A question about prime gaps











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I have been reading the Wikipedia article about prime gaps (http://en.wikipedia.org/wiki/Prime_gap) and I came across the following:




Hoheisel was the first to show that there exists a constant $theta<1$ such that $$pi(x+x^theta)-pi(x)simfrac{x^theta}{log x}$$ as $x$ tends to infinity.




I have a few questions:




  1. What's the meaning of the $sim$ symbol?


  2. Does the symbol $sim$ mean that $pi(x+x^theta)-pi(x)$ is asymptotic to $x^theta/log x$? If this is the case, does this mean that $pi(x+x^theta)-pi(x)$ is always greater than $x^theta/log x$ at least for sufficiently large $x$? Is $pi(x+x^theta)-pi(x)$ the same as $pi[x,x+x^theta]$, that is to say, the amount of primes in the interval $[x,x+x^theta]$?



I'm not sure what the above expression means, and I'm not sure what it means when they say that a certain function is "asymptotic" to another function or when they say that a certain asymptotic formula holds.




  1. Also, in the above expression, is $log$ the natural logarithm or the logarithm in base $10$?






number-theory functions prime-numbers asymptotics prime-gaps







share|cite|improve this question




















  • 4




    $sim$ means asymptotic equality, so $$lim_{xtoinfty} (pi(x+x^theta)-pi(x))frac{log x}{x^theta} = 1.$$ And $log$ is the natural logarithm.
    – Daniel Fischer♦
    Mar 13 '14 at 23:35










  • Thank you for your comment. But does this mean that $pi(x+x^theta)-pi(x)$ is always greater than $x^theta/log x$? And is $pi(x+x^theta)-pi(x)$ the same as $pi[x,x+x^theta]$?
    – User X
    Mar 14 '14 at 1:28










  • No, it need not always be greater. It cannot be too much smaller in the long run, however. If $pi[x,x+x^theta]$ means the number of primes between $x$ and $x+x^theta$, then it means the same as $pi(x+x^theta)-pi(x)$.
    – Daniel Fischer♦
    Mar 14 '14 at 1:49










  • I think I'm kind of confused and maybe I'm mixing up things. For example, if consider the function $f(x)=1/x$, then its asymptotes are the $x$-axis and the $y$-axis. If consider the positive values for $x$, then as $x$ increases $f(x)=1/x$ gets smaller and smaller but never reaches $0$ (the graph of the function never touches the $x$-axis). This is why I thought that if a function $f(x)$ was asymptotic to another function $g(x)$, then $f(x)$ was always greater than $g(x)$ (or the other way round). What can I do to correctly understand this specific topic? I'm confused.
    – User X
    Mar 14 '14 at 2:19












  • $f sim g$ means $$lim_{xtoinfty} frac{f(x)}{g(x)} = 1.$$ That is something different from (but not entirely unrelated to) the asymptotes of functions like the axes for $1/x$.
    – Daniel Fischer♦
    Mar 14 '14 at 2:23















up vote
2
down vote

favorite












I have been reading the Wikipedia article about prime gaps (http://en.wikipedia.org/wiki/Prime_gap) and I came across the following:




Hoheisel was the first to show that there exists a constant $theta<1$ such that $$pi(x+x^theta)-pi(x)simfrac{x^theta}{log x}$$ as $x$ tends to infinity.




I have a few questions:




  1. What's the meaning of the $sim$ symbol?


  2. Does the symbol $sim$ mean that $pi(x+x^theta)-pi(x)$ is asymptotic to $x^theta/log x$? If this is the case, does this mean that $pi(x+x^theta)-pi(x)$ is always greater than $x^theta/log x$ at least for sufficiently large $x$? Is $pi(x+x^theta)-pi(x)$ the same as $pi[x,x+x^theta]$, that is to say, the amount of primes in the interval $[x,x+x^theta]$?



I'm not sure what the above expression means, and I'm not sure what it means when they say that a certain function is "asymptotic" to another function or when they say that a certain asymptotic formula holds.




  1. Also, in the above expression, is $log$ the natural logarithm or the logarithm in base $10$?






number-theory functions prime-numbers asymptotics prime-gaps







share|cite|improve this question




















  • 4




    $sim$ means asymptotic equality, so $$lim_{xtoinfty} (pi(x+x^theta)-pi(x))frac{log x}{x^theta} = 1.$$ And $log$ is the natural logarithm.
    – Daniel Fischer♦
    Mar 13 '14 at 23:35










  • Thank you for your comment. But does this mean that $pi(x+x^theta)-pi(x)$ is always greater than $x^theta/log x$? And is $pi(x+x^theta)-pi(x)$ the same as $pi[x,x+x^theta]$?
    – User X
    Mar 14 '14 at 1:28










