问:
Probability of three disjoint events
I have a question which I haven't been able to figure out.

Given three disjoint events $X,Y,$ and $Z$ find

a) the probability that exactly one event occurs

b) the probability that exactly two of the events occur.

So far, the progress I have made on a) is $$P(\text{exactly one}) = [P(X)(1 - P(Y))(1 - P(Z))] + [P(Y)(1 - P(X))(1 - P(Z))] + [P(Z)(1 - P(X))(1 - P(Y))].$$

I took a similar approach for the second one, too. However, given that these sets are disjoint, I think that I might be headed in the wrong direction.

答:
(a) The probability you are interested in is given by:
\begin{align*}
\mathbb{P}(A\cap B^{c}\cap C^{c}) + \mathbb{P}(A^{c}\cap B\cap C^{c}) + \mathbb{P}(A^{c}\cap B^{c}\cap C)
\end{align*}

Since $A\cap B = \varnothing$, we conclude that $A\subseteq B^{c}$.

Moreover, since $A\cap C = \varnothing$, we conclude that $A\subseteq C^{c}$.

Therefore $A\subseteq B^{c}\cap C^{c}$, whence we get that $A\cap B^{c}\cap C^{c} = A$.

Similarly, $A^{c}\cap B\cap C^{c} = B$ and $A^{c}\cap B^{c}\cap C = C$, and we are done.

(b) As @lulu has mentioned, such probability equals zero because $A$, $B$ and $C$ are pairwise disjoint.

More precisely, one has
\begin{align*}
\mathbb{P}(A\cap B\cap C^{c}) + \mathbb{P}(A\cap B^{c}\cap C) + \mathbb{P}(A^{c}\cap B\cap C) = 3\times \mathbb{P}(\varnothing) = 0
\end{align*}

Hopefully this helps!

Title: Probability of three disjoint events


问:
Probability of selecting at least one from each group probability
A team of 5 managers is to be selected from a group of 10 managers - 5 from company A, 3 from company B, and 2 from company C. In how many ways can this be done if the team must contain at least one manager from each company?

I tried selecting one from each company first:
$$
5C1 * 3C1 * 2C1 = 30
$$
Then, I selected 2 from the remaining 7 people:
$$
7C2=21
$$
So I got $ 30 * 21 = 630 $, but the answer is 175. What did I do wrong?

答:
I'm late to the party, but here is why your answer is incorrect:

Lets simplify the question so it's easier to follow. Imagine "Sam" and "Dean" work in company A,"Mike" works in company B and we are asked to select 3 managers in total in a way that it contains at least one manager from each company.

We have three managers in total and we want to select three managers so there is only one way to do so.

