Summation over multiple arguments












0














This might seem stupid, but I'm really stuck. I don't understand how to calculate the following explicitly:



$$sum_{s_1=pm1} sum_{s_2=pm1} sum_{s_3=pm1} e^{-{s_1s_2}}e^{-{s_2s_3}}$$



(it's the Ising model for 3 lattice sites).



I don't understand how I can calculate this in a brute force way, since the sums for $s_1$ and $s_3$ only apply to one part of the equation to be summed over.



(I know you can simplify it and end up with a much nicer expression in terms of cosh)



Help! Thanks.






summation exponential-sum







share|cite|improve this question





























    0














    This might seem stupid, but I'm really stuck. I don't understand how to calculate the following explicitly:



    $$sum_{s_1=pm1} sum_{s_2=pm1} sum_{s_3=pm1} e^{-{s_1s_2}}e^{-{s_2s_3}}$$



    (it's the Ising model for 3 lattice sites).



    I don't understand how I can calculate this in a brute force way, since the sums for $s_1$ and $s_3$ only apply to one part of the equation to be summed over.



    (I know you can simplify it and end up with a much nicer expression in terms of cosh)



    Help! Thanks.






    summation exponential-sum







    share|cite|improve this question



























      0












      0








      0







      This might seem stupid, but I'm really stuck. I don't understand how to calculate the following explicitly:



      $$sum_{s_1=pm1} sum_{s_2=pm1} sum_{s_3=pm1} e^{-{s_1s_2}}e^{-{s_2s_3}}$$



      (it's the Ising model for 3 lattice sites).



      I don't understand how I can calculate this in a brute force way, since the sums for $s_1$ and $s_3$ only apply to one part of the equation to be summed over.



      (I know you can simplify it and end up with a much nicer expression in terms of cosh)



      Help! Thanks.






      summation exponential-sum







      share|cite|improve this question















      This might seem stupid, but I'm really stuck. I don't understand how to calculate the following explicitly:



      $$sum_{s_1=pm1} sum_{s_2=pm1} sum_{s_3=pm1} e^{-{s_1s_2}}e^{-{s_2s_3}}$$



      (it's the Ising model for 3 lattice sites).



      I don't understand how I can calculate this in a brute force way, since the sums for $s_1$ and $s_3$ only apply to one part of the equation to be summed over.



      (I know you can simplify it and end up with a much nicer expression in terms of cosh)



      Help! Thanks.






      summation exponential-sum




      summation exponential-sum






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Nov 21 '18 at 13:36









      Jean-Claude Arbaut

      14.7k63464




      14.7k63464










      asked Nov 21 '18 at 13:20









      Learn4life

      906




      906






















          2 Answers
          2






          active

          oldest

          votes


















          4














          Proceeding from right to left:



          $$begin{align*}sum_{s_1=pm1} sum_{s_2=pm1} sum_{s_3=pm1} e^{-{s_1s_2}}e^{-{s_2s_3}}& =sum_{s_1=pm1} sum_{s_2=pm1}e^{-{s_1s_2}}left[ e^{{s_2}}+e^{-{s_2}} right]\
          &=sum_{s_1=pm1}left( e^{{s_1}}left[ e^{{-1}}+e^{1} right]+e^{-{s_1}}left[ e^{1}+e^{-{1}} right]right)\
          &=sum_{s_1=pm1}left( [e^{{s_1}}+e^{-{s_1}}]left[ e^{{-1}}+e^{1} right]right)\
          &=[e^{{-1}}+e^{1}]left[ e^{{-1}}+e^{1} right]+[e^{{1}}+e^{-1}]left[ e^{{-1}}+e^{1} right]\
          &=2[e^{{-1}}+e^{1}]^2\
          &=4(cosh(2)+1)
          end{align*}$$






          share|cite|improve this answer































            0














            A slightly alternate approach than in the other answer (dividing up the sum differently):



