Linear First Order Difference Equations (Iterative/General Method)

In summary, to solve this problem, you need to use the iterative method and rearrange the equations to get the general solution.
  • #1
rabbitchannel
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Homework Statement


I am almost done with a chapter all about this topic and this type of question is the only one I can't get. This is linear first order difference equations. The question is:

Given the unemployment Ut equation:

Ut = [tex]\alpha[/tex] + [tex]\beta[/tex] Ut-1
[tex]\alpha[/tex], [tex]\beta[/tex] > 0

b. Suppose that there are occasional shocks to the demand for labor causing shifts in Ut. The modified equation for Ut becomes:

Ut = [tex]\alpha[/tex] + [tex]\beta[/tex] Ut-1 + et

where et varies over time. Show that the solution to the modified equation is:

Ut = [tex]\beta[/tex]tU0 + [tex]\frac{\alpha(1-\betat)}{1-\beta}[/tex] + e1[tex]\beta[/tex]t-1 + e2[tex]\beta[/tex]t-2 + ... + et-1[tex]\beta[/tex] + et

Don't know how to fix that there. It should be (1-[tex]\beta[/tex]t)

Homework Equations



General Method:
Pc + Pp = General method

Yt = (Y0 - [tex]\frac{c}{1+a}[/tex])(-a)t + [tex]\frac{c}{1+a}[/tex]

I've also got the derived formula for supply and demand but that requires two functions.

The Attempt at a Solution


Ok, I can't get the iteration. This is what I've tried:

Ut = [tex]\alpha[/tex] + [tex]\beta[/tex]Ut-1 + et

Ut+1 = [tex]\alpha[/tex] + [tex]\beta[/tex]Ut + et+1

After this point I don't know what to do. I tried to do this:

Ut+1 = [tex]\beta[/tex]([tex]\alpha[/tex] + [tex]\beta[/tex]Ut + et+1) + [tex]\alpha[/tex] + et+1

Basically multiplying the whole equation by [tex]\beta[/tex] then adding: [tex]\alpha[/tex] + et+1. Once I do it for 3 periods I can determine the general function but it is different from the given one. I lack the 1-[tex]\beta[/tex] on that denominator. I can solve other equations but have trouble when something else, such as et+1 is added. I've also used the general method but it also turns out different. I was under the impression that I can use the iterative and general solutions for any first order linear difference equation. Am I wrong?

Any help would be greatly appreciated. Thanks!

P.S. I would like to thank the system for auto-logging me out while trying to preview my first ever post, thereby deleting a chunk of what I wrote. Good thing I saved. :\
 
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  • #2
The solution to this problem is actually relatively simple. Start with the given equation:Ut = \alpha + \beta Ut-1 + etThen, subtract Ut-1 from both sides to get:Ut - Ut-1 = \alpha + \beta Ut-1 - \beta Ut-1 + etSimplifying, we have:Ut - Ut-1 = \alpha + etNow, let's rearrange and solve for Ut:Ut = Ut-1 + \alpha + etNow, let's use the same technique to get the next period's equation:Ut+1 = Ut + \alpha + et+1We can continue this pattern of rearranging and substituting to get the general equation:Ut = \betatU0 + \frac{\alpha(1-\betat)}{1-\beta} + e1\betat-1 + e2\betat-2 + ... + et-1\beta + etThis is the solution to the modified equation.
 

FAQ: Linear First Order Difference Equations (Iterative/General Method)

1. What is a linear first order difference equation?

A linear first order difference equation is a mathematical equation that describes the relationship between the current value of a variable and its previous value. It can be written in the form of xn+1 = axn + b, where a and b are constants and xn represents the value of the variable at time n.

2. What is the iterative method for solving linear first order difference equations?

The iterative method for solving linear first order difference equations involves using the previous value of the variable to calculate the next value. This process is repeated until the desired number of values is obtained. It is an approximate method and can be time-consuming for large numbers of iterations.

3. What is the general method for solving linear first order difference equations?

The general method for solving linear first order difference equations involves finding the general solution to the equation by solving for the constants a and b. This involves using algebraic techniques such as substitution and elimination. Once the general solution is found, specific solutions can be obtained by substituting in initial conditions.

4. What is the difference between a linear first order difference equation and a nonlinear first order difference equation?

A linear first order difference equation has a constant coefficient, while a nonlinear first order difference equation has a variable coefficient. This means that the relationship between the current value and the previous value of the variable is not constant, making it more difficult to solve.

5. What are some real-world applications of linear first order difference equations?

Linear first order difference equations can be used to model population growth, interest rates, and chemical reactions. They are also commonly used in economics and finance to model changes in prices and stock market trends. Additionally, they can be used in engineering to model the behavior of electrical circuits and mechanical systems.

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