问:
Differentiability Implies Continuity (Multivariable Calculus)
I am reading Hubbard and Hubbard's Vector Calculus, Linear Algebra, and Differential Forms, and in it they provide the following theorem and proof. My issue is not with the theorem, but more so with the proof.
Note that, in the theorem below, Equation 1.7.20 refers to the basic definition of the (multivariable) derivative, and $L$ is the linear transformation which defines the derivative.

In particular, they seem to implicitly be distributing the limit in order to make the desired conclusion. That is, they seem to be using the theorem below (bullet point 1).

But this theorem of course requires the hypothesis that both $f$ and $g$ have the given limit. Translating to our case in the proof of Prop. 1.7.11, use of Theorem 1.5.26 in the proof would require a priori knowledge that
$$ \lim_{\mathbf{h}\to \mathbf{0}}\mathbf{f}(\mathbf{a}+\mathbf{h})-\mathbf{f}(\mathbf{a})$$
exists (the second term in the limit of course exists, as justified). But how can we know this without resorting to the epsilon-delta definition of the limit? Indeed, that this limit exists (and is 0) is precisely what we are trying to prove. Thus, is this one of those instances where the authors intend for you to fill in the details, or is there really some way to deduce the result without resorting to epsilons and deltas (I imagine it's quite simple).

答: No, it is not an argument by contradiction. If$$\lim_{h\to0}\frac{f(a+h)-f(a)-L(h)}{\|h\|}=0,$$then$$\lim_{h\to0}\bigl(f(a+h)-f(a)-L(h)\bigr)=\lim_{h\to0}\|h\|.\frac{f(a+h)-f(a)-L(h)}{\|h\|}=0$$and therefore, since $\lim_{h\to0}L(h)=0$ (since $L$ is linear and continuous), we have$$\lim_{h\to0}f(a+h)=\lim_{h\to0}\bigl(f(a+h)-f(a)\bigr)+f(a)=0+f(a)=f(a).$$


问:
Is this derivation of $\frac{dy}{dx}$ correct?
Derivatives show us how fast something is changing at any point. For example; the gradient of the graph of $y = x^2$ at any point is twice the value of $x$ thereat. The process of finding the derivation of a gradient / slope of a function $y=f(x)$ or $y = x^2$ is as follow.

Pick any two points $A$ and $B$ close to each other on the curve of $y =x^2$. The coordinates of $A$ on the curve are $(x, y)$ or $(x, x^2)$. Add $Δx$ at $A$ as usual. When $x$ increases by $Δx$, then $y$ increases by $Δy$. The $x$ changes from $x$ to $(x +Δx)$ while $y$ changes from $y$ to $(y + Δy)$ or $f(x)$ to $(x+Δx)^2$. Thus the $x$ and $y$ coordinates of $B$ on the curve are $(x + Δx, y + Δy)$ or $([x+Δx, (x+Δx)^2].$ Now the instantaneous rate of change is given by

$$\frac{Δy}{Δx} = \frac{[(x + Δx)^2 – x^2]}{[x + Δx - x]}$$

$$\frac{Δy}{Δx} = \frac{[x^2 + Δx^2+2xΔx − x^2]}{ Δx}$$

$$\frac{Δy}{Δx} = \frac{[2x + Δx]}{ 1}$$

Reduce $Δx$ close to zero by taking limit ($Δx$ to $dx$ and $Δy$ to $dy$)

$$\frac{Δy}{Δx} = 2x + dx$$

$$\frac{Δy}{Δx} = 2x \tag1\label{eq1} $$
OR

$$dy = 2x.dx \tag2\label{eq2}$$

ABC is an infinitesimal triangle made by $dx, dy$, and hypotenuse or slope of tangent where point $A$ and $C$ are always on the curve. Length of $AB$ = Base = $dx$, Length of $BC$ = Perpendicular= $dy$ and Length of Hypotenuse = $AC$. $\angle CAB$ or $\angle BAC$ is the slope of a tangent

According to the aforementioned $\eqref{eq1}$ or $\eqref{eq2}$

•$\frac{dy}{dx}$ is directly proportional to $x$ or $\angle CAB$ is directly proportional to $x$.

• $dx$ is indirectly proportional to $x$ OR $x$ is inversely proportional to $dx$.

• $dy$ is directly proportional to $x.dx$ or $dx$

The length of $dx > dy$ when $\angle CAB < 45$ degrees The length of $dx = dy$ when $\angle CAB = 45$ degrees The length of $dx < dy$ when $\angle CAB > 45$ degrees.

The proportionality of both the $\angle CAB$ and $dy$ with $x$ are in contradiction with the proportionality of $x$ and $dx$ in the triangle $ABC$ after probing the equation of $\frac{dy}{dx} = 2x$ beyond its derivation on a graph of $y = x^2$. When $x$ increases; $dx$ decreases, $dy$ increases, and $\angle CAB$ increases. This means $AC$ also increases and ultimately SECANT when $x$ increases. Our goal is to bring $dx, dy$ and $AC$ to zero (not away from zero either positively or negatively - Point $C$ has to be on the curve) or secant to tangent by reducing them close to zero but here $dx$ heads toward zero but $dy$ and $AC$ increases when $x $ increases on axis mathematically.

Although the difference in the length of $dx$ and $dy$ can be noticeable clearly on the graph if we examine the triangle $ABC$ at two different points for a gradient ($\frac{dy}{dx}$), say when an $\angle BAC = 0.1$ degrees and 89.9 degrees on the curve but UNIT CIRCLE is the best example for observing the change in an $\angle CAB$ (say 0.1 and 89.9 degrees) of a triangle $ABC$ for $dy$ and $dx$ and the comparison of their lengths.

$RISE = dy = 2x$ and $RUN = dx = 1$ (always constant) in a GRADIENT of 1 in $2x$ which we obtained from the $\eqref{eq1}$ of $\frac{dy}{dx}=\frac{2x}{1}$ at any point on the curve when there is no difference between secant and tangent – No idea how do we get $\frac{dy}{dx} = 2x.dx$ but above said contradiction may be due to the introduction of another curve of $y =(x+dx)^2$ at a point where we seize $x$ or $y=x^2$  deliberately and introduce delta $x$ OR when function $y = f(x)$ changed to $y=f(x+Δx)^2$. The value of $x$ has reached to its maximum value instead of unlimited when a curve $y=x^2$ doesn’t continue anymore at a point where we introduce delta $x$ or $dx$ as $y=x^2$ and $y =(x+dx)^2$ are two different types of curve (two diffrent functions).

Further, integration is the reverse process of differentiation. Although delta $x$ or $dx$ is ignored during the process of derivation of $\frac{dy}{dx}$ because of their small values but we can’t ignore them in the process of integration which makes a lot of difference in summation. They can’t be disappeared forever and should resurface during the process of integration or summation.

Similarly, $dy$ is the small vertical change in $y$, therefore, we take the sum of all the small vertical lengths [dy(s)] not the whole slice or y-coordinate(s) from zero to its value on the curve when we integrate both sides of the equation of $dy = 2xdx$ but it turns into function of $x^2$ or area under the graph – no idea how but summation of vertical lengths on a graph gives vertical length only not curve?

The derivation of the natural relationship of a gradient of 1 in $2x$ at any point with $y=x^2$ or $\frac{dy}{dx}=2x$ is still unbeknownst to illuminates - Anyone who is in agreement with all above unless satisfied by logic.

Below figure may help with the above.

Image

答: As $\dfrac{\Delta y}{\Delta x}\approx\dfrac{dy}{dx}=2x$, for small $x$, $\Delta x\gg\Delta y$ and for large $x$, $\Delta x\ll\Delta y$. Really nothing weird.

ContentLicense: CC BY-SA 4.0


问:
Differentiation Operator a Contraction Mapping
Let $C^{\infty}[a,b]$ be the space of all infinitely differentiable functions on [a,b] with norm $$ || f || = \max _{[0,1]} | f(x) | , f \in C^{\infty}[a,b]$$

Is the differentiation operator $\frac{d}{dx}$ a contraction mapping on $C^{\infty}[a,b]$?

I'm confused. Operators are not until later in the textbook, and the contraction mappings I've worked with are using the Mean Value Theorem which are referencing a derivative? Ah!

答:
It's not a contraction.

Let $f(x)=\sin nx$, $g(x)=\cos nx$ ($n$ is a number we need to find). Then $f,g\in C^{\infty}[a,b]$. And
\begin{align*}
\left\|f-g\right\|=&\underset{[a,b]}{\max}|\sin nx-\cos nx|=\sqrt{2}\underset{[a,b]}{\max}|\sin(nx-\frac{\pi}{4})|
\\
\left\|\frac{df}{dx}-\frac{dg}{dx}\right\|=&\left\|n\cos nx+n\sin nx\right\|=n\underset{[a,b]}{\max}|\cos nx+\sin nx|=\sqrt{2}n\underset{[a,b]}{\max}|\sin(nx+\frac{\pi}{4})|
\end{align*}
If we take $n$ large enough that $n\ge\frac{2\pi}{b-a}$, then
\begin{align*}
\left\|f-g\right\|=\sqrt{2},\qquad\left\|\frac{df}{dx}-\frac{dg}{dx}\right\|=\sqrt{2}n.
\end{align*}
If we also make $n\ge1$,
$$\left\|\frac{df}{dx}-\frac{dg}{dx}\right\|\ge\left\|f-g\right\|.$$
Therefore the operator $d/dx$ is not contraction.

ContentLicense: CC BY-SA 3.0


问:
Finding appropriate values for $a$ and $b$ in bisection method
When you use the bisection method to find the roots of a function $f(x)$, you have to start with values $a$ and $b$ such that $f(a)$ and $f(b)$ have different signs- so (by the intermediate value theorem) you can find a root between $a$ and $b$. Only thing is... I'm automating the process with some code, and the code is supposed to be very general- so any combinations of regular algebraic arithmetic, trigonometry, logarithms, and hyperbolic functions.

This makes it pretty hard to estimate $a$ and $b$ on the spot such that they have opposite signs. Is there any way I can do this (without randomizing $a$ and $b$ and then running values through the same function to hone them down until they have separate signs)?

Thanks

Edit: I'm using an algorithm called Lehmer-Schur which is generalized to complex values using disks rather than a specific interval. I realize that it would be a lot simpler to pick really big bounds or try using some kind of randomizer, but this seems really crude and I was hoping there'd be a better solution. This is a big ask since (within the restrictions on composition mentioned above) there are no limits on the complexity of the function- so I understand that this may not be possible at all. I'm just keen to know what other people are doing or if there are any smart solutions.

答:
There is no issue with choosing very large brackets for a function, because the bracket will close in on the root very rapidly. To accelerate initial convergence to find the order of magnitude of the root, see here and here. The given routines will guarantee convergence in roughly 70 iterations over the largest finite bracket using doubles (somewhat useless if you don't care about relative accuracy).

If the values at infinity give the same sign (i.e. $f(\infty)=f(-\infty)=\pm\infty$), then you can try using optimization methods on that bracket to attempt to find how close it gets to the opposing infinity (i.e. $f(x)\to\mp\infty$). If there is a sign change, then you have two brackets (from the relative extrema to each infinity), and if there is no sign change then you really have no guarantee on the existence of a root. You may also wish to consider the case when $|f(x_\text{relative-extrema})|$ is very small as a double (or higher) root.

If the root is expected to be simple, then superlinear convergence may be obtained by using better bracketing methods, such as Chandrupatla's method. Feel free to look into the source code on GitHub and take out the portions that fit your needs.


问:
Am I determining the Image correctly?
basically we have

$f: \mathbb{Z} \rightarrow\mathbb{N_0} $
with
$f (n) = \left\{
\begin{array}{ll}
-2n & n \leq 0 \\
2n-1 & \, \textrm{n > 0} \\
\end{array}
\right. $

And I should determine $Im(2\mathbb{Z})$ the image of f under $2\mathbb{Z}$ with $2\mathbb{Z} := \{2x | x\in \mathbb{Z}\}$.

(case 1) $n \leq 0$

We put in numbers like $\{...,-8,-6,-4,-2,0 \}$ So $f(-8) = 16, f(-6) = 12, f(-4) = 8, f(-2) = 4, f(0) = 0$

So we have $Im(2\mathbb{Z}) = \{...,16,12,8,4,0\}$ in the case $n \leq 0$

You can see that this set is basically multiplies of $4$ so it can be written as $4n$.

(case 2) $n > 0$

We put in numbers like $\{2,4,6,8,... \}$ So $f(2) = 3, f(4) = 7, f(6) = 11, f(8) = 15$

So we have $Im(2\mathbb{Z}) = \{3,7,11,15,...\}$ in the case $n > 0$

You can see that this set is basically multiplies of $4$ subtracted by 1 so it can be written as $4n-1$.

Can we conclude that $Im(2\mathbb{Z}) = \{  4n \lor 4n-1 \forall n \in \mathbb{N_0}\}$ ?

答:
We want to proof equality of sets. You can seperate cases. Then for $n>0$ you have $2\mathbb{Z}=2\mathbb{N}$, and for $n\leq 0$ we have $-2\mathbb{Z}=\mathbb{N}_0$. So in the first case, try to proof $f(2\mathbb{N})=\{4n-1;n\geq 1\}$. For equality of sets you have to proof "$\subseteq$" and "$\supseteq$".

So lets proof $f(2\mathbb{N})\subseteq \{4n-1;n\geq 1\}$.

For that we have to proof that every $m\in f(2\mathbb{N})$ has the form $4n-1$ where $n$ is a natural number strict greater $0$.

We have $m\in f(2\mathbb{N})$. So there is an $x\in2\mathbb{N}$ with $f(x)=m$. So $x$ is even, and we can write $x=2i$. Furthermore we know that $2i>0$. So $f(2i)=2(2i)-1=4i-1$.

Which is exactly what we had to show. So for $i=n$, we have proved that $m=4n-1$.

The other set equality can then be proved similar.
Also there should be a more direct proof, but splitting it up like this might be helpful at the beginning.

This might look intimidating, as this is a little bit "back and forth", but it is kinda clear what we have to do in each step.

Now try and proof $\supseteq$ similar to the given proof.

Start with: Let $4n-1\in\{4n-1; n\geq 1\}$. Then we have to find a $m\in2\mathbb{N}$ with $f(m)=4n-1$.

