Can $a(n) = frac{n}{n+1}$ be written recursively?












2












$begingroup$


Take the sequence $$frac{1}{2}, frac{2}{3}, frac{3}{4}, frac{4}{5}, frac{5}{6}, frac{6}{7}, dots$$



Algebraically it can be written as $$a(n) = frac{n}{n + 1}$$



Can you write this as a recursive function as well?



A pattern I have noticed:




  • Take $A_{n-1}$ and then inverse it. All you have to do is add two to the denominator. However, it is the denominator increase that causes a problem here.


I am currently in Algebra II Honors and learning sequences






sequences-and-series recursion







share|cite|improve this question









New contributor




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







$endgroup$

















    2












    $begingroup$


    Take the sequence $$frac{1}{2}, frac{2}{3}, frac{3}{4}, frac{4}{5}, frac{5}{6}, frac{6}{7}, dots$$



    Algebraically it can be written as $$a(n) = frac{n}{n + 1}$$



    Can you write this as a recursive function as well?



    A pattern I have noticed:




    • Take $A_{n-1}$ and then inverse it. All you have to do is add two to the denominator. However, it is the denominator increase that causes a problem here.


    I am currently in Algebra II Honors and learning sequences






    sequences-and-series recursion







    share|cite|improve this question









    New contributor




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







    $endgroup$















      2












      2








      2


      1



      $begingroup$


      Take the sequence $$frac{1}{2}, frac{2}{3}, frac{3}{4}, frac{4}{5}, frac{5}{6}, frac{6}{7}, dots$$



      Algebraically it can be written as $$a(n) = frac{n}{n + 1}$$



      Can you write this as a recursive function as well?



      A pattern I have noticed:




      • Take $A_{n-1}$ and then inverse it. All you have to do is add two to the denominator. However, it is the denominator increase that causes a problem here.


      I am currently in Algebra II Honors and learning sequences






      sequences-and-series recursion







      share|cite|improve this question









      New contributor




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







      $endgroup$




      Take the sequence $$frac{1}{2}, frac{2}{3}, frac{3}{4}, frac{4}{5}, frac{5}{6}, frac{6}{7}, dots$$



      Algebraically it can be written as $$a(n) = frac{n}{n + 1}$$



      Can you write this as a recursive function as well?



      A pattern I have noticed:




      • Take $A_{n-1}$ and then inverse it. All you have to do is add two to the denominator. However, it is the denominator increase that causes a problem here.


      I am currently in Algebra II Honors and learning sequences






      sequences-and-series recursion




      sequences-and-series recursion






      share|cite|improve this question









      New contributor




      Levi K 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




      Levi K 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 yesterday









      Jyrki Lahtonen

      110k13172390




      110k13172390






      New contributor




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









      asked yesterday









      Levi KLevi K

      435




      435




      New contributor




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





      New contributor





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






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






















          5 Answers
          5






          active

          oldest

          votes


















          6












          $begingroup$

          begin{align*}
          a_{n+1} &= frac{n+1}{n+2} \
          &= frac{n+2-1}{n+2} \
          &= 1 - frac{1}{n+2} text{, so } \
          1 - a_{n+1} &= frac{1}{n+2} text{, } \
          frac{1}{1 - a_{n+1}} &= n+2 &[text{and so } frac{1}{1 - a_n} = n+1]\
          &= n+1+1 \
          &= frac{1}{1- a_n} +1 \
          &= frac{1}{1- a_n} + frac{1-a_n}{1-a_n} \
          &= frac{2-a_n}{1- a_n} text{, then } \
          1 - a_{n+1} &= frac{1-a_n}{2- a_n} text{, and finally } \
          a_{n+1} &= 1 - frac{1-a_n}{2- a_n} \
          &= frac{2-a_n}{2- a_n} - frac{1-a_n}{2- a_n} \
          &= frac{1}{2- a_n} text{.}
          end{align*}






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Brilliant! I got the same answer, but I appreciate all the work you put in showing how to get there. I just found it by noticing a pattern with 2's, but your solution is a great mathematical approach. Thank you!
            $endgroup$
            – Levi K
            15 hours ago



















          3












          $begingroup$

          After some further solving, I was able to come up with an answer



          It can be written $${A_{n + 1}} = frac{1}{2 - A_{n}}$$ where $$A_1 = frac{1}{2}$$






          share|cite|improve this answer










          New contributor




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






          $endgroup$





















            1












            $begingroup$

            Just by playing around with some numbers, I determined a recursive relation to be



            $$a_n = frac{na_{n-1} + 1}{n+1}$$



            with $a_1 = 1/2$. I mostly derived this by noticing developing this would be easier if I negated the denominator of the previous term (thus multiplying by $n$), adding $1$ to the result (giving the increase in $1$ in the denominator), and then dividing again by the desired denominator ($n+1$).






