问:
Using the Intermediate Value Theorem and Rolle's theorem to determine number of roots

Use the Intermediate Value Theorem and Rolle's Theorem to show the that the polynomial $$p(x) =  x^{5} + x^{3} + 7x - 2$$ has a unique real root.

Can someone please give some hints on how to do this question. Thanks in advance.

答:
The Intermediate Value Theorem establishes existence:  there is at least one real root.

Notice that $p(0) = -2 < 0$ and $p(1) = 7 > 0$.  Since $p$ is continuous, the I.V.T. guarantees a number $c$ such that $p(c) = 0$.  (In fact, we know that $0 < c < 1$.)

Rolle's Theorem establishes uniqueness:  there is at most one real root.
Why?  Suppose that there were two roots $a, b \in \mathbb{R}$.  Since $p$ is differentiable, Rolle's Theorem guarantees a number $c \in (a, b)$ where $p'(c) = 0$.  What's wrong with that?  The derivative
$$
p'(x) = 5x^4 + 3x^2 + 7 > 0
$$
for all $x \in \mathbb{R}$.  Why?  It's quadratic in $x^2$ and its discriminant $(3)^2 - 4(5)(7) < 0$.


问:
Let $R$ be a finite ring, and $S$ its subring. Then $|S|$ divides $|R|$
I'm supposed to answer if this statement is true or false. I am extremely confused... I don't even know where to start. Does anybody have any hints?</p>

答: Subrings may not form a group with respect to multiplication, but subrings ALWAYS by definition form a group with respect to addition. Hence this statement is a direct application of Lagrange's theorem. The cosets will be additive ones!</p>

ContentLicense: CC BY-SA 4.0


问:
Problem finding the limit of a random variable sequence ( attempt included )
Let $X$ be a real random variable defined in the probability space $(\Omega, \mathcal A, P)$

Suppose we have a set $B_k = \{|X|:|X|\in [\frac{1}{k}, k]\}$, with $k ∈ \mathbb N$, prove that $B_k\in \mathcal A.$

Let $(Y_k), k\in\mathbb N$, be a sequence of positive real random variables such that  $Y_k= |X|\mathbf{1}_{B_k}$. Prove that $P(\lim_{k\to \infty} Y_k = |X|) = 1.$

($\mathbf{1}_{B_k}$ is the indicator function).

my attempt :

For the first question, we have X a measurable function and  $[\frac{1}{k}, k]$ is a borel set therefore the results.

2.Now here is where I have a problem, I tried to prove that $\lim_{k\to \infty} Y_k = |X|$ using the fact that $\lim_{k\to \infty}\frac{1}{k}$=0 and when $\lim_{k\to \infty} K=+\infty$ therefore we have the indicator function on the positive real line always equals to 1 therefore we get that equality but I don't know where to go with that, I don't know how the probability equals to 1 and it kinda feels wrong to reason like that because what I get is P(X) with no assigned value to X which doesnt make sense I guess?

Any help or hint would be appreciated :).

答:
Hint

$B_k$ has no sense as written ! It should rather be $$B_k:=\left\{\omega \in \Omega \mid |X(\omega )|\in\left[\frac{1}{k},k\right]\right\}.$$

But the idea is correct. For the second question, since $$B_k\nearrow \{|X|>0\},$$

i.e. $B_k\subset B_{k+1}$ for all $k$, and $$\bigcup_{k\in\mathbb N}B_k=\{|X|>0\},$$

you have that $$\boldsymbol 1_{B_k}\nearrow \boldsymbol 1_{\{ |X|>0\}}.$$


问:
Let V be a vector space over $Z_3$ and dim(V)=3 ,then V has how many elements
The following question is part of a mock test which I am trying to solve and I need help in this question.

Let V be a vector space over $Z_3$ and dim(V)=3 ,then V has how many elements?

I used formula $(3^3 -1) (3^3 -3) (3^3 -3^2)$ but answer is given 125 elements.

I am asking here because I am not good in this type of questions so that you can tell me whether I am wrong and what is the right approach.

thnaks!!

答: Since there is a basis consisting of three elements, and the base field has order three, there will be $3^3=27$ total linear combinations.


问:
Proof of "infinite" series
This question comes from an exercise in programming ( the code is NOT the question)
Meta code

pick a composite natural number $N_1$

pick a comisite natural number $N_2 , N_2> N_1$

compute difference $N_2-N_1 =:d$

compute highest prime factor of $N_1 =:P_1$

compute highest prime factor of $N_2 =:P_2$

compute difference $P_2-P_1 =: D$

if $d=D$ compute product of $P_1 \cdot P_2 =: NN$

that is the exercise and now the question how to prove $NN$ series is infinite

or prove that there are infinitely many natural numbers which might be returned by step 7

答:
Let $p_1,p_2$ be consecutive primes. As they are coprime, there are integers $u,v$ such that $up_1+vp_2=1$. We can assume that $0>u>-p_2$, for if $u\ge 0$ or $u\le -p_2$, we can replace $(u,v)$ with $(u\mp p_2, v\pm p_1$), perhaps repeatedly. Then clearly $0<v<p_1$. So $-u$ and $v$ are positive integers divisible only by primes $<p_2$.
We also know that $p_2-p_1$ is divisible only by primes $<p_2$.
Therefore, if we let $n_1=-(p_2-p_1)up_1$ and $n_2=(p_2-p_1)vp_2$, then
$$ n_2-n_1 = (p_2-p_1)(vp_2+up_1)=p_2-p_1.$$
Therefore, $p_1p_2$ is a possible output of step 7.

Example: With $p_1=23$, we let $p_2=29$. We find $-5p_1+4p_2=1$, so $6=-30p_1+24p_2$, which leads to $n_1=690$, $n_2=696$ where $n_2-n_1=p_2-p_1=6$.

Title: Proof of "infinite" series


答:
First of all, I think you meant $B_k:=\{\omega\in \Omega: |X(\omega)|\in [\frac{1}{k}, k]\}.$ The way you have defined the set $B_k,$ it is a subset of $\mathbb{R}.$

Now for your second problem. The question is essentially saying that $Y_k\to |X|$ almost surely. Consider the set $Z:=\{\omega: 0\le |X(\omega)|<\infty\}$. Observe that $P(Z)=1.$

We now want to show that if  $\omega\in Z,$ then $Y_k(\omega)\to |X|.$ To see this, firs consider the $\omega\in Z$ Such that $|X(\omega)|=x\neq 0.$ Then there exists $k_0$ such that $\frac{1}{k}\le |X(\omega)|\le k$ for all $k\ge k_0.$ In other words, $Y_{k}(\omega)=|X|(\omega)$ for all $k\ge k_0.$ This proves the claim.

The case when $|X(\omega)|=0$ need to be handled separately. But it is easy. Note that if $|X(\omega)|=0,$ then $Y_k(\omega)=0$ for all $k.$

Title: Problem finding the limit of a random variable sequence ( attempt included )

Body:
Let $X$ be a real random variable defined in the probability space $(\Omega, \mathcal A, P)$

Suppose we have a set $B_k = \{|X|:|X|\in [\frac{1}{k}, k]\}$, with $k ∈ \mathbb N$, prove that $B_k\in \mathcal A.$

Let $(Y_k), k\in\mathbb N$, be a sequence of positive real random variables such that  $Y_k= |X|\mathbf{1}_{B_k}$. Prove that $P(\lim_{k\to \infty} Y_k = |X|) = 1.$

($\mathbf{1}_{B_k}$ is the indicator function).

my attempt :

For the first question, we have X a measurable function and  $[\frac{1}{k}, k]$ is a borel set therefore the results.

2.Now here is where I have a problem, I tried to prove that $\lim_{k\to \infty} Y_k = |X|$ using the fact that $\lim_{k\to \infty}\frac{1}{k}$=0 and when $\lim_{k\to \infty} K=+\infty$ therefore we have the indicator function on the positive real line always equals to 1 therefore we get that equality but I don't know where to go with that, I don't know how the probability equals to 1 and it kinda feels wrong to reason like that because what I get is P(X) with no assigned value to X which doesnt make sense I guess?

Any help or hint would be appreciated :).


答:
We have $p(0)\lt 0$ and $p(1)\gt 0$, so by the Intermediate Value Theorem there is an $r$ between $0$ and $1$ such that $p(r)=0$.

We now show there is only root. For suppose that $a$ and $b$ are two distinct roots. Then by Rolle's Theorem, there is a $c$ between $a$ and $b$ such that $p'(c)=0$. However. $p'(x)=5x^4+3x^2+7$, so $p'(x)\gt 0$ for all real $x$.

ContentLicense: CC BY-SA 3.0

Title: Using the Intermediate Value Theorem and Rolle's theorem to determine number of roots

Body:
Use the Intermediate Value Theorem and Rolle's Theorem to show the that the polynomial $$p(x) =  x^{5} + x^{3} + 7x - 2$$ has a unique real root.

Can someone please give some hints on how to do this question. Thanks in advance.


问:
How to derive Lobachevsky's formula for the angle of parallelism
I'm lightly studying some non-Euclidean Geometry and in the book I am reading there is no proof or derivation from where the Lobachevsky formula for angle of parallelism comes from:

$$\Pi(x)=2\tan^{-1}\left(e^{-x}\right)$$

Any help? Thanks

P.S. I couldn't find anything by google search either.

答:
This formula is defining hyperbolic geometry that originated at first definitively from Bolyai, Lobachevsky and Gauss, so they cannot be derived from any other earlier results.

The angle of parallelism is more clearly seen in the Beltrami 3d model compared to the other models.

There is no need to to derive afresh imho but there is a need to recognize/reconcile with earlier models with the Beltrami 3d view.

Let $$ x=\theta, \Pi (x)= \psi(\theta)\;, \Pi(x)= 2 \tan^{-1}(e^{-x})\tag1 $$

$$  \rightarrow \psi(\theta)/2= \psi/2= \tan^{-1}(e^{-\theta})\tag2$$

Take reciprocals, add and simplify using trigonometric and hyperbolic function relations.

$$ \sin \psi = \text{sech} (\theta) \tag3$$

We are here introducing Beltrami's interpretation of hyperbolic geometry.

$ \psi$ is the angle of parallelism when the asymptotic line is everywhere parallel to the axis of rotation.

$$ \sin \psi = \text{sech} (\theta)=\dfrac{r}{a} \tag4$$

Recognize that $\psi$ is angle made by asymptotic line or hyperbolic geodesic to the meridian of a pseudo-sphere of cuspidal radius $a$, the $r$ is radius at any point in cylindrical coordinates as

$$ (r, \theta,z) \tag5$$

where $\theta$ is polar angle reckoned from cuspid equator, $\phi$ angle tangent makes to axis of symmetry.

we have

$$ \psi= \phi \; ; \tag 6$$

Since

$$ \tan \psi= \dfrac{r\;d\theta}{dz} \tag7$$

we get by integration 3d parameterization:

$$ r/a= \sin \psi = \text{sech} (\theta) ;\; z/a = \theta- \tanh(\theta)\; ; \tag 8 $$

that define the geodesic parallel lines shown in red for one half of the pseudosphere. Note that at the cuspidal equator ( maximum radius)

$$ \psi= \pi/2 \tag 9 $$

and as $ \theta \rightarrow \infty, r\rightarrow 0\;\tag{ 10 }$

The above are in agreement with Wiki: Angle of Parallelism.

There can be two parallels through one point, with another one symmetrical to the plane of meridian.

The angles of parallelism $  \Pi (x)$ or $ \psi(\theta)$ vary from point to point between $ 0 - \pi$, being $\pi/2$ at cusp.


问:
Gentzen Calculus: Proving $\vdash^G_\Sigma (E3_\cong),(E1),\Gamma\rightarrow \Delta, \forall x_1\forall x_2(x_1\cong x_2\supset x_2\cong x_1)$
I am trying to prove some theorems using Gentzen calculus and have a few questions. For example:

Show that
$$\vdash^G_\Sigma (E3_\cong),(E1),\Gamma\rightarrow \Delta, \forall x_1\forall x_2(x_1\cong x_2\supset x_2\cong x_1),$$ where $(E1)$ is the sentence
$$\forall x(x\cong x)$$ and $(E3_p )$ is the sentence
$$\forall x_1...\forall x_{n+n}((x_1\cong x_{n+1}\wedge...\wedge x_n\cong x_{n+n})\supset p(x_1,...,x_n)\supset p(x_{n+1},...,x_{n+n})).$$
My attempt:

$1\quad (E3_\cong),(E1),\Gamma\rightarrow \Delta, \forall x_1\forall x_2(x_1\cong x_2\supset x_2\cong x_1)\quad R\forall:2$

$2\quad (E3_\cong),(E1),\Gamma\rightarrow \Delta,\forall x_2(x_2\cong x_2\supset x_2\cong x_2)\quad R\forall :3$

$3\quad (E3_\cong),(E1),\Gamma\rightarrow \Delta, x_2\cong x_2\supset x_2\cong x_2\quad R\supset:4$

$4\quad x_2\cong x_2,(E3_\cong),(E1),\Gamma\rightarrow \Delta,x_2\cong x_2\quad \text{Ax}$

From step 1 to step 2 I changed $x_1$ to $x_2$ because $x_2$ is a fresh variable in the first sequent. Is this derivation correct?

Furthermore, there is another exercise:

Show that
$$\vdash^G_\Sigma (E3_\cong),(E1),\Gamma\rightarrow \Delta,\forall x_1\forall x_2\forall x_3((x_1\cong x_2\wedge x_2\cong x_3)\supset x_1\cong x_3).$$
Here I probably have to use $(E1)$ and $(E3_\cong)$.

