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Limit of absolute value of sequence | By definition, if $$\lim_{n \rightarrow \infty} a_n=a$$ then for all $\varepsilon>0$ there exists a $N \in \mathbb{N}$ such that $n \geq N$ implies $|a_n-a| \leq \varepsilon$.
We may also verify case-by-case that $$\big||x|-|y|\big| \leq |x-y|$$ for all $x,y \in \mathbb{R}$. In particular,$$\big||a_n|-|a|\big| \leq |a_n-a|.$$
Combining these yields the proof. |
Geodesics are minimizing in a simply connected manifold without conjugate points? | The fact that it is the universal cover of a compact manifold is not essential, what you need is completeness of the Riemannian manifold $\tilde M$ (the universal cover of a compact manifold is always complete) and its simple connectivity. Assuming these two properties and the NCP property, just follow the proof of the Cartan-Hadamard theorem to conclude that for each $p\in \tilde M$ the exponential map $\exp_p: T_p \tilde M\to \tilde M$ is a diffeomorphism. From this, you see that any two points are connected by a unique geodesic. This implies that geodesics are global length minimizers. |
Probability that one value from one exponential is smaller than the value from another | Since we assume independence:
$$P(Y_1<Y_2)=\mathbb{E}_{Y_2}[P(Y_1<Y_2|Y_2)]=\int_{[0,\infty)}P(Y_1<y|Y_2=y)f_{Y_2}(y)dy=$$
$$=\int_0^\infty\left(\int_0^y\lambda_1e^{-\lambda_1x}dx\right)\lambda_2e^{-\lambda_2 y}dy=\int_0^\infty\left(1-e^{-\lambda_1 y}\right)\lambda_2e^{-\lambda_2 y}dy=$$
$$=1-\int_0^{\infty}\lambda_2e^{-y(\lambda_1+\lambda_2)}dy=1-\frac{\lambda_2}{\lambda_1+\lambda_2}=\frac{\lambda_1}{\lambda_1+\lambda_2}$$ |
Is $\mu$ complete measure? | Let $A$ be a null-set and alet $f$ be its indicator function. Then every indicator function $g$ of a subset $B$ of $A$ equals $f$ $\mu$-almost everywhere. So $B$ is measurable and $\mu$ complete.
Of course, this has to hold for all measurable functions $f$, not just a given one. |
Throwing a dice and add the digit that appears to a sum and stopping when sum $\geq 100$. | Basic approach. Imagine drawing a tree, with a root labelled $0$. The running count of each node is the label on that node, plus the sum of the labels of all of its direct ancestors. We build on the tree as follows: Under any node whose running count is not yet $100$, we add six more nodes, labelled $1$ through $6$. We repeat until there are no nodes left whose running count is less than $100$.
At the end of this process, we obviously have a finite tree. How many $1$s are there? How many $6$s? Was there any time when we added a $1$ but not a $6$, or vice versa? |
Proving that $[\mathbb{Q}(\sqrt{\sqrt{p+q}+\sqrt{q}},\sqrt{\sqrt{p+q}-\sqrt{q}}):\mathbb{Q}]=8$. | We have $\sqrt {p+q}=n\in\Bbb N$, so i will use this $n$ below.
Let us consider the tower of fields:
$\require{AMScd}$
\begin{CD}
{} @. L=\Bbb Q\left(\ \sqrt {n\pm\sqrt q}\ \right)\\
@. @AAA\\
{} @. K=\Bbb Q(\ \sqrt p, \sqrt q\ )\\
@.\nearrow @.\nwarrow\\
\Bbb Q(\sqrt p) @. {} @. \Bbb Q(\sqrt q )\\
@.\nwarrow @.\nearrow\\
{} @. \Bbb Q @.{}
\end{CD}
Some remarks first:
The extension $K=\Bbb Q(\sqrt p,\sqrt q):\Bbb Q$ has degree four, else $\sqrt p$, $\sqrt q$ would differ by a rational factor, but $p\ne q$.
The vertical arrow is an extension of fields.
First $\sqrt q\in L$, since $n\pm\sqrt q\in L$.
Also, because the product of the two numbers $\sqrt{n\pm\sqrt q}$
is $\sqrt{n^2-q}=\sqrt{(p+q)-q}=\sqrt p$, we also have $\sqrt p\in L$.
For short, $L=
\Bbb Q\left(\ \sqrt {n+\sqrt q},\ \sqrt p,\ \sqrt q\ \right)
=
\Bbb Q\left(\ \sqrt {n-\sqrt q},\ \sqrt p,\ \sqrt q\ \right)
=K(\sqrt {n+\sqrt q})
=K(\sqrt {n-\sqrt q})
$.
It remains to show that the extension $L:K$ has degree two.
If not, then we would have a linear relation over $\Bbb Q$ of the shape:
$$
\sqrt{n+\sqrt q}=A+B\sqrt q+\sqrt p(C+D\sqrt q)\in K\ .
$$
Apply now the Galois morphism $\sqrt p\to -\sqrt p$, $\sqrt q\to+\sqrt q$ of $K=\Bbb Q(\sqrt p,\sqrt q)$, to get parallely
$$
\begin{aligned}
\sqrt{n+\sqrt q} &=A+B\sqrt q+\sqrt p(C+D\sqrt q)\in K\ ,\\
\pm \sqrt{n+\sqrt q} &=A+B\sqrt q-\sqrt p(C+D\sqrt q)\in K\ .
\end{aligned}
$$
(To be pedant and avoid any questions that i may put myself, i added that $\pm$ in the last relation, imposed by the minimal polynomial condition over $K$, the L.H.S being a root of $X^2 -n-\sqrt q\in \Bbb Q(\sqrt q)\ [X]$.)
The representation is unique, so we have either
$$
\begin{aligned}
\sqrt{n+\sqrt q} &=A+B\sqrt q\ ,\text{ or}\\
\sqrt{n+\sqrt q} &=\sqrt p(C+D\sqrt q)\ .
\end{aligned}
$$
We use now the other Galois morphism, $\sqrt p\to \sqrt p$, $\sqrt q\to-\sqrt q$, getting either
$$
\begin{aligned}
\pm\sqrt{n-\sqrt q} &=A-B\sqrt q\ ,\text{ or}\\
\pm\sqrt{n-\sqrt q} &=\sqrt p(C-D\sqrt q)\ .
\end{aligned}
$$
We multiply, so $\pm \sqrt{n+\sqrt q} \cdot \sqrt{n-\sqrt q}=\pm \sqrt{n^2-q}=\pm\sqrt p$ is either $A^2-qB^2\in\Bbb Q$ or $p(C^2-qD^2)\in \Bbb Q$, thus a contradiction.
The linear relation cannot hold. So the degree of the vertial field extension is two.
$\square$ |
Evaluating $\lim\limits_{x \to 0} \frac{\ln(1+x^{144})-\ln^{144}(1+x)}{x^{145}}$ | A simple series expansion $$\log (1+z) = z - \frac{z^2}{2} + O(z^3)$$ followed by a binomial expansion will evaluate the limit quite easily. |
some subsets of $Gl_n$ that are path connected | Let A and B be two matrices in $GL_n(\mathbb{C})$ Define $$g(t)=At+(1-t)I$$ let $t_1,\dots,t_n$ be the roots of the polynomial $$det[g(t)=At+(1-t)I]=0$$ Now choose $t\in \mathbb{C}\setminus\{t_1,\dots,t_n\}$, As $\mathbb{C}\setminus\{t_1,\dots,t_n\}$ is path connected(why?) I can choose a path $$f:[0,1]\rightarrow \mathbb{C}\setminus\{t_1,\dots,t_n\}$$ such that $$f(0)=0$$ and $$f(1)=1$$ Now choose $$G(t)=f(t)A+(1-f(t))I$$ So clearly $G(t)$ is a path from A to I so in the same way there will be path from I to B and hence A to B, This shows $GL_n(\mathbb{C})$ is path connected.
Use the fact that for every matrix $U\in U(n)$ there is an
invertible matrix $g$ such that $U = gDg^{−1}$ where $D = diag(e^{ix_1},\dots,e^{ix_n})$
(here $x_1,\dots,x_n$ are real numbers) to show that the topological group $U(n)$
is connected.show that every point of $U(n)$ can be joined by a path
with the identity matrix. |
Rearrange $y = xa-zc$ so that $a-c$ is on one side of the equation. | Not if you want $a-c$ to be the only appearances of $a$ and $c$ in the rewritten equation.
Otherwise, it's easy, of course:
$$ y-xa+zc+a-c = a-c $$
You can achieve that one of $a$ and $c$ appear only as part of $a-c$, at the cost of assuming that $x$ or $z$ is nonzero:
$$ \frac{y - (x-z)c}{x} = a-c $$ |
Proof for a graph distance | I would approach this using a proof by contradiction. Suppose $d(x, y) + d(y, z) < d(x, z)$. By definition, $d(x, y)$ is the length of the shortest path from $x$ to $y$, and $d(y, z)$ is the length of the shortest path from $y$ to $z$. However, $d(x, z)$ is the length of the shortest path from $x$ to $z$, so $d(x, y) + d(y, z)$ can't be less than $d(x, z)$. So we have a contradiction.
Notice how I generalize here, rather than using specific examples. Remember that example is not valid proof, unless you are using a counterexample to disprove something. |
the convergence of the infinite product of matrices composed by matrces with the spectral radius less than 1 | For a counterexample, let $A,B$ be given by
\begin{align*}
A&=
\pmatrix{
0&\frac{1}{4}\\
-1&1\\
}
\\[4pt]
B&=
\pmatrix{
0&-\frac{1}{4}\\
1&1
}
\end{align*}
Then the eigenvalues of $A$ are ${\large{\frac{1}{2}}},{\large{\frac{1}{2}}}$, and the eigenvalues of $B$ are also ${\large{\frac{1}{2}}},{\large{\frac{1}{2}}}$, so each of $A,B$ has spectral radius less than $1$.