            share|cite|improve this answer









            $endgroup$





















              0












              $begingroup$

              You can, and in multiple ways. Such as:$$a_n=frac{1}{a_{n-1}+frac{2}{n}}tag{1}$$



              or



              $$a_n=frac{na_{n-1}+1}{n+1}tag{2}$$



              etc.



              equation 1, is simply noting: $$frac{n-1}{n}+frac{2}{n}=frac{n+1}{n}=frac{1}{a_n}$$ Where the first fraction in the sum is $a_{n-1}$



              and equation 2, simply notes:



              $$n=na_{n-1}+1$$



              etc.






              share|cite|improve this answer









              $endgroup$





















                0












                $begingroup$

                Perhaps a bit simpler is to note that
                $$
                overbrace{ frac1{1-a_n} }^{n+1}+1=overbrace{frac1{1-a_{n+1}}}^{n+2}\
                $$
                solving for $a_{n+1}$ yields
                $$
                frac1{2-a_n}=a_{n+1}\
                $$






                share|cite|improve this answer









                $endgroup$














                  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
                  });


                  }
                  });






                  Levi K 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%2f3176633%2fcan-an-fracnn1-be-written-recursively%23new-answer', 'question_page');
                  }
                  );

                  Post as a guest















                  Required, but never shown

























                  5 Answers
                  5






                  active

                  oldest

                  votes








                  5 Answers
                  5






                  active

                  oldest

                  votes









                  active

                  oldest

                  votes






                  active

                  oldest

                  votes









                  6












                  $begingroup$

                  begin{align*}
                  a_{n+1} &= frac{n+1}{n+2} \
                  &= frac{n+2-1}{n+2} \
                  &= 1 - frac{1}{n+2} text{, so } \
                  1 - a_{n+1} &= frac{1}{n+2} text{, } \
                  frac{1}{1 - a_{n+1}} &= n+2 &[text{and so } frac{1}{1 - a_n} = n+1]\
                  &= n+1+1 \
                  &= frac{1}{1- a_n} +1 \
                  &= frac{1}{1- a_n} + frac{1-a_n}{1-a_n} \
                  &= frac{2-a_n}{1- a_n} text{, then } \
                  1 - a_{n+1} &= frac{1-a_n}{2- a_n} text{, and finally } \
                  a_{n+1} &= 1 - frac{1-a_n}{2- a_n} \
                  &= frac{2-a_n}{2- a_n} - frac{1-a_n}{2- a_n} \
                  &= frac{1}{2- a_n} text{.}
                  end{align*}






                  share|cite|improve this answer









                  $endgroup$













                  • $begingroup$
                    Brilliant! I got the same answer, but I appreciate all the work you put in showing how to get there. I just found it by noticing a pattern with 2's, but your solution is a great mathematical approach. Thank you!
                    $endgroup$
                    – Levi K
                    15 hours ago
















                  6












                  $begingroup$

                  begin{align*}
                  a_{n+1} &= frac{n+1}{n+2} \
                  &= frac{n+2-1}{n+2} \
                  &= 1 - frac{1}{n+2} text{, so } \
                  1 - a_{n+1} &= frac{1}{n+2} text{, } \
                  frac{1}{1 - a_{n+1}} &= n+2 &[text{and so } frac{1}{1 - a_n} = n+1]\
                  &= n+1+1 \
                  &= frac{1}{1- a_n} +1 \
                  &= frac{1}{1- a_n} + frac{1-a_n}{1-a_n} \
                  &= frac{2-a_n}{1- a_n} text{, then } \
                  1 - a_{n+1} &= frac{1-a_n}{2- a_n} text{, and finally } \
                  a_{n+1} &= 1 - frac{1-a_n}{2- a_n} \
                  &= frac{2-a_n}{2- a_n} - frac{1-a_n}{2- a_n} \
                  &= frac{1}{2- a_n} text{.}
                  end{align*}






                  share|cite|improve this answer









                  $endgroup$













                  • $begingroup$
                    Brilliant! I got the same answer, but I appreciate all the work you put in showing how to get there. I just found it by noticing a pattern with 2's, but your solution is a great mathematical approach. Thank you!
                    $endgroup$
                    – Levi K
                    15 hours ago














