Past exam of the mathematics course of the University of Cambridge 2013 iii Paper 11 3 2 Solution Created 2026-10-03 Updated 2026-10-07
Write and . Choose unit complex numbers so thatis real and nonnegative. Cauchy-Schwarz, using , bounds byAfter setting , each summand is a unit phase times the Fourier coefficient on a finite abelian group of at frequency .
Let count pairs with and . Grouping terms, another Cauchy-Schwarz and the Parseval identity yieldwhere is the additive energy of a frequency graph. The last inequality uses for each .
ThusThe bound proves that derivative correlations force additive frequency energy. This energy counts exactly the ordered quadruples satisfying the two requested additive relations, after relabelling the difference equality as a sum equality. Shift equalities initially hold modulo , but the shifts lie in and , so they are also integer equalities. Frequency equalities hold in , as required. Consequently a common exponent works for both conclusions, after adjusting absolute implicit constants and using .
For the quadratic phase, the multiplicative derivative on the overlap is . Hence an explicit choice isTake . The correlation magnitude is , and every additive quadruple of shifts satisfies the frequency relation. The interval of shifts has such quadruples by Cauchy-Schwarz. These particular real frequencies need not belong to the grid used to prove existence for general ; evaluating them at the interval's integer coordinates gives the stated exact formula.
A genuinely nonlinear example is the bracket-linear frequency functionIt has exact additive quadruples, yet agrees with any affine function at only shifts, uniformly in the affine function. The question permits giving this example without proof, but the mechanism is useful. For a pair with sum , the value has only two possibilities, or . There are pair classes, so Cauchy-Schwarz produces collisions with both additive relations.
To see why no long affine agreement occurs, three agreement points force the corresponding lattice points to be collinear: eliminating the affine slope gives times an integer determinant equal to an integer, and irrationality makes that determinant zero. A rational line of slope contains agreement shifts in one residue class modulo , and closeness of to bounds their span by . Lines with large have arbitrarily small possible density; for bounded , irrationality of bounds the number uniformly. Thus the maximum agreement is .
Past exam of the mathematics course of the University of Cambridge 2013 iii Paper 11 4 Solution Created 2026-10-03 Updated 2026-10-07
Use the phase-distance convention for a Bohr set:where the norm is distance to the nearest integer. This is the Bohr set in phase-distance convention; the alternative condition has a different radius and corresponding constants. Assume , as requires.
For each residue , form in the -dimensional torus. Around these points place translates of a box of side length in each coordinate. For , their volumes sum to . If , two boxes overlap, and their distinct indices differ by a nonzero with every phase distance less than . Thus the Bohr set contains a nonzero element at the stated threshold. For larger radii the entire group is already present.
Letting the box width decrease to and using the finite number of possible nonzero , obtain a nonzero with for every . By the triangle inequality, lies in whenever . Since is prime, these multiples are distinct up to length . Therefore, in the usual radius range , the arithmetic progressionhas the requested length. For literally unrestricted , the correct lower bound is : a progression of distinct residues cannot exceed . For example, , makes the uncapped printed bound impossible. If is empty, the Bohr set is the whole group and has a progression of length ; the displayed exponent is simply inapplicable.
Now transfer the dense integer set to a larger cyclic group. Choose a prime , using Bertrand's postulate, and regard as a subset of . Its density is at least . Normalize Fourier coefficients on a finite abelian group and convolution byThe Fourier inversion and Parseval identity formulas are and ; convolution turns into multiplication of Fourier coefficients on a finite abelian group.
For , set . Then Parseval identity boundsAlso, with , the nonnegative convolution is supported exactly on the modular set , andThe frequencies outside contribute in absolute value at most . For , the real part of each phase from is at least ; moreover and . Hence . We have proved the cyclic Bogolyubov lemma with explicit phase radius:The previous Bohr set argument gives a modular arithmetic progression of length at least . The remaining step is lifting a short modular progression to an integer progression. Every one of its elements has a unique integer lift belonging to . Successive lifted differences lie in and are congruent to the same residue. Since , at most one integer in this difference interval represents that residue, so every successive difference is the same. The lifts form a genuine integer arithmetic progression, not merely a modular one.
For sufficiently large , the constants can be absorbed into half the exponent. ThusNo attempt to optimize this absolute constant is needed.
A counterexample comes from dense integer sets with only logarithmic-length progressions. To show that itself need not have such a progression, choose a random subset by independent inclusion with probability . Its size has mean and variance at most , so Chebyshev's inequality shows . There are at most increasing arithmetic progressions of any fixed length , and each belongs to with probability . Taking makes the union bound at most . For large , some realization has at least elements and no length- arithmetic progression. Take a subset of exactly elements. It retains that avoidance. Such a set has no progression longer than , and therefore no progression of length for any fixed , eventually. Interpret the prescribed exact cardinality on values of where it is integral, or use its integer part.
