答:
Hint:

In the diagram, you will notice by Pythagoras that

$(r_3 + r_2)^2 = (r_3 -r_2)^2 + (t_{23})^2$ where $t_{23}$ is the length of the tangent.

So $t_{23} = 2 \sqrt{r_3 r_2}$

Similarly $t_{12} = 2 \sqrt{r_1 r_2}, t_{13} = 2 \sqrt{r_1 r_3}$

Now $t_{23} = t_{12} + t_{13}$ and that leads you to the answer.

Title: Three circle touching each other and having a common tangent

Body:
Three circle with radius $r_1$, $r_2$ and $r_3$ where ($r_1$<$r_2$<$r_3$) touch each other externally . If they have a common tangent then the value of $\sqrt{\frac{r_1}{r_2}}+\sqrt{\frac{r_1}{r_3}}$=__________

My approach is illustrated in the diagram

How will I proceed from here


问:
Given a ring homomorphism $f: \mathbb C \to \mathbb C$, prove it is either the identity mapping or complex conjugation
I am struggling to round off this homework question and require a gentle push in the right direction.

Here's what I have so far:

If $f$ is a homomorphism from the complex numbers to the complex numbers, then I would have the following: $$ f(z+w)=f(z)+f(w) \; \forall (z,w) \in \mathbb(C)$$
$$f(zw)=f(z)f(w) \; \forall (z,w) \in \mathbb(C)$$

and so if I represent a complex number $x+yi$ as $(x+0i)+(0+yi)$ then I would arrive at the following:
$$f(x+yi) = x+yf(i) = u(x,y)+v(x,y)i$$

The issue I have is that I am not sure now how to establish that the real component of this function is $x$. I attempted to look at particular cases such as $x=x , y=0$ and $x=0, y=y$ etc, but cannot establish even the real component is definitively $x$.

One such idea I had was to show that for certain, $f(i)$ is purely imaginary and so $yf(i)$ has no contribution to the real component, but again cannot arrive at a conclusion here.

Any help would be greatly appreciated!

答:
Continuity is essential. There exist additive, multiplicative,

one-to-one discontinuous functions on $%
%TCIMACRO{\U{2102} }%
%BeginExpansion
\mathbb{C}
%EndExpansion
!.$ However, continuity of $f$ can be replaced by the condition that $f$
maps $%
%TCIMACRO{\U{211d} }%
%BeginExpansion
\mathbb{R}
%EndExpansion
$ into $%
%TCIMACRO{\U{211d} }%
%BeginExpansion
\mathbb{R}$

We have $f(rz)=rf(z)$ for all rational $r.$ Since $(f(1))^{2}=f(1)$ we get $%
f(1)=0$ or $1.$ If $f(1)=0$ then $f(r)=0$ for all $r$ rational, hence for
all $r$ real. Note that $[f(i)]^{2}=f(-1)=-f(1)=0$ so $f(i)=0.$ We now get $%
f(a+ib)=f(a)+f(i)f(b)=0.$ Thus $f\equiv 0$ in this case. Let $f(1)=1.$ Then $%
[f(i)]^{2}=f(-1)=-f(1)=-1$ so $f(i)=i$ or $f(i)=-i.$ In the first case $%
f(a+ib)=f(a)+f(i)f(b)=a+ib$ and in the second case $%
f(a+ib)=f(a)+f(i)f(b)=a-ib.$ Conclusion: $f\equiv 0$ or $f(z)=z$ $\forall z$
or $f(z)=\overset{-}{z}$ $\forall z.$

ContentLicense: CC BY-SA 4.0


答:
This is quite false. Actually there are $2^{2^{\aleph_0}}$ ring automorphisms $\mathbb{C} \to \mathbb{C}$ other than the identity and complex conjugation, although they all require some choice to construct and they are all non-measurable. See, for example, this MO thread and this math.SE thread.

There are even ring homomorphisms that aren't automorphisms, which can be constructed as follows: $\mathbb{C}$ embeds into $\mathbb{C}(t)$ and hence into the algebraic closure $\overline{\mathbb{C}(t)}$. It's a general fact about algebraically closed fields of characteristic $0$ that an uncountable such field is determined up to isomorphism by its cardinality, so $\overline{\mathbb{C}(t)}$ is actually isomorphic to $\mathbb{C}$, and hence we have an injection $\mathbb{C} \to \mathbb{C}$ exhibiting $\mathbb{C}$ as a proper subfield of itself (in fact we have many such injections, acted on by a group at least as large as the absolute Galois group of $\mathbb{C}(t)$).

What you can show about an arbitrary ring homomorphism $f : \mathbb{C} \to \mathbb{C}$ is that $f(i)^2 = f(i^2) = f(-1) = -1$, so $f(i) = i$ or $-i$. You can also show that $f(\mathbb{Q}) = \mathbb{Q}$ without much difficulty, so $f$ at least restricts to a map $\mathbb{Q}(i) \to \mathbb{Q}(i)$ which is either the identity or complex conjugation. The problem is that you have no control over the image $f(\mathbb{R})$ of the reals, and in particular no guarantee that it's $\mathbb{R}$ again. In fact the constructions above imply that $\mathbb{C}$ has a bunch of (non-measurable!) subfields abstractly isomorphic to $\mathbb{R}$ other than the obvious one.

This is despite the fact that (and this is a nice exercise) $\mathbb{R}$ itself has no nontrivial endomorphisms.

Title: Given a ring homomorphism $f: \mathbb C \to \mathbb C$, prove it is either the identity mapping or complex conjugation

Body:
I am struggling to round off this homework question and require a gentle push in the right direction.

Here's what I have so far:

If $f$ is a homomorphism from the complex numbers to the complex numbers, then I would have the following: $$ f(z+w)=f(z)+f(w) \; \forall (z,w) \in \mathbb(C)$$
$$f(zw)=f(z)f(w) \; \forall (z,w) \in \mathbb(C)$$

and so if I represent a complex number $x+yi$ as $(x+0i)+(0+yi)$ then I would arrive at the following:
$$f(x+yi) = x+yf(i) = u(x,y)+v(x,y)i$$

The issue I have is that I am not sure now how to establish that the real component of this function is $x$. I attempted to look at particular cases such as $x=x , y=0$ and $x=0, y=y$ etc, but cannot establish even the real component is definitively $x$.

One such idea I had was to show that for certain, $f(i)$ is purely imaginary and so $yf(i)$ has no contribution to the real component, but again cannot arrive at a conclusion here.

Any help would be greatly appreciated!


问:
Computing $\int_0^{\pi\over2} \frac{dx}{1+\sin^2(x)}$?
How would you compute$$\int_0^{\pi\over2} \frac{dx}{1+\sin^2(x)}\, \, ?$$</p>

答:
HINT:

$$\int_0^{\pi\over2} \frac{dx}{1+\sin^2(x)}= \int_0^{\pi\over2} \frac{\csc^2xdx}{\csc^2x+1}=\int_0^{\pi\over2} \frac{\csc^2xdx}{\cot^2x+2}$$

Put $\cot x=u$

ContentLicense: CC BY-SA 3.0


问:
How to visualize $\frac{a}{b} \div \frac{c}{d} \equiv \frac{a}{b} \times \frac{d}{c}$?
I read https://matheducators.stackexchange.com/q/7837, but it doesn't answer my question because the Multiplicand is a whole integer there. The trick  is to rationalize the denominator, to enable me to compare the numerator. I'm just using  $\dfrac{4}{6} \div \dfrac{1}{2} \equiv \dfrac{4}{6} \div \color{red}{\dfrac{3}{6}}$. I can see that $4/6 = \color{red}{1}$
of the top row (each row representing $3/6$) + $\color{green}{1/3}$ of the bottom row = 4/3.

Now   $a \neq b \neq c \neq  d \neq 0, 1$. I rationalize the denominator. $\dfrac{\dfrac{a}{b}}{\dfrac{c}{d}} \equiv \dfrac{\dfrac{ad}{bd}}{\dfrac{cb}{db}} $. Now what do I do? How can I compare $ad$ with $cb$?

答:
Imagine you have the fraction $\frac{a}{b}$.  This can be readily made by a set of segments b long, a coloured.

Beside it is a second set of coloured rods. The length of $a$ is made identical to $c$ new segments, the total length of the rod is $d$ segments.

The lengths of the two rulers then stand in the ratio of $\frac{b}{d}$ of these two rulers, since we have made $a=c$.

The other way is to resort to a graph.

In the first part, we make a line $b$ long and divide it to $a$:

o-------a---b.

In the second part, we make a line o------c---d.

Geometric division is performed by drawing a line between the divisor and $1$.  Here, $1$ means $b$ and the line is drawn from $c$ to $b$.

Continue $oab$ outwards and draw a second line from $d$ parallel to $cb$, to intersect with $ab$.  You will see that it is on the other side of $a$.


问:
Evaluate $\lim_{x\to \infty} \frac{ (\log x)^2}{x^n}, n>0$
It’s a $\frac{\infty}{\infty}$ form

Here is what I tried
$$\lim_{x\to \infty} \frac{ (\log (1+(x-1))^2}{(x-1)^n}$$

I tried reducing it to the standard from, but the $n$ i hindering the process. How do I proceed?

答:
Hint:

Let $\log x=y, x=e^y$

$$\lim_{x\to\infty}\dfrac{(\ln x)^2}{x^n}=\lim_{y\to\infty}\dfrac{y^2}{e^{ny}}$$

Now $e^{ny}=\sum_{r=0}^\infty\dfrac{(ny)^r}{r!}$

Now divide the numerator & the denominator by $y^2$


问:
$\Delta ABC$, $AC = 2BC$ and $\angle C = 90^\circ$ and $D$ is the foot of the altitude from $C$ onto $AB$. Find $AE : EC$ .

$\Delta ABC$, $AC = 2BC$ and $\angle C = 90^\circ$ and $D$ is the foot of the altitude from $C$ onto $AB$. A circle with diameter $AD$ intersects the segment $AC$ at $E$. Find $AE : EC$.

What I Tried: Here is a picture :-

The first thing I did was use similarity. For example one can find multiple similar triangles like $\Delta ABC \sim \Delta ACD \sim \Delta CBD$. After assuming $AC = x$ and $AD = y$ we get everything there in the picture.

Next from Pythagorean Theorem we will get only one equation involving $x$ and $y$ :-
$$5y^2 = x^2$$

The problem is, how will I proceed finding $AE$ and $ED$ ? I am thinking that $ED \parallel BC$ , but I have no proof of it and neither I cannot proceed further.

Can anyone help me? Thank You.

答:
$\angle DEA = 90º$ by Thales' theorem, so $\angle CED = 90º$.

Thus $\Delta DCE \sim \Delta CBD$ by AA. This gives:

$$\frac{EC}{CD} = \frac{DB}{DC} \Rightarrow \frac{EC}{2y} = \frac{y}{x}.$$

Write out $AC:EC$ for convenience in terms of $x,y$. Now using the fact that $5y^2 =x^2$:

$$\frac{AC}{EC} = \frac{2x}{2y^2/x} = \frac{x^2}{y^2} = 5.$$

Therefore $AE:EC$ is:

$$\frac{AC-EC}{EC} = \frac{5-1}{1} = \boxed{4}.$$


答:
It might be simpler to rewrite this as
$$\lim_{x\rightarrow\infty}\left(\dfrac{\log(x)}{x^{n/2}}\right)^2$$
and use theorems on limit comparisons to show that the function in the brackets tend to $0$.

Title: Evaluate $\lim_{x\to \infty} \frac{ (\log x)^2}{x^n}, n>0$

Body:
It’s a $\frac{\infty}{\infty}$ form

Here is what I tried
$$\lim_{x\to \infty} \frac{ (\log (1+(x-1))^2}{(x-1)^n}$$

I tried reducing it to the standard from, but the $n$ i hindering the process. How do I proceed?


问:
why $D\setminus f(D)$ is open?
Suppose $f$ is C-analytic in $D(0; 1)$ and maps the unit circle into itself. Show then that $f$ maps the
entire disc onto itself.

My attempt :  I  got the  answer  given below but i have  some confusion in the given below answer  mark in red box

My doubt: why $D\setminus f(D)$ is open ?

My thinking : Here $D(0,1)=D$ is an open disk and $f(D)$  is also open since contnious image of open is open . But this will contradicts the theorem complement of open set is closed because $f(D)$ is not closed

I think it may be $D\setminus f(D)$$=$ (open) $-$(open)$=$ open

答: Let $0<r<1-|w|$. Then the disk of radius $r$ around $w$ is contained in $D\setminus f(D)$ because if $\zeta $ is in this disk then $|f(z)-\zeta| \geq |f(z)-w|-|w-\zeta| >1-|w| -r >0$ (so $f(z) \neq w$). We have proved that any point $w$ of $D\setminus f(D)$ is an interior point.

Title: why $D\setminus f(D)$ is open?


Body:
Since, $\log x < x$ for all $x>0$,
$$
\log x = \frac{4}{n}\log(x^{n/4})  < \frac{4}{n}x^{n/4}
$$
and therefore,
$$
0<\frac{{(\log x)^2 }}{{x^n }} < \frac{{16}}{{n^2 }}\frac{1}{{x^{n/2} }} \to 0.
$$


问:
$x^{5}+y=t , x^{2}+y^{3}=t^{2}$ and $ f=x^{3}y $ then$ \frac{df}{dt}=?$
$x^{5}+y=t , x^{2}+y^{3}=t^{2}$ and $  f=x^{3}y $ then$  \frac{df}{dt}=?$

I use chain rule but i couldnt result?

$\frac{df}{dx}\frac{dx}{dt}+\frac{df}{dy}\frac{dy}{dt}$ am i wrong?

答:
Implicitly differentiate the first 2 equations

$$ 5x^4\frac{dx}{dt} + \frac{dy}{dt} = 1 $$

$$ 2x\frac{dx}{dt} + 3y^2\frac{dy}{dt} = 2t = 2(x^5+y) $$

Can you now solve for $\frac{dx}{dt}$ and $\frac{dy}{dt}$ and get the answer?


答:
Note that $\angle AED$ must be $90^{\circ}$ as it is inscribed in a semi-circle. Thus $\triangle ADE \sim \triangle ABC$ and so

$$\frac{AE}{EC} = \frac{AD}{DB} = 4$$

Title: $\Delta ABC$, $AC = 2BC$ and $\angle C = 90^\circ$ and $D$ is the foot of the altitude from $C$ onto $AB$. Find $AE : EC$ .

