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Confluent hypergeometric functions solve the Kummer differential equation $z{y}^{\prime \prime }+(b-z)y^{\prime }-ay=0$. The regular solution is the confluent hypergeometric function of the first kind; a second standard solution is conventionally denoted $U(a,b,z)$. The family arises by coalescing singular points of the Gauss hypergeometric equation.

Multiply a solution with parameters $(c-a,c-b;c)$ by $(1-z{)}^{c-a-b}$. Its local exponents become exactly those of the Gauss hypergeometric equation with parameters $(a,b;c)$: those at one shift by $c-a-b$, and those at infinity shift by its negative. Direct differentiation gives the same differential equation. Near zero, choose the factor with value one; uniqueness of the analytic normalized solution, for $c$ not a nonpositive integer, gives the identity. Compatible analytic continuation extends it to other domains and parameters where the normalized functions are defined.

For compatible local branches near $x=0$,It follows by transforming the harmonic oscillator equation $y^{\prime \prime }+k^{2}y=0$ into the Gauss hypergeometric equation and selecting the solution with value one and derivative zero at the origin.

For the Gauss hypergeometric equation with parameters $\alpha ,\beta ,\gamma$, a Frobenius solution $y=x^{\sigma }{\sum }_{n\ge 0}{a}_n{x}^n$, $a_0\ne 0$, has indicial roots $\sigma =0$ and $\sigma =1-\gamma$. Matching each successive power gives the displayed undivided recurrence. The usual ratio formula is valid only when its left coefficient is nonzero. At integer-separated roots, a zero coefficient requires a compatibility check as in an undetermined coefficient at Frobenius resonance; otherwise a logarithmic companion may be necessary. For example, $\gamma =0,\sigma =0$ gives $0=\alpha \beta a_0$ at $n=1$, obstructing that exponent-zero series when $\alpha \beta \ne 0$.

The Gauss hypergeometric function is the normalized analytic solution at zero,initially for $|z|<1$ and $c\notin \{0,-1,-2,\dots \}$, followed by analytic continuation where possible.

The ordinary hypergeometric function is ${}_2{F}_1(a,b;c;z)={\sum }_{n\ge 0}(a{)}_n(b{)}_n{z}^n/((c{)}_{n}n!)$ near zero for admissible $c$. It solves the Gauss hypergeometric equation and extends by analytic continuation.

A Papperitz symbol records the three regular singular points of a second-order Fuchsian differential equation and the two local exponents at each. Exponents at infinity refer to solutions behaving as $z^{-\gamma }$, that is, to powers of the local coordinate $1/z$. Their sum satisfies the Fuchs relation: the six exponents add to one. The symbol represents the two-dimensional solution space, not one normalized solution by itself.

The specified exponents give the Gauss hypergeometric equationSet $d=c-a-b$ and let $v=F(c-a,c-b;c;z)$. Its equation isFor $u=(1-z{)}^{d}v$, differentiation gives $u^{\prime }=(1-z{)}^d[v^{\prime }-dv/(1-z)]$ and $u^{\prime \prime }=(1-z{)}^d[v^{\prime \prime }-2d{v}^{\prime }/(1-z)+d(d-1)v/(1-z{)}^2]$. Substitution into the first equation leaves $(1-z{)}^d$ times exactly the second equation: its $v^{\prime }$ coefficient is $c-(a+b+1+2d)z$, and its $v$ coefficient simplifies to $-(c-a)(c-b)$. Hence $u$ solves the same equation as $F(a,b;c;z)$.

The symbol gives the same check geometrically: multiplication by $(1-z{)}^d$ shifts the exponents at one from $\{0,-d\}$ to $\{d,0\}$ and those at infinity from $\{c-a,c-b\}$ to $\{b,a\}$; it does not change the exponents at zero. Choose the branch with $(1-0{)}^d=1$. Then $u$ is analytic at zero and $u(0)=1$. For $c\notin \{0,-1,-2,\dots \}$ the analytic recurrence uniquely determines this normalized solution, provingThe identity continues wherever both sides are defined with compatible branches. At exceptional parameters where the stated normalization does not uniquely define an analytic function, a parameter-continuation convention is needed; the generic proof does not silently assume uniqueness there. This is Euler's hypergeometric transformation.

Write the Papperitz symbol by listing pairs of exponents at $0,1,\infty$. For $F(a,c-b;c;w)$ they are $(0,1-c)$, $(0,b-a)$, and $(a,c-b)$. Under $w=z/(z-1)$ the points $0,1,\infty$ in the $z$-plane correspond respectively to $0,\infty ,1$ in the $w$-plane. Thus the pulled-back exponents are $(0,1-c)$ at zero, $(a,c-b)$ at one, and $(0,b-a)$ at infinity.

