问:
$\mathbb F$ is a field and $\phi:\mathbb Z \to \mathbb F$ is a surjective ring homomorphism,then the number of elements in $\mathbb F$ is prime
Assume $\mathbb F$ is a field and $\phi:\mathbb Z \to \mathbb F$ is a surjective ring homomorphism,prove that the number of elements in $\mathbb F$ should be a prime number.

Based on my knowledge

$\phi(a+b)=\phi(a)+\phi(b)$ for all $a,b \in \mathbb Z$

$\phi(ab)=\phi(a)\phi(b)$ for all $a,b \in \mathbb Z$

$\phi(1)=\phi(1_{\mathbb F})$

Moreover surjectivity of $\phi$ implies that for every element  $x \in \mathbb F$ there is some $a \in \mathbb Z:\phi(a)=x$

But I don't understand how this is related to the number of elements in $\mathbb F$ and how it should be a prime number,maybe the surjectivity of such a mapping implies that $\mathbb F$ is a finite field and hence the order of $\mathbb F$ should be of the form $p^n$ ,where $n$ is a natural number and $p$ is a prime number,but even if it happens then $\text{ord}(\mathbb F) $ is not necessarily a prime number.

答: Hint: By the homomorphism theorem $\phi$ induces an isomorphism $\mathbb Z/\operatorname{ker}(\phi)\to \mathbb F$. What are the possible ideals $I\subseteq \mathbb Z$, quotients $\mathbb Z/I$ and when is such a quotient a field?


问:
An example of tensor product
Let L be field extension of K. Can any one show that Hom($K^n,K^m$)${\otimes}_K$L=Hom($L^n,L^m$).Thanks.</p>

答:
First, note that $\hom(K^n,K^m)\cong K^{m\times n}$, by arranging the images of the standard basis vectors in the columns of a matrix.
We can then naturally define a $K$-bilinear map $K^{m\times n}\times L\,\to\,L^{m\times n}$, namely
$$(A,\lambda)\, \mapsto\lambda\cdot A$$
where $A$ is an $m\times n$ matrix with coefficients in $K$. This mapping induces a linear $\phi:K^{m\times n}\otimes L\,\to\,L^{m\times n}$, which has an inverse:

Let $E_{ij}$ be the matrix which has $1$ at the $(i,j)$th entry and 0 everywhere else. These form a basis of $L^{m\times n}$ over $L$. It means that any matrix $B\in L^{m\times n}$ can be (uniquely) written in the form
$$B=\sum_{ij}\lambda_{ij}\,E_{ij}\,,\quad \ \lambda_{ij}\in L$$
where anyways, $B=(\lambda_{ij})_{ij}$. Since $\{0,1\}\subseteq K$, we have $E_{ij}\in K^{m\times n}$, and then consider
$$\psi:= \ B\mapsto \sum_{ij}(E_{ij}\otimes \lambda_{ij})\,.$$
Prove that this is indeed the inverse of $\phi$.

ContentLicense: CC BY-SA 3.0

Title: An example of tensor product


答:
For each integer $n$, one has $\phi(n) = \phi(n1) =  n\phi(1) = n1_F$.
Since $\phi$ is onto, $F = \{n1_F\mid n\in\Bbb Z\}$.

If $F$ has characteristic 0, it contains the rational numbers and so the mapping cannot be onto.

Thus $F$ must have characteristic $p>0$ and $p$ is prime.
In $F$,  $n1_F=0$ holds only if $n$ is prime. Otherwise, it would contain zero divisors.

ContentLicense: CC BY-SA 4.0

Title: $\mathbb F$ is a field and $\phi:\mathbb Z \to \mathbb F$ is a surjective ring homomorphism,then the number of elements in $\mathbb F$ is prime

Body:
Assume $\mathbb F$ is a field and $\phi:\mathbb Z \to \mathbb F$ is a surjective ring homomorphism,prove that the number of elements in $\mathbb F$ should be a prime number.

Based on my knowledge

$\phi(a+b)=\phi(a)+\phi(b)$ for all $a,b \in \mathbb Z$

$\phi(ab)=\phi(a)\phi(b)$ for all $a,b \in \mathbb Z$

$\phi(1)=\phi(1_{\mathbb F})$

Moreover surjectivity of $\phi$ implies that for every element  $x \in \mathbb F$ there is some $a \in \mathbb Z:\phi(a)=x$

But I don't understand how this is related to the number of elements in $\mathbb F$ and how it should be a prime number,maybe the surjectivity of such a mapping implies that $\mathbb F$ is a finite field and hence the order of $\mathbb F$ should be of the form $p^n$ ,where $n$ is a natural number and $p$ is a prime number,but even if it happens then $\text{ord}(\mathbb F) $ is not necessarily a prime number.


答:
Hint: Since $\phi$ is a ring homomorphism, $\phi(n)=n\phi(1)$. Since the field is finite (otherwise $\phi$ is injective -- what goes wrong in this case?), there must be some integer $m$ for which $\phi(m)=0$, and thus $\mathbb F$ consists exactly of
$$\{0,\phi(1),\dots,(m-1)\phi(1)\}.$$
What does this mean about $m$, and thus the size of the field?


问:
How to find integration without any sort of modification to the current function?
I want to integrate a difficult single variable function but I don't want to modify the given function i.e. factorising, rationalizing, etc.
The question is:
$$f(x)= (x^3-4x^2+4x)(x^2-4)^{-1}$$

The answer turns out to be equals to:
$g(x)= (x^2/2)+4\ln(|x-4|)+C$ after factorising the question. But I don't want this!
If you can see, there is something special about the question itself.
At $x=2$ the $f(x)=0/0$. But when we calculate derivative of $g(x)$ we get,
$h(x)= (x^2-4x+4)(x-4)^{-1}$ but $h(x)$ is not equals to $0/0$ at $x=2$...
So I want a $F(X)$ whose derivative will give "exactly" my original given $f(x)$- no factorization, no rationalization, etc...
(You can use u-substitution but be careful my objective shall be achieved)
and I'm well aware that $h(x)$ is equivalent to $g(x)$...
In short, I want a F(X) such that its derivative is exactly equals to my given f(x)- it doesn't matter what techniques you use! You can even use technique other than integration if possible.

Thank you for spending your precious time
Respected Regards,
Swayam Jha

答:
The trouble is that you are treating the indeterminate form $f(2)=0/0$ as a number, while $f(2)$ is well defined, as you can verify by looking at the graph. Applying L'Hopital (or by whatever method you like$^\dagger$), we see that $f(2)=0$. As expected, $g'(2)=0$, so $g'(2)=f(2)$. If you really wanted, I suppose you could write

$$ g'(x)=\frac{4}{x-4}+x=\frac{x^2-4x+4}{x-4}$$

Multiply top and bottom by $x-2$

$$ g'(x)=\frac{(x^2-4x+4)(x-2)}{(x-4)(x-2)}=\frac{x^3-6x^2+12x-8}{x^2-6x+8}$$

Which also has the indeterminate form $0/0$ at $x=2$

EDIT: actually, just multiply $g'$ by $1=x/x$ and we recover your original function

$$ g'(x)=\frac{(x^2-4x+4)(x)}{(x-4)(x)}=\frac{x^3-4x^2+4x}{x^2-4x}=f(x)$$

$\dagger$ It is simplest by reversing the last step (dividing by $1=x/x$)

$$f(x)= \frac{x^3-4x^2+4x}{x^2-4x}=\frac{x^2-4x+4}{x-4}$$

Evaluating the right hand side at $x=2$ gives $0/2=0$


问:
question on five number summary & quantile.
I know that in five number summary :

25% of a data set lies between Min & 1st quartile.

50% of a data set lies between Min & 2nd quartile, that is, Median.

75% of a data set lies between Min & 3rd quartile.

& 100% of a data set lies between Min & Max.

But my cocept is contradicting as i run a data set in R Programming Language.

while i wrote the command fivenum(x) & quantile(x) the percentiles are not matching that is, "the 1st quartile of fivenum(x)" is not "the the value below which 25% of the data fall in quantile(x)"

