问:
What is the Op Category?
By what is, I do not mean the construction, the construction is easy. What I mean is what does the Op category really represent to us in terms of the original? Oh sure, it represents the dual... but that is circular reasoning.

For every category $\textbf{C}$ there is a mirror image category $\textbf{C}^{op}$. When we "walk" through $\textbf{C}$ we are walking through $\textbf{C}^{op}$ "contravariantly" BUT $f^{op}$ does something different for us.

So the question really is, what is $f^{op}$ actually doing to the internal structure? If I have some category $\textbf{C}$ and follow some path of arrows from a to b then in $\textbf{C}^{op}$ I have to take the path of arrows from b to a and this somehow tells me something relevant. In some sense it seems like an inverse. It is almost as if $f^{op}$ is some "generalized" inverse that takes us back from where we came. If that sorta is true then $\textbf{C}^{op}$ is sort of an "inverse" category that tracks all changes, a sort of "undo" class that allows us, in theory, to go back.

If that is true then it would not be much different than just saying that for every arrow in a category there is an op-arrow that sort of acts like a generalized inverse that in theory lets us reverse any path of arrows.

I guess this would make total since if a category that has inverse for all arrows is self-dual(hence the generalized inverses are just normal inverses and).

Anyways, I'm trying to get some type of intuitive understanding of the opposite category so that when it is talked about I can actually reason from it in a concrete way. This inverse analogy seems like it would work but it also seems that $f^{op}$ doesn't necessarily, by definition, have to have any relation to $f$, it just has to be any arrow that reverses direction.

This leads me to another problem I have, in the definition of the opposite category it seems that we can actually have many opposite categories. Only the arrow has to be reverses, but the definition doesn't say anything about it having to have any other relationship to $f$ other than exchanging the domain and codomain. Of course, it is clear that this is not true, I haven't been able to figure out way from the book.

I guess this is really ultimately asking how does $f^{op}$ really relate to $f$ rather than just saying it is the opposite?

答:
Here is how I regard the opposite category. If $\mathscr{A}$ is a category, then $\mathscr{A}^{op}$ is obtained by reversing arrows (as you probably know already). What does an arrow $f^{op}$ in $\mathscr{A}^{op}$ mean? Actually,... it means almost nothing. For example, take the category $\textbf{Set}$. A morphism in $\textbf{Set}^{op}$ means what? It is definitely not an ordinary function, because there is a morphism $\S\to\emptyset$ for any set $S$, which makes no sense to call this a function. And let's be clear, the construction of an opposite category is not meant to give some concrete applications. In the end, we first invent it to capture the notion of duality. Given a statement in category theory, you can get the dual statement immediately by invoking the opposite category. But we often rewrite the dual statement to eliminate the opposite category. A basic example is a covariant/contravariant functor. A contravariant functor is a covariant functor from the opposite category. By defining this way, we don't even have to explicitly define what a contravariant functor really is. But it offers no contextual meaning. In the end, we still have to translate it back to the version without using the opposite category. The opposite category is meant to be a tool only, concrete interpretation is almost non-existing.

It also doesn't stop you to learn more category theory without really getting this concept fully. I suggest you continue without worrying too much about this. As you learn more, you probably will understand what opposite category "really is"

Title: What is the Op Category?


问:
Algorithm to solve $p(x)f''(x) + q(x)f(x) = 0$ where $p(x), q(x)$ are polynomials
Consider the following ordinary differential equation.

$$ p(x)f''(x) + q(x) f(x) = 0$$

Here, $p(x), q(x)$ are given polynomials in $x$. For example, in my physics education, I am interested in solving the following.

$$ 4x^2 f''(x) +(-15-8x^2-4x^4)f(x) = 0 $$

What are the general algorithms to solve these kind of differential equations? Any references are also very appreaciated. One of the ways that I know of would be assuming series solution and solve (recursively) for coefficients, but is there something else?

答:
There are difficult to have general algorithms, just solve it case by case.

Let $r=x^2$ ,

Then $\dfrac{df}{dx}=\dfrac{df}{dr}\dfrac{dr}{dx}=2x\dfrac{df}{dr}$

$\dfrac{d^2u}{dx^2}=\dfrac{d}{dx}\left(2x\dfrac{df}{dr}\right)=2x\dfrac{d}{dx}\left(\dfrac{df}{dr}\right)+2\dfrac{df}{dr}=2x\dfrac{d}{dr}\left(\dfrac{df}{dr}\right)\dfrac{dr}{dx}+2\dfrac{df}{dr}=2x\dfrac{d^2f}{dr^2}2x+2\dfrac{df}{dr}=4x^2\dfrac{d^2f}{dr^2}+2\dfrac{df}{dr}=4r\dfrac{d^2f}{dr^2}+2\dfrac{df}{dr}$

$\therefore4r\left(4r\dfrac{d^2f}{dr^2}+2\dfrac{df}{dr}\right)+(-15-8r-4r^2)f=0$

$16r^2\dfrac{d^2f}{dr^2}+8r\dfrac{df}{dr}-(4r^2+8r+15)f=0$

Let $f=r^au$ ,

Then $\dfrac{df}{dr}=r^a\dfrac{du}{dr}+ar^{a-1}u$

$\dfrac{d^2f}{dr^2}=r^a\dfrac{d^2u}{dr^2}+ar^{a-1}\dfrac{du}{dr}+ar^{a-1}\dfrac{du}{dr}+a(a-1)r^{a-2}u=r^a\dfrac{d^2u}{dr^2}+2ar^{a-1}\dfrac{du}{dr}+a(a-1)r^{a-2}u$

$\therefore16r^2\left(r^a\dfrac{d^2u}{dr^2}+2ar^{a-1}\dfrac{du}{dr}+a(a-1)r^{a-2}u\right)+8r\left(r^a\dfrac{du}{dr}+ar^{a-1}u\right)-(4r^2+8r+15)r^au=0$

$16r^2\dfrac{d^2u}{dr^2}+32ar\dfrac{du}{dr}+16a(a-1)u+8r\dfrac{du}{dr}+8au-(4r^2+8r+15)u=0$

$16r^2\dfrac{d^2u}{dr^2}+8(4a+1)r\dfrac{du}{dr}-(4r^2+8r-16a^2+8a+15)u=0$

Choose $-16a^2+8a+15=0$ , i.e. $a=\dfrac{5}{4}$ or $-\dfrac{3}{4}$ , the ODE becomes

$16r^2\dfrac{d^2u}{dr^2}+48r\dfrac{du}{dr}-(4r^2+8r)u=0$ or $16r^2\dfrac{d^2u}{dr^2}-16r\dfrac{du}{dr}-(4r^2+8r)u=0$

$4r\dfrac{d^2u}{dr^2}+12\dfrac{du}{dr}-(r+2)u=0$ or $4r\dfrac{d^2u}{dr^2}-4\dfrac{du}{dr}-(r+2)u=0$

The latter is simpler, since it has trivial particular solution $u=e^{-\frac{r}{2}}$

Let $u=e^{-\frac{r}{2}}v$ ,

Then $\dfrac{du}{dr}=e^{-\frac{r}{2}}\dfrac{dv}{dr}-\dfrac{e^{-\frac{r}{2}}v}{2}$

$\dfrac{d^2u}{dr^2}=e^{-\frac{r}{2}}\dfrac{d^2v}{dr^2}-\dfrac{e^{-\frac{r}{2}}}{2}\dfrac{dv}{dr}-\dfrac{e^{-\frac{r}{2}}}{2}\dfrac{dv}{dr}+\dfrac{e^{-\frac{r}{2}}v}{4}=e^{-\frac{r}{2}}\dfrac{d^2v}{dr^2}-e^{-\frac{r}{2}}\dfrac{dv}{dr}+\dfrac{e^{-\frac{r}{2}}v}{4}$

$\therefore4r\left(e^{-\frac{r}{2}}\dfrac{d^2v}{dr^2}-e^{-\frac{r}{2}}\dfrac{dv}{dr}+\dfrac{e^{-\frac{r}{2}}v}{4}\right)-4\left(e^{-\frac{r}{2}}\dfrac{dv}{dr}-\dfrac{e^{-\frac{r}{2}}v}{2}\right)-(r+2)e^{-\frac{r}{2}}v=0$

$4r\dfrac{d^2v}{dr^2}-4r\dfrac{dv}{dr}+rv-4\dfrac{dv}{dr}+2v-(r+2)v=0$

$4r\dfrac{d^2v}{dr^2}-4(r+1)\dfrac{dv}{dr}=0$

$r\dfrac{d^2v}{dr^2}=(r+1)\dfrac{dv}{dr}$

$\dfrac{\dfrac{d^2v}{dr^2}}{\dfrac{dv}{dr}}=1+\dfrac{1}{r}$

$\ln\dfrac{dv}{dr}=r+\ln r+c$

$\dfrac{dv}{dr}=Cre^r$

$v=C_1(r-1)e^r+C_2$

$e^\frac{r}{2}u=C_1(r-1)e^r+C_2$

$r^\frac{3}{4}f=C_1(r-1)e^\frac{r}{2}+C_2e^{-\frac{r}{2}}$

$f=C_1(r-1)r^{-\frac{3}{4}}e^\frac{r}{2}+C_2r^{-\frac{3}{4}}e^{-\frac{r}{2}}$

$f=C_1(x^2-1)x^{-\frac{3}{2}}e^\frac{x^2}{2}+C_2x^{-\frac{3}{2}}e^{-\frac{x^2}{2}}$

ContentLicense: CC BY-SA 4.0


问:
Example of a continuous function $f$ into the unit circle, such that there is no continuous $g$ with $f=e^{{\rm i}g}$
Let $E$ be a metric space, $$U:=\left\{z\in\mathbb C:\left|z\right|=1\right\}$$ and $f\in C^0(E,U)$. Say that $f$ is inessential if $$f=e^{{\rm i}g}\tag1$$ for some $g\in C^0(E,\mathbb R)$.

I was able to show that if $f(E)\ne U$, then $f$ is inessential. Now I wonder what can possibly go wrong if $f(E)=U$.

The most trivial example of $f(E)=U$ should be $$E=\{(x,y)\in\mathbb R^2:x^2+y^2=1\},$$ $g$ being the inverse of the polar coordinate map $$(-\pi,\pi]\to E\;,\;\;\;\phi\mapsto\left(\cos\phi,\sin\phi\right)\tag2$$ and $f:=e^{{\rm i}g}$. Now $(1)$ is trivially satisfied by the very definition of $f$.

Now, clearly, this is a very special construction. Does anyone know a counterexample of $(1)$ when $f(E)=U$?

答: If I understand correctly, you are looking for a continuous surjective map $f:E\longrightarrow U$ such that there exists no continuous map $g:E\to\mathbb R$ with $f=\mathrm e^{\mathrm ig}$. An example is the projection $f:\mathbb C\backslash\{0\}\longrightarrow U,~r\mathrm e^{\mathrm i\varphi}\mapsto\mathrm e^{\mathrm i\varphi}$. A fitting $g:\mathbb C\backslash\{0\}\longrightarrow \mathbb R$ would be a continuous branch of the argument function $\operatorname{arg}$ on the punctured plane. But it is known that such a branch does not exist. Otherwise $\mathrm i \operatorname{arg}(z)+\ln\vert z\vert$ would be locally an inverse of $\exp$, and thus locally an antiderivative of $\frac1z$, and since it's defined on the maximal domain of $\frac1z$, it would be a global antiderivative, but then all contour integrals over $\frac1z$ would vanish, which we know to be false.


问:
Find $AH$, where $H$ is the orthocenter of $\Delta ABC$

In an acute-angled triangle $\Delta ABC$ , points $D,E,F$ are feet of the perpendiculars drawn from $A,B,C$ onto $BC,CA,AB$ respectively. Suppose $\sin A = \frac{3}{5}$ and $BC = 39$. Find $AH$, where $H$ is the orthocenter of $\Delta ABC$ .

What I Tried: Here is a picture :-

I am a bit weak at trigonometry, I had learnt the use of $\sin,\cos,\tan$ and all in right-angled triangles. Next I came to know that we have the Law of sines and cosines. The Law of sines tell us :-
$$\frac{\sin A}{BC} = \frac{\sin B}{AC} = \frac{\sin C}{AB} = \frac{3}{195}$$

From here, I don't know how to proceed. Some other ideas include that $H$ must be inside $\Delta ABC$, as the triangle is acute-angled. But is that intended?

Can anyone help? Thank You.

答:
See here.

Use formula $$AH = 2R \cos A = BC \cot A $$

 $$\cot A = \dfrac{4}{3} $$
 $$ AH = 52$$

Summary of proof for $AH=2R\cos A$  in the link:

$BFCH$ as constructed is a parallelogram. Opposite sides are equal hence we have this formula.


问:
Dirichlet's/Abel's test like convergence to $0$ for absolutely convergent series
Consider the series $S_N = \sum_{k=0}^N a_k b_k$.
Let:

$a_k > 0$

$\sum_{k=0}^N a_k < M$ for any $N$

$a_k \rightarrow 0$ as $N \rightarrow \infty$

$b_k = \beta^{N-k}$ with $\beta \in (0,1)$.

I am pretty sure that this implies convergence of $\sum_{k=0}^N a_k b_k$ to $0$ as $N \rightarrow \infty$, because $a_k$ is finite and $b_k$ tends to $0$ for increasing $N$. But unfortunately, I'm not able to prove it.

The problem is that all available tests/theorems aim to prove convergence to a finite value instead of to zero. I tried a Abel's test/Dirichlet's test like approach (with summation of parts) but this alsways ends up in convergence to a nonzero value.

Any help and hint is appreciated. Thanks!

答:
Just write
$$\sum_{k=0}^N a_kb_k = \sum_{k=0}^N a_k \beta^{N-k}=\frac{\sum_{k=0}^N a_k \beta^{-k}}{\beta^{-N}}$$

and apply Kronecker's lemma.


答:
Welcome in MSE.

This is tricky! The main tool we use is the summation by parts, which is an analogue of the integration by parts in the discrete world. Setting

$$T_n := \sum_{k=0}^n \frac{a_k}{\beta^k} $$

You are asking whether $\beta^n T_n$ tends to zero. Notice that $(T_p - T_{p-1})\beta^p = a_p$. Summing from $p=0$ to $p=N$ and using the second condition we have

$$ M > \sum_{p=0}^N (T_p - T_{p-1}) \beta^p $$

And applying summation by parts

$$ M > \beta^N T_{N+1} - T_0 - \sum_{p=1}^N T_p (\beta^p - \beta^{p-1} ) = \beta^n T_{n+1} - T_0 + (1-\beta)\sum_{p=1}^N T_p \beta^{p-1} $$

$$ > - T_0 + (1-\beta) \sum_{p=1}^{N+1} T_p \beta^{p-1} $$

Which implies

$$ \frac{\beta(M+T_0)}{1-\beta} > \sum_{p=1}^{N+1} T_p \beta^p $$

Now the right hand side is a bounded and positive series in $N$, hence it will converge for $N \to \infty$. Since it converges, the general term $T_p \beta^p$ must converge to zero, which was our thesis!

