Mixing
Divergent products.
$begingroup$
My question are about divergent products.
I'm a Dutch student so i may lack the skil to write it down in the correct notation and forgive my spelling errors.
A thing i've found on the internet was that $$frac{sqrt{2 pi}}{Gamma(n)} = prod_{c=0}^{infty} (c+n) $$
according to my own numerical foundings should be true but i've been unable to find these results anywhere nor the logic so i'm hoping someone could point out to me where to look. I rewrite the product as sums and calculate them the "normal" way for diveregent sums.
only for positive n
(1)$$n!= (n/e)^n sqrt{2npi} prod_{c=1}^{infty}(1-c/n)$$
(2)$$n!prod_{c=1}^{infty}(n+c)=(n/e)^n sqrt{2pi} $$
And if i continue with my logic i get:
(3)$$prod_{c=1}^{infty}(1-c/n) prod_{c=1}^{infty}(1+c/n)=1$$
(4)$$prod_{c=1}^{infty}( 1+frac{-c^2+c-n-1}{(n+1)^2})=frac{e}{((n+1)/n)^n}$$
Is there someone who can explain of point out if i'm wrong or right and where i make the mistake.
So to sum it up, I simpely split the product into sums and calculate it that way. I've calculate the first/second statement numericaly but I've been unable to calculate the 4th yet i was able to check the points as n tends to infinity and 0. Yet i wonder if this is correct and what is the difference with the form at the beginning.
I'm awere that this first formula can be writing more detailed if you also wish to make it fit on the negative side, but i don't know exactly how, I would be eager to know.
Edit:
Because it raised some questions here i analyse the first product
$$prod_{c=1}^{infty}(1-c/n)=$$
$$1+sum_{c=1}^{infty}-c/n + $$
$$1/2!((sum_{c=1}^{infty}-c/n)^2-sum_{c=1}^{infty}(-c/n)^2)+$$
$$1/3!((sum_{c=1}^{infty}-c/n)^3-3*(sum_{c=1}^{infty}-c/n)*sum_{c=1}^{infty}(-c/n)^2+2*sum_{c=1}^{infty}(-c/n)^3)+ ...$$
The pattern i've posted before but it's the refined stirling numbers etc. just how one would look to a convergent product and or write a factorial as polynomal.
But in this case =
$$prod_{c=1}^{infty}(1-c/n)=1-1/(-12n)+1/(2*(-12n)^2)-1/6*(1/(-12n)^3+2*1/(120*n^3))+...$$
To calculate these negative values of the zeta function i've my own way, but again i'm terrible at the notation and i'm sure there are errors in it, but the idea is the following:
For every d>1, most easy is d=2, I take the constant part of the left handed side, and i try to isolate the sum i'm looking for in the right handed part.
$$sum_{n=1}^{dp} f(n)sum_{k=1}^{d-1} (e^{frac{2ipi k}{d}})^{n}=sum^p_{n=1} (d f(nd)-f(n))$$
But please ignore this, cause it's so wrong formulated a whole story on it's own, and not the topic now, cause i can find similair results on the web which conform my results, it's only for me a way to find the zeta values which are essential here.
gamma-function divergent-series infinite-product
asked Jan 30 '15 at 1:19
GerbenGerben
813
$endgroup$
|
show 8 more comments
$begingroup$
My question are about divergent products.
I'm a Dutch student so i may lack the skil to write it down in the correct notation and forgive my spelling errors.
A thing i've found on the internet was that $$frac{sqrt{2 pi}}{Gamma(n)} = prod_{c=0}^{infty} (c+n) $$
according to my own numerical foundings should be true but i've been unable to find these results anywhere nor the logic so i'm hoping someone could point out to me where to look. I rewrite the product as sums and calculate them the "normal" way for diveregent sums.
only for positive n
(1)$$n!= (n/e)^n sqrt{2npi} prod_{c=1}^{infty}(1-c/n)$$
(2)$$n!prod_{c=1}^{infty}(n+c)=(n/e)^n sqrt{2pi} $$
And if i continue with my logic i get:
(3)$$prod_{c=1}^{infty}(1-c/n) prod_{c=1}^{infty}(1+c/n)=1$$
(4)$$prod_{c=1}^{infty}( 1+frac{-c^2+c-n-1}{(n+1)^2})=frac{e}{((n+1)/n)^n}$$
Is there someone who can explain of point out if i'm wrong or right and where i make the mistake.
So to sum it up, I simpely split the product into sums and calculate it that way. I've calculate the first/second statement numericaly but I've been unable to calculate the 4th yet i was able to check the points as n tends to infinity and 0. Yet i wonder if this is correct and what is the difference with the form at the beginning.
I'm awere that this first formula can be writing more detailed if you also wish to make it fit on the negative side, but i don't know exactly how, I would be eager to know.
Edit:
Because it raised some questions here i analyse the first product
$$prod_{c=1}^{infty}(1-c/n)=$$
$$1+sum_{c=1}^{infty}-c/n + $$
$$1/2!((sum_{c=1}^{infty}-c/n)^2-sum_{c=1}^{infty}(-c/n)^2)+$$
$$1/3!((sum_{c=1}^{infty}-c/n)^3-3*(sum_{c=1}^{infty}-c/n)*sum_{c=1}^{infty}(-c/n)^2+2*sum_{c=1}^{infty}(-c/n)^3)+ ...$$
The pattern i've posted before but it's the refined stirling numbers etc. just how one would look to a convergent product and or write a factorial as polynomal.
