Asymptotics of a solution to an equation.

Question

I have derived an error probability \begin{align*} \frac{1}{\log p}Q\Big((1-\tau)\sqrt{\log p}\Big)+\Big(1-\frac{1}{\log p}\Big)Q\Big(\tau\sqrt{\log p}\Big), \end{align*} where $Q(\cdot)$ in the Q-function defined here. I want to know the scaling of $\tau$ in terms of $p$ that would minimize the expression above (I don't care for the exact solution). What I thought of was to equate the two terms above and solve for $\tau$ but this turns out to be intractable. More precisely, I want to solve (not exactly but just to get the scaling of $\tau$ in terms of $p$) \begin{align} \frac{1}{\log p}Q\Big((1-\tau)\sqrt{\log p}\Big)=\Big(1-\frac{1}{\log p}\Big)Q\Big(\tau\sqrt{\log p}\Big). \end{align} While I'm aware of the approximation $Q(x)\approx e^{-x^2/2}$ for large $x$, I didn't manage to get a scaling of $\tau$ with just that. Is there any other way that I am missing? Thanks.

Answer 1

Using this approximation, you get the (different, approximate) equation $$ \frac{1}{\log p} e^{-\frac{1}{2}(1-\tau)^2\log p} = \left(1-\frac{1}{\log p} \right)e^{-\frac{1}{2}\tau^2\log p} $$ which becomes, reorganizing, $$ e^{-\frac{1}{2}\log p}e^{\tau\log p}e^{-\frac{1}{2}\tau^2\log p} = \left(\log p-1 \right)e^{-\frac{1}{2}\tau^2\log p} $$ i.e., $$ e^{\tau\log p} = e^{\frac{1}{2}\log p}\left(\log p-1 \right) $$ Taking logs and reorganizing again, we get $$ \tau = \frac{1}{2} + \frac{\log \log p}{\log p} + \frac{\log \left(1-\frac{1}{\log p}\right)}{\log p} = \boxed{\frac{1}{2} + \frac{\log \log p}{\log p} - \frac{1}{\log^2 p} + o\!\left(\frac{1}{\log^2 p}\right)} $$ Of course, the third (or even second) term in the expansion might not be meaningful to keep, given that we started with an approximation to begin with.

Answer 2

Hint

For a given $p$, just solve the equation in $\tau$. Generate a table and curve fit it.

For example, using $p=10^k$, you would have $$\left( \begin{array}{cc} k & \tau_k \\ 1 & 0.565213 \\ 2 & 0.688279 \\ 3 & 0.689201 \\ 4 & 0.677200 \\ 5 & 0.664213 \\ 6 & 0.652361 \\ \end{array} \right)$$

Edit

Let $p=e^{2 x^2}$ and the equation to be solved becomes $$\left(1-2 x^2\right) \text{erfc}(\tau x)+\text{erfc}((1-\tau) x)=0$$ Perform a series expansion around $t=\frac 12$ and use series reversion $$t=\frac 12+\frac{\sqrt{\pi } e^{\frac{x^2}{4}} \left(x^2-1\right) \text{erfc}\left(\frac{x}{2}\right)}{2 x^3}+\frac{\pi e^{\frac{x^2}{2}} \left(x^2-1\right)^3 \text{erfc}\left(\frac{x}{2}\right)^2}{8 x^6}+$$ $$\frac{\pi ^{3/2} e^{\frac{3 x^2}{4}} \left(x^2-1\right)^3 \left(2 x^4-4 x^2+3\right) \text{erfc}\left(\frac{x}{2}\right)^3}{48 x^9}+$$ $$\frac{\pi ^2 e^{x^2} \left(x^2-1\right)^5 \left(6 x^4-16 x^2+15\right) \text{erfc}\left(\frac{x}{2}\right)^4}{384 x^{12}}+\cdots$$

$$\left( \begin{array}{ccc} x & \text{estimate} & \text{solution} \\ 2 & 0.683872 & 0.683949 \\ 3 & 0.631964 & 0.634414 \\ 4 & 0.591784 & 0.597100 \\ 5 & 0.565749 & 0.572688 \\ 6 & 0.548799 & 0.556343 \\ 7 & 0.537402 & 0.544968 \\ 8 & 0.529463 & 0.536758 \\ 9 & 0.523751 & 0.530643 \\ 10 & 0.519521 & 0.525966 \\ 11 & 0.516311 & 0.522307 \\ 12 & 0.513821 & 0.519389 \\ 13 & 0.511855 & 0.517022 \\ 14 & 0.510276 & 0.515075 \\ 15 & 0.508990 & 0.513454 \\ 16 & 0.507929 & 0.512087 \\ 17 & 0.507044 & 0.510925 \\ 18 & 0.506299 & 0.509928 \\ 19 & 0.505665 & 0.509065 \\ 20 & 0.505122 & 0.508313 \\ \end{array} \right)$$

If we add a systematic $\sim 0.003$, it seems to be quite decent.

