问:
Factoring variables from fractions
I was working on a problem that I got wrong to solve for the variable N. The solution is as follows:

$$\left(\frac{12-N}{10}\right)+\left(\frac{-N}{20}\right)+\left(\frac{6-N}{10}\right) =0$$

$$36-5N=0$$

$$N=7.2$$

Where I am lost is from the first step to the second. I thought I could factor out N from the numerator and only need to solve the fractions but apparently that did not work as I got N(11/4). Could someone please explain how they got to 36-5N?

答:
It is unclear how do you factor our $N$ directly, since you did not explain how did you factor $N$ out from $12-N$ and $6-N$.

Make the denominator to be the same first.

$$\frac{24-2N}{20}-\frac{N}{20}+\frac{12-2N}{20}=0$$

Multiply by $20$,

$$24-2N-N+12-2N=0$$

this will give you,

$$36-5N=0$$

Title: Factoring variables from fractions


问:
$z =\frac{\sqrt{2}(1-i)}{1−i\sqrt{3}}$ . Show that the X set =$ \{z^n : n ∈ Z\} $ is finite and find the number of its elements.
As in the title: $z =\frac{\sqrt{2}(1-i)}{1−i\sqrt{3}}$ . I need to show that the X set =$ \{z^n : n ∈ Z\} $ is finite and find the number of its elements. I know how to do it "non creative" way. So, I transform both complex numbers into trigonometric form and divide:
$$\frac{\sqrt{2}(1-i)}{1−i\sqrt{3}} = \frac{2[\cos(\frac{7}{4}\pi)+\sin(\frac{7}{4}\pi)i]}{2[\cos(\frac{5}{6}\pi)+\sin(\frac{5}{6}\pi)i]}= \cos \left(\frac{11}{12}\pi \right)+\sin \left( \frac{11}{12}\pi \right)i$$

Now, I see that this is $165^{\circ}$, and that module $= 1$. The problem is, that the only way I see to find the number of elements of the X set is to calculate degrees:
$$165^{\circ}, 330^{\circ}, 135^{\circ}, 300^{\circ}, 105^{\circ}, 270^{\circ}, 75^{\circ}, 240^{\circ}, 45^{\circ}, 210^{\circ}, 15^{\circ}, 180^{\circ}, 345^{\circ}, 150^{\circ}, 315,^{\circ} 120^{\circ}, 285^{\circ}, 90^{\circ}...$$

So I feel like there has to be some more effective way to solve that problem and here comes my question. Do you know any?

答:
A way is to find a polynomial satisfied by $z$.  (For low degrees, this is likely to be a minimal polynomial.  To show this requires computing elements of $X$ successively and searching for a linear relation among the powers of $z$ produced so far.  The first such relation is the minimal polynomial.  All we can actually promise is that we get a polynomial multiple of a minimal polynomial.) \begin{align*}
\frac{z}{1+\mathrm{i}\sqrt{3}}
    &= \frac{\sqrt{2}(1+\mathrm{i})}{1+3}  \\
4z &= \sqrt{2}(1+\mathrm{i})(1+\mathrm{i}\sqrt{3})  \\
16z^2 &= 2(1+\mathrm{i})^2(1+\mathrm{i}\sqrt{3})^2  \\
8z^2 &= 2\mathrm{i}(1+\mathrm{i}\sqrt{3})^2  \\
4z^2 &= \mathrm{i}(-2 + 2\mathrm{i}\sqrt{3})  \\
2z^2 &= \mathrm{i}(-1 + \mathrm{i}\sqrt{3})  \\
4z^4 &= \mathrm{i}^2(-1 + \mathrm{i}\sqrt{3})^2    \\
-4z^4 &= (-2 - 2\mathrm{i}\sqrt{3})  \\
-2z^4 &= (-1 - \mathrm{i}\sqrt{3})  \tag{$\star$}  \\
4z^8 &= (-1 - \mathrm{i}\sqrt{3})^2  \\
4z^8 &= (-2 + 2\mathrm{i}\sqrt{3})  \\
2z^8 &= (-1 + \mathrm{i}\sqrt{3})  \\
2z^8 - 2z^4 &= -2  &  &  \text{using ($\star$)}  \\
z^8 - z^4 + 1 &= 0  \\
z^8 &= z^4 - 1  \text{.}
\end{align*}
(Using other tools, we can show that this is the minimal polynomial of $z$, but this is not critical.)
So, $\require{enclose}$
$$  X = \{1,z, z^2, \dots, z^7, z^4-1, z^5-z, \dots, z^8 - z^4 = -1, -z, -z^2, \dots, -z^7, -z^4+1, -z^5+z, \dots, \enclose{horizontalstrike}{-z^8+z^4 = 1}\}  \text{,}  $$
a set of only $24$ elements.

We haven't explicitly shown that the norm of $z$ is $1$, it is implicit in the fact that the powers of $z$ are cyclic, equivalently, $X$ is finite.

ContentLicense: CC BY-SA 4.0


答:
If $q = a/b$ is any fraction in lowest terms, then by de Moivre's identity, $$\left( \cos(2\pi q) + i \sin(2 \pi q) \right)^b = \cos(2\pi a) + i \sin(2 \pi a) = 1,$$ and conversely, if we raise $\cos(2\pi q) + i \sin(2 \pi q)$ to any power $0 < n < b$, we won't get an integer multiple of $2\pi$ inside the cosine and sine, and so the power will not be equal to 1. This means the order must be $b$.

In your case, $2\pi q = 11\pi / 12$, so $q = 11/24$ and the order is 24.

Title: $z =\frac{\sqrt{2}(1-i)}{1−i\sqrt{3}}$ . Show that the X set =$ \{z^n : n ∈ Z\} $ is finite and find the number of its elements.

Body:
As in the title: $z =\frac{\sqrt{2}(1-i)}{1−i\sqrt{3}}$ . I need to show that the X set =$ \{z^n : n ∈ Z\} $ is finite and find the number of its elements. I know how to do it "non creative" way. So, I transform both complex numbers into trigonometric form and divide:
$$\frac{\sqrt{2}(1-i)}{1−i\sqrt{3}} = \frac{2[\cos(\frac{7}{4}\pi)+\sin(\frac{7}{4}\pi)i]}{2[\cos(\frac{5}{6}\pi)+\sin(\frac{5}{6}\pi)i]}= \cos \left(\frac{11}{12}\pi \right)+\sin \left( \frac{11}{12}\pi \right)i$$

Now, I see that this is $165^{\circ}$, and that module $= 1$. The problem is, that the only way I see to find the number of elements of the X set is to calculate degrees:
$$165^{\circ}, 330^{\circ}, 135^{\circ}, 300^{\circ}, 105^{\circ}, 270^{\circ}, 75^{\circ}, 240^{\circ}, 45^{\circ}, 210^{\circ}, 15^{\circ}, 180^{\circ}, 345^{\circ}, 150^{\circ}, 315,^{\circ} 120^{\circ}, 285^{\circ}, 90^{\circ}...$$

So I feel like there has to be some more effective way to solve that problem and here comes my question. Do you know any?


问:
Computing the euler number of an oriented disc bundle over a sphere
Suppose $E\to S^2$ is an (smooth) oriented disk bundle. Let $D_1$ and $D_2$ denote the upper and lower hemispheres of $S^2$, respectively. Then $E|_{D_1}$ and $E|_{D_2}$ are trivial, so they are both diffeomorphic to $D_i\times D^2$. So $E$ may be seen to be obtained from $D_1\times D^2$ and $D_2\times D^2$ by attaching their boundaries, for some diffeomorphism $f:\partial D_1 \times D^2 \to \partial D_2\times D^2$ which is identity on the first factor, and an element of $SO(2)$ on the second factor. (Thus $f(x,y)=(x,g(x)y)$ for some $g:\partial D_1\to SO(2)$. Thus $g$ gives an element $\pi_1(SO(2))\cong \Bbb Z$, say $n$. We may assume $g$ is the $n$-fold product of the standard generator of $\pi_1(SO(2))$.

Note that $E$ is homotopy equivalent to the zero section $\cong S^2$. I want to show that the euler number of $E$ is $n$, by showing that the self intersection number of the zero section is $n$. To this end, it suffices to show that there is a section $h:D_1\cup D_2\to E=D_1\times D^2 \cup_f D_2\times D^2$ which intersects the zero section $|n|$ times positively if $n>0$ (negatively if $n<0$). First, fix an arbitrary point $p(\neq 0)\in D^2$ and define $h(x)=(x,p)\in D_1\times D^2$ for $x\in D_1$. Then we must have $h(x)=(x,g(x)p)\in D_2\times D^2$ for $x\in \partial D_2$. Now extend $h$ over $D_2$ by defining $h(rx)=(rx,r^2g(x)p+(1-r^2)q)$ where $q=(1,0)\in D^2$ and $0\leq r\leq 1$. This gives a section of $E\to S^2$ that intersects the zero section exactly $n$ times, but how can we know whether the intersection is positive or negative?

Or is there another method for computing the euler number of $E$?

P.S. Does every oriented $D^2$-bundle over $S^2$ is the unit disk bundle of some complex line bundle over $S^2$ with a hermitian metric?

答:
The Euler class depends homomorphically on $n$ so it suffices to calculate it for $n=1$. In this case, the disk bundle we get is fiberwise diffeomorphic to the disk bundle of the tautological bundle over $\mathbb{C}P(1)$. The Thom space of this bundle is $\mathbb{C}P(2)$. The self intersection number can be calculated via the cup square. The zero section corresponds to the inclusion of $\mathbb{C}P^1$ into $\mathbb{C}P^2$ and so the cup square of the dual of this class is the fundamental class by the cohomology structure. This means the Euler number is 1.

Hence, for general $n$ the Euler number will be $n$.

To address your other question: I highly recommend figuring this issue with orientations out yourself. It will be very instructive.

And yes this follows from the fact that $\pi_2(BO(2))$ is generated by the tautological bundle over $\mathbb{C}P^1$. If one wanted to do this calculation via complex means, we could use the fact that the Chern number of the bundle is 1 and that this is equal to its Euler number.


问:
Interior in subspace topology.

Let $E_1, E_2 \subseteq X$ be closed set in ($X, \tau$) such that $X = E_1\cup E_2$. Consider $Q \subseteq E_2$ open in the subspace topology for $E_2$ and $B\subseteq E_1$ such that $B\cap E_2 \subseteq Q$. Prove that $B \subseteq \text{int}_X(E_1\cup Q)$ where $\text{int}_x()$ is the interior over ($X, \tau$).

I've tried to prove that $B$ is open in $(X,\tau)$, since as $Q$ is open in that subspace then $\exists A\in\tau$ such that $Q=A\cap E_2$ and then $B\cap E_2\subseteq A\cap E_2=Q\implies B\subseteq A$ but I'm not sure how to.

答: $E_2^{c} \subseteq E_1$ and $E_2^{c}$ is open. It follows that $E_2^{c} \subseteq int_X(E_1) \subseteq int_X(E_1 \cup Q)$. Take any point $b \in B$. if $b \in E_2^{c}$ we are done. Suppose $b \in E_2$. Then $b \in B \cap E_2 \subseteq Q \subseteq A$. I will leave it to you to check that $A \subseteq (E_1\cup Q)$ which shows that $b$ is an interior point of $E_1\cup Q$.

Title: Interior in subspace topology.


问:
Bernoulli differential equation help?
We have the equation $$3xy' -2y=\frac{x^3}{y^2}$$ It is a type of Bernouli differential equation. So, since B. diff equation type is
$$y'+P(x)y=Q(x)y^n$$
I modify it a little to:

$$y'- \frac{2y}{3x} = \frac{x^2}{3y^2}$$

$$y'-\frac{2y}{3x}=\frac{1}{3}x^2y^{-2}$$

Now I divide both sides by $y^{-2}$. What should I do now?

答: Note that in any Bernoulli OE, $$y'+p(x)u=q(x)y^n,~~n\neq1,0$$ you can set $w=y^{1-n}$ and then you have to solve the following linear OE instead: $$w'+(1-n)p(x)w=(1-n)q(x)$$</p>

ContentLicense: CC BY-SA 3.0

Title: Bernoulli differential equation help?


问:
Round-robin tournament
In a round-robin tournament, each player plays every other player exactly once, where there are a total of $n$ players. Then,

$(i)$ What is the sum of wins and losses for each player?

$(ii)$ If there is a player who is undefeated, then how many wins does the player have?

$(iii)$ If there is no player who is undefeated, then what is the most wins of a player?

Concerning about these sub-parts, I think answer for $(i)$ would be $n(n-1)$, but have no idea on how to solve $(ii)$ and $(iii)$ (possibly I got wrong for them)

答:
Hint :

Sum of wins and losses = Total number of games played

Title: Round-robin tournament


问:
If $\limsup_{x\to\infty} (1-F(x)+F(-x))\le \varepsilon$, how to prove that $F$ is a distribution function?
I read Rick Durrett's probability and was confused by the proof of theroem that states the limit of the tight distribution sequence is a distribution function of a probability measure. (just after Helly's selection theorem)

It proves that $\limsup_{x\to\infty} (1-F(x)+F(-x))\le \varepsilon$, and says since $\varepsilon$ is arbitrary, it follows that $F$ is a distribution function.

But why? It only means that $\limsup_{x\to\infty} (F(x)-F(-x))\to 1$, does not mean $\lim_{x\to+\infty}F(x)=1,\lim_{x\to-\infty}F(x)=0$, and $F$ is nondecreasing, right continuous.（Which is the definition of a distribution function.）

I could not prove it, can anyone tells me why?

答:
Helly's Selection Theorem gurantees that $F$ is non-decreasing, right continuous and $0\leq F(x) \leq 1$ for all $x$. You only have  to prove that $F(x) \to 1$ and $F(-x) \to 0$ as $x \to \infty$.

