For $p\geq 0$ the Legendre differential equation is $$(1-t^2)y''-2ty'+p(p+1)y=0.$$ Two linearly independent solutions that I have found for this diff equation are $$y_1(t)=1+\sum_{n=1}^\infty(-1)^n\frac{[(p-2n+2)...(p-2)p][(p+1)(p+3)...(p+2n-1)]}{(2n)!}t^{2n}$$and $$t+\sum_{n=1}^\infty(-1)^n\frac{[(p-2n+1)...(p-3)(p-1)][(p+2)(p+4)...(p+2n)]}{(2n+1)!}t^{2n+1}.$$ How do I show that for $p=m$ (with $m$ an integer) the Legendre diff equation has a polynomial solution of degree $m$?
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1You have used the variable $n$ as an index; so your question is rather "if $m$ is an integer, how do I show that for $p=m$" ...... Hint: if $m$ is even, put $m=2n_0-2$, and compute the coeffcient of $t^{2n}$ in the first series for $n\geq n_0+1$ – Kelenner Oct 21 '15 at 18:48
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@Kelenner You were right about the m/n stuff, I'm not sure what to do with the hint. Do you mind making an answer of that? – Oct 21 '15 at 18:50
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1It is only a hint, not an answer. The factor $(p-2n+2)...(p-2)$ is zero for what values of $p$ ? – Kelenner Oct 21 '15 at 18:53
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@Kelenner for p=2? – Oct 21 '15 at 19:05
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1$p=2,4,...2n-2$. What is the product if $p=m=2n_0-2$ with $n\geq n_0$ ? – Kelenner Oct 21 '15 at 19:08
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@Kelenner then the coefficients are zero? – Oct 21 '15 at 19:15
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1Yes, you are true; and then the first series is a polynomial. – Kelenner Oct 21 '15 at 19:17
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@Kelenner but what if p isnt $2n_0-2$? – Oct 21 '15 at 19:19
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1If $n$ is even, then one series terminates and gives you a polynomial. If $n$ is odd, the other series terminates and gives you a polynomial. You always get one polynomial solution if $n$ is an integer. The other solution is unbounded near $x = \pm 1$. – Disintegrating By Parts Oct 22 '15 at 02:57
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See my post About the Legendre differential equation for details of why the equation has a polynomial solution $P_n(x)$ of degree n (the p in your notation). You can find the other solution there also, for which the expansion does not terminate and which is normally denoted as $Q_n(x)$.
By the way the closed form you included in the question is not really practical in my opinion; better remember Rodrigues' formula $$P_n(x) = \frac{1}{2^nn!}[(x^2-1)^n]^{(n)}$$ which clearly is a polynomial of degree n.
