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









$endgroup$













    Your Answer





    StackExchange.ifUsing("editor", function () {
    return StackExchange.using("mathjaxEditing", function () {
    StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
    StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
    });
    });
    }, "mathjax-editing");

    StackExchange.ready(function() {
    var channelOptions = {
    tags: "".split(" "),
    id: "69"
    };
    initTagRenderer("".split(" "), "".split(" "), channelOptions);

    StackExchange.using("externalEditor", function() {
    // Have to fire editor after snippets, if snippets enabled
    if (StackExchange.settings.snippets.snippetsEnabled) {
    StackExchange.using("snippets", function() {
    createEditor();
    });
    }
    else {
    createEditor();
    }
    });

    function createEditor() {
    StackExchange.prepareEditor({
    heartbeatType: 'answer',
    autoActivateHeartbeat: false,
    convertImagesToLinks: true,
    noModals: true,
    showLowRepImageUploadWarning: true,
    reputationToPostImages: 10,
    bindNavPrevention: true,
    postfix: "",
    imageUploader: {
    brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
    contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
    allowUrls: true
    },
    noCode: true, onDemand: true,
    discardSelector: ".discard-answer"
    ,immediatelyShowMarkdownHelp:true
    });


    }
    });














    draft saved

    draft discarded


















    StackExchange.ready(
    function () {
    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3032691%2fclosed-form-for-sum-n-1-infty-logn-xn%23new-answer', 'question_page');
    }
    );

    Post as a guest















    Required, but never shown

























    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






























            draft saved

            draft discarded




















































            Thanks for contributing an answer to Mathematics Stack Exchange!


            • Please be sure to answer the question. Provide details and share your research!

            But avoid …



            • Asking for help, clarification, or responding to other answers.

            • Making statements based on opinion; back them up with references or personal experience.


            Use MathJax to format equations. MathJax reference.


            To learn more, see our tips on writing great answers.




            draft saved


            draft discarded














            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3032691%2fclosed-form-for-sum-n-1-infty-logn-xn%23new-answer', 'question_page');
            }
            );

            Post as a guest















            Required, but never shown





















































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown

































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown







            -

            Popular posts from this blog

            Different font size/position of beamer's navigation symbols template's content depending on regular/plain...

            Image
            3 Let's suppose I'd like to divert the navigation symbols template (e.g. in order replace them by the frame numbers, see this bug report ;) I would be facing troubles regarding the font size/position, depending on regular/plain frames, as shown by the following MCE. How could I get rid of these troubles? documentclass{beamer} usetheme{PaloAlto} setbeamertemplate{navigation symbols}{insertframenumber/inserttotalframenumber} begin{document} begin{frame} frametitle{A regular frame} end{frame} begin{frame}[plain] frametitle{A plain frame} end{frame} end{document} beamer header-footer share | improve this question asked Dec 10 '18 a
            Read more

            Berounka

            Image
            Berounka.mw-parser-output .entete.map{background-image:url("//upload.wikimedia.org/wikipedia/commons/7/7a/Picto_infobox_map.png")} Caractéristiques Longueur 139  km Bassin 8 861  km 2 Bassin collecteur Elbe Débit moyen ? Cours Source Confluence de la Mže et de la Radbuza · Localisation Pilsen · Altitude 305  m · Coordonnées 49° 45′ 13″ N, 13° 23′ 24″ E Confluence Vltava · Localisation Prague · Altitude 187  m · Coordonnées 49° 59′ 43″ N, 14° 24′ 04″ E Géographie Pays traversés   République tchèque modifier   La Berounka est une rivière de la République tchèque. Elle porte le nom de Mže depuis sa source en Allemagne près de la frontière jusqu'à sa confluence avec la Radbuza à Plzeň. Elle continue ensuite sous le nom de Berounka jusqu'à sa confluence avec la Vltava, près de Prague. La rivière est une destination prisée pour les amateurs de canoë, qui appréc
            Read more

            I want to find a topological embedding $f : X rightarrow Y$ and $g: Y rightarrow X$, yet $X$ is not...

            Image
            0 0 $begingroup$ This is related to a problem of showing that none of the intervals $(0,1], (0,1), [0,1]$ are homeomorphic to another, since it's in the same problem block. I've tried using $f(x) = 1 - x , f(x) = 1/(x-1)$ between varying pairs of those intervals to no avail. Do I need a function as complicated as the topologist's sine to show this? general-topology connectedness share | cite | improve this question edited Dec 10 '18 at 3:46 user1101010 775 1 7 30 asked Sep 24 '13 at 23:15
            Read more