Computing $AB$, we get
$$
AB
=
\pmatrix{
\frac{1}{4}&\frac{1}{4}\\
1&\frac{5}{4}\\
}
$$
which has eigenvalues
$$
\frac{3}{4}+\frac{\sqrt{2}}{2},\;\frac{3}{4}-\frac{\sqrt{2}}{2}
$$
so $AB$ has spectral radius greater than $1$.
It follows that $(AB)^n$ doesn't approach $0$ as $n$ approaches infinity. |
Coin flipping game- probability of winning | Let $B$ be the event that Bob wins. Assuming that the coins are fair, conditioning on the first flips yields
$$ \mathbb{P}(B)=\frac{1}{4}\mathbb{P}(B\mid HT)+\frac{1}{4}\mathbb{P}(B\mid TH)+\frac{1}{2}\mathbb{P}(B\mid HH,TT)=\frac{1}{4}+\frac{1}{2}\mathbb{P}(B\mid HH,TT)$$
But $\mathbb{P}(B\mid HH,TT)=\mathbb{P}(B)$ since if the flips are $HH$ or $TT$ then the game starts over. Hence
$$ \frac{1}{2}\mathbb{P}(B)=\frac{1}{4} $$
or $\mathbb{P}(B)=\frac{1}{2}$. |
Differential equations and linear algebra way | One way is separation of variables. One can prove that the solution with $u(0) = x$ (in your notation) of $\frac{\mathrm{d}}{\mathrm{d}t}u(t) = u(t)$ fulfills
$$
\int^{u(t)}_x \frac{1}{g(s)}~\mathrm{d}s= \int^t_0 h(s)~\mathrm{d}s,
$$
where $g(s) = s$ and $h(s) = 1$. This comes from the ODE being in the form
$$
\frac{\mathrm{d}}{\mathrm{d}t} u(t) = g(u)h(t).
$$
I'm sure you can find the solution yourself. |
Inverse of a linear transformation | It is the reflexion itself.
What is the image of the image of a point by a reflexion? |
are all transition matrix for symmetric matrices orthogonal | No, that is not the case, because the vectors you pick in your transition matrix need not have norm $1$.
Take $A=\left(\begin{array}{cc}2&3\\3&2\end{array}\right)$. Then one transition matrix is given by
$P=\left(\begin{array}{cr}1&1\\1&-1\end{array}\right)$. But then $P^{-1} = \left(\begin{array}{cr}\frac{1}{2} & \frac{1}{2}\\ \frac{1}{2} & -\frac{1}{2}\end{array}\right)$, $D=\left(\begin{array}{cr}5&0\\0&-1\end{array}\right)$. Note that $P^{-1}\neq P^T$.
You can always pick a $P$ for which $P^T=P^{-1}$ (by normalizing the columns), but not every matrix you pick will have that property. |
Is precomputation of the prime number decomposition feasible? | The problem is storage space. The total amount of harddrives ever produced by humanity can store around $10^{21}$ numbers.* But twenty-one-digit numbers are easy to factorise. So we could never store enough factorisations to make this plan worthwhile.
*For citation see here: https://www.zdnet.com/article/what-is-the-worlds-data-storage-capacity/ That's an old article, but it doesn't really matter. Thirty-digit numbers are also easy to factorise. |
Lebesgue integration by substitution | The integrand $f$ can (and should) essentially be ignored here.
It's probably clearer to prove a more general substitution rule. If $g: X\to Y$ is a map between measure spaces and $\mu$ is a measure on $X$, then its image measure $g\mu$ is defined by $(g\mu)(B)=\mu(g^{-1}(B))$ (on a suitable $\sigma$-algebra on $Y$). This comes with a substitution rule
$$
\int_Y f(y)\, d(g\mu)(y) = \int_X f(g(x))\, d\mu(x) ;
$$
to prove this, start out by checking it for $f=\chi_B$ (this holds by the definition of $g\mu$) and then for general measurable $f\ge 0$ by approximating by simple functions, and then we have it for general $f\in L^1$.
Now back to your case: it suffices to show that Lebesgue measure on $Y=[c,d]$ is the image measure of $d\mu=|g'|\, dt$ on $X=[a,b]$, under the map $g$ (let's call this measure $\nu$). This is obvious because it holds for intervals:
$$
\nu(p,q) = \mu(g^{-1}(p,q)) = \int_{g^{-1}(p)}^{g^{-1}(q)} g'(t)\, dt = g(g^{-1}(p)) - g(g^{-1}(q)) = p-q,
$$
as desired (here I assumed $g'>0$, but of course the calculation is analogous in the other case). |
Anayltical Solution for this differential equation? | You can solve the second equation to find $v_2(t) = C e^{t^2/2}$ with $C$ is constant, and then substitute $v_2$ into the first equation to find $v_1$. |
Let $B:=\{B_{\epsilon}(x):\forall x\in \Bbb{R}^3,\epsilon>1\}$. Does $B$ form a basis for a topology? | No, it doesn't form the basis of a topology. Take several balls $B_\varepsilon(x)$ such that their intersection $I$ is small; more precisely, such the diameter if $I$ is smaller than $2$. Then $I$ would be an open set, but $I$ contains no ball whose radius is greater than $1$. |
change the order of $\limsup$ in sequence | No, here's a counterexample: On $(0,1),$ let $u(x) = x^{-1/2}, v(x) = x^{-1/4}.$ Then $uv\in L^1(0,1).$ Let $u_n(x) = u(x) + x^{-3/4}/n.$ Then $u_n \to u$ in $L^1(0,1).$ However $\int_0^{1} u_nv = \infty$ for all $n.$
NOTE: I changed my previous answer; I think the above is easier. |
Dirac Delta limiting representation | Distributions are functionals which act linearly and continuously on a space of some class of test functions, such as compactly supported, or rapidly decreasing, or just all continuous functions as above. By analogy with the Riesz representation lemma, which says that certain classes of linear functionals are actually integrals, we may pretend that some other linear functionals, like the Dirac delta function, are also integrals.
The main reason to prefer smooth functions of compact support is that it makes integration by parts particularly easy to use, and therefore weak derivatives of distributions is easy to define.
For example, I would like to know what the derivative of the Dirac delta function (distribution) is.
$$\int^\infty_\infty g(x)\delta'(x-x_0)\,dx = g(x)\delta(x-x_0)\big|^\infty_\infty-\int^\infty_\infty g'(x)\delta(x-x_0)\, dx
$$
Since a priori $\delta$ is not a function, I do not know how to define its derivative. The left-hand side of the above equation is not defined, though I would like it to be true. So I simply define it to be true. I define the derivative of the delta function to be that functional which makes the above equation true. This is called a weak derivative.
Now if I chose $g(x)$ to be of compact support, or to at least converge sufficiently rapidly to zero at infinity, then I can conclude that the boundary term in my integration by parts vanishes. So I can define the derivative $\delta'(x-x_0)$ to be the functional which sends $g$ to $-g'(x_0).$
But if I assume that $g(x)$ is just any continuous function, then I have to carry that boundary term, and my weak derivative loses some good properties. This is a good reason to use test functions of compact support. |
Graph skeleton for thorus $S_1$ | This can be done. You know that $K_7$ can be embedded on $S_1$ and in fact can be a triangulation of $S_1$. It will be regular of degree 6 and will have 7 vertices, 21 edges, and 14 triangular faces. Now consider its dual graph; it will be regular of degree 3, and will have 14 vertices, 21 edges, and 7 hexagonal faces and by construction will be a skeleton of $S_1$. |
Case where an orthogonal projection matrix becomes a diagonal matrix? | In order for your formula to make sense, $A$ must have linearly independent columns. We therefore assume that this is the case.
$H$ will be diagonal if and only if $A$ has an invertible $n \times n$ submatrix and $A$ has zeros in all entries outside this submatrix. |
The rate at which the vector force is acting | Hint: It's simply asking $F.v$ To get that, all you need to do is $F=ma$ and scalar multiply with $v$. Mass will be constant |
What is the probability of being in a "run" of length $k$? | One is in a run of length $k$ if, for some $i$, the following events happen:
$i$ entries after the current position coincide with the entry at the current position
$k-i-1$ entries before the current position coincide with the entry at the current position
the next entries before and after the current position do not coincide with it.
These events are disjoint for different values of $i$, each has probability $1/2^{k+1}$, and exactly the integers $i$ such that $0\leqslant i\leqslant k-1$ are valid hence there are $k$ of them. Finally, for every $k\geqslant1$, $$P(\text{run of length exactly}\ k)=k/2^{k+1}.{}$$ |
What's the $z$-derivative of $|g|^2$ for $g(z)$ analytic? | By Leibniz's rule, $$\dfrac{\partial}{\partial z} (g \overline{g}) = \dfrac{\partial g}{\partial z} \overline{g} + g \dfrac{\partial \overline{g}}{\partial z} = g' \overline{g} + 0 = g' \overline{g}$$ |
Is it true that $ \sup_x (f(x) - g(x)) \geq \sup_x f(x) - \sup_x g(x)? $ | Hint: Ask the (easier) question, how $\sup F+\sup G$ compares to $\sup(F+G)$. Then let $F=f-g$ and $G=g$. |
is it possible to use induction to prove the following? | The two recurrences (for $\det A_n$ and $\det B_n$) are clearly correct, so it’s just a matter of proving the two closed forms, and you can use an ordinary induction argument to prove them simultaneously.
