(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

Show, by direct examination of the Frobenius series solution to Legendre's differential equation that;

[tex] P_n(x) = \sum_{k=0}^{N} \frac{(-1)^k(2n-k)!} {2^n k! (n-k)! (n-2k)!}x^{n-2k} ;\ N=\frac{n}{2}\ \mathrm{n\ even,}\

N=\frac{n-1}{2}\ \mathrm{n\ odd}[/tex]

Write down the first five Legendre polynomials due to this formula. Show also that [tex]P_n(1)=1[/tex] and [tex]P_n(-1)=(-1)^n[/tex]

2. Relevant equations

The Frobenius/Puiseux/Fuchs series solution is: [tex]y= \sum_{n=0}^{\infty} a_{n}x^{k+n}[/tex]

Legendre's Diff Eq is: [tex](1-x^{2})y''-2xy'+l(l+1)=0[/tex]

3. The attempt at a solution

[tex]y= \sum_{n=0}^{\infty} a_{n}x^{k+n}

y'= \sum_{n=0}^{\infty} (k+n)a_{n}x^{k+n-1}

y''= \sum_{n=0}^{\infty} (k+n)(k+n-1)a_{n}x^{k+n-2}[/tex]

Setting the [tex]x^k[/tex] power to zero yields:

[tex] a_2 = \frac{k(k+1)-l(l+1)}{(k+2)(k+1)} a_0[/tex]

Setting the [tex]x^{k+1}[/tex] power to zero yields:

[tex] a_3 = \frac{(k+1)(k+2)+l(l+1)}{(k+3)(k+2)}a_1[/tex]

I believe setting the [tex]x^{k-2}\ \mathrm{and}\ x^{k-1} [/tex] powers yield:

[tex] k(k-1)a_0 = 0

k(k+1)a_1 = 0 [/tex]

although this would seem to imply my first two constants are zero, which would make all higher order constants zero as well since each constant can be written in terms of an earlier constant. This is the first part I am stuck on, and the second is getting an answer in terms of [tex]P_n(x)[/tex] instead of [tex]y(x)[/tex]. Thank you for any direct help or links to reading material!

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# Legendre's Diff Eq using Frobenius

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