But in this case =
$$prod_{c=1}^{infty}(1-c/n)=1-1/(-12n)+1/(2*(-12n)^2)-1/6*(1/(-12n)^3+2*1/(120*n^3))+...$$
To calculate these negative values of the zeta function i've my own way, but again i'm terrible at the notation and i'm sure there are errors in it, but the idea is the following:
For every d>1, most easy is d=2, I take the constant part of the left handed side, and i try to isolate the sum i'm looking for in the right handed part.
$$sum_{n=1}^{dp} f(n)sum_{k=1}^{d-1} (e^{frac{2ipi k}{d}})^{n}=sum^p_{n=1} (d f(nd)-f(n))$$
But please ignore this, cause it's so wrong formulated a whole story on it's own, and not the topic now, cause i can find similair results on the web which conform my results, it's only for me a way to find the zeta values which are essential here.
gamma-function divergent-series infinite-product
asked Jan 30 '15 at 1:19
GerbenGerben
813
$endgroup$
4
$begingroup$
My guess: All of your formulas are nonsense.
$endgroup$
– GEdgar
Jan 30 '15 at 1:36
2
$begingroup$
I'm pretty sure none of my formulas are nonsense, actualy i find this insulting the way you state it. What i do is the same done to find divergent series. I found these "myself" and the first formula i entered at wolfram product calcualator and in some documents at google. The prouduct formulas i gave are founded when i wanted to calculate a product, i splited them into a lot of divergent sums. Maybe i should had have added this one $$prod_{n=1}^{infty} (1+c)=sum_{n=0}^{infty} (2n)!/(n!)^2*(c/4)^n*(-1)^n= 1/(1+c)^{1/2}$$
$endgroup$
– Gerben
Jan 30 '15 at 1:39
2
$begingroup$
@GEdgar : lets be nice :)
$endgroup$
– Arjang
Jan 30 '15 at 1:42
1
$begingroup$
The way i did split the product up in sums, see my post at math.stackexchange.com/questions/460579/… Also here's another topic where someone did the trivial product above math.stackexchange.com/questions/871254/… But i've no clue how he did it and it looks so much more confusing to me.
$endgroup$
– Gerben
Jan 30 '15 at 1:58
1
$begingroup$
One way to treat convergence of infinite products (the limit of the truncations to finite products, often called partial products) is by converting them to infinite sums (series) by taking logarithms. In any case I think it evident that your first infinite product diverges, assuming your index $c$ steps through the nonnegative integers, unless $n$ happens to be a nonpositive integer (in which case the partial products are eventually zero).
$endgroup$
– hardmath
Jan 30 '15 at 3:16
|
show 8 more comments
$begingroup$
My question are about divergent products.
I'm a Dutch student so i may lack the skil to write it down in the correct notation and forgive my spelling errors.
A thing i've found on the internet was that $$frac{sqrt{2 pi}}{Gamma(n)} = prod_{c=0}^{infty} (c+n) $$
according to my own numerical foundings should be true but i've been unable to find these results anywhere nor the logic so i'm hoping someone could point out to me where to look. I rewrite the product as sums and calculate them the "normal" way for diveregent sums.
only for positive n
(1)$$n!= (n/e)^n sqrt{2npi} prod_{c=1}^{infty}(1-c/n)$$
(2)$$n!prod_{c=1}^{infty}(n+c)=(n/e)^n sqrt{2pi} $$
And if i continue with my logic i get:
(3)$$prod_{c=1}^{infty}(1-c/n) prod_{c=1}^{infty}(1+c/n)=1$$
(4)$$prod_{c=1}^{infty}( 1+frac{-c^2+c-n-1}{(n+1)^2})=frac{e}{((n+1)/n)^n}$$
Is there someone who can explain of point out if i'm wrong or right and where i make the mistake.
So to sum it up, I simpely split the product into sums and calculate it that way. I've calculate the first/second statement numericaly but I've been unable to calculate the 4th yet i was able to check the points as n tends to infinity and 0. Yet i wonder if this is correct and what is the difference with the form at the beginning.
I'm awere that this first formula can be writing more detailed if you also wish to make it fit on the negative side, but i don't know exactly how, I would be eager to know.
Edit:
Because it raised some questions here i analyse the first product
$$prod_{c=1}^{infty}(1-c/n)=$$
$$1+sum_{c=1}^{infty}-c/n + $$
$$1/2!((sum_{c=1}^{infty}-c/n)^2-sum_{c=1}^{infty}(-c/n)^2)+$$
$$1/3!((sum_{c=1}^{infty}-c/n)^3-3*(sum_{c=1}^{infty}-c/n)*sum_{c=1}^{infty}(-c/n)^2+2*sum_{c=1}^{infty}(-c/n)^3)+ ...$$
The pattern i've posted before but it's the refined stirling numbers etc. just how one would look to a convergent product and or write a factorial as polynomal.
But in this case =
$$prod_{c=1}^{infty}(1-c/n)=1-1/(-12n)+1/(2*(-12n)^2)-1/6*(1/(-12n)^3+2*1/(120*n^3))+...$$
To calculate these negative values of the zeta function i've my own way, but again i'm terrible at the notation and i'm sure there are errors in it, but the idea is the following:
For every d>1, most easy is d=2, I take the constant part of the left handed side, and i try to isolate the sum i'm looking for in the right handed part.
$$sum_{n=1}^{dp} f(n)sum_{k=1}^{d-1} (e^{frac{2ipi k}{d}})^{n}=sum^p_{n=1} (d f(nd)-f(n))$$
But please ignore this, cause it's so wrong formulated a whole story on it's own, and not the topic now, cause i can find similair results on the web which conform my results, it's only for me a way to find the zeta values which are essential here.
gamma-function divergent-series infinite-product
asked Jan 30 '15 at 1:19
GerbenGerben
813
$endgroup$
My question are about divergent products.