the commands that i have run in R console:

```<code>> y <- c(7,12,9,1)

> fivenum(y)

 [1]  1.0      4.0     8.0    10.5    12.0

 > quantile(y)

   0%   25%   50%   75%  100%

 1.00  5.50  8.00  9.75 12.00
</code>```

by which points am i being misleading?

答:
$\frac{n+1}{2}\times 25\% =1.25\\
\frac{n+1}{2}\times 75\% =3.75$

The fivenum command takes the $25^{th}$ percentile as the number half way between $1$ and $7$  and the $75^{th}$ as the number half way between $9$ and $12$, whereas the quantile command may have been an attempt to produce the numbers $0.25$ of the way between $1$ and $7$ and $0.75$ of the way between $9$ and $12$ that contained an error and ended up giving the numbers $0.25$ of the way between 7 and 1 and $0.75$ of the way between $12$ and $9$.


问:
Properties of Measuring Points in perspective drawing
In his Complete Guide to Perspective Drawing, Craig Attebery defines a central concept in two-point perspective: Measuring Points, which can be used to determine depth in lines going away from the viewer.

Attebery states, without proof, that the distance from Vanishing Points to Measuring Points should be the same as the distance from Vanishing Points to the Station Point (artist’s eye). That is, Measuring Points can be obtained using a compass:

A different theory, still, comes from Joseph D’Amelio’s Perspective Drawing Handbook. From the latter’s definition, Measuring Points would be Vanishing Points themselves. As each set of parallel lines has its Vanishing Point, this would imply each Vanishing Point to have not a single, but infinite Measuring Points:

Both theories, as interpreted above, cannot be right: Vanishing Points have either a single or infinite measuring points. The question is, thus:

How many Measuring Points does each Vanishing Point has? If only one, why should Attebery’s statement hold: that the distance from Vanishing Points to Measuring Points  be the same as the distance from Vanishing Points to the Station Point?

答:
Each direction has a specific vanishing point.

Guide lines that have the same vanishing point, correspond to the same direction.

You can use any direction for guide lines, because the angle between the guide lines and the object edge only affects the scale.

Here is visual proof:

The black line represents a line divided into four equal length sections.
Red guide line corresponds to one measurement point, and blue guide line to another measurement point.
Both red lines and blue lines intersect the horizontal measuring line at regular intervals; only the exact scale varies.

Note that both Craig Attebery and Joseph D'Amelio both provide a way to choose a good measuring point.  However, any measuring point not too close to the vanishing point of the line that is to be measured, will work.

The method shown by D'Amelio at makes the measurements along the measuring line easy, so that one can use a rule/scale without having to do any math.

How many measuring points does each vanishing point have?

An infinite number.  The choice of the vanishing point only affects the scale between the measuring line and the object (line to be measured).

Since this is math.stackexchange.com, I think it is pertinent to explore the actual projective math.  It is surprisingly simple, you see, if we choose a particular coordinate system.

Let Station Point (eye) be at origin $(0, 0, 0)$, and the picture plane perpendicular to the $z$ axis, at distance $d$.  (You are then gazing directly along the positive $z$ axis.)

Any point $(x, y, z)$ in 3D space corresponds to point $(x^\prime, y^\prime, z^\prime)$ on the picture plane via
$$\left\lbrace \; \begin{aligned}
x^\prime &= x \frac{d}{z} \\
y^\prime &= y \frac{d}{z} \\
z^\prime &= z \frac{d}{z} = d \\
\end{aligned} \right.$$
Note that $x=0$, $y=0$ corresponds to the sight line; the line that passes through the station point (eye) and is perpendicular to the picture plane.

Vanishing points are points at infinite distance.  You pick any line, say one that passes through $(x_0, y_0, z_0)$ in direction $(x_\Delta, y_\Delta, z_\Delta)$, $z_\Delta \gt 0$, so that the line parametrized using $t \in \mathbb{R}$ is
$$\left\lbrace ~ \begin{aligned}
x &= x_0 + t x_\Delta \\
y &= y_0 + t y_\Delta \\
z &= z_0 + t z_\Delta \\
\end{aligned} \right.$$
and the vanishing point corresponding to that line is $(x^\prime_v ,\; y^\prime_v ,\; z^\prime_v = d)$:
$$\left\lbrace \; \begin{aligned}
x^\prime_v &= \lim_{t \to \infty} \left( \bigr( x_0 + t x_\Delta \bigr) \left( \frac{d}{z_0 + t z_\Delta} \right) \right) = d \frac{ x_\Delta }{ z_\Delta } \\
y^\prime_v &= \lim_{t \to \infty} \left( \bigr( y_0 + t y_\Delta \bigr) \left( \frac{d}{z_0 + t z_\Delta} \right) \right) = d \frac{ y_\Delta }{ z_\Delta }  \\
z^\prime_v &= \lim_{t \to \infty} \left( \bigr( z_0 + t z_\Delta \bigr) \left( \frac{d}{z_0 + t z_\Delta} \right) \right) = d \\
\end{aligned} \right.$$
This means that every direction $(x_\Delta, y_\Delta, z_\Delta)$ in to the picture, $z_\Delta \gt 0$, has its own explicit vanishing point at $(d x_\Delta / z_\Delta, \; d y_\Delta / z_\Delta, \; d)$.

In other words, all parallel lines do indeed have the same vanishing point.

Note that the above is an exact mathematical description of what vanishing points are.  This is not any kind of approximation.  This is what happens in reality, if we think of the image plane as e.g. a window, or the wall in a camera obscura with origin at the pinhole and the wall perpendicular to the $z$ axis at $(0, 0, -d)$.  (The origin is technically the focal point of the lens used.)

For clarity, I use $\;^\prime$ to denote picture coordinates, although they are in the same coordinate system as everything else.  $z^\prime = d$ for all points on the picture plane, so it is okay to just drop the $z$ coordinate.

We can furthermore define the rotation of the image plane around the $z$ (depth) axis, by defining horizontal parallel to the $x$ axis ($x z$ plane), and vertical parallel to the $y$ axis ($y z$ plane).  Horizon plane is then $y = 0$.  If we choose a left-handed coordinate system, $x$ increases right and $y$ up.  If our eye is at height $h$, then the ground plane is $y = -h$ (because our eye is at origin, $(0, 0, 0)$).

Both guides OP referenced involve lines at a specific height, the same plane $y$. Because all lines on such planes have $y_\Delta = 0$, their vanishing points are on the horizon line at $x^\prime = d x_\Delta / z_\Delta$, $y^\prime = 0$.  The exact height $y = h$ where the plane is, does not matter.

If we use angle $\theta$ to describe the direction of a line on such a plane (say, on the ground, or any other horizontal plane), with $\theta = 0$ being directly along the sight line, $\theta \gt 0$ right, $\theta \lt 0$ left, we can use $x_\Delta = \sin \theta$, $z_\Delta = \cos \theta$, and the vanishing point $x^\prime$ coordinate on the horizon line is
$$x^\prime = d \frac{\sin\theta}{\cos\theta} = d \tan\theta$$
and inversely, if we know the $x$ coordinate $x^\prime$ (and the distance from station point to picture plane $d$),
$$\theta = \arctan\left(\frac{x^\prime}{d}\right)$$

The vanishing point corresponding to a perpendicular line on the same plane ("ground level"), $\theta \pm 90°$, is
$$x^\prime_\perp = d \tan(\theta \pm 90°) = -d \cot\theta = \frac{-d}{\tan\theta}$$
and therefore the distance $L$ between vanishing points corresponding to perpendicular directions is
$$L = \lvert x^\prime - x^\prime_\perp \rvert = d \lvert \tan\theta + \cot\theta \rvert$$
For symmetric two-point perspective, $\theta = 45°$ and $L = 2 d$.  If $\theta = 0$, the other vanishing point is at ±infinity, and vice versa.  Here is a small table of angles, so you can have an idea of how $L$ varies as a function of $\theta$:
$$\begin{array}{c|c}
\theta & L \\
\hline
45° & 2.000 \; d \\
40°, 50° & 2.031 \; d \\
35°, 55° & 2.128 \; d \\
30°, 60° & 2.309 \; d \\
25°, 65° & 2.611 \; d \\
20°, 70° & 3.111 \; d \\
15°, 75° & 4.000 \; d \\
\end{array}$$

If we want to choose the vanishing point so that the distances from the point where the measuring line and the object edge intersect match along both (the measuring line and the object edge), we do need to choose a specific vanishing point. Looking from above,

we want to choose a direction $\varphi$ (and corresponding vanishing point) for the blue guide lines so that measurements along the black measuring line match the ones along the red object line.  (Specifically, that distances from the point where the measuring line intersects the object line, match.)
If the object line direction is $\theta$, then the guide line direction $\varphi$ is
$$\varphi = \begin{cases} - \frac{90° - \theta}{2} = \frac{1}{2}\theta - 45°, & \theta \ge 0 \\
\frac{90° + \theta}{2} = \frac{1}{2}\theta + 45°, & \theta \lt 0 \\
\end{cases}$$
If $x^\prime$ is the $x$ coordinate for the object line vanishing point (or true vanishing point), we need to calculate $\theta$ from
$$\theta = \arctan\left(\frac{x^\prime}{d}\right)$$
and then we can calculate the measuring point $x$ coordinate $x^\prime_m$,
$$x^\prime_m = d \tan\varphi$$
Here is a small table of various object line directions, and the distance $L$ from the object line vanishing point to the optimal measurement point:
$$\begin{array}{c|c|c|c|c}
\theta & x^\prime & \varphi & x^\prime_m & L \\
\hline
0°  & 0.000 \; d & -45°   & -1.000 \; d & 1.000 \; d \\
5°  & 0.087 \; d & -42.5° & -0.916 \; d & 1.004 \; d \\
10° & 0.176 \; d & -40°   & -0.839 \; d & 1.015 \; d \\
15° & 0.268 \; d & -37.5° & -0.767 \; d & 1.035 \; d \\
20° & 0.364 \; d & -35°   & -0.700 \; d & 1.064 \; d \\
25° & 0.466 \; d & -32.5° & -0.637 \; d & 1.103 \; d \\
30° & 0.577 \; d & -30°   & -0.577 \; d & 1.155 \; d \\
35° & 0.700 \; d & -27.5° & -0.521 \; d & 1.221 \; d \\
40° & 0.839 \; d & -25°   & -0.466 \; d & 1.305 \; d \\
45° & 1.000 \; d & -22.5° & -0.414 \; d & 1.414 \; d \\
50° & 1.192 \; d & -20°   & -0.364 \; d & 1.556 \; d \\
55° & 1.428 \; d & -17.5° & -0.315 \; d & 1.743 \; d \\
60° & 1.732 \; d & -15°   & -0.268 \; d & 2.000 \; d \\
65° & 2.145 \; d & -12.5° & -0.222 \; d & 2.366 \; d \\
70° & 2.747 \; d & -10.0° & -0.176 \; d & 2.924 \; d \\
75° & 3.732 \; d & -7.5°  & -0.132 \; d & 3.864 \; d \\
80° & 5.671 \; d & -5°    & -0.087 \; d & 5.759 \; d \\
85° & 11.430 \; d & -2.5° & -0.044 \; d & 11.474 \; d \\
\end{array}$$


问:
Finite order operators has pure point spectrum
I have to prove that the spectrum of an operator $T:H\rightarrow H$, $H$ Hilbert, s.t. $T^k=id_H$ has pure point spectrum. Now, it is obvious that $\sigma_0(T)\subseteq\{e^{i\mu2\pi/k}\}_{\mu=0}^{k-1}$ is the point spectrum of $T$ and $\sigma_0(T)\subseteq\sigma(T)$. However, I cannot find out how to prove that the two spectra coincide. I guess one has to prove that $T-\lambda{id_H}$ is invertible whenever $\lambda\notin\sigma_0(T)$. I thought I can do it in this way: $T-\lambda id_H$ is clearly injective under the assumptions on $\lambda$ (otherwise $\lambda\in\sigma_0(T)$). To prove that it's surjective, one may use the fact that $H=ker(T-\lambda id_H)\oplus Im(T-\lambda id_H)=\{0\}\oplus Im(T-\lambda id_H)=Im(T-\lambda id_H)$ (the equalities are isometric equivalences). Then, the open mapping theorem would imply the continuity of $(T-\lambda id_H)^{-1}$. Is it correct?

However, I'd like to avoid to invoke the open mapping theorem. Is there an easy way to do so?

答: Hint: You can directly write down the inverse of $T -\lambda I$: Compute $(I+S+S^{2}+\cdots +S^{k-1})) (T-\lambda I)$ where $S=\frac T {\lambda}$. You will see that $T-\lambda I$ is invertible whenever $\lambda ^{k} \neq 1$ and $\lambda \neq 0$.


问:
Calculate how $\mathbb{E}[V|X]$ distributed.
Given $X,Y$ i.i.d where $\mathbb{P}(X>x)=e^{-x}$ for $x\geq0$ and $\mathbb{P}(X>x)=1$ for all $x<0$
and $V=\min(X,Y)$

Calculate how $\mathbb{E}[V|X]$ distributed.

I've found that $F_{V|X=x}(v)=\left\{\begin{array}{rcl} 0&t\leq0\\1-e^{-t}&0\leq t\leq x\\1&else\end{array}\right.$

And I've tried using the formula $\mathbb{E}[V|X]=\int_{\infty}^{\infty}vf_{V|X=x}dv$ and I got that $\mathbb{E}[V|X]=-xe^{-x}-e^{-x}+1$ and in the answer I had to compare with they got $\mathbb{E}[V|X]\sim U(0,1)$

Not sure how to get to this distribution any help?

答:
$$E[V|X] = E[X|X, Y\ge X]P(Y\ge X|X) + E[Y|X, Y<X]P(Y<X|X) = X\int_{X}^{\infty}e^{-y}dy + \int_{0}^Xye^{-y}dy = Xe^{-X} + 1 - e^{-X}(1+X) = 1 - e^{-X}$$
Which means that $E[V|X]\sim U(0, 1)$ since $F_{X}(X)$ is always distributed $U(0, 1)$ if $F_X(x)$ is a CDF of r.v. $X$ which obviously is the case here.


问:
$\mu(E_{\epsilon})<+\infty$ such that if $F \cap E_{\epsilon} = \emptyset$, then $\left \| f \chi_{F} \right \|_{p} < \epsilon$.
Let $f \in L_{p}(X,\Sigma,\mu)$, $1 \leq p < \infty$ and $\epsilon>0$. There exists a set $E_{\epsilon} \in \Sigma$ with $\mu(E_{\epsilon})<+\infty$ sucht that if $F \cap E_{\epsilon} = \emptyset$, then $\left \| f \chi_{F}  \right \|_{p} < \epsilon$.

I tried to construct the set $E_{\epsilon}=\left \{ X \setminus A : A \in \Sigma, \left \| f \chi_{A}  \right \|_{p} < \epsilon   \right \}$, then it satisfies the above property, but wasn't able to show that this set has finite measure. Actually, $E_{\epsilon}$ defined as above may not even belong to the $\sigma$-algebra $\Sigma$ when there is an uncountable quantity of sets $A$.

