linear difference eqn-scary exams

Dear Mathematicians,

I have a question for you.thanks.

Lets say that you had to use a linear difference equation to solve some population problem.

How do you determine how many orders it is ?ie.first order,second order,3rd order etc.

eg.so how would I determine (if I am finding for n+1 or not)the long term chance of survival after contracting a deadly virus.

Re: linear difference eqn-scary exams

A linear difference equation can be written in terms of "differences", $\displaystyle \Delta^n a(i)$ ($\displaystyle \Delta a(i)= a(i+1)- a(i)$, $\displaystyle \Delta^2 a(i)= \Delta a(i+1)- \Delta a(i)$, etc.) or in terms of different subscripts: $\displaystyle a(i+ n)$, $\displaystyle a(i+ n- 1)$, etc.. The "order" of a difference equation is, in the first case, the largest power of $\displaystyle \Delta$ and, in the second, the largest "n".

For example $\displaystyle \Delta^3 a(n)- 3\Delta^2 a(n)+ 2\Delta a(n)- a(n)= 0$ is a third order equation because of the $\displaystyle \Delta^3$.

Since $\displaystyle \Delta a(n)= a(n+1)- a(n)$, $\displaystyle \Delta^2(x)= \Delta (a(n+1)- a(n))= a(n+2)- a(n+1)- (a(n+1)- a(n))= a(n+2)- 2a(n+1)+ a(n)$, and $\displaystyle \Delta^3 a(n)= \Delta (a(n+2)- 2a(n+1)+ a(n))= (a(n+3)- 2a(n+2)+ a(n+1))- (a(n+2)- 2a(n+1)+ a(n))= a(n+3)- 3a(n+2)+ 3a(n+1)- a(n)$ that same equation can be written as $\displaystyle a(n+3)- 6a(n+2)+ 11a(n+1)-4a(n)= 0$. That, again, has order 3 because of the "n+3".

Quote:

eg.so how would I determine (if I am finding for n+1 or not)the long term chance of survival after contracting a deadly virus.

I don't understand the "e.g.". Knowing the order does not greatly help in solving the equation. There are a large variety of ways of solving such equations.