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Solve $x^x=2x$ where $x\in\mathbb C$.

Obviously, one solution is $x=2$. By WA, another solution is $x=0.346...$. How to solve it analytically, e.g. using Lambert W function?

Thank you.

JSCB
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  • What is WA please ? – Ewan Delanoy Jun 18 '13 at 08:21
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    WA is Wolfram Alpha. – JSCB Jun 18 '13 at 08:26
  • I can show you how I started:

    The Lambert W function is given as $z=W(z)*e^{W(z)}$. What you are trying to obtain is a expression on x which matches this function.

    You can start with $x^x=2x$ using the logarithm you obtain $x\ln{x}=\ln{2x}=\ln{2}+\ln{x}$. Then: $$x\ln{x}-\ln{x}=\ln{2}$$ $$(x-1)\ln{x}=\ln{2}$$ $$e^{\ln{x-1}}\ln{x}=\ln{2}$$ Now we almost arrived at the desired form. However, $x-1$ is still causing troubles...

     – Karl Hardr Jun 18 '13 at 09:12
  • What about http://www.wolframalpha.com/input/?i=plot+log%28abs%28%282%2Bx%2Biy%29^%282%2Bx%2Biy%29-2%282%2Bx%2Biy%29%29%29+for+x%3D-4+to+4 – Gottfried Helms Oct 09 '13 at 19:24
  • @GottfriedHelms: your link broke, could you write out the function another way so we can read it? – abiessu Oct 09 '13 at 19:45
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    Hmm, I don't see how I can do that. It is just calling WA with the command "plot log(abs((2+x+yi)^(2+x+yi)-2(2+x+yi))) for x=-4 to +4" – Gottfried Helms Oct 09 '13 at 20:26
  • @GottfriedHelms: that's what I was hoping for, thank you – abiessu Oct 09 '13 at 20:49
  • @GottfriedHelms: so what function is being plotted there and how does it relate to this question? – abiessu Oct 09 '13 at 20:51
  • @abiessu:??? it's just plotting the absolute value of the function $ g(x)= (2+x)^(2+x)-2(2+x) = 0$ over the complex numbers. Because we know apriori, that one solution of the original function $f(z)$ is at $z_0=2$ I shift the function to let the center of the coordinate-system at $z=2$. After that, the solutions are, where the absolute value equals zero (the real coordinate must then be recentered). After that, the log enhances the display of the zeros to make them better visible in the plot. – Gottfried Helms Oct 09 '13 at 21:09
  • @GottfriedHelms: thank you, I was missing the "recenter due to known solution" piece. – abiessu Oct 09 '13 at 21:12
  • let us continue this discussion in chat – abiessu Oct 09 '13 at 21:23

1 Answers1

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$$x^x=2x$$

$$=e^{x\ln x}=e^{(a+bi)(\ln|a+bi|+i(\arg(a+bi)+2k\pi))}$$

$$=e^{a\ln|a+bi|-b(\arg(a+bi)+2k\pi)}\cdot e^{bi\ln|a+bi|+ai(\arg(a+bi)+2k\pi)}$$

$$\implies {e^{a\ln|a+bi|}\over e^{b(\arg(a+bi)+2k\pi)}}=2|a+bi|$$

$$\text{and}\qquad e^{bi\ln|a+bi|+ai(\arg(a+bi)+2k\pi)}={a+bi\over|a+bi|}$$

$$\implies (a^2+b^2)^{a-1\over 2}=2e^{b\arg(a+bi)+2bk\pi}\tag 1$$

$$\text{and}\qquad\text{cis}(b\ln\sqrt{a^2+b^2}+a\arg(a+bi)+2ak\pi)=\text{cis}(\arg(a+bi)+2l\pi)\tag 2$$

$$(2)\implies \frac b2\ln(a^2+b^2)+(a-1)\arg(a+bi)=2l\pi-2ak\pi\tag 3$$

This is almost to the point of evaluating all-real numbers. Fortunately, many graphing programs will evaluate $\arctan(\tfrac ba)$ using all branches and correctly evaluating at e.g. $x=0$ such that $\arg(a+bi)$ can be graphed as $\arctan(\tfrac ba)$. Here are some graphs with $a$ along the $x$-axis and $b$ along the $y$-axis:

Wide angle view of magnitude (red) vs. angle (green) functions

Closeup of real solutions and first complex solution

As demonstrated in the pictures, there are the three real solutions at $x\approx -0.658, 0.346, 2$, and there are many complex solutions, one of which is near the location $3+3i$.

Also, we have two equations to work with, so we can continue the direct analysis:

$$(1)\implies a^2+b^2=(2e^{b\arg(a+bi)+2bk\pi})^{2\over a-1}$$

Plugging into $(2)$:

$$\frac b2\ln((2e^{b\arg(a+bi)+2bk\pi})^{2\over a-1})+(a-1)\arg(a+bi)=2l\pi-2ak\pi$$

$$=\frac b{a-1}(b\arg(a+bi)+2bk\pi)\ln2+(a-1)\arg(a+bi)$$

$$=\frac {b^2\ln 2+(a-1)^2}{a-1}\arg(a+bi)+\frac{2b^2k\pi\ln 2}{a-1}$$

Unfortunately, this doesn't lend itself to any further breakdown that I am aware of. The graph is not easy to create, since the branches appear to be a near-full covering of 2D space, and even setting e.g. $k=1$ only makes for confusing graphs that do not correspond to the solutions identified above.

abiessu
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