问:
Determine the distribution of $Y_i$
Let $X_1, \ldots , X_n$ be independent, identically distributed random variables with $$P(X_i=-1)=P(X_i=1)=\frac{1}{2}$$
We consider the random variables $Y_i=\max \{X_i,X_{i+1}\}$, $i=1,\ldots , n-1$.

(a) Determine the distribution of $Y_i$, $i=1,\ldots , n-1$.

(b) Calculate the expected value of $Y_i$, $i=1,\ldots , n-1$.

(c) Calculate the covariance of $Y_i$ and $Y_j$, i.e. $\text{Cov}(Y_i, Y_j)=E(Y_iY_j)-E(Y_i)E(Y_j)$, $i,j=1,\ldots , n-1$.

$$$$

For question (a) the r.v. $Y_i$ has the same distribution as the r.v.s $X_i$, or not?

答:
For question (a) the distribution of $Y_i$ is not the same as the distribution of $X_i$ because we have four possible outcomes for $(X_i, X_{i+1})$ which all have the probability $\frac{1}{4}$:
$$(-1,-1), (1,-1), (-1,1), (1,1)$$
Taking the max of both values we get 1 in three of four cases and -1 in only one case. Therefore
$$P(Y_i = 1) = \frac{3}{4}, \ P(Y_i=-1) = \frac{1}{4}$$

Calculating the expected value with this should be no problem.

I hope I could help you.

Title: Determine the distribution of $Y_i$


问:
Substituition of the integral limit in the first fundemantal theorem of calculus
As I was reading G. Strang's Calculus book, I came across an example problem I couldn't understand. The topic is the First Fundamental Theorem of Calculus. The form proposed for the theorem is (Picture): $$F( x) \ =\ \int ^{x}_{a} f( t) dt\\$$ After proving this, there is an example problem. $$Let\ F( x) \ =\ \int ^{\sqrt{x}}_{1} \sin(t)dt.
Find\ F'( t)$$Book solves the question by (Picture) $$u(x) = \sqrt{x}\ then\ F(x) = \int ^{u(x)}_{1}\sin(t)dt\\F'( x) \ = \sin( u( x))\frac{du}{dx}\\
=\sin( u( x)) \cdot \left(\frac{1}{2} x^{-\frac{1}{2}}\right)\\
=\frac{\sin\sqrt{x}}{2\sqrt{x}} \cdot \\$$My question is that, in it's original form, there is an $x$ in the upper limit of the integral. When we substitute it with $u(x)$, shouldn't we change $F(x)$ with $F(u(x))$ also? How does it stay unchanged?

答: Well, it’s exactly what you said, just write it in the following way: $F(x) = \int_1^{\sqrt{x}} \sin t \, dt =: G(\sqrt{x})$. Then by the Fundamental Theorem of Calculus, you get that $G’(\sqrt{x}) = \sin(\sqrt{x})$. At the same time, by the chain rule we get that $F’(x) = G’(\sqrt{x}) \frac{d \, \sqrt{x}}{d \, x} = \frac{\sin \sqrt{x}}{2 \sqrt{x}}.$ I hope this sheds some light on the issue. :)

ContentLicense: CC BY-SA 4.0


问:
The number of groups of order 32
There are 51 groups of order ($32=2^5$). My question is how this number was computed.

Graham Higman and Charles Sims gives an estimate for the number of $p$-goups (i.e. groups of order $p^n$ where $p$-is a prime number).

I will be very grateful for any suggestion.

答:
Just to put this discussion in context, the groups of order $32$ were first enumerated by G.A. Miller in the paper

Miller, G. A., The regular substitution groups whose orders is less than 48, , 28, (1896), 232–284,

which solves the problem for all order up to $48$. This was of course done by hand. Although plenty of mistakes were made in subsequent calculations of this type, as far as I know Miller's work is accurate. But unfortunately I don't have any easy access to the paper (the University of Warwick library does not go back that far).

The lists of groups of order $2^n$ for $n \le 6$ were published in a printed book in 1964:

Hall, Marshall, Jr.; Senior, James K. The groups of order $2^n\,(n\leq 6)$. The Macmillan Co., New York; Collier-Macmillan, Ltd., London 1964 225 pp.

I remember looking through that many years ago, but I expect the printed version is regarded as redundant now!

Title: The number of groups of order 32


问:
Proof of Leibniz Rule for formal derivatives
Is there a method to prove Leibniz's Rule, in the context of formal derivatives of polynomials over Finite Fields?

If I have two polynomials $a(x), b(x) \in GF(q)[x]$, How can I prove that $D(a(x)b(x)) = a(x)D(b(x)) + D(a(x))b(x)$

The Wikipedia page for formal derivatives indicates that this can be proven. However, it does not reference any proofs and I could not find any.

答: Hint: Distribution of multiplication over addition, and linearity of the formal derivative, means you can start with monomials and build from there.</p>


问:
The Bayes Error, Book: "A Probabilistic Theory of Pattern Recognition, Bayes decision"
The book is really hard to understand, could someone, please, explain this part from Example 2:

The Bayes classifier decides $1$ if $T + B + E < 7$ and $0$ otherwise. $E$ and $B$ are intangible, and not available to the observer, we have access only to $T$. Let us assume that $T, B$ and $E$ are i.i.d exponential random variables (thus, they have density $e^{-u}$ on $[0, \infty)$).

Could you please explain how we derive this part:

\begin{equation}
    P(Y=1\mid T) = P(E + B < 7 - T\mid T) = \max\left(0, 1 - (1 + 7 - T)e^{-(7-T)}\right)
\end{equation}

I am interested in $\max\left(0, 1 - (1 + 7 - T)e^{-(7-T)}\right)$. I understand that there is an integration of $E + B$ which are exponentially distributed, but how is it calculated? What are the steps and over what are we integrating ?

答:
If $T$, $E$ and $B$ are i.i.d. $\sim \exp(1)$ random variables, the derivation of $\mathbb{P}(Y=1|T)$ can be done in different ways. For instance, conditioning on $T=t$ :

$\begin{align}
\mathbb{P}(E + B\le 7 - t) &= \int_0^\infty \mathbb{P}(E+B\le 7-t|B=b)f_{B}(b)db \\
&= \int_0^{7-t}\mathbb{P}(E+B\le 7-t|B=b)\text{e}^{-b}db \\
&= \int_0^{7-t} \mathbb{P}(E\le 7-t-b)\text{e}^{-b}db \\
&= \int_0^{7-t}\left(1-\text{e}^{-(7-t-b)}\right) \text{e}^{-b}db\\
&= \int_0^{7-t}\text{e}^{-b}db - \int_0^{7-t} \text{e}^{-(7-t)}db \\
&=1-e^{-(7-t)} - (7-t) e^{-(7-t)} \\
&= 1 - (1 + 7 -t)e^{-(7-t)} \end{align}$

This is how the expression is obtained. Now the $\max$ comes from the fact that there are values of $t$ for which the expression is negative. These values are all $t\ge 7$, for which you see that the probability is actually $0$. So the correct expression is actually
$$\mathbb{P}(E + B\le 7 - t) = \max\{0,1 - (1 + 7 -t)e^{-(7-t)}\} \, \forall t\ge 0$$

Since this holds for all $t$, this proves that the conditional probability is indeed
$$  \mathbb{P}(Y=1|T) = \max\{0,1 - (1 + 7 -t)e^{-(7-t)}\} $$
as expected.


问:
Classification of groups of order $56$.
The particular problem that I am facing is that when $H$ is a normal subgroup of order $7$ and $K$ is a subgroup of order $8$ .Now,  we can consider the map $\theta :K \to Aut(H)$ we see that $\theta(k) \to khk^{-1}$ and $khk^{-1}=h^{-1}$.

I am unable to proceed after this .I need to prove that :

$a)$ when $K$ is isomorphic to $Z_2 \times Z_2 \times Z_2$ there is only one group which is non Abelian.

$(b) $ when $K = Z_4 \times Z_2$ there are two non isomorphic  groups.

$(c) $ when $K=Z_8$ there is only one group

答:
${\rm Aut}(H) \cong C_6$ is cyclic of order $6$. So in all cases, the non-trivial homomorphisms $\theta$ all have image of order $2$ and kernel $L$ of order $4$.

So you just need to find the normal subgroups $L$ of $K$ of order $4$, and then determine which of these give rise to isomorphic groups $H \ltimes K$.

In Case a), we have $L \cong C_2 \times C_2$, and all such $K$ give rise to isomorphic groups $H \ltimes K$ (note that all such $L$ are equivalent under ${\rm Aut}(K)$).

In Case b), we can have $L \cong C_4$ or $L \cong C_2^2$. The $C_2^2$ is unique. There are two copies of $C_4$, but they give isomorphic $H \ltimes K$.

In Case c) we must have $L=C_4$.

Title: Classification of groups of order $56$.


问:
Why does this Infinite Series have contradictory convergences?
In my Calculus II homework, I encountered the following exercise:

If the $n$th partial sum of a series $\sum_{n=1}^\infty a_n$ is
$$s_n = \frac {n-1}{n+1}$$
find  $a_n$ and $\sum_{n=1}^\infty a_n$.

I solved the exercise this way: ( I took $S_n=(n-1)/(n+1)$ to be equation (1))

Equations (2) and (3) answer the exercise's questions. However, when I tried to corroborate my answer in equation (3) by using the found $a_n$ in equation (2), I encountered an inconsistency that I haven't yet been able to harmonize. This is what I tried to do:

Equations (3) and (4) should have yielded the same answer but, this disparity I have been so far unable to harmonize. Your kind comments on what to do will be greatly appreciated.

答:
This is not an answer, just a small addition to the post of player3236:

As already in the post of player3236 mentioned $s_n$ is only definied for all $n\in\mathbb{N}$. So you can't calculate $s_n-s_{n-1}$ for all $n\in\mathbb{N}$. But you can calculate
$$s_{n+1}-s_n=a_{n+1}$$
without changing the range of n. By a similar calculation as you did above you will receive $a_{n+1}=\frac{2}{(n+1)(n+2)}$ for all $n\in\mathbb{N}$. Now you still have to find $a_1$, which can either be done directly by the given formular $a_1=s_1=\frac{1-1}{1+1}=0$ or by using the formular for $a_{n+1}$
$$s_2=\frac{1}{3}=\frac{2}{(1+1)(1+2)}=a_2
$$
and thus $a_1=0$ since $s_2=a_2+a_1$.


问:
Check if Cantor’s Theorem is applicable in the following examples. Also, find $\bigcap_{n\in\Bbb{N}} F_n$ in each case
So we have to find whether Cantor's Theorem is applicable to the following
(i) $F_n = [n, \infty)$ (ii) $F_n = (0, \frac{1}{n})$ (iii) $F_n = [1 − \frac{1}{n} , 2 + \frac{1}{n}]$

Now I've tried doing these as

$F_n$ is non-empty and closed and $[n+1,\infty)\subset[n,\infty)$. But since $n\to\infty$, the length of $F_n$ does not tend to $0$. So not applicable

$F_n$ is an open interval so not applicable

$F_n$ is non-empty and closed. But as $n\to\infty$, $F_n\to [1,2]$ and length $[1,2]$ is not equal to $0$. So not applicable.

Is my method correct or am I doing something wrong??

答:
Note that the exact statement of Cantor’s Intersection Theorem is the following:

The intersection of any decreasing sequence of non-empty nested compact sets is non-empty.

Thus, as you correctly stated, it is not applicable in cases a) and b). However, it is clearly applicable in case c), since that sequence satisfies all of the assumptions of Cantor’s Theorem.


问:
Classifying map
Let $\xi=(E,p,B)$ a principal $G$-bundle and $\eta=(P,\pi,Q)$ a real vector bundle such that $\operatorname{rank}(\eta)=n$. We can consider a classificant space $BG$. What is the classifying map $f:X \rightarrow BG$? Why the name "classifying"? How can build classifying map for $\eta$? </p>

答:
Given any principal $G$-bundle $E' \to X'$, and a map $X \to X'$, the pullback $E' \times_X' X \to X$ is a map that's easily seen to be a principal $G$-bundle.  The idea of the classifying space is that it's a space $BG$, with a principal $G$-bundle $EG \to BG$, such that any principal $G$-bundle over any space $X$ is a pullback of this bundle.  Thus, the principal $G$-bundles over $X$ (up to isomorphism) are in bijective correspondence with the maps $X \to BG$ (up to homotopy).  That is, maps to $BG$ 'classify' principal $G$-bundles.

In many cases, once you prove that such a space exists (which isn't at all obvious), you don't pay much attention to the explicit construction of classifying maps.  Indeed, part of the point of the thing is to replace a somewhat opaque theory -- the study of vector bundles and principal bundles -- with a theory which has more machinery developed for it -- the study of homotopy classes of maps.  This is just my opinion, though -- your mileage may vary.

One way to construct $BG$ is to let $EG$ be any contractible space with a free $G$-action, and $BG = EG/G$.  Clearly $EG \to BG$ is a principal $G$-bundle, and the contractibility of and freeness of the $G$-action on $EG$ implies that there are no obstructions to defining a $G$-map from a principal $G$-bundle to $EG$, which descends to a map $X \to BG$.  For more details, see e.g. these notes.

Now, this answers your questions about principal $G$-bundles.  The secret to the last question is that real vector bundles really are principal bundles as well -- they're principal $O(n)$-bundles!  To turn a vector bundle $E \to X$ into a principal $O(n)$-bundle, replace it by the frame bundle $\mathrm{Fr}(E) \to X$, which is a bundle whose fiber over a point $x \in X$ is the set of orthonormal bases for $E_x$.  (You have to pick a metric on the vector bundle first, but this always exists if $X$ is, for example, paracompact.)  Conversely, given a principal $O(n)$-bundle $P \to X$, you can replace it by the bundle whose fibers are $P_x \times \mathbb{R}^n$ mod its diagonal $O(n)$-action, which is a real vector bundle.  These mappings are mutually inverse, so when we talk about real vector bundles, we might as well be talking about principal $O(n)$-bundles.

