问:
How can we compute $\int_{0}^{\pi}{-\frac{2\cos^2(x)}{(1+t\cos(x))^3}}dx$
I tried to compute the following integral but it got really messy .. ( we have $|t| < 1$)

$$\int_{0}^{\pi}{-\frac{2\cos^2(x)}{(1+t\cos(x))^3}}dx$$

I tried integration by parts, substitution. There is maybe a trick I don't see. Thanks in advance for your help.

答:
First, try:
$$I(t)=\int_0^\pi\frac{1}{1+t\cos(x)}dx$$
notice that:
$$I'(t)=-\int_0^\pi\frac{\cos(x)}{\left[1+t\cos(x)\right]^2}dx$$
$$I''(t)=\int_0^\pi\frac{2\cos^2(x)}{\left[1+t\cos(x)\right]^3}dx$$
so you just want: $-I''(t)$

$$I(t)=\int_0^\pi\frac{1}{1+t\cos(x)}dx$$
now if we use the substitution $u=\tan(x/2)\Rightarrow dx=\frac{2}{1+u^2}du,\cos(x)=\frac{1-u^2}{1+u^2}$ and we will have the bounds $(0,\infty)$:
$$I(t)=\int_0^\infty\frac{2}{1+t\frac{1-u^2}{1+u^2}}\frac{du}{1+u^2}=\int_0^\infty\frac{2du}{(1+u^2)+t(1-u^2)}=\int_0^\infty\frac{2du}{(1-t)u^2+(1+t)}$$
now I think you can take it from here

We can rewrite this as:
$$I(t)=\frac{2}{1+t}\int_0^\infty\frac{du}{\frac{1-t}{1+t}u^2+1}$$
now if we make the substitution: $\sqrt{\frac{1-t}{1+t}}u=\tan v$ we will get:
$$I(t)=\frac{2}{1+t}\sqrt{\frac{1+t}{1-t}}\int_0^{\pi/2}\frac{\sec^2v dv}{\tan^2v+1}=\frac{\pi}{\sqrt{1-t^2}}$$
Which finally will give us:
$$\mathbb{I}(t)=-I''(t)=-\frac{2\pi t^2+\pi}{(1-t^2)^{5/2}}$$


问:
The degree-2 curves through three points is given by $\gamma L_1L_2+\lambda L_2L_3+\mu L_3L_1=0$, where the $L_i$ are the lines joining those points
I came across this equation while reading on conics and circles.

The family of degree-$2$ curves passing through three points is given by
  $$\gamma L_1L_2 + \lambda L_2L_3 + \mu L_3L_1 = 0$$
  where $L_1$, $L_2$, and $L_3$ are the equations of lines joining the three points, and $\lambda$, $\mu$ and $\gamma$ are parameters.

I can see that the three points definitely satisfy this equation , but I am unable to say much about rest of the points.

How would you prove this relation? Explain, please. :-)

答:
Simple explanation for pairwise satisfying the assembly of straight lines and conic generation from lines .

We know that

$$ \lambda' L_2 + \mu' L_3 = 0$$

represents a straight line passing through the intersection point of $L_2,L_3.$

Divide given relation by $L_1L_2L_3$

$$\gamma/ L_3 + \lambda/ L_1 + \mu /L_2 = 0$$

Extending the same logic for a pair we have a single curve passing through all three intersection points with another combination  parameter set.

Setting one parameter zero three pair-wise solutions for three sets of straight lines gives three points for the vertices of a triangle.

Given second degree equation is a conic which must satisfy/pass through these  three points. That is the triangle is inscribable or escribable in the conic... in a sense as being concurrent.

ContentLicense: CC BY-SA 4.0


问:
Question in Example 2.6.4 Page 57 S. G. Krantz Topology
This example is from S. G. Krantz ( A guide to Topology) and I have a question in an example on page 57 in section: Quotient Topology.

I am unable to understand why this map is 2-to-1 and why 2 points are mapped to same point under π iff a=-b.

答: That $\pi : S^N \to \mathbb P^N$ is $2$-to-$1$ means that there are exactly two points $a,b \in S^N$ such that $\pi(a) = \pi(b)$. Notice that the last equation tells us that $b = \lambda a$ for some non-zero real number $\lambda$, but $$1=\|b\| = \|\lambda a\| = |\lambda|\|a\| = |\lambda|,$$ so $\lambda = -1$ since $a \neq b$. Of couse, if $a=-b$, then $a \sim b$ and $\pi(a) = \pi(b)$.


问:
Is it "legal" to switch $\forall x$ with $\forall s\forall t$?
In the book The Foundations of Mathematics by Kenneth Kunen, in a chapter about set theory formalizations, the following first order logic sentence is provided (referred to as the Axiom of Comprehension/Separation Schema):

$\forall z \Big (\exists y \color{red}{\forall x} \big ( \color{red}{x} \in y \leftrightarrow x \in z \land \varphi(\color{red}{x}) \big ) \Big )$. I have colored in $\color{red}{\text{red}}$ the variable that I will be focusing on for the remainder of this question.

Consider the following two sets:

$A = \{ x \in \mathbb N \ |\  x \lt 5\}$.

$B = \{ \langle s, t \rangle \in \mathbb N \times \mathbb N \ |\ s \lt 5 \land t\lt 4 \}$

Now, let's say I wanted to look at set $A$ and confirm that it exists through the perspective of the Axiom of Comprehension Schema:

$\exists y \forall x (x \in y \ \leftrightarrow \ x \in \mathbb N \land x \lt 5)$ where $\varphi(x):= x \lt 5$. Cool. Set A exists.

To confirm set $B$'s existence, however, I run into an issue when it comes to inserting the predicate $\varphi$:

$\exists y \forall x (x \in y \ \leftrightarrow \ x \in \mathbb N \times \mathbb N \ \land \text{something} )$

I use the phrase "$\text {something}$" because "$\varphi (x)$" does not appear to decompose correctly into a statement about two variables...i.e. $\varphi(x)$ cannot be used to represent "$s \lt 5 \land t \lt 4$" because $s$ and $t$ are not necessarily equal, and therefore do not represent one object $x$.

A quick fix seems to be to switch $\forall x$ with $\forall s \forall t$ in order to generate the following statement:

$\exists y \forall s \forall t (\langle s,t \rangle \in y \ \leftrightarrow \ \langle s,t \rangle \in \mathbb N \times \mathbb N \ \land s \lt 5 \land t \lt 4  )$

I am uncertain why this is something that is allowed. Specifically, this seems to suggest that $\forall x  \ \varphi(x)$ is 'semantically' (regular English definition) equivalent to $\forall s \forall t \ \varphi (s,t)$.

From what I understand about first order logic, $\forall x \ \varphi (x)$ essentially means "For all singular objects in the domain of discourse, the following sentence about a singular object is true". Comparatively, I interpret $\forall s \forall t \ \varphi(s,t)$ as saying "For any two objects in the domain of discourse, the following sentence is true about two objects".

Is the syntax "$\ s,t\ $" a singular object in the domain of discourse? It doesn't strike me as one...hence my confusion.

Any input is greatly appreciated! Thank you.

答:
Here is a possible workaround in order to show that $B$ is a set:

You could use a single variable $w$ (to encode the pair $\langle s,t\rangle$) and then use
$$\phi(w):=\exists s\exists t\;\, w=\,\langle s,t\rangle \land s<5 \land t<4$$


答:
You can choose $\varphi(x)$ to be $x_0<5\land x_1<4$, if you have the notation $x=\langle x_0,\,x_1\rangle$ for $2$-tuples. Mathematicians often abbreviate this by saying $\varphi(\langle s,\,t\rangle)$ iff $s<5\land t<4$, with it being taken as implicit that $\neg\varphi(x)$ whenever $x$ isn't a $2$-tuple (although that wouldn't matter in the present context).

However, the existence of a Cartesian product $X\times Y$, such as $\Bbb N\times\Bbb N$, is proven with several axioms:

First we convince ourselves ordered pairs exist, e.g. on Kuratowski's definition with unordered-pair (an axiom in Zermelo set theory, a theorem in ZF) repeatedly.

Replacement tells us first that $X\times\{a\}$ exists for each $a\in Y$, then that there is a set of these Cartesian products.

We can now take a union to get $X\times Y$.

If this question presupposes we already accept the above, we can prove $B$ exists by writing $B=X\times Y$ with$$X=\{s\in\Bbb N|s<5\},\,Y=\{t\in\Bbb N|t<4\},$$which lets us use comprehension twice, with very obvious choices of $\varphi$.

Title: Is it "legal" to switch $\forall x$ with $\forall s\forall t$?

Body:
In the book The Foundations of Mathematics by Kenneth Kunen, in a chapter about set theory formalizations, the following first order logic sentence is provided (referred to as the Axiom of Comprehension/Separation Schema):

$\forall z \Big (\exists y \color{red}{\forall x} \big ( \color{red}{x} \in y \leftrightarrow x \in z \land \varphi(\color{red}{x}) \big ) \Big )$. I have colored in $\color{red}{\text{red}}$ the variable that I will be focusing on for the remainder of this question.

Consider the following two sets:

$A = \{ x \in \mathbb N \ |\  x \lt 5\}$.

$B = \{ \langle s, t \rangle \in \mathbb N \times \mathbb N \ |\ s \lt 5 \land t\lt 4 \}$

Now, let's say I wanted to look at set $A$ and confirm that it exists through the perspective of the Axiom of Comprehension Schema:

$\exists y \forall x (x \in y \ \leftrightarrow \ x \in \mathbb N \land x \lt 5)$ where $\varphi(x):= x \lt 5$. Cool. Set A exists.

To confirm set $B$'s existence, however, I run into an issue when it comes to inserting the predicate $\varphi$:

$\exists y \forall x (x \in y \ \leftrightarrow \ x \in \mathbb N \times \mathbb N \ \land \text{something} )$

I use the phrase "$\text {something}$" because "$\varphi (x)$" does not appear to decompose correctly into a statement about two variables...i.e. $\varphi(x)$ cannot be used to represent "$s \lt 5 \land t \lt 4$" because $s$ and $t$ are not necessarily equal, and therefore do not represent one object $x$.

A quick fix seems to be to switch $\forall x$ with $\forall s \forall t$ in order to generate the following statement:

$\exists y \forall s \forall t (\langle s,t \rangle \in y \ \leftrightarrow \ \langle s,t \rangle \in \mathbb N \times \mathbb N \ \land s \lt 5 \land t \lt 4  )$

I am uncertain why this is something that is allowed. Specifically, this seems to suggest that $\forall x  \ \varphi(x)$ is 'semantically' (regular English definition) equivalent to $\forall s \forall t \ \varphi (s,t)$.

From what I understand about first order logic, $\forall x \ \varphi (x)$ essentially means "For all singular objects in the domain of discourse, the following sentence about a singular object is true". Comparatively, I interpret $\forall s \forall t \ \varphi(s,t)$ as saying "For any two objects in the domain of discourse, the following sentence is true about two objects".

Is the syntax "$\ s,t\ $" a singular object in the domain of discourse? It doesn't strike me as one...hence my confusion.

Any input is greatly appreciated! Thank you.


问:
True or False: The line y=x is always at 45 degrees to the x-axis?
I have a doubt.

My teacher has posted a question that Is it always/sometimes/never true that the line y=x is at 45 degrees to the x-axis?

I know that the slope of the line y=x is always 1, and the line must make 45 degrees with x-axis.

I am a bit confused when I am plotting the line y=x at different X-scales (see Figures 1 and 2 attached).
I do not visually see the line at 45 degrees to the x-axis in Figure 2.

What am I missing?

Which answer would be correct and why?
(a) It is always true
(b) It is sometimes true because when we change the scale of x or y axis, graph will distort and not be at 45 degrees.

See images attached.
Fig 1 with equal x and y axis
Fig 2 with different axis

答: Yes, you can give different scales to your x and y axes, and then when you draw the line y=x, it will not be at 45 degrees to the x-axis. However, what someone usually means when they say that the line is at 45 degrees, what they almost always mean is that the line would be at 45 degrees if the x and y axes were scaled equally (and at right angles to each other). It is a convention. And it is a useful convention, because it allows us to uniformly represent slopes as the tangents of angles.


问:
Are Pringles an example of anti-de Sitter space?
Pringles, the potato chip, have the classic saddle shape of anti-de Sitter space.  They seem to be the perfect teaching example of anti-de Sitter space, but I've never heard Pringles used in examples. Is there a reason why Pringles arn't used in teaching anti-de Sitter space?

https://en.wikipedia.org/wiki/Anti-de_Sitter_space

答:
A Pringle is likely closer to a hyperbolic paraboloid (saddle):

which is described by $z=ay^2-bt^2$.

Whereas, anti-deSitter space is an embedding of the form $y^2-t^2=-\alpha^2$, which is a "quasi-sphere". Both have negative curvature, but other than that they are different.


问:
How do I prove that if $n$ is greater or equal to $4$, then $(2n)!$ is greater than or equal to $10^n$?
I used the method of induction to prove this.

For the basis step, when $n=4$, the statement is $[2(4)]!≥10^4$ which is the same as $40320≥10000$ which is true.

Next, for the inductive step, we show $\mathrm{S}_k$ implies $\mathrm{S}_{k+1}$ for some integer $n=k≥1$.

By assuming that $(2k)!≥10^k$ is true. Therefore, we need to show $(2k+2)!≥10^{k+1}$ for $n=k+1$. First, $10^{k+1}=10\cdot10^k$. By the induction hypothesis, $(2k)!≥10^k$, so $10\cdot(2k)!≥10\cdot10^k$.

Since $k≥4$, we know that $2(k+1)≥10$.                So, $2(k+1)\cdot(2k)!≥10\cdot10^k$ by substitution. This simplifies to $(2(k+1))!≥10\cdot10^k$ which equals $(2(k+1))!≥10^{k+1}$ which we needed to show.

I know my expansion from $2(k+1)(2k)!$ to $(2(k+1))!$ is wrong, but I don't know what the correct way to get to $(2(k+1))!$ from $2(k+1)(2k)!$.

答:
Assuming $(2n)! \geq 10^n$,

For your induction step -

$(2n+2)!\geq 10^{n+1} \,$ or $\, (4n^2+6n+2)\frac{(2n)!}{10^n} \geq 10$

Is true if $\, (4n^2+6n+2)\geq 10$ (which seems obvious for $n \geq 1$)

Or you can go on as below,

i.e. if $\, 2n^2 + 3n - 4 \geq 0 \implies n \geq \frac{\sqrt41}{4} - \frac{3}{4}$

i.e. if $n\,  \geq 1$, which is true.


