Talk:Equation xy = yx

Error
Section 1, second paragraph,

if r,n are positive integers, r ≥ 3 or n ≥ 3 then r^(r+n) > (r+n)^r ;

This is trivially false (set r=1, n=3, then 1 < 4). Should it read "r>=3 and n>=3" ? That would certainly seem to be true but I have no idea if it's what Hengel actually pointed out. Salaw (talk) 18:22, 19 December 2017 (UTC)


 * You are correct in that the statement as given above is wrong. However, Hengel proves the result with no restriction on n other than it being a positive integer. I've corrected the statement. --Bill Cherowitzo (talk) 04:31, 20 December 2017 (UTC)

The constraint x=y produces valid solutions for all complex numbers except perhaps x=y=0. — Preceding unsigned comment added by 2605:A601:41FA:8400:4083:5783:1E30:2047 (talk) 18:39, 2 November 2018 (UTC)

Derivative dy/dx of the pairs (x,y) not on the line y=x
Section 2 (Positive real solutions) second paragraph says that, for the $$(x,y)$$ pairs not on the line $$y=x$$, the derivative $$dy/dx$$ is equal to $$-n^2$$ for $$n>0$$ and $$n \neq 1$$. But isn't
 * $$dy/dn = n^\frac{n}{n-1}\left(\left(\frac{1}{n-1}-\frac{n}{(n-1)^2}\right)\ln(n)+\frac{1}{n-1}\right)= \frac{n^\frac{n}{n-1}\left(n-1-\ln(n)\right)}{(n-1)^2}$$

and
 * $$dx/dn = n^\frac{1}{n-1}\left(\frac{1}{n(n-1)}-\frac{\ln(n)}{(n-1)^2}\right) = \frac{n^\frac{1}{n-1}\left(n-1-n\ln(n)\right)}{n(n-1)^2}$$

so
 * $$dy/dx = (dy/dn)/(dx/dn) = n^2\left(\frac{n-1-\ln(n)}{n-1-n\ln(n)}\right)$$

for the $$(x,y)$$ pairs not on the line $$y=x$$? 2A02:1811:BC10:E700:E86A:DB82:177C:B51 (talk) 16:17, 5 February 2022 (UTC)


 * Indeed, seems to be a mistake by the editor. The source given doesn't mention the derivative, but it's a straightforward calculation. I have fixed it. -- Hugo Spinelli (talk) 14:02, 27 September 2023 (UTC)

Removal of redundant proof
I have removed the following proof:

"A general solution to $x^y = y^x$ is obtained by noting that the positive real quadrant can be 'covered' by the intersection of the two equations $y=mx$ and $y=x^n$ ($m>0,n>0$). Requiring that some points also satisfy the equation $x^y=y^x,$ means that $x^{mx}=(x^n)^x=x^{nx},$ and by comparing exponents, $m=n.$ Thus, the 'covering' equations mean that $nx=x^n,$ and by exponentiating both sides by $1/(n-1)$ ($n \neq 1$), $x=n^{1/(n-1)},$ and $y=n^{n/(n-1)}.$ The case of $m=n=1$ corresponds to the solution $y=x$. The full solution thus is $(y=x) \cup \left(n^{1/(n-1)},n^{n/(n-1)}\right) \text{ for } n > 0, n \neq 1 .$"

Reason: The other proof for the parametric solution is very similar, but much more concise. This one adds nothing of value to the article, in my opinion. If anyone disagrees, feel free to add it back to the article. Hugo Spinelli (talk) 05:25, 28 September 2023 (UTC)