Consider this function with the complementary error function $\mathrm{erfc}(x)$ $$e^{2x} \mathrm{erfc}\left(\sqrt{2} x\right) - \frac{1}{2} e^{\frac{1}{2}x} \mathrm{erfc}\left(\frac {x}{\sqrt{2}}\right)$$ I am interested in finding the minimum and maximum of this function for $x \geq 1$, but I am not sure how I should approach this problem. Intuitively I would compute the derivative and set it to zero, but I have no idea how to get values for $x$ out of this thing $$2 e^{2x} \mathrm{erfc}\left(\sqrt{2} x\right) - \frac{4}{\sqrt{2 \pi}}e^{2x - 2x^2} - \frac{1}{4} e^{\frac{1}{2}x} \mathrm{erfc}\left(\frac {x}{\sqrt{2}}\right) + \frac{1}{\sqrt{2 \pi}} e^{\frac{1}{2} x - \frac{1}{2} x^2} = 0$$ I thought, this might be possible by using the bounds for $x > 0$ that are mentioned here, but I got stuck on something like $$4 + \frac{3}{2} x - \frac{3}{2} x^2 + \ln\left(\frac{1}{x + \sqrt{x^2 + 1}} - 1\right) - \ln\left(\frac{1}{x + \sqrt{x^2 + 4}} - 1\right) = 0$$

Would anyone have an idea how to find the minimum or the maximum of this function for $x \geq 1$? I am open both for analytical and numerical approaches.

  • $\begingroup$ for what stands $$erfc$$? $\endgroup$ – Dr. Sonnhard Graubner Dec 23 '16 at 18:43
  • $\begingroup$ @Dr.SonnhardGraubner Complementary error function $\endgroup$ – Mr Tsjolder Dec 23 '16 at 18:53
  • $\begingroup$ @Dr.SonnhardGraubner $ erfc(x) = \frac{1}{\sqrt(2)} \int_{x}^{\infty} e^{-t^2}dt $ $\endgroup$ – Rebellos Dec 23 '16 at 18:55
  • $\begingroup$ @Dr.SonnhardGraubner I edited my question and added a link $\endgroup$ – Mr Tsjolder Dec 23 '16 at 18:57
  • $\begingroup$ is it $$\frac{1}{\sqrt{2}}\int_{\pi}^{\infty}e^{-t^2}dt$$? $\endgroup$ – Dr. Sonnhard Graubner Dec 23 '16 at 18:57

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