问:
Projection of 3D line onto 2D plane
Consider the following picture:

There are two planes $\pi, \pi'$, and two associated (violet) euclidean coordinate frames, with $z-$axis orthogonal to the respective planes.
We are given a point $x' \in \pi'$, in coordinates with respect to the right frame of reference. Thus, $x'=(x'_1,x'_2,1)$.

Task: show that the line $l: d \mapsto d \cdot x'$ (in coordinates of the right frame), when projected into $\pi$, is a 2D line

By projection of the point $(x,y,z)$ (in left coordinates) onto $\pi$ I mean $(x/z, y/z)$.

Here is my idea: find a change of coordinates $(R,T)$ from right to left frame, and write $l$ in left coordinates, i.e. $l: d\mapsto Rx'+T$. This is obviously still a line, but when projecting onto $\pi$ I get a non linearity. What am I doing wrong?

答:
Let's examine the 3D line specified by the vector-valued function
$$\mathbf{p}(\lambda) = (1 - \lambda) \mathbf{p}_0 + \lambda \mathbf{p}_1 = \mathbf{p}_0 + \lambda (\mathbf{p}_1 - \mathbf{p}_0)$$
Without loss of generality, we use a coordinate system where the projection plane is perpendicular to the $z$ axis at $z = 1$.  Then, the projection simplifies to
$$\left\lbrace \begin{aligned}
x^\prime &= \frac{x}{z} \\
y^\prime &= \frac{y}{z} \\
z^\prime &= \frac{z}{z} = 1 \\
\end{aligned} \right.$$
The projection of points
$$\mathbf{p}_0 = \left[ \begin{matrix} x_0 \\ y_0 \\ z_0 \end{matrix} \right], \quad \mathbf{p}_1 = \left[ \begin{matrix} x_1 \\ y_1 \\ z_1 \end{matrix} \right]$$
to the projection plane are
$$\mathbf{q}_0 = \left[ \begin{matrix} \frac{x_0}{z_0} \\ \frac{y_0}{z_0} \\ 1 \end{matrix} \right], \quad \mathbf{q}_1 = \left[ \begin{matrix} \frac{x_1}{z_1} \\ \frac{y_1}{z_1} \\ 1 \end{matrix} \right]$$
and the projection of $\mathbf{p}(\lambda_q)$, the vector-valued function specifying the line, is
$$\mathbf{q}(\lambda_q) = \left[\begin{matrix} \displaystyle\frac{x_0 + \lambda_q (x_1 - x_0)}{z_0 + \lambda_q (z_1 - z_0)} \\ \displaystyle\frac{y_0 + \lambda_q (y_1 - y_0)}{z_0 + \lambda_q (z_1 - z_0)} \\ 1 \end{matrix}\right]$$
whereas the vector-valued function specifying the line between the two projected points is
$$\mathbf{d}(\lambda_d) = \left[\begin{matrix} \displaystyle\frac{x_0}{z_0} \\ \displaystyle\frac{y_0}{z_0} \\ 1 \end{matrix}\right] + \lambda_d \left[\begin{matrix}\displaystyle\frac{x_1}{z_1} - \frac{x_0}{z_0} \\ \displaystyle\frac{y_1}{z_1} - \frac{y_0}{z_0} \\ 0 \end{matrix}\right] = \left[\begin{matrix} \displaystyle\frac{x_0 z_1 + \lambda_d (x_1 z_0 - x_0 z_1)}{z_0 z_1} \\ \displaystyle\frac{y_0 z_1 + \lambda_d (y_1 z_0 - y_0 z_1)}{z_0 z_1} \\ 1 \end{matrix}\right]$$
The two are collinear if their vector cross product is the zero vector:
$$\begin{aligned}
\left( \mathbf{q}(\lambda_q) - \mathbf{q}(0) \right) \times \left( \mathbf{d}(\lambda_d) - \mathbf{d}(0) \right) &= \left[ \begin{matrix} \displaystyle \frac{x_0 + \lambda_q (x_1 - x_0)}{z_0 + \lambda_q (z_1 - z_0)} - \frac{x_0}{z_0} \\ \displaystyle \frac{y_0 + \lambda_q (y_1 - y_0)}{z_0 + \lambda_q (z_1 - z_0)} - \frac{y_0}{z_0} \\ 0 \end{matrix} \right] \times \left[\begin{matrix} \displaystyle \lambda_d \left( \frac{x_1}{z_1} - \frac{x_0}{z_0} \right) \\ \displaystyle \lambda_d \left( \frac{y_1}{z_1} - \frac{y_0}{z_0} \right) \\ 0 \end{matrix}\right] \\
~ &= \left[ \begin{matrix} 0 \\ 0 \\ \displaystyle\frac{\lambda_q ( x_1 y_0 - x_0 y_1 )}{ \lambda_q z_0 z_1 + (1 - \lambda_q) z_0^2 } \end{matrix} \right] \\
~ &= \left[ \begin{matrix} 0 \\ 0 \\ \displaystyle\frac{\lambda_q ( x_1 y_0 - x_0 y_1 )}{ z_0^2 + \lambda_q z_0 ( z_1 - z_0 )} \\ \end{matrix} \right] \\
\end{aligned}$$
i.e., if and only if $x_1 y_0 = x_0 y_1$.

This means that if in the coordinate system where the projection plane is perpendicular to the $z$ axis,

the line intersects with the $z$ axis (in which case there is a parametrisation for the line with $x_0 = y_0 = 0$ or $x_1 = y_1 = 0$),

or

the line is in the $x z$ plane (in which case $y_0 = y_1 = 0$),

or

the line is in the $y z$ plane (in which case $x_0 = x_1 = 0$),

the projected line is also a straight line.  (Technically, the first case also includes the other two cases: the projected line is a straight line if and only if the 3D line intersects the $z$ axis, when the projection plane is perpendicular to the $z$ axis.)

Title: Projection of 3D line onto 2D plane


问:
comparisons of side lengths on a triangle
I am trying to find the relationship between the sides of any triangle as part of a bigger problem.

I was thinking about altering the Pythagorean theorem, but all I got was the useless $a^2+b^2\lesseqgtr c^2$ which is basically completely useless.

What I am after is something like $a^2+b^2\leq c^2$, which will actually mean something!!

答:
The basic triangle inequality states that
$$a+b\ge c$$
if $c$ is the biggest side. Triangle inequalities that are related to Pythagoras theorem are
$$a^2+b^2 < c^2, \quad\text{if angle $C$ is obtuse (greater than 90°)},$$
$$a^2+b^2 > c^2, \quad\text{if angle $C$ is acute (less than 90°)}.$$
In general we may write
$$a^2+b^2>\frac{c^2}{2}.$$
For more triangle inequalities, see here.

ContentLicense: CC BY-SA 4.0


问:
Describing/sketching region of integration of triple integral
I'm having great difficulty with the following problem:

This question concerns the integral $\int_{0}^{2}\int_{0}^{\sqrt{4-y^2}}\int_{\sqrt{x^2+y^2}}^{\sqrt{8-x^2-y^2}}\!z\ \mathrm{d}z\ \mathrm{d}x\ \mathrm{d}y$. Sketch or describe in words the domain of integration. Rewrite the integral in both cylindrical and spherical coordinates. Which is easier to evaluate?

Below is what I believe I have established so far...

The projection of this integral's domain onto the $xy$-plane is the portion of the circle $x^2+y^2=4$ on $0\le x\le2,\ y\ge0$.

The bounds on $z$ correspond to

$z^2=x^2+y^2$ (cone) and $x^2+y^2+z^2=8$ (sphere).

These bounds intersect at

$x^2+y^2=4$.

Below $z=2$ (where the bounds on $z$ intersect), I believe that the cone and cylinder, $x^2+y^2=4$, are completely inside the sphere.

Would it hence be correct to say that the region of integration is the solid lying between the cone and the cylinder, on $x\ge0$, $y\ge0$ and $0\le z\le2$? I'm struggling to visualize this problem.

When I attempt to move on, and evaluate the integral in cylindrical/spherical coordinates, my solutions differ by a factor of 2.

That is, I evaluated this integral as,

$\int_{0}^{\frac{\pi}{2}}\int_{0}^{2}\int_{0}^{\sqrt{8-r^2}}\!z\ r\ \mathrm{d}z\ \mathrm{d}r\ \mathrm{d}\theta=2\pi$

And,

$\int_{0}^{\frac{\pi}{2}}\int_{0}^{\frac{\pi}{2}}\int_{0}^{2\sqrt{2}}\!\rho\ \cos\phi\ \rho^2 \sin \phi\ \mathrm{d}\rho\ \mathrm{d}\theta\ \mathrm{d}\phi=4\pi$

Can you please help me to identify where I am going wrong?

Thank you very much.

答:
You are very close.
The integral in cylindrical coordinates has the right value, but the lower limit for $z$ should be $r$, not $0$.
Notice that in spherical coordinates the angle $\phi$ should be integrated from $0$ to $\pi/4$. This restrict us to the volume inside the cone.

Addendum:
The volume is the region between the cone and the sphere in octant I ($x,y,z\ge 0$).
See the figure below.

In cylindrical coordinates we have
$$\int_{0}^{\frac{\pi}{2}}\int_{0}^{2}\int_{r}^{\sqrt{8-r^2}}\!z\ r\ \mathrm{d}z\ \mathrm{d}r\ \mathrm{d}\theta.$$

In spherical coordinates we have
$$\int_{0}^{\frac{\pi}{4}}\int_{0}^{\frac{\pi}{2}}\int_{0}^{2\sqrt{2}}\!\rho\ \cos\phi\ \rho^2 \sin \phi\ \mathrm{d}\rho\ \mathrm{d}\theta\ \mathrm{d}\phi.$$

ContentLicense: CC BY-SA 3.0


问:
How to prove the mathematical induction is true?
I have no idea about the underlying theory from which the mathematical induction was derived.

How to prove the mathematical induction is true?

答:
Well, in principles induction is, having something holding for $0$, denoted as $h(0)$, and if $h(n)$ implies $h(n')$, where $n'$ is a successor of $n$, i.e. $n+1$, we have $h(n)$ for any natural number of $n$.

On the other hand, natural numbers are defined as the smallest set that contains 0 and closed to successor operation (i.e. if $n$ is natural so is $n'$).

The correctness of induction follows strictly from the fact the natural numbers are the the smallest set satisfying the same condition as induction.

ContentLicense: CC BY-SA 2.5


问:
How to show random walk on semi-infinite line is null-recurrent
The figure shows the Markov chain with the transition probabilities. I need to prove that it is null recurrent by showing
$$ \sum_n p_{00}^{(n)} \to \infty$$
Showing the above is enough as I can show it is not positive recurrent as no stationary distribution exists for this reversible chain. However, I am unable to show that the sum diverges. For a random walk on the infinite line, we could directly write $p_{00}^{(2n)} = ^{2n}C_n (0.5)^{2n}$ but I am confused about how to proceed in this semi-infinite chain.

答:
If you've shown that the random walk on the infinite line is recurrent, you've shown that when you go from state $1$ to state $2$ in that random walk, you will return to state $1$ with probability $1$.

This means that when you go from state $1$ to state $2$ in the semi-infinite random walk, you will also return to state $1$ with probability $1$, because the states $\{2,3,4,\dots\}$ can't distinguish the infinite and semi-infinite line. (Before returning to $1$, a random walk on the infinite line will never visit $\{0,-1,-2,-3,\dots\}$, so it doesn't matter if those states actually exist.)

Therefore the random walk on the semi-infinite line is also recurrent.


问:
What exactly is sheafification?
I have recently learned about the very BASICS of sheaves, but I was wondering is there an easier definition for sheafification? I could not find anywhere an easier definition for sheafification. I kind of compare it to local rings, where in local rings, you collect the local data at a point for a variety or an algebraic set where some function is defined on that regular set. Am I heading in the right direction? If not, what is the rigorous definition of sheafification and what is some intuition one could use to understand it?

I would much more appreciate the intuition and a clear definition, rather than rigour.

答:
Question: "If not, what is the rigorous definition of sheafification and what is some intuition one could use to understand it?"

Answer: If $X$ is a scheme and $\mathcal{E},\mathcal{F}$ are quasi coherent $\mathcal{O}:=\mathcal{O}_X$-modules, with $\phi: \mathcal{E}\rightarrow \mathcal{F}$ a map of $\mathcal{O}$-modules you want to form kernels, images, tensor products exterior products etc of the sheaves $\mathcal{E},\mathcal{F}$ and the map $\phi$. In the case of the tensor product $\otimes$ you define the presheaf

$$ S(U):= \mathcal{E}(U)\otimes_{\mathcal{O}(U)} \mathcal{F}(U).$$

The presheaf $S$ is not a sheaf, and the associated sheaf $S^+ := \mathcal{E}\otimes \mathcal{F}$ (see Hartshorne, Section II.5) is a sheaf with the property that for any open affine set $V:=Spec(A)$ with $\mathcal{E}(V):=E, \mathcal{F}(V):=F$ with $E,F$ left $A$-modules, it follows
there is an isomorphism of $A$-modules

$$ S^+(V) \cong E\otimes_A F$$

Hence the sheaf $S^+$ has the correct behaviour and generalize the notion of tensor product of modules. If $\mathcal{E}, \mathcal{F}$ are locally trivial, it follows the tensor product, exterior product, symmetric product etc is again locally trivial. You get compatibility formulas valid for modules: If $\mathcal{E}$ is locally trivial it follows

$$ Hom_{\mathcal{O}}(\mathcal{F}\otimes \mathcal{E}, \mathcal{G}) \cong Hom_{\mathcal{O}}( \mathcal{F}, Hom_{\mathcal{O}}(\mathcal{E}, \mathcal{G})).$$

As you can see this is similar to what happens for $A$-modules. The sheafification process seems "unintuitive" but it has all the correct functorial properties.

Example 1. If you want to define a morphism of sheaves

$$f^+: \mathcal{E}\otimes \mathcal{F} \rightarrow \mathcal{G}$$

where $\mathcal{G}$ is a sheaf, by the universal property of the $S^+$-construction, it is enough to define a morphism of presheaves

$$f: S \rightarrow \mathcal{G}.$$

If $f$ exists, by the universal property you get a unique morphism $f^+: S^+ \rightarrow \mathcal{G}$. And to define a morphism $f$ you can do this using basic properties of the tensor product.

