What is the value of this expression involving the dilogarithm $operatorname{Li}_2$ and $sqrt{2}$?












1












$begingroup$


I'm carrying out a calculation, and the end result is $$operatorname{Li}_2(1/sqrt2) - operatorname{Li}_2(1 - 1/sqrt2) +
operatorname{Li}_2(2 - sqrt2) - operatorname{Li}_2(sqrt{2} - 1).$$
The value appears to be $pi^2/12$. How does one prove this?






definite-integrals special-functions polylogarithm







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Where we can presume: $operatorname{Li}_2(z)=sum_{k=1}^infty frac{z^k}{k^2}$ the dilogarithm function.
    $endgroup$
    – Mason
    Nov 25 '18 at 7:20








  • 2




    $begingroup$
    Note that $pi^2/12 =zeta(2)/2 =frac{1}{2} sum_{k=1}^infty frac{1}{k^2}$. So this can give you a direction in which to manipulate this expression.
    $endgroup$
    – Mason
    Nov 25 '18 at 7:23






  • 1




    $begingroup$
    the closed from follows from integrating by parts $Li_2(z) = -int frac{log(1-z)}{z}dz$ @Mason
    $endgroup$
    – reuns
    Nov 25 '18 at 7:33








  • 1




    $begingroup$
    Also I think one may be able to arrive there by exploiting the functional equations that appear in the first link I offered.
    $endgroup$
    – Mason
    Nov 25 '18 at 7:34












  • $begingroup$
    @Mason Yes that's the point of integrating by parts $int_1^z frac{log(1-s)}{s}ds$
    $endgroup$
    – reuns
    Nov 25 '18 at 7:48


















1












$begingroup$


I'm carrying out a calculation, and the end result is $$operatorname{Li}_2(1/sqrt2) - operatorname{Li}_2(1 - 1/sqrt2) +
operatorname{Li}_2(2 - sqrt2) - operatorname{Li}_2(sqrt{2} - 1).$$
The value appears to be $pi^2/12$. How does one prove this?






definite-integrals special-functions polylogarithm







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Where we can presume: $operatorname{Li}_2(z)=sum_{k=1}^infty frac{z^k}{k^2}$ the dilogarithm function.
    $endgroup$
    – Mason
    Nov 25 '18 at 7:20








  • 2




    $begingroup$
    Note that $pi^2/12 =zeta(2)/2 =frac{1}{2} sum_{k=1}^infty frac{1}{k^2}$. So this can give you a direction in which to manipulate this expression.
    $endgroup$
    – Mason
    Nov 25 '18 at 7:23






  • 1




    $begingroup$
    the closed from follows from integrating by parts $Li_2(z) = -int frac{log(1-z)}{z}dz$ @Mason
    $endgroup$
    – reuns
    Nov 25 '18 at 7:33








  • 1




    $begingroup$
    Also I think one may be able to arrive there by exploiting the functional equations that appear in the first link I offered.
    $endgroup$
    – Mason
    Nov 25 '18 at 7:34












  • $begingroup$
    @Mason Yes that's the point of integrating by parts $int_1^z frac{log(1-s)}{s}ds$
    $endgroup$
    – reuns
    Nov 25 '18 at 7:48
















1












1








1


0



$begingroup$


I'm carrying out a calculation, and the end result is $$operatorname{Li}_2(1/sqrt2) - operatorname{Li}_2(1 - 1/sqrt2) +
operatorname{Li}_2(2 - sqrt2) - operatorname{Li}_2(sqrt{2} - 1).$$
The value appears to be $pi^2/12$. How does one prove this?






definite-integrals special-functions polylogarithm







share|cite|improve this question











$endgroup$




I'm carrying out a calculation, and the end result is $$operatorname{Li}_2(1/sqrt2) - operatorname{Li}_2(1 - 1/sqrt2) +
operatorname{Li}_2(2 - sqrt2) - operatorname{Li}_2(sqrt{2} - 1).$$
The value appears to be $pi^2/12$. How does one prove this?






