Mixing
Gamma function in Taylor series
The formula I am having some problems is this one
$$
fleft( x right) =sin left(frac { 2(x-3)!-x+1 }{ 2x } pi right) =sin left(frac { 2,Gamma (x-2)-x+1 }{ 2x } pi right)
$$
Although this might seem too general it is a variation of Wilson's theorem which (in my opinion) has slightly better properties than the general Wilson's theorem formula.
If you look at the graph you'll see that the values for $f(x)$ where $x$ is prime is 0 while for $x$ nonprime tends to be converging toward $-1$ and in almost a perfect fashion and rapidly.
My idea is this: turn this into infinite Taylor series and add $x$ to it, that way we could get a good approximation of prime counting function to infinity.
In other words:
$$
lim _{ xrightarrow infty }{ x+sum _{ nepsilon N }^{ }{ f(n) } } =lim _{ xrightarrow infty }{ Pi (x) }
$$
The problem I am having is actually converting it into Taylor series. I really didn't have any experience with this kind of equations so I was wondering is there any trick to do this, or should I just let this one pass
Edited: Changed the formula, forgot to multiply by $pi$
prime-numbers
edited Apr 7 '13 at 17:42
vonbrand
19.8k63058
asked Apr 7 '13 at 17:20
user1398593
364
add a comment |
The formula I am having some problems is this one
$$
fleft( x right) =sin left(frac { 2(x-3)!-x+1 }{ 2x } pi right) =sin left(frac { 2,Gamma (x-2)-x+1 }{ 2x } pi right)
$$
Although this might seem too general it is a variation of Wilson's theorem which (in my opinion) has slightly better properties than the general Wilson's theorem formula.
If you look at the graph you'll see that the values for $f(x)$ where $x$ is prime is 0 while for $x$ nonprime tends to be converging toward $-1$ and in almost a perfect fashion and rapidly.
My idea is this: turn this into infinite Taylor series and add $x$ to it, that way we could get a good approximation of prime counting function to infinity.
In other words:
$$
lim _{ xrightarrow infty }{ x+sum _{ nepsilon N }^{ }{ f(n) } } =lim _{ xrightarrow infty }{ Pi (x) }
$$
The problem I am having is actually converting it into Taylor series. I really didn't have any experience with this kind of equations so I was wondering is there any trick to do this, or should I just let this one pass
Edited: Changed the formula, forgot to multiply by $pi$
prime-numbers
edited Apr 7 '13 at 17:42
vonbrand
19.8k63058
asked Apr 7 '13 at 17:20
user1398593
364
add a comment |
The formula I am having some problems is this one
$$
fleft( x right) =sin left(frac { 2(x-3)!-x+1 }{ 2x } pi right) =sin left(frac { 2,Gamma (x-2)-x+1 }{ 2x } pi right)
$$
Although this might seem too general it is a variation of Wilson's theorem which (in my opinion) has slightly better properties than the general Wilson's theorem formula.
If you look at the graph you'll see that the values for $f(x)$ where $x$ is prime is 0 while for $x$ nonprime tends to be converging toward $-1$ and in almost a perfect fashion and rapidly.
My idea is this: turn this into infinite Taylor series and add $x$ to it, that way we could get a good approximation of prime counting function to infinity.
In other words:
$$
lim _{ xrightarrow infty }{ x+sum _{ nepsilon N }^{ }{ f(n) } } =lim _{ xrightarrow infty }{ Pi (x) }
$$
The problem I am having is actually converting it into Taylor series. I really didn't have any experience with this kind of equations so I was wondering is there any trick to do this, or should I just let this one pass
Edited: Changed the formula, forgot to multiply by $pi$
prime-numbers
edited Apr 7 '13 at 17:42
vonbrand
19.8k63058
asked Apr 7 '13 at 17:20
user1398593
364
The formula I am having some problems is this one
$$
fleft( x right) =sin left(frac { 2(x-3)!-x+1 }{ 2x } pi right) =sin left(frac { 2,Gamma (x-2)-x+1 }{ 2x } pi right)
$$
Although this might seem too general it is a variation of Wilson's theorem which (in my opinion) has slightly better properties than the general Wilson's theorem formula.
If you look at the graph you'll see that the values for $f(x)$ where $x$ is prime is 0 while for $x$ nonprime tends to be converging toward $-1$ and in almost a perfect fashion and rapidly.
My idea is this: turn this into infinite Taylor series and add $x$ to it, that way we could get a good approximation of prime counting function to infinity.
In other words:
$$
lim _{ xrightarrow infty }{ x+sum _{ nepsilon N }^{ }{ f(n) } } =lim _{ xrightarrow infty }{ Pi (x) }
$$
The problem I am having is actually converting it into Taylor series. I really didn't have any experience with this kind of equations so I was wondering is there any trick to do this, or should I just let this one pass
Edited: Changed the formula, forgot to multiply by $pi$
prime-numbers
prime-numbers
edited Apr 7 '13 at 17:42
vonbrand
19.8k63058
asked Apr 7 '13 at 17:20
user1398593
364
edited Apr 7 '13 at 17:42
vonbrand
19.8k63058
asked Apr 7 '13 at 17:20
user1398593
364
edited Apr 7 '13 at 17:42
vonbrand
19.8k63058
edited Apr 7 '13 at 17:42
vonbrand
19.8k63058
edited Apr 7 '13 at 17:42
vonbrand
19.8k63058
19.8k63058
asked Apr 7 '13 at 17:20
user1398593
364
asked Apr 7 '13 at 17:20
user1398593
364
asked Apr 7 '13 at 17:20
user1398593
364
364
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
Well it seems I will answer my own question.
