test series

Question

Arrange the following functions in asymptotically increasing order

f1(n) = n0.999999 log n

f2(n) = 10000000n

Please explain your solution. Thanks

Comments

f1(n) < f2(n)

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why so? f1 is approx nlogn while f2 is n

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@Manu

no, f2(n) > f1(n) is correct, for small values of 'n'...it seems that f1(n) is big but as you go towards infinity f2(n) grow tremendously..

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yes @nitish you're correct. i overlooked the problem but now i have a proof that f2(n)> f1(n)

Answer 1

$f1(n) = n^{0.999999} log n \\ f2(n) = 10000000n$

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Convert both functions in such a way that they are comparable.

$f1(n) = n^{0.999999} log n  =10000000 * n^{0.999999} * logn * \frac{1}{10000000}\\f2(n) = 10000000n =10000000 \ * \ n^{0.999999} \ * \ n^{0.000001} $

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Let $10000000 * n^{0.999999}  = X$

$f1(n) = X * logn * \frac{1}{10000000}\\f2(n) = X \ * \ n^{0.000001} $

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For n=$2^{2^{10}}$

$f1(n) = 2^{10} * \frac{1}{10000000} = 1024 * 10^{-7}\\f2(n) = 2^{0.001024} = 1.000710035 $

$\therefore f2(n)>f1(n) $

Note : Using one value would not ensure the function plot but definetly hint something about function .

Comments

see my comment above, i have given less mathematical proof for this Problem.