Math Problem Statement

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)).

Solution

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. logn\sqrt{\log n}
  2. log(logn)\log(\log n)
  3. 2logn=n2^{\log n} = n (because 2logn=n2^{\log n} = n)
  4. nlognn \log n

Sequence 2:

  1. (logn)100(\log n)^{100}
  2. n1100n^{\frac{1}{100}}
  3. 2log2n32=n322^{\log 2n^{\frac{3}{2}}} = n^{\frac{3}{2}}
  4. nlognn \log n

Sequence 3:

  1. n\sqrt{n}
  2. 2logn2^{\sqrt{\log n}}
  3. n14n^{\frac{1}{4}}
  4. (logn)3(\log n)^3
  5. n2n^2

Sequence 4:

  1. 2logn2^{\sqrt{\log n}}
  2. n\sqrt{n}
  3. n(logn)2n(\log n)^2
  4. ni2n^{\frac{i}{2}} (Note: This term is ambiguous due to the "i", but assuming ii 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(logn)\log(\log n), logn\log n, logn\sqrt{\log n}, (logn)2(\log n)^2, (logn)100(\log n)^{100}
  • Polynomial functions: n\sqrt{n}, n1100n^{\frac{1}{100}}, n14n^{\frac{1}{4}}, nn, nlognn \log n, n2n^2
  • Exponential functions: 2logn=n2^{\log n} = n, 2log2n32=n322^{\log 2n^{\frac{3}{2}}} = n^{\frac{3}{2}}, 2logn2^{\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 logn<log(logn)<n<nlogn\sqrt{\log n} < \log(\log n) < n < n \log n.
  2. Sequence 2:

    • This sequence is not correctly ordered since n1100n^{\frac{1}{100}} grows faster than (logn)100(\log n)^{100}.
  3. Sequence 3:

    • This sequence is also not correctly ordered because (logn)3(\log n)^3 grows slower than n14n^{\frac{1}{4}} and 2logn2^{\sqrt{\log n}}.
  4. Sequence 4:

    • This sequence starts with an exponential term, 2logn2^{\sqrt{\log n}}, followed by a polynomial n\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 nlognn \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.

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Math Problem Analysis

Mathematical Concepts

Big-O notation
Function growth rates
Logarithmic functions
Polynomial functions
Exponential functions

Formulas

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Theorems

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Suitable Grade Level

Advanced