Body:
$\Delta ABC$, $AC = 2BC$ and $\angle C = 90^\circ$ and $D$ is the foot of the altitude from $C$ onto $AB$. A circle with diameter $AD$ intersects the segment $AC$ at $E$. Find $AE : EC$.

What I Tried: Here is a picture :-

The first thing I did was use similarity. For example one can find multiple similar triangles like $\Delta ABC \sim \Delta ACD \sim \Delta CBD$. After assuming $AC = x$ and $AD = y$ we get everything there in the picture.

Next from Pythagorean Theorem we will get only one equation involving $x$ and $y$ :-
$$5y^2 = x^2$$

The problem is, how will I proceed finding $AE$ and $ED$ ? I am thinking that $ED \parallel BC$ , but I have no proof of it and neither I cannot proceed further.

Can anyone help me? Thank You.


问:
Partial derivatives of homogenous functions.
Consider a homogenous function $f(x, y)$ of degree $r$. That is,

$f(tx, ty) = t^{r}f(x,y)$.

In all textbooks,

$\frac{t{\partial f(tx, ty)}}{\partial x} = \frac{t^{r}{\partial f(x, y)}}{\partial x}$.

I think it should be written as
$\frac{t{\partial f(tx, ty)}}{\partial (tx)} = \frac{t^{r}{\partial f(x, y)}}{\partial x}$

because of the chain rule, because $f(tx, ty)$ is first a function of $tx$ and $ty$ (and only 'ultimately' of $x$ and $y$) and then $tx$ and $ty$ are functions of $x$ and $y$ respectively. Please help.

答:
To provide a proof of
$$t\frac{\partial f}{\partial x}(tx,ty)=t^r\frac{\partial f}{\partial x}(x,y)$$
it is sufficient to show $\frac{\partial f}{\partial x}$ is homogeneous of degree $r-1$.

By definition
$$\frac{\partial f}{\partial x}(tx,ty)=\lim_{h\to 0}\frac{f(tx+h,ty)-f(tx,ty)}{h}.$$

Using homogeneity, we can rewrite this as
$$t^r\lim_{h\to 0}\frac{f(x+\frac{h}{t},y)-f(x,y)}{h}.$$

Then, as $t$ is independent of $h$, this is equal to

$$t^r\lim_{h/t\to 0}\frac{1}{t}\frac{f(x+\frac{h}{t},y)-f(x,y)}{h/t}=t^{r-1}\lim_{s\to 0}\frac{f(x+s,y)-f(x,y)}{s}$$

where $s=h/t$. But this is precisely $t^{r-1}\frac{\partial f}{\partial x}(x,y)$. The result follows immediately.


问:
Eigenvalues with positive real parts imply instability
Given an autonomous ode $\dot{x}=F(x)$ with $0$ being its equilibrium point, and all eigenvalues of $DF(0)$ have non-zero real parts.

I have learned that if the real parts of eigenvalues are all negative , then the system is asymptotically stable near $0$. Now I guess that if one of the eigenvalues has positive real part then the system is unstable near $0$, but I can't give a proof, can anyone help me?

Many thanks!

答:
You are comparing two systems of differential equations, the linear system
$$
\dot v=Av
$$
and the non-linear system
$$
\dot x=F(x)
$$
with $F(0)=0$ a stationary point at the origin and $DF(0)=F'(0)=A$ as the Jacobian. If $F$ is sufficiently smooth, say twice differentiable at least, the vector fields look similar close to the origin.

Now if one eigenvalue of $A$ is real and positive, or a conjugate eigenpair has a positive real part, then there are solutions of the linear system that connect the stationary point with infinity moving outwards. The question asks if the non-linear system is similar enough that solutions close-by inherit enough of this divergence to prevent this point from being stable.

The Hartman-Grobman theorem answers this in the most comprehensive way (lesser tools may be able to also answer this with more effort) by the claim that there is some diffeomorphism $\Phi$ mapping a neighborhood of the origin in the linear system to a neighborhood of the stationary point of the non-linear system (here also the origin) with $Φ(x)=x+O(|x|^2)$ so that $$x(t)=Φ(e^{At}v_0)$$ is the solution of $$\dot x=F(x)~~\text{ with }~~x_0=Φ(v_0)\iff v_0=Φ^{-1}(x_0)$$ for $x_0$ close to the fixed point $0$ and $t$ small enough.

Now if $A$ has an eigenvalue $\lambda=\alpha+i\beta$ with positive real part $α$, then there will be an eigenvector $u+iw$ leading to real solutions $e^{αt}(\cos(βt)u-\sin(βt)w)$ for the linear equation and thus
$$
x(t)=Φ(v(t))=Φ\bigl(e^{αt}(\cos(βt)u-\sin(βt)w)\bigr)
$$
(for a real positive eigenvalue just set $β=0$).
Now if you fix some point $x(t_*)$ on this solution that is still inside the range of the diffeomorphism, then the points $x_0=x(t_0)$, $t_0<t_*$ can be selected arbitrarily close to the origin (select $β(t_*-t_0)\in 2\pi\Bbb Z$ to get points on the same ray to the origin). Thus you get initial points arbitrarily close to the origin that have solutions that pass through the same point $x(t_*)$ (relatively) far away from the origin. This is then a contradiction to both definitions of stability and asymptotic stability.


问:
Integrate $\int_0^1 \Bigl(\mathrm{Li}_2\bigl(\frac{x-1}{x}\bigr)\Bigr)^2 \mathrm{d}x$
This question has some relationship to this integral: Let $\mathrm{Li}_2$ be the dilogarithm. Then, numerically,
$$
  \int_0^1 \Bigl(\mathrm{Li}_2\bigl(\frac{x-1}{x}\bigr)\Bigr)^2 \mathrm{d}x
  =
  \frac{17}{180}\pi^4,
$$
as can be checked by Mathematica code like

```<code>NIntegrate[ PolyLog[2, (x-1)/x]^2, {x,0,1}, WorkingPrecision -> 100]
</code>```

and

