Understanding the Floor and Fractional Part Functions

Oct 13, 2025 by talynt.org 149 views

Introduction

In this article, we will delve into the world of mathematics and explore a complex expression involving the floor and fractional part functions. The given expression is: $[x] + \sum_{r=-1}^{100} \{x+r\} \frac{1}{100}$ . Our goal is to simplify and solve this expression, providing a clear and concise understanding of the underlying mathematical concepts.

Understanding the Floor and Fractional Part Functions

Before we dive into the expression, let's briefly discuss the floor and fractional part functions. The floor function, denoted by $[x]$ , gives the greatest integer less than or equal to $x$ . On the other hand, the fractional part function, denoted by $\{x\}$ , gives the decimal part of $x$ .

For example, if $x = 3.7$ , then $[x] = 3$ and $\{x\} = 0.7$ . Similarly, if $x = -2.3$ , then $[x] = -3$ and $\{x\} = 0.7$ .

Breaking Down the Expression

Now that we have a good understanding of the floor and fractional part functions, let's break down the given expression. We have:

$[x] + \sum_{r=-1}^{100} \{x+r\} \frac{1}{100}$

This expression consists of two parts: the floor function $[x]$ and the summation of the fractional part function $\{x+r\}$ .

Simplifying the Expression

To simplify the expression, let's first focus on the summation part. We have:

$\sum_{r=-1}^{100} \{x+r\} \frac{1}{100}$

This summation represents the sum of the fractional parts of $x+r$ for $r$ ranging from $-1$ to $100$ .

Using the Properties of the Fractional Part Function

The fractional part function has some interesting properties that we can exploit to simplify the expression. One such property is that the fractional part function is periodic with a period of $1$ . This means that $\{x+r\} = \{x\}$ for any integer $r$ .

Using this property, we can rewrite the summation as:

$\sum_{r=-1}^{100} \{x\} \frac{1}{100}$

Evaluating the Summation

Now that we have simplified the summation, let's evaluate it. We have:

$\sum_{r=-1}^{100} \{x\} \frac{1}{100}$

This summation represents the sum of the fractional part of $x$ multiplied by $\frac{1}{100}$ for $r$ ranging from $-1$ to $100$ .

Since the fractional part function is periodic with a period of $1$ , the summation can be rewritten as:

$101 \{x\} \frac{1}{100}$

Combining the Terms

Now that we have evaluated the summation, let's combine the terms. We have:

$[x] + 101 \{x\} \frac{1}{100}$

This expression represents the simplified form of the original expression.

Solving the Expression

To solve the expression, we need to find the value of $x$ that satisfies the equation. Let's rewrite the equation as:

$[x] + 101 \{x\} \frac{1}{100} = k$

where $k$ is a constant.

Using the Properties of the Floor and Fractional Part Functions

The floor and fractional part functions have some interesting properties that we can exploit to solve the equation. One such property is that the floor function is a step function, meaning that it takes on integer values at integer points.

Using this property, we can rewrite the equation as:

$[x] + 101 \{x\} \frac{1}{100} = k$

$[x] = k - 101 \{x\} \frac{1}{100}$

Finding the Value of $x$

To find the value of $x$ , we need to find the value of $k$ that satisfies the equation. Let's assume that $k$ is an integer.

Since the floor function is a step function, we know that $[x]$ is an integer. Therefore, we can rewrite the equation as:

$k - 101 \{x\} \frac{1}{100} = [x]$