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I have to resolve with respect $K$ the following inequality \begin{equation*}\frac{1}{1-\frac{e_k}{2}}-1\leq \tau\end{equation*} where $\tau$ is given and \begin{equation*}e_K\triangleq \textrm{erfc}\left(\sqrt{\frac{K}{2}}\right)\end{equation*} and $\textrm{erfc}(\cdot)$ is the complementary error function. The intuition says that the solution must be something like \begin{equation*}K\geq f(\tau)\end{equation*} for some suitable function $f(\cdot)$. The problem is that I endup finding the following solution \begin{equation*} K \leq 2\,\left(\textrm{ierfc}\left(\frac{2\tau}{\tau+1}\right)\right)^2 \end{equation*} where $\textrm{ierfc}(\cdot)$ is the inverse of the complementary error function. The derivation is pretty trivial and I don't really understand where I'm making the mistake.

EDIT I provide more details about my derivation. The starting inequality is equivalent to \begin{equation*}\textrm{erfc}\left(\sqrt{\frac{K}{2}}\right) \leq \frac{2\tau}{\tau+1}\end{equation*} now, applying the inverse erf to both members gives \begin{equation*}\sqrt{\frac{K}{2}} \leq \textrm{ierfc}\left(\frac{2\tau}{\tau+1}\right)\end{equation*} and so \begin{equation*}K\leq 2 \left(\textrm{ierfc}\left(\frac{2\tau}{\tau+1}\right)\right)^2\end{equation*}

Is it possible that when I pass from $\textrm{erfc}(\cdot)$ to $\textrm{ierfc}(\cdot)$ the verse of the inequlity must be changed for some reason?

matteogost
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1 Answers1

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Using the error function, the equality write $$\text{erf}\left(\sqrt{\frac{K}{2}}\right)=\frac{\tau -1}{\tau +1}$$ For a rather good approximation, have a look here $$\text{erf}(x)\sim \sqrt{1-\exp\Big(-\frac 4 {\pi}\,\frac{1+\alpha\, x^2}{1+\beta \,x^2}\,x^2 \Big)}$$ where $$\alpha=\frac{10-\pi ^2}{5 (\pi -3) \pi } \qquad \text{and} \qquad \beta=\frac{120-60 \pi +7 \pi ^2}{15 (\pi -3) \pi }$$

Squaring $$\frac{2 K (\alpha K+2)}{\pi (\beta K+2)}=\log \left(\frac{(\tau +1)^2}{4 \tau }\right)$$ which is a quadratic equation in $K$.

Claude Leibovici
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