Update 4.6.md

New solution giving a simple and exact expression for n_j and a proof of its correctness.

docs/Chap04/4.6.md

@@ -2,8 +2,18 @@
 
 > Give a simple and exact expression for $n_j$ in equation $\text{(4.27)}$ for the case in which $b$ is a positive integer instead of an arbitrary real number.
 
+We state that $\forall{j \ge 0}, n_j = \left \lceil \frac{n}{b^j} \right \rceil$.  
+
+Indeed, for $j=0$ we have from the recurrence's base case that $n_0 = n = \left \lceil \frac{n}{b^0} \right \rceil$.  
+Now, suppose $n_{j - 1} = \left \lceil \frac{n}{b^{j - 1}}  \right \rceil$ for some $j > 0$. By definition, $n_j = \left \lceil \frac{n_{j - 1}}{b} \right \rceil$.  
+It follows from the induction hypothesis that $n_j = \left \lceil \frac{\left \lceil \frac{n}{b^{j - 1}} \right \rceil}{b} \right \rceil$.  
+Since $b$ is a positive integer, equation $(3.4)$ implies that $\left \lceil \frac{\left \lceil \frac{n}{b^{j - 1}} \right \rceil}{b} \right \rceil = \left \lceil \frac{n}{b^j} \right \rceil$.  
+Therefore, $n_j = \left \lceil \frac{n}{b^j} \right \rceil$.
+
+P.S.  
 $n_j$ is obtained by shifting the base $b$ representation $j$ positions to the right, and adding $1$ if any of the $j$ least significant positions are non-zero.
 
+
 ## 4.6-2 $\star$
 
 > Show that if $f(n) = \Theta(n^{\log_b a}\lg^k{n})$, where $k \ge 0$, then the master recurrence has solution $T(n) = \Theta(n^{\log_b a}\lg^{k + 1}n)$. For simplicity, confine your analysis to exact powers of $b$.