Past exam of the mathematics course of the University of Cambridge 2015 ii Paper 4 6B Solution Created 2026-09-24 Updated 2026-10-06
The Papperitz symbol describes a homogeneous second-order Fuchsian differential equation on the Riemann sphere, with three regular singular points and the two indicial exponents at each. At a finite point , an exponent corresponds locally to a factor ; at infinity it corresponds to . Repeated or resonant exponents can require logarithms. The six exponents obey the Fuchs relation that their sum is one. With three singularities there is no independent accessory parameter, so these data determine the normalized equation.
A Möbius transformation of a Papperitz symbol moves the singular points without changing their exponents. Multiplication of the dependent variable by a local power shifts the two exponents by that power, using the reciprocal local coordinate at infinity.
Set and . At , the exponent pairs for are respectively , , and . The Möbius transformation sends to . Before rescaling, therefore has exponent pairs at zero, at one, and at infinity. Multiplication by subtracts at one and adds at infinity. The resulting pairs areexactly those of the original Gauss hypergeometric equation. Near zero choose the branch with at zero. The resulting solution is analytic there and has value one. Uniqueness of the normalized analytic solution yieldsThis is the Pfaff transformation. The normalization presupposes the usual admissible parameters, in particular . Elsewhere the identity holds by analytic continuation with compatible branches, or by parameter limits wherever those are defined.
Past exam of the mathematics course of the University of Cambridge 2017 iii Paper 137 1 Solution Created 2026-10-03 Updated 2026-10-07
Use the Fourier transform normalization . A precise version of the Poisson summation formula is that, for every Schwartz function ,Both series have absolute convergence. Here being a Schwartz function means being smooth with for all nonnegative integers . This hypothesis makes the sums and the termwise operations below legitimate; the formula is not being asserted for arbitrary integrable functions.
Form the periodization of a Schwartz function . Every differentiated series has uniform convergence on , by rapid decay, so is a smooth periodic function of period one. Its th Fourier coefficient isThe interchange follows from absolute convergence uniformly on this interval. Integration by parts shows that these Fourier coefficients decrease faster than every inverse power of . Thus the Fourier series converges absolutely and uniformly to ; for example the standard convergence theorem for twice continuously differentiable periodic functions applies. Evaluating at zero proves the Poisson summation formula.
For , scaling the given Gaussian Fourier transform givesApplying the Poisson summation formula gives the real-parameter Jacobi theta function transformation
For , termwise application of the Mellin transform is justified by integrating absolute values and givesEach positive contributes , and the factor one half removes the equal positive and negative terms. This is the Mellin representation of the completed Riemann zeta function.
Split the integral at one. On , the Jacobi theta function transformation writes . Integrating the first two terms and substituting in the third yields the pole-subtracted theta integral for the completed zeta function:The remaining integral is an entire function of : , and on each compact set of this exponential dominates all powers of and all factors arising from differentiation. It therefore supplies a meromorphic continuation of to the whole plane, with simple poles at one and zero, of residues and , respectively. Its expression is unchanged by .
The reciprocal Gamma function is entire, with simple zeros at its nonpositive integer arguments. Consequentlyis holomorphic everywhere except for a simple pole at , of residue one. The apparent pole at cancels; in fact gives . This proves the analytic continuation of the Riemann zeta function. The completed functional equation of the Riemann zeta function isThese are equalities of meromorphic functions; at apparent singularities they are interpreted by continuation. Equivalently, is an entire function with .
Past exam of the mathematics course of the University of Cambridge 2017 iii Paper 137 2 Solution Created 2026-10-03 Updated 2026-10-07
Extend the Dirichlet character periodically to all integers, putting when . Its Dirichlet L-function, initially on , isThe Euler product follows from unique prime factorization and absolute convergence; a Dirichlet character is completely multiplicative on this extension. Write for the principal Dirichlet character, equal to one on units and zero elsewhere.
For a nonprincipal Dirichlet character , character orthogonality gives . Explicitly, multiplication by a unit with permutes the unit residues and multiplies this sum by , forcing it to vanish. For setThe numerator is at zero and the denominator is , so is bounded, indeed analytic, near zero; it decays exponentially at infinity. Initially for , absolute convergence justifiesThis Mellin transform integral is holomorphic for , locally uniformly in , and division by the Gamma function proves the requested analytic continuation to the left of the line one.
