Question

Let \(\left\\{t_{m}\right\\}_{m=0}^{\infty}\) be a sequence of points such that \(t_{0}=0, t_{m+1}>t_{m},\) and \(\lim _{m \rightarrow \infty} t_{m}=\infty\). For each nonnegative integer \(m\), let \(f_{m}\) be continuous on \(\left[t_{m}, \infty\right)\), and let \(f\) be defined on \([0, \infty)\) by $$ f(t)=f_{m}(t), t_{m} \leq t

Short answer

Based on the given sequence of functions and the definition of a piecewise function, we can show that this function is continuous on the interval [0, ∞) by considering the continuity of each function in the sequence for a specific range. We then prove the step function representation by breaking down the definition of a step function and analyzing its behavior for specific cases. Lastly, we verify that the series in the step function converges for all t in [0, ∞) by considering two separate scenarios: 1) when t is less than t1 and 2) when t is greater than or equal to t1. Both cases yield a series that converges, thus proving that the piecewise function is continuous and has a proper step function representation with converging series.

Step-by-step solution

Prove f(t) is piecewise continuous

Given that each \(f_m(t)\) is continuous on \([t_m, \infty)\), we only need to consider the case where \(t\) lies between two consecutive points \(t_m\) and \(t_{m+1}\). We know that \(t_0 = 0\) and \(\lim_{m\rightarrow \infty} t_m = \infty\). Thus, for \(t \in [t_m, t_{m+1}), m=0,1,\ldots\), we have \(f(t)=f_m(t)\), which is continuous by definition. This ensures the continuity of \(f(t)\) on \([0, \infty)\).

Prove the step function representation

To prove: $$ f(t) = f_0(t) + \sum_{m=1}^{\infty} u(t-t_m)(f_m(t)-f_{m-1}(t)) $$ We'll break this down using the definition of a step function, which is defined as: $$ u(t) = \begin{cases} 0, & t < 0 \\ 1, & t \ge 0 \end{cases} $$ Now, consider a fixed \(t \in [0, \infty)\). There exists a unique \(m\) such that \(t \in [t_m, t_{m+1})\). Therefore, we have: $$ f(t) = f_m(t), $$ and also $$ f_m(t) - f_{m-1}(t) = \begin{cases} f_m(t), & m = 0 \\ f_m(t) - f_{m-1}(t), & m > 0 \end{cases} $$ Applying the step function in this representation: $$ u(t-t_k)(f_k(t)-f_{k-1}(t)) = \begin{cases} f_k(t), & k = m \\ 0, & \text{otherwise} \end{cases} $$ where \(k = 0, 1, 2, \ldots.\) Now, let's substitute everything back into the sum: $$ f(t) = f_0(t) + \sum_{k=1}^{\infty} u(t-t_k)(f_k(t)-f_{k-1}(t)) $$ The only nonzero term in the sum is when \(k = m\). Therefore, $$ f(t) = f_0(t) + u(t-t_m)(f_m(t)-f_{m-1}(t)) = f_m(t), $$ which proves the step function representation of \(f(t)\).

Show the convergence of the series

We now need to show that the series converges for all \(t\in [0, \infty)\). Observe that for any \(t\in [0, \infty)\): if \(t < t_1\), then \(u(t - t_m) = 0\) for \(m \ge 1\), which gives the series $$ f(t) = f_0(t), $$ which converges trivially. if \(t \ge t_1\), then there exists \(M \in \mathbb{N}\) such that \(t_M \le t < t_{M+1}\). Then all the terms of the series vanish when \(m > M\). The resulting finite sum is $$ f(t) = f_0(t) + \sum_{m=1}^{M} u(t-t_m)(f_m(t)-f_{m-1}(t)), $$ which converges for all \(t\in [0, \infty)\). This concludes the proof that the series converges for all \(t\) in \([0, \infty)\).

Key concepts

Step Function Representation

A step function is a piecewise constant function having a finite or infinite number of intervals in which it holds a constant value. The representation of a function as a step function makes it easier to analyze, especially in the context of piecewise continuous functions. Let's delve into how the function \( f(t) \) is expressed using a step function.

The given function, \( f(t) = f_0(t) + \sum_{m=1}^{\infty} u(t-t_m)(f_m(t)-f_{m-1}(t) \), utilizes the unit step function \( u(t) \), which is defined as:

• \( u(t-t_m) = 0 \) when \( t < t_m \)
• \( u(t-t_m) = 1 \) when \( t \geq t_m \)

This allows the function \( f(t) \) to switch between different functions \( f_m(t) \) over their respective intervals \([t_m, t_{m+1})\).

For instance, if \( t \) lies in the interval \([t_m, t_{m+1})\), only the term \( f_m(t) \) is non-zero, because the unit step function makes all previous terms vanish when \( t-t_k < 0 \) for \( k > m \). This characteristic inherently provides each segment of \( f(t) \) a controlled, step-like behavior that simplifies convergence analysis over the partitioned domain.

Convergence of Series

Convergence is a critical concept when considering series, particularly when looking at functions expressed in terms of infinite summations like step functions. A series is said to converge if its terms tend to a finite limit as they increase indefinitely.

In our example, the convergence of the series within the function \( f(t) \) is assured by the way terms beyond a certain point become zero.

Here's how it works for the function \( f(t) \):

• For \( t < t_1 \), the series is simplified to \( f_0(t) \), a straightforward scenario with no infinite terms.
• When \( t \ge t_1 \), a finite integer \( M \) exists where \( t_M \le t < t_{M+1} \). Beyond this point, all additional terms in the series vanish because the step function \( u(t-t_m) \) will be zero for all \( m > M \).

Since only a finite number of non-zero terms contribute for any chosen \( t \), the series is naturally convergent. This behavior guarantees \( f(t) \) can be calculated accurately for any point on its domain, making it a stable piecewise continuous representation.

Limit of Sequence

The concept of the limit of a sequence is pivotal in understanding how a sequence progresses and approaches a particular value as the term index increases indefinitely. In our context, we examine the sequence \( \{t_m\} \) with properties \( t_0 = 0, t_{m+1} > t_m \), and \( \lim_{m \to \infty} t_m = \infty \).

This sequence tells us about the increasing intervals \([t_m, t_{m+1})\) where each partial function \( f_m(t) \) operates. As the indices \( m \) increase, the endpoints \( t_m \) tend toward infinity. Essentially, this property ensures that for every real number \( t \), there exists a sufficiently large \( m \) such that \( t < t_m \).

Here is why this is vital:

• For \( f(t) \), being piecewise defined, this sequence sets up the switching mechanism at each \( t_m \).
• The sequence approaching infinity ensures the entire real line \([0, \infty)\) can be covered using these successive intervals, while ensuring every \( f_m(t) \) is smoothly transitioned onto the next.

By analyzing the limit of the sequence, we observe how the full domain of \( f(t) \) is partitioned into manageable pieces, allowing the function to remain piecewise continuous.