                  6












                  6








                  6





                  $begingroup$

                  begin{align*}
                  a_{n+1} &= frac{n+1}{n+2} \
                  &= frac{n+2-1}{n+2} \
                  &= 1 - frac{1}{n+2} text{, so } \
                  1 - a_{n+1} &= frac{1}{n+2} text{, } \
                  frac{1}{1 - a_{n+1}} &= n+2 &[text{and so } frac{1}{1 - a_n} = n+1]\
                  &= n+1+1 \
                  &= frac{1}{1- a_n} +1 \
                  &= frac{1}{1- a_n} + frac{1-a_n}{1-a_n} \
                  &= frac{2-a_n}{1- a_n} text{, then } \
                  1 - a_{n+1} &= frac{1-a_n}{2- a_n} text{, and finally } \
                  a_{n+1} &= 1 - frac{1-a_n}{2- a_n} \
                  &= frac{2-a_n}{2- a_n} - frac{1-a_n}{2- a_n} \
                  &= frac{1}{2- a_n} text{.}
                  end{align*}






                  share|cite|improve this answer









                  $endgroup$



                  begin{align*}
                  a_{n+1} &= frac{n+1}{n+2} \
                  &= frac{n+2-1}{n+2} \
                  &= 1 - frac{1}{n+2} text{, so } \
                  1 - a_{n+1} &= frac{1}{n+2} text{, } \
                  frac{1}{1 - a_{n+1}} &= n+2 &[text{and so } frac{1}{1 - a_n} = n+1]\
                  &= n+1+1 \
                  &= frac{1}{1- a_n} +1 \
                  &= frac{1}{1- a_n} + frac{1-a_n}{1-a_n} \
                  &= frac{2-a_n}{1- a_n} text{, then } \
                  1 - a_{n+1} &= frac{1-a_n}{2- a_n} text{, and finally } \
                  a_{n+1} &= 1 - frac{1-a_n}{2- a_n} \
                  &= frac{2-a_n}{2- a_n} - frac{1-a_n}{2- a_n} \
                  &= frac{1}{2- a_n} text{.}
                  end{align*}







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered yesterday









                  Eric TowersEric Towers

                  33.6k22370




                  33.6k22370












                  • $begingroup$
                    Brilliant! I got the same answer, but I appreciate all the work you put in showing how to get there. I just found it by noticing a pattern with 2's, but your solution is a great mathematical approach. Thank you!
                    $endgroup$
                    – Levi K
                    15 hours ago


















                  • $begingroup$
                    Brilliant! I got the same answer, but I appreciate all the work you put in showing how to get there. I just found it by noticing a pattern with 2's, but your solution is a great mathematical approach. Thank you!
                    $endgroup$
                    – Levi K
                    15 hours ago
















                  $begingroup$
                  Brilliant! I got the same answer, but I appreciate all the work you put in showing how to get there. I just found it by noticing a pattern with 2's, but your solution is a great mathematical approach. Thank you!
                  $endgroup$
                  – Levi K
                  15 hours ago




                  $begingroup$
                  Brilliant! I got the same answer, but I appreciate all the work you put in showing how to get there. I just found it by noticing a pattern with 2's, but your solution is a great mathematical approach. Thank you!
                  $endgroup$
                  – Levi K
                  15 hours ago











                  3












                  $begingroup$

                  After some further solving, I was able to come up with an answer



                  It can be written $${A_{n + 1}} = frac{1}{2 - A_{n}}$$ where $$A_1 = frac{1}{2}$$






                  share|cite|improve this answer










                  New contributor




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






                  $endgroup$


















                    3












                    $begingroup$

                    After some further solving, I was able to come up with an answer



                    It can be written $${A_{n + 1}} = frac{1}{2 - A_{n}}$$ where $$A_1 = frac{1}{2}$$






                    share|cite|improve this answer










                    New contributor




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






                    $endgroup$
















                      3












                      3








                      3





                      $begingroup$

                      After some further solving, I was able to come up with an answer



                      It can be written $${A_{n + 1}} = frac{1}{2 - A_{n}}$$ where $$A_1 = frac{1}{2}$$






                      share|cite|improve this answer










                      New contributor




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






                      $endgroup$



                      After some further solving, I was able to come up with an answer



                      It can be written $${A_{n + 1}} = frac{1}{2 - A_{n}}$$ where $$A_1 = frac{1}{2}$$







                      share|cite|improve this answer










                      New contributor




                      Levi K 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








                      edited yesterday









                      user1952500

                      833712




                      833712






                      New contributor




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









                      answered yesterday









                      Levi KLevi K

                      435




                      435




                      New contributor




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





                      New contributor





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






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























                          1












                          $begingroup$

                          Just by playing around with some numbers, I determined a recursive relation to be



