# 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.