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I'm trying to solve this recurrence relation using the iterative method and i keep getting the different answer from using the master theorem.

$$\begin{aligned} T(n) &= 5T(n/2) +n^2 \\ &= 5^2 T(n/4) + (5/4)n^2 + n^2 \\ &= 5^3 T(n/8) + (5/4)^2 n^2 + (5/4)n^2 + n^2 \\ &= 5^k T(n/(2^k)) + n^2 \sum_{i=0}^{k-1}(5/4)^i \end{aligned}$$

$n = 2^k => \ln(n) = k$

$$\begin{aligned} &=5^{\ln(n)} T(1) + n^2 * \Theta(5^{\ln(n)}) \\ &= n^{\ln(5)} + n^2 * \Theta(n^{\ln(5)}) \end{aligned}$$

$O(n^2 * n^{\ln(5)}) = O(n^{2+\ln(5)})$

Right?

According to the master Theorem,

$a= 5$; $b=2$; $p=0$; $k=2$; $5>4$

This result in $\Theta(n^{\ln(5)})$ for this problem,

So, do I ignore the $n^2$ in the iterative method? Did I do it incorrectly? If it correct, then why is there a difference between master Theorem and iterative method? Master Theorem have a tighter constraint?

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Check your maths. The sum with (5/4)^i is much smaller than theta(5^ln n).

Intuitively, T(n) = 5 T(n/2) gives you growth of n^(ln 5)= n^2 * n (ln 1.25). Adding another O (n^2) shouldn't make a difference.

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You have been calculating correctly all along, except when it comes to $\sum_{i=0}^{k-1}(\frac54)^i$, where $n=2^k$.

Note that $\sum_{i=0}^{k-1}(\frac54)^i = \frac{1-(\frac54)^k}{1-\frac54} = 4((\frac54)^k - 1) = 4\cdot2^{\ln(\frac54)k} - 4 = 4n^{\ln(\frac54)} - 4$, where $\ln(x)$ means $\log_2(x)$.

So you would have

$$\begin{aligned} T(n) &= 5^k T(n/(2^k)) + 4n^2n^{\ln(\frac54)} - 4n^2 \\ &=n^{\ln(5)} T(1) + 4n^{\ln(5)} - 4n^2 \\ &=(T(1) + 4)n^{\ln(5)} - 4n^2 \\ &= O(n^{\ln(5)}) \end{aligned}$$

The result above obtained by iterative method is, as we should have been expecting, indeed consistent with the result obtained by the master Theorem.

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