Past exam of the mathematics course of the University of Cambridge 2013 iii Paper 63 7 Solution Created 2026-10-03 Updated 2026-10-07
A stability analysis asks whether perturbations in an evolution remain bounded on a fixed physical time interval by a constant independent of the mesh. For a linear update , the basic requirement isThis permits physical growth bounded by ; it need not require every step to be a contraction. Consistency of a numerical method measures how well its stencil reproduces the differential equation, whereas stability of a numerical method controls the accumulated defects. For a well-posed linear problem, the Lax equivalence theorem connects stability and convergence for a consistent approximation in the specified norm.
On the whole line or a periodic grid, constant-coefficient stencils commute with translations. The Fourier modes diagonalize such spatial operators, turning a scalar one-step scheme intoThe Fourier symbol is found simply by replacing a shift by . The Parseval identity makes this modal calculation a norm estimate: if at every resolvable frequency, then the discrete norm cannot increase. More generally, a uniform gives and a fixed-time stability bound. All frequencies matter, including those near the grid scale; a small-wavenumber expansion establishes consistency but cannot test the entire stability range.
For the heat equation, put and . The Forward Euler diffusion scheme hasThe condition for every is exactly . This illustrates a parabolic time-step restriction . The Backward Euler diffusion scheme instead haswhich is contractive for every . The Crank-Nicolson diffusion scheme hasand is likewise unconditionally stable. However, its high-frequency amplification approaches for large , so it need not damp rapidly decaying modes effectively and may produce alternating transients from nonsmooth data. An A-stable time integrator can stabilize a diffusion semidiscretization with left-half-plane spectrum, but damping and accuracy remain separate questions.
For transport , define the hyperbolic Courant number . Forward Euler with a centered spatial difference gives and . It is unstable under a fixed nonzero CFL ratio. For , the upwind finite difference scheme hasThus is the stable range. Upwinding for negative reverses the spatial difference and gives . The Lax-Wendroff advection scheme gives a second-order example:It is stable for , but that result does not imply monotonicity or exclude oscillations at shocks. The Courant–Friedrichs–Lewy condition has a domain-of-dependence interpretation for explicit hyperbolic schemes; the Fourier calculation supplies the sharper algebraic stability criterion for the particular stencil.
For multilevel methods, each frequency gives an amplification polynomial and a companion matrix. Every amplification root must lie in the closed unit disk, with unit-modulus roots simple. Uniformity across frequencies and meshes is essential: one also needs control of the associated matrix powers, for example through a uniformly bounded diagonalization or the uniform power bound from separated amplification roots. A repeated unit root generates a Jordan block and polynomial growth in the number of steps. The same issue occurs for systems: checking only eigenvalue moduli of a nonnormal Fourier amplification matrix is insufficient. For instancehas unit eigenvalues but no fixed-time mesh-uniform bound as . For a hyperbolic system with a constant coefficient matrix admitting a uniformly bounded characteristic diagonalization, scalar modal estimates for each characteristic speed do give a system estimate. A positive energy symmetrizer provides another route. Frozen-coefficient Fourier tests for variable coefficients are useful local diagnostics, but a global estimate must also account for coefficient variation and boundary terms.
Boundary conditions change the allowable modes and can change the conclusion. A periodic calculation is exact only for a periodic closure. On a finite interval with zero Dirichlet boundary conditions, the centered diffusion matrix is diagonalized instead by the discrete sine transform. On interior points its eigenvalues are before scaling by . Thus the exact fixed-grid Forward Euler restriction iswhich tends to the whole-line threshold as the grid is refined. Backward Euler and Crank-Nicolson remain stable for all nonnegative . An appropriate Neumann boundary condition uses cosine-type modes and includes a constant mode with eigenvalue zero. The constant mode must be preserved, rather than forced to decay; depending on the boundary stencil, the natural discrete inner product may have endpoint weights.
For hyperbolic transport, the continuous and discrete boundary closures must respect the direction of propagation. If on , prescribe the left inflow value; prescribing an independent outflow value can overdetermine the problem. The energy method exposes this directly:Homogeneous inflow is dissipative, while forced inflow contributes data to the estimate. A compatible upwind boundary update has the same one-sided information flow. For diffusion,The last term vanishes for homogeneous Dirichlet or Neumann conditions. With Robin boundary conditions of dissipative sign it contributes a further nonpositive term. A non-dissipative boundary can produce genuine PDE growth, which a valid stability estimate should represent rather than incorrectly suppress.
Even a perfectly stable interior symbol cannot detect every defective boundary closure of a difference scheme. As a concrete counterexample, leave a stable explicit heat stencil on all grid points away from the left endpoint, but replace its first interior row by , still keeping . The interior Fourier symbol is unchanged, yet makes the finite-interval update unstable. The faulty boundary-adjacent row creates a mode absent from the periodic interior calculation. Thus the entire finite update matrix, or an energy estimate including its boundary rows, must be examined.
