Find all polynomials of following property












0












$begingroup$


Find all polynomials $P$ with real coefficients, such that for all real $x$:



$1+P(x)={P(x+1)+P(x-1)over 2}$






polynomials







share|cite|improve this question









$endgroup$








  • 2




    $begingroup$
    The second order difference is constant . So the third order difference is 0. So poly is at most quadratic.
    $endgroup$
    – Soumik Ghosh
    Dec 10 '18 at 21:47












  • $begingroup$
    Could you explain it somehow more precisely because I dont quite understand you
    $endgroup$
    – James Caldy
    Dec 10 '18 at 21:50






  • 2




    $begingroup$
    @JamesCaldy Rearrange the given eqn as $$ [P(x+1) - P(x)] - [P(x) - P(x-1)] = 2$$
    $endgroup$
    – rsadhvika
    Dec 10 '18 at 21:51










  • $begingroup$
    What am I supposed to do next though? I cant see yet
    $endgroup$
    – James Caldy
    Dec 10 '18 at 22:01










  • $begingroup$
    I mean for me it looks familliar to $f''(x)-2f'(x)+f(x)=0$ equation... however dont know how to connect those
    $endgroup$
    – James Caldy
    Dec 10 '18 at 22:02
















0












$begingroup$


Find all polynomials $P$ with real coefficients, such that for all real $x$:



$1+P(x)={P(x+1)+P(x-1)over 2}$






polynomials







share|cite|improve this question









$endgroup$








  • 2




    $begingroup$
    The second order difference is constant . So the third order difference is 0. So poly is at most quadratic.
    $endgroup$
    – Soumik Ghosh
    Dec 10 '18 at 21:47












  • $begingroup$
    Could you explain it somehow more precisely because I dont quite understand you
    $endgroup$
    – James Caldy
    Dec 10 '18 at 21:50






  • 2




    $begingroup$
    @JamesCaldy Rearrange the given eqn as $$ [P(x+1) - P(x)] - [P(x) - P(x-1)] = 2$$
    $endgroup$
    – rsadhvika
    Dec 10 '18 at 21:51










  • $begingroup$
    What am I supposed to do next though? I cant see yet
    $endgroup$
    – James Caldy
    Dec 10 '18 at 22:01










  • $begingroup$
    I mean for me it looks familliar to $f''(x)-2f'(x)+f(x)=0$ equation... however dont know how to connect those
    $endgroup$
    – James Caldy
    Dec 10 '18 at 22:02














0












0








0





$begingroup$


Find all polynomials $P$ with real coefficients, such that for all real $x$:



$1+P(x)={P(x+1)+P(x-1)over 2}$






polynomials







share|cite|improve this question









$endgroup$




Find all polynomials $P$ with real coefficients, such that for all real $x$:



$1+P(x)={P(x+1)+P(x-1)over 2}$






polynomials




polynomials






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 10 '18 at 21:41









James CaldyJames Caldy

191




191








  • 2




    $begingroup$
    The second order difference is constant . So the third order difference is 0. So poly is at most quadratic.
    $endgroup$
    – Soumik Ghosh
    Dec 10 '18 at 21:47












  • $begingroup$
    Could you explain it somehow more precisely because I dont quite understand you
    $endgroup$
    – James Caldy
    Dec 10 '18 at 21:50






  • 2




    $begingroup$
    @JamesCaldy Rearrange the given eqn as $$ [P(x+1) - P(x)] - [P(x) - P(x-1)] = 2$$
    $endgroup$
    – rsadhvika
    Dec 10 '18 at 21:51










  • $begingroup$
    What am I supposed to do next though? I cant see yet
    $endgroup$
    – James Caldy
    Dec 10 '18 at 22:01










  • $begingroup$
    I mean for me it looks familliar to $f''(x)-2f'(x)+f(x)=0$ equation... however dont know how to connect those
    $endgroup$
    – James Caldy
    Dec 10 '18 at 22:02














  • 2




    $begingroup$
    The second order difference is constant . So the third order difference is 0. So poly is at most quadratic.
    $endgroup$
    – Soumik Ghosh
    Dec 10 '18 at 21:47












  • $begingroup$
    Could you explain it somehow more precisely because I dont quite understand you
    $endgroup$
    – James Caldy
    Dec 10 '18 at 21:50






