Holonomic function

Remarks

Example: $f(x) = \frac{1}{1 - x} = \sum_{n \ge 0} x^n$ satisfies $(1 - x) f'(x) - f(x) = 0$, since $f'(x) = \frac{1}{(1-x)^2}$; here $r = 1$, $p_1 = 1 - x$ and $p_0 = -1$. Non-example: $\tan x$ is not holonomic. An analytic holonomic function extends along paths avoiding the finitely many zeros of $p_r$, so it has only finitely many singularities, while $\tan$ has poles at all points $\frac{\pi}{2} + k\pi$. Equivalently, $f$ is holonomic if and only if the derivatives $f, f', f'', \ldots$ span a finite-dimensional vector space over the field of rational functions $K(x)$. Another equivalent form: $f = \sum_n a_n x^n$ is holonomic if and only if its coefficients satisfy a recurrence $q_d(n) a_{n+d} + \cdots + q_0(n) a_n = 0$ for all $n$, with polynomials $q_i$ not all zero; such sequences are called P-recursive.