Assuming P(x) is a polynomial, can one deduce that P(x) is a polynomial of degree 0 or 1 from the following condition: 2 P ( x ) = P ( x − 2 k n ) + P ( x + 2 k n ) , ∀ xwhere n is a fixed positive integer and k ∈ N ∪ { 0 }The identity gives a feeling like the values of P(x) is "equally distributed" and its graph illustrates a straight line since P ( x ) = P ( x − 2 k n ) + P ( x + 2 k n ) 2 so its value at x is an arithmetic mean of its values at x − 2 k n and x + 2 k n . I don't have a potential argument though.

Question

Assuming P(x) is a polynomial, can one deduce that P(x) is a polynomial of degree 0 or 1 from the following condition:  2  P  (  x  )  =  P            (        x  −            2      k        n              )        +  P            (        x  +            2      k        n              )        ,        ∀  xwhere n is a fixed positive integer and   k  ∈      N    ∪  {  0  }The identity gives a feeling like the values of P(x) is &quot;equally distributed&quot; and its graph illustrates a straight line since  P  (  x  )  =                                                  P                                    (                                x          −                                    2              k                        n                                              )                                +          P                                    (                                x          +                                    2              k                        n                                              )                                                                                        2                    so its value at x is an arithmetic mean of its values at   x  −            2      k        n   and   x  +            2      k        n  . I don&#039;t have a potential argument though.

latatuy · Accepted Answer

Step 1Yes, even for a single fixed k.Taking   n  =  2 and   k  =  1 as an example, the given equation implies   P  (  x  +  2  )  −  2  P  (  x  +  1  )  +  P  (  x  )  =  0 for all x; the left-hand side is the result of twice applying the finite difference operator to P(x). While formulas for finite differences of polynomials are not quite as slick as their derivatives, it is still the case that the finite difference of a degree-d polynomial is a degree-  (  d  −  1  ) polynomial when   d  ≥  1. It follows that the only polynomials whose second-order finite difference vanishes are linear polynomials.Step 2Other values of n and k follow immediately just by rescaling the variable x by an appropriate constant, or using finite differences with an offset different from 1.

Marlene Brooks · Accepted Answer

Step 1Assuming that P is a polynomial to begin with, the property transfers to the derivative of P. If P has degree   n  ≥  2, after   n  −  2 differentiations we arrive at a polynomial of degree 2. If   Q  (  x  )  =  a      x    2    +  b  x  +  c satisfies the condition, then, with   r  =      2    k        /    n  ≠  0,  2  a      x    2    +  2  b  x  +  2  c  =  a  (  x  −  r      )    2    +  b  (  x  −  r  )  +  c  +  a  (  x  +  r      )    2    +  b  (  x  +  r  )  +  cyielding  a      r    2    =  0Step 2&amp;lt;br&amp;lt;Therefore   a  =  0, a contradiction.

Assuming P(x) is a polynomial, can one deduce that P(x) is a polynomial of degree 0 or 1 from the following condition: 2P(x)=P(x-(2^k)/n)+P(x+(2^k)/n), forall x where n is a fixed positive integer and k in N cup {0}.

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