So real rank $n$ vector bundles are classified by $BO(n)$, which there actually is a nice geometric construction of.  Let $EO(n)$ be the space of sequences of $n$ orthonormal vectors in an infinite-dimensional inner product space.  This clearly has a free $O(n)$-action, and you can check that it's contractible.  The quotient $EO(n)/O(n)$ is the space of $n$-dimensional planes (we've forgotten their bases) in this infinite-dimensional space, also called the Grassmannian $\mathrm{Gr}_n(\mathbb{R}^\infty)$.  In particular, one-dimensional vector bundles are classified by $\mathrm{Gr}_1(\mathbb{R}^\infty) = \mathbb{R}P^\infty$.

Similar remarks apply to other other sorts of vector bundles.  For example, complex rank $n$ vector bundles are the same as principal $U(n)$-bundles, and these are classified by the Grassmannian of copies of $\mathbb{C}^n$ in $\mathbb{C} P^\infty$.

ContentLicense: CC BY-SA 3.0

Title: Classifying map


问:
What is the invariance principle?
Let $X_1,X_2,...,X_n$ be a random sample , $0<p<1$

$p_X(x)=(1-p)^{x-1}p$, $x=1,2,...$

I have worked out the MLE and have shown with further working that it is a maximum, but the next part of the question asks Find the maximum likelihood estimator for $\mathbf{θ=\frac{1}{p}}$. I think the invariance principle is required for this part?

Could someone explain what the invariance principle is? Many Thanks

答:
the invariance principle simply says that

"if $\hat{p}$ is the MLE for $p$ then $g(\hat{p})$ is MLE for $g(p)$"

thus

$$\hat{\theta}_{ML}=\overline{X}_n$$

Title: What is the invariance principle?


问:
Eccentricity of the conic $x^{2}-4xy-2y^{2}+10x+4y=0$?
Q:

How do I find the eccentricity of the conic $x^{2}-4xy-2y^{2}+10x+4y=0$?

Doubt:

How do I solve this, because it is not like the normal hyperbola but also not like the $xy=c^2$ type.

答:
How do I find the eccentricity of the conic
\begin{align} x^2-4xy-2y^2+10x+4y&=0 \tag{1}\label{1} \end{align}

A general equation is of the form

\begin{align}
Ax^2+Bxy+Cy^2+Dx+Ey+F&=0
\tag{2}\label{2}
\end{align}

represents a hyperbola, if the its discriminant $\Delta=B^2-4AC>0$.
In case of \eqref{1} we have

\begin{align}
A&=1
,\quad
B=-4
,\quad
C=-2
,\quad
D=10
,\quad
E=4
,\quad
\text{and }\quad F=0
,
\end{align}

hence

\begin{align}
\Delta&=(-4)^2-4\cdot1\cdot(-2)
=24>0
,
\end{align}
so, the equation \eqref{1} indeed represents a hyperbola
which the eccentricity is given by

\begin{align}
e&=
\sqrt{\frac{2\sqrt{(A-C)^2+B^2}}{\eta(A+C)+\sqrt{(A-C)^2+B^2}}}
\tag{3}\label{3}
,
\end{align}
where $\eta=1$
if the determinant of the matrix
\begin{align}
M&=
\begin{bmatrix}
A         & B/2 & D/2 \\
B/2 & C         & E/2 \\
D/2 & E/2 & F
\end{bmatrix}
\tag{4}\label{4}
\end{align}
is negative,
and $\eta=-1$
if the $\det(M)>0$.

For \eqref{1} we have

\begin{align}
M&=
\begin{bmatrix}
\phantom{-}1 & -2 & 5 \\
-2 & -2 & 2 \\
\phantom{-}5 &  \phantom{-}2 & 0
\end{bmatrix}
\tag{5}\label{5}
,\\
\det{M}&=6>0
,
\end{align}
hence $\eta=-1$ and

\begin{align}
e&=\sqrt{\tfrac53}
\approx 1.29
.
\end{align}


问:
Where does the 3rd power of the denominator of the curvature of the graph of a function come frome?
According to Wikipedia, the curvature of the graph of a function $f$ is given by the following ratio (assuming second-differentiability).

$$\mathscr{k}_f(x) = \frac{f^{\prime\prime}(x)}{(1 + [f^{\prime}(x)]^2)^{\frac{3}{2}}}$$

The denominator of $\mathscr{k}_f(x)$ is nearly of the form $(a^2 + b^2)^{\frac{1}{2}}$ that makes me think about it as the radius of a circle, but this intuition is cut short by taking that "radius" to the third power. Where does this third power come from in the derivation or definition of this functional?

答:
It is the numerator that should be looked at as it influences shape more strongly.

To me it appears that you are thinking more on rotation up from the vicinity's lowest tangent position at bottom position of a parabola, circle, catenary etc.

Curvature is defined as the speed or rate at which slope changes. When the rotation $\phi$ goes all the way up to 90 degrees then  to include same curvature it  is necessary to augment/ modify the parabola curvature  away from the starting “shallow”   situation.

The curvature is not proportional  to $\sec \phi$  but should be proportional to $\sec^3 \phi$

If that change is not made, it keeps getting flatter and flatter.E.g., if we had only the ode

$$ \dfrac{y''}{\sqrt{1 +y^{‘2}}}$$

then on integration we find this to be a catenary that refuses to curl up. With a circle,  curvature is increased to:

$$ \dfrac{y’’}{[1 +y^{‘2}]^\frac32}$$

Similarly leave alone the numerator and looking at the denominator as sum of two squares $ dx^2+dy^2 $  for constant

$$ \dfrac{y’’}{[1 +y^{'2}]}$$

we have a catenary of uniform strength where vertical limbs straighten up fast by this slope

In the more exaggerated case of a Cornu's spiral for example where the curvature is proportional to the arc length

$$ \dfrac{y’’}{[1 +y^{‘2}]^\frac32}= s \text{(=arc length )} $$

it curls up almost indefinitely like a creeper or ivy.

The behavior  of slope / curvature would be clear on seeing graphs of some of these curves very roughly sketched:


问:
$\Omega=\{1,2,3,4\}$. Consider the events $E = \{1,2\}, F = \{1,3\}$ and $G = \{3,4\}$,Which option correct regarding independence of events?
Let $P \ $ be a probability function that assigns the same weight to each of the points of the sample
space $\Omega = \{1,2,3,4\}$. Consider the events $E = \{1,2\}, F = \{1,3\}$ and $G = \{3,4\}$. Which of the following statement(s) is (are) true?

$(1)$ $E$ and $F$ are independent

$(2)$ $E$ and $G$ are independent

$(3)$ $F$ and $G$ are independent

$(4)$ $E$, $F$ and $G$ are independent

I am not sure what they are asking. I know for independence we need to show $P(A \cap B)=P(A)P(B)$
but I am not able to understand events here. Does that mean event $E$ is selecting $1$ and $2$?

$P(\text{selecting $1$} )=\dfrac{1}{4}=P(\text{selecting $2$} )$

Therefore $P(E)=\dfrac{1}{16}=P(F)$

But $E\cap F = \{1\}$ therefore $P(E\cap F)=\dfrac{1}{4}\ne P(E)P(F)$. Is it true that they are not independent?

答:
Let $ \ $ be a probability function that assigns the same weight to each of the points of the sample space $Ω = \{1,2,3,4\}$.

This make me thinking that if $E = \{1,2\}$ then $P(E) = \frac{1}{2}$ because the probability of choosing one element of $E$ under the assumption above is exactly $\frac{1}{2}$.

To be more clear, if you had a subset $K \subseteq \Omega$ such that $|K| = 3$ then I would think that $P(K) = \frac{3}{4}$ and notice that this assumption is reasonable since if $|K| = 4$ then $K = \Omega$ and $P(K) = P(\Omega) = 1$

Now you can go on checking when $P(A\cap B) = P(A)P(B)$ and check which answer is true.


问:
Does $\forall(S) [S\subseteq S$ ] generate an infinite chain of subsets? If yes, is this a problem?
In a discussion I've  had recently regarding basic notions of set theory, my interlocutor pointed out what seemed to him a " regression ad infinitum".

His point was related to the fact that any set is a subset of itself. Say we consider the set $S=\{a,b\}$.

One of $S$'s subsets is $S$ itself : $S\in P(S) = \{ \emptyset, \{a\} , \{a\} , S\}$   And $S$ as subset of itself still has $S$ has subset.

It seems to generate an infinite descending  chain of subsets.

Hence two questions :
(1) is there actually an infinite chain of subsets?
(2) in case there is such a chain, does it pose any problem?

Or, to ask my question differently : suppose you represent the subset relation by a diagram , starting at bottom with the empty set , included in singetons $\{a\}$ and $\{b\}$, which in turn are included in $S$. At the end of the diagram there is an arrow from S to itself ( due to the reflexivity of the subset relation ). Is the whole diagram " duplicated" at its top, indefinitely?

At first I suspected the " objection " arose from a confusion between inclusion and membership relation . But I'm not sure it is the case.

答:
One of $S$'s subsets is $S$ itself $[\dots]$ and $S$ as subset of itself still has $S$ as subset.

You just repeated it twice.


答: Similarly, what's wrong with $1 \le 1 \le 1 \le \dots$ (or where $\le$ is replaced by $\ge$)? If you're happy with this, then you should be happy with the sets.

Title: Does $\forall(S) [S\subseteq S$ ] generate an infinite chain of subsets? If yes, is this a problem?

Body:
In a discussion I've  had recently regarding basic notions of set theory, my interlocutor pointed out what seemed to him a " regression ad infinitum".

His point was related to the fact that any set is a subset of itself. Say we consider the set $S=\{a,b\}$.

One of $S$'s subsets is $S$ itself : $S\in P(S) = \{ \emptyset, \{a\} , \{a\} , S\}$   And $S$ as subset of itself still has $S$ has subset.

It seems to generate an infinite descending  chain of subsets.

Hence two questions :
(1) is there actually an infinite chain of subsets?
(2) in case there is such a chain, does it pose any problem?

Or, to ask my question differently : suppose you represent the subset relation by a diagram , starting at bottom with the empty set , included in singetons $\{a\}$ and $\{b\}$, which in turn are included in $S$. At the end of the diagram there is an arrow from S to itself ( due to the reflexivity of the subset relation ). Is the whole diagram " duplicated" at its top, indefinitely?

At first I suspected the " objection " arose from a confusion between inclusion and membership relation . But I'm not sure it is the case.


问:
What is the probability that the dice drawn was one with $6$ points on all sides?
(a) In a container there are $10$ real dice, and $10$ dice, at which each of the six sides occurs with the same probability in a throw, but at $5$ of these dice two sides each have $2$, $4$ and $6$ points and with the last $5$ dice all sides have $6$ points. One of these $20$ dice is drawn at random from the container. This die is thrown $n$ times and a $6$ is thrown each time (without seeing the other sides). What is the probability that the dice drawn was one with $6$ points on all sides?

(b) How many of the real dice (with $6$ different sides) have to be in the container so that the probability that the dice drawn is a real one is equal to the probability that one with $6$ points on all sides was drawn? (Again under the fact that $6$ was rolled $n$ times.) How many is that for $n = 2$?

$$$$

For question (a) :

We consider the events :

S = A "6" is thrown

A = The "6" was from one of the dice with $6$ points on all sides

B = The "6" was from one of the dice with two sides each have $2$, $4$ and $6$ points

C = The "6" was from one of the normal dice

The probability that the "6" we got is from one of the dice with $6$ points on all sides is equalto
\begin{align*}P(A\mid S)&=\frac{P(A\cap S)}{P(S)}=\frac{1}{P(S\mid A)\cdot P(A)+P(S\mid B)\cdot P(B)+P(S\mid C)\cdot P(C)}\\ & =\frac{1}{1\cdot \frac{5}{20}+\frac{2}{6}\cdot \frac{5}{20}+\frac{1}{6}\cdot \frac{10}{20}}\\ & =\frac{1}{ \frac{1}{4}+\frac{1}{12}+\frac{1}{12}}\\ & =\frac{1}{ \frac{5}{12}}\\ & =\frac{12}{5}\end{align*} I must have done something wrong since the result is greater than  $1$

答:
The formula you used is correct, however you wrote $P(A \cap S) = 1$ in the numerator which is not true. It holds that $P(A \cap S) = P(A)$ in this case because $S$ will automatically occur when $A$ is true (or in "mathematical terms": $A \subset S$). Therefore your numerator should be $\frac{5}{20}$ instead. With that you would get:
$$\frac{\frac{5}{20}}{\frac{5}{12}} = \frac{12}{20} = 0.6$$


问:
What does this bracket notation mean?
I am currently taking MIT6.006 and I came across this problem on the problem set. Despite the fact I have learned Discrete Mathematics before, I have never seen such notation before, and I would like to know what it means and how it works, Thank you:
$$ f_3(n) = \binom n2$$
(Transcribed from screenshot)

答:
$\binom{n}{k}$ usually denotes the Binomial coefficient which is explicitly given as $$\binom{n}{k} = \frac{n!}{k!(n - k)!}$$
for non-negative integers $k \le n$.

Title: What does this bracket notation mean?


问:
How to prove that sequence spaces $l^{p}, l^{\infty}$ and function space $C[a. b]$ are of infinite dimension
I am studying about the sequence space $l^{p}, l^{\infty}$ and function space $C[a. b]$. It is mentioned in the book that all of these spaces are of infinite dimension. I want to prove that these spaces are of infinite dimension. I am not finding a way to proceed.

Any help and suggestions would be helpful to me.