In fact, the same argument proves Mellin continuation of a nonprincipal Dirichlet L-function to the entire plane. If at zero, subtract this Taylor polynomial on and add its explicit integrals:The last integral is holomorphic on . Its possible simple poles at nonpositive integers cancel against zeros of . Letting increase shows that is an entire function, without any primitivity assumption.
For real , use the absolutely convergent Euler product logarithmThe higher-power remainder has the uniform estimateFor a nonprincipal Dirichlet character , invoke the allowed nonvanishing of a nonprincipal Dirichlet L-function at one. Its holomorphy and nonvanishing give a holomorphic logarithm on a small disk about one. On the connected real interval , differs from this logarithm by a fixed element of : the difference is continuous with exponential one. Thus the prime-character sum near one is bounded. This branch argument is needed for complex-valued Dirichlet characters.
For ,The finite product has a positive limit as , and the residue-one pole of the Riemann zeta function givesHere for nonprincipal Dirichlet characters means bounded complex magnitude.
Finally, for a residue class coprime to , Orthogonality of Dirichlet characters givesOnly primes not dividing occur, so the character orthogonality applies to every term. This sum diverges as . A finite collection of primes would give a bounded sum, a contradiction. Every reduced residue class contains infinitely many primes. This is the Dirichlet theorem on primes in arithmetic progressions; the coprimality hypothesis is essential.
Past exam of the mathematics course of the University of Cambridge 2017 iii Paper 304 1 iv Solution Created 2026-10-03 Updated 2026-10-05
The frequency-independent measure factor cancels in the ratio:Upon analytic continuation in frequency, this has poles at and zeros at , for , apart from cancellations when numerator and denominator vanish together. They match those of . To justify equality, rather than merely matching this divisor, use the hyperbolic-sine infinite productIt follows from pairing the Weierstrass product for the reciprocal gamma function at opposite imaginary arguments and using the Gamma reflection formula. Normal convergence on compact sets follows from . Thus, for positive real frequencies,The cutoff ratio determines the frequency dependence, agreeing with the thermal partition function of a quantum harmonic oscillator.
The printed assertion about an absolute needs a normalization qualification: independence of from alone does not ensure existence of a nonzero limit. Set . Then , so if , its limit is . The choice gives the required limit. Conversely, is a positive frequency-independent normalization with no absolute limit, though every ratio above still converges. Matching zeros and poles alone also permits nonvanishing entire functions; the convergent normalized product is what rules them out here.
Past exam of the mathematics course of the University of Cambridge 2017 iii Paper 304 3 v Solution Created 2026-10-03 Updated 2026-10-05
At fixed and , the radial integrands in the scalar shell integral in six dimensions behave as in the ultraviolet. ThusConsequently the proper box quartic vertex is ultraviolet finite. The reducible triangle insertion contains the logarithmic cubic-vertex divergence. The internal bubble contains the mass divergence; its nonzero-momentum expansion additionally has a logarithmic kinetic-term divergence. An unsubtracted tadpole attachment has the quartic one-point divergence and is canceled by tadpole subtraction, or incorporated consistently in the chosen stationary background.
These divergent lower-point subgraphs require counterterms, not a new independent divergent coupling. More generally, cubic graph identities give the superficial degree of divergenceVacuum, one-, two-, and three-point functions need vacuum-energy, linear, mass, wave-function renormalization and cubic-coupling counterterms. Proper higher-point functions have negative superficial degree and are finite after subtraction of divergent subgraphs. Hence the perturbative renormalizability of cubic scalar theory in six dimensions permits a continuum perturbative expansion with finitely many renormalized parameters; induced irrelevant higher-field and derivative interactions are finite predictions at finite scale, not an infinite list of ultraviolet parameters.
This perturbative statement does not establish a nonperturbative real Euclidean measure. For any nonzero real , tends to in one field direction, so even its constant-field integral is divergent. Thus Stability or a specified analytic continuation would be additional input beyond these loop calculations.
Past exam of the mathematics course of the University of Cambridge 2017 iii Paper 306 3 Solution Created 2026-10-03 Updated 2026-10-05
Use a genus-zero Polyakov path integral, target signature and all four momenta incoming. Define the Mandelstam variables by , , . The string mass-shell condition is and momentum conservation is , soThe reduced amplitude omits the overall momentum-conservation Dirac delta distribution and any conventional overall scattering-matrix phase. Interpret each tachyon vertex operator as normal ordered; self-contractions must not be included.