Multiplication by $(1-z{)}^{-a}$ shifts the exponents at one by $-a$ and those at infinity by $+a$, yieldingThis is the original Gauss hypergeometric equation. Both functions are analytic and normalized to one at zero, so uniqueness of that normalized local solution proves the Pfaff transformationContinue the identity using the branch $|\arg (1-z)|<\pi$. As usual, the parameters must admit the normalized hypergeometric solution; excluded singular parameter values are treated only when their limiting solution exists.

The chain rule gives $z_x=\sin 2x$, $z_x^2=4z(1-z)$, and $z_{xx}=2(1-2z)$. Thereforewhich, after division, is the printed equation. It has only three regular singular points, $0,1,\infty$, and is thus a Papperitz equation. At zero and one the indicial roots are $0,1/2$. At infinity, inserting $w\sim z^{-\rho }$ gives $\rho =\pm n/2$. Its Papperitz symbol isThe exponent sum is one, as required for a three-singularity second-order equation.

Multiplying the equation by $z(1-z)$ puts it in Gauss hypergeometric equation form with $a=n/2$, $b=-n/2$, $c=1/2$. The solution analytic at zero and normalized to one is ${}_2{F}_1(a,b;c;z)$. Composing it with $z={\sin }^{2}x$ gives an even analytic solution of $w_{xx}+n^{2}w=0$ with $w(0)=1$, $w_x(0)=0$. Uniqueness of this ordinary differential equation impliesThis Gauss hypergeometric function identity initially holds near zero, and on $|x|<\pi /2$ with the usual real branch; it extends by matching analytic continuation. It is not an identity on every real $x$ if the left side is always reevaluated on its principal $z$ branch: for $n=1$ that principal value is $\sqrt{1-{\sin }^{2}x}=|\cos x|$. Continuing the branch consistently restores the stated cosine.

Let $\delta =1-m\downarrow 0$. This is the complete elliptic integral of the first kind in parameter notation; the modulus used in some definitions is $\sqrt{m}$. Near the endpoint, $s=\pi /2-\vartheta$ givesUse matched asymptotic expansion with an intermediate cutoff $b$ satisfying $\sqrt{\delta }\ll b\ll 1$. 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 $(1-m)\log (1/(1-m))$, rather than merely $1-m$. More explicitly, if $L=\log (4/\sqrt{\delta })$,The logarithmic correction arises from the next terms integrated through the overlap. Its coefficient can also be found by substituting $L+\delta (aL+b)$ into the Gauss hypergeometric equation satisfied here, $m(1-m)K_{mm}+(1-2m)K_m-K/4=0$, giving $a=1/4$, $b=-1/4$.

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 $z_j$, an exponent $r$ corresponds locally to a factor $(z-z_j{)}^r$; at infinity it corresponds to $z^{-r}$. 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 $w=z/(z-1)$ and $v(w)=F(a,c-b,c;w)$. At $w=0,1,\infty$, the exponent pairs for $v$ are respectively $(0,1-c)$, $(0,b-a)$, and $(a,c-b)$. The Möbius transformation sends $z=0,1,\infty$ to $w=0,\infty ,1$. Before rescaling, $v(w(z))$ therefore has exponent pairs $(0,1-c)$ at zero, $(a,c-b)$ at one, and $(0,b-a)$ at infinity. Multiplication by $(1-z{)}^{-a}$ subtracts $a$ at one and adds $a$ at infinity. The resulting pairs areexactly those of the original Gauss hypergeometric equation. Near zero choose the branch with $(1-z{)}^{-a}=1$ 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 $c\notin \{0,-1,-2,\dots \}$. Elsewhere the identity holds by analytic continuation with compatible branches, or by parameter limits wherever those are defined.

Choose a branch of $t=1-z$ with $|\arg t|<\pi$ and write $d=c-a-b\notin \mathbb{Z}$. The Gauss hypergeometric equation isChanging to $t$ turns this into the same equation with third parameter $a+b+1-c=1-d$. Its solution regular at $t=0$ is $y_1=F(a,b,1-d;t)$. Substitution of $y=t^{d}v$ gives the Gauss hypergeometric equation for $v$ with parameters $c-a,c-b,1+d$. Thus $y_2=t^{d}F(c-a,c-b,1+d;t)$ is also a solution. Their distinct nonintegrally separated characteristic exponents at a regular singular point $0,d$ imply linear independence near $z=1$, soThis also follows from the exponent transformation in the Papperitz symbol.

