exact-closed form solutions of recurrences












0














How can I find the exact-closed form solution of this recurrence?



$$f(n) = (1/2)f(n-2) + n/2,$$
where n is even, $f(0)=1$



My answer:



I try to guess a solution by unwinding the formula for f(n)



$f(n) = (1/2)f(n-2) + n/2$$



$= (1/2) ((1/2)f(n-4) + ((n-2)/2) )+n/2$



$=(1/2)^2f(n-4)+n/2^2+n/2 -2/2^2$



$=(1/2)^3f(n-6)+n/2^3+n/2^2+n/2 -2/2^2-4/2^3$



$=(1/2)^4f(n-8)+n/2^4+n/2^3+n/2^2+n/2 -2/2^2-4/2^3-6/2^4$



Continuing in this manner suggest the sum:



$f(n)=1/2^{(n/2)}f(0)+n(1/2^{n/2}+1/2^{(n/2)-1}+...+1/2)-(2/2^2+4/2^3+6/2^4+...)$



$f(n)=1/2^{(n/2)}+n(1-(1/2)^{n/2})-(2/2^2+4/2^3+6/2^4+...)$



How could I solve the expression $(2/2^2+4/2^3+6/2^4+...)$ and find the exact-closed form solution of this recurrence?






discrete-mathematics recurrence-relations







share|cite|improve this question



























    0














    How can I find the exact-closed form solution of this recurrence?



    $$f(n) = (1/2)f(n-2) + n/2,$$
    where n is even, $f(0)=1$



    My answer:



    I try to guess a solution by unwinding the formula for f(n)



    $f(n) = (1/2)f(n-2) + n/2$$



    $= (1/2) ((1/2)f(n-4) + ((n-2)/2) )+n/2$



    $=(1/2)^2f(n-4)+n/2^2+n/2 -2/2^2$



    $=(1/2)^3f(n-6)+n/2^3+n/2^2+n/2 -2/2^2-4/2^3$



    $=(1/2)^4f(n-8)+n/2^4+n/2^3+n/2^2+n/2 -2/2^2-4/2^3-6/2^4$



    Continuing in this manner suggest the sum:



    $f(n)=1/2^{(n/2)}f(0)+n(1/2^{n/2}+1/2^{(n/2)-1}+...+1/2)-(2/2^2+4/2^3+6/2^4+...)$



    $f(n)=1/2^{(n/2)}+n(1-(1/2)^{n/2})-(2/2^2+4/2^3+6/2^4+...)$



    How could I solve the expression $(2/2^2+4/2^3+6/2^4+...)$ and find the exact-closed form solution of this recurrence?






    discrete-mathematics recurrence-relations







    share|cite|improve this question

























      0












      0








      0







      How can I find the exact-closed form solution of this recurrence?



      $$f(n) = (1/2)f(n-2) + n/2,$$
      where n is even, $f(0)=1$



      My answer:



      I try to guess a solution by unwinding the formula for f(n)



      $f(n) = (1/2)f(n-2) + n/2$$



      $= (1/2) ((1/2)f(n-4) + ((n-2)/2) )+n/2$



      $=(1/2)^2f(n-4)+n/2^2+n/2 -2/2^2$



      $=(1/2)^3f(n-6)+n/2^3+n/2^2+n/2 -2/2^2-4/2^3$



      $=(1/2)^4f(n-8)+n/2^4+n/2^3+n/2^2+n/2 -2/2^2-4/2^3-6/2^4$



      Continuing in this manner suggest the sum:



      $f(n)=1/2^{(n/2)}f(0)+n(1/2^{n/2}+1/2^{(n/2)-1}+...+1/2)-(2/2^2+4/2^3+6/2^4+...)$



      $f(n)=1/2^{(n/2)}+n(1-(1/2)^{n/2})-(2/2^2+4/2^3+6/2^4+...)$



      How could I solve the expression $(2/2^2+4/2^3+6/2^4+...)$ and find the exact-closed form solution of this recurrence?






      discrete-mathematics recurrence-relations







      share|cite|improve this question













      How can I find the exact-closed form solution of this recurrence?



      $$f(n) = (1/2)f(n-2) + n/2,$$
      where n is even, $f(0)=1$



      My answer:



      I try to guess a solution by unwinding the formula for f(n)



      $f(n) = (1/2)f(n-2) + n/2$$



      $= (1/2) ((1/2)f(n-4) + ((n-2)/2) )+n/2$



      $=(1/2)^2f(n-4)+n/2^2+n/2 -2/2^2$



      $=(1/2)^3f(n-6)+n/2^3+n/2^2+n/2 -2/2^2-4/2^3$



      $=(1/2)^4f(n-8)+n/2^4+n/2^3+n/2^2+n/2 -2/2^2-4/2^3-6/2^4$



      Continuing in this manner suggest the sum:



      $f(n)=1/2^{(n/2)}f(0)+n(1/2^{n/2}+1/2^{(n/2)-1}+...+1/2)-(2/2^2+4/2^3+6/2^4+...)$



      $f(n)=1/2^{(n/2)}+n(1-(1/2)^{n/2})-(2/2^2+4/2^3+6/2^4+...)$



      How could I solve the expression $(2/2^2+4/2^3+6/2^4+...)$ and find the exact-closed form solution of this recurrence?






      discrete-mathematics recurrence-relations




      discrete-mathematics recurrence-relations






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Nov 26 '18 at 14:28









      user614642

      278




      278






















          3 Answers
          3






          active

          oldest

          votes


















          2














          Hint: Look for a solution in the form $$f(n) = A r^n + B n + C$$






          share|cite|improve this answer





















          • Sorry but I do not know how is this hint related to this question, could you please explain more about it?(There is only one way to solve this kind of recurrences in my lecture notes: unwind the formula and guess for a solution, so I do not know if there are other ways to solve it)
            – user614642
            Nov 26 '18 at 15:26



















          0














          Hint



          Let $g(m)=f(m)+a_0+a_1m+cdots$



          $n=2f(n)-f(n-2)=2g(n)-g(n-2)+a_0(2-1)+a_1{2n-(n-2)}$



          Set $a_i=0$ for $ige2$ and $a_1=1$ and $a_0+2a_1=0$



          so that $g(n)=dfrac{g(n-2)}2=dfrac{g(n-2r)}{2^r}$



          Set $2r=n$






          share|cite|improve this answer





















          • I don’t understand why 2f(n)-2f(n-2)=2g(n)-g(n-2)+a(2-1)+a1(2n-(n-2)),where does the term a(2-1)+a1(2n-(n-2)) come from?How do you know that?
            – user614642
            Nov 27 '18 at 3:36





















          0














          Your direct question is about the sum
          $$
          eqalign{
          & 2/2^{,2} + 4/2^{,3} + 6/2^{,4} + cdots = cr
          & = 1 cdot left( {{1 over 2}} right)^{,1} + 2 cdot left( {{1 over 2}} right)^{,2} + 3 cdot left( {{1 over 2}} right)^{,3} + cdots = cr
          & = sumlimits_{1, le ,k} k left( {{1 over 2}} right)^{,k} cr}
          $$



          Since
          $$
          eqalign{
          & sumlimits_{0, le ,left( {,1, le } right),k} {k,z^{,k} } = zsumlimits_{0, le ,k} {k,z^{,k - 1} }
          = z{d over {dz}}sumlimits_{0, le ,k} {z^{,k} } = cr
          & = z{d over {dz}}{1 over {left( {1 - z} right)}} = {z over {left( {1 - z} right)^{,2} }} cr}
          $$
          then the sum equals
          $$
          sumlimits_{1, le ,k} k left( {{1 over 2}} right)^{,k} = 2
          $$






          share|cite|improve this answer





















          • Sorry.Maybe I did not explain clearly about this question.The sequence 2/2^2 +4/2^2+... is a summation to n, not to infinity,and its first term 2/2^2 first appears when n=4, as you can see in my try to unwind the formula
            – user614642
            Nov 27 '18 at 3:26











          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%2f3014395%2fexact-closed-form-solutions-of-recurrences%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









          2














          Hint: Look for a solution in the form $$f(n) = A r^n + B n + C$$






          share|cite|improve this answer





















          • Sorry but I do not know how is this hint related to this question, could you please explain more about it?(There is only one way to solve this kind of recurrences in my lecture notes: unwind the formula and guess for a solution, so I do not know if there are other ways to solve it)
            – user614642
            Nov 26 '18 at 15:26
















          2














          Hint: Look for a solution in the form $$f(n) = A r^n + B n + C$$






          share|cite|improve this answer





















          • Sorry but I do not know how is this hint related to this question, could you please explain more about it?(There is only one way to solve this kind of recurrences in my lecture notes: unwind the formula and guess for a solution, so I do not know if there are other ways to solve it)
            – user614642
            Nov 26 '18 at 15:26














          2












          2








          2






          Hint: Look for a solution in the form $$f(n) = A r^n + B n + C$$






          share|cite|improve this answer












          Hint: Look for a solution in the form $$f(n) = A r^n + B n + C$$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Nov 26 '18 at 14:37