I'm a Dutch student so i may lack the skil to write it down in the correct notation and forgive my spelling errors.
A thing i've found on the internet was that $$frac{sqrt{2 pi}}{Gamma(n)} = prod_{c=0}^{infty} (c+n) $$
according to my own numerical foundings should be true but i've been unable to find these results anywhere nor the logic so i'm hoping someone could point out to me where to look. I rewrite the product as sums and calculate them the "normal" way for diveregent sums.
only for positive n
(1)$$n!= (n/e)^n sqrt{2npi} prod_{c=1}^{infty}(1-c/n)$$
(2)$$n!prod_{c=1}^{infty}(n+c)=(n/e)^n sqrt{2pi} $$
And if i continue with my logic i get:
(3)$$prod_{c=1}^{infty}(1-c/n) prod_{c=1}^{infty}(1+c/n)=1$$
(4)$$prod_{c=1}^{infty}( 1+frac{-c^2+c-n-1}{(n+1)^2})=frac{e}{((n+1)/n)^n}$$
Is there someone who can explain of point out if i'm wrong or right and where i make the mistake.
So to sum it up, I simpely split the product into sums and calculate it that way. I've calculate the first/second statement numericaly but I've been unable to calculate the 4th yet i was able to check the points as n tends to infinity and 0. Yet i wonder if this is correct and what is the difference with the form at the beginning.
I'm awere that this first formula can be writing more detailed if you also wish to make it fit on the negative side, but i don't know exactly how, I would be eager to know.
Edit:
Because it raised some questions here i analyse the first product
$$prod_{c=1}^{infty}(1-c/n)=$$
$$1+sum_{c=1}^{infty}-c/n + $$
$$1/2!((sum_{c=1}^{infty}-c/n)^2-sum_{c=1}^{infty}(-c/n)^2)+$$
$$1/3!((sum_{c=1}^{infty}-c/n)^3-3*(sum_{c=1}^{infty}-c/n)*sum_{c=1}^{infty}(-c/n)^2+2*sum_{c=1}^{infty}(-c/n)^3)+ ...$$
The pattern i've posted before but it's the refined stirling numbers etc. just how one would look to a convergent product and or write a factorial as polynomal.
But in this case =
$$prod_{c=1}^{infty}(1-c/n)=1-1/(-12n)+1/(2*(-12n)^2)-1/6*(1/(-12n)^3+2*1/(120*n^3))+...$$
To calculate these negative values of the zeta function i've my own way, but again i'm terrible at the notation and i'm sure there are errors in it, but the idea is the following:
For every d>1, most easy is d=2, I take the constant part of the left handed side, and i try to isolate the sum i'm looking for in the right handed part.
$$sum_{n=1}^{dp} f(n)sum_{k=1}^{d-1} (e^{frac{2ipi k}{d}})^{n}=sum^p_{n=1} (d f(nd)-f(n))$$
But please ignore this, cause it's so wrong formulated a whole story on it's own, and not the topic now, cause i can find similair results on the web which conform my results, it's only for me a way to find the zeta values which are essential here.
gamma-function divergent-series infinite-product
gamma-function divergent-series infinite-product
asked Jan 30 '15 at 1:19
GerbenGerben
813
asked Jan 30 '15 at 1:19
GerbenGerben
813
edited Jan 30 '15 at 15:51
Gerben
asked Jan 30 '15 at 1:19
GerbenGerben
813
asked Jan 30 '15 at 1:19
GerbenGerben
813
asked Jan 30 '15 at 1:19
GerbenGerben
813
813
4
$begingroup$
My guess: All of your formulas are nonsense.
$endgroup$
– GEdgar
Jan 30 '15 at 1:36
2
$begingroup$
I'm pretty sure none of my formulas are nonsense, actualy i find this insulting the way you state it. What i do is the same done to find divergent series. I found these "myself" and the first formula i entered at wolfram product calcualator and in some documents at google. The prouduct formulas i gave are founded when i wanted to calculate a product, i splited them into a lot of divergent sums. Maybe i should had have added this one $$prod_{n=1}^{infty} (1+c)=sum_{n=0}^{infty} (2n)!/(n!)^2*(c/4)^n*(-1)^n= 1/(1+c)^{1/2}$$
$endgroup$
– Gerben
Jan 30 '15 at 1:39
2
$begingroup$
@GEdgar : lets be nice :)
$endgroup$
– Arjang
Jan 30 '15 at 1:42
1
$begingroup$
The way i did split the product up in sums, see my post at math.stackexchange.com/questions/460579/… Also here's another topic where someone did the trivial product above math.stackexchange.com/questions/871254/… But i've no clue how he did it and it looks so much more confusing to me.
$endgroup$
– Gerben
Jan 30 '15 at 1:58
1
$begingroup$
One way to treat convergence of infinite products (the limit of the truncations to finite products, often called partial products) is by converting them to infinite sums (series) by taking logarithms. In any case I think it evident that your first infinite product diverges, assuming your index $c$ steps through the nonnegative integers, unless $n$ happens to be a nonpositive integer (in which case the partial products are eventually zero).
$endgroup$
– hardmath
Jan 30 '15 at 3:16
|
show 8 more comments
4
$begingroup$
My guess: All of your formulas are nonsense.