问:
Semi direct product-groups that are isomorphic
The questions asks me to find all groups up to isomorphism of the semi direct product $C_5 \rtimes C_4$

Now I've done the working out to get four groups (Note I've used $X$ as an element of $C_5$ and $Y$ as an element of $C_4$ so in all of these $X^5=1$, $Y^4=1$):

a). $YX=XY$ - which abelian so it is just the group $C_{20}$

b). $YXY^{-1}=X^2$

c). $YXY^{-1}=X^4$

d). $YXY^{-1}=X^3$

Now I have found that by changing the generator $Y^3$ to say $W$ in (b) you get the group (d) so (b) and (d) are isomorphic. I also thought that the same applied to (b) and (c) when you change the generator in (b) from $Y^2$ to say $Z$ but in the answers this isn't the case and it only says that (b) and (d) are isomorphic. Can anyone tell me how to spot which groups are isomorphic and which aren't? My notes aren't very clear and I've tried to search online to no avail. Thanks.

答:
You can't change $y^2$ to a generator since $y^2$ isn't a generator of $\mathbb{Z}_4$.  The group c is the binary dihedral group (the double of the symmetries of a 5 gon).  Note that in this group $y^2$ is central since $yxy^{-1}$ flips $x$ around to its inverse, so if you do that twice you're back where you started.  Meanwhile the other two non abelian guys actually have only the identity in their centers! Note that you now need to do $yxy^{-1}$ not twice, but four times to get back where you started.  Think of $y$ as playing the role of multiplication by $-1$ in $\mathbb{Z}_5$ in c.) while playing the role of multiplication by $2$ in $\mathbb{Z}_5$ in the other two (b and d).  If you multiply by $-1$ twice you get $1$ (${(-1)^2=1 \bmod 5}$), but you need to do it four times since in $\mathbb{Z}_5$, ${2^4=16 = 1 \bmod 5}$, but ${2^2=4 \neq 1 \bmod 5}$.

In general, you can show that for $p$ prime, ${\mathbb{Z}_p \rtimes_a \mathbb{Z}_n}$ (the second generator acts as multiplication by $a$ on the first) and
${\mathbb{Z}_p \rtimes_b \mathbb{Z}_n}$ (the second generator acts as multiplication by $b$ on the first or if you prefer to write things multiplicatively like you did above, acting by raising the first generator to the $a$th power) are isomorphic iff the multiplicative orders of $a$ and $b$ in $\mathbb{Z}_p$ are the same.

For instance, $\mathbb{Z}_{101} \rtimes_{19} \mathbb{Z}_{25}$ and $\mathbb{Z}_{101} \rtimes_{36} \mathbb{Z}_{25}$ are not isomorphic (but both make sense since both when raised to the $25$th power are equal to $1$) since $36^5=1 \bmod 101$, but $19^5 \neq 1 \bmod 101$.

ContentLicense: CC BY-SA 3.0


问:
Finding a non-linear equation from a set of points
I'm trying to figure this out and I need help.

I have a set of points and I know they are quadratic because the differences between the y points aren't the same.

But how do I find the equation of a nonlinear equation?

X  |  Y

-2 | -1

-1 | -4

3 |  4

4 | 11

答:
First you posit a quadratic equation in unknowns a, b, c thus:
Y = a$X^2$ + b$X$ + c.  Now plug in your five X and Y values, yielding five (linear!) equations in a, b, c.  Solve for a, b, c (I assume you know how to solve linear equations in three unknowns).


答:
It is always at $45^{\circ}$. It may not "look" like $45^{\circ}$, but that's because you're "cheating" and measuring the angle with the original coordinate system.

Then again, if you measure the coordinates of a point $(x,y)$ on the line $y=x$ in the scaled system using the original coordinate system, then clearly $y\neq x$.

So is $y=x$ or not? Clearly it is.

This is like drawing a circle on a piece of paper, then placing the paper flat on a table and squatting down near the level of the table and looking at the circle nearly edge-on. It looks squished and no longer looks like a circle but instead looks like a flattened ellipse. But of course, it is a circle.

Title: True or False: The line y=x is always at 45 degrees to the x-axis?

Body:
I have a doubt.

My teacher has posted a question that Is it always/sometimes/never true that the line y=x is at 45 degrees to the x-axis?

I know that the slope of the line y=x is always 1, and the line must make 45 degrees with x-axis.

I am a bit confused when I am plotting the line y=x at different X-scales (see Figures 1 and 2 attached).
I do not visually see the line at 45 degrees to the x-axis in Figure 2.

What am I missing?

Which answer would be correct and why?
(a) It is always true
(b) It is sometimes true because when we change the scale of x or y axis, graph will distort and not be at 45 degrees.

See images attached.
Fig 1 with equal x and y axis
Fig 2 with different axis


问:
If $X_1$ and $X_2$ are independent random variables with distribution $N(0,1)$, then proof the sum of $X_{1}^2 + X_{2}^2$ has gamma distribution.
I know $Y=X^2$ has density function $\frac{1}{\sqrt{2y\pi}}e^{\frac{-y}{2}} $ and this is gamma distribution $ G(\frac{1}{2},\frac{1}{2})$. My question is how can I do the sum of $X_{1}^2 + X_{2}^2$ and that is equal to $G(1, \frac{1}{2})$.

答:
One way is polar coordinates:
\begin{align}
& F_Y(y) = \Pr(Y\le y) \\[8pt]
= {} & \iint\limits_{x_1,x_2\,:\,x_1^2\,+\,x_2^2\,\le\,y} \frac 1 {2\pi} e^{-x_1^2/2} e^{-x_2^2/2} \, d(x_1,x_2) \\[8pt]
= {} & \iint\limits_{x_1,x_2\,:\,x_1^2\,+\,x_2^2\,\le\,y} \frac 1 {2\pi} e^{-(x_1^2+x_2^2)/2} \, d(x_1,x_2) \\[8pt]
= {} & \int_0^{2\pi} \left( \int_0^{\sqrt y} \frac 1 {2\pi} e^{-r^2/2} (r\,dr) \right) \, d\theta
\end{align}
The inner integral does not depend on $\theta,$ i.e. it is constant as $\theta$ goes from $0$ to $2\pi,$ so this expression is the length, $2\pi-0$ of the interval times that constant. Thus it is
$$
\int_0^{\sqrt y} e^{-r^2/2} (r\,dr) = \int_0^{y/2} e^{-u} \, du
$$
(When $r = \sqrt y$ then $u = r^2/2 = y/2.$)

Also it is sometimes useful to know that if independent random variables have distributions
$$
\frac 1 {\Gamma(\alpha_i)} (\lambda y)^{\alpha_i-1} e^{-\lambda y} (\lambda\,dy) \text{ for } y>0\quad \text{for } i=1,2,
$$
then the sum of those two random variables have distribution
$$
\frac 1 {\Gamma(\alpha_1+\alpha_2)} (\lambda y)^{\alpha_1+\alpha_2-1} e^{-\lambda y} (\lambda\,dy) \text{ for } y>0.
$$
In your example, you have $\alpha_1=\alpha_2 = \frac 1 2,$ so $\alpha_1+\alpha_2-1=0.$ And $\lambda=1/2.$


问:
Maximizing Pieces of fabric cut out of a roll
I have a problem I am trying to solve in google sheets.

I have a roll of fabric that is 120 yards long of that I need to cut into 1 yard, 2 yard, 3 yard, 4 yard, and 5 yard pieces. We have varying demand for each size.

Our historic sales ratio is thus:
1 yard 53%
2 yard 24%
3 yard 15%
4 yard 4%
5 yard 5%

I am trying to make a calculator that will take an input of quantity of rolls of fabric and break it up into a piece count that will not exceed 120 Linear yards.

With tweaking numbers around I found an equation that makes it work, but I don't know why it works. Could someone help me understand why my work around works and also perhaps help me design an equation that automatically optimizes the product qty with a maximum constraint?

Example:

For each product (which is the 1, 2, 3, 4, and 5 yard pieces) I made the follow equation

1 yard:    .53120.54 = 34
2 yard:    .24120.54 = 16
3 yard:    .15120.54 = 9
4 yard:    .4120.54 = 2
5 yard:    .5120.54 = 3

then

(1yard * 34 pieces)+(216)+(39)+(42)+(53)= 118

for some reason .54  works on keeping the sum of fabric piece length under 120yard

Is there a function of equation I can use to optimize the piece count while using my historic sales ratios?

Thanks so much!

答:
It's very hard to read what you have typed here.  Please use MathJax if you make any future posts.

One problem is that you sales figures add up to $101\%$  You'd get better results if you carry another decimal place or two.  Anyway, what you're trying to do is to figure out how much of the fabric is sold in each size.

Using $100$ sales as a base, we sell $53$ yards in one-yard pieces, $48$ yards in two-yard pieces, $45$ yards in $3$-yard pieces, $16$ yards in $4$ yard pieces, and $25$ yards in $5$-yard pieces, for a total of $187$ yards.  Now, we only have $120$ yards, so we want to adjust the figures by multiplying by $\frac{120}{187}\approx.6417$.

We have two problems.  The first is that this doesn't give whole numbers, but this is easily dealt with by rounding.  The second is that we don't get numbers that are easily divided into pieces of the right size.  For example, we get that $28.877$ yards of fabric should be cut into $3$-yard pieces.  Should we make $9$ or $10$ pieces?

In short, there isn't a way to solve the problem exactly.  I suppose you could do better if you have a great many rolls, and you're willing to divide them in a number of different ways, but that may be more trouble than it's worth.


问:
L2-limit of sum similar to the Itô integral
We define a sequence of partitions $\Pi_{n} = \{0=t_{0} < t_{1} <\ldots<t_{k_{n}} =T\}$ for $n\geq1.$
For any $\lambda \in[0,1]$ we set $t^{\lambda}_{n,j}=\lambda t_{j}+(1-\lambda)t_{j-1}.$
Then in $L^2$ $$\lim_{n\to\infty} \sum_{j=1}^{k_{n}} W_{t^{\lambda}_{n,j}} \left(W_{t_{j}}-W_{t_{j-1}}\right) = \int_{0}^{T} W_{t}\,dW_{t}+\lambda T,$$
where $\int_{0}^{T}W_{t}\,dW_{t}$ is the Itô integral and $W_t$ the Brownian motion.

I do not know where to start to show this limit. Clearly for $\lambda=0$ we get the usual definition of the Itô integral. I tried to consider $(W_{t^{\lambda}_{n,j}}-W_{t_{j-1}}+W_{t_{j-1}})$, but I could not solve that either. Any idea?

答:
The trick is to rewrite the sum to

$$\sum_{j=1}^{k_{n}} W_{t_{j-1}} \left(W_{t_{j}}-W_{t_{j-1}}\right) +\sum_{j=1}^{k_{n}} (W_{t^{\lambda}_{n,j}}-W_{t_{j-1}})^{2}
   + \sum_{j=1}^{k_{n}} (W_{t^{\lambda}_{n,j}}-W_{t_{j-1}}) \left(W_{t_{j}}-W_{t^{\lambda}_{n,j}}\right).$$

Then the first sum is clear, the second is almost a quadratic variation (gives $\lambda T$) and the last one goes to 0.


问:
Polynomials in the Pancake problem
I noticed something interesting in this table. The columns can be expressed by polynomials of degree k. I toke the first $k+1$ numbers from each column and used Lagrange's interpolation. Surprisingly, I got that this extrapolation gives exact values of all other numbers in the column. I can't check this for $k=7$.
$$k=0: 1$$
$$k=1: n-1$$
$$k=2: n^2-3n+2$$
$$k=3: n^3-5n^2+8n-5$$
$$k=4: n^4-\frac{15}2n^3+\frac{29}2n^2+3n-17$$
$$k=5: n^5-\frac{65}6n^4+\frac{173}6n^3+\frac{148}3n^2-\frac{862}3n+265$$
$$k=6: n^6-\frac{883}{60}n^5+\frac{157}3n^4+\frac{2155}{12}n^3-\frac{4570}3n^2+\frac{42767}{15}n-967$$
$$k=7: \frac{1679}{1680}n^7-\frac{2765}{144}n^6+\frac{21541}{240}n^5+\frac{69163}{144}n^4-\frac{717821}{120}n^3+\frac{1462277}{72}n^2-\frac{10388033}{420}n+5037$$

If I set the $n$ to one where the first positive value occurs, it is:
$$k=0: 1$$
$$k=1: n$$
$$k=2: n^2+n$$
$$k=3: n^3+n^2-1$$
$$k=4: n^4+\frac92n^3+n^2-\frac92n+1$$
$$k=5: n^5+\frac{55}6n^4+\frac{31}2n^3-\frac{14}3n^2-2n+1$$
$$k=6: n^6+\frac{917}{60}n^5+\frac{713}{12}n^4+\frac{565}{12}n^3-\frac{5}{12}n^2+\frac{409}{30}n-3$$
$$k=7: \frac{1679}{1680}n^7+\frac{142}{9}n^6+\frac{192}{5}n^5-\frac{3743}{36}n^4-\frac{18917}{240}n^3+\frac{14119}{36}n^2-\frac{50761}{140}n+100$$

Questions:

Can it be proven that the colums of that table are always polynomials?

Can be the coefficients predicted in a way? There seems to be an order but I can't find it.

答:
This is a very very very partial answer, treating only cases k = 0, 1, 2, 3. But I hope it inspires some further work on the remaining cases.

Think of the numbers as points at distance $k$ from a fixed base point in the pancake graph. We think about the basepoint as the pile of pancakes that is completely sorted in the desired way. Vertices (stacks) that are reachable in $k$ steps from this point (but not in less steps) are exactly the stacks that need $k$ steps to reach the base point - which is what the table is counting. (Of course from the pancake graph perspective, the choice of base point is not important since the group $S_n$ acts transitively on the vertices of the graph by graph isomorphisms.)

The first three columns are easy. Of course there is only point at distance 0 from the basepoint: the base point itself. Also as every vertex in the pancake graph has degree $(n-1)$ (there are at each moment $n-1$ possible flips to make) it is clear that each point has $n-1$ neighbors. This is the $k=1$ column.

