问:
Help proving $\left|e^{-(s-1/2)\log(s)}e^s\right|\ll_\varepsilon e^{|s|^{1+\varepsilon}}$
I'm trying to work through this proof and I'm stuck at the fourth equation down on page 2. Firstly, there is a typo in that the coefficient $\sqrt{2\pi}$ should be $1/(\sqrt{2\pi})$. I believe there is a second typo which is that it should read $\ll_\varepsilon e^{|s|^{1+\varepsilon}}$ not $\ll_\varepsilon e^{|s|^\varepsilon}$ (empirical results using limits support this). The first typo is inconsequential, but assuming the second corrected typo I'm now trying to prove
$$\left|e^{-(s-1/2)\log(s)}e^s\right|\ll_\varepsilon e^{|s|^{1+\varepsilon}},\tag{*}$$
for all $\varepsilon>0$, where $\text{Re}(s)\geq\delta>0$, presumably for some $\delta>0$.

I manipulated the left hand side,
$$|e^{-(s-1/2)\log s+s}|=|e^{-(\sigma-1/2)\log|s|+t\text{Arg}(s)+\sigma}|.$$
So we need to show for some $C>0$ that
$$e^{-(\sigma-1/2)\log|s|+t\text{Arg}(s)+\sigma}<Ce^{|s|^{1+\varepsilon}}.$$
I'm stuck at this point and can't seem to prove the inequality. Any help or a hint would be appreciated.

答:
Write $C = e^c$ to get things down from the exponents. You want to show that for a sufficiently large $c$ (depending on $\varepsilon$) the inequality
$$-(\sigma - 1/2)\log \lvert s\rvert + t\operatorname{Arg}(s) + \sigma \leqslant \lvert s\rvert^{1 + \varepsilon} + c$$
holds throughout the half-plane $\sigma \geqslant \delta$.

For this we can be pretty brutal. Write $x = \lvert s\rvert$ and note that $\sigma \leqslant x$, $\lvert t\rvert \leqslant x$, and the argument is bounded (in absolute value) by $\pi/2$. Thus it suffices to see that
$$(x + 1/2) \lvert\log x\rvert + \pi/2 x + x \leqslant x^{1+\varepsilon} + c$$
holds on the interval $[\delta, +\infty)$ if $c$ is sufficiently large. But
$$\lim_{x \to +\infty} \frac{\log x}{x^{\varepsilon}} = 0$$
implies
$$\lim_{x \to +\infty} \:\bigl((x+1/2)\lvert \log x\rvert + (1 + \pi/2)x - x^{1+\varepsilon}\bigr) = -\infty\,,$$
so the function on the left is bounded from above, as we wanted to show.

To see that it's indeed a typo and $e^{\lvert s\rvert^{\varepsilon}}$ on the right hand side does not work, consider $s = \frac{1}{2} + it$. For large $\lvert t\rvert$ we have
$$\Bigl\lvert e^{-(s-1/2)\log s + s}\Bigr\rvert \geqslant e^{\lvert t\rvert}$$
but
$$e^{\lvert s\rvert^{\varepsilon}} < e^{(2\lvert t\rvert)^{\varepsilon}}\,,$$
which shows the inequality
$$\Bigl\lvert e^{-(s-1/2)\log s + s}\Bigr\rvert \leqslant Ce^{\lvert s\rvert^{\varepsilon}}$$
cannot hold for $\varepsilon < 1$.


问:
Does $u\in L^p(B)$ implies $u_{|\partial B_t}\in L^p(\partial B_t)$ for almost $t\in (0,1]$?
Let $B$ be the unit ball in $\mathbb{R}^N$ with center in origin and consider the space $L^p(B)$ with Lebesgue measure ($1<p<\infty$). Let $B_t\subset B$ be a concentric ball of radius $t\in (0,1]$. For fixed $t$ we can define $L^p(\partial B_t$) in the same sense as $L^p(B)$, by using the surface measure of $\partial B_t$.

My question is: Let $u\in L^p(B)$. Is true that for almost every $t\in (0,1]$ (Lebesgue measure), it does make sense to talk about the (classical) restrictuon of $u$ to $\partial B_t$, to wit, $u_{|\partial B_t}$ and $u_{|\partial B_t}\in L^p(\partial B_t)$?

Thank you

答:
By the change of variables theorem for Lebesgue integration on $\mathbb{R}^n$ we have
$$
\int_{B}|f(x)|^pdx=\int_0^1t^{n-1}\Bigl(\int_{\partial B}|f(t\,x)|^p\,d\sigma\Bigr)\,dt=\int_0^1\Bigl(\int_{\partial B_t}|f(x)|^p\,d\sigma_t\Bigr)\,dt<\infty,
$$
where $d\sigma$ is surface measure on $\partial B$ and $d\sigma_t$ is surface measure on $\partial B_t$. Thus
$$
\int_{\partial B_t}|f(x)|^p\,d\sigma_t<\infty\quad
$$
for almost every $t\in(0,1\,]$.</p>

ContentLicense: CC BY-SA 3.0


问:
Sampling distribution for the mean of a large sample of salaries
Hello I am trying to solve this question. But, I can not find any approach, We know that avg is 600 and the question is asking for 500? and also if we sum the sd and avg we can not come near to the 1000£ on the second part. Could you please help me to solve this question?

Average income of nurses is 600£ and the standard deviation was 266£ . A sample of 1,000 nurses
was randomly chosen to verify if this sample would have a similar mean income as the entire population.
a) Probability that the mean income of the nurses sampled is less than 500£ .
b) Probability that the mean income of the nurses sampled is greater than 1000£ .

答:
Your idea of looking at the number of standard deviations $\sigma$ from the mean $\mu$ is a reasonable approach.

If the distribution of incomes is $\mathsf{Norm}(\mu=600, \sigma = 266),$ then $\bar X_{1000} \sim \mathsf{Norm}(600, 266/\sqrt{1000}),$ where $SD(\bar X_{1000}) = 266/\sqrt{1000} = 8.4117.$

By the Central Limit Theorem the distribution
for $\bar X_{1000}$ is very nearly normal even if the population of nurse incomes is not normally distributed.

In (a) you seek $P(\bar X_{1000} < 500) \approx 0.$  This is asking for the probability in a normal distribution that lies
to the left of about eleven standard deviations below its mean, so you should not be surprised by the nearly zero probability.

For more precision in such computations, you can 'standardize' and look at printed standard normal CDF tables use software (calculator or computer).
Use whatever method is discussed in your course text or lectures. I used R statistical software
to get the value stated above.

```<code>pnorm(500, 600, 266/sqrt(1000))
[1] 6.810586e-33
</code>```

(b) Similar:

```<code>1 - pnorm(1000, 600, 266/sqrt(1000))
[1] 0
</code>```

Note: Maybe a more user-friendly initial problem of this kind would have asked
for $P(\bar X_{1000} < 590) =  0.1173$ and $P(\bar X_{1000} > 615) = 0.0373.$

ContentLicense: CC BY-SA 4.0

Body:
Your idea of looking at the number of standard deviations $\sigma$ from the mean $\mu$ is a reasonable approach.

If the distribution of incomes is $\mathsf{Norm}(\mu=600, \sigma = 266),$ then $\bar X_{1000} \sim \mathsf{Norm}(600, 266/\sqrt{1000}),$ where $SD(\bar X_{1000}) = 266/\sqrt{1000} = 8.4117.$

By the Central Limit Theorem the distribution
for $\bar X_{1000}$ is very nearly normal even if the population of nurse incomes is not normally distributed.

In (a) you seek $P(\bar X_{1000} < 500) \approx 0.$  This is asking for the probability in a normal distribution that lies
to the left of about eleven standard deviations below its mean, so you should not be surprised by the nearly zero probability.

For more precision in such computations, you can 'standardize' and look at printed standard normal CDF tables use software (calculator or computer).
Use whatever method is discussed in your course text or lectures. I used R statistical software
to get the value stated above.

pnorm(500, 600, 266/sqrt(1000))
[1] 6.810586e-33

(b) Similar:

1 - pnorm(1000, 600, 266/sqrt(1000))
[1] 0

Note: Maybe a more user-friendly initial problem of this kind would have asked
for $P(\bar X_{1000} < 590) =  0.1173$ and $P(\bar X_{1000} > 615) = 0.0373.$


问:
What is the result of ${^{i}i}$?
$i+i=2i$, $i\cdot i=-1$, $i^i=e^{-\pi/2}$, but what is the result of ${^{i}i}$? That is, what is the $i$-th tetration of $i$?

答:
In order to answer your question we first need to extend tetration to allow complex bases and heights. To the best of my knowledge there is no currently accepted way to do this, so your question is not really answerable.

Keep in mind that priori there is no reason for a function $\mathbb{N}\rightarrow\mathbb{N}$ to have a "natural" extension to a larger domain like $\mathbb{Q},\mathbb{R}$, or $\mathbb{C}$. Of course many functions do have this property (addition, multiplication, exponentiation with some subtleties, the factorial, ...), but this sort of generalizability should not be assumed at the outset.

There has been some work on extending tetration to complex bases and heights, although like I said above I don't believe that any of it reaches expressions like ${}^ii$. See e.g. here or here.

Title: What is the result of ${^{i}i}$?


问:
Trap Rule for sin(x)

Use the trapezoidal rule with $N=6$ to approximate the arc length of the curve $f(x) = \sin(x)$ from $x=0$ to $x=\pi$.

So I found that $\Delta x = \frac{\pi}{6}$ which means that my interval points are $0,\frac{\pi}{3}, \frac{\pi}{2}, \frac{2\pi}{3}, \frac{5\pi}{6}$ and $\pi$.

Is that right?

And then after that I just use the trap rule forumula computing the value of $\sin$ of all my interval points.

Is my thinking correct?

答:
The formula for step size is given by:

$$\displaystyle h = \frac{b-a}{N} = \frac{\pi - 0}{6} = \frac{\pi}{6}$$

We are also given that $x_0 = 0$.

Title: Trap Rule for sin(x)


问:
Prove the following sequence is convergent: $a(1) = 1$ and $a(n+1) = 1/2 \cdot (a(n) + 3/a(n))$
I know the limit is $\sqrt 3$. And I tried proving that the sequence is decreasing from $a(2)$, yet when I got to this point, I have been stuck:

$a(n+1) - a(n) = (3 - a(n)^2) / 2a(n)$ ?

Thank you very much!

答:
Let’ s prove by induction that for $n \ge 2$ $a_n -\sqrt 3 \gt0$

It’s true for $n=2$ as $a_2=2$. Suppose that the inequality holds for $n$. As

$$\begin{aligned}
a_{n+1} -\sqrt 3 &= \frac{1}{2}\left(a_n +\frac{3}{a_n}\right)-\frac{1}{2}\left(\sqrt 3+\frac{3}{\sqrt 3}\right)\\
&=\frac{1}{2}\left(a_n-\sqrt 3\right)\left(1+\frac{\sqrt 3}{a_n}\right)
\end{aligned}$$

The inequality also holds for $n+1$.

Now using the equality written in your question and the inequality we just proved, you get the desired result that $\{a_n\}$ is decreasing starting at the second term.


答:
At this time, we don't even have a definition of tetration with rational heights, let alone complex ones. (If we did, we might try to treat real heights by limits, then complex heights by some as yet unimaginable generalization.) Why not?

I can define $x^{3/2}$ as the solution of $y^2=x^3$, because the associativity of multiplication ensures this $y$ also satisfies $y^{2n}=x^{3n}$, as we'd expect of $x^{(3n)/(2n)}$. But the same trick doesn't work with ${}^{3/2}x$, since exponentiation doesn't associate (or commute, for that matter).

See also here.

Title: What is the result of ${^{i}i}$?

Body: $i+i=2i$, $i\cdot i=-1$, $i^i=e^{-\pi/2}$, but what is the result of ${^{i}i}$? That is, what is the $i$-th tetration of $i$?


答:
Hint:

On the interval $(\sqrt 3,+\infty)$ the function $\:f(x)=\dfrac12\Bigl(x+\dfrac 3x\Bigr)$ is increasing and $\:f(x)<x$.

Note that $a_2=2\in(\sqrt 3,+\infty)$.

Title: Prove the following sequence is convergent: $a(1) = 1$ and $a(n+1) = 1/2 \cdot (a(n) + 3/a(n))$

Body:
I know the limit is $\sqrt 3$. And I tried proving that the sequence is decreasing from $a(2)$, yet when I got to this point, I have been stuck:

$a(n+1) - a(n) = (3 - a(n)^2) / 2a(n)$ ?

Thank you very much!


问:
Confusion over definition of natural numbers
In my course we have been given the following definition of the natural numbers

The natural numbers $\mathbb{N}$ are the smallest set such that

 $0\in\mathbb{N}$

 if $n\in\mathbb{N}$ then $n+1\in\mathbb{N}$

Surely the rational numbers also satisfy this definition. They are the same size as $\mathbb{N}$, have $0\in\mathbb{Q}$ and we can take any $n\in\mathbb{Q}$ and have $n+1\in\mathbb{Q}$.

Is the definition not as rigorous as it could be or are the naturals the smallest set to satisify the definition and I'm using ideas about cardinality incorrectly?

答: $ 0\in \Bbb N $ and $ 0 $ is the smallest element. There is no element $ n\in \Bbb N $ such that $ n+1=0 $, and this is not true in $ \Bbb Q$.


答: The notion of “smallest set” has nothing to do with cardinality. It refers to set inclusion. No proper subset of the naturals as defined here will satisfy the two conditions.

Title: Confusion over definition of natural numbers

Body:
In my course we have been given the following definition of the natural numbers

The natural numbers $\mathbb{N}$ are the smallest set such that

 $0\in\mathbb{N}$

 if $n\in\mathbb{N}$ then $n+1\in\mathbb{N}$

Surely the rational numbers also satisfy this definition. They are the same size as $\mathbb{N}$, have $0\in\mathbb{Q}$ and we can take any $n\in\mathbb{Q}$ and have $n+1\in\mathbb{Q}$.

Is the definition not as rigorous as it could be or are the naturals the smallest set to satisify the definition and I'm using ideas about cardinality incorrectly?


问:
Understanding Proof of Poisson Summation Formula
Consider a proof from a textbook on Harmonic Analysis:

Note that $\mathcal{S}(ℝ)$ denotes the Schwartz Space.

Question 1: Why does the top left formula in the proof start out as:

$$
\int_0^1 \left( \sum_{m ∈ ℤ} \phi(x+m) \right) \mathrm{e}^{-2 \pi i n x} \,\mathrm{d}x?
$$

Shouldn't the integrand be $\int_{-\infty}^{\infty}$?