```<code>N[ 17/180 * Pi^4, 100]
</code>```

both of which output the same digits

```<code>9.199747486544674627886031420599927173585383 ...
</code>```

Is this a known identity? Some related results appear in Freitas and Laurenzi, but without this specific argument in the dilogarithm.

答:
OK, this follows from Freitas' paper. In Table 3 there

$$\int_0^1 \mathrm{Li}_2^2(x) \mathrm{d}x = 6 - 2\zeta(2) - 4\zeta(3) + \frac{5}{2}\zeta(4),$$
and Formula (2.2.5) of Table of integrals and formulae for Feynman diagram calculations states
$$\mathrm{Li}_2\Bigl(-\frac{y}{1-y}\Bigr) = - \mathrm{Li}_2(y) - \frac{1}{2}\ln^2(1-y),$$
and Mathematica can solve
$$
\frac{1}{4}\int_0^1 \ln^4(1-x)\mathrm{d}x = 6
$$
as well as
$$
\int_0^1 \mathrm{Li}_2(x) \ln^2(1-x) \mathrm{d}x
=
\frac{1}{15} \left(60 (\zeta (3)-3)+5 \pi ^2+\pi ^4\right)
$$

Using these formulae, the claim follows.


问:
Prove that 4 is not a 3-adic cube
I'm trying to show that 4 is not a 3-adic cube.

On the Wikipedia page for Hensel's lemma (https://en.wikipedia.org/wiki/Hensel%27s_lemma) I read that: "4 is not a 3-adic cube since it is not a cube mod 9"

However, I'm not sure why a number not being a cube mod 9 implies it is not a 3-adic cube. Is there a nice proof of this fact?

答:
If $4$ were a cube in $\mathbf Z_3$, it would be the cube of a $3$-adic unit, say $4=u^3$. Reduction mod $3$ shows that actually $u$ is a principal unit, i.e. has the form $u=1+w$, with $w\in 3\mathbf Z_3$. Expanding $(1+w)^3$ shows immediately that $v_3 (u^3 -1) \ge 2$, whereas $v_3 (4-1)=1$, a contradiction.

Actually one could tackle similar questions in the general case of $\mathbf Z_p$, where $p$ is any prime, by using the multiplicative structure of the group of units (resp. principal units) $U$ (resp. $U_1$) of $\mathbf Z_p$, see e.g. Serre's booklet "A course in arithmetic", chap. II, §3. In particular, for $p$ odd and $n \ge 1$, denote $U_n =1+p^n\mathbf Z_p$. If $x\in U_n - U_{n+1}$, then $x^p\in U_{n+1} - U_{n+2}$ (§3.2, lemma). Analogous (but more complicated) results are available for any finite extension of $\mathbf Q_p$.

Title: Prove that 4 is not a 3-adic cube


问:
Finding the coordinate vector

If $v_1$ and $v_2$ are two linearly independent vectors in $\mathbb{R}^4$ ,$U= \text{span} \{v_1,v_2,v_1+v_2\}$ and $B= \{v_1+2v_2,v_2\}$. Prove that $B$ is a basis in $U$ and find coordinate vector of  $v_1$ with respect to $B$

答:
First step: show that $v_1,v_2,v_1+v_2$ are  linearly dependent . Cnclusion: $ \dim U=2.$

Second step: show that $B \subseteq U.$

Third step: show that $v_1+2v_2, v_2$ are  linearly independent.

These steps give:  $\text{span}B=U$, hence $B$ is a basis of $U$.

For the rest make the "Ansatz": $v_1=a(v_1+2v_2)+bv_2$ for $a,b \in \mathbb R.$ Compute $a $ and $b.$

Title: Finding the coordinate vector


问:
If $2^{2k}-x^2\bigm|2^{2k}-1$ then $x=1$
This is the $y=2^k$ case of this question.

Suppose that $k\geq1$ and $0<x<2^k$ and $2^{2k}-x^2\bigm|2^{2k}-1$. Is it necessarily the case that $x=1$?

Equivalently: Suppose that there are two positive divisors of $2^{2k}-1$ which average to $2^k$. Is it necessarily the case that these two divisors are $2^k-1$ and $2^k+1$?

答:
The equation resolves to $(2^k+x)(2^k-x) \mid (2^k+1)(2^k-1)$.  Since we have the first factor larger than the rest, we would look for a common factor in $(x- 1)$ and $(2^k-1)$ or $(x+1)$ and $2^k+1$,

We could suppose that the first is a product of say $(ab)(cd)$, and that the second is a product of $(ac)(bd)e$, where $ac$ divides $2^k+1$ and $bd$ divides $2^k-1$.  The common divisor between the first two and the last two factors, is $x+1$, and between the second and third, and the first and fourth, $x-1$.

But this common factor must also divide $2^k+1$ and $2^k-1$, and so must divide $2$.

So there is no other number which divides pairs of divisors (ie $2^k \pm 1$ and $2^k\pm x$).


问:
Normal covering spaces - equivalent definitions for connected spaces
A covering space $p: Y \to X$ is normal when for all $ x \in X$ and for all $x_1', x_2' \in p^{-1}(x)$ there is a deck transformation $\phi$ with $x_2' = \phi(x_1')$.

I am asked to show that when $X$ is connected, then this condition is equivalent to saying that there exists one $x_0 \in X$ so that for all $x_1', x_2' \in p^{-1}(x_0)$, there exists a deck transformation taking $x_1'$ to $x_2'$.
One direction of this is easy, the other I am having trouble with.

I have thought about defining the set $A$ to be the set of points in $X$ for which there exist points in its fibre that aren't mapped to teach other by any deck transformation.
If I can show that $A$ is both open and closed, then I will be done.
However I am not sure how to show this.
Is this the correct approach, and if so, how should I continue?

答:
Your approach is correct, but as far as I can see yo need further assumptions on $X$.

Call $x  \in X$ a normal point of $p$ if for all $y_1, y_2 \in p^{-1}(x)$ there is a deck transformation $\phi$ with $y_2 = \phi(y_1)$. Let us first prove the following

Lemma. Let $U$ be an evenly covered connected open subset of $X$. If some $\xi \in U$ is a normal point of $p$, then all $x \in U$ are normal points of $p$.

$p^{-1}(U)$ is the disjoint union of open $V_\alpha \subset Y$ which are mapped by $p$ homeomorphically onto $U$ ("sheet decomposition of $p^{-1}(U)$"). The $V_\alpha$ are the connected components of $p^{-1}(U)$. Let $x \in U$ and $y_i \in p^{-1}(x)$. There are unique $\alpha_i$ such that $y_i \in V_{\alpha_i}$. Let $\eta_i \in p^{-1}(\xi)$ be the unique point contained in $V_{\alpha_i}$. There exists a deck transformation $\phi$ such that $\eta_2 = \phi(\eta_1)$. The set $\phi(V_{\alpha_1})$ is a connected component of $p^{-1}(U)$ such that $\eta_2 = \phi(\eta_1) \in \phi(V_{\alpha_1})$. Thus $\phi(V_{\alpha_1}) = V_{\alpha_2}$. Therefore $y_2 = \phi(y_1)$.

Why do we need the connectedness of $U$? In the non-connected case the sheet decomposition of $p^{-1}(U)$ is not unique (see Covering projections: What are the sheets over an evenly covered set?), thus the sheet decomposition $\{\phi(V_\alpha) \}$ of $p^{-1}(U)$ may  differ from $\{V_\alpha \}$ and we cannot conclude that $\phi(V_{\alpha_1}) = V_{\alpha_2}$. Hence we cannot be sure that $y_2 = \phi(y_1)$. Of course there may exist a deck transformation $\phi'$ such that $y_2 = \phi'(y_1)$, but there is no general strategy to find it (and maybe it is different from $\phi$).

You might argue that $p^{-1}(U) \approx U \times F$ with a discrete $F$, thus certainly all $x \in U$ are normal points of the trivial covering $p_U : p^{-1}(U) \to U$. That is, for all $x \in U$ and all $y_i \in p^{-1}(x)$ there exists a deck tranformation $\phi_U$ for $p_U$ with $y_2 = \phi_U(y_1)$. But there is no reason to assume that $\phi_U$ extends to deck transformation for $p$.

Now let us assume that $X$ is locally connected.

Let $N$ denote the set of normal points of $p$. Since each $x \in X$ has an evenly covered connected open neigborhood, the above lemma shows that $N$ and $X \setminus N$ are open in $X$. Thus $N = X$.


问:
Does LIpschitz condition implies boundedness by its constant?
Let $g: \mathbb{R} \to \mathbb{R}$ be a locally Lipschitz function such that $\rho:[t_{0}, t_{1}] \to \mathbb{R}$ is the unique solution of the p.v.i defined by
$$ \begin{cases}
\rho'(t) & = g(\rho(t)) \\
\rho(t_{0}) & = \rho_{0}
\end{cases}. $$
Defining $L := Lip(g\vert_{[t_{0}, M]})$ where $M:= \max_{t \in [t_{0}, t_{1}]}(\rho(t))$, can we say that $\vert \rho'(t) \vert \leq L \vert \rho(t) \vert$? Is obvious that $\rho'(t)$ is bounded, but is that bound correct?

答:
If $$R=\max_{t\in[t_0,t_1]}\|ρ(t)-ρ_0\|$$ and $L$ is a Lipschitz constant on $\bar B(ρ_0,R)$, then indeed you get a bound from the Lipschitz condition, but it should look like
$$
\|ρ'(t)\|\le \|f(ρ_0)\|+L\|ρ(t)-ρ_0\|\le \|f(ρ_0)\|+LR.
$$
Per the Grönwall theorem a bound on the solution evolution follows,
$$
\|ρ(t)-ρ_0\|\le \|f(ρ_0)\|\frac{e^{Lt}-1}L.
$$


Body:
n>0;

Hospital:

1)$ \dfrac{2\log x}{x(nx^{n-1})}= \dfrac{2\log x}{nx^n};$

And again:

2)$\dfrac {2}{x(n^2x^{n-1})}= \dfrac{2}{n^2x^n};$


问:
Proof $X_n/a_n \rightarrow 0$ given $\sum_{i=1}^{n}X_i/a_n \rightarrow 0$ (in probability)
I’m trying to solve the following problem: Given a sequence $(a_n)$ of positive real numbers such that $\lim_{n \to \infty} \frac{a_{n+1}}{a_n}=1$ (could be $a_n=n$ for example) and a sequence of random variables $X_n$, I need to prove that if $(\sum_{i=1}^{n}X_i)/a_n \rightarrow 0$ in probability then $X_n/a_n \rightarrow 0$ also in probability.

I tried to bound $|X_n/a_n|$ by $\sum_{i=1}^{n}|X_i|/a_n$ to apply the definition of convergence in probability, but since the $X_n$ might not be positive this doesn’t work. I also tried using the definition of a Cauchy sequence in probability to prove $X_n/a_n$ is also Cauchy but I got a mess that doesn’t seem to arrive to anything. Finally I tried to extract a subsequence that converges almost surely but again I don’t seem to get anything useful.

Could you give me any hint?

答: Let $Y_n:= a_n^{-1}\sum_{i=1}^nX_i$. Then by assumption, $Y_n\to 0$ in probability and it follows that $Y_{n+1}\frac{a_{n+1}}{a_n}-Y_n\to 0$ in probability (using $a_{n+1}/a_n\to 1$), which gives the wanted conclusion.


问:
Distribution of sum of squares of two normal random variable
Let $X1$ and $X2$ be independent random variables, each following a normal density with mean $0$ and variance $\sigma^2$. Find the distribution of

$R = \sqrt{(X1^2 + X2^2)}$?

Hint: Let X1 = R cos θ and X2 = R sin θ. Obtain the joint pdf of R and θ, then integrate out θ.

答:
The joint density of $(X_1,x_2)$ is the following

$$f_{X_1X_2}(x_1,x_2)=\frac{1}{2\pi\sigma^2}e^{-\frac{x_1+x_2}{2\sigma^2}}$$

now simply passing in polar coordinates you get

$$f_{R\Theta}(r,\theta)=\frac{1}{2\pi}\times \frac{r}{\sigma^2}e^{-\frac{r^2}{2\sigma^2}}=f_{\Theta}(\theta)\times f_R(r)$$

Where

$f_{\Theta}$ is the angle's distribution, uniform in $[0;2\pi]$

$ f_R(r)$ is the "radius distribution", as requested, a Rayleigh

thus not any integration is needed to answer


问:
Finding orthonormal bases with respect to an inner product.

Hi i was wondering if anyone could help me with my linear algebra question as I am a bit stuck as we've only just touched on the idea of orthonormal bases.

So far my knowledge is that an orthonormal base includes vectors which have a size of 1.

So from this my thinking is that the only answer that is incorrect is a) as when you calculate the size of each factor of a you get $1, \sqrt{2}, \sqrt{3}$

and the size of each component of b), c) and d) all give 1.

as for example d. $((1/\sqrt{2}),(1/2),0)^T$

$=\sqrt{(1/\sqrt{2})^2+(1/2)^2*2+0}$ $=1$

So hence I was wondering if I was on the right track or whether I've missed something. Thanks in advance

答:
First of all you can check this simply by checking all the necessary inner products. A basis $\{\mathbf{u}_1, \mathbf{u}_2, \mathbf{u} \}$ for $\mathbb{R}^3$ is orthonormal with respect to the inner product $< \cdot, \cdot >$ if and only if

$$< \mathbf{u}_i, \mathbf{u}_j > = \begin{cases}
1, & i=j \\
0, & i \neq j
\end{cases}.$$

However, there is a more compact way of formulating this for weighted dot products like this using matrices. We consider $\mathbb{R}^n$ equipped with the weighted dot product

$$< \mathbf{x}, \mathbf{y} > = w_1 x_1y_1 + \cdots + w_nx_n y_n,$$

for weights $w_1,..., w_n \neq 0$. Observe that the inner product can be written as

$$< \mathbf{x}, \mathbf{y} > = \mathbf{x}^T W \mathbf{y},$$

where $W = \text{diag}(w_1, ..., w_n).$

We can now reformulate our criterion for orthonormality: A basis $\{ \mathbf{u}_i \}_{i=1}^{n}$ is an orthonormal basis for $\mathbb{R}^n$ if and only if

$$\mathbf{u}_i^T W \mathbf{u}_j = \begin{cases}
1, & i=j \\
0, & i \neq j
\end{cases}$$

Observe that if we define $U = [\mathbf{u}_1 \cdots \mathbf{u}_n]$, this can be written as

$$U^T W U = I_n,$$

which gives you a nice and compactly formulated criterion.


问:
Doubt about discontinuity
I found this exercise:>Let $f$ be a monotone function, then it can have only a countable number of discontinuities.

I've seen around here a lot of posts about the same but I never get to any conclusion. What happens with Identity function restricted to the set of irrationals? It is monotone, but, isn't it discontinuous at every point of its domain? Which, by the way is uncountable.

I know that if $f$ were defined on an interval, it would be easy to prove, but this is not the case...

Could anyone clarify me please?

答: Actually, the identity function, restricted to the irrationals, is continuous at every point of its domain. If you want a $\varepsilon-\delta$ proof, then just pick, for any $\varepsilon>0$, $\delta=\varepsilon$. So, it is not a counter-example.

Title: Doubt about discontinuity


问:
Does $\sum_{k=1}^{\infty} (-1)^k \frac{x^{2k}}{5^k \sqrt k}$ converge?
Let $a_k=(-1)^k \frac{1}{5^k \sqrt k}$.

This is an example in my book, the author claims that according to the root test the limit does not exist, because $a_k=0$ for k odd (is this true?), but proceeds to show that it indeed has a convergence radius of $R=5$, with the argument that limsup for a bounded sequence always exists.

答:
Substitute $t=\frac{x^2}{\sqrt{5}}$ then the inifinite sum is polylog of t.

$$\sum_{k=1}^{∞}(-1)^k\frac{x^{2k}}{5^k \sqrt{k}}=\sum_{k=1}^{∞}(-1)^k\frac{t^k}{\sqrt{k}}=PolyLog(\frac{1}{2},-t)=PolyLog(\frac{1}{2},-\frac{x^2}{\sqrt{5}})$$

Have a look at Polylogarithm. This is granted by the law, that is a new series can be led back to a well known series, it converges like that one modified by the substitution necessary to transform both series into each other.

This is a Dirichlet power series in $s=-\frac{1}{2}$. This definition is valid for arbitrary complex order $s$ and for all complex arguments $z$ with $|z| < 1$. In this question $x$ with $|x| < \sqrt{5}$. It can be extended to $|x| ≥ \sqrt{5}$ by the process of analytic continuation. This follows by the definition of the power series of $PolyLog$.

Central to the prove by using the $k$-root limes superior law is that
$$ \lim_{k\rightarrow∞}k^{\frac{1}{2k}}=1$$

That is not easily seen ad hoc. $k^{\frac{1}{2k}}=\exp (\frac{1}{2 k} \log k)$. We can now take the limit of the argument of $\exp$ since this function is well behaved. We know already that $k$ grows much fast than $\log$ for large value. So the exponent approches $0$ for very large $k$. Some limit that exists properly exist as a quotient to. The limes superior $k$-root law helps and proves the result.

On the real axis

-1 without the substitution" />

-1 with the substitution" />

The result by comparison to the known series provides a rather different convergence behavior than the rules and laws. We get ride of the singularity at $-1$ because the argument is square an the factor is positive. The singularity would be at $1$ if the sign was not explicitly changed.

The radius of convergence of the series on the real axis for $x$ is ℝ. The example shows that substitution and search a suitable well known power series is much mightier than using the laws.

Body:
Substitute $t=\frac{x^2}{\sqrt{5}}$ then the inifinite sum is polylog of t.

$$\sum_{k=1}^{∞}(-1)^k\frac{x^{2k}}{5^k \sqrt{k}}=\sum_{k=1}^{∞}(-1)^k\frac{t^k}{\sqrt{k}}=PolyLog(\frac{1}{2},-t)=PolyLog(\frac{1}{2},-\frac{x^2}{\sqrt{5}})$$

Have a look at Polylogarithm. This is granted by the law, that is a new series can be led back to a well known series, it converges like that one modified by the substitution necessary to transform both series into each other.

This is a Dirichlet power series in $s=-\frac{1}{2}$. This definition is valid for arbitrary complex order $s$ and for all complex arguments $z$ with $|z| < 1$. In this question $x$ with $|x| < \sqrt{5}$. It can be extended to $|x| ≥ \sqrt{5}$ by the process of analytic continuation. This follows by the definition of the power series of $PolyLog$.

