Problem 26
Question
which initial value produces \(y_{1}=y_{0}\) (steady state)? $$ y(t+1)=-\frac{1}{4} y(t)+6 $$
Step-by-Step Solution
Verified Answer
The initial value is \(y_0 = 4.8\).
1Step 1: Understand the Problem
The problem asks for the initial value that produces a steady state, meaning that \(y_{1} = y_{0}\). This indicates that the system does not change over time, so \(y(t + 1) = y(t)\).
2Step 2: Set Up the Steady State Equation
We know that at steady state, \(y(t + 1) = y(t)\). Substitute this equality into the equation: \ \[ y(t) = -\frac{1}{4} y(t) + 6 \].
3Step 3: Solve for the Steady State Value
Rearrange the equation from Step 2 to solve for \(y(t)\): \ \[ y(t) + \frac{1}{4}y(t) = 6 \]. Factor the left-hand side: \ \[ \frac{5}{4} y(t) = 6 \]. Divide each side by \(\frac{5}{4}\) to isolate \(y(t)\): \ \[ y(t) = \frac{6 \cdot 4}{5} = \frac{24}{5} \].
4Step 4: Interpret the Solution
The solution \( y(t) = \frac{24}{5} \) or \(4.8\) is the initial value that produces a steady state. This means that if the initial value \(y_{0}\) is \(4.8\), then \(y_{1} = y_{0} = 4.8\).
Key Concepts
Steady StateInitial ValueRecursive Sequences
Steady State
In the context of difference equations, a steady state refers to a condition where the value of a sequence remains constant over time. Essentially, the sequence does not undergo any changes with each progression. For this to occur, the terms of the sequence must satisfy a particular condition where each subsequent value equals the previous one.
This can be expressed mathematically as:
Identifying a steady state involves setting the equation for the sequence equal to its prior value and solving for the unknown variable, which in this problem is the initial value. Thus, for the given equation, the steady state arises when the initial value is 4.8, ensuring the sequence remains constant.
This can be expressed mathematically as:
- \( y(t+1) = y(t) \)
Identifying a steady state involves setting the equation for the sequence equal to its prior value and solving for the unknown variable, which in this problem is the initial value. Thus, for the given equation, the steady state arises when the initial value is 4.8, ensuring the sequence remains constant.
Initial Value
The initial value in a recursive sequence is the starting point from which the sequence begins to unfold. This initial condition is vital because it can determine the future behavior of the entire sequence.
In our exercise, the initial value is denoted as \( y_0 \) and the task is to find this specific value that leads directly into a steady state situation, where consecutive terms remain unchanged.
In our exercise, the initial value is denoted as \( y_0 \) and the task is to find this specific value that leads directly into a steady state situation, where consecutive terms remain unchanged.
- For the equation \( y(t+1) = -\frac{1}{4} y(t) + 6 \), identifying the proper initial value ensures that the recursive formula holds true, maintaining equal subsequent terms.
Recursive Sequences
Recursive sequences are sequences where the terms are defined in relation to previous terms using a specific formula or rule. This recursive relation relies heavily on initial conditions, as once given, they enable the generation of subsequent terms.
Recursive sequences follow a format such as \( y(t+1) = f(y(t)) \), where each term is calculated by applying a function to the preceding term. This makes them distinct from other sequences where each term is calculated independently of the others.
Recursive sequences follow a format such as \( y(t+1) = f(y(t)) \), where each term is calculated by applying a function to the preceding term. This makes them distinct from other sequences where each term is calculated independently of the others.
- These sequences often represent dynamic systems, modeling real-world phenomena like population growth or financial forecasts.
- They can be simple or quite sophisticated, depending on the functional relation used.
Other exercises in this chapter
Problem 25
Invent three functions \(f, g, h\) such that for \(x>10\) $$(1+1 / x)^{x}
View solution Problem 26
Questions \(20-29\) are about the derivative \(d y / d x=c b^{x}\). Find a base \(b\) for which \(\left(b^{h}-1\right) / h \approx 1\). Use \(h=1 / 4\) by hand
View solution Problem 26
Graph \(e^{x}\) and \(\sqrt{e^{x}}\) at \(x=-2,-1,0,1,2 .\) Another form of \(\sqrt{e^{x}}\) is _______.
View solution Problem 27
Questions \(20-29\) are about the derivative \(d y / d x=c b^{x}\). Find the second derivative of \(y=b^{x}\) and also of \(x=\log _{b} y .\)
View solution