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I'm going through some problems in Theorems, Corollaries, Lemmas, and Methods of Proof, and I'm stuck at a certain problem that seemed very interesting until I couldn't solve it for the life of me.

Let $\circ$ be defined on $\mathbb R^+$ as $a\circ b = a^b b^a$. One can show that $\circ$ is Abelian, that $\mathbb R^+$ is closed under $\circ$, and that there exists identity element $e=1$ such that $a\circ e = e \circ a = a$.

The next part of the problem asks to find $2^{-1}$, which means solving for $x$ where $2^x x^2 = 1$. Because $\circ$ is Abelian, there's no other way to arrange this expression. I tried taking logarithm of both sides to get

$$\begin{align} \log_2(2^x x^2) &= \log_2(1) \\ \log_2(2^x) + \log_2(x^2) &= 0 \\ x + 2\log_2(x) &= 0 \\ x &= -2\log_2(x) \\ 1 &= -2 \frac{\log_2(x)}{x} \\ -\frac{1}{2} &= \log_2(x^{\frac{1}{x}}) \\ 2^{-\frac{1}{2}} &= x^{\frac{1}{x}} \end{align}$$

This doesn't really get me anywhere. I tried other manipulations, but all expressions I got were quite ugly. WolframAlpha doesn't return any meaningful results for positive reals. I keep coming back to this problem, but it looks like I won't be able to solve it. Perhaps I'm missing some algebra I have to use to get the result.

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Looks like a job for Lambert's W function. –  marty cohen Jan 28 '12 at 19:05
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Does one ask to find the inverse or to prove that it exists? –  Did Jan 28 '12 at 19:07
    
@DidierPiau The book asked to find it, so I assume it exists. –  Phonon Jan 28 '12 at 20:37
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As Raymond Manzoni says, $-2$ works. I think it is worth noting, however, that some elements of $\mathbb{R}^+$ may not be invertible. Here's a plot of $(\frac{1}{3})^x x^{\frac{1}{3}}$ which suggests $\frac{1}{3}$ is one such element. –  Dejan Govc Jan 28 '12 at 21:38
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Since nobody mentioned it, let me add that an approximate value of the inverse of $2$ is $0.766665$. Let $u=\mathrm eW(1/\mathrm e)=0.756945...$ Every point in $(1,+\infty)$ has a unique inverse in $(u,1)$, the unique inverse of $\mathrm e$ is $u$ and vice versa, the unique inverse of $1$ is $1$, each point in $(u,1)$ has two inverses, one in $(1,\mathrm e)$ and the other in $(\mathrm e,+\infty)$ (for example $2$ and $4$ have the same inverse), and no point in $(0,u)$ has an inverse. –  Did Jan 29 '12 at 14:05

1 Answer 1

up vote 3 down vote accepted

The equation $2^{-x}=x^2$ admits three real solutions : $x=-2,\;x=-4\;$ and a more sophisticated solution involving the LambertW function as returned by WolframAlpha (the only positive solution will be this last one).

These solutions may be obtained by writing $\,2^{-x}=x^2\,$ as $\,1=x\,e^{x\frac{\ln(2)}2}\,$ or $\,-1=x\,e^{x\frac{\ln(2)}2}\,$ that we may express as :
$\;\displaystyle \frac{\ln(2)}2=\left(x\frac{\ln(2)}2\right)\,e^{\left(x\frac{\ln(2)}2\right)}\;$ or $\;\displaystyle \frac{-\ln(2)}2=\left(x\frac{\ln(2)}2\right)\,e^{\left(x\frac{\ln(2)}2\right)}$

Since the LambertW function is defined implicitly by $\displaystyle z=W(z)e^{W(z)}$ we see that the answers are given by $\;W\left(\frac{\ln(2)}2\right)=x\frac{\ln(2)}2\;$ and $\;W\left(\frac{-\ln(2)}2\right)=x\frac{\ln(2)}2\;$ that is by
$$\tag{1}\boxed{\displaystyle x=\frac2{\ln 2} W\left(\frac{\ln 2}2\right)}$$ and $$\tag{2}\;\boxed{\displaystyle x=\frac2{\ln 2} W\left(\frac{-\ln 2}2\right)}$$

The subtle point is that this second solution $(2)$ will be split in two sub-solutions since the $\rm LambertW$ function admits two branches for $x \in \left(-\dfrac1e,0\right)\;$ (i.e. for $ W\left(\dfrac{-\ln 2}2\right)$) :

  • the upper one gives $W_0\left(\dfrac{-\ln 2}2\right)=-\ln 2\;$ (divided by $\ln 2$ and multiplied by $2$ returns the $x=-2\,$ solution) while
  • the lower branch gives the value noted $W_{-1}\left(\dfrac{-\ln 2}2\right)$ (after multiplication by $\dfrac2{\ln 2}$ this returns the $x=-4\;$ solution).

The solution $(1)$ is the only real and positive solution : $$\boxed{\displaystyle x=\frac2{\ln 2} W\left(\frac{\ln 2}2\right)\approx 0.766664695962123}$$ To clarify further : you probably had to find an approximation of the answer (possibly using the plot of $x\to 2^{-x}$ and $x \to x^2$ on the same graphics as shown on WolframAlpha) or find it numerically (say by iterations or with a Taylor series) or play with the LambertW function...

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OK, and thanks for the edit; I misread the answer missing the positive solution. –  Marc van Leeuwen Oct 29 '13 at 5:05
    
@Marc: this was one of my early posts here and a fine occasion for an edit so thanks for taking care! Cheers, –  Raymond Manzoni Oct 29 '13 at 8:02

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