My question(s): what does $p(x_1,...,x_n)$ mean when $p$ is replaced by $\cong$ ? In this case, I don't really know how to work with $(E3_\cong)$. What I mean is: what happens if, for example, I apply the rule $L\forall$ to $(E3_\cong)$ ?; how many variables $(x_1,x_2,...)$ should I consider for $(E3_\cong)$ in this particular exercise?

I appreciate any answer/insight. And thank you very  much.

答:
See G.Takeuti, Proof Theory, page 37-on, for equality axioms:

$$\to x_1=x_1$$

$$x_1=y_1, \ldots x_n=y_n \to f(x_1,\ldots,x_n)=f(y_1,\ldots,y_n),$$

where $f(...)$ is an $n$-argument places function symbol, and

$$x_1=y_1, \ldots x_n=y_n, P(x_1,\ldots,x_n) \to P(y_1,\ldots,y_n),$$

where $P(...)$ is an $n$-argument places predicate ($=$ included).

Reading your proof bottom-up. the last step seems wrong to me: when we have $∀x_2(x_2 ≅ x_2 ⊃ x_2 ≅ x_2)$ there is no way to recover $x_1$.

To prove Symmetry, we need the Substitution axiom for formulas with formula $(z = x_1)$ as $P(...)$.

Thus, the substitution $P(z,x_1)[x_1/z]$ will get $(x_1=x_1)$ and the substitution $P(z,x_1)[y_1/z]$ will get $(y_1=x_1)$:

$$x_1=y_1, (x_1 = x_1) \to (y_1=x_1).$$

The proof must be:

$$\dfrac{ \dfrac{}{x_1=y_1 , x_1 = x_1 \to y_1 = x_1}}{x_1=x_1,x_1=y_1 \to y_1=x_1},$$

using suitable rules and interchanges.

Now, we have to use Reflexivity axiom: $\dfrac {}{\to x_1=x_1}$ to derive: $x_1=y_1 \to y_1=x_1$, using Cut.

The final result will need $(\text R \supset)$ followed by two $(\text R \forall)$ to get:

$$\to \forall x_1 \forall y_1 (x_1=y_1 \supset y_1=x_1).$$

Similar for second proof. We need the atomic formula $(z = x_3)$ with the following instance of the axiom:

$x_1 = x_2, (z=x_3)[x_2/z] → (z=x_3)[x_1/z]).$


问:
Probability that at least one Ace is dealt in poker with $6$ players
If two cards are dealt to each of the 6 players from a 52 deck of cards what is the probability that there is at least one ace dealt?

I've tried to solve this using the following reasoning:

In total there are $ {52 \choose 12 } \cdot {12 \choose 6}$ ways to deal 12 cards to the 6 players.

Count the number of ways to deal with at least one ace.

To count 2 notice that there is in general ${4 \choose N} \cdot 48 \cdot 47 \cdots (48 - (12 - N)) $ ways to choose $N$ aces in the $12$ dealt cards. For each of these combinations there are also ${12 \choose 6}$ ways to distribute two cards to each player. Thus in general the probability would be, following the above reasoning:

$$ P(\text{at least one A}) = \frac{{4 \choose N} \cdot 48 \cdot 47 \cdots (48 - (12 - N))}{ {52 \choose 12 } \cdot {12 \choose 6}^2 }$$

However, this does not give a sensible answer.

Can anyone see where the flaw in the above reasoning is?

Edit:

Thanks to those who have answered below - I can see that taking the inverse of the question by asking how to deal no aces is much easier. However, can anyone show me how to do it by count the positive cases as I have tried above?

答:
The probability of at least one ace is 1 minus the probability of no ace:
$$1-\frac{\binom{4}{0}\binom{48}{12}}{\binom{52}{12}}=\frac{2759}{4165},$$
so it is about twice as likely to have at least one ace than no ace.

If you prefer to count positive cases:
$$\sum_{k=1}^4 \frac{\binom{4}{k}\binom{48}{12-k}}{\binom{52}{12}}=\frac{2759}{4165}$$


问:
Solve a recurrence relation in two variables
Suppose the function $f : \mathbb{N}\to\mathbb{N}$ is such that $f(1)=1$ and
$$f(2m) + f(2n) = 2(f(m+n) + f(m-n))$$
for all $m>n>0$. I would like to show that this implies $f(n) = n^2$ for all $n\in\mathbb{N}$.

It is easy to show that $f(n)=n^2$ is a solution, just by substituting, but I'm having trouble proving uniqueness. I know about mathematical induction but in this case, the fact that there are two variables $m$ and $n$ is giving me trouble. Any help appreciated.

Added: the part that's giving me trouble is that I cannot assume that $f(0)$ exists. If I could assume that, then $f(0)=0$ by putting $m=n=0$, and then by putting $m=n$ I could show that $f(2m) = 4f(m)$, from which it's not hard to continue.

I guess my question is: am I allowed to extend $f$ by defining $f(0)=0$, and use that in my reasoning? It seems like cheating somehow.

答:
I'm not quite satisfied with this solution, but it's the best I can come up with. All my approaches ultimately depend on finding initial values like $f(2), f(3), f(4)$, for which I had no motivating approach other than trying numerous cases and hoping it eventually works out.

The following 2 equations are the most important that we'd use:

Calculating powers of 2 from smaller powers of 2:
$$f(2^{k+1} ) + f(2^k) = 2 f(3\times 2^{k-1}) + 2 f( 2^{k-1} ) = 4 f(2^k) + 4 f( 2^{k-1} )$$

Calculating multiples of $2^{k-a}$ from multiples of $2^{k-a+1}$:

Where $O$ is an odd number,

$$f( \frac{O+1}{2} \times 2^{k-a+1}) + f( \frac{O-1}{2} \times 2^{k-a+1})           =2 f( O \times 2^{k-a} ) + f( 2^{k-a} ).$$

Claim 1: Let $ f(2) = a, f(4) = b$. Then, the rest of the values could be determined from these unknowns.

Proof of claim 1: We do so by induction, showing that we can determine the values from $f(2^k+1)$ to $f(2^{k+1})$.

We use the first equation to get $2^{k+2}$, then use the second equation to calculate the odd multiples of $2^{k}$ less than $2^{k+2}$, then calculate the odd multiples of $2^{k-1}$ less than $2^{k+2}$, then calculate the odd multiples of $2^{k-2}$ less than $2^{k+2}$, so on and so forth. $_\square$

Note: While we have determined these values through the equations, we still have to ensure that the all of the equations hold in general. It is by verifying these equations that we get more information.

Claim 2: We can show that $f(4) = 3f(2) + 4$, and that $f(2) = 4$.

Corollary: With $f(1) = 1, f(2) = 4, f(4) = 16$, we can repeat the induction at the start (once again working with the powers of 2 then multiples of $2^{k-a}$) and prove that $f(n) = n^2$.

Finally, verify that with $ f(n) = n^2$, both equations hold in all generality. So this is indeed a solution.

Proof of Claim 2: We can list out some of the these values, with the sequence obtained via the above algorithm:

$f(1) = 1$

$f(2) = a$

$f(4) = b$

$f(3) = (a+b-2) / 2 $

$f(8) = 4a+3b$

$f(6) = a+2b$

$f(5) = (a+3b-2) /2$

$f(7) = (5a+5b-2)/2$

With $ f(8) + f(2) = 2f(5) + 2f(3)$, we get that $ b = 3a + 4$. Rewriting the above list, we have

$f(1) = 1$

$f(2) = a$

$f(3) = 2a+1$

$f(4) = 3a + 4$

$f(5) = 5a + 5$

$f(6) = 7a+8$

$f(7) = 10a + 9$

$f(8) = 13a + 12$

There seems to be nothing further we can do, so let's generate more values.

$f(16) = 51a+52$

$f(12) = 29a + 28$

$f(10) = 20a+20$

$f(9) = (33a+30) / 2$

(I skipped listing out the rest)

With $f(16) + f(2) = 2 f(9) + 2 f(7) $, we get that $ a = 4$.

(With this value, you can check that the listed ones work out to $f(n) = n^2$.)$_\square$


问:
Prove that $gcd(u \times a, u \times b) = |u| \times gcd(a,b)$
The question: "Prove $gcd(u \times a, u \times b) = |u| \times gcd(a,b)$"

We have the Theorem which says that for any intergers $a$ and $b$ both not $0$, have a positive gcd which can be expressed as $gcd(a,b) = x\times a + y \times b$.

My attempts:

First, let $gcd(a,b) = x \times a + y \times b$, but let denote $gcd(a,b)$ as $d$, since $d$ divides $a$ and $b$, it divides $u \times a$ and $u \times b$ also, thus it is a positive common divisor of $u \times a$ and $u \times b$.

Now, let $gcd(u \times a, u \times b) = x' \times (u \times a) + y' \times (u \times b)$, but lets denote $gcd(u \times a, u \times b$) as $p$, since $d$ is a positive common divisor of $u \times a$ and $u \times b$, then $d$ must divide $p$, from which follows that $p = k \times d$ for some integer $k$, but it is the same as:
$$gcd(u\times a, u \times b) = k \times gcd(a,b)$$

And since $gcd(u \times a, u \times b) = x'\times (u \times a) + y' \times (u \times b)$, it follows that $gcd(u \times a, u \times b) = u \times (x' \times a + y' \times b)$ then i got:
$$gcd(u\times a, u \times b) = k \times (x \times a + y \times b) = u \times (x' \times a + y' \times b)$$

But I got stucked here, then I choosed to try another approach, first trying to prove the equality is valied to all positive integers $u$ using induction.

Let $B = \{u : u \in \mathbb{N} \land [gcd(u \times a, u \times b) = u \times gcd(a,b)]\}$.

First $1 \in B$ since $gcd(1 \times a, 1 \times b) = gcd(a,b) = 1 \times gcd(a,b)$.

Now assume $u \in B$ that is $gcd(u \times a, u \times b) = u \times gcd(a,b)$, now if we take $(u+1) \times gcd(a,b)$ we got $u \times gcd(a,b) + gcd(a,b)$ and by the assumption it is the same as $gcd(u \times a, u \times b) + gcd(a,b)$.

But I got stucked again, I will like to know if its possible to prove $gcd((u+1) \times a, (u+1) \times b) = gcd(u×a,u×b)+gcd(a,b)$.

I really want to know if its possible to continue from some of these two approaches, but other paths will be appreciated too.

Edit: I just noticed the answer from @blackmirror7, add the information to finish my first attempt, I dont think is needed to provide more information on the first approach, but meanwhile I managed to reach the answer following other path as follows:

First assume $a,b$ and $u$ are not $0$ because these are trivial cases, then we can proceed as follows:

$gcd(a,b) = d, \text{such that } 0<d$

Then $a=dm$ and $b=dn$ for some integers $m$ and $n$, and $gcd(m,n)=1$, for if there is some positive common divisor of $m$ and $n$ bigger than $1$, let say $c$ we got:

$m=cr$ and $n=cs$, for some $r,s$, thus $a=dcr$ and $b=dcs$ and $dc$ divides both $a$ and $b$. but as $1<c$ then $d<dc$, which contradicts the fact that $gcd(a,b)=d$.

Now let $gcd(ua, ub)=k$, since $ua=udm$ and $ub=udn$ we know that $ud$ divides both $ua$ and $ub$ then it must divide $k$ too, thus we have $k = pud$ for some $p$.

Since $k$ divides both $ua$ and $ub$, we have $ua = xk$ and $ub = yk$, for some $x,y$, but that is $udm = xpud$ and $udn=ypud$, from which follows $m=xp$ and $n=yp$, but by our previous result that $gdc(m,n)=1$ and the fact $p$ is a common divisor of $m$ and $n$ we have that $p=1$ or $p=-1$.

Finally, since $ua$ and $ub$ are not $0$, we know that $0<k=gcd(ua,ub)$, then $|k| = k$, but $|k| = |pud| = |p|.|u|.|d|$, and since $|p|=1$ and $0<d$ we got $k = |u|.d$, that is $gcd(ua,ub) = |u|.gcd(a,b)$.

I know that the first approach using the Bézout's identity is much more short, but i think its worth to present another approaches too, Hope it help someone.

答:
Let gcd(ua,ub)=k

gcd(ua,ub)= smallest positive integer of the form uax+uby for some integers x,y.

uax+uay=u(ax+by)=k and by the same property of gcd, we have u(ax+by)=u(gcd(a,b))


问:
Uniformly convergent sequence of holomorphic function in every compact subset converges to holomorphic function

$\{f_n\}^\infty_{n=1}$ is a sequence of holomorphic functions that
converges uniformly to a function $f$ in every compact subset of
$\Omega$, then $f$ is holomorphic in $\Omega$.

We let $D$ be any disc whose closure is contained in $\Omega$. Then for any triangle $T$ contained in $D$, by Goursat's theorem, we have $\int _T f_n(z)dz=0$. It then asserts that

$$\int_T f_n(z)dz\to \int_T f(z)dz\text{ as }z\to \infty$$

in the closure of $D$, because of the uniform convergence of $f_n$. This seems a basic question, but can anybody please elaborate what is happening here?

