Is it possible to represent this function as a polynomial, by removing the ceiling function?












1












$begingroup$


I've been working through a derivation and have arrived at the following expression:



$$E = 1 - frac{x}y left( bigglceil dfrac{x}{y} biggrceil right)^{-1}$$



where $x,y in mathbb{R^+}$.



I would like to know whether this can be reduced further, by converting the floor function to a polynomial to simplify expression?



furthermore, lets say there was a constant k, where $k in mathbb{N^+}$ hence,



$$E = 1 - frac{x}{k.y} left( bigglceil dfrac{x}{k.y} biggrceil right)^{-1}$$



could this be reduced to,



$$E = 1 - frac{x}y left( bigglceil dfrac{x}{y} biggrceil right)^{-1}$$



Kind regards!






polynomials floor-function ceiling-function







share|cite|improve this question











$endgroup$

















    1












    $begingroup$


    I've been working through a derivation and have arrived at the following expression:



    $$E = 1 - frac{x}y left( bigglceil dfrac{x}{y} biggrceil right)^{-1}$$



    where $x,y in mathbb{R^+}$.



    I would like to know whether this can be reduced further, by converting the floor function to a polynomial to simplify expression?



    furthermore, lets say there was a constant k, where $k in mathbb{N^+}$ hence,



    $$E = 1 - frac{x}{k.y} left( bigglceil dfrac{x}{k.y} biggrceil right)^{-1}$$



    could this be reduced to,



    $$E = 1 - frac{x}y left( bigglceil dfrac{x}{y} biggrceil right)^{-1}$$



    Kind regards!






    polynomials floor-function ceiling-function







    share|cite|improve this question











    $endgroup$















      1












      1








      1





      $begingroup$


      I've been working through a derivation and have arrived at the following expression:



      $$E = 1 - frac{x}y left( bigglceil dfrac{x}{y} biggrceil right)^{-1}$$



      where $x,y in mathbb{R^+}$.



      I would like to know whether this can be reduced further, by converting the floor function to a polynomial to simplify expression?



      furthermore, lets say there was a constant k, where $k in mathbb{N^+}$ hence,



      $$E = 1 - frac{x}{k.y} left( bigglceil dfrac{x}{k.y} biggrceil right)^{-1}$$



      could this be reduced to,



      $$E = 1 - frac{x}y left( bigglceil dfrac{x}{y} biggrceil right)^{-1}$$



      Kind regards!






      polynomials floor-function ceiling-function







      share|cite|improve this question











      $endgroup$




      I've been working through a derivation and have arrived at the following expression:



      $$E = 1 - frac{x}y left( bigglceil dfrac{x}{y} biggrceil right)^{-1}$$



      where $x,y in mathbb{R^+}$.



      I would like to know whether this can be reduced further, by converting the floor function to a polynomial to simplify expression?



      furthermore, lets say there was a constant k, where $k in mathbb{N^+}$ hence,



      $$E = 1 - frac{x}{k.y} left( bigglceil dfrac{x}{k.y} biggrceil right)^{-1}$$



      could this be reduced to,



      $$E = 1 - frac{x}y left( bigglceil dfrac{x}{y} biggrceil right)^{-1}$$



      Kind regards!






      polynomials floor-function ceiling-function




      polynomials floor-function ceiling-function






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Jan 9 at 12:57







      Afroeskimo

















      asked Jan 9 at 10:56









      AfroeskimoAfroeskimo

      115




      115






















          1 Answer
          1






          active

          oldest

          votes


















          1












          $begingroup$

          If I understand your question correctly then the answer is No. The function
          $$
          f(x)=frac{x}{lceil x rceil}
          $$
          has infinitely many (jump) discontinuities.



          If we can somehow convert $f(x)$ into a polynomial or a rational function then it would be continuous on the whole $Bbb R$ except for only finitely many points. This is impossible since it'd contradict the fact that $f(x)$ has infinitely many discontinuity.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Understood! I would like to know whether it would be possible to remove a positive integer that resides in the ceil function, in a way that I have mentioned in the edited version of my question?
            $endgroup$
            – Afroeskimo
            Jan 9 at 12:29










          • $begingroup$
            @Afroeskimo In general you cannot pull a constant out of the floor/ceiling function so I'm afraid the answer is also No.
            $endgroup$
            – BigbearZzz
            Jan 9 at 12:49










          • $begingroup$
            even if the constant is an integer?
            $endgroup$
            – Afroeskimo
            Jan 9 at 13:03












          • $begingroup$
            @Afroeskimo Nope, you can consider $x=1.5$ and $k=2$, for example.
            $endgroup$
            – BigbearZzz
            Jan 9 at 13:09












          Your Answer








          StackExchange.ready(function() {
          var channelOptions = {
          tags: "".split(" "),
          id: "69"
          };
          initTagRenderer("".split(" "), "".split(" "), channelOptions);

