Inverse of $f(x) = fracxlog(x)$

Clash Royale CLAN TAG#URR8PPP

up vote
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$pi(x)$ is defined as number of primes less than or equal to x. Gauss showed that $pi(x) approx fracxlog(x)$. I want to calculate the inverse of that approximation because I want to estimate how far should I sieve in order to get the number of primes that I want. Here are the steps that I took.

$$beginaligny &= fracxlog(x)\ x &= y log(x) \ e^x&=e^ylog(x) \ e^x &= (e^log(x))^y \ e^x &= x^yendalign$$

I am stuck at this point. How can I calculate $y$ in terms of $x$?

logarithms inverse-function

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edited Aug 6 at 20:58

gimusi

65.5k73684

asked Aug 6 at 20:42

yasar

1674

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  For an exact formula, use the Lambert W function, but in your case a numerical method like Newton's method will do.
  – Jean-Claude Arbaut
  Aug 6 at 20:46
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  I believe you can't do it analytically.
  – fonini
  Aug 6 at 20:46
  

  
  
  
  

add a comment | 

up vote
2
down vote

favorite

$pi(x)$ is defined as number of primes less than or equal to x. Gauss showed that $pi(x) approx fracxlog(x)$. I want to calculate the inverse of that approximation because I want to estimate how far should I sieve in order to get the number of primes that I want. Here are the steps that I took.

$$beginaligny &= fracxlog(x)\ x &= y log(x) \ e^x&=e^ylog(x) \ e^x &= (e^log(x))^y \ e^x &= x^yendalign$$

I am stuck at this point. How can I calculate $y$ in terms of $x$?

logarithms inverse-function

share|cite|improve this question

edited Aug 6 at 20:58

gimusi

65.5k73684

asked Aug 6 at 20:42

yasar

1674

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  For an exact formula, use the Lambert W function, but in your case a numerical method like Newton's method will do.
  – Jean-Claude Arbaut
  Aug 6 at 20:46
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  I believe you can't do it analytically.
  – fonini
  Aug 6 at 20:46
  

  
  
  
  

add a comment | 

up vote
2
down vote

favorite

up vote
2
down vote

favorite

$pi(x)$ is defined as number of primes less than or equal to x. Gauss showed that $pi(x) approx fracxlog(x)$. I want to calculate the inverse of that approximation because I want to estimate how far should I sieve in order to get the number of primes that I want. Here are the steps that I took.

$$beginaligny &= fracxlog(x)\ x &= y log(x) \ e^x&=e^ylog(x) \ e^x &= (e^log(x))^y \ e^x &= x^yendalign$$

I am stuck at this point. How can I calculate $y$ in terms of $x$?

logarithms inverse-function

share|cite|improve this question

edited Aug 6 at 20:58

gimusi

65.5k73684

asked Aug 6 at 20:42

yasar

1674

$pi(x)$ is defined as number of primes less than or equal to x. Gauss showed that $pi(x) approx fracxlog(x)$. I want to calculate the inverse of that approximation because I want to estimate how far should I sieve in order to get the number of primes that I want. Here are the steps that I took.

$$beginaligny &= fracxlog(x)\ x &= y log(x) \ e^x&=e^ylog(x) \ e^x &= (e^log(x))^y \ e^x &= x^yendalign$$

I am stuck at this point. How can I calculate $y$ in terms of $x$?

logarithms inverse-function

share|cite|improve this question

edited Aug 6 at 20:58

gimusi

65.5k73684

asked Aug 6 at 20:42

yasar

1674

share|cite|improve this question

share|cite|improve this question

edited Aug 6 at 20:58

gimusi

65.5k73684

edited Aug 6 at 20:58

gimusi

65.5k73684

edited Aug 6 at 20:58

gimusi

65.5k73684

65.5k73684

asked Aug 6 at 20:42

yasar

1674

asked Aug 6 at 20:42

yasar

1674

asked Aug 6 at 20:42

yasar

1674

1674

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  For an exact formula, use the Lambert W function, but in your case a numerical method like Newton's method will do.
  – Jean-Claude Arbaut
  Aug 6 at 20:46
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  I believe you can't do it analytically.
  – fonini
  Aug 6 at 20:46
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  For an exact formula, use the Lambert W function, but in your case a numerical method like Newton's method will do.
  – Jean-Claude Arbaut
  Aug 6 at 20:46
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  I believe you can't do it analytically.
  – fonini
  Aug 6 at 20:46
  

