I am stuck in solving this equation: $$\bbox[5px,border:2px solid black]{x+1=x^{x^2}}$$ My approach was like this:
$$x^2=y$$ $$x+1=x^y$$ $$(x+1)^2=x^{2y}$$ $$(x+1)^2=y^y$$ $$(\sqrt {y}+1)^2=y^y$$ $$y+2\sqrt y+1=y^y$$ From here I could not go any further, any help is highly appreciated.
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- 6$\begingroup$ Do you have any reason to believe there exists a closed form solution to this? $\endgroup$Tony Theo– Tony Theo2025-01-26 15:33:28 +00:00Commented Jan 26 at 15:33
- 4$\begingroup$ Unless you want to count $x=0$ as a solution, I'm only seeing one positive real solution, which seems to require numerical methods to specify. $\endgroup$lulu– lulu2025-01-26 15:43:38 +00:00Commented Jan 26 at 15:43
- $\begingroup$ Maybe a Taylor expansion around $\frac 32$ would give an approximation ($1.501728977$ on CAS with degree 1), but at this point just trial and error on a calculator e.g. pluging $f(1.502)$ and refining in way easier. $\endgroup$zwim– zwim2025-01-26 16:09:21 +00:00Commented Jan 26 at 16:09
- $\begingroup$ Please try to make the titles of your questions more informative. For example, Why does $a<b$ imply $a+c<b+c$? is much more useful for other users than A question about inequality. See also the section on titles in How to ask a good question. (The part entitled "Make your title your question" is especially relevant to this.) $\endgroup$Martin Brandenburg– Martin Brandenburg2025-01-27 09:42:15 +00:00Commented Jan 27 at 9:42
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1 Answer
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2 Without looking at the plot
Consider $$f(x)=x^{x^2}-x-1$$ The first derivative $$f'(x)=x^{x^2} (x+2 x \log (x))-1$$ cancels at $x=1$.
Expand $f(x)$ as a very short series around $x=1$ $$f(x)=-1+2 (x-1)^2+2 (x-1)^3+O\left((x-1)^4\right)$$ Use power series reversion to obtain $$x=1+\frac{\sqrt{f(x)+1}}{\sqrt{2}}-\frac{f(x)+1}{4}+O\left((f(x)+1)^{3/2}\right)$$
$$f(x)=0 \quad \implies \quad x_0=\frac{3}{4}+\frac{1}{\sqrt{2}}=1.45711$$
Say that $x_0=\frac 3 2$ and use one single iteration of Newton-like methods of order $n$. You will obtain an explicit result such as
$$ x_{(2)}=\frac 32+\frac{40-18\ 2^{3/4} \sqrt[4]{3}}{-16+27\ 2^{3/4} \sqrt[4]{3}+108\ 2^{3/4} \sqrt[4]{3} \coth ^{-1}(5)}$$
Converting to decimals, the results $$\left( \begin{array}{ccc} n & x_{(n)} & \text{method} \\ 2 & 1.50172897696 & \text{Newton} \\ 3 & 1.50172178173 & \text{Halley} \\ 4 & 1.50172179228 & \text{Householder} \\ 5 & 1.50172179227 & \text{no name} \\ \end{array} \right)$$
- $\begingroup$ Hi Claude ! Cheers ! Very good approximate. For the fun I tried with experimental mathematics method : $$x\simeq \frac32+\frac{G}{\sinh^2(e)}\sin\left( 2G^2-\frac{\pi}{2}\right)\simeq 1.501721792266425...$$ to compare with numerical solving $$x\simeq 1.50172179226759...$$ $G\simeq 0.915965594177219...$ is the Catalan number. Of course there is no mathematic significance behind which isn't so good that with your method. $\endgroup$JJacquelin– JJacquelin2025-01-27 09:13:18 +00:00Commented Jan 27 at 9:13
- $\begingroup$ @JJacquelin. Hi Jean ! This is beautiful. Thanks for posting it $\endgroup$Claude Leibovici– Claude Leibovici2025-01-27 10:02:35 +00:00Commented Jan 27 at 10:02