In many diciplines that utlizes mathematics, we often see the equation $$y=y(x)$$ where $y$ might be other replaced by whichever letter that makes the most sense in ...

May 2, 2013 · 3 You could modify the proof of Sylvester's determinant theorem to show that $$\text {det} \left ( \lambda I_m - XY \right) = \text {det} \left ( \lambda I_n - YX \right)$$ for all $\lambda \neq 0$. …

Aug 3, 2022 · I am showing that if for some group $G$, $xy=yx$ for every $x, y \in G$ then $$ (xy)^n=x^ny^n.$$ I claim this holds by induction on $n$. So base case if $n=1$, we ...

Jul 3, 2024 · We know that $2^4 = 4^2$ and $(-2)^{-4} = (-4)^{-2}$. Is there another pair of integers $x, y$ ($x\\neq y$) which satisfies the equality $x^y = y^x$?

Then $y$ must start with $x$ and have some trailing characters $b$: $xy = yx \rightarrow y = xb$. This tells us a lot; from it we can conclude that both $x$ and $y$ must start and end with $b$.

Nov 30, 2016 · 0 This might be same argument as in previous answer: $$ (x^ {-1}\cdot yx)^2= (yx\cdot x^ {-1})^2 \,\,\,\, \Rightarrow \,\,\,\, x^ {-1}y^2x=y^2.$$

Aug 13, 2015 · We wish to show that $f_ {xy}=f_ {yx}$ for a continuous function $f (x,y)$ of two variables with continuous second partial derivatives. To do this, we use a helpful lemma.

One thought I had is that "whenever $x^ {-1},y^ {-1}$ exist in $R$, then $xxyy=xyxy\implies xy=yx$".

May 8, 2014 · If anyone cares, it seems that the two parts intersect at the point $ (e,e)$.

Nov 25, 2022 · How to prove $u_ {xy}=u_ {yx}$? Ask Question Asked 3 years, 3 months ago Modified 3 years, 3 months ago