I am reading a textbook on differential equations.The chapter I am studying now is about solving differential equations using series.The book solves Legendre's differential equation$$(1-x^{2})y''-2xy'+v(v+1)y=0$$ using the series method. First, it assumes that the equation has a solution in the form of$$y=c_{0}+c_{1}x+c_{2}x^{2}+\cdots+c_{n}x^{n}+\cdots$$ Then takes the first and second derivatives of $y$ and puts them in the equation $(1-x^{2})y''-2xy'+v(v+1)y=0$ to get the coefficients $c_{i}$. we have$$y'=c_{1}+2c_{2}x+3c_{3}x^{2}+\cdots+nc_{n}x^{n-1}\cdots$$ and $$y''=2c_{2}+3\times 2c_{3}x+4\times 3c_{4}x^{2}+\cdots+n(n+1)c_{n}x^{n-2}+\cdots$$ Now put these expressions in the equation$$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 2c_{2}+3\times 2c_{3}x+4\times 3c_{4}x^{2}+\cdots+(n+2)(n+1)c_{n+2}x^{n}+\cdots$$ $$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ -2c^{2}x^{2}-\cdots-n(n-1)c_{n}x^{n}-\cdots$$ $$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ -2c_{1}x-2\times 2c_{2}x^{2}-\cdots-2nc_{n}x^{n}-\cdots$$ $$+v(v+1)c_{0}+v(v+1)c_{1}x+v(v+1)c_{2}x^{2}+\cdots+v(v+1)c_{n}x^{n}+\cdots=0$$ Equating the coefficient of each power of x to zero, we have $$2c_{2}+v(v+1)c_{0}=0\Rightarrow c_{2}=-{{v(v+1)}\over{2}}c_{0}$$ $$3\times 2c_{3}-2c_{1}+v(v+1)c_{1}=0\Rightarrow c_{3}=-{{(v-1)(v+2)}\over{3!}}c_{1}$$ $$\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots$$ $$(n+2)(n+1)c_{n+2}+\left[{-n(n-1)-2n+v(v+1)}\right]c_{n}=0\Rightarrow c_{n+2}=-{{(v-n)(v+n+1)}\over{(n+2)(n+1)}}c_{n}$$ The last relation is a recursive relation. Using the recursive relation, all odd coefficients are expressed in terms of $c_{1}$ and all even coefficients are expressed in terms of $c_{0}$ and $c_{0}$ and $c_{1}$ are arbitrary constants.So we have: $$c_{4}=-{{(v-2)(v+3)}\over{4\times 3}}c_{2}={{(v-2)v(v+1)(v+3)}\over{4!}}c_{0}$$ $$c_{5}=-{{(v-3)(v+4)}\over{5\times 4}}c_{3}={{(v-3)(v-1)(v+2)(v+4)}\over{5!}}c_{1}$$ And the rest of the coefficients are calculated in the same way. By inserting these expressions for the coefficients into $y=c_{0}+c_{1}x+c_{2}x^{2}+\cdots+c_{n}x^{n}+\cdots$ we obtain $$y=c_{0}(1-{{v(v+1)}\over{2!}}x^{2}+{{(v-2)v(v+1)(v+3)}\over{4!}}x^{4}-+\cdots )$$ $$\ \ \ +c_{1}(x-{{(v-1)(v+2)}\over{3!}}x^{3}+{{(v-3)(v-1)(v+2)(v+4)}\over{5!