问:
Let $A ⊆ E ⊆ R$, where $E$ is measurable and $A$ is non-measurable. Prove that $m^∗ (E − A) > 0$
I have been asked to prove a statement-
Let $A ⊆ E ⊆ R$, where $E$ is measurable and $A$ is non-measurable. Prove that $m^∗
(E − A) > 0$.

I used the Caratheordory criterion-
For a measurable set $E$ and any set $A⊆ R$
$$m^∗(A) = m^∗(A ∩ E)+m^∗(A ∩ E^c)$$

Now $m^∗(A) = m^∗(A ∩ E)$ which gives that $m^∗(A ∩ E^c)=0$

What am I missing here?

答:
I assume that measurable here means Lebesgue measurable.

Let $A \subseteq E \subseteq \Bbb R$, where $E$ is measurable and $A$ is non-measurable. Then $m^∗ (E − A) > 0$.

Proof:  Since $A \subseteq E$, we have that $A=E-(E-A)$.

If $m^∗(E − A) = 0$ then $E- A$ is measurable and then $A$ is measurable.

Since  $m^∗ (E − A) \geqslant 0$, by the counter-positive, if $A$ is non-measurable, then $m^∗ (E − A) > 0$.


问:
Formal proof for $A \cup (B-A) = A\cup B$
I am just confused by the whole concept if someone can explain it to me or provide a different example so I can understand this would be much appreciated.

I want to prove that $$A \cup (B-A) = A\cup B$$

答:
Hint:

Usually, showing that two sets are equal (say $C=D$), can be undertaken in two steps:

Show that $C\subseteq D$ (that is, that every element $c\in C$ satisfies $c\in D$)

Show that $D\subseteq C$ (that is, that every element $d\in D$ satisfies $d\in C$.)

This is sufficient to show that the two sets are equal.

So, for your case:

Let $x\in A\cup (B\setminus A)$.  How can you show that $x\in A\cup B$? Start by noting that $x\in A\cup (B\setminus A)$ implies that either $x\in A$ or $x\in B\setminus A$. Can you proceed from there?

If you can do that, then you need to do the reverse: assume you have some $y\in A\cup B$, and prove that $y\in A\cup(B\setminus A)$.

ContentLicense: CC BY-SA 4.0


问:
Computational complexity of unknotting problem?
The Wikipedia article on the unknotting problem says "a major unresolved challenge is to determine [...] whether the problem lies in the complexity class P".  It mentions some work towards this result and a recent claim that it is true.  But then finally it says "the unknotting problem has the same computational complexity as testing whether an embedding of an undirected graph in Euclidean space is linkless," referencing this paper.  Doesn't that statement imply that the unknotting problem is in P, by the graph minor theorem and the existence of a polynomial time algorithm to test for a specific minor in a given graph?  What am I missing here?  What is the current status of this problem?</p>

答:
There is a difference between the following two problems.

(1) Decide if a given abstract graph $G$ admits a linkless embedding in $\mathbb{R}^3$.  [This can be solved in polynomial time, as you observe. The authors of the cited paper call this the "flat and linkless embedding" problem.]

(2) Decide if a given embedding of an abstract graph $G$ into $\mathbb{R}^3$ is linkless.  [This is the subject of Conjecture 1.1 if the cited paper.  The authors prove this polynomial time, relative to an oracle for the "link" problem.]

Basically, in (2) you are given a fixed embedding that you have to understand.  This is a lot more information than just the abstract graph, and appears to make the problem harder.

ContentLicense: CC BY-SA 3.0


问:
Proof of the principle of backwards induction
I have difficulty in neatly writing down a proof for the following from Terence Tao's Analysis I, where $m{+\!+}$ is the successor function:

Let $n$ be a natural number, and let $P(m)$ be a property pertaining to the natural numbers such that whenever $P(m{+\!+})$ is true, then $P(m)$ is true. Suppose that $P(n)$ is also true. Prove that $P(m)$ is true for all natural numbers $m\leq n$; this is know as the principle of backwards induction. (Hint: apply induction to the variable $n$.)

First of all, I am unsure about what the base case should look like. For the induction step, I understand that if we suppose inductively that $P(n)$ is true, that then for a natural number $a$ s.t. $a{+\!+}=n$ it holds that $P(a)$ is true, and then for a natural number $b$ s.t. $b{+\!+}=a$ it holds that $P(b)$ is true etc. Hence for all natural numbers $m\leq n$, $P(m)$ is true.

Could anyone please tell me what the base case should look like, and whether there is a neater way of writing down the induction step?

答:
Same answer as @coffeemath but made more clear. The statement to be proved is this:

$\forall n\in \textbf{N}, \Bigg (P(n) \implies (\forall m \le n, P(m)) \Bigg )$

which we can rewrite as, taking $R(n) := (\forall m \le n, P(m))$

$\forall n\in \textbf{N}, \Bigg (P(n) \implies R(n) \Bigg )$

Define $Q(n):=(P(n) \implies R(n) )$. We want to show $\forall n\in \textbf{N}, Q(n)$ and we shall use the principle of mathematical induction on $Q(n)$.

Step 1: $Q(0):=(P(0) \implies R(0) )\equiv (P(0) \implies P(0) )$. Thus, Q(0) is true.

Step 2: Assume $Q(n)$ is true.

Step 3: Since $P(n{+\!+})\implies P(n)$ and $P(n) \implies R(n)$, we have $P(n{+\!+}) \implies R(n)$. (syllogism). Now $ (P(n{+\!+}) \implies R(n)) \equiv (P(n{+\!+}) \implies R(n)) \land P(n{+\!+})) \equiv (P(n{+\!+}) \implies R(n{+\!+})) \equiv Q(n{+\!+})$.

Hence $Q(n)$ is true $\forall n$. QED.


问:
Lusin’s Theorem for Polish spaces with infinite Radon measure
I’m working on the following exercise in Klenke’s Probability Theory: A Comprehensive Course (Exercise 13.1.3), which asks us to prove the following generalization of Lusin’s Theorem:

Let $\Omega$ be a Polish space, let $\mu$ be a $\sigma$-finite measure on $(\Omega, \mathcal B(\Omega))$, and let $f : \Omega \to \mathbb R$ be a map. Show that the following are equivalent:

There is a Borel measurable map $g : \Omega \to \mathbb R$ with $f = g$ almost everywhere.

For any $\epsilon > 0$, there is a compact set $K_\epsilon$ with $\mu(\Omega \setminus K_\epsilon) < \epsilon$ such that the restricted function $f|_{K_\epsilon}$ is continuous.

As stated, this exercise is wrong when $\mu(\Omega) = \infty$: if $\Omega = \mathbb R$, no compact set has a complement with finite Lebesgue measure, so it should be a closed set $K_\epsilon$.

Furthermore, $\mu$ must be more than just $\sigma$-finite. Let $\Omega = \mathbb R$, and $\mu = \sum_{q \in \mathbb Q} \delta_q$ be the counting measure of the rationals. Then $\mu$ is certainly $\sigma$-finite, but if $f$ is a Borel-measurable map and if $K \subset \mathbb R$ is closed with $\mu(K^c) < \epsilon$ for $\epsilon < 1$, then we must have $\mu(K^c) = 0$, or $K \supset \mathbb Q$. But then since $K$ is closed, $\overline{\mathbb{Q}}= \mathbb R \subset K$, so $f$ must be continuous on $\mathbb R$ in order for the claim to hold. So we need more than $\sigma$-finite.

One way to edit the exercise is to instead assume $\mu$ is Radon and modify Statement 2 like so:

There is a Borel measurable map $ g : \Omega \to \mathbb R$ with $f = g$ $\mu$-a.e.

For any subset $A \subset \Omega$ with $\mu(A) < \infty$, and for any $\epsilon > 0$, there is a compact $K_\epsilon \subset A$ such that $f|_{K_\epsilon}$ is continuous.

These statements may be shown to be equivalent, since one can show Radon measures on Polish are $\sigma$-finite (see the discussion below).

But suppose we want to show the “original” Statement 2:

For any $\epsilon > 0$, there is a closed $K_\epsilon \subset \Omega$ with $\mu(K_\epsilon^c) < \epsilon$ such that the restricted function $f|_{K_\epsilon} : K_\epsilon \to \mathbb R$ is continuous.

What conditions must we impose on the Polish space $\Omega$ with infinite Radon measure $\mu$ in order to guarantee that this is true?

答:
All references I can find have extra assumptions, but I believe they are not needed. Hopefully my argument is correct, but there is a strictly positive probability that I overlooked some tricky details. I will prove the following, which is the most general version I could manage with a Radon measure.

Theorem: Let $X$ be a topological space, $\mu$ be a Radon measure on $X$ and $f\colon X\to Y$ a function, where $Y$ is a second countable topological space. The following are equivalent:

There is a Borel function $g$ such that $f=g$ almost everywhere.

For every $\epsilon>0$ there is a closed $F\subseteq X$ such that $f_{|F}$ is continuous and $\mu(X\setminus F)<\epsilon$.

Proof: $(1)\implies $(2) Fix $\epsilon>0$ and a countable basis $\{Y_n\}_{n\in\Bbb N}$ for $Y$ and, for each $n$, find a closed $F_n$ and an open $V_n$ such that $F_n\subseteq g^{-1}(Y_n)\setminus D\subseteq V_n$ and $\mu(V_n\setminus F_n)<\epsilon/2^{n+1}$, where $D=\{x\in X\mid f(x)\neq g(x)\}$. Let $U=\bigcup(V_n\setminus F_n)$ and note that $U$ is an open set with $\mu(U)<\epsilon$. Let $F=X\setminus U$, and note that $g^{-1}(Y_n)\cap F=V_n\cap F$, hence $g_{|F}$ is continuous, but $f=g$ on $F$ so $f_{|F}$ is also continuous.

$(2)\implies(1)$ For every $n\in\Bbb N$ we can find a closed set $F_n$ such that $f_{|F_n}$ is continuous and $\mu(X\setminus F_n)<1/n$. Let $F=\bigcup F_n$, let $Z=X\setminus F$ and fix a countable basis $\{Y_n\}_{n\in\Bbb N}$ for $Y$. We have $$f^{-1}(Y_m)=(f^{-1}(Y_m)\cap Z)\cup\bigcup_{n\in\Bbb N}f^{-1}(Y_m)\cap F_n,$$
where the set on the right is Borel, since $f_{|F_n}$ is continuous, while the set on the left is contained in the Borel null set $Z$, so if we define $g:X\to Y$ to agree with $f$ on $F$ and to be constant on $Z$ we have the function we were looking for.

As an aside note that to some authors a Radon measure means a measure which is tight rather than just inner regular, meaning that for all measurable $U$ and all $\epsilon>0$ there is a compact $K\subseteq U$ with $\mu(U\setminus K)<\epsilon$. Those authors usually state Lusin's theorem with compact $K_\epsilon$ and then remark that if the measure is Borel regular instead of Radon then the theorem still holds by replacing "compact" with closed", maybe the book you're following is using this convention?


问:
Puzzle: Finding the next number in the sequence of numbers
I'm trying to solve this puzzle and couldn't come up with the next value

答:
Usually these puzzles want you to combine two of the numbers to get the third.

That is, $f(a,b)=c$, where $f$ combines common operations $+,-,\times,\div,\sqrt\space,\dots$

Notice that if $a\in\{93,72\}$ and $b\in\{63,44\}$ decrease, so does $c\in\{39,29\}$.

 My first guess was that addition $+$ was used, but the result was too large.

 To fix this, I added division $\div$ by a constant $4$.

 This turned out to work out. $$\frac{93+63}{4}=39, \qquad \frac{72+44}{4}=29, \qquad \frac{47+13}{4}=15.$$


问:
Does incompressible in the smooth sense imply incompressible in the measure theoretic sense?
A smooth map $T$ is called incompressible if it has zero divergence and a measurable map $T$ is incompressible if $T^{-1}(B) \subset B$ implies $\mu(T^{-1}(B))=\mu(B)$ where $\mu$ is some probability measure.  Are these two versions of incompressibility equivalent in $R^n$?

答:
These two notions are not equivalent, but there is a connection. To avoid confusion, I'll refer to the two cases as divergence-free vector fields and measure-preserving maps respectively.

As stated, any constant map $\mathbb{R}^n\to\mathbb{R}^n$ is divergence-free as a vector field, but clearly not measure preserving. It's easy to come up with lots of other linear counterexamples, some of which are invertible.

Instead, given a smooth vector field $V:\mathbb{R}^n\to\mathbb{R}^n$, we can (at least locally) define the flow of $V$ to be a family of maps $\Theta_V:\mathbb{R}\times\mathbb{R}^n\to\mathbb{R}^n$ which displaces each point along the integral curves of $V$. In other words, it is defined as the solution to the following initial value problem:
$$
\frac{\partial}{\partial t}\Theta_V(t,x)=V(\Theta_V(t,x)),\ \ \ \Theta_V(0,x)=x
$$
One can show with a bit of calculus that if $V$ is incompressible, then its flow $\Theta_V$ preserves the Lebesgue measure on $\mathbb{R}^n$ (i.e. each map $\Theta(t,\_):\mathbb{R}^n\to\mathbb{R}^n$ is measure preserving). This will not be the case for other measures on $\mathbb{R}^n$ (though some will give rise to a similar notion of divergence), and not every measure-preserving map is the flow of a vector field.

This notion generalizes beyond $\mathbb{R}^n$. For instance, a sufficiently nice measure on a smooth $n$-manifold $M$ can be described (locally, up to sign) by a nonvanishing differential $n$-form $\omega$. The divergence of a vector field can then be defined in terms of a Lie derivative of $\omega$.
$$
\operatorname{div}(X)\omega=\mathcal{L}_X\omega
$$
And once again we have that the flow of a divergence-free vector field is measure-preserving. This notion applies, for instance, to other measures on $\mathbb{R}^n$ with smooth nonvanishing density functions and to Riemannian manifolds and their canonical measures.


问:
Showing that $f=0$ a.e. then $\int f=0$.
Let $\left(X,\mathcal{M},μ\right)$ be a measure space and f be integrable over $X$ with respect to $\mu$. I want to show that if $f=0$ almost everywhere on $X$ then $\int_{E}fd\mu=0$ for every $E\in\mathcal{M}$.

