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