问:
How to prove AC = BD?
Triangle ABC has $\widehat{A} = 130^\circ;\, \widehat{B} = 30^\circ.$ D and E are the intersection between the mid-perpendicular of $AC$ and $BC, AB,$ respectively.

a) Prove ADE is an isosceles triangle

b) AC = BD

a) is an easy one, but b) is in a different level (at least with me)

How can I prove AC= BD?

Please help me!

答:
Define $D'$ as the reflection of point $D$ over $EB$.

Here's the rough sketch ( only angle chase is required ) :  show $D'CEB$ cyclic $\rightarrow \Delta AD'B_1$ isosceles $\rightarrow CD=AB_1 \rightarrow \Delta CB_1B$ isosceles

Here's the detailed one.

Claim: $D' \in (CEB)$

The proof is only angle chase.

Note that $\angle DAC=\angle DCA=20 \implies\angle DAE=\angle ADE=70 \implies \angle DEA= 40 \implies\angle EDB=110$ .

Also note that $\Delta ECA$ is isosceles, so $\angle CEA =2\cdot \angle DEA\implies CEA =80 \implies ECB=70$

Now since $D'$ is the reflection of point $D$ over $EB$. So $\angle ED'B=\angle EDB=110$

By cyclic quads angles property we get $D'\in (CEB)$.

Now define $CA\cap (ECB)=B_1$

Claim: $AD'B_1$ is isosceles. Hence $CD=AB_1$.

Again since $D'$ is the reflection of point $D$ over $EB$. So $\angle D'EA=\angle AED=\angle DEC=40\implies \angle D'EC=120\implies CB_1D'=60$( using cyclic quad $CB_1D'E$)

Also note that, since $\Delta ECA$ is isosceles and $\angle CEA=80 \implies \angle ACE=50$

So by cyclic quad $CED'B_1$, we get $\angle ED'B_1=130$ and also $\angle ADE=\angle AD'E=70 \implies \angle AD'B_1=\angle ED'B_1-\angle ED'A=60$.

So $AD'B_1$ is isosceles. Also $AD'=AD$ ( $D'$ is reflected point) , So $AD=AB_1$, but $AD=CD\implies CD=AB_1$ .

Claim:$\Delta CB_1B$ isosceles

Consider cyclic quad $ED'B_1B$. We get $\angle D'EB+\angle D'B_1B=180 \implies \angle CB_1B=80$ . But $\angle ACB=20$. So $\Delta CB_1B$ isosceles  .

Claim:$AC=BD$

So we have $CB_1=CB$ , but we also have $CD=AB_1 \implies AC=BD$

And we are done!

Title: How to prove AC = BD?


问:
When is $A^T A$ a positive definite matrix for non square $A$?
Let $A$ be a $n \times k$ matrix, $A\neq 0$. We know that $(A^T A)$ is positive semi-definite as:

$v^T (A^T A) v = (Av)^T (Av) = ||Av||^2 \geq 0 $

and furthermore, positive definite if $Av \neq 0 \; \forall v \neq 0$ .

Due to $A$ not necessarily being a square matrix the usual arguments of determinants and inverse are out the window, but we can still say that if $Av = 0$ for some $v = (v_1, \cdots, v_k) \neq (0,\cdots,0)$, we must have

$v_1  \begin{pmatrix}
      A_{11}\\\vdots\\A_{n1}
    \end{pmatrix} + \cdots +  v_k \begin{pmatrix}
      A_{1k}\\\vdots\\A_{nk}
    \end{pmatrix} = 0   $

ie. the columns of $A$ are linearly dependent.

So if the columns of $A$ are linearly independent, ie. $A$ has full (column?) rank, this can never happen. Is this a necessary and sufficient condition for $A^T A$ to be positive definite?

答:
Let $A$ be an $m\times n$ matrix, then the symmetric matrix $A^{T}A$ is automatically positive semi-definite
$$
x^{T}A^{T}Ax=(Ax)^{T}Ax=\|Ax\|^{2}\geq 0
$$
Now take note of the following three cases :

$1)$ If $m=n$, then $A$ is a square matrix and $A^{T}A$ is for sure a square matrix and thus $A^{T}A$ is postive-definite if and only if $A$ is of full rank.

$2)$ If $m< n$ then $A^{T}A$ is not-full rank in-fact the rank is at most $m$. Therefore, it is only positive semi-definite.

$3)$ If $m>n$ and this is the case that you should be careful with because the rank of $A$ is at most $n$ and thus it may be positive definite or positive semi-definite.

ContentLicense: CC BY-SA 4.0


问:
How to find the values of $x,y,z$ to give a maximum value of $\min\{ax,by,cz\}/(x+y+z)$.
$a,b,c$ are all known positive rational numbers (e.g. $a=2.1,b=3.5,c=7.4$). They are also always $\geq1$ but I'm not sure if that helps. $\min\{ax,by,cz\}$ means the minimum out of those three products.

This feels like an optimisation problem but I'm not sure how to approach it. I'm hoping to solve this programmatically, so that it can be solved for several thousand sets of coefficients.

答:
One way to model a minimum so it can be solved programmatically is to turn it into a series of optimization problems. In each of these problems you evaluate the set of points for which one of the elements of the minimum is minimal:

$$
\max_{x_1,...,x_n\in\mathbb{R}}\min\{f_1(x_1),...,f_n(x_n)\}/(x_1+...+x_n) =
\max_{i=1,..,.n}
\left(\max_{x_1,...,x_n\in\mathbb{R}\\f_i(x_i)\le f_1(x_1)\\...\\f_i(x_i)\le f_n(x_n)}f_i(x_i)/(x_1+...+x_n)
\right)
$$

The above formula holds in an even more general context, but depending on how complex the $f_i$ are, it may or may not be programmatically feasible.

In your case the constraints create a convex polyhedron, so a nonlinear solver probably can solve even a series of 1000 problems rather quickly.

That said, doing a theoretical analysis of your optimization problem and simplifying it will bring you a possibly tremendous speed up


问:
A bijective function between a square and its side
Is there a simple geometric proof that there exists a continuous bijective function between a square and its side?

And is there some explicit continuous function or formula $f^1(z)\mapsto (x,y)$ and $f(x,y)\mapsto z$, with $(x,y) \in [0,1]\times[0,1]$ and $z \in [0,1]$?

And is there a constructive proof that two sets of the same cardinality have a bijective function?

答: As Chandru1 commented, there are several space filling curves. These on the other hand are usually only surjective. With Cantor-Schröder-Bernstein's Theorem we find a bijection. This is not geometric though and I couldn't imagine to have a geometric proof for this, because $[0,1]$ and $[0,1]^2$ are not homeomorphic. As for the last question, having a bijection between two sets is the definition of having the same cardinality.</p>

ContentLicense: CC BY-SA 3.0


问:
What does the set R/{0} represent?
It looks like it refers to the set R- {0}, but I looked the question (from where I got it) up on the internet, and it's written the same everywhere. The function in consideration has a term with a zero in the denominator, so I'm pretty sure, but I'd like a confirmation.</p>

答:
R/{0} = R$- \{0\} =$ the set of all $x$ such that $x$ belongs to $R$ and $x$ does not belong to {0} = the set of all $x$ such that x belongs to R and  $x\not = 0$.

$R$ is a set, the set of real numbers.

If you want $R$ without $0$ in it, you cannot get this new set by writing : $R - 0$. The reason is that :

(a) If you treat $O$ as a number ( and not a set) the expression does not mean anything ( set operations - here : set difference -  operate only on sets). You cannot " substact" a number from a set.

(b) If you treat $O$ as a set, namely, as the empty set ( as set theory does) you get :

$R-O = R - \emptyset = $ the set of all $x$ such that (1) $x\in R$ and (2) $x\not\in\emptyset = R$.

Note: condition (2) changes nothing, since any object whatever satisfies it.

So if you want $R$ without $0$ in it, you have to "substact" , so to say, the singleton set  $\{0\}$ from the set $R$, that is, you need the set difference : $R - \{0\}$.

Title: What does the set R/{0} represent?


答:
Do you see why $\triangle ADF \cong \triangle DAH$?

That leads to $DF = AH = \frac{AC}{2}$

$BD = 2 \times DF$ (as $ \, \angle FBD = 30^0$)

Title: How to prove AC = BD?

Body:
Triangle ABC has $\widehat{A} = 130^\circ;\, \widehat{B} = 30^\circ.$ D and E are the intersection between the mid-perpendicular of $AC$ and $BC, AB,$ respectively.

a) Prove ADE is an isosceles triangle

b) AC = BD

a) is an easy one, but b) is in a different level (at least with me)

How can I prove AC= BD?

Please help me!


问:
Reducing trigonometry functions in exponents
I need to know a bit about replacing the trig functions in exponents with simpler algebraic equations which can then be reduced with logarithms. For example, an answer to this question said that

$$e^{\sin(x)} = e^{\frac{1}{2i}(e^{ix} - e^{-ix})}$$

applies, which would do the trick. As well as this, using a few basic log rules it could be adjusted to deal with any base as well. I do have a few questions though.

Firstly, how does this work? It looks like it has a fair bit to do with Euler's identity but I don't understand the exact derivation.

Next, how do I generalize it (For example, with the exponent as $4*sin^2(3x)$)? I honestly don't know if it barely involves changing numbers or if it doesn't apply at all anymore.

Finally, is this transferrable to the other trig functions? If it's connected to Euler's identity as I think it is, it should be easy to change it to apply to cosine instead, and then the two together may somehow be used to derive the transformation (if I can call it that) to tan.

Thank you so much for your help!

答:
The derivation of the definitions of $\sin x$ and come from subtracting Euler's formulas.
\begin{align}
e^{ix}&=\cos x+i\sin x\\
e^{-ix}&=\cos x-i\sin x\\
\implies\sin x&=\frac1{2i}(e^{ix}-e^{-ix})
\end{align}
It therefore follows that $$\exp(\sin x)=\exp\left(\frac1{2i}(e^{ix}-e^{-ix})\right).$$

For your example, we can write
\begin{align}
4\sin^2(3x)=4(\sin(3x))^2&=4\left(\frac1{2i}(e^{3ix}-e^{-3ix})\right)^2\\
&=-(e^{3ix}-e^{-3ix})^2\\
&=2-e^{6ix}-e^{-6ix}\\
\implies\exp(4\sin^2(3x))&=\exp(2-e^{6ix}-e^{-6ix}).\\
\end{align}


问:
Internal Bisector of a triangle
In $\Delta ABC$, the bisector of the angle A meets the side BC at D and circumscribed circle at E, then DE equals to

(A) $\frac{{{a^2}\cos \frac{A}{2}}}{{2\left( {b + c} \right)}}$

(B) $\frac{{{a^2}\sec \frac{A}{2}}}{{2\left( {b + c} \right)}}$

(C) $\frac{{{a^2}\sin \frac{A}{2}}}{{2\left( {b + c} \right)}}$

(D) $\frac{{{a^2}\cos ec\frac{A}{2}}}{{2\left( {b + c} \right)}}$

My approach is as follow

Internal Bisector property

$\frac{{AB}}{{AC}} = \frac{{BC}}{{DC}} \Rightarrow \frac{c}{b} = \frac{t}{{a - c}}$

How do I proceed in order to find DE

答:
Since a full derivation is not required, take relative proportions and well known properties of an easily calculable  equilateral triangle of side $2, \angle  BAO =30^{\circ}, $ altitude is divided by center in ratio $2:1$ etc.

$$ OD=DE= \dfrac{1}{\sqrt 3}$$

Only option (B) gives this value.

Title: Internal Bisector of a triangle


问:
Confidence interval of population standard deviation using measurements with uncertainties
Let's say I have a sample of N measurements which I use to calculate the standard deviation of said sample:

$s=\sqrt{\frac{\sum{(x-\bar{x})^2}}{N-1}}$

I can use this value to place a 68% confidence interval on the population standard deviation:

$P(\sqrt{\frac{ks^2}{q_{0.84}}}<\sigma<\sqrt{\frac{ks^2}{q_{0.16}}})=0.68$,

where the coefficients of the confidence interval are calculated using the appropriate chi-squared distribution.

However, my individual measurements have uncertainties, which means that $s$ itself has an associated error (I am using the formulation from Ahn & Fessler 2003 to calculate it, although the choice is not critical here).

My question is: How do I incorporate the error on $s$ into the calculation of the confidence interval for the population standard deviation?

Thank you very much for your help

答:
Here is an example of one kind of bootstrap CI for an unknown population standard deviation. Suppose I have $n=100$ observations.