$endgroup$
– GEdgar
Jan 30 '15 at 1:36
2
$begingroup$
I'm pretty sure none of my formulas are nonsense, actualy i find this insulting the way you state it. What i do is the same done to find divergent series. I found these "myself" and the first formula i entered at wolfram product calcualator and in some documents at google. The prouduct formulas i gave are founded when i wanted to calculate a product, i splited them into a lot of divergent sums. Maybe i should had have added this one $$prod_{n=1}^{infty} (1+c)=sum_{n=0}^{infty} (2n)!/(n!)^2*(c/4)^n*(-1)^n= 1/(1+c)^{1/2}$$
$endgroup$
– Gerben
Jan 30 '15 at 1:39
2
$begingroup$
@GEdgar : lets be nice :)
$endgroup$
– Arjang
Jan 30 '15 at 1:42
1
$begingroup$
The way i did split the product up in sums, see my post at math.stackexchange.com/questions/460579/… Also here's another topic where someone did the trivial product above math.stackexchange.com/questions/871254/… But i've no clue how he did it and it looks so much more confusing to me.
$endgroup$
– Gerben
Jan 30 '15 at 1:58
1
$begingroup$
One way to treat convergence of infinite products (the limit of the truncations to finite products, often called partial products) is by converting them to infinite sums (series) by taking logarithms. In any case I think it evident that your first infinite product diverges, assuming your index $c$ steps through the nonnegative integers, unless $n$ happens to be a nonpositive integer (in which case the partial products are eventually zero).
$endgroup$
– hardmath
Jan 30 '15 at 3:16
4
4
$begingroup$
My guess: All of your formulas are nonsense.
$endgroup$
– GEdgar
Jan 30 '15 at 1:36
$begingroup$
My guess: All of your formulas are nonsense.
$endgroup$
– GEdgar
Jan 30 '15 at 1:36
2
2
$begingroup$
I'm pretty sure none of my formulas are nonsense, actualy i find this insulting the way you state it. What i do is the same done to find divergent series. I found these "myself" and the first formula i entered at wolfram product calcualator and in some documents at google. The prouduct formulas i gave are founded when i wanted to calculate a product, i splited them into a lot of divergent sums. Maybe i should had have added this one $$prod_{n=1}^{infty} (1+c)=sum_{n=0}^{infty} (2n)!/(n!)^2*(c/4)^n*(-1)^n= 1/(1+c)^{1/2}$$
$endgroup$
– Gerben
Jan 30 '15 at 1:39
$begingroup$
I'm pretty sure none of my formulas are nonsense, actualy i find this insulting the way you state it. What i do is the same done to find divergent series. I found these "myself" and the first formula i entered at wolfram product calcualator and in some documents at google. The prouduct formulas i gave are founded when i wanted to calculate a product, i splited them into a lot of divergent sums. Maybe i should had have added this one $$prod_{n=1}^{infty} (1+c)=sum_{n=0}^{infty} (2n)!/(n!)^2*(c/4)^n*(-1)^n= 1/(1+c)^{1/2}$$
$endgroup$
– Gerben
Jan 30 '15 at 1:39
2
2
$begingroup$
@GEdgar : lets be nice :)
$endgroup$
– Arjang
Jan 30 '15 at 1:42
$begingroup$
@GEdgar : lets be nice :)
$endgroup$
– Arjang
Jan 30 '15 at 1:42
1
1
$begingroup$
The way i did split the product up in sums, see my post at math.stackexchange.com/questions/460579/… Also here's another topic where someone did the trivial product above math.stackexchange.com/questions/871254/… But i've no clue how he did it and it looks so much more confusing to me.
$endgroup$
– Gerben
Jan 30 '15 at 1:58
$begingroup$
The way i did split the product up in sums, see my post at math.stackexchange.com/questions/460579/… Also here's another topic where someone did the trivial product above math.stackexchange.com/questions/871254/… But i've no clue how he did it and it looks so much more confusing to me.
$endgroup$
– Gerben
Jan 30 '15 at 1:58
1
1
$begingroup$
One way to treat convergence of infinite products (the limit of the truncations to finite products, often called partial products) is by converting them to infinite sums (series) by taking logarithms. In any case I think it evident that your first infinite product diverges, assuming your index $c$ steps through the nonnegative integers, unless $n$ happens to be a nonpositive integer (in which case the partial products are eventually zero).
$endgroup$
– hardmath
Jan 30 '15 at 3:16
$begingroup$
One way to treat convergence of infinite products (the limit of the truncations to finite products, often called partial products) is by converting them to infinite sums (series) by taking logarithms. In any case I think it evident that your first infinite product diverges, assuming your index $c$ steps through the nonnegative integers, unless $n$ happens to be a nonpositive integer (in which case the partial products are eventually zero).
$endgroup$
– hardmath
Jan 30 '15 at 3:16
|
show 8 more comments
2 Answers
2
active
oldest
votes
$begingroup$
In the equation $frac{sqrt{2 pi}}{Gamma(n)} = prod_{c=0}^{infty} (c+n)$, the right-hand side may be interpreted as a zeta-regularized product. Then by definition, we have $$prod_{c=0}^{infty} (c+n)=exp(-Z'(0))$$ where $Z(s)$ is the function defined by $$Z(s)=sum_{c=0}^infty(c+n)^{-s}$$
for $s$ with large enough real part, and by analytic continuation as necessary. The rest of the computation is given in Example 3 on p.220 of J. R. Quine, S. H. Heydari and R. Y. Song (1993), Zeta regularized products.
answered Jan 30 '15 at 4:00
Chris CulterChris Culter
21.3k43887
$endgroup$
1
$begingroup$
But where is the (e/n)^n element lost? I can write it down all the way but I wasn't used to the gamma function notation until recently, and i prefer to start my product at 1 instead of 0 and i'm not sure if i can chance it that easily. My grip is at the product I computed above since i can actualy check if it's correct. I divide a product of 1/n out there since it should be equal to the root of n cause it can be seen as constant. i'm very happy for your help.