          Robert Israel

          318k23208457




          318k23208457












          • Sorry but I do not know how is this hint related to this question, could you please explain more about it?(There is only one way to solve this kind of recurrences in my lecture notes: unwind the formula and guess for a solution, so I do not know if there are other ways to solve it)
            – user614642
            Nov 26 '18 at 15:26


















          • Sorry but I do not know how is this hint related to this question, could you please explain more about it?(There is only one way to solve this kind of recurrences in my lecture notes: unwind the formula and guess for a solution, so I do not know if there are other ways to solve it)
            – user614642
            Nov 26 '18 at 15:26
















          Sorry but I do not know how is this hint related to this question, could you please explain more about it?(There is only one way to solve this kind of recurrences in my lecture notes: unwind the formula and guess for a solution, so I do not know if there are other ways to solve it)
          – user614642
          Nov 26 '18 at 15:26




          Sorry but I do not know how is this hint related to this question, could you please explain more about it?(There is only one way to solve this kind of recurrences in my lecture notes: unwind the formula and guess for a solution, so I do not know if there are other ways to solve it)
          – user614642
          Nov 26 '18 at 15:26











          0














          Hint



          Let $g(m)=f(m)+a_0+a_1m+cdots$



          $n=2f(n)-f(n-2)=2g(n)-g(n-2)+a_0(2-1)+a_1{2n-(n-2)}$



          Set $a_i=0$ for $ige2$ and $a_1=1$ and $a_0+2a_1=0$



          so that $g(n)=dfrac{g(n-2)}2=dfrac{g(n-2r)}{2^r}$



          Set $2r=n$






          share|cite|improve this answer





















          • I don’t understand why 2f(n)-2f(n-2)=2g(n)-g(n-2)+a(2-1)+a1(2n-(n-2)),where does the term a(2-1)+a1(2n-(n-2)) come from?How do you know that?
            – user614642
            Nov 27 '18 at 3:36


















          0














          Hint



          Let $g(m)=f(m)+a_0+a_1m+cdots$



          $n=2f(n)-f(n-2)=2g(n)-g(n-2)+a_0(2-1)+a_1{2n-(n-2)}$



          Set $a_i=0$ for $ige2$ and $a_1=1$ and $a_0+2a_1=0$



          so that $g(n)=dfrac{g(n-2)}2=dfrac{g(n-2r)}{2^r}$



          Set $2r=n$






          share|cite|improve this answer





















          • I don’t understand why 2f(n)-2f(n-2)=2g(n)-g(n-2)+a(2-1)+a1(2n-(n-2)),where does the term a(2-1)+a1(2n-(n-2)) come from?How do you know that?
            – user614642
            Nov 27 '18 at 3:36
















          0












          0








          0






          Hint



          Let $g(m)=f(m)+a_0+a_1m+cdots$



          $n=2f(n)-f(n-2)=2g(n)-g(n-2)+a_0(2-1)+a_1{2n-(n-2)}$



          Set $a_i=0$ for $ige2$ and $a_1=1$ and $a_0+2a_1=0$



          so that $g(n)=dfrac{g(n-2)}2=dfrac{g(n-2r)}{2^r}$



          Set $2r=n$






          share|cite|improve this answer












          Hint



          Let $g(m)=f(m)+a_0+a_1m+cdots$



          $n=2f(n)-f(n-2)=2g(n)-g(n-2)+a_0(2-1)+a_1{2n-(n-2)}$



          Set $a_i=0$ for $ige2$ and $a_1=1$ and $a_0+2a_1=0$



          so that $g(n)=dfrac{g(n-2)}2=dfrac{g(n-2r)}{2^r}$



          Set $2r=n$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Nov 26 '18 at 14:50









          lab bhattacharjee

          223k15156274




          223k15156274












          • I don’t understand why 2f(n)-2f(n-2)=2g(n)-g(n-2)+a(2-1)+a1(2n-(n-2)),where does the term a(2-1)+a1(2n-(n-2)) come from?How do you know that?
            – user614642
            Nov 27 '18 at 3:36




















          • I don’t understand why 2f(n)-2f(n-2)=2g(n)-g(n-2)+a(2-1)+a1(2n-(n-2)),where does the term a(2-1)+a1(2n-(n-2)) come from?How do you know that?
            – user614642
            Nov 27 '18 at 3:36


















          I don’t understand why 2f(n)-2f(n-2)=2g(n)-g(n-2)+a(2-1)+a1(2n-(n-2)),where does the term a(2-1)+a1(2n-(n-2)) come from?How do you know that?
          – user614642
          Nov 27 '18 at 3:36






