Beta-function recurrence 2026-09-28
The Beta function obeysThese identities follow immediately from the beta--gamma identity and the Gamma function recurrence.
Past exam of the mathematics course of the University of Cambridge 2006 ii Paper 4 14E b Solution Created 2026-09-24 Updated 2026-09-24
Past exam of the mathematics course of the University of Cambridge 2006 ii Paper 4 8E Solution Created 2026-09-24 Updated 2026-09-24
Past exam of the mathematics course of the University of Cambridge 2011 ii Paper 1 8E Solution Created 2026-09-24 Updated 2026-09-24
Past exam of the mathematics course of the University of Cambridge 2012 iii Paper 29 3 Solution Created 2026-10-03 Updated 2026-10-07
For , define the Riemann zeta function by the absolutely convergent Dirichlet seriesThe Euler product follows from unique prime factorization and absolute convergence. The counting-function integral givesBecause the fractional part is bounded, the last integral, and its derivatives with respect to on compact subsets, converge uniformly for . It is a holomorphic function there. This proves the Meromorphic continuation of the Riemann zeta function to the right half-plane, with a simple pole at one of residue one and no other singularity in that half-plane.
The Gamma function isThe Gamma function recurrence extends it meromorphically; its poles are at the nonpositive integers and it has no zeros. The functional equation of the Riemann zeta function can be stated asor, equivalently,These are identities of meromorphic functions; the second also extends the Riemann zeta function to the remaining half-plane. The critical strip is ; its boundary lines will also be treated below, rather than included among possible exceptional zeros.
For , the absolutely convergent reciprocal Euler product gives . For , the factors , , and in the second functional equation of the Riemann zeta function are finite and nonzero. Hence the only zeros there are the zeros of the sine factor:Each is simple. At zero, the sine zero cancels the pole of ; using its residue one gives , not zero. Nonvanishing on the rest of follows from the boundary-line proof below and the functional equation of the Riemann zeta function. Together these facts prove that the trivial zeros of the Riemann zeta function are the only zeros outside the open critical strip.
Here is a Jensen disk proof of the zeta zero-count bound which avoids any unproved left-half-plane growth estimate. Set , a holomorphic function throughout , including at one. The integral continuation formula givesFor each integer , apply Jensen's formula in the disk with centre , outer radius and inner radius . The outer disk stays in , so its maximum modulus is . At its centre,since . Therefore the number of zeros in its inner disk, counted with multiplicity, is at mostThe rectangle , fits in the inner disk because its farthest point has distance . Summing over such disks counts zeros in the right half of the critical strip up to height . The first functional equation of the Riemann zeta function bijects zeros, with multiplicities, in the left half with their reflections in the right half; its Gamma function factors are finite and nonzero in the strip. ThusThe pole of zeta function at one does not count as a zero: .
Finally, for real and , the Euler-product logarithms and the nonnegative trigonometric polynomialgive the three-four-one product proof of zeta boundary nonvanishing:If had a zero of order , its factor would be as . The real zeta function factor has pole order three, and the factor at stays bounded because . The product would tend to zero as , contradicting that it is at least one. ThereforeAt there is a pole, not a zero. Applying the sine-form functional equation of the Riemann zeta function at now proves nonvanishing on the remaining imaginary axis, completing the earlier assertion about all zeros outside the critical strip.
Past exam of the mathematics course of the University of Cambridge 2014 iii Paper 46 4 Solution Created 2026-10-03 Updated 2026-10-06
Take an oriented closed string in flat, critical bosonic string theory, with and a mostly-plus target Minkowski metric. After continuation of the worldsheet to Euclidean signature, the Polyakov path integral sums over embeddings and worldsheet metrics, divided by worldsheet diffeomorphisms and Weyl transformations. In a flat target its kinetic action isTarget-time continuation or analytic continuation of external momenta defines the Lorentzian scattering amplitude; a naive real Euclidean Gaussian for timelike would not be convergent.