Taking more terms and expanding for large $x$ and back to $p$ $$t=\frac12+\frac{86021}{13860 \log (p)}-\frac{12322516}{51975 \log ^2(p)}+$$ $$\frac{2075034539}{207900 \log ^3(p)}-\frac{57417072928}{155925 \log ^4(p)}+\cdots$$

Answer 3

Let $p=e^{2 x^2}$ and $\tau=t+\frac 12$ and we need to find the zero of function $$f(t)=\left(1-2 x^2\right) \text{erfc}\left(\left(\frac{1}{2}+t\right) x\right)+\text{erfc}\left(\left(\frac{1}{2}-t\right) x\right)\tag 1$$ Expanded as a series around $t=0$, it write $$\color{red}{Q}=\frac{\sqrt{\pi } e^{\frac{x^2}{4}} \left(x^2-1\right) \text{erfc}\left(\frac{x}{2}\right)}{2 x^3}=\sum_{n=0}^\infty (-1)^{n}\, \frac{P_n}{n!}\,t^{n+1}$$ where the first polynomials $P_n(x)$ are $$\left( \begin{array}{cc} n & P_n \\ 1 & 1 \\ 2 & x^2-1 \\ 3 & x^2 \left(x^2-2\right) \\ 4 & x^2 (x^2-1) \left(x^2-6\right) \\ 5 & x^4 \left(x^4-12 x^2+12\right) \\ 6 & x^4 (x^2-1) \left(x^4-20 x^2+60\right) \\ 7 & x^6 \left(x^6-30 x^4+180 x^2-120\right) \\ 8 & x^6 (x^2-1) \left(x^6-42 x^4+420 x^2-840\right) \\ 9 & x^8 \left(x^8-56 x^6+840 x^4-3360 x^2+1680\right) \\ 10 & x^8 (x^2-1) \left(x^8-72 x^6+1512 x^4-10080 x^2+15120\right) \\ \end{array} \right)$$

Using the expansion to $O(t^{16})$ and then power series reversion, $$t= Q \left(1+P_1\,Q+\frac {4P_1^2-P_2 }2Q^2+ \frac {30P_1^3-15P_1P_2+P_3 }6Q^3+\cdots \right)$$ which could be extended to infinity using Morse and Feshbach generalization.

Expanding as a series for large values of $x$ $$\tau=\frac{1}{2}+\frac{1195757}{360360\, x^2}-\frac{1053992873}{12612600\, x^4}+\frac{1797048860741}{756756000\, x^6}-\frac{127690048177157}{2270268000\, x^8}+\frac{2201850337465043}{1945944000\, x^{10}}-\frac{40092528910219199}{1945944000\, x^{12}}+O\left(\frac{1}{x^{14}}\right)$$

What needs to notice is the difference in the coefficients when compared to those given in my previous answer. The reason is that each term of the series reversion affects even the lowest order term.

Some results

$$\left( \begin{array}{ccc} x & \text{estimate} & \text{solution} \\ 10 & 0.526730 & 0.525966 \\ 11 & 0.522831 & 0.522307 \\ 12 & 0.519694 & 0.519389 \\ 13 & 0.517139 & 0.517022 \\ 14 & 0.515035 & 0.515075 \\ 15 & 0.513285 & 0.513454 \\ 16 & 0.511816 & 0.512087 \\ 17 & 0.510572 & 0.510925 \\ 18 & 0.509510 & 0.509928 \\ 19 & 0.508598 & 0.509065 \\ 20 & 0.507808 & 0.508313 \\ \end{array} \right)$$

Back to $p$ $$\color{blue}{\tau=\frac 12 +\frac{1195757}{180180 \log (p)}-\frac{1053992873}{3153150 \log ^2(p)}+\frac{1797048860741}{94594500 \log ^3(p)}+\cdots}$$

Solving the equation for $t$

If we want to solve $(1)$ for $t$ by Newton method $$t_0=\frac 12 \qquad \implies \qquad t_1= \frac{\sqrt{\pi } e^{\frac{x^2}{4}} \left(x^2-1\right) \text{erfc}\left(\frac{x}{2}\right)}{2 x^3}$$ which, by Darboux theorem, is an underestimate of the solution since $\forall x$ $$f\left(0\right)~ <~0 \qquad \text{and} \qquad f\left(0\right)\times f''\left(0\right)~ >~0 $$ So, the solution will be reached without any overshoot as shown below vor $x=10$ $$\left( \begin{array}{cc} n & t_n \\ 0 & 0.0000000 \\ 1 & 0.0097128 \\ 2 & 0.0186786 \\ 3 & 0.0247301 \\ 4 & 0.0259520 \\ 5 & 0.0259658 \\ \end{array} \right)$$