A more systematic boundary stability of a finite-difference method analysis transforms time and any tangential spatial variables, then looks for modes with and on a half-line. The interior dispersion relation selects decaying spatial modes; the boundary equations must determine their coefficients without admitting a nonzero growing homogeneous mode. A uniform inverse estimate for the boundary system, expressed by the Uniform Kreiss--Lopatinskii condition, also rules out arbitrarily weak boundary control near limiting frequencies. Merely excluding one obvious unstable mode is weaker than proving a mesh-uniform estimate. Discrete summation by parts and suitable boundary penalties offer an energy-based alternative.
Finally, nonlinear scalar conservation laws can develop discontinuities, so linearized Fourier stability is only one part of the analysis. A monotone conservative scheme, such as the Engquist-Osher method under its CFL restriction, has an invariant state range and L1 contraction of a monotone conservative scheme, with a discrete entropy inequality selecting admissible shocks. These complement the Fourier and boundary estimates. For an evolution problem, the meaningful outcome is a stability bound in a specified norm, uniform over the chosen mesh family and compatible with the actual boundary and initial data.
Past exam of the mathematics course of the University of Cambridge 2014 ib Paper 1 15A iv Solution Created 2026-09-24 Updated 2026-10-06
The Parseval identity for the orthonormal stationary-state expansion is . Splitting the amplitudes into even and odd indices givesMultiply by to obtain the requested identity. Its integer constants areThe series converges absolutely, since both parity contributions are . This derivation uses the normalization of the prepared state and completeness of the well stationary states, rather than a separate summation trick.
Past exam of the mathematics course of the University of Cambridge 2014 ib Paper 4 17D ii Solution Created 2026-09-24 Updated 2026-10-06
The Parseval identity for a complex Fourier series isIt holds for every square-integrable function, with complex conjugation expressed by the absolute squares. It is the Pythagorean theorem extended to the complete orthonormal system of complex exponentials.
Past exam of the mathematics course of the University of Cambridge 2014 ib Paper 4 17D v Solution Created 2026-09-24 Updated 2026-10-06
For real nonintegral , the Parseval identity and the coefficient formula giveSolving for the sum yieldsAt , the sine square is one and the cotangent is zero, giving .
Past exam of the mathematics course of the University of Cambridge 2014 iii Paper 66 5 b Solution Created 2026-10-03 Updated 2026-10-06
Use the Fourier transform for the spatial Cauchy problem, or its periodic analogue, and regard the two starting levels as independently perturbed data. A Fourier mode with spatial factor has amplification roots satisfyingPut and . ThenIf , both roots have modulus one and their separation is bounded below uniformly in frequency:The uniform power bound from separated amplification roots now controls the two-level companion matrix for every time step. Its entries are uniformly bounded, and its eigenvector conditioning is bounded by the reciprocal root gap. The Parseval identity transfers this frequency-uniform bound to the spatial norm. This proves stability, rather than merely checking each root's modulus.
If , the frequency has a root outside the unit disk, so there is exponential instability. If , the two roots at coincide on the unit circle. The companion matrix is not a scalar matrix and has a nontrivial Jordan block; its powers grow linearly in the number of steps. Frequencies arbitrarily near that value produce the same lack of a uniform bound for localized Fourier packets, so this also invalidates Cauchy stability, not only periodic plane-wave stability. At the double amplification root is , and at it is .
Therefore the full two-level stability range for a fixed positive Courant ratio isThe endpoint is moreover not a positive time step. Bounds deteriorate as approaches either endpoint; the displayed range is not a uniform claim over ratios arbitrarily close to one. A prescribed starter that removes one special parasitic component can change behavior for selected initial data, but does not establish the requested unrestricted two-level stability.
Past exam of the mathematics course of the University of Cambridge 2015 iii Paper 68 6 Solution Created 2026-10-03 Updated 2026-10-06
For an evolutionary partial differential equation, stability of a numerical method means that errors in the starting data and forcing stay controlled on each fixed interval , with constants independent of the mesh. Fourier stability analysis is especially effective for a constant-coefficient finite difference method on a uniform infinite or periodic grid: translation invariance makes different Fourier modes evolve independently.
For a scalar one-step scheme on the integer lattice, insert , with . The Fourier symbol of each shift is , and the update reduces toThe amplification factor must be defined for every relevant frequency; for an implicit scheme this includes checking that its denominator is nonzero. The Parseval identity turns a bound into a discrete L2 norm bound. Thus gives contractivity, while the more general bound gives for . Conversely, frequencies with amplification uniformly greater than one produce unstable wave packets or periodic Fourier modes under refinement.
For a system, is an amplification matrix; for a multilevel scheme, a companion matrix evolves the vector of time levels. The actual requirement is a uniform bound on matrix powers, not just their spectral radius. A nontrivial Jordan block at a unit-modulus eigenvalue creates polynomial growth in the time index. Even simple eigenvalues inside the unit disk can fail to give a uniform bound if the eigenvector matrices become ill-conditioned as the mesh changes. Uniformly controlled diagonalization of a matrix, or a suitable quadratic energy estimate, resolves this issue. The power boundedness of a two-level Fourier scheme illustrates why root multiplicities and conditioning matter.