An external tachyon is represented by the tachyon vertex operator , with conformal weights . The physical integrated vertex has , so . Schematically its tree amplitude isThe Faddeev-Popov determinant from conformal gauge is represented by the worldsheet ghost fields. At tree level the worldsheet is a Riemann sphere, whose unpunctured complex structure has no moduli. Its residual conformal automorphisms are the Möbius transformations, . Fix three insertion points, accompanying their unintegrated vertices by the required ghost factors. The remaining complex insertion positions are integrated over the sphere; equivalently one integrates all positions and divides by the residual conformal group.
The embedding fields are free, withTheir zero-mode integral gives momentum conservation, and their nonzero-mode Gaussian integral gives the Koba-Nielsen factorThis explains the sphere tachyon position integral without needing its evaluation. For four tachyons the resulting Virasoro–Shapiro amplitude displays the exchanged string spectrum directly.
Use all-incoming external momenta and introduce alongside the two printed dimensionless Mandelstam variables. Since and , one obtains . The physical center-of-mass energy squared in the channel is . The channel measures the analogous crossed momentum transfer, with sign determined by the mostly-plus convention. Rewriting the Gamma factors in a symmetric form givesThe Gamma function poles imply, at generic fixed values of the other invariant,The denominator Gamma factors can remove residues at special intersecting channel kinematics; the statement concerns a generic single-channel limit. An -channel pole occurs when the intermediate momentum satisfies the mass-shell condition for a closed-string state:The pole at exchanges the ground-state tachyon, the pole at exchanges massless states, and the positive integer poles exchange the infinite massive tower. The -channel interpretation is the crossed version. Factorization means that each residue is a sum of products of couplings to intermediate physical states that couple to the chosen external particles. It need not expose every representation at that mass.
For example, the Gamma function recurrence and Gamma function residue at a nonpositive integer give the dimensionless Virasoro–Shapiro amplitude pole residueThe residue polynomial has degree , consistent with maximum spin in the exchanged level. The amplitude also has the corresponding -channel poles by crossing symmetry.
The massless fields can be treated as target backgrounds rather than separate asymptotic insertions. Write the target metric as , introduce a Kalb–Ramond field , and a dilaton . In conventional Euclidean signs their string nonlinear sigma model action isHere is the antisymmetric tensor density. The three backgrounds correspond to the graviton, antisymmetric tensor and scalar states at closed-string level one. Expanding the vacuum functional in , then Fourier expanding the backgrounds, produces exactly their integrated string vertex operators. Its functional derivatives therefore generate the amplitudes with massless external strings. The connected vacuum functional organizes connected amplitudes; the spacetime effective action organizes the corresponding vertices after treating massless propagation consistently.
At momenta small compared with , massive string propagators can be expanded in powers of momenta over their masses. The analytic part of the amplitudes consequently determines local higher-derivative interactions, ordered by powers of . Massless exchange poles are retained through propagation of the massless fields, rather than expanded into local contact terms. Up to field redefinitions, the leading massless-sector action in the string-frame metric isThe terms denoted contain additional derivatives, including curvature-squared terms in the bosonic theory. The expansion concerns the massless sector around the perturbative bosonic background; the tachyon instability remains and is not cured by omitting its field from this displayed action. Thus this is a formal perturbative effective description, not a claim of a stable bosonic vacuum.
Finally split the dilaton into a constant and its variation, . By the Gauss-Bonnet theorem, its dilaton Euler-characteristic weighting on a connected closed oriented surface follows fromDefine the string coupling by . A genus- path integral is weighted byThe sphere carries , the torus carries , and each extra handle adds . At higher genus one integrates over complex-structure moduli as well as insertion points, with the associated antighost insertions supplying the correct moduli measure. With canonically normalized external vertices an -point genus- amplitude scales as .
