问:
Intuition for meagre sets
So I'm putting together a presentation in topology on BCT and I have the following theorems:

Let $D$ be the set of somewhere differentiable functions $[0,1] \to \mathbb R$. Then $D$ is meagre in $C[0,1]$.

Let $f \colon [0,1] \to \mathbb R$ be a differentiable function with derivative $f'$. Let $X$ be the set of points for which $f'$ is discontinuous. Then $X$ is meagre in $[0,1]$.

My temptation to rationalise these intuitively (as a taster in the introduction) as follows:

Most continuous functions are nowhere differentiable.

Derivatives are continuous at most points.

This is based off my understanding that a set $A$ is meagre in $B$ if it is small relative to $B$ in a precise sense. But there are instances where a meagre set can be quite large (even of full measure), and I've been told that it's not really correct to say "most" in this context. However, I've seen quite a lot of posts on here giving similar intuitive restatements.

I thought I'd ask here for clarification - is it proper/well-understood to say "most" here? I could say "many" but that doesn't really feel as punchy. If it's not proper to say most here, is there any better way to put it? My topological intuition is still fairly weak.

答:
There are multiple notions of "mostness" out there. Measure and category each provide their own notions of "most" - namely, "full measure" and "comeager," respectively - and these generally do not agree: there are null sets which are comeager and there are meager sets which have full measure.

Whether one should use the term "most" in a category-based sense depends on context. Certainly terms like "almost every" are exclusively connected to measure, while "generic" is exclusively connected to category, but "most" is somewhat flexible. I would personally say that in analysis, "most" should be reserved for the measure-theoretic sense, primarily because category plays a significantly lesser role there; however, in general topology I would take the opposite stance, since category makes sense in arbitrary topological spaces while measure doesn't.

Re: what terminology you should use, let me go back to the term "generic" I mentioned above. This is a rather subtle notion; intuitively, a generic point is a point which doesn't lie in any "simple" meager set. Of course this isn't an absolute notion, but it's often used as a weasel-word when a comeager set is "hidden in the background." So for example one might say

A generic function is nowhere differentiable

or

Given a differentiable function $f$, the derivative $f'$ is continuous at a generic point.

Unfortunately, this terminology is somewhat uncommon. It's also frequently grammatically unwieldy: in the second example above it would sound better to say e.g. "The derivative of a differentiable function is generically continuous," but I've only rarely seen this usage outside of logic.

Ultimately since measure gets talked about far more than category in most areas of mathematics (in my experience anyways), it gets the bulk of the "simple language" descriptions.

Title: Intuition for meagre sets


问:
How to solve this question
prove that if $R\leq S$ and $S\leq N$ then $P(N,S)$ is divisible by $P(N,R)$

Let $s = r +k$ since s is greater than r by some value but we don’t we don’t how much so I used k

$\frac{n!}{(n-(r+k))!}$/$\frac{n}{n-r}!$

$\frac{n!}{(n-r-k))!}$/$\frac{n}{n-r}!$ where you can cancel out n!

I am not getting how to solve farther than this.

Now , I also don’t the meaning behind this question . Because $r$ can be just any value. There may be many values of  $P(N,S)$  divisible by $P(N,R)$. Many different values for $N , R$ and $S$. so , from this way of solving . Can we find at least which value is $n,r$ and $s$ also. Because i think just by dividing variables , how can I find the real values when they are just not given.

答:
Start by proving that the quotient
$\frac{P(n,s)}{P(n,r)}$
is indeed an integer.

ContentLicense: CC BY-SA 4.0

Title: How to solve this question


问:
Smallest vector topology
This is with regard to this question: Topology induced by seminorms and initial topology

I saw somwehere that topology $\mathcal{S}$ is the smallest topology with respect to which all the seminorms are continuous and $V$ is a topological vector space. But I don't have any proof of it.

Is the arbitrary intersection of vector topologies a vector topology. How does one show the "smallest" assertion?

答:
If $\tau$ is a vector topology such that the seminorm $p$ is continuous, then
$$x\mapsto p(x-x_0)$$
is also continuous since it is the composition of $p$ with the translation by $x_0$ (the latter being continuous thanks to the fact that $\tau$ is a vector topology).

Therefore the "ball"
$$
  B_{x_0, p, r} =\{x \in V : p(x - x_0) < r \}
  $$
is open relative to $\tau$.

If $\mathscr P$ is a given family of seminorms, and if $\tau$ is a vector topology relative to which all seminorms in $\mathscr P$ are cotinuous, we therefore conclude that the topology induced by $\mathscr P$ is smaller than $\tau$.

Title: Smallest vector topology


问:
Rules or laws regarding about factorials
For any of you that don't know what a factorial is, here's a start:

$x! = x ×(x-1) ×(x-2)  ×... 2  × 1$

So $6!$ would be $6 ×5 ×4 ×3 ×2 ×1$

Now onwards to my question.

If $x$ is in the range $12 >= x >= 48$, How many $x$ values there are such that $(x-1)!$ is not divisible by $x$?

I considered listing out all the numbers but I suspect there is a rule here, is there?

Thanks!

答:
$$n\nmid (n-1)!\iff n=4\lor n\text{ is prime} $$

Proof: A prime divding a product divides one of the factors. Thus if $n$ is prime and $n\mid (n-1)!$, then $n$ must divide one of the factors $1,2,\ldots, n-1$, which is not the case already because these factors are too small.
We observe that  $4\nmid 3!=6$ and $1\mid 0!=1$. Remains the case of $n$ composite and $>4$, so $n $can be written as $n=ab$ with integers $1<a,b<n$. If $a\ne b$, then clearly, $a$ and $b$ appear among the factors $1,2,\ldots, n-1$ in the definition of $(n-1)!$, so $ab\mid (n-1)!$. If $a=b$ then (as $n>4$) $n=ab>2a$ so that $a$ and $2a$ appear among the factors $1,2,\ldots, n-1$, so $n=ab\mid (2\cdot 2a)\mid (n-1)!$.


问:
How do you respond to "I was always bad at math"?
Here in the U.S., it is my experience that over 75% of adults I meet socially will volunteer that phrase or a variation upon learning that I am a mathematician.  I find this frustrating, since almost nobody would  brag about being bad at history or English or really any other subject.   Normally I change the subject, or perhaps say they must have had a bad classroom experience, but that feels unsatisfying and turns the conversation in an ugly direction.

Surely some of the bright minds here on Math.SE have found a good response. Note that I'm not looking to  alienate a new acquaintance, so let's keep it positive.

答: Just ask what the student likes to do or does well. Then show that there is math in it. Then you can use this as a base to elaborate and cultivate an interest.</p>

ContentLicense: CC BY-SA 3.0


问:
Homology groups of spherical $3$-manifolds
Let $G$ be a finite subgroup of $SO(4)$, acting freely on $\mathbb{S}^3$. How can we compute $H_2(\mathbb{S}^3/G; \mathbb{Z})$? Is it zero?

答:
As Steve D suggests in the comments, this follows from Poincaré duality. The quotient $S^3/G$ is orientable because $G$ is a subgroup of $SO(4)$, and therefore

$$H_2(S^3/G; \mathbb{Z}) \cong H^1(S^3/G; \mathbb{Z}) \cong \operatorname{Hom}(\pi_1(M/G), \mathbb{Z}) = \operatorname{Hom}(G, \mathbb{Z}) = 0.$$

More generally, for any finite group $G$ which acts freely on a rational homology sphere $X$ with $\dim X = 2n + 1 \geq 3$, we have $H_{2n}(X/G; \mathbb{Z}) = 0$. To see this, first note that $X/G$ must be orientable. There are two ways to do this:

As $ \pi : X \to X/G$ is a finite covering, $\pi^* : H^*(X/G; \mathbb{Q}) \to H^*(X; \mathbb{Q})$ is injective, so $b_i(X/G) \leq b_i(X)$ for all $i$. As $X$ is a $(2n+1)$-dimensional rational homology sphere, its only non-zero Betti numbers are $b_0(X) = b_{2n+1}(X) = 1$. Note that $b_0(X/G) = 1$, while $b_{2n+1}(X/G) = 1$ if and only if $X/G$ is orientable, and all other Betti numbers are zero. As $\chi(X/G) = 0$, we must have $b_{2n+1}(X/G) = 1$ and hence $X/G$ is orientable.

Let $f : X \to X$ be a fixed-point free homeomorphism. By the Lefschetz fixed point theorem, the alternating sum of traces of $f_*$ on rational homology is zero. As $X$ is a rational homology sphere, it only has non-zero rational homology in degrees $0$ and $2n+1$ in which case it is one dimensional. On degree zero homology, $f_*$ is always the identity, while on degree $2n + 1$ homology, it acts by the identity if and only if $f$ is orientation-preserving. As the alternating sum of traces is zero, we see that $f_*$ must be the identity on $H_{2n+1}(X; \mathbb{Q})$ and hence $f$ is orientation-preserving. In particular, $G$ must be an orientation-preserving action, and hence $X/G$ is orientable.

Now that we know $X/G$ is orientable, we see that $H_{2n}(X/G; \mathbb{Z}) \cong H^1(X/G; \mathbb{Z})$ by Poincaré duality. As mentioned in point 1 above, we have $b_1(X/G) \leq b_1(X) = 0$ so $H^1(X/G; \mathbb{Z})$ has rank zero. As the first cohomology group is always torsion-free, we conclude that $H^1(X/G; \mathbb{Z}) = 0$ and hence $H_{2n}(X/G; \mathbb{Z}) = 0$.


答:
I believe you are so close to the solution. As you suggested let

$$
s=r+k
\\n=s+m=r+k+m
$$

where $k\ge0$ and $m\ge0$. Then

$$
\frac{P(N,S)}{P(N,R)}=\frac{n!}{(n-s)!}\frac{(n-r)!}{n!}=\frac{(k+m)!}{m!}=(m+1)(m+2)...(m+k)
$$

if $k=0$, then the division becomes 1.

Title: How to solve this question

Body:
prove that if $R\leq S$ and $S\leq N$ then $P(N,S)$ is divisible by $P(N,R)$

Let $s = r +k$ since s is greater than r by some value but we don’t we don’t how much so I used k

$\frac{n!}{(n-(r+k))!}$/$\frac{n}{n-r}!$

$\frac{n!}{(n-r-k))!}$/$\frac{n}{n-r}!$ where you can cancel out n!

I am not getting how to solve farther than this.

Now , I also don’t the meaning behind this question . Because $r$ can be just any value. There may be many values of  $P(N,S)$  divisible by $P(N,R)$. Many different values for $N , R$ and $S$. so , from this way of solving . Can we find at least which value is $n,r$ and $s$ also. Because i think just by dividing variables , how can I find the real values when they are just not given.


问:
Polynomials : $p(z)=(z-r)q(z)$
Theorem : Let $p$ be a polynomial of degree $n\geq 1$, with leading co-efficient $1$, and let $r$ be a roof of $p$. Then  $p(z)=(z-r)q(z)$, where $q$ is a polynomial of degree $n-1$, with leading co-efficient $1$. The proof is given : Let $p(z)$ have the form $$p(z)=z^n +a_1z^{n-1}+\cdots a_{n-1}z+a_n$$

and let $c$ be any complex number. Then $$p(z)-p(c)=(z^n - c^n)+a_1(z^{n-1}-c^{n-1})+\cdots+a_{n-1}(z-c)=(z-c)q(z)$$

where $q$ is the polynomial given by $$q(z)=z^{n-1} + cz^{n-2}+c^2z^{n-3}+\cdots+c^{n-1}$$$$+a_1(z^{n-2}+cz^{n-3}+\cdots+c^{n-2})+\cdots+a_{n-1}$$

Where is this formula for $q$ derived from? I think with this theorem, one can factor polynomials and solve their roots - provided we are able to guess at least one root $r$. But the problem is we need to find $q$. How can I compute $q$ for, lets say this polynomial : $z^3 -3z^2 +4$?

答:
Have you not learned polynomial long division? It works almost exactly like long division in arithmetic:

$$\require{enclose}\begin{array}{r}
z^2-\phantom 0z-2\\[-3pt]
z-2\enclose{longdiv}{z^3-3z^2 + 0z + 4}\\[-3pt]
\underline{z^3-2z^2\phantom {\;\;+0z+4}}\\[-3pt]
-z^2+0z+4\\[-3pt]
\underline{-z^2+2z\phantom{+4\;\,}}\\[-3pt]
-2z+4\\[-3pt]
\underline{-2z+4}\\[-3pt]
0
\end{array}$$
So $z^3 -3z^2 + 4 = (z-2)(z^2 -z-2)$.

When you write out the division, put in terms for every power of $z$ in the dividend, as I did by adding "$+0z$". You will need that space eventually.

Take the ratio of the leading term of the dividend to the leading term of the divisor. In this case that is $\frac {z^3}z = z^2$. Write this result on top as the first term of the quotient.

Multiply the divisor by the just-found quotient term and write the result underneath the dividend, lining up the terms by degree.

Subtract and write the result below the line. This is the remainder.

If the remainder is still of equal or higher degree than your divisor, it becomes your new dividend. Repeat the process to get the next term in the quotient.

Continue until the remainder is of lower degree than the divisor ($0$ is of lower degree than anything other than itself).

This works for any polynomial dividend and divisor. But if the divisor is of degree $1$, then there is a shortcut version called synthetic division that does the same thing, but with much less writing.

As for finding that first root: There is a trick called the Rational Root Theorem. This is great for math homework, because the teacher can carefully construct examples where you can use the rational root theorem to find roots and divide them out, thus giving you good practice in larger degree polynomials and their behavior. It also has some number theory implications.

But for "real world" problems (ones that were not carefully constructed to be easy to solve), the rational root theorem is of practically no value, rarely ever giving you an actual root. There for degrees higher than $4$, you generally have to use approximation techniques to find approximate roots (generally this is easier than using the $3$rd and $4$th degree versions of the quadratic formula as well).

Title: Polynomials : $p(z)=(z-r)q(z)$


答: $(x-1)!$ is not divisible by $x$ if $x$ is prime. This is due to the fact that since $x$ is prime, it does not have any factors less than it, other than $1$. Thus, $x$ is does not divide any of the numbers from $x-1$ to $1$, and thus $x$ does not divide $(x-1)!$. Note that this is only ONE condition I found.