The free-boson worldsheet propagator is . Wick theorem then evaluates the normal-ordered exponential correlator as the Koba-Nielsen factorThe worldsheet zero mode supplies momentum conservation. Four vertices give and the sphere contributes through the string genus expansion, leaving .
The sphere's residual Möbius transformations fix three insertion points. In sphere gauge fixing for four string vertices, take . The worldsheet ghost fields, equivalently the Faddeev-Popov determinant for this residual group, supply . At finite , this determinant grows as , while the matter correlator decays as by momentum conservation and the tachyon mass shell; their product has a finite limit. Equivalently the weight- matter operator at infinity is normalized with , while the ghost pair uses the inverse factor. After stripping the fixed-position normalization, onlyremains. Here choose the standard complex-coordinate measure . With instead, the reduced normalization must acquire a factor two to give the same requested amplitude. Absolute vertex/sphere normalization is a convention; this choice fixes it consistently with the displayed prefactor.
Set , , . Then and the integrand is . To evaluate the complex beta integral first work where . This ensures convergence near , and infinity. Set and . Schwinger parameterization givesCompleting the square, the Gaussian integral is . Change variables to and ; their Jacobian determinant is . Integrating gives . The remaining integral is the ordinary beta function . ThusRestoring and substituting the Mandelstam variables proves the Virasoro–Shapiro amplitude:For physical scattering the original position integral generally fails to converge. The formula defines the amplitude by analytic continuation from the convergence domain, with the desired scattering boundary value at real poles; the convergent integral should not be claimed valid for every physical momentum.
The gamma function is a meromorphic function, with simple poles at nonpositive integers and no zeros, while its reciprocal is an entire function. Consequently it is a meromorphic function of the independent invariants and a crossing-symmetric scattering amplitude: permuting leaves it unchanged. Its generic channel poles areand the same tower in and . They represent the exchanged bosonic string mass spectrum: the tachyon, massless states including the graviton, and an infinite sequence of massive closed-string levels. There are no threshold branch cuts at this tree order.
A precise check of factorization is the Virasoro–Shapiro amplitude pole residue. Near , set in the nonsingular factor. The gamma function recurrence givesCombining this with showsThe residue is a degree- polynomial in , consistent with exchange up to spin and scattering-amplitude factorization. At exceptional kinematics reciprocal Gamma zeros can remove apparent poles. In particular, when both and approach nonpositive integers, the zero from cancels the putative double pole, leaving channel simple-pole terms. Thus one must not count products of numerator poles without using .
Past exam of the mathematics course of the University of Cambridge 2017 iii Paper 311 4 Solution Created 2026-10-03 Updated 2026-10-06
Work first in units . Black-hole thermodynamics starts with the laws of black-hole mechanics. For a connected stationary regular Killing horizon, the Zeroth law of black-hole mechanics makes its surface gravity constant when the Einstein field equations and dominant energy condition hold. This parallels uniform equilibrium temperature. For neighboring stationary asymptotically flat four-dimensional Einstein-Maxwell black holes, the First law of black-hole mechanics iswhere is the horizon angular velocity and its electric potential relative to infinity. The last terms are rotational and electromagnetic work, analogous to the work terms in . The second law of black-hole mechanics, expressed by Hawking's area theorem, makes the total future horizon area nondecreasing for classical matter obeying the null energy condition and global assumptions such as strong asymptotic predictability. The third law of black-hole mechanics is an unattainability statement: subject to the usual regularity assumptions, the weak energy condition, and a bounded stress-energy tensor, no finite physical process reduces the surface gravity of a regular horizon to zero. It does not assert that extremal black holes have zero area or zero entropy.
Classical geometry alone fixes an analogy, not a nonzero physical temperature. Quantum field theory supplies Hawking temperature . Comparing the area term of the first law with gives for nonextremal holes, and hence the standard Bekenstein-Hawking entropy. Restoring physical constants,Here restored is the physical acceleration surface gravity and is the squared Planck length. The first-law comparison determines the area coefficient, leaving an additive entropy constant unspecified; the displayed formula is the usual convention. Its area scaling, rather than ordinary volume scaling, signals that the available thermodynamic degrees of freedom of a gravitating system are constrained by the horizon geometry.
