问:
Is $\sum_{n\ge0}(-1)^n\frac{\Gamma(\tfrac{n+1}{2})}{\Gamma(\tfrac{n}2+1)}=\frac{2}{\sqrt{\pi}}$ true?

Prove/disprove $$\sum_{n\ge0}(-1)^n\frac{\Gamma(\tfrac{n+1}2)}{\Gamma(\tfrac{n}2+1)}=\frac{2}{\sqrt{\pi}}.\tag 1$$

As far as I can tell, this is true, although it seems to converge very slowly.

I came up with a proof but I don't know if it's valid.

Let $$J=\int_0^\pi \frac{xdx}{1+\sin x}.$$
On one hand, we have
$$\frac1{1+\sin x}=\sum_{n\ge0}(-1)^n\sin(x)^n,$$
so that
$$J=\sum_{n\ge0}(-1)^n p_n,\tag 2$$
where
$$
\begin{align}
p_n&=\int_0^\pi x\sin(x)^ndx\\
&=\int_\pi^0 -(\pi-x)\sin(\pi-x)^ndx\\
&=\pi\int_0^\pi\sin(x)^ndx-p_n\\
\Rightarrow p_n&=\frac\pi2\int_0^\pi\sin(x)^ndx.
\end{align}
$$
And since $\sin(x)=\sin(\pi-x)$, $$p_n=\pi\int_0^{\pi/2}\sin(x)^ndx=\frac{\pi^{3/2}}{2}\frac{\Gamma(\tfrac{n+1}2)}{\Gamma(\tfrac{n}2+1)}.\tag 3$$

On the other hand, we have $1+\sin x=2\sin^2(\tfrac{x}2-\tfrac\pi4)$, so that
$$\begin{align}
J&=\frac12\int_0^\pi\frac{xdx}{\sin^2(\tfrac{x}2-\tfrac\pi4)}\\
&=2\int_{\pi/4}^{3\pi/4}\frac{tdt}{\sin^2t}-\frac\pi2\int_{\pi/4}^{3\pi/4}\frac{dt}{\sin^2 t}\\
&=2\left(\ln\sin x-x\cot x\right)\bigg|_{\pi/4}^{3\pi/4}-\frac\pi2\left(-\cot x\right)\bigg|_{\pi/4}^{3\pi/4}\\
&=2\pi-\frac\pi2\cdot2=\pi.
\end{align}$$
Then from $(2)$ and $(3)$, we have
$$\frac{\pi^{3/2}}{2}\sum_{n\ge0}(-1)^n\frac{\Gamma(\tfrac{n+1}2)}{\Gamma(\tfrac{n}2+1)}=\pi,$$
which is equivalent to $(1)$. $\square$

Can you come up with any other proofs to $(1)$? Thanks!

Edit (11/12/2020):

Here is a proof that the interchange of the sum and integral in $(2)$ is valid.

The partial sums $$S_M(x)=\sum_{n=0}^M(-1)^n\sin(x)^n$$
form a uniformly convergent sequence of functions for $x$ in $[0,\pi/2)$ or $(\pi/2,\pi]$, and they converge to the limit $$\lim_{M\to\infty}S_M(x)=\frac1{1+\sin x},\qquad x\in[0,\pi]\setminus\{\pi/2\}.$$
Choose $\varepsilon>0$ and notice that
$$J=\int_{0}^{\pi}\frac{xdx}{1+\sin x}=\int_{\pi/2-\varepsilon}^{\pi/2+\varepsilon}\frac{xdx}{1+\sin x}+\int_0^{\pi/2-\varepsilon}\frac{xdx}{1+\sin x}+\int_{\pi/2+\varepsilon}^\pi\frac{xdx}{1+\sin x}.$$
The sums $S_M(x)$ converge uniformly as $M\to\infty$ when $x\in[0,\pi/2-\varepsilon]\cup[\pi/2+\varepsilon,\pi]$, so we can interchange the sum and integral to get
$$J=\int_{\pi/2-\varepsilon}^{\pi/2+\varepsilon}\frac{xdx}{1+\sin x}+\sum_{n\ge0}(-1)^n(a_n(\pi/2-\varepsilon)+b_n(\pi/2+\varepsilon)),$$
where
$$\begin{align}
a_n(t)&=\int_0^t x\sin(x)^ndx\\
b_n(t)&=\int_t^\pi x\sin(x)^ndx.
\end{align}$$
We have $a_n(t)+b_n(t)=p_n$ for all $t\in[0,\pi]$. As $\varepsilon$ approaches $0$, $\int_{\pi/2-\varepsilon}^{\pi/2+\varepsilon}\frac{xdx}{1+\sin x}$ approaches $0$.