They are both clearly correct when $n=1$. Suppose that they hold for some $n\ge 1$. Then
$$\begin{align*}
\det A_{n+1}&=(1-a)\det A_n-n\deg B_n\\
&=(1-a)(-1)^na^{n-1}(a-n)-n(-1)^{n-1}a^{n-1}\\
&=(1-a)(-1)^na^n-n(1-a)(-1)^na^{n-1}-n(-1)^{n-1}a^{n-1}\\
&=(1-a)(-1)^na^n+n(-1)^na^n\\
&=(n+1-a)(-1)^na^n\\
&=(-1)^{n+1}a^n\big(a-(n+1)\big)\;,
\end{align*}$$
and
$$\begin{align*}
\det B_{n+1}&=\det A_n-(n-1)\det B_n\\
&=(-1)^na^{n-1}(a-n)-n(-1)^{n-1}a^{n-1}\\
&=(-1)^na^n\;,
\end{align*}$$
and the result follows by induction on $n$. |
Proof that $\operatorname{ker}(A^{T}A) = \operatorname{ker}(A)$? | For
$\vec v \in \ker A \tag{1}$
we have
$A \vec v = 0, \tag{2}$
whence
$A^TA \vec v = 0, \tag{3}$
showing that $\vec v \in \ker A^TA$; thus $\ker A \subset \ker A^TA$. To go the other way, note that
$A^TA \vec v = 0 \tag{4}$
implies that
$\vec v^T A^TA \vec v = 0, \tag{5}$
or
$(A \vec v)^T(A \vec v) = 0; \tag{6}$
but $(A \vec v)^T(A \vec v)$ is a scalar quantity which if written out in terms of components is
$0 = (A \vec v)^T(A \vec v) = \sum (A \vec v)_i^2, \tag{7}$
where $(A \vec v)_i$ is the $i$-th component of $A\vec v$. Thus we must have
$(A \vec v)_i = 0 \tag{7}$
for each and every $i$, implying
$A \vec v = 0 \tag{8}$
or
$\vec v \in \ker A, \tag{9}$
as desired. QED.
Hope this helps. Cheers,
and as always,
Fiat Lux!!! |
Angles between vectors in 3D | Let me change your notation of the axes in order to keep the standard chirality.
So let's take the pelvis to lie on the $y,z$ plane, with the origin at the hip joint $H$ and with the $x$ axis in the retro-antero direction.
In standing up position the femur will extend for a length $f$ along the negative $z$ axis and so the knee $K$ will be at
$(0,0,-f)$.
We take the reference system $x',y',z'$ to be fixed with the femur, with origin in $K$ and being parallel
to $x,y,z$ when standing up.
We can model the knee joint as a hinge with a single degree of freedom.
Apart from the varus angle $\beta$, the hip is also capable of a moderate "torsion" angle
$\alpha$, which corresponds to rotating the foot in-out ( I don't know the anathomical term for that).
We take $\beta$ to be a right-hand rotation around $x'$, moving $z'$ to $z''$, and
$\alpha$ to be a right hand-rotation around $z'$, (not $z''$).
We have then the flexion, that we individuate with $\phi$, the right-hand angle around $y''$,
which theoretically spans from$0$ to $\pi$.
The tibia will extend a length of $t$ along $-z''$. from the knee $K$ to the ankle $A$.
So the reference system $x'',y'',z''$ is placed with the origin in $K$, is fixed with the tibia extending along
$-z''$ to $(0,0,-t)$ and with the axis $y''$ being the axis of the knee joint.
We pass then to define the movement of the knee in the frame fixed with the pelvis.
We have two possible angular movements, which we choose to define as:
- a rotation $\theta$ around $x$, which defines the lateral inflexion;
- a rotation $\eta$ around $y$, which defines the frontal inflexion, with positive values corresponding to a backward flexion;
Having defined the kinematic model, we pass now to the matematical analysis.
The matrices
$$
{\bf R}_{\,{\bf x}} (\alpha ) = \left( {\matrix{
1 & 0 & 0 \cr
0 & {\cos \alpha } & { - \sin \alpha } \cr
0 & {\sin \alpha } & {\cos \alpha } \cr
} } \right)\quad {\bf R}_{\,{\bf y}} (\beta ) = \left( {\matrix{
{\cos \beta } & 0 & {\sin \beta } \cr
0 & 1 & 0 \cr
{ - \sin \beta } & 0 & {\cos \beta } \cr
} } \right)\quad {\bf R}_{\,{\bf z}} (\gamma ) = \left( {\matrix{
{\cos \gamma } & { - \sin \gamma } & 0 \cr
{\sin \gamma } & {\cos \gamma } & 0 \cr
0 & 0 & 1 \cr
} } \right)
$$
represent the unit column vectors of a reference system, rotated wrt to the base system, around the axis indicated and for an angle
measured according to the right-hand rule.
Formally, the expression
$$
{\bf v'}_b = {\bf R}_{\,{\bf x}} (\alpha )\;{\bf v}_b
$$
returns the vector $\bf {v'}$ which is the vector$\bf v$ rotated by the angle $\alpha$ around the $x$ axis.
Both the vectors are column vectors expressed in the same base system as the axis $x$: that's what the suffix $b$ is going to remind.
Therefore, taking the unit vectors along the axes $ {\bf i} , {\bf j} , {\bf k}$, joined horizontally into a matrix, the above translates into
$$
{\bf U}'_b = \left( {\matrix{
{{\bf i}'} & {{\bf j}'} & {{\bf k}'} \cr
} } \right)_b = {\bf R}\,\;\left( {\matrix{
{\bf i} & {\bf j} & {\bf k} \cr
} } \right)_b = {\bf R}\,{\bf U}_b = {\bf R}\,{\bf I} = {\bf R}
$$
Regarding the composition of the rotations, we have two possible schemes: extrinsic and intrinsic definition.
If we apply a rotation around $x$, followed by a rotation around the $y$ axis, the last being (as the first) relevant to the base system, then we have the extrinsic definition,
and the global rotation, in the base system, will be expressed by the right-to-left product of the matrices
$$
{\bf R}_{\,{\bf T}} = {\bf R}_{\,{\bf y}} \;{\bf R}_{\,{\bf x}}
$$
If instead, the rotation about $x$ is followed by a rotation around the "new" $y'$ axis (intrinsic definition), since in the base system
this is expressed as
$$
{\bf R}_{\,{\bf T}} = {\bf R}_{\,{\bf y}'} \;{\bf R}_{\,{\bf x}} = {\bf R}_{\,{\bf x}} \;{\bf R}_{\,{\bf y}} \;{\bf R}_{\,{\bf x}} ^{\, - \,1} \;{\bf R}_{\,{\bf x}} = {\bf R}_{\,{\bf x}} \;{\bf R}_{\,{\bf y}}
$$
that is with the product left-to-right of the corresponding matrices.
That premised, we can conclude
a) concerning the vector $\vec{HK}$, applying first $\eta$ then $\theta$, both extrinsic
$$
\eqalign{
& {\bf U}'_b = {\bf R}_{\,{\bf x}} (\theta )\;{\bf R}_{\,{\bf y}} (\eta ) \cr
& \vec {HK} = {\bf U}'_b \left( {\matrix{
0 \cr
0 \cr
{ - f} \cr
} } \right) = - f\;{\bf R}_{\,{\bf x}} (\theta )\;{\bf R}_{\,{\bf y}} (\eta )\;{\bf k} \cr}
$$
b) concerning the vector $\vec {KA}$, applying first $\beta$, then $\alpha$ extrinsic, then $\phi$ intrinsic, in the reference $x',y',z'$
$$
\eqalign{
& {\bf U}''_b = {\bf R}_{\,{\bf z}} (\alpha )\;{\bf R}_{\,{\bf x}} (\beta )\;{\bf R}_{\,{\bf y}} (\phi ){\bf U}'_b = \cr
& = {\bf R}_{\,{\bf z}} (\alpha )\;{\bf R}_{\,{\bf x}} (\beta )\;{\bf R}_{\,{\bf y}} (\phi )\;{\bf R}_{\,{\bf x}} (\theta )\;{\bf R}_{\,{\bf y}} (\eta ) \cr
& \vec {KA} = {\bf U}''_b \left( {\matrix{
0 \cr
0 \cr
{ - t} \cr
} } \right) = - t\;{\bf R}_{\,{\bf z}} (\alpha )\;{\bf R}_{\,{\bf x}} (\beta )\;{\bf R}_{\,{\bf y}} (\phi )\;{\bf R}_{\,{\bf x}} (\theta )\;{\bf R}_{\,{\bf y}} (\eta )\;{\bf k} \cr}
$$
c) and finally, we have of course
$$
\eqalign{
& \vec {HK} = \vec {HK} + \vec {KA} = \cr
& = - \left( {f\;{\bf I}\; + t\;{\bf R}_{\,{\bf z}} (\alpha )\;{\bf R}_{\,{\bf x}} (\beta )\;{\bf R}_{\,{\bf y}} (\phi )} \right){\bf R}_{\,{\bf x}} (\theta )\;{\bf R}_{\,{\bf y}} (\eta )\;{\bf k} \cr}
$$ |
Calculating Probabilities for Substitution Ciphers using Frequency Analysis | You might be interested in the paper "The Markov Chain Monte Carlo Revolution" by Diaconis, which you can find online. In this paper, Diaconis shows how Stanford students were able to break a simple substitution cipher by using the Metropolis algorithm, using digraph frequencies. Diaconis says an attempt based on the frequencies of single letters failed. |
Divergence and Curl of a vector field $\vec{F}$ | This information about the vector field can be heuristically determined by looking at how the arrows are distributed over the plane. If you imagine that the plane is covered in fluid, and that each arrow tells a particle passing through it what its velocity has to be, then you may interpret the vector field as a "static visualization" of the motion of the fluid.
Telling the divergence of the vector field at a point is equivalent to telling how much "denser" the fluid is getting there, if it flows according to the arrows. So if the arrows "seem to be directed toward" this point, the fluid particles tend to aggregate around it, and we say that the fluid converges there, or that it has negative divergence. Instead, if the arrows seem to be pointing away from the point, then the fluid is "thinning out", the fluid particles tend to escape from it, and we say that the fluid diverges from there, or that it has positive divergence. If the fluid seems to do neither thing, then you may say that the divergence there is approximately zero, and that the field or the fluid are solenoidal. In other words, if you draw a (small) circle centered at the point, and the arrows seem to always cross the boundary of the circle, you have nonzero divergence there; the divergence is positive if the arrows are directed outward (source point), it is negative if the arrows are directed inward (sink point).
