Questions on a self-made theorem about polynomialsFor any complex polynomial P degree n: ∑ k = 0 n + 1 ( − 1 ) k ( n + 1 k ) P ( a + k b ) = 0 ∀ a , b ∈ C Basically, if P is quadratic, P ( a ) − 3 P ( a + b ) + 3 P ( a + 2 b ) − P ( a + 3 b ) = 0 (inputs of P are consecutive terms of any arithmetic sequence). This can be generalized to any other degrees.- Has this been discovered? If yes, what's the formal name for this phenomenon?- Is it significant/Are there important consequences of this being true?- Can this be generalized to non-polynomials?

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Questions on a self-made theorem about polynomialsFor any complex polynomial P degree n:      ∑          k      =      0              n      +      1        (  −  1      )    k                      (                            n        +        1            k                      )              P  (  a  +  k  b  )  =  0    ∀  a  ,  b  ∈      C  Basically, if P is quadratic,   P  (  a  )  −  3  P  (  a  +  b  )  +  3  P  (  a  +  2  b  )  −  P  (  a  +  3  b  )  =  0 (inputs of P are consecutive terms of any arithmetic sequence). This can be generalized to any other degrees.- Has this been discovered? If yes, what&#039;s the formal name for this phenomenon?- Is it significant/Are there important consequences of this being true?- Can this be generalized to non-polynomials?

Davion Fletcher · Accepted Answer

Step 1What you wrote is the finite difference operator of order   n  +  1, acting on P,      ∑          k      =      0              n      +      1        (  −  1      )    k                      (                            n        +        1            k                      )              P  (  a  +  k  b  )  =      Δ          n      +      1        P  (  a  +  k  b  )  .Notice that by linearity it suffices for the property to hold for all monomials       k    m    ,  m  ≤  n and it is easily explained by the fact that the first order difference of a polynomial is a polynomial of degree one less.  (  k  +  1      )    m    −      k    m    =      k    m    +  m      k          m      −      1        +  ⋯  −      k    m    .Step 2Illustration (  n  =  3):      Δ    4        k    m    =  (  (      4    m    −      3    m    )  −  (      3    m    −      2    m    )  )  −  (  (      3    m    −      2    m    )  −  (      2    m    −      1    m    )  )    −  (  (      3    m    −      2    m    )  −  (      2    m    −      1    m    )  )  −  (  (      2    4    −      1    m    )  −  (      1    4    −      0    m    )  )    =      4    m    −  4  ⋅      3    m    +  6  ⋅      2    m    −  4  ⋅      1    m    +      0    m    .and                    1                          1                          1                          1                          1                                  0                          0                          0                          0                                        0                          0                          0                                              0                          0                                                    0                                0                          1                          2                          3                          4                                  1                          1                          1                          1                                        0                          0                          0                                              0                          0                                                    0                                0                          1                          4                          9                          16                                  1                          3                          5                          7                                        2                          2                          2                                              0                          0                                                    0                                0                          1                          8                          27                          64                                  1                          7                          19                          37                                        6                          12                          18                                              6                          6                                                    0            Final remark:On can show that       Δ          n      +      1            k          n      +      1        =  (  −  1      )    n    n  !, so that for a polynomial of degree   n  +  1 the sum is  (  −  1      )    n    n  !      p          n      +      1            b          n      +      1        ,independently of a.

Christopher Saunders · Accepted Answer

Step 1The key point is that finite differences and derivatives commute:   D      Δ    h    =      Δ    h    D.For   f  ∈      C    1    (      R    ) you can compute      1    h        Δ    h    [  f  ]  (  x  )  =            f      (      x      +      h      )      −      f      (      x      )        h    =      1    h        ∫    0    h    D  [  f  ]  (  x  +      x    1    )    d      x    1  For   f  ∈      C    n    (      R    ), iterating the above formula, you get                              1                      h            n                                    Δ          h          n                [        f        ]        (        x        )                            =                                            1                              h                                  n                  −                  1                                                                    Δ              h                              n                −                1                                      [            f            ]            (            x            +            h            )            −                          1                              h                                  n                  −                  1                                                                    Δ              h                              n                −                1                                      [            f            ]            (            x            )                    h                                                  =                  1          h                          ∫          0          h                          1                      h                          n              −              1                                      D                  Δ          h                      n            −            1                          [        f        ]        (        x        +                  x          1                )                d                  x          1                                                  =                  1                      h            2                                    ∫          0          h                          ∫          0          h                          1                      h                          n              −              1                                                D          2                          Δ          h                      n            −            2                          [        f        ]        (        x        +                  x          1                +                  x          2                )                d                  x          1                d                  x          2                                                  =                ⋯                                          =                  1                      h            n                                    ∫          0          h                ⋯                  ∫          0          h                          D          n                [        f        ]        (        x        +                  x          1                +        ⋯        +                  x          n                )                d                  x          1                ⋯        d                  x          n                .            Step 2The last integral is a weighted average of       D    n    [  f  ] over the segment   [  x  ,  x  +  n  h  ]. More precisely, let       X    1    ,  …  ,      X    n    ∼      U    n    i    f    o    r    m    (  0  ,  h  ) be i.i.d. and   S  =      X    1    +  ⋯  +      X    n  . Then the previous expression can be viewed as      1          h      n            Δ    h    n    [  f  ]  (  x  )  =      E        [          D      n        [    f    ]    (    S    )    ]  The mean value theorem for integrals applies and tells us that there exists       x    ∗    ∈  (  x  ,  x  +  n  h  ) such that      1          h      n            Δ    h    n    [  f  ]  (  x  )  =      D    n    [  f  ]  (      x    ∗    )  .Regarding your proposition, it follows trivially from the fact that if P is a polynomial of degree n, then       D          n      +      1        [  P  ]  =  0 identically, hence also       Δ    h          n      +      1        [  P  ]  =  0.

For any complex polynomial P degree n: sum_{k=0}^{n+1}(-1)^k ((n+1),(k)) P(a+kb)=0 forall a, b in C

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