To get data for a demonstration, I sample from a t distribution with $\nu = 12$ DF, hence $\sigma^2 = \nu/(\nu-2) = 12/10 = 1.2,$ $\sigma = \sqrt{1.2} = 1.0955.$  [You would
substitute your data for my x. Presumably, your data are modeled
to be nearly but not exactly normal, as mentioned in your question.]

```<code>set.seed(1114)
x = rt(100, 12)
s.obs = sd(x);  s.obs
[1] 1.102585

set.seed(2020)
d.re = replicate(5000, sd(sample(x,100,rep=T))/s.obs)
UL = quantile(d.re, c(.975,.025))
s.obs/UL
     97.5%      2.5%
 0.9659272 1.3009683
</code>```

Notice that the 95% nonparametric bootstrap CI $(0.966, 1.301)$ contains
the observed SD $S_{obs} = 1.1026$ of the data x.

Brief rationale for bootstrap code: If we knew the distribution of
$S/\sigma$ then we could get $L$ and $U$ such that
$P\left(L \le \frac{S}{\sigma}\le U\right) = 0.95.$ Then
a 95% CI for $\sigma$ would be of the form
$\left(\frac{S}{U}, \frac{S}{L}\right).$

In order to approximate the unknown distribution of $S/\sigma$ we enter the 'bootstrap world', temporarily using s.obs $(S_{obs})$ as a proxy for $\sigma.$
We take $B = 5000$ samples of size $n=100$ with replacement
from x. This is called re-sampling. Then we obtain standard deviations $S^*$ of these samples and divide by s.obs
to get an idea of the unknown distribution of $D^*\approx S/\sigma.$
We take upper and lower quantiles $U^*, L^*$ of the simulated
distribution of $D^*$s.

Then returning s.obs to its original role as our observerd sample SD,
we obtain the 95% nonparametric bootstrap CI of the form
$\left(\frac{S_{obs}}{U^*},\frac{S_{obs}}{L^*}\right).$ In the R
code the suffix .re replaces the re-sampled quantities denoted above by $*$s.

This is called a nonparametric bootstrap because the bootstrap procedure has not
assumed that the data were sampled from a normal distribution (or another
distribution of known type). We have assumed only that our $n=100$ observations are randomly sampled from
some distribution for which the standard deviation exists.

Body:
Here is an example of one kind of bootstrap CI for an unknown population standard deviation. Suppose I have $n=100$ observations.

To get data for a demonstration, I sample from a t distribution with $\nu = 12$ DF, hence $\sigma^2 = \nu/(\nu-2) = 12/10 = 1.2,$ $\sigma = \sqrt{1.2} = 1.0955.$  [You would
substitute your data for my x. Presumably, your data are modeled
to be nearly but not exactly normal, as mentioned in your question.]

set.seed(1114)
x = rt(100, 12)
s.obs = sd(x);  s.obs
[1] 1.102585

set.seed(2020)
d.re = replicate(5000, sd(sample(x,100,rep=T))/s.obs)
UL = quantile(d.re, c(.975,.025))
s.obs/UL
     97.5%      2.5%
 0.9659272 1.3009683

Notice that the 95% nonparametric bootstrap CI $(0.966, 1.301)$ contains
the observed SD $S_{obs} = 1.1026$ of the data x.

Brief rationale for bootstrap code: If we knew the distribution of
$S/\sigma$ then we could get $L$ and $U$ such that
$P\left(L \le \frac{S}{\sigma}\le U\right) = 0.95.$ Then
a 95% CI for $\sigma$ would be of the form
$\left(\frac{S}{U}, \frac{S}{L}\right).$

In order to approximate the unknown distribution of $S/\sigma$ we enter the 'bootstrap world', temporarily using s.obs $(S_{obs})$ as a proxy for $\sigma.$
We take $B = 5000$ samples of size $n=100$ with replacement
from x. This is called re-sampling. Then we obtain standard deviations $S^*$ of these samples and divide by s.obs
to get an idea of the unknown distribution of $D^*\approx S/\sigma.$
We take upper and lower quantiles $U^*, L^*$ of the simulated
distribution of $D^*$s.

Then returning s.obs to its original role as our observerd sample SD,
we obtain the 95% nonparametric bootstrap CI of the form
$\left(\frac{S_{obs}}{U^*},\frac{S_{obs}}{L^*}\right).$ In the R
code the suffix .re replaces the re-sampled quantities denoted above by $*$s.

This is called a nonparametric bootstrap because the bootstrap procedure has not
assumed that the data were sampled from a normal distribution (or another
distribution of known type). We have assumed only that our $n=100$ observations are randomly sampled from
some distribution for which the standard deviation exists.


问:
Submanifold such that $M-A$ is simply connected then $M$ is simply connected
I am trying to prove the following statement

Let $M$ be a manifold and $A$ a submanifold of codimension greater or equal to $2$. If $M-A$ is Simply connected then $M$ is simply connected.

Now I was able to see that $M$ is path connected. For the part where $\pi_1(M)=\{0\}$ is where I don't know what to do . Does anyone have any suggestions? Thanks in advance.

答: Hint: Given a closed curve $\gamma$ in $M$, problems can only occur if $\gamma$ intersects $A$, i.e., $Y:=\gamma^{-1}(A)$ is a non-empty subset of $[0,1]$. Now if we can deform $\gamma$ such that it avoids $A$, we are in $M-A$ and can contract. For the simplest case, consider an isolated point $t\in Y$. We want to "push" $\gamma(t)$ (i.e., continuously modify $\gamma$ in a small neighbourhood of $t$) in a direction that does not keep $\gamma(t)$ in $A$ and also not in a direction along $\gamma$ because that could "pull" other points of the curve into $A$. By the codimension condition, we have at least one "other" direction available to play with.


问:
I just don't see what I do wrong - number of surjections seems higher than number of functions.
EDIT: Answer added.

I haven't slept much lately and I've been raging on this thing for a couple hours now. I really hope some people here can have the same obsession/rage and will help me out.

I have two sets, A and B.
|A| = m, |B| = n.
I was looking for the number of surjections from A to B, and I found a formula here:
http://www.ma.utexas.edu/users/kbi/COURSES/TERM/11S/325K/L17.pdf

I seem to have understood the following:

You have to calculate the Stirling number of the second kind, and then multiply it with n! due to the fact that Stirling numbers only divide/group (and do not map).
I have the following in maple:

f:=(n,m)->n!/(m!*(n-m)!);

test:=Sum((-1)^if(n,n-i)(n-i)^m,i=0..n);

If I set m to be 500 and n to be 300, then test > n^m, the latter being the total number of functions.
The number of surjective functions is a subset of the number of functions and should therefore not be higher.
My blood is boiling, because I know I'm overseeing something idiotic.
Hoping for help.

EDIT:

Not sure what the protocol is on adding answers, but this question has been answered.
It is most likely a problem with maple.
I added a screenshot of my maple worksheet.
I defined a function f1(n,m) myself, that acted odd.
I googled and found that maple has a default implemented Stirling2(n,m) function.
As the screenshot shows, both functions give the same result for many values of n and m, but not for all.
Apparantly I need 10 rep to post images, so I'll link to imgur:

答: This question has been answered. It is most likely a problem with maple. I added a screenshot of my maple worksheet. I defined a function f1(n,m) myself, that acted odd. I googled and found that maple has a default implemented Stirling2(n,m) function. As the screenshot shows, both functions give the same result for many values of $n$ and $m$, but not for all.


问:
GP and combinations divisibility
Is the number :

${100 \choose 0}\cdot2^0 + {100 \choose 1}\cdot2^1+{100 \choose 2}\cdot2^2+{100 \choose 3}\cdot2^3 \ldots + {100 \choose 100}\cdot2^{100}$

divisible by 3?

I tried looking at the first few terms which can be written as :

$2^0 + \frac{100}{1}\cdot 2^1 + \frac{100 \cdot 99}{2!}\cdot2^2+\frac{100 \cdot 99\cdot98}{3!}\cdot2^3+\ldots$

So I my guess it that the general term

$a_n = \frac{\prod_{i=0}^n 100-i}{n!}\cdot2^n$

but I am not sure how to continue from here to obtain the final value and check if it is divisible by 3. Is there another way?

答:
Hint: write the binomial  expansion of $${(1+2)}^{100}$$
Do you see a similarity?

Title: GP and combinations divisibility


问:
Let $F$ be an infinite field and let $f(x) ∈ F[x]$. If $f(a) = 0$ for infinitely many $a ∈ F$, show that $f = 0$.
Step 1:
Suppose that $F$ is an infinite field and $f(x) \in F[x]$.
To claim the statement,
"If $f(a)=0$ for infinitely many elements $a$ of $F$, then $f(x)=0$".

To prove this statement using proof by contradiction.

Suppose that $f(x)=a_{n}x^{n} + a_{n-1}x^{n-1} + \cdots + a_{1}x+a_{0}$ with $\deg (f(x))=n$.

May I have your idea about the next steps?

答:
We have to understand the background idea at first. A non-zero polynomial $f(x) \in F[x]$ of degree $n$ has at most $~n~$ zeros in the field $F$. However your question says that a polynomial has $\text{infinitely}$ many zeros and hence we can think of some messy things lies inside.

Now suppose in your question the polynomial $f(x)$ is of degree $n$. Assume the $f(x) \neq0$. But it is given that $f(x)$ has infinitely many zeros. So there are $n+1$ elements $a_1,~a_2, \cdots, a_{n},~a_{n+1}$ such that \begin{align}f(x) &=(x-a_1)(x-a_2)\cdots (x-a_n)(x-a_{n+1}) g(x),  \end{align} where $g(x) \in F[x]$. But this means that $\text{deg}(f) \geq n+1$, which is a contradiction [ $\because \text{deg}(f)=n$] unless $f(x) \equiv 0$. Hence $f(x) \equiv 0$.


问:
Integer matrices with determinant 1
Let $A$ be an integer matrix of rank $n$ and let $v_1,\dots,v_n$ be the row vectors of $A$ (which are linear independent). Assume that the convex hull of $S:=\{0,v_1,\dots,v_n\}$ does not contain other integer points than the points in $S$. Is there an elementary way to show that $|\det(A)|=1?$

答: If this condition on lattice points is true, then it's true after left-multiplying or right-multiplying by any matrix in $GL_n(\Bbb Z)$. This means that putting your matrix in Smith normal form preserves this condition. As Smith normal form is diagonal, the conclusion can be reached: if there's a diagonal entry of magnitude 2 or more, there's a lattice point which isn't a $v_i$ in the convex hull.

Title: Integer matrices with determinant 1


问:
Proving that $\frac{1}{1 + \lVert\mathbf{A}\rVert} \leq \lVert (\mathbf{I}_{d} - \mathbf{A})^{-1} \rVert$
Let $\mathbf{A} \in \mathbb{R}^{d \times d}$ such that $\lVert \mathbf{A}\rVert < 1$, where $\lVert \heartsuit\rVert$ is matrix norm. Show that
\begin{equation*}
      \frac{1}{1 + \lVert\mathbf{A}\rVert}
    \leq
      \lVert (\mathbf{I}_{d} - \mathbf{A})^{-1} \rVert
    \leq
      \frac{1}{1-\lVert\mathbf{A}\rVert}.
\end{equation*}

I have shown that $ (\mathbf{I}_{d} - \mathbf{A})$ is invertible, using the fact that the subset of invertiable matrices is open. Then using geometric series, I could show one side of the inequality $$\lVert (\mathbf{I}_{d} - \mathbf{A})^{-1} \rVert  = \lVert\sum_{k=0}^{\infty} \mathbf{A}^{k}\rVert \leq \sum_{k=0}^{\infty}  \lVert\mathbf{A}\rVert^k =  \frac{1}{1-\lVert\mathbf{A}\rVert}$$

Since $\lVert\mathbf{A}\rVert^{n+1} \leq \lVert\mathbf{A}\rVert^{n} $ und $\lim_{n \to \infty} \lVert\mathbf{A}\rVert^n = 0$, we have
$$\sum_{k=0}^{\infty} (-1)^k \lVert\mathbf{A}\rVert^k = \frac{1}{1+\lVert\mathbf{A}\rVert}$$
but I can't show that it bounds $\lVert (\mathbf{I}_{d} - \mathbf{A})^{-1} \rVert $ from below. Any help?

Body:
In general $\|B^{-1}\|\ge\frac{1}{\|B\|}$ since $1=\|BB^{-1}\|\le\|B\|\|B^{-1}\|$.

Also, $\|I-A\|\le1+\|A\|$. Hence combining the two, $$\|(I-A)^{-1}\|\ge\frac{1}{\|I-A\|}\ge\frac{1}{1+\|A\|}$$


问:
A problem about uniformly integrable over E?

Definition：let $\mathcal F$ be a family of functions,if for each $ε＞0$,there is a $δ＞0$ such that for each $f∈\mathcal F$,if $A⊆E$ is measurable and $m(A)＜δ$,then $\int_A |f|＜ε$. We say $\mathcal F$ is uniformly integrable over $E$.

problem:let $\mathcal F$ be a family of functions,each of which is integrable over $E$ then $\mathcal F$ uniformly integrable over $E$ iff  for each $ε＞0$,there is a $δ＞0$ such that for each $f∈\mathcal F$,if $A⊆E$ is measurable and $m(A)＜δ$,then $|\int_A f|＜ε$.

From left to right is easy,only use the defition of uniformly integrable and a inequality, but conversely how to do?
This problem comes from Royden' real Analysis.

答:
I suppose your family consists of real valued measurable functions.

Suppose the second condition is given. If $m(A) <\delta$ then (for any $f \in \mathcal F$) $m(A \cap \{x: f(x) >0\})<\delta$ so $|\int_{A \cap \{x: f(x) >0\}} f |<\epsilon$. But this is same as $\int_{A \cap \{x: f(x) >0\}} |f| <\epsilon$. Similarly $\int_{A \cap \{x: f(x) \leq 0\}} |f| <\epsilon$ Add these two finish the proof. .


问:
If $\mu$ is a complex measure and $\|\mu\| = \mu(X)$, then $\mu$ is a positive measure
Taken from Conway's A course in functional analysis Chapter 3 Section 7 Problem 2

Problem Statement: Let $X$ be a set and $\Omega$ a $\sigma$-algebra of subsets of $X$. Suppose $\mu$ is a complex-valued countably additive measure defined on $\Omega$ such that $\|\mu\| = \mu(X) < \infty$. Show that $\mu(\Delta) \geq 0$ for every $\Delta$ in $\Omega$