$endgroup$
– Gerben
Jan 30 '15 at 4:05
1
$begingroup$
Ah, sorry, I'm afraid I didn't follow that part of the question. All I meant to do was to address the very first part with this answer.
$endgroup$
– Chris Culter
Jan 30 '15 at 4:25
1
$begingroup$
My opinions ... (1) "$frac{sqrt{2 pi}}{Gamma(n)} = prod_{c=0}^{infty} (c+n)$" is simply wrong. (2) "$frac{sqrt{2 pi}}{Gamma(n)}$ is the zeta-regularized value of $prod_{c=0}^{infty} (c+n)$" may be correct. Better would be (3) "$frac{sqrt{2 pi}}{Gamma(n)}$ is a zeta-regularized value of $prod_{c=0}^{infty} (c+n)$" since a given numerical product may be zeta regularized to two or more different values.
$endgroup$
– GEdgar
Jan 30 '15 at 13:47
1
$begingroup$
To be honest i don't doubt the given product as zeta regularized value, although i do not fully understand it. However there's a difference between the zeta-regularized product and my own product. I use a faculty and start at 1 maybe it's in there. I don't belief that one product could nor should give multiple values, so i just discard that one. A more realistic option is that i'm wrong, yet why is it so efficient to calculate these products (and intunitive) and does it actualy work? Where's the mistake.
$endgroup$
– Gerben
Jan 30 '15 at 16:02
add a comment |
$begingroup$
An old maybe outdated answer
$$
(n/e)^n*sqrt{(2npi)}/n!=$$
$$prod_{c=1}^{infty}(1+c/n)=$$
$$exp{(sum_{c=1}^{infty}frac{zeta(-c)}{c*n^c})} $$
answered Feb 9 '15 at 12:27
GerbenGerben
13413
$endgroup$
$begingroup$
Gerben: It seem that you have two accounts (one registered and one unregistered). If you have access to both of them, you should be able to merge them following the instructions here. If you do not have access to the other account, you will probably need some assistance from moderators or SE team to merge those two accounts.
$endgroup$
– Martin Sleziak
Feb 12 '15 at 10:47
$begingroup$
math.stackexchange.com/questions/39378/… seems to have another way of computing the product, atleast at value n=1.
$endgroup$
– Gerben
Jun 21 '15 at 16:20
add a comment |
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$begingroup$
In the equation $frac{sqrt{2 pi}}{Gamma(n)} = prod_{c=0}^{infty} (c+n)$, the right-hand side may be interpreted as a zeta-regularized product. Then by definition, we have $$prod_{c=0}^{infty} (c+n)=exp(-Z'(0))$$ where $Z(s)$ is the function defined by $$Z(s)=sum_{c=0}^infty(c+n)^{-s}$$
for $s$ with large enough real part, and by analytic continuation as necessary. The rest of the computation is given in Example 3 on p.220 of J. R. Quine, S. H. Heydari and R. Y. Song (1993), Zeta regularized products.
answered Jan 30 '15 at 4:00
Chris CulterChris Culter
21.3k43887
$endgroup$
1
$begingroup$
But where is the (e/n)^n element lost? I can write it down all the way but I wasn't used to the gamma function notation until recently, and i prefer to start my product at 1 instead of 0 and i'm not sure if i can chance it that easily. My grip is at the product I computed above since i can actualy check if it's correct. I divide a product of 1/n out there since it should be equal to the root of n cause it can be seen as constant. i'm very happy for your help.
$endgroup$
– Gerben
Jan 30 '15 at 4:05
1
$begingroup$
Ah, sorry, I'm afraid I didn't follow that part of the question. All I meant to do was to address the very first part with this answer.
$endgroup$
– Chris Culter
Jan 30 '15 at 4:25
1
$begingroup$
My opinions ... (1) "$frac{sqrt{2 pi}}{Gamma(n)} = prod_{c=0}^{infty} (c+n)$" is simply wrong. (2) "$frac{sqrt{2 pi}}{Gamma(n)}$ is the zeta-regularized value of $prod_{c=0}^{infty} (c+n)$" may be correct. Better would be (3) "$frac{sqrt{2 pi}}{Gamma(n)}$ is a zeta-regularized value of $prod_{c=0}^{infty} (c+n)$" since a given numerical product may be zeta regularized to two or more different values.
$endgroup$
– GEdgar
Jan 30 '15 at 13:47
1
$begingroup$
To be honest i don't doubt the given product as zeta regularized value, although i do not fully understand it. However there's a difference between the zeta-regularized product and my own product. I use a faculty and start at 1 maybe it's in there. I don't belief that one product could nor should give multiple values, so i just discard that one. A more realistic option is that i'm wrong, yet why is it so efficient to calculate these products (and intunitive) and does it actualy work? Where's the mistake.
$endgroup$
– Gerben
Jan 30 '15 at 16:02
add a comment |
$begingroup$
In the equation $frac{sqrt{2 pi}}{Gamma(n)} = prod_{c=0}^{infty} (c+n)$, the right-hand side may be interpreted as a zeta-regularized product. Then by definition, we have $$prod_{c=0}^{infty} (c+n)=exp(-Z'(0))$$ where $Z(s)$ is the function defined by $$Z(s)=sum_{c=0}^infty(c+n)^{-s}$$
for $s$ with large enough real part, and by analytic continuation as necessary. The rest of the computation is given in Example 3 on p.220 of J. R. Quine, S. H. Heydari and R. Y. Song (1993), Zeta regularized products.
answered Jan 30 '15 at 4:00
Chris CulterChris Culter
21.3k43887
$endgroup$
1
$begingroup$
But where is the (e/n)^n element lost? I can write it down all the way but I wasn't used to the gamma function notation until recently, and i prefer to start my product at 1 instead of 0 and i'm not sure if i can chance it that easily. My grip is at the product I computed above since i can actualy check if it's correct. I divide a product of 1/n out there since it should be equal to the root of n cause it can be seen as constant. i'm very happy for your help.
