Asymptotic expression for the $n$th prime number
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Quoting from the Wikipedia article:
As a consequence of the prime number theorem, one gets an asymptotic expression for the $ n $th prime number, denoted by $ p_n $:
$$ p_n sim n log n.$$
Can you explain how we get this approximate expression? I understand the number of prime numbers less than or equal to an integer $ N $, denoted as $ pi(N) $ is approximately:
$$
pi(N) sim frac{N}{log N}.
$$
If the $ n $ prime number is approximately $ n log n $, it implies that there are approximately $ n $ prime numbers less than or equal to $ n log n $, i.e.,
$$
pi(n log n) sim n.
$$
But substituting $ N $ with $ n log n $ in the formula of $pi(N)$ we get
$$
pi(N) sim frac{N}{log N} = frac{n log n}{log (n log n)} notsim n.
$$
What am I doing wrong? How can I show that the statement quoted from Wikipedia is correct?
prime-numbers
edited Nov 23 at 13:55
Klangen
1,49011232
asked May 9 at 3:17
Lone Learner
249416
add a comment |
up vote
2
down vote
favorite
Quoting from the Wikipedia article:
As a consequence of the prime number theorem, one gets an asymptotic expression for the $ n $th prime number, denoted by $ p_n $:
$$ p_n sim n log n.$$
Can you explain how we get this approximate expression? I understand the number of prime numbers less than or equal to an integer $ N $, denoted as $ pi(N) $ is approximately:
$$
pi(N) sim frac{N}{log N}.
$$
If the $ n $ prime number is approximately $ n log n $, it implies that there are approximately $ n $ prime numbers less than or equal to $ n log n $, i.e.,
$$
pi(n log n) sim n.
$$
But substituting $ N $ with $ n log n $ in the formula of $pi(N)$ we get
$$
pi(N) sim frac{N}{log N} = frac{n log n}{log (n log n)} notsim n.
$$
What am I doing wrong? How can I show that the statement quoted from Wikipedia is correct?
prime-numbers
edited Nov 23 at 13:55
Klangen
1,49011232
asked May 9 at 3:17
Lone Learner
249416
log(n log n) is very close to log(n).
– Jalex Stark
May 9 at 3:24
add a comment |
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Quoting from the Wikipedia article:
As a consequence of the prime number theorem, one gets an asymptotic expression for the $ n $th prime number, denoted by $ p_n $:
$$ p_n sim n log n.$$
Can you explain how we get this approximate expression? I understand the number of prime numbers less than or equal to an integer $ N $, denoted as $ pi(N) $ is approximately:
$$
pi(N) sim frac{N}{log N}.
$$
If the $ n $ prime number is approximately $ n log n $, it implies that there are approximately $ n $ prime numbers less than or equal to $ n log n $, i.e.,
$$
pi(n log n) sim n.
$$
But substituting $ N $ with $ n log n $ in the formula of $pi(N)$ we get
$$
pi(N) sim frac{N}{log N} = frac{n log n}{log (n log n)} notsim n.
$$
What am I doing wrong? How can I show that the statement quoted from Wikipedia is correct?
prime-numbers
edited Nov 23 at 13:55
Klangen
1,49011232
asked May 9 at 3:17
Lone Learner
249416
Quoting from the Wikipedia article:
As a consequence of the prime number theorem, one gets an asymptotic expression for the $ n $th prime number, denoted by $ p_n $:
$$ p_n sim n log n.$$
Can you explain how we get this approximate expression? I understand the number of prime numbers less than or equal to an integer $ N $, denoted as $ pi(N) $ is approximately:
$$
pi(N) sim frac{N}{log N}.
$$
If the $ n $ prime number is approximately $ n log n $, it implies that there are approximately $ n $ prime numbers less than or equal to $ n log n $, i.e.,
$$
pi(n log n) sim n.
$$
But substituting $ N $ with $ n log n $ in the formula of $pi(N)$ we get
$$
pi(N) sim frac{N}{log N} = frac{n log n}{log (n log n)} notsim n.
