# Legendre polynomials, Hypergeometric function

## Homework Statement

$$_2F_1(a,b;c;x)=\sum^{\infty}_{n=0}\frac{(a)_n(b)_n}{(c)_nn!}x^n$$
Show that Legendre polynomial of degree ##n## is defined by
$$P_n(x)=\,_2F_1(-n,n+1;1;\frac{1-x}{2})$$

## Homework Equations

Definition of Pochamer symbol[/B]
$$(a)_n=\frac{\Gamma(a+n)}{\Gamma(a)}$$

## The Attempt at a Solution

$$P_n(x)=\,_2F_1(-n,n+1;1;\frac{1-x}{2})=\sum^{\infty}_{k=0}\frac{\frac{\Gamma(-n+k)}{\Gamma(-n)}\frac{\Gamma(n+k+1)}{\Gamma(n+1)}}{(k!)^2}\frac{(1-x)^k}{2^k}$$
Main problem for me is ##\Gamma(-n)##. Bearing in mind that ##n## is degree of polynomial, ##\Gamma(-n)## diverge. What is solution of this issue?

Yes but I have concrete question. What to do with ##\Gamma(-n), n>0##?

TeethWhitener
Gold Member
Yes but I have concrete question. What to do with ##\Gamma(-n), n>0##?
My point was: you can't do anything with ##\Gamma(-n), n>0##, which diverges. You have to look at ##\frac{\Gamma(-n+k)}{\Gamma(-n)} ##, which doesn't diverge.

But ##\Gamma(-n), n>0## does not diverge. It does not exist. ##\Gamma(-n),n>0## is not defined.

Last edited:
TeethWhitener
Gold Member
Bearing in mind that ##n## is degree of polynomial, ##\Gamma(-n)## diverge. What is solution of this issue?
But ##\Gamma(-n), n>0## does not diverge. It does exist. ##\Gamma(-n),n>0## is not defined.

The point I'm trying to make is that even if a piece of your function is divergent, it might be the case that the whole function actually converges. For a trivial example, let ##\Lambda(x)=\frac{1}{x}##, ##\Omega(x)=\frac{2}{x}##, and build the function ##A(x) = \frac{\Lambda(x)}{\Omega(x)}##. Even though both ##\Lambda(x)## and ##\Omega(x)## diverge, ##A(x)## does not.

With this consideration in mind, my advice remains as before: take a closer look at ##\frac{\Gamma(-n+k)}{\Gamma(-n)}## and supplement with the Mathworld page on Pochhammer symbols.

Yes. I understand you. However, I am not sure what of the two is correct. Does ##\Gamma(-1) \to \infty##, or ##\Gamma(-1)## is not defined?

TeethWhitener
Gold Member
See 2:27.

TeethWhitener
Gold Member
I have no idea what your point is. The gamma function is undefined at negative integers. Its limit doesn't exist at those points; see the plot in post #8 (This is what I meant by divergent, which admittedly is sloppy terminology). The Pochhammer symbol is analytic everywhere in the complex plane.

Ray Vickson
$${n \choose k} = \frac{\Gamma(n+1)}{\Gamma(n-k+1) \Gamma(k+1)} = \frac{n!}{(n-k)! k!},\hspace{3ex}(1)$$ where we have used the definition ##u! = \Gamma(u+1).## For integer ##n \geq k \geq 0##, (1) becomes
$${n \choose k} = \frac{n(n-1) \cdots(n-k+1)}{k!} \hspace{3ex}(2)$$ We can use (2) to define ##{n \choose k}## when ##k \geq 0## is an integer but ##n## is non-integer and/or ##n < 0##. In that case the original definition in (1) no longer makes any sense.