definite-integrals special-functions polylogarithm




definite-integrals special-functions polylogarithm






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 25 '18 at 7:25









Lord Shark the Unknown

102k1059132




102k1059132










asked Nov 25 '18 at 7:18









alphaalpha

233




233








  • 1




    $begingroup$
    Where we can presume: $operatorname{Li}_2(z)=sum_{k=1}^infty frac{z^k}{k^2}$ the dilogarithm function.
    $endgroup$
    – Mason
    Nov 25 '18 at 7:20








  • 2




    $begingroup$
    Note that $pi^2/12 =zeta(2)/2 =frac{1}{2} sum_{k=1}^infty frac{1}{k^2}$. So this can give you a direction in which to manipulate this expression.
    $endgroup$
    – Mason
    Nov 25 '18 at 7:23






  • 1




    $begingroup$
    the closed from follows from integrating by parts $Li_2(z) = -int frac{log(1-z)}{z}dz$ @Mason
    $endgroup$
    – reuns
    Nov 25 '18 at 7:33








  • 1




    $begingroup$
    Also I think one may be able to arrive there by exploiting the functional equations that appear in the first link I offered.
    $endgroup$
    – Mason
    Nov 25 '18 at 7:34












  • $begingroup$
    @Mason Yes that's the point of integrating by parts $int_1^z frac{log(1-s)}{s}ds$
    $endgroup$
    – reuns
    Nov 25 '18 at 7:48
















  • 1




    $begingroup$
    Where we can presume: $operatorname{Li}_2(z)=sum_{k=1}^infty frac{z^k}{k^2}$ the dilogarithm function.
    $endgroup$
    – Mason
    Nov 25 '18 at 7:20








  • 2




    $begingroup$
    Note that $pi^2/12 =zeta(2)/2 =frac{1}{2} sum_{k=1}^infty frac{1}{k^2}$. So this can give you a direction in which to manipulate this expression.
    $endgroup$
    – Mason
    Nov 25 '18 at 7:23






  • 1




    $begingroup$
    the closed from follows from integrating by parts $Li_2(z) = -int frac{log(1-z)}{z}dz$ @Mason
    $endgroup$
    – reuns
    Nov 25 '18 at 7:33








  • 1




    $begingroup$
    Also I think one may be able to arrive there by exploiting the functional equations that appear in the first link I offered.
    $endgroup$
    – Mason
    Nov 25 '18 at 7:34












  • $begingroup$
    @Mason Yes that's the point of integrating by parts $int_1^z frac{log(1-s)}{s}ds$
    $endgroup$
    – reuns
    Nov 25 '18 at 7:48










1




1




$begingroup$
Where we can presume: $operatorname{Li}_2(z)=sum_{k=1}^infty frac{z^k}{k^2}$ the dilogarithm function.
$endgroup$
– Mason
Nov 25 '18 at 7:20






$begingroup$
Where we can presume: $operatorname{Li}_2(z)=sum_{k=1}^infty frac{z^k}{k^2}$ the dilogarithm function.
$endgroup$
– Mason
Nov 25 '18 at 7:20






2




2




$begingroup$
Note that $pi^2/12 =zeta(2)/2 =frac{1}{2} sum_{k=1}^infty frac{1}{k^2}$. So this can give you a direction in which to manipulate this expression.
$endgroup$
– Mason
Nov 25 '18 at 7:23




$begingroup$
Note that $pi^2/12 =zeta(2)/2 =frac{1}{2} sum_{k=1}^infty frac{1}{k^2}$. So this can give you a direction in which to manipulate this expression.
$endgroup$
– Mason
Nov 25 '18 at 7:23




1




1




$begingroup$
the closed from follows from integrating by parts $Li_2(z) = -int frac{log(1-z)}{z}dz$ @Mason
$endgroup$
– reuns
Nov 25 '18 at 7:33






$begingroup$
the closed from follows from integrating by parts $Li_2(z) = -int frac{log(1-z)}{z}dz$ @Mason
$endgroup$
– reuns
Nov 25 '18 at 7:33






1




1




$begingroup$
Also I think one may be able to arrive there by exploiting the functional equations that appear in the first link I offered.
$endgroup$
– Mason
Nov 25 '18 at 7:34






