问:
Looking for a function $f$ that solves $f'(x)= \alpha(x) f(x)$ for a continuous $\alpha$ and starts off positive but then turns negative
I am looking for a continuously differentiable function $f:[-1,1]\to\mathbb R$ (note that I don't care about the interval $[-1,1]$ in particular; it can be replaced by any interval $[a,b]$ with $a<b$ of your choice) that satisfies

$f(t)>0$ for all $-1\le t<0$,

$f(0)=0$,

$f(t)<0$ for all $0<t\le 1$

as well as

$$f'(t)=\alpha(t) f(t)$$

for some continuous $\alpha:[-1,1]\to\mathbb R$.

Essentially I am looking for something like this:

My attempts: The most obvious attempt is  $f(t)=-t^3$. However, one notices that we need to have $\alpha(t)=\frac{3}{t^2}$ for $t\not=0$, which cannot be extended to a continuous function. So $f(t)=t^3$ doesn't work.

In fact, if $f$ is any polynomial with $f(0)=0$, we know that $\frac{f'}{f}$ diverges at $0$ so that $\alpha=\frac{f'}f$ can't be continuous....

Next try, $f(t)=-\exp\left(-\frac1t\right)$ for $t>0$ and $f(-t)=-f(t)$: This fails immediately as for $t\not=0$, $f'(t)=\frac{f(t)}{t^2}$, so that, once again, $\alpha(t)=\frac{f'(t)}{f(t)}=\frac{1}{t^2}$ can't be made continuous at $t=0$.

Final attempt: I define $g(t)=2 \operatorname W(\exp(t))-t-1$, where $\operatorname W$ is the Lambert W function. This is a rotation of the function $t\mapsto\exp(t)$ by 45 degrees and it looks like this:

Then I define $$f(t)=\begin{cases}
g(t+1) & t\le 0,\\ -g(t+1) & t>0.
\end{cases}$$

It looks like this:

However, once again we have that $\lim_{t\to0}\frac{f'(t)}{f(t)}$ is non-existant so that $\alpha$ can't be continuous at $0$.

So my question: Does a function $f$ as above even exist? If yes, can you give me an example? If not, why doesn't such a function exist??

答:
Note that, $$F:[-1,1]\times\mathbb R\to\mathbb R, (t,y)\mapsto \alpha(t)y$$ is Lipschitz-continuous in the second variable: For all $t\in[-1,1]$, $$|\alpha(t) y_1-\alpha(t)y_2|=\lvert\alpha(t)\rvert\cdot\lvert y_1-y_2\rvert \le\lVert\alpha\rVert_{L^\infty([-1,1])}\lvert y_1-y_2\rvert,$$

so that Picard-Lindelöf can immediately be applied to show that the only solution is $y\equiv 0$.

Note: I was so focused on the fact that the function $\alpha$ doesn't have to be Lipschitz continuous that I forgot that Lipschitz-continuity in $t$ isn't required...


问:
Question about the map $S^1\to S^1$ in the context of the real projective plane $\mathbb{R}P^2$
I was recently working on an exercise to compute $H^*(\mathbb{R}P^2\times \mathbb{R}P^2,\mathbb{Z})$ and there was a particular step in the solution for which i would like to get a better intuition.

The solution says

Viewing $\mathbb{R}P^2$ as $D^2/{\sim}$ where we identify $p \in S^1 =
 \partial D^2$ with $−p$, we set $U$ to be a small disk around $0 \in
 D^2$ and $V = \mathbb{R}P^2 − \{0\}$. Thus $U$ is contractible and $V$
deformation retracts onto the subspace $S^1/{\sim}$, which is
homeomorphic to $S^1$ through a loop joining $p$ and $−p$. The intersection $U\cap V$ is an annulus, so the diagram

$\require{AMScd}$ \begin{CD} U\cap V @>{}>> V\\ @AAA @AAA\\ S^1
 @>{f}>> S^1 \end{CD}

in which the lower map $f\colon S^1\to S^1$ is $z\mapsto z^2$,
commutes up to homotopy.

I would love to understand why the lower map needs to be chosen as $z\to z^2$.

My thoughts:

Clearly the annulus $U\cap V$ is homotopy equivalent to $S^1$ and $V$ is homotopy equivalent to $S^1/{\sim}$ which apparently seems to be homeomorphic to $S^1$ as pictured in the solution. The homotopy equivalence of $U\cap V$ and $V$ is also comprehensible by construction.

So up to homotopy i understand the top horizontal arrow and the two vertical arrows. If i imagine the bottom horizontal map $f$ to be the attaching map of the open disk onto the boundary circle $S^1$ of $\mathbb{R}P^2$ i understand why it's $z\mapsto z^2$.

Why can't we, for example, just consider the identity map $\operatorname{id}\colon S^1\to S^1$ as the bottom horizontal map. Why is the attaching map the obvious choice?

And related to that question: Is that the reason why we have chosen the disk $D^2/{\sim}$ identified with antipodal points rather than the unit circle $S^1/{\sim}$?

答:
View $S^1$ as the unit sphere in $\mathbb{C}$ and the disc $D^2$ as the cone on $S^1$. Thus the disc has coordinates $(z,t)$ where $z\in S^1$ and $t\in[0,1]$, and $S^1\times\{0\}$ is identified to a point. Then $\mathbb{R}P^2$ is the quotient of the disc formed by the identification $(z,1)\sim(-z,1)$. I will write the coordinates in $\mathbb{R}P^2=D^2/\sim$ with angular brackets.

We have
$$U=\{\langle z,t\rangle\in\mathbb{R}P^2\mid t\leq 1/2\},\qquad V=\{\langle z,t\rangle\in\mathbb{R}P^2\mid t>0\}.$$
The contraction of $U$ is given by $F_s\langle z,t\rangle=\langle z,(1-s)t\rangle$. The inclusion $S^1\hookrightarrow U\cap V$ is given by $z\mapsto \langle z,\frac{1}{2}\rangle$ and is a homotopy equivalence. The retraction of $V$ onto $\mathbb{R}P^1\cong \{\langle z,1\rangle\in\mathbb{R}P^2\}$ is the projection $\langle z,t\rangle\mapsto\langle z,1\rangle=\langle-z,1\rangle$, which is well-defined since $\langle z,t\rangle\in V\Rightarrow t>0$. The homotopy $G_s\langle z,t\rangle=\langle z,(1-s)t+s\rangle$ is that which is required to show that the map is a deformation retraction.

We identify $\mathbb{R}P^1$ with $S^1$ by the inverse homeomorphisms
$$\langle z,1\rangle\mapsto z^2,\qquad z\mapsto \langle \sqrt z,1\rangle=\langle-\sqrt z,1\rangle.$$
Call $\alpha: V\rightarrow S^1$ the homotopy equivalence which is the composite of the previous projection followed by this identification. Then clearly the composite $S^1\hookrightarrow U\cap V\hookrightarrow V\xrightarrow{\alpha}S^1$ which appears in your diagram is the degree $2$ map
$$z\mapsto z^2.$$

Homefully you agree with this. The point of course is that we never had a chance to consider any other map to put in your square diagram: the map $f$ was what was given to use by the geometry. One thing we should note is that we have made special use of the fact that $\alpha$ is a homotopy equivalence.

If we're feeling brave we can try to see what the maps above look like in homogeneous coordinates. Take $\mathbb{R}P^2$ as a quotient of $S^2$ by the relation $(x,y,z)\simeq(-x,-y,-z)$ and write its coordinates with square brackets as $[x,y,z]=[-x,-y,-z]$, where $(x,y,z)\in\mathbb{R}^3$ satisfy $x^2+y^2+z^2=1$.

The quotient map $\Phi:D^2\rightarrow\mathbb{R}P^2$ is given by
$$\Phi(x,y,t)=[t\cdot x,t\cdot y,\sqrt{1-t^2}],$$
and it's easily seen that this induces a homeomorphism of $D^2/\sim$ as above onto $\mathbb{R}P^2=S^2/\simeq$. The sets which appear are
$$U=\{[x,y,z]\in\mathbb{R}P^2\mid |z|\geq\sqrt{3/4}\},\qquad V=\{[x,y,z]\in\mathbb{R}P^2\mid |z|<1\}$$
(Note the change in signs of the last coordinate!). The inclusion $S^1\hookrightarrow U\cap V$ is the map $(x,y)\mapsto\Phi(x,y,1/2)=[\frac{1}{2}x,\frac{1}{2}y,\sqrt{3/4}]$. Note that this is indeed an inclusion since any coset in its image has a unique representative with positive last coordinate. The projection $V\rightarrow \mathbb{R}P^1$ is the map $[x,y,z]\mapsto[x/\sqrt{x^2+y^2},y/\sqrt{x^2+y^2},0]$ (you might like to down the homotopy $G_s$ from before using these coordinates to show again that this map is a deformation retraction). The identification $\mathbb{R}P^1\cong S^1$ is induced using stereographic projecion. In the direction we need it is
$$[x,y,0]\mapsto (x^2-y^2,2xy).$$
The composite $S^1\hookrightarrow U\cap V\hookrightarrow V\xrightarrow{\alpha} S^1$ is now
$$(x,y)\mapsto(x^2-y^2,2xy).$$
(Notice that this is exactly the same map as before: $x+iy\mapsto (x+iy)^2$). I'll leave you to check any details.

So, to answer your second question: this is why it's better for this question to consider $D^2/\sim$ - everything is much easier!

ContentLicense: CC BY-SA 4.0


问:
Smart method for computing the determinant of the matrix defined by $A_{ii}=a+b$ and $A_{ij}=a$?
We are given the following $n\times n$ matrix, whose terms are:

$a+b$ for the terms in the diagonal ($i=j$).

$a$ for the rest of the elements.

I know we can simply compute this and try to find a pattern, but is there a faster way? Maybe using some properties of the determinant of a matrix?

答:
\begin{align*}
& \begin{vmatrix}
a + b & a     & a     & \cdots & a \\
a     & a + b & a     & \cdots & a \\
a     & a     & a + b & \cdots & a\\
\vdots & \vdots & \vdots & \ddots & \vdots \\
a     & a     & a     & \cdots &a + b
\end{vmatrix} \\
= & \begin{vmatrix}
na + b & na + b   & na + b     & \cdots & na + b \\
a     & a + b & a     & \cdots & a \\
a     & a     & a + b & \cdots & a\\
\vdots & \vdots & \vdots & \ddots & \vdots \\
a     & a     & a     & \cdots &a + b
\end{vmatrix} \\
= & (na + b)\begin{vmatrix}
1 & 1   & 1     & \cdots & 1 \\
a     & a + b & a     & \cdots & a \\
a     & a     & a + b & \cdots & a\\
\vdots & \vdots & \vdots & \ddots & \vdots \\
a     & a     & a     & \cdots &a + b
\end{vmatrix} \\
= & (na + b)\begin{vmatrix}
1 & 1   & 1     & \cdots & 1 \\
0     & b & 0    & \cdots & 0 \\
0     & 0    & b & \cdots & 0 \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
0     & 0    & 0   & \cdots & b
\end{vmatrix} \\
=& (na + b)b^{n - 1}
\end{align*}

Can you figure out what operations I used in each step?


问:
Understanding 0-dim. case in proposition 15.3 in "Introduction to smooth manifolds" by Lee
I am trying to understand the proof of the following proposition. It is taken from Lee's book "Introduction to smooth manifolds":

I have trouble understanding the 0-dim. case. In partucluar why $\omega>0$ implies $\mathcal{O}_{\omega}$ is +1 and the other case as well.

Since it says this case is immediate I suppose I am missing something obvious, but unfortunately I am completely stuck. I am also not sure what "consistently oriented" means in the case of a 0-dim. vector space. There is only a definition for the case $n\geq 1$. In Jänich's book "Vector Analysis" I found something that would make sense to me regarding the definion of orientaion in the 0-dim. case. It is the following:

Then $\mathcal{O}_{\omega}=[\emptyset]$, since the empty set is the only ordered basis in $V=0$?

Thank you very much in advance!

答:
Note that my book gives a separate definition of orientations of zero-dimensional vector spaces, which doesn't have anything to do with consistently oriented bases. The definition appears on page 379, about a third of the way down the page:

For the special case of a zero-dimensional vector space $V$, we define
an orientation of $V$ to be simply a choice of one of the numbers $\pm 1$.

In the statement of Proposition 15.3, the sentence "if $n=0$, then $\mathscr O_\omega$ is $+1$ if $\omega>0$, and $-1$ if $\omega<0$" is not a claim to be proved -- it's the definition of the orientation $\mathscr O_\omega$ determined by a particular choice of $\omega$.

Since $\Lambda^0 V^* = \mathbb R$ (no matter what the dimension of $V$ is), in the case $n=0$, an element $\omega\in \Lambda^n V^*$ is just a real number, and any such number specifies a unique orientation of $V$ by the recipe quoted above.

The only thing to be proved in that case is the claim that two nonzero $0$-covectors (i.e., nonzero real numbers) determine the same orientation if and only if each is a positive multiple of the other, and this is just the statement that two nonzero real numbers have the same sign iff each is a positive multiple of the other.


问:
How do I integrate $(\frac{-1}{\sin^2x})(\frac{\cos x}{\sin x}) d\theta$?
I have a question:

Right before this question, in a previous part, they had me find the derivative of $\frac{1}{\sin^2x}$, which I found to be $\frac{-2\cos\theta}{\sin^3\theta}$.

