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Past exam of the mathematics course of the University of Cambridge 2013 iii Paper 55 1 Solution by
Codex 0 Created 2026-10-07 Updated 2026-10-14
Let and define the cosmological density power spectrum by . The contribution per logarithmic wave-number interval is the dimensionless cosmological power spectrum . The linear cosmological mass variance on mass scale iswith a comoving smoothing radius and the top-hat filter in Fourier space. On halo scales the observed spectrum is consistent with greater variance at smaller masses. Growth therefore brings smaller objects to the collapse threshold earlier, statistically; larger structures assemble by accretion and dark-matter halo mergers. This statistical growth from smaller bound systems to larger ones is hierarchical galaxy formation. It is an ordering of dark-matter assembly, not a rule that every small visible galaxy must precede every large visible galaxy.
For linear modes the linear growth factor gives . During matter domination , so and the characteristic nonlinear mass increases. Once modes become nonlinear, coupling between scales and halo formation change the shape, so the same multiplication cannot be used for the entire late-time spectrum. At low redshift accelerated expansion suppresses linear growth. Galaxy clustering traces the matter spectrum with galaxy bias; it should not be equated directly with an unbiased matter measurement.
The broad linear shape was set by the cosmological transfer function before and around matter-radiation equality, with baryonic acoustic structure also imprinted before recombination. A nearly scale-invariant primordial curvature spectrum has , with close to one. After converting curvature perturbations to matter-density perturbations,Thus nearly scale-invariant primordial curvature does not mean a constant density . Modes with enter the horizon after equality and have . Modes with enter during radiation domination, when radiation controls the expansion and cold-matter perturbations grow only slowly. The cold-dark-matter transfer function behaves approximately as at large . Hence the density spectrum turns over near :The turnover records the equality horizon, while the late nonlinear excess records gravitational clustering. On galactic scales the effective slope is greater than , giving the growing small-scale variance needed for hierarchical galaxy formation.
Cold dark matter has negligible primordial thermal velocities and a very short collisionless free streaming length on galactic scales. Warm dark matter has appreciable residual velocities while structure is being seeded; particles stream across small fluctuations and reduce their contrast. Its cosmological transfer function is consequently cut off below a characteristic length, suppressing low-mass halos and delaying their formation. Above that cutoff its assembly can still be hierarchical. The important distinction is the free-streaming scale, rather than the present temperature or an arbitrary particle-mass label.
Linear evolution assumes . When becomes of order unity, overdense regions depart strongly from the Hubble flow and can turn around and collapse. Collisionless dark matter develops multistream motion after trajectories cross; gravitational mixing redistributes energy and produces a bound dark matter halo. The formal infinite-density collapse of an ideal spherical pressureless solution is not the physical endpoint. A roughly virialized halo has and a characteristic virial velocity . Its gas virial temperature is conventionallyIt measures the thermal energy associated with the gravitational potential; the numerical factor depends on the velocity-dispersion convention. It does not imply that the collisionless dark matter has a thermodynamic gas temperature.
The unheaded request about baryon conversion efficiency of a halo is also answered here. Define using the cosmic baryon fraction . In small halos, shallow potentials let stellar feedback drive outflows or repeatedly heat star-forming gas; supernova energy per stellar mass is roughly fixed while binding energy per gas mass scales as . Photoheating during reionization also prevents very small halos from retaining or accreting cool gas. Molecular/atomic cooling thresholds further reduce star formation in the smallest systems. These effects make fall toward low mass.
Near , gas can cool efficiently and the potential is deep enough to retain more of it, while a long-lived hot atmosphere and maintenance heating are less effective than in larger systems. At high mass, higher virial temperature and lower cooling efficiency let a substantial hot atmosphere persist. Active-galactic-nucleus feedback can prevent that atmosphere from supplying cold gas and can expel some gas; the cooling-time bottleneck alone is not an adequate explanation for the low stellar fractions of massive groups and clusters. The peak reflects a competition between gas supply/cooling and feedback, rather than complete conversion of all baryons at a sharply universal mass. Its exact location and height depend on epoch, metallicity, gas history and the stellar population included.
