Find-the-superimum-of-the-set-n-2-2-n-

Question Number 74554 by shubham90412@gmail.com last updated on 26/Nov/19

F i n d t h e s u p e r i m u m o f t h e s e t {\frac{n^{2}}{2^{n}}}

Answered by MJS last updated on 26/Nov/19

trying

S = {0, \frac{1}{2}, 1, \frac{9}{8}, 1, \frac{25}{32}, \frac{9}{16}, \dots}

calculating

f (x) = \frac{x^{2}}{2^{x}}

f^{'} (x) = \frac{x}{2^{x}} (2 - x \ln 2)

\frac{x}{2^{x}} (2 - x \ln 2) = 0 \Rightarrow x = 0 \lor x = \frac{2}{\ln 2} \approx 2.89

f^{″} (x) = \frac{{(\ln 2)}^{2} x^{2} - 4 x \ln 2 + 2}{2^{x}}

f^{″} (0) > 0 \Rightarrow x = 0 is minimum

f^{″} (\frac{2}{\ln 2}) < 0 \Rightarrow x = \frac{2}{\ln 2} is maximum

\Rightarrow we have to try n = ⌊ \frac{2}{\ln 2} ⌋ = 2 \lor n = ⌈ \frac{2}{\ln 2} ⌉ = 3

\Rightarrow supremium at n = 3

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