My problem is with the integration part (I've separated my variables). I tried u-substitution with $u = \sin^4\theta$, and also integration by parts, but neither got anywhere. I seem to have gotten into an loop with integration by parts, and u-substitution ended when I still had a $\cos\theta$ in my numerator after all substitutions/cancelling etc. So now, having spent a considerable amount of time on this problem, I don't see how to integrate the right hand side after separating variables. The right hand side is:

$$\left(\frac{-1}{\sin^2\theta}\right)\left(\frac{\cos \theta}{\sin \theta}\right) d\theta$$

答: As @player3236 notes, in the problem's original notation $xdx=-\frac{\cos\theta}{\sin^3\theta}d\theta$. With $s=\sin\theta$,$$x^2=\int 2xdx=-\int 2s^{-3}ds=s^{-2}+C=\csc^2\theta+C.$$I'll leave you to find $C$.


问:
If $f$ is continuous and $g$ is of bounded variation then $f$ is integrable with respect to $g$.
Suppose $f$ is continuous on $[a,b]$ and $g$ is of bounded variation on $[a,b]$.

Prove that $f$ is integrable with respect to $g$ on $[a,b]$.

How can I prove this?

答:
You can prove this in the easier case when $g$ is monotone on $[a,b]$, and then use that $g=V-(V-g)$ is a difference of monotone functions, where $V$ is the total variation of $g$ over $[a,x]$, i.e $V_g(a,x)$. You'll have to use that $f$ is uniformly continuous, and that $g$ has variation bounded by $g(b)-g(a)$. Then, choose a partition  with mesh $<\delta$ such that for each $|x-y|<\delta$, we have $$\tag 1 |f(x)-f(y)|<\frac{\epsilon}{A}$$ where $A=2(g(b)-g(a))$.

You'll have to use that being Riemann Steiltjes integrable is equivalent to the Riemann condition (for monotone integrators), that is, that given $\epsilon >0$ there exists a partition $P_\epsilon$  such that $P\supset P_\epsilon$ implies $$U(P,f,\alpha)-L(P,f,\alpha)<\epsilon$$

To this end, you'll have to sue that, because of $(1)$, $$M_k-m_k\leq \epsilon/A$$

where $M_k,m_k$ are the supremum and infimum of $f$ over the $k$th interval. Good luck.

ContentLicense: CC BY-SA 3.0


问:
Correlation of Rolling Two Dice
If A is a random variable responsible for calculating the sum of two independent rolls of a die, and B is the result of calculating the value of first roll minus the value second roll, is is true that A and B have a $cov(A,B)\neq 0$? In other words, is it true that they are correlated?

I've come to the conclusion that they must be correlated because they are not independent, that is, the event of A can have an impact on event B, but I remain stuck due to the fact that causation does not necessarily imply correlation.

I know that independence $->$ uncorrelation, but that the opposite isn't true.

答:
Covariance of independent variables is $0$ but covariance of dependent variables is not necessarily non-zero: it might be $0$ (which is exactly what happens in this case), so your conclusion is untrue.

Let $X_1,X_2$ be independent random variables denoting the number rolled on the two fair die respectively. $X_1,X_2$ are identically distributed. $A=X_1+X_2,B=X_1-X_2$.

$\Bbb E[B]=E[X_1]-E[X_2]=0.\\\Bbb E[AB]=E[X_1^2]-E[X_2^2]=0.$

So the covariance $\Bbb E[AB]-\Bbb E[A]\Bbb E[B]$ is $0$.

Title: Correlation of Rolling Two Dice


问:
Let $ X $ be a topological space, paracompact, $T_2$. Let $S \subset X$, with $S$ is generalized-$F_{\sigma}$ then $ S $ is paracompact.
Let $ X $ be a topological space, paracompact, $T_2$, and $S \subset X$ with $S$ is generalized-$F_{\sigma}$ then $ S $ is paracompact.

A subset S of a topological space $X$ is called a generalized-$F_{\sigma}$ set in $X$ if for all open $G \subset X$  with $S \subset G$, there exists an $F_{\sigma}$-set $F \subset X$ such that $S \subset F \subset G$.

We have to $F$ an $F_{\sigma}$-set in a paracompact space is paracompact, but how can I say that $ S $ is also paracompact, does it inherit the paracompact of $ F $?

答: Let $\mathscr{U}$ be a relatively open cover of $S$. For each $U\in\mathscr{U}$ let $V_U$ be an open set in $X$ such that $U=V_U\cap S$, and let $\mathscr{V}=\{V_U:U\in\mathscr{U}\}$. There is an $F_\sigma$-set $F$ such that $S\subseteq F\subseteq\bigcup\mathscr{V}$, and $F$ is paracompact, so $\{V_U\cap F:U\in\mathscr{U}\}$ has a locally finite $F$-open refinement $\mathscr{R}$ that covers $F$. Now just check that $\{R\cap S:R\in\mathscr{R}\}$ is a locally finite $S$-open refinement of $\mathscr{U}$ that covers $S$.


问:
The Introduction of boundary conditions in physical problems: A problem with a result in a book
I'm asking this question again, since I incurred in an error while copying the relations from the text.\

Returning to an old book about the foundations of physics by Lindsay and Margenau, I wasn't able to follow a result in a section about the introduction of boundary conditions in physical problems.

The imposed conditions on the wave equation solution in a tube of length $l$ are:

$f_{1}(-ct) + f_{2}(ct)=0$

$f_{1}(l-ct)+f_{2}(l+ct)=0$

from this you`ll get (copying the exact relation from the text)

$f_{2}(ct)=-f_{1}(-ct)$

$f_{2}(ct+l)=-f_{1}(l-ct)= f_{2}(ct-l)$

It is the third term of this equation ($f_{2}(ct-l)$) that I cant figure out how it appears from the others two and the above relations, sadly it is a non-trivial result for me. I`ll appreciate your help.

答:
The first condition $f_1(-ct) + f_2(ct)=0$ basically asserts that $f_1(t) = -f_2(t)$.

To see this, do the substitution $t \to -\dfrac t c$.

Hence we have $-f_1(l-ct) = --f_2(-(l-ct)) = f_2(ct-l)$, via the substitution $t \to l-ct$.


Title: The Introduction of boundary conditions in physical problems: A problem with a result in a book

Body:
I'm asking this question again, since I incurred in an error while copying the relations from the text.\

Returning to an old book about the foundations of physics by Lindsay and Margenau, I wasn't able to follow a result in a section about the introduction of boundary conditions in physical problems.

The imposed conditions on the wave equation solution in a tube of length $l$ are:

$f_{1}(-ct) + f_{2}(ct)=0$

$f_{1}(l-ct)+f_{2}(l+ct)=0$

from this you`ll get (copying the exact relation from the text)

$f_{2}(ct)=-f_{1}(-ct)$

$f_{2}(ct+l)=-f_{1}(l-ct)= f_{2}(ct-l)$

It is the third term of this equation ($f_{2}(ct-l)$) that I cant figure out how it appears from the others two and the above relations, sadly it is a non-trivial result for me. I`ll appreciate your help.

Body:
$f_{1}(c(ct+l)) + f_{2}(-c(ct+l))=0$

$f_{1}(l-c(ct+l))+f_{2}(l+c(ct+l))=0$

the same applies to $l-ct$ like above for $ct+l$:

$f_{1}(c(l-ct)) + f_{2}(-c(l-ct))=0$

$f_{1}(l-c(l-ct))+f_{2}(l+c(l-ct))=0$


问:
A permutation problem
Consider all the permutations of the digits $1, 2, \dots, 9$. Find the number
of permutations each of which satisfies all of the following:

the sum of the digits lying between 1 and 2 (including 1 and 2)
is 12

the sum of the digits lying between 2 and 3 (including 2 and 3)
is 23

the sum of the digits lying between 3 and 4 (including 3 and 4)
is 34, and

the sum of the digits lying between 4 and 5 (including 4 and 5)
is 45.

How to solve such problems?

答:
$\{4,5\}$ are sure the first and the last elements. You have the template for permutation:

$$4*******5$$ The sum between $3$ and $4$ is $(34-3-4)=27$. Note that $1$ and $2$ are both between  $3$ to $4$. Then you have the only place to set $3$:
$$4*****365$$
The sum of the numbers which are not chosen yet is $27$. You should reach the sum of $(23-2-3)=18$ between $2$ and $3$. And $(12-1-2)=9$. Again, the only way, because $1$ should be between of $2$ and $3$:
$$472***365$$
Then
$$472918365$$
So there are 2 combinations: $472918365$ and $563819274$.

Title: A permutation problem


问:
Determining the Remainder
Is it possible to calculate the remainder of two given values with merely addition, subtraction, multiplication, and division? Is there an algorithm or formula if it is even possible?

For instance, say we have two values: 100 and 30. If we divide 100 by 30, we get 3.333.... But is there any way to reach 0.333... (without knowing from the get go how many wholes are included)?

To further illustrate what I mean, and sticking with the example above, one way to find the remainder would be the following formula: 100 / 30 / 10 * 30 = 0.333... BUT this obviously does not work for any two given numbers.

Some more examples (The values in the brackets are the values I am after):

100 / 40 = 2.5 (0.5)
450 / 50 = 9 (0)
11 / 4 = 2.75 (75)

Sorry, if this question is not clear. It is based on a programming challenge I have encountered. I want to use only CSS to calculate the remainder of a text's line height, given a specific vertical offset of the text. For instance, the height of the window might be 100, whereas the line height would be 30. But CSS has no function to easily determine the remainder, and I cannot use loops or recursion (i.e. 100 - 30 - 30 - 30 - 30 < 0), nor conditional (i.e. if N < 0, do...).

My math is awful. But I am wondering whether or not it is mathematically possible?

答:
You are looking for the fractional part of a number.

To make use of this, define a function of two variables that gives one output:
$$t=f(x,y)=\frac{x}y-\left\lfloor\frac{x}y\right\rfloor.$$

Using your examples above, we have the following:
\begin{align}
f(100,30)&=\frac{100}{30}-\left\lfloor\frac{100}{30}\right\rfloor=\frac{10}3-3=\frac13\\
f(100,40)&=\frac{100}{40}-\left\lfloor\frac{100}{40}\right\rfloor=\frac52-2=\frac12\\
f(450,50)&=\frac{450}{50}-\left\lfloor\frac{450}{50}\right\rfloor=9-9=0\\
f(11,4)&=\frac{11}{4}-\left\lfloor\frac{11}{4}\right\rfloor=\frac{11}{4}-2=\frac34
\end{align}

Title: Determining the Remainder


问:
Inertia subgroup of finite extension over $\mathbb{Q}_p$ whose ramification index is not divisible by $p$
Let $K$ be an extension of $\mathbb{Q}_p$ and let $L/K$ be a finite extension with $p \nmid e$ where $e = e(L/K)$ is the ramification index of $L/K$. Let $I=I(L/K)$ be the intertia subgroup of $L/K$.

Question Is there a result showing that $I$ must be cyclic?

I still have a vague understanding of things like inertia subgroups, Galois groups over $\mathbb{Q}_p$, etc., so I think a reference to obtain the basics to solve the above question would be the best for me. I don't mind an answer to my question though. Thanks!

答:
Yes, this is standard stuff. A finite, normal, totally tamely ramified extension $L\supset K$ of local fields has a Galois group that injects into $\kappa^\times$, where $\kappa$ is the (common) residue field.

If $\pi$ is a uniformizer of $L$, you send $\sigma\in\text{Gal}^L_K$ to the image of $\frac{\sigma(\pi)}\pi$ in $\mathcal O_L/\pi\mathcal O_L=\kappa$. Show it’s a homomorphism, and injective if the degree is prime to $p$. So the Galois group   is cyclic.


问:
Does there exist a dense bijection on the unit interval
I was looking at the question "Are there any valid continuous Sudoku grids?" and that lead me to this question.

Let $U$ be the unit interval $[0,1]$ or $[0,1)$. (pick which is easier for you)

Does there exist a bijective function $f: U \to U$ such that for every subinterval $I$ of $U$, the set $\{f(x)\ |\ x \in I\}$ is dense in $U$?

答:
Choose open subintervals $I_1,I_2,\dots$ of $U$  such that if $J$ is an open subinterval of $U,$ then $J$ contains $I_n$ for some $n.$ (For example we could let $I_1,I_2,\dots$  be the open subintervals of $U$ with rational endpoints.)

We can then choose pairwise disjoint countable sets $D_n\subset I_n$ such that $D_n$ is dense in $I_n$ for each $n.$

And we can choose pairwise disjoint countable sets $E_n\subset U$ such that $E_n$ is dense in $U$ for each $n.$

Now define $f:U\to U$ as follows: For each $n=1,2,\dots$ let $f$ be a bijection of $D_n$ onto $E_n.$ Then $f$ is a bijection of $\cup D_n$ onto $\cup E_n.$ Now $U\setminus \cup D_n$ has the same cardinality as $U\setminus \cup E_n.$ So to finish the definition of $f,$ let $f$ be any bijection of $U\setminus \cup D_n$ onto $U\setminus \cup E_n.$

It follows that $f:U\to U$ is a bijection. Let $I\subset U$ be a subinterval. Then $I_n\subset I$ for some $n.$ Since $f(I_n) = E_n,$ $f$ is the desired function.


问:
Commutative ring, commutative ring without zero divisors or field?
I'm in my first year of University and struggle with the following exercise in linear algebra.