Summing connected worldsheets of every genus of a surface yields the string-loop effective action expansionEach coefficient has its own low-energy expansion. The two parameters have different roles: resolves finite string size through higher derivatives, while counts additional string loops.
Past exam of the mathematics course of the University of Cambridge 2015 iii Paper 34 4 Solution Created 2026-10-03 Updated 2026-10-06
In the Bühlmann model, a latent risk parameter is drawn from a population distribution. Conditional on , the yearly observations are independent and identically distributed random variables, with conditional expectation and conditional variance . Define the structural parametersHere is the expected process variance, while is the variance of hypothetical means. The Bühlmann credibility premium is the best affine estimate of from the observed claims, under mean squared error. Predicting the next claim gives the same affine estimate: the extra conditional observation noise contributes the constant to the prediction error.
The law of total variance and conditional independence giveAn affine estimate can be written as : for any chosen , optimizing the constant makes its expected value equal to . The normal equations for the linear least-squares projection areFor they force all to agree, with . Thus the credibility factor and premium areThe credibility factor increases with the observation count and between-risk variance, and decreases with within-risk variance. If , the risk mean is known and ; if and , one observation reveals it and . If both vanish, the premium is the fixed value and the factor is immaterial.
In the specified model, the conditional law is a gamma distribution with shape and scale . ThereforeThe prior is an inverse-gamma distribution with shape and scale . To obtain its moments directly, substitute in the defining integral, obtainingThe Gamma function recurrence yieldsThe assumption makes both structural variances finite. HenceIt follows that the model-specific credibility estimate is
For the Bayes estimator under squared error loss, the quantity to estimate is , so the optimum is its posterior mean. The likelihood function, viewed as a function of , is proportional toMultiplication by the prior shows gamma scale inverse-gamma conjugacy:Although the printed hint only mentions integer shapes, the same substitution and Gamma integral normalize this posterior for every positive real shape, so no integrality of is needed. Its posterior mean gives the Bayesian estimate and comparisonThis exact Bühlmann credibility for gamma claims holds for every observed sample, not merely on average. Here the posterior mean is affine in the sample mean, so the best affine Bühlmann credibility premium is also the unrestricted Bayes estimator under squared error loss.
Past exam of the mathematics course of the University of Cambridge 2015 iii Paper 49 3 Solution Created 2026-10-03 Updated 2026-10-06
Independent metric and Hamiltonian parametrization. The Lorentzian Polyakov action isThe worldsheet metric is independent of the embedding. Its equation sets the traceless worldsheet stress tensor to zero, making the metric locally conformal to the induced worldsheet metric; substitution recovers the Nambu–Goto action. Eliminating momentum in the Nambu-Goto phase-space action instead gives . On the positive-lapse branch, choose the independent metric, up to a positive Weyl transformation, asIts determinant is . Multiplication by recovers the reduced density, including the mixed term . This proves the phase-space identification of the Polyakov metric.
Conformal gauge and residual coordinate freedom. In the Hamiltonian formulation, conformal gauge is . The equations become , while and remain as constraints. In the metric formulation, conformal gauge is , or simply after fixing Weyl invariance. The metric parametrization proves these choices equivalent. Gauge-fixing the metric does not discard the Virasoro constraints.
With , independent reparameterizations multiply the flat metric by a conformal factor. A compensating Weyl transformation restores its chosen representative. Infinitesimally the Conformal Killing equation is . These are the residual conformal transformations of two-dimensional Minkowski spacetime, with the periodicity conditions appropriate to a closed string.
Ghost determinant and anomaly. For gauge conditions , the Faddeev-Popov determinant is the determinant of the linearized variationInsert this determinant in the gauge-fixed path integral and exponentiate it with anticommuting ghosts and antighosts: the additional action is proportional to . In metric gauge fixing, the trace is removed by Weyl invariance, and a worldsheet diffeomorphism changes the trace-free metric throughConsequently is a vector ghost and a symmetric trace-free antighost. A conventional normalization of the worldsheet ghost action isA rescaling of the antighost changes only this overall kinetic normalization. In conformal gauge, after normalizing the chiral fields, the kinetic terms areThe closed-string ghosts are periodic, with independent left and right systems. Their conformal weights are for and for , so the supplied bc ghost system expression gives . Each free embedding boson has central charge one per chirality. HenceThis central charge of reparameterization ghosts cancels the string conformal anomaly in each chirality separately. Having two sectors does not double the required target dimension.
