Approximately solve the equation. Find the first two terms of the approximation.

Question

Approximately solve the equation. Find the first two terms of the approximation.

By $a>>1$ and $a<<1$

$\ln x=e^{-ax}$

$x=e^{e^{-ax}}$

$x_{0}\sim1$

$x_{1}=e^{e^{-a}}$

$\left|e^{e^{-a}}-1\right|<<1$

$x=1+e^{-a}+\frac{e^{-2a}}{2}$

Have I considered one case correctly and how to proceed for the second?

Answer 1

I got a bit different answer. When we use the perturbation theory to find an approximate solution, we search for the solution in the form $x=x_0+x_1+x_2+...$. We also have to check at every iteration that $x_0\gg x_1\gg x_2 ...$

1. $\ln x=e^{-ax},\,\, a\gg1$.

We are looking for the solution in the form $x=1+\epsilon; \,\,\epsilon\ll1$ $$\ln(1+\epsilon)=\epsilon-\frac{\epsilon^2}{2}+ ... =e^{-a}e^{-a\epsilon}$$ Let's also suppose that $a\epsilon\ll1$ (we will have to check whether it is true). Decomposing the exponent $$=\epsilon-\frac{\epsilon^2}{2}+ ... =e^{-a}(1-a\epsilon+\frac{a^2\epsilon^2}{2}-...)$$ Taking in turn $\epsilon =e^{-a}+\delta; \,\,\delta\ll e^{-a}$ (we also see that the requirement $a\epsilon\sim ae^{-a}\ll1$ is met for $a$ big enough). $$e^{-a}+\delta-\frac{(e^{-a}+\delta)^2}{2}+...=e^{-a}\Big(1-a(e^{-a}+\delta)+\frac{a^2(e^{-a}+\delta)^2}{2}-...\Big)$$ In this equation we have to keep all terms $\sim e^{-2a}$ $$\delta-\frac{e^{-2a}}{2}-e^{-a}\delta+ ... =-ae^{-2a}-ae^{-a}\delta + ...$$ $\delta \ll e^{-a}$, so we have to drop in the equation all the terms $\sim e^{-a}\delta$ $$\delta-\frac{e^{-2a}}{2}=-ae^{-2a}\,\,\Rightarrow\,\, \delta=e^{-2a}\Big(\frac{1}{2}-a\Big)\ll e^{-a}$$ Therefore, for $\mathbf{a\gg1}$ the solution looks $$\boxed{x=1+e^{-a}+e^{-2a}\Big(\frac{1}{2}-a\Big)+o(e^{-2a})}$$ 2. $\ln x=e^{-ax},\,\, a\ll1$ We are looking for the solution in the form $x=e+\epsilon; \,\,\epsilon\ll1$ $$\ln(e+\epsilon)=\ln e+\ln\Big(1+\frac{\epsilon}{e}\Big)=1+\frac{\epsilon}{e}-\frac{\epsilon^2}{2e^2}+...=1-a(e+\epsilon)+\frac{a^2}{2}(e+\epsilon)^2+...$$ Choosing in turn $\epsilon=-ae^2+\delta,\, \delta\ll a$, and keeping only the terms $\sim a^2$, we get the equation for $\delta$ $$\frac{\delta}{e}-\frac{a^2e^4}{2e^2}=a^2e^2+\frac{a^2}{2}e^2\,\,\Rightarrow\,\, \delta=2e^3a^2$$ For $\mathbf{a\ll1}$ the solution looks $$\boxed{x=e-e^2a+2e^3a^2+o(a^2)}$$

Answer 2

Considering that we look for an approximation of the zero of function $$f(x)=\log(x)-e^{-ax}$$ if $a \gg 1$, build the series expansion around $x=1$ and use series reversion to obtain $$\color{red}{x=1+t+\frac{\left(a^2+e^a\right) }{2 \left(a+e^a\right)}t^2+O\left(t^3\right)}\quad \text{with}\quad \color{red}{t= \frac{1 }{a+e^a}}$$

A few examples $$\left( \begin{array}{ccc} a & \text{estimation} & \text{solution} \\ 1 & 1.305106165 & 1.309799586\} \\ 2 & 1.113387034 & 1.113808277 \\ 3 & 1.044499191 & 1.044524585 \\ 4 & 1.017240818 & 1.017241979 \\ 5 & 1.006542360 & 1.006542410 \\ 6 & 1.002445628 & 1.002445628 \end{array} \right)$$

For sure, if $a$ is really large, we can neglect the terms in $a$ and the above would reduce to $$\color{blue}{x=1+e^{-a}+\frac{1-2 a}{2} e^{-2 a}-\frac{30 a^2+9 a-1}{6} e^{-3 a}+\cdots}$$

Now, for $a\ll 1$ doing the same around $x=e$ leads to $$\color{red}{x=e+t+\frac{\left(e^2 a^2+e^{e a}\right) }{2 e \left(e a+e^{e a}\right)}t^2+O\left(t^{3}\right)}\quad \text{with}\quad \color{red}{t=\frac{e \left(1-e^{e a}\right)}{e a+e^{e a}}} $$

Let $a=5^{-k}$ and some results $$\left( \begin{array}{ccc} k & \text{estimation} & \text{solution} \\ 1 & 1.974782640 & 1.964280299 \\ 2 & 2.474136359 & 2.473869824 \\ 3 & 2.661616514 & 2.661612633 \\ 4 & 2.706561144 & 2.706561108 \\ 5 & 2.715921436 & 2.715921436 \end{array} \right)$$

and,expanding the formula for small value of $a$ $$\color{blue}{x=e-e^2 a+2 e^3 a^2-\frac{14 }{3}e^4a^3+\cdots}$$

The $\color{red}{\text{red formulae}}$ are better than the $\color{blue}{\text{blue formulae}}$

Edit

If you want an almost exact solution, perform one single iteration of Householder method.

Using $x_0=1$, this will give $$x_1=1+\frac{3 \left(a^2+(4 a-1) e^a+2 e^{2 a}\right)}{a^3+2 \left(6 a^2-3 a+1\right) e^a+6(3 a-1) e^{2 a}+6 e^{3 a}}$$which, for $a=1$ gives $x_1=1.309573776$.

Using $x_0=e$, this will give $$x_1=e-\frac{3 e \left(e^{e a}-1\right) \left(e^2 a^2+3 e^{2 e a}+e^{e a} (e a (e a+4)-1)\right)}{e^3 a^3+14 e^{3 e a}+2 e^{e a} (e a (2 e a (e a+3)-3)+1)+e^{2 e a} (e a (e a (e a+6)+24)-10)}$$which, for $a=\frac15$ gives $x_1=1.963434019$.