For the heat equation, put . Centered space with the Forward Euler method hasThe Von Neumann stability analysis condition holds exactly for on the full frequency interval. The Backward Euler diffusion scheme instead has and is stable for every . The Crank-Nicolson diffusion scheme hasIt too is unconditionally stable, but poorly resolved high-frequency modes have for large , giving oscillatory numerical transients. A-stability therefore does not guarantee strong damping; L-stability distinguishes the damping of the Backward Euler method.
For the advection equation , let . Forward time and centered space giveFor nonzero fixed , repeated steps amplify some modes by a fixed factor greater than one, so the method is unstable under the usual refinement . This calculation also shows the refinement qualification: if , its finite-time growth can be bounded by , though the restrictive scaling defeats the usual hyperbolic time-step choice. For , the upwind finite difference scheme hasso it is stable for . The Lax-Wendroff advection scheme instead hasgiving . These examples separate the effects of the spatial stencil, temporal approximation and Courant number; a higher order of a numerical method by itself does not establish stability.
A periodic boundary condition is ideal for this analysis. The discrete Fourier transform diagonalizes the circulant update, with only the discrete frequencies needed on a grid of points. A bound on all is a convenient guarantee over every such grid. On the whole line, the Fourier transform uses a continuous frequency interval and the Parseval identity proves the corresponding square-summable-data estimate.
Homogeneous Dirichlet boundary conditions do not admit arbitrary complex exponential Fourier modes. For the standard centered second difference on interior points, a discrete sine transform diagonalizes the actual boundary-value matrix:The same heat amplification formulas then apply at these sine frequencies. For example, the exact forward-Euler contractivity limit on that fixed grid is ; the mesh-independent sufficient limit follows by including all frequencies. Homogeneous Neumann boundary conditions can similarly permit a discrete cosine transform, provided the endpoint discretization and its weighted inner product are chosen consistently. The constant mode is then present, reflecting conservation of the heat equation's spatial mean. For inhomogeneous boundary conditions, subtract a suitable lifting of the prescribed boundary data and estimate the induced source using the homogeneous evolution's bound and a Duhamel principle estimate. Bounds for the lifting and source must themselves be uniform in the mesh.
General boundary conditions require separate boundary stability of a finite-difference method. For an advection equation, incoming data are prescribed at the inflow boundary, while an outflow closure must respect the outgoing characteristics. An interior periodic Fourier symbol cannot detect a growing mode confined near a boundary. A normal-mode test on a half-line seeks modes with and that satisfy both the interior recurrence and boundary closure. Their existence proves instability, and uniform control requires more than merely excluding isolated growing roots; the Uniform Kreiss--Lopatinskii condition addresses boundary resolvent bounds. Alternatively, a direct energy method using summation by parts can include the boundary terms and establish the required estimate.
Variable coefficients and nonuniform meshes usually destroy exact Fourier transform diagonalization. Frozen-coefficient Fourier stability analysis is then a useful diagnostic, but it is not automatically a proof for the full variable-coefficient boundary problem. The mesh-uniform energy method in question 5 is an example of a direct proof. Also, an L2 norm proof is not automatically a mesh-uniform maximum-norm proof: finite-dimensional norm-equivalence constants may grow with the number of grid points.
Finally, stability measures error propagation; consistency of a numerical method measures the defect of inserting the exact solution. For a well-posed linear initial-value problem in the chosen norm, the Lax equivalence theorem says that a consistent approximation is convergent exactly when it is stable. Boundary discretization and starting data must be included in that assertion. A Fourier calculation proves convergence only after consistency, well-posedness and the actual boundary treatment have also been checked.
Past exam of the mathematics course of the University of Cambridge 2015 iii Paper 7 4 Solution Created 2026-10-03 Updated 2026-10-06
Identify with the Boolean hypercube , with coordinatewise multiplication and normalized Haar measure assigning mass to each point. Its Bernoulli function is a Rademacher random variable. The Walsh functions on a hypercube are the Walsh characters , indexed by all subsets , including . Independence of the sign coordinates gives . There are characters, equal to the dimension of , so they form an orthonormal basis.
Flipping coordinate negates exactly when . Thus the coordinate-flip generator on a hypercube satisfiesFor , Parseval identity gives . Equivalently its Dirichlet form of a Markov chain iswhere flips coordinate .
Set and . Since , its Walsh expansion contains only sets of even size. Every nonconstant such set has , giving the even-function spectral gap on a hypercubeAt each the Hahn-Banach theorem supplies a real supporting linear functional of norm at most one with ; for a complex normed space use the real part of a complex norming functional. At take . Convexity of the norm gives . Since , sum these inequalities to get . Therefore . Combining the two bounds yields the sharp Rademacher second-moment inequalityNo smoothness of the norm is required. The constant is sharp: two equal nonzero real coefficients give modulus or with equal probabilities. This is the second-versus-first moment case of the Kahane-Khintchine inequality for arbitrary normed vector spaces.
