NPTEL Assignment Question

Question

Comments

Answer B

Answer 1

Answer C : $\rightarrow F_1<F_3<F_2$

$\rightarrow$ comparing $ F_1 $and $F_2$

$F_1(n)=n^{\sqrt n}$ ,

$F_2(n)=2^n$

let, $n=K^2$

$F_1(K^2) =  (K^2)^{\sqrt{K^2}} = (K^2)^K=K^{2K} =2^{2Klog_2(K)}$

$F_2(K^2)=2^{(K^2)}$

$K^2 >2Klog_2(K)$

$ \therefore F_2 > F_1$ 

$\rightarrow$ comparing $ F_1 $and $F_3$

$F_1(K^2)=2^{2Klog_2(K)}$

$F_3(K^2)=2^{(\frac {K^2}{2})}.(K^2)^{10}$

here simply, $\frac {K^2}{2}>2.Klog_2(K)$

$ \therefore F_3 > F_1$ 

$\rightarrow$ comparing $ F_2 $and $F_3$

$F_2=2^n=2^{\frac  n 2}\times 2^{\frac  n 2}$

$F_3=2^n=2^{\frac  n 2}\times n^{10}$

$2^{\frac n 2}>n^{10}$ as $exponentiation > polynomial$

$ \therefore F_2 > F_3$ 

 

Thanks @ 

Anuj parashar for rectifying mistake.

 

Alternate way to compare $F_1$ and $F_2$

taking log on both sides,

$log_2(F_1(n))=\sqrt n log_2(n)$

$log_2(F_2(n))=n log_2(2)=n$

now putting $n=K^2$,

$log_2(F_2(n))=K^2$

$log_2(F_1(n))=\sqrt {K^2}log_2(K^2)=2Klog(K)$

$K^2>2Klog_2(K)$

 

Comments

In 2nd line power must be root(n) not n/2

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Thank you for pointing that out,

Under current circumstances entire solution collapses XD

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Between 2^n and n^(root n) 2^n greater

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Intuitively and experimentally yes, but how do I prove the same fact mathematically?

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n^ root n  and 2^n take log both side then compare both functions u will get ur ans