$$
What am I doing wrong? How can I show that the statement quoted from Wikipedia is correct?
prime-numbers
prime-numbers
edited Nov 23 at 13:55
Klangen
1,49011232
asked May 9 at 3:17
Lone Learner
249416
edited Nov 23 at 13:55
Klangen
1,49011232
asked May 9 at 3:17
Lone Learner
249416
edited Nov 23 at 13:55
Klangen
1,49011232
edited Nov 23 at 13:55
Klangen
1,49011232
edited Nov 23 at 13:55
Klangen
1,49011232
1,49011232
asked May 9 at 3:17
Lone Learner
249416
asked May 9 at 3:17
Lone Learner
249416
asked May 9 at 3:17
Lone Learner
249416
249416
log(n log n) is very close to log(n).
– Jalex Stark
May 9 at 3:24
add a comment |
log(n log n) is very close to log(n).
– Jalex Stark
May 9 at 3:24
log(n log n) is very close to log(n).
– Jalex Stark
May 9 at 3:24
log(n log n) is very close to log(n).
– Jalex Stark
May 9 at 3:24
add a comment |
2 Answers
2
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up vote
1
down vote
$$log(nlog n)=log n+loglog n=(1+o(1))log n$$
therefore
$$frac{nlog n}{log(nlog n)}=frac{nlog n}{(1+o(1))log n}
=frac{n}{(1+o(1))}sim n.$$
answered May 9 at 3:25
Lord Shark the Unknown
99.6k958131
add a comment |
up vote
0
down vote
It seems you are very new to prime number theory hence I am not sure if you understand the little $o$ notations. Hence I present @Lord Shark's answer in layman's terms.
$$
log (nlog n) = log n + loglog n
$$
But $loglog n$ is exponentially smaller compared to $log n$ i.e. it can be ignore compared to $log n$. Hence as an approximation, we have
$$
log (nlog n) sim log n
$$
With this approximation
$$
pi (nlog n) sim frac{nlog n}{log (nlog n)} sim frac{nlog n}{log n} sim n
$$
The smallest values of $x$ for which $pi(x) = n$ is $x = p_n$ hence we have
$$
pi (p_n) = n sim pi(nlog n)
$$
or $p_n sim nlog n$.
answered May 9 at 4:18
Nilotpal Kanti Sinha
4,01721437
add a comment |
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2 Answers
2
active
oldest
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2 Answers
2
active
oldest
votes
active
oldest
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active
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up vote
1
down vote
$$log(nlog n)=log n+loglog n=(1+o(1))log n$$
therefore
$$frac{nlog n}{log(nlog n)}=frac{nlog n}{(1+o(1))log n}
=frac{n}{(1+o(1))}sim n.$$
answered May 9 at 3:25
Lord Shark the Unknown
99.6k958131
add a comment |
up vote
1
down vote
$$log(nlog n)=log n+loglog n=(1+o(1))log n$$
therefore
$$frac{nlog n}{log(nlog n)}=frac{nlog n}{(1+o(1))log n}
=frac{n}{(1+o(1))}sim n.$$
answered May 9 at 3:25
Lord Shark the Unknown
99.6k958131
add a comment |
up vote
1
down vote
up vote
1
down vote
$$log(nlog n)=log n+loglog n=(1+o(1))log n$$
therefore
$$frac{nlog n}{log(nlog n)}=frac{nlog n}{(1+o(1))log n}
=frac{n}{(1+o(1))}sim n.$$
answered May 9 at 3:25
Lord Shark the Unknown
99.6k958131
$$log(nlog n)=log n+loglog n=(1+o(1))log n$$
therefore
$$frac{nlog n}{log(nlog n)}=frac{nlog n}{(1+o(1))log n}
=frac{n}{(1+o(1))}sim n.$$
answered May 9 at 3:25
Lord Shark the Unknown
99.6k958131
answered May 9 at 3:25
Lord Shark the Unknown
99.6k958131
answered May 9 at 3:25
Lord Shark the Unknown
99.6k958131
answered May 9 at 3:25
Lord Shark the Unknown
99.6k958131
99.6k958131
add a comment |
add a comment |
up vote
0
down vote
It seems you are very new to prime number theory hence I am not sure if you understand the little $o$ notations. Hence I present @Lord Shark's answer in layman's terms.