Title: Dirichlet's/Abel's test like convergence to $0$ for absolutely convergent series

Body:
Consider the series $S_N = \sum_{k=0}^N a_k b_k$.
Let:

$a_k > 0$

$\sum_{k=0}^N a_k < M$ for any $N$

$a_k \rightarrow 0$ as $N \rightarrow \infty$

$b_k = \beta^{N-k}$ with $\beta \in (0,1)$.

I am pretty sure that this implies convergence of $\sum_{k=0}^N a_k b_k$ to $0$ as $N \rightarrow \infty$, because $a_k$ is finite and $b_k$ tends to $0$ for increasing $N$. But unfortunately, I'm not able to prove it.

The problem is that all available tests/theorems aim to prove convergence to a finite value instead of to zero. I tried a Abel's test/Dirichlet's test like approach (with summation of parts) but this alsways ends up in convergence to a nonzero value.

Any help and hint is appreciated. Thanks!


问:
For anypair of $(x,y)$ is $x+y + xy$ always unique?
I was solving a computer science problem and was thinking of a way to encrypt an unordered pair of $(x,y)$ to a single value, I tried playing with bits a bit but did not get any conclusions But I think for any unordered pair of $(x,y)$ the Value of $x+y + xy$ is always unique for it, It is purely intuition-based, I am looking for a valid proof, If it is wrong Please present a counterexample for it.
For eg a pair of $(2,5)$ no other pair of numbers can give me the value $17$.
If anyone has any other way to encrypt two numbers Please suggest
The numbers are always Positive Numbers.

答:
Since no one has mentioned this yet, an easy way to see that $x+y+xy$ is not unique is to factorize a similar expression:

$$x+y+xy = 1+x+y+xy-1=(x+1)(y+1)-1$$

So given that $(x+1)(y+1)$ is not prime for some $x,y$, we can always find another pair $x', y'$ such that $x+y+xy = x'+y'+x'y'$. The counterexamples discussed stem from $2\times26 = 4\times 13$ and $3\times 6 = 2\times 9$.

In general, we can rewrite for nonzero $w$:

$$ux+vy+wxy=\frac {uv}w+ux+vy+wxy-\frac{uv}w=\frac1w((u+y)(v + xw) - uv)$$

hence for integers $u,v,w$ the value of this expression cannot be unique to every pair.


问:
For a symmetric matrix, the geometric and algebraic multiplicities are equal
Could anyone tell me how to prove the following proposition?

Let $\lambda$ be an eigenvalue of a symmetric matrix. Then, its geometric multiplicity equals its algebraic multiplicity.

答:
Let's review some basic matrix theorems first:

Geometric Multiplicity $\gamma(\lambda)$ is the dimension of $ker(A-
\lambda I)$, and the number of Jordan Blocks for an eigenvalue $\lambda$; Algebraic Multiplicity $\mu(\lambda)$ is the sum of the sizes of all Jordan Blocks for an eigenvalue $\lambda$; and $\gamma\leq \mu$.

Suppose $A$ is symmetric, we can diagonalize it as $A=Q\Lambda Q^{T},\ Q^{T}=Q^{-1}$.

$\Lambda$ is a diagonal matrix with eigenvalues of matrix $A$ on the main diagonal line; $Q$ is an Orthonormal matrix with column space as span of Eigenspaces. We can always construct an Eigenspace for each $\lambda$ with size of Algebraic Multiplicity $\mu(\lambda)$.

For a specific eigenvalue $\lambda$, if Geometric Multiplicity $\gamma(\lambda)$ is equal to Algebraic Multiplicity $\mu(\lambda)$, this means the size of the largest Jordan Block should be 1 and there are $\mu=\gamma$ blocks for $\lambda$. In another words, for any eigenvaule $\lambda$ and its eigevector $v$:  $$(A-\lambda I)^{\mu}v=0\iff (A-\lambda I)v=0$$

Prove:
$$(A-\lambda I)^{\mu}v=(Q\Lambda Q^{T}-\lambda I)^{\mu}v=[Q(\Lambda-\lambda I)Q^{T}]^{\mu}v=[Q(\Lambda-\lambda I)^{\mu}][Q^{T}v]=0$$Since column vectors of $Q$ are constructed as orthonormal basis, we get the equation above. Take an example of $\lambda=\lambda_{i}$ for clarification, assume Algebraic Multiplicity $\mu(\lambda_{i}) = 2$, $v = v_{i_{1}}$ as its eigenvector and construct another base vector $v_{i_2}$ subject to $v_{i_{2}} \bot v_{i_{1}}$, which span eigenspace of $\lambda_{i}$. Rewrite equation above as 2 separate parts considering orthonormality of eigenvectors:
$$ \left(
\begin{array}{cccccc}
  |&\cdots&|&|&\cdots&| \\
  (\lambda_{1}-\lambda)^{2}v_{1}&\cdots&(\lambda_{i}-\lambda)^{2}v_{{i}_{1}}&(\lambda_{i}-\lambda)^{2}v_{i_{2}}&\cdots&(\lambda_{n}-\lambda)^{2}v_{n} \\
  |&\cdots&|&|&\cdots&|
\end{array}
\right)
\left(
\begin{array}{ccc}
  0\\
  \vdots\\
  1\\
  0\\
  \vdots\\
  0
\end{array}
\right) = (\lambda_{i} -\ \lambda)^{2}v=0 $$
$$hence\ for\ \mu\in \mathbb{Z}^{+},\ (A-\lambda I)^{\mu}v=Q(\Lambda-\lambda I)^{\mu}Q^{T}v=0$$
$$(A-\lambda I)v=0$$

vice versa.


问:
Evaluating a convergent improper triple integral over the unit sphere
In Exercise 5 (f) of Angus Taylor's Advanced calculus (p. 659) one is asked to find the value of the following integral if convergent:

$$I:=\underset{R}{\iiint}\dfrac{x^2 y^2 z^2}{r^{17/2}}\mathrm dV$$

where $R$ is the unit sphere $x^2+y^2+z^2\leq 1$ and $r^2=x^2+y^2+z^2$.

Observing that $\dfrac{x^2 y^2 z^2}{r^{17/2}}\leq \dfrac{r^6}{r^{17/2}}=r^{-5/2}$ I proved that $I$ is convergent.

Using spherical co-ordinates $r$, $\theta $, $\phi $ i.e.

$$\begin{align*}x&=r\sin \phi \cos \theta\\y&=r\sin \phi \sin \theta\\z&=r\cos \theta\end{align*}$$

I transformed the integral $I$ into

$$I=\int\nolimits_0^{2\pi }\left(\int_0^{\pi }\left(\lim_{\delta \to 0}\int_{\delta }^1\left(r^2 \sin \phi\right)\dfrac{x^2 y^2 z^2}{r^{17/2}}\;\mathrm dr\right)\;\mathrm d\phi \right)\;\mathrm d\theta$$

$$=\lim_{\delta \to 0}\left( \int_{\delta }^1 r^{-1/2}\mathrm dr\right)\int_0^{2\pi }\cos^4 \theta \sin^2 \theta \mathrm d\theta\int_0^{\pi }\sin^5 \mathrm d\phi $$

$$=2\cdot \dfrac18 \pi \cdot \dfrac{16}{15}=\dfrac4{15}\pi $$

In the solutions the answer is $\dfrac8{105}\pi$. Since sometimes there
are a few book typos (in the exercises) to prevent undue copying, I ask the
following

Question: What is the correct solution, $\dfrac4{15}\pi $ or $\dfrac8{105}\pi $?

UPDATE (Correction): instead of $z=r\cos \theta $ it is

$z=r\cos \phi $

See a comment from whuber.

The integral $I$ is transformed into

$$I=\int_0^{2\pi }\left(\int_0^{\pi }\left(\lim_{\delta \to 0}\int_{\delta }^1\left(r^2 \sin \phi\right)\dfrac{x^2 y^2 z^2}{r^{17/2}}\;\mathrm dr\right)\;\mathrm d\phi \right)\;\mathrm d\theta$$

Since

$$(r^2 \sin \phi )\dfrac{x^2 y^2 z^2}{r^{17/2}}=(r^2\sin \phi )\dfrac1{r^{17/2}}\left( r\sin \phi \cos \theta \right)
^{2}\left( r\sin \phi \sin \theta \right) ^{2}\left( r\cos \phi \right) ^{2}$$

$=r^{-1/2}\cos ^{2}\theta \cdot\sin ^{2}\theta \cdot\cos ^{2}\phi \cdot\sin ^{5}\phi $,

the transformed integral becomes (if I am right):

$$I=\left(\lim_{\delta \to 0} \int_{\delta
}^1 r^{-1/2}\mathrm dr\right)\int_0^{2\pi }\cos^2\theta\cdot\sin^2\theta \;\mathrm d\theta
\int_0^{\pi }\cos^2 \phi \cdot\sin^5 \phi \;\mathrm d\phi$$

$$=2\cdot \dfrac14 \pi \cdot \dfrac{16}{105}=\dfrac8{105}\pi$$

The correct solution will be $\dfrac8{105}\pi $ as in the book.

答:
For this answer, I'll use $\rho$ instead of $r$, I tend to confuse things when dealing with both cylindrical and spherical coordinates

Converting the original integrand into spherical coordinates gives

$$\frac{1}{16}\frac{(\sin(2\theta)\sin(2\phi)\sin(\phi))^2}{\rho^{5/2}}$$

which we multiply by the Jacobian $\rho^2 \sin(\phi)$ for the triple integration to give the final integrand

$$\frac{1}{16}\frac{(\sin(2\theta)\sin(2\phi))^2\sin(\phi)^3}{\sqrt{\rho}}$$

As for setting up the limits of the integration, we can exploit symmetry and cut up the unit sphere into octants to simplify the mathematics, thus:

$$8\int_{0}^{1}\int_{0}^{\pi/2}\int_{0}^{\pi/2}{\frac{1}{16}\frac{(\sin(2\theta)\sin(2\phi))^2\sin(\phi)^3}{\sqrt{\rho}}}\mathrm{d}\theta\mathrm{d}\phi\mathrm{d}\rho$$

and then bring out the constant term to give

$$\frac12\int_{0}^{1}\int_{0}^{\pi/2}\int_{0}^{\pi/2}{\frac{(\sin(2\theta)\sin(2\phi))^2\sin(\phi)^3}{\sqrt{\rho}}}\mathrm{d}\theta\mathrm{d}\phi\mathrm{d}\rho$$

Let's separate things out:

$$\frac12\int_{0}^{1}\frac{\mathrm{d}\rho}{\sqrt{\rho}}\int_{0}^{\pi/2}\sin(2\theta)^2 \mathrm{d}\theta\int_{0}^{\pi/2}{\sin(2\phi)^2\sin(\phi)^3}\mathrm{d}\phi$$

The integral with respect to $\rho$ is 2, which cancels out the multiplicative factor, so we're left with

$$\int_{0}^{\pi/2}\sin(2\theta)^2 \mathrm{d}\theta\int_{0}^{\pi/2}{\sin(2\phi)^2\sin(\phi)^3}\mathrm{d}\phi$$

Evaluating the angular integrals gives:

$$\left.\frac{\theta}{2}-\frac{\sin(4\theta)}{8}\right|_{\theta=0}^{\theta=\pi/2}=\frac{\pi}{4}$$

$$\left.-\frac{5\cos(\phi)}{16}-\frac{\cos(3\phi)}{48}+\frac{3\cos(5\phi)}{80}-\frac{\cos(7\phi)}{112}\right|_{\phi=0}^{\phi=\pi/2}=\frac{32}{105}$$

and we get the result $\frac{8\pi}{105}$.

ContentLicense: CC BY-SA 3.0

ContentLicense: CC BY-SA 2.5


问:
Proving that $\mathbb{Z}[x]/\langle x^2 + 1\rangle$  is isomorphic to $\mathbb{Z}[i]$
I have some general idea of going about this. I'm trying to come up with a suitable mapping. I've come up with $\phi\colon a + \langle x^2 + 1\rangle\longmapsto a$ where $a$ is an irreducible element in $\mathbb{Z}[i]$ but this mapping has some issues. For example take
$5x + x^2 + 1$, which is $5x \pmod{x^2+1}$. $5x$ does not belong to $\mathbb{Z}[i]$. So my question is, is there a better mapping?

I know for a fact that if you add any integer call it $b$ to $\langle x^2 + 1\rangle$, you will get $b$ back which belongs to $\mathbb{Z}[i]$. I'm stuck when $b$ is not an integer.

答:
Here is a perhaps slightly more concrete way of looking at things.  In the longer run, the more abstract approaches are essential.

What are the elements of $\mathbb{Z}[x]/(x^2+1)$?  They are equivalence classes.  Two polynomials are called equivalent if they differ by a multiple of the polynomial $x^2+1$.

In our situation, we can give a more concrete description of the equivalence classes.  Let $P(x)$ be a polynomial with integer coefficients.  Imagine dividing $P(x)$ by $x^2+1$, using the ordinary procedure for dividing polynomials.  We get in general a remainder which is a polynomial of the shape $bx+a$ (one or both of $a$ and $b$ may be $0$).

So we have
$$P(x)=Q(x)(x^2+1) +bx+a$$
where $Q(x)$ is the quotient.  If you look at the above displayed formula, you will see that $P(x)$ and $bx+a$ are equivalent. Formally,
$$P(x) +(x^2+1)= bx+a + (x^2+1)$$
It is easy to see that if $bx+a$ and $dx+c$ are different polynomials, then they determine different equivalence classes, since their difference is not divisible by $x^2+1$.  Thus every equivalence class corresponds to a unique polynomial of the form $bx+a$.

This is analogous to what happens with congruence modulo $m$.  The collection of equivalence classes there is $\mathbb{Z}/(m)$, but we can think of $\mathbb{Z}/(m)$ as (sort of) $0, 1, 2, \dots, m-1$ with a funny addition and multiplication.

Now where should we send $a+bx+(x^2+1)$?  By basic properties of isomorphism, once we know where $1+(x^2+1)$ and $x+(x^2+1)$ are to go, the rest is determined. It is natural, indeed necessary, to send the equivalence class of $1$ to $a+bi$, where $a=1$ and $b=0$.

Where should $x+(x^2+1)$ go?  It is reasonable to send it to $i$.  So then the general object $a+bx +(x^2+1)$ should go to $a+bi$.

