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Show that \(A_{t}=0\) for all \(t\) is a solution to Equation 1.18 $$ A_{t+1}-A_{t}=2 \pi k \sqrt{A_{t}} $$ This means that for every \(t,\) if \(A_{t}=0\) and \(A_{t+1}\) is computed from the equation, then \(A_{t+1}=0\) (and, yes, this is simple).

Short Answer

Expert verified
If \( A_{t} = 0 \), \( A_{t+1} = 0 \) also holds; hence, it's a solution.

Step by step solution

01

Understand the Given Relation

The given equation is \( A_{t+1} - A_{t} = 2 \pi k \sqrt{A_{t}} \). We need to prove that \( A_{t} = 0 \) for all \( t \) is a solution.
02

Substitute A_t with 0

Let's substitute \( A_{t} = 0 \) into the equation: \( A_{t+1} - 0 = 2 \pi k \sqrt{0} \).
03

Simplify the Equation

Since the square root of zero is zero, it simplifies to \( A_{t+1} = 0 \).
04

Conclude the Verification

The equation \( A_{t+1} = 0 \) confirms that if \( A_{t} = 0 \), then \( A_{t+1} = 0 \) as well. Therefore, \( A_{t} = 0 \) for all \( t \) is indeed a solution to the equation.

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Key Concepts

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

Difference Equations
Difference equations are mathematical expressions that relate the difference between successive terms in a sequence. They are often used to model complex systems where changes occur in discrete steps over time. For example, in life sciences, difference equations can help describe how populations grow or shrink from one generation to the next.
One key aspect of difference equations is that they express change in terms of earlier values. This allows us to understand how processes evolve over time. For instance, the given equation \( A_{t+1} - A_{t} = 2 \pi k \sqrt{A_t} \) is a simple difference equation expressing that the change from \(A_t\) to \(A_{t+1}\) is related to the square root of \(A_t\).
  • Useful for modeling discrete-time processes.
  • Illustrative in understanding progression using previous states.
  • Frequently used in various scientific fields, especially biology and ecology.
Solutions of Equations
In solving equations, the goal is to find values for variables that satisfy the given expressions. A solution of an equation means that when the solution is substituted into the equation, the terms balance or hold true. In our specific case, we're asked to show that \( A_t = 0 \) for all \( t \) is a solution.
When solving the difference equation, substituting \( A_t = 0 \) results in both sides of the equation balancing, as shown in the solution steps. This verification fulfills the condition of being a valid solution. For different equations, finding solutions often requires exploring the equation's structure, experimenting with different values, or using techniques like factorization and substitution.
  • Identify the variables that need values.
  • Substitute solutions to verify if they balance the equation.
  • Examine the structure for further insights into possible solutions.
Mathematical Modeling
Mathematical modeling involves using mathematical expressions to represent and study real-world phenomena. It helps scientists and researchers predict behaviors and test hypotheses. In life sciences, mathematical models, like our difference equation, are tools for simulating biological processes or predicting changes in ecosystems.
Models translate observations into a format that can be manipulated and analyzed. For example, in this exercise, the equation models a process where the next state \( A_{t+1} \) depends on the current state \( A_t \). This is particularly impactful in biology where growth, decay, and population dynamics can be modeled with such equations.
  • Convert complex systems into understandable forms.
  • Enable prediction and testing of various scenarios.
  • Bridge the gap between theoretical and applied sciences.
Recurrence Relations
Recurrence relations express sequences where the next terms are calculated from preceding ones. They form the backbone of difference equations and play a crucial role in mathematical modeling. Recurrence relations guide us in understanding how sequences develop over time, which is invaluable for subjects like computer science, physics, and biology.
In our case, \( A_{t+1} = A_t + 2 \pi k \sqrt{A_t} \) is a recurrence relation. It shows how each term \( A_{t+1} \) is derived from its predecessor \( A_t \), providing insight into how changes accrue over each iteration.
  • Define infinite sequences with finite information.
  • Useful for iterative calculations and simulations.
  • Form the backbone of algorithms and numerical methods.

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Most popular questions from this chapter

A one-liter flask contains one liter of distilled water and \(2 \mathrm{~g}\) of salt. Repeatedly, 50 \(\mathrm{ml}\) of solution are removed from the flask and discarded after which \(50 \mathrm{ml}\) of distilled water are added to the flask. Introduce notation and write a dynamic equation that will describe the change of salt in the beaker each cycle of removal and replacement. How much salt is in the beaker after 20 cycles of removal?

Two kilos of a fish poison, rotenone, are mixed into a lake which has a volume of \(100 \times 20 \times 2=4000\) cubic meters. No water flows into or out of the lake. Fifteen percent of the rotenone decomposes each day. a. Write a mathematical model that describes the daily change in the amount of rotenone in the lake. b. Let \(R_{0}, P_{1}, R_{2}, \cdots\) denote the amounts of rotenone in the lake, \(P_{t}\) being the amount of poison in the lake at the beginning of the \(t \underline{h}\) day after the rotenone is administered. Write a dynamic equation representative of the mathematical model. c. What is \(R_{0} ?\) Compute \(R_{1}\) from your dynamic equation. Compute \(R_{2}\) from your dynamic equation. d. Find a solution equation for your dynamic equation.

Determine the doubling times or half-lives of the following exponential equations. (a) \(y=0.5^{t}\) (b) \(y=2^{3 t}\) (c) \(y=0.1^{0.1 t}\) (d) \(y=1000.8^{t}\) (e) \(y=45^{3 t}\) (f) \(y=0.00015^{0.1 t}\) (g) \(y=100.8^{2 t}\) (h) \(y=0.01^{3 t}\) (i) \(y=0.01^{0.1 t}\)

Find a formula for a population that grows exponentially and a. Has an initial population of 50 and a doubling time of 10 years. b. Has an initial population of 1000 and a doubling time of 50 years. c. Has in initial population of 1000 and a doubling time of 100 years.

Equation 1.30 $$ P_{t+1}-P_{t}=b $$ represents a large number of equations $$ \begin{array}{r} P_{1}-P_{0}=b \\ P_{2}-P_{1}=b \\ P_{3}-P_{2}=b \\ \vdots \quad \vdots \quad \vdots \quad \vdots \\ P_{n-1}-P_{n-2}=b \\ P_{n}-P_{n-1}=b \end{array} $$ Add these equations to obtain $$ P_{n}=P_{0}+n b $$ Substitute \(t\) for \(n\) to obtain $$ P_{t}=P_{0}+t b $$

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