问:
What (if anything) can we infer about a first order theory from its category of models?
Does the structure of the category of models of a given first order theory where the arrows are embeddings tell us anything interesting about the theory itself?

Two models $M$ and $M'$ are isomorphic if and only if there exists an embedding $f : M \to M'$ whose inverse $f^{(-1)}$ is an embedding from $M'$ to $M$. Also, embeddings compose. Thus, we have a category.

As a trivial observation, we can rehash the Löwenheim–Skolem theorem in categorical language, if we take some effort to remove the finite models ahead of time.

Picking a satisfiable theory $T$ with no finite models produces a category $C$. If we take $C$ and produce a new category $D$ that identifies all the arrows in every homset in $C$, we get a category that is isomorphic to the category of infinite cardinals, where the homset $\text{Hom}(\kappa, \lambda)$ is a singleton if and only if $\kappa \le \lambda$ and $\text{Hom}(\kappa, \lambda)$ is empty otherwise.

The above statement is a consequence of the Löwenheim–Skolem theorem; it might be equivalent, though, if "cardinal gaps" are always detectable by the structure of the category.

答:
We can say a lot about the first-order theory by just looking at its category of models. Let's use elementary embeddings instead of just embeddings. They are the 'right' arrows to take because they respect first-order logic. You could also work with just embeddings to get some sort of positive logic, which is also interesting. But that is harder and you asked about first-order logic.

So just to remind you: for structures $M$ and $N$ we call a function $f: M \to N$ an elementary embedding if for every first-order formula $\phi(\bar{x})$ and every tuple $\bar{a} \in M$ we have
$$
M \models \phi(\bar{a}) \quad \Longleftrightarrow \quad N \models \phi(f(\bar{a})).
$$
For a first-order theory $T$ let us denote by $\mathbf{Mod}(T)$ the category of models of $T$ with elementary embeddings.

Indeed, we can already detect the Löwenheim-Skolem theorem. For this we have the language of accessible categories. Roughly this works as follows: we can give a purely category-theoretic definition of what a $\lambda$-presentable object is (see here, although there they call it "compact" instead of "presentable"). In $\mathbf{Mod}(T)$, for any $\lambda > |T|$, we have that an object $M$ is $\lambda$-presentable iff $|M| < \lambda$. So this gives us a notion of size. Then we can formulate the Löwenheim-Skolem theorem roughly as "every object is a colimit of small (i.e. $< |T|^+$) objects".

One easy thing that we can see is whether or not $T$ is complete. Namely, $T$ is complete exactly when $\textbf{Mod}(T)$ has the Joint Embedding Property (JEP). JEP means that whenever we have objects $M_1$ and $M_2$ that then there is an object $N$ and arrows $M_1 \to N \leftarrow M_2$. It should be a nice exercise to prove that this is equivalent to being complete.

Another thing that we can easily see from $\mathbf{Mod}(T)$ is whether or not the theory is $\kappa$-categorical for any $\kappa$. This is again done using the notion of presentability. We can detect the 'size' of an object purely category-theoretically and then we can just ask if all the objects of that 'size' are isomorphic.

Just to contrast, let's look at something that $\textbf{Mod}(T)$ cannot detect. It cannot detect whether or not $T$ has quantifier elimination. Indeed, for any theory $T$ we can add a relation symbol $R_\phi(\bar{x})$ for every formula in the language of $T$ and then let $T'$ extend $T$ by saying $\forall \bar{x}(\phi(\bar{x}) \leftrightarrow R_\phi(\bar{x}))$ (we call $T'$ the Morleyisation of $T$). Then $T'$ has quantifier elimination, but $\textbf{Mod}(T)$ and $\textbf{Mod}(T')$ are clearly equivalent categories.

Finally I should say something about classification theory. This is a very deep subject where the aim is to classify theories by how well-behaved they are. The class of stable theories is the most well-known here and the most studied, but there are also more general classes like simple theories. We can detect whether or not a theory is stable or simple (or neither) from its category of models! I think that we can do much more, but this is all very recent research...


问:
Which methods to use to integrate $\int{\frac{x^4 + 1}{x^2 +1}}\, dx$
I have this integral to evaluate:

$$\int{\frac{x^4 + 1}{x^2 +1}}\, dx$$

I have tried substitution, trig identity and integration by parts but I'm just going round in circles.

Can anyone explain the method I need to work this out?

答:
I think lab bhattacharjee's way is probably the best, but since you mentioned you tried it, I believe trig substitution would work as well.  The substitution

$$x=\tan t,dx=\sec^2tdt$$

would result with

$$\int\tan^4t+1=\int(\sec^2t-1)\tan^2t+1dt$$
$$\int\tan^2t\sec^2tdt+\int1-(\sec^2t-1)dt$$

Clear the parentheses, integrate, and backsubstitution should be rather easy.

ContentLicense: CC BY-SA 3.0


问:
Fundamental Group of $S^1$
As is well known, $\pi_1(S^1,x)\cong \mathbb{Z}$ where we can regard the loop $l$ based at $x$ going around $S^1$ counterclockwise once as the generator. Can someone explain why the composition $l$ and $l^{-1}$ is homotopic to the zero element, i.e. the point $x$?

答:
Let the generator $l$ be described as mapping $t$ to $x+(\cos t, \sin t)$ for $0 \le t \le 2 \pi).$ Then doing $l$ and then $l^{-1}$ is homotopic to the constant loop at $x$ given by the  map taking each $t \in [0,4\pi]$ to $x$.

I leave the setting up of the homotopy to you. It works since it doesn't matter what point one stops at before turning around, since only the two endpoints need to stay at $x.$

ContentLicense: CC BY-SA 4.0

Title: Fundamental Group of $S^1$


答:
$\ell \circ \ell^{-1}$ is the following loop: you go all the way around, then retrace your steps going backwards. The homotopy to the constant loop works by "pulling" the end of this loop; you can physically model this quite explicitly using a rubber band. Formally (it is much more annoying to write this out formally than to draw it or build a physical model) if we think of $\ell$ as the path

$$f(x) : [0, 2\pi] \ni t \mapsto e^{it} \in S^1$$

and $\ell^{-1}$ as the path

$$g(t) : [0, 2\pi] \ni t \mapsto e^{-it} \in S^1$$

then their composite is the path

$$h(t) : [0, 2\pi] \ni t \mapsto \begin{cases} e^{2it} & \text{ if } t \in [0, \pi] \\ e^{-2it} & \text{ if } t \in [\pi, 2\pi] \end{cases}$$

and the homotopy $H$ from this path to the constant path is

$$H(t, s) : [0, 2\pi] \times [0, 1] \ni (t, s) \mapsto \begin{cases} e^{2its} & \text{ if } t \in [0, \pi] \\ e^{2i(\pi - t) s} & \text{ if } t \in [\pi, 2\pi] \end{cases}.$$

$s = 1$ is the path $h(t)$ and $s = 0$ is the constant path; decreasing $s$ has the effect of "slowing down" the loop so you don't quite go all the way to the end before turning back, and you keep slowing down more and more until eventually you're not moving at all.

This argument is in no way specific to the circle and generalizes to showing that reversing the direction of a loop always provides the inverse in the fundamental group.

Title: Fundamental Group of $S^1$

Body: As is well known, $\pi_1(S^1,x)\cong \mathbb{Z}$ where we can regard the loop $l$ based at $x$ going around $S^1$ counterclockwise once as the generator. Can someone explain why the composition $l$ and $l^{-1}$ is homotopic to the zero element, i.e. the point $x$?


问:
Currently working through ODE exercise book and troubling question has appeared and i cant find any useful info to help with the it
I'm currently completing a exercise sheet on ODEs and I'm having trouble arriving at the answer.

Any help would be much appreciated.

Given the following differential equations:
\begin{align*}
\begin{cases}
dx/dt = -x + y -1\\\\
dy/dt = -x - y + 3
\end{cases}
\end{align*}

When $t = 0$, $x = 0$ and $y = 3$.

a) Find the general solution for $x$ and $y$ in terms of $t$.

Thanks.

答:
Differentiate both sides of the first equation:

$x'' = -x' + y'$ (3)

Sub (2) into (3):

$x'' = -x' + (-x - y + 3)$

$x'' = -x' - x - y + 3$ (4)

Rearrange (1) and sub into (4):

$x'' = -x' - x - (x' + x + 1) + 3$

$x'' = -x' - x - x' - x + 2$

$x'' + 2x' + 2x - 2 = 0$

Solve for $x(t)$ then sub into (1) to solve for $y$.


问:
When does the isomorphism $G\simeq \ker(\phi)\times \operatorname{im}(\phi)$? hold?
Suppose you have a group isomorphism given by the first isomorphism theorem:

$$G/\ker(\phi) \simeq \operatorname{im}(\phi)$$

What can we say about the group $\ker(\phi)\times \operatorname{im}(\phi)$? In particular, when does the following hold:

$$G\simeq \ker(\phi)\times \operatorname{im}(\phi)?$$

I ask this question because i want to prove that $GL_n^+(\mathbb{R}) \simeq SL_n(\mathbb{R}) \times \mathbb{R}^*_{>0}$, with $GL_n^+(\mathbb{R})$ the group of matrices with positive determinant. I proved that $SL_n(\mathbb{R})$ is a normal subgroup and that $GL_n^+(\mathbb{R})/  SL_n(\mathbb{R}) \simeq \mathbb{R}^*_{>0}$, using the surjective homomorphism $\det(M)$. I tried something with semidirect products but I got stuck.

答: Even if it can't be applied to your example, I would like to point out that in the abelian case (more generally in any abelian category) it's equivalent to have a split exact sequence: $0 \to \ker(\phi) \to G \to Im(\phi) \to0$</p>


答:
Say $S^1$ is the unit circle in the complex plane, with basepoint $1$.

Then the obvious representative of $l \cdot l^{-1}$ is the map from $[0,1] \to S^1$ which is given by

$$
\begin{cases}
e^{2\pi i 2x} & 0 \leq x \leq \frac{1}{2} \\
e^{-2\pi i (2x-1)} & \frac{1}{2} \leq x \leq 1
\end{cases}
$$

We can then consider the homotopy

$$
\begin{cases}
e^{2\pi i (1-t) 2x} & 0 \leq x \leq \frac{1}{2} \\
e^{-2\pi i (1-t) (2x-1)} & \frac{1}{2} \leq x \leq 1
\end{cases}
$$

Notice when $t=0$ this is our original map. However when $t=1$ this is a constant function (which corresponds to the trivial path at the basepoint $1$ (Why doesn't a similar idea work for $l$?). The idea is that as $t$ increases, we don't walk as far along the circle before turning around. At $t=1$, we don't walk at all.

This is summarized in the following picture:

I hope this helps ^_^

Title: Fundamental Group of $S^1$


问:
Bayes' Theorem Solution Verification

An insurance company examines its pool of auto insurance customers and gathers the following information:

All customers insure at least one car

$64\%$ of the customers insure more than one car.

$20\%$ of the customers insure a sports car.

Of those customers who insure more than one car, $15\%$ insure a sports car.

What is the probability that a randomly selected customer insures exactly one car and that car is not a sports car?

My attempt:
Let $S$ denote the event that a policyholder insures a sports car. Let $X$ denote the number of cars insured. We need to find $P(\{X = 1\}\cap S')$. Note that
$$P(\{X = 1\}\cap S') = P(\{X = 1\}| S')\cdot P(S') = [1-P(\{X = 1\}| S)]\cdot (1-0.2) = [1-P(\{X = 1\}| S)]\cdot 0.8$$

Also, $0.2 = [P(S|X=1) \cdot 0.36] + [0.64 \cdot 0.15] \implies P(S|X=1) \approx 0.289$

Using Bayes' Theorem, $P(\{X = 1\}| S) = \dfrac{P(S|X=1) \cdot P(X=1)}{P(S|X=1) \cdot P(X=1) + P(S|X>1) \cdot P(X>1)} = \dfrac{0.289 \cdot 0.36}{(0.289 \cdot 0.36) + (0.15 \cdot 0.64)} \approx 0.52$.

Therefore, $P(\{X = 1\}\cap S') = [1-0.52] \cdot 0.8 \approx 0.384$, which is not correct. Can someone please let me know where I went wrong? Thanks!

答:
$A=$"insures more than a car"

$B=$"insures a sport car"

We are asked to get
$$P(\bar A \cap \bar B)=P(\overline{A\cup B})=1-P(A\cup B)$$

We obtain
$$P(A\cup B)=P(A)+P(B)-P(A\cap B)=0,64+0,20-P(A\cap B)$$

We need
$$P(B/A)=\dfrac{P(A\cap B)}{P(A)}\implies P(A\cap B)=P(B/A)P(A)=0,15\cdot 0,64 =0,096$$

So, finally

$$P(\bar A \cap \bar B)=1-P(A\cup B)=1-(0,64+0,20-P(A\cap B))=1-(0,64+0,20-0,096).$$

That is

$$P(\bar A \cap \bar B)=1-(0,64+0,20-0,096)=0,256.$$

Title: Bayes' Theorem Solution Verification


问:
What is an open set in a topological space?
I recently started learning topology to help me understand limits and continuity better for calculus, and I am struggling with some of the definitions.

What I am getting confused with is why is every set in a topology considered to be open and when talking about sets in the topology we always say the set is open.

My intuitive notion of openness from  previous knowledge of mathematics is an interval that does not contain its endpoints, so there is an infinite sequence at the end points, e.g., $(0,1)$ is an open interval.

However, in topology, for example, the singleton $\{1\}$
is considered an open set—how is this so? Why are sets in a topology always open? And what is the definition of an open set in a topological space?

Thanks in advance.

