Proving x^n = ∑(x!/x!-k!)S(n,k): Why x Must Be >0 - Insights & Explanation

[Punkyc7]

So the question was, Let x > 0.

Prove that x$^{n}$ = $\sum$ $\frac{x!}{x!-k!}$ S(n, k).

Where the sum goes from k = 1 to n and S(n, k) is th Stirling numbers.

I believe I have proven what I needed to, but my question is why does x have to be greater than 0? Couldn't we define a function that maps [n] to {-x, ..., 1}.

 

[Ray Vickson]

So the question was, Let x > 0.

Prove that x$^{n}$ = $\sum$ $\frac{x!}{x-k!}$ S(n, k).

Where the sum goes from k = 1 to n and S(n, k) is th Stirling numbers.

I believe I have proven what I needed to, but my question is why does x have to be greater than 0? Couldn't we define a function that maps [n] to {-x, ..., 1}.

Is this equation accurate? The kth term blows up when x = k! for some k in {1,2,...,n}.

RGV

 

[Punkyc7]

that should be an x! on the bottom... ill fix that