Fractional part distribution












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It is known that the distribution of ${sqrt{n} }$, evaluated over the integer values of $n$, is uniform in the interval $[0,1)$. Let us consider the sum $$S(K)=sum_{n=1}^K left({sqrt{n}}-frac{1}{2} right) sqrt{n}$$ I noted that the average value of $sum_{K=1}^N S(K)$, calculated over all integers $N$ and for $N rightarrow infty$, converges to $1/4$. I suppose that this asymmetric distribution of $S(K)$ with respect to zero dipends on some properties of the distribution of $sqrt{n} mod 1$, but would be happy to better understand it by a formal proof.






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    $begingroup$


    It is known that the distribution of ${sqrt{n} }$, evaluated over the integer values of $n$, is uniform in the interval $[0,1)$. Let us consider the sum $$S(K)=sum_{n=1}^K left({sqrt{n}}-frac{1}{2} right) sqrt{n}$$ I noted that the average value of $sum_{K=1}^N S(K)$, calculated over all integers $N$ and for $N rightarrow infty$, converges to $1/4$. I suppose that this asymmetric distribution of $S(K)$ with respect to zero dipends on some properties of the distribution of $sqrt{n} mod 1$, but would be happy to better understand it by a formal proof.






    number-theory summation fractional-part







    share|cite|improve this question











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      2








      2


      1



      $begingroup$


      It is known that the distribution of ${sqrt{n} }$, evaluated over the integer values of $n$, is uniform in the interval $[0,1)$. Let us consider the sum $$S(K)=sum_{n=1}^K left({sqrt{n}}-frac{1}{2} right) sqrt{n}$$ I noted that the average value of $sum_{K=1}^N S(K)$, calculated over all integers $N$ and for $N rightarrow infty$, converges to $1/4$. I suppose that this asymmetric distribution of $S(K)$ with respect to zero dipends on some properties of the distribution of $sqrt{n} mod 1$, but would be happy to better understand it by a formal proof.






      number-theory summation fractional-part







      share|cite|improve this question











      $endgroup$




      It is known that the distribution of ${sqrt{n} }$, evaluated over the integer values of $n$, is uniform in the interval $[0,1)$. Let us consider the sum $$S(K)=sum_{n=1}^K left({sqrt{n}}-frac{1}{2} right) sqrt{n}$$ I noted that the average value of $sum_{K=1}^N S(K)$, calculated over all integers $N$ and for $N rightarrow infty$, converges to $1/4$. I suppose that this asymmetric distribution of $S(K)$ with respect to zero dipends on some properties of the distribution of $sqrt{n} mod 1$, but would be happy to better understand it by a formal proof.






      number-theory summation fractional-part




      number-theory summation fractional-part






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      edited Dec 7 '18 at 23:31







      Anatoly

















      asked Dec 6 '18 at 15:02









      AnatolyAnatoly

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          Intuitively, we can consider the range of $n$ from $k^2$ to $(k+1)^2-1=k^2+2k$. Over this range $lfloor sqrt n rfloor=k,$ so ${sqrt n}=sqrt n-k$. As $(k=frac 12)^2=k^2+k+frac 14$ the terms from $k^2$ through $k^2+k$ have ${sqrt n}-frac 12lt 0$ and all the terms from $k^2+k+1$ through $k^2+2k$ have ${sqrt n}-frac 12gt 0$. The positive ones get to multiply the larger $sqrt n$'s. On the other hand, there is one less positive term.



          Here is a failed attempt: We can write your sum with the limit of $K to infty$ as
          $$S=sum_{n=1}^infty left({sqrt{n}}-frac{1}{2} right) sqrt{n}
          =sum_{m=1}^inftysum_{n=1}^{2m}left(sqrt{m^2+n}-m-frac 12right)sqrt{m^2+n}$$
          where we have broken the sum into pieces with each integer part of the square root. Designating each term in the outer sum as $S_m$ we have
          $$S_m=sum_{n=0}^{2m}left(sqrt{m^2+n}-m-frac 12right)sqrt{m^2+n}\
          =sum_{n=0}^{2m}(m^2+n)-(m+frac 12)sqrt{m^2+n}\
          =2m^3+frac 12(2m)(2m+1)-sum_{n=0}^{2m}(m+frac 12)sqrt{m^2+n}$$
          I tried turning the sum into an integral, but it is not accurate enough.






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            $begingroup$

            Intuitively, we can consider the range of $n$ from $k^2$ to $(k+1)^2-1=k^2+2k$. Over this range $lfloor sqrt n rfloor=k,$ so ${sqrt n}=sqrt n-k$. As $(k=frac 12)^2=k^2+k+frac 14$ the terms from $k^2$ through $k^2+k$ have ${sqrt n}-frac 12lt 0$ and all the terms from $k^2+k+1$ through $k^2+2k$ have ${sqrt n}-frac 12gt 0$. The positive ones get to multiply the larger $sqrt n$'s. On the other hand, there is one less positive term.