Your answer is incorrect because first it picks one manager from company A (2C1 = 2) and one manager from company B (1C1 = 1). Then it picks one manager from the remaining managers (1C1 = 1) and says there are two ways to do this. This method gives you the following selections:

```<code>- Picking "Sam" from company A, picking "Mike" from company B and picking "Dean" from company A.
- Picking "Dean" from company A, picking "Mike" from company B and picking "Sam" from company A.
</code>```

which are the same selection. That's why your answer is always bigger than the correct answer.

ContentLicense: CC BY-SA 3.0

Body:
I'm late to the party, but here is why your answer is incorrect:

Lets simplify the question so it's easier to follow. Imagine "Sam" and "Dean" work in company A,"Mike" works in company B and we are asked to select 3 managers in total in a way that it contains at least one manager from each company.

We have three managers in total and we want to select three managers so there is only one way to do so.

Your answer is incorrect because first it picks one manager from company A (2C1 = 2) and one manager from company B (1C1 = 1). Then it picks one manager from the remaining managers (1C1 = 1) and says there are two ways to do this. This method gives you the following selections:

- Picking "Sam" from company A, picking "Mike" from company B and picking "Dean" from company A.
- Picking "Dean" from company A, picking "Mike" from company B and picking "Sam" from company A.

which are the same selection. That's why your answer is always bigger than the correct answer.

ContentLicense: CC BY-SA 4.0


问:
Sufficient condition for uniform convergence of $f_{\epsilon}(x, \cdot)$ to $f(x, \cdot)$ implies that $f_{\epsilon} \rightarrow f$ uniformly.
Consider functions $f_{\epsilon}: K_x \times K_u \rightarrow \mathbb{R}$ for all $\epsilon >0 $ and $f: K_x \times K_u \rightarrow \mathbb{R}$. $K_x \subset \mathbb{R}^n$ and $K_u \subset \mathbb{R}^m$ are compact spaces.
Suppose that given $x \in K_x$, $f_{\epsilon}(x, \cdot)$ and $f(x, \cdot)$ are continuous (on $K_u$), and $f_{\epsilon}(\cdot, u)$ and $f(\cdot, u)$ are continuous (on $K_x$).

If $f_{\epsilon}(x, \cdot)$ converges uniformly to $f(x, \cdot)$ as $\epsilon \rightarrow 0^+$, then does it imply that $f_{\epsilon} \rightarrow f$ uniformly?
I guess that from the uniform convergence of $f_{\epsilon}(x, \cdot)$ to $f(x, \cdot)$,
we may choose $\overline{\epsilon}(x) > 0 $ for any $\epsilon' >0$ such that $|f_{\epsilon(x)}(x, u) - f(x, u)| < \epsilon', \forall \epsilon(x) \leq \overline{\epsilon}(x), \forall (x, u) \in K_x \times K_u$. Then, $\epsilon := \min_{x \in K_x}\overline{\epsilon}(x)$ implies the desired result.

Actually, I'm not sure that such $\epsilon$ exists. Is it correct? If not, are there any sufficient conditions to derive similar conclusions?

EDIT: with an additional condition: $f_{\epsilon}$ converges pointwise to $f$ as $\epsilon \rightarrow 0^+$.

答:
For a counterexample, take $f_\epsilon : [0,1] \times [0,1] \to \mathbb{R}$ where

$$f_\epsilon(x,u) = \begin{cases}\frac{\epsilon xu}{x^3+u^3}, &(x,u) \neq (0,0)\\0, & (x,u) = (0,0)\end{cases}$$

The function $f_\epsilon$ is separately continuous in $x$ and $u$ and converges pointwise to the (continuous) zero function as $\epsilon \to 0+$.   For fixed $x$ and all $u \in [0,1]$ we have

$$\left|\frac{\epsilon xu}{x^3+u^3} \right|\leqslant \begin{cases}\frac{\epsilon}{x^2}, & 0 < x \leqslant 1\\0, & x= 0 \end{cases}\underset{\epsilon \to 0+}\longrightarrow 0,$$

and, thus, $f_\epsilon(x,\cdot) \to 0$ uniformly for $u \in [0,1]$.

However, for $\epsilon \in (0,1)$

$$\sup_{(x,u) \in [0,1]^2}\left|\frac{\epsilon xu}{x^3+u^3} \right| \geqslant \frac{\epsilon \cdot \epsilon \cdot \epsilon}{\epsilon^3 + \epsilon^3 } = \frac{1}{2}  \underset{\epsilon \to 0+}\longrightarrow \frac{1}{2} \neq0$$

and the convergence is not uniform on $[0,1]\times[0,1]$.


问:
The simplest $\Delta$-complex structure on $S^2$
I think my reasoning is correct, but I want to run through it here because having the right intuition will make similar problems easier in future.

A 2-simplex is homeomorphic to a closed disc, and a closed disc is homeomorphic to a hemisphere, so we can "build" $S^2$ out of two 2-simplices. However we need to have 3 specified vertices, say $v_0, v_1$ and $v_2$, on the circumference where the two hemispheres meet. This also means three 1-simplices joining these vertices.

Is this the simplest $\Delta$-structure possible?

答:
You need at least two $2$-simplices, but you can glue them up in another way to get an equally simple structure. Take a simplex and glue two of its sides together to get a cone. Do this for another simplex, and then glue the boundary circles together. This is a $\Delta$ complex structure on the sphere with 3 vertices 3 edges and 2 faces, just like yours, but glued together differently.

To see that you can't get away with just one $2$-simplex, you just have to notice that there's no way of gluing the sides of a triangle together to get a sphere. (or even a surface).


问:
Prime Numbers and Elliptic Equations
I came across one elliptic equation of the form $y^2 = x^3 + p^2$ being $p$ prime, and taking $p \neq 3$, I want to have more understanding why there is no rational point $x$, for $y = 3p$ or $y = 3p^2$, such that:

$$y^2 = x^3 + p^2$$

I want to know in terms of equivalence classes approach also if possible.

答:
Let $y^2 = x^3 + p^2$;

take $y=3p$ place into equation; $9p^2 = x^3 + p^2$ then $x^3 = 8p^2$ take the cube-root; $x=2\sqrt[3]{p^2}$, this implies there is no rational solution.

take $y=3p^2$ place into equation; $9p^4 = x^3 - p^2$ then we have $$x^3 - (9p^4 + p^2)$$ Look at the discrimant $\Delta_x = -27 (9p^4-p^2)^2$. It is negative, therefore, one real two complex roots. Complext roots cannot be a solution. The real root is $p^{2/3}\sqrt[3]{9p^2-1}$. This concludes that no rational solution here, too.

Title: Prime Numbers and Elliptic Equations


问:
Statistics - Chebyshev's rule, my answer is wrong for some reason?
I have the following problem :

Determine what age interval will contain at least 95% of the data (Chebyshev's) ?

Now, I have standard deviation of 1.516, mean of 19.211.

The formula is $1-(1/k^2) = .95$ So I solve for k to get $\sqrt 2$.

Now, I calculate the interval by mean + $\sqrt 2$*standard deviation = RIGHT AGE INTERVAL

Why is this wrong? I get approximately $21.3578$.

The left interval, let's just skip it for now.

Is my logic off?

答:
Your calculations do have an error: your $\sqrt{2}$ should be $\sqrt{\frac{1}{1-0.95}}=\sqrt{20}$.

You seem to have started with Chebyshev's inequality of the form $$\mathbb P(|X-\mu|\le k \sigma) \ge 1-\frac1{k^2}$$

The right hand side is $0.95$ when $k=\sqrt{20} \approx 4.472136$ so your interval should be $[\mu-k\sigma, \mu+k \sigma]$ and in this case $[12.431,25.991]$


问:
(simple) Expectation of random variable as a multipart function
Let the random variable $Y \in [0, \infty)$, a real number $\theta >0$, and the random variable $X$ such that $X = \theta - \min(\theta,Y)$, thus, $X \in [0, \theta]$. That is, $X = 0$ if $Y > \theta$ and $X = \theta-Y$ if $Y \leq \theta$. I would like to know if the expectation of $X$, $E[X] = E[\theta-Y|Y \leq \theta] P(Y \leq \theta) = (\theta-E[Y|Y\leq \theta])P(Y \leq \theta)=\theta P(Y \leq \theta)-\int_{0}^{\theta}y f_{Y}(y)dy$ is valid.
If so, since $E[X] >0$, then, $\theta P(Y \leq \theta) > \int_{0}^{\theta}y f_{Y}(y)dy$ (?) Thanks! </p>

答: Yes, this is correct.  It's the same idea as Markov's inequality.</p>


问:
Proof that expectation of $X^T \Sigma^{-1} X = \dim X$
Given a random variable $X \in \mathbb{R}^n$, with variance/covariance matrix $\Sigma$ and mean 0, how do we prove the following identity?
$$ \mathbb{E}(X^T\Sigma^{-1}X) = n = \dim X$$

Does this identity have a name? It was provided as a hint, to be used without proof, to prove that the multivariate normal distribution has maximum differential entropy for a given variance matrix $\Sigma$.

答: $$ \mathbb{E}(X^T\Sigma^{-1}X) = \mathbb{E}(\mathrm{Tr} (X^T\Sigma^{-1}X))=\mathbb{E}(\mathrm{Tr} (XX^T\Sigma^{-1}))=\mathrm{Tr}(\mathbb{E}(X X^T)\Sigma^{-1}) = \mathrm{Tr}(\Sigma \Sigma^{-1}) = n$$


问:
Inequality Induction Proof, how should I proceed?
I have been trying to prove inequalities using induction to no avail.

For example,

Prove the following:
$\frac{1}{2^{1}}+\frac{2}{2^{2}}+\frac{3}{2^{3}}+...+\frac{n}{2^{n}}<2$

Base Case:

$\frac{1}{2^{1}} = 2$

$\frac{1}{2}<2\;$  Which is true.

We assume, for n=k, that

$\frac{1}{2^{1}}+\frac{2}{2^{2}}+\frac{3}{2^{3}}+...+\frac{k}{2^{k}}<2\;\;\;$ is true.

As induction step with n=k+1, we need to prove that

$\frac{1}{2^{1}}+\frac{2}{2^{2}}+\frac{3}{2^{3}}+...+\frac{k+1}{2^{k+1}}<2\;\;\;$

How should I proceed from here?

答:
A case can be made that this response is defective, because I am abandoning the OP's approach, which I consider the wrong approach for this problem.  This is a classic geometric series problem where (for $0 < t < 1$) you have

$$S = t + t^2 + t^3 + \cdots = t \times \frac{1}{1-t}$$

and

$$T = t + 2t^2 + 3t^3 + \cdots =
S(1 + t + t^2 + \cdots) =
t \times \left(\frac{1}{1-t}\right)^2.$$


问:
Borel algebra always strictly smaller than its completion?
[Edited] I am not particularly looking for an answer from a reputable source as the bounty says. Just any clear explanation would suffice.

This question is induced by a remark in Folland's Real Analysis where Lebesgue-Stieltjes measure is discussed. In the chapter, it says one can show that the domain of a Lebesgue-Stieltjes measure $\bar{\mu_f}$ (which is the completion of the Borel measure $\mu_f$ generated by an increasing right-continuous function $f$) is always strictly larger than the Borel sigma algebra on $\mathbb{R}$. I wanted to know how to show this.

More generally, a natural question is that in the Carathéodory extension where we extend a premeasure $\mu$ from a measure-theoretic semi-ring $\mathcal{S}$ to a $\mu$-measurable algebra $\mathcal{M}_{\mu}$, I wanted to know when we could conclude that $\sigma (\mathcal{S})$, the sigma algebra generated by $\mathcal{S}$, is strictly smaller than $\mathcal{M}_{\mu}$.

Here are some (hopefully correct) facts that I know of:

If the premeasure $\mu$ is sigma-finite, its extension to $\mathcal{M}_{\mu}$ is the completion.

$\sigma(\mathcal{S})$ can be described by using transfinite induction indexed by the first uncountable ordinal.

If the cardinality of $\mathcal{S}$ is between that of the natural numbers and the reals, then the cardinality of $\sigma(\mathcal{S})$ is that of the reals.

[Added] Cantor set is uncountable but of Lebesgue measure zero, so the set of Lebesgue measurable sets has the same cardinality as $2^{\mathbb{R}}$. So at least it shows the set of Lebesgue measurable sets is strictly larger than Borel sets on $\mathbb{R}$.

I would appreciate any reference or help.

答:
I'm going to stick with the Lebesgue-Stieltjes part of the question. Hopefully Folland had something simpler in mind, but here's one way to show that the domain of the completion has cardinality $2^{\mathbb{R}}$.

Note: I'm not much of a set theorist, so in my mind the only uncountable subsets of $\mathbb{R}$ have size $|\mathbb{R}|$. If someone wants to go over this argument with a more discerning eye in this regard and make some comments, I'd love to know if some things should be changed.

The main idea, as already noted in the original question and the comments, will be to find an uncountable Borel set $G$ with $\mu_{f}(G) = 0$. By completeness, every one of its $2^{\mathbb{R}}$-many subsets will be in the domain of $\overline{\mu}_{f}$.

A good reference for what I'm about to do is Giovanni Leoni's A First Course in Sobolev Spaces (1st edition). All the theorems and propositions I state will be from that book.

The first step will be to decompose $f$ into two increasing functions:
$$f = f_{AC} + f_{S},$$
where $f_{AC}$ is absolutely continuous and $f_{S}$ is differentiable with derivative $0$ $\mathcal{L}^{1}$-a.e (see Theorem 3.73).

With this decomposition, we have
$$\mu_{f} = \mu_{f_{AC}} + \mu_{f_{S}}.$$
Thus, our goal is to find an uncountable Borel set $S$ with measure $0$ with respect to each of these two measures.

We will start with $f_{S}$. By (the proof of) Theorem 3.72, the set $E = \{x : f_{S}'(x) = 0\}$ satisfies $\mathcal{L}^{1}(\mathbb{R} \backslash E) = 0$ and $\mathcal{L}^{1}(f_{S}(E)) = 0$. Note that $E$ is Borel.

Since $f_{S}$ is increasing, it has only countably many discontinuities. Thus the set
$$F = E \backslash \{x : f_{S} \text{ is discontinuous at } x\}$$
is Borel and satisfies $\mathcal{L}^{1}(\mathbb{R} \backslash F) = 0$, $\mathcal{L}^{1}(f_{S}(F)) = 0$, and $f_{S}$ is continuous on $F$. By Proposition 5.9,
$$\mu_{f_{S}}(F) = \mathcal{L}^{1}(f_{S}(F)) = 0.$$
Now we have to deal with $f_{AC}$. By the absolute continuity of $f_{AC}$, we know that $\mu_{f_{AC}}$ is absolutely continuous with respect to the Lebesgue measure. Since $F$ is an uncountable Borel set, it contains a nonempty perfect set $G \subseteq F$ of Lebesgue measure $0$ (as per bof's answer to this question), which means that $\mu_{f_{AC}}(G) = 0$.

All told, we have
$$\mu_{f}(G) = \mu_{f_{AC}}(G) + \mu_{f_{S}}(G) = 0,$$
which completes the proof.


答:
You may first prove the following equality:

Prove that $\frac 1 {2^1} + \frac 2 {2^2} + \dots + \frac k {2^k} = 2 - \frac {k + 2} {2^k}$.

This can be proved by induction on $k$ (easy exercise).

Once it is proved, the original inequality follows.

Title: Inequality Induction Proof, how should I proceed?

Body:
I have been trying to prove inequalities using induction to no avail.

For example,

Prove the following:
$\frac{1}{2^{1}}+\frac{2}{2^{2}}+\frac{3}{2^{3}}+...+\frac{n}{2^{n}}<2$

Base Case:

$\frac{1}{2^{1}} = 2$

$\frac{1}{2}<2\;$  Which is true.