Here's a video by Grant Sanderson (3Blue1Brown): notice how fluid particles seem to want to get inside the yellow circle, to converge there – the vector field has negative divergence there. In this other video, the situation is reversed.
On the other hand, telling the curl of the vector field at a point is equivalent to telling how much the fluid is rotating counterclockwise around that point. If you draw a (small) circle centered at the point and the arrows seem to tell fluid particles to run along the circle counterclockwise, then the vector field has positive curl there, while if they seem to go in the other direction the vector field has negative curl. If the particles do not seem to be rotating around that point, like in the two videos linked above, then the curl is close to zero (the fluid and the field are irrotational there).
Here's another video by Grant Sanderson, depicting fluid flow according to a vector field. Notice how the fluid particles behave near the small red circle: they seem to rotate counterclockwise, indicating that the curl at the center of the circle should be positive. (The particles also seem to converge, so the divergence there should be negative!) Had the fluid particles rotated in the other direction (clockwise), the sign of the curl would have been reversed (negative). |
A unitarity axiom in the vector space axioms | Canceling $\alpha$ in
$\alpha\cdot(1\cdot\boldsymbol x)=\alpha\cdot\boldsymbol x$
is actually multiplying both sides by $\alpha^{-1}$:
$$
\alpha^{-1}(\alpha\cdot(1\cdot\boldsymbol x))=\alpha^{-1}(\alpha\cdot\boldsymbol x)
$$
Now you need to use associativity of scalar multiplication:
$$
(\alpha^{-1}\alpha)\cdot(1\cdot\boldsymbol x)=(\alpha^{-1}\alpha)\cdot\boldsymbol x
$$
and then
$$
1\cdot(1\cdot\boldsymbol x)=
1\cdot\boldsymbol x
$$
which begs the question.
Perhaps it is easier to see this by trying to prove that $\alpha\cdot \boldsymbol x=\alpha\cdot\boldsymbol y$ implies $\boldsymbol x=\boldsymbol y$. You'll end up with $1\cdot\boldsymbol x=1\cdot\boldsymbol y$. |
Prove that $\frac{\sin(x)}{1+|x|}$ is not Lebesgue integrable | By Dirichlet's test, it is improperly Riemann integrable on $\mathbb{R}^+$. In explicit terms, the inverse Laplace transform gives
$$ \int_{0}^{+\infty}\frac{\sin(x)}{1+x}\,dx = \int_{0}^{+\infty}\mathcal{L}(\sin x)(s)\mathcal{L}^{-1}\left(\frac{1}{1+x}\right)(s)\,ds = \int_{0}^{+\infty}\frac{e^{-s}}{1+s^2}\,ds$$
and by the Cauchy-Schwarz inequality $$\int_{0}^{+\infty}\frac{e^{-s}}{1+s^2}\,ds\leq\sqrt{\frac{\pi}{8}}. $$
Similarly, for any $n\in\mathbb{N}^+$ we have
$$0\leq \int_{0}^{+\infty}\frac{\sin(nx)}{1+x}\,dx \leq\frac{1}{n},\qquad 0\leq \int_{0}^{+\infty}\frac{\cos(nx)}{1+x}\,dx \leq \frac{1}{n^2}$$
and the last inequality can be used for proving $\frac{\sin x}{1+x}\not\in L^1(\mathbb{R}^+)$. Indeed
$$\left|\sin x\right|=\frac{2}{\pi}-\frac{4}{\pi}\sum_{n\geq 1}\frac{\cos(2nx)}{4n^2-1} $$
holds uniformly over any compact subset of the real line, hence
$$ \int_{0}^{M}\frac{\left|\sin x\right|}{1+x}\,dx = \frac{2}{\pi}\log(M+1)+O\left(\sum_{n\geq 1}\frac{1}{n^2(4n^2-1)}\right)=\frac{2}{\pi}\log M+O(1).$$ |
Combinations problem for all possible combinations | Both formulas count the number of subsets of a set of size $2^n$.
In the first formula each of the $2^n$ elements can either be "in" or "out" of the subset. So there are $2$ possibilities for each of $2^n$ elements, giving a total of $2^{2^n}$ subsets.
The second formula sums over the number of subsets of size $i$, so it gives $$\sum_{i=0}^{2^n} \text{nr subsets of size } i=\sum_{i=0}^{2^n} \binom{2^n}{i}.$$ |
Grassmannians and $\mathrm{GL}(n,\mathbb{R})$ | The fact that $\Omega Gr_n \simeq GL(n,\mathbb{R})$ follows from the fact that $Gr_n$ is the classifying space of $GL(n,\mathbb{R})$. In general, $\Omega BG \simeq G$.
Now for your initial question: in order to use this adjunction we need to see that two maps are homotopic, if and only if any basepointed representatives are basepoint homotopic. Recall that isomorphism classes of n-dimensional vector bundles over a basepointed CW complex X correspond to homotopy classes of maps $X \rightarrow Gr_n$. Now if we let the basepoint of $Gr_n$ be $\mathbb{R}^n$ sitting inside $\mathbb{R}^\infty$, then we want to understand basepointed homotopy classes of maps from $X$ to $Gr_n$. These correspond to vector bundles with fiber over $* \in X$ equal to $\mathbb{R}^n$ up to isomorphism of vector bundles that restricts to the identity over $*$.
So given two basepointed vector bundle $V, V'$ and an isomorphism between them, take a nice enough closed trivializing neighborhood $U$ around $*$. Then we can identify our vector bundle in this neighborhood with $U \times \mathbb{R}^n$ so that it is the identity at the basepoint. Then we take a map $U \times \mathbb{R}^n \rightarrow GL(n,\mathbb{R})$ such that it is the the inverse of $V|_* \rightarrow V'|_*$ on $*$ and $0$ on $\partial U \times \mathbb{R}^n$. We can then act on $V'$ via this map to get an isomorphism of vector bundles that restricts to the identity on the basepoint. Hence, basepointed and free homotopy classes of maps into $Gr_n$ coincide. |
What will be the value of $\det(A^2+B^2)$? | the following observation may help:
set $X=A^2+B^2$
then
$$
XA=A^3+B^2A = B^3 + A^2B = XB
$$
so
$$
X(A-B) = 0
$$ |
Topology on the tensor product of two topological vector spaces -- how properties does it maintains? | AFAIK there is no single "canonical" topology on $A \otimes B$. Instead there are several that make sense.
A good source is
Francois Treves: "Topological Vector Spaces, Distributions and Kernels"
chapter 43: "The Two Main Topologies on Tensor Products", where the author defines the $\epsilon$ and the $\pi$ topologies.
The $\pi$-topology is the strongest locally convext topology such that the canonical embedding of $A \times B \to A \otimes B$ is continuous (locally convex by definition).
( The definition of the $\epsilon$ topology is rather involved, so you better look it up in the book I mentioned.) |
Minimal polynomial in a quotient | Note that $x^5=1$ in $M$. Thus
$$\beta^2=2+x^2+x^3$$
so
$$\beta^2+\beta-1=1+x+x^2+x^3+x^4=0$$ |
How to solve this integral: $\int_{-1}^{1} x^k (1-x^2)^{(n/2)-2} \, dx$ | Following Daniel's comments:
Case $1$ --- $\;k\ge 0\;$ is odd: In this case, the integrand $\;x^k(1-x^2)^{\frac n2-2}\;$ is an odd continuous function in a symmetric interval and thus the integral equals zero.
Case $2$ --- $\;k=2m\ge 0\;$ is even: in this case our integral equals twice the integral over half the interval and we can substitute:
$$u:=x^2\implies dx=\frac{du}{2\sqrt u}\implies$$
$$\int\limits_{-1}^1 x^k(1-x^2)^{\frac n2-2}dx=2\int\limits_0^2 (x^2)^m(1-x^2)^{\frac n2-2}dx=$$
$$2\int\limits_0^1u^m(1-u)^{\frac n2-2}\frac{du}{2u^{1/2}}=\int\limits_0^1u^{m-1/2}(1-u)^{\frac n2-2}du=:B\left(\frac{k+1}2\,,\,\color{}{\frac{n-2}2}\right)$$
and, for example using the relation between gamma and beta functions, we have
$$B\left(\frac{k+1}2\,,\,\frac{n-2}2\right)=\frac{\Gamma\left(\frac{k+1}2\right)\Gamma\left(\frac{n-2}2\right)}{\Gamma\left(\frac{k+n-1}2\right)}$$ |
Calculatin differential | Use the chain rule, let $z = y + t (x-y)$:
$$ \frac{\partial f(y + t (x-y))}{\partial t} = f'(z)\frac{\partial z}{\partial t} = f'(z)(x-y)$$
Since you are working in $\mathbb{R}^n$ then actually
$f'(z) (x-y)$ means $(x-y)\cdot \nabla f(z)$ |
How to solve the following differential equation of the form $\frac{k}{2}\frac{d^{2}f}{dx^{2}}-f+f^{3}=0$? | $$\frac{k}{2}f''-f+f^3=0$$
with conditions : $f(0)=f'(0)=0$
$$f''(0)=\frac{2}{k}(f(0)-f(0)^3)=0$$
$\frac{k}{2}f'''-f'+3f^2f'=0$
$$f'''(0)=\frac{2}{k}(f'(0)-3f(0)^2f'(0))=0$$
And so on... By induction, all the successive derivatives are $=0$.
This draw to expect the solution $f(x)=0$.
