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







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$endgroup$








  • 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

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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











    $endgroup$













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      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

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      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






























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