问:
ELMO Shortlist 2012/G3: Geometry involved angle bisector and right angle

Given $\triangle ABC$ with $I$ as its incenter, draw $ID$ perpendicular to $BC$ and $IP$ perpendicular to $AD$. Prove that $\angle BPD=\angle DPC$

It seems like just a normal classic geometry with incenter and some perpendicular, but still I wasn't able to solve it. The solution of this problem involving Pole/Polar or sth like that, that I dun study or know anything about it.

Here is the link to the problem and solution

I'm a highschool student who like to solve geometry and I want to solve it using just normal method without involving higher theorems and here is my approach:

Let $E,F$ be another touching point on $AB$ and $AC$ respectively.

Since $IP\perp AD$ it suggests that $A,E,I,P,F$ lie on the same circle with diameter $AI$

Let $J$ be the midpoint of $AI$ so $IJ\perp EF$

Denote $M,N$ be the point where $PC, PB$ meet $(J)$

So $\angle BPD=\angle DPC$ if and only if $AM=AN$ or $EF\parallel MN$ or $EM=FN$.

Another way around is to prove that
\begin{equation}
\frac{BD}{DC}=\frac{BP}{PC}
\end{equation}
Unfortunately, that's all about that I can do. Please help

To simplify the pic I only take the important part because if I take point $C$ too, the pic looks ugly

答:
This problem is naturally projective, so you are handicapping yourself by forbidding methods like poles/polars/harmonic cross-ratios. The following uses Menelaus' theorem -- you might categorise it as a "higher theorem", but it was known $2000$ years ago so...

Let $X=IP\cap BC$. Then $XI\cdot XP=XD^2$, so $X$ lies on the radical axis of $(AIP)$ and the incircle, which is $EF$. So now apply Menelaus' theorem on $\triangle ABC$ and line $XFE$ (undirecting lengths):
$$1=\frac{BX}{XC}\cdot\frac{CF}{FA}\cdot\frac{AE}{EB}=\frac{BX}{XC}\cdot\frac{s-c}{s-a}\cdot\frac{s-a}{s-b}\implies\frac{BX}{XC}=\frac{s-b}{s-c}=\frac{BD}{DC}.$$
Now by the sine rule, we deduce that
\begin{align*}
1=\frac{BD}{DC}\div\frac{BX}{XC}&=\frac{\sin\angle BPD}{\sin\angle DPC}\div\frac{\sin\angle BPX}{\sin\angle XPC} \\
&=\frac{\sin\angle BPD}{\sin\angle DPC}\div\frac{\cos\angle BPD}{\cos\angle DPC} \\
&=\frac{\tan\angle BPD}{\tan\angle DPC}.
\end{align*}
Hence it follows that $\angle BPD=\angle DPC$.


问:
Would a circle overlap a parabola's bottom by more than just its vertex?
I mean, out of the condition that a circle actually crosses the parabola.
My question is when a circle is "inside" a parabola, would it touch part of the parabola other than just the parabola's vertex at the bottom?

So just the curiosity brought me crazy...but would this be possible?? (There was an image to show but my reputation is not enough since I'm a freshmeat :P)

Thanks xx

答: It depends on the circle, and on the parabola. If the parabola is very narrow, or the circle very large, then they'd meet at two points additionally to the vertex of the parabola. If the parabola is wide, or the circle small, they'd meet only at the vertex. </p>

ContentLicense: CC BY-SA 3.0


问:
If $x=\max(0,y+\alpha)$, how can we write $y$?
If we have $x=\max(0,y+\alpha)$
so we have $x=0$ or $x=y+\alpha$.

How can we write $y$? Only as $y=x-\alpha$, or do we add $y=-\alpha$, too?

答: $x=f(y)=\max(0,y+\alpha)$, $f$ is not strictly monotone so $f^{-1}$ does not exist. Therefore you can't express $y$ as a function of $x$.

ContentLicense: CC BY-SA 4.0


问:
Non homogenous equation- Plotting a graph of y against x

Given the following equations:
$$x'=-x+y-1$$
$$y'=-x-y+3$$
When $t=0$, $x=0$ and $y=3$
Find the general solution for $x$ and $y$ in terms of $t$.

Draw a graph of $y$ against $x$ for values of $t$ >0. Describe what happens as $t$ approaches infinity.

I have found the general solution for $x$ and $y$ to be:
$$x(t)=e^{-t}  (A cost+B sin t)+1$$
$$y(t)=e^{-t} (A cost+B sint)+2$$

Applying the initial conditions the particular solutions are:
$$x(t)=e^{-t}  (- cost+ sin t)+1$$
$$y(t)=e^{-t} ( cost+ sint)+2$$.

Can someone please help verify that this is correct so far and help me with how to proceed in graphing this and explaining the resultant graph.

答: Since you have explicit solutions, you can use one of the many graph plotter websites. To determine the behavior as $t\rightarrow\infty$, it might help to first look into the behaviors of $t\mapsto e^{-t}$, $t\mapsto \cos(t)$, $t\mapsto\sin(t)$ as $t$ approaches infinity and you should be able to figure a behavior for the product of those elementary functions.


问:
Combinatorial proof of $\sum_{k=0}^n \binom{n}k \binom{n-k}{m-k} = 2^m \binom{n}m$
Combinatorial proof of: $$\sum_{k=0}^n \binom{n}k \binom{n-k}{m-k} = 2^m \binom{n}m$$

My answer is:

number of ways to form 2 committees of different members from a group of n people.

For the first method, choose k members of the first committee from n people, and form the other committee with (m-k) members from the remaining (n-k) people, since k members are already in the first committee. The product is the number of ways.

For the second method, choose m members who will be part of the two committees from n people. The number of ways to form committees from m people is the subset of m members, which is $2^m$. The product is the number of ways.

Did I get some things right?

答:
$\newcommand{\Set}[1]{\left\{ #1 \right\}}$You are nearly there. You have $n$ people. You want to select $m$ of them, and then divide these $m$ people in a group labelled $0$, and a group labelled $1$.

On the right-hand side, you choose a subset $A$ of $m$ people in $\dbinom{n}{m}$ ways, and then divide these $m$ people in the two groups, by choosing one of the $2^{m}$ characteristic functions $A \to \Set{0, 1}$, which tells you whether person $a \in A$ goes into group $0$ or $1$.

On the left-hand side, for every $k \le m$ you first choose $k$ people to go in the group labelled $0$, and then $m - k$ people to go in the group labelled $1$.


问:
What is the probability that there are exactly 3 pennies in box number 1?
One hundred pennies are being distributed independently and at random into 30 boxes, labeled 1, 2, ..., 30. What is the probability that there are exactly 3 pennies in box number 1?

I don't know what type distribution this problem involves and also how do I proceed to solve it? Any hint?

答:
There are only two possibilities relevant to the question: either a penny lands in box 1 (probability $\frac1{30}$) or it doesn't. Therefore a binomial distribution is in play and the answer is
$$\binom{30}3(1/30)^3(29/30)^{27}$$


答:
Pick the three pennies that'll go in the first box in $\binom{100}{3}$ ways. Probability that three given pennies actually go into that box (and the other $97$ do not) is $(1/30)^3\cdot(29/30)^{97}$, so the answer is
$$\binom{100}{3}\cdot\left(\frac{1}{30}\right)^3\cdot\left(\frac{29}{30}\right)^{97}$$

Body:
One hundred pennies are being distributed independently and at random into 30 boxes, labeled 1, 2, ..., 30. What is the probability that there are exactly 3 pennies in box number 1?

I don't know what type distribution this problem involves and also how do I proceed to solve it? Any hint?


问:
What does commutativity of partial derivatives imply about geometry about the surface of a function?
Suppose we have a function $f(x,y)$ and it has the property that:

$$  \partial_x \partial_y f(x,y) = \partial_y \partial_x f(x,y)$$

What does this imply about the geometry of the surface described the function?

答:
The second order differential operator commutes for $ f\in \mathcal{C}^2$. Which means $\partial_x\partial_y f$ is defined in the neighborhood of $(x_0, y_0)$ and continuous at $(x_0, y_0)$ for all points in the domain of $f$.

$\mathcal{C}^2$ functions are twice differentiable meaning they are "smooth" in some definitions of the word. You can read more about them here.


问:
Find $[(\sqrt 2 +1)^8]$, where $[.]$ is integer part function
$$2^4 (1+\frac{1}{\sqrt 2})^8$$
$$=16 (1+8\frac{1}{\sqrt 2})$$
$$=106.4$$ which is nowhere close to the actual answer. What’s wrong with my approximation?

Body:
$(1+x)^n$ can be approximated to $1+nx$ only when $x \to 0$

So your method will not work

Let's add $(1 - \sqrt{2})^8$ on both sides

$(1+\sqrt{2})^8 + (1 - \sqrt{2})^8 = 2(1+ \binom{8}{2}2+\binom{8}{4}4+\binom{8}{6}8+16) = 1154$

$(1+\sqrt{2})^8 + (1 - \sqrt{2})^8 = 1154$

$(1+\sqrt{2})^8 = 1154-(1 - \sqrt{2})^8$

$1154$ is an integer and $(1 - \sqrt{2})^8$ is a very small fraction.

Therefore,

$[(1+\sqrt{2})^8] = 1154-1 = 1153$


问:
How to show that the dimension of the output subspace $D_K$ is $m$?
I have some problems with a homework question, and would appreciate if I can get some help.

Consider a single-input-single-output system

$$\dot{x} = Ax + Bu \\
y = Cx$$

and its corresponding transfer function

$$G(s) = C(sI-A)^{-1}B = \frac{s^m + b_{m-1}s^{m-1} + ...+b_0}{s^n + a_1s^{n-1}+...+ a_n},$$
where $A \in {\rm I\!R}^{n \times n}$, $B \in {\rm I\!R}^{n \times 1}$, $C \in {\rm I\!R}^{1 \times n}$. The relative degree $r = n - m$ is the excess degree of the denominator compared to the numerator.

Let $u = Kx$. Then the collection of all initial states such that $y(t) = 0$, denoted by $D_K = \{x_0: Ce^{(A+BK)t}x_0 = 0 \; \forall t\}$ is a subspace in ${\rm I\!R}^n$. Show that there is a solution $K$ such that the dimension of $D_K$ is $m$.

I do not really understand what a dimension means in this case, and how to show the dimension of a output subspace. I have earlier shown that the Markov parameters $CA^kB = 0,\; k= 0,1,2\dots r-2 $ and $CA^{r-1}B \ne0$. Also, I know that the matrix $ \begin{bmatrix}
C\\
CA\\
\vdots \\
CA^{r-1}
\end{bmatrix}$ has full row rank.

Any help is appreciated.

EDIT: The system (A,B) is not necessarily controllable. The solution given to this question is given below. I understand everything except how this implies that $D_K$ has dimension $m$:

"We can compute that $y^{k}(t) = CA^kx(t), \; k = 0,...,r-1$ and $y^r(t) = CA^rx(t) + CA^{r-1}Bu$. Then $y(t) = 0$ implies that $CA^kx(t) = 0, \; k = 0,...,r-1$ and $u = -(CA^{r-1}B)^{-1}CA^rx(t)$, which gives that $D_K$ dimension $n-r = m$."

答:
By using the relation of the matrix exponential with the Cayley-Hamilton theorem it is also possible to write $D_K$ as

$$
D_K = \left\{x\,|\,M\,x = 0\right\},
$$

with

$$
M =
\begin{bmatrix}
C \\ C\,(A + B\,K) \\ \vdots \\ C\,(A + B\,K)^{n-1}
\end{bmatrix}.
$$

The set $D_K$ is thus essentially equivalent to the null-space of $M$. It can also be noted that $M$ is the observability matrix associated with the pair $(A+B\,K, C)$, so for example $D_K$ would be empty if $(A+B\,K, C)$ is observable. The relation between observability (and controllability) of the state space representation and its corresponding transfer function is that if the state space representation is not observability (or controllability) then the corresponding transfer function would have one or more pole-zero-cancellations. For each pole-zero-cancellation either the observability or controllability matrix would reduce its rank by one.

When assuming that $(A,B)$ is controllable then when using pole-placement for choosing $K$ and place a subset of the poles at the zeros of the corresponding transfer function would result in $m$ pole-zero-cancellations. It can be noted that using state feedback does not alter the controllability, thus the rank of closedloop observability matrix has to become $n-m$. Or in other words the null-space of the closedloop observability matrix would be of dimension $m$.

I am not sure if this could also be shown for systems who are not controllable. I think it would still work for systems whose controllability matrix has at least a rank of $m$, but below that I suspect that it would not work. For example consider the extreme case of $B=0$.

Edit:

Given that $y^{(r)}(t) = C\,A^r x(t) + C\,A^{r-1} B\,u$ and $u = -(C\,A^{r-1} B)^{-1} C\,A^r x(t)$, it follows that $y^{(r)}(t) = 0\ \forall\ x(t) \in \mathbb{R}^n$. Here $z^{(n)}(t)$ denotes the $n\text{th}$ derivative of $z(t)$ with respect to $t$ to avoid the confusion with raising to the power. The fact that $y^{(r)}(t) = 0$ does not yet imply that $y(t) = 0$. This would only follow if all the "initial conditions"  of the output are zero, thus $y^{(k)}(t) = 0\ \forall\ k = 0,\dots,r-1$. Those time derivative are also shown to be related to each other using $y^{(k)}(t) = C\,A^k x(t), \forall\ k = 0,\dots,r-1$. Thus the zero "initial conditions" is equivalent to

$$
R =
\begin{bmatrix}
C \\ C\,A \\ \vdots \\ C\,A^{r-1}
\end{bmatrix}, \\
R\,x(0) = 0.
$$

The solutions for $x(0)$ which satisfy this all lie in the null-space of $R \in \mathbb{R}^{r \times n}$. It is stated that $R$ has full row rank or equivalently has a rank of $r$, therefore the null-space of $R$ should have a dimension of $n - r = m$. One might think that having $x(0)$ would only ensure that $y^{(k)}(0) = 0\ \forall\ k = 0,\dots,r-1$, but that fact that $y^{(r)}(t) = 0\ \forall\ x(t) \in \mathbb{R}^n$ ensures that $y(t) = 0$ (since none of $y^{(k)}(t)\ \forall\ k = 0,\dots,r-1$ can become non-zero) which would also imply that $x(t)$ remains in the null-space of $R$.