Of course, in case of initial conditions different from $f(0)=f'(0)=0$ the result would not be so trivial. |
Change of Basis Proof example | There are change of bases matrices $B$ and $C$ such that $BB_1=B_2$ and $CC_1=C_2$. These change of bases matrices have rank $n$ and $m$, respectively. So, you only need to show that the rank of $BB_1$ and $CC_1$ have the same rank as $B_1$ and $C_1$, respectively. |
Convergence in a metric space | Assume that for a certain metric $d$ your sequence converges to $g\in\mathbb{R}$. Then its even subsequence also converges to $g$ but it converges to $0$ thus $g=0$ similarly considering odd subsequence one shows that $g=1$ which gives a contradiction. |
multiplicative semi-norms on $\mathbb{C}[x]$ | Given a seminorm $p$, let $\mathfrak m_p = \{f \in \Bbb C[x] : p(f) = 0\}$. This ideal is prime, as if $fg \in \mathfrak m_p$, $p(f)p(g) = 0$, so one of $p(f)$ or $p(g)$ is zero. So $\mathfrak m_p = \langle x-a\rangle$ for some $a \in \Bbb C$ or $\mathfrak m_p = \{0\}$.
Suppose $\mathfrak m_p = \{0\}$. Then $p$ is in fact a norm on $\Bbb C[x]$ and extends to a norm on $\Bbb C(x)$ by $p'\left(\frac{a}{b}\right) = p(a)/p(b)$. So $\Bbb C(x)$ has the structure of a normed field. Passing to the completion we now have a complete normed field that contains $\Bbb C(x)$, and is thus infinite dimensional over $\Bbb C$; this contradicts Gelfand-Mazur, and thus $\mathfrak m_p$ must have been maximal.
Write $\mathfrak m_p= \langle x-a_p\rangle$. Then for any $f(x)$, $f(x) = g(x)+c$ where $g \in \mathfrak m_p$. Then the reverse triangle inequality gives $|p(f)-p(c)| \leq p(f-c) = p(g) = 0$, so $p(f) = p(c) = |c| = |f(a_p)|$. Hence $p(f) = |f(a_p)|$ for all $f \in \Bbb C[x]$ as desired. |
What does the equation $x^2+24xy+68y^2=0$ represent? | Completing the square w.r.t $x$ i get $$(x+12y)^2-76y^2 \color{red}{=0}$$
This is a good step; now rewrite using $a^2-b^2 = (a-b)(a+b)$ to get:
$$\left(x+12y + \sqrt{76}y \right)\left(x+12y - \sqrt{76}y \right)=0$$
But this implies:
$$x+12y + \sqrt{76}y = 0 \;\vee\; x+12y - \sqrt{76}y = 0$$
And these are equations of...? |
Stirling's approximation for Gamma function | Not a real solution, just a strategy: can you use the fact that the inequality holds for integers to show that it holds for rationals? Because then continuity would do it, right? |
Simple field extension inequality proof | Let $f$ be the minimal polynomial of $\alpha$ over $\mathbb{Q}$ and $g$ the minimal polynomial over $F$.
Note that $f$ is a polynomial with coefficients in $F$ (because $\mathbb{Q}\subset F\subset \mathbb{C}$) with $f(\alpha)=0$.
Since $g$ is the minimal polynomial over $F$, we get that $f$ must divide $g$.
In particular, $[F(\alpha):F]=\deg f\le \deg g=[\mathbb{Q}(\alpha):\mathbb{Q}]$. |
about convection term in the NSE | Your calculation is essentially correct, but be careful in the second to last equality since $\partial u /\partial x$ and $\partial u / \partial y$ are both vectors, so the term $(\partial u /\partial x, \partial u /\partial y)$ is not really a vector, but a matrix. In fact, one can also show (using your calculation) that
$$
(u \cdot \nabla) u =
\begin{pmatrix}
\partial_x u_1 & \partial_y u_1 \\
\partial_x u_2 & \partial_y u_2
\end{pmatrix}
\begin{pmatrix}
u_1 \\ u_2
\end{pmatrix}
= Du u^T
$$
where $D u$ is the Jacobian matrix of partials and $u^T$ is the transpose of the row vector $u$. |
Find $a_n$ via $a_n - a_{n-1}=3^n$ | I generalize a tad, setting
$a_0 = a, \tag 1$
$a_n - a_{n - 1} = b^n; \tag 2$
then
$a_1 - a_0 = a_1 - a = b, \tag 3$
$a_1 = a + b; \tag 4$
$a_2 - a_1 = a_2 - (a + b) = b^2, \tag 5$
$a_2 = a + b + b^2; \tag 6$
I make the inductive hypothesis:
$a_k = a + \displaystyle \sum_1^k b^j; \tag 7$
then
$a_{k + 1} = a_k + b^{k + 1} = a + \displaystyle \sum_1^k b^j + b^{k + 1} = a + \sum_1^{k + 1}b^j; \tag 7$
it follows by induction that for general $n \in \Bbb N$,
$a_n = a + \displaystyle \sum_1^n b^j = a + \dfrac{b^{n + 1} - b}{b - 1}; \tag 8$
taking
$a_0 = a = 1, \; b = 3, \tag 9$
we find
$a_n = 1 + \dfrac{3^{n + 1} - 3}{2} = \dfrac{3^{n + 1} - 1}{2}. \tag{10}$
I'm not sure what the name of this sequence is, but it's called a "geometric something-or-other", I'll warrant. |
What is the logic proof that ¬ (∃x)(Fx) => (∀x)(Fx → Gx) | Before attempting to construct a formal proof of anything, you should have an informal reason to believe it is true. I suggest:
$\lnot \exists \langle x,~ Fx \rangle$ means $F$ is always false
$\forall \langle x,~ Fx \implies Gx \rangle$ follows from "false always implies anything ($G$ in this case)"
A constructed proof should follow that general outline. Here is a natural deduction one:
$$\begin{array} {l}
%
\quad \begin{array} {|l} \lnot \exists \langle x,~ Fx \rangle \\ \hline
%
\quad \begin{array} {|l} Fy \\ \hline
%
\exists \langle x,~ Fx \rangle \\
%
\bot \\
%
Gy \end{array} \\
%
\forall \langle x,~ Fx \implies Gx \rangle \end{array} \\
%
\lnot \exists \langle x,~ Fx \rangle \implies \forall \langle x,~ Fx \implies Gx \rangle \end{array}$$ |
On the existence of a bounded linear functional | You did good.
If $\phi=(\cdot,g)$ then $\|\phi\|=\|g\|$, i.e. Riesz representation is an isometric isomorphism. Indeed, by Cauchy-Schwarz, $\|\phi\|\leq\|g\|$. And $\phi(g)=\|g\|^2$ so $\phi(g/\|g\|)=\|g\|$ hence $\|\phi\|\geq \|g\|$.
Now you see that $g=\frac{h}{\|h\|}$ yields $\|\phi\|=\|g\|=1$ and $\phi(h)=(h,g)=\|h\|^2/\|h\|=\|h\|$. |
Self-adjoint operators and positive | Hint: $\varphi \psi$ is similar to $\varphi^{1/2}\psi\varphi^{1/2}=\varphi^{-1/2}(\varphi \psi)\varphi^{1/2}$. |
Prove that language is non-context free $L=\{a^{n^2}b^n|n\ge 0\}$ | The pumping lemma will work. Show that if $r,s\ge 0$, and at least one of $r$ and $s$ is positive, then there is always a $k\ge 0$ such that
$$p^2+(k-1)r\ne\big(p+(k-1)s\big)^2\;.$$ |
Simplifying expression using Euler's formula | The exponents go from $-0iw$ to $-3iw$, so take the middle point $-3iw/2$, getting
$$
e^{-3iw/2}\cdot[ae^{+3iw/2}+be^{iw/2}+be^{-iw/2}+ae^{-3iw/2}]=\\
e^{-3iw/2}\cdot\left[2a\frac{e^{+3iw/2}+e^{-3iw/2}}{2}+2b\frac{e^{iw/2}+e^{-iw/2}}{2}\right]
$$ |
Norm of linear combination of vectors in the "same general direction" | The claim is not true in general. And I think looks true because your intuitive picture of $x$ and $y$ being in the "same direction" is not what the conditions $||x|| \leq ||x + y||$ and $||y|| \leq ||x + y||$ imply. I'll give a counterexample in $\mathbb{R}^2$ itself:
Take the vectors, in $(r,\theta)$ coordinates, to be $x = (1,0)$ and $y = (1, \frac{2\pi}{3})$. Then $x+y = (1, \frac{\pi}{3})$ satisfies the given conditions, even though $x$ and $y$ are not in the "same general direction". It is easy to show now that for any $c > 0$, we have $||(1 + c)y|| > ||x + (1 + c)y||$. |
Minimum distance of point to origin | $$\boxed{ t = \left( \frac{1}{8} - \frac{\sqrt{78}}{72} \right)^\tfrac{1}{3} + \left( \frac{1}{8} + \frac{\sqrt{78}}{72} \right)^\tfrac{1}{3} }$$
How? I entered $A t^3 -B t = 1$ to a CAS system (like Wolfram Alpha) and it solved the cubic equation. If you want to know more about the roots of a cubic polynomial, look at equations (71), (72) and (73) in this write-up from Wolfram. |
Showing Eigenvalues are non negative in a graph problem | For every vector $f$ in $R^n$ we have:
$$f'Mf=f'Df-f'Wf=\sum_{i=1}^nd_if_i^2-\sum_{i,j=1}^nf_if_jw_{ij}$$
$$\frac{1}2(\sum_{i=1}^n d_if_i^2+\sum_{j=1}^nd_jf_j^2-2\sum_{i,j=1}^nf_if_jw_{ij})=\frac{1}2\sum_{i,j=1}^nw_{ij}(f_i-f_j)^2\geq0$$
This shows you that M is a positive semi-definite matrix, and hence the eigenvalues are non-negative. |
How to show that the triangle is equilateral triangle? | $$\text{Now,}\sum(\cot A-\cot B )^2=2\sum(\cot^2A-\cot A\cot B)$$
$$\text{Again,}\sum\cot^2A=(\sum\cot A)^2-2\sum \cot A\cot B$$
We have $\displaystyle\cot(A+B)=\frac{\cot A\cot B-1}{\cot A+\cot B}$
and as $\displaystyle A+B+C=\pi, \cot(A+B)=\cot(\pi-C)=-\cot C$
Comparing values of $\displaystyle\cot(A+B),$ we get $\displaystyle\sum \cot A\cot B=1$
If we set $\cot A=p$ etc.,
we have $pq+qr+rp=1$ and $p+q+r=\sqrt3$
$\displaystyle\implies(p-q)^2+(q-r)^2+(r-p)^2=2(p^2+q^2+r^2-pq-qr-rp)=2\{(p+q+r)^2-2(pq+qr+rp)-pq-qr-rp\}=2\{(p+q+r)^2-3(pq+qr+rp)\}=2\{(\sqrt3)^2-3\}=0$
As $p,q,r$ are real, we have $\displaystyle p=q=r$
$\displaystyle\implies\cot A=\cot B\implies A=n\pi+B$ where $n$ is any integer.
