What is the value of the Dirichlet Eta Function at s=1/2?











up vote
3
down vote

favorite












Although $eta(1)$ is known to be $ln(2)$, I have not seen an analytically calculated value for $eta(frac{1}{2});$



$$etaleft(frac{1}{2}right) = sum_{n=1}^{infty}frac{(-1)^{(n+1)}}{sqrt{n}}$$



A web calculator gives the value to be 0.6, which seems to be right.










share|cite|improve this question




























    up vote
    3
    down vote

    favorite












    Although $eta(1)$ is known to be $ln(2)$, I have not seen an analytically calculated value for $eta(frac{1}{2});$



    $$etaleft(frac{1}{2}right) = sum_{n=1}^{infty}frac{(-1)^{(n+1)}}{sqrt{n}}$$



    A web calculator gives the value to be 0.6, which seems to be right.










    share|cite|improve this question


























      up vote
      3
      down vote

      favorite









      up vote
      3
      down vote

      favorite











      Although $eta(1)$ is known to be $ln(2)$, I have not seen an analytically calculated value for $eta(frac{1}{2});$



      $$etaleft(frac{1}{2}right) = sum_{n=1}^{infty}frac{(-1)^{(n+1)}}{sqrt{n}}$$



      A web calculator gives the value to be 0.6, which seems to be right.










      share|cite|improve this question















      Although $eta(1)$ is known to be $ln(2)$, I have not seen an analytically calculated value for $eta(frac{1}{2});$



      $$etaleft(frac{1}{2}right) = sum_{n=1}^{infty}frac{(-1)^{(n+1)}}{sqrt{n}}$$



      A web calculator gives the value to be 0.6, which seems to be right.







      number-theory






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Nov 13 at 22:59









      User525412790

      313114




      313114










      asked Nov 13 at 10:08









      Akira Bergman

      214




      214






















          3 Answers
          3






          active

          oldest

          votes

















          up vote
          11
          down vote



          accepted










          Isn't just
          $$etaleft(frac{1}{2}right)=sum_{n=1}^inftyfrac{(-1)^{(n+1)}}{sqrt{n}}=left(1-sqrt{2}right) zeta left(frac{1}{2}right)approx 0.6048986434$$



          Edit



          Remember the general relation
          $$etaleft(sright)=left(1-2^{1-s}right) zeta (s)$$ If you want a quick and dirty shortcut evaluation, for $0 leq s leq 1$, you could use
          $$etaleft(sright)=frac 12+left( log (2)-frac{1}{2}right), s^{0.895}$$






          share|cite|improve this answer






























            up vote
            7
            down vote













            A careful computation shows that the numerical value is
            $$0.6048986434216303702472...$$
            which is not $0.6$. One should be aware that the above series converge really slowly.



            As Claude Leibovici indicates, one can relate its value to the Riemann's Zeta function value at $1/2$. However, as far as I know, there is no analytic formula of $zeta(1/2)$, so this is why you haven't seen an "analytically calculated value for $eta(1/2)$".



            EDIT2: As pointed again in the comments by leftaroundabout, I missread the OEIS link given in the answer of R. J. Mathar. What equals $$gamma/2 + pi/4 - (1/2 + sqrt{2})log(2) + log(pi)/2,$$where $gamma$ is the Euler-Mascheroni constant, is $eta'(1/2)/eta(1/2)$ not $eta(1/2)$.






            share|cite|improve this answer























            • You are very correct ! We start a no-end loop. By the way $to +1$
              – Claude Leibovici
              Nov 13 at 10:51






            • 4




              It does seem to be expressible exactly in terms of the Euler-Mascheroni constant though.
              – leftaroundabout
              Nov 13 at 16:01










            • @leftaroundabout Do you have a reference, I searched, but I didn't find it.
              – Josué Tonelli-Cueto
              Nov 13 at 18:47






            • 1




              @user3059799 Corrected, editing from the phone is hard
              – Josué Tonelli-Cueto
              Nov 13 at 20:11






            • 1




              The formula seems incorrect - I get the wrong answer when calculating it. OEIS seems to say that if $c$ is the OP's constant, and $d$ is the constant described here, then your formula gives $d/c$. The formula on OEIS for $d$ includes $zeta(1/2)$, so I suspect all they are giving is Claude Leibovici's formula in disguise.
              – Mike Miller
              Nov 14 at 1:18




















            up vote
            5
            down vote













            The numerical value of 0.604898... is provided in http://oeis.org/A113024 .






            share|cite|improve this answer





















              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',
              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%2f2996561%2fwhat-is-the-value-of-the-dirichlet-eta-function-at-s-1-2%23new-answer', 'question_page');
              }
              );