We assume, for n=k, that

$\frac{1}{2^{1}}+\frac{2}{2^{2}}+\frac{3}{2^{3}}+...+\frac{k}{2^{k}}<2\;\;\;$ is true.

As induction step with n=k+1, we need to prove that

$\frac{1}{2^{1}}+\frac{2}{2^{2}}+\frac{3}{2^{3}}+...+\frac{k+1}{2^{k+1}}<2\;\;\;$

How should I proceed from here?


问:
Intuition when solving for intersection of two surfaces

Consider the ellipsoid $x^2 +  5y^2 + z^2 = 1$ and the plane $x+2z=1$. Study their intersection

Now, I can substitute into ellipse,

$$ x = 1-2z$$ and get an equation in $z $ and $y$ which is a circle and if we plug in $ \frac{x-1}{-2}=z$ into the equation, we get equation in $x$ and $y$ which is an ellipse.

Now, what does it mean geometrically when we plug reduce the variables with different variable from plane we plug into the surface?

答:
The circle and ellipse are the projections onto the $y,z$- and $x,y$-planes, respectively. The projection onto the $x,z$-plane is a line segment along $x=1-2z$.

Here's a view of these projections. First, the shapes viewed at an angle:

$\hskip{3cm}$

Next, looking straight along the axes, you can see the intersections form a circle, a line segment, an ellipse:

$\hskip{3cm}$
$\hskip{1cm}$
$\hskip{1cm}$


问:
How To: Calculate The Σ / A of Pascal's Triangle To Base 111
How would I calculate the sum or area of all the numbers in Pascal's Triangle stopping at row 111?

Would I refer to this as the sum? Or the area (A) within the triangle?

Also, how would I create a graphical image of Pascal's Triangle in Base 111? Like this
Pascal's Triangle and how do I express a number that is within the row 111?

I have invested hours of research into these questions.  The mathematical literature sends me in an infinite rabbit hole of confusion.  I tend to struggle with basics.

Thank you for your knowledge if you've shared some with me.

答:
Since the formula for calculating the sum of the numbers in row $n$ is $2^n$, the sum of all numbers up to a certain point is the sum of the first $n$ $2^n$ numbers, or in other case, $$2^{n+1} -1$$ In your case with the row being up to 111, the steps to answer this question is:

Step 1: Substitute $n$ with 111 $$2^{111+1} -1$$

Step 2: Use a calculator to calculate the expression because the product is very large.

The answer is $$5.192296858... * 10^{33}$$

Remember to calculate the "Area" of a pascal triangle up to base $n$ is $2^{n+1} -1$ :)


答:
Your computations are valid. The inequality at the end is direct since
$$
\theta P[Y\leqslant\theta]=\int_0^\theta \theta f_Y(y)\mathrm dy\geqslant\int_0^\theta y f_Y(y)\mathrm dy=E[Y;Y\leqslant\theta].
$$</p>

Title: (simple) Expectation of random variable as a multipart function

Body:
Let the random variable $Y \in [0, \infty)$, a real number $\theta >0$, and the random variable $X$ such that $X = \theta - \min(\theta,Y)$, thus, $X \in [0, \theta]$. That is, $X = 0$ if $Y > \theta$ and $X = \theta-Y$ if $Y \leq \theta$. I would like to know if the expectation of $X$, $E[X] = E[\theta-Y|Y \leq \theta] P(Y \leq \theta) = (\theta-E[Y|Y\leq \theta])P(Y \leq \theta)=\theta P(Y \leq \theta)-\int_{0}^{\theta}y f_{Y}(y)dy$ is valid.
If so, since $E[X] >0$, then, $\theta P(Y \leq \theta) > \int_{0}^{\theta}y f_{Y}(y)dy$ (?) Thanks! </p>


问:
Intuitive understanding of a non-full functor
A functor $T : C \to B$ is full if for each $c,c' \in C$ and for each $g : T c \to T c'$,  there is at least a morphism $f : c \to c'$ such that $g = T f$.

What does this tell me about non-full functors?

My understanding is that, as soon as there's a morphism $g : T c \to T c'$ in $B$, this morphism must come from a morphism in $C$; on the other hand, if the functor is not full, <s>i.e.</s> e.g. there is no morphism $c \to c'$ in $C$, then there  must be another pair of objects $d,d' \in C$ such that

$T d = T c$,

$T d' = T c'$,

$d \neq c \vee d' \neq c'$,

$\exists h:d \to d'$ such that $T h = g$.

Is this correct?

答:
From the comments:

Saying that a functor $F : C \to D$ is not full is saying that there are some objects $Fx$ and $Fy$ in $D$ that have ~bonus arrows~ between them that don't come from arrows in $C$.

For instance, look at the category with 2 objects and one arrow, shown below:

$$x \to y$$

Then a functor from $C \to \mathsf{Set}$, say, is a choice of 2 sets $X$ and $Y$, plus a choice of function $f : X \to Y$.
As long as there are multiple choices of $f$, this functor cannot be full.

More generally, saying that a functor is full is saying that we can understand every arrow in $D(Fx,Fy)$ as the image of an arrow that already existed in $C(x,y)$. Saying that a functor is not full is saying that there are some extra arrows between $Fx$ and $Fy$ that exist in $D$ but not $C$.

I hope this helps ^_^


问:
Price of dividend paying stock
Consider a dividend paying stock $S$ and suppose that its value just before a dividend payment of $D >0$ at time $t$ is denoted by $S(t−)$. The price after the payment should be $S(t)=S(t-)-D$?

How can I justify that. Is there an arbitrage argument?

答: If a stock price is the discounted value of future income, then immediately before the dividend, the future income is the amount $S(t-)$, and immediately after the dividend, the future income is the amount $S(t) = S(t-)-D$. It's lessened by $D$ because that's no longer part of the stream of future income. There's nothing more to it than that.

Title: Price of dividend paying stock


问:
Closed form for subgroups of a set
I am working through an algebra text and am struggling to find the positive integer $k$ (in terms of $r$ and $s$ such that every nontrivial subgroup of the group $G=\{nr+ms\mid n,m\in \Bbb Z\}$ with $r$ and $s\in\Bbb Z$ takes the form $k\Bbb Z$.

For example, we know that for $\Bbb Z$ every nontrivial subgroup is of the form $k\Bbb Z$.

So far, I know that $G$ is cyclic as $\gcd(r,s)$ generates $G$. So we know that all subgroups of $G$ are cyclic but I am not sure how to use that or if that is even the correct method to approach this problem.

答: You've already figured out that $G = d\mathbb{Z}$ where $d = \gcd(r,s)$. So subgroups of $G$ correspond to subgroups of $\mathbb{Z}$ under the isomorphism $\mathbb{Z}\to G:x\to dx$. Can you take it from there?

Title: Closed form for subgroups of a set


问:
Base and dimension in a subspace
I need help interpreting the answer to a question about the base and dimension of a subspace within linear algebra. I have a subspace W of $R^5$ that is spanned by the vectors:

$${v_1}=\begin{pmatrix} 1 \\ 2 \\ 3 \\ -1 \\ 1 \end{pmatrix} ,  {v_2}=\begin{pmatrix} 0 \\ 1 \\ 1 \\ -1 \\ 2 \end{pmatrix} ,  {v_3}=\begin{pmatrix} -1 \\ -1 \\ -2 \\ 1 \\ 2 \end{pmatrix}
    \text{and} \,{v_4}= \begin{pmatrix} 1 \\ 0 \\ 1 \\ 0 \\ -4 \end{pmatrix}$$

and I will determine the dimension of W and also specify the base of W.

To do this I put the vectors as columns in a matrix and simplified this using gauss:

$$\begin{pmatrix}1 & 0 & -1 & 1 \\ 2 & 1 & -1 & 0 \\ 3 & 1 & -2 & 1 \\ -1 & -1 & 1 & 0 \\ 1 & 2 & 2 & -4\end{pmatrix}\sim\begin{pmatrix}1 & 0 & 0 & 0 \\ 0 & 1 & 0 & -1 \\ 0 & 0 & 1 & -1 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0\end{pmatrix}.\\$$

Here I concluded that the dimension max can be equal to three as we have three rows of pivot elements, then I also concluded that you can express v4 = -v2-v3 (I have also tested this and it is true) and the two vectors v2 and v3 must then be linearly independent. But what I have a hard time interpreting is if this means that v2 and v3 are the basis of W (and not v1,v2 and v3 as my first thought was)? And if the dimension in that case becomes two and not three (as the maximum can be)?

Thanks in advance!

答:
I would do that  transposing, and performing row operations, to make the whole process:
\begin{align}
&\begin{bmatrix}
1&2&3&-1&1\cr 0&1&1&-1&2\cr -1&-1&-2&1&2\cr 1&0&1&0&-4
\end{bmatrix}\xrightarrow[v_4\leftarrow v_4-v_1]{v_3\leftarrow v_3+v_1}
\begin{bmatrix}
1&2&3&-1&1\cr 0&1&1&-1&2\cr 0&1&1&0&3\cr 0&-2&-2&1&-5
\end{bmatrix}\\[1.5ex]
\xrightarrow[v_4\leftarrow v_4+v_2+v_3]{}
&\begin{bmatrix}
1&2&3&-1&1\cr 0&1&1&-1&2\cr 0&1&1&0&3\cr 0&0&0&0&0
\end{bmatrix}\xrightarrow{v_3\leftarrow v_3-v_2}
\begin{bmatrix}
1&2&3&-1&1\cr 0&1&1&-1&2\cr 0&0&0&1&1\cr 0&0&0&0&0
\end{bmatrix}
\end{align}
Therefore, we obtain that the matrix has rank $3$, and the vectors of the first three rows: $v_1, v_2$ and $v'_3=v_1-v_2+v_3$ are a basis of $W$. As
$\operatorname{Span}(v1,v_2,v'_3)=\operatorname{Span}(v1,v_2,v_3)$, there results that $\{v_1,v_2,v_3\}$ is also a basis of $W$.

Title: Base and dimension in a subspace


问:
Evaluate the flux of $F=\langle\sin(xyz), x^2y, z^2e^{x/5}\rangle$ through surface $S$ ... $4y^2+z^2=4, \space x\in [-2,2]$
I am trying to find the flux of  $\vec F=\langle\sin(xyz), x^2y, z^2e^{x/5}\rangle$ through the surface $S$ where $S$ consists of the elliptical cylinder defined by

$S$ ... $4y^2+z^2=4, \space x\in [-2,2]$

My first instinct is to parametarize the surface to get $\vec F \cdot \vec n$,

where

$$\vec r = \left\langle x,y,\pm \sqrt{1-y^2} \right\rangle$$

I get $$\vec n = \left\langle 0, \pm \frac{2y}{\sqrt{1-y^2}},1\right\rangle$$

so $$\vec F \cdot \vec n = \pm \frac{2x^2y^2}{\sqrt{1-y^2}} + 4(1-y^2)e^{x/5}.$$

I know that $x \in [-2,2]$ and $y \in \left[-\sqrt{1-\left(\frac{z}{2}\right)^2}, \sqrt{1-\left(\frac{z}{2}\right)^2} \right]$, and using a calculator I would be able to evaluate it, but with certain techniques the answer simplifies to

$$\frac{16\pi + 160 \sinh(2/5)}{3}.$$

I am not as familiar with integration involving hyperbolic functions and that is probably the reason why I am not seeing this.

Is there anyone that can confirm this fact?

答:
The easiest way to do this integral would be via divergence theorem:

$$\nabla\cdot \vec{F} = yz\cos(xyz)+x^2+2ze^{\frac{x}{5}}$$

Why does this make life easy? Two of the terms are odd functions of $z$ and the cylinder has $z$ symmetry (i.e. $(x,y,-z)\in$ cylinder $\iff (x,y,z)\in$ cylinder). Thus the integrals of those two terms vanish and we are left with

$$\iiint\limits_V x^2\:dV = \int_{-2}^2x^2dx\cdot\iint\limits_{4y^2+z^2\leq 4}dA = \frac{16}{3}\cdot 2\pi = \frac{32\pi}{3}$$

Now to only get the contribution from the cylinder $S$, we must subtract the integrals over the planes on the "top" and "bottom".

$$\iiint\limits_V \nabla\cdot\vec{F} dV = \iint_S\vec{F}\cdot dS+\iint_{x=2}\vec{F}\cdot dS + \iint_{x=-2}\vec{F}\cdot dS$$

$$\implies \iint_S\vec{F}\cdot dS = \iiint\limits_V \nabla\cdot\vec{F} dV - \iint_{x=2}\vec{F}\cdot dS - \iint_{x=-2}\vec{F}\cdot dS$$

However the integrals simplify to

$$\iint_{x=2}\vec{F}\cdot dS + \iint_{x=-2}\vec{F}\cdot dS = 2\iint\limits_{4y^2+z^2\leq 4}\sin(2yz)dA = 0$$

by the same odd symmetry. Thus the answer to the original surface integral is $\frac{32\pi}{3}$


问:
How to clear $T$ from the formula $vf = vp ( 1 + t \times i )$?
Given the formula $vf = vp ( 1 + t \times i )$, I would like to clear $t$.

I would like to know what is the steps to do this. I have issues with formulas and don't understand. Please help me.

答:
That's elementary algebra.
Given

$$
v_f = v_p(1 + it)
$$

you get
$$
\frac{v_f}{v_p} = 1 + it,
$$
then
$$
\frac{v_f}{v_p} -1 = it
$$
so
$$
t = \frac{1}{i}\left( \frac{v_f}{v_p} -1 \right)
$$


问:
For i.i.d. $\{X_n\}$ with non-constant $X_n > 0$ and $\mathbb{E}X_n = q \leq 1$ the product $\prod\limits_{n=1}^{+\infty}X_n \to 0$ almost surely
Suppose $\{X_n\}$ is i.i.d. with $X_n > 0$ and each $X_n$ is not constant almost surely.

I want to show that if $\mathbb{E}X_n = q \leq 1$, then the product $\prod\limits_{n=1}^{+\infty}X_n \to 0$ almost surely. In the problem statement, there is also a hint that $q < 1$ and $q = 1$ can be considered separately.

I thought about this one for a long time and I think applying logarithm operation to both sides and transform this product into a sum might help, but I couldn't progress from there. I'm still unable to see how I can reduce this problem to using laws of large numbers.

I would appreciate any help!

答:
Since $X_{n}>0$, we have that $q=E[X_{n}]>0$. Define the process
$M=\{M_{n}\mid n\in\mathbb{N}\}$ by $M_{n}=\prod_{k=1}^{n}\frac{X_{k}}{q}.$
Let $\mathbb{F}=\{\mathcal{F}_{n}\mid n\in\mathbb{N}\}$ be the raw-filtration
induced by $M,$ i.e., $\mathcal{F}_{n}=\sigma\{M_{1},M_{2},\ldots,M_{n}\}$.
Note that we also have $\mathcal{F}_{n}=\sigma\{X_{1},X_{2},\ldots,X_{n}\}.$
We go to show that $M$ is a $L^{1}$-bounded $\mathbb{F}$-martingale.
Clearly $M_{n}$ is integrable (recall that the product of two independent
random variables is integrable). Moreover, since $X_{n+1}$ and $\mathcal{F}_{n}$
are independent while $M_{n}$ is $\mathcal{F}_{n}$-measurable, we
have that
\begin{eqnarray*}
 &  & E\left[M_{n+1}\mid\mathcal{F}_{n}\right]\\
 & = & E\left[\frac{X_{n+1}}{q}M_{n}\mid\mathcal{F}_{n}\right]\\
 & = & M_{n}E\left[\frac{X_{n+1}}{q}\mid\mathcal{F}_{n}\right]\\
 & = & M_{n}E\left[\frac{X_{n+1}}{q}\right]\\
 & = & M_{n}.
\end{eqnarray*}
Moreover, since $M_{n}>0$, we clearly have $E\left[|M_{n}|\right]=E\left[M_{n}\right]=E[M_{1}]=E[X{}_{1}/q]=1$.
This shows that $\sup_{n}E\left[|M_{n}|\right]<\infty.$

By Martingale Convergence Theorem, there exists a $\mathcal{F}_{\infty}$-measurable
random variable $\xi$ with $E[|\xi|]<\infty$ such that $M_{n}\rightarrow\xi$
pointwisely a.e. (Note that, in general, we do not have $M_{n}\rightarrow\xi$
in $L^{1}$).

Case 1: $q<1$. We have that $\prod_{k=1}^{n}X_{n}=q^{n}M_{n}\rightarrow0$
a.e. because $q^{n}\rightarrow0$ while $M_{n}\rightarrow\xi$ a.e.

Case 2: Define $a=E\left[\sqrt{X_{n}}\right].$ By Cauchy-Schwarz inequality,
we have that
\begin{eqnarray*}
a & = & \int\sqrt{X_{n}}\cdot1dP\\
 & \leq & \left\{ \int X_{n}dP\right\} ^{\frac{1}{2}}\left\{ \int1^{2}dP\right\} ^{\frac{1}{2}}\\
 & = & 1
\end{eqnarray*}
and the equality holds iff $\sqrt{X_{n}}$ and $1$ are linearly dependent (as elements in $L^2$, which is false since $X_{n}$ is not constant a.e.). Therefore, $0<a<1.$

