Past exam of the mathematics course of the University of Cambridge 2008 iii Paper 73 4 iii c Solution by
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The agreement of the main light-element abundances with a hot early nuclear-burning phase, and the independent baryon-density consistency between nucleosynthesis and the microwave background, gives strong support to the hot Big Bang framework. These measurements connect the first minutes to recombination, despite their very different temperatures, densities and observables. In particular, the large primordial helium mass fraction and residual deuterium are natural outputs of Big Bang nucleosynthesis, rather than arbitrary products of subsequent star formation.
The cosmological lithium problem prevents claiming that the simplest abundance calculation explains every observation without qualification. One must distinguish nuclear-rate uncertainties, stellar depletion, atmosphere and abundance systematics, and possible new particle physics. Thus the broad hot Big Bang framework is strongly supported, while lithium tests the completeness of its simplest nucleosynthesis and stellar-abundance interpretation. An unresolved lithium discrepancy alone does not erase the independent successes of deuterium, helium and the cosmic microwave background.
Past exam of the mathematics course of the University of Cambridge 2008 iii Paper 73 1 vi b Solution by
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Motion toward incoming cosmic microwave background radiation produces a blueshift. The photon occupation function of a blackbody depends on , so the same Doppler factor multiplies both frequency and effective temperature. Looking forward along the Sun's peculiar velocity, one obtainsTaking the mean cosmic microwave background temperature as about givesThe numerical total inherits the rounded mean temperature; the robust estimate is the positive millikelvin dipole increment.
Past exam of the mathematics course of the University of Cambridge 2008 iii Paper 73 1 v b Solution by
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In the local, nonrelativistic Hubble law approximation, . Ignoring the peculiar velocity gives an inferred distance , soThe requested accuracy therefore requiresAt exactly the worst-case error is , so “better than” requires a strictly larger distance. This is the conventional leading-order estimate. If one retains the exact Doppler factor and the product term while treating as exact, the positive-velocity case gives , with . It requires . That small refinement is conditional on the distance-redshift prescription; the simple answer uses the local approximation consistently.
Past exam of the mathematics course of the University of Cambridge 2008 iii Paper 73 1 v a Solution by
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Let be the photon frequency in the local comoving frame just after emission. The kinematic redshift gives , and propagation in the FRW metric gives . Multiplication givesEquivalently, . Both redshifts being small is what permits their approximate addition. Here the observer's own peculiar velocity is assumed already corrected to the Hubble flow frame.
Past exam of the mathematics course of the University of Cambridge 2008 iii Paper 41 5 b ii Solution by
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Converting size to a regression factor allows two unrestricted contrasts rather than imposing a linear trend on numerical size codes. The command relevel with a numeric second argument selects the second existing factor level as its reference level. For sorted codes this is medium, coded . However, the displayed coefficient labels size1 and size3 instead correspond to codes with code as reference. The printed coding and coefficient labels are inconsistent. The actual factor levels and labels must be checked. Under the intended medium-size reference, the two reported contrasts compare small and large businesses with medium ones; with literal coding their printed names would be size0 and size2.
Using B's delayed-entry survival object, the Cox proportional-hazards model isHere is the unspecified baseline hazard for a medium-size village business and is the Cambridge indicator. The fitted hazard ratios are conditional on the other covariates and constant over business age under the proportional hazards assumption. The function fits regression coefficients through partial likelihood; its survival object must actually be B's version, rather than A's earlier object with the same name.
The location coefficient , with standard error , gives and a two-sided Wald test p-value about . Its reported hazard ratio is , with 95% confidence interval . At the same age and size, a Cambridge business therefore has an estimated closure hazard about 55.2% lower than a village business. The reciprocal ratio is . A hazard ratio is not a survival-probability ratio or a proportional increase in lifetime.
The intended small-versus-medium contrast is , with standard error , and p-value . Its hazard ratio is , with 95% confidence interval . The intended large-versus-medium contrast is , with standard error , and p-value ; its hazard ratio is , with interval . Neither size contrast provides convincing evidence of an effect. The wide intervals also show why these results do not establish equality of size-specific hazards.
