问:
Problem with forces

A motor boat with the power to steer across a river at $30$ kmph is moving such that the bow is pointed in a northerly direction. The stream is moving eastward at $6$ kmph. The river is $1$ km wide. Where on the opposite side will the boat meet the land?

I wasted enough time on this problem already without getting any answer. Can I know how its
solution?

All I know is that the position vector of the boat, $t$ hours later, would be: $\vec{r}(t)= (6t)\hat{i}+(30t)\hat{j}$, but I'm not sure where to go from there. In other words, how to take advantage of the info given that the width of the river is $1$ km.

P.S. The book's answer just says: $200$m east of the initial point (and I have no clue what do they mean by east of initial point?)

Thanks in advance.

答:
There is one massive problem with the question. It doesn't tell us where the opposite back is? Is it 1 km to the North of 1 km to the East?

Velocity has the nice property that it is additive. The velocity of the boat is 30${\bf j}$ km/h. The velocity of the stream is 6${\bf i}$ km/h. Hence the resulting velocity is (6${\bf i}$ + 30${\bf j}$) km/h. Since the velocity is constant, the displacement is (6${\bf i}t$ + 30${\bf j}t$) km.

If it is 1 km to the North we need to solve $6t = 1$, i.e. $t=1/6$ and the final position is (${\bf i}$ + 5${\bf j}$) km. If it is 1 km to the East then we need to solve $30t = 1$, i.e. $t = 1/30$ and position is (0.2${\bf i}$+0.5${\bf j}$) km.

If the bank is to the South or to the West, the boat will never make it. If the bearing is strictly between $0^{\circ}$ and $180^{\circ}$ then other answer are possible.

Title: Problem with forces


问:
Finding relations for finite matrix groups in GAP
Given two invertible $n \times n$ integer matrices A and B I want to look at the group $G$ generated by $A$ and $B$ inside $Gl_n(\mathbb{Z})$ with GAP. My goal is the see relations for A and B with the help of GAP in concrete examples.

Here is a concerete example:
$A=[ [ 0, 0, 0, -1 ], [ 0, 0, 1, 1 ], [ 0, 1, 0, 1 ], [ -1, -1, -1, -1 ] ]$
$B=[ [ 0, 0, 0, 1 ], [ 0, 0, 1, 0 ], [ 0, 1, 0, 0 ], [ 1, 0, 0, 0 ] ]$ (I wrote the matrices in GAP notation).

Then I can obtain the group $G$ in GAP via: G:=Group(A,B);
I can even find the isomorphism class of $G$ in GAP via the command  StructureDescription(G); to see that in this example $G$ is isomorphic to $D_{12}$.

Question: How can I find in GAP a description for a given group $G$ of the minimal relations that describe $G$ for the generators $A$ and $B$?

There are relations such as $(AB)^2=1$ but I can not see them in GAP since GAP displays A and B as matrices, but I want to see the relations "abstractly" for example as $(AB)^2=1$ and not written out as matrices where it is hard/impossible to see what the relations between A and B are.

In the example some relations would be $A^6=1$,$B^2=1$ and $(AB)^2=1$ (found by hand) but there can be more and I want to see them all in GAP in this form or at least some minimal relations.

