# ${x}^{y} = {y}^{x}$ Challenge

Recently I posted a calculus problem Is there a closed form?.

In this problem, there is this dispute which interested me:

Here are the main points of the dispute which I want to talk about:

"For any value of $t>0,\quad t\neq 1$, there is a solution set $(x,y)=\left(t^{\frac{1}{t-1}},t^{\frac{t}{t-1}}\right)$

And due to symmetry, $x=y$ is also a solution..."

This dispute impressed me as I did not manage to find the solutions for which $x \neq y$

However,

It happens that this does not include all the solutions and that for some values of $x$, the solution for $y$ can be negative.

So, here is a challenge (which I can't solve):

Find the values of $x$ which has $3$ solutions for $y$

Find a way to find the solutions of $y$ that are negative. (You can also try to find the complex solutions too, if there are)

Good luck! Note by Julian Poon
6 years, 5 months ago

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## Comments

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We get $3$ solutions for $x = 2$. The first $2$ can be found by observation, namely $y = 2$ and $y = 4$. The last one ends up being negative, and since $x$ is both a base and exponent there isn't an algebraic means of finding this third solution. Probably the most straightforward way is to use Newton's Method on the function $f(x) = 2^{x} - x^{2}$ and start with a first guess of $x_{0} = 0$. The iterative formula for the $(n + 1)$st approximation would be

$x_{n+1} = x_{n} - \dfrac{2^{x_{n}} - (x_{n})^{2}}{2^{x_{n}}*\ln(2) - 2x_{n}}$.

We would then get $x_{1} = -1.4427, x_{2} = -0.89707, x_{3} = -0.77347, x_{4} = -0.766685, x_{5} = -0.7666647.$

We could keep going with a few more iterations depending on how many significant figures of accuracy we need.

- 6 years, 5 months ago

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Wow. I feel like I just unlocked a new power.

- 6 years, 5 months ago

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Hahaha. It might not be as powerful as telekinesis or anything but Newton's Method is very useful and efficient. It doesn't always work, however, especially if you make an errant first guess, (you'll know that this is the case when the successive approximations start to fluctuate wildly). Also, if there are multiple real roots then you need to make appropriate first guesses and go through the iterative process for each one individually. For this problem if you were to start with $x_{0} = 1$ the approximations would converge to $2$, and if you were to start with $x_{0} = 3$ the approximations would converge to $4$, giving us all three roots.

Newton's Method can be adapted for the complex plane, but I'd rather leave that mess alone for now. :)

- 6 years, 5 months ago

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However, sometimes Newton's Method does not work, right? For example, $f(x)=\sqrt{x}$.

- 6 years, 5 months ago

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That's right. It often won't work near asymptotes and local extrema. In the case of $f(x) = \sqrt{x}$ the problem is that $f'(x)$ is not defined at the root $x = 0$, which results in the increasing oscillation of the successive approximations. The reasons for failure are numerous, but my rule of thumb for the method is that "it works when it works and doesn't when it doesn't", i.e., it doesn't hurt to give it a try, since more often than not it will work. :)

- 6 years, 5 months ago

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You will always get 3 values of $y$ when x is even & 2 values when x is odd.

- 6 years, 5 months ago

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Mind if you define what you mean by E and O? Im not very good at set notation.

- 6 years, 5 months ago

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Sure I'll clarify it :)

- 6 years, 5 months ago

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Oh Yeah. Never noticed that. Do you happen to have a proof for that?

- 6 years, 5 months ago

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Graphs would help here, sketch the graph of both the expressions

Expression of the form $c^{x}$ will always lie in first and second quadrant, where c is constant

Now Expression of the form $x^{c}$

When c is even, graph will be in $1^{st}$ & $2^{nd}$ quadrant, there will surely be an intersection in $2^{nd}$ quadrant & 2 intersections in $1^{st}$ quad.

When c is odd, graph will be in $1^{st}$ & $3^{rd}$ quadrant so there won't be any negative value satisfying the equation.

Note:- for $x = e$ intersection points in $1^{st}$ quadrant will coincide resulting in only 1 solution.

- 6 years, 5 months ago

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I have yet to prove it but I found this interesting relation:

$y$ has $3$ solutions if

$x=2n+k$ or $x=(2n-1)+g$

Where $n$ is a positive integer $n>0$ and

$g,k<1$ and if you make $k$ or $g$ into an integer by taking $g \times {10}^{p} = g'$ or $k \times {10}^{p} = k'$,

$g'={a}_{2p-1}$ and $k'={a}_{2p}$

where $p$ is an integer $p>0$ and

${a}_{n+2}=2{a}_{n+1}-{a}_{n}$

At least it works until $g'$ or $h'$ reaches $100$. Beyond that, it works only sometimes. (sometimes as to when $k'$ or $g'$ is equal to ${a}_{n} \times {10}^{p}$ where ${a}_{n} < 100$ which is actually expected since it can be bashed to show that it works for all ${a}_{n}$ lesser than $100$)

I don't think this is complete

- 6 years, 5 months ago

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Good point. And in the limit as integer $n \rightarrow \infty$ for $x = 2n$ the roots, besides $y = 2n$, go to $-1$ and $1$, both from the right.

- 6 years, 5 months ago

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Good to know that an invalid dispute raised by me started a valid and interesting discussion.

- 6 years, 5 months ago

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You may want to investigate Lambert's $W$ function (a multiply-valued complex function, akin to the the complex version of $\ln x$), defined as the inverse of $ye^y=x$. In practice, it is computed using Haley's method (a second order version of Newton's), but it yields closed form solutions for this equation and several similar ones. In the case of $x^y=y^x,$ the solutions are of the form $y=\frac{-xW_b(-\frac{\ln x}{x})}{\ln x},$ where $b$ denotes the branch to compute of Lambert's $W$ function. It also can simplify infinite tetration (when it converges): $y=x^{x^{x^{\cdot^{\cdot^{\cdot}}}}},$ to $y=\frac{W(-\ln x)}{\ln x}.$

- 6 years, 4 months ago

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