Define a process $Y=\{Y_{n},\,\,\,n\in\mathbb{N}\}$ by $Y_{n}=\frac{\sqrt{M_{n}}}{a^{n}}.$
It can be proved similarly that $Y$ is a $\mathbb{F}$-martingale.
For, by Cauchy-Schwarz inequality, $M_{n}$ is integrable $\Rightarrow$
$\sqrt{M_{n}}$ is integrable. Clearly $Y_{n}$ is $\mathcal{F}_{n}$-measurable.
Moreover,
\begin{eqnarray*}
 &  & E\left[Y_{n+1}\mid\mathcal{F}_{n}\right]\\
 & = & E\left[\frac{\sqrt{X_{n+1}}}{a}\cdot Y_{n}\mid\mathcal{F}_{n}\right]\\
 & = & Y_{n}E\left[\frac{\sqrt{X_{n+1}}}{a}\mid\mathcal{F}_{n}\right]\\
 & = & Y_{n}E\left[\frac{\sqrt{X_{n+1}}}{a}\right]\\
 & = & Y_{n}.
\end{eqnarray*}
Since $Y$ is non-negative, we have that $E\left[|Y_{n}|\right]=E\left[Y_{n}\right]=E\left[Y_{1}\right].$
It follows that $\sup_{n}E\left[|Y_{n}|\right]<\infty$, i.e., $Y$
is $L^{1}$-bounded. By Martingale Convergence Theorem again, there
exists an integrable random variable $\eta$ such that $Y_{n}\rightarrow\eta$
a.e.. Recall that $M_{n}\rightarrow\xi$ a.e. and notice that $M_{n}=a^{2n}Y_{n}^{2}$.
Letting $n\rightarrow\infty$, we have that $\xi=0$ a.e. because
$a^{2n}\rightarrow0$ while $Y_{n}^{2}\rightarrow\eta^{2}<\infty$
a.e.


问:
Translation and scaling invariance of the Hardy-Littlewood maximal function
I have as an exercise to prove some properties of the Hardy-Littlewood maximal function, namely positivity, sublinearity, homogeneity, translation invariance and scaling invariance. The first three are easy but the last two is what I am struggling with and I can't find any tips online. We assume that $f \in L^1(\mathbb{R}^n)$.

(d) (Translation invariance) $M(\tau_y f) (x) = (\tau_y Mf) (x)$, where $\tau_y f(x) = f(x +y)$

(e) (Scaling invariance) $ M(\eta_a f)(x) = (\eta_a Mf)(x)$, where $\eta_a f(x) = f(ax)$.

I think I have gotten somewhere with (d) since the ball does not really change no matter what you call it's centre so I tried doing some substitution business, but I could not quite piece it out. Any help would be appreciated.
Using this definition:
$$Mf(x)=\sup_{r>0}\frac{1}{|B(x,r)|}\int_{B(x,r)}|f(t)|dt$$

答:
Let $g_r(x)=\frac{1}{|B(0,r)|}\chi_{B(0,r)}(x)$. Then
$$
Mf(x)=\sup_{r>0} M_rf(x),.
$$
where
$$
M_rf(x)=\int_{\mathbb{R}^d}|g_r(x-t)f(t)|dt.
$$

Notice that changing variables $u=t+y$ we get
$$
M_r \tau_y f(x)=\int_{\mathbb{R}^d}|g_r(x-t)\tau_yf(t)|dt=\int_{\mathbb{R}^d}|g_r(x-t)f(t+y)|dt=\int_{\mathbb{R}^d}|g_r(x+y-u)f(u)|du=\tau_y M_r f(x).
$$
Part d) follows by taking supremum over $r>0$.

A similar argument based on the change of variables $u=at$ gives part e).


问:
Fourier analysis questions
Can anyone give me a hand with the proof of this properties?

Prove that:

a) The linear span of the set $\left\{T_bh/b\in\mathbb{R}\right\}$ is dense in $L_2(\mathbb{R})$, where $h(x)=e^{-\pi x^2}$.

b)$\blacksquare$ $V_{\phi}f(t,w)=e^{-2\pi itw}V_{\hat{\phi}}\hat{f}(w,-t)$.

$\blacksquare$ $V_{\phi}(T_{u}E_{\eta}f)(t,w)=e^{-2\pi iwu}V_{\phi}f(t-u,w-\eta)$.

where $V_{\phi}f(t,w)=\int_{\mathbb{R}}f(s)\overline{\phi(s-t)}e^{-2\pi iws}ds$, for $f,\phi\in{L_2(\mathbb{R}})$, with $\phi\neq0$.

($E_a$ is the modulation $E_af(t)=e^{2\pi iat}f(t)$, and $T_a$ is the traslation $T_af(t)=f(t-a)$, and $V_{\phi}f$ is the Gabor tranform of $f$)

Thank you very much in advance.

答:
Once question a) is done, use (sesqui)linearity of the involved operators to do this when $f=T_ae^{-\pi x^2}$ and $\phi=T_be^{-\pi x^2}$.

For question a), what we have to show is that if $f\in L^2$ and for each $c$, we have $\langle f,T_ce^{-x^2}\rangle=0$ then $f=0$.
To do that, use Plancherel identity and $\widehat{T_c(x\mapsto e^{-\pi x^2})}=e^{icx}e^{-\pi x^2}$. The function $x\mapsto \widehat f(x)e^{-\pi x^2}$ is integrable, and its Fourier transform is $0$, hence $\widehat f=0$ and $f=0$.

Title: Fourier analysis questions


问:
Atiyah-MacDonald Ex 2.10

Let $A$ be a ring. Let $\mathfrak a$ be an ideal contained int he Jacobson Radical of $A$. Let $u:M \rightarrow N$ be a morphism. Let $N$ be finitely generated. Show that if $w:M/\mathfrak a M \rightarrow  N/ \mathfrak a N$ is surjective then so is $u$.

Firstly this screams Nakayamma's lemma.
We have the following commutative diagram which we can apply the snake lemma to.

$\DeclareMathOperator{\coker}{coker}$
If we suppose that $w$ is surjective then the $\coker(w)=0$. Then we have $\ker(w)\rightarrow \coker(v)\rightarrow \coker(u)\rightarrow 0$.

I'm not to sure what to do from here. The map $v$ is just the restriction of $u$ so there should be some nice between their cokernels but I cannot see what.

答: You don't even need exact sequences here. $N$ is finitely generated, $u(M)$ is its submodule, and it is easy to show that $N=\mathfrak{a}N+u(M)$. Now just apply corollary $2.7$ from Atiyah-Macdonald.

Title: Atiyah-MacDonald Ex 2.10


问:
How to evaluate $ \sum_{n=r}^{\infty} n^2 \binom{n-1}{r-1} p^r (1-p)^{n-r}$?
I want to evaluate the following sum

$$ \sum_{n=r}^{\infty} n^2  \binom{n-1}{r-1} p^r (1-p)^{n-r} $$

where $r \in \mathbb{N}$ and $0<p<1$.

Using WA I know that the sum evaluates to $\frac{r(r-p+1)}{p^2}$ but I don't know how to show this. I tried simplifying the sum a bit using that
$$
n  \binom{n-1}{r-1} = r  \binom{n}{r}
$$
Which gives me the new equivalent sum
$$
\sum_{n=r}^{\infty} n^2  \binom{n-1}{r-1} p^r (1-p)^{n-r} = r\sum_{n=r}^{\infty} n  \binom{n}{r} p^r (1-p)^{n-r}
$$
but I didn't know how to continue. I have also previously shown that
$$
\sum_{k=s}^{\infty} \binom{k-1}{s-1} (1-t)^{k-s} t^{s} = 1
$$
for $\lvert 1 - t \rvert <1$. I tried to apply this but I couldn't find a way to get rid of the factor of $n$ multiplying the binomial coefficients. Can anyone give me any hints as to how I can show this?

答:
Using an extension of the method described in this answer, it is straightforward to show that

$$n(n+1) \binom{n-1}{r-1} = \frac{(n+1)!}{(r-1)!(n-r)!} = r(r+1) \frac{(n+1)!}{(r+1)!(n-r)!} = r(r+1) \binom{n+1}{r+1},$$ consequently

$$\begin{align}
\sum_{n=r}^\infty n(n+1) \binom{n-1}{r-1} p^r (1-p)^{n-r}
&= r(r+1) \sum_{n=r}^\infty \binom{n+1}{r+1} p^r (1-p)^{n-r} = r(r+1) \\
&= r(r+1) p^{-2} \sum_{n=r+2}^\infty \binom{n-1}{(r+2)-1} p^{r+2} (1-p)^{n-(r+2)} \\
&= \frac{r(r+1)}{p^2}.
\end{align}$$

And since the linked answer also provided $$\sum_{n=r}^\infty n \binom{n-1}{r-1} p^r(1-p)^{n-r} = \frac{r}{p},$$ we combine these results to obtain $$\sum_{n=r}^\infty n^2 \binom{n-1}{r-1} p^r(1-p)^{n-r} = \frac{r(r+1)}{p^2} - \frac{r}{p} = \frac{r(r+1-p)}{p^2}.$$


问:
Vertical Asymptotes of $f(x) = \frac{\ln(|4x+1|)}{\ln(\sqrt{3x+4})}$
I want to make sure I understand the concept correctly. If I'm trying to find the vertical asymptotes of the following:
$$ f(x) = \frac{\ln(|4x+1|)}{\ln(\sqrt{3x+4})}$$

Do we not care of when $ln(x)$ is not defined, and just evaluate for the denominator being equalling $0$ in which case the only vertical asymptote is $x = -1$, or do we need to be careful? If so, for what do I need to look out?

Thanks!

答:
For rational functions, you would be right in thinking that you only need to worry about the denominator.  However, as the logarithmic function carries a vertical asymptote itself (which grows far more quickly compared to the rate of rational functions not near asymptotes), we also have to consider those.

As such, I am sure it would not surprise you to know that the vertical asymptotes of this function are $x=-1$ and $x=-\frac14$.


问:
How to show function is uniformly continous on certain interval
Let $ f\colon \mathbb{R} \to \mathbb{R}$ be  $$x \mapsto \begin{cases} \frac{(x-1)^2}{x^2+1}, &|x| \neq 1 \\2, &|x|=1\end{cases} $$

How can I show that $f$ is uniformly continuous on $[-2,2]\backslash \{ - 1,1\}$?
I thought it could be done with  the $ \epsilon - \delta$ criteria but I am not sure because of the specific interval.

答: $\frac{(x-1)^2}{x^2+1}$ defines a continuous function on $[-2,2]$. Continuous functions on compact sets are uniformly continuous, so this function is uniformly continuous on $[-2,2]$, and thus also on all subsets of $[-2,2]$, including $[-2,2]\backslash\{\pm1\}$, where it's identical to $f$. So $f$ is uniformly continuous on said set.


问:
Extreme random variable sequence independence
During solution of probability problem I came across following:
Let $X_n$ - be a sequence of i.i.d (independent, identically distributed) random variables, uniformly distributed over $[0;1]$. Then is it true that random variables $m_n$, determined as $m_n=min\{X_1, ..., X_n\}$ are also independent or not?

If not, do you have any ideas of proving that such  approaches 0 almost surely then? I was trying to proof this via fact mentioned above, and do not have anymore ideas..

Do you have any ideas about this one? This is very crucial for my solution, would appreciate any help!

答:
(This answer is in response to the comments by OP).

We have $0 \leq m_{n+1} \leq m_m \leq 1$. This implies that $m =\lim_{ n\to \infty} m_m$ exist almost surely. In particular, this gives convergence in probability. But $P(m_n >\epsilon) =(P(X_1>\epsilon)^{n} \to 0$ (since $x^{n} \to 0$ whenver $|x|<1$). It follows that $m_m \to 0$ in probability and $m_n \to m$ in probability. Hence, $m=0$ almost surely and $\lim_{ n\to \infty} m_n=0$ almost surely.


问:
Frobenius coin problem
Suppose that you only have coins worth, say 3 and 5 euros. According to Sylvester result we can find the Frobenius nr $g(3,5)=15-3-5=7$ so 7 is the largest integer that cannot be written as $a_{1}k_{1}+a_{2}k_{2}$ for $k_{1},k_{2}\in\mathbb{N}$ and $a_{1},a_{2}$ are the values of these coins.

a) how do you pay 8€,9€ and 10€ with these coins?

b)use a) to show that it is possible to pay all amounts that are greater than 10€ with the coins 3€ and 5€.

c) show that it is impossible to pay the amount of 7€ with these coins.

I am afraid I do not understand 100% the whole idea behind the Frobenius numbers.

a) can we just take 3€+5€=8€ and 3€+3€+3€=9€ and 5€+5€=10€ this seems suspicious of how easy it is....

b)do I have to use both coins? or just 3€ or 5€?
11€=3€+5€+3€

12€=3€+3€+3€+3€

13€=3€+5€+5€

14€=5€+3€+3€+3€

.

.

c) if we could pay 7€ with these coins we could have written

$7€=k_{1}5€+k_{2}3€$ but this is impossible as $k_{1},k_{2}\in\mathbb{N}$

can someone please explain to me what should be done in this exercise and how?

答:
Your approach to (c) can be made to work. You have the Diophantine equation $3x+5y=7$. One solution is $x=-1,y=2$, so the general solution is

$$\left\{\begin{align*}
x&=-1+5k\\
y&=2-3k\;.
\end{align*}\right.$$

Since we require that $x\ge 0$, we must have $k\ge 1$, but then $y\le-1<0$, so there is no solution in non-negative integers.

However, the numbers are so small that it’s easier to examine cases, unless you’re very comfortable with solving linear Diophantine equations. Since $3+5>7$, you clearly cannot use both denominations to make $7$. But $7$ is not a multiple of $3$, so you can’t make it using only $3$’s, and it’s not a multiple of $5$, so you can’t make it using only $5$’s. Thus, you can’t make it at all.

For (b) you really do need a proof by induction. For your induction step try to prove that if you can make $n,n+1$, and $n+2$, then you can make $n+1,n+2$, and $n+3$; do you see why that would give you the desired result once you know how to make $8,9$, and $10$?

Title: Frobenius coin problem


问:
Calculating the Rolling Variance of a set of numbers
I would like to subscribe to a WebSocket stream which will supply me with many numbers per second. From this data, I would like to calculate the variance of say the last 1000 numbers.

How can I do this in a rolling fashion? That is, I would like some computation comparable to this one for the mean of the last 1000 numbers:

$$\rm{mean}_{i+1} = \rm{mean}_{i} + \frac{1}{1000}\left(x_{i+1}-x_{i-999}\right)$$

Thanks in advance for any help.
Ben

答:
$$\rm{variance}_{i+1} = \frac{1}{1000}\sum_{j=i-999}^{i+1}\left(x_j-\rm{mean}_{i+1}\right)^2\\
=\frac{1}{1000}(x_{i+1}^2-x_{i-999}^2)+\rm{variance}_{i}-\rm{mean}_{i+1}^2+\rm{mean}_{i}^2,$$
where you already computed $\rm{mean}_{i+1}$ according to your equation.
If you wonder why the equality holds, see here.


问:
Prove an greater than 11 can be written as a sum of only 3's and only 7's.
Prove an greater than 11 can be written as a sum of only 3's and only  7's.

Is there a way to do this by strong induction. I would greatly appreciate someone showing me how to apply the classic strong induction technique to this question.

答:
$$12=3+3+3+3,13=3+3+7,14=7+7.$$

Now assume the result to be true for all $n$ such that  $12\le n\le k$, where  $k\ge14$. Then consider $n=k+1$.

$n=3+(k-2)$ where, by assumption $k-2$ can be expressed as an appropriate sum. Hence, by strong induction, every number greater than $12$  can be so expressed.


问:
Why is the derivative of $\arctan(\frac{x}{\sqrt{1-x^2}})$ the same as the derivative of $\arcsin(x)$?
I'm self studying math through MIT opencourseware. Unfortunately the solutions of the problem sets are not very user-friendly and don't contain much explanation. There is a question which is bugging me and I couldn't come up with any explanation (there is no explanation provided in the solutions as well).

The question is:

Why is the derivative of $\arctan(\frac{x}{\sqrt{1-x^2}})$ the same as the derivative of $\arcsin(x)$?

I have solved the derivative of the arctan part and it's obvious to me how to get to the $\frac{1}{\sqrt{1-x^2}}$ answer. But I don't understand how this is related to the $\arcsin(x)$. Why are they the same?

Thank you in advance

答:
For $\alpha$ in the first quadrant, if $\sin(\alpha)=x$, then $\cos(\alpha)=\sqrt{1-x^2}$, and so
$$\tan(\alpha) = \frac{\sin(\alpha)}{\cos(\alpha)} = \frac{x}{\sqrt{1-x^2}}.$$
That means that if $\arcsin(x)=\alpha$, then $\arctan\left(\frac{x}{\sqrt{1-x^2}}\right) = \alpha$ as well. That is, $\arcsin(x) = \arctan\left(\frac{x}{1-x^2}\right)$. They are the same function (at least on $[0,1)$), so they have the same derivative.

