Hey /sci/, I'm working on this problem involving the binomial

>>8692478
>I know extending factorials into the negative numbers involves the Riemann Zeta Function, so I'm wondering if this formula has any relation to that?
extending factorial requires the gamma function, not the zeta function

and you can extend it everywhere except negative integers

>>8692484
Oh, so (-1)! is undefined? I wonder what the meaning of this formula is, and why it popped up from the binomial distribution.

>>8692521
yes its undefined

it's probably just from an algebra mistake you made (there's no negative factorials in the original sum)

>>8692526
>it's probably just from an algebra mistake you made (there's no negative factorials in the original sum)
But when substituting, a negative factorial appears. I don't think this is an algebra mistake, but take a look at my work:
[eqn] \sum _{y=0} ^m \Big[ \frac{m!}{y!(m-y)!} p^y (1-p)^{m-y} \Big] =1 [/eqn]
Plugging in y=x-1 and m=n-1:
[eqn] \sum _{x=1} ^{n-1} \Big[ \frac{(n-1)!}{(x-1)!(n-x)!} p^{x-1} (1-p)^{x-y} \Big] =1 [/eqn]
Multiplying both sides by np:
[eqn] \sum _{x=1} ^{n-1} \Big[ \frac{n!}{(x-1)!(n-x)!} p^x (1-p)^{x-y} \Big] =np [/eqn]
Making the limits of the sum match the form given in the OP:
[eqn] \sum _{x=0} ^{n-1} \Big[ \frac{n!}{(x-1)!(n-x)!} p^x (1-p)^{x-y} \Big] - \frac{n!}{(0-1)!(n-0)!} p^0 (1-p)^{n-0} =np [/eqn]
[eqn] \sum _{x=0} ^{n-1} \Big[ \frac{n!}{(x-1)!(n-x)!} p^x (1-p)^{x-y} \Big] - \frac{(1-p)^{n} }{(-1)!} =np [/eqn]
[eqn] \sum _{x=0} ^{n} \Big[ \frac{n!}{(x-1)!(n-x)!} p^x (1-p)^{x-y} \Big] - \frac{(1-p)^{n} }{(-1)!} - \frac{n!}{(n-1)!(n-n)!} p^n (1-p)^{n-n} =np [/eqn]
[eqn] \sum _{x=0} ^{n} \Big[ \frac{n!}{(x-1)!(n-x)!} p^x (1-p)^{x-y} \Big] - \frac{(1-p)^{n} }{(-1)!} - \frac{n}{0!} p^n =np [/eqn]
Simplifying further:
[eqn] np- \frac{(1-p)^{n} }{(-1)!} -np^n = np [/eqn]
Solving for (-1)!
[eqn] (-1)!=- \frac{(1-p)^n }{np^n } [/eqn]

>>8692649
but the negative factorial doesn't appear when you do a substitution. its appearing when you added and subtracted it to make the limits match, and those terms you added aren't defined so the result you get at the end won't make sense

>>8692677
also note that changing p and n changes your value of (-1)!

>>8692680
>>8692649
also you can just see how to do it here if you want
https://en.wikipedia.org/wiki/Binomial_distribution#Mean

you don't want to just substitute right away, you want to substitute when it makes sense to

>>8692677
>those terms you added aren't defined
But the only reason the terms are undefined is because (-1)! is undefined. If we define it to be [math] - \frac{(1-p)^n }{np^n } [/math] then it is consistent with the binomial theorem.
>so the result you get at the end won't make sense
I'm just not understanding why substituting at the wrong time would yield a different answer. If we define the negative factorial in this way then at least we are being consistent.

>>8692680
I know, that's why I found it so interesting. Do you know if this formula converges to anything as n approaches infinity? It sort of looks like the formula for e^x in that way.

>>8692690
>>8692708
Thanks!

>>8692744
> If we define it to be −(1−p)nnpn then it is consistent with the binomial theorem.
it's not consistent because it depends on n and p, and so your definition of (-1)! changes for every positive integer n and every probability p between 0 and 1, it doesn't lead to anything sensible
if p=0 it definitely doesn't make sense
if p=1/3 and n=1 you'd get (-1)!= -2
if p=1/3 and n=3 you'd get (-1)!=-2.66666...

>I'm just not understanding why substituting at the wrong time would yield a different answer.
you did the substitution fine but your starting point was wrong by multiplying the sum equal to 1 by np

you never actually relate the sum or np to the mean which involves the extra 'x' term in the sum

you cant add it and subtract it in the same way that you cant define 1/0 by adding and subtracting it and grouping one of them into a sum, the thing you're thrown in doesn't exist

>>8692796
(continued) basically by doing what you did you're saying 'if (-1)! exists then it should equal -(1-p)^n/np^n', but this is not well-defined because of the dependence on n and p

>>8692796
>you never actually relate the sum or np to the mean which involves the extra 'x' term in the sum
Shit, I didn't catch that either. Multiplying by x makes that first subtraction 0, which gets rid of the (-1)! term all together. Thanks.

>>8692861
But it also changes that second term to [math] n^2 p^n [/math] and changes the RHS to [math] \sum _{x=1} ^{n-1} xnp [/math]

Dividing both sides by np, we have [eqn] 1-np^{n-1} = \sum _{x=1} ^{n-1} x [/eqn]
Again we can adjust the bounds:
[eqn] \sum _{x=0} ^{n} x -0-n=1-np^{n-1} [/eqn]
Rearranging:
[eqn] \sum _{x=0} ^{n} x=1+n(1-p^{n-1} ) [/eqn] [eqn] - \frac{1}{12}= \lim_{n\to\infty} 1+n(1-p^{n-1} ) [/eqn] [eqn] - \frac{13}{12}= \lim_{n\to\infty} n(1-p^{n-1} ) [/eqn]
Since p<1 [eqn] - \frac{13}{12}= \lim_{n\to\infty} n(1-0) [/eqn]
[eqn] - \frac{13}{12}= \infty [/eqn]
It appears that if the sum of all integers is -1/12, then the classic "infinity" symbol would be -13/12

>>8692478
You made an error, the stuff I say below also reffers to you, because I believe that you made the same awfull things.

>>8692649
There are so many things wrong with that.
I seriously cant belive it.

It is no wonder that you get (-1)! if everything you do is wrong.

[math]\sum _{y=0} ^m \Big[ \frac{m!}{y!(m-y)!} p^y (1-p)^{m-y} \Big] =1[/math]
means
[math]\sum _{x=1} ^{n-1} \Big[ \frac{(n-1)!}{(x-1)!(n-x-1)!} p^{x-1} (1-p)^{n-1-x+1} \Big] =1[/math]
(there is probably still some errors in that)

I can not be bothered to check the rest. What I can tell you is that getting (-1)! means that you have done something wrong.

I dont even want to think about what you did in your 4th equation but I am absolutely certain that whatever you would bring many mathematicians to the verge of suicide.
(Notice that in 3rd equations you never have to calculate (-1)! but in the 4th you do).