Similarly: each of these $n-1$ neighbors has $n-2$ neighbors apart from the base point, so the number of points at distance $k = 2$ should be $(n-1)(n-2)$. Expanding the brackets we recover your result: $n^2 - 3n + 2$.

Alas this strategy breaks down for $k = 3$. The pancake graphs contain a lot of hexagons (isomorphic embeddings of the 3-pancake graph) which means that if we take the naive prediction of $(n-1)(n-2)(n-2)$ we count some, and perhaps many, points twice or perhaps even more often it there are multiple hexagons containing both the base point and the point of interest. So the answer will certainly be less than $(n-1)(n-2)(n-2)$, but by how much?

Well lets first rewrite this crude upper bound in the form you use, so without brackets. We get $n^3 - 5 n^2 + 8 n - 4$. This is very interesting, it is only 1 more than what you find! This means that of the $n^3 - 5 n^2 + 8 n - 4$ points at distance 3 there is only one that is double counted!

Knowing this it is not hard to see which one is it: we already mentioned the 3-pancake subgraph. It consists of moves that only move the top 3 pancakes. Apparently:

Any set of k=3 moves starting at the completely sorted stack that moves a pancake of size 4 or higher to a position it did not have before, reaches a state that cannot be reached in 3 steps in any other way.

Instead of writing 'apparently' we could also just post this as a lemma, prove it and derive the formula for the $k = 3$ case from it.

Now the proof of this lemma is not so hard because 3 is a really really small number. The real question is:

what is the proper generalization of this lemma to other values of $k$?

I will think about that and come back to it later.

Title: Polynomials in the Pancake problem


答:
I'm going to say it's a meaningless question.

If you skew the graph by having the x axis and y axis be on different scales, or by depicting the x, y axis so they are not perpendicular, Or having the scale of the y axis be linear but the x axis be logrithmic or geometric or exponential (in which case $y=x$ won't even be a line) then no, the angle of that drawing of the function is not at $45^\circ$).

But those are skewed interpretations of "the" graph and just interpretive drawings.  The are not "the" graph.

We define the "graph" of $y=x$ as: The set of all $(x,y)$ points in $\mathbb R^2$ where $(x,y)$ satisfy the equation $y=x$.  But that's not a drawing... thats a bunch of points.  What does it mean to say they are a "line" or that they have an angle?  Well, it is presumed that we are depicting them on "THE" Cartesian plane and THE Cartesian plane has perpendicular $x$, $y$ axes on the same scale.  And on that plane with that scale, and those axis, yes, there is only one instance of a graph $y= x$ and it is at an $45^\circ$ angle.

Graphing it on a skewed axis.... is simply not on the cartesian plane.  They are another interpretation.

But it is vague as to what we actually "mean" when we say "$y=x$ is always at a right angle".  Does that mean on the universal Cartesian plane?  In which case the answer is "YES".  Or does it mean any possible depiction? In which case the answer is "NO".


问:
Is there a formula to calculate the minimum height of an n-nary tree with L leaves?
I'm trying to figure out if there is a way to calculate the minimum height of an n-nary tree with L leaves. Is there such a formula?</p>

答:
I found it:

$$ \text{minimum height}= \lceil \log_n L \rceil$$


答:
Alternative proof without induction:

Firstly, if $n=4, (2n)! = 8! = 40320 > 10^4 = 10^n$.

If $n\ge 5$, note that
$$\forall 1\le i \le n, i(2n+1-i)=2n + (i-1)(2n-i) \ge 2n \ge 10.$$
Therefore $$(2n)! = \underset{i=1}{\overset{n}{\Pi}} \left[i(2n+1-i)\right] \ge  10^n. \blacksquare$$

Title: How do I prove that if $n$ is greater or equal to $4$, then $(2n)!$ is greater than or equal to $10^n$?

Body:
I used the method of induction to prove this.

For the basis step, when $n=4$, the statement is $[2(4)]!≥10^4$ which is the same as $40320≥10000$ which is true.

Next, for the inductive step, we show $\mathrm{S}_k$ implies $\mathrm{S}_{k+1}$ for some integer $n=k≥1$.

By assuming that $(2k)!≥10^k$ is true. Therefore, we need to show $(2k+2)!≥10^{k+1}$ for $n=k+1$. First, $10^{k+1}=10\cdot10^k$. By the induction hypothesis, $(2k)!≥10^k$, so $10\cdot(2k)!≥10\cdot10^k$.

Since $k≥4$, we know that $2(k+1)≥10$.                So, $2(k+1)\cdot(2k)!≥10\cdot10^k$ by substitution. This simplifies to $(2(k+1))!≥10\cdot10^k$ which equals $(2(k+1))!≥10^{k+1}$ which we needed to show.

I know my expansion from $2(k+1)(2k)!$ to $(2(k+1))!$ is wrong, but I don't know what the correct way to get to $(2(k+1))!$ from $2(k+1)(2k)!$.


问:
How large should n be to guarantee the error in approximating ln(.5) with ln(1+x) centered at a = 0 is less than .0001?
I am asked to find the number of terms such that the error of approximating ln(.5) < 0.001 using Taylor's inequality. I can do these using the remainder theorem from the alternating series test, but we are required to use Taylor's inequality.

I am struggling right now to find M such that
$ |f^{n+1}(x)| \le M$.

I have found the derivative
$|f^{n+1}(x)|= \frac{n!}{(1+x)^{n+1}}$. My problem is with finding M. I know that $x \in (-.5, 0)$, and the derivative (absolute value) is decreasing. If I plug -.5 into the derivative to get M, M seems to just blow up and isn't bounded.

Is there any other way to find M in this situation?

答:
Since $\ln(1+x)=x-x^2/2+x^3/3+\dots+\mathcal o(x^n)$, we just need to choose $n$ so that $1/2^n\lt1/10000$.  Since $2^{17}\gt10000$, $n=17$ will work.

For a better estimate, note that we actually have the $n$-th term equal to $(-1)^{n+1}x^n/n$ , and so we can get away with $n=13$.


答:
With induction but more clear:

You have proved the base case, so let's continue.

Let $(2n)!\ge 10$. Then:
$$(2(n+1))!=(2n+2)!=(2n+2)(2n+1)(2n)!=(4n^2+6n+2)(2n)!\ge (4n^2+6n+2)\cdot 10^n$$

Clearly $4n^2+6n+2 \ge 10$ for $n \ge 1$ so:
$$(2(n+1))! \ge (4n^2+6n+2)\cdot 10^n \ge 10\cdot 10^n=10^{n+1}$$
$$(2(n+1))! \ge 10^{n+1}$$
And we have proved the inductive step $\blacksquare$

Just to clarify, you were actually almost there. We know that $2(k+1)\cdot (2k)! = (2k+2) \cdot (2k)! \le (2k+2)(2k+1) \cdot (2k)! = (2(k+1))!$

If $2(k+1)\cdot (2k)! \ge 10^{k+1}$ and $(2(k+1))! \ge 2(k+1)\cdot (2k)!$ then $(2(k+1))! \ge 10^{k+1}$


问:
Integral $\int \frac{2x^2-1}{\pm x(2x^2-1)+2x+ \sqrt{1+2x^2}} d x$
Consider the indefinite integral
$$F_{\pm}(x):=\int \frac{2x^2-1}{\pm x(2x^2-1)+2x+ \sqrt{1+2x^2}} \mathbb{d} x.$$
For all $x\in \mathbb{R}$ it is
$$ F_{+}(x)=  \log(-x + \sqrt{1 + 2 x^2}).$$
But what is $F_{-}$?

Note that numerical studies suggest that it does exist in a certain range. Math software already struggles to prove the formula for $F_{+}$.

答:
The problem is $$F_\pm(x)=\int\frac{2x^2-1}{\pm x(2x^2-1)+2x+\sqrt{1+2x^2}}\,\mathrm{d}x.$$ Let $x=2t/(2-t^2)$, then $$\begin{aligned}F_+(t)&=2\int\bigg(\frac{t}{2-t^2}+\frac{1+t}{t^2+2t+2}\bigg)\,\mathrm{d}t,\\F_-(t)&=2\int\bigg(\frac{t}{t^2-2}-\frac{t^3-t^2-6t-2}{t^4-2t^3-12t^2-4t+4}\bigg)\,\mathrm{d}t.\end{aligned}$$ You can integrate $F_+$ directly: $$F_+(t)=-\log(2-t^2)+\log(t^2+2t+2)+C.$$
For $F_-$ you need only solve a quadratic equation to factor the denominator and then integrate directly: $$\begin{aligned}F_-(t)=\log(2-t^2)&+\frac{-17+\sqrt{17}}{34}\log(t^2+(\sqrt{17}-1)t+2)\\&+\frac{-17-\sqrt{17}}{34}\log(-t^2+(\sqrt{17}+1)t-2)+C.\end{aligned}$$
After this, solve a quadratic equation to get a formula for $F_\pm(x)$.


问:
Do endomorphisms of the adjoint representation of a Lie algebra commute?
Given a field $k$ of characteristic $0$ and a finite dimensional simple Lie algebra $\mathfrak{g}$ over $k$. Consider the adjoint representation $(\mathfrak{g}, \mathrm{ad})$ of $\mathfrak{g}$ and let $\mathrm{End}(\mathfrak{g}, \mathrm{ad})$ denote the ring of $\mathfrak{g}$-module endomorphisms with regard to this representation.

The claim is: $\mathrm{End}(\mathfrak{g}, \mathrm{ad})$ is a field extension of $k$ and $\dim_k\mathrm{End}(\mathfrak{g}, \mathrm{ad})$ is equal to the number of simple components of $g \otimes_k \overline{k}$ where $\overline{k}$ denotes the algebraic closure of $k$.

I've come this far: Since $\mathfrak{g}$ is simple, the adjoint representation must be irreducible. Thus, $\mathrm{End}(\mathfrak{g}, \mathrm{ad})$ is a $k$ skew-field. But why do all elements from $\mathrm{End}(\mathfrak{g}, \mathrm{ad})$ commute?
If $A, B \in \mathrm{End}(\mathfrak{g}, \mathrm{ad})$, one can argue that $AB - BA$ is either $0$ or invertible. However, I wasn't able to rule out the latter case.

Also: How are $\mathfrak{g} \otimes_k \overline{k}$ and $\mathfrak{g}$ related with regard to (semi)-simplicity? I know that $\mathfrak{g}$ is semi-simple iff $\mathfrak{g} \otimes_k \overline{k}$ is semi-simple. Is there a corresponding result for the simple case?

答:
$\DeclareMathOperator{\g}{\mathfrak g}$
$\DeclareMathOperator{\ad}{\mathrm{ad}}$
$\DeclareMathOperator{\End}{\mathrm{End}}$

I tried to give a short intro to this theory in section 4.1 of my thesis which generally follows Jacobson, N.: A note on non-associative algebras. Duke Math. J. 3 (1937), no. 3, 544--548. doi:10.1215/S0012-7094-37-00343-0. Here is the part relevant to your question:

Re first question:

For a $k$-Lie algebra $\g$ define

$$K := K(\g) := \{ s \in \End_k(\g): s \circ \ad_{\g}(x) = \ad_{\g}(x) \circ s \text{ for all } x \in \g \}.$$

We view it as associative $k$-algebra and remark that as such it identifies with what you call $\End(\g, \ad)$.

If $\g$ is simple, then (as you remark) $K$ is a skew field by Schur's lemma.

In fact, it is a field; namely, since $\g = [\g, \g]$ it suffices to see that two elements $s, t \in K$ commute on a commutator $[x,y]$ for $x,y \in \g$. But
$$ s(t([x,y])) = s([x, ty]) = [sx, ty] = t([sx, y]) = t(s([x,y])) $$
where we have used, from left to right, that $t$ commutes with $\ad_{\g}(x)$, $s$ with $-\ad_{\g}(ty)$, $t$ with $\ad_{\g}(sx)$ and $s$ with $-\ad_{\g}(y)$.

One calls $K$ the centroid of $\g$ and remarks that $\g$ has a natural structure as Lie algebra over $K$. When viewed as such, write $^K \g$.

Re second question:

First, some notation. For a Lie algebra $\g$ over $k$, let $A(\g)$ be the (associative, unital) $k$-subalgebra of $\End_k(\g)$ generated by all $\ad_{\g}(x)$, $x \in \g$. Remark straightaway that for any field extension $L|k$, $a \otimes \ad_{\g}(x) \mapsto \ad_{\g_L} (a \otimes x)$ defines a natural isomorphism of associative $L$-algebras:

$$(*) \qquad L \otimes_k A(\g) \cong A(\g_L)$$

Also remark that $\g$ is a (left) $A(\g)$-module, and that an ideal of $\g$ is the same as an $A(\g)$-submodule.

Further, the inclusion $A(\g) \subseteq \End_k(\g)$ factors through natural maps $A(\g) \hookrightarrow \End_K(^K\g) \hookrightarrow \End_k(\g)$, and the first arrow is bijective by Jacobson's density theorem. (The theorem is absent from Jacobson's paper I quoted above, as he only proved it eight years later!) Consequently, the following are equivalent:

$\g$ is simple and $K = k$

$A(\g) = \End_k(\g)$.

In this case we call $\g$ central simple. So e.g. $^K\g$ is central simple if $\g$ is simple. It follows from $(*)$ that every scalar extension of a central simple Lie algebra is again central simple, a fortiori absolutely simple (A Lie algebra $\g$ over $k$ is called absolutely simple if $\g_{\bar k} := \g \otimes_k \bar k$ is simple over $\bar k$, or equivalently, $\g_K$ is simple over $K$ for every extension $K|k$.). But we have much more:

Proposition (4.1.2 in my thesis): Let $\g$ be a simple Lie algebra and $L|k$ a Galois extension containing the centroid $K$. Then $\g_L \simeq \g_1 \times ... \times \g_r$ where $r = [K:k]$ and the $\g_i$ are absolutely simple Lie algebras over $L$. In particular, $\g$ is central simple if and only if it is absolutely simple.

