# Is there a way to “recreate” data with scikit-learn?

seed(7)

def inverse_lda(lda, C):

c = np.means_[c]

AB, C = make_blobs(n_samples=333, n_features=2, centers=2) # toy data

A, B = AB.

seed(7)

def inverse_lda(lda, C):

c = np.means_[c]

AB, C = make_blobs(n_samples=333, n_features=2, centers=2) # toy data

A, B = AB.

I have drawn a line between two points A(x,y)---B(x,y)

Now I have a third point C(x,y).The distance dist(AB) can be calculated as:

___________________________

/ 2 2

V (A.

I have a line segment AB made up of two LatLngs, A and B.I have another LatLng C and a bearing x degrees.

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.

In the key of Bb major I came across a chord that contained the notes Ab, C, D and G.This is an Ab major 7 flat 5 (Abmaj7(b5)) chord (if you hear Ab as its root).

If I use fma(a, b, c) in cuda, it means that the formula ab+c is calculated in a single ternary operation.But if I want to calculate -ab+c, does the invoking fma(-a, b, c) take one more multiply operation?

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)).

, 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 $c=1-\lambda^{-1}$.

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.

Good evening,

Consider $x^4+y^4+z^4=2t^4$ where x,y,z,t integer.Ramanujan gave two parametrizations: if $a+b+c=0$ then

$$a^4(b-c)^4+ b^4(c-a)^4+ c^4(a-b)^4= 2(ab+bc+ca)^4$$

and

$$(a^3+2abc)^4(b-c)^4+(b^3+2abc)^4(c-a)^4+(c^3+2abc)^4(a-b)^4=2(ab+ac+bc)^8$$

(listed in Mathworld, equations 144 and 146).

We show that the coefficient of $X^c$ or of $X^{b+c-1}$ of $P(X)$ is negative.Note that $ac>b+c$ and so on, hence

\begin{equation}

P(X)\equiv\frac{(1-X^a)(1-X^b)(1-X^c)(1-X^d)}{(1-X)^2(1-X^{ab})}\pmod{X^{b+c}}.

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?

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.

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.

Is it the case there is a state $\tau$ on $A$ such that $|\tau(ab)|=\|ab\|$?I'm particularly interested in the case when $a,b$ are projections and $\|ab\|=1$.

Then $\mu(3C)=2\mu(C)$, where $aB = \{ ax : x \in B \}$.Thus if $a$ is a power of $3$, $\mu(aC) = a^{\log_3 2} \mu(C)$.

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.

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.

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).

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 $\mathrm{Mor}(\mathrm{Ab})$.