But as $0<A,B<\pi, A=B$ |
How I can find the matrix $A$ of this quadratic form? | Unless I'm misunderstanding something I think this is pretty straightforward:
$$\sum_{1\le i<j\le n}(x_i-x_j)^2=(x_1-x_2)^2+(x_1-x_3)^2+\ldots+(x_1-x_n)^2+\ldots+(x_{n-1}-x_n)^2=$$
$$=x_1^2-2x_1x_2+x_2^2+\ldots+x_1^2-2x_1x_n+x_n^2+\ldots+x_{n-1}^2-2x_{n-1}x_n+x_n^2=$$
$$(n-1)\left[x_1^2+\ldots+x_n^2\right]-2\sum_{1\le i<j\le n}x_ix_j\;\;\ldots\ldots$$ |
The $x$-coordinate of the two points $P$ and $Q$ on the parabola $y^2=8x$ are roots of $x^2-17x+11$. | Note that
$$x_1+x_2=17,\>\>\>\>\>x_1x_2 =11$$
Then, the intersection coordinates of the two tangents are (GM of abscissa, AM of ordinate),
$$x=\sqrt{x_1x_2} = \sqrt{11}$$
$$y = \frac{y_1+y_2}2=\sqrt2(\sqrt{x_1}+\sqrt{x_2})=\sqrt2\sqrt{x_1+x_2+2\sqrt{x_1x_2} }
=\sqrt{34+4\sqrt{11}}$$
Since the focus of $y^2=8x$ is $(2,0)$, the distance is thus,
$$\sqrt{(x-2)^2+y^2} =\sqrt{(\sqrt{11}-2)^2+34+4\sqrt{11}}=\sqrt{49}=7$$ |
Proof that a discrete subgroup of $\mathbb(R)^n$ is isomorphic to an integer lattice | I have found the answer, through proving that $z_k$ is not injective. As it takes finitely many values, there exists $(k,n)\in {(\mathbb{N}^*)}^2$ such that $z_k = z_{k+n}$. Thus,
$$kx-y_k=(k+n)x-y_{k+n} \iff nx = y_{k+n}-y_k\in H$$
Therefore, there exists an $n\in\mathbb{N}^*$ such that $nx\in H$. |
How can I solve probability questions using the lebesgue integral? | Here's how you can utilize Markov chains to solve this problem.
Let $S_0$ be the state that the number isn't divisible by $2$ nor $5$, $S_1$ the state the number is divisible by $2$ but not $5$, $S_3$ the state the number is divisible by $5$ but not $2$, and $S_4$ the (absorbing) state the number is divisible by $10$.
This Markov chain generates the following transition matrix $P$: $$P=\begin{pmatrix}\frac{7}{9}&\frac{1}{9}&\frac{1}{9}&0\\ 0&\frac{8}{9}&0&\frac{1}{9}\\ 0&0&\frac{8}{9}&\frac{1}{9}\\ 0&0&0&1\end{pmatrix}$$ The probability the number is divisible by ten after $n$ selections equals $$p_{14}^{(n)}=e_1^TP^n e_4$$ |
Full House Probability: Why is my answer incorrect? | You are assuming that the Full House, which is of the form $AAABB$ for some $A$ and $B$, is being dealt to you in that very order: i.e. that first you get all the three $A$'s, and then you get the two $B$'s. However, if the cards are being dealt to you in the order $BAABA$ you end up with the same Full House. Indeed, there are $5 \choose 3$ ways in which the three $A$'s are being dealt to you to get three $A$'s and two $B$'s. |
When does a Markov chain not have a steady state? | Every Markov chain on a finite state space has an invariant distribution. As you said, this follows directly from the condition that the rows sum to $1$.
It is possible for a Markov chain on a finite state space to have multiple invariant distributions. However, the Perron-Frobenius theorem tells us that these can be decomposed into distributions which are concentrated on strongly connected components of the state space. Decomposing the process into strongly connected components results in one or more different Markov chains each of which has a unique invariant distribution.
Any transient states of the original chain will not be states in any of these sub-chains, since they are not in any strongly connected component. Confusingly, this means that a chain that is not irreducible is generically not reducible into chains on disjoint subsets of the state space. That's because generically some transient states would have to be in more than one sub-chain, which is because generically one can reach more than one strongly connected component starting from a transient state.
As an important special case, an irreducible finite state Markov chain has a unique invariant distribution which assigns positive probability to all states.
However many invariant distributions the process has, it can happen that no invariant distribution is approached over time. When the state space is finite, it turns out that this only happens when the chain is "periodic" (meaning that there are states $i,j$ and an integer $n>1$ such that all paths from $i$ to $j$ have a length which is a multiple of $n$). In this case the transition matrix has an eigenvalue which is not $1$ and has modulus $1$. If the corresponding eigenvector contributes to the initial condition, then its contribution does not decay, and no invariant distribution is approached. The classic example of a periodic chain is $P=\begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}$, but in general it is not essential that the chain is deterministic for it to be periodic.
It is possible for a Markov chain on an infinite state space to not have any invariant distributions. This is roughly because probability mass can escape to infinity. A simple example is a Markov chain on $\mathbb{Z}$ which deterministically moves one unit to the right at every step. Another example is the simple symmetric random walk: it has an invariant measure which is uniform on $\mathbb{Z}$, but this measure cannot be normalized. |
Characteristic curves : Why do I get different results? | They are a few typos in your answer.
And :
Also, the arctan and arccos must be considered as multivalued in the final equations. With convenient choice of signs $\quad \cos^{-1}\left(\frac{y}{\pm\sqrt{x^2+y^2}}\right)\pm\frac{\pi}{2}= \tan^{-1}\left(\frac{y}{x}\right)$
So, the result is the same for both ways. |
What functions satisfy $f(x)=f(x-1)+1$ | Define $g(x) = f(x)-x$. Then the equation says exactly that $g$ is periodic of period $1$. So the general solution is $f(x) = x+g(x)$, where $g$ is any function of period $1$. |
Notation of integral. | In this case, $f(x)$ would be the probability density function and $F(x)$ would be the cumulative distribution function. They are related by
$$f(x)=\frac{dF(x)}{dx}\ ,$$
so $dF(x)=f(x)\,dx$ and the two integrals are equivalent.
In practice, to evaluate the first integral you would probably begin by transforming it to the second anyway. |
Tools for plotting behavior of differential equations | I'll second Matlab but make specific metion to the Matlab toolkits pplane
and Matcont.
Matcont is particularly useful for investigating bifurcations, which seems to be what you're interested in. |
Closed form for $\sqrt{1+\sqrt[2!]{2^2+\sqrt[3!]{3^3+...}}}$ | Given the fact that neither Somos' quadratic recurrence constant, nor the nested radical constant are known to possess a closed form, I find it highly doubtful that this one will fare any better... Same as to the nature of the number, given that the nature of the afore-mentioned two constants is also unknown. |
A few questions about KSVD algorithm (dictionary learning) in a paper | $x_{T}^{k}$ denotes the $k$th row in $X$
The main idea is that: the authors first defined some dictionary, and then based on them, using K-SVD to denoising the image. After that, the dictionary is updated by OMP (Orthogonal Matching Pursuit). This work will be iteratively done until satisfying convergence condition.
$K-SVD$ is a very original paper on dictionary learning, the authors are initially defined $w_{k}$. But, the following work from other researchers show that it is not quite good. Tons of papers have been trying to improve the performance of $K-SVD$. For better performance, you may read some papers which don't need to initially define the dictionary (dictionary is defined by noisy data):
[1] K. Dabov, A. Foi, V. Katkovnik, and K. Egiazarian, “Image Denoising by Sparse 3D Transform-domain Collaborative Filtering,” IEEE Trans. on Image Process., vol. 16, no. 8, pp. 2080-2095, Aug. 2007
[2] K. Dabov, A. Foi, V. Katkvnik, K. Egiazarian, “BM3D Image Denoising with Shape-Adaptive Principal Component Analysis,” Proc. of Workshop on Signal Process. With Adaptive Sparse Structure Representations, 2009
[3] P. Chatterjee, and P. Milanfar, “Patch-Based Near-Optimal Image Denoising,” IEEE Trans. on Image Process., vol. 21, no. 4, pp. 1635-1649, Apr. 2012 |
Is there any state that satisfies for all $x$, $x$ equals $5$ | What you are calling a state seems to be what it usually called a model. It is a set of elements and relationships between them that satisfy a set of axioms. If you have an axiom $\forall x (x=5)$ the model $\{5\}$ satisfies that axiom. If you have an axiom (your belief) that $\forall x (x \neq 5)$ any model that does not include $5$ satisfies it. |
Finding a plane parallel to a line | Choos any vector not parallel to $(1,-1,3)$ such as $(1,0,0)$
The vector perpendicular to both of these, i.e. The cross product is $(0,3,1)$
This is the normal to the plane.