Example 2. As an example you should verify that there is an isomorphism of sheaves $\mathcal{O}(m)\otimes \mathcal{O}(n) \cong \mathcal{O}(m+n)$ on $C:=\mathbb{P}^1$. Let $C:=Proj(k[x_0,x_1])$ and let $U_i:=D(x_i)$. Let $t:=\frac{x_1}{x_0}, s:=\frac{1}{t}$. Let $L_m:=\mathcal{O}(m)$ for an integer  $m$. Let $S(U):=L_m(U)\otimes_{\mathcal{O}_C(U)}L_n(U)$. Define

$$ f_0: S(U_0):=k[t]x_0^m\otimes k[t]x_0^n \rightarrow k[t]x_0^{m+n}$$

by

$$f_0(f(t)x_0^m\otimes g(t)x_0^n):= fgx_0^{m+n}.$$

Do a similar construction for $U_1$ to get a map

$$f_1: S(U_1) \rightarrow k[s]x_1^{m+n}.$$

It follows $f_i$ glue to induce a canonical map

$$f: S^+:= \mathcal{O}(m)\otimes \mathcal{O}(n) \rightarrow \mathcal{O}(m+n)$$

and since $f_0,f_1$ are isomorphisms, it follows $f$ is an isomorphism. Try to generalize to projective space.

Title: What exactly is sheafification?


问:
Completeness Axiom of $\mathbb{R}$.
I use the following as the axiom of completeness of the reals $\mathbb{R}$:
$$\forall X,Y\in \mathcal{P}(\mathbb{R})\backslash\{\emptyset\}: (\forall x\in X\quad\forall y\in Y: x\leq y) \implies \exists c\in\mathbb{R}\quad\forall x\in X\quad\forall y\in Y: x\leq c\leq y.$$

From this I deduce that $\forall a,b\in\mathbb{R}:a\leq b\implies \exists c \in\mathbb{R}: a\leq c\leq b$ as this is the special case of the axiom above with $X:=\{a\}$ and $Y:=\{b\}$.

My question now is: Can I use this to prove, that the open Interval $(a,b)$ is not empty if $a<b$? And why is the statement $\forall a,b\in\mathbb{R}:a<b\implies(a,b)\neq\emptyset$ not enough to be equivalent to the Axiom of Completeness of $\mathbb{R}$?

答:
As Henry and Vercassivelaunos point out in the comments, the trick to answering your second question is to consider $\mathbb Q$, which you can show is not complete but is dense. (Terminology: A linear order $<$ is said to be dense if whenever $a<b$ there is some $z$ with $a<z<b$.)

My hint for your first question is to also try to prove that the answer is no. In this case, a counterexample would be a linear order $(A,<)$ which satisfies $\forall a,b\in A: a\le b\implies\exists c\in A:a\le c\le b$ but isn't dense. My advise is to start with your favorite non-dense linear order, and check to see if it satisfies this second property.

Edit: Getting back to your second question, not only are there dense linear orders which are not complete (like $\mathbb Q$), there are also complete linear orders which are not dense. Unlike in the previous paragraph I can't promise that your favorite non-dense linear order is a counterexample (maybe you have gross favorites), but I think you can show e.g. $\mathbb N$ is complete (by your definition) but not dense.

Edit 2: Above I took your questions as asking if the implications held over linear orders, since the question doesn't talk about fields. But it's also possible that you meant to ask in the context of ordered fields (which I think is the usual context when talking about the completeness axiom). In that case the answer to your second question is still no because of $\mathbb Q$, but the answer to your first question is yes. In fact:

Exercise 1: Show that every ordered field has the property that $\forall a,b\in A: a\le b\implies\exists c\in A:a\le c\le b$.

Exercise 2: Show that every ordered field is dense.

The first exercise should be easy if you've already checked that your favorite non-dense linear order satisfies said property. My hint for the second is, for $a<b$, to show that $z=\frac{a+b}{1+1}$ is well-defined (which is not obvious from just the field axioms!) and satisfies $a<z<b$.

Title: Completeness Axiom of $\mathbb{R}$.


问:
Strongest increase of a gradient $f(x,y)$ at the position $x_0$
I am trying to figure out how to interpret the gradient of my scalar field at the position $x_0$.

The gradient of the function $$f: \mathbb{R}^n \supset \ U \longrightarrow \mathbb {R}$$ at the point $x_0$ indicates the direction in which the function value of $f$ increases the most, starting from the point $x_0$.

Considering my function: To determine the gradient at the point $f(x, y) = x^3 - 3xy^2$ we must first calculate the partial derivatives with respect to $x$ and $y$ and write them into a column vector.

The partiel derivate of $x$ at the position $x,y$ is:
$$ \frac{\partial f }{\partial x}(x,y) = 3x^2-3y^2 $$
The partiel derivate of $y$ at the position $x,y$ is:
$$ \frac{\partial f }{\partial y}(x,y) = -6xy $$

Therefore I get: $$\nabla f(x,y) = \begin{pmatrix} \frac{\partial f }{\partial x}\\ \frac{\partial f }{\partial y} \end{pmatrix} = \begin{pmatrix} 3x^2-3y^2 \\ -6xy  \end{pmatrix} $$

Plotting the scalar field I get the following:

The point $(0,0)$ is a critical point and a saddle. What does the gradient at $x_0$ tells me exactly when there is no increase?

答: When the gradient of $f(x,y)$ at $x_0$ is $\nabla f =\vec{0}$ it's telling you that your function is approximately constant near $x_0$ with respect to all directions. In the case of your question it means that there exists a neighbourhood of $x_0$, however close to it, such that the function is similar to an horizontal plane, i.e. there is not a direction with respect to which the function increases the most, starting from $x_0$.


问:
Brownian motion on the circle.
Let $(B_t)$ a Brownian motion on $\mathbb R$. The following puzzled. So, I read for example here that $(e^{iB_t})_{t\geq 0}$ is a Brownian motion on the circle.  But using the stereographic projection, $\left(\frac{2B_t}{1+B_t^2},\frac{B_t^2-1}{1+B_t^2}\right)$ should also be a Brownian motion, right ? But it looks that both seem to be very different in the sens that the first one will reach the north pole in infinite time a.s. (and looks more faithful of the Brownian motion on $\mathbb R$), whereas the other one will reach all point of the circle in finitely many time. So, at the end, I'm not so sure why these two brownian motion are so different, and I'm suspicious if $\left(\frac{2B_t}{1+B_t^2},\frac{B_t^2-1}{1+B_t^2}\right)$ is really a Brownian motion of the circle. Can some one explain a bit better how things behave ?

答:
Let $$W_t:=\left(\frac{2B_t}{1+B_t^2},\frac{B_t^2-1}{1+B_t^2}\right),\quad t\geq 0.$$
Then $(W_t)_{t\geq 0}$ is not a Brownian motion on the circle. Nevertheless, it's a Brownian motion on $\mathbb S^{1}\setminus \{(0,1)\}$.

Title: Brownian motion on the circle.


问:
Eigen-Decomposition of Householder operator
Define the eigen decomposition of householder operator $H=I-2ww^*$ where $\vert\!\vert w\vert\!\vert=1$, geometrically! I know that the eigen-values are $[-1, 1 \dots 1]$ but I cannot understand about the eigen-vectors and in particular I cannot depict the solution, geometrically!

Can anybody possibly help me to understand it through drawing a geometrical interpretation?

Thanks alot

答:
Here, $Hv=(I-2ww^*)v=v-2w(w^*v)$

Notice that $w^*v$ is the length of the perpendicular from the $v$ onto the orthogonal complement of $w$.

In the above image, $w=\left(\frac1{\sqrt2}, \frac1{\sqrt2}\right)$, it's orthogonal complement is the line $y=-x$, and $w^*v$ is the length of the line $VP$.

As can be seen in the image, $w(w^*v)$ is the vector $\overrightarrow{PV}$. Subtracting it twice from $v$ gives its image over the line $y=-x$. Thus, $H$ is a reflection map around the orthogonal complement of $w$.

Clearly, all the vectors lying perpendicular to $w$ remain as they are (and correspond to eigenvalue $1$), while the ones parallel to $w$ get inverted (and correspond to eigenvalue $-1$).


问:
How can we show that the set of all polynomials with integer coefficients denumerable?
Is the set of all polynomials with integer coefficients, denumerable?

I know that this set is countable but how can I show that it is denumerable as well.

答:
For completeness sake, I will write down two proofs of the fact that the set $\mathbb{Z}[x]$ of polynomials on integer coefficients is denumerable, meaning countable and infinite. As David Mitra points out in a comment, it is certainly of cardinality at least that of $\mathbb{Z}$, since one can easily present an injective function $f:\mathbb{Z}\rightarrow\mathbb{Z}[x]$: namely,
$$f(n)=n, \quad\text{for all}\quad n\in\mathbb{Z}.$$
One could then consider the set $\mathbb{Z}[x]_{m}$ of polynomials of degree at most $m$, which is easily seem to be equinumerous to $\mathbb{Z}^{m}$, which in turn has the same cardinality as that of $\mathbb{Z}$; then, by noticing that $\mathbb{Z}[x]=\bigcup_{m\in\mathbb{N}}\mathbb{Z}[x]_{m}$, one obtains $\mathbb{Z}[x]$ has at most as many elements as a denumrable union of denumerable sets, which is easily seem to be denumerable as well.

This proof is rather standard, but I like very much an alternative one: consider the map
$$g:\mathbb{Z}[x]\rightarrow\mathbb{Q}\setminus\{0\}$$
such that, for a polynomial $p(x)=m_{0}+m_{1}x+m_{2}x^{2}+\cdots+m_{k}x^{k}$, one has
$$g(p(x))=p_{0}^{m_{0}}\times p_{1}^{m_{1}}\times\cdots\times p_{k}^{m_{k}},$$
where $p_{i}$ is the $i$-th prime number. One can prove, without much work, that $g$ is, in fact, bijective, and since $\mathbb{Q}$ is denumerable, the desired result holds.


问:
Proof of maximum principle theorem using mean value theorem
Let $f:\mathbb{R}^2 \rightarrow \mathbb{R}$ be a function twice differentiable such that $\Delta f = 0$ on an open disc $D$ of center $0$ and radius $\rho > 0$.

We are given the mean value formula:

$$f(0)=\frac{1}{\pi \rho^2}\iint_{D(0, \rho)}f(x,y)dxdy$$

Let $U$ be a connected and bounded set of $\mathbb{R}^2$ and suppose that $\Delta f=0$ on U. Prove that if $f_{\lvert \overline{U}}$ reaches its maximum on $U$ then $f$ is constant on $U$.

I think I have the structure of the proof but fail to formalize it.

My attempt to prove it:

Let $m$ be the maximum of the function.

Since $U$ is connected, $\forall x \in U$ there exists a path $\phi:[0,1]\rightarrow U$ such that $\phi(0)=x$ and $\phi(1)=m$.

Let's take a finite number of points in $U$, $x_1, x_2, ..., x_n$ along the path and let $\epsilon_1, ..., \epsilon_n$ be the radiuses of the open discs $D(x_i, \epsilon_i)$ such that the discs overlap.

Using the mean value formula for $m$, it follows that:

$$f(m)=\frac{1}{\pi \rho^2}\iint_{D(m, \epsilon_{max})}f(x,y)dxdy$$

where $\epsilon_{max}$ is the radius of the disc $D(m, \epsilon_{max})$ such that it overlaps the disc $D(x_n, \epsilon_n)$.

Hence it follows that $f(m)=f(x_n)=...=f(x_1)$.

Hence by generalizing this reasoning to every $x$ in $U$, it follows that $f$ is constant on $U$.

答:
What you've missed is using the fact that $m$ is an interior maximum.

The intuition for the argument begins by considering a local form of the mean value property, meaning $f(x_0)=\frac{1}{\pi r^2} \iint_{B_r(x_0)} f(x) dx$ for some particular $x_0$ and some particular $r>0$. Say you have this property  and $f(x_0)>f(x)$ for some $x$ in that disk, then you must also have $f(x_0)<f(y)$ for some $y$ in that disk to compensate for the contribution to integration from a vicinity of $x$ being too small.

More formally, we know there exists a disk $B_r(m)$ on which $f(m) \geq f(x)$. We want to show $f(m)=f(x)$ for all $x \in B_r(m)$. So suppose by way of contradiction that $f(m)>f(x_0)$ for some $x_0 \in B_r(m)$. Then by continuity there exists a whole neighborhood of $x_0$ such that for all $x$ on that neighborhood, $f(m)>f(x)$. Since $f(m) \geq f(x)$ on all of $B_r(m)$, you conclude that $f(m)> \frac{1}{\pi r^2} \iint_{B_r(m)} f(x) dx$ which contradicts the mean value property. So you conclude that no such $x_0$ exists i.e. $f(m)=f(x)$ for all $x \in B_r(m)$.

You can now propagate this around the domain "manually" by using the idea that you said: pick a point in $B_r(m)$, draw a new disk around that point, and repeat, until you arrive at any given point $x_0$. Alternatively you can use topology, since what we just said implies that $\{ x : f(x)=M \}$ is open and it is automatically closed by continuity, and a nonempty clopen subset of a connected space is the whole space.


问:
Mixing time of a lifted Markov Chain.
Let $(X_t)$ be a Markov chain  and $(Y_t)$ a lifted chain of $(X_t)$, that is, the lifted chain splits each state in several states.

$t_O$ denotes the mixing time of the original chain;

$t_L$ denotes the mixing time of the lifted chain;

How can we related the mixing times of both chains? Intuitively it makes sense that by considering a larger state space, the lifted chain will mix faster, but I don't find a result saying that that is always true.

Is it always the case that $t_L \leq t_O$?

Edit: In fact what is always true is that
$$t_O \leq t_L.$$

答:
If you go by the total variation distance from a stationary distribution, as defined for instance on Wikipedia, then the lifted chain will never mix faster.