Thanks

答:
You can exhibit an infinite linearly independent family:

For $\ell^p$ or $\ell^{\infty}$, consider the sequences of the form $(0,\dots,0,1,0, \dots)$.

For $C([a,b])$, consider the functions $t \mapsto e^{ct}$.

Remark: In fact, the second point shows that $C([a,b])$ has dimension $\mathfrak{c}$.


答:
It's the binomial coefficient, which counts the number of 2-subsets of an $n$-element set.

It's computed as $\frac{n(n-1)}{2}$.

It generalises to $n\choose k$.

Title: What does this bracket notation mean?

Body:
I am currently taking MIT6.006 and I came across this problem on the problem set. Despite the fact I have learned Discrete Mathematics before, I have never seen such notation before, and I would like to know what it means and how it works, Thank you:
$$ f_3(n) = \binom n2$$
(Transcribed from screenshot)


答:
It's the binomial coefficient:
$$
\binom{n}{k}=\frac{n!}{k!(n-k)!)}$$

Title: What does this bracket notation mean?


问:
How do I write $ba$ in the form $a^i b^j\ $?

Let $p$ be a prime number and $q \in \Bbb N$ be such that $q\ |\ (p-1).$ Let $u \in \Bbb Z_p^*$ be an element of order $q.$ Now consider the group $G$ defined by $$G = \left \langle a, b\ \big |\ a^p = b^q = e,\ b^{-1} a b = a^u\ \right \rangle.$$ Prove that any element of $G$ can be written as $a^i b^j,$ for some $0 \leq i \lt p$ and $0 \leq j \lt q$ and hence $|G| = pq.$

My question is $:$ How do I write $ba$ in the form of $a^ib^j\ $? Can anybody please help me in this regard.

Thanks for your time.

Body: Since $b^{-k}ab^k=a^{u^k}$, then we have $ba=b^{-(q-1)}ab^{q-1}b=a^{u^{q-1}}b$.


问:
How to find the length of the projection of $v$ onto $u$ without knowing the dot product formula?
Let's say I have two $2$-D vectors, $v$ and $u$. I want to find the length of the projection,$P$, of $v$ onto $u$ as depicted by the red arrow in the figure below:

Now I know this problem becomes trivial if I use the dot product. However, I'm still sort of struggling to understand the dot product so I want to see if I can arrive to the answer without using it (and maybe the dot product will pop out along the way).

I've been trying to use the Law of Cosines as my starting idea:
$$\vert\vert{u-v}\vert\vert^2 = \vert\vert{u}\vert\vert^2 + \vert\vert{v}\vert\vert^2 -2\vert\vert{u}\vert\vert*\vert\vert{v}\vert\vert\cos{\theta}$$

I also know that:

$$cos{\theta} = \frac{P}{\vert\vert{v}\vert\vert}$$

Therefore:

$$P = \frac{\vert\vert{u}\vert\vert^2 + \vert\vert{v}\vert\vert^2 - \vert\vert{u-v}\vert\vert^2}{2\vert\vert{u}\vert\vert}$$

答:
Let $\theta$ be the angle between $u$ and $v$. Then the projection of $v$ onto $u$ is $v \cos \theta$, and $\cos \theta$ can be found using the law of cosines to get:

$$\text{length of projection} = v \cdot\frac{||u||^2 + ||v^2|| + ||u-v||^2}{2 \times||u||\times||v||}$$


问:
Bolza example like Question
I have to find $u$ minimizing $\int_0^1 F$ with $F(x,u(x),u'(x)) = (1-(u'(x))^2)^2+(u(x))^2$ with $u(0) = 0$ and $u(1) = 1$.

I'm relatively new to CoV and got told i should try Euler-Lagrange-Equation which gives me
$2u(x) + 4u''(x) - 12{(u'(x))}^2 \cdot u''(x) = 0$.

But then I don't have a clue how to solve this ODE. Can you please tell me if this is the right way and I should learn solving (even ugly) ODEs?

Thank you very much in advance,
JohnDoe746

答:
You can use approximation methods such direct variational method if there is no exact solution. Let $\hat u(x)$ be the approximate solution
$$\hat u(x)=a\sin\bigg(\frac{\pi x}2\bigg)$$
which satisfies the boundary conditions $\hat u(0)=0$ and $\hat u(1)=1$. Then
$$I=\int_0^1\bigg((1-\hat u'(x)^2)^2+u(x)^2\bigg)dx=\int_0^1\bigg(\bigg(1- \bigg(\frac{a\pi}2 \cos\frac{x\pi}2 \bigg)^2\bigg)^2+\bigg(a\sin\frac{\pi x}2\bigg)^2\bigg)dx$$

You can use Wolfram Alpha to find the solution of integral such as
$$I=\frac{3\pi^4}{128}a^4-\frac{\pi^2-2}4a^2+1$$
To minimize the integral wrt to $a$
$$\frac{d\,I}{d\,a}=\frac{3\pi^4}{32}a^3-\frac{\pi^2-2}2a=0$$
which has the solution
$$a_1=0\qquad a_{2,3}=\pm \frac 4{\pi^2 \sqrt{3}}\sqrt{\pi^2-2}$$
If you reevaluate the integral with
$$I(0)=1$$
$$I\bigg(\pm \frac 4{\pi^2 \sqrt{3}}\sqrt{\pi^2-2}\bigg)=0.58$$
which means that you can approximate with
$$\hat u(x)=\bigg(\pm \frac 4{\pi^2 \sqrt{3}}\sqrt{\pi^2-2}\bigg)\sin\bigg(\frac{\pi x}2\bigg)$$
to minimize the integral.

Title: Bolza example like Question


问:
Prove that the equation $a^2+ 2a−3 =b^2$ does not have a solution in positive integers $a$ and $b$
Prove that the equation $a^2+ 2a−3 =b^2$ does not have a solution in positive integers.

I have tried proving this by contradiction but I can't get to work. Can someone please help me understand how to prove this and how to write this proof?

答:
Assume that there exist positive integers $a$ and $b$ satisfying your requirements. Then:
$$a^2+2a+1-4 = b^2 \implies (a+1)^2-4=b^2$$
Thus, if $x=a+1$ and $y=b$, then $x^2-y^2=4$ where $x$ and $y$ are positive integers. Can you see why this is impossible?


问:
Limit of $E(X_n-X\vert\mathcal F_n)$ or how to handle the fact that the conditional changes
Consider real RVs $X_n$ that converge to $X$ a.s. and are dominated by an integrable $Y$, so $\lvert X_n\rvert\le Y$ a.s. for all $n$ and $E(\lvert Y\rvert)<\infty$. Now let $\mathcal F_n$ be a filtration. I want to prove that
$$\lim_{n\rightarrow\infty}E(X_n-X\vert\mathcal F_n)=0$$
both in $L^1$ and almost surely. I have no clue how to handle the fact that the $\sigma$-Algebra in the conditional expectation changes as well over $n$. I'm not even sure I can pull in the limit at all in this case.

Is there a tool/trick to work around the fact that the conditional changes as well?

答:
By the triangle inequality, we have

$$|E(X_n-X \mid \mathcal{F}_n)| \leq E(|X_n-X| \mid \mathcal{F}_n).$$

Using the tower property, we find that

$$E\big[  \big|E(X_n-X \mid \mathcal{F}_n)\big| \big] \leq E(|X_n-X|).$$

Because of the assumptions on $X_n$, the dominated convergence theorem shows that the right-hand side converges to $0$, and so $E(X_n-X \mid \mathcal{F}_n) \to 0$ in $L^1$.

To prove almost sure convergence, we use Lévy's upward theorem. Fix $m \in \mathbb{N}$. Clearly,

$$|E(X_n-X \mid \mathcal{F}_n)| \leq E \left( \sup_{k \geq m} |X_k-X| \mid \mathcal{F}_n \right)$$

for $n \gg 1$. Set $\mathcal{F} := \sigma(\mathcal{F}_n; n \in \mathbb{N})$, then Lévy's upward theorem yields

$$\limsup_{n \to \infty} |E(X_n-X \mid \mathcal{F}_n)| \leq E\left( \sup_{k \geq m} |X_k-X| \mid \mathcal{F} \right) \tag{1}$$

almost surely. Since

$$\sup_{k \geq m} |X_k-X| \xrightarrow[]{m \to \infty} 0$$

and

$$\sup_{k \geq m} |X_k-X| \leq 2 |Y| \in L^1,$$

the dominated convergence theorem (for conditional expectations) entails that the right-hand side of $(1)$ converges almost surely to $0$ as $m \to \infty$. This finishes the proof.

In case that you are not familiar with the dominated convergence theorem for conditional expectations, you can look it up for instance, in the probability book by Chow & Teicher (Theorem 2 in Section 7.1) or in the measure theory book by Schilling (Corollary 27.15, 2nd ed).


答:
The reason that a condition in i. fails, is to note that for all $n$, $\operatorname{diam}(F_n)=+\infty$ and it follows from that that $\lim_{n \to \infty} \operatorname{diam}(F_n) =+\infty$ in particular, so the diameters do not tend to $0$ as they should.

Another condition in ii. fails because $F_n$ is not closed (that it's an open interval is not relevant, as open does not always imply it's not closed). In both i. and ii. we have $\bigcap_n F_n = \emptyset$, so the conclusion fails too, showing the necessity of the diamater resp. closedness conditions.

For iii. again $\operatorname{diam}(F_n) = 1+\frac{2}{n} \to 1$, so it does not apply (limit not $0$). But indeed $\bigcap_n F_n = [1,2]$, as you claimed. So the conclusion fails too: the intersection is not a singleton.

You should be more precise in checking the conditions. That the conclusion of the theorem fails indicates that not all conditions of the theorem are fulfilled, but you're asked to say which condition exactly fails how...

Title: Check if Cantor’s Theorem is applicable in the following examples. Also, find $\bigcap_{n\in\Bbb{N}} F_n$ in each case

Body:
So we have to find whether Cantor's Theorem is applicable to the following
(i) $F_n = [n, \infty)$ (ii) $F_n = (0, \frac{1}{n})$ (iii) $F_n = [1 − \frac{1}{n} , 2 + \frac{1}{n}]$

Now I've tried doing these as

$F_n$ is non-empty and closed and $[n+1,\infty)\subset[n,\infty)$. But since $n\to\infty$, the length of $F_n$ does not tend to $0$. So not applicable

$F_n$ is an open interval so not applicable

$F_n$ is non-empty and closed. But as $n\to\infty$, $F_n\to [1,2]$ and length $[1,2]$ is not equal to $0$. So not applicable.

Is my method correct or am I doing something wrong??


问:
Operators embedded in second dual as continuous functions on state spaces
Let $A$ be a unital $\mathrm{C}^*$-algebra. There are some well-known facts about the state space $S(A)$ and the set of pure states $PS(A)$. We might remark that in general $PS(A)$ is not weak$*$-closed but if we take the weak star closure, the pure state space, $\overline{PS(A)}^{w*}$, we have have a weak$*$-compact space.

If $f\in A$ is embedded in the second dual $\imath:A\hookrightarrow A^{**}$ via the Gelfand transform (normally only done for commutative $\mathrm{C}^*$-algebras),

$$\widehat{f}(\rho):=\rho(f)\qquad (\rho\in A^*),$$

then the nature of weak*-convergence of a net of states $\rho_\lambda\rightarrow \rho\in A^*$ implies that $\hat{f}$ is a continuous function from $S(A)\rightarrow \mathbb{C}$. The multiplication $\widehat{f}\otimes \widehat{g}\mapsto \widehat{fg}$ in $\imath(A)\subset C(S(A))$ is not the pointwise multiplication.

As an example consider $A=M_2(\mathbb{C})$, whose state space is parameterised by the closed unit ball in $\mathbb{R}^3$. It turns out that $\imath(A)\subset C(S(A))$ consists of all the affine functions, $(x,y,z)\mapsto ax+by+cz+d$.

Question 1: In general, is there any hope of generally describing
which continuous functions arise in $\imath(A)\subset C(S(A))$?

If we restrict $\widehat{f}$ to $\overline{PS(A)}^{w*}$ we have a continuous function $\tilde{f}\in C(\overline{PS(A)}^{w*})$ on the pure state space.

If we restrict $\tilde{f}$ to $f_0\in C(PS(A))$, then we can extend $f_0$ by linearity to the (not closed) convex hull $\operatorname{co}PS(A)$.

Question 2: Is every continuous function  $\widehat{f}\in\imath(A)$
determined by its restriction to $PS(A)$ (or perhaps
$\overline{PS(A)}^{w*}$), in the sense that it is determined uniquely
by linear/convex extension?

In the finite commutative case, the pure states $PS(A)$ are a finite set, every function $F\in C(S(A))$ is of the form $\widehat{f}$ for some $f\in A$, and all of these functions are got by linearly extending $F_{\left|PS(A)\right.}$ to $S(A)$.

I understand there are studies that try and recover a $\mathrm{C}^*$-algebra from the state and structures on it. I am not trying to be so comprehensive: just trying to describe a way to consider elements of a noncommutative $\mathrm{C}^*$-algebra as continuous functions on a compact Hausdorff space.

答: Thanks to Ruy I was able to find some references such as Landsman (p.18-19) which discusses an answer to Q. 1 and Q. 2 is yes.</p>


问:
how many solutions are possible for the system $x+y+z+w=30$, where $0≤x, 0≤y≤3, 0≤w≤3, 0≤z≤2$?
I want to find the number of solutions for the system $x+y+z+w=30$, where $0≤x, 0≤y≤3, 0≤w≤3, 0≤z≤2$.

I know that we can choose $x$ based on the sum of $w+y+z$, so I tried to calculate to number of solutions to the system $w+y+z=k$
where $0≤k≤8$ but I got stuck.

Any help would be appreciated.

答:
It's $4\times 4\times 3=48$.

This is because once you know what $y$, $w$ and $z$ are, then $x$ is determined. The restrictions on $y$, $w$ and $z$ give you $48$ choices, so there are $48$ in total.