$$
log (nlog n) = log n + loglog n
$$
But $loglog n$ is exponentially smaller compared to $log n$ i.e. it can be ignore compared to $log n$. Hence as an approximation, we have
$$
log (nlog n) sim log n
$$
With this approximation
$$
pi (nlog n) sim frac{nlog n}{log (nlog n)} sim frac{nlog n}{log n} sim n
$$
The smallest values of $x$ for which $pi(x) = n$ is $x = p_n$ hence we have
$$
pi (p_n) = n sim pi(nlog n)
$$
or $p_n sim nlog n$.
answered May 9 at 4:18
Nilotpal Kanti Sinha
4,01721437
add a comment |
up vote
0
down vote
It seems you are very new to prime number theory hence I am not sure if you understand the little $o$ notations. Hence I present @Lord Shark's answer in layman's terms.
$$
log (nlog n) = log n + loglog n
$$
But $loglog n$ is exponentially smaller compared to $log n$ i.e. it can be ignore compared to $log n$. Hence as an approximation, we have
$$
log (nlog n) sim log n
$$
With this approximation
$$
pi (nlog n) sim frac{nlog n}{log (nlog n)} sim frac{nlog n}{log n} sim n
$$
The smallest values of $x$ for which $pi(x) = n$ is $x = p_n$ hence we have
$$
pi (p_n) = n sim pi(nlog n)
$$
or $p_n sim nlog n$.
answered May 9 at 4:18
Nilotpal Kanti Sinha
4,01721437
add a comment |
up vote
0
down vote
up vote
0
down vote
It seems you are very new to prime number theory hence I am not sure if you understand the little $o$ notations. Hence I present @Lord Shark's answer in layman's terms.
$$
log (nlog n) = log n + loglog n
$$
But $loglog n$ is exponentially smaller compared to $log n$ i.e. it can be ignore compared to $log n$. Hence as an approximation, we have
$$
log (nlog n) sim log n
$$
With this approximation
$$
pi (nlog n) sim frac{nlog n}{log (nlog n)} sim frac{nlog n}{log n} sim n
$$
The smallest values of $x$ for which $pi(x) = n$ is $x = p_n$ hence we have
$$
pi (p_n) = n sim pi(nlog n)
$$
or $p_n sim nlog n$.
answered May 9 at 4:18
Nilotpal Kanti Sinha
4,01721437
It seems you are very new to prime number theory hence I am not sure if you understand the little $o$ notations. Hence I present @Lord Shark's answer in layman's terms.
$$
log (nlog n) = log n + loglog n
$$
But $loglog n$ is exponentially smaller compared to $log n$ i.e. it can be ignore compared to $log n$. Hence as an approximation, we have
$$
log (nlog n) sim log n
$$
With this approximation
$$
pi (nlog n) sim frac{nlog n}{log (nlog n)} sim frac{nlog n}{log n} sim n
$$
The smallest values of $x$ for which $pi(x) = n$ is $x = p_n$ hence we have
$$
pi (p_n) = n sim pi(nlog n)
$$
or $p_n sim nlog n$.
answered May 9 at 4:18
Nilotpal Kanti Sinha
4,01721437
answered May 9 at 4:18
Nilotpal Kanti Sinha
4,01721437
answered May 9 at 4:18
Nilotpal Kanti Sinha
4,01721437
answered May 9 at 4:18
Nilotpal Kanti Sinha
4,01721437
4,01721437
add a comment |
add a comment |
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log(n log n) is very close to log(n).
– Jalex Stark
May 9 at 3:24