I am learning how to solve the time complexity for the recurrence relation
$$ T(n) = 2T(n - 1) + n^2\text{, where }T(1) = 1 $$
The solution notes that I should begin by considering the following sequence:
$$ \text{Define }T^k(n) = T^{k - 1}(n + 1) - T^{k - 1}(n) \\ \text{and let }T^{(0)}(n) = T(n) $$
How is the sequence $T^k(n)$ relevant to solving my original recurrence relation? This just seems to be complicating the problem by defining an entirely new sequence.
We make Kaya's nice observations slightly more formal. The sequence $T^k$ is the $k$th difference sequence of the original sequence $T$. You can think of it as a discrete $k$th derivative. For example, if $T$ is a $k$th degree polynomial then $T^k$ is constant. We will see below what happens when $T$ is exponential.
For a sequence $T$ (all sequences range over the positive integers), let $\Delta(T)$ be defined by $$ \Delta(T)(n) = T(n+1)-T(n). $$ So $T^k = \Delta^{(k)}(T)$, that is $T^k$ is obtained by applying $k$ times the operator $\Delta$. The operator $\Delta$ has an inverse, which corresponds to indefinite integration: $$ T(n) = T(1) + \sum_{k=1}^{n-1} \Delta(T)(k). $$ If we are only given the sequence $\Delta(T)$ then we don't know $T(1)$, and we can think of it as a constant of integration. For concreteness, define another operator $\Sigma$ given by $$ \Sigma(T)(n) = \sum_{k=1}^{n-1} T(k). $$ The "fundamental theorem of calculus" then states that $$ \Delta(\Sigma(T)) = T $$ and $$ \Sigma(\Delta(T)) = T - T(1). $$ Another useful property of the operators $\Delta$ and $\Sigma$ is their linearity.
Furthermore, we have the nice formula $$\Delta^{(k)}(n^k) = k!,$$ which can be proved by induction on $k$. This formula also implies that $$\Delta^{(k+1)}(n^k) = 0.$$ In fact, this is true for any polynomial of degree at most $k$. Conversely, $\Sigma(n^k)$ is some polynomial of degree $k$.
We can do the same for exponentials: $$ \Delta(c^n) = (c-1)c^n, \quad \Sigma(c^n) = \frac{c^n-1}{c-1}.$$
Finally, it will be useful to use the shift operator $S$ given by $$S(T)(n) = T(n+1).$$ This operator commutes with $\Delta$: $S(\Delta(T)) = \Delta(S(T))$.
Now to the problem at hand. The sequence $T$ satisfies $$S(T) = 2T + (n+1)^2. $$ Taking the third difference, we get $$S(\Delta^{(3)}(T)) = 2\Delta^{(3)}(T).$$ Here we used the linearity of $\Delta$, the commutativity of $\Delta,S$, and the fact that the third derivative of $(n+1)^2$ vanishes. We can conclude that $$ \Delta^{(3)}(T) = C 2^n, \quad C = \Delta^{(3)}(T)(1)/2. $$ We now want to repeatedly apply the operator $\Sigma$. Each time we apply $\Sigma$, we get some constant of integration. After the first time, we get $$ \Delta^{(2)}(T) = C 2^n + C_1 $$ for some $C_1$, using $\Sigma(2^n) = 2^n-1$. While we can express $C_1$ in terms of $T$, it is better not to. The next time we apply $\Sigma$, $C_1$ turns into a linear terms, and so $$ \Delta^{(1)}(T) = C 2^n + C_1 n + C_2 $$ for some $C_2$. The next time we apply $\Sigma$, $C_1 n$ turns into a quadratic term, and so $$ T = C 2^n + P_2 $$ for some quadratic polynomial $P_2$.
It remains to find $C$ and $P_2$. Recall that $C = \Delta^{(3)}(T)(1)/2$. Kaya calculated $\Delta^{(3)}(T)(1) = 12$ and so $C = 6$. Therefore $$ T = 6 \cdot 2^n + P_2. $$ The first few values in $T$ are $1,6,21$, and so $P_2(1) = -11$, $P_2(2) = -18$, $P_2(3) = -27$. Write $P_2 = B_2 n^2 + B_1 n + B_0$, and so $\Delta^{(2)}(P_2) = 2B_2$. On the other hand, $\Delta(-11,-18,-27,\ldots) = (-7,-9,\ldots)$ and so $\Delta^{(2)}(-11,-18,-27,\ldots) = (-2,\ldots)$, and we conclude that $B_2 = -1$. We can now calculate $B_1 n + B_0 = (-10,-14,\ldots)$, and so $B_1 = -4$ and $B_0 = -6$. We can conclude that $P_2 = -n^2 - 4n - 6$, and we get Kaya's formula.