That's because for every subset $A$ of the state space of $(X_t)$, we have a corresponding subset $B$ of the state space of $(Y_t)$ consisting of all sub-states of every state in $A$. The events $\Pr[X_t \in A]$ and $\Pr[Y_t \in B]$ coincide, and so do the stationary measures $\pi(A)$ and $\pi(B)$, so we have $$|\Pr[X_t \in A] - \pi(A)| < \frac14 \iff |\Pr[Y_t \in B] - \pi(B)| < \frac14.$$ The condition is satisfied for all subsets $A$ at the same time it's satisfied for all "lifted subsets" $B$ of this form.

In particular, at time $t_L$, the second inequality holds for all $B$, in particular for all "lifted subsets" $B$, so the first inequality holds for all $A$; we conclude that $t_O \le t_L$. However, the reverse inequality does not hold: if the first inequality holds for all $A$, we can only conclude the second inequality for sets $B$ of the right form.

For an example where the lifted chain has a much worse mixing time, imagine the following two cases:

The lifted chain has two copies of the original chain, with no transitions between them. In that case, $(Y_t)$ will never mix, because it can never leave the copy it started in to go to the other copy.

The lifted chain has two copies of the original chain, with one very rare transition between them in each direction. In that case, $(Y_t)$'s mixing time will be dominated by the time it takes to see that transition sufficiently many times, rather than the mixing time within each copy.

Title: Mixing time of a lifted Markov Chain.


问:
Basic doubt about exponents.
I have a doubt regarding this:

$\begin{aligned}  \ln y &=\ln \left(x^{x^{x}}\right) \\ &=x \ln \left(x^{x}\right) \\ &=x \cdot x \cdot \ln x \\ &=x^{2} \cdot \ln x \\ &=\ln \left(x^{x^{2}}\right) \neq \ln \left(x^{x^{x}}\right) \end{aligned}$

I KNOW that I did it wrong ( in step 3) but can any body explain me what exactly I did wrong and point out the exact fundamental  understanding that I lack.
Please try to be extremely specific while explaining the error.

If u want more info on the question, feel free to comment.

答:
Your error was due to incorrect association of the exponents:

$$ x^{x^x} = x^\left(x^x\right) \neq \left(x^x\right)^x. $$

As you know,

$$ x \ln(u) = \ln\left(u^x\right). $$

Observe that

$$ x \ln \left(x^x\right) =   \ln \left(\left(x^x\right)^x\right)
 \neq \ln \left(x^{\left(x^x\right)}\right) = \ln\left(x^{x^x}\right). $$

Also observe that

$$ \left(x^x\right)^x = x^\left(x^2\right). $$

And note that you are inconsistent in the way you associate exponentiation, because you treated $x^{x^x}$ as $\left(x^x\right)^x$
but you treated $x^{x^2}$ as $x^\left(x^2\right).$

Title: Basic doubt about exponents.


答:
The problem is that
$$\log(x^{x^x})=x^x\log x\neq x^2\log x.$$
The logarithmic law you wanted to use is
$$\log(a^b)=b\log a.$$
You set $a=x^x$ and $b=x$. The problem is that $a^b\neq x^{x^x}$, $a^b=x^{x^2},$ because $[(a^b)]^c=a^{bc}$. The correct way is to set $a=x$ and $b=x^x$.

Title: Basic doubt about exponents.

Body:
I have a doubt regarding this:

$\begin{aligned}  \ln y &=\ln \left(x^{x^{x}}\right) \\ &=x \ln \left(x^{x}\right) \\ &=x \cdot x \cdot \ln x \\ &=x^{2} \cdot \ln x \\ &=\ln \left(x^{x^{2}}\right) \neq \ln \left(x^{x^{x}}\right) \end{aligned}$

I KNOW that I did it wrong ( in step 3) but can any body explain me what exactly I did wrong and point out the exact fundamental  understanding that I lack.
Please try to be extremely specific while explaining the error.

If u want more info on the question, feel free to comment.


问:
Volume of truncated prism
I needed to find the volume of what Wikipedia calls a  truncated prism, which is a prism (with triangle base) that is intersected with a halfspace such that the boundary of the halfspace intersects the three vertical edges of the prism at heights $h_1, h_2, h_3$.

I was able to find the formula
$$
V=A\frac{h_1+h_2+h_3}{3},
$$
(where $A$ is the area of the triangle base) online, but without proof.

I was also able to prove this formula myself, but with a really nasty proof. (I integrated the area of the horizontal cross-sections; after passing the first intersection with the hyperplane at height $h_1$ these cross-sections have the form of the base triangle minus a quadratically increasing triangle, then after crossing the first intersection at height $h_2$ they have the form of a quadratically shrinking triangle)

Do you know of an elegant proof of the volume formula?

PS: Wikipedia cites 'William F. Kern, James R Bland,Solid Mensuration with proofs, 1938, p.81' for the name truncated prism, but I cannot find this book.

答: The book (Solid mensuration, by Willis F. Kern and James R. Bland. ) that you are looking for can be read on the site https://babel.hathitrust.org/cgi/pt?id=mdp.39015064340733</p>

Title: Volume of truncated prism


问:
Adjoint system associated to a linear system of PDEs
I am reading about adjoint systems of linear partial differential equations and appears the following example of a system over the domain $(a,b)\times (0,T)$, where $T>0$ and $a,\, b\in \mathbb{R}$:

$$
f_t + a_1 g_x =0,
$$
$$g_t - a_2 g_{xx} + a_3 f_x + a_4 h_x =0,  $$
$$
h_t - a_5 h_{xx}+ a_6 g_x =0,
$$
where each $a_i$ is a real number. The adjoint system associated to the system above reads as

$$ - u_t - a_1 v_x =0, $$
$$ - v_t - a_2 v_{xx} - a_3 u_x - a_4 w_x =0, $$
$$ - w_t - a_5 w_{xx} - a_6 v_x =0.$$

How could I obtain this adjoint system ? What is the procedure to get it ?

答:
You can write your original system as $A \vec{f} = \vec{0}$ where $\vec{f} = (f,g,h)$ and

\begin{align}
A &= \begin{pmatrix}
\partial_{t} & a_{1} \partial_{x} & 0 \\
a_{3} \partial_{x} & \partial_{t} - a_{2} \partial_{xx} & a_{4} \partial_{x} \\
0 & a_{6} \partial_{x} & \partial_{t} - a_{5} \partial_{xx}
\end{pmatrix}
\end{align}

is a matrix of disembodied derivative operators. Recall that $M$ is adjoint to $N$ if $\langle M x, y \rangle = \langle x, N y \rangle$ for a suitable notion of $\langle \cdot, \cdot \rangle$. In this case, we want to use integration by parts and so we let $\langle \cdot, \cdot \rangle = L^{1}$, though we will also consider $\langle \cdot, \cdot \rangle$ to be a projection (dot product) so we can solve the entire system simultaneously.

Computing $\langle A \vec{f}, \vec{u} \rangle$ where $\vec{u} = (u, v, w)$, $\Omega = (a, b) \times (0, T)$ and IBP stands for integration by parts

\begin{align}
0 &= \langle A \vec{f}, \vec{u} \rangle \quad \text{(as $A \vec{f} = \vec{0}$)} \\\\
&= \int_{\Omega} (A \vec{f}) \cdot \vec{u} \ dx dt \\\\
&=\int_{\Omega} \left[ (f_{t} + a_{1} g_{x}) u + (a_{3} f_{x} + g_{t} - a_{2} g_{xx} + a_{4} h_{x}) v + (a_{6} g_{x} + h_{t} - a_{5} h_{xx}) w \right] \ dx dt \\\\
&\underbrace{=}_{\text{IBP}} \text{(terms that need to be evaluated)} \\
&- \int_{\Omega} \left[ f u_{t} + g (a_{1} u_{x}) + f (a_{3} v_{x}) + g (v_{t} + a_{2} v_{xx}) + h (a_{4} v_{x}) + g (a_{6} w_{x}) + h (w_{t} + a_{5} w_{xx}) \right] dx dt \\\\
&= \text{(terms that need to be evaluated)} \\
&- \int_{\Omega} \left[ f (u_{t} + a_{3} v_{x}) + g (a_{1} u_{x} + v_{t} + a_{2} v_{xx} + a_{6} w_{x}) + h (a_{4} v_{x} + w_{t} + a_{5} w_{xx}) \right] dx dt \\\\
&= \text{(terms that need to be evaluated)} + \int_{\Omega} \vec{f} \cdot (B \vec{u}) \ dx dt \\\\
&= \text{(terms that need to be evaluated)} + \langle \vec{f}, B \vec{u} \rangle
\end{align}

where

\begin{equation}
B = \begin{pmatrix}
- \partial_{t} & - a_{3} \partial_{x} & 0 \\
-a_{1} \partial_{x} & - \partial_{t} - a_{2} \partial_{xx} & -a_{6} \partial_{x} \\
0 & -a_{4} \partial_{x} & - \partial_{t} - a_{5} \partial_{xx}
\end{pmatrix}
\end{equation}

which is almost the result you have written down, except the coefficients $a_{1}$ and $a_{3}$ are swapped, as are the $a_{4}$ and $a_{6}$ coefficients, which is a typo on your end or the papers. You also need to specify boundary and initial conditions on the functions to determine what happens to the evaluated terms.


问:
How to show that the space is connected but not locally-connected.
Given that $I = [0,1]$, the following is a subspace of $R^2$:

$$X = (\{1\} × I) ∪ \left(I × \left(\{0\} \cup\{ \frac{1}{n} \mid n \in \Bbb N \}\right)\right)$$

How can one show that it is connected but not locally-connected?

答:
It is connected, because it is path connected. Given any point $(x,y)\in X$ there is a straight linear path connecting $(x,y)$ to $(1,y)$. Then there's a straight linear path connecting $(1,y)$ to $(1,1)$. Meaning every point can be connected to $(1,1)$, and so $X$ is path connected.

Not being locally connected is a bit more difficult. Consider $v=(0,0)$ and let $U\subseteq X$ be a small enough neighbourhood of $v$, say $U=\big((-\infty,1/2)\times \mathbb{R}\big)\cap X$. The reason I choose such $U$ is because I don't want $\{1\}\times I$ to intersect with it.

Now let $V\subseteq U$ be any subneighbourhood of $U$. Consider the projection $\pi:V\to\mathbb{R}$, $\pi(x,y)=y$. Since $\{1\}\times I$ does not intersect $U$, then it does not intersect $V$ as well. And thus the image $im(\pi)$ of $\pi$ is some infinite subset of $A=\{0\}\cup\{\frac{1}{n}\}_{n=1}^\infty$. But in $A$ every point except for $0$ is clopen. Let $r\in im(\pi)$ be some non-zero point. Then $\pi^{-1}(r)$ is clopen as well, meaning $V$ is not connected.


问:
Inverse trigonometric substitution for integrals
I am trying to solve this integral by inverse trigonometric substitution.
$$
\int_{0}^1{\sqrt{4kx-k^2x^2}dx}
$$ where k is an arbitrary constant.
I completed the square to get this:$$
\int_{0}^1\sqrt{2^2-(kx-2)^2}dx
$$
Then, I substituted $kx-2=2\sin \theta$ into the integral to obtain:
$$
\int_{-\pi/2}^{\sin^{-1}(\frac{k-2}{2})}2\cos{\theta}\frac{2}{k}\cos\theta d\theta \\
=\frac{2}{k}\int_{-\pi/2}^{\sin^{-1}(\frac{k-2}{2})}2\cos^2{\theta}+1-1d\theta \\
=\frac{1}{k}\int_{-\pi/2}^{\sin^{-1}(\frac{k-2}{2})}2\cos(2\theta)+2d\theta \\
=\frac{1}{k}(\sin(2\theta)+2\theta)|_{-\pi/2}^{\sin^{-1}(\frac{k-2}{2})}
$$
However, when I tried to check by substituting $k=2$, I get$$
\int_{0}^1{\sqrt{4\cdot2\cdot x-2^2x^2}dx}=1.333333754
$$ but my answer returns$$
\frac{1}{k}(\sin(2\theta)+2\theta)|_{-\pi/2}^{-\sin^{-1}(\frac{k-2}{2})}=1.570796327$$
I am not sure where I have gone wrong with my substitution. I thought I merely left out a 2 multiple but that was not the case. Thank you very much for your help.

答:
You want to substitute $kx-2=2\sin\theta$. For the lower boundary, $x=0$, we have
$$-2=2\sin\theta\Rightarrow\theta=-\frac{\pi}{2}.$$
And for $x=1$, we have
$$k-2=2\sin\theta\Rightarrow\theta=\arcsin\left(\frac{k}{2}-1\right).$$


问:
Finding the SDE
Let $X_t = e^{(\mu-\frac{\sigma^2}{2})t + \sigma B_t}$, where $B_t$ is a standard Brownian motion. How do I find the SDE satisfied by $X_t^{-1}?$ I know I must use Ito's formula but not sure how to apply that. Any help would be appreciated.

答:
First, let $Y_t := \ln(X_t^{-1}) = -(\mu-\frac{\sigma^2}{2})t - \sigma B_t$ (which is well defined). We see that $Y_t$ is a solution to the SDE
$$ dY_t = -(\mu-\frac{\sigma^2}{2})dt - \sigma dB_t \tag1 $$
Which means that $Y_t$ is a regular drift-diffusion process, so applying Itô's lemma to $g(t,Y_t) = e^{Y_t} = f(Y_t)$ gives
$$df(Y_t) = f'(Y_t)dY_t + \frac{1}2f''(Y_t)d\langle Y\rangle_t \tag2$$
And the quadratic variation of $Y$ is given by $d\langle Y\rangle_t =(dY_t)^2 =  \sigma^2dt$, so replacing in (2), we get
$$\begin{align} d(X^{-1}_t) = df(Y_t) &= f'(Y_t)dY_t + \frac{1}2f''(Y_t)\sigma^2dt \\
&= f(Y_t)\cdot(dY_t + \frac{\sigma^2}2dt) \\
&= f(Y_t)\cdot\left(-(\mu-\frac{\sigma^2}{2})dt - \sigma dB_t + \frac{\sigma^2}2dt\right) \\
&= f(Y_t)\cdot\left(-(\mu-\sigma^2)dt - \sigma dB_t\right)\\
&= -X_t^{-1}\cdot \left(\sigma dB_t + (\mu-\sigma^2)dt\right)\end{align} $$

Finally, $X^{-1}_t$ is solution to the SDE
$$d(X^{-1}_t) = X_t^{-1}\cdot \left(-\sigma dB_t + (\sigma^2 - \mu)dt\right)\qquad \square$$

Another way to prove it could have been to show that $X_t$ is an Itô diffusion process and applying Itô's lemma to $X_t^{-1} := {1 \over X_t} = f(X_t)$ directly (I haven't tried).