Does this work?  You need to show this mapping is one-to-one and onto, or to use more currently approved language, that the mapping is a bijection.  That is easy, given the prior discussion.

Then we need to show that the mapping "preserves" addition and multiplication.  The fact that addition is preserved is mechanical.  Multiplication is a bit trickier.  The product
$$[P(x)+(x^2+1)][Q(x)+(x^2+1)]$$
is, by definition, $P(x)Q(x)+(x^2+1)$.

Apply this to $P(x)=a+bx$, $Q(x)=c+dx$.
Note that $(a+bx)(c+dx)=ac +bdx^2 +(ad+bc)x$.

But $bdx^2=bd(x^2+1)-bd$, so $bdx^2$ is equivalent to $-bd$.
Putting things together, we find that
$(a+bx)(c+dx)$ is equivalent to $ac-bd +(ad+bc)x$.  This should remind you of the rule for multiplying complex numbers, and help to take care of showing that the mapping behaves well under multiplication.

A bit of intuition The polynomial $x^2+1$ is equivalent to the $0$-polynomial, since they differ by a multiple of $x^2+1$.  So in $\mathbb{Z}[x]/(x^2+1)$, "$x^2+1$" behaves like $0$, so "$x^2$" behaves like "$-1$". Thus the square of "$x$" is "$-1$".

More formally,
$$[x+(x^2+1)]^2 =x^2+(x^2+1) =-1+(x^2+1)$$

Note: There is another mapping that would give an isomorphism, namely the one that takes $a+bx +(x^2+1)$ to $a-bi$.


问:
Use method separation of variables: $\frac{{\partial u}}{{\partial y}} = \frac{{{\partial ^2}u}}{{\partial {x^2}}} - 4u$
I did the dpe:

$$\frac{\partial u}{\partial y} = \frac{\partial ^2 u}{\partial x^2} - 4u$$

$0 < x < \pi $

With boundary conditions: $\begin{array}{l}
 u(0,y) = u(\pi ,y) = 0 \\
 u(x,0) = {x^2} - \pi x \\
 \end{array}$

I used method separation of variables to solve the problem, and i got the solution is:

$$u(x,y) = \sum\limits_{n = \text{even}} \frac{- 8}{\pi n^3} \sin (nx) e^{ - (4 + n^2)y} $$

However, I feel that the solution is not fit with the boundary conditions
Could you give some hint??

答:
Using separation of variables
$$
u = XY \\
u_y = XY' \\
u_{xx} = X''Y \\
XY' = X''Y - XY \\
\frac {Y'}Y = \frac {X''}X - 4
$$
Since LHS and RHS are functions of different variables, they must be equal to some constant
$$
\frac {Y'}Y = \frac {X''}X - 4 = -\lambda^2-4 \\
\frac {Y'}Y = -\lambda^2-4 \\
Y = Y_0 e^{-(\lambda^2+4) y} \\
\frac {X''}X = -\lambda^2 \\
X = X_{01} \cos( \lambda x) + X_{02} \sin (\lambda x) \\
u = Y_0 e^{-(\lambda^2+4) y} \left (X_{01} \cos(\lambda x) + X_{02} \sin (\lambda x) \right )
$$
Applying first and second BCs
$$
u(0, y) = Y_0 e^{-(\lambda^2+4) y} X_{01} = 0 \\
u(\pi, y) = Y_0 e^{-(\lambda^2+4) y} X_{02} \sin (\lambda \pi) = 0
$$
From first equation $X_{01} = 0$, since if $Y_0$ one gets trivial solution. From second
$$
\lambda \pi = \pi n, \quad n \in \mathbb Z \\
\lambda = n\\
u = \sum_{n = 1}^\infty Y_0 e^{-(n^2+4) y} X_{01n} \sin nx
$$
Coefficients $X_{01n}$ can absorb $Y_0$, so without loss of generality, one can say
$$
u = \sum_{n = 1}^\infty X_{01n} e^{-(n^2+4) y} \sin nx
$$
Now apply third BC
$$
u(x, 0) = \sum_{n = 1}^\infty X_{01n} \sin nx = x^2 - \pi x
$$
So, basically you need to find LHS expansion of $x^2-\pi x$.
$$
b_n = \frac 2\pi \int_0^\pi (x^2 - \pi x) \sin n x dx = -\frac 4 {n^3 \pi} (-1+\cos n \pi) = \frac 8{n^3 \pi},\quad n \text{ is odd}
$$
so
$$
u = \frac 8{\pi^3} \sum_{n = odd} \frac {e^{-(n^2+4) y} \sin nx}{n^3}
$$</p>


问:
Prove $\vert e^x - 1 - x \vert \le x^2$ without derivatives
By comparison to the geometric series, prove that $\vert e^x - 1 - x \vert \le x^2$ for all $x \in \mathbb{R}$, $\vert x \vert \le 1$.

I'd like to solve this without calculating any derivatives, but I'm seriously wondering how I can use geometric series here. The best way I could think of is a certain geometric series that is greater or equal to the left, but less than or equal to the right side ... But does such a series even exist?

Any help is greatly appreciated.

答:
If $x \in[0,1]$, we get $$e^x -1 -x = \sum_{n=2}^\infty\frac{x^n}{n!} \le \sum_{n=2}^\infty \frac{x^n}{2^{n-1}}= \frac{x^2}{2-x}\leq x^2$$

Credit and my gratitude goes to @orangeskid for pointing out how to correct my flaw.


问:
Why is the following map $\mu : G_K \to M_d(\Bbb C_K) : g \mapsto a_{ij}(g)$, where $g(w_j) = \sum_{i} a_{ij}(g)w_i$, continuous?
Let $K$ be a p-adic field and let $\Bbb C_K$ be the completion of its algebraic closure. Let $G_K := \operatorname{Gal}(\overline{K}/K)$. Fixing a basis $\lbrace w_1, \dots, w_d\rbrace$ of a $\Bbb C_K$-vector space $W$ we have a map

$$\mu := \mu_{(w_j)} : G_K \to M_d(\Bbb C_K), g \mapsto a_{ij}(g)$$

where the $a_{ij}(g)$ are given by $g(w_j) = \sum_{i}a_{ij}(g)w_i$ for all $j$.

The claim is that this $\mu$ defines a continuous map, but I can't see why this is the case. I assume the topology on $M_d(\Bbb C_K)$ is the product topology (let's say after identifying $M_d(\Bbb C_K)$ with $\Bbb C_K^{d^2}$).

答:
If I'm not wrong, I think you meant $W = \Bbb{C}_K^d$.

Any $g\in G_K= Gal(\overline{K}/K)$ extends to a continuous automorphism of $\Bbb{C}_K$.

For a fixed $w\in W$, $g(w)$ is defined the obvious way. $g\to g(w)$ is continuous because  for all $n$, $w = u_n+p^n v_n$ with $u_n\in \overline{K}$, $v_n\in O_{\Bbb{C}_K}$, $g(p^nv_n)\in p^n O_{\Bbb{C}_K}$, and $g(u_n)$ is one of the finitely many conjugates of $u_n$.

$w\to g(w)$ is not $\Bbb{C}_K$ linear, it is only $K$-linear. So it defines an homomorphism $G_K\to GL_{K-vector space}(W)$.

$(g,w)\to g(w)$ is continuous because if $w'$ is close enough to $w$ then $u_n'=u_n$.

Your function is taking a basis $w_1,\ldots,w_d$ of $\Bbb{C}_K$, putting them in a matrix $M$, and letting $\mu(g)=\pmatrix{g(w_1)\\ \vdots \\ g(w_d)}M^{-1}$, clearly continuous $G_K\to \Bbb{C}_K^{d\times d}$. Not an homomorphism at all.


Title: Prove $\vert e^x - 1 - x \vert \le x^2$ without derivatives

Body:
By comparison to the geometric series, prove that $\vert e^x - 1 - x \vert \le x^2$ for all $x \in \mathbb{R}$, $\vert x \vert \le 1$.

I'd like to solve this without calculating any derivatives, but I'm seriously wondering how I can use geometric series here. The best way I could think of is a certain geometric series that is greater or equal to the left, but less than or equal to the right side ... But does such a series even exist?

Any help is greatly appreciated.


问:
Distribution of Random Matrix by Vector
The question is just a little variant of this post but I thought to be careful based on the minor differences.

Let $A$ be an $n\times n$ random matrix with independent and identically distributed entries sampled from $N(\mu,\sigma)$ and let $x$ be a (deterministic) vector in $\mathbb{R}^n$; here $\mu,\sigma \in \mathbb{R}$ with $\sigma>0$.  What is the distribution of the random vector
$$
Y= Ax.
$$
From the linked post, I know that it is Gaussian $N(\mu',\Sigma')$, since ultimately, $Y$ is an affine transformation of a normal random vector (when mapping $A$ into a vector in $\mathbb{R}^{n^2}$).  However, what are the actual quantities $\mu'$ and $\Sigma'$?

I think it should be $Y\sim \mathcal{N}\left(\boldsymbol{0}_n , x^{\top}x\right)$ however, I'm not sure if I've made a mistake...

答:
Just use the definition.

For any entry $Y_i$ of $Y$ we have
$$
Y_i = \sum_j A_{i,j} x_j
$$
and since the mean is linear,
$$
\mu'_i = \mathbb E[Y_i] = \sum_j \mathbb E[A_{i,j}] x_j = \mu\sum_j  x_j
$$

For the variance, notice that $Y_i$ and $Y_j$ are independent whenever $i\ne j$, so the covariance matrix is diagonal. If you're not convinced, recall that the covariance is bilinear, so
$$
i\ne j\implies \Sigma_{i,j}' = Cov(Y_i,Y_j) = \sum_{s,k} Cov(A_{i,s},A_{j,k}) x_sx_k  = 0
$$
On the diagonal, we have (since the entries of $A$ are independent)
$$
\Sigma_{i,i}' = Cov(Y_i,Y_i) = Var(Y_i) = \sum_j Var(A_{i,j}) x_j^2 = \sigma^2\sum_j x_j^2
$$
so in the end you'd have
$$
Y\sim \mathcal{N}\left(\mu\sum_j  x_j \cdot \boldsymbol{1}_n ,  \sigma^2\sum_j x_j^2\cdot I_n\right)
$$

Title: Distribution of Random Matrix by Vector


问:
Can a $6\times6$ all positive entry matrix have eigenvalues of $\{-1,1,-8,2,6,7\}$?
Is there any $6\times 6$ real square matrix $A$, all of whose entries are positive with the eigenvalues of $\{-1,1,-8,2,6,7\}$?

I am thinking it is related to the trace, determinant, or similar matrices but I am having trouble getting further.

答:
The answer is no, by the Perron-Frobenius theorem. In fact, the spectral radius of a positive matrix is always an eigenvalue of the matrix.

In your case, the matrix has spectral radius 8, that is not an eigenvalue.


问:
This proof that Folner sequences imply amenability
During various travels, I encountered the following page on Wikipedia:

http://en.wikipedia.org/wiki/Folner_sequence

The proof in the "Proof of amenability" section is kind of neat and I'd like to incorporate it into some background material because it's easy to appreciate, but it doesn't appear (at least, as far as I can tell) in the one and only one referenced paper (Folner, 1955). The other books, papers, etc I have all talk in the language of means.

Does somebody have a source for this? Or is it a "made-nice-for-Wikipedia" restatement of some result I failed to appreciate in the paper?

答: I'm certain it appears in Grimmett's book "Percolation".  It's absolutely central to an argument by Burton and Keane that if a graph is amenable there can be at most one infinite cluster in independent bond percolation so there should be a reference in there.</p>


问:
On the number of prime divisors of a product of 4 consecutive integers
In PEN (Problems in Elementary Number Theory) Project, there is a problem A111 as follows

Find $n$ if $n(n+1)(n+2)(n+3)$ has exactly three distinct prime divisors.

It's easy to see that at least 2, 3 should be two of the prime divisors of $n(n+1)(n+2)(n+3)$, and
I think there are infinitely many $n$ satisfying the above property, but I haven't found out the general answer.

Please help me.

Thanks.

答:
First not ethat $n=1$ is only almost a solution as $n(n+1)(n+2)(n+3)$ has only two prime divisors. Hence we ma assume $n>1$.

Let $p$ denote the third prime.

If $n$ is a multiple of $3$, then so is $n+3$. Then $n+1, n+2$ are divisible only by $2$ and $p$, and non of them is divisible by both. In other words, we have either
$$\tag1 n=3^a,\quad n+1=2^b,\quad n+2=p^c,\quad n+3=2^k3^d$$
or
$$\tag2 n=2^k3^a,\quad n+1=p^c,\quad n+2=2^b,\quad n+3=3^d.$$
Note that one of $a,d$ must be $=1$, and one of $b,k$ must be $=1$. This boils down to the following variants:

$(1)$ with $a=1$, so $n=3$, $n+1=4$, $n+2=5$, $n+3=6$, which is indeed a solution

$(1)$ with $b=1$ leads to $n=1$, so no.

$(1)$ with $d=k=1$, so $n=3$ - see above

$(2)$ with $d=1$ or $b=1$ leads to $n=0$, so no

$(2)$ with $a=k=1$, so $n=6$, $n+1=7$, $n+2=8$, $n+3=9$, which is a solution.

Next, assume that $n$ is not a multiple of $3$. Then exactly one of the four numbers is a multiple of $2$, exactly two are multiples of $2$ and exactly one is a multiple of $p$. To cover four numbers this way, we must have that each of the for numbers is a prime power. In particular, the two powers of $2$ must be $2$ and $4$. This leaves us with $n=2$, $n+1=3$, $n+2=4$, $n+3=5$, which is a solution.

In summary, $n$ is a solution iff
$$ n\in\{2,3,6\}.$$


答:
Let $E=U$ and let $f:E\to U$ be the identity map.

Suppose $f=\exp(ig)$ for some continuous function $g:E\to\mathbb{R}$.

Our goal is to derive a contradiction.

Since $f$ is injective, $g$ must be injective.

In particular, $g$ is not constant.

Since $E$ is compact and connected, we must have $g(E)=[a,b]$ for
some $a,b\in\mathbb{R}$ with $a < b$.

But then since $g$ is a continuous bijection from the compact space $E$ to the Hausdorff space $[a,b]$, it would follow that the $g$, when regarded as a map from $E$ to $[a,b]$, is a homeomorphism.

But $E$ is not homeomorphic to $[a,b]$ since for example, removing any point from $E$ always yields a connected space, whereas removing any point from $[a,b]$ other than $a$ or $b$ always yields a space which is not connected.

Title: Example of a continuous function $f$ into the unit circle, such that there is no continuous $g$ with $f=e^{{\rm i}g}$

Body:
Let $E$ be a metric space, $$U:=\left\{z\in\mathbb C:\left|z\right|=1\right\}$$ and $f\in C^0(E,U)$. Say that $f$ is inessential if $$f=e^{{\rm i}g}\tag1$$ for some $g\in C^0(E,\mathbb R)$.

