2 votes 2 votes Algorithms algorithms asymptotic-notation test-series + – dragonball asked Oct 15, 2017 • retagged Jul 13, 2022 by makhdoom ghaya dragonball 938 views answer comment Share Follow See all 2 Comments See all 2 2 Comments reply sourav. commented Oct 15, 2017 reply Follow Share $(A)?$ 0 votes 0 votes dragonball commented Oct 15, 2017 reply Follow Share Plz explain with solution . 0 votes 0 votes Please log in or register to add a comment.
Best answer 2 votes 2 votes $f(n)=n^{\frac{1}{\sqrt{\log n}}}= e^{\log n^{\frac{1}{\sqrt{\log n}}}}=e^{\sqrt{\log n}}$ $g(n)=\sqrt{\log n}=e^{\log \sqrt{\log n}}$ $h(n)=n^{\frac{1}{100}}=e^{\log n^{\frac{1}{100}} }=e^{k \times \log n}$ $g(n)\, < \,f(n) \, <h(n)$ sourav. answered Oct 15, 2017 • selected Oct 15, 2017 by dragonball sourav. comment Share Follow See all 11 Comments See all 11 11 Comments reply dragonball commented Oct 15, 2017 reply Follow Share can u tell me what is the wrong with this solution ? 0 votes 0 votes sourav. commented Oct 15, 2017 reply Follow Share it will be better to derive the asymptotic function in simplest term as you can .after you are done and if you have enough time,you can put large values to verify .don't put larger value to verify the answer in your first attempt, it might be risky and you may lose your marks :) 1 votes 1 votes dragonball commented Oct 15, 2017 reply Follow Share Thanks . But for all other questions my solution didn't went wrong by choosing a large number , but for this question only it went wrong . Why this happened ? Secondly, what is the wrong with my solution as depicted above . Kindly help so that in future I could apply the things judiciously . 0 votes 0 votes sourav. commented Oct 15, 2017 reply Follow Share you have done right, but i am not getting why you have written option $D$? you too are getting $A$ 0 votes 0 votes dragonball commented Oct 15, 2017 reply Follow Share For h(n) it is 10.24 not 1024 . So the larger value is f(n) and not h(n) . Little confused. 0 votes 0 votes sourav. commented Oct 15, 2017 reply Follow Share I thought it as $1024$.As you know if $f(n)=O(g(n))\Rightarrow f(n) \leq c \times g(n) ,\text{where }n > n_0$ sometimes when you put larger value then you may lie the within the range of $n_0$,perfect example is the case you did.That's the reason to break the asymptotic function in simpler terms. Well you try with $2^{65,536}$ you will get you answer 2 votes 2 votes dragonball commented Oct 15, 2017 reply Follow Share Thanks . It helped a lot. But a liitle confusion . I have solved almost more than 95 percent of the problem using this method only and I got correct answer . But this case failed . Could u tell me the best approach so that it must be true in each and every case. 0 votes 0 votes sourav. commented Oct 15, 2017 reply Follow Share i have told you to simplify it in simpler terms see comment above 1 votes 1 votes dragonball commented Oct 15, 2017 reply Follow Share Thank u :) 0 votes 0 votes rajatmyname commented Apr 15, 2018 reply Follow Share But I think if you compare f(n) and h(n), f(n) is greater because it will have 1/(logn)^1/2 term whereas h(n) will contain a constant term 1/100 0 votes 0 votes Ashutosh_k commented Apr 11, 2019 reply Follow Share I am facing problems in logarithmic conversions. Like how you did the last step of simplifying f(n). Please suggest resources to read about these formulae. 0 votes 0 votes Please log in or register to add a comment.