# 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).

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)).

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}}.

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.

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

And of course, a group $G$ acting on $X$ is just a group homomorphism $G \to \text{Aut }X$.In fact, any ordinary category $C$ has a free $\text{Ab}$-enriched category $\mathbb{Z}[C]$ equipped with the universal functor from $C$ to an $\text{Ab}$-enriched category.

The equation $c^2 = a^2 + b^2 + ab$ is the law of cosines for a triangle with integer sides $a$, $b$, and $c$, and a 120 degree angle opposite side $c$.By the substitution $x = (a-b)/2$, $y = (a+b)/2$ it can be transformed to $x^2 + 3y^2 = 4z^2$, which is a more familiar equation, whose solutions are given in parametric form on page 353 (Corollary 6.

Let $C = AB^{\top}$.$I$ is of dimension $d \times m$ and $J$ is of dimension $m \times d$).

$A = N_G(A)$, $B = N_G(B)$, $C = N_G(C)$;

$AB = BC = CA = G$;

$A \cap B = B \cap C = C \cap A = 1$.One is $\Delta(X) = \{ (x,x): x \in X \}$.

Say we have a cohomological functor F from a triangulated category $C$ to the category $Ab$ of abelian groups, e.Given two objects $x, y$ in $C$, can we find a complex that's like $RHom(x,y)$ whose cohomology computes $Ext_C^i(x,y):=Hom_C(x,y[i])$?