I have to determine if the following sets are a commutative ring, a ring without zero dividers or a field.

$
\mathbb{Z}[ \sqrt{3} ] := \{ a+b\sqrt{3} | a,b \in \mathbb{Z} \} \subseteq \mathbb{R}
$

and
$
\mathbb{Q}[ \sqrt{3} ] := \{ a+b\sqrt{3} | a,b \in \mathbb{Q} \} \subseteq \mathbb{R}
$

I know that $\mathbb{(Q,+,\cdot)}$ is a field and am allowed to use that. Therefore I would just say, that $
\mathbb{Q}[ \sqrt{3} ] := \{ a+b\sqrt{3} | a,b \in \mathbb{Q} \} \subseteq \mathbb{R}
$ is a field because it only uses the operations $+$ and $\cdot$ and $a, b\in \mathbb{Q}$ and we know that $\mathbb{Q}$ is a field.

For $\mathbb{Z}[ \sqrt{3} ] $ I would say that $\mathbb{Z}$ an abelian group:

In $\mathbb{Z}$ we can use $(a+b)+c=a+(b+c)$

in $\mathbb{Z}$ we can use  $a+b =b+a$

It is $0 \in \mathbb{Z}$, therefore $a+0 = 0+a = a$

There exists $-a \in \mathbb{Z}$ such that $a+(-a) = a-a= 0$.

We also can say, that $\mathbb{Z}$ is a monoid because

if $a,b,c \in \mathbb{Z}$ then $(a\cdot b)\cdot c = (a \cdot b) \cdot c$

there is $1 \in \mathbb{Z}$ therefore for $a \in \mathbb{Z}$ we have $1 \cdot a = a$

And at last, we can say that $\mathbb{Z}$ is distributive.

I'm not exactly sure if this answer would suffice or if this is missing the goal of the exercise. Can someone help me?

答:
$x,y\in$ $\mathbb{Z}[ \sqrt{3} ] = \{ a+b\sqrt{3} | a,b \in \mathbb{Z} \} \subseteq \mathbb{R}$

$x=a+b\sqrt3$ , $y=c+d\sqrt3$

Let  $xy=0$ then $ac+(ad+bc)\sqrt3+3db=0$

we know that $ad+bc=0$ and $ac+3db=0$

$d(a-3b)=(a-b)c$

$bc^2=3db^2$

$b(c^2-3db)=0$

$b=0$ or $c^2=3db$

$b=0$ $\Rightarrow$ $a=0$ which means $x=0$

$c^2=3db$ $\Rightarrow$ $c^2+ac=0$

$c^2+ac=0$ $\Rightarrow$ $c=0$ or $c=-a$ if $c=0$ $y=0$ or $x=0$

it is same at  $a=-c$ situation so it does not have zero diveder


问:
Is $a\cdot (-1)^a$ monotonic?
I'm having difficulty understanding what is a monotonic sequence? Is this Monotonic? This is because that this function doesn't increase or decrease with consistency. Also does $\dfrac{(-1)^a}{a}$ have a limit? No right?

答:
First of all, there is some ambiguity in the term monotonic.

For example, with respect to the phrase "monotonically increasing",
at least one mathSE reviewer that I have exchanged comments with regards this as synonomous with "strictly increasing".

Alternatively, I have seen old math books that regard "monotonically increasing" as synonymous with "non-decreasing."

Consider (for example) the sequence

$$ (1 \times [-1]^1), ~(2 \times [-1]^2), ~(3 \times [-1]^3) \cdots$$

The even #'d terms are positive and the odd #'d terms are negative.  Therefore, the sequence is bouncing back and forth between greater than zero and less than zero.

Therefore, under any definition of monotonic, the sequence is not monontonic.

Title: Is $a\cdot (-1)^a$ monotonic?


问:
Find a continuous function such that$f:\mathbb{R}\to \mathbb{R}$, $f(0)=0$ and $f(2x)\geq x+f(x)$ and $f(3x)\leq 2x+f(x)$.
Could you give me a hint for this problem?

I have to find all continuous functions $f:\mathbb{R}\to \mathbb{R}$ such that $f(0)=0$ and  $f(2x)\geq x+f(x)$ and $f(3x)\leq 2x+f(x)$.

I think this might be all the $f(x)=|x|$,  but can it be any other functions?

答:
$f(2x)\ge x+f(x)$

rewrite the second as $f(x)+2x\ge f(3x)$

Take $$f(x) =x$$

we have $$2 x\ge x+x;\;x+2x\ge3x$$
verified for any $x\in\mathbb{R}$


答:
It is not monotonic. You need your sequence to obey $a_n\leq a_{n+1}$ for every $n$ or $a_{n+1}\leq a_n.$ If $a_n = n\cdot (-1)^n,$ then $a_{2n-1}\leq a_{2n}\geq a_{2n+1}.$ So there is no consistency in the inequality.

That sequence $(-1)^n/n$ actually converges to $0$ because for $n$ big enough you will find $(-1)^n/n$ as closed as you want to $0.$ This words explain the definition of limit as the $\epsilon -\delta$ arguments.

Title: Is $a\cdot (-1)^a$ monotonic?

Body: I'm having difficulty understanding what is a monotonic sequence? Is this Monotonic? This is because that this function doesn't increase or decrease with consistency. Also does $\dfrac{(-1)^a}{a}$ have a limit? No right?


问:
"Converse" of second Fundamental Theorem of Calculus?
Let $f: \mathbb{R} \to \mathbb{R}$ be continuous. Let $F: \mathbb{R} \to \mathbb{R}$ be such that $F(b) - F(a) = \int_{a}^{b} f(x) \, dx$ for all $a, b \in \mathbb{R}$. Is $F$ differentiable, and is $F’ = f$?

One reason I am asking is in the context of probability. Given the cumulative distribution function $F$ of a continuous random variable $Y$, $F(y) := P(Y \leq y)$, can the probability density function $f := F'$ be (alternatively) characterized by the property $F(b) - F(a) = \int_{a}^{b} f(x) \, dx$ for all $a, b \in \mathbb{R}$?

答: Yes. Fix any $a \in \mathbb{R}$. Define $G: \mathbb{R} \to \mathbb{R}$ by $G(x) = F(x) - F(a) = \int_a^x f(t) \, dt$. Differentiating and applying the Fundamental Theorem of Calculus, we get $G'(x) = F'(x) = f(x)$, so $F' = f$.


问:
Limit of Multivariable Fourier Series
If I have some Fourier Series representation of a function with $x$ period of $2L$

$$G(x,y) = \sum_{n = 1}^{\infty} \left[a_n \sin\left(\frac{n \pi x}{L}\right) + b_n \cos\left(\frac{n \pi x}{L}\right)\right]f_n(y)$$

Where each of the $f_n(y) \to 0 $ as $y \to \infty$ do we necessarily have that

$$\lim_{y \to \infty} G(x,y) = 0$$

I do not know the relevant analysis -- maybe we are not always allowed to take termwise limits of an infinite sum? Is there a general result that applies here? How might I go about calculating limits of sums like this, is there a particular method that is usually applied?

Thank You

答: Well, guess in your case you are correct. You can see for example "Elementary Properties of Limits of Functions" (Ch.6, p.156) in S. Krantz Real Analysis and Foundations.</p>

Title: Limit of Multivariable Fourier Series


问:
Question about proof of closure of set.
Let $\Gamma$ be the Moore Plane Topology and consider the following set:
\begin{equation*}
    A=\bigcup_{q\in\mathbb{Q}\setminus \{ 0\}} \left(\{q\} \times \left[0,\frac{1}{d_q}\right]\right).
\end{equation*}

Where $q = \frac{n_q}{d_q}: n_q,d_q \in \mathbb{Z}$ are relativily prime and $d_q >0$.

I am tring to show that if $x \in \{0\} \cup (\mathbb{R} \setminus \mathbb{Q})$ and $y = 0$, then $(x,0) \in \operatorname{cl}(A)$.

Below is my attempt:

(my) Solution: let $x \in \mathbb{R}$ and consider the tangent disk at $(x,0)$ of radius $\epsilon > 0$. Let $q \in \mathbb{Q}(q = \frac{n_q}{d_q})$ such that $|x-q|<\sqrt{(\frac{1}{d_q})(2\epsilon - \frac{1}{d_q})}$. Then the point $(q, \frac{1}{d_q}) \in A$ is in the tangent disk at $(x,0)$ of radius $\epsilon > 0$, since, the following holds:
\begin{equation*}
    \begin{split}
        (x-q)^2 + (\epsilon - \frac{1}{d_q})^2 &< (\frac{1}{d_q})(2\epsilon - \frac{1}{d_q}) + (\epsilon - \frac{1}{d_q})^2 = \epsilon^2.
    \end{split}
\end{equation*}
Therefore, for any $\epsilon > 0$ we have $D_{(x,0),\epsilon} \cap A \neq \emptyset$, hence, the point
$(x,0) \in \operatorname{cl}_{\tau} (A)$, for any $x \in \mathbb{R}$. Therefore, $\mathbb{R} \times \{0\}$, i.e. the $x$-axis, is in the closure.

答:
Let $\epsilon>0$. There is a prime $p$ such that $\frac2p<\epsilon$, so there is a rational $q\in(x,x+\epsilon)\setminus\Bbb Z$ such that $d_q=p$. Using a prime ensures that $d_q$ really is $p$ unless $q$ is an integer, and taking $p$ large enough so that $\frac2p<\epsilon$ ensures that there are at least two consecutive multiples of $\frac1p$ between $x$ and $x+\epsilon$, so that we can can choose one that isn’t an integer. Then

$$\begin{align*}
|x-q|^2+\left(\epsilon-\frac1{d_q}\right)^2&\le\frac1{p^2}+\epsilon^2-\frac{2\epsilon}p+\frac1{p^2}\\
&=\epsilon^2+\frac{2-2p\epsilon}{p^2}\\
&<\epsilon^2\,,
\end{align*}$$

since $p\epsilon>1$, and therefore $\left\langle q,\frac1{d_q}\right\rangle\in D(\langle q,0\rangle,\epsilon)$.

Title: Question about proof of closure of set.


问:
Number of house subgraphs in a Erdős-Rényi graph?
Considering we have a "house subgraph", which is a graph composed of 5 nodes, $v_1, \ldots v_5$ in which the edges
$\{v_1,v_2\}, \{v_1,v_3\}, \{v_2,v_3\}, \{v_2,v_4\}, \{v_3,v_5\}, \{v_4,v_5\}$ exist, and no other edge between them exists
(see Figure below). What is the expected number of house subgraphs in a random graph according
to $G(n, p)$ (Erdős-Rényi graph with $n$ vertices and edge probability $p$)?

House subgraph figure:

答:
Let's solve a more general problem:

Suppose $\Gamma$ is a graph with $m$ vertices and $k$ edges. What is the expected number of induced subgraphs isomorphic to $\Gamma$ in $G(n, p)$ ($n$-vertex Erdos-Renyi random graph with edge probability $p$)?

Let's solve it first for the case, where $n = m$. Then there is only one induced subgraph of size $m$ in our random graph. What is the probability of it being isomorphic to $\Gamma$?

There are $m!$ bijections from its vertices to the vertices of $\Gamma$, but only $\frac{m!}{|Aut(\Gamma)|}$ are distinct in respect to automorphisms of $\Gamma$. For all those distinct bijections the events of them being isomorphisms are disjoint and probability of each of them equals $p^k (1 - p)^{\frac{m(m-1)}{2} - k}$. Thus the total probability of $G(m, p)$ being isomorphic to $\Gamma$ is $\frac{m! p^k (1 - p)^{\frac{m(m-1)}{2} - k}}{|Aut(\Gamma)|}$.

Now assume $n$ is arbitrary. For any $m$-element set $A$ of vertices of our random graph assume $I_A$ to be the Bernoulli random variable that equals $1$ if induced subgraph on those vertices is isomorphic to $\Gamma$ and $0$ otherwise. Then, due to linearity of expectation, the expected number of induced subgraphs isomorphic to $\Gamma$ is the sum of expected values of all $I_A$. We have already found, that the expectation of each of them (equal to their probability to be $1$) is $\frac{m! p^k (1 - p)^{\frac{m(m-1)}{2} - k}}{|Aut(\Gamma)|}$. And the number of $m$-element subsets of vertices to which they correspond is $C_n^m = \frac{n!}{m!(n-m)!}$.

Thus, the answer is

$$\frac{n!p^k (1 - p)^{\frac{m(m-1)}{2} - k}}{(n-m)!|Aut(\Gamma)|}$$

Now you can calculate the corresponding numbers for your specific case (i.e. the number of vertices, edges and automorphisms of the "house graph") and put them into this formula.


问:
Models and Sets
I'm asked to find models whose universes are subsets of the reals, where the statement is true in the first model and false in the second.
$$(1) \ \forall x \exists y(x<y)$$
An example answer I have is $\mathbb N \models \forall x \exists y(x<y)$ and $\mathbb Z^- \not\models \forall x \exists y (x<y)$
But I'm struggling to understand the reasoning fundamentally. I think I Intuitively understand it. I acknowledge $\mathbb N \subseteq \mathbb R$, and when I think of $\{0, 1, 2, ..., x\}$ as my subset I see that it is always possible for $x<y$ since $y$ can be found between elements of $\mathbb N$. For example, at $x=3$ $y$ may be $3.5$, so $3<3.5$ checks. But I see there are also examples where some $y$ is not greater than $x$, such as $y=2.2$, then $x=3$ and therefore $2.2\not\ge3$. So $\exists y(y>x)$ holds.

