xy=yx{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,t1t>0,\quad t\neq 1, there is a solution set (x,y)=(t1t1,ttt1)(x,y)=\left(t^{\frac{1}{t-1}},t^{\frac{t}{t-1}}\right)

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


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

However,

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


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

Find the values of xx which has 33 solutions for yy

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


Good luck!

Note by Julian Poon
4 years, 10 months ago

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1 vote

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We get 33 solutions for x=2x = 2. The first 22 can be found by observation, namely y=2y = 2 and y=4y = 4. The last one ends up being negative, and since xx 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)=2xx2f(x) = 2^{x} - x^{2} and start with a first guess of x0=0x_{0} = 0. The iterative formula for the (n+1)(n + 1)st approximation would be

xn+1=xn2xn(xn)22xnln(2)2xnx_{n+1} = x_{n} - \dfrac{2^{x_{n}} - (x_{n})^{2}}{2^{x_{n}}*\ln(2) - 2x_{n}}.

We would then get x1=1.4427,x2=0.89707,x3=0.77347,x4=0.766685,x5=0.7666647.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.

Brian Charlesworth - 4 years, 10 months ago

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

Julian Poon - 4 years, 10 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 x0=1x_{0} = 1 the approximations would converge to 22, and if you were to start with x0=3x_{0} = 3 the approximations would converge to 44, 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. :)

Brian Charlesworth - 4 years, 10 months ago

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@Brian Charlesworth However, sometimes Newton's Method does not work, right? For example, f(x)=x3f(x)=\sqrt[3]{x}.

Daniel Liu - 4 years, 10 months ago

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@Daniel Liu That's right. It often won't work near asymptotes and local extrema. In the case of f(x)=x3f(x) = \sqrt[3]{x} the problem is that f(x)f'(x) is not defined at the root x=0x = 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. :)

Brian Charlesworth - 4 years, 10 months ago

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

Krishna Sharma - 4 years, 10 months ago

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

Julian Poon - 4 years, 10 months ago

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@Julian Poon Sure I'll clarify it :)

Krishna Sharma - 4 years, 10 months ago

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

Julian Poon - 4 years, 10 months ago

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@Julian Poon Graphs would help here, sketch the graph of both the expressions

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

Now Expression of the form xcx^{c}

When c is even, graph will be in 1st1^{st} & 2nd2^{nd} quadrant, there will surely be an intersection in 2nd2^{nd} quadrant & 2 intersections in 1st1^{st} quad.

When c is odd, graph will be in 1st1^{st} & 3rd3^{rd} quadrant so there won't be any negative value satisfying the equation.

Note:- for x=ex = e intersection points in 1st1^{st} quadrant will coincide resulting in only 1 solution.

Krishna Sharma - 4 years, 9 months ago

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

yy has 33 solutions if

x=2n+kx=2n+k or x=(2n1)+gx=(2n-1)+g

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

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

g=a2p1g'={a}_{2p-1} and k=a2pk'={a}_{2p}

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

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

At least it works until gg' or hh' reaches 100100. Beyond that, it works only sometimes. (sometimes as to when kk' or gg' is equal to an×10p{a}_{n} \times {10}^{p} where an<100{a}_{n} < 100 which is actually expected since it can be bashed to show that it works for all an{a}_{n} lesser than 100100)

I don't think this is complete

Julian Poon - 4 years, 9 months ago

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Good point. And in the limit as integer nn \rightarrow \infty for x=2nx = 2n the roots, besides y=2ny = 2n, go to 1-1 and 11, both from the right.

Brian Charlesworth - 4 years, 10 months ago

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

Janardhanan Sivaramakrishnan - 4 years, 9 months ago

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You may want to investigate Lambert's WW function (a multiply-valued complex function, akin to the the complex version of lnx\ln x), defined as the inverse of yey=xye^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 xy=yx,x^y=y^x, the solutions are of the form y=xWb(lnxx)lnx,y=\frac{-xW_b(-\frac{\ln x}{x})}{\ln x}, where bb denotes the branch to compute of Lambert's WW function. It also can simplify infinite tetration (when it converges): y=xxx, y=x^{x^{x^{\cdot^{\cdot^{\cdot}}}}}, to y=W(lnx)lnx.y=\frac{W(-\ln x)}{\ln x}.

Joseph Miller - 4 years, 9 months ago

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