$begingroup$
Also I think one may be able to arrive there by exploiting the functional equations that appear in the first link I offered.
$endgroup$
– Mason
Nov 25 '18 at 7:34














$begingroup$
@Mason Yes that's the point of integrating by parts $int_1^z frac{log(1-s)}{s}ds$
$endgroup$
– reuns
Nov 25 '18 at 7:48






$begingroup$
@Mason Yes that's the point of integrating by parts $int_1^z frac{log(1-s)}{s}ds$
$endgroup$
– reuns
Nov 25 '18 at 7:48












1 Answer
1






active

oldest

votes


















2












$begingroup$

The following identity could be found in Russion version of article about dilogarithm:



$$begin{aligned}operatorname{Li}_2(xy)&=operatorname{Li}_2(x)+operatorname{Li}_2(y)-operatorname{Li}_2left(frac{x(1-y)}{1-xy}right)-operatorname{Li}_2left(frac{y(1-x)}{1-xy}right)-\
&-lnleft(frac{1-x}{1-xy}right)lnleft(frac{1-y}{1-xy}right)end{aligned}$$
Therefore
$$begin{aligned}
operatorname{Li}_2(x)+operatorname{Li}_2(y)&=operatorname{Li}_2(xy)+operatorname{Li}_2left(frac{x(1-y)}{1-xy}right)+operatorname{Li}_2left(frac{y(1-x)}{1-xy}right)+\
&+lnleft(frac{1-x}{1-xy}right)lnleft(frac{1-y}{1-xy}right)
end{aligned}$$
Now we can simplify the original expression (let's name it $E$):



$$begin{aligned}E&=left(operatorname{Li}_2frac1{sqrt2}+operatorname{Li}_2left(2-sqrt2right)right)-operatorname{Li}_2left(1-frac1{sqrt2}right)-operatorname{Li}_2(sqrt2-1)=\
&=color{red}{operatorname{Li}_2(sqrt2-1)}+operatorname{Li}_2left(frac{frac1{sqrt2}(1-(2-sqrt2))}{1-(sqrt2-1)}right)+operatorname{Li}_2left(frac{(2-sqrt2)(1-frac1{sqrt2})}{1-(sqrt2-1)}right)+\
&+lnleft(frac{1-(2-sqrt2)}{1-(sqrt2-1)}right)lnleft(frac{1-frac1{sqrt2}}{1-(sqrt2-1)}right)-operatorname{Li}_2left(1-frac1{sqrt2}right)color{red}{-operatorname{Li}_2(sqrt2-1)}=\
&=operatorname{Li}_2left(frac12right)+color{red}{operatorname{Li}_2left(1-frac1{sqrt2}right)}+lnleft(frac1{sqrt2}right)lnleft(frac12right)color{red}{-operatorname{Li}_2left(1-frac1{sqrt2}right)}=\
&=frac{pi^2}{12}-frac12ln^22+frac12ln^22=frac{pi^2}{12}
end{aligned}$$



In the last line special value of $operatorname{Li}_2left(frac12right)$ was used which equals to $frac{pi^2}{12}-frac12ln^22$.






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%2f3012534%2fwhat-is-the-value-of-this-expression-involving-the-dilogarithm-operatornameli%23new-answer', 'question_page');
    }
    );

    Post as a guest















    Required, but never shown

























    1 Answer
    1






    active

    oldest

    votes








    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    2












    $begingroup$

    The following identity could be found in Russion version of article about dilogarithm:



    $$begin{aligned}operatorname{Li}_2(xy)&=operatorname{Li}_2(x)+operatorname{Li}_2(y)-operatorname{Li}_2left(frac{x(1-y)}{1-xy}right)-operatorname{Li}_2left(frac{y(1-x)}{1-xy}right)-\
    &-lnleft(frac{1-x}{1-xy}right)lnleft(frac{1-y}{1-xy}right)end{aligned}$$
    Therefore
    $$begin{aligned}
    operatorname{Li}_2(x)+operatorname{Li}_2(y)&=operatorname{Li}_2(xy)+operatorname{Li}_2left(frac{x(1-y)}{1-xy}right)+operatorname{Li}_2left(frac{y(1-x)}{1-xy}right)+\
    &+lnleft(frac{1-x}{1-xy}right)lnleft(frac{1-y}{1-xy}right)
    end{aligned}$$
    Now we can simplify the original expression (let's name it $E$):



