5
$\begingroup$

Can someone help me finding a lower bound to the function $$f(x)=\frac{x-1}{e^{-1}-xe^{-x^2}},$$ where $x\in[1,+\infty[$?

Taking the derivative and then solve $f'(x)=0$ isn't analytically possible. Then I tried the second best thing, find a lower bound but I don't really know how to start, so any help would be most welcome.

$\endgroup$
  • 2
    $\begingroup$ Are you seeking to find a tight lower bound? $\endgroup$ – user40314 Sep 26 '13 at 15:27
  • $\begingroup$ @user40314 In the sense that the bound should be close to the actual minimum?Well, yes. $\endgroup$ – PML Sep 26 '13 at 15:30
  • $\begingroup$ Alpha finds $\approx 2.33815$ at $x \approx 1.34929$ $\endgroup$ – Ross Millikan Sep 26 '13 at 17:44
  • $\begingroup$ Since $$-\frac{1}{f(x)}=\frac{xe^{-x^2}-e^{-1}}{x-1},$$ we have that if $f$ obtains its lowest value at $x=a$ then the line connecting $(1,e^{-1})$ and $(a,ae^{-a^2})$ is tangent to the graph of $xe^{-x^2}$ at $a$. From here we get $$1-2a^2+2a^3=e^{a^2-1}.$$ That $\approx 1.34929$ is the only root of this equation with $a > 1$. $\endgroup$ – njguliyev Sep 26 '13 at 20:40
  • $\begingroup$ @njguliyev Your last result of $f(\approx 1.34929)$ being the minimum is taken from where? Numerical methods? I was hoping to find a lower bound from analysis. $\endgroup$ – PML Sep 26 '13 at 21:17
2
$\begingroup$

Following njguliyev's comment, consider
$$ \frac{1}{f(x)}=\frac{g(x)-g(1) }{x-1}\tag{1}$$ where $g(x)=-xe^{-x^2}$. By the Mean value theorem, every value of $1/f$ on $(1,\infty)$ is also attained by $g'$ on $(1,\infty)$. So we need an upper bound on $g'$. Since $$g'(x) = (2x^2-1)e^{-x^2}$$ $$g''(x) = 2x(3-2x^2)e^{-x^2}$$ it follows that $g'$ is maximal when $x^2=3/2$. Thus, the maximum of $g'$ on $(1,\infty)$ is $2e^{-3/2}$. This gives an upper bound on (1), and consequently $$f(x) \ge \frac{e^{3/2}}2 ,\quad x\ge 1$$ This lower bound is $\approx 2.24$, not far from the minimum of $\approx 2.34$ found in comments numerically.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.