Question 2: What justifies the leap from

$$
\sum_{m ∈ ℤ} \int_0^1 \phi(x+m) \mathrm{e}^{-2 \pi i n x} \,\mathrm{d}x = \sum_{m ∈ ℤ} \int_m^{m+1} ϕ(y) \mathrm{e}^{-2 \pi i n y} \,\mathrm{d}y ?
$$

答:
there is lot easier than that

consider the fourier series expansion for the 1-periodic function

$$ \lfloor x\rfloor = x - \frac{1}{2} + \frac{1}{\pi}  \sum_{k=1}^\infty \frac{\sin(2 \pi k x)}{k} $$

and get this into the Euler sum formula

and use Euler sum formula plus integration by parts

$$ \sum_{n=-\infty}^{\infty}f(n) = - \int_{-\infty}^{\infty}f'(x)\lfloor x] $$

and you have the Poisson sum formula, very easily


问:
Rank of element in generic extension versus rank of its name
I sometimes see the following fact used in some arguments:

suppose $M[G]$ is a generic extension of $M$ by a forcing $\mathbb P$ and suppose $x\in M[G]$ has rank $<\gamma$, where $\gamma$ is some limit ordinal above rank$(\mathbb{P})$. Then there is a name $\tau\in M^\mathbb{P}$ such that $\tau_G=x$ and $\tau$ has rank $<\gamma$.

For example, this fact is used in Reitz's The Ground Axiom in proving that the ground model is definable, at the end of the first paragraph in the proof of Lemma 7.1.

But I'm not sure how to prove this. Any help appreciated!

Added after edit: if we assume additionally that $\gamma$ is a $\beth$-fixed point (equivalently, $H_\gamma=V_\gamma$. This holds in the special case in Lemma 7.1 referenced above), then I think the following argument works.

By induction on rank, we show that if $x\in (H_\gamma)^{M[G]}$, then there is a name $\sigma\in H_\gamma\cap M^{\mathbb{P}}$ such that $\sigma_G=x$. So suppose this holds for all sets of lower ranks than $x$. Hence each $y\in trcl(x)$ has a name $n(y)$ whose rank is lower than $\gamma$. Now collecting all those names, let $z=\{n(y)\mid y\in trcl(x)\}$. Since $x\in (H_\gamma)^{M[G]}$, we know $|trcl(x)|=\kappa<\gamma$. This means also that $|z|=\kappa$. The preceding cardinality claims are all in the sense of $M[G]$, and we fix a surjection $f:\kappa\to z$ in $M[G]$.

Let $\rho$ be a name for $x$ and $\tau$ be a name for $z$. By the truth lemma, we may fix some $p\in G$ such that
$$
p\Vdash \rho\in (H_\gamma)^{M[G]} \wedge \text{ every element of } \rho \text{ has a name of rank } <\gamma \\ \wedge \tau \subseteq \check{(H_\gamma)} \wedge\dot{f}:\kappa \to \tau \text{ is a surjection}
$$

We then proceed to define our low-rank name $\sigma$ for $x$. For each $\alpha<\kappa$, we let

$$
X_\alpha = \{ q \in \mathbb{P} \mid \ (\exists \pi \in H_{\gamma}\cap M^{\mathbb{P}})~ q \leq p \wedge q \Vdash (\dot{f}(\alpha)=\check{\pi} \wedge \pi \in \rho)\}
$$
In other words, $X_\alpha$ collects those conditions below $p$ that will force (the evaluation of) an element in $z$ to be an element of $x$.

Now for each $X_\alpha$, fix a maximal antichain $A_\alpha$ that it intersects. For each $\alpha<\kappa$ and $q\in X_\alpha\cap A_\alpha$, there is some $\mathbb P$-name $v(\alpha,q)$ such that $q\Vdash v(\alpha,q)\in\rho\wedge \dot f(\alpha)=\check{v(\alpha,q)}$. Now we can define the name $\sigma$ to be
$$
\sigma = \{(\pi,q)\mid (\exists\alpha)( \alpha < \kappa \wedge q \in X_{\alpha} \cap A_{\alpha} \wedge \pi = v(\alpha,q))\}
$$
Then $\sigma$ is a name in $H_\gamma\cap M^{\mathbb P}$, and $p\Vdash \sigma=\rho$.

Second edit: it seems like the special case sketched above has a duplicate (?) Regardless, I'd still be interested in seeing how to argue for the stronger claim quoted.

答:
I will work over $V$ instead of $M$. I think the following proof works over $\mathsf{ZFC^-}$ (i.e., $\mathsf{ZFC}$ without Power Set and with Collection and the well-ordering principle) with the existence of $\mathcal{P}(\mathbb{P})$. (Especially, it holds over $M=H_\theta$ for large regular $\theta$.)

Lemma. Let $x\in V^\mathbb{P}$ be a name such that $\operatorname{rank}x\ge\operatorname{rank}\mathbb{P}$ and $\gamma$ be a ordinal greater than $\operatorname{rank}\mathbb{P}$. If $p\Vdash \operatorname{rank}x=\check{\gamma}$, then there is $\tau\in V^\mathbb{P}$ such that

$p\Vdash x=\check{\tau}^{\dot{G}}$, and

$\operatorname{rank}\tau\le\gamma_0+3n$, where $\gamma=\gamma_0+n$ for some limit $\gamma_0$ and $n\in\omega$.

Let me introduce some notation on ordinals: for each ordinal $\alpha$, $\alpha^*$ and $\alpha^@\in\omega$ be ordinals such that $\alpha=\alpha^*+\alpha^@$ and $\alpha^*$ is a limit ordinal.

I will use induction on the rank of $x$. Without loss of generality, we may assume that

if $(y,q)\in x$ then $q\le p$, and

(Downward closeness) if $(y,q)\in x$ and $r\le q$, then $(y,r)\in x$

by replacing $x$ to
$$x'=\{(y,r)\mid \exists q (y,q)\in x \text{ and } r\le p,q\}.$$
Since $\operatorname{rank} x\ge\operatorname{rank}\mathbb{P}$,  we have $\operatorname{rank}x'\le\operatorname{rank}x$.

Then for each $(y,q)\in x$, $q\Vdash \operatorname{rank}y<\check{\gamma}$. Find a maxinal antichain $A_{y,q}$ below $q$ which decides the value or $\operatorname{rank}y$; that is, if $r\in A_{y,q}$ then there is an ordinal $\beta_{y,q,r}<\gamma$ such that $r\Vdash \operatorname{rank}y=\check{\beta}_{y,q,r}$.

By the inductive hypothesis, we can find $\tau_{y,q,r}$ such that $r\Vdash y=\tau_{y,q,r}^{\dot{G}}$ and
$$\operatorname{rank}\tau_{y,q,r}\le\beta_{y,q,r}^*+3\beta_{y,q,r}^@.$$
Now take
$$\tau=\{(\tau_{y,q,r},r)\mid (y,q)\in x\text{ and }r\in A_{y,q}\}.$$
Then we can prove $p\Vdash x=\check{\tau}^{\dot{G}}$. It remains to check the rank of $\tau$. We can see that
$$\operatorname{rank}(\tau_{y,q,r},r)\le\max(\operatorname{rank}r, \beta_{y,q,r}^*+3\beta_{y,q,r}^@)+2$$

Case 1. If $\gamma$ is a limit ordinal, then the right-hand-side is strictly less than $\gamma$. Hence $\operatorname{rank}\tau\le\gamma$.

Case 2. If $\gamma=\gamma_0+n$ for some limit $\gamma_0$ and $1\le n<\omega$, then
$$p\Vdash \forall y\in x (\operatorname{rank} y\le\check{\gamma}_0+\check{n}-1).$$
Hence the corresponding $\beta_{y,q,r}$ satisfies $\beta_{y,q,r}\le \gamma_0+n-1$, and thus $\tau_{y,q,r}$ satisfies
$$\operatorname{rank}\tau_{y,q,r}\le\gamma_0+3(n-1).$$
The remaining argument is direct, and we have $\operatorname{rank} \tau\le\gamma_0+3n$.


问:
Show that $\frac{1}{\sqrt{1+kx}}\geq 1-\frac{kx}{2}$

Let $k>0$. Show that $$\frac{1}{\sqrt{1+kx}}\geq 1-\frac{kx}{2}$$ for all $x>0$.

In order to prove the requested result I'm trying to prove that $$\frac{1}{\sqrt{1+kx}} -1+\frac{kx}{2}\geq0$$ for positive $x$. However, I'm stuck at proving the last inequality. Thank you in advance

答:
If $x>\frac{2}{k}$, then
$$ 1-\frac{kx}2<0$$
and then the inequality holds automatically. For $x\in(0,\frac{2}{k}]$, by the LMVT for $f(x)=\frac{1}{\sqrt{1+kx}}$ in $[0,x]$, there is $\xi\in(0,x)$ such that
$$ \frac{1}{\sqrt{1+kx}}-1=-\frac{k}{2(\sqrt{1+k\xi})^3}(x-0)\ge -\frac{kx}{2} $$
and so
$$ \frac{1}{\sqrt{1+kx}}\ge1-\frac{kx}{2}. $$
Here
$$ \frac{k}{2(\sqrt{1+k\xi})^3}\le \frac{k}{2} $$
is used.


问:
LTI with all zero eigenvalues
Say we have this LTI:

$$
\dot{x} = Ax + Bu \\
y = Cx + Du
$$

If the eigenvalues of the A matrix are all zero, what does that mean for the stability of the LTI? Is it unstable or marginally stable and why?

答:
If the matrix $A$ is diagonalizable, then the system is (marginally) stable. Indeed, in some variables $z= Px$, $\det P\ne 0$, the system
$\tag{1}
\dot x= Ax
$
takes the form
$\dot z=PAP^{-1}z=0z=0$. This means that $z$ is constant and also $x$. Hence, the solution does not leave any neighborhood of the origin which includes the initial point of the solution.

If $A$ is not diagonalizable, then the system is unstable. In this case, there exists some change of variables $z= Px$, $\det P\ne 0$, which transforms the system (1) into the
form
$$\tag{2}
\dot z= Jz,
$$
where $J$ is a matrix in Jordan normal form:
$$
J=diag(J_1,\ldots,J_k),\qquad
J_j=\left(
\begin{array}{lllll}
0 & 1 & 0 & \dots & 0\\
0 & 0 & 1 & \dots & 0\\
\vdots  & \vdots & \vdots & \ddots & \vdots \\
0 & 0 & 0 & \dots & 1\\
0 & 0 & 0 & \dots & 0\\
\end{array}
\right),\quad j=1\ldots k.
$$
Since $A$ is not diagonalizable, there is at least one Jordan block $J_p$ of some size $k>1$. Consider the equations of (2), which correspond to the rows of the block $J_p$. They are
$$
\begin{array}{c}
\dot z_i= z_{i+1}\\
\dot z_{i+1}= z_{i+2}\\
\vdots\\
\dot z_{i+k-1}= 0;\\
\end{array}
$$
thus, the (part of the) solution is
$$z_{i+k-1}= C_1,$$ $$z_{i+k-2}= C_1 t+C_2,$$ $$z_{i+k-3}= C_1 \frac{t^2}2+C_2t+C_3$$ etc., where $C_1$, $C_2$, $\ldots$ are constants. Now notice that $z_{i+k-2}$ is unbounded when $C_1\ne 0$, thus, $z$ is unbounded, thus, $x$ is unbounded. Hence, the system is unstable.

Title: LTI with all zero eigenvalues


问:
Functional equation problem: $ f \left( y ^ 2 - f ( x ) \right) = y f ( x ) ^ 2 + f \left( x ^ 2 y + y \right) $
This functional equation problem is from the Latvian Baltic Way team selection competition 2019:

Find all functions $ f : \mathbb R \to \mathbb R $ such that for all real $ x $ and $ y $,
$$ f \left( y ^ 2 - f ( x ) \right) = y f ( x ) ^ 2 +  f \left( x ^ 2 y + y \right) \text . \tag 1 \label {eqn1} $$

OK, so I think that the only answer is $ f ( x ) = 0 $.

I just want to see if my proof that it is the only solution is correct.

So we start off by plugging $ y = - y $. We get that
$$ f \left( y ^ 2 - f ( x ) \right) = - y f ( x ) ^ 2 + f \Big( - y \left( x ^ 2 + 1 \right) \Big) \text . $$

Then we add the two equations together getting that
$$ 2 f \left( y ^ 2 - f ( x ) \right) = f \Big( - y \left( x ^ 2 + 1 \right) \Big) + f \Big( y \left( x ^ 2 + 1 \right) \Big) \text . $$

From the above equations we get that
$$ \frac { f \Big( - y \left( x ^ 2 + 1 \right) \Big) + f \Big( y \left( x ^ 2 + 1 \right) \Big) } 2 = y f ( x ) ^ 2 + f \left( x ^ 2 y + y \right) \text. \tag 2 \label {eqn2} $$

Now, if we plug $ x = - x $ then we will get that the LHS is the same and that the RHS is $ y f ( - x ) ^ 2 + f \left( x ^ 2 y + y \right) $.

So we proceed by subtracting the two and getting that
$$ 0 = y f ( x ) ^ 2 - y f ( - x ) ^ 2 \text . $$

So, lets assume that $ y \ne 0 $ getting that
$$ 0 = \big( f ( x ) - f ( - x ) \big) \big( f ( x ) + f ( - x ) \big) \text . $$

Now we do a two case analysis, 1) the function is even and 2) the function is odd.

Lets start with the function being even then from \eqref{eqn2} we get that
$$ 0 = y f ( x ) ^ 2 \text , $$
which of course implies that the function is just $ 0 $.

OK, now the odd case. Since the function is odd, $ f ( 0 ) = 0 $. Then plugging $ x = 0 $ in \eqref{eqn1} we get that
$$ f \left( y ^ 2 \right) = f ( y ) \text , $$
which implies that $ f $ is also an even function. Since $ f $ is both even and odd, it can only be $ 0 $.

Since we got that $ f $ is zero in both cases, the only solution to the equation is $ f ( x ) = 0 $.

答:
You can show that the only function $ f : \mathbb R \to \mathbb R $ satisfying
$$ f \left( y ^ 2 - f ( x ) \right) = y f ( x ) ^ 2 + f \left( x ^ 2 y + y \right) \tag 0 \label 0 $$
for all $ x , y \in \mathbb R $ is the constant zero function, by continuing your own argument. You've established showing
$$ f ( - x ) ^ 2 = f ( x ) ^ 2 \text . \tag 1 \label 1 $$
Now, take the "$ y f ( x ) ^ 2 $" term in \eqref{0} to the left-hand side, square both sides and rearrange the terms to get
$$ f \left( y ^ 2 - f ( x ) \right) ^ 2 + y ^ 2 f ( x ) ^ 4 - f \left( x ^ 2 y + y \right) ^ 2 = 2 y f ( x ) ^ 2 f \left( y ^ 2 - f ( x ) \right) \text . \tag 2 \label 2 $$
Substituting $ - y $ for $ y $ in \eqref{2} and using \eqref{1}, the left-hand side will be equal to the left-hand side of \eqref{2}, while the right-had side will be the opposite of the right-hand side of \eqref{2}. This in particular implies that
$$ y f ( x ) ^ 2 f \left( y ^ 2 - f ( x ) \right) = 0 \text . \tag 3 \label 3 $$
Now, if for some $ a \in \mathbb R $ we have $ f ( a ) \ne 0 $, then setting $ x = a $ and considering all nonzero values for $ y $ in \eqref{3}, we can see that for all $ z > - f ( a ) $ we have $ f ( z ) = 0 $. But then letting $ x = a $ and $ y = 1 + \frac { | f ( a ) | } { 1 + a ^ 2 } $ in \eqref{0} leads to a contradiction with $ f ( a ) \ne 0 $, since we have $ y > 0 $, $ y ^ 2 - f ( a ) > - f ( a ) $ and $ a ^ 2 y + y > - f ( a ) $. Therefore $ f ( x ) $ must be equal to $ 0 $ for all $ x \in \mathbb R $.