Proof:  Writing $K = k[X]/(f)$ where $f$ is a minimal polynomial of a primitive element of $K|k$, we have $L \otimes_k K \cong \prod_{i=1}^r L_i$ (as $L$-algebras) where the $L_i$ are all $L$ but with an $L$-action twisted via certain elements $\sigma_i : L \simeq L_i$ of the Galois group $Gal(L|k)$, permuting the zeros of $f \in L[X]$. In particular, $r = [K:k]$. Then with $(*)$,
\begin{align*}
A(\g_{L}) &\cong L \otimes_k \End_K(^K\g) \cong \End_{L\otimes_k K}((L \otimes_k K) \otimes_K (^K\g) ) \\
&\cong \End_{\prod_{i=1}^r L_i} (\bigoplus_{i=1}^r (^K\g)_{L_i}) \cong \prod_{i=1}^r \End_{L_i}((^K\g)_{L_i}).
\end{align*}
Calling $e_i$ the $i$-th idempotent in the last product, the $A(\g_L)$-module $e_i \cdot \g_L$ is a simple ideal $\g_i$ in $\g_L$, which is in fact the simple $L$-Lie algebra deduced from $(^K\g)_L$ by scalar extension (i.e. twisting the $L$-action) with $\sigma_i$.


问:
Finding largest $r$ and smallest $q$ in $\mathbb{N}$ so that $\exp(r)\lt\int_{1}^{2}\exp(x^3+x)\mathrm dx\lt \exp(q)$ by hand
Initially I was given that $k_{1}\le \int_{1}^{2}\exp(x^3+x)\mathrm dx\le k_{2}$, with few options of what $k_{1},k_{2}$ could be. With the very basic technique of approximation, one can argue that the definite integral in question lies between $\exp(2)$ and $\exp(10)$. But I could not arrive at a direct method for comparing the integral with a different value inside this interval say $\exp(3)$.

So what if we were supposed to find the largest $r$ and smallest $q$ with $r,q\in\mathbb{N}$ such that $\exp(r)\lt\int_{1}^{2}\exp(x^3+x)\mathrm dx\lt\exp(q)$. Any hints are appreciated. Thanks.

Body:
Let $I = \displaystyle\int_{1}^{2}e^{x^3+x}\,dx$. Using integration by parts, we have
\begin{align*}
I &= \int_{1}^{2}\dfrac{(3x^2+1)e^{x^3+x}}{3x^2+1}\,dx
\\
&= \left[\dfrac{e^{x^3+x}}{3x^2+1}\right]_{1}^{2}+\int_{1}^{2}\dfrac{6xe^{x^3+x}}{(3x^2+1)^2}\,dx
\\
&= \left(\dfrac{e^{10}}{13}-\dfrac{e^2}{4}\right)+\int_{1}^{2}\dfrac{6xe^{x^3+x}}{(3x^2+1)^2}\,dx
\end{align*}

You can check that $\dfrac{12}{169} \le \dfrac{6x}{(3x^2+1)^2} \le \dfrac{3}{8}$ and $e^{x^3+x} \ge 0$ for all $1 \le x \le 2$. Hence, $$\dfrac{12}{169}I = \int_{1}^{2}\dfrac{12}{169}e^{x^3+x}\,dx \le \int_{1}^{2}\dfrac{6xe^{x^3+x}}{(3x^2+1)^2}\,dx \le \int_{1}^{2}\dfrac{3}{8}e^{x^3+x}\,dx = \dfrac{3}{8}I.$$

Using the above facts, we have $$\left(\dfrac{e^{10}}{13}-\dfrac{e^2}{4}\right)+\dfrac{12}{169}I \le I \le \left(\dfrac{e^{10}}{13}-\dfrac{e^2}{4}\right)+\dfrac{3}{8}I,$$ which can be rearranged to $$\dfrac{169}{157}\left(\dfrac{e^{10}}{13}-\dfrac{e^2}{4}\right) \le I \le \dfrac{8}{5}\left(\dfrac{e^{10}}{13}-\dfrac{e^2}{4}\right).$$

Using $e^2 \le 8$, the upper bound can be relaxed to $I \le \dfrac{8}{5}\left(\dfrac{e^{10}}{13}-\dfrac{e^2}{4}\right) \le \dfrac{8}{65}e^{10} \le \dfrac{64}{65}e^8 \le e^8$.

Using $e^8 \ge e^3 \ge 20$, the lower bound can be relaxed to $I \ge \dfrac{169}{157}\left(\dfrac{e^{10}}{13}-\dfrac{e^2}{4}\right) \ge \dfrac{e^{10}}{13}-\dfrac{e^2}{4} = \dfrac{e^{10}}{13}\left(1-\dfrac{13}{4e^8}\right) \ge \dfrac{e^{10}}{13}\left(1-\dfrac{13}{80}\right) \ge \dfrac{e^{10}}{20} \ge e^7$.

Hence, $e^7 \le I \le e^8$.


问:
probability to reach a spesific point before other
a frog can jump each time forward or backward (one step) with an equal probability of 0.5,
the frog stands in point 0,
what is the probability that the frog will reach to point 100 before point -1?
tried to look at this answer but not sure that it is the same
The probability of reaching a certain amount of money before going bust in a casino game.</p>

答:
Let $p_k$ be the probability of the frog's reaching $100$ before $-1$, if it is currently at $k$, where $1\leq k\leq100$.  We have $$p_k=\cases{
0,&$k=-1$\\
1,&$k=100$\\
\frac{p_{k-1}}2+\frac{p_{k+1}}2,&otherwise}$$

It is easily seen that $$p_k=\frac{k+1}{101}$$ is the solution, so the desired probability is $$p_0=\frac1{101}$$


问:
Finding the limit when denominator = 0
I'm having trouble solving this limit:

$$\lim_{x \to -2^-} \frac{1}{(x + 2)^2}$$

I can't find a way to rationalize the denominator. Also, is there a way to do it without plugging in -2.001 and stuff or graphing it?

EDIT:

I realized after asking this question that it doesn't matter if you take the above limit from the right, left, or both. It's always $+\infty$. Here's an equation that gives $+\infty$ from the right and $-\infty$ from the left: $$\lim_{x \to 3^+} \frac{x - 4}{x - 3}$$

How do I (algebraically) determine if it is positive or negative?

答: As $x \to -2^{-}$, your fraction becomes of form $\frac{1}{\epsilon}$, where $\epsilon$ is an arbitrarily small number, and 1 divided by a 'small' number is clearly a 'large' number (the smaller the denominator, the larger the value of the fraction). I hope you can see it from here.</p>

Title: Finding the limit when denominator = 0


问:
Determining if function sequence is not uniformly convegent.
We let $f_n(x)=\frac{1}{1+x^{2n}}$

I want to prove that the sequence is not uniformly convergent on $(-1,1)$, but is uniformly convergent on any compact set in $(-1,1)$

One possibility is to use a sequence $x_n$, converging to 1, and show that $|f_n(x_n)-f(x_n)|\geq\epsilon$

If i choose $x_n =(\frac{1}{2})^{1/n}$, will that work?

答:
hint

If the convergence to $f:x\mapsto 1 $ at $ (-1,1) $ was uniform, we would have

$$\lim_{n\to+\infty}\lim_{x\to 1^-}f_n(x)=$$

$$\lim_{x\to 1^-}\lim_{n\to +\infty}f_n(x)$$

Other approach. $ f_n $ is an even function, we study it at $ [0,1)$.

$$g_n(x)=|f_n(x)-1|$$
$$=\frac{x^{2n}}{1+x^{2n}}$$

$$g'_n(x)(1+x^{2n})^2=2nx^{2n-1}$$

$ g_n $ is strictly increasing at $ [0,1) $ and
$$\sup_{0\le x<1}g_n(x)=\lim_{1^-}g_n(x)=\frac 12$$
the convergence is not uniform at $ (-1,1)$

but at $ [0,A] $, $$\sup_{[0,A]}g_n=g_n(A)\to 0$$

which proves the uniform convergence at $ [-A,A] $.


问:
Why is this a valid conclusion?
I have the following argument:

S1: $ (\neg (R \vee Q)) \wedge P $

S2: $ P \rightarrow Q $

S3: $ P \wedge Q $

Conclusion :

S: $R$

When I run it through a logic checker, it’s valid. The premises contradict each other, though, so I’m confused why this is the case.

答:
The premises of this argument are unsatisfiable (there exists no interpretation which makes all of the premises true).

Therefore, any formula is a logical consequence of these premises.

Indeed, this is the case for any set of unsatisfiable premises.

To see why this is the case, note that

$ [((\neg (R \vee Q)) \wedge P) \wedge (P \rightarrow Q) \wedge (P \wedge Q)] \rightarrow R$

is a Tautology as the antecedent of the implication is a Contradiction.

Therefore, $ R $ is a logical consequence of the premises:

$ (\neg (R \vee Q)) \wedge P, P \rightarrow Q, P \wedge Q \vDash R $

Title: Why is this a valid conclusion?


问:
The orthogonal projection of the vector b is not equal to $AA^+b$
Given matrix $A:= \begin{pmatrix}1&2\\3&6\end{pmatrix}$

I am trying to solve this task:

... compute $P:=AA^+$. Finally, evaluate $Pb$ with
$b:=\begin{pmatrix}2\\4\end{pmatrix}$, and show that $Pb$ is indeed
the orthogonal projection of the vector b onto the subspace by the
column vectors of the matrix A.

$A^+=\frac{1}{50}
\begin{pmatrix}1&3\\2&6\end{pmatrix}$

$P=AA^+$

$P=
\begin{pmatrix}1&2\\3&6\end{pmatrix}
\frac{1}{50}
\begin{pmatrix}1&3\\2&6\end{pmatrix}$

$P=
\frac{1}{10}
\begin{pmatrix}1&3\\3&9\end{pmatrix}$

$Pb=
\frac{1}{10}
\begin{pmatrix}1&3\\3&9\end{pmatrix}
\begin{pmatrix}2\\4\end{pmatrix}
=\frac{7}{5}\begin{pmatrix}1\\3\end{pmatrix}$

$proj_A(b)=\frac{u_1b}{u_1u_1}u_1+\frac{u_2b}{u_2u_2}u_2$

$proj_A(b)=\frac{\begin{pmatrix}1\\3\end{pmatrix}\begin{pmatrix}2\\4\end{pmatrix}}{\begin{pmatrix}1\\3\end{pmatrix}\begin{pmatrix}1\\3\end{pmatrix}}\begin{pmatrix}1\\3\end{pmatrix}+\frac{\begin{pmatrix}2\\6\end{pmatrix}\begin{pmatrix}2\\4\end{pmatrix}}{\begin{pmatrix}2\\6\end{pmatrix}\begin{pmatrix}2\\6\end{pmatrix}}\begin{pmatrix}2\\6\end{pmatrix}$

$proj_A(b)=\frac{14}{10}\begin{pmatrix}1\\3\end{pmatrix}+\frac{28}{40}\begin{pmatrix}2\\6\end{pmatrix}$

$proj_A(b)=\frac{14}{10}\begin{pmatrix}1\\3\end{pmatrix}+\frac{14}{10}\begin{pmatrix}1\\3\end{pmatrix}$

$proj_A(b)=\frac{28}{10}\begin{pmatrix}1\\3\end{pmatrix}=\frac{14}{5}\begin{pmatrix}1\\3\end{pmatrix}$

$Pb$ is not equal $proj_A(b)$. What is wrong here? Why are not they equal?

答:
Your solution for $P$ and $Pb$ is correct.

You can simply verify that $Pb$ is in the column space of $A$ and that $b-Pb$ is orthogonal to this column space, which is now the span of $\pmatrix{1\\3}$.

The other method works only if $u_1,u_2$ are orthogonal, and in that case it produces the projection to the plane spanned by $u_1$ and $u_2$, which only makes sense in 3 or more dimensional spaces.

Here the columns of $A$ are not orthogonal, rather they are parallel, so the subspace we project on is one dimensional, a line.


问:
On skolomizing quantifiers
I have a quick question on how skolomizing quantifiers. I understand the following

∀x(Something(x) -> ∃y(Else(x,y))) => ∀x(Something(x) -> (Else(x,f(x)))

What I need help understanding is what to do when it's configured like this:

∃x(Something(x) -> ∀y(Else(x,y)))

If I eliminate the existential quantifier first like what's usually done, do I put down x as f(y) or y as f(x). Do I instead start with the universal quantifier?

答:
You always first pull out all quantifiers to the front. So

$\phantom{\rightsquigarrow} \exists x (Something(x) \to \forall y (Else(x,y)))\\
\rightsquigarrow \exists x \forall y (Something(x) \to Else(x,y))$

Then you eliminate the quantifiers from left to right, so first $\exists x$, then $\forall y$.

There are two slightly different variants of skolemization, and I don't know which variant you learnt:

The first variant eliminates universal quantifiers one by one. In that case, the skolem function for the existential quantifier is applied to all variables occurring free in the remaining formula. Usually these will be the variables left over from eliminated universal quantifiers having occurred left to the existential quantifier.

The second variant first computes the universal closure of the formula to bind all free variables, stepwise eliminates the existential quantifiers but leaves the universal ones in place for the moment, and removes the universal quantifiers all in one batch at the end. In that case, the skolem function for the existential quantifier is applied to all variables bound by universal quantifiers occurring before the existential quantifier.

In both variants, for the present case there are no variables to apply the skolem function to, because the universal quantifier only comes afterwards. So the resulting function is a 0-place function, with no arguments at all:

$\phantom{\rightsquigarrow} \exists x \forall y (Something(x) \to Else(x,y))\\
\rightsquigarrow \forall y (Something(f()) \to Else(f(),y))$

Zero-place functions behave like individual constants, and depending on the preferred notation conventions, you can also write $a$ instead of $f()$.

Then you eliminate the universal quantifier:

$\phantom{\rightsquigarrow} \forall y (Something(f()) \to Else(f(),y))\\
\rightsquigarrow Something(f()) \to Else(f(),y)$

And you're done.

The skolem function never contains the bound variable being eliminated, only the ones on which the existential quantifier is dependent, "dependent" in the sense of bound left of the existential quantifier or free, and also no variables bound after (right of) the existential quantifier.

Title: On skolomizing quantifiers


问:
Solving ODE using frobenius method. 3 coefficients
I'm trying to learn frobenius method by solving some problems (ODEs).

