Kummer's transformation

Remarks

Proof idea: put $w(z) = e^z v(-z)$ into Kummer's equation with parameters $a, b$. A direct computation shows that $w$ is a solution exactly when $v$ solves Kummer's equation with parameters $b - a, b$. So $e^z M(b - a, b, -z)$ is a solution of the first equation that is analytic at $0$ with value $1$. For $b \notin \set{0, -1, -2, \dots}$ there is only one such solution, namely $M(a, b, z)$. Indeed, the other exponent at $0$ is $1 - b$, and a solution with that exponent cannot be analytic with value $1$ there. Example: for $a = b$, the left side is $e^z$ and the right side is $e^z M(0, b, -z) = e^z \cdot 1$. Example: for $a = 1$, $b = 2$, the right side is $e^z \frac{e^{-z} - 1}{-z} = \frac{e^z - 1}{z} = M(1, 2, z)$.

Going deeper

For a student

The identity is often used to move between large positive and large negative arguments, since one side may be much easier to estimate than the other. It also shows that if $a$ or $b - a$ is a nonpositive integer, then $M(a, b, z)$ is a polynomial or $e^z$ times a polynomial. The Tricomi function satisfies a companion identity, $U(a, b, z) = z^{1 - b} U(a - b + 1, 2 - b, z)$, often called Kummer's second transformation.