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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)).
solve this desgin algorithm and analysis problem.

To solve this problem, we'll evaluate each sequence of functions in the image and check if they are arranged in ascending order of their growth rates. 

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

### Analysis:
- \( \sqrt{\log n} \) grows slower than \( \log(\log n) \).
- \( 2^{\log n} = n \) (since \( 2^{\log n} = n \)).
- \( n \log n \) is a combination of linear and logarithmic growth, and grows faster than \( n \).

This sequence is not correctly ordered because \( \sqrt{\log n} \) actually grows slower than \( \log(\log n) \).

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

### Analysis:
- \( (\log n)^{100} \) is a large logarithmic term, but still grows slower than any polynomial.
- \( n^{\frac{1}{100}} \) is a very slowly growing polynomial.
- \( 2^{\log 2n^{\frac{3}{2}}} \) simplifies to \( n^{\frac{3}{2}} \), which grows faster than \( n \log n \).
- \( n \log n \) grows faster than \( n^{\frac{1}{100}} \) but slower than \( n^{\frac{3}{2}} \).

This sequence is not correctly ordered because \( n^{\frac{1}{100}} \) should come after \( (\log n)^{100} \).

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

### Analysis:
- \( \sqrt{n} \) grows faster than \( 2^{\sqrt{\log n}} \), since \( 2^{\sqrt{\log n}} \) is very slow compared to polynomial functions.
- \( n^{\frac{1}{4}} \) grows slower than \( \sqrt{n} \) but faster than \( (\log n)^3 \).
- \( n^2 \) is the fastest growing term here.

This sequence is not correctly ordered because \( 2^{\sqrt{\log n}} \) should come before \( \sqrt{n} \).

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

### Analysis:
- \( 2^{\sqrt{\log n}} \) is very slow compared to \( \sqrt{n} \).
- \( \sqrt{n} \) is slower than \( n(\log n)^2 \).
- \( n^{\frac{i}{2}} \) has an ambiguous term, likely intended to grow as a power of \( n \), but it should be slower than \( n(\log n)^2 \) for any small \( i \).

This sequence is **incorrectly ordered** due to the ambiguity with \( n^{\frac{i}{2}} \).

### Conclusion:

**None of the sequences provided in the image are correctly ordered in ascending order of growth rate.**

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

1. Can you explain how to determine the growth rate of a function using limits?
2. What is the significance of comparing the growth rates of algorithms?
3. How do you simplify complex growth functions like \( 2^{\log n} \)?
4. How do logarithmic and exponential functions compare in terms of growth rate?
5. What are some examples of real-world problems where this type of analysis is critical?
6. How does the base of a logarithm affect its growth rate?
7. Can growth rates of functions be compared graphically?
8. What is the difference between Big-O, Big-Theta, and Big-Omega notation?

**Tip:** When analyzing sequences of functions for growth rate, it's useful to remember the hierarchy: logarithmic functions grow slower than polynomial functions, which grow slower than exponential functions.

Explore how to determine the growth rates of functions and analyze their order using asymptotic notation. This problem examines sequences of functions and determines their relative growth rates, crucial for algorithmic analysis.

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