              Post as a guest















              Required, but never shown

























              3 Answers
              3






              active

              oldest

              votes








              3 Answers
              3






              active

              oldest

              votes









              active

              oldest

              votes






              active

              oldest

              votes








              up vote
              11
              down vote



              accepted










              Isn't just
              $$etaleft(frac{1}{2}right)=sum_{n=1}^inftyfrac{(-1)^{(n+1)}}{sqrt{n}}=left(1-sqrt{2}right) zeta left(frac{1}{2}right)approx 0.6048986434$$



              Edit



              Remember the general relation
              $$etaleft(sright)=left(1-2^{1-s}right) zeta (s)$$ If you want a quick and dirty shortcut evaluation, for $0 leq s leq 1$, you could use
              $$etaleft(sright)=frac 12+left( log (2)-frac{1}{2}right), s^{0.895}$$






              share|cite|improve this answer



























                up vote
                11
                down vote



                accepted










                Isn't just
                $$etaleft(frac{1}{2}right)=sum_{n=1}^inftyfrac{(-1)^{(n+1)}}{sqrt{n}}=left(1-sqrt{2}right) zeta left(frac{1}{2}right)approx 0.6048986434$$



                Edit



                Remember the general relation
                $$etaleft(sright)=left(1-2^{1-s}right) zeta (s)$$ If you want a quick and dirty shortcut evaluation, for $0 leq s leq 1$, you could use
                $$etaleft(sright)=frac 12+left( log (2)-frac{1}{2}right), s^{0.895}$$






                share|cite|improve this answer

























                  up vote
                  11
                  down vote



                  accepted







                  up vote
                  11
                  down vote



                  accepted






                  Isn't just
                  $$etaleft(frac{1}{2}right)=sum_{n=1}^inftyfrac{(-1)^{(n+1)}}{sqrt{n}}=left(1-sqrt{2}right) zeta left(frac{1}{2}right)approx 0.6048986434$$



                  Edit



                  Remember the general relation
                  $$etaleft(sright)=left(1-2^{1-s}right) zeta (s)$$ If you want a quick and dirty shortcut evaluation, for $0 leq s leq 1$, you could use
                  $$etaleft(sright)=frac 12+left( log (2)-frac{1}{2}right), s^{0.895}$$






                  share|cite|improve this answer














                  Isn't just
                  $$etaleft(frac{1}{2}right)=sum_{n=1}^inftyfrac{(-1)^{(n+1)}}{sqrt{n}}=left(1-sqrt{2}right) zeta left(frac{1}{2}right)approx 0.6048986434$$



                  Edit



                  Remember the general relation
                  $$etaleft(sright)=left(1-2^{1-s}right) zeta (s)$$ If you want a quick and dirty shortcut evaluation, for $0 leq s leq 1$, you could use
                  $$etaleft(sright)=frac 12+left( log (2)-frac{1}{2}right), s^{0.895}$$







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Nov 14 at 4:53

























                  answered Nov 13 at 10:43









                  Claude Leibovici

                  117k1156131




                  117k1156131






















                      up vote
                      7
                      down vote













                      A careful computation shows that the numerical value is
                      $$0.6048986434216303702472...$$
                      which is not $0.6$. One should be aware that the above series converge really slowly.



                      As Claude Leibovici indicates, one can relate its value to the Riemann's Zeta function value at $1/2$. However, as far as I know, there is no analytic formula of $zeta(1/2)$, so this is why you haven't seen an "analytically calculated value for $eta(1/2)$".



                      EDIT2: As pointed again in the comments by leftaroundabout, I missread the OEIS link given in the answer of R. J. Mathar. What equals $$gamma/2 + pi/4 - (1/2 + sqrt{2})log(2) + log(pi)/2,$$where $gamma$ is the Euler-Mascheroni constant, is $eta'(1/2)/eta(1/2)$ not $eta(1/2)$.






                      share|cite|improve this answer























                      • You are very correct ! We start a no-end loop. By the way $to +1$
                        – Claude Leibovici
                        Nov 13 at 10:51






                      • 4




                        It does seem to be expressible exactly in terms of the Euler-Mascheroni constant though.
                        – leftaroundabout
                        Nov 13 at 16:01










                      • @leftaroundabout Do you have a reference, I searched, but I didn't find it.
                        – Josué Tonelli-Cueto
                        Nov 13 at 18:47






                      • 1




                        @user3059799 Corrected, editing from the phone is hard
                        – Josué Tonelli-Cueto
                        Nov 13 at 20:11






                      • 1




                        The formula seems incorrect - I get the wrong answer when calculating it. OEIS seems to say that if $c$ is the OP's constant, and $d$ is the constant described here, then your formula gives $d/c$. The formula on OEIS for $d$ includes $zeta(1/2)$, so I suspect all they are giving is Claude Leibovici's formula in disguise.
                        – Mike Miller
                        Nov 14 at 1:18

















                      up vote
                      7
                      down vote













                      A careful computation shows that the numerical value is
                      $$0.6048986434216303702472...$$
                      which is not $0.6$. One should be aware that the above series converge really slowly.



