Expand a function in terms of Legendre polynomials

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SUMMARY

The discussion focuses on expanding a function defined on the interval (a,b) using Legendre polynomials through the transformation u = (2x-a-b)/(b-a), which maps the function onto the interval (-1,1). The key point is to demonstrate that this transformation is onto, meaning that for every point in (-1,1), there exists a corresponding point in (a,b). The confusion arises from interpreting the mapping and the relationship between the functions involved. A clear understanding of the mapping process is essential for successfully applying Legendre polynomial expansions.

PREREQUISITES
  • Understanding of Legendre polynomials and their properties
  • Familiarity with function transformations and mappings
  • Basic knowledge of linear algebra concepts
  • Experience with interval notation and function behavior
NEXT STEPS
  • Study the properties and applications of Legendre polynomials in function approximation
  • Learn about function transformations and their implications in mathematical analysis
  • Explore the concept of onto functions and how to prove mappings between intervals
  • Review linear algebra fundamentals, particularly focusing on mappings and transformations
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Mathematicians, physics students, and anyone involved in numerical analysis or function approximation using polynomial expansions.

Logarythmic
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Problem:
Suppose we wish to expand a function defined on the interval (a,b) in terms of Legendre polynomials. Show that the transformation u = (2x-a-b)/(b-a) maps the function onto the interval (-1,1).

How do I even start working with this? I haven't got a clue...
 
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I do not speak english in everyday life, and so the words "maps the function onto the interval (-1,1)" seems confusing to me. I would normally interpret them as "u is such that [itex]u\circ f[/itex]:(a,b)-->(-1,1)". But that does not seem to make sense, since we do not know the form of f. What does make sense, is to understand it as "u is such that [itex]f\circ u[/itex]:(a,b)-->R is the same as f:(-1,1)-->R". In other words, you need to show that [itex]f\circ u[/itex] is a restriction of f. But that is easy, you only need to show that u is onto from (a,b) to (-1,1).
 
Ok, I'm really not good at linear algebra so I think I'll jump this problem. ;)
 

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