Title: Finding the SDE


问:
Is $a_n = \frac{c^n}{n!}$ decreasing?
For any $n \in \mathbb{N}$, $a_n = \frac{c^n}{n!}$, where $c \ge 0$. A teacher at my class said the sequence {$a_n$} is monotonic, so I suppose it is decreasing, but I am unable to prove it. For any $2 < c < \infty$, {$a_n$} first increases until reaching a point $N \in \mathbb{N}$, $n \ge N$, then {$a_n$} starts to decrease. For example, {$\frac{3^n}{n!}$} $=$ {$3, 4.5, 4.5, 3.375, 2.025, ...$} so it decreases from $n \ge 4$ onwards. If a sequence does not decrease from the very beginning, it is not a decreasing sequence. It seems to me the only choice of $c$ that makes the sequence truly decreasing is $0 \le c \le 2$. So is {$a_n$} really decreasing?

Any help would be greatly appreciated!

答: That sequence is eventually decreasing, which means that, for some $p\in\Bbb N$, the sequence $(a_n)_{n\geqslant p}$ is decreasing. That follows from the fact that $\frac{a_{n+1}}{a_n}<1$ if $n$ is large enough.

Title: Is $a_n = \frac{c^n}{n!}$ decreasing?


问:
Stating that a function is injective using quantifiers
Using quantifiers, we have to state that a function f is injective meaning that for a unique element in the domain, there will be a unique image associated with that specific element and no two pre images can share a same image. So my answer for this would be:

$$\forall a\forall b(f(a)=f(b)\leftrightarrow a=b)$$

This answer seems correct to me as $f(a)$ can only be equal to $f(b)$ when a=b.

But the answer given is:
$$\forall a\forall b(f(a)=f(b)\rightarrow a=b)$$

I don't understand this answer.For all $a$ and for all $b$ if $f(a)=f(b)$ then $a=b$ , i get that part but in the case $f(a)\not=f(b)$ then $a=b$ or $a\not=b$, but we need to ensure that $a\not=b$ and in this statement both options are possible. Please explain.

答:
The formula
$$\forall a\forall b\Big(f(a)=f(b)\rightarrow a=b\Big)$$
states that the function $f$ is injective: meaning that, for any two elements of its domain, if $f(a)=f(b)$, then you actually have that the elements you started with are one and the same, $a=b$.

But your formula is not actually wrong! It just has a redundant part, that is,
$$\forall a \forall b\Big(f(a)=f(b)\leftarrow a=b\Big):$$
what is a function? A function $f:X\rightarrow Y$ is a relation on $X\times Y$ (that is, a subset of $X\times Y$) such that, if both $(a,y_{1})$ and $(a, y_{2})$ are in this relation, then $y_{1}=y_{2}$. But we usually write, to mean that $(x,y)$ is in a function $f$, $f(x)=y$, so this means that a function is a relation satisfying, for any $a$ and $b$ in its domain, that
$$a=b\quad\text{implies}\quad f(a)=f(b),$$
what is exactly the redundant part of your formula. The extra implication you wrote down is already built in in the definition of function, and is therefore not necessary, although not wrong as well.


问:
Weak Stationary Time Series
If $\{X_t : t \in Z\}$ is a weakly stationary time series does that necessarily mean that  $\{X^2_t : t \in Z\}$ weakly stationary time series as well? I can't really think of an example to prove or disprove this statement.

答: No, it does not. Weak stationarity requires finite variance. So let $X_t$ have finite variance, but not finite 3th moment. Then $X_t^2$ does not have finite second moment,i.e. variance does not exist, therefore can not be weakly stationary.

Title: Weak Stationary Time Series


答:
We have $ \frac{a_{n+1}}{a_n}= \frac{c}{n+1}.$

Hence

$$a_{n+1 } \ge a_n \iff n \ge c-1.$$

Title: Is $a_n = \frac{c^n}{n!}$ decreasing?

Body:
For any $n \in \mathbb{N}$, $a_n = \frac{c^n}{n!}$, where $c \ge 0$. A teacher at my class said the sequence {$a_n$} is monotonic, so I suppose it is decreasing, but I am unable to prove it. For any $2 < c < \infty$, {$a_n$} first increases until reaching a point $N \in \mathbb{N}$, $n \ge N$, then {$a_n$} starts to decrease. For example, {$\frac{3^n}{n!}$} $=$ {$3, 4.5, 4.5, 3.375, 2.025, ...$} so it decreases from $n \ge 4$ onwards. If a sequence does not decrease from the very beginning, it is not a decreasing sequence. It seems to me the only choice of $c$ that makes the sequence truly decreasing is $0 \le c \le 2$. So is {$a_n$} really decreasing?

Any help would be greatly appreciated!


问:
Injectivity of $A-\lambda I$
I'm reading a paper on determinants and on one point the author states that:

A complex number $\lambda$ is called an eigenvalue of matrix $A$ if $A-\lambda I$ is not injective.

Why is this? Could someone clarify :)

Thank you! =)

答:
Let's view $A$ as a linear operator from vector spaces $V \to W$.  An eigenvalue $\lambda$ and eigenvector $x$ satisfy

$$Ax = \lambda x$$

So equivalently:

$$( A - \lambda I ) x = 0$$

But $( A - \lambda I ) 0 = 0$ and $0 \ne x$.  This shows that $A - \lambda I$ is not injective.

Title: Injectivity of $A-\lambda I$


问:
Let $f_n$ be a sequence of M-Lipschitz functions. Show that $f$ is M-Lipschitz
Let $f_n$ be a sequence of M-Lipschitz functions on interval $[a,b]$. Suppose $f_n\to f$ pointwise as $n\to\infty$ on $[a,b]$. Show that $f$ is M-Lipschitz too.

I tried everything to prove the statement, starting from the definitions and trying to create a subdivision, but didn't succeed. I don't really see how pointwise convergence could allow us to prove that $f$ is M-Lipschitz, as $\epsilon$ depends on $x$ (so i tried to pass by a subdivision to get the max of $N$). But, it seems to me too difficult as solution for this such of statement. If someone could explain in details (including graphics or examples, please, if possible), I would really appreciate it. Thank you in advance for help!

答:
For $x,y\in [a,b]$ we have:
\begin{align*}
|f(x)-f(y)|&=\left|\lim_{n\to\infty} f_n(x)-\lim_{n\to\infty}f_n(y)\right|\\
&=\lim_{n\to\infty}|f_n(x)-f_n(y)|\\
&\leq \lim_{n\to\infty}M|x-y|\\
&=M|x-y|
\end{align*}
Therefore $f$ is $M$-Lipschitz.


问:
How to compute the multivariable limit of a multivariable function as the variables approach infinity.

Consider a particle in a three-dimensional potential
$$V(x,y,z)=\frac{A \left(x^3+2 y^3+3 z^3+4 a^3\right)}{\left(x^2+y^2+z^2+a^2\right)^2}$$
This particle has scattering states if $E>E_0$, where

$E_0=$

How does one go about calculating the limit as $x,y,z$ approach infinity for a multivariable function? Take my photo for example.

答:
Treat the variables as an $\mathbb{R}^n$ vector, and then extend that vector towards infinity:
$$\lim_{x,y,z \to \infty} f(x,y,z) = \lim_{r \to \infty} f(C_x r, C_y r, C_z r), \quad 0 \ne C_x, C_y, C_z \in \mathbb{R}$$

In other words,
$$\begin{aligned}
V(x, y, z) & = \frac{ A ( x^3 + 2 y^3 + 3 z^3 + 4 a^3 ) }{( x^2 + y^2 + z^2 + a^2 )^2 } \\
\lim_{x,y,z \to \infty} V(x,y,z) & = \lim_{r \to \infty} V(X r, Y r, Z r) \\
~ & = \lim_{r \to \infty} \frac{ A r^3 ( X^3 + 2 Y^3 + 3 Z^3 ) + 4 A a^3 ) }{( (X^2 + Y^2 + Z^2) r^2 + a^2 )^2 } = 0 \\
\end{aligned}$$

When you have a multivariate function, and wish to find the limit when two or more variables approach infinity, you should consider those $N$ variables as an $\mathbb{R}^N$ point, and all other variables constants.

Treat the direction of that point (with respect to origin) as a constant, and the distance from origin the only variable.

In the original question, we have three variables $x$, $y$, and $z$, that all approach infinity.  We use three real constants, say $X$, $Y$, and $Z$ to describe the direction, and $r$ the single variable, so that $x \to r X$, $y \to r Y$, and $z \to r Z$.  This way, we convert the multivariate limit to a limit in $r$ only.

If the direction constants ($X$, $Y$, and $Z$) do not appear in the resulting limit, we know the limit is not dependent on the direction.  If they do, the limit is dependent on the direction, and the limit itself tells us how.

In that sense, the direction constants do not matter, because we expect them to not affect the limit; but we do need to use them in order to be sure.  We never care about their numerical value (other than that they're never all zero, because that is not a valid direction; and we can pretty safely assume none of them are zero if that makes evaluating the limit easier).


问:
Can $a\sin{bx}$ be expressed in matrix form?
(I am trying to determine if given a dataset and this function, can it be solve using the Least Squared Error Method.)
Thanks</p>

答:
yes it can, so lets use your assumption that:
$$y=a\sin(bx)$$
however there will be an associated error for each term, namely:
$$r_i=y_i-y(x_i)$$
$$r_i=y_i-a\sin(bx_i)$$
now you simply define your sum:
$$S=\sum r_i^2=\sum\left[y_i-a\sin(bx_i)\right]^2$$
now you want to optimise for $a,b$ such that $S$ is as small as possible, which would mean:
$$\frac{\partial S}{\partial a}=0$$
$$\frac{\partial S}{\partial b}=0$$
can you take it from here?

$$\frac{\partial S}{\partial a}=-2\sin(bx_i)\left[y_i-a\sin(bx_i)\right]=0$$
$$\therefore y_i\sin(bx_i)-a\sin^2(bx_i)=0$$
$$\frac{\partial S}{\partial b}=-2ax_i\cos(bx_i)\left[y_i-a\sin(bx_i)\right]=0$$
$$\therefore ax_iy_i\cos(bx_i)-a^2x_i\sin(bx_i)\cos(bx_i)=0$$
unfortunately as you can see these two simultaneous equations are hard to solve for $a,b$. Also if it wasn't clear all of the above expressions are summations but I removed the symbols as it can make it look very messy


问:
Prove $\frac1{\sin A}+\frac1{\sin B}+\frac1{\sin C}-\frac12(\tan\frac{A}{2}+\tan\frac{B}{2}+\tan\frac{C}{2}) \ge \sqrt{3}$
Let $ABC$ be a triangle. Prove that: $$\frac{1}{\sin A}+\frac{1}{\sin B}+\frac{1}{\sin C}-\frac{1}{2}\left(\tan\frac{A}{2}+\tan\frac{B}{2}+\tan\frac{C}{2}\right) \ge \sqrt{3} $$
My attempt:
$$P=\frac{1}{\sin A}+\frac{1}{\sin B}+\frac{1}{\sin C}-\frac{1}{2}\left(\tan\frac{A}{2}+\tan\frac{B}{2}+\tan\frac{C}{2}\right) = \frac{1}{2}\left(\cot\frac{A}{2}+\cot\frac{B}{2}+\cot\frac{C}{2}\right)$$
$$\alpha = \cot\frac{A}{2}+\cot\frac{B}{2}+\cot\frac{C}{2}=\cot\frac{A}{2}\cdot\cot\frac{B}{2}\cdot\cot\frac{C}{2}$$
$$\Leftrightarrow \alpha \ge3.\sqrt[3]{\alpha} \Leftrightarrow \alpha^2\ge27\Leftrightarrow \alpha\ge3\sqrt{3}$$
$$\Rightarrow P=\frac{1}{2}\alpha\ge\frac{3\sqrt{3}}{2}$$
Hmm where I was wrong ?

答:
You are not wrong because
$$
\frac32\sqrt{3}\ge \sqrt3.
$$

You just have proved a stronger inequality than required.


问:
Inifinte coins flipping
Suppose there are infinitely many coins labled $1,2,3...$ and define a R.V. $X_n=1$ when the coin $n$ is a head and $X_n=0$ when the coin n is a tail. Also we assume the for any integer m, it is not possible that $X_i=1$ for all $i\ge m$. Now define another new R.V $Z=\sum_{n\ge 1}\frac{X_n}{2^n}$. So question i) What is $P(Z\le0.625)$ and ii)In general, for all $x\in (0,1)$, what is the probability $Pr(Z\le x)$

I have pondered this question quite a long time, but have no idea how to do it. I found that $0.625=\frac{1}{2}+\frac{1}{8}$ and it seems quite like some terms of $Z$ but still not quite work to solve the question.

答:
The terms are such that if all terms after the $n$-th coin are 1, their total contribution to $Z$ is exactly the contribution of the $n$-th coin. You assume that this event has probability $0$ (which is true for independent coin flips), so their contribution will be strictly less than the contribution of the $n$-th coin.

So for the first question: compute first $P(Z > \frac{1}{2} + \frac{1}{8})$: we need $X_1$ to be 1 (or all other terms sum to less than $\frac{1}{2}$, and either $X_2$ or $X_3$ or both to be 1 too: if they are both $0$ the other terms after $X_3$ sum to less than $\frac{1}{8}$.
So the probability $P(Z > \frac{1}{2} + \frac{1}{8}) = \frac{1}{2}\cdot(1-\frac{1}{2}\cdot\frac{1}{2})$, so the asked for probability is its complement.

Try to generalize for the second one.

Title: Inifinte coins flipping


问:
if $f(x,y)=\frac{xy^2}{x^2+y^6}$ and $f(0,0)=0$ show that $f$ is ubounded at any neighborhood of (0,0)

if $f(x,y)=\frac{xy^2}{x^2+y^6}$ and $f(0,0)=0$ show that $f$ is ubounded at any neighborhood of (0,0) but the restritions to straight line in $\mathbb{R}^2$ is cont.

