Equations via Lambert's W function

Question

I'm studying Lambert's W function and I came across the equation $2^x = 2x$. Upon inspection it is easy to see that $x = 1$ and $x = 2$ are the real solutions to the equation. Solving for Lambert's W function, we have: $$ \begin{split} &2^x = 2x \ \Rightarrow \ x2^{-x} = \frac{1}{2} \ \Rightarrow \ -x\log 2 \ e^{-x\log 2} = -\frac{\log 2}{2} \ \Rightarrow \ W(-x \log 2\ e^{-x\log 2}) = W\biggl(-\frac{\log 2}{2}\biggr) \ \Rightarrow\\ &-x\log 2 = W\biggl(-\frac{\log 2}{2}\biggr) \ \Rightarrow \ x = -\frac{1}{\log 2}W\biggl(-\frac{\log 2}{2}\biggr) \end{split} $$ But, $-\frac{1}{e} < -\frac{\log 2}{2} < 0$. Thus, $$ x = \begin{cases} -\frac{1}{\log 2}W_0\biggl(-\frac{\log 2}{2}\biggr) = -\frac{1}{\log 2}W_0[2^{-1}\log(2^{-1})] = -\frac{1}{\log 2}\cdot \log(2^{-1}) = 1\\ -\frac{1}{\log 2}W_-1\biggl(-\frac{\log 2}{2}\biggr) = -\frac{1}{\log 2}W_{-1}[2^{-1}\log(2^{-1})] = -\frac{1}{\log 2}\cdot 2\log(2^{-1}) = 2 \end{cases} $$ Why $W_{-1}[2^{-1}\log(2^{-1})] = -2\log 2$? Where did factor $2$ come from?

Answer 1

You seem to believe that $$ W_{-1} \left( \frac{1}{2} \ln \frac{1}{2} \right) $$ should be $- \ln 2$. The Lambert $W$ function has some identities that might make one suspect this, but none are applicable at $(1/2) \ln (1/2)$.

$$ W_{-1}(x \ln x) = \ln x, x \leq \frac{1}{\mathrm{e}} $$ is such an identity. It does not apply here because $2 < \mathrm{e}$, so $\frac{1}{2} > \frac{1}{\mathrm{e}}$

Let's look at $W_{-1}(x \ln x)$.

The identity applies to the part to the left of the apex and not to the part to the right. You are interested in $x = 1/2$, which is to the right. (The apex is at the branch point of $W$, where $x \ln x = -1/\mathrm{e}$, i.e., where $x = 1/\mathrm{e}$. Since we are crossing the branch point, we should expect some kind of non-smoothness there.)

Ultimately, the answer to your question is just because $$ W_{-1} \left( \frac{1}{2} \ln \frac{1}{2} \right) = -2 \ln 2 \text{.} $$

Answer 2

$\require{begingroup} \begingroup$ $\def\e{\mathrm{e}}\def\W{\operatorname{W}}\def\Wp{\operatorname{W_0}}\def\Wm{\operatorname{W_{-1}}}$

When the argument of $\W(u)$ is in the range $(-\tfrac1\e,0)$, the number $u$ always can be expressed in two equivalent forms: \begin{align} u&=\ln\left(a^{\tfrac 1{1-a}} \right)\cdot a^{\tfrac 1{1-a}} \tag{1}\label{1} \\ &=\ln\left(a^{\tfrac a{1-a}} \right)\cdot a^{\tfrac a{1-a}} \tag{2}\label{2} . \end{align}

Indeed, multiplying \eqref{1} by $1=a\cdot\frac1a$, we have

\begin{align} u&= a\cdot\tfrac 1a\cdot\ln\left(a^{\tfrac 1{1-a}} \right)\cdot a^{\tfrac 1{1-a}} \tag{3}\label{3} \\ &= a\cdot\ln\left(a^{\tfrac 1{1-a}} \right)\cdot a^{\tfrac 1{1-a}}\cdot \tfrac 1a \tag{4}\label{4} \\ &= \ln\left(a^{\tfrac a{1-a}} \right)\cdot a^{\tfrac 1{1-a}-1} = \ln\left(a^{\tfrac a{1-a}} \right)\cdot a^{\tfrac 1{1-a}-\tfrac{1-a}{1-a}} = \ln\left(a^{\tfrac a{1-a}} \right)\cdot a^{\tfrac a{1-a}} , \tag{5}\label{5} \end{align}

that is, \eqref{1}$\equiv$\eqref{2}.

Recall that the expression of the form $\W(t\exp{t})=t$, just by the definition of $\W$, ignoring any questions about branches of $\W$. So, applying $\W$ to both forms of $u$ in \eqref{1}, \eqref{2}, we have two distinct results,

\begin{align} \W\left( \ln\left(a^{\tfrac 1{1-a}} \right)\cdot a^{\tfrac 1{1-a}} \right) &= \ln\left(a^{\tfrac 1{1-a}} \right) =\frac{\ln a}{1-a} \tag{6}\label{6} \\ \text{and } \W\left( \ln\left(a^{\tfrac a{1-a}} \right)\cdot a^{\tfrac a{1-a}} \right) &= \ln\left(a^{\tfrac a{1-a}} \right) =\frac{a\ln a}{1-a} \tag{7}\label{7} . \end{align}

The last two expressions are known as
Parametric representation of the real branches of the Lambert W function.

Note that both values obtained from \eqref{6}, \eqref{7} are negative, and one of them is always greater than $-1$ (belongs to the branch $\Wp$), while the other is less than $-1$ (belongs to the branch $\Wm$).

Note also, that the positive parameter $a$ can be either less, or greater than $1$.

When $a<1$, the expression \eqref{7} is recognized as $\Wp(u)$ while the expression \eqref{6} is recognized as $\Wm(u)$ and $a=\tfrac{\Wp(u)}{\Wm(u)}$.

In context of the original question, we have for $a=\tfrac12$,

\begin{align} u&=\tfrac12\cdot\ln\tfrac12 \tag{8}\label{8} \\ &=\ln\left(a^{\tfrac a{1-a}} \right)\cdot a^{\tfrac a{1-a}} =\ln\left(a^{\tfrac 1{1-a}} \right)\cdot a^{\tfrac 1{1-a}} \tag{9}\label{9} \\ &=\tfrac12\cdot\ln\tfrac12 =\tfrac14\cdot\ln\tfrac14 \tag{10}\label{10} . \end{align}

\begin{align} \W(u)&=\W(\tfrac12\cdot\ln\tfrac12)=\ln\tfrac12 =-\ln2 =\Wp(u)\approx -0.69314718>-1 \tag{11}\label{11} ,\\ \W(u)&=\W(\tfrac14\cdot\ln\tfrac14)=\ln\tfrac14 =-2\ln2 =\Wm(u)\approx -1.38629436<-1 \tag{12}\label{12} . \end{align}

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