}}x^{5}-+\cdots )$$ We notice that the general solution is in the form $$y(x)=c_{0}R_{v}(x)+c_{1}S_{v}(x)$$ where $R_{v}(x)$ and $S_{v}(x)$ are two linearly independent solutions. If $v=n$ (a non-negative integer), then according to the recursive relation $c_{n+2}=-{{(v-n)(v+n+1)}\over{(n+2)(n+1)}}c_{n}$, we have $$c_{n+2}=c_{n+4}=\cdots=0$$ As a result, when n is even, the series $R_{n}(x)$ is a polynomial of degree n and $S_{n}(x)$ will be a series, and when $n$ is odd, the series $S_{n}(x)$ will be a polynomial of degree $n$ and $R_{n}(x)$ will be a series. Therefore, for every non-negative integer n, either $R_{n}(x)$ or $S_{n}(x)$ (and not both) is a polynomial of degree n. These polynomials, after being multiplied by some constants, are called Legendre polynomials and are denoted by $P_{n}(x)$. We write the recursive relation $c_{n+2}=-{{(v-n)(v+n+1)}\over{(n+2)(n+1)}}c_{n}$ in the form $$c_{n}=-{{(n+2)(n+1)}\over{(v-n)(v+n+1)}}c_{n+2}$$ Then we obtain all the non-zero coefficients in terms of $c_{v}$. In this case, $c_{v}$ is an arbitrary constant and it is customary to choose $$c_{v}={{(2v)!}\over{2^{v}(v!)^{2}}}={{1\times 3\times 5\times\cdots\times (2v-1)}\over{v!}}$$ Now, according to relation $c_{n}=-{{(n+2)(n+1)}\over{(v-n)(v+n+1)}}c_{n+2}$, we have: $$c_{v-2}=-{{v(v-1)}\over{2(2v-1)}}c_{v}=-{{v(v-1)}\over{2(2v-1)}}{{(2v)!}\over{2^{v}(v!)^{2}}}$$ $$=-{{v(v-1)(2v)(2v-1)(2v-2)!}\over{2(2v-1)2^{v}v(v-1)!v(v-1)(v-2)!}}$$ $$=-{{(2v-2)!}\over{2^{v}(v-1)!(v-2)!}}$$ And in a similar way $$c_{v-4}=-{{(v-2)(v-3)}\over{4(2v-3)}}c_{v-2}={{(v-2)(v-3)}\over{4(2v-3)}}{{(2v-2)!}\over{2^{v}(v-1)!(v-2)!}}$$ $$={{(2v-4)!}\over{2^{v}2!(v-2)!(v-4)!}}$$ and in the general case for $v-2k\geq 0$ we have $$c_{v-2k}=(-1)^{k}{{(2v-2k)!}\over{2^{v}k!(v-k)!(v-2k)!}}$$ and assuming that $v$ is a non-negative integer like $n$, the resulting solution is called a Legendre polynomial of degree $n$ and we denote it by $P_{n}(x)$ and according to the relation $c_{v-2k}=(-1)^{k}{{(2v-2k)!}\over{2^{v}k!(v-k)!(v-2k)!}}$, we have $$P_{n}(x)=\sum\limits_{k=0}^{\left[{{{n}\over{2}}}\right]}{(-1)^{k}{{(2n-2k)!}\over{2^{n}k!(n-k)!(n-2k)!}}}x^{n-2k}$$ So the general solution of a Legendre equation with $v=n$ is $$y=a_{1}P_{n}(x)+a_{2}Q_{n}(x)$$ where $P_{n}(x)$ is a polynomial of degree n which is calculated by the formula $P_{n}(x)=\sum\limits_{k=0}^{\left[{{{n}\over{2}}}\right]}{(-1)^{k}{{(2n-2k)!}\over{2^{n}k!(n-k)!(n-2k)!}}}x^{n-2k}$ and q is an infinite series. We call $P_{n}(x)$'s Legendre polynomials and $Q_{n}(x)$'s second type Legendre functions.