For example:

$$xy''+(2x+1)y'+(x+1)y=0$$

Let $y=\sum\limits_{n=0}^\infty a_nx^{n+r}$. Then, I took derivatives and put into the equation:
$$\sum\limits_{n=0}^\infty a_n(n+r)^2x^{n+r-1}+2\sum\limits_{n=0}^\infty a_n(n+r+1)x^{n+r}+\sum\limits_{n=0}^\infty a_nx^{n+r+1}=0$$
After I shifted to make their orders same:
$$\sum\limits_{k=-2}^\infty a_{k+2}(k+r+2)^2x^{k+r+1}+2\sum\limits_{k=-1}^\infty a_{k+1}(k+r+2)x^{k+r+1}+\sum\limits_{k=0}^\infty a_kx^{k+r+1}$$
And if I leave first $k=-2, k=-1$ parts, I can find relationship among 3 coefficient:
$$\sum\limits_{k=0}^\infty [x^{k+r+1}(a_{k+2}(k+r+2)^2+a_{k+1}(k+r+2)+a_k)]=0$$
Now, here I could find relationship with $a_{k+2},a_{k+1},a_k$. But, in this method we should find proportionality between 2 coefficients, not 3. For example, this: Frobenius Method to solve $x(1 - x)y'' - 3xy' - y = 0$

Can you, please, suggest a solution?

答:
You state:

After I shifted to make their orders same:
  $$
\sum_{k = -2}^\infty a_{k + 2}(k + r + 2)^2 x^{k + r + 1} + 2 \sum_{k = -1}^\infty a_{k + 1}(k + r + 2)x^{k + r + 1}+\sum_{k = 0}^\infty a_k x^{k + r + 1}
$$

What you do is the following. Given that
$$
\sum_{k = 0}^\infty a_k (k + r)^2 x^{k + r - 1} + \sum_{k = 0}^\infty a_k(2 k + 2 r + 1)x^{k + r} + \sum_{k = 0}^\infty a_k x^{k + r + 1}
$$
(note that your second sum was incorrectly calculated)

you need to separate the necessary terms of the sums in order to group the powers of $x$ correctly, i.e:

\begin{multline}
\sum_{n = 0}^\infty a_n (n + r)^2 x^{n + r - 1} + 2 \sum_{n = 0}^\infty a_n (n + r + 1) x^{n + r} + \sum_{n = 0}^\infty a_n x^{n + r + 1} = \\
a_0 r^2 x^{r-1} + a_1 (r + 1)^2 x^r + \sum_{n=2}^\infty a_n (n + r)^2 x^{n + r - 1} + (2 r + 1) a_0 x^r + \\ \sum\limits_{n = 1}^\infty a_n (2 n + 2 r+ 1)x^{n + r} +\sum_{n = 0}^\infty a_n x^{n + r + 1} = 0
\end{multline}

Regrouping orders, you have

\begin{multline}
a_0 r^2 x^{r-1} + [a_1 (r + 1) + a_0 (2 r + 1)] x^r + \\ \sum_{k = 0}^\infty \left\{ a_{k + 2}(k + r + 2)^2 + 2 a_{k + 1} (k + r + 2) + a_k \right\} x^{k + r + 1} = 0
\end{multline}

Each power of $x$ needs to vanish, hence $r^2 = 0$. This is the indicial polynomial (details here). This means that $r = 0$ and

\begin{align}
a_1 + a_0 &= 0 \\
a_{k + 2}(k + 2)^2 + 2 a_{k + 1} (2 k + 3) + a_k &= 0
\end{align}

which closes the recurrence relation. The first tree terms are
\begin{align}
a_1 &= -a_0\\
a_2 &= \frac{1}{2!}a_0\\
a_3 &= -\frac{1}{3!}a_0
\end{align}
and it's clear that a relationship is forming. By induction, the whole solution can be computed.

Note that, assuming that $y$ is somehow well behaved, for $x \sim 0$,

$$
x y'' + (2x + 1) y' + (x + 1) y = 0 \quad \sim \quad y' + y = 0.
$$

Proposing the anzats $y(x) = e^{-x} z(x)$ and substituting in the original ode,

$$
x y'' + (2x + 1) y' + (x + 1) y = e^{-x}\left(x z'' + z'\right) = 0,
$$

and it's easily verified that $z = c_1 \log x + c_2$. Hence

$$
y(x) = e^{-x}\left(c_1 \log x + c_2\right)
$$

Cool trick ha?


问:
Is the area under the Cantor Lebesgue function a Borel set of $\mathbb{R}^2$?
If $f$ is the Cantor Lebesgue function, then is the set $\{(x,y) \in [0,1]^2 : 0\leq y \leq f(x) \}$ Borel ?

答: Yes: Cantor's function is continuous, therefore its hypograph is closed in $(\operatorname{domain})\times\Bbb R$. Specifically, that set of yours is $([0,1]\times[0,\infty))\cap g^{-1}(-\infty,0]$, where $g:[0,1]^2\to \Bbb R$ is the continuous function $g(x,y)=y-f(x)$.


问:
Calculating $\sin{\pi x}$ in a weird way
Got bored in math class and found out that $$\sin{\pi x} = \frac{imag((-x)^x)}{real(x^x)}$$ for x > 0. Can anybody please proove to me that the above equation is true? (Because I myself can't)

Thanks in advance.

答: express $(-x)^x$ in the form $a+ib$ and you will find that it contains the trig term $sin(\pi x)$. Once you divide this expression by $x^x$ you will arrive at the result you are trying to prove. Tip: You will need to use DeMoivre's Theorem.


问:
Properties of connection principal bundle
I am working on a proof of Cartan's formula
$$
DA(u,v) = dA(u,v) + [A(u),A(v)],
$$
where $A$ is a connection one form on a principal bundle and $u,v$ are two vector fields. $D$ denotes exterior covariant derivative One way of solving this should be using the identity
$$
dA(u,v) = u(A(v)) - v(A(u)) - A([u,v]).
$$
From the definition of the exterior covariant derivative, I am inclined to split my vector fields such that $u = H(u) + V(u)$, where H and V denote the horizontal and vertical components of the vector fields. In this way, I can use the fact that $A(H(u)) = 0$, but I am left with a lot of terms under the form such as $$v(A(V(u))).$$

My question is if there is any (defining) property of how a connection one form acts on a vertical vector field $V(u)$. My intuition/guess is that it should vanish or stay constant at least while going over a horizontal path, but I might be wrong.

答:
Your formula $\require{enclose} \enclose{horizontalstrike}{is}$ [was] incorrect. Notice that if your [original] formula were true, we would have $DA=dA$ on coordinate vector fields, hence on all vector fields. The correct formula is $ DA(u,v) = dA(u,v) + [A(u),A(v)]. $

To prove the formula, we first check it when one of the vector fields is a fundamental vector field. Let $X \in \mathfrak g$ and let $X^\ast_p = \frac{\rm{d}}{\rm{d}t} p e^{tX} |_{t=0} $ be the corresponding fundamental vector field. We have $DA(X^\ast,v) = dA(X^\ast,v) + [A(X^\ast),A(v)] = - \mathrm{ad}_X A(v) + [X,A(v)] = 0$. Since an arbitrary vertical vector field can be expressed as a $C^\infty$ linear combination of fundamental vector fields, this shows that the formula holds when one of the vector fields is vertical.

Next we show that they have the same pullback under horizontal projection $H$: $ H^\ast DA = DA = H^\ast dA = H^\ast (dA+ \frac{1}{2}[A,A]) $ since $H^\ast [A,A] = [H^\ast A,H^\ast A]= 0$. (Perhaps confusingly, $[A,A](u,v) = [A(u),A(v)] - [A(v),A(u)] = 2 [A(u),A(v)]$.)

Now you can decompose $u$ and $v$ into their horizontal and vertical parts and see that the formula holds.


问:
Index of $n\mathbb{Z}$ in $\mathbb{Z}$.
This is problem 1 of Ch.2, Section 6 in Artin's Algebra (Ed.1). It asks to determine the index $[\mathbb{Z} : n\mathbb{Z}]$. I have that if $|n|\geq 1$, then $[\mathbb{Z} : n\mathbb{Z}] = |n|$, but if $n=0$, then $[\mathbb{Z} : n\mathbb{Z}] = \infty$. Is this the case? Also, how would one go about proving the formula by induction for positive integers $n$?

答:
Consider the abelian groups $G=(\mathbb{Z},+)$ and $H=(n\mathbb{Z},+)$.

Answering your first question: yes, this is the case. If $|n|\geq1$, you have exactly $|n|$ cosets of $(n\mathbb{Z},+)$ in $G$ and these are:
$$0+H$$
$$1+H$$
$$...$$
$$(|n|-1)+H$$
Thus, $$[(\mathbb{Z},+):(n\mathbb{Z},+)]=|n|$$
For $n=0$, $0\mathbb{N}=\{0\}$ and the above list of cosets becomes infinite and countable. So,
$$[(\mathbb{Z},+):(0\mathbb{Z},+)]=|\mathbb{N}|$$
as @PeterTamaroff noted.

Addressing your second quiestion, let $n\in\mathbb{N}$.

We could make a proof by induction by defining a set $C_k$, whose elements are the $k$ cosets of the subgroup $H=(n\mathbb{Z},+)$ in $G=(\mathbb{Z},+)$: $$C_k:=\{\quad\bar{H}\subset G \,:\,\bar H=i+H \quad i=0,1,...,k-1\quad\} $$
Starting with $n=1$ we see that $\mathbb{Z}=n\mathbb{Z}$, so
$$ C_1=\{0+1\mathbb{Z} \},\quad|C_1|=1.$$
For $n=2$ $$C_2=\{0+2\mathbb{Z},1+2\mathbb{Z}\}=C_1\cup\{1+2\mathbb{Z} \}, \quad |C_2|=2.$$
Supose then that for $n=k$, $$C_{k}=\bigcup_{i=0}^{k-2}\{i+k\mathbb{Z}\}\cup\{(k-1)+k\mathbb{Z}\}=\bigcup_{i=0}^{k-1}\{i+k\mathbb{Z}\}, \quad|C_{k}|=k.$$ Then for $n=k+1$ we get:
$$ C_{k+1}=\bigcup_{i=0}^{k-1}\{i+k\mathbb{Z}\}\cup\{k+k\mathbb{Z}\}=\bigcup_{i=0}^{k}\{i+k\mathbb{Z}\}, \quad |C_{k+1}|=k+1. \qquad \blacksquare$$

Title: Index of $n\mathbb{Z}$ in $\mathbb{Z}$.


问:
What are the elements of the quadratic ring $\mathbb{Z}[ \sqrt{15}]$?
Hi MathStackExchange,

I am having trouble understanding this quadratic ring: $$\mathbb{Z}[\sqrt{15}]$$ whose elements can be written as $$ \{a+b\sqrt{15} | a,b \in \mathbb{Z}\}$$
If anyone understands what these elements actually look like,and could provide a few examples, that would be very helpful.
Wikipedia describes the set as :
"The ring Z[ω] consists of all roots of all equations $x^2 + Bx + C = 0$whose discriminant $B^2 − 4C$ is the product of D by the square of an integer. In particular √D belongs to Z[ω], being a root of the equation $x^2 − D = 0$, which has 4D as its discriminant." link to page - see Quadratic Rings section

...which doesn't seem to match the other expression I have for the elements.

For example, is something like(just a random example) $999999 + (-2)\sqrt{15}$ in this ring? It seems like it could be from the way the elements can be represented?

My Answer (see approved answer for more detail):

My example $999999 + (-2)\sqrt{15}$ is indeed an element of the quadratic ring. In general, the motivation for mentioning rings of this type in Algebra textbooks/lectures is that they have so many interesting properties. For example, notice that $\mathbb[Z\sqrt{15}]$ is not a Unique Factorization Domain. You can find two different factorizations of irreducible elements for some elements in the ring.

Quadratic integer rings are explicitly defined like this, which makes it clear to see that elements of this ring in particular are indeed of the form $ \{a+b\sqrt{15} | a,b \in \mathbb{Z}\}$

答: Yes it is a subring of $\Bbb{R}$ (or $\Bbb{C}$). For any complex number $t$ then $\Bbb{Z}[t]$ is a subring of $\Bbb{C}$, and when $t$ is the root of a monic integer polynomial of degree $n$ then $\Bbb{Z}[t] = \{ \sum_{j=0}^{n-1} c_j t^j, c_j\in \Bbb{Z}\}$.


问:
Bounds when computing surface integral of a scalar function

Compute the surface integral of the function $x+z$ over the first octant of the plane $x+y+z=1$

My attempt:

$$ \vec{n} \cdot \vec{k} = \sqrt{1 + f_x^2 + f_y^2} = \sqrt{3}$$

Then,

$$ \int_{\partial S} (x+z) dS= \int_{R} (x+z) \frac{dx dy}{\vec{n} \cdot \hat{k}} = \frac{1}{\sqrt{3}} \int_{R} (x+z)dx dy$$

Now, I plug in equation of plane for $z$ and rearrange the order ( I've ommited the root three outside)

$$ \int_{R} (1-y) dy dx = \int_?^? \int_?^? (1-y) dy dx$$

Now, for the inner integral I am running a double integral in the $x-y$ plane so I put $z=0$ in the equation of plane which gives $x+y=1$ or $ y=1-x$ , and the outer integral is from $ 0 \to 1$,

$$ \int_0^1 \int_{0}^{1-x} (1-y) dy dx  = \int_0^1 [x^2 -1 ] dx = -\frac23$$

But the true answer of the original integral $ \frac{1}{\sqrt{3}}$ / the new integral should evaluate to one. What have I done wrong?

答:
$z = f(x,y) = 1 - x - y$ is the parametrization of your surface.

$f_x = -1, f_y = - 1, \sqrt{1 + f_x^2 + f_y^2} = \sqrt3$

$\int_{S} (x+z) dS = \int_{A} (x+z) \sqrt{1 + f_x^2 + f_y^2} \, dA = \sqrt3 \displaystyle  \int_0^1\int_{0}^{1-x} (1-y)  \, dy \, dx = \frac{1}{\sqrt3}$

Also note $(x+z)$ is a scalar function and not a vector field so it is not necessary to find $\, \hat{n}\,$ for your integral.

EDIT:

Specifically to your question on the mistake, when you have a vector field your integral translates to

$\displaystyle \iint_R (\vec{F} \cdot \hat{n}) \frac{dx \, dy}{\hat{n}.\vec{k}}$

Here your surface is $x + y + z = 1$ which has a normal vector $(1, 1, 1)$.

Unit normal  $\hat{n} = (\frac{1}{\sqrt3}, \frac{1}{\sqrt3}, \frac{1}{\sqrt3})$

$\hat{n} \cdot \vec{k} = \frac{1}{\sqrt3}$

So your integral should be

$\sqrt3 \displaystyle \iint_R (x + z) \, dx \, dy$


问:
Find the vector function for two intersecting surfaces
Question: Find the vector function for the two intersecting surface $y=x^2$ and $10y=x^2-z^2+16$.