My attempt:

**Proving continuity **

if $f(x,y)=\frac{xy^2}{x^2+y^6}$  then $y=mx+c$ is equation of any line then we can get $f(x,mx+c)= \frac{x.(mx)^2+c^2.x+2mxc}{x^2+(mx+c)^6}$ then the function is continuous at any point $(x,y)$ which is on the line$y=mx+c$,

We pick any point $(a,ma+c)$ on the line $y=mx+c$,then we try to show that the function is continuous at the point

We know that $f(a,ma+c)= \frac{ma^3+c^2.a+2mac}{a^2+(ma+c)^6}$ .

Now, $\lim_{x \to a}(f(x,mx+c))=\lim_{x \to a}\frac{x^3m^2+c^2x+2mxc}{x^2+(mx+c)^6}=f(a,ma+c)$(by the property of limits).

Proving uboundedness

In this part I did some back calculation . I showed that the function will not be bounded if the numerator becomes infinity or when the denominator becomes infinity.

For the denominator to become $0$ we see that $y^6+x^2=0$ then $(y^2)^3= -x^2$ then $y^2=(-x)^{\frac{2}{3}}$ . So I have shown that if $(x,y) \to (0,0) $ along the given curve $y^2=(-x)^{\frac{2}{3}}$ the limit of the function $f$ doesnt exist.

I am not sure whether this is correct but this has been essentially my attempt.

答:
Equation $y^6+x^2=0$ has only one solution in #\mathbb{R}^2$, and that is $(0,0)$.

To prove that the function is unbounded in the neighbourhood of $(0,0)$ consider (for $y\neq 0$)
$$ f(y^3,y) = \frac{y^3 \cdot y^2}{(y^3)^2+ y^6} = \frac{1}{2y}$$


问:
Calculating the limit $\lim\limits_{\theta \to 0} \frac{\sin \theta}{\tan \theta}$
I'm calculating this limit and would kindly appreciate feedback on my solution

$\lim\limits_{\theta \to 0}\dfrac{\sin \theta}{\tan \theta}$

What I've tried:

given that $\tan \theta = \dfrac{\sin \theta}{\cos \theta}\;,$

then I rearrange the equation like so:

$$\frac{\sin \theta}{\tan \theta} = \frac{\sin \theta \cos \theta }{\sin \theta} = \cos \theta$$

As $\theta$ approaches $0$, then is it true that $\dfrac{\sin \theta}{\tan \theta}=\cos\theta\to1\;?$

答:
Your solution is 100% correct.

An alternative way using the L's hospital rule--

$\lim _{x\rightarrow 0}\left(\frac{\sin x}{\tan x}\right)=\frac{\frac{d}{dx}\left(\sin x\right)}{\frac{d}{dx}\left(\tan x\right)}=\frac{\cos x}{\sec ^2x}=\cos ^3x=1$

Absolutely unnecessary, but since L's hopital rule is taught after direct limits, I thought it might be useful later.... :~)


问:
SAT question - Random Sample/Population
I'm tutoring a student in preparation for the SAT. Below is a question that I think has two good answers - both A and D. However, D is the official answer. If D is indeed the best answer here, would someone please explain why.

In particular, it reads as if option A is a subset of option D, and if the newspaper showed bias (i.e. political bias) then should not A be the answer?

答: They polled from the list of all voters, not from the newspaper readers.  Therefore D is the only correct answer.</p>

Title: SAT question - Random Sample/Population


问:
Calculate the representation matrix of $L$ with respect to the standard bases of $\mathbb{R}^3$ and $\mathbb{R}^2$

The Question

With respect to the bases,

$$a_1 =\begin{pmatrix}
     1 \\
     1 \\
     1 \end{pmatrix},\ a_2 =\begin{pmatrix}
     0 \\
     1 \\
     2 \end{pmatrix},\ a_3 =\begin{pmatrix}
     1 \\
     0 \\
     0 \end{pmatrix} \text{of}\ \mathbb{R}^3$$

$$b_1 =\begin{pmatrix}
     -1 \\
     2  \end{pmatrix},\ b_2 =\begin{pmatrix}
     1 \\
     -1  \end{pmatrix} \text{of}\ \mathbb{R}^2$$

the linear mapping of $L:\mathbb{R}^3 \rightarrow \mathbb{R}^2$ is
represented by the matrix:

$$A =\begin{pmatrix}
     1 & 0 & 3 \\
     -2 & 2 & 0  \end{pmatrix}.$$

a) Calculate the representation matrix of $L$ with respect to the
standard bases of $\mathbb{R}^3$ and $\mathbb{R}^2.$

b) Give a basis of $c_1, c_2, c_3$ of $\mathbb{R}^3$ and a basis $d_1, d_2$ of $\mathbb{R}^2$ such that $L$ is represented by the matrix

$$A_0 =\begin{pmatrix}
     1 & 0 & 0 \\
     0 & 1 & 0  \end{pmatrix}$$

with respect to these bases.

The Understanding

Can someone help me solve this task or at least explain me how it should be done. I don't understand this very well.

答:
Notice that we have a linear transformation from domain $\mathbb{R}^3$ to the co-domain $\mathbb{R}^2$.

(a) It is given that linear transformation L is represented by:

\begin{align}
A =\begin{pmatrix}
     1 & 0 & 3 \\
     -2 & 2 & 0  \end{pmatrix}
\end{align}

Now we want to find the representation matrix for the linear transformation L with respect to the bases standard basis $\beta = \{(1,0,0),(0,1,0),(0,0,1)\}$ and $\gamma = \{(1,0),(0,1)\}$.

\begin{align}
T((1,1,1)) = A\begin{pmatrix}
          1 \\
          1 \\
          1 \\ \end{pmatrix}
          =\begin{pmatrix}
          1 & 0 & 3 \\
          -2 & 2 & 0  \end{pmatrix}
          \begin{pmatrix}
          1 \\
          1 \\
          1 \\ \end{pmatrix}
          = \begin{pmatrix}
          4 \\
          0 \\
          \end{pmatrix}
\end{align}

\begin{align}
T((0,1,2)) = A\begin{pmatrix}
          0 \\
          1 \\
          2 \\ \end{pmatrix}
          =\begin{pmatrix}
          1 & 0 & 3 \\
          -2 & 2 & 0  \end{pmatrix}
          \begin{pmatrix}
          0 \\
          1 \\
          2 \\ \end{pmatrix}
          = \begin{pmatrix}
          6 \\
          2 \\
          \end{pmatrix}
\end{align}

\begin{align}
T((1,0,0)) = A\begin{pmatrix}
          1 \\
          0 \\
          0 \\ \end{pmatrix}
          =\begin{pmatrix}
          1 & 0 & 3 \\
          -2 & 2 & 0  \end{pmatrix}
          \begin{pmatrix}
          1 \\
          0 \\
          0 \\ \end{pmatrix}
          = \begin{pmatrix}
          1 \\
          -2 \\
          \end{pmatrix}
\end{align}

Now expressing these vectors as linear combinations of b1,b2,b3:

T(a1) = 4b1 + 8b2 (verify)

T(a2) = 8b1 + 14b2 (verify)

T(a3) = -1b1 + 0b2 (verify)

So the desired matrix:
\begin{align}
\begin{pmatrix}
          4 & 8 & -1 \\
          8 & 14 & 0
\end{pmatrix}
\end{align}


问:
Trying to figure out where the numbers are coming from for this work/integration problem
There is a cylindrical tank lying on its side filled completely with liquid. The diameter of the tank is 4ft, and the length of the tank is 10 ft.

Here is a terrible drawing of it but close enough:

I have to set up an integral to find the work required to pump all the liquid to a level 1 ft above the top of the tank.

So I have my formula:

Work = π (weight density) ∫ (area)(displacement)(thickness)

Someone else did this problem and obtained a bounds of 2 to -2, a displacement of (3-y), an area of 2x, and a thickness of dy. Where did the 2 to -2, (3-y), and 2x come from? This is probably intuitive but I don't see it.

答:
(Where did $\pi$ come from?)

Look at the tank straight on, so that you only see the circular side. Imagine that circle on the plane, with the center at $(0,0)$. Since the circle has diameter $4$, the bottom of the circle is at $(0,-2)$, and the top of the circle is at $(0,2)$. You want to lift the liquid to one foot above the top of the tank, so you are trying to lift it to the line $y=3$.

Now, take a horizontal slice of that circle of thickness $\Delta y$, at height $y_i$, where $y_i$ is somewhere between $-2$ and $2$. This slice corresponds to a volume of liquid that is approximately equal to the area of this slice, times $10$ (to account for the length of the tank).

To find the area, note that the circle has equation $x^2+y^2 = 4$, so that the $x$ coordinate is given by $x=\pm\sqrt{4-y^2}$. So the width of the slice is going to be $2x=2\sqrt{4-(y_i)^2}$. The thickness is $\Delta y$. So the volume of the slab of liquid is approximately equal to
$$20\sqrt{4 - y_i^2}\Delta y\text{ cubic feet.}$$
Now, you want to lift it all the way to $y=3$; how far from $y=3$ are you? Since the slab is at height $y_i$, the distance to $y=3$ is $3-y_i$.

So the work done in lifting this one slice of liquid is approximately:
$$\begin{align*}
\text{Work}&=\text{Weight}\times\text{Displacement}\\
&= \left(\text{Volume}\times\text{Density}\right)\times\text{Displacement}\\
&\approx \left(20\sqrt{4-y_i^2}\Delta y\right)\times\text{Density}\times(3-y_i)\\
&= \text{Density}\times 20(3-y_i)\sqrt{4-y_i^2}\Delta y.
\end{align*}$$
(Be sure to get the units right; the density should be in pounds per cubic feet, in which case the units of work will be ft-lbs. )

Now, to get the total work, you slice up the tank into these thin slices, figure out the work for each, and add it all up:
$$\text{Work}\approx \sum_{i=1}^n \text{Density}\times 20(3-y_i)\sqrt{4-y_i^2}\Delta y.$$
Taking the limit as $n\to\infty$, this becomes an integral. The integrand is
$$\text{Density}\times 20(3-y)\sqrt{4-y^2}\,dy.$$
(The $\sqrt{4-y^2}$ is the value of $x$ at the slice).

What are the limits of integration? Going back to the picture, what are the possible values of $y$? Because $y$ denotes where you are in the tank, and the tank extends from $y=-2$ to $y=2$, then the limits of integration go from $-2$ to $2$.


答:
\begin{align*}
A-\lambda I\text{ is not injective} & \iff\exists v_{1},v_{2}\colon v_{1}\neq v_{2}\text{ and }(A-\lambda I)v_{1}=(A-\lambda I)v_{2}\\
 & \iff\exists x\colon x\neq0\text{ and }(A-\lambda I)x=0\\
 & \iff A-\lambda I\text{ is singular}\\
 & \iff\det(A-\lambda I)=0
\end{align*}
The last line regarding the determinant is probably the definition
of eigenvalue you are familiar with.</p>

Title: Injectivity of $A-\lambda I$

Body:
I'm reading a paper on determinants and on one point the author states that:

A complex number $\lambda$ is called an eigenvalue of matrix $A$ if $A-\lambda I$ is not injective.

Why is this? Could someone clarify :)

Thank you! =)


问:
Is this statement about n-tuples incomplete?
To prove the following statement is an exercise in my real analysis book.

There are,
$$\frac{r!}{r_1!r_2!...r_n!}$$
ordered n-tuples of integers $(i_1,i_2,i_3,...,i_n)$ that contain $r_1$ ones, $r_2$ twos,...,$r_n$ n's.

What is $r$ here? is it to be assumed that $r_1+r_2+r_3...+r_n=r$? It seems strange however, since $(i_1,i_2,...,i_n)$ contain $n$ integers, is it not the case that $r_1+r_2+r_3...+r_n=n$?

答:
$r$ is the number of symbols. The symbols are going from $1$ to $n.$ For example, take $n=5,$ the string $12321$ has $r=5$ symbols, where one is $r_1=2$ times, two is $r_2=2$ times, $r_3=1$ and $r_4=0,r_5=0$(there are no fours nor fives in the string). So, $r_1+r_2+r_3+r_4=5=r.$

Formally, $$r_j=|\{m\in [n]:i_m=j\}|.$$

It seems then that this is just the case that $r=n$(typo, perhaps) In general, the result is for $(i_1,\cdots ,i_r)$ for general $r.$


问:
Will the perspective projections of an equally sized sphere and circle be congruent?
I am not a mathematician (yet), but rather more of an artist and have a question about perspective which can be described with projective geometry.

The diameter of a sphere and a circle are equal. A station point and picture plane are perpendicular to the center of each shape. Each shape is equal distance from the picture plane and station point.

When we make the projection will the shapes from the sphere and circle on the picture plane be congruent?

Consider this, you cannot see 50% of a sphere unless the projection point is infinitely far away from the sphere (orthographic projection). So the most we can see in finite space is upwards of 49% but not quite 50%, which is where the hemisphere and diameter of the sphere are. If the circular contour of the sphere being projected isn't from the hemisphere, but very close to it, how can the two shapes appear congruent?

答:
In the above figure you see that there are three kinds of rays emanating from the station point: rays that hit the surface of the sphere and illuminate it there, rays that miss the sphere and illuminate the projection plane, and finally the rays that just touch the sphere. These special rays form a circular cone and touch the sphere along the red circle, which is the shadow limit on the sphere. The image of the red circle is a conic section on the projection plane. This conic section is again a circle in my disposition.

Now you have in addition a wire circle of the same radius as the sphere has. If you place this circle properly, which is not on the sphere, you can have an image coinciding with the shadow boundary of the sphere on the projection plane. This is the green circle in my figure.


问:
Prove that $a\mathrel{R}b \iff$ **a** is a solution to the equation $X²-2bX+b²=0$ is transitive
Let $\mathrel{R}$ by a relation defined on the set  $\mathbb{R}$

$a\mathrel{R}b \iff$ a is a solution to the equation $X²-2bX+b²=0$

How do I prove that it's a transitive relation?

