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I came across the Legendre differential equation today and I'm curious about how to solve it. The equation has the form:
$$(1 - x^2)y'' - 2xy' + \nu(\nu +1)y = 0, (1)$$
Where ##\nu## is a constant.
The equation has singularities at ##x_1 = \pm 1## where both ##p## and ##q## are not analytic. The best ordinary point to use would probably be ##x_0 = 0##, which yields the radius of convergence ##R ≥ |x_0 - x_1| = 1## for the series solution:
##y = \sum_{n=0}^{∞} a_nx^n, |x - x_0| < R##.
Taking derivatives and simplifying ##(1)##, I found the recurrence relation to be:
##a_{n+2} = \frac{n(n+1) - \nu(\nu+1)}{(n+1)(n+2)} a_n##
From which I have noticed some interesting properties.
If ##\nu = 0##, then all of terms involving ##n## even are zero. The terms involving ##n## odd can be deduced from: ##a_{n+2} = \frac{n}{n+2}##. When ##n = 1, 3, 5 ...## it can be observed that ##a_{2n+1} = \frac{1}{n+2}##.
The series solution can then be written as:
##y = \sum_{n=0}^{∞} a_nx^n = \sum_{n=1}^{∞} a_{2n} x^{2n} + \sum_{n=1}^{∞} a_{2n+1} x^{2n+1} = a_0 + \sum_{n=1}^{∞} \frac{1}{n+2} x^{2n+1} ##.
If ##\nu = 2m## or ##\nu = 2m +1## for some ##m ≥ 1##, then some other interesting things seem to happen. If ##\nu## is even, all of the even ##n## terms of the recurrence relation are zero after a certain point and there is only an odd sum plus a few even terms. If ##\nu## is odd, all of the odd ##n## terms of the recurrence relation are zero after a certain point and there is only an even sum plus a few odd terms.
Does this sound alright? Finding solutions for ##\nu > 0## has proven slightly difficult to formalize.
$$(1 - x^2)y'' - 2xy' + \nu(\nu +1)y = 0, (1)$$
Where ##\nu## is a constant.
The equation has singularities at ##x_1 = \pm 1## where both ##p## and ##q## are not analytic. The best ordinary point to use would probably be ##x_0 = 0##, which yields the radius of convergence ##R ≥ |x_0 - x_1| = 1## for the series solution:
##y = \sum_{n=0}^{∞} a_nx^n, |x - x_0| < R##.
Taking derivatives and simplifying ##(1)##, I found the recurrence relation to be:
##a_{n+2} = \frac{n(n+1) - \nu(\nu+1)}{(n+1)(n+2)} a_n##
From which I have noticed some interesting properties.
If ##\nu = 0##, then all of terms involving ##n## even are zero. The terms involving ##n## odd can be deduced from: ##a_{n+2} = \frac{n}{n+2}##. When ##n = 1, 3, 5 ...## it can be observed that ##a_{2n+1} = \frac{1}{n+2}##.
The series solution can then be written as:
##y = \sum_{n=0}^{∞} a_nx^n = \sum_{n=1}^{∞} a_{2n} x^{2n} + \sum_{n=1}^{∞} a_{2n+1} x^{2n+1} = a_0 + \sum_{n=1}^{∞} \frac{1}{n+2} x^{2n+1} ##.
If ##\nu = 2m## or ##\nu = 2m +1## for some ##m ≥ 1##, then some other interesting things seem to happen. If ##\nu## is even, all of the even ##n## terms of the recurrence relation are zero after a certain point and there is only an odd sum plus a few even terms. If ##\nu## is odd, all of the odd ##n## terms of the recurrence relation are zero after a certain point and there is only an even sum plus a few odd terms.
Does this sound alright? Finding solutions for ##\nu > 0## has proven slightly difficult to formalize.