答:
There is no such thing as an "open set," only a open subset of a topological space. "The singleton $\{ 1 \}$ is an open set" is not a meaningful statement; the meaningful statement is whether the singleton $\{ 1 \}$ is an open subset of some topological space containing $1$. For example it is not an open subset of $\mathbb{R}$, or of the closed interval $[0, 1]$ (with their usual topologies). However, it is an open subset of itself, considered as a one-point space (which has a unique topology).

Sets in a topology are open by definition; that's what "open" means.

I suggest that when starting to learn topology you completely ignore the ordinary English meaning of the word "open" and just work with the axioms abstractly for awhile. It's not the most fun or intuitive way to do it but at least you aren't letting preconceptions get in the way. Also, I don't really recommend this as a way to better understand calculus; you'd be better off picking up a textbook on real analysis.


问:
Trouble applying the Tan double angle formula for $5\tan(2\theta)=4\cot(\theta)$
I have to solve the following equation

 $$5\tan(2\theta)=4\cot(\theta),\space \space \space 0°<\theta<180°$$

Applying the double angle formula for $\tan(2\theta)$

$$\frac{10\tan(\theta)}{1-\tan^2(\theta)}=4\cot(\theta)$$

I did solve and simplify this further but did not arrive at a correct value for $\theta$ so I decided to confirm where I went wrong on wolfram alpha, and to my surprise my very first step turned out to be wrong.

What mistake have I made here, is the double angle formula I have applied, $\tan(2\theta)=\dfrac{2\tan(\theta)}{1-\tan^2(\theta)}$, wrong or is there some restriction upon applying it?

答:
HINT

The identity which you have applied is correct. More precisely, we have that
\begin{align*}
\tan(2\theta) = \frac{\sin(2\theta)}{\cos(2\theta)}
= \frac{2\sin(\theta)\cos(\theta)}{\cos^{2}(\theta) - \sin^{2}(\theta)}  = \frac{2\tan(\theta)}{1 - \tan^{2}(\theta)}
\end{align*}

Consequently, the proposed equation is equivalent to
\begin{align*}
5\tan(2\theta) = 4\cot(\theta) \Longleftrightarrow \frac{10\tan(\theta)}{1 - \tan^{2}(\theta)} = \frac{4}{\tan(\theta)}
\end{align*}

Can you take it from here?


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:
I started to develop your idea from the OP, now we already have an accepted, good solution, but let me submit one more, since there is in the given constellation the following interesting property, not used or mentioned in all other solutions from AoPS:

Lemma: Define $T=MF\cap NE$. Then the lines $MF$, $NE$, $AI$, $BC$ are concurrent in $T$.

This "coincidence" immediately solves the problem in the OP (respecting the idea from the OP) by symmetry, since from the lemma $M,N$ are symmetric w.r.t. the axis $AJIT$, so $\overset\frown{AM}=\overset\frown{AN}$, and the angles of interest are pointing to these arcs.

Proof of the lemma:

First of all, the lines $EF$, $PI$, $BC$ are concurrent in a point $S$.
Many linked proofs use this step, here is an other (and the same) proof. Consider the inversion $*$ centered in $I$ which invariates $(DEF)$. The the circle $\odot(IPD)$ is transformed in the line perpendicular in $D^*=D$ on the diameter $ID$, which is $BDC$. The circle $\odot(IEF)$ is transformed in the line $EF=E^*F^*$, perpendicular on the diameter $IA$ and passing through the fixed points $E,F$. So $S=P^*$ is on both lines, i.e. $I,P,S=P^*=BC\cap EF$ are collinear.

Consider now the polygon $MFAENP$. Using Pascal's theorem, the following points are collinear.
$$
\begin{aligned}
MF \cap EN &= T\ ,\\
FA \cap NP &= B\ ,\\
AE \cap PM &= C\ .&&\text{ So }T\in BC\ .
\end{aligned}
$$
Further consider the hexagon $IP N EF A$, and apply again
Pascal's theorem, obtaining the colinearity of:
$$
\begin{aligned}
IP\cap EF &= S\ ,\\
PN\cap FA &= B\ ,\\
NE\cap AI &= (\text{ a point on $SB$ and $NE$ }) =T\ .\text{ So }T\in AI\ .
\end{aligned}
$$
$\square$


问:
In a differential equation: Will all rewritings yield a solution?
I've been asked to solve the following differential equation:

$$y\frac{\mathrm{d}y}{\mathrm{d}x}=(y+1)^2$$

In the book, it's solved as follows:

$$\frac{y}{(y+1)^2}\,\mathrm{d}y=\mathrm{d}x$$

And then we integrate both sides. Now, I noticed that this rewriting is a particular choice for it. There are also other rewritings:

$$\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{(y+1)^2}{y} \hspace{1cm} \mathrm{d}y=\frac{(y+1)^2}{y}\,\mathrm{d}x \hspace{1cm} \frac{\mathrm{d}x}{\mathrm{d}y}=\frac{y}{(y+1)^2}$$

These are just some of them. This may be a very silly question but: Will all of those rewritings yield solutions to our original equation? I noticed that for some of them, the solution is going to be different, for example: In the second, it won't be defined for $y=0$, in the fourth it won't be defined for $y=-1$.

I also noticed that in doing this, the "solutions" look very different, for example:

Solving $\frac{y}{(y+1)^2}\,\mathrm{d}y=\mathrm{d}x$ we obtain: $\frac{1}{y+1}+\log{\left(y+1\right)}=x+c$.

Solving $\mathrm{d}y=\frac{(y+1)^2}{y}\,\mathrm{d}x$ we obtain: $x \left(y+1\right)^2+c=y^2$.

答:
Consider the relation

$dy=\frac{(y+1)^2}{y}dx$

We have $\int dy = y + c$

But the second integral is not of a known form. We are blocked for the integral:

$\int \frac{(y(x)+1)^2}{y(x)} dx$

There would be a missing y' for example.

Only the first writing by separating the variables gives a result. This is actually a shortcut using the differential forms. But this can be rigorously justified by:

$\frac{y(x)}{(y(x)+1)^2} y'(x)=1$

which gives by integrating

$\int \frac{y(x)}{(y(x)+1)^2} y'(x) dx =\int 1. dx$

By changing variable z=y(x) in the first integral:

$\int \frac{z}{(z+1)^2} dz =\int 1. dx = x+c$

$\frac{1}{z+1}+\ln|z+1|+d=x+c$

Returning to the old variable

$\frac{1}{y(x)+1}+\ln|y(x)+1|+d=x+c$

etc.


问:
Is this function measurable? Something to do with Bochner space and norms.
Suppose $f:[0,T]\to X$ is a measurable map where $X$ is Hilbert space. Suppose also that $R(t):X \to X^*$ is an isometric isomorphism with
$$\lVert R(t)f(t)\rVert_{X^*} = \lVert f(t) \rVert_X$$
also measurable.

Can we deduce
$$t \mapsto R(t)f(t)$$ is measurable?

答:
As long as $R(t)$ is otherwise arbitrary, the answer is no. Take a non-measurable set $E$ and put
$$
R(t) = \begin{cases}R&t \in E\\-R & t \not\in E\end{cases}
$$
for some isometric isomorphism $R$. Together with a constant $f$, this yields a non-measurable $R(t)\,f(t)$.</p>


答:
You are not applying the correct input to Wolfram Alpha.  The symbol = represents an assignment operator.  The symbol === represents SameQ[] in the Wolfram language, and tests whether two things are identical.  Neither of these is proper.  Do not try using Wolfram Alpha to check your work unless you understand its syntax and limitations.

The identity $$\tan 2\theta = \frac{2 \tan \theta}{1 - \tan^2 \theta}$$ is correct, however, there are some restrictions due to discontinuities and periodicity.  That said, we can deal with this in the normal course of searching for solutions.

For the given equation, let $z = \tan \theta$.  Then $$5 \tan 2\theta = \frac{10z}{1-z^2},$$ and $$4 \cot \theta = \frac{4}{z}.$$  Consequently $$10z^2 = 4(1-z^2),$$ from which we obtain $$z^2 = \frac{2}{7}.$$  Hence $$\tan \theta = \pm \sqrt{\frac{2}{7}}.$$  If we take the positive root, then $\theta$ is in quadrant I or III.  If we take the negative root, then $\theta$ is in quadrant II or IV.

In the first case, we may take $$(\sin \theta, \cos \theta) = \pm \left(\frac{\sqrt{2}}{3},  \frac{\sqrt{7}}{3}\right),$$ where the sign must be the same for each coordinate.  In the second case, we may take $$(\sin \theta, \cos \theta) = \left(\pm \frac{\sqrt{2}}{3},  \mp \frac{\sqrt{7}}{3}\right),$$ where the signs must be opposite for each coordinate.  Now all that is left is to check that these possibilities all work.

Title: Trouble applying the Tan double angle formula for $5\tan(2\theta)=4\cot(\theta)$

Body:
I have to solve the following equation

 $$5\tan(2\theta)=4\cot(\theta),\space \space \space 0°<\theta<180°$$

Applying the double angle formula for $\tan(2\theta)$

$$\frac{10\tan(\theta)}{1-\tan^2(\theta)}=4\cot(\theta)$$

I did solve and simplify this further but did not arrive at a correct value for $\theta$ so I decided to confirm where I went wrong on wolfram alpha, and to my surprise my very first step turned out to be wrong.

What mistake have I made here, is the double angle formula I have applied, $\tan(2\theta)=\dfrac{2\tan(\theta)}{1-\tan^2(\theta)}$, wrong or is there some restriction upon applying it?


问:
Calculating the value of $\sum^\infty_{k=1}\frac{1}{k(9k^2-1)}$
My first thought is to split it up into:

$$\sum^\infty_{k=1}\;\frac{3}{2(3k-1)} + \frac{3}{2(3k+1)} - \frac{1}{k}$$

This is starting to look vaguely like some sort of rearrangement of the alternating harmonic series with some factors thrown in. The way I would evaluate one of those types of sums would be to exploit the fact that $\lim_{n \to \infty} H_n - \log n = \gamma$ exists to find a formula for the partial sum, then take the limit and the answer would just fall out.

Maybe I'm just being stupid, but I can't see exactly how to do it in this case. Alternative methods are also welcome.

答:
Let $q\ge2$ be an integer. Then set
$$f_q(a,b,c)=\sum_{n\ge1}a^{qn}b^{qn-1}c^{qn-2}=\frac{1}{bc^2}\sum_{n\ge1}(abc)^{qn}=\frac{a^{q}b^{q-1}c^{q-2}}{1-(abc)^q},
\qquad |abc|<1.$$
Then we may see
$$\sum_{n\ge1}\frac{1}{n(q^2n^2-1)}=q\int_0^1\int_0^1\int_0^1f_q(a,b,c)dadbdc=q\int_{(0,1)^3}f_q(a,b,c)dA.$$
Here, $dA=da\,db\,dc$.

Thus your sum is given as
$$S=\sum_{n\ge1}\frac{1}{n(9n^2-1)}=3\int_{(0,1)^3}\frac{a^3b^2c}{1-(abc)^3}dA.$$
I'm sure there is a closed form, but the triple integral is very tedious. Complex numbers may be involved.

An alternate form is
$$S=\int_0^1\int_0^x\int_0^u\frac{t}{1-t^3}dt\,du\,dx.$$
This 'simplifies' to
$$S=\int_0^1\left(\frac16J_1(x)-\frac{1}{\sqrt3}J_2(x)-\frac13J_3(x)+\frac{\pi}{2\sqrt3}x\right)dx,$$
where
$$\begin{align}
J_1(x)&=(x+2)\ln(1+x+x^2)+\sqrt3\arctan\tfrac{1}{\sqrt3}(2x+1)-2x-\frac\pi{2\sqrt3}\\
J_2(x)&=-\frac{\sqrt3}{4}\ln(1+x+x^2)+(x+\tfrac12)\arctan\tfrac{1}{\sqrt3}(2x+1)-\frac\pi{12}\\
J_3(x)&=(x-1)\ln(x-1)+\ln(-1)-x.
\end{align}$$


问:
Roots of a matrix equation
Let $X \in \mathbb{C}^{n \times m} $ be a rectangular matrix of full rank, and $X^*$ its hermitian conjugate, let $A \in \mathbb{C}^{m \times m}$ a square matrix, and let $f: \mathbb{C} \to \mathbb{C}$ be defined by
$$
f(z) = \det \left( I_n + X \frac{1}{A - z I_m} X^* \right)
$$
where $I_n$ is the $n \times n$ identity matrix.

Show that for $n \leq m$ solutions of $f(z)= 0$ are given by $z$ equal to an eigenvalue of $B : = A + X^* X$.

My attempt: for $n = m$, $f(z)$ can be straightforwardly rearranged to
$$
f(z) = \det \left( X \frac{1}{B-A} \big( B - I_m z \big) \frac{1}{A - z I_m} X^* \right) = \frac{\det(X) \det(B - I_m z) \det(X^*)}{\det(A - z I_m) \det(B - A)}
$$
where the desired result follows from the factor $\det(B - I_m z)$.

However I cannot factorise the determinant in this way for $n < m$, and I am unsure how to proceed.

答:
With the Sylvester determinant identity, we have
$$
\begin{aligned}
\det\left(I_n + \left( X\frac 1{A - zI_m}\right)X^*\right) &=
\det\left(I_m + X^*\left( X\frac 1{A - zI_m}\right)\right)
\\
& =\det\left(I_m + \big(X^*X \big) \frac 1{A - zI_m}\right)
\\
& =\det\left(\big(A - zI_m + X^*X\big)\frac 1{A - zI_m}\right)
\\
& =\frac{\det\Big(\big(A + X^*X\big) - zI_m)\Big)}{\det(A - zI_m)}.
\end{aligned}
$$

Title: Roots of a matrix equation


问:
A sum of Euler products
Maybe someone have come across a particular sum related to Euler's product in the form:

$$S(x) = \frac{1}{5} (1 - \frac{2}{3}) + \frac{1}{7} (1 - \frac{2}{3})(1 - \frac{2}{5}) + \ ... \ = \sum_{p > 3}^{p < x} \ \frac{1}{p} \; \prod_{p_l > 1}^{p_l < p} \; (1 - \frac{2}{p_l})$$

where the sum and the products are over primes and we are interested in asymptotics at $x \to \infty$.