I was able to show that if $f(E)\ne U$, then $f$ is inessential. Now I wonder what can possibly go wrong if $f(E)=U$.

The most trivial example of $f(E)=U$ should be $$E=\{(x,y)\in\mathbb R^2:x^2+y^2=1\},$$ $g$ being the inverse of the polar coordinate map $$(-\pi,\pi]\to E\;,\;\;\;\phi\mapsto\left(\cos\phi,\sin\phi\right)\tag2$$ and $f:=e^{{\rm i}g}$. Now $(1)$ is trivially satisfied by the very definition of $f$.

Now, clearly, this is a very special construction. Does anyone know a counterexample of $(1)$ when $f(E)=U$?


问:
$2 ^ {\sin ^ 2x}\cos x = 1$
An equation I find to be difficult. $$2 ^ {\sin ^ 2x}\cos x = 1$$. I have tried proving that $\frac{1}{\cos x}$ and $2 ^ {\sin ^ 2x}$ have different monotonies, but they do not. The substitution $\cos x = a$ would make the equation $2 ^ {1 - a ^ 2} a = 1$, but I cannot continue with this either.

One of my friends found a solution. If we let $sin ^ 2 \ x = a$, and square the equation, we get $2 ^ {\frac{2}{x}}(1 - \frac{1}{x}) = 1$. Multiply it with $x$ and, then raise it to the power of x: $4(x - 1) ^ x = x ^ x$, so $$(\frac{x}{x - 1}) ^ x = 4$$

Now, it is quite easy to prove $(\frac{x - 1}{x}) ^ x$ is strictly increasing. So
$(\frac{x - 1}{x}) ^ x = \frac {1}{4}$ has only one solution, which is $x = 2$.
Then, $sin ^ 2 x = 2$, so $x = \frac{\pi}{4} + 2k\pi, \ k \in Z $, since $2k\pi$ is the period.

答:
One graphic approach.
Let me change $x$ by $t$.

Using paramtrical equation, $x=\sin t,\, y=\cos t$ is the unit circle.

But then $2^{\sin^2 t}\cos t=1$ becames $2^{x^2} y=1$ which means $y=2^{-x^2}$.

So if you plot the function $y=2`{-x^2}$ and look for intersections with the unit circle, you'll get that there are 3 solutions: 1 is the point $(0,1)$ and the other two solutions cannot be solved analitically

Finally, go back and set $(\sin t,\cos t)=(0,1)\Rightarrow t=0+2k\pi$, and the same for the other two points.

EDIT:

That is what Mathematica says:
$$\left\{\{x\to 0\},\left\{x\to \fbox{$-\frac{\sqrt{W_{c_1}\left(-\frac{\log (2)}{2}\right)+2
   \log (2)}}{\sqrt{2 \log (2)}}\text{ if }c_1\in \mathbb{Z}$}\right\},\left\{x\to
   \fbox{$\frac{\sqrt{W_{c_1}\left(-\frac{\log (2)}{2}\right)+2 \log (2)}}{\sqrt{2 \log
   (2)}}\text{ if }c_1\in \mathbb{Z}$}\right\}\right\}$$

Title: $2 ^ {\sin ^ 2x}\cos x = 1$


问:
Logical translation of the verb " to be " in " the cat is a mammal ". ( Inclusion or membership?)

Standardly, the sentence " the cat is a mammal " would be translated as

for all x , if x is a cat, then x is a mammal,

meaning that the set of cats is included in the set of mammals : $C\subseteq M$.

There is another interpretation: the species " cat" is a mammals species. Under this alternative interpretation ( admitting that a " species" is some sort of set or collection), we would have : $ C \in M $*

with C = the set of cats and M* = the set of species of mammals.

The distinction I'm referring to is analogous to the medieval distinction between personal supposition and simple supposition of a term. In " horses are animals" the term " horse" supposes personnaly while in " horse is a species" , the term supposes simply.

What are the drawbacks of the second interpretation ( where " is" is interpreted as membership relation holding between a species and a set of species)? Why has the first interpretation become standard?

答: To see the problem with your second approach, consider the sentence "The Siamese cat is a mammal." The first approach translates simply as $S\subseteq M$; but your second approach requires us to define a new set $Q$, the set of sub-species of mammals, before we can say $C\in Q$. You can see how cumbersome this might get.


问:
What is the exact length of the belt, in inches?

As shown below, 2 circular pulleys with centers 8 inches apart are connected with a tight belt. The belt wraps 2/3 of the way around the larger pulley, which has a radius of 5 inches, and 1/3 of the way around the smaller pulley, which has a radius of 1  inch. What is the exact length of the belt, in inches?

答:
Hints:

Add some lines to get something like

You know $AC=5$ and $BD=1$ and $CD=8$ so using similar triangles you can find $DG$ and $CG$.  That will give you $AG$ using Pythagoras and thus $AB$ and by symmetry $EF$.

You can find $\theta = \frac{\pi}{3}$ from the "$2/3$ of the way around the larger pulley ... and $1/3$ of the way around the smaller pulley" or from the sides of the triangles.  You can use this to find the length of these curved arcs.

Add up the lengths of belt


问:
a problem on nonconstant holomorphic function has a zero or no in the closed unit disk
Let $f:D \to \mathbb{C}$ be a non-constant holomorphic function ($D$ is the closed unit disk) such that $|f(z)|=1$ for all $z$ satisfying $|z|=1$ . Then prove that there exist $z_0 \in D$ such that $f(z_0)=0$

My thought:-

By Maximum Modulus Theorem $|f(z)|$ has Maximum value on the curve which is $1$.

By Minimum Modulus Theorem if $f(z)\ne0$ for all $z\in D$, then it has its minimum value on the boundary which is $1$.

Then $|f(z)|=1$ for all $z\in D$.

Hence $|f(z)|$ is constant which is a contradiction.

Is my thinking correct?

答: Suppose $f(z)\ne 0\in D\forall z$, $g(z)={1\over f(z)}$ is Holomorphic on $D$ and $|g(z)|=1$ on $|z|=1$,   and by MMP $|g(z)|$  attains Maxima $1$ on $|z|=1$  so$|g(z)| \le 1\forall z \in D$ but then $|f(z)|>1 \Leftrightarrow$</p>


问:
Rigorous Construction of $e^{x}$
Consider :
$$
E(x)=\sum_{n=0}^{\infty}\frac{x^{n}}{n!}
$$
Based on some derivations from the binomial theorem, we note that $\forall x,y\in\mathbb{R}$, $E(x+y)=E(x)E(y)$ and $E(x)E(-x)=1$. We must show that $E(n)=e^{n}$ for all $n\in\mathbb{N}$ and we must show that $E(k)=e^{k}$ for all $k\in\mathbb{Z}$.

Any hint?

答: Since $E(1)=e=e^1$, you can now prove by induction that $^(\forall n\in\Bbb N):E(n)=e^n$. On the other hand, $E(0)=1$ and therefore $E(x)E(-x)=1$; in other words, $E(x)=\frac1{E(-x)}$. And now, if $k\in\Bbb Z$ and $k<0$,$$E(k)=\frac1{E(-k)}=\frac1{e^{-k}}=e^k.$$

Title: Rigorous Construction of $e^{x}$


答:
(Click on the diagram for better viewing.)

(Took me quite a while to make this diagram. But when I first saw the problem, $2/3$rd, $1/3$rd information wasn't given. Assuming tangency of belt, this information isn't even required. Angle can be calculated as shown.)

Length of belt would be $$ \dfrac{240}{360}\cdot 2\pi\cdot 5 + 2\cdot4\sqrt{3} + \dfrac{120}{360}\cdot 2\pi\cdot 1 \Rightarrow G$$

Body: As shown below, 2 circular pulleys with centers 8 inches apart are connected with a tight belt. The belt wraps 2/3 of the way around the larger pulley, which has a radius of 5 inches, and 1/3 of the way around the smaller pulley, which has a radius of 1  inch. What is the exact length of the belt, in inches?


问:
Let BD bisect $\angle ABC$ in $\Delta ABC$. Given $\angle ABC$ = 60 degrees and AB = BC + CD, find $\angle BAC$

I tried using angle bisector theorem and cosine rule but to no avail. Please help.

答:
Construct the point $E$ on $AB$ such that $EB = BC$. Then $AE = BA - BE = BA - BC = CD$.

Besides $\triangle BED \cong \triangle BCD$ by SAS. Hence $ED = CD = AE$.

We have $\angle EAD = \angle EDA = \alpha$. $\angle EDB = \angle CDB = \frac12(180^\circ - \alpha)$.

We also have $\angle BED = 2\alpha$. Now considering the interior angles of $\triangle BED$,

$$30^\circ + 2\alpha + \frac12(180^\circ - \alpha) = 180^\circ$$


问:
How is $x^2$ related to $x^Tx$
If $x$ is a column vector how is $x^2$ related to $x^\top x$?

Where $x^\top$ is the transpose of $x$.

答:
Let $ x \in \mathbb{R}^n $ and $ \left \| \cdot  \right \| $ be the euclidean norm,

$ \left \| x \right \|^2 = < x, x > = ^tx \cdot x $

Title: How is $x^2$ related to $x^Tx$


答:
The $\Gamma$ function fulfills the symmetry relation
$$ \Gamma(z)\Gamma(1-z) = \frac{\pi}{\sin(\pi z)} $$
hence by applying $\frac{d}{dz}\left(\log(\cdot)\right)$ to both sides
$$\psi(z)-\psi(1-z) =-\pi\cot(\pi z) $$
and by differentiating once again
$$ \psi'(z)+\psi'(1-z) = \frac{\pi^2}{\sin^2(\pi z)}. \tag{1}$$
Since
$$ \psi'(z) = \sum_{n\geq 0}\frac{1}{(n+z)^2}\tag{2} $$
equations $(1)$ and $(2)$ with $z=\frac{1}{10}$ prove the claim.

Title: How $\sum_{n=1}^{\infty} \left( \frac{1}{(10n-9)^2} + \frac{1}{(10n-1)^2} \right)$ equals to $\frac{{\pi}^2}{50} \frac 1 { 1-\cos{\frac {\pi} 5} }$

Body:
Why is
$$
\sum_{n=1}^{\infty} \left( \frac{1}{(10n-9)^2} + \frac{1}{(10n-1)^2} \right) = \frac{\pi^2}{50} \frac{1}{1- \cos \frac{\pi}{5}}
$$
?

Each form of
$$
\sum_{n=1}^{\infty} \frac{1}{(10n-9)^2}
$$
and
$$
\sum_{n=1}^{\infty} \frac{1}{(10n-1)^2}
$$

couldn't be calculated but the sum of them has a closed form


问:
Simplify Boolean expression $A'B'C'D' + A'B'CD + A'B'CD' + AB'CD' + AB'C'D'$
Using karnaugh map, I know that my final answer should be $B'D'+A'B'C$ But I cannot simplify this expression to that.

So far I got...

$A'B'C'D' + A'B'CD + A'B'CD' + AB'CD' + AB'C'D'$

$=A'B'C(D+D')+ A'B'C'D'+ AB'CD' + AB'C'D'$

$= A'B'C + B'C'D'(A'+A) + AB'CD'$

$= A'B'C + B'C'D' + AB'CD'$

$= A'B'C + B'D'(C'+AC)$

$= A'B'C + B'D'(A+C')(C+C')$

$= A'B'C + B'D'(A+C')$

What did I do wrong?

How can I get rid of that $A+C'$?

Body:
$A'B'C'D' + A'B'CD + A'B'CD' + AB'CD' + AB'C'D'$

$= A'B'CD + A'B'C'D' + A'B'CD' + AB'CD' + AB'C'D'$

$ = A'B'CD + A'B'D'(C + C') + AB'D' (C + C') = A'B'CD + A'B'D' + AB'D'$

$ = A'B'C + A'B'D' + AB'D' = A'B'C + B'D' \,$

(using CD + D' = C + D').


问:
Uniform continuity implies existence of increasing continuous function
In the book that I’m reading, the author makes the following assertion, which I was not able to prove:

If $c:\mathbb R \times \mathbb R\to \mathbb R$ is a continuous function on a compact set (i.e $c$ restricted to a compact set) and hence uniformly continuous, then there exists an increasing continuous function $w:\mathbb R_+ \to \mathbb R_+$, with $w(0)=0$ and such that
$$
\mid c(x,y) - c(x’,y’) \mid \leq w(d(x,x’) + d(y,y’))
$$

Can anyone prove that this is in fact true?

P.s: Note that $d$ is a metric.

答:
A friend of  mine showed me the solution, so here it is:

I'll sove this for $\mathbb R$ instead of $\mathbb R^2$ for sake of simplicity, because the proof is the same. So, I'll change $c:\mathbb R \times \mathbb R \to \mathbb R$ for $f:\mathbb R\to \mathbb R$, also continuous.

Let $K$ be the compact and
$$ w(z) := \sup_{x,y:d(x,y)\leq z} |f(x) - f(y)|$$
Hence, $|f(x) - f(y)|\leq w(d(x,y))$. Also, it's clear that $w$ is an increasing function. We have to prove that $w$ is continuous, and $w(0)=0$.

It's clear that $w(0) = 0$, since $d(x,y) =0 \iff x=y$.

The only thing left to prove is the continuity, which we'll prove for $w$ at 0, and a similar proof can be done to other points.

Note that since $f$ is continuous on a compact set, it is then uniformly continuous. Therefore,
$$\forall \epsilon>0, \exists  \delta>0, \quad \text{s.t} \quad  d(x,y) \leq \delta \implies |f(x) - f(y)| \leq \epsilon$$

Taking the $\sup$ , we obtain that:
$$
\sup_{x,y:d(x,y)\leq \delta}|f(x) - f(y)| = w(\delta) \leq \epsilon
$$

This means that $d(x,y)\leq \delta \implies w(\delta) \leq \epsilon$, hence,
$\lim_{\delta \to 0}w(\delta) = 0$.

Which proves that $w(0) =0$m and that $w$ is continuous on $0$. The only thing left to do is to prove the continuity for the rest of the points, which I could not do actually.


问:
Suppose $\Vert\vec v\Vert=1$. Does $\vec v\cdot A\vec v=1$ imply $A=I$?
I'm trying to show that $\vec v\cdot A\vec v=1 \iff A=I$. The reverse implication is straightforward. Somehow I'm unable to proceed for the forward part. I'm not even sure if its true. Any help would be much appreciated.

