Question

Given the following list of functions, determine the order of growth of each using big-Theta notation and put all the functions in order from slowest-growing to fastest-growing. Be sure to put functions of equal growth rate on the same level. Unless otherwise noted, you can assume all logarithms are base-2.

6nlog(2n)+8n

4n2log(log(8n))+8n2+n

500

n3+7nlog(n2) + 4n

2n+2n+1

log(4n2)+3n+1

12

8log(24n)+10

8n2log(5n2)+7n+200

4log(n3)+1000

100log(16n)log(n6)+23

8nlog(log(n4))+6n+32

9log(log(8n))

Answer 1

`Hey,

Note: Brother if you have any queries related the answer please do comment. I would be very happy to resolve all your queries.

6nlog(2n)+8n is theta(n*log(n))

4(n^2)log(log(8n))+8n^2+n=theta(n^2*log(log(n)))

500 is theta(1)

n^3+7nlog(n^2) + 4n=theta(n^3)

2^n+2n+1=theta(2^n)

log(4n^2)+3n+1=theta(n)

12=theta(1)

8log(24n)+10=theta(log(n))

8n^2log(5n^2)+7n+200=theta(n^2*log(n))

4log(n^3)+1000=theta(log(n))

100log(16n)log(n^6)+23=theta(log(n)^2)

8nlog(log(n^4))+6n+32=theta(n*log(log(n)))

9*log(log(8n))=theta(log(log(n)))

theta(1)=theta(1)<theta(log(log(n)))<theta(log(n))=theta(log(n))<theta(log(n)^2)<theta(n)<theta(n*log(log(n)))<theta(n*log(n))<theta(n^2*log(log(n)))<theta(n^2*log(n))<theta(n^3)<theta(2^n)

Kindly revert for any queries

Thanks.