Title: Am I determining the Image correctly?


问:
Locally compact space $X$, $C_b(X)$ is reflexive if and only if $X$ is finite.
Taken From Conway's A Course in Functional Analysis Chapter 3 Section 11 Problem 2

Problem Statement: Show that for a locally compact space $X$, $C_b(X)$ is reflexive if and only if $X$ is finite.

$C_b(X)$ is the set of all continuous functions $f: X \rightarrow \mathbb{F}$ such that $||f|| = \sup\{|f(x)|: x \in X\}< \infty$. However, from here I'm not sure how to prove this is reflexive.

Any tips would be greatly appreciated!

答: Assuming that $X$ is infinite (and Hausdorff), let $\{V_n\}_n$ be a family of pairwise disjoint, non-empty open subsets of $X$.  For each $n$, choose a continuous function $f_n$ of norm one, vanishing off $V_n$.  Next show that the closed subspace generated by the $f_n$ is isometrically isomorphic to $c_0$.  Since $c_0$ is not reflexive, neither is $X$.


问:
Finitely additive measure on $\mathbb R$
Suppose $\mathcal B$ is the Borel $\sigma$-algebra on $\mathbb R$. Let $\mu : \mathcal B \rightarrow [0, \infty ]$ be a finitely additive(but not necessarily countably additive), translation-invariant, ''measure'', with the property that $\mu(K) < \infty$ for every compact $K \subset \mathbb R $.

Let $P(\mathbb R)$ denote the power set of $\mathbb R$.

Does there exist a finitely additive, translation-invariant "measure", $\tilde \mu : P(\mathbb R) \rightarrow [0, \infty ]$, extending $\mu$ ?

In particular, let $\mu$ be the Lebesgue measure on the Borel $\sigma$-algebra. Can something be said about this special case at least?

答: I don't know about the general case. But here Tao states that the Lebesgue case is true, using the Hahn-Banach theorem. </p>


问:
Are all commutative rings with identity Noetherian?
According to the definition of Noetherian ring, a ring is said to be Noetherian if all increasing chains of ideals eventually stabilize.

Now, let $R$ be a commutative ring with identity and we take a collection $C$ of proper ideals of $R$. Consider an increasing chain of ideals
$$I_1\subseteq I_2\subseteq I_3\subseteq ....$$
Now, $I=\cup_{i}I_i$ is a maximal ideal, as any proper ideal containing $I$, is in $C$ and falls in the increasing chain, hence it falls inside $I$ and if, $I=R$, then $1\in I$, therefore, $1\in I_i$ for any $i$, which again contradicts the definition of $I_i$.

So, does all commutative rings with identity falls inside Noetherian rings?

答:
No: for example $F[x_1,x_2,\ldots]$ in countably many variables is commutative, has identity, and is not Noetherian.

There are holes in the reasoning you've given.  The first one is

Now, $I=\cup_{i}I_i$ is a maximal ideal

This does not follow, even if the ring was Noetherian. In a Noetherian ring, that chain is always going to stabilize, and so the union will be the place where it stabilizes. But there's no reason to expect it will be a maximal ideal.  The reason you gave: "Any ideal containing $\cup I_i$ falls in the increasing chain" does not hold. Consider the ring I gave above and the chain of ideals $(x_2)\subseteq (x_2,x_4)\subseteq (x_2,x_4,x_6)\subseteq \cdots$. The ideal $(x_1, x_2,x_3,x_4,\ldots)$ contains the union of the "even variable chain" but it is not one of the members of the chain, and it is not even equal to the union of the chain.

Secondly, even if the union is a maximal ideal, it does not help determine if the ring is Noetherian or not.  One can construct non Noetherian valuation rings for which the union of all proper ideals is indeed the maximal ideal, but the ring is not Noetherian.


问:
The characteristic of the field does not divide the dimension of an irreducible representation
Let $k$ be an algebraically closed field of characteristic $p$, and $G$ be a finite group whose order is not divisible by $p$. I would like to prove the following: if $V$ is an irreducible representation of $kG$, then $\dim V \neq 0$ in $k$, i.e. $p$ does not divide the dimension of $V$.

Here is one explanation of why it is true: for representations over $\mathbb{C}$, the dimension of a representation must divide the order of the group. Then, there is a claim that the dimensions of the irreducible representations of $G$ over $k$ are equal to the dimensions of the irreducible representations of $G$ over $\mathbb{C}$. (I do not know why this claim is true). This then implies that over $kG$, the dimension of any irreducible representation $V$ divides $|G|$, and hence is coprime to $p$.

Is there a shorter argument, which avoids the character theory over $\mathbb{C}$?

答:
The claim that the dimensions of irreducible representations over $k$ are the same as over $\mathbb{C}$ is not particularly hard. It follows from these three general facts about representations in characteristic p:

If an irreducible representation over $k$ lifts to characteristic zero, the lift is irreducible too.

Any representation of a finite group $G$ which is a projective object in the category of $G$ representations over $k$ can be lifted to characteristic zero.

If $p \nmid |G|$ representations of $G$ are semisimple, and therefore every object is projective.

If you insist on not using that fact you could use the general theory of blocks and defect groups in characteristic p, although I'm not sure that's an easier theory. Here are some general facts from that theory that imply what you want:

Each irreducible representation lies in a unique block.

Each block has an associated defect group, a p-subgroup of $G$.

A block contains a representation of dimension prime to $p$ iff the defect group is a Sylow subgroup.

Since $p \nmid |G|$, each irreducible representation forms its own block.

Probably neither of these explanations is quite as simple as you'd like, but given that the characteristic zero result is fairly nontrivial I'm somewhat doubtful a slick proof of this will be found.


问:
Find the greatest integer $k$ for which $1991^k$ divides $1990^{{1991}^{1992}}+1992^{{1991}^{1990}}$
Find the greatest integer $k$ for which $1991^k$ divides $$1990^{{1991}^{1992}}+1992^{{1991}^{1990}}$$

It is easy to see that $k \geq 1$ as $1990 \equiv -1$ and $1992 \equiv 1 \pmod{1991}$
Also, I thought that perhaps as $1991$ is the product of two distinct primes, it would be worth looking at small values of $(pq)^k||(pq-1)^{{pq}^{pq+1}}+(pq+1)^{{pq}^{pq-1}}$ for primes $p$ and $q$. Any help would be really appreciated.

答:
Let $a=1991$, and let $$b=(a-1)^{ a^{a+1} } + (a+1)^{ a^{a-1} }$$

The goal is to compute $\nu_{a}(b)$. To do so, we will evaluate $b$ modulo $a^{a+1}$.

Applying binomial expansions, $$b\equiv (-1)^{a^{a+1}} + 1 + a^{a-1} a\mod{a^{a+1}}$$ (because all the higher order terms vanish).

Since $a^{a+1}$ is odd, this simplifies to $$b \equiv a^a\mod{a^{a+1}}$$
Therefore, $\nu_{a}(b)=a$. To rephrase in the terminology of your question, $k=1991$.


问:
Struggling to find a solution to this improper integral
I am working on a probability exercice and I am trying to calculate E(Y) which comes down to this expression :

$$ E(Y) = \int_{-∞}^{+∞} y\frac{e^{-y}}{(1+e^{-y})^{2}} \, \mathrm{d}y $$

I tried to use integrals by part but it diverges and I can't find a good change of variables.

Any idea ?

答:
This density is an even function, and the first moment integrand is an odd function. The expected value is then zero.

If we denote $f(y)=\dfrac{e^{-y}}{(1+e^{-y})^2}$, you can see that
$$
f(-y)=\frac{e^y}{(1+e^y)^2} = \frac{e^{2y} \cdot e^{-y}}{e^{2y}(e^{-y}+1)} = f(y)
$$

and if you interpret the improper integral as
$$
\int_{\mathbb{R}} y f(y) dy = \lim_{a \to +\infty} \int_{-a}^a y f(y) dy
$$

you see that its value must be zero.


问:
Higher Ext vanish iff module is free
Let $\Lambda$ be a commutative ring with $1$. Let $M$ be a $\Lambda$-module. Is it true that $\mathrm{Ext}^i_\Lambda(M,\Lambda)=0$ for all $i\ge1$ if and only if $M$ is free over $\Lambda$?

The 'if' part of the question is clear: compute the Ext groups by a projective resolution of $M$. Since $M$ is free, we may choose the trivial resolution, which makes all Ext groups vanish.

The question is, does the converse hold?

答: The converse is false; if $M$ is any projective module, then all higher Ext groups vanish, for exactly the same reason (compute the Ext groups by the trivial projective resolution). In general, there are plenty of examples of projective modules that are not free. For example, take a non-principal ideal of a Dedekind domain.

Title: Higher Ext vanish iff module is free


问:
Can every theorem involving countable things necessarily be proven by induction?
Last night I was working on some problems to remember how proof by induction worked, and I began to wonder if it's possible to prove every theorem on countable things by induction.  I understand proof by induction may not be the most elegant or simple way to construct the proof.

If so, is there a proof of this?  If not, is there a counterexample?

BTW, the problem I was struggling with, but finally figured out, was to prove that:

$\sum_{i=1}^n i^3 = (\sum_{i=1}^ni)^2$

答: Nope. The Faulhaber formulas can be obtained by the method of indeterminate coefficients, leading to small linear systems of equations. (And I doubt that your concern has anything to do with countability.)</p>


答: Note that the integrand is an odd function$\big(f(y)=-f(-y)\big)$ $\to$which means that integral is $0$

Title: Struggling to find a solution to this improper integral

Body:
I am working on a probability exercice and I am trying to calculate E(Y) which comes down to this expression :

$$ E(Y) = \int_{-∞}^{+∞} y\frac{e^{-y}}{(1+e^{-y})^{2}} \, \mathrm{d}y $$

I tried to use integrals by part but it diverges and I can't find a good change of variables.

Any idea ?


答:
I shall prove a more general result: Let $n>1$ be an odd positive integer. Then $n^n \|[(n-1)^{n^{n+1}}+(n+1)^{n^{n-1}}]$.

Proof:
\begin{align}
(n-1)^{n^2}+(n+1)=\sum_{i=0}^{n^2}{\binom{n^2}{i}(-1)^{n^2-i}n^i}+(n+1) & \equiv \binom{n^2}{1}n-1+(n+1) \pmod{n^2}\\
& \equiv n \pmod{n^2}
\end{align}

Now applying Lifting the Exponent Lemma on each prime factor of $n$, we have $n^n \|[(n-1)^{n^{n+1}}+(n+1)^{n^{n-1}}]$.

For your special case, $k=1991$.

Title: Find the greatest integer $k$ for which $1991^k$ divides $1990^{{1991}^{1992}}+1992^{{1991}^{1990}}$

Body:
Find the greatest integer $k$ for which $1991^k$ divides $$1990^{{1991}^{1992}}+1992^{{1991}^{1990}}$$

It is easy to see that $k \geq 1$ as $1990 \equiv -1$ and $1992 \equiv 1 \pmod{1991}$
Also, I thought that perhaps as $1991$ is the product of two distinct primes, it would be worth looking at small values of $(pq)^k||(pq-1)^{{pq}^{pq+1}}+(pq+1)^{{pq}^{pq-1}}$ for primes $p$ and $q$. Any help would be really appreciated.


问:
What are the orders of a element $a + H$ of a quotient group $G/H$
I would like somebody to check my answers

$G = (\mathbb{Q, +}), H = (\mathbb{Z, +}), a = \frac{2}{3}$

I think the answer is $\infty$, because $a + H = \{\frac{2}{3} + h: h \in \mathbb{Z}\}$

$G = (\mathbb{Z_{12}, +_{12}}), H = \{0,3,6,9\}, a = 5$

$a +H  = \{5,8,11,2\}$, so the answer is 4.

答:
By definition, the order of the coset $aH$ in the quotient group $G/H$ (so, $H\unlhd G$) is the least positive integer $n$ such that $(aH)^n=H$, being $H$ the identity of $G/H$. But, by normality of $H$ in $G$, it is $(aH)^n=a^nH$ (induction on $n$), so you are looking for the least $n$ such that $a^nH=H$ or, equivalently, such that $a^n\in H$. In your case (both $G$ are abelian, so I use the additive notation):

$na \in H=\Bbb Z\Longrightarrow n=3$

$na \in H=\{0,3,6,9\}\Longrightarrow n=3$


问:
The expected number of heads
Given $10$ fair coins:

In the first round, we toss each coin once which gives us a combination of heads and tails.

In the second round, we only toss those coins that landed on the tail in the first round.

What is the expected number of heads after this experiment $?$

Intuition tells me is $5 + 2.5 = 7.5$.

答:
Intuition is not enough to solve an exercise...

The expected #H in the first round is evidently $10\times\frac{1}{2}=5$

In the second round,

$Y|X\sim Bin\Big(10-x;\frac{1}{2}\Big)$

thus

$$\mathbb{E}[Y]=\mathbb{E}[\mathbb{E}[Y|X]]=\mathbb{E}\Bigg[\frac{10-x}{2}\Bigg]=5-\frac{1}{2}\mathbb{E}[X]=\frac{5}{2}$$

Title: The expected number of heads


答:
$$E=\int_{-\infty}^{\infty} y\frac{e^{-y}}{(1+e^{y})^2}dy=\int_{-\infty}^{\infty} \frac{y}{4} \text{sech}^2(y/2) dy=0.$$
As the integrand is an odd function.

However,  $$\int_{0}^{\infty} \frac{y}{4} \text{sech}^2(y/2) dy$$
is convergent and finite.


答: The order is simply how many times can we do the operation so that the result yields an element of the subgroup. For example, how many times you add $  =\frac{2}{3}$ to become inside $H = Z$, the integer numbers. $\frac{2}{3}+\frac{2}{3}+\frac{2}{3}=\frac{6}{3}=2$ which is an integer number. So the order of $a$ is 3.

Title: What are the orders of a element $a + H$ of a quotient group $G/H$

Body:
I would like somebody to check my answers

$G = (\mathbb{Q, +}), H = (\mathbb{Z, +}), a = \frac{2}{3}$

I think the answer is $\infty$, because $a + H = \{\frac{2}{3} + h: h \in \mathbb{Z}\}$

$G = (\mathbb{Z_{12}, +_{12}}), H = \{0,3,6,9\}, a = 5$

$a +H  = \{5,8,11,2\}$, so the answer is 4.