    $$begin{aligned}E&=left(operatorname{Li}_2frac1{sqrt2}+operatorname{Li}_2left(2-sqrt2right)right)-operatorname{Li}_2left(1-frac1{sqrt2}right)-operatorname{Li}_2(sqrt2-1)=\
    &=color{red}{operatorname{Li}_2(sqrt2-1)}+operatorname{Li}_2left(frac{frac1{sqrt2}(1-(2-sqrt2))}{1-(sqrt2-1)}right)+operatorname{Li}_2left(frac{(2-sqrt2)(1-frac1{sqrt2})}{1-(sqrt2-1)}right)+\
    &+lnleft(frac{1-(2-sqrt2)}{1-(sqrt2-1)}right)lnleft(frac{1-frac1{sqrt2}}{1-(sqrt2-1)}right)-operatorname{Li}_2left(1-frac1{sqrt2}right)color{red}{-operatorname{Li}_2(sqrt2-1)}=\
    &=operatorname{Li}_2left(frac12right)+color{red}{operatorname{Li}_2left(1-frac1{sqrt2}right)}+lnleft(frac1{sqrt2}right)lnleft(frac12right)color{red}{-operatorname{Li}_2left(1-frac1{sqrt2}right)}=\
    &=frac{pi^2}{12}-frac12ln^22+frac12ln^22=frac{pi^2}{12}
    end{aligned}$$



    In the last line special value of $operatorname{Li}_2left(frac12right)$ was used which equals to $frac{pi^2}{12}-frac12ln^22$.






    share|cite|improve this answer









    $endgroup$


















      2












      $begingroup$

      The following identity could be found in Russion version of article about dilogarithm:



      $$begin{aligned}operatorname{Li}_2(xy)&=operatorname{Li}_2(x)+operatorname{Li}_2(y)-operatorname{Li}_2left(frac{x(1-y)}{1-xy}right)-operatorname{Li}_2left(frac{y(1-x)}{1-xy}right)-\
      &-lnleft(frac{1-x}{1-xy}right)lnleft(frac{1-y}{1-xy}right)end{aligned}$$
      Therefore
      $$begin{aligned}
      operatorname{Li}_2(x)+operatorname{Li}_2(y)&=operatorname{Li}_2(xy)+operatorname{Li}_2left(frac{x(1-y)}{1-xy}right)+operatorname{Li}_2left(frac{y(1-x)}{1-xy}right)+\
      &+lnleft(frac{1-x}{1-xy}right)lnleft(frac{1-y}{1-xy}right)
      end{aligned}$$
      Now we can simplify the original expression (let's name it $E$):



      $$begin{aligned}E&=left(operatorname{Li}_2frac1{sqrt2}+operatorname{Li}_2left(2-sqrt2right)right)-operatorname{Li}_2left(1-frac1{sqrt2}right)-operatorname{Li}_2(sqrt2-1)=\
      &=color{red}{operatorname{Li}_2(sqrt2-1)}+operatorname{Li}_2left(frac{frac1{sqrt2}(1-(2-sqrt2))}{1-(sqrt2-1)}right)+operatorname{Li}_2left(frac{(2-sqrt2)(1-frac1{sqrt2})}{1-(sqrt2-1)}right)+\
      &+lnleft(frac{1-(2-sqrt2)}{1-(sqrt2-1)}right)lnleft(frac{1-frac1{sqrt2}}{1-(sqrt2-1)}right)-operatorname{Li}_2left(1-frac1{sqrt2}right)color{red}{-operatorname{Li}_2(sqrt2-1)}=\
      &=operatorname{Li}_2left(frac12right)+color{red}{operatorname{Li}_2left(1-frac1{sqrt2}right)}+lnleft(frac1{sqrt2}right)lnleft(frac12right)color{red}{-operatorname{Li}_2left(1-frac1{sqrt2}right)}=\
      &=frac{pi^2}{12}-frac12ln^22+frac12ln^22=frac{pi^2}{12}
      end{aligned}$$