问:
Solution of System of Linear Differential Equations
$$x' + y' - x - 2y = 2e^t$$
$$x' + y' - 3x - 4y = e^{2t}$$

I didn't do anything for this.
I found $Dx + Dy - 2x - 3y$...(*)..but nothing came.

答:
Substitute $u = x + y: u' = x' + y'$. The equations can be rewritten as
$$u' - 2u + x = 2e^t\\u'-4u+x=e^{2t}$$

Now subtract the second equation from the first.
$$2u=2e^t-e^{2t}\\u=e^t-\dfrac{1}{2}e^{2t}$$

Now plug this value for $u$ into the first equation.
$$(e^t-e^{2t})-(2e^t-e^{2t})+x = 2e^t\\-e^t+x=2e^t\\x=3e^t$$

Now use this to find $y$.
$$3e^t + y = e^t-\dfrac{1}{2}e^{2t}\\y=-2e^t-\dfrac{1}{2}e^{2t}$$


答: You can take advantage of the fact that you have $x'+y'$ in both equations. Subtract both line and identify $x$ as a function of $y$, then work from there to get $y$.

Title: Solution of System of Linear Differential Equations

Body:
$$x' + y' - x - 2y = 2e^t$$
$$x' + y' - 3x - 4y = e^{2t}$$

I didn't do anything for this.
I found $Dx + Dy - 2x - 3y$...(*)..but nothing came.


问:
Supremum of a set given by $\sum\limits_{k=1}^{n}\frac{a_{k}}{a_{k}+a_{k+1}+a_{k+2}}$ (one inequality)
I've found a solution for this problem: $\inf$ and $\sup$ of a set given by $\sum\limits_{k=1}^{n}\frac{a_{k}}{a_{k}+a_{k+1}+a_{k+2}}$ but I'm somehow struggling to understand one inequality, that is
$$\frac{a_{k}}{a_{k}+a_{k+1}+a_{k+2}}<  \frac{\sum_{i=1}^{k}a_{i}+\sum_{i=k+3}^{n}a_{i}}{\sum_{i=1}^{n}a_{i}}$$
(I created a new topic since I cannot comment overthere). Tia.

答:
Hint: If $ \frac{a}{b} < 1$, then for any $ c>0$, $\frac{a}{b} < \frac{ a+c}{b+c}$.

Apply this directly. If you're stuck, show your work and what you've tried.


问:
Uniform distribution on the n-sphere.
I have the next RV:

$$\underline{W}=\frac{\underline{X}}{\frac{||\underline{X}||}{\sqrt{n}}}$$
where $$X_i \tilde \  N(0,1)$$
It's a random vector, and I want to show that it has a uniform distribution on the n-sphere with radius $$\sqrt{n}$$.

I understand that it has this radius, just calculate it. But I don't understand from calculating the CDF how to I arrive at uniform distribution.

Thanks in advance, MP.

答: Just a hazy thought: once the radius question is settled, you have to prove that the direction of the rv is uniformly distributed. What about fixing an arbitrary unit vector $\vec{n}$, and computing the probability that $\langle \vec{n}, X \rangle \geq 0$? If it is $\frac{1}{2}$ for any $\vec{n}$, doesn't it give you the result?

Title: Uniform distribution on the n-sphere.


答:
This is a cute inequality, that comes from the following observation:

$$\frac{\sum_{i=1}^{k}a_{i}+\sum_{i=k+3}^{n}a_{i}}{\sum_{i=1}^{n}a_{i}}=1-\frac{a_{k+1}+a_{k+2}}{\sum a_i}>1-\frac{a_{k+1}+a_{k+2}}{a_k+a_{k+1}+a_{k+2}}=\frac{a_k}{a_k+a_{k+1}+a_{k+2}}$$

Title: Supremum of a set given by $\sum\limits_{k=1}^{n}\frac{a_{k}}{a_{k}+a_{k+1}+a_{k+2}}$ (one inequality)

Body:
I've found a solution for this problem: $\inf$ and $\sup$ of a set given by $\sum\limits_{k=1}^{n}\frac{a_{k}}{a_{k}+a_{k+1}+a_{k+2}}$ but I'm somehow struggling to understand one inequality, that is
$$\frac{a_{k}}{a_{k}+a_{k+1}+a_{k+2}}<  \frac{\sum_{i=1}^{k}a_{i}+\sum_{i=k+3}^{n}a_{i}}{\sum_{i=1}^{n}a_{i}}$$
(I created a new topic since I cannot comment overthere). Tia.


问:
The exact definition of a form
I am studying Fulton's Algebraic Geometry book and I can't understand what a form is! Sometimes it factors a polynomial $F$ to something like:
$$F=f_0 + f_1 + f_2 +...+ f_m$$
and it calls $f_i$  a "form". What is the exact definition of a form?

答:
Putting together the answers in the comments to the question:

Viktor Vaughn writes: This is defined on the very first page of the book: "We call F homogeneous, or a form, of degree $d$, if all coefficients $a_{(i)}$ are zero except for monomials of degree $d$."

Quasi writes: Thus, for example, $x^2+3xy−2yz$ is a form of degree $2$, since all terms have total degree $2$, but $x^2+3x^2y−2yz$ is not a form.

(Note that $3x^2y$ has degree $2+1=3$.)

See also https://en.wikipedia.org/wiki/Homogeneous_polynomial.

Title: The exact definition of a form


问:
proof any finite sequence of real numbers is bounded
For any sequence of real numbers over a finite domain such as,
$ \langle a_n \rangle = a, a+1, \ldots, b $

Can someone show me (in formal terms) how it is I can go about defining a bijection between the range $ \left\{ f(n) : a ≤ n ≤ b \right\} $ and subset of natural numbers so that I can go on to use the fact that any finite set of real numbers is bounded to assert the existence of a maximal and minimal elements of the range.

Or if this is not the way to formally show that maximal and minimal elements of the range exist (obviously they do), can someone show me the way.

Thank you!!

答:
Let $A\subseteq\mathbb{R}$. The real number $\alpha$ is the maximal element of $A$ provided that $\alpha\in A$ and for all $a\in A$ we have $a\leq\alpha$.

For $n\in\mathbb{N}$, define $$\mathbb{N}_n=\{k\in\mathbb{N}:1\leq k\leq n\}$$

By definition, a set $A$ is said to have $n$ elements if there exists a bijection $$f:\mathbb{N}_n\to A$$

We will show that every finite subset of the real numbers has a maximal element. Showing that every finite subset of the real numbers has a minimal element is almost identical.

Define $$S=\{n\in\mathbb{N}:\text{ if }f:\mathbb{N}_n\to A\text{ is a bijection, then $A$ has a maximal element}\} $$ We want to show that $S$ is inductive, that is, if $1\in S$ and if $n\in S$, then $n+1\in S$.

Note: We require $A$ to be non-empty since the empty set cannot have a maximal or minimal element, since there are no elements in the empty set.

Base Case: Let $n=1$ and suppose $f:\mathbb{N}_1\to A$ is a bijection. Then $A$ has $1$ element, call it $f(1)$. Since $f(1)\in A$ and $f(1)\geq a$ for all $a\in A$, then $f(1)$ is the maximal element of $A$. This completes the base case.

Inductive Hypothesis: Let $n=k$. If $f:\mathbb{N}_k\to A$ is bijective, then $A$ has a maximal element.

Inductive Step: Let $n=k+1$. Suppose $f:\mathbb{N}_{k+1}\to A$ is bijective. One can show that the function $\hat{f}:\mathbb{N}_k\to A\setminus\{f(k+1)\}$ defined as $$\hat{f}(n)=f(n)$$is bijective. Then $A\setminus\{f(k+1)\}$ has $k$ elements, so by hypothesis, $A\setminus\{f(k+1)\}$ has a maximal element, call it $\alpha\in A\setminus\{f(k+1)\}$.

One can now distinguish two cases for bijection $f:\mathbb{N}_{k+1}\to A$:

(1) $f(k+1)\leq\alpha$. In this case, the maximal element of $A$ is $\alpha$, since $\alpha\geq a$ for all $a\in A\setminus\{f(k+1)\}$ and $\alpha\geq f(k+1)$.

(2) $f(k+1)>\alpha$. In this case, the maximal element of $A$ is $f(k+1)$ since $\alpha\geq a$ for all $a\in A\setminus\{f(k+1)\}$ and $f(k+1)>\alpha$.

In both cases, $A$ has a maximal element, so $k+1\in S$, which implies $S$ is inductive. This completes the proof.

Also note that this can be generalized to a set $A$ together with a non-strict total order $\leq$ and strict total order $<$, since the Inductive Step required us to compare two elements.


问:
Why is better to work with the spectrum of prime ideals than with the maximal one, for example in the definition of affine scheme.
When we have an algebraic variety we can identify the points of the variety with maximal ideals of the coordinate ring.

I would like to know why it is more natural to define the main structure of the theory of schemes, the affine scheme, with prime ideals and not with maximal ones.

When Grothendieck was creating theory of schemes, why did he decide to work with the normal spectrum instead of the maximal one?

(As you can see I dont have an strong background of Algebraic Geometry, I just want to have some intuition)

In which sense the schemes generalize the notion of variety and why is better to work with this notion?

答:
This is the moment in which I should give a very well documented answer, full of critical examples, very long and well explained. I'll try, but it's been a while I am not in algebraic geometry, so I ask for tolerance :)

If you think of maximal ideals as points in your variety, prime ideals represents irreducible subvarieties: lines, curves, planes...  So when we say that a maximal ideal can be mapped to prime ideal which is non maximal, we are saying that a rational function can, in principle, send a point to a line. What the hell does this mean? Let's see an example.

Take a ring $A$ and a prime $p$ which is not maximal: you can imagine $p$ represent a line. Consider the localization $A_p$. Now $pA_p$ becomes maximal in $A_p$, and the inclusion $i:A\to A_p$ sends via preimage $pA_p$ to the non maximal ideal $p$. See? We sent a "point" to a line, but we all know this is a lie. Let me explain better.

When you take a local ring $R$ there is only one maximal ideal, but we don't want to think this to be a one-point variety: as you have seen above, the richness of this variety is in the "neighborhood" of the maximal ideal.

Also, since we cannot really talk about disks and balls in the euclidean space as we do in classical geometry (this would require using inequalities, which are not defined for arbitrary fields), we need a notion of being close for defining a topology. This is achieved by declaring subvarieties to be the closed sunsets. That's not finer as the classical topology, but we can live with that. For example, here it is a thing that I struggled with: take a plane and remove three general lines; is this connected? In classical topology of course it is not, but in algebraic geometry there is no way you can disconnect it, no matter which (robust) notion of connection you take. This emanates from the fact that you have to use lines and not segments to connect stuff; also, lines can be punctured.

Let me also state a final remark which is crucial as a springboard to schemes. Local to global principles play a crucial role thorough all geometry, and in a special way in algebraic geometry.

Schemes are objects which locally looks like affine algebraic varieties, and this often permits to do local-to-global arguments (i.e. reduce to the case in which you have $Spec A$ and then use commutative algebra). In the same way, noetherian rings locally looks like local rings; this means that a property for a ring $A$ is equivalent to the same property for the ring $A_p$, where $p$ ranges over all primes. Not all criteria holds if you substitutes all primes with maximal.

Of course, not all properties of schemes are local (see connectedness) and not all properties of rings are local, but nevertheless this is probably something that was used a lot and well established at the time schemes were introduced. It would be nice if someone with historical knowledge could comment, though.


问:
Projection and inner product space
Definition: Let $V$ be vector space, and $U$, $W$ be two subspaces such that $V=U\oplus W$.

We know that there exists for each $v \in V$ only one $u \in U$ and only one $w \in W$ such that $v=u+w$. Using this, we define a projection $P_{U,V}\colon V\longrightarrow V$ to be: $P_{U,W}(v)=u$

Now my question is this:

Let $V$ be an inner product space, and let $U$ be subspace of $V$. Let $\left\{e_{1},\ldots ,e_{n}\right\}$ be an orthogonal basis for $U$.

Let us define the orthogonal projection $P_{U}\colon V\longrightarrow V$ as

$$P_{U}(v)=\sum_{i=1}^n \langle v,e_{i}\rangle e_{i}$$

I need to prove that  $P_{U}=P_{U,U^{\perp}}$.

$P_{U,U^{\perp}}$ is according to the definition in beginning.

How do I do it? I am sitting 1 hour on that and I have no clue.

plus I need to proof that $P_{U}$ is self adjoint

Basically, I have to prove that if $v=u+w$, then $P_{U}(v)=\sum_{i=1}^n \langle v,e_{i}\rangle e_{i}$  and

$P_{U,U\perp}(v)=u$ meaning I have to show $u=\sum_{i=1}^n \langle v,e_{i}\rangle e_{i}$.

But how do I do it ?

答:
Hint: you can prove $P_U=P_{U,U^\perp}$ by checking that $P_U(v)=P_{U,U^\perp}(v)$ for every $v\in V$.