And since $a_n(t), b_n(t)$ are continuous, $a_n(\pi/2-\varepsilon)+b_n(\pi/2+\varepsilon)$ approaches $a_n(\pi/2)+b_n(\pi/2)=p_n$, so that
$$J=\sum_{n\ge0}(-1)^np_n$$
as desired. $\square$

Body:
Based on this answer:
$$\frac{2}{\sqrt{\pi }}\int_{0}^{\infty }{\frac{1}{{{\left( 1+{{x}^{2}} \right)}^{\frac{n}{2}+1}}}dx}=\frac{\Gamma \left( \frac{n+1}{2} \right)}{\Gamma \left( \frac{n}{2}+1 \right)}$$
so the sum in question
$$\begin{align}
  & =\sum\nolimits_{n=0}^{\infty }{\left\{ {{\left( -1 \right)}^{n}}\frac{2}{\sqrt{\pi }}\int_{0}^{\infty }{\frac{1}{{{\left( 1+{{x}^{2}} \right)}^{\frac{n}{2}+1}}}dx} \right\}} \\
 & =\frac{2}{\sqrt{\pi }}\int_{0}^{\infty }{\sum\nolimits_{n=0}^{\infty }{\frac{{{\left( -1 \right)}^{n}}}{{{\left( 1+{{x}^{2}} \right)}^{\frac{n}{2}+1}}}}dx} \\
 & =\frac{2}{\sqrt{\pi }}\int_{0}^{\infty }{\frac{dx}{\sqrt{1+{{x}^{2}}}\left( 1+\sqrt{1+{{x}^{2}}} \right)}} \\
 & =\frac{2}{\sqrt{\pi }}\left. \frac{\sqrt{1+{{x}^{2}}}-1}{x} \right|_{0}^{\infty } \\
 & =\frac{2}{\sqrt{\pi }} \\
\end{align}$$


问:
Drawing numbers without replacement
So if I were to draw 11 unique numbers from 0-99, (0 included), what is the probability that the last digit of all numbers drawn contain 0-9. (i.e a good outcome would be 10,11,12,13,14,15,16,17,18,19,20 as the last digits of 0-9 are selected).

Would I need to use the inclusion exclusion principle?
Is the answer to this question $1- \dfrac{\binom {10}{1} \binom {90}{11}- \binom {10}{2} \binom {80}{11}+ \binom {10}{3} \binom {70}{11}- \binom {10}{4} \binom {60}{11}+ \binom {10}{5} \binom {50}{11} - \binom {10}{6} \binom {40}{11} + \binom {10}{7} \binom {30}{11} - \binom {10}{8} \binom {20}{11} + \binom{10}{9} \binom{10}{11}}{\binom{100}{11}}$?

答:
Your answer is correct, but there is really no need to use the devious route of inclusion-exclusion.

There are $10$ ways to choose the digit that will occur twice,

and the rest is just application of the "normal" hypergeometric formula, thus

$10\times\left[\frac{\dbinom{10}2\dbinom{10}{1}^9}{\dbinom{100}{11}}\right]$

ContentLicense: CC BY-SA 4.0

Title: Drawing numbers without replacement


问:
Weak formulation, divergence free
I was trying to find the weak formulation of the following problem

\begin{align}
-\vartheta\Delta u + b \cdot \nabla u &= f \quad \text{in $\Omega$} \\
u &= 0 \quad \text{in $\partial \Omega$}
\end{align}

where $b:\bar{\Omega}\rightarrow \mathbb{R}^2, b\in C^1(\Omega)$ is a divergence free velocity field. I have computed the weak formulation by multiplying both sides with a test function and integrating over the domain. I have

\begin{equation}
a(u,v) = \int_{\Omega} \vartheta \nabla u \cdot \nabla v+ (b \cdot \nabla u)  v
\end{equation}

Is this done? I don't know how to use the fact that $b$ is divergence free.
Next I wanted to show that the bilinear form is bounded. Normally I would try and do this using the $L^2$ Norm and a Sobolev Inequality $\|u\| \leq C\|Du\|$. But I feel I am doing something wrong because it is not working.

答:
Your blf as in the comment, i.e.,

$$a(u,v) = \int_\Omega \vartheta \nabla u \cdot \nabla v -\int_\Omega (bu) \cdot \nabla v $$

for all $v \in H_0^1(\Omega)$ is correct. It is bounded in $H_0^1 \times H_0^1$ since

$$\begin{aligned}|a(u,v)| &\leq \vartheta \|\nabla u\|_{L^2}  \|\nabla v\|_{L^2} + \|b\|_{L^\infty} \|u\|_{L^2(\Omega)} \|\nabla v\|_{L^2(\Omega)} \\
&\leq (\vartheta+\|b\|_{L^\infty}) \|u\|_{H^1} \|v\|_{H^1}.
\end{aligned}$$

Title: Weak formulation, divergence free


问:
Probability Calculations on Highway
I read this Google Interview Question.

Q:If the probability of observing a car in 30 minutes on a highway is 0.95, what is the probability of observing a car in 10 minutes (assuming constant default probability)?

A:The trick here is that .95 is the probability for 1 or more cars, not the probability of seeing just one car.
The prob. of NO cars in 30 minutes is 0.05, so the prob of no cars in 10 minutes is the cube root of that, so the prob of seeing a car in 10 minutes is one minus that, or ~63%

My question is, if The probability of NO cars in 30 minutes is 0.05, why the probability of no cars in 10 minutes is the cube root of that ??

Which algorithm used in this question?

答:
Assuming a Poisson Distribution, we know that the probability mass function is:

$$\text{P}(X=k) = \frac{\lambda^k e^{-\lambda}}{k!} $$

The probability that no cars pass in thirty minutes is $1 - 0.95 = 0.05$:
$$\text{P}(X=0) = 0.05 = \frac{\lambda^0 e^{-\lambda}}{0!} = e^{-\lambda} $$

We can solve for lamda by taking the log of both sides:
$$-\text{ln}(0.05) = \lambda $$

$$\lambda = 2.995 \approx 3 $$

(To answer the OP's question, this is where the cube root comes from, because the $ln$ is the inverse of $e$ and gives us the exponential value.)