The equation of the plane is therefore $3y+z=d$
Now we choose $d$ so that the point $(1,-2,0)$ does not lie on the plane, such as $d=0$ and you are done. |
An uncountable set of reals in which any finite sum of elements is irrational - without choice | Yes: You can take
$$ A=\biggl\{ \sum_{n=0}^\infty 10^{-n^2} f(n) \biggm| f \in \mathbb \{1,2\}^{\mathbb N} \biggr\}$$
This clearly has $2^{\aleph_0}$ elements. And the nonzero digit in the decimal expansions are ever further from each other, so for every finite sum of elements from the set, there will be increasingly long runs of $0$s in the decimal expansion of the sum ...
If we change the exponents in the scaling factor $10^{-n^2}$ to grow fast enough (something like $10^{-n!}$ should suffice), we can even make sure that all of the finite sums will be Liouville numbers and therefore transcendental. |
The characteristic values of the p-th compound matrix | Let $V$ be a complex $n$-dimensional vector space. What I get out of this in modern language is that:
the linear transformations $V\to V$ with $n$ distinct eigenvalues are dense in the space of all linear transformations,
the map taking a linear transformation $A:V\to V$ to its multiset of eigenvalues is continuous (in Hausdorff distance, say),
the map taking a linear transformation $A:V\to V$ to its $p$th exterior power $\bigwedge^p A:\bigwedge^p V \to \bigwedge^p V$ is also continuous (the author uses $\mathfrak{A}_p$ and "compound matrix" for this latter concept), and
if a linear transformation $A:V\to V$ has $n$ distinct eigenvalues, then the eigenvalues of $\bigwedge^p A:\bigwedge^p V\to \bigwedge^p V$ are given by all products of $p$-element subsets of the set of eigenvalues of $A$.
Consequently, the author concludes that for any linear transformation $A$ (whose eigenvalues need not be distinct), the eigenvalues of $\bigwedge^p A:\bigwedge^p V\to \bigwedge^p V$ are given by all products of $p$-element sub-multisets of the multiset of eigenvalues of $A$ (accounting properly for multiplicity). To draw this conclusion, the author uses density and continuity to reduce to the case of $n$ distinct eigenvalues for which the result has already been proven. |
Optimization Calculus Question | Your problem should have said $a > 0$ and $b > 0$ because the function has no maximum value in the interval if either is negative.
That being said,
$$ x(1-x) \frac{d}{dx} x^a (1-x^b) = a(1-x) - bx$$Setting that to zero gives
$$
x = \frac{a}{a+b}$$
and then $y$ can be read off. |
Pick out correct statement | a. is true by the identity theorem.
b. is true by the fundamental theorem of algebra.
c. If $f$ is a non-constant analytic function in the disc $D=\{z: |z| < 1 \}$, then $f(D)$ is open in $\mathbb C$. Hence c. is not true. |
Can you please help with the inscribed angle theorem? | "On the remaining part of the circle" means not on the arc that was mentioned in the first part of the theorem.
The radii drawn in red divide the circle into two arcs.
Arc $A$ is the short arc that sits inside the angle marked $2x$ and is the "arc of a circle" that subtends the angle $2x$ at the center of the circle.
Arc $B$ is the long arc that goes around the top of the circle from one end of the short arc to the other. Part B is "the remaining part of the circle".
From any point on arc $B,$ arc $A$ subtends an angle $x.$
On the other hand, you could take arc $B$ as the "arc of a circle"
and take arc $A$ as the "remaining part",
in which case you find that arc $B$ subtends the angle
$2\pi - 2x$ (that is, $360^\circ - 2x$)
and the yellow angle on arc $A$ is half that angle. |
Calculate the value of the error (only) with two decimal places for: $(5.1 +/- 0.4) / (2.5 +/- 0.2)$ | I get $Z_{max}=5.5/2.3\approx 2.3913$, while $Z_{nom}=5.1/2.5= 2.04$. Your answer has more than two places. |
Understanding a combinatorics problem | Well take that there are $4!$ combinations when $ag_3$ are together(take $n=\{ag_3\}$ so $\{nb_1b_2b_3\}$ has 4! permutations,so the number of exactly two girls together is $5!-4!=96$.Now you can take 2 girls in 3 ways $\{g_1g_2\},\{g_1g_3\},\{g_2g_3\}$ now the number of combinations so that there are exactly two girls together is $3*(5!*2!-96)$ and there are $4!*3!=144$ combinations with exactly 3 girls sitting together now $6!-3*(5!*2!-96)-144=144$ is the number of combinations when girls are setting apart(we substracted the number of all girls sitting together and exactly two girls sitting together) |
Regularity of Laplace equation | In order to obtain a unique weak solution, you additionally need some boundary condition. Let us for sake of exposition assume that we require $u \in W_0^{1, 2}(\Omega)$. Then your statement is true in the sense that for any $r \in (0, 1)$ and $M_1 \gt 0$, there is $M_2 \gt 0$ such that if $f \in L^\infty(\Omega)$ with $\|f\|_{L^\infty(\Omega)} \le M_1$ then the corresponding solution fulfills $\|u\|_{C^{1+r}(\overline \Omega)} \le M_2$.
One way to prove that: For any $p \in (1, \infty)$, the operator $A: W_0^{2, p}(\Omega) \to L^p(\Omega)$, $\varphi \mapsto \Delta \varphi$ is an ismorphism. Thus, if $u$ is a solution to your PDE with right hand side $f$, then $u = A^{-1} f$ and hence $\|u\|_{W^{2, p}(\Omega)} \le \|A^{-1}\| \|f\|_{L^p(\Omega)} \le C \|f\|_{L^\infty(\Omega)}$.
By Morrey's inequality, $W^{2, p}(\Omega) \hookrightarrow C^{1 + r}(\overline \Omega) $ for $r = 1-\frac np$ provided $p > n$, so that we obtain the desired inequality by choosing $p$ large enough.
A more elementary proof can be found in Gilbarg–Trudinger, if I remember correctly.
Finally, a small warning: The argument fails for $p=\infty$ and you do not obtain a bound in $C^2(\overline \Omega)$. |
Prove that there is a point $z \in [a,b]$ such that $f(z)= {f(x_1)+ \dots + f(x_n)\over n} $ | Sketch: let $m=\min_i f(x_i)$ and $M=\max_i f(x_i)$ both of which are attained since $f$ is bounded and the sequence is finite. Let $A$ denote the average of $f$ over the points. Then $m\leq A\leq M$. By the intermediate value theorem there is a point $z$ that achieves the average.
Now figure out why you need $f(a)\neq f(b)$ to make the above precise. I have a feeling you meant to say $z\in (a,b)$. |
Binary relation with a for all (x,y) | This means $x$ is related to $y$ (or $xRy$) if there exists a $z\in\mathbb{Z}$ (that may depend on $x$ and $y$) (and I assume that $Z$ in your question is the set of integers) such that $\frac{x}{y} = 3^z$. So, to check whether or not $1R1$, you have $x=1$, $y=1$, and you ask yourself, "Is there a $z\in\mathbb{Z}$ such that $\frac{1}{1} = 3^z$?." The answer is yes ($z=0$), and you conclude, "Hey $1R1$". How about $1R2$? You ask whether there is a $z\in\mathbb{Z}$ such that $\frac{1}{2} = 3^z$. Now, the answer is no, and you continue like this.. |
Conceptual differences between the notations $\int_{a}^{b}f$ and $\int_{[a,b]}f$ | I'd explain it in the following way:
1. The expression
$$\int_{[a,b]}f(x)\>{\rm d}x$$
denotes a limit of Riemann sums:
$$\int_{[a,b]}f(x)\>{\rm d}x:=\lim_{\ldots}\sum_{k=1}^N f(\xi_k)\>\mu(I_k)\ ,$$
whereby we have partitions of $[a,b]$ into subintervals $I_k$ of length $\mu(I_k)$ in mind. This idea can immediately be transported to a higher-dimensional setting: If, e.g., $B\subset{\mathbb R}^3$ is a ball then it makes sense to consider
$$\int_B f({\bf x})\>{\rm d}({\bf x}):=\lim_{\ldots}\sum_{k=1}^N f(\xi_k)\>\mu(B_k)\ ,$$
where we have partitions of $B$ into small pieces $B_k$ of volume $\mu(B_k)$ in mind. Note that lengths and volumes are just nonnegative numbers and have no inherent "orientation" or "direction".
2. On the other hand we have learnt to differentiate and to "integrate" functions $f:\>{\mathbb R}\to{\mathbb R}$. Given such an $f$ the set of functions $F$ such that $F'=f$ is completely determined by $f$ and is denoted (for whatever reasons$\ldots$) by $$\int f(x)\>dx\ .$$ After we have found one such primitive $F$ of $f$ we know all of them, and we could denote this set as well by $\langle F\rangle$, or similar. It so happens that time and again we have to deal with differences $F(v)-F(u)$ even before we know $F$ explicitly. To handle this in a swift way we introduce the following notation: Given any two points $u$ and $v$ in the domain of $f$ the expression
$$\int_u^v f(x)\>dx$$
is called the definite integral of $f$ from $u$ to $v$, and denotes the difference $F(v)-F(u)$, computed with any primitive $F$ of $f$. Apart from the definition of derivative there are no limits involved in this setup, and formulas like
$$ \int_u^v f(x)\>dx=-\int_v^u f(x)\>dx,\quad \int_{\phi(u)}^{\phi(v)} f(x)\>dx=\int_u^v f\bigl(\phi(t)\bigr)\>\phi'(t)dt$$
are immediate.