And with the example of the false model, my thinking is that if $x\in \mathbb Z^-$ then $x$ is always less than $y$ for ALL $y$.

Is there a better way I can notate a justification for these answers? Are my interpretations correct?

答:
I'm afraid your understanding is not quite right; for instance, in the $<$-structure $\mathbb{N}$, there is no element $3.5$; even though $\mathbb{N}$ is a substructure of $\mathbb{R}$, which does contain $3.5$, by restricting our attention to the (sub)structure $\mathbb{N}$ we are essentially choosing to ignore all the elements of $\mathbb{R}\setminus\mathbb{N}$. In particular, we are choosing to ignore the element $3.5\in\mathbb{R}\setminus\mathbb{N}$.

Also, for example, this sentence

And with the example of the false model, my thinking is that if $x\in\mathbb{Z}^-$ then $x$ is always less than $y$ for ALL $y$.

is not right either; for example, take the element $x=-5\in\mathbb{Z}^-$, and let $y=-7\in\mathbb{Z}^-$. Is $x<y$?

However, your examples are still correct. Let's flesh the question out; what does it mean for the sentence $\forall x\exists y (x<y)$ to hold in an $<$-structure $M$? It means that, for any $x\in M$, we can give a choice of $y\in M$ such that $x<y$.

Let's check whether or not this is true in your examples. First, for $\mathbb{N}$. Let $x\in\mathbb{N}$ be any natural number. Then choose $y=x+1$. Is $y\in\mathbb{N}$? Indeed, for the natural numbers are closed under the successor operation. (For instance, if $x=2$, we can choose $y=3$.) Also, $x<x+1=y$ for any natural number $x$, so this choice of $y$ is a "witness" to the truth of the sentence $\forall x\exists y (x<y)$ in $\mathbb{N}$, and we are done.

Now, what about $\mathbb{Z}^-=\{-1, -2, -3, ...\}$? Well, we wish to show that $\mathbb{Z}^-$ does not satisfy the sentence $\forall x\exists y (x<y)$, and for this it suffices to construct a single counterexample – ie, to give a choice of $x\in\mathbb{Z}^-$ such that, for any $y\in\mathbb{Z}^-$, $x$ is not strictly less than $y$. (Or, restated, a choice of $x\in\mathbb{Z}^-$ such that, for any $y\in\mathbb{Z}^-$, $x\leqslant y$.) I claim $x=-1$ gives such a choice, and is hence a witness to the falseness of the sentence $\forall x\exists y (x<y)$ in $\mathbb{Z}^-$. Can you see why this $x$ is a suitable choice?

Title: Models and Sets


问:
Permutations Integers 1 to 9 all even numbers stay in their natural positions
I have these questions regarding Inclusion-Exclusion Principle two of them have been bothering me. The questions goes:

How many ways can we write the numbers 1 to 9 [1,2,3,4,5,6,7,8,9] if:

All even numbers stay in their natural positions

No even numbers stay in their natural positions

All odd numbers stay in their natural positions

No odd numbers stay in their natural positions

I solved no. 2 and 4 using the principle of inclusion-exclusion but haven't been able to solve 1 and 4. Should the answers of 2 and 4 be of help?

答:
To 1:

You have five places where you can put the five odd numbers (1, 3, 5, 7 and 9):

_ 2 _ 4 _ 6 _ 8 _

So you don't need to bother about the even numbers. The number of different permutations will be: 5! = 120

To 3:

You have four places where you can put the four even numbers (2, 4, 6 and 8):

1 _ 3 _ 5 _ 7 _ 9

So you don't need to bother about the odd numbers. The number of different permutations will be: 4! = 24


问:
Error in brachistichrone derivation
While attempting an independent derivation of the functional to be minimized in the brachistochrone problem, I derived an equation that prevents the known solution to the problem, specifically
$$\left(\frac{dx}{dt}\right)^2\frac{d^2y}{dx^2}=0\tag{1}$$
where $t\in[0, T]$, $x=x(t)$, and $y=y(x)$. This implies that either the particle is stationary or the curve must be (piecewise) linear. I haven't been able to detect where I went wrong.

For those unfamiliar with the brachistochrone problem, the functional to be minimized is the time taken for a particle constrained to a curve to travel from a point A to a lower point B in the presence of a uniform gravitational field. The curve is assumed to be confined to a plane, and I assume point B is to the right of point A, in the conventional sense.

Throughout, $\dot{x}=\frac{dx}{dt}$, $y'=\frac{dy}{dx}$, and this notation is generalized for higher derivatives in the obvious way. I started by deriving the resultant force on the particle at a given point on a somewhat arbitrary curve.  $$F(x)=m(\vec{g}^T\vec{e})\vec{e}\tag{2}$$
where $\vec{g}=\begin{bmatrix}0&-g\end{bmatrix}$ and $\vec{e}=\frac{\tau(x)}{\lVert\tau(x)\rVert}$ for $\tau(x)=\begin{bmatrix}1&y'(x)\end{bmatrix}$, a unit vector tangent to the curve. This should yield the component of the gravitational force tangent to the curve, which I find to be $$F(x)=-\frac{mgy'}{1+(y')^2}\begin{bmatrix}1&y'\end{bmatrix}\tag{3}$$

Next, I set up equations describing Newton's 2nd:
$$\frac{d^2x}{dt^2}=-\frac{gy'}{1+y'^2}\tag{4}$$
$$\frac{d^2y}{dt^2}=-\frac{gy'^2}{1+y'^2}\tag{5}$$

As I've defined $y$ to only implicitly depend on $t$, I rewrite the LHS of $(5)$ through the chain rule, yielding the equations:
$$\ddot{x}=-\frac{gy'}{1+y'^2}\tag{4*}$$
$$\dot{x}^2y''+\ddot{x}y'=-\frac{gy'^2}{1+y'^2}\tag{5*}$$
which should hold for all $t$. The erroneous equation emerges from these two equations. By multiplying $(4^*)$ by $-y'$ and taking the linear combination of this with  $(5^*)$, equation $(1)$ is obtained. What gives? Thanks in advance.

答:
All right, I tracked down my issue. The force is not generally constrained to the tangent of the curve. Rather, this is only the case when the slope is constant. The degenerate case of a stationary particle also satisfies this constraint. This explains equation $(1)$.

Looks like this ended up being more of a physics than a math question.

Title: Error in brachistichrone derivation


问:
Finding the tangents common to two rotated ellipses?
Is there a way to find the four tangents that two rotated ellipses share?

I believe that if two ellipses do not intersect and do not contain one another, they will have four tangents in common and I wish to find these tangents.
I know how to complete the same process with circles and wish to do the same with ellipses.

答:
Let us suppose that the ellipses are given by $f_1(x,y)=c_1$ and $f_2(x,y)=c2$.  We seek points $(x_1,y_1)$ and $(x_2,y_2)$ such that the lines through those points are mutually tangent to the ellipsies.  Clearly, $(x_1,y_1)$ must satisfy $f_1(x_1,y_2)=c_1$ and similiarly for the second point.  Furthermore, the gradients of $f_1$ and $f_2$ must be parallel at their respective points.  This yields $\nabla f_1(x_1,y_1) = \lambda \nabla f_2(x_2,y_2)$.  Finally, we need the line through the points to perpendicular to the gradient vectors.  This yields a total of five equations in the five unkowns $x_1$, $y_1$, $x_2$, $y_2$, and $\lambda$.

Let's apply this in the specific case
$$x^2 - x*y+y^2 = 4$$
and
$$2x^2+x*y+3y^2 = 1.$$

These ellipses do intersect, but I think that's no problem.  The equations are sufficiently complicated that I think I'll do this with Mathematica.

```<code>f1[x_, y_] = x^2 - x*y + y^2;
f2[x_, y_] = 2 x^2 + x*y + 3 y^2;
grad1[x_, y_] = {D[f1[x, y], x], D[f1[x, y], y]};
grad2[x_, y_] = {D[f2[x, y], x], D[f2[x, y], y]};
eqs = Flatten[{
  f1[x1, y1] == 1,
  f2[x2, y2] == 4,
  ({x1, y1} - {x2, y2}).grad1[x1, y1] == 0,
  Thread[grad1[x1, y1] == lambda*grad2[x2, y2]]
}];
pts = {{x1, y1}, {x2, y2}} /. NSolve[eqs, {x1, y1, x2, y2, lambda}]

(* Out: {
  {{1.11372, 0.820906}, {1.37585, 0.122754}},
  {{0.601823, 1.15435}, {-0.245408, 1.17882}},
  {{-1.11372, -0.820906}, {-1.37585, -0.122754}},
  {{-0.601823, -1.15435}, {0.245408, -1.17882}}} *)
</code>```

You can use Solve rather than NSolve but the resulting expressions are quite complicated.  Let's visualize, as well.

```<code>line[{pt1_, pt2_}] := ParametricPlot[t*pt1 + (1 - t) pt2, {t, -2, 3},
  PlotRange -> 2, PlotStyle -> Black];
lines = line /@ pts;
cp = ContourPlot[{
  f1[x, y] == 1,
  f2[x, y] == 4
}, {x, -3, 3}, {y, -3, 3},
ContourStyle -> {{Thick, ColorData[1, 1]}, {Thick,
  ColorData[1, 2]}}];
Show[Flatten[{lines, cp}], PlotRange -> 2]
</code>```

Body:
Let us suppose that the ellipses are given by $f_1(x,y)=c_1$ and $f_2(x,y)=c2$.  We seek points $(x_1,y_1)$ and $(x_2,y_2)$ such that the lines through those points are mutually tangent to the ellipsies.  Clearly, $(x_1,y_1)$ must satisfy $f_1(x_1,y_2)=c_1$ and similiarly for the second point.  Furthermore, the gradients of $f_1$ and $f_2$ must be parallel at their respective points.  This yields $\nabla f_1(x_1,y_1) = \lambda \nabla f_2(x_2,y_2)$.  Finally, we need the line through the points to perpendicular to the gradient vectors.  This yields a total of five equations in the five unkowns $x_1$, $y_1$, $x_2$, $y_2$, and $\lambda$.

Let's apply this in the specific case
$$x^2 - x*y+y^2 = 4$$
and
$$2x^2+x*y+3y^2 = 1.$$

These ellipses do intersect, but I think that's no problem.  The equations are sufficiently complicated that I think I'll do this with Mathematica.

f1[x_, y_] = x^2 - x*y + y^2;
f2[x_, y_] = 2 x^2 + x*y + 3 y^2;
grad1[x_, y_] = {D[f1[x, y], x], D[f1[x, y], y]};
grad2[x_, y_] = {D[f2[x, y], x], D[f2[x, y], y]};
eqs = Flatten[{
  f1[x1, y1] == 1,
  f2[x2, y2] == 4,
  ({x1, y1} - {x2, y2}).grad1[x1, y1] == 0,
  Thread[grad1[x1, y1] == lambda*grad2[x2, y2]]
}];
pts = {{x1, y1}, {x2, y2}} /. NSolve[eqs, {x1, y1, x2, y2, lambda}]

(* Out: {
  {{1.11372, 0.820906}, {1.37585, 0.122754}},
  {{0.601823, 1.15435}, {-0.245408, 1.17882}},
  {{-1.11372, -0.820906}, {-1.37585, -0.122754}},
  {{-0.601823, -1.15435}, {0.245408, -1.17882}}} *)

You can use Solve rather than NSolve but the resulting expressions are quite complicated.  Let's visualize, as well.

line[{pt1_, pt2_}] := ParametricPlot[t*pt1 + (1 - t) pt2, {t, -2, 3},
  PlotRange -> 2, PlotStyle -> Black];
lines = line /@ pts;
cp = ContourPlot[{
  f1[x, y] == 1,
  f2[x, y] == 4
}, {x, -3, 3}, {y, -3, 3},
ContourStyle -> {{Thick, ColorData[1, 1]}, {Thick,
  ColorData[1, 2]}}];
Show[Flatten[{lines, cp}], PlotRange -> 2]


问:
Solving Underdetermined Non-Linear Equation of Probabilities
I'm trying to see if I can solve for g in the set of linear equations:
$$ga+gb = m$$ $$ga+gb+2gab=n$$
where g, a, and b are all probabilities (i.e. between 0 and 1). I know that the problem is underdetermined, but I was wondering if through some clever algebra it is possible to get a unique solution for g or if there is a counterexample showing why it is not (i.e. two sets of g, a, and b that produce identical m and n). I tried squaring equation 1 and combining it with equation 2 to get (ga-gb) on one side of an equation, but that didn't work and I tried generating counterexamples from
$$gab = g'a'b'$$
$$ga+gb = g'a'+g'b'$$
But I can't seem to either definitively prove that g is or isn't universally solvable. My intuition (because it is an underdetermined set of equations) leads me to believe that it isn't... Thanks so much!

答:
Yes, there are several solution with the same $m$ and $n$. For example, taking
$(g,a,b)=(\frac{1}{2},\frac{2}{3},0)$ and $(g,a,b)=(\frac{1}{3},1,0)$ the equations are both times satisfied with $m=n=\frac{1}{3}$.

