Please help in Gamma function to series conversion.

Click For Summary

Discussion Overview

The discussion revolves around finding a series representation for the Gamma function, specifically \Gamma(n-3/2+1), and its relation to Bessel functions. Participants explore various approaches to derive the series, particularly focusing on the values of n from 0 to 3 and the challenges associated with n=0.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant requests assistance in deriving the series representation of \Gamma(n-3/2+1) and provides specific values for n=0 to n=3.
  • Another participant suggests a method of substituting n with n-3 to express \Gamma(n-3/2+1) in terms of \Gamma(n-3) but notes that this approach may only apply for n ≥ 3.
  • The original poster expresses difficulty in obtaining a series solution that includes n=0, indicating that the challenge lies in this specific case.
  • A later post mentions a related problem involving the Bessel function and its solution, suggesting a connection to the original inquiry.
  • Some participants reference external resources, such as a Wikipedia article on Bessel functions, but do not provide specific solutions or insights into the original problem.

Areas of Agreement / Disagreement

Participants do not reach a consensus on how to derive the series representation for \Gamma(n-3/2+1), and multiple competing views and approaches are presented. The discussion remains unresolved regarding the specific challenge of n=0.

Contextual Notes

Participants express uncertainty about the applicability of certain mathematical transformations and the limitations of their approaches, particularly concerning the value of n=0.

yungman
Messages
5,741
Reaction score
291
Can you show me how to get the series representation of \Gamma(n-3/2+1)?


For example \Gamma(n+3/2+1)=\frac{(2n+3)(2n+1)!}{2^{2n+2}.n!}.


I cannot figure out how to write a series with:

n=0 => \Gamma(0-3/2+1)= -2\sqrt{\pi}

n=1 => \Gamma(1-3/2+1)= \sqrt{\pi}

n=2 => \Gamma(2-3/2+1)= 1/2\sqrt{\pi}

n=3 => \Gamma(3-3/2+1)= 4/3\sqrt{\pi}

This is not homework. I have spent 2 days on this and can't figure it out!

Thanks

Alan
 
Physics news on Phys.org
If I understand you correctly, the trick is to replace n by n-3:
\Gamma(n - \frac 3 2 +1) = \Gamma((n-3) + \frac 3 2 + 1) =\frac{(2(n-3)+3)(2(n-3)+1)!}{2^{2(n-3)+2}(n-3)!}=\frac{(2n-3)(2n-5)!}{2^{2n-4}(n-3)!}

At least I think this works for n \ge 3.
 
LCKurtz said:
If I understand you correctly, the trick is to replace n by n-3:
\Gamma(n - \frac 3 2 +1) = \Gamma((n-3) + \frac 3 2 + 1) =\frac{(2(n-3)+3)(2(n-3)+1)!}{2^{2(n-3)+2}(n-3)!}=\frac{(2n-3)(2n-5)!}{2^{2n-4}(n-3)!}

At least I think this works for n \ge 3.

Thanks for trying. But that was where I got really stuck. I can easily for find the series representation for n=1,2,3,4... It is the n=0 that is the problem. The series solution has to cover n=0,1,2...

Actually I am working on the problem is to show

J-3/2(x)=\sqrt{\frac{2}{\pi x}}[\frac{-cos(x)}{x}-sin(x)]

Where this is solution of Bessel function.
 
Please, even if you have any suggestion, I would like to listen to it. I am very despirate! I absolutely ran out of ideas!
 
LCKurtz said:
It has been many years since I looked at Bessel functions, so I can't help you more. I assume you have already looked at resources on the web such as this wikipedia article:

http://en.wikipedia.org/wiki/Bessel_function#Bessel_functions_of_the_first_kind_:_J.CE.B1

If you haven't already looked there you might find something useful. Good luck.

Thanks

I am studying on my own, this question only on the first section, before a lot of the information your link shows. I have to spend some time looking at this. Yes I did look at this before.

The question in the book ask both J(3/2) and J(-3/2). The possitive is relative easy because the gamma function always stay possitive.
 

Similar threads

  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 2 ·
Replies
2
Views
4K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
6
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 7 ·
Replies
7
Views
3K
  • · Replies 15 ·
Replies
15
Views
1K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 2 ·
Replies
2
Views
1K