Can this summation be simplified via factorization?

[humy]

With m being some arbitrary positive natural number constant and n being a positive natural number variable, can the summation of;

∑[k=0, n-1] (n - k)^m (-1)^k n! / ( k! (n - k)! )

be simplified to a NONE summation by it being factorized?
The reason why I think there just might be a way is because I see contained in there, more specifically the "(-1)^k n! / ( k! (n - k)! ) " part, binomial coefficients, which implies to me it just might be a way to simplify it by factorizing it.
I tried to see if I could factorize it but so far I just cannot get past the " (n - k)^m " component of the summation for that because that part to me seems to me to 'get in the way' of factorization.

 

[Deleted member 4993]

With m being some arbitrary positive natural number constant and n being a positive natural number variable, can the summation of;

∑[k=0, n-1] (n - k)^m (-1)^k n! / ( k! (n - k)! )

be simplified to a NONE summation by it being factorized?
The reason why I think there just might be a way is because I see contained in there, more specifically the "(-1)^k n! / ( k! (n - k)! ) " part, binomial coefficients, which implies to me it just might be a way to simplify it by factorizing it.
I tried to see if I could factorize it but so far I just cannot get past the " (n - k)^m " component of the summation for that because that part to me seems to me to 'get in the way' of factorization.

I have not tried it - but I suggest:

Write out the first few terms (say 4 or 5) and see if it is amenable to "telescoping".

 

[Dr.Peterson]

Since m is a constant, I don't think this will relate to the binomial theorem, or simplify by any other obvious method I can see. Writing it out for n=1, 2, 3 doesn't make me hopeful for factorization.

But perhaps if you told us where the series came from, that might suggest a simpler form that could be tried, and proved by induction.

 

[humy]

I have already written out all the terms of this summation from n=1 to n=8 and here are the first 5;

for n=1
sum = 1*1^m
= 1

for n=2
sum = 2^m - 2*1^m
= 2^m - 2

for n=3
sum = 3^m - 3*2^m + 3*1^m
= 3^m - 3*2^m + 3

for n=4
sum = 4^m - 4*3^m + 6*2^m - 4*1^m
= 4^m - 4*3^m + 6*2^m - 4

for n=5
sum = 5^m - 5*4^m + 10*3^m - 10*2^m + 5*1^m
= 5^m - 5*4^m + 10*3^m - 10*2^m + 5

But I didn't see how this helps me.

The summation is just part of a much more complex equation that is for the likelihood probability of a distribution of an unknown number g of none-numerical categories and, trust me, knowing where that summation came from won't help us here! Some of the equations I have derived will more than fill a whole page of a book!

 

[Dr.Peterson]

There's a good chance it can't be simplified.

 

[humy]

There's a good chance it can't be simplified.

If that's true, I wouldn't mind too much.
I am going to put the equation with that summation in a book about statistics I am planning to get published and thought that summation looked a bit ugly and thought it would be nice if it could somehow be simplified to a neater expression but that's is really about esthetics rather than practicality so its not really that important as it wouldn't be a mar to my good work if it doesn't simplify.

 

[Steven G]

With m being some arbitrary positive natural number constant and n being a positive natural number variable, can the summation of;

∑[k=0, n-1] (n - k)^m (-1)^k n! / ( k! (n - k)! )

be simplified to a NONE summation by it being factorized?
The reason why I think there just might be a way is because I see contained in there, more specifically the "(-1)^k n! / ( k! (n - k)! ) " part, binomial coefficients, which implies to me it just might be a way to simplify it by factorizing it.
I tried to see if I could factorize it but so far I just cannot get past the " (n - k)^m " component of the summation for that because that part to me seems to me to 'get in the way' of factorization.

What is a NONE summation. Please be more clear.
You say that you want to factor? It already is factored. It may be able to be simplified but it is still factored. Please read up on what a factor is and how to tell if you have more than one term. In order to factor something you need at least two terms. Do you know what a term is?

 

[Dr.Peterson]

Huh? It's a summation; it has n terms. Why do you say it has one term and is already factored?

I suspect he means "non-summation" (and "non-numerical"), and is struggling for a way to say what he means. I'm not sure what kind of factorization he is hoping for, but probably something like the binomial formula would give (a power of a binomial).

 

[Steven G]

Huh? It's a summation; it has n terms. Why do you say it has one term and is already factored?

I suspect he means "non-summation" (and "non-numerical"), and is struggling for a way to say what he means. I'm not sure what kind of factorization he is hoping for, but probably something like the binomial formula would give (a power of a binomial).

I meant the general term is not factorable.