Channel variables, poles and the vector state. With particles 1,2 incoming and 3,4 outgoing, the Mandelstam variables are and . They measure squared center-of-mass energy and momentum transfer. In the specified units , the external tachyon mass squared is , so .
The Gamma function residue at is . At , its argument varies with the opposite sign to . The Gamma function recurrence givesThus the Veneziano amplitude pole residues areFor generic fixed the simple poles are . A zero of the residue product at an exceptional negative integer makes the corresponding pole removable. Fixed must be away from its own channel poles; fixing it at a channel singularity does not give an ordinary finite meromorphic function of .
At , the residue is . This is proportional to , the contraction of conserved scalar-pair vector currents. Each current is orthogonal to the exchanged momentum because the external masses in its pair are equal. Alternatively, with scattering-angle cosine , one has ; at this pole , a pure spin-one angular dependence. ThereforeThe massless vector pole of the Veneziano amplitude is the massless open-string vector seen in the two-tachyon exchange channel.
Past exam of the mathematics course of the University of Cambridge 2016 iii Paper 304 1 Solution Created 2026-10-03 Updated 2026-10-06
Use the Euclidean path integral with weight and Fourier transform convention . The momentum-space Feynman rules are: an internal scalar propagator ; a quartic vertex together with for incoming momenta; and an integral over each independent loop momentum. Multiply by the Feynman-diagram symmetry factor; external propagators are retained for a full correlation function and removed for an amputated connected correlation function.
To enumerate the requested one-particle-irreducible Feynman diagrams, let be the number of quartic vertices and the number of internal edges. Four external legs imply , while one loop order implies . Thus . Both internal edges must connect the two vertices: an alternative with a tadpole and one connecting edge would disconnect on cutting that edge. The only graphs are therefore the three pairings of labelled external legs, each with Feynman-diagram symmetry factor .
The corresponding channel momenta are , , and , with all external momenta incoming. This exhausts the connected one-loop four-point one-particle-irreducible Feynman diagrams.
For the Euclidean massive loop integral, first take and an integer , where the integral converges. Schwinger parameterization givesThe supplied Gaussian integral then yieldsIntegrating over , and using the Gamma function recurrence to obtain , provesOutside its initial convergence range, the right-hand side defines the meromorphic continuation used in dimensional regularization; the divergent ordinary integral is not being assigned a convergent value.
For the scalar bubble integral, the Feynman parameter identity followed by a translation of loop momentum givesWith , the Gamma function satisfies , and the parameter integral tends to one. Hence the ultraviolet pole is independent of external momentum:One can also see why the same pole occurs without introducing a Feynman parameter: at large , the difference between this integrand and is ultraviolet integrable near four dimensions, so both have the same dimensional regularization pole. The positive mass avoids an infrared ambiguity in this argument.
In the quantum effective action convention, the tree-level four-point vertex is , the negative of the amputated connected correlation function tree vertex. The three bubble corrections giveThus the pole counterterm is . The modified minimal subtraction scheme also subtracts the conventional finite combination, or equivalently absorbs it into the subtraction-scale convention; that does not change the one-loop renormalization-group beta function. Using that scale convention, writeHere is dimensionless, , and the conventional constant scale factor is implicit. The one-loop tadpole is momentum independent, so wave-function renormalization does not contribute at this order. Differentiating at fixed , with , gives . ThereforeFor the stable quartic scalar field theory, with , the one-loop quartic scalar beta function is positive: the running coupling increases toward the ultraviolet and decreases toward the infrared. The theory is not asymptotically free; extrapolating the one-loop flow gives a Landau pole.
