MHB Upper and Lower bound of the recursive relation

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The discussion focuses on determining the asymptotic upper and lower bounds for the recursive relation T(n) = 5T(n/5) + n/ln(n). The master theorem is initially considered for application, but doubts arise regarding its suitability due to the nature of f(n) being asymptotically smaller than n. A substitution method is proposed, transforming the relation using m = log_5(n) to simplify the analysis. The resulting equation leads to a summation involving S(m), which requires further exploration to derive the bounds. The conversation highlights the challenges in applying the master theorem and the potential of substitution for solving the recursive relation.
evinda
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Hello! (Wave)

I have to define an asymptotic upper and lower bound of the recursive relation $T(n)=5 T(\frac{n}{5})+\frac{n}{ \lg n}$.

I thought that I could use the master theorem,since the recursive relation is of the form $T(n)=aT(\frac{n}{b})+f(n)$

$$a=5 \geq 1 , b=5>1 , f(n)=\frac{n}{ \lg n}$$

$$f'(n)=\frac{ \lg n-1}{ \lg^2 n}>0 \Rightarrow \lg n >1 \Rightarrow n>2$$

So, $f(n)$ is asymptotically positive and increasing $\forall n>2$.

$$n^{\log_b a}=n^{\log_5 5}=n$$

We see that $f(n) < n$

$$f(n)=O(n^{ \log_b a- \epsilon})=O(n^{1- \epsilon})$$

But how can we find the $\epsilon$ ? (Thinking) Or can't we apply in this case the master theorem? (Worried)
 
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I think that the master theorem cannot be applied. (Shake)

But..we could do it with substitution,right? (Thinking)

We set $m= \log_5{n} \Rightarrow n=5^m$

Then,we have:

$$T(5^m)=5T(5^{m-1})+\frac{5^m}{m} \\ \Rightarrow \frac{T(5^m)}{5^m}=\frac{T(5^{m-1})}{5^{m-1}}+\frac{1}{m}$$

Let $S(m)=\frac{T(5^m)}{5^m}$

So:

$$S(m)=S(m-1)+\frac{1}{m} \\ S(m-1)=S(m-2)+\frac{1}{m-1} \\ \dots \\ \dots \\ S(2)=S(1)+\frac{1}{2} \\ + --------------- \\ \Rightarrow S(m)=S(1)+\left ( \frac{1}{2}+ \dots + \frac{1}{m}\right )$$

But,how could we continue? (Thinking)
 

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