  
  
  
  

2

2

For an exact formula, use the Lambert W function, but in your case a numerical method like Newton's method will do.
– Jean-Claude Arbaut
Aug 6 at 20:46

For an exact formula, use the Lambert W function, but in your case a numerical method like Newton's method will do.
– Jean-Claude Arbaut
Aug 6 at 20:46

1

1

I believe you can't do it analytically.
– fonini
Aug 6 at 20:46

I believe you can't do it analytically.
– fonini
Aug 6 at 20:46

add a comment | 

2 Answers
2

active

oldest

votes

up vote
3
down vote

Resorting to Lambert's $W$ is a little overkill.

With $$x=yln y,$$ we have

$$frac xlog x=fracyln yln(yln y)=fracyln yln y+lnln yapprox y.$$

Given that the distribution is anyway approximate, this can be a sufficient approximation of the inverse. Or you can use it as a starting value for Newton's iterations (use few of them anyway).

share|cite|improve this answer

edited Aug 6 at 20:59

answered Aug 6 at 20:50

Yves Daoust

111k665205

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $x = y / ln y$ I think? division not multiplication?
  – Claude
  Aug 6 at 20:53
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @Claude: the problem statement says $y=x/log x$, then $xapprox ylog y$.
  – Yves Daoust
  Aug 6 at 20:54
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  oh, I see now what you are doing
  – Claude
  Aug 6 at 20:56
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @yasar Try to plug it: set $x := yln y$. Then $$fracxln x = fracyln yln(yln y)) = fracyln yln y + lnln y = fracy1+fraclnln yln y = y - yfraclnln yln y + oleft(yfraclnln yln y right) = y + o(y)$$
  – Clement C.
  Aug 6 at 21:16
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @ClementC. Indeed, but it does work here: it is known that $n (log n + log log n - 1) < p(n) < n (log n + log log n)$ for $n ge 6$.
  – Robert Israel
  Aug 7 at 4:50
  

  
  
  
  

 | 
show 5 more comments

up vote
2
down vote

To be clear about what is hinted at in the comments,
$$
f^-1(y) = -y,W(-tfrac1y)
$$
where $W$ is the Lambert-$W$ function. (You can verify this in Mathematica.) If you wanted an asymptotic expansion check the wiki article and plug one in.

share|cite|improve this answer

answered Aug 6 at 21:02

cdipaolo

527211

add a comment | 

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2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
3
down vote

Resorting to Lambert's $W$ is a little overkill.

With $$x=yln y,$$ we have

$$frac xlog x=fracyln yln(yln y)=fracyln yln y+lnln yapprox y.$$

Given that the distribution is anyway approximate, this can be a sufficient approximation of the inverse. Or you can use it as a starting value for Newton's iterations (use few of them anyway).

share|cite|improve this answer

edited Aug 6 at 20:59

answered Aug 6 at 20:50

Yves Daoust

111k665205

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $x = y / ln y$ I think? division not multiplication?
  – Claude
  Aug 6 at 20:53
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @Claude: the problem statement says $y=x/log x$, then $xapprox ylog y$.
  – Yves Daoust
  Aug 6 at 20:54
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  oh, I see now what you are doing
  – Claude
  Aug 6 at 20:56
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @yasar Try to plug it: set $x := yln y$. Then $$fracxln x = fracyln yln(yln y)) = fracyln yln y + lnln y = fracy1+fraclnln yln y = y - yfraclnln yln y + oleft(yfraclnln yln y right) = y + o(y)$$
  – Clement C.
  Aug 6 at 21:16
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @ClementC. Indeed, but it does work here: it is known that $n (log n + log log n - 1) < p(n) < n (log n + log log n)$ for $n ge 6$.
  – Robert Israel
  Aug 7 at 4:50
  

  
  
  
  

 | 
show 5 more comments

up vote
3
down vote

Resorting to Lambert's $W$ is a little overkill.