My attempted solution: I know that $x = rcos(t)$ and that $y=rsin(t)$ in this case since the intersection is an elipse shape, however I'm not really sure how to calculate $r$ without $x^2 +y^2$.

I'd be much appreciated if someone could show me how this is done. Thanks!

答:
First of all, if this ellipse is not a circle, then you are not going to have the same "$r$" in both equations.

Second, it may be easier to use such a parametrization of an ellipse for another pair of variables, not necessarily for $x$ and $y$.

Here's what I would do. Since $y=x^2$ according to the first equation, we can substitute $x^2$ for $y$ into the second equation. We'll get $10x^2=x^2-z^2+16$, which indeed is an equation of an ellipse. By simplifying this equation (move $x^2$ and $z^2$ to the left-hand side, and then divide everything by $16$), you will see how to parametrize $x=a\cos(t)$, $z=b\sin(t)$ with appropriate values of $a$ and $b$. And then the parametrization for $y$ comes from $y=x^2$.


问:
move a point along a vector by a given distance
Given a point
$P = [x, y, z]$
and a vector
$v = [v_x, v_y, v_z]$

I want to move the point along the vector by a fixed amount $d$

Thank you

答:
To move a point along a vector, we multiply the vector by a constant ($\epsilon$), and then add it to the point
$P_{moved} = P + \epsilon v = [x + \epsilon v_x, y + \epsilon v_y, z + \epsilon v_z]$

Now, the distance moved $d = \sqrt{(P_{moved} - P)^2} = \epsilon\sqrt{v_x^2 + v_y^2 + v_z^2}$

Therefore $\epsilon = \frac{d}{\sqrt{v_x^2 + v_y^2 + v_z^2}}$

So $P_{moved} = P + \frac{d}{\sqrt{v_x^2 + v_y^2 + v_z^2}}v$


问:
Proof Using Contrapositive
I need help on solving this problem.
Prove that if A x B = {}, then A = {} or B = {}.

(Prove the contrapositive) ( {} = Empty Set )

答:
You want to prove
$$A\times B=\emptyset \Longrightarrow A=\emptyset \mbox{ or } B=\emptyset$$
by contrapositive, that means:
$$A\not=\emptyset \mbox{ and } B\not=\emptyset \Longrightarrow A\times B\not= \emptyset\mbox.$$
This is simply becouse if $A,B$ are not empty, there exist $a\in A$ and $b\in B$, so the element $(a,b)\in A\times B$ which, now, cannot be empty.

Title: Proof Using Contrapositive


问:
Proving first supplement of the quadratic reciprocity law
I'm trying to prove the first supplement of the quadratic reciprocity law in its particular form:

$$-\bar{1} \text{is a square in} \mathbb{Z}/p\mathbb{Z} \iff p \equiv 1 \pmod 4$$

For the forward part, I'm thinking about applying Euler's criterion, since we know $-\bar{1}$ is a square. For the backward part, my starting point is that I have proved that $x^{p-1}=\bar{1}$ and that implies that:

$$\Bigl(x^{\frac{p-1}{2}} - \bar{1} \Bigr)\Bigl(x^{\frac{p-1}{2}} + \bar{1} \Bigr) = \bar{0}$$

答: If $-1$ is a square mod $p$, say $-1 \equiv a^2 \pmod{p}$, then $\overline{a}$ has order $4$ in $\Bbb F_p^\times$. By Lagrange we get that $4 \mid (p-1)=|\Bbb F_p^\times|$, so $p \equiv 1\pmod{4}$


问:
Prove $\sin x < x < \tan x$ for $0 < x < \pi/2$ algebraically
I know there are some proofs showing that, for  $0 < x < \pi/2$:

$\sin x < x < \tan x $

with the use of the unit circle, however, can I prove this algebraically in some way as well?

答: One simple way could be to demonstrate that , over $[0, pi/2)$ $\sin$ is concave while $\tan$ is convex, and that $\sin(0)=0=\tan(0)$ and $\sin '(0) = 1 =\tan'(0)$.


问:
Constructing projective resolution of a chain complex
I am trying to construct the projective resolution in the category of chain complexes of

$\dots \to 0 \to M \to 0 \to \dots$

It seems like it should be possible to do this in terms of the projective resolution of $M$ but I am completely stuck.

I know a projective chain complex is split exact and formed by projectives, so if we think of the resolution as a half plane double complex, the column with $M$ must be a projective resolution of $M$.

I was trying to use the trick of $0 \to P \to P \to 0$ is a projective complex whenever $P$ is projective, but if I put that on top of our complex we don't necessarily get exactness.

答:
If
$$\dots\to P_2\to P_1\to P_0 \to M\to0$$
is a projective resolution of $M$ as a module, then $\dots\to0\to M\to0\to\dots$ has a resolution (by projective chain complexes) in the category of chain complexes of the following form (I'll let you figure out the differentials):

$\require{AMScd}$
\begin{CD}
@.\vdots@.\vdots@.\vdots@.\vdots@.\vdots@.\vdots@.\\
@.@VVV@VVV@VVV@VVV@VVV@VVV\\
\cdots@>>>0@>>>P_2@>>>P_2\oplus P_1@>>>P_1\oplus P_0@>>> P_0@>>>0@>>>\cdots\\
@.@VVV@VVV@VVV@VVV@VVV@VVV\\
\cdots@>>>0@>>> P_1@>>>P_1\oplus P_0@>>>P_0@>>>0@>>>0@>>>\cdots\\
@.@VVV@VVV@VVV@VVV@VVV@VVV\\
\cdots@>>>0@>>>P_0@>>>P_0@>>>0@>>>0@>>>0@>>>\cdots\\
@.@VVV@VVV@VVV@VVV@VVV@VVV\\
\cdots@>>>0@>>>M@>>>0@>>>0@>>>0@>>>0@>>>\cdots\\
@.@VVV@VVV@VVV@VVV@VVV@VVV\\
@.0@.0@.0@.0@.0@.0
\end{CD}


问:
To what extent does the connection one-form on a principal fibre bundle determine the Ehresmann connection?
I have seen a few sources that define the connection on a principal fibre bundle as a choice of "horizontal vector space" $H_pP$ (to my understanding this is an Ehresmann connection) and also as a choice of Lie algebra valued connection 1-form. I have also seen some sources state that there is some relationship between the two (i.e. can determine the other). My question is how uniquely does the connection 1-form determine an Ehresmann connection and vice-versa?

答:
An Ehresmann connection and a connection $1$-form are basically the same thing.

Starting off with an Ehresmann connection, i.e. a "horizontal" subbundle $HP$ of $TP$ complementary to the vertical bundle $VP$, we may also define a projection $TP \to VP$. This projection is just a fiberwise operation coming from basic linear algebra: If $W$ is a  vector space with subspace $W'$, then a projection $W \to W'$ is equivalent to a complement $W''$ to $W'$.

Moreover, on a principal bundle, there is a diffeomorphism $P \times \mathfrak{g} \to VP$ given by $(p,X) \mapsto \frac{\rm{d}}{\mathrm{d}t} pe^{tX} |_{t=0}$. So putting the data together, we get a map $TP \to VP \to P \times \mathfrak{g} \to \mathfrak{g}$ where the last map is just projection onto the second factor. This composition is the connection $1$-form. On the other hand, given a connection $1$-form $\omega$, we can define the horizontal subbundle to be the kernel of $\omega$.


问:
Is there any notation for the undecimal system?
Decimal is
0
1;2;3;4;5;6;7;8;9
10;11;...
But undecimal
0
1;2;3;4;5;6;7;8;9;?
?0;?1

TWas this system used by an ancient civilisation ?

(I don't know what to tag)

答: Hexadecimal traditionally uses 'digits' 0,1,2,3,4,5,6,7,8,9,A,B,C,D,E,F; it looks reasonable to use 0,1,2,3,4,5,6,7,8,9,A for undecimal.</p>


问:
Showing that $|\delta(s,u)-\delta(s,v)|=1$ for every $(u,v)\in E$
Part of a problem in algorithms, I'm trying to prove the following statement:

If $G=(V,E)$ is an undirected graph and $s\in V$ is the the root (has path from it to any other vertex in $G$)  so for every $(u,v)\in E$ we get $|\delta(s,u)-\delta(s,v)|=1$ where $\delta(s,u)$ is the shortest path between $s$ and $u$.

What I did: Split into two independent parts:

If in the shortest path between $s$ and $v$ does not contain $u$ this means that the shortest path between $s$ and $u$ is the shortest path between $s$ and $v$ and the edge $(u,v)$ so $\delta(s,u)=\delta(s,v)+1$.

If in the shortest path between $s$ and $v$ does contain $u$ this means...

Now I'm stuck. I want to say that if shortest path between $s$ and $v$ does contain $u$, then it should be the shortest path between $s$ and $v$ without the edge $(u,v)$. But why can I say it? Maybe the vertex $u$ does not come before the vertex $v$ in that path (something like $s\to u_1 \to \ldots \to u\to u_3\to \ldots \to v$). How to show it?

答:
You are very close.

Argue by contradiction.

Suppose there exists $u,v\in V$ such that $\{u,v\}\in E$ and $\lvert\delta(s,u)-\delta(s,v)\lvert>1$. Assume, without loss of generality, that $\delta(s,v)<\delta(s,u)$.

Let $P:=sw_1w_2\ldots w_kv$ be a shortest path from $s$ to $v$. Note that $u$ cannot be on this path. (Why?)

Then the path $P':=sw_1w_2\ldots w_kvu$ is a path of length $\delta(s,v)+1$ connecting $s$ to $u$... but $\delta(s,v)+1<\delta(s,u)$, which is shorter than $\delta(s,u)$, a contradiction.


问:
$A$ and $B$ are positive self-adjoint matrix such that $AB$ is self-adjoint then $AB$ is positive
Question: $A$ and $B$ are positive self-adjoint matrix such that $AB$ is self-adjoint then prove that $AB$ is positive.

My first try: Here I can see that the same question was there but it has a little amount of typo. What I have to prove is that $(x, ABx)>0$ for all $x>0$.

Now, $(x,ABx)=(Ax,Bx)=(BAx,x)$ now we also have that $AB=BA$ because $AB=(AB)^*=B^*A^*=BA$ because $A,B,AB$ are all self-adjoint and we can also see that $(BAx,Ax)>0$ but I can't get rid of $A$ in the 2nd co-ordinate.

My second try: Then I tried using this result that "if $A$ and $B$ are positive self-adjoint matrix then there exist a basis $x_1,\cdots, x_n$ of $X$ satisfies an eqn of the form $$Ax_j=\lambda_j Bx_j.$$
where $\lambda_j$ is real."
Now $A,B$ are positive $\Rightarrow \lambda_j>0$ using Generalised Raleigh quotient then $$(x_j,ABx_j)=(Ax_j,Bx_j)=(\lambda_j Bx_j,Bx_j)=\lambda_j||Bx_j||^2>0.$$
Hence $AB$ is positive.
Here I am confused that what is the place where I used $AB$ is self-adjoint! Please help. You can also help me with different proofs.

答:
Let us start with the positive operators $A$, $B$ as in the OP. Then $AB$ self-adjoint implies
$$
AB = (AB)^*=B^*A^*=BA\ ,
$$
so that $AB$ are commuting. Using continuous functional calculus w.r.t. the function $f(x)=\sqrt x$ defined on $[0,\infty)$ (and this domain contains the spectrum of $A$, and the spectrum of $B$) we obtain square roots $S=f(A)$, $T=f(B)$, i.e. $A=S^2$, and $T=B^2$.

Explicitly, for $A$ only, let $p_n$ be a sequence of polynomials such that $p_n\to f$ on a compact interval (for instance $[0,\|A\|]$) contaning the spectrum of $A$. Such a sequence is insured by Stone-Weierstraß. Then $$p_n(A)\to f(A)=:S\ .$$
Here, $f(A)$ is defined as $\lim p_n(A)$, the limit exists, Cauchy sequence. (Functional calculus of bounded operators shows this does not depend on $(p_n)$, but we do not need this.)
We denote by $S$ this value.

Then
$$
\begin{aligned}
S^2
&=(f(A))^2=(\lim p_n(A))^2=\lim (p_n(A))^2
=\lim p_n^2(A)
\\
&=(\lim p_n^2)(A)
=(\lim p_n)^2(A)
\\
&=f^2(A)=\operatorname{id}(A)=A\ .
\\[3mm]
SB &=(\lim p_n(A))B=\lim p_n(A)B\\
&=\lim B p_n(A)=B(\lim p_n(A))=BS\ .
\\[3mm]
(S^*x, y)
&=(x,Sy)=(x,(\lim p_n(A))y)=(x,\lim p_n(A)y)=\lim (x,p_n(A)y)\\
&=\lim (p_n(A)x,y)=(\lim p_n(A)x,y)=((\lim p_n(A))x,y)\\
&=(Sx,y)\ ,\qquad\text{ for all $x,y$ in the given Hilbert space.}
\end{aligned}
$$
(We have used $AB=BA$. These properties are basic properties of the functional calculus. Starting from $AB=BA$ we get to $f(A)B=Bf(A)$, so $SB=BS$. Similarly, starting with $SB=BS$ we obtain $Sf(B)=f(B)S$. So $ST=TS$ the two operators $S,T$ also commute.)

Let now $x$ be $\ne 0$. We have in a row:
$$
(ABx,x) =(SSTTx,x)=(TSSTx,x)=(STx,STx)=\|STx\|^2>0\ .
$$
(We have $(STx,STx)=\|STx\|^2\ge 0$, and in case of an equality, from $(STx,STx)=0$ we get first $Tx=0$, since $S>0$, then from $(Tx,Tx)=0$ also $x=0$, since $T>0$. Contradiction, since we started with an $x\ne 0$.)

Alternatively, we could have intoduced only $T$, and have the same argument with $SS$ replaced by $A$, e.g. $(ABx,x)=(ATTx,x)=(TATx,x)=(ATx,Tx)>0$ since $Tx\ne 0$ since $(Tx,Tx)=(TTx,x)=(Bx,x)>0$.

