问:
Is there a Weierstrass function with asymptotic line?
I haven't seen a Weierstrass function with asymptotes.
They are all fractals and the sum of trigonometric functions.

If you know a function that is nowhere-differentiable but is continuous and has asymptotes, please let me know, and if not, please explain why.

答:
Let $f : \mathbb (0,1) \to \mathbb R$ be continuous and nowhere differentiable.  Then
$f_2(x) = \arctan(f(x))$ is continuous, nowhere differentiable, and bounded.  Then
$$f_3(x) = f_2(x)+\frac{1}{x}$$ is continous, nowhere differentiable, and has vertical asymptote at $x=0$.

......

Let $g :\mathbb R \to \mathbb R$ be continuous and nowhere differentiable.  Then $g_2(x) = \arctan(g(x))$ is continuous, nowhere differentiable, and bounded.  Then
$$g_3(x) = e^{-x}g_2(x)$$
is continous, nowhere differentiable, and has horizontal asymptote as $x \to +\infty$.


问:
The formula of matrix calculus
Suppose $A \in \mathbb{R}^{p \times p}$ is a semi-positive defined symmetric matrix. Then $A^{1/2}$ is well-defined. Now I want to know does there exists a formula for
$$\frac{\partial A^{1/2} }{\partial A}  ?$$
Thanks so much!

BTW, if we find an explicit formula for $\frac{\partial A^{1/2} }{\partial A}$, then does the following approximation holds?
$$B^{1/2} - A^{1/2} = \frac{\partial A^{1/2} }{\partial A} (B - A)+ O(\|B-A\|_2^2) $$
for a proper $p \times p$ matrix $B$?

答:
$\def\p#1#2{\frac{\partial #1}{\partial #2}}$
The square root function is
$$F(S) = S^{1/2}$$
For an SPD matrix, it can be squared, differentiated, vectorized, and inverted to yield
$$\eqalign{
S &= F^2 \\
dS &= dF\,F + F\,dF \\
{\rm vec}(dS)
 &= (F\otimes I + I\otimes F)\;{\rm vec}(dF) \\
 &= (F\oplus F)\;{\rm vec}(dF) \\
ds &= (F\oplus F)\,df \\
df &= (F\oplus F)^{-1}\,ds \\
  &= G\,ds \\
\p{f}{s} &= G \\
}$$
where $(\otimes,\oplus)$ denote the Kronecker product and Kronecker sum, respectively.

In vectorized form, the first order Taylor expansion is
$$\eqalign{
df &\approx G\,ds \quad\implies\quad
f(s+ds)-f(s) &\approx G\,ds \\
}$$
The matrix form requires the gradient as a fourth-order tensor
and the double-dot product
$$\eqalign{
dF &\approx \Gamma:dS \quad\implies\quad
dF_{ij} &\approx \sum_{k}\sum_{\ell}\,\Gamma_{ijk\ell}\;dS_{k\ell} \\
}$$
Converting between the matrix/tensor forms of the gradient $(G/\Gamma)$ is simple but tedious.

In general
$$\eqalign{
G &\in {\mathbb R}^{mn\times pq} \quad\iff\quad
\Gamma \in {\mathbb R}^{m\times n\times p\times q} \\
G_{\alpha\beta} &= \Gamma_{ijk\ell} \\
\alpha &= i+(j-1)\,m \\
\beta &= k+(\ell-1)\,p \\
i &= 1+(\alpha-1)\,{\rm mod}\,m \\
j &= 1+(\alpha-1)\,{\rm div}\,m \\
k &= 1+(\beta-1)\,{\rm mod}\,p \\
\ell &= 1+(\beta-1)\,{\rm div}\,p \\
}$$
However, in this particular case  $\;m=n=q=p$

Extend this solution to semi-SPD matrices $(A,B)\,$ by writing them as perturbations $(\lambda,\mu\to 0)$ of $S$ in symmetric matrix directions $(X,Y)$
$$\eqalign{
A &= S + \lambda X, \qquad B &= S + \mu Y \\
}$$
Then we can use the SPD solution, with the gradient evaluated at $\Gamma=\Gamma(S),\,$ to write
$$\eqalign{
F(B) - F(S) &\approx \Gamma:\mu Y \\
F(A) - F(S) &\approx \Gamma:\lambda X \\
}$$
Subtraction then yields the expected formula
$$\eqalign{
F(B) - F(A) &\approx \Gamma:(\mu Y - \lambda X) \\
            &\approx \Gamma:(B-A) \\
}$$

ContentLicense: CC BY-SA 4.0

Title: The formula of matrix calculus


问:
Solve polynomial interpolation $f(x)=\frac{bc(x - b)(x-c)}{(a - b)(a-c)} + \frac{ac(x - a)(x - c)}{(b - a)(b - c)} + \frac{ab(x-a)(x-b)}{(c-a)(c-b)}$
I think the polynomial

$$f(x)=\frac{bc(x - b)(x-c)}{(a - b)(a-c)} + \frac{ac(x - a)(x - c)}{(b - a)(b - c)} + \frac{ab(x-a)(x-b)}{(c-a)(c-b)}$$

may be easy to find, but I am a bit unexperienced. So, I would be glad to get some help!

$$f(a) = bc,\>\>\>f(b) = ac,\>\>\>f(c) = ab$$

which is what the polynomial should satisfy.

答:
Let $f(x)=\alpha x^2 + \beta x + \gamma$. Then
\begin{align}
& f(a)=\alpha a^2 + \beta a + \gamma = bc \tag1\\
 & f(b)=\alpha b^2 + \beta b + \gamma = ca \tag2 \\
 & f(c)=\alpha c^2 + \beta c + \gamma = ab \tag3 \\
\end{align}
Take (1)-(2) and (1)-(3) to get
\begin{align}
& \alpha (a+b) + \beta +c= 0\tag4 \\
 & \alpha (a+c) + \beta +b = 0\tag5\\
\end{align}
Then, take (4)-(5) to get $\alpha =1$. Substitute into (5) to get $\beta = -(a+b+c)$ and, in turn, $\gamma = ab+bc +ca$. Thus
$$f(x) = x^2 -(a+b+c)x + ab+bc+ca$$


问:
Parent and childs of a full d-node tree
i have a full d-node tree (by that mean a tree that each node has exactly d nodes as kids).
My question is, if i get a random k node of this tree, in which position do i get his kids and his parent?

For example, if i have a full binary tree, the positions that i can find the parent,left and right kid of the k node are $\dfrac k2, 2k, 2k+1$ respectively.

Thanks in advance.

答:
If you use zero-based indexing, so that the root is position $0$, then the last node in level $\ell$ of the tree has index $\sum_{i=1}^\ell d^i=\frac{d(d^\ell-1)}{d-1}$. Thus, node $k$ is in level $\ell$ iff

$$\frac{d}{d-1}(d^{\ell-1}-1)<k\le\frac{d}{d-1}(d^\ell-1)\;,$$

or

$$d^{\ell-1}<k\left(1-\frac1d\right)+1\le d^\ell\;,$$

i.e.,

$$\ell=\left\lceil\log_d\left(k\left(1-\frac1d\right)+1\right)\right\rceil\;.$$

Let

$$\ell(k)=\left\lceil\log_d\left(k\left(1-\frac1d\right)+1\right)\right\rceil\;,$$

the level of node $k$. Node $k$ has

$$k-\frac{d(d^{\ell-1}-1)}{d-1}-1$$

predecessors in level $\ell(k)$, each of which has $d$ offspring, so its $d$ offspring have indices from

$$\frac{d(d^\ell-1)}{d-1}+d\left(k-\frac{d(d^{\ell-1}-1)}{d-1}-1\right)+1=dk+1$$

through $dk+d=d(k+1)$.

Now assume that $k>0$, and node $m$ is the parent of node $k$. Then

$$dm+1\le k\le d(m+1)\;,$$

so

$$m+\frac1d\le\frac{k}d\le m+1\;,$$

and therefore $$\left\lceil\frac{k}d\right\rceil=m+1\;,$$

i.e.,

$$m=\left\lceil\frac{k}d\right\rceil-1\;.$$

ContentLicense: CC BY-SA 3.0

Title: Parent and childs of a full d-node tree


问:
numerically integrating $\int_0^{10} \cos x^3 dx $
The integrand is highly oscillating at the right end of the interval.

I thought of this way

$$ I = \int_0^1 \cos x^3 d x + \int_1^{10} \cos x^3 d x = \int_0^1 \cos x^3 d x + \int_1^{1000} \cos y d y^{1/3} . $$

The second integral is free of singularity now. But the problem is that the interval is too big. Moreover, there is much cancellation there.

Do anyone have any better idea?

答:
I'll consider only the last integral of the form
$$
I = \int_1^{1000} f(y) \cos y dy.
$$
Let's split the integration domains into $2\pi$ segments like
$$
I = \int_1^\pi f(y) \cos y dy + \int_{\pi}^{3\pi} f(y) \cos y dy + \dots
+ \int_{315\pi}^{317\pi} f(y) \cos y dy + \int_{317\pi}^{1000} f(y) \cos y dy.
$$
Again, considering only the middle terms we can write
$$
I' = \sum_{k=1}^{158} \int_{(2k-1)\pi}^{(2k+1)\pi} f(y) \cos y dy.
$$
Each integral can now be converted by a variable transformation to an integral for $[-\pi, \pi]$ domain:
$$
I' = \sum_{k=1}^{158} \int_{-\pi}^{\pi} f(y + 2k\pi) \cos z dz.
$$
Moreover, the sum and the integral sign can now be interchanged:
$$
I' = \int_{-\pi}^{\pi} \underbrace{\sum_{k=1}^{158} f(z + 2k\pi)}_{g(z)} \cos z dz.
$$
Not the original problem is reduced to evaluating only five integrals without highly-oscillating features. Instead you need to evaluate $g(z)$ which may be quite expensive.


问:
How to prove that this is a group isomorphism?
The conditions are the following.
Let $(G,\circ)$ be a group, and $a$ an element of $G$.

Prove that this is a group isomorphism:

$$\phi : G \to G : \quad\phi(x) = a^{-1}\circ x\circ a$$

I know that I have to prove it is:

well-defined

bijective

homomorphic

The main question I have is can I switch the e.g. $a$ and $x$? So the function would be $\phi(x)= a^{-1} \circ a \circ x$, then it would be pretty easy, but since $G$ is not abelian I am not quite sure how to proof the 3 axioms for the isomorphism.

Can anyone help me?

答:
1)
$$x=y\Rightarrow a^{-1}x=a^{-1}y \Rightarrow  a^{-1}xa=a^{-1}ya \Rightarrow \varphi(x)=\varphi(y)$$
2)
The implications above are actually equivalences(because of the cancellability in a group), so the function is injective. Let's proof it's surjective:
$g\in G$. Since $g=a^{-1}(aga^{-1})a=\varphi(aga^{-1})$, the function is surjective.

3)
$$\varphi(xy)=a^{-1}xya=a^{-1}x1ya=a^{-1}x(aa^{-1})ya=(a^{-1}xa)(a^{-1}ya)=\varphi(x)\varphi(y)
$$


答:
Hint:

Since $\varphi$  maps element to element, it is well-defined.

Consider $\phi:G\rightarrow G$ defined by $\phi(x)=a\circ x\circ a^{-1}$ for all $x\in G$. Try to show that $\phi\varphi=\text{id}_G=\varphi\phi$. This shows that $\varphi$ is bijective.

Let $x,y\in G$. Check that $$\varphi(x\circ y)=a^{-1}\circ (x\circ y)\circ a=(a^{-1}\circ x\circ a)\circ (a^{-1}\circ y\circ a)=\varphi(x)\circ\varphi(y).$$
This shows that $\varphi$ is a homomorphism.

Title: How to prove that this is a group isomorphism?

Body:
The conditions are the following.
Let $(G,\circ)$ be a group, and $a$ an element of $G$.

Prove that this is a group isomorphism:

$$\phi : G \to G : \quad\phi(x) = a^{-1}\circ x\circ a$$

I know that I have to prove it is:

well-defined

bijective

homomorphic

The main question I have is can I switch the e.g. $a$ and $x$? So the function would be $\phi(x)= a^{-1} \circ a \circ x$, then it would be pretty easy, but since $G$ is not abelian I am not quite sure how to proof the 3 axioms for the isomorphism.

Can anyone help me?


问:
When will two cars pass each other
There was a question in my math text book the other day that stated:

$2$ cars each travelling at a constant velocity around a ring ,
complete exactly $4$ and $7$ rounds in one hour. If they start at exactly the same time from the same place but travel in opposite directions around the track, how many times will they
pass each other in the one hour?

I tried various things using $\displaystyle v = \frac{d}{t}$ but couldnt get very far. Any help is appreciated, thanks.

答: Let's look from the driver in the slower car's point of view. If he is traveling at $4$ laps per hour and the other car is traveling at $7$ laps per hour in the opposite direction, it looks like the other car is traveling at $11$ laps per hour. From the slower car's viewpoint, he seems to be sitting still while the other car has passed him $11$ times.</p>

Title: When will two cars pass each other


答: For the surjectivity consider arbitrary $g\in G$ then we know that $a\circ g \circ a^{-1}\in G $ because $a\in G$ and $G$ is a group but so $\phi(a\circ g \circ a^{-1})=a^{-1}\circ a\circ g\circ a^{-1}\circ a=g$. Because $g$ was arbitrary the map is surjective. Now for the injectivity assume $\phi(g)=\phi(h)$, i.e. $a^{-1}\circ g \circ a=a^{-1}\circ h \circ a$ which is equivalent to $a\circ a^{-1}\circ g \circ a\circ a^{-1}=a\circ a^{-1}\circ g \circ a\circ a^{-1}$, i.e. $g=h$ since we are in a group, so the right and left translation maps are bijective. Now for the homomorphism we have $\phi(g\circ h)=a^{-1}\circ g \circ h\circ a=a^{-1}\circ g\circ a\circ a^{-1} \circ h\circ a=\phi(g)\circ\phi(h)$ so we are done.