A similar result holds for $\alpha$ in the fourth quadrant, which are the angles you get when $x$ is negative: you still have $\cos(\alpha)=\sqrt{1-\sin^2(\alpha)}$, because the cosine is nonnegative for angles in $[-\frac{\pi}{2},\frac{\pi}{2}]$.

Since they are in fact the same function where they are both defined (which is $-1\lt x\lt 1$), it is not a surprise they have the same derivative.

More generally, two functions have the same derivative exactly when (if and only if) they differ by a constant (this is the Constant Function Theorem). So for example, $-\arctan(x)$ and $\arctan(\frac{1}{x})$ (for $x\gt 0$) have the same derivative:
$$\begin{align*}
-\frac{d}{dx}\arctan(x) &= -\frac{1}{1+x^2}\\
\frac{d}{dx}\arctan\left(\frac{1}{x}\right) &= \left(\frac{1}{1+\frac{1}{x^2}}\right)\left(\frac{1}{x}\right)’\\
&= \frac{1}{\quad\frac{x^2+1}{x^2}\quad}\left(-\frac{1}{x^2}\right)\\
&= -\frac{1}{x^2+1}.
\end{align*}$$
so $-\arctan(x)$ and $\arctan\left(\frac{1}{x}\right)$ differ by a constant on $(0,\infty)$. You can figure out the value since at $x=1$ you get $-\arctan(1) = -\frac{\pi}{4}$, and $\arctan(\frac{1}{1}) = \frac{\pi}{4}$, so the two functions differ by $\frac{\pi}{2}$.


问:
Non-convergence of a sequence implies 'consecutive' partial limits
Let $n,d$ be fixed positive integers, and let $X^k=(x_1^k, \dots, x_n^k)$ be a bounded sequence, where $x_i^k \in \mathbb{R}^d$. (Every element of $X^k$ is an $n$-tuple of points in $\mathbb{R}^d$).

Claim: Suppose that $X^k$ has a finite number $N_0 \ge 2$ of accumulation points. Then there exist distinct accumulation points $Y=(y_1, \dots, y_n),Z=(z_1, \dots, z_n)$ and distinct subsequences $X^{k_j},X^{k_l}$ such that $X^{k_j} \to Y, X^{k_l} \to Z$ and $k_l=k_j+1$.

How to prove this claim?

I saw it mentioned in a paper without proof, so I assume there is an elementary solution.

答:
In your claim:

$\{X^k\}_{k\in\mathbb N}$ must be bounded

the number of accumulation points of $\{X^k\}_{k\in\mathbb N}$ must be greater or equal than 2

I define   $\ \forall k \in \mathbb N \ , \ U^k=X^{k+1}-X^k$.

$\{U^k\}_{k\in\mathbb N}$ is bounded and has therefore at least one point of accumulation $U$.

If $U\neq 0$, there exists a subsequence $\{U^{\varphi(k)}\}_{k\in\mathbb N}$ that converges to $U$. I extract from $\{X^{\varphi (k)}\}_{k\in\mathbb N}$ a subsequence $\{X^{\varphi \circ \psi(k)}\}_{k\in\mathbb N}$ that converges to $Y$. We have:
$\ \lim X^{\varphi \circ \psi(k)} = Y \ $ and $\ \lim X^{\varphi \circ \psi(k)+1} = Y +U$.

If the only point of accumulation of $\{U^k\}_{k\in\mathbb N}$ is $0$, then, $\{U^k\}_{k\in\mathbb N}$ converges to $0$. I note $d$ the minimal distance between the points of accumulation of  $\{X^k\}_{k\in\mathbb N}$ and $\varepsilon =d/3$. There exists $N\in \mathbb N$ such that $\forall p\geqslant N$, $X^p$ is in the union of $B(a,\varepsilon)$ where $a$ is a point of accumulation of  $\{X^k\}_{k\in\mathbb N}$ (otherwise, there would be another point of accumulation). There exists $M \geqslant N$ such that $\forall p\geqslant M \ , \|X^{p+1}-X^p\| < \varepsilon $. If we note $b$ the point of accumulation such that $X^M \in B(b,\varepsilon ) $, we can show by induction that $\forall p\geqslant M \ , X^{p} \in B(b,\varepsilon)$. But that would imply that $b$ is the only point of accumulation of  $\{X^k\}_{k\in\mathbb N}$. There is a contradiction.

And the claim is proved .


问:
Price of dividend paying stock
Consider a dividend paying stock $S$ and suppose that its value just before a dividend payment of $D >0$ at time $t$ is denoted by $S(t−)$. The price after the payment should be $S(t)=S(t-)-D$?

How can I justify that. Is there an arbitrage argument?

答:
Not necessarily, because of tax considerations.  Depending on the jurisdiction, capital gains may be taxed at a different rate than dividends.  Depending on these rates, selling a stock before the dividend at $S(t-)$ may be a better or a worse deal than
getting the dividend $D$ and selling at $S(t-)-D$.  If enough investors prefer one to the other, it will affect the stock prices.

Title: Price of dividend paying stock


问:
explaining the derivative of $x^x$
You set the following exercise to your calculus class:

Q1. Differentiate $y(x) = x^x$.

A student submits the following solution:

Let $g(a)=a^x$ and $f(x)=x$.  Then $y(x) = g(f(x))$, so by the chain rule, $y'(x) = f'(x) g'(f(x)) = 1 \cdot x \cdot (x^{x-1}) = x^x$.

How would you explain to the student why their solution is incorrect?

To be clear, I know why this is wrong but am interested in good ways to explain it to undergraduate or high school students.

In this question someone has problems differentiating $x^x$, but they didn't take the approach of my hypothetical student.

答:
Well, an important first step is to convince them that their argument is, in fact, wrong.  A simple appeal to the geometric meaning of the derivative can be very useful in this regard.  Not only does this provide a very clear and easy to understand demonstration of the incorrectness of their result but it also encourages them to think about the derivative in more than one way.  All too often students at this level think purely algebraically, without serious geometric consideration.

So start with a simple plot of the graph of $x^x$, which any reasonable mathematical software should be able to provide:

Now, the graph lies completely above the $x$-axis, so the function value is always positive.  The slope of the graph is clearly negative, however, just to the right of the $y$-axis. Thus, there is no way that this function can be its own derivative.

Title: explaining the derivative of $x^x$


答:
It could help to notice that your function's domain of definition is $\;\left(-\frac43,\,\infty\right)\setminus\{-\frac14\}\;$, and also for these values

$$\log\sqrt{3x+4}=\frac12\log(3x+4)$$

so your function is actually

$$f(x)=\frac{2\log|4x+1|}{\log(3x+4)}$$

Since  $\;\log|4x+1|\xrightarrow[x\to-1/4]{}-\infty\;$  and the denominator equals $\;\log13/4\;$ , then $\;x=-1/4\;$ is a vertical asymptote.

And also when the denominator gets close to zero but the numerator doesn't (as then we'd need to check more carefully), and this happens when $\;x\to-1\;$ , which you already had.

In any other value of the domain the function is just the quotient of two well defined, continuous functions and thus continuous itself.

Title: Vertical Asymptotes of $f(x) = \frac{\ln(|4x+1|)}{\ln(\sqrt{3x+4})}$

Body:
I want to make sure I understand the concept correctly. If I'm trying to find the vertical asymptotes of the following:
$$ f(x) = \frac{\ln(|4x+1|)}{\ln(\sqrt{3x+4})}$$

Do we not care of when $ln(x)$ is not defined, and just evaluate for the denominator being equalling $0$ in which case the only vertical asymptote is $x = -1$, or do we need to be careful? If so, for what do I need to look out?

Thanks!


答:
Conceptually, both $\theta_t(x)=\arctan\frac{x}{\sqrt{1-x^2}}$ and  $\theta_s(x)=\arcsin(x)$ represent angles as functions of $x$. Equivalently, the two expressions are $\tan( \theta_t (x))= \frac{x}{\sqrt{1-x^2}}$ and $\sin(\theta_s(x))=x$, respectively. Then, $\cos (\theta_s(x))= \sqrt{1-x^2}$ and

$$\tan(\theta_s(x) )= \frac {\sin(\theta_s(x))}{\cos(\theta_s(x))}= \frac{x}{\sqrt{1-x^2}}=\tan( \theta_t(x))
$$

Thus, $\theta_t(x)$ and $\theta_s(x)$ are the same. So are their derivatives.

Body:
I'm self studying math through MIT opencourseware. Unfortunately the solutions of the problem sets are not very user-friendly and don't contain much explanation. There is a question which is bugging me and I couldn't come up with any explanation (there is no explanation provided in the solutions as well).

The question is:

Why is the derivative of $\arctan(\frac{x}{\sqrt{1-x^2}})$ the same as the derivative of $\arcsin(x)$?

I have solved the derivative of the arctan part and it's obvious to me how to get to the $\frac{1}{\sqrt{1-x^2}}$ answer. But I don't understand how this is related to the $\arcsin(x)$. Why are they the same?

Thank you in advance


问:
Statistics probability density function
I was given the probability density function

$$f(x) = \begin{cases} 0.75(1 - x^2) & \text{if} -1 < x < 1 \\ 0 & \text{otherwise} \end{cases},$$

and I want to calculate the mean. I know that the definition of mean for a continuous random variable is the integral from $-\infty$ to $\infty$ of $xf(x) dx$.

My question was, since my function does not go to infinity, it only goes from -1 to 1 and is 0 elsewhere, how do I calculate that integral for the mean? Is it zero?

答:
As others have already said:

$\int_{-\infty}^{\infty} xf(x)dx$

$= \int_{-\infty}^{-1} xf(x)dx + \int_{-1}^{1}xf(x)dx + \int_{1}^{\infty}xf(x)dx$

$= 0 + \int_{-1}^{1}xf(x)dx + 0$

$= \int_{-1}^{1}xf(x)dx$

Title: Statistics probability density function


答: Your density is even, symmetric around zero and unimodal thus Mean, Mode and Median are all zero. No further calculation are needed</p>

Title: Statistics probability density function

Body:
I was given the probability density function

$$f(x) = \begin{cases} 0.75(1 - x^2) & \text{if} -1 < x < 1 \\ 0 & \text{otherwise} \end{cases},$$

and I want to calculate the mean. I know that the definition of mean for a continuous random variable is the integral from $-\infty$ to $\infty$ of $xf(x) dx$.

My question was, since my function does not go to infinity, it only goes from -1 to 1 and is 0 elsewhere, how do I calculate that integral for the mean? Is it zero?


问:
Rewriting arguments using quantifiers, variables, and predicate logic
There are two statements which sounds similar to me but their answers are different according to answer sheet.

All birds can fly

Answer: ∀ x [B(x) → F(x)]

Some men are genius

Answer: ∃ x [M(x) ^ G(x)]

In the first part implication is used but in second part AND is used. There is no mention of 'if then' or 'And' in  statements so why the answers are different for similar sounding statements?

答:
In the system you're studying (first-order logic), a language has a fixed "universe" (also called a "domain of discourse"), and the quantifiers $\forall$ and $\exists$ always range over (i.e. "apply to") all the objects in that universe.

If the universe is just birds (i.e. if our language is only for talking about birds and nothing else), then "all birds can fly" can be written as just $\forall x F(x)$, since the "birds" part is built into the $\forall$. Similarly, if the universe is just men (sad!), then "some men are genius" can be written as $\exists x G(x)$. However, if we want write both statements in the same language, then the domain of discourse needs to include both men and birds, so we think of the universe as just "objects" and introduce predicates $B$ and $M$ to identify the birds and men. The quantifiers of this language range over all objects.

It's fairly clear that "some men are genius" is equivalent to "there exists an object that is a man and is genius", so we can write it as $\exists x (M(x)\land G(x))$. "All birds can fly" is trickier: we want to say something about just birds, but $\forall$ is going to give us a statement about all objects. To make this work, we need a formula inside the $\forall$ that says $F(x)$ if $x$ is a bird but says nothing extra about $x$ if $x$ is not a bird. You should convince yourself that such a formula is $\lnot B(x)\lor F(x)$, which is equivalent to $B(x)\to F(x)$ by the definitions of the logical symbols. We think of $\forall x (B(x)\to F(x))$ as saying "for every object x, if x is a bird, then x can fly", and it works because $B(x)\to F(x)$ is true when $x$ is not a bird.


问:
Symmetric Matrices Using Pythagorean Triples

Find symmetric matrices A =$\begin{pmatrix} a &b \\   c&d
\end{pmatrix}$ such that $A^{2}=I_{2}$.

Alright, so I've posed this problem earlier but my question is in regard to this problem.

I was told that $\frac{1}{t}\begin{pmatrix}\mp r & \mp s \\   \mp s & \pm r  \end{pmatrix}$ is a valid matrix $A$ as $A^{2}=I_{2}$, given the condition that $r^{2}+s^{2}=t^{2}$, that is, (r,s,t) is a Pythagorean Triple.

Does anybody know why this works?

答:
It works because $$A^2 = \frac{1}{t^2}\begin{pmatrix}r & s\\s & -r\end{pmatrix}\begin{pmatrix}r&s\\s&-r\end{pmatrix} = \frac{1}{t^2}\begin{pmatrix}r^2+s^2 & 0\\0 & r^2 + s^2\end{pmatrix}.$$
and you want the diagonals to be 1, i.e. $\frac{r^2 + s^2}{t^2} = 1$.  </p>


问:
How do I simplify $\overline{X}\overline{Y} + YZ + \overline{X}Y\overline{Z}$?
$$\overline{X}\overline{Y} + YZ + \overline{X}Y\overline{Z}$$

I'm having a lot of trouble simplifying this boolean expression.

I used commutative property and re-arranged it as my first step:
$$\overline{X}Y\overline{Z} + \overline{X}\overline{Y} + YZ$$
2nd Step: Factored out the common literal, $\overline{X}$.
$$\overline{X}(\overline{Y} + Y\overline{Z}) + YZ$$
3rd Step: ? I first used one of the identities to turn $\overline{Y} + Y$ into 1; I wasn't too sure what to do here and I believed what I did was wrong, so I then used the distributive property for the sum term in the parentheses instead. But, I don't know if this aforementioned step is correct.

答:
$\overline{X}\overline{Y}+YZ+\overline{X}Y\overline{Z}=$

$\overline{X}\overline{Y}+\color{blue}{YZ}+\overline{X}Y\overline{Z}=$

$=\overline{X}\overline{Y}+\color{blue}{YZ\left(1+\overline{X}\right)}+\overline{X}Y\overline{Z}=$

$=\overline{X}\overline{Y}+YZ+\overline{X}YZ+\overline{X}Y\overline{Z}=$

$=\overline{X}\overline{Y}+YZ+\overline{X}Y\left(Z+\overline{Z}\right)=$

$=\overline{X}\overline{Y}+YZ+\overline{X}Y=$

$=\overline{X}\left(\overline{Y}+Y\right)+YZ=$

$=\overline{X}+YZ$

More details about the steps:

$\color{blue}{YZ=YZ\cdot1=YZ\left(1+\overline{X}\right)}$

because $\;1+\overline{X}=1\;$ whatever is the value of $\;X\;,$

indeed, if $\;X=0\;,\;$ then $\;1+\overline{X}=1+1=1\;,$

whereas if $\;X=1\;,\;$ then $\;1+\overline{X}=1+0=1\;.$


问:
Natural deduction: how to prove the argument below
Sorry my title is a bit vague, but I have a specific problem I'm trying to solve.

$$P \rightarrow Q \vdash \lnot(P\land\lnot Q)$$

I need to prove this using natural deduction, I can do it using equivalent symbols in propositional logic, but I can't seem to figure out where to even begin in terms of natural deduction.

I (mostly) understand the introduction and elimination rules for connectives, but I just don't know what to do here. I know I've got to turn the premise into the conclusion step by step using introduction and elimination of these logical connectives, but how?

答:
$P\to Q\vdash\lnot(P\land\lnot Q)$

I (mostly) understand the introduction and elimination rules for connectives, but I just don't know what to do here. I know I've got to turn the premise into the conclusion step by step using introduction and elimination of these logical connectives, but how?

You seek to derive a negated statement.   Negation Introduction is clearly named as the rule to use for this, so what do you need for it?   Why, it requires deriving a contradiction under the assumption of the positive statement.

$$\begin{array}{|l}~~~~\begin{array}{|l}~~\mathrm m.~\phi\\\hline~~~~\vdots\\~~\mathrm n.~~\bot\end{array}\\~~\mathrm o.~~\lnot\phi\qquad{\lnot}\mathsf I~\mathrm {m{-}n}\end{array}$$

So lets set that up

$$\def\fitch#1#2{~~~~\begin{array}{|l}#1\\\hline#2\end{array}}\fitch{~~1.~~P\to Q}{\fitch{~~2.~~P\land\lnot Q}{~~~~\vdots\\~~?.~\bot}\\~~?.~\lnot(P\land\lnot Q)\quad{\lnot}\mathsf I~2{-}?}$$

So, how do you derive a contradiction - or rather, what two contrary statements are you able to derive in order to use Negation Elimination -?   Well, you have that assumption and the premise to work with so Conditional Elimination and Conjunction Elimination rules are indicated.$$\begin{array}{|l}~~\mathrm m.~~\phi\\~~\mathrm n.~~\lnot\phi\\~~\mathrm o.~~\bot\qquad{\lnot}\mathsf E~\mathrm{m,n}\end{array}\qquad\begin{array}{|l}~~\mathrm m.~~\phi\\~~\mathrm n.~~\phi\to\psi\\~~\mathrm o.~~\psi\qquad{\to}\mathsf E~\mathrm {m,n}\end{array}\qquad\begin{array}{|l}~~\mathrm m.~~\phi\land\psi\\~~\mathrm n.~~\phi\qquad{\land}\mathsf E~\mathrm m\\~~\mathrm o.~~\psi\qquad{\land}\mathsf E~\mathrm m\end{array}$$

So, having assumed $P\land\lnot Q$, we can use that infer $P$, and also to infer $\lnot Q$, with Conjunction Elimination.  Having derived $P$ we may use that and the premise $P\to Q$ to infer $Q$ with Conjunction Elimination.  Finally, having derived both $Q$ and $\lnot Q$, we may infer the contradiction with Negation Elimination.

$$\fitch{~~1.~~P\to Q}{\fitch{~~2.~~P\land\lnot Q}{~~3.~\lnot Q\qquad{\land}\mathsf E~2\\~~4.~P\qquad{\land}\mathsf E~2\\~~5.~~Q\qquad{\to}\mathsf E~4,1\\~~6.~\bot\qquad{\lnot}\mathsf E~5,3}\\~~7.~\lnot(P\land\lnot Q)\quad{\lnot}\mathsf I~2{-}6}$$


问:
Metric space - an example
Show that $(\ell^p, d_p)$ is not a metric space for $p<1.$ Here $\ell^p$ and $d_p$ are similarly defined as
$$\ell^p = \left\{ x: \mathbb{N} \to \mathbb{R} \Big| \sum_{i=1}^\infty |x_i|^p < \infty  \right\}, \quad d_p(x,y) = \left( \sum_{i=1}^\infty |x_i - y_i|^p\right)^{\frac{1}{p}}. $$

Assume that $p=-1$, $$d_{-1}(x,y) = \left( \sum_{i=1}^\infty |x_i - y_i|^{-1}\right)^{-1} \quad \forall x, y \in \ell^{-1}$$ and
the definiteness axiom is not met: $d_{-1}(x, y)= 0 \iff x=y$. This is because, if $d_{-1}(x,y) = 0$, then the function has a zero denominator which is undefined and $(\ell^{-1}, d_{-1})$ is not a metric space. My question is how do we prove it generally that $(\ell^{-1}, d_{-1})$ is not a metric, what I proved is just one possible case when $p=-1.$

答:
Your proof for $p = -1$ extends to all negative $p$.  $p = 0$ is also trivial.  So you need $0 < p < 1$.  You might check for violations of the triangle inequality, as that is the most important part of the definition of a metric.  Consider $3$ points $x$, $y$ and $z$ and fill in the formulas. Try small numbers so you can see what is going on.  For example, with $p = 1/2$, try $x = (4,4,0,0,0...)$, $y = (4,0,0,0,0...)$ and $z = (0,0,0...)$.
$$(\sqrt{4-4} + \sqrt{4-0} + 0...)^2 = 4=d(x,y)$$   and
$$(\sqrt{4-0} + 0 + 0 + ...)^2  = 4=d(y,z)$$
$$(\sqrt{4-0} + \sqrt{4-0} + 0 + ...)^2 = 16 = d(x,z)$$

We are supposed to have $d(x,y) + d(y,z)\geq d(x,z)$ but this is not the case.

Title: Metric space - an example


问:
How to price a contract that is denominated in another currency using the Martingale Approach to the Black and Scholes theory?
I am taking a course in asset pricing and I have the following problem at hand:

Suppose that the level of the UK FTSE100 index (in British pounds) evolves according to
$$\frac{\mathrm{d}S_t}{S_t}=\mu_S\mathrm{d}t+\sigma_S\mathrm{d}Z_{S,t}.$$  Assume that the dividend yield on the index is constant and equal to $\delta$. Suppose that the dollar value of a single British pound equals $X_t$, where
$$\frac{\mathrm{d}X_t}{X_t}=\mu_X\mathrm{d}t+\sigma_X\mathrm{d}Z_{X,t},\quad\mathrm{d}Z_{X,t}\mathrm{d}Z_{S,t}=\rho\mathrm{d}t.$$
Suppose that the interest rate on dollar-denominate bonds is constant at $r^{US}$, while the interest rate on the pound-denominated bonds is constant at $r^{UK}$.

Consider a derivative security that pays $S_T$ dollars at time $T$. Derive the market value of this security at time $0$.

In this course, we are emphasizing the ''Martingale Approach''. I usually look for a change of measure that will make the risky asset have the drift equal to the risk free rate and then I try to derive the dynamics of $S_T$ under this new measure. After I have done that, it's all a matter of taking expectations of known functions of something we the distribution of.

But here there are two differences that made me get lost. First, there are dividends. My intuition is that the dynamics for $S_t$ should take into account that it pays dividends. Why isn't the drift $\mu_S - \delta$?

Moreover, I understand that I should be studying the dynamics of $M_t = X_t S_t$, which is the value of the index in dollars. I derived the following dynamics:

$$\frac{dM_t}{M_t} = (\mu_S + \mu_X + \rho \sigma_X \sigma_Z)dt + \sigma_S dZ_{S, t} + \sigma_X dZ_{X, t}$$

But I am stuck here. Should I look for the measure $Q$ that will make $M_t/B_t$ a martingale? And how do I deal with dividends? Any solutions/hints are deeply appreciated! Thanks in advance.

答:
I believe I have figured out a solution myself. Here it goes:

Start noticing that we are in a complete markets world. We have two risky assets and two Brownian motions. Let $Z_t = (Z^S_{t}, Z^X_{t})'$ and $dB = d(B^S_t, B^X_t)' = dZ_t + (\eta_t, \gamma_t)'dt$ be the stacked Brownian motions under the physical measure and under the risk-neutral measure. We will characterize the processes $(\eta_t, \gamma_t)$.

What is really happening is that from the point of view of the US investor he has two risky at assets at his disposal: the British bond $U_t$ converted to American dollars $X_tU_t$ and the British index also converted in American dollars $M_t = X_tS_t$. First, I study the gain process of the British bond and make sure that it will be a martingale. Since it pays no coupons, the discounted gain process has evolution given by:
$$d\left(\frac{X_tU_t}{B_t}\right) = \frac{X_tU_t}{B_t}\left[(\mu_X + r^{UK} - r^{US} - \sigma_X \gamma_t)dt + \sigma_X dB^X_t \right]$$
Under the risk-neutral measure, if we let $\gamma_t = \frac{\mu_X + r^{UK} - r^{US} - \sigma_X}{\sigma_X}$, this process will be a martingale.

Now, we study the discounted gain process for the stock after making the exchange rate conversion. Taking into account the dividend yield, we can write that
$$d\left(\frac{M_t}{B_t}\right) = M_t\left[(\mu_X + \mu_S + \sigma_X\sigma_S \rho + \delta - r^{US} - \sigma_X\gamma_t - \sigma_S \eta_t)dt + (\sigma_S, \sigma_X)dB \right]$$
where the last term entails an inner product. Using our result for $\gamma_t$, if we let
$$\eta_t = \frac{\mu_S + \sigma_X\sigma_S \rho + \delta - r^{UK}}{\sigma_S}$$
we have that $M_t/B_t$ is also a martingale under the risk-neutral measure. Further, we notice that we can drop the subscript $t$ since these are constants and, as such, will trivially satisfy Novikov's condition.

Having fully characterized the risk-neutral measure, it's easy to see that
$$d\left(\frac{S_t}{B_t}\right) = \frac{S_t}{B_t}(r^{UK} - r^{US} - \delta - \sigma_X \sigma_S\rho)dt + \frac{S_t}{B_t}\sigma_s dB^S_t$$

Then, using the fact that Ito's integral is a martingale, the value $V_t$ of this derivative is given by:
$$V_t = S_t e^{(r^{UK} - r^{US} - \delta - \sigma_X \sigma_S\rho)(T-t)}$$


问:
Peak / local maxima notation
I was wondering if there exists a common notation for local maxima / peaks of some function $f: \mathbb{R}\rightarrow\mathbb{R}$. More specifically, I am looking for something similar to $$x'=\operatorname{arg\,\max_x}f(x)$$
but instead of one global maxima, I am interest in the set of all local maxima. I could imaging a notation similar to
$$\{x'\}=\operatorname{arg\, local\max_x}f(x)$$
(I came up with that myself. Not very happy with it.)

Using this notation (stolen from Wikipedia)
$${\displaystyle {\underset {x\in S}{\operatorname {arg\,max} }}\,f(x):=\{x\mid x\in S\wedge \forall y\in S:f(y)\leq f(x)\}.}$$
one could maybe construct something using this local maxima definition, also from Wikipedia.

However, this gets convoluted quite quickly. I am looking for something simple and clean. It doesn't have to be super exact, it just needs to convey the idea.

Thanks

答:
Something like this?

arg local max $f(x)$ = $\{x | \exists r>0$ s.t. $\forall y \in (x-r,x+r), f(y) \leq f(x)  \}$

Title: Peak / local maxima notation


问:
Probability of no matching pairs of shoes
There are 3 different pairs ( i.e. 6 units say aa, bb, cc) of shoes in a lot. Now three person come and pick the shoes randomly (each get 2 units). Let p be the probability that no one is able to wear shoes (i.e. no one gets a correct pair) , then the value of $\frac{13p}{4-p}$ is?

答:
The probability for the first person to pick a bad pair is $\frac 45$ (only one out of five shoes is a good second pick).
After this there is one valid pair left. The probability that the person to pick the left shoe of this pair also picks the right shoe, is $\frac 13$. Hence $p=\frac45\cdot\frac23=\frac{8}{15}$. I hope you can figure out the rest.</p>


问:
Calc 2 visualizing equations in 3D
Is there a way to create and visualize a 3D object given an equation? I am currently studying calc 2 and would really like to be able to interact or view the equations I am given. For example, I have this problem:

I would like to be able to rotate it around whatever axis I want in order to create the 3D shape. Is this possible? I doesn't have to tell me the area, I just want a 3d shape.

Thanks!

答:
GeoGebra is a good place to check visualization.

In particular, you can try this: https://www.geogebra.org/m/zBRtUVfR

Title: Calc 2 visualizing equations in 3D


问:
What is the integral $\;\displaystyle\int\left[x^2(x-2)\right]^{-2/3}\text{d}x\;$?
What is the integral $\;\displaystyle\int\left[x^2(x-2)\right]^{-2/3}\text{d}x\;$?

I have tried solving this integral by using U-substitution and integration by parts but got nowhere. Don't really know what else to try. WolframAlpha gives an answer to the integral but is unable to show steps for the solution. Any help would be appreciated!

答:
We have
$$ I = \int x^{-\frac43}(x-2)^{-\frac23}dx = \int \frac{1}{x(x-2)}\sqrt[3]{\frac{x-2}{x}} d x$$
Integrals of the form $$\int Q(x) \sqrt[n]{\frac{ax+b}{cx+d}}dx $$ where $Q$ is a rational function can be solved using the substitution
$$ t = \sqrt[n]{\frac{ax+b}{cx+d}} $$
In this case, we have
$$ t = \sqrt[3]{\frac{x-2}{x}}$$
that is
$$ x= \frac{2}{1-t^3}$$
It will transform the integral into an integral of a rational function.


问:
Variance for grouped bivariate data.
If for example I have two random variables X and Y. X stands for height and Y for weight. I compute the conditional variance for Y at each X. ( Both are discrete). Then can I compute the total variance by summing each weighted conditional variance where weights would be number of cases the measurements have been taken and then dividing the total amount of measurements taken?</p>

答: Use this. Your calculation would only be the first term.</p>

Title: Variance for grouped bivariate data.


问:
Finding the center and max side length of a square inscribed between two concentric circles

I am attempting to find the center and max side length of an unrotated square inscribed between two concentric circles such that the square stays between the two circles at any arbitrary rotation around the circle. So far, I have gathered:

Given $d$ is the rotation around the circles in radians, $r_1$ is the radius of the inner circle, and $r_2$ is the radius of the outer circle,

$Length(d) = \frac{r_2 - r_1}{\sqrt{2}} + f(d)$

$Center(d) = (\cos(d) * \frac{r_2 + r_1}{2} + g_x(d) * \cos(d), \sin(d) * \frac{r_2 + r_1}{2} + g_y(d) * \sin(d))$

where $Length(d)$ is the side length of the square and $Center(d)$ is the center of the square.

Behaviour:

The min size of the square is $\frac{r_2 - r_1}{\sqrt{2}}$ at $\frac{n\pi}{4}$ where $n\mod 2 = 1$, and the max size is $\frac{2\sqrt{5r_2^2 - r_1^2} - 2r_1}{5}$ at $\frac{n\pi}{2}$ where n is in the set of all integers.

At $\frac{n\pi}{4}$ where $n\mod 2 = 1$; $f(d) = 0$, $g_x(d) = 0$, $g_y(d) = 0$

At $n\pi$ where $n$ is in the set of all integers; $f(d) > 0$, $g_x(d) = 0$, $g_y(d) \ne 0$

At $\frac{n\pi}{2}$ where $n\mod 2 = 1$; $f(d) > 0$, $g_x(d) \ne 0$, $gy(d) = 0$

I'm guessing $f(d)$ and $g(d)$ must be trig expressions but I have no idea how to derive them.