For an example with $0<g,a,b<1$, take $(g,a,b)=(\frac{1}{3},\frac{1}{4},\frac{1}{5})$ and $(g,a,b)=(\frac{1}{4},\frac{9+\sqrt{21}}{30},\frac{9-\sqrt{21}}{30})$. Then both equations are satisfied with
$$
(m,n)=\left(\frac{3}{20},\frac{11}{60}\right).
$$


问:
What does the underscore positive sign mean outside a parentheses?
I know this might be a simple question but in the equation $\sum(|y_i − w^T x_i − b| − \epsilon)_+$ does the underscore positive sign mean $(|y_i − w^T x_i − b| − \epsilon)>=0$ ?

答:
It may mean the "positive part" of the expression.

Generally, one writes $f = f_+-f_-$, where $f_+$ is the positive part of $f$ and $f_-$ is the negative part of $f$. These are the values
$$f_+ = \tfrac12(|f| + f)$$
$$f_- = \tfrac12(|f| - f)$$
Note that these are both nonnegative.

If $f\geq 0$, then $f=f_+$.

If $f\leq0$, then $f = -f_-$.


问:
Derivative of a Complex Function as a $\mathbb{C}$-linear Transformation
I'm trying to show that the derivative of a differentiable complex function is a $\mathbb{C}$-linear mapping of $\mathbb{C}$ to itself, and since every $\mathbb{C}$-linear map is of the form
$$
\begin{bmatrix}
c_1 & c_2 \\
-c_2 & c_1
\end{bmatrix}
$$
then I can deduce Cauchy-Riemann equations.

In particular I think when $f$ is differentiable in a neighborhood of $a$, it's enough to show that $Df (a)$ respects scalar multiplication over $\mathbb{C}$ but my efforts always need Cauchy-Riemann equations to prove the desired which means it doesn't work since we want to show them.

In addition I mean a complex-valued function differentiable at neighborhood of $a$, if the limit $\lim_{h \rightarrow0}\frac{f(x+h)-f(x)}{h}$ exists for every $x$ in that neighborhood.

答:
Differentiability is a pointwise issue, you we only need to worry about what is happening at $a$. Suppose $f:U \to \Bbb{C}$ is a  function defined on an open subset of the complex plane, $a\in U$, and that $f'(a):= \lim\limits_{h\to 0}\frac{f(a+h)-f(a)}{h}$ exists. I hope you can convince yourself that this is equivalent to
\begin{align}
\lim_{h\to 0}\dfrac{f(a+h) - f(a) - f'(a)\cdot h}{h} = 0.
\end{align}
Or equivalently, we can put absolute values everywhere
\begin{align}
\lim_{h\to 0}\dfrac{\left|f(a+h) - f(a) - f'(a)\cdot h\right|}{|h|} = 0.
\end{align}
Look at what we have inside; the function $h\mapsto f'(a)\cdot h$ is a $\Bbb{C}$-linear transformation $\Bbb{C}\to \Bbb{C}$. This means exactly that $Df_a(h)= f'(a)\cdot h$, and hence that $Df_a:\Bbb{C}\to \Bbb{C}$ is $\Bbb{C}$-linear.

Finally, since every $\Bbb{C}$-linear transformation has that matrix representation you describe (relative to the  basis $\{1, i\}$ of the $2$-dimensional vector space $\Bbb{C}$ over the field $\Bbb{R}$), you can deduce the Cauchy-Riemann equations.


问:
Proving that $f(n) \geq n$
Here is the question in full:

Let $f : \mathbb{N} \rightarrow \mathbb{N}$ be any function such that for $i \in \mathbb{N}$, $f(i) < f(i+1)$. Prove that $lim_{n \to \infty} x_n = L$, then $lim_{n \to \infty} x_{f(n)} = L$.

Note that my definition of $\mathbb{N}$ excludes $0$.

Here was my solution:

First we show that $f(n) \geq n$ for all $n \in N$ using induction.

Base case: $n = 1$, so we see $f(1) \geq 1$ because if not, then it would be a contradiction (we would get $f(1) = 0 \text{ or something negative} \notin \mathbb{N}$).

Inductive step: Suppose that $f(n) \geq n$. Prove that $f(n+1) \geq n$.

By assumption $f(n+1) > f(n)$ and by hypothesis $f(n) \geq n$ so this tells us that $f(n+1) \geq n$. Hence $f(n) \geq n$.

By definition of limit, we know there exists $N \in \mathbb{N}$ such that $n \geq N$ implies $|x_n - L| < \epsilon$ for $\epsilon > 0$.

Because $f(n) \geq n$, then $f(n) \geq n \geq N$ which implies, by definition, that $|x_{f(n)} - L| < \epsilon$, hence showing that $lim_{n \to \infty} x_{f(n)} = L$. QED

My question is: is this proof correct?

答:
$f : \mathbb{N} \rightarrow \mathbb{N}$ such that $f(n)<f(n+1)$ means that $f$ is strictly increasing in $\mathbb{N}$ that is $f(n)\to\infty$ as $n\to\infty$.

In other words the sequence $\{x_{f(n)}\}$ is a subsequence of a convergent sequence, therefore it converges.

Title: Proving that $f(n) \geq n$


问:
Fundamental group of a sphere with a handle
Consider a sphere $S^2$ with a circle attached to it by a point. I want to calculate its fundamental group. I will use the following different version of Van Kampen theorem

If $X$ is path connected and $X=X_1 \cup X_2$ with $X_1$ path connected and $X_2$ simply connected and both open and such that $X_1 \cap X_2= X_0= A\cup B$ where $A,B$ are disjoint and both open and simply connected. Then $\pi_1(X)= \pi(X_1)* \mathbb{Z}$

Then what I can do is cut a line from the handle to obtain $X_1$ and take a longer line as $X_2.$

In this way the intersection is given by two lines which are both simply connected.

Moreover the fundamental group of $X_1$ is trivial because if I cut a line from the handle then I can rectract the handle to a point on the sphere which is contractible.

Thus the $\pi_1$ of the sphere with a handle must be $\mathbb{Z}.$

Does it work?

答:
There is no need to use Seifer-van-Kampen.

You just want to show that any path in $X$ is homotopic to a path just in the circle $S^1$ (which we assume to be attached to $S^2$ at a point $p$). So take a path $\gamma:[0,1]\to X$. Then $\gamma^{-1}(X\setminus S^1)$ consists of finitely many open intervals. Consider such an interval $(a,b)\subset[0,1]$ that has $\gamma(a)=\gamma(b)=p$ and $\gamma([a,b])\subset X$. We know that $S^2$ is simply-connected, so $\gamma|_{[a,b]}$ is homotopic to the constant path at $p$. So $\gamma$ is homotopic to a path $\tilde{\gamma}$ with $\tilde{\gamma}^{-1}(X\setminus S^1)$ consisting of one less interval than $\gamma^{-1}(X\setminus S^1)$. By induction $\gamma$ is thus homotopic to a path entirely in $S^1$.

So we conclude $\pi_1(X)\cong\pi_1(S^1)\cong\mathbb Z$.

In some sense we used exactly the idea of the proof of Seifert-van-Kampen.


问:
Solving second order ODE with variable coefficients
How to solve this second order ODE

$$\frac{d^2 u}{dx^2} - k^2\frac{\gamma-3x^2}{\gamma x-x^3}\frac{d u}{d x} + k^3u=0 $$
Where $k, \gamma$ are real constants and $\gamma\neq 0$.

I solved this using frobenius method since it has three regular singular points.
However it gets complicated and wanted to know if there is any other way to solve it!

https://en.m.wikipedia.org/wiki/Frobenius_method

答:
Assume $k\neq0$ for the key cases

Case $1$: $\gamma=0$

Then $\dfrac{d^2u}{dx^2}-\dfrac{3k^2}{x}\dfrac{du}{dx}+k^3u=0$

$x\dfrac{d^2u}{dx^2}-3k^2\dfrac{du}{dx}+k^3xu=0$

Which can reduce to Bessel ODE

Case $2$: $\gamma\neq0$

Then $\dfrac{d^2u}{dx^2}-k^2\dfrac{\gamma-3x^2}{\gamma x-x^3}\dfrac{du}{dx}+k^3u=0$

$\dfrac{d^2u}{dx^2}-\dfrac{k^2(3x^2-\gamma)}{x(x^2-\gamma)}\dfrac{du}{dx}+k^3u=0$

Let $r=x^2$ ,

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

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

$\therefore4r\dfrac{d^2u}{dr^2}+2\dfrac{du}{dr}-\dfrac{2k^2(3r-\gamma)}{r-\gamma}\dfrac{du}{dr}+k^3u=0$

$4r(r-\gamma)\dfrac{d^2u}{dr^2}+(2(r-\gamma)-2k^2(3r-\gamma)\dfrac{du}{dr}+k^3(r-\gamma)u=0$

$4r(r-\gamma)\dfrac{d^2u}{dr^2}-(2(6k^2-1)r-(k^2-1)\gamma))\dfrac{du}{dr}+k^3(r-\gamma)u=0$

Which relates to Heun's Confluent Equation.


问:
An integer $n \geq 2$ is called square-positive- proof?
An integer $n \geq 2$ is called square-positive if there are $n$ consecutive positive integers whose sum is a square. Determine the first four square-positive integers.

So I have found the first four square-positive numbers, but I need to prove that why it $4$ is not a square-positive number and I also need to write a general formula for determining whether a number is square-positive or not. I have tried to write the sum of consecutive positive integers like this $a + a +1 + a + 2 + a+3 \dots a - 1$ and I wrote it like this for all numbers, and part of the proof for why $4$ isn't a square-positive number is that $4a + 6$ is not divisible with $4$. But I haven't got so far.

Here is my answer:

2 : 4 + 5 = 9  which is 3^2

3 : 2 + 3 + 4 = 9 which is 3^2

5: 18 + 19 + 20 + 21 + 22 = 100  which is 10^2

6: 35 + 36 + 37 + 38 + 39 + 40 = 225  which is 15^2

Interesting fact is that for all odd numbers and some even numbers like 6 and 10, you can find out which number is the first (the one you start with and then go forward here like 3, 2, 18 and 35) using this formula :

(I show it in an example because I still can't write it algebraically):

For example: the sum of 95 subsequent numbers is 5n + 10

(10^2 - 10) /5 = 18

So your first number is 18

And if you keep adding, 18 + 19 + 20 + 21 + 22 you get 100 which is 10^2, the same number you squared.

答:
There is a pretty simple test to show that $n=4q$, where q is odd, cannot be a solution.

The sum of the first $a$ integers is:
$$\frac{a(a+1)}{2}$$
so the sum of $n$ consecutive integers starting at $a$ is:
$$\frac{(a+n)(a+n-1)}{2}-\frac{a(a-1)}{2} = \frac{n(2a+n-1)}{2}$$
since this number must be equal to a perfect square:
$$\frac{n(2a+n-1)}{2}=k^2$$
substitute $n=4q$ for some odd integer $q$:
$$2q(2a-1+4q)=k^2$$
it should be apparent that $k$ must be an even integer, therefore substitute $k=2m$:
$$2(2a-1+4q)q=4m^2$$
$$(2a-1+4q)q=2m^2$$
The numbers $2a-1+4q$ and $q$ must be odd, but the RHS must be even, therefore it is not possible for this equation to be solved over the integers, and so there is no solution for $n=4q$.

I know that this isn't a full proof, but I am still working on it


问:
Proving that two matrices are congruent, in order to establish a rank and signature result.
I'm reading Further Linear Algebra by by T. S. Blyth, E. F. Robertson.  I'm somewhat stuck on Problem 9.14.  In particular, they want us to establish that the quadratic form
$$
Q(x_1,\ldots,x_n) = \sum_{r < s} (x_r - x_s)^2
$$
has both rank and signature equal to $n-1$.  The hint they give us is to show that the matrix for $Q$ with respect to the standard basis has $n-1$ on the diagonal, and $-1$ elsewhere.  Then they ask us to show that this matrix is congruent to the diagonal matrix with entries $n-1, n-2, \ldots, 2, 1, 0$. I'm assuming there is some clever basis transformation that gives us $P^t A P = \text{diag}(n-1, n-2, \ldots, 2, 1, 0)$, but I can't figure it out.  Any help?

答:
Thanks all!  Just by brute force algebra, I figured out that if you want to turn
$\sum_{i=1}^n (n-1)x_i^2 - 2 \sum_{1 \leq i < j \leq n}x_ix_j$ into $(n-1)y_1^2+(n-2)y_2^2+\cdots + 1 \cdot y_{n-1}^2 + 0\cdot y_n^2$, then using $y_k = \sqrt{\frac{n}{n-k+1}}\left(x_k - \frac{x_{k+1}+\cdots+x_n}{n-k}\right)$ for $k = 1, \ldots, n-1$ and $y_n = x_n$, does the trick.  Again, I don't know what the pedagogical purpose of this exercise was.


问:
Probability that the sum of three (or any number) of independent but not identical exponential RV is greater than a number?

Suppose $X_1$, $X_2$ are i.i.d. exponential with rate $\lambda = 3$ and  $Y_1, Y_2, Y_3, Y_4$ are i.i.d. exponential with $\lambda = 4$. Find $P(X1+2X2+Y1>3)$.