So far everything is clear but after this book says: As we said before, the general solution of Legendre's equation is in general form $$y(x)=c_{0}\left[{{\rm 1}+\sum\limits_{{\it k}{\rm =1}}^{\infty}{{{{\rm (}-{\rm 1}{\rm )}^{{\it k}}{\rm (}{\it v}-{\rm (}{\rm 2}{\it k}-{\rm 2}{\rm ))...(}{\it v}-{\rm 2}{\rm )}{\it v}{\rm (}{\it v}+{\rm 1}{\rm )(}{\it v}+{\rm 3}{\rm )...(}{\it v}+{\rm (}{\rm 2}{\it k}-{\rm 1}{\rm ))}}\over{{\rm (}{\rm 2}{\it k}{\rm )!}}}}{\it x}^{{\rm 2}{\it k}}}\right]$$ $$\ \ \ \ \ \ \ \ \ +c_{1}\left[{{\rm x}+\sum\limits_{{\it k}{\rm =1}}^{\infty}{{{{\rm (}-{\rm 1}{\rm )}^{{\it k}}{\rm (}{\it v}-{\rm (}{\rm 2}{\it k}-{\rm 1}{\rm ))...(}{\it v}-{\rm 3}{\rm )(v}{\rm -}{\rm 1)(}{\it v}+2{\rm )(}{\it v}+4{\rm )...(}{\it v}+2k{\rm )}}\over{{\rm (}{\rm 2}{\it k+1}{\rm )!}}}}{\it x}^{{\rm 2}{\it k+1}}}\right]$$ $$=c_{0}R_{v}(x)+c_{1}S_{v}(x)$$ and when $v=n$ ($n$ a non-negative integer) either $R_{n}(x)$ or $S_{n}(x)$ is a polynomial of degree $n$ and as it was said before, we denote these polynomials with $P_{n}(x)$. $P_{n}(x)$ can be obtained from the following equation: $$P_{n}(x)=\left\{{\matrix{ {{{R_{n}(x)}\over{R_{n}(1)}},even\ n}\cr {{{S_{n}(x)}\over{S_{x}(1)}},\ odd\ n}\cr }}\right.\ \ \ \ \ \ \ \ \ \ \ \ \ (1)$$ And the other solution, which is an infinite series and is denoted by $Q_{n}(x)$, can be obtained from the following relation: $$Q_{n}(x)=\left\{{\matrix{ {R_{n}(1)S_{n}(x)\ \ \ ,even\ n}\cr {-S_{x}(1)R_{n}(x),\ \ \ odd\ n}\cr }}\right.\ \ \ \ \ \ \ \ \ \ \ \ \ (2)$$ This is where I run into trouble. The book does not explain where formulas 1 and 2 came from.
These formulas are equivalent to say that for even ${\it n}$, Legendre polynomials ${\it P}_{{\it n}}{\rm (}{\it x}{\rm )}$ 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:$${\it P}_{{\rm 4}}{\rm (}{\it x}{\rm )}={{{\rm 1}+\sum\limits_{{\it k}{\rm =1}}^{{\rm 2}}{{{{\rm (}-{\rm 1}{\rm )}^{{\it k}}{\rm (}{\it 4}-{\rm (}{\rm 2}{\it k}-{\rm 2}{\rm ))...(}{\it 4}-{\rm 2}{\rm )}{\it 4}{\rm (}{\it 4}+{\rm 1}{\rm )(}{\it 4}+{\rm 3}{\rm )...(}{\it 4}+{\rm (}{\rm 2}{\it k}-{\rm 1}{\rm ))}}\over{{\rm (}{\rm 2}{\it k}{\rm )!}}}}{\it x}^{{\rm 2}{\it k}}}\over{{\rm 1}+\sum\limits_{{\it k}{\rm =1}}^{{\rm 2}}{{{{\rm (}-{\rm 1}{\rm )}^{{\it k}}{\rm (}{\it 4}-{\rm (}{\rm 2}{\it k}-{\rm 2}{\rm ))...(}{\it 4}-{\rm 2}{\rm )}{\it 4}{\rm (}{\it 4}+{\rm 1}{\rm )(}{\it 4}+{\rm 3}{\rm )...(}{\it 4}+{\rm (}{\rm 2}{\it k}-{\rm 1}{\rm ))}}\over{{\rm (}{\rm 2}{\it k}{\rm )!}}}}}}={{{\rm 1}+{{-{\rm 1}{\rm \times}{\rm 4}{\rm \times}{\rm 5}}\over{{\rm 2}{\rm !}}}{\it x}^{{\rm 2}}+{{{\rm 2}{\rm \times}{\rm 4}{\rm \times}{\rm 5}{\rm \times}{\rm 7}}\over{{\rm 4}{\rm !}}}{\it x}^{{\rm 4}}}\over{{\rm 1}+{{-{\rm 1}{\rm \times}{\rm 4}{\rm \times}{\rm 5}}\over{{\rm 2}{\rm !