Given the section this problem is in, I assume we are supposed to use Banach limits to solve this problem. However, Banach limits are linear functionals acting on $l^\infty$ spaces which makes it hard for me to see how to apply it to this problem.

答:
According to a consequence of Radon-Nikodym Theorem (see Rudin, Real & Complex Analysis, Theorem 6.12), there exists a measurable function $h:X\to \mathbb C$, with $|h(x)|=1$, for all $x\in X$, such that
$$
d\mu=h\,d|\mu|,
$$
and hence
$$
\int_X d|\mu|=\|\mu\|=\mu(X)=\int_X d\mu=\int_X d\mu=\int_X h\, d|\mu|
$$
Hence
$$
0=\int_X (1-h)\, d|\mu|=\int_X (1-\mathrm{Re}\,h)\, d|\mu|,
$$
and hence Re$\,h(x)=1$, almost everywhere, which implies that Im$\,h(x)=0$, almost everywhere, since $|h|=1$. Thus $h(x)=1$, almost everywhere, and hence $\mu$ is a positive measure.


问:
Does a function being not differentiable imply it's continuous?
My textbook defines $p = f'(a^{+})$ and $q= f'({a^-})$, and says;

(I) p=q =>f is differentiable at x = a => f is continuous at x = a

(II)p$\ne$q  => f is not differentiable at x = a, but f is continuous at x=a.

How does the second one work out? It needn't be continuous, right?

答:
If $f(x)$ is defined in some neighbourhood of $a\in\mathbb R$ and both $\lim_{x\to a^-}\frac{f(x)-f(a)}{x-a}$ and $\lim_{x\to a^+}\frac{f(x)-f(a)}{x-a}$ exist (as finite numbers), then $f(x)$ will be continuous at $x=a$.

This is because $f(x)=f(a)+(x-a)\frac{f(x)-f(a)}{x-a}$ so, if, say, $\lim_{x\to a^-}\frac{f(x)-f(a)}{x-a}=p$, then $\lim_{x\to a^-}f(x)=f(a)+0\cdot p= f(a)$, and similar for the limit "from above".


问:
" A 7 is drawn from a deck of cards, then without replacing the card, a 2 is drawn." is NOT an independent event?

Two events are independent if the outcome of one
event does not affect the outcome of the other event.
One of the following statements does NOT describe
independent events. Which one?

F. A coin lands heads up, then a single 6-sided die
lands with a 3 face up.

G. A king is drawn from a deck of cards, then a coin
lands heads up.

H. A 4 is drawn from a deck of cards, then after
replacing the card, a 4 is drawn.

J. A single 6-sided die lands with a 2 face up, then
after being rolled a second time, the die lands with
a1 faceup.

K. A 7 is drawn from a deck of cards, then without
replacing the card, a 2 is drawn.

The answer is K, but I don't think it is not a independent event, since what does drawing a 7 have to do with drawing a 2?

答:
I think your definition of independence is a bit misleading, as it focuses completely on the outcome, not the process. In that vein it is completely understandable to say „drawing a 7 doesn’t affect drawing a 2“.

However, what dependence/independence really tries to describe is, whether the outcome of the first event changes the context to determine the probability of the second event.

Take for example the first two. As the events are taken from completely different activities (throwing dices vs throwing coins) the output of one will certainly not affect the context of the other.

In the last example this is different. Both events come from the same activity and not inserting the first card back does change the context of drawing a 2. Indeed the probability of drawing a 2 goes up to be $4/(n-1)$, which was $4/n$ before removing a 7. Hence the events are dependent.


问:
Integration of $\int \sqrt\frac{x-1}{x^5}\ dx$?
How would you integrate $\int \sqrt\frac{x-1}{x^5}\ dx$?

I've tried to split it up to integrate by parts, with $u = \sqrt{x-1}$ and $v' = \frac{1}{x^5}$. Then I get $-\frac{2}{3}x^{-3/2}(x-1)^{1/2} + \frac{1}{3} \int x^{-3/2}(x-1)^{-1/2}$. It seems to me that I need to further pursue integration by parts on the second integral, but that would seem to throw me into an endless loop. How can I proceed?

I am given that the answer to this is $\frac{2}{3}(1-\frac{1}{x})^{3/2} + C$. What might be the fastest way to achieve this answer?

答:
Hint we have $$\int \frac{1}{x^2}\sqrt{1-\frac{1}{x}}$$

Now put $$t=1-\frac{1}{x}$$
can you end it now?


问:
Find the number of different words that can be formed from the letters of the word ‘TRIANGLE’ so that no vowels are together
I have a question that says: Find the number of different words that can be formed from the letters of the
word ‘TRIANGLE’ so that no vowels are together

I did this in the following way:
Number of ways in which TRIANGLE can be arranged -
Number of ways in which TRIANGLE can be arranged where 3 vowels are together -
Number of ways in which TRIANGLE can be arranged  where 2 vowels are together = 8! - (6!*3!) - (7!*2!*3) = 5760

The correct answer in my book is given to be 14400 calculated as:
xTxRxNxGxLx : x depicts spaces where vowels can be arranged and they are not together.
Therefore, Number of ways in which Consonants can be arranged*Number of ways in which vowels can be arranged  = 5!6C33! = 14400.

Can anyone help me figure out why the method I follow isn't working?

答:
Applying inclusive - exclusive principle

Making group with two vowels$(7!\times 2!\times 3)$ will cover also making group with three vowels $(6!\times3!)$

for example $TR(IA)ENGL$ arrangements of $(IA)$ together also has $E(IA)$  and $(IA)E$

$$8! - (7!\times 2!\times 3) + (6!\times3!)= 14400$$


问:
The limit of the sequence $\,\,a_n=\frac{n(\sqrt[n]{n}-1)}{\operatorname{log} n}$
I’m currently studying a course in real analysis and I’m having some trouble with calculating limits like the following one:

$$   \lim _{n \rightarrow \infty} n \cdot \frac{\sqrt[n]{n}-1}{\operatorname{log} n} $$

The answer would be as follows:

\begin{array}{l}
=\lim _{n \rightarrow \infty} \frac{\sqrt[n]{n}-1}{\frac{\log n}{n}}=\lim _{n \rightarrow \infty} \frac{\sqrt[n]{n}-1}{\log \sqrt[n]{n}} \\
\approx \lim _{n \rightarrow \infty} \frac{\sqrt [n]{n}-1}{\sqrt [n]{n}-1}=1
\end{array}

The problem I have with this possible resolution of the limit is that we have to use an approximation, which is valid since n to the 1/n goes to 1. I suppose that this ‘approximation’ has something to do with Taylor series, but that’s something we haven’t covered yet in class. So I was wondering whether there was another way of computing this limit, which does not require the use of such ‘approximation’. There are a bunch of other limits like this which can be solved by this kind of ‘approximation’.

答:
$$
n\cdot \frac{\sqrt[n]{n}-1}{\log n}=\frac{\mathrm{e^{\frac{\log n}{n}}}-1}{\frac{\log n}{n}}=\frac{\mathrm{e}^{a_n}-1}{a_n}
$$
where $a_n>0$ and $a_n\to 0.$

Next, use the fact that $\lim_{h\to 0}\frac{\mathrm{e}^{h}-1}{h}=1$, to obtain that
$\lim_{n\to\infty}\frac{\mathrm{e}^{a_n}-1}{a_n}=1$.


问:
Prove that $G$ is a finitely generated group under multiplication.

Consider the set of matrices $$G = \left \{\begin{pmatrix} s & b \\ 0 & 1 \end{pmatrix}\ \bigg |\ b \in \Bbb Z,\ s \in \left \{-1,+1 \right \} \right \}.$$ Then show that $G$ is a finitely generated group under multiplication.

It is easy to show that $G$ is a group under multiplication. What I think is that $$G = \left \langle \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}, \begin{pmatrix} -1 & 1 \\ 0 & 1 \end{pmatrix} \right \rangle.$$

The reason behind my claim is that the elements of $G$ are of the form $\begin{pmatrix} 1 & n \\ 0 & 1 \end{pmatrix}$ or $\begin{pmatrix} -1 & n \\ 0 & 1 \end{pmatrix},$ for some $n \in \Bbb Z.$ Now for any $n \in \Bbb Z$ we have $$\begin{pmatrix} 1 & n \\ 0 & 1 \end{pmatrix} = {\begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}}^n.$$ Also if $n$ is an odd integer then $$\begin{pmatrix} -1 & n \\ 0 & 1 \end{pmatrix} = {\begin{pmatrix} -1 & 1 \\ 0 & 1 \end{pmatrix}}^n;$$ and if $n$ is an even integer then $$\begin{pmatrix} -1 & n \\ 0 & 1 \end{pmatrix} =  {\begin{pmatrix} -1 & 1 \\ 0 & 1 \end{pmatrix}}^{n+1} \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix} ;$$ which proves the claim.

Is my above reasoning correct at all? Can anybody please verify it?

Thanks in advance.

答:
Well, you need to consider what happens when the generators multiply:

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


问:
$f(yf(x) + y) = xy + f(y)$ for all $x, y.$ Prove $f$ is surjective
I'm stucking with this problem for days. f is a function from $\mathbb{R}$ to $\mathbb{R}$ satisfies:

$$f(yf(x) + y) = xy + f(y) \qquad \forall x, y \in \mathbb{R}$$

Prove $f$ is surjective, and hence find all the functions satisfy this equation.

I think fixing a $x$ and $y$ as a constant and study the function might be a good idea, but so far I've been only playing with $x=1,0$ and $y=1,0,$ and they didn't seem to help in proving surjectivity.

Any help is appreciated. Thanks!

(Proving injectivity is easy)

答:
Lemma (1):  $f(f(x)+1) = x+f(1)$.

Proof of Lemma (1) : let $y = 1$.

Lemma (2): $f$ is injective.

Proof of Lemma (2) by contradiction: if $x \neq y$ and $f(x) = f(y)$, then

$f(f(x)+1) = f(f(y)+1) = x = y$, which contradicts with $x \neq y$.

Lemma (3): $f$ is surjective:

Proof of Lemma (3): By Lemma (1), $f(f(x-f(1))+1) = x - f(1) + f(1) = x$.

Lemma (4): $f(0) = 0$.

Proof of Lemma (4): By Lemma (1):

$f(f(0)+1) = f(1)$.

By the injectivity of $f$, $f(0) + 1 = 1$, then $f(0) = 0$.

I stuck at proving $f(1) = 1$...


答:
Let $f(1)=a$. Then setting $y=1$, we get
$$
f\big(f(x) + 1\big) = x + a, \quad \text{for all $x\in\mathbb R$},
$$
which implies that
$$
f\big(f(x-a) + 1\big) = x, \quad \text{for all $x\in\mathbb R$}.
$$

Title: $f(yf(x) + y) = xy + f(y)$ for all $x, y.$ Prove $f$ is surjective

Body:
I'm stucking with this problem for days. f is a function from $\mathbb{R}$ to $\mathbb{R}$ satisfies:

$$f(yf(x) + y) = xy + f(y) \qquad \forall x, y \in \mathbb{R}$$

Prove $f$ is surjective, and hence find all the functions satisfy this equation.

I think fixing a $x$ and $y$ as a constant and study the function might be a good idea, but so far I've been only playing with $x=1,0$ and $y=1,0,$ and they didn't seem to help in proving surjectivity.

Any help is appreciated. Thanks!

(Proving injectivity is easy)


问:
Isomorphisms preserve order of elements

If $f:G\rightarrow H$ is an isomorphism, then $\forall g \in G$, we have that the order of $g$ is the same as the order of $f(g)$

I can understand intuitively why this is true, but what would be the clearest way of proving this? Would you use some contradiction or would you define a bijection from the set generated by $g$ to the set generated by $f(g)$?

答:
Let $g\in G$ with order $n$. Then $e_H=f(e_G) = f(g^n) = f(g)^n$.
So the order of $f(g)$ is a divisor of $n$.

Let $d$ be the order of $f(g)$. Then $d$ divides $n$.
Thus $e_H = f(g)^d = f(g^d)$ and hence $f(e_G) =e_H= f(g^d)$.
But $f$ is an isomorphism (one-to-one is required here) and so $e_G=g^d$. By hypothesis ($d$ divides $n$), $d=n$.

Title: Isomorphisms preserve order of elements


问:
In a quaternion, are j and k not just equal to i?
I have been listening to many videos and reading but I am very confused. Firstly, I read that quaternions exist in $\mathbb{R}^4$ which would seem to exclude imaginary numbers completely (I would have guessed). But secondly, if a quaternion is sort of like a complex number, why do we need $j$ and $k,$ why can't we just use $i$ throughout ($a+ib+jc+kd$) is different somehow than ($a+ib+ic+id$) and if so, in what way does $i$ differ from $j$ and $k$?

Also, if quaternions have something to do with rotation is 3D space, why is a quaternion in a hypersphere? If we want to do something with rotations in on the plane, is 3D space involved?

答:
Quaternions build a number system similar, but distinct from the complex numbers. Every quaternion may be written in the form $a + bi + cj + dk$ and $i,k,j$ being pairwise distinct make sure that this representation is in fact unique. If $i=j=k$ then we would have
$$a + bi + cj + dk = a + (b+c+d)i + 0j +0k$$
violating uniqueness. You can think of it as adding three independent imaginary axes to the real line $\Bbb R$, ie making it a 4-dimensional vector space. But the real clue is that, just like the complex numbers have $i^2=-1$, quaternions come with algebraic relations (I think it was along the lines of $i^2 = j^2 = ijk = -1$), which allow you to have a sort of multiplication on that vector space. It turns out this multiplication is very handy to describe composing rotations.

Finally let’s answer the question, why quaternions require to be in $\Bbb R^4$ rather than $\Bbb R^3$. Quaternions mimick specifying an axis of rotation $r\in \Bbb R^3$ as well as an angle $\alpha \in \Bbb R$, which adds up to be a quantity in $\Bbb R^4$. Of cause you can use Euler Angles, which require one dimension less, but those lack most of the cool properties of quaternions...


问:
fraction rased to a power of log base 2
The below expression simplifies to $\frac{y}{n}$,
$$
\left(\dfrac{1}{2} \right)^{\log_2\left(\frac{1}{y/n}\right)}