Title: Rules or laws regarding about factorials

Body:
For any of you that don't know what a factorial is, here's a start:

$x! = x ×(x-1) ×(x-2)  ×... 2  × 1$

So $6!$ would be $6 ×5 ×4 ×3 ×2 ×1$

Now onwards to my question.

If $x$ is in the range $12 >= x >= 48$, How many $x$ values there are such that $(x-1)!$ is not divisible by $x$?

I considered listing out all the numbers but I suspect there is a rule here, is there?

Thanks!


问:
Are there general procedures to find group isomorphisms?
This question is actually a series of related questions that have been bothering me while studying group theory. It seems to me that if we have two groups $G,H$ of same order and $G$ has a subgroup of order $n$ and $H$ has no subgroup of order $n$, then we can't have an isomorphism.

Now, suppose that we have two subgroups $G,H$ and they have the same quantity of subgroups $G_1,\dots,G_n$ and $H_1,\dots,H_n$ and $|G_i|=|H_i|$ does this means we have an isomorphism? Or are there groups where these conditions are true but there is no isomorphism?

I've been thinking about mappings from generating sets of groups, if we can map each generator of $G$ to another generator of $H$ such that each generator from $H$ and $G$ have the same order, do we have an isomorphism? Perhaps this is equivalent to the previous one but I am not really sure.

This is bothering me because in my lectures, we speak about what is an isomorphism but we never go in depth to actually talk about general procedures to find isomorphisms. Perhaps it's too early? I don't know.

答:
Determining whether two (finite) groups are isomorphic is quite difficult in general. There are lots of obstacles to isomorphism - basically, any non-silly property you can think of is preserved by isomorphisms, so any non-silly disagreement between two groups witnesses their non-isomorphicness - but it's difficult to guarantee an isomorphism without explicitly constructing one. (In particular, per the comments above neither of your conditions is sufficient to conclude that the two groups are isomorphic.)

That said, there are some tools we can use.

In the abelian case, group isomorphism is actually easy. The fundamental theorem of finite abelian groups gives a canonical way to describe any finite abelian group such that two finite abelian groups are isomorphic iff they have the same description. And the analysis required to get that description isn't very hard.

Even in the non-abelian case, we can show that a putative isomorphism between two groups $G$ and $H$ must satisfy certain constraints, and so narrow the range of possibilities we have to consider. For example, it has to send Sylow subgroups to Sylow subgroups appropriately, so breaking down the two groups via Sylow theory can make it easier to find an isomorphism if one exists (or prove that there is no isomorphism, for that matter).


问:
Give syntax proof of FOL Formla $\exists x.P(x) \lor \exists x.Q(x)$ $\vdash$ $\exists x.(P(x) \lor Q(x))$
So the question is pretty much described in the title already. I have to show the following result. I have tried it but am failing to do so. Anyone who can please help me in understanding its proof. I have attached the picture of formulas that may prove to be helpful.

$\exists x.P(x) \lor \exists x.Q(x)$ $\vdash$ $\exists x.(P(x) \lor Q(x))$

答:
Let us get you started with the derivation for $\exists x~.P(x)~\lor~\exists x~.Q(x)\vdash \exists x~.(P(x)\lor Q(x))$ using the Gentzen tree system.

$$\dfrac{\exists x~.P(x)~\lor~\exists x~.Q(x)\\\qquad\qquad\vdots}{\exists x~.(P(x)\lor Q(x))}{?}$$

Well, existential statements may be derived either by construction, or by reduction to absurdity.   Which is appropriate here?

We have existences in the premise, so that indicates we may be constructive.   Thus we shall be using existential introduction using a witnesses derived from the premise's existentials.

But wait!   To access those existentials we will first need to eliminate that disjunction.

$$\dfrac{\lower{1.5ex}{\exists x~.P(x)~\lor~\exists x~.Q(x)}\quad\dfrac{[\exists x~.P(x)]^1\\\quad\vdots}{\exists x~.(P(x)\lor Q(x))}\quad\dfrac{[\exists x~.Q(x)]^2\\\quad\vdots}{\exists x~.(P(x)\lor Q(x))}}{\exists x~.(P(x)\lor Q(x))}{\small\lor\textsf{elim}^{1,2}}$$

Now you can look to the existential introduction and elimination rules, and disjunction introduction while we are at it.$$\dfrac{\lower{1.5ex}{\exists z~.\phi(z)}~~\dfrac{[\phi(t)]^1\\\quad\vdots}{\psi}}{\psi}{\small\exists\textsf{elim}^1}\qquad\dfrac{\rho(t)}{\exists z~.\rho(z)}{\small\exists\textsf{intro}}\\~\\~\\\dfrac{\chi(s)}{\chi(s)\lor\varphi(s)}{\small\lor\textsf{intro}}\qquad\dfrac{\varphi(t)}{\chi(t)\lor\varphi(t)}{\small\lor\textsf{intro}}$$

Just put it together.


问:
If $f\colon G\to H$ is a surjective homomorphism, then $|C_G(g)| \geq |C_H(f(g))|$

Let $G$ be finite, $f\colon G\to H$  be a surjective homomorphism (hence $H$ is finite) and $g \in G$. Prove the order of center of $g$ in $G$ is greater than or equal to the order of the center of $f(g)$ in $H$, i.e. $$|C_G(g)| \ge |C_H(f(g))|.$$

Attempts at solution.

If $f$ is an isomorphism this is clear. So if $\ker f=K$ we can consider $G \to G/K \xrightarrow{f'} H$, where $f'$ is the induced map. So we can reduce this problem to the case where $f$ is a projection map.

Also if $f(x)f(g) = f(g)f(x)$ then $f(x^{-1}g^{-1}xg) = e$.

So if $h = f(y)$, $h \in C_H(f(g))$ iff $[h,g] \in K$.

The center of $g$ is the set of fixed points of the conjugation map (by $g$). And the center of $h$ is the set of fixed points under conjugation by $h$. So maybe this can be done by considering how the orbits collapse mod a normal subgroup?

答:
Let us denote by $[g]$ the conjugacy class of $g$ in $G$. The set
$$
[g]\cap Kg=\{g=g_1,g_2,\ldots,g_m\}
$$
is obviously finite, and $m\le |K|$.

On the other hand $f(x)$ commutes $f(g)$ in $H$, iff $xgx^{-1}g^{-1}\in K$, or iff
$$
xgx^{-1}\in Kg.
$$
All the elements $g_i,i=1,2,\ldots,m,$ are conjugates in $G$ so the equation
$$xgx^{-1}=g_i$$ holds for exactly $|C_G(g)|$ different elements $x\in G$. Irrespective of the choice of the index $i$.

Denote the subgroup $f^{-1}(C_H(f(g)))$ by $M$.
Putting all this together we have calculated that
$$
|M|=m\cdot |C_G(g)|.
$$
On the other hand $C_H(f(g))=f(M)\cong M/K$, so
$$
|C_H(f(g))|=\frac{|M|}{|K|}=\frac{m\cdot |C_G(g)|}{|K|}\le |C_G(g)|,
$$
as $m\le |K|$.


问:
A prime ideal of $\mathbb{Z} \times \mathbb{Z}$ that is not maximal.
So, I've just proven that $I = \{(a,0): a \in \mathbb{Z}\}$ is a prime ideal of $\mathbb{Z} \times \mathbb{Z}$ (Operations defined coordinate per coordinate). But I can't get a grip on how to begin to prove that $I$ is not maximal.

The only thing I think about is that, if $I$ is maximal, the quotient ring $\mathbb{Z} \times \mathbb{Z} / I$ is a field, since $1 = (1,1) \in \mathbb{Z} \times \mathbb{Z}$ is it's unity.

I just need a hint on how to approach this problem.

Any help or response is much apreciated.

答: Hint: Your ideal is properly contained in $\mathbb{Z}\times 2\mathbb{Z}$.


问:
Find the volume a solid by triple integration
We need to find the volume of the solid bounded by the $xy$-plane, the cylinder $x^2+y^2=2x$, and the cone $z=\sqrt{x^2+y^2}.$ The volume is given by the triple integral $\int \int \int_S \,dx\,dy\,dz,$where S is the region of the bounded solid.

I am trying to use spherical co-ordinates where the integral becomes $\int \int \int_T r^2 \phi \,dr\,d\theta\,d\phi$  but I am unable to set the correct bounds for the $r,\theta, \phi$ integrals. The order of integration I am trying is $\,dr\,d\theta\,d\phi$.

Can you suggest how to correctly setup the integral in spherical co-ordinates?

答:
We'll set up the volume integration using each coordinate system, with the calculation in cylindrical coordinates serving to check the result from the application of spherical coordinates.

Using cylindrical coordinates, the cylinder of radius $ \ 1 \ $ with symmetry axis $ \ x \ = \ 1 \ $ has the equation $ \ x^2 \ - \ 2x \ + \ 1 \ + \ y^2 \ = \ 1 \ \ \rightarrow \ \ r \ = \ 2 \ \cos \theta \ $ and the cone is described by $ \ r \ = \ z \ $ .  A "slice" through the solid of interest taken parallel to the $ \ xy- $ plane looks like

The slice is symmetrical about the $ \ xz-$ plane ( $ \theta \ = \ 0 $ ) and the intersections of the cone (red) and the cylinder (blue) occur at

$$ z \ = \ 2 \cos \theta \ \ \Rightarrow \ \ \theta \ = \ \pm \arccos \left( \frac{z}{2} \right) \ \ . $$

Thus, applying the symmetry of the slices, we can set up the volume integral as

$ V \ \ = \ \ 2 \ \int_0^2 \ \int_0^{\arccos(z/2)} \ \int_z^{2 \cos \theta} \ \ r \ dr \ \ d\theta \ \ dz $

$ \ \ \ \ = \ \ 2 \ · \ \frac{1}{2}  \ \int_0^2 \ \int_0^{\arccos(z/2)} \ \ (2 \cos \theta)^2 \ - \ z^2  \  \ d\theta \ \ dz   $

$ \ \ \ \ = \ \  \int_0^2 \  \ \left( \ 4 \left[ \frac{1}{2} \left( \theta \ + \sin \theta \cos \theta \right) \right] \ - \ z^2  \theta \  \right) \vert_0^{\arccos(z/2)} \ \ dz   $

[using a right triangle with angle $ \ \theta \ $ , "adjacent side" $ \ z \ $ , and hypotenuse $ \ 2 \ $ ]

$ \ \ \ \ = \ \  \int_0^2 \  \ 2 \arccos(z/2) \ + \ 2  ·  \frac{\sqrt{4 - z^2}}{2}  ·  \frac{z}{2} \ - \ z^2   \arccos(z/2) \ \ \ dz   $

$ \ \ \ \ = \ \  \left[ \ 2 z \arccos(z/2) \ - \ 2 \sqrt{4 - z^2} \ + \ \frac{1}{2} · \left(  -\frac{1}{3} (4-z^2)^{3/2}  \right) \ - \ \frac{z^2}{3}   \arccos(z/2) \ + \ \left( \frac{z^2 + 8}{9} \right) \ \sqrt{4-z^2}\ \right] |_0^2  $

$ \ \ \ \ = \ \  \left[ \ 4 \arccos(1) \ - \ 2 \sqrt{4 - 2^2} \ - \ \frac{1}{6}   · (4-2^2)^{3/2}  \ - \ \frac{2^2}{3}   \arccos(1) \ + \ \left( \frac{2^2 + 8}{9} \right) \ \sqrt{4-2^2} \ \right] $
$ \ \ \ \ \ \ - \ \left[ \ 0 · \arccos(0/2) \ - \ 2 \sqrt{4 - 0^2} \ - \ \frac{1}{6} ·  (4-0^2)^{3/2} \ - \ \frac{0^2}{3}   \arccos(0/2) \ + \ \left( \frac{0^2 + 8}{9} \right) \ \sqrt{4-0^2} \ \right]  $

$ \ \ \ \ = \ \  \left[ \ 0 \ \right]  \ - \ \left[ \ 0  \ - \ 4 \ - \ \frac{4^{3/2}}{6}  \ - \ 0 \ + \  \frac{8}{9}  · 2 \ \right] \ \ = \ \ 4 \ + \ \frac{8}{6} \ - \ \frac{16}{9} \ \ = \ \ \frac{36 \ + \ 12 \ - \ 16}{9} \ \ = \ \ \frac{32}{9} \ \ .   $

For the volume integral in spherical coordinates, the differential form is $ \ \rho^2 \ d\rho \ \ \sin \phi \ d\phi \ \ d\theta \ $ , with $ \ \rho \ $ being distance measured from the origin (rather than $ \ r  \ $ in cylindrical coordinates being measured perpendicularly from the $ \ z-$ axis), and $ \ \phi \ $ being the polar angle measured "downward" from the $ \ z-$ axis.

The surface of the cone is always at $ \ \phi \ = \ \frac{\pi}{4} \ $ , the $ \ xy-$ plane is at $ \ \phi \ = \ \frac{\pi}{2} \ $ , and a "vertical slice" through the solid of our concern at a given $ \ \theta \ $ looks like

The expression for $ \ \rho \ $ is found from $$ \ \sin \phi \ = \ \frac{2 \cos \theta }{\rho} \ \ \Rightarrow \ \ \rho \ = \ \frac{2 \cos \theta }{\sin \phi} \ \ . $$

While this looks like it will be unpleasant to work with, the integration in spherical coordinates is actually simpler than the one above, and proves to be separable.  We can again exploit symmetry, this time integrating over $ \ \theta \ $ from $ \ 0 \ $ to $ \ \frac{\pi}{2} \ $ , the full projection of the cylinder onto the $ \ xy-$ plane.