            Starting with the inside sum of
            $$sum_{s_1=pm1} sum_{s_2=pm1} sum_{s_3=pm1} e^{-{s_1s_2}}e^{-{s_2s_3}},$$
            we can factor out the term that doesn't depend on $s_3$ to get
            $$sum_{s_1=pm1} sum_{s_2=pm1} sum_{s_3=pm1} e^{-{s_1s_2}}e^{-{s_2s_3}}=
            sum_{s_1=pm1} sum_{s_2=pm1} e^{-{s_1s_2}}sum_{s_3=pm1} e^{-{s_2s_3}}.
            $$
            Now, let's look at the exponent of the internal sum, since $s_2$ is either $+1$ or $-1$, then $-s_2$ is either $+1$ or $-1$. Multiplying by $s_3$ which is either $+1$ or $-1$ results in one of each of $+1$ and $-1$. Therefore, the sum simplifies to
            $$
            sum_{s_1=pm1} sum_{s_2=pm1} e^{-{s_1s_2}}sum_{s_3=pm1} e^{-{s_2s_3}}=
            sum_{s_1=pm1} sum_{s_2=pm1} e^{-{s_1s_2}}(e+e^{-1})=(e+e^{-1})sum_{s_1=pm1} sum_{s_2=pm1} e^{-{s_1s_2}}.
            $$
            By applying the same argument as above, we see that the exponent of this $e$ is one of each of $+1$ and $-1$, so we get
            $$
            (e+e^{-1})sum_{s_1=pm1} sum_{s_2=pm1} e^{-{s_1s_2}}=(e+e^{-1})sum_{s_1=pm1} (e+e^{-1})=(e+e^{-1})^2sum_{s_1=pm1}1.
            $$
            Since there are only two values for $s_1$, we get that this simplifies to $2(e+e^{-1})^2$, which can then be simplified in terms of $cosh$.






            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',
              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%2f3007721%2fsummation-over-multiple-arguments%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









              4














              Proceeding from right to left:



              $$begin{align*}sum_{s_1=pm1} sum_{s_2=pm1} sum_{s_3=pm1} e^{-{s_1s_2}}e^{-{s_2s_3}}& =sum_{s_1=pm1} sum_{s_2=pm1}e^{-{s_1s_2}}left[ e^{{s_2}}+e^{-{s_2}} right]\
              &=sum_{s_1=pm1}left( e^{{s_1}}left[ e^{{-1}}+e^{1} right]+e^{-{s_1}}left[ e^{1}+e^{-{1}} right]right)\
              &=sum_{s_1=pm1}left( [e^{{s_1}}+e^{-{s_1}}]left[ e^{{-1}}+e^{1} right]right)\
              &=[e^{{-1}}+e^{1}]left[ e^{{-1}}+e^{1} right]+[e^{{1}}+e^{-1}]left[ e^{{-1}}+e^{1} right]\
              &=2[e^{{-1}}+e^{1}]^2\
              &=4(cosh(2)+1)
              end{align*}$$






              share|cite|improve this answer




























                4














                Proceeding from right to left:



                $$begin{align*}sum_{s_1=pm1} sum_{s_2=pm1} sum_{s_3=pm1} e^{-{s_1s_2}}e^{-{s_2s_3}}& =sum_{s_1=pm1} sum_{s_2=pm1}e^{-{s_1s_2}}left[ e^{{s_2}}+e^{-{s_2}} right]\
                &=sum_{s_1=pm1}left( e^{{s_1}}left[ e^{{-1}}+e^{1} right]+e^{-{s_1}}left[ e^{1}+e^{-{1}} right]right)\
                &=sum_{s_1=pm1}left( [e^{{s_1}}+e^{-{s_1}}]left[ e^{{-1}}+e^{1} right]right)\
                &=[e^{{-1}}+e^{1}]left[ e^{{-1}}+e^{1} right]+[e^{{1}}+e^{-1}]left[ e^{{-1}}+e^{1} right]\
                &=2[e^{{-1}}+e^{1}]^2\
                &=4(cosh(2)+1)
                end{align*}$$






                share|cite|improve this answer


























                  4












                  4








                  4






                  Proceeding from right to left:



                  $$begin{align*}sum_{s_1=pm1} sum_{s_2=pm1} sum_{s_3=pm1} e^{-{s_1s_2}}e^{-{s_2s_3}}& =sum_{s_1=pm1} sum_{s_2=pm1}e^{-{s_1s_2}}left[ e^{{s_2}}+e^{-{s_2}} right]\
                  &=sum_{s_1=pm1}left( e^{{s_1}}left[ e^{{-1}}+e^{1} right]+e^{-{s_1}}left[ e^{1}+e^{-{1}} right]right)\
                  &=sum_{s_1=pm1}left( [e^{{s_1}}+e^{-{s_1}}]left[ e^{{-1}}+e^{1} right]right)\
                  &=[e^{{-1}}+e^{1}]left[ e^{{-1}}+e^{1} right]+[e^{{1}}+e^{-1}]left[ e^{{-1}}+e^{1} right]\
                  &=2[e^{{-1}}+e^{1}]^2\
                  &=4(cosh(2)+1)
                  end{align*}$$






                  share|cite|improve this answer














                  Proceeding from right to left:



                  $$begin{align*}sum_{s_1=pm1} sum_{s_2=pm1} sum_{s_3=pm1} e^{-{s_1s_2}}e^{-{s_2s_3}}& =sum_{s_1=pm1} sum_{s_2=pm1}e^{-{s_1s_2}}left[ e^{{s_2}}+e^{-{s_2}} right]\
                  &=sum_{s_1=pm1}left( e^{{s_1}}left[ e^{{-1}}+e^{1} right]+e^{-{s_1}}left[ e^{1}+e^{-{1}} right]right)\
                  &=sum_{s_1=pm1}left( [e^{{s_1}}+e^{-{s_1}}]left[ e^{{-1}}+e^{1} right]right)\
                  &=[e^{{-1}}+e^{1}]left[ e^{{-1}}+e^{1} right]+[e^{{1}}+e^{-1}]left[ e^{{-1}}+e^{1} right]\
                  &=2[e^{{-1}}+e^{1}]^2\
                  &=4(cosh(2)+1)
                  end{align*}$$







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Nov 21 '18 at 13:38

























                  answered Nov 21 '18 at 13:32









                  smcc

                  4,297517




                  4,297517























                      0














                      A slightly alternate approach than in the other answer (dividing up the sum differently):



                      Starting with the inside sum of
                      $$sum_{s_1=pm1} sum_{s_2=pm1} sum_{s_3=pm1} e^{-{s_1s_2}}e^{-{s_2s_3}},$$
                      we can factor out the term that doesn't depend on $s_3$ to get
                      $$sum_{s_1=pm1} sum_{s_2=pm1} sum_{s_3=pm1} e^{-{s_1s_2}}e^{-{s_2s_3}}=
                      sum_{s_1=pm1} sum_{s_2=pm1} e^{-{s_1s_2}}sum_{s_3=pm1} e^{-{s_2s_3}}.
                      $$
                      Now, let's look at the exponent of the internal sum, since $s_2$ is either $+1$ or $-1$, then $-s_2$ is either $+1$ or $-1$. Multiplying by $s_3$ which is either $+1$ or $-1$ results in one of each of $+1$ and $-1$. Therefore, the sum simplifies to
                      $$
                      sum_{s_1=pm1} sum_{s_2=pm1} e^{-{s_1s_2}}sum_{s_3=pm1} e^{-{s_2s_3}}=
                      sum_{s_1=pm1} sum_{s_2=pm1} e^{-{s_1s_2}}(e+e^{-1})=(e+e^{-1})sum_{s_1=pm1} sum_{s_2=pm1} e^{-{s_1s_2}}.
                      $$
                      By applying the same argument as above, we see that the exponent of this $e$ is one of each of $+1$ and $-1$, so we get
                      $$
                      (e+e^{-1})sum_{s_1=pm1} sum_{s_2=pm1} e^{-{s_1s_2}}=(e+e^{-1})sum_{s_1=pm1} (e+e^{-1})=(e+e^{-1})^2sum_{s_1=pm1}1.
                      $$
                      Since there are only two values for $s_1$, we get that this simplifies to $2(e+e^{-1})^2$, which can then be simplified in terms of $cosh$.






                      share|cite|improve this answer


























                        0














                        A slightly alternate approach than in the other answer (dividing up the sum differently):