                          $$a_n = frac{na_{n-1} + 1}{n+1}$$



                          with $a_1 = 1/2$. I mostly derived this by noticing developing this would be easier if I negated the denominator of the previous term (thus multiplying by $n$), adding $1$ to the result (giving the increase in $1$ in the denominator), and then dividing again by the desired denominator ($n+1$).






                          share|cite|improve this answer









                          $endgroup$


















                            1












                            $begingroup$

                            Just by playing around with some numbers, I determined a recursive relation to be



                            $$a_n = frac{na_{n-1} + 1}{n+1}$$



                            with $a_1 = 1/2$. I mostly derived this by noticing developing this would be easier if I negated the denominator of the previous term (thus multiplying by $n$), adding $1$ to the result (giving the increase in $1$ in the denominator), and then dividing again by the desired denominator ($n+1$).






                            share|cite|improve this answer









                            $endgroup$
















                              1












                              1








                              1





                              $begingroup$

                              Just by playing around with some numbers, I determined a recursive relation to be



                              $$a_n = frac{na_{n-1} + 1}{n+1}$$



                              with $a_1 = 1/2$. I mostly derived this by noticing developing this would be easier if I negated the denominator of the previous term (thus multiplying by $n$), adding $1$ to the result (giving the increase in $1$ in the denominator), and then dividing again by the desired denominator ($n+1$).






                              share|cite|improve this answer









                              $endgroup$



                              Just by playing around with some numbers, I determined a recursive relation to be



                              $$a_n = frac{na_{n-1} + 1}{n+1}$$



                              with $a_1 = 1/2$. I mostly derived this by noticing developing this would be easier if I negated the denominator of the previous term (thus multiplying by $n$), adding $1$ to the result (giving the increase in $1$ in the denominator), and then dividing again by the desired denominator ($n+1$).







                              share|cite|improve this answer












                              share|cite|improve this answer



                              share|cite|improve this answer










                              answered yesterday









                              Eevee TrainerEevee Trainer

                              10k31742




                              10k31742























                                  0












                                  $begingroup$

                                  You can, and in multiple ways. Such as:$$a_n=frac{1}{a_{n-1}+frac{2}{n}}tag{1}$$



                                  or



                                  $$a_n=frac{na_{n-1}+1}{n+1}tag{2}$$



                                  etc.



                                  equation 1, is simply noting: $$frac{n-1}{n}+frac{2}{n}=frac{n+1}{n}=frac{1}{a_n}$$ Where the first fraction in the sum is $a_{n-1}$



                                  and equation 2, simply notes:



                                  $$n=na_{n-1}+1$$



                                  etc.






                                  share|cite|improve this answer









                                  $endgroup$


















                                    0












                                    $begingroup$

                                    You can, and in multiple ways. Such as:$$a_n=frac{1}{a_{n-1}+frac{2}{n}}tag{1}$$



                                    or



                                    $$a_n=frac{na_{n-1}+1}{n+1}tag{2}$$



                                    etc.



                                    equation 1, is simply noting: $$frac{n-1}{n}+frac{2}{n}=frac{n+1}{n}=frac{1}{a_n}$$ Where the first fraction in the sum is $a_{n-1}$



                                    and equation 2, simply notes:



                                    $$n=na_{n-1}+1$$



                                    etc.






                                    share|cite|improve this answer









                                    $endgroup$
















                                      0












                                      0








                                      0





                                      $begingroup$

                                      You can, and in multiple ways. Such as:$$a_n=frac{1}{a_{n-1}+frac{2}{n}}tag{1}$$



                                      or



                                      $$a_n=frac{na_{n-1}+1}{n+1}tag{2}$$



                                      etc.



                                      equation 1, is simply noting: $$frac{n-1}{n}+frac{2}{n}=frac{n+1}{n}=frac{1}{a_n}$$ Where the first fraction in the sum is $a_{n-1}$



                                      and equation 2, simply notes:



                                      $$n=na_{n-1}+1$$



                                      etc.






                                      share|cite|improve this answer









                                      $endgroup$



                                      You can, and in multiple ways. Such as:$$a_n=frac{1}{a_{n-1}+frac{2}{n}}tag{1}$$



                                      or



                                      $$a_n=frac{na_{n-1}+1}{n+1}tag{2}$$



                                      etc.