$ V \ \ = \ \ 2 \ \int_0^{\frac{\pi}{2}} \ \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \ \int_0^{2 \cos \theta \ / \sin \phi} \ \  \rho^2 \ d\rho \ \ \sin \phi \ d\phi\ \ \ d\theta  $

$ \ \ \  \ \ = \ \ 2 \ · \ \frac{1}{3}  \ \int_0^{\frac{\pi}{2}} \ \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \ \ \left( \frac{2 \cos \theta }{\sin \phi} \right)^3  \  \ \sin \phi \ d\phi\ \ \ d\theta   $

$ \ \ \  \ \ =  \ \frac{2}{3} \ · \ 8  \ \int_0^{\frac{\pi}{2}} \ \cos^3 \theta \ \ d\theta \ \ \ \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \  \left( \frac{1 }{\sin \phi} \right)^2  \   \ d\phi\ \ \ = \ \ \frac{16}{3}   \ \int_0^{\frac{\pi}{2}} \ \cos^3 \theta \ \ d\theta \ \ \ \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \  \csc^2 \phi   \   \ d\phi\   $

$ \ \ \  \ \ =  \  \frac{16}{3}   \ \int_0^{\frac{\pi}{2}} \ \cos^3 \theta \ \ d\theta \ \ · \  -\cot \phi   \  |_{\pi/4}^{\pi/2} \ \ = \ \ \frac{16}{3}   \ · \ \frac{2}{3}  \ · \  - ( 0  -  1 ) \ \ = \ \ \frac{32}{9} \ \ , $

with the first definite integral being one ( $ \ \int_0^{\frac{\pi}{2}} \ \sin^n \theta \ \ d\theta \ = \ \int_0^{\frac{\pi}{2}} \ \cos^n \theta \ \ d\theta \ $ ) for which there is a familiar formula.


问:
A criterion for irreducible topological subspace
Let $X$ be a non-empty spectral space and $P$ be a closed subset of $X$. If $U_1,U_2$ are two arbitary quasi-compact open subsets satisfying $P\cap U_1\neq\emptyset$ and $P\cap U_2\neq\emptyset$, then $P\cap U_1\cap U_2\neq\emptyset$.

Can we deduce that $P$ is irreducible from the above conditions?

Thanks in advance.

答: Yes (assuming $P$ is nonempty).  If $P$ were reducible then it would have a pair of disjoint nonempty open subsets, meaning there would exist $V_1$ and $V_2$ open in $X$ such that $P\cap V_1$ and $P\cap V_2$ are nonempty but $P\cap V_1\cap V_2$ is empty.  Now pick a point $p_i\in P\cap V_i$ and let $U_i$ be a compact open neighborhood of $p_i$ contained in $V_i$.


问:
The Sphere of Discrete Metric Space
Let $(X,d)$ denotes the usual discrete metric space. Specify the spheres for $\epsilon = 1, \frac12, 2$.

I strictly used the definition of the sphere such that whenever the metric is not equal to epsilon then we get empty set (all red colors). Can anyone check my result to approve or disprove my results?

We find all the spheres, $\partial{B}_\epsilon(x)$, as follows
\begin{align*}
d(x,y) & = \left.
\begin{cases}
{\color{red}1} {\color{red}\not= {\color{red}2,}} & {\color{red}x \ \not={\color{red}y}} \\ {\color{red}0 \not= {\color{red}2},} &\  {\color{red}x = {\color{red}y}}
\end{cases}
\right\} = \emptyset,  \text{ for } \epsilon = 2, \ \partial B_\epsilon(x) = \emptyset.
 \\
d(x,y) &=  \left.
\begin{cases}
1 = 1, & x \ \not=y \\ {\color{red}{0 \not= 1}}, & {\color{red}{x = y}}
\end{cases}
\right\} =  \{x \not= y\}, \text{ for } \epsilon = 1, \ \partial B_\epsilon(x) = \{x \not= y\} = \{x,y\}.
 \\
d(x,y) = \epsilon & =  \left.
\begin{cases}
{\color{red}1} {\color{red}\not= {\color{red}{1/2},}} & {\color{red}x \ \not={\color{red}y}} \\ {\color{red}0 \not= {\color{red}{1/2}},} &\  {\color{red}x = {\color{red}y}}\end{cases}
\right\} =  \emptyset, \text{ for } \epsilon = 1/2, \ \partial B_\epsilon(x) = \emptyset.
\end{align*}

答:
You should not be limiting yourself to just two points. For each $x\in X$ the sphere of radius $1$ about $x$ is $X\setminus\{x\}$: every point of $X$ except $x$ itself is at distance $1$ from $x$. The spheres of radius $2$ and radius $\frac12$ are therefore empty, since there is no point at distance $2$ or $\frac12$ from $x$.

The closed balls of radii $\frac12$, $1$, and $2$ around $x$, however, are $\{x\}$, $X$, and $X$, respectively, and the corresponding open balls are $\{x\}$, $\{x\}$, and $X$.

Title: The Sphere of Discrete Metric Space


问:
Maximum Y in a rotate ellipse with a, b and phi
We have major axis, minor axis and the phi between major axis and y axis in a rotated ellipse.
How can we find the maximum y?</p>

答: It's easy to see that $\phi=0$ gives the maximum $y$. If you think vectorially, the position vector in direction of the major axis is longest. If you take a different vector and then project it, it gets shorter and shorter again.


问:
What is the quotient group of $O_2$/$SO_2$?
I am working on this as a challenge problem, but I have never worked with a quotient group outside of modular values in the integers.</p>

答: $SO_2$ is, by definition,  the kernel of the homomorphism $O_2\to \Bbb F^\times$,  $A\mapsto \det A$, which happens to be onto.


答: In the discrete metric, the spheres of radii $1$ and $2$ and center $\{x\}$ are equal to $X\setminus\{x\}$ and $\emptyset$ respectively.  The sphere of radius $1/2$ centered at $\{x\}$ is $\emptyset$.

Title: The Sphere of Discrete Metric Space

Body:
Let $(X,d)$ denotes the usual discrete metric space. Specify the spheres for $\epsilon = 1, \frac12, 2$.

I strictly used the definition of the sphere such that whenever the metric is not equal to epsilon then we get empty set (all red colors). Can anyone check my result to approve or disprove my results?

We find all the spheres, $\partial{B}_\epsilon(x)$, as follows
\begin{align*}
d(x,y) & = \left.
\begin{cases}
{\color{red}1} {\color{red}\not= {\color{red}2,}} & {\color{red}x \ \not={\color{red}y}} \\ {\color{red}0 \not= {\color{red}2},} &\  {\color{red}x = {\color{red}y}}
\end{cases}
\right\} = \emptyset,  \text{ for } \epsilon = 2, \ \partial B_\epsilon(x) = \emptyset.
 \\
d(x,y) &=  \left.
\begin{cases}
1 = 1, & x \ \not=y \\ {\color{red}{0 \not= 1}}, & {\color{red}{x = y}}
\end{cases}
\right\} =  \{x \not= y\}, \text{ for } \epsilon = 1, \ \partial B_\epsilon(x) = \{x \not= y\} = \{x,y\}.
 \\
d(x,y) = \epsilon & =  \left.
\begin{cases}
{\color{red}1} {\color{red}\not= {\color{red}{1/2},}} & {\color{red}x \ \not={\color{red}y}} \\ {\color{red}0 \not= {\color{red}{1/2}},} &\  {\color{red}x = {\color{red}y}}\end{cases}
\right\} =  \emptyset, \text{ for } \epsilon = 1/2, \ \partial B_\epsilon(x) = \emptyset.
\end{align*}


问:
Find a matrix $A$ with no zero entries such that $A^3=A$
I took a standard $2 × 2$ matrix with entries $a, b, c, d$ and multiplied it out three times and tried to algebraically make it work, but that quickly turned into a algebraic mess. Is there an easier method to solve this?

Body: $$\begin{pmatrix}7&-6\\8&-7\end{pmatrix} $$


答:
The quantity $d(x,y)$ is a number, not a set. It does not make sense to write $d(x,y)=\cdots =\emptyset$. And the expression $\{x\ne y\}$ does not make any sense.

Recall the definition of sphere  of radius $r$ centered at $x$:
$$
\partial B(x,r)=\{y\in X: d(x,y)=r\}\tag{1}
$$

On the other hand, given any $x,y\in X$, $d(x,y)$ is either $0$ or $1$. So the set in (1) must be empty whenever $r\ne 1$ and $r\ne 0$.

For $r=1$, observe that $d(x,y)=1$ for every $y\ne x$ and thus
$$
\partial B(x,1)=\{y\in X: d(x,y)=1\}=\{y\in X: y\ne x\}=X\setminus\{x\}.
$$

Title: The Sphere of Discrete Metric Space


问:
Projection/Dot Product Q
If we had a vector representing a Boost Pad (like on Mario cart) <4, 2) let's say , and the speed of Mario on the cart let's say, <3, 4> and we said that if you go perpendicular on the Boost Pad (90angle) you won't get any boost, any other angle and you get 3* speed boost...does this mean that any dot product simply gives us information about direction..but not about speed? (at least not directly).

or if someone says that a vector represents the speed of car, let's say, is that just an abstraction that really means it has some quantity/size (i.e. speed) ?? but it doesn't actually mean anything practically until you relate it to some frame of reference (ie. where it started, how long it took, etc)

Hope this makes sense...but for some us non physics people, relating these things to real world things is sometimes quite difficult.

答:
does this mean that any dot product simply gives us information about direction..but not about speed?

Dot product will give you how much one vector is "in the direction" of the other as a value between -1 through 0 to 1 (from opposite direction through perpendicular to same direction).

In game programming, they use speed as the scalar, the magnitude of the velocity vector. They will often make the distinction between a position vectors (or points) and a direction vector. That is, a point has position, can be translated, etc... but a direction doesn't have a position, and will usually remained normalized so its magnitude isn't important either. They can be used in conjunction with each other. For you mario cart example, the process would be something like:

add the direction of the mario cart with the direction of the boost to get the new direction (don't forget to normalize the result)

scale this direction by the boost scalar value

scale again by the timestep value (delta time)

add this to mario carts position vector every update to get the new position with boost applied in the direction of the booster.

So to sum up, you are basically have the right intuition, just that direction vectors and position vectors are both vectors but are used to represent different things, direction vectors are arrows that can be located anywhere (their position is not encoded in the data type but their direction) and position vectors are points just like any other that are displacements from the origin, and don't have a direction, and applying vector addition and scalar multiplication you can move a position in a direction a certain amount.

Hope this clears things up.

This is a good tutorial that might help clear things up too.

Title: Projection/Dot Product Q


问:
Commitees with more of one type of participant
Assume that in a class there are 15 actuarial science students and 12 mathematics students. How many ways are there to form a committee of 7 members with more actuarial science students than mathematics students?

That is the question that I have tried to solve.

First I did ${15\choose 7}+{15\choose 6}{12\choose 1}+{15\choose 5}{12\choose 2}+{15\choose 4}{12\choose 3}$

Can someone please tell me if this is the correct way to go about this question or if I have done it wrong.

答:
This is correct.  You're simply dividing the cases when you have $7$, $6$, $5$, and $4$ actuarial students in the committee.

If you were worried about the multiplication, this is correct because you can pair any two choices of actuarial students with a group of mathematics students.  You choose some $k$ actuarial students and $n$ math students (choose the actuarial students first, and then the math students, and apply the rule of product), where $k+n = 7$ and $k > n$.


答:
An easier approach is noticing that if $A^3 = A$, then $A^2 = I$, which should be much more tractable.  This also means that $A = A^{-1}$ and you can equate the generic abcd with its inverse.  Further, the determinant of A must be 1 or -1, since it squares to 1.  So that simplifies the inverse formula too.

Addendum: if $A^3 = A$, then $det(A)^3 = det(A)$ so $det(A)$ satisfies $x^3 = x$, which has solutions x = 1, 0, and -1. If x = 1 or -1, $A$ is invertible and the above description works.  If $A$ is not invertible, then it must have two rows or columns that are proportional, which is a further simplification for the search.

Title: Find a matrix $A$ with no zero entries such that $A^3=A$

Body: I took a standard $2 × 2$ matrix with entries $a, b, c, d$ and multiplied it out three times and tried to algebraically make it work, but that quickly turned into a algebraic mess. Is there an easier method to solve this?


问:
Prove that $f \times g$ is injective

Consider the maps $f:U_1 \to \mathbb{R}^2$ and $g:U_2 \to \mathbb{R}^2$, where

$f(x^1,x^2)=x^2$ and $g(y^1,y^2)=y^2$

$U_1=\{(x^1,x^2) \ | \ x^1>0\}$ and $U_2=\{(y^1,y^2) \ | \ y^1<0 \}$

Where $x^i$ is the $i$-th coordinate

Prove that $f \times g$ is injective

Suppose that $(f \times g) \big( (x^1,x^2),(y^1,y^2)\big) =(f \times g) \big( (a^1,a^2),(b^1,b^2)\big)$.

Then $(x^2,y^2)=(a^2,b^2)$ which implies that $x^2=a^2$ and $y^2=b^2$.

Now, we have to show that $x^1=a^1$ and $y^1=b^1$, but I couldn't complete this.

Any help would be appreciated. Thanks

Title: Prove that $f \times g$ is injective

Body: We have $(f \times g)((1,0),(-1,0)) = (0,0) = (f \times g)((2,0),(-1,0))$ but $((1,0),(-1,0)) \ne ((2,0),(-1,0))$ so it's not injective.


问:
Confusion with limits with respect to a certain metric
I was looking at some proofs that certain metric spaces are not complete, and I have been confused.

Here is a link to an example of a main source of my confusion:Show that this metric is not complete

Here is the line that drove my confusion:

No, there is a error in the proof. In fact the functions $f_n(x) = x^n$ converge to the zero function in the metric $\rho$ on $C([0,1]):$

$$ \sqrt {\int_0^1 |x^n-0|^2\,dx} = 1/\sqrt {2n+1} \to 0.$$

If I understand this correctly, if we have a metric space equipped with a certain metric $d$, all limits will be computed with respect to this metric. We can pretend no other metrics exist within the space, as we did with limits in $\mathbb{R}$ in the beginning. I do not understand the last line because it seems to me that the user is taking the limit with respect to the metric $d(x,y)=|x-y|$ in $\mathbb{R}$ when he writes $\frac{1}{\sqrt{2n+1}} \rightarrow 0$. I know I am really confused here, and would appreciate if someone could help clear this confusion.