This is the exercise 6.P in "Elements of Integration and Lebesgue Measure", R Bartle.

答:
Let $E_n=\{x\in X: |f(x)|\geq \frac{1}{n}\}$ and observe that:
$$\infty > \int_{E_n} |f|^p d\mu \geq \frac{1}{n} \mu (E_n) $$
thus $\mu(E_n) < \infty$ for each $n\in\mathbb{N}$. We show that:
$$\lim_{n\rightarrow\infty} \int_{E_n} |f|^p d\mu = \int |f|^p d\mu $$
We have that $|f|^p\chi_{E_n}\leq|f|^p$ for each $n\in\mathbb{N}$ and, also, $|f|^p\chi_{E_n}\rightarrow|f|^p$. Thus, by the dominated convergence theorem we get the above expression.
Finally, given $\epsilon>0$, there exists $N\in\mathbb{N}$ such that:
$$\epsilon^p + \int_{E_N} |f|^p d\mu > \int |f|^p d\mu $$
by taking $E_{\epsilon}=E_N$, if $F\cap E_{\epsilon}=\emptyset$, then:
$$\epsilon^p + \int_{E_\epsilon} |f|^p d\mu \geq \int |f|^p d\mu \geq \int_{E_{\epsilon}\cup F} |f|^p d\mu = \int_F |f|^p d\mu + \int_{E_{\epsilon}} |f|^p d\mu \implies$$
$$\epsilon^p > \int_F |f|^p d\mu$$


问:
True or false: $x^{T}A^{T}Ax\geq 0$
I have this problem:

For matrix with $m\times n$ and for every $x\in \mathbb R^{n}$ :

$x^{T}A^{T}Ax\geq 0$

Any tip how to start?

I am thinking about semi-positive matrices, but I am not sure.

My try:

$x^{T}A^{T}Ax=(Ax)^{T}(Ax)$

答:
Hint:

$x^TA^TAx = (Ax)^T Ax$. Now use a well known property of the inner product.

Title: True or false: $x^{T}A^{T}Ax\geq 0$


问:
Cyclotomic polynomial dividing $f\in\mathbb{Q}[X]$
I know that $\varphi (n)\geq \sqrt{\frac{n}{2}}$ for natural numbers $n$, and that the cyclotomic polynomials are irreducible in $\mathbb{Q}[X]$. Let $f\in\mathbb{Q}[X]$ be a polynomial of degree $m$ with root $\alpha\in\mathbb{C}$, where $\alpha ^k=1$ for some $k\in\mathbb{N}\backslash \{ 0\}$. I must show that there is an $l\leq 2m^2$ s.t. $\Phi_l\mid f$.

I understand that if $\Phi_l\mid f$ then $l\leq 2m^2$, but I can't show that $\Phi_l\mid f$. Do you have any hints?

答:
You've pretty much got all the right ingredients -- all you need to do is put them together in the right order.

Since $\mathbb Q[X]$ is a unique factorization domain (well, all we need is that it's a GCD domain), we can find $\gcd(f,\Phi_\ell)$ for various $\ell$. Since $\Phi_\ell$ is irreducible, this is either $\Phi_\ell$ or a constant: what happens if it's a constant for all $\ell$?


问:
Automorphisms of the Petersen graph
I am trying to find out the automorphism group of the Petersen graph. My book carries the hint: "Show that the $\tbinom{5}{2}$ pairs from {1, . . . , 5} can be used to label the vertices in such a way that a simple rule determines when there is an edge. To find the full automorphism group, consider the subgroup that fixes a vertex and its three neighbors."

Accordingly the rule is that there is an edge if 2-sets are disjoint. What I am not getting is to how to use the second sentence of the hint to find the automorphism group. Its not at all clear what the order of such a subgroup will be and how such a subgroup will be the requisite automorphism group.

Also I am not really proficient with group action, so if that is involved I would appreciate a detailed explanation.

答:
Since each vertex is labeled with a subset with two elements of $\{1,2,3,4,5\}$, then any permutation of $\{1,2,3,4,5\}$ is going to induce a permutation of the vertices. Moreover, if $\{a,b\}\cap\{c,d\}=\emptyset$, and $\sigma$ is a permutation of $\{1,2,3,4,5\}$, then $\{\sigma(a),\sigma(b)\}\cap\{\sigma(c),\sigma(d)\} = \emptyset$. That is: this permutation of the vertices sends adjacent vertices to adjacent vertices (and non-adjacent vertices to non-adjacent vertices).

Also, two permutations induce the same permutation of the vertices if and only if they are identical permutations (you should prove this). That means that every element of $S_5$ induces a distinct automorphism of the graph.

Are these all the automorphisms, or are there more? If $\tau$ is any automorphism, then by composing with an appropriate permutation of $\{1,2,3,4,5\}$ you may assume that the map fixes $\{1,2\}$; that means that the vertices adjacent to $\{1,2\}$, $\{3,4\}$, $\{3,5\}$, and $\{4,5\}$, must map to each other. Composing with an appropriate permutation (that fixes $1$ and $2$) you may assume that the permutation also fixes $\{3,4\}$; and composing again by a suitable permutation, you can make it fix also $\{3,5\}$ $\{4,5\}$ (again, you need to prove this). So then you are left with an automorphism that fixes $\{1,2\}$ and its three neighbors. If you can show that this is also given by a permutation of $\{1,2,3,4,5\}$, then you will have shown that every automorphism "comes" from an element of $S_5$ (since composing with suitable automorphism that do give you the identity).

Title: Automorphisms of the Petersen graph


问:
Understanding the (Partial) Converse to Cauchy-Riemann
We have that for a function $f$ defined on some open subset $U  \subset \mathbb{C}$ then the following if true:

Suppose $u=\mathrm{Re}(f), v=\mathrm{Im}(f)$ and that all partial derivatives $u_x,u_y,v_x,v_y$ exists and are continuous on $U$. Suppose further that they satisfy the Cauchy-Riemann equations. Then $f$ is holomorphic on $U$.

The proof for this is readily available, though there is a subtlety that I can't understand.

We essentially want to compute $\lim_{h \rightarrow 0} \dfrac{f(z+h)-f(z)}{h}$ where $h=p+qi \in \mathbb{C}$.

We need a relationship like $u(x+p,y+q)-u(x,y)=pu_x(x,y)+qu_y(x,y)+o(|p|+|q|)$. Why is this relationship true?

答:
One way to look at this is by interpreting $f$ as a function from $\mathbb{R}^2 \to \mathbb{R}^2$. This allows you to apply whatever you know about differentiability in $\mathbb{R}^2$ to $\mathbb{C}$. In other words, you interpret $f$ as  function $$
  f \,:\, (u,v) \to \left(\textrm{Re}(f(u+iv)), \textrm{Im}(f(u+iv)\right) =: (f_1(u,v),f_2(u,v)) \text{.}
$$

If all the partial derivatives $\frac{\partial f_1}{\partial u}$, $\frac{\partial f_1}{\partial v}$, $\frac{\partial f_2}{\partial u}$, $\frac{\partial f_2}{\partial v}$ exist and are continuous, $f$ is differentiable in the $\mathbb{R}^2$-sense, i.e. for every $\mathbf{x}=(u,v)$ there's a linear function $df$ which approximates $f$ around $\mathbf{x}$. In other words, $$\begin{aligned}
  f(\mathbf{x} + \Delta) &= f(\mathbf{x}) + df_{\mathbf{x}}\cdot \Delta + R_{\mathbf{x}}(\Delta)
 \quad\text{where}\quad
  \lim_{\Delta\to 0} \frac{\left|\left| R_{\mathbf{x}}(\Delta) \right|\right|}{||\Delta||} = 0
  \quad\text{and}\quad \\
  df_{\mathbf{x}} &= \left(\begin{matrix}
  \frac{\partial f_1}{\partial u} &\frac{\partial f_1}{\partial v} \\
  \frac{\partial f_2}{\partial u} &\frac{\partial f_2}{\partial v}
  \end{matrix}\right) \text{.}
\end{aligned}$$

If the cauchy-riemann differential equations hold, $df_\mathbf{x}$ is then a scaled rotation, i.e. there are $A_\mathbf{x}$ and $B_\mathbf{x}$ such that $$
  df_\mathbf{x} = \left(\begin{matrix}
  A_\mathbf{x}& B_\mathbf{x} \\
  -B_\mathbf{x}& A_\mathbf{x}
  \end{matrix}\right) \text{.}
$$
This form of $df_\mathbf{x}$ allows you to translate back into $\mathbb{C}$, because its exactly these matrices which correspond to complex multiplication. Simply set $$
  f'(z) = f'(u+iv) = A_{(u,v)} + iB_{(u,v)} \text{.}
$$
and translate the approximation back to $\mathbb{C}$, i.e. $$
  f(z+\Delta) = f(z) + f'(z)\cdot\Delta + R_z(\Delta)
  \quad\text{where}\quad
  \lim_{\Delta\to 0} \frac{\left| R_{\mathbf{x}}(\Delta) \right|}{|\Delta|} = 0 \text{,}
$$
which is exactly what you want to show.


答: Let $y = Ax$. Then, $x^\top A ^\top A x = y^\top y = \Vert y \Vert_2^2 \geq0$

Title: True or false: $x^{T}A^{T}Ax\geq 0$

Body:
I have this problem:

For matrix with $m\times n$ and for every $x\in \mathbb R^{n}$ :

$x^{T}A^{T}Ax\geq 0$

Any tip how to start?

I am thinking about semi-positive matrices, but I am not sure.

My try:

$x^{T}A^{T}Ax=(Ax)^{T}(Ax)$


问:
Convergence of sequence of exponential matrices.
If $\{A_k\}$ is a sequence of $n \times n \hspace{1mm}$  matrices so that $\sum_{i,j=1}^n|a_{k_{ij}}|^2 < \infty$  for every $k$, i.e sum of squares of all the entries is finite for each matrix of the sequence. How do we ensure or what will be the criterion for convergence of $\exp(A_k)$ to say $\exp(A)$?

We may assume the norm as $$||B||=\sum_{i,j=1}^n|b_{ij}|^2$$ where $B=(b_{ij})_{1\leq i,j \leq n}$. The definition I am using for $\exp(A)$ is
$$\exp(A)=I+A+\frac{A^2}{2!}+\cdots+\frac{A^n}{n!} +\cdots $$
The above definition of $\exp(A)$ may complicate the proof of convergence hence feel free to share any other definition of $\exp(A)$.

答:
If $A_k\to A$ then $e^{A_k}\to e^A$.

Proof: (In what follows, any norm that satisfies $\|AB\|\le\|A\|\|B\|$ can be used.) $$B^n-A^n=B^{n-1}(B-A)+B^{n-2}(B-A)A+\cdots+B(B-A)A^{n-1}+(B-A)A^{n-1}$$
\begin{align}\therefore\ \|B^n-A^n\|&\le(\|B\|^{n-1}+\|B\|^{n-2}\|A\|+\cdots+\|A\|^{n-1})\|B-A\|\\
&\le {nc}^{n-1}\|B-A\|\end{align} where $c=\max(\|A\|,\|B\|)$.

Turning to the exponential,$$e^B-e^A=(B-A)+\tfrac{1}{2!}(B^2-A^2)+\cdots$$
$$\therefore\ \|e^B-e^A\|\le\|B-A\|(1+\tfrac{2c}{2!}+\tfrac{3c^2}{3!}+\cdots+\tfrac{c^k}{k!}+\cdots)=e^c\|B-A\|$$ So, given any $0<\epsilon<1$, let $k$ be large enough that $\|A_k-A\|<\epsilon$, so $\|A_k\|<\|A\|+\epsilon\le\|A\|+1$; then the above shows $\|e^{A_k}-e^A\|\le e^{\|A\|+1}\epsilon$, which can be made arbitrarily small.


问:
Prove the polynomial $x^4+4 x^3+4 x^2-4 x+3$ is positive
Given the following polynomial
$$
x^4+4 x^3+4 x^2-4 x+3
$$
I know it is positive, because I looked at the graphics

and I found with the help of Mathematica that the following form
$$
(x + a)^2 (x + b)^2 + c^2(x + d)^2 + e^2
$$
can represent the polynomial with the following values for the constants
$$
\left(x-\frac{1}{2}\right)^2
   \left(x+\frac{5}{2}\right)^2+\frac{5}{2} \left(x+\frac{1}{5}\right)^2+\frac{107}{80}
$$

I suppose there are simpler ways to prove that the polynomial is positive, perhaps by using some inequalities.

Please, advice.

答:
if $x\ge 0$ by AM-GM $$x^4+3=x^4+1+1+1\ge 4x$$  and the inequality $$\color{blue}{x^4+3-4x}+4x^3+4x^2>0$$ is obvious as equality of am-gm and $x=0$ is simultaneously is not achieved.

If $x\le 0$ then replace $x$ by $-t $ and we have to prove $$t^4-4t^3+4t^2+4t+3>0$$which is true as $$t^4+4t^2\ge 4t^3$$ by AM-GM and as  $t\ge 0$ and the rest of terms are postive (note :again equality of both AM-GM and t=0 is not achieved)


答:
The conditional pdf $f_{V\mid X}$ that you write is not defined since the joint density $f_{V,X}$ does not exist wrt Lebesgue measure. This is because $V=X$ has a positive probability.

You can write  $$V=\min(X,Y)=X\mathbf1_{X<Y}+Y\mathbf1_{X>Y}\,,$$

where $\mathbf1_A$ is an indicator variable.

Now

\begin{align}
\mathbb E\left[V\mid X\right]&=\mathbb E\left[X\mathbf1_{X<Y}\mid X\right]+\mathbb E\left[Y\mathbf1_{X>Y}\mid X\right]
\\&=X\mathbb E\left[\mathbf1_{X<Y}\mid X\right]+\mathbb E\left[Y\mathbf1_{X>Y}\mid X\right]
\end{align}

For fixed $x>0$, we have $$\mathbb E\left[\mathbf1_{x<Y}\right]=\mathbb P\left(Y>x\right)=e^{-x}$$ and $$\mathbb E\left[Y\mathbf1_{x>Y}\right]=\int_0^xye^{-y}\,\mathrm{d}y=1-e^{-x}-xe^{-x}$$

This suggests that $$\mathbb E\left[V\mid X\right]=Xe^{-X}+1-e^{-X}-Xe^{-X}=1-e^{-X}$$

You can verify this has a uniform distribution on $(0,1)$.

Body:
Given $X,Y$ i.i.d where $\mathbb{P}(X>x)=e^{-x}$ for $x\geq0$ and $\mathbb{P}(X>x)=1$ for all $x<0$
and $V=\min(X,Y)$

Calculate how $\mathbb{E}[V|X]$ distributed.

I've found that $F_{V|X=x}(v)=\left\{\begin{array}{rcl} 0&t\leq0\\1-e^{-t}&0\leq t\leq x\\1&else\end{array}\right.$

And I've tried using the formula $\mathbb{E}[V|X]=\int_{\infty}^{\infty}vf_{V|X=x}dv$ and I got that $\mathbb{E}[V|X]=-xe^{-x}-e^{-x}+1$ and in the answer I had to compare with they got $\mathbb{E}[V|X]\sim U(0,1)$

Not sure how to get to this distribution any help?


答:
Here's a slightly more motivated way to find a sum-of-squares representation.

We should want the first term to be the square of a quadratic with leading coefficient $1$, and to cancel out the $x^3$ term we need it to be $(x^2+2x+a)^2$ for some real $a$. This will leave us to show that
$$-2ax^2-(4a+4)x+(3-a^2)\geq 0;$$
the simplest case of this is to choose $a=-1$ (to remove the $x$ term entirely), which gives us that
$$x^4+4x^3+4x^2-4x+3=(x^2+2x-1)^2+2x^2+2>0.$$

Title: Prove the polynomial $x^4+4 x^3+4 x^2-4 x+3$ is positive

Body:
Given the following polynomial
$$
x^4+4 x^3+4 x^2-4 x+3
$$
I know it is positive, because I looked at the graphics

and I found with the help of Mathematica that the following form
$$
(x + a)^2 (x + b)^2 + c^2(x + d)^2 + e^2
$$
can represent the polynomial with the following values for the constants
$$
\left(x-\frac{1}{2}\right)^2
   \left(x+\frac{5}{2}\right)^2+\frac{5}{2} \left(x+\frac{1}{5}\right)^2+\frac{107}{80}
$$

I suppose there are simpler ways to prove that the polynomial is positive, perhaps by using some inequalities.

Please, advice.


问:
A question about the set of pairs $(x,y)$ such that $0<x<y<p/2$ and $p\mid x^2+y^2$ for a fixed prime number $p$
Like the title, fix a prime $p$. Consider the pairs of integers $(x,y)$ such that
