I just read a paper 'A pocket calculator determination of energy eigenvalues' by J Killingbeck (1979).
I have some questions about section 2.
Why $R<0$ implies the $E$ estimate was too high and $R>1$ implies too low? And is it possible that $0<R<1$ remains?
Why "if the starting point is chosen at $r=h(l+1)$, then $R(hl)$ can be assigned any finite value without disturbing the calculation at larger $r$ values" ?
How to guess a trial energy $E$ (especially the ground state energy)?
(Actually I'm interested in a later paper by Killingbeck (1982) 'Finite-difference methods for eigenvalues'(Link: http://iopscience.iop.org/0022-3700/15/6/009), in which the author applies the method in 'A pocket calculator determination of energy eigenvalues' to some potentials with highly singularity at origin)
mod note: copying the text of whoplisp's post here since it isn't actually an answer to the question
They convert the Schroedinger equation into a finite difference equation (instead of a variational approach):
$V$ is a central potential
Ansatz with solid harmonic $\xi_l$ of degree $l$ (e.g. $x$ or $xy$)
finite difference approximation
$h^2D^2\phi \rightarrow \phi(r+h)+\phi(r-h)-2\phi(r)$
$2hD\phi \rightarrow \phi(r+h)-\phi(r-h)$
crucial step: introduce ratio variable $R(r)$:
convert Schroedinger equation in recursive equation to calculate $R(r)$:
the corresponding 1D Schroedinger equation is:
if $r=h(l+1)$ is chosen then $R(hl)$ can be assigned any finite value without disturbing the calculation at large $r$ values
to find the ground states with angular momentum $l$ start with an estimate for $E$, some non-zero $R(hl)$ and some small $h$. $R(r)$ as $r$ increases will either become negative or pass through unity from below.
first thing happens when guessed $E$ was too high
second thing indicates $E$ was too low (because $\psi$ increases with $r$)
you just try various values and 'sandwhich' the solution
example for $V=\lambda r-1/r$:
$\lambda=1$, $h=.1$, state=2p: 1.9759