答: A comment that got too long that discuses (locally) uniform convergence vs pointwise convergence. One can use Runge theorem to construct a sequence of polynomials $P_n$ that converge to zero pointwise except at the origin, where $P_n(0)=1$; then although the integral of $P_n$ converges on every closed curve to the integral of $f$ as $f(z)=0, z \ne 0, f(0)=1$, $f$ is not continuous at zero hence not holomorphic there; in general with pointwise convergence, Osgood's theorem ($f_n \to f$ pointwise on $U$ and $f_n$ analytic, then here is an open dense subset $U_0$ of $U$, where $f_n \to f$ locally uniform so $f$ is analytic at least on $U_0$ is the best one can get; a good (fairly elementary) discussion of what happens is in the American Math Monthly paper by Beardon and Minda (pdf LINK)


问:
Distribution $P_{\frac{1}{x^3}}$
How  to show that for $\phi\in D(R)$, $<P_{\frac{1}{x^3}},\phi>=v.p.\int_{-\infty}^{\infty}\frac{\phi(x)-x\phi'(0)}{x^3}$ defines a distribution?

It is easy to check that  $P_{\frac{1}{x^3}}$ is correctly defined and linear on D(R), but how to show that it continuously depends on $\phi$?

Thanks in advance.

答:
One approach is begin with $f(x)=\log|x|$, which is locally integrable and therefore defines a distribution via $\phi\mapsto \int \phi(x)\log|x|\,dx$.

The third distributional derivative $f'''$ is the distribution $\phi\mapsto -\int \phi'''(x)\log|x|\,dx$. Formally, this can be viewed as $(\log |x|)'''=2x^{-3}$. To make this rigorous, we should integrate by parts (taking due diligence around $0$). To simplify the computations, observe that

Every  function $\mathbb R\to\mathbb R$ is the sum of an odd function and an even function.

The distribution $\log|x|$ is even in the sense that it annihilates odd test functions.

The third derivative of an even distribution is odd: it annihilates even test functions.

Therefore, it suffices to work with $-\int \phi'''(x)\log|x|\,dx$ when $\phi$ is odd. Take  $\epsilon>0$:
$$
\begin{split}
-\int_{|x|>\epsilon}\phi'''(x)\log |x|\,dx &= -2\int_{\epsilon}^\infty \phi'''(x)\log x\,dx \\
&= 2(\log \epsilon)\phi''(\epsilon)+2\int_\epsilon^\infty \phi''(x)\frac{dx}{x}
\\
&= 2(\log \epsilon)\phi''(\epsilon)- 2\epsilon^{-1}\phi'(\epsilon) +2\int_\epsilon^\infty \phi'(x)\frac{dx}{x^2}
\\
&= 2(\log \epsilon)\phi''(\epsilon)- 2\epsilon^{-1}\phi'(\epsilon) -2\epsilon^{-2}\phi(\epsilon) + 4\int_\epsilon^\infty \phi(x)\frac{dx}{x^3}
\end{split}
$$
Since $\phi$ is odd, $\phi(0)=\phi''(0)=0$. As $\epsilon\to 0$,
$$\begin{split} 2(\log \epsilon)\phi''(\epsilon) &= o(1) \\
- 2\epsilon^{-1}\phi'(\epsilon) &=  - 2\epsilon^{-1}\phi'(0)+o(1)  \\
-2\epsilon^{-2}\phi(\epsilon)&=-2\epsilon^{-1}\phi'(0)+o(1) \\
\end{split}$$

Add these up, express $-4\epsilon^{-1}\phi'(0)$ as $\int_{|x|>\epsilon} \frac{-2\phi'(0)}{x^2}\,dx$, and there you have it:
$$
-\int_{|x|>\epsilon}\phi'''(x)\log x\,dx = 2\int_{|x|>\epsilon}\frac{\phi(x)-x\phi'(0)}{x^3}\,dx
$$
The limit as $\epsilon\to 0$ is the principal value in question.

Title: Distribution $P_{\frac{1}{x^3}}$


问:
A question about a pointwise convergent sequence $(u_n) \subset W_0^{1,p}(I)$.
Let $I$ be a bounded  open interval, assume for simplicity that  $I=(0,1)$. Let  $(u_n)_n \subset W_0^{1,p}(I)\  \ \  (1<p<\infty)\ $  be a sequence such that
the sequence of weak derivatives $(u_n') \subset L_p(I)$ is bounded and $u_n(t)\to u(t)$ for all $t \in I$. How can one show that $u_n\to u$  $\textbf{uniformly}$ on $\overline I$?

My attempt:
By Poincare's inequality, we have  that $(u_n)_n$ is bounded and since $W_0^{1,p} \hookrightarrow C(\overline I) \ \textbf{compactly}$, we see that $(u_n)_n$ is relatively compact and thus we can find a uniformly convergent sequence:
$$u_{n_k} \to \hat u \  \ \text{ in }  \ C(\overline I) .$$
Since $u_n(t) \to u(t)$ for all $t \in I$, we infer that $u=\hat u$. Can we somehow show that $u_n \to u$ in $C(\overline I)$?

Thanks in advance!

答:
You can use a subsequence-subsequence argument:

Assume that $u_n$ does not converge towards $u$ in $C(\bar I)$.
Then, there is a subsequence $u_{n_k}$ with $\| u_{n_k} \ge u\|_C \ge \varepsilon > 0$. But your argument shows that this subsequence has a subsequence which converges towards $u$ in $C(\bar I)$. This is a contradiction.


问:
Writing $\sqrt[\large3]2+\sqrt[\large3]4$ with nested roots
Let $C\subset\Bbb R$ be the smallest set containing $0$ and closed under whole number addition/subtraction, whole number exponents, and whole number roots. That is, for all $c\in C$ and $n\in\Bbb N$, we have $c\pm n\in C$, $c^n\in C$, and $c^{1/n}\in C$.

We know that $\sqrt2+\sqrt3\in C$, since it's equal to $\sqrt{\sqrt{24}+5}$. Similarly, Ramanujan proved that:
$$\sqrt[\large3]{\sqrt[\large3]{1458}-9}-1=\sqrt[\large3]4-\sqrt[\large3]2$$
so $\sqrt[\large3]4-\sqrt[\large3]2\in C$. (Well, Ramanujan proved something equivalent.)

I found that $\left(\sqrt[\large3]2-1\right)^{-1}-1=\sqrt[\large3]2+\sqrt[\large3]4$, but I'm not allowed to use negative exponents/roots, so this doesn't prove that $\sqrt[\large3]2+\sqrt[\large3]4\in C$. So:

Is it true that $\sqrt[\large3]2+\sqrt[\large3]4\in C$? If not, why not?

Body:
Note
$$(\sqrt[3]2+\sqrt[3]4 -1)^2= 5- \sqrt[3]{4}
$$
which yields
$$\sqrt[3]2+\sqrt[3]4 =\sqrt{5-\sqrt[3]{4}}+1$$


问:
A question on conditional probability (Cambridge Admissions Exam)
I was solving the following problem

and I got to part (b). I thought of approaching this the following way:

There are two types of cartons, one with a 60 per cent probability of being selected and the other with 40 per cent. I split it intwo scenarios and I calculated each of the conditional probabilities and then added them together (alongside with the probability of each carton being selected). Here is what I have done:

First consider the case where he carton is of skimmed milk.

$$P(X>500 | X<505)= \frac{P(500 < X <505)}{P(X<505)}=\frac{b-\frac{1}{2}}{b}=\frac{2b-1}{2b}$$

Case for full-fat milk

$$P(X>500 | X<505)= \frac{P(500 < X <505)}{P(X<505)}=\frac{b-a}{a}$$

Thus the total probability must be

$$=\frac{6}{10} \frac{2b-1}{2b} + \frac{4}{10} \frac{b-a}{a} $$

However, this is wrong according to the markscheme given here:

My question: Why is it wrong to consider the different cases separately for part (i)?

答:
Sorry for attaching an image. I am fairly new to MSE and dont know any formatting as of now.

Coming to the answer its just because the probability of picking skimmed milk given that carton contains less than 505 ml is not 0.6 and similarly not 0.4 for full fat milk. Those probabilities should be calculated considering that there is less than 505 ml in the carton.

Expressions


问:
How can you calculate $\lim _{x\rightarrow 1}\; \frac{\root {m} \of {x}-1}{\root {n} \of {x}-1}$without using L'Hospital's law?
I'm trying to calculate this limit with substitution: $\lim _{x\rightarrow 1}\; \frac{\root {m} \of {x}-1}{\root {n} \of {x}-1}$

I started by multiplying with $\frac{\root{n} \of {x}}{\root{n} \of {x}}$ $\to$ $\lim _{x\rightarrow 1}\; \frac{\root {mn} \of {x^{n+m}}-\root{n} \of {x}}{\root {n} \of {x^2}-\root{n} \of {x}}$

I then substituted $\root{mn}\of{x}$ with $t$, so I got $x=t^{mn}$ with $\lim _{x\rightarrow 1}\;\root{mn}\of{x}=1$

$\lim _{t\rightarrow 1}\;\frac{t^{n+m}-\root{n} \of {t^{mn}}}{\root{n} \of {t^{2mn}}-\root{n} \of {t^{mn}}}$ $\to$ $\lim _{t\rightarrow 1}\;\frac{t^{n+m}-t^m}{t^{2n}-t^m}$ $\to$ $\lim _{t\rightarrow 1}\;\frac{t^m(t^n-1)}{t^m(t^m-1)}$ $\to$ $\lim _{t\rightarrow 1}\;\frac{t^n-1}{t^m-1}$

$\lim _{t\rightarrow 1}\;\frac{t^n-1}{t^m-1}$ $\to$ $\lim _{t\rightarrow 1}\;\frac{(t^{n-1}+t^{n-2}+...+t^1+t^0)}{(t^{m-1}+t^{m-2}+...+t^1+t^0)}$

So if we put $1$ in we will get $\frac{n*1}{m*1}$ which is $\frac{n}{m}$

Is this right?

Thanks

Body:
Set $x=e^y$, and consider $y \rightarrow 0.$

$f(y):=\dfrac{e^{y/m}-1}{e^{y/n}-1}=$

$\dfrac{1+y/m+O((y/m)^2)-1}{1+y/n+O((y/n)^2) -1}.$

Take the limit.


问:
Prove that the Guassian curvature of the given surface is zero

Let $\gamma_1,\gamma_2:I\to\mathbb{R}^3$ be a pair of regular space curves with the same domain, $I$. Define $\sigma:I\times(0,1)\to\mathbb{R}^3$ as $\sigma(t,u) = u\cdot\gamma_1(t)+(1-u)\cdot\gamma_2(t)$. Assume that $\gamma_1,\gamma_2$ are chosen such that $\sigma$ is regular. Prove that $K(t,1/2)=0$ for all $t\in I$.

In the above, $K$ denotes the Gaussian curvature. To solve the above problem,
\begin{align}
\sigma_t &= u\gamma_1'+(1-u)\gamma_2'\\
\sigma_u&= \gamma_1-\gamma_2\\
\sigma_t\times\sigma_u &= (u\gamma_1'+(1-u)\gamma_2')\times (\gamma_1-\gamma_2)\\
&= \frac{1}{2}(\gamma_1'+\gamma_2')\times (\gamma_1-\gamma_2)\\
&\text{in this case ($u=\frac {1}{2}$).}\\
\sigma_{uu} &= 0\\
\sigma_{tu} &= \gamma_1'-\gamma_2'
\end{align}

So that we need to show $<\gamma_1'-\gamma_2',(\gamma_1'+\gamma_2')\times (\gamma_1-\gamma_2)> = 0$ (This is because, $K = \frac{eg-f^2}{EG-F^2}$ where $e,g,f$ from second fundamental form and $E,F,G$ form the first fundamental form. Since $\sigma_{uu} = 0, g = 0$ and now, to show $K = 0$ is equivalent to show $f = 0$.)

Here is where I'm stuck. Why is $<\gamma_1'-\gamma_2',(\gamma_1'+\gamma_2')\times (\gamma_1-\gamma_2)> =0$?

I used the notation in 'Differential geometry in curves and surfaces' by Tapp.

答:
Let's let $a(t) = \gamma_1(t) - \gamma_2(t), b(t) = \gamma_1(t) + \gamma_2(t)$.

Evidently $\gamma_1 = \frac12 (a(t) + b(t)), \gamma_2 = \frac12 (b(t) - a(t))$, so there no loss of generality in this renaming.

Now you're asking why, for any curves $a, b$, we have
$$
\langle a', b' \times a \rangle = 0.
$$

Let's look at $a(t) = (t, t^2, 0), b(t) = (0, 0, t)$. Then the expression above is
\begin{align}
\langle a', b' \times a \rangle
&= \langle (1, 2t, 0), (0,0,1) \times (t, t^2, 0) \rangle \\
&= \langle (1, 2t, 0), (-t^2,t,0) \rangle \\
&= -t^2 + 2t^2
\end{align}
which, for $t = 1,$ for instance, is not zero as desired.

So the thing you're trying to prove is false.

Personally, I suspect that the overall claim might be true, but my instincts aren't great. If they are true, then it means that your computations of the various tangents and normals are wrong somewhere. I don't see where --- but I didn't look too closely --- but I'd recheck carefully if I were you.

I also have the suspicion (using the triple-product identities) that if the $\gamma$s are both unit speed curves, then the proof might be pretty easy. But I leave that to you.

Big picture remark

When you're down in the weeds in a diff'l geometry computation, try an example.


问:
Problem of minimum in physics.