$$
x^2+y^2=0 \bmod p \quad x \leq y \leq (p-1)/2.
$$

I find some property

$p$ needs to be $4k+1$

There exists an element $a$ such that $a^2=-1$

The sum of all set $y$ is twice as large as the sum of all set $x$

for example:$\mathbb Z_{13} $

$1^2+5^2=0$ mod $p$,

$2^2+3^2=0$ mod $p$ ,

$4^2+6^2=0$ mod $p$

then $(1+2+4)\cdot 2=5+3+6$

I know the 1) 2) is about Euler Criterion, but I have no idea about 3).

Please give me some hint for 3). Thanks.

答:
A hint as requested - not a full proof

Consider a pair $(x,y)$ satisfying your opening condition.

Then $(y-x)^2+(y+x)^2=0\mod p$ and $(y-x)^2+(p-y-x)^2=0\mod p$.

Note that  $y-x<y+x$ and $y-x<p-y-x$. We can then choose whichever of $y+x$ and $p-y-x$ is less than or equal to $(p-1)/2$.

We have therefore constructed a pair $(x-y,*)$ satisfying your opening condition.

Now let $X,Y,D$ be the sets of $x,y,y-x$, respectively, such that $(x,y)$ satisfies your opening condition. The above argument suggests that $X$ and $D$ are the same sets and then property 3 will automatically hold.

Example

You can see the $X=D$ result in your example, where both sets consist of $1,2,4$.


问:
Laplacian in a spherical coordinate system question
Q: Consider a function f = r^2 given in the spherical coordinate system. Using the expression for the scale factors, as well as the formula for the Laplacian compute the Laplacian ∇^2 f . The result is a number times r^(n-2) . State this number.

Not sure where I've gone wrong with this, I think I'm close to the answer but I can't seem to get r^(n-2), or get rise of cosec(theta). I'll show my workings:

f = r^2 hr = 1 htheta = r hphi = rsintheta

grad f = 1/hr * ∂f/∂r * er + 1/htheta * ∂f/∂theta * etheta + 1/hphi * ∂f/∂phi * ephi

= 2r er

∇ . A = 1/h1h2h3 [∂(h2h3A1)∂u1 + ∂(h1h3A2)/∂u2 + ∂(h1h2A3)/∂u3]

∇ = 2rer = A1e1 where A1 = Ar = 2r

∇ . A = 1/hrhthetahphi (∂(r * rsintheta * A1)/∂r + ∂(1 * rsintheta * A2)/∂theta + ∂(1 * r * A3)/∂phi)

Now A2 = A3 = 0, A1 = 2r

∇ . A = 1/r^2*sin^2theta * ∂/∂r (r^2 * sin(theta) *2r)

= 1/r^2*sin^2theta * ∂/∂r (2r^3 sin(theta))

= 1/r^2*sin^2 (theta) * 6r^2 sin(theta)

= 6/sin(theta)

Now I know that the correct answer is 6, so am I correct? Is my working correct?

答:
I must admit I find our OP Sam Connell's question a little unclear; nevertheless I feel I can make enough sense out of it to give it a try, so here goes:

We first find $\nabla^2 r^2$ in three dimensions:

By definition we have

$\nabla^2 r^2 = \nabla \cdot (\nabla r^2); \tag 1$

it is both well-known and easy to see (e.g. by working in cartesian coordinates) that

$\nabla r^2 = \dfrac{d(r^2)}{dr} \nabla r = 2r \nabla r; \tag 2$

$\nabla r = \dfrac{\partial r}{\partial r} \hat e_r = \hat e_r, \tag 3$

where $\hat e_r$ denotes the unit vector field tangent to the radial lines and pointing in the outward direction; combining (2) and (3) yields

$\nabla r^2 = 2r \hat e_r; \tag 4$

thus,

$\nabla^2 r^2 = \nabla \cdot (2r \hat e_r) = \nabla (2r) \cdot \hat e_r + 2r \nabla \cdot \hat e_r$
$= 2\nabla r \cdot \hat e_r + 2r \nabla \cdot \hat e_r = 2 \hat e_r \cdot \hat e_r + 2r \nabla \cdot \hat e_r = 2 + 2r \nabla \cdot \hat e_r, \tag 5$

where we have exploited the well-known identity $\nabla \cdot (fX) = \nabla f \cdot X + f \nabla \cdot X$, holding for any differentiable function $f$ and vector field $X$; here $f = 2r$ and $X = \hat e_r$; now,

$\hat e_r = r^{-1} \vec r, \tag 6$

and it is easy to compute that

$\nabla \cdot \vec r = 3, \tag 7$

since $\vec r = (x, y, z) \in \Bbb R^3$; therefore,

$\nabla \cdot \hat e_r = \nabla \cdot (r^{-1} \vec r) = \nabla (r^{-1}) \cdot \vec r + r^{-1} \nabla \cdot r$
$= -r^{-2} \hat e_r \cdot \vec r + r^{-1}\nabla \cdot \vec r$
$= -r^{-2} r + 3r^{-1} = 2r^{-1}; \tag 8$

$\nabla^2 r^2 = 2 + 2r\nabla \cdot \hat e_r = 2 + 2r(2r^{-1}) = 6. \tag 9$


问:
Coin betting problem.
A coin bet is placed between two friends such that the person who wins four tosses first is the winner. What is total number of possible ways in which the bet can play out?</p>

答:
Avoiding elegance, there will end up being $k$ tosses, where $k \in \{4,5,6,7\}.$  Therefore, you want $\sum_{k=4}^7 f(k)$, where $f(k)$ is the number of different coin toss sequences that result in the game ending after exactly $k$ coin tosses.

Clearly, $f(4) = \binom{4}{0} + \binom{4}{4} = 2$.

That is, there are $\binom{4}{0}$ ways of getting zero heads and

$\binom{4}{4}$ ways of getting 4 heads.

For a game to end after exactly 5 coin tosses, two things must happen:

One of the two people must be ahead 3 to 1, after 4 coin tosses.

That person must also win the 5th coin toss.

There are $\binom{4}{3}$ ways that exactly 3 of the 1st 4 coin tosses come up heads.  If this happens, for the game to end after exactly 5 coin tosses, with heads the winner, the 5th coin toss must also be heads.  Therefore, there are $4$ coin toss sequences that allow the game to end after exactly 5 coin tosses, with heads the winner.

By symmetry, there are exactly 4 coin toss sequences that allow the game to end after exactly 5 coin tosses, with tails the winner.

Therefore $f(5) = 2 \times \binom{4}{1} = 8.$

For a game to end after exactly 6 coin tosses, two things must happen.

After 5 coin tosses, someone is ahead 3 to 2.

On the 6th coin tosses, the person who was ahead also wins the 6th toss.

By arithmetic very similar to the computation of $f(5)$, you have that

$f(6) = 2 \times \binom{5}{3} = 20.$

For the game to end on exactly 7 tosses, the game must be tied after 6 tosses, 3 to 3.  This can happen in $\binom{6}{3}$ ways.  In that event, the 7th toss might be either heads or tails.

Therefore, $f(7) = 2 \times \binom{6}{3} = 40.$

Final answer:

$$\sum_{k=4}^7 f(k) = 2 + 8 + 20 + 40 = 70.$$

Addendum

The answer may be re-expressed as

$$2 \times \left[\binom{3}{3} + \binom{4}{3} + \binom{5}{3} + \binom{6}{3}\right].$$

Examining Pascal's triangle, and its rule, re

$\binom{n}{k} + \binom{n}{k+1} = \binom{n+1}{k+1},$

$\binom{3}{3} + \binom{4}{3} = \binom {4}{4} + \binom{4}{3} = \binom{5}{4}.$

Then,

$\binom{5}{4} + \binom{5}{3} = \binom{6}{4}.$

Then,

$\binom{6}{4} + \binom{6}{3} = \binom{7}{4}.$

Addendum-1
As the other responses indicate, it is not a coincidence that the computation yields $2 \times \binom{7}{4}.$
Originally, I wasn't able to grasp the analysis.  Then it hit me.

It is true that you can't simply assume that the contest will go 7 coin flips, since for example, you will never have one person getting 5 heads out of 7.  That is, the contest ends before the person can get the 5th head.

However: assume that the person flipping heads wins.  There is a bijection between each of the winning sequences (which will be of length 4,5,6, or 7) and all the possible ways that exactly 4 of the 7 coin tosses come up heads [i.e. $\binom{7}{4}$].

To see this, pretend that if the game finishes after $k$ coin tosses, with Heads the winner and $k < 7$, then the coin toss sequence is continued with exactly $(7 - k)$ trailing tails.  Assuming that the total number of Heads remains constant at 4, there is only 1 such continuation possible, since the remaining $(7 - k)$ pretended coin tosses must all be tails.

Because of this bijection, the number of sequences with Heads the winner must equal $\binom{7}{4}.$

Title: Coin betting problem.


问:
Question on simple modules and vector spaces
Let $F$ be a field and let $R = F[x]$.

If $V$ is a finite-dimensional $F$-vector space and $f:V \rightarrow V$ is a linear map, then explain how to make $V$ into a left $R$-module via $x\cdot v = f(x)$ for any $v \in V$.

Let $F$ be algebraically closed. A left $R$-module $V$ is called simple if the only submodules are {0} and $V$. Show that if $V$ has dimension 2 or more, then it cannot be simple.

Let $F = \mathbb{R}$ and let $V$ be a 2-dimensional $\mathbb{R}$-vector space. Give an explicit $f:V \rightarrow V$ such that $V_f$ is a simple $R$-module. This shows being algebraically closed in the previous part of the problem is required.

I got (1) done.

For (2) I know that the submodules of $V$ correspond to the $f$-stable subspaces of $V$ i.e. $W \subseteq V$ such that $f(W) \subseteq W$. Would the Rank-Nullity theorem somehow be applicable here?

I'm not sure how to go about starting (3).

I appreciate any help. Thanks in advance.

答:
For (2), let $\lambda$ be an eigenvalue of $f$ (it must exist, since $F$ is algebraically closed) and consider the subspace $Fv$. It's a submodule, distinct from $\{0\}$ and from $V$, and therefore $V$ is not simple.

For (3), consider$$\begin{array}{rccc}f\colon&\Bbb R^2&\longrightarrow&\Bbb R^2\\&(x,y)&\mapsto&(-y,x).\end{array}$$If it had a submodule distinct from $\{0\}$ and from $V$, it would be $1$-dimensional, and therefore it would be an eigenspace. But $f$ has no real eigenvalues.


答:
Let the polynomial be denoted by $f(x)$.
$f'(x)=4(x^3+3x^2+2x-1)$ has only one real root $\approx 0.3247$. At that point $f''(x)=4(3x^2+6x+2)>0$. Thus $f(x)$ has minimum value at $\approx 0.3247$ and $f(0.3247)\approx 2.271$.

$\therefore f(x)>0 \;\forall x\in \mathbb R$.


问:
Basic question with tiling tiles in a rectangle
Find the number of ways of tiling a 2 by 10 rectangle with 1 by 2 tiles which can be placed in any orientation.

I’m just not sure how to approach it begin counting it

答:
Since I already have a tiling program it's a 2 minute job to plug this in and get the answer '51'. If I didn't have a tiling program it would take maybe an hour to write a simple one.

Clarification: There are 89 ways to tile if you count rotations/reflections as distinct tilings. My program generates all 89 tilings, and for each it generates all distinct rotations and reflections, then assigns each a canonical score and discards tilings which aren't the smallest canonically.


问:
How to know a function is integrable or not?
Let say
$$ h(x) = \begin{cases}x^2,& x \in \mathbb {Q}\\-x^2,& x \notin \mathbb{Q}\end{cases}
           $$

Is there a difference between riemann integrable and integrable?

And can I just say contiuous function is integrable and hence the above function is not
integrable in [-1,1]?

答:
A function is Riemann Intrgrable if and only if the set of its points
  of discontinuity has measure zero.

In your case, the function $h$ is everywhere discontinuous in $[-1,1]$, except from zero. The measure of $[-1,1]\setminus \{0\}$ which is the set of the points of discontinuity of $h$, is $1\neq 0$. Therefore the function is not Riemann integrable.

When we refer to an integrable function in the notions of measure theory, we usually mean a measurable function $f$, defined on a measure space $X$ such that $\int_X|f(x)|dx<\infty$ (where the last integral is Lebesgue integral). In your case the function $h$ is measurable, and bounded by $1$. Therefore:
$$\int_{[-1,1]}|h|\leq \int_{[-1,1]}1=2<\infty$$ so the function is Lebesgue integrable


问:
Converting job assignment problem into a 0-1 linear programming problem

There are $n$ people who need to be assigned to execute $n$ jobs, one person per job. (That is, each person is assigned to exactly one job and each job is assigned to exactly one person.) The cost that would accrue if the $i$-th person is assigned to the $j$-th job is a known quantity $C[i, j]$ for each pair $i, j = 1, \dots , n$. The problem is to assign the people to the jobs to minimize the total cost of the assignment. Express the assignment problem as a $0$-$1$ linear programming problem.

Let $M$ be an $n \times n$ matrix where the value at $M_{ij}$ indicates the cost of assigning person $i$ to job $j$, or $C[i, j]$.

Let $x$ be an $n \times n$ matrix where the value at $x_{ij}$ denotes whether person $i$ was assigned to job $j$. If the assignment occurred, $x_{ij}$ is 1; otherwise, it is 0.

I've managed to come up with these definitions, but unfortunately I'm not sure how to proceed with the problem. The canonical form of a linear programming problem is as follows:

maximize $c^Tx$, subject to $Ax\leq b$, and $x\geq 0$

However, it seems the goal of this problem is to minimize, not to maximize. How do I convert this minimization problem into a maximization problem? Furthermore, how would the constraints be defined by this sort of problem? Are my definitions adequate for the problem?

答:
The required LP is

$\begin{align}
\min & \sum_{i=1}^n\sum_{j=1}^n M_{ij} x_{ij} \\
& \sum_{i=1}^n x_{ij} = 1 \quad \forall j \in \{1,\dots, n\} \tag{each job is assigned to exactly one person} \\
& \sum_{j=1}^n x_{ij} = 1 \quad \forall i \in \{1,\dots, n\} \tag{each person is assigned to exactly one job} \\
& x_{ij} \in \{0,1\} \quad \forall i,j \in \{1, \dots, n\}
\end{align}$

To understand the objective function, for each job $j$, there exist unique $i_0 \in \{1, \dots, n\}$ such that $x_{ij} = \begin{cases} 1 &\text{if } i = i_0 \\ 0 &\text{otherwise,} \end{cases}$ so the cost for the $j$-th job is $M_{i_0j} = \sum
\limits_{i=1}^n M_{ij}$.  Sum the previous equality over $j \in \{1,\dots,n\}$ to get the above double sum.

How do I convert this minimization problem into a maximization problem?

To minimize a quantity is equivalent to maximizing its additive inverse.

