Paper 4

www.maths.cam.ac.uk/undergrad/pastpapers/files/2023/paperii_4_2023.pdf

Table of contents 13k 286
- 1G Paper 4 121 1
  - Solution 1G 121
- 2F Paper 4 295 10
  - i 2F 72 1
    - Solution i 72
  - ii 2F 57 1
    - Solution ii 57
  - iii 2F 25 1
    - Solution iii 25
  - iv 2F 64 1
    - Solution iv 64
  - v 2F 77 1
    - Solution v 77
- 3I Paper 4 278 4
  - a 3I 130 1
    - Solution a 130
  - b 3I 148 1
    - Solution b 148
- 4I Paper 4 354 8
  - i 4I 48 1
    - Solution i 48
  - ii 4I 108 1
    - Solution ii 108
  - iii 4I 70 1
    - Solution iii 70
  - iv 4I 128 1
    - Solution iv 128
- 5J Paper 4 250 6
  - i 5J 31 1
    - Solution i 31
  - ii 5J 42 1
    - Solution ii 42
  - iii 5J 177 1
    - Solution iii 177
- 6C Paper 4 91 1
  - Solution 6C 91
- 7E Paper 4 291 1
  - Solution 7E 291
- 8D Paper 4 120 1
  - Solution 8D 120
- 9B Paper 4 160 1
  - Solution 9B 160
- 10D Paper 4 238 4
  - a 10D 108 1
    - Solution a 108
  - b 10D 130 1
    - Solution b 130
- 11G Paper 4 388 11
  - a 11G 74 1
    - Solution a 74
  - b 11G 99 1
    - Solution b 99
  - c 11G 97 4
    - i c 51 1
      - Solution i 51
    - ii c 46 1
      - Solution ii 46
  - d 11G 118 1
    - Solution d 118
- 12F Paper 4 679 8
  - a 12F 107 1
    - Solution a 107
  - b 12F 87 1
    - Solution b 87
  - c 12F 138 1
    - Solution c 138
  - d 12F 347 1
    - Solution d 347
- 13J Paper 4 391 10
  - i 13J 53 1
    - Solution i 53
  - ii 13J 67 1
    - Solution ii 67
  - iii 13J 77 1
    - Solution iii 77
  - iv 13J 90 1
    - Solution iv 90
  - v 13J 104 1
    - Solution v 104
- 14C Paper 4 391 1
  - Solution 14C 391
- 15D Paper 4 250 1
  - Solution 15D 250
- 16H Paper 4 645 8
  - a 16H 119 1
    - Solution a 119
  - b 16H 151 1
    - Solution b 151
  - c 16H 229 1
    - Solution c 229
  - d 16H 146 1
    - Solution d 146
- 17H Paper 4 517 8
  - a 17H 42 1
    - Solution a 42
  - b 17H 172 1
    - Solution b 172
  - c 17H 95 1
    - Solution c 95
  - d 17H 208 1
    - Solution d 208
- 18I Paper 4 449 7
  - a 18I 219 1
    - Solution a 219
  - b 18I 230 4
    - i b 57 1
      - Solution i 57
    - ii b 173 1
      - Solution ii 173
- 19H Paper 4 358 1
  - Solution 19H 358
- 20H Paper 4 478 9
  - a 20H 168 1
    - Solution a 168
  - b 20H 199 4
    - i b 142 1
      - Solution i 142
    - ii b 57 1
      - Solution ii 57
  - c 20H 111 1
    - Solution c 111
- 21G Paper 4 264 1
  - Solution 21G 264
- 22F Paper 4 310 8
  - a 22F 72 1
    - Solution a 72
  - b 22F 131 1
    - Solution b 131
  - c 22F 45 1
    - Solution c 45
  - d 22F 62 1
    - Solution d 62
- 23F Paper 4 427 8
  - a 23F 88 1
    - Solution a 88
  - b 23F 117 1
    - Solution b 117
  - c 23F 124 1
    - Solution c 124
  - d 23F 98 1
    - Solution d 98
- 24G Paper 4 353 1
  - Solution 24G 353
- 25G Paper 4 482 8
  - a 25G 153 1
    - Solution a 153
  - b 25G 132 1
    - Solution b 132
  - c 25G 77 1
    - Solution c 77
  - d 25G 120 1
    - Solution d 120
- 26K Paper 4 158 4
  - a 26K 95 1
    - Solution a 95
  - b 26K 63 1
    - Solution b 63
- 27J Paper 4 247 6
  - a 27J 52 1
    - Solution a 52
  - b 27J 102 1
    - Solution b 102
  - c 27J 93 1
    - Solution c 93
- 28K Paper 4 237 6
  - a 28K 62 1
    - Solution a 62
  - b 28K 61 1
    - Solution b 61
  - c 28K 114 1
    - Solution c 114
- 29K Paper 4 325 8
  - a 29K 129 1
    - Solution a 129
  - b 29K 63 1
    - Solution b 63
  - c 29K 50 1
    - Solution c 50
  - d 29K 83 1
    - Solution d 83
- 30J Paper 4 317 8
  - a 30J 69 1
    - Solution a 69
  - b 30J 80 1
    - Solution b 80
  - c 30J 42 1
    - Solution c 42
  - d 30J 126 1
    - Solution d 126
- 31E Paper 4 174 1
  - Solution 31E 174
- 32A Paper 4 341 10
  - i 32A 44 1
    - Solution i 44
  - ii 32A 50 1
    - Solution ii 50
  - iii 32A 61 1
    - Solution iii 61
  - iv 32A 115 1
    - Solution iv 115
  - v 32A 71 1
    - Solution v 71
- 33B Paper 4 368 13
  - a 33B 200 8
    - i a 32 1
      - Solution i 32
    - ii a 17 1
      - Solution ii 17
    - iii a 77 1
      - Solution iii 77
    - iv a 74 1
      - Solution iv 74
  - b 33B 80 1
    - Solution b 80
  - c 33B 88 1
    - Solution c 88
- 34D Paper 4 258 4
  - a 34D 100 1
    - Solution a 100
  - b 34D 158 1
    - Solution b 158
- 35A Paper 4 449 11
  - a 35A 170 1
    - Solution a 170
  - b 35A 33 1
    - Solution b 33
  - c 35A 51 1
    - Solution c 51
  - d 35A 195 4
    - i d 140 1
      - Solution i 140
    - ii d 55 1
      - Solution ii 55
- 36A Paper 4 319 10
  - a 36A 34 1
    - Solution a 34
  - b 36A 70 1
    - Solution b 70
  - c 36A 75 1
    - Solution c 75
  - d 36A 50 1
    - Solution d 50
  - e 36A 90 1
    - Solution e 90
- 37B Paper 4 262 10
  - a 37B 146 4
    - i a 70 1
      - Solution i 70
    - ii a 76 1
      - Solution ii 76
  - b 37B 116 4
    - i b 51 1
      - Solution i 51
    - ii b 65 1
      - Solution ii 65
- 38C Paper 4 478 12
  - a 38C 66 1
    - Solution a 66
  - b 38C 89 1
    - Solution b 89
  - c 38C 76 1
    - Solution c 76
  - d 38C 62 1
    - Solution d 62
  - e 38C 112 1
    - Solution e 112
  - f 38C 73 1
    - Solution f 73
- 39C Paper 4 225 6
  - a 39C 59 1
    - Solution a 59
  - b 39C 91 1
    - Solution b 91
  - c 39C 75 1
    - Solution c 75
- 40C Paper 4 344 10
  - a 40C 56 1
    - Solution a 56
  - b 40C 69 1
    - Solution b 69
  - c 40C 67 1
    - Solution c 67
  - d 40C 73 1
    - Solution d 73
  - e 40C 79 1
    - Solution e 79

which is continuous, maps the annulus to itself, and has no fixed point because the origin is not in the annulus. If the disc and annulus were homeomorphic, the homeomorphism invariance of the fixed-point property would give the annulus the fixed-point property, a contradiction. Therefore they are not homeomorphic.

Solved by gpt-5.6-sol high.

v

Words: 77 Articles: 1

Solution

Words: 77

No. Take

B = [0, 1] \times {0} \subset R^{2} .

(18)

It contains more than one point and has the fixed-point property of a closed interval, but it is not homeomorphic to the closed unit disc. Indeed, deleting any point

(t, 0)

with

0 < t < 1

disconnects

B

, so such a point is a cut point. Deleting one point from a two-dimensional closed disc leaves a path-connected space. Since being a cut point is preserved by homeomorphisms, the two spaces cannot be homeomorphic. This gives the required counterexample.

Solved by gpt-5.6-sol high.

3I (Coding and Cryptography)

Words: 278 Articles: 4

a

Words: 130 Articles: 1

Solution

Words: 130

For the standard dot product on

F_{2}^{n}

, the dual code is

C^{⊥} = {y \in F_{2}^{n} : y \cdot c = 0 for every c \in C} .

(19)

It is an intersection of kernels of linear functions, and is therefore a linear code.

Let

σ

denote cyclic right shift. If

y \in C^{⊥}

and

c \in C

, then

(σ y) \cdot c = y \cdot (σ^{- 1} c) .

(20)

A cyclic code is closed under both

σ

and

σ^{- 1}

, so

σ^{- 1} c \in C

and the right-hand side vanishes. Hence

σ y \in C^{⊥}

, proving directly that the dual of a cyclic code is cyclic.

Identify words with polynomials in

F_{2} [X] / (X^{n} - 1)

. If the generator polynomial of a cyclic code is the monic divisor

g (X)

and

g (X) h (X) = X^{n} - 1,

(21)

then the generator polynomial of

C^{⊥}

is the monic reciprocal

g^{⊥} (X) = h^{*} (X) = X^{d e g h} h (X^{- 1}) / h (0) .

(22)

Equivalently, the parity-check polynomial of one code becomes, after reversal, the generator polynomial of its dual.

Solved by gpt-5.6-sol high.

b

Words: 148 Articles: 1

Solution

Words: 148

Over

F_{2}

, subtraction equals addition. The polynomial

X^{7} - 1

has the factorization

X^{7} - 1 = (X + 1) (X^{3} + X + 1) (X^{3} + X^{2} + 1) .

(23)

The two cubic factors have no root in

F_{2}

, so they are irreducible, and they are reciprocal to one another. Put

f = X^{3} + X + 1, f^{*} = X^{3} + X^{2} + 1.

(24)

Every cyclic code corresponds to one monic divisor of

X^{7} - 1

. The complete list is

g (X) 1 X + 1 f f^{*} (X + 1) f (X + 1) f^{*} f f^{*} = 1 + X + \dots + X^{6} (X + 1) f f^{*} = X^{7} - 1 parameters [7, 7, 1] [7, 6, 2] [7, 4, 3] [7, 4, 3] [7, 3, 4] [7, 3, 4] [7, 1, 7] [7, 0] code whole-space linear code even-weight binary code Hamming code reversed Hamming code binary simplex code reversed binary simplex code binary repetition code zero code .

(25)

Thus the two cubic generators give the two cyclic coordinate orientations of Hamming's

[7, 4, 3]

code. Their duals have generators

(X + 1) f

and

(X + 1) f^{*}

and are the

[7, 3, 4]

simplex codes. The remaining four are the familiar whole-space, even-weight, repetition, and zero codes. This is the complete classification of binary cyclic codes of length seven.

Solved by gpt-5.6-sol high.

4I (Automata & Formal Languages)

Words: 354 Articles: 8

i

Words: 48 Articles: 1

Solution

Words: 48

A grammar

G = (Σ, V, P, S)

is a right-linear regular grammar when every production has one of the forms

A \to wB or A \to w,

(26)

where

A, B \in V

and

w \in Σ^{*}

. Thus a production has at most one variable on its right-hand side, and that variable is at the right end.

Solved by gpt-5.6-sol high.

ii

Words: 108 Articles: 1

Solution

Words: 108

We prove the claim by induction on the length of the grammar derivation from

S

. At length zero the sentential form is

S = ϵ S

, which has the required form.

Suppose the current sentential form is

w A

. Because

w

is terminal, the only symbol to which a production can apply is

A

. A production

A \to u B

produces

w A \Rightarrow w u B,

(27)

again a terminal word followed by one variable. A production

A \to u

produces the terminal word

w u

. Once a completely terminal word is reached, no further production can apply. The induction proves the sentential form of a right-linear regular grammar: every reachable

α

is either

w A

w

Solved by gpt-5.6-sol high.

iii

Words: 70 Articles: 1

Solution

Words: 70

In every derivation beginning at

w A

, the terminal prefix

w

remains unchanged because no production has a terminal on its left-hand side. By the sentential-form result, each step rewrites the sole variable at the right and appends terminals immediately before its successor variable. Hence

w A \Rightarrow^{*} v B

(28)

forces

v = w u

for some

u \in Σ^{*}

. Removing the inert prefix

w

from every sentential form in the derivation gives

A \Rightarrow^{*} u B .

(29)

Solved by gpt-5.6-sol high.

iv

Words: 128 Articles: 1

Solution

Words: 128

An accepting derivation in a right-linear grammar has the form

S \Rightarrow w_{1} A_{1} \Rightarrow w_{1} w_{2} A_{2} \Rightarrow \dots \Rightarrow w_{1} \dots w_{m},

(30)

with one active variable until the final terminal production.

If no variable repeated on any accepting derivation, an accepting derivation could contain at most

∣ V ∣

variable occurrences. Since the production set is finite, only finitely many such derivations and hence finitely many terminal words would exist. The hypothesis that

L (G)

is infinite therefore supplies an accepting derivation in which some variable

A

occurs twice.

Split this derivation at the two occurrences:

S \Rightarrow^{*} u A, A \Rightarrow^{*} v A, A \Rightarrow^{*} z,

(31)

where

u, v, z \in Σ^{*}

. The first segment makes

A

an accessible variable of a regular grammar, the middle segment makes it a looping variable of a regular grammar, and the last makes it a terminable variable of a regular grammar. Thus

A

has all three properties, proving the accessible looping terminable variable criterion.

Solved by gpt-5.6-sol high.

5J (Statistical Modelling)

Words: 250 Articles: 6

i

Words: 31 Articles: 1

Solution

Words: 31

Use the R linear-model formula with only its implicit intercept:

fit1 <- lm(log(price) ~ 1, data = cars)

This models every log-price by one common mean plus an error term.

Solved by gpt-5.6-sol high.

ii

Words: 42 Articles: 1

Solution

Words: 42

The other quantitative columns are mpg, psngr, length, width, and weight, so use

fit2 <- lm(log(price) ~ mpg + psngr + length + width + weight, data = cars)

This is a normal linear model with an intercept and five slopes.

Solved by gpt-5.6-sol high.

iii

Words: 177 Articles: 1

Solution

Words: 177

Treat make as a categorical predictor, converting it to a factor explicitly if necessary:

fit3 <- lm(log(price) ~ mpg + psngr + length + width + weight + factor(make), data = cars)

The models are nested. The first nested-model F-test compares fit2 with the intercept-only fit1. Adding the five quantitative predictors reduces the residual sum of squares from

8584.0

3349.1

, with

F = 69.7334, p < 2.2 \times 1 0^{- 16} .

(32)

There is overwhelming evidence that these quantitative car properties jointly improve the model.

The second test compares fit3 with fit2. Adding make reduces the residual sum of squares further to

840.8

, with

F = 5.3891, p = 2.541 \times 1 0^{- 8} .

(33)

Thus manufacturer has a highly significant effect even after adjusting for the five quantitative properties, and fit3 is preferred among these nested models.

There are

93

observations because the intercept-only model has

92 = 93 - 1

residual degrees of freedom. The make term uses

31

additional degrees of freedom. By the degrees of freedom of a factor predictor, a factor with

m

represented levels contributes

m - 1

degrees of freedom when an intercept is present. Hence

m - 1 = 31, m = 32

(34)

unique manufacturers occur in the dataset.

Solved by gpt-5.6-sol high.

6C (Mathematical Biology)

Words: 91 Articles: 1

Solution

Words: 91

Put

a = f^{'} (0) > 0

. Linearizing the morphogen reaction-diffusion equation at

C = 0

gives

C_{t} = D C_{xx} + a C .

(35)

Seek separated modes

C (x, t) = X (x) e^{λ t}

. Then

D X^{''} + a X = λ X, X (0) = 0, X^{'} (L) = 0.

(36)

The mixed Dirichlet-Neumann modes on an interval are

X_{n} (x) = sin (k_{n} x), k_{n} = \frac{( n + 1/2 ) π}{L}, n = 0, 1, 2, \dots .

(37)

Indeed, the Dirichlet condition selects sine functions, while the Neumann condition requires

cos (k_{n} L) = 0

. Their growth rates are

λ_{n} = a - D k_{n}^{2} = a - D \frac{( n + 1/2 ) ^{2} π ^{2}}{L ^{2}} .

(38)

The largest growth rate is the lowest mode,

λ_{0} = a - \frac{D π ^{2}}{4 L ^{2}} .

(39)

Linear stability requires

λ_{0} < 0

, after which all higher modes also decay. Therefore the critical length for a linearly growing morphogen is

L < \frac{π}{2} \frac{D}{f ^{'} ( 0 )} .

(40)

Equality gives a neutral lowest mode, while a larger domain is linearly unstable.

Solved by gpt-5.6-sol high.

7E (Further Complex Methods)

Words: 291 Articles: 1

Solution

Words: 291

The Papperitz symbol

P ⎩ ⎨ ⎧ 00 1 - c 10 c - a - b \infty a b z ⎭ ⎬ ⎫

(41)

specifies a second-order Fuchsian differential equation. The first row lists its three distinct regular singular points

0, 1, \infty

;

z

is the independent variable. The two entries below each singular point are its characteristic exponents. Thus local solutions have leading behaviors

1, z^{1 - c} (z \to 0),

(42)

1, (1 - z)^{c - a - b} (z \to 1),

(43)

and, with the usual convention at infinity,

z^{- a}, z^{- b} (z \to \infty) .

(44)

When an exponent difference is an integer, a logarithmic second solution may replace the naive second Frobenius power. The entries obey the Fuchs relation

0 + (1 - c) + 0 + (c - a - b) + a + b = 1.

(45)

For a second-order equation with exactly three regular singular points, these exponent data determine the equation up to multiplication by a nonzero function; there is no accessory parameter. The symbol is therefore the one for the Gauss hypergeometric equation. Its distinguished solution

F (a, b; c; z)

is the exponent-zero solution analytic at zero and normalized to one there. For the ordinary power-series definition one assumes

c \in / {0, - 1, - 2, \dots}

, with exceptional parameter values handled separately or by continuation.

Now put

u = \frac{z}{z - 1}, Y (z) = F (a, c - b; c; u) .

(46)

The hypergeometric equation in

u

has exponent pairs

(0, 1 - c) at u = 0, (0, b - a) at u = 1, (a, c - b) at u = \infty.