Note: In case of finitely dimensional spaces, things are simple. The two commuting self-adjoint operators can be diagonalized simultaneously w.r.t. some ortonormal basis, and if $A=\operatorname{diag}(a_1,\dots,a_n)>0$, then $S=\sqrt A:=
\operatorname{diag}(\sqrt{a_1},\dots,\sqrt{a_n})>0$ is the explicit square root of $A$ (which is positive), it is diagonal w.r.t. the same basis, et caetera.

Note: See also

Functional calculus for bounded operators using continuous functions

Functional calculus for bounded operators using Borel functions


问:
Let $(X,d)$ be a metric space and let $K$ be a dense subset of $X$.
Given that $\{B_r(k)\}_{k \in K}$, is it true that $\{B_r(k)\}_{k \in K}$ is also open cover for $X$.

My goal is to show that $X \subseteq \cup_{k \in K}B_r(k)$. I know since $K$ is a dense subset of $X$, for all $x \in X$, either $x \in K$ or $x$ is limit point of K. The first case is easy but I am struggling with the second case.

答: If $x$ is a limit point of $K$, then in particular there is some $k \in K$ such that $d(x,k) < r$.  Then $x \in B_r(k)$.


答:
The curve is not a plane curve as we can see in the picture below.

From the definition
$$
\begin{cases}
y=x^2\\
10 y=x^2-z^2+16\\
\end{cases}
$$
a first parametrization is $$(t,t^2,\pm\sqrt{16-9t^2})$$
This is not nice because we should use two vectors, so we set $t=\alpha\sin u$
$$z^2=16-9t^2\to z^2=16-9 \alpha^2\sin^2 u$$
we want $9\alpha^2=16\to \alpha=\frac{4}{3}$ thus, setting
$t=\frac{4}{3}\sin u$ we get $z^2=16-16\sin^2 u\to z=4\cos u$
$$\left(\frac{4}{3}\sin u,\frac{16}{9}\sin^2 u,4\cos u \right);\;u\in[0,2\pi]$$

Hope this can help

Title: Find the vector function for two intersecting surfaces

Body:
Question: Find the vector function for the two intersecting surface $y=x^2$ and $10y=x^2-z^2+16$.

My attempted solution: I know that $x = rcos(t)$ and that $y=rsin(t)$ in this case since the intersection is an elipse shape, however I'm not really sure how to calculate $r$ without $x^2 +y^2$.

I'd be much appreciated if someone could show me how this is done. Thanks!


问:
Existence of two measurable functions s.t. $\int_{D}f^3dx = \int_{D}g^4dx = 1$ and $\int_{D}(f+g)^2dx = \pi^3 $
Given $D = \lbrace (x,y) : x^2+y^2 \le 1 \rbrace $, say if we could find two measurable functions $f, g: D \to [0, + \infty]$ s.t. (then give an explicit example) :
$$\int_{D}f^3dx =  \int_{D}g^4dx = 1$$

and

$$\int_{D}(f+g)^2dx = \pi^3 $$

I tried to reason in terms of characteristic functions, but I am stuck on this exercise, some hints will be greatly appreciated. Thank you

答: Note that, if $f,g$ satisfy the first two equalities, then $$\int_Df^2\leq\left(\int_Df^3\right)^{2/3}|D|^{1/3}=\pi^{1/3},$$ and also $$\int_Dg^2\leq\left(\int_Dg^4\right)^{1/2}|D|^{1/2}=\pi^{1/2}.$$ So then $$\int_D(f+g)^2\leq 2\int_D(f^2+g^2)\leq 2\pi^{1/3}+2\pi^{1/2},$$ but this last term is less than $\pi^3$. Therefore, no such functions exist.


问:
Extensions of fields and subfields generated by a field and a set
K ⊆ L ⊆ M be two field extensions and α ∈ M algebraic
over K then it is algebraic over L.

Help me prove this please

答:
Since $\alpha \in M$ is algebraic over $K$, it is a zero of some polynomial

$k(x) \in K[x]; \tag 1$

since

$K \subset L, \tag 2$

we have

$K[x] \subset L[x], \tag 3$

which shows that

$k(x) \in L[x]; \tag 4$

but this by definition implies $\alpha$ is algebraic over $K$.


答:
Make up your own notation.  Or as there is only one symbol to add we can call it $TEN$.

so the numbers are:

$1;2;3;4;5;6;7;8;9;TEN$

$10;11;12;13;14;15;16;17;18;19;1TEN$

$20;21.......$

Although any modern mathematician doing this would probably use $A$.

I'm more interested in how the Maori used it practically.  I imaging they counted to ten and probably had a place holder that the gave as an actually number and a transition to the next set of ten number (and then a second place holder).... Maybe?

Title: Is there any notation for the undecimal system?

Body:
Decimal is
0
1;2;3;4;5;6;7;8;9
10;11;...
But undecimal
0
1;2;3;4;5;6;7;8;9;?
?0;?1

TWas this system used by an ancient civilisation ?

(I don't know what to tag)


问:
Number of Zonotope Edges Parallel to Generator
Suppose we have a zonotope $Z$ that is the Minkowski sum of line segments $U_1+\dots +U_n$. All the edges of $Z$ are parallel to some $U_i$. Is it also true that the number of edges parallel to $U_i$ equals the number of edges parallel to $U_j$ for every $j\neq i$? Surely this question has been asked and probably answered before. I looked through a page of zonohedra and verified that they all answer the question true.

https://www.ics.uci.edu/~eppstein/junkyard/ukraine/ukraine.html

It is known that $Z$ is a zonotope iff $Z$ is the projection of a cube from $\mathbb{R}^n$. Of course, a cube in $\mathbb{R}^n$ is the sum of $n$ orthonormal vectors and has the same number of edges parallel to each of its generators. I suspect there may be a good induction proof down this way by adding one generator, $v$, at a time and  but haven't found one. The new generator divides the old zonotope into two pieces. The edges parallel to $v$ will stretch between the two halves.

Any help is appreciated!
Thanks! Aaron

答:
That is not the case.

Consider a $2n$-gonal prism with $n\ge 3$ (with top and bottom face being regular polygons). It is a zonotope.

Then there are $2n$ edges connecting the top and the bottom face, all parallel. But choose an edge in the top or bottom face: it is parallel to only three other edges.


问:
Is there a geometric relationship between plane geometry and polynomials?
It is well known that the complex plane is algebraically closed: Every polynomial has a zero. The relationship seems, to me, to run deeper: For every complex-differentiable function, there exists a power series which converges to it; power series could be regarded as a sort of a generalized polynomial. Every entire function with a pole at infinity is a polynomial. Etc.

Complex analysis can in some ways be regarded as the successor to Euclidean plane geometry. Many proofs exist of the aforementioned facts, which seem to link the polynomials much more tightly to the complex plane than to the real line, but is there an intuitive, geometric reason as to why these transformations seem so fundamentally linked to what is essentially plane geometry?

答:
Well, nice question.

First thing to mention is probably the Gauss correspondence b/w ruler and compass constructions and roots of quadratic polynomials.

Second connection:
 Every polynomial is a characteristic polynomial of a companion matrix; the number of real roots of a polynomial can be read from the determinants of the corresponding Hurwitz matrix; Jordan blocks of a matrix correspond to multiplicities of eigenvalues, which are roots of its characteristic polynomial, etc... So, theory of polynomials and special matrices are intrinsically connected. Now, matrices can be viewed as 2D/plane objects, because they have rows and columns, and special matrices can be described geometrically... Also, taking limits one gets complex functions defined on the complex plane from both theories to get a unified self-contained theory.


问:
Prove that $\left | f(x) - a \right | \leq \frac{1}{2} \left | x - a \right |$
$f(x) = 1 - \frac{1}{x}(\sqrt{1 + x^2}-1)$

$|f'(x)| \leq \frac{1}{2}$

$a$ is a solution for $f(x) = x$ where  $0.65 < a < 0,7$

The question says:

Prove that $\left | f(x) - a \right | \leq \frac{1}{2} \left | x - a \right |$

I have no idea from where to begin, if I knew I would have posted my attempt at a solution but I didn't even know how to start it.

Any ideas?

答:
Since $x = a$ is a solution to $f(x) = x$ , thus $f(a) = a$ . Now, take any $x \in \mathbb{R}$ . Without loss of generality, let $x \leqslant a$ . Then, by Lagrange's Mean Value Theorem, there exists $h_x \in (x , a)$ such that $$\left\vert \frac{f(x) - a}{x - a}\right\vert = \left\vert \frac{f(x) - f(a)}{x - a}\right\vert = \vert f'(h_x) \vert \leqslant \frac{1}{2}$$ which is what you require to show.

Similarly, if $x \geqslant a$ , then an exactly similar calculation will show the same result. Therefore, for any given $x \in \mathbb{R}$ , we'll have : $$\left | f(x) - a \right | \leq \frac{1}{2} \left | x - a \right |$$

Sorry if I didn't understand your question properly.


问:
Taylor expansion of $\log(z^2-3z+2)$ at z = 0
Find the Taylor series of the function $$f(z) = \log(z^2-3z+2) \quad \forall z \in \mathbb{C} \backslash \left\{1,2 \right\} $$ at $z = 0$ and calculate the radius of convergence for the resultant series.

答: Use $f=\ln2+\ln(1-z)+\ln(1-z/2)$. I'll leave you to work out the Taylor series. Hopefully, you can see which term determines the radius of convergence.


问:
Why is the Hilbert polynomial of projective varieties an invariant?
Suppose $S=k[x_0,...,x_n]$ and $M$ is the homogeneous coordinate ring of some projective variety in $P^n$. Then $M$ is a graded $S$ module. By a theorem of Hilbert-Serre, Hilbert polynomial is defined to be the unique polynomial, $P_M$, such that $P_M(l)=\dim_k (M_l)$.

Is the Hilbert polynomial an invariant under birational map or isomorphism? Why?

答:
No, the Hilbert polynomial is defined as a function of the specific projective embedding. Its degree is an invariant (always the dimension of the projective variety), but the general polynomial isn’t.

For instance, the Hilbert polynomial of the identity inclusion $\mathbb{P}^1 \subset \mathbb{P}^1$ is $H(T)=T+1$. But the Hilbert polynomial of the Veronese embedding $\mathbb{P}^1 \longmapsto \mathbb{P}^n$ is $H(T)=nT+1$ (I think).


问:
In a topological space$X$ prove that $\overline A+\overline B\subseteq\overline {(A+B)}$
Let $X$ be a topological space prove that  $\overline A+\overline B\subseteq\overline {(A+B)}$

where $A+B=\{a+b: a\in A, b\in B\}$

My attempt

Suppose $X$ and $Y$ are topological spaces and $f: X \rightarrow Y$ is a map. Then $f$ is continuous if and only if $f(\overline{A})\subseteq \overline{f(A)}$, where $\overline{A}$ denotes the closure of an arbitrary set $A$.

so I consider $f=+$

would it be possible?

答:
So if we are working in a topological vector space, or a more generally topological (semi)group $X$ where $p: X \times X \to X$ defined by $p(x,y)=x+y$ is continuous, then we can indeed apply the general fact that $f: X \to Y$ is continuous iff $f[\overline{C}] \subseteq \overline{f[C]}$ for all $C \subseteq X$. Note that $A + B = p[A \times B]$ by comparing their resp. definitions.

We apply it to $p$ and use $C= A \times B$ as the subset of its domain. In the product topology we know that $\overline{A \times B} = \overline{A} \times \overline{B}$ and so as $p$ is continuous:

$$\overline{A} + \overline{B} = p[\overline{A} \times \overline{B}] = p[\overline{A \times B}] \subseteq \overline{p[A \times B]} = \overline{A + B}$$

as required.


问:
Number of $n \to p \bmod n$ before getting to 0
There isn't much background context, but is there any estimations on how many iterations of $n \to p\% n$ are needed before $n$ becomes 0? Percentage sign is modulo. ($p$ is fixed, prime in my context but not sure if it matters).

For example, when $p=10^9+7$, the maximum number of iterations needed is $50$. When $p=998244353$ (a common prime in competitive programming) the number of iterations is $45$. I don't see any proof that it requires approximately $\log p$, but it seems like so.

Thank you for your time and help.

答:
Here's a semi-logarithmic plot of the maximum number of iterations required for a given $p$, with $p$ ranging from 5 to 5000, checking all $2 \le n < p-1$. Both the best & worst cases do seem to be growing logarithmically.

And here's a plot of the mean number of iterations, for $1 \le n < p$.

These plots were produced using Sage / Python. You can play with an online interactive version here (the script is actually encoded in the URL, not stored on the server). If you set the 'hi' parameter too high, the server will probably time out.


问:
Every linear operator in irreducible $G$-module can be represented as element of $\mathbb{C}[G]$
Suppose that $G$ is a finite group and $V$ is a (complex) irreducible representation of $G$, that is, for any $g\in G$ we have $gV=V$ (here we use notation $gv$ for the action of element $g$ on vector $v$ instead of, for example, $\rho(g)v$, where $\rho\colon G\to\operatorname{End} V$).

There is a natural embedding of the group algebra $\Phi\colon\mathbb{C}[G]\hookrightarrow\operatorname{End} V$ which is defined by the formula
$$
\Phi\left(\sum_{g\in G}c_g\cdot g\right)(v)=\sum_{g\in G}c_g\cdot gv.
$$

I found somewhere (can't remember now) that $V$ is irreducible iff $\Phi$ is surjective, i. e. every linear operator on $V$ can be realised as an element of $\mathbb{C}[G]$, but I don't know how to prove that.

The "if" part is easy. Indeed, if $V$ is reducible, then there is a $G$-invariant subspace $W\subsetneq V$ and we can take projection $P\in\operatorname{End} M$ onto $U$, where $U\in W$ such that $U\oplus W=V$ (and define $P|_{U}=\operatorname{id}_{U}$ and $P|_{W}=0$). It's clear that for any $a\in\mathbb{C}[G]$ we have $\operatorname{Im}\Phi(a)\subset W$, so there is no $a\in\mathbb{C}[G]$ such that $\Phi(a)=P$.

Thus, it remains to prove the "only if" part. Any ideas? I am a beginner in the representation theory, so  any help would be appreciated.

答:
Proving it with the Jacobson Density Theorem is not so bad, in the sense that Wikipedia has a fully self-contained proof. Or you can prove the Artin-Wedderburn Theorem without it and use that.

