Closed form for $sum_{n=1}^infty log(n) * x^n$












0












$begingroup$


As in the title, I'm in
quest for $sum_{n=1}^infty log(n)cdot x^n$, where $0 le x lt 1$



Wolfram Alpha says: $-operatorname{PolyLog}^{(1, 0)}(0, x)$, but I don't understand what that means. (Of course, PolyLog stays for the polylogarithm).



Background



It's about "how many bits I need to encode a real number $0 < r < 1$ with a tolerance $delta/2$? The "naive" response is $-log_2(delta)$.



Nevertheless (long story short) I need a different approach:




  1. I can encode every positive integer $n$ with approximately $Ccdotlog(n)$ bits


  2. Let $0 < x_i < 1$ be a pseudo-random sequence, and let $N$ be the 1st index so that $r-delta/2 <x_N< r+delta/2$.
    Then let's say that we can transmit $r$ via $N$ (with the tolerance $delta$). So we need $Ccdotlog(N)$ bits...


  3. But then I need the expected value
    $E(Ccdotlog(N)) = sum_{n=1}^infty Ccdotlog(n)cdotdeltacdot(1-delta)^{n-1}$
    $=Ccdot{deltaover1-delta} sum_{n=1}^infty log(n) cdot (1-delta)^n$







sequences-and-series summation logarithms







share|cite|improve this question











$endgroup$












  • $begingroup$
    When you say "log" do you mean log base-10 or natural logarithm?
    $endgroup$
    – R. Burton
    Dec 9 '18 at 18:01










  • $begingroup$
    I suspect it has no closed form better than that. What do you need this for?
    $endgroup$
    – Ethan Bolker
    Dec 9 '18 at 18:01










  • $begingroup$
    Please edit the question to include a link to the Wolfram alpha output.
    $endgroup$
    – Shaun
    Dec 9 '18 at 18:11










  • $begingroup$
    @R.Burton natural logarithm
    $endgroup$
    – giuliolunati
    Dec 9 '18 at 19:19










  • $begingroup$
    @EthanBolker sorry, too long to explain now... maybe I'll edit the question.
    $endgroup$
    – giuliolunati
    Dec 9 '18 at 19:30
















0












$begingroup$


As in the title, I'm in
quest for $sum_{n=1}^infty log(n)cdot x^n$, where $0 le x lt 1$



Wolfram Alpha says: $-operatorname{PolyLog}^{(1, 0)}(0, x)$, but I don't understand what that means. (Of course, PolyLog stays for the polylogarithm).



Background



It's about "how many bits I need to encode a real number $0 < r < 1$ with a tolerance $delta/2$? The "naive" response is $-log_2(delta)$.



Nevertheless (long story short) I need a different approach:




  1. I can encode every positive integer $n$ with approximately $Ccdotlog(n)$ bits


  2. Let $0 < x_i < 1$ be a pseudo-random sequence, and let $N$ be the 1st index so that $r-delta/2 <x_N< r+delta/2$.
    Then let's say that we can transmit $r$ via $N$ (with the tolerance $delta$). So we need $Ccdotlog(N)$ bits...


  3. But then I need the expected value
    $E(Ccdotlog(N)) = sum_{n=1}^infty Ccdotlog(n)cdotdeltacdot(1-delta)^{n-1}$
    $=Ccdot{deltaover1-delta} sum_{n=1}^infty log(n) cdot (1-delta)^n$







sequences-and-series summation logarithms







share|cite|improve this question











$endgroup$












  • $begingroup$
    When you say "log" do you mean log base-10 or natural logarithm?
    $endgroup$
    – R. Burton
    Dec 9 '18 at 18:01










  • $begingroup$
    I suspect it has no closed form better than that. What do you need this for?
    $endgroup$
    – Ethan Bolker
    Dec 9 '18 at 18:01










  • $begingroup$
    Please edit the question to include a link to the Wolfram alpha output.
    $endgroup$
    – Shaun
    Dec 9 '18 at 18:11










  • $begingroup$
    @R.Burton natural logarithm
    $endgroup$
    – giuliolunati
    Dec 9 '18 at 19:19










  • $begingroup$
    @EthanBolker sorry, too long to explain now... maybe I'll edit the question.
    $endgroup$
    – giuliolunati
    Dec 9 '18 at 19:30














0












0








0





$begingroup$


As in the title, I'm in
quest for $sum_{n=1}^infty log(n)cdot x^n$, where $0 le x lt 1$



Wolfram Alpha says: $-operatorname{PolyLog}^{(1, 0)}(0, x)$, but I don't understand what that means. (Of course, PolyLog stays for the polylogarithm).