(47)

The Möbius transformation of a Papperitz symbol sends

u = 0, 1, \infty ⟵ z = 0, \infty, 1,

(48)

Y

has symbol

P ⎩ ⎨ ⎧ 00 1 - c 1 a c - b \infty 0 b - a z ⎭ ⎬ ⎫ .

(49)

On the other hand,

F (a, b; c; z)

has exponent pairs

(0, 1 - c), (0, c - a - b), (a, b)

(50)

0, 1, \infty

. Multiplication by

(1 - z)^{a}

applies the dependent-variable rescaling of a Papperitz symbol: it adds

a

to both exponents at

z = 1

and subtracts

a

from both at infinity. Hence

(1 - z)^{a} F (a, b; c; z)

(51)

has exactly the same three exponent pairs as

Y

Both functions are analytic near

z = 0

and equal one at

z = 0

, so uniqueness of the normalized exponent-zero hypergeometric solution gives the Pfaff transformation

F (a, c - b; c; \frac{z}{z - 1}) = (1 - z)^{a} F (a, b; c; z) .

(52)

The identity first holds near zero with the branch of

(1 - z)^{a}

equal to one there, and then extends by analytic continuation on any domain where compatible branches are chosen.

Solved by gpt-5.6-sol high.

8D (Classical Dynamics)

Words: 120 Articles: 1

Solution

Words: 120

An adiabatic invariant remains constant to leading order when a system parameter changes on a timescale much longer than one period. For a periodic one-degree-of-freedom system, the adiabatic invariance of the action supplies such an invariant.

The momentum magnitude between collisions is

p_{0} = 2 m E

. A complete orbit crosses the gap once in each direction, so the unnormalized action used in the question is

I = \oint p d q = 2 L p_{0} = 2 L 2 m E .

(53)

Consequently

\frac{d I}{d E} = \frac{2 L m}{2 m E} = \frac{2 L}{2 E / m} = T,

(54)

because the speed is

2 E / m

and the round-trip distance is

2 L

When

L (t)

varies adiabatically, conservation of the adiabatic particle between moving parallel walls action gives

E (t) = E (0) (\frac{L ( 0 )}{L ( t )})^{2} .

(55)

A moving wall does work at each collision: an approaching wall raises the particle's energy and a receding wall lowers it. Elasticity holds in the instantaneous rest frame of the wall.

Solved by gpt-5.6-sol high.

9B (Cosmology)

Words: 160 Articles: 1

Solution

Words: 160

The Flatness problem is that the observed cosmological density parameter is close to

Ω = 1

, although in an ordinary decelerating universe any departure from one grows with time. Extrapolation backward therefore requires implausibly precise early cancellation of the curvature term.

The cosmological perfect-fluid continuity equation is

\overset{ρ}{˙} + 3 H (ρ + P) = 0.

(56)

Consequently

\frac{d}{d t} (ρ a^{2}) = a^{2} \overset{ρ}{˙} + 2 a \overset{a}{˙} ρ = - H a^{2} (ρ + 3 P) .

(57)

For an expanding universe

H > 0

. If

ρ + 3 P < 0,

(58)

then

ρ a^{2}

increases. The Friedmann equation implies

Ω^{- 1} - 1 = \frac{constant \times ∣ k ∣}{ρ a ^{2}} .

(59)

It therefore decreases during this period, driving

Ω

toward one. By the Friedmann acceleration equation, the same condition gives

\overset{a}{¨} > 0

when the cosmological constant is included in the effective fluid. This is the inflationary solution of the flatness problem.

For the scalar field, the slow-roll approximation means

\dot{ϕ}^{2} ≪ V (ϕ), ∣ \ddot{ϕ} ∣ ≪ 3 H ∣ \dot{ϕ} ∣.

(60)

The equations reduce to

3 H^{2} ≃ m^{2} ϕ^{2}, 3 H \dot{ϕ} ≃ - 2 m^{2} ϕ .

(61)

Taking the positive-field branch gives

H ≃ m ϕ / 3

, and therefore

\dot{ϕ} ≃ - \frac{2 m}{3} .

(62)

With

ϕ (0) = ϕ_{i}

ϕ (t) = ϕ_{i} - \frac{2 m}{3} t .

(63)

Now integrate

H = \overset{a}{˙} / a ≃ m ϕ / 3

lo g \frac{a ( t )}{a _{i}} = \int_{0}^{t} \frac{m}{3} (ϕ_{i} - \frac{2 m}{3} s) d s = \frac{m ϕ _{i}}{3} t - \frac{m ^{2} t ^{2}}{3} .

(64)

Thus the quadratic-potential slow-roll solution is

a (t) = a_{i} exp [\frac{m ϕ _{i}}{3} t - \frac{m ^{2} t ^{2}}{3}] .

(65)

Finally,

ϕ_{i}^{2} - ϕ (t)^{2} = \frac{4 m ϕ _{i}}{3} t - \frac{4 m ^{2}}{3} t^{2},

(66)

so equivalently

a (t) = a_{i} exp [\frac{ϕ _{i}^{2} - ϕ ( t ) ^{2}}{4}] .

(67)

Solved by gpt-5.6-sol high.

10D (Quantum Information and Computation)

Words: 238 Articles: 4

a

Words: 108 Articles: 1

Solution

Words: 108

Let the unknown qubit be

∣ α ⟩ = a ∣0 ⟩ + b ∣1 ⟩

. Alice holds this qubit and her half of the shared Bell state

∣ Ψ^{+} ⟩

, while Bob holds the other half.

Alice performs a measurement of her two qubits in the Bell basis and sends the two-bit outcome to Bob through the classical channel. For the shared state

∣ Ψ^{+} ⟩

, the teleportation with the psi-plus Bell state correction table is

Alice’s outcome Φ^{+} Φ^{-} Ψ^{+} Ψ^{-} Bob’s state X ∣ α ⟩ XZ ∣ α ⟩ ∣ α ⟩ Z ∣ α ⟩ Bob applies X ZX I Z

(68)

up to physically irrelevant global phases. Here

X

and

Z

are the Pauli X gate and Pauli Z gate. After the indicated correction, Bob's qubit is

∣ α ⟩

. This is quantum teleportation; no physical copy of the unknown qubit is sent.

Solved by gpt-5.6-sol high.

b

Words: 130 Articles: 1

Solution

Words: 130

Let

D

denote Alice's unknown input qubit and

A

her GHZ qubit. Expanding the four-qubit state in the Bell basis of

D, A

gives

∣ α ⟩_{D} ∣ GHZ ⟩_{A BC} = \frac{1}{2} [∣ Φ^{+} ⟩_{D A} (a ∣00 ⟩ + b ∣11 ⟩)_{BC} + ∣ Φ^{-} ⟩_{D A} (a ∣00 ⟩ - b ∣11 ⟩)_{BC}

(69)

+ ∣ Ψ^{+} ⟩_{D A} (b ∣00 ⟩ + a ∣11 ⟩)_{BC} + ∣ Ψ^{-} ⟩_{D A} (b ∣00 ⟩ - a ∣11 ⟩)_{BC}],

(70)

where a global sign in the last term is immaterial.

For outcomes

Φ^{+}

and

Φ^{-}

, Bob and Charlie already have

∣ ϕ_{1} ⟩ = a ∣00 ⟩ + b ∣11 ⟩, ∣ ϕ_{2} ⟩ = a ∣00 ⟩ - b ∣11 ⟩ .

(71)

For either

Ψ

outcome, both apply the Pauli X gate. Since

X \otimes X

exchanges

∣00 ⟩

and

∣11 ⟩

, the

Ψ^{+}

branch becomes

∣ ϕ_{1} ⟩

and the

Ψ^{-}

branch becomes

∣ ϕ_{2} ⟩

, up to a global phase. This proves the Bell measurement on one qubit and one leg of a GHZ state result.

Finally, tracing out Charlie gives

ρ_{B} = ∣ a ∣^{2} ∣0 ⟩ ⟨ 0∣ + ∣ b ∣^{2} ∣1 ⟩ ⟨ 1∣

(72)

for either sign. This reduced state has rank two exactly when

a \neq = 0

and

b \neq = 0

. By the entanglement criterion for a two-term correlated state,

∣ ϕ_{1} ⟩ and ∣ ϕ_{2} ⟩ are entangled iff a \neq = 0 and b \neq = 0.

(73)

Solved by gpt-5.6-sol high.

11G (Number Theory)

Words: 388 Articles: 11

a

Words: 74 Articles: 1

Solution

Words: 74

For an odd prime

p

, the Legendre symbol is

(\frac{a}{p}) = ⎩ ⎨ ⎧ 0, 1, - 1, p ∣ a, a \neq \equiv 0 (mod p) is a quadratic residue, a is a quadratic nonresidue .

(74)

Euler criterion states that

a^{(p - 1) /2} \equiv (\frac{a}{p}) (mod p) .

(75)

p ∣ a

, both sides vanish. Otherwise choose a primitive root

g

and write

a \equiv g^{m}

. Since

g^{(p - 1) /2} \equiv - 1

a^{(p - 1) /2} \equiv (- 1)^{m} .

(76)

The power

g^{m}

is a square exactly when

m

is even, proving the criterion. Taking

a = - 1

gives the first supplementary law for quadratic reciprocity

(\frac{- 1}{p}) = (- 1)^{(p - 1) /2} .

(77)

Solved by gpt-5.6-sol high.

b

Words: 99 Articles: 1

Solution

Words: 99

Because

I

and

A

commute, the matrix binomial theorem gives

(I + A)^{p} = \sum_{j = 0}^{p} (j p) A^{j} .

(78)

For prime

p

, every intermediate coefficient

(j p)

is divisible by

p

, proving the prime-power binomial congruence for a matrix

(I + A)^{p} \equiv I + A^{p} (mod p) .

(79)

For

A = (01 - 1 0), A^{2} = - I,

(80)

one has

(I + A)^{2} = 2 A

and

(I + A)^{4} = - 4 I

. If

p = 4 k + 1

, then

(I + A)^{p} = (- 4)^{k} (I + A) \equiv I + A,

(81)

(- 4)^{k} \equiv 1 (mod p)

. If

p = 4 k - 1

, then

A^{p} = - A

and

(I + A)^{p} = (- 4)^{k} (I + A)^{- 1} = \frac{( - 4 ) ^{k}}{2} (I - A) \equiv I - A,

(82)

(- 4)^{k} \equiv 2 (mod p)

By Euler criterion,

(\frac{2}{p}) \equiv 2^{(p - 1) /2} (mod p)

. In either case the preceding congruence gives

(\frac{2}{p}) = (- 1)^{k}

. Since

(p^{2} - 1) /8

has the same parity as

k

for

p = 4 k \pm 1

, this is the second supplementary law for quadratic reciprocity

(\frac{2}{p}) = (- 1)^{(p^{2} - 1) /8} .

(83)

Solved by gpt-5.6-sol high.

c

Words: 97 Articles: 4

i

Words: 51 Articles: 1

Solution

Words: 51

Write

χ (a) = (\frac{a}{p})

. When

p \equiv 1 (mod 4)

, the first supplementary law for quadratic reciprocity gives

χ (- 1) = 1

, so

χ (p - a) = χ (a)

. Pairing

a

with

p - a

gives

T = p \sum_{a = 1}^{(p - 1) /2} χ (a) .

(84)

The sum of the nontrivial character

χ

over all nonzero residue classes is zero. Because its two half-sums are equal when

χ (- 1) = 1

0 = \sum_{a = 1}^{p - 1} χ (a) = 2 \sum_{a = 1}^{(p - 1) /2} χ (a) .

(85)

Hence

T = 0.

(86)

Solved by gpt-5.6-sol high.

ii

Words: 46 Articles: 1

Solution

Words: 46

By Euler criterion,

T \equiv \sum_{a = 1}^{p - 1} a^{(p + 1) /2} (mod p) .

(87)

Choose a primitive root

g

. For any integer

m

not divisible by

p - 1

\sum_{a = 1}^{p - 1} a^{m} = \sum_{j = 0}^{p - 2} g^{jm} = \frac{g ^{m (p - 1)} - 1}{g ^{m} - 1} \equiv 0 (mod p) .

(88)

For

p > 3

, the exponent

m = (p + 1) /2

satisfies

0 < m < p - 1

. Therefore

T \equiv 0 (mod p) .

(89)

Together with part (i), this is the weighted complete Legendre-symbol sum.

Solved by gpt-5.6-sol high.

d

Words: 118 Articles: 1

Solution

Words: 118

Put

H = (p - 1) /2

. Because

p \equiv 7 (mod 8)

, the supplementary laws give

(\frac{- 1}{p}) = - 1, (\frac{2}{p}) = 1.

(90)

Let

R

and

N

be the quadratic residues and nonresidues, respectively, among

1, \dots, H

, and put

S = \sum_{r \in R} r .

(91)

Since

- 1

is a nonresidue, the complete set

Q

of least positive quadratic residues is

Q = R \cup {p - n : n \in N} .

(92)

u = ∣ N ∣

, its sum is

\sum_{q \in Q} q = S + u p - \sum_{n \in N} n = 2 S + u p - \frac{H ( H + 1 )}{2} .

(93)

Since

2

is a residue, multiplication by two permutes

Q

modulo

p

. Exactly the

u

upper-half residues

p - n

cross

p

when doubled. Equality of the sums before and after reduction therefore gives

\sum_{q \in Q} q = 2 \sum_{q \in Q} q - u p, \sum_{q \in Q} q = u p .

(94)

Comparing the two formulas yields

2 S = \frac{H ( H + 1 )}{2} .

(95)

H = (p - 1) /2

S = \frac{H ( H + 1 )}{4} = \frac{p ^{2} - 1}{16} .

(96)

This is the lower-half quadratic-residue sum for primes congruent to seven modulo eight.

Solved by gpt-5.6-sol high.

12F (Topics In Analysis)

Words: 679 Articles: 8

a

Words: 107 Articles: 1

Solution

Words: 107

For

a, b \in R^{n + 1}

, the triangle inequality and the Cauchy-Schwarz inequality give

∣ S (a) - S (b) ∣ \leq ∥ T (a - b) ∥_{\infty} \leq r = 0 \sum n ∣ a_{r} - b_{r} ∣ \leq n + 1 ∥ a - b ∥_{2} .

(97)

Thus

S

is a continuous function. The unit sphere in the finite-dimensional Euclidean normed vector space

R^{n + 1}

is compact. Moreover,

S (a) = 0

implies that the polynomial

T a

vanishes identically, so every coefficient

a_{r}

is zero. Hence

S

is strictly positive on the unit sphere. By the extreme value theorem, it has a positive minimum

δ = min_{∥ a ∥_{2} = 1} S (a) > 0.

(98)

The supremum norm is homogeneous. For

a \neq = 0

∥ T a ∥_{\infty} = ∥ a ∥_{2} S (\frac{a}{∥ a ∥ _{2}}) \geq δ ∥ a ∥_{2} .

(99)

It follows directly that

∥ T a ∥_{\infty} \to \infty

whenever

∥ a ∥_{2} \to \infty

Solved by gpt-5.6-sol high.

b

Words: 87 Articles: 1

Solution

Words: 87

Write each polynomial of degree at most

n

T a

and define

F (a) = ∥ T a - f ∥_{\infty} .

(100)

This is a continuous function, and the reverse triangle inequality together with part (a) gives

F (a) \geq ∥ T a ∥_{\infty} - ∥ f ∥_{\infty} \geq δ ∥ a ∥_{2} - ∥ f ∥_{\infty} .

(101)

Consequently

F

is a coercive function. Choose

R

so large that

F (a) > F (0)

outside the closed Euclidean ball of radius

R

. That ball is compact, so the extreme value theorem supplies a minimizer

a_{*}

there. Then

P = T a_{*}

is a best uniform approximation to

f

among polynomials of degree at most

n

Solved by gpt-5.6-sol high.

c

Words: 138 Articles: 1

Solution

Words: 138

The Chebyshev alternation theorem says that a polynomial

P

of degree at most

n

is a best uniform approximation to

f \in C ([0, 1])

if and only if there are

n + 2

points

0 \leq t_{0} < t_{1} < \dots < t_{n + 1} \leq 1

(102)

and a sign

ε \in {1, - 1}

for which

f (t_{j}) - P (t_{j}) = ε (- 1)^{j} ∥ f - P ∥_{\infty} .

(103)

This is the equiripple criterion.

To prove sufficiency, put

E = ∥ f - P ∥_{\infty}

and suppose that another polynomial

Q

of degree at most

n

satisfies

∥ f - Q ∥_{\infty} < E

. At every

t_{j}

, the difference

R = Q - P

must have the same sign as

f - P

; hence its signs alternate at the

n + 2

ordered points. The intermediate value theorem then gives at least one distinct root of

R

in every interval

(t_{j}, t_{j + 1})

, for at least

n + 1

roots in all. The Lagrange root bound over a field forces the nonzero polynomial

R

to have degree at least

n + 1

, contradicting

de g R \leq n

. Therefore no

Q

gives a smaller error, and

P

is best.

Solved by gpt-5.6-sol high.

d

Words: 347 Articles: 1

Solution

Words: 347

As printed, this part contains an indexing error. The displayed product ends at

m - 1

, but the hypotheses contain

m + 1

alternating intervals indexed by

0, \dots, m

. On the first

m

intervals the printed polynomial has sign

(- 1)^{i}

, while on the last interval it has sign

(- 1)^{m - 1}

rather than the required

(- 1)^{m}

. For example, when

m = 1

the printed product is the constant

Q = 1

, which cannot reduce a negative extremum of

g

on the final interval. The intended polynomial is

Q (t) = (- 1)^{m} \prod_{j = 1}^{m} (t - w_{j}), w_{j} = \frac{v _{j - 1} + u _{j}}{2} .

(104)

We prove the stated conclusion with this correction.

Each root

w_{j}

lies in the gap between the

(j - 1)

st and

j

th active intervals. Therefore

Q

has sign

(- 1)^{i}

throughout

[u_{i}, v_{i}]

. In particular, wherever

∣ g ∣ = M

, the numbers

Q

and

g

have the same sign, and

Q

is nonzero.

Let

K_{+} = {t : g (t) = M}, K_{-} = {t : g (t) = - M} .