There are three cars every thirty minutes, which is $0.1$ every minute. So, we multiply that by 10 minutes and get $\lambda=1$ for units of 10 minutes.

Then we calculate the probability for exactly 0 cars in 10 minutes:
$$\text{P}(X=0) = \frac{\lambda^0 e^{-1}}{0!} = 0.368 $$

Then, to find the probability of at least one car, we just subtract that from one:

$$\text{P}(X>0) = 1 - 0.368 = 0.632$$

The probabilities (p) for the given number of cars (k) in the 10 minute interval looks like this:

ContentLicense: CC BY-SA 3.0

Title: Probability Calculations on Highway


问:
General Solution of $\sin\theta=3\cos\theta$
I'm a high school level maths student currently working through some exercises for the general solution of trigonometric equations and have come across this one that I am stuck on. Any hints would be much appreciated!

Question:

Determine the general solution of the trigonometric equation:
  $\sin\theta=3\cos\theta$

So the first thing I think of doing when seeing this is making sure either side has the same ratio. I can do this using co-ratios, so:

$\sin\theta=3\sin(90^\circ-\theta)$

$\sin\theta=3\cos\theta$

From here I get stuck; there are no values to compute a reference angle with.

The textbook gives the solution as:

$\theta=71.6^\circ+k\cdot 180^\circ, k\in\mathbb{Z}$

Can anyone give me a hint? Thanks in advance!

-Shaun

Body:
$$\sin \theta=3\cos \theta$$
$$\frac{\sin \theta}{\cos \theta}=3$$
$$\tan \theta=3$$
$$\theta=\tan^{-1} (3)+180^\circ k$$
$$\theta\approx71.56505118^{\,\circ}+180^\circ k $$


问:
Expected values in mixed strategies
If we have two actions $a_1$ and $a_2$ that in an equilibrium mixed strategy should be mixed $70\%/30\%$ with both having an expected value of say $10$. What happens to the expected values of the actions if we $100\%$ of the time do action $a_1$?

My thinking was that the expected value of the action we always do $a_1$ will go down?

Edit: Practical example. In a hand of poker, we calculate the optimal strategy, which is a mixed strategy between betting our hand for size $A$ $70\%$ of the time and size $B$ $30\%$ of the time. The expected value of these two bets is equal since it is a mixed strategy. My question then is by deviating from this optimal strategy, what would happen to the expected values. My intuition was that if we start betting our hand for size $A$ $100\%$ of the time instead of $70\%$, then the expected value of this action would go down? But maybe it stays the same, that's what I'm asking.

Thanks a lot

答:
If you mix actions, it means that you are indifferent between them so, given the other player does not change his strategy, you might as well play just one of them $100\%$ of the time and get the same payoff. However, if the other player responds, then there might be multiple scenarios and it also depends whether you consider a new equilibrium or just the payoff of the new strategy profile.

For example, in the "matching pennies" game, if you play pure then the other player best responds and your payoff is lower. But in such a game, there is no pure NE, and you can best respond to his best response and so on.

On the other hand, in a game such as "Battle of the Sexes", if you play purely and the other player best responds, you both arrive at one of the pure equilibria of the game, where the payoff is strictly higher than the mixed one.

Title: Expected values in mixed strategies


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

Help me! I'm clueless.

答:
Hint Since $f$ is continuous on the closed interval $[a,b]$, it is uniformly continuous on $[a,b]$. Then, there exists some $\delta>0$ so that

$$|x-y|<\delta \Rightarrow  |f(x)-f(y)|<\epsilon \,.$$

Pick some integer $n>0$ so that $\frac{b-a}{n}<\delta$. Split the interval $[a,b]$ in $n$ equal subintervals, $[x_{i-1},x_i]$.

Let $g$ be the piecewise line which connects the points $(x_i,g(x_i)$. Prove that it works.


问:
Bounded sequence in a Hilbert space has weakly convergent subsequence
Suppose $H$ is a Hilbert space with an orthonormal basis $(e_n)$. Let $(x_n)$ be a bounded sequence in $H$.

Prove that $(x_n)$ contains a subsequence $(x_{n_k})$ and $H$ contains an element $x$ so that $\langle x_{n_k},e_n\rangle\to\langle x,e_n\rangle$ for each $n$.

Prove that $\langle x_{n_k},x'\rangle\to\langle x,x'\rangle$ for any $x'\in H$.

I am not sure how to proceed with this question. I am familiar wiht the Riesz representation theorem, which states that for every $f\in H^*$ there exists $y\in H$ so that $f(x)=\langle x,y\rangle$. But I don't know how to apply that here.

I tried defining a linear functional $f(h)=\langle h,e_n\rangle$ and seeing what I can do from there, but I am truly stuck. I have seen some proofs using the Banach-Alaoglu theorem, but I am wondering if there is a proof without using this result. Any help in this question will be highly appreciated - thanks in advance.

答:
I don't even think you need to apply Riesz to get this result.