  • 2




    $begingroup$
    @JamesCaldy Rearrange the given eqn as $$ [P(x+1) - P(x)] - [P(x) - P(x-1)] = 2$$
    $endgroup$
    – rsadhvika
    Dec 10 '18 at 21:51










  • $begingroup$
    What am I supposed to do next though? I cant see yet
    $endgroup$
    – James Caldy
    Dec 10 '18 at 22:01










  • $begingroup$
    I mean for me it looks familliar to $f''(x)-2f'(x)+f(x)=0$ equation... however dont know how to connect those
    $endgroup$
    – James Caldy
    Dec 10 '18 at 22:02








2




2




$begingroup$
The second order difference is constant . So the third order difference is 0. So poly is at most quadratic.
$endgroup$
– Soumik Ghosh
Dec 10 '18 at 21:47






$begingroup$
The second order difference is constant . So the third order difference is 0. So poly is at most quadratic.
$endgroup$
– Soumik Ghosh
Dec 10 '18 at 21:47














$begingroup$
Could you explain it somehow more precisely because I dont quite understand you
$endgroup$
– James Caldy
Dec 10 '18 at 21:50




$begingroup$
Could you explain it somehow more precisely because I dont quite understand you
$endgroup$
– James Caldy
Dec 10 '18 at 21:50




2




2




$begingroup$
@JamesCaldy Rearrange the given eqn as $$ [P(x+1) - P(x)] - [P(x) - P(x-1)] = 2$$
$endgroup$
– rsadhvika
Dec 10 '18 at 21:51




$begingroup$
@JamesCaldy Rearrange the given eqn as $$ [P(x+1) - P(x)] - [P(x) - P(x-1)] = 2$$
$endgroup$
– rsadhvika
Dec 10 '18 at 21:51












$begingroup$
What am I supposed to do next though? I cant see yet
$endgroup$
– James Caldy
Dec 10 '18 at 22:01




$begingroup$
What am I supposed to do next though? I cant see yet
$endgroup$
– James Caldy
Dec 10 '18 at 22:01












$begingroup$
I mean for me it looks familliar to $f''(x)-2f'(x)+f(x)=0$ equation... however dont know how to connect those
$endgroup$
– James Caldy
Dec 10 '18 at 22:02




$begingroup$
I mean for me it looks familliar to $f''(x)-2f'(x)+f(x)=0$ equation... however dont know how to connect those
$endgroup$
– James Caldy
Dec 10 '18 at 22:02










2 Answers
2






active

oldest

votes


















0












$begingroup$

Solving the difference equation



$$
P(x+1)-2P(x)+P(x-1)=2
$$



we obtain



$$
P(x) = x^2-x + C_1 x + C_2
$$






share|cite|improve this answer











$endgroup$





















    0












    $begingroup$

    $$forall x, 1+P(x)=frac12(P(x+1)+P(x-1)) tag{1}$$



    Setting $Q:=P'$, (1) implies, by differentiation



    $$forall x, Q(x)=frac12(Q(x+1)+Q(x-1)) tag{2}$$



    but this is only possible for a degree 1 or degree 0 polynomial (see proof below).



    Thus $P$ is at most a second degree polynomial ;



    Plugging expression $P(x)=ax^2+bx+c$ into (1), we get :



    $$1-a=0$$




    Thus the set of solutions is all polynomials of the form
    $$P(x)=1x^2+bx+c.$$




    Proof that (2) is only possible for polynomial of degree $leq 1$:



    If (2) is verified, it means in particular that



    $$forall nin mathbb{Z} Q(n+1)-Q(n)=Q(n)-Q(n-1)=...=Q(2)-Q(1)=Q(1)-Q(0) tag{3}$$



    Let $a=Q(1)-Q(0)$.



    (3) implies that



    $$Q(n+1)=Q(0)+na$$



    which is verified only for polynomials of degree at most 1 (any polynomial of degree $geq 2$ has $lim_{x to infty}|Q(x)/x|=infty$).