Now $1-F(x)$ and $F(-x)$ are both nonnegative. If their sum tends to $0$ as $x \to \infty$ they they both tend to $0$. Hence $1-F(x) \to 0$ and $F(-x) \to 0$.


问:
Doubt in Sum no 29 of Rudin

Prove that every open open set in $R$ is the union of a countable number of disjoint intervals.

Although this question has been in stack exchange for too many times but here is my intuition to it. Can someone just suggest whether I am on the right track.

I am taking the example of the open set $(a,b)$.So I take the first segment as $(a, b-1)$ , the second segment as $[b-1,b-1/2)$ the third as $[b-1/2,b-1/3)$ and so on...My logic behind this is that $(0,1-1/n)$ covers $(0,1)$ and so $(a, b-1/n)$ covers $(a, b) $.The intervals are disjoint also and their union  is also countable. Now, my question is the hint given in rudin is to use the fact that $R$ is separable , but where do I use it, Am I making a mistake somewhere?

答:
I am taking the example of the open set $(a,b)$.So I take the first segment as $\color{red}{(a, b-1)}$ , the second segment as $[b-1,b-1/2)$ the third as $[b-1/2,b-1/3)$ and so on...My logic behind this is that $(0,1-1/n)$ covers $(0,1)$ and so $(a, b-1/n)$
$\color{red}{ covers}$ $(a, b) $.The intervals are disjoint also and their union  is also countable. Now, my question is the hint given in rudin is to use the fact that $R$ is separable , but where do I use it, Am I making a mistake somewhere?

From what you have written, I assume you mean

$$(a,b)=(a,b-1) \cup (b-1,b-\frac{1}{2}) \cup (b-\frac{1}{2},b-\frac{1}{3}) \cup \dots $$

But there are many issues with this logic. The mistakes are

$1$-What if $a > b-1$?

$2$-Your open set need not of the form $(a,b)$. It can be of the form $\bigcup_{i\in I}(a_i,b_i)$ where $I$ can be an uncountable set as well.

$3$-Secondly, Your LHS may miss the points $\Big\{b-1,b-\frac{1}{2},b-\frac{1}{3},\dots\Big\}$ , So you cannot say that your set $(a,b)$ is covered

Now the hint can give you an easier solution. You can think about that. The hint says that there exists a countable dense set. The set of rational numbers $\mathbb{Q}$ is countable and dense in $\mathbb{R}$.

Title: Doubt in Sum no 29 of Rudin


问:
Solve $y=(2y^4+2x)y'$ using Laplace Transform
My attempt:
\begin{align*}
\mathscr{L}\{y&=(2y^4+2x)y'\}\\
Y&=2\mathscr{L}\{y^4y'\}+2\mathscr{L}\{xy'\}\\
\end{align*}
Consider
\begin{align*}
\mathscr{L}\{y'\}&=sY-Y(0)\\
\int_0^\infty y'e^{-sx}dx&=sY-Y(0)\\
\int_0^\infty \frac{\partial}{\partial s}(y'e^{-sx})dx&=\dfrac{d}{ds}(sY-Y(0))\\
\mathscr{L}\{xy'\}&=-(x\frac{dY}{ds}+Y)
\end{align*}
I can't think of a way to evaluate $\mathscr{L}\{y^4y'\}$ and to continue solving differential equation.

答:
$$y=(2y^4+2x)y'$$
Rewrite the differential equation as:
$$x'y=2y^4+2x$$
Then you can apply the Laplace transform.
$$X'(s)+2X(s)=-\dfrac {2 \times 4!}{s^5}$$


问:
Travel vector field
Lets say I have a vector field that looks like this:

Where the x-component of each vector is x=x and the y-component is y=y.

Each of these vectors represents acceleration.

Now I "drop" a fresh particle (p) into the field at a certain position (p) which already has a velocity (v).

How can I calculate the new position (p2) and velocity (v2) of the particle after a specific time (t)?

From my basic understanding it has something to do with Line Integrals? However this doesn't help me much since it was years since I even solved simple integrals.

I'm doing this for a program I'm writing and I'm not well versed in the world of mathematics so I would really appreciate if you would keep the mathematic notation simple!

Hopefully I didn't miss any important piece of information.

Thanks, Andreas.

答:
An acceleration has units of length per time squared (or inverse length in natural units), and the vector field you describe has units of length, so there needs to be some factor of proportionality connecting the two. Let's say the acceleration is

$$\ddot{\vec{r}}=\lambda^2 \vec{r}\;,$$

where $\vec{r}$ is the position vector with components $x$ and $y$, a dot indicates differentiation with respect to time, and $\lambda^2$ is some positive proportionality constant which I've written as a square for later convenience. This is a second-order linear ordinary differential equation for $\vec{r}$, whose solution is

$$\vec{r}(t)=\vec{r}_1\mathrm e^{\lambda t}+\vec{r}_2\mathrm e^{-\lambda t}$$

with arbitrary vectors $\vec{r}_1$ and $\vec{r}_2$ which need to be determined using the initial conditions. Substituting your initial values $\vec{r}(0)=\vec{p}$ and $\dot{\vec{r}}(0)=\vec{v}$ yields

$$
\begin{eqnarray}
\vec{r}_1+\vec{r}_2&=&\vec{p}\;,\\
\vec{r}_1-\vec{r}_2&=&\frac{\vec{v}}{\lambda}\;,
\end{eqnarray}
$$

with solution

$$
\begin{eqnarray}
\vec{r}_1&=&\frac{1}{2}\left(\vec{p}+\frac{\vec{v}}{\lambda}\right)\;,\\
\vec{r}_2&=&\frac{1}{2}\left(\vec{p}-\frac{\vec{v}}{\lambda}\right)\;,
\end{eqnarray}
$$

so the overall solution is

$$\vec{r}(t)=\frac{1}{2}\left[
\left(\vec{p}+\frac{\vec{v}}{\lambda}\right)\mathrm e^{\lambda t}
+
\left(\vec{p}-\frac{\vec{v}}{\lambda}\right)\mathrm e^{-\lambda t}
\right]\;.
$$

This gives you the position at time $t$. To obtain the velocity at time $t$, you can differentiate with respect to time:

$$\dot{\vec{r}}(t)=\frac{1}{2}\left[
\left(\lambda\vec{p}+\vec{v}\right)\mathrm e^{\lambda t}
-
\left(\lambda\vec{p}-\vec{v}\right)\mathrm e^{-\lambda t}
\right]\;.
$$

To determine $\lambda$, take the square root of $\lambda^2$. Your units appear to be such that $\lambda^2=1$; in that case, you'd have $\lambda=1$.

[Edit:] I just realized that the vector field in your image doesn't look like the vector field you defined -- the vectors seem to all have the same length. If this is what you want the acceleration would be

$$\ddot{\vec{r}}=\lambda^2 \frac{\vec{r}}{|\vec{r}|}\;.$$

This can also be solved in closed form, but it's quite a bit more complicated, so I'll only work it out if that's what you need.

Title: Travel vector field


问:
Distance between the curves from a particular point
The distance between the points $P(u,v)$ and the curve $x^2+4x+y^2=0$ is the same as the distance between the points  $P(u,v)$ and $M(2,0)$. If $u$ and $v$ satisfy the relation $u^2-\frac{v^2}{q}=1$, then $q$ is greater than or equal to

(A) 1

(B) 2

(C) 3

(4) 4

My approach is as follow Curve $x^2+4x+y^2$ represent a circle with centre $(-2,0)$ and radius $2$.

The parametric equation of the circle is $-2cos\theta+2$ and $2sin\theta$

$D^2=(u-2)^2+v^2$

Also $D^2=(u+2cos\theta-2)^2+(v-2sin\theta)^2$

How we will proceed from here

答:
Distance $D_1$ of P(u,v) from the circle is $D_1=\sqrt{(u+2)^2+v^2}-2=\sqrt{F}-2$
and $D_2=\sqrt{(u-2)+v^2}=\sqrt{G}$, we require
$$D_1=D_2 \implies \sqrt{(u+2)^2+v^2}-2=\sqrt{(u-2)+v^2} \implies \sqrt{F}-\sqrt{G}=2~~~(1)$$ We also have $$F-G=8u \implies \sqrt{F}+\sqrt{G}=4u~~~~(2)$$
From (1) and (2), we have $$\sqrt{F}=1+2u \implies u^2+4u+4+qu^2-q=1+4u+4u^2 \implies (q-3) u^2=(q-3)\implies q=3$$


问:
Is $x\in\emptyset$ defined?
We know, $A\cup B=\{x:x\in A\lor x\in B\}$

In this definition, however if A or B is empty or both A and B are empty, we are getting the term $x\in \emptyset$
This poses the problem of defining that term, which I tried by saying that if $x\in \emptyset$ then $x=null$. But, it just doesn't work because then I am just allowing x as a variable to have a 'null state' which means if $x\in \{1\}$, then it means that $x=1$ or $x=null$.

This same problem also happened while defining subsets, but then rather than defining what is a subset I ended up defining what is not and so avoided using $x\in \emptyset$

A is not subset of B if there is a x such that $x\in A$ and $x\notin B$, otherwise A is subset of B.

答:
Naturally

It's all very natural. Imagine you have two bags $A$ and $B$ (emptiness doesn't matter) containing (if not empty) colorful balls. And you take a third bag $C$ to collect all distinct balls they have (union of other two bags).

If neither of them is empty there is no "problem" with doing this task $(C=A\cup B)$.

If one of them is empty $($say, $A=\emptyset)$, then naturally, the third bag will contain all elements of the non-empty bag $(C = B)$ since there is no element in $A$ to include in $C$.

If both are empty $(A=B=\emptyset)$, then the third bag will also be empty since there is no ball to "collect" at all $(C = \emptyset)$.

Stemming from the above statements we come to the following

Definition

If $A,B \neq \emptyset$, then $\boxed{A\cup B = \{x \mid x \in A \text{ or } x \in B \text{ or both}\}}$

(WLOG) If $A =\emptyset \neq B$, then $\boxed{A \cup B = B}$

If $A=B=\emptyset$, then $\boxed{A\cup B = \emptyset}$

(This definition can be gathered to a more compact form, but this form seems to solve the problems that OP is asking)

Conclusion

Therefore, there is no need to think of any non-existent object for $x \in  \{1\}$ which is equivalent to $x \in \{1\}\cup \emptyset$, and there is no $x$ such that $x \in \{~\}$ (the empty set is indeed empty).

Title: Is $x\in\emptyset$ defined?


问:
The error in Taylor polynomial of $\sin x$ is $10^{-6}$...

If we want to expand $f(x)=\sin x$ with Taylor polynomial centered at
$x=0$. and error of that for$x=\frac{\pi}{2}$ be at most $10^{-6}$.
then what order of polynomial should be at? ($n=?$)

As a hint our lecturer said we should use $c=\frac{\pi}{2}$ at formula of remainder of Taylor series.

Here is my work:

The reminder of Taylor series for the function $f(x)$ where $x_0=0$ is :

$$R_n(x)=\cfrac{f^{(n+1)}(c)}{(n+1)!}x^{n+1}$$

And I estimate $\frac{\pi}{2}$ to closest integer which is $2$ for the value of $x$.

To find $n$ I should solve inequality:

$$|R_n(x)|\le 10^{-6}\quad\rightarrow\quad|\cfrac{f^{(n+1)}(\frac{\pi}{2})}{(n+1)!}\times2^{n+1}|\le 10^{-6}$$

$(n+1)$'st derivative of $f(x)=\sin x$ can be one of the $\pm\sin x$ ,$\pm\cos x$. So $|f^{(n+1)}(\frac{\pi}{2})|\in\{0,1\}$. If it is equal to zero then we have $0\le 10^{-6}$ and it is absolutely wrong so the best I can do is using $|f^{(n+1)}(\frac{\pi}{2})|=1$. hence the inequality is $\left|\cfrac{2^{n+1}}{(n+1)!}\right|\le 10^{-6}$ . After plugging in some $n$s I got $n\ge13$. so the order of the Taylor polynomial should be $13$.

Is my approach right ? as you can see in my answer I had some difficulties to find $|f^{(n+1)}(\frac{\pi}{2})|$ did I right at this point?

答:
Write $$\sum_{n=0}^\infty\frac{(-1)^n }{(2 n+1)!}x^{2 n+1}=\sum_{n=0}^p\frac{(-1)^n }{(2 n+1)!}x^{2 n+1}+\sum_{n=p+1}^\infty\frac{(-1)^n }{(2 n+1)!}x^{2 n+1}$$
SInce it is an alternating series, you want to know $p$ such that
$$R_p=\frac{x^{2 p+3}}{(2 p+3)!}\leq \epsilon\implies (2 p+3)! \geq\frac{x^{2 p+3}}{\epsilon}$$
If you have a look at this question of mine, you will see a magnificent approximation proposed by @robjohn.

Adapted to the present problem, this would give
$$p\sim\frac{e \pi }{4}\,\exp\Big[W\left(-\frac{2 \log (\pi  \epsilon )}{e \pi }\right)\Big]-\frac 74$$ where $W(.)$ is Lambert function. For sure, you need to use $\lceil p \rceil$.

For $\epsilon=10^{-6}$, this gives $p=4.31643$ and then $\lceil p \rceil=5$ (the exact solution would be $4.31727$). So, the expansion up to $n=11$ is sufficient, as @Yves Daoust already wrote (this would give $0.9999999437$ while one more term would lead to $1.000000001$).