问:
let $(X,d)$ be a metric space. How I can show that any finite subset of $X$ is closed.

let $(X,d)$ be a metric space. How I can show that any finite subset of $X$ is closed.

Can a finite subset of $X$ be open ?

Definitions:

a set $F\subseteq X$ is closed (in$(X,d)$) if  $\bar F =F$.

a set $U\subseteq X$ is open (in$(X,d)$) if  $U^o=U$

答:
Let {x1,x2,....xn} be a finite subset of the metric space $(X,d)$.

Let 'x' be one of the limit points of the this finite set.

Just observe that for r = minimum{$d(x,x_i)$| i=1,2...n},
the sphere with radius r about x, $S_r$(x) does not contain any point of the finite set.

This simply means there exist no limit points of a finite set.
Which also means that the finite set contains all its limit points(since there are none, lol).

And hope you know that a set which contains all its limit points is a Closed set.

Simple. ;)))


问:
Prove the inequality $\ln\left(\Gamma(x^x+1)\right)< x^x\ln\left(\Gamma(x+1)\right)$ for $0<x<1$
Following some toughts on Approximation of $\Big[\Gamma(1+x)\Big]^{-1}$ for $0 \leq x \leq 1$ (for the art for art's sake). .

I have the inequality :

Claim:

Let $0<x<1$ then we have :

$$\Gamma(x^x+1)< \Big(\Gamma(x+1)\Big)^{x^x}$$

The interesting part is near the value $x=1$.

First I rewrite the inequality as :

$$\ln\left(\Gamma(x^x+1)\right)< x^x\ln\left(\Gamma(x+1)\right)$$

Then I introduce the function :

$$f(x)=\ln\left(\Gamma(x^x+1)\right)-x^x\ln\left(\Gamma(x+1)\right)$$ and now I want to differentiate but it gives nothing good .I have tried also simple refinements without success.

My question :

How to prove the claim ?

Thanks in advance !

答:
Not a final answer but too long for a comment.

Hoping that you do not mind, I shall define
$$f(x)=x^x\log\left(\Gamma(x+1)\right)-\log\left(\Gamma(x^x+1)\right)$$

Close to $x=1$, we have
$$f(x)=\left(\frac{\gamma }{2}+\frac{\pi ^2}{12}-1\right) (1-x)^3+$$ $$\frac{1}{12} (1-x)^4
   \left(8-2 \gamma -\pi ^2-4 \psi ^{(2)}(2)\right)+O\left((1-x)^5\right)$$ which matches the function quite decently for $0.8 \leq x \leq 1.0$.

Close to $x=0$, we have
$$f(x)=x ((\gamma -1) \log (x)-\gamma )+$$ $$\frac{1}{12} x^2 \left(\left(6 \gamma -\pi ^2\right)
   \log ^2(x)-12 \gamma  \log (x)+\pi ^2\right)+O\left(x^3\right)$$which matches the function quite decently for $0.0 \leq x \leq 0.1$.

Using the first term only, this predicts a maximum value at $$x_*=e^{\frac{1}{\gamma -1}}\approx 0.0939237$$ and $f(x_*)=0.0330039$ while a rigorous optimization would give a maximum value of $0.0330818$ at $x=0.103099$.


问:
Computing Relative Homology
I'm sort of uncertain about how to compute relative homologies, so I would like some feedback on my work. As I understand it, you entirely use the idea that they form long exact sequences.

Suppose $X = S^1 \times [0, 1]$, the cylinder. And let $A = S^1 \times \{0, 1\}$ the upper and lower boundary circles. We want to compute $H_n(X,A)$ for every $n$.

So we use the idea that the homology groups form a long exact sequence $$...H_n(A) \rightarrow H_n(X) \rightarrow H_n(X, A) \rightarrow H_{n-1}(A) \rightarrow H_{n-1}(X) .....$$

Next since $X$ deformation retracts to a circle, $H_n(X) = \mathbb{Z}$ for $n = 1, 2$ and trivial otherwise. Similarly, as $A$ is a disjoint union of two circles, $H_n(A) = \mathbb{Z} \oplus\mathbb{Z}$ for $n = 1,2$ and trivial otherwise. This means that $H_{n}(X,A) = 0$ if $n \geq 4$. For $n = 3$,

$$.. 0 \rightarrow H_3 (X, A) \rightarrow \mathbb{Z} \oplus \mathbb{Z} \rightarrow \mathbb{Z}$$

is what we obtain. Which means that $H_3(X, A)$ is a subgroup of $\mathbb{Z} \oplus \mathbb{Z}$. Moreover, the map $i_{*} : \mathbb{Z} \oplus \mathbb{Z} \rightarrow \mathbb{Z}$ is induced by the inclusion $i : C_2(A) \rightarrow C_2(X)$.

However, I'm not sure which properties of $i_{*}$ are to be used to get what $H_3(X, A)$ is exactly. I'm facing similar problems for the other $H_n(X,A)$ ($n = 0,1,2$).

I think I could get it done after looking at a concrete example but such examples are surprisingly hard to find. Most examples, like the ones in Hatcher, are trivial and quick. I'm sure this one is easy as well and I'm just being an idiot but I would still appreciate the help.

答:
For $n \ge 3$ we have an exact sequence
$$0 = H_n(X) \to H_n(X,A) \to H_{n-1}(A) = 0$$
which shows $H_n(X,A) = 0$.

Let $S^1_k = S^1 \times \{ k \}$ for $k= 0,1$. Let $i : A \to X$ and $i_k : S^1_k \to A$ denote inclusions. The maps $u_k : S^1 \to S^1_k, u_k(x) = (x,k)$, are homeomorphisms. The projection $p : X \to S^1$ and the maps $j_k : S^1 \to X, j_1(x) = (x,k)$, are homotopy equivalences.

Since the $S^1_k$ are the path components of $A$, the map
$$h  : H_n(S^1_0) \oplus H_n(S^1_1) \to H_n(A), h(a,b) =  (i_1)_*(a) + (i_2)_*(b)$$ is an isomorphism. Moreover
$$\psi : H_n(S^1) \oplus H_n(S^1) \to H_n(S^1_0) \oplus H_n(S^1_1), \psi(c,d) = ((u_1)_*(c),(u_2)_*(d))$$
is an isomorphism.

For $n = 0,1$ let us compute $\phi = p_* i_*h \psi: H_n(S^1) \oplus H_n(S^1) \to H_n(S^1)$. Let $g$ one of the two generators of $H_n(S^1)$. Obviously $\phi(g,0) = p_*i_*h\psi(g,0) = p_*i_*(i_1)_*(u_1)_*(g) = g$. Similarly $\phi(0,g) = g$. Thus $\phi(ag,bg) = (a+b)g$. We conclude that $\phi$ is onto with $\ker \phi \approx \mathbb Z$. Thus also $i_*$ is onto with $\ker i_* \approx \mathbb Z$. Hence $j_* : H_n(X)\to H_n(X,A)$ is the zero map because $\ker j_* =\operatorname{im} i_* = H_n(X)$.

The exact sequence
$$0 = H_2(X) \to H_2(X,A)  \stackrel{\partial}{\to} H_1(A) \stackrel{i_*}{\to} H_1(X)$$
shows that $H_2(X,A) \approx \operatorname{im} \partial = \ker i_*  \approx \mathbb Z$.

The exact sequence
$$H_1(X)  \stackrel{0}{\to} H_1(X,A) \stackrel{\partial}{\to} H_0(A) \stackrel{i_*}{\to} H_0(X)$$
shows that $\ker \partial = \operatorname{im} 0 = 0$, i.e. $\partial$ is injective. But $\operatorname{im} \partial = \ker i_*  \approx \mathbb Z$, thus $H_1(X,A) \approx \mathbb Z$.

The exact sequence
$$H_0(X) \stackrel{j_*}{\to}  H_0(X,A) \to 0$$
shows that $j_*$ is surjective. But $j_*$ is the zero map, thus $H_0(X,A) = 0$.

Title: Computing Relative Homology


问:
Affine transformation mapping circle to ellipse
I'm trying to find an affine transformation that maps the unit circle to an ellipse centered at $(1,3)$ such that points P$(-3,-1)$ and R$(5,7)$ are at the greatest distance from the centre of the ellipse along the major axis while points Q$(0,4)$ and S$(2,2)$ are at the greatest distance from the centre of the ellipse along the minor axis.

I think we need to rotate the unit disc followed by scaling and then another rotation. But I don't know how to find the matrix that does that. Ultimately, I think there will be translation by $(1,3)^T.$

Any help appreciated.

答:
Hints :

$(5,7), (1,3), (-3,-1)$ lie on line $y=mx+c$

$(0,4), (1,3), (2,2)$ lie on line $y=\ldots$

Conclude this is rotation with $\theta = \tan^{-1} m$

Use rotation matrix
$$ \begin{bmatrix}
\cos \theta & -\sin \theta \\
\sin \theta & \cos \theta \\
\end{bmatrix}
$$

Find lengths of axes of ellipse. Hence find scaling matrix. $$ \begin{bmatrix}
a & 0 \\
0 & b \\
\end{bmatrix}
$$


问:
Let $T:[0,1]\to[0,1]$ be a monotone function, and suppose that $T(x)=\frac{f(x)+1-f(1-x)}{2}$. Can we find $f$?
Let $T:[0,1]\to[0,1]$ be a monotone and continuous function, and suppose that $\forall x\in[0,1]:T(x)=\frac{f(x)+1-f(1-x)}{2}$.

Is there a unique monotone and continuous function $f$ that satisfies the above?

Can we express $f(x)$ explicitly (as a function of $T$)?

For example, if $T(x)=ax+\frac{1-a}{2}$ for some $a>0$, then $f(x)=T(x)-T(0)=ax$ works.

As @Clement answered, if $f$ satisfies the condition, then so will $f+c$ for some constant $c$.

Is $f$ unique up to an additive constant?

答:
This is definitely not unique, as for any $a \in \mathbb{R}$, we have that if $f(x) = x + a$ then $T(x) = x$.

It is not unique up to additive constant either. The idea is that if $f(x) = f(1 - x)$, then $T(x) = \frac{1}{2}$ necessarily. For instance, consider:
$$
f(x) = ax^n + a(1 - x)^n
$$
where $a \in \mathbb{R}$ and $n \in \mathbb{N}$. Then for any of such $a,n$, we have that $T(x) = \frac{1}{2}$.

EDIT: There also exists $T$ such that no such $f$ exists. One example is the case where $T(x) = 0$. Then this reduces to the problem of solving for:
$$
f(1 - x) = f(x) + 1
$$
But letting $x = 0$ gives $f(1) = f(0) + 1$, and letting $x = 1$ gives $f(1) = f(0) - 1$, a contradiction. This probably suggests that $f$ cannot be expressed in terms of $T$.


问:
Matrix factorization with a semi-orthogonality constraint
Given $X \in \mathbb{R}^{n \times d}$, I have an optimization problem with an orthogonality constraint:

$$\min_{_{P \in \mathbb{R}^{n \times k}, L \in \mathbb{R}^{k \times d}}}\left \| X - PL \right \|_{F}^{2} \quad \text{s.t.} \quad \:P^{T}P=I_{k} \tag{1}$$

I have tried to solve problem (1) by alternating between solving problem (2) and problem (3) as follows.

Solving P:

$$\min_{_{P \in \mathbb{R}^{n \times k}}}(\left \| X - PL \right \|_{F}^{2})  \quad \text{s.t.} \quad  \:P^{T}P=I_{k} \tag{2}$$

Problem (2) is the Orthogonal Procrustes problem.

By applying compact SVD on $XL^{T}$, we obtain:
$XL^{T}=U\Sigma V^{T}$ where $U$ is of size $n$ by $k$, $\Sigma$ is of size $k$ by $k$, and $V$ is of size $k$ by $k$.
Then the solution to problem (2) is :

$P=UV^{T}$.

Solving L:

$$\min_{_{L \in \mathbb{R}^{k \times d}}}(\left \| X - PL \right \|_{F}^{2}) \quad \text{s.t.} \quad P^{T}P=I_{k} \tag{3}$$

Solution of problem (3) is straightforward:

$L = P^{T}X$

Question

My optimizer always converges at a local minimum. Is there another way to solve problem (1) ?

答:
Yes, there is another way. Note that with the existence of the rank-revealing $QR$ factorization, we find that a matrix $M$ can be written in the form $M = PL$ with $P$ of size $n \times k$, $L$ of size $k \times d$, and $P^TP = I$
if and only if $M$ has rank at most $k$. With that, we can equivalently frame problem (1) as
$$
\min_{M}\|X - M\|_F \quad \text{s.t.} \quad \operatorname{rank}(M) \leq k. \tag{0}
$$
In other words, this is simply a low-rank approximation problem. As is explained in the linked page, the EYM theorem guarantees that the minimizer $M_*$ can be obtained from the truncated SVD of $X$. That is, let
$$
X = U\Sigma V^{\top} \in \mathbb{R}^{n\times d}
$$
be a thin singular value decomposition of $X$ and partition $U$, $\Sigma=:\operatorname{diag}(\sigma_1,\ldots,\sigma_{\min\{n,d\}})$, and $V$ as follows:
$$
U =: \begin{bmatrix} U_1 & U_2\end{bmatrix}, \quad
\Sigma =: \begin{bmatrix} \Sigma_1 & 0 \\ 0 & \Sigma_2 \end{bmatrix}, \quad\text{and}\quad
V =: \begin{bmatrix} V_1 & V_2 \end{bmatrix},
$$
where $U_1$ is $n\times k$, $\Sigma_1$ is $k\times k$, and $V_1$ is $d\times k$. Then the rank-$k$ matrix, obtained from the truncated singular value decomposition
$$
M_* = U_1 \Sigma_1 V_1^{\top},
$$
minimizes the objective function for (0). If we take $P = U_1$ and $L = \Sigma_1V_1^\top$, then we find that $P$ and $L$ satisfy the required constraints.