Particle production explains the quantum input. For a real scalar field obeying the massless covariant wave equation, the Klein-Gordon inner product on solutions isThe wave equation makes its current divergence-free, so the product is independent of a Cauchy hypersurface when the boundary flux vanishes. Use suitably normalized wave packets or the appropriate continuum distributions. Asymptotic past and future Minkowski spacetimes give preferred positive-frequency solution mode bases and , normalized by , and . Between those regions, a nonstationary geometry generally has no preferred positive-frequency splitting.
The mode bases are related by a Bogoliubov transformation,The last identities follow from conservation of the Klein-Gordon inner product and are the canonical identities for a bosonic Bogoliubov transformation. Expanding the field in either complete basis and extracting its positive-frequency coefficient givesThus the state annihilated by every hasThis particle number from Bogoliubov coefficients is nonzero when time evolution mixes positive and negative frequencies. It describes one state as vacuum in the early particle basis and populated in the late basis; it does not require an arbitrary choice of a vacuum at each intermediate time. For a finite implementable transformation the state remains a squeezed pure quantum state. In infinitely many modes, a common unitary bosonic Fock space implementation requires additional conditions, such as the beta map being a Hilbert-Schmidt operator; finite wave packet observables need not share every global divergence of an ideal continuum calculation.
For gravitational collapse, the future is not globally Minkowski: it contains a black hole. The relevant late out-basis includes modes reaching future null infinity together with modes entering the future event horizon. Modes on infinity alone are not a complete Cauchy basis. Nevertheless the same mode-mixing calculation determines the outgoing particle flux. Near a nonextremal horizon, the logarithm in the tortoise coordinate converts regular early null coordinates into the Hawking exponential ray mapwhere is the early affine null coordinate and is late retarded time at infinity. Backward propagation of a late outgoing mode gives a factor for . Its positive- and negative-frequency Fourier integrals differ by analytic continuation around the logarithmic branch. With a convergence regulator they reduce toTaking gives the thermal ratio of Hawking Bogoliubov coefficients, . Combining this with the canonical normalization yields the bosonic occupation . Late-time wave packets turn formal continuum coefficients into a finite number flux.
A Schwarzschild black hole has , so this spectrum has . The exterior curvature potential partly reflects the outgoing modes. Its transmission probabilities are the greybody factors, giving, for one massless scalar species in the stationary late-time approximation,Thus the horizon temperature is universal, while the spectrum received at infinity is not a perfect featureless blackbody radiation spectrum. The derivation assumes the near-horizon quantum state inherited from a regular collapse vacuum and ignores rapid backreaction over the timescale of the packets. Exponentially blueshifted precursor frequencies indicate the usual short-distance assumption in the semiclassical calculation, not an independently demonstrated quantum-gravity description of the endpoint.
The outgoing positive energy flux drives black-hole evaporation. At leading order for a large isolated Schwarzschild black hole, area scales as and temperature as , giving and a lifetime of order , with the coefficient depending on species and transmission factors. Its negative heat capacity of a Schwarzschild black hole, , means that it heats up as it loses mass and cannot be in stable canonical equilibrium with an unlimited thermal reservoir. The approximation fails when quantum-gravitational scales are reached and does not fix whether the endpoint is complete evaporation, a remnant, or something else.
Area loss does not contradict the classical area theorem, because the quantum stress-energy tensor need not obey the classical null energy condition; negative horizon energy accompanies the positive outgoing flux. The appropriate thermodynamic statement is the generalized second law, involving . Finally, Hawking radiation raises the black hole information paradox: if a pure quantum state completely evaporates into an exactly thermal mixed quantum state and its partners disappear, the result conflicts with ordinary unitary time evolution. Early outgoing radiation can be mixed simply because of its entanglement with interior modes while the complete state remains a pure quantum state; that fact alone is not information loss. Resolving the fate of all correlations through the evaporation endpoint requires more than the leading semiclassical flux calculation.
Past exam of the mathematics course of the University of Cambridge 2017 ii Paper 1 13E Solution Created 2026-09-24 Updated 2026-10-05
For , expand . Absolute integrability, bounded near zero by a constant times , permits termwise integration. The Gamma function integral then givesSpecify the Hankel contour to run from the negative real axis below the cut towards zero, circle zero counterclockwise with fixed radius , and return above the cut. Take . The circle excludes every nonzero pole of the denominator. For its radius can shrink to zero, and the lower and upper rays give respectively and . By the Gamma reflection formula,away from integer in that half-plane, with limiting equality at the removable points.