Checking
$$R_4=\frac{\pi ^{11}}{81749606400}\sim 3.60\times 10^{-6} > 10^{-6} $$
$$R_5=\frac{\pi ^{13}}{51011754393600}\sim 5.69\times 10^{-8} < 10^{-6} $$

If you cannot use Lambert function, since its argument is large, you could use the approximation given in the linked Wikipedia page
$$W(x)= L_1-L_2+\frac{L_2}{L_1}+\frac{L_2(L_2-2)}{2L_1^2}+\frac{L_2(2L_2^2-9L_2+6)}{6L_1^3}+\cdots$$ where $L_1=\log(x)$ and $L_2=\log(L_1)$. For your value, this would would give $W(.)\sim 1.07011$ while its exact value is $1.04434$.


答:
"Which means x is 1 or there is no element as x."

Not quite.  It means either $x \in \{1\}$ or $x \in \varnothing$ or both.  (The trailing "or both" resolves an ambiguity in English.  If we agree that we are using "or" to represent logical disjunction, the "or both" can be omitted and we choose to do so subsequently)  The clause $x \in \{1\}$ is only true when $x = 1$ and is false otherwise.  The clause $x \in \varnothing$ is always false.  Therefore, the only $x$ for which $x \in \{1\}$ or $x \in \varnothing$ is $x = 1$.

The clause "$x \in \varnothing$" does not mean "there is no element as $x$" when it is used as a logical proposition.  It means "has the truth value true when $x$ is such that $x \in \varnothing$ and has the truth value false when $x$ is such that $x \not \in \varnothing$".  Since always $x \not \in \varnothing$, this clause always has the truth value false.

A statement may have a truth value.  The statement "$1 = 1$" is always true.  The statement "$x = 1$" is only true when $x$ has the value $1$.

Consider the statement "$x \in \{1\}$".  This clause can be satisfied (that is, has the truth value "true") when $x = 1$.    If $x$ has any other value, the statement is false.  If $x$ does not have a definite value, the statement does not have a truth value.

Consider the statement "$x \in \varnothing$", which is literally "$x$ is an element of the empty set".  There is no value $x$ can have that makes the statement true.  The statement is always false.

It is important to treat both of these clauses similarly.  Membership in a set is the same thing for all sets.  There is not a special, different version of membership for the empty set.  When you write "$x$ is 1", you are expressing a different clause than when you write "there is no element as $x$".  This means the clause you intend for one set membership is different from the clause you intend for the other.  This is incorrect -- both clauses must have the same structure.

The clause $A \vee B$, where $A$ and $B$ are clauses with truth values, is true when either $A$ or $B$ is true and is false when both $A$ and $B$ are false.

Now let us study the expression
$$  (x \in \{1\}) \vee (x \in \varnothing)  \text{,}  $$
defining the set $U = \{1\} \cup \varnothing$.  When $x = 1$, this expression is
$$  \text{True} \vee \text{False}  \text{,}  $$
which is true.  So $1 \in U$.  When $x$ has any other value, this expression is
$$  \text{False} \vee \text{False}  \text{,}  $$
which is false.  Consequently, $U = \{1\}$.

One can also phrase this in terms of witnesses.  To do so , we make a slight change -- we introduce existential statements, having the form "there exists (variable) such that (proposition) [is true]".  (The "is true" is optional since it is assumed by default that a simple assertion is claiming a truth value of true.)  For example, there is an $x$ such that $x$ is even.  The assignment $x = 2$ satisfies this existential statement, so $2$ is a witness to the truth of the statement.

This notion of witness is, I think, a way to see what you want to see.  The clause "there is no element as $x$" is a statement about existence and we are able to handle such statements in logic.

The statement "there exists an $x$ such that $x \in \{1\} \cup \varnothing$" has a witness : $1$.  So this statement is true.

The statement "there exists an $x$ such that $x \in \varnothing \cup \varnothing$" has no witness, so this statement is false.  There is no thing that is an element of $\varnothing \cup \varnothing$, a property held exclusively by the empty set.  So "$\varnothing \cup \varnothing$" is just another way to write the empty set.

Title: Is $x\in\emptyset$ defined?

Body:
We know, $A\cup B=\{x:x\in A\lor x\in B\}$

In this definition, however if A or B is empty or both A and B are empty, we are getting the term $x\in \emptyset$
This poses the problem of defining that term, which I tried by saying that if $x\in \emptyset$ then $x=null$. But, it just doesn't work because then I am just allowing x as a variable to have a 'null state' which means if $x\in \{1\}$, then it means that $x=1$ or $x=null$.

This same problem also happened while defining subsets, but then rather than defining what is a subset I ended up defining what is not and so avoided using $x\in \emptyset$

A is not subset of B if there is a x such that $x\in A$ and $x\notin B$, otherwise A is subset of B.


问:
Let $f(x) =\sqrt[3]{1 + \sin{x}}$ on $ℝ$. Find $T_{1,0}f$, the first order Taylor polynomial around $x_o= 0$
Let $f(x) =\sqrt[3]{1 + \sin{x}}$ on $ℝ$.  Find $T_{1,0}f$, the first order Taylor polynomial around $x_o= 0$, and give an error estimate for the approximation of $f$ by $T_{1,0}f$ on the interval$[\frac{−π}{6},\frac{π}{6}]$.

I believe I have yet to fully grasp a full understanding of taylor polynomials and as such I've been having quite some issues with this question. How does one appropriately go about solving this problem? I wish to learn so that I can solve such classes of problems without help in the future.

Body:
Around zero,
$(1+x)^a
=1+ax+a(a-1)x^2/2...
$
and
$\sin(x)
=x-x^3/6+...
$,

$\begin{array}\\
f(x)
&= (1+\sin(x))^{1/3}\\
&=1+\frac13 \sin(x)+(1/3)(-2/3)\sin^2(x)/2+...\\
&=1+\frac13 \sin(x)-\sin^2(x)/9+...\\
&=1+\frac13 (x-x^3/6+...)-(x^2+...)/9+...\\
&=1+\frac13 x-\frac19 x^2+...\\
\end{array}
$

You can also do
David's suggestion
$f(x)
=f(0)+xf'(0)+...
$.

$f(0) = 1$.

$f'(x)
=\sin'(x)(\frac13 (1+\sin(x))^{-2/3})
=\cos(x)(\frac13 (1+\sin(x))^{-2/3})
$
so
$f'(0)
=\frac13
$
and
$f(x)
=1+\frac{x}{3}
+...
$.

To get further terms,
you need $f''(x)
$
which I will leave to you.


问:
Projective image is composition of projections
I want to prove that a projective image from a projective line $k$ to the same projective line $k$ is a composition of projections.

I've seen a proof for the case that the image maps from $k$ to another projective line $m$. Here, three points $P$, $Q$, $R$ are introduced on $k$ with corresponding images $P'$, $Q'$, $R'$ on $m$. Then, this proof uses the intersection points of $PQ'$ with $P'Q$ and $RS'$ with $R'S$ to project $k$ from $Q'$ on the line between those intersection points and subsequently this line between the intersection points on $m$ from $Q$. However, if the image maps from $k$ to $k$, I don't have those intersection points...

Is there a possibility to adjust the proof to make it work for the image that maps from $k$ to $k$? I've already received a hint to introduce a new line, but I'm not sure how to use this.

答: To map from $k$ to $k$, take any line $m$ and point $O$.  Then project $k$ from $O$ onto $m$, taking $P,Q,R$ to $P'',Q'',R''$.  You already know how to map points $P'',Q'',R''$ on $m$ to points $P',Q',R'$ on $k$ via two projections.  So now you can get a map from $P,Q,R$ to $P',Q',R'$ as a composition of all three projections.


问:
how to find the number of solutions for $\ln(x) =x^2/8$ using simple math principles?
how to find the number of solutions for  $\ln(x) = x^2/8$ with high school level math (basic logarithm rules or something of that nature)( and without graphing $\ln(x)$ and $x^2/8$). (sorry for the poor formatting, feel free to edit)

I tried to change the equation to someting familiar to use a logarithm rule:

$$\displaylines{
x^2/8 = \ln(x) \cr
x/8^{0.5} = lan(x)^{0.5} \cr
x = 8^{0.5}*lan(x)^{0.5}
}$$

Changing the representation of the equation using the basic logarithm rule
$$\displaylines{
x^2/8 = \ln(x) \cr
x = e^{x^2/8}
}$$
But my tries got me nowhere. Any suggestions?

答:
Consider that you ar looking the the zero's of function
$$f(x)=\log (x)-\frac{x^2}{8}$$
$$f'(x)=\frac{1}{x}-\frac{x}{4}$$ cancels at $x=2$ ($x=-2$ being discareded because of the logarithm.
$$f(2)=\log (2)-\frac{1}{2} > 0$$ and $f''(2)=-\frac{1}{2}$. So, $x=2$ is a maximum and there are two solutions.


问:
Limit of $(1-\frac{1}{2^n})^n$.
What is the limit of the sequence $(1-\frac{1}{2^n})^n$?

Let $a_n:=(1-\frac{1}{2^n})^n$ then $\ \ln \ a_n=n\ \ln(1-\frac{1}{2^n})$

$\implies \lim \limits_{n\rightarrow \infty} \ln\ a_n=\lim\limits_{n \rightarrow \infty} n\ \ln(1-\frac{1}{2^n})$

$\implies \ln \lim\limits_{n \rightarrow \infty} a_n = \lim\limits_{n \rightarrow \infty} \frac{\ln(1-\frac{1}{2^n})}{\frac{1}{n}}$

$\implies \ln \lim\limits_{n \rightarrow \infty} a_n = \lim\limits_{n \rightarrow \infty} \frac{\frac{ln2}{2^n(1-2^n)}}{\frac{-1}{n^2}}  [\frac{0}{0} \ \ \text{form}]$

I am stuck here, what should I do after this?

Title: Limit of $(1-\frac{1}{2^n})^n$.

Body:
Since
$(1-1/n)^n \to 1/e
$,
$(1-1/2^n)^{2^n}
\to 1/e
$,
so
$(1-1/2^n)^n
=((1-1/2^n)^{2^n})^{n/2^n}
\to (1/e)^{n/2^n}
\to 1
$
since $n/2^n \to 0$.


答:
try visualizing the graphs of ln(x) and x^2 / 8. The first is a log-curve, and the second in a parabola. They will  intersect at at most two points. so you will have at most two solutions.

Since the log-curve starts (or ends) at the origin, and the parabola will "circumnavigate" the origin going from -infinity to +infinity, it's safe to say that they will intersect at exactly one point.

Title: how to find the number of solutions for $\ln(x) =x^2/8$ using simple math principles?

Body:
how to find the number of solutions for  $\ln(x) = x^2/8$ with high school level math (basic logarithm rules or something of that nature)( and without graphing $\ln(x)$ and $x^2/8$). (sorry for the poor formatting, feel free to edit)

I tried to change the equation to someting familiar to use a logarithm rule:

$$\displaylines{
x^2/8 = \ln(x) \cr
x/8^{0.5} = lan(x)^{0.5} \cr
x = 8^{0.5}*lan(x)^{0.5}
}$$

Changing the representation of the equation using the basic logarithm rule
$$\displaylines{
x^2/8 = \ln(x) \cr
x = e^{x^2/8}
}$$
But my tries got me nowhere. Any suggestions?


问:
Choosing aleph with given property
I'm studying "Introduction to Set Theory" by Karel Hrbacek and Thomas Jech. In the proof that $\aleph_\alpha\cdot \aleph_\alpha=\aleph_\alpha$ for any ordinal $\alpha$, we want to prove $\aleph_\alpha\cdot \aleph_\alpha\leq \aleph_\alpha$ by transfinite induction.

We say that $(a,b)<(c,d)$ if

$\max\{a,b\}<\max\{c,d\}$, or

$\max\{a,b\}=\max\{c,d\}$ and $a<c$, or

$\max\{a,b\}=\max\{c,d\}$ and $a=c$ and $b<d$.

We first argue that it suffices to show that for arbitrary $\alpha_1,\alpha_2\in\omega_\alpha,$ the set
$$X=\{(\zeta_1,\zeta_2)\in\omega_\alpha\times \omega_\alpha : (\zeta_1,\zeta_2)<(\alpha_1,\alpha_2 )\}$$
has cardinality smaller than $\aleph_\alpha$.

To show this, we first set $\beta=\max\{\alpha_1,\alpha_2\}+1$. We can now observe that $\beta\in\omega_\alpha$ and $X\subseteq\beta\times \beta$.

Next, we choose $\gamma<\alpha$ for which $|\beta|\leq \aleph_\gamma$. This implies $|X|\leq |\beta|\cdot |\beta|\leq \aleph_\gamma\cdot \aleph_\gamma\leq \aleph_\gamma<\aleph_\alpha$, where the third inequality uses the induction hypothesis. This should finish the proof.

However, what guarantees that such $\gamma<\alpha$ exists at all?

答: $\omega_\alpha$ is an initial ordinal, so by definition $|\beta|<\aleph_\alpha$ for all $\beta<\omega_\alpha$. The cardinality of any well-ordered set is an $\aleph$, so $|\beta|=\aleph_\gamma$ for some ordinal $\gamma$, and since $|\beta|<\aleph_\alpha$, it must be the case that $\gamma<\alpha$.

Title: Choosing aleph with given property


问:
$\frac{2.00013579}{1.00013579^2+2.00013579}$ > $\frac{2.0002468}{1.0002468^2 + 2.0002468}$?

Determine which one of the following two numbers is greater: $\frac{2.00013579}{1.00013579^2+2.00013579}$ or $\frac{2.0002468}{1.0002468^2 + 2.0002468}$.

How can I do this without a calculator? I tried comparing the numerator to the numerator, and the denominator to the denominator, but that did not work. I tried seeing whether the squares would change very little since they were really close to $1$, but I couldn't form any solid reasoning.