Added: You've got the start of the right strategy, but let me modify it it a bit. Let's start with your $v=u+w$ with $u\in U$ and $w\in U^\perp$ (I think you might be forgetting about this last fact.)

You know that $P_{U,U^\perp}(v)=u$

Since the $e_i$ are a basis for $U$, you can write $u=\sum \alpha_ie_i$ so that $v=\sum \alpha_ie_i +w$ where $w\in U^\perp$. Now, compute $P_{U}(v)=P_{U}(\sum\alpha_ie_i +w)=\_\_\_\_$.

Next hint:
$P_{U}(\sum\alpha_ie_i +w):=\sum_j \langle \sum_i\alpha_ie_i +w,e_j\rangle e_j$

Title: Projection and inner product space


问:
Variance-Covariance matrix for Weighted Least Squares
For ordinary least squares (OLS), the solution to the system $X\beta = y$ is

$\hat{\beta} = (X^T X)^{-1} X^T y$

and the variance on the solution parameters is

$Var(\hat{\beta}) = \sigma^2 (X^T X)^{-1}$

where the vector y denoted our observables and $\sigma$ are the errors on these observables.

If I instead would obtained the solution from the weighted least squares as

$\hat{\beta} = (X^T W X)^{-1} X^T W y$

where $W_{ii} = 1 / \sigma^2_i$, what would be the corresponding variance-covariance matrix of $\hat{\beta}$? Is it the same as in the OLS case?

答:
In weighted least squares, you have $y=X\beta+\varepsilon$ where $\operatorname E(\varepsilon)=0$ and $\operatorname{Var}(\varepsilon)=\Omega=\operatorname{diag}(\sigma_1^2,\ldots,\sigma_n^2)$ is positive definite.

Weighted least squares estimator of $\beta$ is then $$\hat\beta_{\text{WLS}}=(X^T\Omega^{-1}X)^{-1}X^T\Omega^{-1}y=Py \quad(\text{say})$$

Therefore,

\begin{align}
\operatorname{Var}\left(\hat\beta_{\text{WLS}}\right)&=
P\cdot\operatorname{Var}(y)\cdot P^T
\\&=(X^T\Omega^{-1}X)^{-1}X^T\Omega^{-1}\cdot\operatorname{Var}(y)\cdot((X^T\Omega^{-1}X)^{-1}X^T\Omega^{-1})^T
\\&=(X^T\Omega^{-1}X)^{-1}X^T\Omega^{-1}\cdot\Omega\cdot\Omega^{-1}X(X^T\Omega^{-1}X)^{-1}
\\&=(X^T\Omega^{-1}X)^{-1}
\end{align}

You can see that this matches with $\operatorname{Var}\left(\hat\beta_{\text{OLS}}\right)$ when $\sigma_i^2=\sigma^2$ for each $i$.


问:
Problems with understanding rotor's conception
By definition rotor is:
$$ rot \vec A =
    \begin{bmatrix}
    (\frac{\partial A_z}{\partial y} - \frac{\partial A_y}{\partial z})  \\
    (\frac{\partial A_x}{\partial z} - \frac{\partial A_z}{\partial x})  \\
    (\frac{\partial A_y}{\partial x} - \frac{\partial A_x}{\partial y})  \\
    \end{bmatrix}
$$

And I'm trying to visualize it. So if I have a vector field, take one vector, and find a projection, and I don't understand how we can take derivative on, for example for $\hat i$, $A_z$ if there is no y axis anymore (see pic.1) and for $A_y$ there is no z axis anymore

答:
Take a point $(x,y,z)$. We can evaluate $\vec{A}(x,y,z)=(A_x,A_y,A_z)$ which is the vector of the vector field at $(x,y,z)$.

We may reinterpret this as being three separate functions $A_x,A_y$ and $A_z$, which take points of the space as input and give single numbers as output. As such, we can take partial derivatives of these three functions entirely without issue. These are probably the projections you talk about. Note that it is only the output that is projected. The input is still 3-dimensional.

Alternately, consider the partial derivatives of $\vec A$ itself. For instance,
$$
\frac{\partial \vec A}{\partial x}=\lim_{h\to 0}\frac{\vec A(x+h,y,z)-\vec A(x,y,z)}{h}
$$
which is the instantaneous change of $\vec A$ as you move in the $\vec i$ direction. Similarly with the two other partial derivatives. These three partial derivatives are vectors, and as such they each have $x,y$ and $z$ components.

Both these approaches give the same result. It is up to you which one you like best.


问:
idempotent matrix over local ring is similar to a diagonal matrix with elements $0$ and $1$
I want to know why any idempotent matrix $P$ over local ring is similar to a diagonal matrix with elements $0$ and $1$.

I saw a proof in lemma 3.3 of ncatlab, but I don't understand why $P$ is similar to a diagonal matrix if it has an invertible $r$-minor, and why such a matrix can be chosen with entries $0$ and $1$.

Could somebody explain more?

答:
$\newcommand{\Set}[1]{\left\{ #1 \right\}}$Let $V$ be the the underlying free module over the commutative local ring $A$ with unity.

Consider the submodules
$$
V_{0}
=
\Set{ v \in V : P v = 0},
$$
and
$$
V_{1}
=
\Set{ v \in V : P v = v}.
$$
Clearly $V_{0} \cap V_{1} = \Set{0}$.

For $v \in V$ one has
$$
v = (v - P v) + P v,
$$
where

$v - P v \in V_{0}$, as $P (v - P v) = P v - P^{2} v = P v - P v = 0$, and

$P v \in V_{1}$, as $P (P v) = P^{2} V = P v$.

Therefore $V = V_{0} \oplus V_{1}$, and $P$ is zero on $V_{0}$ and the identity on $V_{1}$.

Addendum The point of $A$ being local is that the direct summands $V_{0}, V_{1}$ are also free,


问:
$(f_n)$ converges uniformly $\iff \|f_n-f\|_\infty=\sup\{|f_n(x)-f(x)|;\ x\in I\}$
I am trying to prove the following statement : $(f_n)$ converges uniformly $\iff \sup\{|f_n(x)-f(x)|;\ x\in I\}=\|f_n-f\|_\infty\to 0.$
$\implies$ :
We have $\forall \epsilon \leq 0$ $\exists N$ $\forall n \leq N$ $\forall x \in I$ : $|f_n(x)-f(x)| \leq \epsilon$.

We also have that $|f_n(x)-f(x)| \leq \sup\{|f_n(x)-f(x)|;\ x\in I\}$ but then I don't really see how to properly link it to the $\epsilon$.

For $\leftarrow$, I fix an $\epsilon\geq 0$. Then there exists $n_0\in \mathbb{N}$ such that $$\forall n\geq n_0, \sup \{|f(x) - f_n(x)| : x \in I\}\leq \epsilon $$
Which implies, $$\forall n \leq n_0 |f(x) - f_n(x)|\leq \sup \{|f(x) - f_n(x)| : x \in I\}\leq \epsilon , \text{ for any } x\in I.$$
Hence, $f_n\rightarrow f $ as $n\rightarrow\infty$ uniformly. I don't know if I did it correctly though...

答:
$\Longrightarrow:$ Suppose that $(f_n)_{n\in\Bbb N}$ converges uniformly to some function $f$; you want to prove$$\lim_{n\to\infty}\sup\left\{\bigl|f_n(x)-f(x)\bigr|\,\middle|\,x\in I\right\}=0.\tag1$$Take $\varepsilon>0$. Since $(f_n)_{n\in\Bbb N}$ converges uniformly to $f$, there is some $N\in\Bbb N$ such that$$(\forall x\in I)(\forall n\in\Bbb N):n\geqslant N\implies\bigl|f_n(x)-f(x)\bigr|<\frac\varepsilon2.$$Therefore,$$(\forall n\in\Bbb N).n\geqslant N\implies\sup\left\{\bigl|f_n(x)-f(x)\bigr|\,\middle|\,x\in I\right\}\leqslant\frac\varepsilon2<\varepsilon,$$and this proves that we indeed have $(1)$.

$\Longleftarrow:$ If we have $(1)$, take $\varepsilon>0$ and take $N\in\Bbb N$ such that$$n\geqslant N\implies\sup\left\{\bigl|f_n(x)-f(x)\bigr|\,\middle|\,x\in I\right\}<\varepsilon.$$So, if $n\in\Bbb N$ and if $x\in I$, then$$n\geqslant N\implies\bigl|f_n(x)-f(x)\bigr|\leqslant\sup\left\{\bigl|f_n(x)-f(x)\bigr|\,\middle|\,x\in I\right\}<\varepsilon.$$


问:
Let $U=\Bbb{Z} \times (-\frac12,\frac12) = \{(m,y) \mid m\in \Bbb{Z}, |y| < \frac12 \}$. Show that $U$ is an open set in the metric space $(A, d_A).$

Let $A= \Bbb{Z} \times \Bbb{R} \subset \Bbb{R}^2$ and let $d_A$ be the Euclidean metric restricted to $A$. Let $U=\Bbb{Z} \times (-\frac12,\frac12) = \{(m,y) \mid m\in \Bbb{Z}, |y| < \frac12 \}$. Show that $U$ is an open set in the metric space $(A, d_A).$

My idea was to define $U_0 = \{(0,y) \mid |y| < \frac12\}$ and show that this is open and since an arbitary union of open sets are open I would have that $U$ is open since $U = \bigcup_{i\in \Bbb{Z}}U_i$. I cannot seem to be able to show that $U_0$ would be open.

If I pick a point $x=(0, x_0) \in U_0$ and construct a ball $B(x,r)$ with $r = \frac12-x_0 >0$ I should be using this to show that $B(x,r) \subset U_0$? I'm getting very confused about what should be a subset of what etc in order to get this to work...

答:
The metric space $(A,d_A)$ has the associated subspace topology $$\mathcal{T}_A=\{A\cap V|V\text{ is open in }\mathbb{R}^2\} $$ Take $$V=\mathbb{R}\times(-\frac{1}{2},\frac{1}{2}) $$ and show that $V$ is open in $\mathbb{R}^2$. One can do this by showing that for any $z\in V$, there exists an $r>0$ such that $\{w\in\mathbb{R}^2:d(z,w)<r\}\subseteq V $ where $d$ is the usual Eucidean metric. Alternatively, you could observe that both sets are open in the usual topology on the reals, so their product is open in the usual topology on the plane.

Then $$A\cap V=(\mathbb{Z}\times\mathbb{R})\cap(\mathbb{R}\times(-\frac{1}{2},\frac{1}{2}))=(\mathbb{Z}\cap\mathbb{R})\times(\mathbb{R}\cap(-\frac{1}{2},\frac{1}{2}))=\mathbb{Z}\times(-\frac{1}{2},\frac{1}{2})=U$$ so $U$ is in $\mathcal{T}_A$.


问:
Nonempty finite subsets of $\mathbb{N}$ have at least one maximal element.
I am trying to prove the following proposition:

If $X$ is a finite, nonempty subset of $\mathbb{N}$, then there exists some $x_0 \in X$ such that $x \leq x_0$, for all $x \in X$, i.e., $X$ has a maximal element.

Here's my attempt:

We induct on $n$ - the number of elements of $X$. If $X$ is empty, then the proposition is vacuously true. The base case $n = 1$ is trivial. Now suppose inductively that if $|X| = n$ for some natural number $n$, then $X$ has a maximal element. We now show that if $|X| = n + 1$, then $X$ has a maximal element as well. Since $X$ is a nonempty subset of $\mathbb{N}$, the well-ordering principle tells us that $X$ has a minimal element - call it $a$. The set $X \setminus \{a\}$ has $n$ elements and therefore - by the inductive hypothesis - it has a maximal element - call it $x_0$. Since $a \leq x$ for all $x \in X$, then in particular $a \leq x_0$. Since $x_0$ is the maximal element of $X \setminus \{a\}$, it follows that $x \leq x_0$ for all $x \in X \setminus \{a\}$. This implies that $x \leq x_0$ for all $x \in X$ (since $X = X \setminus \{a\} \cup \{a\}$), and therefore $x_0$ is a maximal element of $X$. This closes the induction, and the desired result follows.

Could anyone tell me if this is correct/well-written? If not, can anyone point me in the right direction? Thanks in advance.

答:
Looks good.  You used induction and the well ordering principle.  Since these are equivalent statements, you can expect there is a way to do this using only one of these.

Only induction: if $X$ has $n+1$ elements, choose an arbitrary element $a$ in $X$.  Then $X \backslash \{a\}$ has $n$ elements, and by induction it has a maximal element $m$.  You can argue that either $m$ or $a$ is a maximal element of $X$.

Only well ordering: Let $Y = \{ -x : x \in X\}$.  By the well ordering principle, $Y$ has a least element $m$.  Then $-m$ must be a maximal element of $X$.


问:
Double inequality - two or three separate inequalities?
If I have the double inequality
$a\geq b \geq c$, does it mean I have two separate inequalities
\begin{align}
a&\geq b \\
b&\geq c
\end{align}
Or is it three separate inequalities
\begin{align}
a&\geq b \\
b&\geq c \\
a & \geq c
\end{align}
?

答:
All three expressions are equivalent. Some times expression like $$\begin{align}
a&\geq b \\
b&\geq c
\end{align}$$ is useful and some times $$\begin{align}
a&\geq b \\
b&\geq c \\
a & \geq c
\end{align}$$
For example in convex optimization when we are going to apply duality any of this two expressions may cause completely different dual problems, one solvable and another not!


问:
$R=(3+\sqrt 5)^{2n}$ and $f=R-[R]$, where $[.]$ is an integer part function. Then find $R(1-f)$
$f$ basically stands for the fractional part of $R$, but since value of $n$ isnt know, I don’t understand how it’s possible to derive it, because it is tricky to write in general terms. How do I solve it?