问:
Changing probabilities after the fact
I have a state which is a list of probabilities for next states as follows:

$\left[0, \frac {1}{2}, 0, 0, \frac{1}{2}\right]$

I realise that the second element here actually has a $\left(\frac 2 7\right)$ chance of looping back to this state. So I can remove it like so:

$\frac 1 2 -\left(\frac 1 2 \times \frac 2 7\right)$

Which gives me $\frac 5 {14}$, which is correct in this case, but I might still have the wrong method for others

I want the other element to equal $\frac 9 {14}$, which I have been able to do, but not in the general case.

Hope this is clearer.

答:
I am taking quite the long shot here, but how about

$ p_j' = (1-p_1') \cdot \dfrac{p_j}{\displaystyle\sum_{k=2}^n p_k} $,

where $ p_1, \dots , p_n $ are the original probabilities and $ p_j' $ the new ones. It keeps the sum at 1 and the "extra" probability is divided somewhat analogously.

Title: Changing probabilities after the fact


问:
How to show that $A′A$ is a positive deﬁnite matrix given rank$(A) = k$?
I have difficulty in trying to prove the following question.Let $A = n \times k, k ≤ n$. Show that, if rank$(A) = k$, then $A′A$ is a positive deﬁnite matrix ($A'$ denotes the transpose of $A$). Do you have any idea how it can be solved? Thanks a lot.

答:
Show that $A'A$ is positive semidefinite, i.e. its eigenvalues are nonnegative. (Hint: if $v$ is a $\lambda$-eigenvector of $A'A$, then $\lambda \|v\|^2 = v^\top A^\top A v = \|Av\|^2 \ge 0$.)

Thus it suffices to show that the eigenvalues of $A'A$ are nonzero, i.e. that $A'A$ is invertible. It may help to note that $A'A$ is a $k \times k$ matrix with the same rank as $A$.


问:
Divergence of Curl for a vector space of dimension $n$
It is a fact that the divergence of curl, once the conditions of continuity $f (x, y,z)$ are satisfied, is equal to zero. However, all the demonstrations I found focus on three-dimensional space and use a lot of algebra. Is there any way to generalize the result to dimension $n$ more succinctly?

答:
Curl is defined as an operation on 3-dimensional vector fields.

In three dimensions the algebra is not difficult at all.

$\nabla \cdot (\nabla \times F) = 0$

Which is just an identity of linear algebra.

To generalize to higher dimensional spaces we need a more abstract tool.  We get into the world of differential m-forms and exterior derivatives.  For an explanation of these, you will need to do your own research.  Michael Penn has a series of lectures on Youtube you might want to check out.  But the algebra of these operators becomes surprisingly simple.

On the left are the definitions in standard vector calculus.  On the right are the equivalent expressions in notation of exterior derivatives and Hodge operators.

$d$ is the exterior derivative.

$\omega_F$ expresses the Vector valued function $F$ as a differential form.

$\star$ is the Hodge operator.

Gradient:

$\nabla F \sim df$

Curl:

$\nabla \times F = \star d\omega_F$

Divergence:

$\nabla \cdot F = \star d\star \omega_F$

The algebra of these forms when you have $\star\star$ the stars cancel.  And $dd=0$ always

Curl of the gradient of a function, $\nabla \times (\nabla F)$

$\star dd F = 0$

and the Gradient of the curl of a function $\nabla\cdot (\nabla \times F)$

$\star d\star \star d \omega_F = \star dd \omega_F = 0$


问:
Power series raised to a non integer power
I would like to know whether there is a general formula to raise a power series to a non integer exponent:

$$
\left(\sum_{k = 0}^{+\infty} a_k x^k\right)^\alpha, \text{ where } \alpha \in \mathbb{R}.
$$

I have seen previous posts related to an integer power: integer power 1, integer power 2, integer power 3. I have also come accross to the following couple of posts that might be closely related to my question: power series^power series and fractional exponent.

It seems that an answer to this problem could involve a generalisation of the multinomial theorem, maybe in the like of this.

There might already be some literature about this, I would be happy to be referred to papers, books, etc.!

EDIT : I am also interested in the corresponding questions of convergence!

答:
There is a slight ambiguity in your question since you do not explain what kind of formula you would like to see: however, since it seems, from the references you cited, that you mean a power series formula, you can build it by using Newton's generalization of the binomial theorem
$$
t\mapsto (1+t)^\alpha = \sum_{k=0}^\infty \binom{\alpha}{k} t^k.\label{1}\tag{1}
$$
By using it, you have that
$$
\left(\sum_{k=0}^\infty a_kx^k\right)^{\alpha}=
\begin{cases}
\displaystyle \sum_{n=0}^\infty \binom{\alpha}{n} \left[\sum_{k=1}^\infty a_kx^k -1\right]^n & a_0=0\\
\\
\displaystyle a_0^\alpha\sum_{n=0}^\infty \binom{\alpha}{n} \left[\sum_{k=1}^\infty a_kx^k\right]^n & a_0\neq0\\
\end{cases}\label{2}\tag{2}
$$
and you could further expand this formula by using the generalization of Cauchy Product formula. However, the computational work involved in the whole procedure seems daunting.

The convergence properties of \eqref{2} can be inferred from the convergence properties of \eqref{1}, listed in the related wikipedia entry, and from the properties of the power series $s(x) = \sum_{k=1}^\infty a_k x^k$: call $D(R)$ the convergence disk (or interval, if $x, a_k\in\Bbb R$ for all $n\in\Bbb N$) of $s(x)$, and $C\big(s^\alpha(x)\big)$ the convergence region of \eqref{2}, i.e the largest open set in $D$ where \eqref{2} converges. Since \eqref{1} converges absolutely for any real $\alpha$ if $|t|<1$ then
$$
C\big(s^\alpha(x)\big) =
\begin{cases}
\big\{x\in D(R) : |s(x)-1|<1\big\} & a_0=0 \\
\\
\big\{x\in D(R) : |s(x)-a_0|<{a_0}\big\} & a_0\neq0 \\
\end{cases}
$$
In a similar fashion, you can analyze the boundary of $C\big(s^\alpha(x)\big)$ by the convergence properties of \eqref{1} on the boundary $|t|=1$ of its convergence disk.

Edit: comments and further notes

Following a comment of the Asker, let's prove the following formula
$$
s^n(x)=\left(\sum_{k=1}^{\infty} a_k x^k\right)^n = \sum_{k=1}^{\infty} \sum_{\substack{0< k_1, \ldots, k_n\le k \\
k_1+\ldots+k_n=k}}a_{k_1} \cdots a_{k_n} x^k \quad\forall n\in\Bbb N,\, n\ge 2\label{3}\tag{3}
$$
Let's proceed by induction

For $n=2$ equation \eqref{3} is simply the Cauchy Product of the power series $s(x)$ by itself.

Assuming that \eqref{3} holds true for a general $n\in\Bbb N$ with $n\ge 2$, lets see if it holds for $n+1$. By using Cauchy's product formula,
$$
\begin{split}
\left(\sum_{k=1}^{\infty} a_k x^k\right)^{n+1} & = \left(\sum_{k=1}^{\infty} a_k x^k\right)\cdot\left(\sum_{k=1}^{\infty} a_k x^k\right)^n \\
&= \Bigg( \sum_{k=1}^{\infty} a_k x^k\Bigg)\Bigg(\sum_{k=1}^{\infty} \sum_{\substack{0< k_1, \ldots, k_n\le k \\
k_1+\ldots+k_n=k}}a_{k_1} \cdots a_{k_n} x^k\Bigg)\\
&= \sum_{k=1}^{\infty} \Bigg(\sum_{m=1}^{k} a_m \sum_{\substack{0< k_1, \ldots, k_n\le k-m \\
k_1+\ldots+k_n=k-m}}a_{k_1} \cdots a_{k_n} \Bigg)x^k\\
&= \sum_{k=1}^{\infty} \Bigg(\sum_{\substack{0< k_1, \ldots, k_n, k_{n+1}\le k \\
k_1+\ldots+k_n + k_{n+1}=k}}a_{k_1} \cdots a_{k_n} a_{k_{n+1}}\Bigg)x^k
\end{split}
$$
and the validity of the formula for $n$ implies its validity for $n+1$.

Thus formula \eqref{3} is proven true for every $n\in\Bbb N$ by induction.

A final note. There are several ways of representing the composition of functions $\left(\sum_{k=0}^\infty a_kx^k\right)^{\alpha}$ for a real $\alpha$: for example it is possible to express it by using Faa di Bruno's formula in its various flavors. As stated in the comments, If you know the derivative of the composition of two analytic functions you know the Taylor series of their composition since this is likewise an analytic function. In our specific case we have
$$ s^\alpha(x)=\sum_{k=0}^\infty \frac{1}{k!}\frac{\mathrm{d}^k}{\mathrm{d}x^k}s^\alpha(x)\big|_{x=0} x^k $$
and, as remarked also in the Wikipedia entry on the said formula, you can calculate the Taylor expansion coefficients directly.

What is the "most convenient" way of proceeding? It depends heavily on the structure of the power series $s(x)$ and on on your purpose: in a word it depends on the context of your application.


答:
The linear mapping $T : \mathbb{R}^{k} \to \mathbb{R}^{n}$ with $k \le n$ induced by the matrix $A$ is injective. This is because the dimension of the image of $T$ equals $k$ (because rank $A = k$) so the dimension of the kernel of $T$ equals $0$. So for all $x \in \mathbb{R}^{k}$ with $x \ne 0$ we have $Ax \ne 0$. This implies

$x^{T}(A^{T}A)x = (Ax)^{T}(Ax) = \|Ax\|^{2} > 0$

therefore the quadratic form $A^{T}A$ is positive definite by definition.

Title: How to show that $A′A$ is a positive deﬁnite matrix given rank$(A) = k$?

Body: I have difficulty in trying to prove the following question.Let $A = n \times k, k ≤ n$. Show that, if rank$(A) = k$, then $A′A$ is a positive deﬁnite matrix ($A'$ denotes the transpose of $A$). Do you have any idea how it can be solved? Thanks a lot.


问:
Question about calculation of sets
Here I want to prove of find a counter example of the statements:

if $A, B, C$ are all sets, then
$$
(A\setminus(B∩C))\setminus(B\setminus C)=A\setminus B
$$
To prove these I draw a Venn diagram

I think it's true but how can I prove it.

答:
Let us do it by parts. To begin with, one has that
\begin{align*}
A - (B\cap C) & = A\cap(\overline{B\cap C}) = A\cap(B^{c}\cup C^{c}) = (A\cap B^{c})\cup(A\cap C^{c})
\end{align*}

Now we can approach the whole expression:
\begin{align*}
(A - B\cap C)-(B - C) & = ((A\cap B^{c})\cup(A\cap C^{c}))-(B\cap C^{c})\\\\
& = ((A\cap B^{c})\cup(A\cap C^{c}))\cap (B^{c}\cup C)\\\\
& = ((A\cap B^{c})\cap(B^{c}\cap C))\cup((A\cap C^{c})\cap(B^{c}\cup C))\\\\
& = (A\cap B^{c}\cap C)\cup((A\cap B^{c}\cap C^{c})\cup\varnothing)\\\\
& = (A\cap B^{c}\cap C)\cup(A\cap B^{c}\cap C^{c})\\\\
& = (A\cap B^{c})\cap(C\cup C^{c})\\\\
& = A\cap B^{c} = A - B
\end{align*}

Hopefully this helps!

Title: Question about calculation of sets


答:
Well your going after arc length here so you either wanna switch sin(x) to sqrt(1 + cos(x)^2)

Since dLength^2 = dx^2 + dy^2

Or:

You can keep in mind it is the length of the top sides of your trapezoids, Not their area, that you want to sum up.

Besides that you should be good to go.

Title: Trap Rule for sin(x)

Body:
Use the trapezoidal rule with $N=6$ to approximate the arc length of the curve $f(x) = \sin(x)$ from $x=0$ to $x=\pi$.

So I found that $\Delta x = \frac{\pi}{6}$ which means that my interval points are $0,\frac{\pi}{3}, \frac{\pi}{2}, \frac{2\pi}{3}, \frac{5\pi}{6}$ and $\pi$.

Is that right?

And then after that I just use the trap rule forumula computing the value of $\sin$ of all my interval points.

Is my thinking correct?


问:
How to determine if this isomorphism is a linear transformation?
I am presented with the following isomorphism:

$$\begin{bmatrix}x & y\end{bmatrix}\longrightarrow\begin{bmatrix}x\\
y
\end{bmatrix}$$

Is this a linear transformation and why?

Thank you for your help.

答:
The mapping $T$ shows that $T(x,y)^{r} = (x, y)^{c}$, where the superscripts $\textbf{r}$ and $\textbf{c}$ represent the row and column vectors.

Then by definition of a linear transformation we have: $T(x, y)^{r} + T(u,w)^{r} = (x, y)^{c} + (u, w)^{c} = (x + u, y + w)^{c} = T(x+u, y + w)^{r}$.

Now, you can prove the second condition, namely: $T(\alpha x, \alpha y)^{r} = \alpha T(x, y)^{r}$.


答: Observe that $a_n > 0, \forall n \ge 1$, and by using the AM-GM inequality: $\forall n \ge 1 \implies (a_{n+1})^2 \ge \left(\dfrac{1}{2}\cdot 2\sqrt{a_n\cdot \dfrac{3}{a_n}}\right)^2 = \left(\sqrt{3}\right)^2 = 3 \implies a_{n+1} \ge \sqrt{3}, \forall n \ge 1$. Thus: $a_{n+1} - a_n = \dfrac{1}{2}\left(a_n+\dfrac{3}{a_n}\right)-a_n= \dfrac{1}{2}\left(\dfrac{3}{a_n}-a_n\right)= \dfrac{3-a_n^2}{2a_n}< 0\implies a_{n+1} < a_n$. The sequence is decreasing and bounded below by $\sqrt{3}$ hence converges to $L$ that satisfies the equation: $L = \dfrac{1}{2}\left(L+\dfrac{3}{L}\right)\implies L^2 = 3\implies L = \sqrt{3}$ since $L \ge \sqrt{3}$.


