Many months ago in class I came up with the problem: $$x^{(x+1)} = (x+1)^x$$ Using the solve function on my calculator I have found that the answer is around 2.29... This is backed up by the graph. However I was determined to find the inverse function where: $$x = f(y) $$ or find the answer algebraically $$ x = _-$$ Being a lowly first year A-level student this has been pretty much impossible. So far doing some simple rearranging the equation looks like: $$ x^{\frac 1 x} - \frac 1 x - 1=0$$I've tried many methods and had a good look online. So far I have just about been able to solve the equation $x^{\frac 1 x} = y$ by finding the inverse of $x^x$ graphically so $$ x^x = y $$ $$ x = P(y)$$ where P is the inverse function of $x^x$, then doing $$x^{\frac 1 x} = y$$ $$e^{ln(x^{\frac 1 x})}=y$$ $$\frac1 x (ln(x))=ln(y)$$ $$xln(x)=\frac 1 {ln(y)}$$ $$x^x = e^{\frac1 {ln(y)}}$$ $$x=P\biggl(e^{\frac1 {ln(y)}}\biggr)$$I don't know how to fit this in to my original equation to have just x on one side and no x's on the other side ... I do not want to use any guesswork methods or methods where you work your way to the answer slowly. I have tried using methods where you go one step up above powers so $x^x$ becomes something like $x@2$ where @ is used like + or X then trying to find the inverse of this like - is to + and / is to X and $\sqrt x$ is to $x^2$, to help you bridge the barrier between the $x^x$ and the $\frac 1 x$ but I couldn't find any way of doing this. Thank you for the help.
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4Many (one might even say "almost all") equations cannot be solved analytically; we can only approximate solutions numerically. This appears to be one of those, although I don't have a proof of that claim. – G Tony Jacobs Dec 09 '17 at 17:42
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2There might be no closed form solution, but you can iterate $x_{n+1}=(1+1/x_n)^{x_n}$ with $x_0=2$ to converge to the solution. – M. Winter Dec 09 '17 at 18:46
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1Welcome to stackexchange. This is a good question. Unfortunately (as the comments say) it does not have a good answer. That happens more often than mathematicians might wish. Don't let that discourage you from asking. – Ethan Bolker Dec 09 '17 at 20:51
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The search "Foias Constant" and references at wikipedia may help. – Raymond Manzoni Dec 13 '17 at 11:00
3 Answers
HINT.- It is illusory to want to have a closed algebraic form of the solution because the equation is transcendental so we search the solution with numerical methods.
Since $2^3=8\lt 9=(2+1)^2$ and $3^4=81\gt 64=(3+1)^3$ we deduce that the answer $x_0$ is such that $2\lt x_0\lt3$. The number $x_0$ is the only zero of the function $f(x)=x^{x+1}-(x+1)^x$. We have $$f(x)=x^{x+1}-(x+1)^x=x^{x+1}(x+1)^x\left(\frac{1}{(x+1)^x}-\frac{1}{x^{x+1}}\right)$$ Hence $x_0$ is the zero of the function $$g(x)=\frac{1}{(x+1)^x}-\frac{1}{x^{x+1}}$$ which converges to $0$ faster than $f(x)$ in the neighborhood of $x_0$ being negative at the left of $x_0$.
A way so is to approximate $g(x)$. For example one has $$g(2.3)\approx 0.0001652....\gt 0\Rightarrow\color{red}{2\lt x_0\lt2.3}$$ and thus we can approximate more.
ADDITIONAL NOTE.-Another optional way could be maybe as follows:
$$f(x)=x^{x+1}\left(1-\frac{1}{x}\left(1+\frac{1}{x}\right)^x\right)$$ which gives $x_0$ as the only fixed point of the function $h(x)=\left(1+\dfrac 1x\right)^x$
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It seems that the problem of finding the inverse function is at least as hard as this, or exactly hard as this:
Find $z$ as a function of $w$:
$$\sum_{n=1}^{+\infty} \frac {(-1)^{n+1}}{n}(z^n-(w^n +\sum_{k=0}^{n} \binom {n} {k}w^{n+1-k}(-1)^k))=0$$
We can approximate the solution building around $x=2$ the Taylor expansion of function $$f(x)=(x+1)\log(x)-x \log(x+1)$$ the first derivatives of which being$$f'(x)=\frac{1}{x}+\frac{1}{x+1}+\log (x)-\log (x+1)$$ $$f''(x)=-\frac{x^2+x+1}{x^2 (x+1)^2}\qquad f'''(x)=\frac{(2 x+1) \left(x^2+x+2\right)}{x^3 (x+1)^3}$$ This would give $$f(x)=-\log \left(\frac{9}{8}\right)+ \left(\frac{5}{6}-\log \left(\frac{3}{2}\right)\right)(x-2)-\frac{7}{72} (x-2)^2+\frac{5}{162} (x-2)^3+O\left((x-2)^4\right)$$ Using the expansion to $O\left((x-2)^2\right)$ would give a solution which is $$x=2+\frac{6 \log \left(\frac{9}{8}\right)}{5-\log \left(\frac{729}{64}\right)}\approx \color{red}{2.2}7528$$ I shall not write the solution of the quadratic corresponding to the expansion up to $O\left((x-2)^3\right)$ but the solution will be $x\approx \color{red}{2.29}506$.
Solving the cubic corresponding to the expansion up to $O\left((x-2)^4\right)$ would give $x\approx \color{red}{2.29}297$.
We could even do better building the simplest $[1,1]$ Padé approximant of function $f(x)$ $$f(x) \approx \frac{f(2)+\left(f'(2)-\frac{f(2) f''(2)}{2 f'(2)}\right)(x-2) } {1-\frac{ f''(2)}{2 f'(2)}(x-2) }$$ and solving for $x$ the numerator get an explicit expression which evalutes $x\approx \color{red}{2.293}65$
Building the simplest $[1,2]$ Padé approximant would lead to $x\approx \color{red}{2.2931}3$ still a the price of a linear equation.
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