Summation of $frac{cos n theta}{2^n}$












3












$begingroup$


I would like to compute the following sum:



$$sum_{n=0}^{infty} frac{cos ntheta}{2^n}$$



I know that it involves using complex numbers, although I'm not sure how exactly I'm supposed to do so. I tried using the fact that $cos theta = {e^{itheta} + e^{-itheta}over 2}$. I'm not sure how to proceed from there though. A hint would be appreciated.






complex-numbers summation







share|cite|improve this question











$endgroup$

















    3












    $begingroup$


    I would like to compute the following sum:



    $$sum_{n=0}^{infty} frac{cos ntheta}{2^n}$$



    I know that it involves using complex numbers, although I'm not sure how exactly I'm supposed to do so. I tried using the fact that $cos theta = {e^{itheta} + e^{-itheta}over 2}$. I'm not sure how to proceed from there though. A hint would be appreciated.






    complex-numbers summation







    share|cite|improve this question











    $endgroup$















      3












      3








      3


      1



      $begingroup$


      I would like to compute the following sum:



      $$sum_{n=0}^{infty} frac{cos ntheta}{2^n}$$



      I know that it involves using complex numbers, although I'm not sure how exactly I'm supposed to do so. I tried using the fact that $cos theta = {e^{itheta} + e^{-itheta}over 2}$. I'm not sure how to proceed from there though. A hint would be appreciated.






      complex-numbers summation







      share|cite|improve this question











      $endgroup$




      I would like to compute the following sum:



      $$sum_{n=0}^{infty} frac{cos ntheta}{2^n}$$



      I know that it involves using complex numbers, although I'm not sure how exactly I'm supposed to do so. I tried using the fact that $cos theta = {e^{itheta} + e^{-itheta}over 2}$. I'm not sure how to proceed from there though. A hint would be appreciated.






      complex-numbers summation




      complex-numbers summation






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Nov 16 '15 at 7:26









      Mike Pierce

      11.7k103585




      11.7k103585










      asked Nov 16 '15 at 7:25









      Gummy bearsGummy bears

      1,89311532




      1,89311532






















          4 Answers
          4






          active

          oldest

          votes


















          3












          $begingroup$

          Consider the series
          $$S=sum_{n=0}^{infty}left(frac{e^{itheta}}{2}right)^n.$$
          This is a geometric series whose sum is
          $$S=frac{2}{2-e^{itheta}}.$$
          Now the real part of $S$ is the sum you are looking for.






          share|cite|improve this answer









          $endgroup$









          • 1




            $begingroup$
            Aha...... So I replace $e^{itheta}$ with $cos theta + isin theta$ and find the real part?
            $endgroup$
            – Gummy bears
            Nov 16 '15 at 7:38



















          0












          $begingroup$

          If you are only looking for a hint, write your sum as $$sum_{n=0}^inftyfrac{e^{intheta}+e^{-intheta}}{2cdot2^n}$$ Break it up as two sums, each of which are geometric, so you can use the geometric series formula.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Crap.... How did I miss that..... -_- Thanks though.
            $endgroup$
            – Gummy bears
            Nov 16 '15 at 7:37










          • $begingroup$
            Anurag's solution is better, once you get used to thinking of $cos(theta)$ as first and foremost the real part of $e^{itheta}$. This answer goes the opposite way, thinking of $cos(theta)$ as being built out of $e^{itheta}$ instead of the other way round.
            $endgroup$
            – alex.jordan
            Nov 16 '15 at 7:45



















          0












          $begingroup$

          I know I'm late but there's a slightly different solution I want to present that doesn't involve any exponentiation.