答:
I assume you found something like$$\left(\sum_{i=1}^ni\right)^2-\left(\sum_{i=1}^{n-1}i\right)^2=\left(\sum_{i=1}^ni-\sum_{i=1}^{n-1}i\right)\left(\sum_{i=1}^ni+\sum_{i=1}^{n-1}i\right)=n\cdot n^2=n^3.$$I think whether all theorems on countable sets are provable by induction is an open question; the hard part would be checking some ordering exists for which a proof of the inductive step must exist. Whether I'm right or wrong, this question is probably a better fit for MathOverflow than for math.se, because it looks very, very nontrivial.

But I suspect far more theorems are thereby provable than first seem to be. There's a very complicated inductive argument in Wiles's proof of (the semistable case of) the modularity theorem, for which he needed not only a sensible ordering of E- & M-series, but also a way to make the inductive step work for said ordering. (I'm not an expert on it, but I expect he worked backwards from needing the inductive step to work to inspire him in how he'd order them.)

Title: Can every theorem involving countable things necessarily be proven by induction?

Body:
Last night I was working on some problems to remember how proof by induction worked, and I began to wonder if it's possible to prove every theorem on countable things by induction.  I understand proof by induction may not be the most elegant or simple way to construct the proof.

If so, is there a proof of this?  If not, is there a counterexample?

BTW, the problem I was struggling with, but finally figured out, was to prove that:

$\sum_{i=1}^n i^3 = (\sum_{i=1}^ni)^2$


问:
A simple concept question about charaterization of closeness of a space.
For first countable topological space  X we know if $x\in \bar{A} \iff \text{exist}\ x_n\to x$  with $x_n\in A$

Hence if $A$ is closed ,$x\in A $ there must exist a sequence converge to it.

My question is does the following statement holds?:

if for any $x\in A$ exist $x_n\to x$ with $x_n\in A$ then $A$ is
closed?

It seems not,since closed iff contains all the limit point,there may exist some sequence $y_n\to y\notin A$. correct?

答: No. The sequence "for any $x \in A$, exists $x_n \rightarrow x$ with $x_n \in A$" is always true, since you can consider the constant sequence $x_n = x$.


答: An alternate solution is as follows: imagine you flipped all the coins twice.  Then any coin that gave you heads on the first flip or the second flip would be one of the ones you want to count.  The probability of getting at least one head in two flips is $3/4$, so the expected number of coins that get at least one head is $10 \times 3/4 = 7.5$.

Title: The expected number of heads

Body:
Given $10$ fair coins:

In the first round, we toss each coin once which gives us a combination of heads and tails.

In the second round, we only toss those coins that landed on the tail in the first round.

What is the expected number of heads after this experiment $?$

Intuition tells me is $5 + 2.5 = 7.5$.


问:
Generalized eigenspace - Construction algorithm proof
Suppose that $A$ is an $n\times n$ nilpotent matrix.

Let $x_1,\ldots,x_m$ be eigenvectors of $A$ forming a basis for the eigenspace of $A$.

For $j=1,\ldots,m$, form a Jordan chain $$C\left(x_j\right)=\left\{v_{j,1},\ldots,v_{j,k_j}\right\}$$ by "going backward", that is, start from the eigenvector $v_{j,1}=x_j$ and iteratively solve $$Av_{j,k}=v_{j,k-1}$$ while the system has a solution (until you find a system that has no solution).

Several sources I found on the web, as well as an old textbook, seem to suggest that $$C=C\left(x_1\right) \cup \ldots \cup C\left(x_m\right)$$  is a basis for the generalized eigenspace of $A$, that is, for the null space of $A^n$, without offering a formal proof that the algorithm works. How would you prove that? Any reference would be highly appreciated.

答:
This approach does not work. Consider the matrix
$$
A=\pmatrix{0&0&1\\0&0&0\\0&0&0}
$$
with eigenvectors $\pmatrix{1\\1\\0}$ and  $\pmatrix{1\\-1\\0}$. None of them is in the range of $A$, only their sum is.

Instead  one has to work from the other direction: Assume $A^dv=0$ but $A^{d-1}v\ne 0$. Then $(A^{d-1}v \dots Av,v)$ are linear independent and part of the Jordan basis. A reference for this approach is the book by Axler.


问:
show that $f(x) = \frac{x^2}{ 1 + \vert x \vert} $ is uniformly continuous on $\mathbb{R}$
I started proving it by saying that on [-1;1] the map is continuous and so we have uniform continuity.
Then, I want to show that f is uniform continuous on [1; $\infty$[ and ]-$\infty$ ; -1].
I wanted to do it with the definition, I have the following expression: $\forall x,y \in [1; \infty$[ \
$\mid f(x) - f(y) \mid = \mid$ $\frac{x^2}{1 + \mid x \mid} - \frac{y^2}{ 1 + \mid y \mid} \mid =
\mid \frac{x^2 - y^2 + x^2y -y^2x}{(1+x)(1+y)} \mid = \mid \frac{(x - y)( x+ y + xy)}{(1+x)(1+y)} \mid$
I don't know how to conclude. I also have to show that the result stays true for $x \in [1; \infty$[ and
y $\in  ]-\infty ; -1]$ ?

答:
You are almost there as $f$ is even and for $x,y \ge 0$

$$\begin{aligned}\left\vert f(x) - f(y) \right\vert &= \left\vert \frac{x^2}{1 + \vert x \vert} - \frac{y^2}{ 1 + \vert y \vert} \right\vert\\
&=
\left\vert \frac{x^2 - y^2 + x^2y -y^2x}{(1+x)(1+y)} \right\vert\\
&= \left\vert \frac{(x - y)( x+ y + xy)}{(1+x)(1+y)}\right\vert\\
&\le \vert x - y \vert \end{aligned}$$

knowing that $0 \le x+ y + xy \le (1+x)(1+y)$


答:
First, as the function is even, we may suppose $x,y\ge 0$. In this case,
$$|f(x)-f(y)|=|x-y|\frac{x+y+xy}{(1+x)(1+y)}<|x-y|\frac{1+x+y+xy}{(1+x)(1+y)}=|x-y|.$$

Title: show that $f(x) = \frac{x^2}{ 1 + \vert x \vert} $ is uniformly continuous on $\mathbb{R}$

Body:
I started proving it by saying that on [-1;1] the map is continuous and so we have uniform continuity.
Then, I want to show that f is uniform continuous on [1; $\infty$[ and ]-$\infty$ ; -1].
I wanted to do it with the definition, I have the following expression: $\forall x,y \in [1; \infty$[ \
$\mid f(x) - f(y) \mid = \mid$ $\frac{x^2}{1 + \mid x \mid} - \frac{y^2}{ 1 + \mid y \mid} \mid =
\mid \frac{x^2 - y^2 + x^2y -y^2x}{(1+x)(1+y)} \mid = \mid \frac{(x - y)( x+ y + xy)}{(1+x)(1+y)} \mid$
I don't know how to conclude. I also have to show that the result stays true for $x \in [1; \infty$[ and
y $\in  ]-\infty ; -1]$ ?


问:
To find a specfic equation $z-\frac{1}{s\ln(s)}-n\ln(s)==0$
Here I want to find the solution to the following equation
$$z-\frac{1}{s\ln(s)}-n\ln(s)=0,$$
where $z$ and $n$ are constants.
Here the aim is to find the solution $s=?$.

Any suggestion is welcome.

答:
Some thoughts that may help.

Similar equations can often be solved, at least numerically, in 'general' by the application of a numerical method as have been developed in the field of Numerical Analysis.

See, for example, this Wikipedia discussion on Newton's method, also known as the Newton–Raphson method.

This may not be the direct algebraic solution that you are seeking, however, plotting the solution as a function of the underlying parameter values, does, at least, present a graphical-based general solution.

While not likely, you may be able to also identify the underlying functional forms composing the solution. In this regard, try, for example, examing plots where one of the parameters is set at a constant value.

Relatedly, take the derivative of the target equation and work with that (where the solution then becomes a differential equation exercise), or try applying some variable transformations. In this case, I would use y = Exp (s), and then relook at the equation and perhaps employing series expansions.

Can proceed again to examine a graphical solution for aid.

OwnerDisplayName: Blueka


问:
Direct product and commutativity of factors
If $H$ and $K$ are both normal subgroups of $G$, and if every element of $G$ can be written uniquely as $hk$ with $h\in H$ and $k\in K$ (so $G$ is the direct product of $H$ and $K$), does it follow that $H$ and $K$ commute, i.e. that $kh=hk$ always?

答:
Yes, that is always true. Note that $G$ being the direct product of $H$ and $K$ means that the map $\Phi\colon G\to H\times K$ given by $hk\mapsto (h,k)$ is an isomorphism, where $H\times K$ carries the group structure given by $(h_1,k_1)(h_2,k_2)=(h_1h_2,k_1k_2)$. Under this isomorphism, note that
$$
\Phi(kh) = \Phi((1k)(h1)) = \Phi(1k)\Phi(h1) = (1,k)(h,1) = (1h,k1) = (h,k) = \Phi(hk).
$$
So $\Phi(kh)=\Phi(hk)$ and hence $kh=hk$.

To elaborate on the direct proof suggested by @AugustLiu in the comments: We have $H\cap K = \{1\}$, since otherwise $1\neq g\in H\cap K$ would have the two different expressions
$$
g = \underbrace{g}_{\in H} \cdot \underbrace{1}_{\in K} = \underbrace{1}_{\in H} \cdot \underbrace{g}_{\in K}.
$$
Now note that since $H$ and $K$ are normal subgroups, we have
$$
H \ni \underbrace{(khk^{-1})}_{\in H} \, \underbrace{h^{-1}}_{\in H} = \underbrace{k}_{\in K} \underbrace{(hk^{-1}h^{-1})}_{\in K} \in K
$$
so that $khk^{-1}h^{-1}\in H\cap K=\{1\}$. Hence $khk^{-1}h^{-1}=1$, which is equivalent to $kh=hk$.

Yet another proof goes like this: The element $kh\in G$ must have a unique expression $kh=h'k'$ with $h'\in H$ and $k'\in K$. However, this yields
$$
h = k^{-1} h' k' = \underbrace{(k^{-1} h' k)}_{\in H} \underbrace{(k^{-1} k')}_{\in K}
$$
and the unique expression of $h$ as a product of an element in $H$ and an element in $K$ is $h=h\cdot 1$, so that $1=k^{-1} k'$ and hence $k'=k$. By a similar argument $h'=h$ and hence $kh=hk$.


问:
Proving A Propositional Logic Statement Using Only Equivalence Laws
I'm trying to figure out how to prove

$$ \neg (p \vee q) \vee (\neg p \wedge q) \quad \Leftrightarrow \quad \neg p \tag{*} $$

using purely equivalence laws in propositional logic, the equivalence laws being here: link.

What I've done so far is to use de Morgan's law on one of the brackets, then use the distributive law. However, I wasn't able to get $\neg p$ from doing so. Could someone explain how to show $\text{(*)}$ using purely equivalence laws?

Thanks in advance.

答:
You're almost there:

$$\neg(p \lor q) \lor (\neg p \land q)$$

$$ \overset{DeMorgan}{\Leftrightarrow}$$

$$(\neg p \land \neg q) \lor (\neg p \land q)$$

$$\overset{Distribution}{\Leftrightarrow}$$

$$\neg p \land (\neg q \lor q)$$

$$\overset{Law of Excluded Middle}{\Leftrightarrow}$$

$$\neg p \land \top$$

$$\overset{Identity}{\Leftrightarrow}$$

$$\neg p$$


问:
Triangle Area Problem with intersecting lines
I recently saw this geometry problem and it's been killing me. Given $\overline{AB}=\overline{AC}=20$, $\overline{AD}=\overline{AE}=12$, and $[ADFE]=24$ with $[\cdot]$ denoting area, find the area of $\triangle BCF$ (see diagram)

I started by considering line $\overline{DE}$. It's pretty clear to see that $\triangle DEF \sim \triangle CBF$. We also have $\triangle {ADE}\sim\triangle{ABC}$, with $9[ABC]=25[ADE]$. But once i've figured out these parts, I get stuck. What would be the next steps in this situation? I assume I'd gave to take advantage of the area ratios I've gotten, but I can't workout the details of that argument.

答:
Connect $AF$. By symmetry, $[ADF] = [AEF]= 12$.

Notice that $\triangle DEF \sim \triangle CBF$ in the same ratio $3:5$, so we have $\dfrac {EF}{FC} = \dfrac {DF}{FB} = \dfrac35$.

Hence $\dfrac {[AEF]}{[AFC]} = \dfrac35 = \dfrac {12}{12+ [DFC]}$, giving $[DFC] = 8$.

We also have $\dfrac {[CDF]}{[FBC]} = \dfrac 35$, giving $[FBC] = \dfrac{40}3$.


答:
$CD = BE = 8$

Join $AF$. $[ADF] = [AEF] = 12$

Now

$$ \dfrac{[AFD]}{[FDC]} = \dfrac{AD}{CD} = \dfrac{12}{8}$$

$$ \Rightarrow [FDC] = 8 = [FEB]$$

Also

$$ \dfrac{AD}{CD} = \dfrac{[ABD]}{[BDC]} =  \dfrac{[ADFE]+[FEB]}{[FDC]+[BFC]} = \dfrac{24+8}{8+[BFC]}$$

$$ \Rightarrow [BFC] = \dfrac{40}{3}$$

Title: Triangle Area Problem with intersecting lines

Body:
I recently saw this geometry problem and it's been killing me. Given $\overline{AB}=\overline{AC}=20$, $\overline{AD}=\overline{AE}=12$, and $[ADFE]=24$ with $[\cdot]$ denoting area, find the area of $\triangle BCF$ (see diagram)

I started by considering line $\overline{DE}$. It's pretty clear to see that $\triangle DEF \sim \triangle CBF$. We also have $\triangle {ADE}\sim\triangle{ABC}$, with $9[ABC]=25[ADE]$. But once i've figured out these parts, I get stuck. What would be the next steps in this situation? I assume I'd gave to take advantage of the area ratios I've gotten, but I can't workout the details of that argument.


问:
Filippov's Inductive proof for Jordan Canonical form

Let $A$ be an $n×n$ complex matrix. Then there exists an invertible matrix $T$ such that $T^{−1}AT=J$ where $J$ is a Jordan form matrix having the eigenvalues of $A$. Equivalently, the columns of $T$ consist of a set of independent vectors $x_1, . . . , x_n$ such that $Ax_k=λ_kx_k$ , or $Ax_k=λ_kx_k+x_{k−1}$.

