Problem 14 - Calculus

Question

Take three steps of \(y(t+1)=2 y(t)+1\) from \(y_{0}=0\).

Step-by-Step Solution

Answer

The values are \( y_1 = 1 \), \( y_2 = 3 \), \( y_3 = 7 \).

Step 1: Understand the Initial Value

The initial value is provided as \( y_0 = 0 \). This means at time \( t = 0 \), the value of \( y \) is 0. We will use this initial value to find the subsequent values of \( y(t) \) for the next three steps.

Step 2: Calculate \( y_1 \)

The equation is \( y(t+1) = 2y(t) + 1 \). To find \( y_1 \), substitute \( t = 0 \) into the equation: \( y_1 = 2y_0 + 1 \). Given that \( y_0 = 0 \), this simplifies to \( y_1 = 2(0) + 1 = 1 \).

Step 3: Calculate \( y_2 \)

Now, substitute \( t = 1 \) into the equation to find \( y_2 \): \( y_2 = 2y_1 + 1 \). From the previous step, we found \( y_1 = 1 \). Thus, \( y_2 = 2(1) + 1 = 3 \).

Step 4: Calculate \( y_3 \)

Finally, substitute \( t = 2 \) into the equation to find \( y_3 \): \( y_3 = 2y_2 + 1 \). From the previous step, we found \( y_2 = 3 \). Therefore, \( y_3 = 2(3) + 1 = 7 \).

Key Concepts

Initial Conditions

Initial conditions are the foundation of solving recurrence relations. They provide a starting point for the sequence that the relation defines. In our exercise, the initial condition is given as \( y_0 = 0 \). This tells us that at the time \( t = 0 \), the value of \( y \) is 0. Initial conditions are crucial because they allow us to calculate future values in the sequence.

In a recurrence relation like \( y(t+1) = 2y(t) + 1 \), the initial condition helps determine the

1. starting point, which directly affects the outcome of the entire sequence,
2. ensures the sequence is uniquely defined, and
3. helps predict future values by providing the necessary base.

Iterative Process

The iterative process involves repeatedly applying the same operation until the desired number of steps is completed. Here, to find subsequent terms, we apply the given recurrence relation iteratively. This means we use each new result to find the next value, as shown in the step-by-step solution.

For example, starting with the initial condition \( y_0 = 0 \), we calculate the next few terms:

• Calculate \( y_1 \): Substitute \( y_0 \) into the equation to find \( y_1 = 2(0) + 1 = 1 \).
• Calculate \( y_2 \): Use \( y_1 \) to get \( y_2 = 2(1) + 1 = 3 \).
• Calculate \( y_3 \): Use \( y_2 \) to find \( y_3 = 2(3) + 1 = 7 \).

Through this iterative process, we systematically build upon each previous result to determine the next value in the sequence.

Linear Recurrence Relations

Linear recurrence relations are equations that express each term of a sequence as a linear combination of its previous terms, often involving some fixed coefficients and constants.Our given relation, \( y(t+1) = 2y(t) + 1 \), is an example of a linear recurrence relation. This formula establishes that each new term is twice the previous term plus one.

Key aspects of linear recurrence relations include:

• Coefficient: The number multiplying the previous term (\(2\) in our case).
• Constant: The number added after scaling (\(1\) here).

Linear recurrence relations are common because they can model various real-world phenomena and often have straightforward solutions. Each step in the sequence depends linearly on the previous term, making calculations systematic and predictable.