Let's carefully analyze the proposed recursive definition: **Recursive definition:** - \( f(0) = 2 \) - \( f(n) = 4 - n \) for \( n > 0 \) **Step 1: Is this a valid recursive definition?** A proper recursive definition of a function from nonnegative integers (\( \mathbb{N}_0 \)) to integers (\( \mathbb{Z} \)) must define: - A base case (usually for 0), and - A rule to find \( f(n) \) in terms of **previous values** of the function, typically \( f(n-1) \) or earlier. **Examination of this case:** - *Base case:* \( f(0) = 2 \) **(OK)** - *Recursive step:* \( f(n) = 4 - n \) for \( n > 0 \) - **But:** This does **not** use \( f(n-1) \) or any prior values of \( f \). It defines \( f(n) \) directly for each \( n > 0 \). **Conclusion:** This is **not** a recursive definition—it's a **piecewise definition**. --- **Step 2: Is the function well-defined? What is its formula?** The function **is** well-defined for all nonnegative integers because for any \( n \), \( f(n) \) is unambiguously defined: - \( f(0) = 2 \) - For \( n > 0 \), \( f(n) = 4 - n \) So, the formula is: \[ f(n) = \begin{cases} 2 & \text{if } n = 0 \\ 4-n & \text{if } n > 0 \end{cases} \] Or, using the Iverson bracket notation: \[ f(n) = 2 \cdot \mathbf{1}_{n = 0} + (4 - n) \cdot \mathbf{1}_{n > 0} \] --- **Step 3: Does there exist a closed formula for all \( n \)?** Notice that for \( n=0 \), \( f(0)=2 \), but for \( n>0 \), \( f(n)=4-n \). You could write: \[ f(n) = \begin{cases} 2 & \text{if } n = 0 \\ 4-n & \text{if } n > 0 \end{cases} \] Or, if you prefer a single expression: \[ f(n) = (4-n) + (n-2)\cdot \mathbf{1}_{n=0} \] But in general, the piecewise form is simplest. --- **Step 4: Is it a recursive definition?** - **No.** The value of \( f(n) \) does **not** depend on \( f(n-1) \) or any previous value. - A valid recursive definition must use previous values. --- **Step 5: Multiple choice selection** If the question asks whether this is a valid recursive definition, the answer is **No**. If the question asks for the formula for \( f(n) \), then the answer is: \[ f(n) = \begin{cases} 2 & \text{if } n = 0 \\ 4-n & \text{if } n > 0 \end{cases} \] --- **Summary Table:** | Option | Is it recursive? | Well-defined? | Formula for \( f(n) \) | |-------------------------------|------------------|---------------|-------------------------------| | \( f(0) = 2; f(n) = 4-n \) | No | Yes | \( f(n) = 2 \) if \( n=0 \), \( 4-n \) if \( n>0 \) | --- ### **Explanation** - **Not recursive:** Because a recursive definition must define \( f(n) \) in terms of earlier values, and this one does not. - **Well-defined:** Yes, because for every \( n \), we know what \( f(n) \) is. - **Formula:** \( f(n) = 2 \) if \( n=0 \), \( f(n) = 4-n \) if \( n>0 \). --- **In summary:** This is a valid function definition, but **not** a recursive one. The correct formula for \( f(n) \) is the piecewise definition above.