Central to the prove by using the $k$-root limes superior law is that
$$ \lim_{k\rightarrow∞}k^{\frac{1}{2k}}=1$$

That is not easily seen ad hoc. $k^{\frac{1}{2k}}=\exp (\frac{1}{2 k} \log k)$. We can now take the limit of the argument of $\exp$ since this function is well behaved. We know already that $k$ grows much fast than $\log$ for large value. So the exponent approches $0$ for very large $k$. Some limit that exists properly exist as a quotient to. The limes superior $k$-root law helps and proves the result.

On the real axis

The result by comparison to the known series provides a rather different convergence behavior than the rules and laws. We get ride of the singularity at $-1$ because the argument is square an the factor is positive. The singularity would be at $1$ if the sign was not explicitly changed.

The radius of convergence of the series on the real axis for $x$ is ℝ. The example shows that substitution and search a suitable well known power series is much mightier than using the laws.


问:
The Probability of Getting X More Heads Than Tails in Y Tosses of a Coin
You toss a coin 30 times. A head is a win, a tails is a loss. What's the probability of getting two heads ahead of the cumulative tails in these 30 tosses. You stop tossing the coin once you have achieved two heads more than tails.

For example:

tails, heads, heads, heads

is a win because you have two more heads than tails.

tails, tails tails, tails, heads, tails, heads, heads, tails, heads, tails, heads, heads, heads, heads, heads

is a win because you have 7 tails, and 9 heads.

What is the probability of having x heads ahead of the cumulative tails in y number of tosses?

Hopefully that makes sense and thanks in advance, and please provide a formula for how this is calculated.

答:
Thank you for your gratitude, always happy to help a chemist! The question is very valid, asking questions is often much more important than answering them in mathematics. For what it is worth your intuition is very nearly perfect.

Let us ask ourself another question to answer yours: Why did we select 16? This is because 16 heads (and thus 30-16 = 14 ) tails was the absolute threshold to ensure we ended with a net +2 heads. Hence for $y$ total tosses to ensure a net of $+x$ we would require $\frac{y+x}{2}$ heads and $\frac{y-x}{2} $ tails. That is where the 16 came from. Hence if $x+y \equiv 0 $ mod 2. I.e are both even or both odd then the final solution is:

$ \frac{{y \choose (\frac{x+y}{2})} + \sum\limits_{i=(\frac{x+y}{2})+1}^{y}2{y \choose i}}{2^{y}}$

Now what about the annoying case of $x+y \equiv 1 $ mod 2. I.e one is odd and the other even. Funnily enough this is soooooo much easier. To ensure a net of at least $+x$ we must have $\frac{y+x+1}{2}$ heads and $\frac{y-x-1}{2}$ tails. Notice these are integers, and sum to $y$ and have difference $x+1$. As having a diffirence of just $x$ after $y$ tosses is not possible. I'd recommend playing around with these numbers using $y=29$ and $x= 2$

Now let $\Omega_{\geq\frac{y+x+1}{2}}$ and $\Omega_{<\frac{y+x+1}{2}}$ Be defined as before and clearly everything in $\Omega_{\geq\frac{y+x+1}{2}}$ eventually has a difference of $+ (x+1)$ heads and thus has is in our "successful set" $\mathcal{S} $

There is again a 1 to 1 correspondence as before by our flipping after we first have net $+x$  heads from $\Omega_{\geq\frac{y+x+1}{2}}$ to $\Omega_{<\frac{y+x+1}{2}}$

Hence the size of our successful set is just double the size of $\Omega_{\geq\frac{y+x+1}{2}}$ and the final answer is : $ \frac{  \sum\limits_{i=(\frac{x+y+1}{2})}^{y}2{y \choose i}}{2^{y}}$

Very happy to answer any further clarity, would definitely recommend you read page $6$ of this article on reflecting as it is exactly what we did.

Oskar


问:
Product of N independent uniform random variables
We have N independent random variables of $X_1 , X_2 , ... X_N$ with uniform distribution $U(0,1)$. the distribution of $Y$ as
$$
Y = \frac{1}{X_1X_2X_3 ... X_N}
$$
is desired.

To solve this problem, I write:
$$
T = \ln(Y) = - \sum_{i=1}^N \ln(X_i) = \sum_{i=1}^N Z_i
$$
where $ Z_i =-\ln(X_i) $. Then the distribution of $Z_i$ random variables could be written as
$$
f_{Z_i}(Z_i) = \frac{f_{X_i}(e^{-Z_i})}{|e^{Z_i}|} = e^{-Z_i}f_{X_i}(e^{-Z_i})
$$
But I do not know how to solve it any further. Would be helpful to utilize its Characteristic function in order to rewrite the sum of natural-log sum of random variables? Or Should I solve it in a different way?

Thanks for your help in advance.

答:
I've found another answer for this question, for whom get here by search.

If we have
$$
\phi_T(s) = E\{e^{sT}\} = E\{e^{s\sum_{i=0}^N Z_i}\} = \prod_{i=0}^NE\{e^{sZ_i}\}
$$
$$
= \prod_{i=0}^N\phi_{Z_i}(s) = \left( \phi_{Z_i}(s)\right)^N = (\frac{1}{1-s})^N
$$
in Laplace transform, there is this duality
$$
f_T(t) \leftrightarrow L^{-1}\{\phi_T(-s)\} \rightarrow f_T(t) = \frac{1}{(N-1)!}t^{N-1}e^{-t}u(t)
$$
So finally we could write this
$$
T = \ln(Y) \rightarrow Y = e^T
$$
$$
f_Y(y) = \frac{f_T(\ln(y))}{|e^{\ln(y)}|} =  \frac{1}{(N-1)!}\frac{(\ln(y))^{N-1}}{y^2}u(y-1)
$$


问:
A question about the arctangent addition formula.
In the arctangent formula, we have that:

$$\arctan{u}+\arctan{v}=\arctan\left(\frac{u+v}{1-uv}\right)$$

however, only for $uv<1$. My question is: where does this condition come from? The situation is obvious for $uv=1$, but why the inequality?

One of the possibilities I considered was as following: the arctangent addition formula is derived from the formula:

$$\tan\left(\alpha+\beta\right)=\frac{\tan{\alpha}+\tan{\beta}}{1-\tan{\alpha}\tan{\beta}}.$$

Hence, if we put $u=\tan{\alpha}$ and $v=\tan{\beta}$ (which we do in order to obtain the arctangent addition formula from the one above), the condition that $uv<1$ would mean $\tan\alpha\tan\beta<1$, which, in turn, would imply (thought I am NOT sure about this), that $-\pi/2<\alpha+\beta<\pi/2$, i.e. that we have to stay in the same period of tangent.

However, even if the above were true, I still do not see why we have to stay in the same period of tangent for the formula for $\tan(\alpha+\beta)$ to hold. I would be thankful for a thorough explanation.

答:
A simpler explanation  of why the identity  $$\arctan u+\arctan v=\arctan\dfrac{u+v}{1-uv}$$
is valid if & only if  $uv <1$.

By definition of the arctan function, as both sides have the same tangent, we only need to check under which condition
$$-\frac\pi 2<\arctan u+\arctan v < \frac\pi2.$$
Let $\alpha=\arctan u$, $\beta=\arctan v$. A priori, $\alpha+\beta\in(-\pi,\pi)$, and on this interval,
$$\alpha+\beta \in\Bigl(-\frac\pi 2,\frac\pi 2\Bigr)\iff \cos(\alpha+\beta)>0$$
Now $$\cos(\alpha+\beta)=\cos\alpha\cos\beta-\sin\alpha\sin\beta = \cos\alpha\cos\beta(1-\tan\alpha\tan\beta)\\=\underbrace{\cos\alpha\cos\beta}_{>0}\,(1-uv)$$
has the sign of $1-uv$.


问:
Different definitions for the same function space
today my teacher ask me: but I do not how is possible it

What is the difference or relation between the following two spaces:

$$\{ u \in L^2 \ :\ \exists g\in L^2 \mbox{ such that }\int u\phi' = \int g\phi \mbox{ for all  }  \phi \in C_c^1 \}$$
and
$$\{ u \in L^2 \ :\  \mid \mid u \mid \mid_{L^2}  + \mid \mid u' \mid \mid_{L^2}<\infty \} ?$$

is it true? how is possible?

答:
In the following I'll refer to your first space as X and to the second as Y.

As the question is set in Lebesgue-Spaces $L^2$, $u'$ means the weak derivative of $u$ which is defined by $u'\in L^2$ and
$$ \int u(t)\phi'(t) dt = - \int u'(t)\phi(t)dt$$
for all $\phi\in C_c^1$.
I'll just assume, that you forgot the '-' in your equation.

So let $u$ be in X. Therefore there exists a weak derivative $u'\in L^2$. We get $||u'||_{L^2}<\infty$ and obviously already have $||u||_{L^2}<\infty$ because of $u\in L^2$. Hence $u$ is in Y.

Now let $u$ be in Y. This yields the existence of a weak derivative $u'$ of $u$ which is in $L^2$. In consequence in fulfills the equation and we get $u\in X$.

Consequently, $X=Y$ and the claim is proven.


答:
The identity function restricted to the set of irrationals, $f:\mathbb{Q}^c\to \mathbb{R}$ defined by $x\mapsto x$, is continuous on $\mathbb{Q}^c$ and discontinuous on $\mathbb{Q}$, which is countable. I don't see how this is a counter-example to the statement.

To prove this, let $a\in \mathbb{Q}^c$. For any $\varepsilon >0$, choose $\delta =\varepsilon >0$. For any $x \in \mathbb{Q}^c$ such that $|x-a|<\delta $, we have $$|f(x)-f(a)|=|x-a|<\delta =\varepsilon .$$ Thus, we have that $f$ is continuous at the irrationals, i.e. on its domain. Also see that $f$ has removable discontinuity at the rationals since it is not defined at those points. Note that $\mathbb{Q}$ is countable.

Title: Doubt about discontinuity

Body:
I found this exercise:>Let $f$ be a monotone function, then it can have only a countable number of discontinuities.

I've seen around here a lot of posts about the same but I never get to any conclusion. What happens with Identity function restricted to the set of irrationals? It is monotone, but, isn't it discontinuous at every point of its domain? Which, by the way is uncountable.

I know that if $f$ were defined on an interval, it would be easy to prove, but this is not the case...

Could anyone clarify me please?


问:
How to find that derivative of this function is invertible
The following question was part of my real analysis assignment and I having a hard time solving it.

Let $ f : \mathbb{R}^2 \to \mathbb{R}^2 $ be given by the formula $f(x,y)=( 3x+2y+y^2 +|xy| , 2x+3y +x^2+|xy|) $.

I have proved it to be differentiable at $(0,0)$ but I am unable to think whether Df(0,0) is invertible or not?

Jacobian at $(0,0)$ is $5$ so its invertible by inverse function theorem but the answer is it's not invertible.

what I am doing wrong ?

please help.

答:
Let $f_1(x,y)\equiv (f_1(x),f_2(y))$ for each $(x,y)\in\Bbb R^2$. Then the derivative $f’(x,y)$ of the function $f(x,y)$ is a matrix
$\begin{pmatrix}\frac{\partial{f_1}}{\partial{x}} &  \frac{\partial{f_1}}{\partial{y}}\\
\frac{\partial{f_2}}{\partial{x}} &  \frac{\partial{f_2}}{\partial{y}}\end{pmatrix}$,
and $f’(0,0)=\begin{pmatrix} 3 & 2 \\ 2 & 3\end{pmatrix} $ is invertible.

But the conditions of the inverse function theorem does not applicable to $f$ at $(0,0)$, because there are no neighborhood of $(0,0)$ where $f$ is continuously differentiable.

Nevertheless, the function $f$ is invertible on some open neighborhood of $(0,0)$. Let us show this.

Consider auxiliary functions $f_+$ and $f_-$ from $\mathbb{R}^2$ to $\mathbb{R}^2$ such that for each $(x,y)\in\Bbb R^2$ we have

$$f_+(x,y)=(3x+2y+y^2 +xy , 2x+3y +x^2+xy),$$
$$f_-(x,y)=(3x+2y+y^2 -xy , 2x+3y +x^2-xy).$$

Each of these two functions is continuously differentiable and has non-zero Jacobian at $(0,0)$. So, by the inverse function theorem, there exists $0<r<1$ such that these functions are invertible on a closed square $S=\{(x,y)\in\Bbb R^2:|x|+|y|\le r\}$. In particular, the restictions of these functions onto $S$ are injective.

Lemma. The restriction $f_S$ of $f$ onto $S$ is injective.

Therefore,  by the invariance of domain, the restriction of $f_S$ (and so of $f$) onto the interior of $S$ is a homeomorphic embedding, and so invertible.

Proof of Lemma.  The boundary $B$ of $S$ consists of four segments. We claim that $f|B$ is injective. For this suppose $(x,y)\in B$ and consider the following cases.

I)) $x>0$, $y>0$, $y=-x+r$. Then $f(x,y)=(x-xr+r^2+2r , xr-x+3r) $. Thus $f_1(x,y)+f_2(x,y)=r^2+5r$.

II)) $x<0$, $y>0$, $y=x+r$. Then $f(x,y)=(5x+xr+r^2+2r, 5x-xr +3r) $. Thus $f_1(x,y)+f_2(x,y)=10x+r^2+5r$.

III)) $x<0$, $y<0$, $y=-x-r$. Then $f(x,y)= (x+xr+r^2-2r , -x-xr-3r) $. Thus $f_1(x,y)+f_2(x,y)=r^2-5r$.

IV)) $x>0$, $y<0$, $y=x-r$. Then $f(x,y)=(5x-rx+r^2-2r, 5x+xr-3r) $. Thus $f_1(x,y)+f_2(x,y)=10x+r^2-5r$.

Using the above equations for $f_1(x,y)+f_2(x,y)$ and a condition $0<|x|<r$ it is easy to check that $f|B$ is injective.

Also it is easy to check that $f(B)$ is a quadrilateral with vertices $f(0,r)=r(2+r,3)$, $f(r,0)=r(3,2+r)$, $f(0,-r)=r(-2+r,-3)$, and $f(-r,0)=r(-3,-2+r)$, which is symmetric with respect to a line $x=y$ and a point $f(0,0)=(0,0)$ is contained in the bounded component of $\Bbb R^2\setminus f(B)$.

Let $G$ be a geometric graph with vertices $(0,0)$, $(0,r)$, $(r,0)$, $(0,-r)$, and (eight) edges connecting them induced from $B$ and diagonals of $S$. It is easy to check that $G$ is three-connected.

Let $L$ be any of latin numbers from $I$ to $IV$ and $f_L$ be the restriction of $f$ onto $L$-th (closed) quadrant $Q_L$ of $\Bbb R^2$. Then $f_L$ coincides with the restriction onto $L$ of $f_+$ (if $L=I$ or $L=III$) or of $f_-$ (if $L=II$ or $L=IV$).

Let $B^*$ be the “support” of $G$, that is a union of $B$ with the diagonals. We claim that $f|G$ is injective. Indeed, let $z$ and $z’$ be distinct points of $B^*$. If $z, z’\in B$ then $f(z)\ne f(z’)$, because $f|B$ is injective. If $z\in B^*\setminus B$ then $z$ belongs to one of diagonals of $S$. Then $z\in (Q_I\cup Q_{III})\cap (Q_{II}\cup Q_{IV})$, so $f(z)=f_+(z)=f_-(z)$. If $z’\in Q_I\cup Q_{III}$ then $f(z’)=f_+(z’)\ne f_+(z)=f(z)$. If $z’\in Q_{II}\cup Q_{IV}$ then $f(z’)=f_-(z’)\ne f_-(z)=f(z)$.

Therefore $f$ induces a plane embedding $\hat f$ of $G$.  By Whitney's theorem, all plane embeddings of a three-connected planar graph are equivalent, that is, obtainable from one another by a plane homeomorphism up to the choice of outer face. Since $f(0,0)=(0,0)$ is contained in the bounded component of $\Bbb R^2\setminus f(B)$, both  $G$ and $\hat f(G)$ have as the outer face the unique face with four vertices. Therefore there exists a homeomorphism $h$ of $\Bbb R^2$ such that $hf|B^*$ is an identity map.

Let $L$ be any of Latin numbers from $I$ to $IV$. Since a triangle $\Delta_L=S\cap Q_L$ is compact and the function $hf_L$ is continuous and injective, $hf_L|\Delta_L$ is a homeomorphic embedding.  Let $B_L$ be the boundary of  $\Delta_L$. Since a boundary of a compact subset $K$ of $\Bbb R^2$ consists of points $x$ of $K$ having no neighborhood in $K$ homeomorphic to an open disk, $hf_L(B_L)=B_L$ is the boundary of  $hf_L(\Delta_L)$. The set $\Bbb R^2\setminus B_L$  consists of two connected components  $\Delta_L\setminus B_L$ and $\Bbb R^2\setminus\Delta L$, so a connected set $C=hf_L(\Delta_L\setminus B_L)$ is contained in one of them. It is easy to construct a continuous map $r$ from a set $(\Bbb R^2\setminus\Delta L) \cup B_L$ onto its boundary $B_L$ such that $r|B_L $ is the identity map, that is, $r$ is a retraction. If $C\subset \Bbb R^2\setminus\Delta_L$ then $r|hf_L(\Delta_L)$ is a retraction of the set $ hf_L(\Delta_L)$ onto its boundary $B_L$, which contradicts the drum theorem, since $ hf_L(\Delta_L)$ is homeomorphic to a closed disk. Therefore $C=hf_L(\Delta_L\setminus B_L)\subset \Delta_L\setminus B_L$. Since this holds for each $L$, we have that the map $hf_S$ is injective, and so is $f_S$. $\square$


问:
If in a category $\mathcal{C}$ we have that $\mathrm{Ob}(\mathcal{C})$ is a set, is then $\mathrm{Mor}(\mathcal{C})$ a set?
in the definition of a small category $\mathcal{C}$ we have that both $\mathrm{Ob}(\mathcal{C})$ and  $\mathrm{Mor}(\mathcal{C})$ have to be sets, so my question is if $\mathrm{Ob}(\mathcal{C})$ is a set, isn't $\mathrm{Mor}(\mathcal{C})$ also a set?

I've tried to prove it in ZFC axiomatic but I'm not able to do it.

I think what we essentially need to proof is that
$$A=\left\{ \mathcal{C}(X,Y): X,Y \in \mathrm{Ob}(\mathcal{C}) \right\}$$
is a set, beacuse then we just have that $\mathrm{Mor}(\mathcal{C})=\bigcup A$.

