Math Problem Statement
\item You are given a function F : $\mathbb{Z} -> \mathbb{Z}^+$\ where F ( n ) =\begin{cases} 1 & \text{if n is 1} \\ F$(n/2)$ & \text{if n is even }\\ 1+ F(3n-8)& \text{if n is odd and n$ > $1 } \end{cases}\ for each integer n $\ge 1$. \ \item If F well defined?
Solution
To determine if the function is well-defined, we need to check whether the function produces a unique output for each input .
The function is defined as follows:
\begin{cases} 1 & \text{if } n = 1 \\ F\left(\frac{n}{2}\right) & \text{if } n \text{ is even} \\ 1 + F(3n - 8) & \text{if } n \text{ is odd and } n > 1 \end{cases}$$ ### Analysis of Each Case 1. **Base Case ($$n = 1$$):** - When $$n = 1$$, $$F(1) = 1$$. This is a fixed value and poses no issue. 2. **Even $$n$$:** - If $$n$$ is even, $$F(n) = F\left(\frac{n}{2}\right)$$. - This means that the function value for even $$n$$ is determined by the function value at $$\frac{n}{2}$$. - Since $$n$$ is halved each time, eventually $$n$$ will reach 1, where $$F(1) = 1$$ is already defined. Thus, for even numbers, the function is well-defined and finite because the halving process will always reach 1. 3. **Odd $$n > 1$$:** - If $$n$$ is odd and greater than 1, $$F(n) = 1 + F(3n - 8)$$. - Here, we need to consider the sequence produced by repeatedly applying this function. - The sequence $$3n - 8$$ for odd $$n$$ can either increase or decrease depending on the value of $$n$$. The critical point is whether or not this sequence will eventually reach an even number or 1, which would then lead the sequence to eventually terminate. ### Behavior of $$3n - 8$$ for Odd $$n > 1$$ - **If $$n = 3$$:** - $$F(3) = 1 + F(3(3) - 8) = 1 + F(9 - 8) = 1 + F(1) = 1 + 1 = 2$$. - **If $$n = 5$$:** - $$F(5) = 1 + F(3(5) - 8) = 1 + F(15 - 8) = 1 + F(7)$$. - Continuing this process for $$n = 7$$, $$n = 9$$, etc., it can be complex. But in each case, we need to check if the function eventually leads to an even $$n$$ or reaches 1. ### Conclusion The function $$F(n)$$ will be well-defined if the sequences generated by the rules for odd $$n > 1$$ either eventually reach 1 or reach an even $$n$$ which then leads to 1. The behavior of these sequences can be intricate, but for most small values of $$n$$, the sequences do reach 1. However, without a general proof that $$3n - 8$$ always reduces to an even number or 1 for all $$n$$, we cannot guarantee $$F(n)$$ is well-defined for all $$n \geq 1$$. Therefore, **$$F$$ is well-defined for many cases, but it requires a deeper analysis to prove definitively for all $$n$$.** The general behavior of the sequence $$3n - 8$$ for arbitrary odd $$n$$ would need to be thoroughly explored to ensure that it does not produce an infinite loop without reaching an even number or 1. --- Would you like to explore this behavior for specific values or go deeper into proving the general case? Here are some related questions: 1. What happens to $$F(n)$$ when $$n$$ is a large odd number? 2. Can you show that $$F(n)$$ reaches 1 for all small values of $$n$$? 3. How does the behavior of $$3n - 8$$ differ for different values of $$n$$? 4. What would be the function values for specific $$n$$, such as $$F(7)$$ or $$F(9)$$? 5. How can we construct a proof to determine if the function is well-defined for all integers $$n \geq 1$$? **Tip:** Analyzing recursive functions often involves understanding the behavior of sequences they generate. Working through small examples can offer valuable insights.Ask a new question for Free
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Math Problem Analysis
Mathematical Concepts
Recursive Functions
Number Theory
Formulas
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Theorems
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Suitable Grade Level
Advanced High School to College
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