Proof of a lower bound of the recurrence relation (the CLRS's 4.6-2 exercise)

Question

I am trying to find a solution to the ex. 4.6-2 of the Introduction to Algorithms by Cormen, Leiserson, Rivest, Stein (the third edition). It requires, for recurrence relations $T(n)=aT(n/b)+f(n)$ where $a\geq 1, b > 1$, $n$ is an exact power of $b$ and $f(n)$ is an asymptotically positive function, to prove that if $f(n) = \Theta(n^{\log_ba}lg^{k}n)$, where $k\geq0$, then $T(n)=\Theta(n^{\log_ba}lg^{k+1}n)$.

Since $T(n) = n^{\log_ba} + g(n)$ where $g(n) = \sum_{j=0}^{\log_b n - 1} a^{j}f(n/b^{j})$, I decided to consider the $g(n)$ function at the first. And I have shown $g(n) = O(n^{\log_ba}lg^{k+1}n)$ already (I think so).

But the doing the proof $g(n) = \Omega(n^{\log_ba}lg^{k+1}n)$ became a challenge for me.

Below is my research (with the simplified assumption $k$ is an integer). By condtition, there is such constant $c$ that

$g(n) \geq$

$c \sum_{j=0}^{\log_b n - 1} a^{j}(n/b^{j})^{log_ba}log^{k}(n/b^{j}) =$

$cn^{\log_ba}\sum_{j=0}^{\log_b n - 1} log^{k}(n/b^{j}) =$

$cn^{\log_ba}\sum_{j=0}^{\log_b n - 1}(logn - logb^{j})^{k} =$

$cn^{\log_ba}\sum_{j=0}^{\log_b n - 1}\sum_{i=0}^{k} {k \choose i}log^{k-i}n(-logb^{j})^{i} =$

$cn^{\log_ba}log^{k}n\sum_{j=0}^{\log_b n - 1}\sum_{i=0}^{k} {k \choose i}(-logb^{j}/logn)^{i} =$

$cn^{\log_ba}log^{k}n \biggl(log_bn + \sum_{j=0}^{\log_b n - 1}\sum_{i=1}^{k} {k \choose i}(-logb^{j}/logn)^{i} \biggr) \geq$

$c'n^{\log_ba}log^{k+1}n - cn^{\log_ba}log^{k}n\sum_{j=0}^{\log_b n - 1}\sum_{i=1}^{k} {k \choose i}(logb^{j}/logn)^{i} =$

$A(n) - B(n) = \Theta(n^{\log_ba}lg^{k+1}n) - B(n)$

Actually I am stuck with it. I can not show that $B(n)$ grows slower than $A(n)$. For instance, since $(logb^{j}/logn)^{i} \le 1$ we are able to enhance our $\geq$ condition by the substitution $B(n)$ to some fucntion $B'(n)$ with the sums of binominal coefficients only. But then finally $B'(n)$ has $n^{\log_ba}log^{k+1}n$.

So how to prove $g(n) = \Omega(n^{\log_ba}lg^{k+1}n)$ ?

Answer 1

All we need to prove is the following, where $b>1, k\ge 0$ and $k$ may or may not be an integer. $$\sum_{j=0}^{\lfloor\log_b n\rfloor} \left(\log_b\left(\frac n{b^{j}}\right)\right)^k=\Omega((\log_b n)^{k+1})$$

In order to show the idea of the proof clearly, assume that $n = b^m$. The above estimate is $$\sum_{j=0}^{m} (m-j)^k=\Omega(m^{k+1})$$

Here is the proof.

$$\sum_{j=0}^{m} (m-j)^k=\sum_{i=1}^m i^k\ge\sum_{i=1}^m \int_{i-1}^{i} u^k du=\int_{0}^{m} u^k du = \left.\frac{u^{k+1}}{k+1}\right |^m_0= \frac{m^{k+1}}{k+1} $$