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

[Void123]

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.

 

[lanedance]

is m an integer?

maybe try showing your working & what special properties of the gamma function can you use to help?

 

[Void123]

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.

 

[Void123]

I think I figured it out.

 

[lanedance]

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
$$\Gamma (n+1) = n \Gamma(n)$$