1. **State the problem:** We want to solve the recurrence relation $$T(n) = T\left(\frac{n}{2}\right) + T\left(n^{\frac{1}{2}}\right) + n$$ for large $n$. 2. **Understand the recurrence:** This recurrence expresses $T(n)$ in terms of two smaller subproblems: one of size $\frac{n}{2}$ and another of size $n^{1/2}$, plus a linear cost $n$. 3. **Approach:** The recurrence is non-standard because of the two different recursive calls with different sizes. We analyze the dominant term by considering the sizes of the subproblems. 4. **Intuition:** The term $T(n/2)$ reduces the problem size by half each time, while $T(n^{1/2})$ reduces it to the square root. The $T(n/2)$ term will dominate the recursion depth because it shrinks slower than $n^{1/2}$. 5. **Estimate the recursion depth for $T(n/2)$:** The depth is about $\log_2 n$ because each step halves $n$. 6. **Estimate the recursion depth for $T(n^{1/2})$:** The depth is about $\log_2(\log_2 n)$ because repeatedly taking square roots reduces $n$ very quickly. 7. **Dominant cost:** The $T(n/2)$ branch dominates the recursion tree height, so the total cost is roughly the sum over $\log n$ levels of $n$ work at each level. 8. **Summation:** The total work is approximately $$\sum_{k=0}^{\log_2 n} \frac{n}{2^k} = n \sum_{k=0}^{\log_2 n} \frac{1}{2^k} \approx 2n$$ because the geometric series sums to about 2. 9. **Conclusion:** The solution to the recurrence is $$T(n) = \Theta(n)$$. This means the function grows linearly with $n$ for large $n$.