What are the Different Ways to Define Legendre Polynomials?

  • Thread starter Thread starter JM00404
  • Start date Start date
Click For Summary
SUMMARY

This discussion focuses on the various methods to define Legendre polynomials, specifically addressing four problems related to their properties and applications. The first problem involves demonstrating the nth Legendre polynomial using a binomial expansion and differentiation. The second problem seeks the degree three Legendre approximation of a piecewise function. The third problem utilizes the generating function for Legendre polynomials to derive the Henyey-Greenstein phase function. The fourth problem establishes a relationship between the leading term of Legendre polynomials and any polynomial of the same degree.

PREREQUISITES
  • Understanding of Legendre polynomials and their properties
  • Familiarity with binomial expansion techniques
  • Knowledge of generating functions in mathematical analysis
  • Basic concepts of polynomial approximation and orthogonality
NEXT STEPS
  • Study the derivation of Legendre polynomials using the Rodrigues' formula
  • Explore the applications of Legendre polynomials in physics, particularly in solving potential problems
  • Learn about the properties of orthogonal polynomials and their implications in approximation theory
  • Investigate the Henyey-Greenstein phase function and its significance in scattering theory
USEFUL FOR

Mathematicians, physicists, and engineers interested in polynomial approximations, mathematical analysis, and applications of Legendre polynomials in various fields.

JM00404
Messages
7
Reaction score
0
Evening. I am having difficulty solving the the problems that have been included below. For the first problem, I essentially followed what the hint suggested that I do and I still cannot "see" the solution. I am honestly not sure how to even go about solving the second problem. For the third problem, I differentiated the generating function for the legendre polynomial, which gave a solution that was somewhat similar to the "Henyey-Greenstein phase function" but deviates from it slightly (the answer expression that I have right now is
[tex]2[r^2-1](\frac{d}{dx})[1+r^2-2cos(\theta)]^{-1/2}[/tex].
Finally, for the fourth problem, I really don't know where to begin. Any and all assistance/guidance/hints would be very much appreciated. Thank you.

Regards

I. Show that the nth Legendre polynomial is given by
[tex]P_n(x)=\sum_{k=0}^n\left(\stackrel{n}{k}\right)\left(\stackrel{n+k}{k}\right)\left(\stackrel{x-1}{2}\right)^k.[/tex]
Hint: Write [tex](x^2-1)^n=(x-1)^n[(x-1)+2]^n[/tex], apply the binomial expansion to the term in [\cdots], and differentiate n-times.

II. Find the degree three Legendre approximation of the function
[tex]f(x)=\left\{\stackrel{0 (-1\leq x<0)}{1 (0\leq x<1)}.[/tex]

III. Use the formula
[tex]\frac{1}{\sqrt{1+r^2-2rx}}=\sum_{n=0}^\infty P_n(x)r^n[/tex]
to derive the formula for the ``Henyey-Greenstein phase function''
[tex]\frac{1-r^2}{(1+r^2-2rcos(\theta))^{3/2}}=\sum_{n=0}^\infty (2n+1)P_n(cos(\theta))r^n[/tex].

IV. Let [tex]c_n[/tex] be the leading term of [tex]P_n[/tex] and set
[tex]\tilde{P}_n=c^{-1}P_n=x^n+[/tex] (lower powers of x).
Prove that if [tex]Q=x^n+\cdots[/tex] is any polynomial of degree n with leading coefficient
one, then
[tex]<Q,Q>\geq<\tilde{P}_n,\tilde{P}_n>[/tex]
with equality only if [tex]Q=\tilde{P}_n[/tex].
(Hint: Write [tex]Q=\tilde{P}_n+h[/tex], where h is a linear
combination of [tex]P_0,P_1,\ldots P_{n-1}[/tex], and note that [tex]<Pn,h>=0[/tex].)
 
Last edited:
Physics news on Phys.org
There are several equivalent ways of defining the Legendre polynomials. It is impossible to know how YOU should do these without know what definitions or formulas you have available to you.
 

Similar threads

  • · Replies 7 ·
Replies
7
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K
Replies
5
Views
2K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 14 ·
Replies
14
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 2 ·
Replies
2
Views
3K
Replies
5
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K