Furthermore, how would the constraints be defined by this sort of problem?

Use the standard technique of transforming an equality constraint to a set of two inequality constraints:

$$x = a \iff \begin{cases} - x &\le -a \\ x &\le a. \end{cases}$$

Are my definitions adequate for the problem?

They are more than enough.  You don't need to define $M$, since you already have $C$, which represents the costs.

To write the LP is a more concise matrix form, denote

$\mathbf{1} \in \{0,1\}^n$ the $n \times 1$ vector filled with $1$.

$C = (C[i,j])_{ij} = (c_{ij})_{ij} \in \mathbb{R}^{n \times n}$: the matrix $C$ is defined by $C[i,j]$ as the $(i,j)$-th element, so that

$(C^T)_{ij} = c_{ji}$: the $(i,j)$-th element of $C^T$ is the $(j,i)$-th element of $C$.

$X = (x_{ij})_{ij} \in \{0,1\}^{n \times n}$

Remarks: I prefer using capital letters for matrices, and use small letters with a subscript to denote the corresponding matrix elements.

Note that the cost function is
\begin{align}
\sum_{i=1}^n \sum_{j=1}^n C[i,j] x_{ij} &= \sum_{i=1}^n \sum_{j=1}^n (C^T)_{ji} x_{ij} \tag{take transpose of $C$} \\
&= \sum_{j=1}^n \left( \sum_{i=1}^n (C^T)_{ji} x_{ij} \right) \tag{change order of summation} \\
&= \sum_{j=1}^n \left( C^T X \right)_{jj} \tag{defintion of matrix product} \\
&= \mathop{\mathrm{tr}}(C^T X). \tag{definition of trace}
\end{align}
$\sum\limits_{i=1}^n x_{ij} = 1$ can be rewritten as $\mathbf{1}^T X = \mathbf{1}^T$; $\sum\limits_{j=1}^n x_{ij} = 1$ can be rewritten as $X \mathbf{1} = \mathbf{1}$.
Hence the LP in matrix form is $\min\{\mathop{\mathrm{tr}}(C^T X) \mid \mathbf{1}^T X = \mathbf{1}^T, X \mathbf{1} = \mathbf{1}, X \in \{0,1\}^{n \times n} \}$


问:
Divisor Arithmetic
The product of the proper positive integer factors of $n$ can be written as $n^{(ax+b)/c}$, where $x$ is the number of positive divisors $n$ has, $c$ is a positive integer, and the greatest common factor of the three integers $a$, $b$, and $c$ is 1. What is $a+b+c$?

I know that this works with the divisors of 24
24 has 8 divisors
So x=8, a=1, b=10, c=6
$24^3=13824=24^{(8*1+10)/6}$

$\gcd(1,6,10)$

$a+b+c=1+10+6=17$

But it also works with 60
which with 60, $a+b+c=19$
I'm not sure what would be my final answer could someone help me? (also it seems like this could have infinite solutions)

答:
You just showed one case where $n=24$ where it will naturally have infinite solutions. Look for a solution in all cases. Spoiler below.

$a=1,b=-2,c=2$ solves all the cases. So $a+b+c=1$ is the solution.

Title: Divisor Arithmetic


问:
Why is $\sum_{k=1}^{\infty} \frac{(-1)^{k}}{(2 k) !}=\left.\left(\sum_{k=0}^{\infty} \frac{(-1)^{k}}{(2 k) !} x^{2 k}-1\right)\right|_{x=1}$?
In order to sum up the series we have to realize that it is almost the expansion of the cosine function. But we need $k=0$ in the sum. I don't know how to do this. The solution is   $\sum_{k=1}^{\infty} \frac{(-1)^{k}}{(2 k) !}=\left.\left(\sum_{k=0}^{\infty} \frac{(-1)^{k}}{(2 k) !} x^{2 k}-1\right)\right|_{x=1}$, but where does $-1$ on the right hand side come from and how did we achieve $k=0$?

My approach was to do the following, which leads to a dead end...(right?)

$\sum_{k=1}^{\infty} \frac{(-1)^{k}}{(2 k) !}=-\sum_{k=1}^{\infty} \frac{(-1)^{k-1}}{(2 k) !}=\cdots$

答:
For a convergent series $ \sum_{k=0}^{\infty}a_k$ we have

$$\sum_{k=0}^{\infty}a_k=a_0+\sum_{k=1}^{\infty}a_k,$$

hence

$$\sum_{k=1}^{\infty}a_k= \sum_{k=0}^{\infty}a_k-a_0.$$


问:
Stochastik of triangle Angle
As an electrical engineer I am interested in the phase shift of two signals. At one specific moment I can identify the phasor of a signal as a vector in the complex plane. To transfer the problem into math:

I've got two 2D-vectors $\vec{a}$ and $\vec{b}$, which both change random in time, but are somehow dependend of each other. I am able to measure the average and the maximum absolute value of the length of both vectors for a certain period of time. Additionally, I am able to measure the average and the maximum of the length of the geometric sum $|\vec{a}+\vec{b}|$ of the two vectors.

Now, I want to use the law of cosines $|\vec{a}+\vec{b}|^2 = |\vec{a}|^2+|\vec{b}|^2-2|\vec{a}||\vec{b}|cos(\varphi)$ to tell something of the statistic of the phase $\pi - \varphi$ between both vectors.

I am kind of worried whether the average of the phase is possible to determine in that way since I killed the phase information previously by taking the absolute values. Is it as simple as that?

答:
If you know $|a+b|$, $|a|$, and $|b|$ at a particular timestep, you can certainly calculate the cosine of the angle between them using (essentially; I think you made a typo) the method you described:
$$\cos(\varphi)=-\frac{|a+b|^2-|a|^2-|b|^2}{2|a||b|}.$$
Unfortunately, if all you know are the average (and maximum) lengths of $a$, $b$, and $a+b$, you can't get any distributional information on $\cos(\varphi)$ (or any other function of $\varphi$) -- it's not necessarily true that
$$\mathbb E[f(x,y,z)] = f(\mathbb E[x], \mathbb E[y], \mathbb E[z])$$
if $f$ is some "complicated" (non-linear) function.

If, however, you are able to make this calculation at every timestep, then you can compute $\cos(\varphi)$ for every timestep, and thus whatever else you want involving the distribution of $\varphi$.

Title: Stochastik of triangle Angle


问:
Surjective map from $\mathbb{Z}^+$ to $\mathbb{Z}$.
Surjective map between $\mathbb{Z^{+}}$ and $\mathbb{Z}$.
I am confused as to how this is even possible to create.

答:
\begin{align}
f(n)=\begin{cases}
 \frac n2 & \text{if n is even}\\
-{n-1\over 2} & \text{if n is odd}
\end{cases}
\end{align}

$f:\mathbb Z^+\mapsto \mathbb Z$ is a bijection and thus a surjection also.


Body:
Just another solution (similar to Albus Dumbledore's).

$$x^4+4x^3+4x^2-4x+3=x^2(x+2)^2-4x+3 = x^4+4x^3+(2x-1)^2 +2$$
For $x<0$ we have $-4x+3>0$ and therefore $x^2(x+2)^2-4x+3>0$.

For $x>0$ it follows from $x^3>0$ that $x^4+4x^3+(2x-1)^2 +2>0$.


问:
Finite order operators has pure point spectrum
I have to prove that the spectrum of an operator $T:H\rightarrow H$, $H$ Hilbert, s.t. $T^k=id_H$ has pure point spectrum. Now, it is obvious that $\sigma_0(T)\subseteq\{e^{i\mu2\pi/k}\}_{\mu=0}^{k-1}$ is the point spectrum of $T$ and $\sigma_0(T)\subseteq\sigma(T)$. However, I cannot find out how to prove that the two spectra coincide. I guess one has to prove that $T-\lambda{id_H}$ is invertible whenever $\lambda\notin\sigma_0(T)$. I thought I can do it in this way: $T-\lambda id_H$ is clearly injective under the assumptions on $\lambda$ (otherwise $\lambda\in\sigma_0(T)$). To prove that it's surjective, one may use the fact that $H=ker(T-\lambda id_H)\oplus Im(T-\lambda id_H)=\{0\}\oplus Im(T-\lambda id_H)=Im(T-\lambda id_H)$ (the equalities are isometric equivalences). Then, the open mapping theorem would imply the continuity of $(T-\lambda id_H)^{-1}$. Is it correct?

However, I'd like to avoid to invoke the open mapping theorem. Is there an easy way to do so?

答:
Using the hint I was given, I prove that $T$ is a pure point spectrum operator. Clearly, $0\notin\sigma(T)$, since $T^k=id_H \ \Longrightarrow \ T^{k-1}=T^{-1}$, that is $T$ is invertible and its inverse is obviously a continuous mapping (as a composition of bounded operators from $H$ into itself). Let $\lambda\in\mathbb{C}\setminus(\{0\}\cup\sigma_0(T))$. Set $T^0=id_H$.
\begin{equation}
\underbrace{\sum_{j=0}^{k-1}\frac{T^j}{\lambda^j}}_{$=:S$}(Tx-\lambda x)=\sum_{j=0}^{k-1}\frac{T^{j+1}x}{\lambda^j}-\frac{T^j(\lambda x)}{\lambda^j}=\frac{1-\lambda^{k}}{\lambda^{k-1}}x=x.
\end{equation}
That is, $\frac{\lambda^{k-1}}{1-\lambda^{k}}S=T^{-1}$ and the assumptions on $\lambda$ imply the well-posedness of the factor $\frac{\lambda^{k-1}}{1-\lambda^{k}}$ and the fact that it does not vanish.


问:
Prove that $\int_{-\infty}^{\infty} \sin x \, dx = 0 $
$$\int_{-\infty}^{\infty} \sin x \, dx$$
When I am doing the proof for this, why do i have to split it into
$\int_{-\infty}^a \sin x \, dx + \int_a^\infty \sin x \, dx $?
where a is a constant </p>

答:
The assertion that the integral is $0$ doesn't really make sense: the convergence of this improper integral requires that both

\begin{gather}
\lim_{a\to-\infty}\int_{a}^{0}\sin x \, dx \\
\lim_{b\to\infty}\int_{0}^{b}\sin x \, dx
\end{gather}

exist and are finite and neither does. The "break point" $0$ is arbitrary and can be any real number. This is by definition of improper integral, at least the most common definition one finds.

The second limit doesn't exist, because if you compute it on the sequences $2n\pi$ or $2n\pi + \pi/2$ you get different limits:

$$
\lim_{n\to\infty}\int_{0}^{2n\pi}\sin x \, dx=
\lim_{n\to\infty}[-\cos x]_{0}^{2n\pi}=0
$$
$$
\lim_{n\to\infty}\int_{0}^{2n\pi+\pi/2}\sin x \, dx=
\lim_{n\to\infty}[-\cos x]_{0}^{2n\pi+\pi/2}=1
$$

In the same way you show that the first limit doesn't exist (just change the variable with $x=-y$).

Therefore, we can't say that $\displaystyle\int_{-\infty}^{\infty}\sin x\,dx$ is equal to a number, much less that it's zero, unless we give the symbol some other meaning than an improper integral.

If you are considering the principal value, but you should clearly specify it, because it's a different thing than an improper integral, in general, you indeed get $0$, because the sine function is odd: $\sin(-x)=-\sin x$, so, for $a>0$,

$$
\int_{-a}^{a}\sin x\,dx = 0
$$

hence

$$
\mathrm{p.v.}\!\!\int_{-\infty}^{\infty}\sin x\,dx=
\lim_{a\to\infty}\int_{-a}^{a}\sin x\,dx = 0
$$

The first equality in the above line is the definition of the principal value integral.


答:
The function $\varphi\colon\mathbb{Z}^+\to\mathbb{Z}$ given by
\begin{align}
\varphi(m)=\begin{cases}
-\frac{m}{2},& m \text{ is even}\\
\frac{m-1}{2}, & m \text{ is odd}
\end{cases}
\end{align}
is such an example. Remark $\varphi=-f$ where $f$ is defined in Saikai Prime's answer.

Title: Surjective map from $\mathbb{Z}^+$ to $\mathbb{Z}$.

Body:
Surjective map between $\mathbb{Z^{+}}$ and $\mathbb{Z}$.
I am confused as to how this is even possible to create.


问:
Solutions of $f(k,s)=0$ as $k\to\infty$
Let $f(k,s) = D(s) + kN(s)$ where $D(s)$ and $N(s)$ are polynomials of $s \in \mathbb{C}$ such that $\text{Deg}(D) = n, \ \text{Deg}(N) = m$ and $n\ge m$. Also $D(s)$ and $N(s)$ doesn't have common factor. Find the values of $s$ such that $$\lim_{k\to \infty}f(k,s) = 0  \tag{1}$$ My try: Assume that $N(s) \not = 0$. Obviously, in this case $\lim_{k\to \infty}f(k,s)$ is divergent for any $s$. Now suppose $N(s) = 0$ and we have $\lim_{k\to \infty}f(k,s) = D(s)$. So the answer is $D(s) = N(s) = 0$ which is impossible since $D(s)$ and $N(s)$ doesn't have common factor. What's wrong in my reasoning?

Also is $(1)$ equivalent to find $s$ such that $f(k,s) = 0$ and then $\lim_{k \to \infty} s(k)$? I mean that we first find the roots of $f(k,s)$ with respect to $k$ and see what happens to the roots when $k \to \infty$.

Edit: The correct answer is as follows. Divide the equation $D(s) + kN(s) = 0$ by $k$ and let $k \to \infty$. So we have $$\lim_{k \to \infty}(\frac{D(s)}{k} + N(s)) = N(s) = 0$$ I don't know why my argument is flawed and whether the mentioned equivalence holds.

Edit2: Thanks to @Alex Ravsky, it turns out that the true question is

What are the solutions of $f(k,s)=0$ as $k\to\infty$?

答:
What are the solutions of $f(k,s)=0$ as $k\to\infty$?

It can happen that the only root $s(k)$ of $f(k,s)$ is $k$ (for instance, when $D(s)=s$ and $N(s)=-1$), then  $\lim_{k \to \infty} s(k)=\infty$.

But in order to develop the argument from the proposed answer we suppose that there exists $R\ge 0$ such that for each $k$ the equation $f(k,s)=0$  has a root $s(k)$ contained in a set $K=\{s\in\Bbb C:|z|\le R\}$. Then for any $k\ne 0$ we have $\frac{D(s(k))}{k} + N(s(k))=0$. Since the set $K$ is compact and a function $D$ is continuous, there exists $M>0$ such that $|f(s)|\le M$ for each $s\in K$. This follows $\lim_{k \to \infty} N(s(k))=0$. Let $S=\{s\in K: N(s)=0\}$.

We can describe an asymptotic behavior of $s(k)$ as follows. For each $\varepsilon>0$ there exists natural $k_\varepsilon$ such that for any $k> k_\varepsilon$ there exists $s\in S$ with $|s(k)-s|<\varepsilon$. Indeed, suppose for a contradiction that
there exist $\varepsilon>0$ and a sequence $(k_n) _{n\in\Bbb N}$, tending to infinity, such that for each $n$,
$$s(k_n)\in K_\epsilon=\{s\in K: |s(k_n)-s|\ge \varepsilon\mbox{ for each }s\in S\}.$$

A sequence $(s(k_n))_{n\in\Bbb N}$ of points of a compact set $K_\epsilon$ has a subsequence converging to a point $s\in K_\epsilon$. Since $\lim_{n \to \infty} N(s(k_n))=0$ and the function $N$ is continuous, we have $N(s)=0$. But then $s\in S$ and
so $|s(k_n)-s|<\varepsilon$ for some $n$. Then $s(k_n)\not\in K_\varepsilon$, a contradiction.

Update. Let $D(s)=a_ns^n+a_{n-1}s^{n-1}+\dots+a_0$ and $N(s)=b_ms^n+b_{m-1}s^{n-1}+\dots+b_0$. Let $s(k)$ be a root of $f(k,s)$. If $N(s(k))=0$ then $D(s(k))=0$, so polynomials $N(s)$ and $D(s)$ have a common root, which is impossible. Therefore $N(s(k))\ne 0$ and $k=-D(s(k))/N(s(k))$.

Suppose that there exists a sequence $k_j$ tending to infinity such that $s(k_j)$ tends to infinity. Then  for each $j$,