          I don’t understand why 2f(n)-2f(n-2)=2g(n)-g(n-2)+a(2-1)+a1(2n-(n-2)),where does the term a(2-1)+a1(2n-(n-2)) come from?How do you know that?
          – user614642
          Nov 27 '18 at 3:36













          0














          Your direct question is about the sum
          $$
          eqalign{
          & 2/2^{,2} + 4/2^{,3} + 6/2^{,4} + cdots = cr
          & = 1 cdot left( {{1 over 2}} right)^{,1} + 2 cdot left( {{1 over 2}} right)^{,2} + 3 cdot left( {{1 over 2}} right)^{,3} + cdots = cr
          & = sumlimits_{1, le ,k} k left( {{1 over 2}} right)^{,k} cr}
          $$



          Since
          $$
          eqalign{
          & sumlimits_{0, le ,left( {,1, le } right),k} {k,z^{,k} } = zsumlimits_{0, le ,k} {k,z^{,k - 1} }
          = z{d over {dz}}sumlimits_{0, le ,k} {z^{,k} } = cr
          & = z{d over {dz}}{1 over {left( {1 - z} right)}} = {z over {left( {1 - z} right)^{,2} }} cr}
          $$
          then the sum equals
          $$
          sumlimits_{1, le ,k} k left( {{1 over 2}} right)^{,k} = 2
          $$






          share|cite|improve this answer





















          • Sorry.Maybe I did not explain clearly about this question.The sequence 2/2^2 +4/2^2+... is a summation to n, not to infinity,and its first term 2/2^2 first appears when n=4, as you can see in my try to unwind the formula
            – user614642
            Nov 27 '18 at 3:26
















          0














          Your direct question is about the sum
          $$
          eqalign{
          & 2/2^{,2} + 4/2^{,3} + 6/2^{,4} + cdots = cr
          & = 1 cdot left( {{1 over 2}} right)^{,1} + 2 cdot left( {{1 over 2}} right)^{,2} + 3 cdot left( {{1 over 2}} right)^{,3} + cdots = cr
          & = sumlimits_{1, le ,k} k left( {{1 over 2}} right)^{,k} cr}
          $$



          Since
          $$
          eqalign{
          & sumlimits_{0, le ,left( {,1, le } right),k} {k,z^{,k} } = zsumlimits_{0, le ,k} {k,z^{,k - 1} }
          = z{d over {dz}}sumlimits_{0, le ,k} {z^{,k} } = cr
          & = z{d over {dz}}{1 over {left( {1 - z} right)}} = {z over {left( {1 - z} right)^{,2} }} cr}
          $$
          then the sum equals
          $$
          sumlimits_{1, le ,k} k left( {{1 over 2}} right)^{,k} = 2
          $$






          share|cite|improve this answer





















          • Sorry.Maybe I did not explain clearly about this question.The sequence 2/2^2 +4/2^2+... is a summation to n, not to infinity,and its first term 2/2^2 first appears when n=4, as you can see in my try to unwind the formula
            – user614642
            Nov 27 '18 at 3:26














          0












          0








          0






          Your direct question is about the sum
          $$
          eqalign{
          & 2/2^{,2} + 4/2^{,3} + 6/2^{,4} + cdots = cr
          & = 1 cdot left( {{1 over 2}} right)^{,1} + 2 cdot left( {{1 over 2}} right)^{,2} + 3 cdot left( {{1 over 2}} right)^{,3} + cdots = cr
          & = sumlimits_{1, le ,k} k left( {{1 over 2}} right)^{,k} cr}
          $$



          Since
          $$
          eqalign{
          & sumlimits_{0, le ,left( {,1, le } right),k} {k,z^{,k} } = zsumlimits_{0, le ,k} {k,z^{,k - 1} }
          = z{d over {dz}}sumlimits_{0, le ,k} {z^{,k} } = cr
          & = z{d over {dz}}{1 over {left( {1 - z} right)}} = {z over {left( {1 - z} right)^{,2} }} cr}
          $$
          then the sum equals
          $$
          sumlimits_{1, le ,k} k left( {{1 over 2}} right)^{,k} = 2
          $$






          share|cite|improve this answer












          Your direct question is about the sum
          $$
          eqalign{
          & 2/2^{,2} + 4/2^{,3} + 6/2^{,4} + cdots = cr
          & = 1 cdot left( {{1 over 2}} right)^{,1} + 2 cdot left( {{1 over 2}} right)^{,2} + 3 cdot left( {{1 over 2}} right)^{,3} + cdots = cr
          & = sumlimits_{1, le ,k} k left( {{1 over 2}} right)^{,k} cr}
          $$



