Looking for infinite series resembling an exponential












3












$begingroup$


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









$endgroup$

















    3












    $begingroup$


    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









    $endgroup$















      3












      3








      3





      $begingroup$


      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




      share|cite|improve this question










      asked 3 hours ago









      Mike BattagliaMike Battaglia

      1,036523




      1,036523






















          1 Answer
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          $begingroup$

          You can solve this system of equations explicitly in terms of the function
          $$F(z)=sum_{m=1}^{infty} mu(m)z^m=z-z^2-z^3-z^5+z^6+cdots$$
          which 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
          $$sum_{n=1}^{infty} f(kn)=sum_{n=1}^{infty} F(r^{kn})=sum_{m,n=1}^{infty} mu(m)r^{nmk}=sum_{m=1}^{infty}sum_{d|m}mu(d)r^{mk}=r^k$$
          as desired.






          share|cite|improve this answer









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            $begingroup$

            You can solve this system of equations explicitly in terms of the function
            $$F(z)=sum_{m=1}^{infty} mu(m)z^m=z-z^2-z^3-z^5+z^6+cdots$$
            which 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
            $$sum_{n=1}^{infty} f(kn)=sum_{n=1}^{infty} F(r^{kn})=sum_{m,n=1}^{infty} mu(m)r^{nmk}=sum_{m=1}^{infty}sum_{d|m}mu(d)r^{mk}=r^k$$
            as desired.






            share|cite|improve this answer









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              4












              $begingroup$

              You can solve this system of equations explicitly in terms of the function
              $$F(z)=sum_{m=1}^{infty} mu(m)z^m=z-z^2-z^3-z^5+z^6+cdots$$
              which 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
              $$sum_{n=1}^{infty} f(kn)=sum_{n=1}^{infty} F(r^{kn})=sum_{m,n=1}^{infty} mu(m)r^{nmk}=sum_{m=1}^{infty}sum_{d|m}mu(d)r^{mk}=r^k$$
              as desired.






              share|cite|improve this answer









              $endgroup$
















                4












                4








                4





                $begingroup$

                You can solve this system of equations explicitly in terms of the function
                $$F(z)=sum_{m=1}^{infty} mu(m)z^m=z-z^2-z^3-z^5+z^6+cdots$$
                which 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
                $$sum_{n=1}^{infty} f(kn)=sum_{n=1}^{infty} F(r^{kn})=sum_{m,n=1}^{infty} mu(m)r^{nmk}=sum_{m=1}^{infty}sum_{d|m}mu(d)r^{mk}=r^k$$
                as desired.






                share|cite|improve this answer









                $endgroup$



                You can solve this system of equations explicitly in terms of the function
                $$F(z)=sum_{m=1}^{infty} mu(m)z^m=z-z^2-z^3-z^5+z^6+cdots$$
                which 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
                $$sum_{n=1}^{infty} f(kn)=sum_{n=1}^{infty} F(r^{kn})=sum_{m,n=1}^{infty} mu(m)r^{nmk}=sum_{m=1}^{infty}sum_{d|m}mu(d)r^{mk}=r^k$$
                as desired.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered 3 hours ago









                Gjergji ZaimiGjergji Zaimi

                62.8k4163310




                62.8k4163310






























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