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







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    up vote
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    down vote

    favorite
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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







    share|cite|improve this question
























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      up vote
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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







      share|cite|improve this question













      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






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      asked yesterday









      Larry Freeman

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      3,26321239






















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          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 $pi(x^{1/v})$.






          share|cite|improve this answer























          • the product of these numbers is the least common multiple of $x$.
            – Larry Freeman
            yesterday












          • Now, I am clear. Thanks!
            – Larry Freeman
            yesterday











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          1 Answer
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          up vote
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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 $pi(x^{1/v})$.






          share|cite|improve this answer























          • the product of these numbers is the least common multiple of $x$.
            – Larry Freeman
            yesterday












          • Now, I am clear. Thanks!
            – Larry Freeman
            yesterday















          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 $pi(x^{1/v})$.






          share|cite|improve this answer























          • the product of these numbers is the least common multiple of $x$.
            – Larry Freeman
            yesterday












          • Now, I am clear. Thanks!
            – Larry Freeman
            yesterday













          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 $pi(x^{1/v})$.






          share|cite|improve this answer














          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 $pi(x^{1/v})$.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited yesterday

























          answered yesterday









          ajotatxe

          52.1k23688




          52.1k23688












          • the product of these numbers is the least common multiple of $x$.
            – Larry Freeman
            yesterday












          • Now, I am clear. Thanks!
            – Larry Freeman
            yesterday


















          • the product of these numbers is the least common multiple of $x$.
            – Larry Freeman
            yesterday












          • Now, I am clear. Thanks!
            – Larry Freeman
            yesterday
















          the product of these numbers is the least common multiple of $x$.
          – Larry Freeman
          yesterday






          the product of these numbers is the least common multiple of $x$.
          – Larry Freeman
          yesterday














          Now, I am clear. Thanks!
          – Larry Freeman
          yesterday




          Now, I am clear. Thanks!
          – Larry Freeman
          yesterday


















           

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