# linear difference eqn-scary exams

• Jun 11th 2013, 08:05 AM
n22
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.
• Jun 11th 2013, 10:29 AM
HallsofIvy
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.