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









New contributor




Akira Bergman is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
























    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









    New contributor




    Akira Bergman is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
    Check out our Code of Conduct.






















      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









      New contributor




      Akira Bergman is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.











      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









      New contributor




      Akira Bergman is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.











      share|cite|improve this question









      New contributor




      Akira Bergman is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.









      share|cite|improve this question




      share|cite|improve this question








      edited 2 days ago









      User525412790

      313114




      313114






      New contributor




      Akira Bergman is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.









      asked Nov 13 at 10:08









      Akira Bergman

      214




      214




      New contributor




      Akira Bergman is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.





      New contributor





      Akira Bergman is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.






      Akira Bergman is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.






















          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
              2 days ago




















            up vote
            5
            down vote













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






            share|cite|improve this answer








            New contributor




            R. J. Mathar is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
            Check out our Code of Conduct.


















              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
              });


              }
              });






              Akira Bergman is a new contributor. Be nice, and check out our Code of Conduct.










               

              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 2 days ago

























                  answered Nov 13 at 10:43









                  Claude Leibovici

                  116k1156131




                  116k1156131






















                      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
                        2 days ago

















                      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
                        2 days ago















                      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 2 days ago

























                      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
                        2 days ago




















                      • 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
                        2 days ago


















                      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
                      2 days ago






                      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
                      2 days ago












                      up vote
                      5
                      down vote













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






                      share|cite|improve this answer








                      New contributor




                      R. J. Mathar is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                      Check out our Code of Conduct.






















                        up vote
                        5
                        down vote













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






                        share|cite|improve this answer








                        New contributor




                        R. J. Mathar is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                        Check out our Code of Conduct.




















                          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








                          New contributor




                          R. J. Mathar is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                          Check out our Code of Conduct.









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







                          share|cite|improve this answer








                          New contributor




                          R. J. Mathar is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                          Check out our Code of Conduct.









                          share|cite|improve this answer



                          share|cite|improve this answer






                          New contributor




                          R. J. Mathar is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                          Check out our Code of Conduct.









                          answered Nov 13 at 14:29









                          R. J. Mathar

                          511




                          511




                          New contributor




                          R. J. Mathar is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                          Check out our Code of Conduct.





                          New contributor





                          R. J. Mathar is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                          Check out our Code of Conduct.






                          R. J. Mathar is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                          Check out our Code of Conduct.






















                              Akira Bergman is a new contributor. Be nice, and check out our Code of Conduct.










                               

                              draft saved


                              draft discarded


















                              Akira Bergman is a new contributor. Be nice, and check out our Code of Conduct.













                              Akira Bergman is a new contributor. Be nice, and check out our Code of Conduct.












                              Akira Bergman is a new contributor. Be nice, and check out our Code of Conduct.















                               


                              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

                              Сан-Квентин

                              8-я гвардейская общевойсковая армия

                              Алькесар