      In the last line special value of $operatorname{Li}_2left(frac12right)$ was used which equals to $frac{pi^2}{12}-frac12ln^22$.






      share|cite|improve this answer









      $endgroup$
















        2












        2








        2





        $begingroup$

        The following identity could be found in Russion version of article about dilogarithm:



        $$begin{aligned}operatorname{Li}_2(xy)&=operatorname{Li}_2(x)+operatorname{Li}_2(y)-operatorname{Li}_2left(frac{x(1-y)}{1-xy}right)-operatorname{Li}_2left(frac{y(1-x)}{1-xy}right)-\
        &-lnleft(frac{1-x}{1-xy}right)lnleft(frac{1-y}{1-xy}right)end{aligned}$$
        Therefore
        $$begin{aligned}
        operatorname{Li}_2(x)+operatorname{Li}_2(y)&=operatorname{Li}_2(xy)+operatorname{Li}_2left(frac{x(1-y)}{1-xy}right)+operatorname{Li}_2left(frac{y(1-x)}{1-xy}right)+\
        &+lnleft(frac{1-x}{1-xy}right)lnleft(frac{1-y}{1-xy}right)
        end{aligned}$$
        Now we can simplify the original expression (let's name it $E$):



        $$begin{aligned}E&=left(operatorname{Li}_2frac1{sqrt2}+operatorname{Li}_2left(2-sqrt2right)right)-operatorname{Li}_2left(1-frac1{sqrt2}right)-operatorname{Li}_2(sqrt2-1)=\
        &=color{red}{operatorname{Li}_2(sqrt2-1)}+operatorname{Li}_2left(frac{frac1{sqrt2}(1-(2-sqrt2))}{1-(sqrt2-1)}right)+operatorname{Li}_2left(frac{(2-sqrt2)(1-frac1{sqrt2})}{1-(sqrt2-1)}right)+\
        &+lnleft(frac{1-(2-sqrt2)}{1-(sqrt2-1)}right)lnleft(frac{1-frac1{sqrt2}}{1-(sqrt2-1)}right)-operatorname{Li}_2left(1-frac1{sqrt2}right)color{red}{-operatorname{Li}_2(sqrt2-1)}=\
        &=operatorname{Li}_2left(frac12right)+color{red}{operatorname{Li}_2left(1-frac1{sqrt2}right)}+lnleft(frac1{sqrt2}right)lnleft(frac12right)color{red}{-operatorname{Li}_2left(1-frac1{sqrt2}right)}=\
        &=frac{pi^2}{12}-frac12ln^22+frac12ln^22=frac{pi^2}{12}
        end{aligned}$$



        In the last line special value of $operatorname{Li}_2left(frac12right)$ was used which equals to $frac{pi^2}{12}-frac12ln^22$.






        share|cite|improve this answer









        $endgroup$



        The following identity could be found in Russion version of article about dilogarithm:



        $$begin{aligned}operatorname{Li}_2(xy)&=operatorname{Li}_2(x)+operatorname{Li}_2(y)-operatorname{Li}_2left(frac{x(1-y)}{1-xy}right)-operatorname{Li}_2left(frac{y(1-x)}{1-xy}right)-\
        &-lnleft(frac{1-x}{1-xy}right)lnleft(frac{1-y}{1-xy}right)end{aligned}$$
        Therefore
        $$begin{aligned}
        operatorname{Li}_2(x)+operatorname{Li}_2(y)&=operatorname{Li}_2(xy)+operatorname{Li}_2left(frac{x(1-y)}{1-xy}right)+operatorname{Li}_2left(frac{y(1-x)}{1-xy}right)+\
        &+lnleft(frac{1-x}{1-xy}right)lnleft(frac{1-y}{1-xy}right)
        end{aligned}$$
        Now we can simplify the original expression (let's name it $E$):



