Question

Determine whether the given function is periodic. If so, find its fundamental period. $$ f(x)=\left\\{\begin{array}{ll}{(-1)^{n},} & {2 n-1 \leq x<2 n,} \\ {1,} & {2 n \leq x<2 n+1 ;}\end{array} \quad n=0, \pm 1, \pm 2, \ldots\right. $$

Short answer

If so, find its fundamental period. Answer: The given function f(x) is not periodic.

Step-by-step solution

Understand the function

The given function is defined as $$ f(x)=\left\{\begin{array}{ll}{(-1)^{n},} & {2 n-1 \leq x<2 n,} \\\ {1,} & {2 n \leq x<2 n+1.}\end{array} \quad n=0, \pm 1, \pm 2, \ldots\right. $$ We can tabulate the function with some specific n's for better understanding. | n | x range | f(x) | |--- |--------------- |------ | | 0 | -1 ≤ x < 0 | 1 | | 1 | 0 ≤ x < 1 | -1 | | -1 | 1 ≤ x < 2 | 1 | | 2 | 3 ≤ x < 4 | 1 | | -2 | 5 ≤ x < 6 | -1 | From the table, we can see that the function repeats itself after a certain interval.

Check for periodicity

A function is periodic if there exists a positive number T such that f(x+T) = f(x) for all x in the domain of the function. By looking at the table we created in Step 1, we can observe that f(x) alternates between 1 and -1 for each consecutive integer interval. Therefore, the function seems to have a period of 2. To confirm this, let's check if f(x+2) = f(x) for all x in the domain of the function. 1. Case 1: If \((2n-1) \leq x < 2n\), then \(f(x) = (-1)^n\). $$ f(x+2)=\left\{\begin{array}{ll}{(-1)^{n+1},} & {2(n+1)-1 \leq x+2<2(n+1),} \\\ {1,} & {2(n+1) \leq x+2<2(n+1)+1,}\end{array}\right. $$ If \((2n-1) \leq x < 2n\), then \((2(n+1)-1) \leq x+2 < 2(n+1)\), i.e., \(f(x+2) = (-1)^{n+1}\). So, we have \(f(x+2) = (-1)^{n+1} = -(-1)^n = -f(x)\). 2. Case 2: If \(2n \leq x < 2n+1\), then \(f(x) = 1\). $$ f(x+2)=\left\{\begin{array}{ll}{(-1)^{n+1},} & {2(n+1)-1 \leq x+2<2(n+1),} \\\ {1,} & {2(n+1) \leq x+2<2(n+1)+1.}\end{array}\right. $$ If \(2n \leq x < 2n+1\), then \(2(n+1) \leq x+2 < 2(n+1)+1\), i.e., \(f(x+2) = 1\). So, we have \(f(x+2) = 1 = f(x)\). In one hand, we have f(x)=-f(x+2); in the other hand f(x)=f(x+2). Therefore, there is not a sufficient condition to prove periodicity.

Conclusion

The given function, $$ f(x)=\left\{\begin{array}{ll}{(-1)^{n},} & {2 n-1 \leq x<2 n,} \\\ {1,} & {2 n \leq x<2 n+1,}\end{array} \quad n=0, \pm 1, \pm 2, \ldots\right. $$ is not periodic since we couldn't prove the condition f(x+T) = f(x) for all x in the domain of the function, with a fixed positive T.

Key concepts

Fundamental Period

Understanding the fundamental period of a function is crucial in mathematical function analysis. The fundamental period is defined as the smallest positive value of T for which the function repeats itself, that is, for any value x in the function's domain, it holds that the function value at x is the same as at x+T. This concept is essential in the study of periodic functions which often arise in fields such as physics with wave motions, or in engineering with signal processing.

In exploring periodicity, it's important to remember that the fundamental period must be consistent for all values of x within the domain of the function. Moreover, the existence of a fundamental period allows the simplification of complex functions by studying them over one period instead of their entire domain.

Mathematical Function Analysis

Mathematical function analysis involves understanding the characteristics of a function, including its behavior, continuity, and in this context, its periodicity. Analyzing functions allows mathematicians and scientists to predict and model real-world phenomena. When examining a function to determine its periodic nature, one not only looks for repetition in values but also scrutinizes the rule that governs the function for clues about its behavior over different intervals.

For instance, with the given exercise function, the tabulation of specific values provides an insight into the function's behavior at various intervals. This early step is at the core of mathematical function analysis and lays the ground for further exploration of the function's properties.

Periodicity Check

A periodicity check is a process of determining whether a function exhibits periodic behavior. This is done by applying the definition of a periodic function: there must exist a positive constant T, such that for all x within the domain, the equality f(x) = f(x+T) holds. The steps to prove or disprove this involve picking arbitrary points within the domain and examining if the function assumes the same values at x and x+T intervals. The function may show periodic behavior over several intervals but to be considered periodic, the pattern must be consistent across its entire domain.

The ultimate goal of this analysis is to establish a consistent period T, or to demonstrate its absence. Our exercise demonstrates a scenario where despite having an alternating behavior, the function doesn't satisfy the conditions for periodicity for a constant value T. This occurrence underlines the importance of thoroughness in periodicity checks.