Prime clasfication by some constructive function












3














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.










share|cite|improve this question
























  • @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
















3














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.










share|cite|improve this question
























  • @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














3












3








3


0





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.










share|cite|improve this question















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






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share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 21 '18 at 12:02









amWhy

192k28224439




192k28224439










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


















  • @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










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