$$k_j=-\frac{D(s(k_j))}{N(s(k_j))}=\frac{a_n s(k_j)^n+o(|s(k_j)|^n)} {b_m s(k_j)^m+o(|s(k_j)|^m)}=\frac{a_n}{b_m}s(k_j)^{n-m} +o(|s(k_j)^{n-m}|).$$

Title: Solutions of $f(k,s)=0$ as $k\to\infty$


问:
How few $(42^\circ,60^\circ,78^\circ)$ triangles can an equilateral triangle be divided into?
This is the parallel question to this other post with many answers already, in the sense that the $(42^\circ,60^\circ,78^\circ)$-similar triangles form the only non-trivial rational-angle tiling of the equilateral triangle (and the regular hexagon), modulo a real conjugation of the coordinate field (a subfield of $\mathbf{Q}(\zeta_{60})$) which transforms between $(42^\circ,60^\circ,78^\circ)$-similar triangles and $(6^\circ,60^\circ,114^\circ)$-similar triangles. (Reference: M. Laczkovich's Tilings of triangles.)

My attempt has been the following:

Since $\sin(42^\circ)$ and $\sin(78^\circ)$ have nested radicals, I tried to get rid of them by restricting my basic tiling units to only the $60^\circ$-angled isosceles trapezoids and parallelograms that are a single row of the triangular tiles. They have shorter-base-to-leg ratios of the form
$$m\cdot\frac{9-3\sqrt{5}}{2}+n\cdot\frac{11-3\sqrt{5}}{2}\quad\left(m,n\ge 0\right)$$
which are automatically algebraic integers. Any potential tiling of the equilateral triangle from these quadrilateral units corresponds to some integer polynomial relation of the above algebraics, whose polynomial degree correlates with the number of quadrilateral pieces in the tiling.

Unfortunately all the above algebraics have large norms, so a blind search for the desired polynomial is out of the question, and I had to reduce the pieces' proportions again to the rationals. I was able to find a $60^\circ$-angled isosceles trapezoid with shorter-base-to-leg ratio of $10$ using $79$ tiles, and a $60^\circ$-angled parallelogram with neighboring sides' ratio of $11$ using $80$ tiles. Thus a few more tiles produce a $60^\circ$-angled rhombus, and another few more tiles produce a $60^\circ$-angled isosceles trapezoid with shorter-base-to-leg ratio $1$, three of which tile an equilateral triangle, using a total of $121\,170$ triangular tiles. While I was at it, I found this less related post that might reduce my number of tiles to a bit below a hundred thousand.

Meanwhile, I also did a quick computer search through some conceptually simple configurations that attempt to tile the equilateral triangle using less than about $50$ tiles, and I found nothing at all.

I get this feeling that about a hundred thousand tiles is not the optimal amount for such a tiling, so I'm asking to see if people have better ideas. I'm unable to provide cash incentive as the parallel post did, but anyone who tries this puzzle will sure have my gratitude.

Edit suggested by RavenclawPrefect:

To get to the quadrilateral tiling units that I used, the first thing is to de-nest the radicals as I mentioned above. As $\mathbf{Q}(\zeta_{60})$ is Galois over $\mathbf{Q}(\sqrt{3})$ (the base field here should not be $\mathbf{Q}$ but instead the coordinate field of the equilateral triangle), if we can geometrically construct any length $\ell$ (or techinically, ratio $\ell$), such that when we perform the same geometric construction but with all the $42^\circ$ angles and $78^\circ$ angles swapped with each other, we still provably construct the same $\ell$, then it must hold that $\ell\in\mathbf{Q}(\sqrt{5})$, so that $\ell$ doesn't contain any nested radicals.

There were a couple of ideas on what $\ell$ should specifically be, most of them parallel ideas that can all be found in the parallel question for the square. I settled on the above $\mathbf{Q}(\sqrt{5})$-quadrilaterals (the ones that are a single row of triangular tiles) because they had the smallest numerator norms among others. As a non-example, there was a double-decker idea using $9$ tiles that resulted in a trapezoid with ratio a rational multiple of $889-321\sqrt{5}$, yuck. There were also some non-triviality in which way the triangles should be oriented when being put into a single row, but some more calculation showed that the above $(m,n)$ form are all we really get. More precisely, a trapezoid also can't have $m=0$, and a parallelogram also can't have $n=0$.

After all that work, the rest has really been a matter of trial-and-error. Among all the $(m,n)$ form, I picked a parallelogram with the smallest norm, which is an $(m,n)=(0,1)$ parallelogram with $4$ tiles, and rotated it so that it becomes a $\frac{11+3\sqrt{5}}{38}$-parallelogram. Then $19$ of those make a $\frac{11+3\sqrt{5}}{2}$-parallelogram with $76$ tiles, and obviously I combined it to a $(1,0)$-trapezoid and a $(0,1)$-parallelogram to get to the rational quadrilaterals.

So the process was more like "I frankly don't know what else to do" rather than "I see potential simplifications but I don't know the optimum". It's also why I'm seeking for completely new ideas (see above) that aren't found in the parallel question about the square.

RavenclawPrefect also asked a well-motivated question for if the same tiling could be performed but with congruent tiles. M. Laczkovich proved this is impossible in a subsequent paper Tilings of Convex Polygons with Congruent Triangles.

答:
Here’s a trapezoid of ratio $1$ tiled by $195$ triangles, found in a brute-force search.  Using three of these to build an equilateral triangle takes $3 \cdot 195 = \mathbf{585}$ triangles.

Old answer

This basic $60^\circ$ trapezoid of ratio $\frac{9 - 3\sqrt 5}{2}$ uses three triangles, and this basic $60^\circ$ parallelogram of ratio $\frac{11 - 3\sqrt 5}{2}$ uses four triangles:

Any number $r \in \mathbb Q[\sqrt 5]$ can be decomposed as $r = \frac{11 - 3\sqrt 5}{2}u + \frac{2}{11 - 3\sqrt 5}v$ with $u, v \in \mathbb Q$.  If $u, v \ge 0$, then we can tile a parallelogram of ratio $r$ using basic parallelograms by combining affine transformations of the tilings of rectangles of ratios $u$ and $v$ using squares.  For example, here’s a 72-triangle parallelogram of ratio $1 = \frac{11 - 3\sqrt 5}{2}\cdot\frac{1}{11} + \frac{2}{11 - 3\sqrt 5}\cdot\frac{19}{11}$, derived from square tilings of $1 × 11$ and $19 × 11$ rectangles.

Using this idea in a version of the “staircase” construction yields much more efficient tilings of the trapezoid of ratio $1$.  Here’s one with $45 \cdot 4 + 10 \cdot 3 + 44 \cdot 4 = 386$ triangles.  (I’m now using a dynamic programming algorithm that tiles the green region and the blue region each in one pass rather than slicing them into parallelograms.  For clarity, divisions of basic trapezoids/parallelograms into three/four triangles are not pictured.)

Using three of these to build an equilateral triangle takes $3 \cdot 386 = \mathbf{1158}$ triangles.

Probably a yet more efficient tiling could be constructed by cutting a small number of basic trapezoids out of the equilateral triangle more or less arbitrarily until a single parallelogram remains, solving for its ratio $r \in \mathbb Q[\sqrt 5]$, and applying the above rectangle-tiling construction once.  Finding a way to do this such that $u, v \ge 0$ has been trickier than I expected though.


问:
Do the same results of a discrete metric space apply to a metric space $d(x,y)$ = $1/L$, $x\neq y$, $0$, $x=y$
What if a metric is the same as a discrete metric, but instead of being $1$ when $x\neq y$, it equals some real number $1/M$ instead?

For example would the metric space $(\mathbb{R},d)$ where $d$ is the discrete metric, but instead of $1$ is a real number $1/M$, be compact if and only if $\mathbb{R}$ is finite?

答: If, on a set $X$ you define $d_K(x,y)=K$, for some $K\in(0,\infty)$, then you get a metric (which is the discrete one if $K=1$), and, for any value of $K$, the topology induced by that metric is the discrete one, that is, it's the topology for which every subset of $X$ is open. So, yes, $X$ endowed with such a metric is compact if and only if $X$ is finite.


问:
Quasi-regular elements which are not in the Jacobson radical
Suppose $R$ is a commutative ring. An element $x \in R$ is called quasi-regular if $1-x$ is a unit. Are there examples of $R$ and $x$ such that $x$ is quasi-regular, but $yx$ is not, for some $y \in R$? In other words, examples of quasi-regular $x$ which is not in the Jacobson radical of $R$.

答: Take $R=\mathbb{Z}$ and $x=2$.

Body: Take $R=\mathbb{Z}$ and $x=2$.


答:
This is an improper integral, so, you have to define it as a limit. You have
$$
\int_{-\infty}^\infty \sin(x)dx=\lim_{R\rightarrow\infty}\int_{-R}^R \sin(x)dx=0.
$$
If you now take a different $a$, you have
$$
\int_{-\infty}^a+\int_{a}^{\infty}\sin(x)dx=\lim_{R\rightarrow\infty}\int_{-R}^{-a}+\int_{-a}^{a}+\int_{a}^{R}\sin(x)dx=\int_{-a}^a\sin(x)dx=0
$$
and nothing changes. Thus, I think that you don't need to split it as $\int_{-\infty}^a+\int_{a}^{\infty}$.

PD: I consider the integral as Principal Value integral (when I pass to the limit the domain is symmetric). See the answer by @egreg

Title: Prove that $\int_{-\infty}^{\infty} \sin x \, dx = 0 $

Body:
$$\int_{-\infty}^{\infty} \sin x \, dx$$
When I am doing the proof for this, why do i have to split it into
$\int_{-\infty}^a \sin x \, dx + \int_a^\infty \sin x \, dx $?
where a is a constant </p>


答:
As @Chaos & @Fred said, it's because of first term added to right hand side ($k=0$). Also notice the benefit of relation:

$$f(x) = \sum_{k=0}^{\infty} \frac{(-1)^k}{(2k)!}x^{2k} \Longrightarrow
f''(x) = -f(x) \Longrightarrow f(x) = a \sin (x) + b \cos (x)$$

May this is the purpose of relation.

Title: Why is $\sum_{k=1}^{\infty} \frac{(-1)^{k}}{(2 k) !}=\left.\left(\sum_{k=0}^{\infty} \frac{(-1)^{k}}{(2 k) !} x^{2 k}-1\right)\right|_{x=1}$?

Body:
In order to sum up the series we have to realize that it is almost the expansion of the cosine function. But we need $k=0$ in the sum. I don't know how to do this. The solution is   $\sum_{k=1}^{\infty} \frac{(-1)^{k}}{(2 k) !}=\left.\left(\sum_{k=0}^{\infty} \frac{(-1)^{k}}{(2 k) !} x^{2 k}-1\right)\right|_{x=1}$, but where does $-1$ on the right hand side come from and how did we achieve $k=0$?

My approach was to do the following, which leads to a dead end...(right?)

$\sum_{k=1}^{\infty} \frac{(-1)^{k}}{(2 k) !}=-\sum_{k=1}^{\infty} \frac{(-1)^{k-1}}{(2 k) !}=\cdots$


问:
Is this argument for uniform continuity good?
Consider the function $f:(-\infty, 1)\to \mathbb{R}$, $f(x)=\frac{1}{x-1}$. I would like to prove that this function is uniformly continuous.

My argument is as follows: the function is continous because it is the division of two continuous functions. Now, $\displaystyle \lim_{x\to -\infty}f(x)=0$, so we are done.

I want to make sure that this is a correct way of proving uniform continuity. I know that this works for functions defined on an interval that contains $\infty$, but I want to make sure that this also works for $-\infty$.

Furthermore, what if our function's domain is no longer an interval, but a union of intervals that contains $\infty$ or $-\infty$? Does the conclusion still hold?

答:
No, this is not true. The problem here is not $-\infty$, the problem happens near the point $1$. For example, $1-\frac{1}{n}$ is a Cauchy sequence of elements in $(-\infty,1)$. However, $f(1-\frac{1}{n})=-n$ is clearly not a Cauchy sequence. So $f$ is not uniformly continuous. (otherwise it would map Cauchy sequences to Cauchy sequences)

Another way to see this: in general, if $a,b$ are real numbers (not $\infty$ or $-\infty$) and a function $f$ is uniformly continuous at $(a,b)$ then it can be extended to a continuous function $[a,b]$. However, your function can't be continuously extended to the point $x=1$, as the limit $\lim\limits_{x\to 1^-} f(x)$ doesn't exist.


问:
Integrals identity
By numerical integration I found the identity
$$ \int_{-\infty}^{0}\frac{db}{\sqrt{R_4(b,W)}} = 2\int_{(W+1)^2}^{\infty}\frac{db}{\sqrt{R_4(b,W)}} $$
where $R_4(b,W)= b \ (b-4) \ (b-(W-1)^2) \ (b-(W+1)^2) $ and $W>3$.
Now I would like to prove that formally but I can't find the right way. I started by substituting
$$ b \rightarrow t=-b $$ and
$$ t \rightarrow x=t+(W+1)^2 $$
to have the same integration range on both sides, but $R_4(b,W)$ becomes quite strange.

答:
Let the roots of $R_4$ be $a=(W+1)^2,b=(W-1)^2,c=4,d=0$. Byrd and Friedman 251.00 and 258.00 resolves both sides in terms of elliptic integrals as follows:
$$gF(\varphi,m)=2gF(\psi,m)$$
where $g=\frac2{\sqrt{(a-c)(b-d)}}$, $m=k^2=\frac{(b-c)(a-d)}{(a-c)(b-d)}=\frac{(W-3)(W+1)^3}{(W+3)(W-1)^3}$, $\sin^2\varphi=\frac{a-c}{a-d}=1-\frac4{(W+1)^2}$ (hence $\cos\varphi=\frac2{W+1}$) and $\sin^2\psi=\frac{b-d}{a-d}=\frac{(W-1)^2}{(W+1)^2}$ (hence $\cos\psi=\frac{2\sqrt W}{W+1}$). Thus it remains to show $F(\varphi,m)=2F(\psi,m)$. Applying the $F$ addition formula to the RHS (see DLMF 19.11.14) gives
$$\sin\varphi=\frac{2\sin\psi\cos\psi\sqrt{1-m\sin^2\psi}}{1-m\sin^4\psi}=\frac{\sqrt{(W-1)(W+3)}}{W+1}$$
$$\sin^2\varphi=\frac{(W-1)(W+3)}{(W+1)^2}=1-\frac4{(W+1)^2}$$
which matches the expression we obtained for $\sin^2\varphi$ earlier. So the equality is true.

Title: Integrals identity


问:
Laplacian 2D kernel - is it separable?
I'm wondering if the 2D laplacian kernel

