Legendre polynomials, Hypergeometric function

[LagrangeEuler]

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?

 

[TeethWhitener]

Some useful properties of Pochhammer symbols can be found here:
http://mathworld.wolfram.com/PochhammerSymbol.html

 

[LagrangeEuler]

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

 

[TeethWhitener]

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.

 

[LagrangeEuler]

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

 

[TeethWhitener]

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.

You're contradicting yourself.

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.

 

[LagrangeEuler]

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]

https://en.wikipedia.org/wiki/Gamma_function#/media/File:Gamma_plot.svg

 

[LagrangeEuler]

See 2:27.

 

[TeethWhitener]

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]

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

When these things occur in appropriate ratios, we can get meaningful results even if the numerator and denominator are both undefined. The reason is that we start by defining the ratio in the region where no problems occur, express the ratio in another way, then use that other way to extend the ratio into regions where the original definition would be nonsense. For example, the binomial coefficient for general $n \geq 0$ and integer $k$ ($n \geq k \geq 0$) is
$${n \choose k} = \frac{\Gamma(n+1)}{\Gamma(n-k+1) \Gamma(k+1)} = \frac{n!}{(n-k)! k!},\hspace{3ex}(1)$$ 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.