It is necessary to go from a point $A(0,0)$ to a point $B(a,b)$ walking from A to $P(x,0)$ with a speed $v_1$ and then until the point B with a speed $v_2$.

Find where is the point in which it is necessary to abandon x axis in order to have the minimum time to complete the path.

I've called $x$ the distance of A from P.
The function is $$t(x)=\frac{x}{v_1}+\frac{\sqrt{(a-x)^2+b^2}}{v_2}$$
$$\frac{dt}{dx}=\frac{1}{v_1}+\frac{(a-x)*(-1)}{v_2* \sqrt{(a-x)^2+b^2}}=0 \Rightarrow v_2*\sqrt{(a-x)^2+b^2}=v_1*(a-x) \Rightarrow x^2(v_2^2-v_1^2)+x(-2av_2^2+2av_1^2)+(a^2v_2^2+b^2v_2^2-v_1^2a^2)=0$$

The solutions of this equation are $x_{1/2}  =a \pm \frac{v_2 b}{\sqrt{v_1^2-v_2^2}}  $
Putting $\frac{dt}{dx}>0$ it is verified for $0<x- \frac{v_2 b}{\sqrt{v_1^2-v_2^2}}$ and for $x>a + \frac{v_2 b}{\sqrt{v_1^2-v_2^2}}$.

So the minimum point is for $a + \frac{v_2 b}{\sqrt{v_1^2-v_2^2}}$
and  the minimum time is

$$t(a + \frac{v_2 b}{\sqrt{v_1^2-v_2^2}})= \frac{b (v_1^2+v_2^2)}{v_1v_2 \sqrt{v_1^2-v_2^2}}+\frac{a}{v_1}$$

I'm not sure that I haven't done mistakes, because in the solution on the book it indicates
$a  - \frac{v_2 b}{\sqrt{v_1^2-v_2^2}}  $  as the solution of minimum.

and  the minimum time

$$t(a - \frac{v_2 b}{\sqrt{v_1^2-v_2^2}})= \frac{b\sqrt{v_1^2-v_2^2}}{v_1v_2}+\frac{a}{v_1}$$

but, in particular , it distinguishes the case of $\frac{v_2^2}{v_1^2}\ge \frac{a^2}{a^2+b^2}$
in which the minimum time is possible walking from A to B directly.

答:
You are correct that
$$ \frac{dt}{dx} = \frac{1}{v_1} - \frac{a-x}{v_2 \sqrt{(a-x)^2+b^2}} \tag1$$
and that $\frac{dt}{dx} = 0$ implies
$$ v_2 \sqrt{(a-x)^2+b^2} = v_1 (a-x), \tag2$$
which (I think--I did not check your arithmetic here) implies
$$ x^2(v_2^2-v_1^2)+x(-2av_2^2+2av_1^2)+(a^2v_2^2+b^2v_2^2-v_1^2a^2) = 0. \tag3$$

But you seem to be assuming (incorrectly) that Equation $(3)$ implies Equation $(2)$.
It does not. Equation $(3)$ does imply
$$ v_2 \sqrt{(a-x)^2+b^2} = \pm v_1 (a-x), $$
but this is not the same thing.
Also, the region where the left side of Equation $(3)$ is positive does not correspond to the region where $\frac{dt}{dx}$ is positive.

What Equation $(3)$ gives you is two possible points at which $\frac{dt}{dx}$ might be zero. To determine which points are actual zeros, compare each of them against Equation $(2)$.
Assuming that $v_1$ and $v_2$ both are positive,
in Equation $(2)$ we see that the left side is positive and that
therefore $a - x$ must also be positive.
That is, $x < a.$ Hence of the two roots of Equation $(3)$, the only one that can be a zero of $\frac{dt}{dx}$ is the one that is less than $a,$ namely

$$ x_0 = a - \frac{v_2 b}{\sqrt{v_1^2-v_2^2}}. $$

To determine whether this is a local minimum or a local maximum,
you must examine the sign of $\frac{dt}{dx}$ for values of $x$
less than $x_0$ or greater than $x_0$,
not the values of the left-hand side of Equation $(3).$

Title: Problem of minimum in physics.


问:
Can't understand definition of system $P$ given by Kurt Godel in his "On formally undecidable propositions..."
I am reading On formally undecidable propositions of Principia Mathematica and related systems by Kurt Godel. And I am not quite sure I am getting his definitions right. See the screenshot beneath:

...and so on for every natural number as type.

What does it mean for natural number to define a type? As far as I understand $x_1$ could reference to any natural number, for example $x_1 = 12$ is correct, but it can't reference set of natural numbers, so $x_1 = \{ 12, 1 \}$ is not correct.

But what about $x_2$? What is "class of individuals"? Is it the class from the set theory (i.e. a collection of items that looks like set but can not be a set because it may lead to contradiction)? Or Kurt Godel uses term "class" to mean something different? To my understanding, $x_2 = \{ 12, 1 \}$ is correct, but $x_2 = 12$ is not.

Why do we ever need to distinct between variables?

答:
It is High-Order Logic.

Unfortunately, the formal language of Gödel's paper is today quite unused :

By a sign of type $1$ we understand a combination of signs that has [anyone of] the forms

$$a, fa, ffa, fffa, \ldots,$$

and so on, where $a$ is either $0$ or a variable of type $1$. In the first case, we call such a sign a numeral. [...] A combination of signs that has the form $a(b)$, where $b$ is a sign of type $n$ and a a sign of type $n + 1$, will be called an elementary formula.

In modern terms, Type-1 is the collections of (individual) terms: the constant $0$, and terms $f(a)$ where $f$ is the successor function.

The intended model is $\mathbb N$; thus the above terms are numerals:

$0, f(0) [i.e. 1], ff(0) [i.e.2]$, and so on.

See axiom I.1: $\lnot (fx_1=0)$. It is the first Peano axioms: number zero is not a successor.

There are no predicate constants because, being Hihg-Order Logic, we can define equality $x_1=y_1$ as follows:

$x_2 \Pi (x_2(x_1) \to x_2(y_1))$

that reads : $\forall P [P(x_1) \to P(y_1)]$.

In Gödel's symbols, $x_1,y_1$ are variables of Type-1 (ranging over individuals, i.e. numbers) while $x_2$ is a variable of Type-2 (ranging over predicates, i.e. "classes of individuals").

See axiom I.3 (Induction): $x_2(0) \cdot x_1 \Pi (x_2(x_1) \supset x_2(f(x_1)) \supset x_1 \Pi (x_2(x_1))$, that is exactly:

$P(0) \land \forall x_1 (P(x_1) \to P(f(x_1)) \to \forall x_1 (P(x_1)).$

$x_3$ is a variable of Type-3 that ranges over predicates of predicates: "classes of classes of individuals", "and so on, for every natural number as a type".

See axiom IV.1: $\exists x_{n+1}(x_n \Pi (x_{n+1}(x_n) \equiv a))$, that reads: $\exists \Psi \forall P(\Psi(P) \equiv \varphi)$.

It expresses the fact for every suitable formula ranging over Type-$n$ "objects" there is the corresponding predicate of Type-(n+1) [Comprehension axiom].


问:
An alternative description of the first Stiefel-Whitney class
I've heard this description of the first Stiefel-Whitney class of a vector bundle but I don't know why this is true. Can anyone help me with this, please?

The first Stiefel-Whitney class of a vector bundle is an element in the first cohomology group of the base space $H^1(B,Z_2)$. This element can be seen as a map $w_1 : \pi_1(B) \rightarrow Z_2$
(because of the universal coefficient theorem and the fact that $H_1$ is the abelianization of $\pi_1$), so it assigns to each loop based at a fixed point an element of $Z_2$. Now this number is $0$ if and only if that loop is orientation preserving (locally we have an orientation and when we go around a loop and come back to our first place, we can ask if orientation has changed or not)

答:
If you have a triangulated manifold then it is orientable when its n simplices can be given signs (+/-1) so that the signed sum is an integer cycle.

If the manifold is not orientable then any signed sum of the n simplices is not a cycle and it turns out that what goes wrong is that some of the n-1 simplices in the boundary get counted twice rather than cancelling each other out.

This boundary which is a sum of doubles of n-1 simplices can be divided by two to give a 2 torsion Z homology class. This n-1 dimensional cycle represents the obstruction to orienting the manifold. I would strongly suspect that the first Stiefel Whitney class of the tangent bundle is the Poincare dual to this homology class. Check it out.


问:
Extension of a bounded linear map
Let $K \subset H$ both be Hilbert spaces. I'm trying to prove that for any bounded linear map $T: K \to X$ with $X$ a normed space, there exists an extension $\tilde{T}: H \to X$ such that the norm is preserved ($||T|| = ||\tilde{T}||$). Many of the answers I find on the internet about this have the extra condition that $X$ is Banach, I'm not able to come up with a solution when $X$ is not. What I tried so far is the following:

Extend $T: K -> X$ to $\bar{T}: \bar{K} \to \tilde{X}$ where $\bar{K}$ is the closure of the Hilbert subspace $X$ and $\tilde{X}$ is the completion of the normed space $X$ as we can always associate a Banach space to a normed space. Recalling that $K$ is dense in $\bar{K}$ this gives us an extension via the BLT.

Project $H$ onto the closed convex subset $\bar{K}$; $P_{\bar{K}}: H \to \bar{K}$

Taking the composition of these two maps we get a map $\bar{T} \circ P_\bar{K}: H \to \tilde{X}$.

I think I have to add a map to this to go from $\tilde{X} \to X$ or somehow be able to prove that any complete space gets mapped to a complete space under $T$ (which is not true I think).

Besides proving this I am also trying to find a pair of normed spaces $Y \subset X$, a two-dimensional normed space $Z$ and a bounded linear map $S: Y \to Z$ such that any extension $\tilde{S}: X \to Z$ satisfies $||\tilde{S}|| > ||S||$ indicating that the norm preservation relies on both $K \subset H$ being Hilbert spaces and not just normed spaces. I am also unable to come up with this but I think being able to complete the proof will help with this part too.

答:
We start out with the assumption that $K$ is Hilbert, which includes completeness in its definition. Therefore $K$ is already a closed convex subset of $H$ and step 1 is not needed. Just compose projection of $H$ onto $K$ with $T$ to get the desired map.

$$H \stackrel{P_K}{\to} K\stackrel{T}{\to} X$$

Title: Extension of a bounded linear map


问:
Homomorphisms between $ \mathbb{Z} $ modules.

Calculate $\newcommand\Hom{\operatorname{Hom}}\Hom(\mathbb Z \oplus \mathbb Z_{p^\infty},\mathbb Z \oplus \mathbb Z_{p^\infty})$. Where $ \mathbb Z_{p^\infty}= \bigcup_{k=1}^{\infty}\bar{\langle\frac1{p^k}\rangle}$ for $p$ prime.

As $\mathbb Z$ and  $\mathbb Z_{p^\infty}$ are $\mathbb Z$-modules then applying the theorem $\Hom_A ( \bigoplus_{i \in I} M_i, \prod_{j \in J} N_j ) \cong\prod_{(i,j) \in I \times J} \Hom(M_i,N_j)$ because for finite indices have $\bigoplus_{i=1}^{n}M_i = \prod_{i=1}^n M_1$. Then the problem reduces to find those $\Hom$ where

$$ \Hom(\mathbb{Z}, \mathbb{Z}) \cong \mathbb{Z} $$
$$ \Hom(\mathbb{Z},\mathbb{Z}_{p^\infty}) \cong \mathbb{Z}_{p^\infty}$$
but  $\Hom(\mathbb{Z}_{p^\infty},\mathbb{Z})$ and  $\Hom(\mathbb{Z}_{p^\infty},\mathbb{Z}_{p^\infty})$ not how to calculate them.

答:
$\def\Z{\mathbb{Z}}\def\Hom{\mathrm{Hom}}$We have $\Hom(\Z_{p^\infty},\Z)=0$ because $\Z_{p^\infty}$ is divisible and $\Z$ is reduced.

On the other hand $\Hom(\Z_{p^\infty},\Z_{p^\infty})$ is the ring of $p$-adic integers. You find its description in any Algebra book (Cohn, for instance, or Atiyah-McDonald), or http://en.wikipedia.org/wiki/P-adic_number


问:
What is this numerical approximation for definite integrals?
I've recently been reading a programming textbook that I've seen recommended to undergraduates. In it, the following approximation appears:

What is this approximation? It seems elementary enough to have a name, but I do not believe that I have encountered it before. In fact, I can't even recall having seen this notation.

To be clear, I just want to know what the name of this trick is. Once I've got that, I'm happy to do any further reading myself.

答: This is the "midpoint rule" approximate Riemann sum/"method of Riemann summation". Note that they are using $dx$ to stand for a small real number (usually denoted $h$ or $\Delta x$), rather than something more calculus-y.


问:
Solve the differential equation $\frac{dx}{dy} -\frac{x \log x}{1+\log x} =\frac{e^y}{1+\log x}$, if $y(1)=0$
I don’t know what form this. It’s probably a reducible to linear form equation, but I don’t know which term to substitute or how to rearrange. Can I get a hint?</p>

答:
$$\frac{dx}{dy} -\frac{x \log x}{1+\log x} =\frac{e^y}{1+\log x}$$
$$({1+\log x}){dx} -(x \log x)dy=e^ydy$$
Note that:
$$d(x \log x)=(1+\log x) dx$$
So that we have:
$$d(x \log x) -x \log xdy={e^y}dy$$
Multiply by $e^{-y}$:
$$e^{-y}d(x \log x)+ x \log xde^{-y}=dy$$
$$d(e^{-y}x \log x)=dy$$
Integrate:
$$e^{-y}x \log x-y=C$$


问:
Convergents of an infinite continued fraction oscillate around irrational number
If $\alpha$ is an irrational number and $\frac{p_n}{q_n}$ for $n\geqslant 0$ are the convergents to $\alpha$.