答:
Let $A:\vec{i}\mapsto\vec{i}+\vec{j}, \vec{j}\mapsto -\vec{i}+\vec{j}$ is given on an orthonormal basis $(\vec{i}, \vec{j})$ of a $2$-dimensional dot product space. Let $|\vec{v}|=1$ for a vector $\vec{v}$, and express $\vec{v}=x\vec{i}+y\vec{j}$ in co-ordinates in this basis. Now:

$$\vec{v}\cdot A(\vec{v})=(x\vec{i}+y\vec{j})\cdot(x(\vec{i}+\vec{j})+y(-\vec{i}+\vec{j}))=x(x-y)+y(x+y)=x^2+y^2=1$$

but $A\ne I$.

Update: In an orthonormal base $e$, the value of $\vec{v}\cdot A\vec{v}$ is given as $\vec{v}_e^TA_e\vec{v}_e$ where $\vec{v}_e$ is the column of co-ordinates of $\vec{v}$ and $A_e$ is the matrix of $A$ in $e$. However, note that, for any column $x$ and square matrix $A$ of appropriate size we have $x^TAx=x^TA^Tx=x^T(\frac{1}{2}(A+A^T))x$ so we can always replace matrix $A$ with its symmetric component $\frac{1}{2}(A+A^T)$.

Now, let's go back: we know that $\vec{v}\cdot I\vec{v}=1$ for $|\vec{v}|=1$, however, the previous consideration makes it clear that we can just start with another operator $A\ne I$ if we can somehow make $\frac{1}{2}(A_e+A_e^T)=I_e=I$ (identity matrix). This is the motivation for me to build a counterexample which will, as a matrix, have something like $\begin{bmatrix}1&-1\\1&1\end{bmatrix}$.

Update 2: One can see that the chosen map $A$ above is the rotation by $45^\circ$ followed by the multiplication by the factor $\sqrt{2}$. In other words, $A\vec{v}$ is rotated from $\vec{v}$ by $45^\circ$, which reduces the dot product by the factor $\sqrt{2}=\sec(45^\circ)$ but then is multiplied by $\sqrt{2}$ to get to the initial value of the dot product. You may take any angle instead of $45^\circ$.


问:
Explain proof that any positive definite matrix is invertible

If an $n \times n$ symmetric A is positive definite, then all of its eigenvalues are positive, so $0$ is not an eigenvalue of $A$. Therefore, the system of equations $A\mathbf{x}=\mathbf{0}$ has no non-trivial solution, and so A is invertible.

I don't get how knowing that $0$ is not an eigenvalue of $A$ enables us to conclude that $A\mathbf{x}=\mathbf{0}$ has the trivial solution only. In other words, how do we exclude the possibility that for all $\mathbf{x}$ that is not an eigenvector of $A$, $A\mathbf{x}=\mathbf{0}$ will not have a non-trivial solution?

答:
Note that $A^TA$ is symmetric and positive definite and we have
$$(\det A)^2=\det A\det A^T= \det (A^TA)\neq 0 $$

this reduces the question to the symmetric case.

Now if A is symmetric. Assume that is $0$ is an eigenvalue, then we have  $Ax=0$ with $x\neq 0$. By positive definite, we get $$0=(Ax,x)>0$$ which is impossible. So $x=0$ is the only solution to $Ax=0$.

NB: note that by definition,  an eigenvector is never zero


问:
Finding a rational number in between two rationals
Can anyone give me some hints or show me how to solve this problem?

Problem Find a rational number in between $\frac{9}{10}$ and $\frac{10}{11}$ which may be written in the form $\frac{m}{2^n}$ where $m$ is an integer and $n$ is a non negative integer.

I just do not know where to even start or what theorems I need to use to approach this. I would appreciate hints or guidance to solving these types of problems.

答:
Hint

Some ingredients:

$1/2^n$ is positive and can be as small as you want providing $n$ is large enough.

Then $1/2^n, 2/2^n, ...$ is an increasing sequence with constant difference between two terms.

If an interval, like $(9/10,10/11)$ is of length larger than $1/2^n$, then one element of the sequence above has to belong to it.

You should be done!

Note: making a drawing of all that will help you a lot to understand what happens.


问:
Is 1331 a porous number?
Porous numbers are numbers $k$ which are not multiples of 10 such that every m with sum of digits = $k$ and $k$ a divisor of both $m$ and rev($m$) has a zero in its digits. rev($m$) is the digit reversal of $m$ (e.g. rev(123) = 321).

Below 1000 there are only 11, 37, 74, 101 and 121 which fulfill these requirements (see OEIS sequence A337832).

Since $11$ and $11^2$ are porous, I wonder if $11^3$ and eventually all $11^n$ might be porous as well.

Title: Is 1331 a porous number?

Body:
By creating random numbers , I finally found a number showing that $1331=11^3$ is NOT porous.

This is the $252$-digit number :

$$353825666971232183215917592547993744495841357998656914595823152857793685258835251136486926388153276487648939825745571369191664461982939148175161852945169978959436475151898734955879246341597631697429537824325715778878473864688738548526257969637457226557$$

No idea whether $11^4,11^5,\cdots$ are porous, but my guess is they are not.


问:
How to solve expectation / first moment of Gaussian Integral?
How can I solve the following integral?

$E[I] = \int_{I=0}^\infty I \frac{1}{\sqrt{2\pi}\sigma}\exp(-\frac{(\mu-f-I)^2}{2\sigma^2}) dI$

The result is supposed to be:

$\sigma[\frac{\mu - f}{\sigma}\Phi(\frac{\mu - f}{\sigma}) + \phi(\frac{\mu - f}{\sigma})]$, where $\Phi(.)$ and $\phi(.)$ are the Gaussian probability density function and the cumulative probability function.

I tried integration by parts and variable substitution like done here.

But neither approach gets me the solution. I would like to know the steps to get to the result.

This whole thing is from Bayesian Optimisation, like here: p.23 E. Brochu, PhD Thesis 2010: Interactive Bayesian Optimization, https://circle.ubc.ca/bitstream/handle/2429/30519/ubc_2011_spring_brochu_eric.pdf?sequence=3

答:
Using the change of variable $\sigma x=\mu-f-I$ and the shorthand $\alpha=(\mu-f)/\sigma$, one sees that the integral to be computed is
$$
J=\int_{-\infty}^\alpha\sigma\cdot(\alpha-x)\cdot\varphi(x)\mathrm dx.
$$
The last piece of information you might need to conclude is that, since $x\varphi(x)=-\varphi'(x)$,
$$
\int_{-\infty}^\alpha(-x)\cdot\varphi(x)\mathrm dx=\varphi(\alpha).
$$</p>


问:
Probability of a random set of binary numbers XORing to $0$
Question - Given a randomly generated set of binary numbers, what is the probability that all of them, when XORed with each other, yield $0$?

Additional information -

All of the binary numbers in the set are different from each other. As an example, $\{0001, 0010, 0011\}$ is a valid set, but not $\{0001, 0010, 0010\}$ or $\{0010, 0010, 0010\}$.

All of the binary numbers have equal length $n$ which is some given. For example, one could have a given length of 4, meaning numbers such as $00001$ and $010$ are not allowed.

The length of the set $m$ is a given, its members are randomly generated.

Each possible valid binary number has an equal likelihood of being generated.

Context - For a school math project, I decided to analytically investigate the average percentage of errors an error correction code can correct given a data set and error rate. This question I am asking was a part of said project that I am unable to solve. If necessary, I can provide additional context.

I already know how to do this problem if restriction 1 is lifted, but did not really know how to proceed. I visualized the randomly generated data set as being one number stacked on top of another, kind of how we write when adding large numbers on paper by hand, and then taking my answer to be the probability that each column of bits had an even number of $1$s.

Updated context - Looking at the answers already provided and doing some of the math myself, it is clear that using this in my project will take me well over the page count in all of the explanations. I have come up with a simpler path for my investigation, but am leaving this question up here as I am still interested in this problem and I see no reason to stop anyone interested in answering or seeing answers to this question from doing so.

1rst edit - I have assigned $m$ as the length of the set and $n$ as the length of the binary sequences

2nd edit - I have added additional context

答:
This isn't a quick solution, so others may improve on this; however, it is a good start.

You can use conditional probability. Let's say you want to generate $n$ random $k$-bit binary numbers. Let $B \in \{0, 1\}^{n \times k}$ be a random matrix where $b_{ij} = 1$ if the $j$th bit of the $i$th number is $1$, otherwise $b_{ij} = 0$. Let $P \in [0, 1]^{n \times k}$ represent a matrix, where $p_{ij} = \mathbb{P}(b_{ij} = 1)$. The first row of $P$ is $0.5$ because of condition $(4)$ from your question. From the second row on, you need to rely on conditional probabilities, e.g.:

$$p_{21} = \mathbb{P}(b_{21} = 1) = \mathbb{P}(b_{21} = 1 | b_{11} = 1)p_{11} + \mathbb{P}(b_{21} = 1 | b_{11} = 0)(1 - p_{11})$$

In the case of $p_{21}$, the probability doesn't depend on $b_{1i}$ for $i > 1$. This isn't true for all the entries, so you could condition on all the previous variables, it doesn't hurt, but a bit more tedious to write it out.

You can write down expressions for the conditional probabilities, e.g.:

$$\mathbb{P}(b_{21} = 1 | b_{11} = 1) = \frac{2 ^ {k - 1} - 1}{2 ^ k - 1}$$

This is because you already selected a $k$-bit number for the first row, so there are $2 ^ k - 1$ numbers left. Of those, there are $2 ^ {k - 1}$ which start with a zero and ${2 ^ {k - 1} - 1}$ which start with a one, because the condition is $b_{11} = 1$.

Probably, you can come up with a general formula for the entries of $P$, I haven't checked this, though.

Once you have $P$ defined, you can calculate the probability of the $i$th column XOR'ing to zero. Take an even-length combination of row indices, multiple those entries in $P$ together with $1 - P$ for the indices you didn't select. If you add up the probabilities for all these combinations, you'll get the probability of the $i$th column XOR'ing to zero.

The difficulty of this problem is that there is a non-zero covariance between the rows as they have to be unique. Also consider whether the first restriction is meaningful, you can start with a valid set $\{0100, 0010, 0110\}$ and XOR'ing the first two element creates an invalid set $\{0110, 0110\}$ but doesn't actually affect the final outcome of the whole set XOR'ing to zero. You can go the other way around as well, you can start with an invalid set $\{0110, 0110\}$ and convert it by expanding to $\{0100, 0010, 0110\}$. There can be cases when this expansion leads to invalid sets again. Those map to problems where you have too many numbers selected, for example, for 3-bit numbers, you can select at most 8 unique numbers, selecting the 9th will create an invalid set.


问:
Struggling to find solution: question 1
Can someone help with this problem ?
$$\cos(x)\cos(y)dx + (\sin(x)\cos(y) − \sin(x)\sin(y) + y) dy = 0$$

I have tried to find an integrating factor but in the book shows only how to find the integrating factors dependent on one variable (either x or y): so this fails both theorems and is seeming to be dependent on both the variables. I've been struggling with this for a while now, so any help is appreciated

答:
$$\cos(x)\cos(y)dx + (\sin(x)\cos(y) − \sin(x)\sin(y) + y) dy = 0$$
Without the $y$ term:
$$\cos(x)\cos(y)dx + (\sin(x)\cos(y) − \sin(x)\sin(y)) dy = 0$$
$$\cos(y)d(\sin x) + \sin(x)d(\sin(y)) + \sin(x)d(\cos(y)) = 0$$
$$\cos(y)d(\sin x)  + \sin(x)d(\cos(y))+ \sin(x)d(\sin(y)) = 0$$
$$d(\cos(y) \sin (x)) + \sin(x)d(\sin(y)) = 0$$
$$\dfrac {d(\cos(y) \sin x)}{\cos(y) \sin (x)} + \dfrac {d(\sin(y))}{\cos(y) } = 0$$
Integrate both sides.
$$\ln {|\cos(y) \sin (x)|} +\int  \dfrac {du}{\sqrt {1-u^2} } = C$$
Where $u=\sin y$.

Maybe the $y$ term is a trigonometric function and there is an error in the differential equation.

Title: Struggling to find solution: question 1


问:
How to prove the laguerre polynomial has n zero point?
Here is an equation:

$$
e^x\frac{d^n(x^ne^{-x})}{dx^n}=0
$$

Now I want to prove that this equation has $n$ different roots.

I tried to convert the equation to this form:

$$
\sum\limits_{k=0}^{n}C_n^k(-1)^{n-k}\frac{n!}{k!}x^{n-k}=0
$$

But I don't know how to prove that it has exactly $n$ different roots. So what should I do?

A hint is enough!

答:
Rolle's theorem is enough. $f(x)=x^n e^{-x}$ has a root with multiplicity $n$ at the origin and it vanishes at $+\infty$.

$f'(x)$ has a root with multiplicity $n-1$ at the origin, a simple root at some $x_0\in\mathbb{R}^+$ (local maximum for $f$) and it vanishes at $+\infty$. $f''(x)$ has a root with multiplicity $n-2$ at the origin, a simple root somewhere between $0$ and $x_0$ and a simple root somewhere after $x_0$. Use induction and you are done.

The same argument also applies to (shifted) Legendre polynomials, essentially given by $\frac{d^n}{dx^n}\left(x^n(1-x)^n\right)$.

The largest root of a Laguerre polynomial can be estimated using the techniques shown here for Hermite polynomials.


问:
how to solve the integral with terms involve exponentiation
I have problem of solving or approximating the following integral:

$$ \int_{0}^{\infty} \frac{(1- (ax+1)^{(1-n)})^{(m-1)}}{(ax+1)^n} dx$$

I tried substitution or simplification, but it did not work. It was not successful.

Can anyone suggest any tips please?

Thank you.

答:
I am assuming $a>0,n>1$.

Keep $1-(ax+1)^{1-n}=u,du=a(n-1)dx/(ax+1)^n$ which gives$$I=\int_{0}^1\frac{u^{m-1}}{a(n-1)} du=\begin{cases}\frac1{am(n-1)},&m>0\\\infty,&m\le0\end{cases}$$

For $n=1$ the integrand is $0$. For $n<1$ we get $-\infty$ for $m\le0$ but the same result as above for $m>0$.


问:
Does $\sum_{n=1}^\infty \frac{\sqrt[n]e-1}{n}$ converge?
Does $$\sum_{n=1}^\infty \frac{\sqrt[n]e-1}{n}$$ converge or diverge?

Could you please help me out with this one? I tried the ratio test, but it's inconclusive...

I also tried to use the inequality $e^x \ge x+1$, but it doesn't get me anywhere.

答:
You can apply the direct comparision test if you construct a suitable series which forms an upper bound.