问:
How to prove the following: Assume A infinite,B denumerable show that A~$A\cup B$
Assume A is infinite,B is denumerable.
Prove A$\approx (A \cup B)$(rewrite)

Reference:

7.18 Theorem A is an infinite set iff A is equipotent with a proper subset of itself.

1.20 Theorem If A and B are any classes, then

i) A⊆A∪B and B⊆A∪B.

ii) A∩B⊆A. and A∩B⊆B

My thoughts on it are as follows :
By 7.18 if A is infinite then it is equipotent
with a proper subset of itself, so by
1.20(i),A$\subseteq A\cup B$

Then there exist an injective function f:$\omega\mapsto A\cup B$
Check f is bijective once defined.

If A is denumerable ,we know there is a bijective
function f:$\omega\mapsto A\cup B$ and we
have a function g:$A\cup B\mapsto\omega$
Once defined show it is onto

I don’t know how to create brackets for
two functions thus I state “once it is created “
Then ?

I don’t think I can conclude somehow?

答:
If $B\subseteq A$, there is nothing to prove, since in that case $A\cup B=A$, so assume that $B\nsubseteq A$. $A$ is infinite, so it has a denumerable subset $D$.

Show that $B\cup D$ is denumerable.

Let $f:\omega\to D$ and $g:\omega\to B\cup D$ be bijections. Define

$$h:A\cup B\to A:x\mapsto\begin{cases}
a,&\text{if }a\in A\setminus(B\cup D)\\
f\left(g^{-1}(x)\right),&\text{if }x\in B\cup D\,,
\end{cases}$$

and show that $h$ is a bijection.

(Some of the discussion in my answer to this related question may prove helpful.)


问:
Uniqueness of the sum of roots of unity
Everyone knows that the sum of all roots of unity $r_k$ equals to zero:
$$\sum^{n-1}_{k=0} r_k=\sum^{n-1}_{k=0} e^{\frac{i2k\pi}{n}} =0$$

Does anyone know if it possible to prove that the equation below:

$$\sum^{n-1}_{k=0} a_k r_k=\sum^{n-1}_{k=0} a_k e^{\frac{i2k\pi}{n}} =0$$ where $a_i\in N$ (non-negative integers)

has only one one solution: $a_0=a_1=...=a_{n-1}=constant$

Probably you can suggest me which direction to look in? May be Kummer rings theory can help or the simple prove exist.

答:
I'll make the assumption the coefficients are real (I think this is slightly more interesting posing an additional constraint), and that $n\ge 3$.

The idea linearity. Let
\begin{align*}
{\bf s} &= (1,\cos(2\pi /n),\dots, \cos(2\pi (n-1)/n))\\
{\bf i} &= (0,\sin (2\pi/n),\dots, \sin (2\pi (n-1)/n))
\end{align*}

represent the real vectors consisting of the real and imaginary parts of $(1, \omega, \dots, \omega^{n-1})$ where $\omega = e^{2\pi i /n}$.

Next define $T:{\mathbb R}^n \to {\mathbb R}^2$ by letting

$$ T {\bf a}  = ({\bf a}\cdot {\bf s} ,{\bf a} \cdot {\bf i}),$$

where  ${\bf a} = (a_0,\dots, a_{n-1})$. Then $T$ is linear and  $\sum_{j=0}^{n-1} a_j \omega^j=0$ if and only if $T{\bf a}=(0,0)$.

Since $n\ge 3$, it follows that $T$ is onto. Indeed: $Te_1 =(1,0)$ and $T e_2 = (\cos 2\pi/n,\sin 2\pi/n)$, which is linearly independent of $T e_1$.

Therefore by the rank + nullity Theorem, the null space of $T$ has dimension $n-2$.

To find a basis for the null space, you can choose  $n-2$  vectors of the form
$(x_2,y_2,1,0,\dots,0)$, $(x_3,y_3,0,1,0,\dots,0)$, ...,  $(x_{n-1},y_{n-1},0,\dots,0,1)$, where for each $j=2,\dots,n-2$, the constants  $x_j,y_j$ are chosen so that the resulting vector is orthogonal to both ${\bf s}$ and ${\bf i}$. Solving the equations we have

\begin{align*} &y_j = -\frac{\sin (2\pi j/n)}{\sin (2\pi /n)}\\
& x_j = -\cos(2\pi j/n) +\cos(2\pi/n)\frac{\sin (2\pi j/n)}{\sin (2\pi /n)}.
\end{align*}

In the case $n=4$, this gives the vectors
$(1,0,1,0),(0,1,0,1)$

Title: Uniqueness of the sum of roots of unity


问:
Find the angle $\angle MNP$ from the circle with centre C
Let us consider a circle with centre at C, and diameter MCN. Consider a point P on the circumference of the circle and join MP and NP such that MPN is a triangle with $\angle MPN =π/2$. Now consider a point A on the same semicircle, I mean on the arc MP and two points B and D on the side MP such that ABCD is a square, recall that C is the centre of the circle. Then find the $\angle MNP$.

I mean I can not find a particular value of the required angle. If I assume radius as $r$, then somehow using triangles law I found a relation but not so close. Can you please help me to solve this question.

Thanks in advance.

答:
Since $ABCD$ is a square and the diagonal $BD$ lies on the line segment $MP$, so does the midpoint of $BD$ (let's call it $O$), which is also the midpoint of $AC$.

We also have $AC\perp BD$. Hence $OC \perp MP$, and thus $MO = OP$.

Using $MC = CN$ and the midpoint theorem, we see that $OC$ is half the length of $PN$. But $OC$ is half the length of $AC$, which is the radius.

Hence $PN$ has equal length as the radius. This gives $\angle MNP = 60^\circ$.


问:
Understanding the values of Möbius function
I want to understand how did he get the values for this intersection poset:

So I know that he used the Möbius function but I don't understand how he get the values only using the following :

But how did he end up with negative values? If someone can explain to me thanks in advance and why for $tuv$ he gets $2$ not $3$

答:
If you transpose the $\mu(x,y)$ term, the condition that

$$\sum_{z\in[x,y]}\mu(x,z)=0$$

for $x<y$ becomes

$$\mu(x,y)=-\sum_{z\in[x,y)}\mu(x,z)\,.$$

In his example, for instance,

$$\mu(u)=\mu(\hat 0,u)=-\sum_{z\in[\hat 0,u)}\mu(\hat 0,z)=-\mu(\hat 0,\hat 0)=-1\,,$$

since $\hat 0$ is the only element of $[\hat 0,u)$.


问:
Prove the following sequence is convergent: $a(1) = 1$ and $a(n+1) = 1/2 \cdot (a(n) + 3/a(n))$
I know the limit is $\sqrt 3$. And I tried proving that the sequence is decreasing from $a(2)$, yet when I got to this point, I have been stuck:

$a(n+1) - a(n) = (3 - a(n)^2) / 2a(n)$ ?

Thank you very much!

Body:
$$\frac{a_{n+1}+\sqrt3}{a_{n+1}-\sqrt3} = \frac{a_n+\frac{1}{a_n}+2\sqrt3}{a_n+\frac{1}{a_n}-2\sqrt3}=\left(\frac{a_n+\sqrt3}{a_n-\sqrt3}\right)^2 \\\implies \frac{a_n+\sqrt3}{a_n-\sqrt3}=\left(\frac{a_1+\sqrt3}{a_1-\sqrt3}\right)^{2^{n-1}}= \left(\frac{1+\sqrt3}{1-\sqrt3}\right)^{2^{n-1}}\\ \implies
a_n = \sqrt 3 \cdot \frac{(1+\sqrt3)^{2^{n-1}}+ (1-\sqrt3)^{2^{n-1}}}{(1+\sqrt3)^{2^{n-1}}-(1-\sqrt3)^{2^{n-1}}} \to \sqrt 3, n \to \infty.$$


问:
Need help with logical expression involving existential and universal quantifier
P is the set of "all exams at university". E(x) denotes the predicate "x is easy". Translate the following sentence to formal logic. "Only one exam is easy." Solution:

I understand the first part, that there exists a test "x" for which the predicate "is easy" holds. But I don't understand the second part. I read it as such that for all y,z element of the set of exams, if y is easy and z is easy and there exist an exam x that is easy it follows that y = z.

答:
The second part says that if you pick any two easy exams, then they will turn out to be one and the same exam. This will rule out that there are two (or more) different easy exams. Indeed, you could equivalently write that second part as:

$$\neg \exists y, z \in P: (E(y) \land E(z) \land y \neq z)$$

Finally, here's an efficient way to say that there is exactly one easy exam:

$$\exists x \in P: \forall y \in P: (E(y) \leftrightarrow x = y)$$


问:
Let ${\rm Hom}:G\to G'$ be a homomorphism. Prove that if $H\unlhd G$ then ${\rm Hom}(H)\unlhd G'$.
Let ${\rm Hom}:G\to G'$ be a homomorphism. Prove that if $H\unlhd G$ then ${\rm Hom}(H)\unlhd G'$.

My current approach is to show that ${\rm Hom}(gHg^{-1}) = g'{\rm Hom}(H)g'^{-1}$ using the properties of homomorphisms and normal subgroups but it seems too simple for this to be correct.

答:
This is not true. Take any $H$ that is not a normal subgroup of a group $G$.

Then the inclusion of $H$ in $G$ is a group homomorphism and the image of $H$ is obviously not normal in $G$.

This is however true if the homomorphism is supposed to be onto.


答:
Counterexample: consider any example of a non-normal subgroup $G$ of a group $G'$. The identity map $\varphi : G \hookrightarrow G'$ is a homomorphism. $H=G$ is a normal subgroup of $G$, while $\varphi(H) = G$ is a non-normal subgroup of $G$ by assumption.

However, if the homomorphism is surjective, then the claim is true. To show $g' \text{Hom}(H) g'^{-1} = \text{Hom}(H)$ write $g'$ as the image of some element $g$ of $G$ and use your work.

Body:
Let ${\rm Hom}:G\to G'$ be a homomorphism. Prove that if $H\unlhd G$ then ${\rm Hom}(H)\unlhd G'$.

My current approach is to show that ${\rm Hom}(gHg^{-1}) = g'{\rm Hom}(H)g'^{-1}$ using the properties of homomorphisms and normal subgroups but it seems too simple for this to be correct.


问:
Factoring $−x^{3}−3x^{2}+4$
I have this polynomial:

$$−x^{3}−3x^{2}+4$$

and I know it factors to this:

$$−(x−1)(x+2)(x+2)$$

But how do I do this formally and in a step by step way?

答:
HINT

Let $f(x) = -x^{3} - 3x^{2} + 4$. Then notice the sum of its coefficients equals zero. This means that $f(1) = 0$.

Now you can factor $x - 1$ out, whence you are going to obtain a second degree polynomial.

Can you take it from here?

Title: Factoring $−x^{3}−3x^{2}+4$


问:
Recover the stalk at a certain point along the base.
Let $X$ be a topological space; its open sets form a directed set by the rule
 $$U\prec V\quad\textrm{iff}\quad V\subseteq U$$
If $\mathscr F$ is a sheaf, for example of rings, then by definition, for every $x\in X$ we have that
$$\mathscr F_x=\varinjlim_{U\ni x}\mathscr F(U)$$
where clearly $U$ is varying along all open neighborhoods of $x$.

Now if $\mathcal B$ is a base for the topology, intuitively one should have that

$$\mathscr F_x=\varinjlim_{\substack {B\in\mathcal B\\B\ni x}}\mathscr F(B)$$

because $\{\mathscr F(B)\}$ looks like as a sort of subsequence of $\{\mathscr F(U)\}$. However I cannot make the above intuitive statement in a formal proof.

答:
The key is that "only the really small open sets matter". Let then $F_x$ be the usual stalk at $x$ with canonical inclusions $i_U:F(U)\to F_x$, whereas $F_{(x)}$ is the "basic" stalk at $x$ has inclusions $j_B:F(B)\to F_{(x)}$.

On the one hand basic open sets are open sets and therefore we have for each $B\in \mathcal B$ a map $i_B:F(B)\to F_x$, so the universal property of the colimit gives us a map $F_{(x)}\to F_{x}$.

On the other hand, since $\mathcal B$ is a basis for the topology, for any $U$, there exists some $B\subseteq U$ a basic open set.

Now consider $\mathcal B_U$ the family of basic open sets contained in $U$, then for each $B\in \mathcal B_U$ we have a map $\mu_B:F(U) \to F(B)$ that naturally commutes with restrictions (that is $\mu_C = r_{B,C}\circ\mu_B$). This gives a map $F(U)\to F_{(x)}$ defined by $j_B\circ\mu_B$ which does not depend on the choice of $B$, this family gives rise to a map $F_x\to F_{(x)}$ (by the universal property pf the colimit).

Lastly I couldn't piece out how to do this in general but we are left to show that the maps $\alpha:F_x\rightleftarrows F_{(x)}:\beta$ are inverse to each other, the setting I chose is when $F$ is a sheaf of sets.

In this setting we have a representation of the stalks as the set of germs/"basic" germs at $x$

$$F_x = \coprod_{U\ni x} F(U)\big/\sim \qquad
F_{(x)} = \coprod_{B\ni x} F(B)\big/\sim $$

where $s_U \sim s_V$ iff there exists $W\subseteq U\cap V$ such that $r_{U,W}(s_U) = r_{V,W}(s_V)$ (in both cases where $U,V,W$ are general open sets or just basic open sets).

Thus take an element $s\in F_x$ it is a germ of some element $s_U\in F(U)$ therefore it restricts to an element $s_B\in F(B)$ for some basic open set $B$ containing $x$, thus determining a "basic germ", since we can take $U$ to be a basic open set itself, this map $F_x\to F_{(x)}$ is surjective. Now take two germs $s,t\in F_x$ so that $\alpha(s) = \alpha (t)$, furthermore let $s_U$ and $t_V$ be the representatives of $s$ and $t$ respectively, then $\alpha(s)$ and $\alpha(t)$ have representatives $s_{B}$ and $t_{B'}$ in $F_{(x)}$, and since they are the same, itstands to rason to conclude there exists some basic open set $D\subseteq B\cap B'$ so that $r_{B,D}s_B = r_{B',D}t_{B'}$ but $s_B = r_{U,B}s_U$ and $t_{B'}=r_{V,B'}t_V$ thus $r_{U,D}s_U= r_{V,D}t_V$ and so $t=s$. Therefore $\alpha$ is injective and surjective which is an isomorphism in $\mathbf{Set}$.


答:
Given $ABCD$ is a square, $AC \perp MP, CE = \frac{r}{2}, CE \parallel NP$

$\triangle MCE \sim \triangle MNP, PN = r$

So, $\angle MNP = 60^0$.

