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Please write your answers in a legible way and submit a single pdf file. 1. Consider functions \( f_{1}, f_{2}: \mathbb{N} \rightarrow \mathbb{R} \) which are asymptotically positive. Using the definitions show that if \( f_{1}=O\left(f_{2}\right) \) then \( f_{1}+f_{2}=\Theta\left(f_{2}\right) \).

Ask by Fleming Boone. in the United States
Jan 23,2025

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If \( f_{1} = O(f_{2}) \) and both functions are asymptotically positive, then \( f_{1} + f_{2} = \Theta(f_{2}) \).

Solución

**Title:** Proof that \( f_{1} + f_{2} = \Theta\left(f_{2}\right) \) Given \( f_{1} = O\left(f_{2}\right) \) --- **Introduction** In the analysis of algorithms and asymptotic behavior of functions, understanding the relationship between different growth rates is crucial. This proof demonstrates that for two asymptotically positive functions \( f_{1}, f_{2}: \mathbb{N} \rightarrow \mathbb{R} \), if \( f_{1} = O\left(f_{2}\right) \), then the sum \( f_{1} + f_{2} \) is \( \Theta\left(f_{2}\right) \). --- **Definitions** 1. **Big-O Notation (\( O \))**: - **Formal Definition**: \( f_{1} = O\left(f_{2}\right) \) if there exist positive constants \( c \) and \( n_{0} \) such that for all \( n \geq n_{0} \), \[ f_{1}(n) \leq c \cdot f_{2}(n) \] 2. **Big-Theta Notation (\( \Theta \))**: - **Formal Definition**: \( f = \Theta\left(g\right) \) if and only if \( f = O\left(g\right) \) and \( f = \Omega\left(g\right) \). Formally, there exist positive constants \( c_{1}, c_{2} \) and \( n_{0} \) such that for all \( n \geq n_{0} \), \[ c_{1} \cdot g(n) \leq f(n) \leq c_{2} \cdot g(n) \] 3. **Little-O Notation (\( o \))** (for completeness): - **Formal Definition**: \( f = o\left(g\right) \) if for every positive constant \( c \), there exists a positive integer \( n_{0} \) such that for all \( n \geq n_{0} \), \[ 0 \leq f(n) < c \cdot g(n) \] --- **Proof** **Given:** - \( f_{1}, f_{2}: \mathbb{N} \rightarrow \mathbb{R} \) are asymptotically positive functions. - \( f_{1} = O\left(f_{2}\right) \). **To Prove:** - \( f_{1} + f_{2} = \Theta\left(f_{2}\right) \). **Proof Steps:** 1. **Upper Bound (\( f_{1} + f_{2} = O\left(f_{2}\right) \)):** Since \( f_{1} = O\left(f_{2}\right) \), by definition, there exist constants \( c > 0 \) and \( n_{0} \) such that for all \( n \geq n_{0} \), \[ f_{1}(n) \leq c \cdot f_{2}(n) \] Adding \( f_{2}(n) \) to both sides: \[ f_{1}(n) + f_{2}(n) \leq c \cdot f_{2}(n) + f_{2}(n) = (c + 1) \cdot f_{2}(n) \] Therefore, \[ f_{1} + f_{2} \leq (c + 1) \cdot f_{2} \] This satisfies the definition of \( f_{1} + f_{2} = O\left(f_{2}\right) \) with constants \( c' = c + 1 \) and the same \( n_{0} \). 2. **Lower Bound (\( f_{1} + f_{2} = \Omega\left(f_{2}\right) \)):** Since both \( f_{1} \) and \( f_{2} \) are asymptotically positive, for all \( n \) sufficiently large, \( f_{2}(n) > 0 \). Consider the sum: \[ f_{1}(n) + f_{2}(n) \geq f_{2}(n) \] This inequality holds because \( f_{1}(n) \geq 0 \) (as \( f_{1} \) is asymptotically positive). Thus, \[ f_{1} + f_{2} \geq 1 \cdot f_{2}(n) \] This satisfies the definition of \( f_{1} + f_{2} = \Omega\left(f_{2}\right) \) with constant \( c_{1} = 1 \) and the same \( n_{0} \). 3. **Combining Upper and Lower Bounds:** From steps 1 and 2, we have: \[ 1 \cdot f_{2}(n) \leq f_{1}(n) + f_{2}(n) \leq (c + 1) \cdot f_{2}(n) \] Therefore, by definition: \[ f_{1} + f_{2} = \Theta\left(f_{2}\right) \] --- **Conclusion** By demonstrating both the upper and lower bounds, we have shown that if \( f_{1} = O\left(f_{2}\right) \) and both functions are asymptotically positive, then the sum \( f_{1} + f_{2} \) grows asymptotically at the same rate as \( f_{2} \). Hence, \( f_{1} + f_{2} = \Theta\left(f_{2}\right) \).

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To show that \( f_{1} + f_{2} = \Theta(f_{2}) \) under the condition that \( f_{1} = O(f_{2}) \), we first recall the definitions involving Big O and Big Theta notation. By definition, \( f_{1} = O(f_{2}) \) means that there exist positive constants \( C \) and \( n_{0} \) such that for all \( n \geq n_{0} \): \[ f_{1}(n) \leq C \cdot f_{2}(n). \] This indicates that \( f_{1} \) grows at a rate that is bounded above by some constant multiple of \( f_{2} \) for sufficiently large \( n \). Now, when we consider \( f_{1} + f_{2} \), we can express it as follows: \[ f_{1}(n) + f_{2}(n) \leq C \cdot f_{2}(n) + f_{2}(n) = (C + 1) \cdot f_{2}(n). \] This shows that \( f_{1} + f_{2} \) is \( O(f_{2}) \). To satisfy the other requirement for \( \Theta(f_{2}) \), we need to establish the opposite inequality. Notice that since \( f_{2} \) is asymptotically positive, for sufficiently large \( n \), \( f_{2}(n) > 0 \). Thus: \[ f_{1}(n) + f_{2}(n) \geq f_{2}(n) \quad \text{(because \( f_{1} \) is always non-negative)}. \] This means \( f_{1} + f_{2} \) is lower bounded by \( f_{2}(n) \). Combining both results, we have: \[ f_{1} + f_{2} = O(f_{2}) \quad \text{and} \quad f_{1} + f_{2} \geq f_{2}, \] which implies: \[ f_{1} + f_{2} = \Theta(f_{2}). \] Thus, we have shown that if \( f_{1} = O(f_{2}) \), then it indeed follows that \( f_{1} + f_{2} = \Theta(f_{2}) \).

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