Can anyone help out this question....?
Arrange the following functions in increasing order of growth rate. ie. if g(n) follows f(n) in your list, then f(n) is necessarily O(g(n)).

a)2^2n
b)2^n^2
c)n^2log(n)
d)n
e)n^2n

Question

Can anyone help out this question....?
Arrange the following functions in increasing order of growth rate. ie. if g(n) follows f(n) in your list, then f(n) is necessarily O(g(n)).

a)2^2n
b)2^n^2
c)n^2log(n)
d)n
e)n^2n

ganeshie8 · Answer

put n=10000 and see

anonymous · Answer

i am confused by the last part

saranya · Answer

increasing order : d>a>b>c>e
is this correct ?

anonymous · Answer

some should be easy 
$f(n)=n$ is first on the list

anonymous · Answer

i think you have it right, but i would have written the inequalities the other way

anonymous · Answer

hmm maybe i am wrong

anonymous · Answer

looks like $2^{x^2}$ grows faster than $x^{2x}$

anonymous · Answer

http://www.wolframalpha.com/input/?i=limit+x+to+infinity+%282^%28x^2%29%29%2F%28x^%282x%29%29

saranya · Answer

can yu give me ur ans. i will work out n c to t :)

saranya · Answer

give ur ans in increasing order :)

anonymous · Answer

i didn't do any work, i just checked with wolfram
if $\lim_{x\to\infty}\frac{2^{x^2}}{x^{2x}}=\infty$ then $2^{x^2}$ grows faster

anonymous · Answer

here is another comparison http://www.wolframalpha.com/input/?i=limit+x+to+infinity+%282^%282x%29%29%2F%28x^%282log%28x%29%29%29

saranya · Answer

yup so thats what i mentioned ryt ? 
:)

anonymous · Answer

a)2^2n
b)2^n^2
c)n^2log(n)
d)n
e)n^2n

least is d
then looks like c
from the last wolfram link

anonymous · Answer

then my guess is a 
then e 
then b
but i was cheating and going fast
check each
gotta run

saranya · Answer

okay :)

saranya · Answer

yu were correct dude :)