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I am trying to polish my second answer to this question in Mathematics Stack Exchange.

The problem is to find the asymptotics of $t$, solution of the implicit equation $$\color{blue}{\left(1-2 x^2\right) \text{erfc}\left(\left(\frac{1}{2}+t\right) x\right)+\text{erfc}\left(\left(\frac{1}{2}-t\right) x\right)=0}\tag 1$$ for large $x >0 $ (then small values of $t$).

I had no problem to arrive to the fact that it is equivalent to find $t$ solution of $$Q=\frac{\sqrt{\pi }\,\, e^{\frac{x^2}{4}} \left(x^2-1\right) \text{erfc}\left(\frac{x}{2}\right)}{2 x^3}=\sum_{n=0}^\infty (-1)^{n}\, \frac{P_n}{n!}\,t^{n+1}\tag 2$$ where $P_n$ is a polynomial of degree $2n$ in $x$.

Now, the problem is related to the power series reversion of $(2)$ truncated to $O(t^{p+1})$

$$t_{(p)}=Q\sum_{n=0}^p T_n\,Q^n \tag 3$$

which would be followed by the series expansion for large $x$ to have $$t_{(p)}=\sum_{n=1}^p \frac {a_n}{x^{2n}}\tag 4$$

The problem is that the leading order of $T_n$ is $x^{2n}$ which makes that $$x^{2n}\,Q^n=1-\frac{3 n}{x^2}+O\left(\frac{1}{x^4}\right)$$ This means that all coefficients $a_n$ depend on $p$. For example, as a function of $p$, coefficient $a_1$ form the sequence $$\left\{1,\frac{3}{2},\frac{11}{6},\frac{25}{12},\frac{137}{60 },\frac{49}{20},\frac{363}{140},\frac{761}{280},\frac{7129 }{2520},\cdots\right\}$$ which seems to correspond to the harmonic number $H_n$. The problem is much worse with the next coefficients.

It seems that I would need to use very large $p$ to obtain a good asymptotics of the asymptotics if I stay with this procedure.

Similarly, using the first iterate of Newton-like methods of order $n$ and continuing witge series expansion $$\left( \begin{array}{cccc} n & a_1 & a_2 & a_3\\ 2 & 1 & -3 & 14 \\ 3 & 2 & -14 & 118 \\ 4 & 3 & -45 & 834 \\ 5 & 4 & -140 & 6604 \\ 6 & 5 & -455 & 59510 \\ 7 & 6 & -1530 & 576450 \\ 8 & 7 & -5201 & 5759978 \\ \end{array} \right)$$

Is there any way to do it even totally changing this process ? Any idea or suggestion would be very welcome.

Edit

Using the approximation $$ \text{erfc}(x)\sim \frac{e^{-x^2}}{x\sqrt{\pi }}$$ we face the problem of solving for $t$ $$\color{blue}{e^{-2 x^2 t}=\frac 1{1-2x^2}\,\,\frac{2t+1}{2t-1}}\tag 5$$ which has an explicit solution in terms of the generalized Lambert function.

This could hide a logarithmic contribution somewhere.

As shown below, the solution of $(5)$ is a quite good approximation of the exact solution. $$\left( \begin{array}{ccc} x & \text{sol. of }(5)& \text{sol. of }(1)\\ 3 & 0.1282499 & 0.1344144 \\ 4 & 0.0952576 & 0.0971005 \\ 5 & 0.0720334 & 0.0726883 \\ 6 & 0.0560754 & 0.0563435 \\ 7 & 0.0448454 & 0.0449680 \\ 8 & 0.0366964 & 0.0367576 \\ 9 & 0.0306099 & 0.0306428 \\ 10 & 0.0259471 & 0.0259658 \\ \end{array} \right)$$

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  • $\begingroup$ How did you get formula two, which separates t,x? I would expect additional assumptions concerning O[x], O[t] $\endgroup$ Jan 2, 2023 at 9:47
  • $\begingroup$ @UlrichNeumann. Just expanding $(1)$ as a series around $t=0$. Term $Q$ is just the first term coefficient divided by the second. $\endgroup$ Jan 2, 2023 at 9:53
  • $\begingroup$ Thanks. Assuming t->0 and x->Infinity I'm missing information concerning Asymptotic of x t. $\endgroup$ Jan 2, 2023 at 9:59
  • $\begingroup$ @UlrichNeumann. $(3)$ is simple and rigorous. Now, the problem comes when going from $(3)$ to $(4)$. May be, I took a wrong approach. Cheers :-) $\endgroup$ Jan 2, 2023 at 10:07
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    $\begingroup$ Try: Reduce[Exp[-2*x^2 *t] == 1/(1 - 2 x^2)*(Series[(2 t + 1)/(2 t - 1), {t, 0, 1}] // Normal), t] then we have; $$t=-\frac{1}{4}+\frac{W\left(\frac{1}{2} e^{\frac{x^2}{2}} x^2 \left(-1+2 x^2\right)\right)}{2 x^2}$$ $\endgroup$ Jan 3, 2023 at 13:36