                        Starting with the inside sum of
                        $$sum_{s_1=pm1} sum_{s_2=pm1} sum_{s_3=pm1} e^{-{s_1s_2}}e^{-{s_2s_3}},$$
                        we can factor out the term that doesn't depend on $s_3$ to get
                        $$sum_{s_1=pm1} sum_{s_2=pm1} sum_{s_3=pm1} e^{-{s_1s_2}}e^{-{s_2s_3}}=
                        sum_{s_1=pm1} sum_{s_2=pm1} e^{-{s_1s_2}}sum_{s_3=pm1} e^{-{s_2s_3}}.
                        $$
                        Now, let's look at the exponent of the internal sum, since $s_2$ is either $+1$ or $-1$, then $-s_2$ is either $+1$ or $-1$. Multiplying by $s_3$ which is either $+1$ or $-1$ results in one of each of $+1$ and $-1$. Therefore, the sum simplifies to
                        $$
                        sum_{s_1=pm1} sum_{s_2=pm1} e^{-{s_1s_2}}sum_{s_3=pm1} e^{-{s_2s_3}}=
                        sum_{s_1=pm1} sum_{s_2=pm1} e^{-{s_1s_2}}(e+e^{-1})=(e+e^{-1})sum_{s_1=pm1} sum_{s_2=pm1} e^{-{s_1s_2}}.
                        $$
                        By applying the same argument as above, we see that the exponent of this $e$ is one of each of $+1$ and $-1$, so we get
                        $$
                        (e+e^{-1})sum_{s_1=pm1} sum_{s_2=pm1} e^{-{s_1s_2}}=(e+e^{-1})sum_{s_1=pm1} (e+e^{-1})=(e+e^{-1})^2sum_{s_1=pm1}1.
                        $$
                        Since there are only two values for $s_1$, we get that this simplifies to $2(e+e^{-1})^2$, which can then be simplified in terms of $cosh$.






                        share|cite|improve this answer
























                          0












                          0








                          0






                          A slightly alternate approach than in the other answer (dividing up the sum differently):



                          Starting with the inside sum of
                          $$sum_{s_1=pm1} sum_{s_2=pm1} sum_{s_3=pm1} e^{-{s_1s_2}}e^{-{s_2s_3}},$$
                          we can factor out the term that doesn't depend on $s_3$ to get
                          $$sum_{s_1=pm1} sum_{s_2=pm1} sum_{s_3=pm1} e^{-{s_1s_2}}e^{-{s_2s_3}}=
                          sum_{s_1=pm1} sum_{s_2=pm1} e^{-{s_1s_2}}sum_{s_3=pm1} e^{-{s_2s_3}}.
                          $$
                          Now, let's look at the exponent of the internal sum, since $s_2$ is either $+1$ or $-1$, then $-s_2$ is either $+1$ or $-1$. Multiplying by $s_3$ which is either $+1$ or $-1$ results in one of each of $+1$ and $-1$. Therefore, the sum simplifies to
                          $$
                          sum_{s_1=pm1} sum_{s_2=pm1} e^{-{s_1s_2}}sum_{s_3=pm1} e^{-{s_2s_3}}=
                          sum_{s_1=pm1} sum_{s_2=pm1} e^{-{s_1s_2}}(e+e^{-1})=(e+e^{-1})sum_{s_1=pm1} sum_{s_2=pm1} e^{-{s_1s_2}}.
                          $$
                          By applying the same argument as above, we see that the exponent of this $e$ is one of each of $+1$ and $-1$, so we get
                          $$
                          (e+e^{-1})sum_{s_1=pm1} sum_{s_2=pm1} e^{-{s_1s_2}}=(e+e^{-1})sum_{s_1=pm1} (e+e^{-1})=(e+e^{-1})^2sum_{s_1=pm1}1.
                          $$
                          Since there are only two values for $s_1$, we get that this simplifies to $2(e+e^{-1})^2$, which can then be simplified in terms of $cosh$.