The connection between these two streams of thought is the following: If $a<b$ then
$$\int_{[a,b]}f(x)\>{\rm d}x\ =\ \int_a^b f(x)\>dx\ .$$
This is not a tautology, but a total miracle, and is called the Fundamental Theorem of Calculus. |
Surface Integral over scalar field | Total mass of S is given by
$$\mathrm{
\iint_{S}m\;\color{blue}{dS}}
$$
Using spherical coordinates (physics convension) this integral can be written as
$$\mathrm{
\int_0^{2\pi}\int_0^{\pi\over2} R^2\sin^2\theta\;\color{blue}{R^2\sin\theta\;d\theta\;d\phi}=\frac{4}{3}\pi R^4
}$$ |
Which solution is the right one?? | The Fundamental Theorem of Algebra can not be extended to power series.
Example:
$$e^z\neq 0$$
that is
$$1+z+\frac{z^2}{2!}+\frac {z^3}{3!}+\cdots\neq 0$$ |
Angle between two 3D lines | You can think of the formula as giving the angle between two lines intersecting the origin. So just "move" the intersection of your lines to the origin, and apply the equation.
Shifting lines by $( -1,-1,-1 )$ gives us:
Line $1$ is spanned by the vector $\vec{u} = ( 2,1,-6 )$
Line $2$ is spanned by the vector $\vec{v} = (0,-5,5)$
Now calculating the angle between the lines is a direct application of the equation you gave. |
Limit of $\frac{1}{e^3}$ | For the first one, take logarithms and note that
$$\lim_{n \to \infty} n \log(1 - \frac{3}{n-2}) = \lim_{n \to \infty} \frac{n}{n-2} \cdot \lim_{n \to \infty} (n-2) \log(1 - \frac{3}{n-2}).$$
The second limit on the right-hand side can be computed using your provided fact.
For the other question, Arthur has already given you a hint in the comments. |
Question trying to physically understand Algorithm run time as proportion | Since no one really answered this, this is what I think is some physical intuition behind the situation.
It takes $\sqrt{n}\ microseconds$ to solve a given problem, so no matter what it is true that it took $\frac{\sqrt{n}\ microseconds}{problem\ of\ any\ size\ n}$. So if the problem took 1 second to solve (for whatever n), the fraction would be $\frac{1\ seconds}{problem\ of\ size\ n}$. And since we know the first fraction is always true we can set up a proportion $\frac{\sqrt{n}\ microseconds}{problem\ of\ any\ size\ n} = \frac{1\ seconds}{problem\ of\ size\ n}$. So now we can solve the proportion for $n$. |
If $d_G(v)≤k \; \forall \; v \in V(G)$, then there exists a matching with $≥\frac{|V(G)|}{2k}$ edges | Consider a maximal matching, and suppose it has $s$ edges, or also said $2s$ vertices. The matching is maximal, so any other vertex is connected to one of the $2s$ vertices, otherwise there would be another edge that can be added to the matching.
This means that the total number of vertices is at most $2ks$ since every vertex can be connected to at most $k$ other vertices. It means
$$|V|\le 2ks \implies s\ge \frac{|V|}{2k}$$ |
Radius of convergence of a power series | You have
$$\lim_{n\to \infty} \sqrt[n]{a_n} = \lim_{n\to \infty} {3}\frac{1}{3}\sqrt[n]{a_n}= \lim_{n\to \infty} {3}\sqrt[n]{\frac{a_n}{3^n}} = 3$$
So, with root test for power series, $R=\frac{1}{3}$ |
Does the order preserve in different norm spaces? | Probably not, let $b^{(1)} = b^{(2)} = (1/4, 1/4, 1/4, 1/4)$, $a^{(1)} = b^{(1)} + (-\epsilon, -2\epsilon, -3\epsilon, 6\epsilon)$ and $a^{(2)} = b^{(2)} + (-3\epsilon, -3\epsilon, 3\epsilon, 3\epsilon)$ for some small $\epsilon$, then
$$\sum_i |a_i^{(1)} - b_i^{(1)}| = 12\epsilon = \sum_i |a_i^{(2)} -b_i^{(2)}|,$$ but
$$\sum_i |a_i^{(1)} - b_i^{(1)}|^2 = (1^2 + 2^2 + 3^2 + 6^2)\epsilon^2 > 4\times 3^2\epsilon = \sum_i |a_i^{(2)} -b_i^{(2)}|$$
More generally, your claim would imply that
$$\sum_i |a_i^{(1)} - b_i^{(1)}|= \sum_i |a_i^{(2)} - b_i^{(2)}|$$ implies
$$\sum_i |a_i^{(1)} - b_i^{(1)}|^2 = \sum_i |a_i^{(2)} - b_i^{(2)}|^2$$
which immediately seems wrong. |
Why should there is a $c\in [0,1]$: $f(c)=f(c+1)$. | $g(0) = f(1)-f(0)$ while $g(1)=f(2)-f(1) = f(0)-f(1)=-g(0)$.
So if $g(0)$ is positive, $g(1)$ is negative and vice versa (or both are $0$ already).
The intermediate value theorem (IVT) implies that a continuous $g$ must have a point where it assumes $0$ inbetween those points. |
Consider a set $U$ of $23$ different compounds in a chemistry lab | Consider this as a graph problem with $23$ vertices corresponding to $23$ different compounds. Whenever a compound reacts with other, we have an edge in between them. So, according to the question, we have $9$ vertices with exactly $3$ edges. Therefore, total degree of these $9$ vertices (subset S) is $27$ which is odd. However, we know that total degree of a graph is always even, because each edge contributes degree $2$. Therefore, total degree of the vertices in $U \setminus S$ must be odd to make the total degree of $U$ even.
Now, Statement I is not always true, because to make the total degree of $U \setminus S$ odd, each vertex of $U \setminus S$ don't need to have odd degree; only one vertex in $U \setminus S$ having odd degree is enough which makes Statement II always true. If all the vertices in $U \setminus S$ have even degrees, then the total degree of $U \setminus S$ cannot be odd. So, Statement III is always false. Hope it helps. |
Describe the complex numbers that satisfy these sums | Suppose the complex numbers $\beta_0,\ldots,\beta_{N-1}$ are completely arbitrary except for the property that they are not all zero. Define $$\beta_N = -\beta_1 - \cdots - \beta_{N-1}$$ and $$B = \sum_{i=0}^N |\beta_i|^2.$$
The numbers $\alpha_i = \dfrac{\beta_i}{\sqrt B}$ satisfy the two conditions. If these numbers are special they can't be too special since it is trivial to find examples. |
How to apply Leibniz rule for this integral? | $$\int\limits_{0}^{a(t)}{(a(t)-s)^{\alpha-1}}f(s)ds = \left|-\frac{(a(t)-s)^\alpha}{\alpha}f(s) \right|_{0}^{a(t)}+\frac{1}{\alpha}\int\limits_{0}^{a(t)}{(a(t)-s)^\alpha}{f^{\prime}(s)}ds$$ |
Why if a set is equal to its dual, then its complementary is not L3. | False. Let $\mathcal{A} = \{A \in [n] : 1 \in A\}$. Then $\mathcal{A}$ is increasing and $$\mathcal{A}^* = \{B \in [n] : [n]\setminus B \not \in \mathcal{A}\} = \{B \in [n] : [n] \setminus B \not \ni 1\} = \{B \in [n] : 1 \in B\} = \mathcal{A}.$$ However, $\mathcal{A}^c = \{A \in [n] : 1 \not \in A\}$ is $L^3$: for any $A_1,A_2,A_3 \in \mathcal{A}^c$, $1 \not \in A_1\cup A_2 \cup A_3$ and thus $A_1\cup A_2 \cup A_3 \not = [n]$. |
Verifying a Closure Operation | From $N\cap\overline{E}\neq\emptyset$ you know that $N$ is a neighborhood of some element of $\overline{E}$, and thus $N\cap E\neq\emptyset$ by your definition of $\overline{E}$. |
How could one write $Aut(C_{30})$ as a product of cyclic groups? | First of all let us choose a generator $g\in C_{30}$, and view it as $\mathbb{Z}/{30\mathbb{Z}}$.
Any automorphism is determined by the image of the class $[1]$, so the endomorphisms are given by
$$
\text{End}(\mathbb{Z}/30\mathbb{Z}) = \{[1]\mapsto[n]\ \mid n\in \mathbb{Z}\}.
$$
The multiplication structure of composing endomorphisms comes down to multiplying the integers $n$, so the automorphisms are exactly the endomorphisms such that the class $[n]$ is invertible mod 30.
With this knowledge we find
$$
\text{Aut}(\mathbb{Z}/30\mathbb{Z}) \cong (\mathbb{Z}/30\mathbb{Z})^*.
$$
By the chinese remainder theorem,
$$
(\mathbb{Z}/30\mathbb{Z})^* \cong (\mathbb{Z}/5\mathbb{Z})^* \times (\mathbb{Z}/2\mathbb{Z})^* \times (\mathbb{Z}/3\mathbb{Z})^* \cong (\mathbb{Z}/4\mathbb{Z})^* \times (\mathbb{Z}/2\mathbb{Z})^*.
$$
EDIT: replaced automorphism with endomorphism. |
A question about covariant functor | I assume that by $S$ you mean the category of sets and functions of sets.
Now first, if you have a function $\phi\colon A \to B$ then for $h_X$ to be covariant this means $\phi$ should induce a function $\hom(X, A) \to \hom(X, B)$. And indeed it does. This function maps $X \overset{\psi}{\to} A$ to the composition $X \overset{\psi}{\to} A \overset{\phi}{\to} B$. I recommend that you check that the appropriate axioms are satisfied.
Second, if $X$ and $Y$ are sets then the cartesian product $X \times Y$ is certainly a set, so yes $X \times Y$ is an object in $S$.
Now for $(f, h_X)$ to be an adjoint pair you need an isomorphism
$$\hom(X \times A, B) \simeq \hom(A, \hom(X, B))$$
which is functorial in $A$ and $B$. So you need to take a function $\phi\colon X \times A \to B$ and construct a function $A \to \hom(X, B)$. How about sending $a \in A$ to the function $\phi(-, a)\colon X \to B$? I'll leave you to check that this gives an isomorphism between those hom sets and is appropriately functorial in $A$ and $B$. |
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