Math Problem Statement
Consider the following function f(n). The argument is an integer, and negative integers are allowed. f(n) = { n+1 (n≤0), f(f(n-2)) (n > 0) } For example, f(1) can be calculated as follows f(1) = f(f(1-2)) = f(f(-1)) = f(0) = 1 In this case, the number of function calls in the calculation of f(1) is 3.
Q1. Find the number of function calls in computing f(2). Q2. Prove that f(n) = 1 for all nonnegative integers n. Q3. Let n be a nonnegative integer. when computing f(n), express the number of function calls T(n) in terms of n.
Solution
Let's carefully analyze the given recursive function and address each part of the problem one by one.
Given function:
\begin{cases} n + 1 & \text{if } n \leq 0 \\ f(f(n-2)) & \text{if } n > 0 \end{cases}$$ ### Q1: Find the number of function calls in computing $$f(2)$$. To compute $$f(2)$$: $$f(2) = f(f(2-2)) = f(f(0)) = f(1)$$ Now, to compute $$f(1)$$: $$f(1) = f(f(1-2)) = f(f(-1)) = f(0) = 1$$ Thus, we have: $$f(2) = f(1) = 1$$ The number of function calls is as follows: 1. Call to $$f(2)$$ initiates a call to $$f(0)$$. 2. The call to $$f(0)$$ results in a call to $$f(1)$$. 3. The call to $$f(1)$$ results in two calls: one to $$f(-1)$$, which leads to $$f(0)$$, and one to $$f(0)$$. Thus, the total number of function calls is 5. ### Q2: Prove that $$f(n) = 1$$ for all nonnegative integers $$n$$. We will prove this by induction. **Base case:** For $$n = 0$$, we have: $$f(0) = 0 + 1 = 1$$ So, $$f(0) = 1$$. **Inductive step:** Assume that for all $$k \leq n$$, $$f(k) = 1$$. We need to prove that $$f(n+1) = 1$$. For $$n+1 > 0$$, the function definition tells us: $$f(n+1) = f(f(n-1))$$ By the inductive hypothesis, $$f(n-1) = 1$$. Therefore, this simplifies to: $$f(n+1) = f(1)$$ We have already computed that $$f(1) = 1$$. Therefore, $$f(n+1) = 1$$. By the principle of mathematical induction, $$f(n) = 1$$ for all nonnegative integers $$n$$. ### Q3: Let $$n$$ be a nonnegative integer. When computing $$f(n)$$, express the number of function calls $$T(n)$$ in terms of $$n$$. We can derive a recursive relation for the number of function calls $$T(n)$$ based on the recursive definition of $$f(n)$$. For $$n \leq 0$$, the function is computed directly: $$T(n) = 1 \quad \text{for } n \leq 0$$ For $$n > 0$$, we have: $$f(n) = f(f(n-2))$$ Thus, the number of function calls is: $$T(n) = T(n-2) + T(n-2) + 1 = 2T(n-2) + 1$$ This recursive relation can be solved using iteration. For example: - $$T(0) = 1$$ - $$T(2) = 2T(0) + 1 = 2(1) + 1 = 3$$ - $$T(4) = 2T(2) + 1 = 2(3) + 1 = 7$$ In general, the number of function calls $$T(n)$$ grows exponentially in $$n$$, and can be expressed as: $$T(n) = 2^k - 1 \quad \text{where } n = 2k$$ Thus, $$T(n)$$ is a geometric series with growth rate $$2^k$$. --- Would you like further details on this explanation or have any questions? ### 5 Follow-up Questions: 1. How would the function $$f(n)$$ behave for negative values of $$n$$? 2. Can you derive a closed form for the recursive formula for $$T(n)$$? 3. How does the growth rate of the number of function calls compare to the Fibonacci sequence? 4. What are the computational implications of a function like this in real-world applications? 5. How would memoization affect the performance of computing $$f(n)$$? ### Tip: Recursive functions like this one can often be optimized using **memoization** to store previously computed results, drastically reducing redundant function calls.Ask a new question for Free
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Math Problem Analysis
Mathematical Concepts
Recursion
Mathematical Induction
Number Theory
Formulas
f(n) = { n+1 (n≤0), f(f(n-2)) (n > 0) }
T(n) = 2T(n-2) + 1
Theorems
Mathematical Induction
Suitable Grade Level
Grades 11-12
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