                                      equation 1, is simply noting: $$frac{n-1}{n}+frac{2}{n}=frac{n+1}{n}=frac{1}{a_n}$$ Where the first fraction in the sum is $a_{n-1}$



                                      and equation 2, simply notes:



                                      $$n=na_{n-1}+1$$



                                      etc.







                                      share|cite|improve this answer












                                      share|cite|improve this answer



                                      share|cite|improve this answer










                                      answered 5 hours ago









                                      Roddy MacPheeRoddy MacPhee

                                      720118




                                      720118























                                          0












                                          $begingroup$

                                          Perhaps a bit simpler is to note that
                                          $$
                                          overbrace{ frac1{1-a_n} }^{n+1}+1=overbrace{frac1{1-a_{n+1}}}^{n+2}\
                                          $$
                                          solving for $a_{n+1}$ yields
                                          $$
                                          frac1{2-a_n}=a_{n+1}\
                                          $$






                                          share|cite|improve this answer









                                          $endgroup$


















                                            0












                                            $begingroup$

                                            Perhaps a bit simpler is to note that
                                            $$
                                            overbrace{ frac1{1-a_n} }^{n+1}+1=overbrace{frac1{1-a_{n+1}}}^{n+2}\
                                            $$
                                            solving for $a_{n+1}$ yields
                                            $$
                                            frac1{2-a_n}=a_{n+1}\
                                            $$






                                            share|cite|improve this answer









                                            $endgroup$
















                                              0












                                              0








                                              0





                                              $begingroup$

                                              Perhaps a bit simpler is to note that
                                              $$
                                              overbrace{ frac1{1-a_n} }^{n+1}+1=overbrace{frac1{1-a_{n+1}}}^{n+2}\
                                              $$
                                              solving for $a_{n+1}$ yields
                                              $$
                                              frac1{2-a_n}=a_{n+1}\
                                              $$






                                              share|cite|improve this answer









                                              $endgroup$



                                              Perhaps a bit simpler is to note that
                                              $$
                                              overbrace{ frac1{1-a_n} }^{n+1}+1=overbrace{frac1{1-a_{n+1}}}^{n+2}\
                                              $$
                                              solving for $a_{n+1}$ yields
                                              $$
                                              frac1{2-a_n}=a_{n+1}\
                                              $$







                                              share|cite|improve this answer












                                              share|cite|improve this answer



                                              share|cite|improve this answer










                                              answered 4 hours ago









                                              robjohn♦robjohn

                                              270k27313642




                                              270k27313642






















                                                  Levi K is a new contributor. Be nice, and check out our Code of Conduct.










                                                  draft saved

                                                  draft discarded


















                                                  Levi K is a new contributor. Be nice, and check out our Code of Conduct.













                                                  Levi K is a new contributor. Be nice, and check out our Code of Conduct.












                                                  Levi K is a new contributor. Be nice, and check out our Code of Conduct.
















                                                  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%2f3176633%2fcan-an-fracnn1-be-written-recursively%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

                                                  GameSpot

                                                  Image
                                                  body.skin-minerva .mw-parser-output table.infobox caption{text-align:center} GameSpot 戰地風雲：惡名昭彰2在Gamespot的評論 网站类型 新聞 持有者 CBS 创始人 Pete Deemer Vince Broady Jon Epstein 网站 http://www.gamespot.com/ 注册 Optional (free and paid) 推出时间 1996年5月1日 [1] GameSpot （中国大陆：游戏基地），於1996年5月由Pete Deemer和Vince Broady創立，是一個提供新聞、評論、預告片、下載及其他的相關資訊的電子遊戲網站。GameSpot被一間後來被CNET收購的企業ZDNet所收購。根據Alexa，GameSpot.com是200個網路擁擠最嚴重的網站之一。 除了由GameSpot員工創作的內容，網站還允許用戶寫評論、網誌、之後在網路論壇分享。一些在CNET旗下的GameFAQs分享。 2004年， GameSpot被Spike TV的觀眾選上「電子遊戲賞節目」贏得「最傑出遊戲網站。 [2] 其他的遊戲網站還有IGN、1UP.com、GameSpy是它最大的競爭對手。2008年，根據Compete.com的統計，「gamespot.com」吸引了最少6000萬人的點擊率。 [3] GameSpot的主頁鏈結了到最近新聞、評論、預告、和一些有關遊戲機的入口：Wii、任天堂DS、電腦遊戲、Xbox 360、PSP、PlayStation 2、PlayStation 3。它還有一列「最受歡迎遊戲名單」，還有給用戶快速獲得遊戲資訊的搜尋器。GameSpot 還包括一些小範圍的遊戲機：任天堂64、GameCube、Game Boy Color、Game Boy Advance、Xbox、PlayStation、SEGA Saturn、Dreamcast、Neo Geo Pocket Color、N-Gage、手機遊戲。 目录 1 歷史 1.1 國際歷史 1.2 著名的員工 2 評論和分...
                                                  Read more