          We can write the summation as the real part of



          $$sum_{n=0}^{infty} frac{cos ntheta + isin mtheta}{2^n}$$



          Using De Moivre's theorem:



          $$sum_{n=0}^{infty} left(frac{cos theta + isin theta}{2}right)^n$$



          We can then calculate the real portion of the infinite sum.



          begin{align*}
          operatorname{Re}left(frac{1}{1-r}right)
          &= operatorname{Re}left(frac{1}{frac{2-(cos theta + isin theta)}{2}}right) \
          &= operatorname{Re}left(frac{2}{(2-costheta) - isintheta}right) \
          &= operatorname{Re}left(frac{2((2-costheta) + isintheta)}{4-4costheta+1}right) \
          &= frac{4-2costheta}{4-4costheta+1}
          end{align*}



          Hope this helps!






          share|cite|improve this answer











          $endgroup$





















            -1












            $begingroup$

            begin{align*}
            sum_{n=0}^{infty} frac{cos ntheta}{2^n}
            &= sum_{n=0}^infty frac{e^{i n theta} + e^{- i n theta}}{2^{n+1}} \
            &= sum_{n=0}^infty frac{(e^{i theta})^n + (e^{- i theta})^n}{2^{n+1}} \
            &= sum_{n=0}^infty frac{(e^{i theta})^n}{2^{n+1}} + sum_{n=0}^infty frac{(e^{- i theta})^n}{2^{n+1}} \
            &= sum_{n=0}^infty frac12 left(frac{e^{i theta}}{2}right)^n
            + sum_{n=0}^infty frac12 left(frac{e^{-i theta}}{2}right)^n .
            end{align*}



            These last two sums are geometric series. Can you finish it?






            share|cite|improve this answer









            $endgroup$














              Your Answer








              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%2f1531525%2fsummation-of-frac-cos-n-theta2n%23new-answer', 'question_page');
              }
              );

              Post as a guest















              Required, but never shown

























              4 Answers
              4






              active

              oldest

              votes








              4 Answers
              4






              active

              oldest

              votes









              active

              oldest

              votes






              active

              oldest

              votes









              3












              $begingroup$

              Consider the series
              $$S=sum_{n=0}^{infty}left(frac{e^{itheta}}{2}right)^n.$$
              This is a geometric series whose sum is
              $$S=frac{2}{2-e^{itheta}}.$$
              Now the real part of $S$ is the sum you are looking for.






              share|cite|improve this answer









              $endgroup$









              • 1




                $begingroup$
                Aha...... So I replace $e^{itheta}$ with $cos theta + isin theta$ and find the real part?
                $endgroup$
                – Gummy bears
                Nov 16 '15 at 7:38
















              3












              $begingroup$

              Consider the series
              $$S=sum_{n=0}^{infty}left(frac{e^{itheta}}{2}right)^n.$$
              This is a geometric series whose sum is
              $$S=frac{2}{2-e^{itheta}}.$$
              Now the real part of $S$ is the sum you are looking for.






              share|cite|improve this answer









              $endgroup$









              • 1




                $begingroup$
                Aha...... So I replace $e^{itheta}$ with $cos theta + isin theta$ and find the real part?
                $endgroup$
                – Gummy bears
                Nov 16 '15 at 7:38














              3












              3








              3





              $begingroup$

              Consider the series
              $$S=sum_{n=0}^{infty}left(frac{e^{itheta}}{2}right)^n.$$
              This is a geometric series whose sum is
              $$S=frac{2}{2-e^{itheta}}.$$
              Now the real part of $S$ is the sum you are looking for.






              share|cite|improve this answer









              $endgroup$



              Consider the series
              $$S=sum_{n=0}^{infty}left(frac{e^{itheta}}{2}right)^n.$$
              This is a geometric series whose sum is
              $$S=frac{2}{2-e^{itheta}}.$$
              Now the real part of $S$ is the sum you are looking for.







              share|cite|improve this answer












              share|cite|improve this answer



              share|cite|improve this answer










              answered Nov 16 '15 at 7:30









              Anurag AAnurag A

              26.4k12351




              26.4k12351








              • 1




                $begingroup$
                Aha...... So I replace $e^{itheta}$ with $cos theta + isin theta$ and find the real part?
                $endgroup$
                – Gummy bears
                Nov 16 '15 at 7:38














              • 1




                $begingroup$
                Aha...... So I replace $e^{itheta}$ with $cos theta + isin theta$ and find the real part?
                $endgroup$
                – Gummy bears
                Nov 16 '15 at 7:38








              1




              1




              $begingroup$
              Aha...... So I replace $e^{itheta}$ with $cos theta + isin theta$ and find the real part?
              $endgroup$
              – Gummy bears
              Nov 16 '15 at 7:38




