Problem 56
Question
Compute \(N_{t}\) and \(N_{t} / N_{t-1}\) for \(t=2,3,4, \ldots .20\) when $$ N_{t+1}=N_{t}+2 N_{t-1} $$ with \(N_{0}=1\) and \(N_{1}=1\).
Step-by-Step Solution
Verified Answer
Calculate the sequence and compute ratios for each \( t \). Continue until \( N_{20} \).
1Step 1: Understanding the Recurrence Relation
We are given the recurrence relation \( N_{t+1} = N_{t} + 2N_{t-1} \), with the initial conditions \( N_{0} = 1 \) and \( N_{1} = 1 \). We need to compute \( N_{t} \) for \( t=2 \) to \( 20 \) and the ratios \( \frac{N_{t}}{N_{t-1}} \).
2Step 2: Computing Initial Values
Using the initial conditions:\- \( N_{0} = 1 \) \- \( N_{1} = 1 \) \We will start computing \( N_{t} \) from \( t=2 \).
3Step 3: Apply Recurrence for \( N_2 \)
Calculate for \( t=2 \):\\[ N_{2} = N_{1} + 2N_{0} = 1 + 2 \times 1 = 3 \]
4Step 4: Compute Ratio for \( N_2 \)
Calculate the ratio for \( t=2 \): \\[ \frac{N_{2}}{N_{1}} = \frac{3}{1} = 3 \]
5Step 5: Continue with \( N_3 \)
Calculate for \( t=3 \):\\[ N_{3} = N_{2} + 2N_{1} = 3 + 2 \times 1 = 5 \]
6Step 6: Compute Ratio for \( N_3 \)
Calculate the ratio for \( t=3 \): \\[ \frac{N_{3}}{N_{2}} = \frac{5}{3} \approx 1.67 \]
7Step 7: Repeat for \( N_4 \)
Calculate for \( t=4 \):\\[ N_{4} = N_{3} + 2N_{2} = 5 + 2 \times 3 = 11 \]
8Step 8: Compute Ratio for \( N_4 \)
Calculate the ratio for \( t=4 \): \\[ \frac{N_{4}}{N_{3}} = \frac{11}{5} = 2.2 \]
9Step 9: Continue Iteratively
Continue calculating \( N_t \) using \( N_{t+1} = N_{t} + 2N_{t-1} \) up to \( t=20 \), and each time compute the ratio \( \frac{N_{t}}{N_{t-1}} \).
10Step 10: Summarize Results
After calculations, summarize the sequence values: for example \( N_5 = 21 \), \( N_{6} = 43 \), etc., continuing until \( N_{20} \). For each \( t \), compute \( \frac{N_{t}}{N_{t-1}} \).
Key Concepts
SequenceMathematical InductionRatio Analysis
Sequence
In mathematics, a sequence is an ordered list of numbers that usually follows a specific pattern or rule. Each number in the list is called a term. For this exercise, the sequence is defined by a recurrence relation: \( N_{t+1} = N_{t} + 2N_{t-1} \) with initial conditions \( N_0 = 1 \) and \( N_1 = 1 \).
This means each term can be calculated based on the previous two terms. To determine the sequence from \( N_2 \) to \( N_{20} \), you start by using the initial given values and apply the recurrence relation iteratively. The calculated values form a sequence that captures how each number evolves from those that precede it. This sequential approach helps recognize patterns, predict future values, and analyze trends within the set.
This means each term can be calculated based on the previous two terms. To determine the sequence from \( N_2 \) to \( N_{20} \), you start by using the initial given values and apply the recurrence relation iteratively. The calculated values form a sequence that captures how each number evolves from those that precede it. This sequential approach helps recognize patterns, predict future values, and analyze trends within the set.
- The beginning terms \( N_0 \) and \( N_1 \) set the stage for the sequence.
- Each subsequent term relies on the ones before it, showing the interconnectedness inherent in sequences.
- Understanding the rule gives insight into how the sequence behaves over time.
Mathematical Induction
Mathematical induction is a powerful technique used in mathematics to prove statements that are true for all natural numbers. It works on the principle that if something is true for the initial case, and if assuming it true for one case implies its truth for the next, it must be true for all cases.
In the context of sequences defined by recurrence relations, such as \( N_{t+1} = N_{t} + 2N_{t-1} \), induction can be used to prove properties about the entire sequence. Here’s how you might approach it:
In the context of sequences defined by recurrence relations, such as \( N_{t+1} = N_{t} + 2N_{t-1} \), induction can be used to prove properties about the entire sequence. Here’s how you might approach it:
- Base Step: Verify the statement for an initial value, such as \( t = 0 \)
- Inductive Step: Assume the statement is true for \( t = k \) (inductive hypothesis), and prove it for \( t = k+1 \).
- This method confirms the result by showing its consistency over an infinite set of cases, building on the logic link between each one.
Ratio Analysis
Ratio analysis in this context refers to examining the ratio \( \frac{N_t}{N_{t-1}} \) for the sequence defined by our recurrence relation. This ratio helps in understanding how quickly the sequence grows compared to its previous values.
By calculating these ratios, you gain insight into the nature and behavior of the sequence. For our sequence defined as \( N_{t+1} = N_{t} + 2N_{t-1} \), ratios show how one term relates proportionally to the previous term.
By calculating these ratios, you gain insight into the nature and behavior of the sequence. For our sequence defined as \( N_{t+1} = N_{t} + 2N_{t-1} \), ratios show how one term relates proportionally to the previous term.
- Beginning with the initial terms \( N_0 \) and \( N_1 \), calculating \( \frac{N_2}{N_1}, \frac{N_3}{N_2}, \ldots, \frac{N_{20}}{N_{19}} \) can reveal trends or patterns.
- High ratios indicate rapid growth, while lower ratios suggest slower increase.
- Analyzing these ratios provides a clearer picture of how the sequence evolves over time.
Other exercises in this chapter
Problem 55
$$ \text { In Problems } \text { , write } N_{t} \text { as a function of } t \text { for each recursion } $$ $$ N_{t+1}=\frac{1}{3} N_{t} \text { with } N_{0}=
View solution Problem 55
Formal Definition of Limits: \lim _{n \rightarrow \infty} a_{n}=a .\( Find the limit \)a\(, and determine \)N\( so that \)\left|a_{n}-a\right|N\( for the given
View solution Problem 56
$$ \text { In Problems } \text { , write } N_{t} \text { as a function of } t \text { for each recursion } $$ $$ N_{t+1}=\frac{1}{3} N_{t} \text { with } N_{0}=
View solution Problem 56
Formal Definition of Limits: \lim _{n \rightarrow \infty} a_{n}=a .\( Find the limit \)a\(, and determine \)N\( so that \)\left|a_{n}-a\right|N\( for the given
View solution