Prime clasfication by some constructive function
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How to prove or justify the following:
$$ f(g)= frac1{1-g^2} prod_{k=1}^{infty} left(frac{sin(pi frac gk)}{pi frac gk} cdot frac 1{1-frac{g^2}{k^2}}right).
$$
The above statement can illustrate the following facts:
(1) if $f(g) = 0$, then $g$ is composite number
(2) If $f(g)$ is not equal to $0$, the $g$ is prime number
(3) if $f(1+g) = 0$, then $g$ is prime.
Please justify.
number-theory prime-numbers conjectures
edited Nov 21 at 12:02
amWhy
191k27223438
asked Nov 16 '12 at 10:26
vmrfdu123456
1427
|
show 2 more comments
up vote
3
down vote
favorite
How to prove or justify the following:
$$ f(g)= frac1{1-g^2} prod_{k=1}^{infty} left(frac{sin(pi frac gk)}{pi frac gk} cdot frac 1{1-frac{g^2}{k^2}}right).
$$
The above statement can illustrate the following facts:
(1) if $f(g) = 0$, then $g$ is composite number
(2) If $f(g)$ is not equal to $0$, the $g$ is prime number
(3) if $f(1+g) = 0$, then $g$ is prime.
Please justify.
number-theory prime-numbers conjectures
edited Nov 21 at 12:02
amWhy
191k27223438
asked Nov 16 '12 at 10:26
vmrfdu123456
1427
@martini! Thank you so much for your editing work.
– vmrfdu123456
Nov 16 '12 at 10:32
2
Note that $1/(1-g^2)$ is redundant
– Norbert
Nov 16 '12 at 10:34
1
I am confused. What is $f$? It is given by the formula, then the formula is true by definition. And if not, how is it defined?
– Harald Hanche-Olsen
Nov 16 '12 at 10:40
f is given by the formula. of course, formula is true by definition.
– vmrfdu123456
Nov 16 '12 at 10:54
2
For me it seems that in this form, (3) contradicts (1) and (2).
– Berci
Nov 16 '12 at 11:02
|
show 2 more comments
up vote
3
down vote
favorite
up vote
3
down vote
favorite
How to prove or justify the following:
$$ f(g)= frac1{1-g^2} prod_{k=1}^{infty} left(frac{sin(pi frac gk)}{pi frac gk} cdot frac 1{1-frac{g^2}{k^2}}right).
$$
The above statement can illustrate the following facts:
(1) if $f(g) = 0$, then $g$ is composite number
(2) If $f(g)$ is not equal to $0$, the $g$ is prime number
(3) if $f(1+g) = 0$, then $g$ is prime.
Please justify.
number-theory prime-numbers conjectures
edited Nov 21 at 12:02
amWhy
191k27223438
asked Nov 16 '12 at 10:26
vmrfdu123456
1427
How to prove or justify the following:
$$ f(g)= frac1{1-g^2} prod_{k=1}^{infty} left(frac{sin(pi frac gk)}{pi frac gk} cdot frac 1{1-frac{g^2}{k^2}}right).
$$
The above statement can illustrate the following facts:
(1) if $f(g) = 0$, then $g$ is composite number
(2) If $f(g)$ is not equal to $0$, the $g$ is prime number
(3) if $f(1+g) = 0$, then $g$ is prime.
Please justify.
number-theory prime-numbers conjectures
number-theory prime-numbers conjectures
edited Nov 21 at 12:02
amWhy
191k27223438
asked Nov 16 '12 at 10:26
vmrfdu123456
1427
edited Nov 21 at 12:02
amWhy
191k27223438
asked Nov 16 '12 at 10:26
vmrfdu123456
1427
edited Nov 21 at 12:02
amWhy
191k27223438
edited Nov 21 at 12:02
amWhy
191k27223438
edited Nov 21 at 12:02
amWhy
191k27223438
191k27223438
asked Nov 16 '12 at 10:26
vmrfdu123456
1427
asked Nov 16 '12 at 10:26
vmrfdu123456
1427
asked Nov 16 '12 at 10:26
vmrfdu123456
1427
1427
@martini! Thank you so much for your editing work.
– vmrfdu123456
Nov 16 '12 at 10:32
2
Note that $1/(1-g^2)$ is redundant
– Norbert
Nov 16 '12 at 10:34
1
I am confused. What is $f$? It is given by the formula, then the formula is true by definition. And if not, how is it defined?
– Harald Hanche-Olsen
Nov 16 '12 at 10:40
f is given by the formula. of course, formula is true by definition.
– vmrfdu123456
Nov 16 '12 at 10:54
2
For me it seems that in this form, (3) contradicts (1) and (2).
– Berci
Nov 16 '12 at 11:02
|
show 2 more comments
@martini! Thank you so much for your editing work.
– vmrfdu123456
Nov 16 '12 at 10:32
2
Note that $1/(1-g^2)$ is redundant
– Norbert
Nov 16 '12 at 10:34
1
I am confused. What is $f$? It is given by the formula, then the formula is true by definition. And if not, how is it defined?
– Harald Hanche-Olsen
Nov 16 '12 at 10:40
f is given by the formula. of course, formula is true by definition.
– vmrfdu123456
Nov 16 '12 at 10:54
2
For me it seems that in this form, (3) contradicts (1) and (2).
– Berci
Nov 16 '12 at 11:02
@martini! Thank you so much for your editing work.
– vmrfdu123456
Nov 16 '12 at 10:32
@martini! Thank you so much for your editing work.
– vmrfdu123456
Nov 16 '12 at 10:32
2
2
Note that $1/(1-g^2)$ is redundant
– Norbert
Nov 16 '12 at 10:34
Note that $1/(1-g^2)$ is redundant
– Norbert
Nov 16 '12 at 10:34
1
1
I am confused. What is $f$? It is given by the formula, then the formula is true by definition. And if not, how is it defined?
– Harald Hanche-Olsen
Nov 16 '12 at 10:40
I am confused. What is $f$? It is given by the formula, then the formula is true by definition. And if not, how is it defined?
– Harald Hanche-Olsen
Nov 16 '12 at 10:40
f is given by the formula. of course, formula is true by definition.
– vmrfdu123456
Nov 16 '12 at 10:54
f is given by the formula. of course, formula is true by definition.
– vmrfdu123456
Nov 16 '12 at 10:54
2
2
For me it seems that in this form, (3) contradicts (1) and (2).
– Berci
Nov 16 '12 at 11:02
For me it seems that in this form, (3) contradicts (1) and (2).
– Berci
Nov 16 '12 at 11:02
|
show 2 more comments
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@martini! Thank you so much for your editing work.
– vmrfdu123456
Nov 16 '12 at 10:32
2
Note that $1/(1-g^2)$ is redundant
– Norbert
Nov 16 '12 at 10:34
1
I am confused. What is $f$? It is given by the formula, then the formula is true by definition. And if not, how is it defined?
– Harald Hanche-Olsen
Nov 16 '12 at 10:40
f is given by the formula. of course, formula is true by definition.
– vmrfdu123456
Nov 16 '12 at 10:54
2
For me it seems that in this form, (3) contradicts (1) and (2).
– Berci
Nov 16 '12 at 11:02