This was only part of the question, but this was a part that I had no idea how to do. Obviously, $X_1 + 2X_2 + Y_1$ are all independent so this ends up being an $\exp(3)+\exp(3/2)+\exp(4)$, and we are trying to find the probability that the sum of three independent exponential variables is greater than $3$. My initial thought was to use some kind of order statistics, where I find that the minimum of the three variables is greater than $3$. However, I then realized that was completely wrong since it is the sum of three exponential random variables. Does anybody know how to proceed with such a question?

答:
If $X\sim\mathsf{Expo}(\lambda)$ and $Y\sim\mathsf{Expo}(\mu)$ are independent where $\lambda\ne\mu$, then the density of $Z:=X+Y$ can be found by convolution:
\begin{align}
f_Z(t) &= \int_{\mathbb R}f_X(s)f_Y(t-s)\ \mathsf ds\\
&= \int_0^t \lambda e^{-\lambda s}\mu e^{-\mu(t-s)}\ \mathsf ds\\
&= \lambda\mu e^{-\mu t} \int_0^t e^{-(\lambda +\mu)s}\ \mathsf ds\\
&= \frac{\lambda\mu}{\mu-\lambda} (e^{-\lambda t}-e^{-\mu t}).
\end{align}
Moreover, if $c>0$ then for all $t>0$ we have
$$
\{cX\leqslant t\} = \left\{X\leqslant \frac tc\right\},
$$
and a quick computation shows that $cX$ is exponentially distributed with rate $\frac\lambda c$. So we can use convolution to find the distribution of $Z:=X_1+2X_2+Y_1$; assuming $\mathbb E[X_1] = \frac1\lambda$ and $\mathbb E[Y_1]=\frac1\mu$, we find that
$$
f_Z(z) = \frac{\lambda  \mu  e^{-(\lambda+\mu)t} \left(\lambda  e^{\mu  t}-2 \lambda  e^{\frac{\lambda  t}{2}+\mu  t}+2 \mu  e^{\frac{\lambda  t}{2}+\mu  t}+\lambda  e^{\lambda  t}-2 \mu  e^{\mu  t}\right)}{(\lambda -2 \mu ) (\lambda -\mu )}.
$$
From this we find
$$
\mathbb P(Z>3) = \int_3^\infty f_Z(t)\ \mathsf dt = \frac{e^{-3 (\lambda +\mu )} \left(e^{3 \lambda } \lambda ^2+e^{3 \mu } \mu  \left(-4 e^{\frac{3 \lambda }{2}} (\lambda -\mu )+\lambda -2 \mu \right)\right)}{(\lambda -2 \mu ) (\lambda -\mu )}.
$$
Substituting $\lambda=3$ and $\mu=4$, the above is
$$
\frac15 e^{-12}\left(4 e^3 \left(4 e^{9/2}-5\right)+9\right)\approx 0.0350662.
$$
I sincerely hope your instructor does not expect you to do computations like these by hand...


问:
Proving a f is continuous at 1/3
Show that $f(x) = \frac{1}{5x}$ is continuous at $x = \frac{1}{3}$. I have to use an $\varepsilon - \delta$ proof? I am having trouble choosing my delta because I am confused as to how $|\frac{1}{x} - \frac13|$ can get to $|x - \frac{1}{3}|$?

答:
A function is continuous at a given point $x_0$ if for every $ϵ>0$ you can find some $δ>0$ such that $|x−x_0|<δ⟹|f(x)−f(x_0)|<ϵ.$

What is useful with these problems is to "solve them backwards". Start with
$$\left|\frac{1}{5x} - \frac{1}{15}\right| < \epsilon$$
What happens when $0<x<3$?
$$\frac{1}{5x} - \frac{1}{15} < \epsilon$$
$$\frac{1}{x} - \frac{1}{3} < 5 \epsilon$$
$$\frac{1}{x} < \frac{1}{3} + 5 \epsilon = \frac{1+15 \epsilon}{3}$$
$$x > \frac{3}{1+15\epsilon}$$

What happens when $x>3?$
$$-\frac{1}{5x} + \frac{1}{15} < \epsilon$$
$$-\frac{1}{x} + \frac{1}{3} < 5\epsilon$$
$$  \frac{1}{3} - 5\epsilon <\frac{1}{x}$$
Now we're gonna want to guarantee that the left side of this inequality is positive, so choose $\epsilon <\frac{1}{15}.$
$$  \frac{1-15\epsilon}{3} <\frac{1}{x}$$
$$  \frac{3}{1-15\epsilon} >x$$.

In conclusion, we see that when $\epsilon <\frac{1}{15}$ and $x>0,$ then $x$ lies in the range
$$\frac{3}{1+15\epsilon} < x < \frac{3}{1-15\epsilon}.$$
So
$$\frac{3}{1+15\epsilon} -3 < x -3 < \frac{3}{1-15\epsilon} -3.$$
Now think about how you can choose $\delta$ so that $|x-3| < \delta$ implies the above condition. (Hint: just make sure $\delta$ is the minimum of the two). Do you need to worry about $x<0$? (Hint: you could just choose $\delta$ to guarantee this doesn't happen).

Title: Proving a f is continuous at 1/3


问:
Does there exist a countably infinite extension ring of the rational numbers?
This question came up as I was thinking about whether there could be a ring with the conventional addition and multiplication defined on it that contains all of $\mathbb{Q}$ and at least one other element and is countable. Intuitively, a set that is related to $\mathbb{Q}$ in the same way $\mathbb{Q}$ is related to $\mathbb{Z}$. I tried to attack the question using the fact that if a ring $B$ contains all of $A$ and at least one other element $c$, it must contain every element of the form $a_0 +a_1c +...  +a_nc^n$. i.e. every element of the polynomial ring $A[c]$. But unfortunately couldn't get anywhere that way. There is some related stuff on the internet. Notably, this and this (the second one seems to be evidence for the existence of such a set). The first link, however, refers to power sets which are different to 'polynomial sets' at least in the fact that a 'polynomial set' doesn't contain infinite polynomials (sequences of coefficients). Cantor's diagonal argument can't be applied for the same reason.

答:
Yes, $\Bbb Q[x]$, the ring of polynomials. If you want to have something "nicer", look at $\Bbb Q[\sqrt 2]$, or perhaps $\Bbb Q^{\text{alg}}$, the field of all algebraic numbers (which is a countable subfield of $\Bbb C$), or its intersection with $\Bbb R$.

All of these are countable. $\Bbb Q[x]$ is countable because the set of finite sequences from a countable set is itself countable, and $\Bbb Q[\sqrt 2]$ can be either seen as a quotient of $\Bbb Q[x]$ (so it is countable as the image of a countable set), or directly as $\{p+q\sqrt 2\mid p,q\in\Bbb Q\}\cong\Bbb Q\times\Bbb Q$. And finally, $\Bbb Q^{\text{alg}}$ is countable because each number is the root of a polynomial in $\Bbb Q$, and each polynomial has finitely many roots, so it is a countable union of finite sets, and therefore countable.

Similarly, you can consider $\Bbb Q(\pi)$, or even just adjoin any new symbol with some equation, $\langle\Bbb Q,x\mid x+x=0, x\neq 0\rangle$, so we add an element of additive order $2$, and the ring is no longer a field.


问:
Find $a$ and $b$ such that probability is the biggest
Basketball player A hit the basket with probability $\frac{1}{2}$, and B with $\frac{1}{10}$. Player A can shoot $a$ times, and B $b$ times, $a+b=20$. Find $a$ and $b$ such that probability to make at least one point each is the biggest.

I just begging my course in probability and I have a problem with this. I tried to find max value of $(1-\frac{1}{2^{a}})(1-\frac{9^{20-a}}{10^{20-a}})$(complement of event that some of players doesn't make a point), and get solution a=1, b=19, but solution in book is a=5,b=15.

Sorry for my bad English, any help?

答:
Your formula looks completely correct.

I did the lazy person approach and calculated the probabilities for $a=0, \ldots, 20$. I got a maximal probability at $a=5$. So you must have found the maximum incorrectly, with numbers this small its worth just having a computer run through all 21 possibilities.


问:
Randomly restarted Lévy process is again a Lévy process
Let

$(\Omega,\mathcal A,\operatorname P)$ be a probability space;

$(\mathcal F_t)_{t\ge0}$ be a filtration on $(\Omega,\mathcal A,\operatorname P)$;

$E$ be a $\mathbb R$-Banach space;

$(L_t)_{t\ge0}$ be a $E$-valued process on $(\Omega,\mathcal A,\operatorname P)$.

Remember that $L$ is called $\mathcal F$-Lévy if

$L$ is $\mathcal F$-adapted;

$L_0=0$;

$L_{s+t}-L_s$ and $\mathcal F_s$ are independent for all $t\ge s\ge0$;

$L_{s+t}-L_s\sim L_s$ for all $t\ge s\ge0$.

Assume $L$ is $\mathcal F$-Lévy. Let $\tau$ be a $\mathcal F$-stopping time, $\tilde\Omega:=\{\tau<\infty\}$, $\tilde{\mathcal A}:=\left.\mathcal A\right|_{\tilde\Omega}$, $\tilde{\operatorname P}:=\left.\operatorname P\right|_{\tilde\Omega}$, $$\mathcal G_t:=\mathcal F_{\tau+t}\;\;\;\text{for }t\ge0$$ and $$X_t(\omega):=L_{\tau+t}(\omega)-L_\tau(\omega)\;\;\;\text{for }(\omega,t)\in\tilde\Omega\times[0,\infty).$$

How can we show that $X$ is a $\mathcal G$-Lévy process on $(\tilde\Omega,\tilde{\mathcal A},\tilde{\operatorname P})$?

(1.) and (2.) are clearly trivial.

I think the easiest way to show (3.) and (4.) is to approximate $\tau$ in a suitable way. So, let's first assume that $\tau$ is finite and $k:=\left|\tau(\Omega)\right|\in\mathbb N$. Then, $$\tau(\Omega)=\{t_1,\ldots,t_k\}\tag1$$ for some $0\le t_1<\cdots<t_k$. Since $\{\tau=t_i\}\in\mathcal F_{t_i}\subseteq\mathcal F_{t_i+s}$, we easily obtain \begin{equation}\begin{split}\operatorname P\left[X_{s+t}-X_s\in B\right]&=\sum_{i=1}^k\operatorname P\left[\tau=t_i,L_{t_i+s+t}-L_{t_i+s}\in B\right]\\&\sum_{i=1}^k\operatorname P\left[\tau=t_i\right]\operatorname P\left[L_{t_i+s+t}-L_{t_i+s}\in B\right]\\&\operatorname P\left[L_t\in B\right]\sum_{i=1}^k\operatorname P\left[\tau=t_i\right]=\operatorname P\left[L_t\in B\right]\end{split}\tag2\end{equation} for all $B\in\mathcal B(E)$ and $s,t\ge0$; which is (4.).

Analogously¹, since $L_{t_i+s+t}-L_{t_i+s}$ and $\mathcal F_{t_i+s}$ are independent and $\{\tau=t_i\}\in\mathcal F_{t_i}\subseteq\mathcal F_{t_i+s}$ for all $i\in\{1,\ldots,k\}$, \begin{equation}\begin{split}\operatorname P\left[X_{s+t}-X_s\in B\mid\mathcal G_s\right]&=\sum_{i=1}^k\operatorname P\left[\tau=t_i,L_{t_i+s+t}-L_{t_i+s}\in B\mid\mathcal F_{\tau+s}\right]\\&=\sum_{i=1}^k1_{\left\{\:\tau\:=\:t_i\:\right\}}\operatorname P\left[L_{t_i+s+t}-L_{t_i+s}\in B\mid\mathcal F_{t_i+s}\right]\\&=\sum_{i=1}^k1_{\left\{\:\tau\:=\:t_i\:\right\}}\operatorname P\left[L_{t_i+s+t}-L_{t_i+s}\in B\right]\\&=\sum_{i=1}^k\operatorname P\left[\tau=t_i,L_{t_i+s+t}-L_{t_i+s}\in B\right]\\&=\operatorname P\left[X_{s+t}-X_s\in B\right]\end{split}\tag3\end{equation} almost surely for all $B\in\mathcal B(E)$; which is (3.).

Can we derive the general case by approximating $\tau$ with $\mathcal F$-stopping times of the formerly considered form?

EDIT 1: Both, $(2)$ and $(3)$, should hold line by line when $\tau$ is finite and $\tau(\Omega)$ is countable. We simply need to use the sums $\sum_{r\in\tau(\Omega)}\operatorname P\left[\tau=r,L_{r+s+t}-L_{r+s}\in B\right]$ and $\sum_{r\in\tau(\Omega)}\operatorname P\left[\tau=r,L_{r+s+t}-L_{r+s}\in B\mid\mathcal F_{\tau+s}\right]$ instead.

EDIT 2: Now assume $\tau$ is only finite. Let $\tau_n$ be a $\mathcal F$-stopping time² on $(\Omega,\mathcal A,\operatorname P)$ for $n\in\mathbb N$, such that $\tau_n(\Omega)$ is countable and $$\tau_n\ge\tau_{n+1}\tag4$$ for all $n\in\mathbb N$ and $$\tau_n\xrightarrow{n\to\infty}\tau\tag5.$$

Let $X^{(n)}_t:=L_{\tau_n+t}-L_{\tau}$ and $\mathcal G^{(n)}_t:=\mathcal F_{\tau_n+t}$ for $t\ge0$. By $(4)$, $$\mathcal G^{(n)}_t\supseteq\mathcal G^{(n+1)}_t\;\;\;\text{for all }t\ge0\tag6$$ for all $n\in\mathbb N$. Now assume $L$ is right-continuous. Then, by $(4)$ and $(5)$, $$X^{(n)}_t\xrightarrow{n\to\infty}X_t\;\;\;\text{for all }t\ge0\tag7.$$

Let $B\in\mathcal B(E)$ and $s,t\ge0$. By what we've already shown, $$\operatorname P\left[X^{(n)}_{s+t}-X^{(n)}_s\in B\mid\mathcal G^{(n)}_s\right]=\operatorname P\left[X^{(n)}_{s+t}-X^{(n)}_s\in B\right]\tag8$$

Using $(7)$ and the dominated convergence theorem, the right-hand side of $(8)$ should converge to $\operatorname P\left[X_{s+t}-X_s\in B\right]$.