There exists an inductive proof by Filippov,

When $n=1$ the Jordan canonical form of the matrix $[a]$ is $[a]$ itself.

Assume the existence of a Jordan canonical form for all $r\times r$ matrices, $r=1,2,...,n-1$, ie. $A_{r\times r}$ is simlar to a Jordan matrix.

Consider an $n\times n$ matrix

Assume that $\lambda=0$ is an eigenvalue, ie., $A$ is singular. So the dimention of $C(A)$ is $r<n$.

Then it says that, by the induction hypothesis A (more precisely the linear operator associated with A) restricted by its range has a Jordan canonical form.

How do I make sense of this statement ?

I think that means,

thanks @ancientmathematician for pointing into the right direction.

If we think of another tansformation, $T:range(A)\to range(A)$ and the corresponding matrix is $B_{r\times r}$ associated with it, then from the induction hypothesis there exists a Jordan canonical basis $(w_1,\cdots,w_r)$ for the $range(A)$ such that $Bw_k=λ_kw_k$ , or $Bw_k=λ_kw_k+w_{k−1}$.

The book seems to use same $A$ for the $B_{r\times r}$, might be the source of confusion here !

Ok.

Even if that is the case, how can one advance the proof from here and find the conditions for the Jordan basis for the original matrix $A_{n\times n}$ ?

How do I get from $Bw_k=λ_kw_k$ , or $Bw_k=λ_kw_k+w_{k−1}$.  to $Aw_k=λ_kw_k$ , or $Aw_k=λ_kw_k+w_{k−1}$ for the original $n\times n$ matrix $A$ ?

Please check A  Primer of Abstract Mathematics By Robert B. Ash.

答:
Thanks @ancientmathematician for the hint.

If we think of another transformation, $T:range(A)\to range(A)$ and the corresponding matrix is $B_{r\times r}$ associated with it, ie., $B_{r\times r}$ represents the same $range(A)\to range(A)$ transformations as $A_{n\times n}$, just the space becomes smaller,

then from the induction hypothesis there exists a Jordan canonical basis $(w'_1,\cdots,w'_r)$ for the $range(A)$ such that $Bw'_k=λ_kw'_k$ , or $Bw'_k=λ_kw'_k+w'_{k−1}$.

Step 1

This implies there exists a Jordan canonical basis $(w_1,\cdots,w_r)$ for the $range(A)$ such that $Aw_k=λ_kw_k$ , or $Aw_k=λ_kw_k+w_{k−1}$. The vectors $w_i$ and $w'_i$ are the same except that $w_i$ is a vector in the larger $\mathbb{C}^n$ space and $w'_i$ is the exact same vector represented in the smaller subspace $\mathbb{C}^n$.

Step 2

Let the subspace $N(A)\cap R(A)$ has dimension $p$. The subspace $N(A)\cap R(A)$ is the eigenspace corresponding to the eigenvalue $\lambda=0$. So among the basis vectors $(w_1,\cdots,w_r)$ there are $p$ linearly independent eigenvectors that has eigenvalue $\lambda=0$. Since $w_i\in R(A)$ we have $w_i=Ay_i$ for some $y_i$. Since $Ay_i=0y_i+w_i$ we can place each $y_i$ after each corresponding $w_i$ in the string for all $p$ members in $N(A)\cap R(A)$.

Step 3

Since $dim[N(A)]=n-r$ there are $n-r-p$ vectors $z_i\in N(A)$ that do not belong to $N(A)\cap R(A)$ and $Az_i=0$, ie., there are $n-r-p$ linearly independent eigenvector $z_i\in N(A)$ but not a member of $R(A)$.

Since we have $n$ vectors in $\mathbb{C}^n$ that are in the form $Ax_k=λ_kx_k$ , or $Ax_k=λ_kx_k+x_{k−1}$, we just have to verify the linear independence of $w_i$, $y_i$, $z_i$ to prove the existence of the Jordan canonical basis for any singular square matrix.


问:
Does $c^n(n!+c^n)\lt (n+c^2)^n$ hold for all positive integers $n$ and $c\gt 0$?
I am  not sure whether the following inequality is true? Some small $n$ indicates it is true. </p>

<p>Let $n$ be a positive integer and $c\gt0$, then
$$c^n(n!+c^n)\lt(n+c^2)^n.$$ </p>

答: Consider $n = 1$ and $c = 10$. Then, $c^n(n!+c^n) = 110$ while $(n+c^2)^n = 101$. You will be able to show a number of other similar counterexamples when $n=1$.</p>


问:
Is there a name for relations with this property?
Is there a name for relations $\rho : X \rightarrow Y$ such that for all $x,x' \in X$ and all $y,y' \in Y$ we have that the following conditions $$xy \in \rho$$ $$x'y \in \rho$$ $$xy' \in \rho$$ imply that $$x'y' \in \rho?$$

Here's a couple of alternative characterizations.

Define $\rho(x) = \{y\in Y \,|\, xy \in \rho\}$ for all $x \in X$. Then for all relations $\rho : X \rightarrow Y,$ we have that $\rho$ has the property of interest iff for all $x,x' \in X$ it holds that either $\rho(x) = \rho(x')$ or $\rho(x) \cap \rho(x') = \emptyset$.

Call a relation $\kappa : X \rightarrow Y$ Cartesian iff there exist $A \subseteq X$ and $B \subseteq Y$ such that $\kappa = A \times B$. Call two relations $\kappa$ and $\kappa$' strongly disjoint iff their images are disjoint, and their "left-images" are also disjoint. Then for all relations $\rho : X \rightarrow Y,$ we have that $\rho$ has the property of interest iff it can be expressed as a strongly disjoint union of Cartesian relations $X \rightarrow Y$.

A few observations:

If a relation has the property of interest, so too does its converse.

Every function has the property (and thus, so too does its converse).

If two relations have the property, their composition does, too; thus, we obtain a category.

The property is preserved under arbitrary strongly disjoint unions.

答:
If the relation is seen as an undirected bipartite graph with edges joining $X$ and $Y$, the axiom is that a path of length 3 closes to a cycle of length 4.

In this form it is easy to check that "are both related to the same element in the other set (or equal)" is an equivalence relation on $X$ and on $Y$, and that if $x$ is related to $y$, all elements of $X$ equivalent to $x$ are related to all elements of $Y$ equivalent to $y$.  This is the condition for there to be quotients of the sets $X$ and $Y$ on which the relation descends to a one-to-one correspondence.

This can be expressed by saying that $\rho$ is/induces an "identification between quotients" of $X$ and $Y$.  That property is not closed under composition as far as I can see, but the other properties are true.


问:
Subspace contains the nullspace of a linear function, then subspace is itself either $V$ or the nullspace
For a linear functional $\phi$ on the vector space $V$ I claim that if $U$ is a subspace such that $\mathrm{null}{\phi} \subseteq U$ then it must be that either $U = \mathrm{null}{\phi}$ or that $U = V$.

I know that $\text { null } \phi =\{f \in V: \phi f=0\}$ and i know how to prove that $\mathrm{null}{\phi}$ is itself a subset of $V$. I'm having trouble applying any of that to prove my claim. Thanks in advance for the help.

答:
Assume that $U \ne null(\phi)$. Then there is $u_0 \in U$ such that $\phi(u_0)\ne 0$.  Set $u:= \frac1{\phi(u_0)}u_0$.  Then $u_1 \in U$ and $\phi(u_1)=1$.
Now let $v\in V$.
Then
$$
v = (v - \phi(v)u_1) + \phi(v)u_1
$$
with $v - \phi(v)u_1\in null(\phi)$ and $ \phi(v)u_1\in U$. So $v\in U$ and $U=V$.


问:
probability question convergence
Let $(\Omega, \mathcal{F}, \mathbb{P})$ be a probability space.

Define an rv such that $X_n=1$ with $\mathbb{P}(X_n=1)=1$ and 0 with $\mathbb{P}(X_n=0)=0$ for all n.

Then, $X_n\rightarrow 1$ in $\mathbb{P}$

I am confused as to how to prove this. I am aware that we must show that

$\mathbb{P}(|X_n-1|>\epsilon)\rightarrow 0$ and $n\rightarrow \infty$ but i'm having issues in constructing suitable cases.

Attempt:

Let $0<\epsilon<1$ then

$\{$ $X_n=0$ $\}$ $=$ $\{$ $|X_n-1|>\epsilon$ $\}$

thus, $\mathbb{P}(X_n=0 )=\mathbb{P}(|X_n-1|>\epsilon)=0$

is this correct?

答:
If you ignore $\mathbb{P}(X_n=0)$ then your proof could become:

For all $\epsilon >0$ you have

$0 \le \mathbb{P}(|X_n-1|>\epsilon) \le \mathbb{P}(|X_n -1| \not=0) = 1-\mathbb{P}(X_n=1)=1-1=0$

so $\mathbb{P}(|X_n-1|>\epsilon) =0$

Title: probability question convergence


问:
Variation vs Differential
I am a bit confused about the difference between variation $\delta u$ and differential $du$.

I saw them in terms of minimizing a functional. The way the variation operates seems very similar to derivatives (chain-rule and so on).

I have a hunch that variations talk about small deviations in function (throughout its domain) whereas differentials talk about small changes in discrete quantity. Am I right about this? An answer in Layman's terms would be much appreciated.

答:
In simplistic terms, a differential relates to the increase in the value of a function, an object taking a scalar as argument and returning a scalar, for a "small" variation in the independent variable.
A variation relates to the increase in the value of a functional, and object taking a function as argument and returning a scalar, for a small variation in the argument function. Your intuition seems correct, although I am not so sure about the reference to "discrete quantity".</p>

Title: Variation vs Differential


问:
Does exactness of a sequence of groups imply exactness of the dual?
If $A \rightarrow B \rightarrow C$ is exact, with morphisms a and b respectively, that is $Im(a)=ker(b)$, then $C^*\rightarrow B^* \rightarrow A^*$ is also exact?

$A^*=Hom(A,G)$

A, B, C and G are groups, but I guess it might be also a good question in the case of modules.

The inclusion $Im(b^*)\subset ker(a^*)$ is easy, but the other inclusion is quite difficult to prove (maybe not even true)

Any ideas?

Edit: I was able to prove the statement in the case b has a right inverse. Maybe this is a necessary condition, I will check it later.

Edit 2: The splitting condition is sufficient but not necessary. As was mentioned in the answers, one could choose G to be an injective object and the sequence would always be exact.

答:
In general not, you need that $G$ is injective, as that is precisely one of the definitions of an injective objects, for abelian groups this is equivalent to being divisible.

you can find a lot of examples of $Hom(_,G) failing to be exact even on this page.
http://www-users.math.umn.edu/~garrett/m/repns/notes_2014-15/04b_adjoints_exactness.pdf


问:
$a \in \mathbb{R}$ and $a^7, a^{12} \in \mathbb{Q}$. Prove that $a \in \mathbb{Q}$.
Let $a \in \mathbb{R}$ and $a^7, a^{12} \in \mathbb{Q}$. Prove that $a \in \mathbb{Q}$.

my solution:

$a^7 \in \mathbb{Q} \Rightarrow a^{14} \in \mathbb{Q}$

$a^{14}=a^{12}\cdot a^{2}\in \mathbb{Q}$, but $a^{12} \in \mathbb{Q} \Rightarrow$ $a^2 \in \mathbb{Q}$.

Using the same rule:
$a^{21} \in \mathbb{Q}$ and $a^{24}\in \mathbb{Q} \Rightarrow$ $a^{24}=a^{2} \cdot a \cdot a^{21} \Rightarrow a \in \mathbb{Q} $.

Is there any other solution?

答: Well, we could the same things more directly: write $a$ as $\frac{(a^{12})^3}{(a^7)^5} = \frac{a^{36}}{a^{35}}$, or as $\frac{(a^7)^7}{(a^{12})^4} = \frac{a^{49}}{a^{48}}$, and conclude that it's in $\mathbb Q$. Your work probably simplifies to one of these after some substitutions.


答:
Hint: $\gcd(7, 12) = 1$

so done.

Body:
Let $a \in \mathbb{R}$ and $a^7, a^{12} \in \mathbb{Q}$. Prove that $a \in \mathbb{Q}$.

my solution:

$a^7 \in \mathbb{Q} \Rightarrow a^{14} \in \mathbb{Q}$

$a^{14}=a^{12}\cdot a^{2}\in \mathbb{Q}$, but $a^{12} \in \mathbb{Q} \Rightarrow$ $a^2 \in \mathbb{Q}$.

Using the same rule:
$a^{21} \in \mathbb{Q}$ and $a^{24}\in \mathbb{Q} \Rightarrow$ $a^{24}=a^{2} \cdot a \cdot a^{21} \Rightarrow a \in \mathbb{Q} $.

Is there any other solution?

Body:
Hint: $\gcd(7, 12) = 1$

so done.


问:
Sum of a series with
I am trying to compute the sum of the following series:
$$
\sum_{n=1}^\infty 3^{n-1}\sin^{3}\left(a \over \,3^{n}\right)
$$
The series is clearly convergent by the ratio test, as well as by comparison with $(\frac{a}{3^n})^3$. If I take the derivative with respect to $a$, I am eventually left with $\cos(\frac{a}{3^n})-\cos^3(\frac{a}{3^n})$, which doesn't seem very helpful, plus I'm not sure it's okay to take the derivative. Multiplying by sine functions to try to get a telescopical sum didn't lead me anywhere. Any ideas will be appreciated. Thanks!

Title: Sum of a series with

Body:
As a hint:$$\sin(3x)=3\sin x-4\sin^3x \to \color{red}{\sin^3(x)=\frac14\Big(3\sin x-\sin(3x)\Big)}$$so
$$\sum_{n=1}^\infty 3^{n-1}\sin ^3{\frac{a}{3^n}}=\\
\frac14\sum_{n=1}^\infty 3^{n-1}(3\sin {\frac{a}{3^n}}-\sin {\frac{3a}{3^n}})=\\
\frac14\sum_{n=1}^\infty (3^n\sin {\frac{a}{3^n}}-3^{n-1}\sin {\frac{a}{3^{n-1}}})=\\
\frac14\sum_{n=1}^\infty (f(n)-f(n-1))$$


答:
Your solution is fine. You could do it in one line ...
\begin{eqnarray*}
a=\frac{(a^{12})^3}{(a^7)^5}.
\end{eqnarray*}

Title: $a \in \mathbb{R}$ and $a^7, a^{12} \in \mathbb{Q}$. Prove that $a \in \mathbb{Q}$.