(105)

These are compact sets. On their union, the continuous function

∣ Q ∣

has a positive minimum. Hence, throughout some open neighbourhood

U

K_{+} \cup K_{-}

, subtracting a sufficiently small positive multiple

η Q

moves

g

strictly towards zero and gives

∣ g - η Q ∣ < M

. On the compact complement

[0, 1] ∖ U

, continuity gives a uniform margin

∣ g ∣ \leq M - ϵ

for some

ϵ > 0

. Taking also

η ∥ Q ∥_{\infty} < ϵ

(106)

shows that

∣ g - e t a Q ∣ < M

there. Thus

∥ η Q - g ∥_{\infty} < M

(107)

for every sufficiently small

η > 0

It remains to prove necessity in the Chebyshev alternation theorem. Let

P

be a best approximation of degree at most

n

, put

g = f - P

, and let

M = ∥ g ∥_{\infty}

. If

g

has no sequence of

n + 2

extrema with alternating signs, its sets of positive and negative extrema can be collected, from left to right, into

m + 1

alternating compact groups with

m \leq n

. Choose disjoint intervals

[u_{i}, v_{i}]

containing those groups and separated by regions on which

∣ g ∣ < M

. After replacing

g

- g

if necessary, they satisfy the displayed hypotheses. The corrected construction produces a polynomial

Q

of degree

m \leq n

and a small

η > 0

such that

∥ f - (P + η Q) ∥_{\infty} = ∥ g - η Q ∥_{\infty} < M .

(108)

This contradicts the assumed optimality of

P

. Therefore every best polynomial has the required

n + 2

alternating extrema.

Solved by gpt-5.6-sol high.

13J (Statistical Modelling)

Words: 391 Articles: 10

i

Words: 53 Articles: 1

Solution

Words: 53

Let

Y_{i}

be the first team's goal count in match

i

, let

x_{i}^{T}

be row

i

of the design matrix, and let

β

be the regression coefficient vector. The fitted Poisson regression assumes that the

Y_{i}

are independent random variables with

Y_{i} \sim Pois (μ_{i}), lo g μ_{i} = x_{i}^{T} β .

(109)

The likelihood function maximized by glm is therefore

L (β; y) = \prod_{i = 1}^{64} \frac{μ _{i}^{y_{i}} e ^{- μ_{i}}}{y _{i} !} = \prod_{i = 1}^{64} \frac{e x p { y _{i} x _{i}^{T} β - e ^{x_{i}^{T} β} }}{y _{i} !} .

(110)

Solved by gpt-5.6-sol high.

ii

Words: 67 Articles: 1

Solution

Words: 67

None of the displayed one-coefficient Wald tests for team1_form rejects at the

5%

significance level. That does not establish that formation has no effect: the relevant null hypothesis sets all nine displayed team1_form coefficients to zero simultaneously, and several individually imprecise estimates may still be jointly significant. One should compare the fitted model with the nested model omitting team1_form by an analysis of deviance for nested generalized linear models, equivalently a likelihood-ratio test of those joint restrictions.

Solved by gpt-5.6-sol high.

iii

Words: 77 Articles: 1

Solution

Words: 77

Writing

Y_{i}

for team1_goal and

x_{i}

for team1_xg, the first hypothesis concerns only the expected value:

H_{1} : E (Y_{i} ∣ x_{i}) = x_{i} for every i .

(111)

The second specifies the entire conditional probability distribution:

H_{2} : Y_{i} ∣ x_{i} \sim Pois (x_{i}) for every i,

(112)

with independence between matches as assumed in the question. The Poisson limit theorem, often called the law of small numbers in this setting, suggests a Poisson distribution when a goal count is the sum of many approximately independent rare scoring opportunities.

Solved by gpt-5.6-sol high.

iv

Words: 90 Articles: 1

Solution

Words: 90

Under

H_{1}

, the slope in this through-origin linear regression is

1

. Its observed Wald statistic is

t = \frac{1.15790 - 1}{0.08643} = 1.83.

(113)

The corresponding two-sided p-value is about

0.07

, so the fit does not give evidence against

H_{1}

at the

5%

significance level.

The usual least-squares standard error assumes constant conditional variance. Goal counts are counts, and under the natural Poisson distribution model their variance equals their mean

x_{i}

, so observations with different expected goals are heteroscedastic. The reported homoscedastic standard error is therefore doubtful; a heteroscedasticity-robust calculation or a suitable count model would be more credible.

Solved by gpt-5.6-sol high.

v

Words: 104 Articles: 1

Solution

Words: 104

The fitted Poisson regression has

Y_{i} ∣ x_{i} \sim Pois (μ_{i}), lo g μ_{i} = β_{0} + β_{1} lo g x_{i},

(114)

so that

μ_{i} = e^{β_{0}} x_{i}^{β_{1}}

. Hypothesis

H_{2}

requires both restrictions

β_{0} = 0, β_{1} = 1,

(115)

as well as the adequacy of the assumed Poisson distribution. The displayed interval addresses only the single slope restriction

β_{1} = 1

. Even if the separate interval for the intercept also contains zero, separate one-parameter intervals do not implement the relevant joint hypothesis test, because the two coefficient estimators can be correlated. One should test

(β_{0}, β_{1}) = (0, 1)

jointly, for example by a likelihood-ratio test, and separately assess the Poisson goodness of fit. Thus the slope interval alone cannot decide whether

H_{2}

is rejected at the

5%

level.

Solved by gpt-5.6-sol high.

14C (Mathematical Biology)

Words: 391 Articles: 1

Solution

Words: 391

This is a birth-death process with birth and death rates

b_{n} = γ + β n, d_{n} = α n (n - 1) .

(116)

Its transition diagram has the two arrows

n γ + β n n + 1, n α n (n - 1) n - 1.

(117)

The gain into state

n

comes from a death in state

n + 1

or a birth in state

n - 1

, while the loss is the sum of both rates out of state

n

. The birth-death master equation is consequently

\frac{\partial P ( n , t )}{\partial t} = α n (n + 1) P (n + 1, t) - [α n (n - 1) + γ + β n] P (n, t) + [γ + β (n - 1)] P (n - 1, t) .

(118)

Multiplying by

n

, summing over the nonnegative integers, and shifting the summation indices shows that each birth contributes

+ 1

and each death contributes

- 1

. Thus the first-moment equation of a birth-death process gives

\frac{d ⟨ n ⟩}{d t} = ⟨ b_{n} - d_{n} ⟩ = - α ⟨ n^{2} ⟩ + (α + β) ⟨ n ⟩ + γ .

(119)

At a stationary state, use the variance identity

⟨ n^{2} ⟩ = ⟨ n ⟩^{2} + (Δ n)^{2}

. Solving the resulting quadratic equation gives

⟨ n ⟩ = \frac{α + β}{2 α} \pm \frac{( α + β ) ^{2}}{4 α ^{2}} + \frac{γ}{α} - (Δ n)^{2} .

(120)

The minus root is inadmissible whenever it is negative, namely when

(Δ n)^{2} < γ / α

. Equality would give zero mean, which is also incompatible with a positive immigration rate

γ

, so for

γ > 0

the minus branch requires

(Δ n)^{2} > γ / α

even to be a possible mean.

For a continuum approximation, write

b (n) = γ + β n

and

d (n) = α n (n - 1)

. Applying the Kramers-Moyal expansion to the two gain terms and retaining derivatives through second order gives the Fokker-Planck equation

\frac{\partial P}{\partial t} = \frac{\partial}{\partial n} [g (n) P] + \frac{1}{2} \frac{\partial ^{2}}{\partial n ^{2}} [h (n) P],

(121)

where the negative drift and infinitesimal jump variance are

g (n) h (n) = d (n) - b (n) = α n^{2} - (α + β) n - γ, = d (n) + b (n) = α n^{2} + (β - α) n + γ .

(122)

This truncation requires the typical population and the scale on which

P

varies to be much larger than the unit jump size, with the rates varying smoothly across that scale.

The positive zero of

g

n_{*} = \frac{α + β + s q r t ( α + β ) ^{2} + 4 α γ}{2 α} = \frac{β}{α} + 1 + \frac{γ}{β} + O (α),

(123)

so in particular

n_{*} \sim β / α

α / β \to 0

. Put

x = n - n_{*}

. The linear noise approximation uses

g (n) = g_{*}^{'} x + O (x^{2}), h (n) = h_{*} + O (x),

(124)

where

g_{*}^{'} = 2 α n_{*} - (α + β) = (α + β)^{2} + 4 α γ > 0

(125)

and, because

b (n_{*}) = d (n_{*})

h_{*} = h (n_{*}) = 2 b (n_{*}) = 2 (γ + β n_{*}) .

(126)

In a stationary state with zero Fokker-Planck probability current,

g_{*}^{'} x P + \frac{h _{*}}{2} \frac{d P}{d x} = 0.

(127)

Its normalized solution is the normal distribution

P (x) ≃ \frac{g _{*}^{'}}{π h _{*}} exp (- \frac{g _{*}^{'} x ^{2}}{h _{*}}) .

(128)

It follows that

⟨ n ⟩ ≃ n_{*} \sim \frac{β}{α}, (Δ n)^{2} ≃ \frac{h _{*}}{2 g _{*}^{'}} \sim \frac{β}{α} .

(129)

These estimates agree with the exact stationary first-moment relation on its plus branch to leading order: inserting

(Δ n)^{2} \sim β / α

gives

⟨ n ⟩ \sim β / α

. Moreover,

\frac{Δ n}{n _{*}} s im \frac{α}{β} ≪ 1, Δ n \sim \frac{β}{α} ≫ 1.

(130)

Thus the stationary mass lies far from the boundary

n = 0

, is narrow relative to its mean, yet changes across many lattice sites. These are precisely the large-population and slow-variation conditions needed for the diffusion approximation.

Solved by gpt-5.6-sol high.

15D (Classical Dynamics)

Words: 250 Articles: 1

Solution

Words: 250

A transformation of phase space is canonical when it preserves the symplectic form, or equivalently all Poisson brackets. For one degree of freedom this is

d Q \land d P = d q \land d p,

(131)

equivalently

{Q, P}_{q, p} = 1

Suppose a type-two generating function for a canonical transformation

F (q, P)

defines

p = F_{q}, Q = F_{P} .

(132)

Then

d q \land d p d Q \land d P = F_{qP} d q \land d P, = F_{Pq} d q \land d P .

(133)

Equality of the mixed partial derivatives gives

d q \land d p = d Q \land d P

, so the transformation is canonical wherever it is locally invertible.

For

F_{0} (q, P) = \int_{0}^{q} 2 P - u^{2} d u,

(134)

the fundamental theorem of calculus and differentiation under the integral sign give

p = 2 P - q^{2}, Q = \int_{0}^{q} \frac{d u}{2 P - u ^{2}} = arcsin \frac{q}{2 P} .

(135)

Thus

q = 2 P sin Q, p = 2 P cos Q,

(136)

and in particular

P = (p^{2} + q^{2}) /2

For the unit-frequency simple harmonic motion, the transformed Hamiltonian is simply

H = P

. Hamilton's equations become

\dot{Q} = \frac{\partial H}{\partial P} = 1, \dot{P} = - \frac{\partial H}{\partial Q} = 0.

(137)

Therefore

P (t) = E, Q (t) = t - t_{0},

(138)

and transformation back gives the familiar solution

q (t) = 2 E sin (t - t_{0}), p (t) = 2 E cos (t - t_{0}) .

(139)

Now consider the weak quartic oscillator

H (q, p) = \frac{1}{2} (p^{2} + q^{2}) + ϵ q^{4} .

(140)

Choose the modified generating function

F_{ϵ} (q, P) = \int_{0}^{q} 2 P - u^{2} - 2 ϵ u^{4} d u .

(141)

It gives

p = 2 P - q^{2} - 2 ϵ q^{4},

(142)

so the transformed Hamiltonian is again

H = P

. Hence

P = E

and

Q = t - t_{0}

. Expanding the square root with the Taylor series gives

F_{ϵ} (q, P) = F_{0} (q, P) - ϵ \int_{0}^{q} \frac{u ^{4}}{2 P - u ^{2}} d u + O (ϵ^{2}) .

(143)

Differentiation with respect to

P

therefore yields

Q = arcsin \frac{q}{2 P} + ϵ I (q, P) + O (ϵ^{2}),

(144)

where

I (x, y) = \int_{0}^{x} \frac{u ^{4}}{( 2 y - u ^{2} ) ^{3/2}} d u .

(145)

Set

θ = t - t_{0}

and

q_{0} = 2 E sin θ, p_{0} = 2 E cos θ .

(146)

Writing

q = q_{0} + ϵ q_{1} + O (ϵ^{2})

and expanding the inverse sine relation at

q_{0}

gives

0 = \frac{q _{1}}{p _{0}} + I (q_{0}, E), q_{1} = - p_{0} I (q_{0}, E) .

(147)

Expanding

p^{2} = 2 E - q^{2} - 2 ϵ q^{4}

similarly gives

p = p_{0} + ϵ p_{1} + O (ϵ^{2}), p_{1} = q_{0} I (q_{0}, E) - \frac{q _{0}^{4}}{p _{0}} .

(148)

Consequently

\frac{q}{p} = \frac{q _{0}}{p _{0}} + ϵ (\frac{q _{1}}{p _{0}} - \frac{q _{0} p _{1}}{p _{0}^{2}}) + O (ϵ^{2}) = tan θ - ϵ I (q_{0}, E) (1 + tan^{2} θ) + ϵ q_{0}^{2} tan^{3} θ + O (ϵ^{2}) .

(149)

Since

θ = t - t_{0}

, this is the required expression. The formula is understood away from turning points where

p_{0} = 0

and the ratio

q / p

itself is singular.

Solved by gpt-5.6-sol high.

16H (Logic and Set Theory)

Words: 645 Articles: 8

a

Words: 119 Articles: 1

Solution

Words: 119

Assume the axiom of choice. By the well-ordering theorem, every set is bijective with an ordinal, and hence with an initial ordinal. Any two ordinals are comparable, so for any sets

K, L

either

K

injects into

L

L

injects into

K

. Transitivity is immediate, and antisymmetry on cardinalities follows from the Cantor-Schröder-Bernstein theorem. Thus cardinal comparison is a total order.

Conversely, assume the cardinal comparability principle. Given any set

X

, let

h (X)

be its Hartogs ordinal. By Hartogs theorem, there is no injection

h (X) \to X

. Comparability therefore supplies an injection

X \to h (X)

. Pulling the ordinal order back to

X

gives a well-order of

X

. Hence every set can be well-ordered, and the well-ordering theorem implies the axiom of choice.

Solved by gpt-5.6-sol high.

b

Words: 151 Articles: 1

Solution

Words: 151

Under the axiom of choice, identify every infinite cardinal

κ

with its initial ordinal. We prove

κ^{2} = κ

by transfinite induction on infinite cardinals. The countable case follows from the usual diagonal enumeration of

N \times N

For the induction step, well-order the Cartesian product

κ \times κ

by first comparing

max {α, β}

(150)

and then using lexicographic order among pairs with the same maximum. The predecessors of

(α, β)

lie in

(η + 1) \times (η + 1)

, where

η = max {α, β} < κ

. Put

μ = ∣ η + 1∣ < κ

. If

μ

is infinite, the induction hypothesis makes this predecessor set have cardinal at most

μ^{2} = μ < κ

; if

μ

is finite, the same conclusion is immediate.

Let

θ

be the order type of this well-order. Every proper initial segment of

θ

therefore has cardinal less than

κ

. Hence

θ < κ^{+}

, since otherwise its initial segment of order type

κ

would have cardinal

κ

. Thus

∣ κ \times κ ∣ = ∣ θ ∣ \leq κ

. The map

α \mapsto (α, 0)

gives the reverse injection, and the Cantor-Schröder-Bernstein theorem yields

κ^{2} = κ .

(151)

Solved by gpt-5.6-sol high.

c

Words: 229 Articles: 1

Solution

Words: 229

Let

i : K \times L ⟶ K ⊔ L

(152)

be the injection witnessing

κλ \leq κ + λ

, and fix

(k_{*}, l_{*}) \in K \times L

. Extend the inverse of

i

to a surjection

p : K ⊔ L ⟶ K \times L

(153)

by setting

p (z) = {i^{- 1} (z), (k_{*}, l_{*}), z \in i [K \times L], z \in / i [K \times L] .

(154)

This construction uses only the fixed pair, not the axiom of choice.

Restrict

p

to the

K

-summand and take its second coordinate:

r : K ⟶ L, r (k) = π_{L} (p (k, 0)) .

(155)

r

is onto, it is the required surjective function from

K

L

Otherwise choose one

l_{0} \in L ∖ r [K]

. For each

k \in K

, surjectivity of

p

gives a preimage of

(k, l_{0})

. No such preimage lies in the

K

-summand, by the choice of

l_{0}

, so it lies in the

L

-summand. Unless

(k, l_{0}) = (k_{*}, l_{*})

, this preimage is exactly

i (k, l_{0})

and is unique: all points outside the range of

i

map only to the default pair. Distinct

k

give distinct preimages because

i

is injective.

l_{0} \neq = l_{*}

, these unique preimages therefore define an injective function

K \to L

. The only exceptional case is

l_{0} = l_{*}

, where they initially define an injection

j : K ∖ {k_{*}} ⟶ L .

(156)

j

is onto, extend it arbitrarily at

k_{*}

to obtain a surjection

K \to L

. If it is not onto, choose one

l^{'} \in L ∖ j [K ∖ {k_{*}}]

and set

j (k_{*}) = l^{'}

; this gives an injection

K \to L

. These cases prove the product-sum comparison lemma entirely in ZF.

Solved by gpt-5.6-sol high.

d

Words: 146 Articles: 1

Solution

Words: 146

Part (b) proves one implication. Conversely, assume in ZF that

κ^{2} = κ

(157)

for every infinite cardinal

κ

. Let

X

be any set and let

γ = h (X)

be its Hartogs ordinal. Finite

X

is already well-orderable, so suppose

X

is infinite and form the disjoint union

A = X ⊔ γ .

(158)

By hypothesis,

A \times A

is bijective with

A

. Since

γ \times X

injects into

A \times A

, we obtain

∣ γ ∣ ∣ X ∣ \leq ∣ γ ∣ + ∣ X ∣.

(159)

Apply part (c), with

K = γ

and

L = X

. It gives either an injection

γ \to X

or a surjection

s : γ \to X

. The first alternative contradicts Hartogs theorem. In the second, every fiber

s^{- 1} ({x})

is a nonempty set of ordinals and therefore has a least member. The map

x ⟼ min s^{- 1} ({x})

(160)

is an injection of

X

into the ordinal

γ

and pulls its well-order back to

X

. Thus every set is well-orderable, so the well-ordering theorem gives the axiom of choice. This proves the Tarski cardinal-square theorem.

Solved by gpt-5.6-sol high.

17H (Graph Theory)

Words: 517 Articles: 8

a

Words: 42 Articles: 1

Solution

Words: 42

The vertex form of Menger theorem states that if

x

and

y

are distinct nonadjacent vertices of a finite graph, then the maximum number of pairwise internally vertex-disjoint paths from

x

y

equals the minimum cardinality of an

x

y

vertex separator.