What can we do with the left-hand side? Maybe $(6)$ is the crucial ingredient which allows us to obtain convergence to $\operatorname P\left[X_{s+t}-X_s\in B\mid\mathcal G_s\right]$ as desired ...

Now, in order to conclude for general finite $\tau$, I guess we need to assume right-continuity, but how do we then need to argue exactly?

Remark: I'm also unsure whether we need to impose further assumptions on $(\mathcal F_t)_{t\ge0}$ like completeness or right-continuity.

¹ If $Y\in\mathcal L^1(\operatorname P;E)$, then $$\operatorname E\left[1_{\left\{\:\tau\:=\:t\:\right\}}Y\mid\mathcal F_\tau\right]=1_{\left\{\:\tau\:=\:t\:\right\}}\operatorname E\left[X\mid\mathcal F_t\right]\;\;\;\text{almost surely}.$$

² We could, for example, take

答:
First of all, note that, say, $Y_n \to Y$ almost surely does not imply $\mathbb{P}(Y_n \in B) \to \mathbb{P}(Y \in B)$. The latter holds only if $B$ is such that $\mathbb{P}(Y \in \partial B)=0$. You can easily see this issue if you consider for instance $Y_n := \frac{1}{n}$ and $B=\{0\}$. As a consequence, we cannot take, in general, the limit on the right-hand side of (8).

Let $f$ be a bounded continuous function. Since $X_t^{(n)} \to X_t$, we have, by the dominated convergence theorem,

$$\mathbb{E}(f(X_{s+t}-X_s) \mid \mathcal{G}_s) = \lim_{n \to \infty} \mathbb{E}(f(X_{s+t}^{(n)}-X_s^{(n)}) \mid \mathcal{G}_s). \tag{1}$$

From $\tau_n \geq \tau$ we see that $\mathcal{G}_s^{(n)} = \mathcal{F}_{\tau_n+s} \supseteq \mathcal{F}_{\tau+s}=\mathcal{G}_s$. Consequently, by the tower property of conditional expectation,

$$ \mathbb{E}(f(X_{s+t}^{(n)}-X_s^{(n)}) \mid \mathcal{G}_s) = \mathbb{E} \bigg [ \mathbb{E}(f(X_{s+t}^{(n)}-X_s^{(n)}) \mid \mathcal{G}_s^{(n)}) \mid \mathcal{G}_s \bigg].$$

By your earlier considerations for stopping times taking only finitely many values, we can compute the right-hand side:

\begin{align*} \mathbb{E}(f(X_{s+t}^{(n)}-X_s^{(n)}) \mid \mathcal{G}_s) &= \mathbb{E} \bigg[ \mathbb{E}(f(X_{s+t}^{(n)}-X_s^{(n)})) \mid \mathcal{G}_s\bigg] \\ &= \mathbb{E}(f(X_{s+t}^{(n)}-X_s^{(n)})). \end{align*}

Plugging this into $(1)$ and using once more the dominated convergence theorem and the right-continuity of the sample paths, we arrive at

$$\mathbb{E}(f(X_{s+t}-X_s) \mid \mathcal{G}_s) = \mathbb{E}(f(X_{t+s}-X_s)),$$

which should be all you need. (At least for the case $\mathbb{P}(\tau<\infty)=1$, which you were considering.)

Proving the assertion for stopping times which may take the value $+\infty$ requires some more work. Define $\tau_n := \min\{\tau \wedge n\}$ and denote by $X^{(n)}$ the corresponding restarted Lévy process with filtration $\mathcal{G}^{(n)}$. Then, by the previous step of the proof,

$$\mathbb{E}(f(X_{t+s}^{(n)}-f(X_s^{(n)}) \mid \mathcal{G}_s^{(n)}) = \mathbb{E}(f(X_{t+s}^{(n)}-X_s^{(n)}))= \mathbb{E}(f(L_t)).$$

Since $\{\tau \leq n\} \in \mathcal{F}_{\tau} \cap \mathcal{F}_n = \mathcal{F}_{\tau \wedge n} \subseteq \mathcal{G}_s^{(n)}$, we can multiply both sides by $1_{\{\tau \leq n\}}$ to obtain that

$$\mathbb{E}(1_{\{\tau \leq n\}} f(X_{t+s}^{(n)}-f(X_s^{(n)}) \mid \mathcal{G}_s^{(n)})=1_{\{\tau \leq n\}} \mathbb{E}(f(L_t)). \tag{2}$$

We would like to let $n \to \infty$. To this end, we first show that

$$\mathbb{E}(1_{\{\tau \leq n\}} f(X_{t+s}^{(n)}-f(X_s^{(n)}) \mid \mathcal{G}_s^{(n)}) \xrightarrow[]{L^1} \mathbb{E}(1_{\{\tau<\infty\}} f(X_{t+s}-X_s) \mid \mathcal{G}_s). \tag{3}$$

By the triangle inequality,

\begin{align*} &\mathbb{E}\bigg|\mathbb{E}(1_{\{\tau \leq n\}} f(X_{t+s}^{(n)}-f(X_s^{(n)}) \mid \mathcal{G}_s^{(n)})- \mathbb{E}(1_{\{\tau <\infty\}} f(X_{t+s}-X_s) \mid \mathcal{G}_s)\bigg| \\&\leq \mathbb{E}\bigg|\mathbb{E}(1_{\{\tau \leq n\}} f(X_{t+s}^{(n)}-f(X_s^{(n)}) \mid \mathcal{G}_s^{(n)})- \mathbb{E}(1_{\{\tau<\infty\}} f(X_{t+s}-X_s) \mid \mathcal{G}_s^{(n)})\bigg|\\ &\quad +\mathbb{E}\bigg| \mathbb{E}(1_{\{\tau<\infty\}} f(X_{t+s}-X_s) \mid \mathcal{G}_s^{(n)})- \mathbb{E}(1_{\{\tau<\infty\}}f(X_{t+s}-X_s) \mid \mathcal{G}_s)\bigg| \\ &=: \Delta_1+\Delta_2. \end{align*}

For the first term we see, using the tower property,

$$\Delta_1 \leq \mathbb{E}(|1_{\{\tau \leq n\}} f(X_{t+s}^{(n)}-X_s^{(n)})-1_{\{\tau<\infty\}} f(X_{t+s}-X_s)|).$$

If $\omega \in \{\tau<\infty\}$, then $X_{r}^{(n)}(\omega)=X_r(\omega)$ for $n=n(\omega)$ sufficiently large, and so the dominated convergence theorem yields $I_1 \to 0$ as $n \to \infty$. On the other hand, $\mathcal{G}_s = \sigma(\bigcup_n \mathcal{G}_s^{(n)})$, see this question, and so Lévy's upwards theorem yields $I_2 \to 0$ as $n \to \infty$. This then proves $(3)$. Because of $(3)$, we can choose an almost surely convergent subsequence

$$\mathbb{E}(1_{\{\tau \leq n_k\}} f(X_{t+s}^{(n_k)}-f(X_s^{(n_k)}) \mid \mathcal{G}_s^{(n_k)}) \xrightarrow[]{\text{a.s.}} \mathbb{E}(1_{\{\tau<\infty\}} f(X_{t+s}-X_s) \mid \mathcal{G}_s). \tag{4}$$

Letting $n \to \infty$ in (2) now gives

$$ \mathbb{E}(1_{\{\tau<\infty\}} f(X_{t+s}-X_s) \mid \mathcal{G}_s) = 1_{\{\tau<\infty\}} \mathbb{E}(f(L_t)). \tag{5}$$

Taking expectation on both sides yields

$$ \mathbb{E}(1_{\{\tau<\infty\}} f(X_{t+s}-X_s)) = \mathbb{P}(\tau<\infty) \mathbb{E}(f(L_t)),$$

i.e.

$$\mathbb{E}(f(L_t)) = \frac{1}{\mathbb{P}(\tau<\infty)}  \mathbb{E}(1_{\{\tau<\infty\}} f(X_{t+s}-X_s)).$$

Plugging this into $(5)$ shows that

$$ \mathbb{E}(1_{\{\tau<\infty\}} f(X_{t+s}-X_s) \mid \mathcal{G}_s) = 1_{\{\tau<\infty\}} \frac{1}{\mathbb{P}(\tau<\infty)}  \mathbb{E}(1_{\{\tau<\infty\}} f(X_{t+s}-X_s)).$$

If we define a probability measure $\tilde{P}(A) := \frac{\mathbb{P}(A \cap \{\tau<\infty\})}{\mathbb{P}(\tau<\infty)}$ on $\tilde{\Omega} := \{\tau<\infty\}$, then this is equivalent to

$$\mathbb{E}_{\tilde{\mathbb{P}}}(f(X_{t+s}-X_s) \mid \mathcal{G}_s) = \mathbb{E}_{\tilde{\mathbb{P}}}(f(X_{t+s}-X_s)).$$


问:
Smart method for computing the determinant of the matrix defined by $A_{ii}=a+b$ and $A_{ij}=a$?
We are given the following $n\times n$ matrix, whose terms are:

$a+b$ for the terms in the diagonal ($i=j$).

$a$ for the rest of the elements.

I know we can simply compute this and try to find a pattern, but is there a faster way? Maybe using some properties of the determinant of a matrix?

答:
Let  $\mathbf e_i$ denote the $i$th standard base vector and $\mathbf j=\sum\mathbf e_i$ denote the "all-ones" vector.

Clearly, $A\mathbf j=(b+na)\mathbf j$, so $\mathbf j$ is an eigenvector of eigenvalue $b+na$.

Also $A\mathbf e_i=b\mathbf e_i+a\mathbf j$ so that for $2\le i\le n$, $A(\mathbf e_i-\mathbf e_1)=b(\mathbf e_i-\mathbf e_1)$, i.e., we know $n-1$ linearly independent eigenvectors of eigenvalue $b$.

All in all, we have found a basis consisting of eigenvectors an know their eigenvalues, and can
conclude that $$\det A=(b+na)\cdot b^{n-1}.$$


问:
Prove that all meromorphic bijections are Möbius transformations
With the notation $\overline{\mathbb C}= \mathbb C\cup \{\infty\}$, I want to show that every meromorphic bijection $f:\overline{\mathbb C}\rightarrow \overline{\mathbb C}$ is a Möbius transformation.  A Mobius transformation is any

$$ \frac{az+b}{cz+d} , \quad a,b,c,d\in\mathbb C$$

with $ad-bc\ne 0$.  I have that the meromorphic functions in $\overline {\mathbb C}$ are the rational functions.  I can see that, because $f$ is rational then it has a partial fraction decomposition.  But so does $z^2$ and that's a rational function ... so I can guess that somehow the assumption of being bijective comes in right about now.  And somehow it should tell us that the maximum degree of the numerator and denominator is 1.

I'm also guessing that somehow I should first inspect the poles and argue that the largest order of any pole is 1, and then make some kind of argument about $f(1/z)$.

So ok, to make an attempt on some of these guesses: $f$ has the form

$$ f(z) = {P(z)\over Q(z)} $$

If $\deg P \geq \deg Q$ then we can perform long division  to separate this into

$$ {P(z)\over Q(z)} = R(z)+{S(z)\over Q(z)} $$

Where $R$ and $S$ are polynomials and $\deg S < \deg Q$.  I'm not sure why the degree of $R$ should have to be 0 but it seems like it must, because I'm guessing we'll turn the ${S(z)\over Q(z)}$ into a sum of polynomials in $1/(z-z_k)$ for the $z_k$ roots of $Q$.  Then with each of these being a Mobius transformation, we want to argue that their sum is a Möbius transformation ...

Well, I'm actually now unsure that their sum is a Möbius transformation now that I think of it.  Maybe this whole path was a giant dead-end.  Should I be trying to do something entirely different from partial fractions?

答: You can compose any such bijection $\varphi:\overline{\mathbb C}\to\overline{\mathbb C}$ with a Möbius transformation $\mu$ to get a meromorphic bijection mapping $\infty\mapsto\infty$. Restricting to $\mathbb C$, this is an entire bijection $\mathbb C\to\mathbb C$, with a pole at $\infty$. The pole at $\infty$ makes it a polynomial, and bijective polynomials are exactly the linear ones, which are Möbius transformations. So $\varphi\circ\mu$ is a Möbius transformation. And since Möbius transformations are a group under composition, $\varphi$ is one, too.


问:
What is the "dimension" of a locally ringed space?
Let $(X,\mathscr{O}_X)$ be a locally ringed space. If it is a scheme, the natural notion of dimension is the dimension of the subjacent topological space (the size of the biggest chain of irreducible closed subsets). But if $X$ is a manifold, I think that the natural notion of dimension is perhaps the dimension of the Zariski tangent space.