答:
If $f_n(x)=x^n$ and $\mathbf{0}$ is the constant $0$ function, then

$$\rho(f_n,\mathbf{0})=\sqrt{\int_0^1|x^n-0|^2\,dx}\,,$$

so

$$\rho(f_n,\mathbf{0})=\frac1{\sqrt{2n+1}}$$

for each $n$. This means that $\lim\limits_{n\to\infty}\rho(f_n,\mathbf{0})=0$, which in turn easily implies that the sequence $\langle f_n:n\in\Bbb N\rangle$ is $\rho$-Cauchy: given $\epsilon>0$, there is an $n_\epsilon\in\Bbb N$ such that $\rho(f_n,\mathbf{0})<\frac{\epsilon}2$ whenever $n\ge n_\epsilon$, so

$$\rho(f_m,f_n)\le\rho(f_m,\mathbf{0})+\rho(\mathbf{0},f_n)<\frac{\epsilon}2+\frac{\epsilon}2=\epsilon$$

whenever $m,n\ge n_\epsilon$.


答:
If $ad-bc=1,$ with $a,b,c,d$ non-zero, you can use:

$$\begin{pmatrix}-bc&-bd\\ac&ad\end{pmatrix}$$

For example, $a=2,b=1,c=1,d=1$ gives:

$$\begin{pmatrix}-1&-1\\2&2\end{pmatrix}$$

I got this by taking:

$$\begin{pmatrix}a&b\\c&d\end{pmatrix}^{-1}\begin{pmatrix}0&0\\0&1\end{pmatrix}\begin{pmatrix}a&b\\c&d\end{pmatrix}$$

Where $B=\begin{pmatrix}0&0\\0&1\end{pmatrix}$ satisfies $B^2=B$ and hence $B^3=B.$

You can also choose $B=\begin{pmatrix}-1&0\\0&1\end{pmatrix}$ or $B=\begin{pmatrix}0&0\\0&-1\end{pmatrix}.$


问:
Properties of Logarithms (evaluate)
use $\log 4= 0.602$ and $\log 12=1.079$ to evaluate the logarithm

$\log 3$

I'm very confused on how I will evaluate this one I've tried other things but I'm not sure

答: Are you familiar with the logarithm property $log(\frac ab) = log(a) - log(b)$? If not, here's a proof https://www.khanacademy.org/math/algebra2/x2ec2f6f830c9fb89:logs/x2ec2f6f830c9fb89:log-prop/a/justifying-the-logarithm-properties. Now that you know how it works, apply the rule. We know the quanitities of $log(12)$ and $log(4)$ and we need to find $log(3)$. We can see that $3 = \frac{12}4$, so we can try to use our quotient rule. We get $log(3) =log(\frac{12}{4}) = log(12) - log(4) = 1.079 - .602 = .477$. Thus, $log(3) = .477$.

Title: Properties of Logarithms (evaluate)


答:
I'm not sure why no one has yet posted

$$(1)^3=(1)$$


答:
One approach is to notice that if $A^2 = A$, then $A^3 = A  A^2 = A^2 = A$. So it suffices to find a matrix with $A^2 = A$. Since $A$ is square these are matrices associated to projections (not neccesarily orthogonal projections). It's much easier to algebraicly solve $A^2 = A$ for the entires. As an example,

$$A = \frac{1}{2}\begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix}$$

will do what you want.


答:
It is useful to think about a problem like that in geometric terms.

We want to have $ A^3=A $ which could be thought as some kind of rotation of the plane.

Along these lines what we want is if we rotate a vector by $\theta $ under $A$ to be equal to a rotation by $3 \theta$ corresponding to $A^3$.

A rotation by $\theta$ is given by $A_\theta=
\left (\begin{array}{cc}
\cos \theta & -\sin \theta\\
\sin \theta  &  \cos \theta
\end{array} \right )$.

So if we choose $\theta = \pi = 3\theta  ~\text{mod} 2 \pi $

We get
$$A_{\pi/2}=
\left (\begin{array}{cc}
-1 & 0\\
0  &  1
\end{array} \right ).$$

Which should satisfy $$ A_{\pi}^3 = A_{\pi} $$

We can also change the coordinates by $\pi/3$ so to eliminate the zeros.
So we define
\begin{align*}
A &= A_{\pi/3}  A_{\pi} A_{-\pi/3} \\
&=  \left (\begin{array}{cc}
\frac{1}{2} & -\frac{\sqrt{3}}{{2}}\\
\frac{\sqrt{3}}{{2}} &  \frac{1}{2}
\end{array} \right ) \cdot  \left (\begin{array}{cc}
-1 & 0\\
0  &  1
\end{array} \right ) \cdot \left (\begin{array}{cc}
\frac{1}{2} & \frac{\sqrt{3}}{{2}}\\
-\frac{\sqrt{3}}{{2}} &  \frac{1}{2}
\end{array} \right )
\end{align*}


答: $A^3=A$ can be written as $A(A-I)(A+I)=0$, which is satisfied by any matrix that is similar to a $2x2$ diagonal matrix with diagonals in the set $\{0,1,-1\}$.


问:
Trying to fit an equation to a graph
I'm trying to fit an equation to the scatter plot below. Does anyone know what equation can produce data points like this?

Scatter plot

答:
If those spikes are actually infinite, then
$$y = \dfrac{x(x-150)(x-300)(x-700)(x-850)(x-1000)}{(x-30)^2(x-220)^2(x-410)^2(x-590)^2(x-780)^2}$$
would produce something similar, where the factors in the numerator are the places where the function is $0$, while the factors in the denominator are where the spikes are at.
The numerator factors are all linear because at each of the zeros, the function actually changes sign. The factors in the denominator are all squared because at each spike, the function does not change sign.

(I had to divide everything by 100 to not overflow Desmos)

As you can see, it has the right general shape but isn't a great match. To improve it,
multiply your data by the denominator above (adjusted to line up the factors to the actual spike locations in your curve). Plot the result. Does it still have spikes? If so, increase the exponent for any factor corresponding to a remaining spike by $2$ and try again. Keep doing this until all the spikes are gone.

The resulting curve can be approximated by a polynomial. Once you obtain a good polynomial approximation of that curve, divide it by the factors that were multiplied earlier to get a better approximation of your curve.

Title: Trying to fit an equation to a graph


问:
Why is the sum of multiplicity lengths the same as the cumulative sum of partitions up to the previous integer?
Let $\lambda$ be a partition of $N$ ($\lambda\vdash N$). In the multiplicity representation, $\lambda=(a_{1},a_{2},\ldots a_{m(\lambda)})$, such that
$$\sum_{k=1}^{m(\lambda)}{ka_{k}}=N,$$
where $m(\lambda)$ is the largest $k$ with $a_k>0$ within $\lambda$.

Heuristically, I've found that
$$\sum_{\lambda \vdash N}s(\lambda)=\sum_{n=0}^{N-1}p(n),$$
where $s(\lambda)$ is the number of non-zero $a_k$ within $\lambda$, and $p(n)$ is the number of partitions of integer $n$.

Is there a formal proof, or better yet, an intuitive understanding of this identity?

I'd also appreciate any suggestions to improve the terminology I used so I can find the appropriate resources on my own in the future.

答:
Suppose that $1\le n\le N$; $k$ is a part of a partition $\lambda$ of $N$ if and only if the rest of $\lambda$ is a partition of $N-k$. Conversely, every partition $\mu$ of $N-k$ can be extended to a partition of $N$ with a part $k$. Note that even if $k$ is also a part of $\mu$, $\mu$ can be extended to a unique partition of $N$ with a part $k$. Thus, there is one partition of $N$ with a part $k$ for each partition of $N-k$. It follows that the number of partitions of $N$ in which $a_k\ne 0$ is $p(N-k)$ and hence that

$$\sum_{\lambda\vdash N}s(\lambda)=\sum_{k=1}^Np(N-k)=\sum_{n=0}^{N-1}p(n)\,.$$


问:
Is the set theory (ZF) a structure?
According to the definition, generally speaking, a structure $\langle A;R;F,C\rangle$ is such that $A$ is a non-empty set, $R$ is the set of relations, $F$ is the set of functions, and $C$ is a set of constants. For example $\langle\mathbb{R};; +,\cdot, ^{-1};0,1\rangle$ would be the field of the real numbers. Now, in the construction of the set theory (ZF), we need to make a structure. But then $A$ would be the set of sets, which is impossible... What am I missing?</p>

答:
This is a very delicate point in set theory.

First of all we need to point out that the concept of "set" is not an absolute one. Different models of set theory, will think that different mathematical objects are indeed sets.

This is an external point of view. We consider the universes of set theory as sets, from the outside. The Russell paradox, and indeed all the classical paradoxes of set theory, state that the universe cannot be a set from its own point of view. This is where internal and external points of view are important.

Furthermore, if in a universe of set theory which satisfies the axioms of $\sf ZF$ and we can find a set $M$ and a relation $E$ such that $\langle M,E\rangle$ is a model of $\sf ZF$ then by the completeness theorem we have proved that $\sf ZF$ is consistent - in that particular universe. But from the incompleteness theorem we know that a theory like $\sf ZF$ cannot prove its own consistency, therefore we can never even prove that such set structure exists.

If $V$ is the universe of all sets that mathematics have, and suppose that it satisfies all the axioms of $\sf ZF$, then we know that (1) it is not a set itself; (2) we cannot prove that there is a set which is a structure satisfying the axioms of $\sf ZF$; (3) if there is such set, then we can talk about objects which are in that structure, which that particular structure thinks of as sets, and we can contrast them to actual sets.

This is close to the Skolem paradox, by the way.

In any case, as I said, this is a very delicate point and one has to study quite a lot of logic and set theory in order to understand it completely and become comfortable with it.

Title: Is the set theory (ZF) a structure?


问:
Vector parametric form for 2 equations and 3 variables
I need help to find the vector parametric form of the following linear system.

I have tried doing it on my own but I can't seem to get the system in the right form to get a solution.

\begin{align*}
\begin{cases}
x - 2y - z = -4\\\\
-2x + 4y + 3z =  1
\end{cases}
\end{align*}

答:
If we add twice the first row to the second row, we get that $z = -7$, as you have mentioned.

Based on it, we obtain the relation $x - 2y + 7 = -4$. That is to say, $x = 2y - 11$.

Consequently, the solution set to this system of linear equations is given by
\begin{align*}
S = \{(2a - 11,a,-7)\in\mathbb{R}^{3} \mid a\in\mathbb{R}\}
\end{align*}

Hopefully this helps!


问:
Under which domain transformations are the eigenvalues of $-\Delta$ invariant?
Let $d\in\mathbb N$, $\Omega\subseteq\mathbb R^d$ be bounded and open, $$\mathfrak a(u,v):=\langle\nabla u,\nabla v\rangle_{L^2(\Omega,\:\mathbb C^d)}\;\;\;\text{for }u,v\in H_0^1(\Omega)$$ and $$P:=-\Delta:C_c^\infty(\Omega)\to C_c^\infty(\Omega).$$ By the Lax-Milgram theorem, $$\forall f\in L^2(\Omega):\exists !v_f\in H_0^1(\Omega):\forall u\in H_0^1(\Omega):\mathfrak a(u,v_f)=\mathcal l_f(u):=\langle u,f\rangle_{L^2(\Omega)}.$$ The map $$S:L^2(\Omega)\to H_0^1(\Omega)\;,\;\;\;f\mapsto v_f$$ is

a bounded linear operator from $L^2(\Omega)$ to $H_0^1(\Omega))$;

a compact linear operator on $L^2(\Omega)$;

a self-adjoint bounded linear operator on $L^2(\Omega)$.

Remember that if $f\in L^2(\Omega)$, then $u\in L^2(\Omega)$ is a weak solution of the symbolic equation $Pu=f$ if $$\langle u,P\varphi\rangle_{L^2(\Omega)}=\langle f,\varphi\rangle_{L^2(\Omega)}\;\;\;\text{for all }\varphi\in C_c^\infty(\Omega).\tag1$$ It's easy to see that $u\in H_0^1(\Omega)$ is a weak solution of $Pu=f$ iff $$Sf=u\tag2.$$

By the spectral theorem, $$S=\sum_{i\in I}\mu_ie_i\otimes e_i,$$ where the sum converges in $\mathfrak L(L^2(\Omega))$ and $I:=\mathbb N\cap[0,\operatorname{rank}S]$, for some $(\mu_i)_{i\in I}\subseteq(0,\infty)$ nonincreasing and some orthonormal basis $(e_i)_{i\in I}$ of $\overline{\mathcal R(S)}^{L^2(\Omega)}$. Moreover, if $$\mu_n:=0\;\;\;\text{for }n\in\mathbb N\setminus I,$$ then $$\mu_n\xrightarrow{n\to\infty}0\tag3.$$ As usual, $(e_i)_{i\in I}$ can be supplemented to an orthonormal basis $(e_n)_{n\in\mathbb N}$ of $L^2(\Omega)$ by an orthonormal basis $(e_n)_{n\in\mathbb N\setminus I}$ of $\mathcal N(S)$.

If $$\lambda_n:=\left.\begin{cases}\displaystyle\frac1{\mu_i}&\text{, if }\mu_i\ne 0\\\infty&\text{, otherwise}\end{cases}\right\}\;\;\;\text{for }n\in\mathbb N,$$ we immediately see that $e_i\in H_0^1(\Omega)$ and $e_i$ is the unique weak solution of $Pu=\lambda_ie_i$ for all $i\in I$.

Let $\tilde\Omega\subseteq\mathbb R^d$ be bounded and open. Assume $\tilde\Omega=\iota(\Omega)$ for some $\iota:\Omega\to\tilde\Omega$. Under which assumption on $\iota$ will the eigenvalues $\tilde{\lambda_i}$ of $$\tilde P:=-\Delta:C_c^\infty(\tilde\Omega)\to C_c^\infty(\tilde\Omega),$$ obtained in the same way as the $\lambda_i$, coincide with the eigenvalues $\lambda_i$ of $P$?

Clearly, if $\iota$ is a $C^1$-diffeomorphism with $$\left|\det{\rm D}\iota\right|=1\tag4,$$ we easily see that $$\tilde e_n:=e_n\circ\iota^{-1}\;\;\;\text{for }n\in\mathbb N$$ is an orthonormal basis of $L^2(\tilde\Omega)$. I guess that's the least we need to assume.

