f(n)=4n 2 +5n 2 ∗log(n) , g(n) = n^3g(n)=n 3 Now to show f(n)=O(g(n))f(n)=O(g(n)) when number of operation increases i.e., n\rightarrow \inftyn→∞ then \frac{f(n)}{g(n)}=\frac{4n^3+5n^2.log\ n}{n^3}=4+\frac{5.log\ n}{n} g(n) f(n) = n 3 4n 3 +5n 2 .log n =4+ n 5.log n Now, \lim_{n\to\infty}lim n→∞ \frac{f(n)}{g(n)}=4+\lim_{n\to\infty}\frac{5.log\ n}{n} g(n) f(n) =4+lim n→∞ n 5.log n =4+5\lim_{n\to\infty} \frac{1/n}{1}=4+5lim n→∞ 1 1/n [Using L'Hospital Rule] =4+5\times0=4=4+5×0=4 So, |\frac{f(n)}{g(n)}|\rightarrow 4∣ g(n) f(n) ∣→4 as n\rightarrow\inftyn→∞ So, f(n)=O(g(n))f(n)=O(g(n))
The Answer to the Question
is below this banner.
Can't find a solution anywhere?
NEED A FAST ANSWER TO ANY QUESTION OR ASSIGNMENT?
Get the Answers Now!You will get a detailed answer to your question or assignment in the shortest time possible.
Here's the Solution to this Question
Write down a few first terms of a sequence:
it is easy to see that the even terms of a sequence are zero, so