My attempt: By the Simple Approximation Theorem, there is an increasing sequence $\{\phi_n\}$ of non-negative simple functions that converge pointwise to $f$. Therefore, by the Monotone Convergence Theorem,

$$\int_{E}fd\mu = \lim_{n\to\infty}\int_{E}\phi_nd\mu = \lim_{n\to\infty}\phi_n\int_{E}d\mu = 0\cdot\mu(E) = 0.$$

答:
You  cannot pull $\phi_n$ out of the integral since it is not a  constant.

Observe that $f^{+}=\max \{f,0\}=\max \{0,0\}=0$ almost evrywhere. Similarly, $f^{-}=0$ almost everywhere. This allows us to assume that $f$ is non-negative.

There exist non-negative simple functions $\phi_n$ increasing to $f$. It follows that each $\phi_n=0$ almost everywhere. I will let you check that if $\phi$ is simple and $0$ almost everywhere then its integral is $0$. [Use definition of integral for  a simple function]. Now $\int f d\mu=\lim \int \phi_n  d\mu=\lim 0=0$.


问:
When can we affirm this equality about integral?
Suppose we have $f(x) = \int G(x_{0},x_{1},...)\,dx_{0}\,dx_{1}\dotsb dx_{n}$
When can we affirm that $$df = G(x_{0},x_{1},...)? \tag1$$
Basically, I am having trouble to understand how to deal with differentials of functions, intuitively I thought that we can do that:

$$\delta f(x) = \int \sum\left(\frac{\partial G}{\partial dx_{i}}\right)\,dx_{0}\,dx_{1}
\dotsb dx_{n} \tag2$$

But I am not sure how $(2)$ reduces to $(1)$.

答: In some notations this is true; for example, the volume element can be written $dV=A \ du \ dv$ for some $A$. In the case of your integral, the proper way of dealing with it is probably $df=Gdx_0dx_1\cdots dx_{n-1}dx_n$, for which its usefulness depends on the context.


问:
Integrating the complex function $f(z) = x^2 + iy^2$
Suppose $f(z) = x^2 + iy^2$ (where $z = x + iy$ ) and consider
$C:z(t) = t + it,0 \leq t \leq 1$

Evaluate
$\int_{C} f(z)dz$

I have tried solving this,
I know that
$\int_{C} f(z)dz = \int_a^b f(z(t))z'(t)dt$ for $a \leq t \leq b$

From $z(t)$ I can easily get $z'(t)$
But I struggle to obtain $f(z(t))$ because $f(z)$ is not expressed in terms of z.

Please help

答:
Hint.

Note that $f(x+iy)=x^2+iy^2$. What do you have for
$
f(t+it)
$?


问:
How to find the value of $g'(2)$?
If we have
$$
\lim_{x\to1} \cfrac{f(x)-f(1)}{x^2-1} = 4\;,
\quad
g(x)+g(2)=2\;
\quad \text{and } \quad
(f \circ g)'(2) = 8,
$$
what is the value of $g'(2)$?

To solve the problem I rewrote the limit as
$$
\lim_{x\to1} \cfrac{f(x)-f(1)}{x-1} \times \cfrac{1}{x+1} = \cfrac{f'(1)}{2} = 4
$$
so $f'(1)=8$. And we have $(f \circ g)'(2) = g'(2) \times f'(g(2)) = 8$. So in order to solving the problem I need to find $f'(g(2))$. I don't know how to do it. and also how can I use $g(x)+g(2)=2$?

答:
HINT: $g(2)+g(2)=2$ and therefore $g(2)=?$

However as $g$ is a constant function, so does $f\circ g,$ which contradicts to the given data.

Title: How to find the value of $g'(2)$?


问:
Question about improper integral
Does $\forall \epsilon \in (0,a]: \int _{\epsilon}^{a} f(x)dx =0$ imply that $\lim_{h \rightarrow 0} \int _h^a f(x)dx=0$. I guess it is somehow the definition of it, but I need to know this exactly. </p>

答:
It's even stronger. The limit only says that the integral goes to zero when $h$ goes to zero, whereas the first statement says that it is zero for all $h$.

In fact, your first statement implies that $f = 0$ a.e. on $(0,a]$, I think.

Title: Question about improper integral


问:
Fermat's Little Theorem Problem with an exponent of an exponent
How do you solve this?

$$2^{5^{2017}} \bmod 11$$

I tried my own method, and I got that:

$$2^{5^{2017}} \equiv 10\pmod{11}$$

But I'm not sure it's the correct answer.

Thanks!

答: Note that since $11$ is a prime, $2^{10}\overset{11}{\equiv}1$ Hence because $5^{2017}\overset{10}{\equiv}5$, we have $2^{5^{2017}}\overset{11}{\equiv}2^5\overset{11}{\equiv}10$ and you are right!


答:
When someone writes $f(z) = x^2 + iy^2$ is is usually understood that $z = x + yi$, i.e. $x$ is the real part of $z$ and $y$ is the imaginary part of $z$.

So if $z = t +it$, can you find the real and imaginary parts of $z$?

Title: Integrating the complex function $f(z) = x^2 + iy^2$

Body:
Suppose $f(z) = x^2 + iy^2$ (where $z = x + iy$ ) and consider
$C:z(t) = t + it,0 \leq t \leq 1$

Evaluate
$\int_{C} f(z)dz$

I have tried solving this,
I know that
$\int_{C} f(z)dz = \int_a^b f(z(t))z'(t)dt$ for $a \leq t \leq b$

From $z(t)$ I can easily get $z'(t)$
But I struggle to obtain $f(z(t))$ because $f(z)$ is not expressed in terms of z.

Please help


问:
Integral of two error functions times a Gaussian
I am trying to find some expression for this integral, ideally as a function of the constants. I can't put the full integral in the title as it's too long.

I doubt a closed form exists, but I'm even stuck with how I would go about it.

$$\frac{1}{4\sqrt{2\pi \sigma_3}}\int_{-\infty}^\infty \exp\left(-\frac{(\mu_3-z)^2}{2\sigma_3}\right)\left(1+\text{Erf}\left[\frac{\mu_2-z}{\sqrt{2}\sqrt{\sigma_2}}\right]\right)\left(1-\text{Erf}\left[\frac{\mu_1-z}{\sqrt{2}\sqrt{\sigma_1}}\right]\right)dz$$

This was derived when considering $P(Y<Z<X)$ where $Y,Z,X$ are random variables that all follow a normal distribution, but with distinct parameters.

答:
Partial Answer: If we assume independence, then you have
$$
U := X-Z \sim N(\mu_X-\mu_Z, \sigma^2_X + \sigma^2_Z)
$$
and
$$
V := Z-Y \sim N(\mu_Z-\mu_Y, \sigma^2_Z + \sigma^2_Y)
$$
and further,
$$
\text{Cov}(U, V) = -\text{Cov}(Z,Z) = -\sigma^2_Z
$$
so that
$$
\begin{bmatrix}
U\\V
\end{bmatrix} \sim N
\left (
\begin{bmatrix}
\mu_X-\mu_Z\\\mu_Z-\mu_Y
\end{bmatrix},
\begin{bmatrix}
\sigma^2_X+\sigma_Z^2  & -\sigma^2_Z\\-\sigma^2_Z & \sigma^2_Y+\sigma_Z^2
\end{bmatrix}
\right )
$$
We therefore have
$$
P(Y<Z<X) = P(U > 0, V>0).
$$
If $U,V$ were correlated standard Gaussians with correlation $\rho$, then we would have
$$
P(U>0, V>0) = \frac{1}{4} + \frac{1}{2\pi} \arcsin \rho
$$
based on the result in this question. So perhaps you can use the result there to work out something nicer for your problem?


问:
Is it possible to find a basis of $\mathbb{R}^4$ with only two vectors?
One of the exercises I did recently is the following:

Find a basis of the subspace of $\mathbb{R}^4$ consisiting of all vectors of the form:
$$ \left[
\begin{array}{c}
  x_1\\
  -5x_1 + x_2\\
  5x_1 - 8x_2\\
  -9x_1 + 7x_2
\end{array}
\right] $$

The solution to this problem is $\{(1,-5,5,-9), (0,1,-8,7)\}$.

How is it possible to have a basis of $\mathbb{R}^4$ with only 2 vectors? Don't I need 4 of them? With 2 vectors, the most I can get is a basis for $\mathbb{R}^2$, because they only span a plane, or am I completely wrong?

答: That's not a basis of all of $\Bbb{R}^4$.  That's a basis of the 2-dimensional subspace that the given vector traces out as $x_1$ and $x_2$ are varied.  It's a plane, a copy of $\Bbb{R}^2$, sitting in an ambient $\Bbb{R}^4$.


问:
Finding All Combinations
Question

Consider the following prices on balls:

\begin{alignat}{2}
  &\text{Red (R):}   &       15&\\
  &\text{Green (B):} &        6&\\
  &\text{Blue (G):}  &\qquad  8&
\end{alignat}

Question

Is there an "elegant" way of finding all the possible ways of bying balls for a total sum of $90$?

My own approach

By a brute force search, I have found the following possibilities, but can I somehow find this in a more "elegant" way?

\begin{alignat}{3}
  (0\,&\text{R},{} & 3\,&\text{G},{} &9\,&\text{B})\\
  (0\,&\text{R},   & 7\,&\text{G},   &6\,&\text{B})\\
  (0\,&\text{R},   &11\,&\text{G},   &3\,&\text{B})\\
  (0\,&\text{R},   &15\,&\text{G},   &0\,&\text{B})\\
  (2\,&\text{R},   &10\,&\text{G},   &0\,&\text{B})\\
  (2\,&\text{R},   & 2\,&\text{G},   &6\,&\text{B})\\
  (2\,&\text{R},   & 6\,&\text{G},   &4\,&\text{B})\\
  (4\,&\text{R},   & 1\,&\text{G},   &3\,&\text{B})\\
  (4\,&\text{R},   & 5\,&\text{G},   &0\,&\text{B})\\
  (6\,&\text{R},   & 0\,&\text{G},   &0\,&\text{B})
\end{alignat}

Update

$(2\,\text{R},{} 6\,\text{G},{} 4\,\text{B}) \to (2\,\text{R},{} 6\,\text{G},{} 3\,\text{B})$

答:
We are looking for values $r,g,b$ such that
$$15r +6g + 8b = 90.$$
Immediately note that since $15$ is odd, $r$ must be even, let $r = 2R$. Then we want

$$30R + 6g + 8b = 90.$$

Next, observe that $6|\;30R,\;6g,\;90 \implies 6|8b,$ so $b$ must be a multiple of 3, let $b=3B$, now we want

$$30R + 6g + 24B = 90$$

$$\iff 5R + g + 4B = 15$$

Now I would consider the cases $R=0,1,2,3$, there is still some case by case work to do but perhaps it's easier.

Title: Finding All Combinations


答: No.  $\Bbb R^4$ is four dimensional, so requires a basis consisting of four vectors.  As was stated, what you have is a basis for a subspace of $\Bbb R^4$.  A two dimensional subspace.

Title: Is it possible to find a basis of $\mathbb{R}^4$ with only two vectors?

Body:
One of the exercises I did recently is the following:

Find a basis of the subspace of $\mathbb{R}^4$ consisiting of all vectors of the form:
$$ \left[
\begin{array}{c}
  x_1\\
  -5x_1 + x_2\\
  5x_1 - 8x_2\\
  -9x_1 + 7x_2
\end{array}
\right] $$

The solution to this problem is $\{(1,-5,5,-9), (0,1,-8,7)\}$.

How is it possible to have a basis of $\mathbb{R}^4$ with only 2 vectors? Don't I need 4 of them? With 2 vectors, the most I can get is a basis for $\mathbb{R}^2$, because they only span a plane, or am I completely wrong?


问:
Find $c\in\mathbb{R}$, and $f$ such that $\int_c^xt\cdot f(t)\,\mathrm{d}x=x\sin (x)+\cos (x)+\frac{x^3}3$

Find a number and a function that satisfies:
$$\displaystyle\int_{c}^{x}t\cdot f(t)\,\mathrm{d}x=x\sin (x)+\cos (x)+\dfrac{x^3}{3}$$

I tried taking derivative both sides so I got:

$$\left(\displaystyle\int_{0}^{x}t\cdot f(t)\,\mathrm{d}t-\displaystyle\int_{0}^{c}t \cdot f(t)\,\mathrm{d}t\right)^\prime=x \cos(x)+x^2 $$

It's clear that the right part of the integral will be $0$ because it's constant:

$$\left( \displaystyle\int_{0}^{x} t \cdot f(t)\,\mathrm{d}t \right)^{\prime}=x \cos(x)+x^2$$

But I didn't do rest. Because I don't know how to derive the integral with two functions. I know with one function like $\int_{0}^{x} f(t)\,\mathrm{d}t$. Thanks!

答:
Let's define:
$$g(t)=tf(t)$$
$$F(x)=\int_0^xg(t)\,dt$$
so we have:
$$F(x)-F(c)=x\sin(x)+\cos(x)+\frac{x^3}3$$
now if we differentiate both sides it should be clear that:
$$g(x)=xf(x)=\left(x\sin(x)+\cos(x)+\frac{x^3}3\right)'$$
now calculate the derivative and divide through by $x$:
$$xf(x)=\sin(x)+x\cos(x)-\sin(x)+x^2$$
$$f(x)=x+\cos(x)$$

Now if we sub back in:
$$\int_c^x\left[t^2+t\cos(t)\right]dt=x\sin(x)+\cos(x)+\frac{x^3}{3}$$
$$x\sin(x)+\cos(x)+\frac{x^3}{3}-\frac{3c\sin(c)+3\cos(c)+c^3}{3}=x\sin(x)+\cos(x)+\frac{x^3}3$$
and now you are left with:
$$3c\sin(c)+3\cos(c)+c^3=0$$
$$c\sin(c)+\cos(c)+\frac{c^3}{3}=0$$
and now there isn't an analytical way to solve this so a numerical method is the best you can do


答:
Take $x_1= 1$ and $x_2= 0$ so that $\begin{bmatrix}0 \\-5 \\ 5 \\ -9 \end{bmatrix}$.  Take $x_1= 0$ and $x_2= 1$ so that $\begin{bmatrix} 0 \\ 1 \\ -8 \\ 7\end{bmatrix}$.

Those form a basis for this two dimensional space.