                      As Claude Leibovici indicates, one can relate its value to the Riemann's Zeta function value at $1/2$. However, as far as I know, there is no analytic formula of $zeta(1/2)$, so this is why you haven't seen an "analytically calculated value for $eta(1/2)$".



                      EDIT2: As pointed again in the comments by leftaroundabout, I missread the OEIS link given in the answer of R. J. Mathar. What equals $$gamma/2 + pi/4 - (1/2 + sqrt{2})log(2) + log(pi)/2,$$where $gamma$ is the Euler-Mascheroni constant, is $eta'(1/2)/eta(1/2)$ not $eta(1/2)$.






                      share|cite|improve this answer























                      • You are very correct ! We start a no-end loop. By the way $to +1$
                        – Claude Leibovici
                        Nov 13 at 10:51






                      • 4




                        It does seem to be expressible exactly in terms of the Euler-Mascheroni constant though.
                        – leftaroundabout
                        Nov 13 at 16:01










                      • @leftaroundabout Do you have a reference, I searched, but I didn't find it.
                        – Josué Tonelli-Cueto
                        Nov 13 at 18:47






                      • 1




                        @user3059799 Corrected, editing from the phone is hard
                        – Josué Tonelli-Cueto
                        Nov 13 at 20:11






                      • 1




                        The formula seems incorrect - I get the wrong answer when calculating it. OEIS seems to say that if $c$ is the OP's constant, and $d$ is the constant described here, then your formula gives $d/c$. The formula on OEIS for $d$ includes $zeta(1/2)$, so I suspect all they are giving is Claude Leibovici's formula in disguise.
                        – Mike Miller
                        Nov 14 at 1:18















                      up vote
                      7
                      down vote










                      up vote
                      7
                      down vote









                      A careful computation shows that the numerical value is
                      $$0.6048986434216303702472...$$
                      which is not $0.6$. One should be aware that the above series converge really slowly.



                      As Claude Leibovici indicates, one can relate its value to the Riemann's Zeta function value at $1/2$. However, as far as I know, there is no analytic formula of $zeta(1/2)$, so this is why you haven't seen an "analytically calculated value for $eta(1/2)$".



                      EDIT2: As pointed again in the comments by leftaroundabout, I missread the OEIS link given in the answer of R. J. Mathar. What equals $$gamma/2 + pi/4 - (1/2 + sqrt{2})log(2) + log(pi)/2,$$where $gamma$ is the Euler-Mascheroni constant, is $eta'(1/2)/eta(1/2)$ not $eta(1/2)$.






                      share|cite|improve this answer














                      A careful computation shows that the numerical value is
                      $$0.6048986434216303702472...$$
                      which is not $0.6$. One should be aware that the above series converge really slowly.



                      As Claude Leibovici indicates, one can relate its value to the Riemann's Zeta function value at $1/2$. However, as far as I know, there is no analytic formula of $zeta(1/2)$, so this is why you haven't seen an "analytically calculated value for $eta(1/2)$".



                      EDIT2: As pointed again in the comments by leftaroundabout, I missread the OEIS link given in the answer of R. J. Mathar. What equals $$gamma/2 + pi/4 - (1/2 + sqrt{2})log(2) + log(pi)/2,$$where $gamma$ is the Euler-Mascheroni constant, is $eta'(1/2)/eta(1/2)$ not $eta(1/2)$.







                      share|cite|improve this answer














                      share|cite|improve this answer



                      share|cite|improve this answer








                      edited Nov 14 at 21:30

























                      answered Nov 13 at 10:48









                      Josué Tonelli-Cueto

                      3,6521027




                      3,6521027












                      • You are very correct ! We start a no-end loop. By the way $to +1$
                        – Claude Leibovici
                        Nov 13 at 10:51






                      • 4




                        It does seem to be expressible exactly in terms of the Euler-Mascheroni constant though.
                        – leftaroundabout
                        Nov 13 at 16:01










                      • @leftaroundabout Do you have a reference, I searched, but I didn't find it.
                        – Josué Tonelli-Cueto
                        Nov 13 at 18:47






