HOME

TheInfoList



OR:

In mathematics, Hermite's identity, named after
Charles Hermite Charles Hermite () FRS FRSE MIAS (24 December 1822 – 14 January 1901) was a French mathematician who did research concerning number theory, quadratic forms, invariant theory, orthogonal polynomials, elliptic functions, and algebra. Herm ...
, gives the value of a
summation In mathematics, summation is the addition of a sequence of any kind of numbers, called ''addends'' or ''summands''; the result is their ''sum'' or ''total''. Beside numbers, other types of values can be summed as well: functions, vectors, m ...
involving the
floor function In mathematics and computer science, the floor function is the function that takes as input a real number , and gives as output the greatest integer less than or equal to , denoted or . Similarly, the ceiling function maps to the least ...
. It states that for every
real number In mathematics, a real number is a number that can be used to measurement, measure a ''continuous'' one-dimensional quantity such as a distance, time, duration or temperature. Here, ''continuous'' means that values can have arbitrarily small var ...
''x'' and for every positive
integer An integer is the number zero (), a positive natural number (, , , etc.) or a negative integer with a minus sign ( −1, −2, −3, etc.). The negative numbers are the additive inverses of the corresponding positive numbers. In the language ...
''n'' the following identity holds:. : \sum_^\left\lfloor x+\frac\right\rfloor=\lfloor nx\rfloor .


Proof

Split x into its
integer part In mathematics and computer science, the floor function is the function that takes as input a real number , and gives as output the greatest integer less than or equal to , denoted or . Similarly, the ceiling function maps to the least in ...
and
fractional part The fractional part or decimal part of a non‐negative real number x is the excess beyond that number's integer part. If the latter is defined as the largest integer not greater than , called floor of or \lfloor x\rfloor, its fractional part ca ...
, x=\lfloor x\rfloor+\. There is exactly one k'\in\ with :\lfloor x\rfloor=\left\lfloor x+\frac\right\rfloor\le x<\left\lfloor x+\frac\right\rfloor=\lfloor x\rfloor+1. By subtracting the same integer \lfloor x\rfloor from inside the floor operations on the left and right sides of this inequality, it may be rewritten as :0=\left\lfloor \+\frac\right\rfloor\le \<\left\lfloor \+\frac\right\rfloor=1. Therefore, :1-\frac\le \<1-\frac , and multiplying both sides by n gives :n-k'\le n\, \ Now if the summation from Hermite's identity is split into two parts at index k', it becomes : \begin \sum_^\left\lfloor x+\frac\right\rfloor & =\sum_^ \lfloor x\rfloor+\sum_^ (\lfloor x\rfloor+1)=n\, \lfloor x\rfloor+n-k' \\ pt& =n\, \lfloor x\rfloor+\lfloor n\,\\rfloor=\left\lfloor n\, \lfloor x\rfloor+n\, \ \right\rfloor=\lfloor nx\rfloor. \end


Alternate proof

Consider the function :f(x) = \lfloor x \rfloor + \left\lfloor x + \frac \right\rfloor + \ldots + \left\lfloor x + \frac \right\rfloor - \lfloor nx \rfloor Then the identity is clearly equivalent to the statement f(x) = 0 for all real x. But then we find, : f\left(x + \frac \right) = \left\lfloor x + \frac \right\rfloor + \left\lfloor x + \frac \right\rfloor + \ldots + \left\lfloor x + 1 \right\rfloor - \lfloor nx + 1 \rfloor = f(x) Where in the last equality we use the fact that \lfloor x + p \rfloor = \lfloor x \rfloor + p for all integers p. But then f has period 1/n. It then suffices to prove that f(x) = 0 for all x \in [0, 1/n). But in this case, the integral part of each summand in f is equal to 0. We deduce that the function is indeed 0 for all real inputs x.


References

{{DEFAULTSORT:Hermite's Identity Mathematical identities Articles containing proofs