Existence of a real valued function satisfying $f'(x)=f(x+1)$ where $f(x)ne0$












2












$begingroup$


I am wondering about the existence of a real valued function that satisfies $f'(x)=f(x+1)$ other than the trivial solution $f(x)=0$.
I thought a likely solution would be a repeating (perhaps sinusoidal) function, where $f(0)=0, f'(0)=1,f'(1)=0,f'(2)=-1,f'(3)=0$, etc.



I tried to find such a function on Desmos, however I failed after realizing that for a Sin function the derivative at $0$ is always greater than the value of the maximum since between the zero and the maximum the function is strictly decreasing.



In my exploration, I was able to find a solution to the similar differential equation, $f'(x)=f(x+1)*{piover2}$, with the solution $f(x)=sin{({piover2}x)}$. This was easier to find due to the coefficient of $piover2$, which removes the obstacle of the previous inequality. Perhaps this provides hope for a solution to the other one.



Overall, the question is, is there any functional solution for the equation $f'(x)=f(x+1)$ where $f(x)ne0$?






ordinary-differential-equations derivatives functional-equations delay-differential-equations







share|cite|improve this question









$endgroup$

















    2












    $begingroup$


    I am wondering about the existence of a real valued function that satisfies $f'(x)=f(x+1)$ other than the trivial solution $f(x)=0$.
    I thought a likely solution would be a repeating (perhaps sinusoidal) function, where $f(0)=0, f'(0)=1,f'(1)=0,f'(2)=-1,f'(3)=0$, etc.



    I tried to find such a function on Desmos, however I failed after realizing that for a Sin function the derivative at $0$ is always greater than the value of the maximum since between the zero and the maximum the function is strictly decreasing.



    In my exploration, I was able to find a solution to the similar differential equation, $f'(x)=f(x+1)*{piover2}$, with the solution $f(x)=sin{({piover2}x)}$. This was easier to find due to the coefficient of $piover2$, which removes the obstacle of the previous inequality. Perhaps this provides hope for a solution to the other one.



    Overall, the question is, is there any functional solution for the equation $f'(x)=f(x+1)$ where $f(x)ne0$?






    ordinary-differential-equations derivatives functional-equations delay-differential-equations







    share|cite|improve this question









    $endgroup$















      2












      2








      2





      $begingroup$


      I am wondering about the existence of a real valued function that satisfies $f'(x)=f(x+1)$ other than the trivial solution $f(x)=0$.
      I thought a likely solution would be a repeating (perhaps sinusoidal) function, where $f(0)=0, f'(0)=1,f'(1)=0,f'(2)=-1,f'(3)=0$, etc.



      I tried to find such a function on Desmos, however I failed after realizing that for a Sin function the derivative at $0$ is always greater than the value of the maximum since between the zero and the maximum the function is strictly decreasing.



      In my exploration, I was able to find a solution to the similar differential equation, $f'(x)=f(x+1)*{piover2}$, with the solution $f(x)=sin{({piover2}x)}$. This was easier to find due to the coefficient of $piover2$, which removes the obstacle of the previous inequality. Perhaps this provides hope for a solution to the other one.



      Overall, the question is, is there any functional solution for the equation $f'(x)=f(x+1)$ where $f(x)ne0$?






      ordinary-differential-equations derivatives functional-equations delay-differential-equations







      share|cite|improve this question









      $endgroup$




      I am wondering about the existence of a real valued function that satisfies $f'(x)=f(x+1)$ other than the trivial solution $f(x)=0$.
      I thought a likely solution would be a repeating (perhaps sinusoidal) function, where $f(0)=0, f'(0)=1,f'(1)=0,f'(2)=-1,f'(3)=0$, etc.



      I tried to find such a function on Desmos, however I failed after realizing that for a Sin function the derivative at $0$ is always greater than the value of the maximum since between the zero and the maximum the function is strictly decreasing.



      In my exploration, I was able to find a solution to the similar differential equation, $f'(x)=f(x+1)*{piover2}$, with the solution $f(x)=sin{({piover2}x)}$. This was easier to find due to the coefficient of $piover2$, which removes the obstacle of the previous inequality. Perhaps this provides hope for a solution to the other one.



      Overall, the question is, is there any functional solution for the equation $f'(x)=f(x+1)$ where $f(x)ne0$?






      ordinary-differential-equations derivatives functional-equations delay-differential-equations




      ordinary-differential-equations derivatives functional-equations delay-differential-equations






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Jan 31 at 0:27









      volcanrbvolcanrb

      682216




      682216






















          3 Answers
          3






          active

          oldest

          votes


















          2












          $begingroup$

          This is an example of delay differential equations. Make an ansatz that $f(x) = e^{ax}cos(bx)$ is a solution. Plugging this to the equation,



          $$ e^{ax}(acos(bx) - bsin(bx)) = e^{ax+a}(cos(bx)cos(b)-sin(bx)sin(b)). $$



          So it suffices to find a pair of numbers $(a, b)$ for which



          $$ a = e^a cos(b) quad text{and} quad b = e^a sin (b). $$



          Or, if we write $alpha = a + ib$, then $alpha = e^{alpha}$. It can be proved that this equation has many solutions. (This is an instance of the method using characteristic equation.) One such solution is numerically given by



          $$ (a, b) = (0.576412723031435283cdots, 0.374699020737117494cdots).$$



          although it is likely impossible to provide a closed form in terms of elementary functions.