It is possible to approach it with estimating the products first (as $\sim \frac{1}{log^2 p_l}$) then summation over $p$ but it looks messy to keep track of the terms. Wonder if there can be alternative strategies?

答:
With some transformations, I found a solution with an exact analytical expression:

$$S(p) = \frac{1}{3} - \Pi_2(p_{pr})$$

where $p_{pr}$ is the previous prime, $\Pi_2 = \prod (1 - \frac{2}{p})$. It follows immediately from the relation between two consecutive members of the sum $N(p_{pr}), \ N(p)$ that can be obtained from the definition of the sum: $N(p) = (p_{pr} - 2) \ N(p_{pr})$.

Title: A sum of Euler products


答:
The whole point of having a general topology is that you get to define which sets are and aren't "open", to make the rules of the game, and then get to see what that does and how things are different in the "world" so created versus the usual real numbers. If we take the idea that an open set "doesn't contain its own boundary", which is what you are after but how I originally heard it phrased, being able to define open sets to be whatever you want them to be (so long as you meet the rules for how they must be structured under union and intersection) means, in effect, you get to define what constitutes a "boundary" and what doesn't. You get to make what is and isn't an "end point".

To see why that has an impact, note that the only reason that $0$ and $1$ are "boundaries" of $(0, 1)$ is because of the ordering on the reals, which ensures that $0 < x < 1$ whenever $x \in (0, 1)$, and also, there's nothing in between 0 and 1 and the set $(0, 1)$, i.e. no points $y$ such that $0 < y < x$ for every $x \in (0, 1)$, and similarly for $1$.

But suppose we re-ordered the reals, so that both points $0$ and $1$ came before the points we consider to be in $(0, 1)$ (in the usual definition.). E.g. suppose we ordered the reals to look like

$$(\text{stuff}) < 0 < 1 < 2 < (\text{stuff}) < (\text{the numbers in $(0, 1)$}) < (\text{more stuff})$$

Now, suddenly, $(0, 1)$ no longer has boundary points $0$ and $1$. So there is no absolute notion of a "boundary point". It depends on the order, and we just redefined what the boundary was by redefining the order.

And topology is even more flexible than that. And orders are just one source, but far from the only one, of topologies.

Title: What is an open set in a topological space?

Body:
I recently started learning topology to help me understand limits and continuity better for calculus, and I am struggling with some of the definitions.

What I am getting confused with is why is every set in a topology considered to be open and when talking about sets in the topology we always say the set is open.

My intuitive notion of openness from  previous knowledge of mathematics is an interval that does not contain its endpoints, so there is an infinite sequence at the end points, e.g., $(0,1)$ is an open interval.

However, in topology, for example, the singleton $\{1\}$
is considered an open set—how is this so? Why are sets in a topology always open? And what is the definition of an open set in a topological space?

Thanks in advance.


问:
How many ways can I put down two indistinguishable pieces on an ordinary $8 \times 8$ chessboard if they must either be in the same row or column?
I am a student in middle school and I was wondering if anyone could help me with the following problem:

How many ways can I put down two indistinguishable pieces on an ordinary $8\times 8$ chessboard, if the pieces must either be in the same row or be in the same column?

I think there would be 64 options for the first piece since it can go anywhere on the chessboard, but I'm not sure about the second piece.

答: The first piece can go in any of the $64$ squares. The second piece can then be in any of $14$ positions, since there are $7$ unoccupied squares in the row of the first piece, as well as $7$ unoccupied squares in the column of the first piece. This would seem to give us $64\cdot 14$ choices for the placement of the two pieces. However, order doesn't matter (we said the pieces are indistinguishable), so the actual number of choices is $(64\cdot 14)/2$, which is $\boxed{448}$.


答:
Consider the set $\mathbb{R}$. A topology on $\mathbb{R}$ is a collection* of subsets on $\mathbb{R}$ that we declare to be open.

For example, the usual topology on $\mathbb{R}$ consists of all open intervals like $(0,1)$ and $(a,b)$ and all unions of such intervals. In this topology, the singleton set $\{1\}$ is not open, since it cannot be written as a union of open intervals $(a,b)$.

There are other topologies on $\mathbb{R}$. We could, for example, declare that every subset of $\mathbb{R}$ is open. In other words, the topology consists of all subsets of $\mathbb{R}$. In this topology, $\{1\}$ is open.

As another example, we could declare that the only open sets are $\mathbb{R}$ and $\emptyset$. For this topology, $\{1\}$ is not open. Also, $(0,1)$ is not open.

There are many other topologies that we can put on $\mathbb{R}$. Of course, some are more useful and interesting than others.

*Note: We can't just grab an arbitrary collection of subsets $\mathbb{R}$ and call it a topology. The collection needs to contain $\mathbb{R}$ and the empty set and it needs to be closed under arbitrary unions and finite intersections.

There is nothing special about $\mathbb{R}$ here. Given an arbitrary set $X$, a topology on $X$ is a collection of subsets of $X$ (that contains $X$ and $\emptyset$ and is closed under arbitrary unions and finite intersections). The sets in the topology are called open sets of that topology.


问:
Identity: volume integral of sum of norms
I am trying to show $\int_D\left(|\nabla\cdot\vec{u}|^2+|\nabla\times\vec{u}|^2\right)dV=\int_D\left(|\nabla u_x|^2+|\nabla u_y|^2+|\nabla u_z|^2\right)dV$ but am having no luck.

Here $D$ is a smooth, bounded domain in $\mathbb{R}^3$, and $\vec{u}$ is a smooth vector field in $\mathbb{R}^3$, such that $\vec{u}=\vec{0}$ on the boundary $∂D$.

I have managed to show $\int_D\left(|\nabla\cdot\vec{u}|^2+|\nabla\times\vec{u}|^2\right)dV= \int_D\left(2|\nabla\cdot\vec{u}|^2-∂_iu_j∂_ju_i\right)dV$. I have tried a tonne of things from here but haven't made any step towards the solution.

A hint I am given is that $∂_{x_i} ∂_{x_j} f= ∂_{x_j} ∂_{x_i} f(x)$, but I have no idea why the fact that partial derivatives commute relates to this problem.

答:
We have

$$\tag{1}|\nabla\cdot\vec{u}|^2 = \left(\frac{\partial u_x}{\partial x} + \frac{\partial u_y}{\partial y} +\frac{\partial u_z}{\partial z} \right)^2 \\ =A + 2 \frac{\partial u_x}{\partial x}\frac{\partial u_y}{\partial y}+ 2 \frac{\partial u_x}{\partial x}\frac{\partial u_z}{\partial z}+ 2 \frac{\partial u_y}{\partial y}\frac{\partial u_z}{\partial z}\\ = A + 2 \frac{\partial }{\partial x}\left(u_x\frac{\partial u_y}{\partial y}\right)+ 2 \frac{\partial }{\partial y}\left(u_y\frac{\partial u_z}{\partial z}\right)+ 2 \frac{\partial }{\partial z}\left(u_z\frac{\partial u_x}{\partial x}\right) - 2u_x \frac{\partial ^2 u_y}{\partial x \partial y}- 2u_y \frac{\partial ^2 u_z}{\partial y \partial z}- 2u_z \frac{\partial ^2 u_x}{\partial z \partial x},$$

where

$$A= \left(\frac{\partial u_x}{\partial x} \right)^2 + \left(\frac{\partial u_y}{\partial y} \right)^2+\left(\frac{\partial u_z}{\partial z} \right)^2, $$

and

$$\tag{2}|\nabla\times\vec{u}|^2 =\left(\frac{\partial u_z}{\partial y} - \frac{\partial u_y}{\partial z} \right)^2 + \left(\frac{\partial u_z}{\partial x} - \frac{\partial u_x}{\partial y} \right)^2+ \left(\frac{\partial u_y}{\partial x} - \frac{\partial u_y}{\partial z} \right)^2\\= B -  2 \frac{\partial u_z}{\partial y}\frac{\partial u_y}{\partial z}-  2 \frac{\partial u_z}{\partial x}\frac{\partial u_x}{\partial z}-  2 \frac{\partial u_y}{\partial x}\frac{\partial u_x}{\partial y}= \\ B - 2 \frac{\partial }{\partial x}\left(u_z\frac{\partial u_x}{\partial z}\right)- 2 \frac{\partial }{\partial y}\left(u_x\frac{\partial u_y}{\partial x}\right)- 2 \frac{\partial }{\partial z}\left(u_y\frac{\partial u_z}{\partial y}\right) + 2u_z \frac{\partial ^2 u_x}{\partial x \partial z}+ 2u_x \frac{\partial ^2 u_y}{\partial y \partial x}+ 2u_y \frac{\partial ^2 u_z}{\partial z \partial y},$$

where

$$B = \left(\frac{\partial u_z}{\partial y} \right)^2 + \left(\frac{\partial u_y}{\partial z} \right)^2 + \left(\frac{\partial u_z}{\partial x} \right)^2 + \left(\frac{\partial u_x}{\partial z} \right)^2 + \left(\frac{\partial u_y}{\partial x} \right)^2 + \left(\frac{\partial u_x}{\partial y} \right)^2  $$

The last three terms on the RHS of (1) cancel the last three terms on the RHS of (2) (using the hint), and

$$A+B = |\nabla u_x|^2 +|\nabla u_y|^2+ |\nabla u_z|^2$$

Thus, adding (1) and (2), we get

$$|\nabla\cdot\vec{u}|^2+ |\nabla\times\vec{u}|^2= |\nabla u_x|^2 +|\nabla u_y|^2+ |\nabla u_z|^2 + C,$$

where

$$C = 2 \frac{\partial }{\partial x}\left(u_x\frac{\partial u_y}{\partial y}\right)+ 2 \frac{\partial }{\partial y}\left(u_y\frac{\partial u_z}{\partial z}\right)+ 2 \frac{\partial }{\partial z}\left(u_z\frac{\partial u_x}{\partial x}\right)- 2 \frac{\partial }{\partial x}\left(u_z\frac{\partial u_x}{\partial z}\right)- 2 \frac{\partial }{\partial y}\left(u_x\frac{\partial u_y}{\partial x}\right)- 2 \frac{\partial }{\partial z}\left(u_y\frac{\partial u_z}{\partial y}\right) $$

It follows from the divergence theorem that  $\int_D C \,dV = 0$, since each term of $C$ is the divergence of a vector field that vanishes on the boundary $\partial D$ where the velocity $\vec{u}$ vanishes.

For example, since $u_x = 0$ on $\partial D$, we have

$$ \int_D\frac{\partial }{\partial x}\left(u_x\frac{\partial u_y}{\partial y}\right)\, dV = \int_{\partial D}u_x\frac{\partial u_y}{\partial y} n_x \,dS = 0$$

Therefore,

$$\int_D \left(|\nabla\cdot\vec{u}|^2+ |\nabla\times\vec{u}|^2\right)\, dV = \int_D \left(|\nabla u_x|^2 +|\nabla u_y|^2+ |\nabla u_z|^2\right)\, dV$$


问:
prove that a function is a diffeomorphism
anyone can help me to prove that $f(\alpha,x,y,z) = (\sinh(\alpha) x, \sinh(\alpha)y,\cosh(\alpha)z)$ is a diffeomorphism? In fact, i'm not sure if it's a diffeomorphism.</p>

答: At every point $p$, the derivative is a linear map $df_p:\mathbb{R}^4\longrightarrow \mathbb{R}^3$. Such a map can never be invertible for dimension reasons. So your function is not a diffeomorphism, not even local. Of course, it is not even a homeomorphism, due to the invariance of domain. Not even local again.</p>


问:
How to solve the coupled differential equations with 3 variables?
I have the following set of differential equations,
\begin{aligned}
& x' - \alpha(yz' - zy') =  -\beta(yP_z -zP_y) \\
& y' - \alpha(zx' - xz') =  -\beta(zP_x -xP_z) \\
& z' - \alpha(xy' - yx') =  -\beta(xP_y -yP_x)
\end{aligned}
I know how to implement runga kutta 4th order or Euler or Heun's method...what I can't figure out is how to update the $x'$ or $y'$ or $z'$ term while I am solving it. Any help or direction towards possible resources would be appreciated.

答:
Let $\vec{r} = (x,y,z)^T$ and $\hat{r}$ be corresponding unit vector.
Let $\vec{P} = (P_x,P_y,P_z)^T$ and $\vec{Q} = \vec{P} - (\vec{P}\cdot\hat{r})\hat{r}$. The ODE at hand can be rewritten as

$$\vec{r}' = \vec{r}\times( \alpha \vec{r}' - \beta \vec{P})
= \vec{r}\times(\alpha \vec{r}' - \beta \vec{Q})\tag{*1}
$$
Taking dot product with $\vec{r}$ on both sides, we find

$$\vec{r}\cdot\vec{r}' = 0 \implies |\vec{r}|^2 = R^2$$
for some constant $R$. Taking cross product with $\vec{r}$ on both sides of $(*1)$, we get

$$\require{cancel}
\begin{align}
\vec{r} \times \vec{r}' &= \vec{r}\times(\vec{r} \times(\alpha \vec{r}' - \beta \vec{Q}))\\
&= (\color{red}{\cancelto{0}{\color{gray}{\vec{r}\cdot(\alpha \vec{r}' - \beta \vec{Q})}}})\vec{r} - R^2 (\alpha \vec{r}' - \beta \vec{Q})\\
&= -R^2(\alpha \vec{r}' - \beta \vec{Q})
\end{align}
$$
Substitute this back into RHS of ($*1$) , we get

$$\begin{align}
\vec{r}' &= -\alpha R^2(\alpha \vec{r}' - \beta \vec{Q}) - \beta \vec{r} \times \vec{Q}\\
\iff \vec{r}' &=
\frac{\beta}{1 + \alpha^2R^2}(\alpha R^2 \vec{Q} - \vec{r} \times \vec{Q})
\end{align}
$$
In this form, the definition of $\vec{r}'$ is free of itself and you can throw it to any ODE solver you like.