答: If the square does not rotate as it moves around the circle, you can imagine a reference frame in which it is rotating, as you sweep around the circles.  So no matter how the square rotates, it must not go past the circles.  If you trace the paths of the corners of the squares, you get another circle, of diameter r2-r1.  The square must remain in that circle, so its longest "diameter" is the diagonal.  Therefore $(r2-r1) / \sqrt(2)$ is the edge length.


问:
Irreducible polynomial with integer coefficients of degree $> 1$ has no rational roots
In the middle of a homework problem I realized I proved this lemma: an irreducible polynomial $q \in \mathbb{Z}[x]$ with $\deg q > 1$ has no rational roots. This seems... odd, since many well-known theorems on irreducibility and existence of roots for polynomials with integer coefficients (like the rational root theorem and Eisenstein's criterion) do not go this far. What went wrong in this proof? Or otherwise, am I missing a super easy corollary of the definition of an irreducible polynomial that makes the lemma trivial? Any hints are appreciated.

Here is the proof:

Assume $\beta$ is a root of $q$. Since $\mathbb{Z}[x]$ is a PID we have that $I = (q)$ is a maximal ideal of $\mathbb{Z}[x]$ and so $K = \mathbb{Z}[x]/I$ is a field. Define $f: K \to \mathbb{Q}$ by $f(r + I) = r(\beta)$; this map is well defined because if $r + I = s + I$ then $r - s \in I = (p)$, so $r(x) - s(x) = t(x) p(x)$ for some $t \in \mathbb{Z}[x]$ and thus $r(\beta) = s(\beta) + t(\beta) p(\beta) = s(\beta)$. This map is also a field homomorphism since $(r + s)(\beta) = r(\beta) + s(\beta)$ and $(rs)(\beta) = r(\beta) s(\beta)$. Being a homomorphism between fields, $f$ is injective. If $r(\beta) = 0$ then $r + I \in \ker f = \{I\}$ and so $r \in I = (q)$; this means, in particular, that $\deg r \geq \deg q > 1$ (I'm adopting the convention that $\deg 0 = \infty$), so any polynomial in $\mathbb{Z}[x]$ with $\beta$ as a root has degree greater than $1$. But if $\beta = a/b \in \mathbb{Q}$ then the polynomial $r(x) = bx - a \in \mathbb{Z}[x]$ has $r(\beta) = 0$ and $\deg r = 1$, a contradiction. Hence $\beta \notin \mathbb{Q}$ and so $q$ has no rational roots.

答:
“Since $\mathbb{Z}[x]$ is a PID...”

Ehr... no, it isn’t. $(2,x)$ is not principal.

It’s also not true that $\mathbb{Z}[x]/(q)$ is a field when $q$ is irreducible: $\mathbb{Z}[x]/(x)\cong\mathbb{Z}$ is not a field (this because $\mathbb{Z}[x]$ is not a PID, so $(q)$ need not be maximal). I don’t think your proof can be salvaged at this point.

Note that maximal ideals of $\mathbb{Z}[x]$
have the form $(q(x),p)$, where $q$ is an irreducible polynomial of degree greater than or equal to $1$, and $p$ is a rational prime. The quotient will be a field of positive characteristic.

Your result is true: it follows from Gauss’s Lemma: Gauss’s Lemma says that if $p(x)\in\mathbb{Z}[x]$ is such that the gcd of its coefficients is $1$, and $p(x) = r(x)s(x)$ with $r(x),s(x)\in\mathbb{Q}[x]$, then there exist polynomials $R(x),S(x)\in\mathbb{Z}[x]$, $\deg(R)=\deg(r)$, $\deg(S)=\deg(s)$, such that $p(x) = R(x)S(x)$.

In particular, if a polynomial $p(x)$ is irreducible in $\mathbb{Z}[x]$, then it remains irreducible in $\mathbb{Q}[x]$. In particular, it cannot have rational roots if $\deg(p)\gt 1$.

Also, note that

The Rational Root Theorem is not about irreducible polynomials; it has nothing to say about it, and it just gives you a finite list of possible roots. I don’t think it is related to the result.

Eisenstein’s Criterion allows you to deduce irreducibility in $\mathbb{Q}[x]$: as such, it necessarily implies no roots when the polynomial has degree greater than $1$. So I do not understand why you say its conclusion “does not go as far” as the result at hand. Having no roots in $\mathbb{Q}[x]$ is weaker than being irreducible.


问:
why is -1 mod 3 = 2?
Title says it all. Google doesn't have anything. Trying to get it to let me submit by typing more words. There is nothing really to be added but it wont let me submit without this.</p>

答:
If you ignore the meaning of $ -1 $ and consider it an arbitrary symbol, what properties does it have? By definition, it is the value such that $(-1) + 1 = 0$.

But, working mod 3, $2+1= 3 \equiv 0.$ But from earlier, we have $0 = (-1) + 1$. If you combine the two into $2+1 \equiv (-1)+1 \mod 3$ and drop the $+1$ from both sides, you get $2 \equiv -1 \mod 3$.

Title: why is -1 mod 3 = 2?


问:
Does a $\mathrm{C}^*$-algebra generated by projections contain support projections
I think we have for a (normal?) state $\varphi$ on a von Neumann algebra $\mathcal{M}$ a projection $p_\varphi$ with some nice properties called its support. It arises as follows:

Define the following null space:

$$N_\varphi=\left\{g\in \mathcal{M}\,|\,\varphi(|g|^2)=0\right\}.$$

This set is a $\sigma$-weakly closed left ideal. Therefore there exists a projection $q_\varphi$ such that $N_\varphi=\mathcal{M}q_\varphi$. Some properties include the fact that $g\in N_\varphi$ if and only if $gq_\varphi=g$. Also, for all $f\in \mathcal{M}$ we have
$$\varphi(q_\varphi)=\varphi(fq_\varphi)=\nu(q_\varphi f)=0.$$

Also if we define the projection $p_\varphi:=1_{\mathcal{M}}-q_\varphi$. We have
$$\varphi(f)=\varphi(fp_\varphi)=\varphi(p_\varphi f)=\varphi(p_\varphi fp_\varphi),$$
and $\varphi(p_\varphi)=1$.

I understand that a $\mathrm{C}^*$-algebra generated by projections is not necessarily a von Neumann algebra... but

Question: Does a $\mathrm{C}^*$-algebra generated by projections have support projections?

答:
Here is a simple example. Let $A=K(H)$, the compact operators on a separable Hilbert space. It is generated by its projections, which are the finite-rank projections in $B(H)$.

Fix an orthonormal basis $\{e_n\}$, and let $\phi:A\to\mathbb C$ be the state
$$
\phi(x)=\sum_n\frac1{2^n}\,\langle xe_{2n},e_{2n}\rangle.
$$
This state is "normal", since it is the restriction of a normal state in $B(H)$ (the same definition above works for any $x\in B(H)$). The support projection of $\phi$ is
$$
p=\sum_n \langle \cdot,e_{2n}\rangle\,e_{2n},
$$
which is not in $A$.


问:
Why do certain relatively free groups have perfect commutator subgroups?
I am trying to understand an argument in a group theory, which is not my strong suit (I mostly work with semigroups and rarely delve into the actual structure of groups).

Let $\boldsymbol{\mathcal{V}}$ be a non-Abelian variety of groups such that every solvable group in $\boldsymbol{\mathcal{V}}$ is Abelian. Apparently, the relatively free group on two generators, $F_2$, has a perfect commutator subgroup; that is $F_2^{(1)} = F_2^{(2)}$. This is presented as an obvious fact, but I have no idea of how to prove it. I would like to know why it is true. For my purposes, I can assume that the variety has bounded exponent, but this doesn't seem to be needed.

Thanks!

答:
In fact, this holds for every group $G\in\mathcal{V}$.

Let $G\in\mathcal{V}$, and let $K=G/G’’$. This is solvable, since $K’’=G’’/G’’$ is trivial. By assumption, this means that $K$ is abelian, and hence that $K’=\{e\}$. But $K’=G’/G’’$, hence $G’=G’’$.

(The fact that $K’’=G’’/G’’$ and $K’=G’/G’’$ follows because for any variety $\mathfrak{W}$ and any surjective group homomorphism $H\to M$, we have that $\mathfrak{W}(H)$ maps onto $\mathfrak{W}(M)$. Thus, $G’$ maps onto $K’$, and $G’’$ onto $K’’$; using the variety of abelian groups, and of solvable groups of length at most $2$.)


问:
Prove $n = (1 + a_n)^n ≥ n(n − 1)a_n^2 /2]$
Show that $n^{1/n} → 1$. [Hint: let $a_n := n^{1/n} − 1$. Then $n = (1 + a_n)^n ≥ n(n − 1)a_n^2 /2]$.

I know how to prove this in using the hint and the squeeze theorem, but how is it possible to prove $n = (1 + a_n)^n ≥ n(n − 1)a_n^2 /2]$? I thought using a taylor series, but I can't use it.

答:
By Binomial Theorem:

$$(1+a_n)^n = \sum_{k=0}^n \binom nk a_n^k \ge \binom n 2 a_n^2 = n(n-1)a_n^2/2$$


问:
Proving that $P\left(\bigcup_{j=1}^n E_j^{(n)}\right)\sim\sum_{j=1}^n P(E_j^{(n)})$ for independent events $E_j^{(n)}$

For arbitrary events $\{E_j, 1\le j\le n\}$, we have

$$P\left(\bigcup_{j=1}^n E_j\right)\ge\sum_{j=1}^n P(E_j) - \sum_{1\le j < k \le n} P(E_jE_k)$$

If $\forall n: \{E_j^{(n)}, 1\le j\le n \}$ are independent events, and

$$P\left(\bigcup_{j=1}^n E_j^{(n)} \right)\to 0 \quad \text{as} \quad n\to\infty$$

then $P\left(\bigcup_{j=1}^n E_j^{(n)} \right) \sim \sum_{j=1}^n P(E_j^{(n)})$.

My attempt at showing this last claim:

I am going to guess that the ~ symbol in the problem means that

$$\lim_{n\to\infty}\frac{P\left(\bigcup_{j=1}^n E_j^{(n)} \right)}{\sum_{j=1}^n P(E_j^{(n)})}=1$$

From

$$\sum_{j=1}^n P(E_j^{(n)})\ge P\left(\bigcup_{j=1}^n E_j^{(n)}\right)\ge\sum_{j=1}^n P(E_j^{(n)}) - \sum_{1\le j < k \le n} P(E_j^{(n)}E_k^{(n)})$$

and dividing we have

$$1\ge \frac{P\left(\bigcup_{j=1}^n E_j^{(n)}\right)}{\sum_{j=1}^n P(E_j^{(n)})} \ge 1 - \frac{\sum_{1\le j < k \le n} P(E_j^{(n)}E_k^{(n)})}{\sum_{j=1}^n P(E_j^{(n)})}$$

so then the problem reduces to showing

$$\lim_{n\to\infty}\frac{\sum_{1\le j < k \le n} P(E_j^{(n)}E_k^{(n)})}{\sum_{j=1}^n P(E_j^{(n)})} = 0$$

But how to make progress after that?

答:
A quantitative version of the result, proved by a recursion over the number of events involved: for every independent events $E_i$,
$$
1-P\left(E\right)\leqslant\frac{P\left(E\right)}{\sum\limits_iP(E_i)}\leqslant1,\qquad E=\bigcup\limits_iE_i.
$$
Thus, for independent events $E_i^{(n)}$, the events $E^{(n)}=\bigcup\limits_iE_i^{(n)}$ are such that
$$
P\left(E^{(n)}\right)\to0\implies P\left(E^{(n)}\right)\sim\sum\limits_iP(E_i^{(n)}).
$$</p>


答:
Integral-calculator.com gives "steps". I copy below, but not sure if they are "useful"-

Problem:
$$
\int \frac{1}{\left((x-2) x^{2}\right)^{\frac{2}{3}}} d x
$$
Rewrite/simplify:
$$
=\int \frac{1}{(x-2)^{\frac{2}{3}} x^{\frac{4}{3}}} \mathrm{~d} x
$$ Substitute $u=\sqrt[3]{x} \longrightarrow \frac{\mathrm{d} u}{\mathrm{~d} x}=\frac{1}{3 x^{\frac{2}{3}}} \longrightarrow \mathrm{d} x=3 x^{\frac{2}{3}} \mathrm{~d}u$:
$$\tiny x=u^{3} \\
\tiny \frac1{x^{2/3}} =\frac1{u^2}\\
=3 \int \frac{1}{u^{2}\left(u^{3}-2\right)^{\frac{2}{3}}} \mathrm{~d} u
$$
Substitute $v=\frac{\sqrt[3]{u^{3}-2}}{u} \rightarrow \frac{\mathrm{d} v}{\mathrm{~d} u}=\frac{u}{\left(u^{3}-2\right)^{\frac{2}{3}}}-\frac{\sqrt[3]{u^{3}-2}}{u^{2}} \longrightarrow \mathrm{d} u=\frac{1}{\frac{u}{\left(u^{3}-2\right)^{\frac{2}{3}}}-\frac{\sqrt[3]{u^{3}-2}}{u^{2}}} \cdot \mathrm{d}  v$
$$=\int \frac{1}{2} \mathrm{~d} v$$
(from here it is obvious so I stop)

Body:
What is the integral $\;\displaystyle\int\left[x^2(x-2)\right]^{-2/3}\text{d}x\;$?

I have tried solving this integral by using U-substitution and integration by parts but got nowhere. Don't really know what else to try. WolframAlpha gives an answer to the integral but is unable to show steps for the solution. Any help would be appreciated!


问:
Open in subspace topologies implies open
Let $X$ be a topological space and let $X_1, X_2 \subseteq X$ two subsets with
$X = X_1 \cup X_2$. We view $X_1$ and $X_2$ as topological spaces equipped with the subspace topology inherited from $X$. Let $U \subseteq X_1 \cap X_2$ be a subset with the property that $U$ is open as a subset of $X_1$ and as a subset of $X_2$.

Since $U$ is open w.r.t. $X_1$ we can find an open $V_1 \subseteq X$ such that $U = V_1 \cap X_1$. Analogously, since $U$ is open w.r.t. $X_2$ we can find an open $V_2 \subseteq X$ such that $U = V_2 \cap X_2$.

I have to show that $U$ is open w.r.t $X$, but I do not really see how to continue. I thought about considering $U  = (V_1 \cap X_1) \cup (V_2 \cap X_2)$ and somehow find a way to write this as $A \cap (X_1 \cup X_2)$ for some $A$ expressed in $V_1$ and $V_2$, but it didn't really work out well. Could someone help me with this?

答:
Let $V=V_1\cap V_2$. Then

$$\begin{align*}
V\cap X_1&=(V_1\cap V_2)\cap X_1\\
&=V_2\cap(V_1\cap X_1)\\
&=V_2\cap U\,.
\end{align*}$$

$U=V_2\cap X_2$, so $U\subseteq V_2$, and therefore

$$V\cap X_1=V_2\cap U=U\,.$$

A similar argument shows that $V\cap X_2=U$. Can you finish it from here?


问:
Strang's proof of SVD and intuition behind matrices $U$ and $V$
In lecture 29 of MIT 18.06, Professor Gilbert Strang "proves" the singular value decomposition (SVD) by assuming that we can write $A = U\Sigma V^T$ and then deriving what $U$, $\Sigma$, and $V$ must be based on the eigendecomposition of
$$ AA^T = U\Sigma ^2 U^T$$
and
$$ A^TA = V\Sigma ^2 V^T$$

My intuition tells me there's something wrong with first assuming that we can write $A$ in this form. As in, we are finding $U$, $\Sigma$ and $V$ for those matrices that have this form, but what if some matrices couldn't be written in this form in the first place?

Also, is there some intuition to be found about $U$ and $V$ by only thinking about them as eigenvectors of $AA^T$ and $A^TA$? In the sense that we should be able to know why $V$ will have its properties (be mapped orthogonally by $A$) by just looking at it as an eigenvector of $A^TA$ (which somehow implicitly "encodes" the basis $U$). Not sure if this intuition makes sense either.

答:
I guess, intuitively, the existence of SVD can be proofed by contradiction. Assume that SVD doesn't exists for some matrix $$X \neq USV^T$$ for $U$ and $V$ being orthogonal and $D$ is diagonal. Using the fact that $XX^T$ forms a symmetric matrix $A$; and the eigendecomposition $A = WDW^T$ exists; then
$$A = WDW^T = WD^{1/2}I (WD^{1/2}I)^T $$
and that implies there exists such matrix $X = WD^{1/2}I$. Since the 3-tuple $(W,D^{1/2},I)$ satisfies the conditions of the SVD 3-tuple $(U,S, V)$, this leads to a contradiction on $X \neq USV^T$.

Am I correct...?


问:
Using theorem of logical equivalences to show $p \land (\sim q \lor p) \equiv p$
I'm new to the whole discrete math thing, and I'm having trouble finding any laws to start breaking the statement $p \land (\sim q \lor p) \equiv p$ down into its equivalences laws.

Can I have some help please. here's the chart I'm using

heres an example of the format I need to answer in.

statements       | reason

~[~p^(pvq)]vq=t  | given

p^(pvq)vq=t      | distributive law