2 Answers 2

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Solving for $t$ $$\color{blue}{e^{-2 x^2 t}=\frac 1{1-2x^2}\,\,\frac{2t+1}{2t-1}}\tag 5$$ We use the approximation $\frac{2t+1}{2t-1}$ at zero.

aprox = Series[(2 t + 1)/(2 t - 1), {t, 0, 1}] // Normal

R = Reduce[Exp[-2*x^2 *t] == 1/(1 - 2 x^2)*aprox, t]

R[[3, 4]] /. C[1] -> 0 // Expand

(*t == -(1/4) + ProductLog[2 E^(x^2/2) x^2 (-(1/4) + x^2/2)]/(2 x^2)*)
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  • $\begingroup$ Great solution ! For the last table in my answer, this gives $$\{0.133613,0.0970609,0.0727293,0.0563794,0.0449921,0.0367731,0.0306 528,0.0259723\}$$ $\endgroup$ Jan 3, 2023 at 14:33
  • $\begingroup$ You are welcome. :) $\endgroup$ Jan 3, 2023 at 14:37
  • $\begingroup$ Equation (5) might be solved analytically (exact) too : x == Sqrt[(t - ProductLog[-1, -((E^t t (1 + 2 t))/(1 - 2 t))])/(2 t)] $\endgroup$ Jan 3, 2023 at 16:11
  • $\begingroup$ @UlrichNeumann . OP want solving for t not for x ? $\endgroup$ Jan 3, 2023 at 16:14
  • $\begingroup$ @MariuszIwaniuk That is of course true, but your solution doesn't work if you take higher order series expansion for aprox! $\endgroup$ Jan 3, 2023 at 16:21
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After @MariuszIwaniuk's comment and answer

$$e^{-2 x^2 t}=\frac 1{1-2x^2}\,\,\frac{2t+1}{2t-1}\sim \frac {1+4t}{2x^2-1}\quad \implies \quad t=-\frac 14+\frac 1{2x^2}W\left(\frac{1}{2} e^{\frac{x^2}{2}} x^2 \left(2 x^2-1\right)\right)$$ Using the expansions for large argument $$L_1=\log\left(\frac{1}{2} e^{\frac{x^2}{2}} x^2 \left(2 x^2-1\right)\right)=\frac {x^2}2+4 \log (x)-\frac{1}{2 x^2}+O\left(\frac{1}{x^4}\right)$$ $$L_2=\log \left(\log\left(\frac{1}{2} e^{\frac{x^2}{2}} x^2 \left(2 x^2-1\right)\right)\right)=2\log(x)-\log(2)+\frac{8 \log (x)}{x^2}+O\left(\frac{1}{x^4}\right)$$ $$L_1-L_2+\frac {L_2}{L_1}=\frac{x^2}2+2\log(x)+\log(2)-\frac{8 \log (x)+4 \log (2)+1}{2 x^2}$$

$$\color{blue}{t=\frac{\log (x)}{x^2}+\frac{\log (2)}{2 x^2}-\frac{2 \log (x)}{x^4}-\frac{4 \log (2)+1}{4 x^4}+\cdots}$$ Using the exact solutions of the original equation $$\left(1-2 x^2\right) \text{erfc}\left(\left(\frac{1}{2}+t\right) x\right)+\text{erfc}\left(\left(\frac{1}{2}-t\right) x\right)=0$$ and curve fitting the model $$t=a\frac{\log (x)}{x^2}+\frac{b}{2 x^2}-c \frac{ \log (x)}{x^4}-\frac{d}{4 x^4}$$ for $10 \leq x \leq 100$

$$\begin{array}{l|lll} \text{} & \text{Estimate} & \text{Std Error} & \text{Confidence Interval} \\ \hline a & \color{red}{+0.99945} & 0.000010 & \{+0.99943,+0.99947\} \\ b & \color{red}{+0.69808} & 0.000084 & \{+0.69791,+0.69824\} \\ c & +2.57442 & 0.003419 & \{+2.56770,+2.58114\} \\ d & -2.19752 & 0.024778 & \{-2.24625,-2.14879\} \\ \end{array}$$

The mean and maximum absolute errors are $1.29\times 10^{-8}$ and $1.09\times 10^{-7}$ $\color{red}{\large (!!)}$.

Using

$$t_0=\frac{\log (x)}{x^2}+\frac{\log (2)}{2 x^2}$$ the first iterate of Newton method applied to the original equation gives (again) $$t_1=\frac{\log (x)}{x^2}+\frac{\log (2)}{2 x^2}-\frac{2 \log (x)}{x^4}-\frac{4 \log (2)+1}{4 x^4}$$ which, by Darboux theorem, is an underestimate of the solution (for $x=10$, $t_1=0.0259368$ while the solution is $t=0.0259658$) while $t_0$ is an overestimate of it.

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