$$

I have tried to change bases but haven't figured out how to come up with the solution $y/n$.

Thanks in advance.

Title: fraction rased to a power of log base 2

Body:
let $y/n = x$
$$(2^{-1})^{\log_2 x^{-1}} =2^{\log_2 x}=x=\frac{y}{n}$$


答:
Let $f \in \mathrm{Hom}_{\mathbf{Gr}}(G, G')$ be an injective morphism between the groups $G$ and $G'$. It is a fundamental property of morphisms (not necessarily injective) that $f\left[\langle X \rangle\right]=\langle f[X] \rangle$ for any subset $X \subseteq G$, where $\langle X \rangle$ denotes the subgroup of $G$ generated by $X$.

Let us fix $x \in G$. By definition, the order of $x$ is $\left|\langle x \rangle\right|$, the order of the monogenous subgroup generated by $x$. Taking into account the general property mentioned above, we have $\langle f(x) \rangle=f\left[\langle x \rangle\right]$. As $f$ is injective, it follows by elementary cardinal theory that $|\langle f(x) \rangle|=\left|f\left[\langle x \rangle\right]\right|=|\langle x \rangle |$ (in general $|f[T]|=|T|$ for any subset $T$ of the domain of definition of the injection $f$). This means nothing else than the relation you seek to prove: the order of $x$ and the order of its image $f(x)$ coincide.

Title: Isomorphisms preserve order of elements

Body:
If $f:G\rightarrow H$ is an isomorphism, then $\forall g \in G$, we have that the order of $g$ is the same as the order of $f(g)$

I can understand intuitively why this is true, but what would be the clearest way of proving this? Would you use some contradiction or would you define a bijection from the set generated by $g$ to the set generated by $f(g)$?


问:
Image of an injective complex function of an open set
Let $U$ be an open subset of $\mathbb C$ and let $f:U\to \mathbb C$ be an injective complex function. Is it true that $f(U)$ is open? I guess not, but I can not find a counter example. What if $U$ is also connected? Does then $f(U)$ is open? Thanks.

答: Each open subset of $\Bbb{C}$ has cardinality equal to the continuum, and hence there will be an injective function mapping $U$ (injectively and surjectively) to the real line. Such a map will not be particularly simple to define, but $f(U) = \Bbb{R}$ is not open in $\Bbb{C}$ (regardless of whether $U$ is connected or not).


问:
Closed form for $\sum_{n=1}^\infty\frac{1}{2^n\left(1+\sqrt[2^n]{2}\right)}$
Here is another infinite sum I need you help with:
$$\sum_{n=1}^\infty\frac{1}{2^n\left(1+\sqrt[2^n]{2}\right)}.$$
I was told it could be represented in terms of elementary functions and integers.</p>

Body:
Note that

$$\frac{2^{-n}}{2^{2^{-n}}-1}-\frac{2^{-(n-1)}}{2^{2^{-(n-1)}}-1}  = \frac{2^{-n}}{2^{2^{-n}}+1} $$

Thus we have a telescoping sum.  However, note that

$$\lim_{n \to \infty} \frac{2^{-n}}{2^{2^{-n}}-1} = \frac{1}{\log{2}}$$

Therefore the sum is

$$a_1-a_0 + a_2-a_1 + a_3-a_2 + \ldots + \frac{1}{\log{2}} = \frac{1}{\log{2}}- a_0$$

where

$$a_n = \frac{1}{2^n \left ( 2^{2^{-n}}-1\right)}$$

or

$$\sum_{n=1}^{\infty} \frac{1}{2^n \left ( 1+ \sqrt[2^n]{2}\right)}= \frac{1}{\log{2}}-1$$


问:
What is the sum of the following equation?
Let $\lambda_1>\lambda_2\cdots>\lambda_m>0$ be a decreasing sequence. Now, consider the following equation:
$$\sum_{j=1}^m\frac{1}{\prod\limits_{k\neq j}(1-\frac{\lambda_k}{\lambda_j})}$$
I have verified that the result of the above equation is $1$ when $m=2,3$.

I want to know how to get a general proof that the above conclusion holds for any positive integer $m$.

Thanks a lot for any help!

答:
Hint:

Consider the Lagrange interpolating polynomials


问:
Series $\sum_{n=1}^{\infty}\frac{\cos(nx)}{n^2}$
Is it true that for $x\in[0,2\pi]$ we have
$$\sum_{n=1}^{\infty}\frac{\cos(nx)}{n^2}=\frac{x^2}{4}-\frac{\pi x}{2}+\frac{\pi^2}{6}$$
How can I prove it?
For other intervals what is the value of above series if is convergent?</p>

答: Another way to solve this is by finding the fourier series of x^2, then solving for (x+pi)^2 with the cosine addition formula.</p>


问:
How to use matrix language to describe the solution process?
I see that the linear equations $\left\{\begin{array}{l}
2 x+y-t=-2 \\
3 y+z+2 t=3
\end{array}\right.$ can be solved in the following way:

I have uploaded a picture to give the details of the algorithm for the mathematical solution. It is obvious that this method is quite different from the calculation method commonly used in linear algebra textbooks. I want to explain the principle of this algorithm clearly, but I don't even understand the principle at present.

The above similar contents are from page 206 of this textbook:

$$\left(\begin{array}{cccc|c}
2 & 1 & 0 & -1 & 2 \\
0 & 3 & 1 & 2 & -3 \\
\hline 1 & 0 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 & 0 \\
0 & 0 & 1 & 0 & 0 \\
0 & 0 & 0 & 1 & 0
\end{array}\right) \rightarrow\left(\begin{array}{cccc|c}
1 & 0 & 0 & 0 & 2 \\
3 & -6 & 1 & 5 & -3 \\
\hline 0 & 1 & 0 & 0 & 0 \\
1 & -2 & 0 & 1 & 0 \\
0 & 0 & 1 & 0 & 0 \\
0 & 0 & 0 & 1 & 0
\end{array}\right)\rightarrow\left(\begin{array}{cc|cc|c}
1 & 0 & 0 & 0 & 0 \\
3 & 1 & 0 & 0 & -9 \\
\hline 0 & 0 & 1 & 0 & 0 \\
1 & 0 & -2 & 1 & -2 \\
0 & 1 & 6 & -5 & 0 \\
0 & 0 & 0 & 1 & 0
\end{array}\right)\rightarrow\left(\begin{array}{cc|cc|c}
1 & 0 & 0 & 0 & 0 \\
3 & 1 & 0 & 0 & 0 \\
\hline 0 & 0 & 1 & 0 & 0 \\
1 & 0 & -2 & 1 & -2 \\
0 & 1 & 6 & -5 & -9 \\
0 & 0 & 0 & 1 & 0
\end{array}\right)$$

Then we can get the following relationship:

$$\left(\begin{array}{l}
x \\
y \\
z \\
t
\end{array}\right)=\left(\begin{array}{c}
0 \\
-2 \\
9 \\
0
\end{array}\right)+\left(\begin{array}{cc}
1 & 0 \\
-2 & 1 \\
6 & -5 \\
0 & 1
\end{array}\right)\left(\begin{array}{l}
c_{1} \\
c_{2}
\end{array}\right)$$

But I want to use matrix language to describe and prove the solution process:

$$\begin{array}{c}
A.X = B\\
M.A.X = M.B\\
M.A.X.N = M.B.N\\
\text{If }M.B.N = 0\\
\text{then }M.A.X.N = 0\\
\end{array}$$

How can I simply describe and prove the solution process?

答:
I don't understand your approach in the picture for to solve this problem. However, this problem in the matricix language is easy. Note that  if $$(S.E.L): \left\{\begin{aligned} 2x+y+0z-t=-2 \\ 0x+3y+z+2t=3 \end{aligned} \right.$$
so, we can write the S.E.L in the form $$AX=B \quad \text{and} \quad (A|B)$$
Taking the coefficients of the system of linear equations, we construct the matrix $A$, this's $$A=\begin{pmatrix} 2 & 1 & 0 & -1 \\ 0 & 2 & 1 & 2 \end{pmatrix}$$
and taking the unknows of the S.E.L, we have $$X=\begin{pmatrix} x \\ y \\ z \\ t \end{pmatrix}$$
and finally, $$B=\begin{pmatrix} -2 \\ 3 \end{pmatrix}$$
Therefore, you have $$\left\{\begin{aligned} 2x+y+0z-t=-2 \\ 0x+3y+z+2t=3 \end{aligned} \right. \iff \underbrace{\begin{pmatrix} 2 & 1 & 0 & -1 \\ 0 & 2 & 1 & 2 \end{pmatrix}}_{A}\underbrace{\begin{pmatrix} x \\ y \\ z \\ t \end{pmatrix}}_{X}=\underbrace{\begin{pmatrix} -2 \\ 3 \end{pmatrix}}_{B}$$
Now, you can use  Gaussian elimination here $$(A|B) \iff \begin{pmatrix} 2 & 1 & 0 & -1 & | & -2 \\ 0 & 2 & 1 & 2 & | & 3\end{pmatrix}$$


问:
Find bijection $\phi:[n]\times S_{n-1}\to S_n$ and conclude that $\vert S_n\vert = n!$
$[n] = \{1,...,n\}$

I am struggling with finding the bijection.

$\underline{\text{Assume } \mathbf{n=3:}}$

Than we have $A:=[n]\times S_{n-1} = \{(1,\pi_1), (2,\pi_1),(3,\pi_1),(1,\pi_2), (2,\pi_2),(3,\pi_2)\}$,

with $\pi_1 = id$ and $\pi_2 = (12)$

$S_3 = \{(123), (12)(3), (13)(2), (23)(1), (132), id\}$

So we are looking for some $\phi:A\to S_3$. The mapping for $a\in A \mapsto \sigma\in S_3$ is not obvious to me.

Has someone a hint?

The conclusion should be easy if the bijection is found.

Edit:

Due to your hints and further research I found

$\phi:[n]\times S_{n-1}\to S_n, (i,\pi_j)\mapsto \pi_j\circ (in)$, $i\in[n], j\in [n-1]$

My current status on $\phi^{-1}$ is:

$\phi^{-1}:S_n\to [n]\times S_{n-1}, \pi \mapsto (\pi^{-1}(n), ???)$

I am still struggling with $\phi^{-1}$. Does someone has a hint?

答:
Well, we have $\pi_1 = \pmatrix{1 & 2\\ 1&2}$ and $\pi_2 = \pmatrix{1 & 2 \\ 2 & 1}$ or as sequences of the images $\pi_1 = 12$ and $\pi_2=21$ without delimiters.

Rule: in $(i,\pi)$ write $n$ in the $i$th position of $\pi$. Here:

$(1,p_1) = 312$, $(2,p_1)=132$ and $(3,\pi_1)= 123$ or $(1,\pi_1) = \pmatrix{1 & 2 & 3\\ 3&1&2}$, $(2,\pi_1) = \pmatrix{1 & 2 & 3\\ 1&3&2}$ and $(3,\pi_1) = \pmatrix{1 & 2 & 3\\ 1&2&3}$.
This demonstrates also  the general situation.


问:
Are all limits solvable without L'Hôpital Rule or Series Expansion
Is it always possible to find the limit of a function without using L'Hôpital Rule or Series Expansion?

For example,

$$\lim_{x\to0}\frac{\tan x-x}{x^3}$$

$$\lim_{x\to0}\frac{\sin x-x}{x^3}$$

$$\lim_{x\to0}\frac{\ln(1+x)-x}{x^2}$$

$$\lim_{x\to0}\frac{e^x-x-1}{x^2}$$

$$\lim_{x\to0}\frac{\sin^{-1}x-x}{x^3}$$

$$\lim_{x\to0}\frac{\tan^{-1}x-x}{x^3}$$

答: In general, $ \lim_{x \to 0} \frac{f(x) - \sum_{k = 1}^{n - 1} \frac{f^{(k)}(0)\cdot x^k}{k!}}{x^n} = \frac{f^{(n)}(0)}{n!} $. This can be proven using the Mean Value Theorem $n$ times and induction.</p>


答:
This is not a very elegant way of recursively deriving the factorial expression for the order of the finite symmetric groups, but nevertheless let us attempt a suggestion.

Fix $n \in \mathbb{N}$ and consider the map:
$$\begin{align*}
\Phi \colon \Sigma_{n+1} &\to [1, n+1] \times \Gamma\\
\Phi(\sigma)&=\left(\sigma(n+1), (n+1\ \sigma(n+1)) \circ \sigma\right),
\end{align*}$$
where $\Gamma=\mathrm{Stab}_{\Sigma_{n+1}}(n+1)$ is the subset (subgroup actually) of permutations of the natural interval $[1, n+1]$ which fix $n+1$, subset which is clearly isomorphic (as a group, even!) to $\Sigma_n$.

The map $\Phi$ is the bijection you are looking for.

Title: Find bijection $\phi:[n]\times S_{n-1}\to S_n$ and conclude that $\vert S_n\vert = n!$

Body:
$[n] = \{1,...,n\}$

I am struggling with finding the bijection.

$\underline{\text{Assume } \mathbf{n=3:}}$

Than we have $A:=[n]\times S_{n-1} = \{(1,\pi_1), (2,\pi_1),(3,\pi_1),(1,\pi_2), (2,\pi_2),(3,\pi_2)\}$,

with $\pi_1 = id$ and $\pi_2 = (12)$

$S_3 = \{(123), (12)(3), (13)(2), (23)(1), (132), id\}$

So we are looking for some $\phi:A\to S_3$. The mapping for $a\in A \mapsto \sigma\in S_3$ is not obvious to me.

Has someone a hint?

The conclusion should be easy if the bijection is found.

Edit:

Due to your hints and further research I found

$\phi:[n]\times S_{n-1}\to S_n, (i,\pi_j)\mapsto \pi_j\circ (in)$, $i\in[n], j\in [n-1]$

My current status on $\phi^{-1}$ is:

$\phi^{-1}:S_n\to [n]\times S_{n-1}, \pi \mapsto (\pi^{-1}(n), ???)$

I am still struggling with $\phi^{-1}$. Does someone has a hint?


Title: The limit of the sequence $\,\,a_n=\frac{n(\sqrt[n]{n}-1)}{\operatorname{log} n}$

Body:
I’m currently studying a course in real analysis and I’m having some trouble with calculating limits like the following one:

$$   \lim _{n \rightarrow \infty} n \cdot \frac{\sqrt[n]{n}-1}{\operatorname{log} n} $$

The answer would be as follows:

\begin{array}{l}
=\lim _{n \rightarrow \infty} \frac{\sqrt[n]{n}-1}{\frac{\log n}{n}}=\lim _{n \rightarrow \infty} \frac{\sqrt[n]{n}-1}{\log \sqrt[n]{n}} \\
\approx \lim _{n \rightarrow \infty} \frac{\sqrt [n]{n}-1}{\sqrt [n]{n}-1}=1
\end{array}

The problem I have with this possible resolution of the limit is that we have to use an approximation, which is valid since n to the 1/n goes to 1. I suppose that this ‘approximation’ has something to do with Taylor series, but that’s something we haven’t covered yet in class. So I was wondering whether there was another way of computing this limit, which does not require the use of such ‘approximation’. There are a bunch of other limits like this which can be solved by this kind of ‘approximation’.

Body:
Set $y := n^{1/n} >1$, and consider $y \rightarrow 1$.

$\log y= \log (1+(y-1))=$

$(y-1)+O((y-1)^2);$

$\lim_{y \rightarrow 1}\dfrac{y-1}{(y-1)+O((y-1)^2)}=1.$


问:
Computing the Fourier transform of three distributions - one last part
In my analysis class, we are now covering distribution theory. We take the Schwartz space of functions $S(\mathbb{R})$ and its continuous dual space, $S'(\mathbb{R})$, the space of tempered distributions. For $T \in S'(\mathbb{R})$ and $\phi \in S(\mathbb{R})$, we use the notation
$$\langle T,\phi \rangle = T(\phi)$$
and when $T(x)$ is a function, we define $$\langle T,\phi \rangle = \int_{-\infty}^{\infty} T(x)\phi(x)dx$$ For the Foruier transform on tempered distributions, we define $$ \langle \mathcal{F}T,\phi \rangle = \langle T,\mathcal{F}\phi \rangle $$ We are asked to compute the Fourier transform of the following functions as distributions

$H(x)=\chi_{[0,\infty)} (x)$, the Heaviside step function.

$xH(x)$.

$x^2 \sin(x)$