问:
Integral of fractional expression $\int^3_0 \frac{dx}{1+\sqrt{x+1}}$
I want to solve this integral and think about call $\sqrt{x+1} = t \rightarrow t^2 = x+1$
$$\int^3_0 \frac{dx}{1+\sqrt{x+1}}$$
Now the integral is : $$\int^3_0 \frac{2tdt}{1+t}$$ now I need your suggestions.

Thanks.

答:
The substitution is a good idea. The details are not quite right. We get $dx=2t\,dt$, and therefore our definite integral is equal to
$$\int_{t=1}^2 \frac{2t}{1+t}\,dt.$$
Now can you take over? It may be useful to note that $\frac{2t}{1+t}=2-\frac{2}{1+t}$.</p>


问:
How to show this conclusion which concerns two distribution functions?
If $F$ and $G$ are two distribution functions, how to prove that $$\int_{0}^{1} \lvert F^{-1}(t)-G^{-1}(t)\rvert dt=\int_{-\infty}^{+\infty}\lvert F(x)-G(x)\rvert dx$$ with $$F^{-1}(p)=\inf\{x\in\mathbb{R},F(x)\geq p\}$$ I have no idea to deal with the integral of the absolute value of the difference of two distribution functions. The left side is in fact equal to $\mathbb{E}[\lvert F^{-1}(U)-G^{-1}(U)\rvert]$ where $U$ is in the uniform distribution on $[0,1]$.\
This is an important exercice in my course and I really need someone who can help me solve it.

答:
I can answer this question under his help.

Let $\Phi(x,t)=\mathbb{1}_{\{F(x)\geq G(x),t\in [G(x),F(x)]\}}+\mathbb{1}_{\{F(x)<G(x),t\in [F(x),G(x)]\}}$, then we have $$\lvert F(x)-G(x)\rvert=\int_{0}^{1}\Phi(x,t)dt$$
so using the Fubini's theorem, we have
\begin{equation}
\begin{split}
\int_{-\infty}^{+\infty}\lvert F(x)-G(x)\rvert dx&=\int_{-\infty}^{+\infty}\int_{0}^1 \Phi(x,t)dtdx\\
&=\int_0^1(\int_{-\infty}^{+\infty}\Phi(x,t) dx)dt\\
&=\int_0^1\lvert F^{-1}(t)-G^{-1}(t)\rvert dt
\end{split}
\end{equation}
because $$\int_{-\infty}^{+\infty}\Phi(x,t) dx=\lvert F^{-1}(t)-G^{-1}(t)\rvert,\forall t\in[0,1]$$

Is it correct? I want to ask if it is correct.


答: Using the Bezout relation $1=12\times3-7\times5$, we have $(a^{12})^3/(a^7)^5=a\in\mathbb Q$.


问:
Integral using Gamma and Beta function
the integral I have found this integral on past exams papers for a course in Mathematical Methods of Physics. I really cant find any material on the textbook which could be helpfull for the solution. Do you have any idea how to solve this or can you suggest any material I could use? I post the question in the picture bellow. I have to express the solution using Beta and Gamma functions. Any help would be valuable. Thanks in advance!

Thanks for the edit, its the 1st time I post here, sorry if my post was wrong

$$\int_{\mathbb{R}^n}(a+|x|)^m\exp\left(-\lambda\left[\sum_{i=1}^nx_i^2\right]^k\right)d^nx$$
$$x=\mathbb{R}^n,k>0,m\in\mathbb{N},\Re(\lambda)>0$$

答:
Deriving the Surface Area of a Sphere

Start with the $1$ dimensional integral
$$
\int_{-\infty}^\infty e^{-\pi x^2}\,\mathrm{d}x=1\tag1
$$
and take the product of $n$ copies:
$$
\int_{-\infty}^\infty\int_{-\infty}^\infty\cdots\int_{-\infty}^\infty e^{-\pi \left(x_1^2+x_2^2+\cdots+x_n^2\right)}\,\mathrm{d}x_1\,\mathrm{d}x_2\cdots\,\mathrm{d}x_n=1\tag2
$$
Since the integrand is constant over spherical shells, we can easily write it as a spherical integral
$$
\int_0^\infty e^{-\pi r^2}\omega_{n-1}r^{n-1}\,\mathrm{d}r=1\tag3
$$
where $\omega_{n-1}$ is the surface area of the $n-1$ dimensional unit sphere.

Thus, we get
$$
\begin{align}
1
&=\int_0^\infty e^{-\pi r^2}\omega_{n-1}r^{n-1}\,\mathrm{d}r\tag{4a}\\
&=\frac{\omega_{n-1}}2\int_0^\infty e^{-\pi r^2}r^{n-2}\,\mathrm{d}r^2\tag{4b}\\
&=\frac{\omega_{n-1}}2\int_0^\infty e^{-\pi r}r^{n/2-1}\,\mathrm{d}r\tag{4c}\\
&=\frac{\omega_{n-1}}{2\pi^{n/2}}\int_0^\infty e^{-r}r^{n/2-1}\,\mathrm{d}r\tag{4d}\\
&=\frac{\omega_{n-1}}{2\pi^{n/2}}\Gamma(n/2)\tag{4e}
\end{align}
$$
Explanation:

$\text{(4a)}$: copy $(3)$

$\text{(4b)}$: $r\,\mathrm{d}r=\frac12\mathrm{d}r^2$ and pull the constant out front

$\text{(4c)}$: substitute $r\mapsto r^{1/2}$

$\text{(4d)}$: substitute $r\mapsto r/\pi$

$\text{(4e)}$: definition of the Gamma function

Therefore, we get
$$
\omega_{n-1}=\frac{2\pi^{n/2}}{\Gamma(n/2)}\tag5
$$

Application to the Question

We can write the integral from the question as an integral on spherical shells:
$$
\begin{align}
&\int_0^\infty\overbrace{(a+r)^me^{-\lambda r^{2k}}}^{\substack{\text{integrand on a}\\\text{shell of radius $r$}}}\overbrace{\omega_{n-1}r^{n-1\vphantom{r^2}}}^{\substack{\text{surface area}\vphantom{g}\\\text{of the shell}}}\,\mathrm{d}r\tag{6a}\\
&=\frac{\omega_{n-1}}{2k}\sum_{j=0}^m\binom{m}{j}\int_0^\infty a^{m-j}r^{\frac{j+n}{2k}-1}e^{-\lambda r}\,\mathrm{d}r\tag{6b}\\
&=\frac{\omega_{n-1}}{2k}\sum_{j=0}^m\binom{m}{j}a^{m-j}\lambda^{-\frac{j+n}{2k}}\int_0^\infty r^{\frac{j+n}{2k}-1}e^{-r}\,\mathrm{d}r\tag{6c}\\
&=\frac{\omega_{n-1}}{2k}\sum_{j=0}^m\binom{m}{j}a^{m-j}\lambda^{-\frac{j+n}{2k}}\Gamma\!\left(\frac{j+n}{2k}\right)\tag{6d}
\end{align}
$$
Explanation:

$\text{(6a)}$: integral from the question

$\text{(6b)}$: substitute $r\mapsto r^{\frac1{2k}}$ and apply the Binomial Theorem

$\text{(6c)}$: substitute $r\mapsto r/\lambda$

$\text{(6d)}$: definition of the Gamma function

Title: Integral using Gamma and Beta function


问:
The largest possible $m$ so that you are guaranteed to have a blue copy of $K_m$.
Let $G$ be a complete graph on $25$ vertices with each edge coloured red or blue. If there are no red triangles, what is the largest possible $m$ so that you are guaranteed to have a blue copy of $K_m$?

I tried using Ramsey's Theorem but could not proceed. Need some help in this.

答:
Since you know Ramsey theory ...

Hint: $R(3,7) = 23. R(3,8) = 28$

So the answer is $m=7$.

If you're not allowed to quote the result, find a way to show that $ R(3,7) \leq 25$, likely by induction on:

$$R(k, l) \leq \begin{cases} R(k-1,l) + R(k, l-1) -1 & \text{if } R(k-1,l), R(k, l-1) \text{are both even}    \\ R(k-1,l) + R(k, l-1)  & \text{otherwise} \end{cases}$$


问:
Does $\left\vert\mathrm{sech}^2(x)\mathrm{tanh}(x)\mathrm{tanh}\left(\dfrac{x}{10a}\right)\right\vert\leq\dfrac{10}{a^2}$ hold?
Playing a little bit with wolfram I "found" an interesting (?) inequailty
$$
\forall x\in\mathbb{R}, \qquad \left\vert\mathrm{sech}^2(x)\mathrm{tanh}(x)\mathrm{tanh}\left(\dfrac{x}{10a}\right)\right\vert\leq\dfrac{10}{a^2},
$$
for all parameters $a\gg 1$. Here, when I say that I "found" this inequality, I mean that I check it for several huge values of $a$, so I was wondering if it actually holds for all $a$ sufficiently large, let's say, $a>10$. What confuses me is that, at first order, the left-hand should behave as $\tfrac{1}{a}x^2$, so it doesn't looks that easy to obtain a $a^{-2}$ on the right-hand side. Does anyone knows how to prove something like this?

答:
It is false.

For $a\ge 313$ let's say, $a=400$ we have that LHS
$$\tanh \left(\frac{x}{4000}\right) \tanh (x) \text{sech}^2(x)$$
has a max $0.00008$ at $x=1.0095$, larger than RHS value $\frac{10}{400^2}=0.0000625$


问:
Creating subgroups in GAP
Suppose we have a group G:

```<code>F := FreeGroup(3);
G := F / [F.1^3, F.2^4, F.3^5, F.1*F.2*F.1^-1];
</code>```

(the relations don't matter). How do we get a subgroup of G generated by G.1 and G.2?

答: According to the manual, simply use Subgroup(G, [G.1, G.2]);.

Title: Creating subgroups in GAP


答:
Your post was full of typos, which might account for your problems. You can put:

```<code>H := Subgroup(G, [G.1,G.2]);
</code>```

Or, for more general situations:

```<code>gens := GeneratorsOfGroup(G);;
H := Subgroup(G, [gens[1], gens[2]] );
</code>```

Title: Creating subgroups in GAP

Body:
Suppose we have a group G:

```<code>F := FreeGroup(3);
G := F / [F.1^3, F.2^4, F.3^5, F.1*F.2*F.1^-1];
</code>```

(the relations don't matter). How do we get a subgroup of G generated by G.1 and G.2?

Body:
Your post was full of typos, which might account for your problems. You can put:

H := Subgroup(G, [G.1,G.2]);

Or, for more general situations:

gens := GeneratorsOfGroup(G);;
H := Subgroup(G, [gens[1], gens[2]] );


问:
Spivak's Calculus: Chapter 12, Problem 26 - Follow Up
Chapter 12 - Problem 26)

Suppose that $f(x) > 0$ for all $x$, and that $f$ is decreasing. Prove that there is a continuous decreasing function $g$ such that $0 < g(x) \le f(x)$ for all $x$.

So this question has already been asked and "solved" on MSE 9 years ago here, but the accepted answer isn't very detailed, and I think it is in fact flawed (or I've just misunderstood it). I tried commenting to open up the question again, but it seems pretty dead now, hence why I'm making this follow up.

The answer says to "make $g$ piecewise linear with $g(n) = f(n+1)$". Can someone explain what this means exactly? I will write my thoughts below, but it's a lot, so feel free to skip.

My thoughts: Notice that if we try to simply let $g(x) = f(x+1)$, then it works perfectly except for the fact that $g$ may not be continuous, because $f$ need not be continuous (otherwise we could just let $g(x) = f(x)$ in that case!). So if we could just modify this $g$ to make it continuous somehow then we're done.

Fortunately, $f$ is decreasing on $\mathbb{R}$, which means the left and right limits do exist, however they might disagree. This means $f$ can only have jump discontinuities that jump downwards.

So what if we took all the points in $\mathbb{R}$ where $f$ has a jump discontinuity, and just joined lines between them? (I think this is what the answer meant by piecewise linear function?) This would guarantee that $g$ is continuous, however, this approach has some fixable flaws.

First flaw, for starters, it isn't necessarily true that this $g$ would be always smaller than $f$! For example, consider this picture, where $f$ is the red function, and $g$ is the black function:

Sure $g$ is continuous now, but we've lost the $g(x) \leq f(x)$ property! We can fix this easily by letting $g$ be the smaller of the piecewise linear function and $f$. Then the picture becomes like this:

To make this a bit more rigorous, first we need the set of all the points where $f$ is discontinuous:

$S = \big\{x: \lim_{y \rightarrow x^-} (f(y)) = \lim_{y \rightarrow x^+} (f(y)) \big\}$

Then let $l(x)$ be the piecewise linear function joining all points $\big(x, \lim_{y \rightarrow x^+} [f(y)] \big)$, where $x \in S$.

Then finally let $g(x) = \text{Min}\big(f(x), l(x)\big)$.

Now this would work fine, so long as $l(x)$ is well defined. But must it necessarily be so? I'm not sure, and this is where I'm stuck. For example, what if the set $S$ contains not isolated points, but an entire interval of points? For example, what if $f$ is a function that has a jump discontinuity at every point in $[0,1]$? Then to construct $l(x)$, we'd need to join all these jump discontinuity points in $[0,1]$, of which it isn't obvious at all we can do that.

Now you might say that an interval of jump discontinuities is impossible, and you'd be right. However the proof of that comes much much later in the book and is certainly beyond the knowledge of this chapter. But more importantly, even if $f$ doesn't have an interval of jump discontinuities, there are other ways $l(x)$ can be questionable.