问:
The crumpled paper problem
I saw a theorem recently (and have seen similarly phrased problems before), which I found quite surprising. Unfortunately, I'm not sure what its name is (or if it has one). I'll refer to it as the "crumpled paper problem". It goes something like this:

Suppose you were to take two pieces of paper, identical in size and shape, and place one on top of another such that all 4 corners and all 4 edges line up. Now, every point along the top sheet can be "mapped" to its corresponding point on the bottom sheet. If you were to then crumple up the top sheet of paper, and toss it on top of the bottom sheet, there necessarily exists a point on the top sheet that lies precisely above its "mapping point" on the bottom sheet, no matter how it was crumpled or where it was placed.

This result, to me, at least, is very surprising. Intuitively, it seems to me that it must be that there is some way to crumple the paper, or location to toss it, such that no points line up. I can sort've imagined progressively folding the paper (analogous to crumpling), and kind of see how this might be, however, it's still quite challenging.

My question is, how is this proven, does the problem have a name, and to what field of mathematics does this belong? I'm very curious how this can be shown.

I've added those tags which I thought may be relevant. Unfortunately, I have no clue to what precise area of mathematics this belongs to. Please do correct tags as necessary.

答:
It is all about continuity of the crumpling transform. The 1D case is easier to understand.

If you plot the abscissa of the points of the crumpled line vs. the abscissa of the corresponding point in the flat line, you get a continuous function in the range $[0,1]$. As the graph of the function splits the unit square horizontally and is continuous, it must meet the diagonal of the square.

(In other terms, the crumpled abscissa starts larger and ends smaller than the straight abscissa, so it must be equal somewhere.)

In the 2D case, you can plot both the crumpled abscissa and ordinate as a function of the flat abscissa and ordinate, which yield two surfaces in a unit cube. These surfaces meet the oblique planes $x=u$ and $y=v$ along two continuous curves, and these curves meet in at least one point.

This works even if the paper is elastic and can be stretched (i.e. topological matter).

Title: The crumpled paper problem


答: Hint: You need to find infinite linearly independent familiy in each of these spaces. I suggest you to search them among functions with disjoint supports.

Title: How to prove that sequence spaces $l^{p}, l^{\infty}$ and function space $C[a. b]$ are of infinite dimension

Body:
I am studying about the sequence space $l^{p}, l^{\infty}$ and function space $C[a. b]$. It is mentioned in the book that all of these spaces are of infinite dimension. I want to prove that these spaces are of infinite dimension. I am not finding a way to proceed.

Any help and suggestions would be helpful to me.

Thanks


问:
Support Projection for Characters in $\mathrm{C}^*$-Algebra with finitely many characters
Related to this question...

For a normal state $\varphi$ on a von Neumann algebra $\mathcal{M}$ there is a projection $p_\varphi$ with some nice properties called its support. Some properties include that for all: $f\in \mathcal{M}$:
$$\varphi(f)=\varphi(fp_\varphi)=\varphi(p_\varphi f)=\varphi(p_\varphi fp_\varphi),$$
and $\varphi(p_\varphi)=1$. Also, for $q_\varphi:=1_{\mathcal{M}}-p_\varphi$, and all $f\in \mathcal{M}$ we have
$$\varphi(q_\varphi)=\varphi(fq_\varphi)=\nu(q_\varphi f)=0.$$
EDIT: It is the smallest such projection with these properties.

Now suppose that $A$ is a (infinite dimensional, separable, generated by projections) $\mathrm{C}^*$-algebra with finitely many characters. Is there a projection $p_C$ such that for all characters (non-zero *-homomorphisms $\varphi:A\rightarrow \mathbb{C}$) and all $f\in A$  we have

$$\varphi(f)=\varphi(fp_C)=\varphi(p_C f)=\varphi(p_C fp_C),$$
and $\varphi(p_C)=1$. Also, for $q_C:=1_{A}-p_C$, and all $f\in A$ we have
$$\varphi(q_C)=\varphi(fq_C)=\nu(q_C f)=0?$$

答:
A correspondent wrote to say that  working with the enveloping von Neumann algebra $A^{**}:=\pi_{\text{GNS}}(A)''$, like any state, a character $\varphi\in S(A)$ extends to a normal state $\omega_\varphi$ on $A^{**}$, and thus has a support projection $p_\varphi\in A^{**}$ such that $\omega_\varphi(p_\varphi)=1$. Furthermore, if $\omega_\varphi(p)=1$, then $p_\varphi\leq p$, and also for all $f\in A^{**}$:
$$\omega_\varphi(f)=\omega_\varphi(p_\varphi f)=\omega_\varphi( fp_\varphi)=\omega_\varphi(p_\varphi fp_\varphi).$$
If the character space $\Omega(A)$ is empty, define $p_C=0$. Otherwise, define:
$$p_C=\sup\{\omega_\varphi\,:\,\varphi\in \Omega(A)\}.$$


问:
Asymptotic Expansion of Digamma Function
While reading the wikipedia page of the Digamma function (https://en.wikipedia.org/wiki/Digamma_function#Asymptotic_expansion)

I noticed that it said the asymptotic expansion for the digamma function ($\psi(z)$) can be obtained from using

\begin{equation} \psi(z+1) = -\gamma + \sum_{n=1}^{\infty} (\frac{1}{n} - \frac{1}{n+z}) \end{equation} (where $\gamma$ is the Euler–Mascheroni constant) combined with Euler–Maclaurin formula to conclude

\begin{equation} \psi(z) \approx \log(z) - \frac{1}{2z} \end{equation}   My main confusion about this is that when I tried to use the Euler–Maclaurin formula to approximate the sum as an integral, I could not figure out how to obtain a $\gamma$ to cancel out the $\gamma$ term in the series expansion. Any  help on how to obtain the above relationship using the formula or just where the $\gamma$ comes from when I apply the Euler-Maclaurian formula would be greatly appreciated!

答:
I shall assume that $|\arg (z+1)|<\pi$. By the Euler–Maclaurin formula,
\begin{align*}
\psi (z + 1) = &  - \gamma  + \int_1^{ + \infty } {\left( {\frac{1}{t} - \frac{1}{{t + z}}} \right)dt}  + \frac{1}{2}\left( {1 - \frac{1}{{1 + z}}} \right) \\ & + \int_1^{ + \infty } {\left( {\left\{ {1 - t} \right\} - \frac{1}{2}} \right)\left( {\frac{1}{{t^2 }} - \frac{1}{{(t + z)^2 }}} \right)dt}
\\  = & \; \log (z + 1) - \gamma  + \int_1^{ + \infty } {\frac{{\left\{ {1 - t} \right\}}}{{t^2 }}dt}  - \frac{1}{{2(z + 1)}} - \int_1^{ + \infty } {\frac{{\left\{ {1 - t} \right\} - \frac{1}{2}}}{{(t + z)^2 }}dt}.
\end{align*}
Thus, it remains to show that
$$
\int_1^{ + \infty } {\frac{{\left\{ {1 - t} \right\}}}{{t^2 }}dt}  = \gamma
$$
and
$$
\int_1^{ + \infty } {\frac{{\left\{ {1 - t} \right\} - \frac{1}{2}}}{{(t + z)^2 }}dt}  = \mathcal{O}\!\left( {\frac{1}{{(z + 1)^2 }}} \right).
$$
The first one can be shown as follows:
\begin{align*}
\int_1^{ + \infty } {\frac{{\left\{ {1 - t} \right\}}}{{t^2 }}dt} & =\sum\limits_{k = 1}^\infty  {\int_k^{k + 1} {\frac{{\left\{ {1 - t} \right\}}}{{t^2 }}dt} }  = \sum\limits_{k = 1}^\infty  {\int_0^1 {\frac{{1 - t}}{{(k + t)^2 }}dt} } \\ & = \sum\limits_{k = 1}^\infty  {\left[ {\frac{1}{k} - \log \left( {1 + \frac{1}{k}} \right)} \right]}  = \mathop {\lim }\limits_{n \to  + \infty } \sum\limits_{k = 1}^n {\left[ {\frac{1}{k} - \log \left( {1 + \frac{1}{k}} \right)} \right]}
\\ &
 = \mathop {\lim }\limits_{n \to  + \infty } \left(\sum\limits_{k = 1}^n {\frac{1}{k}}  - \log (n + 1) \right)= \gamma .
\end{align*}
For the second claim, we use integration by parts and some estimations to obtain
\begin{align*}
& \left| {\int_1^{ + \infty } {\frac{{\left\{ {1 - t} \right\} - \frac{1}{2}}}{{(t + z)^2 }}dt} } \right| = \left| {\int_1^{ + \infty } {\frac{{\left\{ t \right\} - \left\{ t \right\}^2 }}{{(t + z)^3 }}dt} } \right| \\ & \le \frac{1}{4}\int_1^{ + \infty } {\frac{1}{{\left| {t + z} \right|^3 }}dt}  = \frac{1}{4}\int_0^{ + \infty } {\frac{1}{{\left| {t + (z + 1)} \right|^3 }}dt}
\\ &
 \le \frac{1}{4}\int_0^{ + \infty } {\frac{1}{{(t + \left| {z + 1} \right|)^3 }}dt} \sec ^3 \left( {\frac{{\arg (z + 1)}}{2}} \right) = \frac{1}{{8\left| {z + 1} \right|^2 }}\sec ^3 \left( {\frac{{\arg (z + 1)}}{2}} \right).
\end{align*}
In summary, replacing $z+1$ by $z$,
$$
\psi (z) = \log z - \frac{1}{{2z}} + R(z)
$$
where
$$
\left| {R(z)} \right| \le \frac{1}{{8\left| z \right|^2 }}\sec ^3 \left( {\frac{{\arg z}}{2}} \right)
$$
and $|\arg z|<\pi$. With more careful considerations, the constant $\frac{1}{8}$ can be replaced by $\frac{1}{12}$.


问:
Span of subsets and span of their union.
Let $S$ and $T$ be two subsets of a vector space $E$. I want to prove

$span(S) + span(T) = span(S \cup T)$.

Here is my approach: let $span(S) = span(v_1, \dots, v_m)$ and $span(T) = span(u_1, \dots, u_n)$.

$span(S) + span(T) \subseteq span(S \cup T)$. Let $w \in span(S) + span(T)$. Then $w = x+y$, where $x \in span(S)$ and $y \in span(T)$. Since $x \in span(S)$, $x = \alpha_1 v_1 + \dots + \alpha_m v_m$ for some $\alpha_i \in \mathbb{R}$, $i=1, \dots, m$. Similarly, $y = \beta_1 u_1 + \dots + \beta_n u_n$ for some $\beta_i \in \mathbb{R}$, $i = 1, \dots, n$. Therefore, since $w = \alpha_1 v_1 + \dots + \alpha_m v_m + \beta_1 u_1 + \dots + \beta_n u_n$, $w \in span(v_1, \dots, v_m, u_1, \dots, u_n$) i.e. (not sure if the following step is right) $w \in span(S\cup T)$.

$span(S \cup T)\subseteq span(S) + span(T)$. Let $w \in span(S \cup T)$, then $w = \alpha_1 v_1 + \dots + \alpha_m v_m + \beta_1 u_1 + \dots + \beta_m u_m = x+y$, where $x \in span(S)$ and $y \in span(T)$. Therefore, $w \in span(S) + span(T)$.

Im not sure if those steps are right, mainly the ones involving the fact that a vector belongs to the union of subsets. Is this correct or is there another way, more elegant maybe, to prove this? Thank you!

答:
That is not correct. You are assuming that $S$ and $T$ are finite, but that is not part of the assumptions of the problem.

Since $S\subset S\cup T$, $\operatorname{span}(S)\subset\operatorname{span}(S\cup T)$. For the same reason, $\operatorname{span}(T)\subset\operatorname{span}(S\cup T)$, and therefore$$\operatorname{span}(S)+\operatorname{span}(T)\subset\operatorname{span}(S\cup T).$$

On the other hand, if $v\in\operatorname{span}(S\cup T)$, then there are vectors $v_1,\ldots,v_k\in S$, there are vectors $v_{k+1},\ldots,v_l\in T$ and there are scalars $\lambda_1,\ldots,\lambda_l$ such that$$v=\overbrace{\lambda_1v_1+\cdots+\lambda_lv_k}^{\phantom{\operatorname{span}(S)}\in\operatorname{span}(S)}+\overbrace{\lambda_{k+1}v_{k+1}+\cdots+\lambda_lv_l}^{\phantom{\operatorname{span}(T)}\in\operatorname{span}(T)},$$and therefore $v\in\operatorname{span}(S)+\operatorname{span}(T)$.


问:
Heat inflow where $u(x,y,z) = 2y^2+2z^2$ , $K=6.5$ through surface $S$ $y^2+z^2=6, \space x\in[0,4]$
I just want to confirm that my calculation is right with the following problem.

I want to find the rate of heat flow where $u(x,y,z) = 2y^2+2z^2$ , $K=6.5$ through surface $S$... $y^2+z^2=6, \space x\in[0,4]$

I know that

$$\int \int_S -K\nabla \vec u\cdot d\vec S$$

is the usual heatflow equation, but the problem asks us to find the heatflow inwards the surface so the answer will simply be the opposite of this expression.

Without nice shortcuts or anything, I parametarized the surface using $y = \sqrt{6}\cos{\theta}$, $z = \sqrt{6}\sin{\theta}$ and $x=x$ so

$$\nabla \vec u = <0,4\sqrt{6}\sin{\theta},4\sqrt6\cos{\theta}>$$

and

$$\vec n = <0,\sqrt6\cos{{\theta},} \sqrt{6} \sin{\theta}>$$

where the dot product is simply $24$.

So, the inflow simply becomes

$$K(24)A$$

where $A$ is the lateral area of the cylinder, which is $8\sqrt{6}\pi$ .