  • No, it need not always be greater. It cannot be too much smaller in the long run, however. If $pi[x,x+x^theta]$ means the number of primes between $x$ and $x+x^theta$, then it means the same as $pi(x+x^theta)-pi(x)$.
    – Daniel Fischer♦
    Mar 14 '14 at 1:49










  • I think I'm kind of confused and maybe I'm mixing up things. For example, if consider the function $f(x)=1/x$, then its asymptotes are the $x$-axis and the $y$-axis. If consider the positive values for $x$, then as $x$ increases $f(x)=1/x$ gets smaller and smaller but never reaches $0$ (the graph of the function never touches the $x$-axis). This is why I thought that if a function $f(x)$ was asymptotic to another function $g(x)$, then $f(x)$ was always greater than $g(x)$ (or the other way round). What can I do to correctly understand this specific topic? I'm confused.
    – User X
    Mar 14 '14 at 2:19












  • $f sim g$ means $$lim_{xtoinfty} frac{f(x)}{g(x)} = 1.$$ That is something different from (but not entirely unrelated to) the asymptotes of functions like the axes for $1/x$.
    – Daniel Fischer♦
    Mar 14 '14 at 2:23













up vote
2
down vote

favorite









up vote
2
down vote

favorite











I have been reading the Wikipedia article about prime gaps (http://en.wikipedia.org/wiki/Prime_gap) and I came across the following:




Hoheisel was the first to show that there exists a constant $theta<1$ such that $$pi(x+x^theta)-pi(x)simfrac{x^theta}{log x}$$ as $x$ tends to infinity.




I have a few questions:




  1. What's the meaning of the $sim$ symbol?


  2. Does the symbol $sim$ mean that $pi(x+x^theta)-pi(x)$ is asymptotic to $x^theta/log x$? If this is the case, does this mean that $pi(x+x^theta)-pi(x)$ is always greater than $x^theta/log x$ at least for sufficiently large $x$? Is $pi(x+x^theta)-pi(x)$ the same as $pi[x,x+x^theta]$, that is to say, the amount of primes in the interval $[x,x+x^theta]$?



I'm not sure what the above expression means, and I'm not sure what it means when they say that a certain function is "asymptotic" to another function or when they say that a certain asymptotic formula holds.




  1. Also, in the above expression, is $log$ the natural logarithm or the logarithm in base $10$?






number-theory functions prime-numbers asymptotics prime-gaps







share|cite|improve this question















I have been reading the Wikipedia article about prime gaps (http://en.wikipedia.org/wiki/Prime_gap) and I came across the following:




Hoheisel was the first to show that there exists a constant $theta<1$ such that $$pi(x+x^theta)-pi(x)simfrac{x^theta}{log x}$$ as $x$ tends to infinity.




I have a few questions:




  1. What's the meaning of the $sim$ symbol?


  2. Does the symbol $sim$ mean that $pi(x+x^theta)-pi(x)$ is asymptotic to $x^theta/log x$? If this is the case, does this mean that $pi(x+x^theta)-pi(x)$ is always greater than $x^theta/log x$ at least for sufficiently large $x$? Is $pi(x+x^theta)-pi(x)$ the same as $pi[x,x+x^theta]$, that is to say, the amount of primes in the interval $[x,x+x^theta]$?



I'm not sure what the above expression means, and I'm not sure what it means when they say that a certain function is "asymptotic" to another function or when they say that a certain asymptotic formula holds.




  1. Also, in the above expression, is $log$ the natural logarithm or the logarithm in base $10$?






number-theory functions prime-numbers asymptotics prime-gaps




number-theory functions prime-numbers asymptotics prime-gaps






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 22 at 9:52









Klangen

1,43711232




1,43711232










asked Mar 13 '14 at 23:33









User X

114110




114110








  • 4




    $sim$ means asymptotic equality, so $$lim_{xtoinfty} (pi(x+x^theta)-pi(x))frac{log x}{x^theta} = 1.$$ And $log$ is the natural logarithm.
    – Daniel Fischer♦
    Mar 13 '14 at 23:35










  • Thank you for your comment. But does this mean that $pi(x+x^theta)-pi(x)$ is always greater than $x^theta/log x$? And is $pi(x+x^theta)-pi(x)$ the same as $pi[x,x+x^theta]$?
    – User X
    Mar 14 '14 at 1:28










  • No, it need not always be greater. It cannot be too much smaller in the long run, however. If $pi[x,x+x^theta]$ means the number of primes between $x$ and $x+x^theta$, then it means the same as $pi(x+x^theta)-pi(x)$.
    – Daniel Fischer♦
    Mar 14 '14 at 1:49