For the complex-circle assertion, write each independent Steinhaus random variable as , where is uniform on and all quadrant signs are independent. Conditional on the angles, is a Rademacher sum with complex coefficients and . Its conditional second moment is , independent of the angles. The preceding inequality gives its conditional first moment at least . Average over the angles and square to obtain the Steinhaus first-moment lower bound
Past exam of the mathematics course of the University of Cambridge 2015 iii Paper 7 5 Solution Created 2026-10-03 Updated 2026-10-06
On the polynomial subspace of , the Gaussian creation and annihilation operators are and . Gaussian integration by parts makes them adjoints there and gives . The Gaussian number operator isThe Probabilists' Hermite polynomials are . Their generating function is , giving . The commutator also yieldsThe orthogonality follows by repeated integration by parts; the leading coefficient is one, so these polynomials span every polynomial. By the allowed density assumption is an orthonormal basis. Define the closed number operator by on . This diagonal multiplication operator is self-adjoint and positive semidefinite; polynomial truncations show it is the closure of the polynomial operator.
The Ornstein-Uhlenbeck semigroup is , or . It also has the Mehler formula for the Ornstein-Uhlenbeck semigroupTo verify the formula, apply its right side to : the Gaussian moment-generating function makes the result , proving the eigenvalue identity on every Hermite polynomial. Positivity and invariance of Gaussian measure give contraction by Jensen inequality, so polynomial density extends the equality to all . The coefficient expansion givesFor continuously differentiable with bounded derivative, differentiate the Mehler expectation by the dominated convergence theorem to get the Ornstein-Uhlenbeck gradient commutation identityThe same identity extends to the Gaussian Sobolev form domain. Mehler's formula also gives pointwise convergence to the Gaussian mean for such , since bounded derivative permits at most linear growth.
The Gaussian Dirichlet energy is the energy form of :For polynomials this follows from , and closure extends it to the form domain. For the given continuously differentiable , Gaussian integration by parts yields , so Parseval identity proves the derivative-energy equality directly. Bounded derivative puts in the Gaussian Sobolev space, hence in that form domain; it need not be in the full operator domain, so is not always a legitimate initial definition.
Use the entropy functional . Begin with bounded smooth bounded away from zero, and put . Invariance and Gaussian integration by parts give the Ornstein-Uhlenbeck entropy dissipation identityThe gradient identity and the permitted weighted Cauchy-Schwarz inequality, applied with weight , implyThe entropy tends to zero as by Mehler's formula and bounded convergence. Integrating in time and using invariance yields . Set to obtain the sharp Gaussian logarithmic Sobolev inequalityFor the stated possibly unbounded , apply the argument to smooth positive clipped approximations with a common positive lower bound and uniformly bounded derivatives. They can converge pointwise together with their derivatives and have a common linear-growth bound. Since Gaussian measure integrates every polynomial moment, dominated convergence theorem passes both the entropy and the derivative energy to the limit. This also justifies use of the supplied inequality when itself is unbounded. The normalization concerns , and does not imply ; the second term in the entropy definition must be retained.
Past exam of the mathematics course of the University of Cambridge 2016 ib Paper 2 15C c Solution Created 2026-09-24 Updated 2026-10-06
With unit constant bending coefficients and unit length, the quadratic elastic energy isTo find the first loss of positivity, put . The clamped conditions give and . Extend periodically with period one. The sharp periodic Wirtinger inequality gives . For completeness, expand the mean-zero periodic in its Fourier series . Parseval identity giveswhich proves the inequality and its equality condition. The same holds for . ConsequentlyFor , the straight filament is the unique minimum. Equality in the periodic Wirtinger inequality has ; the endpoint value eliminates . Integration and the position boundary conditions therefore give the first buckling load and modes:At this load every such pair has zero quadratic elastic energy and solves the Euler-Lagrange equations. For these modes give negative energy, signaling instability of the straight configuration. The quadratic small-slope model does not select a finite post-buckling amplitude; nonlinear terms would be needed for that. This is Euler buckling of an elastic filament.
Past exam of the mathematics course of the University of Cambridge 2017 iii Paper 129 1 Solution Created 2026-10-03 Updated 2026-10-05
Use the usual convention that a three-term arithmetic progression must be nonconstant. There is a genuine small-case defect in the printed statement: for , , and , neither conclusion can hold with . We prove the intended Roth density-increment step for every odd integer , including the small values of , rather than silently treating as covered. Counting constant arithmetic progressions would make the first alternative vacuous for every nonempty .
Suppose that has no nonconstant three-term arithmetic progression, and let be its actual density of a finite subset. Thus . Put , , and embed in the cyclic group . A congruence with is an equality over the integers, since . Thus this embedding creates no extra three-term arithmetic progressions. Since is odd, multiplication by two permutes .