                          share|cite|improve this answer












                          A slightly alternate approach than in the other answer (dividing up the sum differently):



                          Starting with the inside sum of
                          $$sum_{s_1=pm1} sum_{s_2=pm1} sum_{s_3=pm1} e^{-{s_1s_2}}e^{-{s_2s_3}},$$
                          we can factor out the term that doesn't depend on $s_3$ to get
                          $$sum_{s_1=pm1} sum_{s_2=pm1} sum_{s_3=pm1} e^{-{s_1s_2}}e^{-{s_2s_3}}=
                          sum_{s_1=pm1} sum_{s_2=pm1} e^{-{s_1s_2}}sum_{s_3=pm1} e^{-{s_2s_3}}.
                          $$
                          Now, let's look at the exponent of the internal sum, since $s_2$ is either $+1$ or $-1$, then $-s_2$ is either $+1$ or $-1$. Multiplying by $s_3$ which is either $+1$ or $-1$ results in one of each of $+1$ and $-1$. Therefore, the sum simplifies to
                          $$
                          sum_{s_1=pm1} sum_{s_2=pm1} e^{-{s_1s_2}}sum_{s_3=pm1} e^{-{s_2s_3}}=
                          sum_{s_1=pm1} sum_{s_2=pm1} e^{-{s_1s_2}}(e+e^{-1})=(e+e^{-1})sum_{s_1=pm1} sum_{s_2=pm1} e^{-{s_1s_2}}.
                          $$
                          By applying the same argument as above, we see that the exponent of this $e$ is one of each of $+1$ and $-1$, so we get
                          $$
                          (e+e^{-1})sum_{s_1=pm1} sum_{s_2=pm1} e^{-{s_1s_2}}=(e+e^{-1})sum_{s_1=pm1} (e+e^{-1})=(e+e^{-1})^2sum_{s_1=pm1}1.
                          $$
                          Since there are only two values for $s_1$, we get that this simplifies to $2(e+e^{-1})^2$, which can then be simplified in terms of $cosh$.







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered Nov 21 '18 at 13:40









                          Michael Burr

                          26.6k23262




                          26.6k23262






























                              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%2f3007721%2fsummation-over-multiple-arguments%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

                              MongoDB - Not Authorized To Execute Command

                              Image
                              .everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty{ height:90px;width:728px;box-sizing:border-box; } 0 I have successfully enabled authorization on MongoDB and I have created an account on the admin database and then I created an account for my database called test. The following connection string to connect to my test database works successfully: mongo --host 192.168.17.52 --port 27017 -u user1 -p password --authenticationDatabase test Only problem I have now is, I cannot execute commands such as: show dbs. I get the following error when I try to do so: "errmsg" : "not authorized on admin to execute command { listDatabases: 1.0, lsid: { id: UUID("a1d5bc0d-bc58-485e-b232-270758a89455") }, $db: "admin...
                              Read more

                              in spring boot 2.1 many test slices are not allowed anymore due to multiple @BootstrapWith

                              Image
                              1 I tried to upgrade a yummy sandwich made of two test slices (@JsonTest and @JdbcTest in my case, crunchy test code in between) adding spring boot 2.1 flavour to it. But it seems it was not much of a success. I cannot annotate my tests with many @...Test since they are now each bringing their own XxxTestContextBootstrapper. It used to work when they all used same SpringBootTestContextBootstrapper. @RunWith(SpringRunner.class) @JdbcTest @JsonTest public class Test { @Test public void test() { System.out.printn("Hello, World !"); } } The error I get from BootstrapUtils is illegalStateException : Configuration error: found multiple declarations of @BootstrapWith for test class I understand I might be doing something wrong here but is there an easy way I could load both Json and Jdbc contex...
                              Read more

                              How to fix TextFormField cause rebuild widget in Flutter

                              Image
                              .everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty{ height:90px;width:728px;box-sizing:border-box; } 0 I've got the same issue at #9280. I tried all solutions from that topic with/without GlobalKey for Scaffold and Form but the problem wasn't fixed. I tried to build a authentication form with login and signup. But the keyboard doesn't show when I tap the TextFormField at the first time. Then I try to input some letter from physical keyboard, the widget is rebuilt. Like this: login screen Nothing to be log when I run flutter run --verbose, so I logged manually. Every tapping on field, the widget called dispose -> init -> build. [√] Flutter (Channel beta, v1.0.0, on Microsoft Windows [Version 10.0.17763.195], ...
                              Read more