Title: Find the angle $\angle MNP$ from the circle with centre C

Body:
Let us consider a circle with centre at C, and diameter MCN. Consider a point P on the circumference of the circle and join MP and NP such that MPN is a triangle with $\angle MPN =π/2$. Now consider a point A on the same semicircle, I mean on the arc MP and two points B and D on the side MP such that ABCD is a square, recall that C is the centre of the circle. Then find the $\angle MNP$.

I mean I can not find a particular value of the required angle. If I assume radius as $r$, then somehow using triangles law I found a relation but not so close. Can you please help me to solve this question.

Thanks in advance.


问:
Prove that the subsets $[-2, 2] \times [-2, 2]$ and $[-3, 3] \times [-1, 1]$ of $\mathbf{R}^2$ are homeomorphic
In order to prove homeomorphism, I would have to define a function in $\mathbf{R}^2$ which is a continuous invertible bijection whose inverse is also continuous.

Plotting the cartesian product on Desmos looks as follows:

Is there an intuitive way to think about this? Also how should I go about finding such a function? I'm very new to topology and metric spaces so apologies for the vague question. Any hints would be appreciated.

答: Yes, there is an intuitive way to think about this. In $[-2,2]\times[-2,2]$, you want to expand the first interval by a factor $\frac32$ in order to get $[-3,3]$ and you want to expand the second interval by a factor $\frac12$ in order to get $[-1,1]$. So, consider the map$$\begin{array}{ccc}[-2,2]\times[-2,2]&\longrightarrow&[-3,3]\times[-1,1]\\(x,y)&\mapsto&\left(\frac32x,\frac12y\right).\end{array}$$Can you take it from here?


答:
Hint

Find two homeomorphisms
$$\begin{cases}
F : [-2,2] \to [-3,3]\\
G : [-2,2] \to [-1,1]
\end{cases}$$
Then consider
$$H :(x,y) \mapsto (F(x),G(y)).$$

Title: Prove that the subsets $[-2, 2] \times [-2, 2]$ and $[-3, 3] \times [-1, 1]$ of $\mathbf{R}^2$ are homeomorphic

Body:
In order to prove homeomorphism, I would have to define a function in $\mathbf{R}^2$ which is a continuous invertible bijection whose inverse is also continuous.

Plotting the cartesian product on Desmos looks as follows:

Is there an intuitive way to think about this? Also how should I go about finding such a function? I'm very new to topology and metric spaces so apologies for the vague question. Any hints would be appreciated.


问:
Use an inverse matrix to find [x]b for the given x and B.
My linear algebra textbook has asked me to find [x]_b for the given x and basis B using an inverse matrix. The answer key does not provide anything but a solution, which doesn't help me learn the process to complete a problem like this seeing that I could have one like it on my final exam.

A similar problem with different values is worked through on Chegg; however, there are pieces missing from the explanation that leave me unable to figure out where certain numbers come from.

Could someone walk me through the process so that I can try it with other similar homework problems?

B = {[3 -5],[-4 6]}, x = [2 -6]

答:
In the regular basis, suppose you have vector $\vec x=(x_0\ y_0)$. What this means is that you can get $\vec x$ by going in the $(1\ 0)$ direction $x_0$ units and then $y_0$ units in the $(0\ 1)$ direction. So you write $$\vec x=x_0(1\ 0)+y_0(0\ 1)$$
Let's suppose now that you write this vector in a similar fashion using a basis $B$, with vectors $(a\ b)$ and $(c\ d)$. Then $$\vec x=(x_0^B\ y_0^B)_B=x_0^B(a\ b)+y_0^B(c\ d)$$
Here $x_0^B$ is just a number, not $(x_0)$ to power $B$.

So how would you write this in a matrix form? Write the matrix created by putting the basis vectors vertically next to each other
$$B=\begin{pmatrix}a&c\\b&d\end{pmatrix}$$
Now you know the vector $\vec x$ is the same, no matter the representation. Then write all vectors as vertical matrices. You get:
$$B\begin{pmatrix}x_0^B\\y_0^B\end{pmatrix}=\begin{pmatrix}a&c\\b&d\end{pmatrix}\begin{pmatrix}x_0^B\\y_0^B\end{pmatrix}=\begin{pmatrix}1&0\\0&1\end{pmatrix}\begin{pmatrix}x_0\\y_0\end{pmatrix}$$
Now multiply both sides on the left with $B^{-1}$. Using $B^{-1}B=I$ and $B^{-1}I=B^{-1}$ you get $$\begin{pmatrix}x_0^B\\y_0^B\end{pmatrix}=B^{-1}\begin{pmatrix}x_0\\y_0\end{pmatrix}$$


问:
Infinite shortest paths in graphs
From Wikipedia:

"If there is no path connecting two
  vertices, i.e., if they belong to
  different connected components, then
  conventionally the distance is defined
  as infinite."

This seems to negate the possibility that there are graphs with vertices connected by an infinite shortest path (as opposed to being not connected).

Why is it that for every (even infinite) path between two vertices there is a finite one?

Note that infinite paths between vertices do exist - e.g. in the infinite complete graph -, but they are not the shortest.

答:
To expand on my comment: It's clear that if an infinite path is defined as a map from $\mathbb N$ to the edge set such that consecutive edges share a vertex, then any vertices connected by such an infinite path are in fact connected by a finite section of the path. To make sense of the question nevertheless, one might ask whether it is possible to use a different ordinal than $\omega$, say, $\omega\cdot2$, to define an infinite path. But that doesn't make sense either, since there's no way (at least I don't see one) to make the two parts of such a path have anything to do with each other -- at each limit ordinal, the path can start wherever it wants, since there's no predecessor for applying the condition that consecutive edges share a vertex.

Note that the situation is different in infinite trees, which can perfectly well contain infinite paths connecting the root to a node. This is because the definition of a path in an infinite tree is different; it explicitly attaches the nodes on levels corresponding to limit ordinals to entire sequences of nodes, not to individual nodes; such a concept doesn't exist in graphs.

Title: Infinite shortest paths in graphs


问:
Which of the following are subspaces of $M$?
Let $M$ be a vector space of all $3\times 3$ real matrices and let $$A=\begin{pmatrix}2&3&1\\0&2&0\\0&0&3\end{pmatrix}.$$ Which of the followings are subspaces of $M?$

$\{X\in M:XA=AX\}$

$\{X\in M:X+A=A+X\}$

$\{X\in M:\text{trace}(XA)=0\}$

$\{X\in M:\det(XA)=0\}$

答: only number 1 is subspace becuase for each $x_1 ,x_2 \in M$ we have $x_1A=Ax_1$ and $x_2A=Ax_2$ so for each $c\in F$ we will have $(cx_1+x_2)A=cx_1A+x_2A=Acx_1+Ax_2=A(cx_1+x_2)$</p>


问:
Domain of composite of function and its inverse
I'm looking for the domain of definition of the  composite function of one function and its inverse.

Let $$H(x)= \frac{e^x -1}{e^x +1} $$ and its inverse $$G(x)=\ln{ \frac{x+1}{1-x}}$$

Is $-1 <x <1 $ the domain of definition of $H(G(x))$ ?

What is weird is that by going through the computer algebra tool Mathematica, I get $\mathbb{R}$ as the domain.
I think the tool does a simplification and gets the identity function, and that explains why I get $\mathbb{R}$ as the domain.
Am I right?

The graph of $$ \frac{x+1}{1-x}: $$

The graph of $H(x)$ in blue and $G(x)$ in orange:

答:
$G(x)$ is defined for $\frac{x+1}{x-1} \gt 0$, i.e. for  $x \in (-\infty, -1) \cup (1,\infty)$. Which is also the domain of $H \circ G$.

So you made a mistake in $H$ domain and the software you used too.


问:
How to find a bound for Fourier coeffiecients
Let $ -\pi=x_{0}<x_{1}<...<x_{m}=\pi $

be a partition of the interval $ [-\pi,\pi) $

And let $ s\left(x\right)=\sum_{j=1}^{m}\alpha_{j}1_{\left(x_{j-1},x_{j}\right)}\left(x\right)$ be a step function such that $ \alpha_{1}\leq\alpha_{2}\leq...\leq\alpha_{m}\in\mathbb{R} $

Where $ 1_{\left(x_{j-1},x_{j}\right)}\left(x\right)=\begin{cases}
1 & x\in\left(x_{j-1},x_{j}\right)\\
0 & x\in[-\pi,\pi)\setminus\left(x_{j-1},x_{j}\right)
\end{cases} $

I have to find a constant $ C $ such that $ C $ does not depend on $ s $ and the following inequality holds:

$ |\hat{s}\left(k\right)|\le C\frac{\max\left\{ -\alpha_{1},\alpha_{m}\right\} }{|k|} $

I already calculated the Fourier coefficients of the indicators, and it is given by:

$ \hat{1}_{\left(a,b\right)}\left(k\right)=\begin{cases}
\frac{b-a}{2\pi} & k=0\\
\frac{1}{2\pi ik}\left(e^{-ika}-e^{-ikb}\right) & k\ne 0
\end{cases} $

I actually did find such $ C $, I managed to prove that for $ C=\frac{m}{2\pi} $ the inequality holds. Is it possible to find such $ C $ that would not depend on $ s $ or on the partition $ x_1<x_2<...<x_m $ ?

This is part of a guided exercise and I need such $ C $ for the next steps.

Thanks in advance.

答:
We want to bound $ |\sum_{j=1}^{m}\alpha_{j}\left(e^{-ikx_{j-1}}-e^{ikx_{j}}\right)| $

Let us denote $ B_{i}=\sum_{j=1}^{i}\left(e^{-ikx_{j=1}}-e^{-ikx_{j}}\right) $

So by Abel's summation formula we have:

$ |\sum_{j=1}^{m}\alpha_{j}\left(e^{-ikx_{j-1}}-e^{ikx_{j}}\right)|=|\alpha_{m}\left(e^{-ik_{m-1}}-e^{-ikx_{m}}\right)+\sum_{i=1}^{m-1}B_{i}\left(\alpha_{i+1}-\alpha _{i}\right)| $

$ \underset{a_{i}\leq a_{i+1}}{\leq}2|\alpha_{m}|+\sum_{i=1}^{m-1}|B_{i}|\left(\alpha_{i+1}-\alpha_{i}\right) $

Now notice that $ B_i $ is just a telescopic sum and thus $ |B_i| < 2 $ So:

$ \underset{a_{i}\leq a_{i+1}}{\leq}2|\alpha_{m}|+\sum_{i=1}^{m-1}|B_{i}|\left(\alpha_{i+1}-\alpha_{i}\right)\leq2|\alpha_{m}|+2\sum_{i=1}^{m-1}\left(\alpha_{i+1}-\alpha_{i}\right)\underset{telescopic\thinspace\thinspace\thinspace sum}{=}2|\alpha_{m}|+2\left(\alpha_{m}-\alpha_{1}\right)\leq4|\alpha_{m}|+4|\alpha_{1}| $

So all in all

$ |\hat{s}\left(k\right)|=|\frac{1}{2\pi}\intop_{-\pi}^{\pi}\sum_{j=1}^{m}\alpha_{j}1_{\left(x_{j-1},x_{j}\right)}e^{-ikx}dx|=|\frac{1}{2\pi}\sum_{j=1}^{m}\alpha_{j}\hat{1}_{\left(x_{j-1},x_{j}\right)}\left(k\right)|=\frac{1}{2\pi|k|}|\sum_{j=1}^{m}\alpha_{j}\left(e^{-ikx_{j-1}}-e^{ikx_{j}}\right)|\leq\frac{4}{2\pi|k|}\left(|\alpha_{1}|+|\alpha_{m}|\right) $

And since $ a_1\leq a_2 \leq ... \leq a_m $

we have

$ |\hat{s}\left(k\right)|\leq\frac{8}{2\pi|k|}\max\left\{ -a_{1},a_{m}\right\}  $


问:
$I$ a maximal ideal in ring $R$ if and only if for all $a \in R\setminus I, \exists r\in R, b \in I$ such that $b+ar=1$.
Let $R$ be a commutative Ring with $1\neq 0$. Let $I$ be a proper ideal of $R$. Show that $I$ is maximal $\iff$ for all $a \in R\setminus I$ there are $r\in R$ and $ b \in I$ such that $b+ar=1$.

I'm not sure how to approach proving this.

答:
Hint:

To show $\Rightarrow$, let $I$ maximal and $a \in R \setminus I$.

Then $\langle I, a \rangle \supsetneq I$. Since $I$ is maximal, what must $\langle I, a \rangle$ be? Do you see why this shows $1 = b  +ar$?

To show $\Leftarrow$, we use a slick trick: recall $I$ is maximal if and only if $R/I$ is a field (see here for a proof).

Then for some $a \in R \setminus I$, saying $b+ar = 1$ is saying exactly that $ar = 1 \text{ (mod $I$) }$. Do you see how to show that $R/I$ must be a field, and thus $I$ must be maximal?

I hope this helps ^_^


问:
Five years ago, Tim was twice as old as Jim... Math Question
Five years ago, Tim was twice as old as Jim. In three years, Jim’s age will be $2/3$ that of Tim. What are the ages of Tim and Jim now?

答:
I found the following property of ratios useful for quick calculation: if $\frac{a}{b} = \frac{c}{d}$ then $\frac{a}{b}=\frac{c}{d}=\frac{a+kc}{b+kd}$ as long as $b+kd \ne 0$.

Now,
$T-5=2(J-5), \frac 23 (T+3) = J+3$, we have

$$1=\frac{3J+9}{2T+6} = \frac{2J-10}{T-5} = \frac{3J+9-2(2J-10)}{2T+6-2(T-5)}=\frac{29-J}{16}$$

Then you have $J=13$ immediately, and $T=2(J-5)+5=21.$


问:
Truth set with two variables
So I have no idea how to go about this question; it asks for the truth set of the following:

$x \in \mathbb{R}$ and $5 \in \{y \in \mathbb{R} \mid x^2 + y^2 < 50 \}$

Can anyone enlighten me on this one?

Title: Truth set with two variables

Body: In other words, $\{x\in\Bbb R|x^2+5^2<50\}=(-5,\,5)$.


答:
Hints:

$X+A=A+X$ for all $X\in M$, since matrix addition is commutative.

The trace function is additive.

Since $\det(A)\neq 0$, $\det(XA)=0$ iff $\det(X)=0$.

Title: Which of the following are subspaces of $M$?

