I'm solving non-homogeneous linear recurrences for my combinatorics class and my teacher skipped a bunch of steps in his notes, so I am trying to make sense of a particular "step" he took.
We were given based operations on sequences where $a_n$ is a sequence $a_0,a_1,a_2,\cdots $ and $E$ is the operator that shifts a sequence, e.g. $E(a_n)=a_{n+1}$:$$c(a_n)=(c\cdot a_n)$$$$(A+B)(a_n)=A(a_n)+B(a_n)$$$$(AB)(a_n)=A(B(A-n))$$So, for example:$$(E+2)(a_n)=(a_{n+1})+(2a_n)$$$$E^3(a_n)=E^2(a_{n+1})=E(a_{n+2})=a_{n+3}$$So I am given the non-homogeneous recurrence relation that I need to solve:$$F_n=2a_{n-1}+1$$The first step is to express the recurrence of the homogeneous part of the recurrence using sequence operators, but immediately there is where he loses me. In my notes it says $(E-2)(a_n)$, but I do not understand where he gets this from, because according to the sequence operations, he is claiming:$$2a_{n-1}\rightarrow (E-2)(a_n)=E(a_n)-2(a_n)=a_{n+1}-2a_n$$And unless I am doing my math incorrectly,$$2a_{n-1}\neq a_{n+1}-2a_n$$So where exactly does he get $(E-2)(a_n)$ from?
My mistake was in not understanding that he switched the inital relation from $$a_n=2a_{n-1}+1$$To:$$a_n-2a_{n-1}=1$$So he is not claiming that $2a_{n-1}=(E-2)(a_n)$ but instead claiming that$$a_n-2a_{n-1}\rightarrow (E-2)(a_n)=E(a_n)-2(a_n)=a_{n+1}-2a_n$$Which is true.