Here's what I tried :

Let a, b, c in R such that $a\mathrel{R}b$ and $b\mathrel{R}c$

$a\mathrel{R}b \iff$ $a²-2ba+b²=0$

$b\mathrel{R}c \iff$ $b²-2cb+c²=0$

This is where I am stuck, I don't know how to advance any further to prove that $a\mathrel{R}c$

答:
Hint: Factor $x^2-2bx+b^2$

 $(x-b)^2$

Recognize that if $a$ is a solution to $x^2-2bx+b^2=0$ then

 $a=b$

Your relation can then be rewritten as

 $a\mathcal{R} b\iff a=b$


问:
Is it correct to have understanding of logic with the use of set theory, to make learning it easier?
My teacher taught us mathematical logic by relating it to set theory (not entirely).
He asked us to relate the propositional variable p to a set P, and similarly q to a set Q.
The intuitive relations were-
$$
p ∧ q\equiv P \cap Q ,
p ∨ q\equiv P \cup Q,
p\rightarrow q\equiv \overline{P}∪Q
$$

He says that we can relate the value of a propositional variable to a set like-

If p has a truth value of False then take it as an empty set and if it has a truth value of True then take it as the universal set S or U.

This literally helps in simplifying long mathematical statements easily by converting them to sets and then to converting them again to propositional variables. ( rather than writing long and hard truth tables)

Is this an easy way to learn logic or is it because it just simplifies thing fast? I want to know if this approach for simplifying things out is sound, or will i run into some roadblocks later.

Any suggestions and comments are welcome if they make this intuitive understanding more concrete.

Just a heads-up I'm just a high-school student.

答: In some sense, the answer to your question is yes. It gives a geometric viewpoint on Boolean algebra (the algebra of propositions), known as Stone duality. A proposition is thought of as some subspace and it can be true or false depending on your position. Even if you don't learn Stone duality, this viewpoint is helpful and sound.</p>


问:
Show $\cosh x$ and $\sinh x$ are continious using $\varepsilon - \delta$ proof
I have to prove that $\sinh x$ and $\cosh x$ are continuous functions. I have to use the hyperbolic addition formula, and the inequalities:

$|\sinh x| \leq 3|x|, \, |x|<\frac{1}{2}$

$|\cosh x -1| \leq 3|x|, \, |x|<\frac{1}{2}$

I've tried doing an $\varepsilon-\delta$ proof, but I am only able to show it for $|x|<1/2$ or not at all. I believe it's beneficial to first look at $x\in[0,\infty )$ and then use symmetry to extend the argument to all of $R$. Here's what I've tried:

$\cosh x \,\mathrm{is\, continious \,on} \, R_+ \Leftrightarrow \forall x\in R_+ \forall \epsilon >0\exists\delta>0: |\cosh x-\cosh y|<\varepsilon\, \mathrm{ such\, that}\, |x-y|<\delta$

$|\cosh x - \cosh y| = |\cosh x -1 + 1 - \cosh y| = |(\cosh x -1) - (\cosh y -1)|$

$\leq |3|x|-3|y|| = 3||x|-|y||\leq3|x-y|$

And since $|x-y|<\delta$, one can pick a $\delta = \frac{\varepsilon}{3}$. The problem is, this is only valid for $0<x<\frac{1}{2}$ (and thus not on the desired the interval) because of the inequality I used. Also, I haven't used the hyperbolic addition formula.

This thread: Prove cosh(x) and sinh(x) are continuous. deals with the same problem, but my problem must not be solved simply by stating that $\sinh x$ and $\cosh x$ consists of continious functions, as that thread does.

答:
If you have to, the addition formula $\cosh(x+h)=\cosh(x)\cosh(h)+\sinh(x)\sinh(h)$ and $\sinh(x+h)=\cosh(x)\sinh(h)+\sinh(x)\cosh(h)$ gives
$$
\begin{aligned}
\cosh(x+h)-\cosh(x)&=\cosh(x)[\cosh(h)-1]+\sinh(x)\sinh(h)\\
\sinh(x+h)-\sinh(x)&=\sinh(x)[\cosh(h)-1]+\cosh(x)\sinh(h).
\end{aligned}
$$
So by triangle inequality, the given bound, and $\cosh(x)+\sinh(x)=e^x$, we have
$$
\begin{aligned}
\lvert\cosh(x+h)-\cosh(x)\rvert&\leq\cosh(x)\overbrace{[\cosh(h)-1]}^{\leq 3\lvert h\rvert}+\lvert\sinh(x)\overbrace{\sinh(h)}^{\leq 3\lvert h\rvert}\rvert\leq 3 e^x\lvert h\rvert
\\
\lvert\sinh(x+h)-\sinh(x)\rvert&=\lvert\sinh(x)\rvert\underbrace{[\cosh(h)-1]}_{\leq 3\lvert h\rvert}+\cosh(x)\underbrace{\lvert\sinh(h)\rvert}_{\leq 3\lvert h\rvert}\leq 3e^x\lvert h\rvert\\
\end{aligned}
$$
Hence if we choose $\delta<\min(\frac13\epsilon e^{-x},\frac12)$, we get $\lvert h\rvert<\delta\implies \lvert f(x+h)-f(x)\rvert\leq\epsilon$ for both $f=\cosh$ and $f=\sinh$.


问:
How to solve this nested feedback loop

I have a nested feedback loop that looks like this. There is an input $X(s)$, output $Y(s)$ and a plant with a function of
$$F(s)=\frac{R^3+R^2sL}{s^2R^2L-R^3s}$$

The input $X(s)$ will subtract the output $Y(s)$, and the result $X1=X(s)-Y(s)$ will be multiplied by the plant function above into $X2=F(s) \cdot X1$, then $X2$ will add $X$ to get $Y$ ($X2+X(s)=Y(s)$). I can't see where I can move things left or right to "unnest" them. Could you guys help me?

Title: How to solve this nested feedback loop

Body:
Let us write $$\begin{aligned}X_1(s) &= X(s) - Y(s) \\ X_2(s) &= F(s)X_1(s) = F(s)X(s) - F(s)Y(s) \\ Y(s) &= X(s) + X_2(s) \\&= X(s) + F(s)X(s) - F(s)Y(s)\end{aligned}.$$
Then
$$
\left(1+F(s)\right)Y(s) = \left(1+F(s)\right)X(s)
$$
and $Y(s) = X(s)$.


问:
Prove that the tangent space of a submanifold is a subspace of the tangent space of the manifold
So I ask if $M$ and $N$ are $k$-manifold and $h$-manifold of $\Bbb R^n$ respectively such that $N\subseteq M$ then the tangent space in $N$ at $y\in N$ is a subspace of the tangent space in $M$. So if $N=\partial M$ the result is surely true but otherwise I do not know.

So to prove it I covered $M$ by carefully-chosen coordinate patches. As a first case for $y\in M\setminus\partial M$ we choose a coordinate patch $\alpha:U\rightarrow V$ about $y_0$ such that $U$ is open in $\Bbb R^k$ and contains the unit cube $(0,1)^k$ and such that $\alpha$ carries a point of $(0,1)^{h}\times\{0\}^{k-h}$ to $y_0$: if we begin with an arbitrary coordinate patch $\alpha:U\rightarrow V$ about $y_0$ we can obtain one of the desired type be preceding $\alpha$ by a translation and a stretching in $\Bbb R^k$. As second case for $y_0\in\partial M$ we choose a coordinate patch $\alpha:U\rightarrow V$ such that $U$ is open in $H^k$ and $U$ contains $(0,1)^k$ and such that $\alpha$ carries a point of $(0,1)^{h}\times\{0\}^{k-h}$ to the point $y_0$. Exactly we can do this choosing a close rectangle $C:=\prod_{i=1}^k[a_i,b_i]$ about $\alpha^{-1}(y_0)$ contained in $U$ and so defining a diffeomorphism $\phi:\Bbb R^k\rightarrow\Bbb R^k$ in any $x\in\Bbb R^k$ such that $\phi[C]=[0,1]^k$ and $\phi(x_0)\in\Bbb R^h\times\{0\}^{k-h}$. Indeed if $\psi$ is the inverse of $\phi$ then $\alpha\circ\psi$ is the coordinate patch of the type described above.

So in any case if $\alpha$ is a coordinate patch of $M$ about $y\in N$ then $y=\alpha(x)$ for $x\in U\cap\Bbb R^h\times\{0\}^{k-h}$ and thus if $U_0\subseteq\Bbb R^h$ is an open set such that $U\cap\Bbb R^h\times\{0\}^{k-h}=U_0\times\{0\}^{k-h}$ then the map $\alpha\circ\iota$ is a coordinate patch of $N$ at $y$ where $\iota$ is the natural embedding of $\Bbb R^h$ in $\Bbb R^k$. So we observe that the column of the derivative of $\alpha\circ\iota$ are exactly the first $h$ column of the derivative of $\alpha$ and thus the statement follows.

So as you can see I did not able to construct the map $\phi$. In particular I tried to define it at $x\in\Bbb R^k$ trough the equation
$$
\big[\phi(x)\big](i)\:=\frac{x_i-a_i}{b_i-a_i}
$$
for each $i=1,...,k$ but generally $\phi(x_0)\notin\Bbb R^h\times\{0\}^{k-h}$ where $x_0=\alpha^{-1}(y_0)$, although $\phi[C]=[0,1]^k$. So could someone help me, please?

答:
This answer is using another, less computational method, but here's another/simpler way to see it:

$N$ is a submanifold of $M$, so there is an inclusion map $i : N \hookrightarrow M$. We can use that map to "push forward" tangent vectors of $N$ to tangent vectors of $M$.

This can simply be done by considering the jacobian induced by $i$ (call it $Di|$), which, at any $x \in N$, is a smooth map from $T_xN$ to $T_xM$. All you need to show
then is that it is injective, which will imply that $Di|_x(T_xN) \leq T_xM$.

The last step is identifying $Di|_x(T_xN)$ with $T_xN$, because both spaces are isomorphic by construction.


答: Here is a coordinate-free proof. Consider the inclusion $i : N \to M$. As $N$ is a submanifold, $i$ is an embedding, and for each $p \in N$, its differential $\mathrm{d}i_p : T_p N \to T_pM$ is injective, and identifies $T_pN$ as a linear subspace $\mathrm{d}i_p(T_pN) \subset T_pM$.

Title: Prove that the tangent space of a submanifold is a subspace of the tangent space of the manifold

Body:
So I ask if $M$ and $N$ are $k$-manifold and $h$-manifold of $\Bbb R^n$ respectively such that $N\subseteq M$ then the tangent space in $N$ at $y\in N$ is a subspace of the tangent space in $M$. So if $N=\partial M$ the result is surely true but otherwise I do not know.

So to prove it I covered $M$ by carefully-chosen coordinate patches. As a first case for $y\in M\setminus\partial M$ we choose a coordinate patch $\alpha:U\rightarrow V$ about $y_0$ such that $U$ is open in $\Bbb R^k$ and contains the unit cube $(0,1)^k$ and such that $\alpha$ carries a point of $(0,1)^{h}\times\{0\}^{k-h}$ to $y_0$: if we begin with an arbitrary coordinate patch $\alpha:U\rightarrow V$ about $y_0$ we can obtain one of the desired type be preceding $\alpha$ by a translation and a stretching in $\Bbb R^k$. As second case for $y_0\in\partial M$ we choose a coordinate patch $\alpha:U\rightarrow V$ such that $U$ is open in $H^k$ and $U$ contains $(0,1)^k$ and such that $\alpha$ carries a point of $(0,1)^{h}\times\{0\}^{k-h}$ to the point $y_0$. Exactly we can do this choosing a close rectangle $C:=\prod_{i=1}^k[a_i,b_i]$ about $\alpha^{-1}(y_0)$ contained in $U$ and so defining a diffeomorphism $\phi:\Bbb R^k\rightarrow\Bbb R^k$ in any $x\in\Bbb R^k$ such that $\phi[C]=[0,1]^k$ and $\phi(x_0)\in\Bbb R^h\times\{0\}^{k-h}$. Indeed if $\psi$ is the inverse of $\phi$ then $\alpha\circ\psi$ is the coordinate patch of the type described above.

So in any case if $\alpha$ is a coordinate patch of $M$ about $y\in N$ then $y=\alpha(x)$ for $x\in U\cap\Bbb R^h\times\{0\}^{k-h}$ and thus if $U_0\subseteq\Bbb R^h$ is an open set such that $U\cap\Bbb R^h\times\{0\}^{k-h}=U_0\times\{0\}^{k-h}$ then the map $\alpha\circ\iota$ is a coordinate patch of $N$ at $y$ where $\iota$ is the natural embedding of $\Bbb R^h$ in $\Bbb R^k$. So we observe that the column of the derivative of $\alpha\circ\iota$ are exactly the first $h$ column of the derivative of $\alpha$ and thus the statement follows.

So as you can see I did not able to construct the map $\phi$. In particular I tried to define it at $x\in\Bbb R^k$ trough the equation
$$
\big[\phi(x)\big](i)\:=\frac{x_i-a_i}{b_i-a_i}
$$
for each $i=1,...,k$ but generally $\phi(x_0)\notin\Bbb R^h\times\{0\}^{k-h}$ where $x_0=\alpha^{-1}(y_0)$, although $\phi[C]=[0,1]^k$. So could someone help me, please?


问:
How do you find all ideals of the R := K x K, where K is a field?
Consider the unitary commutative ring R := K x K where K is a field. The sum and product are defined as $$(a,b) + (c,d) = (a + c, b + d)$$   and $$(a,b) . (c,d) = (ac, bd)$$

How do you find all ideals of this ring?

For a field K, I know that if I < K and I is an ideal of K then I = 0 or I = K. Does it suffice to show that for a,b,c,d $\in$ K, (a,b), (c,d) $\in$ R, since K is closed so (ac, bd) $\in$ R ? So R is a field, or do I need to show every (ac, bd) is a unit? so it's only ideals are itself and 0?

答: Hint: $K \times K \cong K[x]/(x^2-x)$

Body: Hint: $K \times K \cong K[x]/(x^2-x)$


答:
Yes, the path you used in your attempt (as given in the main comments) works.

Below I outline a strategy for discovering such a path . . .