Past exam of the mathematics course of the University of Cambridge 2018 iii Paper 137 1 Solution Created 2026-10-03 Updated 2026-10-05
The Mellin transform iswhere this integral converges. Rapid decay at infinity makes the integral over an entire function of , but alone gives no control near zero. With the additional expansion, the integral initially converges absolutely for .
For , expand . Absolute convergence justifies termwise integration, using the Gamma integral, and gives the scaled Bose integral
There is a missing hypothesis in the general continuation claim: one needs . The original PDF, like the TeX, only says that the sequence increases. Under the intended additional hypothesis, split the integral at one and subtract terms of the asymptotic expansion:Continuity of at zero makes it bounded on . The last integral is holomorphic on : on every compact subset, its integrand and all its derivatives are dominated by an integrable power of times a power of . The expressions for successive agree on their common initial domain and hence everywhere they overlap by the identity theorem. As , these half-planes cover . Taking with isolates the term , while all other terms are holomorphic near . Thus the corrected claim isThis is meromorphic continuation of a Mellin transform from an asymptotic expansion.
To show why the correction matters, set , , andwhere is continuous, equals one for and vanishes for . On , the remainder after division by is a uniformly convergent series of nonnegative powers of , so it extends continuously to zero with value . Away from zero, define by the required remainder quotient; it is continuous on the rest of as well. Thus all the printed hypotheses hold. Its Mellin transform equalsThe series continues meromorphically on and has genuine poles at , accumulating at . A meromorphic function on cannot have such an accumulation of poles. This is the accumulating asymptotic exponents obstruct Mellin continuation counterexample.
For the Gamma function, the Gamma function recurrence givesAt the numerator is and the product of the nonzero denominator factors is . Therefore the residues of the Gamma function are
The generating function of the Bernoulli numbers gives the convergent expansion near zeroThe subtraction proof above applies to this expansion, including its zero coefficients, which give no pole. At , for , the residue of its transform is , while the residue of is . Their quotient defines a holomorphic continuation of the Riemann zeta function at this point, including when the first residue is zero. Dividing gives the Bernoulli formula for zeta values at nonpositive integers:In particular the generating-series convention is , so the formula includes .
Past exam of the mathematics course of the University of Cambridge 2018 iii Paper 306 4 Solution Created 2026-10-03 Updated 2026-10-05
The closed string already contains the particle required for quantum gravity. For bosonic string theory in its critical dimension of string theory, the mass-shell condition and closed-string level matching areAt , the states are massless. After imposing the Virasoro constraints and quotienting null string states, the transverse polarization tensor decomposes into symmetric trace-free, antisymmetric and scalar pieces under . They describe a graviton, the Kalb–Ramond field and the dilaton. In particular, the symmetric trace-free sector is a massless spin-two graviton. Its number of degrees of freedom is .
The state–operator correspondence associates each physical string state with a string vertex operator. For example, the matter part of a massless closed-string vertex operator isIts conformal weights are , so its integrated form is invariant under changes of string worldsheet coordinates. At fixed insertion positions it is accompanied by from the worldsheet ghost fields. Physical vertices represent BRST cohomology classes; longitudinal changes of the polarization tensor are BRST-exact operators and decouple from physical scattering amplitudes. For the graviton, this string-state gauge redundancy becomes the linearized target-space diffeomorphism .
The Polyakov path integral computes a closed-string scattering amplitude by inserting the external string vertex operators, integrating their unfixed positions and the inequivalent worldsheet moduli, and including the worldsheet ghost fields. On a Riemann sphere three positions are fixed by the residual conformal group. For example, contractions of tachyon vertex operators produce the Koba-Nielsen factor, whose position integral gives the Virasoro–Shapiro amplitude.
