where y might be other replaced by whichever letter that makes the most sense in context. My question is what does y mean in this case. I think that y means both a function, since y(x), but also a variable whose value is equall to the output of function y. Is that correct?

May 2, 2013 · It is very easy to check that XY X Y and YX Y X have the same nonzero eigenvalues. Just apply Y Y to the identity XYv = λv X Y v = λ v and note Yv ≠ 0 Y v ≠ 0 if λv ≠ 0 λ v ≠ 0.

Nov 26, 2024 · It looks fine to me; the only subtle bit is in the first equality, where you have to subsititute yx y x for xy x y and vice-versa. But because you're computing this on the curve defined by that equality, it looks valid to me.

Nov 9, 2010 · Say xy = yx, and x> y> 0. Taking logs, ylogx = xlogy; rearranging, (logx) / x = (logy) / y. Let f(x) = (logx) / x; then this is f(x) = f(y). Now, f′(x) = (1 − logx) / x2, so f is increasing for x <e and decreasing for x> e. So if xy = yx has a solution, then x> e> y. So y must be 1 or 2. But y = 1 doesn't work. y = 2 gives x = 4. (I've always thought of this as the ``standard'' solution to ...

Feb 6, 2015 · For more on solving word equation see chapter 9 in Lothaire's Combinatorics on Words (1983/1997), which actually starts with xy = yx x y = y x as the motivating example.

Jan 22, 2015 · The reduced result of x1 ⋅ … ⋅xk ⋅y1 … ⋅yk x 1 ⋅ … ⋅ x k ⋅ y 1 … ⋅ y k will be a multiplication (or sum) of elements, each belong to a different Pi P i. Therefore they are commutative and we can get xy = yx x y = y x.

Dec 2, 2016 · Let G be a group and (xy)2 = (yx)2 (x y) 2 = (y x) 2 for all x, y ∈ G x, y ∈ G. Show that xy2 =y2x x y 2 = y 2 x for all x, y ∈ G x, y ∈ G. I understand the solution for this question but I have seen some people solve it in this way: Choose y =x−1y y = x − 1 y and substitute in for y y

xy =yx y log x = x log y log xdy + y xdx = log y dx + x ydy x y = y x y log x = x log y log x d y + y x d x = log y d x + x y d y

Nov 20, 2019 · Suppose that x ∈ Rn x ∈ R n and y ∈Rn y ∈ R n. The matrices xyT x y T and yxT y x T are both of rank 1 1. Is the sum of these two matrices of rank 1 1, i.e. is the matrix xyT + yxT x y T + y x T of rank 1 1? I think the answer is positive and I am trying to come up with an argument that shows that this is true. In general,

Nov 30, 2016 · I'm not quite sure if I did this right but I had another solution that made no sense to me: This is the solution I do not understand. I did not solve it this way. $=&gt;(xy)^{-1}(xy)^2(yx)^{-1} ...