Is there a way to show that the convergents alternate between being $<\alpha$ and $>\alpha$?

答:
Suppose that $\frac{a}b<\frac{c}d$, where $a,b,c$, and $d$ are positive integers. For any $k\ell\in\Bbb Z^+$ we have

$$\frac{ka+\ell c}{kb+\ell d}-\frac{a}b=\frac{\ell(bc-ad)}{kb^2+\ell bd}>0$$

and

$$\frac{ka+\ell c}{kb+\ell d}-\frac{c}d=\frac{k(ad-bc)}{kbd+\ell d^2}<0\,,$$

so that

$$\frac{a}b<\frac{ka+\ell c}{kb+\ell d}<\frac{d}d\,.$$

In particular, it follows from this that $\frac{p_{k+2}}{q_{k+2}}$ always lies between $\frac{p_k}{q_k}$ and $\frac{p_{k+1}}{q_{k+1}}$. Use this to show that the even convergents are increasing, the odd convergents are decreasing, and all of the even convergents are less than any of the odd convergents.


问:
Solving partial differential equation $u_x + 2xu_y + (2x - 6)u = 0$
I'm trying to solve the differential equation above by separating the variables. Thus, I let $u(x,y) = X(x)Y(y)$. As such, $u_x = X'(x)Y(y)$ and $u_y = Y'(y)X(x)$.

The differential equation above can then be rewritten as:
$X'(x)Y(y) + 2xX(x)Y'(y) + (2x - 6)X(x)Y(y) = 0$

However, I find myself stuck after this step as I'm unable to separate the variables and shift all the functions/variables with x to 1 side of the equation and all the functions/variables with y to the other side of the equation.

Some help will be greatly appreciated. Thanks a lot :)

答:
$$u_x + 2xu_y = (6-2x)u$$
Why not the method of characteristics ?

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

The Charpit-Lagrange system of characteristic ODEs is :
$$\frac{dx}{1}=\frac{dy}{2x}=\frac{du}{(6-2x)u}$$
A first characteristic equation comes from solving $\frac{dx}{1}=\frac{dy}{2x}$ :
$$y-x^2=c_1$$
A second characteristic equation comes from solving $\frac{dx}{1}=\frac{dy}{2x}=\frac{6dx-dy}{6-2x}=\frac{du}{(6-2x)u}\quad;\quad d(6x-y)=\frac{du}{u}$
$$u\:e^{y-6x}=c_2$$
The general solution of the PDE expressed on the form of implicit equation $c_2=F(c_1)$ is :
$$u\:e^{y-6x}=F(y-x^2)$$
$F$ is an arbitrary function.
$$\boxed{u(x,y)=e^{6x-y}F(y-x^2)}$$
If some boundary condition was specified in the wording of the problem one could determine the function $F$ in order to find the particular solution satisfying both the PDE and the boundary condition.


问:
Question regarding the notation of $\equiv$ and $\iff$
What difference does the notation in those two terms make?
$$
=:⟺∀:(∈\iff∈)
$$
$$
=:⟺∀:(∈\equiv∈)
$$

答:
The formula can be  read as

$X=Y \equiv_{Def}\forall(x)  ( x\in X \leftarrow\rightarrow x\in Y)$

( In words : saying that X = Y is logically equivalent ( by definiiton) to saying that X and Y happen to have exactly the same elements).

with

$\equiv_{Def}$ denoting logical equivalence ( more precisely, equivalence-by-definition)

and

$\leftarrow\rightarrow$ denoting material equivalence or material bi-implication, which is a truth fonctional operator.

The first relation is a metalogical relation; the second belongs to the object language.

The relation between logical equivalence and material equivalence is as follows : formulae $\phi$ and $\psi$  are logically equivallent when the material conditional $ (\phi\leftarrow\rightarrow\psi)$ is true in all logically possible cases.

The iff that is in the middle is a logical equivalence, more precisely, an equivalence-by-definition. equivalence-by-definition works in the same way as ordinary logical equivalence ( that is, two propositions are equivalent just in case it is impossible for them not to have different truth values, whatever possible case is considered).

Note : equivalence is interesting since it allows to substitute the LHS for the RHS ( and vice versa).

The iff that is on the left side is not a logical equivalence, but a material  bi-implication. Two propositions are materially equivalent just in case it factually happens that they have the same truth value, or if you prefer, just in case it factually happens that we do not have the first true and the second false, and reciprocally) .

Consider this application of the extensionality principle.

Let H be the set of animal that have a heart and K the set of animal that have kidneys.

The material conditional  $\forall(x)  ( x\in H \leftarrow\rightarrow x\in K)$ is true.

By the definition of set equality, the formula just above  is logically equivalent to saying that the two sets are equal, that is, it is a logicall impossibility ( once the definition is set forth) that the material conditional holds while $H=K$ does not, and vice versa.

But this is not to claim that having a heart is logically equivalent to having kidneys. It simply happens factually that , as a matter of fact, the two sets have exactly the same elements, but a world in which an animal has a heart without having kidneys ( or the other way round) is still logically possible.

To put it briefly: set identity is logically equivalent to co-extensionality; but , by itself, coextensionality holds even when it is only factual or contingent. Hence the material conditional on the LHS.


问:
How did Bessel functions come to be denoted by $J_n$?
The $n$th Bessel function of the first kind is usually denoted $J_n(x)$.

Where did the use of the letter $J$ to indicate the Bessel function come from?

答:
" Bessel defined the function now known by his name by the
following definite integral:
$$\int \cos (h\epsilon-k \sin \epsilon) d\epsilon=2\pi I_k^h$$
where $h$ is an integer and the limits of integration are $0$ and $2\pi$. His
$I_k^h$ is the same as the modern $J_h(k)$, or rather $J_n$(x).

O. Schlomilch
following P. A. Hansen explained the notation $J_{\lambda±n}$ where $\lambda$ signifies
the argument and $\pm$ the index of the function. Schlomilch usually
omits the argument. Watson points out that Hansen and Schlomilch
express by $J_{\lambda,n}$ what now is expressed by $J_n(2\lambda)$. Schlafly marked it $^nJ(x)$. Todhunter uses the sign $J_n(x)$. $J_n(x)$ is known as the "Bessel function of the first kind of order $n$," while $Y^n(x)$, an allied function
introduced in 1867 by Karl Neumann is sometimes called "Newmann's Bessel function of the second kind of order $n$." It is sometimes marked $Y_n(x)$.

Watson says: "Functions of the types $J \pm (n+1/2)(z)$ occur with
such frequency in various branches of Mathematical Physics that
various writers have found it desirable to denote them by a special
functional symbol. Unfortunately no common notation has been
agreed upon and none of the many existing notations can be said to
predominate over the others. He proceeds to give a summary of
the various notations
in his Theory of Bessel Functions, pages 789, 790, Watson gives a
list of 183 symbols used by him as special signs pertaining to that
subject. "

This, in outermost quotes is a paragraph taken from Cajori's Book "A history of Mathematical Symbols" , clause 664, pg 279, vol-II explaining the symbol chosen for Bessel Function of the first kind. Hope it helps.
Cajori, F., A history of mathematical notations. Vol. II: Notations mainly in higher mathematics., XVIII + 367 p., 20 fig. Chicago, The Open Court Publishing Company (1929). ZBL55.0002.02.f


答:
$\mathbb{Z}/p^{\infty}$ is divisible, this implies $\hom(\mathbb{Z}/p^{\infty},\mathbb{Z})=0$.

We have $\mathbb{Z}/p^{\infty}  \cong \mathrm{colim}_n ~\mathbb{Z}/p^n$, hence $\hom(\mathbb{Z}/p^{\infty},\mathbb{Z}/p^{\infty}) \cong \lim_n  \hom(\mathbb{Z}/p^n,\mathbb{Z}/p^{\infty})$

Now $\hom(\mathbb{Z}/p^n,\mathbb{Z}/p^{\infty})$ is isomorphic to the $p^n$-torsion of $\mathbb{Z}/p^{\infty}$, which is isomorphic to $\mathbb{Z}/p^n$. Hence, $\hom(\mathbb{Z}/p^{\infty},\mathbb{Z}/p^{\infty}) = \mathbb{Z}_p$, the ring of $p$-adic integers.

Title: Homomorphisms between $ \mathbb{Z} $ modules.

Body:
Calculate $\newcommand\Hom{\operatorname{Hom}}\Hom(\mathbb Z \oplus \mathbb Z_{p^\infty},\mathbb Z \oplus \mathbb Z_{p^\infty})$. Where $ \mathbb Z_{p^\infty}= \bigcup_{k=1}^{\infty}\bar{\langle\frac1{p^k}\rangle}$ for $p$ prime.

As $\mathbb Z$ and  $\mathbb Z_{p^\infty}$ are $\mathbb Z$-modules then applying the theorem $\Hom_A ( \bigoplus_{i \in I} M_i, \prod_{j \in J} N_j ) \cong\prod_{(i,j) \in I \times J} \Hom(M_i,N_j)$ because for finite indices have $\bigoplus_{i=1}^{n}M_i = \prod_{i=1}^n M_1$. Then the problem reduces to find those $\Hom$ where

$$ \Hom(\mathbb{Z}, \mathbb{Z}) \cong \mathbb{Z} $$
$$ \Hom(\mathbb{Z},\mathbb{Z}_{p^\infty}) \cong \mathbb{Z}_{p^\infty}$$
but  $\Hom(\mathbb{Z}_{p^\infty},\mathbb{Z})$ and  $\Hom(\mathbb{Z}_{p^\infty},\mathbb{Z}_{p^\infty})$ not how to calculate them.


问:
The three-dimensional bundle underlying a quaternionic line bundle
Consider a principal $Sp(1)$-bundle $P$ over a compact space $X$, i.e. a quaternionic line bundle. We can push $P$ forward along the double cover $Sp(1) \cong SU(2) \to SO(3)$ to get a three-dimensional real oriented vector bundle $E$ over $X$.

Is there a nice interpretation of $E$ in terms of $P$? For example, can we say something about its characteristic class $w_2(E)$ in terms of $p_1(P)$?

答:
If the $SU_2$-bundle $P\rightarrow X$ is classified by a map $f:X\rightarrow BSU_2$, then the pushforward $E$ is classified by the composite $X\xrightarrow{f} BSU_2\xrightarrow{B\theta} BSO_3$, where $B\theta$ is the classifying map of the projection $\theta:SU_2\rightarrow SO_3$. Thus $P\cong f^*ESU_2$ and $E\cong(B\theta\circ f)^*ESO_3\cong f^*(B\theta^*ESO_3)$. Since $H^2(BSU_2;\mathbb{Z}_2)=0$ we get $w_2(B\theta^*ESO_3)=0$, so
$$w_2(E)=f^*w_2(B\theta^*ESO_3)=0.$$
On the other hand the Pontryagin class $p_1(E)$ is a more interesting computation. We'll proceed again by means of the universal example.

Notice that the double cover $SU_2\cong Spin_3\rightarrow SO_3$ is exactly the adjoint representation of $SU_2$. Thus if $Q$ is the real vector bundle associated to $B\theta^*ESO_3$, then we can write $Q\cong ESU_2\times_{Ad}\mathfrak{su}_2$, where $\mathfrak{su}_2$ is the Lie algebra of $SU_2$. Now $p_1(B\theta^*ESO_3)=p_1(Q)$, and $p_1(Q)=-c_2(Q^\mathbb{C})$ by definition, where $Q^\mathbb{C}$ is the complexification of $Q$. By the above this is $Q^\mathbb{C}\cong ESU_2\times_{Ad}(\mathfrak{su}_2)^\mathbb{C}$ and we'll return to this is a second. First observe that the Chern classes are stable, so if $\epsilon$ is a trivial complex line bundle, then $c_2(Q)=c_2(Q\oplus\epsilon)$.

Now, the complexification of $\mathfrak{su}_2$ is the complex Lie algebra $\mathfrak{sl}(2;\mathbb{C})$ of traceless $2\times2$ complex matrices, and there is an isomorphism $\mathfrak{sl}(2;\mathbb{C})\oplus\mathbb{C}\cong\mathfrak{gl}(2;\mathbb{C})\cong Hom(\mathbb{C}^2,\mathbb{C}^2)$, where the trivial $\mathbb{C}$ factor corresponds to the central subgroup of $\mathfrak{gl}(2;\mathbb{C})$ generated by complex multiples of the identity. This gives us
$$Q^\mathbb{C}\cong ESU_2\times_{Ad}\mathfrak{sl}(2;\mathbb{C}),\qquad\qquad Q^\mathbb{C}\oplus\epsilon\cong Hom(EU_2,EU_2)|_{BSU_2}\cong \overline{EU_2}\otimes EU_2|_{BSU_2}$$
where we identify the conjugate and dual bundles of $EU_2$ with the aid of a suitable Hermitian metric.

