Evaluating Summation of Floor Function \(\left[ {\frac{1}{4} + \frac{{\rm{n}}}{{1000}}} \right]\)

The question asks us to find the value of the summation \(\mathop \sum \limits_{{\rm{n}} = 1}^{1000} {\rm{f}}\left( {\rm{n}} \right)\), where the function \({\rm{f}}\left( {\rm{n}} \right)\) is defined as \({\rm{f}}\left( {\rm{n}} \right) = \left[ {\frac{1}{4} + \frac{{\rm{n}}}{{1000}}} \right]\). The notation [x] represents the integral part of x, which is the greatest integer less than or equal to x. This is also known as the floor function.

Understanding the Function \({\rm{f}}\left( {\rm{n}} \right)\)

The function \({\rm{f}}\left( {\rm{n}} \right)\) takes the input n, calculates the value of \(0.25 + \frac{{\rm{n}}}{{1000}}\), and then finds the integral part of that value. We need to sum these integral parts for n ranging from 1 to 1000.

Let's examine how the value of \(0.25 + \frac{{\rm{n}}}{{1000}}\) changes as n increases from 1 to 1000:

• When n = 1, \(0.25 + \frac{1}{1000} = 0.25 + 0.001 = 0.251\).
• When n = 1000, \(0.25 + \frac{1000}{1000} = 0.25 + 1 = 1.25\).

So, the expression inside the floor function, \(0.25 + \frac{{\rm{n}}}{{1000}}\), ranges from 0.251 (for n=1) up to 1.25 (for n=1000).

Determining the Value of \({\rm{f}}\left( {\rm{n}} \right)\)

Since the expression \(0.25 + \frac{{\rm{n}}}{{1000}}\) is between 0.251 and 1.25 for \(1 \le n \le 1000\), the integral part \(\left[ {0.25 + \frac{{\rm{n}}}{{1000}}} \right]\) can only be 0 or 1.

Case 1: \({\rm{f}}\left( {\rm{n}} \right) = 0\)

The integral part is 0 when the number inside the brackets is less than 1 (and greater than or equal to 0). So, \({\rm{f}}\left( {\rm{n}} \right) = \left[ {0.25 + \frac{{\rm{n}}}{{1000}}} \right] = 0\) when \(0 \le 0.25 + \frac{{\rm{n}}}{{1000}} < 1\).

Since n is from 1 to 1000, \(\frac{{\rm{n}}}{{1000}}\) is always positive, so \(0.25 + \frac{{\rm{n}}}{{1000}}\) is always greater than 0. We only need to consider the upper bound:

\(0.25 + \frac{{\rm{n}}}{{1000}} < 1\)

Subtracting 0.25 from both sides:

\(\frac{{\rm{n}}}{{1000}} < 1 - 0.25\)

\(\frac{{\rm{n}}}{{1000}} < 0.75\)

Multiplying by 1000:

\(n < 750\)

So, \({\rm{f}}\left( {\rm{n}} \right) = 0\) for n values from 1 up to 749. The number of terms where \({\rm{f}}\left( {\rm{n}} \right) = 0\) is \(749 - 1 + 1 = 749\).

Case 2: \({\rm{f}}\left( {\rm{n}} \right) = 1\)

The integral part is 1 when the number inside the brackets is greater than or equal to 1 (and less than 2). So, \({\rm{f}}\left( {\rm{n}} \right) = \left[ {0.25 + \frac{{\rm{n}}}{{1000}}} \right] = 1\) when \(1 \le 0.25 + \frac{{\rm{n}}}{{1000}} < 2\).

We need to satisfy \(1 \le 0.25 + \frac{{\rm{n}}}{{1000}}\):

\(1 - 0.25 \le \frac{{\rm{n}}}{{1000}}\)

\(0.75 \le \frac{{\rm{n}}}{{1000}}\)

Multiplying by 1000:

\(750 \le n\)

We also need to satisfy \(0.25 + \frac{{\rm{n}}}{{1000}} < 2\):

\(\frac{{\rm{n}}}{{1000}} < 2 - 0.25\)

\(\frac{{\rm{n}}}{{1000}} < 1.75\)

Multiplying by 1000:

\(n < 1750\)

Combining the conditions \(750 \le n\) and \(n < 1750\), and considering that n goes up to 1000 in the summation, \({\rm{f}}\left( {\rm{n}} \right) = 1\) for n values from 750 up to 1000.

The number of terms where \({\rm{f}}\left( {\rm{n}} \right) = 1\) is \(1000 - 750 + 1 = 251\).

Calculating the Total Summation

The summation \(\mathop \sum \limits_{{\rm{n}} = 1}^{1000} {\rm{f}}\left( {\rm{n}} \right)\) can be split into two parts based on the value of \({\rm{f}}\left( {\rm{n}} \right)\):

\(\mathop \sum \limits_{{\rm{n}} = 1}^{1000} {\rm{f}}\left( {\rm{n}} \right) = \mathop \sum \limits_{{\rm{n}} = 1}^{749} {\rm{f}}\left( {\rm{n}} \right) + \mathop \sum \limits_{{\rm{n}} = 750}^{1000} {\rm{f}}\left( {\rm{n}} \right)\)

• For \(n = 1, 2, \dots, 749\), \({\rm{f}}\left( {\rm{n}} \right) = 0\). The sum for these terms is \(749 \times 0 = 0\).
• For \(n = 750, 751, \dots, 1000\), \({\rm{f}}\left( {\rm{n}} \right) = 1\). The sum for these terms is \(251 \times 1 = 251\).

The total sum is the sum of these two parts:

Total Sum = \(0 + 251 = 251\).

Adding the contributions from both ranges gives the final sum: \(0 + 251 = 251\).

Thus, the value of \(\mathop \sum \limits_{{\rm{n}} = 1}^{1000} {\rm{f}}\left( {\rm{n}} \right)\) is 251.

Revision Table: Summation and Floor Function

Additional Information: Properties of the Floor Function

The floor function, denoted as \(\lfloor x \rfloor\) or [x], is a fundamental concept in mathematics and computer science. Here are some key properties:

• For any real number x, \(\lfloor x \rfloor\) is an integer.
• \(\lfloor x \rfloor \le x < \lfloor x \rfloor + 1\).
• \(\lfloor n \rfloor = n\) if and only if n is an integer.
• \(\lfloor x + n \rfloor = \lfloor x \rfloor + n\) for any integer n.
• \(\lfloor x \rfloor = \lfloor y \rfloor\) if and only if there exists an integer k such that \(k \le x < k+1\) and \(k \le y < k+1\).

In this problem, we used the property that \(\lfloor x \rfloor = k\) for some integer k if and only if \(k \le x < k+1\). We applied this to determine the range of n for which \({\rm{f}}\left( {\rm{n}} \right)\) takes the values 0 and 1.

Calculating summations involving floor functions often requires identifying intervals where the expression inside the floor function falls between consecutive integers, as the floor function's value is constant within such intervals.