          share|cite|improve this answer









          $endgroup$





















            1












            $begingroup$

            Let $f$ be a $C^{infty} $ function with compact support in $(0,1)$ and define $f(0)$ and $f(1)$ to be $0$. For $n leq x leq n+1$ define $f(x)= f^{(n)}(x-n)$.






            share|cite|improve this answer









            $endgroup$





















              1












              $begingroup$

              This is an example of a linear delay differential equation. Given a finite set of solutions, then any linear combination is also a solution. Now define
              $, f_k(x) := exp(Re(t_k)x) cos(Im(t_k)x),$ where $, t_k := -W_k(-1) ,$
              and where $, W_k(z) ,$ is the $k$-th branch of the Lambert W function. It is easy to check that $, f_k(x), ,$ for any integer $,k,,$ is a solution of $, f'(x) = f(x+1).,$






              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
                });


                }
                });














                draft saved

                draft discarded


















                StackExchange.ready(
                function () {
                StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3094323%2fexistence-of-a-real-valued-function-satisfying-fx-fx1-where-fx-ne0%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












                $begingroup$

                This is an example of delay differential equations. Make an ansatz that $f(x) = e^{ax}cos(bx)$ is a solution. Plugging this to the equation,



                $$ e^{ax}(acos(bx) - bsin(bx)) = e^{ax+a}(cos(bx)cos(b)-sin(bx)sin(b)). $$



                So it suffices to find a pair of numbers $(a, b)$ for which



                $$ a = e^a cos(b) quad text{and} quad b = e^a sin (b). $$



                Or, if we write $alpha = a + ib$, then $alpha = e^{alpha}$. It can be proved that this equation has many solutions. (This is an instance of the method using characteristic equation.) One such solution is numerically given by



                $$ (a, b) = (0.576412723031435283cdots, 0.374699020737117494cdots).$$



                although it is likely impossible to provide a closed form in terms of elementary functions.






                share|cite|improve this answer









                $endgroup$


















                  2












                  $begingroup$

                  This is an example of delay differential equations. Make an ansatz that $f(x) = e^{ax}cos(bx)$ is a solution. Plugging this to the equation,



                  $$ e^{ax}(acos(bx) - bsin(bx)) = e^{ax+a}(cos(bx)cos(b)-sin(bx)sin(b)). $$



                  So it suffices to find a pair of numbers $(a, b)$ for which



                  $$ a = e^a cos(b) quad text{and} quad b = e^a sin (b). $$



                  Or, if we write $alpha = a + ib$, then $alpha = e^{alpha}$. It can be proved that this equation has many solutions. (This is an instance of the method using characteristic equation.) One such solution is numerically given by



                  $$ (a, b) = (0.576412723031435283cdots, 0.374699020737117494cdots).$$



                  although it is likely impossible to provide a closed form in terms of elementary functions.






                  share|cite|improve this answer









                  $endgroup$
















                    2












                    2








                    2





                    $begingroup$

                    This is an example of delay differential equations. Make an ansatz that $f(x) = e^{ax}cos(bx)$ is a solution. Plugging this to the equation,



                    $$ e^{ax}(acos(bx) - bsin(bx)) = e^{ax+a}(cos(bx)cos(b)-sin(bx)sin(b)). $$



                    So it suffices to find a pair of numbers $(a, b)$ for which



                    $$ a = e^a cos(b) quad text{and} quad b = e^a sin (b). $$



                    Or, if we write $alpha = a + ib$, then $alpha = e^{alpha}$. It can be proved that this equation has many solutions. (This is an instance of the method using characteristic equation.) One such solution is numerically given by



                    $$ (a, b) = (0.576412723031435283cdots, 0.374699020737117494cdots).$$



                    although it is likely impossible to provide a closed form in terms of elementary functions.






                    share|cite|improve this answer









                    $endgroup$



                    This is an example of delay differential equations. Make an ansatz that $f(x) = e^{ax}cos(bx)$ is a solution. Plugging this to the equation,



                    $$ e^{ax}(acos(bx) - bsin(bx)) = e^{ax+a}(cos(bx)cos(b)-sin(bx)sin(b)). $$



                    So it suffices to find a pair of numbers $(a, b)$ for which



                    $$ a = e^a cos(b) quad text{and} quad b = e^a sin (b). $$



                    Or, if we write $alpha = a + ib$, then $alpha = e^{alpha}$. It can be proved that this equation has many solutions. (This is an instance of the method using characteristic equation.) One such solution is numerically given by



                    $$ (a, b) = (0.576412723031435283cdots, 0.374699020737117494cdots).$$



                    although it is likely impossible to provide a closed form in terms of elementary functions.