          Since
          $$
          eqalign{
          & sumlimits_{0, le ,left( {,1, le } right),k} {k,z^{,k} } = zsumlimits_{0, le ,k} {k,z^{,k - 1} }
          = z{d over {dz}}sumlimits_{0, le ,k} {z^{,k} } = cr
          & = z{d over {dz}}{1 over {left( {1 - z} right)}} = {z over {left( {1 - z} right)^{,2} }} cr}
          $$
          then the sum equals
          $$
          sumlimits_{1, le ,k} k left( {{1 over 2}} right)^{,k} = 2
          $$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Nov 26 '18 at 16:57









          G Cab

          18k31237




          18k31237












          • Sorry.Maybe I did not explain clearly about this question.The sequence 2/2^2 +4/2^2+... is a summation to n, not to infinity,and its first term 2/2^2 first appears when n=4, as you can see in my try to unwind the formula
            – user614642
            Nov 27 '18 at 3:26


















          • Sorry.Maybe I did not explain clearly about this question.The sequence 2/2^2 +4/2^2+... is a summation to n, not to infinity,and its first term 2/2^2 first appears when n=4, as you can see in my try to unwind the formula
            – user614642
            Nov 27 '18 at 3:26
















          Sorry.Maybe I did not explain clearly about this question.The sequence 2/2^2 +4/2^2+... is a summation to n, not to infinity,and its first term 2/2^2 first appears when n=4, as you can see in my try to unwind the formula
          – user614642
          Nov 27 '18 at 3:26




          Sorry.Maybe I did not explain clearly about this question.The sequence 2/2^2 +4/2^2+... is a summation to n, not to infinity,and its first term 2/2^2 first appears when n=4, as you can see in my try to unwind the formula
          – user614642
          Nov 27 '18 at 3:26


















          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%2f3014395%2fexact-closed-form-solutions-of-recurrences%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

          Plaza Victoria

          Image
          Plaza Victoria Plaza Victoria en 2017 Otros nombres Plaza de La Victoria Localización El Almendral, Valparaíso, Chile Vías adyacentes Chacabuco, Molina, Plaza Victoria y Edwards Creación 1841 Metro Bellavista Ubicación 33°02′46″S 71°37′11″O  /  -33.04624167, -71.61980833 Coordenadas: 33°02′46″S 71°37′11″O  /  -33.04624167, -71.61980833 [editar datos en Wikidata] La Plaza Victoria , también llamada Plaza de la Victoria , [ 1 ] ​ es una plaza ubicada en el sector El Almendral del plan de la ciudad de Valparaíso, Chile, de importancia patrimonial, que incluye valiosas estatuas y especies vegetales. [ 2 ] ​ La plaza se creó como tal en la primera mitad del siglo XIX, tomando previamente los nombres de Plaza Nueva y Plaza de Orrego . Si bien desde sus inicios tuvo relevancia histórica, no fue sino a partir de fines de los años 1860 cuando comenzó a cobrar mayor importancia y se convirtió en uno de los principales centros sociales de l
          Read more

          In PowerPoint, is there a keyboard shortcut for bulleted / numbered list?

          Image
          .everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty{ height:90px;width:728px;box-sizing:border-box; } 2 OneNote has two useful keyboard shortcuts for bulleted and numbered lists, namely Ctrl + . and Ctrl + - . Is there something similar for PowerPoint where the work with lists is also very common? Haven't found anything here: Use keyboard shortcuts in PowerPoint. keyboard-shortcuts microsoft-powerpoint share | improve this question edited Aug 14 '14 at 14:49 Linger 2,826 10 28 40
          Read more

          How to put 3 figures in Latex with 2 figures side by side and 1 below these side by side images but in...

          Image
          1 0 I want to put 3 images in LateX such as 2 figures are side by side horizotally and 3rd figure below these 2 side by side figures but in middle. I have following code: begin{figure}[H] centering begin{minipage}{.5textwidth} centering includegraphics[width=.50linewidth]{Text/Images/Genelec_8010_AP.jpg} captionof{figure}{Genelec 8010 AP} label{fig:Genelec 8010} end{minipage}% begin{minipage}{.5textwidth} centering includegraphics[width=.69linewidth]{Text/Images/Genelec_8020_CPM.jpg} captionof{figure}{Genelec 8020 CPM} label{fig:Genelec 8020} end{minipage}% end{figure} %%Here is my 3rd picture code begin{figure} centering includegraphics[width=5cm]{Text/Images/Genelec_8030_BPM.jpg} caption{Genelec 8030 BPM} label{fig:Genelec 8030} end{figure} I want to ask ho
          Read more