问:
Find the adjoint of an operator $T\colon \ell^1\to\ell_0^\infty$
Let $\ell_0^\infty=\{x=(x_k)_{k\in\Bbb{N}};x_k\to0\}$. I already proved that the operator $T\colon\ell^1\to\ell_0^\infty$ given by $$T(x_1,x_2,x_3,...):=\left(\sum_{j=1}^\infty x_j,\sum_{j=2}^\infty x_j,\sum_{j=3}^\infty x_j,...\right)$$ is well defined and belongs to $B(\ell^1;\ell_0^\infty)$. Of course $(Tx)_k=\sum_{j=k}^\infty x_j$ is the k-th coordinate of the vector $Tx$.

Now I need to find the k-th coordinate of the adjoint of $T$, in the sense that, if $T^\prime$ denotes the adjoint of $T$, it has the form $T^\prime\colon(\ell_0^\infty)^\prime\to(\ell^1)^\prime$. But we already konw that we can make the identification $$(\ell^1)^\prime=\ell^\infty\qquad\text{and}\qquad(\ell_0^\infty)^\prime=\ell^1,$$ hence our adjoint operator is $T^\prime\colon\ell^1\to\ell^\infty$.

The thing is, I never face this kind of problem in my life before, and in the references that my teacher is usying don't have anything like it.

Thanks for any atention or help! :)

答:
$T'((a_n)) (e_i)=(a_n) (Te_i)=(a_n) (1,1,...,1,0,0...)=a_1+a_2+...+a_i$. Here there are $i$ one's in $(1,1,...,1,0,0...)$ and $e_i$ is the sequnce with $1$ in the $i-$ th place and $0$ elsewhere.

I am writing $(a_n)(b_n) $ for the value of $(a_n) \in \ell^{1}=(\ell_0^{\infty})'$ at $(b_n) \in \ell_0^{\infty}$ so $(a_n)(b_n) =\sum_n a_nb_n$.


问:
Find the line equation $ax + by+ c =0$ from point and angle.
Given a point $P$ and an angle $\alpha$ (see image) how does one find the corresponding line equation on the form $ax + by + c = 0$?

An image of a point $P$ and an angle $\alpha$

答:
I did manage to solve it myself. Here is my solution:

Finding a line on the form $ax+by+c=0$ from a point and an angle:

Let $P_1:(x_1,y_1)$ be an arbitrary point and $\alpha$ be an arbitrary angle which together define a line in the plane. Then a new point $P_2$ on the line can be constructed by taking a unit step along the direction given by $\alpha$ hence $P_2:(x_1+cos(\alpha),y_1+sin(\alpha))$.

Then from this post we know a point $P:(x,y)$ lies on the line connecting $P_1$ and $P_2$ if and only if the area of the parallellogram with sides $P_1P_2$ and $P_1P$ is zero. This can be expressed using the determinant as
$$
\begin{align}
\begin{vmatrix}
(x_1+cos(\alpha))-x_1 & x-x_1 \\
(y_1+sin(\alpha))-y_1 & y-y_1
\end{vmatrix} = 0 \\ \Longleftrightarrow
(y_1-(y_1+sin(\alpha)))x+((x_1+cos(\alpha))-x_1)y+x_1(y_1+sin(\alpha))-(x_1+cos(\alpha))y_1=0 \\ \implies
-sin(\alpha)x+cos(\alpha)y+x_1(y_1+sin(\alpha))-y_1(x_1+cos(\alpha))=0
\end{align}
$$
Hence (up to scale) $a =-sin(\alpha)$, $b=cos(\alpha)$ and $c=x_1(y_1+sin(\alpha))-y_1(x_1+cos(\alpha))$.


问:
Theorem. (Lindenbaum)
I have found the following Lindenbaum theorem (here, page7)

It states that:

Suppose $X$, $Y$ are linear orders. If $X$ is isomorphic to
an initial segment of $Y$ (with $f$) , and $Y$ is isomorphic to a final segment of $X$ (with $g$), then X $\simeq$ Y.

I can't seem to prove it. I have the bijection in the constructed out of $f$, $g$
as in the classical proof of the Schroeder-Bernstein theorem, but I can't show that it's  order embedding. Any tips ?

Edit: In the article it's written " the hypotheses(f, g are isomorphisms) guarantee that this bijection is order-preserving"

But how does it guarantee it ?

答:
Disclaimer: I'm not an expert, just want to try solving this problem.

Let $X_f$ be final segment of $X$ and $X_i = X \setminus X_f$, i.e. $X = X_i \sqcup X_f$.
Let $Y_i$ be initial segment of $Y$ and $Y_f = Y \setminus Y_i$, i.e. $Y = Y_i \sqcup Y_f$.
Let's construct a chain
$$
\cdots \rightarrow  f^{-1}(g^{-1}(a)) \rightarrow g^{-1}(a) \rightarrow   a  \rightarrow  f(a) \rightarrow  g(f(a)) \rightarrow \cdots
$$
This sequence may terminate on the left either in set $X$, or in set $Y$, or it may never terminate.
Thus we can split all elements into $X$-stoppers, $Y$-stoppers, and never-stoppers.

We use $f$ as bijection for $X$-stoppers and never-stoppers, obviously we can compare any two of them because $f$ preserves order.

We use $g$ as bijection for $Y$-stoppers, obviously we can compare any two of them because $g$ preserves order.

We have to prove that $X$-stopper can be compared to $Y$-stopper.

Statement 1 All $X$-stopper chains originate in $X_i$.
Proof. Consider the last $f^{-1}$ in the chain above that yielded an element $x \in X$. By the definition of $X$-stopper there is no such $y \in Y$ that $g(y)=x$. But $Y$ is isomorphic to $X_f$, thus $x \notin X_f \rightarrow x \in X_i$.

Statement 2 All $Y$-stopper chains originate in $Y_f$.

Final Statement
All $X$-stoppers are smaller than $Y$-stoppers.

Proof
Consider two elements $x_X$ and $x_Y$ from $X$-stopper and $Y$-stopper chains.
Let me apply $f^{-1}$ and $g^{-1}$ until exhaustion.
Note that sign $<$ or $>$ between $x_X$ and $x_Y$ remains unchanged at every step since we are applying the same function to both elements simultaneously.
Suppose, the $X$-chain was the first one to exhaust, which means we can draw our chains as
$$
\begin{aligned}
      &x'_X \xrightarrow[]{f} \dots \xrightarrow[]{g} x_X \xrightarrow[]{f} \dots \\
y''_Y \xrightarrow[]{g} \dots \xrightarrow[]{g} &x'_Y \xrightarrow[]{f} \dots \xrightarrow[]{g} x_Y \xrightarrow[]{f} \dots \\
\end{aligned}
$$
By the statement 1 $x'_X \in X_i$.
The other chain did not stop since it is a $Y$-chain, thus there exists $g^{-1}(x'_Y)$ and $x'_Y \in X_f$.
Thus $x'_X < x'_Y$ and as a result we get $x_X < x_Y$.
If we stop in $Y$, similar reasoning applies.

Title: Theorem. (Lindenbaum)


问:
Taylor expansion in elementary symmetric polynomials
Assume that I have an analytic, symmetric function $F(x,y,z)$. Such a function should have an expansion of the form
$$  F(x,y,z)  = \sum_{k,m,n\geq 0} c_{k,m,n} e_1^ne_2^m e_3^k$$

where $e_1,e_2,e_3$ are the elementray symmetric polynomials. How would I go about calculating these $c_{m,n,k}$ in an efficient way (by efficient I mean something better than rewriting the polynomials $$ P_{n_1,n_2,n_3}(x_1,x_2,x_3) = \sum_{\sigma \in S_3} x^{n_{\sigma(1)}}y^{n_{\sigma(2)}}z^{n_{\sigma(3)}}$$
 in terms of elementary polynomials by hand...).

More concretely I would be interested in knowing whether the function
$$ G(x_1,x_2,x_3)  =  \sum_{\sigma \in S_3} \operatorname{exp}\Big[ Ax_{\sigma(1)}+Bx_{\sigma(2)}+Cx_{\sigma(3)}\Big] $$
can be rewritten as a function in elementary symmetric polynomials in a nice form.

答:
Elementary polynomial for the elementary symmetric function
$$f\left( t \right) = {t^3} - {e_1}{t^2} + {e_2}t - {e_3} = \left( {t - x} \right)\left( {t - y} \right)\left( {t - z} \right)$$
Symmetric function
$$F\left( {x,\;y,\;z} \right) = \sum\limits_{{\alpha _1} + 2{\alpha _2} + 3{\alpha _3} = {\beta _x} + {\beta _y} + {\beta _z} = 0}^\infty  {K\left( {{\alpha _1},\;{\alpha _2},\;{\alpha _3},\;{\beta _x},\;{\beta _y},\;{\beta _z}} \right)\frac{{e_1^{{\alpha _1}}e_2^{{\alpha _2}}e_3^{{\alpha _3}}}}{{{\beta _x}!{\beta _y}!{\beta _z}!}}\frac{{{\partial ^{{\beta _x} + {\beta _y} + {\beta _z}}}F}}{{\partial {x^{{\beta _x}}}\partial {y^{{\beta _y}}}\partial {z^{{\beta _z}}}}}(0,0,0)} $$
Coefficient $K$ has a generating function $J$;

$$K\left( {{\alpha _1},\;{\alpha _2},\;{\alpha _3},\;{\beta _x},\;{\beta _y},\;{\beta _z}} \right) = \mathop {\lim }\limits_{\left( {p,\,q,\,x,\,y,\,z} \right) \to \left( {0,\;0,\;0,\;0,0} \right)} \frac{{{{\left( { - 1} \right)}^{{\beta _x} + {\beta _y} + {\beta _z}}}}}{{{\alpha _2}!{\alpha _3}!{\beta _x}!\left( {{\beta _x} + {\beta _y}} \right)!\left( {{\beta _x} + {\beta _y} + {\beta _z}} \right)!}}\frac{{{\partial ^{{\alpha _2} + {\alpha _3} + 3{\beta _x} + 2{\beta _y} + {\beta _z}}}}}{{\partial {p^{{\alpha _2}}}\partial {q^{{\alpha _3}}}\partial {x^{{\beta _x}}}\partial {y^{{\beta _x} + {\beta _y}}}\partial {z^{{\beta _x} + {\beta _y} + {\beta _z}}}}}J_{{\beta _x},{\beta _y}}\left( {p,\,q,\,x,\,y,\,z} \right)$$

$$J_{u,v}\left( {p,\,q,\,x,\,y,\,z} \right) = \frac{{{y^{u + 1}}{z^{u + v + 2}}\left( {y + z + yz} \right)\left( {1 + z + p{z^2}} \right)}}{{\left( {xy + xz + yz + xyz} \right)\left( {y\left( {y + z} \right)\left( {1 + z} \right) + {z^2} + p{y^2}{z^2}} \right)\left( {1 + z + p{z^2} + q{z^3}} \right)}}$$
It is possible to write down a closed-form expression for the coefficient $J$. We would recommend the use of a math formula processing software like Mathematica to get a concrete number. {I am very old and terminally ill, I do not have enough vigor to complete the calculation, sorry. I tested some examples using Mathematica though.}

Full proof  is lengthy for the generating function and not having enough space here, so we give a sketch of proof:
In this proof, we switch notation $x,y,z$. Now, ${x_1},{x_2},{x_3}$ are three roots of the elementary polynomial $f$ assuming, without loss of generality, they are all distinct. It is a standard assumption, especially when we deal with symmetric expressions.
Review that the Lagrangean interpolation formula is the same as polynomial remainder.
$$\sum\limits_{i = 1}^3 {g\left( {{x_i}} \right)\frac{{f\left( x \right)}}{{\left( {x - {x_i}} \right)f'\left( {{x_i}} \right)}}}  = g\left( x \right)\mathop {\bmod }\limits_x f\left( x \right)$$
where $f'$ is the derivative of $f$. Notice that the remainder of the cubic polynomial is a quadratic polynomial, where the constant term is a symmetric function. Apply this observation three times to our symmetric function of interest, if $F$ is a symmetric function, then
$$F\left( {{x_1},{x_2},{x_3}} \right) = \left\{ {\left\{ {F\left( {x,y,z} \right)\mathop {\bmod }\limits_{x \to 0} \frac{{f\left( x \right)}}{{\left( {x - y} \right)\left( {x - z} \right)}}} \right\}\mathop {\bmod }\limits_{y \to 0} \frac{{f\left( y \right)}}{{y - z}}} \right\}\mathop {\bmod }\limits_{z \to 0} f\left( z \right)$$
Now remember the Cauchy integration result on a symmetric sum
$$\sum\limits_{i = 1}^3 {g\left( {{x_i}} \right)}  = \frac{1}{{2\pi \sqrt { - 1} }}\oint {g\left( x \right){{\left\{ {\log f} \right\}}^\prime }\left( x \right)dx} $$
this contour contains all roots of the elementary polynomial $f$, we apply this Cauchy formula to the Lagrangean interpolation/polynomial remainder calculation. The equivalent sequential remainder expression for the symmetric function has a slight problem that divisors are rational functions, but this problem can be easily fixed replacing them by difference polynomials, also do not forget replacing similarly $g$‘s coefficients in the original Lagrange formula in the Cauchy integration form. Consequently, our symmetric function $F$ is equivalently expressed as three-dimensional multiple complex integration in ${\mathbb{C}^3}$. After we replace integration variables to their reciprocal and so does each contour reverse direction accordingly, we find an aesthetic candidate.
$$\hat J\left( {p,\,q,\,x,\,y,\,z} \right) = \left( {1 - \frac{{qx\left( {x - y} \right)\left( {x - z} \right)}}{{1 + x + p{x^2} + q{x^3}}}} \right)\left( {1 - \frac{{q{y^2}\left( {y - z} \right)}}{{1 + y + p{y^2} + q{y^3}}}} \right)\left( {1 - \frac{{q{z^3}}}{{1 + z + p{z^2} + q{z^3}}}} \right)$$
it is not yet a generating function unfortunately, so we prepare a re-modification of complex modulo integrations, we find the result. QED

The Symmetric sum of exponential functions
$$\begin{array}{c}
G\left( {{x_1},\,{x_2},\,{x_3}} \right) = \sum\limits_{\sigma  \in {S_3}} {\exp \left( {{C_1}{x_{\sigma \left( 1 \right)}} + {C_2}{x_{\sigma \left( 2 \right)}} + {C_3}{x_{\sigma \left( 3 \right)}}} \right)} \\
 = \sum\limits_{{\alpha _1} + 2{\alpha _2} + 3{\alpha _3} = {\beta _1} + {\beta _2} + {\beta _3} = 0}^\infty  {K\left( {{\alpha _1},\;{\alpha _2},\;{\alpha _3},\;{\beta _1},\;{\beta _2},\;{\beta _3}} \right)\frac{1}{{{\beta _1}!{\beta _2}!{\beta _3}!}}\left\{ {\sum\limits_{\sigma  \in {S_3}} {C_{\sigma \left( 1 \right)}^{{\beta _1}}C_{\sigma \left( 2 \right)}^{{\beta _2}}C_{\sigma \left( 3 \right)}^{{\beta _3}}} } \right\}e_1^{{\alpha _1}}e_2^{{\alpha _2}}e_3^{{\alpha _3}}}
\end{array}$$


问:
A question on countability of isolated points of a subset of R
The question is to prove that with respect to the euclidean metric on the Real numbers prove that if A is any subset of R, isoA is countable and hence deduce  that if A is uncountable the A' is uncountable.