Is there a "good" notion of dimension in locally ringed spaces? If so, how does this notion relates to the dimension of the subjacent topological space, the dimension of the Zariski tangent space and the Krull dimension of the stalks?

答:
I think the answer is no, because "all locally ringed spaces" is a really broad assortment of objects - it contains too many different types of things for only one concept to really get everything done. This isn't really a proof, though, just an explanation of how most common options do not work. (If you can come up with some axioms you'd want dimension to satisfy, perhaps you can edit them in to your post and we'll see about getting a real proof together.)

Purely algebraic notions defined only in terms of data from the local rings can't be compatible with our topological expectations: given any local ring $R$, the one-point space with $R$ as its sheaf of rings gives a locally ringed space. In any morally upright notion of dimension, this space would have dimension zero: it's a point! But this means that any invariant of local rings you pick to give "dimension" should return zero on all local rings, which isn't great.

Purely topological notions don't work across all of things that can be considered locally ringed spaces either. Krull dimension doesn't work for Hausdorff spaces as every irreducible closed set is a singleton, so we always get dimension 0. All the usual notions of topological dimension (Lebesgue covering, small inductive, large inductive) all fail on schemes with a single closed point because any open cover of such a scheme must contain the whole space. This immediately implies that any such notion of dimension must return zero.


问:
Given $\sum_{n=0}^{\infty }x^n$ converges to $(1-x)^{-1}$ on $(-1,1)$, find a power series that converges to $(1-x)^{-2}$ on $(-1,1)$.
How would I go about solving this problem thus far I was thinking to use the Differentiation Theorem or I may use the Power Series Theorem.

I am trying to show the first few non zero terms but not sure if this is the right approach to this question.

答:
As the derivative of $\frac{1}{1-x}$ is $\frac{1}{(1-x)^2}$

if we sum the derivatives of $x^n$, that is $nx^{n-1}$ we get
$$\sum _{n=1}^{\infty } n x^{n-1}=\frac{1}{(1-x)^2}$$


问:
Uniform continuity of functions and restrictions.
Hi I have this exercise on my real analysis course, I haven´t seen topology course:

Let $A_{i} \subset \mathbb{R}$ an interval for all $i \in \mathbb{N}$. Is it true or false?

a) If $f_{1}: A_{1} \longrightarrow \mathbb{R}$ and $f_{2}: A_{2} \longrightarrow \mathbb{R}$ are uniformly continous and $f: A_{1} \cup A_{2} \longrightarrow \mathbb{R}$ is continous, then $f$ is uniformly continous.

b) Let $f_{i}: A_{i} \longrightarrow \mathbb{R}$ uniformly continous for all $i \in \mathbb{N}$ and $f: \displaystyle{\bigcup_{i \in \mathbb{N}}{A_{i}}} \longrightarrow \mathbb{R}$ continous, then $f$ is uniformly continous.

Since they seem very similar I suppose I need induction on $i \in \mathbb{N}^{\geq 2}$ for both of them. However I am not abble to prove any case. For the fisrst $i = 2$ I do not have any theorem to use or a definition wich implies union of intervals on my textbook. Do you have any hint?

答: The first statement is false. Take $A_1=[0,1)$, $A_2=[1,2]$, $f_1(x)=0$, and $f_2(x)=1$. Then $f_1$ and $f_2$ are uniformly continuous, but $f$ isn't even continuous.


问:
Does the series $\sum_{n=1}^{\infty}|x|^\sqrt n$ converge pointwise? If it then what would be the sum?
Does the series $\sum_{n=1}^{\infty}|x|^\sqrt n$ converge pointwise?</p>

答:
If $|x|\geqslant1$ then $|x|^\sqrt n\geqslant1$ does not converge to zero hence the series $\sum\limits_{n\geqslant1}|x|^\sqrt{n}$ diverges. If $|x|\lt1$ then for every $n$ in the slab $k^2\leqslant n\lt(k+1)^2$, $|x|^\sqrt{n}\leqslant|x|^k$ hence $\sum\limits_{n\geqslant1}|x|^\sqrt{n}\leqslant\sum\limits_{k\geqslant1}(2k+1)|x|^k$, which proves that the series $\sum\limits_{n\geqslant1}|x|^\sqrt{n}$ converges.
As Harald Hanche-Olsen mentioned in a comment, this can be viewed as a case of Cauchy's condensation test.

The sum $S(x)=\sum\limits_{n\geqslant1}|x|^\sqrt{n}$ has no simple expression when $|x|\lt1$. However, $\sum\limits_{k\geqslant1}r^k=\frac{r}{1-r}$ and $\sum\limits_{k\geqslant1}kr^k=\frac{r}{(1-r)^2}$ for every $|r|\lt1$ hence the upper bound explained above shows that $S(x)\leqslant|x|\frac{3-|x|}{(1-|x|)^2}\leqslant\frac{2}{(1-|x|)^2}$.
Likewise, for every $k^2\leqslant n\lt(k+1)^2$, $|x|^\sqrt{n}\geqslant|x|^{k+1}$ hence $S(x)\geqslant\sum\limits_{k\geqslant1}(2k+1)|x|^{k+1}=|x|^2\frac{3-|x|}{(1-|x|)^2}$.
In particular, $S(x)\sim\frac2{(1-|x|)^2}$ when $|x|\to1$.


问:
Connection between exponential function as repeated multiplication and $\lim_{n \to \infty} (1 + \frac{1}{n})^n$ (also known as compund interest)
I cannot find intuitive explanation for formula $e^x = \lim_{n \to \infty} (1 + \frac{1}{n} \cdot x)^n.$ I will write down my understanding of the exponential function so you can understand the question better.

Multiplication.
For hole numbers, multiplication is repeated addition (subtraction). Let's say we want to find $5 \cdot 3 = x.$ We know (or better, we defined this to be) $x = 3 + 3 + 3 + 3 + 3 = 15.$

For rational numbers it's simillar but it has two steps: $\frac{3}{2}7 = x.$ Problem on the left tells us to find number $y$ such that when I add it to itself gives me $7$ then I have to add that number to zero 3 times. So $y = 3.5.$ Now, I have to find $y + y + y.$ So it's kinda back and fort game, but it still can be desribed using repeated addition (subtraction).

Finally for irrational numbers multiplication can be approximated with error of $\epsilon \rightarrow 0$ using arithmetic in $\mathbb{Q}.$

Exponentiation.
Simillary as before we can do repeated multiplication. This arithmetically maybe isn't clear as before, but we can still do it (and for example doubling of something can easly be understood). Reasoning is simillar as before, just turn addition into multiplication. For example, number $7^{\frac{3}{2}} = x$ we can find as follows: find a number $y$ which multiplied by itself gives $7$ (so $y = 2.645...$). Then find $y \cdot y \cdot y.$ Thus $x = 18.520...$ For irrational exponent (base) we can  approximated it very well.

Now, we can investigate doubling of a money in hours, for example (suppose "money" can be any real number). In other words we are investigating function $f(x) = 2^x$ where $x$ is time passed in hours after 0th second.

Let's say we are starting from $2^0 = 1$ dollar. Right after 1 hour we have $2$ dollars, etc. Also 1 hour before starting moment we had $0.5$ dollars and so on. This starting from 1 (*) will be crucial.

We can observe that if more time passes doubling become larger, of course. Take intervals of one hour and see resulting change in amount of money, for example $ \frac{2^{0} - 2^{- 1}}{1 \ \text{hour}} <  \frac{2^1 - 2^0}{1 \ \text{hour}} <\frac{2^{10} - 2^9}{1 \ \text{hour}}.$

Of course, change in money over small time interval (dx) is proportional to amount of money we already have, $2^x$ for some fixed $x.$ After hypothetical calculation we get $\frac{d}{dx}2^x \approx 0.6931 \cdot 2^x$ and simillary for any positive base, $\frac{d}{dx}b^x = kb^x.$ We want to find base $a$ such that $k = 1$ (then obviously $\frac{d}{dx}a^x = 1 \cdot a^x = a^x$). After algebraic manipulation we can find that magical number and denote it by "e": $e:= a = \lim_{n \to \infty} (1 + \frac{1}{n})^n.$

I am interested in intuition why this is true. Simply: why we add to starting moment, 1 (*), this fraction $\frac{1}{n}$ and then multiply it by itself $n$ times for large $n$?

Simillary, why we add $x$ times this fraction $\frac{1}{n}$ and then again multilpy that $n$ times (I understand $e^x = e^{1 \cdot x}$ but I do not see it clearly.

Of course if you give good argument for $e^x = \lim_{n \to \infty}(1 + \frac{x}{n})^n$ then setting $x = 1$ we get basic formula for $e.$ Please, I am looking for intuition and logic, not just algebra. Thanks!

Also, is this limit well behaved, i. e. is inequality
$|e - \lim_{n \to \infty}(1 + (\frac{1}{n + 1})^{n + 1}| < |e - \lim_{n \to \infty} ((1 + \frac{1}{n})^n|$ true?

答:
$1$ is not important, it's just a matter of scale. You can start with million dollars and call it "1", meaning "1 million". It's just what you have at the beginning.

Consider that you take interest $\alpha$, that is, after $1$ time unit, you would have $1+\alpha$ money, after two time-units, you would have $(1+\alpha)^2$ and so on -- if it is compounded once per time unit.

However, if you split the time unit into $n$ smaller time-intervals and take the interest more often, you will have $(1+\frac{\alpha}{n})$ money after the first $\frac{1}{n}$-th time, then $(1+\frac{\alpha}{n})^2$ after $\frac{2}{n}$ time, and so on. After the full time unit, you would have $(1 + \frac{\alpha}{n})^n$.

Algebraically speaking, if you want to prove that $(1+\frac{x}{n})^n$ is similar to $\big((1+\frac{1}{n})^n\big)^x$ (for large $n$), you can think of it as
$$
(1+\frac{x}{n})^n = (1+\frac{1}{\frac{n}{x}})^n = (1+\frac{1}{\frac{n}{x}})^{x\frac{n}{x}} = \big((1+\frac{1}{\frac{n}{x}})^{\frac{n}{x}}\big)^x \simeq \big((1 + \frac{1}{M})^M\big)^x
$$
for some large $M$.


问:
Find a continuous function such that$f:\mathbb{R}\to \mathbb{R}$, $f(0)=0$ and $f(2x)\geq x+f(x)$ and $f(3x)\leq 2x+f(x)$.
Could you give me a hint for this problem?

I have to find all continuous functions $f:\mathbb{R}\to \mathbb{R}$ such that $f(0)=0$ and  $f(2x)\geq x+f(x)$ and $f(3x)\leq 2x+f(x)$.

I think this might be all the $f(x)=|x|$,  but can it be any other functions?

答:
From $f(2x)\ge x+f(x)$ we obtain, repeatedly replacing $x$ by $x/2$,
$$f(x)\ge x/2+f(x/2)\ge x/2+x/4+f(x/4)\ge\ldots\ge\sum^n_{k=1}x/2^k+f(x/2^n).$$
By continuity and $f(0)=0$, we have $\lim_{n\to\infty}f(x/2^n)=0$, and thus,
$$f(x)\ge\sum^\infty_{k=1}x/2^k=x.$$
From the second inequality, we get in the same way
$$f(x)\le2x/3+f(x/3)\le2x/3+2x/9+f(x/9)\le\ldots\le\sum^\infty_{k=1}2x/3^k=x.$$
So $x\ge f(x)\ge x$, i.e. $f(x)=x$.


问:
Is this claim about limit of multivariable functions correct?
let $ f:\mathbb{R}^{n}\to\mathbb{R}^{m} $.

define a projection function $ \pi^{i}$ by:
$ \pi^{i}\left(x_{1},...,x_{n}\right)=x_{i} $.

Is it true that $ \underset{x\to x_{0}}{\lim}f\left(x\right)=\overline{l} $ if and only if for and $ 1\leq i\leq n $ it follows that $ \pi^{i}\left(f\left(x\right)\right)\underset{\pi^{i}\left(x\right)\to\pi^{i}\left(x_{o}\right)}{\longrightarrow}\pi^{i}\left(\overline{l}\right) $

At first I thought I proved this theorem (because the equvivalent theorem holds for sequences), but I though that maybe iterated limits contradict this statement so Im not sure that my proof is correct (maybe I made some assumptions werent correct).

So I'll be glad if someone can tell if this statement is true or not. Thanks in advance

答:
it is true. just observe that
$|\pi^j(f(x)) - \ell_j| \leq \|f(x) -\ell\| \leq \sqrt{m} \max\limits_{j=1\ldots m}\{|\pi^j(f(x)) - \ell_j|\}$


问:
Growth-rate vs totality
How can one prove the statement, "If a function grows fast enough, it cant be proven total in PA, unless PA is inconsistent"? How fast must it grow to be not provably total? </p>

答:
The fast-growing hierarchy (or Wainer  hierarchy) has the property that a recursive function is $\sf{PA}$-provably total iff it is dominated by $f_\alpha$ for some $\alpha < \varepsilon_0$.  (Similarly with the Hardy hierarchy.)

For example, the Goodstein function grows asymptotically with $f_{\varepsilon_0}$, and so is not $\sf{PA}$-provably total (this is essentially Cichon's and Caicedo's proofs of the Kirby-Paris result that Goodstein's theorem is not $\sf{PA}$-provable).

Title: Growth-rate vs totality