But what do we need to assume further? Do we need $\iota$ to extend to a linear isometry on $\mathbb R^d$ or something like that?

答:
The question when two domains have the same Laplacian eigenvalues has been popularized as "Can one hear the shape of a drum" by Kac and has received a lot of research interest.

Some (increasingly less) simple observations:

It is clearly sufficient that $\iota$ is an isometry, because in this case $f\mapsto f\circ \iota$ is a unitary operator from $L^2(\tilde\Omega)$ to $L^2(\Omega)$.

You cannot expect to extract too much information on the map $\iota$ itself, because the input data only fix the range of $\iota$. The better question would be when there exists a map between $\Omega$ and $\tilde \Omega$ with nice properties.

To get an interesting result, you should make at least some minimal assumptions on the boundaries of the domains. Otherwise, you can always just remove a set of capacity $0$ from $\Omega$ without changing the eigenvalues of $\Delta$. For example, the Laplacian on the disk and the Laplacian on the disk with countably many points removed have the same eigenvalues. In this case, the domains are not even homeomorphic.

Even for Lipschitz domains, there does not need to exist an isometry between $\Omega$ and $\tilde \Omega$. The existence of isospectral domains that are not isometric was first proven by Gordon, Webb and Wolpert building on previous work most notably by Sunada. You can find pictures of a pair of such domains in the wikipedia article linked above.

To the best of my knowledge, the question whether isospectral domains are are necessarily isometric is open for some very natural classes such as convex domains and domains with $C^\infty$-boundary.

$\Delta$ and $\tilde \Delta$ have the same eigenvalues if and only if there exists a unitary $U\colon L^2(\Omega)$ to $L^2(\tilde \Omega)$ such that $U\Delta=\tilde \Delta U$. The reason why this question often has a negative answer is that there are just too many isometries of $L^2$. If in contrast $V\Delta_p=\tilde \Delta_p V$ for the Laplacians acting on $L^p$ and a surjective isometry $V\colon L^p(\Omega)\to L^p(\tilde \Omega)$, then $\Omega$ and $\tilde \Omega$ must be isometric under minor regularity assumptions by work of Arendt.


答:
$$
\left(
\begin{array}{cc}
\cos t & \sin t \\
\sin t & - \cos t
\end{array}
\right)
$$

with eigenvectors as columns of

$$
\left(
\begin{array}{cc}
\cos \frac{t}{2} & -\sin  \frac{t}{2} \\
\sin  \frac{t}{2} &  \cos \frac{t}{2}
\end{array}
\right) \; \; , \; \;
$$

this confirmed by the identity

$$
\left(
\begin{array}{cc}
\cos t & \sin t \\
\sin t & - \cos t
\end{array}
\right)
\left(
\begin{array}{cc}
\cos \frac{t}{2} & -\sin  \frac{t}{2} \\
\sin  \frac{t}{2} &  \cos \frac{t}{2}
\end{array}
\right) =
\left(
\begin{array}{cc}
\cos \frac{t}{2} & \sin  \frac{t}{2} \\
\sin  \frac{t}{2} & - \cos \frac{t}{2}
\end{array}
\right)
$$

It does not need to be symmetric, although the reflection is what first came to mind for $M^2 = I,$  which happens when $M^3 = M$ and $M$ is nonsingular. For real matrices, we need just trace zero and determinant $-1$

As  long as $$ a^2 + bc = 1 $$  we can use
$$
\left(
\begin{array}{cc}
a & b \\
c & - a
\end{array}
\right)
$$

for example

$$
\left(
\begin{array}{cc}
7 & 8 \\
-6 & - 7
\end{array}
\right)
$$

We see the eigenvectors as columns

$$
\left(
\begin{array}{cc}
a & b \\
c & - a
\end{array}
\right)
\left(
\begin{array}{cc}
1+a & -b \\
c & 1+ a
\end{array}
\right) =
\left(
\begin{array}{cc}
1+a & -b \\
c & -1- a
\end{array}
\right)
$$


问:
Why define vector spaces over fields instead of a PID?
In my few years of studying abstract algebra I've always seen vector spaces over fields, rather than other weaker structures. What are the differences of having a vector space (or whatever the analogous structure is called[module?]) defined over a principal ideal domain which is not also a field? What properties of vector spaces break down when defining it over a PID? Please give an example (apart from letting the vector space be the field/PID itself) where the vector space over a field has properties that the module over a PID does not.</p>

答:
Actually, someone has already asked a question about Pathologies in module theory, whose solutions will show you a lot about how module theory differs (in interesting ways) from vector space theory. In short, when you begin to lose nice properties, you begin to appreciate them more :)

After learning about vector spaces and their usefulness in linear algebra, they serve as a baseline for intuition about module theory. When you pass from a field to a less-nice ring, things begin to break down. This isn't really a bad thing: we just learn more about how the nice properties come about.

$\Bbb F[x]$ is an important principal ideal domain, and it turns out a lot of linear algebra can be recovered from module theory over these polynomial rings.

$\Bbb Z$ is also an important PID, and the study of its modules is the study of abelian groups.

But there are still lots of other questions to ask. For example: What rings have modules who are isomorphic to copies of the ring? (answer: division rings.)

For what rings are the modules direct sums of simple modules? (answer semisimple rings.

For what rings are the modules all flat? (answer: von Neumann regular rings)

So you can see that moving from fields to PIDs is just one step in a big web of moving outward from fields, exploring what their modules look like. You don't have to stop at PID's!


问:
How did they get from (5/3)^(log3(n)) to n^log3(5/3)
Here's the note I got from class:

So I was wondering how my prof got from
$$\biggr(\frac{5}{3}\biggl)^{\log_3n}$$
to
$$n^{\log_3(5/3)}.$$

答:
If we take $\ln$ of both sides we have

$$\ln\left(\frac{5}{3}\right)\cdot \log_3n$$

which can be rewritten as

$$\ln\left(\frac{5}{3}\right)\cdot \frac{\ln(n)}{\ln(3)} = \ln(n) \cdot \frac{\ln\left(\frac{5}{3}\right)}{\ln(3)}$$

$$= \ln(n) \cdot \log_3\left(\frac{5}{3}\right) \implies e^{\ln(n)\cdot\log_3\left(\frac{5}{3}\right)} = n^{\log_3(5/3)}$$


问:
A prime ideal of $\mathbb{Z} \times \mathbb{Z}$ that is not maximal.
So, I've just proven that $I = \{(a,0): a \in \mathbb{Z}\}$ is a prime ideal of $\mathbb{Z} \times \mathbb{Z}$ (Operations defined coordinate per coordinate). But I can't get a grip on how to begin to prove that $I$ is not maximal.

The only thing I think about is that, if $I$ is maximal, the quotient ring $\mathbb{Z} \times \mathbb{Z} / I$ is a field, since $1 = (1,1) \in \mathbb{Z} \times \mathbb{Z}$ is it's unity.

I just need a hint on how to approach this problem.

Any help or response is much apreciated.

答:
$\;\;\;\;\;$ From the OP then $I=\Bbb Z\times\{0\}$. By the first homomorphism theorem then  $$\frac{\Bbb Z\times\Bbb Z}{\Bbb Z\times\{0\}}\approx\Bbb Z$$ as $\Bbb Z\times\{0\}$ is the kernel of the ring epimorphism  $\pi:\Bbb Z\times\Bbb Z\to\Bbb Z$ defined by $\pi(x,y)=y$.

$\;\;\;\;\;$ $\Bbb Z$ is an integral domain but not a field. Therefore, $\Bbb Z\times\{0\}$ is a prime $\Bbb Z\times\Bbb Z$ ideal but not a maximal one.


问:
Evaluating $\int\frac{1}{5\cos x+\sin x+7}~dx$
Evaluating  $$\int\frac{1}{5\cos x+\sin x+7}~dx.$$

This can be done by substituting $$\sin x = \frac{2t}{1 + t^2}$$ and $$\cos x = \frac{1 - t^2}{1 + t^2}.$$

However after I substitute it I cannot simplify it to get anything easier to integrate.

After substituting I got: integral $$\frac{1 + t^2}{2(t + 2)(t + 3)}$$ or $$\frac{1 + t^2}{12 + 2t^2 + 10t}.$$

Could someone give me a hint?

Many thanks.

答:
Let $t = x - \tan^{-1}\frac15$ to integrate

\begin{align}
& \int\frac{1}{5\cos x+\sin x+7}~dx
= \int \frac1{\sqrt{26}\cos t +7}dt\\
=& \int \frac{2d(\tan\frac t2)}{(7+\sqrt{26})+(7-\sqrt{26})\tan^2\frac t2}\\
=& \frac2{\sqrt{23}}\tan^{-1} \left( \sqrt{\frac{ 7-\sqrt{26}}{7+\sqrt{26}}}\tan\frac t2 \right)+C
\end{align}


问:
Prove that $16(a\sin a + \cos a - 1)^2 \le 2a^4 + a^3 \sin 2a, \ \forall a\ge 0$

Problem 1: Prove that
$$16(a\sin a + \cos a - 1)^2 \le 2a^4 + a^3 \sin 2a, \ \forall a\ge 0.\tag{1}$$

This is the stronger version of the following Prove that $12(a\sin a+\cos a-1)^2\le 2a^4+a^3\sin(2a)$,$\forall a\in (0,\infty)$:

Problem 2: Prove that
$$12(a\sin a + \cos a - 1)^2 \le 2a^4 + a^3 \sin 2a, \ \forall a\ge 0. \tag{2}$$

For Problem 2, there is a very nice solution using Cauchy-Bunyakovsky-Schwarz inequality for integral Prove that $12(a\sin a+\cos a-1)^2\le 2a^4+a^3\sin(2a)$,$\forall a\in (0,\infty)$.
Indeed, the inequality (2) is nothing but
$$12\left(\int_0^a x\cos x \mathrm{d} x \right)^2 \le 12\left(\int_0^a x^2 \mathrm{d} x\right)
\left(\int_0^a \cos^2 x \mathrm{d}x\right).$$

For Problem 2, I gave a complicated proof. See Prove that $12(a\sin a+\cos a-1)^2\le 2a^4+a^3\sin(2a)$,$\forall a\in (0,\infty)$

Are there any nice solutions for Problem 1?

Any comments and solutions are welcome and appreciated.

答:
$\color{brown}{\textbf{Preliminary notes.}}$

Dividing of the given inequality $(1)$ to $\;4a^4\;$ allows to present it in the equivalent form of
$$\left(2\text{ sinc } a - \text{ sinc}^2\,\dfrac a2\right)^2
\le \dfrac{1+\text{sinc }2a}2,\tag1$$
or $\;f_1(a)\le f_2(a),\;$ where
$$f_1(x) = \left(2\text{ sinc } x - \text{ sinc}^2\,\dfrac x2\right)^2,\quad
f_2(x)\le \dfrac{1+\text{sinc }2x}2.\tag2$$

It suffices to prove the inequalities
$$f_1(x) \le (1-\,^1\!/_6\,x^2)^2\le f_2(x),\quad \text{if}\quad x\in\left[0,\dfrac\pi2\right],\tag3$$
and
$$f_1(x)\le \dfrac{4\pi-3}{8\pi}\le f_2(x),\quad \text{if}\quad x\in\left(\dfrac\pi2,\infty\right).\tag4$$

We can obtain the derivative
$$f_1'(x)= -\dfrac{8}{x^5}\,\sin x\left(\tan \dfrac x2-x\right)\bigg((x^2-2) \cos x - 2 x \sin x+2\bigg)\tag5$$

The root of the derivative ar $\;x=r\approx\dfrac73\;$ corresponds to the root of $\;f_1.$

The root of the derivative near $\;x=m\approx 4\;$ corresponds to the secondary maximum of $\;f_1.$

$\color{brown}{\mathbf{\text{Interval }\left[0,\dfrac\pi2\right].}}$

Taking in account inequality $\;x^2<3,\;$ one can get
$$2\text{ sinc } x-\text{ sinc }^2\,\dfrac x2
 = 2\left(1-\dfrac16x^2+\dfrac1{5!}x^4-\dfrac1{7!}x^6+\dots\right)
 - \left(1-\dfrac16\left(\dfrac x2\right)^2+\dfrac1{5!}\left(\dfrac x2\right)^4+\dots\right)^2
$$$$
\le 2-\dfrac13x^2+\dfrac{x^4}{60} - \left(1-\dfrac{x^2}{24}\right)^2
\le 1-\dfrac14x^2+\dfrac1{36}x^4 \le 1-\dfrac14x^2+\dfrac1{12}x^2= 1-\dfrac{x^2}6,
$$

$$f_1(x)\le(1-\,^1\!/_6 x^2)^2.$$

Om the other hand,
$$\dfrac12\big(1+\text{ sinc } 2x\big)- (1-\,^1\!/_6 x^2)^2
 = 1-\dfrac1{12}\,(2x)^2+\dfrac1{2\cdot5!}(2x)^4-\dfrac1{2\cdot7!}(2x)^6+\dots -1 + \dfrac13x^2 - \dfrac1{36}x^4
$$$$
\ge\left(\dfrac1{15}-\dfrac1{36}\right)x^4\ge 0,
$$$$
f_2(x)\ge (1+\,^1\!/_6x^2)^2.
$$
Therefore, inequality $(3)$ is proved.

