Decomposing ∑ i = 0 2 n x i as a simple sum of squares

Question

Decomposing       ∑          i      =      0              2      n            x    i   as a simple sum of squares

Keyon Fitzgerald · Accepted Answer

Note that                              ∑                      k            =            0                                2            n                                    x          k                                    =                                            x                              2                n                +                1                                      −            1                                x            −            1                                                            =        (        2        n        +        1        )                  ∫          0          1                (        1        −        t        +        t        x                  )                      2            n                                    d                t                                          =                              2            n            +            1                                2                          2              n              +              1                                                ∫                      −            1                    1                (        x        +        1        +        (        x        −        1        )        t                  )                      2            n                                    d                t                                          =                  ∑                      k            =            0                                2            n                                                                                            2                n                                                    C              k                        (            2            n            +            1            )                                2                          2              n              +              1                                                ∫                      −            1                    1                (        x        +        1                  )                      2            n            −            k                          (        x        −        1                  )                      k                                    t          k                          d                t                                          =                  ∑                      k            =            0                                n                                                                                            2                n                                                    C                              2                k                                      (            2            n            +            1            )                                              2                              2                n                                      (            2            k            +            1            )                          (        x        +        1                  )                      2            n            −            2            k                          (        x        −        1                  )                      2            k                          ;            whence                                          ∑                          k              =              0                                      2              n                                            x            k                    =                      ∑                          k              =              0                                      n                                                                                                            2                  n                                                            C                                  2                  k                                            (              2              n              +              1              )                                                      2                                  2                  n                                            (              2              k              +              1              )                                (          (          x          +          1                      )                          n              −              k                                (          x          −          1                      )                          k                                            )            2                                .Consequently, if                                       2          n                            C                  2          k                    (      2      n      +      1      )                      2                  2          n                    (      2      k      +      1      )       is a sum of squares (all nonnegative rationals are), then       ∑          k      =      0              2      n            x    k   is a sum of squares with rational coefficients.

Gael Gardner · Accepted Answer

Let     ω   be a primitive     (  2  n  +  1      )          t      h         root of unity, so         ω          2      n      +      1        =  1   and         ω          2      n      +      1      −      k        =            1              ω        k              =            ω      k        ¯    Then:  P  (  x  )  =      ∑          k      =      0              2      n            x    k    =      ∏          k      =      1              2      n        (  x  −      ω    k    )  =      ∏          k      =      1        n              (        (  x  −      ω    k    )  (  x  −            ω      k        ¯    )            )        =                                          ∏                          k              =              1                        n                    (          x          −                      ω            k                    )                ⏟                    Q      (      x      )                                                  ∏                          k              =              1                        n                    (          x          −                                    ω              k                        ¯                    )                ⏟                                                Q                          ¯            (      x      )      Q is a polynomial with complex coefficients which can be written as     Q  (  x  )  =  U  (  x  )  +  i  V  (  x  )   where U,V are polynomials with real coefficients. It follows that:  P  (  x  )  =  Q  (  x  )        Q    ¯    (  x  )  =            (        U  (  x  )  +  i  V  (  x  )            )                  (        U  (  x  )  −  i  V  (  x  )            )        =      U    2    (  x  )  +      V    2    (  x  )

Decomposing <munderover> ∑ <mrow class="MJX-TeXAtom-ORD"> i

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