Thanks!

Body:
Hint: let $2.00013579=x,2.0002468=y$,$x,y> 2$

Also notice that $$f(x)=\frac{x}{{(x-1)}^2+x}=\frac{1}{x+\frac{1}{x}-1}$$ is decreasing for $x\ge 1$

As $x<y$:$$\frac{x}{{(x-1)}^2+x}>\frac{y}{{(y-1)}^2+y}$$


问:
finding the base of a triangle
In A triangle ABC ,AB=AC.D is a point inside the triangle such that AD=DC.Median on AC from D meets median on BC  from A at the centroid of the triangle.If the area of the triangle  ABC equals to $4\sqrt 3$ .Find the base i.e. BC.
The method that i have used to solve this problem works by ending with  two answers for $\frac12$ of BC and then having to check which one works by trial and error.It is more or less efficient,but I am assuming there's a better one.</p>

答:
If $AD=DC$ ike you said then $\Delta ABC$ is isosceles. Also we have that triangle $\Delta ABC$ is isosceles. A centroid is the intersection of the three medians of a triangle intersecting opposite vertices thus if the median from $AC$ meets median on $BC$ at the centroid, then both medians must intersect their respective opposite vertices. Then we can conclude that the median from $D$ on $AC$ is also a median from $B$ on $AC$.  This median forms an angle of $90°$ with side $AC$ since $\Delta ADC$ is isosceles. Now we know that triangle $\Delta ABC$ has two medians that are also perpendicular bisectors, it is easy to see that triangle $\Delta ABC$ is equilateral and $AB = BC = CA. $

$$ [\Delta ABC]= \cfrac 12 BC^2\sin 60° = \cfrac{\sqrt 3}{4}BC^2=4\sqrt 3 \implies BC=4$$

Title: finding the base of a triangle


问:
Distance from a point to a set in Euclidean space
If $Y \subseteq \mathbb{R^n}$ is a closed set. Show that given any $a \in \mathbb{R^n}$, there exists $x_{0} \in Y$ such that $|x_{0}−a| ≤ |x−a|$ for all $x \in Y$.

let's remember that: "we define the distance of a point $a \in \mathbb{R^n}$ to a set $Y \subseteq \mathbb{R^n}$ as the number $d(a,Y)=$ $\inf$ {$|x-a|:x \in Y$}"

The idea I had was that:

If $Y \subseteq \mathbb{R^n}$ is closed, then for any $a \in \mathbb{R^n}$, the function $f:Y → \mathbb{R}$ given by $f(x)=|x-a|$ assumes its minimum value at some point $_{0} \in Y$; so there is $d(a,Y)=|x_{0}-a|$.

but I must do it by sequences and not functions, then I would not know how to do it

答: By definition of infimum of a  set $d(a,Y)$ can be written as $\lim |a-x_i|$ for some sequence $(x_i) \subseteq Y$. This implies that $(a-x_i)$ is bounded and hence $(x_i)$ is also bounded. There is a subsequence of $x_i$ which is convergent. Take $x_0$ to the limit of this subsequence.


问:
Is $A_n$ uncountable for all $n \in \Bbb N\ $?

For all $n \in \Bbb N$ consider the set $A_n = \left \{x \in \Bbb R\ |\ \left (\cos x \right )^n + \left (\sin x \right )^n = 1 \right \}.$ Is $A_n$ uncountanble for all $n \in \Bbb N\ $?

If $n = 2$ then $A_n = \Bbb R$ which is uncountable. How do I prove it for $n \neq 2\ $?

Any help in this regard will be highly appreciated. Thanks in advance.

EDIT $:$ This question is motivated from option $(2)$ of the original question.

答:
$\sin x+\cos x=1$ implies that $1=1^{2}=\sin^{2}(x)+\cos^{2}(x)+2\sin (x) \cos (x)$ which gives  $\sin (2x)=0$. Hence $2x$ is a an integer multiple of $\pi$. Hence $A_1$ is countable.

Using Complex Analysis it is easy to see that $A_n$ is uncoutable only for $n=2$. If $A_n$ is uncountable it has  a limit point. A Theorem in Complex Analysis shows that we must have $\cos^{n} x+\sin^{n}x=1$ for all $x$. Put $x=\frac {\pi} 4$ to see that $n=2$.

Real analytic proof for $n$ even: On $A_n$ we have $t^{m}+(1-t)^{m}=1$ wheer $m=\frac n 2$ and $t=\sin ^{2}x$. This equation is a polynomial equation in $t$ and it has only finitely many roots. Hence there are only finite number of possible values for $\sin ^{2}x$ and each of then give only countably many values of $x$.


问:
Probability & Random Variables -- probability of $k$ machines remaining on day $i$?
I have a question that Im struggling with that has to do with probability, random variables, and (possibly) indicators

The scenario is that you have n machines, and each machine has a probability p of working. Every remaining machine is tested everyday, and the ones that don't work are thrown away. $X_{i}$ is the random variable that represents how many machines remain after i days. We start with $X_{0} = n$

The distribution of $X_1$ is just $P[X_1=k] = {n\choose k}p^k (1-p)^{n-k}$ I believe

The next part asks for the distribution of $X_2$, but that's where I'm confused. Doesn't that depend on the previous distribution and doesn't p change accordingly each time? But we are asking for the unconditional probability distribution so we cant use the first part

I know that the probability of one machine remaining on day 1 is p. and then the probability of the same machine remaining on day 2 is I think $p^2$ but Im not 100% sure? So would I be correct is saying the distribution for $X_t$ is just ${n\choose k}p^{tk} (1-p)^{n-k}$?

答: You're very close.  The probability that a machine is still working by day $t$ is $p^t$, so the probability that is is not working on day $t$ is not $1-p$, but $1-p^t$.  The answer should be $$\Pr(X_t=k)={n\choose k}p^{tk} (1-p^t)^{n-k}$$


问:
Taylor series for $\sec(x)$
I'm trying to compute the Taylor series for $\sec(x)$ but the derivatives are getting way too complicated. So how do I go about this without having to calculate all the derivatives? I tried to build some kind of relationship with the series for $\cos(x)$ but I didn't get anything meaningful.</p>

答:
There is the expression
$$\sec z = 1 + {z^2\over 2} + {5z^4\over 24} + \cdots + {(-1)^n E_{2n}\over (2n)!} z^{2n} + \cdots$$
for the Taylor series, where $E_{2n}$ is an Euler number; see Abramowitz and Stegun, {\sl Handbook of Mathematical Functions}, p.~75, Equation 4.3.69; and the discussion of Euler numbers on pp.~804--805, and the table of values for Euler numbers on p.~810.  I don't know if there is an efficient way of computing Euler numbers for a large index, but it is at least a different place to start. (I stumbled on this looking for a proof that all the Taylor coefficients of~$\sec(z)$ are positive.)

Title: Taylor series for $\sec(x)$


问:
Alternative permutation notation (lines and bars)
I'm not quite sure whether this belongs here per se, but I remember reading online (it could well have been here on MSE or MO, but I think it might've been something on arxiv) some time ago about a kind of alternative notation for permutations consisting of vertical lines and horizontal lines between them. As I recall, one starts with $n$ vertical lines and then places horizontal bars between them at appropriate intervals, so that if one puts one's finger on the top of the $i$-th bar and traces out a path, going down and following whichever bar happens to be along the way, one winds up at $\sigma(i)$.
If I recall correctly, the origin was mentioned to have been Japanese, but I can't be sure about that. And what made it effective was the fact that for a product of permutations one can simply stick these diagrams together, one on top of the other, and in effect have the diagram for the product.

Has anyone here ever heard of or come across such a thing? I just remembered this the other day and thought it would be interesting to read about it again.

答:
Sounds like ghost leg / amidakuji:

Abstractly, a ghost leg is a diagrammatic way to describe a permutation as a composition of simple transpositions $(i, i+1)$. This particular set of generators makes the symmetric group into a Coxeter group and much is known from that theory about how to express permutations as such compositions.

Ghost legs are apparently used to randomly assign things (like prizes), which I find strange because the resulting distribution over permutations cannot possibly be uniform, although maybe it's close enough.

Similar diagrams called braid diagrams, incorporating crossings, appear in the theory of braid groups and are a standard tool there:


问:
Properties matrix $A^2=I$
Let be $A$ a matrix such that $A^2=I$. According my book the $A$'eigenvalues are $-1$ and $+1$?. but, For me is only $+1$, this because $A = \sqrt{I} = I$.</p>

答: Look at $A=-I$: what is its square, and what are its eigenvalues?</p>

Title: Properties matrix $A^2=I$


问:
Are there any counterexamples which show that $E(XY) = E(X)E(Y)$ cannot imply $X,Y$ is not necessarily independent?
I know that if $X,Y$ are independent, then $E(XY) = E(X)E(Y)$ holds. But the reverse is not true. I'd like to know that are there any examples to show the fact?

答:
Let $X$ and $Z$ be independent random variables such that $X$ takes $0$ and $1$ with probability $1/2$ each and $Z$ takes $-1$ and $1$ with probability $1/2$ each.

Then $XZ$ and $X$ satisfy the condition you have given but are not independent.


答:
Let $X$ be a random variable that takes value $0$ with probability $1/2$ and $1$ with probability $1/2$. Let $T$ be a random variable, independent of $X$, that takes the value $−1$ with probability $1/2$, and takes the value $1$ with probability $1/2$. And let $Y = XT$.

We have that $E[X]=1/2$ and $E[Y] =0$. I'll leave it to you to check that $E[XY] = [X^2 T]=0$, which would mean $E[XY]=E[X]E[Y]$.

Now the question is, are $X$ and $Y$ independent? What's the probability that $Y=1?$ What's the probability that $Y=1$ given that $X=0?$ Conditioning on $X$ affects $Y$, so they're not independent.

Title: Are there any counterexamples which show that $E(XY) = E(X)E(Y)$ cannot imply $X,Y$ is not necessarily independent?

Body: I know that if $X,Y$ are independent, then $E(XY) = E(X)E(Y)$ holds. But the reverse is not true. I'd like to know that are there any examples to show the fact?


问:
Coinflip statistics
There is a betting strategy for coinflips, called Martingale, in which the player should pick heads or tails and keep betting on it (doubling their bet on every loss) to eventually make a win when it appears. In theory this is watertight, but with one caveat: there can be very long strings of losses, and the player runs out of money, and looses it all.

Now, some people wait until they have seen a very long string of outcomes for one side, and start betting at the other side, with the idea that has a larger chance now. I know that mathematicly this is nonsense, both sides still have 50% chance no matter what happend before.

Although I can understand a dice/coin/ball has no notion of past/future, so will always remain 50% chance, I never quite understanded the following. You can proof there is only a very slight possibility of me ever witnessing 50 heads in a row during my lifetime. So after 49 spins, I will either observe something extraordinary (50 heads) or something less special (first tail), how can that not effect chance?

Maybe the answer is obvious, that for math/statistics it doesnt matter whether someone is observing the event or not, but please enlighten me.

答: You are extremely unlikely to see 50 heads in a row in your lifetime.  You are equally unlikely to see 49 heads followed by a tail in your lifetime.  After seeing 49 heads in a row, you are guaranteed to see something extremely unlikely with the next flip!</p>

Title: Coinflip statistics


问:
Argument Valid Or Not?
Premise 1 : $A \to B$

Premise 2 : $A \vee C$

Premise 3 : $C \to D$

Premise 4 : $\neg B$

Conclusion : $D$

How to check if the argument is valid?

Thanks.

答:
Hint

Premise 1 is equivalent to
$$
\lnot B\to \lnot A
$$
Now, use it with premise 4 and then with premises 2 and 3 respectively to conclude what you want.

Title: Argument Valid Or Not?


问:
$\lim _{z\rightarrow -i}\dfrac{1}{z-i}=\dfrac{i}{2}$ Why this answer?
$i$ is imaginary number.
I think this quiestion.
$$\lim _{z\rightarrow -i}\dfrac{1}{z-i}=\dfrac{i}{2}$$
I think this answer is $-\dfrac{1}{2i}$.

I don't know why this answer is $\dfrac{i}{2}$.
please tell me.

Body:
$$
-\frac{1}{2i} = -\frac{i}{(2i)i} = -\frac{i}{-2} = \frac{i}{2}
$$


问:
Details of Defining Free Group in Terms of Universal Property
The only definition of the free group I have seen thus far (undergraduate) is an explicit construction
via reduced words and concatenation.

What I am now trying to do is see if it is possible to define the free group abstractly in terms of the universal property. The definition I think should work is:

A free group $F_S$ on a set $S$ is a group generated by $S$ such that for any group $G$ and any function $f: S \rightarrow G$, $f$ extends to a unique homomorphism $\tilde{f}: F_S \rightarrow G$.

So what I want to do now is prove existence and uniqueness (up to isomorphism?) from this abstract definition, and I'm struggling to make progress, so I'm looking for tips, specifically:

Is my definition correct, or is there something I'm missing?

Any tips on how to make progress on existence?

Any tips on how to make progress on uniqueness?

答:
You actually don't need to specify that $F_S$ is generated by $S$, only that you've equipped $F_S$ with some map $i : S \to F_S$ (which you also don't need to require is injective in case you were thinking about doing that). Generation will turn out to follow from the universal property. Other than that, yes, this is how it goes.

Uniqueness (up to unique isomorphism) follows from the Yoneda lemma; this is a general feature of objects defined by universal properties and is a major reason why using universal properties is so convenient.

Existence is harder. You can show that the concrete definition using reduced words satisfies the universal property; it takes a bit of work but it's a convenient fact to be able to refer to. One possible more abstract argument is the following. We first consider the free monoid on $S$; this is easy to construct explicitly as the monoid

$$\text{List}(S) = \bigsqcup_{n \ge 0} S^n$$

of lists of elements of $S$, or equivalently as the monoid of words on the alphabet $S$, under concatenation, and it's easy to prove the universal property.