问:
Show that $\int_0^{2 \pi}\vert rie^{it}\vert dt=2 \pi r$
I am struggling with an normally simple integral calculation. I wanna show:
$$\int_0^{2 \pi}\vert rie^{it}\vert dt=2 \pi r$$
I tried the following:
$\int_0^{2\pi}\vert rie^{it}\vert dt=\vert ri \vert \int_0^{2\pi}\vert e^{it}\vert dt = $r$ *(\vert e^{(2 \pi i)} \vert- \vert e^{0} \vert)=0$...

Where is my mistake? Thanks for your help.

Body: since $|r|=r, |i|=1, |e^{it}|=1$, $$I=\int_{0}^{2\pi} |ri e^{it}| dt=\int_{0}^{2\pi} r dt=2\pi r.$$


问:
Construct a linear programming problem for which both the primal and the dual problem has no feasible solution
Construct (that is, find its coefficients) a linear programming problem with at most two variables and two restrictions, for which both the primal and the dual problem has no feasible solution.

For a linear programming problem to have no feasible solution it needs to be either unbounded or just not have a feasible region at all I think. Therefore, I know how I should construct a problem if it would only have to hold for the primal problem. However, could anyone tell me how I should find one for which both the primal and dual problem have no feasible solution? Thank you in advance.

答: Let $A=\left(\begin{smallmatrix} -1&0\\0&1\end{smallmatrix}\right)$, $b=\left(\begin{smallmatrix}1\\1\end{smallmatrix}\right)=-c$.  $Ax\ge b$ and $A^Ty\le c$ cannot both be satisfied with positive $x,y$.</p>


问:
Coordinate system change and Jacobi matrix
Assume we want to change variables from the cartesian coordinates by the following transformation:

$ x=u,\thinspace\thinspace\thinspace y=v $

Then Jacobi matrix of this transformation would be

$ \begin{pmatrix}\frac{\partial x}{\partial u} & \frac{\partial x}{\partial v}\\
\frac{\partial y}{\partial x} & \frac{\partial y}{\partial v}
\end{pmatrix} $

I cannot understand what is actually the transformation. As far as Im familier with, Jacobi matrix represent the differential of a function, that is, given a differentiable function, say, $ f: \mathbb{R}^2 \to \mathbb{R}^2 $

Jacobi matrix of $ f $ in some point $ x $ where $ f $ is differentiable at, would be   $ \begin{pmatrix}\frac{\partial f^{1}}{\partial x} & \frac{\partial f^{1}}{\partial x}\\
\frac{\partial f^{2}}{\partial y} & \frac{\partial^{2}}{\partial y}
\end{pmatrix} $

So what exactly plays the role of $ f $ in the changing of the coordinate systen $ x=u $ and $y=v $ ?

I hope my question is clear.

Thanks in advance

答:
It's easy to get confused with notation when dealing with co-ordinate transformations. A co-ordinate transformation of $\mathbb R^2$ is by definition a differentiable map $f:\mathbb R^2 \to \mathbb R^2$. This can be written, using $(x,y)$ to denote a point in $\mathbb R^2$, as $(x,y) \to (f_1(x,y),f_2(x,y))$, or alternatively we can denote the transformed co-ordinates as $(u,v)$ in which case $(x,y) \to (u(x,y),v(x,y))$.

Thinking about $f$ as the polar co-ordinate transformation, the transformation from cartesian $(x,y)$ to polar $(r,\theta)$ co-ordinates is given by $f(x,y) = (r,\theta)$ where $r$ and $\theta$ are functions of $x$ and $y$ defined as
$$ r(x,y) = \sqrt{x^2 + y^2}$$
$$\theta(x,y) = arctan(y/x)$$
whereas the inverse transformation from polar to cartesian is given by $f^{-1}(r,\theta) = (x,y)$ where
$$x(r,\theta) = r \cos \theta$$
$$y(r,\theta) = r \sin \theta$$


问:
How to find all planar squares with respect to the taxicab norm or the uniform norm?
Excuse me if this is a silly question. Let $\|\cdot\|$ be any norm on $\mathbb R^2$. We say that four points $a,b,c,d\in\mathbb R^2$ form a square with respect to the norm $\|\cdot\|$ if $\|a-b\|=\|b-c\|=\|c-d\|=\|d-a\|$ and $\|a-c\|=\|b-d\|$. Conceptually, this means the four sides of the polygon with $a,b,c,d$ as vertices have equal lengths and so do the two diagonals.

How to characterize all squares w.r.t. the $1$-norm or the $\infty$-norm? It isn't hard to see that all "Euclidean squares" (i.e. squares w.r.t. to the Euclidean norm) are squares w.r.t. $\|\cdot\|_1$ or $\|\cdot\|_\infty$, but is the converse true?

答:
Your definition of square is somewhat problematic.  An example of a Euclidean square, by your definition, would be $ (0,0), (1,0), (0,0), (1,0) $.  However if you insist that the points be distinct, I think the definition works as intended.

Counter-example
However, $ l^\infty $ seems to admit some additional squares.  A set of examples is $ (0,0), (1,a), (0,a), (1,0) $ for $ -1 < a < 1 $.  This at least shows that not all $ l^\infty $ squares are Euclidean squares.  Moreover, since $ l^\infty(\mathbb{R}^2) $ and $ l^1(\mathbb{R}^2) $ are isometric (via the transformation $ l^1 \ni (x,y) \leftrightarrow (x+y, x-y) \in l^\infty $), anything we say about $ l^\infty $ squares carries over to $ l^1 $ squares.

Possible generalizations I don't think a proof would be terribly interesting, but I am fairly confident that every non-Euclidean-square $ l^\infty $-square (by the definition you gave) looks approximately like the set of examples I just gave, up to rescaling and 90 degree rotations.  If you really want to write down a proof, begin by noting that after translation, rescaling, and 90 degree rotations (and cyclic relabeling of vertices), any example of a square can be put in the standard position $ OPQR $ where
$$
O=(0,0), P=(1,\alpha), |\alpha| \leq 1
$$
Then one does some case analysis on the possible locations of $ Q $ which must be a distance $ 1 $ from $ P $, and hence must be on one of the four edges of the (square-shaped) sphere of radius 1 centered at $ P $.

Salvaging the conjecture You could thus salvage your conjecture if you additionally stipulate that squares be non-self-intersecting.  In that case, all $ l^\infty $ squares really would be Euclidean squares.  After selecting $ O = (0,0)  $ and $ P = (1,\alpha), |\alpha| \leq 1 $, the non-self-intersection assumption forces $ Q = ( 1 + \beta, \pm 1 ) $ where $ |\beta| \leq 1 $.  Assume $ Q = (1 + \beta, 1) $ without loss of generality.  Split into four cases by $ \alpha \leq 0 $, $ \alpha > 0 $, $ \beta \leq 0 $ and $ \beta > 0 $.  In each case we find we are forced by the restrictions to either take $ \alpha = \beta = 0 $ or $ \beta = -\alpha $, and in all cases we recover that $ OPQR $ is a Euclidean square.


问:
finding an example of an equivalence relation
Find an example of an equivalence relation on the set $\mathbb{N}$.

What I know so far:

An equivalence relation is reflexive, symmetric and transitive. However I have no idea of an example.

答:
I thought to add a few more things which might help see equivalence relation through a different lens and will help you easily generate equivalence relation not just on $\mathbb{N}$ but on any set you come across.

Actually, you can see equivalence relations everywhere around you. Consider your classroom for example:

Students having a black pen of the same color is an equivalence relation.

Students having a pen of the same color is, not an equivalence relation(why?)

Students having the same gender is an equivalence relation.

Students with the same eye color is an equivalence relation.

Well, now do you realize the work of an equivalence relation? It actually generates a partition (i.e it breaks the whole set into disjoint classes) on the set it is defined. See the examples above.

In 1. The classroom gets divided into 2 disjoint classes, one having a black pen and the other not having a black pen.

Now can you see why "2" is not an equivalence relation? Remember: Equivalence relation generates partition. What if a student has 1 black and 1 blue pen?

Note: Given a set, there is a one-to-one correspondence between equivalence relations and partitions defined on that set. Can you find this bijection?

Reflexivity ensures that every element of the set is present in one of the classes.

Symmetry and Transitivity together ensure the classes produced are disjoint.

Remark: An equivalence relation is actually a generalization of the notion of equality.
You can explicitly construct a definition of $"="$ on a set using equivalence relations on that set.

i.e For any a,b in a set. One can define $a=b$ if for any given equivalence relation, a and b belong to the same class of the partition the relation generates.


问:
Check polynomial $p_n(x)$ is always divisible by $x^2 - 2x\cos (t) +1$
For any real $t$ and $n \in\mathbb{N}$, I want to show that $p_n(x)=x^{n+1}\cos [(n-1)t] - x^n\cos (nt) - x\cos (t) +1$ is divisible by $q(x) = x^2 - 2x\cos (t) +1$. As a hint I have: complex numbers.

I don't know how to attack the problem using complex numbers. In the case $n=1$, $p_1(x)=q(x)$. Also I compute and check by hand the case $n=2$. Then I start thinking about induction, but the idea failed.

Any idea how this can be solved via complex numbers? Thank you in advance.

答:
Here is a rolling start.  Use the quadratic formula on the quadratic to get the roots

$$x =  {\cos(t)\pm\sqrt{4\cos^2(t) - 4}\over 2}
=\cos(t)\pm i\sin(t) = e^{\pm it}. $$

Are these roots of the $n$th degree polynomial?


问:
How to simplify $\frac{(\sec\theta -\tan\theta)^2+1}{\sec\theta \csc\theta -\tan\theta \csc \theta} $
How to simplify the following expression :

$$\frac{(\sec\theta -\tan\theta)^2+1}{\sec\theta \csc\theta -\tan\theta \csc \theta} $$

Body:
Here is a detailed solution.
Ready, set, go!
$$\require{cancel}\begin{align}\frac{\left(\sec\theta-\tan\theta\right)^2+1}{\sec\theta\csc\theta-\tan\theta\csc\theta}\\&=\frac{\sec^2\theta-2\sec\theta\tan\theta+\color{blue}{\tan^2\theta+1}}{\sec\theta\csc\theta-\tan\theta\csc\theta}\\&=\frac{\sec^2\theta-2\sec\theta\tan\theta+\color{blue}{\sec^2\theta}}{\sec\theta\csc\theta-\tan\theta\csc\theta}\tag{1}\label{ko-eq1}\\&=\frac{2\sec^2\theta-2\sec\theta\tan\theta}{\sec\theta\csc\theta-\tan\theta\csc\theta}\\&=\frac{2\sec\theta\cancel{\left(\sec\theta-\tan\theta\right)}}{\csc\theta\cancel{\left(\sec\theta-\tan\theta\right)}}\\&=\frac{2\sec\theta}{\csc\theta}\\&=\frac{\frac{2}{\cos\theta}}{\frac{1}{\sin\theta}}\\&=\frac{2}{\cos\theta}\sin\theta\\&=\frac{2\sin\theta}{\cos\theta}\\&=2\tan\theta\end{align}$$
In equation $\eqref{ko-eq1}$, $\tan^2\theta+1=\sec^2\theta$.
I hope this helps.


问:
Let $z \in \mathbb C$, show using de Moivre's that the product of the roots of $z^n=1$ is $1$ when $n$ is odd.
I am taking a course which introduces complex numbers and some of the theorems relating to them. I wish to prove the following result:

Prove, using de Moivre's Theorem, that the product of the roots of $$z^n=1$$ is $1$ when $n$ is odd for $z \in \mathbb C$.

I do not see how de Moivre is very helpful here. Any ideas, since De Moivre only introduces a sine and cosine, I do not see how this helps met get farther.
$$\cos(n\pi) + \sin(n\pi)i =1. $$

答:
By looking at $z^n=1$ and taking the complex norm on both sides we see $|z^n|=|1| \implies |z|^n =1 \implies |z|=1$ we see that all solutions lie on the unit circle. This is why they are called the complex roots of unity. All that we need to determine are the arguments that satisfy this equation, so for $z= \cos(\theta) + i \sin (\theta)$ we want to know which $\theta$ solve the equation $z^n =1$?

Now it turns out that the roots are $\{ e^{\pi i k/n} | k \in \{1, \dots, n \} \}$ a result that follows from de Moivre's by considering the numbers of solutions of the real sine and cosine on the interval $[0,2\pi]$. Look for instance here. Essentially one wishes to look at:
$$ z^n =1 \implies(\cos(n \theta)+ \sin(n\theta)i)^n=1.$$
Or using de Moivre we get:
$$  \cos(n\theta) + \sin(n\theta)i=1+0i.$$
This gives us two very real equations, we pick one of them for instance  $\cos(n\theta)=1$, this is equivalent to $\sin(n\theta)=0$.
We see that we get equality when $n \theta= 2\pi k $ for $k\in \mathbb Z$, or equivalently $\theta= \frac{2 \pi k i}{n} $. The corresponding arguments on the interval of $[0,2\pi]$ are then given by:
$$ \frac{2 \pi}{n}, \frac{4 \pi}{n}, \frac{6 \pi}{n}, \dots, \frac{2n \pi}{n}=1.$$
Each argument corresponds to the vertex of a regular $n$-gon that represents the solutions of the $n$-roots unity.  We now have all information about the argument and the modulus/length of these numbers. See the picture below for $n=6$.

Since $n=6$ is even the product of these numbers in polar form becomes:
$$ e^{\frac{2\pi i}{6}} e^{\frac{4\pi i}{6}}e^{\frac{6\pi i}{6}}e^{\frac{8\pi i}{6}}e^{\frac{10\pi i}{6}}e^{\frac{12\pi i}{6}}= e^{\frac{42\pi i}{6}}=e^{7 \pi i}=-1.  $$
We will now look at the general product of complex roots of unity.
We again use the exponential law $e^{a+b}=e^a \cdot e^b$, this conveniently changes a product into a sum of the arguments as observed above for the case $n=6$.
$$ \prod_{k=1}^{n} e^{ 2\pi k i/ n}= e^{\sum_{k=1}^n 2\pi k i/ n }=e^{ 2 \pi i/n\sum_{k=1}^n k  } .$$
Now observe that for this last term we have $\sum_{k=1}^n k = \frac{n(n+1)}{2}$ which gives:
$$ \prod_{k=1}^{n} e^{ 2\pi k i/ n}=e^{ \frac{2 \pi i}{n} \frac{n (n+1)}{2} }= e^{\pi(n+1)i }= \left( e^{\pi \ i}\right) ^{n+1}=(-1)^{n+1}.$$
If $n$ is odd, this gives $1$, if $n$ is even this gives $-1$.


