MHB Reversing recurrence relationships

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Reversing recurrence relationships can be a valuable technique, especially when boundary conditions are inaccurate or unknown. Evaluating equations like the Legendre polynomial can be approached both forwards and backwards, with the forward direction solving for Pl+1 and the reverse for Pl-1. The arithmetic involved in manipulating these equations is straightforward, but clarity on the implications of each direction is essential. Numeric methods with clear step points allow for easy evaluation in both directions, while those with variable step sizes can also be reversed by adjusting the step. Overall, the discussion confirms that reversing relationships is valid and effective, provided the methods are understood correctly.
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A couple of times I have come across the suggestion that numerically evaluating a recursive relation in reverse can be a valuable approach. I can see this where, for example, the boundary conditions at one 'end' are inaccurate or undiscoverable. However, while the arithmetic of manipulating such equations seems simple, I wonder if I am missing something?
One example is a Legendre polynomial, given by (l+1)Pl+1 + lPl - (2l+1)xPl=0
Should I evaluate this in the 'forward' direction, by solving for Pl+1, and in the reverse direction by solving for Pl-1? I am also struggling for some intuition as to what the difference(s) may be?
 
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After writing a few programs to compare, I am satisfied that when a numeric method has clear '3 points', then I can evaluate this in the 'forward' direction, by solving for Pl+1, and in the reverse direction by solving for Pl-1. In some other methods where steps are only visible as +(some step size, like h), then we can 'reverse' direction by using -h. This was all intuitively obvious, I just wanted confirmation I wasn't missing anything else ...
 
Thread 'Erroneously  finding discrepancy in transpose rule'
Obviously, there is something elementary I am missing here. To form the transpose of a matrix, one exchanges rows and columns, so the transpose of a scalar, considered as (or isomorphic to) a one-entry matrix, should stay the same, including if the scalar is a complex number. On the other hand, in the isomorphism between the complex plane and the real plane, a complex number a+bi corresponds to a matrix in the real plane; taking the transpose we get which then corresponds to a-bi...

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