Flag Differential Calculus> f(x)=[x] + [2x] + [3x] +........+[nx] – n...
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f(x)=[x] + [2x] + [3x] +........+[nx] – nx(n+1)/2

where [.] denotes greatest integer function. Find the period of the function above.
0 clue on how to begin with this one, would appreciate some tips on how to approach periodic functions within series, I constantly do badly at these type of problems

Arjun Panchmatia , 10 Years ago
Grade
anser 1 Answers
Askiitians Tutor Team

To determine the period of the function \( f(x) = [x] + [2x] + [3x] + \ldots + [nx] - \frac{nx(n+1)}{2} \), where \([.]\) denotes the greatest integer function, we need to analyze how the function behaves as \( x \) changes. The greatest integer function, also known as the floor function, outputs the largest integer less than or equal to \( x \). This characteristic will play a crucial role in our analysis.

Understanding the Components of the Function

Let's break down the function into its components:

  • Greatest Integer Function: The term \([kx]\) for \( k = 1, 2, \ldots, n \) will change its value at each integer multiple of \( \frac{1}{k} \). This means that each term \([kx]\) is piecewise constant, changing at specific intervals.
  • Summation of Terms: The sum \([x] + [2x] + \ldots + [nx]\) accumulates these piecewise constant values, which will also change at the same intervals.
  • Linear Adjustment: The term \(-\frac{nx(n+1)}{2}\) is a continuous linear function that does not affect the periodicity directly but shifts the overall function vertically.

Identifying the Periodicity

To find the period of \( f(x) \), we need to identify the smallest positive \( T \) such that \( f(x + T) = f(x) \) for all \( x \). Given the nature of the greatest integer function, we can observe the following:

  • Each term \([kx]\) will repeat its values every \( \frac{1}{k} \).
  • Thus, the function \( f(x) \) will repeat its behavior at the least common multiple (LCM) of the periods of each individual term.

Calculating the Least Common Multiple

The periods of the individual terms \([kx]\) are \( \frac{1}{1}, \frac{1}{2}, \frac{1}{3}, \ldots, \frac{1}{n} \). The least common multiple of these fractions can be calculated as follows:

  • The LCM of the denominators \( 1, 2, 3, \ldots, n \) is simply \( n! \) (n factorial).
  • Since the LCM of the fractions \( \frac{1}{k} \) is given by \( \frac{\text{LCM of the numerators}}{\text{GCD of the denominators}} \), and here the numerators are all 1, the period is \( n! \).

Conclusion on the Periodicity

Therefore, the period of the function \( f(x) \) is \( n! \). This means that if you shift \( x \) by \( n! \), the function will return to its original value. Understanding periodic functions often involves recognizing the behavior of their components, especially when dealing with piecewise functions like those involving the greatest integer function. By analyzing how each term contributes to the overall function, we can effectively determine the period.

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