(possible) Alternative solution to a difference equation

In summary, the conversation discusses a difference equation and its solution, which alternates between a(0) for even values of n and 1 - a(0) for odd values of n. The solution also holds true for all values of n and is written as a(n) = a(0)*-1^n + (-1^n - 1)*(-1/2). It is important to use parentheses correctly in mathematical equations. Additionally, initial values are necessary to fully solve a difference equation, just like
  • #1
Nerd-ho
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Hi, I'm a high school graduate who's been going through some UNISA (University of South Africa) mathematical modelling material and I've come across the following difference equation -

a(n+1) = 1 - a(n).

The answer given in the material was that the solution alternates between a(0) for n even and 1 - a(0) for n odd.

I think I've come up with a solution that holds true for all n, namely:

a(n) = a(0)*-1^n + (-1^n - 1)*(-1/2)

Am I correct?

Regards, Travis
 
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  • #2
Nerd-ho said:
Hi, I'm a high school graduate who's been going through some UNISA (University of South Africa) mathematical modelling material and I've come across the following difference equation -

a(n+1) = 1 - a(n).

The answer given in the material was that the solution alternates between a(0) for n even and 1 - a(0) for n odd.

I think I've come up with a solution that holds true for all n, namely:

a(n) = a(0)*-1^n + (-1^n - 1)*(-1/2)

Am I correct?

Regards, Travis

Sure, that's correct. It's exactly the same solution written in a different way.
 
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  • #3
Nerd-ho said:
Hi, I'm a high school graduate who's been going through some UNISA (University of South Africa) mathematical modelling material and I've come across the following difference equation -

a(n+1) = 1 - a(n).

The answer given in the material was that the solution alternates between a(0) for n even and 1 - a(0) for n odd.

I think I've come up with a solution that holds true for all n, namely
:

a(n) = a(0)*-1^n + (-1^n - 1)*(-1/2)

Am I correct?

Regards, Travis

It is vital for you to learn to use parentheses. What you wrote is
$$a(n) = a(0) \times -1^n + (-1^n -1 ) \times (-1/2),$$
which evaluates to ##a(n) = -a(0) + (-2) \times (-1/2) = -a(0) + 1## always. You probably want
$$ a(n) = a(0) \times (-1)^n + ((-1)^n -1) \times (-1/2),$$
which produces ##a(0)## for even ##n## and ##1-a(0)## for odd ##n##.

So, written properly, your formula is a(n) = a(0) * (-1)^n + ((-1)^n - 1) * (-1/2), which, of course, is the same as a(0)*(-1)^n + (1/2)*(1 - (-1)^n)).
 
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  • #4
  • #5
Have we been given the full actual question?

Doesn’t a0, a1, a2, ... = 6.8, -5.8, 6.8, -5.8,... also satisfy your equation?

In other words, for the solution of a difference equation you need initial values, as many as the order of the equation, just like differential equations
 
  • #6
epenguin said:
Have we been given the full actual question?

Doesn’t a0, a1, a2, ... = 6.8, -5.8, 6.8, -5.8,... also satisfy your equation?

In other words, for the solution of a difference equation you need initial values, as many as the order of the equation, just like differential equations

##a(0)## IS the initial value. For your solution ##a(0)=6.8##.
 
  • #7
OK a0 is a general initial value, then the general solution is a0. 1-a0, a0, 1-a0,...
 

1. What is a difference equation?

A difference equation is a mathematical equation that describes the relationship between the current value of a variable and its previous values. It is commonly used in fields such as economics, physics, and engineering to model systems that change over time.

2. Why would an alternative solution to a difference equation be needed?

An alternative solution to a difference equation may be needed if the current solution is not accurate enough or if it is too complex to calculate. It can also be useful for comparing different methods of solving the same equation.

3. What are some possible alternative solutions to a difference equation?

Some possible alternative solutions to a difference equation include numerical methods such as Euler's method, finite difference methods, and numerical integration techniques. Other approaches may involve transforming the equation into a different form or using approximations.

4. How do you determine which alternative solution is the best?

The best alternative solution for a difference equation depends on the specific problem and the desired level of accuracy. It is important to consider factors such as computational efficiency, ease of implementation, and the trade-off between accuracy and complexity.

5. Can alternative solutions to difference equations be used in real-world applications?

Yes, alternative solutions to difference equations can be used in real-world applications. They are often used to model complex systems and can provide valuable insights and predictions for various fields such as economics, physics, and biology.

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