答:
The following GAP computation works, although it has just come up with the same relations as the ones that you had already calculated. To be honest, I am
not completely sure what is that you are trying to compute. What do you mean by a "minimal relation" for example?</p>
```<code>gap> A:=[[0,0,0,-1], [0,0,1,1], [0,1,0,1],[-1,-1,-1,-1]];
  [ [ 0, 0, 0, -1 ], [ 0, 0, 1, 1 ], [ 0, 1, 0, 1 ], [ -1, -1, -1, -1 ] ]
gap> B:=[[0,0,0,1],[0,0,1,0],[0,1,0,0], [1,0,0,0]];
  [ [ 0, 0, 0, 1 ], [ 0, 0, 1, 0 ], [ 0, 1, 0, 0 ], [ 1, 0, 0, 0 ] ]
gap> G:=Group(A,B);
  <matrix group with 2 generators>
gap> Size(G);
  12
gap> isfp := IsomorphismFpGroup(G);
  CompositionMapping( [ (2,4)(3,6), (1,2,3,5,6,4), (1,3,6)(2,5,4) ] ->
  [ F1, F2, F3 ], CompositionMapping(
  [ (1,2)(3,4)(5,6,7,9,10,8), (3,4)(6,8)(7,10) ] ->
  [ (1,2,3,5,6,4), (2,4)(3,6) ], <action isomorphism> ) )
gap> F := Image(isfp);
  <fp group on the generators [ F1, F2, F3 ]>
gap> FF := Subgroup(F, [Image(isfp,A), Image(isfp,B)]);
  Group([ F2, F1 ])
gap> pres := PresentationSubgroupMtc(F, FF);
  <presentation with 2 gens and 3 rels of total length 12>
gap> TzPrintRelators(pres);
  #I  1. _x2^2
  #I  2. (_x2*_x1^-1)^2
  #I  3. _x1^6
</code>```

ContentLicense: CC BY-SA 4.0

Body:
The following GAP computation works, although it has just come up with the same relations as the ones that you had already calculated. To be honest, I am
not completely sure what is that you are trying to compute. What do you mean by a "minimal relation" for example?

gap> A:=[[0,0,0,-1], [0,0,1,1], [0,1,0,1],[-1,-1,-1,-1]];
  [ [ 0, 0, 0, -1 ], [ 0, 0, 1, 1 ], [ 0, 1, 0, 1 ], [ -1, -1, -1, -1 ] ]
gap> B:=[[0,0,0,1],[0,0,1,0],[0,1,0,0], [1,0,0,0]];
  [ [ 0, 0, 0, 1 ], [ 0, 0, 1, 0 ], [ 0, 1, 0, 0 ], [ 1, 0, 0, 0 ] ]
gap> G:=Group(A,B);
  <matrix group with 2 generators>
gap> Size(G);
  12
gap> isfp := IsomorphismFpGroup(G);
  CompositionMapping( [ (2,4)(3,6), (1,2,3,5,6,4), (1,3,6)(2,5,4) ] ->
  [ F1, F2, F3 ], CompositionMapping(
  [ (1,2)(3,4)(5,6,7,9,10,8), (3,4)(6,8)(7,10) ] ->
  [ (1,2,3,5,6,4), (2,4)(3,6) ], <action isomorphism> ) )
gap> F := Image(isfp);
  <fp group on the generators [ F1, F2, F3 ]>
gap> FF := Subgroup(F, [Image(isfp,A), Image(isfp,B)]);
  Group([ F2, F1 ])
gap> pres := PresentationSubgroupMtc(F, FF);
  <presentation with 2 gens and 3 rels of total length 12>
gap> TzPrintRelators(pres);
  #I  1. _x2^2
  #I  2. (_x2*_x1^-1)^2
  #I  3. _x1^6


问:
Are these 3 Matrices definable as products
I have 3 Matrices:

$$ A)
\left(
\begin{matrix}
    2 & 1 & -1 \\
    0 & -1 & 1 \\
\end{matrix}
\right)
$$

$$ B)
\left(
\begin{matrix}
    0 & 1  \\
    -1 & 0 \\
    1 & 1 \\
\end{matrix}
\right)
$$

$$ C)
\left(
\begin{matrix}
    0 & 1 & 0 \\
    1 & 0 & 2 \\
    -1 & 1 & 1 \\
\end{matrix}
\right)
$$

And I am trying to figure out if these matrices are definable as a product.

I believe there are some requirements that are set for the dimensions that determine wether I can define these matrices as products.

From what I've come to assume so far, I can create a product of AB by multiplying the top row of A with the first column of B.

This leaves me with:

$$ AB)
\left(
\begin{matrix}
    2 * 0 + 1 * (-1) + (-1) * 1 & 2 * 1 + 1 * 0 + (-1) * 1 \\
    0 * 0 + (-1) * (-1) + 1 * 1 & 0 * 1 + (-1) * 0 + 1 * 1 \\
\end{matrix}
\right)
$$

And if I am performing these steps correctly the product of AB would be:

$$
\left(
\begin{matrix}
    -2 & 1 \\
    2 & 1 \\
\end{matrix}
\right)
$$

Is this correct so far?

Also, are there any of these matrices that I cannot define as a product?

I'm wondering if I can create products from BA , AC , CA , BC & CB in the same way I did here?

答:
You're asking which matrix products are defined. It all comes down to the dimensions of the matrices. If $X$ is a $x_1×x_2$ matrix and $Y$ is a $y_1×y_2$ matrix, $XY$ is only defined if $x_2=y_1$, and the result is a $x_1×y_2$ matrix.

Thus the products $AB,BA,AC,CB$ are defined; $CA,BC$ are not.


问:
Evaluating limits of floor functions
I have 2 limits to calculate that I'm not sure if I've "proven" correctly.

(a) $\lim_{x\to 0}$ $x\lfloor{x}\rfloor$.

My guess is that since this has to be continuous, (however not sure if I have to prove this), I can just plug in 0 and I get

$\lim_{x\to 0}$ $x\lfloor{x}\rfloor$ $=$ $0\lfloor{0}\rfloor$ $=$ $0$

How can this be shown without just trying $0$?

(b) $\lim_{x\to \infty}$ $\frac{\lfloor{x}\rfloor}{x}$

My guess is that since $\lfloor{x}\rfloor$ is defined as the biggest number $\leq$ $x$, the limit has to be $1$. But I'm not quite sure what to prove here since my answer is merely a guess.

How do approach this? Is there a simple  way to calculate these?

答:
For $h\to0^+,\lfloor h\rfloor=0$ so the RHL is $0$. For $h\to0^-,\lfloor h\rfloor=-1$ so LHL$=\lim_{h\to0^-}-h=0$. So the limit is $0$. You can't say that $f(x)$ is continuous at $0$ and just substitute $0$ to get the limit since that is what you are supposed to show here.

Note that $\lfloor x\rfloor=x-\{x\}$ where $\{x\}\in[0,1)$ is the fractional part of $x$. The limit becomes$$\lim_{x\to\infty}\frac{x-\{x\}}x=1-0=1.$$

Title: Evaluating limits of floor functions


问:
Is a euclidic relation not asymmetric?
The relation ${\sim}\subseteq A^2$ is called euclidic if for all $x,y\in A$ it follows that:
$$x\sim y ~~\wedge ~~ x\sim z \implies y\sim z$$
The relation ${\sim}\subseteq A^2$ is called asymmetric if for all $x,y \in A$ it follows that:
$$x\sim y \implies y\not \sim x$$

The task asks to check if: $\sim $ euclidic $\implies \sim$ not asymmetric.

Attempt:
Suppose $\sim$ is both euclidic and asymmetric, for some $x,y\in A$ such that $x\sim y$ it follows that $x\sim y$ which by euclidic property implies that $y\sim y$, but then from asymmetric property it follows that $y\not \sim y$. So we arrive at $y\sim y$ and $y\not \sim y$, which is a contradiction.

However, if we take that $\sim$ is empty relation, then it is both euclidic and asymmetric, since both premises are trivially achieved. Now I wonder if empty relations count as relations.

答: The empty relation is indeed a relation,  Your argument works assuming there exist $x,y\in A$ such that $x\sim y$.  But, it does not apply to the empty relation, since in that case there do not exist any such $x$ and $y$.  So indeed, the implication you are investigating is not always true: the empty relation is a counterexample.

Title: Is a euclidic relation not asymmetric?


问:
How to show the following limit
Let $f:\mathbb{N^*} \to \mathbb{N^*}$ an injective function. Show that $\displaystyle \sum _{n=0}^{+\infty}\frac{f(n)}{n^2}=+\infty$. I tried to :

Show that $f(n) \ge n$ for all $n$, but i think i've found that's false. ($f(1)=2$ $f(2)=1$ for example)

Let $(S_n)_n$ a sequence such that $S_n= \frac{f(n)}{n^2}$, then using some criterion like Alembert criterion's, yet it didn't work.

Any help, please !

答:
For all $n \geq 1$, let $$S_n = \sum_{k=1}^n \frac{f(k)}{k^2}$$

You have that
$$S_{2n}-S_n = \sum_{k=n+1}^{2n} \frac{f(k)}{k^2} \geq \frac{1}{(2n)^2} \sum_{k=n+1}^{2n} f(k) \geq \frac{1}{4n^2}(1+2+...+n)=\frac{n(n+1)}{8n^2}\geq \frac{1}{8}$$

(justification of $\sum_{k=n+1}^{2n} f(k) \geq 1+2+...+n$ : the sum is made by $n$ distinct integers - because $f$ is injective -, so it is greater or equal to the smallest sum of $n$ distinct integers you can have, which is $1+2+...+n$.)

So $S_{2n}-S_n$ does not tend to $0$, so $(S_n)$ does not converge.

Title: How to show the following limit


问:
Playing a game of SCORE!
There are 4 people playing a game of SCORE!. In the game of SCORE!, every player rolls 2 dice, and adds their sum. The players win if the sum of all 8 dice is more than 17. What is the probability the players win?

I was thinking of using casework, but I felt that it was perhaps too bashy. Is there a better method?

答: For an exact answer, I would make a spreadsheet.  Start with some blank lines, then put the numbers $1$ through $48$ in column A.  The table will have the number of ways to make each sum from a number of dice.  The number of dice will increase from $1$ in column B to $8$ in column I.  Put $1$ in the rows from $1$ to $6$ in column B as there is $1$ way to get each sum from one die.  To the right, each cell is the sum of the cells one column to the left and one to six rows up, corresponding to the fact that you can start with each of those sums and have the last die make the number on the row.  Copy down and copy right make ths easy.  Finally add all the numbers in the rows $18$ through $48$ and divide by $6^8$

Title: Playing a game of SCORE!


答:
(a) The left-hand limit is that of $-x$, since $\lfloor x\rfloor=-1$ in $[-1,0)$. The right-hand limit is that of $0$, since $\lfloor x\rfloor=0$ in $[0,1)$. The two one-sided limits agree at $0$, so the ordinary limit is also $0$.

(b) Rewrite this as $\frac{x-\{x\}}x=1-\frac{\{x\}}x$, where $\{x\}$ is the fractional part of $x$ (always in $[0,1)$). $\frac{\{x\}}x$ can be squeezed to $0$, so the limit is $1$.

Title: Evaluating limits of floor functions

Body:
I have 2 limits to calculate that I'm not sure if I've "proven" correctly.

(a) $\lim_{x\to 0}$ $x\lfloor{x}\rfloor$.

My guess is that since this has to be continuous, (however not sure if I have to prove this), I can just plug in 0 and I get

$\lim_{x\to 0}$ $x\lfloor{x}\rfloor$ $=$ $0\lfloor{0}\rfloor$ $=$ $0$

How can this be shown without just trying $0$?

(b) $\lim_{x\to \infty}$ $\frac{\lfloor{x}\rfloor}{x}$

My guess is that since $\lfloor{x}\rfloor$ is defined as the biggest number $\leq$ $x$, the limit has to be $1$. But I'm not quite sure what to prove here since my answer is merely a guess.

How do approach this? Is there a simple  way to calculate these?


问:
Linear operator is trace non-increasing if dual operator is sub-unital
I am dealing with the following problem and hope some of you can help me.

Consider two finite dimensional Hilbert spaces $A$ and $B$ and a linear map $L \in \mathcal{L}(A) \rightarrow \mathcal{L}(B)$. The dual map $L^*$ is defined via the Hilbert-Schmidt inner product, namely $\forall M \in \mathcal{L}(A) ~\forall N \in\mathcal{L}(B): ~Tr[N^* L(M)] = Tr[(L^*(N))^* M]$.
Furthermore, we define

$A \leq B ~:\Leftrightarrow B-A$ is positive semi-definite

I want to show the following statement.

(i) $\forall M \in \mathcal{L}(A): ~ Tr[L(M)] \leq Tr[M]$ $\Leftrightarrow$ $L^*(I^B) \leq I^A$

So, I have to show two directions. I cannot relate the positivity to a condition under the trace (and following the definition of the dual, I only have statements under the trace).

e.g. for (i) $"\Rightarrow"$ I tried the following:

I started with setting $N=I^B$ and obtained from the definition of the dual

$Tr[I^B L(M)] = Tr[(L^*(I^B))^*M]$

$\Leftrightarrow Tr[M] \geq Tr[L(M)] = Tr[(L^*(I^B))^*M]$

This has to hold for arbitrary $M \in \mathcal{L}(A)$, so

$\Leftrightarrow \forall M \in \mathcal{L}(A): ~ Tr[(I^A - L^*(I^B))M] \geq 0 $.

Let $\{ v_1,...,v_n\}$ be the set of eigenvectors of $(I^A - L^*(I^B))$, where $n = dim(A)$.
Now I can choose $M$ to be the identity matrix and take the trace with respect to the eigenbasis  $\{ v_1,...,v_n\}$, then I find that all eigenvalues of $(I^A - L^*(I^B))$ are positive, hence we have shown this direction.

Unfortunately, I do not know anything about the geometric multiplicity of the eigenvalues of $(I^A - L^*(I^B))$, hence we do not even know whether we find an eigenbasis. So, in general, my argument seems to be wrong.

Does anyone know how to prove this statement?
I am grateful for any help/hint!

答:
Your statement (i) as it stands is wrong. To see this consider the map
$$
L:\mathbb C^{2\times 2}\to \mathbb C^{2\times 2}\quad \begin{pmatrix}a&b\\c&d\end{pmatrix}\mapsto \begin{pmatrix}1&0\\0&0\end{pmatrix}^*\begin{pmatrix}a&b\\c&d\end{pmatrix}\begin{pmatrix}1&0\\0&0\end{pmatrix}=\begin{pmatrix}a&0\\0&0\end{pmatrix}
$$
which is obviously

linear

completely positive (due to Kraus form)

self-dual, that is, $L=L^*$ (if $L(A)=K^*AK$ then $L^*(B)=KBK^*$, but here the Kraus operator is self-adjoint: $K=K^*$).

In particular, $L^*$ is sub-unital because $L$ is (self-duality). However---despite $L^*$ being sub-unital---$L$ violates your definition of "trace non-increasing":
$$
\operatorname{tr}\Big(L \begin{pmatrix}-1&0\\0&-1\end{pmatrix} \Big)=\operatorname{tr}\begin{pmatrix}-1&0\\0&0\end{pmatrix}=-1\ {\color{red}>}-2=\operatorname{tr}\begin{pmatrix}-1&0\\0&-1\end{pmatrix}\,.
$$

What you really want to show is the following statement:
$$
\forall_{M\in\mathcal L(A){\bf, M\geq 0}}\ \operatorname{tr}(L(M))\leq\operatorname{tr}(M)\ \Leftrightarrow\ L^*(I^B)\leq I^A\,,\tag*{(1)}
$$
that is, one only checks the trace condition for $M$ positive semi-definite. Then the proof of (1) becomes rather simple:
\begin{align*}
\forall_{M\geq 0}\ \operatorname{tr}(L(M))\leq\operatorname{tr}(M)&\ \Leftrightarrow\ \forall_{M\geq 0}\ \operatorname{tr}(ML^*(I^B))\leq\operatorname{tr}(M)\\
&\ \Leftrightarrow\  \forall_{M\geq 0}\ 0\leq \operatorname{tr}\big(M\big(I^A-L^*(I^B)\big)\big)\\
&\ \Leftrightarrow\ 0\leq I^A-L^*(I^B)\ \Leftrightarrow\ L^*(I^B)\leq I^A
\end{align*}
The only non-trivial (but not too difficult, either) step here is to show that a matrix $A$ is positive semi-definite if and only if $\operatorname{tr}(AB)\geq 0$ for all $B\geq 0$.


问:
Use the compound interest formula to calculate the interest rate for an account that was compounded semi-annually and ...
Use the compound interest formula to calculate the interest rate for an account that was compounded semi-annually, had an initial deposit of \$9,000 and was worth \$13,373.53 after 10 years.

Formula: $A(t)=P(1+\frac{r}{n})^{nt}$

The solution provided in my book is 4% whereas I get an absurd answer -94%. I'd like to understand where I went wrong. My working:

$n=2$
$P=9000$
$A(t)=13373.53$
$t=10$

Plugging the above into the formula with granular steps to isolate where I went wrong:
$$13373.53=9000(1+\frac{r}{2})^{2*10}$$
$$13373.53=9000(1+\frac{r}{2})^{20}$$
$$\frac{13373.53}{9000}=(1+\frac{r}{2})^{20}$$
$$1.48595=(1+\frac{r}{2})^{20}$$
$$2(1.48595)=2(1+\frac{r}{2})^{20}$$
$$2.972=(2+r)^{20}$$
$$2.972^{\frac{1}{20}}=2+r$$
$$r=2.972^{\frac{1}{20}}-2$$
$$r=-0.944$$

Where did I go wrong and how can I arrive at 4%?

答:
The wrong step is:

$$2(1.48595) = 2(1+\frac r2)^{20}$$
$$2.972 = (2+r)^{20}$$

Since $$(2+r)^{20} = (2(1+\frac r2))^{20} = 2^{20}(1+\frac r2)^{20} \ne 2(1+\frac r2)^{20}$$

Multiplying by $2$ in the previous step is unnecessary; we can take the root right then and there.


问:
Let $G$ be a simple planar graph in which every vertex has the same degree $k$. Prove that $k \leq 5$
Let $G$ be a simple planar graph in which every vertex has the same degree $k$. Prove that $k \leq 5$. This is a problem I was given by my professor, but I am struggling to see why it would be true, let alone prove it. It seems to me that any simple graph regular on degree $5$ would have a subgraph of $K_5$, which is not planar. Therefore, the graph itself could not be planar. Any direction on how to disprove my supposed counterexample and start this proof would be extremely helpful!

答:
Due to Euler's formula we know that edges <= 3n-6 in a planar graph. In your graph which is a regular graph the number of edges = kn/2.
Now compare 3n-6 to kn/2. k < 6 - 12/n which tends to 6 as n tends to infinity. So k < 6.</p>


答:
The problem is just after $$1.48595=\left(1+\frac{r}{2}\right)^{20}$$ You have now
$$\sqrt[20]{1.48595}=1+\frac{r}{2}\implies r=0.04$$

Title: Use the compound interest formula to calculate the interest rate for an account that was compounded semi-annually and ...

Body:
