Some Special Functions of Mathematical Physics Legendre Polynomials

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Some Special Functions of Mathematical Physics
Legendre Polynomials
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There are many special functions that are being
used by Physicists in the solution of physical
problems, like, Legendre Polynomials, Hermite
Polynomials, Bessels Functions, etc. to name a
few. In this lecture we define Legendre
Polynomials and study their properties. Much of
the material covered here is taken from the book
An Introduction to Linear Analysis - By Kreider,
Kuller, Ostberg Perkins (Addison Wesley)
Chapter 11.
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We define the Legendre polynomials, determine
the differential equations satisfied by them,
show the orthogonality of the Legendre
polynomials, and find the recurrence relation
satisfied by them and the zeros of these
polynomials.
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Definition The Legendre polynomials, Pn(x),
n0,1,2, are defined by the Rodrigues formula
Thus
etc.
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• It is clear that
• Pn(x) is a polynomial of degree n.
• If n is even, Pn (x) contains only even powers of
  x and if n is odd, Pn(x) contains only odd powers
  of x.
• The coefficient of xn in Pn(x)

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• The coefficient of xn-2 in Pn(x)

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Leibniz Rule for the nth derivative of product of
two functions
Let u, v be two functions of x. Then
where superscripts denote the order of the
derivative.
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The differential equation satisfied by Pn(x)
Let
Hence
Multiplying both sides by (x2-1), we get
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Differentiating both sides (n1) times w.r.t x
(using Leibniz formula), we get
i.e
Dividing throughout by 2nn!, we get
where
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Thus y Pn(x) satisfies the so-called Legendres
d.e. of order n viz
Theorem
Pn(1) 1, Pn(-1) (-1)n
Proof
Let w (x2-1)n (x1)n(x-1)n Differentiating
both sides n times using Leibniz rule, we get
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We note that in RHS, except for the first and
last term, every other term contains a factor of
(x1) and a factor of (x-1).
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Noting that w(n)2nn!Pn(x), putting x 1, we get
We also note for future reference, for k lt n,
(as every term in w(k), ( k lt n ), contains a
factor of (x1) and a factor of (x-1)).
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Orthogonality of the Legendre Polynomials
We show
Proof
For any function f(x), consider
On integrating by parts, we get
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Now by the remark made earlier
at x ?1
Hence

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We note that if f(x) is a polynomial of degree lt
n, f(n)(x) 0 Hence if m lt n,
0, as m lt n and so Pm(n)(x)0
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Since Pn(x) is a polynomial of degree n, Pn(n)
(x) n!leading coefficient of Pn(x)
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Thus
(as (-1)n(x2-1)n (1-x2)n)
Now look at
Put
We get
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Thus