I am still not that comfortable with proving things like countability so any answers or help is appreciated

thanks

答:
Seirios has given a fine answer directed at your specific exercises; mine is just additional commentary that may add some insight or give you another way of thinking about the problems.

Note that one can prove the second result without explicitly proving the first. Suppose that $A$ is an uncountable subset of $\Bbb R$. Let $\mathscr{B}$ be the set of open intervals with rational endpoints; $\Bbb Q$ is countable, so $\mathscr{B}$ is countable as well. If $x\in\Bbb R\setminus A'$, there is some $B_x\in\mathscr{B}$ such that $x\in B_x$ and

$$B_x\cap A=\begin{cases}
\varnothing,&\text{if }x\notin A\\
\{x\},&\text{if }x\in A\;.
\end{cases}$$

(The cases can be combined into the single statement that $B_x\cap A\subseteq\{x\}$, but I thought that it might be clearer to separate the two possibilities.)

Let $\mathscr{B}_0=\{B_x:x\in\Bbb R\setminus A'\}$, and let $U=\bigcup\mathscr{B}_0$, and let $A_0=A\setminus U$. $\mathscr{B}_0\subseteq\mathscr{B}$, so $\mathscr{B}_0$ is countable, and each member of $\mathscr{B}_0$ contains at most one point of $A$, so $U\cap A$ is countable. If $A_0$ were countable, then $A=A_0\cup(U\cap A)$ would be the union of two countable sets and would therefore be countable, so $A_0$ must be uncountable. Thus, if we can show that $A_0\subseteq A'$, it will follow immediately that $A'$ is uncountable.

Suppose that $x\in A_0$. Then $x\notin U$, so $x\notin B_y$ for any $y\in\Bbb R\setminus A'$. In particular, $x\notin\Bbb R\setminus A'$, i.e., $x\in A'$, and we’re done: $A'$ is uncountable.

Your first result is hidden inside the proof above: it’s essentially the observation that $U\cap A$ is countable.

Almost the same argument actually proves the following stronger result:

Let $A$ be an uncountable subset of $\Bbb R$. Then there is an uncountable $A_0\subseteq A$ such that if $x\in A_0$, and $U$ is open neighborhood of $x$, then $U\cap A_0$ is uncountable (and of course then $U\cap A$ is also uncountable). In other words, $A$ has uncountably many points that are very far indeed from being isolated: their open nbhds not only have infinite intersection with $A$, but actually have uncountable intersection with $A$.

The details are in this answer.


问:
Alternative approach to prove that if a sequence is a Cauchy Sequence, then it converges.
I am familiar with the traditional proof that if a sequence of real numbers is a Cauchy sequence, then it is convergent. However, I am stuck on a problem that asks me to prove the implication using the fact that if all the subsequences of a sequence converge to a real number then the sequence itself converges.

Question: (i) Let $(a_n)$ be a bounded sequence of real numbers and let $x \in \mathbb{R}$ have the property that every subsequence of $(a_n) \neq (a_n)$ converges to $x$. Show that $(a_n) \to x$. (ii) Furthermore, use this result to prove that if a sequence is a Cauchy sequence, then it converges.

Proving part (i) was fairly easy for me. Outlining a proof for part (ii) is tripping me up. I am having a hard time applying the properties of a Cauchy sequence to the proof other than the fact that this implies the sequence is bounded.

Any help is greatly appreciated.

答:
Background of this question: Some lecturers in analysis use this method to prove that Cauchy implies convergence. Generally, these lecturers teach it after teaching Bolzano-Weierstrass Theorem. Now let's prove it.

Suppose $\{a_j\}$ is Cauchy, then we know $\{a_j\}$ is bounded. (Recall that every Cauchy sequence is bounded.) By B-W theorem, there exists a subsequence $\{a_{j_k}\}$ which converges to $c$. Notation: $\lim\limits_{k \to \infty} a_{j_k}=c$

Our goal is to show $\lim\limits_{j \to \infty} a_{j}=c$

$\forall \epsilon >0$, we need to show $\exists N \in \mathbb{N}$, s.t. whenever $j>N$, we have $|a_j-c|<\epsilon$.

But we know $\{a_j\}$ is Cauchy, thus given$\frac{\epsilon}{2}>0$, $\exists N \in \mathbb{N}$ s.t. whenever $j>k>N$, we have $|a_j-a_k|<\frac{\epsilon}{2}$.

Now when $j>N$, $|a_j-c| \leq |a_j-a_{j_k}+a_{j_k}-c|\leq |a_j-a_{j_k}|+|a_{j_k}-c|<\frac{\epsilon}{2}+\frac{\epsilon}{2}=\epsilon$, where we choose $k$ so big s.t. $j_k > N$ and $|a_{j_k} - c|<\frac{\epsilon}{2}$.


问:
$A_1 \times ... \times A_n$ is countable if $A_1, ..., A_n$ are countable
Suppose that $A_1, ..., A_n$ are countable sets. Show that the cartesian product $A := A_1 \times ... \times A_n$ is countable.

My attempt:

Sets are said to be countable if they are finite or if they have the same cardinality as some subset of $\mathbb{N}$ (i.e. we can find some bijection $f: A \rightarrow S$ or $f: S \rightarrow A$ where $S \subset \mathbb{N}$).

Assume that $A_1, ..., A_n$ are countable sets. Then, there exists bijections $fi: \mathbb{N} \rightarrow A_i$ for $i = 1, ..., n$.

Define $g: \mathbb{N} \rightarrow A$ as follows

My issue arises here in finding such a bijective function without it being too complicated. How would I go about finding one? I am also open to any suggestions. Any assistance is welcomed.

答:
It suffices to show that the cartesian product of two countable sets is countable, because an inductive argument will yield the general result.

According to your definition of countable sets, if at least one set is finite, the result is immediate. So the only interesting case is when both sets are countably infinite.

In particular, it is enough to prove that $\mathbb{N}\times\mathbb{N}$ is countable. Define a function $f:\mathbb{N}\to\mathbb{N}\times\mathbb{N}$ as follows. If $n\in\mathbb{N}$ then $n$ can be written uniquely as $2^{m-1}\cdot(2k-1)$, where $m,k\in\mathbb{N}$. Define $f(n)=(m,k)$. By the uniqueness of such expressions, we see that $f$ is one-to-one. On the other hand, if $(m,k)$ is any pair of $\mathbb{N}\times\mathbb{N}$ then setting $n=2^{m-1}\cdot(2k-1)$ yields an integer so that $f(n)=(m,k)$, so $f$ is onto. Thus $f$ is a bijection, and this shows that $\mathbb{N}\times\mathbb{N}$ has the same cardinality as $\mathbb{N}$, i.e. it is countable.


答:
It is sufficient to prove this in the case where $A_i=\mathbb{N}$ for all $i$. So we have to show that $\mathbb{N^n}$ is a countable set. Let $p_1,p_2,...,p_n$ be the first $n$ prime numbers, and define a map $f:\mathbb{N^n}\to\mathbb{N}$ by:

$f(m_1,m_2,...,m_n)=p_1^{m_1}p_2^{m_2}...p_n^{m_n}$

It is well known that prime decomposition is unique, and so $f$ is injective. So we showed that there is an injection from $\mathbb{N^n}$ to $\mathbb{N}$, and so $\mathbb{N^n}$ is countable. (it is in bijection with the image of $f$, which is a subset of $\mathbb{N}$)

Title: $A_1 \times ... \times A_n$ is countable if $A_1, ..., A_n$ are countable

Body:
Suppose that $A_1, ..., A_n$ are countable sets. Show that the cartesian product $A := A_1 \times ... \times A_n$ is countable.

My attempt:

Sets are said to be countable if they are finite or if they have the same cardinality as some subset of $\mathbb{N}$ (i.e. we can find some bijection $f: A \rightarrow S$ or $f: S \rightarrow A$ where $S \subset \mathbb{N}$).

Assume that $A_1, ..., A_n$ are countable sets. Then, there exists bijections $fi: \mathbb{N} \rightarrow A_i$ for $i = 1, ..., n$.

Define $g: \mathbb{N} \rightarrow A$ as follows

My issue arises here in finding such a bijective function without it being too complicated. How would I go about finding one? I am also open to any suggestions. Any assistance is welcomed.


问:
In how many ways can they form a committee of 6 members that does not include both Chloe and George?
The Data Science Club has 18 members, including Chloe and George who are currently not in good terms. In how many ways can they form a committee of 6 members that does not include both Chloe and George? (Note: Either Chloe or George can be on the committee but not both of them.)</p>

答: In case you haven’t figured it out yet, it’s ${18\choose 6}-{2\choose2}{16\choose4}= {18\choose 6}-{16\choose4} $. There are only 16 people to choose from after forcing Chloe and George to be in the group. That is, you subtract the number of groups with Chloe and George (the undesired groups) from the total number of possible groups to get the number of groups that are permissible.


答:
Do it by induction.

Prove that if $A,B$ are countable then $A\times B$ are countable (use the "diagonal line" argument).

$1\mapsto (a_1,b_1); 2 \mapsto (a_2,b_1); 3\mapsto (a_1, b_2); 4\mapsto (a_3, b_1); 5\mapsto (a_2,b_2); 6\mapsto (a_1,b_3);7\mapsto (a_4, b_1).... etc. $

Then if we assume $A_1 \times A_2..... \times A_k$ is countable then $A_1\times A_2\times..... \times A_k \times A_{k+1} = (A_1\times A_2\times..... \times A_k) \times A_{k+1}$ is the product of TWO countable sets and is therefore countable.

So by indcuction $A_1 \times A_2..... \times A_n$ is countable for all $n$.


问:
Difference between multiplying and dividing numbers by percentages?
Basically, what is the difference between $1000\times1.03$ and $1000/.97$?

For some reason I feel like both should result in the same number. I only ask because I'm working a problem with a percentage of waste added in. For $3\%$ waste, I would think that you could multiply the amount by $1.03$ to add $3\%$. However, my professor divided by $.97$.

答:
Just multiply

$$1.03 \times 0.97=0.9991$$ to see the difference.

Something that increases 3% and then decreases 3%... has slightly decreased. This is because the decrease of 3% is not measured compared to the original value, but to the increased value.


答:
$$
\newcommand{\bv} {{\mathbf v}}
\newcommand{\bu} {{\mathbf u}}
$$
If the line direction is $\alpha$, then a vector pointing along the line is $\pmatrix{\cos \alpha \\ \sin \alpha}$. Rotating counterclockwise by 90 degrees, we get that the vector $\bv = \pmatrix{-\sin \alpha \\ \cos \alpha}$ is perpendicular to the line. That means that if $\bu$ is any vector pointing along the line, we know that $\bv \cdot \bu = 0$. Hold that thought.

Now we have a point $(x_1, y_1)$ on the line, and if $(x,y)$ is an arbitrary point on the line, then $\bu = \pmatrix{x-x_1 \\ y - y_1}$ is a vector pointing along the line, so $\bv \cdot \bu = 0$. Writing that out, we get
\begin{align}
0
&= \bv \cdot \bu \\
&= \pmatrix{-\sin \alpha \\ \cos \alpha} \cdot \pmatrix{x-x_1 \\ y - y_1} \\
&= -\sin (\alpha) (x-x_1)+ \cos (\alpha) (y - y_1) \\
\end{align}
Separating out the coefficients of $x$ and $y$, we get
$$
-\sin(\alpha) x + \cos(\alpha) y + (\sin(\alpha)x_1 - \cos(\alpha)y_1) = 0
$$
which has the required form.

Notice that my form for $c$ is simpler than yours --- that's because you have both $x_1y_1$ and $-x_1 y_1$ in yours.

Title: Find the line equation $ax + by+ c =0$ from point and angle.

Body:
Given a point $P$ and an angle $\alpha$ (see image) how does one find the corresponding line equation on the form $ax + by + c = 0$?

An image of a point $P$ and an angle $\alpha$


答: You could use an approach similar to the Cantor proof that the algebraic numbers are countable.  Order the $A_k=(a_k(j))$  For an element of $A$,let $B=(a_1(j_1),....a_n(j_n))$. Define $b=\sum_{k=1}^n j_k$.  For each $b$ there are only a finite number of $B's$ with $b$ as the sum.  These can be ordered in any way.  The totality for all (integer) values of $b$ is countable and includes all members of $A$.


问:
Using back substitution to solve the matrix system with fractions
I was asked to use back substitution to solve the system:
$$ \left[
\begin{array}{ccc|c|c}
   1 & 3/2 &1/2&a&9/4\\
   0 & 1&1&b&-1/2\\
0&0&1&c&0
\end{array}
\right] $$

The answer =

$$ \left[
\begin{array}{c}
   3\\
   -1/2\\
0
\end{array}
\right] $$

I understand how to perform the back substitution and get the first two answers: $c = 0$ and $b = -\frac{1}{2}$

But how do we get $a = 3$?

The fractional back substitution must be getting me confused.

