Loading ...
Sorry, an error occurred while loading the content.
 

Re: Sum of x/log(x) x=2..n

Expand Messages
  • David Broadhurst
    ... f(x,lx) = x/lx - 1/lx + a/lx^2 - b/lx^3; modify at will a=1; b=1; m=2; chosen values p50 terms=10;trunc=10^4;g=f(x,lx);d=[];bv=bernvec(terms+1);
    Message 1 of 8 , Aug 8, 2009
      --- In primenumbers@yahoogroups.com,
      "Cino Hilliard" <hillcino368@...> wrote:

      > I want to examine
      > sm(m,n)=sum(x=m,n,x/log(x)-1/log(x)+a/log(x)^2-b/log(x)^3)
      > a=1,b=1,m=2

      f(x,lx) = x/lx - 1/lx + a/lx^2 - b/lx^3; \\ modify at will
      a=1; b=1; m=2; \\ chosen values

      \p50
      terms=10;trunc=10^4;g=f(x,lx);d=[];bv=bernvec(terms+1);

      {for(n=1,2*terms-1, \\ store odd derivatives
      g=deriv(g,x)+1/x*deriv(g,lx);if(n%2,d=concat(d,g)));}

      fx(x)=f(x,log(x));
      dx(y)=subst(subst(d,x,y),lx,log(y));

      {emac(m,n)=local(dm,dn);
      dm=dx(m);dn=dx(n);intnum(x=m,n,fx(x))+(fx(n)+fx(m))/2
      +sum(k=1,terms,bv[k+1]/((2*k)!)*(dn[k]-dm[k]));}

      s=sum(k=2,trunc-1,fx(k));

      {sm(m,n) = \\ as requested
      if(n>trunc&&trunc>m&&m>1,s-sum(k=2,m-1,fx(k))+emac(trunc,n));}

      for(k=10,22,print([k,sm(m,sqrtint(10^k))]))

      realprecision = 57 significant digits (50 digits displayed)
      [10, 455051173.78013348859945302791104662093667330309988]
      [11, 4118034570.4638268343263084222227560024582402009128]
      [12, 37607913583.517014991610186636613681414323919745603]
      [13, 346065399465.03248684994266161913298076743032790216]
      [14, 3204941752475.7266213978929589359007763992788467339]
      [15, 29844569450368.625194208305114011697818820369920647]
      [16, 279238341514801.87986838501363369093160129071417051]
      [17, 2623557157209703.2566630532132281009952796686474077]
      [18, 24739954285430064.167128134429584525396645626800130]
      [19, 234057667279245072.15948934429135642246573100181670]
      [20, 2220819602565566257.9962162533153260118661266449421]
      [21, 21127269485065195786.472528704950924409440892097567]
      [22, 201467286689267167012.50527750985870920146256019519]

      Comment: This is an inefficient method of fiddling with Li(x).
      Cino is trying to avoid integration. Yet it is needed for
      Euler-Maclaurin summation.

      David
    • Cino Hilliard
      Hi David, You did a lot of work here to develop emac(m,n).I appreciate it as it saved me a lot of time and possibly money. Thank you. ... Maybe. Anyway I don t
      Message 2 of 8 , Aug 8, 2009
        Hi David,
        You did a lot of work here to develop emac(m,n).I appreciate
        it as it saved me a lot of time and possibly money.
        Thank you.

        David Broadhurst wrote:
        > Comment: This is an inefficient method of fiddling with Li(x).

        Maybe. Anyway I don't mind inefficiencies as I have to use them
        everyday like the QWERTY keyboard and all the right handed stuff
        around for this lefty.

        Surprizingly, the most efficient combination in the real world for moving a mass from point A to point B with the least expenditure
        of energy, is a human riding a bicycle on a flat surface. Then
        there is the hummingbird which is one of the the least efficient systems that has been around for a while.

        Nevertheless, Your emac(m,n) gets good mileage and is more
        accurate than Li(x) =-eint1(log(1/x)) and
        li(x,n) = lg=log(x);x*sum(k=1,n,(k-1)!/lg^k)

        gp > p23=1925320391606818006727
        gp > em23=emac(2,sqrt(10^23))
        %116 = 1925320391608063705941.14436225628
        gp > Li23=Li(10^23)
        %117 = 1925320391614054155138.78012956636
        gp > R23=R(10^23)
        %118 = 1925320391607837268776.09905063742
        gp > 1-em23/p23
        %122 = -6.4700878855011600657703221947361 E-13
        gp > 1-Li23/p23
        %123 = -3.7584125963269129092185767109212 E-12
        gp > 1-R23/p23
        %124 = -5.2939866712178918250597351383218 E-13

        For 10^22, emac is better than R(x).

        The actual sum of the primes < sqrt(n) ~ Pi(n) is not as
        accurate as emac but it is quite good wit RE below.

        For sum of primes < 10^n, See the b-file at
        http://www.research.att.com/~njas/sequences/A046731

        sump11 = 201467077743744681014
        gp > p22 = 201467286689315906290
        %137 = 201467286689315906290
        gp > 1. - sump11/p22
        %139 = 0.00000103711..
        This is magnitudes better than RE of 10^22/log(10^22)-1) =
        0.000423348356239

        I find this sum of primes <= sqrt(n) ~ primes <= n to be an
        amazing property of prime numbers and the integers.

        Bottomline, thanks to David, we have an eulermac function in
        Pari.

        Cheers and Roebuck,
        Cino
      • David Broadhurst
        ... Glad to help. ... Not so amazing when you realize that the integrals I1(N) = intnum(x=2, sqrt(N), x/log(x)) I2(N) = intnum(y=2, N , 1/log(y)) differ
        Message 3 of 8 , Aug 9, 2009
          --- In primenumbers@yahoogroups.com,
          "Cino Hilliard" <hillcino368@...> wrote:

          > Hi David,
          > You did a lot of work here to develop emac(m,n).
          > I appreciate it as it saved me a lot of time and
          > possibly money. Thank you.

          Glad to help.

          > I find this sum of primes <= sqrt(n) ~ primes <= n to be
          > an amazing property of prime numbers and the integers.

          Not so amazing when you realize that the integrals

          I1(N) = intnum(x=2, sqrt(N), x/log(x))
          I2(N) = intnum(y=2, N , 1/log(y))

          differ only by a constant, no matter what the value of N.

          Proof: Transform the latter by setting y = x^2. Then
          I2(N) - I1(N) = - I1(2) = 1.92242131492155809316615998937951547...

          David
        Your message has been successfully submitted and would be delivered to recipients shortly.