Asymptotic expansion of $x\exp(1/x)=\exp(\lambda)$ as $\lambda\to\infty$

Question

I am struggling to find a three-term expansion of the following equation $x\exp(1/x)=\exp(\lambda)$ as $\lambda\to\infty$ for each of the solutions for $x$. I graphed the functions $1/x$ and $\log(x)$ after taking the logarithm on both sides of the equation. I know, correct me if I am wrong, that for small root I get the $x\sim 1/\lambda$ balance, while for the large root the balance is $\log(x)\sim\lambda$. Please guide me on how to proceed from here.

Answer 1

We first can solve the equation exactly as follows: $$ xe^{1/x}=z\implies \frac{1}{x}e^{-1/x}=\frac{1}{z}\implies -\frac{1}{x}e^{-1/x}=-\frac{1}{z}. $$ It follows that $$ x=-\frac{1}{W_0(-1/z)}\ \text{or}\ x=-\frac{1}{W_{-1}(-1/z)}, $$ where $W_0(\cdot)$ is the principal branch of the Lambert $W$-function. In your question you have $z=e^\lambda$, where $\lambda\to\infty$ and so we want the solution for $x$ in terms of $W_0(\cdot)$. As such, we can use the Taylor series for $W_0(\cdot)$ to obtain the desired asymptotic expansion. First note $$ W_0(s)=s\sum_{n=1}^\infty\frac{(-(n+1))^n}{(n+1)!}s^n. $$ Using the formula for the multiplicative inverse of formal power series we write $$ \frac{1}{W_0(s)}=\frac{1}{s}\left(1+s-\frac{1}{2}s^2+\mathcal O(s^3)\right). $$ Substituting $s=-1/z$ we then obtain as $z\to\infty$: $$ x=-\frac{1}{W_0(-1/z)}=z-1-\frac{1}{2z}+\mathcal O(z^{-2}). $$ Substituting $z=e^\lambda$ then gives us the final asymptotic result for $\lambda\to\infty$: $$ x\sim e^\lambda-1-\frac{1}{2}e^{-\lambda}. $$

Edit:

If we want the second solution $x\ll 1$ then the asymptotic series for $z\to\infty$ can be derived using the same process as above and an appropriate series expansion for $W_{-1}(\cdot)$.

Answer 2

After @Aaron Hendrickson's answer, for the second branch $$x=-\frac{1}{W_{-1}\left(-e^{-\lambda }\right)}$$ we could use the asymptotic expansion given here $$W_{-1}(t)\approx L_1-L_2+\frac{L_2}{L_1}+\frac{L_2(L_2-2)}{2L_1^2}+\frac{L_2(2L_2^2-9L_2+6)}{6L_1^3}+\cdots$$ where $L_1=-\lambda$ and $L_2=\log(\lambda)$ to obtain $${W_{-1}\left(-e^{-\lambda }\right)}=-\lambda -\log (\lambda )-\frac{\log (\lambda )}{\lambda }+\frac{(\log (\lambda )-2) \log (\lambda )}{2 \lambda ^2}-$$ $$\frac{\log (\lambda ) (2\log ^2(\lambda ) -9\log (\lambda )+6)}{6 \lambda ^3}+\cdots$$ and then $$x=\frac{1}{\lambda }-\frac{\log (\lambda )}{\lambda ^2}+\frac{\log (\lambda )(\log (\lambda )-1) }{\lambda ^3}-\frac{\log (\lambda )(2 \log ^2(\lambda )-5 \log (\lambda )+2) }{2\lambda ^4}+\cdots$$

or, more compact, $$x=\frac 1 \lambda- \frac{\log (\lambda )}{2 \lambda ^4}\Big[2 \left(\lambda ^2+\lambda +1\right)-(2 \lambda -\log (\lambda )+5) \log (\lambda ) \Big]+\cdots$$ which is quite accurate.

Using $\lambda=5^k$, some results $$\left( \begin{array}{ccc} k & \text{approximation} & \text{solution} \\ 1 & 0.147920235019603 & 0.144157704688610 \\ 2 & 0.035322286425834 & 0.035280299526838 \\ 3 & 0.007700440302570 & 0.007700214802757 \\ 4 & 0.001583662506572 & 0.001583661640015 \\ 5 & 0.000319177824945 & 0.000319177822221 \\ 6 & 0.000063960468364 & 0.000063960468356 \\ 7 & 0.000012798154410 & 0.000012798154410 \end{array} \right)$$

Edit

We can improve the approximation making one single iteration of Newton method. This would give $$x_1=\frac {x_0}{1-x_0} \left(1-e^{\lambda -\frac{1}{x_0}}\right)$$

Trying for $\lambda=5$

• $x_0=\frac 1 \lambda-\frac{\log (\lambda )}{\lambda ^2}=0.135622$ gives $x_1=0.142285$

• $x_0=\frac 1 \lambda-\frac{\log (\lambda )}{\lambda ^2}+\frac{\log (\lambda )(\log (\lambda )-1) }{\lambda ^3}=0.143469$ gives $x_1=0.1441445$