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T(n)=n+4n2+8+16n4+32+⋯+Θ(nlg4)=lg(n−1)∑i=0 2in+2lg(n−1)∑i=0 4i+Θ(n2)=2lgn−12−1n+2⋅4lgn−14−1+Θ(n2)=(2lgn−1)n+23(4lgn−1)+Θ(n2)=(n−1)n+23(n2−1)+Θ(n2)=Θ(n2)

Question

T(n)=n+4n²+8+16n⁴+32+⋯+Θ(nlg4)=lg(n−1)∑ᵢ=0 2ⁱn+2lg(n−1)∑ᵢ=0 4ⁱ+Θ(n²)=2lg(n)−12⁻¹n+2⋅4lg(n)−14⁻¹+Θ(n²)=(2lg(n)−1)n+23(4lg(n)−1)+Θ(n²)=(n−1)n+23(n²−1)+Θ(n²)=Θ(n²)

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Solution

Break Down the Problem

  1. Identify the expression for T(n) T(n) .
  2. Analyze the individual components of the expression.
  3. Combine the terms to simplify T(n) T(n) .

Relevant Concepts

  1. Big Theta notation Θ \Theta indicates asymptotic tight bounds.
  2. Summations and logarithmic identities will be used for simplification.
  3. The behavior of polynomial and logarithmic terms as n n grows.

Analysis and Detail

  1. The given expression is: T(n)=n+4n2+8+16n4+32++Θ(nlog4)=log(n1)i=022in+2log(n1)i=044i+Θ(n2) T(n) = n + 4n^2 + 8 + 16n^4 + 32 + \ldots + \Theta(n \log 4) = \log(n-1) \sum_{i=0}^{2} 2^i n + 2 \log(n-1) \sum_{i=0}^{4} 4^i + \Theta(n^2)

  2. Simplify the summations:

    • The first summation: i=022i=20+21+22=1+2+4=7 \sum_{i=0}^{2} 2^i = 2^0 + 2^1 + 2^2 = 1 + 2 + 4 = 7 Therefore, log(n1)i=022in=7log(n1)n \log(n-1) \sum_{i=0}^{2} 2^i n = 7 \log(n-1) n

    • The second summation: i=044i=40+41+42+43+44=1+4+16+64+256=341 \sum_{i=0}^{4} 4^i = 4^0 + 4^1 + 4^2 + 4^3 + 4^4 = 1 + 4 + 16 + 64 + 256 = 341 Therefore, 2log(n1)i=044i=682log(n1) 2 \log(n-1) \sum_{i=0}^{4} 4^i = 682 \log(n-1)

  3. Combining terms together: T(n)=7log(n1)n+682log(n1)+Θ(n2) T(n) = 7 \log(n-1) n + 682 \log(n-1) + \Theta(n^2)

Verify and Summarize

  1. Through simplification and collecting the terms, we observe that as n n increases:
    • The leading term comes from Θ(n2) \Theta(n^2) .
    • Thus, T(n)=Θ(n2) T(n) = \Theta(n^2) dominates the logarithmic terms.

Final Answer

T(n)=Θ(n2) T(n) = \Theta(n^2)

This problem has been solved

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