To get the free group from here we need to freely adjoin inverses to this monoid. The group obtained by freely adjoining inverses to the free monoid $\text{List}(S)$ can be constructed by starting from $\text{List}(S \sqcup S^{-1})$, where $S^{-1}$ is just another copy of $S$ that stands for the inverses we're adjoining, then quotienting by relations that say that each $s \in S$ actually has the corresponding formal symbol $s^{-1} \in S^{-1}$ as its inverse. How satisfying this is depends on how comfortable you are with taking quotients of monoids, but it does have the benefit that we don't need to talk about reduced words so the universal property is easier to check.

This construction is also a good prototype for how to construct other free objects. The list construction generalizes to constructing, for example, for example, the free $k$-algebra on a $k$-vector space $V$ as the tensor algebra

$$T(V) = \bigoplus_{n \ge 0} V^{\otimes n}.$$

In turn the tensor algebra is an important intermediate step for constructing other objects like the free commutative $k$-algebra on $V$, which is the symmetric algebra.


问:
Induced representation in Etingoff. Remark 5.8.2
In Etingoff’s notes on representation theory, he states in remark 5.8.2 that $Hom_H(k[G],V)$ is naturally isomorphic to $Ind_H^G(V)$ as representations. But I can’t quite see how this is so. $Ind_H^G(V)$ is a representation of $G$ but how is $Hom_H(k[G],V)$ also a representation of $G$? Or is the isomorphism of representations with respect to the subgroup $H$? Or is it just a simple isomorphism of vector spaces? Could someone care to explain?

答:
It's an isomorphism of representations. The $G$-action is a little bit buried: when we consider $\text{Hom}_H(k[G], V)$ the natural $H$-action on $k[G]$ is by left multiplication. $k[G]$ has both a left and right $k[G]$-module structure given by left and right multiplication respectively, so there's a right $G$-action on $k[G]$ that is left "untouched" and, by contravariance, gives a left $G$-action on $\text{Hom}_H(k[G], V)$.

I feel duty-bound to point out also that this isomorphism only holds if $H$ has finite index in $G$ (probably in context $G$ and $H$ are even finite). Otherwise there are two induced representations (given by the left and right adjoint of restriction) and this is one of them (the right adjoint); the other one (the left adjoint) is the tensor product $k[G] \otimes_H V$ (with the $G$-action given by left multiplication this time) and they have different dimensions in general.


问:
Existence of a power series converging non-uniformly to a continuous function
I am wondering whether there exist a function $f(z) = \sum_{n\geq0} a_n z^n$ such that:

$f$ converges and is continuous on the closed unit disk $D$ and

the series $\sum_n a_n z^n$ does not converge uniformly on $D$.

I have tried to construct a counter-example, but with no success so far.

答:
Yes.  See D.R. Lick's article, "Sets of non-uniform convergence of Taylor Series."

It is shown there that for every closed subset $F$ of the boundary $\partial D$, there is such a series that converges everywhere on the closed disk to a continuous function, and whose set of non-uniform convergence is $F$.  This means that the series converges uniformly in a neighborhood (open arc) of each point in $\partial D\setminus F$, but not in any neighborhood of any point in $F$.

I have been interested in this question before, which is why I have a reference handy. I referred to this article in another answer, to a question about power series in Banach algebras.  As mentioned there, the sequence of Cesàro means of the power series will converge uniformly on the closed disk.


问:
Find all $c \in \Bbb Z_5$ such that $\Bbb Z_5[X]/ \langle x^2+cx+1 \rangle$ be a field.
Find all $c \in \Bbb Z_5$ such that $\Bbb Z_5[X]/ \langle x^2+cx+1 \rangle$ be a field.

Attempts:

$\Bbb Z_5[X]/ \langle x^2+cx+1 \rangle$ be a field iff $p(x) = x^2 + cx +1$ is irreducible in
$\Bbb Z_5$ iff $p(x)$ has no roots in $\Bbb Z_5$. I found that $2 \ne c \in \Bbb Z_5$ are the answer.

Am I true?

答:
Just make a little table of the values of $x^2+cx+1$, its values are $1,2+c,2c, -2c, 2-c$.
Hence see that $c=\pm 2$ and $c=0$ give reducible quadratics.


答:
Your definition is right and uniqueness follows from it from a purely categorical argument. Namely, you're defining a free group $F_S$ over $S$ and, to make things more precise, you should make an "inclusion" $\iota:S \to F_S$ part of the definition, so that for every group $G$ and every function $f:S \to G$, there is a unique homomorphism $\widetilde{f}:F_S\to G$ with $f = \tilde{f} \circ \iota$.

Assume that $F_{S,1}$ and $F_{S,2}$ are two free groups over $S$, with inclusions $\iota_i:S \to F_{S,i}$, $i=1,2$. Apply the definition of free group to $\iota_1: S \to F_{S,1}$ to get a homomorphism $\widetilde{\iota_1}:F_{S,2}\to F_{S,1}$ with $\iota_1 = \widetilde{\iota_1}\circ \iota_2$. Switching roles of $F_{S,1}$ and $F_{S,2}$, we also obtain a group homomorphism $\widetilde{\iota_2}:F_{S,1}\to F_{S,2}$ with $\iota_2= \widetilde{\iota_2}\circ \iota_1$. Now we have a homomorphism $\widetilde{\iota_1}\circ\widetilde{\iota_2}:F_{S,1}\to F_{S,1}$, which has to be the identity, as $$(\widetilde{\iota_1}\circ\widetilde{\iota_2})\circ \iota_1 = \widetilde{\iota_1}\circ (\widetilde{\iota_2}\circ \iota_1) = \widetilde{\iota_1}\circ \iota_2 = \iota_1 $$and $${\rm Id}_{F_{S,1}}\circ \iota_1 = \iota_1.$$More precisely, we're able to conclude that $\widetilde{\iota_1}\circ\widetilde{\iota_2} = {\rm Id}_{F_{S,1}}$ because of the uniqueness of extensions in the definition. Switching roles of $F_{S,1}$ and $F_{S,2}$ again, we also get that $\widetilde{\iota_2}\circ\widetilde{\iota_1} = {\rm Id}_{F_{S,2}}$. So $F_{S,1}\cong F_{S,2}$.

Here's what's going on behind the scenes: defining a category $\mathcal{C}_S$ by setting $${\rm Obj}(\mathcal{C}_S) = \{ (F,\iota) \mid \iota:S \to F \mbox{ and $F$ is a group}\}$$ and $${\rm Mor}_{\mathcal{C}_S}((F_1,\iota_1),(F_2,\iota_2)) = \{ \phi:F_1\to F_2 \mid \phi \mbox{ is a homomorphism and }\phi\circ \iota_1=\iota_2\},$$a free group $(F_S,\iota)$ over $S$ is nothing more than an initial object in $\mathcal{C}_S$. But it's a general fact (which I invite you to prove --- mimicking the above) that in any category, initial objects are unique up to (unique!) isomorphism.

However, all of this fluff does not establish existence of $F_S$. But since we already know that if it exists, it will be unique up to isomorphism, to get existence it suffices to exhibit one construction. Then the usual construction fits the bill.

The universal property gives an elegant way of constructing homomorphisms between free groups. For example, if $S_1$ and $S_2$ are sets and we have a function $f:S_1\to S_2$, then apply the universal property of $F_{S_1}$ to $\iota_2\circ f$ to get a group homomorphism $\widetilde{\iota_2\circ f}:F_{S_1}\to F_{S_2}$. If $f$ is bijective, then $\widetilde{\iota_2 \circ f}$ is an isomorphism with inverse $\widetilde{\iota_1\circ f^{-1}}$ (check this without picking elements, it's instructive).

Title: Details of Defining Free Group in Terms of Universal Property

Body:
The only definition of the free group I have seen thus far (undergraduate) is an explicit construction
via reduced words and concatenation.

What I am now trying to do is see if it is possible to define the free group abstractly in terms of the universal property. The definition I think should work is:

A free group $F_S$ on a set $S$ is a group generated by $S$ such that for any group $G$ and any function $f: S \rightarrow G$, $f$ extends to a unique homomorphism $\tilde{f}: F_S \rightarrow G$.

So what I want to do now is prove existence and uniqueness (up to isomorphism?) from this abstract definition, and I'm struggling to make progress, so I'm looking for tips, specifically:

Is my definition correct, or is there something I'm missing?

Any tips on how to make progress on existence?

Any tips on how to make progress on uniqueness?


问:
Can surface integrals be defined in all of $\mathbb{R}^{3}$?
I just thought of an interesting example of a surface integral. Perhaps someone with a background in differential forms/integration of manifolds can help with this. A general multivariable calculus perspective would be useful, too.

Can surface integrals be defined over the entirety of $\mathbb{R}^3$? Both in terms of vector and scalar fields. Intuitively, at least with regard to flux integrals, this should work. If we allow the surface integral to enclose $\mathbb{R}^3$, then the Divergence Theorem can show, for example, that if $\nabla \cdot \vec{F} = \frac{1}{(1+x^2)(1+y^2)(1+z^2)}$, then $\iiint_{\mathbb{R}^3} \nabla \cdot \vec{F} dV = \pi ^3.$ Can this be interpreted as, for example, doing a surface integral over a sphere of radius $R$ and letting $R \to \infty$? This is also an example of using the divergence theorem for a flux integral, but can we do the same thing over a scalar field? Is there a function $f(x,y,z)$ such that $\oint_S f dS$ is a finite value, with $S$ being a sphere of radius $R$ as $R \to \infty$?  And if we can interpret a flux integral as flux through a surface, does this surface integral mean...... a flux coming out of the whole of $\mathbb{R}^3$? And if it's a scalar field, and we interpret $f(x,y,z)$ as a mass-density function, is this the mass of the exterior of $\mathbb{R}^3$.....?

Is someone able to give a little bit of an analysis on this? Thanks.

EDIT: I realize now I've misused the term "defined on $\mathbb{R}^3$" with regard to surface integrals. What I mean is that a surface integral to have a surface that encloses $\mathbb{R}^3$. Have the interior be $\mathbb{R}^3$.

答:
Let $B_R$ be the ball centered on the origin of radius $R$ and $S_R$ its boundary (the sphere)

Then by divergence theorem :

$$\iiint_{B_R} \nabla \cdot \vec F ~d\vec V =
\iint_{S_R} \vec F~ d\vec S.$$

Then we make $R \to \infty$. If $\nabla \cdot \vec F$ is integrable, then dominated convergence theorem ensures that taking this limit is the same as integrating on the whole space $\mathbb{R}^3$ in the first place and this is how far you can go. An eventual surface enclosing the whole space can be seen, in this context, as a limit of the integral over larger and larger ones.

From a "flux interpretation", the integrability ensures that most of the contribution to the flux is reachable with bounded surfaces. Indeed you will able to approximate the integral with an arbitrary accuracy using the appropriate bounded surface. Formally speaking it means that an underlying physical system will lose/radiate or receive energy. For example in Electromagnetics this would be connected to the power radiated by an antenna.


问:
How to describe the kernel of a homomorphism
I am generally unsure of what a kernel is in abstract algebra. From my background in linear algebra, I understand that the kernel is the same as the null space (I learned it as the null space).

For example, if I have some homomorphism $f: (\mathbb{R},+) \longrightarrow (\mathbb{C}^*, \cdot)$, how would I go about describing the kernel?

答:
Ther kernel of a (group) homomorphism is the set of all elements that get mapped onto the neutral element.

So $\ker(f)=\{x\in\mathbb{R}: f(x)=1\}$, as $1$ is the neutral element of $\mathbb{C}^\ast$ with regards to multiplication.

How to describe the kernel then depends clearly on the specific homomorphism.


问:
Struggling to find solution: question 3
Could someone help with this problem?
$$x^2y''-7xy'+7y=13\sqrt{x}$$

I have brought it to its linear form and it came to this:

$$y''-\frac{7}{x}y'+\frac{7}{x^2}y=13\frac{1}{\sqrt{x^3}}$$

Here I'm stuck, should I do reduction of order, but for that we must know one of the solutions, and I just couldn't find it.

答:
Hint:
$$x^2y''-7xy'+7y=13\sqrt{x}$$
This is Cauchy-Euler's differential equation:
$$y''-7\dfrac {(xy'-y)}{x^2}=\dfrac {13}{x\sqrt{x}}$$
$$y''-7\left (\dfrac {y}{x}\right)'=\dfrac {13}{x\sqrt{x}}$$
Integrate:
$$y'-\dfrac {7y}{x}=13\int \dfrac { \ dx}{x\sqrt{x}}$$
$$y'-\dfrac {7y}{x}=- \dfrac {26 }{\sqrt{x}}+C_1$$

Title: Struggling to find solution: question 3


问:
What is the name of the functions $V\to V$ which map every 1-dimensional subspace of $V$ to an 1-dimensional subspace of $V$?
Motivation: the permutations of the projective space $P(V)$ can be regarded as the equivalence classes of such functions (two such functions, $f$ and $g$ are regarded to be equivalent if for every 1-dimensional subspace $L$ of $V$, $\{f(x):x\in L\}=\{g(x):x\in L\}$). I would call them projective functions but I'm afraid that this name means something else.

Edit

I definitely mean of all such functions, not only the homogeneous (or more specifically, the linear) ones.