  • I think I'm kind of confused and maybe I'm mixing up things. For example, if consider the function $f(x)=1/x$, then its asymptotes are the $x$-axis and the $y$-axis. If consider the positive values for $x$, then as $x$ increases $f(x)=1/x$ gets smaller and smaller but never reaches $0$ (the graph of the function never touches the $x$-axis). This is why I thought that if a function $f(x)$ was asymptotic to another function $g(x)$, then $f(x)$ was always greater than $g(x)$ (or the other way round). What can I do to correctly understand this specific topic? I'm confused.
    – User X
    Mar 14 '14 at 2:19












  • $f sim g$ means $$lim_{xtoinfty} frac{f(x)}{g(x)} = 1.$$ That is something different from (but not entirely unrelated to) the asymptotes of functions like the axes for $1/x$.
    – Daniel Fischer♦
    Mar 14 '14 at 2:23














  • 4




    $sim$ means asymptotic equality, so $$lim_{xtoinfty} (pi(x+x^theta)-pi(x))frac{log x}{x^theta} = 1.$$ And $log$ is the natural logarithm.
    – Daniel Fischer♦
    Mar 13 '14 at 23:35










  • Thank you for your comment. But does this mean that $pi(x+x^theta)-pi(x)$ is always greater than $x^theta/log x$? And is $pi(x+x^theta)-pi(x)$ the same as $pi[x,x+x^theta]$?
    – User X
    Mar 14 '14 at 1:28










  • No, it need not always be greater. It cannot be too much smaller in the long run, however. If $pi[x,x+x^theta]$ means the number of primes between $x$ and $x+x^theta$, then it means the same as $pi(x+x^theta)-pi(x)$.
    – Daniel Fischer♦
    Mar 14 '14 at 1:49










  • I think I'm kind of confused and maybe I'm mixing up things. For example, if consider the function $f(x)=1/x$, then its asymptotes are the $x$-axis and the $y$-axis. If consider the positive values for $x$, then as $x$ increases $f(x)=1/x$ gets smaller and smaller but never reaches $0$ (the graph of the function never touches the $x$-axis). This is why I thought that if a function $f(x)$ was asymptotic to another function $g(x)$, then $f(x)$ was always greater than $g(x)$ (or the other way round). What can I do to correctly understand this specific topic? I'm confused.
    – User X
    Mar 14 '14 at 2:19












  • $f sim g$ means $$lim_{xtoinfty} frac{f(x)}{g(x)} = 1.$$ That is something different from (but not entirely unrelated to) the asymptotes of functions like the axes for $1/x$.
    – Daniel Fischer♦
    Mar 14 '14 at 2:23








4




4




$sim$ means asymptotic equality, so $$lim_{xtoinfty} (pi(x+x^theta)-pi(x))frac{log x}{x^theta} = 1.$$ And $log$ is the natural logarithm.
– Daniel Fischer♦
Mar 13 '14 at 23:35




$sim$ means asymptotic equality, so $$lim_{xtoinfty} (pi(x+x^theta)-pi(x))frac{log x}{x^theta} = 1.$$ And $log$ is the natural logarithm.
– Daniel Fischer♦
Mar 13 '14 at 23:35












Thank you for your comment. But does this mean that $pi(x+x^theta)-pi(x)$ is always greater than $x^theta/log x$? And is $pi(x+x^theta)-pi(x)$ the same as $pi[x,x+x^theta]$?
– User X
Mar 14 '14 at 1:28




Thank you for your comment. But does this mean that $pi(x+x^theta)-pi(x)$ is always greater than $x^theta/log x$? And is $pi(x+x^theta)-pi(x)$ the same as $pi[x,x+x^theta]$?
– User X
Mar 14 '14 at 1:28












No, it need not always be greater. It cannot be too much smaller in the long run, however. If $pi[x,x+x^theta]$ means the number of primes between $x$ and $x+x^theta$, then it means the same as $pi(x+x^theta)-pi(x)$.
– Daniel Fischer♦
Mar 14 '14 at 1:49




No, it need not always be greater. It cannot be too much smaller in the long run, however. If $pi[x,x+x^theta]$ means the number of primes between $x$ and $x+x^theta$, then it means the same as $pi(x+x^theta)-pi(x)$.
– Daniel Fischer♦
Mar 14 '14 at 1:49












I think I'm kind of confused and maybe I'm mixing up things. For example, if consider the function $f(x)=1/x$, then its asymptotes are the $x$-axis and the $y$-axis. If consider the positive values for $x$, then as $x$ increases $f(x)=1/x$ gets smaller and smaller but never reaches $0$ (the graph of the function never touches the $x$-axis). This is why I thought that if a function $f(x)$ was asymptotic to another function $g(x)$, then $f(x)$ was always greater than $g(x)$ (or the other way round). What can I do to correctly understand this specific topic? I'm confused.
– User X
Mar 14 '14 at 2:19