```<code>0  1  0
1 -4  1
0  1  0
</code>```

is also a separable kernel. How can I find that out?

答:
In general, you need to verify the rank of the kernel (considered as a matrix). If rank is one, than (by SVD decomposition) you can find two vectors whose outer product is the kernel.
In this case, the rank of the laplacian is 2, hence it is not separable.</p>

Title: Laplacian 2D kernel - is it separable?


问:
$2 \int_{0}^{1/2} x^{p-1} (1-x)^{p-1}dx $ in terms of gamma function
This question was asked in my mid term exam which has been concluded and I was unable to solve it and couldn't do it again when I tried to do at home. Hence , I am posting it here for help.

Prove that $\frac{ \Gamma(p) \Gamma(p)} { \Gamma(2p)} = 2 \int_{0}^{1/2} x^{p-1} (1-x)^{p-1}dx$.

I am not able to think how this problem should be attempted as I am not able to get any intution regarding it. So, I request you to give some hints so that I am able to start it.

答:
Obviously the integrand is symmetric about $x=\frac12$, so the factor of $2$ may be merged into the integral bounds giving
$$\int_0^1x^{p-1}(1-x)^{p-1}\,dx$$
which by definition of the beta function is
$$\mathrm B(p,p)=\frac{\Gamma(p)^2}{\Gamma(2p)}$$


问:
Inverse image of a non-zero prime ideal under a surjective ring homomorphism
Let $R, S$ be commutative rings with unity . Let $f:R\to S$ be a  surjective ring homomorphism

$Q\subseteq S$ be a non-zero pime ideal . Which of the following statements are true?

$(a)f^-(Q)$ is a non-zero prime ideal in $R$

$(b)f^-(Q)$ is a maximal  ideal in $R$ if $R$ is a PID

$(c)f^-(Q)$ is a maximal  ideal in $R$ if $R$ is a finite commutative ring with unity .

$(d)f^-(Q)$ is a maximal ideal in $R$ if $x^5=x$ for all $x\in R$

$(a)$ True.

Let $r_1r_2 \in  f^-(Q)$

Then $f(r_1r_2 )=f(r_1)f(r_2) \in Q$. Since $Q$, is prime ideal, we have either of  $ f(r_1),f(r_2)$ belongs to $Q$ .

So either $r_1 \in f^-(Q)$ or $r_2 \in f^-(Q)$

Also $f^-(Q)$ is non-zero and proper since $Q$ is non-zero and $f$ is surjective.

This proves the result.

$(b)$ True since every non-zero prime ideal in a PID is maximal.

$(c)$ True.

$f^-(Q)$ is non-zero prime .

$\Rightarrow R/f^-(Q)$ is finite ID (since $R$ is finite)
and hence a field .

$\Rightarrow f^-(Q)$ is maximal ideal.

$(d)$ I am not sure about this.

Can it be said that order (multiplicative) of every element is $4$ ?

Can it be said that the ring is finite (since order of every element is finite) and then considering the logic in $(c)$ ?

Can it be said that the ring is PID under the given conditions ?

Please help me complete by appropriate hints . Thank you.

答:
Hint for (d): If you have an integral domain $A$ with the property $x^5=x$ for all $x\in A$, then for any $x\ne0$ we have $x^4=1$, so ...

Apply this to $R/f^{-1}(Q)$

I think $R$ might neither be finite nor a PID with the given assumptions, consider e.g. $$R=\Bbb F_5[(x_n)_{n\in\Bbb N}]/\langle (x_n^5-x_n)_{n\in\Bbb N})\rangle$$


问:
In which points is the following function complex differentiable?
Let $a,b,c,d \in \Bbb C$ s.t. $cd \neq 0$. We define
$$f(z)=\frac{az+b}{cz+d}$$
I am looking at the points, where $f$ is complex differentiable. I though to use the Cauchy-Riemann equations, but I don't know how to use it here...

Many thanks for some help!

答: $f$ is the fraction of two complex differentiable maps. So $f$ is complex differentiable where the denominator doesn't vanish, i.e. for $z \in \mathbb C\setminus\{-\frac{d}{c}\}$.


问:
Show that two angles are equal

Acute-angled triangle ABC is inscribed in circle $c$. In the smaller arc BC we choose a random point F and draw a line parallel to AC which intersects side AB in point Q. Then from point Q we draw a parallel to FC which intersects side AC to point H. From vertex B we draw a line parallel to FC which intersects the circle at point S. Show that angle SHC is equal to BAC.

I have tried the following: I have extended line SH until it meets the circle at point which I name G. Clearly G lies also on the line QF but we don't know it yet.

Since FC and BS are parallel chords of the circle, then BFCS must be an isosceles trapezoid, hence arcs BC and FS are equal. Therefore angle CAB is equal to angle SGF (inscribed angles of the same circle, having equal arcs). Then obviously SHE is equal to SGF.

But how can I prove that point G also lies on the parallel from F?

Or, is there any other way to prove that the required angles are equal?

Thank you in advance.

答:
Extend $FQ$ to meet the circle at point $L$.Let the points at which $FL$ and $CA$ intersect $BS$ be $T$ and $U$ respectively. Draw $AL$, $LS$ and $LH$$[$We haven't yet proved that points $L$,$H$ and $S$ are collinear.$]$

$AC\parallel FL$   $\Rightarrow$  Quadrilateral $ACFL$ is an isosceles trapezoid.

Since $QH\parallel CF$, quadrilateral $ALQH$ is an isosceles trapezoid as well.

$\Rightarrow$   $\angle ALH=\angle AQH=\angle ABU=\angle ABS=\angle ALS$

$\Rightarrow$   $\angle ALH=\angle ALS$

Hence, points $L$, $H$ and $S$ are collinear.

Thus, $\angle SHC=\angle SLF=\angle HLQ=\angle QAH=\angle BAC$

Title: Show that two angles are equal


问:
The definability of set of theorems of a theory T which is a consistent extension of Q
I have difficulty understanding lemma 17.2 in page 221 while reading "Computability and Logic" fifth edition by George Boloos.

Lemma 17.2: Let T be a consistent theory extending Q. Then the set of Godel numbers of theorems of T is not definable in T.
Here a definable set S means there is a predicate F(x) such that F(a) is correct under the standard interpretation of arithmetic if and only if a belongs to S , so F(x) and S have a similar extension, so to speak.

The proof of Lemma 17.2: Suppose F(y) defines the set S of Godel numbers of theorems of T. By diagonal lemma there is a sentence G such that
$$T\vdash G \leftrightarrow \lnot F(g)$$
with g is the Godel number of G. Then G is a theorem of T. For if we assume G is not a theorem of T, then g is not in S, and since F(y) defines S, we have $T\vdash \lnot F(g)$; but then since $T\vdash G \leftrightarrow \lnot F(g)$, we have $T\vdash G$ after all. But since G is a theorem of T, g is in S, and so we have $T\vdash F(g)$, but then since $T\vdash G \leftrightarrow \lnot F(g)$, we have $T\vdash \lnot G$ , and T is inconsistent.

I have 2 problems regarding the above proof.

I don't quite understand the bold text in the proof. If g is not in S, $\lnot F(g)$ is true, but this does not entail $T\vdash \lnot F(g)$.

While reading the "Introduction to Godel's incompleteness theorems" by Peter Smith, I also came through the assertion that the set of theorems T is definable (or expressible in the author's terms) by the predicate Prov(x), while the set of true sentences is not even definable (Tarski's theorem). Does this assertion contradict the lemma?

Thank you!

答:
No; see page 207:

If $T$ is a consistent theory in the language of arithmetic, we say a set $S$ is defined
in $T$ by $D(x)$ if for all $n$, if $n$ is in $S$, then $D(\overline n)$ is a theorem of $T$ [i.e. $T \vdash D(\overline n)$], and if $n$ is not in $S$, then $\lnot D(\overline n)$ is a theorem of $T$.

Note: I've used $\overline n$ for the numeral corresponding to number $n$.

Thus, the argument is: "we assume G is not a theorem of T", i.e. $T \nvdash G$, and thus "g is not in S".

But "F(y) defines S",  i.e. $T \vdash \lnot F(\overline g)$ iff $g \notin S$.

Thus, from "g is not in S" it follows that $T \vdash \lnot F(\overline g)$.

Unfortunately, the name is the same but the definition is different from the Definable set of model theory:

a set $A\subseteq M$ is definable in $\mathcal {M}$ if and only if there exists a formula $\varphi$ such that for all $a\in M$

$$a \in A \text  { iff } \mathcal {M}\models \varphi [a].$$


问:
not understanding a question about IVT

Hello, I do not understand why the first answer is correct : in fact the intermediate value theorem only yields that there exists a $u$ such that $\frac{\pi \sqrt{2}}{8}=f'(u)...$

答:
On $]0,\pi/4[$ we have $f(u)=u \cos u$ and $f'(u)=\cos u-u \sin u$

As $f'(u)$ is continuous, $f'(0)=1$ and $f'(\pi/4)\approx 0.15$, fot the IVT there exists at least one $u\in]0,\pi/4[$ such that $f'(u)=\frac{\sqrt 2}{2}$

Title: not understanding a question about IVT


问:
Generalization of degree of a map (reference request)
Let $f:M\to N$ be a continuous map between two oriented manifolds of dimension $n$. There are a lot of alternative definitions of degree, but I'm only interested in the following one. Since $M,N$ are oriented, that means that $H_n(M)=H_n(N)=\mathbb{Z}.$ So the induced map in homology
$$
f_\ast:H_n(M)\cong \mathbb{Z}\xrightarrow{d}\mathbb{Z}\cong H_n(N)
$$
is multiplication by an integer $d$, which we call the degree.

Next, I would like to generalize this definition to manifolds that are not oriented. There's an obvious way to do this. Let $p\in \mathbb{Z}$ be a prime, and let $f:M\to N$ be a map between manifolds such that $H_n(M)= H_n(N)= \mathbb{Z}/p\mathbb{Z}.$ Then
$$
f_\ast:H_n(M)\cong \mathbb{Z}/p\mathbb{Z}\xrightarrow{[d]}\mathbb{Z}/p\mathbb{Z}\cong H_n(N)
$$
is multiplication by some $[d]\in \mathbb{Z}/p\mathbb{Z},$ which I'm calling mod p degree.

This seems like such an immediately useful idea that somebody has to have done it before. But I don't know where in the literature to look.  Does anybody have any suggestions?

Edit: Cross-posted to Math Overflow here.

答:
This is only interesting for $p=2$, and this is indeed the mod-2 degree. Searching for this should give you many references.

For a map $f:M\rightarrow N$ between non-orientable manifolds you can sometimes refine the mod-2 degree and orient the map $f$ instead of the manifolds. You get a notion of twisted degree. I think this is due to Paul Olum, perhaps in the paper below.

MR0058212 (15,338a) Reviewed
Olum, Paul
Mappings of manifolds and the notion of degree.
Ann. of Math. (2) 58 (1953), 458–480.
56.0X


问:
Find the period of $f(x)=\sin^2(x)$
$f(x+p)=f(x)$

$\sin^2(x+p)=\sin^2(x)$

$\sin^2(x+p)-\sin^2(x)=0$

$[\sin(x+p)-\sin(x)][\sin(x+p)+\sin(x)]=0$

$\sin(x+p)-\sin(x)=0$

$\sin(x+p)=\sin(x)$

This is the part iI don't get:

$x+p=x+2kπ$

$p=2kπ$

My question: how did we get $x+p=x+2kπ$ from $\sin(x+p)=\sin(x)$

And yeah I know the exercise doesn't end there, but I only want to understand this part first.

答:
To answer the question directly, suppose $\sin^2x $ is periodic with period $p$.  We need to find the smallest $p > 0$ such that
\begin{equation}
\sin^2(x+p) = \sin^2 x \tag{1}\label{eq1}
\end{equation}
for all $x$, if such exists.

Method 1:
$\cos 2x = \cos^2 x - \sin^2 x = 1 - 2\sin^2 x$.  Therefore, $\sin^2x = \frac{1}{2}\Big( 1 - \cos2x \Big)$.  If you accept that $\sin$ and $\cos$ are both periodic with period $2\pi$ it is thus apparent that $\sin^2x$ is also periodic with period $\pi$.

Method 2:
Suppose $\sin^2x$ is periodic with period $p$.  Then taking $x=0$ in equation \eqref{eq1} gives $\sin^2 p = 0$.  Now we are only interested in the smallest positive $p$ and therefore we must have $p \geqslant \pi$ since $\pi$ is the smallest positive zero of $\sin x$.  Now take $p = \pi$.

\begin{align}
\sin^2(x+\pi) -\sin^2 x &= \Big(\sin(x+\pi)-\sin(x)\Big)\Big(\sin(x+\pi) +\sin (x)\Big) \\
&= \Big(\sin x \cos \pi + \cos x \sin \pi - \sin x\Big)\Big(\sin x \cos \pi + \cos x \sin \pi + \sin x\Big) \\
&=\Big(-\sin x - \sin x\Big)\Big(-\sin x + \sin x\Big) \\
&=0
\end{align}
So we have $\sin^2(x+\pi) = \sin^2(x) $ for all $x$ and $\pi$ is the smallest number with this property so $\sin^2 x$ is periodic with period is $\pi$.

Title: Find the period of $f(x)=\sin^2(x)$


问:
Basic question with tiling tiles in a rectangle
Find the number of ways of tiling a 2 by 10 rectangle with 1 by 2 tiles which can be placed in any orientation.

I’m just not sure how to approach it begin counting it

答:
I am not going to be providing a full solution but just a hint;

let $a_n$ denote number of ways to cover a $2*n$ rectangle with given tiles.

I'd suggest you to draw and see how $a_n$ = $a_{n-1}$ + $a_{n-2}$. (HINT; you can place the $2*1$ tiles in two ways, horizontally and vertically) and then find $a_0$ , $a_1$ and $a_2$ to get the rest or study some methods to solve recusrsion


问:
Question about limits at infinity.
if for example I have fucntions $f$ and $g$, we've prove in class that if
$$\lim _{x\to a}\left(f\left(x\right)-g\left(x\right)\right)=L$$

$$or$$

$$\lim _{x\to a}\left(f\left(x\right)\cdot g\left(x\right)\right)=L$$

then both limits of $f$ and $g$ exist, or both don't exist.

but what about

$$\lim _{x\to ∞/-∞}\left(f\left(x\right)-g\left(x\right)\right)=L$$

$$or$$

$$\lim _{x\to ∞/-∞}\left(f\left(x\right)\cdot g\left(x\right)\right)=L$$

Does it still hold that both of f,g must exist or both must not exist?
If so, how do I prove it?