Body:
Let $M$ be a vector space of all $3\times 3$ real matrices and let $$A=\begin{pmatrix}2&3&1\\0&2&0\\0&0&3\end{pmatrix}.$$ Which of the followings are subspaces of $M?$

$\{X\in M:XA=AX\}$

$\{X\in M:X+A=A+X\}$

$\{X\in M:\text{trace}(XA)=0\}$

$\{X\in M:\det(XA)=0\}$


问:
Is a closed set in a metric space necessarily an infinite intersection of open sets? If so, why?
This has been on my mind for a few days now. I know that an infinite intersection of open sets can be closed, but is a closed set necessarily an infinite intersection of open sets?</p>

答:
Let $(X, d)$ a metric space and $F \subseteq X$ be closed. If we define

$$ \rho(x) = \operatorname{dist}(x, F) := \inf\{ d(x, x') : x' \in F\}, $$

then it is routine to check that $\rho$ is a continuous function with $F = \rho^{-1}(\{0\})$. Then

$$ F = \rho^{-1}(\{0\}) = \bigcap_{\epsilon > 0} \rho^{-1}((-\infty, \epsilon)) $$

shows that $F$ is an intersection of open sets.

Remarks.

$\rho$ is in fact $1$-Lipschitz, meaning that $\left| \rho(y) - \rho(x) \right| \leq d(y, x)$ for all $x, y \in X$.

For a general subset $F \subseteq X$, the inverse image $\rho^{-1}(\{0\})$ is the closure of $F$.

The set $\rho^{-1}((-\infty, \epsilon)) = \{x \in X : \operatorname{dist}(x,F) < \epsilon\}$ is often called the $\epsilon$-neighborhood of $F$ for an obvious reason.


问:
Proving that there is no function that satisfies the conditions
Prove that there can't be a function $g$ such that $g(x) \neq 0$ and $\lim_{x\to a}(\frac{1}{g(x)}+g(x)) = 0$

I tried to prove this by contradiction.

I assumed that such function exists, and from the definition of limit there exists a $\delta > 0$ so that for all $0<|x-a|<\delta$ :

$|\frac{1}{g(x)}+g(x) - 0| = |\frac{1+g^2(x)}{g(x)}| < ... < \epsilon$

But I couldn't get to a contradiction.

Can I get a lead?

答: Let $\epsilon = 2$, and for any $\delta > 0$,  $\forall c \in  (a - \delta, a + \delta)$, either $f(c) > 0$ or $f(c) < 0 \implies \dfrac{1}{f(c)} + f(c) \ge 2$ or $\dfrac{1}{f(c)} + f(c) \le -2\implies \left|\dfrac{1}{f(c)} + f(c)-0\right| = \left|\dfrac{1}{f(c)} + f(c)\right|\ge 2= \epsilon$. This shows the limit cannot be $0$, and this means $g(x)$ can't exist.


问:
Power of a point tangents
Two circles $A$ and $B$ with centers $P$ and $Q$, respectively, are externally tangent to each other. The power of point $P$ with respect to circle $B$ is $8$. The power of point $Q$ with respect to circle $A$ is $15$. What is the ratio of the areas of circle $A$ and circle $B$?

How do I go about solving this problem? Is  there perhaps a formula for ratios of areas of circles given different powers of points?

Title: Power of a point tangents

Body:
$PT^2=8,QU^2=15$ for the secant-tangent theorem
$$
\begin{cases}
r(r+2R)=8\\
R(R+2r)=15\\
\end{cases}
$$

$$
\begin{cases}
r^2+2rR=8\to -15r^2-30rR=-120\\
R^2+2rR=15\to8R^2+16rR=120\\
\end{cases}
$$
Add the two equations
$$8R^2-14rR-15r^2=0$$
Divide all terms by $r^2$ and set $r/R=z$
$$8-14z-15z^2=0\to z=\frac{2}{5};\;\left(z= -\frac{3}{4}\right)$$

$$\frac{r}{R}=\frac{2}{5}$$
$$\frac{Area_A}{Area_B}=\frac{4}{25}$$


问:
Convergence of the derivative of a power series on the boundary
I'm wondering if the series $\sum a_nx^n$ converges at $x=R$, the radius of convergence, can we conclude that it's derivative also converges when x=R ?

答: No, we cannot. The power series $\sum_{n=1}^\infty\frac{x^n}{n^2}$ has radius of convergence $1$ and it converges on $[-1,1]$. But the power series $\sum_{n=1}^\infty\frac{x^{n-1}}n$ only converges on $[-1,1)$; it diverges when $x=1$.


问:
All local cohomology modules being zero
Let $R$ be a Noetherian ring with unit, $I$ be an ideal of $R$ and let $M$ be a finitely generated $R$-module. Suppose $H_{I}^j(M)=0$ for all $j$, then how can one show that $M=IM$?

The converse of the above is trivial.

答: This one is easy, too. Assume that $IM\neq M$. Then $\operatorname{grade}(I,M)=n<\infty$. Now prove by induction on $n$ that $\operatorname{grade}(I,M)=\min\{i:H_I^i(M)\neq 0\}$ and get a contradiction.</p>

Title: All local cohomology modules being zero


答:
Yes it is.

Let $(X,d)$ be a metric space and $C$ a closed subset of $X$. Then $X \setminus C$ is an open set.

For all $x \in X \setminus C$ there exists a radius $r_x>0$ such that the open ball $B(x,r)$ is contained in $X \setminus C$. In particular
$$\overline{B(x, r_x/2)} \subseteq B(x,r_x) \subseteq X \setminus C$$
where $\overline{B(x, r_x/2)}$ denotes the closed ball centered at $x$.

This implies that
$$X \setminus C = \bigcup_{x \in X \setminus C} \overline{B(x, r_x/2)}$$

Using De Morgan's law, this is equivalent to
$$C = \bigcap_{x \in X \setminus C} X \setminus \overline{B(x, r_x/2)}$$
This shows that $C$ is an intersection of open sets.

Title: Is a closed set in a metric space necessarily an infinite intersection of open sets? If so, why?

Body: This has been on my mind for a few days now. I know that an infinite intersection of open sets can be closed, but is a closed set necessarily an infinite intersection of open sets?</p>


答:
Note that
$$ \bigg(\frac{1}{g(x)}+g(x)\bigg)^2=\frac{1}{g^2(x)}+g^2(x)+2\ge2 $$
and hence if
$$ \lim_{x\to a}(\frac{1}{g(x)}+g(x)) = 0$$
we would have
$$ 0\ge2$$
which is absurd.

Title: Proving that there is no function that satisfies the conditions

Body:
Prove that there can't be a function $g$ such that $g(x) \neq 0$ and $\lim_{x\to a}(\frac{1}{g(x)}+g(x)) = 0$

I tried to prove this by contradiction.

I assumed that such function exists, and from the definition of limit there exists a $\delta > 0$ so that for all $0<|x-a|<\delta$ :

$|\frac{1}{g(x)}+g(x) - 0| = |\frac{1+g^2(x)}{g(x)}| < ... < \epsilon$

But I couldn't get to a contradiction.

Can I get a lead?


问:
Asymptotically $n(1-\frac{a\log n}{n})^{n/2-1}(1-\frac{b\log n}{n})^{n/2}\sim n^{1-\frac{a+b}{2}}$
Suppose $a,b>0$ and $\frac{a+b}{2}<1$

Why is the following relation true as $n\rightarrow\infty $ $$n(1-\frac{a\log n}{n})^{n/2-1}(1-\frac{b\log n}{n})^{n/2}\sim n^{1-\frac{a+b}{2}}$$

The $1$ in the exponent on the right hand side value is obviously clear... So Why does $(1-\frac{a\log n}{n})^{n/2}(1-\frac{b\log n}{n})^{n/2}$ reduce to $n^{\frac{a+b}{2}}$?

答:
Hint:

Use $\left ( 1+\frac{x}{n} \right)^n \sim e^x$.

I hope this helps ^_^


问:
Differentiability of $f(x) = x^2 \sin{\frac{1}{x}}$ and $f'$

Let $f(x) = x^2 \sin{\frac{1}{x}}$ for $x\neq 0$ and $f(0) =0$.

(a) Use the basic properties of the derivative, and the Chain Rule to show that $f$ is differentiable at each $a\neq 0$ and calculate $f'(a)$.

You may use without proof that $\sin$ is differentiable and that
$\sin' =\cos$.

Not even sure what this is asking.

(b) Show that $f$ is differentiable at $0$ and that $f'(0) =0$.

$\frac {f(x)-f(0)}{x-0} \to \lim_{x \to 0} x \sin(1/x)$.

$x \sin(1/x) \leq |x|$ and $\lim_{x \to 0} |x|=0$.

Thus $f(x)$ is differentiable at $0$; moreover $f^{'}(0)=0$.

(c) Show that $f'$ is not continuous at $0$.

$f{'}(x)=x^{2} \cos(1/x) (-x^{-2}) + 2x \sin (1/x)$.

In pieces: $\lim_{x \to 0} \cos (1/x)$.

$f^{'}(0-)$ nor $f{'}(0+)$ exists as $x \to 0$ $f^{'}(x)$ oscillates infinity between $-1$ and $1$ with ever increase frequency as $x \rightarrow 0$ for any $p>0$ $[-p,0]$, $[-p,p]$ or $[0,p]$ $f$ is not continuous.

Question: How to show more rigorously?

答:
$x^2$ is continuous and differentiable over $\mathbb{R}$

$\sin(x)$ is continuous and differentiable over $\mathbb{R}$

$\frac 1 x$ is continuous and differentiable over all $\mathbb{R}$ except $0$. And is a function of $\mathbb{R \to R}$

$\displaystyle \sin\left(\frac 1 x\right)$ is therefore continuous and differentiable for all $\mathbb R$ except $0$ where it is undefined.

$\displaystyle x^2\sin\left(\frac 1 x\right)$ is therefore continuous and differentiable for all $\mathbb R$ except possibly $0$.

To compute $f'(a)$ use product rule followed by chain rule to find:

$$F'(a) = 2a\sin\left(\frac 1 a\right) - \cos\left(\frac 1 a\right)$$


问:
A countable subset of R without isolated points
Is there a countable subset of $\mathbb{R},$ which doesn't have any isolated points?

It is an easy consequence of the Baire Category theorem that any closed countable set of reals has an isolated point. I am just wondering, if we can drop the condition about the set being closed.

答: How about $\Bbb Q$?  I don't think it has any isolated points.


问:
If $f$ is a function differentiable at $a$ find: $\underset{h\rightarrow 0}{\lim} \frac{f(a+7h)-f(a-9h^2)}{h}$
If $f$ is a function differentiable at $a$ find:
$\underset{h\rightarrow 0}{\lim} \frac{f(a+7h)-f(a-9h^2)}{h}$

I am struggling to understand what to do. I tried brute forcing this question and I get $\frac{f(a)-f(a)}{0}$ which makes no sense to me. Any clarification would be appreciated.

Body: Hint: $\dfrac{f(a+7h)-f(a-9h^2)}{h} = \dfrac{(f(a+7h) -f(a)) + (f(a) - f(a - 9h^2))}{h} = 7\cdot\dfrac{f(a+7h) - f(a)}{7h} + 9h\cdot \dfrac{f(a) - f(a - 9h^2)}{9h^2}$


问:
Is there a name for the "opposite" of continuity?
Recently I was working on a proof where I eventually wanted to show that for a continuous function $f$ we have the following property at $x$:

For all $\delta > 0$ there exists some $\epsilon > 0$ such that for every $x' \in N_\delta(x)$:

$$|f(x') - f(x)| \leq \epsilon$$

This statement looks superficially like a statement about the continuity of $f$, but it seems to reduce to a claim about $f$ being bounded on a neighborhood at $x$:

$$\sup_{x' \in N_\delta(x)} f(x') < \infty$$

Is there more to this "opposite" definition of continuity? Does it have another name? Is it really just that $f$ is bounded on a neighborhood?

答:
Hölder continuity is defined as :

$$\forall x,y \in \mathbb R^d , |f(x)-f(y)| \leq C|x-y|^\alpha$$

Maybe if you set $\alpha=0$ you get what you are working with?


问:
Similar cones - volumes and lateral areas
Two similar cones have volumes 9$\pi$ and 72$\pi$. If the lateral area of the larger cone is 32$\pi$, what is the lateral area of the smaller cone?

I did the following...

$\frac {(9\pi)^3} {(32\pi)^2} = \frac {x}{(32\pi)^2}$

resulting in a lateral area of $4\pi$. Is this right?

答:
If you scale lengths by a factor $f$, then areas scale by $f^2$ and the volume scales by $f^3$. So the factor between the larger and the smaller cone is

$$ f = \sqrt[3]{\frac{72\pi}{9\pi}} = \sqrt[3]8 = 2 $$

Now you can scale the area with that. Going from larger to smaller, you have to divide by $f^2$.

$$ \frac{32\pi}{f^2} = \frac{32\pi}4 = 8\pi $$

So your result appears to be incorrect. The equation from which you derive it looks like a copy and paste error: both sides are completely equal for $x=(9\pi)^3$, which has nothing to do with the answer you gave.


问:
Why is $\int \sum \limits_{n=1}^{\infty}\frac{(-1)^{n+1}}{n}\chi_{[n-1,n)}(x) \, \mathrm{d}x=\sum \limits_{n=1}^{\infty}\frac{(-1)^{n+1}}{n}$?
I have to show that for $f: [0,\infty)\to \mathbb{R}$, we have$$\int \limits_{0}^{\infty}\sum \limits_{n=1}^{\infty}\frac{(-1)^{n+1}}{n}\chi_{[n-1,n)}(x) \, \mathrm{d}x=\log(2)$$ I don't understand why $$\int \limits_{0}^{\infty}\sum \limits_{n=1}^{\infty}\frac{(-1)^{n+1}}{n}\chi_{[n-1,n)}(x) \, \mathrm{d}x=\sum \limits_{n=1}^{\infty}\frac{(-1)^{n+1}}{n}$$
Could someone please explain?

Body:
For any integer $N>0$, if $x>N$, $\chi_{[n-1,n)}=0$ for $x\le N$ and hence
$$ \int \limits_{0}^{N}\sum \limits_{n=1}^{\infty}\frac{(-1)^{n+1}}{n}\chi_{[n-1,n)}(x) \, \mathrm{d}x=\int \limits_{0}^{N}\sum \limits_{n=1}^{N}\frac{(-1)^{n+1}}{n}\chi_{[n-1,n)}(x) \, \mathrm{d}x=\sum \limits_{n=1}^{N}\frac{(-1)^{n+1}}{n}.$$
Then letting $N\to\infty$, we have
$$\int \limits_{0}^{\infty}\sum \limits_{n=1}^{\infty}\frac{(-1)^{n+1}}{n}\chi_{[n-1,n)}(x) \, \mathrm{d}x=\sum \limits_{n=1}^{\infty}\frac{(-1)^{n+1}}{n}.$$


问:
Why does the pair $(c,d)\in\mathbb Z/N\mathbb Z$ always has a coprime lift?
If we know $c,d\in\mathbb Z/N\mathbb Z$ and $\gcd(c,d,N)=1$. How do we prove there always exist $k_1, k_2\in\mathbb Z$ such that $\gcd(c+k_1N, d+k_2N)=1$ ?

It looks simple but I've been stuck for long. Thanks.

