We have the recurrence relation:

$T(2^n) = T(2^{n-1}) + 2^n$

Step 1: Change Variables

Let $N = 2^n$, so that $n = \log_2 N$. Rewriting the recurrence in terms of $N$:

$T(N) = T(N/2) + N$

This is now in the standard recurrence form:

$T(N) = T(N/b) + f(N)$

where:

• $b = 2$,
• $f(N) = N$.

Step 2: Apply the Master Theorem

The Master Theorem applies to recurrences of the form:

$T(N) = aT(N/b) + f(N)$

where:

• $a = 1$,
• $b = 2$,
• $f(N) = N$.

Now, we compare $f(N) = N$ with $N^{\log_b a} = N^{\log_2 1} = N^0 = 1$.

Since $f(N) = N$ grows asymptotically faster than $N^0 = 1$, we use Case 3 of the Master Theorem, which states that if $f(N) = \Omega(N^c)$ for some $c > \log_b a$, and if the regularity condition holds ($a f(N/b) \leq c f(N)$ for some $c < 1$ and sufficiently large $N$), then:

$T(N) = \Theta(f(N))$

Since $N = \Omega(N^0)$ (where $c = 1 > 0$), and the regularity condition holds, we conclude:

$T(N) = \Theta(N)$

Conclusion

Since we initially substituted $N = 2^n$, we get:

$T(2^n) = \Theta(2^n)$

Would you like a more detailed breakdown of any step?

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Related Questions:

1. How does the Master Theorem apply to different forms of recurrence relations?
2. What are the conditions for using each of the three Master Theorem cases?
3. How do we verify the regularity condition in Case 3 of the Master Theorem?
4. How does changing variables help in solving recurrence relations?
5. What are some real-world applications of recurrence relations in algorithms?

Tip: The Master Theorem is great for divide-and-conquer recurrences, but it doesn’t work for all cases (e.g., when subproblems have varying sizes).