问:
Why is this function not differentiable?
Why is the function

$$\begin{align}
f(x,y)=\begin{cases}\frac{y^3}{x^2+y^2},&(x,y)\ne(0,0)\\\\
0,&(x,y)=(0,0)
\end{cases}
\end{align}$$

not differentiable at $(0,0)$?

By setting $x=0$ or $y=0$ I would get  $f_x(0,0)=0$, $f_y(0,0)=-2$ and when I look at the 3-D Surface plot, $f(x,y)$ looks quite "normal". I gaze already hours on it and cannot identify the problem.

Obviously the partial derivates exist, but are not continuous, but I cannot see where.

Is there an analytic way, I can obtain the result from? How can I come to the conclusion, without gazing at the curve?

答:
You can show that the partial derivatives are not continuous:
$$
\nabla f =
\begin{pmatrix}
-\frac{2 x y^{3} }{(x^{2} + y^{2})^2} \\
\frac{3 x^2 y^2 -y^4}{(x^2+y^2)^2}
\end{pmatrix}
$$

We convert this into polar coordinates:

$$
\nabla f =
\begin{pmatrix}
-2 \cos{\alpha} \sin{\alpha}^3 \\
\sin{\alpha}^2 (\cos{2 \alpha}+2)
\end{pmatrix}
$$

You see that both components depend on the angle, so the partial derivatives are not steady and the function is not differentiable in $(0, 0)^\top$.


答: This is equivalent to $z^{\pm1}$ being roots, with $z:=e^{it}$. We only need check $\pm=+$, since $p_n$ has real coefficients, so its nonreal roots are in conjugate pairs. So just simplify$$p_n(z)=z^{n+1}(z^{n-1}+z^{1-n})/2-z^n(z^n+z^{-n})/2-z(z+z^{-1})/2+1.$$

Title: Check polynomial $p_n(x)$ is always divisible by $x^2 - 2x\cos (t) +1$

Body:
For any real $t$ and $n \in\mathbb{N}$, I want to show that $p_n(x)=x^{n+1}\cos [(n-1)t] - x^n\cos (nt) - x\cos (t) +1$ is divisible by $q(x) = x^2 - 2x\cos (t) +1$. As a hint I have: complex numbers.

I don't know how to attack the problem using complex numbers. In the case $n=1$, $p_1(x)=q(x)$. Also I compute and check by hand the case $n=2$. Then I start thinking about induction, but the idea failed.

Any idea how this can be solved via complex numbers? Thank you in advance.


问:
A value at a point of continuity is determined by dense subset?
I know that given topological space $(X,\tau_X)$ and $(Y,\tau_Y)$, such that $D\subseteq X$ is a dense subset and both spaces are Hausdorff, a continuous function $f:X\to Y$ is determined by its values on $D$. Is the remark also remains true when dealing with not necessarily continuous functions on the whole space, but rather at point. More concretely, if $f$ is continuous at a point $x_0\in X$, is the value of $f(x_0)$ determined by $f\vert_D$?
I know that determing continuity of a function at a point, is saying that $x_0$ is in the interior of $f^{-1}[V]$ for all $f(x_0)\in V\in \tau_Y$, but lately I have been confusing myself alot.

答:
If $Y$ is Hausdorff and if $D$ is a dense subset of  $X$:

Let $f:X\to Y$ and $g:X\to Y$ be continuous at $x_0$ with $f|_D=g|_D.$

By contradiction, suppose $f(x_0)\ne g(x_0).$ Let $V_f, V_g$ be disjoint open subsets of $Y$ with $f(x_0)\in V_f$ and $g(x_0)\in V_g.$

By definition of continuity at a point, there are open subsets $U_f, U_g$ of $X$, each containing $x_0$, such that $f[U_f]\subseteq V_f$ and $g[U_g]\subseteq V_g .$

Now $U_f\cap U_g$ is open in $X,$ and non-empty (because it contains $x_0$) so there exists $d\in D\cap U_f\cap U_g.$

So $f(d)\in V_f$ but also $f(d)=g(d)\in V_g,$ which is absurd because $V_f,V_g$ are disjoint.


问:
Area of Triangle in n-Dimensional Euclidean Space
Given 3 points $x_1,x_2,x_3\in\mathbb{R}^n$ in Euclidean space. What is the area of the triangle they span?

Of course the formulas for $\mathbb{R}^2$ and $\mathbb{R}^3$ are well known. But how do the formulas generalize to higher dimensions?

My motivation is to generalize the formula stated in the publication Polygon Laplacian Made Simple by Bunge et al. Please, note that I am not looking for the volume of a "simplex". In my problem, the number of points is always 3 regardless of the dimension. Thank you!

答:
Hint:
You can use Heron's formula
Remember that the distance between two points $x,y$ in euclidean standard space is:
$$d(x,y)=\sqrt{\sum_{i=1}^n (x_i-y_i)^2}$$


问:
Write a polynomial of degree $5$ that is divisible by $x^2−2$
I'm stuck with this problem and I can't think of how to do it.

I tried this:

$$x^2(ax^3+bx^2+cx+d)$$

Then I did this:

Sorry for my English, I'm an Argentinian teenager

答:
Its $f(x) = (x^2-2)\cdot (ax^3+bx^2+cx+d)$.

Dividing a general 5th degree polynomial into $x^2-2$ is fine.


问:
Determinant (Cauchy?)
I have to show that

${det}\left ( \begin{bmatrix}
\sum_{i=1}^{n}t_i^2 & \sum_{i=1}^{n}t_i\\
\sum_{i=1}^{n}t_i & n
\end{bmatrix} \right )>0$

My solution so far

$n \cdot (\sum_{i=1}^{n}t_i^2)-(\sum_{i=1}^{n}t_i)^2>0\Leftrightarrow$

$n \cdot (\sum_{i=1}^{n}t_i^2)>(\sum_{i=1}^{n}t_i)^2\Leftrightarrow \\$

$n(t_{1}^2+t_{2}^2+...+t_{n}^2)>(t_{1}+t_{2}+...+t_{n})^2$

All the $t$'s are not the same. But my conclusion seems wrong. Can I use Cauchy or any other method to make it obvious?

Thanks!

Title: Determinant (Cauchy?)

Body:
Let $e=(1,...,1)^T\in\mathbb{R}^n,t=(t_1,..,t_n)^T$ then we have $|e|^2=1^2+...+1^2=n$ hence by Cauchy Schwarz we have $$|<e,t>|^2=|t_1+...+t_n|^2\leq |t|^2|e|^2=(t_1^2+...t_n^2)n$$
Since $<e,t>=e_1t_1+...e_nt_n=t_1+...+t_n$.


问:
Existence of bounded derivative inverse (Deuflhard Exercise 1.1)
The following exercise is from Deuflhard's "Newton Methods for Nonlinear Problems" :

Given a nonlinear $C^1$-mapping $F:X\to Y$ over some domain $D\subset X$ for Banach spaces $X$, $Y$, each endowed with some norm $||\cdot||$. Assume a Lipschitz condition of the form
\begin{align*} ||F'(x)-F'(y)||\leq \gamma||x-y||\end{align*}
for $x,y\in D$. Let the derivative at some point $x^0$ have a bounded inverse with
\begin{align*} ||F'(x^0)^{-1}||\leq \beta_0.\end{align*}
Show that then, for all arguments $x\in D$ in some open ball $S(x^0,\rho)$ with $\rho=\frac{1}{\beta_0\gamma}$, there exists a bounded derivative inverse with
\begin{align*} ||F'(x)^{-1}||\leq \dfrac{\beta_0}{1-\beta_0\gamma||x-x_0||}.\end{align*}

I am fairly new to operator theory. I understand the hypotheses, but I don't really understand the relationship between the norms of operators and their inverses. Any hints at how to proceed with this exercise will be much appreciated.

答:
As a consequence of the Banach Perturbation lemma, one has

If $A$ and $B$ are matrices such that $A$ nonsingular, $||A^{-1}||<\theta$, and $||A||^{-1}\,||B||<1$, then $A+B$ is nonsingular and

$||B^{-1}||\leq \dfrac{\theta}{1-\theta||B||}$.

In the case above, take $A=F'(x_0)$ and $B=F'(x)-F'(x^0)$. Then, we have
$||A^{-1}||\,||B||\leq \beta_0\gamma||x-x^0||$. If $x\in S(x^0,\rho)$ where $\rho=1/\beta_0\gamma$, one has $||A^{-1}||\,||B||<1$, therefore $A+B$ is nonsingular with bound given above. Substituting in $F'$ for $A$ and $B$, we have

$||F'(x)^{-1}||\leq \dfrac{\beta_0}{1-\beta_0\gamma||x-x^0||}$.


问:
$F(x,y)=f(x)g(y)$, $F(r\cos\theta,r\sin\theta)=h(\theta)$, find $F(x,y)$
My progress:
$$F_x(x,y)=f'(x)g(y)$$  $$F_y(x,y)=f(x)g'(y)$$ also, $$F(r\cos\theta,r\sin\theta)=h(\theta)$$
taking partial derivative with respect to $r$ and using chain rule:
$$F_x(x,y)\cos\theta+F_y(x,y)\sin\theta=0$$
$$F_x(x,y)r\cos\theta+F_y(x,y)r\sin\theta=0$$
$$F_x(x,y)x+F_y(x,y)y=0$$
$$f'(x)g(y)x+f(x)g'(y)y=0$$
And......I got stuck. if only the positive sign were negative.....then that would be easy. Any help would be appreciated!

答:
Write your last equation in the form
${x f'(x)\over f(x)}+{y\, g'(y)\over g(y)}=0$. This implies that
$$-{x f'(x)\over f(x)}={y\, g'(y)\over g(y)}=\lambda$$
for some real (or complex) constant $\lambda$. The ODE
$$g'(y)={\lambda\over y}g(y)$$
has the solutions
$$g(y)=C' y^\lambda\qquad(y>0)\ ,$$
and similarly you obtain
$$f(x)=C'' x^{-\lambda}\qquad(x>0)\ .$$
It follows that in the first quadrant $I:=\bigl\{(x,y)\bigm|x>0, \ y>0\bigr\}$
you have $$F(x,y)=C\left({y\over x}\right)^\lambda\qquad\bigl((x,y)\in I\bigr)\ .$$
This $F$ is obviously constant on rays emanating from the origin.

One can  also produce an $F$ with a larger domain. E.g., when $\lambda>0$ you can put $$F(x,y):=\left|{y\over x}\right|^\lambda\ .$$
This is defined on all of ${\mathbb R}^2\setminus\{y$-axis$\}$.


问:
What (fractional) exponent rule is being used here?
Here is the example we were given:

Given that $32√2 = 2^a$, find the value of $a$.

Here's the working out that was supplied:

$$32√2 = 2^5 \times 2^\frac{1}{2} = 2^\frac{11}{2}$$
$$a = \frac{11}{2}$$

I'm confused about how this:

$$2^5\times2^\frac{1}{2}$$

Was converted into this:

$$2^\frac{11}{2}$$

What happened?

Thank you.

Body:
For all $ x \in \mathbb{R}^+,\ y \in \mathbb{Q},\ z \in \mathbb{Q}$

$$ x^yx^z = x^{y+z}. $$

Therefore, $2^5\times2^{\small\frac{1}{2}} = 2^{5 +\small\frac{1}{2}} = 2^{\small\frac{10}{2} + {\small\frac{1}{2}}} = 2^{\small\frac{11}{2}}$


答: How about $x^5-2x^3$?  The other three roots, besides $\pm\sqrt2$, are zero.

Title: Write a polynomial of degree $5$ that is divisible by $x^2−2$

Body:
I'm stuck with this problem and I can't think of how to do it.

I tried this:

$$x^2(ax^3+bx^2+cx+d)$$

Then I did this:

Sorry for my English, I'm an Argentinian teenager


问:
Prove $1^r+2^r+.....\equiv 0\pmod p$ , given that p is odd prime and $0<r<p-1$
An exercise reads as follows:
If $p$ is an odd prime and $0<r<p-1$ then prove that $s=1^r+2^r+...(p-1)^r\equiv 0\mod p$

I know one clever way to prove it is to use a primitive root $g$ and write the given sum as: $g^r+g^{2r}+g^{3r} +...+g^{(p-1)r}$ which is a geometric process easy to handle and deduce the result.

But I am looking for a more primitive way right now (suppose I know nothing about primitive roots). My thought is: the polynomial $$f=x^{p-1}-1$$ has $p-1$ roots distinct roots in the field $\Bbb{Z_p}$ so one can also write $$f= (x-1)(x-2)(x-3)...(x-(p-1))$$. By inspection of the coefficients of the 2 equal polynomials we get

$t_1=1+2+3+...+(p-1)\equiv0 \mod p$

$t_2=1\cdot 2+1\cdot 3+...+(p-2)\cdot( p-1)\equiv 0 \mod p$

$t_3=1\cdot 2\cdot3+1\cdot 2\cdot4+...+(p-3)\cdot (p-2)\cdot(p-1)\equiv 0\mod p$

and so on ....

now suppose $r=2$ then $s=t^2_1 -2t_2$, where we deduce that $p$ divides $s$

But how in general to compute $s$ by the $t_i$ mentioned above? Is there an algorithm for that? Thanks a lot.

答:
I do not know how in general to compute $s$ by your $t_i$'s, but here is a useful fact for your question.

Denote $s_\ell:=\sum_{i=1}^{p-1}i^\ell$, then one can prove that
$$p^{r+1}-1=\sum^p_{k=2}((k)^{r+1}-(k-1)^{r+1})=\sum_{\ell=0}^r {r+1\choose \ell}s_\ell$$
by binomial theorem. It follows that
\begin{align*}
s=s_r&=\frac{1}{r+1}(p^{r+1}-1-\sum_{\ell=0}^{r-1} {r+1\choose \ell}s_\ell)\\
&=\frac{1}{r+1}(p^{r+1}-1-(p-1)-\sum_{\ell=1}^{r-1} {r+1\choose \ell}s_\ell)\\
&=\frac{1}{r+1}(p^{r+1}-p-\sum_{\ell=1}^{r-1} {r+1\choose \ell}s_\ell)
\end{align*}

Relevant post:

Prove $\sum_{n = 1}^{p - 1} n^{p - 1} \equiv (p - 1)! + p \pmod {p^2}$ for $p$ being an odd prime


问:
What is the numerator of the denominator (5) of the prime factorization of $6/30$
I am simplifying the fraction $6/30$ which I know is $1/5$.