There is a similar formula that clearly says the nature of prime numbers and it goes like this:
$$pi (m)=sum _{ j=2 }^{ m }{ frac { sin ^{ 2 }{ (pi frac { { ((j-1)!) }^{ 2 } }{ j } ) } }{ sin ^{ 2 }{ (frac { pi }{ j } ) } } } $$
This function really is a prime counting function but you can see that it is a little tricky to develop it into a working formula since it should be integrated over ${ 1 }_{ N }$ .
However since I'm already answering I'm going to show what I planned on doing with the formula:
$$pi (x)quad approx quad sum _{ n=1 }^{ n=x }{ [quad cos { (frac { 2*(n-3)!+1 }{ 2n } pi ) } -cos { (frac { 2*(n-3)!+1 }{ n } pi ) } quad ] } $$
answered Apr 10 '13 at 12:04
user1398593
364
furthermore it converges toward $π(x)$
– user1398593
Apr 10 '13 at 19:03
add a comment |
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1 Answer
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1 Answer
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oldest
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Well it seems I will answer my own question.
There is a similar formula that clearly says the nature of prime numbers and it goes like this:
$$pi (m)=sum _{ j=2 }^{ m }{ frac { sin ^{ 2 }{ (pi frac { { ((j-1)!) }^{ 2 } }{ j } ) } }{ sin ^{ 2 }{ (frac { pi }{ j } ) } } } $$
This function really is a prime counting function but you can see that it is a little tricky to develop it into a working formula since it should be integrated over ${ 1 }_{ N }$ .
However since I'm already answering I'm going to show what I planned on doing with the formula:
$$pi (x)quad approx quad sum _{ n=1 }^{ n=x }{ [quad cos { (frac { 2*(n-3)!+1 }{ 2n } pi ) } -cos { (frac { 2*(n-3)!+1 }{ n } pi ) } quad ] } $$
answered Apr 10 '13 at 12:04
user1398593
364
furthermore it converges toward $π(x)$
– user1398593
Apr 10 '13 at 19:03
add a comment |
Well it seems I will answer my own question.
There is a similar formula that clearly says the nature of prime numbers and it goes like this:
$$pi (m)=sum _{ j=2 }^{ m }{ frac { sin ^{ 2 }{ (pi frac { { ((j-1)!) }^{ 2 } }{ j } ) } }{ sin ^{ 2 }{ (frac { pi }{ j } ) } } } $$
This function really is a prime counting function but you can see that it is a little tricky to develop it into a working formula since it should be integrated over ${ 1 }_{ N }$ .
However since I'm already answering I'm going to show what I planned on doing with the formula:
$$pi (x)quad approx quad sum _{ n=1 }^{ n=x }{ [quad cos { (frac { 2*(n-3)!+1 }{ 2n } pi ) } -cos { (frac { 2*(n-3)!+1 }{ n } pi ) } quad ] } $$
answered Apr 10 '13 at 12:04
user1398593
364
furthermore it converges toward $π(x)$
– user1398593
Apr 10 '13 at 19:03
add a comment |
Well it seems I will answer my own question.
There is a similar formula that clearly says the nature of prime numbers and it goes like this:
$$pi (m)=sum _{ j=2 }^{ m }{ frac { sin ^{ 2 }{ (pi frac { { ((j-1)!) }^{ 2 } }{ j } ) } }{ sin ^{ 2 }{ (frac { pi }{ j } ) } } } $$
This function really is a prime counting function but you can see that it is a little tricky to develop it into a working formula since it should be integrated over ${ 1 }_{ N }$ .
However since I'm already answering I'm going to show what I planned on doing with the formula:
$$pi (x)quad approx quad sum _{ n=1 }^{ n=x }{ [quad cos { (frac { 2*(n-3)!+1 }{ 2n } pi ) } -cos { (frac { 2*(n-3)!+1 }{ n } pi ) } quad ] } $$
answered Apr 10 '13 at 12:04
user1398593
364
Well it seems I will answer my own question.
There is a similar formula that clearly says the nature of prime numbers and it goes like this:
$$pi (m)=sum _{ j=2 }^{ m }{ frac { sin ^{ 2 }{ (pi frac { { ((j-1)!) }^{ 2 } }{ j } ) } }{ sin ^{ 2 }{ (frac { pi }{ j } ) } } } $$
This function really is a prime counting function but you can see that it is a little tricky to develop it into a working formula since it should be integrated over ${ 1 }_{ N }$ .
However since I'm already answering I'm going to show what I planned on doing with the formula:
$$pi (x)quad approx quad sum _{ n=1 }^{ n=x }{ [quad cos { (frac { 2*(n-3)!+1 }{ 2n } pi ) } -cos { (frac { 2*(n-3)!+1 }{ n } pi ) } quad ] } $$
answered Apr 10 '13 at 12:04
user1398593
364
answered Apr 10 '13 at 12:04
user1398593
364
answered Apr 10 '13 at 12:04
user1398593
364
answered Apr 10 '13 at 12:04
user1398593
364
364
furthermore it converges toward $π(x)$
– user1398593
Apr 10 '13 at 19:03
add a comment |
furthermore it converges toward $π(x)$
– user1398593
Apr 10 '13 at 19:03
furthermore it converges toward $π(x)$
– user1398593
Apr 10 '13 at 19:03
furthermore it converges toward $π(x)$
– user1398593
Apr 10 '13 at 19:03
add a comment |
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