Background



It's about "how many bits I need to encode a real number $0 < r < 1$ with a tolerance $delta/2$? The "naive" response is $-log_2(delta)$.



Nevertheless (long story short) I need a different approach:




  1. I can encode every positive integer $n$ with approximately $Ccdotlog(n)$ bits


  2. Let $0 < x_i < 1$ be a pseudo-random sequence, and let $N$ be the 1st index so that $r-delta/2 <x_N< r+delta/2$.
    Then let's say that we can transmit $r$ via $N$ (with the tolerance $delta$). So we need $Ccdotlog(N)$ bits...


  3. But then I need the expected value
    $E(Ccdotlog(N)) = sum_{n=1}^infty Ccdotlog(n)cdotdeltacdot(1-delta)^{n-1}$
    $=Ccdot{deltaover1-delta} sum_{n=1}^infty log(n) cdot (1-delta)^n$







sequences-and-series summation logarithms







share|cite|improve this question











$endgroup$




As in the title, I'm in
quest for $sum_{n=1}^infty log(n)cdot x^n$, where $0 le x lt 1$



Wolfram Alpha says: $-operatorname{PolyLog}^{(1, 0)}(0, x)$, but I don't understand what that means. (Of course, PolyLog stays for the polylogarithm).



Background



It's about "how many bits I need to encode a real number $0 < r < 1$ with a tolerance $delta/2$? The "naive" response is $-log_2(delta)$.



Nevertheless (long story short) I need a different approach:




  1. I can encode every positive integer $n$ with approximately $Ccdotlog(n)$ bits


  2. Let $0 < x_i < 1$ be a pseudo-random sequence, and let $N$ be the 1st index so that $r-delta/2 <x_N< r+delta/2$.
    Then let's say that we can transmit $r$ via $N$ (with the tolerance $delta$). So we need $Ccdotlog(N)$ bits...


  3. But then I need the expected value
    $E(Ccdotlog(N)) = sum_{n=1}^infty Ccdotlog(n)cdotdeltacdot(1-delta)^{n-1}$
    $=Ccdot{deltaover1-delta} sum_{n=1}^infty log(n) cdot (1-delta)^n$







sequences-and-series summation logarithms




sequences-and-series summation logarithms






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 9 '18 at 22:49







giuliolunati

















asked Dec 9 '18 at 17:58









giuliolunatigiuliolunati

1013




1013












  • $begingroup$
    When you say "log" do you mean log base-10 or natural logarithm?
    $endgroup$
    – R. Burton
    Dec 9 '18 at 18:01










  • $begingroup$
    I suspect it has no closed form better than that. What do you need this for?
    $endgroup$
    – Ethan Bolker
    Dec 9 '18 at 18:01










  • $begingroup$
    Please edit the question to include a link to the Wolfram alpha output.
    $endgroup$
    – Shaun
    Dec 9 '18 at 18:11










  • $begingroup$
    @R.Burton natural logarithm
    $endgroup$
    – giuliolunati
    Dec 9 '18 at 19:19










  • $begingroup$
    @EthanBolker sorry, too long to explain now... maybe I'll edit the question.
    $endgroup$
    – giuliolunati
    Dec 9 '18 at 19:30


















  • $begingroup$
    When you say "log" do you mean log base-10 or natural logarithm?
    $endgroup$
    – R. Burton
    Dec 9 '18 at 18:01










  • $begingroup$
    I suspect it has no closed form better than that. What do you need this for?
    $endgroup$
    – Ethan Bolker
    Dec 9 '18 at 18:01










  • $begingroup$
    Please edit the question to include a link to the Wolfram alpha output.
    $endgroup$
    – Shaun
    Dec 9 '18 at 18:11










  • $begingroup$
    @R.Burton natural logarithm
    $endgroup$
    – giuliolunati
    Dec 9 '18 at 19:19










  • $begingroup$
    @EthanBolker sorry, too long to explain now... maybe I'll edit the question.
    $endgroup$
    – giuliolunati
    Dec 9 '18 at 19:30
