$endgroup$
– Gerben
Jan 30 '15 at 4:05
1
$begingroup$
Ah, sorry, I'm afraid I didn't follow that part of the question. All I meant to do was to address the very first part with this answer.
$endgroup$
– Chris Culter
Jan 30 '15 at 4:25
1
$begingroup$
My opinions ... (1) "$frac{sqrt{2 pi}}{Gamma(n)} = prod_{c=0}^{infty} (c+n)$" is simply wrong. (2) "$frac{sqrt{2 pi}}{Gamma(n)}$ is the zeta-regularized value of $prod_{c=0}^{infty} (c+n)$" may be correct. Better would be (3) "$frac{sqrt{2 pi}}{Gamma(n)}$ is a zeta-regularized value of $prod_{c=0}^{infty} (c+n)$" since a given numerical product may be zeta regularized to two or more different values.
$endgroup$
– GEdgar
Jan 30 '15 at 13:47
1
$begingroup$
To be honest i don't doubt the given product as zeta regularized value, although i do not fully understand it. However there's a difference between the zeta-regularized product and my own product. I use a faculty and start at 1 maybe it's in there. I don't belief that one product could nor should give multiple values, so i just discard that one. A more realistic option is that i'm wrong, yet why is it so efficient to calculate these products (and intunitive) and does it actualy work? Where's the mistake.
$endgroup$
– Gerben
Jan 30 '15 at 16:02
add a comment |
$begingroup$
In the equation $frac{sqrt{2 pi}}{Gamma(n)} = prod_{c=0}^{infty} (c+n)$, the right-hand side may be interpreted as a zeta-regularized product. Then by definition, we have $$prod_{c=0}^{infty} (c+n)=exp(-Z'(0))$$ where $Z(s)$ is the function defined by $$Z(s)=sum_{c=0}^infty(c+n)^{-s}$$
for $s$ with large enough real part, and by analytic continuation as necessary. The rest of the computation is given in Example 3 on p.220 of J. R. Quine, S. H. Heydari and R. Y. Song (1993), Zeta regularized products.
answered Jan 30 '15 at 4:00
Chris CulterChris Culter
21.3k43887
$endgroup$
In the equation $frac{sqrt{2 pi}}{Gamma(n)} = prod_{c=0}^{infty} (c+n)$, the right-hand side may be interpreted as a zeta-regularized product. Then by definition, we have $$prod_{c=0}^{infty} (c+n)=exp(-Z'(0))$$ where $Z(s)$ is the function defined by $$Z(s)=sum_{c=0}^infty(c+n)^{-s}$$
for $s$ with large enough real part, and by analytic continuation as necessary. The rest of the computation is given in Example 3 on p.220 of J. R. Quine, S. H. Heydari and R. Y. Song (1993), Zeta regularized products.
answered Jan 30 '15 at 4:00
Chris CulterChris Culter
21.3k43887
answered Jan 30 '15 at 4:00
Chris CulterChris Culter
21.3k43887
answered Jan 30 '15 at 4:00
Chris CulterChris Culter
21.3k43887
answered Jan 30 '15 at 4:00
Chris CulterChris Culter
21.3k43887
21.3k43887
1
$begingroup$
But where is the (e/n)^n element lost? I can write it down all the way but I wasn't used to the gamma function notation until recently, and i prefer to start my product at 1 instead of 0 and i'm not sure if i can chance it that easily. My grip is at the product I computed above since i can actualy check if it's correct. I divide a product of 1/n out there since it should be equal to the root of n cause it can be seen as constant. i'm very happy for your help.
$endgroup$
– Gerben
Jan 30 '15 at 4:05
1
$begingroup$
Ah, sorry, I'm afraid I didn't follow that part of the question. All I meant to do was to address the very first part with this answer.
$endgroup$
– Chris Culter
Jan 30 '15 at 4:25
1
$begingroup$
My opinions ... (1) "$frac{sqrt{2 pi}}{Gamma(n)} = prod_{c=0}^{infty} (c+n)$" is simply wrong. (2) "$frac{sqrt{2 pi}}{Gamma(n)}$ is the zeta-regularized value of $prod_{c=0}^{infty} (c+n)$" may be correct. Better would be (3) "$frac{sqrt{2 pi}}{Gamma(n)}$ is a zeta-regularized value of $prod_{c=0}^{infty} (c+n)$" since a given numerical product may be zeta regularized to two or more different values.
$endgroup$
– GEdgar
Jan 30 '15 at 13:47
1
$begingroup$
To be honest i don't doubt the given product as zeta regularized value, although i do not fully understand it. However there's a difference between the zeta-regularized product and my own product. I use a faculty and start at 1 maybe it's in there. I don't belief that one product could nor should give multiple values, so i just discard that one. A more realistic option is that i'm wrong, yet why is it so efficient to calculate these products (and intunitive) and does it actualy work? Where's the mistake.
$endgroup$
– Gerben
Jan 30 '15 at 16:02
add a comment |
1
$begingroup$
But where is the (e/n)^n element lost? I can write it down all the way but I wasn't used to the gamma function notation until recently, and i prefer to start my product at 1 instead of 0 and i'm not sure if i can chance it that easily. My grip is at the product I computed above since i can actualy check if it's correct. I divide a product of 1/n out there since it should be equal to the root of n cause it can be seen as constant. i'm very happy for your help.
