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The solutions of this DE are called Legendre Functions. If n is a non-negative integer, then the Legendre Functions are called Legendre Polynomials, Pn(x).
Legendre's DE is a second order ordinary differential equation, so two sets of functions are needed to form the general solution. Legendre Polynomials of the second kind are called Qn(x). Then, the general solution to this DE is
y(x) = AnPn(x)+BnQn(x)
However, Qn(x) has singularities at ±1, so Bn is forced to be zero if the solution is meant to have a meaningful physical interpretation.
Pn(x)= 1
——
2nn!· dn
——
dxn(x2−1)n
(n+1)Pn+1(x)=(2n+1)xPn(x)−nPn−1(x)
(2n+1)Pn(x)=Pn+1'(x)−Pn−1'(x)
(x2−1)Pn'(x)=nxPn(x)−nPn−1(x)
Pn−1'(x)=xPn'(x)−nPn(x)
Pn+1'(x)=xPn'(x)+(n+1)Pn(x)
Pn(x) Qn(x) 0 P0(x)=1 Q0(x)=(1/2)ln((x+1)/(x−1)) 1 P1(x)=x Q1(x)=−1+(1/2) x ln((x+1)/(x−1)) 2 P2(x)=(1/2)(3x2−1) Q2(x)=(−3/2)x+(1/4)(−1+3x2) ln((x+1)/(x−1)) 3 P3(x)=(1/2)(5x3−3x) Q3(x)=(2/3)−(5/2)x2−(3/4)x(1−(5/3)x2) (ln((x+1)/(x−1)) To find the recurrence relation for coefficients of Pn, begin with the first recurrence relation, above, reproduced here:
(n+1)Pn+1(x)=(2n+1)xPn(x)-nPn-1(x)
Now, by replacing n+1 with n, and then divide by n to get
Pn(x)=((2n-1)/n) xPn-1(x) + ((1-n)/n) Pn-2(x)
This gives the triangle of coefficients (using Microsoft Excel), emphasizing the formula in cell D5 (which is replicated to all the cells below the n=0 row) as follows:
A B C D E F G H I J K L M N O 1 n Coefficients, constant first, increasing degree 2 0 1 3 1 0 1 4 2 -0.5 0 1.5 5 3 0 -1.5
=(2*$A5-1)/$A5*C4+
(1-$A5)/$A5*D30 2.5 6 4 0.375 0 -3.75 0 4.375 7 5 0 1.875 0 -8.75 0 7.875 8 6 -0.313 0 6.563 0 -19.688 0 14.438 9 7 0 -2.188 0 19.688 0 -43.313 0 26.813 10 8 0.273 0 -9.844 0 54.141 0 -93.844 0 50.273 11 9 0 2.461 0 -36.094 0 140.766 0 -201.094 0 94.961 12 10 -0.246 0 13.535 0 -117.305 0 351.914 0 -427.324 0 180.426 13 11 0 -2.707 0 58.652 0 -351.914 0 854.648 0 -902.129 0 344.449 14 12 0.226 0 -17.596 0 219.946 0 -997.09 0 2029.79 0 -1894.471 0 660.194 Now, A060818 gives the denominators of the coefficients, so we can make a new table, multiplying each coefficient by the appropriate denominator to give a table of numerators. Here, the formulas in cells B5 and D5 are emphasized to aid your understanding:
A B C D E F G H I J K L M N O 1 n A060818
(denom.)A100258:
Numerators of Coefficients, constant first, increasing degree2 0 1 1 3 1 1 0 1 4 2 2 -1 0 3 5 3 2
=2^(FLOOR($A5/2,1)+
FLOOR($A5/4,1)+
FLOOR($A5/8,1))0 -3
=Sheet1!D5*
$B50 5 6 4 8 3 0 -30 0 35 7 5 8 0 15 0 -70 0 63 8 6 16 -5 0 105 0 -315 0 231 9 7 16 0 -35 0 315 0 -693 0 429 10 8 128 35 0 -1260 0 6930 0 -12012 0 6435 11 9 128 0 315 0 -4620 0 18018 0 -25740 0 12155 12 10 256 -63 0 3465 0 -30030 0 90090 0 -109395 0 46189 13 11 256 0 -693 0 15015 0 -90090 0 218790 0 -230945 0 88179 14 12 1024 231 0 -18018 0 225225 0 -1021020 0 2078505 0 -1939938 0 676039
|Pn(x)|≤1
Pn(x) is an odd function if n is odd, and so in particular Pn(0)=0 if n is odd.
Pn(x) is an even function if n is even.
If n is even, Pn(0)=(−1)n/2 (1·3·5·...·(n−1)) / (2·4·6·...·n).
The proof by induction begins with the recurrence relation for the Pn in terms of Pn-1 and Pn-2.
Pn(x)=((2n-1)/n) xPn-1(x) + ((1-n)/n) Pn-2(x)
Now, this recurrence relation gets a lot simpler when x=0:
Pn(0)= ((1-n)/n) Pn-2(0)
and using recursion starting with P0(0)=1, P1(0)=0, the result follows.
Pn(1)=1
Pn(−1)=(−1)n
Pn(x) = (1/π) π
∫
0(x+sqrt(x2−1)cosφ)n dφ
1
∫
−1f(x)Pn(x)dx = (−1/2)n/n! 1
∫
−1(x2−1)nf(n)(x) dx, where f(x) is differentiable n times on −1≤x≤1. f(n)(x) denotes the nth derivative of f(x).
Engineering Fundamentals: Legendre Polynomial
Wikipedia: Legendre polynomials
Chebyshev's Differential Equations: First kind (1−x²)y''−xy'+n²y=0, Second kind (1−x²)y''−3xy'+n(n+2)y=0; Chebyshev's Polynomials
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