Body:
I am adding it as an answer because it was way too clumsy for a comment.

Let $(3+\sqrt{5})^{2n} + (3 - \sqrt{5})^{2n} = I$ where $I$ is an integer

$[(3+\sqrt{5})^{2n}] = I-1$(because $(3-\sqrt{5})^{2n}$ is a small fraction)

$f = (3 + \sqrt{5})^{2n} - [(3 + \sqrt{5})^{2n}] = (3+\sqrt{5})^{2n}-I+1=(3+\sqrt{5})^{2n}-(3+\sqrt{5})^{2n} - (3 - \sqrt{5})^{2n}+1$

$f=-(3 - \sqrt{5})^{2n}+1$

$1-f = (3-\sqrt{5})^{2n}$


问:
Can a set $S$ of subsets of $\mathbb{N}$ such that either $P \subset Q$ or $Q \subset P$ for all $P,Q \in S$ be uncountable?
I've tried a bunch of different approaches to this question (and would particularly appreciate some hints that don't give the whole thing away immediately).

I've noticed that if $S$  contains some finite subsets, then for each $n \in \mathbb{N}$ there is at most one subset of size $n$ in the collection $S$ as $S$ can't contain duplicates (in order to satisfy the given condition). Certainly if $A_i$ is a finite subset contained in $S$ of size $i$, then $A_i \subset A_j$ if $i > j$.

I've tried considering the possibilities where there are some finite subsets in the collection $S$, and then we can also add in infinite-size subsets that are each a superset of all of the finite subsets. However it seems rather difficult to visualise what happens when you consider infinite subsets of $\mathbb{N}$ that can't be obtained by removing finitely many elements from $\mathbb{N}$. What are good approaches to take to figure out whether such a set can be uncountable - is there an element of construction here?

答: Giving just a hint is hard because the idea is the whole thing. The point is that $(\Bbb R,\le)$ can be constructed by identifying $\Bbb R$ with the family of intervals of $\Bbb Q$ which are bounded above, unbounded below and without maximum, with the ordering being $\subseteq$. And $\mathcal P(\Bbb Q)$, as a poset, is essentially the same as $\mathcal P(\Bbb N)$.


问:
How to find 3 x 3 matrix inverses
Is there a way of finding the inverse of a $3 \times 3$ matrix without forming an augmented matrix with the identity matrix? Also, is there a quick way of checking that a  $3 \times 3$ matrix's inverse exists, without trying to compute it? Thanks.</p>

答:
Starting with a $3\times 3$ matrix
$$
A=\left(\begin{matrix}
a & b & c \\
d & e & f \\
g & h & i
\end{matrix}\right)
$$
1. Check that the matrix is invertible. First we need to check that we can actually invert the matrix. This amounts to showing that its determinant is non-zero.
$$
\det A=a\det\left(\begin{matrix}e & f \\ h & i\end{matrix}\right)-b\det\left(\begin{matrix}d & f \\ g & i\end{matrix}\right)+c\det\left(\begin{matrix}d & e \\ g & h\end{matrix}\right)
$$
2. Find the cofactors of the matrix. For each of the $(i,j)$-th entries of the matrix $A$, we denote the cofactor $C_{ij}$ by $(-1)^{i+j}M_{ij}$, where $M_{ij}$ is the determinant of the $2\times 2$ matrix obtained from $A$ by removing the $i$-th row and the $j$-th column. For example
$$
M_{23}=\det\left(\begin{matrix}
a & b & - \\
- & - & - \\
g & h & -
\end{matrix}\right)
=\det\left(\begin{matrix}a & b \\ g & h\end{matrix}\right)=ah-bg
$$
So $C_{23}=(-1)^{2+3}(ah-bg)=bg-ah$.

3. Find the adjugate matrix. After finding all the cofactors, the adjugate of $A$ is
$$
\text{adj }A=\left(\begin{matrix}
C_{11} & C_{12} & C_{13} \\
C_{21} & C_{22} & C_{23} \\
C_{31} & C_{32} & C_{33}
\end{matrix}\right)^T=\left(\begin{matrix}
C_{11} & C_{21} & C_{31} \\
C_{12} & C_{22} & C_{32} \\
C_{13} & C_{23} & C_{33}
\end{matrix}\right)
$$
That is you make a matrix out of the cofactors, and then you transpose the matrix.

4. We're done! Finally, we can can write the inverse as
$$
A^{-1}=\frac{1}{\det A}\text{adj }A
$$

Title: How to find 3 x 3 matrix inverses


问:
Sum of two squares theorem
I am working on the problem: I need to quickly check if positive number $n$ can be expressed as $n^2=a^2+b^2$. I found this theorem: https://en.wikipedia.org/wiki/Sum_of_two_squares_theorem

But it seems I am missing something. For example number $4$ falls under theorem condition: its prime decomposition doesn't have any prime $p\equiv 3{\pmod {4}}$. But $4$ cannot be expressed as sum of 2 squares.

答:
Too big for a comment

The wiki presentation is wrong on at least one point. It says,

"The prime decomposition of the number $3430$ is $2 \cdot 5 \cdot 7^3$. This time, the exponent of $7$ in the decomposition is $3$, an odd number. So $3430$ cannot be written as the sum of two squares."

However $\qquad (2058^2+2744^2=3430^2)\qquad$
because $\qquad(2058,2744,3430)=686\times (3,4,5)$

To find if some square can be the sum of two squares, we begin with Euclid's formula shown here as

$$ \quad A=m^2-k^2,\quad B=2mk,\quad C=m^2+k^2\quad$$

We solve the C-function for $k$ and try a range of m-values to see which yield integers. Here is an example

\begin{equation}
C=m^2+k^2\implies k=\sqrt{C-m^2}\qquad\text{for}\qquad \bigg\lfloor\frac{ 1+\sqrt{2C-1}}{2}\bigg\rfloor \le m \le \lfloor\sqrt{C-1}\rfloor
\end{equation}
The lower limit ensures $m>k$ and the upper limit ensures $k\in\mathbb{N}$.

$$C=65\implies \bigg\lfloor\frac{ 1+\sqrt{130-1}}{2}\bigg\rfloor=6 \le m \le \lfloor\sqrt{65-1}\rfloor=8\quad\land \quad m\in\{7,8\}\Rightarrow k\in\{4,1\}\\$$
$$F(7,4)=(33,56,65)\qquad \qquad F(8,1)=(63,16,65) $$

This "formula" will not find $3430$ directly but, after factoring $3430$ we find that it is $686\cdot 5$ and this formula will find a triple for $C=5$. Multiplying each term by $686$ then yields $\quad (2058,2744,3430)$.

Title: Sum of two squares theorem


问:
Find kernel of $([x+y]_2, 2x-y)$
In my practice problem I have:

$\phi=\mathbb{Z}\times \mathbb{Z} \rightarrow (\mathbb{Z}/2\mathbb{Z})\times \mathbb{Z}$ by homomorphism

$\phi(x,y)=([x+y]_2, 2x-y)$

Find $\ker(\phi)$ and show it is isomorphic to $\mathbb{Z}$

My answer:

$\ker(\phi)=\{ (x,2x) \in \mathbb{Z}^2 | x \in 2\mathbb{Z} \}$ as this ensures $[x+y]=[0]$ and $2x=y$

Let $f=x$ for $x \in 2\mathbb{Z}$. It is injective because $f=(0)$ iff $x=0$

Is my answer correct and how do I show it is surjective? It seems strange that $\mathbb{Z} \cong 2\mathbb{Z}$ because what would the preimage of $13 \in \mathbb{Z}$ be in $2\mathbb{Z}$

答:
As additive groups we indeed have $\Bbb Z\cong 2\Bbb Z$ by $x\mapsto 2x$, so the correspondent element of $13\in\Bbb Z$ is $26\in2\Bbb  Z$.

Your solution for the kernel is fine, we can write it alternatively as
$$\ker\phi=\{(2k,\, 4k):k\in\Bbb Z\}$$
and then define $f:\ker\phi\to\Bbb Z,\ f((2k,4k))=k$.

If you wish, you can express the same by $f((x,2x))=x/2$ and noting $x\in 2\Bbb Z$ which ensures $x/2\in\Bbb Z$.

Its inverse is $g(k)=(2k,\, 4k)$.

Title: Find kernel of $([x+y]_2, 2x-y)$


问:
Amoeba extinction probability
This has been asked several times on this website but I am trying to come up with a formal approach to justify why the extinction probability cannot be 1.

The problem statement is as follows. You start off with one amoeba in a petri dish. Every minute every amoeba in the dish either dies, does nothing, splits into two or splits into three. All these events have equal probability of $1/4$. What happens to amoebas are independent of each other. What is the probability that the amoebas eventually die out?

Let $A_n$ be the number of amoebas at minute $n$. $A_0 = 1$. Define $\tau := \inf\{n \geq 1: A_n = 0\}$ and $p := P\{\tau < \infty\}$. By conditioning on what happens to the first amoeba, we can write

$$p = \frac{1}{4}(1 + p + p^2 + p^3)$$
This has two positive solutions, namely $\sqrt{2}-1$ and $1$. Based on some intuitive reasoning, we rule out $1$ as the solution. I want to make that intuition formal. I can write

$$E[A_{n+1} \mid \mathcal{F_n}] = \frac{3}{2}A_n$$
$$E[A_n] = \left(\frac{3}{2}\right)^n$$

I don't see how these contradict $\tau < \infty$ a.s. $A_n$ is a submartingale. I tried stopping it at $\tau$ but I don't see where to go from there.

答:
Let $q_n=\mathbb P(A_n=0)$, so $p=\lim_{n\to\infty}q_n$. We can show that the PGF $G_n$ for the distribution of $X_n$ is the $n$-fold composition $G_n(t)=G^{(n)}(t)$, where $G(t)=\left(1+t+t^2+t^3\right)/4$ is the PGF for $A_1$.

Now $q_{n+1}=\mathbb P(X_{n+1}=0)=G_{n+1}(0)=G(G_n(0))=G(q_n)$, so taking $n\to\infty$ implies $p=G(p)$.

Claim: $p$ is the smallest non-negative root of $t=G(t)$.

Indeed, suppose the smallest root is $r$. Then $q_0=0\leq r$. Now if $q_n\leq r$, we have $$q_{n+1}=G(q_n)\leq G(r)=r.$$
So $q_n\leq r$ always, hence $p=r$.

Title: Amoeba extinction probability


答:
I agree with you for the kernel.

We have $\text{ker}(\phi)=\{(2u,4u)|u\in \mathbb{Z}\}$

To find an isomorphism, we can for example consider the application::

$\varphi: \mathbb{Z}\to \text{ker}(\phi)$ such that $\varphi(x)=(2x,4x)$

I pass on the check of homomorphism. The injectivity is clear...

Regarding surjectivity, it's pretty obvious, but you have to write it down.

Let $f\in \text{ker}(\phi)$. By definition of $\text{ker}(\phi)$, there exists $u\in\mathbb{Z}$ such that $f=(2u,4u)$ and we remark that $f=\varphi(u)$.

For your last remark, the isomorphism Between $\mathbb{Z}$ and $2\mathbb{Z}$ (quite natural) is the multiplication by 2 and the surjectivity comes from the fact that we find an antecedent of an element of $2\mathbb{Z}$ by dividing by 2.

Title: Find kernel of $([x+y]_2, 2x-y)$

Body:
In my practice problem I have:

$\phi=\mathbb{Z}\times \mathbb{Z} \rightarrow (\mathbb{Z}/2\mathbb{Z})\times \mathbb{Z}$ by homomorphism

$\phi(x,y)=([x+y]_2, 2x-y)$

Find $\ker(\phi)$ and show it is isomorphic to $\mathbb{Z}$

My answer:

$\ker(\phi)=\{ (x,2x) \in \mathbb{Z}^2 | x \in 2\mathbb{Z} \}$ as this ensures $[x+y]=[0]$ and $2x=y$

Let $f=x$ for $x \in 2\mathbb{Z}$. It is injective because $f=(0)$ iff $x=0$

Is my answer correct and how do I show it is surjective? It seems strange that $\mathbb{Z} \cong 2\mathbb{Z}$ because what would the preimage of $13 \in \mathbb{Z}$ be in $2\mathbb{Z}$


问:
Meaning of "Algebra of all functions separates points on unit circle"

Let $K$ be the unit circle in the complex plane, and let $\mathcal{A}$
be the algebra of all functions of the form
$$f(e^{i\theta})=\sum\limits^{N}_{n=0}c_ne^{tn\theta}\text{ ($\theta$
 real)}$$ Then $\mathcal{A}$ separates points on $K$ and $\mathcal{A}$
vanishes at no point of $K$, but nevertheless there are continuous
functions on $K$ which are not in the uniform closure of
$\mathcal{A}$.

Could anybody explain what this question means? Is there any specific math term of "algebra"? And I don't understand how we say it separates the unit circle. I'd appreciate any explanation. Thanks!

答:
Being an algebra means that if $f_1$ and $f_2$ are two such functions and if $\alpha$ is a scalar, then $f_1+f_2$, $f_1\cdot f_2$ and $\alpha f_1$ all belong to it also.

And asserting that this set separates the points means that if $x,y\in K$ and $x\ne y$, then there's some function $f$ from that set such that $f(x)\ne f(y)$.


答:
An algebra is a vector space (or, more generally, a module) equipped with a multiplication that plays nicely with the vector space operations. In this case, the vector space operations are all pointwise, and the multiplication is just pointwise multiplication of functions.