Past exam of the mathematics course of the University of Cambridge 2013 iii Paper 5 2 Solution by
Codex 0 2026-10-07
For an integer tuple , the monomial alternant isNegative exponents require nonzero coordinates; all exponents in the character expansion below are nonnegative. The power-sum symmetric polynomial is . Put , so is the Vandermonde determinant.
Let a conjugacy class of have cycles of length , with , and put . The product is an alternating polynomial homogeneous of degree . In an alternating polynomial, a monomial with two equal exponents has zero coefficient, since interchanging those variables fixes the monomial and reverses its sign. Grouping the remaining monomials by their permutation orbits gives a unique expansion in alternants with .
Such tuples of the indicated total degree are exactly for partitions of an integer of with at most parts. Define the class functionThe displayed monomial occurs with coefficient in and in no other ordered alternant, soThis proves the expansion and explicitly defines its coefficients. They depend only on the cycle counts and hence are class functions. Identifying these coefficients with Specht module characters is the Frobenius alternant character formula, which the remaining parts allow us to assume.
Past exam of the mathematics course of the University of Cambridge 2013 iii Paper 5 1 Solution by
Codex 0 2026-10-07
We take algebras to be unital and modules to be unital. A finite-dimensional semisimple algebra over the complex numbers is one whose regular left module is a semisimple module, that is, a direct sum of simple modules. We use the basic finite-dimensional equivalences: this is equivalent to a zero Jacobson radical, and the Jacobson radical of any finite-dimensional algebra is a nilpotent ideal.
The Artin–Wedderburn theorem here says that there are positive integers withThe nonisomorphic simple modules are the standard column modules of the factors, of dimensions , and the factor sizes are unique up to reordering.
Here is a proof. Decompose the regular left module as with pairwise nonisomorphic simple modules . These are all the simple modules: any simple module is generated by a nonzero vector and is therefore a quotient of the regular module. The Schur lemma gives for and . For the latter assertion, an endomorphism has an eigenvalue over , and the kernel of its difference from that scalar is a nonzero submodule, hence the whole simple module. ConsequentlyEvery regular-module endomorphism is right multiplication by its value at , so . Taking opposite algebras and using matrix transposition gives the asserted decomposition with .
For a matrix algebra , the matrix units show that its only simple module is : the spaces are isomorphic through , and a nonzero vector in one of them generates one copy of the column module. Simplicity makes that copy all of . In a product algebra, the mutually orthogonal central idempotents decompose every module into its factor modules; a simple module uses exactly one factor. This proves the module assertion and also uniqueness, since the primitive central idempotents and the dimensions of their simple modules determine the factors.
Past exam of the mathematics course of the University of Cambridge 2013 iii Paper 47 1 Solution by
Codex 0 2026-10-07
Use the Minkowski metric with signature , so the Sine-Gordon equation is . The light-cone coordinates here satisfyIn particular, there is no extra factor of four with this coordinate normalization. The Bäcklund transformation requires , because one of its equations contains .
Put and . In the parameter convention of this paper the Sine-Gordon Bäcklund transformation gives and . Differentiating, for a twice differentiable transformed field, givesSince , the addition formula yields . Therefore the transformed field satisfies the same Sine-Gordon equation:Subtracting the two differentiated equations also gives , showing the compatibility with the seed equation. This proves the unheaded request before the numbered parts, without assuming a particular soliton form.
Past exam of the mathematics course of the University of Cambridge 2013 iii Paper 44 3 Solution by
Codex 0 2026-10-07
A regulator and a renormalization condition introduce the reference mass scale , even when the classical theory has no mass. Loop amplitudes contain dimensionless logarithms of momentum or distance ratios involving . The resulting running coupling and field normalization compensate changes of this arbitrary reference scale.
Let and . Hold the bare parameters fixed and define and . Assuming multiplicative field renormalization and no mixing or additive contact terms for the correlator, differentiating gives the Callan-Symanzik equationFor the dimensionless two-point factor put . Its equation is . LetThe characteristic solution of the multiplicative Callan-Symanzik equation isTo check the sign, equals . The characteristic flow and its accumulated multiplier give precisely this evolution. Changing changes the dimensionless momentum and renormalized ; the same bare two-point correlation function is recovered after the compensating field normalization. An unnormalized renormalized correlator need not remain numerically identical under that change, but physical predictions do.