$begingroup$
When you say "log" do you mean log base-10 or natural logarithm?
$endgroup$
– R. Burton
Dec 9 '18 at 18:01




$begingroup$
When you say "log" do you mean log base-10 or natural logarithm?
$endgroup$
– R. Burton
Dec 9 '18 at 18:01












$begingroup$
I suspect it has no closed form better than that. What do you need this for?
$endgroup$
– Ethan Bolker
Dec 9 '18 at 18:01




$begingroup$
I suspect it has no closed form better than that. What do you need this for?
$endgroup$
– Ethan Bolker
Dec 9 '18 at 18:01












$begingroup$
Please edit the question to include a link to the Wolfram alpha output.
$endgroup$
– Shaun
Dec 9 '18 at 18:11




$begingroup$
Please edit the question to include a link to the Wolfram alpha output.
$endgroup$
– Shaun
Dec 9 '18 at 18:11












$begingroup$
@R.Burton natural logarithm
$endgroup$
– giuliolunati
Dec 9 '18 at 19:19




$begingroup$
@R.Burton natural logarithm
$endgroup$
– giuliolunati
Dec 9 '18 at 19:19












$begingroup$
@EthanBolker sorry, too long to explain now... maybe I'll edit the question.
$endgroup$
– giuliolunati
Dec 9 '18 at 19:30




$begingroup$
@EthanBolker sorry, too long to explain now... maybe I'll edit the question.
$endgroup$
– giuliolunati
Dec 9 '18 at 19:30










2 Answers
2






active

oldest

votes


















1












$begingroup$

Definition of Polylogarithm:
http://mathworld.wolfram.com/Polylogarithm.html



No closed form exists in terms of elementary functions (addition, multiplication, powers, etc.), at least not in terms of real functions. You might be able to write it as a complex-valued function or improper integral.



Given that the polylogarithm is already a special function, I suspect that any closed form will be in terms of special functions rather than something nice.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Yes, I read that, so I'd understand what $PolyLog(0,q)$ means. What I don't understand is the "exponent" $^{(1,0)}$
    $endgroup$
    – giuliolunati
    Dec 9 '18 at 19:38












  • $begingroup$
    After working on it for a bit, I don't think that your sum has anything to do with polylogarithms at all. I have no idea what "$-PolyLog^{(1,0)}(0,x)$" is supposed to mean; it might be an error. Either way, The closest I can get to approximating the curve without spending more time on it is approximately $frac{x^2}{1.2-x^2}$.
    $endgroup$
    – R. Burton
    Dec 9 '18 at 21:32










  • $begingroup$
    Thank you for spending time on that! Where that approximation come from?
    $endgroup$
    – giuliolunati
    Dec 9 '18 at 22:09












  • $begingroup$
    Wild guess based on the form of the graph (which is similar visually similar to to $frac{x^2}{(a-x)^2}$, then adjusting $a$ until I got as close to $sum_{n=1}^{1000 }log(n)x^n$ as possible.
    $endgroup$
    – R. Burton
    Dec 9 '18 at 23:30





















0












$begingroup$

Ok, I found here what the notation $f^{(1,0)}$ means in Wolfram Alpha: it's the derivative wrt the 1st variable.



So the response is $-frac{partial operatorname{Li}(s,t)}{partial s}bigg|_{(s=0, t=x)}$






share|cite|improve this answer









$endgroup$













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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    1












    $begingroup$

    Definition of Polylogarithm:
    http://mathworld.wolfram.com/Polylogarithm.html



    No closed form exists in terms of elementary functions (addition, multiplication, powers, etc.), at least not in terms of real functions. You might be able to write it as a complex-valued function or improper integral.



    Given that the polylogarithm is already a special function, I suspect that any closed form will be in terms of special functions rather than something nice.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Yes, I read that, so I'd understand what $PolyLog(0,q)$ means. What I don't understand is the "exponent" $^{(1,0)}$
      $endgroup$
      – giuliolunati
      Dec 9 '18 at 19:38












    • $begingroup$
      After working on it for a bit, I don't think that your sum has anything to do with polylogarithms at all. I have no idea what "$-PolyLog^{(1,0)}(0,x)$" is supposed to mean; it might be an error. Either way, The closest I can get to approximating the curve without spending more time on it is approximately $frac{x^2}{1.2-x^2}$.
      $endgroup$
      – R. Burton
      Dec 9 '18 at 21:32