答:
In case you haven’t figured it out yet, your system of equations is

$a+3/2b+1/2c=9/4\\
b+c=-1/2\\
c=0$

Now you simply work your way up, “back substituting” what you get from each rung of the ladder. Third line already gives you $c=0$ in plain sight. So now you have

$b+c=-1/2\\
b+0=-1/2\\
b=-1/2$

Now the first line tells you
$a+3/2b+1/2c=9/4\\
a+3/2(-1/2)+0=9/4\\
a-3/4=9/4\\
a=12/4=3$

So your answers are $a=3, b=-1/2, c=0$. A system of equations in rref or row echelon form are especially conducive to being solved via back substitution, because what else would you do? Add equation 1 to equation 3? Then you lose the information that c already equals 0.

Note you don’t have to back substitute one line at a time. You could substitute what you get from the bottommost equation into all the equations above it at once.


问:
Which one result in mathematics has surprised you the most?
A large part of my fascination in mathematics is because of some very surprising results that I have seen there.

I remember one I found very hard to swallow when I first encountered it, was what is known as the Banach Tarski Paradox. It states that you can separate a ball $x^2+y^2+z^2 \le 1$ into finitely many disjoint parts, rotate and translate them and rejoin (by taking disjoint union), and you end up with exactly two complete balls of the same radius!

So I ask you which are your most surprising moments in maths?

Chances are you will have more than one. May I request post multiple answers in that case, so the voting system will bring the ones most people think as surprising up. Thanks!

答:
How the "inverse" of area under a curve is the slope.

(I mean: $\frac{d}{dx} \int x {\mathrm dx} = x$)

LastEditorDisplayName: anon


问:
Clarification on $L^1$ convergence
I am asked to prove the following:

Let $f_h(x)=f(x-h)$.

$(1)$ If $f$ is a continuous function of bounded support, then $\vert \vert f_h-f \vert \vert_{L^1} \rightarrow 0,$ as $h$ goes to $0$.

$(2)$ If $f \in L^1(\mathbb{R})$, then $\vert \vert f_h-f \vert \vert_{L^1} \rightarrow 0,$ as $h$ goes to $0$.

Now I know how to do the first, but is there any work to do for the latter? since continuous function of bounded support are dense in $L^1$ is there much work for $(2)$ or? So can I approximate $f$ by a continuous function of bounded support for $(2)$?

答:
Your intuition is correct. We need the following observation: for any $f\in L^1(\mathbb{R})$ we have that $\|f\|_1=\|f_h\|_1$ for all $h\in\mathbb{R}$. Prove this first for simple functions using the fact that the Lebesgue measure is invariant under translation; then prove this for positive, measurable functions by an approximation argument and finally prove this for $L^1$ functions by breaking them down to their real and imaginary parts first and then positive and negative parts too.

Now let $f\in L^1$ and $\varepsilon>0$. Find a function of bounded support so that $\|f-g\|_1<\varepsilon$. Then
$$\|f-f_h\|_1\leq\|f-g\|_1+\|g-g_h\|_1+\|g_h-f_h\|_1=\|f-g\|_1+\|g-g_h\|_1+\|(f-g)_h\|_1=$$ $$=2\|f-g\|_1+\|g-g_h\|_1<2\varepsilon+\|g-g_h\|_1\xrightarrow{h\to0}2\varepsilon.$$
Since $\varepsilon>0$ is arbitrarily small, this proves that $\|f-f_h\|_1\to0$.

Title: Clarification on $L^1$ convergence


问:
Number of doublings in a bounded submartingale
Let $1\leq X_n \leq N$ be a submartingale with respect to a filtration $(\mathcal{F}_n)_{n\geq 1}$, i.e., $\mathbb{E}[X_{n+1}\mid\mathcal{F}_{n}] \geq X_n$ and $N \geq 0$ is a scalar. Suppose $X_0 = 1$ and let $n_k$ denote the index at which the process doubles for the $k$-th time, i.e., $n_0 = 0$ and for $k \geq 1$,
$$
n_k = \min\{n > n_{k-1}: X_n > 2X_{n-1}\}.
$$
Let $\kappa_N$ be the set of all these indices until time $N$ and $K_N = |\kappa_N|$ be the total number of doublings until time $N$. Can we bound $\mathbb{E}[K_N]$?

My Approach: In the special case where $X_{n+1} \geq X_n$ monotonically, we claim that $K_N \leq 1 + \log N$. because otherwise,
$$
X_{n_{K_N}} = \prod_{2 \leq n \leq n_{K_T}}\frac{X_{n}}{X_{n-1}} \cdot X_1 \geq \prod_{n \in \kappa_N, \\n \neq 1}\frac{X_{n}}{X_{n-1}} > 2^{K_N-1} > N,
$$
which is a contradiction with $X_{n_{K_N}} \leq N$. Can we prove something similar in the more general case of submartingales?

Edit: Originally the question was with $0 \leq X_i \leq N$. But, now it is changed to $1 \leq X_i \leq N$ to avoid the counter example provided by @Peter Morfe.

答:
Here is an explicit submartingale for which $E[K_N] \ge cN$ for a constant $c>0$. (We will not try to optimize this constant).

Suppose that $2^\ell-1 \le N <2^{\ell+1}-1$. Consider the Markov chain $\{X_t\}$ (a variant of the "greasy ladder") that from 1 jumps to 3, and for each $j \in [2,\ell-1]$ jumps from $2^j-1$  to either $1$ or $2^{j+1}-1$ with equal probability. The chain is started at $X_0=1$ and is absorbed at $2^\ell-1$.
This chain is indeed a submartingale.

Let $\tau$ be the first time $t$ when $X_t=2^\ell-1$. Then $$P(\tau \ge N/8)\ge (1-2^{2-\ell})^{N/8-1} \ge 1/e$$ so $E[\tau \wedge N] \ge \frac{N}{8e}$. Let $I_t=1$ if $X_t \ge X_{t-1}$ and $I_t=0$  otherwise. Then $S_T=\sum_{t=1}^T (I_t-1/2)$ is a submartingale, so $$0 \le E[S_{\tau \wedge N}] \le E[K_N]- E[\tau \wedge N]/2 \le
E[K_N]-\frac{N}{16e} \,.$$


问:
Compute $f(f(1))$ if $f(x)=3x+2$
Question :

Let the function $f$ be defined by $f(x)=3x+2$.

Then $f(f(1))=$

$f^2(1)$=

and $f^2(f(1))=$

for this first one I replaced the $x$ by $1$ so it gives $5$, the second one $=25$ and $100$ for the last
but I am not sure if it is the right way to resolve it.
Thank you,

Title: Compute $f(f(1))$ if $f(x)=3x+2$

Body:
if $f(x)=3x+2$ then $f(f(x))=f(3x+2)=3(3x+2)+2=9x+6+2=9x+8$.

So you can calculate $f(f(1))$ in either of two ways:

$$f(f(1))= 9(1)+ 8= 9+ 8= 17$$ or

$$f(1)=  3(1)+ 2= 3+ 2= 5$$ so
$$f(f(1))= f(5)= 3(5)+ 2= 15+ 2= 17.$$


答: Well you see $1/.97\approx 1.0309$. So that’s actually about a 3.1% increase. Your professor was taking liberties with saying a 3% increase. That is $x/.97=1.031x$, slightly more than 3%. It is unknown whether the professor intentionally did something opposite of tradition.

Title: Difference between multiplying and dividing numbers by percentages?

Body:
Basically, what is the difference between $1000\times1.03$ and $1000/.97$?

For some reason I feel like both should result in the same number. I only ask because I'm working a problem with a percentage of waste added in. For $3\%$ waste, I would think that you could multiply the amount by $1.03$ to add $3\%$. However, my professor divided by $.97$.


问:
Logic problem that i am struggling with
$1$-   The universal set for this problem is the set of students attending Miskatonic University. Let

-$M$ denote the set of Math majors.

-$CS$ denote the set of Computer Science majors.

-$T$ denotes the set of students who had a test on Friday.

-$P$ denotes those students who ate pizza last Thursday.

Using only the set theoretical notation we have following assertions

$1$. Computer Science majors had a test on Friday $$CS \subseteq T$$

$2$. No Math majors ate pizza last Thursday-----> $$M^c \subseteq P$$ or $$M \cap P=\varnothing$$ (The book said that the answer is $M \cap P=\varnothing$, idk why?)

$3$. Some Math majors did not eat pizza last Thursday$$C^c \cap P=\varnothing?$$ I am struggling with translating context to symbols.

$4$. Those Computer Science majors who did not have a test on Friday ate pizza on Thursday
$$CS^c \in T \subseteq P?$$
$5$. Math or Computer Science major who ate pizza on Thursday did not have a test on Friday
$$M \cup CS \subseteq P \land T = \varnothing?$$

I need help please. I am trying my best to learn the logic and operations on sets.

I understand that $A \cup B$ reads "$x \in A$ or $x \in B$"; $A \cap B$" means $x \in A$ and $x \in B$" but when it comes to question like this, I have no clue how to write. For instance, $(2)$, how is $M \cap P = \varnothing$? Why can't it be $M^C  \subseteq P$? How does "and" or" "E" used to write symbols?

答:
$(1)$ is correct.

$(2)$ is $M\cap P=\varnothing $ Since $M\cap P$ are the Math Majors who ate Pizza last Thursday.

$(3)$ is $M\cap P^c\ne\varnothing $ Since $M\cap P^c$ are the Math Majors who did not eat Pizza last Thursday.

$(4)$ is $(CS\cap T^c)\subseteq P $ Since $CS\cap T^c$ are the Computer Science Majors who did not have test on Friday.

$(5)$ is $((M\cup CS)\cap P)\subseteq T^c $ Since$((M\cup CS)\cap P)$ are Math or Computer Science Majors who ate Pizza last Thursday.

Title: Logic problem that i am struggling with


问:
Formal proof for the limit of $\frac{\tanh(x)-1}{e^{-2x}}$ as $x \rightarrow \infty$
Formal proof for the limit of $\frac{\tanh(x)-1}{e^{-2x}}$ as $x \rightarrow \infty$.

So far

Keep in mind I have to use the definition for a limit. I.e for this would be a proof for the limit at $\infty$ with a finite value (I have checked this to be $-2$). I would have to use the definition that states $\lim_{x \ \to \  \infty} f(x) = L \Leftrightarrow
  \forall \ \epsilon>0\; (\exists \ \delta : (\;x>\delta\implies |f(x) - L|\leq\epsilon)).
$ i.e I start off by by finding that $$\left | \frac{\tanh{x}-1}{e^{-2x}}-(-2) \right | \iff \left | \frac{2}{1+e^{2x}} \right |$$ Or just
$\frac{2}{1+e^{2x}}$ since we will assume $\epsilon >0$. Now I would have to to start my proof and this is where I struggle.

I know that
"for alle $\epsilon>0$ let $\delta=\frac{2}{1+e^{2\epsilon}}>0$ and $x>\delta>0$....."

*Remark: I know we use $N,M,K$ and not $\delta$ normally.

答:
$$\frac{\tanh(x)-1}{e^{-2x}}=\bigg(\frac{e^{2x}-1}{e^{2x}+1}-1\bigg)\cdot e^{2x}=\frac{-2e^{2x}}{{e^{2x}+1}}$$

So $$\bigg|\frac{-2e^{2x}}{e^{2x}+1}-(-2)\bigg|=\bigg|2-\frac{2e^{2x}}{e^{2x}+1}\bigg|=\frac{2}{e^{2x}+1}$$

Now if $\varepsilon>0$ find $\delta>0$ such that when $x>\delta$ we have that $e^{2x}+1>\frac{2}{\varepsilon}$. We can do that because the function $\phi(x)=e^{2x}+1$ tends to $\infty$ as $x\to\infty$.
Then for $x>\delta$ we have that
$$e^{2x}+1>\frac{2}{\varepsilon}\iff\frac{2}{e^{2x}+1}<\varepsilon. $$


问:
Is there a Math symbol that means "associated"
I am looking for a Math symbol that means "associated" and I don't mean "associated" as something as complicated as isomorphism or anything super fancy.

I am looking for a symbol that means something like "$\triangle ABC$ [insert symbol] $A_{1}$" (as in triangle ABC "associated" with area_{1}) Or want to say something like "The eigenvector associated with the eigenvalue"

You get the idea.

答:
You can always make up a symbol (I've done so on several occasions), but, like user4594 said, whenever you first use it in a work, you need to always precisely define it. If you find yourself needing to use the word association or correlation a lot, and you don't want to use existing notation (e.g. corr(A,B)), then you can insert a symbol and either make a footnote that defines the symbol or include a legend/key at the beginning or end of what you are creating.

Using a symbol for every relationship between arguments often unnecessarily burdens your reader, but sometimes, I find that it helps me remember things better if I can assign symbols to certain words, especially if I end up using the words repeatedly in a single work.

Here is an example of a group of symbols I have created (not designed, but assigned):
"⌽A, ⦵B, ⍉C, ⦵D"
...meaning:
"If A, then B, else if C, then D."
I know that "A⟶B" ("If A, then B") is more commonly used, but in my particular document, a ⟶ meant something else entirely.
If you want to make a novel symbol but don't want to personally design one (font editors can get really complicated), there are some fonts you can download that consist mostly or entirely of symbols (many of which you will probably have never seen before). Some examples, at least in Windows (I don't own a Mac or Linux machine), are Cambria Math, Wingdings (including 2 and 3), Segoe UI Symbol is quite extensive (it even has symbols of playing cards and dominoes among other things), as is Segoe UI Emoji, and Symbol. The font Segoe UI Historic has a lot of random characters that were probably part of some ancient languages, like    (click here if you can't see them). There are plenty of options to choose from. If you use an obscure symbol, and you want to send the document to someone else, either send them the font file you used (impractical) or save the document as a *.pdf or image file, because if the other person doesn't have the font installed on their device, they will probably just see boxes.

Usually, though, I only do things like that when I'm making notes for myself (I use plenty of custom shorthands when I take notes). Neuropsychological research has suggested that the human mind has a rough limit, reported as anywhere from 3-7 distinct items, to what can be held in working memory at a time (Cowan, 2010), so it may be counterproductive to use too many symbols, as your ideas will be lost to your readers while they try to keep all the symbols straight. It's probably a good idea to apply the adage: when in doubt, just write it out.


问:
Is there a symbol for set operation $A \cup B^\complement$?
Many of the 16 binary set operations (corresponding to the 16 logical connectives) don’t seem to have common dedicated symbols. Instead, they are often simply expressed in terms of $\cup, \cap, \setminus$, etc.

