Apostol calculus 2- Legendre polynomials

Question

I'm trying to understand this formula that Apostol uses to define Legendre polynomials.

Suppose $$\alpha=2m, m\in \mathbb{Z}$$ Then:

$$(2m+1)(2m+3)\dotsb(2m+2n-1)=\frac{(2m+2n)!m!}{2^{n}(2m)!(m+n)!}$$ I don't understand the right hand side of this equality. Can anyone help me, please?

Answer 1

I will assume that you are familiar with following equation $$(m)!\;(m+1)=(m+1)!$$

Multiply numerator and denominator by $(2m)!\;(2m+2)(2m+4)\cdots(2m+2n)$ in L.H.S . You get

$(2m+1)(2m+3)\cdots(2m+2n-1)\left(\displaystyle\frac{(2m)!\;(2m+2)(2m+4)\cdots(2m+2n)}{(2m)!\;(2m+2)(2m+4)\cdots(2m+2n)}\right)$ .

Rearrange numerator as $$\displaystyle\frac{(2m)!\;(2m+1)(2m+2)(2m+3)\cdots(2m+2n)}{(2m)!\;(2m+2)(2m+4)\cdots(2m+2n)}$$ Numerator is then you see is $(2m+2n)!$ by applying the assumed equation succesively .So LHS becomes $$\displaystyle\frac{(2m+2n)!}{(2m)!\;(2m+2)(2m+4)\cdots(2m+2n)}$$In denominator factor out all $2$'s and club them together$$\displaystyle\frac{(2m+2n)!}{(2m)!\;2(m+1)2(m+2)2(m+3)\cdots2(m+n)}$$ since there are $n$ $2$'s clubbing them makes it$$\displaystyle\frac{(2m+2n)!}{(2m)!\;2^n\;(m+1)(m+2)(m+3)\cdots(m+n)}$$ Now multiply $(m)!$ in both numerator and denominator$$\displaystyle\frac{(2m+2n)!\;(m)!}{(2m)!\;2^n\;(m+1)(m+2)(m+3)\cdots(m+n)(m)!}$$Note that $(m+1)(m+2)(m+3)\cdots(m+n)(m)!=(m+n)!$

This gives us$$\displaystyle\frac{(2m+2n)!\;(m)!}{(2m)!\;2^n\;(m+n)!}$$

Answer 2

Note that:

$$(2m+2)(2m+4)\dotsb(2m+2n)=2^n(m+1)(m+2)\dotsb(m+n)$$ Now

$$\frac{(2m)!(2m+1)(2m+3)\dotsb(2m+2n-1)\cdot(2m+2)(2m+4)\dotsb(2m+2n)}{(2m)!(2m+2)(2m+4)\dotsb(2m+2n)}=\frac{(2m+2n)!}{2^{n}(2m)!(m+1)(m+2)\dotsb(m+n)}=\frac{(2m+2n)!m!}{2^{n}(2m)!m!(m+1)(m+2)\dotsb(m+n)}=\frac{(2m+2n)!m!}{2^{n}(2m)!(m+n)!}$$

Answer 3

It’ll help you to understand double factorials.

$$(n)!!:=(n)(n-2)\cdots$$

Now notice that

$$(2n+1)!=(2n+1)!!(2n)!!$$

Finally $(2n)!!=(2n)(2n-2)\ldots =2^nn!$. These relations will help you derive your result.