        $$begin{aligned}E&=left(operatorname{Li}_2frac1{sqrt2}+operatorname{Li}_2left(2-sqrt2right)right)-operatorname{Li}_2left(1-frac1{sqrt2}right)-operatorname{Li}_2(sqrt2-1)=\
        &=color{red}{operatorname{Li}_2(sqrt2-1)}+operatorname{Li}_2left(frac{frac1{sqrt2}(1-(2-sqrt2))}{1-(sqrt2-1)}right)+operatorname{Li}_2left(frac{(2-sqrt2)(1-frac1{sqrt2})}{1-(sqrt2-1)}right)+\
        &+lnleft(frac{1-(2-sqrt2)}{1-(sqrt2-1)}right)lnleft(frac{1-frac1{sqrt2}}{1-(sqrt2-1)}right)-operatorname{Li}_2left(1-frac1{sqrt2}right)color{red}{-operatorname{Li}_2(sqrt2-1)}=\
        &=operatorname{Li}_2left(frac12right)+color{red}{operatorname{Li}_2left(1-frac1{sqrt2}right)}+lnleft(frac1{sqrt2}right)lnleft(frac12right)color{red}{-operatorname{Li}_2left(1-frac1{sqrt2}right)}=\
        &=frac{pi^2}{12}-frac12ln^22+frac12ln^22=frac{pi^2}{12}
        end{aligned}$$



        In the last line special value of $operatorname{Li}_2left(frac12right)$ was used which equals to $frac{pi^2}{12}-frac12ln^22$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Nov 26 '18 at 5:45









        Mikalai ParshutsichMikalai Parshutsich

        473315




        473315






























            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%2f3012534%2fwhat-is-the-value-of-this-expression-involving-the-dilogarithm-operatornameli%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

            How to change which sound is reproduced for terminal bell?

            Image
            3 Recently upgrade to 18.10 from 18.04, and using the terminal is becoming annoying: the new alert sound is a drip ( /usr/share/sounds/gnome/default/alerts/drip.ogg ), and the previous one is still located at ( /usr/share/sounds/ubuntu/stereo/bell.ogg ). My issue is that at the gnome-terminal, when autocompleting or pressing left BOTH sounds are reproduced, at the same time. I want to just have ubuntu/stereo/bell.ogg active, not both . At the moment I have been only able to reduce the volume of the alerts (but still both are played), or turn the alerts down (but I dislike both "solutions"). gnome-terminal 18.10 teminal-bell share | improve this question
            Read more

            Title Spacing in Bjornstrup Chapter, Removing Chapter Number From Contents

            Image
            up vote 5 down vote favorite 2 I would like to make it so that the vertical space below the title of each chapter is the same when using the Bjornstrup package (i.e. so that the descenders do not affect the spacing - as shown by the red arrow in the picture below). I think I'm fairly happy with the spacing around "A chapter", so I'd really like titles without descenders (e.g. "Contents") to have the same vertical spacing as "A chapter" below the title. I would also like to remove the large "0" in the contents title. Any help appreciated, as always. documentclass[11pt,a4paper]{book} % Packages[![enter image description here][1]][1] usepackage[left=3.5cm, right=3.5cm, top=4.3cm, bottom=4.25cm]{geometry} usepackage[utf8]{inputenc} usepackage{libertine} usepackage[libertine]{newtx
            Read more

            Can I use Tabulator js library in my java Spring + Thymeleaf project?

            Image
            0 I want to add Tabulator js to my java project. I added tabulator.min.js, tabulator.min.css to resources folder. After that added to my html page these dependencies as recommending in readme.md. <link rel="stylesheet" th:href="@{/css/tabulator.min.css}"> <script type="text/javascript" th:src="@{/js/tabulator.min.js}"></script> Now my tables view all elements, that I have in the html page, but not correct. In the documentation of Tabulator is example how to create header and table, var table = new Tabulator("#example-table", { columns:[ {title:"Name", field:"name", sorter:"string", width:200, editor:true}, {title:"Age", field:"age", sorter:"
            Read more