ

is constant on the disk and

ψ_{r} = 0

there. Choose the arbitrary constant to be zero:

ψ (a, θ) = 0, ψ_{r} (a, θ) = 0 .

(465)

The imposed shear

u_{\infty} = γ y e_{x}

has streamfunction

ψ_{\infty} = \frac{γ}{2} y^{2} = \frac{γ r ^{2}}{4} (1 - cos 2 θ) .

(466)

The far-field condition is therefore

\nabla ψ - \nabla ψ_{\infty} ⟶ 0 (r \to \infty) .

(467)

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ii

Words: 99 Articles: 1

Solution

Words: 99

Only the axisymmetric and

cos 2 θ

modes are needed. The exterior biharmonic forms compatible with the far-field velocity are

ψ_{0} (r) = \frac{γ}{4} r^{2} + C_{0} lo g r + D_{0},

(468)

ψ_{2} (r, θ) = (- \frac{γ}{4} r^{2} + C_{2} + D_{2} r^{- 2}) cos 2 θ .

(469)

Terms

r^{2} lo g r

and

r^{4} cos 2 θ

are excluded because they would dominate the imposed shear.

The two conditions at

r = a

applied separately to each Fourier mode give

C_{0} = - \frac{γ a ^{2}}{2}, D_{0} = - \frac{γ a ^{2}}{4} + \frac{γ a ^{2}}{2} lo g a,

(470)

C_{2} = \frac{γ a ^{2}}{2}, D_{2} = - \frac{γ a ^{4}}{4} .

(471)

Hence

ψ (r, θ) = \frac{γ}{4} [r^{2} - a^{2} - 2 a^{2} lo g \frac{r}{a} - (r^{2} - 2 a^{2} + \frac{a ^{4}}{r ^{2}}) cos 2 θ] .

(472)

Direct differentiation gives

ψ = ψ_{r} = 0

r = a

. At infinity the remaining disturbance velocity is

O (r^{- 1})

, so the imposed shear is recovered. This is the Biharmonic stream function for a fixed disk in planar shear.

Solved by gpt-5.6-sol high.

iii

Words: 98 Articles: 1

Solution

Words: 98

The polar shear stress in a Newtonian fluid is

σ_{r θ} = μ [r \frac{\partial}{\partial r} (\frac{u _{θ}}{r}) + \frac{1}{r} \frac{\partial u _{r}}{\partial θ}] .

(473)

r = a

, no slip holds for every

θ

, so

u_{θ} = 0

and

\partial_{θ} u_{r} = 0

. Therefore

σ_{r θ} (a, θ) = μ \partial_{r} u_{θ} (a, θ) = - μ ψ_{rr} (a, θ) .

(474)

Differentiating the solution from part (ii) gives

ψ_{rr} (a, θ) = γ - 2 γ cos 2 θ,

(475)

and hence

σ_{r θ} (a, θ) = - μ γ + 2 μ γ cos 2 θ .

(476)

The moment arm and line element are both

a

, so the torque exerted by the fluid on the disk per unit axial length is

T_{z} = a^{2} \int_{0}^{2 π} σ_{r θ} (a, θ) d θ = - 2 π μ γ a^{2} .

(477)

For

γ > 0

the negative sign denotes the clockwise sense of the ambient shear rotation. This agrees with the hydrodynamic torque on a fixed disk in planar shear.

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40C (Waves)

Words: 400 Articles: 11

a

Words: 97 Articles: 1

Solution

Words: 97

Let

ρ^{'}

and

p^{'}

be the density and pressure perturbations and let

u

be the velocity perturbation. The linear acoustics equations about a uniform quiescent state are

ρ_{t}^{'} + ρ_{0} \nabla \cdot u = 0, ρ_{0} u_{t} = - \nabla p^{'}, p^{'} = c_{0}^{2} ρ^{'} .

(478)

The last relation is the homentropic pressure perturbation law.

Taking the scalar product of the momentum equation with

u

gives

\frac{\partial}{\partial t} (\frac{1}{2} ρ_{0} ∣ u ∣^{2}) = - u \cdot \nabla p^{'} = - \nabla \cdot (p^{'} u) + p^{'} \nabla \cdot u .

(479)

The continuity and pressure relations imply

p^{'} \nabla \cdot u = - \frac{p ^{'} p _{t}^{'}}{ρ _{0} c _{0}^{2}} = - \frac{\partial}{\partial t} (\frac{p ^{'2}}{2 ρ _{0} c _{0}^{2}}) .

(480)

Therefore the acoustic energy conservation law is

\frac{\partial}{\partial t} (K + W) + \nabla \cdot I = 0,

(481)

with

K = \frac{1}{2} ρ_{0} ∣ u ∣^{2}, W = \frac{p ^{'2}}{2 ρ _{0} c _{0}^{2}}, I = p^{'} u .

