In lecture, we covered 1st order recurrence relations and came up with a solution by inspection. I sort of see that we're finding the next term in the sequence by multiplying the initial condition by a power of lambda.
Previously we talked about the forward difference, $\Delta = (E-I)$ which looks somewhat similar to $(E- \lambda I )$, although I'm not sure if that is significant.
This also looks similar to finding eigenvalues.
Doing the algebra after substitution I see that:
$(E- \lambda I ) C\lambda^n = 0$
$CE\lambda^n - C\lambda^{n+1} I=0$
$C\lambda^{n+1} - C\lambda^{n+1} =0$
On my own, I don't see how I would come up with the solution.
Edit: After some thought, I see that you can get the next term in the sequence by starting at an initial condition, $u_0$, and multiplying by $\lambda$ an n number of times. At each step in the sequence, you're multiplying by lambda to get to the next step.
For example, if we're at 8 and we want the next term in the sequence $2^k$, we can either
1) multiply by $u_n=8$ by 2
or
2) pick somwhere to start at, $u_0=C=2$, and multiply by $\lambda^n=2^3$
I still wonder if there's a meaning behind the operator $(E- \lambda I )$
Notes from lecture:
A linear 1st order recurrence relation has the form
$u_{n+1} = \lambda u_n$
$Eu = \lambda u$
$(E- \lambda I ) u = 0$
$\lambda$ is a constant E is the left shift operator $C = u_0 $ is the constant initial condition for our "differential equation"
The solution to this recurrence relation is $u = C \lambda^n$