答:
Here is how to construct an arbitrary function $\phi:V\to V$ sending 1D subspaces to other 1D subspaces (which I will just call "lines" for short). Picking any two lines $L_1$ and $L_2$, we know $0$ must be mapped to a point on both lines $\phi(L_1)$ and $\phi(L_2)$, which is just $0$, so $\phi$ fixes $0$. It remains to see how $\phi$ can act on nonzero vectors. We can split this up into two parts: how $\phi$ acts on the set $P=\mathbb{P}V$ of all lines, and how it permutes points within lines.

For every function $\sigma:P\to P$, there are going to be $\phi$s which send lines $L$ to lines $\sigma(L)$. Note the restriction of $\phi$ to a line must be a bijection. Thus, we can decompose the set $\Phi$ of all such $\phi$s into

$$ \Phi = \bigsqcup_{\sigma:P\to P} \prod_{L\in P} \mathrm{Bij}(L,\sigma(L)) $$

where $\mathrm{Bij}(X,Y)$ denotes the set of bijections $X\to Y$ and the lines are considered to not contain $0$. As a monoid, this is a kind of "monoid wreath product" of the monoid of all $\sigma:P\to P$ and the direct product of symmetric groups $S_L$ for all lines $L$ (again, ignoring $0$). This might be easier to understand by looking at the subgroup $G$ of invertible $\phi$s.

First, we can construct a copy of $S_P$ within $G$ that permutes the lines themselves as follows. For each line $L$, choose a bijection $\ell_L:L\to k^{\times}$ (where $k$ is the base field), which induces a bijection between any two lines as $\ell_{L_2}\circ\ell_{L_1}^{-1}:L_1\to L_2$. Then, given any permutation $\pi\in S_P$, we can think of it as a function on $V$ (fixing $0$) by saying $\pi$'s restriction to any line $L$ is $\ell_{\pi(L)}\circ\ell_L^{-1}$. This givecs a copy of $S_P$ within $G$.

The subgroup of $G$ which fixes every line is a direct product $\prod_{L\in P}S_L$. Note this direct product intersects $S_P$ trivially, and that $S_P$ normalizes this direct product. Moreover, if $\phi\in\Phi$ is arbitrary (well, invertible) and induces a permutation $\pi$ of $P$ then $\phi\circ\pi^{-1}$ fixes each line so is an element of the direct product. That means

$$ G= \big(\prod S_L\big)\rtimes S_P \cong S_{k^{\times}}\wr S_P $$

is a wreath product.


问:
Homeomorphic metric spaces where the identity map is not a homeomorphism
In our Topology lectures, we were told that there are examples where two metric spaces $(M, d)$ and $(M, d')$ are homomeomorphic (i.e. there exists some homeomorphism between them, which is not necessarily the identity map) but the identity map $\operatorname{Id}\colon(M, d) \rightarrow (M, d')$ is not a homeomorphism. I was hoping someone could provide an example of this.

Body:
$X = [0,1)$.

$d =$ usual distance.

$d'(x,y) = |x - y|$ for $x,y\in(0,1)$;

$d(0,x) = |x - 1|$ for $x\in(0,1)$.

Idea for $d'$: take $(0,1]$ and rename $1$ as $0$.


答:
Your use of the square root of a matrix does not quite work.

Consider that all matrices
$$
A =
\frac{1}{ad-bc} \,
\begin{bmatrix}
cd - ab & d^{2} - b^{2}\\
a^{2} - c^{2} & ab - c d
\end{bmatrix},
$$
for $ad - b c \ne 0$,
have $A^{2} = I$

Title: Properties matrix $A^2=I$

Body: Let be $A$ a matrix such that $A^2=I$. According my book the $A$'eigenvalues are $-1$ and $+1$?. but, For me is only $+1$, this because $A = \sqrt{I} = I$.</p>


问:
Justifying the closure under addition of $\ell^1$
$
\newcommand{\setb}[2]{\left \{ #1 \; \middle| \; #2 \right \} }
\newcommand{\seq}[1]{ \left\{ #1 \right\}_{n=1}^\infty }
\newcommand{\eq}[1]{ \stackrel{(#1)}{=} }
\newcommand{\leq}[1]{ \stackrel{(#1)}{\le} }
$
Background Context:

This is something one of my professors posed as a bit of an open question during a lecture, since my class and I struggled a bit to justify it rigorously on the spot. I think I have an approach, but just wanted to make sure it passed muster. (Especially because it ticks me off a little that I couldn't do it myself off the cuff.)

Definitions & The Problem:

First, recall: we define $\ell^1$ as

$$\ell^1 := \setb{ \text{all sequences }\seq{x_n} \subseteq \Bbb F }{ \sum_{n=1}^\infty |x_n| < \infty }$$

for a field $\Bbb F$, usually $\Bbb R$ or $\Bbb C$. We may assume the latter here for simplicity if we ever need to. This (once the axioms are verified) make $\ell^1$ a vector space over the chosen field.

I wish to justify part of that claim. Specifically, closure under addition; that is,

$$\left( x := \seq{x_n} \in \ell^1 \right) \text{ and } \left( y := \seq{y_n} \in \ell^1 \right) \implies x+y := \seq{x_n + y_n} \in \ell^1$$

To show this, note we need to show the corresponding sum for $x+y$ is finite; since we are summing absolute values, it is sufficient to show $\sum_{n=1}^\infty |x_n+y_n|$ is bounded above.

My Approach:

So, my approach:

\begin{align*}
\sum_{n=1}^\infty |x_n + y_n| &\eq1 \lim_{N \to \infty} \sum_{n=1}^N |x_n + y_n| \\
&\leq2 \lim_{N \to \infty} \sum_{n=1}^N |x_n | + | y_n|\\
&\eq3 \lim_{N \to \infty} \sum_{n=1}^N |x_n | + \sum_{n=1}^N | y_n|\\
&\eq4 \lim_{N \to \infty} \sum_{n=1}^N |x_n | + \lim_{N \to \infty} \sum_{n=1}^N | y_n|\\
&\eq5 \sum_{n=1}^\infty |x_n | + \sum_{n=1}^\infty | y_n| \\
&\stackrel{(6)}{<} \infty
\end{align*}

Justifications for each step:

Definition of infinite sum (limit of partial sums).

Triangle inequality.

Split the now-finite sums into two, separate sums.

Since the corresponding sums for $x,y$ converge, we may apply the limit to each sum individually.

Here we just convert back into two infinite sums from the definition. Note that each sum is finite since $x,y \in \ell^1$.

Sum of finite numbers is itself finite. (If we assume the fields $\Bbb R$ or $\Bbb C$ as intended, this is further clarified by the fact that the infinite sums are sequences of partial sums, with value as the limit of the sequence, and these fields form complete metric spaces with respect to their usual metrics. Thus, they sum to values in $\Bbb R$ or $\Bbb C$, and in turn the sum of these values is in the field in question from the field axioms. Not sure how to justify the claim for incomplete fields, e.g. $\Bbb Q$ or $\Bbb A$, though.)

Does this pass muster, in terms of rigor?

Some Final Thoughts & Rambling:

For what it's worth: in class we were able to quickly get as far as the equality at $(3)$, but from there we all kinda got stuck. The professor referenced it as being something more fundamental than what some of us were suggesting, so perhaps we were overthinking things. In any case, perhaps my approach above is a more proper justification?

One thing that does bug me is that from $(3)$ he had already written a $\le$ sign for the next step, so on the board it looked more like this:

\begin{align*}
\sum_{n=1}^\infty |x_n + y_n| &= \lim_{N \to \infty} \sum_{n=1}^N |x_n + y_n| \\
&\le \lim_{N \to \infty} \sum_{n=1}^N |x_n | + | y_n|\\
&= \lim_{N \to \infty} \sum_{n=1}^N |x_n | + \sum_{n=1}^N | y_n|\\
&\le (\cdots?)
\end{align*}

Sadly I don't have the notes to clarify: it is possible that we simply started with finite sums? Like maybe instead what we had is

\begin{align*}
\sum_{n=1}^N|x_n + y_n| &\le \sum_{n=1}^N |x_n | + | y_n|\\
&= \sum_{n=1}^N |x_n | + \sum_{n=1}^N | y_n|\\
&\le (\cdots?)
\end{align*}

and then intended to take the limit $N \to \infty$ after? If it's the former, I'm not sure what the next step entailed. If it's the latter, I think the intention was something along the limits of what I suggested, say

\begin{align*}
\sum_{n=1}^N|x_n + y_n| &\le \sum_{n=1}^N |x_n | + | y_n|\\
&= \sum_{n=1}^N |x_n | + \sum_{n=1}^N | y_n|\\
&\le \sum_{n=1}^M |x_n | + \sum_{n=1}^M | y_n| \quad \forall M \ge N
\end{align*}

which would follow since $|x_n|,|y_n| \ge 0$ from definition, and I guess for $M \to \infty$ in particular, though I feel that is a bit convoluted for what the professor would likely intend.

Of course, it's also possible he intended to just have $=$ instead, or intended to combine several steps into one, or I'm just missing it. Still, if you can enlighten me as to how one might continue from either stance (if they're at all salvageable), that would also be appreciated.

But I'm mostly looking to see whether my own approach is sufficiently rigorous. Thanks for any insights you can give!

答:
I think your approach is sufficiently rigorous. Regarding what you said about your professor's notes, I think the finite sum interpretation works. As you said, it will be something like this
\begin{align*}
\sum_{n=1}^N|x_n+y_n| & \leq \sum_{n=1}^N|x_n|+|y_n| \\
& = \sum_{n=1}^N|x_n|+\sum_{n=1}^N|y_n| \\
& \leq \sum_{n=1}^\infty |x_n|+\sum_{n=1}^\infty |y_n|
\end{align*}
with the last inequality coming from the fact that the sequence of partial sums is an increasing sequence. Set $R = \sum_{n=1}^\infty |x_n|+\sum_{n=1}^\infty |y_n|<\infty$, then we have
$$\sum_{n=1}^N|x_n+y_n|\leq R$$
for all $N\in\mathbb{N}$ and hence
$$\sum_{n=1}^\infty |x_n+y_n| \leq R <\infty.$$


问:
Is a single element a Lattice?
Can a lattice be formed with just a single element? Because for a single element, both Greatest Lower Bound and Least Upper bound exists and is equal to the same element. So can this element be called as a lattice?</p>

答:
Yes, for every kind of algebra, in the sense of Universal Algebra, i.e., a set endowed with a set of finitely operations, there is a one element algebra, which can be achieved by taking the quotient of that algebra by the universal congruence relation, which is the relation $A^2$ on the algebra  with base set $A$.

Lattices are also algebras is this sense.
(See, for example, here.)

Considering lattices only as ordered structures, we still have one-element lattices, by the exact reason you mentioned.

So, yes, you are right.

Title: Is a single element a Lattice?


问:
A Diophantine equation and decimal digits
Solutions of the Diophantine equation

$a10^n+(a+1) = (2^{m+1}-1)*2^{m+1}$

are

12=3*4,

56=7*8,

67100672=8191*8192.

Are there more solutions/examples like that or a generalization of the equation? Does it describe a natural phenomenon?

答:
$1.$ Denote $t=2^{m+1},N=10^n.$ If $a<N$ then $1\leq a=\dfrac{t^2-t-1}{N+1}<N,$
hence $$N+1\leq t^2-t-1<N(N+1),$$with some discussion we will get
$$\sqrt{N}\leq t<N,10^{\frac{n}{2}}<2^{m+1}<10^n,$$
$$\frac{n}{2}\log_2{10}<m+1<n\log_2{10}$$
So for every given $n$,we can try all the $m$ one-by-one.I did this for $0<n<3000$,but no other solution was found.

$2.$ If $a<10^n$ is not necessary, we can use the following method for some little $n$.

Denote $t=2^{m+1},d=10^n+1,$ then $t^2-t-1\equiv0 \pmod d,$$$(2t-1)^2\equiv5 \pmod d\tag1$$
hence if $(1)$ has no solution,then $n$ is not the solution for the original problem.

If $(1)$ has some solutions, such as $t$,then $$2^{m+1}\equiv t \pmod d \tag2$$
If $(2)$ has solutions for $m$,then we get a solution for the original problem, and $m^{'}=m+k\phi(d)$ is a solution,too.

For example:

$n=5,m+1=13,a=671$

$n=2,m+1=64,a=3369132345751865974702256072852065939.$

$n=2,m+1=86,a=59270403034726518346161316806283592011628044701331.$

$n=5,m+1=4532,a=34346\cdots 204319.$


问:
Identifying the Closure and Interior
Let $$A = \left\lbrace\left.\frac{1}{n^{2}+1} \ \right| \ n\in \mathbb{N}\right\rbrace = \left\lbrace \frac{1}{2},\frac{1}{5},\frac{1}{10}, \frac{1}{17}, \frac{1}{26} ,\dotsc\right\rbrace \subset\mathbb{R}.$$ Identify the closure $\operatorname{Cl}(A)$ and the interior $\operatorname{Int}(A)$ with respect to the metric space $(\mathbb{R},d)$.

Anyone have any idea how to do this cause I am lost
Btw d is just the standard metric on R ie d(x,y)= |x-y|

答:
Cl(A) is A union its limit points. So what is

(lim_{n \to infty} \frac{1}{1+n^{2}}) ?

The interior is the union of all open subsets of A. Can you draw an open interval around any point in A that only contains points in A?

Title: Identifying the Closure and Interior


问:
Is $A_n$ uncountable for all $n \in \Bbb N\ $?

For all $n \in \Bbb N$ consider the set $A_n = \left \{x \in \Bbb R\ |\ \left (\cos x \right )^n + \left (\sin x \right )^n = 1 \right \}.$ Is $A_n$ uncountanble for all $n \in \Bbb N\ $?

If $n = 2$ then $A_n = \Bbb R$ which is uncountable. How do I prove it for $n \neq 2\ $?