More generally, if $T$ is a sequence satisfying $T(n+1) = c T(n) + P_k(n)$ for some polynomial $P_k$ of degree $k$, then the very same method shows that $$ T(n) = C c^n + P_{k+1}(n) $$ for some constant $C$ and polynomial $P_{k+1}$ of degree $k+1$. In order to find $C$ and $P_{k+1}$, we can either follow the route used above - express $C$ in terms of $c$, $k$ and $\Delta^{(k+1)}(T)(1)$ - or calculate $T(1),\ldots,T(k+2)$ and solve linear equations in the unknowns $C$ and the coefficients of $P_{k+1}$.
Even more generally, we can handle recurrences of the form $T(n) = c_1 T(n-1) + \cdots + c_l T(n-l) + P_k(n)$. As before, after applying $\Delta^{(k+1)}$, the inhomogeneous terms $P_k$ disappears, and we can solve the resulting recurrence relation. Applying $\Sigma^{(k+1)}$, we will get an "error term" of the form $P_{k+1}$ as before.
General recurrence relations have solutions which are sums of expressions of the form $n^r c^n$ for integer $r \geq 0$, so we also need to calculate the (iterated) discrete integrals of these functions; the result of applying $\Sigma$ once is $P_{r+1}(n) c^n$ for some polynomial $P_{r+1}$ of degree $r+1$, and so applying it $k+1$ time we obtain $P_{r+k+1}(n) c^n$ for some polynomial $P_{r+k+1}$ of degree $r+k+1$. As before, we can find the coefficients of all polynomials by solving linear equations, given enough values of the sequence $T$.
First we write out recurrence relationships for the first few $T^{k}$ functions $k\in\{0,1,2,3\}$. $T^{0}$ is given. $$ T^{0}(n)=2T^{0}(n-1)+n^{2} \\ T^{1}(n)=2T^{1}(n-1)+2n+1 \\ T^{2}(n)=2T^{2}(n-1)+2 \\ T^{3}(n)=2T^{3}(n-1) $$
Then we compute initial conditions for $k>0$. These aren't too tricky as they only require computing up $T^{k-1}(2)$ for finding what $T^{k}(1)$ equals. $$ T^{1}(1)=5 \\ T^{2}(1)=10 \\ T^{3}(1)=12 \\ $$
Now the function $T^{3}$ is trivial to express in closed form: $$ T^{3}(n)=3\cdot 2^{n+1} $$
I admit I do not know why this next step is valid formally, however an ansatz would lead me to try working backwards like this: $$ T^{2}\overset{?}=3\cdot 2^{n+1}+c \\ 3\cdot 2^{n+1}=T^{2}(n+1)-T^{2}(n) \\ 3\cdot 2^{n+1}=3\cdot 2^{n+2}+c -(3\cdot 2^{n+1}+c) \\ 3\cdot 2^{n+2}=3\cdot 2^{n+2} $$
Since this seems to confirm that the guess was right, we look to the initial conditions and solve for $c=-2$. For the next step, $T^{1}(n)$, the auxiliary function would be 1 degree higher in $n$ so again we verify that $T^{1}(n)\overset{?}=3\cdot 2^{n+1}-2+(an+b)$ fits the definition of $T^{2}$, and since it does the value of $T^{1}(1)$ requires that $b=-3$ and $T^{1}(2)$ requires that $a=-2$. Thus we find that $T^{1}(n)=3\cdot 2^{n+1}-2n-5$. Finally for $T^{0}$ the guess is something of the form $T^{0}(n)=T^{1}(n)+(dn^{2}+en+f)$. Again, after verification with definitions of $T^{k}$, looking at the first few values of $T$ we discover that $d=-1$, $e=-2$, and $f=-1$. This leads to the conclusion that: $$ T(n)=3\cdot 2^{n+1}-n^{2}-4n-6 \\ $$
Some quick checks to verify correctness: $$ T(1)=3\cdot 2^{2}-1^{2}-4-6=12-1-4-6=1 \\ 2T(n-1)+n^{2}=2(3\cdot 2^{n}-(n-1)^{2}-4(n-1)-6)+n^{2} \\ = 3\cdot 2^{n+1}-2(n-1)^{2}-8(n-1)-12+n^{2} \\ = 3\cdot 2^{n+1}-2n^{2}+4n-2-8n+8-12+n^{2}\\ = 3\cdot 2^{n+1}-n^{2}-4n-6 = T(n) \\ $$
