me test

Question

Comments

you take, $n=10^{50}$ then tell me which one is larger.

the method you are following to compare asymptotic complexity is not reliable.

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@joshi_nitish

$1)n^{\sqrt{n}}$=$10^{50}$$^{\sqrt{10^{50}}}$

=$\left ( 10^{50} \right )^{50/2}$=$\left ( 10^{50} \right )$$^{25}$=10¹²⁵⁰
$2)$ $2^{n}$ =$2^{10}$$^{50}$=2⁵⁰⁰

Now, see which one greater

1 st one

right?

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@ srestha 

for n=10⁵⁰

n^n/2=n²⁵=10¹²⁵⁰

2ⁿ = ${{2}^{10}}^{50}$=(2^{10........(50 times)} )

2nd one is greater

Answer 1

$1)$ $n^{\sqrt{n}}=2^{\sqrt{n}.logn}$

$2)$ $2^{n}$

$3)$ $n^{10}.2^{n/2}=2^{(10.logn+n/2)}$

$4)$ $\frac{n.(n+1)}{2}-1$

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Last one is trivial so comparing the exponent part of first $3$.

$\Rightarrow \sqrt{n}logn$

$\Rightarrow n=\sqrt{n}.\sqrt{n}$

$\Rightarrow 10.logn+\frac{n}{2} \approx \frac{n}{2}=\frac{1}{2}(\sqrt{n}.\sqrt{n})$

$\therefore logn<\frac{\sqrt{n}}{2}<\sqrt{n}$

$\Rightarrow$ $1<3<2$

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$\color{Red}{4<1<3<2}$

Comments

@saxena sir

how you converted n^sqrt(n)   to    2^sqrt(n).logn

please explain

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$\Large\rightarrow n^{\sqrt{n}}\rightarrow2^{log_2{n^{\sqrt{n}}}}\rightarrow2^{\sqrt{n}.log_2n}$

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@saxena

u can see this algebric formula

$\left ( a^{x} \right )^{y}=a^{xy}$

I know u used right associativity as per C precedence

but according to algebric formula multiplication of power should applicable

That point what do ur view??