Use the compound interest formula to calculate the interest rate for an account that was compounded semi-annually, had an initial deposit of \$9,000 and was worth \$13,373.53 after 10 years.

Formula: $A(t)=P(1+\frac{r}{n})^{nt}$

The solution provided in my book is 4% whereas I get an absurd answer -94%. I'd like to understand where I went wrong. My working:

$n=2$
$P=9000$
$A(t)=13373.53$
$t=10$

Plugging the above into the formula with granular steps to isolate where I went wrong:
$$13373.53=9000(1+\frac{r}{2})^{2*10}$$
$$13373.53=9000(1+\frac{r}{2})^{20}$$
$$\frac{13373.53}{9000}=(1+\frac{r}{2})^{20}$$
$$1.48595=(1+\frac{r}{2})^{20}$$
$$2(1.48595)=2(1+\frac{r}{2})^{20}$$
$$2.972=(2+r)^{20}$$
$$2.972^{\frac{1}{20}}=2+r$$
$$r=2.972^{\frac{1}{20}}-2$$
$$r=-0.944$$

Where did I go wrong and how can I arrive at 4%?


问:
Hyperbola from Pell's equation
Could you please explain the proof of finding infinitely many solutions are existing for $a^2 - 10b^2 = 1$ or $4$ or $9$. Also, discuss, is there any relation between hyperbola and Pell's equation? If yes, how to find solution of $a$ and $b$ of $a^2 -10b^2 = 1$ by reducing into hyperbola form?

Thanks in advance.

答:
The main ingredient is to note that
$$(a^2-10b^2)^n = (a+\sqrt{10}b)(a-\sqrt{10}b)^n = \left(A+\sqrt{10}B \right)\left(A-\sqrt{10}B \right) = A^2-10B^2$$
Once you have this, you can generate infinite solutions.

EDIT

Let us see how to get solutions for $a^2-10b^2 = 1 \tag{$\star$}.$ From guessing, we have $19^2-10 \cdot 6^2 = 1$, i.e., $(a,b) = (19,6)$. Square $(\star)$ to get,
$$1^2 = (19 + 6\sqrt{10})^2(19 - 6\sqrt{10})^2 = \left( 721 + 228\sqrt{10}\right)\left( 721 - 228\sqrt{10}\right) = 721^2 - 10 \cdot 228^2$$
This gives us another solution as $(721,228)$. Keep proceeding like, i.e., instead of squaring, cube it, raise it to the fourth power and so on to get infinite solutions.

To get solutions for $a^2-10b^2 = 4 \tag{$\perp$},$ note that $a$ and $b$ have to be both even, i.e., let $a = 2a_1$ and $b=2b_1$ and find the infinite solutions to $a_1^2 - 10 b_1^2 = 1$ from above. Similarly, for $a^2-10b^2 = 9 \tag{$\dagger$},$ set $a = 3a_2$ and $b=3b_2$ and find infinite solutions to $a_2^2 - 10b_2^2 = 1$.

ContentLicense: CC BY-SA 3.0

Title: Hyperbola from Pell's equation


问:
How to prove this fact about the argmax of a function given properties of partial derivatives
I need to prove that assuming differentiability etc that the following holds. If we have a function $f(x,y)$ and we know that $\frac{\partial ^ 2 f}{\partial x \partial y} < 0$, then this implies that $g(y) = \arg \max_x f(x,y)$ is a decreasing function of $y$. i.e. as $y$ increases, $g(y)$ decreases.

My intuition is that this is because $\frac{\partial ^ 2 f}{\partial x \partial y} = \frac{\partial}{ \partial y} \left( \frac{\partial f}{\partial x} \right) < 0$ and thus the function $h(x,y) = \frac{\partial f}{\partial x}(x,y)$ is decreasing with respect to y. I am not sure this makes a conclusive argument, however.

Any help will be appriecated!

答:
I assume that $f$ is defined such that for all $y$, a unique $\arg \max_x f(x,y)$ exists, and that the domain over which $f$ is defined is convex.

Here is an argument you could use.  Consider any $y_0 < y_1$. Let $x_0 = g(y_0), x_1 = g(y_1)$. Suppose for the purpose of contradiction that $x_0 \leq x_1$.

By the definition of $g$, we see that $f(x_0,y_0) > f(x_1,y_0)$, and $f(x_1,y_1) > f(x_0,y_1)$. Now, define $h(y) = f(x_1,y) - f(x_0,y)$. We see that
$$
\begin{align}
h(y_0) &= f(x_1,y_0) - f(x_0,y_0) < f(x_1,y_1) - f(x_0,y_0)
\\ & < f(x_1,y_1) - f(x_0,y_1) = h(y_1).
\end{align}
$$
That is, $h$ increases from $y_0$ to $y_1$. It follows from the mean value theorem that there must exist a $y_*$ with $y_0 < y_* < y_1$ for which $\frac{dh}{dy}(y_*) > 0$, which is to say that
$$
f_y(x_1,y_*) > f_y(x_0,y_*).
$$
Now, by the mean value theorem, there must exist an $x_*$ with $x_0 < x_* < x_1$ for which
$$
f_{xy}(x_*,y_*) = \frac{\partial }{\partial x} f_y|_{(x,y) = (x_*,y_*)} = \frac{f_y(x_1,y_*) - f_y(x_0,y_*)}{x_1 - x_0} > 0,
$$
which contradicts our premise that $f_{xy} < 0$ over the domain of $f$.


问:
Two methods of Linearization of Second Order ODE are not matched
Take a look at this nonlinear differential equation and linearizing it about $(x,\dot{x})=(0.5,1)$
$$
\ddot{x} + 2x^2\dot{x} + 3\dot{x}^2 + x = 0
$$
In the book I'm reading, the author solved it as letting $y=\dot{x}$ and $z=2x^2\dot{x}+3\dot{x}^2$
$$
\ddot{x} + 2x^2y + 3y^2 + x = 0
$$
And he computed
$$
\frac{dz}{dx}\Big\vert_{0.5,1} = 4xy = 2, \qquad \frac{dz}{dy}\Big\vert_{0.5,1} = 2x^2+6y = 6.5 \\
(z-z_0) = 2(x-x_0) + 6.5(y-y_0) \\
z = 2x + 6.5y + 4
$$
Finally, he got the linearized ODE as
$$
\ddot{x} + 6.5\dot{x} + 3x - 4 = 0
$$

Now, I'm trying to reach same result using Jacobain matrix. If I let $x_1=x, x_2=\dot{x}$, I get
$$
\begin{align}
\dot{x}_1 &= x_2 \\
\dot{x}_2 &= -2x_1^2x_2 - 3x_2^2 - x_1
\end{align}
$$
The jacobain matrix about (0.5, 1) is
$$
J=
\begin{bmatrix}
0&1\\
-3&-6.5
\end{bmatrix}
$$
I got this linearized ODE as
$$
\ddot{x} +6.5\dot{x} + 3x = 0
$$
It is missing the constant -4. What is the problem?

答:
The linearization of $F(u)$ is
$$
F(u_0)+J_F(u_0)(u-u_0)=J_F(u_0)u+F(u_0)-J_F(u_0)u_0.
$$
In your second method you are missing the last two terms. ($u=[x,\dot x]$)


答:
Introduce $S_n=\sum_{k=1}^n \frac{f(k)}{k^2}$. Notice that the first $n$ values of $f(n)$ must be different because of injection. The minimum value of $S_n$ is obtained when the first $n$ values of $f(n)$ are $1,2,3\dots n$.

First show that for any $k,l$ such that $k<l\le n$:

$$\frac{k}{k^2}+\frac{l}{l^2}<\frac{l}{k^2}+\frac{k}{l^2}$$

This is obvious, because you can easily show that it is equivalent to:

$$kl^2+lk^2<l^3+k^3$$

$$0<l^2(l-k)-k^2(l-k)$$

$$0<(l^2-k^2)(l-k)$$

...which is true for $l>k$. It means that the minimum value of $S_n$ is achieved when $f(k)=k$ for $k=1,2,\dots,n$

It means that:

$$S_n\ge S_{n_{min}}=\sum_{k=1}^n \frac{k}{k^2}=\sum_{k=1}^n \frac{1}{k}=H_n$$

...where $H_n$ is the $n$-th harmonic number.

It means that:

$$\lim_{n\to \infty}S_n\ge\lim_{n\to \infty}H_n=\infty$$

$$\sum_{n=1}^{\infty}\frac{f(n)}{n^2}=\infty$$

Title: How to show the following limit

Body:
Let $f:\mathbb{N^*} \to \mathbb{N^*}$ an injective function. Show that $\displaystyle \sum _{n=0}^{+\infty}\frac{f(n)}{n^2}=+\infty$. I tried to :

Show that $f(n) \ge n$ for all $n$, but i think i've found that's false. ($f(1)=2$ $f(2)=1$ for example)

Let $(S_n)_n$ a sequence such that $S_n= \frac{f(n)}{n^2}$, then using some criterion like Alembert criterion's, yet it didn't work.

Any help, please !


问:
The role of the topology in a topological space.
Consider the topological space $(X, T)$ where $X$ is a set and $T$ is a topology which is a family of open subsets of $X$. I learned that the continuity of a function is determined by the topology. Say, the continuity of a function from $(X,T_{1})$ to $(X,T_{2})$ is determined by the relation between $T_{1}$ and $T_{2}$ even though its domain and codomain remain the same. $T_{1}$ and $T_{2}$ are families of the open subsets of $X$ by definition.

My question: Some open sets in $T_{1}$ need not be contained in $T_{2}$ because a set can have different topologies. In that case, are such sets not considered as open sets in $(X,T_{2})$?

答: There is some confusion when you say that “$T$ is a topology which is a family of open subsets of $X$”. This is true, but the whole truth is that $T$ is the family of all open subsets of $X$. So, if $T_1\not\subset T_2$, and if $O\in T_1\setminus T_2$, then $O$ is not an open subset of $(X,T_2)$.


问:
Putnam and Beyond 2e Section 1.5 Example
Question, from Putnam and Beyond, 2E, Section 1.5, first example: There are $n$ markers, each with one side white and the other side black, aligned in a row with their white sides up. At each step, if possible, we choose a marker with the white side up (but not one of the outermost markers), remove it, and reverse the two neighboring markers. Prove that one can reach a configuration with only two markers left if and only if $n-1$ is not divisible by 3.

I claim this example has a counterexample, namely n=6.
My approach is a combinatoric approach: we can count all the valid approaches and show that we don't have any.

Start with our starting arrangement $$WWWWWW$$.
We cannot choose either of the $W$s in the middle because if we do, we would be unable to remove either the second, or the 5th tile. So we must start with tile 2 or tile 5.
$$BOBWWW $$.

We still may not remove tile 4 otherwise we will be unable to remove tile 5. So we remove tile 5. $$BOBBOB$$

And we complete, with 4 remaining tiles. $\square$

Please tell me if there is something wrong with my solution. If not then is this a glaring error or omission (say the condition $n>6$ or something)?

I think the only ambiguity I that potentially could be is if we didn't want to count the outermost markers. If that's the case, this is a poorly worded example.

答:
I think the removed markers leave no actual, empty spots behind. Or, equivalently, "reverse the two neighbouring markers" looks past these open spots to the next remaining marker. Otherwise, specifying that there are two of them seems a bit strange.

We have the following sequence of moves:
$$\begin{array}{llllll}
W&W&W&W&W&W\\
W&W&B&O&B&W\\
B&O&W&O&B&W\\
W&O&O&O&W&W\\
B&O&O&O&O&B\end{array}
$$
where $W$ is a marker with the whie side up, $B$ is a marker with the black side up, and $O$ is a removed marker.


问:
Noncyclic subgroups of multiplicative group of integers mod n
I want to find an $n$ such that $\mathbb{Z}/n\mathbb{Z}^{\times}$ has a noncyclic subgroup, and I'm struggling to think of an example of such a subgroup. How can I construct such a subgroup without thinking about the cyclic subgroups generated by the elements of $\mathbb{Z}/n\mathbb{Z}^{\times}$?

答:
If $(\Bbb Z /n \Bbb Z)^\times$ contains a non-cyclic subgroup then $(\Bbb Z /n \Bbb Z)^\times$ itself must be non-cyclic.

A non-cyclic group must have at least $4$ elements.

Since $(\Bbb Z /8 \Bbb Z)^\times = \{1,3,5,7\}$ contains $4$ elements, we can check if it is also non-cyclic:

$\; 1^2 \equiv 1 \pmod8$

$\; 3^2 \equiv 1 \pmod8$

$\; 5^2 \equiv 1 \pmod8$

$\; 7^2 \equiv 1 \pmod8$

So $(\Bbb Z /8 \Bbb Z)^\times$ is a non-cyclic group.


问:
Adjoint Property of a matrix
$\DeclareMathOperator{\adj}{adj}$

If $|\adj(A)| = |A|^{n - 1} ;\bigl| \adj\bigl(\adj(A)\bigr) \bigr| = {\left| A \right|^{{{\left( {n - 1} \right)}^2}}}$, then $\bigl| {\underbrace {\adj\dots\adj\bigl(\adj(A) \bigr)}_{t \text{ - times}}} \bigr| = {\left| A \right|^{{{\left( {n - 1} \right)}^t}}}$

How do we prove it?

Upto $\bigl|\adj\bigl(\adj(A)\bigr)\bigr| = | A |^{(n - 1)^2}$, it is mentioned in the book but not after that.

答: $\DeclareMathOperator{\adj}{adj}$ $$A(\adj A)=|A|I$$ Put $A=\adj\adj A$ $$(\adj\adj A)(\adj \adj\adj A)=|\adj\adj A|I$$ Now, $\adj\adj A=|A|^{n-2}A$. And, $|\adj\adj A|=|A|^{(n-1)^2}$. Putting this, we get, $$|A|^{n-2}A(\adj \adj\adj A)=|A|^{(n-1)^2}I$$ $$\implies \adj\adj\adj A=|A|^{n^2+1-2n-n+2}A^{-1}$$ $$=|A|^{n^2-3n+3}\frac{\adj A}{|A|}$$ $$=|A|^{n^2-3n+2}A$$ Using the fact that $|kA|=k^n|A|$, we get $|\adj\adj\adj A|=|A|^{(n-1)^3}$. You see the pattern now?

Title: Adjoint Property of a matrix


问:
Proving a block matrix is psd.
I have a special matrix $X \in \mathbb R^{2n \times 2n}$ whose all elements are non-negative.

The matrix is in a block structure: $X = \begin{pmatrix} X^1 & X^2 \\ X^3 & X^4  \end{pmatrix}$ where $X^1 = \mathrm{diag}(x_1, x_2, \ldots,x_n)$ for some $x \in \mathbb{R}^n_+$ and $X^4 = \mathrm{diag}(y_1, y_2,\ldots,y_n)$ for some $y \in \mathbb{R}^{n}_+$, e.g., the diagonal blocks are diagonal matrices. Moroever, $X^2$ is a matrix whose $i$-th row sums to $x_i$, and $j$-th column sums to $y_j$. Similarly, $i$-th row of $X^3$ sums to $y_i$ and $j$-th column of $X^3$ sums to $x_j$.

An example matrix for $n= 2$ would be: $\begin{pmatrix} 0.3061 & 0  & 0 &  0.3061 \\
   0 & 0.6939  &  0.0826  &  0.6112 \\
   0 & 0.0826  &  0.0826  &  0 \\
    0.3061 &   0.6112   & 0 &   0.9174 \end{pmatrix}.$

I would like to show that such an $X$ (in general, not this example above) is positive semi-definite. I don't know if it helps, but I have $\sum_i x_i = 1, \sum_j y_j = 1$.

答:
One method is to use Gershgorin Circle theorem.

Thm: For a symmetric $m\times m$ matrix $A$ with diagonal elements $A_{ii}$, $i = 1,...,m$ the eigenvalues of $A$ lie in the union of sets $[D_{ii}-R_i, D_{ii}+R_i]$ where $R_i = \sum_{i\neq j}|A_{ij}|$.

So, if you can guarantee that for all rows $i$, the quantity $R_i \leq D_{ii}$, then you can ensure that $$\text{eigenvalues}(A) \geq \min_{i=1,...,m} D_{ii}-R_i \geq 0$$ hence, the matrix is PSD!

Title: Proving a block matrix is psd.


问:
$dH_x(v) = dF_x(v)G(x)+F(x)dG_x(v)$
Let $F : M \to \mathbb{R}^{m\times k}$ (where $\mathbb{R}^{m\times k}$ are $m\times k$ matrices with real coefficients)  and $G : M \to \mathbb{R}^{k\times n}$.

Define $H : M \to \mathbb{R}^{m\times n}$ by $H(x)=F(x)G(x)$. I want to show that $dH_x(v) = dF_x(v)G(x)+F(x)dG_x(v)$ but I am not sure how to start.

答:
Take $\gamma:(-\epsilon,\epsilon)\rightarrow M$ such that $\gamma(0)=x$ and $\gamma'(0)=v$ then the chain rule said that
$$
dH_x(v)=(H\circ \gamma)'(0)
$$

but $H\circ \gamma$ it's now a curve on $\mathbb{R}^{m\times n}$ given by
$$
F(\gamma(t))G(\gamma(t))
$$.
if $F=(F_{ij})$ and $G=(G_{ij})$ then
$$
(H\circ \gamma(t))_{ij}=\sum_{k}F_{ik}(\gamma(t))G_{kj}(\gamma(t))
$$
then
$$
\frac{d}{dt}(H\circ \gamma(t))_{ij}=\sum_{k}\frac{d}{dt}(F_{ik}(\gamma(t)))G_{kj}(\gamma(t))+\sum_{k}F_{ik}(\gamma(t))\frac{d}{dt}(G_{kj}(\gamma(t)))
$$

by the chain rule on $\mathbb{R}$, thus
$$
\frac{d}{dt}|_{t=0}(H\circ \gamma(t))_{ij}=\sum_k d(F_{ik})_x(v)G_{kj}(x)+\sum_k F_{ik}(x)d(G_{kj})_x(v)
$$

thus
$$
dH_x(v)=dF_x(v)G(x)+F(x)dG_x(v)
$$

Title: $dH_x(v) = dF_x(v)G(x)+F(x)dG_x(v)$


答:
The answer to your question is yes, such sets need not be "considered as open in $T_2$. To put it more precisely, there are elements of $T_1$ which are not elements of $T_2$ (recall that elements of $T_1$ and $T_2$ are themselves subsets of $X$).

Consider $(\mathbb{R},T_1)$, $(\mathbb{R},T_2)$ and $(\mathbb{R},T_3)$ where $T_1$ is the usual Euclidean topology, $T_2$ is the discrete topology, (so every subset $S$ of $\mathbb{R}$ is in $T_2$), and $T_3$ is the coarse topology (the only elements of $T_3$ are $\mathbb{R}$ and $\emptyset$).

Now, for example, any function $f:(\mathbb{R},T_2) \rightarrow (\mathbb{R},T_1)$ is continuous. Here there are many elements of $T_2$ which are not contained in $T_1$.

The point here is that the topology is an entirely separate thing from the underlying set. I start with the set $\mathbb{R}$, and this does not come with a topology. It is just a set, in the same way that $\{a,b,c\}$ is just a set (despite the fact that $\mathbb{R}$ is a lot bigger). I must equip it with a topology. Once I've done that, I have a topological space. I can create a different topological space, with the same underlying set, by equipping $\mathbb{R}$ with a different topology (and that's what I've done above). Now that I have these two topological spaces, I can start talking about continuous maps from one of them to the other.

Title: The role of the topology in a topological space.

Body:
Consider the topological space $(X, T)$ where $X$ is a set and $T$ is a topology which is a family of open subsets of $X$. I learned that the continuity of a function is determined by the topology. Say, the continuity of a function from $(X,T_{1})$ to $(X,T_{2})$ is determined by the relation between $T_{1}$ and $T_{2}$ even though its domain and codomain remain the same. $T_{1}$ and $T_{2}$ are families of the open subsets of $X$ by definition.