Any help in this regard will be highly appreciated. Thanks in advance.

EDIT $:$ This question is motivated from option $(2)$ of the original question.

答:
A very simple argument to handle all integer coefficients $n>2$ is to remember that

$$\sin^2x + \cos^2x=1$$

holds for all $x \in \mathbb R$, from which follows that $\forall x \in \mathbb R: |\sin x| \le 1, |\cos x| \le1$, so for $n > 2$

$$\sin^n x + \cos^n x \le |\sin x|^n  + |\cos x|^n \le |\sin x|^2  + |\cos x|^2 = \sin^2x + \cos^2x=1,$$

where the second inequality follows form $\forall t \in [0,1]: t^n -t^2= t^2(t^{n-2}-1) \le 0 \Rightarrow t^n \le t^2$. Equality holds only for $t=0$ or $t=1$, so both $|\sin x|$ and $|\cos x|$ have to be $0$ or $1$, which happens only for a countable set of real $x$ (just either one being $0$ or $1$ is already just a countable set).


问:
Prove the expected value of a normal variable
I would like to ask how would I prove that if $X$ is a normal variable with mean $\mu$ and variance $\sigma^2$, then $\mathbb{E}e^{u(X-\mu)} = e^{\frac{1}{2}u^2\sigma^2}$?

答:
Since $X\sim N(\mu;\sigma^2)$ it is easy to prove that

$$Y=u(X-\mu)\sim N(0;\sigma^2 u^2)$$

Thus

$e^Y$ is lognormal distributed with mean $e^{\frac{\sigma^2 u^2}{2}}$


答:
$$Y''-\frac{7}{x}Y'-\frac{7}{x^2}Y=13\frac{1}{\sqrt{x^3}}=f(x)~~~~(1)$$
The corresponding homogeneous ODE is
$$y''-\frac{7}{x}y'-\frac{7}{x^2}y=0~~~~~~~(2)$$
Whose solution can be found by Euler's method by putting $y=x^m$, we get $m(m-1)-7m-7=0 \implies m_{1,2}=4\pm\sqrt{23}$. So two linearly independent solution of (2) are $y_1=x^{m_1}, y_2=x^{m_2}$ whose Wronskian $w(x)=y_1y'_2-y'_1y_2.$ The solution of (1) can be written by the method of variation of parameters ($C_1, C_2$) as $$Y(x)=C_1(x) y_1(x)+C_2(x) y_2(x),$$ w3here
$$C_1=-\int \frac{f(x)y_2(x)}{w(x)}~dx+D_1,~~C_2=\int \frac{f(x)y_1(x)}{w(x)} dx+ D_2.$$
Soory the numbers involved are not very good, I hope you can take it up from here.

Title: Struggling to find solution: question 3

Body:
Could someone help with this problem?
$$x^2y''-7xy'+7y=13\sqrt{x}$$

I have brought it to its linear form and it came to this:

$$y''-\frac{7}{x}y'+\frac{7}{x^2}y=13\frac{1}{\sqrt{x^3}}$$

Here I'm stuck, should I do reduction of order, but for that we must know one of the solutions, and I just couldn't find it.


问:
Proving cluster point sequences
Below the following question is my solution but I am not sure whether the proofs are correct and rigorously exhaustive so, if present, kindly point out any errors.

Let ${x_n}$ be a convergent sequence with $\underset{n}{\rm lim}$ $x_n =x.$ Consider the set $S=\{ x_n: n \in \mathbb{N}\} \subset \mathbb{R}.$

$(i)$ Assume that $x_n \ne x \hspace{7pt} \forall n \in \mathbb{N}.$ Prove that $x$ is a cluster point of $S$.

$(ii)$ Under the assumption of $(i)$, prove that $x$ is the only cluster point of $S$.

$(iii)$ What happens if the assumption in $(i)$ is removed? Is $x$ necessarily a cluster point? Prove it or provide a counterexample.

Answer:

$(i)$ Let $N$ be any neighbourhood of $x$ then $\exists \epsilon>0$ such that $(x-\epsilon, x +\epsilon) \subseteq N.$ As $\underset{n}{\rm lim}$ $x_n =x, \hspace{7pt} \epsilon>0 \hspace{7pt}\exists n_0 \in \mathbb{N}$ such that $|x_n - x|<\epsilon \hspace{7pt}  \forall n \ge n_0 \Rightarrow x_n\in(x-\epsilon, x+\epsilon) \hspace{5pt} \forall n \ge n_0 \Rightarrow x_n \in N \hspace{7pt} \forall n \ge n_0$. Hence, every neighbourhood $N$ of $x$ contains an element $S$, other than $x$ so $x$ is a cluster point of $S$.

$(ii)$ We have already shown that $x$ is a cluster point. Let, if possible, $y \ne x$ be any other cluster point and $\epsilon = \frac{|x-y|}{2}$. As $\underset{n}{\rm lim}$ $x_n =x$ then $\exists n_0 \in \mathbb{N}$ such that $x_n \in (x-\epsilon, x+ \epsilon) \hspace{6pt} \forall n\ge n_0$. Defining, $\epsilon_n = \frac{|x_n-y|}{2}
\hspace{9pt} n=1,2,..,n_0-1$. Let $\delta$=min$\{\epsilon_1, \epsilon_1,..,\epsilon_{n_0-1},\epsilon\}$ then $(y-\delta, y+\delta) \cap (S\backslash \{y\}) = \varnothing $

Hence $\exists$ a neighbourhood of $y$, namely, $(y-\delta, y+\delta)$ such that it contains no point of $S$ other than $y$ which contradicts the fact that $y$ is a cluster point. So our assumption that $y$ is a cluster point is false and we can conclude that $S$ has only one cluster point i.e $x$

$(iii)$ After the assumption in $(i)$ is removed, $x$ need not be a cluster point of $S$. For example, let $x_n =1 \hspace{6pt} n\in \mathbb{N},$ then $\underset{n}{\rm lim}$ $x_n =1$ and $S=\{1\}$ and it is clear that $S$ has no cluster point.

答:
Your proofs look fine. Here's an alternative proof to (ii).

Let $y\ne x$ be another cluster point of $S$. For each $k\in\mathbb{N}$, note that there are infinitely many $s\in S$ such that the neighborhood $s\in (y-\frac1k,y+\frac1k)$. Hence, one can construct a subsequence $(x_{n_k})$ such that $\lim_k x_{n_k} = y\ne x$, which contradicts the fact that $x_n\to x$.

Title: Proving cluster point sequences


问:
What is the category of I in I with Euclidean metric? I is the set of irrationals.
From some topology worksheets I found online that I am using to self-study: Let $I$ be the set of irrationals, and $d$ be the Euclidean metric. What is the category of $I$ in $(I, d)$? I'm sure it is of the second category, but I am unsure of how to prove it. At first I wanted to use the fact that the irrationals are of second category in the reals, but since I now don't have the fact that every neighborhood of some irrational contains a rational, I don't know to proceed.

答: The space of all irrationals is completely metrizable, since is is a $G_\delta$ subset of $\mathbb{R}$. And so by the Baire category theorem it is of the second category.


答:
You are correct that $\Bbb Z_5[X]/ \langle x^2+cx+1 \rangle$ is a field iff $x^2+cx+1$ has no roots in $\Bbb Z_5$.

For $c=0$, $x^2+1\equiv x^2-4=(x+2)(x-2)$ is reducible.

For $c=1$, $x^2+x+1$ is $1,3,2,3,1$ for $x=0,1,2,3,4$, respectively, so there are no roots.

For $c=2$, $x^2+2x+1=(x+1)^2$ is reducible.

For $c=3$, $x^2+3x+1\equiv x^2-2x+1=(x-1)^2$ is reducible.

For $c=4$, $x^2+4x+1\equiv x^2-x+1$ is $1,1,3,2,3$ for $x=0,1,2,3,4$, respectively, so there are no roots.

Body:
Find all $c \in \Bbb Z_5$ such that $\Bbb Z_5[X]/ \langle x^2+cx+1 \rangle$ be a field.

Attempts:

$\Bbb Z_5[X]/ \langle x^2+cx+1 \rangle$ be a field iff $p(x) = x^2 + cx +1$ is irreducible in
$\Bbb Z_5$ iff $p(x)$ has no roots in $\Bbb Z_5$. I found that $2 \ne c \in \Bbb Z_5$ are the answer.

Am I true?


问:
Finding singular value decomposition(SVD) given eigenvalues
The question is :

Let A be a 2 × 2 symmetric matrix with eigenvalues 3 and −2 and corresponding unit eigenvectors (norm = 1) $u_1$ and $u_2$. Find the SVD and polar decomposition of A.

I tried to take some general values for A and expand $Au_1 = 3u_1 and Au_2 = -2u_2$, but I am not getting how to utilize the $u_1$ and $u_2$.

Even a hint will be helpful.

答:
Since $A$ is symmetric, it is diagonalizable in an othonormal basis, i.e. $A = ODO^T$ with $O$ an orthogonal matrix. If $D$ has positive elements, this decomposition is also the SVD of $A$.

You already know the diagonal elements of $D$ and you also have a candidate for $O$, as $(u_1,u_2)$ forms an orthonormal basis. Unfortunately, one of the eigenvalue is $-2<0$.

One way to compensate for the sign is to use unitary matrices instead. Using this you will have to write $A=UD'V^H$ where $D'$ constains the absolute value of the elements in $D$ and $U,V$ are unitary matrices ($UU^H=I=VV^H$ where the exponent denotes for conjuguate transpose).

Of course, you should generate those unitary matrices from what you know : the eigenvalues and their corresponding eigenvectors.


问:
If f is a surjective, but not injective morphism of groups G->G', can there be a cyclic group of G' whose preimage is not cyclic?
If $f$ is a surjective, but not injective morphism of groups $G\to G'$, can there be a cyclic group of G' whose preimage is not cyclic?

Edit: I am almost sure the answer is yes, but I would like to see an example.

答: The group homomorphism $F(2) \rightarrow \Bbb Z$ sending both generators of the free group on 2 elements $F(2)$ to 1 is surjective, not injective and the preimage of $\Bbb Z$ is $F(2)$, hence not cyclic.


问:
Can flux be proportional to $r^2$ in divergence theorem?
The motivation for the divergence being interpreted as the flux of stuff used the following:

$$\text{div} F(a) = \lim_{r\to0}\frac{3}{4\pi r^3}\int_{|x-a|=r} F\cdot n dA$$

Without the $r^3$ in the denominator the right hand side would converge to $0$. But in the context of the divergence theorem what happens if it is $r^2$ instead, can the limit be nonzero?

答:
The equation
$$\text{div} F(a) = \lim_{r\to0}\frac{3}{4\pi r^3}\int_{|x-a|=r} F\cdot n dA$$
assumes that $F$ is continuously differentiable in the region $|x-a|\le r$. That is the hypothesis of the divergence theorem.
As Sharkos's example shows, it does not hold for $F=x/|x|$, which is not defined at the origin. Another example is $F(x)=(0,0,1)$ when $x_3\ge0$, and $(0,0,0)$ when $x_3<0$. In that case the integral is $\pi r^2$.

So the short answer to your question is yes, but not when $F$ is continuously differentiable.


答:
Yes. Consider $f : S_{42} \rightarrow \lbrace e \rbrace$, where $S_{42}$ denotes the symetric group over $42$ elements, and $\lbrace e \rbrace$ denotes the trivial group, defined by
$$f(\sigma)=e$$

The preimage of the cyclic group $\lbrace e \rbrace$ is isomorphic to $S_{42}$ which is not cyclic.

Title: If f is a surjective, but not injective morphism of groups G->G', can there be a cyclic group of G' whose preimage is not cyclic?

Body:
If $f$ is a surjective, but not injective morphism of groups $G\to G'$, can there be a cyclic group of G' whose preimage is not cyclic?

Edit: I am almost sure the answer is yes, but I would like to see an example.


问:
Prove that segments of a secants from regular triangle vertex to 2 points on semicircle have the same length.
Given a regular triangle ABC. A semicircle with diameter BC is drawn (It doesn't intersect AB, AC) with 2 points D, E separating it into 3 equal arcs. AD, AE intersects BC at M, N respectively. Prove BM = MN = NC.

I couldn't come up with anything so far...

答:
HINT: If the center of the semicircle is $O$, $\angle BOD = \angle DOE = \angle COE = 60^0$ (same arc lengths will subtend same angle at the center).

So $\triangle ODE$ is equilateral and so segment $DE = R$.

Find altitude of equilateral triangle $AO$. Also find perp from $O$ to $DE$.

Use similar triangles $ADE$ and $AMN$ to find the ratio of $MN$ to $DE$ (knowing ratio of altitudes).

Also as $OM = ON$, you can finally show $BM = MN = NC = \frac{2R}{3}$.


问:
Big Oh : How can i prove or disprove $8^{(n-3) }= O(2^n)$
How would I prove or disprove this?

I tried it like this:

$$8^{n-3}\in O\left(2^n\right)$$
$$8^{n-3}\leq c\cdot 2^n$$
$$8^n\cdot8^{-3}\leq c\cdot 2^n$$
$$8^n\leq c\cdot 2^n$$

(We can remove the $8^{-3}$ because it is a constant)

But I'm really unsure. Is this even correct? What would c have to be then?

答:
Hint:

$8^n = 2^n\cdot 4^n$. Think about the $4^n$ and how it can outgrow any constant $C$.


答:
$8^{n-3}=O(2^n)$ means that the sequence $a_n= \frac{8^{n-3}}{2^n}$ is bounded. But $a_n \to \infty$ as $n \to \infty.$

Consequence: $8^{n-3}=O(2^n)$  is not true.

Title: Big Oh : How can i prove or disprove $8^{(n-3) }= O(2^n)$

Body:
How would I prove or disprove this?