My question is whether there is any known precedent of a symbol for the operation $A \cup B^\complement$. This would complete the following (where ? is a placeholder for the operation in question):

$A \setminus B = A \cap B^\complement$

$A \operatorname{?} B = A \cup B^\complement$

For more context (based on discussion in the comments below):

In everyday work, of course it makes sense to use just a few symbols (e.g. complement, $\cup, \cap, \setminus$) and express other operations in terms of these. This approach represents a practical tradeoff between two extremes:

At one extreme, only one symbol is needed (just NAND or NOR, cf. functional completeness). The drawback is that most expressions will be quite large when written this way.

At the other extreme, we would have a unique symbol for each of the 16 operations. Expressions may be written more compactly with these symbols, but then there is more to memorize.

(Why are the complement, $\cup, \cap, \setminus$ operations in particular given special status? My only guess is that they correspond with concepts in spoken language: NOT, OR, AND, WITHOUT, which, in turn, may correspond to which operations are most naturally computed by our brains, but I digress...)

A comment below suggests simply using $(B \setminus A)^\complement$ to represent $A \cup B^\complement$, but the same could be said of any operation, e.g. why do we need a symbol for set difference $A \setminus B$ when we can just write $A \cap B^\complement$? Or why do we need a special logical connective for implication (which actually corresponds to the set operation in question here)?

My point is that sometimes things can be written more concisely with dedicated symbols for the uncommon cases. It may be rare that these symbols are needed, but in those instances where one of the uncommon set operations is heavily utilized ($A \cup B^\complement$ in my case here), it would be nice to have some agreed-upon symbol. And here I simply want to know if there is any precedent in the literature for those weird cases.

The diagram below (originally found on Wikipedia here or here) shows that all 16 logical connectives have dedicated symbols. Why not extend this luxury to the corresponding set operations?

答:
In lattice theory, this is nothing but the Heyting arrow in the lattice $\mathcal P(S)$ of subsets of a set $S$:

$$(A\succ B) = A^c\cup B.$$

More generally, if $L$ is a meet-semilattice (meaning a partially ordered set with finite infima $\wedge$), and $a,b\in L$ are any two elements of $L$. An implication from $a$ to $b$ is an element $(a\succ b)\in L$ such that
$$ (x\wedge a) \leq b \hspace{5mm} \text{if and only if} \hspace{5mm} x\leq(a\succ b)$$
for every $x\in L$.
If the implication from $a$ to $b$ exists, then it is unique.

A distributive lattice where every pair of elements has an implication is called a Heyting algebra.

The set $\mathcal P(S)$ of subsets of a set $S$ has an order structure given by inclusion ($\leq=\subseteq$). With this order, $\mathcal P(S)$ is a meet-semilattice, since infima are given by intersection ($\wedge=\cap$). Moreover, $\mathcal P(S)$ is a Heyting algebra, and I claim that for every pair of subsets $A,B\subseteq S$, the implication $A\succ B$ is given by the formula
$$(A\succ B) = A^c\cup B.$$

Proof: take any subset $X\in\cal P(S)$:

If $X\cap A\subseteq B$, then every $x\in X$ which is not in $B$ is in $A^c$. Hence $X\subseteq A^c\cup B$.

Conversely, if $X\subseteq A^c\cup B$, then $X\cap A \subseteq (A^c\cup B)\cap A=B\cap A\subseteq B$.

EDIT:
You can find the Heyting arrow denoted by $A\succ B$ in the works of Harold Simmons. Books like that of Picado and Pultr use $A\to B$, while some sources also use $A\Rightarrow B$ or $A\multimap B$. At some point, Birkhoff's (in his book on lattice theory) uses the symbol $(A:B)$ to denote what he calls the "relative complement of $A$ in $B$", which coincides with what I write as $B\succ A$ (notice the change of order).

Another common Heyting algebra is a topology: i.e. the family of open subsets of a topological space. In this case you have $A\succ B = \text{Int}(A^c\cup B)$.

The intention of this last paragraph is to remind you that $A\succ B=A^c\cup B$ only in the context of the power set $\mathcal P(S)$ of a set $S$ (or, more generally, in the context of a Boolean algebra). If you are working with a different lattice, then the implication may be given by something different, as in the case $A\succ B = \text{Int}(A^c\cup B)$ of the family of open subsets of a topological space.

Then my answer to your last comment is no, $B\succ A$ is not commonly used to denote the set operation $B^c\cup A$, because there may be contexts where the set $B^c\cup A$ has no special importance.


问:
$f\cos(\phi) + g\sin(\phi)=1$ and $f\sin(\phi)-g\cos(\phi)=0 \implies f = \cos\phi$ and $g = \sin(\phi)$
Let $f, g$ be smooth functions such that $f\cos(\phi) + g\sin(\phi)=1$ and $f\sin(\phi)-g\cos(\phi)=0$.  Prove $f = \cos\phi$ and $g = \sin(\phi)$.  It seems easy but I'm not sure where to begin

答: We have that $f(x) \sin x \cos x = \sin x - g(x) \sin^2 x = g(x) \cos^2(x)$ for all $x \in \mathbb{R},$ so that $g(x) = \sin x.$ We deduce that $f(x) \sin(x) = \sin(x) \cos(x)$ for all $x \in \mathbb{R},$ so $f(x) = \cos x$ whenever $x \neq k \pi$ for some $k \in \mathbb{Z}.$ Since $f$ is smooth, it must be the case that $g(x) = \cos(x)$ for all $x \in \mathbb{R},$ which finishes the proof.


问:
Find the line equation $ax + by+ c =0$ from point and angle.
Given a point $P$ and an angle $\alpha$ (see image) how does one find the corresponding line equation on the form $ax + by + c = 0$?

An image of a point $P$ and an angle $\alpha$

答:
Use trig.

If we imagine a bug traveling along the line for $1$ unit.  The number of units in the $x$ direction will be $\cos \alpha$ units, and the number of units in the $y$ direction will be $\sin \alpha$ units.

For any given point $(x,y)$ then change in vertical direction, ($y-y_0$) will be in proportion to the change in horizontal direction, $(x-x_0)$ as $\sin \alpha$ will be to $\cos \alpha$.

That is $(y-y_0):: (x-x_0) \sim \sin \alpha :: \cos \alpha$ so

$\cos\alpha(y-y_0) = \sin \alpha (x-x_0)$.

You might note that using the old idea of "slope/point" formula where $y = mx + b$ or $y-y_0 = m(x-x_0)$ or $\frac {y-y_0}{x-x_0} = m$.  This is  quite simply, $m = \frac {\sin \alpha}{\cos \alpha}=\tan \alpha$.

As you point out in a comment, the fails if the line is vertical and the slope $m$ is infinite and $\alpha = \frac {\pi}2$ and $\tan \alpha = \infty$.

But our old slope/point formula failed in those cases anyway.

..... cut the below ......

Using trig we get $\cos \frac \pi 2(y - y_0) = \sin\frac \pi 2(x-x_0)$

$0 = x-x_0$ and $x = x_0$ is the formula.

SO the slope of the line is $\frac {\sin \alpha}{\cos \alpha} = \tan \alpha$.

And that's it. Once we have the slope and a point we are done.

If the point is $x_0, y_0$ then we have the equation is $(y-y_0) = m (x-x_0) = \tan \alpha (x-x_0)$.

That's all.


问:
Morse Function Definition: Does it implies Morse function is $C^2$?
In my understanding, Morse function just means the determinant of Hessian matrix is nonsingular at critical points.

So my claims are:

the function itself should be continuous

the reference to Hessian matrix in the definition implies Morse functions are twice differentiable - T/F?

But Morse functions are not $C^2$ since the definition does not require continuity after twice differential - T/F?

Thank you very much!

n.b.1. A morse function is defined to be "A function for which all critical points are nondegenerate and all critical levels are different."

n.b.2. $f$ is of class $\mathscr{C}^k$ on $U$ if all iterated partial derivatives of $f$, of order at most $k$, exist and are continuous on $U$.

n.b.3. Differentiable functions are continuos. But the derivative of a differentiable function may not be continuous:
$$f(x) = \left\{\begin{matrix}
x^2 \sin \frac{1}{x} & x \neq 0\\
0 & x=0
\end{matrix}\right.$$

答:
In most introductory texts, a Morse function is defined to be smooth ($C^\infty$), since most such texts do not wish to bother themselves with the minutiae of degrees of regularity. However, the assertion that a Morse function is $C^2$ suffices, at least for the basics of the subject, i.e. results like the Morse Lemma do not require smoothness.

Although you can define the Hessian without $f$ being $C^2$, I can not find any text that develops the theory assuming only twice differentiability. If you read some proofs of the Morse Lemma, they use the fact that the entries of the Hessian are continuous, so I suspect that one must proceed with care when some, but not all, double partials are continuous.


问:
Find the adjoint of an operator $T\colon \ell^1\to\ell_0^\infty$
Let $\ell_0^\infty=\{x=(x_k)_{k\in\Bbb{N}};x_k\to0\}$. I already proved that the operator $T\colon\ell^1\to\ell_0^\infty$ given by $$T(x_1,x_2,x_3,...):=\left(\sum_{j=1}^\infty x_j,\sum_{j=2}^\infty x_j,\sum_{j=3}^\infty x_j,...\right)$$ is well defined and belongs to $B(\ell^1;\ell_0^\infty)$. Of course $(Tx)_k=\sum_{j=k}^\infty x_j$ is the k-th coordinate of the vector $Tx$.

Now I need to find the k-th coordinate of the adjoint of $T$, in the sense that, if $T^\prime$ denotes the adjoint of $T$, it has the form $T^\prime\colon(\ell_0^\infty)^\prime\to(\ell^1)^\prime$. But we already konw that we can make the identification $$(\ell^1)^\prime=\ell^\infty\qquad\text{and}\qquad(\ell_0^\infty)^\prime=\ell^1,$$ hence our adjoint operator is $T^\prime\colon\ell^1\to\ell^\infty$.

The thing is, I never face this kind of problem in my life before, and in the references that my teacher is usying don't have anything like it.

Thanks for any atention or help! :)

答:
$T'$ sends an $f\in (\ell_0^\infty)'$ to a $T'f\in (\ell^1)'$ given by $T'f(x)=f(Tx).$ To each such $f$ there is a unique $y\in \ell^1$ such that $f(z)=\sum y_kz_k.$ On the other hand, to $T'f$ there is a unique $w\in \ell^\infty$ such that $T'f(x)=\sum w_kx_k.$

To determine $w$ all we need to do is set $x=j^{th}$ canonical basis vector for $\ell^1$ and put the information in the previous paragraph together. We get $w_j=(f(Tx))_j=(f\overset{k\ \text{times}}{\overbrace{(1,1,\cdots, 1}},0,0,0,\cdots ,))_j=\sum_{k=1}^jy_k$.

Finally, using the identifications $(\ell^1)^\prime=\ell^\infty\ \text{and}\ (\ell_0^\infty)^\prime=\ell^1,$ we have a formula for the adjoint, namely $y_j\mapsto \sum_{k=1}^jy_k.$


问:
Finding a function as a solution to a system of integral equations
In my research, I reached a point where I need to find a nonnegative function, that satisfies the following:

For any $\phi\in[0,\pi]$:

$[\int^\pi_\phi f(\theta)d\theta][ \int^{\pi-2\phi}_{-\phi} f(\theta)d\theta]+[\int^0_{\pi-\phi} f(\theta)d\theta ][\int^{2\phi}_{\pi+\phi} f(\theta)d\theta]-[\int^\phi_0 f(\theta)d\theta]^2-[\int^{\pi-\phi}_\pi f(\theta)d\theta]^2 = \pi^2\cos^2\frac{\phi}{2}$

I would be very glad if somebody can help me, and thank you in advance for your help!

答:
Unfortunately, this is inconsistent:

If I put $\phi = 0$, I get

$\int^\pi_0 f(\theta)d\theta=\frac{\pi}{\sqrt[]{2}}$

while if I put $\phi=\pi$, I get

$\int^\pi_0 f(\theta)d\theta=0$


问:
If the principle invariants of two second order tensors $A$ and $B$ are equal, then the eigenvalues of $A$ and $B$ are equal. How to prove?
Problem is in the title. I found this problem in an a continuum mechanics textbook intended for applied mathematicians, so naturally everything in this problem is defined over $\mathcal{R}^3$.

The principle invariants of a second order tensor $M$ are defined as
$$
\def\tr{\operatorname{tr}}
\begin{align*}
I_1 &= \tr(M) \\
I_2 &= \frac{1}{2}\bigl[ (\tr(M))^2 - \tr(M^2) \bigr] \\
I_3 &= \det(M)
\end{align*}
$$

Defining the principle invariants in terms of eigenvalues and assuming the matrix is symmetric (which is reasonable given the context of continuum mechanics), we can write
$$
\begin{align*}
a_1 + a_2 + a_3 &= b_1 + b_2 + b_3 \\
a_1a_2 + a_1a_3 + a_2a_3 &= b_1b_2 + b_1b_3 + b_2b_3 \\
a_1a_2a_3 &= b_1b_2b_3,
\end{align*}
$$
where $a_i$ represents the eigenvalues of $A$ and the same for $B$. Now, my plan to show the equality of these eigenvalues was to expand the equation
$$
(a_1 - b_1)(a_2 - b_2)(a_3 - b_3) = 0
$$
and represent the product in terms of the prior equations. However, the algebra has taken up many pages and hours so far and I'm no longer so certain that I'm doing this right. Is my strategy flawed? Perhaps there's a theorem in linear algebra that I've forgotten that might help out?

Thank you for reading! :)

答: Note that $a_{1,2,3}$ are the three roots of the polynomial $$f_a(x)=(x-a_1)(x-a_2)(x-a_3),$$ and similarly for the $b_i$.  Can you show that the two polynomials are equal, and therefore have the same roots?