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Jan 31 at 0:50









                    Sangchul LeeSangchul Lee

                    96.3k12172283




                    96.3k12172283























                        1












                        $begingroup$

                        Let $f$ be a $C^{infty} $ function with compact support in $(0,1)$ and define $f(0)$ and $f(1)$ to be $0$. For $n leq x leq n+1$ define $f(x)= f^{(n)}(x-n)$.






                        share|cite|improve this answer









                        $endgroup$


















                          1












                          $begingroup$

                          Let $f$ be a $C^{infty} $ function with compact support in $(0,1)$ and define $f(0)$ and $f(1)$ to be $0$. For $n leq x leq n+1$ define $f(x)= f^{(n)}(x-n)$.






                          share|cite|improve this answer









                          $endgroup$
















                            1












                            1








                            1





                            $begingroup$

                            Let $f$ be a $C^{infty} $ function with compact support in $(0,1)$ and define $f(0)$ and $f(1)$ to be $0$. For $n leq x leq n+1$ define $f(x)= f^{(n)}(x-n)$.






                            share|cite|improve this answer









                            $endgroup$



                            Let $f$ be a $C^{infty} $ function with compact support in $(0,1)$ and define $f(0)$ and $f(1)$ to be $0$. For $n leq x leq n+1$ define $f(x)= f^{(n)}(x-n)$.







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Jan 31 at 0:39









                            Kavi Rama MurthyKavi Rama Murthy

                            72.6k53170




                            72.6k53170























                                1












                                $begingroup$

                                This is an example of a linear delay differential equation. Given a finite set of solutions, then any linear combination is also a solution. Now define
                                $, f_k(x) := exp(Re(t_k)x) cos(Im(t_k)x),$ where $, t_k := -W_k(-1) ,$
                                and where $, W_k(z) ,$ is the $k$-th branch of the Lambert W function. It is easy to check that $, f_k(x), ,$ for any integer $,k,,$ is a solution of $, f'(x) = f(x+1).,$






                                share|cite|improve this answer











                                $endgroup$


















                                  1












                                  $begingroup$

                                  This is an example of a linear delay differential equation. Given a finite set of solutions, then any linear combination is also a solution. Now define
                                  $, f_k(x) := exp(Re(t_k)x) cos(Im(t_k)x),$ where $, t_k := -W_k(-1) ,$
                                  and where $, W_k(z) ,$ is the $k$-th branch of the Lambert W function. It is easy to check that $, f_k(x), ,$ for any integer $,k,,$ is a solution of $, f'(x) = f(x+1).,$






                                  share|cite|improve this answer











                                  $endgroup$
















                                    1












                                    1








                                    1





                                    $begingroup$

                                    This is an example of a linear delay differential equation. Given a finite set of solutions, then any linear combination is also a solution. Now define
                                    $, f_k(x) := exp(Re(t_k)x) cos(Im(t_k)x),$ where $, t_k := -W_k(-1) ,$
                                    and where $, W_k(z) ,$ is the $k$-th branch of the Lambert W function. It is easy to check that $, f_k(x), ,$ for any integer $,k,,$ is a solution of $, f'(x) = f(x+1).,$






                                    share|cite|improve this answer











                                    $endgroup$



                                    This is an example of a linear delay differential equation. Given a finite set of solutions, then any linear combination is also a solution. Now define
                                    $, f_k(x) := exp(Re(t_k)x) cos(Im(t_k)x),$ where $, t_k := -W_k(-1) ,$
                                    and where $, W_k(z) ,$ is the $k$-th branch of the Lambert W function. It is easy to check that $, f_k(x), ,$ for any integer $,k,,$ is a solution of $, f'(x) = f(x+1).,$







                                    share|cite|improve this answer














                                    share|cite|improve this answer



                                    share|cite|improve this answer








                                    edited Jan 31 at 3:59

























                                    answered Jan 31 at 1:50









                                    SomosSomos

                                    14.7k11337




                                    14.7k11337






























                                        draft saved

                                        draft discarded




















































                                        Thanks for contributing an answer to Mathematics Stack Exchange!


                                        • Please be sure to answer the question. Provide details and share your research!

                                        But avoid …



                                        • Asking for help, clarification, or responding to other answers.

                                        • Making statements based on opinion; back them up with references or personal experience.


                                        Use MathJax to format equations. MathJax reference.


                                        To learn more, see our tips on writing great answers.




                                        draft saved


                                        draft discarded














                                        StackExchange.ready(
                                        function () {
                                        StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3094323%2fexistence-of-a-real-valued-function-satisfying-fx-fx1-where-fx-ne0%23new-answer', 'question_page');
                                        }
                                        );

                                        Post as a guest















                                        Required, but never shown





















































                                        Required, but never shown














                                        Required, but never shown












                                        Required, but never shown







                                        Required, but never shown

































                                        Required, but never shown














                                        Required, but never shown












                                        Required, but never shown







                                        Required, but never shown







                                        Popular posts from this blog

                                        MongoDB - Not Authorized To Execute Command

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

                                        How to fix TextFormField cause rebuild widget in Flutter

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

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

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