To show that $g$ is unbounded in a punctured neighborhood of $(0,0)$, it suffices to
find a path terminating at $(0,0)$ such that $g(x,y)$ approaches infinity as $(x,y)$ approaches $(0,0)$ along that path.

Consider a path in the first quadrant with equation $y=x^a$, where $a$ is a positive constant, left unknown for now.

Then for $(x,y)$ on that path with $(x,y)\ne (0,0)$, we have
$$
g(x,y)=\frac{xy^2}{x^2+y^6}=\frac{x^{2a+1}}{x^2+x^{6a}}
$$
so we would like to find $a > 0$ such that
$$
\lim_{x\to 0^+}\frac{x^{2a+1}}{x^2+x^{6a}}=\infty
$$
To get that to happen, we want $a > 0$ to be such that
$$
\left\lbrace
\begin{align*}
&2a+1 < 2\\[4pt]
&2a+1 < 6a\\[4pt]
\end{align*}
\right.
$$
or equivalently, ${\large{\frac{1}{4}}} < a < {\large{\frac{1}{2}}}$.

Thus we can choose $a={\large{\frac{1}{3}}}$ to achieve the desired result.

Body:
if $f(x,y)=\frac{xy^2}{x^2+y^6}$ and $f(0,0)=0$ show that $f$ is ubounded at any neighborhood of (0,0) but the restritions to straight line in $\mathbb{R}^2$ is cont.

My attempt:

**Proving continuity **

if $f(x,y)=\frac{xy^2}{x^2+y^6}$  then $y=mx+c$ is equation of any line then we can get $f(x,mx+c)= \frac{x.(mx)^2+c^2.x+2mxc}{x^2+(mx+c)^6}$ then the function is continuous at any point $(x,y)$ which is on the line$y=mx+c$,

We pick any point $(a,ma+c)$ on the line $y=mx+c$,then we try to show that the function is continuous at the point

We know that $f(a,ma+c)= \frac{ma^3+c^2.a+2mac}{a^2+(ma+c)^6}$ .

Now, $\lim_{x \to a}(f(x,mx+c))=\lim_{x \to a}\frac{x^3m^2+c^2x+2mxc}{x^2+(mx+c)^6}=f(a,ma+c)$(by the property of limits).

Proving uboundedness

In this part I did some back calculation . I showed that the function will not be bounded if the numerator becomes infinity or when the denominator becomes infinity.

For the denominator to become $0$ we see that $y^6+x^2=0$ then $(y^2)^3= -x^2$ then $y^2=(-x)^{\frac{2}{3}}$ . So I have shown that if $(x,y) \to (0,0) $ along the given curve $y^2=(-x)^{\frac{2}{3}}$ the limit of the function $f$ doesnt exist.

I am not sure whether this is correct but this has been essentially my attempt.


问:
Probability distribution of expression containing random variables
Two independent random variables $A$ and $B$ share the same probability distribution
$$p_{A}(x) = p_{B}(x) = \frac{1}{2^{x+1}} $$
where $x$ is a non negative integer.

N is a random variable given by:
$$N = 4 + 2A + 3B$$

What is the distribution of N? Is there a general method for generating a new probability distribution from an expression containing random variables with given distributions?

答:
First observe that if you want to compute the distribution of a function of $A,B$, given that you have no the joint density, the only alternative is that you assume independence between A and B

Second, note that A an B have the following pmf

$$P(A=x)=\frac{1}{2}\cdot\left(\frac{1}{2}\right)^x$$

where $X=0,1,2,3,...$

this is a Geometric distribution

to find the law of N you have to combine events of A and B

As @Henry alrady commented, the support of N is

$$N \in \{4,6,7,8,9,10,11,...\}$$

Thus

$$P(N=4)=P(A=0;B=0)=\frac{1}{4}$$

$$P(N=6)=P(A=1;B=0)=\frac{1}{8}$$

etc etc

at the end ouf your process (calculating the first 10 outcomes you will get the solution by induction finding that the support is

$$\{4,6,7,8,9,10,11,12,13,14,15,...\}$$

with probabilities

$$\{\frac{1}{4},\frac{1}{8},\frac{1}{8},\frac{1}{16},\frac{1}{16},\frac{1}{16},\frac{1}{32,}\frac{1}{32},\frac{1}{32},\frac{1}{32},\frac{1}{64},...\}$$


问:
Find an equation of the line tangent to the graph of  at the point where =1
So I was given the following prompt when studying for a test:

"A curve $C$ is defined by the parametric equations $x(t)=3+t^2$ and $y(t)=t^3+5t$. Find an equation of the line tangent to the graph of $C$ at the point where $t=1$."

I know this is a bit similar to a question that was asked a while ago, but I was able to work through a bit more of the problem. I ended up reducing the problem down to $y-6=\frac{dy}{dx}(x-4)$, but I'm a bit confused about what $\frac{dy}{dx}$ would actually look like here. I ended up finding an equation for it, but I thought that it had to be a number in the context of this equation? (The equation I found was as follows: $\frac{3t^2+5}{2t}$) Any help would be appreciated!

答: The question wants the tangent line where $t = 1$, so you should plug in $t = 1$ into your (correct) formula for $dy/dx$.


问:
Relation between negative part of function and another function.

I'm wondering how works this fact.

Let be $f:= f^+ - f^-$ and $g$ a function such that $g \le f$.

Then $f^- \le |g|$.

Thank you.

答:
If $f(x) \geq 0$, then $f^-(x)=0$ so $f^-(x) \leq |g(x)|$.

If $f(x) \leq 0$, then $f^-(x)=-f(x) \leq -g(x) = |g(x)|$ because $g(x) \leq f(x) \leq 0$.

If both cases, you get $f^-(x) \leq |g(x)|$, so
$$f^- \leq |g|$$


问:
$f(1/n)=\frac{2n}{3n+1}$, Then
Given that $f:D\rightarrow \mathbb{C}$ is analytic, $D=\{z:|z|<1\}$, analytic at $0$ and satisfies $f(1/n)=\dfrac{2n}{3n+1}$, Then

$f(0)=2/3$

$f$ has a simple pole at $z=-3$

$f(3)=1/3$

No such $f$ exists.

considered $g(z)=f(z)-\dfrac{2}{3+z}$ and zero set of $g$ is $\{\dfrac{1}{n}\}$ has limit point $0\in D$, so can I say $f(z)=\dfrac{2}{3+z}$. so $1$, $3$ are true for sure, for $2$ If our function were defined on $\mathbb{C}$ or a domain containing the point $-3$ then $-3$ is a simple pole thats for true, so in our case $2$ is false am I right? $4$ is false

答:
This is a very poorly formulated question.

Your idea is correct, but it is possible to argue for both 2. and 3. to be either true or false. Notice that the formulation of the problem states that $f$ is defined on the unit disc.

In other words, question 2 and 3 are meaningless, since $z=-3$ and $z=3$ are outside the specified domain of definition. On the other hand a (the) function described by the conditions has a unique meromorphic extension to the entire complex plane, and for this extension, question 2 and 3 make sense. Strictly speaking, the extended function is not the $f$ described in the question, though.

Title: $f(1/n)=\frac{2n}{3n+1}$, Then


问:
computing the limit $\lim_{\theta \to \frac{\pi}{2}} (\sec \theta - \tan \theta)$
I'm trying to compute the following limit and would greatly appreciate your heartening feedback on my solution.

The limit:

$\lim_{\theta \to \frac{\pi}{2}} (\sec \theta - \tan \theta)$

My steps in deriving the solution:

Preliminary identities:

$\sec \theta = \frac{1}{\cos \theta}$

$\tan \theta = \frac{\sin \theta}{\cos \theta}$

$\frac{1}{\cos \theta}-\frac{\sin\theta}{\cos\theta} = \frac{1-\sin\theta}{\cos\theta} = \frac{1-\sin^2\theta}{(1+\sin\theta)\cos\theta} = \frac{\cos^2\theta}{\cos\theta}\cdot \frac{1}{1+\sin\theta} = \frac{\cos\theta}{1+\sin\theta} = \frac{0}{1+1}$

When $\theta \to \frac{\pi}{2}$ then $\cos(\frac{\pi}{2}) = 0$ and $\sin(\frac{\pi}{2}) = 1$

The answer being:

$\lim_{\theta \to \frac{\pi}{2}} (\sec \theta - \tan \theta) = 0$

答:
Your approach is indeed the best way to tackle such problems, here is alternative way using L's hospital rule after applying the preliminary identities...

$\lim _{\theta\rightarrow 0}\frac{\left(1-\sin \theta \right)}{\cos \theta }=\lim _{\theta\rightarrow 0}\frac{\frac{d}{d \theta}\left(1-\sin \theta \right)}{\frac{d}{d \theta}\left(\cos \theta \right)}=\lim _{\theta\rightarrow 0}\frac{\left(-\cos \theta \right)}{\left(-\sin \theta \right)}=\lim _{\theta\rightarrow 0}\tan \theta =0$

Again, that was absolutely unnecessary, yet just in case it would be useful later...... :~)


答:
There is indeed a relation between logic expressions and set theory expression. For given set of basic logic sentences $p, q, \dots$ you can create space $\Omega$ of all possible possibilite situations (that is all possible combinations of logical values that your sentences may have). You can relate every logical sentence to the set of situations in which this sentence is true, that is
$$ p \text{ is true in situation } x \Leftrightarrow x \in P$$
Then you have for example
\begin{align} (p \land q) \text{ is true in situation } x &\Leftrightarrow (p \text{ is true in situation } x )\land (q \text{ is true in situation } x) \Leftrightarrow  \\
& \Leftrightarrow  (x\in P) \land (x\in Q) \Leftrightarrow \\
& \Leftrightarrow  x \in (P \cap Q) \end{align}
which means that the set $P \cap Q$ is the set related to the sentence $p \land q$. Analogously you can find the set operations related to other logic operations. The construction is solid, and if you already know set theory it can be useful to learning logic. For example it can visualise various logic laws (like the de Morgan's laws for example) or allow to immediately teanscribe some theorems about set on the theorems about logic sentences. For example, if you know that
$$ P \cap (Q \cup R) = (P \cap Q) \cup (P \cap R)$$
then you can immediately tell that sentences
$$ p \land (q \lor r) $$
and
$$ (p \land q) \lor (p \land r)$$
are equivalent.

Title: Is it correct to have understanding of logic with the use of set theory, to make learning it easier?

Body:
My teacher taught us mathematical logic by relating it to set theory (not entirely).
He asked us to relate the propositional variable p to a set P, and similarly q to a set Q.
The intuitive relations were-
$$
p ∧ q\equiv P \cap Q ,
p ∨ q\equiv P \cup Q,
p\rightarrow q\equiv \overline{P}∪Q
$$

He says that we can relate the value of a propositional variable to a set like-

If p has a truth value of False then take it as an empty set and if it has a truth value of True then take it as the universal set S or U.

This literally helps in simplifying long mathematical statements easily by converting them to sets and then to converting them again to propositional variables. ( rather than writing long and hard truth tables)

Is this an easy way to learn logic or is it because it just simplifies thing fast? I want to know if this approach for simplifying things out is sound, or will i run into some roadblocks later.

Any suggestions and comments are welcome if they make this intuitive understanding more concrete.

Just a heads-up I'm just a high-school student.


问:
How do one solve a nonlinear combinatoric problem?
I am an undergraduate CS student and I am struggling with a problem.

$Qx = b$ where $Q$ is a constant $m \times n$ matrix (with $m>n$), $x$ is a $n \times 1$ vector and $b$ is a $m\times 1$ vector.

I want to maximize the number of zeros in vector $b$.

subject to: $x(i)>0$ for $i=1,2, \ldots,n$

How do one tackle a problem like this ?

答:
Here's a slow way to do it.

Find a parametrization for $\operatorname{Nul} Q$, and see if it contains a positive vector (which is a matter of examining some inequalities that fall out of the parametrization). If so, this is such an $x$.

Failing that, one at a time, check if each standard basis vector $e_i$ is in $\operatorname{Col} Q$. If it is, Find a parametrization of $Q^{-1}e_i$ (the preimage of $e_i$ - not implying that $Q$ is invertible here), and again see if it contains a positive vector. If so, this is such an $x$.

Failing that, one at a time, find a parametrization of $Q^{-1}\operatorname{Span}(e_i,e_j)$, and again see if it contains a positive vector. If so, this is such an $x$.

Continue with $Q^{-1}\operatorname{Span}(e_i,e_j,e_k)$, etc.


问:
Is it possible to prove that surfaces with compact boundary are homeomorphic by glueing disks to the boundary components?
Let $S_1$ and $S_2$ be two surfaces with compact boundary and the same number of boundary components. Let $M_1$ and $M_2$ be the surfaces obtained by glueing closed disks to the boundary of $S_1$ and $S_2$, respectively. Is there a direct proof that if $M_1$ and $M_2$ are homeomorphic then $S_1$ and $S_2$ are also homeomorphic? I know that this can be done in the compact case using normal forms, as in Massey's book. And if the surfaces are not compact?

答:
If $M_1$ is homeomorphic to $M_2$, then it is homeomorphic via a map preserving the attached disks because we may compose the homeomorphism of $M_1$ and $M_2$ with a homeomorphism that takes the image of the disks attached $M_1$ to the disks attached to $M_2$ (such a homeomorphism exists by isotopy extension). By removing the interior of the attached disks of $M_1$, we have $S_1$ and the restriction homeomorphically maps onto $S_2$.

A similar result will be true in all higher dimensions topologically, but more care is needed in the smooth case. For example, in dimensions above 4 you can attach disks to disks along their boundary and arrive at nondiffeomorphic spheres, so the notions of diffeomorphism and diffeomorphism after capping off boundary components needs refinement (you need to specify how boundary components are capped off).


