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!

## Comments

Sort by:

TopNewestWe 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. – Brian Charlesworth · 2 years, 2 months ago

Log in to reply

– Krishna Sharma · 2 years, 2 months ago

You will always get 3 values of \(y\) when x is even & 2 values when x is odd.Log in to reply

– Brian Charlesworth · 2 years, 2 months ago

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.Log in to reply

– Julian Poon · 2 years, 2 months ago

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

– Krishna Sharma · 2 years, 2 months ago

Sure I'll clarify it :)Log in to reply

– Julian Poon · 2 years, 2 months ago

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

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. – Krishna Sharma · 2 years, 2 months ago

Log in to reply

\(y\) 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 – Julian Poon · 2 years, 2 months ago

Log in to reply

– Julian Poon · 2 years, 2 months ago

Wow. I feel like I just unlocked a new power.Log in to reply

Newton's Method can be adapted for the complex plane, but I'd rather leave that mess alone for now. :) – Brian Charlesworth · 2 years, 2 months ago

Log in to reply

– Daniel Liu · 2 years, 2 months ago

However, sometimes Newton's Method does not work, right? For example, \(f(x)=\sqrt[3]{x}\).Log in to reply

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 · 2 years, 2 months ago

That's right. It often won't work near asymptotes and local extrema. In the case of \(f(x) = \sqrt[3]{x}\) the problem is that \(f'(x)\) is not defined at the root \(x = 0\), which results in the increasing oscillation of the successive approximations. TheLog in to reply

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 \(y=\frac{W(-\ln x)}{\ln x}.\) – Joseph Miller · 2 years, 2 months ago

Log in to reply

Good to know that an invalid dispute raised by me started a valid and interesting discussion. – Janardhanan Sivaramakrishnan · 2 years, 2 months ago

Log in to reply