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 Apr 6 at 4:39









      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 Apr 6 at 3:25









      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
            2 days 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









            $endgroup$





















              1












              $begingroup$

              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









              $endgroup$













              • $begingroup$
                I like the way you thought about it!
                $endgroup$
                – Levi K
                yesterday



















              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









              $endgroup$














                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
                  2 days 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
                  2 days 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 Apr 6 at 3:52









                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
                  2 days 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
                  2 days 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
                2 days 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
                2 days 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 Apr 6 at 4:36









                    user1952500

                    917712




                    917712






                    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 Apr 6 at 3:33









                    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 Apr 6 at 3:32









                            Eevee TrainerEevee Trainer

                            10.2k31742




                            10.2k31742























                                1












                                $begingroup$

                                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









                                $endgroup$













                                • $begingroup$
                                  I like the way you thought about it!
                                  $endgroup$
                                  – Levi K
                                  yesterday
















                                1












                                $begingroup$

                                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









                                $endgroup$













                                • $begingroup$
                                  I like the way you thought about it!
                                  $endgroup$
                                  – Levi K
                                  yesterday














                                1












                                1








                                1





                                $begingroup$

                                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 2 days ago









                                robjohn♦robjohn

                                270k27313642




                                270k27313642












                                • $begingroup$
                                  I like the way you thought about it!
                                  $endgroup$
                                  – Levi K
                                  yesterday


















                                • $begingroup$
                                  I like the way you thought about it!
                                  $endgroup$
                                  – Levi K
                                  yesterday
















                                $begingroup$
                                I like the way you thought about it!
                                $endgroup$
                                – Levi K
                                yesterday




                                $begingroup$
                                I like the way you thought about it!
                                $endgroup$
                                – Levi K
                                yesterday











                                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 2 days ago









                                    Roddy MacPheeRoddy MacPhee

                                    742118




                                    742118






















                                        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

                                        數位音樂下載

                                        Image
                                        數位音樂下載 為非以「實體」方式來販賣音樂的相關產品，而使用數位格式如mp3、AAC等文件格式來進行銷售販賣的一种音乐产业形式。著名的數位音樂下載服務商有iTunes Store、RecoChoku、Google Play音乐和微软Zune等。数字音乐下载在二十一世纪初一度风靡全球， [1] [2] 使得传统唱片产业走向衰落，但其自身也在2010年代流媒体音乐服务如Apple Music、Spotify等的挤压下迅速失去市场份额。 [3] [4] 数字音乐下载有便于携带、传播迅速、制作简单成本低等优点，与流媒体音乐相比也能更好保障音乐人权益； [5] 但它同时很可能带来侵犯版权和滥用的情形， [6] 也给音乐产业带来了诸多不利影响。数字音乐下载为音乐人和乐迷提供了更加便捷直接的交流渠道，也一定程度上推动了网络音乐人的出现和独立音乐的迅速发展。 [7] 目录 1 历史 1.1 技术积淀（1987-2000） 1.2 出现与兴起（2000-2004） 1.3 唱片市场的颠覆者（2004-2007） 1.4 巅峰和衰落（2005-2013） 1.5 摇摇欲坠的现状与不可预知的明天（2014至今） 2 主要文件格式 2.1 PCM编码格式 2.1.1 WAV 2.1.2 APE 2.1.3 FLAC 2.2 苹果推出的无损文件格式 2.2.1 AIFF 2.2.2 ALAC 2.3 有损压缩格式 2.3.1 MP3 2.3.2 AAC 3 主要市场 3.1 iTunes Store 3.2 Google Play音乐商店 3.3 亚马逊音乐商店 3.4 部分其他音乐商店 4 影响与评价 4.1 对音乐产业的影响 4.2 对科技行业的影响 4.3 正面评价 4.4 负面评价 4.4.1 侵权问题 4.4.2 经营模式不明晰 4.4.3 不适应移动时代 4.5 文化现象 5 延伸阅读 6 参考资料 历史 苹果iTunes+iPod的软硬件搭配让数字音...
                                        Read more

                                        When can things happen in Etherscan, such as the picture below?

                                        Image
                                        2 I don't know why balance of the address is 0. when can this situation? etherscan balances share | improve this question asked yesterday Ru Hi Ru Hi 11 1 New contributor Ru Hi is a new contributor to this site. Take care in asking for clarification, commenting, and answering. Check out our Code of Conduct. add a comment  |  ...
                                        Read more

                                        格利澤436b

                                        Image
                                        格利泽436b 太陽系外行星 太陽系外行星列表 艺术家笔下的格利泽436b 母恆星 母恆星 Gliese 436 星座 獅子座 赤經 ( α ) 11 h  42 m  11.0941 s [1] 赤緯 ( δ ) +26° 42′ 23.652″ [1] 距離 33.4 ly (10.2 pc) 光譜類型 M2.5 V [1] 軌道參數 半长轴 ( a ) 0.0291±0.0004 [2] AU 軌道離心率 ( e ) 0.150±0.012 [2] 公轉週期 ( P ) 2.643904±0.000005 [3] d (0.00723849 y) 軌道傾角 ( i ) 85.8 +0.21 −0.25 [3] ° 角距 ( θ ) 2.794 mas 近星點時間 ( T 0 ) 2,451,551.716 ±0.01 JD 半振幅 ( K ) 18.68±0.8 m/s 物理性质 质量 ( m ) 22.2±1.0 [2] M ⊕ 半徑 ( r ) 4.327±0.183 [2] [4] R ⊕ 密度 ( ρ ) 1.51 g cm -3 表面重力 ( g ) 1.18 g 溫度 ( T ) 712±36 [2] K 發現 發現時間 2004 發現者 巴特勒、沃格特、 馬西 et al. 發現方法 徑向速度、凌日法 發現地點 美國加利福尼亞州 發表論文 出版 其他名稱 Ross 905 b, GJ 436 b, [5] LTT 13213 b, GCTP 2704.10 b, LHS 310, AC+27:28217 b, Vyssotsky 616 b, HIP 57087 b, GEN# +9.80120068 b, LP 319-75 b, G 121-7 b, LSPM J1142+2642 b, 1RXS J114211.9+264328 b, ASCC 683818 b, G 147-68 b, UCAC2 41198281 b, BP...
                                        Read more