The point is that we can compute $c_2(Q^\mathbb{C})$ by computing $c_2(\overline{EU_2}\otimes EU_2)$ and resticting from $BU_2$ to $BSU_2$, and this is not so difficult to do. Indeed, introduce a formal factorisation of the total Chern class by writing $c(EU_2)=\prod^2_{i=1}(1+x_i)$. Then $c(\overline{EU_2})=\prod^2_{i=1}(1-x_i)$ and
$$c(\overline{EU_2}\otimes EU_2)=\prod_{i,j=1,2}(1+(x_i-x_j))=\prod_{1\leq j<i\leq 2}(1-(x_i-x_j)^2)=1-(x_1+x_2)^2+4x_1x_2.$$
The formulas here are lifted straight from Hirzebruch's book Topological Methods in Algebraic Geometry, Th. 4.3.3, pg. 64. In any case this gives us $c_1(\overline{EU_2}\otimes EU_2)=0$ and $c_2(\overline{EU_2}\otimes EU_2)=-c_1^2+4c_2$. Pulling this back to $BSU_2$ kills $c_1^2$ and we obtain
$$p_1(Q)=-c_2(Q^\mathbb{C})=-c_2(Q^\mathbb{C}\oplus\epsilon)=-4c_2.$$

Finally returning to the $SO_3$bundle $E\rightarrow X$ we can compute
$$p_1(E)=f^*p_1(Q)=-4\cdot f^*c_2=-4\cdot c_2(P)$$
where of course $P\rightarrow X$ is the $SU_2$-bundle we started with.


问:
Proof $-(x+y) = (-x) + (-y) = -x + -y$
This was left as an exercise in Apostol's Calculus Vol II and I'm not very sure how to proceed and have my doubts on whether the following would be sufficient due to step $(1)$:
$$-(x+y) \stackrel{(1)}{=} -1(x+y) \stackrel{(2)}{=} -1x + -1y \stackrel{(3)}{=} (-x) + (-y) = -x-y$$

Where $(2)$ is allowed by the distributive multiplication property for addition in V and $(3)$ is allowed because $(-a)x = -(ax)$, which is proven by:

Let $z = (-a)x$
$$ z + ax = (-a)x + ax \stackrel{(a)}{=} (-a + a)x = 0x \stackrel{(b)}{=} O $$
Where $(a)$ is allowed by the distributive multiplication property for scalars addition and $(b)$ is true because $0x = O$ (I'm not providing this proof here, but it can be assumed already proven). We find that $z$ is the negative of $ax$ and so $z = -(ax)$.

Would this be a sufficient proof?

答:
You can simply say

-(x+y)=-(x+y)+x-x+y-y=-(x+y)+(x+y)-x-y (since we add x-x and y-y there is no problems)

which then would imply -(x+y)=-x-y

Title: Proof $-(x+y) = (-x) + (-y) = -x + -y$


问:
Alternative ways to say "if and only if"?
There are some scenarios about which I would like to get some confirmation:

when defining a concept A,

We call A, if ... [definition of
  concept A]

Does "if" here mean equivalence
instead of just sufficiency? Is it
incorrect to replace "if" with "if
and only if"?

For precisely what condition is
   B satisfied?

Does "precisely" mean asking for
necessary and sufficient condition
for B?

Some other ways to say "if and only
if" / "necessary and sufficient"?

Some references that summarize some standard terminologies in Mathematics  such as this one?

Thanks and regards!

答:
It distracts the reader to use 'if and only if' in a definition, because it's so obvious that 'only if' is implied.

Definition: We say that $S$ is a snargle if all its corticles are open.

None of us would ask: "What if some of its corticles are not open? Is $S$ still a snargle?" We are not politicians, or lawyers. (I'm not, anyway.)

The phraseology 'if and only if' is best reserved to statements of lemmas, propositions, and theorems, where it plays an important role.


问:
integer programming with indicators
I have the following question, and I need to write it as an integer programming problem:

A manager of a company wants to give presents to his 1000 employers. He can buy the presents from two different suppliers:

Every present costs 110\$. For any present above 500, the manager would have to pay 5\$ extra for insurance (for example, for 502 presents, there will be 10\$ extra for insurance).

Every present costs 120$. For any present above 300, there is a discount of 30% (each present above 300 costs 84\$).

I need to find how many presents he should buy from each supplier, in order to spend the minimal amount of money.

I defined:

$x_1, x_2$ - The number of presents to buy from each supplier.

$z_1$ - Indicator which represents whether or not $x_1 \ge501$

$z_2$ - Indicator which represents whether or not $x_2 \ge301$

I know how to represents the indicators using linear functions, but I don't know how to find a linear objective function.

My best suggestion is:

$$ 110x_1+5z_1(x_1-500)+120x_2-36z_2(x_2-300) $$

However, this is not a linear function, because I multiply the decision variables.

How can I turn it into a linear function?

答:
Introduce nonnegative integer decision variables $y_1$ and $y_2$ to represent "extra" presents.  The problem is to minimize
$$110x_1+5y_1+120x_2-36y_2$$
subject to
\begin{align}
x_1 &\le 500 + y_1 \tag1 \\
y_2 &\le (1000-300) z_2 \tag2 \\
x_2 &\ge 300 z_2 \tag3
\end{align}
Constraint $(1)$ and the objective enforce $y_1 = \max\{x_1-500, 0\}$.  You don't need $z_1$.
Constraint $(2)$ enforces $y_2 > 0 \implies z_2 = 1$.
Constraint $(3)$ enforces $z_2 = 1 \implies x_2 \ge 300$.

Title: integer programming with indicators


问:
Probability that at least one Ace is dealt in poker with $6$ players
If two cards are dealt to each of the 6 players from a 52 deck of cards what is the probability that there is at least one ace dealt?

I've tried to solve this using the following reasoning:

In total there are $ {52 \choose 12 } \cdot {12 \choose 6}$ ways to deal 12 cards to the 6 players.

Count the number of ways to deal with at least one ace.

To count 2 notice that there is in general ${4 \choose N} \cdot 48 \cdot 47 \cdots (48 - (12 - N)) $ ways to choose $N$ aces in the $12$ dealt cards. For each of these combinations there are also ${12 \choose 6}$ ways to distribute two cards to each player. Thus in general the probability would be, following the above reasoning:

$$ P(\text{at least one A}) = \frac{{4 \choose N} \cdot 48 \cdot 47 \cdots (48 - (12 - N))}{ {52 \choose 12 } \cdot {12 \choose 6}^2 }$$

However, this does not give a sensible answer.

Can anyone see where the flaw in the above reasoning is?

Edit:

Thanks to those who have answered below - I can see that taking the inverse of the question by asking how to deal no aces is much easier. However, can anyone show me how to do it by count the positive cases as I have tried above?

答:
The number $X$ of Aces in twelve draws from a standard
deck has a hypergeometric distribution.

You can make a probability table in R as shown below.
(Ignore row numbers in [ ]s.)

```<code>x = 0:4;  pdf = dhyper(x, 4,48, 12)
cbind(x, pdf)

     x         pdf
[1,] 0 0.337575030
[2,] 1 0.437935174
[3,] 2 0.190156062
[4,] 3 0.032505310
[5,] 4 0.001828424
</code>```

This matches @RobPratt's Answer (+1):

```<code>phyper(0, 4,48, 12)
[1] 0.337575
choose(48,12)/choose(52,12)
[1] 0.337575
1 - 2759/4165
[1] 0.337575
</code>```

