Question

Prove by induction that if \(T(n) \leq 4 T(n / 2)+n^{2}\), then \(T(n)=O\left(n^{2} \log n\right)\) (assuming \(n\) is a power of 2 ).

Short answer

The recurrence relation resolves to \(T(n) = O(n^{2} \log n)\) by induction, verifying the hypothesis.

Step-by-step solution

Understanding the Problem

We need to show that the recursive inequality \(T(n) \leq 4 T(n / 2)+n^{2}\) leads to the conclusion that \(T(n)=O(n^{2} \log n)\). This means we will demonstrate that the asymptotic behavior of \(T(n)\) is bounded above by a function \(Cn^{2} \log n\) for some constant \(C\).

Base Case Verification

For induction, we first check the base case. Typically, for \(T(n)\) where \(n=1\), assume \(T(1)\) is a constant \(c\), which satisfies the base case since \(c \leq 4c/2 + 1^{2}\), simplifying to \(c \leq 2c + 1\). This holds true when \(c\) is positive, establishing the base case.

Inductive Hypothesis

Assume that for some \(k\), \(T(k) \leq Ck^{2} \log k\). This is our inductive hypothesis, which we will use to prove that the statement also holds for \(n = 2k\).

Inductive Step Application

Using the recurrence relation, \(T(2k) \leq 4T(k) + (2k)^{2}\). By substituting the inductive hypothesis, we have:\[ T(2k) \leq 4(Ck^{2} \log k) + 4k^{2} \]which simplifies to:\[ T(2k) \leq 4Ck^{2} \log k + 4k^{2} \]

Simplification Using Logarithmic Properties

Recognize that \(\log(2k) = \log 2 + \log k\), hence:\[ T(2k) \leq 4Ck^{2}(\log 2 + \log k) + 4k^{2} \].Which expands to:\[ T(2k) \leq 4Ck^{2} \log k + 4Ck^{2}\log 2 + 4k^{2} \].

Combine and Compare Terms

Group terms involving \(k^{2} \log k\) and constants:\[ T(2k) \leq 4C k^{2} \log k + (4C \log 2 + 4)k^{2} \]. This shows: \( T(2k) \leq 4Ck^{2} \log(2k) \). Thus, \( T(n) = O(n^{2}\log n) \).

Conclusion of Inductive Proof

Since the base case holds and the inductive step is proven, by induction, for \(n = 2k\), \(T(n) \leq Cn^{2} \log n\) holds for all powers of 2. Therefore, \(T(n) = O(n^{2} \log n)\) is valid.

Key concepts

Base Case

In mathematical induction, establishing the base case is like laying the foundation for a building; if it's solid, the structure built upon it is likely to hold strong. The base case of our problem involves verifying the inequality for the smallest instance, typically where \(n=1\). Here, we're assuming \(T(1) = c\), a constant that satisfies the inequality \(c \leq 4c/2 + 1^2\). Simplified, this becomes \(c \leq 2c + 1\). This condition holds true when \(c\) is positive, thus successfully establishing our base case. This step is crucial because it shows that the initial premise holds, allowing the hypothetical 'domino effect' of induction to commence.

Inductive Hypothesis

The inductive hypothesis is the stage in mathematical induction where we assume that our statement holds true for an arbitrary case \(k\). It's like assuming, "If it's true for one, it's true for the next." In our scenario, we assume that \(T(k) \leq Ck^2 \log k\) for some constant \(C\). This assumption serves as a stepping stone to demonstrate why it must also hold for the subsequent integer \(n = 2k\). Think of this as middle ground reasoning, where proving this assumption helps us extend the truth to larger cases, ensuring that our inequality holds across various values.

Inductive Step

The inductive step is the critical link in proving our statement beyond the base case, using the inductive hypothesis. Here, our task is to prove the statement \(T(n) \leq Cn^2 \log n\) for \(n = 2k\), based on the assumption it holds for \(k\). Through substitution and simplification, we use the recurrence relation \(T(2k) \leq 4T(k) + (2k)^2\). By inserting the inductive hypothesis, we get:

• \(T(2k) \leq 4(Ck^2 \log k) + 4k^2\)

Simplifying further and incorporating logarithmic properties, we derive:

• \(T(2k) \leq 4Ck^2(\log 2 + \log k) + 4k^2\)

Finally, grouping terms reveals the final form matches \(T(n) = O(n^2 \log n)\). This shows that if it's valid for \(k\), it must be valid for \(2k\), completing the proof by induction.

Asymptotic Analysis

Asymptotic analysis is a handy tool in mathematics and computer science, particularly useful for understanding the behavior of algorithms as inputs grow large. In our current context, we use it to describe the time complexity of the function, indicating how \(T(n)\) scales with increasing \(n\). The statement \(T(n) = O(n^2 \log n)\) predicts the upper bound of \(T(n)\), implying that while it increases at that rate, it doesn't do so significantly faster. This doesn't necessarily reflect the precise value of \(T(n)\), but offers a big-picture view, primarily focusing on the dominant terms that affect growth rates. Such analysis aids in comparing algorithm efficiency, especially as \(n\) becomes very large.