So, can someone proof that fact? And if it is not true, can someone give and example of a category where $\mathrm{Ob}(\mathcal{C})$ is a set, but  $\mathrm{Mor}(\mathcal{C})$ is not?

答: Every monoid $M$ can be regarded as a category with a single element $\star$ and morphisms $\{f_m\colon\star\to\star\mid m\in M\}$ whose compositions beheave as in the monoid $M$, namely $f_{m_1}\circ f_{m_2}=f_{m_1m_2}$, so in order to find a category with set many objects and class many morphisms we only need a class sized monoid. Natural examples of such objects are cardinal/ordinals under addition.


答:
Recall that we call a category $\mathcal{C}$ locally small if for any two objects $X$ and $Y$ in $\mathcal{C}$ we have that $\operatorname{Hom}(X, Y)$ is a set. If $\operatorname{Obj}(\mathcal{C})$ is a set then $\operatorname{Mor}(\mathcal{C})$ is a set if and only if $\mathcal{C}$ is locally small.

From left to to right is immediate: since $\operatorname{Hom}(X, Y) \subseteq \operatorname{Mor}(\mathcal{C})$ for any $X$ and $Y$ in $\mathcal{C}$ we have that $\operatorname{Hom}(X, Y)$ must be a set. The converse is like your idea. We have that $\operatorname{Hom}(X, Y)$ is a set for all objects $X$ and $Y$ in $\mathcal{C}$. Since $\operatorname{Obj}(\mathcal{C})$ is a set, we have that $\operatorname{Obj}(\mathcal{C}) \times \operatorname{Obj}(\mathcal{C})$ is a set. So
$$
\{\operatorname{Hom}(X, Y) : (X, Y) \in \operatorname{Obj}(\mathcal{C}) \times \operatorname{Obj}(\mathcal{C}) \}
$$
is a set, and so is its union, which is exactly $\operatorname{Mor}(\mathcal{C})$.

If we do not require $\mathcal{C}$ to be locally small then it is not hard to come up with a counterexample. We can define a category $\mathcal{C}$ as follows: it has only one object, say $\star$. We define the arrows of $\mathcal{C}$ to be the class of all sets, so this is a proper class. They all have domain and codomain $\star$, since that is the only object. Composing arrows $A$ and $B$ is done by taking their union. It is easy to check that this is associative and the empty set forms the identity arrow.

Title: If in a category $\mathcal{C}$ we have that $\mathrm{Ob}(\mathcal{C})$ is a set, is then $\mathrm{Mor}(\mathcal{C})$ a set?

Body:
in the definition of a small category $\mathcal{C}$ we have that both $\mathrm{Ob}(\mathcal{C})$ and  $\mathrm{Mor}(\mathcal{C})$ have to be sets, so my question is if $\mathrm{Ob}(\mathcal{C})$ is a set, isn't $\mathrm{Mor}(\mathcal{C})$ also a set?

I've tried to prove it in ZFC axiomatic but I'm not able to do it.

I think what we essentially need to proof is that
$$A=\left\{ \mathcal{C}(X,Y): X,Y \in \mathrm{Ob}(\mathcal{C}) \right\}$$
is a set, beacuse then we just have that $\mathrm{Mor}(\mathcal{C})=\bigcup A$.

So, can someone proof that fact? And if it is not true, can someone give and example of a category where $\mathrm{Ob}(\mathcal{C})$ is a set, but  $\mathrm{Mor}(\mathcal{C})$ is not?


问:
How to integrate $\int_0^\infty e^{-ty^2} \sin t dt$
My book suggests that I do some sort of limiting

$\lim_{A \to \infty} \int_0^A e^{-ty^2} \sin t d t$

But I'm not getting anywhere.

答: you can use Euler formula $e^{i\pi}=\cos(x)+i\sin(x)$ </p>


问:
Finding absolute extreme values on a region D
I am unable to understand how to use lagrange multipliers in this question. Can you lead me a way to solve this question?
I've always used lagrange multipliers with circles; however, in this case, it is an ellipsis. I am unsure about how to apply lagrange multipliers on ellipsis.

Question: Function: $f(x,y)=\frac{1}{3}\cdot \:x^3-\frac{1}{4}\cdot \:x+y^2$

Find the absolute extreme values on the region $D$:
$$D=\left\{\left(x,y\right)\in\mathbb{R}^2:\left(x-1\right)^2+4y^2\le 4 \right\}$$

答:
First step: determine the critical points of $f$ in the interior of $D$ and investigate their nature (lokal max, local min, saddle point).

Second step: investigate the function $F$ on $ \partial D.$ The Lagrange function is:

$$L(x,y, \mu)=f(x,y) - \mu( (x-1)^2+4y^2-4).$$


问:
Second degree linear derivate with exponential function, which general solution?
Suppose $y''-4y=xe^{2x}$.

Solution of the homogenous equation is $y_H(x) = C_{1} e^{2x} + C_{2} e^{-2}$, after solving the characteristic equation with the guess $y=e^{rx}$.

Now my instructor insists that the one solution is either

$y(x) = Ax^{2} e^{2x} + Bxe^{2x}$

or

$y(x) = Ax^{2}e^{2x} + Bxe^{2x}+Ce^{2x}$

...and then I get the solution $y(x) = y_{H}(x) +\frac{1}{8}x^{2}e^{2x}-\frac{1}{16}xe^{2x}$

but I cannot understand the choices. Why do just the forms work? I have tried many different forms there and get lost/wasted a lot time -- and my instructor just says that it is a "guess" and smiles. How can I find the one solution without guessing? For one solution, things such as $y(x)=e^{2x}$ or $y(x)=xe^{2x}$ work so I am a bit lost why the more complex form is "the solution" and why it provides all solutions.

For example, the case $y=e^{2x}$. We get $y'=2e^{2x}$ and $y''=4e^{2x}$ so

$y''-4y = 4e^{2x} - 4 e^{2x} = 0$

so $RHS = xe^{2x} = 0$ so $x=0$, a solution?! If not why? Ok it is naive, it does not contain all cases but how can I know that some form contains all?

答:
Another way to do this is using the Exponential Shift Theorem (see e.g. http://www.math.ubc.ca/~israel/m215/coco/coco.html): For any polynomial $P$, constant $k$ and function $u$, $P(D) e^{kx} u = \exp(k x) P(D+k) u$ (where $D$ stands for derivative).  In your case $P(t) = t^2 - 4$, $k=2$,
and $P(t+2) = (t+2)^2 - 4 = t^2 + 4 t$ so if $y = e^{2x} u$
we have $y'' - 4 y = e^{2x} (u'' + 4 u')$.  Now you want $y'' - 4 y = x e^{2x}$
so $u'' + 4 u' = x$.  Writing $v = u'$, we have the first-order linear equation in $v$: $v' + 4 v = x$, which has a solution $v = \frac{x}{4} - \frac{1}{16}$.  An antiderivative of this is $u = \frac{x^2}{8} - \frac{x}{16}$, corresponding to the particular solution $y = e^{2x} \left(\frac{x^2}{8} - \frac{x}{16}\right)$ of your original equation.  </p>


答:
An alternative way is to recognize that

$$\int_0^{\infty} dt \, e^{-b t} = \frac{1}{b}$$

when $\Re{b} \gt 0$.  Then write

$$\int_0^{\infty} dt \, e^{-t y^2} \, \sin{t} = \Im{\left [ \int_0^{\infty} dt \, e^{-(y^2 - i) t} \right]} = \Im{\left[\frac{1}{y^2-i}\right]} = \frac{1}{y^4+1}$$

Title: How to integrate $\int_0^\infty e^{-ty^2} \sin t dt$

Body:
My book suggests that I do some sort of limiting

$\lim_{A \to \infty} \int_0^A e^{-ty^2} \sin t d t$

But I'm not getting anywhere.


问:
Is there a graphical proof that $9n+1$ is triangular if $n$ is?
It is straightforward to prove that if $n$ is a triangular number then $9n+1$ must be.

Is there a systematic decomposition of a triangle made of $9n+1$ dots into nine triangles of $n$ dots plus one left over to illustrate this fact visually?

答:
Visual proof. ${}{}{}{}{}{}{}{}{}{}{}$

Of course it doesn't depend on the base triangle, here is the same with $T_4$:

With regard to the pattern mentionned by Teresa Lisbon in a comment above, here is how $25T_n+3$ shows up:

The pattern may be repeated with a number of "holes" that is itself a triangle number, for instance:

In light red, the repeated pattern.

This pattern yields the general result: $8T_pT_q+T_p+T_q=T_n$ with $n=2pq+p+q$, and of course $T_n=\frac12n(n+1)$. It's not difficult to prove algebraically.


答: I think it may help your curiosity to prove that any monotonic function from any order topologies to another order topology can have only countably many discontinuities. But as you suggest this isn't always true. Consider the long line $\omega_1\times[0,1)$ with the lexigraphical ordering. Define a monotonic map from it to itself as $f(x,y)=(x,\frac{y}{2})$ if $y<\frac{1}{2}$ and $f(x,y)=(x,\frac{y+1}{2})$ otherwise. In this case you can check $f(x,y)$ is monotonic, but has uncountable discontinuities. Now try and find conditions which guarantee the statement to be true, maybe something to do with paracompact, second countable, Lindelof, or seperability.

Title: Doubt about discontinuity


问:
When is an Integer a Rational Number, and are All Ratios Rational, Even $\frac{\sqrt{7}}{2}$?

$$\Bbb{Q} = \left\{\frac ab \mid \text{$a$ and $b$ are integers and $b \ne 0$} \right\}$$

In other words, a rational number is a number that can be written as one integer over another.

For an integer, the denominator is $1$ in that case. For example, $5$ can be written as $\dfrac 51$.

Is $5$ a rational number? Or is $\dfrac 51$  a rational number? I'm not able to figure out what the definition is actually saying. What are the numbers that cannot be written as one integer over another?

Irrational numbers are the numbers that cannot be written as one integer over another. Roots of numbers that are not perfect squares are examples of irrational numbers.

However, what is this then: $\dfrac {\sqrt 7} {2}$?

答:
Any number for which it is possible to express as the ratio or quotient of integers is a rational number. So yes, $5$ is rational, because it is possible to express this as $\frac 51, \frac {10}{2}...$.

$5$ is also an integer. Every integer is a rational number, but not all rational numbers, e.g. $\frac 12$, is an integer. We know $\frac 12$ is rational, because it is the quotient of two integers.

However, $\dfrac {\sqrt 7}{2}$ is not a ratio of integers. It is a ratio of a non-integer, namely $\sqrt 7$, over an integer. So $\dfrac{\sqrt 7}{2}$ is not rational.


答: Another one, based on the fact that centered hexagonal numbers can be expressed as $6T_n+1$:

Title: Is there a graphical proof that $9n+1$ is triangular if $n$ is?

Body:
It is straightforward to prove that if $n$ is a triangular number then $9n+1$ must be.

Is there a systematic decomposition of a triangle made of $9n+1$ dots into nine triangles of $n$ dots plus one left over to illustrate this fact visually?


问:
Showing that if $P,Q \in \mathbb{S^2}$ are equidistant from $A,B,C \in\mathbb{S^2}$, then $P=Q$
We are given that the equidistant set of two points on $\mathbb{S^2}$ is a "line" (great circle) on $\mathbb{S^2}$. Using this fact I must show that:

If $P,Q \in \mathbb{S^2}$ are equidistant from $A,B,C \in\mathbb{S^2}$, the $P=Q$. Deduce from this result that an isometry of $\mathbb{S^2}$ is determined by the images of three points $A,B,C$ not in a "line".

My work:

"the equidistant set of two points on $\mathbb{S^2}$ is a "line" (great circle) on $\mathbb{S^2}$ "

I am able to understand this result. My understanding is that the reason that this is true is because any two points
on the sphere are simply points in $\mathbb{R^3}$ that are a distance of $1$ unit from the origin, and the set of points equidistant to this pair of points forms a plane in $\mathbb{R^3}$ that passes through the origin. In $\mathbb{S^2}$, this is simply the intersection of this plane with $\mathbb{S^2}$, so the set of points equidistant to the given pair is a great circle or "line" on $\mathbb{S^2}$. But how do I use this to answer the above question?

I am having a hard time visualizing this, and I started out by just writing equations of distance between $P$ and $Q$ and $A,B,C$ and trying to equate them but that gets confusing and messy fast and i'm not sure if it's going in the right direction.

Another idea that I had was first considering the set of equidistant points between $A$ and $B$, which would be a great circle, which we will call $I$. Then consider the set of equidistant points between $A$ and $C$, which would be a great circle, which we will call $J$. Finally, we consider the set of equidistant points between $B$ and $C$, which would again be a great circle, which we will call $K$. Then the points at the intersection of the great circles $I,J,K$ would be equidistant to $A,B,C$. Intuitively I think (I don't know this for sure), these three great circles would intersect in two points. One point would be in the 'triangle' (not sure what to call this, but I mean the curved triangle) $ABC$, and one would be outside $ABC$ on the other side of the sphere. But this would mean that that there are two unique points that are equidistant from $A,B,C$, so i'm not sure if this line of thinking is correct either.

As for the second part about the isometries of $\mathbb{S^2}$, I have no idea where to even begin. Any help would be great and I would really appreciate if it is dumbed down a little bit because I am a complete beginner at non-Euclidean geometry. Thank you!

答:
The set of points equidistant from $A$ and $B$ can be visualized as a (tilted) equator with $A$ and $B$ symmetrical like approximately Porto Rico (Northern H.) and Bogota (Southern H.) are with respect to our usual equator. Let us call this circle $M_{AB}$ ($M$ like "mediator").

You have seen where the problem is : the issue is that the intersection of the 3 "mediators" $M_{AB}, M_{BC}, M_{CA}$ is not a single point, equivalent for a spherical triangle to the circumcenter for a plane triangle, but 2 antipodal points.


问:
I have trouble understanding indexed family
Define $f:\!R → \!R$ by $f(x) = cos(2x)$.
a) $f(\{\frac{-\pi}{3}, \frac{\pi}{2}\})$

ANS: $\{-0.5, 1\}$

b) $f([0, \frac{\pi}{4}])$

ANS: $[0, 1]$

c) $f(\!R)$

ANS: $[-1, 1]$

d) $f^{-1}(\{0\})$

ANS: $\{\frac{\pi}{8}\}$

e)$f(\cup_{n \in \!Z} [4n, 4n + 3])$

ANS: ???

I really have trouble with the last question, and my effort:

ANS: $[4, 7]$, but I am pretty sure I am wrong, how to understand the concept of family index?

Thank you for your time!

答:
$$f\left([0,\pi/4]\right)=\left[\frac{1}{\sqrt{2}},1\right]\tag{b}$$

$$f^{-1}(0)=\frac{\pi}{2}+k\pi;\;\forall k\in\mathbb{Z}\tag{d}$$

For question e) remember that $f(x)$ is periodic with period $\pi$

$$f([0,3]=[-1,1]$$

as you can see from the graph below the interval $[-1,1]$ is already covered when $0\le x\le \pi/2$.

BTW, the union in question e) is simply $$\ldots\cup[-8,-5]\cup [-4,-1]\cup[0,3]\cup[4,7]\ldots$$


问:
Show that the sequence $X_n$ converges to a limit $Y$
Let $A$ be a positive definite $n\times n$ matrix. We use the iteration of the mapping $$f(X)=0.5(X^2+B), \ X\in \mathbb{R}^{n\times n}$$ where $B=I-A$.

Show that the sequence $X_0:=0$ (zero matrix), $X_1=f(X_0), X_2=f(f(X_0)), \ldots $ converges to a limit $Y$ if $\|B\|<1$ and show that $(I-Y)(I-Y)=A$.

We want to show that for every $\epsilon>0$ there is a natural number $N$ such that $|X_n-Y|<\epsilon$, right? How could we do that?

答:
We have $$f(X) = \frac 12X^2 + \frac 12 B$$
Let's try to see what initial terms of $(X_n)$ look like (with $X_0=0$):
$$\begin{align}X_1 &= f(0) = \frac 12 0^2 + \frac 12 B = \frac 12 B\\[2mm]
X_2 &= f(f(0)) = \frac12 \bigg(\frac 12 B\bigg)^2 + \frac 12 B = \frac{1}{2^3}B^2 + \frac{1}{2}B \\[2mm]
X_3 &=f(f(f(0)))= \frac 12\bigg( \frac{1}{2^3}B^2 + \frac{1}{2}B \bigg)^2 + \frac 12 B = \frac1{2^7}B^4 + \frac{1}{2^5}B^3 + \frac {1}{2^3}B^2 + \frac{1}{2} B\end{align}$$
We now notice that (which can be proven by induction):
$$X_n = \frac{1}{2} B + \frac {1}{2^3}B^2 + \dots + \frac{1}{2^{2a_n - 1}}B^{a_n} = \sum_{k=1}^{a_n}\frac{1}{2^{2k - 1}}B^{k}$$
where $a_n = a_{n-1}^\text{th}~ \text{triangular number} + 1$, or explicitly
$$\quad a_{n+1} = \frac{a_n(a_n+1)}{2} + 1$$
for all $n \in \mathbb N$ and $a_0 = 1$.

Why exactly this number? This is because the expression $(c_1 + \dots + c_n)^2$ has $\frac{n(n+1)}{2}$ (which is $n^\text{th}$ triangular number) summands, and we add $1$ to  this number since the function $f$ is adding one extra "constant" term.

Now, since $\|B\| < 1$, the following series
$$\lim_{n\to\infty}X_n = \sum_{n=0}^\infty \frac{1}{2^{2k - 1}}B^{k}$$
is convergent.

Finally, let the limit be $Y$. So, we have
$$X_{n+1} = f(X_n)$$
Now "limiting" both sides, we get
$$Y = f(Y) \\
Y = \frac 12 Y^2 + \frac 12 B \\
Y^2 - 2Y + B = 0$$
By the original substitution $B = I-A$:
$$Y^2 - 2Y + I = A \\