A set $S$ of functions $X \to Y$ separates points of $X$ if for all $a,b \in X$, there is some $f \in S$ such that $f(a) \neq f(b)$.

What the question is getting at is that all of the conditions of the Stone-Weierstrass Theorem are necessary. That theorem says (in one of its forms):

Theorem: If $X$ is a compact Hausdorff space and $\mathcal{A}$ is a subalgebra of $C(X,\mathbb{R})$ containing a non-zero constant function, then $\mathcal{A}$ is dense in $C(X,\mathbb{R})$ if and only if it separates points on $X$.

Here, we have a compact Hausdorff space (the unit circle $K$) and a subalgebra $\mathcal{A}$ of $C(K,\mathbb{R})$ which separates points, but $\mathcal{A}$ is not dense in $C(K,\mathbb{R})$. That is: the requirement for $\mathcal{A}$ to contain a non-zero constant is necessary for the theorem to hold.

Title: Meaning of "Algebra of all functions separates points on unit circle"

Body:
Let $K$ be the unit circle in the complex plane, and let $\mathcal{A}$
be the algebra of all functions of the form
$$f(e^{i\theta})=\sum\limits^{N}_{n=0}c_ne^{tn\theta}\text{ ($\theta$
 real)}$$ Then $\mathcal{A}$ separates points on $K$ and $\mathcal{A}$
vanishes at no point of $K$, but nevertheless there are continuous
functions on $K$ which are not in the uniform closure of
$\mathcal{A}$.

Could anybody explain what this question means? Is there any specific math term of "algebra"? And I don't understand how we say it separates the unit circle. I'd appreciate any explanation. Thanks!


问:
Can we really ignore negative term on complex integration?
So, my problem is shown here .
In the second last integral we have :

$$\frac{M}{R^{k-1}} \int_{\theta_0}^{\pi/2} e^{(R\cos{\theta})t} \Bbb d\theta = \frac{M}{R^{k-1}} \int_{0}^{\phi_0} e^{(R\sin{\theta})t} \Bbb d\phi \tag{Eq. 1}$$

As we know from the screenshot, we set $$\theta = \pi/2 - \phi\tag{Eq. 2}$$. So the left hand side of Eq. $1$ should be:

$$\frac{M}{R^{k-1}} \int_{\pi/2 - \theta_0}^{0} e^{(R\sin{\phi})t} \Bbb d\phi\tag{Eq. 3}$$

Then, from Eq. $2$ we can substitute the lower bound of integral Eq. $3$ and it becomes:

$$\frac{M}{R^{k-1}} \int_{\phi_0}^{0} e^{(R\sin{\phi})t} \Bbb d\phi$$

But as we know in real integration if we want to change lower and upper bound we should multiply the integrand with $(-1)$. So:

$$\frac{M}{R^{k-1}} \int_{0}^{\phi_0} -e^{(R\sin{\phi})t} \Bbb d\phi$$

But wait! Let's see what's shown on the screenshot, it's clear that there's no negative term there. And I believe $M$, $R$, and $k$ have nothing to do with the integral since they're all positive. Or maybe the negative sign vanishes because of absolute value? But we've already removed the absolute value.

So, my question is Where did the negative sign go or maybe I missed something very obvious? Hope you can help me. I spent 2 hours to read this over and over again but still don't understand.

Thanks in advance!

答: At first glance, since $ \theta = \pi/2 - \phi $, you have an extra minus, since $ d\theta = - d\phi $. So the two minus signs cancel out.


问:
What are cohomology rings good for?
I am studying some concepts of algebraic topology myself, and I read lately a bit about cohomology rings (created by the direct sum of cohomology groups) but besides all definitions I could not find any information.

What are they good for?

How do we use them?

Could someone give me a very short descriptive example?

Thank you

答:
The main idea behind the cohomology ring is that you have an extra structure that allows you to say more about your space.  In some instances, you can use it to say that two spaces which have isomorphic (co)homology groups are different because they have different cohomology rings, and in some instances you can infer information about the cohomology groups by knowing that there is an additional structure.

For an example of the first case, the cohomology of real projective space is $H^*(\mathbb{R}P^n;\mathbb{Z}/2\mathbb{Z})\cong \mathbb{Z}/2\mathbb{Z}[x]/(x^{n+1})$.  This distinguishes projective space from a wedge product $\bigvee_{i\leq n} S^i$, because the multiplication in the corresponding cohomology ring is trivial.

For an example of the second case, suppose that a space has a product on it satisfying some mild conditions, making it an H-space.  Then the cohomology ring will be a Hopf algebra,.  If the space is also a connected, compact manifold, then (taking cohomology over a field of characteristic zero) one can use the classification of connective, finite type, commutative hopf algebras (the Milnor-Moore theorem) to say what the ring looks like. In particular, it must be the alternating algebra on odd-degree generators.  This is enough to show that there are no odd-dimensional analogues of the quaternions.

I am blanking on truly simple examples which exemplify the power of cohomology rings, but it should at least be plausible that, just as you can say useful things by knowing that a set has a group structure, you can say useful things by knowing that a graded vector space has a graded-commutative ring structure.

One other application worth mentioning is that the cohomology ring (plus Poincare duality) allows us to study intersections in an algebraic manner.

Suppose that $M$ is an $n$ dimensional compact oriented manifold, and that $X$ and $Y$ are oriented submanifolds with $\dim X + \dim Y = n$.  If we perturb $X$ and $Y$ slightly (so that the intersection is transversal, they will intersect at discrete points, and if we count orientation and multiplicity correctly, the number of intersection points is independent of our perturbation.  We denote this intersection number by $I_M(X,Y)$.  This gives interesting topological invariants.  For example, if we embed $X$ in $X\times X$ via the diagonal embedding, then $I_{X\times X}(X,X)=\chi(X)$ is the Euler characteristic of $X$.

It turns out that we can describe the intersection numbers in terms of the cohomology ring.  Given $X$ and $Y$, there are corresponding cohomology classes $[X]$ and $[Y]$ (called fundamental classes), and pairing the $[X]\cup [Y]$ with the orientation cycle in the top degree homology spits out a number, which is exactly the intersection number of $X$ and $Y$.

This has also been applied to algebraic geometry with great success.  The algebraic geometer in you may want to look into intersection theory/intersection cohomology.

Title: What are cohomology rings good for?


问:
Solve PDE: $~u_x +u_y =u^2~$ Initial condition $~u(x,-x)=x~.$
I need help this: Solve the following initial value problem

$u_x+u_y=u^2,~~ u(x,-x)=x$

Here what I did: The characteristic equations: $~dx/dt=1,~~ dy/dt=1~$ and $~du/dt=u^2~.$

$du/u^2= dx$
Integrate:
$-1/u+ C_1=x$ ;      $C_1 =1/u+ x$;    $ u=1/(C_1-x)$

$dx = dy$
Integrate:
$y + C_2=x$ ; $C_2 = x - y$

Therefore, the complete integral of the equation:
$F ( 1/u + x,x - y) = 0$ or in another form:
$u =1/(C_1-x)=1/(f(C_2)-x)= ( f (x- y)- x )^{-1}$ where $f$ - some differentiable function.
The initial condition $u(x,-x) = x$;
$( f (x+x)- x )^{-1}=x$;
$1=x(f(2x)-x)\implies f(2x)=(1+x^2)/x$

But I don't know what do now...

Thanks

答:
So you get, $$-1/u+ C_1=x\implies C_1=x+1/u\\~~~~~~~~~~~~~~~~~~~~~~\text{and}~~~~~ C_2 = x - y$$
Hence the solution of the given PDE is of the form $~x+1/u=f(x - y)~,$ where $~f~$ is arbitrary function.

Now initial condition gives, $~u(x,-x)=x~,$ so the above solution becomes
$$x+\dfrac 1x=f(2x)~~~~~\text{i.e.,}~~~ f(y)=\dfrac y2+\dfrac 2y$$ $($Putting $y=2x)$

So $$x+\dfrac 1u=f(x - y)\implies x+\dfrac 1u=\dfrac {x-y}2+\dfrac 2{x-y}\implies \dfrac 1u=\dfrac {x-y}2+\dfrac 2{x-y}-x$$ $$\implies \dfrac 1u=\dfrac{y^2-x^2+4}{2(x-y)}\implies (y^2-x^2+4)u=2(x-y)$$
This is the required solution of the given initial valued PDE.


问:
Understanding change of variables in double integrals
In this libre text chapter, in example 15.7.1B, the author illustrates the change of variables using the following example:

The statement near it is written:

For the vertical length  $A:u=0,0≤v≤1$  transforms to  $x=−v^2,y=0$  so this is the horizontal length $A′ $ that joins  $(−1,0)$  and $ (0,0)$

I find it confusing that regardless of whatever function of $x$ could have been on $v$, it seems that all of them would only have a straight line of $A'$ like suppose it was $x=-v^3$, that'd also denote the same line

If I understood correctly, we derive $A'$ from $A$ by feeding constraints of $A$ into the dependencies of the new variables on the old. Under that idea, does the kind of function $x$ is on the old coordinates not matter in sketching the region(only the bounds does)?

答:
When you transform a range to another range via a transformation, the relation between old and new variables are fixed by that transformation.

When you transform a range to another range via a transformation, the only thing you need is to find the boundary of new range.

When you transform a range to another range via a transformation, it has not yet involve any functions.

Let us see a function $f(v,u)$ defined on a domain $D$, your only task is to transform the domain of the function $f(v,u)$ to a new function $\tilde f(x,y)$ on a new domain $\tilde D$. You do not need to know the exact form of the function, only thing you need is to find the boundary of $\tilde D$, based on knowing the boundary of $D$ and the transformation.


问:
Rudin: Showing at least one of the trichotomy holds for dedekind cuts
Let $\alpha ,\beta$ be dedekind cuts, define $\alpha < \beta$ to mean $\alpha$ is a proper subset of $\beta$.

It is clear from the definition of proper subset that at most one of:
$\alpha < \beta$, $\alpha = \beta$, $\alpha > \beta$ can hold.

But we also need to show at least one of the 3 cases must hold to finish trichotomy proof. Rudin only seems to proves that if $\alpha < \beta$, $\alpha = \beta$ are both false, then $\alpha > \beta$ case must be true. This I understand why.

The problem is what if $\alpha > \beta$, $\alpha < \beta$ are false, how to we prove $\alpha = \beta$? He didn't tell the reader to try it themselves, so I'd assume it's really obvious but I can't think of an obvious reason.

The approach I have is first to show that $\alpha$ and $\beta$ cannot both have elements exclusive to the other set otherwise it contradicts definition of dedekind cuts, then naturally I'd get $\alpha \subset \beta, \beta \subset \alpha$ so $\alpha = \beta$. Is this a valid way to prove this?

Because Rudin didn't include it, is there a trivial reasoning I'm missing?

答: If $ \alpha = \beta $ was false, you would get a contradiction. This is because you would have $ \alpha < \beta $ and $ \alpha = \beta $ false. You have proven that this implies $ \alpha > \beta $. In terms of logic showing that $ \neg p $ and $ \neg q $ imply $ r $, is equivalent to showing that one of $ p, q $ and $ r $ is true.


问:
Showing that the right and left regular representations are equivalent
Let $G$ be a finite group. Let $C(G,\mathbb{C})$ be the complex vector space of  all functions from $G$ to $\mathbb{C}$. We define two representations of G on $C(G,\mathbb{C})$: the left regular representation $\ell$ given by $(\ell(g)f)(x)=f(g^{-1}x)$ an the right regular representations $r$ given by $(r(g)f)(x)=f(xg)$. I try to show that the are equivalent. So we need to find an invertible intertwining operator $A$ such that $A\ell(g)=r(g)A$ for all $g\in G$. My book says $(Af)(x)=f(x^{-1})$ works and it is really easy to check that it really works. But I want to find the equation $(Af)(x)=f(x^{-1})$ explicitly.

Attempt: We need to find $A$ such that for any $f\in C(G,\mathbb{C})$ , $$(A\ell(g))(f)=(r(g)A)(f)$$ i.e. $\big((A\ell(g))(f)\big)(x)=\big((r(g)A)(f)\big)(x)$ for all $x\in G$

But $$\big((r(g)A)(f)\big)(x)=r(g)\big((Af)(x)\big)=(Af)(xg)$$
Similarly $$\big((A\ell(g))(f)\big)(x)=(Af)(g^{-1}x)$$

So the equality $(Af)(x)=f(x^{-1})$ given by the book does not imply $(Af)(xg)=(Af)(g^{-1}x)$. I couldn't see my mistake. Can anyone help? And how can we find the equality $(Af)(x)=f(x^{-1})$ for all $x\in G$? Thanks.

答:
Mistake can be traced to the second equality in "But":

$$[r(g)Af] (x) = [r(g)f] (x^{-1}) = f(x^{-1}g),$$

i.e. you have implicitly assumed $r(g)A = Ar(g)$, hence your error.

For completeness, here is the calculation for the left regular repr:

$$[Al(g)f] (x) = [Af] (g^{-1}x) = f((g^{-1}x)^{-1}) = f(x^{-1}g).$$

Aside: I was also trying to check unitary equivalence of $(L,L^2(K))$ and $(R,L^2(K))$, where the group $G=K$ is now compact and the repr space is $L^2(K)$. Since $K$ is compact, the modular function $\Delta$ is identically $1$, and so the operator $A$, now viewed as an operator on $L^2(K)$, is unitary.


问:
Recurrence relations of the form $T(n) = kT(n-1) + f(n)$
What can be said about the asymptotic behaviour of recurrence relations of the form
$$
T(n) = kT(n-1) + f(n), \text{where } k \in \mathbb{N}^{\ge2}
$$
which often occur in recursive algorithm analysis?