问:
Prove that the real root of $X^3-X^2+2X-1$ is the square of the real root of $X^3+X^2-1$.
Context: I'm studying the family of functions of the form:
$$y=a \ln(x) + b \ln(1-x)$$
for $0 \leq x \leq 1$ and $(a,b) \in \mathbb{N} \times \mathbb{N}$ where $\mathbb{N} = \{0, 1, 2,\dots\}$.
When "all" the functions of the family are drawn on the same picture, one gets (click to enlarge):

What I'm calling "pillars" make their appearance. They are located on precise abscissas $x_{c/d}$ which are algebraic numbers, roots of $X^c-(1-X)^d$ polynomials for some positive rational $c/d$ (in irreducible form). To each pillar is associated a vertical wavelength (clearly visible on the picture) which is equal to $-\frac{\ln(x_{c/d})}{d}=-\frac{\ln(1-x_{c/d})}{c}$. I compared the wavelength of the $(c/d=5)$ pillar (for which $x_5$ is the multiplicative inverse of $\rho$, the plastic number) with the wavelength of the $(c/d=3/2)$ pillar (marked green), and found numerically that they were equal. So, the asked question is to prove that, in any of its forms:

$$ -\frac{\ln(x_{3/2})}{2} = -\frac{\ln(x_{5})}{1} $$

$$ x_{3/2} = x_{5}^2 $$

the real root of $X^3-(1-X)^2$ is the square of the real root of $X^5-(1-X)^1$

the real root of $X^3-X^2+2X-1$ is the square of the real root of $X^3+X^2-1$

Body:
Let $f(x)=x^3-x^2+2x-1, g(x)=x^3+x^2-1$. We have

$$-g(x)\cdot g(-x)=-(x^3+x^2-1)((-x)^3+(-x)^2-1)\\
=(x^3+x^2-1)(x^3-x^2+1) = x^6 - (x^2-1)^2\\
=x^6-x^4+2x^2-1=f(x^2)$$

Now if $g(x)=(x-r)(x-s)(x-t)$, then

$$f(x^2)= - (x-r)(x-s)(x-t) (-x-r)(-x-s)(-x-t)\\=(x^2-r^2)(x^2-s^2)(x^2-t^2)
\\\implies f(x)=(x-r^2)(x-s^2)(x-t^2)$$
and we are done.


问:
Computing the value of $\zeta(6)$ and $\zeta(8)$
I know the closed form formula for calculating the values of Zeta function at even integers. I was able to derive it from the coefficients of the power series of $-\dfrac{\pi x}{2}\cot(\pi x)$. Now, I am looking at the way with which Euler was able to derive the values of $\zeta(2n)$.

I have seen this "trick" in Brilliant:

So I have:

$$\dfrac{\sin(x)}{x}=1-\dfrac{x^2}{3!}+\dfrac{x^4}{5!}-\dfrac{x^6}{7!}+\dfrac{x^8}{9!}...\tag{1}$$

$$\dfrac{\sin(x)}{x}=\left(1-\dfrac{x^2}{\pi^2}\right)\left(1-\dfrac{x^2}{4\pi^2}\right)\left(1-\dfrac{x^2}{9\pi^2}\right)\left(1-\dfrac{x^2}{16\pi^2}\right)...\tag{2}$$

$$\dfrac{\sin(ix)}{ix}=1-\dfrac{(ix)^2}{3!}+\dfrac{(ix)^4}{5!}-\dfrac{(ix)^6}{7!}...\tag{3}$$

$$\dfrac{\sin(ix)}{ix}=1+\dfrac{x^2}{3!}+\dfrac{x^4}{5!}+\dfrac{x^6}{7!}...\tag{4}$$

$$\dfrac{\sin(ix)}{ix}=\left(1-\dfrac{(ix)^2}{\pi^2}\right)\left(1-\dfrac{(ix)^2}{4\pi^2}\right)\left(1-\dfrac{(ix)^2}{9\pi^2}\right)\left(1-\dfrac{(ix)^2}{16\pi^2}\right)...\tag{5}$$

$$\dfrac{\sin(ix)}{ix}=\left(1+\dfrac{x^2}{\pi^2}\right)\left(1+\dfrac{x^2}{4\pi^2}\right)\left(1+\dfrac{x^2}{9\pi^2}\right)\left(1+\dfrac{x^2}{16\pi^2}\right)...\tag{6}$$

Multiply $(6)$ and $(2)$, I have:

$$\dfrac{\sin(x)\sin(ix)}{ix^2}=1+\left(\dfrac{x^2}{3!}-\dfrac{x^2}{3!}\right)+\left(\dfrac{x^4}{5!}+\dfrac{x^4}{5!}-\dfrac{x^4}{(3!)^2}\right)+...\tag{7}$$

$$\dfrac{\sin(x)\sin(ix)}{ix^2}=\left(1+\dfrac{x^2}{\pi^2}\right)\left(1-\dfrac{x^2}{\pi^2}\right)\left(1+\dfrac{x^2}{4\pi^2}\right)\left(1-\dfrac{x^2}{4\pi^2}\right)\left(1+\dfrac{x^2}{9\pi^2}\right)\left(1-\dfrac{x^2}{9\pi^2}\right)...\tag{8}$$

$$\dfrac{\sin(x)\sin(ix)}{ix^2}=\left(1-\dfrac{x^4}{\pi^4}\right)\left(1-\dfrac{x^4}{16\pi^4}\right)\left(1-\dfrac{x^4}{81\pi^4}\right)...\tag{9}$$

Comparing the coeffcients of $x^4$ in the infinite Maclaurin series and the infinite product, we will arrive at $\zeta(4)=\dfrac{\pi^4}{90}$.

Now that I wish to continue this process to see if I can get $\zeta(6)$, $\zeta(8)$, $\zeta(10)$, I don't know how to continue. The trick above works fine for $\zeta(4)$, but what about higher $\zeta(2n)$.

I look around and find this interesting post of user17762:

In his answer, he wrote the function on the LHS as $\dfrac{i\sin(z)\sin(\dfrac{z}{i})}{z^2}$. A little bit of manipulation shows that it is exactly like $\dfrac{\sin(x)\sin(ix)}{ix^2}$  or $\dfrac{\sin(x)\sinh(x)}{x^2}$

I have tried to multiply again $\dfrac{\sin(x)\sinh(x)\sin(x/i^2)}{x^2(\frac{x}{i^2})}=\dfrac{\sin^2(x)\sinh(x)}{x^3}$, using the hint found in the answer of the member above. Then I use Wolfram to check the series expansion of this function but the third term of $x^6$ is $\dfrac{5}{3024}$. I thought that I should a find a function where its coefficient of $x^6$ should be $\dfrac{1}{945}$. I am not sure if my reasoning is problematic.

Is there an algorithm to find the function on the LHS?

答:
The generalisation you are looking for is given by the functions
$$f_n \colon \mathbb{R} \to \mathbb{R}, \, f_n (x) = \prod \limits_{k=0}^{n-1} \operatorname{sinc}\left(\mathrm{e}^{\mathrm{i} \pi \frac{k}{n}} x\right) \, , $$
for $n \in \mathbb{N}$ ($\operatorname{sinc}(z) = \frac{\sin(z)}{z}$ with $\operatorname{sinc}(0) = 1$). Since the product representation of $\operatorname{sinc}$ converges absolutely, we can rearrange factors to obtain
$$ f_n(x) = \prod \limits_{k=0}^{n-1} \prod \limits_{l=1}^\infty \left(1 - \mathrm{e}^{2 \pi \mathrm{i} \frac{k}{n}} \frac{x^2}{\pi^2 l^2} \right) = \prod \limits_{l=1}^\infty \prod \limits_{k=0}^{n-1} \left(1 - \mathrm{e}^{2 \pi \mathrm{i} \frac{k}{n}} \frac{x^2}{\pi^2 l^2} \right) = \prod \limits_{l=1}^\infty \left(1 - \left(\frac{x}{\pi l}\right)^{2n} \right)$$
for $x \in \mathbb{R}$ and $n \in \mathbb{N}$ (see this question for a simple proof of the final step). This representation confirms that these functions are real at real arguments. Clearly, the coefficient in front of $x^{2n}$ in the series expansion of $f_n$ is $[x^{2n}] f_n (x) = - \frac{\zeta(2n)}{\pi^{2n}}$ for $n \in \mathbb{N}$. Now simply multiply the power series representations, i.e.
$$ f_n (x) = \prod \limits_{k=0}^{n-1}  \sum \limits_{l_k = 0}^\infty \frac{(-1)^{l_k} \mathrm{e}^{2 \pi \frac{k l_k}{n} \mathrm{i}} x^{2l_k}}{(2l_k+1)!} \, ,$$
and compute the coefficient this way to obtain the zeta values by comparison. As you can easily check using CAS, this works for every $n \in \mathbb{N}$ ($[x^6]f_3(x) = - \frac{1}{945}$ or $[x^8]f_4(x) = - \frac{1}{9450}$, for example), but I have not yet managed to prove $[x^{2n}] f_n(x) = \frac{(-1)^{n} \mathrm{B}_{2n} 2^{2n-1}}{(2n)!}$ analytically.


问:
Stationary Distribution of Tsetlin Library/Move to Front Markov chain
Consider $R_1, R_2, ... R_N$, N items arranged in a sequence. With probability $p_i > 0$, item $R_i$ is moved to the front of the sequence. This defines a finite Markov Chain of $N!$ states.

What is the stationary distribution of this Markov Chain?

Note: I am looking for a solution to this problem with elementary probability that also shows how one would derive the distribution, and not just a proof by induction as is presented here: https://www.jstor.org/stable/3212655?seq=1#metadata_info_tab_contents

答:
Let's generalize the description of the problem slightly, so that item $R_i$ has weight $w_i$, and is moved to the front with probability $p_i = \frac{w_i}{w_1 + w_2 + \dots + w_N}$. Let $f(w_1, w_2, \dots, w_N)$ be the stationary probability of state $(R_1,R_2,\dots,R_N)$: by symmetry, that's the only probability we need to find.

Start the Markov chain in the stationary distribution and then make one more step from there. This will keep the distribution stationary, but make it easy to see that item $R_1$ is in front with probability $\frac{w_1}{w_1 + w_2 + \dots + w_n}$.

Moreover, given that item $R_1$ is in front, the order of $R_2, \dots, R_N$ is distributed exactly as it was in the stationary distribution, because it did not change. If we forget about $R_1$ entirely, we get a size-$(N-1)$ problem with weights $w_2, \dots, w_N$; therefore the probability that $R_2, \dots, R_N$ appear in that order is $f(w_2, \dots, w_N)$.

Multiplying, we get
$$
   f(w_1, w_2, \dots, w_N) = \frac{w_1}{w_1 + \dots + w_n} f(w_2, \dots, w_N)
$$
from which the formula
$$
   f(w_1, w_2, \dots, w_N) = \prod_{i=1}^N \frac{w_i}{w_i + \dots + w_N}
$$
immediately follows.


问:
SAT question - plotting and linear model
I am tutoring someone in preparation for the SAT. Below are two SAT questions that I find bothersome. I have the answers but disagree with the official answers and would appreciate understanding the SAT train of thought if anyone can conjure it up.

For problem #5, the SAT answer is C. I calculated using all points between $x=3$ and $x=4$ and found the percent increase to be around 6.4%.

For problem #7, the SAT answer is A. However, it appears the the greatest difference occurs right after $x=2$ - the beginning of month 3.

答:
Let $x$ denote the number of months from the first class was offered and $y$ the number of people attending to the class. Also, let $\widehat{y}$ denote the estimated number of people attending to the class.

For problem $5$, observe that the predicted number of people attending to the class at month $4$ (that is to say, $x=4$) is slightly smaller than $800$, and the observed number of people who attented is $600$. Then, the percent greater that the actual number of people is
$$\frac{\widehat{y}- y}{y} = \frac{800-600}{600}=33.33\%.$$
Since the predicted value was a bit lower than $800$, I would choose option $C$ to approximate it.

For problem $7$, you have to calculate, for each month $x=0,1,2,3,4,5$, what is the biggest difference $y - \widehat{y}$, which obviously occurs at $x=2$, that is to say, for month $2$.


问:
Tangent space of the pre-image of identity via the Lie product
Let $G$ be a Lie group and denote by $\mu:G\times G\rightarrow G$ its product. I am trying to understand the submanifold $\mu^{-1}(\{e\})=\{(g,g^{-1}):g\in G\}$ in the Lie group product $G\times G$. More precisely, I would like to characterize the tangent space
$$ T_{(g, h)}(\mu^{-1}(\{e\}))\subset T_gG\oplus T_{h}G .$$
I have tried to use the fact that $\mu$ is a submersion and so such tangent space is given by $\ker d\mu_{(g,h)}$. Therefore, if $$(X,Y)\in T_{(g, h)}(\mu^{-1}(\{e\})),$$ then, using that $d\mu_{(g,h)}(X,Y)=d(L_g)_h Y+d(R_h)_gX=0$, we get
$$
X=-d(R_{h^{-1}}\circ L_g)_h Y,
$$
and since $h=g^{-1}$, it follows that
$$
X=-d(R_g\circ L_g)_{g^{-1}}Y,$$
but I can't see if there is a better characterization for this problem. Any ideas would be appreciated!

答:
There are the canonical projection maps $p,q: G\times G \rightarrow G$ defined by

$$p(g,h):=g, q(g,h):=h.$$

Let $Z:=\mu^{-1}(e) \subseteq G\times G$.

There is a canonical map

$$s: G \rightarrow Z \subseteq G \times G$$

defined by $s(g):=(g,g^{-1})$. We get induced map $p,q: Z \rightarrow G$ defined by
$p(g,g^{-1}):=g, q(g,g^{-1}):=g^{-1}$. It follows

$$sp(g,g^{-1})=s(g)=(g,g^{-1})$$

and

$$ps(g)=p(g,g^{-1})=g$$

hence $s: G \rightarrow Z$ seems to be an "isomorphism of manifolds". The set $Z$ is not a subgroup of $G\times G$ since

$$(g,g^{-1})(h,h^{-1}):=(gh, g^{-1}h^{-1}) \notin Z$$

and the map $s$ is not a map of Lie groups. If you give $Z$ the following product

$$(g,g^{-1})(h,h^{-1}):=(gh, h^{-1}g^{-1})=(gh,(gh)^{-1})$$

it follows $s$ is an isomorphism of Lie groups.

Question: "More precisely, I would like to characterize the tangent space
$T_{(g,h)}(μ^{−1}({e})) \subseteq T_g(G)\oplus T_h(G)$."

Answer: It seems to me the tangent space of $Z$ at $x:=(g,g^{-1})$ should be as follows

$$T_x(Z) \cong T_g(G).$$

This is beacause $s$ is an isomorphism and $p(x)=g$.