Asymptotic analysis

Question

Arrange them in increasing order

Comments

$f1<f2 <f4 <f3$

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@Anand how you compared f1,f3 & f4?

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sorry it will be $1,2,4,3$

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Can you please explain how this order come?

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$$f_1=e^{\log (n^{0.99999} \log n)}$$

$$f_1=e^{0.99999 \log n+\log  \log n}$$

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$$f_2=e^{\log (10000000n)}$$

$$f_2=e^{\log 10000000+\log n }$$

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$$f_3=e^{\log (1.000001^{n})}$$

$$f_3=e^{n\log (1.000001)}$$

 

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$$f_4=e^{\log (n^{2})}$$

$$f_4=e^{2 \log (n)}$$

 

Now comparing the powers of $e$,

you will  get the solution.

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@anand yes but between 1 and 2 i think 1 is greater as it has extra log(logn) func..as in your soln..so order shld be 2,1,4,3 is not it..?

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@shubham, yes $1$ has extra $\log \log n$ ,but it does not have complete $\log n$ which $2$ has.

 

$1$ has $0.999999 \log n $.it will become lesser than $log n +\log 1000000$ for very large value of $n$

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ok anand

Answer 1

logarithmic function<polynomial functions<exponential functions(comparison)

f₁(n)=n^0.999999logn

f₂(n)=10000000n

f₃(n)=1.000001ⁿ

f₄(n)=n²

1-we can clearly see that f₃(n) is exponential function and all other are polynomial function so f₃(n) is greatest between all.

2-f₂(n) is less than f₄(n) for sure for larger value of n .

3-now come to f₁(n) and f₂(n) ,cancel the common terms from each now f₁(n)=logn  and f₂(n)=n^.000001 we can clearly see that f₁(n) is logarithmic and f₂(n) is polynomial so f₂(n)>f₁(n) (for large values of n)

from the above discussion we can say that f₁(n)<f₂(n)<f₄(n)<f₃(n).

Comments

chill bhai :D

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@ Anil  between f₁(n)=logn  and f₂(n)=n^.000001 which one is greater? I think due to that fractional power .000001 $f_2$ will give value around 1 (slightly greater) . So, f₂(n)<f₁(n)

 

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f2(n) is greater than  F1(n) because f2(n) is polynomial and F1(n) is logarithmic and polynomial >logarithmic (for large value of n)
We always talk about larger value in asymptotic notation take n=2^(10000000000000000000000000000000000000000 ) like values so this fraction will return into something interesting .