答:
Let $$s = \prod_{p\ |\ c,p\ \nmid\ N} p$$

Take $u$ such that $$uN=1-d\bmod s$$

$d+uN=d\bmod \gcd(c,N)$ is a unit and $d+uN=1\bmod s$ is a unit thus
$$\gcd(c,d+uN)=1$$


问:
Finding the expected value of four dice when only the Lowest three count
Four standard six-sided dice are rolled and any side is equally likely to land face up. The highest value is dropped and the remaining three values are summed. What is the expected value of the result?
If all four dice have the same number, then the value from any one die can be dropped.

I have never seen the question asked in this way before, but I have seen a number of similar problems where the lowest of the four dice is dropped, so I would assume this problem can be solved in the same way. Here is my solution:

Start by finding the expected value of the highest die roll, which I will call $E(H)$:

```<code>There is only one way one can be the highest die roll, which is when all dice show 1.
There are 2^4 – 1^4 = 15 ways where 2 can be the highest
There are 3^4 – 2^4 = 65 ways where 3 can be the highest
There are 4^4 – 3^4 = 175 ways where 4 can be the highest
There are 5^4 – 4^4 = 369 ways where 5 can be the highest
There are 6^4 – 5^4 = 671 ways where 6 can be the highest.

E(H) = (1/1296 * 1) + (15/1296 * 2) + (65/1296 * 3) +  (175/1296 * 4) +  (369/1296 * 5) + (671/1296 * 6)
E(H) = 5.24
</code>```

The expected value of the sum would be $$E(X_1 + X_2 + X_3 - H) = E(X_1) + E(X_2) + E(X_3) - E(H).$$
Since we know the expected value of a die is $3.5$, this comes out to $$3.5 + 3.5 + 3.5 +3.5 - 5.24 = 8.74.$$

I would appreciate it if someone could check my reasoning (and my math) to help me solve this question. Thank you.

答:
I think you meant $\ldots = 8.76$, which is indeed correct.

You're missing an $X_4$ in your sum:

$$E(X_1 + X_2 + X_3 + X_4 - H) = E(X_1) + E(X_2) + E(X_3) + E(X_4) - E(H),$$

although you did account for it in your calculation.


问:
Changing the probabilities of state transitions
I have a state which is a list of probabilities for next states as follows:

$\left[0, \frac {1}{2}, 0, 0, \frac{1}{2}\right]$

i.e. There is a 50% chance that this state will go to state 2 and a 50% chance it will go to state 5

I calculated that if state 2 is taken, it will loop back to this state with a $\frac 2 7$ probability. I know it will only loop back once. How do I alter the probabilities so that the $\frac 2 7$ chance of looping back is taken into account. I know that the correct answer in this case is:

$\left[0, \frac {5}{14}, 0, 0, \frac{9}{14}\right]$

And I also know that

$\left[0, \frac {2}{3}, 0, 0, \frac{1}{3}\right]$

With the second element having a $\frac {1}{2}$ chance of looping back, should end up as:

$\left[0, \frac {1}{2}, 0, 0, \frac{1}{2}\right]$

答:
In this case for state 2 you have ($p_l$ is the chance to loop after going to state 2)

$ p_2 \cdot (1-p_l) + p_2 \cdot p_l \cdot p_2 $

because you either change to state 2 and stay there ($p_2 \cdot (1-p_l)$), or loop back and choose state 2 again ($p_2 \cdot p_l \cdot p_2$). For the other states you have

$ p_j + p_2 \cdot p_l \cdot p_j $

Because you either change to state j (chance $p_j$), you either loop back from state 2 (chance $p_2 \cdot p_l$) and then change to state j.

This gives different answers from the ones in your examples (in the first one I get 6/14 and 8/14 instead of 5/14 and 9/14).

After the comment about calculating the "loop" probability using geometric series sums, I would like to point out that the "usual" approach  in these situations is the following:

(edited for clarity by adding a dice game equivalent example)

Say five people $A_1, A_2, \dots , A_5$ play a game. On phase 0, they roll a 3-sided die and if it lands on 1 or 2 they go to phase $A_2$. If it lands on 3 they go to phase $A_5$.

So $ [p_1,\, p_2,\, p_3,\, p_4,\, p_5] = [0,\frac{2}{3},0,0, \frac{1}{3} ]$ describes the probability to go to phase $A_1, A_2, \dots , A_5$ while we are on phase 0.

On phase $A_2$ they toss a coin. Heads they return to phase 0. Coins and player $A_2$ wins! So on phase $ A_2 $ player $ A_2 $ wins with a probability of $ p_l = \frac{1}{2} $. With probability $ 1 - p_l $ we are back at the beginning (phase 0).

If we reach any other phase $A_i$ - player $A_i$ wins.

To find the probability that $A_2$ wins:

$ P(A_2) = p_2 \cdot (1-p_l) + p_2 \cdot p_l \cdot P(A_2) $,

For the other players:

$ P(A_j) = p_j + p_2 \cdot p_l \cdot P(A_j) $.

Simply solve these equations to get the desired probabilities.


问:
Proof of Gallai Theorem for factor critical graphs
I am trying to find the proof of the following theorem:

Definition 1.2. A vertex $v$ is essential if every maximum matching of $G$ covers $v$ (or $\nu(G-v) = \nu(G)-1$). It is avoidable if some maximum matching of $G$ exposes $v$ (or $\nu(G-v) = \nu(G)$). A graph $G$ is factor-critical if $G-v$ has a perfect matching for any $v \in V(G)$.

Theorem 1.3. Let $G$ be a connected graph. Then $G$ is factor-critical if and only if every vertex is avoidable.

I don't think this is something I can prove on my own. However, I can't seem to find the proof of it online either.

It makes sense intuitively. If for every vertex $v$, in graph $G$, there exists a perfect matching that doesn't include $v$, then removing $v$ and its incident edges would result in a perfect matching because one existed that didn't include $v$ in the first place.

Could someone provide a more formal proof for this theorem, or provide a link to where I might find the proof?

Thank You

答:
We prove the following:

THM: Let $G$ be a connected graph where for each vertex $v \in G$
there is a maximum matching that does not contain $v$. Then $G \setminus \{v\}$
has a perfect matching for each
$v \in V(G)$.

First of all, the graph must be connected for this to be true. Indeed,
2 vertex-disjoint copies of a graph where all vertices are avoidable
also has this property, and the maximum matching leaves at least
2 vertices uncovered. Secondly, this seems challenging to prove. I
apologize for the length, I don't know how to make this any shorter.

We make the following claims:

Claim 1: Let $M^*$ be any maximum matching. Then for any vertex $v$ matched
by $M^*$ at most 1 of $v$'s neighbors are not matched by $M^*$.

Proof of Claim 1: Let us suppose that there is a vertex $v$ and a
matching $M^*$ not covering at least 2 of $v$'s neighbours which we
write as $u_1$, $u_2$. Then let $M^*_v$ be a maximum-cardinality
matching not covering $v$. Then $|M^*_v|=|M^*|$ Then as
$M^*_v$ and $M^*$ are both maximum matchings and $v$ is covered by $M^*$ but not by $M^*_v$, the component $P$ of $M^*_v \cup M^*$ that contains $v$ is a path with (i)
one of its endpoints $v$, and (ii) $|P \cap M^*| = |P \cap M^*_v|$.
Furthermore, each of $u_1,u_2$ is also the endpoint of a path
[or is an isolated vertex] of a component of $M^*_v \cup M^*$. So at
least one of
$u_1,u_2$, say $u_1$, is not in $P$ with $v$. So let $M'_v$ be the set of edges where

(a) $M'_v \cap P$ is the $|M^*_v \cap P|$ edges of $M^*_v \cap P$, plus

(b) $M'_v \setminus P^*$ is the set of $|M^*|-|M^* \cap P|$ $=$ $|M^*|-|M^*_v \cap P|$ edges of $M^*$ that are not in $P$.

Then $|M'_v|=|M^*|$ and leaves exposed both $v$ and $u_1$.
Thus $M'_v + \{u_1v\}$ is a matching with cardinality strictly larger
than $M^*$, contradicting that $M^*$ is a maximum matching. So Claim 1
follows. $\surd$

Now let $v$ be a vertex covered by $M^*$.
We now define an $(v,M^*)$-alternating path to be a path
$P$ where

(i) one endpoint of $P$ is $v$,

(ii) the edge in $P$ incident to $v$ is not
in $M^*$, and

(iii) every alternating edge in $P$ is in $M^*$ [so the edge in
$P$ incident to $v$ is not in $M^*$, but the next edge is in
$M^*$, but the edge after that is not in $M^*$, and so on and so forth].

Claim 2: Let $v$ be a vertex matched by $M^*$. Then for some vertex
$y(v)$ not covered by $M^*$ there is a $(v,M^*)$-alternating path
from $v$ to $M^*$.

Proof of Claim 2: Let $M^*_v$ be a maximum matching that leaves $v$
uncovered. Then the component $P$ of $M^*_v \cap M^*$ is a path with one
endpoint $v$. As $M^*_v$ has the same cardinality as $M^*$ and does not
contain the edge in $M^*$ covering $v$ it follows that alternating edges
of $P$ are in $M^*$ that the other endpoint of $P$ is
not covered by $M^*$. $\surd$

Claim 3: Let $M^*$ be a maximum matching in $G$.
For each $v$ matched by $M^*$ there is exactly one vertex
$y(v)$ not matched by $M^*$ there is a $(v,M^*)$-alternating path
with one endpoint in $v$ and the other endpoint in $y(v)$.

Proof of Claim 3: Let us suppose otherwise. Let
$P_1$ be a $(v,M^*)$-alternating path
where the other endpoint is in $y_1$; where $y_1$ not covered by $M^*$;
a path $P_1$ exists by Claim 2.
And let $P_2$ be a $(v,M^*)$-alternating path
where the other endpoint is in $y_2$; where $y_2$ not covered by $M^*$.
Let us assume that $P_1$ and $P_2$ do not intersect except at $v$.
[Otherwise there is a vertex $v'$ on $P_1 \cap P_2$
that is covered by $M^*$ so that there is a $(v',M^*)$-alternating path
$P'_1$ where the other endpoint is in $y_1$, and a
$(v',M^*)$-alternating path $P'_2$ where the other endpoint is in $y_2$, and $P'_1$ and $P'_2$ intersect only at $v'$.
Namely $P'_1$ is $P_1$ from $v'$ to $y_1$ and $P'_2$ is $P_2$
from $v'$ to $y_2$.]

Then let $M'$ be the matching formed from $M^*$ where (i) the edges in
$M^* \cap P_1$ are replaced by those in $P_1 \setminus M^*$ except
for the one edge in $P_1$ incident to $v$, and (ii) the edges in
$M^* \cap P_2$ are replaced by those
in $P_2 \setminus M^*$ except for the one edge in $P_2$ incident
to $v$. Then because $P_1$ and $P_2$ intersect only at $v$, it follows that $M'$ is indeed a matching and has as many edges as
$M^*$ so $|M'| \ge |M^*|$; and furthermore $v$ is still covered by $M'$.
Also, 2 of $v$'s neighbours $u_1$ and $u_2$ [namely $u_1$ is $v$'s
neighbor on $P_1$ and $u_2$ is $v$'s neighbour on $P_2$] are not
covered by $M'$. So finish by Claim 1. $\surd$

Claim 4: Let $uv$ be an edge in $M^*$. Then $y(v)=y(u)$.

Proof of Claim 4: Let $P_1$ be a $(u,M^*)$-alternating path that ends
in $y_1$; where $y_1$ not covered by $M^*$, and let $P_2$ be a
$(v,M^*)$-alternating path that ends in $y_2$; where $y_2$ not covered
by $M^*$; and $y_1 \not = y_2$. If $P_1$ and $P_2$ intersect then there
exists a $(v,M^*)$ path from $v$ to both $y_1$ and $y_2$;
finish as in the proof of Claim 3.

If $P_1$ and $P_2$ do not intersect then $P=P_1 \cup \{uy\} \cup P_2$
is a path where alternating edges in $P$ are in $M^*$ and
where the endpoints of $P$ are not covered by $M^*$. So the matching
$M'$ formed from $M^*$ by replacing the edges in $P \cap M^*$ with those
in $P \setminus M^*$ is indeed a matching and has one more edge than $M^*$
does. $\surd$

We now finish the proof by showing that $M^*$ leaves only 1 vertex uncovered. Let us suppose that there are at least 2 unmatched
vertices $y_1$ and $y_2$ be $M^*$. Then as $G$ is connected,
let $P$ be a path where one endpoing is an unmatched vertex $y_1$
and the other endpoint is another unmatched vertex $y_2$ and where
every other vertex is matched by $M^*$. Now write $P=y_1v_1\ldots v_ry_2$
where $v_1,\ldots, v_r$ are covered by $M^*$. Then defining $y(v)$ as in
Claim 3, we note the following: $y(v_r)=y_2$
and $y(v_1)=y_1$ and each vertex $v_i$ has exactly one such $y(v_i)$
by Claim 2 so there exists an $i$ such that $y_1=y(v_i) \not = y(v_{i+1})=y_2$
So let $P_{i}$ be a $(v_i,M^*)$-alternating
path from $v_i$ to $y_1$ and let $P_{i+1}$ be a $(v_{i+1},M^*)$-alternating
path from $v_{i+1}$ to $y_2$.

If $v_iv_{i+1}$ is in $M^*$, then finish as in Claim 4.

If $v_iv_{i+1}$ is not in $M^*$ then writing as
$u_{i+1}v_{i+1}$ be the edge in
$M^*$ incident to $v_{i+1}$, by Claim 4 it follows
that $y(u_{i+1})=y(v_{i+1})=y_2$. But then
it follows that there is a $(v_i,M^*)$-alternating path $P$
from $v_i$ to $y_2$ as well as $y_1$; namely the first two
edgres in $P$ are $v_iv_{i+1}$ and $v_{i+1}u_{i+1}$. So finish
via Claim 3.

Thus $M^*$ indeed leaves only 1 vertex unmatched, and as we are given that for each vertex $v$ there is a matching $M^*_v$ that does not cover $v$ and that satisfies $|M^*|=|M^*_v|$, the theorem follows. $\surd$


问:
Factoring $−x^{3}−3x^{2}+4$
I have this polynomial:

$$−x^{3}−3x^{2}+4$$

and I know it factors to this:

$$−(x−1)(x+2)(x+2)$$

But how do I do this formally and in a step by step way?

Title: Factoring $−x^{3}−3x^{2}+4$

Body:
Note
$$−x^{3}−3x^{2}+4= −x^{3}+x^2 - 4x^{2}+4
= −x^{2}(x-1)- 4(x^{2}-1)\\
= -(x-1)(x^2+4x+4)=-(x-1)(x+2)^2
$$


问:
3 white balls, 2 green and 1 red ball
There are $3$ white, $2$ green, and $1$ red balls in a bag.

$2$ balls are drawn from the bag.

What is the probability of drawing $1$ white and $1$ green ball?

I've come up with the following:
Probability to draw a white ball, then green ball is
$\frac{3}{6} \times \frac{2}{5} = \frac{1}{5}$
Also, you can draw green then white, so
$\frac{2}{6} \times \frac{3}{5} = \frac{1}{5}$

The probability is $\frac{1}{5}+\frac{1}{5}=\frac{2}{5}$

