Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 312 / 3

Table of contents
- a 3
  - Solution a
- b 3
  - Solution b
- c 3
  - Solution c
- d 3
  - Solution d
- e 3
  - Solution e

a

Solution

For a Gaussian smoothed density contrast, the Press-Schechter factor of two gives the collapsed mass fraction

F (> M) = 2 \int_{δ_{c}}^{\infty} \frac{d δ}{2 π σ} e^{- δ^{2} / (2 σ^{2})} = erfc (\frac{ν}{2}), ν = \frac{δ _{c}}{σ ( M )} .

The fraction in the interval

[M, M + d M]

is

- (d F / d M) d M = (M / \overset{ρ}{ˉ}) (d n / d M) d M

. Since

d ν / d M = - ν d lo g σ / d M

,

\frac{d n}{d M} = - \frac{2}{π} \frac{ρ ˉ}{M ^{2}} ν e^{- ν^{2} /2} \frac{d lo g σ}{d lo g M}

or, equivalently, the same expression with

∣ d lo g σ / d lo g M ∣

. The minus sign is needed because

σ (M)

decreases with

M

; this is the positive Press-Schechter halo mass function.

Solved by gpt-5.6-sol high.

b

Solution

In the peak-background split, a long-wavelength overdensity changes the local threshold from

δ_{c}

to

δ_{c} - δ_{l}

. At fixed mass, the Press-Schechter halo mass function depends on the threshold through

ν e^{- ν^{2} /2}

. Therefore

b_{L} = \frac{1}{n} \frac{\partial n}{\partial δ _{l}}_{δ_{l} = 0} = - \frac{\partial l o g n}{\partial δ _{c}} = - \frac{1}{δ _{c}} \frac{\partial}{\partial l o g ν} lo g (ν e^{- ν^{2} /2}) .

It follows that the linear Lagrangian halo bias is

b_{L} (M) = \frac{ν ^{2} - 1}{δ _{c}} .

Solved by gpt-5.6-sol high.

c

Solution

On wavelengths much larger than halos, all matter is partitioned among halos. The mass-weighted halo overdensity must therefore equal the matter overdensity. Since

δ_{h} (M) = b (M) δ_{m}

, mass conservation requires

\int_{0}^{\infty} d M \frac{d n}{d M} \frac{M}{ρ ˉ} b (M) = 1.

For the Press-Schechter formalism, the mass-fraction measure becomes

f (ν) d ν = \frac{2}{π} e^{- ν^{2} /2} d ν, ν \geq 0.

It is normalized and is a half-normal distribution, so

\int_{0}^{\infty} f (ν) d ν = 1, \int_{0}^{\infty} ν^{2} f (ν) d ν = 1.

Using the linear Eulerian halo bias

b = 1 + b_{L} = 1 + \frac{ν ^{2} - 1}{δ _{c}}

therefore gives

\int_{0}^{\infty} f (ν) b (ν) d ν = 1 + \frac{1 - 1}{δ _{c}} = 1,

which verifies the consistency relation.

Solved by gpt-5.6-sol high.

d

Solution

Write the matter density as a sum over normalized halo profiles,

ρ (x) = \sum_{i} M_{i} u (x - x_{i} ∣ M_{i}), \int d^{3} x u (x ∣ M) = 1.

For

k \neq = 0

, its density contrast is

δ (k) = \frac{1}{ρ ˉ} \sum_{i} M_{i} u (k ∣ M_{i}) e^{- i k \cdot x_{i}} .

If halo locations form an uncorrelated Poisson process, only equal-halo terms survive after subtracting the homogeneous contribution. Replacing the sum per unit volume by the halo abundance gives the one-halo term

P (k) = P_{1 h} (k) = \int_{0}^{\infty} d M \frac{d n}{d M} (\frac{M}{ρ ˉ})^{2} ∣ u (k ∣ M) ∣^{2} .

Solved by gpt-5.6-sol high.

e

Solution

For distinct halos, insert their linearly biased correlation

ξ_{hh} (r ∣ M_{1}, M_{2}) = b (M_{1}) b (M_{2}) ξ_{lin} (r)

into the double sum. Fourier transformation converts

ξ_{lin}

into

P_{lin}

, while the two independent mass integrals factorize. The result is the halo model

P (k) = P_{1 h} (k) + P_{2 h} (k),

where

P_{2 h} (k) = [\int_{0}^{\infty} d M \frac{d n}{d M} \frac{M}{ρ ˉ} b (M) u (k ∣ M)]^{2} P_{lin} (k) .

Thus the requested function is

f (M, \overset{ρ}{ˉ}) = \frac{M}{ρ ˉ} .

At small

k

, profile normalization gives

u \to 1

and the bias consistency relation makes the square bracket tend to one, so the two-halo term approaches the linear matter spectrum.

Solved by gpt-5.6-sol high.

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