Confluent hypergeometric function 2026-10-05
Confluent hypergeometric functions solve the Kummer differential equation . The regular solution is the confluent hypergeometric function of the first kind; a second standard solution is conventionally denoted . The family arises by coalescing singular points of the Gauss hypergeometric equation.
Hypergeometric cosine identity 2026-09-29
For compatible local branches near ,It follows by transforming the harmonic oscillator equation into the Gauss hypergeometric equation and selecting the solution with value one and derivative zero at the origin.
Past exam of the mathematics course of the University of Cambridge 2005 ii Paper 1 8A Solution Created 2026-09-24 Updated 2026-09-24
Past exam of the mathematics course of the University of Cambridge 2006 ii Paper 3 14E a Solution Created 2026-09-24 Updated 2026-09-24
Past exam of the mathematics course of the University of Cambridge 2008 ii Paper 1 14C Solution Created 2026-09-24 Updated 2026-09-24
Past exam of the mathematics course of the University of Cambridge 2014 iii Paper 74 2 b Solution Created 2026-10-03 Updated 2026-10-06
Let . This is the complete elliptic integral of the first kind in parameter notation; the modulus used in some definitions is . Near the endpoint, givesUse matched asymptotic expansion with an intermediate cutoff satisfying . Away from the endpoint the leading integral iswhile the endpoint integral isAdding them removes the arbitrary cutoff and gives the logarithmic endpoint asymptotic of the complete elliptic integralThe order of the next term is therefore , rather than merely . More explicitly, if ,The logarithmic correction arises from the next terms integrated through the overlap. Its coefficient can also be found by substituting into the Gauss hypergeometric equation satisfied here, , giving , .
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 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 2019 ii Paper 1 14A c Solution Created 2026-09-24 Updated 2026-09-29
A Möbius transformation of a Papperitz symbol changes the independent variable by a Möbius transformation. In particular,maps to , respectively. Because a Möbius map is locally biholomorphic away from its pole, it carries each exponent pair with its singular point and givesNow define a dependent-variable rescaling of a Papperitz symbol by writing the old dependent variable asso that . This dependent-variable rescaling of a Papperitz symbol subtracts from both exponents at zero, subtracts from both exponents at one, and adds to both exponents at infinity. Hence has symbolDefineThe Fuchs relation gives and . After renaming as , the symbol is thereforewhich is the Gauss hypergeometric equation. Renaming as gives exactly the displayed symbol in the question.
Past exam of the mathematics course of the University of Cambridge 2020 ii Paper 1 14E ii Solution Created 2026-09-24 Updated 2026-09-29
The solution of with and is . Near , the map has the local inverse , so this solution is analytic in , corresponds to exponent zero at , and has value one there. Uniqueness of the normalized local solution of the Gauss hypergeometric equation therefore gives the hypergeometric cosine identityor, when on the chosen local branch,
Past exam of the mathematics course of the University of Cambridge 2020 ii Paper 1 14E i Solution Created 2026-09-24 Updated 2026-09-29
Put with . The chain rule givesand henceAfter substitution and division by , the ordinary differential equation becomesComparison with the Gauss hypergeometric equation gives, up to interchanging and ,
At and , the coefficient of has at most a simple pole and that of has at most a double pole, so both are regular singular points. The regular singular point at infinity criterion gives the same conclusion at infinity. The corresponding Papperitz symbol isIndeed, the exponent pairs are at zero, at one, and at infinity; their sum also satisfies the Fuchs relation.
Past exam of the mathematics course of the University of Cambridge 2022 ii Paper 4 7E Solution Created 2026-09-23 Updated 2026-09-25
The Papperitz symboldescribes the two-dimensional solution space of a second-order homogeneous Fuchsian differential equation on the Riemann sphere with exactly three regular singular points . Near , its two independent Frobenius behaviours are locallysubject to logarithmic modifications in resonant cases. The six characteristic exponents obey the Fuchs relation
For the Gauss hypergeometric equation, the exponents at infinity are and . Put . Transforming the Papperitz symbol from to shows thatandare local solutions near , with respective leading behaviours and . When , these behaviours are distinct and the two solutions are linearly independent. They therefore form a basis of the solution space on any simply connected common domain with compatible branch choices.
The solution normalized at zero can be analytically continued into that domain. Since it solves the same second-order equation, it must be a constant linear combination of the basis at infinity:The constants depend on and the branch convention, but not on . This is the hypergeometric connection formula at infinity.
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.