I get the solution $1248\sqrt{6}\pi$, but can anyone else confirm that I got the right number?

答:
$\nabla \vec u = <0,4\sqrt{6}\cos{\theta},4\sqrt6\sin{\theta}>$. You just have a typo as your reversed the order of $y$ and $z$ component when you wrote but your dot product is correct.

$\vec n = \ <0, \sqrt6\cos{{\theta},} \sqrt{6} \sin{\theta}>$ is radially outward vector but given the question we will choose $\vec n = <0, - \sqrt6\cos{{\theta},} - \sqrt{6} \sin{\theta}>$.

If you do your calculation using outward pointing vector, you will have to just reverse the sign in the end as you rightly said.

Now this is where you have a mistake. Your normal vector is not normalized  so multiplying again by surface area will give a wrong result.

Since you have parametrized the surface as $r(x,\theta) = <x, \sqrt6 \cos\theta, \sqrt6 \sin\theta>$

$dS = \vec{n}\ dx \ d\theta $ ($\vec{n} = r'_{x} \times r'_{\theta}$). So your integral is simply,

$\displaystyle 24K \int_0^{2\pi} \int_0^{4} dx \ d\theta = 1248 \pi$


问:
When does the Frobenius Norm equal normalized inner product?
Assume you need to design an image retrieval system where you want
to find similar images to a query image $X \in \Bbb{R}^{N \times M}$ among the database
images $Y_{k} \in \Bbb{R}^{N \times M}$

You have two options, using the Frobenius norm: $\operatorname{dis}(X,Y_{k}) = \left \| X-Y_{k} \right \|_{F}$
or the normalized inner product:
$$\operatorname{sim}(X,Y_{k}) = \frac{\operatorname{tr}(X^{T}Y_{k})}{\left \|X  \right \|_{F}\left \|Y_{k}  \right \|_{F}}$$

My Question is:

Under which condition will a system using the Frobenius norm
yield the same result than a system that uses the normalized inner
product?

Any help is greatly appreciated!

答: Both systems will return the same results if the “image matrices” would already be normalized to unit norm. In this case, one can easily show that the $|| X – Y_k||_F^2 = 2 – 2*tr({X^T * Y_k})$, i.e., they will result in the same ranking.


问:
Upper bound on expected value of $\frac{1}{(1+X)^2}$ where $X$ is binomial
As observed in Find the expected value of $E[\frac{1}{\left(X+1\right)^2}]$ where X is binomial, there exists a closed form solution to $$\mathbb{E}\left[\frac{1}{(1+X)^2}\right]$$ where $X$ is binomial random variable i.e $X \sim \textrm{Bin}(n,p)$. I am interested in proving an upper bound which I expect to be $$\mathbb{E}\left[\frac{1}{(1+X)^2}\right] \leq \frac{c(p)}{n^2}$$ for $n>1$ where $c(p)$ is some constant dependent on $p$. Any ideas on how to approach this?

EDIT: After @NN2's answer, I changed the question from $O(1/n^2)$ to $c(p)/n^2$ for $n>1$.

答:
As $X$ follows binomial distribution, we can write $X=\sum_{i=1}^n Z_i$ with $Z_i$ follows the Bernoulli distribution. Then, according to central limit theorem, we have

$$\frac{1}{n}X =\frac{1}{n}\sum_{i=1}^n Z_i \xrightarrow{n\rightarrow +\infty}N(p,p(1-p))$$

So,

\begin{align}
\mathbb{E}\left[\frac{1}{(1+X)^2}\right] &= \mathbb{E}\left[\frac{1}{n^2}\frac{1}{(\frac{1}{n}+\frac{X}{n})^2}\right] \\
&= \frac{1}{n^2}\mathbb{E}\left[\frac{1}{(\frac{1}{n}+\frac{X}{n})^2}\right] \xrightarrow{n\rightarrow +\infty}\frac{1}{n^2}\mathbb{E}\left[\frac{1}{N^2(p,p(1-p))}\right]\\
\end{align}


问:
One square per person
There are n persons.

Each person draws k interior-disjoint squares.

I want to give each person a single square out of his chosen k, so that the n squares I give are interior-disjoint.

What is the minimum k (as a function of n) for which I can do this?

NOTES:

For n=1, obviously k=1.

For n=2, obviously k must be more than 2, since with 2 squares per person, it is easy to think of situations where both squares of person 1 intersect both squares or person 2. It seems that k=3 is enough, but I couldn't prove this formally.

If we don't limit ourselves to squares, but allow general rectangles, then even for n=2, no k will be large enough, as it is possible that every rectangle of player 1 intersects every other rectangle of player 2. So, the sqauare limitation is important.

EDIT: The problem has two versions: in one version, the squares are all axis-aligned. In the second version, the squares may be rotated. Solutions to any of these versions are welcome.

EDIT: Here is a possibly useful claim, relevant for the axis-aligned version:

Claim 1: If two axis-aligned squares, A and B, intersect, then one of the following 3 options hold:

At least 2 corners of A are covered by B, and B is as large or larger than A;

One corner of A is covered by B, and one corner of B is covered by A,

At least 2 corners of B are covered by A, and A is as large or larger than B.

Thus, if A intersects B, then, out of the 8 corners of A and B, at most 6 corners remain uncovered.

答:
I found an upper bound for the axis-aligned version. It is based on the following claim:

Claim 2: a square A can intersect at most 4 interior-disjoint squares that are as large or larger that A.

Proof: If A intersects B and B is as large or larger than A, then at least one corner of A is covered by B (See Claim 1 at the end of the question). Since A has 4 corners, there can be at most 4 such squares that are interior-disjoint.

Now we can give each person a square of his choice, using the following algorithm:

Find the smallest square out of the $nk$ available squares. If there are more than one - choose one arbitrarily.

Give the smallest square to its owner, send that owner home, and remove all squares of that owner.

Remove all squares that are intersected by the smallest square. By Claim 2, there are at most 4 such squares per person. Thus, each of the n-1 persons has at least k-4 squares. Return to step 1.

If $k = 4n-3$, then the algorithm will eventually terminate with 1 person and 1 square.

This is only half a solution. Can we do better than $4k-3$ in the axis-aligned version?

Title: One square per person


问:
Continuous Matrix-Valued Functions as Bounded Operators
In Theorem 1.3, the following $\mathrm{C}^*$-algebra is considered:

$$A:=\left\{f\in C([0,1],M_2(\mathbb{C})):f(0),\,f(1)\text{ diagonal}\right\}.$$

I want to identify a Hilbert space that these these act on. I want to say that via left multiplication, $g\mapsto fg$, the elements of $A$ are bounded operators on:

$$H:=\left\{f\in L^2([0,1],M_2(\mathbb{C})):f(0),\,f(1)\text{ diagonal}\right\}.$$

Define matrix elements:
$$f(x)=\left(\begin{array}{cc}f_{11}(x) & f_{12}(x)\\ f_{21}(x) & f_{22}(x)\end{array}\right),$$
I want to suggest that each of the $f_{ij}$ are in $L^2([0,1])$ where the norm is got from, e.g.
$$\langle f_{ij},g_{k\ell}\rangle_{L^2([0,1])}=\int_0^1 \overline{f_{ij}(x)}g_{k\ell}(x),$$

and define the inner product on $H$ by:

$$\langle f,g\rangle =\sum_{i,j=1}^2 \langle f_{ij},g_{ij}\rangle_{L^2([0,1])}.$$

Does this actually check out? As in:

Is the representation of $A$ (via left-multiplication) as bounded operators on $H$ faithful?

Or are there some subtleties and confusions that I am missing out on?

Relevant: $A$ is generated by projections
$$p(x)=\left(\begin{array}{cc}1 & 0 \\ 0 & 0\end{array}\right)$$
and
$$q(x)=\left(\begin{array}{cc}x & \sqrt{x(1-x)}\\ \sqrt{x(1-x)} & 1-x\end{array}\right)$$

答: I went with $\pi(A)\subset B(L^2([0,1],\mathbb{C}^2))$ in the end.


问:
Existence of a local maximum on an open ball
Let $f: \mathbb{R}^2 \rightarrow \mathbb{R}$ be a function of class $\mathcal{C}^2$. Let $D$ be an open ball in $\mathbb{R}^2$. We suppose there exists $m_0 \in D$ such that $\forall m \in \partial D, f(m_0) \geq f(m)$.

Show that there exists a local maximum $m_1$ of $f$ in $D$.

My proof so far:

I have shown that $cl(D)$ is a compact set.

Since $f$ is continuous and $cl(D)$ is a compact set, then $f:cl(D) \rightarrow \mathbb{R}$ reaches its supremum.

I guess what I have left to prove is that this supremum or upper bound is in D., but I don't know how.

答: There exist $x$ in $\overline D$ such that $f(x) \geq f(y)$ for all $y \in \overline D$. If $x \in D$ we are done. If $x \in \partial D$ the $f(m_0) \geq f(x) \geq f(y)$ for all $y \in \overline D$ si $m_0$ so a local maximum.


答:
So you already know that $f:cl(D)\to\mathbb{R}$ achieves global supremum. Recall that $\partial D$ and $D$ form a partitioning of $cl(D)$. Now $f(m_0)\geq f(m)$ for any $m\in\partial D$. And so if that global supremum is in $\partial D$ then $m_0\in D$ is a global supremum as well (by our inequality). Otherwise if the global supremum is not in $\partial D$ then it has to lie in the other piece, i.e. $D$.

And so we have that $f:cl(D)\to\mathbb{R}$ has a global supremum inside $D$. Now $D$ is open in $\mathbb{R}^2$ and thus the global supremum is a local supremum of the entire $f:\mathbb{R}^2\to\mathbb{R}$.

Title: Existence of a local maximum on an open ball

Body:
Let $f: \mathbb{R}^2 \rightarrow \mathbb{R}$ be a function of class $\mathcal{C}^2$. Let $D$ be an open ball in $\mathbb{R}^2$. We suppose there exists $m_0 \in D$ such that $\forall m \in \partial D, f(m_0) \geq f(m)$.

Show that there exists a local maximum $m_1$ of $f$ in $D$.

My proof so far:

I have shown that $cl(D)$ is a compact set.

Since $f$ is continuous and $cl(D)$ is a compact set, then $f:cl(D) \rightarrow \mathbb{R}$ reaches its supremum.

I guess what I have left to prove is that this supremum or upper bound is in D., but I don't know how.


问:
Does $\forall(S) [S\subseteq S$ ] generate an infinite chain of subsets? If yes, is this a problem?
In a discussion I've  had recently regarding basic notions of set theory, my interlocutor pointed out what seemed to him a " regression ad infinitum".

His point was related to the fact that any set is a subset of itself. Say we consider the set $S=\{a,b\}$.

One of $S$'s subsets is $S$ itself : $S\in P(S) = \{ \emptyset, \{a\} , \{a\} , S\}$   And $S$ as subset of itself still has $S$ has subset.

It seems to generate an infinite descending  chain of subsets.

Hence two questions :
(1) is there actually an infinite chain of subsets?
(2) in case there is such a chain, does it pose any problem?

Or, to ask my question differently : suppose you represent the subset relation by a diagram , starting at bottom with the empty set , included in singetons $\{a\}$ and $\{b\}$, which in turn are included in $S$. At the end of the diagram there is an arrow from S to itself ( due to the reflexivity of the subset relation ). Is the whole diagram " duplicated" at its top, indefinitely?

At first I suspected the " objection " arose from a confusion between inclusion and membership relation . But I'm not sure it is the case.

答: The confusion, as you suspected, is between $\in$ and $\subseteq$. The axiom of foundation states that no set satisfies $x\in x$. Is such a set was allowed, then it would lead to an infinite membership chain $\dots \in x\in x\in x\in x\dots $. Set inclusion, on the other hand, always satisfies $x\subseteq x$ simply since it states that every member of $x$ is a member of $x$ --- clearly true. On this note one should mention that sets that do satisfy $x\in x$ are studied. This is known as non well-founded set theory.


问:
How do I rewrite the vector equation of a line in symmetric form?
The problem given was "Given a the vector equation $r(t)=(−4−1t)i+(−3+1t)j+(3+5t)k$, rewrite this in terms of the symmetric equations for the line."

(quotient involving $x$)$=\quad -x- 4$

(quotient involving $y$)$=\quad\frac{y+3}{1}$

(quotient involving $z$) ?

I am having trouble figuring out the quotient involving $z$. I intially got $\frac{z-3}{5}$ but that was wrong. I googled how to do problems similiar to this one and it said to write them as parametric equations and then solve for $t$ which worked for the other two. I don't know what I am doing wrong and any help would be appreciated. Thank you !

答:
Re-write $\vec r=(-4\vec i-3\vec j+3\vec k)+t(-1\vec i+\vec j+ 5\vec k)$
So this line passes through the point $(-4,-3,3)$ having direction ratios as $(-1,1,5).$ The equation of liner is:
$$\frac{x+4}{-1}=\frac{y+3}{1}=\frac{z-3}{5}=t.$$


问:
Eigenvalues and eigenvectors of AB and BA, proof.
$A$ is an $n \times k$ matrix and $B$ is an $k \times n$ matrix.

If $v_1, ..., v_l$ are linearly independent eigenvectors of $BA$ corresponding to a single nonzero eigenvalue $c$, then $Av_1, ..., Av_l$ are linearly independent eigenvectors of $AB$ corresponding to $c$.

I need help proving this.

答: Hint: What happens if you left-multiply the equation $BAv_i=kv_i$ with $A$?


问:
Simple limit question involving binomial coefficient $\lim \binom{n}{n-k} (p)^n (1-p)^n$ where $0 \leq p \leq 1$
The question is pretty simple, how does one compute $\lim_{n \rightarrow \infty} \binom{n}{n-k} (p)^n (1-p)^n$ where $0 \leq p \leq 1$ and $0 \leq n-k \leq n$?