Initially assume $\Re c>\Re b>0$, so the Euler integral for the hypergeometric function applies. If $\Re d>0$, taking $z\to 1$ gives an integral evaluated by the beta function and determinesIndeed $y_1\to 1$ and $y_2\to 0$ there. To find $B$, instead work in the nonempty parameter region $\Re d<0$, put $z=1-ϵ$ and scale the integration variable by $1-s=ϵu$. The singular contribution isThe defining integral of the beta function evaluates this integral to $\Gamma (c-b)\Gamma (-d)/\Gamma (a)$. Since ${ϵ}^{-d}{y}_1\to 0$ and ${ϵ}^{-d}{y}_2\to 1$, it determines $B$. Analytic continuation in the parameters then yields the hypergeometric connection formulaThe usual power-series definition requires $c\notin \{0,-1,-2,\dots \}$; 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 $z$.

A Möbius transformation of a Papperitz symbol changes the independent variable by a Möbius transformation. In particular,maps $a,b,c$ to $0,1,\infty$, 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 $v=u^{-\alpha }(1-u{)}^{-\beta }y$. This dependent-variable rescaling of a Papperitz symbol subtracts $\alpha$ from both exponents at zero, subtracts $\beta$ from both exponents at one, and adds $\alpha +\beta$ to both exponents at infinity. Hence $v$ has symbolDefineThe Fuchs relation gives ${\alpha }^{\prime }-\alpha =1-C$ and ${\beta }^{\prime }-\beta =C-A-B$. After renaming $u$ as $z$, the symbol is thereforewhich is the Gauss hypergeometric equation. Renaming $A,B,C$ as $a,b,c$ gives exactly the displayed symbol in the question.

The solution of $y^{\prime \prime }+k^{2}y=0$ with $y(0)=1$ and $y^{\prime }(0)=0$ is $y(x)=\cos (kx)$. Near $x=0$, the map $z={\sin }^{2}x$ has the local inverse $x=\arcsin \sqrt{z}$, so this solution is analytic in $z$, corresponds to exponent zero at $z=0$, and has value one there. Uniqueness of the normalized local solution of the Gauss hypergeometric equation therefore gives the hypergeometric cosine identityor, when $z={\sin }^{2}x$ on the chosen local branch,

Put $y(x)=w(z)$ with $z={\sin }^{2}x$. The chain rule givesand henceAfter substitution and division by $4z(1-z)$, the ordinary differential equation becomesComparison with the Gauss hypergeometric equation gives, up to interchanging $A$ and $B$,

At $z=0$ and $z=1$, the coefficient of $w^{\prime }$ has at most a simple pole and that of $w$ 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 $(0,1-C)=(0,1/2)$ at zero, $(0,C-A-B)=(0,1/2)$ at one, and $(A,B)=(k/2,-k/2)$ at infinity; their sum also satisfies the Fuchs relation.

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 $z_1,z_2,z_3$. Near $z_i$, 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 $a$ and $b$. Put $w=z^{-1}$. Transforming the Papperitz symbol from $z$ to $w$ shows thatandare local solutions near $z=\infty$, with respective leading behaviours $z^{-a}$ and $z^{-b}$. When $a-b\notin \mathbb{Z}$, 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 $F(a,b;c;z)$ 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 $a,b,c$ and the branch convention, but not on $z$. This is the hypergeometric connection formula at infinity.

The Papperitz symbolspecifies a second-order Fuchsian differential equation. The first row lists its three distinct regular singular points $0,1,\infty$; $z$ 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 $F(a,b;c;z)$ is the exponent-zero solution analytic at zero and normalized to one there. For the ordinary power-series definition one assumes $c\notin \{0,-1,-2,\dots \}$, with exceptional parameter values handled separately or by continuation.

Now putThe hypergeometric equation in $u$ has exponent pairsThe Möbius transformation of a Papperitz symbol sendsso $Y$ has symbol

On the other hand, $F(a,b;c;z)$ has exponent pairsat $0,1,\infty$. Multiplication by $(1-z{)}^a$ applies the dependent-variable rescaling of a Papperitz symbol: it adds $a$ to both exponents at $z=1$ and subtracts $a$ from both at infinity. Hencehas exactly the same three exponent pairs as $Y$.

Both functions are analytic near $z=0$ and equal one at $z=0$, so uniqueness of the normalized exponent-zero hypergeometric solution gives the Pfaff transformationThe identity first holds near zero with the branch of $(1-z{)}^a$ equal to one there, and then extends by analytic continuation on any domain where compatible branches are chosen.