}}}+{{{\rm 2}{\rm \times}{\rm 4}{\rm \times}{\rm 5}{\rm \times}{\rm 7}}\over{{\rm 4}{\rm !}}}}}={{{\rm 1}-{\rm 10}{\it x}^{{\rm 2}}+{{{\rm 35}}\over{{\rm 3}}}{\it x}^{{\rm 4}}}\over{{\rm 1}-{\rm 10}+{{{\rm 35}}\over{{\rm 3}}}}}={{{\rm 3}}\over{{\rm 8}}}{\rm (}{\rm 1}-{\rm 10}{\it x}^{{\rm 2}}+{{{\rm 35}}\over{{\rm 3}}}{\it x}^{{\rm 4}}{\rm )}={{{\rm 1}}\over{{\rm 8}}}{\rm (}{\rm 3}-{\rm 30}{\it x}^{{\rm 2}}+{\rm 35}{\it x}^{{\rm 4}}{\rm )}$$ and for odd ${\it n}$, Legendre polynomials ${\it P}_{{\it n}}{\rm (}{\it x}{\rm )}$ are obtained from this formula: $${{{\rm x}+\sum\limits_{{\it k}{\rm =1}}^{\left[{{{n}\over{2}}}\right]}{{{{\rm (}-{\rm 1}{\rm )}^{{\it k}}{\rm (}{\it n}-{\rm (}{\rm 2}{\it k}-{\rm 1}{\rm ))...(}{\it n}-{\rm 3}{\rm )(n}{\rm -}{\rm 1)(}{\it n}+2{\rm )(}{\it n}+4{\rm )...(}{\it n}+2k{\rm )}}\over{{\rm (}{\rm 2}{\it k+1}{\rm )!}}}}{\it x}^{{\rm 2}{\it k+1}}}\over{{\rm 1}+\sum\limits_{{\it k}{\rm =1}}^{\left[{{{n}\over{2}}}\right]}{{{{\rm (}-{\rm 1}{\rm )}^{{\it k}}{\rm (}{\it n}-{\rm (}{\rm 2}{\it k}-{\rm 1}{\rm ))...(}{\it n}-{\rm 3}{\rm )(n}{\rm -}{\rm 1)(}{\it n}+2{\rm )(}{\it n}+4{\rm )...(}{\it n}+2k{\rm )}}\over{{\rm (}{\rm 2}{\it k+1}{\rm )!}}}}}}={\it P}_{{\it n}}{\rm (}{\it x}{\rm )}$$(provided that n be odd) In this formula, $\left[{{{n}\over{2}}}\right]$ is the largest integer smaller than ${{n}\over{2}}$. for example:$${\it P}_{{\it 5}}{\rm (}{\it x}{\rm )}{\rm =}{{{\rm x}+\sum\limits_{{\it k}{\rm =1}}^{2}{{{{\rm (}-{\rm 1}{\rm )}^{{\it k}}{\rm (}{\it 5}-{\rm (}{\rm 2}{\it k}-{\rm 1}{\rm ))...(}{\it 5}-{\rm 3}{\rm )(5}{\rm -}{\rm 1)(}{\it 5}+2{\rm )(}{\it 5}+4{\rm )...(}{\it 5}+2k{\rm )}}\over{{\rm (}{\rm 2}{\it k+1}{\rm )!}}}}{\it x}^{{\rm 2}{\it k+1}}}\over{{\rm 1}+\sum\limits_{{\it k}{\rm =1}}^{2}{{{{\rm (}-{\rm 1}{\rm )}^{{\it k}}{\rm (}{\it 5}-{\rm (}{\rm 2}{\it k}-{\rm 1}{\rm ))...(}{\it 5}-{\rm 3}{\rm )(5}{\rm -}{\rm 1)(}{\it 5}+2{\rm )(}{\it 5}+4{\rm )...(}{\it 5}+2k{\rm )}}\over{{\rm (}{\rm 2}{\it k+1}{\rm )!}}}}}}={{x+{{-1\times 4\times 7}\over{3!}}x^{3}+{{2\times 4\times 7\times 9}\over{5!}}x^{5}}\over{1+{{-1\times 4\times 7}\over{3!}}+{{2\times 4\times 7\times 9}\over{5!}}}}={{x-{{14}\over{3}}x^{3}+{{21}\over{5}}x^{5}}\over{1-{{14}\over{3}}+{{21}\over{5}}}}={{15}\over{8}}(x-{{14}\over{3}}x^{3}+{{21}\over{5}}x^{5})={{1}\over{8}}(15x-70x^{3}+63x^{5})$$ which is also true. I want to prove these two formulas(1 & 2).If you can, please help me or if you know a textbook that mentions these two formulas, please introduce it to me. thanks