    • $begingroup$
      Thank you for spending time on that! Where that approximation come from?
      $endgroup$
      – giuliolunati
      Dec 9 '18 at 22:09












    • $begingroup$
      Wild guess based on the form of the graph (which is similar visually similar to to $frac{x^2}{(a-x)^2}$, then adjusting $a$ until I got as close to $sum_{n=1}^{1000 }log(n)x^n$ as possible.
      $endgroup$
      – R. Burton
      Dec 9 '18 at 23:30


















    1












    $begingroup$

    Definition of Polylogarithm:
    http://mathworld.wolfram.com/Polylogarithm.html



    No closed form exists in terms of elementary functions (addition, multiplication, powers, etc.), at least not in terms of real functions. You might be able to write it as a complex-valued function or improper integral.



    Given that the polylogarithm is already a special function, I suspect that any closed form will be in terms of special functions rather than something nice.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Yes, I read that, so I'd understand what $PolyLog(0,q)$ means. What I don't understand is the "exponent" $^{(1,0)}$
      $endgroup$
      – giuliolunati
      Dec 9 '18 at 19:38












    • $begingroup$
      After working on it for a bit, I don't think that your sum has anything to do with polylogarithms at all. I have no idea what "$-PolyLog^{(1,0)}(0,x)$" is supposed to mean; it might be an error. Either way, The closest I can get to approximating the curve without spending more time on it is approximately $frac{x^2}{1.2-x^2}$.
      $endgroup$
      – R. Burton
      Dec 9 '18 at 21:32










    • $begingroup$
      Thank you for spending time on that! Where that approximation come from?
      $endgroup$
      – giuliolunati
      Dec 9 '18 at 22:09












    • $begingroup$
      Wild guess based on the form of the graph (which is similar visually similar to to $frac{x^2}{(a-x)^2}$, then adjusting $a$ until I got as close to $sum_{n=1}^{1000 }log(n)x^n$ as possible.
      $endgroup$
      – R. Burton
      Dec 9 '18 at 23:30
















    1












    1








    1





    $begingroup$

    Definition of Polylogarithm:
    http://mathworld.wolfram.com/Polylogarithm.html



    No closed form exists in terms of elementary functions (addition, multiplication, powers, etc.), at least not in terms of real functions. You might be able to write it as a complex-valued function or improper integral.



    Given that the polylogarithm is already a special function, I suspect that any closed form will be in terms of special functions rather than something nice.






    share|cite|improve this answer









    $endgroup$



    Definition of Polylogarithm:
    http://mathworld.wolfram.com/Polylogarithm.html



    No closed form exists in terms of elementary functions (addition, multiplication, powers, etc.), at least not in terms of real functions. You might be able to write it as a complex-valued function or improper integral.



    Given that the polylogarithm is already a special function, I suspect that any closed form will be in terms of special functions rather than something nice.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Dec 9 '18 at 18:09









    R. BurtonR. Burton

    45619




    45619












    • $begingroup$
      Yes, I read that, so I'd understand what $PolyLog(0,q)$ means. What I don't understand is the "exponent" $^{(1,0)}$
      $endgroup$
      – giuliolunati
      Dec 9 '18 at 19:38












    • $begingroup$
      After working on it for a bit, I don't think that your sum has anything to do with polylogarithms at all. I have no idea what "$-PolyLog^{(1,0)}(0,x)$" is supposed to mean; it might be an error. Either way, The closest I can get to approximating the curve without spending more time on it is approximately $frac{x^2}{1.2-x^2}$.
      $endgroup$
      – R. Burton
      Dec 9 '18 at 21:32










    • $begingroup$
      Thank you for spending time on that! Where that approximation come from?
      $endgroup$
      – giuliolunati
      Dec 9 '18 at 22:09












    • $begingroup$
      Wild guess based on the form of the graph (which is similar visually similar to to $frac{x^2}{(a-x)^2}$, then adjusting $a$ until I got as close to $sum_{n=1}^{1000 }log(n)x^n$ as possible.
      $endgroup$
      – R. Burton
      Dec 9 '18 at 23:30




















    • $begingroup$
      Yes, I read that, so I'd understand what $PolyLog(0,q)$ means. What I don't understand is the "exponent" $^{(1,0)}$
      $endgroup$
      – giuliolunati
      Dec 9 '18 at 19:38












