- 06 Nov '17 08:59 / 8 editsis there any way to simplify either

( ∑[k=c, p] (k – 1)! / ( k^m (k – c)! ) ) / ∑[k=c, p] ( (k – 1)! / k^(m-1) (k – c)! )

where

c, m, p ∈ ℕ,

1≤c≤m,

c≤p

or

( ∑[k=c, ∞] (k – 1)! / ( k^m (k – c)! ) ) / ∑[k=c, ∞] (k – 1)! / ( k^(m-1) (k – c)! )

where

c, m ∈ ℕ,

1≤c<m>2,

c ∈ [1, m–2],

m ∈ [3, ∞ )

?

I really hope there is as I use them often in my book I am writing + I find their evaluation is currently horribly computatively inefficient.

In each case the numerator and denominator are almost identical except m in the sum for the numerator is replaced with m-1 in the sum for the denominator. - 08 Nov '17 18:37

Not offhand, the ratio of factorials is k-1 permute c and if you were dividing by k! then it would be the binomial expansion coefficient. This might be tractable using some sort of generating function, I'll play around with it, but my intuition is it's not going to simplify.*Originally posted by @humy***is there any way to simplify either**

( ∑[k=c, p] (k – 1)! / ( k^m (k – c)! ) ) / ∑[k=c, p] ( (k – 1)! / k^(m-1) (k – c)! )

where

c, m, p ∈ ℕ,

1≤c≤m,

c≤p

or

( ∑[k=c, ∞] (k – 1)! / ( k^m (k – c)! ) ) / ∑[k=c, ∞] (k – 1)! / ( k^(m-1) (k – c)! )

where

c, m ∈ ℕ,

1≤c<m>2,

c ∈ [1, m–2],

m ∈ [3, ∞ )

?

I really hope there is as I use ...[text shortened]... tical except m in the sum for the numerator is replaced with m-1 in the sum for the denominator. - 08 Nov '17 18:47 / 7 edits

Fortunately, even if it doesn't simplify, I have worked out an efficient work around of evaluating the second one and I have just tested it today and it works albeit at the price of requiring an awful lot of computer memory needed to hold a huge 2D array of new maths constants that I call the "nac constants". It works around ~1000 times faster than using the simple but inefficient 'brute force' method. I will explain this efficient method in my book.*Originally posted by @deepthought***... but my intuition is it's not going to simplify.**

As for the first one; I found no reasonable method to improve its evaluation but concluded that isn't a problem because each sum is at least finite.

And I did find the first one at least find it simplifies for these special case;

( ∑[k=1, p] 1/p ) / p for m=1 (which implies we must also have c=1)

( ∑[k=1, p] 1/k^m ) / ∑[k=1, p] k^(1–m) for c=1 - 12 Nov '17 16:25 / 6 edits

my field is NOT specifically maths in particular (try AI) and never was and I am NOT a fully qualified mathematician and never will be.*Originally posted by @eladar***Here you go again asking for some smart person in your field to do your math for you.**

I do most of my work myself including any required maths but I don't have the arrogance to think no one is smart enough to help me when I just occasionally get stuck. What is wrong with just occasionally asking smart person for help (on something) when I get stuck and then do all the rest of my research work ( ~99% ) myself? If they contribute, like they sometimes actually do, I will cite due credit to their help when I publish my results in my book. I already have a list of people who I intend to cite credit in my book thanks to their contributions -what is wrong with that? - 12 Nov '17 16:51

So you are trying to do something beyond your abilities and need others to pull it off.*Originally posted by @humy***my field is NOT specifically maths in particular (try AI) and never was and I am NOT a fully qualified mathematician and never will be.**

I do most of my work myself including any required maths but I don't have the arrogance to think no one is smart enough to help me when I just occasionally get stuck. What is wrong with just occasionally asking smart person ...[text shortened]... who I intend to cite credit in my book thanks to their contributions -what is wrong with that?

Usually people have to pay money for consulting fees. - 12 Nov '17 19:27 / 3 edits

nope, but that doesn't mean nobody is allowed to help me on the rare occasion when I ask for help required.*Originally posted by @eladar***So you are trying to do something beyond your abilities**

Don't understand you problem with that. - 13 Dec '17 14:10 / 2 editsI have totally given up on simplifying those expressions but now got one more similar one which I hope can somehow simplify;

( ∑[k=c, ∞] (k + m – 1)! / ( k^2 (k + 1) (k + 2m – 1)! ) ) / ∑[k=c, ∞] (k + m – 1)! / ( k (k + 1) (k + 2m – 1)! )

where

c, k, m ∈ ℕ

1≤c≤k

1≤m

I have tried and failed to simplify it and my feeling is that, despite such an apparently small visual difference between the numerator and the denominator, it probably doesn't but still hope there's something I've missed here. - 15 Dec '17 10:59 / 1 edit

I have just made some new applied-math discoveries that renders that above formula almost obsolete so no longer quite so concerned with it as I was.*Originally posted by @humy***I have totally given up on simplifying those expressions but now got one more similar one which I hope can somehow simplify;**

( ∑[k=c, ∞] (k + m – 1)! / ( k^2 (k + 1) (k + 2m – 1)! ) ) / ∑[k=c, ∞] (k + m – 1)! / ( k (k + 1) (k + 2m – 1)! )

where

c, k, m ∈ ℕ

1≤c≤k

1≤m

I have tried and failed to simplify it and my feeling is that, despite such ...[text shortened]... ator and the denominator, it probably doesn't but still hope there's something I've missed here.