# Find perturbation solutions to the real roots of $xe^{-x^3} = \epsilon$

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!

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 $$x_0=\epsilon$$, $$\ln(x_{n+1})=\ln(\epsilon)+x_n^3\Rightarrow x_{n+1}=\epsilon e^{x_n^3}.$$

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,

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

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 $$x_1 = y+y^4+O(y^7),\;\; x_2 = y+y^4+7x^7/2!+O(y^{10}),\;\;\dots.$$