Our textbook says that, in the sense of distributions, one has the derivative $H'(x)=\delta_0(x)$ so I thought I could take the Fourier transform on both sides and get, using the derivative identity $ip \hat{H}=\frac{1}{\sqrt{2\pi}}$ but I do not know if I can "divide" with distributions, given other answers on this site and elsewhere, when dividing we need to take a constant multiple of $\delta$ which I do not know why, and I do not how to compute the constant. For the second part, I thought I could do the derivative and get using the product rule (which I think holds for distributions) $(xH)'=H+xH'=H+x\delta_0$ but I do not know how to compute the Fourier transform of $x\delta_0$. I have managed part 3. I would therefore appreciate any help on 1 and 2, and I thank all helpers.

************* Progress: I have managed part 3 but parts 1 and 2 are still a mystery to me. I do not know how to "divide" in the sense of distributions. I thank all helpers.

答:
I'm using
$$
\mathcal{F}\{f(x)\} = \int_{-\infty}^{\infty} f(x)e^{-ix\xi} \, dx.
$$

First we take the Fourier transform of the $\operatorname{sign}$ function using $\operatorname{sign}'=2\delta$:
$$
2
= \mathcal{F}\{2\delta\}
= \mathcal{F}\{\operatorname{sign}'\}
= i\xi \mathcal{F}\{\operatorname{sign}\}
.
$$
From this we can conclude that
$$
\mathcal{F}\{\operatorname{sign}\}
= \operatorname{pv}\frac{2}{i\xi} + C\delta(\xi).
$$
Here I have used the facts that $u(x)=\operatorname{pv}\frac{1}{x}$ is the odd solution to $xu(x)=1$ and that $x\delta(x)=0.$

But $\operatorname{sign}$ is an odd function, and so must be its Fourier transform, so $C=0.$ Thus,
$$
\mathcal{F}\{\operatorname{sign}\}
= \operatorname{pv}\frac{2}{i\xi}.
$$

Now, $H=\frac12(\operatorname{sign}+1),$ so
$$
\mathcal{F}\{H\}
= \mathcal{F}\{\frac12(\operatorname{sign}+1)\}
= \frac12(\mathcal{F}\{\operatorname{sign}\}+\mathcal{F}\{1\})
= \frac12(\operatorname{pv}\frac{2}{i\xi} + 2\pi\delta(\xi))
= \operatorname{pv}\frac{1}{i\xi} + \pi\delta(\xi)
.
$$

Then,
$$
\mathcal{F}\{xH\}
= i\frac{d}{d\xi}\mathcal{F}\{H\}
= i\frac{d}{d\xi}(\operatorname{pv}\frac{1}{i\xi} + \pi\delta(\xi))
= -\operatorname{fp}\frac{1}{\xi^2} + i\pi\delta'(\xi)
.
$$

Finally,
$$
\mathcal{F}\{x^2\sin x\}
= (i\frac{d}{d\xi})^2 \mathcal{F}\{\sin x\}
= -\frac{d^2}{d\xi^2} \mathcal{F}\{ \frac{1}{2i}(e^{ix}-e^{-ix}) \}
= -\frac{1}{2i} \frac{d^2}{d\xi^2} 2\pi(\delta(\xi-1)-\delta(\xi+1)) \\
= i\pi (\delta''(\xi-1)-\delta''(\xi+1))
.
$$


问:
Difference between binomial and hypergeometric distributions
I am having trouble when it comes to picking which distribution to use to solve a probability problem, let's take the following exercise as an example:

Suppose that out of $500$ lottery tickets sold, $200$ pay off at least
the cost of the ticket. Now suppose that you buy $5$ tickets. Find the
probability that you will win back at least the cost of $3$ tickets.

I used the hypergeometric distribution while solving it but the solution manual indicates a binomial distribution. The reason I chose the hypergeometric distribution is that because I don't think these trials are independent with fixed probability, so for example I have $1/200$ chance of picking the first ticket that win back its cost but $1/199$ for the second one, etc...
This reasoning got me question almost all probability problems with binomial distribution I did before and I am lost no. Any help would be appreciated.

答:
Which model you use depends on whether the $500$ tickets represents the population of all tickets that were sold, or whether the problem is merely stating that out of every $500$ tickets sold, $200$ of these will pay off at least the cost of the ticket, and that the total number of tickets is very much larger than $500$.  Now, I tend to favor the former interpretation, because it is rather odd to state the proportion of "non-losing" tickets as $200$ out of $500$, when $2$ out of $5$ would have sufficed.

Let's look at the hypergeometric model.  We have a population of $N = 500$ tickets of which $n = 200$ are "non-losing."  You choose $m = 5$ tickets from this population and are interested in the random number $X$ of non-losing tickets occurring among the five.  Thus $$\Pr[X \ge 3] = \sum_{x=3}^5 \Pr[X = x] = \sum_{x=3}^5 \frac{\binom{n}{x}\binom{N-n}{m-x}}{\binom{N}{n}} = \sum_{x=3}^5 \frac{\binom{200}{x}\binom{300}{5-x}}{\binom{500}{5}} = \frac{1010589063}{3190558595} \approx 0.316744.$$

Now, consider the binomial model.  We would have $n = 5$ tickets drawn with non-losing probability $p = 2/5 = 0.4$, hence $$\Pr[X \ge 3] = \sum_{x=3}^5 \binom{n}{x} p^x (1-p)^{n-x} = \sum_{n=3}^5 \binom{5}{x} (0.4)^x (0.6)^{5-x} = \frac{992}{3125} \approx 0.31744.$$

These are very similar, but the hypergeometric model has slightly smaller probability.  The two models would diverge more substantially if $N$ were smaller or $m$ were larger.

As a more advanced exercise, consider the limiting case where $N \to \infty$ and $n = Np$ for some fixed $p \in (0,1)$.  What happens to the hypergeometric PMF?


问:
A function that is both open and closed but not continuous
This does not have to be a very extravagant example just something that I can wrap my head around to have a concrete idea.

I was thinking that this could be satisfied by the function

$f: [0, 2\pi) \rightarrow B_1 = \{(x,y): x^2 + y^2 = 1\}$

$$
f(x) = (\cos(x),\sin(x))
$$

The point of discontinuity is obviously at $(1,0)$.

I was hoping to get a little help convincing myself that this is both an open and closed mapping. (Still getting use to the definition). So, my definition as stated follows.

Let $f:X \rightarrow Y$ be a function on the indicated spaces. Then $f$ is an $\bf{open}$ $\bf{function}$ or $\bf{open}$ $\bf{mapping}$ if for each open set $O$ in $X$, $f(O)$ is open in $Y$. The function $f$ is a $\bf{closed}$ $\bf{function}$ or $\bf{closed}$ $\bf{mapping}$ if for each closed set $C$ in $X$, $f(C)$ is closed in $Y$.

My idea was along the lines of the following.

Let $O$ be an open set in the interval $[0, 2\pi)$. Then $O = (a,b)$, $a>0, b<2\pi$.  Suppose $U = (0, 2\pi) = \cup O$. Where $U$ is the collection of all open sets $O$. Then $f(U)$ is the unit circle excluding the point $(1,0)$ Therefore an open set.

Let $C$ be a closed set in the interval $[0, 2\pi)$. Let $\epsilon > 0$ be given. Then $C = [a, b]$ such that $a \geq 0$, $b \leq 2\pi- \epsilon$. Suppose $H = [0, 2\pi- \epsilon]= \cup C$. $H$ is the collection of all closed sets $C$. Then $f(H)$ is the approximately the unit circle radius 1 starting at the point $(1,0)$ ending and including a point before the point $(\cos(2\pi- \epsilon), \sin(2\pi- \epsilon))$. Therefore a closed set.

Thus we have found a discontinuous, closed and open map.

答:
You can find your answer even in very simple functions defined on $R$!
for example consider $f:R\to A$ where $A = (-\infty , 1]\cup(2,\infty)$ and
$$ f(x) =
 \begin{cases}
x   & x \in (-\infty , 1]\\
x+1 & x \in (1,\infty)
\end{cases}
$$
$f$ is clearly discontinuous but it's open and closed. to verify just derive $f^{-1}(x)$!


问:
What's the bijection between scalar/inner products and (certain) almost complex structures (on $\mathbb R^2$)?
Asked on maths overflow here.

What's the bijection between (equivalence classes of) scalar products (I guess 'scalar product' is the same as 'inner product') and a.c.s. (almost complex structure/s) on $\mathbb R^2$?

From Example 1.2.12 of Daniel Huybrechts - Complex Geometry An Introduction.

Assumptions and notation:

I just pretend $V = \mathbb R^2$ literally instead of just an isomorphism.

Let $\Phi(V)$ be the set of real symmetric positive definite $2 \times 2$ matrices. This set is in bijection with inner products on $V$, I believe. We have according to this,

$$\Phi(V) = \{\begin{bmatrix}
h & f\\
f & g
\end{bmatrix} \ | \ h+g, hg-f^2 > 0 \}_{h,f,g \in \mathbb R}$$

Let $\Gamma(V)$ be the (matrix representations of) a.c.s. on $V$. We have, according to this,

$$\{\begin{bmatrix}
a & b\\
\frac{-1-a^2}{b} & -a
\end{bmatrix}\}_{a,b \in \mathbb R, b \ne 0}=: \Gamma(V) \subseteq Auto_{\mathbb R}(V) \subseteq End_{\mathbb R}(V)$$

I understand that the 'rotation' matrices in $V$ are $SO(2) := \{R(\theta) := \begin{bmatrix}
\cos(\theta) & -\sin(\theta)\\
\sin(\theta) & \cos(\theta)
\end{bmatrix}\}_{\theta \in \mathbb R}$, though I'm not sure that Huybrechts has the same usage of the term 'rotation'. (I ask about this later.)

Questions:

A. For injectivity (except for the equivalence class):

Given (equivalence class of) scalar product ($[M]$ of) $M$, choose unique $I$ that assigns $v$ to the one described. I'll call this map $\gamma: \Phi(V) \to \Gamma(V)$, $\gamma(M)=I$. (Later, $\tilde \gamma: \frac{\Phi(V)}{\tilde{}} \to \Gamma(V)$, $\tilde \gamma([M])=I$.)

It's 'rotation by $\pi/2$' or something. In what way? For $M=I_2$ (2x2 identity), then $I$ is indeed 'rotation by $\pi/2$', in the sense that it's $\begin{bmatrix}
0 & 1\\
-1 & 0
\end{bmatrix} \in SO(2) \cap \gamma(V)$, which is the '$R(\theta)$', for $\theta = \pi/2$.

What exactly is the formula for $I=\begin{bmatrix}
a & b\\
\frac{-1-a^2}{b} & -a
\end{bmatrix} \in \Gamma(V)$ given $M = \begin{bmatrix}
h & f\\
f & g
\end{bmatrix} \in \Phi(V)$?

I'm asking because

2a - I would exceed wolfram computation time

2b - I notice for a different $M$ I tried, $I$ isn't a 'rotation matrix' in the sense of $SO(2)$. In fact, I believe the only 'rotation' matrices that are also a.c.s. are $\pm \begin{bmatrix}
0 & 1\\
-1 & 0
\end{bmatrix}$, i.e. $SO(2) \cap \gamma(V) = \{\pm \begin{bmatrix}
0 & 1\\
-1 & 0
\end{bmatrix}\}$. However, I think $I$ kind of 'rotates by $\pi/2$' in some other sense.

2c - I think $SO(2) \cap \gamma(V)$ isn't meant to be the image of $\gamma$

B. For surjectivity:

I'll call whatever map we would have as $\phi: \Gamma(V) \to \Phi(V)$, $\phi(I)=M$

Given an a.c.s. $I$, what are some possible scalar products $M$?

There's a comment that goes choosing the unique $M_v$ such that for some $v \in V \setminus 0$, we have $\{v,I(v)\}$ as an orthonormal basis. I tried this out (long to type!), and the only thing missing was the positively oriented. I guess either $\{v,I(v)\}$ or $\{v,-I(v)\}$ is positively oriented though. So I'll let $M_v$/$N_v \in \Phi(V)$ correspond to $\{v,I(v)\}$/$\{v,-I(v)\}$. Then by fixing $v$ (I ask about non-fixing of $v$ later), we have $\phi(I)=M_v$ or $N_v$, whichever corresponds to positively oriented basis. I'll just call this $\phi(I)=L_v$ Is this right?

Is $\phi$ supposedly an inverse (or right inverse or left inverse or whatever) to $\gamma$ (or $\tilde \gamma$ or whatever), in the sense that $\gamma(\phi(I)) = I$ for all (a.c.s.) $I \in \Gamma(V)$?

This whole thing about the $v$ makes me think there's another equivalence relation going on here. Is there?

This seems like we can have maps parametrised by the nonzero $v$, namely $\phi_v: \Gamma(V) \to \Phi(V)$. In this case, we might investigate if $\phi_v(I)=L_v=L_w=\phi_w(I)$ or at least if $[L_v]=[L_w]$ under the old equivalence relation of positive scalar $\lambda$, i.e. $L_v = \lambda L_w$. If this investigation turns out negative, then I think there's some problem like if 2 inner products are equivalent if they are from the same a.c.s. $I$ under $\phi_{\cdot}$, but for possibly different $v$ and $w$, then I think the equivalence class of $L_v$ under this new relation, which is $\{L_w\}_{w \ne 0}$, might not be the same as the equivalence class of $L_v$ under the old relation, which is $\{\lambda L_v\}_{\lambda > 0}$.

Ideas:

Perhaps there's some matrix thing here about how scalar products are in bijection with positive definite symmetric matrices and then almost complex structures are rotation matrices or something that are square roots of $-I_2$. Like given pos def symmetric $B$, there exists unique a.c.s. $J$ such that (something something).

Perhaps this is related, but I'd rather not further analyse the question or read through the answer given that I've spent over a month on almost complex structures BEFORE we even put inner products on vector spaces. Please consider spoon-feeding me here.

答:
Fix a two-dimensional real vector space $V$. There are three kinds of extra data we can impose upon $V$:

An orientation, a function $\omega$ which measures a basis $(v_1, v_2)$ and outputs $\omega(v_1, v_2) \in \{\pm 1\}$.

A complex structure, an $\mathbb{R}$-linear operator $I \colon V \to V$ satisfying $I^2 = -\operatorname{id}_V$.

A scalar product $B \colon V \times V \to \mathbb{R}$, which is bilinear, symmetric, and positive-definite.

For example, when $V = \mathbb{R}^2$ and $(e_1, e_2)$ is the standard basis, then we have the standard structures:

The orientation of a basis $(v_1, v_2)$ is the sign of the determinant of the change-of-basis matrix from $(e_1, e_2)$ to $(v_1, v_2)$.

The complex structure is a rotation counter-clockwise by $\pi/2$, the linear operator defined by $I e_1 = e_2$ and $I e_2 = -e_1$.

The dot product $B(e_1, e_1) = B(e_2, e_2) = 1$ and $B(e_1, e_2) = 0$.