    • $begingroup$
      After working on it for a bit, I don't think that your sum has anything to do with polylogarithms at all. I have no idea what "$-PolyLog^{(1,0)}(0,x)$" is supposed to mean; it might be an error. Either way, The closest I can get to approximating the curve without spending more time on it is approximately $frac{x^2}{1.2-x^2}$.
      $endgroup$
      – R. Burton
      Dec 9 '18 at 21:32










    • $begingroup$
      Thank you for spending time on that! Where that approximation come from?
      $endgroup$
      – giuliolunati
      Dec 9 '18 at 22:09












    • $begingroup$
      Wild guess based on the form of the graph (which is similar visually similar to to $frac{x^2}{(a-x)^2}$, then adjusting $a$ until I got as close to $sum_{n=1}^{1000 }log(n)x^n$ as possible.
      $endgroup$
      – R. Burton
      Dec 9 '18 at 23:30


















    $begingroup$
    Yes, I read that, so I'd understand what $PolyLog(0,q)$ means. What I don't understand is the "exponent" $^{(1,0)}$
    $endgroup$
    – giuliolunati
    Dec 9 '18 at 19:38






    $begingroup$
    Yes, I read that, so I'd understand what $PolyLog(0,q)$ means. What I don't understand is the "exponent" $^{(1,0)}$
    $endgroup$
    – giuliolunati
    Dec 9 '18 at 19:38














    $begingroup$
    After working on it for a bit, I don't think that your sum has anything to do with polylogarithms at all. I have no idea what "$-PolyLog^{(1,0)}(0,x)$" is supposed to mean; it might be an error. Either way, The closest I can get to approximating the curve without spending more time on it is approximately $frac{x^2}{1.2-x^2}$.
    $endgroup$
    – R. Burton
    Dec 9 '18 at 21:32




    $begingroup$
    After working on it for a bit, I don't think that your sum has anything to do with polylogarithms at all. I have no idea what "$-PolyLog^{(1,0)}(0,x)$" is supposed to mean; it might be an error. Either way, The closest I can get to approximating the curve without spending more time on it is approximately $frac{x^2}{1.2-x^2}$.
    $endgroup$
    – R. Burton
    Dec 9 '18 at 21:32












    $begingroup$
    Thank you for spending time on that! Where that approximation come from?
    $endgroup$
    – giuliolunati
    Dec 9 '18 at 22:09






    $begingroup$
    Thank you for spending time on that! Where that approximation come from?
    $endgroup$
    – giuliolunati
    Dec 9 '18 at 22:09














    $begingroup$
    Wild guess based on the form of the graph (which is similar visually similar to to $frac{x^2}{(a-x)^2}$, then adjusting $a$ until I got as close to $sum_{n=1}^{1000 }log(n)x^n$ as possible.
    $endgroup$
    – R. Burton
    Dec 9 '18 at 23:30






    $begingroup$
    Wild guess based on the form of the graph (which is similar visually similar to to $frac{x^2}{(a-x)^2}$, then adjusting $a$ until I got as close to $sum_{n=1}^{1000 }log(n)x^n$ as possible.
    $endgroup$
    – R. Burton
    Dec 9 '18 at 23:30













    0












    $begingroup$

    Ok, I found here what the notation $f^{(1,0)}$ means in Wolfram Alpha: it's the derivative wrt the 1st variable.



    So the response is $-frac{partial operatorname{Li}(s,t)}{partial s}bigg|_{(s=0, t=x)}$






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      Ok, I found here what the notation $f^{(1,0)}$ means in Wolfram Alpha: it's the derivative wrt the 1st variable.



      So the response is $-frac{partial operatorname{Li}(s,t)}{partial s}bigg|_{(s=0, t=x)}$






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        Ok, I found here what the notation $f^{(1,0)}$ means in Wolfram Alpha: it's the derivative wrt the 1st variable.



        So the response is $-frac{partial operatorname{Li}(s,t)}{partial s}bigg|_{(s=0, t=x)}$






        share|cite|improve this answer









        $endgroup$



        Ok, I found here what the notation $f^{(1,0)}$ means in Wolfram Alpha: it's the derivative wrt the 1st variable.



        So the response is $-frac{partial operatorname{Li}(s,t)}{partial s}bigg|_{(s=0, t=x)}$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 9 '18 at 22:47









        giuliolunatigiuliolunati

        1013




        1013






























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