(possible) Alternative solution to a difference equation

[Nerd-ho]

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

 

[Dick]

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.

 

[Ray Vickson]

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)).

 

[Nerd-ho]

Thanks @Dick and @Ray Vickson

@Ray Vickson Apologies, I used spacing to perform that function. I'll use parenthesis in future

 

[epenguin]

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

 

[Dick]

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 $a(0)=6.8$.

 

[epenguin]

OK a₀ is a general initial value, then the general solution is a₀. 1-a₀, a₀, 1-a₀,...