I need help with the following problem:
f(n) = n^2 + 4n is O(n^2)
Are you the same person who asked this question earlier today??
http://www.jiskha.com/display.cgi?id=1207150648
Perhaps we are in the same class.
http://www.jiskha.com/display.cgi?id=1207186401
To determine if f(n) = n^2 + 4n is O(n^2), we need to find a constant value c and a positive integer value n0 such that for all values of n greater than or equal to n0, f(n) ≤ c * n^2.
Let's break down the problem step by step:
Step 1: Simplify the function.
f(n) = n^2 + 4n
Step 2: Express f(n) in terms of n^2.
f(n) = n^2 + 4n = (1 + 4/n) * n^2
Step 3: Set an inequality.
(1 + 4/n) * n^2 ≤ c * n^2
Step 4: Simplify the inequality.
1 + 4/n ≤ c
Step 5: Find the maximum value of c.
To find the maximum value of c, we need to let n approach infinity. As n approaches infinity, 4/n approaches 0, so the inequality simplifies to:
1 ≤ c
Therefore, any value of c that is greater than or equal to 1 satisfies the inequality.
Step 6: Choose a value for n0.
Since we have found that c ≥ 1 satisfies the inequality, we can choose any positive integer value for n0.
In conclusion, f(n) = n^2 + 4n is O(n^2) with c ≥ 1 and any value of n0.