It's pretty self-explanatory.

However, when I use prime factorization to simplify it, I am left with a denominator of $5$ with nothing above.

$6/30$ :

$6 = 2 \cdot 3$

$30 = 2 \cdot 3 \cdot 5$

$2$ and $3$ cancel each other out.

You are left with $5$ as the denominator with nothing above as the numerator. What above it makes it $1/5 = 20\% = 20/100$.

答:
You forgot that

$$6=3\times2\times1$$

You always have a $1$ in this cases.


问:
$O$ is not dense in $\mathbb R^n$
$A$ is a matrix of order $n$, and $x$ is an $n$-dimensional column vector. The set
$$
O=\{x,Ax,A^2x,\cdots,A^mx,\cdots\}
$$
Prove that, $O$ is not dense in $\mathbb R^n$.

This claim seems to be trivial: it just looks like $\mathbb N$ in $\mathbb R$. I recalled how we verify the denseness of $\mathbb Q$ in $\mathbb R$, but it seeems that the same method doesn't work. Can anyone help?

答:
I think the hints in the comments provide you enough help for a simple proof.

You consider two cases: the norm sequence $(||A^n x||)_{n\in\mathbb N}$ either diverges to infinity, or is eventually bounded by some real number $M$ (it could still oscillate and not converge, but we'll see that we don't care).

If it diverges, it means that any open bounded subspace of $\mathbb R^n$ (e.g., $B=\{x\in\mathbb R^n : ||x||<M\}$) contains at most a finite number of elements of $O$, which means that $O$ can't be dense in $\mathbb R^n$ (because its closure doesn't even "fill" $B$).

If the norm is (eventually) bounded ($\leq M$), you can use the same argument for $\mathbb R^n\smallsetminus \overline{B}$ (it contains at most a finite number of elements of $O$...).

In either case, $O$ cannot be dense in $\mathbb R^n$.

Title: $O$ is not dense in $\mathbb R^n$


问:
Show that an integral function is a Borel function
I've been trying to prove that the Hardy-Littlewood Maximal function is a Borel Function for a while now (and was hoping to get some help) and have been stuck on the final step for a while, which is to show that, fixing $r>0$ and letting $(X,d)$ be a metric space then

$$I(x) =  \int_{B_r(x)} |f| d\mu$$ is a Borel function. Where $f\in L^1(\mu)$ and $\mu$ is a outer Borel measure (which means Borel sets are $\mu$ measurable) with the property that $\mu(B_r(x)) < \infty$ for all $r$ and $x$.

I've boiled the problem down to showing that the above holds if one can show that the function $\mu(B_r(x)\cap A)$ is a Borel function. Where $A$ is any arbitrary $\mu$-measurable set.

I have tried to come up with counter examples just in case but any weird measure that I come up with it tends to break one of the two conditions imposed on $\mu$.

$\underline{EDIT}$
Here are some definitions that may not be standard and hence have added them to clear things up. (Also in the course I am doing we call outer measures, measures so this might explain the weird naming I the following definitions)

A set $A \subset X$ is said to be $\mu$ measurable if $\forall E \subset X$ the
$$\mu(E) = \mu(E\cap A) +\mu(E\setminus A)$$

$\mu$ is said to be Borel if every Borel set is $\mu$ measurable. e.g. $\delta _0$ is Borel when $X=\mathbb{R}$ but also the Vitali set is $\delta _0$ measuable but not a Borel set

A function $f:X \to \mathbb{R}$ is said to be $\mu$ Borel if for all $a \in \mathbb{R}, f^{-1}((a,\infty)$ is a Borel set.

Many thanks in advance.

答:
If you mean that $\mu$ is the measure induced by an outer measure on the Borel sets, then we can take simple functions $s_n\uparrow |f|$ and apply the Monotone or Dominated Convergence Theorem: if we define $I_n$ by

$I_n(x):=\int_X s_n(x) \chi_{B_r}(x)d\mu$ then the $I_n$ are  $\mu-$measurable and satisfy

$\lim I_n(x)=\int_X \lim s_n(x) \chi_{B_r}(x)d\mu=\int_X|f(x)|\chi_{B_r}(x)d\mu =\int _{B_r(x)}|f(x)|d\mu=I(x)$

so $I$ is measurable being the limit of $\mu-$ measurable functions.


问:
Prime decomposition in ring extensions
Let $R\subseteq R'$ be Dedekind domains, let $\mathfrak{p}$ be a nonzero prime ideal of $R$. Then $\mathfrak{p}R'$ is an ideal of $R'$ and it has a factorization
$$\mathfrak{p}R'=\mathfrak{P}_1^{e_1}\cdots\mathfrak{P}_g^{e_g}$$
in which $\mathfrak{P}_1,\ldots\mathfrak{P}_g$ are distinct prime ideals of $R'$ and $e_1,\ldots,e_g$ are positive integers.

At this point, my textbook says something obscure (in my opinion). It says: note thet $\mathfrak{P_i}\cap R=\mathfrak{p}$ for each $i$. $\textbf{Thus}$  the integer $e_i$ is completely determined by $\mathfrak{P}_i$.

I can't understand the sentence in itself: what does it mean that the exponent is completely determined by the ideal?? Secondly, how the previous argument should show this?

Any suggestion/comment/help would be appreciated

答:
For an ideal $\mathfrak{P}$ appearing the in the decomposition of $\mathfrak{p}$, one often writes the suggestive notation $\mathfrak{P} | \mathfrak{p}$. Note that this means $\mathfrak{p}$ is contained in the ideal $\mathfrak{P}$. Then the power $e$ appearing in the decomposition is the highest power of $\mathfrak{P}$ in which $\mathfrak{p}$ is contained. In other words,

$\mathfrak{p} \in \mathfrak{P}^e$ but $\mathfrak{p} \notin \mathfrak{P}^{e+1}$.

This is the sense in which the $e$ is determined by this decomposition.

Title: Prime decomposition in ring extensions


答: $1$ has the key property that $1×n=n$ for all $n$, just as $0$ has the key property that $0+n=n$.  When you add no numbers, you get $0$.  When you multiply no numbers, you are left with $1$

Title: What is the numerator of the denominator (5) of the prime factorization of $6/30$

Body:
I am simplifying the fraction $6/30$ which I know is $1/5$.

It's pretty self-explanatory.

However, when I use prime factorization to simplify it, I am left with a denominator of $5$ with nothing above.

$6/30$ :

$6 = 2 \cdot 3$

$30 = 2 \cdot 3 \cdot 5$

$2$ and $3$ cancel each other out.

You are left with $5$ as the denominator with nothing above as the numerator. What above it makes it $1/5 = 20\% = 20/100$.


答:
Sum of squares is non-negative. So$$\sum\left(t_i-\frac1n\sum t_i\right)^2\ge0$$

Open the outer sum,$$\begin{align*}\sum t_i^2+\frac1{n^2}\sum\left(\sum t_i\right)^2-\frac2n\sum\left(t_i\sum t_i\right)&\ge0\\\sum t_i^2+\frac1n\left(\sum t_i\right)^2-\frac2n\left(\sum t_i\right)^2&\ge0\\\sum t_i^2&\ge \frac{(\sum t_i)^2}n\end{align*}$$The equality occurs iff $t_i=\frac1n\sum t_i$ i.e. $t_i$ are constant.

Title: Determinant (Cauchy?)

Body:
I have to show that

${det}\left ( \begin{bmatrix}
\sum_{i=1}^{n}t_i^2 & \sum_{i=1}^{n}t_i\\
\sum_{i=1}^{n}t_i & n
\end{bmatrix} \right )>0$

My solution so far

$n \cdot (\sum_{i=1}^{n}t_i^2)-(\sum_{i=1}^{n}t_i)^2>0\Leftrightarrow$

$n \cdot (\sum_{i=1}^{n}t_i^2)>(\sum_{i=1}^{n}t_i)^2\Leftrightarrow \\$

$n(t_{1}^2+t_{2}^2+...+t_{n}^2)>(t_{1}+t_{2}+...+t_{n})^2$

All the $t$'s are not the same. But my conclusion seems wrong. Can I use Cauchy or any other method to make it obvious?

Thanks!


问:
Prove $i\frac{d}{dx}$ is not essentially self-adjoint.
Consider $T=i\frac{d}{dx}$ on $C_c^{\infty}(0,\infty)\subset L^2(0,\infty)$, the infinitely differentiable functions with compact support away from the origin. Prove that $T$ is not essentially self-adjoint.

My guess is to show that $(T^*-i)$ or $(T^*+i)$ has non trivial kernel, but I dont know how to do this. For instance, I don't know what $T^*$ is in the first place.

答:
By the definition of the adjoint, for any $f \in C^\infty_c(0,\infty)$ and $g \in L^2(0,\infty)$, we must have
$$
\langle Tf, g \rangle = \langle f, T^*g \rangle.
$$
Whenever $g$ is differentiable and within the domain of $T^*$, integration by parts yields the following:
$$
\begin{align}
\langle Tf,g \rangle &= \int_0^\infty i \frac d{dt}f(t) \bar g(t)\,dt
\\ & =
\left.if(t)\bar g(t)\right|_0^\infty - \int_0^\infty f(t) \cdot \overline{-i\frac{d}{dt}g(t)}\,dt =
 \int_0^\infty f(t) \cdot \overline{i\frac{d}{dt}g(t)}\,dt
\end{align}
$$
so that $T^*g = i \frac{d}{dx} g$. With this is established, consider the function $g(x) = e^{-x}$. It suffices to note that

$g \in \mathrm{dom}(T^*)$ (i.e. $f \mapsto \langle Tf,g \rangle$ is continuous over $\mathrm{dom}(T)$),

$g \in \ker(T^* + i)$.


问:
Simplify $\frac {\sec^2\theta - \cos^2\theta}{\tan^2\theta}$

My question is to simplify:$$\frac {\sec^2\theta - \cos^2\theta}{\tan^2\theta}$$

My workings:

$$\frac {\sec^2\theta - \cos^2\theta}{\tan^2\theta}=\frac {\frac {1}{\cos^2\theta}- \cos^2\theta} {\tan^2\theta}$$

Then I am unable to continue, can someone please show me some working outs? Thank you.

答:
Here is a detailed solution.
Move on!
$$\require{cancel}
\begin{align}
\frac{\sec^2\theta-\cos^2\theta}{\tan^2\theta}&=\frac{\frac{1}{\cos^2\theta}-\cos^2\theta}{\frac{\sin^2\theta}{\cos^2\theta}}\tag{1}\label{ko-eq1}\\
&=\frac{\frac{1-\cos^4\theta}{\cancel{\cos^2\theta}}}{\frac{\sin^2\theta}{\cancel{\cos^2\theta}}}\tag{2}\label{ko-eq2}\\
&=\frac{1-\cos^4\theta}{\sin^2\theta}\tag{3}\label{ko-eq3}\\
&=\frac{\cancel{\left(1-\cos^2\theta\right)}\left(1+\cos^2\theta\right)}{\cancel{1-\cos^2\theta}}\tag{4}\label{ko-eq4}\\
&=1+\cos^2\theta\tag{5}\label{ko-eq5}
\end{align}$$
\eqref{ko-eq1} Expand $\sec^2\theta$ as $\frac{1}{\cos^2\theta}$ and $\tan^2\theta$ as $\frac{\sin^2\theta}{\cos^2\theta}$.

\eqref{ko-eq2} Get a common denominator of $\frac{1}{\cos^2\theta}-\cos^2\theta$ and cancel both $\cos^2\theta$. The expression $\frac{\frac{1-\cos^4\theta}{\cos^2\theta}}{\frac{\sin^2\theta}{\cos^2\theta}}$ is equal to $\frac{1-\cos^4\theta}{\cos^2\theta}\cdot\frac{\cos^2\theta}{\sin^2\theta}$.

\eqref{ko-eq3} After cancelling both $\cos^2\theta$.

\eqref{ko-eq4} Expand $1-\cos^4\theta$ as $\left(1-\cos^2\theta\right)\left(1+\cos^2\theta\right)$, change $\sin^2\theta$ as $1-\cos^2\theta$ and cancel the same terms on numerator and denominator.

\eqref{ko-eq5} Final result.

I hope this helps.


问:
When does a variety have a point over a finite field for sufficiently large primes p?
Let $X$ be an algebraic variety over the rational numbers. Suppose that $X$ has positive dimension. I would like to say that $X(\mathbb{F}_p)$ is non-empty for sufficiently large primes $p$. One idea is to use the Weil conjectures, but that seems like overkill. So my two questions are:

When can one deduce that a system of polynomial equations has a solution over $\mathbb{F}_p$ for sufficiently large primes $p$? I might have guessed that $\dim(X) > 0 $ is sufficient, but it seems like the affine variety $X^2+Y^2 = 0$ minus the point $(0,0)$ (which has no solutions for $p = -1 \mod 4$) suggests that extra conditions are necessary.

Is there an elementary proof of the (correct) version of part 1?

答:
If $k$ is the field of rational numbers and $A$ is a finitely generated $k$-algebra with $X:=Spec(A)$, it follows $X(\mathbb{F}_p)$ is empty for any prime number $p>0$. If $\mathfrak{m}\subseteq A$ is a maximal ideal it follows $k \subseteq \kappa(\mathfrak{m})$ is a finite field extension of $k$, but $k$ has characteristic zero and hence this is impossible.

