Looking for infinite series resembling an exponential

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












6












$begingroup$


I'm looking for some $f(x)$ that has the following property:



$sum_x=1^infty f(kx) = r^k$



for some real $0 < r < 1$, and at least for strictly positive integer $k$.



Does such an $f(x)$ exist?



This could also be thought of in terms of some sequence of real numbers $f[n]$.



I posted this at MSE but got no answer, so thought I would try MO.






real-analysis sequences-and-series power-series







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  • $begingroup$
    The m.se post is math.stackexchange.com/questions/3078889/…
    $endgroup$
    – Gerry Myerson
    Jan 23 at 19:59















6












$begingroup$


I'm looking for some $f(x)$ that has the following property:



$sum_x=1^infty f(kx) = r^k$



for some real $0 < r < 1$, and at least for strictly positive integer $k$.



Does such an $f(x)$ exist?



This could also be thought of in terms of some sequence of real numbers $f[n]$.



I posted this at MSE but got no answer, so thought I would try MO.






real-analysis sequences-and-series power-series







share|cite|improve this question









$endgroup$











  • $begingroup$
    The m.se post is math.stackexchange.com/questions/3078889/…
    $endgroup$
    – Gerry Myerson
    Jan 23 at 19:59













6












6








6





$begingroup$


I'm looking for some $f(x)$ that has the following property:



$sum_x=1^infty f(kx) = r^k$



for some real $0 < r < 1$, and at least for strictly positive integer $k$.



Does such an $f(x)$ exist?



This could also be thought of in terms of some sequence of real numbers $f[n]$.



I posted this at MSE but got no answer, so thought I would try MO.






real-analysis sequences-and-series power-series







share|cite|improve this question









$endgroup$




I'm looking for some $f(x)$ that has the following property:



$sum_x=1^infty f(kx) = r^k$



for some real $0 < r < 1$, and at least for strictly positive integer $k$.



Does such an $f(x)$ exist?



This could also be thought of in terms of some sequence of real numbers $f[n]$.



I posted this at MSE but got no answer, so thought I would try MO.






real-analysis sequences-and-series power-series




real-analysis sequences-and-series power-series






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




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asked Jan 22 at 0:26









Mike BattagliaMike Battaglia

1,153623




1,153623











  • $begingroup$
    The m.se post is math.stackexchange.com/questions/3078889/…
    $endgroup$
    – Gerry Myerson
    Jan 23 at 19:59
















  • $begingroup$
    The m.se post is math.stackexchange.com/questions/3078889/…
    $endgroup$
    – Gerry Myerson
    Jan 23 at 19:59















$begingroup$
The m.se post is math.stackexchange.com/questions/3078889/…
$endgroup$
– Gerry Myerson
Jan 23 at 19:59




$begingroup$
The m.se post is math.stackexchange.com/questions/3078889/…
$endgroup$
– Gerry Myerson
Jan 23 at 19:59










1 Answer
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$begingroup$

You can solve this system of equations explicitly in terms of the function
$$F(z)=sum_m=1^infty mu(m)z^m=z-z^2-z^3-z^5+z^6+cdots$$
where $mu(m)$ is the Möbius function.
The function $F(z)$ can be shown to be transcendental (see the paper "Transcendence of Power Series for Some Number Theoretic Functions", by Coons and Borwein). With this notation your sequence can be succinctly expressed as
$$f(n)=F(r^n).$$
Indeed, one can check that
$$sum_n=1^infty f(kn)=sum_n=1^infty F(r^kn)=sum_m,n=1^infty mu(m)r^nmk=sum_m=1^inftysum_mmu(d)r^mk=r^k$$
as desired.






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    1 Answer
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    active

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    1 Answer
    1






    active

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    active

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    14












    $begingroup$

    You can solve this system of equations explicitly in terms of the function
    $$F(z)=sum_m=1^infty mu(m)z^m=z-z^2-z^3-z^5+z^6+cdots$$
    where $mu(m)$ is the Möbius function.
    The function $F(z)$ can be shown to be transcendental (see the paper "Transcendence of Power Series for Some Number Theoretic Functions", by Coons and Borwein). With this notation your sequence can be succinctly expressed as
    $$f(n)=F(r^n).$$
    Indeed, one can check that
    $$sum_n=1^infty f(kn)=sum_n=1^infty F(r^kn)=sum_m,n=1^infty mu(m)r^nmk=sum_m=1^inftysum_mmu(d)r^mk=r^k$$
    as desired.






    share|cite|improve this answer











    $endgroup$

















      14












      $begingroup$

      You can solve this system of equations explicitly in terms of the function
      $$F(z)=sum_m=1^infty mu(m)z^m=z-z^2-z^3-z^5+z^6+cdots$$
      where $mu(m)$ is the Möbius function.
      The function $F(z)$ can be shown to be transcendental (see the paper "Transcendence of Power Series for Some Number Theoretic Functions", by Coons and Borwein). With this notation your sequence can be succinctly expressed as
      $$f(n)=F(r^n).$$
      Indeed, one can check that
      $$sum_n=1^infty f(kn)=sum_n=1^infty F(r^kn)=sum_m,n=1^infty mu(m)r^nmk=sum_m=1^inftysum_mmu(d)r^mk=r^k$$
      as desired.






      share|cite|improve this answer











      $endgroup$















        14












        14








        14





        $begingroup$

        You can solve this system of equations explicitly in terms of the function
        $$F(z)=sum_m=1^infty mu(m)z^m=z-z^2-z^3-z^5+z^6+cdots$$
        where $mu(m)$ is the Möbius function.
        The function $F(z)$ can be shown to be transcendental (see the paper "Transcendence of Power Series for Some Number Theoretic Functions", by Coons and Borwein). With this notation your sequence can be succinctly expressed as
        $$f(n)=F(r^n).$$
        Indeed, one can check that
        $$sum_n=1^infty f(kn)=sum_n=1^infty F(r^kn)=sum_m,n=1^infty mu(m)r^nmk=sum_m=1^inftysum_mmu(d)r^mk=r^k$$
        as desired.






        share|cite|improve this answer











        $endgroup$



        You can solve this system of equations explicitly in terms of the function
        $$F(z)=sum_m=1^infty mu(m)z^m=z-z^2-z^3-z^5+z^6+cdots$$
        where $mu(m)$ is the Möbius function.
        The function $F(z)$ can be shown to be transcendental (see the paper "Transcendence of Power Series for Some Number Theoretic Functions", by Coons and Borwein). With this notation your sequence can be succinctly expressed as
        $$f(n)=F(r^n).$$
        Indeed, one can check that
        $$sum_n=1^infty f(kn)=sum_n=1^infty F(r^kn)=sum_m,n=1^infty mu(m)r^nmk=sum_m=1^inftysum_mmu(d)r^mk=r^k$$
        as desired.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Jan 23 at 17:58

























        answered Jan 22 at 1:00









        Gjergji ZaimiGjergji Zaimi

        63.2k4165313




        63.2k4165313



























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