答:
The point to which the limit tends is indeed irrelevant. By a change of variable such that $t:=a+\dfrac1x$, you turn a limit to infinity in a limit at any $a$.

Anyway, your notation $\lim_{x\to\infty/-\infty}$ is ambiguous.

Title: Question about limits at infinity.


问:
How do I solve for $x$ in this polynomial function?
The marketing manager of a new soap has figured that her monthly profit
$P$ (in thousands of pesos) is determined by the monthly expense $x$ (in terms of thousands of pesos) according to the function:
$$
(x) = x^3 − 15x^2 + 90x \qquad\text{ for } 0 \leq x \leq 5
$$
For what value of $x$ does the manager get Php 184,000.00 in profits?

答:
With minimal effort of will,

$$\begin{array}{ccccc}0&1&2&3&4&5\\\hline0&
76&
128&
162&
184&
200
\end{array}$$


问:
Classification of infinity as a singular point.
i have the followings problems, i must find and classify the singular points of the followings functions:

$a) \frac{e^{z}}{1+z^{2}}\\$.

$b) \frac{e^{z}}{z(1-e^{-z})}$.

$c) \frac{1}{z^{3}(2-\cos{z})}$.

$d) e^{\frac{z}{1-z}}$.

$e) \cot{z}- \frac{1}{z}$.

$f) \cot{z}- \frac{2}{z}$.

$g) \sin{\frac{1}{1-z}}$.

I have classified the obvious points and they have reviewed it, but they tell me that i must classify infinity as well, but i do not know how to do it, i would greatly appreciate your help please.

答: Not completely sure if this is rigorous, but my approach would be to make the substitution $ w = 1/z $, and rearrange the functions to be in terms of $w$. Classifying $z = ∞$ will then be equivalent to classifying $w = 0$.


问:
Integral of $(-1/(1 + x^2))$
I had a doubt that the integral of $\frac{-1}{1+x^2}$ is given as negative tan inverse $x$. Isn't it equal to $\cot$ inverse ($x$) ?

答:
The answers are completely equivalent:
$$
\int-\frac{1}{1+x^2}\,dx=-\arctan x+c
$$
but also
$$
\int-\frac{1}{1+x^2}\,dx=\operatorname{arccot} x+c
$$
because the functions $-\arctan x$ and $\operatorname{arccot} x$ differ by a constant: indeed
$$
\operatorname{arccot} x+\arctan x=\dfrac{\pi}{2}
$$

Title: Integral of $(-1/(1 + x^2))$


问:
Probability to win a solo card game
I discovered a simple game with cards:

You have a deck of 52 French Cards, this deck has four sets of cards with values from 1 to 10 and J Q K (which values are 11, 12, 13 respectively)

The game is very simple: you reveal the first card and say ONE, reveal the second card and say TWO and so on... After saying THIRTHEEN you restart the voice counting

If the revealed card value is the same as the said number, you have lost. If you finish the deck you have win.

Now comes my question: how can I compute manually the probability of winning to this game? (I already computed with programming so I simply want to know how to compute this probability by hand :) )

Here I attach some example of games:

Drawed cards:
2 4 6 8 5... -> lost because of 5 is the fifth card

4 2... -> lost because of 2 is the second card

13 12 11 10 9 8 7 6 5 4 3 2 2 1 -> lost because 1 is the 14th card (but after 13 we have to restart the count, so 1 correspond to saying ONE)

I hope that some of you can answer to my question :)

答:
This same problem occurred to me when I was an undergraduate, $50$-odd years ago.  I wasn't able to solve it it, though I thought about it off and on for about $15$ years.  Then I ran across rook polynomials, and was surprised how easy it was.

There's a good article on rook polynomials here.

In actually doing the problem, it's easier if you think about calling the cards in the order $1,1,1,1,2,2,2,2,...$ which obviously doesn't change the probability of success.  Then the dark chessboard of forbidden positions consists of $13$ $4\times4$ chessboards, arranged in block diagonal fashion.  Since the rook polynomial of a  $4\times4$ board is $$24x^4+96x^3+72x^2+16x+1,$$ (see the wiki article,) the rook polynomial of the entire dark chessboard is  $$\left(24x^4+96x^3+72x^2+16x+1\right)^{13}$$

Then just apply the formula.

If we have $s$ suits and $r$ ranks (so in the question $s=4, r=13$) then as $s$ remains fixed and $r\to\infty$ one can show that the probability of success approaches $e^{-s}$.  This is the basis of my comment that the probability is approximately $e^{-4},$ though frankly, I can't remember how good that approximation is when $r=13$.  I seem to remember that it came out to be about $1/60$, but I wouldn't bet on it.

Note that when $s=1$, this is just the problem of derangements, and it's well-known that the probability of success approaches $e^{-1}$.

EDIT

One other thing I'd like to mention is that friends I showed the problem to often answered $$\left(\frac{12}{13}\right)^{52},$$ incorrectly treating the events as independent, but this doesn't give a bad first approximation, since
$$\left(\frac{12}{13}\right)^{52}=\left(\left(1-\frac1{13}\right)^{13}\right)^4\approx e^{-4}$$

EDIT

Couldn't resist it.  I wrote this python script

```<code>from sympy import Symbol, factorial

x = Symbol('x')
p = 24*x**4+96*x**3+72*x**2+16*x+1
c = (p**13).as_poly().coeffs()[::-1]
n = sum((-1)**k*c[k]*factorial(52-k) for k in range(53))

prob = n/factorial(52)
</code>```

The probability of success is

$$\frac{4610507544750288132457667562311567997623087869}{284025438982318025793544200005777916187500000000}\\\approx0.016232727467194636$$

Title: Probability to win a solo card game

Body:
This same problem occurred to me when I was an undergraduate, $50$-odd years ago.  I wasn't able to solve it it, though I thought about it off and on for about $15$ years.  Then I ran across rook polynomials, and was surprised how easy it was.

There's a good article on rook polynomials here.

In actually doing the problem, it's easier if you think about calling the cards in the order $1,1,1,1,2,2,2,2,...$ which obviously doesn't change the probability of success.  Then the dark chessboard of forbidden positions consists of $13$ $4\times4$ chessboards, arranged in block diagonal fashion.  Since the rook polynomial of a  $4\times4$ board is $$24x^4+96x^3+72x^2+16x+1,$$ (see the wiki article,) the rook polynomial of the entire dark chessboard is  $$\left(24x^4+96x^3+72x^2+16x+1\right)^{13}$$

Then just apply the formula.

If we have $s$ suits and $r$ ranks (so in the question $s=4, r=13$) then as $s$ remains fixed and $r\to\infty$ one can show that the probability of success approaches $e^{-s}$.  This is the basis of my comment that the probability is approximately $e^{-4},$ though frankly, I can't remember how good that approximation is when $r=13$.  I seem to remember that it came out to be about $1/60$, but I wouldn't bet on it.

Note that when $s=1$, this is just the problem of derangements, and it's well-known that the probability of success approaches $e^{-1}$.

EDIT

One other thing I'd like to mention is that friends I showed the problem to often answered $$\left(\frac{12}{13}\right)^{52},$$ incorrectly treating the events as independent, but this doesn't give a bad first approximation, since
$$\left(\frac{12}{13}\right)^{52}=\left(\left(1-\frac1{13}\right)^{13}\right)^4\approx e^{-4}$$

EDIT

Couldn't resist it.  I wrote this python script

from sympy import Symbol, factorial

x = Symbol('x')
p = 24*x**4+96*x**3+72*x**2+16*x+1
c = (p**13).as_poly().coeffs()[::-1]
n = sum((-1)**k*c[k]*factorial(52-k) for k in range(53))

prob = n/factorial(52)

The probability of success is

$$\frac{4610507544750288132457667562311567997623087869}{284025438982318025793544200005777916187500000000}\\\approx0.016232727467194636$$


问:
calculate the distribution and density of the variable $Y=X^2$
$$ \mbox{Let X have the normal distribution: } N(0,\sigma^{2}) \to E(X)=0, Var(X)=\sigma^{2}. \mbox{ Let's define a new variable: }Y=X^2. \mbox{  Calculate the distribution and density of Y} $$
Is that solution correct?
$$f_{X}(x)= {{1}\over{(2\pi)^{1/2} \sigma}}e^{{-x^2}\over{2\sigma^2}}$$ $$F_{X}(x)= {\phi({{x}\over{\sigma}})} $$ $$ F_{Y}(y)=P(Y<y)=P(X^2<y)=P(-{y}^{1/2} < X < {y}^{1/2})=F_{X}({y}^{1/2})-F_{X}(-{y}^{1/2})=\phi({{y}^{1/2}\over\sigma})-\phi({{-y}^{1/2}\over\sigma})= 2\phi({{y}^{1/2}\over{\sigma}}) $$ But I still don't know how to calculate the density of $Y$.

答:
Let's start form

$$\frac{X}{\sigma}\sim N(0;1)$$

It is well known, and anyway easy to prove that

$$W=\Bigg(\frac{X}{\sigma}\Bigg)^2\sim \chi_{(1)}^2=Gamma(1/2;1/2)$$

Thus

$$Y=\sigma^2\cdot W=Gamma(1/2;1/(2\sigma^2))$$


问:
What is the derivative of $(\mathbf{u}(x))^TA\mathbf{v}(x)$ w.r.t. $x$?
What is $\frac{d}{dx}(\mathbf{u}^TA\mathbf{v})$ where $\mathbf{u}$ and $\mathbf{v}$ are column-vector-valued functions of $x$, and $A$ is a matrix?

Is it $\frac{d\mathbf{u}^T}{dx}A\mathbf{v}+\mathbf{u}^TA\frac{d\mathbf{v}}{dx}$?

答:
Yes, as @Hermis14 said, it's true.

Notice: it's based on hypothesis of "$A$ is a constant matrix".


问:
Subset $A$ of segment $(0,1]$ with Euclidean topology is compact $\Leftrightarrow$ $X(A)$ is compact on the Euclidean plane
Let:

$I(p,q)$ - closed segment with ends $p,q\in \mathbb R^2$

$A\subset (0,1]$

$X(A)=\bigcup \{I \Big( (0,\frac 1x),(x,-\frac 1x) \Big):x\in A \}$

Prove that: subset $A$ of segment $(0,1]$ with Euclidean topology is compact $\Leftrightarrow$ $X(A)$ is compact on the Euclidean plane $\Leftrightarrow$ $X(A)$ is closed on the Euclidean plane

I know the definition of compactness and various theorems, but I don't know how to go about such tasks. Would there be anyone who would be willing to provide a sketch of the reasoning or provide tips that might help me? This is not a homework or an exam. I just prepare for the exam session by doing the tasks from previous years, and because I have very poor lessons, I can't handle them too much and I would like to have at least one model solution to try to draw conclusions for other tasks.

答:
The closed segment $I(p,q)$ has a closed definition. Let $I=[0,1]$ be the standard unit interval. Then

$$I(p,q)=\big\{t\cdot p+(1-t)\cdot q\ \big|\ t\in[0,1]\big\}\subseteq\mathbb{R}^2$$

With that we have:

1) $A$ is compact $\Rightarrow$ $X(A)$ is compact

For $x\neq 0$ denote $v(x)=(0,\frac{1}{x})$ and $w(x)=(x,-\frac{1}{x})$. Both are continuous functions over $\mathbb{R}\backslash\{0\}$. This implies that $X(A)$ is a continuous image of $A\times I$ via $$(x,t)\mapsto t\cdot v(x)+(1-t)\cdot w(x)$$ Since $A$ is compact, then so is $A\times I$ and thus its image $X(A)$.

2) $X(A)$ is compact $\Rightarrow$ $X(A)$ is closed

Obvious. Every compact subset of a Hausdorff space is closed.

3) $X(A)$ is closed $\Rightarrow$ $A$ is compact

First of all there is $\epsilon >0$ such that if $(0,1/x)\in X(A)$ then $x>\epsilon$. Because otherwise $(0,0)$ would be a limit point of $\{(0,1/x)\}_{x\in A}$. But then $(0,0)$ would have to belong to $X(A)$ since we assumed it is closed, which is impossible by the definition of $X(A)$.

This implies that $A\subseteq [\epsilon, 1]$ and so $X(A)$ is bounded. And therefore compact as a closed and bounded subset of Euclidean space.

Now the subset $Y=\big\{(x,y)\in A\ \big|\ x=0\big\}$ is a closed subset of $X(A)$. Furthermore $A$ is the image of $Y$ via $(x,y)\mapsto y$. Since $X(A)$ is compact then so is $Y$ and therefore $A$.


答:
The approach suggested by @user872642 does work, and indeed works well for many functions. However, for some function other methods may lead to simpler calculations. For example, you can look at the behaviour of the function near infinity, i.e. for large $|z|$. The following rules apply:

If the function is bounded near $\infty$, it is said to have a removable singularity in $\infty$.

If $|f(z)| \rightarrow \infty$ as $|z| \rightarrow 0$, the function is said to have a pole. If you want to find the order of that pole, you might have to do the substitution $w=\frac{1}{z}$.

If the function neither vanishes nor blows up as $|z| \rightarrow \infty$, it has an essential singularity in $\infty$.

You can also look at the Laurent series of $f(\frac{1}{z})$, which is what is sometimes (for example by WolframAlpha) referred to as the Laurent expansion at $\infty$.

Title: Classification of infinity as a singular point.

Body:
i have the followings problems, i must find and classify the singular points of the followings functions:

$a) \frac{e^{z}}{1+z^{2}}\\$.

$b) \frac{e^{z}}{z(1-e^{-z})}$.

$c) \frac{1}{z^{3}(2-\cos{z})}$.

$d) e^{\frac{z}{1-z}}$.

$e) \cot{z}- \frac{1}{z}$.

$f) \cot{z}- \frac{2}{z}$.

$g) \sin{\frac{1}{1-z}}$.

I have classified the obvious points and they have reviewed it, but they tell me that i must classify infinity as well, but i do not know how to do it, i would greatly appreciate your help please.