One has taken some positive numbers and built a sequence: $a_1$ is the sum of the numbers; $a_2$ is the sum of the squares of the numbers, $a_3$ is the sum of the cubes of the numbers, etc. Could be so that before $a_5$ $ a_1>a_2>a_3>a_4 > a_5$ and after $a_5$ $a_5 < a_6 < a_7 <...$?

  • $\begingroup$ I would take two numbers $1-\epsilon _1, 1+\epsilon _2$ and try to tune the values of $\epsilon$s to meet the criteria. $\endgroup$ – didgogns Mar 12 '17 at 14:11
  • $\begingroup$ And if it is not possible? $\endgroup$ – idliketodothis Mar 12 '17 at 14:19
  • 3
    $\begingroup$ See this link. $\endgroup$ – didgogns Mar 12 '17 at 14:20
  • $\begingroup$ Your answer generalizes nicely. $\endgroup$ – marty cohen Mar 12 '17 at 16:18
  • $\begingroup$ @didgogns, how did you realise that these numbers are true? $\endgroup$ – idliketodothis Mar 12 '17 at 17:03

Generalizing didgogns' nice solution, if $f(x) =e^{-ax}+e^{bx} =(e^{-a})^x+(e^b)^{x} $, where $a,b > 0$, $f'(x)=-ae^{-ax}+be^{bx}$ so that $f(x)$ has a unique minimum when $ae^{-ax}=be^{bx}$ or $\ln a-ax=\ln b+bx$ or $x =\frac{\ln a-\ln b}{a+b} =\frac{\ln (a/b)}{a+b} =\frac1{b}\frac{\ln (a/b)}{a/b+1} $.

By choosing $a$ and $b$ properly, we can get $x$ to take on any desired value so that $f$ decreases to there and increases after.

For example, if $a/b=e$ then $x=\frac1{b}\frac{1}{e+1}$ so choosing $b=\frac1{x(e+1)}$ and $a=be$ works.


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