Solved by gpt-5.6-sol high.

b

Words: 172 Articles: 1

Solution

Words: 172

The Hall marriage theorem says that a bipartite graph with classes

A, B

has a matching saturating

A

if and only if

∣ N (S) ∣ \geq ∣ S ∣ for every S \subseteq A,

(161)

where

N (S)

is the graph neighbourhood of

S

. Necessity follows because the matching sends the vertices of

S

to distinct vertices of

N (S)

For sufficiency, add vertices

x, y

, join

x

to every vertex of

A

, and join every vertex of

B

y

. Let

W

be an

x

y

vertex separator, and write

W_{A} = W \cap A

and

W_{B} = W \cap B

. If an edge joined a vertex of

A ∖ W_{A}

to a vertex of

B ∖ W_{B}

, it would give an

x

y

path avoiding

W

. Hence

N (A ∖ W_{A}) \subseteq W_{B} .

(162)

Hall's condition gives

∣ W_{B} ∣ \geq ∣ N (A ∖ W_{A}) ∣ \geq ∣ A ∖ W_{A} ∣,

(163)

and therefore

∣ W ∣ \geq ∣ A ∣

. By Menger theorem, there are

∣ A ∣

internally vertex-disjoint

x

y

paths. Each has the form

x ab y

with

a \in A

and

b \in B

; disjointness makes all the

a

and all the

b

distinct. Their middle edges form a matching saturating

A

Solved by gpt-5.6-sol high.

c

Words: 95 Articles: 1

Solution

Words: 95

Construct a bipartite graph with left vertices

A_{1}, \dots, A_{k}

, right vertices

B_{1}, \dots, B_{k}

, and an edge

A_{i} B_{j}

exactly when

A_{i} \cap B_{j} \neq = \emptyset

. For

S \subseteq [k]

, every point of

⋃_{i \in S} A_{i}

lies in some

B_{j}

adjacent to one of those

A_{i}

. Thus

⋃_{i \in S} A_{i} \subseteq ⋃_{j \in N (S)} B_{j} .

(164)

Using finite additivity of Lebesgue measure and the equal-volume hypotheses,

\frac{∣ S ∣}{k} = vol (⋃_{i \in S} A_{i}) \leq vol (⋃_{j \in N (S)} B_{j}) = \frac{∣ N ( S ) ∣}{k} .

(165)

Hence

∣ N (S) ∣ \geq ∣ S ∣

for every

S

. The Hall marriage theorem supplies a perfect matching, which has the form

A_{i} B_{σ (i)}

for a permutation

σ

[k]

. Every matched edge means precisely that

A_{i} \cap B_{σ (i)} \neq = \emptyset

Solved by gpt-5.6-sol high.

d

Words: 208 Articles: 1

Solution

Words: 208

Let

T = {x_{1}, \dots, x_{k}, y_{1}, \dots, y_{k}}

(166)

be the terminal set, and let

C

be the vertex set of the given complete graph

K_{2 k}

. Any vertex set

S

meeting every

T

C

path has size at least

2 k

: if

∣ S ∣ < 2 k

, choose

t \in T ∖ S

and

c \in C ∖ S

. Since

G

2 k

-connected,

G - S

is connected, so it contains a

t

c

path avoiding

S

, a contradiction.

The Set version of Menger theorem therefore gives

2 k

pairwise vertex-disjoint

T

C

paths. Truncate them at their first and last vertices in

T \cup C

. Since both sets have

2 k

vertices, every terminal is the endpoint of one path and the other endpoints are distinct vertices of

C

. Denote the path from a terminal

z

to its clique endpoint by

R_{z}

, and call that endpoint

c_{z}

For each

i

, the clique contains the edge

c_{x_{i}} c_{y_{i}}

. Concatenating

R_{x_{i}}, c_{x_{i}} c_{y_{i}}, R_{y_{i}}

(167)

gives an

x_{i}

y_{i}

path in a graph. The linkage paths are mutually vertex-disjoint, their clique endpoints are all distinct, and the joining clique edges pair those endpoints without introducing a new vertex. The resulting paths

P_{1}, \dots, P_{k}

are therefore mutually vertex-disjoint. In particular, under the stated clique hypothesis, the

2 k

-connected graph is

k

-linked for these terminals.

Solved by gpt-5.6-sol high.

18I (Galois Theory)

Words: 449 Articles: 7

a

Words: 219 Articles: 1

Solution

Words: 219

If a monic polynomial

f

has roots

α_{1}, \dots, α_{n}

in a splitting field, its polynomial discriminant is

disc (f) = \prod_{1 \leq i < j \leq n} (α_{i} - α_{j})^{2} .

(168)

It is a symmetric polynomial in the roots, so it belongs to the base field; it is nonzero exactly when

f

is a separable polynomial.

Put

D = \prod_{i < j} (α_{i} - α_{j}),

(169)

the Vandermonde determinant. A root permutation

σ

sends

D

sgn (σ) D

. Hence the Galois group of a polynomial is contained in the alternating group

A_{n}

exactly when every Galois automorphism fixes

D

, which by the fixed-field property is equivalent to

D \in K

. This implies that

disc (f) = D^{2}

is a square in

K

. Conversely, if

disc (f) = d^{2}

for

d \in K

, then

D^{2} = d^{2}

, so

D = \pm d \in K

. The Galois group therefore fixes

D

and consists of even permutations. This proves the discriminant criterion for an alternating Galois group.

Now let

f (T) = T^{3} - 2 T + 2.

(170)

The rational root theorem shows that none of

\pm 1, \pm 2

is a root, so this cubic is irreducible over

Q

. The discriminant of a depressed cubic is

disc (f) = - 4 (- 2)^{3} - 27 (2)^{2} = 32 - 108 = - 76 = - 4 \cdot 19.

(171)

This is not a square in

Q

, so the Galois group of an irreducible cubic is

S_{3}

over

Q

Over

K = Q (- 19)

, the discriminant is the square

- 76 = (2 - 19)^{2} .

(172)

The cubic remains irreducible: a root in the degree-two extension

K / Q

would generate over

Q

both a degree-three field, by irreducibility, and a subfield of a degree-two field, contradicting the tower law for field extensions. Its Galois group over

K

is consequently

A_{3} ≅ C_{3}

Solved by gpt-5.6-sol high.

b

Words: 230 Articles: 4

i

Words: 57 Articles: 1

Solution

Words: 57

In characteristic two, the formal derivative in positive characteristic of

f (T) = T^{3} + a T + b

(173)

f^{'} (T) = T^{2} + a .

(174)

A polynomial is separable exactly when it and its derivative have no common root. If

r

is a root of

f^{'}

, then

f (r) = r^{3} + a r + b = r (r^{2} + a) + b = b .

(175)

Thus

f

and

f^{'}

have a common root exactly when

b = 0

. Therefore

f

is a separable polynomial if and only if

b \neq = 0

Solved by gpt-5.6-sol high.

ii

Words: 173 Articles: 1

Solution

Words: 173

Let

x_{1}, x_{2}, x_{3} \in L

be the distinct roots of

f

, and define the two cyclic sums

r = x_{1}^{2} x_{2} + x_{2}^{2} x_{3} + x_{3}^{2} x_{1}, s = x_{2}^{2} x_{1} + x_{3}^{2} x_{2} + x_{1}^{2} x_{3} .

(176)

Vieta's relations in characteristic two are

x_{1} + x_{2} + x_{3} = 0, x_{1} x_{2} + x_{2} x_{3} + x_{3} x_{1} = a, x_{1} x_{2} x_{3} = b .

(177)

A direct symmetric expansion using these identities gives

r + s = b, rs = a^{3} + b^{2} .

(178)

Consequently

r

and

s

are the roots of the characteristic-two cubic resolvent

g (T) = T^{2} + b T + a^{3} + b^{2} .

(179)

They are distinct because

r + s = b \neq = 0

. Thus

g

splits into distinct linear factors in

L [T]

A three-cycle of the roots fixes each cyclic sum, whereas any transposition interchanges

r

and

s

. The induced action on

{r, s}

is therefore exactly the sign action of the symmetric group

S_{3}

S_{3} / A_{3}

. If the Galois group

G

is contained in

A_{3}

, it fixes

r

and

s

, so both lie in the fixed field

K

and

g

splits over

K

. Conversely, if

g

splits in

K [T]

, its distinct roots

r, s

belong to

K

and every element of

G

fixes them individually. No element of

G

can then act as an odd root permutation, so

G \subseteq A_{3}

Solved by gpt-5.6-sol high.

19H (Representation Theory)

Words: 358 Articles: 1

Solution

Words: 358

Schur lemma says that an intertwining linear map between irreducible finite-dimensional complex representations is either zero or an isomorphism. In particular, every endomorphism of an irreducible complex representation is a scalar multiple of the identity.

A continuous representation of a topological group is a continuous homomorphism

ρ : G ⟶ GL (V)

(180)

for a finite-dimensional complex vector space

V

. It is a unitary representation if

V

has a positive-definite Hermitian inner product for which

⟨ ρ (g) v, ρ (g) w ⟩ = ⟨ v, w ⟩

(181)

for every

g \in G

and

v, w \in V

For

G = S^{1}

, start with any positive-definite Hermitian form and average it using normalized Haar measure:

⟨ v, w ⟩_{S^{1}} = \int_{S^{1}} ⟨ ρ (z) v, ρ (z) w ⟩ d z .

(182)

Translation invariance makes this form

S^{1}

-invariant, and positivity is preserved, proving unitarity. Since

S^{1}

is abelian, the operators

ρ (z)

commute; since they are unitary, they are normal. Simultaneous diagonalization therefore decomposes

V

into common one-dimensional eigenspaces. Thus every representation of the circle group is a direct sum of one-dimensional representations.

Write

g (x, y, z) = 100 x 10 z y 1 .

(183)

The group law and inverse are

g (x, y, z) g (x^{'}, y^{'}, z^{'}) g (x, y, z)^{- 1} = g (x + x^{'}, y + y^{'}, z + z^{'} + x y^{'}), = g (- x, - y, - z + x y) .

(184)

A calculation gives

g (x, y, z)^{- 1} g (x^{'}, y^{'}, z^{'})^{- 1} g (x, y, z) g (x^{'}, y^{'}, z^{'}) = g (0, 0, x y^{'} - x^{'} y) .

(185)

Every element of the centre

Z

occurs by taking, for example,

x^{'} = y = 0

and

y^{'} = 1

. Hence the commutator subgroup is exactly

Z

. The image of a one-dimensional representation is abelian, so the one-dimensional representation kills the commutator subgroup and its kernel contains

Z

Now let

(ρ, V)

be a complex representation of

G / Z_{0}

. Its restriction to the central subgroup

Z / Z_{0} ≅ R / Z ≅ S^{1}

(186)

is a representation of the circle group. Decompose it into its distinct weight spaces:

V = V_{1} \oplus \dots \oplus V_{d}, ρ (z) ∣_{V_{i}} = θ_{i} (z) id_{V_{i}} .

(187)

Because

Z / Z_{0}

is central, every

ρ (g)

commutes with its action and preserves every common eigenspace. Thus the

V_{i}

are

G / Z_{0}

-subrepresentations, as asserted by the central circle weight-space decomposition.

Let

n_{i} = dim V_{i}

. The map

g ⟼ det (ρ (g) ∣_{V_{i}})

(188)

is a one-dimensional representation of

G / Z_{0}

; after composition with

G \to G / Z_{0}

, its kernel contains

Z

. On

z \in Z / Z_{0}

its value is

det (θ_{i} (z) I_{n_{i}}) = θ_{i} (z)^{n_{i}},

(189)

θ_{i}^{n_{i}} = 1

. A continuous character of

S^{1}

has the form

z \mapsto z^{m}

; the displayed identity forces

m = 0

. Therefore every

θ_{i}

is trivial. Since the

θ_{i}

were distinct,

d = 1

and

θ_{1} = 1

It follows that every finite-dimensional complex representation of

G / Z_{0}

kills the entire nontrivial central circle

Z / Z_{0}

. Its kernel is therefore nontrivial, so no such representation is faithful. This is the real Heisenberg quotient has no faithful finite-dimensional representation.

Solved by gpt-5.6-sol high.

20H (Number Fields)

Words: 478 Articles: 9

a

Words: 168 Articles: 1

Solution

Words: 168

Minkowski's lemma states that if

Λ

is a full lattice in

R^{n}

and

C \subset R^{n}

is measurable, convex, and centrally symmetric, then

vol (C) > 2^{n} covol (Λ)

(190)

implies that

C

contains a nonzero point of

Λ

Let

σ_{1}, σ_{2} : K \to R

be the two real embeddings. The Minkowski embedding of a real quadratic field identifies

O_{K}

with a lattice

Λ = {(σ_{1} (α), σ_{2} (α)) : α \in O_{K}}

(191)

of covolume

∣ D_{K} ∣

. Choose a constant

c

satisfying

2 ∣ D_{K} ∣ < c < 4 ∣ D_{K} ∣ .

(192)

For

R > 0

, consider the closed diamond

C_{R} = {(x, y) : \frac{∣ x ∣}{R} + \frac{∣ y ∣}{c / R} \leq 1} .

(193)

It is convex and centrally symmetric, and its area is

vol (C_{R}) = 2 R \frac{c}{R} = 2 c > 4 ∣ D_{K} ∣ .

(194)

Minkowski's lemma supplies a nonzero

α_{R} \in O_{K}

whose two embeddings lie in

C_{R}

For

(x, y) \in C_{R}

, the arithmetic-geometric mean inequality gives

∣ x y ∣ l e q \frac{1}{4} R \frac{c}{R} = \frac{c}{4} < ∣ D_{K} ∣ .

(195)

Since the coordinate product is the field norm,

∣ N_{K / Q} (α_{R}) ∣ < ∣ D_{K} ∣ .

(196)

Finally,

∣ σ_{2} (α_{R}) ∣ \leq c / R \to 0

R \to \infty

. No fixed nonzero algebraic integer can occur for arbitrarily large

R

, because its second embedding is nonzero. Consequently the elements

α_{R}

obtained along an unbounded sequence of

R

contain infinitely many distinct values. Here

N (α)

denotes the absolute norm, as usual in this inequality.

Solved by gpt-5.6-sol high.

b

Words: 199 Articles: 4

i

Words: 142 Articles: 1

Solution

Words: 142

For

K = Q (10)

, the ring of integers of a quadratic field is

O_{K} = Z [10] .

(197)

The element

ε = 3 + 10

(198)

has field norm

9 - 10 = - 1

, so it is a unit. We claim that

O_{K}^{\times} = {\pm ε^{n} : n \in Z} .

(199)

Let

u

be any unit. Its norm is

\pm 1

. After changing its sign and replacing it by its inverse if necessary, its first real embedding satisfies

u \geq 1

. Choose

n \in Z

so that

1 \leq v = u ε^{- n} < ε .

(200)

Write

v = a + b 10

with

a, b \in Z

. Its conjugate is

v^{'} = \pm v^{- 1}

, so

∣ v^{'} ∣ \leq 1

. Therefore

∣ a ∣ = \frac{∣ v + v ^{'} ∣}{2} < \frac{ε + 1}{2} < 4, ∣ b ∣ = \frac{∣ v - v ^{'} ∣}{2 10} < \frac{ε + 1}{2 10} < 2.

(201)

Thus

∣ a ∣ \leq 3

and

∣ b ∣ \leq 1

. The equation

a^{2} - 10 b^{2} = \pm 1

now shows directly that the only value in

[1, ε)

v = 1

: if

b = 0

then

a = \pm 1

, while if

∣ b ∣ = 1

then

a = \pm 3

and the positive possibilities are either below

1

or equal to

ε

. Hence

u = ε^{n}

after normalization, proving the claimed description of the units of Q of square root ten.

Solved by gpt-5.6-sol high.

ii

Words: 57 Articles: 1

Solution

Words: 57

For

K = Q (- 3)

, put

ω = \frac{- 1 + - 3}{2} .

(202)

Then the ring of integers of a quadratic field is

O_{K} = Z [ω]

, and

N (a + bω) = a^{2} - ab + b^{2} .

(203)

A unit must have norm

1

. Multiplying by four gives

(2 a - b)^{2} + 3 b^{2} = 4,

(204)

b \in {0, \pm 1}

, and checking these cases gives exactly

O_{K}^{\times} = {\pm 1, \pm ω, \pm ω^{2}} ≅ C_{6} .

(205)

These are precisely the sixth roots of unity, proving the description of the units of Q of square root minus three.

Solved by gpt-5.6-sol high.

c

Words: 111 Articles: 1

Solution

Words: 111

Assume that

ζ

is a primitive fifth root of unity. The cyclotomic field

K = Q (ζ)

has degree four and signature

(r_{1}, r_{2}) = (0, 2)

. The Dirichlet unit theorem therefore gives free rank

r_{1} + r_{2} - 1 = 1.

(206)

The roots of unity in a rational cyclotomic field are the ten elements

\pm ζ^{j}

, so

O_{K}^{\times} ≅ C_{10} \times Z .

(207)

For explicit units,

- ζ

has order ten, while

N_{K / Q} (1 + ζ) = \prod_{j = 1}^{4} (1 + ζ^{j}) = Φ_{5} (- 1) = 1.

(208)

Hence

1 + ζ

is a unit. It has infinite order because in the standard complex embedding

∣1 + ζ ∣ = 2 cos (π /5) > 1,

(209)

whereas every root of unity has modulus one. Its image in the free

Z

factor is therefore nonzero and generates a finite-index subgroup of that factor. Consequently

H = ⟨ - ζ, 1 + ζ ⟩

(210)

has finite index in

O_{K}^{\times}

, as described by the unit group of the fifth cyclotomic field.

Solved by gpt-5.6-sol high.

21G (Algebraic Topology)

Words: 264 Articles: 1

Solution

Words: 264

Let

f : (C, d) \to (C^{'}, d^{'})

be a chain map, so

d_{i}^{'} f_{i} = f_{i - 1} d_{i} .

(211)

x \in C_{i}

is a cycle, then

d^{'} f_{i} (x) = f_{i - 1} d (x) = 0

, so

f_{i} (x)

is a cycle. If

x = d_{i + 1} y

is a boundary, then

f_{i} (x) = f_{i} d_{i + 1} y = d_{i + 1}^{'} f_{i + 1} y

(212)

is a boundary. Hence

f_{*} : H_{i} (C) ⟶ H_{i} (C^{'}), [x] ⟼ [f_{i} (x)]

(213)

is a well-defined induced map on homology.

The maps

f

and

g

are chain homotopic if there are homomorphisms

h_{i} : C_{i} ⟶ C_{i + 1}^{'}

(214)

such that

f_{i} - g_{i} = d_{i + 1}^{'} h_{i} + h_{i - 1} d_{i} .

(215)

For a cycle

x

, this gives

f_{i} (x) - g_{i} (x) = d_{i + 1}^{'} h_{i} (x),

(216)

which is a boundary. Thus

f_{*} [x] = g_{*} [x]

and

f_{*} = g_{*}

on homology.