Some checking with wolfram (and the context of the problem that this is involved in) tells me that this should go to zero, but I'm not sure how to show this. I have tried bounding this above by $\lim_{n \rightarrow \infty} n! (p)^n (1-p)^n$, but I think this diverges to infinity.

Any help would be appreciated. Thank you!

答:
$\lim_{n \rightarrow \infty} n! (p)^n (1-p)^n$ diverges to infinity, as the factorial grows faster than the exponential function (with base $<1$) decreases to zero. But you don't have just the factorial, you have $$\binom{n}{n-k}=\binom{n}{k}=\prod_{j=1}^k\frac{n+1-j}{j}=\mathcal O(n^k)\qquad\text{for }n\to\infty.$$
And you know, that the exponential function (with base $<1$) decreases faster to zero than any polynomial.

If your $k$ is not fixed, i.e. $k=k(n)$, you could use that the binomial coefficient is taking it's maximum in "the middle":
$$\binom{n}{k(n)}\leq\binom{n}{\lfloor n/2\rfloor}=\mathcal O(2^n)\qquad\text{for }n\to\infty.
$$
But the last fact "$=\mathcal O(2^n)$" is a thing of its own and I don't know if it's easy to proof. One way could be to proof $$\forall n\in\mathbb N:\binom{n}{\lfloor n/2\rfloor}\leq 2^n$$ by mathematical induction.


问:
Find the general solution to the ODE:
I was asked to find the general solution to these ODEs.

$$\frac{dy}{dx}=\frac{(-8x+3y-31)}{(-3x+y-11)}$$
I tried by rearranging to the form $M(x,y)+N(x,y)\frac{dy}{dx}=0$ and let $y=vx$, but what I got was $$(v^2x-6vx-11v+8x+31)dx+x(-3x+vx-11)dv=0$$
and I could not separate $x$ and $v$ for integration.

$$\frac{dy}{dx}=\frac{y}{x}+\frac{9}{2}xexp(-\frac{2y}{x})+\frac{9}{2}xexp(\frac{2y}{x})$$
I have totally no idea which method to be used in Question 2.

May I know if I'm on the wrong track, and how should I solve them? Thanks.

答:
Your equation is $\dfrac{dy}{dx}=\dfrac{(-8x+3y-31)}{(-3x+y-11)}=-\dfrac{8x-3y+31}{-3x+y-11}$ which is an exact differential equation.

The differential form associated to it is $\omega=(8x-3y+31)dx+(-3x+y-11)dy$ and its domain is $\mathbb R$ which is simply connected hence we have the relation $\text{exact}\iff\text{closed}$.

Let's call the fisrt coefficient of the form $A(x,y)$ and the second coefficient $B(x,y)$, then
$$\dfrac{\partial}{\partial_y}A(x,y)=-3=\dfrac{\partial}{\partial_x}B(x,y)\implies\omega\text{ is closed and exact}.$$
The only thing we have to do now is calculating the potential $U(x,y)$ of the form:
$$\int A(x,y)dx=4x^2-3xy+31x+g(y)$$
$$\dfrac{d}{dy}(4x^2-3xy+31x+g(y))=-3x+g'(y)=-3x+y-11\implies g(y)=\dfrac{y^2}{2}-11y+c$$
$$U(x,y)=4x^2-3xy+31x+\dfrac{y^2}{2}-11y+c, c\in\mathbb R.$$
You can now write the solution explicitly or you could study the implicit function defined by the relation $U(x,y)=0$.

Title: Find the general solution to the ODE:


问:
Prove that the equation $a^2+ 2a−3 =b^2$ does not have a solution in positive integers $a$ and $b$
Prove that the equation $a^2+ 2a−3 =b^2$ does not have a solution in positive integers.

I have tried proving this by contradiction but I can't get to work. Can someone please help me understand how to prove this and how to write this proof?

答:
As has already been mentioned, one finds that the problem can be reduced to a statement regarding $x,y$ as being positive integers, and $x^{2}-y^{2}=4$.

Then $(x+y)(x-y)=4$. If $x+y,x-y = 2$, then $x+y=x-y \implies y=-y \implies y = 0 $ which is not positive.

Clearly $x+y > x-y$ so that the other case is $x+y =4$ and $x-y = 1$.

Subtracting these equations $(x+y)-(x-y)= 2y =3 \implies y = \frac{3}{2}.$

Therefore there are no positive integer solutions.


问:
Method of characteristics for $u_t + (1-2u) u_x = 0$ and shocks
My teacher just went over the method of characteristics and we did an example with shocks then drew a picture, but I wanted to clarify some things for myself.

The example was
$$u_t + (1-2u) u_x = 0$$
such that
$u(x,0 )=
\begin{cases}
      0 & x\leq 0 \\
      0.5 & \geq 1 \\
      x/2 & \text{otherwise}
   \end{cases}
$

So from my understanding due to the method of characteristics I get two ODEs
$\frac{dx}{dt}= 1-2u$ and $\frac{du}{dt} = 0$ where $u = u(x(t),t)$. Solving the second equation first I get $u(x(t),t) = u_0$. Now depending at $x(t)$ I have three different options $0,0.5$, and $x/2$.

Where I am confused is he drew the picture, I am attaching. I get the characteristics for $x<0$ and $x>1$. However, I do not understand how he obtained the characteristics in the middle region. That ODE should be
$\frac{dx}{dt} = 1-x$
Which I can use sep of variables to get
$\frac{1}{1-x}dx = 1dt$ and from there , I get $-ln|1-x| = t + x_0$ so from my understanding the characteristics should look like logarithms. I would greatly appreciate some clarification on this, thank you.

答:
The initial derivation is correct. From $\frac{du}{dt} = 0$ we deduce that $u(t)=u_0$ is constant along the characteristic curves, and from $\frac{dx}{dt} = 1-2u$ we deduce that these curves are straight lines with equation $x(t) = (1-2u_0) t + x_0$ where $x(0) = x_0$. Their slope $1-2u_0$ is deduced from the initial value $u(x_0,0) = u_0$ at the abscissa $x_0$, i.e.
\begin{equation}
x(t) = \left\lbrace\begin{aligned}
& t + x_0, & & x_0 \leq 0\\
& x_0, & & x_0 \geq 1\\
& (1-x_0) t + x_0 , & &\text{otherwise}
\end{aligned}\right.
\end{equation}
In the $x$-$t$ plane, the first family consists of lines parallel to the identity function, the second family produces vertical lines, and the last one produces lines with variable slope. The formation of shocks is described in this post and related ones.


问:
Mathematical Induction prime help
I recently obtained "What is Mathematics?" by Richard Courant and I am having trouble understanding what is happening with the Prime Number Unique Factor Composition Proof (found on Page 23).

The first part:

I have looked over it many times but I just don't understand what he is doing and why, as  an example, if you remove the first factor from either side of the equation you end up with two essentially different compositions that make up a new smaller integer.

I'm sure it is a simple error on my behalf but I have been stuck on this for a long time so I would appreciate a walkthrough explained clearly or some pointers in the right direction. Thank you.

答:
We reframe the logic of the argument,  using a variant of induction called Fermat's Method of Infinite Descent.

Call a positive integer bad if it has (apart from order) more than one factorization as a product of prime powers. Note that $1$ is not bad.

We want to show that there are no bad positive integers.  Suppose to the contrary that there is a bad positive integer $n_1$.

By the argument in the book, there is then a bad positive integer $n_2\lt n_1$.

By the argument in the book, there is then a bad positive integer $n_3\lt n_2$.

And so on. So if there is a bad positive integer, there is an infinite descending sequence
$$n_1\gt n_2\gt n_3\gt n_4\gt \cdots$$
of (bad) positive integers.

However, there is no infinite descending sequence of positive integers!

Instead of using the language of descent. the proof quoted above used the fact that if there is a bad positive integer, there is a smallest bad. The two proofs are completely equivalent, they are really the same proof.  However, descent feels more concrete to me.

Remark: The proof that is used in  What is Mathematics is a clever one, in that it avoids using "Euclid's Lemma." This lemma essentially says that if a prime $p$ divides $ab$, then $p$ divides $a$ or $p$ divides $b$. The standard proof of the Fundamental Theorem uses Euclid's Lemma. Thus in a sense the proof you quoted uses less machinery than the standard proof, it is, sort of, more "elementary." It pays for that by being quite a bit more complicated, and harder to understand. It is the sort of proof that might be appreciated for its cleverness if one is already familiar with the standard proof.

Title: Mathematical Induction prime help


答:
Rewrite as

$$(-3x+y-11)\,dy-(-8x+3y-31)\,dx=0$$

or

$$-3(x\,dy+y\,dx)+(y-11)\,dy+(8x+31)\,dx=0.$$

This is an exact differential, integration is immediate.

Title: Find the general solution to the ODE:

Body:
I was asked to find the general solution to these ODEs.

$$\frac{dy}{dx}=\frac{(-8x+3y-31)}{(-3x+y-11)}$$
I tried by rearranging to the form $M(x,y)+N(x,y)\frac{dy}{dx}=0$ and let $y=vx$, but what I got was $$(v^2x-6vx-11v+8x+31)dx+x(-3x+vx-11)dv=0$$
and I could not separate $x$ and $v$ for integration.

$$\frac{dy}{dx}=\frac{y}{x}+\frac{9}{2}xexp(-\frac{2y}{x})+\frac{9}{2}xexp(\frac{2y}{x})$$
I have totally no idea which method to be used in Question 2.

May I know if I'm on the wrong track, and how should I solve them? Thanks.


问:
What are the bounds on the frequency of divisions by 2 in the Collatz cycles of natural numbers?
If $T^n(x):\Bbb N\times\Bbb N\to\Bbb N$ is the $n^{th}$ iterate of the Collatz function $T(x)=\frac12(3x+1)$ if $x$ odd and $x/2$ if $x$ even.   Then let $Q(x):\Bbb N\to2^{\Bbb N}$ be the parity vector function given by $$Q(x)=\sum_{n=0}^\infty2^n(T^n(x)\pmod2)$$

$Q(x)$ is an infinite binary string containing a $1$ every time $x$'s Collatz sequence is odd and a $0$ every time it is even, for any natural number $x$.

As an aside, $Q$ is a partial function of a larger $\Bbb Z_2\to\Bbb Z_2$ bijection which is a 2-adic isometry.

Let $0\leq r(x)\leq1$ denote the proportion of digits of $Q(x)$ which are zeroes, and equivalently the proportion of steps in $x$'s Collatz sequence which are even.  Therefore for all integers whose Collatz sequences stabilise finitely to the well known cycle $(2,1\ldots)$ we have $r(x)=\frac12$

Let $\overline {\Bbb N}$ indicate the subset of natural numbers whose Collatz sequences eventually stabilise to a cyclic orbit. Now suppose $x\in\overline {\Bbb N}$ is a natural number greater than $2$ and $T(x)$ has an eventually cyclic orbit which is not $(2,1\ldots)$.  Under these suppositions I have fairly easily shown that $r(x)=\frac12$ is an upper bound for $r(\overline {\Bbb N})$ - the proof is not too long to give, if required.  In fact $$\log_6(2)<r(x)<\frac12$$

Question

Since the proof was so easy and my talent is so little, I presume this is known and that there are already sharper bounds on $r(\overline {\Bbb N})$.  What bounds on $r(\overline {\Bbb N})$ are known or provable with current techology?

The larger the $x$, the greater the cycle length and the sharper the upper bound will need to be, with $r(x)$ approaching its lower bound $\log_6(2)$ as $x\to\infty$.

答:
Well, for me the massaging of the Collatz-formulae is easier if we assume the "Syracuse"-style notation of the Collatz-iteration
$$ a_{k+1}= {3a_k+1\over 2^{A_1}} \qquad a_k \small \text{ from the odd integers}
$$ and for a $N$-fold iterated transformation the short, vectorial, notation:
$$ a_{N+1}=T(a_1;[A_1,A_2,...,A_N])
$$
So let $N$ denote the (N)umber of steps $3x+1$ and $S$ denote the (S)um of the exponents $A_k$, which is also the number of $x/2$-steps.

Then, to convert this into the more common version of $(3x+1)/2$ and $x/2$ -stepping, we introduce $E$ the number of even steps without the $(3x+1)/2$ steps, so $E=S-N$.

With that, I understand your $r(a_1)$ as $r(a_1)=E/(N+E) = (S-N)/S = 1- N/S$.

We can observe,

that the trival cycle $1 = T(1;[2,2,2,...2])$ to any length $N$, has the values $S=2N$ and $E=N$ and $r(a_1)= 1-N/S = 1-N/(2N)=1/2 $

that the first cycle in the negative numbers $-1= T(-1;[1,1,1,1,...,1])$ to any length $N$, has  $S=N$, $E=0$, and $r(a_1)= 0$

that the second cycle in the negative numbers $-5=T(-5;[1,2,1,2,1,2,...,1,2])$ to any even length $N=2n$, has $S=3n$, $E=n$ and $r(a_1) = E/(N+E)= n/(3n) = 1/3 $

Now to have a cycle of any length, and other than $T(a_1;[2,2,2,...,2])$ we can use the well known multiplication-formula for the $N$ members of an expected cycle $a_k$ ($k=1..N$)
$$ a_2 \cdot a_3 \cdot ... \cdot a_N \cdot a_1 = \left({3a_1+1\over 2^{A_1}}\right)
\left({3a_2+1\over 2^{A_2}}\right) \cdots \left({3a_N+1\over 2^{A_N}}\right) \tag 1$$
This can be rearranged to
$$ 2^S = 2^{A_1+A_2+...A_N} =\left(3+{1\over a_1}\right)
\left(3+{1\over a_2}\right) \cdots \left(3+{1\over a_N}\right) \tag 2$$
We see, that the rhs must be at least as large as the smallest perfect power of $2$ larger than $3^N$, but at most as $4^N = 2^{2N}$ so we get for the lhs (writing $\gamma=\log_2(3)$, and further below $\gamma_1=\log_2(3)-1$):
$$ 2^{\lceil N \cdot \gamma \rceil} \le 2^S \le 2^{2N} \tag 3$$
which in terms of $S$ means
$$ \lceil N \cdot \gamma \rceil \le S \le 2N \qquad \text{where } S \in \mathbb N^+ \tag 4 $$ and in terms of $E$ instead
$$ \lceil N \cdot \gamma \rceil -N =\lceil N \cdot \gamma_1 \rceil \le E \le N \tag 5$$

From this you have bounds for your $r(a_1)$-parameter:
$$  r(a_1) = { E \over N+E } \implies \\
{\lceil N \cdot \gamma_1 \rceil \over N+E} \le \frac E{N+E} = r(a_1) \le \frac N{N+E} \tag 6
      $$

Using (5) we denote $E_\min = \lceil N \gamma_1 \rceil$ and $E_\max = N$.

