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Often in CLRS, when proving recurrences via substitution, $\Theta(f(n))$ is replaced with $cf(n)$.

For example, on page 91, the recurrence

$$ T(n) = 3T(⌊n/4⌋) + \Theta(n^2) $$

is written like so in the proof

$$ T(n) \le 3T(⌊n/4⌋) + cn^2. $$

But can't $\Theta(n^2)$ stand for, let's say, $cn^2 + n$? Would that not make such a proof invalid?

Further in the proof, the statement

\begin{align} T(n) &\le (3/16)dn^2 + cn^2 \\ &\le dn^2 \end{align}

is reached. But if $cn^2 + n$ was used instead of $cn^2$, that step would instead be the following

$$ T(n) \le (3/16)dn^2 + cn^2 + n $$

Can it still be proven that $T(n) \le dn^2$ if this is so? Do such lower order terms not matter in proving recurrences via substitution?

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By definition, a function $f(n)$ is $\Theta(n^2)$ if there exist constant $b,c,N$ such that for all $n \geq N$, $$ bn^2 \leq f(n) \leq cn^2.$$ In particular, as long as $n \geq N$, we can always replace $f(n)$ by $cn^2$ when upper-bounding an expression.

In your example of $cn^2 + n$, when $n \geq 1$ we have $n \leq n^2$ and so $cn^2 \leq (c+1)n^2$.

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Let me also suggest the following vision of the situation: $\Theta(f)$ is set of functions, which satisfy well know definition. So, when you wrote $T(n) = 3T(⌊n/4⌋) + \Theta(n^2)$, then this means $T(n) \in 3T(⌊n/4⌋) + \Theta(n^2)$, where right side is again set $g+\Theta(f) = \{g+\phi \colon \phi \in \Theta(f) \}$.

So, $T \in g+\Theta(f)$ mean that exists element of $g+\Theta(f)$ which is equal to $T$, but we cannot say, that $T$ equals to any member of $g+\Theta(f)$.

Same is if we write $$T(n)- 3T(⌊n/4⌋) \in \Theta(n^2)$$ Now, knowing the definition of belonging to a set, we just use it and write, that $\exists c_1, c_2 \gt 0, N\in \mathbb{N}$, such that for $n \gt N$ $$c_1 n^2 \leqslant T(n)- 3T(⌊n/4⌋) \leqslant c_1 n^2$$ saying exactly we do not "replace" $T(n)- 3T(⌊n/4⌋)$ in equality, but we "estimate" it in inequality.

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