Now consider the proposed mapping cone

M (f)

. Applying its differential twice gives

(d_{f})_{i - 1} (d_{f})_{i} = (d_{i - 2} d_{i - 1} (- 1)^{i - 1} f_{i - 2} d_{i - 1} + (- 1)^{i} d_{i - 1}^{'} f_{i - 1} 0 d_{i - 1}^{'} d_{i}^{'}) .

(217)

The diagonal entries vanish because

C

and

C^{'}

are chain complexes, and the lower-left entry is

(- 1)^{i - 1} (f_{i - 2} d_{i - 1} - d_{i - 1}^{'} f_{i - 1}) = 0

(218)

by the chain-map identity. Therefore

d_{f}^{2} = 0

and

M (f)

is a chain complex.

There is a short exact sequence of chain complexes

0 ⟶ C^{'} j M (f) p C [- 1] ⟶ 0,

(219)

where

j_{i} (y) = (0, y), p_{i} (x, y) = x, C [- 1]_{i} = C_{i - 1} .

(220)

Its long exact sequence in homology is

\dots ⟶ H_{i} (C^{'}) ⟶ H_{i} (M (f)) ⟶ H_{i - 1} (C) δ_{i} H_{i - 1} (C^{'}) ⟶ \dots .

(221)

To identify the connecting map, represent a class in

H_{i - 1} (C)

by a cycle

x

and lift it to

(x, 0) \in M (f)_{i}

. Then

(d_{f})_{i} (x, 0) = (0, (- 1)^{i} f_{i - 1} x),

(222)

δ_{i} = (- 1)^{i} f_{*} .

(223)

At the preceding occurrence the degree is

i + 1

, giving

(- 1)^{i + 1} f_{*}

. Hence the sequence is exactly

\dots ⟶ H_{i} (C) (- 1)^{i + 1} f_{*} H_{i} (C^{'}) ⟶ H_{i} (M (f)) ⟶ H_{i - 1} (C) (- 1)^{i} f_{*} H_{i - 1} (C^{'}) ⟶ \dots .

(224)

Finally suppose

f - g = d^{'} h + h d

. Define

Φ_{i} : M (f)_{i} ⟶ M (g)_{i}, Φ_{i} (x, y) = (x, y + (- 1)^{i} h_{i - 1} x) .

(225)

A direct calculation gives

(d_{g})_{i} Φ_{i} (x, y) Φ_{i - 1} (d_{f})_{i} (x, y) = (d x, (- 1)^{i} gx + d^{'} y + (- 1)^{i} d^{'} h x), = (d x, (- 1)^{i} f x + d^{'} y + (- 1)^{i - 1} h d x) .

(226)

These are equal precisely because

f - g = d^{'} h + h d

. Thus

Φ

is a chain map. Replacing the plus sign in its definition by a minus sign gives its inverse, so

M (f)

and

M (g)

are isomorphic as chain complexes.

Solved by gpt-5.6-sol high.

22F (Linear Analysis)

Words: 310 Articles: 8

a

Words: 72 Articles: 1

Solution

Words: 72

For each

n

, define the bounded linear functional

T_{n} : H ⟶ C, T_{n} (y) = ⟨ x_{n}, y ⟩ .

(227)

The hypothesis is precisely that

x_{n}

converges weakly to

x

. In particular, for each fixed

y \in H

, the scalar sequence

T_{n} (y)

converges and is therefore bounded. Thus the family

{T_{n}}

is pointwise bounded.

The Uniform boundedness principle gives

sup_{n} ∥ T_{n} ∥ < \infty.

(228)

By the Riesz representation theorem,

∥ T_{n} ∥ = ∥ x_{n} ∥_{H}

. Hence

sup_{n} ∥ x_{n} ∥_{H} < \infty,

(229)

so the sequence is bounded.

Solved by gpt-5.6-sol high.

b

Words: 131 Articles: 1

Solution

Words: 131

Put

z_{n} = x_{n} - x

, so

z_{n} ⇀ 0

. We show that the norm closure of the convex hull of every tail

A_{k} = {z_{n} : n \geq k}

(230)

contains zero. If it did not, the Hahn-Banach separation theorem would give a bounded linear functional

ϕ

, a real number

c > 0

, and, after multiplying

ϕ

by a complex scalar if necessary,

Re ϕ (z) \geq c

(231)

for every

z

in that closed convex hull. In particular

Re ϕ (z_{n}) \geq c

for all

n \geq k

. By the Riesz representation theorem,

ϕ (z) = ⟨ z, y ⟩

for some

y \in H

, contradicting

z_{n} ⇀ 0

Consequently, for every

k

there is a finite convex combination

w_{k} = \sum_{n = k}^{N_{k}} λ_{k, n} z_{n}, λ_{k, n} \geq 0, \sum_{n = k}^{N_{k}} λ_{k, n} = 1,

(232)

with

∥ w_{k} ∥_{H} < 1/ k

. Define

x_{k} = x + w_{k} = \sum_{n = k}^{N_{k}} λ_{k, n} x_{n} .

(233)

Then every

x_{k}

is a convex combination of terms of the original sequence and

∥ x_{k} - x ∥_{H} = ∥ w_{k} ∥_{H} < \frac{1}{k} ⟶ 0.

(234)

This is Mazur lemma in the present Hilbert-space setting.

Solved by gpt-5.6-sol high.

c

Words: 45 Articles: 1

Solution

Words: 45

Apply part (b). Since every

x_{n}

belongs to the convex set

C

, every convex combination

x_{k}

also belongs to

C

. Since

C

is norm closed and

x_{k} \to x

in norm, it follows that

x \in C

. Thus a norm-closed convex set is weakly closed.

Solved by gpt-5.6-sol high.

d

Words: 62 Articles: 1

Solution

Words: 62

No. Take the Hilbert space

H = ℓ^{2}

and its standard orthonormal sequence

(e_{n})

. Let

C = {u \in ℓ^{2} : ∥ u ∥_{2} = 1},

(235)

the unit sphere. This set is norm closed but is not convex. By the Bessel inequality, for every

y = (y_{j}) \in ℓ^{2}

⟨ e_{n}, y ⟩ = \overline{y_{n}} ⟶ 0,

(236)

e_{n} ⇀ 0

; equivalently, this is a weakly null orthonormal sequence. Every

e_{n}

lies in

C

, but the weak limit

0

does not. Hence convexity is essential.

Solved by gpt-5.6-sol high.

23F (Analysis of Functions)

Words: 427 Articles: 8

a

Words: 88 Articles: 1

Solution

Words: 88

Choose a nonzero

ϕ \in C_{c}^{\infty} (R^{n})

supported in a ball of radius

1

, and let

ϕ_{k} (x) = ϕ (x - 3 k e_{1}) .

(237)

Translation preserves both the

L^{2}

norm and the weak-derivative norms, so

∥ ϕ_{k} ∥_{H^{1}} = ∥ ϕ ∥_{H^{1}}

(238)

for every

k

. Thus

(ϕ_{k})

is bounded in the Sobolev space

H^{1} (R^{n})

Distinct translates have disjoint supports. Hence, for

j \neq = k

∥ ϕ_{j} - ϕ_{k} ∥_{2}^{2} = ∥ ϕ_{j} ∥_{2}^{2} + ∥ ϕ_{k} ∥_{2}^{2} = 2 ∥ ϕ ∥_{2}^{2} .

(239)

No subsequence is Cauchy in

L^{2}

, so no subsequence converges there. The embedding

H^{1} (R^{n}) ↪ L^{2} (R^{n})

is therefore not a compact operator. This is the noncompactness of a Sobolev embedding by translation caused by loss of compactness at infinity.

Solved by gpt-5.6-sol high.

b

Words: 117 Articles: 1

Solution

Words: 117

Regard the space of continuous functions vanishing at infinity

C_{0} (R^{n})

as a subspace of

L^{\infty} (R^{n})

. Point evaluation at the origin,

ℓ_{0} (f) = f (0),

(240)

is a bounded linear functional of norm one on this subspace because a continuous function's supremum and essential supremum agree. By the Hahn-Banach theorem,

ℓ_{0}

extends to a bounded linear functional

ℓ : L^{\infty} (R^{n}) ⟶ C .

(241)

Suppose that some

g \in L^{1} (R^{n})

represented this extension:

ℓ (f) = \int_{R^{n}} f (x) g (x) d x .

(242)

Choose

φ \in C_{c} (R^{n})

with

0 \leq φ \leq 1

φ (0) = 1

, and support in the unit ball, and set

φ_{k} (x) = φ (k x)

. Then

ℓ (φ_{k}) = 1

for every

k

. On the other hand,

φ_{k} (x) \to 0

for almost every

x

and

∣ φ_{k} g ∣ \leq ∣ g ∣

. The dominated convergence theorem gives

\int φ_{k} (x) g (x) d x ⟶ 0,

(243)

a contradiction. Thus

ℓ

is a singular functional on L infinity and has no

L^{1}

density.

Solved by gpt-5.6-sol high.

c

Words: 124 Articles: 1

Solution

Words: 124

We prove the estimate first for

u

in the Schwartz space. By the Fourier inversion theorem and the Cauchy-Schwarz inequality,

∣ u (x) ∣ \leq C \int_{R^{n}} ∣ u (ξ) ∣ d ξ \leq C (\int_{R^{n}} (1 + ∣ ξ ∣^{2})^{- n} d ξ)^{1/2} ∥ u ∥_{H^{n}} .

(244)

The integral is finite because

n > n /2

For

0 < α \leq 1

, the elementary bound

∣ e^{i t} - 1∣ \leq C_{α} ∣ t ∣^{α}

(245)

gives

∣ u (x) - u (y) ∣ \leq C ∣ x - y ∣^{α} \int_{R^{n}} ∣ ξ ∣^{α} ∣ u (ξ) ∣ d ξ \leq C ∣ x - y ∣^{α} (\int_{R^{n}} ∣ ξ ∣^{2 α} (1 + ∣ ξ ∣^{2})^{- n} d ξ)^{1/2} ∥ u ∥_{H^{n}} .

(246)

The last integral is finite near zero for every

α > 0

and at infinity exactly when

2 α < n

. Choose, for example,

α = 1/4

, which works for every

n \geq 1

. We obtain

∥ u ∥_{C^{0, 1/4}} \leq C_{n} ∥ u ∥_{H^{n}} .

(247)

Density of the Schwartz space in

H^{n}

extends the estimate and supplies a unique Hölder-continuous representative. Therefore

H^{n} (R^{n}) ↪ C^{0, 1/4} (R^{n})

(248)

continuously. This is the Fourier proof of Hölder regularity from a Sobolev norm.

Solved by gpt-5.6-sol high.

d

Words: 98 Articles: 1

Solution

Words: 98

Use the centered Hardy-Littlewood maximal function

Mθ (x) = sup_{r > 0} \frac{1}{2 r} \int_{x - r}^{x + r} θ (t) d t .

(249)

x > 0

, choosing

0 < r < x

makes the whole interval lie in the positive half-line, so

Mθ (x) = 1

. If

x = 0

, every centered interval has exactly half its length in the positive half-line, so

Mθ (0) = 1/2

x < 0

, the average is zero for

r \leq - x

. For

r > - x

, it is

\frac{x + r}{2 r} = \frac{1}{2} + \frac{x}{2 r} < \frac{1}{2},

(250)

and these values tend to

1/2

r \to \infty

. Hence

Mθ (x) = {1, \frac{1}{2}, x > 0, x \leq 0.

(251)

For the uncentered maximal-function convention, intervals extending arbitrarily far to the right make the supremum equal to

1

at every

x

Solved by gpt-5.6-sol high.

24G (Algebraic Geometry)

Words: 353 Articles: 1

Solution

Words: 353

Two irreducible varieties

X

and

Y

are birational if they contain nonempty Zariski-open subsets

U \subseteq X

and

V \subseteq Y

that are isomorphic. Equivalently, their function fields are isomorphic:

k (X) ≅ k (Y) .

(252)

For an irreducible variety,

dim X = trdeg_{k} k (X) .

(253)

An isomorphism of function fields preserves transcendence degree, so

dim X = dim Y

. This is the dimension from the function field.

Now let

K / C

be a finitely generated field extension. Choose field generators

a_{1}, \dots, a_{m}

and set

A = C [a_{1}, \dots, a_{m}] \subseteq K .

(254)

Then

A

is a finitely generated integral

C

-algebra and

Frac (A) = K

. The affine irreducible variety

X_{0} = Spec A

(255)

has function field

K

. Embed

X_{0}

as a closed affine subvariety of some affine space

A^{N}

, identify that affine space with the standard open chart of projective space

P^{N}

, and take the projective closure

X

X_{0}

. The closure of an irreducible topological space is irreducible, and

X_{0}

is dense and open in

X

. Thus

X

is a projective variety and

k (X) = k (X_{0}) = K .

(256)

This constructs the projective model of a finitely generated field.

For the curve

X = V (y^{2} - x (x - 1)^{2}),

(257)

put

t = \frac{y}{x - 1} .

(258)

In its function field, the defining equation gives

t^{2} = x, y = t (x - 1) = t (t^{2} - 1) .

(259)

Hence

C (X) = C (t)

, so

X

is rational. Equivalently, this is the normalization of y squared equals x times x minus one squared, whose smooth projective normalization is

P^{1}

and has geometric genus zero.

A smooth projective curve in the projective plane of degree

d

has, by the genus of a smooth plane curve formula,

g = \frac{( d - 1 ) ( d - 2 )}{2} .

(260)

Its genus can be zero only for

d = 1

d = 2

. Conversely, a projective line and every smooth conic over

C

are rational, so each is birational to

X

. Therefore the required degrees are exactly

d = 1 or d = 2 .

(261)

Finally, in

A_{x, y, z}^{3}

take

S = V (F), F (x, y, z) = y^{2} - x (x - 1)^{2} .

(262)

The polynomial

F

is irreducible: as a quadratic in

y

, it could factor only if

x (x - 1)^{2}

were a square polynomial, which it is not. Thus

S

is an irreducible affine hypersurface of dimension two. Its function field is

C (S) = C (t, z), x = t^{2}, y = t (t^{2} - 1),

(263)

S

is birational to

A^{2}

The partial derivatives are

F_{x} = - (x - 1) (3 x - 1), F_{y} = 2 y, F_{z} = 0.

(264)

Solving

F = F_{x} = F_{y} = 0

gives

x = 1

y = 0

, with

z

arbitrary. By the Jacobian criterion,

Sing (S) = {(1, 0, z) : z \in C} ≅ A^{1} .

(265)

This is an irreducible subvariety of dimension one, giving the requested singular cylinder over a nodal curve.

Solved by gpt-5.6-sol high.

25G (Differential Geometry)

Words: 482 Articles: 8

a

Words: 153 Articles: 1

Solution

Words: 153

For a positively oriented local parametrization

X (u, v)

with first fundamental form

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

(266)

the area element of a surface is

d A = EG - F^{2} d u \land d v .

(267)

Under an orientation-preserving coordinate change, the Jacobian from

d u \land d v

cancels the inverse Jacobian in the square root of the metric determinant. Hence these local expressions agree and define a global two-form.

The Euler characteristic may be defined from any finite triangulation by

χ (S) = V - E + F,

(268)

or equivalently by the alternating sum of the dimensions of the rational homology groups. Subdivision leaves

V - E + F

unchanged, and the homological formula shows that it is a topological invariant, so the definition does not depend on the triangulation.

Give the boundary its induced orientation and parametrize it by arc length. If

T

is its unit tangent and

N

is the chosen unit normal to the surface, its signed geodesic curvature is

k_{g} = ⟨ D_{s} T, N \times T ⟩,

(269)

where

D_{s}

is the surface covariant derivative. The Gauss-Bonnet theorem for a compact oriented surface with smooth boundary and no corners is

\int_{S} K d A + \int_{\partial S} k_{g} d s = 2 π χ (S),

(270)

where

K

is the Gaussian curvature.

Solved by gpt-5.6-sol high.

b

Words: 132 Articles: 1

Solution

Words: 132

Because

S_{1}

and

S_{2}

are isometric,

χ (S_{1}) = χ (S_{2}), \int_{S_{1}} K d A = \int_{S_{2}} K d A .

(271)

The two induced orientations on their common boundary

γ

are opposite. If

B = \int_{γ} k_{g} d s

(272)

is computed with the boundary orientation from

S_{1}

, then its value for

S_{2}

- B

. Applying the Gauss-Bonnet theorem to the two halves gives

\int_{S_{1}} K d A + B = 2 π χ (S_{1}), \int_{S_{2}} K d A - B = 2 π χ (S_{2}) .

(273)

The equalities above imply

B = 0

. This is the isometric halves give zero total boundary geodesic curvature argument.

Every ambient isometry

ϕ_{x, y}

preserves the magnitude of geodesic curvature and sends

x

y

. Hence

∣ k_{g} ∣

is constant along the connected curve

γ

by the transitive curve symmetry makes geodesic-curvature magnitude constant. If that constant were positive, continuity would force the signed

k_{g}

to have one fixed sign, making its integral nonzero. Therefore

k_{g} \equiv 0

. A curve has zero geodesic curvature exactly when it is a geodesic, so

γ

is geodesic.

Solved by gpt-5.6-sol high.

c

Words: 77 Articles: 1

Solution

Words: 77

No. On the round sphere, take a latitude circle other than the equator. Rotations about the polar axis preserve the circle and act transitively on its points, so all the isometries

ϕ_{x, y}

exist. The latitude separates the sphere into two caps of unequal area, so those caps are not isometric. It is not a great circle and therefore is not a geodesic; equivalently, its geodesic curvature is nonzero. This is a homogeneous nongeodesic latitude.

Solved by gpt-5.6-sol high.

d

Words: 120 Articles: 1

Solution

Words: 120

Again, no. On the unit round sphere and for a sufficiently small nonzero

ε

, consider

γ (θ) = \frac{( c o s θ , s i n θ , ε s i n 3 θ )}{1 + ε ^{2} s i n ^{2} 3 θ}, 0 \leq θ \leq 2 π .

(274)

This is a smooth embedded closed curve and may be reparametrized by arc length. It obeys

γ (θ + π) = - γ (θ),

(275)

so the antipodal isometry preserves the curve and exchanges the two complementary discs. Its restriction therefore gives an isometry

S_{1} \to S_{2}

The curve is not contained in any plane through the origin: its third-harmonic height cannot satisfy a nontrivial linear relation with

cos θ

and

sin θ

. It is therefore not a great circle. Since the closed geodesics of the round sphere are precisely the great circles,

γ

is not geodesic. This antipodally symmetric nongeodesic spherical curve shows that the point-transitivity assumption cannot be omitted.

