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)). solve this desgin algorithm and analysis problem.

Solution

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. logn\sqrt{\log n}
  2. log(logn)\log(\log n)
  3. 2logn2^{\log n}
  4. nlognn \log n

Analysis:

  • logn\sqrt{\log n} grows slower than log(logn)\log(\log n).
  • 2logn=n2^{\log n} = n (since 2logn=n2^{\log n} = n).
  • nlognn \log n is a combination of linear and logarithmic growth, and grows faster than nn.

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

Sequence 2:

  1. (logn)100(\log n)^{100}
  2. n1100n^{\frac{1}{100}}
  3. 2log2n322^{\log 2n^{\frac{3}{2}}}
  4. nlognn \log n

Analysis:

  • (logn)100(\log n)^{100} is a large logarithmic term, but still grows slower than any polynomial.
  • n1100n^{\frac{1}{100}} is a very slowly growing polynomial.
  • 2log2n322^{\log 2n^{\frac{3}{2}}} simplifies to n32n^{\frac{3}{2}}, which grows faster than nlognn \log n.
  • nlognn \log n grows faster than n1100n^{\frac{1}{100}} but slower than n32n^{\frac{3}{2}}.

This sequence is not correctly ordered because n1100n^{\frac{1}{100}} should come after (logn)100(\log n)^{100}.

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

Analysis:

  • n\sqrt{n} grows faster than 2logn2^{\sqrt{\log n}}, since 2logn2^{\sqrt{\log n}} is very slow compared to polynomial functions.
  • n14n^{\frac{1}{4}} grows slower than n\sqrt{n} but faster than (logn)3(\log n)^3.
  • n2n^2 is the fastest growing term here.

This sequence is not correctly ordered because 2logn2^{\sqrt{\log n}} should come before n\sqrt{n}.

Sequence 4:

  1. 2logn2^{\sqrt{\log n}}
  2. n\sqrt{n}
  3. n(logn)2n(\log n)^2
  4. ni2n^{\frac{i}{2}}

Analysis:

  • 2logn2^{\sqrt{\log n}} is very slow compared to n\sqrt{n}.
  • n\sqrt{n} is slower than n(logn)2n(\log n)^2.
  • ni2n^{\frac{i}{2}} has an ambiguous term, likely intended to grow as a power of nn, but it should be slower than n(logn)2n(\log n)^2 for any small ii.

This sequence is incorrectly ordered due to the ambiguity with ni2n^{\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 2logn2^{\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.

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Math Problem Analysis

Mathematical Concepts

Asymptotic Notation
Growth Rate Analysis

Formulas

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Theorems

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Suitable Grade Level

Advanced