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I am trying to take the following limit $$\lim_{x\to \infty } \, \frac{2 x \operatorname{erfc}\left[\frac{x}{\sqrt{2} t}\right]}{t \operatorname{erfc}\left[-\frac{x}{\sqrt{2}}\right]}$$ my first thoughts were to use LHospital's rule after making the top complementary error function as the denominator of the denominator. Upon differentiating I got the following $$\lim_{x\to \infty } \, \frac{\sqrt{2 \pi } \exp\left[\frac{\left(t^2+1\right) x^2}{2 t^2}\right] \operatorname{erfc}\left[\frac{x}{\sqrt{2} t}\right]^2}{\exp\left[\frac{x^2}{2}\right] \left(\operatorname{erf}\left[\frac{x}{\sqrt{2}}\right]+1\right)+t \exp\left[\frac{x^2}{2 t^2} \right]\operatorname{erfc}\left[\frac{x}{\sqrt{2} t}\right]}$$ which seems even more complicated. Maybe using an identity or some form of expansion for the complementary error function may lead to a simpler expression.

Any help will be greatly appreciated.

Thank you.

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If you look here and here, you will see the asymptotic expansion

$$\operatorname{erfc}(z) = \frac{e^{-z^2}}{z\sqrt{\pi}}\sum_{n=0}^\infty (-1)^n \frac{(2n - 1)!!}{(2z^2)^n}$$ $$\operatorname{erfc}(-z) = 2-\frac{e^{-z^2}}{z\sqrt{\pi}}\sum_{n=0}^\infty (-1)^n \frac{(2n - 1)!!}{(2z^2)^n}$$So, for very large $z$, keeping the first term only makes $$\operatorname{erfc}\left(\frac x {t \sqrt 2}\right) \approx\frac{\sqrt{\frac{2}{\pi }} t e^{-\frac{x^2}{2 t^2}}}{x}$$ $$\operatorname{erfc}\left(-\frac x { \sqrt 2}\right) \approx 2-\frac{\sqrt{\frac{2}{\pi }} e^{-\frac{x^2}{2}}}{x}$$

$$\frac{2 x \operatorname{erfc}\left(\frac{x}{\sqrt{2} t}\right)}{t \operatorname{erfc}\left(-\frac{x}{\sqrt{2}}\right)}\approx \frac{2 \sqrt{\frac{2}{\pi }} e^{-\frac{x^2}{2 t^2}}}{2-\frac{\sqrt{\frac{2}{\pi }} e^{-\frac{x^2}{2}}}{x}} \to \sqrt{\frac{2}{\pi }} e^{-\frac{x^2}{2 t^2}}$$

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By exploiting the continued fraction representation for the complementary error function $$\frac{2x\int_{\frac{x}{t\sqrt{2}}}^{+\infty}e^{-z^2}\,dz}{t\int_{-\frac{x}{\sqrt{2}}}^{+\infty}e^{-z^2}\,dz}\approx \sqrt{\frac{2}{\pi}}e^{-\frac{x^2}{2t^2}}$$ hence the wanted limit is simply zero. You may also use concentration (tail) inequalities for the normal distribution (thanks to M.Spivey) or Hoeffding's inequality.

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  • $\begingroup$ Thank you, i do not quite understand how the continued fractions were cancel. $\endgroup$ – Comic Book Guy Sep 7 '16 at 7:36
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The denominator tends to $t$. Depending on the sign of $t$, the numerator tends to $\infty$ or $0$, because the complementary error function tends to $1$ or $0$, with a super-exponential rate.

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