A couple of months ago, I asked this question, and received a solid answer which provided a thorough level of detail and analysis on the logic behind the initial proof for $A(n) = f(n)$.
To clarify, his goal is to show that this recurrence:
$$f(1) = \alpha;$$ $$f(2n) = 2f(n) + \beta;$$ $$f(2n + 1) = 2f(n) + \gamma;$$
is equivalent to the following closed form:
$$f(n) = A(n)\alpha + B(n)\beta + C(n)\gamma;$$
Deducing that,
$$A(n) = 2^m;$$ $$B(n) = 2^m - l - 1;$$ $$C(n) = l$$
However, this requires a proof to be certain.
In order to do this, he first proves that $A(n) = 2^m$ through the following recurrence:
$$A(1) = 1;$$ $$A(2n) = 2A(n) + \beta$$ $$A(2n + 1) = 2A(n) + \gamma$$
for $(\alpha, \beta, \gamma) = (1, 0, 0)$
Then, using $(\alpha, \beta, \gamma) = (1, 0, 1)$ he provides:
$$1 = \alpha;$$ $$2n = 2 \times 1 + \beta$$ $$2n + 1 = 2 \times 1 + \gamma$$
Which satisfies $f(n) = n$.
For $f(n) = 1$, he provides:
$$1 = \alpha;$$ $$1 = 2\times1 + \beta$$ $$1 = 2\times1 + \gamma$$
Which gives us $(\alpha, \beta, \gamma) = (1, -1, -1)$
What I'm trying to figure out is the big picture behind all of this. My understanding is that this technique of generalization for $f$ can be used as a framework for solving different kinds of problems using the fact that we can essentially represent any value $n$ as $2^m + l$, and then derive a recurrence of the given form, using arbitrary constants $(\alpha, \beta, \gamma)$ for some arbitrary purpose.
What I don't understand is... how exactly is it that these proofs actually show this is true? I understand, given the question I provided and its corresponding answer, that they're designed to fill gaps in areas where representing $f(n)$ through some form of these constants requires that $f(n)$ be defined for all n, and that this is a means of providing such proof.
How is it that these statements show this fact when they require that $(\alpha, \beta, \gamma)$ be defined to specific values for each case?