From this you have bounds for your $r(a_1)$-parameter:

$ \displaystyle \lim_{N \to \infty} r(a_1)_{E_{min}} =  1-{ 1 \over (1-\{N \gamma_1 \} + N \gamma_1 )/N+1 } = 1-\frac 1{\gamma_1+1} \\
\qquad  = 1- \log_3(2) = \log_3(1.5) \approx 0.369070246429 \qquad \qquad \tag {7a}$

$ \displaystyle \lim_{N \to \infty} r(a_1)_{E_{max}} =  { E_{max} \over N + E_{max} } = { N  \over 2N  }  = \frac 12 \qquad \qquad \qquad \qquad \qquad \tag {7b} $
The latter is the ratio for the trivial cycle $r(1)$; the ratio $0$ is for $r(-1)$ and (7a) is the limit for any other cycle.

P.S: How did you get the sharper lower bound $\log_6(2)$?


问:
Collatz Conjecture, why a rate of change of $*4$ in the following?
Following the rules of the Collatz Conjecture, in this experiment I have created a list of all odd numbers until $33333$. The list includes 3 columns, such as in the following sample:

A) Starting Odd $(X)$
B) $(X * 3) +1$
C) $X/2$ repeat until odd

1
4
2, (1)

3
10
(5)

5
16
8, 4, 2, (1)

7
22
(11)

9
28
14, (7)

11
34
(17)

I have then checked what kind of numbers in Column A) lead to a specific odd number in Column C) marked in ():

When Column C) ends in 5, We have the following numbers in Column A):

$3, 13, 53, 213, 853, 3413...$ Growing as $+10, +40, +160, +640, +2560..$ at a rate of $*4$

When Column C) ends in 7, We have the following numbers in Column A):

$9, 37, 149, 597, 2389...$ Growing as $+28, +112 , +448, +1792..$ at a rate of $*4$

When Column C) ends in 11, We have the following numbers in Column A):

$7, 29, 117, 469, 1877, 7509, 30037...$ Growing as $+22, +88 , +352, +1408, +5632, 22528 ...$ at a rate of $*4$

When Column C) ends in 13, We have the following numbers in Column A):

$17, 69, 277, 1109, 4437, 17749...$ Growing as $+52, +208 , +832, +3328, +13312 ...$ at a rate of $*4$

When Column C) ends in 17, We have the following numbers in Column A):

$11, 45, 181, 725, 2901, 11605...$ Growing as $+34, +136, +544, +2176, +8704 ...$ at a rate of $*4$

My question is:

How can the rate of change of $*4$ be explained for causing these increments to lead to the same ending numbers in column C)?

答:
Firstly, note that the first number in the $+x$ sequence is always double the column C) number. So, in your first example, $13-3=10$ and $10=2\times5$. That also seems to work, so I would frase your question like this:

Let $n$ be an odd number, if the next odd number in it's Collatz sequence is $m$, then $\exists k\in\mathbb{N}$ such that the number $n+4^k2m$, also leads to $m$

Now, I don't know if there is a better phrasing for that, but that is what I've attempted to prove.

Say n leads to $n$, then $3n+1=2^lm$ for some $l$, so $2m=2^{-j}(3n+1)$, where $l=j+1$. Starting the sequence, $n+4^k2m$ would take you to $3(n+4^k2m)+1=(3n+1)+3·4^k2m=(3n+1)+3·4^k2^{-j}(3n+1)=(3n+1)(3·4^k2^{-j}+1)=$ $(3n+1)(3·2^{2k-j}+1)$, so if we take $k=j/2$, we get $(3n+1)(3·2^{2k-j}+1)=4(3n+1)=2^{l+2}m$.

Now, for this to work, we need $j$ to be even, so $l$ needs to be odd. I think you can prove that if $n$ is odd, $3n+1$ is an odd power of $2$ times the next odd number (unless the next odd number is $1$). Or, what is the same, that if $m$ is odd, $2^{2a}m-1$ is never a multiple of 3. I think you can prove this quite easily with modular congruences, but I'm a bit rusty on those.


问:
How to prove that sequence spaces $l^{p}, l^{\infty}$ and function space $C[a. b]$ are of infinite dimension
I am studying about the sequence space $l^{p}, l^{\infty}$ and function space $C[a. b]$. It is mentioned in the book that all of these spaces are of infinite dimension. I want to prove that these spaces are of infinite dimension. I am not finding a way to proceed.

Any help and suggestions would be helpful to me.

Thanks

答: You just need to construct infinitely many linearly independent vectors. For $l_p$ and $l_{\infty}$ take vectors $(0,0,0,...0,1.....).$ For $C[a,b]$ polynomials $x^n$ will do. </p>


问:
$\chi(G - u-v) = \chi(G) -2,$ for any two vertices $u$ and $v$ then the graph is complete.
Let $G$ be a graph with the property that on the deletion of any two vertices from the graph the chromatic number of the graph decreases by $2$, i.e.
$$\chi(G - u-v) = \chi(G) -2,$$ for any two vertices $u$ and $v$. We have to prove that the graph $G$ is none other than a complete graph.

I tried starting with the graph $G -u-v$. Then adding a vertex should increase the chromatic number by $1$. I am stuck here..

答: You want to show that there is an edge between every pair of vertices. For contradiction, suppose that this is not the case for some $u,v$. Then from any valid coloring of $G - u - v$ with $k$ colors you can get a coloring of $G$ with $k+1$ colors by coloring $u$ and $v$ with the same new color. This shows that $\chi(G) \leq \chi(G-u-v) + 1$, a contradiction.


问:
Definition of non-singular curve
A path $c:[a,b]→R$ is non singular if $c'(t)≠0$ for all $a≤t≤b$. This is the definition which was taught.However I am not sure if $c'(t)$ needs to be finite in the definition. Any help is appreciated.

答: A path is a continuous function, right? But if for some $\tau<\infty$ $$\lim_{t\to\tau} c'(t) = \infty,$$ it fails to be continuous.

Title: Definition of non-singular curve


问:
The crumpled paper problem
I saw a theorem recently (and have seen similarly phrased problems before), which I found quite surprising. Unfortunately, I'm not sure what its name is (or if it has one). I'll refer to it as the "crumpled paper problem". It goes something like this:

Suppose you were to take two pieces of paper, identical in size and shape, and place one on top of another such that all 4 corners and all 4 edges line up. Now, every point along the top sheet can be "mapped" to its corresponding point on the bottom sheet. If you were to then crumple up the top sheet of paper, and toss it on top of the bottom sheet, there necessarily exists a point on the top sheet that lies precisely above its "mapping point" on the bottom sheet, no matter how it was crumpled or where it was placed.

This result, to me, at least, is very surprising. Intuitively, it seems to me that it must be that there is some way to crumple the paper, or location to toss it, such that no points line up. I can sort've imagined progressively folding the paper (analogous to crumpling), and kind of see how this might be, however, it's still quite challenging.

My question is, how is this proven, does the problem have a name, and to what field of mathematics does this belong? I'm very curious how this can be shown.

I've added those tags which I thought may be relevant. Unfortunately, I have no clue to what precise area of mathematics this belongs to. Please do correct tags as necessary.

答:
The mathematical description is this: You have a rectangle $R = [a,b] \times [c,d]$. Crumpling a copy of $R$ and tossing it on the top of the bottom sheet $R$ means that you have an injection $\phi : R \to R \times [0,\infty) \subset \mathbb R^3$. Here each point $(u,v,w)$ in the crumpled image $C = \phi(R)$ is uniquely identified by its"coordinates" $(r,s) \in R$ which means that $(u,v,w) = \phi(r,s)$. Your question is

Does there exists a point $x \in R$ such $\pi(\phi(x)) = x$? Here $\pi : \mathbb R^3 \to \mathbb R^2, \phi(u,v,w) =(u,v)$ is the orthogonal projection onto the $(u,v)$-plane.

In other words, does the function
$$f : R \to R, f(x) = \pi(\phi(x))$$
have a fixed point?

Note that whether or not the crumpling function $\phi$ is continuous is hard to say. If crumpling causes ruptures, even if they are microscopically small, it will be not. But let us assume that $\phi$ is continuous. Then your question is answered in the affirmative by the Brouwer fixed point theorem in dimension $2$. I am not going to give a statement or a proof, search for "Brouwer" in this forum or look here.

Title: The crumpled paper problem


问:
Someone please explain to me quotient rings
Specifically what does stuff like $\mathbb{Q}[x] / \langle x^{2} - 1\rangle$ generally means? I thought it meant the set of polynomials with rational coefficients divided by $x^{2} - 1$, but when I see stuff like $\mathbb{Z}[x]/\langle x^{2} + 1, 5\rangle$ I have no idea what it means.

Also, how do you generally find what $R/I$ is isomorphic to?

答:
Usually, you'd learn that quotient rings are sets of certain residue classes with a certain ring structure imposed. But this really obscures the vision on what quotient rings really are: If $R$ is a commutative ring and $A\subseteq R$ some subset, then conceptually speaking, $R/\langle A\rangle$ is a ring with essentially the same structure as $R$, but with the additional assumption that every element of $A$ is equal to $0$. Of course, if $a_1$ and $a_2$ are $0$, their sum should also be $0$. And if $a$ is $0$ and $r$ is some other element of $R$, then their product should also be $0$. So if every element of $A$ is $0$, then all the elements of the ideal $\langle A\rangle$ should also be $0$, since those elements are exactly those which can be generated by the mentioned operations (adding elements toeach other or multiplying them by some random element). So essentially, $R/I$ with some ideal $I\subseteq R$ is a ring with the structure of $R$, but the additional assumption that every element of $I$ is $0$, and no other elements are $0$.

We can also deduce more relations from this fact. For instance, if $a-b$ is in $I$, then $a-b$ is $0$, so $a$ and $b$ are equal. Such relations must hold in the ring $R/I$ if elements of $I$ are supposed to be $0$. But these forced relations are the only relations that differentiate $R$ from $R/I$. Every single rule about how to do algebra in $R/I$ follows from the two rules that 1. everything works like in $R$, except for the fact that 2. if something is in $I$, it's $0$. You don't need to know more than this to do algebra in $R/I$.

For instance, consider $\mathbb Q[X]/\langle X^2+1\rangle$. I will add a prime to elements of the quotient to differentiate them from elements of the additional ring. At first, this quotient ring is made up of elements which look like those of $\mathbb Q[X]$, so polynomials of the form $a_0'+a_1'X'+\dots+a_n'X'^n$. But we get the additional rule that $X'^2+1'=0'$, or in other words, $X'^2=-1'$, from which we can deduce $X'^{2k}=1'$ if $k$ is even, and $X'^{2k}=-1'$ if $k$ is odd. So those polynomials reduce to
$$a_0'+\dots+a_n'X'^n=(a_0'-a_2'+a_4'-\dots) + (a_1'-a_3'+a_5'-\dots)X'.$$
Those sums can be replaced by their results, say $b'$ and $c'$, to make them into
$$b'+c'X',$$
where $b'$ and $c'$ behave like $b$ and $c$, and $X'^2=-1'$. So it behaves exactly like $a+b\mathrm i$ should in $\mathbb Q[\mathrm i]$, the ring of rationals with $\mathrm i$ added. In fact, it's a nice way to define $\mathbb Q[\mathrm i]$.

Now as an afterthought, we might want to guarantee that such a quotient ring actually exists for all rings and all their ideals. We do that by constructing them explicitly using residue classes. But residue classes are not what quotient rings are about, it's just a tool to guarantee their existence.


问:
If asked to prove that a set $S$ is closed, is it sufficient to prove that the given set is closed in a particular metric space?
To elaborate, if I were asked to prove that a set $S$ is closed, and I know that there exists a particular metric space $(X,d)$ such that $S\subset X$, is it then sufficient to prove that $S$ is closed in $X$, using the particular metric $d$?

答: If no metric is given, then the question makes no sense, even if you are aware of the existence of such a metric space. If someone asks you whether or not $(0,1)$ is closed, what's your answer? Yes, $(0,1)\subset\Bbb R$, but it is closed with respect to the discrete metric, whereas it is not closed with respect to the usual one.


问:
An alternate proof of Egorov's Theorem
I came up with a proof of Egorov's Theorem which I can't find in my books or on the net, which makes me think it's wrong, which means I have a misunderstanding somewhere which needs to be hammered out.

Egorov's Theorem: Let $(X,M,\mu)$ be a finite measure space and $f_n$ a sequence of measurable functions on $X$ that converges pointwise a.e. on $X$ to a function $f$ that is finite a.e. on $X$.  Then for each $\epsilon > 0$, there is a measurable subset $X_{\epsilon}$ of $X$ for which

$f_n\rightarrow f$ uniformly on $X_{\epsilon}$ and $\mu(X\backslash X_{\epsilon})<\epsilon$

Proof:

Let $\epsilon > 0$.

