Math Problem Statement
Consider the following recurrence relation. P(n) =
0 if n = 0 [P(n − 1)]2 − n if n > 0 Use this recurrence relation to compute P(1), P(2), P(3), and P(4). P(1) = P(2) = P(3) = P(4) =
Solution
To compute based on the recurrence relation:
\begin{cases} 0 & \text{if } n = 0, \\ \left[P(n-1)\right]^2 - n & \text{if } n > 0, \end{cases}$$ we will evaluate $$P(1), P(2), P(3),$$ and $$P(4)$$ step by step. ### Step-by-Step Computation: 1. **Base Case: $$n = 0$$:** $$P(0) = 0.$$ 2. **For $$n = 1$$:** $$P(1) = \left[P(0)\right]^2 - 1 = 0^2 - 1 = -1.$$ 3. **For $$n = 2$$:** $$P(2) = \left[P(1)\right]^2 - 2 = (-1)^2 - 2 = 1 - 2 = -1.$$ 4. **For $$n = 3$$:** $$P(3) = \left[P(2)\right]^2 - 3 = (-1)^2 - 3 = 1 - 3 = -2.$$ 5. **For $$n = 4$$:** $$P(4) = \left[P(3)\right]^2 - 4 = (-2)^2 - 4 = 4 - 4 = 0.$$ ### Final Results: $$\begin{aligned} P(1) &= -1, \\ P(2) &= -1, \\ P(3) &= -2, \\ P(4) &= 0. \end{aligned}$$ --- Would you like me to explain any step further? ### Related Questions: 1. What is the general behavior or trend of this recurrence relation for large $$n$$? 2. Can $$P(n)$$ ever become positive again after it becomes negative? 3. How does squaring $$P(n-1)$$ influence the recurrence? 4. Is there a closed-form expression for $$P(n)$$ or must it always be computed recursively? 5. What would happen if $$P(0)$$ were initialized to a value other than 0? ### Tip: When working with recurrence relations, always carefully evaluate the base case and compute terms step by step, as each term depends heavily on the previous ones!Ask a new question for Free
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Math Problem Analysis
Mathematical Concepts
Recurrence Relations
Algebra
Formulas
P(n) = [P(n-1)]^2 - n for n > 0
Base Case: P(0) = 0
Theorems
-
Suitable Grade Level
Grades 10-12
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