              $begingroup$
              Aha...... So I replace $e^{itheta}$ with $cos theta + isin theta$ and find the real part?
              $endgroup$
              – Gummy bears
              Nov 16 '15 at 7:38











              0












              $begingroup$

              If you are only looking for a hint, write your sum as $$sum_{n=0}^inftyfrac{e^{intheta}+e^{-intheta}}{2cdot2^n}$$ Break it up as two sums, each of which are geometric, so you can use the geometric series formula.






              share|cite|improve this answer









              $endgroup$













              • $begingroup$
                Crap.... How did I miss that..... -_- Thanks though.
                $endgroup$
                – Gummy bears
                Nov 16 '15 at 7:37










              • $begingroup$
                Anurag's solution is better, once you get used to thinking of $cos(theta)$ as first and foremost the real part of $e^{itheta}$. This answer goes the opposite way, thinking of $cos(theta)$ as being built out of $e^{itheta}$ instead of the other way round.
                $endgroup$
                – alex.jordan
                Nov 16 '15 at 7:45
















              0












              $begingroup$

              If you are only looking for a hint, write your sum as $$sum_{n=0}^inftyfrac{e^{intheta}+e^{-intheta}}{2cdot2^n}$$ Break it up as two sums, each of which are geometric, so you can use the geometric series formula.






              share|cite|improve this answer









              $endgroup$













              • $begingroup$
                Crap.... How did I miss that..... -_- Thanks though.
                $endgroup$
                – Gummy bears
                Nov 16 '15 at 7:37










              • $begingroup$
                Anurag's solution is better, once you get used to thinking of $cos(theta)$ as first and foremost the real part of $e^{itheta}$. This answer goes the opposite way, thinking of $cos(theta)$ as being built out of $e^{itheta}$ instead of the other way round.
                $endgroup$
                – alex.jordan
                Nov 16 '15 at 7:45














              0












              0








              0





              $begingroup$

              If you are only looking for a hint, write your sum as $$sum_{n=0}^inftyfrac{e^{intheta}+e^{-intheta}}{2cdot2^n}$$ Break it up as two sums, each of which are geometric, so you can use the geometric series formula.






              share|cite|improve this answer









              $endgroup$



              If you are only looking for a hint, write your sum as $$sum_{n=0}^inftyfrac{e^{intheta}+e^{-intheta}}{2cdot2^n}$$ Break it up as two sums, each of which are geometric, so you can use the geometric series formula.







              share|cite|improve this answer












              share|cite|improve this answer



              share|cite|improve this answer










              answered Nov 16 '15 at 7:30









              alex.jordanalex.jordan

              39.7k660122




              39.7k660122












              • $begingroup$
                Crap.... How did I miss that..... -_- Thanks though.
                $endgroup$
                – Gummy bears
                Nov 16 '15 at 7:37










              • $begingroup$
                Anurag's solution is better, once you get used to thinking of $cos(theta)$ as first and foremost the real part of $e^{itheta}$. This answer goes the opposite way, thinking of $cos(theta)$ as being built out of $e^{itheta}$ instead of the other way round.
                $endgroup$
                – alex.jordan
                Nov 16 '15 at 7:45


















              • $begingroup$
                Crap.... How did I miss that..... -_- Thanks though.
                $endgroup$
                – Gummy bears
                Nov 16 '15 at 7:37










              • $begingroup$
                Anurag's solution is better, once you get used to thinking of $cos(theta)$ as first and foremost the real part of $e^{itheta}$. This answer goes the opposite way, thinking of $cos(theta)$ as being built out of $e^{itheta}$ instead of the other way round.
                $endgroup$
                – alex.jordan
                Nov 16 '15 at 7:45
















              $begingroup$
              Crap.... How did I miss that..... -_- Thanks though.
              $endgroup$
              – Gummy bears
              Nov 16 '15 at 7:37




              $begingroup$
              Crap.... How did I miss that..... -_- Thanks though.
              $endgroup$
              – Gummy bears
              Nov 16 '15 at 7:37












              $begingroup$
              Anurag's solution is better, once you get used to thinking of $cos(theta)$ as first and foremost the real part of $e^{itheta}$. This answer goes the opposite way, thinking of $cos(theta)$ as being built out of $e^{itheta}$ instead of the other way round.
              $endgroup$
              – alex.jordan
              Nov 16 '15 at 7:45




              $begingroup$
              Anurag's solution is better, once you get used to thinking of $cos(theta)$ as first and foremost the real part of $e^{itheta}$. This answer goes the opposite way, thinking of $cos(theta)$ as being built out of $e^{itheta}$ instead of the other way round.
              $endgroup$
              – alex.jordan
              Nov 16 '15 at 7:45











              0












              $begingroup$

              I know I'm late but there's a slightly different solution I want to present that doesn't involve any exponentiation.