问:
Negation of specific statement
Definition: We say that $X\subset \mathbb{R}$ is bounded above if $\exists C\in \mathbb{R}$ s.t. $\forall x\in X$ we have $x\leq C$.

$(X \ \text{is bounded above}):=\exists C\in \mathbb{R}((\forall x\in X)\Rightarrow (x\leq C))$

So I want to take the negation of it: $$\neg(X \ \text{is bounded above})\equiv \neg(\exists C\in \mathbb{R}((\forall x\in X)\Rightarrow (x\leq C)))\equiv$$ $$\equiv\forall C\in \mathbb{R} \neg((\forall x\in X)\Rightarrow (x\leq C))$$
But we know that $\neg (P\Rightarrow Q)\equiv P\land \neg Q$ which implies that $$\neg(X \ \text{is bounded above})\equiv \forall C\in \mathbb{R} ((\forall x\in X)\land (x> C)) $$

But I know that the correct negation should be the following: $X\subset \mathbb{R}$ is not bounded above if $\forall C\in \mathbb{R}$ $\exists x\in X$ s.t. $x>C$.

What am I doing wrong in the above?

答:
Right in the begining on the second line. The big quantifier $\forall$ is not part of the implication. It is:

$\exists C\in R (\forall x\in X (x\leq C))$

Edit: Expression $(\forall x\in X) \rightarrow (x\leq C) $ isn't meaningful expression because $(\forall x\in X)$ has no logical value (it is not true-false statement) it is just quantification.

Title: Negation of specific statement


问:
Solve the Lagrangian dual problem associated to $\max_{x_1,x_2\ge0}(3x_1+4x_2)$ such that $x_1^2+x_2^2\le25$
Consider the (non-linear) optimization problem ($P$)

$$max \quad3x_1 + 4x_2$$

$$s.t. \quad x_1^2 + x_2^2 \leq 25$$

$$ \quad x_1,x_2 \geq 0$$

Solve the Lagrangian dual problem.

I don't have a clue how the dual problem is even obtained since the constraint is nonlinear. Could anyone please help me?

答:
gt6989b's work isn't correct. The dual problem involves minimizing over the Lagrange multipliers, not maximizing over $x$. Furthermore, to contruct the Lagrangian dual problem, you need Lagrange multipliers not just for the quadratic constraint but also for the two nonnegativity constraints.

Note that most texts that talk about convex duality assume the primal problem is a minimization. So the derivations below are the negatives of what you'd do if you constructed the Lagrangian from the equivalent minimization problem (with the negative objective $-3x_1-4x_2$).

The Lagrangian is
$$L(x_1,x_2,\lambda_1,\lambda_2,\lambda_3) = 3x_1+4x_2+\lambda_1(25-x_1^2-x_2^2)+\lambda_2x_1+\lambda_3x_2.$$
The Lagrange multipliers $\lambda_1$, $\lambda_2$, and $\lambda_3$ are all nonnegative.
The dual function is
$$g(\lambda_1,\lambda_2,\lambda_3) = \max_{x_1,x_2} 3x_1+4x_2+\lambda_1(25-x_1^2-x_2^2)+\lambda_2x_1+\lambda_3x_2$$
and the dual problem is
$$\begin{array}{ll} \text{minimize} & g(\lambda_1,\lambda_2,\lambda_3) \\
\text{subject to} & \lambda_1,\lambda_2,\lambda_3 \geq 0
\end{array}$$
To complete the construction of the dual problem, you'll want to eliminate $x_1$ and $x_2$ from the dual function $g(\cdot)$ using simple calculus. This will give you an expression for $g$ that is independent of $x_1$ and $x_2$. There might be a couple of division by zero issues there you'll want to be careful with.

Alternatively, instead of constructing the full dual problem, you can determine the correct values of $\lambda_1$, $\lambda_2$, and $\lambda_3$ from complementary slackness conditions. This requires solving for the optimal values of $x_1$ and $x_2$, and examining which constraints are active at the solution.


答:
If we write definition as
$$(\exists C\in \mathbb{R})(\forall x\in X)(x\leq C)\quad (1)$$
Then negation gives
$$(\forall C\in \mathbb{R})(\exists x\in X)(x> C)\quad (2)$$
Now step by step: using, that $(\forall x\in X)Q(x)$ is same as $(\forall x)(x\in X \Rightarrow Q(x))$ and $(\exists x\in X)Q(x)$ is same as $(\exists x)(x\in X \land Q(x))$, definition $(1)$ can be decoded as
$$(\exists C)\big(C\in \mathbb{R} \land (\forall x)(x\in X \Rightarrow x\leq C)\big)\quad (1)$$
Taking negation gives
$$(\forall C)\big(C\notin \mathbb{R} \lor (\exists x)(x\in X \land x> C)\big)$$
which is
$$(\forall C)\big(C\notin \mathbb{R} \lor (\exists x\in X)( x> C)\big)$$
$$(\forall C)\big(C\in \mathbb{R} \Rightarrow (\exists x\in X)( x> C)\big)$$
$$(\forall C\in \mathbb{R})(\exists x\in X)(x> C)\quad (2)$$

Title: Negation of specific statement

Body:
Definition: We say that $X\subset \mathbb{R}$ is bounded above if $\exists C\in \mathbb{R}$ s.t. $\forall x\in X$ we have $x\leq C$.

$(X \ \text{is bounded above}):=\exists C\in \mathbb{R}((\forall x\in X)\Rightarrow (x\leq C))$

So I want to take the negation of it: $$\neg(X \ \text{is bounded above})\equiv \neg(\exists C\in \mathbb{R}((\forall x\in X)\Rightarrow (x\leq C)))\equiv$$ $$\equiv\forall C\in \mathbb{R} \neg((\forall x\in X)\Rightarrow (x\leq C))$$
But we know that $\neg (P\Rightarrow Q)\equiv P\land \neg Q$ which implies that $$\neg(X \ \text{is bounded above})\equiv \forall C\in \mathbb{R} ((\forall x\in X)\land (x> C)) $$

But I know that the correct negation should be the following: $X\subset \mathbb{R}$ is not bounded above if $\forall C\in \mathbb{R}$ $\exists x\in X$ s.t. $x>C$.

What am I doing wrong in the above?


问:
In an abelian category, an object $G$ is a generator iff for any nonzero object $X$, $\operatorname{Hom}(G, X) \neq 0$
In an abelian category, an object $G$ is a generator iff for any nonzero object $X$, $\operatorname{Hom}(G, X) \neq 0$

Would you give me a proof or some references?

I could show the “only if” part.

I don’t know that at which I use the abelian-ness of the category.
In the proof of “only if” part, I didn’t use it.

答:
Here is a proof under the additional hypothesis that $G$ is projective; I don't know either a proof or a counterexample without this assumption. (Edit: In the comments Jeremy Rickard gives the example of $\mathbb{Z}/2\mathbb{Z}$ in the category of $\mathbb{Z}/4\mathbb{Z}$-modules).

Let $G$ be a weak generator, meaning that if $\text{Hom}(G, X) = 0$ then $X = 0$ (so $\text{Hom}(G, -)$ reflects zero objects), which is also projective. We want to show that $G$ is a generator, meaning that $\text{Hom}(G, -)$ is faithful, or equivalently that if $f : X \to Y$ is a morphism such that $\text{Hom}(G, f) = 0$ then $f = 0$ (so $\text{Hom}(G, -)$ reflects zero morphisms).

We have that $f = 0$ iff $\text{im}(f) = 0$. Since $G$ is projective, $\text{Hom}(G, -)$ is exact, so preserves kernels and cokernels, and hence preserves images. . This means that if $f : X \to Y$ is a morphism, then

$$\text{Hom}(G, \text{im}(f)) \cong \text{im}(\text{Hom}(G, f)).$$

So we have equivalences

$$\text{Hom}(G, f) = 0 \Leftrightarrow \text{im}(\text{Hom}(G, f)) = 0 \Leftrightarrow \text{Hom}(G, \text{im}(f)) = 0 \Leftrightarrow \text{im}(f) = 0 \Leftrightarrow f = 0$$

which is the desired result.

Without the hypothesis that $G$ is projective you get only that $\text{Hom}(G, -)$ preserves kernels and so reflects monomorphisms (using $\text{ker}(f)$ instead of $\text{im}(f)$ above).


问:
Normal Distribution and Dice game
Suppose I use the following rules to generate random numbers. I roll a fair $D_6$ twice. The first roll I call $\mu$ and the second $\sigma^2$. I then Generate a normal random variable with mean $\mu$ and variance $\sigma^2$. I call this $X$

In more concise notation: $X \sim N(U_1,U_2)$ where $U_1,U_2$ are identical independent discrete uniform R.V's on $\{1,2,3,4,5,6\}$

Find an $x \in \mathbb{R} $ such that $\mathbb{P}[X\leq x] = \frac{1}{2}$

I think $x=3.5$.

Reasoning: Symmetry. More precisely: $\mathbb{P}[X \leq x] = \sum\limits_{i=1}^6\mathbb{P}[X\leq x|U_1=i]\cdot\mathbb{P}[U_1=i]=\sum\limits_{i=1}^6\mathbb{P}[X\leq x|U_1=i]\cdot\frac{1}{6}$

Now examine each term in the sum, $\mathbb{P}[X\leq x|U_1=i]\cdot\mathbb{P}[U_1=i]$. Let us take $i=1 $ and $i=6$ as $3.5$ is in the middle of these two values we have that $\mathbb{P}[N(1,U_2)\leq3.5]=\mathbb{P}[N(6,U_2)\geq3.5]$ from symmetry of normal dist. And then finally $\mathbb{P}[N(6,U_2)\leq3.5] = 1- \mathbb{P}[N(6,U_2)\geq3.5] = 1 - \mathbb{P}[N(1,U_2)\leq3.5]$

Hence: $\mathbb{P}[N(1,U_2)\leq3.5]+\mathbb{P}[N(6,U_2)\leq3.5] = \mathbb{P}[N(1,U_2)\leq3.5] + 1- \mathbb{P}[N(1,U_2)\leq3.5]=1 $

So if we split the sum into the $3$ pairs of integers that have $3.5$ in the middle, i.e sum to $7$ i.e $\{(1,6) , (2,5) , (3,4) \}$ we end up with:

$\mathbb{P}[X \leq x] = \sum\limits_{i=1}^6\mathbb{P}[X\leq x|U_1=i]\cdot\frac{1}{6} = \frac{1}{6}(1+1+1)=\frac{1}{2}$

Is this correct? :)

The second part of the question is to now have $X \sim N(U_1^2,U_2)$ where $U_1,U_2$ are identical independent discrete uniform R.V's on $\{1,2,3,4,5,6\}$ Find an $x \in \mathbb{R} $ such that $\mathbb{P}[X\leq x] = \frac{1}{2}$

And then finally to have  $X \sim N(U_1^n,U_2)$ for some $n \in \mathbb{N}$
and where $U_1,U_2$ are identical independent discrete uniform R.V's on $\{1,2,3,4,5,6\}$ Find an $x \in \mathbb{R} $ such that $\mathbb{P}[X\leq x] = \frac{1}{2}$

答:
Consider the following slightly more general situation.
We fix some finite index set $K$, in the OP it is the set of all tuples $(i,j)$, with uniform probability on it, and consider a family $(X_k)$ of independent gaussian random variables, indexed by $k\in K$, so that $X_k\in\mathcal N(\mu_k,\sigma^2_k)$. We consider a two-steps process, in the first step we pick $k$ uniformly, (well, this can be easily generalized using some weights,) in the second step we generate a random number by $X_k$.

(I think, i managed to rewrite equivalently the modelling intention from the OP, the stressed independence is essential below.)

This is equivalent to considering the gaussian random variable $X$ in one step,
which models the same experiment,
$$
X=\frac 1{|K|}\sum_{k\in K} X_k\ .
$$
Its  mean is
$$
\mu =\Bbb E [X]=\frac 1{|K|} \sum_k\Bbb E[X_k]=\frac 1{|K|} \sum_k \mu_k\ ,
$$
and the variance is not needed, since we already know that
$$
\Bbb P[X\le \mu]=\frac 12\ .
$$
For the case "$N(U_1^n,U_2)$" the mean is explicitly
$$
\mu=\frac 16(1^n+2^n+3^n+4^n+5^n+6^n)\ .
$$
(I may insert some simulation, if this is really a plus.)

Title: Normal Distribution and Dice game


问:
$|f(x)|$ has 2020 divisors for infinately many $x\in\Bbb N$

Let $f(X)\in\Bbb Z[X]$ be a polynomial of degree $n \ge 1$. Prove that there exist infinately many $x\in\Bbb N$ such that $|f(x)|$ has more than $2020$ distinc prime divisors.

Firstly I want to announce that this is how I remember the problem, so the problem statement might be false, but I thought about it and it seems legit.
My approach was to firstly prove that the mentioned polynomial is inifnately often a composite number.
So let $a\in\Bbb N$ and $f(a) = p$ for some prime $p$. Now let's consider a sequence $a_i = a + kp$ and let $i \in {0,1,2,...}$. It is easy to see $a_i\equiv a\pmod p$ and hence $f(a_i)\equiv f(a)\pmod p$, but we know that a polynimial of degree $n$ can achive a certain value at most $n$ times. So in at most $3n$ cases $f(a) = 0,-p,p$ and as we have inifnately many choices of $i$ we will eventually get a value that is multiple of $p$ but not equal to $0,-p,p$. Now my idea was to prove the problem statement using the above idea but for numbers with $2$ distinct prime divisors and extend this idea for numbers with any number of distinct prime divisors ($2020$ included). But I got stucked. I smilarily proved that there are only finite numbers of cases in which the function $f(x)$ is $0,-pq,pq$ for some primes $p,q$ but how do I know if I'm not getting the multiples of the form $p^xq^y$ for some nonnegative $x,y$? How should I proceed? Or maybe I should change strategy? Any help appreciated.