$endgroup$
– Gerben
Jan 30 '15 at 4:05
1
$begingroup$
Ah, sorry, I'm afraid I didn't follow that part of the question. All I meant to do was to address the very first part with this answer.
$endgroup$
– Chris Culter
Jan 30 '15 at 4:25
1
$begingroup$
My opinions ... (1) "$frac{sqrt{2 pi}}{Gamma(n)} = prod_{c=0}^{infty} (c+n)$" is simply wrong. (2) "$frac{sqrt{2 pi}}{Gamma(n)}$ is the zeta-regularized value of $prod_{c=0}^{infty} (c+n)$" may be correct. Better would be (3) "$frac{sqrt{2 pi}}{Gamma(n)}$ is a zeta-regularized value of $prod_{c=0}^{infty} (c+n)$" since a given numerical product may be zeta regularized to two or more different values.
$endgroup$
– GEdgar
Jan 30 '15 at 13:47
1
$begingroup$
To be honest i don't doubt the given product as zeta regularized value, although i do not fully understand it. However there's a difference between the zeta-regularized product and my own product. I use a faculty and start at 1 maybe it's in there. I don't belief that one product could nor should give multiple values, so i just discard that one. A more realistic option is that i'm wrong, yet why is it so efficient to calculate these products (and intunitive) and does it actualy work? Where's the mistake.
$endgroup$
– Gerben
Jan 30 '15 at 16:02
1
1
$begingroup$
But where is the (e/n)^n element lost? I can write it down all the way but I wasn't used to the gamma function notation until recently, and i prefer to start my product at 1 instead of 0 and i'm not sure if i can chance it that easily. My grip is at the product I computed above since i can actualy check if it's correct. I divide a product of 1/n out there since it should be equal to the root of n cause it can be seen as constant. i'm very happy for your help.
$endgroup$
– Gerben
Jan 30 '15 at 4:05
$begingroup$
But where is the (e/n)^n element lost? I can write it down all the way but I wasn't used to the gamma function notation until recently, and i prefer to start my product at 1 instead of 0 and i'm not sure if i can chance it that easily. My grip is at the product I computed above since i can actualy check if it's correct. I divide a product of 1/n out there since it should be equal to the root of n cause it can be seen as constant. i'm very happy for your help.
$endgroup$
– Gerben
Jan 30 '15 at 4:05
1
1
$begingroup$
Ah, sorry, I'm afraid I didn't follow that part of the question. All I meant to do was to address the very first part with this answer.
$endgroup$
– Chris Culter
Jan 30 '15 at 4:25
$begingroup$
Ah, sorry, I'm afraid I didn't follow that part of the question. All I meant to do was to address the very first part with this answer.
$endgroup$
– Chris Culter
Jan 30 '15 at 4:25
1
1
$begingroup$
My opinions ... (1) "$frac{sqrt{2 pi}}{Gamma(n)} = prod_{c=0}^{infty} (c+n)$" is simply wrong. (2) "$frac{sqrt{2 pi}}{Gamma(n)}$ is the zeta-regularized value of $prod_{c=0}^{infty} (c+n)$" may be correct. Better would be (3) "$frac{sqrt{2 pi}}{Gamma(n)}$ is a zeta-regularized value of $prod_{c=0}^{infty} (c+n)$" since a given numerical product may be zeta regularized to two or more different values.
$endgroup$
– GEdgar
Jan 30 '15 at 13:47
$begingroup$
My opinions ... (1) "$frac{sqrt{2 pi}}{Gamma(n)} = prod_{c=0}^{infty} (c+n)$" is simply wrong. (2) "$frac{sqrt{2 pi}}{Gamma(n)}$ is the zeta-regularized value of $prod_{c=0}^{infty} (c+n)$" may be correct. Better would be (3) "$frac{sqrt{2 pi}}{Gamma(n)}$ is a zeta-regularized value of $prod_{c=0}^{infty} (c+n)$" since a given numerical product may be zeta regularized to two or more different values.
$endgroup$
– GEdgar
Jan 30 '15 at 13:47
1
1
$begingroup$
To be honest i don't doubt the given product as zeta regularized value, although i do not fully understand it. However there's a difference between the zeta-regularized product and my own product. I use a faculty and start at 1 maybe it's in there. I don't belief that one product could nor should give multiple values, so i just discard that one. A more realistic option is that i'm wrong, yet why is it so efficient to calculate these products (and intunitive) and does it actualy work? Where's the mistake.
$endgroup$
– Gerben
Jan 30 '15 at 16:02
$begingroup$
To be honest i don't doubt the given product as zeta regularized value, although i do not fully understand it. However there's a difference between the zeta-regularized product and my own product. I use a faculty and start at 1 maybe it's in there. I don't belief that one product could nor should give multiple values, so i just discard that one. A more realistic option is that i'm wrong, yet why is it so efficient to calculate these products (and intunitive) and does it actualy work? Where's the mistake.
$endgroup$
– Gerben
Jan 30 '15 at 16:02
add a comment |
$begingroup$
An old maybe outdated answer
$$
(n/e)^n*sqrt{(2npi)}/n!=$$
$$prod_{c=1}^{infty}(1+c/n)=$$
$$exp{(sum_{c=1}^{infty}frac{zeta(-c)}{c*n^c})} $$
answered Feb 9 '15 at 12:27
GerbenGerben
13413
$endgroup$
$begingroup$
Gerben: It seem that you have two accounts (one registered and one unregistered). If you have access to both of them, you should be able to merge them following the instructions here. If you do not have access to the other account, you will probably need some assistance from moderators or SE team to merge those two accounts.
$endgroup$
– Martin Sleziak
Feb 12 '15 at 10:47
$begingroup$
math.stackexchange.com/questions/39378/… seems to have another way of computing the product, atleast at value n=1.
