For each part below, determine whether: • S(r) is O(g(x)) g(r) is O(f(x)) • f(r) is O(g(x)) and g(r) is O(f(x)) none of the above. (a) f(x) = 3r+log(r), g(x) = x + (log(x))² (b) f(x)=r!, g(x) = 1²¹ (c) f(x) = g(x) = log(r) (d) f(x)=x²0 +¹.9 +¹8..., g(x) = r².1
For each part below, determine whether: • S(r) is O(g(x)) g(r) is O(f(x)) • f(r) is O(g(x)) and g(r) is O(f(x)) none of the above. (a) f(x) = 3r+log(r), g(x) = x + (log(x))² (b) f(x)=r!, g(x) = 1²¹ (c) f(x) = g(x) = log(r) (d) f(x)=x²0 +¹.9 +¹8..., g(x) = r².1
Database System Concepts
7th Edition
ISBN:9780078022159
Author:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan
Publisher:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan
Chapter1: Introduction
Section: Chapter Questions
Problem 1PE
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Question
5
This is Discreet math please answer questions with proper formatting for proofs
![For each part below, determine whether:
f(x) is O(g(x))
• g(x) is O(f(x))
• J(r) is O(g(x)) and g(x) is O(f(x))
none of the above.
(a) f(x) = 3r+log(r), g(x) = x + (log(x))²
(b) f(x)=r!, g(x) = 1²
(c) f(x) = g(x) = log(r)
(d) f(x)=x20 +¹.9 +¹8..., g(x) = x².1](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F37ffeb4c-90e5-4cb9-a6b6-d76475059d8f%2F07dcec61-55d6-4084-a900-830d18bc9961%2F9hu4wu9_processed.jpeg&w=3840&q=75)
Transcribed Image Text:For each part below, determine whether:
f(x) is O(g(x))
• g(x) is O(f(x))
• J(r) is O(g(x)) and g(x) is O(f(x))
none of the above.
(a) f(x) = 3r+log(r), g(x) = x + (log(x))²
(b) f(x)=r!, g(x) = 1²
(c) f(x) = g(x) = log(r)
(d) f(x)=x20 +¹.9 +¹8..., g(x) = x².1
Expert Solution
![](/static/compass_v2/shared-icons/check-mark.png)
Step 1 forrg for (a) f(x) = 3x + log(x), g(x) = x + (log(x))^2
(a) f(x) is O(g(x)) and g(x) is not O(f(x)) To show that f(x) is O(g(x)), we need to find positive constants c and k such that f(x) <= c*g(x) for all x > k. Taking c = 4 and k = 1, we have:
3x + log(x) <= 4*(x + (log(x))^2) for all x > 1
which is true. Therefore, f(x) is O(g(x)).
To show that g(x) is not O(f(x)), we need to show that for any positive constant c and any k, there exists an x > k such that g(x) > c*f(x). Let c = 1 and k = 1. Then for any x > e, we have:
x + (log(x))^2 > 3x + log(x)
Therefore, g(x) is not O(f(x)).
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