(Y - I)^2 = A \\
(I-Y)(I-Y) = A$$
as desired.


问:
Lagrangian and equations of motion on a pseudospherical surface
A particle of mass m oscillates up and down on a smooth surface of revolution parameterized as:

$$(x= a  \sin  \phi \cos t, y= a \sin  \phi  \sin t,z= a(\cos  \phi+ \log \tan  \phi/2 )\;) $$

The  $z$-axis is vertical up. $t$ is polar angle, $\phi$ is angle of tangent to z-axis.

A 3d spiral motion on this pseudospherical horn ensues without friction under the action of gravity. The trajectories look somewhat like this when viewed along $z$ axis:

Depending on the kinetic energy imparted at start  other orbits can be seen.

Please  help find the Lagrangian of the system and its time period of oscillation. Comments for clarification/ solution appreciated.

答:
With

$$
p = (x,y,z)=\left(a \sin (\phi (\tau )) \cos (t(\tau )),a \sin (\phi (\tau )) \sin (t(\tau )),a \left(\cos (\phi (\tau
   ))+\ln \left(\tan \left(\frac{\phi (\tau )}{2}\right)\right)\right)\right)
$$

we have

$$
v = \dot p = \left(a \left(\phi '(\tau ) \cos (\phi (\tau )) \cos (t(\tau ))-\sin (\phi (\tau )) t'(\tau ) \sin (t(\tau
   ))\right),a \left(\phi '(\tau ) \cos (\phi (\tau )) \sin (t(\tau ))+\sin (\phi (\tau )) t'(\tau ) \cos (t(\tau
   ))\right),a \phi '(\tau ) \cos (\phi (\tau )) \cot (\phi (\tau ))\right)
$$

so the Lagrangian reads

$$
L = \frac 12 m v\cdot v - m g z(\phi(\tau)) = \frac{1}{2} a m \left(a \phi '(\tau )^2 \cot ^2(\phi (\tau ))+a \sin ^2(\phi (\tau )) t'(\tau )^2-2 g \left(\cos
   (\phi (\tau ))+\ln \left(\tan \left(\frac{\phi (\tau )}{2}\right)\right)\right)\right)
$$

the movement equations

$$
\left\{
\begin{array}{rcl}
 \phi ''(\tau )& = & \sin (\phi (\tau )) \left(\sin (\phi (\tau )) \tan (\phi (\tau )) t'(\tau
   )^2-\frac{g}{a}\right)+\phi '(\tau )^2 \csc (\phi (\tau )) \sec (\phi (\tau )) \\
 t''(\tau )& = & -2 \phi '(\tau ) \cot (\phi (\tau )) t'(\tau ) \\
\end{array}
\right.
$$

Solving those DEs for $a = 1, m = 1, g = 10, \phi(0) = 0.5, t(0) = 0.5, \dot\phi(0) = 1,\dot t(0)= 2$ we obtain the corresponding orbit in blue, shown in the attached plot

Follows the orbit projected into the $XY$ plane

Follows the plot for $z$ (black), $t(\tau)/10$ (red), and $r = \sqrt{x^2+y^2}$ (blue)


问:
(Borel) Normal number problem

Problem: Given an integer $b \ge 2$, a number $x \in [0,1)$ is (Borel) normal base-b, if its b-ary representation as a sequence $(\sigma_1,\sigma_2,...)$-where $\sigma_i \in \{0,1,...,b-1\}$ are such that $x=\sum_{i \ge 1} \sigma_i b^{-i}$-obeys the following: For eaach $r \ge 1$ and each $\overline{\sigma_1},...,\overline{\sigma_r} \in \{0,1,...,b-1\}, \frac{1}{n}\#\{k=1,...,n:(\sigma_{k+1},...,\sigma_{k+r})=(\overline{\sigma_1},...,\overline{\sigma_r})\}$ converges to $b^{-r}$ as $n \rightarrow \infty$. A number x is absolutely normal if it is normal base b for every integer $b \ge 2$. Prove if $\mathcal{U}=\textbf{Uniform}([0,1])$, then $\mathcal{U}$ is absolutely normal almost surely.

Theorem 1: Almost every real number is absolutely normal.

Proof: A number $x \in (0,1)$ is not absolutely normal if there exist $b \ge 2$, $L \in \mathbf{N}$ and $a_1,...,a_L \in \{0,...,b-1\}$ such that if $x=\sum_{n=1}^{\infty} \frac{c_n}{b^n}$ then $$\textbf{lim inf}_{n \rightarrow \infty} \frac{\# \{n \le N-L |c_{n+i}=a_i,i=1,...,L\}}{N}<\frac{1}{b^L}$$. Then there exists $s \in \{0,...,L-1\}$ such that $$\textbf{lim inf}_{n \rightarrow \infty} \frac{\# \{n \le N-L, n \equiv s (\textbf{mod L}) |c_{n+i}=a_i,i=1,...,L\}}{N}<\frac{1}{Lb^L}$$. Hence for some rational $\beta<\frac{1}{Lb^L}$, $$\textbf{lim inf}_{n \rightarrow \infty} \frac{\# \{n \le N-L, n \equiv s (\textbf{mod L}) |c_{n+i}=a_i,i=1,...,L\}}{N} \le \beta$$. Denote by $R_{b,La,s,\beta}$ the set of all $x=\sum_{n=1}^{\infty} \frac{c_n}{b^n}$ satisfying $$\textbf{lim inf}_{n \rightarrow \infty} \frac{\# \{n \le N-L, n \equiv s (\textbf{mod L}) |c_{n+i}=a_i,i=1,...,L\}}{N} \le \beta$$
. The result of the prvious paragraph is that the set of not absolutely normal number is a subset of $$\bigcup_{b=2}^{\infty} \bigcup_{L=1}^{\infty} \bigcup_{a_1,...,a_L}^{b-1} \bigcup_{s=0}^{L-1} \bigcup_{\beta \in (0, \frac{1}{Lb^L}\cap \mathcal{Q}} R_{b,L,a,s,\beta}$$.
It is sufficient to prove every set $R_{b,L,a,s,\beta}$ has zero measure. Then the set of not absolutely normal numbers is a subset of a countable union of null sets, hence it is a null set.

Let $b \ge 2, L \in \mathcal{N}, a_1,...,a_L \in \{0,...,b-1\}, s \in \{0,...,L-1\}$ and $\beta \in (0,\frac{1}{Lb^L}$. Put $A=a_1b^{L-1}+a_2b^{L-2}+...+a_L$. Let $$x=\sum_{n=1}^{\infty} \frac{c_n}{b^n}=\sum_{n=1}^s \frac{d_n}{b^n}+\frac{1}{b^s}\sum_{n=1}^{\infty} \frac{d_{s+n}}{b^{Ln}} \in R_{b,L,a,s\beta}$$. Then obviously $$\# \{n \le N-L,n \equiv s(\textbf{mod L})|c_{n+i}=a_i,i=1,...,L\}=\# \{s<n \le [\frac{N-s}{L}]|d_n=A\}$$. Hence $$\textbf{lim inf}_{M \rightarrow \infty} \frac{\# \{s<n \le M| d_m=A}{M}=\textbf{lim inf}_{N \rightarrow \infty} \frac{\# \{s<n \le [\frac{N-s}{L}]|d_n=A\}}{[\frac{N-s}{L}]} \le L\beta$$. From this we obtain $R_{b,L,a,s\beta} \subseteq P$, where $$P=\{x=\sum_{n=1}^s \frac{d_n}{b^n}+\frac{1}{b^s}\sum_{n=1}^{\infty} \frac{d_{s+n}}{b^{Ln}}|\textbf{lim inf}_{N \rightarrow \infty} \frac{\# \{s<n \le N |d_n=A\}}{N} \le L \beta\}$$. Then it is sufficient to prove that the set P has zero measure.

The complete version of the proof is found here.
An elementary proof that almost all real numbers are normal

My question is how am I supposed to mimic the proof to solve the problem? Thanks a lot!

答:
I found [5] in the article you mentioned particularly useful. We only need to prove a simpler version of theorem 8.1 (and corollary 8.2) in the book. So here is my answer:

Suppose $X\sim U(0,1)$. Given $b\geq 2$, and fixed $r\geq 1$. Consider the $b-$ary expansion of $x = \sum_{i=1}^\infty \sigma_ibi^{-1}$. We are interested in the probability of a given string: $(\bar{a}_1,\bar{a}_2,...\bar{a}_r)$, where $a_i = 0,1,...b-1$.

If the target string matches $x$ at index $m$, we can rewrite $x$ by replacing the digits from index $m$ to $m+r-1$:
$$
\begin{split}
    x &= \sum^\infty_1 \frac{\sigma_i}{b_i}\\
    &= \sum^{m-1}_{1} \frac{\sigma_i}{b^i} + \sum^{m+r-1}_{m} \frac{a_i}{b^i} + \sum_{m+r}^\infty \frac{\sigma_i}{b^i}
\end{split}
$$
Multiply both sides by $b^{m-1}$:
$$
b^{m-1}x = (b^{m-1}\sum^{m-1}_{1} \frac{\sigma_i}{b^i}) + \frac{a_mb^{r-1}+a_{m+1}b^{r-2}+ ...a_{m+r-1}}{b^r} + \sum_{r+1}^\infty \frac{\sigma_{m+i-1}}{b^i}
$$
The fractional part $\{b^{m-1}x\}$ doesn't contain $(b^{m-1}\sum^{m-1}_{1} \frac{\sigma_i}{b^i})$, and we have $\sum_{r+1}^\infty \frac{\sigma_{m+i-1}}{b^i}< 1b^{-r}$ since they are the trailing digits of this expansion. Therefore,
$$
\{b^{m-1}x\}\in [\frac{a_mb^{r-1}+a_{m+1}b^{r-2}+ ...a_{m+r-1}}{b^r}, \frac{a_mb^{r-1}+a_{m+1}b^{r-2}+ ...a_{m+r-1} + 1}{b^r}) = I_r
$$
Whenever there is a matching sequence, we have $\{b^{m-1}x\}\in I_r$.

Since $X\sim U(0,1)$, we have that $\{b^{m-1}X\}\sim U(0,1)$ because any fractional part is uniformly distributed across each interval $[n,n+1)$, where $n = 0,1,...b^{m-1}-1$. Since $b^{m-1}X$ is uniform, each interval has probability $1/b^{m-1}$ and therefore the marginal distribution of $\{b^{m-1}X\}$ is uniform $(0,1)$.

Given the first $N$ digits of the expansion of $x$, we need to check $N-r+1$ possible starting point of the desired sequence $b_1b_2...b_r$ and this is equivalent to $\sum^{N-r+1}_1X1_{I_r}$, where $X\sim U(0,1)$. Hence by SLLN,
$$
\lim_n \frac{\{\# k = 1,2,...n: (\sigma_k...\sigma_{k+r}) = (b_1,b_2...b_r)\}}{n} = \lim \frac{\sum^{n-r+1}_1X1_{I_r}}{n-r+1}\frac{n-r+1}{n}\xrightarrow[]{}\frac{1}{b^r}
$$
almost surely.

This is true for any $r\geq 1$, so $X$ is normal almost surely. We can do the same for every $b\geq 2$, and then take the countable intersection of the above events so that $X$ is normal with respect to every base $b\geq 2$ with probability 1. We are done.

See theorem 8.1 in https://web.maths.unsw.edu.au/~josefdick/preprints/KuipersNied_book.pdf

Title: (Borel) Normal number problem


答:
$5$ is an integer and a rational number. It is convention that we omit the denominator when it is $1$.

$\frac{\sqrt{7}}{2}$ is not a rational number as it does not conform to the definition i.e., the numerator is not an integer.

Title: When is an Integer a Rational Number, and are All Ratios Rational, Even $\frac{\sqrt{7}}{2}$?

Body:
$$\Bbb{Q} = \left\{\frac ab \mid \text{$a$ and $b$ are integers and $b \ne 0$} \right\}$$

In other words, a rational number is a number that can be written as one integer over another.

For an integer, the denominator is $1$ in that case. For example, $5$ can be written as $\dfrac 51$.

Is $5$ a rational number? Or is $\dfrac 51$  a rational number? I'm not able to figure out what the definition is actually saying. What are the numbers that cannot be written as one integer over another?

Irrational numbers are the numbers that cannot be written as one integer over another. Roots of numbers that are not perfect squares are examples of irrational numbers.

However, what is this then: $\dfrac {\sqrt 7} {2}$?


问:
Why is the space of bounded functions complete?
Let $(X,d_X),(Y, d_Y)$ be two metric spaces. We define:

\begin{equation}
B(X,Y)=\{f:X\rightarrow Y, f\text{ bounded}\}
\end{equation}
Introducing the metric (of uniform convergence):
\begin{equation}
d(f,g)=\underset{x\in X}{\sup}d_Y(f(x),g(x)) \qquad\forall f,g\in B(X,Y)
\end{equation}

Now, if $Y$ is complete, you can prove that $B(X,Y)$ is complete. I found this proof on a book, but there's a part I don't understand.

Proof

Let $\{f_n\}_{n\geqslant1}$ be a Cauchy sequence in $B(X,Y)$ and $x\in X$. The sequence $\{f_n(x)\}$ is a sequence in $Y$.
$\{f_n(x)\}$ is trivially (so I'll be omitting some steps about this) a Cauchy sequence by definition of $d$:
\begin{equation}
d_Y(f_n(x),f_m(x))\leqslant\underset{x\in X}{\sup}d_Y(f_n(x),f_m(x))=d(f_n,f_m)
\end{equation}

Since $Y$ is complete, $\{f_n(x)\}$ is convergent
\begin{equation}
\lim_{n\to+\infty} f_n(x)=f(x) \tag{1}
\end{equation}

Let's write down the definition of Cauchy sequence for $\{f_n\}$:
\begin{equation}
\forall\epsilon>0 \quad \exists\nu\in\mathbb{N}: \forall n,m>\nu\quad n,m\in\mathbb{N}\implies d(f_n,f_m)<\epsilon
\end{equation}
Thus, we have:
\begin{equation}
\forall x\in X\quad\forall\epsilon>0 \quad \exists\nu\in\mathbb{N}: \forall n,m>\nu\quad n,m\in\mathbb{N}\implies d_Y(f_n(x),f_m(x))<\epsilon \tag{2}
\end{equation}

Well, this is the critical point. The book now takes the limit as $m\to+\infty$ and using $(1)$:

\begin{equation}
d_Y(f_n(x), f(x))\leqslant\epsilon \quad \forall x\in X \quad \forall n>\nu \tag{3}
\end{equation}

Since this is valid $\forall x\in X$, it follows that $d(f_n, f)\leqslant\epsilon$ and the theorem is proved.

As I said, I don't understand the step between $(2)$ and $(3)$. It looks like we are assuming $d$ is continuos when we take the limit. My guess is that he used the reverse triangle inequality and then took the limit but I'm not sure about that. So how do we get from $(2)$ to $(3)$?

答:
Yes, they used that $d$ is continuous.
In a metric space, the metric itself is always continuous.

This fact can be shown (as you suspected) with the reverse triangle inequality.


答: As $5$ can be written as $\frac 51$, yes, $5$ is a rational number. All integers are also rational. The fraction $\frac{\sqrt 7}2$ does not denote a rational number because the numerator $\sqrt 7$ is not an integer.


问:
Can $\mathbb{Q}(x^3,y^3,x+y)$ be generated by only two elements?
Let $\mathbb{Q}(x,y)$ be the field of rational functions in the variables $x, y$ with rational coefficients, and consider its subfield $K=\mathbb{Q}(x^3,y^3,x+y)$. Do there exist $p, q \in K$ such that $K=\mathbb{Q}(p,q)$? If the answer is positive, can you explicitly find such two elements?

This question was asked by reuns in his remarkable answer to my previous post Subextensions of Finitely Generated Fields. If I should follow my intuition, I would bet that the answer is negative, but I have no idea of a possible proof. Any help is welcome.

Body:
We have $K={\bf Q}(x,y)$.

Because $(x+y)^3-x^3-y^3=3xy(x+y)$ so that $xy\in K$ and $x=(x^3+xy(x+y))/((x+y)^2-xy)$.


问:
On the proof that continuous images map compact subsets to compact subsets
I'm trying to understand the proof of the fact that if I have a continuous map between two topological spaces $X$ and $Y$, then the image of a compact subset $A \subseteq X$ is compact in $Y$. The proof for this statement is already provided at this link Proving continuous image of compact sets are compact.

Now what I don't understand is why if a set $\{V_i \} $ is an open cover for $f(A) \subseteq Y$, then $A \ \cap \ f^{-1}(V_i)$ forms an open cover for $A$. I'm aware that the $\{f^{-1}(V_i)\}$ must be open because $f$ is continuous but don't see the connection. I mean what guarantees that these open sets cover the entire subset $A$.

答: If $a \in A$, $f(a) \in f[A]$ so is covered by some $V_j$, so by definition, $a \in A \cap f^{-1}[V_j]$. This holds for all $a \in A$, so $A$ is covered by such open sets.


问:
Type Theory Rules For The Empty Type
I would like some help choosing the rules for the empty type. I am trying to setup a typed lambda calculus with sums like in

Extensional Normalisation and Type-Directed Partial Evaluation for Typed Lambda Calculus with Sums

Except, I am being explicit about type formation (that is, I am keeping track of a context, and applying rules to form types, in a similar way to page 554 of the homotopy type theory book). For example, here is one of my introduction rules for sum types:

$$\begin{array}{c}
\Gamma\vdash x:A\hspace{1em}\Gamma\vdash B:\text{type}\\
\overline{\Gamma\vdash\text{inl}_{A+B}(x):A+B}
\end{array}$$

My questions are (1) whether the following is appropriate for being an elimination rule for the empty type:

$$\begin{array}{c}
\underline{\Gamma\vdash A:\text{type}\hspace{1em}\Gamma\vdash e:0}\\
\Gamma\vdash\text{abort}_{A}(e):A
\end{array}$$

And (2) whether the following is appropriate for being a computation rule for the empty type:

$$\begin{array}{c}
\Gamma,e:0\vdash x:A\\
\overline{\Gamma,e:0\vdash\text{abort}_{A}(e)=x:A}
\end{array}$$

And if not, then what rules should I use ? I do not think my situation is helped by the fact that I do not understand the equational rule for the empty type in the above reference.

答:
Your elimination rule is correct.
In a simple type theory (i.e. without dependent types), you could write the computation rule for the empty type as the following (note that this matches the rule in the paper you link, though the notation is a little different, and I've been explicit about the type $A$).
$$\frac{\Gamma \vdash e : 0 \qquad \vdash A\ \mathrm{type} \qquad \Gamma \vdash a : A}{\Gamma \vdash \mathrm{abort}_A(e) \equiv a : A}$$
You will usually not have contexts of the form $\Gamma, x : A$ (or $\Gamma, e : 0$) in a simple type theory unless you're describing a variable-binding operator (like $\lambda$-abstraction). In the HoTT book, there are more rules of this form because they have dependent types (where you have to take more care about variable binding).

Title: Type Theory Rules For The Empty Type


问:
Solve this logic problem, or justify why it can't be solved?
Five sticks (S1, S2, S3, S4 and S5) are placed next to each other, in a circular way so that S1 is next to S2, S2 is next to S3 and so on, and S5 is next to S1.

The sticks need to be painted black or white, such that each stick can be only painted one colour, and if two sticks are painted the same colour, they must be next to each other.

Solve this problem using propositional logic, or prove that it cannot be solved.

Here is my attempt:

These are the formulas I have come up with:

```<code>S1B → S2B (If S1 is painted black, S2 is also painted black)
S1W → S2W (If S1 is painted white, S2 is also painted white)
S2B → S3B
S2W → S3W
S3B → S4B
S3W → S4W
S4B → S5B
S4w → S5w
S5B → S1B
S5W → S1W
</code>```

I am not sure how to proceed from here. I think I need to combine these formulas into a single formula and show that it is a tautology (or not), but I seem to be stuck here.

Please help.

答:
Imagine that these $5$ letters wrap around like a circle as per your problem. Consider $3,4,5$ black sticks.

If there are $5$ black sticks $BBBBB$, clearly if you put these in a circle there are two that are the same color but they are not next to each other.

Same for the case of $4$ $B$'s, $BBBBW$. Just consider the two $B$'s separated by the $W$.

For the case of $3$ $B$'s, either the $B$'s are together like this: $BBBWW$, or the $W$'s are separated like this: $BBWBW$.

In the first case, clearly two $B$'s are separated by the middle $B$, so it does not satisfy the second condition. In the second case, the $W$'s are separated by a $B$ and the same issue occurs.

It has been shown that this does not work for the cases for $3,4,5$ black sticks. If one considers the cases for $0,1,2$ black sticks, this is the same as the cases for $3,4,5$ white sticks. As this argument is exactly the same as for $3,4,5$ black sticks, there is no such configuration that satisfies the required conditions.


问:
Show that these interpretations describe the same set
Let $f: A \rightarrow B$ be bijective with inverse function $f^{-1}: B \rightarrow A$, and let $B' \subset B$.

Then the symbol $f^{-1}(B')$ has two interpretations.

The preimage of $B'$ under $f$.

Image of $B'$ under $f^{-1}$.

Show that both interpretations are the same set.

From the first interpretation we get the set $f^{-1}(B')=\{x\in A \mid f(x)\in B'\}$.

From the second interpretation we get the set $f^{-1}(B')=\{f^{-1}(y) \mid y\in B'\}$.

Can we just say that these sets are equal? Or do we have to show also something?

答:
Interpretation 1 (The inverse image of $B'$ under $f$): Define $S_1$ in such a way
$$
\forall x\bigg(\big(x \in A \rightarrow (f(x) \in B' \equiv x \in S_1)\big) \wedge \big(x \notin A \rightarrow x \notin S_1\big)\bigg)
$$
Interpretation 2 (The image of $B'$ under $f^{-1}$): Define $S_2$ in such a way
$$
\forall x\bigg(x \in S_2 \equiv \exists y\big(y \in B' \wedge f^{-1}(y)=x\big)\bigg)
$$

Now we show that $S_1=S_2$.

On one hand, if $x \in S_1$, then $f(x) \in B'$. $f^{-1}(f(x)) = x \rightarrow x \in S_2$.

On the other hand, conversely, if $x \in S_2$, then $\exists y(y \in B' \wedge f^{-1}(y)=x)$. $f(x)=f(f^{-1}(y))=y \in B' \rightarrow x \in S_1$.

Therefore, $S_1=S_2$, as desired. $\square$


问:
Group of order $15$ has element of order $5$
I need to show that group of order $15$ has element of order $5$. I need to do that without using Sylow theorems and their consequences and Cauchy theorem.
Basically i need to show that Group of order $15$ can't be consisted of $e$ and elements of order $3$.

答:
We only need to appeal to Lagrange's Theorem and the Class Equation.

Assume that $G$ only has elements of order $3$. Note that this implies a non-trivial proper subgroup must have order $3$: if there would be such a subgroup $H$ of $G$ order $5$, then an element $x$ inside this subgroup must have order $3$, violating the fact that the order of $x$ must divide $|H|$.

Let $x \in G-Z(G)$, then the cardinality of the conjugacy class $\#Cl_G(x)=|G:C_G(x)|=5$ since by assumption $|C_G(x)|=3$. Hence the Class Equation yields $15=|Z(G)|+5k$ $(*)$, for some non-negative integer $k$.

If $k=0$, then $G$ would be abelian. Pick $x,y \in G-\{1\}$, $x \neq y$, then ($x$ and $y$ commute) the subgroup $\{1,x,x^2, y, y^2, xy, xy^2, x^2y, x^2y^2\}$ would have order $9$, not dividing $15$.

Hence $k \gt 0$ and $G$ is not abelian and so $|Z(G)|=1 $ or $3$, but that contradicts equation $(*)$, since $12$ nor $14$ are divisible by $5$.


答: A and B have to be integers. Square root 7 is not an integer. It is irrational so it does not qualify as being a rational number. Yes 5/1 is rational. It perfectly satisfies the definition. </p>


问:
Inequality : definition interval
We have the following parameters :

$y \in [0,1]$

$z \in [0,1]$

and I have now to proof the following thing:

$(y+z-2yz) \in [0,1]$

$0 \leq y \leq 1$

$0 \leq z \leq 1$

Therefore  $\Rightarrow 0 \leq y + z \leq 2 \tag{1}$

Then, $0 \leq yz \leq 1$ $\Rightarrow 0 \leq 2yz \leq 2 $ $\Rightarrow -2 \leq - 2yz \leq 0 \tag{2}$

But $(1) + (2)$ :

$-2 \leq y + z - 2yz \leq 2 $
and not $0 \leq y + z - 2yz \leq 1 $.

答:
Your original method is correct, but is crude because when $y$ increases for example, the first inequality increases but the second decreases.

This solution uses differentiation to find the functions minima and maxima.

Let $y+z=k$, then the function becomes:

$$y+(k-y)-2y(k-y)$$

$$=k-2ky+2y^2$$

Treat $k$ as a constant, and differentiate wrt $y$:

$$-2k+4y$$

This is zero when $y=\frac{k}{2}$, and this is a minimum from the second differential.

Similarly for $z$, and so the line $y=z$ contains the minimum value for the function, which is $0$.

As the original function is a quadratic in $y$, the maximum is obtained on the border of the domain, and on the border the function is linear, hence the maximum value is $1$.

Title: Inequality : definition interval


问:
dimension of eigenvalue 1 subspace of a symmetric zero-diagonal matrix
I am working over the finite field of two elements $\mathbb{F}_2$. Let $A\in \mathbb{F}_2^{n\times n}$ be a symmetric zero-diagonal matrix. I am trying to understand the eigenvalue 1 subspace of such matrices. In particular, what is the maximal dimension of the eigenvalue 1 subspace of such matrices?  It can never be $n$, as that would imply that $A$ is the identity matrix. But how much smaller can it be? Thanks.

