Gauss hypergeometric equation

The Gauss equation has singularities at $0,1,\infty$ and normalized exponent-zero solution $F(A,B;C;u)$ at the origin.

Put $d=c-a-b\notin \mathbb{Z}$ and choose a branch of $(1-z{)}^d$. The solution regular at zero isThe two characteristic exponents at a regular singular point are $0,d$. The Euler integral for the hypergeometric function determines the first coefficient by a finite endpoint limit and the second by scaling the singular endpoint; analytic continuation extends the parameter ranges.

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.

For $\Re c>\Re b>0$ and a fixed branch away from the cut,Expand $(1-zs{)}^{-a}$ for $|z|<1$ and integrate termwise using the beta function to recover the Gauss hypergeometric function series. Analytic continuation extends the identity in $z$.

With compatible branches and initially near $z=0$,It follows by applying a Möbius transformation and a dependent-variable rescaling to the same Papperitz symbol, then matching the normalized analytic solution at zero.

When $C$ is not an integer, a second local solution at zero is $u^{1-C}F(A-C+1,B-C+1;2-C;u)$.

Away from resonant parameter cases, two independent local solutions at infinity areand the expression obtained by exchanging $a$ and $b$. Every analytic continuation of a hypergeometric solution into their common domain is a constant linear combination of this basis.

The binomial series gives $F(A,C;C;u)=(1-u{)}^{-A}$ wherever the defining series converges, followed by analytic continuation.

Special parameter choices can identify a hypergeometric solution with an elementary solution of the same second-order differential equation by matching its local normalization.

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 $k\ne 0$ and compatible local branches near $x=0$,This is obtained from the second local hypergeometric solution at zero and its normalization.