It is well-known that $T(n) \in \Theta(k^n)$ when $f(n) \in O(n)$. But is there a result for general $f$? Or at least for polynomial $f(n) \in \Theta(n^P)$ or other standard classes of $f$?

答:
This reccurence relation can be solved :

Let $ n\in\mathbb{N} $, and $ p\leq n $, we have : \begin{aligned}T\left(n-p\right)&=kT\left(n-p-1\right)+f\left(n-p\right)\\ \iff\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ k^{p}T\left(n-p\right)&=k^{p+1}T\left(n-p-1\right)+k^{p}f\left(n-p\right)\\ \Longrightarrow\sum_{p=0}^{n-1}{\left(k^{p}T\left(n-p\right)-k^{p+1}T\left(n-p-1\right)\right)}&=\sum_{p=0}^{n-1}{k^{p}f\left(n-p\right)}\\ \iff\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ T\left(n\right)-k^{n}T\left(0\right)&=\sum_{p=1}^{n}{k^{n-p}f\left(p\right)}\\ \iff \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ T\left(n\right)&=k^{n}\left(T\left(0\right)+\sum_{p=1}^{n}{\frac{f\left(p\right)}{k^{p}}}\right)\end{aligned}

Thus, $ \left(\forall n\in\mathbb{N}\right) $ : $$T\left(n\right)=k^{n}\left(T\left(0\right)+\sum_{p=1}^{n}{\frac{f\left(p\right)}{k^{p}}}\right)$$


问:
List of quadratic field with the UFD property
Let D be a square free integer let $K=\mathbb{Q}(\sqrt{D})$ and let
$\mathcal{O_{K}}$ be the ring of integer of $K$

My question: where I can find a list of of the value of $D$ which makes the ring of the  quadratic field
$K=\mathbb{Q}(\sqrt{D})$ have the UFD property for $D<300$ ?

Can we check whether any value of $D$ make the the ring of a quadratic field have the UFD property using sage or magma ? if yes what is the command?

答:
The broken one-liner

```<code>L = [n for n in [2..300]
     if n.is_squarefree() and QuadraticField(n, 'a').class_number() == 1]
</code>```

computes in sage the list of all squarefree $n\le 300$, so that $\Bbb Q(\sqrt n)$ has class number one, this corresponds to the property of being UFD. Some first few entries in L are:

```<code>sage: L[:21]
[2, 3, 5, 6, 7, 11, 13, 14, 17, 19, 21, 22, 23, 29, 31, 33, 37, 38, 41, 43, 46]
</code>```

Further checks in sage:

```<code>sage: K.<a> = QuadraticField(19)
sage: K
Number Field in a with defining polynomial x^2 - 19 with a = 4.358898943540674?
sage: K.class_number()
1
sage: K.OK()
Maximal Order in Number Field in a with defining polynomial x^2 - 19 with a = 4.358898943540674?
sage: K.OK().class_number()
1
</code>```

In contrast:

```<code>sage: L.<b> = QuadraticField(34)
sage: L
Number Field in b with defining polynomial x^2 - 34 with b = 5.830951894845300?
sage: L.class_number()
2
sage: LOK = L.OK()
sage: LOK
Maximal Order in Number Field in b with defining polynomial x^2 - 34 with b = 5.830951894845300?
sage: LOK.class_number()
2
sage: L.class_group()
Class group of order 2 with structure C2 of Number Field in b with defining polynomial x^2 - 34 with b = 5.830951894845300?
sage: L.class_group().gens()
(Fractional ideal class (3, b + 1),)
sage: L.ideal(3)
Fractional ideal (3)
sage: L.ideal(3).factor()
(Fractional ideal (3, b + 1)) * (Fractional ideal (3, b + 2))
sage: L.ideal(29).factor()
(Fractional ideal (29, b + 11)) * (Fractional ideal (29, b + 18))
</code>```

Body:
The broken one-liner

L = [n for n in [2..300]
     if n.is_squarefree() and QuadraticField(n, 'a').class_number() == 1]

computes in sage the list of all squarefree $n\le 300$, so that $\Bbb Q(\sqrt n)$ has class number one, this corresponds to the property of being UFD. Some first few entries in L are:

sage: L[:21]
[2, 3, 5, 6, 7, 11, 13, 14, 17, 19, 21, 22, 23, 29, 31, 33, 37, 38, 41, 43, 46]

Further checks in sage:

sage: K.<a> = QuadraticField(19)
sage: K
Number Field in a with defining polynomial x^2 - 19 with a = 4.358898943540674?
sage: K.class_number()
1
sage: K.OK()
Maximal Order in Number Field in a with defining polynomial x^2 - 19 with a = 4.358898943540674?
sage: K.OK().class_number()
1

In contrast:

sage: L.<b> = QuadraticField(34)
sage: L
Number Field in b with defining polynomial x^2 - 34 with b = 5.830951894845300?
sage: L.class_number()
2
sage: LOK = L.OK()
sage: LOK
Maximal Order in Number Field in b with defining polynomial x^2 - 34 with b = 5.830951894845300?
sage: LOK.class_number()
2
sage: L.class_group()
Class group of order 2 with structure C2 of Number Field in b with defining polynomial x^2 - 34 with b = 5.830951894845300?
sage: L.class_group().gens()
(Fractional ideal class (3, b + 1),)
sage: L.ideal(3)
Fractional ideal (3)
sage: L.ideal(3).factor()
(Fractional ideal (3, b + 1)) * (Fractional ideal (3, b + 2))
sage: L.ideal(29).factor()
(Fractional ideal (29, b + 11)) * (Fractional ideal (29, b + 18))


问:
predicate logic question| equality between two sentences
I have two sentences:

$A=\forall x\left[\left(\exists y\left(S\left(x,y\right)\right)\right)\longrightarrow\left(\forall z\exists y\left(f\left(x,y\right)=z\right)\right)\right]$

$B=\forall x\forall z\left[(\forall y\left(f\left(x,z\right)\neq y\right))\longrightarrow(\exists y\left(\lnot S\left(x,y\right)\right))\right]$

The question is if they are both logical equaled.

答:
Converting the implications in both statements to disjunctions, we have:

$$\forall x[\exists y(\neg S(x,y)) \vee\forall z \exists y (f(x,y) = z)]\\
\forall x \forall z [\exists y (f(x,z) = y) \vee \exists y(\neg S(x,y)]$$

Suppose $S(x,y) = TRUE$ for all $x,y$, and $f(x,y) = \sin(y)$.

If $z = 2$, then there is no $y$ such that $\sin(y) = 2$, so we can show that the first sentence is false for these functions.

However, $\sin(z)$ is defined for all real numbers $z$, so there must always exist a $y$ such that $f(x,z) = y$. Using this we can show the second sentence is true for these functions.


答:
Hint:

You can transform the equation with

$$T(n)=k^nU(n),$$ giving

$$k^nU(n)=k^nU(n)+f(n)$$ or

$$U(n)=U(n-1)+k^{-n}f(n).$$

Then,

$$U(n)=U(0)+\sum_{i=1}^n k^{-i}f(i)$$ and $$T(n)=k^nT(0)+k^n\sum_{i=1}^n k^{-i}f(i)$$

Title: Recurrence relations of the form $T(n) = kT(n-1) + f(n)$

Body:
What can be said about the asymptotic behaviour of recurrence relations of the form
$$
T(n) = kT(n-1) + f(n), \text{where } k \in \mathbb{N}^{\ge2}
$$
which often occur in recursive algorithm analysis?

It is well-known that $T(n) \in \Theta(k^n)$ when $f(n) \in O(n)$. But is there a result for general $f$? Or at least for polynomial $f(n) \in \Theta(n^P)$ or other standard classes of $f$?


问:
How do linear transformation properties preserve vector space structure?
Currently I'm reading the Mathematics for Machine Learning book, and in the beginning of section 2.7 of the book, the following is given:

2.7 Linear Mappings

In the following, we will study mappings on vector spaces that preserve their structure, which will allow us to
deﬁne the concept of a coordinate. In the beginning of the chapter, we
said that vectors are objects that can be added together and
multiplied by a scalar, and the resulting object is still a vector. We
wish to preserve this property when applying the mapping:  Consider
two real vector spaces V,W. A mapping Φ : V → W preserves the
structure of the vector space if

Φ(x + y) = Φ(x) + Φ(y) (2.85)

Φ(λx) = λΦ(x) (2.86)

for all x, y ∈ V and λ ∈ R.

Generally, when vectors are added together, or multiplied by a scalar, the result is also a vector (the same type of object).

How are the above properties of a linear transformation preserving this?

As long as the LHS and RHS of both given properties belong to the same vector space, can I not consider the vector properties preserved? Why do they have to be equal?

What does it mean by they preserve the structure of a vector space?

Is there a particular reason Φ(λx + ψy) = λΦ(x) + ψΦ(y) has a form similar to a linear combination (y = a1.x1 + a2.x2 +...)?

答:
Once upon a time there was a  Halloween party to which all vectors of $\mathbb R^n$ were invited.  Of course it was a
costume party and the vectors did their best to choose the most creative costumes.

Besides drinking, eating and dancing, the vectors also did what they love to do most, which is to be multiplied by scalars
and added together.  People had a lot of fun when a prize was announced for whoever could prove or disprove that there is a scalar
$\lambda $ such that
$$
  \lambda \times \text{Robin} = \text{Batman},
  $$
but the eventual revelation that
$$
  \text{Frankenstein} +   \text{Dracula} =   \text{Frankenstein}
  $$
was a big embarrassment for Dracula who thought his expensive costume would not give away his true identity!

In any case,  the function
$$
  T:\mathbb R^n \to  \text{Party},
  $$
assigning to each vector $v$, the party guest played by the vector $v$  is a linear function  and so it preserves the
vector operations in the sense that, e.g. the sum of two vectors doesn't care about which costume  each vector is
wearing.

The expression of this fact is nothing but
$$
  T(v+u)=T(v)+T(u),
  $$
and I encourage you to read the above formula aloud, replacing "$T$" for "the character represented by".


问:
3rd grade exercise: "make your own turning pattern"
My 8 year old has been given a worksheet of numeric sequences, e.g. "what are the next numbers in the sequence 11, 12, 14, 15, 17, ..." and "make your own number pattern" and "make your own colour repeating pattern". I've had no problem helping her with these.

The final exercise is "Make your own turning pattern". No other explanation is given.

Apart from sending her back to her teacher to ask "what's a "turning pattern"?", what would you say this is?

答:
My guess would be they want something like this (as examples): the "turning pattern"
$$\underbrace{\fbox{$\,\uparrow\,\strut$}\;\fbox{$\rightarrow\strut$}\;\fbox{$\,\downarrow\,\strut$}\;\fbox{$\rightarrow\strut$}}_{\large\mathtt{part\, that\, repeats}}\;\fbox{$\,\uparrow\,\strut$}\;\fbox{$\rightarrow\strut$}\;\fbox{$\,\downarrow\,\strut$}\;\fbox{$\rightarrow\strut$}\;\cdots$$
corresponds to walking in a snaking line, and
$$\underbrace{\fbox{$\,\uparrow\,\strut$}\;\fbox{$\,\uparrow\,\strut$}\;\fbox{$\rightarrow\strut$}}_{\substack{\large\mathtt{part\, that}\\ \large\mathtt{repeats}}}\;\fbox{$\,\uparrow\,\strut$}\;\fbox{$\,\uparrow\,\strut$}\;\fbox{$\rightarrow\strut$}\;\cdots$$
corresponds to repeatedly moving like a knight can in chess.

It's worth pointing out that "guess what comes next" exercises like

Find the next numbers in the sequence: 11, 12, 14, 15, 17, ...

are really not very good problems (in my opinion). All they judge is whether you can guess what the teacher is thinking of; they do not ask a real mathematical problem. This is because any finite sequence can be extended in literally any way you want; there is always a justification that can be made for it. There is no answer for what the sequence "should" be.

There is a rigorous sense in which what I've said above is true, but sometimes it can be a little hard to get one's head around for someone not mathematically inclined; it might be easier to note that there is a similar problem with analogies. If you were given an analogy to complete,

$\mathsf{grass} : \mathsf{green}\, ::\, \mathsf{ocean} : \mathord{?}$

you can guess that the answer should be "blue", because the teacher is likely thinking of the relation "grass is green", but there is just as much reason to say any other color. For example, "orange" works, because for all we know, the intended relationship between green and grass is "they start with the same letter" (of course, other colors may require more convoluted explanations).


问:
Using contour integration to evaluate $\int_{0}^{\infty} \frac{2 \cos (x) \ln (x) + \pi \sin (x)}{x^2+4} \, \mathrm dx$
I want to use contour integration to show that $$\int_{0}^{\infty} \frac{2 \cos (x) \log (x) + \pi \sin (x)}{x^2+4} \, \mathrm  dx = \frac{\pi \log (2)}{2e^{2}}.$$

The recommendation in the textbook I'm reading is to integrate the function $$f(z) = \frac{e^{iz} \log(-iz)}{z^{2}+4}$$ around an indented contour that consists of the real axis from $-R$ to $R$, along with the upper half of the circle $|z|=R$.

The fact that $f(z)$ involves $\log(-iz)$ and not simply $\log (z)$ makes me a bit uneasy.

I think we would want to place the branch cut along the negative imaginary axis so that it doesn't intersect the contour.

Would that mean $- \pi  < \arg(-iz) \le  \pi$?

EDIT:

I converted the rest of my question into an answer.

答:
Revisiting this question, I see no reason why we couldn't simply integrate the function $$f(z) = \frac{e^{iz} \log (z)}{z^{2}+4}$$ around the same contour with the branch cut in the same location.