Solved by gpt-5.6-sol high.

26K (Probability and Measure)

Words: 158 Articles: 4

a

Words: 95 Articles: 1

Solution

Words: 95

Write

S_{n} = \sum_{j = 1}^{n} Y_{j}

for the partial sum. Fix an integer

M \geq 1

. The tail-sum formula for expectation, applied to the nonnegative random variable

∣ Y_{1} ∣

, gives

\sum_{n = 1}^{\infty} P (∣ Y_{n} ∣ > M n) = \sum_{n = 1}^{\infty} P (∣ Y_{1} ∣ > M n) = \infty,

(276)

because

E ∣ Y_{1} ∣ = \infty

. The events

{∣ Y_{n} ∣ > M n}

are independent, so the second Borel--Cantelli lemma shows that

∣ Y_{n} ∣/ n > M

infinitely often with probability one. Taking the countable intersection over

M \in N

yields

lim sup_{n \to \infty} \frac{∣ Y _{n} ∣}{n} = \infty

(277)

almost surely.

Now

Y_{n} = S_{n} - S_{n - 1}

, and the triangle inequality gives

\frac{∣ Y _{n} ∣}{n} \leq \frac{∣ S _{n} ∣}{n} + \frac{n - 1}{n} \frac{∣ S _{n - 1} ∣}{n - 1} .

(278)

If the limit superior of

∣ S_{n} ∣/ n

were finite, the right-hand side would have finite limit superior, contradicting the preceding conclusion. Therefore

lim sup_{n \to \infty} \frac{∣ Y _{1} + \dots + Y _{n} ∣}{n} = \infty

(279)

almost surely.

Solved by gpt-5.6-sol high.

b

Words: 63 Articles: 1

Solution

Words: 63

Let

(Y_{n})

be independent and identically distributed random variables with the standard Cauchy distribution, whose probability density function is

f (y) = \frac{1}{π ( 1 + y ^{2} )} .

(280)

Its first absolute moment diverges since

E ∣ Y_{1} ∣ = \frac{2}{π} \int_{0}^{\infty} \frac{y}{1 + y ^{2}} d y = \infty.

(281)

On the other hand, its characteristic function is

φ (t) = e^{- ∣ t ∣}

. The characteristic function of a sum of independent variables therefore gives

E exp (i t \frac{S _{n}}{n}) = φ (t / n)^{n} = e^{- ∣ t ∣} .

(282)

Thus the stability of the Cauchy distribution implies that

S_{n} / n

has the standard Cauchy distribution for every

n

. In particular,

\frac{Y _{1} + \dots + Y _{n}}{n} d Z,

(283)

where

Z

is standard Cauchy and

d

denotes convergence in distribution.

Solved by gpt-5.6-sol high.

27J (Applied Probability)

Words: 247 Articles: 6

a

Words: 52 Articles: 1

Solution

Words: 52

A Non-homogeneous Poisson point process with intensity function

λ

is a random counting measure

N

such that, for every finite family of pairwise disjoint bounded Borel sets

A_{1}, \dots, A_{k}

, the counts

N (A_{1}), \dots, N (A_{k})

are independent random variables and

N (A_{i}) \sim Poisson (\int_{A_{i}} λ (x) d x) .

(284)

The assumed local integrability makes each Poisson parameter finite. This count law and independence characterize the process.

Solved by gpt-5.6-sol high.

b

Words: 102 Articles: 1

Solution

Words: 102

Let

N

be the homogeneous Poisson point process in

R^{3}

with intensity measure

λ d x

, and map each point

x

to its radius

g (x) = ∣ x ∣

. For an interval

(a, b] \subseteq R_{+}

, the pushforward measure of the intensity is

λ vol {x : a < ∣ x ∣ \leq b} = \frac{4 πλ}{3} (b^{3} - a^{3}) .

(285)

The Mapping theorem for Poisson point processes therefore shows that the radii form a Non-homogeneous Poisson point process with intensity function

ρ (r) = \frac{d}{d r} (\frac{4 πλ r ^{3}}{3}) = 4 πλ r^{2}, r > 0.

(286)

R

is the distance to the closest star, the event

{R > r}

says that the ball of radius

r

contains no points. The zero-count probability of a Poisson distribution gives

P (R > r) = exp (- \frac{4 πλ r ^{3}}{3}) .

(287)

Differentiating its cumulative distribution function yields the probability density function

f_{R} (r) = 4 πλ r^{2} exp (- \frac{4 πλ r ^{3}}{3}), r \geq 0.

(288)

Solved by gpt-5.6-sol high.

c

Words: 93 Articles: 1

Solution

Words: 93

For one visitor, let

T_{0} = 0, T_{j} = Z_{1} + \dots + Z_{j},

(289)

and define

p_{j} (u) = P (T_{j - 1} \leq u < T_{j}), 1 \leq j \leq 10.

(290)

A visitor arriving at time

s \leq t

is in room

j

at time

t

exactly when its independent service-time mark satisfies

T_{j - 1} \leq t - s < T_{j}

. Thus it receives mark

j

with probability

p_{j} (t - s)

; a further mark records that it has already left the gallery.

The Independent marking theorem for Poisson point processes makes the point processes carrying the different room marks independent. Consequently

V_{1} (t), \dots, V_{10} (t)

are independent, and Poisson thinning gives

V_{j} (t) \sim Poisson (m_{j} (t)), m_{j} (t) = \int_{0}^{t} λ (s) p_{j} (t - s) d s .

(291)

Equivalently,

m_{j} (t) = \int_{0}^{t} λ (s) P (Z_{1} + \dots + Z_{j - 1} \leq t - s < Z_{1} + \dots + Z_{j}) d s .

(292)

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28K (Principles of Statistics)

Words: 237 Articles: 6

a

Words: 62 Articles: 1

Solution

Words: 62

Assume that

f (x) = 0

wherever

h (x) = 0

on the part of the space relevant to

g

. The importance sampling algorithm draws independent and identically distributed random variables

Y_{1}, \dots, Y_{m}

from

h

, forms the weights

w (Y_{i}) = \frac{f ( Y _{i} )}{h ( Y _{i} )},

(293)

and returns

μ_{m} = \frac{1}{m} \sum_{i = 1}^{m} g (Y_{i}) w (Y_{i}) .

(294)

A change of measure gives

E_{h} [g (Y) w (Y)] = \int_{X} g (y) \frac{f ( y )}{h ( y )} h (y) d y = E_{f} [g (X)] .

(295)

Hence the estimator is unbiased whenever the expectation exists, and, if

E_{h} ∣ g (Y) w (Y) ∣ < \infty

, the strong law of large numbers gives

μ_{m} \to E_{f} [g (X)]

almost surely.

Solved by gpt-5.6-sol high.

b

Words: 61 Articles: 1

Solution

Words: 61

Conditionally on

Y_{1}, \dots, Y_{m}

, the cumulative distribution function of

X^{*}

is the weighted empirical distribution

F_{m}^{*} (t) = P (X^{*} \leq t ∣ Y_{1}, \dots, Y_{m}) = \frac{m ^{- 1} \sum _{i = 1}^{m} w ( Y _{i} ) 1 _{{Y_{i} \leq t}}}{m ^{- 1} \sum _{i = 1}^{m} w ( Y _{i} )}, w (y) = \frac{f ( y )}{h ( y )} .

(296)

For fixed

t

, the strong law of large numbers gives, almost surely,

\frac{1}{m} \sum_{i = 1}^{m} w (Y_{i}) 1_{{Y_{i} \leq t}} ⟶ \int_{(- \infty, t]} f (y) d y = F (t),

(297)

while

\frac{1}{m} \sum_{i = 1}^{m} w (Y_{i}) ⟶ \int f (y) d y = 1.

(298)

The ratio therefore converges almost surely to

F (t)

. This is precisely the claimed convergence in distribution of the self-normalized importance sampling resample.

Solved by gpt-5.6-sol high.

c

Words: 114 Articles: 1

Solution

Words: 114

Apply part (b) simultaneously at every rational

t

; the intersection of these probability-one events still has probability one. On that event, monotonicity and continuity extend the convergence from the rationals to every real

t

More directly, fix

ε > 0

. Since

F

is a continuous cumulative distribution function, choose rational points

t_{0} < t_{1} < \dots < t_{k}

(299)

so that the two tails have

F (t_{0}) < ε

and

1 - F (t_{k}) < ε

, while every increment

F (t_{i}) - F (t_{i - 1}) < ε

. Part (b) makes

∣ F_{m}^{*} (t_{i}) - F (t_{i}) ∣ < ε

at all these finitely many points for all sufficiently large

m

. If

t_{i - 1} \leq t \leq t_{i}

, monotonicity gives

F_{m}^{*} (t_{i - 1}) \leq F_{m}^{*} (t) \leq F_{m}^{*} (t_{i}),

(300)

∣ F_{m}^{*} (t) - F (t) ∣ < 2 ε

; the same estimate follows in the tails. This proves the uniform convergence of distribution functions to a continuous limit:

sup_{t \in R} ∣ F_{m}^{*} (t) - F (t) ∣ ⟶ 0

(301)

almost surely.

Solved by gpt-5.6-sol high.

29K (Stochastic Financial Models)

Words: 325 Articles: 8

a

Words: 129 Articles: 1

Solution

Words: 129

The stock tree has

S_{0} = 5

, first-period values

6, 4

, and terminal successors

7, 5

from

6

and

5, 3

from

4

. When

r = 1/6

, the risk-free growth factor is

R = 7/6

. At the upper time-one node, investing

6

in the risk-free asset produces

7

at time two, while one share bought for

6

produces only

7

5

Use the following predictable self-financing portfolio. Hold nothing initially. If

S_{1} = 6

, short one share and invest the proceeds

6

in the risk-free asset; if

S_{1} = 4

, continue to hold nothing. Its terminal payoff is

7 - S_{2} = {0, 2, S_{2} = 7, S_{2} = 5

(302)

on the upper branch, and zero on the lower branch. It costs zero, is never negative, and is positive with positive probability, so it is an arbitrage.

Solved by gpt-5.6-sol high.

b

Words: 63 Articles: 1

Solution

Words: 63

Now

R = 1 + r = 9/8

. The local risk-neutral up probabilities solve

RS = q S_{u} + (1 - q) S_{d}

. They are

q_{0} = \frac{( 9/8 ) 5 - 4}{6 - 4} = \frac{13}{16}, q_{u} = \frac{( 9/8 ) 6 - 5}{7 - 5} = \frac{7}{8}, q_{d} = \frac{( 9/8 ) 4 - 3}{5 - 3} = \frac{3}{4} .

(303)

The terminal payoff of the European put option with strike

5

0, 0, 2

at stock prices

7, 5, 3

. Backward option pricing gives time-one values

V_{u} = 0, V_{d} = R^{- 1} (\frac{1}{4} \cdot 2) = \frac{4}{9},

(304)

and hence

V_{0} = R^{- 1} (\frac{3}{16} \frac{4}{9}) = \frac{2}{27} .

(305)

The first-period stock holding in the replicating portfolio in a binomial market is

Δ_{0} = \frac{V _{u} - V _{d}}{6 - 4} = - \frac{2}{9} .

(306)

Thus the hedge initially shorts

2/9

of a share.

Solved by gpt-5.6-sol high.

c

Words: 50 Articles: 1

Solution

Words: 50

The European call option with strike

5

pays

2, 0, 0

at terminal stock prices

7, 5, 3

. Its lower-node value is zero, while its upper-node value is

V_{u} = R^{- 1} (\frac{7}{8} \cdot 2) = \frac{14}{9} .

(307)

A second step of backward option pricing gives

V_{0} = R^{- 1} \frac{13}{16} \frac{14}{9} = \frac{91}{81} .

(308)

By the fundamental theorem of asset pricing, this discounted expectation under the unique risk-neutral probabilities is the unique no-arbitrage price.

Solved by gpt-5.6-sol high.

d

Words: 83 Articles: 1

Solution

Words: 83

For the American put option, compare exercise with continuation at time one. At

S_{1} = 6

, both values are zero. At

S_{1} = 4

, immediate exercise pays

1

, whereas the continuation value is the European value

4/9

, so exercise is optimal. At time zero, immediate exercise pays zero and continuation is

R^{- 1} (\frac{13}{16} \cdot 0 + \frac{3}{16} \cdot 1) = \frac{1}{6} .

(309)

The optimal policy is therefore to continue at time zero, exercise at time one if

S_{1} = 4

, and otherwise continue to maturity. This nodewise maximum is the Snell envelope value.

Solved by gpt-5.6-sol high.

30J (Mathematics of Machine Learning)

Words: 317 Articles: 8

a

Words: 69 Articles: 1

Solution

Words: 69

A regression tree chooses a threshold

s

and constant predictions

c_{L}, c_{R}

to minimize the squared-error criterion

min_{s, c_{L}, c_{R}} {\sum_{i : X_{i} \leq s} (Y_{i} - c_{L})^{2} + \sum_{i : X_{i} > s} (Y_{i} - c_{R})^{2}},

(310)

where

s

ranges over thresholds producing two nonempty groups. For a fixed split, the least-squares minimizers are the two sample means,

c_{L} = \frac{1}{n _{L}} \sum_{X_{i} \leq s} Y_{i}, c_{R} = \frac{1}{n _{R}} \sum_{X_{i} > s} Y_{i} .

(311)

Thus the first split is the candidate threshold with the smallest sum of the two within-node residual sums of squares.

Solved by gpt-5.6-sol high.

b

Words: 80 Articles: 1

Solution

Words: 80

With

X_{1} < \dots < X_{n}

, it suffices to put the threshold between

X_{m}

and

X_{m + 1}

for

m = 1, \dots, n - 1

. Form the prefix sums

A_{m} = \sum_{i = 1}^{m} Y_{i}, B_{m} = \sum_{i = 1}^{m} Y_{i}^{2} .

(312)

They take

O (n)

operations to compute. The minimized squared errors for the two children are

L_{m} = B_{m} - \frac{A _{m}^{2}}{m},

(313)

and

R_{m} = (B_{n} - B_{m}) - \frac{( A _{n} - A _{m} ) ^{2}}{n - m} .

(314)

Each candidate value

L_{m} + R_{m}

therefore takes constant time once the prefix sums are available. Scanning all

n - 1

candidates takes

O (n)

operations in the sense of Big O notation, and retaining the minimizing

m

gives the optimal split.

Solved by gpt-5.6-sol high.

c

Words: 42 Articles: 1

Solution

Words: 42

Write

\overset{ˉ}{X}_{m} = \frac{1}{m} \sum_{i = 1}^{m} X_{i}, \overset{ˉ}{Y}_{m} = \frac{1}{m} \sum_{i = 1}^{m} Y_{i} .

(315)

The ordinary least squares estimators give

f_{m} (x) = α_{m} + β_{m} x = \overset{ˉ}{Y}_{m} + β_{m} (x - \overset{ˉ}{X}_{m}),

(316)

where

β_{m} = \frac{\sum _{i = 1}^{m} ( X _{i} - X ˉ _{m} ) ( Y _{i} - Y ˉ _{m} )}{\sum _{i = 1}^{m} ( X _{i} - X ˉ _{m} ) ^{2}} = \frac{\sum _{i = 1}^{m} X _{i} Y _{i} - m X ˉ _{m} Y ˉ _{m}}{\sum _{i = 1}^{m} X _{i}^{2} - m X ˉ _{m}^{2}} .

(317)

The denominator is positive because the sorted inputs are distinct and

m \geq 2

Solved by gpt-5.6-sol high.

d

Words: 126 Articles: 1

Solution

Words: 126

For each

m

, the displayed objective is the sum of the minimized linear-regression residual sums of squares on the prefix

1 : m

and suffix

(m + 1) : n

. Precompute prefix sums of

X_{i}, q u a d Y_{i}, q u a d X_{i}^{2}, q u a d X_{i} Y_{i}, q u a d Y_{i}^{2} .

(318)

This requires

O (n)

operations. The corresponding five sums over a suffix are obtained by subtracting a prefix sum from the total.

For any block of

k

observations, these raw sums give in constant time

S_{xx} = \sum x_{i}^{2} - \frac{( \sum x _{i} ) ^{2}}{k}, S_{x y} = \sum x_{i} y_{i} - \frac{( \sum x _{i} ) ( \sum y _{i} )}{k}, S_{yy} = \sum y_{i}^{2} - \frac{( \sum y _{i} ) ^{2}}{k} .

(319)

By the formula for the residual sum of squares in simple linear regression, its minimized loss is

S_{yy} - \frac{S _{x y}^{2}}{S _{xx}} .

(320)

Hence both losses for a proposed

m

take constant time. Evaluating their sum for

m = 2, \dots, n - 2

and keeping the minimum is another

O (n)

scan, so the complete minimization uses

O (n)

computations.

Solved by gpt-5.6-sol high.

31E (Asymptotic Methods)

Words: 174 Articles: 1

Solution

Words: 174

Write the phase and amplitude as

ϕ (z) = cosh z, f (z) = \frac{1}{z ^{2} + 16} .

(321)

The saddle point relevant to the contour is

z = 0

, since

ϕ^{'} (z) = sinh z

and

ϕ (0) = 1, ϕ^{''} (0) = 1.

(322)

The local steepest-descent direction is therefore vertical: with

z = i y

ϕ (i y) = cos y = 1 - \frac{y ^{2}}{2} + O (y^{4}) .

(323)

The given contour approaches the lines

Im z = - π

at its left end and

Im z = π

at its right end. The poles of

f

are at

z = \pm 4 i

, outside the closed strip

∣ Im z ∣ \leq π

. The Cauchy integral theorem therefore permits deformation to the contour consisting of the lower horizontal ray from

- \infty - iπ

- iπ

, the vertical segment from

- iπ

iπ

, and the upper horizontal ray from

iπ

\infty + iπ

. On either horizontal ray,

cosh (p \pm iπ) = - cosh p,

(324)

so those two integrals are exponentially small, of order

e^{- x} x^{- 1/2}

The vertical segment, oriented upward, contributes

i \int_{- π}^{π} \frac{e ^{x c o s y}}{16 - y ^{2}} d y .

(325)

Its real phase has a unique maximum at

y = 0

. Since

\frac{1}{16 - y ^{2}} = \frac{1}{16} + O (y^{2}),

(326)

the simple-saddle contribution in steepest descent, equivalently the local Gaussian integral, gives

I (x) \sim \frac{i e ^{x}}{16} \int_{- \infty}^{\infty} e^{- x y^{2} /2} d y = \frac{i e ^{x}}{16} \frac{2 π}{x} (x \to \infty) .

(327)

The factor

i

comes from the upward tangent

d z = i d y

of the deformed contour.

Solved by gpt-5.6-sol high.