With $$x=yln y,$$ we have

$$frac xlog x=fracyln yln(yln y)=fracyln yln y+lnln yapprox y.$$

Given that the distribution is anyway approximate, this can be a sufficient approximation of the inverse. Or you can use it as a starting value for Newton's iterations (use few of them anyway).

share|cite|improve this answer

edited Aug 6 at 20:59

answered Aug 6 at 20:50

Yves Daoust

111k665205

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $x = y / ln y$ I think? division not multiplication?
  – Claude
  Aug 6 at 20:53
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @Claude: the problem statement says $y=x/log x$, then $xapprox ylog y$.
  – Yves Daoust
  Aug 6 at 20:54
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  oh, I see now what you are doing
  – Claude
  Aug 6 at 20:56
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @yasar Try to plug it: set $x := yln y$. Then $$fracxln x = fracyln yln(yln y)) = fracyln yln y + lnln y = fracy1+fraclnln yln y = y - yfraclnln yln y + oleft(yfraclnln yln y right) = y + o(y)$$
  – Clement C.
  Aug 6 at 21:16
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @ClementC. Indeed, but it does work here: it is known that $n (log n + log log n - 1) < p(n) < n (log n + log log n)$ for $n ge 6$.
  – Robert Israel
  Aug 7 at 4:50
  

  
  
  
  

 | 
show 5 more comments

up vote
3
down vote

up vote
3
down vote

Resorting to Lambert's $W$ is a little overkill.

With $$x=yln y,$$ we have

$$frac xlog x=fracyln yln(yln y)=fracyln yln y+lnln yapprox y.$$

Given that the distribution is anyway approximate, this can be a sufficient approximation of the inverse. Or you can use it as a starting value for Newton's iterations (use few of them anyway).

share|cite|improve this answer

edited Aug 6 at 20:59

answered Aug 6 at 20:50

Yves Daoust

111k665205

Resorting to Lambert's $W$ is a little overkill.

With $$x=yln y,$$ we have

$$frac xlog x=fracyln yln(yln y)=fracyln yln y+lnln yapprox y.$$

Given that the distribution is anyway approximate, this can be a sufficient approximation of the inverse. Or you can use it as a starting value for Newton's iterations (use few of them anyway).

share|cite|improve this answer

edited Aug 6 at 20:59

answered Aug 6 at 20:50

Yves Daoust

111k665205

share|cite|improve this answer

share|cite|improve this answer

edited Aug 6 at 20:59

edited Aug 6 at 20:59

edited Aug 6 at 20:59

answered Aug 6 at 20:50

Yves Daoust

111k665205

answered Aug 6 at 20:50

Yves Daoust

111k665205

answered Aug 6 at 20:50

Yves Daoust

111k665205

111k665205

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $x = y / ln y$ I think? division not multiplication?
  – Claude
  Aug 6 at 20:53
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @Claude: the problem statement says $y=x/log x$, then $xapprox ylog y$.
  – Yves Daoust
  Aug 6 at 20:54
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  oh, I see now what you are doing
  – Claude
  Aug 6 at 20:56
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @yasar Try to plug it: set $x := yln y$. Then $$fracxln x = fracyln yln(yln y)) = fracyln yln y + lnln y = fracy1+fraclnln yln y = y - yfraclnln yln y + oleft(yfraclnln yln y right) = y + o(y)$$
  – Clement C.
  Aug 6 at 21:16
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @ClementC. Indeed, but it does work here: it is known that $n (log n + log log n - 1) < p(n) < n (log n + log log n)$ for $n ge 6$.
  – Robert Israel
  Aug 7 at 4:50
  

  
  