First note $\mathbb{C}[G]\to\bigoplus\mathrm{End}(V)$ (over all irreps $V$) must be one-to-one: any element of the kernel acts as $0$ on any irrep, hence on any rep, hence sends $1\in\mathbb{C}G$ to $0$, hence is $0$. To see that it is in isomorphism it then suffices to check dimensions match. For this, write $\mathbb{C}G\cong\bigoplus nV$ (where $nV:=V\oplus\cdots\oplus V$) for some multiplicities $n$. Then $\hom_G(\mathbb{C}G,V)\cong V$ ($1\mapsto v$ arbitrarily) so the multiplicities are $n=\dim V$ (recall $\dim\hom_G$ acts as a sort of "inner product" for the multiplicities of irreps in reps). This implies the dimension of $\mathbb{C}[G]$ is $\sum n^2$, same as $\bigoplus\mathrm{End}(V)$.

The map $\mathbb{C}[G]\to\mathrm{End}(U)$ for an irrep $U$ corresponds to a projection map on $\bigoplus\mathrm{End}(V)$. On an arbitrary rep decomposing as $W=\bigoplus mV$ the map $\mathbb{C}[G]\to\mathrm{End}(W)$ lands in the block diagonal subalgebra $\bigoplus m\mathrm{End}(V)$ within (in fact, the diagonal copy of $\mathrm{End}(V)$ in $m\mathrm{End}(V)$) so it is onto iff $W$ is an irrep.


问:
Finding the Sum of a Geometric Series in Finance
I have been stuck on these two problems forever, I have tried using the Sum of a Finite Geometric Series formula, but I cannot get to the intended final product of 200(1.005^(n)-1). I even double checked to make sure the -1 did not have to be in the exponent, I am just at wits end! For clarification, I need help with solving Questions 5 and 6. Mathematics of Loans Q5 and Q6</p>

答:
Note that's only the part in paretheses from (4), not the full answer.

Are you using $1 + r + r^2 + \cdots + r^k = \dfrac{r^{k+1}-1}{r-1}$ ?

In this case, $r=1.005$, so $\dfrac1{r-1} = \dfrac1{0.005} = 200$.

You would also use $k=n-1$, so $r^{k+1}-1 = r^n - 1$.

Putting it together, you do indeed get the answer they suggest.


问:
Trouble understanding why Moore's Plane is not normal
This question was already asked here(Show that the Moore plane is not normal)  but I couldn't understand that answer and also I have another question.

On page 13 of Dolciani Expository text in Topology by S.G Krantz author gives an outline why Moore's plane is not normal. But I am unable to deduce it.

Adding image
Firstly, Author gives two sets $A= \{(x,0)\in P \mid x\in \mathbb{Q}\}$ and $B=\{(x,0)\in P\mid x\in \mathbb{R}\setminus \mathbb{Q}\}$, here $P$ is the Moore plane.

Author tells that both of these sets are closed. In my opinion Closure of both $A$, $B$ should be set $\{(x,0)\in P\mid x\in \mathbb{R}\}$.

Then it it clear that they are disjoint sets.

But can you please tell how to prove that they cannot be seperated by open sets?

I can see it intutively but I am afraid that I am wrong as my intution was wrong in question 1 asked above which was very similar.

I am asking for your help as I am bad in topology as course was marred by a bad instructor and wasn't aware of existence of MSE network then.

答:
The set $L:=\{(x,0): x \in \Bbb R\}$ is discrete in its subspace topology with resp. to the Moore plane. This holds as for a basic subset of $(x,0)$, which is $\{(x,0)\}$ unioned with a tangent ball inside the upper half plane $P :=\{(x,r):x \in \Bbb R, r>0\}$ only intersects $L$ in $\{(x,0)\}$, making it an isolated point in the subspace topology on $L$.

Also, $L$ is closed in the Moore plane (its complement is $P$ which is open by definition of the topology on the Moore plane), so all subsets of $L$ are closed in $L$ (as $L$ is discrete) and so also closed in the Moore plane ("closed in closed is closed") as $L$ is closed.

The fact that we use these two sets specifically has other reasons that will become clear later in the proof.

I personally think that the Jones' lemma proof (see my answer to the linked question in your first line) is the nicer way to go with the proof.


问:
A ship is being pulled by two tugboats problems
A ship is being pulled by two tugboats. The larger tugboat exerts a force 25% greater
than the smaller tugboat and at an angle of 20 degrees west of north.
In what direction should the smaller tugboat pull in order for the ship to head
north?
So far, I assume the smaller force is $x$, so the The larger tugboat exerts a force is $1.25x
\tan(20)=1.25 x/m$, so $m=1.25x/tan20$. also, $x/(1.25x/\tan20)=tan\theta$ which $theta$ is the direction should the smaller tugboat pull in order for the ship to head north. $\theta=\arctan(\tan(20/1.25))$, so am I right of the track? </p>

答:
Let the magnitude of the larger force be $1.25$. So the smaller one has magnitude  $1$.

The west component of the force exerted by the larger tug is $1.25 \sin(20^\circ)$. This must be balanced by the east component of the force exerted by the little tug, which is $\sin(\theta)$. I expect you can now find $\theta$.

Remark: The answer reached in the OP does not make physical sense, for it is clear that the little tug must head more than $20^\circ$ east of north to compensate for the extra power of the big tug.


问:
Would like help with a contour integral.
Disclaimer: the knowledge I have about contour integration is solely from the book "Mathematical Methods in the Physical Sciences" by Mary L. Boas.

I am trying to understand how the following function is derived:
\begin{equation}
\displaystyle\lim_{\epsilon \to 0^+} \int\limits^\infty_{-\infty} \frac{e^{-ixt}}{x+i \epsilon} \; \mathrm{d} x = - 2 \pi i \Theta(t)
\end{equation}
where $\Theta(t)$ denotes the unit step function.

In order to do this, I start by considering the integral:
\begin{equation}
\int\limits_{-\infty}^\infty \frac{e^{-i x t}}{x + i \epsilon} \; \mathrm{d} x
\end{equation}
where $t>0$. In order to evaluate this integral I believe we can use the ``contour integration'' technique and thus I consider:
\begin{equation}
\oint_C \frac{e^{-i z t}}{z + i \epsilon} \; \mathrm{d} z \tag{1}
\end{equation}
where $C$ is the (clockwise) contour as shown in the figure:

Clearly, there is a simple pole at $z= -i \epsilon$ and the residue can be easily found:
\begin{equation}
\mathrm{Res}(-i \epsilon) = \displaystyle\lim_{z \to -i \epsilon}\left\{ (z+i \epsilon) \frac{e^{-i z t}}{z + i \epsilon} \right\}= e^{-\epsilon t}
\end{equation}
Therefore, by the residue theorem, we can evaluate equation $(1)$ as:
\begin{equation}
\oint_C \frac{e^{-i z t}}{z + i \epsilon} \; \mathrm{d} z = - 2 \pi i e^{-\epsilon t}
\end{equation}
where the minus sign is due to the fact that the contour is clockwise. Subsequently, by taking $\epsilon$ to be a very small positive constant this becomes:
\begin{equation}
\oint_C \frac{e^{-i z t}}{z + i \epsilon} \; \mathrm{d} z = - 2 \pi i
\end{equation}
Letting $z=\rho e^{i \theta}$, we can write:
\begin{equation}
\oint_C \frac{e^{-i z t}}{z + i \epsilon} \; \mathrm{d} z = \int\limits^\rho_{-\rho} \frac{e^{-i x t}}{x + i \epsilon} \; \mathrm{d} x + \int\limits^{-\pi}_{0} \frac{e^{-i z t}}{\rho e^{i \theta} + i \epsilon} \rho i e^{i \theta} \; \mathrm{d} \theta \tag{2}
\end{equation}
At this point I am stuck. I believe the second term on the right-hand side of the above equation should tend to zero as $\rho \rightarrow \infty$, however I do not see why this would be true. Any help would be much appreciated.

答:
$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\on}[1]{\operatorname{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
$\ds{\bbox[5px,#ffd]{}}$

\begin{align}
&\bbox[5px,#ffd]{\int_{-\infty}^{\infty}
{\expo{-\ic xt} \over x + \ic 0^{+}}\,\dd x} =
\mrm{P.V.}\int_{-\infty}^{\infty}
{\expo{-\ic xt} \over x}\,\dd x +
\int_{-\infty}^{\infty}
\expo{-\ic xt}\bracks{-\ic\pi\,\delta\pars{x}}\,\dd x
\\[5mm] = &\
\int_{0}^{\infty}
{\expo{-\ic xt} - \expo{\ic xt} \over x}\,\dd x - \ic\pi =
-2\ic\int_{0}^{\infty}
{\sin\pars{xt} \over x}\,\dd x - \ic\pi
\\[5mm] = &\
-2\ic\on{sgn}\pars{t}\int_{0}^{\infty}
{\sin\pars{x} \over x}\,\dd x - \ic\pi =
-\pi\ic\on{sgn}\pars{t} - \ic\pi
\\[5mm] = &\
\bbx{-2\pi\,\ic\,\Theta\pars{t}} \\ &
\end{align}


问:
What does "$\min\limits_x - \cos x$" mean in this optimization question?
I have an assignment with different methods of optimization for the following:

Consider optimization problem
$$\min_x -\cos x$$

I feel dumb but does this mean to use "$\cos x$"? What does $\min\limits_x$ mean?

答:
$\min_\limits{x}f(x)$ means "find the minimum value of $f(x)$ over all possible values of $x$"

$\cos x$ is the trigonometric cosine function, and $-\cos x$ is the same function, multiplied by $-1$.

So the question is asking "what is the smallest value that $-\cos x$ can take?" which is the same as asking "find the largest value $\cos x$ can take, and multiply that by $-1$".


问:
If I flip four coins, what is the probability that the first coin is heads AND there are more heads than tails?
I don't know why this answer is eluding me, but I had to figure it out somehow. I know that the probability of flipping heads first is 1/2. 8/16 outcomes have H first. The probability of having more heads than tails is 5/16: HHHH, HHHT, HHTH, HTHH, THHH. I realize that I can count it and that four of these results have H first, meaning that the answer is 4/16 or 1/4. However, how is this calculated mathematically? You can not simply multiply the two probabilities as they are inclusive. I know that they are independent as well. The first coin flip doesn't matter to having more heads than tails as it is still possible regardless. What probability formula would I need to solve this?</p>

答:
There is nothing "un-mathematical" about your approach; it is perfectly fine.

Another approach is to realize that after the first flip is heads, you need at least 2 heads in the last 3 flips, which has probability $4/8=1/2$. So,
\begin{align}
&P(\text{first flip is heads, more heads than tails in 4 flips})
\\
&= P(\text{first flip is heads}) P(\text{more heads than tails in 4 flips} \mid \text{first flip is heads})
\\
&= P(\text{first flip is heads}) P(\text{at least two heads in the last 3 flips})
\\
&= \frac{1}{2} \cdot \frac{1}{2} = \frac{1}{4}.
\end{align}


问:
Diffeomorphism of product manifolds given diffeomorphisms on the factors.
Given $\mathcal{M} = \mathcal{M_1} \times \dots \times \mathcal{M_k}$ where $\mathcal{M_i}$ are smooth manifolds and $\phi_i:\mathcal{M_i} \to \mathcal{M_i}$ diffeomorphisms for each $i$.

Is $\phi: \mathcal{M} \to \mathcal{M}$ where $\phi = (\phi_1 , \dots , \phi_k) $ a diffeomorphism?

答:
Here is my answer:

$\phi$ has derivative given by $\phi' : T\mathcal{M_1} \times \dots \times T\mathcal{M_k}$ where

$$
\phi' = (\phi'_1,\dots , \phi'_k)
$$

as each $\phi_i$ is diffeomorphic each $\phi'_i$ is continuous. Thus by the Characteristic Property of Product Topology $\phi'$ is also continuous.

The inverse is given by $\phi^{-1} = (\phi_1^{-1} , \dots , \phi_k^{-1})$. A similiar argument shows $\phi'^{-1}$ is continuous.

Thus $\phi$ is a differentiable function with differentiable inverse. So a diffeomorphism.


问:
Integrating $\int_0^\infty\frac{1}{1+x^6}dx$
$$I=\int_0^\infty\frac{1}{1+x^6}dx$$

How do I evaluate this?

答:
$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\on}[1]{\operatorname{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
$\ds{\bbox[5px,#ffd]{}}$

\begin{align}
&\bbox[5px,#ffd]{\int_{0}^{\infty}{\dd x \over 1 + x^{6}}} =
{1 \over 6}\int_{0}^{\infty}{x^{1/6 - 1} \over 1 + x}\,\dd x =
{1 \over 6}\,\
\overbrace{\Gamma\pars{1 \over 6}\Gamma\pars{1 - {1 \over 6}}}
^{\ds{\substack{\ds{Ramanujan's} \\[0.5mm] \ds{Master}\\[0.75mm] \ds{Theorem}}}}
\\[5mm] = &\
{1 \over 6}\,{\pi \over \sin\pars{\pi/6}} =
{1 \over 6}\,\pi\ \underbrace{\csc\pars{\pi \over 6}}_{\ds{2}}\
=\ \bbx{\pi \over 3} \approx 1.0472 \\ &
\end{align}


问:
How to solve these differential equations?
I need to find that W which corresponds to X = 0.6.
It's given that at

1)W= $0$, at X =$0$

2)y=1, at X = 0

\begin{align*}
\frac{dX}{dW} &= 0.024  y(\frac{1-X}{1-0.1478X}) \\
\frac{dy}{dW} &= \frac{-0.0083 (1-0.1478X) }{y}
\end{align*}

I divided both the equation and used separation of variables technique to integrate the equations
\begin{equation}
 \frac{dX}{dy } = \frac{0.024  y(\frac{1-X}{1-0.1478X}) }{ \frac{-0.0083 (1-0.1478X) }{y}}
\end{equation}
Integrating y from 1 -> y  and x from 0 -> 0.6, I got y = 0.52334. Is there any way to find the value of corresponding W?
I know one approach might be to find y as a function of x like this and then replace it and solve for W but the expression has a $y^3$ which would make the integration impossible to solve.

答:
$$\frac{(1-bx)^2}{1-x}\,dx=-ay^2\,dy$$

can be integrated analytically, in the form $y=f(x)$.

Then

$$dw=\frac c{f(x)}\frac{1-bx}{1-x}dx.$$

requires just an antiderivative, but an analytical solution is highly improbable.