My question: Some open sets in $T_{1}$ need not be contained in $T_{2}$ because a set can have different topologies. In that case, are such sets not considered as open sets in $(X,T_{2})$?


问:
If $X\sim \textbf{Laplace}(0,\lambda)$. Prove that $|X|\sim \textbf{Exp}(\lambda)$
How can I prove this statement?

If $X\sim \textbf{Laplace}(0,\lambda^{-1})$. Prove that $|X|\sim \textbf{Exp}(\lambda)$

My approach: Since that $X\sim \textbf{Laplace}(0,\lambda)$, so the density function is $$f_{X}(x)=\frac{1}{2\lambda^{-1}}e^{-\frac{|x-0|}{\lambda^{-1}}}, x\in \mathbb{R}, \lambda>0$$
Now, let $Y=|X|$, so we have that $Y$ is $\nearrow$ in $[0,\infty)$ and $y=x$, $x\in [0,\infty)$.

So, by transformation theorem and taking $h^{-1}(y)=y$, we have that $$f_{Y}(y)=\frac{1}{2\lambda^{-1}}e^{-\frac{|y|}{\lambda^{-1}}},\quad x\geq 0 $$
and similar, we have that $$f_{Y}(y)=\frac{1}{2\lambda^{-1}}e^{-\frac{|-y|}{\lambda^{-1}}},\quad x<0$$
So, we have that $$f_{|X|}(x)=\frac{1}{2\lambda^{-1}}e^{-\frac{|x|}{\lambda^{-1}}}, x\in \mathbb{R}, \lambda>0$$

But, I think that I can't conclude with that information that $|X|\sim \textbf{Exp}(\lambda)$.

答:
The error in your approach is that $h^{-1}$ is not well-defined: for every input $x$, there are two values $y$ (they are $\pm x$) such that $h(y) = x$. If I remember correctly, whether or not $|X|$ is increasing is not really relevant; instead, what is relevant is whether the transformation $h^{-1}$ is increasing (because this is a sufficient condition for invertibility).

Think of $f_{X} (x) dx = P(X \in [x - dx / 2, x + dx / 2])$ for some infinitesimal $dx$. Then the event $|X| \in [x - dx / 2, x + dx / 2]$ corresponds to $X \in [-x - dx / 2, -x + dx / 2]$ or $X \in [x - dx / 2, x + dx / 2]$, which we know have probabilities $f_X(-x)dx$ and $f_X(x)dx$, respectively. Thus,
$$f_{|X|} (x) = \begin{cases}
2 f_X(x) & x \geq 0 \\
0 & x < 0
\end{cases}$$
You should notice that this is exactly the density function of $\text{Expon}(\lambda)$.


问:
What is the easiest way to define a complex number in exponential form in maple?
What is the easiest way to define a complex number in exponential form in maple?
Is there a built-in function?

eg: $\underline{Z} = 600 \cdot e^{-j45^\circ}$

答:
Maple works in radians.

Maple is very happy to work with expressions of the form $\tt 5*exp(I*Pi)$.

For example:

$$\tt Re(5*exp(I*Pi));$$

$$\tt -5$$


问:
If a bijection of a plane to it self takes a rectangle to a rectangle then it takes a square to a square

Let $f$ be a bijection of a plane to it self which takes rectangle to a rectangle.

Prove it takes square to a square.

My idea was to  put whole thing in the coordinate system...?

This is a problem from international math olympiad longlist 1987 and it does not assumes that preseves area.

答:
Such an $f$ maps lines $\ell$ to lines and parallel lines $\ell$, $\ell'$ to parallel lines.

Proof. Given two parallel lines $\ell$, $\ell'$ at distance $d$ you can inscribe a sequence of adjacent rectangles $R_i$ into the infinite strip the two lines form. These $R_i$ will be mapped to adjacent rectangles (of maybe different lengths). The $f(R_i)$ therefore again form an infinite strip, bounded by  two parallel lines. These lines are nothing else but  $f(\ell)$, $f(\ell')$.

Such a map $f$ (even non-area-preserving) maps squares to squares.

Proof. Consider a square $Q$ and split it into four congruent quarters by drawing symmetry lines $\ell_1$, $\ell_2$. The image $R:=f(Q)$ will be a rectangle that is divided into four smaller rectangles by the images   $f(\ell_1)$, $f(\ell_2)$. These images are parallel to the edges of $R$. Now $\ell_1$, $\ell_2$ intersect the edges of $Q$ in four points which form the vertices of a square $Q'$. The images of these four points are lying on the edges of $R$. They can only be the vertices of a rectangle when $R$ is in fact a square.


问:
Expectation and Variance of the sequence $HHH$ in $n$ tosses

Toss a fair coin $n \ge 5$ times, and let $X$ be the number of occurrences
of the pattern $HHH$ in the sequence of tosses. For example, if
$n = 9$ and the outcome is $\omega = HHTHHHHTH$ then $X(\omega) = 2$.
Compute E[$X$] and Var ($X$).

My idea so far is to start with $n=5$, where there are three possible outcomes that give us at least three consecutive heads, namely $HHHTT,THHHT,TTHHH,HHHHT,THHHH,HHHHH$, in the first three outcomes we get $X(\omega)=1$, in the third and fourth outcomes we get $X(\omega)=2$, and for the final outcome we get $X(\omega)=3$.

When we move on to $n=6$ and assess all outcomes we get $4$ occurrences of $X(\omega)=1$, $3$ occurrences of $X(\omega)=2$, $2$ occurrences of $X(\omega)=3$, and $1$ occurrence of $X(\omega)=4$.

I don't know exactly what to do with this information or if I'm going in the right direction, but any help is appreciated.

答:
Indicator random variables are your friend here. There are $n-2$ windows of three tosses in the sequence, and each window has a $\frac18$ chance to come up $HHH$ (i.e. we consider each window as a random variable, and these add to make $X$). Therefore the expected value of $X$ is $\frac{n-2}8$.

For the variance of $X$ we sum up the following:

Variance of each window: $n-2$ times $\frac18×\frac78$, or $\frac7{64}(n-2)$

Covariance of windows one toss apart: twice $n-3$ times $\frac1{16}-\frac1{8^2}$, or $\frac6{64}(n-3)$

Covariance of windows two tosses apart: twice $n-4$ times
$\frac1{32}-\frac1{8^2}$, or $\frac2{64}(n-4)$

Thus the variance of $X$ is
$$\frac1{64}(7(n-2)+6(n-3)+2(n-4))=\frac{15n-40}{64}$$


问:
Which its the subgroup generated by -1 respect to +?
I have the question:

Which its the cyclic subgroup of $\mathbb{Z}$ generated by -1 respect to +? I know that its all $\mathbb{Z}$ but I don´t understad why?

Help, please.

答:
Hints:

The subgroup generated $-1$ is all integer multiples of $-1.$ Do you see why that's the case? If you cannot see it, please read the definition again.


问:
Sum of two squares in a $\Bbb Z/p\Bbb Z$
I need to show that every element in $\Bbb Z/p\Bbb Z$ can be written as a sum of two squares. The case $p=2$ is trivial and $0$ is always $0^2 + 0^2$. So all I have to do is show that every element of $(\Bbb Z/p\Bbb Z)^\times$ (the group of units) can be expressed as a sum of two squares. The question hints that I should consider the set of elements of $(\Bbb Z/p\Bbb Z)^\times$ expressible as a sum of two squares.

Associativity is trivial, the identity element $1 = 1^2 + 0^2$ exists.
Closure holds by the identity $(a^2 + b^2)(c^2 + d^2) = (ac + bd)^2 + (ad - bc)^2$.
I don't know how to show that inverses exist.

答: For finite groups, to show that a subset is a subgroup if you check closure and non-emptyness you don't need to check inverses because each element has finite order: $g^{-1} = g^{o(g) - 1}$ is in the subset by closure assumption.</p>


问:
convergence of a series using the root test
I want to find whether this series diverges or converges using the root test
$$ \sum_{n=1}^\infty \arccos^n \left(\frac{1}{n^2}\right)  $$
$$ \sqrt[n] {\arccos^n \left(\frac{1}{n^2}\right)}  $$
$$  \arccos \sqrt[n] {\left(\frac{1}{n^2}\right)}  $$
$$  \arccos  1 = 0  $$
and since its 0 the series should converge but I am not sure about the calculations there.

答:
The series diverges. Indeed
$$\arccos^n\left(\frac{1}{n^2}\right)\sim \left(\frac{\pi }{2}\right)^n+O\left(\frac{1}{n^2}\right);\;n\to\infty$$
The first $50$ terms sum is about $1.7\times 10^{10}$.


问:
What is the Fourier transformation of $y=exp(-λ|x|)$
What is the Fourier transformation $y=exp(-λ|x|)$

Using the identity: $y ̃(k)=∫_{-∞}^{∞}\exp(ikx)y(x)\,dx$

$y ̃(k)=∫_{-∞}^{∞}\exp(ikx) exp(-λ|x|)\,dx$

$y ̃(k)=∫_{0}^{∞}\exp(ix(k-λ))+∫_{-\infty}^{0}\exp(ix(k+λ))$

I do not know how to continue this as when i try to do it i can't define $exp(\infty)$, is there a identity that i can use

Body:
$f(x)=e^{-a|x|}$, then $$\sqrt{2\pi} F(k)=\int_{-\infty}^{\infty} e^{-a|x|} e^{-ikx} dx=\int_{-\infty}^{0} e^{ax} e^{-ikx} dx+ \int_{0}^{\infty} e^{-ax} e^{-ikx}$$
$$\implies \sqrt{2\pi} F(k)=\frac{1}{a-ik}-\frac{1}{-a-ik}=\frac{2a}{a^2+k^2}$$


问:
Diagonalisation of a linear operator
Given the following linear operator
$$T(x,y,z)=(2x+z,-3y+z,-3z)$$
with $T\colon\mathbb{R}^{3}\to\mathbb{R}^{3}$, I found that its matrix in the canonical basis is
$$[T]=\begin{bmatrix} 2 & 0 & 1 \\ 0 & -3 & 1 \\ 0 & 0 & -3 \end{bmatrix}$$
such that its eigenvalues and eigenvectors are:
$$\lambda_{1}=2,\vec{v_{1}}=(1,0,0),\quad\lambda_{2}=-3,\vec{v_{2}}=(0,1,0)$$
How do I conclude that $T$ is not diagonalisable?

答:
If $T$ is diagonalizable, then the sum of the dimension of the nullspace of $T - \lambda I$ over all eigenvalues $\lambda$ of $T$ must be equal to the dimension of the space, which is in this case $3$.

However, for this case, we find that $\dim \ker(T - 2 I) = \dim \ker (T + 3I) = 1$, which means that the sum $1 + 1 = 2$ is too small.

Title: Diagonalisation of a linear operator


问:
Does every character $\phi$ on a Banach algebra satisfy $\|\phi\|=1$?
A character on a not necessarily unital Banach algebra is a non-zero homomorphism $\phi\colon A\to\mathbb{C}$. If $r(a)$ denotes the spectral radius of some element $a\in A$, then one has $|\phi(a)|\leq r(a)\leq\|a\|$. This shows that $\phi$ is bounded with $\|\phi\|\leq1$. If $A$ is unital then one can easily verify that $\phi(1)=1$, thus $\|\phi\|=1$. But what if $A$ is not unital? Do we still have $\|\phi\|=1$? Or do there exist characters with $0<\|\phi\|<1$ in that case?

EDIT: I just noticed that I also implicitly assumed the existence of a unit for proving that $\|\phi\|\leq1$. But I think one can uniquely extend $\phi$ to a (unital) character $\hat{\phi}$ on the unitization $\widetilde{A}$ here if necessary, and use the prove above to conclude that $\|\phi\|\leq\|\hat{\phi}\|\leq1$.

答:
It can happen that the norm of a character is $< 1$ in a Banach algebra. Here is an explicit example:

Norm of a character in a non-unital Banach algebra without approximate identity

It turns out that the absence of an approximate identity is problematic here. Indeed, if $A$ admits an approximate unit $(u_i)_i$ (this is for example the case when $A$ is a $C^*$-algebra) and $\chi: A \to \mathbb{C}$ is a character on $A$, then choose $a \in A$ with $\chi(a) \neq 0$. Then
$$\lim_i\chi(u_i)= \lim_i \frac{\chi(u_ia)}{\chi(a)} = \frac{\chi(a)}{\chi(a)}=1$$ which shows that $\Vert \chi \Vert \ge 1$. The other inequality is always true, as you already observed, so when the Banach algebra admits an approximate unit everything works out nicely.


问:
$\varepsilon$-$\delta$ proof that $\lim_{x\to 0} \frac{{\cos}^2(x)-1}{x^4} = -\infty$
As the title says, I'm trying to find a suitable $\delta$ to complete the proof of $$\lim_{x\to 0} \frac{{\cos}^2(x)-1}{x^4} = -\infty.$$

Starting out with $$\forall N<0, \exists \delta >0 \text{ s.t. } 0<\lvert x-0\rvert \implies \frac{{\cos}^2(x)-1}{x^4}<N,$$ I thought setting $\delta \leq 1$ could be a good idea, leading to the restriction $-1<x<1$. Then I tried some basic algebraic manipulation like $$\frac{{\cos}^2(x)-1}{x^4}=-\frac{{\sin}^2(x)}{x^4} < - \frac{{\sin}^2(x)}{\delta^4},$$ but that lead me nowhere. So any help or hints would be appreciated.

Body:
Since
$$ \lim_{x\to0}\frac{\sin x}{x}=1 $$
one can choose $\delta_1>0$ such that when $|x|<\delta_1$
$$ \frac{\sin x}{x}\ge \frac12. $$
So when $|x|<\delta_1$, one has
$$\frac{{\cos}^2(x)-1}{x^4}=-\frac{{\sin}^2(x)}{x^4} =-\frac{{\sin}^2(x)}{x^2}\frac{1}{x^2}< - \frac{1}{2x^2}. $$
For any given $N>0$, letting $- \frac{1}{2x^2}<-N$ gives $|x|<\frac{1}{\sqrt{2N}}$. Define $\delta=\min\{\delta_1, \frac{1}{\sqrt{2N}}\}$. Thus when $0<|x|<\delta$,
$$\frac{{\cos}^2(x)-1}{x^4}<-N. $$
Namely
$$\lim_{x\to0}\frac{{\cos}^2(x)-1}{x^4}=-\infty. $$


问:
Applying the IVT on closed intervals
I'm not exactly sure how the intermediate value theorem can be applied to the following question especially with the defined function $g(x)$ which is provided below.

Using the intermediate value theorem with $g(x)=f(x)-x$ and suppose $f:[0,b] \to [0,b]$ is continuous. Prove there is at least one $c \in [0,b]$ such that $f(c)=c$.

Thanks

答:
If $f(0)=0$, then $g(0)=f(0)-0=0$, so $x=0$ is the root of $g(x)=0$,. Next if $f(b)=b$
then $g(b)=b-b=0$, so $x=b$ is a root of $g(x)=0$. In other cases
$g(0)=f(0)-0 >0, g(b)=f(b)-b <0$, so by IVT it follows that $g(x)=0$ has a root in (0,b)$.

Title: Applying the IVT on closed intervals


问:
Solution verification of $\lim_{n\to\infty}\left(1+\left(\sum_{k=1}^{n-1}\frac{k}{n}\sin\left(\frac{k\pi}{n}\right)\right)^{-1}\right)^{n}$
$$\lim_{n\to\infty}\left(1+\left(\sum_{k=1}^{n-1}\frac{k}{n}\sin\left(\frac{k\pi}{n}\right)\right)^{-1}\right)^{n}$$

Source: Romanian Mathematical Magazine Spring edition 2022, problem UP. 354.

I work out in following manner.

Clearly, $\displaystyle \frac{k\pi}{n}>0$  for all $k\in[ 1, n-1]$ so due to    Taylor series around $x=0$ we have $\sin x \leq x$ for all $ x>0$. Choose $\displaystyle x=\frac{k\pi}{n}$  we have then $$\sin\left(\frac{k\pi }{n}\right)\leq \frac{k\pi}{n}\Rightarrow S(n)= \sum_{k=1}^{n-1}\frac{k}{n}\sin\left(\frac{k\pi}{n}\right)\leq \pi\sum_{k=1}^{n-1}\frac{k^2}{n^2}=\frac{\pi}{6}\left[\frac{(n-1)(2n-1)}{n}\right]\cdots(1)$$ Since  the partial sum of the later sum diverges as $n\to \infty$ so $S(n)\to \infty$ (due to comparison test). Therefore,$$L=\lim_{n\to \infty} \left(1+\left(S(n)\right)^{-1}\right)^{n}=1^{\infty}$$ and hence we have $$L= \lim_{n\to\infty}\exp\left(\left(S(n)\right)^{-1}\right)=\exp\left(\lim_{n\to \infty}n\left(\sum_{k=1}^{n-1}\frac{k}{n}\sin\left(\frac{k\pi}{n}\right)\right)^{-1}\right)$$ since latter expression is Riemann integrable function on $[0,1]$ so the limit reduces to $$L= \exp\left(\int_0^1 x\sin(\pi x) dx\right)^{-1} \underbrace {=}_{IBP}e^{\left(\pi^{-1}\right)^{-1}}=e^{\pi}$$

The reason of $(1)$ in above is to show that for large value  of $n$ the original sum is divergent so that its reciprocal is $0$ giving us $1^{\infty}$ form.

Question:
Is my work and inequality $(1)$ supports the valid proof?

I manage  to find the limit of problem  in other way by showing that $$\sum_{k=1}^{n-1}\frac{k}{n}\sin\left(\frac{k\pi}{n}\right)=\frac{1}{2}\cot\left(\frac{\pi}{2n}\right)$$ giving us $e^{\pi}$.

答:
From $S(n) \leq \pi\sum_{k=1}^{n-1}\frac{k^2}{n^2}$ we can't conclude for divergenge, we can proceed as follows

$$\frac{k}{n}\sin\left(\frac{k\pi}{n}\right) =\frac{k}{n}\left(\frac{k\pi}{n}-\frac{\pi^3}{3!}\frac{k^3}{n^3}+\frac{\pi^5}{5!}\frac{k^5}{n^5}+\ldots\right)=\sum_{i=1}^\infty (-1)^{i-1}\frac{\pi^{2i-1}}{(2i-1)!}\left(\frac{k}{n}\right)^{2i}$$

and

$$\sum_{k=1}^{n-1}\frac{k}{n}\sin\left(\frac{k\pi}{n}\right) =\sum_{i=1}^\infty\left( (-1)^{i-1}\frac{\pi^{2i-1}}{(2i-1)!}\sum_{k=1}^{n-1}\left(\frac{k}{n}\right)^{2i}\right)=$$

$$=\sum_{i=1}^\infty\left( (-1)^{i-1}\frac{\pi^{2i-1}}{(2i-1)!}\left(\frac {n}{2i+1}-\frac12+O\left(\frac1n\right)\right)\right)=$$

$$=n\sum_{i=1}^\infty\left( (-1)^{i-1}\frac{\pi^{2i-1}}{(2i+1)(2i-1)!}\right)+O\left(\frac1n\right)=\frac n \pi+O\left(\frac1n\right)$$

indeed

$$\sum_{i=1}^\infty\left( (-1)^{i-1}\frac{x^{2i-1}}{(2i+1)(2i-1)!}\right)=\frac{\sin x -x\cos x}{x^2}$$

then

$$\lim_{n\to\infty}\left(1+\left(\sum_{k=1}^{n-1}\frac{k}{n}\sin\left(\frac{k\pi}{n}\right)\right)^{-1}\right)^{n}=\lim_{n\to\infty}\left(1+\frac {1}{\frac n \pi+O\left(\frac1n\right)}\right)^{n} =e^\pi$$


问:
Line passing through edges proof

Prove that there cannot be a line such that it passes through all the edges of a closed and odd-sided figure made of line segments. The line cannot contain any of the vertices of the figure as one of its points.