So, note that $j\mapsto \langle x_j, e_n\rangle$ is a bounded, complex sequence for every $n$. Hence, there is some subsequence $x_{j,1}$ of the $x_j$ such that $\langle x_{j,1},e_1\rangle$ is convergent. Inductively, let $x_{j,l+1}$ be a subsequence of the $x_{j,l}$ such that $\langle x_{j,l+1},e_{l+1}\rangle$ is convergent. Defining $x_{n_k}=x_{k,k},$ we get a subsequence such that $\langle x_{n_k},e_n\rangle$ is convergent for every $n$. Denote the limit $\alpha_n$.

We wish to argue that $\sum_{n=1}^{\infty} |\alpha_n|^2<\infty$. Indeed, note that for every $N>0$
$$
\sum_{n=1}^N |\alpha_n|^2=\lim_{k\to\infty} \sum_{n=1}^N |\langle x_{n_k},e_n\rangle|^2\leq \limsup_{k\to\infty} \|x_{n_k}\|^2<\infty,
$$
and the right-hand side is independent of $N$.

Now, define $x=\sum_{n=1}^{\infty} \alpha_n e_n$. This an absolutely convergent $\ell^2$-series and hence, $x$ defines an element of $H$.

Now, given $x'=\sum_{n=1}^{\infty} \beta_n e_n$ and $\varepsilon>0$, fix $N$ so large that $\sum_{n=N+1}^{\infty} |\beta_n|^2,\sum_{n=N+1}^{\infty} |\alpha_n|^2<\varepsilon^2$ and we get, defining $\tilde{x}=\sum_{n=1}^N \alpha_n e_n$ and $\tilde{x}'=\sum_{n=1}^N \beta_n e_n$,

\begin{align}
|\langle x_{n_k},x'\rangle-\langle x,x'\rangle|&\leq |\langle x_{n_k},x'-\tilde{x}'\rangle|+|\langle x_{n_k}-\tilde{x},\tilde{x}'\rangle|+|\langle \tilde{x}-x,x'\rangle|\\
&\leq \sup_{k}\|x_{n_k}\| \varepsilon+|\langle x_{n_k}-\tilde{x},\tilde{x}'\rangle|+\varepsilon\|x'\|
\end{align}
Now, $|\langle x_{n_k}-\tilde{x},\tilde{x}'\rangle|$ goes to $0$ by our construction of the $\alpha_n$, and so, since $\varepsilon>0$ was arbitrary and the $x_{n_k}$ are bounded, we're done.


问:
How much additional area does a ridge and furrow system create?
Various parts of Europe used a Ridge and furrow system for their fields. One side effect is it provides additional land area.

Approximating the ridges and furrows using parts of a circle with radius, r, and, given a depth / height of h from the mean height, how much additional area does it create in comparison to flat land with surface area, A?

E.g. flat land, where r is infinite and h is zero, would give a surface area of A.

At the other extreme, land with half circles for the ridges and furrows, where r = h, would give a surface area of A x (circumference of semi circle / distance across semi circle) = A x (πr / 2r) = Aπ/2.

What is the general solution?

答:
We denote the area before ploughing and the area after ploughing as $A_{\mathrm{flat}}$ and $A_{\mathrm{tilled}}$ respectively. We shall write,
$$A_{\mathrm{flat}}= b\times\big[\mathrm{segment}\space AC\big]\quad \mathrm{and}\quad A_{\mathrm{tilled}}= b\times\big[\mathrm{arc}\space ABC\big],$$
where $b$ is the the length of the ridgelet.

Let the height and the radius of the ridgelet be $h$ and $r$ respectively. It follows from the diagram, that
$$BD=h, \quad OA=OB=r, \quad \mathrm{and}\quad OD=r-h.$$
$$\therefore\quad AD=\sqrt{OA^2-OD^2}=\sqrt{r^2-\left(r-h\right)^2}=\sqrt{h\left(2r-h\right)}\quad\rightarrow\quad AC=2AD=2\sqrt{h\left(2r-h\right)}$$

Using this result, we can express $A_{\mathrm{flat}}$ as
$$A_{\mathrm{flat}}= 2b\sqrt{h\left(2r-h\right)}.$$
$$\therefore\quad b=\frac{A_{\mathrm{flat}}}{2\sqrt{h\left(2r-h\right)}},\quad\mathrm{where}\space h\gt 0. \tag{1}$$

If the angle $\measuredangle AOB=\phi$ (Please note that $\phi$ is measured in radians), then we have,
$$\cos\left(\phi\right)=\frac{r-h}{r}.$$
$$\therefore\quad ABC=2r\phi=2r\times\cos^{-1}\left(\frac{r-h}{r}\right) \tag{2}$$

Using (1) and (2), we can express $A_{\mathrm{tilled}}$ in term of $A_{\mathrm{flat}}$ as shown below.
$$A_{\mathrm{tilled}}= 2br\phi=\frac{rA_{\mathrm{flat}}}{\sqrt{h\left(2r-h\right)}}\cos^{-1}\left(\frac{r-h}{r}\right),\quad\mathrm{where}\space h\gt 0$$

Now, if we denote the additional area created by ploughing as $\Delta A$, we can express it as,
$$\Delta A=A_{\mathrm{tilled}}-A_{\mathrm{flat}}=\left(\frac{r}{\sqrt{h\left(2r-h\right)}}\cos^{-1}\left(\frac{r-h}{r}\right)-1\right)A_{\mathrm{flat}},\quad\mathrm{where}\space h\gt 0. \tag{3}$$