                                                  日野市

                                                  Image
                                                  body.skin-minerva .mw-parser-output table.infobox caption{text-align:center} 日野市 日野市 日文轉寫  • 日文 日野市  • 平假名 ひのし  • 罗马字 Hino-shi 金剛寺（關東三十六不動尊靈場第9號）的不動堂 市旗 徽章 日野市在東京都的位置 日野市 日野市在日本的位置 坐标： 35°40′17″N 139°23′42″E  /  35.6714°N 139.395°E  / 35.6714; 139.395 国家   日本 地方 關東地方 都道府縣 東京都 接鄰行政區 府中市、國立市、立川市、昭島市、八王子市、多摩市 政府  • 市長 大坪冬彦 面积  •  总计 27.53 平方公里（10.63 平方英里） 人口 （2014年12月1日）  • 總計 183,323  • 密度 6,660/平方公里（17,200/平方英里） 象徵  • 市樹 樫  • 市花 菊花  • 市鳥 普通翠鳥 时区 日本標準時間 （UTC+9） 地方公共團體編號 13212-8 邮政编码 〒 191-8686 市役所地址 日野市神明一丁目12番1號 電話號碼 +81-42-585-1111 法人編號 1000020132128 網站 http://www.city.hino.lg.jp/ 人口：日野市官方網頁 日野市 （日语： 日野市 ／ ひのし   Hino shi   * / ? ）為一位于東京都（不含島嶼部分）中央地帶的城市。從東京站乘坐中央線特別快速列車45分鐘即可到該市。面積27.53km²。1963年（昭和38年）11月3日，該市開始實行市制，為全日本第559個市。汽車企業日野自動車的總部位於這裡。 往東京都特別區部的通勤率為20.9%，往八王子市的通勤率為12.0%（平成22年國勢調査）。 目录 1 概要 ...
                                                  Read more

                                                  Tu-95轟炸機

                                                  Image
                                                  body.skin-minerva .mw-parser-output table.infobox caption{text-align:center} 图-95 Ту–95 北约代号：熊（Bear） 一架在安格斯空军基地的图-95MS 概觀 類型 战略轰炸机 代號 北约代号： Bear （熊） 乘員 7名 駕駛員2名，機尾炮手1名，其他人員4名 首飛 1952年11月12日 服役 1956年 設計 圖波列夫設計局 產量 超过500架 現況 现役 主要用戶   蘇聯   俄羅斯 衍生機型 图-114、图-119、Tu-95 技术数据 長度 49.50米（162呎5吋） 翼展 51.10米（167呎8吋） 高度 12.12米（39呎9吋） 翼面積 310平方米（3,330平方呎） 空重 90,000公斤（198,000磅） 最大起飛重量 188,000公斤（414,500磅） 發動機 4具库兹涅佐夫NK-12MV型渦輪螺旋槳發動機 功率 4×11,000千瓦（14,800馬力） 性能數據 最大速度 925公里／時（500節，575哩／時） 爬升率 10米／秒（2,000呎／分） 最大升限 12,000米（39,000呎） 最大航程 15,000公里（8,100海浬，9,400哩） 翼負荷 606公斤／平方米（124磅／平方呎） 推重比 235 W／公斤（0.143馬力／磅） 武器装备 機炮 1具/2具AM-23型23公釐雷達控制機砲（機尾） 飛彈 空对地导弹： Kh-20、Kh-22 Kh-26、Kh-55 炸彈 60枚FAB-250炸彈，或 30枚FAB-500炸彈。 其他 载弹量（包括飞弹）：最多15,000公斤（33,000磅） Tu-95熊式MR型機 图-95 （俄语： Ту–95 ， 英语： Tu-95 ），北約代號： Bear （ 熊 ），是蘇聯圖波列夫設計局所研製，是全世界唯一服役的大型四渦輪螺旋槳發動機...
                                                  Read more