Use normalized Fourier analysis on a finite abelian group, with and . For the normalized trilinear arithmetic progression count, character orthogonality givesIndeed, expanding each function in its inverse Fourier transform, averaging in forces , and averaging in then forces . The Parseval identity in this normalization is .
Let , , and , the balanced indicator function of a finite subset. Then . Only constant arithmetic progressions contribute to , whereas summing over centers counts ordered three-term arithmetic progressions in . HenceSet . Telescope by changing one factor at a time:In the Fourier transform formula for each term, bound the Fourier coefficient of by and apply the Cauchy-Schwarz inequality to the other two factors. The permutation preserves the required squared sums. Since and , this givesIf , then , so the difference in the other direction is at least . Consequently Fourier detection of a progression-free subset of an interval gives a nonzero frequency with
To turn this into a density increment, take . The Dirichlet approximation theorem, proved here by placing into equal intervals modulo one, gives such that . DefineFor , the inequalities and show thatUse a progression partition with nearly constant linear phase: every residue chain of modulo has at least points. Split each chain into blocks of consecutive chain points, merging any remainder into its last full block. This partitions into arithmetic progressions with and common difference .
On a block with first point , the linear phase satisfiesLet . Freezing the linear phase on each block introduces error at most . ThereforeThe block sums total zero. Their positive part consequently totals at least . Since all block lengths total , some block has . Its density of a finite subset is at least .
For the remaining , divide into disjoint consecutive triples and a remainder of at most two points. A three-term-arithmetic progression-free uses at most two points of each triple, so ; the largest ratio occurs at . Any singleton set from then has density of a finite subset one, which is at least . We may therefore use the explicit constantsIn the large case the constructed arithmetic progression has length at least ; in the small case its length one does too. Thus these constants prove the intended result for all odd integers .
Past exam of the mathematics course of the University of Cambridge 2017 iii Paper 303 3 b Solution Created 2026-10-03 Updated 2026-10-05
Compute the coupling flow at zero magnetic field, or assume the source has support only in retained Fourier modes, so that the shell Gaussian measure is centered. First justify the shell propagator. For a positive quadratic kernel on the shell, the Gaussian functional integral gives covariance equal to the inverse kernel,with both momenta restricted to . Inserting the Fourier transforms and using the Dirac delta function yields the Gaussian shell covarianceChanging to gives the negative-exponent convention as well. The dependence is on the separation, as required by translation invariance. The printed numerator uses alone: it is correct only if that symbol means the separation or if . For example at , the covariance must equal the constant coincident value , while the literal printed integral generally depends on . Thus the missing separation is a real source qualification, not a change of Fourier-sign convention.
Integrating out the shell gives , up to a field-independent constant. At first order in , the cumulant expansion of a coarse-grained free energy retains . Odd moments of the centered Gaussian measure vanish and Wick theorem gives . Writing ,The last term affects only the constant; the second is a momentum-independent tadpole diagram correction. Before rescaling it changes to , leaves the quartic coefficient at , and produces no gradient correction. Combining with part (a) gives the leading coupling recursionThis is one-loop shell mass renormalization in scalar quartic theory. Here . The shell Gaussian is well defined if there; near this follows from . A negative slow-mode mass may be stabilized by the retained quartic term and does not require pretending that the full unconstrained quadratic measure at negative mass is normalizable.
Linearizing about givesIts mass and quartic eigenvalues are and . The off-diagonal term is an additive critical-mass shift: setting the bare is not generally the critical tuning when . The mass remains relevant for every ; at the two linear eigenvalues coincide and the matrix can have a Jordan block, but both perturbations are still relevant.
For , the quartic interaction is an irrelevant operator, and the Gaussian fixed point attracts weak quartic perturbations after the mass and magnetic field have been tuned. The quartic can nevertheless be a dangerously irrelevant coupling: its positive value stabilizes the ordered phase. For , it is a relevant operator, so the Gaussian description is unstable toward interactions. At it is a marginal operator; first-order perturbation theory alone does not decide its fate.
To settle that borderline case, retain the leading quartic contribution from the second cumulant, at zero external momentum in the local expansion. The two-fast-field part of is . Since , its connected second cumulant contributesKeeping the local quartic term and using Parseval identity gives , where . Thus the one-loop shell quartic renormalization isThe coefficient follows from multiplying the quartic density correction by . This local, zero-external-momentum coupling extraction is not a claim that the exact finite-shell effective action retains only its original polynomial; further operators are generated.