I think I'm kind of confused and maybe I'm mixing up things. For example, if consider the function $f(x)=1/x$, then its asymptotes are the $x$-axis and the $y$-axis. If consider the positive values for $x$, then as $x$ increases $f(x)=1/x$ gets smaller and smaller but never reaches $0$ (the graph of the function never touches the $x$-axis). This is why I thought that if a function $f(x)$ was asymptotic to another function $g(x)$, then $f(x)$ was always greater than $g(x)$ (or the other way round). What can I do to correctly understand this specific topic? I'm confused.
– User X
Mar 14 '14 at 2:19














$f sim g$ means $$lim_{xtoinfty} frac{f(x)}{g(x)} = 1.$$ That is something different from (but not entirely unrelated to) the asymptotes of functions like the axes for $1/x$.
– Daniel Fischer♦
Mar 14 '14 at 2:23




$f sim g$ means $$lim_{xtoinfty} frac{f(x)}{g(x)} = 1.$$ That is something different from (but not entirely unrelated to) the asymptotes of functions like the axes for $1/x$.
– Daniel Fischer♦
Mar 14 '14 at 2:23










1 Answer
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$f(x)sim g(x)$ just means that $lim_{xtoinfty}f(x)/g(x)=1.$ Informally, $g(x)$ is a good approximation for $f(x)$ (and vice versa) when $x$ is large.



$pi(x+x^theta)-pi(x)sim x^theta/log x$ mean that $pi(x+x^theta)-pi(x)$ is greater than $0.99x^theta/log x$ for sufficiently large $x$, and this would remain true (for a different value of "large enough") if you replace 0.99 with any number less than 1. ($log$ is the natural logarithm, yes.)



$pi(x+x^theta)-pi(x)$ is the number of primes in the interval $(x,x+x^theta]$, yes.






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    $f(x)sim g(x)$ just means that $lim_{xtoinfty}f(x)/g(x)=1.$ Informally, $g(x)$ is a good approximation for $f(x)$ (and vice versa) when $x$ is large.



    $pi(x+x^theta)-pi(x)sim x^theta/log x$ mean that $pi(x+x^theta)-pi(x)$ is greater than $0.99x^theta/log x$ for sufficiently large $x$, and this would remain true (for a different value of "large enough") if you replace 0.99 with any number less than 1. ($log$ is the natural logarithm, yes.)



    $pi(x+x^theta)-pi(x)$ is the number of primes in the interval $(x,x+x^theta]$, yes.






    share|cite|improve this answer

























      up vote
      2
      down vote













      $f(x)sim g(x)$ just means that $lim_{xtoinfty}f(x)/g(x)=1.$ Informally, $g(x)$ is a good approximation for $f(x)$ (and vice versa) when $x$ is large.



      $pi(x+x^theta)-pi(x)sim x^theta/log x$ mean that $pi(x+x^theta)-pi(x)$ is greater than $0.99x^theta/log x$ for sufficiently large $x$, and this would remain true (for a different value of "large enough") if you replace 0.99 with any number less than 1. ($log$ is the natural logarithm, yes.)



      $pi(x+x^theta)-pi(x)$ is the number of primes in the interval $(x,x+x^theta]$, yes.






      share|cite|improve this answer























        up vote
        2
        down vote










        up vote
        2
        down vote









        $f(x)sim g(x)$ just means that $lim_{xtoinfty}f(x)/g(x)=1.$ Informally, $g(x)$ is a good approximation for $f(x)$ (and vice versa) when $x$ is large.



        $pi(x+x^theta)-pi(x)sim x^theta/log x$ mean that $pi(x+x^theta)-pi(x)$ is greater than $0.99x^theta/log x$ for sufficiently large $x$, and this would remain true (for a different value of "large enough") if you replace 0.99 with any number less than 1. ($log$ is the natural logarithm, yes.)



        $pi(x+x^theta)-pi(x)$ is the number of primes in the interval $(x,x+x^theta]$, yes.






        share|cite|improve this answer












        $f(x)sim g(x)$ just means that $lim_{xtoinfty}f(x)/g(x)=1.$ Informally, $g(x)$ is a good approximation for $f(x)$ (and vice versa) when $x$ is large.



        $pi(x+x^theta)-pi(x)sim x^theta/log x$ mean that $pi(x+x^theta)-pi(x)$ is greater than $0.99x^theta/log x$ for sufficiently large $x$, and this would remain true (for a different value of "large enough") if you replace 0.99 with any number less than 1. ($log$ is the natural logarithm, yes.)



        $pi(x+x^theta)-pi(x)$ is the number of primes in the interval $(x,x+x^theta]$, yes.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Mar 14 '14 at 17:38









        Charles

        23.6k451113




        23.6k451113






























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