$endgroup$
– Gerben
Jun 21 '15 at 16:20
add a comment |
$begingroup$
An old maybe outdated answer
$$
(n/e)^n*sqrt{(2npi)}/n!=$$
$$prod_{c=1}^{infty}(1+c/n)=$$
$$exp{(sum_{c=1}^{infty}frac{zeta(-c)}{c*n^c})} $$
answered Feb 9 '15 at 12:27
GerbenGerben
13413
$endgroup$
$begingroup$
Gerben: It seem that you have two accounts (one registered and one unregistered). If you have access to both of them, you should be able to merge them following the instructions here. If you do not have access to the other account, you will probably need some assistance from moderators or SE team to merge those two accounts.
$endgroup$
– Martin Sleziak
Feb 12 '15 at 10:47
$begingroup$
math.stackexchange.com/questions/39378/… seems to have another way of computing the product, atleast at value n=1.
$endgroup$
– Gerben
Jun 21 '15 at 16:20
add a comment |
$begingroup$
An old maybe outdated answer
$$
(n/e)^n*sqrt{(2npi)}/n!=$$
$$prod_{c=1}^{infty}(1+c/n)=$$
$$exp{(sum_{c=1}^{infty}frac{zeta(-c)}{c*n^c})} $$
answered Feb 9 '15 at 12:27
GerbenGerben
13413
$endgroup$
An old maybe outdated answer
$$
(n/e)^n*sqrt{(2npi)}/n!=$$
$$prod_{c=1}^{infty}(1+c/n)=$$
$$exp{(sum_{c=1}^{infty}frac{zeta(-c)}{c*n^c})} $$
answered Feb 9 '15 at 12:27
GerbenGerben
13413
edited Jan 14 at 14:39
answered Feb 9 '15 at 12:27
GerbenGerben
13413
answered Feb 9 '15 at 12:27
GerbenGerben
13413
answered Feb 9 '15 at 12:27
GerbenGerben
13413
13413
$begingroup$
Gerben: It seem that you have two accounts (one registered and one unregistered). If you have access to both of them, you should be able to merge them following the instructions here. If you do not have access to the other account, you will probably need some assistance from moderators or SE team to merge those two accounts.
$endgroup$
– Martin Sleziak
Feb 12 '15 at 10:47
$begingroup$
math.stackexchange.com/questions/39378/… seems to have another way of computing the product, atleast at value n=1.
$endgroup$
– Gerben
Jun 21 '15 at 16:20
add a comment |
$begingroup$
Gerben: It seem that you have two accounts (one registered and one unregistered). If you have access to both of them, you should be able to merge them following the instructions here. If you do not have access to the other account, you will probably need some assistance from moderators or SE team to merge those two accounts.
$endgroup$
– Martin Sleziak
Feb 12 '15 at 10:47
$begingroup$
math.stackexchange.com/questions/39378/… seems to have another way of computing the product, atleast at value n=1.
$endgroup$
– Gerben
Jun 21 '15 at 16:20
$begingroup$
Gerben: It seem that you have two accounts (one registered and one unregistered). If you have access to both of them, you should be able to merge them following the instructions here. If you do not have access to the other account, you will probably need some assistance from moderators or SE team to merge those two accounts.
$endgroup$
– Martin Sleziak
Feb 12 '15 at 10:47
$begingroup$
Gerben: It seem that you have two accounts (one registered and one unregistered). If you have access to both of them, you should be able to merge them following the instructions here. If you do not have access to the other account, you will probably need some assistance from moderators or SE team to merge those two accounts.
$endgroup$
– Martin Sleziak
Feb 12 '15 at 10:47
$begingroup$
math.stackexchange.com/questions/39378/… seems to have another way of computing the product, atleast at value n=1.
$endgroup$
– Gerben
Jun 21 '15 at 16:20
$begingroup$
math.stackexchange.com/questions/39378/… seems to have another way of computing the product, atleast at value n=1.
$endgroup$
– Gerben
Jun 21 '15 at 16:20
add a comment |
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$begingroup$
My guess: All of your formulas are nonsense.
$endgroup$
– GEdgar
Jan 30 '15 at 1:36
2
$begingroup$
I'm pretty sure none of my formulas are nonsense, actualy i find this insulting the way you state it. What i do is the same done to find divergent series. I found these "myself" and the first formula i entered at wolfram product calcualator and in some documents at google. The prouduct formulas i gave are founded when i wanted to calculate a product, i splited them into a lot of divergent sums. Maybe i should had have added this one $$prod_{n=1}^{infty} (1+c)=sum_{n=0}^{infty} (2n)!/(n!)^2*(c/4)^n*(-1)^n= 1/(1+c)^{1/2}$$
$endgroup$
– Gerben
Jan 30 '15 at 1:39
2
$begingroup$
@GEdgar : lets be nice :)
$endgroup$
– Arjang
Jan 30 '15 at 1:42
1
$begingroup$
The way i did split the product up in sums, see my post at math.stackexchange.com/questions/460579/… Also here's another topic where someone did the trivial product above math.stackexchange.com/questions/871254/… But i've no clue how he did it and it looks so much more confusing to me.
$endgroup$
– Gerben
Jan 30 '15 at 1:58
1
$begingroup$
One way to treat convergence of infinite products (the limit of the truncations to finite products, often called partial products) is by converting them to infinite sums (series) by taking logarithms. In any case I think it evident that your first infinite product diverges, assuming your index $c$ steps through the nonnegative integers, unless $n$ happens to be a nonpositive integer (in which case the partial products are eventually zero).
$endgroup$
– hardmath
Jan 30 '15 at 3:16