In mathematics, Hermite's identity, named after Charles Hermite, gives the value of a summation involving the floor function. It states that for every real number x and for every positive integer n the following identity holds:
                              ∑                      k            =            0                                n            −            1                                    ⌊          x          +                                    k              n                                ⌋                =        ⌊        n        x        ⌋        .                Split                     x                 into its integer part and fractional part,                     x        =        ⌊        x        ⌋        +        {        x        }                . There is exactly one                               k          ′                ∈        {        1        ,        …        ,        n        }                 with
                    ⌊        x        ⌋        =                  ⌊          x          +                                                                      k                  ′                                −                1                            n                                ⌋                ≤        x        <                  ⌊          x          +                                                    k                ′                            n                                ⌋                =        ⌊        x        ⌋        +        1.                By subtracting the same integer                     ⌊        x        ⌋                 from inside the floor operations on the left and right sides of this inequality, it may be rewritten as
                    0        =                  ⌊          {          x          }          +                                                                      k                  ′                                −                1                            n                                ⌋                ≤        {        x        }        <                  ⌊          {          x          }          +                                                    k                ′                            n                                ⌋                =        1.                Therefore,
                    1        −                                            k              ′                        n                          ≤        {        x        }        <        1        −                                                            k                ′                            −              1                        n                          ,                and multiplying both sides by                     n                 gives
                    n        −                  k          ′                ≤        n                {        x        }        <        n        −                  k          ′                +        1.                Now if the summation from Hermite's identity is split into two parts at index                               k          ′                        , it becomes
                                                                                          ∑                                      k                    =                    0                                                        n                    −                    1                                                                    ⌊                  x                  +                                                            k                      n                                                        ⌋                                                                            =                                  ∑                                      k                    =                    0                                                                              k                      ′                                        −                    1                                                  ⌊                x                ⌋                +                                  ∑                                      k                    =                                          k                      ′                                                                            n                    −                    1                                                  (                ⌊                x                ⌋                +                1                )                =                n                                ⌊                x                ⌋                +                n                −                                  k                  ′                                                                                                                  =                n                                ⌊                x                ⌋                +                ⌊                n                                {                x                }                ⌋                =                                  ⌊                  n                                    ⌊                  x                  ⌋                  +                  n                                    {                  x                  }                  ⌋                                =                ⌊                n                x                ⌋                .