$\color{brown}{\mathbf{\text{Interval }\left(\dfrac\pi2,\infty\right).}}$

Since
$$f_1\left(\dfrac\pi2\right)=\left(\dfrac4\pi-\dfrac8{\pi^2}\right)^2
<\dfrac14<\dfrac{4\pi-3}{8\pi},$$
then
$$f_1(x)\le \dfrac{4\pi-3}{8\pi},\quad \text{if}\quad x\in\left(\dfrac\pi2,r\right].$$

At the same time,
$$f'_1(4.0)>0,\;f'_1(4.1)<0\quad\Rightarrow\quad m\in(4.0,4.1).$$
Then
$$f_1(m) - \dfrac{4\pi-3}{8\pi} < (\text{ sinc}^2 2 - 2\text{ sinc }4.1)^2 - \dfrac{4\pi-3}{8\pi} < 0,$$
and
$$f_1(x)\le \dfrac{4\pi-3}{8\pi},\quad \text{if}\quad x\in\left(\dfrac\pi2,\infty\right].\tag6$$

The least value of $\;\text{ sinc } x\;$ is situated near $\;x=\dfrac{3\pi}2,\;$ wherein
$$\text{ sinc }\dfrac{3\pi}2 = -\dfrac2{3\pi},\quad \text{ sinc }\dfrac{4\pi}3 = -\dfrac{3\sqrt3}{8\pi}>\text{ sinc }\dfrac{3\pi}2,\quad \text{ sinc }\dfrac{5\pi}3 = -\dfrac{3\sqrt3}{10\pi}>\text{ sinc }\dfrac{3\pi}2.$$
Taking in account unimodality of $\;\text{ sinc }x\;$ in$\;\left[\dfrac{4\pi}3,\dfrac{5\pi}3\right],\;$ the least value of $\;\text{ sinc }x\;$
belongs to this interval.
Therefore,
$$\text{ sinc }x>-\dfrac{\sin\dfrac{3\pi}2}{\dfrac{4\pi}3}=-\dfrac3{4\pi},$$
$$f_2(x)>\dfrac{4\pi-3}{8\pi }.\tag7$$
Therefore, inequality $(4)$ is proved.

$\color{brown}{\textbf{Done!}}$


问:
What method can be used for finding Green function for Fokker-Planck equation?
Let's have an equation

$$
u_{t} - (xu)_{x} - \frac{1}{2}u_{xx} = 0, \quad u(x, 0) = g(x), \quad -\infty < x < \infty , \quad 0 < t < \infty .
$$
I need to find a Green function for it. So,
$$
u_{t} - (xu)_{x} - \frac{1}{2}u_{xx} = \delta (x - x_{0})\delta (t - t_{0}).
$$
I tried to use Laplace transform for it, but then I got
$$
su(x, s) - g(x) - u(x, s) - xu_{x}(x, s) - \frac{1}{2}u_{xx}(x, s) = \delta (x - x_{0})e^{-t_{0}s}.
$$

I don't know what to do with $xu_{x}(x, s)$. Can you help me with this problem or with method?

Maybe, some substitution like $u(x, p) = v(x, p)w(x)$ can help?

答:
The ordinary differential equation in $x$ that you have is a form of the Hermite ODE with an inhomogeneous term, so you can solve it using the standard theory of inhomogeneous second-order linear ODEs.

Note, by the way, that if you do a Fourier transform over $x$, the term $xu_x$ transforms into $$(ik)(i\partial_k) u_k(t),$$
so a Fourier transform will turn the second-order PDE w.r.t. $(x,t)$ into a first-order linear PDE w.r.t. $(k,t)$. You can then use, for example, the method of characteristics to solve it. This might be easier than using the Laplace transform.


问:
Find the dimensions of some kernel subspaces of $\operatorname{Hom}(V, W)$.
I am a bit unconfident with the following proof that I made to find the dimensions of the subspace below,

$$
S =\{T\in\operatorname{Hom}(V, W): T(v)=0\}
$$
where $V$ and $W$ are finite-dimensional vector spaces over F of dimensions $n$ and $m$ respectively and $v\in V$ be a non-zero vector. What I did is to define the following,
$$
U:\operatorname{Hom}(V,W)\rightarrow W
$$
where $T\mapsto T(v)$. And show $U$ is a linear transformation, then using the Dimension Theorem to find the $\operatorname{dim}(S)$ since $S=\operatorname{ker}(U)$. To show $U$ is linear is not so difficult, but I was confused at the end that was whether if $\operatorname{Im}(U)=W$. I know $\operatorname{Im}(U)\subset$ W, but if $\operatorname{Im}(U)\neq W$, then I can't derive the conclusion such that $\operatorname{dim}(S)= nm-m$.

Additionally, I also have the following question that is to find the possible dimensions of the subspace below,
$$
C =\{T\in\operatorname{Hom}(V,W):T(v_1)=0,T(v_2)=0,T(v_3)=0\}
$$
where $v_1,v_2,v_3\in V$. Basically, I show this in a similar way to the previous one. Let
$$
\Phi:\operatorname{Hom}(V, W)\rightarrow W\times W\times W
$$
which is defined differently depending on the independence of $\{v_1, v_2, v_3\}$, but to show all of them are linear is not so difficult either. For example, if $v_1,v_2,v_3$ are linearly independent then $\Phi:T\mapsto (T(v_1),T(v_2),T(v_3))$. Since $C=\operatorname{ker}(\Phi)$ and if $\operatorname{Im}(\Phi)=W\times W\times W$, then using Dimension Theorem $\operatorname{dim}(C)=nm-3m$. Similarily, depending on the dependence between $v_1, v_2$ and $v_3$, I can also derive two possible dimensions that are $\operatorname{dim}(C)=nm-2m$ or $\operatorname{dim}(C)=nm-m$.

I wonder how to end the proof (e.g. how to tell $\operatorname{Im}(U)=W$?), and whether if my guesses to the second question is correct or not.

PS: I am a newcomer to linear algebra so if I made anything unclear or wrong in the proof, please help me to point it out since this is helpful for me to improve my skill when writing a proof.

Thanks in advance.

答:
Take a basis $v_1,\dots,v_n$ for $V$ such that $v_1=v$ (this can be done since $v \neq 0$). Then, for each $w \in W$, let $T_w \in \operatorname{Hom}(V,W)$ be the linear map such that $$T_w(a_1v_1+\cdots+a_nv_n) = a_1w$$ for every  $(a_1,\dots,a_n) \in F^n$ (this is well-defined, since each element of $V$ can be written in a unique way as a linear combination of $v_1,\dots,v_n$).

Then $w = T_w(v) = U(T_w) \in \operatorname{im} U$, and since $w$ was arbitrary, it follows that $W = \operatorname{im} U$.

Now, use a similar argument to finish the second problem!


问:
Prove or disprove: if $\Sigma$ is special then $\left\{ \lnot\alpha\,:\,\alpha\in\Sigma\right\} $ is special
Practicing for exam in Logic and trying to solve the following question:

Let $\Sigma\subseteq WFF_{\left\{ \lnot,\to\right\} }$ be set of propositions. $\Sigma$ is called special if for every $\alpha,\beta\in WFF_{\left\{ \lnot,\to\right\} }$ we get $\Sigma\vdash\alpha\to\beta$ or $\Sigma\vdash\beta\to\left(\lnot\alpha\right)$. Prove or disprove: if $\Sigma$ is special then $\left\{ \lnot\alpha\,:\,\alpha\in\Sigma\right\} $ is special.

I'm suspecting that this statment is not true. I choose $\Sigma$ to be $WFF_{\left\{ \lnot,\to\right\} }$ and I proved that $\Sigma$ is special because you can show a proof sequence (because $\left(\alpha\to\beta\right)\in\Sigma$). But how do I disprove $\left\{ \lnot\alpha\,:\,\alpha\in\Sigma\right\} $ to be special? I know the theorem that for every $\Sigma\subseteq WFF_{\left\{ \lnot,\to\right\} }$ we get $\Sigma\vDash\alpha$ iff $\Sigma\vdash\alpha$. I think it might help here.

答:
By $WFF_{\{\neg, \rightarrow\}}$ I assume all of your formulas have only $\neg$ and $\rightarrow$ as connectives? This will make things a bit less clear...

I believe you can find a special set $\Sigma$ such that $\neg\Sigma=\{\neg\alpha : \alpha\in\Sigma\}$ is not special: consider the propositional variables $p$ and $q$ and define
$$\Sigma=\{\neg p, \neg(p\rightarrow\neg q)\},$$
but notice that $\neg(p\rightarrow \neg q)$ is equivalent to $p\wedge q$; this means that the logical closure of $\Sigma$ contains all (well-formed) formulas, and so $\Sigma$ is special.

(As you noticed, you can test this by using truth tables: if both $\neg p$ and $\neg(p\rightarrow\neg q)$ are true, $p$ and $p\rightarrow \neg q$ are false; but if $p\rightarrow \neg q$ is false, $p$ is true and $\neg q$ is false, what would lead to a contradiction. This means $\Sigma$ is contradictory, and in particular can derive any formula, meaning it is special.)

Now, $\neg\Sigma=\{p, p\rightarrow\neg q\}$ (up to equivalences), and notice that $p\rightarrow\neg q$ is equivalent to $\neg p\vee \neg q$. We can easily find formulas $\alpha$ and $\beta$ such that neither $\neg\Sigma\vdash \alpha\rightarrow \beta$ nor $\neg\Sigma\vdash\beta\rightarrow\neg\alpha$: take, for example, new propositional variables $r$ and $s$, $\alpha=r$ and $\beta=s$.

If $p$ and $r$ are true but $q$ and $s$ are false, $\neg\Sigma$ is satisfied but not $r\rightarrow s$, meaning $\neg\Sigma$ does not prove $\alpha\rightarrow\beta$.

If $p$, $r$ and $s$ are true but $q$ is false, $\neg\Sigma$ is satisfied but not $s\rightarrow\neg r$, meaning $\neg\Sigma$ does not prove $\beta\rightarrow\neg\alpha$.


问:
Prove that $e^a e^b = e^{a+b}$
I've read the argument in Rudin, but I think I need a little clarification

\begin{align}
e^a e^b &= \sum_{k=0}^{\infty} \frac{a^k}{k!} \sum_{m=0}^{\infty} \frac{b^m}{m!}\\ &= \sum_{n=0}^{\infty} \frac{ n!}{n!}  \sum_{k=0}^{n} \frac{a^k}{k!} \frac{b^{n-k}}{(n-k)!} \\
&= \sum_{n=0}^{\infty} \frac{(a+b)^n}{n!}\\
&=e^{a+b}\\
\end{align}

I'd like to understand how we get from the first line to the second. Of course $m =n-k$, but how do we ensure that we haven't missed any addends in the limit changing process?

Changing limits: $m=0 \rightarrow n-k=0 \rightarrow k=n$ and $m=\infty \rightarrow n-k=\infty \rightarrow k=0$, assuming $n \to \infty$.

\begin{align}
\frac{b^m}{m!} \sum_{k=0}^{\infty} \frac{a^k}{k!}
&= \sum_{k=0}^{\infty} \frac{a^k}{k!} \frac{b^{m}}{m!} \\
&=\sum_{k=0}^{?} \frac{a^k}{k!} \frac{b^{n-k}}{(n-k)!} \\
\end{align}

how to justify that "?"$=n$ ?

答: Yeah, you're not thinking of this right. Think about taking the first quadrant in $mk$-space and filling it up with (anti-)diagonal lines $n=k+m=\text{constant}$. As you  let $n$ vary from $1$ to $\infty$, you hit every $(m,k)$.</p>

Title: Prove that $e^a e^b = e^{a+b}$


问:
Why null hypothesis in t-tests can be numbers other than zero?
This may sound stupid but please bear with me because it's hard to explain also hypothesis testing is a totally new concept to me and cannot really wrap my head around it

Concern:

Should not T-distributions be used ONLY to test $H_o: \Delta \mu = 0$ ?

My reasoning:

Because basically that's what these T-distributions are only able to test

Let me illustrate , say we ran a paired t-test  with 3 degrees of freedom and confidence interval of 95% and $H_o: \Delta \mu = 0$ , $H_a: \Delta \mu \ne 0$ and we get t-score = -3.908 which corresponds to p-value = 0.015 and according to my "understanding" that means 1.5% of times "our null hypothesis" which says that $\Delta \mu = 0$ is true so we can reject the null hypothesis with a 95% of confidence , right?

BUT and here's what i am trying to say:

We could have got the same T-score (because t-scores are simply the no. of standard deviations away from the hypothesized mean in the sampling distribution) but for a totally "different" null hypothesis (say $H_o: \Delta \mu = 3$) , so should not we construct a totally new T-distrubtion with it's own degress of freedom to test the new null hypothesis ($H_o: \Delta \mu = 3$) and same goes to all other null hypotheses, Why are we using a single disturbtion to test all possible null hypotheses ?

NOTE : I might be making wrong assumptions and of course feel free me to correct me but that's what i was able to grasp

Thanks in advance.

答:
We know that $\frac{\bar x-\mu}{s/\sqrt n}$ follows a t-distribution with n-1 degrees of freedom. Under the null hypothesis, we specify a particular value of $\mu$ that we think is the true population mean. Call it $\mu_0$. By extension, since this is assumed to be the true population mean, therefore $\frac{\bar x-\mu_0}{s/\sqrt n}$ follows a t distribution. If the sample mean $\bar x$ is far from the $\mu_0$, it is unlikely that we would have observed such an extreme sample mean under this t distribution, because the shape of the density is bell-shaped, meaning there is less density on the tails. It is not the fact that it is far from the $\mu_0$ that means that it is incorrect. The mean is only the expected value; it is possible that under another distribution, the farther you are away from $\mu_0$, the more density there is, and so the closer to $\mu_0$, the more unlikely you are to observe that value - for example, take a bimodal normal distribution. Anyway, since there is less density on the tails, we don't think that this distribution is correct. But what were we specifying for this distribution? Everything is fixed except for the mean. Therefore it must be the mean that is incorrect, and we reject the null mean.

There is no particular reason why you must test the hypothesis that $\mu_0=0$. It can be any value that you believe is the true population mean (see the very first sentence).


问:
For which $f$ does $f*f=\sqrt{2\pi}f$ hold?
Im facing the following problem: Find all Elements $f$ of $L^1(\mathbb{T})$ with $f*f=\sqrt{2\pi}f$.

$\mathbb{T}$ being the unit circle. $*$ stands for convolution.

So my first idea was to use the convolution theorem: The convolution theorem states that

$\mathcal{F}\{f * g\} = k\cdot \mathcal{F}\{f\}\cdot \mathcal{F}\{g\}$

where $\mathcal{F}\{f\}$ denotes the Fourier transform of $f$, and $k$ is a constant that depends on the specific Normalizing constant of the Fourier transform.

But I cannot get rid  of the Fourier transform. In fact I have no idea what the necessary condition should look like. Any ideas on this?