Let be $n>1$, then by definition we know that $e^{x}:=\sum\limits_{j=0}^{\infty}\frac{x^j}{j!}=\frac{x^0}{0!}+\frac{x^1}{1!}+\frac{x^2}{2!}+\cdots$.
Hence, $$e^{\frac{1}{n}}-1=\lim\limits_{k\to\infty}\sum\limits_{j=0}^{k}\frac{\left({\frac{1}{n}}\right)^j}{j!}-1=\sum\limits_{j=1}^{\infty}\frac{\left({\frac{1}{n}}\right)^j}{j!}.$$ As
$
\Big|\frac{\left({\frac{1}{n}}\right)^j}{j!}\Big|\leq \left({\frac{1}{n}}\right)^j$ for all $j$, we can conclude that
$$
\sum\limits_{j=1}^{k}\frac{\left({\frac{1}{n}}\right)^j}{j!}\leq \sum\limits_{j=1}^{k}\left({\frac{1}{n}}\right)^j=\frac{n}{n-1}\left(\frac{1}{n}-\left(\frac{1}{n}\right)^{k+1}\right).
$$
Hence, $e^{\frac{1}{n}}-1=\lim\limits_{k\to\infty}\sum\limits_{j=1}^{k}\frac{\left({\frac{1}{n}}\right)^j}{j!}\leq \frac{1}{n-1}$. This yields the following bound for each summand of $\sum\limits_{n=2}^{k}\frac{e^{\frac{1}{n}}-1}{n}$:
$$
\Big|\frac{e^{\frac{1}{n}}-1}{n}\Big|\leq \frac{1}{n(n-1)}.
$$
We know that the series $\left(2+\sum\limits_{n=2}^{\infty}\frac{1}{n(n-1)}\right)$ converges because $\sum\limits_{n=2}^{k}\frac{1}{n(n-1)}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}\cdots -\frac{1}{k}=1-\frac{1}{k}$ and therefore $\sum\limits_{n=2}^{\infty}\frac{1}{n(n-1)}=1$. So from the direct comparision test it follows that $\sum\limits_{n=1}^{\infty}\frac{e^{\frac{1}{n}}-1}{n}$ converges (absolutely).


问:
How can I see if this integral is convergent or not $\int_0^\infty \ \frac{1}{1 + x^4\sin x} \,dx $
I think the integral is convergent, but I don't know how to prove it.

$\int_0^\infty \ \frac{1}{1 + x^4\sin x} \,dx $

答:
The integrand has an order $1$ singularity at each root of $1+x^4\sin(x)$. The real roots are located, for $n\ge1$, on the real axis at
$$
x_n\approx n\pi-\frac{(-1)^n}{n^4\pi^4}\tag{1}
$$
There is one simple root at $x_0=0.79474586313959135719 + 0.61883482670901662817 i$.

The residue at $x_n$ is
$$
\frac1{4x_n^3\sin(x_n)+x_n^4\cos(x_n)}\approx\frac{(-1)^n}{n^4\pi^4}\text{ for }n\ge1\tag{2}
$$
The contour we will use goes from $+i\infty$ to $0$ then to $+\infty$ with infinitesimal clockwise semi-circles to avoid the real singularities. By contour integration, the principal value of integral along the real axis from $0$ to $\infty$ is equal to the integral along the imaginary axis from $0$ to $i\infty$ plus $\pi i$ times the sum of the residues at $x_n$ for $n\ge1$ plus $2\pi i$ times the residue at $x_0$. The sum of the residues obviously converges.

Combining the integral along the imaginary axis, the residues along the real axis (which only combine an imaginary part), and the residue at $x_0$, we get the Cauchy Principal Value for the integral:
$$
\begin{align}
\mathrm{PV}\int_0^\infty\frac{\mathrm{d}x}{1+x^4\sin(x)}
&=\int_0^\infty\frac{i\,\mathrm{d}x}{1+ix^4\sinh(x)}\\
&+\pi i\sum_{n=1}^\infty\frac1{4x_n^3\sin(x_n)+x_n^4\cos(x_n)}\\
&+2\pi i\frac1{4x_0^3\sin(x_0)+x_0^4\cos(x_0)}\\
&=\int_0^\infty\frac{x^4\sinh(x)\,\mathrm{d}x}{1+x^8\sinh^2(x)}\\[9pt]
&+0.85233885641320594757\tag{3}
\end{align}
$$
Thus, although the integrand is not absolutely integrable, it does have a convergent Cauchy Principal Value. Using $(3)$, Mathematica 8 gives the numerical approximation
$$
\mathrm{PV}\int_0^\infty\frac{\mathrm{d}x}{1+x^4\sin(x)}=1.14619893142224184361\tag{4}
$$


答:
Express the fractions in binary:

$
\frac{9}{10} = 0.111001100\cdots{}_2
$

$
\frac{10}{11} = 0.111010001\cdots{}_2
$

So try
$
0.11101000{\color{red}0}{}_2 = \frac{29}{32}
$.

This is the solution with smallest denominator.

Title: Finding a rational number in between two rationals

Body:
Can anyone give me some hints or show me how to solve this problem?

Problem Find a rational number in between $\frac{9}{10}$ and $\frac{10}{11}$ which may be written in the form $\frac{m}{2^n}$ where $m$ is an integer and $n$ is a non negative integer.

I just do not know where to even start or what theorems I need to use to approach this. I would appreciate hints or guidance to solving these types of problems.


问:
Partitioning a set of cardinal $np$
I wish to compute the number of possible partitions $S$ of a set of cardinal $np$ into $n$ subsets of cardinal $p$.
It is easy to obtain the formula :
$\displaystyle S=\sum_{k=0}^n {{pk}\choose{p}}$.
However i am getting stucked here without recouring to algebra
. I have in fact used generating functions to compute the sum, but i am looking for a non-algebraic argument (for example double couting ?) that could help me to compute this sum directly as i am sure there's a way to.

答:
You are closed, but not quite. Just put your $n\cdot p$ objects in a line and permute them. Then divide them from left to right taking $p$ at a time and divide by the order in each group and for the order of the pairings, you will get
$$\frac{(n\cdot p)!}{n!\cdot (p!)^n}$$

Title: Partitioning a set of cardinal $np$


问:
General solution of trigonometric functions
What is the significance of the general solution of trigonometric equations?

$\sin 5x+\sin x= \sin 4x+\sin 2x$ on simplification becomes $\cos 2x=\cos x$ which can be transformed to $\sin 2x=\sin x$.
The general solution of  both forms are different.

for $\cos 2x=\cos x$ , it is $2n\pi /3$ and for $\sin 2x=\sin x$, it is $(2n+1)\pi /3$.

Please tell me  the significance of general solution.

答:
$$\sin 5x+\sin x=\sin 4x +\sin 2x$$

$$2\sin 3x\cos 2x=2\sin 3x \cos x$$

$$\cos 2x=\cos x$$

$$2x=2n\pi \pm x$$

$$x=\frac{2n\pi}{3},2n\pi$$

You would have also to consider the other case when

$$\sin 3x=0$$

$$3x=n\pi$$

$$x=\frac{n\pi}{3}$$

where $n \in Z$

$\sin \theta$ and $\cos \theta$ are periodic functions. They regularly repeat after an interval.

This arose the need of general solution.


问:
How to model the vaccination of healthcare and frontline workers?
Let us consider the standard SEIR model with additional vaccination compartment:

\begin{eqnarray*}
\frac{dS}{dt}&=& -\beta \frac{SI}{N} - bS \\
\frac{dV}{dt}&=& bS - \epsilon(\beta \frac{VI}{N}) \\
\frac{dE}{dt}&=& \beta \frac{SI}{N} + \epsilon(\beta \frac{VI}{N}) - \alpha E \\
\frac{dI}{dt}&=& \alpha E - \gamma I\\
\frac{dR}{dt}&=& f\gamma I\\
\end{eqnarray*}

Here I assume that $V$ is the number of healthcare and frontline workers who are vaccinated. Now, in reality, there are only a limited number of such individuals. Thus, I want to alter the parameter $b$ so that the number of vaccinated persons cannot exceed a specific number, say $K.$ Your suggestions are so much appreciated.

答: This is an answer to my question: Take $$b = r \left( 1 - \frac{V}{K} \right)$$ where $r$ is some constant.


问:
On the nearest-square function - Part 2 and the quantity $m^2 - p^k$ where $p^k m^2$ is an odd perfect number
(Note:  This question has been cross-posted to MO.)

This question is an offshoot of this earlier one and this other question.

Let $n = p^k m^2$ be an odd perfect number with special prime $p$ satisfying $p \equiv k \equiv 1 \pmod 4$ and $\gcd(p,m)=1$.

It was conjectured in Dris (2008) and Dris (2012) that the inequality $p^k < m$ holds.

Brown (2016) showed that the Dris Conjecture (that $p^k < m$) holds in many cases.

It is trivial to show that $m^2 - p^k \equiv 0 \pmod 4$.  This means that $m^2 - p^k = 4z$, where it is known that $4z \geq {10}^{375}$.  (See this MSE question and answer, where the case $m < p^k$ is considered.)  Note that if $p^k < m$, then $$m^2 - p^k > m^2 - m = m(m - 1),$$
and that
$${10}^{1500} < n = p^k m^2 < m^3$$
where the lower bound for the magnitude of the odd perfect number $n$ is due to Ochem and Rao (2012).  This results in a larger lower bound for $m^2 - p^k$.  Therefore, unconditionally, we have
$$m^2 - p^k \geq {10}^{375}.$$
We now endeavor to disprove the Dris Conjecture.

Consider the following sample proof argument:

Theorem If $n = p^k m^2$ is an odd perfect number satisfying $m^2 - p^k = 8$, then $m < p^k$.

Proof

Let $p^k m^2$ be an odd perfect number satisfying $m^2 - p^k = 8$.

Then $$(m + 3)(m - 3) = m^2 - 9 = p^k - 1.$$

This implies that $(m + 3) \mid (p^k - 1)$, from which it follows that
$$m < m + 3 \leq p^k - 1 < p^k.$$
We therefore conclude that $m < p^k$.

QED

So now consider the equation $m^2 - p^k = 4z$.  Following our proof strategy, and the formula in the accepted answer to the first hyperlinked question, we have:

$$m^2 - \bigg(\lfloor{\sqrt{m^2 - p^k} + \frac{1}{2}}\rfloor\bigg)^2 = p^k + \Bigg(4z - \bigg(\lfloor{\sqrt{m^2 - p^k} + \frac{1}{2}}\rfloor\bigg)^2\Bigg).$$

So the only remaining question now is whether it could be proved that
$$\Bigg(4z - \bigg(\lfloor{\sqrt{m^2 - p^k} + \frac{1}{2}}\rfloor\bigg)^2\Bigg) = -y < 0$$
for some positive integer $y$?

In other words, is it possible to prove that it is always the case that
$$\Bigg((m^2 - p^k) - \bigg(\lfloor{\sqrt{m^2 - p^k} + \frac{1}{2}}\rfloor\bigg)^2\Bigg) < 0,$$
if $n = p^k m^2$ is an odd perfect number with special prime $p$?

(Additionally, note that it is known that $m^2 - p^k$ is not a square, if $p^k m^2$ is an OPN with special prime $p$.  See this MSE question and the answer contained therein.)

If so, it would follow that
$$\Bigg(m + \lfloor{\sqrt{m^2 - p^k} + \frac{1}{2}}\rfloor\Bigg)\Bigg(m - \lfloor{\sqrt{m^2 - p^k} + \frac{1}{2}}\rfloor\Bigg) = p^k - y$$
which would imply that
$$\Bigg(m + \lfloor{\sqrt{m^2 - p^k} + \frac{1}{2}}\rfloor\Bigg) \mid (p^k - y)$$
from which it follows that
$$m < \Bigg(m + \lfloor{\sqrt{m^2 - p^k} + \frac{1}{2}}\rfloor\Bigg) \leq p^k - y < p^k.$$

Update (November 11, 2020 - 10:21 PM Manila time) Please check out the recently posted answer for a minor adjustment to the logic that should make the general proof argument work.

答:
Let me illustrate what I have in mind for a small value of $z$, say $z=10$.

Then we have
$$m^2 - p^k = 4z = 40$$
$$m^2 - 49 = p^k - 9$$
$$(m+7)(m-7) = p^k - 9.$$
This implies that
$$(m+7) \mid (p^k - 9)$$
from which it follows that
$$m < m+7 \leq p^k - 9 < p^k.$$

Note that $49$ is not the nearest square to $40$ ($36$ is), but rather the nearest square larger than $40$.

With this minor adjustment in the logic, I would expect the general proof argument to work.


答:
In general, by the definition on the unit circle and for symmetry we have that

$$\cos \theta = \cos \alpha \iff \theta=\alpha +2k\pi \quad \lor \quad \theta=-\alpha +2k\pi$$

therefore in this case

$$\cos 2x = \cos x \iff 2x=x +2k\pi \quad \lor \quad 2x=-x +2k\pi$$

which leads to the solutions

$$x= \frac23k\pi \quad \lor \quad x= 2k\pi$$

which corresponds to $x= \frac23k\pi$ with $k\in \mathbb Z$.

Refer to the related

Solving $\cos(3x) = \cos(2x)$

As an alternative, in this case we can also use that

$$\cos 2x=\cos x \iff 2 \cos^2 x- \cos x -1=0$$

which leads to $\cos x=-\frac12$ and $\cos x=1$.

Title: General solution of trigonometric functions

Body:
What is the significance of the general solution of trigonometric equations?

$\sin 5x+\sin x= \sin 4x+\sin 2x$ on simplification becomes $\cos 2x=\cos x$ which can be transformed to $\sin 2x=\sin x$.
The general solution of  both forms are different.

for $\cos 2x=\cos x$ , it is $2n\pi /3$ and for $\sin 2x=\sin x$, it is $(2n+1)\pi /3$.

Please tell me  the significance of general solution.


问:
How to find all elements of $\mathbb{Z}_{4} \times\mathbb{Z}_{4}/\langle(1,1)\rangle$?
I am studying factor groups, and I saw an example that says

Find all the elements of the factor group $\mathbb{Z}_{4}   \times \mathbb{Z}_{4}/\langle(1,1)\rangle$.

I know that the order of  $\mathbb{Z}_{4}  \times \mathbb{Z}_{4}$  is $16$, and the order of $\langle(1,1)\rangle$ is $4$, so we must have $4$ elements, but how can I find them?

Edit: Actually, this is not a homework, I have the answer saying that those elements are  $\{(1,1),(2,2),(3,3),(0,0)\} , \{(1,2),(2,3),(3,0),(0,1)\} , \{(2,1),(3,2),(0,3),(1,0)\},\{(1,3),(2,0),(3,1),(0,2)\}$.

But I did not understand how they are found, can anyone please explain the steps to find them?

Thanks.