I first set the condition that the points so chosen from the edges must be collinear and tried a bunch of cases but could not actually prove it for any specific case. Also I have no idea as to the generalized proof.

Any help would be appreciated.

Thank you.

答: Every time the line you're drawing crosses an edge of the closed figure, it goes from outside to inside and vice versa. As there are an odd number of edges in the figure, the line drawn starts inside and ends outside and vice versa, which is impossible (the drawn line starts and ends outside). Thus the statement is proved.</p>

Title: Line passing through edges proof


问:
Counting the numbers between $1$ and $1,000,000$ whose digits sum to $30$

What's the number of numbers between $1$ and $1,000,000$ whose digits sum is $30$?

So I thought of this as a stars and sticks problem, so in the case you have $35\choose 5$ numbers whose sum is $30$. But obviously this number of numbers includes many overlaps (if for instance the requested sum was $8$, then $80=800=8000=134=143=431$ etc.). How do I get rid of the overlaps (inclusion-exclusion)?

Thanks in advance for any assistance!

答:
One looks for the number $N$ of integer solutions of
$$
\sum_{i=1}^6n_i=30,\qquad 0\leqslant n_i\leqslant9.
$$
Thus, $N$ is the coefficient of $x^{30}$ in the polynomial
$$
\prod_{i=1}^6\sum_{n_i=0}^9x^{n_i}=\left(\frac{1-x^{10}}{1-x}\right)^6.
$$
To compute $N$, note that
$$
(1-x^{10})^6=\color{red}{1}-\color{red}{6}x^{10}+\color{red}{15}x^{20}-\color{red}{20}x^{30}+o(30),
$$
and that
$$
\frac{1}{(1-x)^6}=\sum_{k=0}^{30}\color{green}{{5+k\choose5}}x^k+o(30),
$$
where $o(30)$ denotes terms of order at least $31$, hence
$$
N=\color{red}{1}\cdot\color{green}{{5+30\choose5}}-\color{red}{6}\cdot\color{green}{{5+20\choose5}}+\color{red}{15}\cdot\color{green}{{5+10\choose5}}-\color{red}{20}\cdot\color{green}{{5+0\choose5}}=50877.
$$</p>


问:
Question about prime ideals in commutative rings
A commutative ring $R$ can have the property that if $I\subset R$ is an ideal such that $\sqrt I=\mathfrak p$, with $\mathfrak p\subset R$ a prime ideal, then $I=\mathfrak p^n$. The notes that I'm using prove that $\mathbb Z$ has this property, and say that "this is equivalent to show that every zero-divisor in $\mathbb Z /\mathfrak p^n$, with $\mathfrak p\subset \mathbb Z$ a prime ideal, is nilpotent". Since in $\mathbb Z$ this is immediately clear, I suppose that the notes mean that if $R$ is a ring with the property above, then every zero-divisor in $R/\mathfrak p^n$ is nilpotent. Can you give me an outline of a proof of this fact? Thanks in advance

答:
What they are alluding to is the notion of a primary ideal.

An ideal $Q$ of a commutative ring $R$ is called primary if,

for all $a,b\in R$, if $ab\in Q$, then either $a\in Q$ or $b^n\in Q$ for some positive integer $n$.

This is easily equivalent to

Every nonzero zero divisor in $R/Q$ is nilpotent.

In some rings (but not all) the primary ideals coincide with powers of prime ideals.


答:
Let's compute eigenvalues of $T.$ They are $2$ and $-3.$

Notice that the geometric multiplicity of $-3$ is $1,$ but the algebraic multiplicity is $2$. This shows that $T$ is Not diagonalizable.

Title: Diagonalisation of a linear operator

Body:
Given the following linear operator
$$T(x,y,z)=(2x+z,-3y+z,-3z)$$
with $T\colon\mathbb{R}^{3}\to\mathbb{R}^{3}$, I found that its matrix in the canonical basis is
$$[T]=\begin{bmatrix} 2 & 0 & 1 \\ 0 & -3 & 1 \\ 0 & 0 & -3 \end{bmatrix}$$
such that its eigenvalues and eigenvectors are:
$$\lambda_{1}=2,\vec{v_{1}}=(1,0,0),\quad\lambda_{2}=-3,\vec{v_{2}}=(0,1,0)$$
How do I conclude that $T$ is not diagonalisable?


问:
How to prove that $\frac{|x+y+z|}{1+|x+y+z|} \le \frac{|x|}{1+|y|+|z|}+\frac{|y|}{|1+|x|+|z|}+\frac{|z|}{1+|x|+|y|}$
I need to prove $\frac{|x+y+z|}{1+|x+y+z|} \le \frac{|x|}{1+|y|+|z|}+\frac{|y|}{1+|x|+|z|}+\frac{|z|}{1+|x|+|y|}$. I've tried to use triangle inequality or to explore the form of $(a+b+c)^2$ but it won't get me anywhere. I would be grateful for some suggestions.

答:
We will prove a stronger statement.

Lemma: $f(x) = \frac{x}{1+x}$ is an increasing function on $ x \geq 0$.

This is obvious by setting $ f(x) = 1 - \frac{1}{1+x}$, which is increasing on $ x \geq -1$.

Lemma: $ |x+y+z| \leq |x| + |y| + |z| $.

This is obvious by the basic properties of absolute values.

Corollary:

$$ \frac{ |x+y+z| } { 1 + |x+y+z|} \leq \frac{|x|+|y|+|z| } { 1 + |x| + |y| + |z| } \leq \sum \frac{ |x| } { 1 + |y| + |z| }$$


答:
Recall that

$$\arccos(x) = \frac{\pi}{2} - \arcsin(x) =\frac{\pi}{2}-x+O(x^2) \to \frac \pi 2 $$

therefore the series diverges since $a_n \not \to 0$.

Title: convergence of a series using the root test

Body:
I want to find whether this series diverges or converges using the root test
$$ \sum_{n=1}^\infty \arccos^n \left(\frac{1}{n^2}\right)  $$
$$ \sqrt[n] {\arccos^n \left(\frac{1}{n^2}\right)}  $$
$$  \arccos \sqrt[n] {\left(\frac{1}{n^2}\right)}  $$
$$  \arccos  1 = 0  $$
and since its 0 the series should converge but I am not sure about the calculations there.


问:
Prove that certain operation is commutative
Suppose we have a set $X$ and a binary operation $\circ : X \times X \to X$, such that $\forall x,y \in X$ the following equalities hold $$y \circ (y\circ x) = x, (x\circ y)\circ y = x.$$

How can prove that this operation is commutative?

答:
Goal: $x \circ y = y \circ x$.

Let $x \circ y = z$. Notice that $(x \circ y) \circ y = z \circ y = x$. Also notice that $z \circ (z \circ y) = z \circ x = y$. Finally we calculate $y \circ x = (z \circ x) \circ x = z = x \circ y$.


问:
Non isomorphic principal $G$-bundles
Remark: throughout this post we work in the smooth category, so that all manifolds, bundles, maps etc are assumed to be smooth.

An exercise asks me to show that there exists no principal $S^1$-bundle isomorphism between the Hopf bundle $S^1 \to S^3 \xrightarrow\pi S^2$ (where $S^1$ acts on $S^3$ by multiplication as usual), and the bundle with the same fiber structure but equipped with the "inverted" action $(w_1,w_2) \cdot z = (w_1z^{-1}, w_2z^{-1})$.

Aiming to reach a contradiction I suppose an $S^1$-bundle isomorphism between these two bundles exists. This is a diffeomorphism $\Phi: S^3 \to S^3$ satisfying:
\begin{array}{rc}
\pi \circ \Phi = \pi, & (*)\\[0.2cm]
\Phi(w_1 z,w_2 z) = \Phi(w_1,w_2)z^{-1}. & (**)
\end{array}

As far as I can see, the conditions $(*)$ and $(**)$ alone do not violate each other, since $\pi$ is invariant under multiplication by elements of $S^1$ (and hence their inverses) anyway. So I guess there should be issues with the diffeomorphy of $\Phi$. More particularly, I would guess smoothness/continuity of $\Phi$ (or its inverse) is an issue, since again I don't see why the equivariance condition $(**)$ would contradict the bijectivity of $\Phi$.

But I'm failing to see how to derive a contradiction here. Am I missing something simpler or am I on the right track?

EDIT:  I just realized that the book has solutions to the exercises in the appendix (I don't remember seeing this in an advanced math book so I didn't even notice they were there). I will try to add an outline of the solution to this post once I figure the details out.

EDIT 2: As promised, here is an outline of a solution. Firstly, since $\Phi$ respects the fibers by $(*)$, it must be of the form
$$ \Phi(w) = w \, \phi(w) $$
for some smooth $\phi:S^3 \to S^1$. By $(**)$ we then see that
$$ w\,\phi(wz)z = \Phi(wz) = \Phi(w)z^{-1} = w\,\phi(w)z^{-1}, $$
or in other words,
$$ \phi(w) = \phi(wz)z^2 $$
since the action is free. This is the condition that we are going to use to derive a contradiction. Since $S^3$ is simply connected, we may write
$$ \phi = e^{i\psi} $$
for some smooth $\psi : S^3 \to \mathbb R$. Thus, taking $z = z(\alpha) = e^{i\alpha}$ with $\alpha \in \mathbb R$, we get
$$ e^{i\psi(w)} = \phi(w) = \phi(wz)z^2 = e^{i\psi(we^{i\alpha})+2i\alpha}. $$
This implies that
$$ \psi(we^{i\alpha}) - \psi(w) + 2\alpha = 2\pi n(\alpha) $$
for some integer $n = n(\alpha)$. But the left hand side depends smoothly on $\alpha$, so the right hand side must as well, i.e. $n(\alpha)$ must be constant (since it is integer-valued). In fact, $n(\alpha) = n(0) = 0$. But then
$$ \psi(we^{i\alpha}) - \psi(w) = -2\alpha, $$
which is absurd, since the left hand side is bounded (since $\psi(S^3)$ is compact), while the right hand side is unbounded.

答:
I apologize in advance for the long post, but, as Qiaochu mentioned, I wanted to learn something from the bare hands approach.

To set notation, I'm going to view $S^3\subseteq \mathbb{C}^2$ as the pairs $(z_1,z_2)$ of unit length vectors, so $|z_1|^2 + |z_2|^2 = 1$.  I'm going to view $S^2\subseteq  \mathbb{C}\oplus \mathbb{R}$ consisting of unit length vectors.

Then the Hopf map $\pi:S^3\rightarrow S^2$ is (according to Wikipedia), given by $\pi(z_1,z_2) = (2z_1\overline{z}_2, |z_1|^2 - |z_2|^2)$.

Let $U = \{(w,t)\in S^2: t > -\epsilon\}$ and $V = \{(w,t)\in S^2: t < \epsilon\}$ where I'm thinking of $\epsilon$ as some fixed very small positive number.  Intuitively, $U$ is the northern hemisphere of $S^2$, except extended slightly below the equation, and $V$ is likewise essentially the southern hemisphere.

Proposition 1:  The open sets $U$ and $V$ form a trivializing cover for the Hopf bundle.

Proof: Let's start with $\pi^{-1}(U)$.  We'll start by finding a section $s_U:U\rightarrow \pi^{-1}(U)$.  So, given $(w,t)\in U$, we want to associate to it $(z_1,z_2)\in \pi^{-1}(U)$.  Let's try making the simplifying assumption that $z_1$ is real and positive.

So, we are solving $(2z_1 \overline{z}_2, z_1^2 - |z_2|^2) = (w, t)$ for $(z_1,z_2)$ under the assumption that $z_1$ is real.  The first equation $2z_1 \overline{z}_2 = w$ can be solved for $z_2$, getting $z_2 = \frac{\overline{w}}{2z_1}$.  Substituting this into the equation $z_1^2 - |z_2|^2 = t$, clearing denominators, we get a quadratic in $z_1^2$.  Using the quadratic formula, together with the fact that $z_1 > 0$, we find $$z_1 = \sqrt{\frac{t+1}{2}}, \text{ and } z_2 =\frac{\overline{w}}{\sqrt{2t + 2}}.$$  Thus, our section $s_U$ is given by $s_U(w,t) = (z_1,z_2)$, with formulas for $z_1$, $z_2$ defined above. I will leave it to you to verify it's a section.

Armed with this section $s_U$, we define $f_U: U\times S^1\rightarrow \pi^{-1}(U)$ by $f_U(w,t,z) = s_U(w,t)z$.  I'll leave it to you to verify that $f_U$ is an $S^1$-equivariant diffeomorphism with inverse $f^{-1}(z_1,z_2) = \left(\pi(z_1, z_2), \frac{z_1}{|z_1|}\right).$

In a similar fashion, we have a section $s_V:V\rightarrow \pi^{-1}(V)$ given by $s_V(w,t) = (z_1,z_2)$ with $$z_1 = \frac{w}{\sqrt{2-2t}} \text{ and } z_2 = \sqrt{\frac{1-t}{2}}.$$  This gives a trivialization $f_V:V\times S^1\rightarrow \pi^{-1}(V)$ given by $f_V(w,t,z) = s_V(w,t)z$ with inverse $f_V^{-1}(z_1,z_2) = \left(\pi(z_1,z_2), \frac{z_2}{|z_2|}\right).$  $\square$

Proposition 2:  The composition $f_V^{-1}\circ f_U$ maps $(w,t,z)$ to $(w,t, \frac{\overline{w}}{|w|} z$.

Proof:  We compute.  \begin{align*} f_V^{-1}(f_U(w,t,z)) &= f_V^{-1}( s_U(w,t)z) \\ &= f_V^{-1}\left(\sqrt{\frac{t+1}{2}}z, \frac{\overline{w}}{\sqrt{2t+2}}z\right) \\ &= \left(\pi\left(\sqrt{\frac{t+1}{2}}z,\frac{\overline{w}}{\sqrt{2t+2}}z\right) , \frac{\frac{\overline{w}}{\sqrt{2t+2}}z}{\left|\frac{\overline{w}}{\sqrt{2t+2}}z\right|}\right) \\&= \left(w,t, \frac{\overline{w}}{|w|} z\right).\end{align*}   $\square$

Using this, we can view $S^3$ as $(U\times S^1) \coprod (V\times S^1)/\sim$ where $(w,t,z)\in U\times S^1$ is identified with $(w,t, \frac{\overline{w}}{|w|} z)\in V\times S^1$ for any $t\in(-\epsilon,\epsilon)$.   Using this description, the projection map $\pi$ is simply projection onto the $w$ and $t$ coordinates.

Now, let's show $\Phi$ cannot exist.  To that end, let's assume $\Phi$ does exist.  Note $(\ast)$ implies that $\Phi$ maps fibers to fibers.  In particular, in our above description, $\Phi$ is given by a pair of maps $\Phi_U:U\times S^1\rightarrow U\times S^1$, $\Phi_V:V\times S^1\rightarrow V\times S^1$ which respect $\sim$.

Since $\Phi$ preserves each fiber, $\Phi_U(w,t,z) = (w,t,\phi_U(w,t,z))$ for some function $\phi_U$.

Proposition 3:  The function $\phi_U$ has the property that $\phi_U(w,t,z) = \phi_U(w,t,1)z^{-1}$.

Proof:  Using $(\ast\ast)$, we know that $$\Phi_U(w,t,z) = \Phi_U((w,t,1)z) = \Phi_U(w,t,1)z^{-1} = (w,t,\phi_U(w,t,1))z^{-1} = (w,t,\phi_U(w,t,1)z^{-1}).$$  On the other hand, $\Phi_U(w,t,z) = (w,t,\phi_U(w,t,z))$.  Thus $\phi_U(w,t,1)z^{-1} = \phi_U(w,t,z)$ as claimed.  $\square$

Of course, the above discussion applies equally well to  $\Phi_V$.  In particular, Proposition 3 is also true for $\phi_V$.

We'll now use the fact that $\Phi$ is well defined to find a relationship between $\phi_U$ and $\phi_V$.

Proposition 4:  We have $\phi_U(w,0,1)\overline{w} = \phi_V(w,0,1)w$.

Proof:  Since $(w,t,z)\in U\times S^1$ is identified with $(w,t,\frac{\overline{w}}{|w|}z)$ in $V\times S^1$, we must have $[\Phi_U(w,t,z)] = [\Phi_V(w,t,\frac{\overline{w}}{|w|} z]$ for all $(w,t,z)$ with $t\in (-\epsilon,\epsilon).$  Set $t=0$ (so $|w| = 1$) and set $z=1$.

Now, $\Phi_U(w,0,1) = (w,0,\phi_U(w,0,1))\in U\times S^1$, and so $(w,0,\phi_U(w,0,1))\sim (w,0, \overline{w} \phi_U(w,0,1))\in V\times S^1$.  Since $\Phi_V(w,0, \overline{w} ) = (w,0, \phi_V(w,0, \overline{w}))$, the condition that $\Phi$ respect $\sim$ implies that $$(w,0,\overline{w}\phi_U(w,0,1)) = (w,0,\phi_V\left(w,0,\overline{w})\right).$$

Using Proposition 3 on the last coordinate, we conclude $\phi_U(w,0,1)\overline{w} = \phi_V (w,0,1)w$ as claimed.  $\square$

We are now ready to reach a contradiction.  Namely, we claim that $\phi_U(w,0,1)\overline{w} =\phi_V(w,0,1)w$ is contradictory.  Viewing $\phi_U(\cdot,0,1):S^1\rightarrow S^1$, the degree of this map must be $0$ because $\phi_U$ extends to the disk $U$.  Likewise, the degree of $\phi_V(\cdot,0,1)$ is $0$.  Thus, the maps $\phi_U(\cdot, 0,1)$ and $\phi_V(\cdot,0,1)$ are homotopy to constants.  It now follows from the equation $\phi_U(w,0,1)\overline{w} = \phi_V(w,0,1)w$ that the maps $w\mapsto \overline{w}$ and $w\mapsto w$ are homotopic.   This is absurd since one has degree $1$ while the other has degree $-1$.  This contradiction establishes that $\Phi$ cannot exist.

Title: Non isomorphic principal $G$-bundles


问:
Function of a matrix - intuition
In my lecture notes there is a topic on functions of matrices. Let $A\in \mathbb{C}^{n\times n}$, and let $f$ be a "nice" function (the lecture notes didn't specify what "nice" means). We define $f(A)$ as in the following Wikipedia article:

https://en.wikipedia.org/wiki/Matrix_function

My question is, why do we choose to define $f(A)$ in such  a way? What do we get from it?

答:
I assume you refer to the extension of a function $f:\Bbb C \to \Bbb C$ so that $f(A)$ can be interpreted, as I mention in my comments.

One motivation comes from linear differential equations. Suppose that we are interested in solving the differential equation $\frac{dx}{dt} = Ax$ and initial condition $x(0) = x_0$ ($x$ is a vector and $A$ is a matrix).  In the scalar (or "$1 \times 1$") case, solving this is straightforward: we arrive at the solution
$$
x(t) = e^{At} \cdot x_0.
$$
This solution works because $\frac d{dt}e^{At} = A e^{At}$.

Now, if we could define the function $e^X$ for a matrix input $X$ in a "natural" way, then we could do the same thing for our more complicated system. In particular, if $e^X$ is defined in such a way that $e^0 = I$ and $\frac d{dt} e^{At} = Ae^{At}$, then it follows that $x(t) = e^{At}x_0$ satisfies
$$
\frac d{dt} x(t) = \frac{d}{dt} e^{At}x_0 = Ae^{At}x_0 = Ax(t), \quad x(0) = e^{0}x_0 = x_0,
$$