EDIT: I found this to be somekind of similar problem: http://sebastiaanjanssens.nl/math/calculating-an-auto-convolution-integral-by-fourier-transforms/

Here the result is $f*f=\pi f$. But in fact I need to work "backwords" in order to get my $f$.

答:
With $f(\theta) = f_0 + f_1\cos\theta + f_2\cos(2\theta) + \cdots$, and assuming appropriate
convergence for exchange of the sums and integral,
\begin{align*}
 (f * f)(\theta) &= \int_0^{2\pi}\, f(\theta - \phi)f(\phi)\, d\phi \\
 &= \int_0^{2\pi}\sum_{m=0}^\infty \sum_{n=0}^\infty\, f_m f_n\cos(m(\theta - \phi)) \cos(n\phi)\, d\phi \\
 &= \sum_{m=0}^\infty \sum_{n=0}^\infty\, f_m f_n\int_0^{2\pi}\cos(m(\theta - \phi)) \cos(n\phi)\, d\phi \\
 &= 2\pi f_0^2 + \sum_{m=1}^\infty \sum_{n=1}^\infty\, \pi f_m^2 \cos(m\theta).
\end{align*}
Equate the coefficients to see what's required for $f*f = Cf$.


问:
Why is it valid to use squared Euclidean distances in high dimensions in multiple regression?
Euclidean distance is not linear in high dimensions. However, in multiple regression the idea is to minimize square distances from data points to a hyperplane.

Other data analysis techniques have been considered problematic for their reliance on Euclidean distances (nearest neighbors), and dimensionality reduction techniques have been proposed.

Why is this not a problem in multiple regression?

答:
I'm not sure how regression applies to this question but Euclidean distance is valid in any number of dimensions as shown by the Pythagorean theorem where:
$$A^2+B^2+C^2+\cdots+X^2=Y^2\quad\text{where}\quad X,Y\space\text{are arbitrary variables}$$
For example, we know that $3^2+4^2=5^2$ can be a diagonal on the front of a box. If the depth of the box is $12$, then we can also know that the distance between opposite corners is shown by the equation
$3^2+4^2+12^2=13^2\space$ because the $5^2$ in $(5,12,13)$ calculations can be replaced with $3^2+4^2$.

Likewise, if we have a $4$D box $\quad 3^2+4^2+12^2+84^2=85^2\quad$ because the triples  $\quad (3,4,5)\quad (5,12,13)\quad (13,84,85)\quad$ can all be similarly joined into a quintuple. The process can be reversed for dimensional reduction.
For example $\quad 3^2+4^2+12^2+84^2=\quad 3^2+12.64911064^2+84^2=85$

At least one form of regression works on minimizing distances and the
corner-to-corner distances here are the shortest [straight-line] distance. The example has shown integer solutions but works the same with non-integers such as those found by $\space A=x_1-x_0\quad B=y_1-y_0\quad C=z_1-z_0\quad \cdots $

There are methods of finding missing values if one is unknown and I have been compiling many of these in a paper. If these would be useful, I can probably translate them to your application if I can understand what your application is.


问:
Results on the Riemann Hypothesis
Are there any papers that are significant to the proof of the RH? Are there also any common methods used to prove it? Were there any fundamental papers, or known results regarding this hypothesis? Work being done on it?

I am curious to see how other people would attack such a question, so that maybe I could understand higher level concepts more easily. And it is interesting.

答:
While there is certainly a lot of work (to put it mildly) around it, I don't think anyone really has a good idea for a line of attack. There are various ideas about what such a line might look like broadly - see e.g. here for one of them (with the caveat that "field with one element" is not what one might expect) - but no actual concrete approach is known, to the best of my knowledge.

Moreover there are known obstacles to resolving RH in either direction, most strikingly "near analogies" which are known to be true or known to be false. Such near analogues provide important negative information in that they demonstrate that certain approaches can't possibly work. So, for example, we can immediately discard any claimed proof of RH which only uses the Euler product, meromorphic continuation, and functional equation.

Title: Results on the Riemann Hypothesis


问:
I can't wrap my head around Peano's 5'th axiom
I'm currently reading about Peano's axiom and the construction of the natural numbers. The fifth Axiom of Peano does not make sense to me:

If a subset $T$ contains $0$ as an element and for all $n \in T$, $s(n)$ is also in $T$, then $T$ is the set of Natural numbers $\mathbb{N}$.

I know that this axiom exists in order to exclude elements which could have the following property:
$$s(a) = b$$ $$s(b) = a$$

These elements would not violate axioms $1$ to $4$ however they don't have the desired "natural number" properties hence axiom $5$ has been created to allegedly clean these elements up. I also know that axiom $5$ is supposed to work as follows:

If $0$ is in our set $T$ then $s(0)$ is also in $T$ but if $s(0)$ is in $T$ then $s(s(0))$ is also in $T$ and so on. Thus we declare this set to be the natural numbers.

Now to my actual question: suppose we throw in $a$ and $b$ and $0$ into $T$. Since we do not yet know what the natural numbers are, we can't exclude $a,b$ either (if my thinking is correct). Then for all $n$ in $T$ the succesor $s(n)$ is also in $T$. As previously stated $s(0), s(s(0)), ...$ are all in in $T$ (the actual natural numbers) but if such elements $a$ and $b$ are also in T then their successor $s(a)$ and $s(b)$ are also in $T$ which is  a true statement since $s(a) = b$ and $s(b) =a$.

Axiom $5$ should exclude such elements but the way I am stating it, it seems like these elements do not violate axiom $5$.

答:
Axiom 5, in conjunction with the other four axioms, tells you that:

Every natural number other than $0$ is a successor; and

You can “reach” every natural number by starting with $0$ and applying the successor function finitely many times.

To see that the five axioms prevent you from having “cycles” with the successor function, note first that we cannot have $s(a)=a$ for any $a$. To see this, consider
$$T = \{n\in\mathbb{N}\mid n\neq S(n)\}.$$
By Peano’s third axiom, $0\in T$ (since $0$ is not a successor, so $0\neq S(0))$.

Assume $k\in T$. Then $k\neq S(k)$. By Peano’s fourth axioms, this implies $S(k)\neq S(S(k))$; hence $S(k)\in T$.

By Peano’s Fifth Axiom, $T=\mathbb{N}$.  In particular, $S(a)\neq a$ for all $a\in\mathbb{N}$.

Similarly, if $S(a)=b$ and $S(b)=a$, then $S(S(a))=a$. Consider
$$T=\{a\in\mathbb{N}\mid a\neq S(S(a))\}.$$
Again, $0\in T$, since $0$ is not a successor. Assume $k\in T$. Then $k\neq S(S(k))$, hence $S(k)\neq S(S(S(k))$, so $S(k)\in T$. Thus, $T=\mathbb{N}$, and you cannot have the two elements you propose.

You can similarly verify that there cannot exist elements $a_1,\ldots,a_m$ with $S(a_i) = a_{i+1}$ for $i=1,\ldots,m-1$, and $S(a_m)=a_1$.


问:
If $f\colon G\to H$ is a surjective homomorphism, then $|C_G(g)| \geq |C_H(f(g))|$

Let $G$ be finite, $f\colon G\to H$  be a surjective homomorphism (hence $H$ is finite) and $g \in G$. Prove the order of center of $g$ in $G$ is greater than or equal to the order of the center of $f(g)$ in $H$, i.e. $$|C_G(g)| \ge |C_H(f(g))|.$$

Attempts at solution.

If $f$ is an isomorphism this is clear. So if $\ker f=K$ we can consider $G \to G/K \xrightarrow{f'} H$, where $f'$ is the induced map. So we can reduce this problem to the case where $f$ is a projection map.

Also if $f(x)f(g) = f(g)f(x)$ then $f(x^{-1}g^{-1}xg) = e$.

So if $h = f(y)$, $h \in C_H(f(g))$ iff $[h,g] \in K$.

The center of $g$ is the set of fixed points of the conjugation map (by $g$). And the center of $h$ is the set of fixed points under conjugation by $h$. So maybe this can be done by considering how the orbits collapse mod a normal subgroup?

答:
An alternative way by character theory:

Lifting all the irreducible characters of $H$ to a possibly proper subset of irreducible characters of $G$ and using column orthogonality relations for $g$ and $f(g)$, we have the result.

So, the equality holds if and only if the other irreducible characters of $G$ has value $0$ at $g$.


问:
Finding formula from given table
I'm sucks at finding formula, can you guys help me? As we can see below, there is a table with x, y, z, and i values. My question, what is formula to get the value of i, if the values of x, y, z are given?

x
y
z
i

1
275
50
137,50

2
275
50
206,25

5
275
50
247,50

I thought the formula as simple as

y-(y*(z/100))

but the formula only apply for first row data, looks like it should involve x value somewhere,  I don't know how

答:
I think your math teacher was trying to be too clever. The simplest formula is just:

$137.50 * (1+\frac{x-1}{x})$

No need to use $y$ or $z$.

Title: Finding formula from given table


问:
$D_n = \{x\in\mathbb R: F(x) - F(x^-) \ge 1/n\}$ is finite for $n\in\mathbb N$ if $F$ is a distribution function

Let $D$ be the set of discontinuity points of a distribution function $F$. $D_n = \{x\in\mathbb R: F(x) - F(x^-) \ge 1/n\}$ for $n\in\mathbb N$. Show that each $D_n$ is finite. Hence, show that a distribution function cannot have an uncountable number of discontinuities.

So, I know that a distribution function $F$ must be such that:

$F$ is an increasing function on $\mathbb R$.

$\lim_{t\to\infty} F(t) = 1$ and $\lim_{t\to -\infty} F(t) = 0$

$F$ is right continuous.

If $F$ has no jump discontinuities in $[a,b]$, then $F(a) = F(b)$.

How do I show the required statement now? I am trying to construct a proof by contradiction. Suppose $D_m$ is not finite for some $m\in\mathbb N$. What now?

I was able to do the second part of the proof anyway, i.e. if the $D_n$'s are all finite then $D$ is countable. Here's how: $$D = \{a\in\mathbb R: F(a) - F(a^-) > 0\} = \bigcup_{n=1}^\infty D_n$$

Assuming that all $D_n$'s are finite, hence countable - we see that $D$ is countable (a countable union of countable sets is countable).

Thanks!

答: If $D_m$ is infinite for some $m$ your function would jump up by at least $1/m$ infinitely many times so it would increase to infinity, violating the fact that the limit at infinity is $1$.


问:
Is the order of a quotient of the multiplicative group of integers a prime power if the group is cyclic?
I am looking at the multiplicative group of integers modulo $n$, denoted $\mathbb{Z}_n^*$, and its subgroups/quotient groups. For a number $p \in \mathbb{Z}$ we denote $\bar{p} \in \mathbb{Z}_n^*$. We are given that $\mathbb{Z}_n^*/H$ is cyclic, where $H=\langle \bar{p} \rangle$ for a prime number $p$ which does not divide $n$. Say $|\mathbb{Z}_n^*|=l\cdot d$ where $|H|=d$. What can we say about the number $l$?

I know if $n = p_1^{k_1}\dots p_r^{k_r}$ then $\mathbb{Z}_n^* \simeq \mathbb{Z}_{p_1^{k_1}}^*\times \dots \times \mathbb{Z}_{p_1^{k_r}}^*$. For odd primes each of these factors are cyclic (as well as for $p_i=2, k_i = 0,1,2$). We know that $l$ isn't necessarily prime, since we could get $n=p_1p_2^k, k>1$ and that $|H|=p_1$, but can $l$ be something else than a prime power? Maybe it is easier to prove that $d$ and $l$ are coprime?

答: No. The size of the quotient need not be a prime power. Consider the field $\Bbb F_{31}:=\frac{\Bbb Z}{31\Bbb Z}$ and denote by $\mathbf x$ the residue class $x+31\Bbb Z$. As presecribed in the OP the prime $2$ is a representative of the residue class $\mathbf 2$ but $$\mathbf 2^5=\mathbf{32}=\mathbf 1\;\;\;\;\;\therefore\;|\langle\mathbf 2\rangle_{*}|=5$$ where $\Bbb F_{31}^*:=\Bbb F_{31}\setminus\{\mathbf 0\}$ is the $\Bbb F_{31}$cyclic units group of size $30$ and $\langle\mathbf 2\rangle_*$ is the $\Bbb F_{31}^*$ subgroup generated by $\mathbf 2$. However, $|\frac{\Bbb F_{31}^*}{\langle\mathbf 2\rangle_{*}}|=\frac{30}{5}=6$ but $6$ is not a prime power.


问:
Contest problem involving circles geometry

Consider a circle with center $O$. the chord $AB$ divide the circle
into two areas $R_1$ and $R_2$ ( $AB$ is not diameter). suppose $S_1$
be a circle which its center is inside $R_1$ and be tangent to $AB$
and be tangent inside $S$. also suppose $S_2$ be a circle which its
center is inside $R_2$ and be tangent to $AB$ and be tangent inside
$S$. we know $S_2$ passes through $O$ and $X$ is contact point of $AB$
and $S_1$. suppose $O$ and $X$ and center of $S_2$ lie on a straight
line. if radius of $S_2$ be equal to $100$. what is the radius of
$S_1$?

$1)30\sqrt3\quad\quad2)20(1+\sqrt3)\quad\quad3)60\quad\quad4)72\quad\quad5)\text{none}$

I don't know how to approach this problem. as first step I should draw this but I'm not sure how to do it ( I can't draw a circle passes through $O$ and be tangent to $AB$ and also tangent inside the circle $S$ and have $O$ and $X$ and center of this circle on a straight line) . any help will be appreciated

答: I believe that this is a correct drawing of the problem.  I added a second figure showing another depiction of these geometric constraints.  It appears that $S_1$ is under-constrained.


问:
$f(x,y)=\frac12y^2e^{−x}$ if $ 0<y<x<∞$ Find $P(Y<1\mid X=3)$
Let $(X,Y)$ be the probability density function:

$$f(x,y)= \begin{cases} \dfrac{1}{2}y^2e^{-x} & \text{if } 0<y<x<\infty, \\[8pt] 0 & \text{otherwise.} \end{cases}$$

Find $P(Y<1\mid X=3)$

Which is $\dfrac{P(Y<1 \cap X=3)}{P(X=3)}$

I have read that probability at a particular point for a continuous distribution is $0$. So the denominator is $0$  right?