              We can write the summation as the real part of



              $$sum_{n=0}^{infty} frac{cos ntheta + isin mtheta}{2^n}$$



              Using De Moivre's theorem:



              $$sum_{n=0}^{infty} left(frac{cos theta + isin theta}{2}right)^n$$



              We can then calculate the real portion of the infinite sum.



              begin{align*}
              operatorname{Re}left(frac{1}{1-r}right)
              &= operatorname{Re}left(frac{1}{frac{2-(cos theta + isin theta)}{2}}right) \
              &= operatorname{Re}left(frac{2}{(2-costheta) - isintheta}right) \
              &= operatorname{Re}left(frac{2((2-costheta) + isintheta)}{4-4costheta+1}right) \
              &= frac{4-2costheta}{4-4costheta+1}
              end{align*}



              Hope this helps!






              share|cite|improve this answer











              $endgroup$


















                0












                $begingroup$

                I know I'm late but there's a slightly different solution I want to present that doesn't involve any exponentiation.



                We can write the summation as the real part of



                $$sum_{n=0}^{infty} frac{cos ntheta + isin mtheta}{2^n}$$



                Using De Moivre's theorem:



                $$sum_{n=0}^{infty} left(frac{cos theta + isin theta}{2}right)^n$$



                We can then calculate the real portion of the infinite sum.



                begin{align*}
                operatorname{Re}left(frac{1}{1-r}right)
                &= operatorname{Re}left(frac{1}{frac{2-(cos theta + isin theta)}{2}}right) \
                &= operatorname{Re}left(frac{2}{(2-costheta) - isintheta}right) \
                &= operatorname{Re}left(frac{2((2-costheta) + isintheta)}{4-4costheta+1}right) \
                &= frac{4-2costheta}{4-4costheta+1}
                end{align*}



                Hope this helps!






                share|cite|improve this answer











                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  I know I'm late but there's a slightly different solution I want to present that doesn't involve any exponentiation.



                  We can write the summation as the real part of



                  $$sum_{n=0}^{infty} frac{cos ntheta + isin mtheta}{2^n}$$



                  Using De Moivre's theorem:



                  $$sum_{n=0}^{infty} left(frac{cos theta + isin theta}{2}right)^n$$



                  We can then calculate the real portion of the infinite sum.



                  begin{align*}
                  operatorname{Re}left(frac{1}{1-r}right)
                  &= operatorname{Re}left(frac{1}{frac{2-(cos theta + isin theta)}{2}}right) \
                  &= operatorname{Re}left(frac{2}{(2-costheta) - isintheta}right) \
                  &= operatorname{Re}left(frac{2((2-costheta) + isintheta)}{4-4costheta+1}right) \
                  &= frac{4-2costheta}{4-4costheta+1}
                  end{align*}



                  Hope this helps!






                  share|cite|improve this answer











                  $endgroup$



                  I know I'm late but there's a slightly different solution I want to present that doesn't involve any exponentiation.



                  We can write the summation as the real part of



                  $$sum_{n=0}^{infty} frac{cos ntheta + isin mtheta}{2^n}$$



                  Using De Moivre's theorem:



                  $$sum_{n=0}^{infty} left(frac{cos theta + isin theta}{2}right)^n$$



                  We can then calculate the real portion of the infinite sum.



                  begin{align*}
                  operatorname{Re}left(frac{1}{1-r}right)
                  &= operatorname{Re}left(frac{1}{frac{2-(cos theta + isin theta)}{2}}right) \
                  &= operatorname{Re}left(frac{2}{(2-costheta) - isintheta}right) \
                  &= operatorname{Re}left(frac{2((2-costheta) + isintheta)}{4-4costheta+1}right) \
                  &= frac{4-2costheta}{4-4costheta+1}
                  end{align*}



                  Hope this helps!







