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For each group of functions, sort the functions in increasing order of asymptotic (big-O) complexity

To sort the functions in increasing order of asymptotic complexity (Big-O notation) for each group, let's analyze each function from the three groups.

### Group 1:

1. \( f_1(n) = n^{0.999999} \log n \)
2. \( f_2(n) = 10000000n \)
3. \( f_3(n) = 1.000001^n \)
4. \( f_4(n) = n^2 \)

**Analysis:**

- \( f_1(n) = n^{0.999999} \log n \) grows slower than \( n^1 \), but the logarithmic factor increases the growth slightly.
- \( f_2(n) = 10000000n \) is linear.
- \( f_3(n) = 1.000001^n \) is exponential, where the base is slightly more than 1.
- \( f_4(n) = n^2 \) is quadratic.

**Order:**
\[ f_1(n) < f_2(n) < f_4(n) < f_3(n) \]

### Group 2:

1. \( f_1(n) = 2^{2^{1000000}} \)
2. \( f_2(n) = 2^{100000n} \)
3. \( f_3(n) = \binom{n}{2} \)
4. \( f_4(n) = n \sqrt{n} \)

**Analysis:**

- \( f_1(n) = 2^{2^{1000000}} \) is a constant value since it does not depend on \( n \).
- \( f_2(n) = 2^{100000n} \) is exponential in \( n \).
- \( f_3(n) = \binom{n}{2} \) is quadratic \( \frac{n(n-1)}{2} \sim \frac{n^2}{2} \).
- \( f_4(n) = n \sqrt{n} = n^{3/2} \), which is polynomial.

**Order:**
\[ f_1(n) < f_4(n) < f_3(n) < f_2(n) \]

### Group 3:

1. \( f_1(n) = n^{\sqrt{n}} \)
2. \( f_2(n) = 2^n \)
3. \( f_3(n) = n^{10} \cdot 2^{n/2} \)
4. \( f_4(n) = \sum_{i=1}^{n}(i+1) = \frac{n(n+1)}{2} + n \)

**Analysis:**

- \( f_1(n) = n^{\sqrt{n}} \) grows extremely fast, faster than any polynomial or exponential of lower order.
- \( f_2(n) = 2^n \) is exponential.
- \( f_3(n) = n^{10} \cdot 2^{n/2} \) is exponential, but with a polynomial factor that slows it down compared to \( 2^n \).
- \( f_4(n) = \frac{n(n+1)}{2} + n \sim \frac{n^2}{2} + n \) is quadratic.

**Order:**
\[ f_4(n) < f_3(n) < f_2(n) < f_1(n) \]

These are the orders of the functions in increasing order of asymptotic complexity for each group.

Do you want more details on any of the specific functions or groups? Here are some related questions:

1. What is Big-O notation and how is it used to compare the growth rates of functions?
2. How does the logarithmic factor in \( f_1(n) = n^{0.999999} \log n \) influence its growth rate?
3. Can you explain why \( 2^{2^{1000000}} \) is considered a constant function in this context?
4. Why does \( f_4(n) = \frac{n(n+1)}{2} + n \) behave quadratically?
5. How do different types of growth rates (linear, quadratic, exponential) compare in practice?
6. What are the implications of using exponential functions like \( 1.000001^n \) in algorithms?
7. How can \( \binom{n}{2} \) be simplified and what does it represent combinatorially?
8. How do you determine which function grows faster when comparing two functions with different exponents?

**Tip:** When comparing functions of different types, always start by identifying whether they are polynomial, exponential, or logarithmic, as this often provides a quick insight into their growth rate relative to one another.

Explore how functions are sorted by asymptotic complexity using Big-O notation. Understand the growth rates of functions across three groups with detailed analysis and comparisons. Suitable for advanced learners and anyone interested in algorithmic complexity.

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