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Chapter 9: Problem 1

Starting with the given initial conditions, write down the first four terms of each of the following sequences. By expressing these as an appropriate power, write down a formula for the general term, \(Y_{t}\) in terms of \(t\). (1) (a) \(Y_{t}=3 Y_{t-1} ; Y_{0}=1\) (b) \(Y_{t}=3 Y_{t-1} ; Y_{0}=7\) (c) \(Y_{t}=3 Y_{t-1} ; Y_{0}=A\) (2) (a) \(Y_{t}=\frac{1}{2} Y_{t-1} ; Y_{0}=1\) (b) \(Y_{t}=\frac{1}{2} Y_{t-1} ; Y_{0}=7\). (c) \(Y_{t}=\frac{1}{2} Y_{t-1} ; Y_{0}=A\) (3) \(Y_{t}=b Y_{t-1} ; Y_{0}=A\)

Short Answer

Expert verified
The general formulas for the sequences are: \(Y_t = 3^t\), \(Y_t = 7 \times 3^t\), \(Y_t = A \times 3^t\), \(Y_t = \frac{1}{2^t}\), \(Y_t = 7 \times \frac{1}{2^t}\), \(Y_t = A \times \frac{1}{2^t}\), \(Y_t = A \times b^t\).

Step by step solution

01

Identify and Write Down First Four Terms

To find the first four terms of each sequence, you need to use the given initial condition and recurrence relation. Start with the initial value and apply the recurrence relation to find subsequent terms.
02

Part 1(a): First Four Terms

Given: \(Y_{0}=1\) and \(Y_{t}=3 Y_{t-1}\), the sequence is:\[Y_0 = 1\]\[Y_1 = 3Y_0 = 3 \times 1 = 3\]\[Y_2 = 3Y_1 = 3 \times 3 = 9\]\[Y_3 = 3Y_2 = 3 \times 9 = 27\]
03

General Formula for 1(a)

Observing the pattern, the general term formula is \(Y_t = 3^t\).
04

Part 1(b): First Four Terms

Given: \(Y_{0}=7\) and \(Y_{t}=3 Y_{t-1}\), the sequence is:\[Y_0 = 7\]\[Y_1 = 3Y_0 = 3 \times 7 = 21\]\[Y_2 = 3Y_1 = 3 \times 21 = 63\]\[Y_3 = 3Y_2 = 3 \times 63 = 189\]
05

General Formula for 1(b)

Observing the pattern, the general term formula is \(Y_t = 7 \times 3^t\).
06

Part 1(c): First Four Terms

Given: \(Y_{0}=A\) and \(Y_{t}=3 Y_{t-1}\), the sequence is:\[Y_0 = A\]\[Y_1 = 3Y_0 = 3 \times A = 3A\]\[Y_2 = 3Y_1 = 3 \times 3A = 9A\]\[Y_3 = 3Y_2 = 3 \times 9A = 27A\]
07

General Formula for 1(c)

Observing the pattern, the general term formula is \(Y_t = A \times 3^t\).
08

Part 2(a): First Four Terms

Given: \(Y_{0}=1\) and \(Y_{t}=\frac{1}{2} Y_{t-1}\), the sequence is:\[Y_0 = 1\]\[Y_1 = \frac{1}{2}Y_0 = \frac{1}{2} \times 1 = \frac{1}{2}\]\[Y_2 = \frac{1}{2}Y_1 = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4}\]\[Y_3 = \frac{1}{2}Y_2 = \frac{1}{2} \times \frac{1}{4} = \frac{1}{8}\]
09

General Formula for 2(a)

Observing the pattern, the general term formula is \(Y_t = \frac{1}{2^t}\).
10

Part 2(b): First Four Terms

Given: \(Y_{0}=7\) and \(Y_{t}=\frac{1}{2} Y_{t-1}\), the sequence is:\[Y_0 = 7\]\[Y_1 = \frac{1}{2}Y_0 = \frac{1}{2} \times 7 = \frac{7}{2}\]\[Y_2 = \frac{1}{2}Y_1 = \frac{1}{2} \times \frac{7}{2} = \frac{7}{4}\]\[Y_3 = \frac{1}{2}Y_2 = \frac{1}{2} \times \frac{7}{4} = \frac{7}{8}\]
11

General Formula for 2(b)

Observing the pattern, the general term formula is \(Y_t = 7 \times \frac{1}{2^t}\).
12

Part 2(c): First Four Terms

Given: \(Y_{0}=A\) and \(Y_{t}=\frac{1}{2} Y_{t-1}\), the sequence is:\[Y_0 = A\]\[Y_1 = \frac{1}{2}Y_0 = \frac{1}{2} \times A = \frac{A}{2}\]\[Y_2 = \frac{1}{2}Y_1 = \frac{1}{2} \times \frac{A}{2} = \frac{A}{4}\]\[Y_3 = \frac{1}{2}Y_2 = \frac{1}{2} \times \frac{A}{4} = \frac{A}{8}\]
13

General Formula for 2(c)

Observing the pattern, the general term formula is \(Y_t = A \times \frac{1}{2^t}\).
14

