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My professor taught us the series solution to ODE's method today in class, and one of our homework problems was to solve the Legendre Equation.

$$\text{Legendre Equation:} \frac{d^2y}{dx^2}(1-x^2) -2x\frac{dy}{dx} + \alpha(\alpha + 1)y = 0$$

By making a series expansion at $k = 0$:

$$y = \sum_{k=0}^\infty{a_nx^n}$$ $$y' = \sum_{k=0}^\infty{na_nx^{n-1}} $$ $$ y'' = \sum_{k=0}^\infty{n(n-1)a_nx^{n-2}}$$ Plugging in: $$\sum_{k=0}^\infty{n(n-1)a_nx^{n-2}}(1-x^2) -2x\sum_{k=0}^\infty{na_nx^{n-1}} + \alpha(\alpha + 1)\sum_{k=0}^\infty{a_nx^n} = 0$$ $$\sum_{n=0}^\infty{n(n-1)a_nx^{n-2}}-\sum_{n=0}^\infty{n(n-1)a_nx^n} -2x\sum_{n=0}^\infty{na_nx^{n-1}}+\alpha(\alpha+1)\sum_{n=0}^\infty{a_nx^n}=0 $$

$$\sum_{n=0}^\infty{n(n-1)a_nx^{n-2}}-\sum_{n=0}^\infty{n(n-1)a_nx^n} -2\sum_{n=0}^\infty{na_nx^{n}}+\alpha(\alpha+1)\sum_{n=0}^\infty{a_nx^n}=0$$

$$\sum_{n=0}^\infty{(n+2)(n+1)a_{n+2}x^{n}}-\sum_{n=0}^\infty{n(n-1)a_nx^n} -2\sum_{n=0}^\infty{na_nx^{n}}+\alpha(\alpha+1)\sum_{n=0}^\infty{a_nx^n}=0$$

$$ \sum_{n=0}^\infty{(n+1)(n+2)a_{n+2}+[-n(n-1)-2n+\alpha(\alpha+1)]a_n}=0, $$

Each term must cancel so: $$(n+1)(n+2)a_{n+2} + [-n(n+1) + \alpha(\alpha +1)]a_n = 0$$

$$a_{n+2} = \frac{n(n+1)-\alpha(\alpha+1)}{(n+1)(n+2)}a_n$$ $$ = \frac{[\alpha + (n+1)](\alpha -n)}{(n+1)(n+2)}$$

Therefore: $$a_2 = \frac{-\alpha(\alpha+1)}{1·2}a_0 $$ $$a_4 = -\frac{(\alpha-2)(\alpha+3)}{3·4}a_2$$ $$ a_4= (-1)^2\frac{[(\alpha-2)\alpha][(\alpha+1)(\alpha+3)]}{1·2·3·4}a_0$$

So the even solution is: $$ y_1(x)=1+\sum_{n=1}^\infty{(-1)^n\frac{[(\alpha-2n+2)...(\alpha-2)\alpha][(\alpha+1)(\alpha+3)...(\alpha+2n-1)]}{(2n)!}x^{2n}}. $$ Then, the odd solution is: $$ y_2(x)=x+\sum_{n=1}^\infty{(-1)^n\frac{[(\alpha-2n+1)...(\alpha-3)(\alpha-1)][(\alpha+2)(\alpha+4)...(\alpha+2n)]}{(2n+1)!}x^{2n+1}}. $$

Does this look right? This is my first attempt at the series solution method, so I would really appreciate any help. Thanks in advance!

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I do not understand the concept of odd and even solutions.

Since the equation is of second order, there will be two arbitraty constants $a_0$ and $a_1$. You properly established that $$a_{n+2} = \frac{n(n+1)-\alpha(\alpha+1)}{(n+1)(n+2)}a_n\tag 1$$ and to me, this is enough.

The solution is then $$y=a_0+a_1x+\sum_{n=2}^\infty a_n x^n$$ For sure, you could write it as $$y=a_0+a_1x+\sum_{n=1}^\infty a_{2n} x^{2n}+\sum_{n=1}^\infty a_{2n+1} x^{2n+1}$$ and using $(1)$ make the coefficients totally explicit solving the recurrence equation given by $(1)$.

Even if this will look more elegant, it would not be very efficient from a computing point of view. You would have quite complex formulae (I wrote them) including gamma functions (as you wrote) that you will need to recompute every time while using the recurrence equation will make the evaluation very unexpensive.

In any manner, you did a good job and $\to +1$.

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