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Arrange the following functions in increasing asymptotic order:

  1. $n^{1/3}$
  2. $e^n$
  3. $n^{7/4}$                                                                              
  4. $n \log^9n$
  5. $1.0000001^n$

 

  1. a, d, c, e, b
  2. d, a, c, e, b
  3. a, c, d, e, b
  4. a, c, d, b, e
asked in Algorithms by Veteran (20.8k points)   | 1k views

1 Answer

+19 votes
Best answer
A < C and A < D

E < B

and

C, D < E as E is exponential function.

Now, we just need to see if C or D is larger.

In C we have a term $n^{3/4}$ and correspondingly in D we have $\log^9n$ (after taking $n$ out).

$n^{3/4}$ is asymptotically larger than $\log^9n$ as when $n = 10^{100}, \log^9 n$ gives $100^9$, while $n^{3/4}$ gives $10^{75} > 100^{37}$ a much higher value and this is true for all higher values of $n$. So, D < C.

Thus A is correct.
answered by Veteran (294k points)  
edited by
$10^{75} = 10.{10^{2}}^{37}= 10.100^{37}$
@arjun sir for any large value of n value of option e is less than option c.please correct me where am i going wrong
not getting how E be in 4 position

 i caluculated its value to large n

which still at 1.000..?
if you have doubt you can try plotting the graph and see. Large value means some large value - can be even one with 100 digits - for the same reason we cannot always find this by substituting values.
it is very confusing sir any other way of solving this question


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