    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%2f3034524%2ffind-all-polynomials-of-following-property%23new-answer', 'question_page');
      }
      );

      Post as a guest















      Required, but never shown

























      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      0












      $begingroup$

      Solving the difference equation



      $$
      P(x+1)-2P(x)+P(x-1)=2
      $$



      we obtain



      $$
      P(x) = x^2-x + C_1 x + C_2
      $$






      share|cite|improve this answer











      $endgroup$


















        0












        $begingroup$

        Solving the difference equation



        $$
        P(x+1)-2P(x)+P(x-1)=2
        $$



        we obtain



        $$
        P(x) = x^2-x + C_1 x + C_2
        $$






        share|cite|improve this answer











        $endgroup$
















          0












          0








          0





          $begingroup$

          Solving the difference equation



          $$
          P(x+1)-2P(x)+P(x-1)=2
          $$



          we obtain



          $$
          P(x) = x^2-x + C_1 x + C_2
          $$






          share|cite|improve this answer











          $endgroup$



          Solving the difference equation



          $$
          P(x+1)-2P(x)+P(x-1)=2
          $$



          we obtain



          $$
          P(x) = x^2-x + C_1 x + C_2
          $$







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 10 '18 at 22:28

























          answered Dec 10 '18 at 22:16









          CesareoCesareo

          8,6393516




          8,6393516























              0












              $begingroup$

              $$forall x, 1+P(x)=frac12(P(x+1)+P(x-1)) tag{1}$$



              Setting $Q:=P'$, (1) implies, by differentiation



              $$forall x, Q(x)=frac12(Q(x+1)+Q(x-1)) tag{2}$$



              but this is only possible for a degree 1 or degree 0 polynomial (see proof below).



              Thus $P$ is at most a second degree polynomial ;



              Plugging expression $P(x)=ax^2+bx+c$ into (1), we get :



              $$1-a=0$$




              Thus the set of solutions is all polynomials of the form
              $$P(x)=1x^2+bx+c.$$




              Proof that (2) is only possible for polynomial of degree $leq 1$:



              If (2) is verified, it means in particular that



              $$forall nin mathbb{Z} Q(n+1)-Q(n)=Q(n)-Q(n-1)=...=Q(2)-Q(1)=Q(1)-Q(0) tag{3}$$



              Let $a=Q(1)-Q(0)$.



              (3) implies that



              $$Q(n+1)=Q(0)+na$$



              which is verified only for polynomials of degree at most 1 (any polynomial of degree $geq 2$ has $lim_{x to infty}|Q(x)/x|=infty$).






              share|cite|improve this answer











              $endgroup$


















                0












                $begingroup$

                $$forall x, 1+P(x)=frac12(P(x+1)+P(x-1)) tag{1}$$



                Setting $Q:=P'$, (1) implies, by differentiation



                $$forall x, Q(x)=frac12(Q(x+1)+Q(x-1)) tag{2}$$



                but this is only possible for a degree 1 or degree 0 polynomial (see proof below).



                Thus $P$ is at most a second degree polynomial ;



                Plugging expression $P(x)=ax^2+bx+c$ into (1), we get :



                $$1-a=0$$




                Thus the set of solutions is all polynomials of the form
                $$P(x)=1x^2+bx+c.$$




                Proof that (2) is only possible for polynomial of degree $leq 1$:



                If (2) is verified, it means in particular that



                $$forall nin mathbb{Z} Q(n+1)-Q(n)=Q(n)-Q(n-1)=...=Q(2)-Q(1)=Q(1)-Q(0) tag{3}$$



                Let $a=Q(1)-Q(0)$.



                (3) implies that



                $$Q(n+1)=Q(0)+na$$



                which is verified only for polynomials of degree at most 1 (any polynomial of degree $geq 2$ has $lim_{x to infty}|Q(x)/x|=infty$).






                share|cite|improve this answer











                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  $$forall x, 1+P(x)=frac12(P(x+1)+P(x-1)) tag{1}$$



                  Setting $Q:=P'$, (1) implies, by differentiation



                  $$forall x, Q(x)=frac12(Q(x+1)+Q(x-1)) tag{2}$$



                  but this is only possible for a degree 1 or degree 0 polynomial (see proof below).



                  Thus $P$ is at most a second degree polynomial ;



                  Plugging expression $P(x)=ax^2+bx+c$ into (1), we get :



                  $$1-a=0$$




                  Thus the set of solutions is all polynomials of the form
                  $$P(x)=1x^2+bx+c.$$




                  Proof that (2) is only possible for polynomial of degree $leq 1$:



                  If (2) is verified, it means in particular that



                  $$forall nin mathbb{Z} Q(n+1)-Q(n)=Q(n)-Q(n-1)=...=Q(2)-Q(1)=Q(1)-Q(0) tag{3}$$



                  Let $a=Q(1)-Q(0)$.