The sample has businesses. The likelihood-ratio test, Wald test and score test jointly test all three coefficients being zero; their statistics are , and on three degrees of freedom, with p-values , and . None rejects at 5%. This is compatible with the significant one-coefficient location test: testing a specified one-dimensional contrast and testing three coefficients together are different questions. The location result offers some adjusted evidence, but the joint output does not demonstrate a strong overall improvement, and interpretation should acknowledge the uncertainty and the number of comparisons.
The reported is a Cox–Snell likelihood pseudo-R-squared, here . It measures likelihood improvement rather than a literal fraction of variance in lifetimes explained. Its attainable maximum is reported as . It does not validate the proportional hazards assumption or establish useful predictive performance. Adjusted survival predictions would require estimating the baseline survival function as well: . The coefficient summary alone does not supply adjusted five- or ten-year survival probabilities.
Before reporting results, check the size coding discrepancy, the precise sampling dates and the survival records. The stated January 2000–December 2005 span is six calendar years, despite being called five years. Check nonnegative ages, entry no later than exit, correctly recorded closure indicators and the same opening-based clock throughout. Establish whether inclusion and censoring are plausibly independent of lifetime given the covariates, and whether different opening cohorts or calendar environments can reasonably share the fitted lifetime model. Simply including delayed entry does not cure informative selection, omitted confounders or dependence between businesses.
Assess proportional hazards using Schoenfeld residuals against time and a proportional hazards assumption test, supplemented by groupwise log-minus-log survival plots. Investigate influential observations with coefficient-deletion diagnostics and deviance residuals, and inspect overall fit using Cox–Snell residuals. Consider a city-by-size interaction or time-varying effects if supported by the available events, rather than relying on the displayed three main effects. Finally, report the uncertainty and declining late risk sets—only six remain at the final listed event—and distinguish a conditional association from a causal explanation or a prediction for all future businesses.
Past exam of the mathematics course of the University of Cambridge 2007 iii Paper 62 1 b ii Solution by
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As , use near zero. ConsequentlyThe flat dust-and-vacuum expansion tends to , the Einstein-de Sitter universe.
Past exam of the mathematics course of the University of Cambridge 2007 iii Paper 62 1 b i Solution by
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As , use to obtainAt fixed Hubble constant, the age of a flat matter-Lambda universe diverges logarithmically. The limiting purely vacuum-dominated de Sitter spacetime has an exponential scale factor and no Big Bang at a finite value of cosmic time in this flat slicing.
Past exam of the mathematics course of the University of Cambridge 2007 iii Paper 41 1 a ii Solution by
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If , let be a nearest point of the closed convex cone to . Existence follows by minimizing the continuous squared distance on a sufficiently large compact ball. The first-order minimization inequality is for every . Set . Taking and , both in the cone, gives . ConsequentlyThe cone's test-weight characterization again yields almost surely. This is separation from a closed convex cone with the sign needed forThe strict benefit is already available at the initial date, even if the terminal payoff happens to be zero.
Past exam of the mathematics course of the University of Cambridge 2007 ia Paper 2 7B i b Solution by
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For the Heaviside step function forcing, the integral begins at zero when :Before zero it vanishes. ThereforeThis causal response of a first-order relaxation equation is continuous at zero and approaches the new constant equilibrium exponentially.
Past exam of the mathematics course of the University of Cambridge 2007 ia Paper 2 7B i a Solution by
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Inserting the Dirac delta forcing into the integral gives the impulse responseIt is zero before the impulse and obeys the homogeneous equation afterward. Integrating the equation across zero gives , exactly the jump in this formula. Thus the formula also verifies the differential equation as a distribution; assigning a particular value at the jump does not affect that distribution.
Past exam of the mathematics course of the University of Cambridge 2007 ia Paper 1 5B ii b Solution by
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Substitute the signed displacement into the Pythagorean theorem relation from part (a):A genuine circle requires . Equality gives a tangent plane and a single point, interpreted as radius zero; a larger distance gives no intersection. Under the stipulated circle intersection, the radicand is positive. Its centre is .