The only extra step would be extracting the real part of the result.

By Jordan's lemma (or the estimation lemma), $ \int f(z) \, \mathrm dz $ vanishes along the upper half of the circle $|z|=R$ as $R \to \infty$.

Therefore,

$$ \begin{align} &\int_{-\infty}^{0} \frac{e^{ix} (\log |x| + i \pi)}{x^{2}+4} \, \mathrm  dx + \int_{0}^{\infty} \frac{e^{ix} \log (x)}{x^{2}+4} \, \mathrm  dx \\ &= \int_{0}^{\infty} \frac{e^{-iu} (\log (u) + i \pi)}{u^{2}+4} \, \mathrm du + \int_{0}^{\infty} \frac{e^{ix} \log (x)}{x^{2}+4} \, \mathrm dx \\ &= \int_{0}^{\infty} \frac{(\cos (x) - i \sin (x)) (\log (x) + i \pi)+ (\cos (x) + i \sin (x))\log (x)}{x^{2}+4}  \, \mathrm dx \\ &= \int_{0}^{\infty} \frac{2 \cos (x) \log (x)+ \pi \sin (x)}{x^{2}+4} \, \mathrm  dx + i \pi \int_{0}^{\infty} \frac{\cos (x)}{x^{2}+4} \, \mathrm dx \\ &= 2 \pi i \ \text{Res}[f(z),2i] \\ &= 2 \pi i \left(\frac{e^{-2} (\log (2) + \frac{i \pi}{2})}{4i} \right) \\&= \frac{\pi \log (2)}{2e^{2}} + i \frac{\pi^{2}}{4e^{2}} .\end{align}$$

By equating the real parts on both sides of the equation,we arrive at the answer.


问:
Let $U\subset\mathbb{R}^3$ be defined by the basis set $B=\{[1,1,1]^T,[1,-1,1]^T\}$. Formulate a system of equations to find the...
orthogonal complementary subspace $W$ of $U$. Use the orthogonality of $U$ and $W$ to propose a fast method to express the vector $[2,2,1]^T$ as a sum of vectors from these complementary subspaces.

Let $B_W=[b_1,b_2,b_3]^T$ be the orthogonal complementary subspace of $B$. In order for me to find it, each of the spaces must be orthogonal to eachtoher, therefore the dot product between them must be $0$: $$[1,1,1]^T\cdot[b_1,b_2,b_3]^T=0\longrightarrow1\cdot b_1+1\cdot b_2+1\cdot b_3=0$$ for the first basis vector of $B$ and $$[1,-1,1]^T\cdot[b_1,b_2,b_3]^T=0\longrightarrow1\cdot b_1-1\cdot b_2+1\cdot b_3=0$$ for the second basis vector of $B$. But this gives me a system of equations of 2 equations and 3 unknowns, I am not sure where I can get the third equation. The last condition I know for $B_W$ to be an orthogonal complementary subspace of $B$ is that the union of the two basis sets must form a basis for $\mathbb{R}^n$.

I am not sure how to do the second part of the problem either, but I think once I get the first part I can attempt the second one too, so I just need a hint, thanks.

SOURCE: Linear Algebra and Optimization for Machine Learning: A Textbook (page 63, problem 2.3.4)

答:
You don't need a third equation. The subspace you're looking for has dimension $1$, so it has infinitely many vectors. In other words, your system of equations must be indeterminate.

For solving the system, if you add both equations you'll get
$$2b_1+2b_3=0\Longrightarrow b_3=-b_1.$$
And substituting in any of the two equations, you can see that $b_2=0$. So your subspace is
$$W=\big\{[b_1,0,-b_1]^T:\ b_1\in\mathbb{R}\big\}=\big\langle[1,0,-1]^T\big\rangle.$$

For the second question, suppose that $\vec v=[2,2,1]^T=\vec u+\vec w$, with $\vec u\in U$ and $\vec w=b[1,0,-1]^T\in W$. Then
$$[2,2,1]^T\cdot[1,0,-1]^T=\underbrace{\vec u\cdot[1,0,-1]^T}_0+b[1,0,-1]^T\cdot[1,0,-1]^T\Longrightarrow$$
$$\Longrightarrow1=2b\Longrightarrow b=\frac{1}{2}.$$
So you have
$$\vec w=\frac{1}{2}[1,0,-1]^T=\left[\frac{1}{2},0,-\frac{1}{2}\right]^T\in W;$$
$$\vec u=[2,2,1]^T-\left[\frac{1}{2},0,-\frac{1}{2}\right]^T=\left[\frac{3}{2},2,\frac{3}{2}\right]^T.$$


问:
How many descending order number between 100 and 600?
Example of valid number are:

(1) 431

(2) 543

(3) 210

(4) 321

Example of not valid number are:

(1) 221

(2) 500

(3) 654

The answer is 120 but I don't know why.

My attempt:

I have to construct a string of length $n=3$ with some constraints.
I split the problem into 5 small problems.

$D_1 = \{\text{all valid number staring with 1}\}$

$|D_1| = 0$

$D_2 = \{\text{all valid number staring with 2}\}$

$|D_2| = 1 \times {2\choose 2}$, because I can choose only from $\{0,1\}$

$D_3 = \{\text{all valid number staring with 3}\}$

$|D_3| = 1 \times {3\choose 2}$, because I can choose only from $\{0,1,2\}$

$D_4 = \{\text{all valid number staring with 4}\}$

$|D_4| = 1 \times {4\choose 2}$, because I can choose only from $\{0,1,2,3\}$

$D_5 = \{\text{all valid number staring with 5}\}$

$|D_5| = 1 \times {5\choose 2}$, because I can choose only from $\{0,1,2,3,4\}$

My solution: $$\sum_{i \in \{1,2,3,4,5\}}|D_i| = 35$$ I can sum directly because there is no double counting but it's wrong...

Editing from the future:

I made a mistake on the summation, the solution is 20. And from the textbook 120 were not the correct answer...

$$\sum_{i \in \{1,2,3,4,5\}}|D_i| = 20$$ I can sum directly because each set is disjoint (there is no double counting). In conclusion my long not-necessary aproach was right but (if you are reading this post) I recommend you to read all the comments below and the answers because they helped me a lot.

答:
Summarizing information from the comments,
a decreasing number of three digits requires three different digits from the set
$\{0,1,2,3,4,5,6,7,8,9\}.$
Once you have chosen the three digits they must appear in the number in decreasing order, so from any one subset of three digits there is only one way to form a decreasing number. So we get a one-to-one correspondence between the three-element subsets of a set of ten elements and the three-digit decreasing numbers.
There are
$$ \binom{10}{3} = 120 $$
three-digit decreasing numbers.

If you put the restriction that the number must be between $100$ and $600,$
the digits must be selected from the six-element set
$\{0,1,2,3,4,5\},$
so there are only
$$ \binom{6}{3} = 20 $$
three-digit decreasing numbers less than $600.$

Note that in the calculation for numbers less than $600,$ we have
$$ \binom 63 = \frac{6!}{3!3!} = \frac{6\cdot 5\cdot 4}{3!} = \frac{120}{6},$$
so if you work out the numerator but forget to divide by $6$
you could mistakenly say the answer is $120$
(which is the correct answer for the number of permutations
of three digits out of six).


问:
How to organize columns and rows of a matrix mathematically
I have a matrix M whose size i.e $8$x$8$ , as follows:

where $x$ can be any non-zero value. . I need to organize this matrix $M$ to move the elements $M$(3:4,3:4) to be in positions $N$(7:8,7:8). On the other hand, the elements of $M$(5:8,5:8) to be moved into $N$(3:6,3:6). It means that the matrix $N$ should be as follows:

My question, how should it be the permutation matrix to be multiplied by $M$ in order to get $N$? Or if there is another mathematical expression we can use to get matrix $N$ based on M. I need it in general case, it means whatever the size of matrix $M$, we can organize it to get the other matrix as in matrix $N$.

EDIT:

What I need exactly is to move the elements in red into last rows and column and then move the elements in blue into their places as follows:

and the resultant matrix is as below:

答:
Hint: If you want to find the matrix that swap  two rows or columns just swap that rows or columns in the Identity matrix. For example if you want to swap columns $3\leftrightarrow 7$ then you should multiply (from left side or right?) your matrix to the following matrix:

$$\begin{pmatrix}1&0&\color{red}{0}&0&0&0&\color{blue}{0}&0\\
0&1&\color{red}{0}&0&0&0&\color{blue}{0}&0\\
0&0&\color{red}{1}&0&0&0&\color{blue}{0}&0\\
0&0&\color{red}{0}&1&0&0&\color{blue}{0}&0\\
0&0&\color{red}{0}&0&1&0&\color{blue}{0}&0\\
0&0&\color{red}{0}&0&0&1&\color{blue}{0}&0\\
0&0&\color{red}{0}&0&0&0&\color{blue}{1}&0\\
0&0&\color{red}{0}&0&0&0&\color{blue}{0}&1\\\end{pmatrix}\to \begin{pmatrix}1&0&\color{blue}{0}&0&0&0&\color{red}{0}&0\\
0&1&\color{blue}{0}&0&0&0&\color{red }{0}&0\\
0&0&\color{blue}{0}&0&0&0&\color{red}{1}&0\\
0&0&\color{blue}{0}&1&0&0&\color{red}{0}&0\\
0&0&\color{blue}{0}&0&1&0&\color{red}{0}&0\\
0&0&\color{blue}{0}&0&0&1&\color{red}{0}&0\\
0&0&\color{blue}{1}&0&0&0&\color{red}{0}&0\\
0&0&\color{blue}{0}&0&0&0&\color{red}{0}&1\\\end{pmatrix}$$

Example: Verify the following:
$$\begin{pmatrix}1&2&3\\
4&5&6\\
7&8&9 \end{pmatrix}\times \begin{pmatrix}1&\color{red}{0}&\color{blue}{0}\\
0&\color{red}{0}&\color{blue}{1}\\
0&\color{red}{1}&\color{blue}{0} \end{pmatrix}=\begin{pmatrix}1&3&2\\
4&6&5\\
7&9&8 \end{pmatrix}$$


问:
Can a set containing non-standard elements only, have a legal set formation?
Is it possible to form a set that contains only non-standard elements legally? A legal set formation means that only internal formulas can be used to form the set.

Here's what I thought: A set containing non-standard elements must be a non-standard set. Otherwise, by transfer principle, it has to be empty. Thus it can't be uniquely described by internal mathematics. It can't take the form as $\{x : A(x)\}$.

答:
A set containing a single non-standard element is perfectly legal.

If the "standard" predicate cannot be used, then translated into NSA you can only define sets that are natural extensions of real sets $X$.  The natural extension of the empty set $X=\emptyset$ is still empty.  If $X$ is inhabited, then its natural extension, which contains all the elements of $X$, will necessarily contain standard elements.


问:
Infinite Non-abelian group with every non-identity has infinite order.
Is there any infinite Non-abelian group whose every non-identity element has infinite order?</p>

答:
Yes. For instance, the so-called free group on two elements $F_2$.

It consists of all finite words (including the empty word) using the four letters $a,b,a^{-1}$ and $b^{-1}$, where the group operation is word concatenation, and with the obvious inverse cancelling (e.g. $abb^{-1}a=aa$).

This is, in some sense, the "most general" group generated with two elements, in the sense that given any group $G=\langle x,y\rangle$ generated by two elements, there is a unique homomorphism $F_2\to G$ that maps $a\mapsto x,b\mapsto y$.

For those interested in algebraic topology, this is the fundamental group of the figure-8.


问:
How many faces does an $n$-dimensional slice of $[-1,1]^m$ have?
How many faces does an $n$-dimensional slice through the center of $[-1,1]^m$ have?

I read that:

"...any $n$-dimensional slice of $[-1,1]^m$ is a body with at most $m$ pairs of faces."

How should I prove this? How do we define a slice here?

My thoughts:

I took the example of $m=n=3$, and it made sense, i.e. cutting a biscuit sort of a shape from a cube in $3D$. I don't know how to generalize though!

Could someone please provide hints or any visuals that might aid my imagination?

答:
I would take "slice" to mean the intersection of a linear subspace of $\mathbf{R}^m$ with the cube. If $m=n$ then your subspace is all of $\mathbf{R}^m$ and your slice of the cube is the entire cube. (So I'm not sure what you mean by "biscuit sort of shape".) If you take $m=3$ and $n=2$, you are slicing the cube with a plane. If the plane is parallel to a pair of faces of the cube, the slice will be a square, but tilting the plane can also give rectangular and hexagonal slices.

The boundary of the slice will be formed from the intersection of the linear subspace and the boundary of the cube, which has $2m$ faces (of dimension $m-1$), so the slice won't have more than $2m$ faces. Since the cube is convex, its intersection with a linear subspace is also convex, so we don't have to worry that a face of the cube will contribute more than one face to the slice. Hence $2m$ really is the maximum. The hexagonal slice of the $3$-dimensional cube shows that this maximum is attainable. On the other hand, if $n=1$, the slice is a line segment, and the number of faces is always $2$ (the two endpoints of the line segment) no matter what $m$ is.

If you take the $4$-dimensional cube, its intersection with a plane  will be a polygonal region with up to $8$ sides. Its intersection with a $3$-dimensional subspace will be a polytope with up to $8$ faces.


问:
Does $\frac{\text{clockrate}(P_2)}{\text{clockrate}(P_1)}=1.37$ say that $P_2$ is $37\%$ faster than $P_1$, or the opposite?
Sorry for the trivial question, but ...