            Here is a failed attempt: We can write your sum with the limit of $K to infty$ as
            $$S=sum_{n=1}^infty left({sqrt{n}}-frac{1}{2} right) sqrt{n}
            =sum_{m=1}^inftysum_{n=1}^{2m}left(sqrt{m^2+n}-m-frac 12right)sqrt{m^2+n}$$
            where we have broken the sum into pieces with each integer part of the square root. Designating each term in the outer sum as $S_m$ we have
            $$S_m=sum_{n=0}^{2m}left(sqrt{m^2+n}-m-frac 12right)sqrt{m^2+n}\
            =sum_{n=0}^{2m}(m^2+n)-(m+frac 12)sqrt{m^2+n}\
            =2m^3+frac 12(2m)(2m+1)-sum_{n=0}^{2m}(m+frac 12)sqrt{m^2+n}$$
            I tried turning the sum into an integral, but it is not accurate enough.






            share|cite|improve this answer









            $endgroup$


















              0












              $begingroup$

              Intuitively, we can consider the range of $n$ from $k^2$ to $(k+1)^2-1=k^2+2k$. Over this range $lfloor sqrt n rfloor=k,$ so ${sqrt n}=sqrt n-k$. As $(k=frac 12)^2=k^2+k+frac 14$ the terms from $k^2$ through $k^2+k$ have ${sqrt n}-frac 12lt 0$ and all the terms from $k^2+k+1$ through $k^2+2k$ have ${sqrt n}-frac 12gt 0$. The positive ones get to multiply the larger $sqrt n$'s. On the other hand, there is one less positive term.



              Here is a failed attempt: We can write your sum with the limit of $K to infty$ as
              $$S=sum_{n=1}^infty left({sqrt{n}}-frac{1}{2} right) sqrt{n}
              =sum_{m=1}^inftysum_{n=1}^{2m}left(sqrt{m^2+n}-m-frac 12right)sqrt{m^2+n}$$
              where we have broken the sum into pieces with each integer part of the square root. Designating each term in the outer sum as $S_m$ we have
              $$S_m=sum_{n=0}^{2m}left(sqrt{m^2+n}-m-frac 12right)sqrt{m^2+n}\
              =sum_{n=0}^{2m}(m^2+n)-(m+frac 12)sqrt{m^2+n}\
              =2m^3+frac 12(2m)(2m+1)-sum_{n=0}^{2m}(m+frac 12)sqrt{m^2+n}$$
              I tried turning the sum into an integral, but it is not accurate enough.






              share|cite|improve this answer









              $endgroup$
















                0












                0








                0





                $begingroup$

                Intuitively, we can consider the range of $n$ from $k^2$ to $(k+1)^2-1=k^2+2k$. Over this range $lfloor sqrt n rfloor=k,$ so ${sqrt n}=sqrt n-k$. As $(k=frac 12)^2=k^2+k+frac 14$ the terms from $k^2$ through $k^2+k$ have ${sqrt n}-frac 12lt 0$ and all the terms from $k^2+k+1$ through $k^2+2k$ have ${sqrt n}-frac 12gt 0$. The positive ones get to multiply the larger $sqrt n$'s. On the other hand, there is one less positive term.



                Here is a failed attempt: We can write your sum with the limit of $K to infty$ as
                $$S=sum_{n=1}^infty left({sqrt{n}}-frac{1}{2} right) sqrt{n}
                =sum_{m=1}^inftysum_{n=1}^{2m}left(sqrt{m^2+n}-m-frac 12right)sqrt{m^2+n}$$
                where we have broken the sum into pieces with each integer part of the square root. Designating each term in the outer sum as $S_m$ we have
                $$S_m=sum_{n=0}^{2m}left(sqrt{m^2+n}-m-frac 12right)sqrt{m^2+n}\
                =sum_{n=0}^{2m}(m^2+n)-(m+frac 12)sqrt{m^2+n}\
                =2m^3+frac 12(2m)(2m+1)-sum_{n=0}^{2m}(m+frac 12)sqrt{m^2+n}$$
                I tried turning the sum into an integral, but it is not accurate enough.






                share|cite|improve this answer









                $endgroup$



                Intuitively, we can consider the range of $n$ from $k^2$ to $(k+1)^2-1=k^2+2k$. Over this range $lfloor sqrt n rfloor=k,$ so ${sqrt n}=sqrt n-k$. As $(k=frac 12)^2=k^2+k+frac 14$ the terms from $k^2$ through $k^2+k$ have ${sqrt n}-frac 12lt 0$ and all the terms from $k^2+k+1$ through $k^2+2k$ have ${sqrt n}-frac 12gt 0$. The positive ones get to multiply the larger $sqrt n$'s. On the other hand, there is one less positive term.



                Here is a failed attempt: We can write your sum with the limit of $K to infty$ as
                $$S=sum_{n=1}^infty left({sqrt{n}}-frac{1}{2} right) sqrt{n}
                =sum_{m=1}^inftysum_{n=1}^{2m}left(sqrt{m^2+n}-m-frac 12right)sqrt{m^2+n}$$
                where we have broken the sum into pieces with each integer part of the square root. Designating each term in the outer sum as $S_m$ we have
                $$S_m=sum_{n=0}^{2m}left(sqrt{m^2+n}-m-frac 12right)sqrt{m^2+n}\
                =sum_{n=0}^{2m}(m^2+n)-(m+frac 12)sqrt{m^2+n}\
                =2m^3+frac 12(2m)(2m+1)-sum_{n=0}^{2m}(m+frac 12)sqrt{m^2+n}$$
                I tried turning the sum into an integral, but it is not accurate enough.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Dec 6 '18 at 22:06









                Ross MillikanRoss Millikan

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                294k23198371






























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