So first I let $E_0$ be the set of points at which $f_n$ does not converge to $f$, union the set of points at which $f$ is infinite.  Then $\mu(E_0)=0$.  Now define

$$A_N = \{x\in X\backslash E_0 : |f_N(x)-f(x)|<\epsilon\},$$

and set $E_k = \bigcap_{N=k}^{\infty}A_N$.  Then since the $A_N$ are measurable and $E_k$ is the countable intersection of measurable sets, then the $E_k$ are measurable.  Furthermore the $E_k$ are ascending, since as $k$ grows one is intersecting fewer and fewer of the $A_N$.

Thus since $\mu(E_0)=0$ (to justify the first equality) and by the continuity of measure (to justify the third), we have:

$$\mu(X)=\mu(X\backslash E_0)=\mu(\bigcup_{k=1}^{\infty}E_k)=\lim_{k\rightarrow\infty}\mu(E_k).$$

Where the second equality holds since for every $x\in X- E_0$ there exists some $K$ such that for all $n > K$, $|f_n(x)-f(x)| < \epsilon$ which is what it means for $x\in E_K$; the reverse inclusion being trivial.

Therefore since $\lim_{k\rightarrow\infty}\mu(E_k)=\mu(X)$, we can find a $K$ such that $\mu(X\backslash E_K)=\mu(X)-\mu(E_K)<\epsilon$, with $f_n\rightarrow f$ uniformly on $E_K$.

Is this correct?

答:
I don't think this is correct, no.  I don't see why $f_n\to f$ uniformly on $E_K$.  Certainly it's true that on $E_K$, for $n\ge K$ and any $x$, $|f_n(x) - f(x)| < \varepsilon$, but $\varepsilon$ was fixed at the beginning; this says nothing about convergence.  Because what happens if you make epsilon smaller?  If this just raised $K$, then you'd have uniform converge; but the problem is that changing $\varepsilon$ also changes all your sets and so changes the set $E_K$ -- you're not talking about convergence on the same set anymore!

So no, this doesn't appear to be right.

(A note: A warning sign that made me suspect it was probably wrong was when you used $\varepsilon$ to define how close $f_n$ and $f$ had to be -- $\varepsilon$ is a measure, so taking it as a distance is unlikely to be meaningful.  Certainly clever arguments that involve doing things like that sometimes work, but it's definitely a warning sign.)

Title: An alternate proof of Egorov's Theorem


问:
Equivalence of two definitions of continuous-time dynamical system
I'm new to dynamical systems. I've found out that there are at least two possible approaches to defining continuous-time dynamical systems but they are equivalent: One is a tuple $(\mathbb{R},X,\phi)$(of course it can be more general than that but I've restricted our discussion to this) such that $X$ is a metric space and $\phi$ is a continuous function from $\mathbb{R} \times X$ to $X$ that satisfy two properties(in other words $\phi$ is a left group action of $\mathbb{R}$ on $X$), and the other is a family of functions $(\phi_t)_{t \in \mathbb{R}}$ from $X$ to $X$ such that $\phi_0 = $ id$_X$ and $\phi_{t + s} = \phi_t \circ \phi_s$. I can prove that the first one induces a family of functions $(\phi^t)_{t \in \mathbb{R}}$ satisfying the given properties. So one direction is OK. And I can prove that the second one induces a function $\phi$ satisfying all the stated properties other than one: Continuity. So the question is how to prove that $\phi$ is continuous. I've defined $\phi$ to map $(t,x)$ to $\phi_t(x)$. Thanks.

答:
This is false in general, if you use rhe second definition you would need to explicitly require  continuity, here is a counter example:

Let $R$ be the real numbers with the discrete topology (0,1 metric), and define $\phi_t : R \rightarrow R$ by $\phi_t(r) = r+t$. All these functions are continuous since $R$ has the discrete topology. These induce the function
$$\phi : \mathbb{R} \times R \rightarrow R$$
$$\phi(t,r) = \phi_t(r)$$
Now, note that $\phi$ is not continuous, since it carries connected components into disconnected ones: to see this just observe that  $\phi$ restricted to $\mathbb{R}\times \{0\}$ is surjective.


问:
Hessian matrix as derivative of gradient
From a text:

For a real-valued differentiable function $f:\mathbb{R}^n\rightarrow\mathbb{R}$, the Hessian matrix $D^2f(x)$ is the derivative matrix of the vector-valued gradient function $\nabla f(x)$; i.e., $D^2f(x)=D[\nabla f(x)]$.

$\nabla f(x)$ is just an $n\times 1$ matrix consisting of $\partial f/\partial x_1,\partial f/\partial x_2,\ldots,\partial f/\partial x_n$.

Then $D[\nabla f(x)]$ must be a $1\times n$ matrix.

But I know that the Hessian matrix is an $n\times n$ matrix consisting of $\partial ^2f/\partial x_i\partial x_j$. How can the given definition be consistent with this?

答:
This is an old discussion and I would not intent to correct anything.
Just in mind that this had caused myself long-time confusion so that I wish to accept some simple rules to make it clear.

If one allows to define the operators of the function $f(x):\mathbb{R}^n\rightarrow\mathbb{R}$:

Gradient operator $\nabla$: defined as $n\times1$ column vector.

Derivative operator $\nabla^T$: defined as a row vector (i.e., $1\times n$),

Hessian operator $\mathbf{H}$: defined as the gradient of the derivative of $f(x)$.
$$\mathbf{H}=\left(\begin{matrix}\frac{\partial}{\partial x_1}\\
\vdots\\\frac{\partial}{\partial x_n}\end{matrix}\right)\left(\begin{matrix}\frac{\partial}{\partial x_1}&
\ldots&\frac{\partial}{\partial x_n}\end{matrix}\right)=\left(\begin{matrix}\frac{\partial}{\partial x_1^2} & \ldots & \frac{\partial^2}{\partial x_1\partial x_n}\\\vdots &\ddots&\vdots \\
\frac{\partial^2}{\partial x_n\partial x_1}& \ldots & \frac{\partial^2}{\partial x_n^2}\end{matrix}\right).$$


问:
Chi square distribution.
Let $X$ be a random variable which follows $N(0,1)$. It is clear that $X^2$ follows $\chi^{2}(1)$. Let $k\in\mathbb{R}^{+}$ be a positive real number. I want to know the distribution of the random variable $\sqrt{X^{2} + k}$.

Thanks in adnvance.

答:
You can start defining analytically the density of $Y=X^2$ that is

$$f_Y(y)=\frac{1}{\sqrt{2\pi y}}e^{-y/2}$$

$y>0$

and then proceed by transforming your rv in the desired $Z=\sqrt{Y+k}$

It results to me

$$f_Z(z)=\frac{2z}{\sqrt{2\pi}\sqrt{z^2-k}}e^{-(z^2-k)/2}\cdot\mathbb{1}_{(\sqrt{k};+\infty)}(z)$$

this is the density of your rv.

Title: Chi square distribution.


问:
If |X| $\le$|Y| and X$\in$ $\mathcal H$ then $\mathcal H$ is uniformly integrable
If $Y$ is some integrable random variable, such that, |X|$\le$|Y| for all $\mathcal H$, then $\mathcal H$ is uniformly integrable. How can I prove this using only the definition of Uniformly integrable. But converse is not true please give a complete example for this. I know that $\mathcal H$ is uniformly integrable if and only if $\mathcal H$ is $L_1$ bounded and $\mathcal H$ is uniform absolute continuity.

答:
What does $\vert X\vert\le\vert Y\vert$ for all $\mathcal H$ mean? I suppose you mean that $\mathcal H$ is a family of random variables, and you have $\vert X\vert\le\vert Y\vert$ for all $X\in\mathcal H$.

As a reminder, $\mathcal H$ is uniformly integrable iff $\sup_{X\in\mathcal H}\mathbb E[\vert X\vert1_{\{\vert X\vert\ge R\}}]$ vanishes as $R$ goes to $+\infty$. If $\vert X\vert\le \vert Y\vert$ for all $X\in\mathcal H$ then
$$
\sup_{X\in\mathcal H}\mathbb E[\vert X\vert1_{\{\vert X\vert\ge R\}}]\le\sup_{Y\in\mathcal H}\mathbb E[\vert Y\vert1_{\{\vert Y\vert\ge R\}}],
$$
where the right-hand side vanishes as $R$ goes to $+\infty$ by dominated convergence. Hence $\mathcal H$ is uniformly integrable.

The converse is not true, see for instance Example of a family of uniformly integrable random variables that is not dominated?


问:
Tell me the ideal selling price to get back a specified number
If I am buying something at xxx,

what is the price to sell it if I want a profit of $2.50 after minus-ing  0.63% (broker fee) from the selling price?

I need to make this into an excel formula, but this is the first step... if you know the excel formula that would be a great help as well!

Thanks!

答:
$$\begin{align*}
\mathsf{profit}&=(\mathsf{selling\;price}) - (\mathsf{broker\;fee})-(\mathsf{price\;you\;bought\;it\;for})\\
\end{align*}$$
So you want to solve
$$\begin{align*}
2.50&=(\mathsf{selling\;price}) - (0.63\%\mathsf{\;of\;selling\;price})-(\mathsf{price\;you\;bought\;it\;for})\\\\
2.50&=(\mathsf{selling\;price})-(0.0063)(\mathsf{selling\;price})-(\mathsf{price\;you\;bought\;it\;for})\\\\
2.50&=(0.9937)(\mathsf{selling\;price})-(\mathsf{price\;you\;bought\;it\;for})\\\\
(0.9937)(\mathsf{selling\;price})&=2.50+(\mathsf{price\;you\;bought\;it\;for})\\\\
(\mathsf{selling\;price})&=\frac{2.50+(\mathsf{price\;you\;bought\;it\;for})}{0.9937}\\\\
\end{align*}$$</p>


问:
How to factor and simplify algebraic expressions with fractional or negative exponents
Im trying to factor:

$(x+3)^{1/2}-(x+3)^{3/2}$

The answer is $-(x+3)^{1/2}(x+2)$

Could someone explain how I would factor the expression?

My answer is

$(x+3)^{1/2}[1-(x+3)]$

答:
$(1-(x+3)) = (1-x-3) = (-x-2) = -(x+2)$

Now isolate the negative sign and you get the answer.


问:
Literature on the PDE $\partial_t f + \partial_x(f(1-f))=0$
I am looking for literature on this variant of the transport PDE:

$$\partial_t f + \partial_x(f(1-f))=0.$$

Do you have any suggestions?

答:
This is a special case of a conservation law that can be (and has been) written in divergence form
$$ \partial_t f + \partial_x F(f) = 0$$
where $F(x) = x(1-x)$. You can find some information about this class of equations in e.g. Evans' Partial Differential Equations, Chapter 3.4. IMO, a softer introduction can be found in Alinhac's book Hyperbolic Partial Differential Equations, Chapter 4. One small point, the theory at some points assumes $F$ is convex, or even uniformly convex, which we can achieve by setting $u = -f$ so that we have the equation
$$ u_t + G(u)_x = 0,\quad G(x) = x(x+1)$$


问:
How to solve $\int_\mathbb{R}\frac{\varepsilon}{(\mu-\lambda)^2+\varepsilon^2}f(\mu)\,d\mu$?
I try to solve the following integral, $$-\int_\mathbb{R}\frac{\varepsilon}{(\mu-\lambda)^2+\varepsilon^2}f(\mu)\,d\mu, $$
with $\varepsilon >0$, $\lambda\in\mathbb{R}$, $\mu\in\mathbb{R}$ and $f\in\mathcal{C}^\infty_c(\mathbb{R})$. I tried to use integration by parts with $$v'=\frac{\varepsilon}{(\mu-\lambda)^2+\varepsilon^2} \Rightarrow v = \arctan\left(\frac{\mu-\lambda}{\varepsilon}\right) $$
$$ u=f(\mu)\Rightarrow u'=f'(\mu) .$$
With this I got, $$-\int_\mathbb{R}\frac{\varepsilon}{(\mu-\lambda)^2 + \varepsilon^2} f(\mu)\,d\mu = -\left(\left[\arctan\left(\frac{\mu-\lambda}{\varepsilon}\right)f(\mu)\right]-\int_{\mathbb{R}} \arctan\left(\frac{\mu-\lambda}{\varepsilon}\right)f'(\mu) \, d\mu\right) $$
Now the first part of the RHS is zero because $f$ is compactly supported. My problem is that I don't know how to proceed with the remaining part. I tried integration by parts here as well but failed.
Now my question is wether I have overlooked something or my previous calculations are wrong.

答:
just some tips

If I understand correctly you want to evaluate $I(\lambda)=\lim_{\epsilon \to 0^+} \int_{\mathbb R}  \frac \epsilon {(\mu - \lambda)^2 + \epsilon^2} f(\mu) \, d\mu$ (i.e. $\epsilon$ is infinitesimal). You can rewrite integral in the form:
$$I(\lambda)=\lim_{\epsilon \to 0^+} \frac{-i}{2}\int_{\mathbb R}  f(\mu) \, d\mu \left(\frac 1 {(\mu - \lambda) -i \epsilon}-\frac 1 {(\mu - \lambda) +i \epsilon}\right) $$

So, $\epsilon$ in fact define the way around the pole $\mu=\lambda$ - which of poles "plays". If you close your contour in upper-half on the complex plane the first term contributes (and will give $\frac{-i}{2}2\pi{i}f(\lambda)=\pi{f}(\lambda)$ - we go counter clockwise); if in the lower half-plane - the second (and will give $\frac{i}{2}(-2\pi{i})f(\lambda)=\pi{f}(\lambda)$ - we go clockwise).

How you close the contour (integration over big half-circle) depends on behaviour of $f(\mu)$ - which is often defined by some physical considerations.