When I say "the" rotation by $\pi/2$, I am really using both the orientation and the scalar product implicitly. An algebraic rotation by $\pi/2$ is simply an operator $I$ squaring to $I^2 = - \operatorname{id}_V$, and there are many operators of this form. For example, I could define $J e_1 = 2 e_1 + 3e_2$ and $J(2e_1 + 3e_2) = -e_1$ and $J$ would be an algebraic rotation by $\pi/2$.

Keep in mind that if $V$ is just a two-dimensional real vector space with no more data, we cannot possibly say if something preserves lengths or angles, think for example of the two-dimensional vector space of functions $f \colon \mathbb{R} \to \mathbb{R}$ spanned by $e^x$ and $\sin x$: is the operator $I(e^x) = \sin x$ and $I(\sin x) = -e^x$ a true "rotation"? We cannot possibly say before we define an inner product on the space, but it is certainly an algebraic rotation since it squares to minus one.

Things brings us to standard notions for "compatibility" of a complex structure with the above:

A complex structure $I$ is compatible with the scalar product $B$ if it is an isometry: $B(Iv_1, Iv_2) = B(v_1, v_2)$ for all $v_1, v_2 \in V$.

A complex structure $I$ is compatible with the orientation if $(v, Iv)$ is positively oriented for any $v \in V$.

Lemma: If $(V, \omega, B)$ is a two-dimensional real vector space equipped with an orientation $\omega$ and scalar product $B$, then there is a unique compatible complex structure $I \colon V \to V$.

Proof: Since $I$ is an isometry it preserves lengths: $B(v, v) = B(Iv, Iv)$ for all $v \in V$. Furthermore, we have $B(v, Iv) = B(Iv, I^2 v) = -B(v, Iv)$ and hence $v$ and $Iv$ are perpendicular for all $v \in V$. Therefore $Iv$ lies in the one-dimensional subspace perpendicular to $v$, and must be one of the two vectors on this line which have the same length as $v$. Out of these two possibilities for $Iv$ we take the one where $\omega(v, Iv) = 1$.

Now, fix an oriented two-dimensional vector space $(V, \omega)$. Define
$$S = \{B \colon V \times V \to \mathbb{R} \mid B \text{ a scalar product}\},$$ $$C = \{I \colon V \to V \mid I^2 = -\operatorname{id}_V \text{ and } \omega(v, Iv) = 1 \text{ for all } v \in V \}$$
$$ \Phi \colon S \to C, \quad \Phi(B) = I_B $$
where $I_B$ is the unique complex structure compatible with the data $(V, \omega, B)$. We want to show that $\Phi$ is surjective, and that whenever $\Phi(B) = \Phi(D)$ then $B = \lambda D$ for some $\lambda \in \mathbb{R}_{>0}$.

Surjectivity: Let $I$ be a complex structure on $V$ compatible with $\omega$. Pick any vector $v \in V$, then $(v, Iv)$ is a positively oriented basis. Define a scalar product $B$ by setting $B(v, v) = B(Iv, Iv) = 1$ and $B(v, Iv) = 0$, in other words $B$ is defined so that $(v, Iv)$ is an orthonormal basis. Since $I$ is compatible with both $\omega$ and $B$, we have that $I = I_B = \Phi(B)$.

"Injectivity": Suppose that $I_B = I_D$ for two scalar products $B, D$. Then $(v, I_B v)$ is a positively oriented orthogonal basis for both $B$ and $D$. Hence there are positive scalars $\lambda, \mu$ such that $(\lambda v, \lambda I_B v)$ and $(\mu v, \mu I_B v)$ are positively oriented orthonormal bases for  $B$ and $D$ respectively, and therefore $\frac{1}{\lambda} B = \frac{1}{\mu} D$. (If this does not convince you, do the simple exercise: a scalar product is entirely determined by an orthonormal basis).

Hopefull that is enough: it is a very drawn out explanation. Intuitively, complex structures are algebraic rotations. A scalar product defines a circle in the space (vectors of unit length) and angles in the space, and an orientation tells you which way around the circle is the "positive" way, hence you get a unique compatible complex structure in the presence of an scalar product and orientation. Scaling up/down the scalar product (making the "unit" circle larger or smaller) doesn't change angles or rotations.


问:
biholomorphically mapping upper unit disk to unit disk
I have to find a biholomorphic map from $\{z \in \mathbb{c} : |z|<1, $Im$ z>0\}$ to the unit disk.

My idea: If I can somehow map the upper half disk into the upper half plane (biholomorphically) I am done, since I can then compose it with Caley transform to get into the unit disk. If I use $f(z)=1/z$ it will only map it to outside of the upper half disk in the half plane, so can I edit it a bit to capture the interior of half disk as well?

答:
First,
$$
f_1(z)=\frac{1+z}{1-z}
$$
takes the upper half of the unit disk to the first quadrant.

Then
$$
f_2(z)=z^2
$$
takes the first quadrant to the upper half plane.

Finally,
$$
f_3(z)=\frac{z-i}{z+i}
$$
maps the upper half plane to the unit disk.

Altogether
$$
f(z)=(f_3\circ f_2\circ f_1)(z)=
\frac{\left(\frac{1+z}{1-z}\right)^2-i}{\left(\frac{1+z}{1-z}\right)^2+i}=
-i\cdot\frac{z^2+2iz+1}{z^2-2iz+1}.
$$


问:
Integration of $\int \sqrt\frac{x-1}{x^5}\ dx$?
How would you integrate $\int \sqrt\frac{x-1}{x^5}\ dx$?

I've tried to split it up to integrate by parts, with $u = \sqrt{x-1}$ and $v' = \frac{1}{x^5}$. Then I get $-\frac{2}{3}x^{-3/2}(x-1)^{1/2} + \frac{1}{3} \int x^{-3/2}(x-1)^{-1/2}$. It seems to me that I need to further pursue integration by parts on the second integral, but that would seem to throw me into an endless loop. How can I proceed?

I am given that the answer to this is $\frac{2}{3}(1-\frac{1}{x})^{3/2} + C$. What might be the fastest way to achieve this answer?

Body:
Another approach:

Let $$I=\int \sqrt{\frac{x-1}{x^{5}}}dx$$so
\begin{eqnarray*}
I&=&\int \frac{\sqrt{x-1}}{x^{5/2}}dx\\
&=&2 \int \frac{\sqrt{u^{2}-1}}{u^4}du, \quad u=\sqrt{x}, du=\frac{1}{2\sqrt{x}}dx\\
&=&2\int\sin^{2}(s)\cos(s)ds, \quad u=\sec(s), du=\tan(s)\sec(s)ds\\
&=&2\int p^{2}dp, \quad p=\sin(s), dp=\cos(s)ds\\
&=&\frac{2p^{3}}{3}+C\\
&=&\frac{2}{3}\left(\frac{x-1}{x^{5}} \right)^{3/2}x^{6}+C
\end{eqnarray*}
Therefore, $$\boxed{\int \sqrt{\frac{x-1}{x^{5}}}dx=\frac{2}{3}\left(\frac{x-1}{x^{5}} \right)^{3/2}x^{6}+C}$$


问:
About the divergence of an infinite series.
Why is the series $\sum \frac{1}{n}$ not convergent, even though $$
\lim_{n \to \infty}\frac{1}{n}=0?$$

Body:
$$
1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{2^n}=\\
=1+\frac{1}2+\left(\frac{1}{3}+\frac{1}{4}\right)+\left(\frac{1}{5}+\cdots+\frac{1}{8}\right)+\cdots+\left(\frac{1}{2^{n-1}+1}+\cdots+\frac{1}{2^n}\right)
$$
But
$$
\frac{1}{3}+\frac{1}{4}\ge \frac{2}{4}=\frac{1}{2}, \\
\frac{1}{5}+\cdots+\frac{1}{8} \ge \frac{4}{8}=\frac{1}{2},
$$
and in general
$$
\frac{1}{2^{n-1}+1}+\cdots+\frac{1}{2^n}\ge \frac{2^{n-1}}{2^n}=\frac{1}{2}
$$
and hence
$$
1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{2^n}\ge 1+\frac{n-1}{2}.
$$


问:
A bijective function between a square and its side
Is there a simple geometric proof that there exists a continuous bijective function between a square and its side?

And is there some explicit continuous function or formula $f^1(z)\mapsto (x,y)$ and $f(x,y)\mapsto z$, with $(x,y) \in [0,1]\times[0,1]$ and $z \in [0,1]$?

And is there a constructive proof that two sets of the same cardinality have a bijective function?

答:
As edited, the answer to the first question is that such a continuous bijection doesn't exist.

The fact that sets of the same cardinality are in bijective correspondence is my definition of "same cardinality", so I'm not sure what the second question is really asking.


问:
REVISITED $^2$: Does a solution in $\mathbb{R}^n$ imply a solution in $\mathbb{Q}^n$?
Let $A ∈ M_{m\times n}(\mathbb{Q})$ and $B ∈ \mathbb{Q}^m$. Suppose that the system of linear
equations $AX = B$ has a solution in $\mathbb{R}^n$. Does it necessarily have a solution
in $\mathbb{Q}^n$?

Where do I start?

I feel I can use this to help:

$$x_i=\frac{1}{a'_{ii}}\left(b'_i - \sum_{j=i+1}^{k} a'_{ij} x_j \right).$$

Here is a thought that just crossed my mind, but I'm not sure if this is legal:

Suppose $Y\in \{\mathbb{R}\setminus \mathbb{Q}\}^n$ and $Y'\in\mathbb{Q}^n$ are solutions to $AX=B$. Then $AY=AY'=B$, or $A(Y-Y')=A(Y+(-Y'))=0$, but  addition between $\{\mathbb{R}\setminus \mathbb{Q}\}^n$ and $\mathbb{Q}^n$ is . . .

答:
No, think about $A=I$, $B=\pi\mathbb{1}$, where $\mathbb{1}$ denotes the vector of ones.

EDIT: I'm assuming $F$ is $\mathbb{R}$ or $\mathbb{C}$.

EDIT 2: For revised question - Then yes (although I can't think of an elegant argument why).

Think about using Gaussian elimination to solve for some solution (the fact that there is a solution in $\mathbb{R}^n$ implies that the algorithm generates a solution). The augmented matrix $[A|B]\in\mathbb{Q}^{m\times(n+1)}$. Gaussian elimination only requires us to swap rows around, add rows together and multiply a row by another element in matrix. $\mathbb{Q}$ is closed under addition and multiplication, thus the end product of the algorithm still belongs to $\mathbb{Q}^{m\times(n+1)}$. If at the end you have some "free variables" (that is, if there is multiple solutions) then just substitute any rational numbers into them.

EDIT 3: Just noticed that what I wrote in the second edit had already been pointed out by @GEdgar and @JonasMeyer in the comments.


问:
Show that ${\left(\frac{1+\sin{\frac{\pi}{8}}+i{\cos{\frac{\pi}{8}}}}{1+\sin{\frac{\pi}{8}}-i{\cos{\frac{\pi}{8}}}}\right)}^{\frac{8}{3}}=-1$

Show that $${\left(\frac{1+\sin{\frac{\pi}{8}}+i{\cos{\frac{\pi}{8}}}}{1+\sin{\frac{\pi}{8}}-i{\cos{\frac{\pi}{8}}}}\right)}^{\frac{8}{3}}=-1$$

My Try

I know that I have to apply de Moivre's formula to simplify this further. But it can only be used for integers. I tried simplifying this expression but its going no where,

${\left(\frac{\sin^2{\frac{\pi}{16}}+\cos^2{\frac{\pi}{16}} +2\cos{\frac{\pi}{16}}\sin{\frac{\pi}{16}}+i{\cos{\frac{\pi}{8}}}}{\sin^2{\frac{\pi}{16}}+\cos^2{\frac{\pi}{16}} +2\cos{\frac{\pi}{16}}\sin{\frac{\pi}{16}}-i{\cos{\frac{\pi}{8}}}}\right)}^{\frac{8}{3}}$

Can someone please give me a hint to work this out? Thank you!

答:
Hint

Let $\dfrac\pi8=\dfrac\pi2-2t$

$$S=\left(\dfrac{1+\cos2t+i\sin2t}{1+\cos2t-i\sin2t}\right)^{8/3}$$

$$=\left(\dfrac{\cos t+i\sin t}{\cos t-i\sin t}\right)^{8/3}$$ as $\cos t\ne0$

Either utilise Proof for de Moivre's Formula

Or using
Intuition behind euler's formula,

$$S=(e^{2it})^{8/3}$$

So, one of the three values of $S$ is

$$e^{16it/3}$$

Can you take it from here?


问:
Why $0$ is not irreducible?
We took the following definition of irreducible:

If $R$ is a commutative ring, $r \in R$ is irreducible if $r = ab$ for $a,b \in R$ implies $a \in R^{*}$ or $b \in R^{*}.$

Then my professor said: "so $0$ is not irreducible" but I do not understand that as I know that $0 = 1.0$ and so we have $1$ is a unit and so $0$ can be irreducible. Could anyone explain that to me, please?

答: In any ring, $0\cdot0 = 0$, so the given implication does not hold (since 0 is not invertible).

Title: Why $0$ is not irreducible?


问:
Simple algebra simplification question?
Hello everyone I have the following question.

I have the following fraction

$f(x)=-\frac{4}{x^2}+\frac{1}{(x-1)^2}$

But how would I reduce it I know I have to use the multiply the opposite numerator by denominator and I got.

$(x-1)^2(4)=4x^2-8x+4$

$1(x^2)=x^2$

So I got got $f(x)=-\frac{4x^2-8x+4+x^2}{x^2(x-1)^2}$

But this is incorrect.What am I doing wrong.

答:
You are close.
Your mistake was applying the "-"
to both terms instead of just one.

You wrote
$f(x)=-\frac{4x^2-8x+4+x^2}{x(x-1)^2}$.

What you should have is
$f(x)=\frac{-(4x^2-8x+4)+x^2}{x(x-1)^2}
=\frac{-4x^2+8x-4+x^2}{x(x-1)^2}
=\frac{-3x^2+8x-4}{x(x-1)^2}
$.

Title: Simple algebra simplification question?


问:
Way to find: $a^n \equiv -1 \mod p$
Fermat's Little Theorem states that if $a$ is an integer and $p$ a prime:

$$a^{p-1} \equiv 1 \mod p$$

This is particularly helpful while calculating congruences.

Is there such a theorem/way to find $n$ such that:

$$a^n \equiv -1 \mod p$$

答:
$a^n+1\equiv 0 \ mod p$

If n is odd we can write:

$a^n+1=(a+1)(a^{n-1}-a^{a-2}+a^{a-3}-
\cdot\cdot\cdot +1)$

p can be a primes factor of RHS of above relation, for example:

$2^5+1=33=3\times 11$

p can be 3 or 11 such that :

$2^5=32\equiv -1 \mod33\equiv -1 \mod 3\equiv -1 \mod 11$

Also due to Euler's criterion if $n=\frac {p-1}2$ and a is co-prime with p we have:

$a^{\frac {p-1}2}=-1 \mod p$

For example p=7 , then:

$n=\frac{7-1}2=3$

Now take a=13 we have:

$13^3=2197=313\times 7+6\equiv 6 \ mod 7\equiv -1 \ mod 7$

a must not have this relation with p:

$a \equiv k^2\mod p $

for example

$11=4+7$ and $4=2^2$

But $13=6+7$ and 6 is not perfect square.

Title: Way to find: $a^n \equiv -1 \mod p$


问:
Finding permutations of 3 types of instance that cannot appear as a block
There are 3 types of cars, there are 6 blue cars, 7 pink cars, 8 white cars.

What is the number of permutations that those cars can be sorted in a parking lot where you cannot have all the cars from the same colour appear as a block.

e.g:

```<code>*,b,b,b,b,b,b,* <-- invalid
*,p,p,p,p,p,p,p,* <-- invalid
*,w,w,w,w,w,w,w,w,* <-- invalid
b,b,b,b,b,p,p,p,p,p,p,w,w,w,w,w,w,w,b,p,w <-- valid
</code>```

I was thinking of doing it on the negative and maybe "combining" blocks of each type as one, and then dividing the solution to 3 ways, but the more I got into thinking the more complex it got.

答: Hint: Try to calculate total no. of ways to arrange them all, no. of ways where cars of one/two/three color(s) are together and try to use Inclusion-Exclusion principle</p>


答:
HINT:

The simplest thing I could think of is using the given relation:

$$1+e^{i\alpha} = 2\cos(\frac{\alpha}{2})e^{i\frac{\alpha}{2}}$$

Simple $\frac{\pi}{2}$ addition/subtration in $\sin{x}$ or $\cos{x}$ can bring both the terms in the numerator and denominator into either

$$1+e^{ix}$$
or
$$1+e^{-ix}$$

form. From there you can apply the relation I have given above.

Title: Show that ${\left(\frac{1+\sin{\frac{\pi}{8}}+i{\cos{\frac{\pi}{8}}}}{1+\sin{\frac{\pi}{8}}-i{\cos{\frac{\pi}{8}}}}\right)}^{\frac{8}{3}}=-1$

Body:
Show that $${\left(\frac{1+\sin{\frac{\pi}{8}}+i{\cos{\frac{\pi}{8}}}}{1+\sin{\frac{\pi}{8}}-i{\cos{\frac{\pi}{8}}}}\right)}^{\frac{8}{3}}=-1$$

My Try

I know that I have to apply de Moivre's formula to simplify this further. But it can only be used for integers. I tried simplifying this expression but its going no where,

${\left(\frac{\sin^2{\frac{\pi}{16}}+\cos^2{\frac{\pi}{16}} +2\cos{\frac{\pi}{16}}\sin{\frac{\pi}{16}}+i{\cos{\frac{\pi}{8}}}}{\sin^2{\frac{\pi}{16}}+\cos^2{\frac{\pi}{16}} +2\cos{\frac{\pi}{16}}\sin{\frac{\pi}{16}}-i{\cos{\frac{\pi}{8}}}}\right)}^{\frac{8}{3}}$

Can someone please give me a hint to work this out? Thank you!


问:
Showing that $f$ is $C^{\infty}$ using the equation $\frac{d^2f}{dx^2} = f\frac{df}{dx}$.
I'm trying to answer Tu's Introduction to Manifolds exercise 1.3, which says in the first item:

Let $U \subset \mathbb{R}^n$ and $V \subset \mathbb{R}^n$ be open subsets. A $C^\infty$ map $F \colon U \to V$ is called a diffeomorphism is it is bijective and has a $C^\infty$ inverse $F^{-1} \colon V \to U$. Show that the function $f \colon ]\frac{-\pi}2, \frac\pi2[ \to \mathbb{R}$ given by $f(x) = \tan x$ is a diffeomorphism.

By doing some calculations, I arrived at the equality in the title: $\frac{d^2f}{dx^2} = f\frac{df}{dx}$. Note that both $f$ and $\frac{df}{dx}$ are continuous and defined everywhere in the given domain. Is it possible to justify formally that this implies that $f$ is in fact $C^\infty$? It is pretty clear to me that this is in fact true, because differentiating both sides gives $\frac{d^3f}{dx^3}$ in function of $\frac{df}{dx}$ and $f$ again, and this keeps happening forever in a way which preserves continuity for all derivatives. The problem is that I really don't know how to justify this formally.

答:
From the product rule, we see that if $g$ and $h$ are differentiable functions, then so it the product $gh$.

Thus the equation
$$ f''=ff' $$
also shows that $f''$ is differentiable, since $f$ and $f'$ are. Repeat this logic to show that each derivative $f^{(n)}$ of $f$ exists and is differentiable.


问:
vertical bar on vector
I am reading a paper and I have problem to understand the equation (this is the full paper)

Assume that a one-dimensional discrete-time signal s of length N
exhibits sparsity in certain orthonormal basis Ψ defined by the basis
vectors $\Psi = [\Psi_{1}|\Psi_{2}|\Psi_{3}|...\Psi_{N}]$. Therefore, the signal s can be
represented using its sparse transform domain vector x as follows:

What is the meaning of the vertical bar in $\Psi = [\Psi_{1}|\Psi_{2}|\Psi_{3}|...\Psi_{N}]$?

答:
The context is:

Assume that a one-dimensional discrete-time signal $s$ of length $N$ exhibits sparsity in certain orthonormal basis $\Psi$ defined by the basis vectors $\Psi=[\Psi_1∣\Psi_2∣\Psi_3∣\ldots\Psi_N]$.

So, it seems that $\Psi$ is an orthonormal basis, that the vectors of that basis are $\Psi_1,\Psi_2,\Psi_3,\ldots,\Psi_N$ and that $[\Psi_1∣\Psi_2∣\Psi_3∣\ldots\Psi_N]$ means $\{\Psi_1,\Psi_2,\Psi_3,\ldots,\Psi_N\}$.

Title: vertical bar on vector


问:
Convergence of $f(g_n(x_n)) \rightarrow f(g(x))$
Let $(x_n)_{n\in \mathbb{N}}$ be a sequence in $\mathbb{R}^d$ such that $x_n \rightarrow x \in \mathbb{R}^d$ ($n\rightarrow \infty$). And $f,g,g_n\in C^1(\mathbb{R}^d,\mathbb{R}^d)$ such that :
$$f(g_n(y))\rightarrow f(g(y)), \,\, \forall y \in \mathbb{R}^d.$$
My question is the following :
Can we deduce that :
$$f(g_n(x_n)) \rightarrow f(g(x)).\,\, (n\rightarrow \infty)$$
Or we'll have to verify other hypothesis ?

答: Without more assumption, this may be false. If you add the hypothesis that $f\circ g_n$ uniformly converges to $f \circ g$, then it would be correct.


问:
Why is the free pro-c-group on an infinite set not the pro-c-completion of the free group?
The set-up is the following: $\mathfrak c$ is a collection of finite groups closed under subgroups, homomorphic images, and extensions. For any group $G$, the pro-$\mathfrak c$-completion $G(\mathfrak c)$ is defined as the topological group $\varprojlim G/U$ where $U$ ranges over all normal subgroups of $G$ such that $G/U\in\mathfrak c$, with $G/U$ treated as discrete topological groups. Note that we have a group homomorphism $\eta\colon G\to G(\mathfrak c)$.

If I understand correctly, the pro-$\mathfrak c$-completion is the left adjoint to the forgetul functor from pro-$\mathfrak c$-groups to groups, as it satisfies the universal property that any group homomorphism $f\colon G\to H$ where $H$ is pro-$\mathfrak c$-group factors as $f=h\circ\eta$ where $h\colon G(\mathfrak c)\to H$ is a continuous group homomorphism (so a morphism of pro-$\mathfrak c$-groups).

It seems to me that the above implies that the free pro-$\mathfrak c$-group on a set $S$ should be the pro-$\mathfrak c$-completion of the free group on $S$. I am reading, however, that this is the case only when $S$ is finite, but not when $S$ is infinite. What am I missing?

答:
Adjoint functors compose: if $F_1, F_2$ are composable functors with left adjoints $G_1, G_2$, then $F_1 \circ F_2$ has left adjoint $G_2 \circ G_1$, and this is straightforward to prove. In this situation $F_1$ is the forgetful functor from groups to sets and $F_2$ is the forgetful functor from pro-c-groups to groups. Hence $G_1$ is the free group functor and $G_2$ is the pro-c-completion.

So you are right if you interpret "free pro-c-group on an infinite set" in the obvious way. However, a cursory google search found Wilson's Profinite Groups which gives a different universal property from the obvious one.


问:
In a quaternion, are j and k not just equal to i?
I have been listening to many videos and reading but I am very confused. Firstly, I read that quaternions exist in $\mathbb{R}^4$ which would seem to exclude imaginary numbers completely (I would have guessed). But secondly, if a quaternion is sort of like a complex number, why do we need $j$ and $k,$ why can't we just use $i$ throughout ($a+ib+jc+kd$) is different somehow than ($a+ib+ic+id$) and if so, in what way does $i$ differ from $j$ and $k$?

Also, if quaternions have something to do with rotation is 3D space, why is a quaternion in a hypersphere? If we want to do something with rotations in on the plane, is 3D space involved?

答:
I think your question reflects some basic misunderstandings about what mathematical objects are. In particular the misunderstanding that there is such a thing as "what a mathematical object is".

Namely, two mathematical objects can be exactly the same even if they are completely different. Look at some examples:

(a) Remainders when dividing by $12$, with addition modulo $12$ (e.g. $6+7=1$); (b) Rotations around origin by a multiple of $30^\circ$ with respect to composition of rotations; (c) Multiplication of nonzero remainders modulo $13$. Those three are different objects, yet they obey the same arithmetic rules - all three have a structure of a group, all three have $12$ elements, and in all three there is one element ("generator") which produces all elements when you use the operation repeatedly. (In (c) you can take the remainder $2$ as such element, as $2^0,2^1,\ldots 2^{11}$ are all different $\pmod{13}$.)

(a) Complex numbers and (b) Pairs of real numbers: $\mathbb R\times\mathbb R=\{(a,b)\mid a,b\in\mathbb R\}$, with addition given as $(a,b)+(c,d)=(a+c,b+d)$ and multiplication given by $(a,b)(c,d)=(ac-bd, ad+bc)$. Obviously, in the second case $(a,0)$ "behaves like" the real complex number $a$ and $(0,b)$ "behaves like" a pure imaginary number $bi$. (In fact, $(0,1)$ "behaves like" $i$.)

What is a moral of this? It is this: we just don't want to distinguish two mathematical objects if they are exactly the same, provided that we forget the nature of the objects they are made of.  In algebra, we formalize this notion of being the same "up to the nature of the constituents" using the definition of a isomorphism. An isomorphism is a bijection between the underlying sets for two algebraic structures, which maps results of operations into results of operations. If there is an isomorphism between two algebraic structures, we call those two structures "isomorphic". This means that you can replace the elements of one with the elements of the other and all the calculation rules remain the same - in other words, they are "the same up to the nature of the constituents". Thus, in mathematics, we don't particularly care which one example of an algebraic structure (out of many mutually isomorphic ones) we get to analyze.

For example, if you make a bijection $f:\mathbb R\times\mathbb R\to\mathbb C$, given by $f(a,b)=a+bi$ - this is one isomorphism between $\mathbb R\times \mathbb R$ (with addition and multiplication as given above) and $\mathbb C$ (with complex number addition and multiplication). In my first example, identify the remainder $x\pmod{12}$ with a rotation by $x\cdot 30^\circ$ and with the remainder of $2^x\pmod{13}$.

Now back to quaternions. There is no point in musing about what they really are. You just need one possible definition of quaternions and the operations between them. Anything that is isomorphic to that example can equally be called "quaternions", and anything that is not isomorphic to that example is, well, not the quaternions.