Graph of distribution of $X:$

```<code>plot(x, pdf, type="h", lwd=3,
     main="Hypergeometric Dist'n of Aces")
 abline(h=0, col="green2")
 abline(v=0, col="green2")
</code>```

Body:
The number $X$ of Aces in twelve draws from a standard
deck has a hypergeometric distribution.

You can make a probability table in R as shown below.
(Ignore row numbers in [ ]s.)

x = 0:4;  pdf = dhyper(x, 4,48, 12)
cbind(x, pdf)

     x         pdf
[1,] 0 0.337575030
[2,] 1 0.437935174
[3,] 2 0.190156062
[4,] 3 0.032505310
[5,] 4 0.001828424

This matches @RobPratt's Answer (+1):

phyper(0, 4,48, 12)
[1] 0.337575
choose(48,12)/choose(52,12)
[1] 0.337575
1 - 2759/4165
[1] 0.337575

Graph of distribution of $X:$

plot(x, pdf, type="h", lwd=3,
     main="Hypergeometric Dist'n of Aces")
 abline(h=0, col="green2")
 abline(v=0, col="green2")


问:
Are there non-Hausdorff spaces that satisfy the Galois correspondence for covering spaces?
The Galois correspondence for covering spaces requires that a space $X$ be path-connected, locally path-connected, and semilocally simply connected. These seem like pretty strict requirements, so I was wondering how pathological a space can be while still fulfilling them. More specifically, I was wondering if anyone had an example of a space $X$ fulfilling all three that is not Hausdorff.

Is this even possible? Could we go further and find a space fulfilling these conditions which is not $T_1$?

答:
These conditions are pretty orthogonal to the separation axioms.  For instance, any nonempty indiscrete space satisfies these properties.  Indeed, any map to an indiscrete space is continuous, so any indiscrete space is contractible (every homotopy you might possibly define is continuous), and locally contractible.

For a less trivial example, consider the space $X=\{a,b,c,d\}$, with the topology generated by the sets $\{a\},\{b,a,c\},\{c\},$ and $\{d,a,c\}$.  Each of these basic open sets is contractible: to see that $\{b,a,c\}$ is contractible, for instance, consider the map $H:\{b,a,c\}\times[0,1]\to\{b,a,c\}$ given by $H(x,t)=x$ for $t<1$ and $H(x,1)=b$ (exercise: check that this is continuous).  It follows that $X$ is path-connected and locally contractible, so the usual theory of covering spaces applies to $X$.  Interestingly, $X$ is not simply connected: in fact, $X$ is weak homotopy equivalent to $S^1$, by a map $S^1\to X$ that sends two points to $b$ and $d$ and the two open arcs between them to $a$ and $c$.  So $X$ has a lot of interesting covering spaces.  For more discussion see Asphericity of pseudo circle (where I construct the universal cover of $X$) and An interesting topological space with $4$ elements.


问:
Picking a point on a sphere uniformly at random
I wanted to read a bit about the Borel-Kolmogorov paradox on Wikipedia, but I don't understand the following: to me it seems that picking a point uniformly at random on the sphere amounts to pick a longitude $\lambda$ uniformly at random from $[-\pi,\pi]$ and picking a latitude $\phi$ uniformly at random from $[-\pi/2,\pi/2]$. However, on Wikipedia it's written we pick the latitude $\phi$ from $[-\pi/2,\pi/2]$ with density $\frac 1 2 \cos(\phi)$. Why is that?

答:
A picture is worth a thousand words:

The density in the "naive" approach is proportional to $\frac{1}{\cos \phi}$...


问:
Is second order logic even a logic?
Second order logic is a language, but, is it a logic?

My understanding is that a logic (or "logical system") is an ordered pair; it is a formal system together with a semantics. However, the language of second-order logic is associated with a variety of inequivalent formal systems and a variety of semantics. So, is it a logic?

I think not. Supposing not, what do sentences like the following (copied from wikipedia) even mean?

Second-order logic, for example, does not have a completeness theorem
  for its standard semantics (but does have the completeness property
  for Henkin semantics).

答:
This is a very interesting philosophical question! There are philosophers on both sides of the issue. A prominent figure who spoke against SOL as logic was W.V.O. Quine. His argument mostly focuses on the point that SOL, which quantifies over sets explicitly, seems to just be "set theory in disguise", and hence not logic proper (see his Philosophy of Logic for more details). People have also put forth the argument you presented, viz. any putative logic must have a sound and complete deduction system in order to be a genuine logic, and since SOL doesn't have one, it follows that it can't be a logic.

On the other side, a prominent defender of SOL as logic proper is George Boolos. See his "On Second-Order Logic" and his "To Be is to Be the Value of a Variable (Or to Be Some Values of Some Variables)". One argument put forward in favor of SOL from a practical standpoint is that it has a way of really shortening proofs (see "A Curious Inference"). Boolos also suggests that we could use monadic SOL to model talk of plurals, i.e. to model sentences like "Some critics only admire one another". Another great defender of SOL is Stewart Shapiro, whose book Foundations Without Foundationalism not only presents a wonderful technical treatment of SOL but also some good philosophical defenses of it. In particular, Shapiro argues that SOL is needed for everyday mathematical discourse. I would highly recommend this book as an introduction to the subject, both mathematically and philosophically.

The status of completeness theorems in this debate is fascinating, but difficult to pin down. After all, many philosophers might just say "Look, so what if SOL is incomplete? That just means that the set of validities doesn't coincide with the set of provable sentences, i.e. there are some sentences which are logically true, but not provable. And while this is unfortunate, why shouldn't we expect this? Why shouldn't we just accept that logic didn't turn out the way we hoped?" On the other hand, many philosophers might say "Look, logic is about making inferences. If there are truths which don't come from some deduction or inference, then how could they be truths of logic?"

I personally don't find the view that genuine logics must have complete proof systems convincing (or at least convincing enough), but I certainly do feel the pull to search for/work with logics with complete proof systems. Ultimately, it comes down to what you think "logic" and "logical consequence" are, which are highly contentious matters in the philosophical literature, and which have no widely accepted answers.

Title: Is second order logic even a logic?


问:
Linearizing the product of a binary and a continuous variable
I have an MIP optimization problem that has a constraint $p\geq xy$, where $x$ is a binary variable, $p$ and $y$ are non-negative continuous variables.

I tried the Big-M method. However, the upper bound for $y$ is quite large ($M \geq 10^{15}$). All the MIP solver I tried have numerical problems and cannot solve the problem correctly. I tried to set the solver parameters and the numerical problem still exists.

Since the value range of $y$ is from $10^{-5}$ to $10^{14}$, it seems the values cannot be scaled directly in the problem.

Is there any other way to linearize the product $xy$ which can avoid numerical problems?

答:
Some solvers allow you to directly express indicator constraints $x=1 \implies p \ge y$.  Behind the scenes the solver might introduce big-M constraints or might branch on the indicator constraints.  In the latter case, the constraint is ignored in the root node linear programming relaxation (weakening the relaxation but avoiding numerical difficulties from big-M); if the constraint is violated (that is, if $x=1$ and $p < y$), the $x=1$ branch imposes linear constraint $p \ge y$.

Alternatively, if you have only one such constraint, just solve two problems, effectively doing this branching yourself.


答:
Visualize the points on the sphere given by every combination of $\lambda \in\{\pi, \pi -\frac{2\pi}{n}, \pi - 2\frac{2\pi}{n}, ..., -\pi + \frac{2\pi}{n}\}$ and $\phi\in\{\frac{\pi}{2},\frac{\pi}{2}-\frac{\pi}{m}, \frac{\pi}{2} - 2\frac{\pi}{m},...,-\frac{\pi}{2} + \frac{\pi}{m}\}$. According to your hypothesis, these points are distributed uniformly across the sphere (i.e, equal density per unit area). However, as you approach the poles, the latitudinal distance between the points will remain constant, but the longitudinal distance will decrease, meaning there will be more point density at the poles.

To offset this, we will need to make points more sparse towards the poles. It is not hard to see that the number of points on a given circle of latitude will have to be proportional to the circumference of that circle. However, a circle at latitude $\phi$ has circumference $2\pi r\cos(\phi)$ (where $r$ is the radius of the sphere), leading to the factor of $\cos(\phi)$ in the latitude distribution.

Title: Picking a point on a sphere uniformly at random

Body: I wanted to read a bit about the Borel-Kolmogorov paradox on Wikipedia, but I don't understand the following: to me it seems that picking a point uniformly at random on the sphere amounts to pick a longitude $\lambda$ uniformly at random from $[-\pi,\pi]$ and picking a latitude $\phi$ uniformly at random from $[-\pi/2,\pi/2]$. However, on Wikipedia it's written we pick the latitude $\phi$ from $[-\pi/2,\pi/2]$ with density $\frac 1 2 \cos(\phi)$. Why is that?


问:
Does the "field" over which a vector space is defined have to be a Field?
I was reviewing the definition of a vector space recently, and it occurred to me that one could allow for only scalar multiplication by the integers and still satisfy all of the requirements of a vector space.

Take for example the set of all ordered pairs of integers. Allow for scalar multiplication over the integers and componentwise vector addition as usual. It seems to me that this is a perfectly well-defined vector space.

The integers do not form a Field, which begs the question: Is there any reason that the "field" over which a vector space is defined must be a mathematical Field? If so, what is wrong with the vector field I attempted to define above? If not, what are the requirements for the scalars? (For instance, do they have to be a Group - Abelian or otherwise?)

答:
While the other answers (and comments) implicitly address the question stated in the title of the OP, I thought it may be useful to include an explicit answer, as well.

Does the “field” over which a vector space is defined have to be a Field?

Yes, a vector space is defined over a field; i.e. if the scalars do not refer to a field, the resulting object is not by definition a vector space.

For completeness: as pointed out in the other answers and comments, there are objects with analogous definitions, in the case that the scalars belong to a ring (as in the example provided in the OP), and the other axioms are met, the resulting object is called a "module." As the subsection on modules on the Wikipedia vector space page says (emphasis added):

Modules are to rings what vector spaces are to fields: the same axioms, applied to a ring R instead of a field F, yield modules. The theory of modules, compared to that of vector spaces, is complicated by the presence of ring elements that do not have multiplicative inverses. For example, modules need not have bases, as the Z-module (that is, abelian group) Z/2Z shows; those modules that do (including all vector spaces) are known as free modules. Nevertheless, a vector space can be compactly defined as a module over a ring which is a field, with the elements being called vectors. Some authors use the term vector space to mean modules over a division ring.[105] The algebro-geometric interpretation of commutative rings via their spectrum allows the development of concepts such as locally free modules, the algebraic counterpart to vector bundles.


问:
Coloring R2 with two colors such that there exists no monochromatic curve
Suppose you have two colors, red and blue. Find a coloring scheme for points in $R^2$ so that there is no line segment or curve in the plane, no matter how short, that is all one color.

I believe there is a theorem that shows this cannot be done (i.e. any coloring scheme will have some line segment or curve that is all one color), but I don't know what the theorem is or the logic behind it.

答:
Assuming the Axiom of Choice, it can be done.

The cardinality of the  continuous curves in the plane (single points excepted) is $\mathfrak c$, the cardinality of the continuum.  Let these be indexed as $C_\alpha$, $\alpha < \beta$, where $\beta$ is the first ordinal of cardinality $\mathfrak c$.  By transfinite induction, we produce disjoint subsets $A = \{a_\alpha: \alpha < \beta\}$ and $B = \{b_\alpha: \alpha < \beta\}$ of the plane such that $a_\alpha \in C_\alpha$ and $b_\alpha \in C_\alpha$ for each $\alpha$.  Colour $A$ white and $B$ black; any points not in $A \cup B$ can be coloured arbitrarily.  Then there are no monochromatic curves.


问:
Showing a sequence of real numbers converges to a supremum
I am trying to prove that any non-empty set $S \subset \mathbb{R}$ which is bounded above has a sequence of numbers $s_n \in S, n = 1, 2, 3, ...$ such that $s_n \rightarrow sup(S)$

My proposed method is to create the set of upper bounds and list them in reverse order i.e. largest upper bound is denoted $u_1$ where $u_1 > u_2>...>sup(S)$. And we list the elements of S in ascending order, i.e. $s_1 <s_2 <  ...<s_n$. Since $|u_n|>|s_n|$ for some $s \in S$ we can say that as $n \rightarrow \infty$ $ u_n \rightarrow supS$ and due to the previous inequality we get that $s_n \rightarrow supS$.

My only problem with the proof is that I have not used the exact formal definition of when a sequence of numbers converges to a value. Any suggestions?

答:
If $s=\sup S$, then for every $n\in\mathbb N$, then number $s-1/n$ is not an upper bound of $S$, since it is smaller than the least upper bound $s$. Thus there exists an $s_n\in S$,
such that
$$
s-\frac{1}{n}<s_n\le s.
$$
Clearly the sequence $\{s_n\}\subset S$ converges to $s=\sup S$.

Note. If $s=\sup S\not\in S$, then the sequence $\{s_n\}\subset S$ can be constructed to be strictly increasing.


答:
You cannot list the upper bounds: there are uncountably many of them. Let $a=\sup S$. You should now distinguish two cases: $a\in S$, and $a\notin S$. If $a\in S$, there is a trivial sequence in $S$ converging to $a$; if you don’t see it immediately, consider Tito Eliatron’s example in the comments.

If $a\notin S$, you will have to use the definition. Use the fact that $a=\sup S$ to show that for each $n\in\Bbb Z^+$ there is an $s\in S$ such that $a-s<\frac1n$, and then use this fact to get a sequence in $S$ converging to $a$.

Title: Showing a sequence of real numbers converges to a supremum

Body:
I am trying to prove that any non-empty set $S \subset \mathbb{R}$ which is bounded above has a sequence of numbers $s_n \in S, n = 1, 2, 3, ...$ such that $s_n \rightarrow sup(S)$

My proposed method is to create the set of upper bounds and list them in reverse order i.e. largest upper bound is denoted $u_1$ where $u_1 > u_2>...>sup(S)$. And we list the elements of S in ascending order, i.e. $s_1 <s_2 <  ...<s_n$. Since $|u_n|>|s_n|$ for some $s \in S$ we can say that as $n \rightarrow \infty$ $ u_n \rightarrow supS$ and due to the previous inequality we get that $s_n \rightarrow supS$.

My only problem with the proof is that I have not used the exact formal definition of when a sequence of numbers converges to a value. Any suggestions?


问:
How to convert a differential equation to an exact form with use of an integrating factor?
I need to solve the differential equation $1+(x/y-\sin y)y'=0$.
The equation is not exact so I try to use an integrating factor to make it exact.

$u(x) =$ integrating factor, only dependent on $x$ (because the my book told me to do it like this to make it easier)
$$M(x,y) = u(x)$$ and $$N(x,y) = u(x)(1+(x/y-\sin y))$$
The answer for $u(x) = y$.
The integrating factor is visible just by inspection but I would like to see how to derive the integrating factor.

I end up with the differential equation $$\frac{u'(x)}{u(x)}= -\frac{1}{x-y\sin y}$$
from here I don't know how to get to $u(x) = y$?

ps. I don't know how to write the equations nicely. If someone wants to send me a link here they explain how to do it, I'll write my next question as a nice formula.