Part 3: First Four Terms

Given: \(Y_{0}=A\) and \(Y_{t}=b Y_{t-1}\), the sequence is:\[Y_0 = A\]\[Y_1 = bY_0 = b \times A = bA\]\[Y_2 = bY_1 = b \times bA = b^2A\]\[Y_3 = bY_2 = b \times b^2A = b^3A\]
15

General Formula for Part 3

Observing the pattern, the general term formula is \(Y_t = A \times b^t\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Recurrence Relations
A recurrence relation is a formula that connects each term in a sequence with previous terms. This relationship helps to generate the sequence step-by-step, starting from the initial condition. For example, in the exercise, we see the recurrence relation in the form of:
  • Part 1(a): \(Y_t = 3Y_{t-1}\)
  • Part 2(a): \(Y_t = \frac{1}{2}Y_{t-1}\)
These recurrence relations indicate how each term is calculated based on the previous one. Applying these steps repeatedly, starting from the initial term, allows us to find the entire sequence.
General Term Formula
The general term formula gives a direct way to calculate any term in the sequence without needing to know all previous terms. It provides a closed form expression for the terms. For instance, after observing the patterns in the sequences, we derived these general term formulas:
  • Part 1(a): \(Y_t = 3^t\)
  • Part 2(a): \(Y_t = \frac{1}{2^t}\)
These formulas help to understand the behavior and properties of the sequence as a whole. By using the general term formula, we can jump directly to any term in the sequence without repeatedly applying the recurrence relation.
Geometric Progression
A geometric progression is a sequence of numbers where each term is found by multiplying the previous one by a fixed, non-zero number called the common ratio. Recognizing a geometric progression allows us to apply known formulas and techniques. In the exercise, we see geometric progressions such as:
  • Part 1(a): \(1, 3, 9, 27, ...\)
  • Part 2(a): \(1, \frac{1}{2}, \frac{1}{4}, \frac{1}{8}, ...\)
These sequences grow or decay exponentially based on the common ratio. This progression forms the basis for the recurrence relations and general term formulas identified earlier.
Initial Conditions
Initial conditions specify the starting point(s) of a sequence, often denoted as the first term or terms. When given an initial condition, such as:
  • Part 1(a): \(Y_0 = 1\)
  • Part 2(a): \(Y_0 = 7\)
These initial terms are combined with recurrence relations to determine the first few terms of the sequence. Initial conditions are crucial because they provide the starting point from which the rest of the sequence is developed. Without these, the sequence cannot be generated accurately.

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Most popular questions from this chapter

Use integration to solve each of the following differential equations subject to the given initial conditions. (a) \(\frac{\mathrm{d} y}{\mathrm{~d} t}=2 t ; y(0)=7\) (b) \(\frac{\mathrm{d} y}{\mathrm{~d} t}=\mathrm{e}^{-3 t} ; y(0)=0\) (c) \(\frac{\mathrm{d} y}{\mathrm{~d} t}=t^{2}+3 t-5 ; y(0)=1\)

(Maple) Solve the differential equation $$ \frac{\mathrm{d} y}{\mathrm{~d} t}=\frac{t y}{t^{2}+1} $$ with initial condition \(y(0)=1\). Plot a graph of this solution on the range \(0 \leq t \leq 5\). Hence, or otherwise, write down a simple expression for the approximate solution when \(t\) is large.

Consider the market model $$ \begin{aligned} &Q_{\mathrm{s}}=2 P-2 \\ &Q_{\mathrm{D}}=-P+4 \\ &\frac{\mathrm{d} P}{\mathrm{~d} t}=\frac{1}{3}\left(Q_{\mathrm{D}}-Q_{\mathrm{S}}\right) \end{aligned} $$ Find expressions for \(P(t), Q_{s}(t)\) and \(Q_{D}(t)\) when \(P(0)=1\). Is this system stable or unstable?

Consider the supply and demand equations $$ \begin{aligned} &Q_{S t}=0.4 P_{t-1}-12 \\ &Q_{D t}=-0.8 P_{t}+60 \end{aligned} $$ Assuming that equilibrium conditions prevail, find an expression for \(P_{t}\) when \(P_{0}=70\). Is the system stable or unstable?

(Excel) An economic growth model is based on three assumptions: (1) Aggregate output, \(y_{t}\), in time period \(t\) depends on capital stock, \(k_{t}\), according to $$ y_{t}=k_{t}^{0.6} $$ (2) Capital stock in time period \(t+1\) is given by $$ k_{t+1}=0.99 k_{t}+s_{t} $$ where the first term reflects the fact that capital stock has depreciated by \(1 \%\), and the second term denotes the output that is saved during period \(t\). (3) Savings during period \(t\) are one-fifth of income so that $$ s_{t}=0.2 y_{t} $$ (a) Use these assumptions to write down a difference equation for \(k_{t}\). (b) Given that \(k_{0}=7000\), find the equilibrium level of capital stock and state whether \(k_{t}\) displays uniform or oscillatory convergence. Do you get the same behaviour for other initial values of capital stock?
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