(p^p)v(p^q)vq=t  | distributive law

pv(p^q)vq        | Idempotent law

pvq=t            | Absorption law

答: $\begin{aligned}p\land(\neg q\lor p)&\equiv p&\text{given}\\(p\land\neg q)\lor(p\land p)&\equiv p&\text{distributive law}\\(p\land\neg q)\lor p&\equiv p&\text{idempotent law}\\(p\land\neg q)\lor(p\land T)&\equiv p&\text{identity law}\\p\land(\neg q\lor T)&\equiv p&\text{distributive law}\\p\land T& \equiv p&\text{domination law}\\p&\equiv p&\text{identity law}\end{aligned}$


问:
Does this change make a nonlinear differential equation into linear differential equation?

is a non linear differential equation.

But if this read

then it would be linear.

Is this correct?

答: That's right. Moreover, you can solve the second one just by integrating it twice.</p>


问:
How do you prove that $m\in\Bbb Z$ implies that $m^2$ is even with ZFC?
On this page is a very simple theorem (Theorem 1.1)
$$\text{If }\;m\in\mathbb{Z}\;\text{ is even, then }\;m^2\text{ is even}$$

How would you prove this theorem using ZFC set theory? What would it look like? How is the theorem represented using sets?

Or does ZFC not apply here?

I am trying to learn how to write proofs and I've read that ZFC is the modern foundation for math now-a-days. But I am having trouble understanding how to use it.

Thanks!

答:
Asking how to prove a theorem about $m^2$ in $\Bbb Z$ using $\sf ZFC$ is similar to asking how to write a JavaScript interpreter from complete scratch, in machine code. It's not that it's not possible, but it's just that this is not what you're supposed to do if you want to write a JavaScript interpreter. For one, it's a lot easier to use an operating system that someone else already wrote, and maybe a different programming language (although you can definitely still insist on writing machine code instructions by hand if you really want to).

When we say that $\sf ZFC$ is a foundation of mathematics, we don't mean that we want to, or even should be able to prove every theorem of mathematics in the language of set theory. For one, much like machine code, the language of set theory is very minimal and is untyped. In particular, the symbol $\Bbb Z$, the concept of "even integer", and the notion of multiplication are not part of this language.

So what do we mean, and how do we do it? Well, we show that in $\sf ZFC$ we can interpret the integers $\Bbb Z$, as some set, and indeed there are many different ways to arrive to such a set. We should prove that any two interpretations of $\Bbb Z$ are isomorphic, and that if we treat $\Bbb Z$ as a ring (i.e. with $0,1,+$ and $\cdot$, and maybe even $\leq$) then this isomorphism is unique. This means that now it doesn't matter which copy of $\Bbb Z$ we used for our proof, we can translate it to any other interpretation.

Finally, we show that all the rules of logic that we are used to naively, or at least "pre-set theory" and are not obviously contradictory can be interpreted in $\sf ZFC$. This allows us to take any proof that we write in the usual language of rings, and turn it into a proof from $\sf ZFC$.

So how do you prove that? Well, you first prove all of those other things. You build an operating system, a compiler for higher-level language, and with that another even higher-level language, and finally, we write an interpreter, and we compile it. And then we have an interpreter. We didn't write the machine code from scratch, we didn't even write the machine code inside the operating system. But we still ended up with an interpreter.

Similarly, you can write a proof that shows pretty much all the things that I wrote, but then, at the very end, your proof will just be a lot of machinery, and the same proof that you know from every other freshman course, or whatever. Because $2$ is a prime number, and prime numbers satisfy $p\mid nm\iff p\mid n$ or $p\mid m$:

$$2\mid m^2\iff 2\mid m\text{ or }2\mid m\iff 2\mid m.$$

(Or some ad-hoc proof of this fact.)


问:
Find the Maximum Trigonometric polynomial coefficient $A_{k}$

Let $n,k$ be given positive integers and $n\ge k$. Let $A_i, i=1, 2, \cdots, n$ be given real numbers. If for all real numbers $x$ we have $$A_{1}\cos{x}+A_{2}\cos{(2x)}+\cdots+A_{n}\cos{(nx)}\le 1$$
Find the maximum value of $A_{k}$.

I don't know if this question has been studied

If  $n=2$ it is easy to solve it.

答:
Since $\cos{kx} =\frac { e^{ikx} + e^{-ikx}}{2}$ we can restate the condition in this form
let $$P(x) = 1 - \sum_{k=-n}^{k=n} \frac{A_{k}}{2}*e^{ikx}$$ where $A_{-k} = A_{k}$.
If $P(x) >= 0$ find the upper bound on $A_{k}$.
Now, there is a lemma due to Riesz which says that a real positive polynomial is a square. In words, that there is a trigonometric polynomial $Q(x) = \sum_{k=-n}^{k=n}\frac{1}{2\pi}a_{k}e^{ikx}$ such that $P(x) = |Q(x)|^{2} = Q(x)\overline Q(x)$
Multiplying out and comparing coefficients, one gets
$$\sum |a_{k}|^{2} = (2\pi)^{2}$$ and $$\frac{1}{2}A_{k} = (\frac{1}{2\pi})^{2}\sum a_{l}*\overline a_{l + k}$$
Then, if one applies Cauchy-Schwartz inequality to the last expression and uses the previous estimate on sum of squares of $a_{k}$'s one gets:
$$A_{k} <= 2$$
Not the optimal estimation (as the case n=1 shows) but a uniform one with respect to $n$.


问:
Unable to understand a part of Harris' proof of Chavalley's Theorem: the image of a constructible set contains a non-empty open set
I am reading Harris' Algebraic Geometry, A First Course and could not understand a part of Theorem 3.16.

The setting is this:
$X\subset\mathbb{P}^m$ is a quasi-projective variety, $f:X\to\mathbb{P}^n$ is a regular map,
and $U\subset X$ is a constructible set.

The part of the proof needs to establish the claim $f(U)$ contains a nonempty open subset $V\subset\overline{f(U)}$ of the closure of $f(U)$.

I am copying the part I don't understand from Harris:

To begin with, we may replace $U$ by an open subset, and so may assume that it is affine; restricting to a smaller affine
open, we may assume the target space is also affine space. After replacing $U$ by the
graph of $f$ we may realize the map $f$ as the restriction to a closed subset $U\subset\mathbb{A}^n$ of
a linear projection $\mathbb{A}^n\to\mathbb{A}^m$, so that it is enough to prove the claim for a locally
closed subset $U\subset\mathbb{A}^n$ under the projection

$$\pi:\mathbb{A}^n\to \mathbb{A}^{n-1}$$

$$(z_i,...,z_n)\mapsto(z_1,...,z_{n-1})$$

Finally, we can replace $\mathbb{A}^{n-1}$ by the Zariski closure Y = $\overline{\pi(U)}$ of the image of $U$
and $\mathbb{A}^{n}$ by the inverse image $\pi^{-1}(Y) = Y \times \mathbb{A}^{1}$. It will thus suffice to establish the
claim for a locally closed subset $U$ of a product $Y\times\mathbb{A}^{1}$ (or, equivalently, a locally
closed subset $U\subset Y\times\mathbb{P}^{1}$) and the projection map $\pi: Y \times\mathbb{P}^{1}\to Y $on the first
factor, with the further assumption that $\pi(U)$ is dense in $Y$.

Basically I have trouble understanding what exactly he is trying to say in "we may realize the map $f$ as the restriction to a closed subset $U\subset\mathbb{A}^n$ of
a linear projection $\mathbb{A}^n\to\mathbb{A}^m$", I have no idea what he is trying to do, particularly, is the new closed subset $U$ related to the old constructible $U$ in some way? What is this linear projection from $\mathbb{A}^n$ to $\mathbb{A}^m$, explicitly?

I am a beginner in algebraic geometry and could speak the language of scheme yet. Any help is really appreciated.

答:
At that stage, you have a regular map from an affine to another affine $f:U\subset \mathbb{A}^n\to f(U)\subset \mathbb{A}^m$. The graph of $f$ is defined to be the closed subset
$$G:=\{(x,f(x))\mid x\in U\}\subset \mathbb{A}^n\times\mathbb{A}^m=\mathbb{A}^{n+m}.$$
It is also canonically isomorphic to $U$ by projecting to the first factor $(x,f(x))\mapsto x$. Under this isomorphism, the map $f$ becomes the projection to the second factor, i.e. it now maps a closed subset $G\subset \mathbb{A}^{n+m}$ under this linear projection $\mathbb{A}^{n+m}\to\mathbb{A}^m$.


问:
Is the following set empty?
$$
sp\left \{
\begin{pmatrix}
1 \\
-1 \\
1 \\
-1
\end{pmatrix} , \begin{pmatrix}
4\\
-2 \\
4 \\
-2
\end{pmatrix} , \begin{pmatrix}
1\\
1\\
1\\
1
\end{pmatrix}
 \right \} \bigcap \left \{ \begin{pmatrix}
x_1\\
x_2\\
x_3\\
x_4
\end{pmatrix} | \begin{matrix}
x_1 + x_2 = 0\\
x_3 = 2x_4
\end{matrix} \right \}$$

I was asked to find the basis of the following set but the vectors of the LHS set never fit to both requirements on the RHS set , or maybe I'm totally wrong and the binary relation between the requirements is OR?

答: I would interpret it as you did. It’s not empty, though: it’s the trivial subspace of $\Bbb R^4$ containing only the zero vector. (And while these objects are groups under vector addition, the fact that you’re talking about the span in the first term makes it clear that they are to be understood as vector spaces, not as groups.)</p>

Title: Is the following set empty?


答:
Building $\Bbb Z$ and establishing basic properties of arithmetic starting with just the axioms of ZFC is a complex task (e.g. one rigorous formal proof that $2+2=4$ is almost 3,000 steps long)!

This process includes introducing notation and terminology for numbers and numeric operations (defined in terms of set operations) and proving useful theorems (e.g. commutativity of addition) using those new symbols. This effectively "hides" the fact that we're "technically" still talking about sets, so ultimately we get to write proofs in a natural, intuitive style like the one you referenced. By unpacking many layers of definitions, we could "trace it all back" to sets and the ZFC axioms, and knowing this reassures us that our arguments are valid as long as ZFC is consistent, but it's rarely useful to actually do the translation--the resulting statements in the language of set theory would be huge, unrecognizable symbolic expressions.

Title: How do you prove that $m\in\Bbb Z$ implies that $m^2$ is even with ZFC?

Body:
On this page is a very simple theorem (Theorem 1.1)
$$\text{If }\;m\in\mathbb{Z}\;\text{ is even, then }\;m^2\text{ is even}$$

How would you prove this theorem using ZFC set theory? What would it look like? How is the theorem represented using sets?

Or does ZFC not apply here?

I am trying to learn how to write proofs and I've read that ZFC is the modern foundation for math now-a-days. But I am having trouble understanding how to use it.

Thanks!


问:
Change vectors condition and still make the equation true
For a equation with random vector, this is true: $(v\times w)\times z=c_{1}v+c_{2}v+c_{3}(v\times w)$. The symbol $ \times $ is the cross product, and $v, w, z \in \mathbb{R^{3}}$, $c_{1}, c_{2}, c_{3} \in \mathbb{R}$. Then suppose that $v$ is changed to $v=0$. But every vector not $v$ is not $0$. Then how do I make sure that the equation is still true?

Similarly, if I suddenly change $v=0$ and $w=0$, $v=mw$, $m \not =0$(the vectors is a multiple,) then how do I still make the equation true?

答:
By my understanding, you are trying to prove that the following statement:

For any $  v ,w,z \in \mathbb{R}^3 $, there are real scalars $ c_1, c_2, c_3$ such that
$$ \tag{*} \label{eq} (v \times w) \times z = c_1 v + c_2 w + c_3 (v \times w).  $$

To prove this, you may separate into several cases first, as you suggested.

Case 1. $v$ and $w$ are linearly dependent.

In this case, one may write $ v = kw$ for some real scalar $k$. Notice that $ v \times w $ = 0 in this case. So the left hand side of \eqref{eq} is zero. One can then pick $c_1 = c_2 = c_3 = 0 $.

Case 2. $v$ and $w$ are linearly independent.

In this case, $v \times w \neq 0$. In addition, $ v \times w $ is orthogonal to both $ v $ and $w$. In particular, the set of vectors $ \lbrace v, w, v \times w \rbrace$ is linearly independent. Since the dimension of $\mathbb{R}^3 $ is 3, the set $ \lbrace v, w, v \times w \rbrace$ forms a basis of $\mathbb{R}^3$. It follows that $ \lbrace v, w, v \times w \rbrace $ spans $\mathbb{R}^3$. \eqref{eq} follows by the definition of a spanning set.


问:
When function A implies function B, when is it correct to say that A is the inverse of B?
I suppose my question is of more philosophical than mathematical interest. Take the natural logarithm as an example. We can say that $$y = \ln{x}$$ implies $$e^{y} = x$$ due to being inverse functions. I guess we take "implies" to mean biconditionality, which is met when we're looking at inverse functions. But are there any examples where we'd say one function implies another, but they're not inverses? I think not, but looking for confirmation.

I know this is a silly question, but I hope it can be useful or interesting to some people.

答:
If $y=|x|$ then $y^2=|x|^2$.

We can't conclude the reverse direction. It is possible that $y=-|x|$.


问:
Quine's notation in Three Grades of Modal Involvement
In his Three Grades of Modal involvement, Quine uses a notation that is not familiar for me. Does someone know how he intends to be using them?

What 'p' represents is a statement, hence true or false (and devoid of free 'x'). If 'p' is true, then the conjunction '$x=\Lambda.p$' is true of one and only one object x, viz., the empty class $\Lambda$; whereas if 'p' is false the conjunction '$x=\Lambda.p$' is true of no object x whatever. The class $\hat{x}(x=\Lambda.p$)  therefore, is the unit class $\iota\Lambda$ or $\Lambda$ itself according as 'p' is true or false. Moreover, the equation: $$\hat{x}(x=\Lambda.p)=\iota\Lambda $$
is by the above considerations, logically equivalent to 'p'.

Does someone know what he means by a conjunction of a statement and a set--'$x=\Lambda.p$'--and what '$\hat{x}$' stands for?
from Three Grades of Modal Involvement

答: $x=\Lambda.p$ means $(x=\Lambda).p$, the conjunction of two statements. For any statement $\phi(x)$ about $x$, the  notation $\hat x\,\phi(x)$ means what we'd nowadays write as $\{x:\phi(x)\}$.