答:
You're definitely on the right track. See if you can prove the following statement by induction on $n$.

For all $n$, there exists some integer $x$ for which $|f(x)|$ has at least $n$ divisors.

You've essentially gone through one level of the inductive step (from "there exists a value that isn't $\pm 1$" to "there exist infinitely many composite values") -- the rest of the approach should generalize similarly. Start with a value of $x$ for which $|f(x)|$ has at least $n$ divisors, and try to find some other $x'$ for which $|f(x')|$ has more divisors than $|f(x)|$.


答:
By Schur's theorem, the set of primes that divide some $f(n)\neq0$ (for $n\in\mathbb N$) is infinite.

So pick $2021$ primes $p_1,p_2,\dots,p_{2021}$ and positive integers $a_1,\dots,a_{2021}$ such that $p_i\mid f(a_i)$. Then by the Chinese Remainder Theorem pick $y$ solving the system
$$y\equiv a_i\pmod{p_i},$$ for each $1\leq i\leq2021$. Then we have infinitely many such $y$, and
$$f(y)\equiv f(a_i)\equiv0\pmod{p_i},$$
so each such $f(y)$ has at least $2021$ prime divisors.

Title: $|f(x)|$ has 2020 divisors for infinately many $x\in\Bbb N$

Body:
Let $f(X)\in\Bbb Z[X]$ be a polynomial of degree $n \ge 1$. Prove that there exist infinately many $x\in\Bbb N$ such that $|f(x)|$ has more than $2020$ distinc prime divisors.

Firstly I want to announce that this is how I remember the problem, so the problem statement might be false, but I thought about it and it seems legit.
My approach was to firstly prove that the mentioned polynomial is inifnately often a composite number.
So let $a\in\Bbb N$ and $f(a) = p$ for some prime $p$. Now let's consider a sequence $a_i = a + kp$ and let $i \in {0,1,2,...}$. It is easy to see $a_i\equiv a\pmod p$ and hence $f(a_i)\equiv f(a)\pmod p$, but we know that a polynimial of degree $n$ can achive a certain value at most $n$ times. So in at most $3n$ cases $f(a) = 0,-p,p$ and as we have inifnately many choices of $i$ we will eventually get a value that is multiple of $p$ but not equal to $0,-p,p$. Now my idea was to prove the problem statement using the above idea but for numbers with $2$ distinct prime divisors and extend this idea for numbers with any number of distinct prime divisors ($2020$ included). But I got stucked. I smilarily proved that there are only finite numbers of cases in which the function $f(x)$ is $0,-pq,pq$ for some primes $p,q$ but how do I know if I'm not getting the multiples of the form $p^xq^y$ for some nonnegative $x,y$? How should I proceed? Or maybe I should change strategy? Any help appreciated.


问:
What is an open set in a topological space?
I recently started learning topology to help me understand limits and continuity better for calculus, and I am struggling with some of the definitions.

What I am getting confused with is why is every set in a topology considered to be open and when talking about sets in the topology we always say the set is open.

My intuitive notion of openness from  previous knowledge of mathematics is an interval that does not contain its endpoints, so there is an infinite sequence at the end points, e.g., $(0,1)$ is an open interval.

However, in topology, for example, the singleton $\{1\}$
is considered an open set—how is this so? Why are sets in a topology always open? And what is the definition of an open set in a topological space?

Thanks in advance.

答:
An open set is an abstraction.  So an "open" set, for the most part, whatever we want it to be.  And a Topology is nothing more or less than a list of sets we say "These ones; these are the sets that we are going to be considered open.

We can make the list really really short by having nothing in the list (except the empty set  and the universal set); or we can make the list really really long by having every possible set in the list.  Or we could make it completely arbitrary by saying sets with only elements wearing purple socks will be "open".

The only rules are:  1) The empty set and the universal set must both be open. 2) Any finite intersection of open sets is open; 3) any union (finite or infinite) of open sets is open.

So if for instance our universal set is $\{Heidi, Mary, Sam, Bill\}$ and our topology is $\{\{Mary,Bill\}, \{Heidi, Sam\}, \{Heidi, Bill\}\}$ then you we have $16$ sets and of them:

$\emptyset$ is open because that's the law.

$\{Heidi\}$ is open because it is the intersection of $\{Heidi,Sam\}$ and $\{Heidi,Bill\}$

$\{Mary\}$ is not open because it's not the union or intersection of any open sets.

$\{Sam\}$ is not open because it's not the union or intersection either.

$\{Bill\}$ is open because it's the intersection of $\{Mary,Bill\}$ and $ \{Heidi, Bill\}$

$\{Heidi, Mary\}$ is not open because we just can't get it from the rules.

$\{Heidi, Sam\}$ is open because we said it was.

$\{Heidi, Bill\}$ is open because we said it was.

$\{Mary, Sam\}$ is not open because we can't get it from the rules.

$\{Mary, Bill\}$ is open because we said it was.

$\{Sam,Bill\}$ is not open because we can't get there.

$\{Heidi, Mary, Sam\}$ is not open.

$\{Heidi, Mary, Bill\}$ is open because it is the union of $\{Heidi, Mary\}$ and $\{Heidi, Bill\}$

$\{Heidi, Sam, Bill\}$ is open because it is the union of $\{Heidi, Sam\}$ and $\{Heidi, Bill\}$.

$\{Mary, Sam, Bill\}$ is not  open.

And $\{Heidi, Mary, Sam, Bill\}$ is open because it is the unversal set and it's the law that it must be open.

....

That's it.  Open means what we say it does.

.....

The real question is if in a metric space to claim "open" means "every point is an interior point" we have to ask: Does that follow the rules?  That is:

Are the empty set and the universal set open?  (Empty set is vacuously open because it has no points so all points are interior points; universal set is open as every open ball is a subset of the universal set so every point is an interior point of the universal set.)

Is the finite intersection of any sets where every point is an interior point itself has that property?  (It does.... It requires a proof though)

Is every union of sets with that property have that property? (ditto).

And finally, do no other set have that property?.... (yeah, it's tedious be it can be proven).

.....

I recently started learning topology to help me understand limits and continuity better for calculus

If it helps you, you will probably be the first.

Limits and real numbers and distances in a metric space make for a very very intuitive topology.  But An abstract topology will probably be hard to "see" off hand.  (Does my example of four people and some groups are open and others are not make any sense?  No, it does not... but it follows the rules!)  What one hopes to get from this is how to verify solutions based entirely on the logic of rules and not instinct or (much worse) assumptions (which can often be dead wrong.)


答:
There are a number of things about topology that are idiosyncratic. It is typical to refer to singleton sets with the same symbol as used for the element of the singleton. This should act as an indication that one is actually working within the power set of the domain of definition.

A topology is given as a partition on the power set. Provided that a partition is compatible with the axioms, the partition will consist of open, closed, clopen (both open and closed) and none of the above. The axioms can be given in terms of either open sets or closed sets. So, these classifications reside in the power set of the power set of the domain of definition. Since one can have clopen sets, this classification does not produce disjoint collections.

As a set, the domain of definition has a sort of special relationship with the topology in which every singleton is open. This is called the discrete topology. Every set is open by arbitrary unions. So every set is closed as well by complementation.

Importantly, one can have topologies where not every set is open. The topology with only the empty set and the domain of definition as open sets satisfies the axioms. It is called the trivial topology.

From these remarks, you should surmise that "what is open" in a topology is a matter of stipulation.  For real analysis and calculus the stipulation comes from our naive conceptions of length and measurement. What becomes important are inequalities. The "in between" mathematical object is called a metric space. This governs how we understand distances between points.

Whatever topology book you are using, it could be highly useful to obtain a copy of "Counterexamples in Topology" by Steen and Seebach (reprints from Dover Books are inexpensive).  It is full of examples, including some simple ones accessible to beginners. It also summarizes what you will learn from main textbook. So, it will become a reference resource as you learn the subject better.


问:
Under what conditions is $\frac{dF(x)}{dx}\Big|_{x-x_0} = \frac{dF(x-x_0)}{dx}\Big |_x$?
I have a very complicated function for which I need to evaluate

$$ \frac{dF(x)}{dx}\Bigg|_{x-x_0}.$$

How can I tell when this will be equivalent to evaluating

$$ \frac{dF(x-x_0)}{dx}\Bigg|_x ?$$

Under what conditions on $F$ is this equivalent? Would this condition change for higher derivatives?

答: Assuming $x_0$ is a constant, the Chain Rule gives $\frac{dF(x-x_0)}{dx}\Bigg|_x =\frac{dF(x-x_0)}{dx}=F'(x-x_0)\cdot \frac{d(x-x_0)}{dx} =\frac{dF(x)}{dx}\Bigg|_{x-x_0}$


问:
Convergence almost surely question
Suppose you have a sequence of independent random variables {$X_i, i\geq1$}, such that
$$\Bbb P(X_i=i^2 -1)=i^{-2},$$ $$ \Bbb P(X_i=-1)=1-i^{-2}.$$
Then, $\frac1n \sum_{i=1}^n X_i$ converges almost surely to a constant c. Find the value of c.

For this question, I thought of applying the strong law of large numbers which says:
$$ \frac1n \sum_{i=1}^n X_i$$ converges almost surely to $\mu$, where {$X_i, i\geq1$} is a sequence of IID random variables whose mean $\mu$ exists.

However, for the question above, $\mu=0$. But c is not equal to $0$. c is, in fact, equal to -1.

Any clarification on this would greatly be appreciated, as I'm not sure how $c=-1$.

答: Use the Borel-Cantelli lemma.  Since $\sum_{n \geq 1} \mathbb{P}(X_n \neq -1) = \sum_{n \geq 1} n^{-2} < \infty$, we know that almost surely, $X_n = -1$ for all $n$ sufficiently large.   Then the desired result follows easily: if $(a_n)$ is a sequence that is eventually constant, it's straightforward to verify that $\frac{1}{N} \sum_{n=1}^{N} a_n$ converges to that constant as $N \to \infty$.

Title: Convergence almost surely question


问:
Is the following series involving $\zeta(s)$ asymptotic to $\frac{1}{2s}$?
Context:

While working on Keiper's  $\tau_s$-function, I found that it could be split into a smooth first term and an oscillating second term:

$$\tau_s=\sum_{k=1}^{\infty} \left( {s+k-1\choose k-1}\cdot \frac{\zeta(k+1)}{(-2)^{k+1}}- {-s+k-1\choose k-1} \sigma_{k+1}^*\right)$$

with $\sigma_k^*$ defined here. I found that numerically the first term appears asymptotic to $\dfrac{1}{2s}$ for $\Re(s) \gt 0$.

Question:

With $s \in \mathbb{C}, \Re(s) > 0$, is the following asymptotic true ?

$$\sum_{k=1}^{\infty} {s+k-1\choose k-1}\cdot \frac{\zeta(k+1)}{(-2)^{k+1}}\asymp \frac{1}{2s} \qquad \text{for} \quad \Re(s) \rightarrow \infty$$

Body:
Use
$${n+\alpha\choose n} = (-1)^n {-\alpha-1\choose n}$$
Then
$$\sum_{k=0}^{\infty} {s+k\choose k}\frac{\zeta(k+2)}{(-2)^{k+2}} = \sum_{n\ge 1} \sum_{k=0}^{\infty} {-s-1\choose k} (2n)^{-k-2} = \sum_{n\ge 1}(2n)^{-2}(1+\frac1{2n})^{-s-1}$$ $$\sim \frac12 \int_1^\infty x^{-2} (1+\frac1{x})^{-s-1}dx=\frac12 \int_0^1 (1+y)^{-s-1}dy=\frac{1-2^{-s}}{2s}$$


问:
Which methods to use to integrate $\int{\frac{x^4 + 1}{x^2 +1}}\, dx$
I have this integral to evaluate:

$$\int{\frac{x^4 + 1}{x^2 +1}}\, dx$$

I have tried substitution, trig identity and integration by parts but I'm just going round in circles.

Can anyone explain the method I need to work this out?

答:
Whenever you have a rational polynomial with a numerator of equal or larger degree than the denominator, try to factor the numerator if possible, or simply, use polynomial long division.

lab bhattacharjee noticed a nice way to simplify the rational integrand by manipulating the numerator to make life easier.

But suppose you're tired and not feeling particularly creative; polynomial long division will give you the same result:

Dividing numerator by denominator using simple but quick polynomial long division gives us:

$$\frac{x^4 + 1}{x^2 +1}=(x^2-1)+\frac2{x^2+1}$$

So, we can express our integral as follows: $$\int \frac{x^4 + 1}{x^2 +1} \, dx\;= \;\;\int x^2 \,dx \;\;- \;\;\int \,dx \;\;+\;\; 2\int \frac 1{x^2+1}\,dx$$

No doubt, you can handle the first two integrals. And for the third: notice that the third integral is in perfect form which integrates nicely to $$2\int \frac 1{x^2+1}\,dx\;\; = \;\;\;2\tan^{-1}(x) + C$$


答:
In fact, you are very close to the solution, by I'm afraid you've made a mistake. Using dumb indexes (summation over repeated indices is implied) we can write: $|\nabla\times\vec{u}|^2=∂_iu_j∂_iu_j-∂_iu_j∂_ju_i$ and $|\nabla\cdot\vec{u}|^2=∂_iu_i∂_ju_j $. In the same fashion $\int_D\left(|\nabla u_x|^2+|\nabla u_y|^2+|\nabla u_z|^2\right)dV =  \int_D\left(∂_iu_j∂_iu_j\right)dV$. Next, integrate by parts the term:  $\int_D\left(-∂_iu_j∂_ju_i\right)dV = \int_D(-∂_i(u_j∂_ju_i)+u_j∂_i∂_ju_i)dV$. Integral $\int_D(-∂_i(u_j∂_ju_i))dV$ vanishes, because $\vec{u}=\vec{0}$ on the boundary of the region of integration.