(482)

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b

Words: 303 Articles: 8

i

Words: 57 Articles: 1

Solution

Words: 57

Use the acoustic velocity potential convention

u = \nabla ϕ, p^{'} = - ρ_{0} ϕ_{t} .

(483)

For a spherically symmetric outgoing wave with time dependence

e^{iω t}

and

k = ω / c_{0}

, write

ϕ (r) = C \frac{e ^{- ik (r - a)}}{r} .

(484)

Linearizing the boundary condition at the mean surface

r = a

gives

ϕ^{'} (a) = iω ϵ .

(485)

Since

ϕ^{'} (a) = - \frac{C}{a ^{2}} (1 + ika),

(486)

we find

C = - \frac{iω ϵ a ^{2}}{1 + ika} .

(487)

Thus the outgoing acoustic field of a pulsating sphere is

ϕ (r, t) = Re [- \frac{iω ϵ a ^{2}}{1 + iωa / c _{0}} \frac{e ^{iω t - i (ω / c_{0}) (r - a)}}{r}] .

(488)

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ii

Words: 92 Articles: 1

Solution

Words: 92

For real harmonic quantities

f (t) = Re (f e^{iω t}), g (t) = Re (g e^{iω t}),

(489)

define the period average by

⟨ f g ⟩ = \frac{ω}{2 π} \int_{0}^{2 π / ω} f (t) g (t) d t = \frac{1}{2} Re (f g^{*}) .

(490)

This is the time average of harmonic power.

r = a

, put

x = ωa / c_{0}

. The velocity and pressure amplitudes are

u_{r} (a) = iω ϵ, p (a) = - i ρ_{0} ω ϕ (a) = - \frac{ρ _{0} ω ^{2} ϵ a}{1 + i x} .

(491)

Hence the mean outward work rate per unit area is

⟨ p^{'} u_{r} ⟩ = \frac{1}{2} Re (p u_{r}^{*}) = \frac{ρ _{0} ω ^{4} ϵ ^{2} a ^{2}}{2 c _{0} ( 1 + x ^{2} )} .

(492)

Multiplication by the area

4 π a^{2}

yields

⟨ P ⟩ = \frac{2 π ρ _{0} ω ^{4} ϵ ^{2} a ^{4} / c _{0}}{1 + ω ^{2} a ^{2} / c _{0}^{2}} = 2 π a^{2} ρ_{0} ω^{2} ϵ^{2} c_{0} \frac{ω ^{2} a ^{2}}{c _{0}^{2} + ω ^{2} a ^{2}} .

(493)

This is the mean acoustic power of a pulsating sphere.

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iii

Words: 45 Articles: 1

Solution

Words: 45

For a harmonic field, the time-averaged energy densities at

r = a

are

⟨ K ⟩ = \frac{ρ _{0}}{4} ∣ u_{r} ∣^{2} = \frac{ρ _{0} ω ^{2} ϵ ^{2}}{4},

(494)

⟨ W ⟩ = \frac{∣ p ∣ ^{2}}{4 ρ _{0} c _{0}^{2}} = \frac{ρ _{0} ω ^{2} ϵ ^{2}}{4} \frac{x ^{2}}{1 + x ^{2}} .

(495)

From part (ii),

∣ ⟨ I_{r} ⟩ ∣ = \frac{ρ _{0} ω ^{2} ϵ ^{2} c _{0}}{2} \frac{x ^{2}}{1 + x ^{2}} .

(496)

Consequently

\frac{c _{0} ⟨ K + W ⟩}{∣ ⟨ I _{r} ⟩ ∣} = \frac{1 + 2 x ^{2}}{2 x ^{2}} = 1 + \frac{c _{0}^{2}}{2 ω ^{2} a ^{2}} .

(497)

This is the surface acoustic energy-to-flux ratio of a pulsating sphere.

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iv

Words: 109 Articles: 1

Solution

Words: 109

c_{0} ≪ ωa

, then

x ≫ 1

. The sphere is acoustically large, and

⟨ K ⟩ \sim ⟨ W ⟩, ∣ ⟨ I_{r} ⟩ ∣ \sim c_{0} ⟨ K + W ⟩ .

(498)

Pressure and velocity are nearly in phase, so the near-surface field behaves locally like a progressive radiating wave. The power approaches

2 π a^{2} ρ_{0} ω^{2} ϵ^{2} c_{0} .

(499)

c_{0} ≫ ωa

, then

x ≪ 1

. The sphere is acoustically compact, and

\frac{⟨ W ⟩}{⟨ K ⟩} \sim x^{2} ≪ 1, \frac{c _{0} ⟨ K + W ⟩}{∣ ⟨ I _{r} ⟩ ∣} \sim \frac{1}{2 x ^{2}} ≫ 1.

(500)

The surface field is mainly reactive kinetic energy: pressure and velocity are nearly in quadrature, energy is stored and returned during each cycle, and only a small mean acoustic energy flux is radiated. The mean power scales as