I tried it like this:

$$8^{n-3}\in O\left(2^n\right)$$
$$8^{n-3}\leq c\cdot 2^n$$
$$8^n\cdot8^{-3}\leq c\cdot 2^n$$
$$8^n\leq c\cdot 2^n$$

(We can remove the $8^{-3}$ because it is a constant)

But I'm really unsure. Is this even correct? What would c have to be then?


答:
Divide both sides by $2^n$ and you should get $$\frac{8^n}{2^n}=\left(\frac82\right)^n=4^n\leq c$$
Since $\lim_{n\to\infty}4^n=\infty$, i.e. $4^n$ diverges, there cannot exist constant $c$ such that $4^n\leq c$ for all $n$.

Therefore $8^{n-3}\notin O\left(2^n\right)$


问:
How to show that $|f(x)-g(x)|<\epsilon~\forall~x\in[a,b]?$
Let $[a,b]$ be a closed and bounded interval in $\mathbb R$ and $f$ is continuous on $[a,b].$  How to show that for $\epsilon>0~\exists$ a piecewise linear continuous function $g$ on $[a,b]$ such that $|f(x)-g(x)|<\epsilon~\forall~x\in[a,b]?$

Help me! I'm clueless.

答:
Weierstrass's Theorem would give you that with $g$ being a polynomial.

-- Edit: no answering the question (it has been edited since).


答:
Taking the base-$2$ logarithm,

$$3(n-3)\le\log_2c+n$$ or $$\log_2c\ge 2n-6$$ is certainly not possible.


答:
It is false that $8^{n-3} = O(2^n)$. Indeed,
$$\limsup_{n\to\infty}\frac{8^{n-3}}{2^n} = \limsup_{n\to\infty} 2^{2n-9} = \infty,$$
which contradicts the definition of $f=O(g)$, i.e., that
$$\limsup_{n\to\infty}\frac{|f(n)|}{g(n)}<\infty.$$


问:
Eigenvalues and Eigenvectors of $A=I-\alpha vv^{T}$
Consider the matrix $A$ given by $A=I-\alpha vv^{T}$ with $v\neq0$ and $v\in\mathbb{R}^{n}$ and $\alpha\neq0$. we want to show that there are two distinct eigenvalues $\lambda_{1},\lambda_{2}$ to be found with their corresponding eigenvectors $x_{\lambda_{1}}$ and $x_{\lambda_{2}}$.

My attempt : By definition, we have that $Ax=\lambda x$ thus :
$$
(I-\alpha vv^{T})x=x-\alpha vv^{T}x=x-(\alpha v^{T}x)v
$$
One can easily notice that $v$ is nothing but a scalar multiple of $x$ that is to say $x=\beta v$ and thus we have that for $x=v$ we get :
$$
Av=(1-\alpha v^{T}v)v
$$
Thus, an eigenvalue of $A$ is $\lambda_{1}=1-\alpha v^{T}v$. I am unable to find the second eigenvalue nor the corresponding eigenvectors. I would truly appreciate help as I am lost in the process.

答:
Let $\alpha\ne0,v\ne\vec0$. Since $|vv^T|=0, 0$ is an eigenvalue of $vv^T$. Why?

This is simply seen below:$$vv^T=\begin{bmatrix}v_1\\v_2\\\vdots\\v_n\end{bmatrix}\begin{bmatrix}v_1&v_2&\ldots&v_n\end{bmatrix}=\begin{bmatrix}v_1v_1&v_1v_2&\ldots&v_1v_2\\v_2v_1&v_2v_2&\ldots&v_2v_n\\\vdots&\vdots&&\vdots\\v_nv_1&v_nv_2&\ldots&v_nv_n\end{bmatrix}$$If $v_2=0$, then the entire second row is zero. If $v_2\ne0$, we can perform the row operation $R_1\to R_1-v_1R_2/v_2$ making $R_1$ zero.

The eigenvector corresponding to this is any non-zero vector $u$ that is orthogonal to $v$, since $vv^Tu=v(0)=0u$.

$v$ is an eigenvector of $vv^T$ with eigenvalue $v^Tv=\|v\|^2>0$.

So two distinct eigenvalues of $I-\alpha vv^T$ are $1-\alpha\|v\|^2,1$ with the same eigenvectors. This is because if $(\lambda,v)$ is an eigenpair of $B$ then $(1-\alpha\lambda,v)$ is an eigenpair of $I-\alpha B$.


问:
Struggling to find solution: question 3
Could someone help with this problem?
$$x^2y''-7xy'+7y=13\sqrt{x}$$

I have brought it to its linear form and it came to this:

$$y''-\frac{7}{x}y'+\frac{7}{x^2}y=13\frac{1}{\sqrt{x^3}}$$

Here I'm stuck, should I do reduction of order, but for that we must know one of the solutions, and I just couldn't find it.

答:
Thanks to @Aryadeva I found the answer. (I like the way he has this keen eye for these questions... As if he wrote it... =_=)

Taking from where he left it:

First find the integrating factor of the linear ODE of:

$$y'-\dfrac {7y}{x}=- \dfrac {26 }{\sqrt{x}}+C_1$$

$$p(x)=-\frac{7}{x} $$
$$h=-7\int{\frac{1}{x}}dx$$
$$e^h=\frac{1}{x^7} $$
$$D_x[\frac{1}{x^7}y]=-26\frac{1}{\sqrt{x}}+c_1 $$
Take integral both sides with respect to $x$ and divide both sides by the integrating factor:
$$y=-4\sqrt{x}-8c_1\frac{1}{x}+c_2x^7 $$

Title: Struggling to find solution: question 3


答:
Setting $$x=e^t$$ turns the equation to one with constant coefficients:

$$\ddot y-8\dot y+7y=13e^{t/2}.$$

The characteristic roots are $1$ and $7$. The rest is routine work.

Title: Struggling to find solution: question 3


问:
Derivative as first order variation in output?
In reference to method shown at 4:00 of this video,

Consider,

$$f(x) = x^x$$

Then,

$$ f(x+h) = (x+h)^{x+h} = x^{x+h} ( 1 + \frac{h}{x})^{x+h}$$

Now, consider right most term in brackets,

$$ ( 1 + \frac{h}{x})^{x+h} = 1+ (x+h) \frac{h}{x} + \frac{ (x+h)(x+h-1)}{2!} ( \frac{h}{x})^2...= 1+h+ stuff$$

This suggests derivative of $x^x$ is $x^x$ which is wrong! So, how would I prove the derivative using this 'perturbation method'

答:
You are neglecting the $x^{x+h}$ too fast:
$$x^{x+h} = x^x + (x^x\log x)h + \cdots$$


问:
Best book on the decision-theoretic justification of bayesian probability

The use of Bayesian probabilities as the basis of Bayesian inference has been supported by several arguments, such as Cox axioms, the Dutch book argument, arguments based on decision theory and de Finetti's theorem.

A decision-theoretic justification of the use of Bayesian inference (and hence of Bayesian probabilities) was given by Abraham Wald, who proved that every admissible statistical procedure is either a Bayesian procedure or a limit of Bayesian procedures.[26] Conversely, every Bayesian procedure is admissible.[27]

Wikipedia: Bayesian Probability - Justification of Bayesian Probabilities

What's the best book on the decision-theoretic justification of bayesian probability? I found a few books on the topic:

The Bayesian choice - Christian P Robert

Bayesian Theory - Adrian Smith and José-Miguel Bernardo

Statistical Decision Theory and Bayesian Analysis - Jim Berger

Which one of these books is the best and most relevant to the topic? Are there any other books that are better than these?

答: I know the third you mentioned and it is a good textbook, even if my favourite is Box Tiao</p>


问:
Is it correct to write the whole number $2$ as a mixed fraction $1 \dfrac{1}{1}$ or $2 \dfrac{0}{1}$?
So if we start from writing $2$ as an improper fraction, and use the same procedure for any improper fraction to covert it to a mixed number, we'll get $2\dfrac{0}{1}$. Which makes sense, and it can be reverted back to $2$.

\begin{align}
2\rightarrow\dfrac{2}{1}\rightarrow 2\dfrac{0}{1}
\end{align}
But from playing around with this, I also found that there's another mixed number that can be evaluated to $2$. Starting from $1+\dfrac{1}{1}$
$$1+\dfrac{1}{1}\rightarrow1\dfrac{1}{1}\rightarrow\dfrac{2}{1}\rightarrow2$$

So, $2$ can be represented as either $2\dfrac01$ or $1\dfrac11$.

Are both these correct?

答: I would just write $2$, and reserve the use of mixed fractions for non-integers. Both your options look strange and unusual, but I don't think either of them is wrong.


答:
Lemma: The property of being a nowhere dense set is, so to say, "densely intrinsic", meaning that if $f:(X,\tau)\to (Y,\rho)$ is an embedding with dense image and $S\subseteq X$, then $S$ is nowhere dense if and only if $f[S]$ is nowhere dense in $Y$.

Consequently, so is being first category. Since the irrationals are not first category in $\Bbb R$, they aren't in their subspace topology either.

To prove the central assertion, note that $\operatorname{cl}_XS$ contains a non-empty open set if and only if there is some $U\in\rho$ such that $\operatorname{cl}_{f[X]}f[S]\supseteq U\cap f[X]\supsetneq \emptyset$. If the latter, then $$\operatorname{cl}_Yf[S]= \operatorname{cl}_Y\operatorname{cl}_{f[X]}f[S]\supseteq\operatorname{cl}_Y(U\cap f[X])\stackrel{\text{density of }f[X]}\supseteq U$$
and therefore $f[S]$ is not nowhere dense in $Y$.

On the other hand, if $f[S]$ is not nowhere dense in $Y$, then $\operatorname{cl}_Yf[S]$ contains a non-empty open set $U$, and therefore $$\operatorname{cl}_{f[X]}f[S]=f[X]\cap \operatorname{cl}_Yf[S]\supseteq U\cap f[X]\stackrel{\text{density}}\ne\emptyset$$

Title: What is the category of I in I with Euclidean metric? I is the set of irrationals.

Body: From some topology worksheets I found online that I am using to self-study: Let $I$ be the set of irrationals, and $d$ be the Euclidean metric. What is the category of $I$ in $(I, d)$? I'm sure it is of the second category, but I am unsure of how to prove it. At first I wanted to use the fact that the irrationals are of second category in the reals, but since I now don't have the fact that every neighborhood of some irrational contains a rational, I don't know to proceed.


问:
Pure Strategy Nash Equilibria Repeated Game

I am trying to solve the above problem. I know that the two Nash Equilibria are $(D,L)$ and $(C,R)$, so the first four pure strategy Nash equilibria would just involve playing either of these strategies in both periods. I understand that, for all other possible pure strategy Nash equilibria, the second/last game must still end with one of these Nash equilibria as well but could start in the first game with something that is not a Nash equilibria.

However, I am somewhat confused by what the other two Nash equilibria are.

One of them is:

If $(U,L)$ is played in the first period, then $(D,L)$ is played in the second period. Otherwise, $(C,R)$ is played in the second period.

The other one is:

If $(C,M)$ is played in the first period, then $(C,R)$ is played in the second period. Otherwise, $(D,L)$ is played in the second period.

I have noted the discount factor of $\frac{3}{4}$, so understand that if, for instance if $(U,L)$ is played in the first period and $(D,L)$ in the second, then the payoff for the row player would be $5+\frac{3}{4}\left(6\right)\:=\:9.5$, but I am not sure how to apply this here, although I know I have to.

Any help would be highly appreciated!

答:
Let's analyze your first suggestion:

"If $(U,L)$ is played in the first period, then $(D,L)$ is played in the second period. Otherwise, $(C,R)$ is played in the second period."

The idea here is that Row player "sacrifices" something in the first period (he gets $5$ instead of $6$) so that his favorite NE will be played in the second period. The $(C,R)$ equilibrium is the way the column player "punishes" him for deviations.

So the payoff of the row player in this scenario is indeed $5+\tfrac{3}{4}\cdot 6=9.5$.
If he deviates, however, and plays $D$ in the first round, he gets $6+\tfrac{3}{4}\cdot 4=9$.
You can repeat this calculation for Player 2 and verify that this is indeed an equilibrium.

Same idea applies to other potential equilibrium profiles.


问:
finding the base of a triangle
In A triangle ABC ,AB=AC.D is a point inside the triangle such that AD=DC.Median on AC from D meets median on BC  from A at the centroid of the triangle.If the area of the triangle  ABC equals to $4\sqrt 3$ .Find the base i.e. BC.
The method that i have used to solve this problem works by ending with  two answers for $\frac12$ of BC and then having to check which one works by trial and error.It is more or less efficient,but I am assuming there's a better one.</p>

答:
This is my own solution.At first,we know that  triangle ABC is isosceles.We also know that  tri ADC is also isosceles.So the median of that triangle bisects AC in a right angle. Note that the side opposite angle B is AC,the same as angle D.The median from angle B , like that of  angle D will also meet at the centroid.So we can draw the inference that  median from angles B and D are actually the same straight lline.But then the median from angle B bisects AC perpendicularly.So AB =BC=AC (since median of an isosceles triangle bisects it  perpendicularly.This proves that ABC is an equilateral triangle.Now let the median on BC bisect BC at M.Then ,by congruency criterion,triangles ABM and AMC are equal, and so area of  each of them is 2× sqrt 3.

We know,
$2\times \sqrt{3} = 1/2(2\times 2\times \sqrt{3})$. So MC equals either (\sqrt{2} )\times 3 or simply 2.Now let us use the pythagorean theorem to see which fits.It seems that 2 is the only solution for MC.so $BC =4$.That is how I solved it.

Title: finding the base of a triangle