          StackExchange.using("externalEditor", function() {
          // Have to fire editor after snippets, if snippets enabled
          if (StackExchange.settings.snippets.snippetsEnabled) {
          StackExchange.using("snippets", function() {
          createEditor();
          });
          }
          else {
          createEditor();
          }
          });

          function createEditor() {
          StackExchange.prepareEditor({
          heartbeatType: 'answer',
          autoActivateHeartbeat: false,
          convertImagesToLinks: true,
          noModals: true,
          showLowRepImageUploadWarning: true,
          reputationToPostImages: 10,
          bindNavPrevention: true,
          postfix: "",
          imageUploader: {
          brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
          contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
          allowUrls: true
          },
          noCode: true, onDemand: true,
          discardSelector: ".discard-answer"
          ,immediatelyShowMarkdownHelp:true
          });


          }
          });














          draft saved

          draft discarded


















          StackExchange.ready(
          function () {
          StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3067319%2fis-it-possible-to-represent-this-function-as-a-polynomial-by-removing-the-ceili%23new-answer', 'question_page');
          }
          );

          Post as a guest















          Required, but never shown

























          1 Answer
          1






          active

          oldest

          votes








          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          1












          $begingroup$

          If I understand your question correctly then the answer is No. The function
          $$
          f(x)=frac{x}{lceil x rceil}
          $$
          has infinitely many (jump) discontinuities.



          If we can somehow convert $f(x)$ into a polynomial or a rational function then it would be continuous on the whole $Bbb R$ except for only finitely many points. This is impossible since it'd contradict the fact that $f(x)$ has infinitely many discontinuity.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Understood! I would like to know whether it would be possible to remove a positive integer that resides in the ceil function, in a way that I have mentioned in the edited version of my question?
            $endgroup$
            – Afroeskimo
            Jan 9 at 12:29










          • $begingroup$
            @Afroeskimo In general you cannot pull a constant out of the floor/ceiling function so I'm afraid the answer is also No.
            $endgroup$
            – BigbearZzz
            Jan 9 at 12:49










          • $begingroup$
            even if the constant is an integer?
            $endgroup$
            – Afroeskimo
            Jan 9 at 13:03












          • $begingroup$
            @Afroeskimo Nope, you can consider $x=1.5$ and $k=2$, for example.
            $endgroup$
            – BigbearZzz
            Jan 9 at 13:09
















          1












          $begingroup$

          If I understand your question correctly then the answer is No. The function
          $$
          f(x)=frac{x}{lceil x rceil}
          $$
          has infinitely many (jump) discontinuities.



          If we can somehow convert $f(x)$ into a polynomial or a rational function then it would be continuous on the whole $Bbb R$ except for only finitely many points. This is impossible since it'd contradict the fact that $f(x)$ has infinitely many discontinuity.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Understood! I would like to know whether it would be possible to remove a positive integer that resides in the ceil function, in a way that I have mentioned in the edited version of my question?
            $endgroup$
            – Afroeskimo
            Jan 9 at 12:29










          • $begingroup$
            @Afroeskimo In general you cannot pull a constant out of the floor/ceiling function so I'm afraid the answer is also No.
            $endgroup$
            – BigbearZzz
            Jan 9 at 12:49










          • $begingroup$
            even if the constant is an integer?
            $endgroup$
            – Afroeskimo
            Jan 9 at 13:03












          • $begingroup$
            @Afroeskimo Nope, you can consider $x=1.5$ and $k=2$, for example.
            $endgroup$
            – BigbearZzz
            Jan 9 at 13:09














          1












          1








          1





          $begingroup$

          If I understand your question correctly then the answer is No. The function
          $$
          f(x)=frac{x}{lceil x rceil}
          $$
          has infinitely many (jump) discontinuities.



          If we can somehow convert $f(x)$ into a polynomial or a rational function then it would be continuous on the whole $Bbb R$ except for only finitely many points. This is impossible since it'd contradict the fact that $f(x)$ has infinitely many discontinuity.






          share|cite|improve this answer









          $endgroup$



          If I understand your question correctly then the answer is No. The function
          $$
          f(x)=frac{x}{lceil x rceil}
          $$
          has infinitely many (jump) discontinuities.



          If we can somehow convert $f(x)$ into a polynomial or a rational function then it would be continuous on the whole $Bbb R$ except for only finitely many points. This is impossible since it'd contradict the fact that $f(x)$ has infinitely many discontinuity.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 9 at 11:11









          BigbearZzzBigbearZzz

          9,07421753




          9,07421753












          • $begingroup$
            Understood! I would like to know whether it would be possible to remove a positive integer that resides in the ceil function, in a way that I have mentioned in the edited version of my question?
            $endgroup$
            – Afroeskimo
            Jan 9 at 12:29










          • $begingroup$
            @Afroeskimo In general you cannot pull a constant out of the floor/ceiling function so I'm afraid the answer is also No.
            $endgroup$
            – BigbearZzz
            Jan 9 at 12:49










          • $begingroup$
            even if the constant is an integer?
            $endgroup$
            – Afroeskimo
            Jan 9 at 13:03