You can check the validity of this formula by substituting $r$ in place of $h$. If you do so, you will get the value of $\Delta A$ corresponding to the semicircular ridgelet, i.e. $\left(\frac{\pi}{2}-1\right)A_{\mathrm{flat}}.$

If you have trouble comprehending trigonometric identities, we give below another formula to determine $\Delta A$, which, however, only yields an approximate value. Besides, the values of $\Delta A$ you obtain from this equation are good if and only if $h$ is almost equal to $r$.

$$\Delta A \approx \left(\frac{3\pi+9k-3k^2+k^3-7}{6\sqrt{k\left(2-k\right)}}-1\right)A_{\mathrm{flat}},\quad\mathrm{where}\quad k=\frac{h}{r}\gt 0 \tag{4}$$


问:
Fields of sets and sigma-algebras of natural numbers
I have the following sets
$$ S=\{\{n\}:n \in \mathbb{N} \}$$
$$ F=\{A \subseteq \mathbb{N}:|A|=2  \}$$
$$ C=\{A \subseteq \mathbb{N}:|A| \leq \aleph_0  \}$$
I denoted by $ c(\cdot)$ field of set generated by this set and by $\sigma(\cdot)$ sigma-algebra generated by this set.

Now, I think that $c(S)=c(F)$ and $c(S)=c(C)$, although I have concerns about the last one - on the other hand $S\subset c(C)$ and so $c(S)\subset c(C)$. Also if $X \in C, |X|=\aleph_0 $ then $|X^{C}|=0$ so it is in $c(S)$ and hence $X\in c(S)$ so $C \subset c(S)$ and $c(C)=c(S)$.
This means that $\sigma(S)=\sigma(F)=\sigma(C)$. I'm not really sure about this. Something tells me that $\sigma(C)=c(C)=P(\mathbb{N})$ (power set) but I have troubles proving it so I don't know if this is the right path or not.

答:
It's not true that $|X| = \aleph_0 \implies |X^C| = 0$. For instance, let $X$ denote the set of even numbers. Then $|X^C|$ is the set of odd numbers, so $|X| = |X^C| = \aleph_0$. Also, if you notice $C$ is just $\mathcal{P}(\mathbb{N})$, since every subset of $\mathbb{N}$ has cardinality $\leq \aleph_0$.

It's not true that $c(S) = c(C)$. Define:
$$
\mathcal{F} := \{X \subseteq \mathbb{N} : \text{$X$ is finite or $\mathbb{N} \setminus X$ is finite}\}
$$
It's not hard to check that $\mathcal{F}$ is indeed a field of sets over $\mathbb{N}$. Then since $S \subseteq \mathcal{F}$, $c(S) \subseteq \mathcal{F}$, but $\mathcal{F} \subsetneq c(C) = \mathcal{P}(\mathbb{N})$.

However, it is true that $\sigma(S) = \sigma(F) = \sigma(C)$. Can you prove it yourself?


问:
Calculate variance of polynomial expression (continuous correlated random variables)
I have the following problem: $f_{(X,Y)}(x, y)$ as a PDF of a continuous random variable. I had to calculate the marginal PDF/CDFs, and the means and vars of these ($E(X), E(Y), Var(X), Var(Y)$).

Now I have another random variable $Z = (X-2)^2 + Y$, and I need to calculate $E(Z)$ and $Var(Z)$.

I had no problems with $E(Z)$, but I do have some trouble with $Var(Z)$.
$Var(Z) = Var(X^2 - 2X + 3Y)$. How do I need to proceed from here? Even if $X$ and $Y$ were not correlated, I still wouldn't know how to calculate $Var(X^2)$. (I did google this, but there were many confusing solutions). Is it $E(X^4) - [E(X^2)]^2$?

And if it is, could you still help me figure out how to calculate the whole $Var(X^2 - 2X + 3Y)$ part?

答:
Use the following formula:

$$V(X+Y+Z)=V(X)+V(Y)+V(Z)+2Cov(X,Y)+2Cov(X,Z)+2Cov(Y,Z)$$

and treat $X^2$ as a RV, so

$$V(X^2)=E((X^2)^2)-(E(X^2))^2$$


问:
Prove $9^n \equiv 9 \pmod{24}\quad \forall n \in \Bbb N^*$
I accidentally found that $9^n\equiv 9\pmod{24} \quad \forall n \in \Bbb N^*$

But don't know how to prove it in an exact way.

Please help

Body:
$$\begin{array}{rcl}9^n-9&=&9(9^{n-1}-1)\\&=&9(9-1)(9^{n-2}+9^{n-3}+\ldots+9+1)\\&=&72(9^{n-2}+9^{n-3}+\ldots+9+1)\end{array}$$

so $9^n-9$ is divisible not only by $24$ but by $72$ as well.


答:
You can prove it using induction:

Initial case: It's trivial that $9\equiv 9 \text{ mod(24)}.$

Induction: We suppose $9^n\equiv 9 \text{ mod(24)}$ for some $n\in\mathbb{N}.$
Then, we'll see if it verifies for $9^{n+1}$:
$$9^{n+1}=9^n\cdot 9,$$
and using modular arithmetic properties we see that
$$(9^n\cdot 9) \text{ mod(24)}\equiv(9^n \text{ mod(24)}) \cdot(9\text{ mod(24)}),$$
so
$$(9^n\cdot 9) \text{ mod(24)}\equiv 9\cdot 9=81 \equiv 9 \text{ mod(24)}.$$
So we have proved that $9^{n+1}\equiv \text{ mod(24)}$, and by induction the statement is proved.