With the circle radius kept fixed, the contour integral is an entire function of : the rays decay exponentially and converge locally uniformly, and the circular part stays away from zero. Multiplication by is analytic for and meromorphic globally. The apparent singularities at positive integers are removable because the integral vanishes there and it already agrees with the original analytic zeta function nearby. Only remains a simple pole. Thus this constructs the meromorphic continuation to the whole plane and an analytic continuation on . Keeping the circular segment is essential when the integral is not locally integrable at zero.
At the two rays cancel, since is single-valued. The Laurent expansionshows that the residue of is . Since , the residue theorem gives
Past exam of the mathematics course of the University of Cambridge 2017 ii Paper 2 12E Solution Created 2026-09-24 Updated 2026-10-05
Choose a branch of with and write . The Gauss hypergeometric equation isChanging to turns this into the same equation with third parameter . Its solution regular at is . Substitution of gives the Gauss hypergeometric equation for with parameters . Thus is also a solution. Their distinct nonintegrally separated characteristic exponents at a regular singular point imply linear independence near , soThis also follows from the exponent transformation in the Papperitz symbol.
Initially assume , so the Euler integral for the hypergeometric function applies. If , taking gives an integral evaluated by the beta function and determinesIndeed and there. To find , instead work in the nonempty parameter region , put and scale the integration variable by . The singular contribution isThe defining integral of the beta function evaluates this integral to . Since and , it determines . Analytic continuation in the parameters then yields the hypergeometric connection formulaThe usual power-series definition requires ; exceptional parameters need a separately specified analytic limit. Reciprocal gamma functions make vanishing coefficients at denominator poles meaningful. The branch must remain fixed when continuing in .
Past exam of the mathematics course of the University of Cambridge 2018 iii Paper 137 2 ii Solution Created 2026-10-03 Updated 2026-10-05
Extend each Dirichlet character periodically to all integers by setting when . Its Dirichlet L-function is initially the absolutely convergent Dirichlet seriesThe group has elements. Part (i) gives, for every integer , including nonunits,For a nonunit both sides are zero, because the fixed class is a unit. Interchanging the finite character sum and the absolutely convergent Dirichlet series therefore provesThis is Orthogonality of Dirichlet characters applied to a residue-class indicator. Any meromorphic continuations of the two sides agree by uniqueness of analytic continuation.
Past exam of the mathematics course of the University of Cambridge 2018 ii Paper 1 7B a Solution Created 2026-09-24 Updated 2026-10-03
In the integral defining the Beta function, put and then . Symmetry givesThis proves the identity for . Both sides extend by analytic continuation to the same meromorphic function, equivalently by the beta--gamma identity and the Gamma duplication formula. The identity holds meromorphically for every ; both sides are finite for and have matching poles at the excluded points.
Past exam of the mathematics course of the University of Cambridge 2019 ii Paper 2 13A a Solution Created 2026-09-24 Updated 2026-10-03
Take the Hankel contour about the negative real axis: it starts at just below the cut, circles the origin counterclockwise, and returns to just above the cut. The notation records this return side of the branch point. Choose the branch for .
For , collapse the contour onto the two banks of the cut. Their phase difference is determined byand the resulting sine factor, together with the Gamma reflection formula , recovers the integral representation (‡). The Hankel integral converges exponentially at its unbounded end and, after the standard local subtraction of the singular part at zero, depends holomorphically on . The apparent poles of are cancelled by zeros of the contour integral except at . It therefore gives the analytic continuation of the Riemann zeta function to .
Past exam of the mathematics course of the University of Cambridge 2019 ii Paper 4 33B b Solution Created 2026-09-24 Updated 2026-10-03
Write . Sincethe outgoing-to-incoming coefficient ratio is . Hence the Partial-wave S-matrix element and scattering phase shift areThe scattering length from a partial-wave S-matrix is
For the stated S-wave scattering solution, as , soUsing the same incoming/outgoing convention,Therefore
The analytic continuation of has a bound-state pole of the scattering amplitude at in the upper half-plane. Since , this givesTo extract the bound-state wavefunction, multiply the scattering solution by and take the pole limit:Thus the unnormalized exterior solution isThese data are collected in the Hyperbolic-tangent S-wave exterior solution.
Past exam of the mathematics course of the University of Cambridge 2020 ii Paper 1 7E ii Solution Created 2026-09-24 Updated 2026-09-29
Given any that is not a nonpositive integer, choose large enough that and defineThe recurrence shows that definitions from different sufficiently large agree, so this gives the unique analytic continuation from the right half-plane. It is analytic except where a denominator factor vanishes, namelyThese points are simple poles. More precisely, at ,because one may writeand use .
