I am reading a textbook on differential equations. This book is not written in English and surely you have not heard of it. The chapter I am studying now is about solving differential equations using series. My question is about Legendre differential equation. The book defines Legendre's differential equation as follows:
differential equation$$(1-x^{2})y''-2xy'+v(v+1)y=0$$ where $v$ is a real number is called Legendre's differential equation. This equation has an answer of the form$$y=c_{0}+c_{1}x+c_{2}x^{2}+\cdots+c_{n}x^{n}+\cdots$$Now we calculate the first and second derivatives of $y$ and put them in the equation $(1-x^{2})y''-2xy'+v(v+1)y=0$ and get the coefficients $c_{i}$...
After a lot of calculations, it is concluded that the general solution of Legendre's equation is in the following form:$$y(x)=c_{0}\left[{1-{{v(v+1)}\over{2!}}x^{2}+{{(v-2)v(v+1)(v+3)}\over{4!}}x^{4}-+\cdots}\right]$$$$\ \ \ \ \ \ \ \ \ \ \ +c_{1}\left[{x-{{(v-1)(v+2)}\over{3!}}x^{3}+{{(v-3)(v-1)(v+2)(v+4)}\over{5!}}x^{5}-+\cdots}\right]$$or$$y(x)=c_{0}\left[{1+\sum\limits_{k=1}^{\infty}{{{(-1)^{k}v(v-2)\cdots\left({v-\left({2k-2}\right)}\right).(v+1)(v+3)\cdots (v+(2k-1))}\over{(2k)!}}x^{2k}}}\right]$$ $$\ \ \ \ \ \ \ \ \ \ \ +c_{1}\left[{x+\sum\limits_{k=1}^{\infty}{{{(-1)^{k}(v-1)(v-3)\cdots\left({v-\left({2k-1}\right)}\right).(v+2)(v+4)\cdots (v+2k))}\over{(2k+1)!}}x^{2k+1}}}\right]$$ $$\ \ \ \ \ \ \ \ \ =c_{0}R_{v}(x)+c_{1}S_{v}(x)$$ and When $v=n$($n$ is a non-negative integer), either $R_{v}(x)$ or $S_{v}(x)$ is a polynomial of degree $n$.In this case, Legendre's polynomial $P_{n}(x)$ can be obtained from the following relation:$$P_{n}(x)=\left\{{\matrix{ {{{R_{n}(x)}\over{R_{n}(1)}}\ \ even\ n}\cr {{{S_{n}(x)}\over{S_{n}(1)}}\ \ odd\ n}\cr }}\right.$$ And the Legendre function of the second kind, which is an infinite series and is denoted by $Q_{n}(x)$, can be obtained from the following relation:$$Q_{n}(x)=\left\{{\matrix{ {R_{n}(1)S_{n}(x)\ \ even\ n}\cr {-S_{n}(1)R_{n}(x)\ \ odd\ n}\cr }}\right.$$The book does not explain where the last two formulas for $P_{n}(x)$ and $Q_{n}(x)$ came from. I want to prove these two formulas.If you can, please help me or if you know a textbook that mentions these two formulas, please introduce it to me. thanks