Past exam of the mathematics course of the University of Cambridge 2007 ia Paper 1 5B ii a Solution by
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Set , where . Then , so lies in the plane. For any position in that plane, is perpendicular to . The Pythagorean theorem therefore givesOn the sphere the left side is . Thus the intersection is the circle in the plane centred at , with squared radius . Its centre must therefore be , andIn terms of the radii, , with the sign fixed by . In the central-plane case , both signs give zero. This is the center and radius of a sphere-plane intersection formula with its orientation made explicit.
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The vector is the perpendicular component of relative to the axis , so its length is the distance from that axis. Hence the surface is an infinite circular cylinder of radius about the line through the origin parallel to . Here is the radius and fixes the axis direction; reversing leaves the same circular cylinder.
If are an orthonormal basis of the plane perpendicular to , a parameterization isThis verifies that all heights and all angular positions, but no interior points, belong to the surface.
Past exam of the mathematics course of the University of Cambridge 2007 ia Paper 1 5B i a Solution by
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For , the dot product equation says that the angle between and the oriented unit vector satisfies . Thus is fixed. More explicitly, using the orthogonal projection from the introductory solution,Consequently the surface is one infinite nappe of a right circular cone, with apex at the origin, including the apex itself. The cone's axis points along , and its acute semi-angle is . Equivalently, is the cosine of the angle from the oriented axis. Squaring the original equation without retaining the sign of would incorrectly add the opposite nappe of a double circular cone. No base disk is included in this infinite lateral surface.
Past exam of the mathematics course of the University of Cambridge 2006 ia Paper 1 7B i b Solution by
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If , then . On the perpendicular plane, vanishes and is a signed right-angle rotation, whose square is minus the identity. Hence its complex eigenvalues are , and
Past exam of the mathematics course of the University of Cambridge 2006 ia Paper 1 7B i a Solution by
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If , use the positively oriented orthonormal basis . The cross product identities give , and . In this basis,Its characteristic polynomial is , so
Past exam of the mathematics course of the University of Cambridge 2003 iii Paper 66 4 a iii Solution by
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Gravitational lensing probes the cluster's projected mass through the deflection of background light. Strong gravitational lensing produces arcs and multiple images; weak gravitational lensing measures coherent distortions of many background galaxies. The lensing convergence is , withThus the image geometry and shear constrain the projected surface density without assuming the cluster is in dynamical or hydrostatic equilibrium. The reconstructed total mass again exceeds the visible stellar and gas contribution. Source-distance errors, line-of-sight structures and the mass-sheet degeneracy affect the reconstruction, but the independent agreement of lensing, galaxy motions and gas confinement is strong evidence for cluster dark matter.
Past exam of the mathematics course of the University of Cambridge 2003 iii Paper 66 4 a ii Solution by
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The hot intracluster gas emits X-rays, which constrain its temperature and density profile. If its pressure is predominantly thermal and it is in hydrostatic equilibrium,Eliminating the pressure givesHere is the mean mass per gas particle. A gas at the observed cluster temperatures would not remain confined without a deep gravitational potential; the inferred total mass exceeds the gas and stellar mass together. Nonthermal pressure, mergers, gas clumping and temperature-profile errors can bias this estimate, but do not turn X-ray luminosity itself into a direct measurement of total mass.
Past exam of the mathematics course of the University of Cambridge 2003 iii Paper 53 2 d Solution by
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The three-gamma matrix is antisymmetric, and for three distinct indices it is nonzero. A pair of nonzero Grassmann components therefore produces a nonzero term exactly as in the scalar case. Consequently this three-form bilinear is not identically zero. Equivalently it is dual to the axial-vector Majorana spinor bilinear, which is allowed; it must not be confused with the vanishing vector in part (a).
Past exam of the mathematics course of the University of Cambridge 2003 iii Paper 53 2 c Solution by
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Four-dimensional Hodge duality identifies with a nonzero convention-dependent multiple of . Every is symmetric by the four-dimensional Majorana bilinear vanishing criterion; taking this linear combination preserves symmetry. HenceThe phase in the duality formula depends on the definition of and orientation, but does not affect the vanishing result.
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