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Which of the following arrangements of functions is in ascending order of growth rate. That is, if g(n) follows f(n) than it should be the case that f(n) is O(g(n)).

To determine which of the given sequences of functions is in ascending order of growth rate, we'll need to analyze the growth rate of each function in each sequence.

### Sequence 1:
1. \( \sqrt{\log n} \)
2. \( \log(\log n) \)
3. \( 2^{\log n} = n \) (because \(2^{\log n} = n\))
4. \( n \log n \)

### Sequence 2:
1. \( (\log n)^{100} \)
2. \( n^{\frac{1}{100}} \)
3. \( 2^{\log 2n^{\frac{3}{2}}} = n^{\frac{3}{2}} \)
4. \( n \log n \)

### Sequence 3:
1. \( \sqrt{n} \)
2. \( 2^{\sqrt{\log n}} \)
3. \( n^{\frac{1}{4}} \)
4. \( (\log n)^3 \)
5. \( n^2 \)

### Sequence 4:
1. \( 2^{\sqrt{\log n}} \)
2. \( \sqrt{n} \)
3. \( n(\log n)^2 \)
4. \( n^{\frac{i}{2}} \) (Note: This term is ambiguous due to the "i", but assuming \(i\) could be a positive integer or even complex, it might not fit well with others in a real number context.)

### Comparing Growth Rates:

- **Logarithmic functions:** \( \log(\log n) \), \( \log n \), \( \sqrt{\log n} \), \( (\log n)^2 \), \( (\log n)^{100} \)
- **Polynomial functions:** \( \sqrt{n} \), \( n^{\frac{1}{100}} \), \( n^{\frac{1}{4}} \), \( n \), \( n \log n \), \( n^2 \)
- **Exponential functions:** \( 2^{\log n} = n \), \( 2^{\log 2n^{\frac{3}{2}}} = n^{\frac{3}{2}} \), \( 2^{\sqrt{\log n}} \)

### General Rules:
1. **Logarithmic growth** (very slow growth)
2. **Polynomial growth** (depends on the degree)
3. **Exponential growth** (fast growth)

Now, based on this:

1. **Sequence 1**: 
   - Growth rates are correctly ordered as \( \sqrt{\log n} < \log(\log n) < n < n \log n \).

2. **Sequence 2**:
   - This sequence is not correctly ordered since \( n^{\frac{1}{100}} \) grows faster than \( (\log n)^{100} \).

3. **Sequence 3**:
   - This sequence is also not correctly ordered because \( (\log n)^3 \) grows slower than \( n^{\frac{1}{4}} \) and \( 2^{\sqrt{\log n}} \).

4. **Sequence 4**:
   - This sequence starts with an exponential term, \( 2^{\sqrt{\log n}} \), followed by a polynomial \( \sqrt{n} \), which grows faster. The sequence is not in the correct order.

### Conclusion:
**Sequence 1** is in ascending order of growth rate.

Would you like further details or have any questions? Here are some related questions:

1. What is Big-O notation, and how is it used to classify functions?
2. How does logarithmic growth compare to polynomial growth?
3. What are common examples of polynomial time algorithms?
4. Why does exponential growth outpace polynomial growth?
5. How can we visually compare the growth rates of different functions?
6. What is the significance of analyzing function growth in computer science?
7. Can you explain the \( n \log n \) term and where it commonly appears?
8. What are some practical applications of these growth rate concepts in algorithms?

**Tip:** When comparing growth rates, remember that logarithmic functions grow the slowest, followed by polynomials, and then exponentials grow the fastest.

Understand function growth rates and how to determine their order of growth. Analyze sequences of functions to identify which grow faster or slower than others. Learn about logarithmic, polynomial, and exponential growth in the context of Big-O notation.

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