For one possible definition, take $\mathbb R^4$ (set of quadruplets of real numbers) and define operations:

$$(a,b,c,d)+(p,q,r,s):=(a+pb+q,c+r,d+s)$$
$$(a,b,c,d)(p,q,r,s):=(ap-bq-cr-ds,aq+bp+cr-ds,ar-bs+cp+dq,as+br-cq+dp)$$

What you get are quaternions. You may want to call the elements $(0,1,0,0),(0,0,1,0),(0,0,0,1)$ the names $i,j,k$, respectively, and you may want to call the element $(a,0,0,0)$ just $a$ (for $a\in \mathbb R$) - silently identifying $\mathbb R\times\{0\}\times\{0\}\times\{0\}$ with $\mathbb R$ via isomorphism $a\mapsto (a,0,0,0)$, in which case you can prove $(a,b,c,d)=a+bi+cj+dk$ and go from there. Of course, you can identify some rotations of $3D$ space with quaternions - again via an isomorphism of the set of rotations (with respect to composition) to a particular subset of quaternions (with respect to multiplication).

Isomorphic structures have all the properties identical, which now lets you prove that quaternions are not isomorphic to complex numbers. (Using tools of linear algebra: quaternions are of dimension $4$ over reals, while complex numbers are of dimension $2$.) Also, in quaternions "of the kind explained above" (i.e. over $\mathbb R^4$) we have $i=(0,1,0,0)\ne(0,0,1,0)=j$) so in no isomorphic structure you can ever have $i=j$ because isomorphisms are bijections - i.e. they must be "one to one".

My bigger point is: do not spend time thinking what quaternions really are. Study their properties. The same properties will be the properties of any particular instance of quaternions - i.e. of any one of the many mutually isomorphic structures of quaternions. A part of a training of a mathematician is to be able to silently and seamlessly switch from one structure to another isomorphic structure, we do it all the time, and you should be able to do so as well.


问:
Line integral with given midpoint and radius of a circle. Daugmans algorithm
I am trying to realize the daugman algorithm in java code. While reviewing the formula I found a line integral with a mid point and a radius as parameters. Now I am trying to understand how to calculate this integral.
$$
\oint\limits_{x0, y0, r}^{}  \frac{I(x,y)}{2πr}  ds
$$
(x0,y0) is the center and r the radius. I(x,y) is the Image with pixelvalues.
Can anyone explain how to calculate this?

Body:
$$
\mathop{\oint}_{s(t)=(x_0+i y_0)+re^{it}}\frac{I(x,y)}{2\pi r}ds=\\
\frac{1}{2\pi r}\int_0^{2\pi} I\bigg(g(t),h(t)\bigg)\sqrt{(g')^2+(h')^2}dt
$$
where $g(t)=\text{Re}[s(t)]$ and $h(t)=\text{Im}[s(t)]$.


问:
Contraction Mapping Theorem proof of uniqueness and convergence
The Contraction Mapping Theorem states the following:

[Quoting from the book: "An Introduction to Numerical Analysis" by Suli and Mayers]

Let $g$ be a real-valued function, defined and continuous on a bounded closed interval $[a,b]$ of the real line, and assume that $g(x)\in[a,b]$ for all $x\in[a,b]$. Suppose, further, that $g$ is a contraction on $[a,b]$. Then, $g$ has a unique fixed point $\xi$ in the interval $[a,b].$ Moreover, the sequence $(x_k)$ defined by $x_{k+1}=g(x_k)$ for $k=0,1,2,...,$, converges to $\xi$ as $k\to\infty$ for any starting value $x_0$ in $[a,b].$

I've been able to prove the existence of a fixed-point $\xi$ as it is a consequence of Brouwer's Fixed Point Theorem. However, I'm unable to prove 1) the uniqueness of $\xi$ in the interval $[a,b].$; and 2) that the sequence $(x_k)$ converges to $\xi$ as $k\to\infty$ for any starting value $x_0$ in $[a,b].$

答:
The function $g$ is a contraction mapping if there exists a $k\in (0,1)$, such that
$$
|g(x)-g(y)|\le k|x-y|,
$$
for all $x,y\in[a,b]$.

Then
$$
|x_{n+1}-x_n|=|g(x_n)-g(x_{n-1})|\le k |x_n-x_{n-1}|
$$
and inductively
$$
|x_{n+1}-x_n|\le k |x_n-x_{n-1}|\le \cdots\le k^n |x_1-x_0|
$$
In particular,
$$
|x_{m}-x_n|\le |x_{m}-x_{m-1}|+\cdots+|x_{n+1}-x_n|\le (k^{m-1}+\cdots+k^{n-1})|x_1-x_0|
\le |x_1-x_0|\sum_{j=n-1}^\infty k^j=\frac{k^{n-1}}{1-k}|x_1-x_0|,
$$
and hence $\{x_n\}$ is a Cauchy sequence, and thus convergent.

If $x_n\to x$, then $x_{n+1}=g(x_n)\to g(x)$, and thus $g(x)=x$.


问:
Neat way to prove that $x^2\gt (1+x)(\ln(1+x))^{2}, ~\forall x\gt 0$
I am supposed to prove the following inequality. One standard method that can be put to use is considering a function $f(x)=x^2-(1+x)(\ln(1+x))^2$ and take its derivative to comment on the behaviour of its increase and arrive at concluding the inequality using the fact that $f(0)=0$.

$$x^2\gt (1+x)(\ln(1+x))^2,~~ \forall x\gt 0$$

I was wondering if there was a neater way to come up with the proof, especially as this is a contest-practice problem. Any hints are appreciated. Thanks.

答:
Hint : It's equivalent to :

$$x+1-1\geq \sqrt{x+1}\ln(x+1)$$

Putting $x+1=y$ it becomes :

$$y-1\geq\sqrt{y}\ln(y)$$

Now we substitute : $y=e^u$ and introduce the function :

$$f(u)=e^u-1-e^{\frac{u}{2}}u$$

The derivative is :

$$f'(u)=e^u-e^{\frac{u}{2}}-\frac{1}{2}e^{\frac{u}{2}}u$$

Now factorize and use $e^x> x+1\quad \forall x>0$

Conclude.


问:
Why is a circle with a cut not a Lipschitz domain?
I know the zero-centered circle with Radius $R$ and with a cut $\Omega = \{x \in \mathbb{R}^n | |x| < R, x \neq \lambda e_1 \textrm{ for } 0 \leq \lambda < R\}$ is not a Lipschitz domain.

The reason that was given to me is, that $\Omega$ does not lie to one side of $\partial \Omega$. However I don't really see how that implies that the domain is non-Lipschitz.

答: Near any point $x_0=\lambda e_1$ with $0\le \lambda<R$, you cannot find a neighbourhood $U$ of $x_0$ such that $U\cap \Omega$ is  the set of points in $U$ that are "below" (relative to some hyperplane through $x_0$) the graph of a Lipschitz function. This is simply because such a (small) neighbourhood would lack points that are $\notin\Omega$ and "above" the function graph.


问:
No. of Interesection of the graph of $y=\sin x$ and $y=x/n$
Let $a_n$ denotes the number of times the line $y=x/n$ cuts the graph of $y=\sin x$. Then which of the following are true?

$(a)a_n$ is a Cauchy sequence.

$(b) a_n$ has a convergent subsequences.

$(c) \inf\{a_n\}$ is $1$.

$(d)$ All terms of $a_n$ are distinct.

My attempt

I am going to show $a_n$ has unbounded subsequence and so $a_n$ is unbounded.

Let $n_k=\lceil (2k+1)\pi\rceil $ where $k\in \mathbb{N}$

Then $a_{n_k}\ge 2(2k+1)+1=4k+3$ since the origin is always there and $\sin x$ is odd function.

So $(a)$ is false.

We have $a_1=1, a_2=a_3=a_4=3$

So $(c)$ is true and $(d)$ is false

Now, intuitively it's clear to me $a_n$ is non-decreasing  and
so if a subsequence is convergent ,it must be bounded above.

I think there is no convergent
subsequence and for that I need to prove every subsequence of $\{a_n\}$ is unbounded above.Am I right?

Please help me complete the problem. Also I was thinking whether a general formula exist for $a_n$.?

Thanks for your time

答: What you did is fine. Note that the sequence $(a_n)_{n\in\Bbb N}$ is increasing (although, as you noticed, not strictly increasing). And every subsequence of an increasing unbounded sequence is also increasing and unbounded. Therefore, it cannot be convergent.


问:
How to show that if two Platonic solids have the same number of edges, vertices, and faces, then they are similar in $\mathbb{R}^{3}$?
Note: It appears that some of the terms here do not have standardized definitions, so some sources may give conflicting info.

I was looking into the proof that there are only five Platonic solids in Basic Concepts of Algebraic Topology by F.H. Croom at page 29, Theorem 2.7. To clarify,

We define a Platonic solid as a simple, regular polyhedron homeomorphic to $S^{2}$.

We define a simple polyhedron to be a polyhedron that does not self-intersect itself.

We define a regular polyhedron to be a polyhedron whose faces are regular polygons all congruent to each other and whose local regions near the vertices are all congruent to each other.

Using homology theory, one can prove that the Euler formula $V-E+F=2$ must hold for Platonic solids. Then by using Euler's formula and invoking a counting argument, we find that there are five possible tuples $(V, E, F)$.
This is a beautiful proof, but I am unsatisfied with a question: How do we know there can't be two non-similar Platonic solids that have the same $(V, E, F)$-tuple?

Almost all sources I've looked at seem to assume it is obvious that two Platonic solids with the same $(V, E, F)$-tuple are similar, and it is not obvious to me.

Does anyone have any suggestions for how to prove this? Alternatively, does anyone know of a reference where this is proved rigorously?

Edit 1: It's not completely clear, but it seems like the definition I used for "regular polyhedra" is different than the one commonly used. Note that I am not assuming any global symmetry, so if any global symmetry is to be invoked, it needs to be proven.

Edit 2: I've been made aware of Cauchy's rigidity theorem, which is proven in, e.g., Proofs From the BOOK by Aigner & Zeigler. One can show that any two Platonic solids that have the same $(V, E, F)$-tuple must be combinatorically equivalent. However, in order for the theorem to apply, we need to show that our Platonic solids are convex. I can't seem to think of any rigorous argument for why the Platonic solids have to be convex.

And actually, you don't need to show that the entire polyhedron is convex. If I'm not mistaken, the proof for Cauchy's rigidity theorem only relies on the vertices of the polyhedron being locally convex. So really it suffices to show the vertices are convex.

答:
If I understand you correctly, then you want your polyhedron to be an embedding of the sphere with a well-defined interior. So we can consider it as the boundary of this interior (this will be important later).

We can now speak about the (generalized) Gaussian curvature of this boundary. By the Gauss-Bonnett theorem, the total curvature of this surface is $2\pi$ times the Euler characteristic of the sphere, hence positive. But in a polyhedron, the curvature is concentrated in the vertices (i.e. is zero everywhere else). The "curvature" at a vertex is better know as its angular defect, and since all vertices are locally the same, they all must have the same angular defect. But since these values add up to a positive value, each such angular defect must be positive.

In sum, the curvature is non-negative everywhere. Now there seems to be the following theorem: the boundary of a shape is convex if and only if it has non-negative curvature everywhere (see the answer to this question). Hence your polyhedron is convex.

Once we know this (and since you already know that the combinatorics is unique) you can apply Cauchy's rigidity theorem to conclude the uniqueness of the Platonic solids.


问:
Centralizer of a given element in $S_n$?
It is known that any two disjoint cycles in $S_n$ commutes. Therefore, any $\pi\in S_n$ which is disjoint with $\sigma$ is in the centralizer of $\sigma$: $C_{S_n}(\sigma)$. Also
$$
\sigma^i\pi\in C_{S_n}(\sigma), 0\leq i\leq|\sigma|. \tag{*}
$$
Here are my questions:

Are these all the elements in $C_{S_n}(\sigma)$? If the answer is no, is there an method to find all the elements in $C_{S_n}(\sigma)$ other than check the elements one by one?

A quick search returns several relevant questions:

Order of the centralizer of a permutation

Centralizer of a specific permutation

Is there a systematic way of finding the conjugacy class and/or centralizer of an element?

The method ($*$) is what I learned so far from these questions and their answers. An immediate example in which ($*$) does not work that I can come up with is $\sigma=(12)(34)(56)$ in $S_6$. No disjoint $\pi$ can be found in this case and
$$\langle(12),(34),(56)\rangle\subset C_{S_6}(\sigma).$$

答:
I will outline a way in which you (or somebody else who reads this thread in the future) can seek an answer to the originally asked question.

Step 1 : If you have a $k$ cycle $\tau$ in $S_n$ then, what can you say about the number of elements in the centralizer of $\tau$ ?

Answer: $k \cdot (n-k)!$ is the number of elements in the centralizer of a $k$ cycle in $S_n$, and it is rather easy to show this. Leave a comment if you want this to be described, but I am not sure when I will find time to reply to a comment.

Step 2: Next, if you denote by $T_{\tau}$ the set $T_{\tau} = \{ \tau ^i \cdot \sigma ~ : 0 \le i \le k-1, \sigma \in S_{n-k} \},$ then how many elements are there in $T_{\tau}$ ?

Answer: $k \cdot (n-k)!$ is the number of elements in $T_{\tau}$ and this is so straightforward that you should be able to show this yourself.

Step 3: Finally notice that $T_{\tau}$ defined as above is a subset of the centralizer $C_{\tau}$ of $\tau$.

All these would tell you that $C_{\tau}$ is in fact isomorphic (as a group) to $\mathbb{Z}_k \times S_{n-k}.$


问:
Why do we study prime ideals?
I hope this isn't an inappropriate question here!

I'd like to ask the following (perhaps slightly ill-posed) question: why do we study prime ideals in general (commutative or non-commutative) rings? The situation is somewhat clear to me in some nice cases such as Dedekind rings, but even in slightly more general settings (e.g. general commutative Noetherian rings) the theory becomes rather horrible. What benefit do we gain from an understanding of the prime ideals of a ring?

Thank you!

答:
My answer cannot - and doesn't want - to keep pace with the other learned answers already given, but I'd like to make the point, that prime ideals are dual to ultrafilters, and that the ultrafilters of Boolean algebras (= Boolean rings) correspond to complete theories, which makes them interesting in their own right.

How this could make them interesting in the case of general algebras/rings: I have no idea.

Title: Why do we study prime ideals?