答:
HINT 1 The elements of the factor group are only uniquely presented up to an element of $\langle 1,1\rangle$, meaning that $(a,b)$ and $(a,b)+(1,1)$ represent the same element in the factor group.

HINT 2 When are two elements $(a,b)$ and $(c,d)$ different in the factor group? (precisely when $(a-c,b-d) \neq c(1,1)$ for some $c$)


答:
To keep the arithmetic simple, let's look for a number of the form $\frac{m}{2^n}$ between $\frac{1}{11}$ and $\frac{1}{10}$. Then just subtract it from $1$ for your answer. So we have:
$$\frac{1}{11}<\frac{m}{2^n}<\frac{1}{10}$$
which is the same as
$$10m<2^n<11m$$
Now look for $m$ such that the interval $(10m,11m)$ contains a power of $2$. You don't have to look for long.


答:
Another possibility is to use the following well-known fact:

$$\frac{a}{c} < \frac{b}{d} \Longrightarrow \frac{a}{c} < \frac{a+b}{c+d} < \frac{b}{d}$$

A simple consequence of this is that

$$\frac{a}{c} < \frac{b}{d} \Longrightarrow \frac{a}{c} < \frac{ma+nb}{mc+nd} < \frac{b}{d} \quad (m,n > 0) $$

This means that for your question, we simply have to find multiples of $10$ and $11$ which add to a power of 2. The simplest one would be $1(10) + 2(11) = 32 = 2^5$, which gives

$$
\frac{9}{10} < \frac{1(9) + 2(10)}{1(10) + 2(11)} < \frac{10}{11} \\
\color{white}{\text{empty}}\\
\frac{9}{10} < \frac{29}{32} < \frac{10}{11}
$$

Trying for $64$ or $128$ just yields fractions equivalent to the above, but we can use $19(10) + 6(11) = 256 = 2^8$ to find another one:

$$
\frac{9}{10} < \frac{19(9) + 6(10)}{19(10) + 6(11)} < \frac{10}{11} \\
\color{white}{\text{empty}}\\
\frac{9}{10} < \frac{231}{256} < \frac{10}{11}
$$


问:
How to prove by induction that given a set $S$ of $n$ elements, the number even subsets of $S$ equals to the number of odd subsets?
How do I prove by induction that given a positive integer $n\ge1$ and a set $S$ of $n$ elements, the number even subsets of $S$ equals to the number of odd subsets? For example, for $S=\{g,c\}$ there are two subsets with even cardinality: the empty set $\varnothing$ and $\{g,c\}$, as well as two subsets with odd cardinality: $\{g\}$ and $\{c\}$.

答:
Since $n\ge 1$ and is finite, you know there is at least one element of $S$.  Suppose one such is $a$.

Then the number of odd subsets of $S$ which contain $a$ is equal to the number of even subsets which do not contain $a$, and  the number of even subsets of $S$ which contain $a$ is equal to the number of odd subsets which do not contain $a$.

Adding the numbers together implies the number of odd subsets of $S$ equals the number of number of even subsets of $S$.

(This will not work when $n=0$, and the proposition would not be true as $\emptyset$ has one even subset but no odd subset).

The argument above does not use induction.  If you must use induction then

The proposition is true when $n=1$ since there is one odd subset and one even subset (the empty set)

Then the number of odd subsets of $S$ which contain $a$ is equal to the number of even subsets which do not contain $a$ which (by the inductive hypothesis) is equal to the number of odd subsets which do not contain $a$ and that is equal to the number of even subsets of $S$ which contain $a$.  Adding the pairs up proves the induction step.


问:
An Application of Inclusion Exclusion Principle.
Let $A$ be a finite set and let $A_{1},A_{2},\ldots,A_{n}$ be subsets of $A$. And let $C$ be a subset of $\{1,2,\ldots,n\}$. Let $B$ be the set of all $x\in A$ such that $x\in A_{i}$ for all $i$ and $x\not \in A_{i}$ if $i\in\{1,2,\ldots,n\}-C$. I have to prove that $$|B|=\sum_{C\subseteq D\subseteq \{1,2,\ldots,n\}}(-1)^{|D-C|}\cdot \bigcap_{d\in D}|A_{d}|$$ where $\bigcup_{d\in D}A_{d}$ is taken as $A$ if $D=\emptyset$. Definitely I know this is an application of Inclusion-Exclusion, but I am not sure how to go about applying the result.

答:
Hint: First notice that $x\in B$ iff $x\in A_i$ for $i\in C$ but $x\not \in A_j$ for $j\not \in [n]\setminus C$ so you want to take the following difference of the ones that are all in $C$ minus the ones that are not in $C.$ Show then that you want
$$B=\left (\bigcap _{i\in C}A_i\right )\setminus \left ( \bigcup _{j\not \in C}A_j\right)=\left (\bigcap _{i\in C}A_i\right )\setminus \left ( \bigcup _{j\not \in C}\left (A_j\cap \bigcap _{i\in C}A_i\right )\right),$$
the last equality is just by definition of difference.
Then take the size of the set and use normal inclusion-exclusion (notice that $C$ is always going to be in the intersection of the I-E and so you are taking sets that contain $C,$ be careful with the sign). Can you finish?


问:
How can I solve $y'' + 8y' + 15y = 35 \theta(2-x)e^{2x}$ using Laplace transform?
I have the following homework problem. Using Laplace transform, I need to solve the following differential equation
$$
y'' + 8y' + 15y = 35 \theta(2-x)e^{2x}
$$
where $\theta(x)$ is the Heaviside step function. I evaluated this differential equation using WolframAlpha, but the solution it presents is for the homogeneous equation, which is not the same as the equation I have since for $2-x>0$ the R.H.S. is $35 e^{2x}\neq 0$. Since I believe there's no general formula for the Laplace transform of a product, I explicitly calculated the Laplace transform of the R.H.S. which gives
$$
\mathcal{L}\left\{35 \theta(2-x)e^{2x} \right\} = 35 \int_{0}^{2}e^{-px}e^{2x} \ dx = \frac{35\left(1-e^{4-2s}\right)}{s-2} \tag{1}
$$
Which would then imply, by taking the Laplace transform on both sides of the original equation, that
$$
Y = \frac{1}{(s+3)(s+5)}\left[\frac{35\left(1-e^{4-2s}\right)}{s-2} +y(0)(s+8) +y'(0) \right]\tag{2}
$$
Now, for the last $2$ summands I know I can apply partial fractions to get the inverse Laplace transform, but for the first term, I have no idea how to inverse Laplace it. I plugged it into WA and got a really long expression for the answer. This last part seemed odd to me and I started doubting if my approach was correct.

I then had the idea that maybe I should try to solve the equation piece-wise, getting $2$ "simpler" equations for $x<2$ and $x>2$, the latter being the homogeneous case. The problem with this is that I don't think this results in a solution to the original problem since in both cases I would get different functions, and I need a function that satisfies both conditions ($x>2$ and $x<2$) simultaneously.

Am I on the right path to solving this problem? Or is my approach wrong? And in any case, is there a simple way to obtain the inverse Laplace transform of equation $(2)$? Thank you!

Body:
$$Y = \frac{1}{(s+3)(s+5)}\left[\frac{35\left(1-e^{4-2s}\right)}{s-2} +...... \right]$$
For the first term you can use the formula:
$$\mathcal {L^{-1}}\{ e^{-cs}F(s)\}=H(t-c)f(t-c)$$
You have that:
$$e^{4-2s}=e^4e^{-2s} \implies c=2$$
And:
$$F(s) =- \frac{35e^4}{(s+3)(s+5)(s-2)}$$


问:
Prove that every eulerian graph with odd number of vertices contains 3 vertices of the same degree.
I'm trying to prove the following problem:

Prove that every eulerian graph with odd number of vertices
contains 3 vertices of the same degree.

I know that eulerian graph must have every vertices of even degree, but I am not sure how to continue from that.

Could you please help me?

Thanks

答: First, I assume our graph has at least $3$ vertices. By definition, a graph $G$ is Eulerian if $G$ is connected and every vertex in $G$ has even degree. Now, suppose $|V(G)| = n$ odd. Then, $n-1$ is even so the possible degrees of $G$ are $\{2,4,6,...,n-1\}$ (We do not include $0$ since $G$ is connected). So, there are $\frac{n-1}{2}$ elements in this set. But we have $n$ vertices so by Pigeonhole Principle, at least $3$ of these have the same degree.


答: Note that taking a quotient by $\langle 1,1\rangle$ takes any element of the form $(x,x)$ to the identity coset.  Why not look at an element like $(1,0)$?</p>

Title: How to find all elements of $\mathbb{Z}_{4} \times\mathbb{Z}_{4}/\langle(1,1)\rangle$?

Body:
I am studying factor groups, and I saw an example that says

Find all the elements of the factor group $\mathbb{Z}_{4}   \times \mathbb{Z}_{4}/\langle(1,1)\rangle$.

I know that the order of  $\mathbb{Z}_{4}  \times \mathbb{Z}_{4}$  is $16$, and the order of $\langle(1,1)\rangle$ is $4$, so we must have $4$ elements, but how can I find them?

Edit: Actually, this is not a homework, I have the answer saying that those elements are  $\{(1,1),(2,2),(3,3),(0,0)\} , \{(1,2),(2,3),(3,0),(0,1)\} , \{(2,1),(3,2),(0,3),(1,0)\},\{(1,3),(2,0),(3,1),(0,2)\}$.

But I did not understand how they are found, can anyone please explain the steps to find them?

Thanks.


问:
Finding a rational number in between two rationals
Can anyone give me some hints or show me how to solve this problem?

Problem Find a rational number in between $\frac{9}{10}$ and $\frac{10}{11}$ which may be written in the form $\frac{m}{2^n}$ where $m$ is an integer and $n$ is a non negative integer.

I just do not know where to even start or what theorems I need to use to approach this. I would appreciate hints or guidance to solving these types of problems.

答:
I'll show you a method that works for any base. I'll do the question but with base $5$ instead of $2$, and leave base $2$ as an exercise.

$\frac{10}{11} - \frac{9}{10} = \frac{100}{110} - \frac{99}{110} = \frac{1}{110}$, i.e. the distance between $\frac{10}{11}$ and $\frac{9}{10}$ is $\frac{1}{110}$.

$\frac{1}{5^2} > \frac{1}{110}\ $, but $\frac{1}{5^3} < \frac{1}{110}$.

So $ \exists k \in \mathbb{Z}$ such that $\frac{k}{5^3} \in (\frac{10}{11}, \frac{9}{10})$, otherwise $\exists j \in \mathbb{Z}\ $ such that both $\frac{j}{5^3} < \frac{99}{110}$ and $\frac{j+1}{5^3} > \frac{100}{110}$, contradicting the fact that $\frac{1}{5^3} < \frac{1}{110}$.

$\frac{9}{10} = \frac{112.5}{125}\ \text{and} \ \frac{10}{11} = \frac{113.6...}{125} ,\ \therefore \frac{113}{125} \ $ does the trick.


问:
For a continuous function on closed bounded interval is absolute minimum value, absolute maximum values are same as inf, sup of image set
I was reading theorem “5.3.9” of “Introduction to real analysis Robert G. Bartle and Donald R. Sherbert”. I didn’t not understand one step in the proof of theorem. In a second line of proof author writes “ by “Maximum and minimum theorem 5.3.4” we have $m,M\in f(I)$. (Where, $m=\inf f(I)$, and $M=\sup f(I)$) How? Can anybody explain?

My opinion: By maximum minimum theorem, as $f$ is continuous on closed and bounded interval $I$ hence, $f$ has absolute minimum and absolute maximum on $I$. Hence by definition of absolute minimum, absolute maximum values we have, there exists points $c_*,c^*\in I$ Such that,

$f(c_*)≤f(x)$ for all $x\in I$ and $f(c^*)≥ f(x)$ for all $x\in I$ .

So how author concluded $m=\inf f(I),M=\sup f(I)\in f(I)$?

Is $m=\inf(f(I))=f(c_*)$ and $M=\sup(f(I))=f(c^*)$ and hence $m,M\in f(I)$?

Am I correct? Please help.

答:
You just need to apply the following fact to the set $A=f[I]$.$\newcommand{\R}{\mathbb R}$

Fact. If $A$ is subset of $\R$ and if $m=\min A$, $M=\max A$ then $m=\inf A$, $M=\sup A$.

This is true for any partially ordered set, but since you are asking this in context of real analysis, I have only mentioned $A\subseteq\R$.

This simply says that if a set has maximum, then it is equal to supremum. (The same is true for minimum and infimum.) Notice that there are sets which have supremum, but do not have maximum - for example, $A=(0,1)$.

Why this claim is true - it depends a bit on how you defined maximum and minimum. But for any reasonable definition you should be able to see that if $M=\max A$, then

$M$ is an upper bond of $A$, i.e., for any $a\in A$ you have $a\le M$.

$M$ belongs to $A$, and thus for any upper bound $U$ of the set $A$ we have $M\le U$. (We have $a\le U$ for every $a\in A$. Since $M\in A$, the inequality is true for $M$, too.)


问:
Group structure on set of homotopy classes
Let $\text{Top}_*$ be the category of pointed topological spaces, and fix some $(X,x)\in\text{Top}_*$. Suppose that for all $Y\in\text{Top}_*$ we can give a group structure to the set $[X,Y]$ of homotopy classes of maps fixing the base point, in such a way that there exists a functor
$$
\mathcal F:\text{Top}_*\to\text{Gr},\;
(Y,y)\mapsto[(X,x),(Y,y)]
$$
where $\text{Gr}$ is the category of groups.

Is it true that the identity of $[(X,x),(Y,y)]$ is the class of the constant map?

If the functor $\mathcal F$ happens to be the most natural one, i.e., the one given by composition: for a continuous map $f:Y\to Z$,

$$\mathcal F(f):[X,Y]\to[X,Z],\quad[h]\mapsto[f\circ h],$$
then this is easily seen to be true. This is, for example, what happens for the fundamental group. However, what if $\mathcal F$ is an arbitrary functor (assuming, of course, that a functor not of this form exists, which for all I know it might not at all)? Is it possible that the identity of group is not the constant map?

答:
Apologies, I have not fully read the question. This answer presumes that the functor lifts the normal Hom functor on both objects and morphisms.

Clearly the identity element of $[X,*]$ is constant. The image of the identity of a group under a group homomorphism is the identity of the codomain. Since, $*$ is initial and the composition of any constant map with another map is constant, we conclude that all identities are represented by constant.


答: Just take a second linearly independent and indivisible element of your group and look what it generates. The elements in the generated group will be a solution to your problem. </p>


问:
Is it true that $|x + y| \geq |x| - |y|$?
Is the above inequality true? I feel that it is, but I am having a tough time convincing myself (or coming up with an appropriate counter-example).</p>

答: $|x|=|x+y-y|\le|x+y|+|y|$ by the triangle inequality.