Integrate by parts the second time and you prove the equality.

Title: Identity: volume integral of sum of norms

Body:
I am trying to show $\int_D\left(|\nabla\cdot\vec{u}|^2+|\nabla\times\vec{u}|^2\right)dV=\int_D\left(|\nabla u_x|^2+|\nabla u_y|^2+|\nabla u_z|^2\right)dV$ but am having no luck.

Here $D$ is a smooth, bounded domain in $\mathbb{R}^3$, and $\vec{u}$ is a smooth vector field in $\mathbb{R}^3$, such that $\vec{u}=\vec{0}$ on the boundary $∂D$.

I have managed to show $\int_D\left(|\nabla\cdot\vec{u}|^2+|\nabla\times\vec{u}|^2\right)dV= \int_D\left(2|\nabla\cdot\vec{u}|^2-∂_iu_j∂_ju_i\right)dV$. I have tried a tonne of things from here but haven't made any step towards the solution.

A hint I am given is that $∂_{x_i} ∂_{x_j} f= ∂_{x_j} ∂_{x_i} f(x)$, but I have no idea why the fact that partial derivatives commute relates to this problem.


问:
Closed form for $\prod_{n=0}^\infty (1-z^{2^n})$
Is there a closed form for the product $$f(z) = \prod_{n=0}^\infty 1-z^{2^n}$$
either as a formal power series or as an analytic function in the disk $|z| < 1$? It's not hard to see that Taylor series coefficients of $f$ about 0 are all $\pm 1$:
$$f(z) = 1 - z - z^2 + z^3 - z^4 + z^5 + z^6 - z^7 + \dotsb$$
and that they form a pattern in blocks, so to speak:
$$+- \quad\to\quad +--+ \quad\to\quad +--+-++- \quad\to\quad \dotsb$$
But I don't know much else. I came across this infinite product when I incorrectly transcribed an exercise from a book, wasting a good deal of time before returning to the book to ask it, "are you sure?". (The exercise was to prove that $\prod_{n=0}^\infty 1 + z^{2^n} = (1-z)^{-1}$, which is easy.)</p>

答:
Notice that if $$f(z)=\prod_{n\ge0}(1-z^{2^n}),$$
then
$$f(z)=(1-z)f(z^2).\tag1$$
Then define the sequence $a_n$ to satisfy $$f(z)=\sum_{n\ge0}a_nz^n.$$
Then from $(1)$ we have
$$\sum_{n\ge0}a_nz^n=(1-z)\sum_{n\ge0}a_nz^{2n}.$$
This is equivalent to
$$\sum_{n\ge0}a_nz^n=b_0+\sum_{n\ge1}(b_n-b_{n-1})z^n,$$
where $b_n=[2|n]\cdot a_{n/2}$, with $[2|n]=\frac{1+(-1)^n}{2}$ being the Iverson bracket. Since $f(0)=1$, we have $a_0=b_0=1$, and for $n\ge1$,
$$a_n=a_{n/2}[2|n]-a_{(n-1)/2}[2|n-1].\tag2$$
For $n=2k$, $$a_{2k}=a_k[2|2k]-a_{k-1/2}[2|2k-1]=a_k.$$
Similarly, for $n=2k+1$, $$a_{2k+1}=a_{k+1/2}[2|2k+1]-a_{k}[2|2k]=-a_k.$$
As you saw, it is then easy to see that $|a_k|=1$. It then follows directly that $a_{2^n}=-1$.

We can find some special cases of $a_m$ where $m=2^k$, such as
$$\begin{align}
a_{2m+1}&=1\\
a_{4m+2}&=1\\
a_{4m+3}&=-1\\
a_{8m+6}&=-1\\
a_{8m+7}&=1\\
a_{16m+14}&=1\\
a_{16m+15}&=-1\\
a_{32m+30}&=-1\\
a_{32m+31}&=1\\
&...
\end{align}$$
which all follow from $a_m=-1$.

Clearly, $a_n$ expresses something about the ways in which one can write $n$ as $a2^k+b$, for very specific $a,b$, but other than that, it is proving exceedingly difficult to find a general closed form for $a_n$. I doubt a closed form for $f(z)$ exists.


问:
Finding the measure of an angle

Let E be a point outside of the square $ABCD$ such $\triangle ABE$ is an equilateral triangle.  What is the measure of $\angle CED$, in degrees?

I need help with this problem. I made a diagram beforehand to help me with this problem. From doing this, I find that $m(\angle CAE)$ is the same as $m(\angle DBE)$. I think these angles are $150^\circ$ because the angles of a square are each $90^\circ$, plus when we have point $E$ makes an equilateral triangle (which has three $60^\circ$ angles). So $60^\circ + 90^\circ = 150^\circ$. This is all I have so far.

答:
Correction: I labeled points of square differently. But our solutions mirror one another.

The statement of the problem is below the image.

Note that both $\angle EBC$ and $\angle EAD$ are $150^\circ$ ($90^\circ$ (measure of an angle of the square + $60^\circ$ (measure of each of three angles in an equilateral triangle)). Note also that $AE=AD=EB=BC$, since the lengths of all sides of square and triangle are equal (for triangle $ABE$ to be equilateral, both $EA$ and $EB$ must be the same length as $AB$.)

Since $AE=AD$, triangle $ADE$ is isosceles (and likewise, triangle $EBC$ is isosceles), and so the angles opposite the equal sides are equal. Since we have that $\angle EBC$ and $\angle EAD$ are $150^\circ$, $m(\angle ADE) + m(\angle AED) = m(\angle CEB) + m(\angle BCE) = 180 - 50 = 30^\circ$.  So, in particular, $m(\angle AED) = m(\angle CEB) = 15^\circ$. So we have that $m(\angle CED) = m(\angle AEB) - 2m(\angle AED) = 60 - 2(15) = 30^\circ$.

Title: Finding the measure of an angle


问:
Finding the range of a vector valued function
For a single valued function, I can infer if the function is monotone from its derivative.

For a vector valued function, is it possible to infer monotonicity from the directional derivative?

For example, define
$$
D=[1,2]\times[1,2],
$$
and
$$
f(x,y)=\left( \frac{2}{1/x+1/y},\sqrt{xy} \right).
$$
Is it possible to show that $f$ maps $D$ to $D$ from its gradient $\nabla f$?

The gradient is
$$
\nabla f = \begin{pmatrix}
\frac{2}{\left(1+x/y\right)^2} & \frac{2}{\left(1+y/x\right)^2} \\
\frac{y^{1/2}}{2x^{1/2}} & \frac{y^{1/2}}{2x^{1/2}}
\end{pmatrix},
$$
whence, for $(x,y)$ in $D$, the directional derivative
$$
\left(\nabla f(x,y)\right)\begin{pmatrix} x \\ y \end{pmatrix},
$$
is always positive and I would like to conclude that, on $D$,
$$
\text{$f$ is minimal at $(1,1)$},\\
\text{$f$ is maximal at $(2,2)$}.
$$

Is it the right way to proceed?

The graph of each component of $f$ looks like this

答:
Note that the Jacobian determinant is
$$\det J_f(x,\, y)=\det\begin{bmatrix}
\frac{2}{\left(1+x/y\right)^2} & \frac{2}{\left(1+y/x\right)^2} \\
\frac{y^{1/2}}{2x^{1/2}} & \frac{y^{1/2}}{2x^{1/2}}
\end{bmatrix}=\\
=\dfrac{1}{{\left(\dfrac{1}{x} + \dfrac{1}{y}\right)}^{2} \sqrt{x y} x} - \dfrac{1}{{\left(\dfrac{1}{x} + \dfrac{1}{y}\right)}^{2} \sqrt{x y} y}= \\
=\dfrac{y-x }{{\left(\dfrac{1}{x} + \dfrac{1}{y}\right)}^{2}  (xy)^{\frac{3}{2}}},$$ therefore $\det J_f(x,\, y)=0$ on the line $y=x.$

Addition:

Denote $$G_1=\{(x, \ y)\colon \;\; {1} < {x} < {2},\;\; x < {y} < {2} \}, \\
G_2=\{(x, \ y)\colon \;\; {1} < {y} < {2},\;\; y < {x} < {2} \}$$ Then $J_f (x,\ y)\ne{0}, \;\;\; \forall(x, \ y)\in{G_1\cup G_2},$
thus $f(x, \ y)$ is a diffeomorphism on each $G_1$ and $G_2$  and

$$f(\partial{G_1})=\partial{f(G_1 )}, \\
f(\partial{G_2})=\partial{f(G_2 )}.$$
As noted by Ted Shifrin, $$\gamma_1=\left\{ f(1,\ y),\;\;  1 < y < 2\right\}=\left\{\left(\dfrac{2y}{y+1},\sqrt y\right)\colon \;\;  1 < y < 2\right\}, \\
\gamma_2=\left\{ f(x,\ 2),\;\;  1 < x < 2\right\} = \left\{\left(\frac{4x}{x+2},\sqrt{2x}\right)\colon \;\;  1 < x < 2\right\}, \\
\gamma_3=\left\{ f(x,\ x),\;\;  1 < x < 2\right\} = \left\{\left(x,x\right)\colon \;\; 1 < x < 2\right\}.$$
Therefore the interior of $\partial{G_1}$ is mapped by $f$ into the interior of $f(\partial{G_1}) = \gamma_1\cup\gamma_2\cup \gamma_3.$
Due to the symmetry of  $f$ with respect to $x$ and $y$ the interior of $\partial{G_2}$ is mapped by $f$ also into the interior of $\gamma_1\cup\gamma_2\cup \gamma_3.$

Different signs of the Jacobian determinant on both sides of the diagonal indicate the opposite orientation of parts of the image of $D.$


问:
A definable set in $(\mathbb{R}, +,-,*,0,1,<)$ that is not definable without parameters.
Consider the structure $(\mathbb{R},+,-,*,0,1,<)$. Can someone exhibit a subset of $\mathbb{R}$ that is definable with parameters in that structure, but not definable without parameters?

答:
Let $\mathcal{R}$ be the field of real numbers. I claim that the parameter-freely-definable elements of $\mathcal{R}$ are exactly the algebraic reals; consequently, sets like $\{\pi\}$ are definable with parameters but are not definable without parameters.

The parameter-free-definability of the algebraic reals follows from elementary algebra. Suppose $f$ is a nonconstant polynomial with rational coefficients, and $a$ is a solution to $f(x)=0$. Since nonconstant polynomials have finite zero sets, there are rationals $p,q$ such that $a$ is the unique zero of $f$ in $(p,q)$. But each rational is obviously definable in $\mathcal{R}$, so "the unique zero of $f$ in $(p,q)$" is in fact a parameter-free definition of $a$ in $\mathcal{R}$.

The hard direction is the converse, and this is where model theory properly enters the picture. We know that the theory of real closed fields admits quantifier elimination. It's easy to check that every real definable in $\mathcal{R}$ by a quantifier-and-parameter-free formula is in fact algebraic, so we're done. Alternatively, quantifier elimination implies model completeness, so since the algebraic reals are a real closed field we get by elementarity that every parameter-freely-definable element of $\mathcal{R}$ is algebraic.


问:
Understanding the arc length parameter notation
I've been struggling with the notion of the arc length parameter and I was wondering if this was due to notations. In several text books I often see this occur. A regular curve $\sigma:[a,b]\rightarrow \mathbb{R}^n$ which is not parametrized according to arc length and so takes in as variable $t$ or rather has domain over $[a,b]$ the reparametrization of $\sigma$ is then $\sigma \circ s^{-1}$ where $s(t)= \int_{a}^{t}\lVert \sigma' (x)\rVert \,dx$ so $s:[a,b]\rightarrow[0,L(\sigma)]$. From what I understand a function that is evaluating the parameter $s$ has domain over $[0,L(\sigma)]$. But then there appears expression of the form $\sigma(s(t))$ which generally don't make sense
Example:

My suspicion is that there is an abuse in notation where the reparametrization of $\sigma$ is just written as $\sigma$ is this the case. Also if this is the case is there a reason behind this use of notation?

答:
In my experience, yes, it is typical for there to be abuse of notation when it comes to parametrization by arc length.

I think it typically comes with the use of a so-called "dummy variable" $t$ in the calculation of the integral $s(t)=\int_0^t||\sigma'(t)||\rm{dt}$.

I think it is often done in integrals of this type, and is just the nature of the beast.


问:
Does anyone know some theorem to solve this integral?
Recently I have been studying calculus and I got stuck with this problem:

I already know that the functions $f$ and $g$ are both real functions, defined and continuous in $[a,b]$, such that
\begin{equation}
\int_{a}^{b}f(t)dt=2\int_{a}^{b}g(t)dt.
\end{equation}

I need to show that exists $c\in[a,b]$, such that $f(c)=2g(c)$.