If $B:=\mathbb{Z}[x_1,..,x_n]/I$ where $\mathbb{Z}$ is the ring of integers and $Y:=Spec(B)$, we get a canonical map $\pi: Y \rightarrow S:=Spec(\mathbb{Z})$.
If you accept Zorns's Lemma/the "Axiom of Choice", it follows there is a maximal ideal $\mathfrak{m} \subseteq B$. The residue field $\kappa(\mathfrak{m})\cong \mathbb{F}_{q^r}$ is finite with characteristic $q>0$. Hence $\mathfrak{m}\cap \mathbb{Z}=(q)$, and it follows the fiber $\pi^{-1}(q)\subseteq Y$ is non-empty. Hence Zorn's Lemma  proves that $Y(\mathbb{F}_{q^r})$ is always non-empty for some $q,r$. The fact that the residue field $\kappa(\mathfrak{m})$ of a maximal ideal in $B$ is finite, is proved in the exercises in Atiyah-Macdonald's book "Commutative Algebra", hence this argument is in some sense "elementary". I have also seen simple online proofs that the residue field $\kappa(\mathfrak{m})$ is finite as well.


问:
How many rearrangements of an equation
How many different rearrangements of the equation $a-b=c+d$ are there with two terms on each side?

I count $-b+a=c+d$ and $a-c=b+d$ as different rearrangements for example, however, $b-a=-c-d$ is not as it involves multiplying both sides by $-1$.

答:
Based on your edit, here are the assumptions to my answer,

i) $a + b = c + d \,$ is same as $\, b + a = d + c$

ii) $a + b = c + d \,$ is same as $\, c + d = a + b$

iii) $a - b = c - d \,$ is same as $\, -a + b = -c + d \,$ (both sides multiplied by $-1$)

iv) But $\,a − b = c + d$ and $a − c = b + d$ are different, even though both are rearrangement of the same mathematical equation.

Keeping two variables on each side, number of ways to arrange variables

$ = \displaystyle \frac{1}{2} {4 \choose 2}$

Now we have two signs $+ \,$ and $- \,$, and there are $4$ places to assign signs to -

$\_ a \_ b = \_ c \_ d \,$ (even though the $+$ sign will not show before the first variable on both sides, it exists).

Now to make sure we do not count cases $(a - b = c - d) \,$ and $ \,(-a + b = -c + d)$ twice, we would assign signs to only $3$ places. It will cover all unique equations.

So number of ways to choose signs = $2^3$.

So total number of ways to make equations with given condition $ = \frac{1}{2} {4 \choose 2} \times 2^3 = 24$

Now please note condition $(iv)$. If we consider them the same, instead of different equations, all arrangements of $(a, b) = (c, d)$ we counted above would suffice. We do not need to count $(a,c) = (b,d)$ and $(a,d) = (b,c)$ separately. They can all be made from the arrangements of $(a, b) = (c, d)$. So in that case, we have only $\frac{24}{3} = 8$ different arrangements. They are -

$\begin{cases}
a + b = c + d \\
a + b = c - d \\
a + b = - c + d \\
a + b = - c - d \\
a - b = c + d \\
a - b = - c - d \\
a - b = c - d \\
a - b = - c + d \\
\end{cases}$

Lastly if you do not care about the order of variables in these equations, there will be only $3$ arrangements.

Title: How many rearrangements of an equation


问:
Probability question related to independent/dependent event and conditional probability
An investment company estimates that the probability that stocks a and b go up is 61 % and 75%,
respectively. Let A be the event that stock a goes up, and let B be the event that stock b goes up.
Assume for simplicity that a stock can only go up or go down (i.e. ignore the case where the stock
maintains exactly the same value).

(a) Assuming that the events A and B are independent, compute the probability that both
stocks go up.

(b) Show that the probability that both stocks go up cannot be smaller than 36%, no matter
whether events A and B are independent or not.

(c) Suppose that it is impossible for both stocks to go down. Compute the probability that
both stocks go up.

So I am dealing with question of probability, I manage to figure the a) and b) sub question by understand  the meaning of dependent event and independent event but I am kind of the stuck in the third sub question c)

a) is simply multiply 61% and 75% together

b) is the probability the stock a goes up multiply the probability the stock b goes up given that stock a goes up

How about c)?
I am guessing that I need to use conditional probability with the statement (impossible for both stock to go down) but I have no idea how to translate the statement in to math

答:
For both $(b)$ and $(c)$ parts, use the inclusion-exlcusion principle:$$P(A\text{ and }B)=P(A)+P(B)-P(A\text{ or }B)$$For $(b)$, note that $P(A\text{ or }B)\le1$ so $P(A\text{ and }B)\ge P(A)+P(B)-1=0.36$.

For $(c)$ you are given $P(A'\text{ and }B')=0\implies P(A\text{ or }B)=1-P(A'\text{ and }B')=1$ giving $P(A\text{ and }B)=0.36$.


答: There are two cases, either $O$ is bounded, or not. If $O$ is bounded, then it cannot be dense. If it is unbounded, then any subsequence is also unbounded and hence no subsequence has a limit in $\mathbb{R}^n$. This implies that $O$ is closed and hence, not dense in $\mathbb{R}^n$.

Title: $O$ is not dense in $\mathbb R^n$

Body:
$A$ is a matrix of order $n$, and $x$ is an $n$-dimensional column vector. The set
$$
O=\{x,Ax,A^2x,\cdots,A^mx,\cdots\}
$$
Prove that, $O$ is not dense in $\mathbb R^n$.

This claim seems to be trivial: it just looks like $\mathbb N$ in $\mathbb R$. I recalled how we verify the denseness of $\mathbb Q$ in $\mathbb R$, but it seeems that the same method doesn't work. Can anyone help?


问:
Does $\sum_{n=1}^\infty(-1)^{n}\frac{n^n}{n!}$ converge or diverge
$\sum_{n=1}^\infty(-1)^{n}\frac{n^n}{n!} $

I got that it diverges but I am not sure

答: Simply expand the numerator and the denominator as products.  Can you find a lower bound for $\frac{n^n}{n!}$? Does this lower bound tend to $0$ as $n\rightarrow\infty$? What does this say about the terms of $(-1)^n \frac{n^n}{n!}$?</p>


问:
Puzzle: Maximum amount of rice
Farmer Felix has just had a bountiful harvest with 1 ton of rice, and now he wants to sell all of it. Before getting his rice to the market, his rice needs to go through inspection. Unfortunately, the inspector is a greedy tyrant, and he wants a part of Felix's harvest. The inspector's price is as follow:

The first inspection costs all the rice.

If Felix gives the inspector x ton of rice, then the next inspection costs x less.

For example, if the inspector is given 0.2, 0.3, 0.5 ton portions in that order, then finally the total amount of rice gets to the market is:
$$0.2\times0 + 0.3\times0.2 + 0.5\times(0.2 + 0.3) = 0.31$$
What is the maximum amount of rice Felix can get to the market (given that the inspector doesn't mind even when he's requested to inspect infinite times)?

答:
I haven't been able to solve it yet. Below is my thoughts about the puzzle. I hope my English is enough to convey my thinking.

At first, I intended to let the inspector to accept a limit amount of times only (2, 3, 4,..., n inspections) but it escalates quickly into finding maximum value of a multi-variable function. So, maybe it'll be a question for another time.

I've been thinking about the answer, but it turns out to be harder than I imagine it to be. Initially, I let Felix giving rice according to f(x) = x, where:

$$ \int_0^{1/2}{xdx} = 1 $$

Felix gives a tiny bit of rice xdx at a time and 1 is the total amount of rice Felix has after the harvest. So the amount of rice to the market after all the inspections is:

$$ \int_0^{1/2}{x({\int_0^xudu}) dx} = \int_0^{1/2}{x\frac{x^2}{2}  dx} = \frac{1}{128}$$

where $${\int_0^xudu}$$ is the total amount of inspected rice at some point (inspection cost discounted) and $$ xdx $$ is the tiny amount of rice Felix is giving at some instant.

Now, when the idea of calculus first rose in my mind, I thought "If I make Felix inspect rice in an increasing amount, he'll get more rice in the end." But calculation proved otherwise, 1/128 is definitely less than 0.51. Am I doing something wrong here?

Also, what I'm curious is what function f(x) will give the maximum amount of rice, thus I try to generalize it as:

$$ \int^a_b{f(x)dx} = 1$$

Also, the amount of rice that gets to market:

$$ \int^a_b{f(x)({\int^x_b{f(u)du})}dx}$$

The train of reasoning is similar to above. And let's assume we can find F(x) such that the derivative F'(x) = f(x), then:

$$ \int^a_b{f(x)({\int^x_b{f(u)du})}dx} =
 \int^a_b{f(x)(F(x) - F(b))dx}$$

Since F(b) is a constant, and the integration of f(x) over (a, b) is 1, I finally arrive at:

$$\int^a_b{f(x)F(x)dx} - F(b)$$

which is not helpful at all. Could you give me some clues about what to do next?

Title: Puzzle: Maximum amount of rice


问:
Regarding the Bolzano-Weierstrass proof
I have doubts regarding the Bolzano-Weierstrass proof My textbook proof same as
Bolzano-Weierstrass proof explanation here my problem is why isn't it $n_1 \geq 1$? instead of  $n_1>1$

Can anyone give me a simple explanation?

Thank you!

答:
It's not so crucial that $n_1 > 1$, but the same idea is used later to make $n_{m+1} > n_m$, which is crucial.

The point is just that $I_1$ contains infinitely many terms of your sequence.  When you're picking $n_1$ you're asking for $x_{n_1} \in I_1$; it can't be the case that the only term of the sequence in $I_1$ is $x_1$, so we're free to ask for $n_1 > 1$.

A similar idea, but a bit more complicated, is what happens later in the proof: Because $I_{m+1}$ contains infinitely many terms of your sequence, the terms $x_1, x_2, \ldots, x_{n_m}$ don't use up all of the terms of the sequence that are in $I_{m+1}$, so we can find $n_{m+1} > n_m$ such that $x_{n_{m+1}} \in I_{m+1}$.

Title: Regarding the Bolzano-Weierstrass proof


问:
You must choose a committee of 8 people containing at least 2 men. There are 9 men and 11 women to choose from. How many committees are there?
You must choose a committee of 8 people containing at least 2 men.

There are 9 men and 11 women to choose from.

How many committees are there?

I reached this answer but it seems a little too simple for what was a ($5$) mark question in a previous exam

${9\choose 8}\cdot{11\choose 6}$

Am I missing something?

答:
It seems that you aren't considering that the committee could be all men, or $7$ men and $1$ women, or ... .

The easiest way to do this I think is to subtract the inadmissible committees from the total.  There are $\binom{20}{8}$ ways to choose $8$ people from $20$.  We must subtract the $\binom{11}8$ committees that comprise only women, and the $\binom91\binom{11}7$ with exactly one man.  The answer is $$\binom{20}{8}-\binom{11}8 -\binom91\binom{11}7$$


问:
What is the proper way to determine error in my measurements?
How do I know what the error is in measurements I took using an oscilloscope?
In the image below you will see on channel #1 of the oscilloscope (yellow) there is a pattern.  I manually measured the time (delta t) it took to fall from the maximum to the minimum point.  I used a selection tool that comes with the oscilloscope but its accuracy depends on the user's selection points.

How do I estimate my error?

答: The time duration you're looking for appears to be the same as the time duration that the 4th signal is LOW.  You can use the cursor to see how much time it takes to go from HIGH to LOW on the 4th channel to find your error bounds.  So if your cursor moves in .02ms increments, and you find that the signal is HIGH at time $t$ and LOW at time $t+.04$, then the drop started somewhere in that .04ms time period.  Thus, your error for the initial time is at most .04ms.</p>


问:
Spiral function with constantly increasing separation between its loops
Spirals following the polar equation $r=a+bθ$ are known to have constant separation between each of its loops, I show this by transforming the polar equation into Cartesian equation by using the transformations $r=(x^2+y^2)^{0.5}$ and $ θ=tan^{-1}\left(\frac{y}{x}\right)$. For computing the separation between loops $n$ and $n+1$, I get the following formula

$x_{n+1}-x_{n}=b π [(n+1)-(n)]= b π$

As shown, the separation between the loops is a constant. When I use the equations $r=a+bθ^p$ or $r=a+bp^θ$, and then compute the separation between loops, for instance, $n$ and $n+1$, I get the following results;

$x_{n+1}-x_{n}=b 2π^p [(n+1)^p-(n)^p]$ for $r=a+bθ^p$

$x_{n+1}-x_{n}=b(p^{2π}-1)p^{n 2π }$ for $r=a+bp^θ$

As one would realize, the separation between loops of those two spirals increase, but not with a constant rate. I tried many more formulae, but I didn’t get any spiral with constant increase rate between its loops, i.e., stratifying the formula;

$x_{n+1}-x_{n}=An$

Wherein $A$ is a constant. Does anyone here know which equation would give me the spiral shape I’m looking for? Thank you so much guys.

答:
You can use $r = a + b\left(\left(\frac\theta\pi\right)^2 - \frac\theta\pi\right).$

This way,

$$\begin{equation}
\begin{split}
x_{n+1}-x_{n} &= b\left[\left(\frac{(n+1)\pi}{\pi}\right)^2 - \frac{(n+1)π}{\pi}-\left(\frac{nπ}\pi\right)^2 - \frac{nπ}{\pi} \right] \\
&=b[n^2 + 2n + 1 - n - 1 - n^2 + n] \\
&= b[2n]
\end{split}
\end{equation}$$

I think this was what you were looking for.


问:
Derivative of $\sqrt{x^TAx}$
Let $x\in \mathbb{R}^n$ and $A\in\mathbb{R}^{n\times n}$ be a positive semi-definite matrix.

Is there a way to express in closed form the following derivative?

$$\frac{\partial}{\partial x} \sqrt{x^TAx}$$

答:
First of all, using the chain rule, we have
$$
\frac{\partial }{\partial x} \sqrt{x^TAx} = \frac{1}{2 \sqrt{x^TAx}}\cdot  \frac{\partial }{\partial x} x^TAx.
$$
One approach to this partial derivative is to write the expression $f(x + h)$ in the form $f(x) + g(x)^Th + o(h)$; by definition, the $g(x)$ for which this holds is equal to $\frac{\partial f}{\partial x}$. With that in mind,
$$
(x+h)^TA(x + h) = x^TAx + x^TAh + h^TAx + h^TAh\\
= x^TAx + x^TAh + [h^TAx]^T + o(h)\\
= x^TAx + x^TAh + x^TA^Th + o(h)\\
= x^TAx + [(A + A^T)x]^Th + o(h).
$$
With that, the derivative of $x \mapsto x^TAx$ is $(A + A^Tx)$, and we have
$$
\frac{\partial }{\partial x} \sqrt{x^TAx} = \frac{1}{2 \sqrt{x^TAx}}\cdot  (A + A^T)x.
$$

Title: Derivative of $\sqrt{x^TAx}$


问:
Complex number such that affixes of itself, its squared and its cube form a right triangle. $z,z^2,z^3$ form a right triangle in argand plane
$\DeclareMathOperator{\aff}{Aff}$
Here is a funny problem  I stumbled upon recently

I call the affix of a complex number the corresponding point in the argand plane, that is

$$\aff(\mathrm{i})=(0;1)$$

Find all $z \in \mathbf{C}$ such that the affixes of $z,z^2\ \& \ z^3$ form a right triangle.

