Hermite's identity is an identity that computes the value of a summation involving floor functions. It is named after Charles Hermite, a French mathematician who did research during the nineteenth century. It is useful in solving equations, proving inequalities, computing sums etc. involving floor functions.
Let be a real number, and let be a positive integer. Then,
Notice that there is no restriction on the sign of i.e. it can be positive or negative.
Before presenting the formal proof, one idea of the proof has to be noted. The key fact here is that can take only two values, namely and . Showing this is not too hard. Note that
Hence, the minimum value of the above terms is . The maximum value is
since is either or .
This was the key observation and it's going to help us in our proof. As you will see later, the key fact also helps us in problem solving.
The case where is an integer is obviously true. So we assume that is not an integer. This means . Since
there exists an such that
Now note that satisfies
Together, these two inequalities give us
Using this, which implies that .
which completes the proof.
There are a couple of things to keep in mind when using Hermite's identity:
- can take only two values, namely and . This idea is very important.
- Substitute wherever you can. This idea isn't specific to Hermite's identity, but it is incredibly useful and deserves a mention here.
Now that we are done talking about some main aspects of Hermite's identity, let's solve some problems.
Prove that for any real , where is the function that rounds a number to its nearest integer.
First note that . Hence by Hermite's identiity.
A good exercise to do at home is to see what Hermite's identity looks like for different values of . This makes it easier to sometimes spot Hermite's identity.
Suppose is a real number for which
The given sum has exactly terms, each of which equals or . If all of the terms equal , then the total sum is which is less than . If all of the terms equal , then the total sum is , which is greater than . It follows that some of the terms take the value while the others take . This means that . Suppose the first terms take the value whie the other terms take the value . Then, . Solving this, we get . This means that and . Therefore,
Prove that .
It must be admitted that there are easier ways to prove this inequality.
Use the fact that and the given problem reduces to proving the inequality . This can be proven by substituting and and considering what the fractional parts of and can be. This part is left to the reader.