Does this ratio say $P_2$ is $37\%$ faster than $P_1$, or the opposite? $$\require{cancel}
\frac{ClockRate(P_2)}{ClockRate(P_1)} =  \left(\frac{(1.2 \times 10^9) \cdot 1.25 }{\cancel{\text{CPUTime}}}\right) \cdot \left(\frac{\cancel{\text{CPUTime}}} { 10^9 \times 1.1}\right)= 1.37$$

Thanks

答: Since: $$P_2 = 137\%  P_1$$  and we are talking about rates, this means that rate $P_2$ is faster!


问:
Does homotopy equivalence to a subspace imply (weak) deformation retract?
Here is a nearly same question, but with a slight difference in interpretation. We use the definition of wikipedia, which is different from that of Hatcher or Munkres (where deformation retract refers to strong deformation retract on wikipedia).

My question is :

Suppose $X$ is a path-connected topological space, and $A$ is its subspace. If $X$ and $A$ are homotopic equivalent, is it true that $A$ is a (weak) deformation retract of $X$?

Both Hatcher and Munkres have provided counterexamples for the case of strong deformation retract, but not the one above. Please help.

答:
This community wiki solution is intended to clear the question from the unanswered queue.

Noel Lundström has sketched a counterexample in a comment. Let $X = \bigvee_{i=1}^\infty S^1_i$ be the countably infinite wedge of copies of the circle $S^1$. This is a $CW$-complex with one $0$-cell and countably infinitely many $1$-cells. The space $A = \bigvee_{i=2}^\infty S^1_i$ is a subcomplex of $X$ which is homeomorphic to $X$; it is also a retract of $X$. However, the inclusion $i : A \to X$ is not a homotopy equivalence, thus $A$ is not a (weak) deformation retract of $X$.

To prove this, assume $i$ has a homotopy inverse $f : X \to A$. Then $i \circ f \simeq id_X$. Let $i_1$ be the inclusion of $S^1_1$ into $X$ and $r_1 : X \to S^1_1$ be the retraction which maps $A$ to the wedge point. We get $r_1 \circ f \circ i_1 \simeq r_1 \circ id_X \circ i_1 = id$. But $r_1 \circ f \circ i_1$ is constant, thus the identity on $S^1_1$ must be inessential. This is a contradiction.


问:
No determinant of a homomorphism even when $\dim V = \dim W$?

Why is there no determinant of a homomorphism $f: V \to W$, even when $\dim V = \dim W$?

I don't know how the concepts in linear algebra are related to each other and I am new to it.

Thanks

答:
Without attempting to discredit any one of the very instructive answers already given, let me just say that

It is indeed possible to define a determinant for all linear transformations between vector spaces of the same finite dimension.

In order to do this one may choose, once and for all, a basis for each and every vector space.  Then, given a linear transformation $f:V\to W$, one may define $\text{det}(f)$ to be the determinant of the matrix representing $f$ relative to the chosen bases of $V$ and $W$.

Whether or not such a concept is useful is another question altogether.
Vector spaces often occur in applications without any preferred basis, or perhaps with one which might not be the one we have chosen in advance, so the above definition of determinant is likely to be totally useless!


问:
What is the point of model theory?
Wikipedia defines model theory as ...

... the study of the relationship between formal theories (a collection of sentences in a formal language expressing statements about a mathematical structure), and their models, taken as interpretations that satisfy the sentences of that theory.

Why does this field need to exist? If we have proof theory, then we can discuss any formal system $\Gamma$ and statements within $\Gamma$ like, say, some arbitrarily chosen sentences $\phi$ and $\psi$, and even theorems like $\Gamma \vdash \phi \to \psi$, assuming that $\psi$ is derivable from $\phi$ in $\Gamma$, of course. Why should anyone care whether $\Gamma \models \phi$ if one already knows that $\Gamma \vdash \phi$?

Thanks for any advice.

答:
Today we have many logics which are axiomatized. And, we understand the term "axiomatic" in a very general way. This had not been true of Hilbert. He used the term in contradistinction with "genetic." He recognized that logicism as practiced by Cantor and Dedkind was inherently constructive. Because constructive mathematics could be interpreted in terms of temporal succession, he wanted to divorce mathematics from constructive approaches.

This relates to platonism in the following sense. Aristotle proceeded from syntactic analysis on the basis of use cases. He distinguishes between demonstrative argumentation and dialectical argumentation. Demonstrative argumentation relates to pedagogy, definition, and "first principles." To a large extent the "exactitude" of ordinary mathematics arises from the idiosyncratic use of definition in mathematics. But, demonstrative  argumentation is described in terms of "ordered knowledge claims." It is genetic. Hilbert's view of mathematics breaks this. In the modern first-order paradigm this divergence from Aristotle's account lies with Beth's definability theorem.

Dialectical argumentation is subdivided into other use cases. It covers legal argumentation (forensic use), political argumentation (rhetorical use), and counterfactual argumentation (analytical use). When people ask whether or not truth in mathematics can be distinguished from simple belief, they are asking if analytical use can really be distinguished from rhetorical use. One place they differ lies with the relationship to definition. Analytically, one cannot prove a definition, and, any counterexample destroys a definition. Unfortunately, Hilbert's approach obfuscates this distinction as well.

What remains is a logic with both existential import and the necessary truth of reflexive equality statements. And, this is precisely what is needed for a platonist to study mathematical objects as if they were objects of experience. These two features admit interpretation of the law of identity as a metaphysical presupposition. One can weaken this by distinguishing metaphysics and ontology. Ontologically, the law of identity can be understood as arising from the formal condition that two occurrences of a denoting symbol denote uniformly.

To understand the significance of Goedel's platonism, look at the SEP entry on "Recursive Functions" at

Plato-dot-Stanford-dot-edu

In Section 1.2 you will find discussion of Skolem's paper on primitive recursive arithmetic. When speaking of Skolem's assumptions on the body of that section, Item ii) will explain how Skolem assumed that provably equal descriptive functions could be substituted for one another. Let me emphasize the expression "provably equal." This speaks to the very description theory that the Hilbert school worked to eliminate from mathematics.

I have worked out non-propositional inference rules by which Skolem's use of description theory is realized. It cannot be done using the usual first-order paradigm. This is because it must manage existential import in a manner similar to negative free logic. And, in particular , numbers are "ordered" according to existential priority as described by Aristotle in his book "Categories." Aristotle describes this on the same section where he explains how demonstrative argumentation is characterized by the ordering of knowledge statements according to what is prior and what is posterior.

If you scroll down to Goedel's contribution to the study of recursive functions, you can see that his formulation had been immediately effective for his questioning of the Hilbert program. The SEP entry goes on to explain that primitive recursive arithmetic had to be extracted from the Goedellian approach. Apparently this clarification had been left to Church and Rosser.

For what it is worth, Goedel's platonism is well known by his own admission. But, I think that comparing the  effectiveness of Goedel's development of recursion to Skolem's careful criticism of how quantifiers were being understood in classical logic exemplifies its significance.

With regard to your question about "temporal verbs," you can pick up any translation of Euclid's "Elements." He commonly speaks of "constructing" elements of diagrams as his proofs proceed. Any variant of the word 'produce' would qualify. Proposition 16 reads:

"In any triangle, if one of the sides be produced, the exterior angle is greater than either of the interior and opposite angles."

Good luck with your studies. For what it is worth, Eric Schecter's book on mathematical propositions is an excellent resource for understanding the application of mathematics to the study of logic independent from foundational programs. You will not, for example, have your study of logic undermined by the ulterior motive of teaching you Goedel's incompleteness theorems before you learn about a plurality of logics. As Schecter points out, ordinary mathematical practice can apply different logics in different circumstances. The desire to exclude irrelevant assumptions speaks to relevance logic rather than classical logic. And, the need to have a relation in recursion theory that distinguishes "defined expressions" from "undefined expressions" is clearly related to the principle of indiscernibility of non-existents from negative free logic.

Title: What is the point of model theory?


问:
If $(1+x+x^2)^n = C_0 + C_1x + C_2 x^2...$, then find $C_0C_1 -C_1C_2 + C_2C_3...$
$$(1+x+x^2)^n = C_0 + C_1 x + C_2 x^2...C_n x^n$$

And
$$(1-\frac 1x +\frac{1}{x^2})^n = C_0 -\frac{C_1}{x}+\frac{C_2}{x^2}...$$

So if we multiply, the two equations, the given series will be obtained by the coefficient of $x^{-1}$ on the LHS.

The LHS$$-(\frac{x^4+x^2 +1}{x^2})^n$$
The given answer is 0, but I don’t understand how $x^{-1} $ can have coefficient 0. If it is, how do I prove it?

答:
$$
 (1+x+x^2)(1-\frac 1x +\frac{1}{x^2}) = x^2 + 1 + \frac{1}{x^2}
$$
so that
$$
(1+x+x^2)^n(1-\frac 1x +\frac{1}{x^2})^n = (x^2 + 1 + \frac{1}{x^2})^n
$$
contains only even powers of $x$. In particular, the coefficient of $x^{-1}$ is zero.


问:
Double inequality - two or three separate inequalities?
If I have the double inequality
$a\geq b \geq c$, does it mean I have two separate inequalities
\begin{align}
a&\geq b \\
b&\geq c
\end{align}
Or is it three separate inequalities
\begin{align}
a&\geq b \\
b&\geq c \\
a & \geq c
\end{align}
?

答:
When you speak of double inequality, $a\ge b\ge c$ generally implies $a\ge c$, as well as $a\ge b$ and $b\ge c$. This is transitivity of (partial) order: $a\ge b$ and $b\ge c$ imply $a\ge c$. Still, you should check the definition of $\ge$ in your context and whether your $\ge$ indeed satisfies transitivity whenever possible, although this is known to be the case as far as integers, rational numbers, and real numbers with standard $\ge$ (which are primary subjects of precalculus) are concerned, and checking these may sometimes be beyond the scope of precalculus.

Let us have a look at the case of nonnegative rational numbers for slightly nontrivial example. We usually define $a/b\ge c/d$ by $$a\times d-b\times c\ge_{\mathbb{Z}} 0,$$
where $a,b>0,c,d>0$ is some nonnegative integers and $\ge_{\mathbb{Z}}$ is the standard order of integers (e.g. $4\ge_{\mathbb{Z}} 1$, $3\ge_{\mathbb{Z}}-5$). Then, suppose $a/b\ge c/d$ and $c/d\ge e/f$ hold. Then it follows $$a\times d-b\times c\ge 0,\quad c\times f-d\times e\ge 0.$$
Multiplying $f$ to both sides of the first inequality yields
$$a\times d\times f-b\times c\times f\ge 0,$$
while $b$ to the second yields
$$b\times c\times f-b\times d\times e\ge 0.$$
Hence $$a\times d\times f-b\times d\times e=(a\times d\times f-b\times c\times f)+(b\times c\times f-b\times d\times e)\ge 0.$$
This indicates $d\times(a\times f-b\times e)\ge 0$ and thus $a\times f-b\times e$ cannot be negative, as $d>0$. Now we can say that $a/b\ge c/d$ and $c/d\ge e/f$ imply $a/b\ge e/f$. In other words, this $\ge$ satisfies transitivity. Then it is often convenient to compactly write $a/b\ge c/d\ge e/f$ to express such an order.

Important points twice: Double inequality $a\ge b\ge c$, generally implies $a\ge c$, in addition to $a\ge b$ and $b\ge c$. When you write it, make sure your $\ge$ satisfies transitivity; otherwise it does not make much sense and is likely to confuse readers.


答:
Another pattern I have recognised is, if we take the first digit as $n$, we have $\frac{(n)(n-1)}{2}$ possibilities for that particular value of $n$. Example: if n=$4$, the possible numbers are {$432,431,430,421,420,410$} which is six in number. Which is also equal to $\frac{4\times3}{2}$.

The reason this works is, suppose if n=$4$, we have three numbers having the second digit as $3$,  two numbers having the second digit as $2$ and one number having the second digit as $1$. Similarly, this works for other values of $n$ too.

so we can say that the total possibilities are: $$\frac{1}{2}\sum_{n=1}^{5}(n)(n-1)$$

Which turns out to be $20$

Title: How many descending order number between 100 and 600?

Body:
Example of valid number are:

(1) 431

(2) 543

(3) 210

(4) 321

Example of not valid number are:

(1) 221

(2) 500

(3) 654

The answer is 120 but I don't know why.

My attempt:

I have to construct a string of length $n=3$ with some constraints.
I split the problem into 5 small problems.

$D_1 = \{\text{all valid number staring with 1}\}$

$|D_1| = 0$

$D_2 = \{\text{all valid number staring with 2}\}$

$|D_2| = 1 \times {2\choose 2}$, because I can choose only from $\{0,1\}$

$D_3 = \{\text{all valid number staring with 3}\}$

$|D_3| = 1 \times {3\choose 2}$, because I can choose only from $\{0,1,2\}$

$D_4 = \{\text{all valid number staring with 4}\}$

$|D_4| = 1 \times {4\choose 2}$, because I can choose only from $\{0,1,2,3\}$

$D_5 = \{\text{all valid number staring with 5}\}$

$|D_5| = 1 \times {5\choose 2}$, because I can choose only from $\{0,1,2,3,4\}$

My solution: $$\sum_{i \in \{1,2,3,4,5\}}|D_i| = 35$$ I can sum directly because there is no double counting but it's wrong...

Editing from the future:

I made a mistake on the summation, the solution is 20. And from the textbook 120 were not the correct answer...

$$\sum_{i \in \{1,2,3,4,5\}}|D_i| = 20$$ I can sum directly because each set is disjoint (there is no double counting). In conclusion my long not-necessary aproach was right but (if you are reading this post) I recommend you to read all the comments below and the answers because they helped me a lot.