More explicitly, for a thin shell , let , and . With increasing length scale , the leading flow isThis convention has the opposite scale direction to a renormalization-group beta function defined using increasing momentum. On the tuned critical surface at , and , so weak positive is marginally irrelevant, tending to zero with logarithmic corrections rather than remaining an exactly marginal parameter. Just below four dimensions, with , the same calculation yields the Wilson-Fisher fixed point , . It does not justify extrapolating a small- expansion to every lower dimension. The ordinary scalar quartic upper critical dimension is therefore
Past exam of the mathematics course of the University of Cambridge 2017 iii Paper 340 2 a Solution Created 2026-10-03 Updated 2026-10-06
Fix the ordering of the orthonormal basis and write , taking indices from . The linear N-term approximation retains the first coefficients:Its index set is independent of , so the map is an orthogonal projection and is linear.
For the nonlinear N-term approximation, let be any set of indices with greatest , resolving ties arbitrarily and padding with zero coefficients if necessary. Since , such a selection exists. ThenThe Parseval identity shows that for any fixed index set the displayed coefficients minimize the Hilbert space error. Choosing the largest squared magnitudes minimizes the omitted sum over every set of at most indices. This is the best N-term approximation; dependence of the selected indices on makes the operation nonlinear.
Past exam of the mathematics course of the University of Cambridge 2017 iii Paper 340 2 d Solution Created 2026-10-03 Updated 2026-10-06
Let bound the piecewise Hölder norms, and let . At scale , classify a wavelet as bad when its localized support of a function meets a discontinuity, and as good otherwise. Compact support and bounded overlap imply at most bad wavelets per scale and at most total wavelets. A good support lies in a single smooth piece, so the preceding wavelet coefficient decay for Hölder functions gives . The permitted bounded-function estimate gives for bad coefficients.
Consider a specific N-term approximation: retain all coarse scaling functions and all wavelets through scale , and also retain the bad coefficients at scales , where . Since , for large . Its term count is bounded byBy the Parseval identity, its squared L2 norm error is the sum of omitted squared coefficients. The good tail satisfiesand the remaining bad tail satisfiesChoose with a fixed sufficiently large ; then the extra terms fit the remaining budget for all sufficiently large , while is comparable to . Padding to terms does not increase the error. The best N-term approximation is at least as accurate as this constructed selection, soThe constant may depend on the finite number of jumps, the piecewise Hölder norms, and the fixed wavelet family. Values exactly at the jumps are immaterial in . The improvement comes from retaining only a bounded number of jump-crossing coefficients at each finer scale, as in best N-term wavelet approximation of piecewise Hölder functions.
Past exam of the mathematics course of the University of Cambridge 2017 iii Paper 341 7 Solution Created 2026-10-03 Updated 2026-10-06
For a fully discrete linear partial differential equation evolution, write in a specified discrete norm. Finite-time stability of a numerical method meanswith independent of the allowed meshes. For a multistep method, augment the state with its time-history values and include a stable starter. Bounds may grow as when the continuous problem has growth; demanding decay for all time would be a stronger assertion. The treatment of initial data, forcing, Dirichlet boundary conditions, and the norm is part of the hypothesis, not a detail supplied by an interior calculation.
For constant coefficients on an infinite or periodic uniform grid, Von Neumann stability analysis substitutes the Fourier mode . The discrete Fourier transform and the Parseval identity turn a uniform multiplier estimate into a discrete L2 norm estimate. In a one-step scalar scheme, proves contractivity, while gives finite-time stability. In a multilevel scheme, every root of a polynomial in its amplification polynomial of a multilevel finite difference scheme matters, as does uniform control of the associated companion matrix.
For the Forward Euler diffusion scheme with ,Thus for all frequencies exactly whenSufficiency follows since ; necessity follows from the highest-frequency mode on even periodic grids. This gives the usual parabolic mesh restriction . For the Backward Euler diffusion scheme,so every is stable. The Crank--Nicolson method hasagain of modulus at most one for all , but stiff modes approach rather than zero. These are PDE manifestations of the A-stability and L-stability distinctions for time integration.
For advection , assume and set . The upwind finite difference scheme is , withIt is contractive when . This also has a direct maximum-norm proof: each newest value is a convex combination of two old values. The resulting Courant–Friedrichs–Lewy condition expresses that the numerical domain of dependence covers the physical one. For negative , the upwind direction must be reversed. By contrast, Forward Euler method time stepping with the centered first finite difference has and for nonzero . Under fixed nonzero refinement it is unstable, since a fixed nontrivial mode grows geometrically over steps. This is not a claim of instability under every imaginable mesh coupling: instead bounds its spurious finite-time growth, since .
The leapfrog advection scheme illustrates a multilevel subtlety. Its amplification polynomial of a multilevel finite difference scheme isFor , both roots of a polynomial have unit modulus and separation at least . A uniformly conditioned eigenbasis of the companion matrix then proves power boundedness of a two-level Fourier scheme for arbitrary history data. At and a grid admitting , the roots coincide on the unit circle, and a Jordan block produces growth. Thus the endpoint fails the ordinary arbitrary-history root condition for a multistep method; checking only that both moduli equal one misses the instability. Restricting the starter or filtering the parasitic mode is an additional hypothesis.