32A (Dynamical Systems)

Words: 341 Articles: 10

i

Words: 44 Articles: 1

Solution

Words: 44

For

x \in [0, 1]

, the nonnegative function

x (1 - x^{2})

has derivative

1 - 3 x^{2}

, so its maximum occurs at

x = 1/ 3

and equals

\frac{1}{3} (1 - \frac{1}{3}) = \frac{2}{3 3} .

(328)

The map preserves the interval exactly when this maximum is at most one:

λ \frac{2}{3 3} \leq 1.

(329)

Thus the upper limit is

λ_{m a x} = \frac{3 3}{2} .

(330)

Solved by gpt-5.6-sol high.

ii

Words: 50 Articles: 1

Solution

Words: 50

A fixed point satisfies

x = λ x (1 - x^{2}),

(331)

so either

x = 0

x^{2} = 1 - 1/ λ

. When

0 < λ < 1

, the latter equation has no real solution, and the origin is the only fixed point. Since

F_{x} (x, λ) = λ (1 - 3 x^{2}), F_{x} (0, λ) = λ,

(332)

the criterion for fixed point stability for an iteration shows that the origin is stable for

λ < 1

Solved by gpt-5.6-sol high.

iii

Words: 61 Articles: 1

Solution

Words: 61

For

λ > 1

, the new fixed point in

[0, 1]

x^{*} = 1 - \frac{1}{λ} .

(333)

The origin has multiplier

F_{x} (0, λ) = λ > 1

and is therefore unstable. At the new fixed point,

F_{x} (x^{*}, λ) = λ (1 - 3 (1 - \frac{1}{λ})) = 3 - 2 λ .

(334)

Hence

∣ F_{x} (x^{*}, λ) ∣ < 1 ⟺ 1 < λ < 2.

(335)

The point

x^{*}

is stable in that interval and unstable for

λ > 2

. At

λ = 2

its multiplier is

- 1

, the threshold for a period-doubling bifurcation.

Solved by gpt-5.6-sol high.

iv

Words: 115 Articles: 1

Solution

Words: 115

Let

x^{*} (λ) = 1 - 1/ λ

and center the map at this parameter-dependent fixed point:

G (X, μ) = F (x^{*} (2 + μ) + X, 2 + μ) - x^{*} (2 + μ) .

(336)

A Taylor polynomial in

X

, with the coefficients expanded at

(x, λ) = (1/ 2, 2)

, gives

G (X, μ) = - X + αμ X + β X^{2} + γ X^{3} + O (μ^{2})

(337)

under the period-doubling scaling

X = O (μ)

. The coefficients are

α = \frac{d}{d λ} F_{x} (x^{*} (λ), λ)_{λ = 2} = - 2,

(338)

β = \frac{1}{2} F_{xx} (\frac{1}{2}, 2) = - 32, γ = \frac{1}{6} F_{xxx} (\frac{1}{2}, 2) = - 2.

(339)

Here we used

F_{x} (x^{*} (λ), λ) = 3 - 2 λ

F_{xx} = - 6 λ x

, and

F_{xxx} = - 6 λ

Composing the local map with itself and retaining the terms of order

μ X

and

X^{3}

gives

G (G (X, μ), μ) - X = - 2 αμ X - 2 (γ + β^{2}) X^{3} + o (μ X + X^{3}) .

(340)

Besides the fixed solution

X = 0

, the two leading solutions therefore satisfy

X^{2} = - \frac{αμ}{γ + β ^{2}} .

(341)

Thus the period-two points born at

λ = 2

are

x_{\pm} = x^{*} \pm \frac{- αμ}{γ + β ^{2}}

(342)

to leading order. With the coefficients above, the radicand is positive for

μ > 0

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v

Words: 71 Articles: 1

Solution

Words: 71

The exact second-iterate equation factors as

F (F (x)) - x = x (λ x^{2} - λ - 1) (λ x^{2} - λ + 1) (λ^{2} x^{4} - λ^{2} x^{2} + 1) .

(343)

The positive nonfixed period-two points come from the final factor. If

y = x^{2}

, their squared values are

y_{\pm} = \frac{1 \pm 1 - 4/ λ ^{2}}{2}, y_{+} + y_{-} = 1, y_{+} y_{-} = \frac{1}{λ ^{2}} .

(344)

They are real and distinct for

λ > 2

The multiplier of a periodic orbit of an iteration is

F^{'} (x_{+}) F^{'} (x_{-}) = λ^{2} (1 - 3 y_{+}) (1 - 3 y_{-}) = λ^{2} (1 - 3 (y_{+} + y_{-}) + 9 y_{+} y_{-}) = 9 - 2 λ^{2} .

(345)

For

λ = 2 + μ

with

0 < μ ≪ 1

9 - 2 λ^{2} = 1 - 8 μ + O (μ^{2}),

(346)

whose modulus is less than one. The new two-cycle is therefore locally asymptotically stable.

Solved by gpt-5.6-sol high.

33B (Principles of Quantum Mechanics)

Words: 368 Articles: 13

a

Words: 200 Articles: 8

i

Words: 32 Articles: 1

Solution

Words: 32

On the tensor-product Hilbert space, addition of angular momentum gives

J = J_{1} \otimes I_{2} + I_{1} \otimes J_{2} .

(347)

Suppressing the identity operators, its components and ladder operators are

J_{x} = J_{1 x} + J_{2 x}, J_{y} = J_{1 y} + J_{2 y}, J_{z} = J_{1 z} + J_{2 z},

(348)

J_{\pm} = J_{x} \pm i J_{y} = J_{1 \pm} + J_{2 \pm} .

(349)

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ii

Words: 17 Articles: 1

Solution

Words: 17

The Clebsch-Gordan decomposition permits

j = ∣ j_{1} - j_{2} ∣, ∣ j_{1} - j_{2} ∣ + 1, \dots, j_{1} + j_{2} .

(350)

Thus a total-angular-momentum singlet

j = 0

occurs exactly when

j_{1} = j_{2}

Solved by gpt-5.6-sol high.

iii

Words: 77 Articles: 1

Solution

Words: 77

Put

J = j_{1} + j_{2}

. The only state with

m = J

is the maximal highest-weight state

∣ J, J ⟩ = ∣ j_{1}, j_{1} ⟩_{1} ∣ j_{2}, j_{2} ⟩_{2} .

(351)

Applying

J_{-} = J_{1 -} + J_{2 -}

and normalizing gives

∣ J, J - 1 ⟩ = \frac{j _{1}}{J}, ∣ j_{1}, j_{1} - 1 ⟩_{1} ∣ j_{2}, j_{2} ⟩_{2} + \frac{j _{2}}{J}, ∣ j_{1}, j_{1} ⟩_{1} ∣ j_{2}, j_{2} - 1 ⟩_{2} .

(352)

The orthonormal vector in the same two-dimensional

m = J - 1

subspace is

∣ J - 1, J - 1 ⟩ = \frac{j _{2}}{J}, ∣ j_{1}, j_{1} - 1 ⟩_{1} ∣ j_{2}, j_{2} ⟩_{2} - \frac{j _{1}}{J}, ∣ j_{1}, j_{1} ⟩_{1} ∣ j_{2}, j_{2} - 1 ⟩_{2} .

(353)

It is annihilated by

J_{+}

, so it is the highest-weight state of the

j = J - 1

multiplet. These are all states with

m \geq J - 1

. The last state exists when both subsystem spins are nonzero.

Solved by gpt-5.6-sol high.

iv

Words: 74 Articles: 1

Solution

Words: 74

A product state factors as

∣ ψ_{1} ⟩ \otimes ∣ ψ_{2} ⟩

; an entangled state admits no such factorization. The state

∣ J, J ⟩

is a product state. When

j_{1}, j_{2} > 0

, each state with

m = J - 1

displayed in part (iii) has two nonzero terms with orthogonal states in both subsystems, so its Schmidt rank is two and it is entangled. If one subsystem has spin zero, only the

j = J

multiplet remains and its states factor through the unique state of that subsystem.

Solved by gpt-5.6-sol high.

b

Words: 80 Articles: 1

Solution

Words: 80

For

∣ J, J ⟩

, tracing out subsystem 2 gives the reduced density matrix

ρ_{1} = ∣ j_{1}, j_{1} ⟩ ⟨ j_{1}, j_{1} ∣,

(354)

which is pure, so its entanglement entropy is zero.

For

∣ J, J - 1 ⟩

, the two subsystem-2 factors in part (iii) are orthogonal. The partial trace therefore removes the cross terms and gives

ρ_{1} = \frac{j _{1}}{J} ∣ j_{1}, j_{1} - 1 ⟩ ⟨ j_{1}, j_{1} - 1∣ + \frac{j _{2}}{J} ∣ j_{1}, j_{1} ⟩ ⟨ j_{1}, j_{1} ∣.

(355)

Its nonzero eigenvalues are

j_{1} / J

and

j_{2} / J

, so the Von Neumann entropy is

S = - \frac{j _{1}}{J} lo g \frac{j _{1}}{J} - \frac{j _{2}}{J} lo g \frac{j _{2}}{J} .

(356)

When

j_{1} = j_{2}

, both eigenvalues are

1/2

and

S = lo g 2

, the maximal entropy for a rank-two reduced state.

Solved by gpt-5.6-sol high.

c

Words: 88 Articles: 1

Solution

Words: 88

j = 0

state exists only when

j_{1} = j_{2} =: j

. Since its total magnetic quantum number is zero, every uncoupled basis term must have

m_{1} + m_{2} = 0

, so

∣0, 0 ⟩ = \sum_{m = - j}^{j} α_{m} ∣ j, m ⟩_{1} ∣ j, - m ⟩_{2} .

(357)

Apply the total raising operator

J_{+} = J_{1 +} + J_{2 +}

. The coefficient of

∣ j, m + 1 ⟩_{1} ∣ j, - m ⟩_{2}

(j - m) (j + m + 1) (α_{m} + α_{m + 1}) .

(358)

The angular momentum singlet state is annihilated by

J_{+}

, hence

α_{m + 1} = - α_{m}, - j \leq m < j .

(359)

All coefficients consequently have the same modulus and alternating signs. Normalization fixes that modulus to

(2 j + 1)^{- 1/2}

, and an arbitrary overall phase may be chosen so that

α_{m} = \frac{( - 1 ) ^{j - m}}{2 j + 1} .

(360)

Thus

∣0, 0 ⟩ = \frac{1}{2 j + 1} \sum_{m = - j}^{j} (- 1)^{j - m} ∣ j, m ⟩_{1} ∣ j, - m ⟩_{2} .

(361)

Solved by gpt-5.6-sol high.

34D (Applications of Quantum Mechanics)

Words: 258 Articles: 4

a

Words: 100 Articles: 1

Solution

Words: 100

Take

B > 0

and use the Landau gauge

A = (0, B x, 0)

. With zero

z

-momentum, the Hamiltonian becomes

H = \frac{1}{2 m} [p_{x}^{2} + (p_{y} + e B x)^{2}] .

(362)

For a state

e^{i k_{y} y} χ (x)

, this is a quantum harmonic oscillator in

x

, centred at

x_{0} = - \frac{ℏ k _{y}}{e B},

(363)

with cyclotron frequency

ω_{c} = \frac{∣ e B ∣}{m} .

(364)

The Landau levels are therefore

E_{n} = ℏ ω_{c} (n + \frac{1}{2}), n = 0, 1, 2, \dots .

(365)

Periodicity in

y

quantizes

k_{y} = 2 π ℓ / R_{y}

. Requiring the corresponding oscillator centre to lie in an

x

interval of length

R_{x}

leaves

D = \frac{∣ e B ∣ R _{x} R _{y}}{2 π ℏ}

(366)

allowed centres for each

n

. This is the degeneracy of a Landau level, equal to the magnetic flux through the rectangle divided by the magnetic flux quantum

h /∣ e ∣

. Consistent magnetic periodic boundary conditions make

D

an integer.

Solved by gpt-5.6-sol high.

b

Words: 158 Articles: 1

Solution

Words: 158

Assume

e B > 0

; reversing the sign only exchanges the spin labels. Since

B \cdot σ = B σ_{3}

, the Pauli term shifts the spinless energies by

\pm ℏ ω_{c} /2

. Thus

E_{n, σ} = ℏ ω_{c} (n + \frac{1}{2} + \frac{σ}{2}), σ = \pm 1.

(367)

After collecting equal energies, the spin splitting of Landau levels is

E_{ℓ} = ℓ ℏ ω_{c}, ℓ = 0, 1, 2, \dots .

(368)

The level

ℓ = 0

contains only

(n, s i g ma) = (0, - 1)

and has degeneracy

D

. Every level

ℓ \geq 1

contains

(ℓ, - 1)

and

(ℓ - 1, + 1)

and has degeneracy

2 D

For

N

noninteracting electrons, the Pauli exclusion principle fills these states from the bottom. If

0 \leq N \leq D

, every electron fits in the zero-energy level and

E_{gs} (N) = 0.

(369)

For

N > D

, write

N - D = 2 D k + r, k \in Z_{\geq 0}, 0 \leq r < 2 D .

(370)

Then the first

k

positive levels are full and

r

states in level

k + 1

are occupied. The ground-state energy is

E_{gs} (N) = ℏ ω_{c} [D k (k + 1) + (k + 1) r] .

(371)

Equivalently, on

D (2 k + 1) \leq N \leq D (2 k + 3),

(372)

E_{gs} (N) = D ℏ ω_{c} k (k + 1) + (k + 1) ℏ ω_{c} [N - D (2 k + 1)] .

(373)

The graph is continuous and piecewise linear: it is flat up to

N = D

, then has slopes

ℏ ω_{c}, 2ℏ ω_{c}, 3ℏ ω_{c}, \dots

, with kinks at

N = D, 3 D, 5 D, \dots .

(374)

Solved by gpt-5.6-sol high.

35A (Statistical Physics)

Words: 449 Articles: 11

a

Words: 170 Articles: 1

Solution

Words: 170

The Clausius statement of the second law says that no cyclic device can have as its sole effect the transfer of heat from a colder reservoir to a hotter one. The Kelvin-Planck statement of the second law says that no cyclic device can have as its sole effect the extraction of heat from one reservoir and its complete conversion into work.

Suppose first that the Clausius statement fails. Use the forbidden device to return to a hot reservoir the heat

Q_{C}

rejected by an ordinary heat engine. The combined device extracts

Q_{H} - Q_{C}

from the hot reservoir and converts all of it into work, violating the Kelvin--Planck statement.

Conversely, suppose the Kelvin--Planck statement fails. Use the work produced by the forbidden one-reservoir engine to drive an ordinary refrigerator. The engine's heat withdrawal and the refrigerator's heat discharge cancel by the work-energy balance, leaving only a transfer of heat from the cold reservoir to the hot reservoir with no external work. This violates the Clausius statement. Each statement therefore implies the other, so the two forms of the Second law of thermodynamics are equivalent.

Solved by gpt-5.6-sol high.

b

Words: 33 Articles: 1

Solution

Words: 33

For

N

particles, the ideal gas equation of state is

p V = N k_{B} T .

(375)

The internal energy of an ideal gas depends only on temperature and obeys

d E = C_{V} d T .

(376)

For constant heat capacity and a convenient choice of energy zero,

E = C_{V} T .

(377)

Solved by gpt-5.6-sol high.

c

Words: 51 Articles: 1

Solution

Words: 51

An adiabatic process has no heat transfer,

δ Q = 0

. With work done by the gas taken as positive, the first law of thermodynamics gives

d E = δ Q - p d V = - p d V .

(378)

Using

d E = C_{V} d T

and

p = N k_{B} T / V

C_{V} \frac{d T}{T} = - N k_{B} \frac{d V}{V} .

(379)

Integration, followed by the Mayer relation

N k_{B} / C_{V} = γ - 1

, yields

T V^{γ - 1} = constant .

(380)

Eliminating

T

with the equation of state gives the reversible ideal-gas adiabat

p V^{γ} = constant, γ = \frac{C _{p}}{C _{V}} .

(381)

Solved by gpt-5.6-sol high.

d

Words: 195 Articles: 4

i

Words: 140 Articles: 1

Solution

Words: 140

This is the ideal Diesel cycle. In the

(p, V)

plane,

A \to B

is an adiabatic curve running up and left,

B \to C

is a horizontal constant-pressure segment running right,

C \to D

is an adiabatic curve running down and right, and

D \to A

is a vertical constant-volume segment running down. The traversal is clockwise, so the enclosed area is positive work output.

In the

(T, S)

plane, the two reversible adiabats are vertical because entropy is constant:

A \to B

rises at

S_{A} = S_{B}

, and

C \to D

falls at

S_{C} = S_{D}

. Along

B \to C

, the heat increment at constant pressure is

δ Q = C_{p} d T = T d S,

(382)

S - S_{B} = C_{p} lo g \frac{T}{T _{B}}, T = T_{B} e^{(S - S_{B}) / C_{p}} .

(383)

Along the constant-volume curve

D \to A

δ Q = C_{V} d T = T d S,

(384)

which gives

S - S_{A} = C_{V} lo g \frac{T}{T _{A}}, T = T_{A} e^{(S - S_{A}) / C_{V}} .

(385)

Thus

B \to C

runs up and right and

D \to A

runs down and left between the two vertical adiabats.

Solved by gpt-5.6-sol high.

ii

Words: 55 Articles: 1

Solution

Words: 55

Heat enters only during the constant-pressure step, so the heat capacity at constant pressure gives

Q_{1} = C_{p} (T_{C} - T_{B}) .

(386)

The positive magnitude of the heat rejected during constant-volume cooling is

Q_{2} = C_{V} (T_{D} - T_{A}) .

(387)

Because the working gas returns to its initial state, the thermodynamic cycle has zero net change in internal energy and

W = Q_{1} - Q_{2}

. Its thermal efficiency is therefore

η = \frac{W}{Q _{1}} = 1 - \frac{C _{V} ( T _{D} - T _{A} )}{C _{p} ( T _{C} - T _{B} )} = 1 - \frac{1}{γ} \frac{T _{D} - T _{A}}{T _{C} - T _{B}} .

(388)

Solved by gpt-5.6-sol high.

36A (Electrodynamics)

Words: 319 Articles: 10

a

Words: 34 Articles: 1

Solution

Words: 34

The macroscopic Maxwell equations in matter are

\nabla \cdot D = ρ, \nabla \cdot B = 0,

(389)

\nabla \times E = - \frac{\partial B}{\partial t}, \nabla \times H = ma t hb fJ + \frac{\partial D}{\partial t} .

(390)

Here

ρ

and

J

are the free charge and current densities.

Solved by gpt-5.6-sol high.

b

Words: 70 Articles: 1

Solution

Words: 70

Dot the Ampère-Maxwell equation with

E

and Faraday's law with

H

E \cdot \frac{\partial D}{\partial t} = E \cdot (\nabla \times H) - E \cdot J,

(391)

H \cdot \frac{\partial B}{\partial t} = - H \cdot (\nabla \times E) .