I don't feel this is right, but the solution in my textbook says $\frac{2}{5}$.

答:
Let P = The probability of drawing 1 white ball and 1 green ball (in any order), it is, naturally, equal to the sum of all possible ways you can draw these 2 balls, which are:

Way 1: Drawing a green ball, then a white ball, for which, the probability is: $$P_1=\frac{3}{6}\times\frac{2}{5}=\frac{1}{5}$$

Way 2: Drawing a white ball, then a green ball, for which, the probability is: $$P_2=\frac{2}{6}\times\frac{3}{5}=\frac{1}{5}$$

And then: $$P=P_1+P_2=\frac{1}{5}+\frac{1}{5}=\frac{2}{5}$$

You could also compute P using the combinatoric function C, like so:

The number of ways to pick 1 green ball from 3 total green balls = $3C1$

The number of ways to pick 1 white ball from 2 total white balls = $2C1$

The number of ways to pick 2 balls from 6 total balls = $6C2$

Therefore:
$$P=\frac{3C1\times2C1}{6C2}=\frac{3\times2}{15}=\frac{2}{5}$$

Title: 3 white balls, 2 green and 1 red ball


问:
Given that $g$ is only continuous at $0$, not on $[0,1]$, show $\lim_{n \to \infty} \int_0^1 g(x^n) dx = g(0)$.
Important Notice: Observe that although this question is very similar to many other questions such as the one in proving $\lim\limits_{n\to\infty} \int_{0}^{1} f(x^n)dx = f(0)$ when f is continous on [0,1], this problem only assumes continuity at $0$, but not on $[0,1]$.

Here is the question:
Assume $g$ is (Riemann) integrable on $[0, 1]$ and continuous at $0$. Show
$\lim_{n \to \infty} \int_0^1 g(x^n) dx = g(0)$.

My attempt at this question: I split up the integral into two parts from $[0,1-\alpha]$ and $[1-\alpha,1]$. For $[0,1-\alpha]$, there is uniform continuity, so the limit $n \to \infty$ can be shifted inside the integral to get $g(0)$.

But how do I keep the integral over $[1-\alpha,1]$ to be small?

Note: This is Exercise $7.4.10$ in Abbott, Understanding Analysis, 2nd edition.

答:
If $g$ is assumed to be Riemann-integrable on $[0,1]$, then it is bounded on $[0,1]$, and we have:

$$\begin{array}{rcl}\left|\int_{1-\alpha}^1 g(x^n)dx\right|&\le&\int_{1-\alpha}^1|g(x^n)|dx\\&\le&\left(\sup_{0\le x\le 1}|g(x)|\right)\int_{1-\alpha}^1 dx\\&=&\alpha\sup_{0\le x\le 1}|g(x)|\\&\to&0\end{array}$$

as $\alpha\to 0$.


问:
How to see that the strategy $\{(1,0), (0,1)\}$ is a Nash equilibrium in this simple game?
Consider the payoff matrix,
$$\begin{bmatrix} (0,0) & (-1,1)\\
(1,-1) & (-10,-10)
\end{bmatrix}$$

Assume players want to maximize payoff.

If the row player choses strategy $(1,0)$ i.e., strategy 1. Then the column player must choose strategy $(0,1)$ to maximize his payoff  ($0$ vs $1$)

But then the row player wants to change to strategy $2$ to change his payoff from $-1$ to $1$. At this point, the column player can only change to strategy $1$ to avoid a payoff of $-10$.

This means the pure Nash of this game is $\{(0,1), (1,0)\}$ and the solution $\{(1,0), (0,1)\}$ will be forever unstable (one player will always want to change).

Why is that $\{(1,0), (0,1)\}$ is a solution when computed? https://intranet.csc.liv.ac.uk/cgi-bin/cgiwrap/rahul/input.py

Is the algorithm for computing pure NE:

First consider case when row player move first?

Next consider case when column player move first?

I think this 2 is what I am missing

答:
If the row player chooses strategy 1, the column player can choose between payoff 0 or 1, and will therefore go with 1 (i.e. strategy 2). If the column player chooses strategy 2, the row player can choose between payoff -1 and -10, and will therefore choose -1 (i.e. strategy 1)

So row 1, column 2 is a Nash equilibrium. As is row 2, column 1 for analogous reasons.


答:
So here is a quick answer because I kinda wanna stay out of any possible discussion at the moment.

Answer

WLOG: $g(0)=0$

I'll start by giving some straight point:
$$ \int_{0}^1 g(x^n)dx = \int_{0}^1 g(t) \frac{1}{n} t^{-1+1/n}dt$$
As $g$ is continuous at $0$, for any $\delta>0$, there is a $1>\epsilon>0$ such that:
$$ |g(x)| \le \delta \quad \forall  \quad |x| \le \epsilon $$
Thus
$$ \left| \int_{0}^1 g(t) \frac{1}{n} t^{-1+1/n}dt \right| \le \underbrace{ \left( \int_{0}^{\epsilon} \delta\frac{1}{n} t^{-1+1/n}dt \right)}_{ \le \delta}+\underbrace{  \left( \int_{\epsilon}^1 |g(t)|\frac{1}{n} \epsilon^{-1+1/n}dt  \right) }_{ \le \frac{1}{n}\epsilon^{-1} \int_{0}^1 |g(t)|dt}$$
Thus ,
$$ \limsup_n  \left| \int_{0}^1 g(t) \frac{1}{n} t^{-1+1/n}dt \right|  \le \delta$$
Hence the conclusion

Comment

This is okay whether the integrability is understoond in which sense, Riemann or Lesbeque.

The formula of changing variables in the very beginning is not really necessary. A similar argument can be constructed without it, but surely more details need to be handle. Fortunately, they are just technical. The idea stays the same.

Title: Given that $g$ is only continuous at $0$, not on $[0,1]$, show $\lim_{n \to \infty} \int_0^1 g(x^n) dx = g(0)$.

Body:
Important Notice: Observe that although this question is very similar to many other questions such as the one in proving $\lim\limits_{n\to\infty} \int_{0}^{1} f(x^n)dx = f(0)$ when f is continous on [0,1], this problem only assumes continuity at $0$, but not on $[0,1]$.

Here is the question:
Assume $g$ is (Riemann) integrable on $[0, 1]$ and continuous at $0$. Show
$\lim_{n \to \infty} \int_0^1 g(x^n) dx = g(0)$.

My attempt at this question: I split up the integral into two parts from $[0,1-\alpha]$ and $[1-\alpha,1]$. For $[0,1-\alpha]$, there is uniform continuity, so the limit $n \to \infty$ can be shifted inside the integral to get $g(0)$.

But how do I keep the integral over $[1-\alpha,1]$ to be small?

Note: This is Exercise $7.4.10$ in Abbott, Understanding Analysis, 2nd edition.


问:
In $\triangle PQR$ with $QR^2+PR^2=5PQ^2$, medians $PM$ and $QN$ meet at $G$. Find $\angle QGM$.
This problem becomes much simpler if we solve it using coordinate geometry but i am seeking a solution which involves pure geometry.

Consider a $\triangle PQR$ in which the relation
$$QR^2+PR^2=5PQ^2$$
holds. Let $G$ be the point of intersection of medians $PM$ and $QN$. Then find the value of $\angle QGM$.

Body:
$QR^2+PR^2=5PQ^2$

median lengths

$$PM=\frac{1}{2}\sqrt{2(PQ^2+PR^2)-QR^2}$$
$$PM=\frac{1}{2}\sqrt{2\left(\frac{x^2+y^2}{5}+y^2\right)-x^2}=\frac{1}{2} \sqrt{\frac{12 y^2}{5}-\frac{3 x^2}{5}}$$
$$QN=\frac{1}{2}\sqrt{2\left(\frac{x^2+y^2}{5}+x^2\right)-y^2}=\frac{1}{2} \sqrt{\frac{12 x^2}{5}-\frac{3 y^2}{5}}$$

$$GM=\frac{1}{3}PM=\frac{1}{6} \sqrt{\frac{12 y^2}{5}-\frac{3 x^2}{5}}$$
$$GQ=\frac{2}{3}QN=\frac{1}{3} \sqrt{\frac{12 x^2}{5}-\frac{3 y^2}{5}}$$
$$GM^2+GQ^2=\frac{x^2}{4}=QM^2\to \widehat{QGM}=90°$$


问:
What is the probability that three babies are boys, given that at least one is a boy?
I know this is a simple problem but I am arguing with a friend about its solution, so I want to show him an "official" proof!
Suppose that in any birth, the probability to have a boy is $48.5\%$.

If we have three persons expecting to deliver, what is the probability that at least one of them gives birth to a boy?

If we know that at least one will give birth to a boy (suppose we have accurate ultra-sound results), what is the probability all three will have a boy?

For the first question, we calculate the probability of one NOT having a boy, which is $1-0.485 = 0.515$ and then the required probability of all three not having a boy is $0.515^3 = 0.1365$ so the probability that at least one will have a boy is $1-0.1365 = 0.8634 = 86.34\%$.

For the second question, since the three events are independent, the probability that all three will have a boy given that at least one will have a boy is equal to the probability that the other two will have a boy. Is it $0.485^2$? I am not sure about the second one.

答:
As others have suggested, you can "just" apply Bayes' theorem.  However, this problem is a relatively simple one, and so it might help to draw out a probability tree to help organize your thoughts.  I will not claim that this is the "best" or most efficient way to approach the problem, but I find that it can help to give some insight into why, as you are relying on something a little more concrete than a statement of a theorem.  For this problem, the tree is something like the following:

I have used blue nodes to indicate boys, and purple nodes to indicate girls.

If we have three persons expecting to deliver, what is the probability that at least one of them gives birth to a boy?

There are two ways to approach this from the diagram.  Either find all of the leaves (terminal nodes) in which there is at least one boy, compute the probability of each such outcome by multiplying along the path leading to this leaf (this is Bayes' theorem in disguise), then add up those probabilities (which we can do, because the events are independent).

Alternatively, we can look for all of the outcomes in which no boys are born, determine the probability of each, and then subtract that probability from $1$ (as all the probabilities must add to $1$).  This latter option seems simpler, as there is only one node to consider.  Therefore
\begin{align}
P(\text{at least one boy})
&= 1 - P(\text{no boys}) \\
&= 1 - 0.515^3 \\
&\approx 1 - 0.1366 \\
&= 0.8634,
\end{align}
which is the answer found in the question.

If we know that at least one will give birth to a boy (suppose we have accurate ultra-sound results), what is the probability all three will have a boy?

In general, the probability of an outcome $A$, given some event $B$, is
$$ P(A \mid B) = \frac{P(A)}{P(B)}.$$
This is, basically, the Law of Total Probability.  Take $A$ to be the event "three boys are born" and $B$ to be the event "at least one boy is born".  The probability of at least one boy was computed in the first part of the question:  $P(B) \approx 0.8634$.  Following the same kind of argument as in the first part (i.e. multiply the probabilities along the path from the root node to the leaf labeled "3 boys"),
$$ P(A) = P(\text{three boys}) = (0.485)^3 \approx 0.1141. $$

Then
\begin{align} P(\text{three boys} \mid \text{at least one boy})
&= \frac{P(\text{three boys})}{P(\text{at least one boy})} \\
&\approx \frac{0.1141}{0.8634} \\
&\approx 0.1322.
\end{align}
In other words, if you know that one of the babies is a boy, then the probability that all three are boys is about 13.2%.


问:
Calculate the volume bounded by surface
Calculate the volume bounded by the surface
$$(x^2+y^2+z^2)^2 = a^2(x^2+y^2-z^2)$$

Using the spherical coordinates
$$\begin{cases} x = rcos\varphi cos\theta & \\ y = rsin\varphi cos\theta \\z = rsin\theta \end{cases}$$
and substituting those into the original equation we get
$$r^2 = a^2(cos^2\theta - sin^2\theta) = a^2cos2\theta$$
and from that
$$0 \leq r \leq a\sqrt{cos2\theta}$$

Calculating Jacobian gives us
$$ J = r^2cos\theta$$

Given all that the target volume could be calculated as

$$V = \int_{0}^{2\pi}\,d\varphi \int_{0}^{\pi}\,d\theta \int_{0}^{a\sqrt{cos2\theta}}\,r^2cos\theta dr$$

But this yields incorrect result, moreover the supposed answer should be calculated given the following integral
$$V = 8\int_{0}^{\frac{\pi}{2}}\,d\varphi \int_{0}^{\frac{\pi}{4}}\,d\theta \int_{0}^{a\sqrt{cos2\theta}}\,r^2cos\theta dr $$

But I have trouble understanding where do the integrating boundaries for $\varphi$ and $\theta$ come from.

I undesrtand that given the fact that the surface and therefore target solid are symmetrical, we can integrate over a certain part of the solid and then multiply the result by a proper constant,
but if we use the following bounds for $\varphi$ and $\theta$

$$0 \leq \varphi \leq \pi/2 \\ 0 \leq \theta \leq \pi/4$$

how come we multiply by 8 and not by 16?

Any tips on what I'm doing wrong ?

答:
The reason is that $$\int_{0}^{\frac{\pi}{2}}d\varphi\int_{0}^{\frac{\pi}{4}} d\theta\int_{0}^{a\sqrt{\cos (2\theta)}} r^2\cos\theta dr$$

Is only in the first octant. We multiply by 8 to cover all 8 octants. The screenshot of the graph will make it obvious that your $\theta$ bounds run from $0$ to $\frac{\pi}{4}$ if you run through the first octant, and the bounds you specify are all 8 octants.

https://gyazo.com/fb0003801540888e101442e4cc56a320

Title: Calculate the volume bounded by surface


问:
Am I missing an equilibrium point?
I am trying to identify all the equilibrium points of this system, and am only concerned with the first quadrant:

$$x'=x(a-x-ay),
 \\ y'=y(b-bx-y)$$

Where a and b are parameters greater than 0. When I calculate the equilibrium points by setting x'=y'=0, I get (0,0), (a,0), (0,b). The problem is, when I plot the nullclines with varying values of a and b, I can get a fourth equilibrium point, as an x-nullcline and y-nullcline intersect. Why is this equilibrium point not showing up when I initially calculate them in terms of my parameters? Thank you for the help.

答:
$$
\begin {cases}
x'=x(a-x-ay)
 \\ y'=y (b-bx-y)
\end{cases}
$$
You didn't solve this system:

$$
\begin {cases}
(a-x-ay)=0
 \\ (b-bx-y)=0
\end{cases}
$$
Where $x \ne 0,$ and $y \ne 0$

Title: Am I missing an equilibrium point?