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+24 votes

Consider the following functions from positive integers to real numbers:

$10$, $\sqrt{n}$, $n$, $\log_{2}n$, $\frac{100}{n}$.

The CORRECT arrangement of the above functions in increasing order of asymptotic complexity is:

- $\log_{2}n$, $\frac{100}{n}$, $10$, $\sqrt{n}$, $n$
- $\frac{100}{n}$, $10$, $\log_{2}n$, $\sqrt{n}$, $n$
- $10$, $\frac{100}{n}$, $\sqrt{n}$, $\log_{2}n$, $n$
- $\frac{100}{n}$, $\log_{2}n$, $10$, $\sqrt{n}$, $n$

+31 votes

Best answer

$10$ is constant. $\therefore$ Growth rate is $0$.

$\sqrt{n}$ grows slower than linear but faster than $\log$. (Consider $\frac {\sqrt {n^2}}{\sqrt n} =\sqrt n$, whereas $\frac {\log \left(n^2\right)}{\log n} = 2$)

$n$ : Growth rate is linear.

$\log_{2}n$ : Growth rate is logarithmic. For asymptotic growth, the base does not matter.

$\frac{100}{n}$ : Growth rate decreases with $n$.

So, correct answer is **$(B)$**.

NOTE: Please never substitute large values of $n$ in such questions. If ever you do, at least do for $2$ such values and take ratio to get the growth rate or plot a graph. Remember ${1.01}^{n} \neq O\left(n^{100}\right)$.

+24 votes

- $10$,Constant
- $\sqrt{n}$,Square root
- $n$, polynomial
- $\log_{2}n$, Logorithmic
- $\frac{100}{n}$. Constant division by polynomial
**(clearly less than constant for every value of n >100)**

Now we know Constant division by polynomial < Constant < Logorithmic <Square root < polynomial

So correct order is $\frac{100}{n}$, $10$, $\log_{2}n$, $\sqrt{n}$, $n$

+13 votes

+10 votes

if we take a very large value of n suppose =2^{1024}

then we can follow the sequence with result

100/2^{1024 }<10 < log1024=1024 <sqrt(2^{1024) } < 2^{1024}

so following we see B is the correct answer

+7 votes

100/n grows inversely with n so for very large values of n, it will become close to zero.

10 IS CONSTANT.

100/n<10

among rest of choices it is very clear that

log n < sqrt(n)<n

thus 100/n<10< log n < sqrt(n)<n

**CHOICE B**

+5 votes

let take a very large N value of 2^{1024}

Now putting this in the respective terms we get

B is the correct answer here

NOTE: More editing coming up

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