                      • 1




                        @user3059799 Corrected, editing from the phone is hard
                        – Josué Tonelli-Cueto
                        Nov 13 at 20:11






                      • 1




                        The formula seems incorrect - I get the wrong answer when calculating it. OEIS seems to say that if $c$ is the OP's constant, and $d$ is the constant described here, then your formula gives $d/c$. The formula on OEIS for $d$ includes $zeta(1/2)$, so I suspect all they are giving is Claude Leibovici's formula in disguise.
                        – Mike Miller
                        Nov 14 at 1:18




















                      • You are very correct ! We start a no-end loop. By the way $to +1$
                        – Claude Leibovici
                        Nov 13 at 10:51






                      • 4




                        It does seem to be expressible exactly in terms of the Euler-Mascheroni constant though.
                        – leftaroundabout
                        Nov 13 at 16:01










                      • @leftaroundabout Do you have a reference, I searched, but I didn't find it.
                        – Josué Tonelli-Cueto
                        Nov 13 at 18:47






                      • 1




                        @user3059799 Corrected, editing from the phone is hard
                        – Josué Tonelli-Cueto
                        Nov 13 at 20:11






                      • 1




                        The formula seems incorrect - I get the wrong answer when calculating it. OEIS seems to say that if $c$ is the OP's constant, and $d$ is the constant described here, then your formula gives $d/c$. The formula on OEIS for $d$ includes $zeta(1/2)$, so I suspect all they are giving is Claude Leibovici's formula in disguise.
                        – Mike Miller
                        Nov 14 at 1:18


















                      You are very correct ! We start a no-end loop. By the way $to +1$
                      – Claude Leibovici
                      Nov 13 at 10:51




                      You are very correct ! We start a no-end loop. By the way $to +1$
                      – Claude Leibovici
                      Nov 13 at 10:51




                      4




                      4




                      It does seem to be expressible exactly in terms of the Euler-Mascheroni constant though.
                      – leftaroundabout
                      Nov 13 at 16:01




                      It does seem to be expressible exactly in terms of the Euler-Mascheroni constant though.
                      – leftaroundabout
                      Nov 13 at 16:01












                      @leftaroundabout Do you have a reference, I searched, but I didn't find it.
                      – Josué Tonelli-Cueto
                      Nov 13 at 18:47




                      @leftaroundabout Do you have a reference, I searched, but I didn't find it.
                      – Josué Tonelli-Cueto
                      Nov 13 at 18:47




                      1




                      1




                      @user3059799 Corrected, editing from the phone is hard
                      – Josué Tonelli-Cueto
                      Nov 13 at 20:11




                      @user3059799 Corrected, editing from the phone is hard
                      – Josué Tonelli-Cueto
                      Nov 13 at 20:11




                      1




                      1




                      The formula seems incorrect - I get the wrong answer when calculating it. OEIS seems to say that if $c$ is the OP's constant, and $d$ is the constant described here, then your formula gives $d/c$. The formula on OEIS for $d$ includes $zeta(1/2)$, so I suspect all they are giving is Claude Leibovici's formula in disguise.
                      – Mike Miller
                      Nov 14 at 1:18






                      The formula seems incorrect - I get the wrong answer when calculating it. OEIS seems to say that if $c$ is the OP's constant, and $d$ is the constant described here, then your formula gives $d/c$. The formula on OEIS for $d$ includes $zeta(1/2)$, so I suspect all they are giving is Claude Leibovici's formula in disguise.
                      – Mike Miller
                      Nov 14 at 1:18












                      up vote
                      5
                      down vote













                      The numerical value of 0.604898... is provided in http://oeis.org/A113024 .






                      share|cite|improve this answer

























                        up vote
                        5
                        down vote













                        The numerical value of 0.604898... is provided in http://oeis.org/A113024 .






                        share|cite|improve this answer























                          up vote
                          5
                          down vote










                          up vote
                          5
                          down vote









                          The numerical value of 0.604898... is provided in http://oeis.org/A113024 .






                          share|cite|improve this answer












                          The numerical value of 0.604898... is provided in http://oeis.org/A113024 .







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered Nov 13 at 14:29









                          R. J. Mathar

                          511




                          511






























                              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.





                              Some of your past answers have not been well-received, and you're in danger of being blocked from answering.


                              Please pay close attention to the following guidance:


                              • 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.


                              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%2f2996561%2fwhat-is-the-value-of-the-dirichlet-eta-function-at-s-1-2%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?

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

                              Title Spacing in Bjornstrup Chapter, Removing Chapter Number From Contents