What is the Proof for J(-m) = [(-1)^m][J(m)]?

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Homework Help Overview

The discussion revolves around proving the relationship J(-m) = [(-1)^m][J(m)], where J represents a function defined in terms of a summation involving the gamma function. The context involves properties of the gamma function and its behavior with negative integers.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the implications of m being an integer and explore the properties of the gamma function, particularly its behavior at negative integers. There are attempts to manipulate the given formulas and clarify the significance of certain terms in the summation.

Discussion Status

Some participants have provided insights into the properties of the gamma function and its implications for the problem. There is an acknowledgment of the need to eliminate terms that lead to infinity, and one participant claims to have made progress in understanding the problem.

Contextual Notes

Participants note that the gamma function has specific behaviors at negative integers, which is crucial for the discussion. There is mention of a constant 'a' that arises from simplifying terms in the equation, indicating that certain assumptions or simplifications are being considered.

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Homework Statement



Prove J(-m) = [(-1)^m][J(m)]

(Note: by "J(-m)" I mean "subscript (-m)")

Homework Equations



J(-m) = sum [((-1)^n) * (x/2)^(2n-m)]/[n! \Gamma(n - m + 1)]

J(m) should be obvious.

The Attempt at a Solution



I tried just plugging in the above formulas hoping to get a simplified answer, but I know I'm missing something in that denominator.
 
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is m an integer?

maybe try showing your working & what special properties of the gamma function can you use to help?
 
Yes, m is an integer. I am also told that \Gamma (-k) = \infty and that I should, therefore, eliminate the terms from the sum that equal zero. Also, 1/[(2^m)(\Gamma (m + 1))] is reduced to some 'a' constant.

That's all I got.

Not entirely sure what to do.
 
I think I figured it out.
 
cool, yeah its true that for negative integers
|\Gamma (-k)| \rightarrow \infty
the other handy ones were for integers:
\Gamma (n) = (n-1)!
and so
\Gamma (n+1) = n \Gamma(n)
 

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