which means that we have a solution to our system. That brings us to exactly the situation in your lecture notes: we have a "nice" function $f(x) = e^x$, and it would be useful if we could make sense of $f(X)$ for a matrix $X$.

Note that this $f$ is very nice in that it has a power series that converges for all inputs. However, it is often useful to apply this analysis to functions that are less nice such as $g(x) = \log(x)$. In some applications, it is also useful consider $\sin(x)$ and $\cos(x)$ in this way.

In general, an approach to extending such functions is referred to as a functional calculus.

Title: Function of a matrix - intuition


问:
Borel operator calculus of compound function
I am studying operator calculus of complex-valued bounded Borel measurable function. In our textbook it is induced by Gelfand representation over $A_N$, which is the smallest C* algebra generated by normal operator $N$. It turns out that
$$\phi (\psi(N))=(\phi \circ \psi) (N), \forall \phi, \psi \text{ continuous.}$$
which can be shown using Gelfand representation. Does it still correct in the case $\phi, \psi$ is bounded Borel measurable? I have seen the spectrum decomposition in this case, but don’t know if it can helps.
Thanks in advance.

答:
The answer is affirmative for all Borel functions  $f$ and $g$.  Here is the reason.

First of all let me say that
the most concrete form of the Spectral Theorem I know  asserts that,
given any normal operator $T$ on a separable  Hilbert space $H$,  there exists a $\sigma $-finite measure space $(X,
\mathscr A, \mu )$,
and a unitary operator $U:H\to L^2(X)$, such that
$$
  T = U^{-1}M_\varphi U,
  \tag{1}
  $$
for a certain  $\varphi \in L^\infty (X)$,  where $M_\varphi $ refers to the pointwise multiplication operator
$$
  \xi \in L^2(X)\mapsto \varphi \xi \in L^2(X).
  $$
In other words, every normal operator is unitarily equivalent to a multiplication operator.

This result is perhaps not so popular because its uniqueness part is a bit messy (see section (3.5) in Sunder, V. S., Functional analysis: spectral theory, Birkhäuser Advanced Texts. Basel: Birkhäuser. ix, 241 p. (1997). ZBL0919.46002.) but it is pretty useful, e.g. here.

Next let $B(\sigma (T))$ denote the algebra of all bounded Borel functions on $\sigma (T)$ and consider the *-homomorphism$^{\dagger}$
$$
  f\in   B(\sigma (T)) \mapsto  U^{-1}M_{f\circ \varphi }U \in  \mathscr B(H).
  \tag{2}
  $$

It is easy to prove that this satisfies all of the properties of the Borel functional calculus, and hence this is
the Borel calculus!

Given any $f$ in $B(\sigma (T))$, we thus have that
$$
  f(T) = U^{-1}M_{f\circ \varphi }U.
  \tag{3}
  $$

Observe that  (3) is precisely the expression of (1) for $T'=f(T)$ and $\varphi '=f\circ \varphi $, so the same reasoning above implies
that the Borel functional calculus for $f(T)$ is
$$
  g\in   B(\sigma (f(T))) \mapsto  U^{-1}M_{g\circ f\circ \varphi }U \in  \mathscr B(H).
  $$

We then conclude that
$$
  g(f(T)) = (g\circ f)(T),
  $$
for all $g$, as desired.

$^{(\dagger)}$
It should be noted that   the spectrum of $T$ coincides with the essential range of the above function $\varphi $.  Moreover,
it is well known that $\varphi (x)$ lies in its essential range for almost all $x$, so the composition $f\circ \varphi $ in (1) is defined a.e. on
$X$, and hence the multiplication operator $M_{f\circ \varphi }$ is well defined.


问:
Prove a limit in two variables is $0$
I have the limit

\begin{align}
\lim_{(x,y) \rightarrow(0,0)} \frac{1-\cos(xy)}{\sqrt{x^2+y^2}(x^2+y^2)}
\end{align}

First thing I did is to use Taylor, so $$1-\cos(xy)=\frac{(xy)^2}{2} -\frac{(xy)^4}{4} + o((xy)^4)$$

Therefore the limit is $$\lim_{(x,y) \rightarrow (0,0)} \frac{1}{2} \frac{x^2y^2}{\sqrt{x^2+y^2}(x^2+y^2)}$$

Now, $\frac{y^2}{x^2+y^2} \leq 1$, therefore I have

$$\left|\frac{1}{2} \frac{x^2y^2}{\sqrt{x^2+y^2}(x^2+y^2)}\right| \leq \left|\frac{x^2}{\sqrt{x^2+y^2}}\right| \leq \frac{x^2}{|x|} = |x|$$

Since $|x| \rightarrow 0 $ as $(x,y) \rightarrow (0,0)$, the limit is $0$, correct?

答:
It seems correct, to avoid Taylor we can use that

$$\frac{1-\cos(xy)}{\sqrt{x^2+y^2}(x^2+y^2)}=\frac{1-\cos(xy)}{(xy)^2}\frac{(xy)^2}{\sqrt{x^2+y^2}(x^2+y^2)}$$

with $\frac{1-\cos(xy)}{(xy)^2} \to \frac12$ by standard limits.

For the conclusion by squeeze theorem, we should also consider a part the case $x=0$.

Title: Prove a limit in two variables is $0$


问:
Can you derive the exponential property $ [f(x+y) = f(x)\cdot f(y)]$ from $y' = y$?
There are a variety of different ways to define the natural exponential function, but the definition that feels the most conceptually elegant to me is to define it as the function that is its own derivative, and that $\text{exp}(0)=1$. I am currently trying to figure out if it is possible to derive all the properties of $e^x$/$\text{exp}(x)$ merely from this fact. In addition, I have defined the natural logarithm to simply be the inverse of $\text{exp}(x)$, which was enough to determine that the derivative of $\ln{x}$ is $\frac{1}{x}$. I have been trying in vain to prove from this definition that $\forall{k}\in\mathbb{R},\text{exp}(x+k)=\text{exp}(x)\space\cdot\space\text{exp}(k)$.

答:
Both expressions solve the initial value problem
$$y'(x)=y(x)
,\quad
y(0)=\exp(k),$$
which has a unique solution. So they must be equal.

(More conceptually, that identity is a reflection of the fact that solutions of first-order ODEs are flows.)


问:
What have I done wrong in solving this problem with indices rules?
The question asks to simplify:

$$\left(\dfrac{25x^4}{4}\right)^{-\frac{1}{2}}.$$

So I used $(a^m)^n=a^{mn}$ to get

$$\dfrac{25}{4}x^{-2} = \dfrac{25}{4} \times \dfrac{1}{x^2} = \frac{25}{4x^2} = \frac{25}{4}x^{-2}$$

However, this isn't the answer, and I can't see what I've done wrong.

This is what the mark scheme says:

$$\left(\dfrac{25x^4}{4}\right)^{-\frac{1}{2}} = \left[\left(\frac{4}{25x^4}\right)^{\frac{1}{2}} \text{ or } \left(\frac{5x^2}{2}\right)^{-1} \text{ or } \frac{1}{\left(\dfrac{25x^4}{4}\right)^{\frac{1}{2}}}\right] = \frac{2}{5}x^{-2}$$

To me, my answer looks more simple than theirs, and I can't see what I've done wrong.

答:
Just to point out the "rule" you forgot:

It is correct that $$(a^m)^n=a^{mn}$$

But you have more than $(x^4)^{-1/2}$ to consider. You need to apply the fact that when a product and/or quotient of numbers/variables, enclosed in parentheses, is raised to a power, you need to distribute that power across the product in parentheses:

$$(ab)^n = a^nb^n\quad \text{or given,} \quad \left(\frac{ab}c\right)^n = \dfrac{a^nb^n}{c^n}$$

So in your case, you need to combine the two rules: $$(ab^m)^n = a^n\cdot \left(b^{m}\right)^n = a^n \cdot b^{mn}$$

So for $$\left(\dfrac{25x^4}{4}\right)^{-\frac{1}{2}}$$ I would simplify matters to express this as $$\left(\frac{25}{4} x^4\right)^{-1/2} = \left(\frac{25}{4}\right)^{-1/2} (x^4)^{-1/2} = \left(\frac{5^2}{2^2}\right)^{-1/2} (x^4)^{-2} = \frac{5^{-1}}{2^{-1}}x^{-2} = \frac 25 x^{-2} = \frac 2{5 x^2}$$


答:
Note that (provided $\exp(k)\ne0$), we have for $f(x):=\frac{\exp(x+k)}{\exp(k)}$,
$$f(0)=\frac{\exp(0+k)}{\exp(k)}=1,\qquad f'(x)=\frac{\exp'(x+k)}{\exp(k)}=\frac{\exp(x+k)}{\exp(k)}=f(x)$$
so that by uniqueness of the solution,  $f(x)=\exp(x)$ for all $x$.
Hence,
$$ \exp(x+k)=\exp(x)\exp(k)\quad \text{if }\exp(k)\ne0.$$
As $\exp(0)=1$, we have $\exp(k)\ne0$ in a neighbourhood of $0$, hence if $\exp(x)=0$, then $\exp$ is also $=0$ in a neighbourhood of $x$. It follows that
$$\{\,x\in\Bbb R\mid \exp(x)=0\,\} $$
is open - but it trivially is also closed, and is non-empty. This means that it is all of $\Bbb R$. Hence we can strike our caveats and have
$$ \exp(x+k)=\exp(x)\exp(k)\quad \text{for all }x,k\in \Bbb R.$$

Title: Can you derive the exponential property $ [f(x+y) = f(x)\cdot f(y)]$ from $y' = y$?

Body: There are a variety of different ways to define the natural exponential function, but the definition that feels the most conceptually elegant to me is to define it as the function that is its own derivative, and that $\text{exp}(0)=1$. I am currently trying to figure out if it is possible to derive all the properties of $e^x$/$\text{exp}(x)$ merely from this fact. In addition, I have defined the natural logarithm to simply be the inverse of $\text{exp}(x)$, which was enough to determine that the derivative of $\ln{x}$ is $\frac{1}{x}$. I have been trying in vain to prove from this definition that $\forall{k}\in\mathbb{R},\text{exp}(x+k)=\text{exp}(x)\space\cdot\space\text{exp}(k)$.


问:
Distribution Function of ln(x)
I am trying to find the probability density of ln(1/X) which can be rewritten as -ln(X) so to find the probability density I first want to find the distribution function. I am not sure how to do so, can somebody help me with determining this? I am looking at when x is uniformly distributed with 0 < X < 1.</p>

答: If a continuous random variable $X$ whose support is all $\ge0$ has CDF $F_X$ and PDF $f_X=F_X^\prime$, then $Y=-\ln X$ is a real-valued continuous random variable of CDF$$F_Y(y)=P(Y\le y) =P(X\ge e^{-y})=1-F_X(e^{-y})$$and PDF$$f_Y(y)=F_Y^\prime(y)=e^{-y}f_X(e^{-y}).$$I leave as an exercise working out the support of $Y$ in terms of that of $X$.

Title: Distribution Function of ln(x)


答:
$$F_X(x)=x$$

You are looking for the density of

$$Y=-log X$$

Use the definition

$$F_Y(y)=\mathbb{P}[Y\leq y]=\mathbb{P}[-logX \leq y]=\mathbb{P}[X\geq e^{-y}]=1-F_X(e^{-y})=1-e^{-y}$$

In other words

$$Y\sim Exp(1)$$

Title: Distribution Function of ln(x)

Body: I am trying to find the probability density of ln(1/X) which can be rewritten as -ln(X) so to find the probability density I first want to find the distribution function. I am not sure how to do so, can somebody help me with determining this? I am looking at when x is uniformly distributed with 0 < X < 1.</p>


问:
How to solve $2^\text{nd}$ order ODE $y'' + 4y' + 8y = 145\cos(3x)$
I tried using the method of undetermined coefficients to find the particular solution for this equation. However, I got stuck halfway through.

I'm not sure if this is the correct way to approach this equation. Help will be greatly appreciated.

答:
You have solved it correctly. You get $U=145/(12i-1)=-1-12i$, so the paeticular solution of $y''+4y'+8y=145e^{i3x}$ is $Ue^{i3x}=(-1-12i)e^{i3x}=12\sin3x-\cos3x+i(\cdot\cdot\cdot)$. Clearly the real part of this particular solution generates the real part of the ODE it satisfies i.e. $145\cos3x$, so the particular solution of original ODE is $12\sin3x-\cos 3x$.

Suppose $f(x)+ig(x)$ is a solution of $\mathcal Ly=h(x)+ik(x)$, where $\mathcal L$ is a linear differential operator with real-valued coefficients (in your case $D^2+4D+8)$. Then$$\mathcal L(f+ig)=\cal Lf+i~\cal Lg=h+ik$$Now equate the real and complex parts. This means $f$ satisfies $\cal Lf=h$ and $g$ satisfies $\cal Lg=k$.

In your case, $f(x)=12\sin3x-\cos 3x$ and $h(x)=145\cos3x$.


答:
Let's think through this backwards.

"there is at least one $c \in [0,b]$ such that $f(c)=c$." is certainly the correct form of the conclusion of the IVT.  So to obtain this, we need to satisfy the IVT's hypotheses.

If there is a $c$ such that $f(c) = c$, then $f(c) - c = 0$.  So, we want to know if $f(x) - x$ has a root, say at $x = c$.  Conveniently, we have already been told to study this function, $g(x) = f(x) - x$.

The hypotheses of the IVT require that $g$ is continuous and that $g$ takes at least one positive value and and at least one negative value somewhere in the interval $[0,b]$.  (Notice that $g$ has a third option: $g$ can be zero throughout the interval.  We won't have to think about that explicitly in what follows.)

First we check that $g$ is continuous.  We have: $f(x)$ is continuous (on its domain, $[0,b]$), the identity function, $x \mapsto x$, is continuous (on $[0,b]$) (It is, at least, a polynomial, which we should know is continuous.), and the difference of continuous functions is continuous.  Therefore, $g$ is continuous (on $[0,b]$).

Next we need to see that $g(x)$ takes positive and negative values.  We don't have enough insight into what $f$ does to prove that directly, so we proceed by contradiction.

Suppose $g(x) \geq 0$ for all $x \in [0,b]$.  That is, suppose that there is no point in $[0,b]$ where $g$ takes a negative value.  Then, in particular, $g(b) \geq 0$, which forces $f(b) \geq b$.  Since the image of $f$ is upper bounded by $b$, we must have $f(b) = b$.  (The easiest way to see this is to draw the square $[0,b]\times[0,b]$, draw the diagonal from lower-left to upper-right (the $f(x) = x$ line).  Then, try to draw the graph of a function starting anywhere on the left edge and ending anywhere on the right edge, without ever intersecting the diagonal.  It's easy enough to stay above the diagonal right up until the right edge, at which you have no choice.)  Therefore, either $c = b$ and the result holds or $g$ takes a negative value at least once on $[0,b]$.

Alternatively, suppose $g(x) \leq 0$ everywhere.  Then, in particular, $g(0) \leq 0$.  This forces $f(x) \leq 0$, which either forces $f(0) = 0$ and the result holds or $g$ takes a positive value at least once on $[0,b]$.

So that's how we arrive at how we work this problem.  When we write it up, we should shuffle these around a little.

Start with: $f(c) = c$ means we should study $g$.

Show that $g$ is continuous.

Show that $g$ takes at least one positive and at least one negative value.

Apply the IVT.

Title: Applying the IVT on closed intervals

Body:
I'm not exactly sure how the intermediate value theorem can be applied to the following question especially with the defined function $g(x)$ which is provided below.

Using the intermediate value theorem with $g(x)=f(x)-x$ and suppose $f:[0,b] \to [0,b]$ is continuous. Prove there is at least one $c \in [0,b]$ such that $f(c)=c$.

Thanks


问:
If a solution differential equation is constant on a interval then it is constant everywhere
Suppose we have a solution to a differential equation that is constant on some interval will it be constant everywhere?

How does one proof this if this is the case?

答: No, for example the function defined by $y(x) = 0$ for $x \le 0$ and $y(x) = x^2$ for $x>0$ satisfies the ODE $y'(x) = 2 \sqrt{|y(x)|}$ for all $x$ (including $x=0$).


答:
It doesn't need to be the case.

Consider the differential equation: $y'=f(t,y)$, where $f:\Bbb R^2\to \Bbb R$ is defined as follows

$f(t,y)=\begin{cases}2t&\text{if }t\ge0\\0&\text{otherwise}\end{cases}$.

We have that $f$ is continuous on $\Bbb R^2$, so the ODE is perfectly valid.

Then for every $a\in\Bbb R$ we have a solution, which would be the function $\phi(t)=\begin{cases}t^2+a&\text{if }t\ge0\\a&\text{otherwise}\end{cases}$, since

$\phi'(t)=\begin{cases}2t&\text{if }t\ge0\\0&\text{otherwise}\end{cases}$.

Title: If a solution differential equation is constant on a interval then it is constant everywhere

Body:
Suppose we have a solution to a differential equation that is constant on some interval will it be constant everywhere?

How does one proof this if this is the case?


问:
maximizing $p_1p_2\cdots p_n$ where $p_j$ are probabilities
Let $V = p_1 p_2 \cdots p_n$ where $p_i \in [0,1]$ and $\sum p_i = 1$, so the $p_i$ define a discrete probability distribution. Now I'd like to find $p$ that maximizes $V$. Intuitively it makes sense that the maximum must be at $p_i = 1/n$ for all $i$, and it is easy to verify that the gradient of $V$ vanishes there. But now I'd like to prove the reverse: If $V$ is maximal then $p_i = 1/n$.

With Lagrangian mulitpliers it is rather straightforward, but I've tried solving the system you get by setting the gradient to zero but without much success. Can anyone spot a simplification or a nice argument that would let me continue there?

