Reversing recurrence relationships

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SUMMARY

Reversing recurrence relationships can be an effective numerical approach, particularly when boundary conditions are inaccurate or unknown. The discussion highlights the evaluation of Legendre polynomials using the recurrence relation (l+1)Pl+1 + lPl - (2l+1)xPl=0. It confirms that both forward (solving for Pl+1) and reverse (solving for Pl-1) evaluations yield valid results, especially when a numeric method has clear step points. The author emphasizes that reversing direction can also be achieved by adjusting the step size, reinforcing the validity of this approach in numerical methods.

PREREQUISITES
  • Understanding of recurrence relations and their applications
  • Familiarity with Legendre polynomials and their properties
  • Basic knowledge of numerical methods and step size manipulation
  • Experience with programming for numerical evaluations
NEXT STEPS
  • Study the properties and applications of Legendre polynomials in detail
  • Learn about numerical methods for solving recurrence relations
  • Explore advanced techniques in numerical analysis, such as finite difference methods
  • Investigate software tools for numerical evaluation, such as MATLAB or Python libraries
USEFUL FOR

Mathematicians, numerical analysts, and programmers interested in advanced numerical methods and the evaluation of recurrence relationships.

ognik
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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 ...
 

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