Recurring Relationship – Why Can We Ignore the Constant Factor in Weis's Proof of the Master Theorem?

Suppose your function is always within range $ cN ^ k $ to $ CN ^ k $, Consider the three repetitions
$$
s (N) = such as (N / b) + cN ^ k, S (N) = aS (N / b) + CN ^ k, R (N) = aR (N / b) + N ^ k.
$$

For suitable initial conditions $ s (N) leq T (N) leq S (N) $ will hold, as a clear proof by induction would show. On the other hand, for suitable initial conditions, $ s (N) = cR (N) $ and $ S (N) = CR (N) $as another simple induction shows. This explains Weis's remark.

What's going on here? As you complete the recursion, add different terms that are different $ Theta (N ^ k) $for different values ​​of $ N $, The contribution of the big theta to the final sum is at most a constant factor. Instead of $ a ^ m sum_ {i = 0} ^ m ( frac {b ^ k} {a}) ^ i $, you get $ Theta bigl (a ^ m sum_ {i = 0} ^ m ( frac {b ^ k} {a}) ^ i bigr) $, The decisive factor is the additional constant factor global and does not accumulate during recursion.

Consider generic repetition as an example $ s (N) = 2s (N-1) + f (N) $with basic housing $ s (0) = 0 $, The solution is $$ s (N) = f (N) + 2f (N-1) + 4f (N-2) + cdots + 2 ^ {N-1} f (1). $$ When we multiply $ f (N) $ With a constant factor, the solution of this repetition is multiplied by the same constant factor, while the constant $ 2 $ remains the same.