问:
Limit of $(x^3 - 3x^2 + 2x)$ as $x \to 1$ using epsilon delta definition
I need to prove the following limit using epsilon delta method. I have come up this on my own just to practice skills.

$$\lim \limits_{x \to 1} \,(x^3 - 3x^2 + 2x) = 0 $$

So, I need to come up with some $\delta >0$ such that given $0 < \lvert x - 1 \rvert <\delta$, I need to prove that

$$ |(x^3 - 3x^2 + 2x) - 0| < \varepsilon \tag{1}$$

I can simplify eq $1$ as follows.

$$ |x| |x-1||x-2| < \varepsilon $$

When we have a non linear term, then we restrict $x$ to some distance from $1$. So, suppose we have
$ 0 < |x-1| < 10$. This leads to $-10 < x-1 < 10$. Doing some algebra, we can get inequalities

$$ 0 \leqslant |x| < 11 \tag{2}$$

$$ 0 \leqslant |x-2| < 11 \tag{3}$$

From eq 2 and 3, it follows that

$$ 0 \leqslant |x||x-2| < 121 $$

And since $0 < |x-1| $, it follows that

$$ |x||x-1||x-2| < 121 |x-1| \tag{4}$$

To show that $|x||x-1||x-2| < \varepsilon$ , its sufficient to show that $ 121 |x-1| < \varepsilon$. And this will happen if $ |x-1| < \frac{\varepsilon}{121} $. So, this is another restriction on $|x-1|$. So, we can let

$$\delta = \text{min}\left(10, \frac{\varepsilon}{121}\right) $$

So, we can let this be our choice of $\delta$. Its clear that $\delta > 0$. So, now the official proof will follow.

Let $\varepsilon > 0$ be some arbitrary real. Suppose $0 < |x-1| < \delta$. With the choice of $\delta$ we have done, this means that $ 0 < |x-1| < 10$. As demonstrated above, equation 4 follows from this inequality

$$ |x||x-1||x-2| < 121 |x-1| $$

But now, we also have $ |x-1| < \frac{\varepsilon}{121} $. It means that $ 121 |x-1| < \varepsilon $. Using this, we get that

$$ |x||x-1||x-2| < \varepsilon  $$

which can be rewritten as

$$ | (x^3 - 3x^2 + 2x) - 0 | < \varepsilon  $$

Since, $\varepsilon > 0$ was arbitrary to begin with, its proven that

$$\lim \limits_{x \to 1} \,(x^3 - 3x^2 + 2x) = 0 $$

Is the proof correct ?

答:
Other methode which easier

$f(x) =x^3 - 3x^2 +2x$

Suppose $|f(x)-l|<\epsilon $

Let's show $|x-1|<\alpha$

$|f(x)-l|=|f(x)-0|=|x^3 - 3x^2 +2x|<\epsilon $

$|x^3 - 3x^2 +2x|=|x^2-2x|×|x-1|<\epsilon $

Suppose $x\in [\frac{1}{2}, \frac{3} {2}]$

$\Rightarrow $$   \frac{1} {4}\leq x^2 \leq \frac{9}{4}$

And

$-3\leq-2x\leq - 1$

So :

$\frac{-11} {4}\leq x^2-2x \leq \frac{5}{4}<\frac{11}{4}$$\Rightarrow $$|x^2 - 2x|\leq\frac{11}{4}$

$\Rightarrow $$|x^2 - 2x||x-1|\leq\frac{11}{4}|x-1|$

We know

$|x^2-2x|×|x-1|<\epsilon$

So $\frac{11}{4}|x-1|<\epsilon $$\Rightarrow $$|x-1|<\frac{4\epsilon }{1 1} $

Pose

$\alpha =\frac{4\epsilon }{1 1} $

So we can see $|x-1|<\alpha$

Finally:

After the definition of limite we showed $lim_{x\to 1} x^3 - 3x^2 +2x=0$


问:
Finding Modulus and Argument for Complex Number $z$
Assume the Complex Number $z$ has argument $$\pi $$ and modulus $$\sqrt{3} $$ how would i find argument and modulus for $$-2iz$$ i already know the correct answer for this(it's from an old test) but i don't know how its calculated. I know how to find modulus and argument but im not sure about this exact math problem. Any help is greatly appreciated, please tell me if i should provide more info.

答:
Since we have the modulus and argument, we know that $z=-\sqrt3$, hence we get $-2iz=2\sqrt3i$.  The argument and modulus for that complex number are $\pi/2$ and $2\sqrt3$.

For explanation, the key is Euler's formula:  $e^{i\theta}=\cos\theta+i\sin\theta$.  This allows us to identify every point $z=x+iy$ in the complex plane with a pair $(r,\theta)$, via $z=re^{i\theta}$..


答:
Choose a basis so that the tensor is represented by a matrix. The characteristic polynomial doesn't depend on the choice of basis, and can be represented in two ways. Expanded, its coefficients are the principal invariants. Factored, its roots are the eigenvalues.
$$
p_A(t) = t^3 - I_1 t^2 + I_2 t - I_3 = (t - \lambda_1)(t - \lambda_2)(t - \lambda_3).
$$
It's clear that if two tensors have same invariants, then they have same eigenvalues.

By the way, you can interpret all of the principal invariants in terms of determinants of principal minors (square submatrices formed by keeping a subset of the rows and the same subset of the columns):

$I_1$ is sum of determinants of $1 \times 1$ principal minors, i.e. the trace

$I_2$ is sum of determinants of $2 \times 2$ principal minors, which is equivalent to given expression

$I_3$ is sum of determinants of $3 \times 3$ principal minor, i.e. the determinant of the entire matrix

Title: If the principle invariants of two second order tensors $A$ and $B$ are equal, then the eigenvalues of $A$ and $B$ are equal. How to prove?

Body:
Problem is in the title. I found this problem in an a continuum mechanics textbook intended for applied mathematicians, so naturally everything in this problem is defined over $\mathcal{R}^3$.

The principle invariants of a second order tensor $M$ are defined as
$$
\def\tr{\operatorname{tr}}
\begin{align*}
I_1 &= \tr(M) \\
I_2 &= \frac{1}{2}\bigl[ (\tr(M))^2 - \tr(M^2) \bigr] \\
I_3 &= \det(M)
\end{align*}
$$

Defining the principle invariants in terms of eigenvalues and assuming the matrix is symmetric (which is reasonable given the context of continuum mechanics), we can write
$$
\begin{align*}
a_1 + a_2 + a_3 &= b_1 + b_2 + b_3 \\
a_1a_2 + a_1a_3 + a_2a_3 &= b_1b_2 + b_1b_3 + b_2b_3 \\
a_1a_2a_3 &= b_1b_2b_3,
\end{align*}
$$
where $a_i$ represents the eigenvalues of $A$ and the same for $B$. Now, my plan to show the equality of these eigenvalues was to expand the equation
$$
(a_1 - b_1)(a_2 - b_2)(a_3 - b_3) = 0
$$
and represent the product in terms of the prior equations. However, the algebra has taken up many pages and hours so far and I'm no longer so certain that I'm doing this right. Is my strategy flawed? Perhaps there's a theorem in linear algebra that I've forgotten that might help out?

Thank you for reading! :)


问:
How can I prove that $B_0 \cup X$ generates $V$?
Let be $T:V \rightarrow W$ a linear transformation with $dim(Ker(T))<\infty$ and $dim(Im(T))<\infty$. Let be $B_0$ a basis of $Ker(T)$. Let be $Y \subset W$ a basis of $Im(T)$, and we have that for every $y_i \in Y$ there exists $x_i \in X \subset V$ such that $T(x_i)=y_i$.

I have to prove that $B_{0} \cup X$ is a basis of $V$. I've already proved that $B_0 \cup X$ is linearly independent but I don't know how to prove that $B_0 \cup X$ generates $V$, can you help me please? I would be very greatful.

答:
If $T(x) =0$, then $x\in Ker(T) $ so $x\in span(B_0)$.
If $T(x) \neq 0$, then $$T(x) = \sum c_iy_i = \sum c_iT(x_i) = T\left(\sum c_ix_i\right) $$

This implies that $x - \sum c_ix_i$ is in the Kernel, which makes $$x - \sum c_ix_i \in span(B_0) \implies x \in span(B_0\cup X) $$


答:
So starting with $f(x) = 3x + 2$ we can say $f(1) = 3\cdot 1 + 2 = 5$

And because $f(1) = 5$, we can substitute $f(f(1)) = f(5)$

Then $f(5) = 3\cdot 5 + 2 = 17$, therefore $f(5) = f(f(1)) = 17$

And so on, just keep recursively calculating step by step until you've worked your way up the whole chain

Title: Compute $f(f(1))$ if $f(x)=3x+2$

Body:
Question :

Let the function $f$ be defined by $f(x)=3x+2$.

Then $f(f(1))=$

$f^2(1)$=

and $f^2(f(1))=$

for this first one I replaced the $x$ by $1$ so it gives $5$, the second one $=25$ and $100$ for the last
but I am not sure if it is the right way to resolve it.
Thank you,


问:
Exchangeable Random Variables and independence between their indicators

$X_1,X_2$ are two exchangeable random variables. Show that the random variable $M = \max\{X_1, X_2\}$ is independent of the indicator random variable $Z = 1_{X_1>X_2}$

I am really struggling to show this: right now my approach is to find the distribution of $P(M>m,Z=1)$ and try to show it equals to $P(M>m)P(Z=1)$. However when I evaluate $P(M>m,Z=1)=P(\max\{X_1, X_2\}>m,X_1>X_2)$, I do not think I can assume the joint density between $M,Z$ which makes it impossible to calculate this probability. How can I proceed with this question?

答:
This is not true. Consider $X_1,X_2$ i.i.d. (so a fortiori exchangeable) Bernoulli with parameter $1/2$.

Then $\Pr[M=0] = \Pr[X_1=0,X_2=0] = 1/4$, and $\Pr[Z=1] = \Pr[X_1=1,X_2=0] = 1/4$. However,
$$
\Pr[M=0,Z=1] = 0 \neq \Pr[M=0]\cdot \Pr[Z=1]
$$
since $\Pr[M=0,Z=1] = \Pr[X_1=0,X_2=0,X_1>X_2]$


问:
prove 3rd derivative of a function in a point
Let $f$ be twice differentiable and $f'''$ exists in one point $x\in D$. I want to show for this one point $$f'''(x)=\lim_{h\rightarrow0}\frac{f(x+3h)-3f(x+2h)+3f(x+h)-f(x)}{h^3}$$

I did the following: I know $\lim_{h_1\rightarrow0}\frac{f(x+h_1)-f(x)}{h_1}=f'(x)$ and so follows by pluging in $$f''(x)=\lim_{h_2\rightarrow0}\frac{f'(x+h_2)-f'(x)}{h_2}=...=\lim_{h_1\rightarrow0}\frac{f(x+h_1)+f(x-h_1)-2f(x)}{h_1^2}$$

Doing this one more time I'll get ($f'''$ exists in $x$)

$$f'''(x)=\lim_{h_3\rightarrow0}\frac{f''(x+h_3)-f''(x)}{h_3}=...=\lim_{h_1\rightarrow0}\frac{f(x+3h_1)-3f(x+2h_1)+3f(x+h_1)-f(x)}{h_1^3}$$

First question: Can you do it like that?

And my second question: Why can we always choose 'the same $h$' for the limits? Formally for e.g. $f''$ it's

$$\lim_{h_2\rightarrow0}\frac{\lim_{h_1\rightarrow0}\frac{f(x+h_1+h_2)-f(x+h_2)}{h_1}-\lim_{h_1\rightarrow0}\frac{f(x+h_1)-f(x)}{h_1}}{h_2}$$ and I have $h_1$ and $h_2$ in it.

Edit
Using L'Hospital twice on the right side I get
$$\dots=\lim_{h\rightarrow0}\frac{3f''(x+3h)-4f''(x+2h)+f''(x+h)}{2h}$$

And so we get $\begin{align}&\dots\\&=\lim_{h\rightarrow0}\left\{\frac16\frac{3f''(x+h)-3f''(x)}{h}-\frac13\frac{12f''(x+2h)-12f''(x)}{2h}+\frac12\frac{9f''(x+3h)-9f''(x)}{3h}\right\}\\&=\frac36\lim_{h\rightarrow0}\frac{f''(x+h)-f''(x)}{h}-\frac{12}3\lim_{h\rightarrow0}\frac{f''(x+2h)-f''(x)}{2h}+\frac92\lim_{h\rightarrow0}\frac{f''(x+3h)-f''(x)}{3h}\\&=f'''(x)\end{align}$

答:
This is not a direct answer to your questions but a simple answer to the problem

By Taylor's theorem  we have
$$f(x+h)=f(x)+hf'(x)+\frac{h^2}{2}f''(x)+\frac{h^3}{6}f'''(x)+o(h^3)$$
$$f(x+2h)=f(x)+2hf'(x)+{2h^2}f''(x)+\frac{4h^3}{3}f'''(x)+o(h^3)$$
$$f(x+3h)=f(x)+3hf'(x)+\frac{9h^2}{2}f''(x)+\frac{9h^3}{2}f'''(x)++o(h^3)$$

so we can see easily (it's a simple calculation) that
$$f'''(x)=\lim_{h\rightarrow0}\frac{f(x+3h)-3f(x+2h)+3f(x+h)-f(x)}{h^3}$$


问:
Partitioning 18 into odd number of even components
I need to find a total number of ways $18$ can be partitioned into odd number of even components.

Does the function below satisfy the "odd number of components" requirement?
$$
P_{1,3,5,7,9}(X) = \frac{1}{1-x} \cdot \frac{1}{1-x^3} \cdot \frac{1}{1-x^5} \cdot \frac{1}{1-x^7} \cdot \frac{1}{1-x^9}
$$

Given that $n$ is even, then every partition of $n$ has an even number of odd parts.
Can it be said that if $n$ is even, then every partition of $n$ has an odd number of even parts?

How can I add to that the requirement that components are supposed to be even?
I have spent 6 hours reading about self-conjugation, odd and distinct parts and generating functions and I'm clueless.

答:
It may be quickest to just write the numbers down: you should be able to find $16$ cases.

$$18 =2+2+14=2+4+12=2+6+10=2+8+8=4+4+10 \\=4+6+8=6+6+6  =2+2+2+2+10=2+2+2+4+8\\=2+2+2+6+6=2+2+4+4+6
=2+4+4+4+4 \\= 2+2+2+2+2+2+6=2+2+2+2+2+4+4 \\
=2+2+2+2+2+2+2+2+2$$

One approach could be to say that, by dividing each part by $2$, the number of ways $18$ can be partitioned into an odd number of even parts is equal to the number of ways $9$ can be partitioned into an odd number of parts of any parity.  That just pushes the issue further down to a question which has been considered slightly more.  OEIS A027193 enumerates these but does not in my view give a simple expression for finding the number of such partitions.