  
  

 | 
show 5 more comments

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $x = y / ln y$ I think? division not multiplication?
  – Claude
  Aug 6 at 20:53
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @Claude: the problem statement says $y=x/log x$, then $xapprox ylog y$.
  – Yves Daoust
  Aug 6 at 20:54
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  oh, I see now what you are doing
  – Claude
  Aug 6 at 20:56
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @yasar Try to plug it: set $x := yln y$. Then $$fracxln x = fracyln yln(yln y)) = fracyln yln y + lnln y = fracy1+fraclnln yln y = y - yfraclnln yln y + oleft(yfraclnln yln y right) = y + o(y)$$
  – Clement C.
  Aug 6 at 21:16
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @ClementC. Indeed, but it does work here: it is known that $n (log n + log log n - 1) < p(n) < n (log n + log log n)$ for $n ge 6$.
  – Robert Israel
  Aug 7 at 4:50
  

  
  
  
  

$x = y / ln y$ I think? division not multiplication?
– Claude
Aug 6 at 20:53

$x = y / ln y$ I think? division not multiplication?
– Claude
Aug 6 at 20:53

1

1

@Claude: the problem statement says $y=x/log x$, then $xapprox ylog y$.
– Yves Daoust
Aug 6 at 20:54

@Claude: the problem statement says $y=x/log x$, then $xapprox ylog y$.
– Yves Daoust
Aug 6 at 20:54

oh, I see now what you are doing
– Claude
Aug 6 at 20:56

oh, I see now what you are doing
– Claude
Aug 6 at 20:56

1

1

@yasar Try to plug it: set $x := yln y$. Then $$fracxln x = fracyln yln(yln y)) = fracyln yln y + lnln y = fracy1+fraclnln yln y = y - yfraclnln yln y + oleft(yfraclnln yln y right) = y + o(y)$$
– Clement C.
Aug 6 at 21:16

@yasar Try to plug it: set $x := yln y$. Then $$fracxln x = fracyln yln(yln y)) = fracyln yln y + lnln y = fracy1+fraclnln yln y = y - yfraclnln yln y + oleft(yfraclnln yln y right) = y + o(y)$$
– Clement C.
Aug 6 at 21:16

1

1

@ClementC. Indeed, but it does work here: it is known that $n (log n + log log n - 1) < p(n) < n (log n + log log n)$ for $n ge 6$.
– Robert Israel
Aug 7 at 4:50

@ClementC. Indeed, but it does work here: it is known that $n (log n + log log n - 1) < p(n) < n (log n + log log n)$ for $n ge 6$.
– Robert Israel
Aug 7 at 4:50

 | 
show 5 more comments

up vote
2
down vote

To be clear about what is hinted at in the comments,
$$
f^-1(y) = -y,W(-tfrac1y)
$$
where $W$ is the Lambert-$W$ function. (You can verify this in Mathematica.) If you wanted an asymptotic expansion check the wiki article and plug one in.

share|cite|improve this answer

answered Aug 6 at 21:02

cdipaolo

527211

add a comment | 

up vote
2
down vote

To be clear about what is hinted at in the comments,
$$
f^-1(y) = -y,W(-tfrac1y)
$$
where $W$ is the Lambert-$W$ function. (You can verify this in Mathematica.) If you wanted an asymptotic expansion check the wiki article and plug one in.

share|cite|improve this answer

answered Aug 6 at 21:02

cdipaolo

527211

add a comment | 

up vote
2
down vote

up vote
2
down vote

To be clear about what is hinted at in the comments,
$$
f^-1(y) = -y,W(-tfrac1y)
$$
where $W$ is the Lambert-$W$ function. (You can verify this in Mathematica.) If you wanted an asymptotic expansion check the wiki article and plug one in.

share|cite|improve this answer

answered Aug 6 at 21:02

cdipaolo

527211

To be clear about what is hinted at in the comments,
$$
f^-1(y) = -y,W(-tfrac1y)
$$
where $W$ is the Lambert-$W$ function. (You can verify this in Mathematica.) If you wanted an asymptotic expansion check the wiki article and plug one in.

share|cite|improve this answer

answered Aug 6 at 21:02

cdipaolo

527211

share|cite|improve this answer

share|cite|improve this answer

answered Aug 6 at 21:02

cdipaolo

527211

answered Aug 6 at 21:02

cdipaolo

527211

answered Aug 6 at 21:02

cdipaolo

527211

527211

add a comment | 

add a comment | 

 

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