答: $A$ can have a unit eigenvalue of geometric multiplicity $n-1$. Let $\{e_1,e_2,\ldots,e_n\}$ be the standard basis of $\mathbb F_2^{n\times n}$. For $i=1,2,\ldots,n-1$, define $v_i=e_i+e_{i+1}$. These $v_i$s are linearly independent. When $A$ is the hollow matrix whose off-diagonal entries are all equal to $1$, we have $Av_i=v_i$ for each $i$.


问:
Prove that $e^n\bmod 1$ is dense in $[0,1]$
I just noticed that I have left unanswered one part of an old multi-part question and so decided to re-ask it separately:

Consider the sequence $e^n\bmod 1$, $n\in\Bbb N$. Show that it is dense in $[0,1]$.

This apparently does require specific (approximation?) properties of $e$, as for example replacing $e$ with any integer leads to a non-dense sequence. On the other hand, for every sequence of numbers $a_n\in(0,1)$, it is not hard to find $\alpha$ such that $|\alpha^{2^n}\bmod 1- a_n|<\frac1n$ for all $n$, or $\beta$ such that $|\beta^n\bmod 1-a_n|<\frac1{1000}$. Hence there exist (irrational) bases that lead to a dense sequence and others that lead to a non-dense sequence.
Other than that I'm a bit at a dead end.

答:
I was asked to turn a comment into an answer, even though it was mainly a quote, not any work of my own. It's well known that the fractional parts $\{\theta^n\}$ are not just dense, but uniformly distributed for almost all $\theta$. The irony is, that for any individual $\theta$, we don't know nearly as much. Let's quote http://www.kurims.kyoto-u.ac.jp/~kenkyubu/bessatsu/open/B34/pdf/B34_009.pdf

For instance, we cannot disprove that $\displaystyle \lim\{e^{n}\}=0$,where
$\{x\}$ is the fractional part of a real number $x$. In the case where
$\alpha$ is a transcendental number, it is generally difficult to
prove that the sequence $\{\alpha^{n}\}(n=0,1, \ldots)$ has two
distinct limit points.

So there's little hope concerning transcendental numbers like $e$, and the results for algebraic $\theta$ aren't exactly mind-boggling, either. We can but hope that there will be some progress, soon.


答:
The statement is false.

Let $a>0$ and define $f : [-a,a] \rightarrow \mathbb R : x \mapsto (a^2 - x^2) = (a-x)(a+x).$

Then clearly $f \in C^2[-a,a]$ with $f(a) = f(-a) = 0$.

Since $f'(x) = -2x$ and $f''(x) = -2$ we have

$$ \max_{x \in [-a,a]} \vert f'(x)\vert = \max_{x \in [-a,a]} \vert -2x \vert = 2a$$

and $$ \max \vert f''(x)\vert = \max \vert -2 \vert = 2. $$

But then we cannot have
$$ 2a = \max \vert f' \vert \leq \frac{(a - (-a))^2}{ 2} \max \vert f''\vert  = \frac{(2a)^2}{2} 2 = (2a)^2$$

for $0< \frac{1}{2}.$

Title: Show that $\max_{x \in [a, b]} |f'(x)| \leq \frac{(b-a)^2}{2} \max_{x \in [a, b]}|f''(x)|$.

Body:
Suppose $f(x) \in C^2([a, b])$ with $f(a) = f(b) = 0$. Show that $$\max_{x \in [a, b]} |f'(x)| \leq \frac{(b-a)^2}{2} \max_{x \in [a, b]}|f''(x)|.$$

My work: Suppose that $f'(x_0) = \max_{x \in [a, b]} |f'(x)|$. One can expand $f(x)$ at $x_0$ as follows:
$$f(a) - f(x_0) = f'(x_0) (a - x_0) + \frac{1}{2}f''(\xi_1) (a-x_0)^2,~\xi_1\in[a, x_0]$$
$$f(b) - f(x_0) = f'(x_0) (b - x_0) + \frac{1}{2}f''(\xi_2) (b-x_0)^2,~\xi_2\in[x_0, b]$$
The difference of the two equations leads to
$$f'(x_0) (b-a) = \frac{1}{2}\Big[f''(\xi_1)(a-x_0)^2 - f''(\xi_2)(b-x_0)^2\Big].$$
I have no idea how to reach the term $\max |f''(x)|$.

Body:
The statement is false.

Let $a>0$ and define $f : [-a,a] \rightarrow \mathbb R : x \mapsto (a^2 - x^2) = (a-x)(a+x).$

Then clearly $f \in C^2[-a,a]$ with $f(a) = f(-a) = 0$.

Since $f'(x) = -2x$ and $f''(x) = -2$ we have

$$ \max_{x \in [-a,a]} \vert f'(x)\vert = \max_{x \in [-a,a]} \vert -2x \vert = 2a$$

and $$ \max \vert f''(x)\vert = \max \vert -2 \vert = 2. $$

But then we cannot have
$$ 2a = \max \vert f' \vert \leq \frac{(a - (-a))^2}{ 2} \max \vert f''\vert  = \frac{(2a)^2}{2} 2 = (2a)^2$$

for $0<a < \frac{1}{2}.$


问:
What is the formal defintion of $\sigma(X_1,X_2,\ldots)$ for a sequence of random variables $X_1,X_2,\ldots$?
In probability theory one often seeks to construct the $\sigma$-field $ \sigma(X_{1},X_{2},\ldots) $ for a sequence of R.V.'s $\{X_n\}_{n\in \mathbb{N}}$ (Assumption: $X_n:(\Omega, \mathcal{F}) \rightarrow (\Omega',\mathcal{F'}) )$.

I realized that I'm still not 100% sure what $\sigma(X_1,X_{2},\ldots)$ means. My recollection is the following definition: $\sigma(X_1,X_{2},\ldots) = \sigma\left\{ \bigcap_{n=1}^{\infty} X_n^{-1}(A_n')  \mid A_1',A_2',\ldots \in \mathcal{F}'  \right\}$. Is this correct? If not, what is the correct formal definition?

答: The definition of $\sigma(X_1,X_2,..)$ is it is the smallest sigma algebra which makes each $X_i$ measurable. It is the intersection of all sigma algebras which make each $X_i$ measurable. It is equal to the smallest sigma algebra which contains sets of the form $X_1^{-1}(A_1)\cap X_2^{-1}(A_2)\cap...\cap X_n^{-1}(A_n)$ where $n$ is a positive integer and $A_i$'s are in $\mathcal  F'$. It is possible replace the finite intersections here by infinite intersections $X_1^{-1}(A_1)\cap X_2^{-1}(A_2)\cap...$ so your description is also correct.


问:
Proof of specific $n\times n$ matrix determinant result
I need to prove that for $n \times n$ matrix it holds that $D_n = n$, where
$$
D_n =
\begin{vmatrix}
1       & -1    & -1    & \dots     & -1    \\
1       &  1    &       &           &       \\
1       &       &  1    &           &       \\
\vdots  &       &       & \ddots    &       \\
1       &       &       &           & 1
\end{vmatrix}
$$
and all blank spaces are $0$.

I have calculated $D_n$ for $n = 1, 2, 3, 4$ using the method of minors and cofactors in order to see whether $D_n = n$ is true (it is) and I noticed that determinants of minors, as well as cofactors, can be either $1$ or $-1$.

Furthermore, using the method I mentioned, we have a sign pattern $+-+-+\dots$ for next cofactors. We always have then
$$\text{sign} \cdot \text{cofactor} \cdot \text{determinant of minor} = 1$$
which means that for $n\times n$ matrix and $(n-1) \times (n-1)$ minors, we have $D_n = \underbrace{1+\dots + 1}_{n} = n$, which is what I need to prove.

There are two problems with my (partial) solution: it is based on the observation for small $n$ values, which might not be true for all $n \in \mathbb{N}$, and even if it was, the solution/proof is not formal. How could I generalize it for all $n$ and make it formally?

答:
Just add the 2nd till the nth row to the first one and develop along the first row.
So, you get

$$D_n =
\begin{vmatrix}
n       &  0   & 0    & \dots     & 0    \\
1       &  1    &       &           &       \\
1       &       &  1    &           &       \\
\vdots  &       &       & \ddots    &       \\
1       &       &       &           & 1
\end{vmatrix} = n$$


答:
Let me present a correct proof that you would never love, which is adapted from A Proof of Cauchy's theorem.

Assume $G$ is a group of order $15$ with unit $e$. We consider a set $X = \{(x_1,x_2,x_3,x_4,x_5)|\ x_i \in G,\ x_0x_1x_2x_3x_4x_5=e\}$.
Note that the number of elements in $X$, which we denote as $\#X$, is $15^4$, since the first four components determine an element in $X$. Now consider $C_5 = $, the cyclic group of order $5$ with a fixed generator $a$, acts on $X$ as cycles by the following rule: $a$ carries $(x_1,x_2,x_3,x_4,x_5)$ to $(x_2,x_3,x_4,x_5,x_1)$.

Observation 1. According to Orbit-stabilizer theorem, the orbits under the action of $C_5$ should be of length either $1$ or $5$.

Observation 2. Orbits of length $1$ is of the form $\{(x,x,x,x,x)\}$, where $x^5 =e$. In other words, they correspond 1-1 exactly to elements of order $5$ in $G$.

Now by the obvious relation
$$|X| = \#\{\mathrm{orbits\ of\ length}\ 1 \} + \#\{\mathrm{orbits\ of\ length}\ 5 \} * 5,$$ we conclude that $5$ divides $\#\{\mathrm{orbits\ of\ length}\ 1 \}$. Finally, since the unit $e$ is an element of order $5$, i.e. $\{(e,e,e,e,e)\}$ is such an orbit of length $1$, there must be some other length-$1$ orbit, i.e. nontrivial element of order $5$.

Title: Group of order $15$ has element of order $5$

Body:
I need to show that group of order $15$ has element of order $5$. I need to do that without using Sylow theorems and their consequences and Cauchy theorem.
Basically i need to show that Group of order $15$ can't be consisted of $e$ and elements of order $3$.

Body:
Let me present a correct proof that you would never love, which is adapted from A Proof of Cauchy's theorem.

Assume $G$ is a group of order $15$ with unit $e$. We consider a set $X = \{(x_1,x_2,x_3,x_4,x_5)|\ x_i \in G,\ x_0x_1x_2x_3x_4x_5=e\}$.
Note that the number of elements in $X$, which we denote as $\#X$, is $15^4$, since the first four components determine an element in $X$. Now consider $C_5 = <a>$, the cyclic group of order $5$ with a fixed generator $a$, acts on $X$ as cycles by the following rule: $a$ carries $(x_1,x_2,x_3,x_4,x_5)$ to $(x_2,x_3,x_4,x_5,x_1)$.

Observation 1. According to Orbit-stabilizer theorem, the orbits under the action of $C_5$ should be of length either $1$ or $5$.

Observation 2. Orbits of length $1$ is of the form $\{(x,x,x,x,x)\}$, where $x^5 =e$. In other words, they correspond 1-1 exactly to elements of order $5$ in $G$.

Now by the obvious relation
$$|X| = \#\{\mathrm{orbits\ of\ length}\ 1 \} + \#\{\mathrm{orbits\ of\ length}\ 5 \} * 5,$$ we conclude that $5$ divides $\#\{\mathrm{orbits\ of\ length}\ 1 \}$. Finally, since the unit $e$ is an element of order $5$, i.e. $\{(e,e,e,e,e)\}$ is such an orbit of length $1$, there must be some other length-$1$ orbit, i.e. nontrivial element of order $5$.


问:
Area of image under Gauss map
I am trying to solve the following question

Let $\mathbf r(u,v)$ be a parametrised smooth surface in $\Bbb R^3$ with $(u,v) \in U$, a connected open subset of $\Bbb R^2$. Let $\mathbf n: U \to S^2$ be the Gauss map that assigns to each point $\mathbf r(u,v)$ of the surface its unit normal. Suppose that $\mathbf n$ is bijective onto its image and that the Gaussian curvature K is nowhere zero in U. Show that the area of $\mathbf n(U)$ is equal to the absolute value of $\int K dA$

I am not really sure where to start, so any hint would be appreciated.

答:
In case you want an explicit computation.
Recall $$dA=\mid r_u \wedge r_v\mid dudv= ((r_u \wedge r_v) \cdot(r_u \wedge r_v))^{1/2}dudv=\sqrt{EG-F^2}dudv$$ by the scalar quadruple product. Then:
$$Area(n(U))=\int\int \mid n_{u} \wedge n_{v}\mid dudv$$$$=\int\int \mid\frac{LN-M^2}{ \sqrt{EG-F^2}} n\mid du dv$$$$=\int\int \sqrt{EG-F^2}\mid K \mid du dv=\int\int\mid K\mid dA$$
(That $n_{u} \wedge n_{v}=\frac{LN-M^2}{ \sqrt{EG-F^2}} n$  can be found on page 65 here: https://courses.maths.ox.ac.uk/node/view_material/48840 and is a result of some simple identities and the fact that a unit vector is orthogonal to its derivatives)

Title: Area of image under Gauss map


问:
Can we say $C_G(x)=C_G(x^2)$
Let $(G,.)$ be a group with $e$ as an identity element. If $x\in G$ and $C_G(x)=\left\{g \in G: gx=xg\right\}$ and $|x|=5$. In the process of proving $C_G(x)=C_G(x^3)$, i proved $C_G(x)=C_G(x^2)$ as follows:

Let $g \in C_G(x)$

So $gx=xg$

Now $gx^2=g(xx)=(gx)x=(xg)x=x(gx)=x(xg)=(xx)g=x^2g$

So all $g \in C_G(x)$ satisfies $gx^2=x^2g$

So can we conclude $$C_G(x)=C_G(x^2)$$

答:
In general: if $x \in G$ and $o(x)=n$, then for every non-zero integer $k$ with gcd$(k,n)=1$ it holds that $C_G(x)=C_G(x^k)$.

Proof by Bézout's Theorem, $a,b \in \mathbb{Z}$ exist with $ak+bn=1$. Hence $x=x^{ak+bn}=x^{ak}x^{bn}=(x^k)^a \cdot 1=(x^k)^a$, so any element centralizing $x^k$ centralizes $x$ and the reverse is obvious.

Title: Can we say $C_G(x)=C_G(x^2)$