With a mass, write and define its running mass by , . The dimensionless equation becomesIts flow is thereforeThe renormalization-group mass suppression criterion is , for example an eventual bound with . Then the mass argument on the right tends to zero. A regular massless limit, uniform along the limiting coupling trajectory, makes the mass negligible at high energies. Merely calling small without controlling this integrated exponent is insufficient.
Past exam of the mathematics course of the University of Cambridge 2013 iii Paper 43 2 Solution by
Codex 0 2026-10-07
Use the usual normalization of the Super-Poincaré algebra, with and under the corresponding index convention:The supercharges are odd operators: they turn bosonic states into fermionic states and vice versa. If is fermion parity, this statement is , soThe same relation holds for the conjugate supercharges.
For a finite-dimensional physical supermultiplet at fixed four-momentum with energy , let count physical bosonic and fermionic states. Cyclicity of the ordinary trace and the parity anticommutation implySumming over the two spinor indices givesThis supertrace pairing at positive energy proves boson-fermion degeneracy in a supermultiplet for massive as well as massless positive-energy representations. It counts on-shell polarizations, not merely the names of fields. The requirement matters: a zero-energy supersymmetric vacuum can be a bosonic singlet without a paired fermionic vacuum.
If supersymmetry-breaking operators are explicitly added to the Lagrangian, the original supercharges generally no longer commute with the full Hamiltonian. They are not conserved symmetries generating finite fixed-energy physical supermultiplets; their original anticommutator does not equal the full translation generator with the breaking terms included. Thus the step replacing the parity-weighted anticommutator by on a closed physical representation fails. The odd parity relation alone does not force energy degeneracy or an equal number of physical states at each mass.
This is explicit versus spontaneous supersymmetry breaking. In spontaneous breaking the action still has conserved supercharges, but the vacuum is not annihilated by them. Acting on particle excitations about that vacuum involves the broken-vacuum/Goldstino sector, so an ordinary finite particle multiplet above an invariant vacuum is no longer the correct pairing argument. The vacuum-energy statements below refer to an exact globally supersymmetric Hamiltonian, including the spontaneously broken case; they are not positivity claims for an arbitrary explicitly broken Hamiltonian.
Past exam of the mathematics course of the University of Cambridge 2013 iii Paper 32 7 Solution by
Codex 0 2026-10-07
Begin by defining the scientific target: prediction for a new patient, or estimation of adjusted covariate effects. Check event definitions, follow-up origins, right-censoring codes and any delayed entry. The analysis assumes independent patients and censoring that is noninformative conditional on modeled covariates. For fixed explanatory variables , the Cox proportional-hazards model iswith an unrestricted baseline hazard and coefficients constant over time. Report as the conditional hazard ratio for a one-unit change in the coded variable, not as a risk ratio or automatically a causal effect.
For untied event times, fit the coefficients by maximizing the Cox partial likelihoodThe baseline cancels within each event risk set. Use an appropriate tied-event method, such as an Efron approximation or an exact method for a genuinely discrete event scale; the Breslow approximation for tied event times is another explicit approximation. Numerical score/information methods fit the model, and inverse information gives conventional covariance estimates for independent subjects. Report coefficient estimates, hazard ratios, uncertainty and the coding that makes their interpretation meaningful. After fitting, the Breslow estimator gives , from which provides predicted survival.
Adequacy is broader than significance of coefficients. Inspect influential observations, deviance residuals and data errors. Cox–Snell residuals should have approximately unit-exponential survival under a well-fitting model, retaining the original censoring indicators; a cumulative-hazard plot of these residuals should be near the 45-degree line. This is an overall diagnostic, not a substitute for the functional-form and proportionality checks below. For prediction, assess held-out discrimination and survival prediction calibration at prespecified time horizons with methods accounting for censoring. A high concordance index alone does not establish good calibration or correct hazard structure.
Use bootstrap or held-out cross-validation that repeats the complete modeling procedure, including imputation, variable selection and tuning. Validation of only the final fitted coefficients understates overfitting from earlier decisions. Report any substantive lack of fit rather than presenting a single time-independent hazard ratio when the data do not support that representation.