答:
$\displaystyle 1 + (\frac{x}{y} - \sin y) y'= 0$.

Actually integrating factor that you have is $u(y)$ (you can try with both, first with $u(x)$ and if that doesn't work, with $u(y)$)

$\displaystyle u(y) + u(y)(\frac{x}{y} - \sin y) y'= 0$

$\displaystyle P = u(y), Q = u(y)(\frac{x}{y} - \sin y)$

$\displaystyle P_y = \frac{\partial u}{\partial y}, Q_x = \frac{u(y)}{y}$

To make ODE exact, $P_y = Q_x$ so $u(y) = y$ is a possible integrating factor.


问:
Prove that $AM = MF$ where $M$ is the intersection between the perpendicular of circle $ω$ and line $AF$

Given is circle $ω$ and line $k$ disjoint with circle $ω$. Point $M$ is
the foot of the perpendicular from the center of the circle onto line $k$. A line
intersects line $k$ at point $A$ and circle $ω$ at points $B$ and $C$. Lines $\overline{BM}$ and
$\overline{CM}$ intersect circle $ω$ for the second time at points $D$ and $E$. Line $\overline{DE}$ meets
line $k$ at point $F$. Prove that $\overline{AM} = \overline{MF}$.

Once again, I've been assigned another set of geometry problems, and this is one of the harder ones that I haven't been able to figure out yet. I've asked my instructor what I could use for this problem, and he told me that Power of a Point would be useful here. But so far, I've tried applying Power of a Point, but I'm not getting anywhere with the few points that currently exist. Can anyone help out?

答:
Indeed, this is just one of generalizations of the Butterfly theorem (see my comment above).

Consider point $B'$ and $C'$ such that $M$ is the midpoint of the segments $BB'$ and $CC'$. Clearly, $BCB'C'$ is a parallelogram with center at $M$. Thus, in order to prove that $AM=MF$ (that is, $A$ and $F$ are symmetric with respect to $M$), we only need to prove that $B'C'$ passes through $F$ (since $A\in BC$).

Firstly, it's easy to see that $B'C'DE$ is a cyclic quadrilateral (because $BC\parallel B'C'$ and $BDCE$ is cyclic).

Secondly, denote the circle symmetric to $\omega$ with respect to $M$ as $\omega'$. Then, $B',C'\in\omega'$ and $k$ is the radical axis of $\omega$ and $\omega'$ (due to the symmetry; note that $OM\perp k$). As we have proved before, the quadrilateral $B'C'DE$ is cyclic. Thus, the radical axes of these three circles are concurrent, i. e. lines $DE$, $k$ and $B'C'$ are concurrent.

Finally, recall that $F=DE\cap k$, so $F\in B'C'$, as desired.


问:
Evaluating limit with $\frac{1}{x}$ exponent
How do you evaluate this limit without L'hopital rule?

$$\lim_{x \rightarrow \infty} \frac{1}{x}\log((e^{\pi x} -1)$$

From that: $\lim_{x \rightarrow \infty} \log((e^{\pi x} -1)^{\frac{1}{x}})$ but now I'm really stuck.

答:
Notice that
$$\log(e^{\pi x}-1)=\log\left(e^{\pi x}\left(1-\frac{1}{e^{\pi x}}\right)\right)=\log(e^{\pi x})+\log\left(1-\frac{1}{e^{\pi x}}\right)=\pi x+\log\left(1-\frac{1}{e^{\pi x}}\right)$$


问:
Fréchet Derivative of $\frac{1}{\|x\|}$
In a Hilbert Space $V$, for function $f:V\to\mathbb{R}$, if $f$ is Fréchet differentiable at $x_0$, the Fréchet Derivative $\nabla f(x_0)$ is  $v$ such that $$ \lim_{x\to x_0} \frac{|f(x) - f(x_0) - \langle v, x-x_0 \rangle|}{\|x-x_0\|} = 0$$

As an example, for $f(x) = \|x\|$, the Fréchet derivative is $\nabla f(x) = \frac{x}{\|x\|}$, as shown in this Wikipedia proof.

Now my question is, what is the Fréchet derivative of $f(x) = \frac{1}{\|x\|}$. My gut says something like $-\frac{x}{\|x\|^3}$ following normal derivative rules, but I can't seem to be able to solve it.

答:
There are two ways to solve this.

The first way is to use the chain rule.
This requires the knowledge that a chain rule holds also in Hilbert spaces for Fréchet
derivatives and not just in $\mathbb R^n$.
Or you could just prove the chain rule yourself.

The second way is to guess the Fréchet derivative (you already have a conjecture)
and then try to verify that the $\lim$ in the definition is indeed $0$.

Title: Fréchet Derivative of $\frac{1}{\|x\|}$


答:
We have this density on the latitude essentially because there is more land near the equator, and thus we want to give the latitudes near the equator more weight.

To find the density yourself, I like to use $[0,\pi]$ for the latitude coordinate, where $\phi = 0$ or $\phi = \pi$ corresponds to the $z$-axis, since it corresponds with the polar coordinate system I was taught in multivariable calculus. Then you'll recall from multivariable calculus that $\rho^2\sin\phi\,d\rho\,d\phi\,d\lambda$ is the volume element of $\mathbb R^3$, written in these coordinates, where $(\rho,\phi,\lambda)\in (0,\infty)\times[0,\pi]\times [-\frac\pi2,\frac\pi2]$.

Evaluating at $\rho = 1$ gives us $\sin\phi\,d\phi\,d\lambda$. Dividing this by $4\pi$, which is the area of the sphere, we get the uniform distribution on the sphere: $\frac{1}{4\pi}\sin\phi\,d\phi\,d\lambda = (\frac{1}{2}\sin\phi\,d\phi)(\frac{1}{2\pi}\,d\lambda)$. Or in other words, to sample a point $(\phi,\lambda)$ uniformly at random from the sphere, we sample the latitude $\phi$ from $[0,\pi]$ from a distribution with density $\frac{1}{2}\sin\phi$ and we sample the longitude $\lambda$ from $[-\pi,\pi]$ uniformly, i.e., from a distribution with density $\frac{1}{2\pi}$.

When $\phi \approx \pi/2$, which corresponds to points near the equator, $\sin\phi\approx 1$, so these points have the most weight.

When $\phi\approx 0$, or $\phi\approx \pi$, $\sin\phi\approx 0$, so these points have much less weight.

Both of the last two bullet points make sense if you think about how there is so much more land in a latitudinal strip of width $d\phi$ near the equator, versus near the poles.


问:
Proof about the kernel of a matrix
Prove or disprove that if $\ker A \neq \{ 0\}$ where $A \in \mathbb{K}^{n\times m}$, then $\ker (BA) \neq\{ 0\}$ for all Matrices $B \in \mathbb{K}^{k\times n}$

I proved that this does not hold for $\ker A = \{ 0\}$, but I am struggling here.

Thanks in advance.

答:
If the kernel of $A$ is not just the zero vector, $\vec{0}$, that means there is a nonzero vector $v$ such that $Av=\vec{0}$. Now check what happens to $(BA)v$.

$(BA)v= B(Av)= B\vec{0} = \vec{0}$.

Title: Proof about the kernel of a matrix


答:
Here is what I did.

Let $\vec{k} \in \ker A/\{0\}$

We have $A\vec{k} = \vec{0}$. We simply right multiply by $B$. We now get $BA\vec{k} = B\vec{0}$, giving us $BA\vec{k} = \vec{0}$. Hence $\vec{k} \in \ker(BA)$ and by the definition of $\vec{k}$ we have that $\ker(BA) \neq \{0\}$

Title: Proof about the kernel of a matrix

Body:
Prove or disprove that if $\ker A \neq \{ 0\}$ where $A \in \mathbb{K}^{n\times m}$, then $\ker (BA) \neq\{ 0\}$ for all Matrices $B \in \mathbb{K}^{k\times n}$

I proved that this does not hold for $\ker A = \{ 0\}$, but I am struggling here.

Thanks in advance.


问:
Commuting square of functors
Let $\mathcal{E}$ be a complete and cocomplete category. Given a functor $i: \mathcal{C} \to \mathcal{D}$ between small categories, there is a triple of adjoint functors between their respective categories of presheaves with values in $\mathcal{E}$:
$$Lan_i =: i_!: [\mathcal{C}, \mathcal{E}] \to [\mathcal{D}, \mathcal{E}]$$
$$(-)\circ i =: i^\ast: [\mathcal{D}, \mathcal{E}] \to [\mathcal{C}, \mathcal{E}]$$
$$Ran_i =: i_\ast: [\mathcal{C}, \mathcal{E}] \to [\mathcal{D}, \mathcal{E}]$$

Suppose we are further given functors $p: \mathcal{C} \to \mathcal{C}$ and $q: \mathcal{D} \to \mathcal{D}$ such that $i \circ p = q \circ i$. It is then obvious that $p^\ast \circ i^\ast\simeq i^\ast \circ q^\ast$. Using the unit $\mathrm{id}_\mathcal{C} \to i^\ast \circ i_!$ and counit $i_! \circ i^\ast \to \mathrm{id}_\mathcal{D}$ of the first adjunction, we obtain a natural transformation of functors $\mathcal{C} \to \mathcal{D}$:
$$i_! \circ p^\ast \to i_! \circ p^\ast \circ i^\ast \circ i_! \simeq i_! \circ i^\ast \circ q^\ast \circ i_! \to q^\ast \circ i_!$$
Is this a natural isomorphism in general? What about if $i$ is fully faithful (in which case the unit $\mathrm{id}_\mathcal{C} \to i^\ast \circ i_!$ is a natural isomorphism)?

答:
You are asking about the Beck–Chevalley conditions for Kan extensions, which is a fairly non-trivial question. Here is one generic situation where things work. Let $u : \mathcal{A} \to \mathcal{C}$ and $v : \mathcal{B} \to \mathcal{C}$ be functors between small categories, let $(u \downarrow v)$ be the comma category, let $p : (u \downarrow v) \to \mathcal{A}$ and $q : (u \downarrow v) \to \mathcal{B}$ be the projections, and let $\theta : u p \Rightarrow v q$ be the canonical natural transformation. Then, the Beck–Chevalley transformations
$$q_! v^* \Rightarrow q_! p^* u^* u_! \Rightarrow q_! q^* v^* u_! \Rightarrow v^* u_!$$
$$u^* v_* \Rightarrow p_* p^* u^* v_* \Rightarrow p_* q^* v^* v_* \Rightarrow p_* q^*$$
are natural isomorphisms. This is Proposition 4.1.19 in my notes.

More generally, there is a notion of exact square, and there is an exact square condition characterising fully faithful functors.

Title: Commuting square of functors


问:
Notation for fields and the elements of the sets used to define them.
Context:

While trying to figure out how to prove that the characteristic of a field $\mathcal{F}$ is either $0$ or prime, I realized that I was conflating the elements of the field with integers, because of the notation.

Question:

Because articulating this question abstractly is beyond my ability, I'll illustrate my confusion with an example:

If $\mathbb{Z}_5$ is the set $\{0,1,2,3,4\}$, along with modular addition and multiplication (i.e. in this case, after performing the standard addition or multiplication, the least remainder after ordinary division by 5; e.g. $4\cdot4 = 1$ and $3+4 = 2$) defines a field. Here,
$$
4 + 1 = 0,
$$
i.e. $1$ is the additive inverse of $4$.

Now, all the classical definitions I've seen usually define the additive inverse as follows (this example is taken from Wikipedia):

for every a in F, there exists an element in F, denoted −a, called the additive inverse of a, such that a + (−a) = 0.

Does the phrase "denoted -a" imply that, in the example provided above, we should denote "$1 = -4$"? Or are such definitions simply stating what they mean when they use the symbol (i.e. it is simply saying we should think of $1$ as the additive-inverse-of-4)? In other words, is "$-4$" an element of this field, even if it is not an element of the set used to define the field (does it even make sense to talk of "an element of a field" in this way)?

Just to be clear, if we were to instead denote the additive inverse of a by na, is it then appropriate to write something like $1 = n4$, for the above field? While I can understand what is meant by this, what bothers me is that $n4$ is not an element of any set described or defined, elsewhere.

Final Considerations

Up until here, I'd be tempted to more or less ignore the whole issue; however, while trying to work out a proof that a field's characteristic is either zero or prime, I found myself wanting to use integers to describe the operations, regardless of whether or not the elements of the set were, themselves, numbers. For example, given the field $\mathbb{F}$ defined by:
$$
\left\{
  \begin{array}{lll}
    \mathbb{S} =  \{a, b\},
    &
     \begin{array}{c|c|c}
       \pmb{+} & \pmb{a} & \pmb{b} \\ \hline
       \pmb{a} & a       & b       \\ \hline
       \pmb{b} & b       & a
     \end{array},
    &
     \begin{array}{c|c|c}
       \pmb{\cdot} & \pmb{a} & \pmb{b} \\ \hline
       \pmb{a} & a       & a       \\ \hline
       \pmb{b} & a       & b
     \end{array},
  \end{array}
\right.
$$

it seems natural and useful to define an expression such as $a + a + a$ as $3a$; however, I'm not sure how to think about the relationship of the symbols introduced by this notation to the field itself (this becomes even more confusing if I use "1" to denote the multiplicative identity). This seems -- at least to my confused mind -- somehow related to the confusion expressed in the first example, above.

Is the use of 3 in "3a" and -4 in "1 = -4" nothing more than a convenient notation to describe the elements of the field? While I feel like that is all they are, I also feel that they carry with them some sort of implied structure (perhaps that of an abelian group?), that isn't accounted for, formally / explicitly.

答:
The notation $-a$ means "the element $b$ that fits the equation $a + b = 0$". So in the integers modulo $5$, $-1 = 4$, $-2 = 3$, $-3 = 2$, $-4 = 1$, and $-0 = 0$.

It is perfectly fine to write $na$, for a positive integer $n$, to mean the sum of $a$ with itself $n$ times.  There is no inconsistency between the "$3$" in $3a$ as $a + a + a$ and the "$3$" in $3a$ as the product of the element $3$ in the field and $a$ because
$$
a + a + a = 1\cdot a + 1 \cdot a + 1 \cdot a = (1 + 1 + 1)\cdot a = 3a.
$$

By the way, it is a terrible idea to limit your viewpoint of the integers modulo 5 as being only the symbols $0,1,2,3,4$ with a wraparound addition and multiplication rule modulo $5$.  You should accept $7$, $11$, and $-46$ as perfect reasonable integers modulo $5$ that happen to be the same thing in the integers modulo $5$ as $2$, $1$, and $4$, respectively.

After all, you should be perfectly comfortable with writing the same fraction in many ways: $1/2 = 2/4 = 3/6 = (-6)/(-12)$, and so on.  It would be extremely inconvenient if you were only allowed to work with fractions in reduced form (try to define addition and multiplication when the inputs and the output can only be written as reduced form fractions). Similarly, it is terrible to restrict yourself to only working with the integers modulo $5$ as $0,1,2,3,4$. The integers modulo $5$ can be represented by all integers, but there is a specific rule about when they get identified: $a$ and $b$ are the "same integer mod $5$" when $a - b$ is a multiple of $5$, just like the fractions $a/b$ and $c/d$ are the "same fraction" when $ad = bc$ as integers.


问:
If $f$ is one-to-one function has horizontal asymptote at $y- a$ , then the function $f^{-1}(x)$
The chooses are :

A- Has a vertical asymptote at x = a

B- Has a vertical asymptote at x = -a

C- Has a horizontal asymptote at y = a

D- Has horizontal asymptote at y= -a

I've tried to find function like this but i couldnt find anything other $f(x)=1/x$ which is the same as its inverse

答: If a function has horizontal asymptote at $y=a$ then its inverse has a vertical asymptote at $x=a$ because inversing a function is like mirroring it about the line $y=x$.


问:
In how many ways can I distribute 5 distinguishable balls into 4 distinguishable boxes such that no box is left empty.
So what I think is the way to solve this question is to first count the total number of ways of putting all the balls into boxes such that the restriction isn't satisfied (i.e. the total number of ways of putting the balls into the boxes). Then using the inclusion-exclusion principle for situations where the boxes are empty.

I think the number of total ways that you can distribute them would be $4^5$ but I'm unsure how to do the second part of the question.

答:
Your first part is correct. The number of the total ways is $4^5$.

Then you might consider the in-exclusion principle: $A_1$: $Box_1$ is empty; $A_2$: $Box_2$ is empty; $A_3$: $Box_3$ is empty; $A_4$: $Box_4$ is empty.

Use the same method that you got the number of total ways. For each of these cases,  distributing $5$ distinguishable balls to $3$ distinguishable boxes, number of ways is $3^5$. $\binom{4}{1}$ shows it has $4$ cases.

$\vert A_1 \bigcap A_2\vert$ (Both Box 1 and Box 2 are empty) $= 2^5$. $\binom{4}{2}$ (from $A_1$ - $A_4$, choose $2$) shows how many cases in this situation.

$\vert A_1 \bigcap A_2 \bigcap A_3\vert$ (Boxes 1,2,3 are empty) $= 1^5$. $\binom{4}{3}$ cases.

$\vert A_1 \bigcap A_2 \bigcap A_3 \bigcap A_4 \vert$ (Boxes 1,2,3,4 are empty)= $0^5$. $\binom{4}{4}$ case.

Then apply inclusion-exclusion principle, $\vert A_1 \bigcup A_2 \bigcup A_3 \bigcup A_4 \vert$(at least one box is empty)
$$=\sum_{k=1}^4  (-1)^{k+1}(4-k)^5  \binom{4}{k}$$

Finally, use the total number $4^5$ minus the number we got from the last step, then we get the number of ways that no box is empty.

To be honest, this method is much more complicated than other approaches. However, I guess you might want to generalise it to distribute $r$ distinguishable balls into $n$ distinguishable boxes(?) Then it definitely should be a good way to think about this question.