To solve that, I've started doing:
\begin{equation}
\int_{a}^{b}f(t)dt=F(b)-F(a)=2[G(b)-G(a)]=2\int_{a}^{b}g(t)dt,
\end{equation}
considering $F(t)$ and $G(t)$ antiderivatives for $f(t)$ and $g(t)$. Multiplying all terms by $\frac{1}{(b-a)}$, we have:

\begin{equation}
\frac{F(b)-F(a)}{b-a}=2\frac{[G(b)-G(a)]}{b-a}.
\end{equation}
I don't know the theorem, but I surely know that exists $c\in[a,b]$ such that
\begin{equation}
F'(c)=\frac{F(b)-F(a)}{b-a}\Leftrightarrow f(c)=\frac{F(b)-F(a)}{b-a}=2\frac{[G(b)-G(a)]}{b-a},
\end{equation}
whitch gives me
\begin{equation}
f(c)=2\frac{[G(b)-G(a)]}{b-a}.
\end{equation}

I'm almost there, but I feel that I'm missing something. Please, help me.

答:
$$\int_{a}^{b}f(t)dt=2\int_{a}^{b}g(t)dt \iff \int_{a}^{b}(f(t)-2g(t))dt=0.
$$

The mean value theorem of integral calculus states that there exists $c\in [a,b]$ such that $$\int_a^b h(t)dt= h(c)(b-a).$$

Taking $h(t)=f(t)-2g(t)$ we are done.


答:
The notation does not look too badly abused.

$\sigma' = \frac {d\sigma}{dt} = \frac {d\sigma}{ds}\frac{ds}{dt} = \dot\sigma\frac{ds}{dt}$ is just the chain rule and the given definitions.

$\frac{ds}{dt} = \|\sigma'(t)\|$ comes from the definition of $s$ and the fundamental theorem of calculus.

Title: Understanding the arc length parameter notation

Body:
I've been struggling with the notion of the arc length parameter and I was wondering if this was due to notations. In several text books I often see this occur. A regular curve $\sigma:[a,b]\rightarrow \mathbb{R}^n$ which is not parametrized according to arc length and so takes in as variable $t$ or rather has domain over $[a,b]$ the reparametrization of $\sigma$ is then $\sigma \circ s^{-1}$ where $s(t)= \int_{a}^{t}\lVert \sigma' (x)\rVert \,dx$ so $s:[a,b]\rightarrow[0,L(\sigma)]$. From what I understand a function that is evaluating the parameter $s$ has domain over $[0,L(\sigma)]$. But then there appears expression of the form $\sigma(s(t))$ which generally don't make sense
Example:

My suspicion is that there is an abuse in notation where the reparametrization of $\sigma$ is just written as $\sigma$ is this the case. Also if this is the case is there a reason behind this use of notation?


问:
Property of probabilities to add up to 1 : conditional case
I was wondering in the case of conditional distributions, where let's say we have $P(X|Y)$, if we fix $Y = a$, would the probabilities of $P(X|Y=a)$ for all values of X necessarily all add up to 1? If so then why?

Additionally, our professor also told us that if we had the joint distribution of $X$ and $Y$, then when we are trying to derive $f_{X|Y}(x|y)$ we are essentially fixing y, and looking at a slice of our pdf, and scaling that so that the integral adds up to 1. I did not quite understand this, and if someone could give me some more geometric intuition that would be super helpful. Thank you!

答:
The conditional distribution of $X$ given some event $A$ is still a probability distribution, so yes, the "probabilities must sum to $1$." For instance, if $X$ is a discrete random variable, then
$$\sum_x P(X=x \mid A) = 1$$
where the sum is over all values $X$ can take. Here, $A$ can be any event, such as $\{Y=a\}$ for some other random variable $Y$.

In the case where $(X,Y)$ has a joint density $f_{X,Y}(x,y)$, what your professor is saying is that the conditional density of $X$ given $Y=y_0$ looks kind of like the function $g(x) := f_{X,Y}(x,y_0)$ (this is the "slice of the joint PDF" where the second component is fixed at $y_0$). However, we do not have $\int_{-\infty}^\infty g(x) \, dx = 1$ necessarily since $\int_{-\infty}^\infty f_{X,Y}(x,y_0) \, dx = f_Y(y_0)$. But if we divide $g$ by this quantity (this is the "rescaling") we can make the integral equal to one. The conditional density ends up being this resulting function: $f_{X \mid Y=y_0}(x) = \frac{f_{X,Y}(x,y_0)}{f_Y(y_0)}$ which does satisfy $\int_{-\infty}^\infty f_{X \mid Y=y_0}(x) \, dx = 1$.

[Proving why this actually is the conditional density is a bit technical, but I guess your professor is just trying to give you intuition. It resembles the discrete case $P(X=x \mid Y=y_0) = \frac{P(X=x, Y=y_0)}{P(Y=y_0)}$.]


问:
How can I show that for all $v\in V$ there's $F\in V^{**} $ such that $F(f)=f(v)$ for all $f\in V^*$?
I'm trying to show that the map $\Phi :V\to V^{**}$ given by $\Phi (v)(f):=f(v)$ is well-defined. To show that I need to prove that for all $v\in V$ there's $F\in V^{**}
$ such that $F(f)=f(v)$ for all $f\in V^*$. However I don't know how to prove that in the infinite dimensional case!

My question: Suppose that $V$ is an infinite dimensional vector space. How can I show that for all $v\in V$ there's $F\in V^{**}
$ such that $F(f)=f(v)$ for all $f\in V^*$?

Below there's a lemma that I'm trying to use to prove the question above.

Lemma: Let $V$ be a vector space. Suppose that $\{v_i\}_{i\in I}$ is a base of $V$. Then there's a unique linearly independent subset $\{v^i\}_{i\in I}$ of $V^*$ such that $v^i(v_j)=\delta _{ij}$ for all $i,j\in I$ in which $\delta $ is the Kronecker delta.

EDIT:

I'll clarify my question:

Given two sets $A$ and $B$, we say that $f\subseteq A\times B$ is a map from $A$ to $B$ (which we denote by $f:A\to B$) if $(\forall x\in A)(\exists !y\in B)\left((x,y)\in f\right)$ is a true proposition.

So, my question is to prove that $\Phi :=\big\{(v,F)\in V\times V^{**}:(\forall  f\in V^*)\left(F(f)=f(v)\right)\big\}$ is a map from $V$ to $V^{**}$. That is, I want to prove that $(\forall v\in V)(\exists! F\in V^{**})\left((v,F)\in \Phi \right)$ is a true proposition.

答: Suppose that $V$ is a vector space over a field $k$. Take $v \in V$. Then, if you define a map $\Phi(v) : V^* \to k$ by $$(\forall f \in V^*) \quad \Phi(v)(f) = f(v)$$ note that $\Phi(v)$ is well-defined ($f \in V^* \implies \Phi(v)(f) \in k$) and is linear since if $f_1,f_2 \in V^*$ and $\lambda \in k$, $$\Phi(v)(\lambda f_1+f_2) = (\lambda f_1+f_2)(v) = \lambda f_1(v)+f_2(v) = \lambda \Phi(v)(f_1)+\Phi(v)(f_2).$$ Thus, $\Phi(v) \in (V^*)^* =: V^{**}$, and then $\Phi : V \to V^{**}$ is well-defined!


问:
Solving $y'=f(x,y)$ when $f$ is a rational function $f(x,y)=\frac{y^2+2xy}{x^2y-x^2}$
Solve the initial value problem $$\frac{dy}{dx}=\frac{y^2+2xy^2}{x^2y-x^2},~~y(1)=1$$

My attempt: I checked whether $(x^2y-x^2)dy-(y^2+2xy)dx=0$ is exact, but it was not exact. Also the substitution $y=ux$ makes $$x\frac{du}{dx}=\frac{u^2+ux-u}{ux-1}$$ and stucked.

Body:
$\frac{dy}{dx}=\frac{y^2+2xy^2}{x^2y-x^2}=\frac{(1+2x)y^2}{(y-1)x^2}\\
$

Your diff eq is separable.

$\int \frac {y-1}{y^2}\ dy = \int \frac {1+2x}{x^2} \ dx$


问:
Show $z_\alpha = -z_\alpha$.
Given $Z\sim N(0,1)$ and the standard normal upper quantile $z_\alpha = \Phi^{-1}(1-\alpha)$ I need to show that $$ z_\alpha=-z_\alpha$$
The hint given was to use the fact that the density function ($\phi$) is an even function and show $\Phi(x)=1-\Phi(-x)$.

Showing that was not too difficult:
$$\begin{align*}
\Phi(x)=&\int_{-\infty}^{\infty}\phi(x)dx -\int_{x}^{\infty}\phi(x)dx\\
=&\int_{-\infty}^{\infty}\phi(x)dx -\int_{-\infty}^{-x}\phi(x)dx\quad\text{because $\phi$ is even}\\
=& 1-\Phi(-x)
\end{align*}
$$

The next hint was to now substitute $x$ for $z$ but I do not see how to arrive at $ z_\alpha=-z_\alpha$

The only other conclusions I came to is that $\Phi(z_\alpha) = 1-\alpha $ an then $\Phi(-z_\alpha )=\alpha$?

Any push in the right direction would be appreciated!

Edit:

I realized substituting it into what was proven does show something that is:

$\Phi(z_\alpha) = 1-\Phi(-z_\alpha)$ which gives us  $1-\alpha = 1-\alpha$

But that did not really achieve any result.

答:
Assuming the problem is asking you to show $z_\alpha = -z_{1-\alpha}$:

By the definition of $z_\alpha$ and $z_{1-\alpha}$, you have $\Phi(z_{1-\alpha}) = \alpha$ and $\Phi(z_\alpha) = 1-\alpha$.

By the work that you have done,  you have $\Phi(z_{1-\alpha}) = 1 - \Phi(-z_{1-\alpha})$.

Thus $\Phi(-z_{1-\alpha}) = 1-\alpha$, and finally $-z_{1-\alpha} = z_\alpha$.

Title: Show $z_\alpha = -z_\alpha$.


问:
Product of matrices; MAPLE giving a strange answer
Either my brain is seriously fried up right now or the computer is wrong.

If I have a matrix $\begin{bmatrix}
4 & -2\\
2 & -1 \\
0 &  0
\end{bmatrix}$ multiply by its transpose $\begin{bmatrix}
4 &2  &0 \\
-2&-1 &0
\end{bmatrix}$, I should get a $3 \times 3$ $\begin{bmatrix}
20 &10  &0 \\
10 &5  &0 \\
 0&0  &0
\end{bmatrix}$

For some reason Maple is giving me a $2 \times 2$ $\begin{bmatrix}
20 &10 \\
10 &5   \\
  \end{bmatrix}$

Why did they delete the last row and column of 0s? You can't do that

答: You have multiplied the second matrix with the first matrix. Sice matrix multiplication is not commute the answer is wrong. And you have also ignored the negative signs. They should not be ignored.</p>


问:
Uniform Topology on $\mathbb{R}^{\Gamma} $ is different from Product Topology on $\mathbb{R}^{\Gamma}$
This question is from My topology assignment and i was unable to solve it and unfortunately I have no intution on how this question should be approached.

Question:If $\Gamma$ is an infinite set then prove that Uniform Topology on $\mathbb{R}^{\Gamma} $ is different from Product   Topology on $\mathbb{R}^{\Gamma}$.

Definition of uniform Topology: Let $\Gamma$ be a non empty set and let $\bar{d}$ be the standard bounded metric on $\mathbb{R}$ . Define a metric $\bar{\mathbb{\rho}}$ on $\mathbb{R}^{\Gamma}$ by  $\bar{\mathbb{\rho}}(x,y)$= {lub${\bar{d}(x_{\alpha} ,y_{\alpha})}: \alpha \in \Gamma$}.Then $ \bar{\mathbb{\rho}}$ is a metric called uniform metric and topology induced by $\bar{\mathbb{\rho}}$ is called uniform topology on $\mathbb{R}^{\Gamma}$.

Can you please tell me on how should this question should be solved.  Also, I am really sorry that I would not be able to Get any attemp as I don't have anything worthy to show.

Thank you!

答:
The "canonical" base (basis) $B_P$ for the Product Topology is the set of all $\prod_{g\in \Gamma} V_g$ such that (i) each $V_g$ is open in $\Bbb R,$ and (ii) $\{g\in \Gamma :V_g\ne \Bbb R\}$ is finite.

Let $U$ be an  open $\bar \rho$-ball of radius $1/3$ in the Uniform Topology. For some set $\{W_g: g\in \Gamma \}$ of non-empty subsets of $\Bbb R,$ we have $U\subseteq V=\prod_{g\in \Gamma}W_g$ where $\sup W_g -\inf W_g\le 2/3$ for each $g.$

No non-empty member of $B_P$ is a subset of $V.$ So $U$ is not a union of members of $B_P.$ So $U$ is not open in the Product Topology.


问:
In grouped discrete frequency distribution, Why should I subtract 0.5 from L in the formula for the estimated median?
I have found different sources calculating the estimated median in a discrete grouped frequency distribution differently.
Some subtracted from L 0.5 in the formula:

L+((N/2–F)/f)C

And some did not.

L means lower boundary of the median class

N means sum of frequencies

F means cumulative frequency before the median class.
Meaning that the class before the median class what
is the frequency

f means frequency of the median class

C means the size of the median class

Here is the formula in text:

lower boundary of the median class+((sum of frequencies/2–cumulative frequency before the median class) / frequency of the median class) * size of the median class

答:
Suppose you saw count data as follows, and tried to estimate the median:

```<code> 0 <= x < 20:    10
20 <= x < 30:    15
30 <= x < 80:    10
</code>```

If the numbers being grouped in this way were real numbers from a continuous distribution, you might intuitively guess that the median was about $25$, while if they were integers (you say "discrete") then you might guess the median was about $24.5$ (using the symmetry to go halfway between $20$ and $29$).

Your formula $L+\dfrac{\frac N2 -F}{f}C$ would suggest  $20+\dfrac{\frac{35}{2}-10}{15}\times (30-20) =25$

The subtraction of $0.5$ is an attempt to handle the ranges when the numbers are integers and the bottom end of the median's range is inclusive while the top end is exclusive; if it had been the other way round then you might instead add $0.5$.

If the integer data had been presented differently, for example in either of the two following ways, there would be less need for this kind of adjustment as your formula would automatically give a lower figure for the estimate of the median, in the former case because $C$ is smaller and in the latter case because $L$ is lower.  The suggested subtraction of $0.5$ is equivalent to trying to translate the original data into the last case.