Past exam of the mathematics course of the University of Cambridge 2013 iii Paper 32 6 Solution by
Codex 0 2026-10-07
The episodes from one patient share treatment, biological susceptibility and prior history, so they are not independent observations. Treating every row as an unrelated subject can substantially understate uncertainty. Two approaches address this within-patient recurrent-event dependence:
- A patient-clustered sandwich covariance matrix retains a suitable working Cox mean/intensity model but groups score contributions by patient. If is the patient's score contribution and the observed information, its form is . Independent patients, rather than independent rows, determine the sampling units. This corrects uncertainty for within-patient dependence when the working estimating equation is appropriate; it does not repair an incorrect mean model.
- A shared frailty model introduces a common latent positive multiplier for all of patient 's episode hazards, often with a specified gamma or lognormal distribution. Given frailty and the relevant history, the event mechanism is modeled through that patient's intensity. Estimate treatment and frailty parameters by integrating or profiling the latent effect. This models persistent heterogeneity directly, but relies on the frailty assumptions and generally gives a conditional treatment-effect interpretation.
Past exam of the mathematics course of the University of Cambridge 2013 iii Paper 32 5 Solution by
Codex 0 2026-10-07
The log-rank test compares survival experience through the allocation of events within the successive risk sets. Its null is equal hazard functions between the groups over follow-up, with independent individuals and noninformative censoring within groups. At each distinct event time , let be the numbers at risk immediately before that time, and let be event counts. Under the null, conditional on the total , the expected number of A events is .
The signed log-rank statistic is the observed-minus-expected scoreA positive score indicates relatively more A events than expected, hence a higher event hazard for A; B's score is its negative. Censoring times do not produce score terms, but remove people from all later risk sets.
At each event time the conditional null allocation is hypergeometric, as if the events were sampled without replacement from the risk set. Its variance gives the score-increment variance, with the finite-population correction for ties. Summing these conditional variances gives , since distinct-time martingale score increments have zero cross covariance under the null. For sufficiently many informative events, is approximately standard normal and approximately chi-squared with one degree of freedom. In the no-tie case, each variance contribution is . A time at which only one group remains at risk contributes no comparative information.
The numerical parts below report signed-score and variance contributions after day 160. They cannot be added as separate chi-squared statistics to the earlier follow-up: add the scores and variances first, and standardize once for the complete dataset. The log-rank test is particularly effective for proportional hazards alternatives; crossing hazards can produce cancelling score contributions.
Past exam of the mathematics course of the University of Cambridge 2013 iii Paper 29 1 Solution by
Codex 0 2026-10-07
For an autoregressive moving-average model, the driving-to-output transfer function is , and the inverse transfer function is . The causality and invertibility root criteria for an ARMA model require these respective rational functions to have power series about zero converging on a disk larger than the unit disk. If the two polynomials have no common factor, the conditions becomeIndeed, outside-disk roots leave a radius of convergence greater than one, so the coefficients decay geometrically and are absolutely summable. Conversely an uncancelled pole inside or on the unit disk prevents the required stable power series. If factors are common, apply the criterion after cancellation, to the noise-driven solution rather than additional homogeneous components. The backshift operator translates these power series into the desired one-sided filters.
The original representation has and , with roots and . There is no cancellation. Hence it is neither causal nor invertible relative to its specified driving noise. Stationarity is nevertheless possible through a two-sided solution: expanding the autoregressive inverse in negative powers givesThis is an anticausal time series representation with square-summable coefficients.
To establish the alternative representation on the same process, defineThe inverse of is a stable one-sided filter. For , the identities and giveThe time-series spectral density of is ; therefore the defined has constant time-series spectral density . Its mean is zero, its variance is , and all its nonzero-lag autocovariances vanish. It is thus weak white noise. Its definition directly givesThis root reflection of an ARMA representation has roots , so is causal and invertible. A constant spectrum proves whiteness, not independence of non-Gaussian coordinates; no Gaussian assumption is needed for the required white noise representation.
For best linear prediction from an infinite past, let be the closed linear span of with . The causal representation expresses every such in present and past values, while invertibility puts every with in . In particular is orthogonal to . The representation at time one then gives the orthogonal projectionConsequently the best linear predictor and its error variance areOrthogonality proves optimality among linear predictors in the closed past span, without asserting that the predictor must be the conditional mean for a non-Gaussian process.
Pinned article: Introduction to the OurBigBook Project
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