The energy method handles variable coefficients and finite boundaries for which Fourier stability analysis may be unavailable. For the Backward Euler diffusion scheme with homogeneous Dirichlet boundary conditions, define . The discrete summation by parts identity isTake the inner product of with . The elementary identity givesThis proves unconditional contractivity including the actual boundary treatment. More generally, a dissipative operator gives of operator norm at most one: with , dissipativity implies , and the Cauchy-Schwarz inequality gives . In finite dimensions this also proves invertibility. For the second-order backward differentiation formula, the BDF2 discrete energy identity used in question 4 controls both time levels and proves unconditional diffusion stability. A repeated root strictly inside the disk is harmless here; the energy argument supplies the uniform bound without a singular eigenvector formula.
A second technique is eigenvalue stability analysis of a finite difference method. Under the method of lines, a time integrator advances by . If is a normal matrix, the scalar linear stability domain criterion on all controls its powers exactly in the corresponding L2 norm. For the centered Dirichlet discrete Laplacian, its eigenvalues lie between and zero; intersecting this interval with a time integrator's stability interval determines its mesh restriction. A uniform bound on the diagonalization of a matrix is needed for nonnormal diagonalizable systems. Eigenvalues alone can be misleading. For example,has both eigenvalues inside the disk for , but the upper-right entry of is . Taking , , this grows like . Hence stable scalar eigenvalues do not give mesh-uniform stability. Boundary closures can introduce precisely the extra growth that an interior Fourier symbol overlooks, so boundary stability of a finite-difference method must be checked separately.
For nonlinear spatial discretizations, a useful replacement for Fourier stability analysis is a convexity argument. If the Forward Euler method map is nonexpansive in a chosen norm for , the second-order strong stability preserving Runge-Kutta methodis also nonexpansive under the same restriction. Indeed, for two inputs, the triangle inequality gives . Its Taylor expansion is , proving order two. Such strong stability preserving Runge-Kutta methods transfer suitable forward-step bounds without relying on a linear spectral calculation.
Finally, consistency of a numerical method explains what stability buys. For a linear Hadamard well-posed problem, the Lax equivalence theorem equates convergence of a consistent discretization with its stability, under the stated approximation-space and norm hypotheses. Directly, if the error satisfies , iteration gives the discrete Duhamel principleThus vanishing normalized local truncation error and initial error yield convergence. With a nonlinear Lipschitz step estimate, the discrete Gronwall inequality plays the same role. A stable but inconsistent stencil, such as the literal defective formulas in questions 3 and 4 under their usual refinement, is not rescued by any amplification bound. The decisive checks are a mesh-uniform evolution bound, a valid boundary closure, and consistency with the actual PDE.
Past exam of the mathematics course of the University of Cambridge 2018 iii Paper 105 3 b iii Solution Created 2026-10-03 Updated 2026-10-05
First construct the Dirichlet eigensystem. By part (i) and the Lax-Milgram theorem, the inverse exists from into . Elliptic regularity gives , and the Rellich-Kondrachov compactness theorem makes a compact operator on . Symmetry givesso is a self-adjoint operator. It is injective, and for . The spectral theorem for compact Hermitian operators supplies an orthonormal basis with , , and . Put . ThenRepeated elliptic regularity makes each smooth up to the boundary.
As with part (ii), the stated characterization omits boundary compatibility. At , its right-hand side is finite for every by Parseval identity, whereas its left-hand side would require . At higher orders, the preceding polynomial example is also decisive for the Dirichlet eigensystem: on ,Although this function belongs to every ordinary , the weighted sum diverges at .
The correct spectral characterization of elliptic Dirichlet domains iswith defined in part (ii). For finite eigenfunction sums the squared norm is exactly the weighted sum: even orders follow from , and odd orders also use . These sums are dense in : if is -orthogonal to every , then , so . They are dense in by construction, and the isomorphisms propagate density to every . Completion in the equivalent norms from part (ii) proves both directions of the corrected equivalence. In particular, the printed characterization is correct for .
Past exam of the mathematics course of the University of Cambridge 2018 iii Paper 105 3 c Solution Created 2026-10-03 Updated 2026-10-05
Take , the natural initial-data assumption needed for the requested convergence but not explicitly stated in this subpart. Use the genuine Dirichlet eigensystem from part (b)(iii) and put . The spectral construction of a parabolic solution isEach finite sum has zero Dirichlet boundary condition and satisfies .
To justify all derivatives, fix . For any integers , the squared norm of a tail of the th time derivative isuniformly for . Use the corrected Sobolev domains of powers of an elliptic Dirichlet operator, not the inaccurate unqualified assertion in the PDF. Their norm equivalence gives convergence in every ordinary . The Sobolev embedding theorem then gives convergence of every desired spatial derivative by choosing large enough. Thus termwise differentiation is justified, , and for every . The trace remains zero, and holds pointwise.
Finally Parseval identity givesby the dominated convergence theorem, since each factor tends to zero and is bounded by one. No boundary compatibility of the initial data is required for positive-time smoothness.