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Feb 3 at 1:02









                  Hanul Jeon

                  17.7k42881




                  17.7k42881










                  answered Feb 3 at 0:48









                  QiLin XueQiLin Xue

                  11




                  11























                      -1












                      $begingroup$

                      begin{align*}
                      sum_{n=0}^{infty} frac{cos ntheta}{2^n}
                      &= sum_{n=0}^infty frac{e^{i n theta} + e^{- i n theta}}{2^{n+1}} \
                      &= sum_{n=0}^infty frac{(e^{i theta})^n + (e^{- i theta})^n}{2^{n+1}} \
                      &= sum_{n=0}^infty frac{(e^{i theta})^n}{2^{n+1}} + sum_{n=0}^infty frac{(e^{- i theta})^n}{2^{n+1}} \
                      &= sum_{n=0}^infty frac12 left(frac{e^{i theta}}{2}right)^n
                      + sum_{n=0}^infty frac12 left(frac{e^{-i theta}}{2}right)^n .
                      end{align*}



                      These last two sums are geometric series. Can you finish it?






                      share|cite|improve this answer









                      $endgroup$


















                        -1












                        $begingroup$

                        begin{align*}
                        sum_{n=0}^{infty} frac{cos ntheta}{2^n}
                        &= sum_{n=0}^infty frac{e^{i n theta} + e^{- i n theta}}{2^{n+1}} \
                        &= sum_{n=0}^infty frac{(e^{i theta})^n + (e^{- i theta})^n}{2^{n+1}} \
                        &= sum_{n=0}^infty frac{(e^{i theta})^n}{2^{n+1}} + sum_{n=0}^infty frac{(e^{- i theta})^n}{2^{n+1}} \
                        &= sum_{n=0}^infty frac12 left(frac{e^{i theta}}{2}right)^n
                        + sum_{n=0}^infty frac12 left(frac{e^{-i theta}}{2}right)^n .
                        end{align*}



                        These last two sums are geometric series. Can you finish it?






                        share|cite|improve this answer









                        $endgroup$
















                          -1












                          -1








                          -1





                          $begingroup$

                          begin{align*}
                          sum_{n=0}^{infty} frac{cos ntheta}{2^n}
                          &= sum_{n=0}^infty frac{e^{i n theta} + e^{- i n theta}}{2^{n+1}} \
                          &= sum_{n=0}^infty frac{(e^{i theta})^n + (e^{- i theta})^n}{2^{n+1}} \
                          &= sum_{n=0}^infty frac{(e^{i theta})^n}{2^{n+1}} + sum_{n=0}^infty frac{(e^{- i theta})^n}{2^{n+1}} \
                          &= sum_{n=0}^infty frac12 left(frac{e^{i theta}}{2}right)^n
                          + sum_{n=0}^infty frac12 left(frac{e^{-i theta}}{2}right)^n .
                          end{align*}



                          These last two sums are geometric series. Can you finish it?






                          share|cite|improve this answer









                          $endgroup$



                          begin{align*}
                          sum_{n=0}^{infty} frac{cos ntheta}{2^n}
                          &= sum_{n=0}^infty frac{e^{i n theta} + e^{- i n theta}}{2^{n+1}} \
                          &= sum_{n=0}^infty frac{(e^{i theta})^n + (e^{- i theta})^n}{2^{n+1}} \
                          &= sum_{n=0}^infty frac{(e^{i theta})^n}{2^{n+1}} + sum_{n=0}^infty frac{(e^{- i theta})^n}{2^{n+1}} \
                          &= sum_{n=0}^infty frac12 left(frac{e^{i theta}}{2}right)^n
                          + sum_{n=0}^infty frac12 left(frac{e^{-i theta}}{2}right)^n .
                          end{align*}



                          These last two sums are geometric series. Can you finish it?







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered Nov 16 '15 at 7:34









                          60056005

                          37.1k752127




                          37.1k752127






























                              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%2f1531525%2fsummation-of-frac-cos-n-theta2n%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

                              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

                              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