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I'm trying to derive the solutions to $x*e^{-x^3} = \epsilon$ using perturbation method. From the equation, I got

$x^3 = \ln{1/{\epsilon}} + \ln{x}$

Set $L_1 = \ln{1/{\epsilon}}$,

it turned to be $x^3 = L_1 + \ln{x}$

I know I can solve $x = L_1 + \ln{x}$ using iteration method, but how can I deal with $x^3$?

Thanks!

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When your write $\ln(x)-x^3=\ln(\epsilon)$ there are two natural cases to consider, $\ln(x)\approx\ln(\epsilon)$ and $x^3\approx-\ln(\epsilon)$, which give $x=\epsilon$ and $x=-\ln(\epsilon)^{1/3}$.

You can verify that if $x=\epsilon$ then $|x^3|\ll|\ln(x)|$, and that if $x=-\ln(\epsilon)^{1/3}$ then $|\ln(-\ln(\epsilon)^{1/3})|\ll|\ln(\epsilon)|$ as $\epsilon\rightarrow0$.

So this gives the leading terms in the approximations for each root, $\epsilon$ and $-\ln(\epsilon)^{1/3}$.

For the iteration for the root around $-\ln(\epsilon)^{1/3}$, you can use almost the same iteration as for $xe^{-x}$ case, $$ x^3=\ln(x)+\ln(1/\epsilon)=\ln(x/\epsilon)\Rightarrow x_{n+1}=\big(\ln(x_n/\epsilon)\big)^{1/3}.$$

For the other root near 0, you can use, starting with $x_0=\epsilon$, $$ \ln(x_{n+1})=\ln(\epsilon)+x_n^3\Rightarrow x_{n+1}=\epsilon e^{x_n^3}.$$

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You may not transform it to the log function since it diverges for small values. Instead, let

$$f(x) = xe^{-x^3} - \epsilon$$

Assume $\epsilon$ is small, so is the root. So approximate the function as

$$f(x) = x(1-x^3)-\epsilon$$

The leading order solution is $x=\epsilon$. The perturbed solution including the next order is,

$$x=\frac{\epsilon}{1-\epsilon^3} =\epsilon + \epsilon^4 $$

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One way to proceed to find solutions of $\, x\,e^{-x^3}=y\,$ is to transform the equation into a known equation whose solution is the Lambert $W$ function. Thus, $\,-(-3x^3)e^{(-3x^3)} = -3y^3.\,$ The solution of this is $\,-3x^3 = W(-3y^3).\,$ Thus, $\, x = (-\frac13 W(-3y^3))^{1/3}$ The known theory of the Lambert $W$ function can supply the needed solutions. For example, for $\,y\,$ near $\,0\,$ the series expansion is $$ x = y + y^4 + 7y^7/2! + 100 y^{10}/3! + \cdots + (3n+1)^{n-1} y^{3n+1}/n! + \cdots. $$ The OEIS sequence A052752 is the sequence of coefficients and the connection with Lambert $W$ is mentioned in the OEIS entry.

You can use iteration with the equation $\, x = y\,e^{x^3}\,$ with $\,x_0 := y.\,$ The recursion $\,x_{n+1} := y e^{x_n^3},\,$ gives $$ x_1 = y+y^4+O(y^7),\;\; x_2 = y+y^4+7x^7/2!+O(y^{10}),\;\;\dots. $$

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