The easiest one to use that I know is that the number of ways $n$ can be partitioned into an odd number of parts is $p(n-1^2)-p(n-2^2)+p(n-3^2)-\cdots$ where $p(n)$ is the total number of partitions of $n$.  So here in the case of $n=9$ you get $p(8)-p(5)+p(0) = 22-7+1=16$.


问:
How to understand the Grothendieck group as an adjoint?
My main goal is to understand the Grothendieck group construction and its generalization to non-commutative setting. I understand its explicit construction via an equivalence relation, but I want to fully understand as an adjoint functor. Here is what I know about this so far: There is a forgetful functor $U: \mathbf{Ab}\to\mathbf{CMon}$ from the category of abelian groups to the category of commutative monoids, which has a "free" left adjoint $Gr.$ If I understood correctly we can write a Kan extension formula for it as
$\require{AMScd}$
\begin{CD}
\mathbf{Ab} @>\text{Id}>> \mathbf{Ab}\\
@V U VV @|\\
\mathbf{CMon} @>>\text{Lan}_U(\text{Id})> \mathbf{Ab}
\end{CD}
By the pointwise formula $Gr(M)=\text{Lan}_U(\text{Id})(M)=\underset{\mathcal{D}}{colim}\,\text{Id}(G),$ where $\mathcal{D}$ is the diagram whose objects are $G\in\mathbf{Ab}$ with a commutative monoid homomorphism $U(G)\to M,$ and morphisms are commutative triangles. Until this point we did not use any structure of these categories. Now I wonder,

how to translate this into the equivalence relation that we know? and

what changes as we go to non-abelian groups?

Thank you in advance for your help.

答:
The forgetful functor from the category of groups to the category of monoids has a left adjoint, which may be considered a non-commutative generalisation of the Grothendieck group construction.
The colimit formula is not very useful for obtaining an explicit description – as you observed, it doesn't use anything special about the categories or functors in question.
Here is a concrete construction.

Let $M$ be a monoid, written multiplicatively with unit $1$.
We consider the group $G$ defined as follows.
The elements of $G$ are finite words of the form $a b^{-1} c d^{-1} \cdots$ where $a, b, c, d, \ldots$ are elements of $M$, subject to the following equivalence relations:

$1^{-1} a \cdots = a \cdots$ for all $a \in M$.

$1 b^{-1} \cdots = b^{-1} \cdots$ for all $b \in M$.

$\cdots c 1^{-1} = \cdots c$ for all $c \in M$.

$\cdots d^{-1} 1 = \cdots d$ for all $d \in M$.

$\cdots a 1^{-1} c \cdots = \cdots (a c) \cdots$ for all elements $a$ and $c$ in $M$.

$\cdots a^{-1} 1 c^{-1} \cdots = \cdots (c a)^{-1} \cdots$ for all elements $a$ and $c$ in $M$.

$\cdots (a b) (c b)^{-1} \cdots = \cdots a c^{-1} \cdots$ for all elements $a, b, c$ in $M$.

$\cdots (a b)^{-1} (a c) \cdots = \cdots b^{-1} c \cdots$ for all elements $a, b, c$ in $M$.

The group operation of $G$ is essentially concatenation: given words $a b^{-1} \cdots$ and $c d^{-1} \cdots$, the product is $a b^{-1} \cdots c d^{-1} \cdots$, with a $1$ or $1^{-1}$ inserted in the middle if necessary.
The inverse of $a b^{-1} c d^{-1} \cdots$ is $\cdots d c^{-1} b a^{-1}$.
It is straightforward, if tedious, to verify that this is indeed a group such that every monoid homomorphism $M \to H$ factors as the obvious map $M \to G$ followed by a group homomorphism $G \to H$, and that this factorisation is unique.


问:
Argument $\int_{-\infty}^{t}\int_{-\infty}^{x_2}f_{X_1,X_2}dx_1dx_2=\int_{-\infty}^{t}\int_{-\infty}^{x_1}f_{X_1,X_2}dx_2dx_1$
Right now I have two integrals
$$
\int_{-\infty}^{t} \int_{-\infty}^{x_2} f_{X_1,X_2}(x_1,x_2) \, dx_1 dx_2 =
\int_{-\infty}^{t} \int_{-\infty}^{x_1} f_{X_1,X_2}(x_1,x_2) \, dx_2 dx_1
$$
with the property that $f_{X_1,X_2}(x_1,x_2)=f_{X_1,X_2}(x_2,x_1)$ for all possible $(x_1,x_2)$. This intuitively make sense if I just draw the area bounded by these two integrals, but I am writing a proof right now and really don't know how can I formally explain it. Some hint would be really appreciated!

答:
By swapping the names of the variables $(x_1, x_2) \leftrightarrow (x_2, x_1)$ in the first expression, you almost get the second equation. But the arguments of the function are now swapped. Fortunately, you have the hypothesis that your function is symmetric in those arguments, so
$$
\def\cb#1{\color{blue}{#1}}
\def\cg#1{\color{green}{#1}}
\begin{align*}
\cb{\int_{-\infty}^{t}} \cg{\int_{-\infty}^{x_2}}
f_{X_1,X_2}(\cg{x_1}, \cb{x_2}) \, \cg{dx_1} \cb{dx_2}
&= \cg{\int_{-\infty}^{t}} \cb{\int_{-\infty}^{x_1}}
f_{X_1,X_2}(\cb{x_2}, \cg{x_1}) \, \cb{dx_2} \cg{dx_1} \\
&= \cg{\int_{-\infty}^{t}} \cb{\int_{-\infty}^{x_1}}
f_{X_1,X_2}(\cg{x_1}, \cb{x_2}) \, \cb{dx_2} \cg{dx_1}.
\end{align*}
$$


问:
In Fisher’s discriminant for multiple classes, How do you manage when $(Sw)$ is singular matrix (so you cant get $(Sw)^{-1}$)?
I am trying to use Fisher’s discriminant for multiple classes to reduce the Dimension of the MNIST data set, similar to this post: https://towardsdatascience.com/an-illustrative-introduction-to-fishers-linear-discriminant-9484efee15ac

But in my Implementation in Python, when I compute $Sw$ it is a singular matrix so I can not get  $(Sw)^{-1}$ to obtain the eigenvectors from  $(Sw)^{-1}Sb$. Am I doing something wrong or what can I do to solve this problem?

I am following the steps given in Pattern Recognition and Machine Learning By Bishop.

Thanks.

答:
You are right, it could be singular. This is known as the small sample size (SSS) problem.

A very typical way is to use the pseudoinverse to estimate the required directions.

There have been extensive research to this problem. Here, you can see a survey of various approaches such as regularized LDA, Orthogonal LDA, Uncorrelated LDA, and Null Space LDA.


问:
Distribution of summation in RSS derivation
I am learning how to derive the RSS formula and there is a step that is not making sense to me:

$\sum_{i=1}^n (x_i-\bar x)(y_i-\bar y)   +   \underbrace{\sum_{i=1}^n   \bar x(y_i+\hat \beta x_i)}_{Q} - \underbrace{ \sum_{i=1}^n \bar x(\bar y-\hat\beta \bar x)}_M-\hat \beta \sum _{i=1}^n(\bar x-x_i)^2$

$Q=n\bar x\bigg(\frac{1}{n}\sum_{i=1}^ny_i-\hat \beta \frac{1}{n}\sum_{i=1}^n x_i\bigg)=n\bar x(\bar y-\hat \beta \bar x)=M$

And for this reason, Q and M cancel out. My question is that I do not understand step 2. I  don't understand how distributing the summation term in the beginning, turns the inside components into "bar" terms...

答:
I believe there's a typo, and $Q = \sum_{i=1}^n \bar{x}(y_i-\hat{\beta}x_i) $ in step 1.

From there, note that $\bar{x}$ is a constant, so it can be factored out of the sum:

$$ Q = \bar{x} \sum_{i=1}^n{(y_i+ \hat{\beta}x_i)} .$$

Then, we split the summation into two separate sums:

$$ Q = \bar{x} \left( \sum_{i=1}^n{y_i} + \sum_{i=1}^n {\hat{\beta}x_i} \right).$$

Then, the same thing happens with $\hat{\beta}$ since it's a constant too:

$$ Q = \bar{x} \left( \sum_{i=1}^n{y_i} - \hat{\beta}\sum_{i=1}^n {x_i} \right).$$

The final step comes from the fact that $\bar{x}=\frac{1}{n}\sum_i {x_i} \Longrightarrow n\bar{x} = \sum_i {x_i}$:

$$ Q = \bar{x} \left( n\bar{y} - \hat{\beta}n\bar{x} \right).$$

$$ \Rightarrow Q = n\bar{x} \left( \bar{y} - \hat{\beta}\bar{x} \right).$$


问:
Why can't we have a third number line for dividing by zero?
Originally, taking the square root of a negative number wasn't possible, but then Rafael Bombelli added an imaginary number axis to allow that. Why, then, is there no 3rd dimensional number line to allow division by 0? If there was, you could create a unit z, where z is 1/0. Therefore, 5/0 could be 5z, etc., and multiplying by its reciprocal is a 90° rotation to the real number axis, just like squaring i causes a 90° rotation to the real axis.

答:
Interesting question! Although I’m not sure Rafael Bombelli created the imaginary axis since his life largely predates the widespread use of Cartesian coordinate systems.

Before postulating a division by zero axis, it may help to revisit why x/0 has consensus as an undefined concept. It’s not just a void of mathematics, its use leads to contradictions within our other established axioms:

It follows from the properties of the number system we are using (that is, integers, rationals, reals, etc.), if b ≠ 0 then the equation
a
/
b
= c is equivalent to a = b × c. Assuming that
a
/
0
is a number c, then it must be that a = 0 × c = 0. However, the single number c would then have to be determined by the equation 0 = 0 × c, but every number satisfies this equation, so we cannot assign a numerical value to
0
/
0

Lots more contradictions in the wiki, over many different areas of math https://en.wikipedia.org/wiki/Division_by_zero

Which then comes around to, what are we hoping to postulate a new axis or number line to achieve? Exploratory math, sure go ahead, but I’d like to tickle your brain more on why you’d even want to do this.


问:
Is the converse of Tychonoff's theorem true?
Tychonoff's theorem states that the product of any collection of compact topological spaces is compact with respect to the product topology.

I would like to know that if the product of some topological spaces is compact with respect to the product topology, does it imply that these spaces are compact?

答: Suppose there is $j$ such that $X_j$ is non-compact. Then, there exists an open covering of X, say $\{U_i\}_{i \in J}$, that has no finite subset covering $X_j$. Then $\{\pi_j^{-1}(U_i) \hphantom{/}/ \hphantom{/} i \in J\}$ is an open covering of $\Pi X_i$, which has no finite subset covering it.


问:
Intuition behind intersection of subspaces with common basis vectors

Let $u, v, w$ be vectors in $\mathbf{R^3}$ such that $V = span\{u, v\}$ and $W = span\{u, w\}$ are planes in $\mathbf{R^3}$. Find $V \cap W$ if

(a) $u, v, w$ are linearly independent.

(b) $u, v, w$ are not linearly independent.

For both the parts in this question, I present the intuition behind my method and some workings I have attempted to layout.

(a) $u, v, w$ are linearly independent.

Intuition

At a glance, we know from this fact that none of the vectors in $\{u, v, w\}$ are linear combinations of the other vectors in the same set. Geometrically, this translates to having two planes in $\mathbf{R^3}$ which intersect along a line. Hence, intuitively we know that $\dim(V \cap W) = 1$.

Working

Take any $a \in V \cap W$. It is clear that

$$
a \in V \cap W \Rightarrow a \in V \ \land a \in W
$$

Since $V = span\{u, v\}$ and $W = span\{u, w\}$, we have

\begin{align}
a &= c_1u + c_2v \\
a &= d_1u + d_2v
\end{align}

for $c_1, c_2, d_1, d_2 \in \mathbf{R}$.

Since $u, v, w$ are linearly independent, this means that

$$
(c_1 - d_1)u + c_2v - d_2w = 0
$$

has only the trivial solution $\Rightarrow c_1 - d_1 = c_2 = d_2 = 0$.

Hence $a = c_1u = d_1u$, so

$$
V \cap W = span\{u\}
$$

(b) $u, v, w$ are not linearly independent.

Intuition

(I am more unclear on this portion of the question, please do correct my thinking if there are any flaws/mistakes)

Given that $u, v, w$ are not linearly independent, by the definition, this means that every vector in $\{u, v, w\}$ can be expressed as a linear combination of the other vectors. In other words, $v = c_1u + c_2w$ for some $c_1, c_2 \in \mathbf{R}$. By this logic, geometrically we should obtain two planes who lie on each other, that is, $\dim(V \cap W) = 2$.

Working

Suppose $u, v, w$ are not linearly independent. By the definition, the system

$$
c_1u + c_2v + c_3w = 0
$$

has non-trivial solutions (not all $c_1, c_2, c_3 = 0$) and $c_1, c_2, c_3 \in \mathbf{R}$.

Expressing $v$ in terms of $u, w$,

$$
v = - \frac {c_3} {c_2} w - \frac {c_1} {c_2} u
$$

Hence

\begin{align}
span\{u, v\} &= span\{u, - \frac {c_3} {c_2} w - \frac {c_1} {c_2} u\} \\
&= span\{u, w\}
\end{align}

So

\begin{align}
V \cap W &= span\{u, w\} \\
&= span\{u, v\} \\
&= span\{v, w\}
\end{align}

Are my workings/ideas correct? Any advice would be greatly appreciated! (I recognize that my  expression of ideas is rather hand-wavy, and hence would appreciate clearer ways of thinking about the particular concepts involved).

答:
Reminds me of the time Bill Jacob drew hands waving on the board.

At first glance I think you got the first part.

As for the second, what if $v=w$. Then your last equality doesn't hold.  I think you are right about the dimension though, the intersection must have dimension two.

If $\{u,v,w\}$ is not linearly independent, then we must have that $\{v,w\}$ is dependent.  That is, $v$ and $w$ are multiples of each other.