I decided to Set $A=\aff(z),B=\aff(z^2),C=\aff(z^3)$

I have proven that :

If $z$ is a solution, then $z\neq 0,\  z\neq 1, \ \& \ z\neq-1$

If the triangle is right angled at A, $z$ must be a non zero imaginary number.

If the triangle is rigth angled at B, $\aff(z)$ has to lie on the vertical line defined by $x=(-1)$, except for $z=(-1)$.

(I'll post my proof when I get some time to write it in $\mathrm{\LaTeX}$ soon.)

I'm down to the third case : I have proven that if the triangle is right angled at $C$, then :
$$\Re \left(\frac{1+z}{z}\right)=0$$

Thanks to some experimentation, I know that $\aff(z)$ should lie on the circle centered at $(-\frac12;0)$ with radius $\frac12$, with $z$ again respecting the first condition above.

Now,how do I prove this ?

Thanks for the help.

Here is an animation.

答:
Partial Solution:

If $z=\cos t+i\sin t,$

$$m_{AB}=\dfrac{\sin2t-\sin t}{\cos 2t-\cos t}=-\cot\dfrac{3t}2\text{ if }\sin\dfrac t2\ne0$$

Similarly,
$$ m_{BC}=-\cot\dfrac{5t}2, m_{CA}=-\cot2t$$

Now we need at exactly one of the products $m_{AB}\cdot m_{BC},m_{BC}\cdot m_{CA},m_{CA}\cdot m_{AB}$  to be $=-1$


问:
Why is this function not differentiable?
Why is the function

$$\begin{align}
f(x,y)=\begin{cases}\frac{y^3}{x^2+y^2},&(x,y)\ne(0,0)\\\\
0,&(x,y)=(0,0)
\end{cases}
\end{align}$$

not differentiable at $(0,0)$?

By setting $x=0$ or $y=0$ I would get  $f_x(0,0)=0$, $f_y(0,0)=-2$ and when I look at the 3-D Surface plot, $f(x,y)$ looks quite "normal". I gaze already hours on it and cannot identify the problem.

Obviously the partial derivates exist, but are not continuous, but I cannot see where.

Is there an analytic way, I can obtain the result from? How can I come to the conclusion, without gazing at the curve?

答:
A few comments too long to be put in the comment section.

1: Quote from here: However, the existence of the partial derivatives (or even of all the directional derivatives) does not in general guarantee that a function is differentiable at a point.

2: In your case $f_y(0,0) = 1$, not $-2$.

3: Just because the partial derivatives $f_x$ and $f_y$ are not continuous at $(0,0)$, does not mean $f$ is not differentiable at $(0,0)$. You can use the definition of differentiability to prove it. You can follow this example.


问:
Derivative of $\big(f(x+dx)-f(x)\big)$
Am I correct to state that taking the derivative of $f(x+dx) - f(x)$ with respect to $x$ does not equal $df/dx$? Or in another words:

$$\frac{d}{dx}\big(f(x+dx) - f(x)\big) \neq \frac{df(x)}{dx}$$

I've reasoned this out by example. If we take $f(x) = x^2$, then:

$$\frac{d}{dx}\big(f(x+dx) - f(x)\big) = \frac{d}{dx}\left((x+dx)^2\right) - \frac{d}{dx}\left(x^2\right) = 2\,dx.$$

The only thing I'm uncertain of is that in the limit as $dx\to0$, the tangents at $x$ and $x+dx$ should be approximately the same (given that the function is smooth for all $x$).

答:
Well, I assume you're working with standard analysis. In this context $dx$ is meaningless unless you define it as one kind of object called differential form, and I'll not talk about these objects now to avoid more confusion. You just have to know that treating $dx$ as "infinitesimal amount added to $x$" is not allowed in standard analysis greatly because this brings many troubles with it. The framework that allows this kind of reasoning is non-standard analysis that many people really like. For example it's easy to see that if you try defining $dx$ as $\Delta x \to 0$ you'll fail drastically because by the usual definition of limit, the limit of $\Delta x$ when $\Delta x$ goes to zero is zero.

I understand your preocupation because many people insist in teaching calculus using those objects, and that is an approach that confuses people very much. My suggestion is that you pick a book like Apostol's Calculus Vol. 1 or Spivak's Calculus (personally I prefer Spivak's book). They will teach you how to deal with calculus without recoursing to infinitesimals, and after you see it you'll understand why people moved from this thought to the modern you: the modern one is just clearer and simpler.

Good luck with your studies!

Title: Derivative of $\big(f(x+dx)-f(x)\big)$


答:
If $A$ is symmetric, you can rewrite it as
$$A = B^T B$$
for some matrix $B$ (see more information on how to find $B$ here).

Hence we have
$$x^TAx = x^TB^TBx = (Bx)^TBx = \|Bx\|^2$$

Taking the square root and differentiating should yield a result.

Where naturally $\|Bx\|^2$ is a scalar.

Hope that helps!

Title: Derivative of $\sqrt{x^TAx}$

Body:
Let $x\in \mathbb{R}^n$ and $A\in\mathbb{R}^{n\times n}$ be a positive semi-definite matrix.

Is there a way to express in closed form the following derivative?

$$\frac{\partial}{\partial x} \sqrt{x^TAx}$$


问:
Proof to show distance between objects in a cube.
Consider a cube of side $2$ feet containing 65 flies. Show that at a given point in time, there are at least $9$ flies such that the distance between any $2$ of them is no larger than $1.74$ feet.

So, I know that for $9$ of the flies to line up across the longest line of the cube would be it's diagonal $=\sqrt{2^2+2^2}$ or $2.828...$. But I'm not sure if this is something I'll need to determine the answer $1.74$ feet. Would someone be able to give me some pointers as to what to do to do this proof?

答: Tile the cube into $2\times 2\times 2$ smaller cubes of side length $1$'. Since $8\cdot 8<65$, one of the cubes must contain at least $9$ flies. The diagonal of such a cube is $\sqrt 3$'.


问:
Corollary of Borsuk-Ulam theorem (equivalent statement?)
I heard that the following are equivalent:

(1) (Borsuk-Ulam) If $f:S^n\to\Bbb R^n$, there is $x\in S^n$ such that $f(x)=f(-x)$.

(2) If $A_{i}\subset S^n$ is closed for $i=1,2,...,n+1$ such that $A_1\cup\cdots A_{n+1} = S^n$, then at least one of these sets must contain a pair of antipodal points

(3) If $A_{i}\subset S^n$ is open for $i=1,2,...,n+1$ such that $A_1\cup\cdots A_{n+1} = S^n$, then at least one of these sets must contain a pair of antipodal points

Well I know $(1)\Rightarrow(2)$. But don't know the above is equivalent. How can I show? By the way, are those equivalent?

答:
$(2)\Rightarrow(3)$ This is just topology. Suppose first that $S^n=A_1\cup A_2$  with each $A_i$ open. We have $(S^n\setminus A_1)\cap (S^n\setminus A_2)=S^n\setminus(A_1\cup A_2)=\emptyset$, so since each $S^n\setminus A_i$ is closed we can use normality to find an open set $U$ with $S^n\setminus A_1\subseteq U\subseteq \overline U\subseteq A_2$. Putting $B_1=S^n\setminus U$ and $B_2=\overline U$ we have $S^n=B_1\cup B_2$ with each $B_i$ a closed set satisfying $B_i\subseteq A_i$.

Now suppose $S^n=A_1\cup\dots\cup A_{n+1}$ with each $A_i$ open. Inductively repeating the above procedure we find closed $B_1,\dots,B_{n+1}\subseteq S^n$ with $B_i\subseteq A_i$ and $S^n=B_1\cup\dots\cup B_{n+1}$. By hypothesis some $B_i$ contains an antipodal point, and hence so does some $A_i$. $\square$

$(3)\Rightarrow(1)$ Let $f:S^n\rightarrow \mathbb{R}^n$ be given and put $g(x)=f(x)-f(-x)$ so that $g(x)=0$ if and only if $f(x)=-f(x)$. Note that $g(-x)=-g(x)$ for all $x\in S^n$. To establish a contradiction we suppose that $0\not\in g(S^n)$.

Take the $n$ open sets $U_i=\{x_i>0\}\subseteq\mathbb{R}^n$, $i=1,\dots,n$, and add to this collection the set $U_{n+1}=\{x_i<0\;\text{for all}\;i=1,\dots,n\}$. Because of the supposition we have $g(S^n)\subseteq\bigcup^{n=1}_{i=1}U_i$.

Letting $A_i=g^{-1}(U_i)$ we can now write $S^n=A_1\cup\dots\cup A_{n+1}$ as a union of $n+1$ open sets. By assumption one of the sets contains an antipodal pair. But this means that there is $x\in S^n$ such that $g(x),g(-x)=-g(x)\in U_i$ for some $i$, and this is absurd. Thus we have a contradiction to the assumption that $g$ is nonvanishing. $\square$.


问:
How to express an interval in a cost function
I'm trying to best express a cost function and I'm unclear on how to properly add an interval-based constraint. The cost function is used to minimize a cubic function.

My cost function has 3 components:

Quadratic $\frac{1}{N} \sum_{i=1}^N{(\hat{y}-y})^2$

L2 Regularization $\lambda\sum_{i=2}^kw_i^2$

Interval-based constraint $\sigma_{f(x)}, x\epsilon(80,\infty)$

Is the following a clear way to express this? I'm concerned that the notation is not clear.

$min \frac{1}{N} \sum_{i=1}^N{(\hat{y}-y})^2 + \lambda\sum_{i=2}^kw_i^2 + \sigma_{f(x)}, x\epsilon(80,\infty)$

答:
This is indeed not clear (to me). This is what I think you mean.

Call your three terms $A$, $B$ and $C$. Then
$$
\text{cost}(x) = \min\begin{cases}
A + B \quad\quad\quad \text{ if } x  \le 80 \\
A+B+C  \quad \text{ if } x > 80
\end{cases}
$$

For clarity, I would in fact define $A$, $B$ and $C$ first rather than inlining them in the \cases statement.


问:
Showing that if $x = \sec\theta + \tan\theta$, then $x + \frac{1}{x} = 2\sec\theta$
$x = \sec\theta + \tan\theta.$

Show that $x+\frac1x = 2\sec\theta$.

Thanks.

I used a few simple trig identities but get nowhere. I am confused about this question a lot. I do not see where the theta and x can be in one.

None are squared so I cannot used the identities I would think. The only thing I can think to do is

$x=\frac{1}{\sin\theta}+ \frac{\cos\theta}{\sin\theta}  $ ...and then put them together.

答:
Here is a detailed answer.
Let's rock!
$$\require{cancel}
\begin{align}
\sec\theta+\tan\theta+\frac{1}{\sec\theta+\tan\theta}&=\frac{\sec\theta\left(\sec\theta+\tan\theta\right)+\tan\theta\left(\sec\theta+\tan\theta\right)+1}{\sec\theta+\tan\theta}\\
&=\frac{\left(\sec\theta+\tan\theta\right)\left(\sec\theta+\tan\theta\right)+\color{red}1}{\sec\theta+\tan\theta}\\
&=\frac{\left(\sec\theta+\tan\theta\right)\left(\sec\theta+\tan\theta\right)+\color{red}{\sec^2\theta-\tan^2\theta}}{\sec\theta+\tan\theta}\\
&=\frac{\sec^2\theta+2\sec\theta\tan\theta\cancel{+\tan^2\theta}+\sec^2\theta\cancel{-\tan^2\theta}}{\sec\theta+\tan\theta}\\
&=\frac{2\sec^2\theta+2\sec\theta\tan\theta}{\sec\theta+\tan\theta}\\
&=\frac{2\sec\theta\cancel{\left(\sec\theta+\tan\theta\right)}}{\cancel{\sec\theta+\tan\theta}}\\
&=2\sec\theta
\end{align}
$$
We substituted $\color{red}1$ with $\color{red}{\sec^2\theta-\tan^2\theta}$.

I hope this helps.


问:
Determinant (Cauchy?)
I have to show that

${det}\left ( \begin{bmatrix}
\sum_{i=1}^{n}t_i^2 & \sum_{i=1}^{n}t_i\\
\sum_{i=1}^{n}t_i & n
\end{bmatrix} \right )>0$

My solution so far

$n \cdot (\sum_{i=1}^{n}t_i^2)-(\sum_{i=1}^{n}t_i)^2>0\Leftrightarrow$

$n \cdot (\sum_{i=1}^{n}t_i^2)>(\sum_{i=1}^{n}t_i)^2\Leftrightarrow \\$

$n(t_{1}^2+t_{2}^2+...+t_{n}^2)>(t_{1}+t_{2}+...+t_{n})^2$

All the $t$'s are not the same. But my conclusion seems wrong. Can I use Cauchy or any other method to make it obvious?

Thanks!

答:
Hint : $n=1+1+ \cdots +1$ and use the Cauchy-Schwartz inequality.

The CS inequality states
\begin{eqnarray*}
(a_1^2+a_2^2+ \cdots +a_n^2)(b_1^2+b_2^2+ \cdots +b_n^2) \geq (a_1 b_1+a_2 b_2+ \cdots +a_n b_n)^2.
\end{eqnarray*}
For your question you need $a_i=1$  and $b_i=t_i$.

Title: Determinant (Cauchy?)