Title: Prove $9^n \equiv 9 \pmod{24}\quad \forall n \in \Bbb N^*$

Body:
I accidentally found that $9^n\equiv 9\pmod{24} \quad \forall n \in \Bbb N^*$

But don't know how to prove it in an exact way.

Please help


问:
Steps to obtain integer solution to 2 equations sharing 3 variables?
I want to calculate possible T natural numbers for which the following two equations are true:

$$T=8x+7 $$
$$T=6y+5$$

Wolfram Alpha gives me the function:

Integer solution:

$T = 24 n + 23$, where $x = 3 n + 2$ and $y = 4 n + 3$, $\forall n\in\mathbb{Z}$

I have not done real math since High School, decades ago. What steps does one follow to get to that function?

P.S.: Symbolab returns something that I cannot see how it maps to WA’s answer. How are both results related?

答:
Equating the two expressions for $T$, and simplifying:
$$8x+7=6y+5\\
8x+2=6y\\
4x+1=3y$$
The RHS is a multiple of $3$, so the left hand side needs to be a multiple of $3$ as well. If you just try some values $0,1,2,3..$ for $x$, you will find that the only ones that make $4x+1$ a multple of $3$ are $2,5,8,11,...$. It is easy to see that this pattern continues, as adding a multiple of $3$ to $x$ will not affect whether the left hand side is divisible by $3$. So we get that $x=3n+2$ for some integer $n$.

A more formal way to solve this is by modular arithmetic, where all multiples of $3$ are ignored:
$$4x+1\equiv3y \pmod 3\\
3x+x+1\equiv3y \pmod 3\\
x+1\equiv0 \pmod 3\\
x\equiv-1 \pmod 3\\
x\equiv-1+3=2 \pmod 3\\
x=3n+2$$

Subtituting this into $4x+1=3y$ we get
$$4(3n+2)+1=3y\\
12n+8+1=3y\\
12n+9=3y\\
4n+3=y$$

So we get that $y=4n+3$ and $x=3n+2$. Substituting either of these into one of the expressions for $T$ gives that $T=24n+23$.

Symbolab gives a different answer because it does not know that you want $x,y$ to be integers. It expresses $T$ and $y$ in terms of $x$, but the expression for $y$ is as a fraction. To make that fraction simplify and get an integer $y$, you would still need to restrict the values of $x$ to $x=3n+2$ for integer $n$.


问:
Differential equation proving identity
I consider the differential equation $ u'(t)+a(t)\cdot u(t)=b(t) $.
I want to show that $$ u(t)=\underbrace{\exp \left( -\int_{t_0}^t a(s)\ ds\right )\cdot c}_{=f(t)}+\underbrace{\int_{t_0}^t \exp \left( -\int_{t_0}^t a(r)\ dr\right )\cdot b(s)\ ds}_{=g(t)} \ ,\quad c\in \mathbb{R}$$
solves this differential equation.

My attemp:

$$ \begin{align}f'(t)&=\frac {d}{dt}\left(\exp \left( -\int_{t_0}^t a(s)\ ds\right )\cdot c \right)\\&=-c\cdot a(t)\cdot \exp \left( -\int_{t_0}^t a(s)\ ds\right )\\[20pt]g'(t)&=\frac{d}{dt}\left(\int_{t_0}^t \exp \left( -\int_{t_0}^t a(r)\ dr\right )\cdot b(s)\ ds\right)\\&=\int_{t_0}^t \frac{d}{dt}\left(\exp \left( -\int_{t_0}^t a(r)\ dr\right )\cdot b(s)\ \right)ds\\&=\int_{t_0}^t \left[\frac{d}{dt}\left(\exp \left( -\int_{t_0}^t a(r)\ dr\right )\right)\cdot b(s)+\exp \left( -\int_{t_0}^t a(r)\ dr\right )\cdot b'(t)\ \right]ds\\&=\int_{t_0}^t \left[-a(t)\cdot \exp \left( -\int_{t_0}^t a(r)\ dr\right)\cdot b(s)+\exp \left( -\int_{t_0}^t a(r)\ dr\right )\cdot b'(t)\ \right]ds \\&=\exp \left( -\int_{t_0}^t a(r)\ dr\right )\cdot \int_{t_0}^t (-a(t)\cdot b(s)+b'(t))\ ds\\&=\exp \left( -\int_{t_0}^t a(r)\ dr\right )\cdot\left(b(t)-a(t)\cdot \int_{t_0}^t b(s)\ ds\right)\\[20pt]a(t)\cdot u(t)&=a(t)\cdot \exp \left( -\int_{t_0}^t a(s)\ ds\right )\cdot c+a(t)\cdot \int_{t_0}^t \exp \left( -\int_{t_0}^t a(r)\ dr\right )\cdot b(s)\ ds\\&=a(t)\cdot \exp \left( -\int_{t_0}^t a(s)\ ds\right )\cdot c+a(t)\cdot \exp \left( -\int_{t_0}^t a(r)\ dr\right )\cdot \int_{t_0}^t b(s)\ ds\\[20pt]u'(t)+a(t)\cdot u(t)&=f'(t)+g'(t)+a(t)\cdot u(t)\\&=\exp \left( -\int_{t_0}^t a(r)\ dr\right )\cdot\left(b(t)-a(t)\cdot \int_{t_0}^t b(s)\ ds\right)+a(t)\cdot \exp \left( -\int_{t_0}^t a(r)\ dr\right )\cdot \int_{t_0}^t b(s)\ ds\\&=\exp \left( -\int_{t_0}^t a(r)\ dr\right )\cdot b(t)\end{align}$$.