答:
You are looking for integers $m,n$ such that $$\frac9{10}<\frac m{2^n}<\frac{10}{11}$$

or

$$\frac9{10}2^n<m<\frac{10}{11}2^n.$$

That means that the left and right expression must have different integer parts,

$$\left\lfloor\frac9{10}2^n\right\rfloor\ne\left\lfloor\frac{10}{11}2^n\right\rfloor.$$

We will be on the safe side if the numbers themselves differ by at least one unit,

$$\frac9{10}2^n\le\frac{10}{11}2^n-1,$$

which gives us

$$2^n\ge\frac1{\dfrac{10}{11}-\dfrac9{10}}=110.$$

So a solution is $n=7,m=116$. Note that this is not the tightest, as $n=5,m=29$ also works.


问:
Two ways to calculate the same sum where the harmonic number pops up.
Can someone help me with this equality?

Prove that
$$\sum_{i=1}^n \frac{x^i}{i} = \sum_{i=1}^n {n \choose i}\frac{(x-1)^i}{i} + H_n$$
where $H_n$ is the harmonic number.

答: Hint: let the LHS be $L(x)$ and the RHS be $R(x)$. For proving $L(x)=R(x)$, it's sufficient to check that $L'(x)=R'(x)$ (done using the binomial theorem) and $L(1)=R(1)$ (done immediately).


问:
A geometry question involving a parallelogram within a triangle.

The marked lengths are equal, what is the value of x?

Apologies if this problem is too easy for this website, yet I find this question to seriously challenge my mathematical capacity.

答:
Let $\angle ABC=a$, then $\angle ACB=180-x-a$.

$BC\|DE$ so $\angle D=a,\angle E=180-x-a$.

Since $\triangle BDF,\triangle CEG$ are isosceles, $\angle DBF=a,\angle GCE=180-x-a$.

$\angle CGF=2(180-x-a)$ since it is an exterior angle of $\triangle CEG$.

Opposite angles of parallelogram are equal so $\angle FBC=\angle CGF$.

Now at $B,a+a+2(180-x-a)=180$, giving $x=90$.


问:
Integral using long division
I have the integral

$$\int\frac{x^2+1}{6x-9x^2}\,\mathrm{d}x$$

$$\rightarrow \left(-9x^2+6x\right):\left(1+x^2\right)\ $$
$$+9x^2+0x+9$$
$$\rightarrow6x+9$$

Im tumped here, because I dont know how to continue with the longdivision to get rid of the 'x'-part. It's not like I can multiply $(1+x^2)$ with a square root, or can I (?), if so, how?

Title: Integral using long division

Body:
$$\frac{x^2+1}{6x-9x^2}=-\frac{x^2+1}{9x^2-6x}=-\frac19\frac{9x^2+9}{9x^2-6x}=-\frac19\left(1+\frac{9+6x}{9x^2-6x}\right)$$
$$=-\frac19-\frac19\frac{3+2x}{3x^2-2x}=-\frac19-\frac{(3+2x)}{9x(3x-2)}$$
can you finish it from here?


答:
Here is a diagram:

Let $\angle DEF = x$, so $\angle DEB = 180^{\circ} - x$. Then, $\angle GFE = 180^{\circ} - x$, which means $\angle GFA = x$. However, since $\triangle DBE$ is isosceles, we have that $\angle DEB = \frac{180 - \angle DEB}{2} = \frac{x}{2}$. Similarly, $\angle GAF = \frac{180 -\angle GFA}{2} = 90^{\circ} - \frac{x}{2}$. Thus, $\angle BCA = 180^{\circ} - \angle CBA - \angle CAB = 180^{\circ} - \angle DBE$$ - \angle GAF = \boxed{90^{\circ.}}$

Title: A geometry question involving a parallelogram within a triangle.

Body:
The marked lengths are equal, what is the value of x?

Apologies if this problem is too easy for this website, yet I find this question to seriously challenge my mathematical capacity.


问:
Understanding the proof of Beck's Theorem
I am trying to understand the proof of Theorem 4.19 in this paper, describing when does the comparison functor $\textbf{D}\to\textbf{C}^T$ gives an isomorphism. However, I am not quite sure what the author really trying to prove though. The argument he gives seems like a copy of Mac Lane's Theorem 1 in Section VI.7, from the "Categories for the Working Mathematicians".

The most puzzling part for me is $(i)\implies (ii)$. We need to show that whenever we have a pair of parallel morphism $f,g:C\to D$ whose image under $G$ has a absolute coequalizer, then $G$ (strictly) creates coequalizers for $f,g$. So we need to start with an absolute coequalizer $q:G(D)\to X$ of $Gf,Gg$ in $\textbf{C}$ and end up with a unique morphism $p:D\to Y$ in $\textbf{D}$ such that $Gp=q$ and $p$ gives a coequalizer of $f,g$. However, when I look at the proof (either the one in the link or Mac Lane's), it looks like that's not what it is doing. For example, the proof in the link starts with an absolute coequalizer in $\textbf{D}$, the thing that is supposed to be obtained at the end of the proof.

Can someone tell me what is the logic of the proof?

答:
I think the proof is written in a slightly ambiguous way. The important point is that when the comparison functor $\textbf{D}\to\textbf{C}^T$ is an isomorphism, it is enough to prove that the forgetful functor $\textbf{C}^T\to\textbf{C}$ creates the good coequalizers. In other words, you can act like $\textbf{D}=\textbf{C}^T$.

The problem in that linked paper is that the identification of $\textbf{D}$ and $\textbf{C}^T$ is not done very explicitly, so there is some confusion about whether $f,g,q$ are in $\textbf{D}$ or $\textbf{C}$.

What happens is that the maps $f,g$ are in $\textbf{D}=\textbf{C}^T$, so they are really maps in $\textbf{C}$ which are compatible with the algebra structures of their domain and codomain; we assume that the corresponding maps in $\textbf{C}$ have an absolute coequalizer $q:D\to Q$ in $\textbf{C}$, and the goal is to prove that there is an algebra structure on $Q$ that makes $q$ the coequalizer of $f$ and $g$ in $\mathbf{C}^T$.


问:
The only factor that is common to 200 and an integer n is 1. When $n/200$ what is the minimum number of digits to the right of the decimal point?
An ACT math problem:

The only factor that is common to 200 and an integer $n$ is 1. When $\frac{n}{200}$ is written as a decimal number, what is the minimum number of digits to the right of the decimal point?

What are some quicker methods to get around this as opposed to trying out a bunch of assumed $n$?

答:
Since $\dfrac 1{200} = 0.005$, we can say that $3$ decimal places are possible at maximum. If the decimal places to the right have to reduce, the ending digits must be zero and it achieved by multiplying $0.005$ with a multiple of $2$, but that violates the condition that $gcd(200,n) = 1$ as $1$ is the only common divisor for $200$ and $n$.

Hence, the minimum number of decimal places is also $3$.


问:
How does $axy+bx+cy+d=0$ represent a rectangular hyperbola?
I'm told a curve of this form $$axy+bx+cy+d=0$$ is a rectangular hyperbola.

But I don't seem to understand where are it's $x^2$ and $y^2$ terms since $x^2-y^2=a$ is also a rectangular hyperbola.

Sorry for the boring question, but did I learn something wrong here?

答:
$$x^2 -y^2 =a^2$$

when rotated by $45^{\circ}$, becomes

$$ xy = a^2/2 $$

Our curve $$ xy + \dfrac{b}{a}x + \dfrac{c}{a}y \color{blue}{+ \dfrac{bc}{a^2}} = \dfrac{-d}{a} \color{blue}{+\dfrac{bc}{a^2}}$$

is

$$ \Big(x+\dfrac{c}{a}\Big)\Big(y+\dfrac{b}{a}\Big)=\dfrac{bc-ad}{a^2}$$

in the form $$ XY = K$$

So it is also a rectangular hyperbola.


问:
Large Deviation, Optimal Transport and Machine Learning Reference
I am looking for references (books/sites/articles) on the following three subjects: Large Deviation, Optimal Transport and Machine Learning References. I would like works which involve any of them individually, or jointly. My motivation for this is the following course description which used to be taught at my school:

Two mathematical frameworks: optimal transport and large deviation
  theory, can be combined to provide tools and methods of analysis for
  some central tasks in data science: classification, clustering and
  regression, dimensionality reduction, conditional probability
  estimation, prediction of rare events. Recurring themes will include
  the power of the composition of simple functions to capture complex
  behavior - as demonstrated by neural networks - and of the
  characterization of both the available observations of the system
  under study and the parameters tuned to model it in terms of
  underlying probability distributions.

If you know of any core books on either Large Deviation or Optimal Trasport that can be grasped at a beginning graduate level, I would much appreciate it!

thank you!

答:
Just in order to add also one particular and comparably famous reference combining aspects of OT and large deviations:

S Adams, Nicolas Dirr, M Peletier, J Zimmer;
From a large-deviations principle to the Wasserstein gradient flow: a new micro-macro passage

https://researchportal.bath.ac.uk/en/publications/from-a-large-deviations-principle-to-the-wasserstein-gradient-flo


问:
Proving Sequent using Natural Deduction
RTP: $P \to Q, R \to \neg Q, (S\to \neg P)\to R \vdash (\neg T ∨ P)\to(T \to S)$ using primitive rules of natural deduction. I've attempted this question multiple times but keep getting stuck on trying to get $S$ so I can use a couple arrow introductions to get the conclusion.

Attempt:

1 (1) $P \to Q$ Assumption

2 (2) $R \to \neg Q$ Assumption

3 (3) $(S\to \neg P)\to R$ Assumption

4 (4) $(\neg T ∨ P)$ Assumption

5 (5) $T$ Assumption

4,5 (6) $P$ $4,5∨$E

1,4,5 (7) $Q$ $1,6\to$E

8 (8) $S$ Assumption

9 (9) $S\to \neg P$ Assumption

8,9 (10) $\neg P$ Assumption

4,5,9 (11) $\neg T$ 6,10RAA(8)

12 (12) R Assumption

2,12 (13) $\neg Q$ $2,12\to$E

1,2,4,5 (14) $~R$ 7,13RAA(12)

I realise now that I've eliminated almost all the necessary assumptions with only line 9 left over (those at lines 4 and 5 which will be eliminated via arrow introduction for the conclusion once I have $S$), but I don't know how I can proceed from here. Any help would be greatly appreciated.

答:
The premise is a negation $\neg(S \to \neg P)$, so it is likely you need to perform negation elimination. For that you need to intermediately derive $S \to \neg P$, resulting in a contradiction from which you may infer anything.

That anything will be $S$, so you prove the conclusion by contradiction, assuming $\neg S$ earlier on and dropping that assumption in the last step.

Now you want to derive $S \to \neg P$; this is an implication, so you have to derive $\neg P$ from an assumption $S$. Given the other assumptions you have at your disposal, how do you think that subderivation can be plugged together?

Title: Proving Sequent using Natural Deduction


答:
Here is a proof using "primitive rules" of Natural Deduction.

From line (4) $(¬T ∨ P)$ --- Assumption, we apply $(\lor \text E)$ with two sub-proofs:

The first one is straightforward:

$\lnot T$ --- assumed [a] for $(\lor \text E)$

$T$ -assumed [b]

$\bot$ --- from 5) and 6) using $(\lnot \text E)$

$S$ --- from 7) using $(\bot \text E)$

$(T \to S)$ --- from 6) and 8) by $(\to \text I)$, discharging [b].

The second sub-proof:

$P$ --- assumed [d] for $(\lor \text E)$

$Q$ --- from 10) and 1) by $(\to \text E)$

$R$ --- assumed [e]

$\lnot Q$ --- from 12) and 2) by $(\to \text E)$

$\bot$ --- from 11) and 13) by $(\lnot \text E)$

$\lnot R$ --- from 12) by $(\lnot \text I)$, discharging [e]

$(S \to \lnot P)$ --- assumed [f]

$R$ --- from 16) and 3) by $(\to \text E)$

$\bot$ --- from 15) and 17) by $(\lnot \text E)$

$\lnot (S \to \lnot P)$ --- from 16) and 18) by $(\lnot \text I)$, discharging [f]

$\lnot S$ --- assumed [g]

$S$ --- assumed [h]

$\bot$ --- from 20) and 21) by $(\lnot \text E)$

$\lnot P$ --- from 22) by $(\bot \text E)$

$(S \to \lnot P)$ --- from 21) and 23) by $(\to \text I)$, discharging [h]

$\bot$ --- from 19) and 24) by $(\lnot \text E)$

$S$ --- from 20) and 25) by $(\neg \neg \text E)$: if $\Gamma , \neg \varphi \vdash \bot$, then $\Gamma \vdash \varphi$, discharging [g]

$(T \to S)$ --- from 26) by $(\to \text I)$.

Now we have derived $(T \to S)$ in the two sub-proofs and we can close the $(\lor \text E)$:

$(T \to S)$ --- from 4), 9) and 27) discharging assumptions [a] and [c]

$(\lnot T \lor P) \to (T \to S)$ --- from 4) and 28) by $(\to \text I)$.

Here is an alternative proof that uses Disjunctive syllogism:

$(¬T ∨ P)$ --- Assumption

$T$ --- assumed [a]

$P$ --- from 4) and 5)

Now we insert lines 10)-26) above to get:

$S$

and we conclude with:

$T \to S$ --- from 5) and 27) by $(\to \text I)$, discharging [a]

Title: Proving Sequent using Natural Deduction

Body:
RTP: $P \to Q, R \to \neg Q, (S\to \neg P)\to R \vdash (\neg T ∨ P)\to(T \to S)$ using primitive rules of natural deduction. I've attempted this question multiple times but keep getting stuck on trying to get $S$ so I can use a couple arrow introductions to get the conclusion.

Attempt:

1 (1) $P \to Q$ Assumption

2 (2) $R \to \neg Q$ Assumption

3 (3) $(S\to \neg P)\to R$ Assumption

4 (4) $(\neg T ∨ P)$ Assumption

5 (5) $T$ Assumption

4,5 (6) $P$ $4,5∨$E

1,4,5 (7) $Q$ $1,6\to$E

8 (8) $S$ Assumption

9 (9) $S\to \neg P$ Assumption

8,9 (10) $\neg P$ Assumption

4,5,9 (11) $\neg T$ 6,10RAA(8)

12 (12) R Assumption

2,12 (13) $\neg Q$ $2,12\to$E

1,2,4,5 (14) $~R$ 7,13RAA(12)

I realise now that I've eliminated almost all the necessary assumptions with only line 9 left over (those at lines 4 and 5 which will be eliminated via arrow introduction for the conclusion once I have $S$), but I don't know how I can proceed from here. Any help would be greatly appreciated.