答:
HINT

Angle between vectors $\vec{KM}, \vec{LM},$ where the points have complex coordinates $z_K, z_L, z_M,$ is equal to
$$\arg{\frac{z_M-z_L}{z_M-z_K}}.$$
In the present exercise we want

$$\arg{\frac{z-z^2}{z-z^3}}=\frac{\pi}{2}+k\pi,\;k\in \mathbb{Z}$$
or
$$\arg{\frac{z^2-z}{z^2-z^3}}=\frac{\pi}{2}+k\pi,\;k\in \mathbb{Z}$$
or
$$\arg{\frac{z^3-z^2}{z^3-z}}=\frac{\pi}{2}+k\pi,\;k\in \mathbb{Z}$$

Title: Complex number such that affixes of itself, its squared and its cube form a right triangle. $z,z^2,z^3$ form a right triangle in argand plane

Body:
$\DeclareMathOperator{\aff}{Aff}$
Here is a funny problem  I stumbled upon recently

I call the affix of a complex number the corresponding point in the argand plane, that is

$$\aff(\mathrm{i})=(0;1)$$

Find all $z \in \mathbf{C}$ such that the affixes of $z,z^2\ \& \ z^3$ form a right triangle.

I decided to Set $A=\aff(z),B=\aff(z^2),C=\aff(z^3)$

I have proven that :

If $z$ is a solution, then $z\neq 0,\  z\neq 1, \ \& \ z\neq-1$

If the triangle is right angled at A, $z$ must be a non zero imaginary number.

If the triangle is rigth angled at B, $\aff(z)$ has to lie on the vertical line defined by $x=(-1)$, except for $z=(-1)$.

(I'll post my proof when I get some time to write it in $\mathrm{\LaTeX}$ soon.)

I'm down to the third case : I have proven that if the triangle is right angled at $C$, then :
$$\Re \left(\frac{1+z}{z}\right)=0$$

Thanks to some experimentation, I know that $\aff(z)$ should lie on the circle centered at $(-\frac12;0)$ with radius $\frac12$, with $z$ again respecting the first condition above.

Now,how do I prove this ?

Thanks for the help.

Here is an animation.


问:
A quadrilateral $ABCD$ with $AB = 10$ , $BC = 6$ , $CD = 8$ and $DA = 2$ has diagonals $AC$ and $BD$ meeting at $O$.

A quadrilateral $ABCD$ with $AB = 10$ , $BC = 6$ , $CD = 8$ and $DA = 2$ has diagonals $AC$ and $BD$ meeting at $O$, such that $\angle COB = 45^\circ$. Find the area of $ABCD$.

What I Tried: Here is a picture :-

The only thing missing to find the area is one of the diagonals, and I can find the area easily by Heron's Formula. Also I thought that since some angles are there, I might have to use trigonometry, and I tried but without success.

We have that $\angle DOC = \angle AOB = 135^\circ$ . I know that the area of a triangle can be found from the formula :- $$\rightarrow \frac{1}{2}ab\sin \gamma$$

Where $\gamma$ is the angle between $a$ and $b$, but I cannot use it sadly as I don't know any of the lengths of $AO$ , $BO$ , $CO$ and $DO$. Next I thought of the formula :-
$$\rightarrow \frac{a}{\sin a} = \frac{b}{\sin b} = \frac{c}{\sin c}$$

But this dosen't work here either due to the same reason.

I think that Law of Cosines might work, but I am actually new to Trigonometry and not sure if it will really work or not, and then I have to consider $4$ equations for $4$ sides, which will be complicated.

Can anyone help me? Thank You.

Note:- Solutions without Trigonometry and mostly welcome. I hate using Trigonometry.

答:
Hint :

$$[ABCD] = \dfrac{1}{2} AC  \cdot BD \cdot \sin 45$$

$$[BAD] = \dfrac{1}{2} \cdot OA \sin 45 \cdot BD$$
$$[BCD] = \dfrac{1}{2} \cdot OC \sin 45 \cdot BD$$

Hint 2 :
Use cosine-rule in all four small triangles and $$\cos (180 - \theta) = -\cos \theta$$.

$$OD^2+OA^2-2\cdot OD\cdot OA\cos 45 = AD^2$$
$$OA^2+OB^2+2\cdot OA\cdot OB\cos 45 = AB^2$$
$$OB^2+OC^2-2\cdot OB\cdot OC\cos 45 = BC^2$$
$$OC^2+OD^2+2\cdot OC\cdot OD\cos 45 = CD^2$$

Obtain

$$4[ABCD] = -AD^2+AB^2-BC^2+CD^2$$


问:
Integration of a Weibull pdf
I am trying to calculate the following integral.

$$\int_0^{\infty} x^{k-1} e^{-Ax^k} dx$$

This is very similar to the Weibull pdf integrated from $0$ to infinity (which would clearly be $1$).

What I've actually done is taken the expected value of a function of $x$ where $x$ follows a Weibull distribution. The integral above shows what is left after I pull out all the constants and combine some constants into a single parameter, $A$.

$$f(x) = \frac{\lambda_1}{2}e^{-(x/\lambda_1)^k}$$
where $X$ ~ $\operatorname{Weibull}(k,\lambda_2)$.

答:
Call $u= Ax^{k}$, I will assume that $k>0$, so when $x=0$ then $u=0$,

$$
{\rm d}u = k A x^{k-1}~{\rm d}x \tag{1}
$$

and your integral becomes

$$
\int_0^{+\infty} {\rm d}x~ x^{k-1} e^{-Ax^k} = \frac{1}{kA}\int_0^{+\infty} \underbrace{{\rm d }x~ x^{k-1} k A}_{{\rm d}u} e^{-\underbrace{(Ax^k)}_{u}} = \frac{1}{kA} \int_0^{+\infty}  {\rm d}u~ e^{-u} = \cdots
$$

Title: Integration of a Weibull pdf


问:
Prove that $\ell^2(\mathbb R)$ is complete
Let $$X:=\ell^2(\mathbb R)=\left\{x=(x_n) _ {n\in\mathbb N}\in \mathbb R^{\mathbb N}\mid\sum_{n=1}^\infty x_n^2<\infty \right\}$$

and $$\|x\| _ X=\left(\sum_{k=1}^\infty x_k^2\right)^\frac{1}{2}.$$

$(X,\|\cdot \|_X)$ is a normed space. Prove that it's complete.

Let $\varepsilon >0$ and $(x^n)_{n\in\mathbb N}\in X^{\mathbb N}$ a Cauchy sequence where $x^n:=(x^{n} _k) _ {k\in\mathbb N}$.

Since $(x^n)$ is a Cauchy sequence, then so is $(x_k^n)_{n\in\mathbb N}$ for all $k\in\mathbb N$. In particular, there is $x_k\in \mathbb R$ s.t. $x_k^n\to x_k$ whenever $n\to \infty $.

Since for all $n\in\mathbb N$, $$\sum_{k=N}^\infty (x^n_k-x_k)^2\underset{N\to \infty }{\longrightarrow }0,$$

if $n\in\mathbb N$ is fixed, there is $N_n\in\mathbb N$, s.t. $$\sum_{k=N_n+1}^\infty (x_k^n-x_k)^2<\frac{\varepsilon }{2}.$$

We have that \begin{align*}
\|x^n-x\| _ X^2&=\sum_{k=1}^{N_n}(x_k^n-x)^2+\sum_{k=N_n+1}^\infty (x_k^n-x_k)^2\\
&\leq \sum_{k=1}^{N_n}(x_k^n-x)^2+\frac{\varepsilon }{2}.
\end{align*}

Question Since $N_n$ depend on $n$, I can't find $M\in\mathbb N$ s.t. $|x^n_k-x_k|<\frac{\varepsilon }{2N_n}$ for all $k=1,...,N_n$. Can someone give a hint on how to solve this problem ?

答:
Fix $\varepsilon > 0$. Let $N = N(\varepsilon)$ be such that
$$n, m \ge N \implies \sum_{k=1}^\infty (x^n_k - x^m_k)^2 < \frac{\varepsilon^2}{4}.$$
This implies that
$$n, m \ge N \implies (x^n_k - x^m_k)^2 < \frac{\varepsilon^2}{4},$$
for any given $k$. This proves $(x^n_k)_n$ is Cauchy, and hence convergent, as you have already detailed.

I claim that $\sum_{k=1}^\infty (x_k^n - x_k)^2$ exists, and is less than or equal to $\frac{\varepsilon^2}{4}$. Note that it simply suffices to show that the partial sums are bounded above by $\frac{\varepsilon^2}{4}$; the fact that each term is non-negative means that the series must converge, by the monotone convergence theorem.

For any $M \in \Bbb{N}$, and $n \ge N$, we have,
$$\sum_{k=1}^M (x_k^n - x_k)^2 = \sum_{k=1}^M \lim_{m \to \infty} (x^n_k - x^m_k)^2 = \lim_{m \to \infty} \sum_{k=1}^M (x^n_k - x^m_k)^2,$$
where the latter equality is the algebra of limits on a (finite) sum. But then,
$$m \ge N \implies \sum_{k=1}^M (x^n_k - x^m_k)^2 \le \sum_{k=1}^\infty (x^n_k - x^m_k)^2 < \frac{\varepsilon^2}{4},$$
hence when taking the limit as $m \to \infty$,
$$\sum_{k=1}^M (x_k^n - x_k)^2 \le \frac{\varepsilon^2}{4},$$
for all $M \in \Bbb{N}$. Then, taking the limit as $M \to \infty$,
$$\sum_{k=1}^\infty (x_k^n - x_k)^2 \le \frac{\varepsilon^2}{4}.$$
Therefore,
$$n \ge N \implies \|x^n - x\| \le \frac{\varepsilon}{2} < \varepsilon.$$


问:
GCD questions. How do I determine whether its true or false? (answer is true for both)
1.Let u and v be two positive integers with u/3<vv and let gcd(u,v) represent the greatest common divisor of u and v. If  u and v are both odd, then gcd(u,v)=gcd(u−v,v)

How do I solve these questions? Can you share the steps to these quiz problems?

Asked my prof and he says:In general, exercising in some rudimentary basic number theory, e.g. to see how modp behaves, helps.
The simplest way is of course just to use the definition of remainder and quotient:
For any two positive integers u and v there exist unique non-negative integers q and r
such that u=qv+r and 0≤r<v.
Just manipulating these q and r alone one can accomplish a lot (personal experience).

答:
So for the first question. You have $\frac{u}{3}<v<\frac{u}{2}$.

Split this into two inequalities.

$u/3<v$ which implies $u<3v$. The second inequality is $u/2>v$. Which implies $u>2v$. So you have $$2v<u<3v$$

If you can show that $0≤u−2v<v$ that proves it is the remainder when you divide u by v.

2.You're asked to show whether or not. $gcd(u,v) = gcd(u-v,v)$

So let $a=gcd(u,v)$. let $b=gcd(u-v,v)$.

Show that $a|u-v$. Then since $a|(u-v)$ and $a|v$, then we know that $a|b$.

Similarly prove that $b|a$. Since a and b are positive and since $a|b$ and $b|a$ we know that $a=b$. (There are a couple of ways to show this. One is using inequalities. Since $a|b$, $a \le b$. Since $b|a$, $b \le a$. $a \le b$ and $b \le a$ means $a=b$)


问:
Transportation Problem - Handling case of infinite capacity to supply
The problem is - In a standard transportation problem, with destination's demand limited, but each factory has unlimited supply to cater to the needs of destination. How should we modify the problem in such cases. I have the following idea.

Assign each factory the capacity of aggregate demand of destinations

Add a dummy destination with transportation cost zero and demand equal to the unbalanced part( Aggregate demand * (no of factories - 1)

Then solve the problem.

The initial solution - One of factories will be supplying its entire supply(modified new supply) to the dummy destination.

Also, Can I use the same method with modification, in case destination can handle unlimited supply but the factories has limited capacity to supply.

I have added a sample question for better understanding of the problem.

答: Yes, setting the supplies to a small upper bound and introducing dummy arcs and a dummy demand node is the standard approach.  Yes, you can apply the same idea with a dummy supply node if the supplies are fixed but the demands are unspecified.</p>


问:
Solve differential equasion $(y'')^2-5y'+6=0$
Solve differential equation  $(y'')^2-5y'+6=0$

For there is no "x" in equation, I made substitution $y'=p(y)$ and $y''=p'p$

$(p'p)^2-5p+6=0$

What can I do with the squared member?

Body:
Let $y'=p$, then $$(y'')^2-5y'+6=0 \implies p'^2-5p+6=0 \implies p'= \sqrt{6-5p}. $$$$\int\frac{dp}{\sqrt{6-5p}}=\int dx+A \implies -\frac{2}{5}\sqrt{6-5p}=x+A.$$
$$\implies p=\frac{dy}{dx}=\frac{6}{5}-\frac{5}{4}(x+A)^2 \implies y=\frac{6x}{5}-\frac{5}{12}(x+A)^3+B$$


问:
Epimorphism from $3\mathbb{Z}$ to $U_8$
I'm trying to disprove/prove that an epimorphism exists from $3\mathbb{Z} $ to $U_8$

$U_8$ is acyclic, and is composed out of $\{1,3,5,7\}$. Therefore I can't map every $3k \in 3\mathbb{Z}$ to a cycle, so I'm thinking of creating a composition $f(x) = h(g(x))$ such that:
$$g(x) = x \mod 12 \\ h(0) = 1, \: h(3) = 3,\: h(6) = 5,\: h(9) = 7 $$

However I'm stuck proving this is even an homomorphism.

答:
Since all the elements of $U_8$ have order $2$, all the even elements of $\mathbb{Z}$ must map to $1$:$$f(2k)=f(k+k)=f(k)\cdot f(k)=1$$
Suppose that $f(3)=x$.  Then $f(-3)=x^{-1}$  Now all the positive odd elements of $3\mathbb{Z}$ map to $x$,
since $$f(3(2k+1))=f(6k)f(3)=x,$$and all the negative odd elements map to $x^{-1}=x$, so $f$ cannot be onto.

Title: Epimorphism from $3\mathbb{Z}$ to $U_8$


问:
Use of exponentiation in topos logic
I'm learning about topos theory and its many uses in (intuitionistic) logic. The definition of an elementary topos $\mathbf{C}$ requires the category to have exponentiation, that is, given $\mathbf{C}$-objects $A$ and $B$, there exists another $\mathbf{C}$-object $B^A$ which is "the object of all $A\to B$ morphisms".