I don't see what went wrong here.

答:
You got the  solution wrong. The correct solution is  $$ u(t)=\underbrace{\exp \left( -\int_{t_0}^t a(s)\ ds\right )\cdot c}_{=f(t)}+e^{-\int_{t_0}^t a(r)\ dr}\underbrace{\int_{t_0}^t \exp \left( \int_{t_0}^s a(r)\ dr\right )\cdot b(s)\ ds}_{=g(t)} \ ,\quad c\in \mathbb{R}$$

Use product rule to write down $g'(t)$ and you will see that $u'(t)+a(t)u(t)=b(t)$.

Title: Differential equation proving identity


问:
Probability of breast cancer
I'm having trouble with a probability problem I've been trying to solve for a while. It's about the accuracy of breast cancer testing. Relevant probabilities are listed below, where:

"$\text{cancer}$" is the event "has breast cancer".

"$+$" is the event "tests positive for breast cancer".

$P(\text{cancer}) = \frac{12}{1000}$

$P(+|\text{cancer}) = \frac{11}{12}$

$P(+) = \frac{31}{1000}$

$P(\text{cancer}|+) = 0.355$

This last line is a result from a previous problem. The next part involves updating the probability of having cancer, but I'm having trouble figuring out what the answer is.

In the next part of the question, there's a woman who has tested positive and her doctor says she's part of a population for which there's a 40% chance of breast cancer.

I need to find the probability that the woman has cancer.

I am confused by this update to the cancer probability, but i will assume that this means $P(\text{cancer})$ has changed.

I also assume this means I need to find a new value for $P(\text{cancer}|+)$, but I'm not getting this right.

$P(+ | \text{cancer}) = \frac{11}{12} = \frac{P(\text{cancer} | +)\cdot P(+)}{P(\text{cancer})} = \frac{P(\text{cancer} | +) \cdot \frac{31}{1000}}{0.40}$

$P(\text{cancer} | +) = \frac{11}{12} \cdot 0.40 \cdot \frac{1000}{31} = 11.828$

The result can't be correct because it's way over 1.

How can i fix this? Thank you in advance for any insight.

答:
I'll try to make this easy, by "normalizing" things. Let's say there are $12000$ people (this number is going to make things be integers--for this first part anyway). Here's what we know:

$\frac{12}{1000}$ have cancer: $144$ people

11 of 12 people with cancer test positive for cancer: $132$ positive tests and $12$ false negatives

$\frac{31}{1000}$ test positive for cancer: $372$ positives (means $240$ false positives)

Means $p(\text{cancer}|+) = \frac{132}{372} = \frac{11}{31} \approx 0.355$ (and this result is from Bayes' Theorem: $p(\text{cancer}|+) = \frac{p(+ | \text{cancer})p(\text{cancer})}{p(+)} = \left(\frac{11}{12}\cdot \frac{12}{1000}\right)\cdot\frac{1000}{31}$)

So now you're telling me you fall into a group that isn't the average ($1.2\%$) but a much larger group: $40\%$. There is one definite, real world, assumption we'll have to make and that is that the accuracy of the test doesn't change. That may not be the case in the real world. Different groups are likely to get different accuracies for a test.

So what do I mean by "same accuracy". There are two possible outcomes to a test:

$\text{# positive results} = \text{# true positives} +\text{# false positives}$

$\text{# negative results} = \text{# true negatives} + \text{# false negatives}$

The rates of false negatives and false positives should remain the same (the number of true negatives and true positives will be determined by the population). This is the assumption that the test will have the same accuracy.

So now let's proceed as above, except we need to find the number of positive tests instead of it being a given:

$\frac{4}{10}$ have cancer: $4800$ people

$11$ of $12$ people with cancer still test positive: $4400$ positive tests ($400$ false negatives--note the false negative rate remains the same through this assumption).

The next part, we need to predict how many false positives we'll get. Assuming it happens at the same rate. Now think about this: where do false positives come from? They come from people that should be negative. In the original example, there were $11856$ people that did not have cancer, of those tested, $240$ came back positive, so the false positive rate was: $\frac{240}{11856} = \frac{5}{247}$. This represents $p(+|\neg \text{cancer})$. This can be slightly tricky to solve (although the above should give some insight):

\begin{align*}
p(+) = p(+|\text{cancer})p(\text{cancer}) + p(+|\neg\text{cancer})p(\neg\text{cancer})\\
\frac{31}{1000} = \frac{11}{12}\frac{12}{1000} + x*\left(1 - \frac{12}{1000}\right)
\end{align*}

Let's just get rid of the $1000$:

\begin{align*}
31 = 11 + 1000x - 12x \leadsto 20 = 1000x - 12x &&\text{divide everything by 4}\\
5 = 250x - 3x \leadsto x = \frac{5}{247} && \text{q.e.d.}
\end{align*}

