Results for query "  ab  x c   "

Boolean algebra: (A'+B)(A+C)

I have a little problem with this expression:

x = (A'+B)(A+C)


I know it can be simplified to:

A'C+AB


since ive used some software to simplify it, but i simply can't see how it is done.This is what i've done so far:

(A'+B)(A+C) =>
A'A + AB + A'C + BC =>
0 + AB + A'C + BC =>
AB + A'C + BC


I just fail to see how i can do this differently and get to the correct result.

Examples of enriched categories which are (co)powered or (co)tensored

We have for each $G: C \to \textbf{Ab}$ an isomorphism $\textbf{Ab}^C(G, F^A) \cong \textbf{Ab}(A, \textbf{Ab}^C(G, F))$ since

$$\begin{array}{rcl}
\textbf{Ab}^C(G, F^A) & \cong & \int_c \textbf{Ab}(Gc, F^A(c)) \\
& \cong & \int_c \textbf{Ab}(Gc, \textbf{Ab}(A, Fc)) \\
& \cong & \int_c \textbf{Ab}(Gc \otimes A, Fc) \\
& \cong & \int_c \textbf{Ab}(A, \textbf{Ab}(Gc, Fc)) \\
& \cong & \textbf{Ab}(A, \int_c \textbf{Ab}(Gc, Fc) \\
& \cong & \textbf{Ab}(A, \textbf{Ab}^C(G, F)).For we have a universal element $\mathbb{Z} \to \textbf{Ab}(A, \textbf{Ab}^C(F^A, F))$ and a map

$$\textbf{Ab}^C(F^A, F) \to \textbf{Ab}^C(\sigma(F^A), \sigma(F)).

Estimates of eigenvalues

, all eigenvalues/vectors of $A$, $B$ and $A+B$) or techniques that could be helpful?Assume that $(I+A+B+AB)^{-1}(A+B)z=\lambda z$, then $(I+c(A+B)+AB)z=0$, where $c=1-\lambda^{-1}$.

Bounding the non-multiplicativity of isometric projection

Let $A=O_AP_A,B=O_BP_B$, then $AB=O_AO_B(O_B^*P_AO_B)P_B$ and we are left with showing that if the product of two positive definite self-adjoint operators $X=O_B^*P_AO_B$ and $Y=P_B$ is $\delta$-close to a unitary operator $V$, then it is $C\delta$-close to $I$.Now, we do not really know much about $XY$ except that it is conjugate to $X^{1/2}YX^{1/2}$, so all eigenvalues are real positive.

Symmetries of non-Riemannian curvature tensor

The curvature tensor, $R_{ab}{}^c{}_d$, can be obtained from a connection which not necessarily is a metric connection.Question

Under what conditions $R_{ab}{}^c{}_c$ vanishes for a general curvature?

Triangles in rigid Riemann surfaces

Is there a triangle $A'B'C'$ such that $AB=A'B',\;\;AC=A'C',\;\;BC=B'C'$ but there is no an isometry which carries the first triangle to the second one.It is given that $|AB|=|A'B'|, \, |BC|=|B'C'|, \, |CA|=|C'A'|$, where $|AB|$ is the hyperbolic length of the geodesic segment $AB$, etc.

An example for a construction on monads/operads?

A monad $(X, A)$ in $C$ is defined as a monoid object $(A : X \rightarrow X, p^A : AA \rightarrow A, e^A: 1_X \rightarrow A)$ in $C(X, X)$.Given two monads $(X, A)$ and $(X, B)$, a distibutive law is a 2-cell $d : BA \rightarrow AB$ satisfying few axioms.

An Entropy Inequality

As I mentioned in the comment, the key to the solution is the inequality
$$
(1+a)\log(1+b)+(1+b)\log(1+a)\ge 2(1+c)\log(1+c)
$$
where $a,b,c>0, ab=c^2$.Dividing by $XY$ and putting $X^{-1}=1+x, Y^{-1}=1+y$, so that $\frac{1-X}X=x,\frac{1-Y}Y=y$, we see that this case reduces exactly to the above inequality.

Finding null-homologous curves via the matrix equation $AB^iC^jx=0$

The automorphisms give rise to automorphisms $B,C$ of the homology group, while the covering gives rise to a (non-injective) homomorphism $A$ from one homology group to the other.Let $A$ be a fixed $m\times n$ matrix, $B,C$ fixed invertible $n\times n$ matrices and $x$ a fixed vector.

power log distance between matrices

Then their singular values satisfy the majorization
\begin{equation*}
\log \sigma(AB) - \log \sigma(B) \prec \log\sigma(A),
\end{equation*}
where $s(X)$ denotes the vector of singular values of an operator $X$.Then,
\begin{equation*}
d(A,B) \le d(A,C) + d(B,C).

Injective objects in Mor(Ab)

Consider the abelian (Grothendieck) category $\mathcal{C} := \mathrm{Fun}(\{0<1\},\mathrm{Ab}) = \mathrm{Mor}(\mathrm{Ab})$.I will use notation $A_0 \to A_1$ for objects of $\mathrm{Mor}(\mathrm{Ab})$.