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

The name of the pictureThe name of the pictureThe name of the pictureClash Royale CLAN TAG#URR8PPP











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

















  • 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














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

















  • 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












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













$pi(x)$ 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












share|cite|improve this question




share|cite|improve this question








edited Aug 6 at 20:58









gimusi

65.5k73684




65.5k73684









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












  • 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










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























  • $x = y / ln y$ 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

















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





















    Your Answer




    StackExchange.ifUsing("editor", function ()
    return StackExchange.using("mathjaxEditing", function ()
    StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix)
    StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
    );
    );
    , "mathjax-editing");

    StackExchange.ready(function()
    var channelOptions =
    tags: "".split(" "),
    id: "69"
    ;
    initTagRenderer("".split(" "), "".split(" "), channelOptions);

    StackExchange.using("externalEditor", function()
    // Have to fire editor after snippets, if snippets enabled
    if (StackExchange.settings.snippets.snippetsEnabled)
    StackExchange.using("snippets", function()
    createEditor();
    );

    else
    createEditor();

    );

    function createEditor()
    StackExchange.prepareEditor(
    heartbeatType: 'answer',
    convertImagesToLinks: true,
    noModals: false,
    showLowRepImageUploadWarning: true,
    reputationToPostImages: 10,
    bindNavPrevention: true,
    postfix: "",
    noCode: true, onDemand: true,
    discardSelector: ".discard-answer"
    ,immediatelyShowMarkdownHelp:true
    );



    );








     

    draft saved


    draft discarded


















    StackExchange.ready(
    function ()
    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2874297%2finverse-of-fx-fracx-logx%23new-answer', 'question_page');

    );

    Post as a guest

















    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























    • $x = y / ln y$ 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














    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























    • $x = y / ln y$ 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












    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















    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















    share|cite|improve this answer



    share|cite|improve this answer








    edited Aug 6 at 20:59


























    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
















    • $x = y / ln y$ 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










    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

























      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























        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













        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













        share|cite|improve this answer



        share|cite|improve this answer











        answered Aug 6 at 21:02









        cdipaolo

        527211




        527211






















             

            draft saved


            draft discarded


























             


            draft saved


            draft discarded














            StackExchange.ready(
            function ()
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2874297%2finverse-of-fx-fracx-logx%23new-answer', 'question_page');

            );

            Post as a guest
































































            Comments

            Post a Comment

            Popular posts from this blog

            What is the equation of a 3D cone with generalised tilt?

            Image
            Clash Royale CLAN TAG #URR8PPP up vote 2 down vote favorite What is the equation of a 3D cone with generalized tilt? I've noticed that in most equations given to represent a cone, there is no parameter which defines the tilt of the cone in 3D space and that most of them have their point at the origin $(0,0,0)$ - I was wondering if anyone could give me a more generalized cone equation for a cone in any position in 3D space 3d share | cite | improve this question edited Jul 25 '16 at 0:05 ervx 9,425 3 13 36 asked Jul 24 '16 at 15:38 Charlene 11 2 2 Welcome to Math.SE! Could you please explain a few things to help people give an answer useful to you: 1. What do you mean by "generalized tilt"? 2. Are you asking about a right circular cone? Does the angle at the vertex matter? 3. What form of answer (implicit, parametric...?) are you looking for? 4. If this is homework, what tools do you have available, an...
            Read more

            Relationship between determinant of matrix and determinant of adjoint?

            Image
            Clash Royale CLAN TAG #URR8PPP up vote 2 down vote favorite We are studying adjoints in class, and I was curious if there is a relationship between the determinant of matrix A, and the determinant of the adjoint of matrix A? I assume there would be a relationship because finding the adjoint requires creating a cofactor matrix and then transposing it. linear-algebra determinant adjoint-operators share | cite | improve this question asked Nov 17 '17 at 17:32 CluelessCoder 32 5 For a square matrix $A$ of order $n$, we have $A(operatornameadj A)=(operatornameadj A)A=|A|I_n$ where $I_n$ is the identity matrix of order $n$. This should tell you about the determinant of the adjoint in terms of that of $A$ (use multiplicative property and other elementary properties of determinants). – Prasun Biswas Nov 17 '17 at 17:35 1 @CluelessCoder: see here: math.stackexchange.com/questions/516127/…...
            Read more

            Color the edges and diagonals of a regular polygon

            Image
            Clash Royale CLAN TAG #URR8PPP up vote 5 down vote favorite 1 Here is the problem: For what $n$ is it possible to color the edges and diagonals of an $n$-side regular polygon with $dfracbinomn23$ colors, such that you use every color exactly three times and for every color the three segments (edges or diagonals) with that color form a triangle? Trivially $nequiv 0 text or 1 pmod3$. I can also prove that if the statement is true for $k$ than it is true for $3k$ as well. How to finish? Please help! Thanks combinatorial-geometry share | cite | improve this question edited Aug 6 at 21:57 alcana 118 4 asked Aug 6 at 21:15 Leo Gardner 356 11 Even assuming "polyhpn" in the title is a typo for polygon , it is unclear what you mean by "the edges and sides". Aren't the side of a regular polygon the same as the edges? A regular $n$-sided polygon has $n$ edges. – hardmath Aug 6 at 21:20 3 $n$ ne...
            Read more