UPDATE: I simplified this question. Just prove that for even n the Legendre polynomials are obtained from this formula: $${{{\rm 1}+\sum\limits_{{\it k}{\rm =1}}^{{{{\it n}}\over{{\rm 2}}}}{{{{\rm (}-{\rm 1}{\rm )}^{{\it k}}{\rm (}{\it n}-{\rm (}{\rm 2}{\it k}-{\rm 2}{\rm ))...(}{\it n}-{\rm 2}{\rm )}{\it n}{\rm (}{\it n}+{\rm 1}{\rm )(}{\it n}+{\rm 3}{\rm )...(}{\it n}+{\rm (}{\rm 2}{\it k}-{\rm 1}{\rm ))}}\over{{\rm (}{\rm 2}{\it k}{\rm )!}}}}{\it x}^{{\rm 2}{\it k}}}\over{{\rm 1}+\sum\limits_{{\it k}{\rm =1}}^{{{{\it n}}\over{{\rm 2}}}}{{{{\rm (}-{\rm 1}{\rm )}^{{\it k}}{\rm (}{\it n}-{\rm (}{\rm 2}{\it k}-{\rm 2}{\rm ))...(}{\it n}-{\rm 2}{\rm )}{\it n}{\rm (}{\it n}+{\rm 1}{\rm )(}{\it n}+{\rm 3}{\rm )...(}{\it n}+{\rm (}{\rm 2}{\it k}-{\rm 1}{\rm ))}}\over{{\rm (}{\rm 2}{\it k}{\rm )!}}}}}}={\it P}_{{\it n}}{\rm (}{\it x}{\rm )}$$ (provided that ${\it n}$ be even) For example: $${\it P}_{{\rm 4}}{\rm (}{\it x}{\rm )}={{{\rm 1}+\sum\limits_{{\it k}{\rm =1}}^{{\rm 2}}{{{{\rm (}-{\rm 1}{\rm )}^{{\it k}}{\rm (}{\it 4}-{\rm (}{\rm 2}{\it k}-{\rm 2}{\rm ))...(}{\it 4}-{\rm 2}{\rm )}{\it 4}{\rm (}{\it 4}+{\rm 1}{\rm )(}{\it 4}+{\rm 3}{\rm )...(}{\it 4}+{\rm (}{\rm 2}{\it k}-{\rm 1}{\rm ))}}\over{{\rm (}{\rm 2}{\it k}{\rm )!}}}}{\it x}^{{\rm 2}{\it k}}}\over{{\rm 1}+\sum\limits_{{\it k}{\rm =1}}^{{\rm 2}}{{{{\rm (}-{\rm 1}{\rm )}^{{\it k}}{\rm (}{\it 4}-{\rm (}{\rm 2}{\it k}-{\rm 2}{\rm ))...(}{\it 4}-{\rm 2}{\rm )}{\it 4}{\rm (}{\it 4}+{\rm 1}{\rm )(}{\it 4}+{\rm 3}{\rm )...(}{\it 4}+{\rm (}{\rm 2}{\it k}-{\rm 1}{\rm ))}}\over{{\rm (}{\rm 2}{\it k}{\rm )!}}}}}}={{{\rm 1}+{{-{\rm 1}{\rm \times}{\rm 4}{\rm \times}{\rm 5}}\over{{\rm 2}{\rm !}}}{\it x}^{{\rm 2}}+{{{\rm 2}{\rm \times}{\rm 4}{\rm \times}{\rm 5}{\rm \times}{\rm 7}}\over{{\rm 4}{\rm !}}}{\it x}^{{\rm 4}}}\over{{\rm 1}+{{-{\rm 1}{\rm \times}{\rm 4}{\rm \times}{\rm 5}}\over{{\rm 2}{\rm !}}}+{{{\rm 2}{\rm \times}{\rm 4}{\rm \times}{\rm 5}{\rm \times}{\rm 7}}\over{{\rm 4}{\rm !}}}}}={{{\rm 1}-{\rm 10}{\it x}^{{\rm 2}}+{{{\rm 35}}\over{{\rm 3}}}{\it x}^{{\rm 4}}}\over{{\rm 1}-{\rm 10}+{{{\rm 35}}\over{{\rm 3}}}}}={{{\rm 3}}\over{{\rm 8}}}{\rm (}{\rm 1}-{\rm 10}{\it x}^{{\rm 2}}+{{{\rm 35}}\over{{\rm 3}}}{\it x}^{{\rm 4}}{\rm )}={{{\rm 1}}\over{{\rm 8}}}{\rm (}{\rm 3}-{\rm 30}{\it x}^{{\rm 2}}+{\rm 35}{\it x}^{{\rm 4}}{\rm )}$$ which is true.