```<code> 0 <= x <= 19:    10
20 <= x <= 29:    15
30 <= x <= 79:    10
</code>```

or

```<code>-0.5 < x < 19.5:    10
19.5 < x < 29.5:    15
29.5 < x < 79.5:    10
</code>```

Body:
Suppose you saw count data as follows, and tried to estimate the median:

 0 <= x < 20:    10
20 <= x < 30:    15
30 <= x < 80:    10

If the numbers being grouped in this way were real numbers from a continuous distribution, you might intuitively guess that the median was about $25$, while if they were integers (you say "discrete") then you might guess the median was about $24.5$ (using the symmetry to go halfway between $20$ and $29$).

Your formula $L+\dfrac{\frac N2 -F}{f}C$ would suggest  $20+\dfrac{\frac{35}{2}-10}{15}\times (30-20) =25$

The subtraction of $0.5$ is an attempt to handle the ranges when the numbers are integers and the bottom end of the median's range is inclusive while the top end is exclusive; if it had been the other way round then you might instead add $0.5$.

If the integer data had been presented differently, for example in either of the two following ways, there would be less need for this kind of adjustment as your formula would automatically give a lower figure for the estimate of the median, in the former case because $C$ is smaller and in the latter case because $L$ is lower.  The suggested subtraction of $0.5$ is equivalent to trying to translate the original data into the last case.

 0 <= x <= 19:    10
20 <= x <= 29:    15
30 <= x <= 79:    10

or

-0.5 < x < 19.5:    10
19.5 < x < 29.5:    15
29.5 < x < 79.5:    10


答:
$$(x^2y-x^2)dy-(y^2+{2xy})dx=0$$
You forgot the square for $y$:
$$(x^2y-x^2)dy-(y^2+\color {red}{2xy^2})dx=0$$
$$\dfrac {(y-1)}{y^2}dy-\dfrac {(1+2x)}{x^2}dx=0$$
The differential is exact.
$$Mdx+Ndy=0$$
$$\partial_y M =\partial_x N=0$$
Note that the DE in the title and the DE in the body are different since you wrote:
$$f(x,y)=\frac{y^2+2xy}{x^2y-x^2}$$

Title: Solving $y'=f(x,y)$ when $f$ is a rational function $f(x,y)=\frac{y^2+2xy}{x^2y-x^2}$

Body:
Solve the initial value problem $$\frac{dy}{dx}=\frac{y^2+2xy^2}{x^2y-x^2},~~y(1)=1$$

My attempt: I checked whether $(x^2y-x^2)dy-(y^2+2xy)dx=0$ is exact, but it was not exact. Also the substitution $y=ux$ makes $$x\frac{du}{dx}=\frac{u^2+ux-u}{ux-1}$$ and stucked.


问:
How do you solve for $x$ when $x = c^x$ where c is to a constant?
How do I solve x = c^x for let us say, c = 5? I'm basically a beginner at math, so is there any solutions other than this lambert w function? I thought of this and  want to solve this because I don't know how to solve it.</p>

答:
Complex number solution:

As said by @achille hui, you can rewrite the equation to $we^w = z$. Then we have $w = W(z)$, where $W$ is branch of Lambert W function.

Real number solution:

Anyways, this equation has no solution in real numbers because $f(x) = x$ and $f(x) = 5^x$ do not intersect.

Edit: Seriously? A downvote?


问:
Showing inversion function for invertible matrices is differentiable
Consider the inversion function $f:GL_n( \mathbb{R}) \rightarrow GL_n (\mathbb{R})$ , $f(X)=X^{-1}.$
Where $GL_n( \mathbb{R})$ denotes the set of invertible $ n \times n$ matrices over the reals.

The question wants me to show that it is a differentiable function and then to calculate its derivative. It says to think of the set as a subset of $\mathbb{R} ^{n^{2}}$.

I know that if the partials exist and are continuous then it is differentiable, I can't calculate the partials explicitly though since it seems too difficult, just thinking about it I know if I were to change 1 entry in the matrix keeping all others constant (this is how I interpret partial derivative of this function, is this correct?), I could find a neighbourhood around that entry such that the matrix is still invertible (since $det:\mathbb{R}^{n \times n} \rightarrow \mathbb{R}$ is continuous? - this has been shown in my lecture notes) Is this the correct way to go about it? I have no solutions available to me so just seeking some clarification on here to make sure my understanding isn't completely all wrong, thanks!

答:
I'd like to add a few remarks addressing the questions of OP as comments to Tsemo's answer. It is likely that said questions are resolved by now for the OP it might be useful to someone else.

Here are some further discussions on the same topic (some quite old): [1], [2], [3].

To address the first question-as-comment, inversion, in the case of finite-dimensional underlying space, ought to be rational (which means it is the ratio of two polynomials), as each entry of the inverse matrix will be a polynomial up to scaling by the determinant. (In this regard it is trivial to compare the $n=1$ case and meditative to compute the derivative of inversion of $2\times 2$ matrices.)

The second question-as-comment I think is more substantial. In that, indeed, the two paragraphs of Tsemo's answer can be distinguished by realizing that the first paragraph works in the case of finite dimensions, while the second paragraph works even when $\mathbb{R}^n$ is replaced by an arbitrary Banach space (which still is not the most general case of this, see e.g. this discussion). Of course this is not a problem as the question was stated for finite dimensions. Still, I think the algebraic shortcut that is guaranteed by finite dimensionality hides something fundamental of the object we are dealing with.

Indeed, from a categorical/Lie theoretic point of view inversion, and inversion being $C^\infty$, are definitional for $GL_n(\mathbb{R})$. Out of $M_n(\mathbb{R})$ we can cut out $GL_n(\mathbb{R})$ by considering it to be the domain of inversion (which makes sense without restrictions in arbitrary dimensions, as opposed to taking determinant to be nonzero).

From this point of view, $GL_n(\mathbb{R})$ being open, inversion being continuous and inversion being $C^\infty$ can all be easily derived from the well-known power series

$$(I+H)^{-1}=\sum_{n\geq0}(-H)^n\mbox{ for }\Vert H\Vert <1, $$

$H$ being a linear operator on $\mathbb{R}^n$ and $I=\operatorname{id}_{\mathbb{R}^n}$. Given the power series, as in Tsemo's answer we can say that for a fixed $A\in GL_n(\mathbb{R})$ and for $H$ with $\Vert HA^{-1}\Vert <1$ (e.g. for $H$ with $\Vert H\vert <\dfrac{1}{\Vert A^{-1}\Vert}=\operatorname{conorm}(A)$, which one may interpret as a byproduct of translating the derivative at $A$ to a derivative at $I$), we have that

$$(A+H)^{-1}
= (A+HA^{-1}A)^{-1}
=A^{-1}(I+HA^{-1})^{-1}
=A^{-1}\sum_{n\geq0}(-HA^{-1})^n,$$

thus

$$\dfrac{\Vert(A+H)^{-1}- [A^{-1}- A^{-1}HA^{-1}]\Vert}{\Vert H\Vert}
\leq \dfrac{\Vert H\Vert \Vert A^{-1} \Vert^2}{1- \Vert H \Vert \Vert A^{-1}\Vert}
\xrightarrow{H\to 0} 0.
$$

In the previous paragraph I believe I made it clear how structural principles hint at the fact that a power series expansion ought to play a role in the question. Even without adhering to the structural point of view though one is lead to the power series. Indeed, trying to take the derivative of the (nonlinear) operator that is inversion means that we're trying to find a linear $\lambda:=\left((\cdot)^{-1}\right)'(A):M_n(\mathbb{R})\to M_n(\mathbb{R}) $ such that

$$\lim_{H\to 0}\dfrac{\Vert (A+H)^{-1}-[A^{-1}+\lambda H] \Vert}{\Vert H \Vert} = 0.$$

Somehow we need to be able to write down $(A+H)^{-1}$ more explicitly so that some terms cancel out. By definition $(A+H) (A+H)^{-1} = I$, so that $(A+H)^{-1} = A^{-1}(I-H(A+H)^{-1})$. Using this recursive formula twice in a row we have:

$$(A+H)^{-1} = A^{-1}-A^{-1}HA^{-1}+A^{-1}HA^{-1}H(A+H)^{-1}.$$

This gives us what we want, granted that we have already guaranteed that inversion is continuous (which would guarantee that the rightmost summand on RHS decays quadratically in $H$). So another way to differentiate inversion would go through a separate proof of inversion being continuous (for a proof of this that doesn't use power series explicitly, see Rudin's Principles of Mathematical Analysis (3e) Theorem 9.8. on p. 209). Still, I hope it's clear by now that even this argument uses the power series, albeit implicitly. (Observe that the same argument, with the recursive formula used arbitrarily many times, establishes the power series we used.)

As a final comment I'd like to add that the fact that inversion is $C^\infty$ is not irrelevant to general purpose mathematics. One straightforward application is a corollary to the Inverse Function Theorem: if $f:U\to \mathbb{R}^n$ is $C^1$ and $f'(x_0)$ is a linear isomorphism, then $f$ is a $C^1$ diffeomorphism near $x_0$, and the formula for the derivative of the inverse of $f$ is:

$$\left(f^{-1}\right)'=(\cdot)^{-1}\circ f'\circ f^{-1}. $$

The regularity of a composition is determined by the worst composant, so that if $f$ is $C^r$, $r\in \mathbb{Z}_{\geq1}$, to begin with, we would end up with a $C^r$ diffeomorphism near any point $x_0$ where $f'(x_0)$ is an linear isomorphism. (For reference purposes this is the content of the Inverse Function Theorem in Lang's Fundamentals of Differential Geometry, pp. 15-16.)


问:
If $\int_0^\infty f(x)\, dx <\infty$, then does $\lim\limits_{n\rightarrow \infty} \int_n^{n+1}\sqrt{f(x)}\, dx = 0$?
Here, $f$ is a non-negative continuous function.

I first noted that

$$
\sum_{n=0}^\infty \int_n^{n+1} f(x) \, dx = \int_0^\infty f(x)\, dx <\infty \implies \lim\limits_{n\rightarrow \infty} \int_n^{n+1}f(x)\, dx = 0
$$

From here, it seems intuitive that $\lim\limits_{n\rightarrow \infty} \int_n^{n+1}\sqrt{f(x)}\, dx = 0$, but I can't make any progress.

I'm looking for a hint on how to proceed next?

答: Try the inequality $$\int_{n}^{n+1} \sqrt{f(x)} \, dx \le \left( \int_n^{n+1} f(x) \, dx \right)^{1/2}.$$


问:
If Else Statement in Linear Program, making it Mixed Integer Program
How to write the following if-else condition in Linear Programming or in Mixed Integer Program?

If a <= b then c = a

else ( a > b) then c = b

a, b, c are all variables. Is it possible to rewrite this without doing it nonlinear?

a,b,c >0

答:
You want to enforce $c=\min(a,b)$.  If $c$ naturally (because of the objective or otherwise) satisfies $c\ge\min(a,b)$, it suffices to enforce $c\le\min(a,b)$:
\begin{align}
c&\le a \\
c&\le b
\end{align}

Otherwise, see
https://or.stackexchange.com/questions/1160/how-to-linearize-min-function-as-a-constraint


问:
Let $z$ be the complex number $7-3i$, find a complex number $w$ such that $zw$ is a negative real number
I feel a bit bad asking this question but here it goes:

My initial idea was to rotate the complex number $z$ onto the negative real axis and then compute what complex number was used to do that but that feels a bit too long-winded for this type of problem, I was wondering if anyone has a shortcut.

答:
HINT

Let $w = a + bi$. Then their product can be written as
\begin{align*}
(7-3i)(a + bi) = 7a + 3b + (7b - 3a)i
\end{align*}

Since $zw$ is a negative number, its imaginary part must equal zero and its real part must be negative.

More precisely, we have the following system
\begin{align*}
\begin{cases}
7b - 3a = 0\\\\
7a + 3b < 0
\end{cases}
\end{align*}

Can you take it from here?


问:
Property of bisectors of right triangle
In triangle $ABC$ $\angle C=90^\circ$, $AA'$ and $BB'$ are angle bisectors intersecting at $I$ ($A'\in BC$, $B'\in AC$). What would be the easiest way to prove that projection of $I$ onto $AB$ lies in equal distances from projections of $A'$ and $B'$ onto $AB$?

The reason I’m asking about the easiest way rather than about any solution is that I think it’s quite clear all the lengths can be calculated either in terms of sides of triangle or trigonometrically via its angles. Both ways seem to be too ugly. So I’m hoping there is smart geometrical solution.

答:
Draw the incircle (with center $I$ and radius $r$) of $\triangle ABC$, and let it meet $BC$ at $P$. Since $IP \perp BC$, we have $IP \parallel AC$. (Note: This is why the right angle at $C$ matters.) Also, $|IP|=r$.

Let $Q$ be the point "near" $A^\prime$ on the incircle such that $IQ \parallel AB$; then, $\angle PIQ = \angle A$ and $\angle A^\prime I Q = \angle A^\prime IP = \frac{1}{2}\angle A$. By SAS, $\triangle IPA^\prime \cong \triangle IQA^\prime$, whence $A^\prime Q \perp IQ$, which is to say: the radius $IQ$ is the projection of $IA^\prime$ onto a line parallel to $AB$. This projection is congruent to the projection of $IA^\prime$ onto $AB$ itself.

The same argument applies to the corresponding projection of $IB^\prime$, so that the projections of $A^\prime$ and $B^\prime$ onto $AB$ each lie at distance $r$ from the projection of $I$.

Title: Property of bisectors of right triangle