Consider this monstrous example: $f(x) =
\begin{cases}
1-\frac{x}{2^{\lfloor 1 - \log_2(|x|)-1 \rfloor}}  & :x \leq \frac{1}{2} \\
\frac{6}{6x+5} & :x > \frac{1}{2}
\end{cases} \Biggr\}$

Looks something like this (click here to view in Desmos):

As it turns out, this $f$ satisfies the questions premises, but it also has the cool property that it contains an infinite number of jump discontinuities in any neighbourhood around 0! As such, in order to construct the piecewise linear function for it, you'd have to join lines between an infinite number of points, and still have a function, which might be possible? But for sure it isn't obvious that it is so...

Those are my thoughts on the problem. So Q1) Is my approach so far in the right direction? Or could it be that I've missed some super simple trick that will make the problem trivial and everything I've said above redundant? Q2) If I am in the right direction, how can I justify taking a linear piecewise function for an infinite number of points in a given interval?

答: Another point of view : define $$g(x)=\int_{x}^{x+1} f(t) dt$$


问:
tightness and convergence for Hilbert spaces
Let $(X_n,n\geqslant 1)$ and $(Y_n,n\geqslant 1)$ be sequences of  stochastic processes taking values in some Hilbert space $H$ and defined on the same probability space. Assume that $(X_n,n\geqslant 1)$ converges weakly to some limit process.

Next, suppose that $\lVert Y_n\rVert_{H} \leq \lVert X_n\rVert_{H}$ for all $n$.

Is $(Y_n,n\geqslant 1)$ tight?

In general, I am confused with the following: Assume me have a majorating  weakly convergent process. Then, another process, majorated by convergent one, is not necessarily convergent, but it it not (is it?) tight either.

答:
Answer

In general, your conjecture is not true. You can see it by the obversation that:  One direct implication from your conjecture is that

"Every sequence of random variables bounded in $H$  is tight"

Which is clearly wrong for any infinite-dimensional normed vector space (not hibert space alone).

For example,taking $H= l_2$ ; $Y_n= e_n$ . The sequence $(e_n)$ is bounded, while it is clearly not a relatively compact subset of $H$.

(Though it converges in weak topology, but this topology is not the canonical topology when we regard $H$ as a normed vector space)

Comments

My two-cent comment: If I'm not wrong, what you want to search is essentially the relation between the tightness and the norm of value space (the codomain). As I showed in the previous example, the first step you need to overcome is to form the bridge between the compacity and norm. However, while that relationship is clear in finite dimension, it is extremely weak when you deal with infinite-dimensional space. Much work needs to be done, some change in approach might be nessessary.


问:
How can all matrices have a _unique_ reduced row echelon form?
Wikipedia states (and I suspect the answer to my question could be that wikipedia should never have been my source) that

every matrix has a unique reduced row echelon form.

I am wondering how this can possibly be a unique matrix when any nonsingular matrix is row equivalent to the identity matrix, which is also their reduced row echelon form.

I am considering two possible interpretations that would result in Wikipedia's statement holding, which are (1) two row equivalent matrices are considered the same matrix$^1$ or (2) when Wikipedia says unique they mean only a single, rather than different from all others$^2$.

Because I am only a sophomore it seemed unwise for me to guess by myself which, if any, of these explanations would be the correct one, which is why I am once again asking for your help: what is meant when Wikipedia states "every matrix has a unique reduced row echelon form."

$^1$ if this were so than the conclusion that all matrices have a unique reduced row echelon form would be trivial since matrixes that have the same reduced row echelon form would just be defined as being the same matrix

$^2$ this would imply that they mean that for every matrix there exists one and only one reduced row echelon form, rather than two matrices have the same row echelon form if and only if they are the same matrix

答:
I think that you have misinterpreted the statement. The statement "every matrix has a unique row-echelon form" can be restated as follows:

For every matrix $A$, there exists exactly one matrix $B$ such that $A$ is row-equivalent to $B$ and $B$ is in reduced row-echelon form (rref).

As an example, consider the matrices
$$
A_1 = \pmatrix{1&2\\0&3}, \quad A_2 = \pmatrix{5&-1\\-1 & 7}, \quad I = \pmatrix{1&0\\0&1}.
$$
Both $A_1$ and $A_2$ are invertible, so $I$ is the rref of both of these matrices. $I$ is the only matrix that is in rref and row-equivalent to $A_1$, so it is the rref of $A_1$. $I$ is also the only matrix that is in rref and row-equivalent to $A_2$, so it is the rref of $A_2$. The fact that $A_1$ and $A_2$ have the same rref does not contradict the fact that they have exactly one rref, i.e. "a unique" rref.


问:
Trace of an Inverse Matrix
I want to know if there is a way to simplify, or a closed form solution of $tr(\Sigma^{-1})$ where $\Sigma$ is a symmetric positive definite matrix.</p>

答: Let $A$ be symmetric positive definite matrix hence $\exists$ a diagonal matrix $D$ whose diagonal entries are nonzero and $A=P D P^{-1}$ so $A^{-1} = P D^{-1} P^{-1}$ and $Tr(A^{-1})= Tr(D^{-1})$.  Now $D$ being diagonal matrix with non zero diagonal entries $D^{-1}$ has diagonal entries reciprocal of the diagonal entries of $D$ so $Tr(D^{-1})$ is sum of the inverses of the diagonal entries of $D$. </p>

Title: Trace of an Inverse Matrix


问:
What is the limit of sum of split Wiener process?
I wish to find the limit in $L^2$ for $n\rightarrow\infty$ of

$A_n = \sum_{i=1}^nW^a(it/n)-W^a((i-1)t/n))(W^b(it/n)-W^b((i-1)t/n))$

How can I go about that?

答:
I'm sorry for not doing this in TeX-form but it is easier to understand in handwritten form.

Fill free to ask if you don't understand smth in my solution.


问:
If $f,g$ are continuous at $a$ and $f(a)<g(a)$, then $f(x)<g(x)$ for $\left|x-a\right|<\delta$
If $f,g\colon X\to \mathbb{R}$ are continuous at $a \in X$ and $f(a)<g(a)$, then there is a $\delta>0$ such that $f(x)<g(x)$ for all $x\in X$ with $\left|x-a\right|<\delta$.

If $f$ and $g$ are continuous at $a\in X$, then for all $\varepsilon>0$ there is a $\delta_f>0$ and a $\delta_g>0$ such that for all $x\in X$ we have

$$\left|x-a\right|<\delta_f\implies \left|f(x)-f(a)\right|<\varepsilon$$
as well as $$\left|x-a\right|<\delta_g\implies \left|g(x)-g(a)\right|<\varepsilon$$

Choose $\varepsilon=g(a)-f(a)$, there there'll be $\delta_f,\delta_g>0$ and let's define $\delta=\min{(\delta_f,\delta_g)}$.

Then we'll have for all $x\in X$ that $$\left|x-a\right|<\delta\implies \left|g(x)-g(a)\right|<\varepsilon \ \text{and}\ \left|f(x)-f(a)\right|<\varepsilon$$

Since $\varepsilon=g(a)-f(a)$, we have

$$\left|g(x)-g(a)\right| <g(a)-f(a) \implies f(a)<g(x)$$

and

$$\left|f(x)-f(a)\right| <g(a)-f(a) \implies f(x)<g(a)$$

But from now I don't really know how to arrive at $f(x)<g(x)$, as $f(a)<g(x)$ and $f(x)<g(a)$ don't really imply much.

答:
You can simplify a lot the proof by denoting $h(x) = g(x) - f(x)$. You have $h(a) >0$ and have to prove that $h(x) >0$ in a neighborhood of $a$.

By continuity of $h$ at $a$, it exists $\delta > 0$ such that $\vert h(x) -h(a)\vert < \frac{ h(a)  }{2}$ for $\vert x-a \vert < \delta$.

As $\vert h(x) -h(a)\vert < \frac{ h(a)  }{2}$ implies $h(x) > \frac{h(a)}{2} >0$, you're done.


问:
Why six Trigonometric Ratios
Why should we use six trigonometric ratios when all the three sides of a right angled triangle (opposite, adjacent and hypotenuse) are connected with the three rations sine, cosine and tangent?
what is the significance of inverses of these ratios(cosec, sec and cot)?</p>

答:
It is not particularly helpful to have $6$ trigonometric functions when you only consider them in the context of right-angled triangles. If you simply want to compute the lengths of the sides of a triangle, and the angles in between them, then $\sin$, $\cos$, and $\tan$ are all you need. However, the reciprocal functions have nice geometric interpretations, one of which is given here:

The 'co-functions' have another interpretation. $\cos \theta$ is sine of the co-angle of $\theta$. So, for example, in a right-angled triangle with angles $\{60,30,90\}$, $\cos 30 = \sin 60$, since $60$ is the co-angle of $30$ (the two angles have to add up to 90).

(Incidentally, this is how the function tangent got its name. It is the length of the line segment that is tangent to a point $P$, with the end points of the line segment being the $x$-axis and $P$.)

That being said, we could do just fine without the reciprocal functions. 3Blue1Brown calls them tattoos on math.

Title: Why six Trigonometric Ratios


问:
Creating a version of conical helix (parametric) about the border of a paraboloid.
Im just getting to know a little more about helix's and i thought it was cool to visualize these helix's in 3D as regular helix's that extend uniformly across the z axis but what was far more interesting was the visualizations of conical as well as spherical spirals. They usually operate by making the radius of the spiral a function of Phi but naturally after examining this i began to wonder what a spiral about the edge of a paraboloid would look like and how you would represent this parametrically. I should start by asking if these equations exist somewhere where i could see them and if not what would they be.

