Math Problem Statement

T(n)=2T(n-1) +(n+5)3^n, n>=1 solve this recurrence relation.

Solution

To solve the recurrence relation T(n)=2T(n1)+(n+5)3nT(n) = 2T(n-1) + (n + 5)3^n for n1n \geq 1, we can use the method of iteration (also known as the unfolding method). This approach involves expanding the recurrence relation step by step until we can identify a pattern or series.

Given: T(n)=2T(n1)+(n+5)3nT(n) = 2T(n-1) + (n + 5)3^n

Let's start unfolding this recurrence relation:

Iteration 1: T(n)=2T(n1)+(n+5)3nT(n) = 2T(n-1) + (n + 5)3^n

Iteration 2: T(n1)=2T(n2)+(n1+5)3n1T(n-1) = 2T(n-2) + (n-1 + 5)3^{n-1}

Substitute T(n1)T(n-1) into T(n)T(n): T(n)=2[2T(n2)+(n1+5)3n1]+(n+5)3nT(n) = 2[2T(n-2) + (n-1 + 5)3^{n-1}] + (n + 5)3^n =4T(n2)+2(n+4)3n1+(n+5)3n= 4T(n-2) + 2(n + 4)3^{n-1} + (n + 5)3^n

Iteration 3: T(n2)=2T(n3)+(n2+5)3n2T(n-2) = 2T(n-3) + (n-2 + 5)3^{n-2}

Substitute T(n2)T(n-2) into the expression for T(n)T(n): T(n)=4[2T(n3)+(n2+5)3n2]+2(n+4)3n1+(n+5)3nT(n) = 4[2T(n-3) + (n-2 + 5)3^{n-2}] + 2(n+4)3^{n-1} + (n+5)3^n =8T(n3)+4(n+3)3n2+2(n+4)3n1+(n+5)3n= 8T(n-3) + 4(n + 3)3^{n-2} + 2(n + 4)3^{n-1} + (n + 5)3^n

Pattern Recognition: After unfolding the recurrence relation several times, we notice a pattern forming. For the general kk-th iteration: T(n)=2kT(nk)+i=0k12i(ni+5)3niT(n) = 2^kT(n-k) + \sum_{i=0}^{k-1} 2^i (n - i + 5) 3^{n-i}

To find the particular solution, we continue unfolding until k=nk = n, where T(nn)=T(0)T(n-n) = T(0). For the base case, we assume T(0)T(0) is given or can be derived from the context.

Final Step (When k=nk = n): T(n)=2nT(0)+i=0n12i(ni+5)3niT(n) = 2^n T(0) + \sum_{i=0}^{n-1} 2^i (n - i + 5) 3^{n-i}

To find an explicit solution, we'd need to know T(0)T(0). The sum involves terms with increasing powers of 3 and decreasing powers of 2, so the dominant term as nn \to \infty will be (n+5)3n(n + 5) 3^n. Thus, the growth of T(n)T(n) can be approximated by the term with the highest growth rate, which is the last term of the sum, giving us:

T(n)(n+5)3nT(n) \approx (n + 5) 3^n

Conclusion:

The exact solution involves the series sum, but asymptotically, the solution grows as Θ(n3n)\Theta(n 3^n). Therefore, the time complexity of the given recurrence relation is T(n)=Θ(n3n)T(n) = \Theta(n 3^n).

Would you like more details or have any questions?

Follow-up Questions

  1. What is the method of iteration in solving recurrence relations?
  2. How can you verify the correctness of the obtained solution?
  3. What are the limitations of the method of iteration?
  4. Can you derive the exact solution for T(n)T(n) if the base case T(0)T(0) is given?
  5. How does the presence of 3n3^n affect the growth rate of the solution?
  6. What is the difference between Θ(n3n)\Theta(n 3^n) and O(n3n)O(n 3^n)?
  7. How would the solution change if T(n)=3T(n1)+(n+5)3nT(n) = 3T(n-1) + (n + 5)3^n?
  8. Can you explain why the largest term in the sum determines the asymptotic growth rate?

Tip

When using the method of iteration, carefully keep track of each term's contribution to ensure an accurate pattern identification.

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Math Problem Analysis

Mathematical Concepts

Recurrence Relations
Algorithm Analysis
Asymptotic Notation

Formulas

Unfolding Method for Recurrence Relations

Theorems

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Suitable Grade Level

Graduate Level