Here's my attempt so far:
First I just plugged in the $\sum p_i = 1$ (and for ease of writing $m:= n-1$, and we assume that the index now goes up to $m$). So

$$V = \underbrace{\Big(\prod_i p_i \Big)}_{=:P}\Big(1 - \underbrace{\sum_i p_i}_{=:S}\Big)$$

Actually it should be possible to prove that we have only one local maximum even without the restriction of $p_i \in [0,1]$:

$$0 \overset{!}{=} \frac{\partial V}{\partial p_j} = \frac{P}{p_j} (1-S) + P(p_j-S) $$

assuming all $p_i \neq 0$ (if some $p_i=0$ we can discard it an restart the problem with less variables) we can simplify that to

$$ S-1 = p_j(p_j-S)$$

(Note that this is a system of equations that holds for all $j=1,\ldots,m$, and here $S$ and $P$ depend on $p_j$.)

Now at that point I thought maybe we can use some fixed point iteration argument but didn't manage to find anything. So I'd love to hear if anyone could expand this argument, but I'm also interested in other elegant ways of solving this.

答:
The AM-GM inequality tells you that
$$(p_1 p_2 \cdots p_n)^{1/n} \leq \frac 1 n \sum_{i = 1}^n p_i$$
and crucially, that equality is achieved if and only if $p_1 = p_2 = \cdots = p_n$. This fact should suffice to prove both directions (the inequality itself tells you that $p_1 = p_2 = \cdots = p_n$ yields an optimal solution, and the information about when equality is achieved tells you that this is the only optimal solution).


问:
Why in proximal gradient descent, "proximal" is referred to as "Backward" and the "gradient" is "Forward"?
I am not into monotone operators theory (but hope someday I can get a hang of it).

Sorry for asking probably stupid question.

Why in proximal gradient descent, "proximal" is referred to as "Backward" and the "gradient" is referred to as "Forward"?

I can imagine why "gradient" is referred to as "Forward" perhaps because it looks for the "downhill" and move forward. But I am not sure. However, I have no clue why "proximal" is referred to as "Backward". Can someone enlighten me? Thank you.

答:
The gradient descent (with fixed step size $t > 0$), i.e.,
$$
x_{k+1} = x_k - t \, \nabla f(x_k)$$
can be rewritten as
$$
\frac{x_{k+1} - x_k}{t} = -\nabla f(x_k),$$
i.e., it is a explicit Euler discretization of the gradient flow ODE
$$
\dot x(t) = -\nabla f(x(t)).
$$
On the other hand, the proximal point method reads
$$
x_{k+1} = \operatorname{prox}_{t \, f}(x_k).$$
This implies
$$
\frac{x_{k+1} - x_k}{t} \in -\partial f(x_{k+1}).$$
This amounts to an implicit Euler discretization of the gradient flow.

Finally, the explicit Euler method is also called 'forward Euler', whereas implicit Euler is called 'backward Euler'.


问:
Question about combinatorics in case the arrangement is not important.
So, I know that the number of combinations of $n$ objects taken $r$ at a time (where the order does not matter) is given by $\frac{n!}{(n - r)!}$ divided by $r!$. Based on my understanding, I would divide the product of given formula $\frac{n!}{(n-r)!}$ by just $r$. I don't understand why it should be divided by $r!$. Can someone explain? For example, if we were to select two managers from five people then the possible ways to do it would be $5\cdot4 \cdot \frac{1}{2} = 10$. Also, I only partly understand the reason you have to divide it by anything when the order of the selected objects is not important. How to better understand the case when the order is not important?

答:
To answer your question, let us look at a very simple example.

Imagine we were asked to find the number of different ways to arrange the letters in the word 'SHEEP' (the words formed do not have to be meaningful). How would this be done?

Since we were asked to arrange the letters in the word 'SHEEP', we will have to use permutations (since the order of the letters in the word matters).

Therefore, the number of arrangements, assuming that all the letters are distinct, is given by:

 $$^5P_5 = \frac {5!}{(5 - 5)!} = \frac {5!}{0!} = \frac {5!}{1} = 5! = 120$$

However, all the letters in the word 'SHEEP' are not distinct; the letter 'E' is repeated twice. Therefore, we will have to divide the number of arrangements by $2!$. This is because, in the above calculation, we have assumed that all the letters are distinct.

Note: We divide by $2!$ not $2$. This is more evident with more repeated letters, such as with the word 'DELETE'. If we notate each 'E' differently, like '$DE_1LE_2TE_3$', then the calculation above sees '$DE_1LE_2TE_3$' differently from '$DE_1LE_3TE_2$' and '$DE_3LE_2TE_1$'. This is clearly not the case, as the 'E' is the same no matter where it is written. The same concept can be applied to words like 'INTERNET' and 'BOOKSHELF', where more than one letter is repeated.

Therefore, the number of arrangements of the letters in the word 'SHEEP' is given by:

 $$\frac {^5P_5}{2!} = \frac{\frac {5!}{(5 - 5)!}}{2!} = \frac{\frac {5!}{0!}}{2!} = \frac{\frac {5!}{1}}{2!} = \frac {5!}{2!} = 5\cdot 4 \cdot 3 = 60$$

Let us now extend this concept with combinatorics where the arrangement does not matter.

Let us take your example - selecting two managers from five people.

Let us first look at arranging two people from a group of five i.e. the order does matter. This is given by:

 $$^5P_2 = \frac {5!}{(5 - 2)!} = \frac {5!}{3!} = 5\cdot 4 = 20$$

Let us now look at selecting two people from a group of five i.e. the order does not matter.

Let us first notate the people who are selected as '$S$' and the people who are not selected as '$N$'. Therefore, the group of people can be rewritten as:

 $$S \; S \; N \; N \; N$$

To solve this problem, all we have to do is find out the number of ways we can arrange the letters above. There are $5!$ total ways to arrange the letters above. However, the letter 'S' is repeated twice and the letter 'N' is repeated thrice. With that in mind, we have to look back at our problem with the word 'SHEEP'. When we were asked to arrange the letters in the word 'SHEEP', we noticed that the letter 'E' was repeated twice. We, therefore, divided the total number of arrangements by $2!$ i.e. the factorial of the number of times a letter was repeated.

Thus, we can apply the same concept here.

The number of ways to select 2 managers from 5 people is given by:

 $$\frac {5!}{2!\cdot 3!} = \frac {5\cdot4}{2\cdot 1} = 10$$

This explains why we have to divide $\frac {5!}{(5 - 2)!}$ by $2!$ i.e. divide $\frac {n!}{(n - r)!}$ by $r!$. It is because of that extra $r!$ group which we are taking note of (the second repeated group). This also explains why $^5C_2$ $=$ $^5C_3$, since selecting 2 from 5 is the same as selecting 3 from 5; the complementary group is left behind.

I hope this helps! I'm new to MSE, so I would really appreciate any and all feedback. Thank you!


问:
Sport Venn Diagram

21% of the students play soccer.

34% play basketball.

30% do track.

2% do soccer and basketball.

6% soccer and track.

9% basketball and track.

36% do none or other.

What percentage of the students play all three sports?
  Display the data in a Venn Diagram.

What percentage of the students only play soccer?

There are 900 students at the school.  Show a Venn Diagram displaying the data as numbers of students.

I cannot figure out how to make these percentages work. Help!

答:
There are a couple of ways to approach the problem; I’ll start with a very straightforward one from first principles. First, here’s a picture:

The blue part of the diagram represents those students who do not participate in soccer, basketball, or track. We’re interested in $x$, the percentage who play all three sports. One way to solve the problem is to fill in the Venn diagram in terms of $x$. I’ve started: we know, for instance, that $2\%$ play soccer and basketball, and $x\%$ play all three sports, so $2-x$ must be the percentage playing soccer and basketball but not track. The figures $6-x$ and $9-x$ were obtained by similar reasoning. Now go after the percentage playing just soccer, just basketball, and just track. We know that $21\%$ play soccer, and $(2-x)+x+(6-x)$ percent play soccer and at least one other of the three sports, so what percentage (in terms of $x$) play just soccer? Repeat this procedure with basketball and track to fill in the last three regions of the Venn diagram. Finally, add up the percentages of all of the white regions; that total, which will involve $x$, must be the $100-36$ percent who play at least one of the three sports, and you’ll have a simple equation to solve for $x$.

If you know something about calculating the cardinalities of unions and intersections of sets, you can short-circuit this calculation. Let $S,B$, and $T$ be the sets of students playing soccer, basketball, and track, respectively. Pretend that there are $100$ students altogether, so that percentages are simply numbers of students. Then you want $|S\cap B\cap T|$, and you know $|S|$, $|B|$, $|T|$, $|S\cap B|$, $|S\cap T|$, $|B\cap T|$, and the cardinality of the complement of $S\cup B\cup T$. That last figure gives you $|S\cup B\cup T|$. Now, do you know a formula that relates all of these quantities? If so, you’re in business, because you know all but one of them.

Added: However, the problem is faulty, because if you actually carry out the computations, which I unfortunately did not bother to do when I originally posted the answer, you’ll find that $x$ is negative!

Title: Sport Venn Diagram


问:
The jumping times $\tau_k$ of a $\mathcal F$-poisson process $N$ are $\mathcal F$-stopping times and $N_t=\sum_k1_{\left\{\:\tau_k\:\le\:t\:\right\}}$
Let $(\Omega,\mathcal A,\operatorname P)$ be a probability space, $(\mathcal F_t)_{t\ge0}$ be a filtration on $(\Omega,\mathcal A,\operatorname P)$ and $(N_t)_{t\ge0}$ be a $\mathcal F$-Poisson process on $(\Omega,\mathcal A,\operatorname P)$ with intensity $\lambda>0$, i.e.

$N$ is an $\mathbb N_0$-valued $\mathcal F$-adapted process on $(\Omega,\mathcal A,\operatorname P)$;

$N_0=0$;

$N_t-N_s$ and $\mathcal F_s$ are independent for all $t\ge s\ge0$;

$N_t-N_s$  is Poisson distributed with parameter $\lambda(t-s)$ for all $t\ge s\ge0$.

Assuming $N$ is almost surely right-continuous, I was able to show that

$N$ is almost surely nondecreasing

and assuming that $N$ is almost surely càdlàg, I was able to show that$^1$

$\operatorname P\left[\forall t\ge0:\Delta N_t\in\{0,1\}\right]=1$.

Assume $N$ is (surely) càdlàg. Let $\tau_0:=0$ and $$\tau_k:=\inf\left\{t>\tau_{k-1}:\Delta N_t\ne0\right\}$$ for $k\in\mathbb N$.

Are we able to show that

$\operatorname P\left[\forall n\in\mathbb N_0:\exists t\ge0:N_t=n\right]=1$;

$\tau_k$ is a $\mathcal F$-stopping time for all $k\in\mathbb N$;

$\tau_k$ is almost surely finite for all $k\in\mathbb N$.

$N_t=\sum_{k\in\mathbb N_0}1_{\left\{\:\tau_k\:\le\:t\:\right\}}$ for all $t\ge0$ almost surely.

All these claims are intuitively trivial, but I really worry about some of the technical details we need to prove them rigorously. For example, does (8.) really hold or do we need to replace $\mathcal F$ with the right-continuous filtration generated by it?

For (7.), I've tried to consider $$\operatorname P\left[N_t<n\right]=\sum_{k=0}^{n-1}\operatorname P\left[N_t=k\right]=e^{-\lambda t}\sum_{k=0}^{n-1}\frac{(\lambda t)^k}{k!}\tag{11}$$ for all $t\ge0$ and $n\in\mathbb N$. $(11)$ would tend to $0$ as $t\to\infty$ if $t^ke^{-\lambda t}\xrightarrow{t\to\infty}0$, but does this really hold? We clearly got $t^ke^{-\lambda t}=e^{k\ln t-\lambda t}$, but now the only useful inequality I'm aware of is $\ln t\le t-1$ for all $t>0$, which is not enough to conclude $k\ln t-\lambda t\xrightarrow{t\to\infty}-\infty$.

$^1$ As usual, if $x:[0,\infty)\to\mathbb R$ is càdlàg, then $x(t-):=\lim_{s\to t-}x(s)$ and $\Delta x(t):=x(t)-x(t-)$ for $t\ge0$.

答:
$\{\tau_1\le t\} = \{N_t\ge 1\}\in\mathcal F_t$.  Likewise for subsequent $\tau_k$.

${\rm P}[\tau_1<\infty] = \lim_{t\to\infty}{\rm P}[\tau_1\le t]=\lim_{t\to\infty}{\rm P}[N_t\ge 1]=\lim_{t\to\infty}1-e^{-\lambda t}=1$. Similarly  for subsequent $\tau_k$.

This should read $N_t = \sum_{k\ge 1}1_{\{\tau_k\le t\}}$. Both sides of the equality-to-be are non-decreasing, cadlag, with jumps of size 1, and initial values $0$. To see that they agree you can check that they have the same jumps at each time $t$. But this is clear because $\Delta N_t=1$ if and only if $t=\tau_k$ for some $k\ge 1$.

This follows from 6. and 8.


问:
Range of function $a \sin(mx) + b \cos(nx)$
What's the range of function $a \sin(mx) + b \cos(nx)$ where $a,b,m,n \in R$?

Not hard to solve for the case where $m=n$. We can let $m=n=1$ WLOG

$a \sin(x) + b \cos(x) = \sqrt{a^2 + b^2} \left( \frac{a}{\sqrt{a^2 + b^2}} \sin(x) + \frac{b}{\sqrt{a^2 + b^2}} \cos(x) \right)$

Then we can substitute $\sin\theta = \frac{b}{\sqrt{a^2 + b^2}}$

How do we handle the case when they are different?

答:
A comment that got too long with some observations:

Assume wlog $abmn \ne 0$ as those cases are easy to consider and let $f(x)=a\sin mx +b \cos nx$; the range of $f$ is always an interval by continuity.

When $m/n$ is irrational the solution is straightforward as we can use Kronecker theorem to find $x$ st $\sin mx =\pm \frac{a}{|a|}(1-\epsilon) , \cos nx =\pm \frac{b}{|b|}(1-\epsilon_1)$, so the range of $f$ being an interval, must be the open maximal one $(-|a|-|b|, |a|+|b|)$ as the ends ar never taken by an easy check ($\sin nx= \pm 1, \cos mx= \pm 1$ implies $m/n$ rational, while obviously $-|a|-|b| \le f(x) \le |a|+|b|$

When $m/n$ is rational, we can reduce by changing variables to $m,n \in \mathbb Z, (m,n)=1$ but that case seems hard in general and I am not sure if there is an explicit way to express the solution (to me it seems that the answer depends on divisibility properties of $m,n$ - definitely on parity - but I could be mistaken of course).

Now $f$ is periodic so the range must be a closed interval and in the case, $m=2k+1, n=2p$ it is easy to see that we can get one maximal end but not the other as we can get $\sin mx =\pm 1$ while $\cos nx =(-1)^p, x =\pm \pi/2$, while when $m$ is even or $m,n$ both odd we cannot reach the end points of the maximal interval. I tried to look at the roots of $f'(x)=0$ as some of those will give the maximum and minimum of $f$ but went nowhere.


问:
When does improper Riemann integration fail for computing expected values?
Riemann integration requires bounded intervals of integration. However, some continuous random variables (like those from the normal or gamma distribution) have unbounded support sets.

For these 2 distributions, would improper Riemann integrals be sufficient to get their expected values? If not, why not?

答:
Usually, improper Riemann integration will correctly provide the expectation of a continuous random variable with an unbounded support. Where you will usually encounter problems with improper Riemann integrals is when the integral of interest does not converge absolutely. In this case the improper integral may have some value, but this value has limited probabilistic meaning; for example the law of large numbers doesn't necessarily hold.

As an example of this to get a feel for it, you can look at the evaluation of the expected value of $X \sin(X)$ where $X$ has PDF $\frac{1}{x^2}$ on $[1,\infty)$ and $0$ otherwise. This expectation is formally $\int_1^\infty \frac{\sin(x)}{x} dx$ which has a value in improper Riemann integration. But actually this random variable will not satisfy the law of large numbers, as you can observe numerically, so it doesn't make a whole lot of sense to say that it has an expected value.

Indeed, if you have Matlab or Octave, try running the following:

```<code>x=1./rand(1000,1);
plot(cumsum(x.*sin(x))./(1:1000)');
</code>```

This shows sample means from this distribution for progressively larger samples, and you see that they don't converge. (Off-topic: the weird 1/rand trick being used there is called the probability integral transformation, which is extremely useful for numerical work in probability.)

Note that this will never be a problem for a random variable of one sign such as your example of the Gamma distribution. It is also not a problem for the normal distribution.

The exception to the second "usually" above is when the improper Riemann integral doesn't exist but the Lebesgue integral does. This is a pretty uncommon situation in practice.

Body:
Usually, improper Riemann integration will correctly provide the expectation of a continuous random variable with an unbounded support. Where you will usually encounter problems with improper Riemann integrals is when the integral of interest does not converge absolutely. In this case the improper integral may have some value, but this value has limited probabilistic meaning; for example the law of large numbers doesn't necessarily hold.

As an example of this to get a feel for it, you can look at the evaluation of the expected value of $X \sin(X)$ where $X$ has PDF $\frac{1}{x^2}$ on $[1,\infty)$ and $0$ otherwise. This expectation is formally $\int_1^\infty \frac{\sin(x)}{x} dx$ which has a value in improper Riemann integration. But actually this random variable will not satisfy the law of large numbers, as you can observe numerically, so it doesn't make a whole lot of sense to say that it has an expected value.

Indeed, if you have Matlab or Octave, try running the following:

x=1./rand(1000,1);
plot(cumsum(x.*sin(x))./(1:1000)');

This shows sample means from this distribution for progressively larger samples, and you see that they don't converge. (Off-topic: the weird 1/rand trick being used there is called the probability integral transformation, which is extremely useful for numerical work in probability.)

Note that this will never be a problem for a random variable of one sign such as your example of the Gamma distribution. It is also not a problem for the normal distribution.

The exception to the second "usually" above is when the improper Riemann integral doesn't exist but the Lebesgue integral does. This is a pretty uncommon situation in practice.