Reference

答:
The intended approach is probably to compute the conditional density of $Y$ given $X=3$ using the formula $f_{Y \mid X=x}(y) = \frac{f_{X,Y}(x,y)}{f_X(x)}$ where $f_X(x) = \int_{-\infty}^\infty f_{X,Y}(x,y) \, dy$, and then compute $P(Y < 1 \mid X=3)$ as $\int_{-\infty}^1 f_{Y \mid X=3}(y) \, dy$.

You've touched on an interesting technical point about conditioning on an event with zero probability. Using the usual definition of conditional probability fails because of division by zero. This Wikipedia section discusses this a little bit with an approximation argument that works for joint densities, but is problematic more generally; see regular conditional probabilities for the general way this is usually handled. But all this is probably out of scope of your class if this is an introductory class to probability; you are likely only expected to know how to handle the case with a joint density as discussed above.


问:
Assuming that the ball bounces indefinitely, find the total distance that the ball travels.
A ball has the property that each time it falls a distance onto a hard, level surface it rebounds to a height $rh$ where $0< r <1$. Suppose that the ball is dropped from an initial height of $H$ meters. Assuming that the ball bounces indefinitely, find the total distance that the ball travels.
Write your answer both with sigma notation and as a closed formula

This is the work I've done to reach sigma notation and the closed formula:

$a_{1}=H$

$a_{2}=2Hr$

$a_{3}=2Hr^{2}$

$a_{4}=2Hr^{3}$

$a_{n}=2Hr^{n-1}$

$s_{n}=H+2Hr+2Hr^{2}+...+2Hr^{n-1}$

Is the sigma notation:

$H+\sum_{n=2}^{\infty}2Hr^{n-1}=2Hr+2Hr^{2}+2Hr^{3}+...+2Hr^{n-1}$

geometric series: $\frac{a_{1}}{1-r}$

$a = 2Hr$

$r = r$

$H + 2Hr$

$H + \frac{2Hr}{1-r}$

$\frac{H(1-r)+2Hr)}{1-r}$

$\frac{H-rH+2Hr}{1-r}$

$H\frac{1-r+2r}{1-r}$

Closed form:
$Distance = H(\frac{1+r}{1-r})$

Does this all look correct?

答:
It seems your difficulty arises because you are doing the parts of the problem in the wrong order. Sigma notation is supposed to be the easy part of the problem, the part that you can do as soon as you understand how to write the $n$th term of the sum,
before you figure out what the closed form of the sum is.

If you can write

$$ s_{n}=H+2Hr+2Hr^{2}+...+2Hr^{n-1} $$

then you should be able to write a $\sum$ notation relevant to the question.
(The first term doesn't fit the right pattern for the $\sum$ notation,
but based on your later work I think you know this and you know how to deal with it.)

The equation
$$ar^{n-1} \stackrel?= \frac{a}{1-r}$$
is plainly and clearly false; just try a few different choices of numbers,
for example $a = 1,$ $r = \frac12,$ $n = 4$ and see how wrong it is.
Remember that when you write $ar^{n-1}$ you are saying to just take $r$ to the $n-1$ power, multiply by $a,$ and then stop and that's your answer.

If you wrote the left side of the equation with $\frac{a}{1-r}$ correctly, it would already be in $\sum$ notation.

Forget about everything you wrote after $\frac{a}{1-r}$ and just try to write the summations for these two places. You might have to look up the  $\frac{a}{1-r}$ to remind yourself this time.


问:
How to find the limit of the following question when x approaches 0 from positive side?
\begin{equation}
\lim _{x \rightarrow 0^{+}}(\sin 2 x)^{\tan 2 x}
\end{equation}

答:
Letting $u = x$:

$$L = \lim_{x\to 0^{+}}\sin(2x)^{\tan(2x)} = \lim_{u\to 0^{+}}\sin(u)^{\tan(u)}$$

Then, we have, noting that $\sin(u)^{\tan(u)} > 0$ for $u > 0$:

$$\ln(L) = \ln\big(\lim_{u\to 0^{+}}\sin(u)^{\tan(u)}\big) =\lim_{u\to 0^{+}}\ln(\sin(u)^{\tan(u)}) =\lim_{u\to 0^{+}}\frac{\ln(\sin(u))}{\cot(u)}$$

Using l'Hopital:

$$\ln(L) = \lim_{u\to 0^{+}}\frac{\frac{d}{du}\ln(\sin(u))}{\frac{d}{du}\cot(u)} = \lim_{u\to 0^{+}}\frac{\cot(u)}{-\csc^{2}(u)} = \lim_{u\to 0^{+}}-\cos(u)\sin(u) = 0$$

Thus, $L = e^{0} = \boxed{1}$

You can also solve this by noting that $\sin(u)\approx \tan(u)\approx u$ for $u\approx 0$, and from there finding $\displaystyle\lim_{u\to 0^{+}}u^{u} = 1$.


答:
Around $0^+$, $\sin(2x)>0$, therefore we can write
$$ \sin(2x)^{\tan(2x)}=\exp(\tan(2x)\log\sin(2x)) $$
But $\log\sin(2x)=\log\left(2x+o(x^2)\right)=\log(2x)+\log(1+o(x))=\log(2x)+o(x)=\log(2x)+o(\log x)$, therefore $\log(\sin(2x))\sim\log(2x)$. Moreover, $\tan(2x)\sim 2x$ thus $\tan(2x)\log\sin(2x)\sim2x\log(2x)\underset{x\rightarrow 0^+}\longrightarrow 0$ and finally, $\lim\limits_{x\rightarrow 0^+}\sin(2x)^{\tan(2x)}=1$.

Title: How to find the limit of the following question when x approaches 0 from positive side?

Body:
\begin{equation}
\lim _{x \rightarrow 0^{+}}(\sin 2 x)^{\tan 2 x}
\end{equation}


问:
Prove $\int_0^\infty\frac{\ln x}{x^3-1}\mathrm{d}x=\frac{4\pi^2}{27}$
Proof of the integral
$$\int_0^\infty\frac{\ln x}{x^3-1}\mathrm{d}x=\frac{4\pi^2}{27}$$
I try to substitute $u = \ln x$. Then $x = e^u,\>\mathrm{d}x = e^u\mathrm{d}u$ and the limits $(0,\infty)\to (-\infty,\infty)$.
The integral becomes $$\int_{-\infty}^\infty \frac{ue^u}{e^{3u}-1}\mathrm{d}u.$$

答:
Split the integral to simplify as follows

\begin{align}\int_0^\infty\frac{\ln x}{x^3-1}{d}x
= &\int_0^1\frac{\ln x}{x^3-1}{d}x + \int_1^\infty\overset{x\to \frac1x}{\frac{\ln x}{x^3-1}}{d}x
=\int_0^1\frac{(1+x)\ln x}{x^3-1}{d}x \\
= &\int_0^1\frac{(x^2+x+1)\ln x}{x^3-1}{d}x
 - \int_0^1\overset{x^3\to x}{\frac{x^2\ln x}{x^3-1}}{d}x\\
=&\int_0^1\frac{\ln x}{x-1}{d}x-\frac19 \int_0^1\frac{\ln x}{x-1}{d}x
=\frac89 \int_0^1\frac{\ln x}{x-1}{d}x\\
=&\frac89\cdot \frac{\pi^2}6
=\frac{4\pi^2}{27}
\end{align}

Integrate $\int_0^1\frac{\ln x}{x-1}{d}x=\frac{\pi^2}6$


问:
Prove that $QBYZ$ is cyclic
Circles $\omega_1$ and $\omega_2$intersect each other at points $A$ and $B$. Point $C$ lies on
the tangent line from $A$ to $\omega_1$ such that $\angle ABC = 90^{°}$. Arbitrary line
passes through $C$ and cuts $\omega_2$ at points $P$ and $Q$. Lines $AP$ and $AQ$ cut
$\omega_1$ for the second time at points $X$ and $Z$ respectively. Let $Y$ be the foot
of altitude from $A$ to $PQ$.

Prove that $QBYZ$ is cyclic.

I proved that $AYBC$ and $PYBX$ are cyclic but I have no idea how to prove that $QBYZ$ is cyclic.

答:
$ \angle 1 = \angle 2$

$ \angle 2 = \angle 3$

$ \angle 3 =\angle 4$  because AYBC is also cyclic.

Title: Prove that $QBYZ$ is cyclic


问:
How do you respond to "I was always bad at math"?
Here in the U.S., it is my experience that over 75% of adults I meet socially will volunteer that phrase or a variation upon learning that I am a mathematician.  I find this frustrating, since almost nobody would  brag about being bad at history or English or really any other subject.   Normally I change the subject, or perhaps say they must have had a bad classroom experience, but that feels unsatisfying and turns the conversation in an ugly direction.

Surely some of the bright minds here on Math.SE have found a good response. Note that I'm not looking to  alienate a new acquaintance, so let's keep it positive.

答:
I was never good at singing till a stranger at a wedding turned to my wife after the service and asked her "who was that bloke with the beautiful voice standing next to you?"

I once helped a person with their last chance at GCSE (high school) - they needed to pass to become a teacher. Looking at their work, they could do what they needed - indeed, over the three exams they'd failed, they'd shown that they could answer every question they'd been asked. They got a 'B' when they needed a 'C' (this was a fortnight before the exam - so really last chance saloon). I said "bottle that feeling and give it your your pupils".

People have been educated to believe they are not capable - most are. The most powerful thing is to catch them doing it, because that has the potential to break the spell.


问:
$3^n+4^n<5^n$ for all $n>2$
I'm doing the following induction proof and wanted to know if this was valid. I think it is, but I'm seeing more complicated solutions than what I did. What I did seems much easier.

Prove that $3^n+4^n<5^n$ for all $n>2$.

When $n=3$ we get $91<125$. No problem, now assume the result is true from $k<n$, $(5^k>3^k+4^k)$ and consider $5^{k+1}=5 \times 5^k>5(3^k+4^k)=5\times 3^k + 5\times 4^k>3\times 3^k+4\times 4^k=3^{k+1}+4^{k+1}$ since $5\times 3^k>3\times 3^k$ and $5\times 4^k> 4\times 4^k$.

Title: $3^n+4^n<5^n$ for all $n>2$

Body:
Here's a method I like
for showing
$f(n) > g(n)$
for $n \ge n_0$.

Show that
$f(n_0) > g(n_0)$.

Show that,
if $n \ge n_0$
and $f(n) > g(n)$ then
$f(n+1)-f(n)
\ge g(n+1)-g(n)
$.

Then $f(n) > g(n)$
for $n \ge n-0$.

In this case,
$f(n) = 5^n,
g(n) = 3^n+4^n$.

Choose $n_0 = 3$.
$f(n_0) = 5^3=125,
g(n_0) = 3^3+4^3 = 91
\lt f(n_0)$.
So (1) holds.

If $n \ge 3$
and $f(n) > g(n)$ then

$f(n+1)-f(n)
=5^{n+1}-5^n
=4\cdot 5^n
\gt 4(3^n+4^n),\\
g(n+1)-g(n)
=2\cdot 3^n+3\cdot 4^n
\lt 3(3^n+4^n)
\lt f(n+1)-f(n)
$.


问:
How to prove that for any two natural numbers a, b, there exist natural numbers c,d with a + d = b +c
I understand the intuition for how to do this, but I don't know how to formally prove it. For example, I know that if you take (a,b) to be (1,2), then a sample (c,d) could be (1,2) since 1 + 2 = 2 + 1. But how do I formally prove that such numbers c,d exist for an arbitrary c,d such that a,b,c,d are all in the naturals.

Thanks in advance!

答:
You need to be careful with the statement that you are trying to prove.  As stated in the title of your question it is false because $c,d$ are chosen before $a,b$.  You need $c,d$ to be chosen after $a,b$, so it should be that for any two natural numbers $a,b$ there exists a pair $c,d$ such that $a+d=b+c$.  Having stated it correctly, a good way to prove it is an algorithm to find a $c,d$ pair.  One such (assuming you accept $0$ as a natural) is

If $a=b, c=d=0$

Else if $a \gt b, d=0, c=a-b$

Else $c=0, d=b-a$


问:
Regular and non regular parametrizations
A proposition from Pressley’s book says, ‘Any reparameterization of a regular curve is regular’ and then gives an example to show that a curve can have both regular and non regular reparameterizations. How does one reconcile this? Does this mean that finding a parameterization is not the same as doing a reparameterization? My question is for some given curve suppose we find two parameterizations $\gamma_1$ and $\gamma_2$, is there always a reparameterization map between these two?
He gives two parameterizations for parabola, $(t,t^2)$ and $(t^3,t^6)$. Is there a reparameterization map between these? If so, doesn’t this contradict the proposition?

答:
This is probably not going to answer your question, but I am going to focus in on: My question is for some given curve suppose we find two parameterizations $\gamma_1$ and $\gamma_2$, is there always a reparameterization map between these two?

I would argue the answer is yes if they are both monotonic (and continuous--it doesn't need to be monotonic to be one-to-one if it's discontinous) and no, in general. Very simple example, let's take the line $y = x$ for $x \in \mathbb{R}: x \geq 0$. An obvious parameterization is $(x, y) \mapsto (t, t)$ (obviously $t\in\mathbb{R}: t \geq 0$). What about $(x, y) \mapsto \left(t^2, t^2\right)$ where $t \in \mathbb{R}$? Surely both parameterize the original curve. However, there isn't a clear map between the two.

OK, so that one is easily resolved by limiting the $\left(t^2, t^2\right)$ to $t \in \mathbb{R}: t \geq 0$, but what about $(x, y) \mapsto \left(2t^3 - 9t^2 + 12t, 2t^3 - 9t^2 + 12t\right)$ where $t \in \mathbb{R}: t \geq 0$? This is not a one-to-one function within the domain $t \geq 0$ (it has two local extrema at $t = 1$ and $t = 2$, respectively):

So while both parameterize the curve $y = x$, we cannot recover the non-monotonic parameterization ($2t^3 - 9t^2 + 12t$) from the monotonic one ($t$)--literally because $2t^3 - 9t^2 + 12t$ isn't one-to-one.