Now, when developing the propositional logic, the valuations, the meet/join/negation/implication operations in the algebra of subobjects I haven't seen the exponential construction at all.

I'm wondering why is it crucial that a categroy has exponential objects (or power objects) in order to "do" strong enough internal logic, or similar enough to the $\mathbf{Set}$ topos internal logic.

What construction, rule of inference, quantification, formula or whatever, needs exponentiation/power objects?

答:
The exponential $B^A$ corresponds to the implication (or function space) $A \Rightarrow B$. The reason you have not seen exponentiation explicitly appear in the treatment of propositional logic is that in that setting you can define implication in terms of negation and disjunction, i.e. as $\neg A \lor B$. However, in other settings (such as constructive logic), this is not possible, and implication must be given separately.

There is a distinction to be drawn, though, between the logic of the lattice of subobjects (giving the propositional logic definable within the topos) and the internal language of the topos itself (which is a higher-order logic). We may have exponentials of the subobjects and not general exponentials in the category itself, or vice versa. The exponential is important in toposes for giving constructions of function spaces (like in the category of sets), which is exactly what makes the internal language "higher-order" logic.

If you want to learn more about exponentiation and implication (or, more generally, function types in a type theory), the place to start is the correspondence between the simply-typed $\lambda$-calculus and cartesian-closed categories. The internal language of toposes subsumes the simply-typed $\lambda$-calculus, so this is a simpler starting point.

This other answer about the distinction between $\vdash$ and $\Rightarrow$ in categorical logic might also be helpful.

Title: Use of exponentiation in topos logic


答:
Instead of using theorems about the relationship between differentiability and properties of partial derivatives, let's just go back to the definition of differentiability:

A function $f(x,y)$ is differentiable at $(a,b)$ if and only if there is a linear function $L(x,y)=cx+dy+e$ such that

$$\lim_{(x,y)\to(a,b)}{f(x,y)-L(x,y)\over||(x,y)-(a,b)||}=0$$

For the function $f(x,y)=y^3/(x^2+y^2)$ and $(a,b)=(0,0)$, polar coordinates $x=r\cos\theta$ and $y=r\sin\theta$ gives

$${f(x,y)-L(x,y)\over||(x,y)-(a,b)||}={r\sin^3\theta-cr\cos\theta-dr\sin\theta-e\over r}=\sin^3\theta-c\cos\theta-d\sin\theta-{e\over r}$$

We can get a limit as $r\to0$ by setting $e=0$, but there is no trigonometric identity of the form $\sin^3\theta=c\cos\theta+d\sin\theta$, so the limit as $(x,y)\to(0,0)$ is not only not $0$, it doesn't exist.

Remark: It's not strictly necessary to do so here, but it might help to keep in mind that the choice of $\theta$ can itself vary with $r$, writing it as $\theta_r$ or $\theta(r)$.  Doing so helps emphasize that the limit as $(x,y)\to(0,0)$ is indeterminate.

Title: Why is this function not differentiable?

Body:
Why is the function

$$\begin{align}
f(x,y)=\begin{cases}\frac{y^3}{x^2+y^2},&(x,y)\ne(0,0)\\\\
0,&(x,y)=(0,0)
\end{cases}
\end{align}$$

not differentiable at $(0,0)$?

By setting $x=0$ or $y=0$ I would get  $f_x(0,0)=0$, $f_y(0,0)=-2$ and when I look at the 3-D Surface plot, $f(x,y)$ looks quite "normal". I gaze already hours on it and cannot identify the problem.

Obviously the partial derivates exist, but are not continuous, but I cannot see where.

Is there an analytic way, I can obtain the result from? How can I come to the conclusion, without gazing at the curve?


问:
Continuous RV: PDF/CDF (Geometric)
Let vector $(X, Y)$ be distributed uniformly on a support of triangular shape $ABC$, where $A=(0;0), B=(2;0), C=(1,1)$. Find $PDF$ and $CDF$ of random variables $X$ and $Y$. Are those random variables independent?

Could you help me with this one? I do have some experience with the basic geometric probability, however I literally don't know what and how to apply here..

答:
When they say $Z \sim U(\Omega), \Omega \subset \mathbb{R}^n$, this means that the PDF is the same in every point, if it is in the support $\Omega$ and zero otherwise:

$$ p(z) = \begin{cases}C, \text{ if } z \in \Omega \\ 0, \text{ if } z \notin \Omega \end{cases} $$,

where the constant C is chosen so that $\int_\Omega C \mathrm{d} z = 1$.

In your case $n = 2$, $z = (x,y)$ and $\Omega = \Delta ABC$.

For the integration purposes the support of $(X,Y)$ can be expressed as the following constraint on values of $Y$:

$$
\Delta ABC = \left\{(x,y) \in [0,2] \times \mathbf{R} \bigg| \begin{cases} y \in (0,x) &\text{ if } x \in [0,1],\\ y \in (0,2-x) &\text{ if } x \in (1,2] \end{cases}\right\}.
$$

So

$$ C = \left(\int_{\Delta ABC} 1 \mathrm{d}y\mathrm{d}x\right)^{-1} = \frac{1}{S_{\Delta ABC}} = 1,$$

therefore

$$ p(x,y) = 1 \text{ if } (x,y) \in \Delta ABC, \text{ and } 0 \text{ otherwise}.$$

To obtain PDF of $X$ in a certain point $X=x$, you must integrate the joint PDF over all values of $Y$ that are consistent with $x$:

$$ p(x) = \begin{cases} \int_0^x 1 \mathrm{d} y = x, \text{ if } x\in [0,1], \\
           \int_0^{2-x} 1 \mathrm{d} y = 2 - x, \text{ if } x \in (1,2],\\
0 \text{ otherwise}.
    \end{cases} $$

The CDF is computed from the PDF according to the definition:

$$ P(x) = \int_0^x p(\xi) \mathrm{d} \xi = \begin{cases} 0, \text{ if } x < 0, \\ \frac{x^2}{2}, \text{ if } x \in [0,1], \\ 2x - \frac{x^2}{2}, \text{ if } x \in (1,2], \\ 1, \text{ if } x > 2.   \end{cases}$$

By the same token, you can easily find the PDF and CDF of $Y$.

Title: Continuous RV: PDF/CDF (Geometric)


问:
What is the ideal $\sum_{i \in I} \mathfrak{a}_i$?
I encountered the ideal $\sum_{i \in I} \mathfrak{a}_i$, where the $\mathfrak{a}_i$'s are ideals of an entire ring. However, no explanation was given about this ideal.

Is it the ideal $\lbrace \sum_{i \in I} a_i \mid a_i \in \mathfrak{a}_i \rbrace$?

Is it possible to express this ideal as the ideal generated by some set, i.e. the intersection of all ideals containing some set?

答:
The sum of a set of ideals is simply the smallest ideal containing all elements in each of the ideals. If your set is finite then you have
$$
I_1+I_2+...+I_n = \{r_1+r_2+...+r_n\mid r_i \in I_i\}
$$
And if you are considering the sum of an infinite set of ideals, elements are simply sums of elements from any finite subset of the ideals. I.e. if you have a set of ideals indexed by the natural numbers then:
$$
\sum_{i \in \mathbb{N}}I_i = \{r_1+r_2+...+r_n\mid r_i \in I_i\, n \in \mathbb{N}\}
$$
In terms of generating set, it's hard to say in general, but you certainly have that the sum of a set of ideals is generated by the union of the generators of each ideal in the sum.


答: It is given by $\sum_{i\in I}\mathfrak{a}_i=\{\sum_{i\in I} a_i\mid \textrm{all but finitely many }a_i=0\}$ as stated in the comments.

Title: What is the ideal $\sum_{i \in I} \mathfrak{a}_i$?

Body:
I encountered the ideal $\sum_{i \in I} \mathfrak{a}_i$, where the $\mathfrak{a}_i$'s are ideals of an entire ring. However, no explanation was given about this ideal.

Is it the ideal $\lbrace \sum_{i \in I} a_i \mid a_i \in \mathfrak{a}_i \rbrace$?

Is it possible to express this ideal as the ideal generated by some set, i.e. the intersection of all ideals containing some set?


问:
What do $L$-functions of curves over $\mathbb Q$ tell us about the curve
Following up this thread: $L$-function of an elliptic curve and isomorphism class

I'd like to ask some more questions for the case of smooth projective curves $C$ over $\mathbb Q$

To be more precise, take this definition ( https://webusers.imj-prg.fr/~marc.hindry/Notes_rev_Brasilia.pdf, (2.2)):

$$L(\rho,s)=\prod_{\mathfrak p}\det(1-\rho(\mathrm{Frob}_p)p^{-s}\mid V^{I_{\mathfrak p}})^{-1}$$
where $\mathfrak p$ runs over the primes of $\mathbb Q$, or more generally: the number field $K$. And $V=H^1_{et}(C,\mathbb Q_\ell)$.

Is there a similar statement for curves of higher genus? I.e. if two $L$-functions $L(C,s)$ and $L(C',s)$ coincide, what can be said about $C$ and $C'$?

Are there further interesting properties that can be read off the $L$-function $L(C,s)$, and not the curve $C$ itself?

Assuming the Generalized Riemann Hypothesis holds for $L$-functions of  curves over $\mathbb Q$, what does that imply about the curve?

Best,

Dan

答: Let $A:=\mathbb{Z}$ be the ring of integers and let $S$ be a scheme of finite type over $A$. If $\pi:X \rightarrow S$ is a scheme of finite type over $S$, you may define the Hasse-Weil L-function $L(X,s):=\prod_{x \in X^{cl}} \frac{1}{1-N(x)^{-s}}$. Here $N(x)$ is the number of elements in in the residue field of $x$. Since $x$ by definition is a closed point it follows $N(x)$ is an integer. If $\pi':Y\rightarrow S$ is another scheme of finite type and $X_t\cong Y_t$ are isomorphic for any closed point $t \in S$, it follows $L(X,s)=L(Y,s)$. In particular if $E,F$ are vector or projective bundles of the same rank over $S$ it follows $L(E,s)=L(F,s)$. Hence if $E$ and $F$ have the same L-function this does not imply that $E$ and $F$ are isomorphic. Hence if $S=C$ is a "curve" it follows vector or projective bundles over $C$ are not determined by their Hasse-Weil L-functions.


问:
Inverse of a particular Toeplitz matrix
I have the following matrix

$$A=\begin{pmatrix}
b & a & 0 & 0 & \cdots & 1\\
a & b & a & 0 & \cdots & 0\\
0 & a & b & a & \cdots & 0\\
0 & 0 & a & b & \cdots & 0\\
\vdots & \vdots & \vdots & \vdots & \ddots & \vdots \\
0 & 0 & a & b & a & 0\\
0 & 0 & 0 & a & b & a\\
1 & 0 & 0 & 0 & a & b
\end{pmatrix}$$

That can be written as $A=T+u_{1}v_{1}^{T}+u_{2}v_{2}^{T}$, where T is tridiagonal. I have to compute the inverse of this matrix in terms of $T^{-1}$. I had thought that I could extend the Sherman-Morris formula, but the procedure for obtaining it cannot be applied to this case. In the statement of the problem there is a hint. It says that it is useful writting $A=L(\mathbb{I}+stuff)U$ [sic]. Where L (U) is a lower (upper) triangular matrix. I have thought that those LU could be the LU factorization of A, but when U compute $L^{-1}(u_{1}v_{1}^{T}+u_{2}v_{2}^{T})U^{-1}$ the stuff term becomes a singular matrix. Do you have any tip to solve this problem?.

答:
You have reduced the problem to that of finding the inverse of $I + L^{-1}(u_1v_1^T + u_2v_2^T)U^{-1}$. With that in mind, apply the Woodbury matrix identity to compute $(I + PQR)^{-1}$, where
$$
P = \pmatrix{L^{-1}u_1 & L^{-1}u_2}, \quad Q =  I_{2 \times 2}, \quad R =  \pmatrix{v_1^TU^{-1}\\v_2^TU^{-1}}.
$$

Title: Inverse of a particular Toeplitz matrix


问:
Simple Harmonic Motion

A particle moving with simple harmonic motion starts from rest at a distance $3m$ from the center of oscillation. If the period is 2$\pi$ seconds:

a) Find the time taken to move to a point $2m$ from the origin

b) Find the velocity and acceleration at this point

答:
A particle moving as a simple harmonic motion is described by the equations
$$x(t) = A \cos(2\pi t/T + \phi_0)$$
and
$$v(t) = -\frac{2\pi A}{T} \sin(2\pi t/T + \phi_0).$$
Here, $A$ is the amplitude, $T$ the period, and $\phi_0$ a "phase-shift" of the motion.

Given your information $T=2\pi$. Additionally you have $v(0)=0$ and $x(0)=3A$. This should suffice to solve your problem.

Title: Simple Harmonic Motion


问:
"Spherical Plot" of a 2D function
I'have seen this graph of a 2D function of $\theta$ and  $\phi$ (let's call it $F(\theta,\phi)$):

You may see that it's not represented as a conventional surface plot. In fact, there aren't $\theta$ and  $\phi$ on x and y axes, but there is $\theta$ at both x and y axis. Along x axis we have $F(\theta,\phi=0°)$, along y axis we have $F(\theta,\phi=90°)$.

Which kind of graph is this? If I plot $F(\theta,\phi=90°)$ with respect to $\theta,\phi$, I get a completely different result.

I need some formal clarifications about it so that I can make a similar graph on MATLAB or other tools.

答:
It's a Cartesian plot of a scalar function $F(\theta, \phi)$ with
$$\left\lbrace ~ \begin{aligned}
x &= \frac{\theta}{2 \pi} \cos \phi \\
y &= \frac{\theta}{2 \pi} \sin \phi \\
z &= F(\theta, \phi) \\
\end{aligned} \right .$$
If
$$\begin{array}{rccl}
0 & \le & F(\theta, \phi) & \le & 1 \\
0 & \le & \theta & \lt & 2 \pi \\
0 & \le & \phi & \lt & 2 \pi \\
\end{array}$$
then
$$\begin{array}{rccl}
-1 & \le & x & \le & 1 \\
-1 & \le & y & \le & 1 \\
 0 & \le & x^2 + y^2 & \le & 1 \\
 0 & \le & z & \le & 1 \\
\end{array}$$

Title: "Spherical Plot" of a 2D function