In other-words for a paraboloid with the equations

```<code>x(θ,φ) = cos(φ) * θ
y(θ,φ) = sin(φ) * θ
z(θ,φ) = θ^2
</code>```

what are the equations for a spiral whos radius is about the border of the paraboloid

答:
Okay i ended up finding a paper exploring spirals on surfaces of revolutions which describes the derivation for these equations

Equations


问:
If $A$ is invertible show that $\det(A) \neq 0$
If $A \in M_{n\times n}(F)$ is not invertible then the rank of $A$ is less than $n$,

thus $\det(A) =0$.

I proved that way, but looks like too simple so I think maybe there is a trick that I missed.

Or I proved it right?

Modification:

($\Rightarrow$) If $A$ is invertible, $AB=I$, $\det(AB)=\det(A)\det(B)=1$ so $\det(A) \neq 0$

($\Leftarrow$) If $\det(A) \neq 0$ , by Cramer's rule, $Ax=b$ has a unique solution and it means $A$ is invertible ($x=A^{-1}b$)

Now I think I complete the proof.

答:
Your proof of the $\Rightarrow$ implication is right. For the other one, do you know about the Adjugate Matrix? To each $A\in M_{n\times n}(F)$ we can associate a matrix $adj(A)$ defined as the tranpose of the cofactor matrix of $A$ such that one always has $A\cdot adj(A)=\det (A) I_{n\times n}$. If $\det(A)\neq 0$ then $\det(A)$ is invertible and therefore you get $A^{-1}=(\det(A))^{-1}adj(A)$.

Notice that the result can be actually strengthened: given a unitary commutative ring $R$, a matrix $A\in M_{n\times n}(R)$ is invertibile if and only if its determinant is invertible in $R$. The proof of this fact goes exactly as the former one just by noticing that what we really used previously was the fact that any non zero element in a field is invertible, as I underlined.


问:
$a \in \mathbb{R}$ and $a^7, a^{12} \in \mathbb{Q}$. Prove that $a \in \mathbb{Q}$.
Let $a \in \mathbb{R}$ and $a^7, a^{12} \in \mathbb{Q}$. Prove that $a \in \mathbb{Q}$.

my solution:

$a^7 \in \mathbb{Q} \Rightarrow a^{14} \in \mathbb{Q}$

$a^{14}=a^{12}\cdot a^{2}\in \mathbb{Q}$, but $a^{12} \in \mathbb{Q} \Rightarrow$ $a^2 \in \mathbb{Q}$.

Using the same rule:
$a^{21} \in \mathbb{Q}$ and $a^{24}\in \mathbb{Q} \Rightarrow$ $a^{24}=a^{2} \cdot a \cdot a^{21} \Rightarrow a \in \mathbb{Q} $.

Is there any other solution?

答:
Here is general way. You need to solve this linear equation

$$\dfrac {a^{12x}}{a^{7y}}=a \Longrightarrow 12x-7y=1$$
where $x,y\in \mathbb {Z^+}$

Now, $\gcd (12,7) =1$ , then we always have a solution.  Thus, it is not even necessary to solve this equation with the Euclidean algorithm.


问:
Prove that $\arctan(\operatorname{arctanh}(x))\geq \operatorname{arctanh}(\arctan(x))$ for $0\leq x <1$
It's inspired by a general result proposed in a book (Dictionary of inequalities second edition by Peter Bullen p.101) related to composition of  function :

In fact we have $x\in[0,1)$:

$$\arctan(\operatorname{arctanh}(x))\geq \operatorname{arctanh}(\arctan(x))\quad (1)$$

To prove it I have introduced a function :

$$f(x)=\arctan(\operatorname{arctanh}(x))-\operatorname{arctanh}(\arctan(x))$$

We differentiate to get:

$$f'(x)= -\frac{1}{((x^2-1)(\operatorname{arctanh}^2(x)+1))}+\frac{1}{((x^2+1)(\arctan^2(x)-1))}$$

And then I use :

Let $0\leq x <1$ then we have :

$$\operatorname{arctanh}(x)\geq x \geq \arctan(x)$$

Then I cannot conclude directly .

Question :

How to prove $(1)$ ?

Thanks in advance !

答:
MacLaurin expansion
$$\arctan\left(\operatorname{arctanh} x\right)=x+\frac{x^5}{15}+\frac{x^7}{45}+\frac{64 x^9}{2835}+\frac{71 x^{11}}{4725}+\frac{5209 x^{13}}{405405}+\frac{2203328 x^{15}}{212837625}+\ldots$$
All coefficients of the series are positive
$$\operatorname{arctanh}\left(\arctan x\right)=x+\frac{x^5}{15}-\frac{x^7}{45}+\frac{64 x^9}{2835}-\frac{71 x^{11}}{4725}+\frac{5209 x^{13}}{405405}-\frac{2203328 x^{15}}{212837625}+\ldots$$
Coefficients have the same absolute value as the previous series, but alternating signs from $5$th degree on.

Thus for $x\in[0,1)$
$$\arctan\left(\operatorname{arctanh} x\right)\ge \operatorname{arctanh}\left(\arctan x\right)$$
while for $x\in(-1,0]$ the opposite happens.


问:
What is the homotopy type of $S^3 \times S^3$ - 4 points?
What is the homotopy type of $S^3 \times S^3$ - 4 points?
Is there a compact CW-complex of $S^3 \times S^3$ - 4 points,
which is a deformation retract of $S^3 \times S^3$ - 4 points and has dimension < 6?

答:
The homology of $X=S^3 \times S^3- \{1,2,3,4\}$ is nontrivial only for $H_3(X)=\mathbb{Z}\oplus\mathbb{Z}$ coming from the wedge of 3-spheres and $H_5(X)=\mathbb{Z}^3$ coming from the spheres enclosing each of the first three points  (this can also be seen with Poincare duality for manifolds with boundary). The sphere enclosing the fourth point is homologous to a combination of the first three.

Let's assume $\{1,2,3\}$ are chosen near the wedge of 3-spheres. Then we may take it so that each sphere enclosing one of the $\{1,2,3\}$ touches the wedge exactly once. Let $A$ be the union of the wedge of the two 3-spheres and the three 5-spheres.

The inclusion $A \rightarrow X$ is clearly a homology equivalence, so it is a homotopy equivalence since $A$ and $X$ are simply connected. Since $X$ can be given a CW structure with $A$ a subcomplex, we can apply the result that if the inclusion of a subcomplex is a homotopy equivalence, the subcomplex is a deformation retract.

Hence, $Y$ is what you're looking for. It is homotopy equivalent to a bouquet of two 3-spheres and three 5-spheres. This is the lowest dimensions CW complex possible, since the homology of $X$ is nontrivial in dimension 5.

I recommend reproducing this argument explicitly for $S^1 \times S^1$ and removing two points.


问:
Could this be the subgroup test described by one axiom?
Proposition: Let $H$ be a non-empty subset of a group $G$ w.r.t. some binary operation, $*$. We have:

$H$ is a subgroup of $G$ $\iff$ $(\forall h_1, h_2 \in H) (h_1 * h_2 ^{-1} \in H)$.

I start the first part of the proof by the supposition that $H$ is a subgroup of $G$ w.r.t. the binary operation $*$. We already know that if $H$ is a subgroup of $G$, then for all $h_1, h_2 \in H$, $h_1 * h_2 \in H$, and $h_1 \in H \implies h_1 ^{-1} \in H$ and $h_2 \in H \implies h_2 ^{-1} \in H$ so that $h_1 * h_1 ^{-1} = e \in H$ and $h_2 * h_2 ^{-1} = e \in H$, where $e$ is the unique identity element in $H$.

How could one relate this information to the fact that $(\forall h_1, h_2 \in H) (h_1 * h_2 ^{-1} \in H)$?

Kind regards.

答:
Hint: By definition, a nonempty subset $H$ of $G$ forms a subgroup of $G$ iff
(1) for all $h,h'\in H$, we have $h\cdot h'\in H$, and (2) for each $h\in H$, we have $h^{-1}\in H$.

These two conditions are equivalent to the subgroup criterion: for all $h,h'\in H$, we have $h\cdot h'^{-1}\in H$.

This is easy to validate.


问:
Given $\phi=(12)(34)(56)$, $\delta=(34)(56)$ and $\lambda=(126)$, is $\langle\phi\delta,\lambda\rangle $ cyclic in $S_6$?
Let $\,\,\,\sigma$, $\tau$ $\in S_8\,\,\,\,$ s.t. $\,\,\,\sigma = (123)(456)\,\,\,$ and $\,\,\,\tau=(14)(25)(36)(78)$

Let $\,\,\,\phi$, $\delta$, $\lambda \in S_6\,\,\,$ s.t. $\,\,\,\phi = (12)(34)(56)$, $\,\,\,\delta = (34)(56)\,\,\,$ and $\,\,\,\lambda = (126)$

Say if $\,\,\,\langle \sigma, \tau\rangle \,\,\,$ is cyclic in $S_8$ and if $\,\,\,\langle\phi \delta, \lambda\rangle \,\,\,$ is cyclic in $S_6$ and prove that.

About $\,\,\langle \sigma,\tau\rangle \,$, it is easy to observe that  $\,\,\langle \sigma\rangle\langle \tau\rangle \,\, = \,\,\,\langle\sigma, \tau\rangle \,$, furthermore $\,\,\,\langle \tau\rangle \leqslant N_{S_8}(\langle\sigma\rangle)\,\,\,$ and viceversa, eventually $\,\,\,(|\sigma|, |\tau|)=1\, \,\,\,\Rightarrow \,\,\,\,\langle\sigma\rangle \cap\langle \tau\rangle \,\,= \{1\}\,\,\,$; therefore I can say: $\,\,\,\langle\sigma, \tau\rangle \,\,\, = \,\,\,\langle\sigma\rangle\langle\tau\rangle \,\,\, \simeq \,\,\,\langle\sigma\rangle \times\langle\tau\rangle \,\,\, \simeq \,\,\,C_3\times C_2\,\,\, \simeq\,\,\, C_6$

Regarding $\,\,\,\langle\phi \delta, \lambda\rangle \,\,\,$ I would say also yes, but I could not find a proper proof, if I am right, any suggestion?

答:
Note that $\phi\delta=(1\ 2)$, so $\phi\delta$ and $\lambda$ only permute the three elements $1$, $2$ and $6$. In this way we see that $\langle\phi\delta,\lambda\rangle$ is isomorphic to a subgroup of $S_3$. Can you now determine whether this group is cyclic or not?

As for the subgroup $\langle\sigma,\tau\rangle$, a more explicit argument would be to show that
$$\sigma^3=\operatorname{id},\qquad\tau^2=\operatorname{id},\qquad\sigma\tau=\tau\sigma.$$
It follows that $(\sigma\tau)^4=\sigma$ and $(\sigma\tau)^3=\tau$, so $\langle\sigma,\tau\rangle=\langle\sigma\tau\rangle$.


问:
Second degree Taylor Polynomial Questions

Hi so I was wondering if anyone could help me as I'm still not quite grasping the concept of these questions and I'm consistently getting the wrong answer.

So far this is what I have done:

$f(x,y)=e^{xy-2x+5y-6},  f(-1,1)=e^{0}=1$

$f_{x}(x,y)=(y-2)e^{xy-2x+5y-6},  f_{x}(-1,1)=-1$

$f_{y}(x,y)=(x+5)e^{xy-2x+5y-6},  f_{y}(-1,1)=4$

$f_{xx}(x,y)=(y-2)^2e^{xy-2x+5y-6},  f_{xx}(-1,1)=1$

$f_{yy}(x,y)=(x+5)^2e^{xy-2x+5y-6},  f_{yy}(-1,1)=16$

$f_{xy}(x,y)=(x+5)(y-2)e^{xy-2x+5y-6},  f_{xy}(-1,1)=-4$

so then I put my answer together and obtain:

$P_{2}(x,y)= 1-(x+1)+4(y-1)+(1/2)(x+1)^2-4(x+1)(y-1)+(16/2)(y-1)^2$

which I have found to be incorrect as the given answer is:

$P_{2}(x,y)=-4-x+4y+8(y-1)^2+1/2(x+1)^2-3(x+1)(y-1)$

So no I'm completely stuck on where I went wrong. Any help would be great.

答: The actual discrepancy between the two expressions is just the value of $f_{xy}$, otherwise it is just a matter of simplification (and I would say your answer is actually the "better" answer in that aspect). As for why that happened, there is a product rule application required to find $f_{xy}$.


答:
There is no trick in "make $g$ piecewise linear with $g(n)=f(n+1)$". It means that we define for each $n \in \mathbb Z$
$$g_n(x) : [n,n+1] \to \mathbb R, g_n(x) = f(n+1) + (f(n+2) - f(n+1))(x-n).$$
This is a linear function on the interval $[n,n+1]$ with slope $f(n+2) - f(n+1) \le 0$. In particular it is continuous and decreasing. For all $x \in [n,n+1]$ we have

$f(x) \ge f(n+1)$ because $f$ is decreasing.

$f(n+1) \ge g_n(x) \ge f(n+2)$ by linearity.

$f(n+2) > 0$ by assumption.

This shows $f(x) \ge g_n(x) > 0$ for all $x \in [n,n+1]$.

Now define
$$g : \mathbb R \to \mathbb R, g(x) = g_n(x) \text{ for } x \in [n,n+1]. $$
This is a well-defined continuous function because $g_n(n+1) = f(n+2) = g_{n+1}(n+1)$. Clearly $f \ge g > 0$.

See the green curve below:

Title: Spivak's Calculus: Chapter 12, Problem 26 - Follow Up

Body:
Chapter 12 - Problem 26)

Suppose that $f(x) > 0$ for all $x$, and that $f$ is decreasing. Prove that there is a continuous decreasing function $g$ such that $0 < g(x) \le f(x)$ for all $x$.

So this question has already been asked and "solved" on MSE 9 years ago here, but the accepted answer isn't very detailed, and I think it is in fact flawed (or I've just misunderstood it). I tried commenting to open up the question again, but it seems pretty dead now, hence why I'm making this follow up.

The answer says to "make $g$ piecewise linear with $g(n) = f(n+1)$". Can someone explain what this means exactly? I will write my thoughts below, but it's a lot, so feel free to skip.

My thoughts: Notice that if we try to simply let $g(x) = f(x+1)$, then it works perfectly except for the fact that $g$ may not be continuous, because $f$ need not be continuous (otherwise we could just let $g(x) = f(x)$ in that case!). So if we could just modify this $g$ to make it continuous somehow then we're done.

Fortunately, $f$ is decreasing on $\mathbb{R}$, which means the left and right limits do exist, however they might disagree. This means $f$ can only have jump discontinuities that jump downwards.

So what if we took all the points in $\mathbb{R}$ where $f$ has a jump discontinuity, and just joined lines between them? (I think this is what the answer meant by piecewise linear function?) This would guarantee that $g$ is continuous, however, this approach has some fixable flaws.

First flaw, for starters, it isn't necessarily true that this $g$ would be always smaller than $f$! For example, consider this picture, where $f$ is the red function, and $g$ is the black function:

Sure $g$ is continuous now, but we've lost the $g(x) \leq f(x)$ property! We can fix this easily by letting $g$ be the smaller of the piecewise linear function and $f$. Then the picture becomes like this:

To make this a bit more rigorous, first we need the set of all the points where $f$ is discontinuous:

$S = \big\{x: \lim_{y \rightarrow x^-} (f(y)) = \lim_{y \rightarrow x^+} (f(y)) \big\}$

Then let $l(x)$ be the piecewise linear function joining all points $\big(x, \lim_{y \rightarrow x^+} [f(y)] \big)$, where $x \in S$.

Then finally let $g(x) = \text{Min}\big(f(x), l(x)\big)$.

Now this would work fine, so long as $l(x)$ is well defined. But must it necessarily be so? I'm not sure, and this is where I'm stuck. For example, what if the set $S$ contains not isolated points, but an entire interval of points? For example, what if $f$ is a function that has a jump discontinuity at every point in $[0,1]$? Then to construct $l(x)$, we'd need to join all these jump discontinuity points in $[0,1]$, of which it isn't obvious at all we can do that.

Now you might say that an interval of jump discontinuities is impossible, and you'd be right. However the proof of that comes much much later in the book and is certainly beyond the knowledge of this chapter. But more importantly, even if $f$ doesn't have an interval of jump discontinuities, there are other ways $l(x)$ can be questionable.

Consider this monstrous example: $f(x) =
\begin{cases}
1-\frac{x}{2^{\lfloor 1 - \log_2(|x|)-1 \rfloor}}  & :x \leq \frac{1}{2} \\
\frac{6}{6x+5} & :x > \frac{1}{2}
\end{cases} \Biggr\}$

Looks something like this (click here to view in Desmos):

As it turns out, this $f$ satisfies the questions premises, but it also has the cool property that it contains an infinite number of jump discontinuities in any neighbourhood around 0! As such, in order to construct the piecewise linear function for it, you'd have to join lines between an infinite number of points, and still have a function, which might be possible? But for sure it isn't obvious that it is so...

Those are my thoughts on the problem. So Q1) Is my approach so far in the right direction? Or could it be that I've missed some super simple trick that will make the problem trivial and everything I've said above redundant? Q2) If I am in the right direction, how can I justify taking a linear piecewise function for an infinite number of points in a given interval?


问:
To find a specfic equation $z-\frac{1}{s\ln(s)}-n\ln(s)==0$
Here I want to find the solution to the following equation
$$z-\frac{1}{s\ln(s)}-n\ln(s)=0,$$
where $z$ and $n$ are constants.
Here the aim is to find the solution $s=?$.

Any suggestion is welcome.

答:
Does it help to substitute $x=\ln s$? Then the equation becomes
$$zx - e^{-x} -nx^2=0$$
which can be solved approximately in the limits of large $x$ (then $e^{-x}\approx 0$ and small $x$ (then $e^{-x}\approx 1-x$).


问:
How many 5-digit numbers have distinct digits and are divisible by 3?
Problem: How many 5-digit numbers have distinct digits and are divisible by 3?

My direction: I have tried to find all numbers with 5 digits that are divisible by 3 (30000 numbers) and then I will subtract from it numbers that do not have 5 distinct digits but I can't find it.

答: Hint: first consider how many numbers there are between $10 \, 000$ and $99 \, 999$ such that all of the digits are distinct. The first digit is between $1$ and $9$, so there are $9$ possible choices. Then, there are $9$ choices for the second digit, $8$ choices for the third digit, etc.


问:
Good source to study complex analysis
I am taking a course on Analytic Number Theory. For that, as a pre-requisite, complex analysis is needed. Can anyone give me a video lecture series to start complex analysis? I need to know about holomorphic functions and their singularities, and contour integration(Mentioned by my course instructor).

If anyone can refer a good book, then also it will be helpful.

答:
I think a good way to go about learning complex analysis would be to use this book:
https://www.fing.edu.uy/~cerminar/Complex_Analysis.pdf

along with these lectures:

https://www.youtube.com/watch?v=bfrIk13rAJ4&list=PL71JUoXcec_mmLm9psjMKo1FYBXS9WHb7&ab_channel=StevenMiller

Hope this helps!

Title: Good source to study complex analysis