Unclear on step #2 of the MathWorld definition of the Reimann Prime Counting Function
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I was reading through the MathWorld article on the Reimann Prime Counting Function.
The first step in the definition is clear to me:
$$f(x) = sum_{p^v < x text{ and p prime}} frac{1}{v}$$
Here is the second step:
$$=sum_{n}frac{pi(x^{1/n})}{n}$$
It is not clear to me how the sum of the reciprocal of the power is equal to the sum of the prime counting function divided by all values of $n$.
I would appreciate if someone could show how the second step follows from the first.
prime-numbers proof-explanation
asked Nov 19 at 17:00
Larry Freeman
3,24721239
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up vote
0
down vote
favorite
I was reading through the MathWorld article on the Reimann Prime Counting Function.
The first step in the definition is clear to me:
$$f(x) = sum_{p^v < x text{ and p prime}} frac{1}{v}$$
Here is the second step:
$$=sum_{n}frac{pi(x^{1/n})}{n}$$
It is not clear to me how the sum of the reciprocal of the power is equal to the sum of the prime counting function divided by all values of $n$.
I would appreciate if someone could show how the second step follows from the first.
prime-numbers proof-explanation
asked Nov 19 at 17:00
Larry Freeman
3,24721239
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I was reading through the MathWorld article on the Reimann Prime Counting Function.
The first step in the definition is clear to me:
$$f(x) = sum_{p^v < x text{ and p prime}} frac{1}{v}$$
Here is the second step:
$$=sum_{n}frac{pi(x^{1/n})}{n}$$
It is not clear to me how the sum of the reciprocal of the power is equal to the sum of the prime counting function divided by all values of $n$.
I would appreciate if someone could show how the second step follows from the first.
prime-numbers proof-explanation
asked Nov 19 at 17:00
Larry Freeman
3,24721239
I was reading through the MathWorld article on the Reimann Prime Counting Function.
The first step in the definition is clear to me:
$$f(x) = sum_{p^v < x text{ and p prime}} frac{1}{v}$$
Here is the second step:
$$=sum_{n}frac{pi(x^{1/n})}{n}$$
It is not clear to me how the sum of the reciprocal of the power is equal to the sum of the prime counting function divided by all values of $n$.
I would appreciate if someone could show how the second step follows from the first.
prime-numbers proof-explanation
prime-numbers proof-explanation
asked Nov 19 at 17:00
Larry Freeman
3,24721239
asked Nov 19 at 17:00
Larry Freeman
3,24721239
asked Nov 19 at 17:00
Larry Freeman
3,24721239
asked Nov 19 at 17:00
Larry Freeman
3,24721239
asked Nov 19 at 17:00
Larry Freeman
3,24721239
3,24721239
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
up vote
1
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accepted
Hint:
How many $v$-th powers of primes are there between $1$ and $x$?
Answer: if $p$ is a prime then $p^v<x$ iff $p<x^{1/v}$. Then the number of $v$-th powers of primes between $1$ and $x$ is $pi(x^{1/v})$.
answered Nov 19 at 17:05
ajotatxe
52.3k23789
the product of these numbers is the least common multiple of $x$.
– Larry Freeman
Nov 19 at 17:09
Now, I am clear. Thanks!
– Larry Freeman
Nov 19 at 23:54
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Hint:
How many $v$-th powers of primes are there between $1$ and $x$?
Answer: if $p$ is a prime then $p^v<x$ iff $p<x^{1/v}$. Then the number of $v$-th powers of primes between $1$ and $x$ is $pi(x^{1/v})$.
answered Nov 19 at 17:05
ajotatxe
52.3k23789
the product of these numbers is the least common multiple of $x$.
– Larry Freeman
Nov 19 at 17:09
Now, I am clear. Thanks!
– Larry Freeman
Nov 19 at 23:54
add a comment |
up vote
1
down vote
accepted
Hint:
How many $v$-th powers of primes are there between $1$ and $x$?
Answer: if $p$ is a prime then $p^v<x$ iff $p<x^{1/v}$. Then the number of $v$-th powers of primes between $1$ and $x$ is $pi(x^{1/v})$.
answered Nov 19 at 17:05
ajotatxe
52.3k23789
the product of these numbers is the least common multiple of $x$.
– Larry Freeman
Nov 19 at 17:09
Now, I am clear. Thanks!
– Larry Freeman
Nov 19 at 23:54
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Hint:
How many $v$-th powers of primes are there between $1$ and $x$?
Answer: if $p$ is a prime then $p^v<x$ iff $p<x^{1/v}$. Then the number of $v$-th powers of primes between $1$ and $x$ is $pi(x^{1/v})$.
answered Nov 19 at 17:05
ajotatxe
52.3k23789
Hint:
How many $v$-th powers of primes are there between $1$ and $x$?
Answer: if $p$ is a prime then $p^v<x$ iff $p<x^{1/v}$. Then the number of $v$-th powers of primes between $1$ and $x$ is $pi(x^{1/v})$.
answered Nov 19 at 17:05
ajotatxe
52.3k23789
edited Nov 19 at 17:12
answered Nov 19 at 17:05
ajotatxe
52.3k23789
answered Nov 19 at 17:05
ajotatxe
52.3k23789
answered Nov 19 at 17:05
ajotatxe
52.3k23789
52.3k23789
the product of these numbers is the least common multiple of $x$.
– Larry Freeman
Nov 19 at 17:09
Now, I am clear. Thanks!
– Larry Freeman
Nov 19 at 23:54
add a comment |
the product of these numbers is the least common multiple of $x$.
– Larry Freeman
Nov 19 at 17:09
Now, I am clear. Thanks!
– Larry Freeman
Nov 19 at 23:54
the product of these numbers is the least common multiple of $x$.
– Larry Freeman
Nov 19 at 17:09
the product of these numbers is the least common multiple of $x$.
– Larry Freeman
Nov 19 at 17:09
Now, I am clear. Thanks!
– Larry Freeman
Nov 19 at 23:54
Now, I am clear. Thanks!
– Larry Freeman
Nov 19 at 23:54
add a comment |
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