Past exam of the mathematics course of the University of Cambridge 2021 iii Paper 150 2 b Solution Created 2026-09-28 Updated 2026-10-07
For a proof, letThe Poisson summation formula applied to a Gaussian function gives the theta transformationThe standard Gamma function integral and termwise integration initially give, for ,Split the integral at one, substitute in the lower half, and use the theta transformation. The result isThe integral is an entire function of because decays exponentially. The right side is visibly invariant under , proving both the analytic continuation and the functional equation of the Riemann zeta function. This is the Mellin representation of the completed Riemann zeta function.
Past exam of the mathematics course of the University of Cambridge 2021 ii Paper 1 14E b Solution Created 2026-09-24 Updated 2026-10-03
An analytic continuation of from to is a holomorphic function on whose restriction to the nonempty overlap agrees with .
Suppose and are two such continuations. Both equal on , soon that nonempty open set. Since is connected, the identity theorem givesThus analytic continuation to is unique.
Past exam of the mathematics course of the University of Cambridge 2021 ii Paper 1 14E d Solution Created 2026-09-24 Updated 2026-09-25
If , the same upper-half-plane contour calculation givesIf , the pole lies below the real integration contour. Closing upward encloses no pole, so Jordan lemma and the Cauchy integral theorem giveHence at a real point the upper and lower boundary values satisfyThus crossing from the upper half-plane to the lower half-plane produces the negative of this jump.
An analytic continuation must be holomorphic, hence continuous, across every point where it is defined. The undeformed real-axis formula for has the nonzero jump above, while the continuation constructed in part (c) retains below the axis. Therefore cannot be the analytic continuation of .
Past exam of the mathematics course of the University of Cambridge 2023 iii Paper 337 1 g Solution 2026-09-28
The analytic continuation from bosonic imaginary frequency to a retarded frequency is . ThusWriting and using the Sokhotski–Plemelj formula gives, in this sign convention,The opposite overall convention for the retarded Green function reverses this sign; the corresponding spectral function is conventionally chosen positive at positive frequency.
Near the quantum critical point, is a function only of . The Green function has the scaling formwithThis is quantum critical scaling with dynamical critical exponent and leading large- anomalous dimension .
Past exam of the mathematics course of the University of Cambridge 2023 ii Paper 2 13E a i Solution Created 2026-09-24 Updated 2026-09-25
PutThis slit complex plane is a simply connected domain. On there is one branch of that equals at zero, and its reciprocal is holomorphic. The primitive of a holomorphic function on a simply connected domain therefore makes the integral from zero path-independent, so is single-valued.
For unrestricted paths, going once around either branch point changes the sign of the square root. The resulting analytic continuation therefore need not return to the original germ, and the value of can depend on the path.
Past exam of the mathematics course of the University of Cambridge 2023 ii Paper 4 7E Solution Created 2026-09-23 Updated 2026-09-25
The Papperitz symbolspecifies a second-order Fuchsian differential equation. The first row lists its three distinct regular singular points ; is the independent variable. The two entries below each singular point are its characteristic exponents. Thus local solutions have leading behaviorsand, with the usual convention at infinity,When an exponent difference is an integer, a logarithmic second solution may replace the naive second Frobenius power. The entries obey the Fuchs relationFor a second-order equation with exactly three regular singular points, these exponent data determine the equation up to multiplication by a nonzero function; there is no accessory parameter. The symbol is therefore the one for the Gauss hypergeometric equation. Its distinguished solution is the exponent-zero solution analytic at zero and normalized to one there. For the ordinary power-series definition one assumes , with exceptional parameter values handled separately or by continuation.
Now putThe hypergeometric equation in has exponent pairsThe Möbius transformation of a Papperitz symbol sendsso has symbol
On the other hand, has exponent pairsat . Multiplication by applies the dependent-variable rescaling of a Papperitz symbol: it adds to both exponents at and subtracts from both at infinity. Hencehas exactly the same three exponent pairs as .
Both functions are analytic near and equal one at , so uniqueness of the normalized exponent-zero hypergeometric solution gives the Pfaff transformationThe identity first holds near zero with the branch of equal to one there, and then extends by analytic continuation on any domain where compatible branches are chosen.