(392)

Adding and using the vector identity

\nabla \cdot (E \times H) = H \cdot (\nabla \times E) - E \cdot (\nabla \times H)

(393)

gives

E \cdot \frac{\partial D}{\partial t} + H \cdot \frac{\partial B}{\partial t} + \nabla \cdot (E \times H) = - E \cdot J .

(394)

This is the local Poynting theorem before specializing the constitutive relations.

Solved by gpt-5.6-sol high.

c

Words: 75 Articles: 1

Solution

Words: 75

Because

ε_{ij}

and

μ_{ij}

are symmetric and time-independent, the linear anisotropic dielectric relations give

E_{i} \partial_{t} D_{i} = ε_{ij} E_{i} \partial_{t} E_{j} = \frac{1}{2} \partial_{t} (ε_{ij} E_{i} E_{j}),

(395)

and similarly

H_{i} \partial_{t} B_{i} = \frac{1}{2} \partial_{t} (μ_{ij} H_{i} H_{j}) .

(396)

Integrate the identity from part (b) over

V

and apply the divergence theorem to obtain

\frac{\partial}{\partial t} \int_{V} \frac{1}{2} (ε_{ij} E_{i} E_{j} + μ_{ij} H_{i} H_{j}) d V + \int_{S} (E \times H) \cdot d S = - \int_{V} E \cdot J d V .

(397)

When

J = 0

, this is conservation of electromagnetic energy. The Poynting theorem in a linear anisotropic medium identifies

u = \frac{1}{2} (ε_{ij} E_{i} E_{j} + μ_{ij} H_{i} H_{j})

(398)

as the energy density and

N = E \times H

(399)

as the energy flux, or Poynting vector.

Solved by gpt-5.6-sol high.

d

Words: 50 Articles: 1

Solution

Words: 50

Substitution of the sinusoidal plane waves into Faraday's law gives

k \times e = ω b .

(400)

The source-free Ampère-Maxwell equation, with

H = B / μ

and

D_{i} = ε_{ij} E_{j}

, gives

k \times \frac{b}{μ} = - ω ε e .

(401)

Eliminating

b = (k \times e) / ω

yields the wave equation for a plane wave in an anisotropic dielectric:

[k \times (k \times e)]_{i} + ω^{2} μ ε_{ij} e_{j} = 0 .

(402)

Solved by gpt-5.6-sol high.

e

Words: 90 Articles: 1

Solution

Words: 90

Suppose

ε_{ij} e_{j} = ε_{e} e_{i}

. Taking the scalar product of the equation in part (d) with

k

gives

k \cdot e = 0

. The vector triple product then reduces the equation to

(- k^{2} + ω^{2} μ ε_{e}) e = 0,

(403)

k^{2} = ω^{2} μ ε_{e}

. From part (d),

H = \frac{B}{μ} = \frac{k \times e}{μ ω} sin (k \cdot x - ω t) .

(404)

Consequently

N = E \times H = \frac{s i n ^{2} ( k \cdot x - ω t )}{μ ω} e \times (k \times e) = \frac{∣ e ∣ ^{2} s i n ^{2} ( k \cdot x - ω t )}{μ ω} k .

(405)

Thus the instantaneous flux, except at its zeros, and its time average both point in the wave-propagation direction

k

Solved by gpt-5.6-sol high.

37B (General Relativity)

Words: 262 Articles: 10

a

Words: 146 Articles: 4

i

Words: 70 Articles: 1

Solution

Words: 70

The Lorenz gauge in linearized gravity gives

0 = \partial^{μ} \overset{ˉ}{h}_{μν} = i k^{μ} H_{μν} e^{ik \cdot x},

(406)

so the symmetric amplitude is transverse:

k^{μ} H_{μν} = 0 .

(407)

Under this condition the supplied linearized Einstein tensor reduces to

G_{μν} = - \frac{1}{2} □ \overset{ˉ}{h}_{μν} .

(408)

The vacuum Einstein field equations and the plane-wave ansatz then imply

k^{ρ} k_{ρ} = 0 .

(409)

Thus the wavevector is null and the perturbation is a plane gravitational wave in linearized gravity. Lorenz gauge alone does not require the trace of

H_{μν}

to vanish; that is a further gauge choice.

Solved by gpt-5.6-sol high.

ii

Words: 76 Articles: 1

Solution

Words: 76

Under an infinitesimal coordinate transformation generated by

ξ_{μ}

, the trace-reversed metric perturbation changes by

δ \overset{ˉ}{h}_{μν} = - \partial_{μ} ξ_{ν} - \partial_{ν} ξ_{μ} + η_{μν} \partial_{ρ} ξ^{ρ} .

(410)

Taking a divergence gives

\partial^{μ} δ \overset{ˉ}{h}_{μν} = - □ ξ_{ν} .

(411)

The transformation therefore preserves Lorenz gauge precisely when

□ ξ_{ν} = 0

. For a plane-wave parameter

ξ_{μ} = a_{μ} e^{ik \cdot x}

, this holds because

k^{2} = 0

. Absorbing a factor of

i

into an arbitrary constant vector

ϵ_{μ}

, the amplitude transformation is

H_{μν} \mapsto H_{μν} + k_{μ} ϵ_{ν} + k_{ν} ϵ_{μ} - η_{μν} k \cdot ϵ .

(412)

This four-parameter freedom is the residual gauge symmetry of linearized gravity.

Solved by gpt-5.6-sol high.

b

Words: 116 Articles: 4

i

Words: 51 Articles: 1

Solution

Words: 51

Use the total mass

M ≃ 60 M_{⊙} ≃ 1.2 \times 1 0^{32} kg .

(413)

With the stated last-orbit estimate,

r = \frac{6 GM}{c ^{2}} ≃ 5 \times 1 0^{5} m .

(414)

The Kepler third law then gives

T = 2 π \frac{r ^{3}}{GM} = 12 π 6 \frac{GM}{c ^{3}} ≃ 3 \times 1 0^{- 2} s .

(415)

A binary's dominant quadrupole radiation has two cycles per orbit, so the last-orbit frequency of a compact binary is

f_{GW} ≃ \frac{2}{T} ≃ 7 \times 1 0^{1} Hz .

(416)

Only one significant figure is meaningful under the approximations.

Solved by gpt-5.6-sol high.

ii

Words: 65 Articles: 1

Solution

Words: 65

A suitable Spatially flat FLRW metric is

d s^{2} = - c^{2} d t^{2} + a^{2} (t) (d x^{2} + d y^{2} + d z^{2}) .

(417)

Both source and detector are comoving, so cosmological redshift stretches the wave period in proportion to the scale factor. If

f_{e}

is the local source frequency at emission and

f_{o}

is the frequency measured in Earth's local frame, then

f_{o} = f_{e} \frac{a ( t _{e} )}{a ( t _{o} )} = \frac{f _{e}}{1 + z}, 1 + z = \frac{a ( t _{o} )}{a ( t _{e} )} .

(418)

The same relation applies to gravitational waves propagating on the FLRW background in the geometric-optics regime.

Solved by gpt-5.6-sol high.

38C (Fluid Dynamics II)

Words: 478 Articles: 12

a

Words: 66 Articles: 1

Solution

Words: 66

The unperturbed fluid wake is the horizontal strip

- h < y < h

: the fluid is stationary there and moves to the right with velocity

U e_{x}

above and below it. Its two shear layers are the straight lines

y = \pm h

. After the symmetric disturbance, draw the upper layer at

y = h + η (x, t)

and the lower one as its reflection at

y = - h - η (x, t)

, so the local wake width is

2 (h + η)

Solved by gpt-5.6-sol high.

b

Words: 89 Articles: 1

Solution

Words: 89

Write the velocity potentials in the upper, middle and lower domains as

ϕ_{+}

ϕ_{0}

and

ϕ_{-}

. Irrotational flow and incompressible flow imply the Laplace equation

\nabla^{2} ϕ_{j} = 0, j \in {+, 0, -} .

(419)

The far-field boundary conditions are

\nabla ϕ_{+} ⟶ U e_{x} (y \to + \infty), \nabla ϕ_{-} ⟶ U e_{x} (y \to - \infty) .

(420)

Every shear layer is a material interface. The exact kinematic boundary condition at

y = h + η

ϕ_{+, y} = η_{t} + ϕ_{+, x} η_{x}, ϕ_{0, y} = η_{t} + ϕ_{0, x} η_{x},

(421)

and at

y = - h - η

it is

ϕ_{-, y} = - η_{t} - ϕ_{-, x} η_{x}, ϕ_{0, y} = - η_{t} - ϕ_{0, x} η_{x} .

(422)

Here each equation is evaluated on the indicated moving interface.

Finally, pressure continuity is the dynamic boundary condition for an inviscid interface. The Unsteady Bernoulli equation gives

ϕ_{+, t} + \frac{1}{2} ∣\nabla ϕ_{+} ∣^{2} = ϕ_{0, t} + \frac{1}{2} ∣\nabla ϕ_{0} ∣^{2} (y = h + η),

(423)

ϕ_{-, t} + \frac{1}{2} ∣\nabla ϕ_{-} ∣^{2} = ϕ_{0, t} + \frac{1}{2} ∣\nabla ϕ_{0} ∣^{2} (y = - h - η),

(424)

after absorbing spatially constant functions of time into the potentials.

Solved by gpt-5.6-sol high.

c

Words: 76 Articles: 1

Solution

Words: 76

The displacement must have small relative amplitude and small slope,

\frac{∣ η ∣}{h} ≪ 1, ∣ η_{x} ∣ ≪ 1.

(425)

Write the potentials as perturbations of the base flow,

ϕ_{\pm} = Ux + φ_{\pm}, ϕ_{0} = φ_{0},

(426)

with

∣\nabla φ_{\pm} ∣/ U

and

∣\nabla φ_{0} ∣/ U

small. A linearization evaluates the conditions at the unperturbed lines

y = \pm h

and discards products of perturbations. At

y = h

this gives

φ_{+, y} = η_{t} + U η_{x}, φ_{0, y} = η_{t}, φ_{+, t} + U φ_{+, x} = φ_{0, t},

(427)

whereas at

y = - h

it gives

φ_{-, y} = - η_{t} - U η_{x}, φ_{0, y} = - η_{t}, φ_{-, t} + U φ_{-, x} = φ_{0, t} .

(428)

The perturbation potentials remain harmonic functions, and their gradients decay at the corresponding far fields.

Solved by gpt-5.6-sol high.

d

Words: 62 Articles: 1

Solution

Words: 62

The linearized equations are homogeneous and have constant coefficients in

x

and

t

. Consequently, spatial Fourier modes evolve independently, and the superposition principle reconstructs arbitrary sufficiently regular initial disturbances from them. It is therefore enough to seek the normal modes

η (x, t) = Re {η_{0} e^{ik x + σ t}} .

(429)

Here the real

k

is the wavenumber,

σ

is the possibly complex temporal exponent, and taking the real part produces a physical displacement.

Solved by gpt-5.6-sol high.

e

Words: 112 Articles: 1

Solution

Words: 112

Put

q = ∣ k ∣

. Decay at infinity and the symmetry of the disturbance suggest the mode potentials

φ_{+} = A e^{- q (y - h)} e^{ik x + σ t}, φ_{-} = A e^{q (y + h)} e^{ik x + σ t},

(430)

φ_{0} = B cosh (q y) e^{ik x + σ t} .

(431)

The middle potential is an even function of

y

because its vertical derivative must take opposite values at the two interfaces. The upper kinematic conditions give

- q A = (σ + ik U) η_{0}, qB sinh (q h) = σ η_{0} .

(432)

The lower conditions give the same two relations. The upper dynamic condition is

(σ + ik U) A = σ B cosh (q h) .

(433)

Eliminating

A

B

and

η_{0}

yields the dispersion relation

(σ + ik U)^{2} + σ^{2} coth (q h) = 0 .

(434)

Writing

C = coth (q h)

, its two roots are

σ = \frac{- ik U \pm q U C}{1 + C} .

(435)

For every real

k \neq = 0

C > 0

and one root has the positive growth rate

Re σ = \frac{∣ k ∣ U c o t h ( ∣ k ∣ h )}{1 + c o t h ( ∣ k ∣ h )} > 0.

(436)

Thus every nonzero wavenumber is linearly unstable in this inviscid vortex sheet model;

k = 0

is neutral.

Solved by gpt-5.6-sol high.

f

Words: 73 Articles: 1

Solution

Words: 73

When

∣ k ∣ h ≫ 1

, the two shear layers are separated by many disturbance decay lengths. Since

coth (∣ k ∣ h) \to 1

, the roots reduce to

σ \sim - \frac{ik U}{2} \pm \frac{∣ k ∣ U}{2} .

(437)

This is exactly the Kelvin-Helmholtz instability of one isolated vortex sheet separating equal-density streams of speeds

U

and

0

: the pattern travels at their mean speed

U /2

, while its unstable amplitude grows at rate

∣ k ∣ U /2

. The exponentially weak interaction between the two layers disappears in this short-wavelength limit.

Solved by gpt-5.6-sol high.

J_{m} = ρ u, J_{p} = p + ρ u^{2} .

(438)

In a steady flow, the conservation laws make both the mass flux and the given energy flux

J_{E} = u (p + W + \frac{1}{2} ρ u^{2})

(439)

constant. Wherever

J_{m} \neq = 0

, their ratio is therefore constant:

\frac{J _{E}}{J _{m}} = \frac{p + W}{ρ} + \frac{1}{2} u^{2} = \frac{γ}{γ - 1} \frac{p}{ρ} + \frac{1}{2} u^{2} = C,

(440)

where

W = p / (γ - 1)

is the internal energy per unit volume of the ideal gas. This is the required steady-flow energy integral.

Solved by gpt-5.6-sol high.

b

Words: 91 Articles: 1

Solution

Words: 91

Work in the shock frame, and let the positive upstream and downstream speeds be

w_{0}

and

w_{1}

. The Rankine-Hugoniot conditions for a perfect gas give conservation of mass and momentum:

ρ_{0} w_{0} = ρ_{1} w_{1}, p_{0} + ρ_{0} w_{0}^{2} = p_{1} + ρ_{1} w_{1}^{2} .

(441)

Put

r = \frac{ρ _{1}}{ρ _{0}}, \frac{p _{1}}{p _{0}} = 1 + β .

(442)

Mass conservation gives

w_{1} = w_{0} / r

, and momentum conservation then gives

w_{0}^{2} = \frac{p _{0}}{ρ _{0}} \frac{β r}{r - 1} . (1)

(443)

Applying the energy integral from part (a) on the two sides gives

\frac{γ}{γ - 1} \frac{p _{0}}{ρ _{0}} + \frac{1}{2} w_{0}^{2} = \frac{γ}{γ - 1} \frac{p _{0} ( 1 + β )}{ρ _{0} r} + \frac{1}{2} \frac{w _{0}^{2}}{r ^{2}} . (2)

(444)

Substitution of (1) into (2), followed by cancellation of

p_{0} / ρ_{0}

, yields

\frac{β ( r + 1 )}{2} = \frac{γ}{γ - 1} (1 + β - r) .

(445)

Solving this linear equation for the density ratio gives

\frac{ρ _{1}}{ρ _{0}} = r = \frac{2 γ + ( γ + 1 ) β}{2 γ + ( γ - 1 ) β} .

(446)

Solved by gpt-5.6-sol high.

c

Words: 75 Articles: 1

Solution

Words: 75

The Taylor series of the shock density ratio at

β = 0

\frac{ρ _{1}}{ρ _{0}} = 1 + \frac{β}{γ} - \frac{γ - 1}{2 γ ^{2}} β^{2} + O (β^{3}) . (1)

(447)

For a reversible adiabatic process,

p / p_{0} = (ρ / ρ_{0})^{γ}

, so the density ratio predicted by the ideal-gas adiabat is

(\frac{p _{1}}{p _{0}})^{1/ γ} = (1 + β)^{1/ γ} = 1 + \frac{β}{γ} - \frac{γ - 1}{2 γ ^{2}} β^{2} + O (β^{3}) . (2)

(448)

Equations (1) and (2) agree through order

β^{2}

. Thus a weak shock satisfies the adiabatic pressure-density relation with an error of order

O (β^{3})

, which in particular establishes the requested accuracy through

O (β^{2})

Solved by gpt-5.6-sol high.

x

, the Rayleigh quotient of the matrix

A

R_{A} (x) = \frac{x ^{T} A x}{x ^{T} x} .

(449)

The Rayleigh quotient iteration starts from a unit vector

v_{0}

. Given

v_{k}

, compute the shift

μ_{k} = R_{A} (v_{k}),

(450)

solve the linear system

(A - μ_{k} I) w_{k} = v_{k},

(451)

and normalize in the Euclidean norm:

v_{k + 1} = \frac{w _{k}}{∥ w _{k} ∥ _{2}} .

(452)

The next eigenvalue estimate is

μ_{k + 1} = R_{A} (v_{k + 1})

. The process stops when the eigenpair residual is sufficiently small.

Solved by gpt-5.6-sol high.

b

Words: 69 Articles: 1

Solution

Words: 69

Differentiate the eigenvalue equation

A (t) v (t) = λ (t) v (t)

(453)

t = 0

, where

A^{'} (0) = E

. Suppressing the argument

0

gives

E v + A v^{'} = λ^{'} v + λ v^{'} .

(454)

Left-multiplication by the transpose of the unit left eigenvector

u

and use of

u^{T} A = λ u^{T}

cancel the terms involving

v^{'}

u^{T} E v = λ^{'} u^{T} v .

(455)

Because the eigenvalue is simple,

u^{T} v \neq = 0

, and hence the first-order perturbation of a simple eigenvalue is

λ^{'} (0) = \frac{u ( 0 ) ^{T} E v ( 0 )}{u ( 0 ) ^{T} v ( 0 )} .

(456)

The definition of the operator norm, followed by the Cauchy-Schwarz inequality, gives

∣ u^{T} E v ∣ \leq ∥ u ∥_{2} ∥ E ∥_{2} ∥ v ∥_{2} = ∥ E ∥_{2} .

(457)

Therefore

∣ λ^{'} (0) ∣ \leq \frac{∥ E ∥ _{2}}{∣ u ( 0 ) ^{T} v ( 0 ) ∣} .

(458)

Solved by gpt-5.6-sol high.

c

Words: 67 Articles: 1

Solution

Words: 67

A

is a real symmetric matrix, the finite-dimensional spectral theorem lets us choose the normalized left eigenvector and right eigenvector to be the same vector. Thus

∣ u^{T} v ∣ = 1, s (λ) = 1 .

(459)

More generally, the unitary diagonalization of a normal matrix shows that a real normal matrix also has coincident normalized left and right eigenspaces for each real eigenvalue. Since the question assumes