          • $begingroup$
            @Afroeskimo Nope, you can consider $x=1.5$ and $k=2$, for example.
            $endgroup$
            – BigbearZzz
            Jan 9 at 13:09


















          • $begingroup$
            Understood! I would like to know whether it would be possible to remove a positive integer that resides in the ceil function, in a way that I have mentioned in the edited version of my question?
            $endgroup$
            – Afroeskimo
            Jan 9 at 12:29










          • $begingroup$
            @Afroeskimo In general you cannot pull a constant out of the floor/ceiling function so I'm afraid the answer is also No.
            $endgroup$
            – BigbearZzz
            Jan 9 at 12:49










          • $begingroup$
            even if the constant is an integer?
            $endgroup$
            – Afroeskimo
            Jan 9 at 13:03












          • $begingroup$
            @Afroeskimo Nope, you can consider $x=1.5$ and $k=2$, for example.
            $endgroup$
            – BigbearZzz
            Jan 9 at 13:09
















          $begingroup$
          Understood! I would like to know whether it would be possible to remove a positive integer that resides in the ceil function, in a way that I have mentioned in the edited version of my question?
          $endgroup$
          – Afroeskimo
          Jan 9 at 12:29




          $begingroup$
          Understood! I would like to know whether it would be possible to remove a positive integer that resides in the ceil function, in a way that I have mentioned in the edited version of my question?
          $endgroup$
          – Afroeskimo
          Jan 9 at 12:29












          $begingroup$
          @Afroeskimo In general you cannot pull a constant out of the floor/ceiling function so I'm afraid the answer is also No.
          $endgroup$
          – BigbearZzz
          Jan 9 at 12:49




          $begingroup$
          @Afroeskimo In general you cannot pull a constant out of the floor/ceiling function so I'm afraid the answer is also No.
          $endgroup$
          – BigbearZzz
          Jan 9 at 12:49












          $begingroup$
          even if the constant is an integer?
          $endgroup$
          – Afroeskimo
          Jan 9 at 13:03






          $begingroup$
          even if the constant is an integer?
          $endgroup$
          – Afroeskimo
          Jan 9 at 13:03














          $begingroup$
          @Afroeskimo Nope, you can consider $x=1.5$ and $k=2$, for example.
          $endgroup$
          – BigbearZzz
          Jan 9 at 13:09




          $begingroup$
          @Afroeskimo Nope, you can consider $x=1.5$ and $k=2$, for example.
          $endgroup$
          – BigbearZzz
          Jan 9 at 13:09


















          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%2f3067319%2fis-it-possible-to-represent-this-function-as-a-polynomial-by-removing-the-ceili%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

          Aardman Animations

          Image
          Aardman Animations, Ltd. Aardman Animations Tipo privada Atividade Animação em stop-motion , Animação em CGI Fundação 1972 Fundador(es) Peter Lord David Sproxton Sede Bristol, Inglaterra  Reino Unido Proprietário(s) Dreamworks Pessoas-chave Peter Lord David Sproxton Nick Park Divisões Aardman Features Aardman Digital Aardman Commercials Aardman Broadcast Aardman International Aardman Rights Aardman Effects Aardman 3-D Systems Aardman Nathan Love AardBoiled Website oficial aardman.com
          Read more

          Are they similar matrix

          Image
          0 $begingroup$ Do $begin{bmatrix} 0&i&0\0&0&1\0&0&0 end{bmatrix} $ and $begin{bmatrix} 0&0&0\-i&0&0\0&1&0 end{bmatrix} $ are similar.Is this True/false Clearly both are nilpotent and one is conjucate transpose of other but how to know if they are similar.i'm stuck. Please help me linear-algebra vector-spaces eigenvalues-eigenvectors generalizedeigenvector share | cite | improve this question edited Dec 10 '18 at 6:38 Vasanth Kris asked Dec 10 '18 at 6:19
          Read more

          “minimization” problem in Euclidean space related to orthonormal basis

          Image
          3 $begingroup$ I have stumbled upon an exercice for second year undegraduate student majoring in economics which I find quite demanding. I have an idea for the solution, but it seems awfully complicated, and I am wondering if there was a simpler way to solve it. This goes as follows: Let $(E, langle, rangle)$ be a euclidean space of dimension $n geq 2$ . 1) For any $x in E$ , show that $||x|| = sqrt{n}$ if and only if there exists $(e_1, dots, e_n)$ an orthonormal basis of $E$ such that $x =e_1+ ldots + e_n$ . 2) For any $(x,y) in E$ , show that $||x|| = sqrt{n}$ , $||y|| = sqrt{frac{n(n+1)(2n+1)}{6}}$ and $langle x,y rangle = frac{n(n+1)}{2}$ if and only if there eixsts $(e_1, ldots, e_n)$ an orthonormal basis of $E$ such that $x = e_1 + ldots + e_n$ and $y = e_1 + 2e_2 + ldots + ne_n$ The first q
          Read more