The string dual resonance property means that one crossing-symmetric scattering amplitude has equivalent expansions in the different channels: its poles exhibit intermediate string states in each channel, rather than separate channel contributions being added again. The scattering-amplitude factorization at a pole identifies the intermediate particle and its couplings. Set , so and the closed-string tachyon has . For four such external states the Mandelstam variables obey . The Virasoro–Shapiro amplitude has generic -channel poles at , exactly the bosonic string mass spectrum in these units.
The massless pole gives an especially direct check. With overall normalization and , the Gamma function recurrence givesThusThe quadratic residue in contains a spin-two exchange. A scalar exchange alone could not give this angular dependence. Combined with the known massless spectrum and scattering-amplitude factorization, it identifies exchange of the graviton, with possible scalar contributions from the dilaton; the antisymmetric Kalb–Ramond field does not couple to two identical scalar tachyons here. Consequently the graviton participates in interactions, rather than being an isolated free state.
Decoupling longitudinal graviton polarizations forces a universal coupling to the conserved stress-energy tensor. Consistency of this massless spin-two gauge invariance extends the linearized coupling to the nonlinear dynamics of general relativity. At distances large compared with , the gravitational sector of the effective action begins with the Einstein-Hilbert action, alongside dilaton and Kalb–Ramond field terms and higher-derivative string corrections. The infinite tower of string states supplies the short-distance completion of this quantum gravity expansion.
Interactions are organized by string worldsheet topology. A constant dilaton gives the string coupling , and a connected oriented surface of genus has Euler characteristic . Its weight is ; with normalized external vertices,The sphere is tree order, the torus is one loop, and each additional handle adds a factor . Each coefficient is an integral over worldsheet moduli, so this is string perturbation theory in , with an independent low-energy expansion in .
For the one-loop vacuum contribution, torus modular invariance identifies with , . Integration is over the standard fundamental domain of the modular group,The potential short-proper-time region , responsible for a point-particle ultraviolet divergence, is absent. It would count metrics already represented elsewhere in . Torus modular invariance is therefore the geometric reason for one-loop ultraviolet finiteness. More explicitly, in the vacuum integrand is proportional to , with the Dedekind eta function; the complete expression is invariant under the modular group.
Ultraviolet finiteness does not make the bosonic one-loop vacuum energy finite. The remaining long-tube region is an infrared divergence from the tachyon: . It signals instability of the bosonic vacuum. Tachyon-free consistent backgrounds remove this particular obstruction, though other infrared effects must still be treated. The ultraviolet improvement comes from the extended string and the complete spectrum together with the worldsheet gauge identifications; the genus expansion remains a perturbative description of quantum gravity.
Past exam of the mathematics course of the University of Cambridge 2019 ii Paper 1 7A b Solution Created 2026-09-24 Updated 2026-09-29
Use the suggested substitutionIt maps to itself and satisfiesHencewhere is the complete elliptic integral of the first kind. Part (a) and the Gamma function recurrence now give
Past exam of the mathematics course of the University of Cambridge 2020 ii Paper 1 7E i Solution Created 2026-09-24 Updated 2026-09-29
Past exam of the mathematics course of the University of Cambridge 2021 ib Paper 3 17B c Solution Created 2026-09-24 Updated 2026-09-25
For monic orthogonal polynomials, taking the inner product of the recurrence with givesUsing part (b) and the Gamma function recurrence,
The Rodrigues expansion also shows that the coefficient of in isComparing the coefficients of intherefore givesand hence
Past exam of the mathematics course of the University of Cambridge 2022 iii Paper 320 3 a Solution 2026-09-28
At fixed radius, use spherical coordinates in velocity space with polar angle measured from the radial direction:The galactic distribution function and volume element give angular weightAll dependence on , , and cancels from ratios of second moments. Symmetry in the tangential plane givesWriting the angular integrals as beta function integrals and using the Gamma function recurrence,Therefore, for every admissible energy factor ,Equivalently, the velocity-anisotropy parameter is the constant .