                  (3) implies that



                  $$Q(n+1)=Q(0)+na$$



                  which is verified only for polynomials of degree at most 1 (any polynomial of degree $geq 2$ has $lim_{x to infty}|Q(x)/x|=infty$).






                  share|cite|improve this answer











                  $endgroup$



                  $$forall x, 1+P(x)=frac12(P(x+1)+P(x-1)) tag{1}$$



                  Setting $Q:=P'$, (1) implies, by differentiation



                  $$forall x, Q(x)=frac12(Q(x+1)+Q(x-1)) tag{2}$$



                  but this is only possible for a degree 1 or degree 0 polynomial (see proof below).



                  Thus $P$ is at most a second degree polynomial ;



                  Plugging expression $P(x)=ax^2+bx+c$ into (1), we get :



                  $$1-a=0$$




                  Thus the set of solutions is all polynomials of the form
                  $$P(x)=1x^2+bx+c.$$




                  Proof that (2) is only possible for polynomial of degree $leq 1$:



                  If (2) is verified, it means in particular that



                  $$forall nin mathbb{Z} Q(n+1)-Q(n)=Q(n)-Q(n-1)=...=Q(2)-Q(1)=Q(1)-Q(0) tag{3}$$



                  Let $a=Q(1)-Q(0)$.



                  (3) implies that



                  $$Q(n+1)=Q(0)+na$$



                  which is verified only for polynomials of degree at most 1 (any polynomial of degree $geq 2$ has $lim_{x to infty}|Q(x)/x|=infty$).







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Dec 10 '18 at 23:46

























                  answered Dec 10 '18 at 22:46









                  Jean MarieJean Marie

                  29.5k42050




                  29.5k42050






























                      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%2f3034524%2ffind-all-polynomials-of-following-property%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

                      Ellipse (mathématiques)

                      Image
                      Pour les articles homonymes, voir Ellipse. L'ellipse est le lieu des points dont la somme des distances à deux points fixes, dits foyers, est constante. Section du cône ou projection du cercle. En géométrie, une ellipse est une courbe plane fermée obtenue par l’intersection d’un cône de révolution avec un plan, à condition que celui-ci coupe l'axe de rotation du cône ou du cylindre : c'est une conique d'excentricité strictement comprise entre 0 et 1. On peut également la définir comme le lieu des points dont la somme des distances à deux points fixes, dits foyers, est constante (sa construction par la méthode du jardinier est très simple). Dans la vie courante, l’ellipse est la forme qu'on perçoit en regardant un cercle en perspective, ou la figure formée par l’ombre d'un disque sur une surface plane. On retrouve aussi, en première approximation [ 1 ] , des ellipses dans les trajectoires des corps célestes (planètes, comètes ou satellites arti...
                      Read more

                      Quarter-circle Tiles

                      Image
                      up vote 0 down vote favorite I have $99$ identical square tiles, each with a quarter-circle drawn on it like this: [asy] size(1.5cm); draw(Arc((2,0),1,90,180),red+1); draw((0,0)--(2,0)--(2,2)--(0,2)--(0,0)); [/asy] When I arrange the tiles in a $9times 11$ rectangular grid, each with a random orientation, what is the expected value of the number of full circles I form? I think this problem has to do with finding the chance any given 2x2 square has a circle, but I can't find it. expected-value share | cite | improve this question asked Nov 20 at 15:03 6minecraftninja 1 2 ...
                      Read more

                      Mont Emei

                      Image
                      Mont Emei.mw-parser-output .entete.map{background-image:url("//upload.wikimedia.org/wikipedia/commons/7/7a/Picto_infobox_map.png")} Vue du sommet Wanfo du mont Emei Géographie Altitude 3 099  m , Wanfo Coordonnées 29° 31′ 11″ nord, 103° 19′ 57″ est Administration Pays   Chine Province Sichuan Ville-préfecture Leshan Géolocalisation sur la carte : Sichuan Mont Emei Géolocalisation sur la carte : Chine Mont Emei modifier   Paysage panoramique du mont Emei, incluant le paysage panoramique du grand Bouddha de Leshan *   Patrimoine mondial de l'UNESCO Passerelle en bois sur le versant ouest Pays   Chine Subdivision Sichuan Type Mixte Critères (iv) (vi) (x) Superficie 15 400 ha Numéro d’identification 779 Zone géographique Asie et Pacifique  ** Année d’inscription 1996 (20 e session) * Descriptif officiel UNESCO ** Classification géog...
                      Read more