OK, back to the example. We know $\frac{5}{247}$ of the people without cancer will test (falsely) positive, so now we find how many there are:

$$
\frac{5}{247}\cdot 7200 \approx 145.75\ \text{false positives}
$$

Giving the total number of positives at approximately $4400 + 145.75 = 4545.75$. So now we do just as we did above: we have $4400$ true positives out of $4545.75$ total positives:

$$
p(\text{cancer}|+) \approx \frac{4400}{4545.75} \approx 96.79\%
$$

But you can see from how we found the false positive rate, how we would solve this using the probabilities:

\begin{align*}
p(+) =&\ p(+|\text{cancer})p(\text{cancer}) + p(+|\neg\text{cancer})p(\neg\text{cancer}) \\
x =&\ \frac{11}{12}\frac{2}{5} + \frac{5}{247}\frac{3}{5} \\
=&\ \frac{11}{30} + \frac{3}{247} = \frac{2807}{7410}
\end{align*}

And finally, using Bayes' Theorem again:

\begin{align*}
p(\text{cancer}|+) =&\ \frac{p(+|\text{cancer})p(\text{cancer})}{p(+)} = \frac{\frac{11}{12}\cdot\frac{2}{5}}{\frac{2807}{7410}} \\
=&\ \frac{11}{30}\cdot\frac{7410}{2807} = \frac{11\cdot 247}{2807} \\
=&\ \frac{2717}{2807} \approx 96.79 \%
\end{align*}

A note on False Positives/Negatives

If you think about a terminal (or rare) medical diagnosis, you might think that $p(\text{cancer}|+) = 35.5\%$ is an inaccurate test. But you can see that it's actually not (when you take mostly cancer patients, it identifies them with a much higher probability). The false positive rate is "high", but what we should actually care about is the false negative rate. Which if you calculate (from my normalized example), there were $12$ false negatives out of a total of $11,628$ negatives ($12000 - 372$ negatives) giving a false negative rate of $\approx 0.10\%$...

\begin{align*}
p(\text{cancer}) =&\ p(\text{cancer}|+)p(+) + p(\text{cancer}|-)p(-) \\
p(\text{cancer}|-) =&\ \frac{p(\text{cancer}) - p(\text{cancer}|+)p(+)}{1 - p(+)} \\
=&\ \frac{\frac{12}{1000} - \frac{11}{31}\cdot\frac{31}{1000}}{\frac{969}{1000}}\\
=&\ \frac{1}{969} \approx 0.103199174407\%
\end{align*}

This is important because it should be less dangerous to investigate further (false positive) than it is more dangerous to be wrong and let the disease continue (false negative).

Title: Probability of breast cancer


问:
$\frac{z_1+z_2}{1+z_1z_2} \in \mathbb{R}$. Is my solution correct? Why?
I have to show that if numbers:

$z_1$ and $z_2 \in \mathbb{C} $

$ |z_1| = |z_2| = 1 $

$z_1*z_2 \neq -1$

then:
$$\frac{z_1+z_2}{1+z_1z_2} \in \mathbb{R}$$

So I thought I can transform that expression like that:
\begin{align*}
\frac{z_1+z_2}{1+z_1z_2}
&= \frac{(z_1+z_2)(1-z_1z_2)}{(1+z_1z_2)(1-z_1z_2)}
= \frac{z_1-z_1*z_1z_2+z_2-z_2*z_1z_2}{|1+z_1z_2|^2}
\\
&= \frac{z_1-|z_1|^2*z_2+z_2-|z_2|^2*z_1}{|1+z_1z_2|^2}
\\
&= \frac{z_1-z_2+z_2-z_1}{|1+z_1z_2|^2} = \frac{0}{|1+z_1z_2|^2}
\\
&= 0
\end{align*}

Is that correct? Could you tell me exactly why $(1+z_1z_2)(1-z_1z_2) = |1+z_1z_2|^2$ ?

I know that $z * \bar z = |z|^2$. But that equality gets me confused (if it is even right).

答:
This is not correct: $$(1+z_1z_2)(1-z_1z_2)=|1+z_1z_2|^2$$

it should be $$(1+z_1z_2)(\overline {1+z_1z_2})=|1+z_1z_2|^2$$

as $|w|^2 =w\cdot \overline{w}$. (Notice further that $\overline {1+z_1z_2} = \overline{1}+\overline {z_1z_2} =1 + \overline {z_1}\cdot \overline{z_2})$

Also, you could easyl check it is not correct if you take $z_1=z_2=1$, you don't get $0$ at the end.


问:
Prove that $-\sqrt{2}\leq \sin\theta+\cos\theta\leq\sqrt{2}$
We have to do it without calculus or any complex inequality. Level of complexity is that we cannot even use the AM-GM inequality.

So I tried,
$$(\sin\theta-\cos\theta)^2\geq0$$
$$1-2\sin\theta\cos\theta\geq0$$
$$\frac12\geq\sin\theta\cos\theta$$
Reverting back to the previous step, $$(\sin\theta+\cos\theta)^2\geq4\sin\theta\cos\theta$$
I am stuck here, please help.

Edit: Sorry, but we can only use knowledge upto class 10th. Which includes,
 $\sin^2\theta+\cos^2\theta=1$ etc.

$\sin(90^{\circ}-\theta)=\cos\theta$ etc.

 And basic trig ratios.

答:
$$-\sqrt{a^2+b^2} \le a\sin \theta+b\cos \theta \le \sqrt{a^2+b^2}$$

Plug in $a=b=1 $ in above formula.
You get:

$$-\sqrt{2} \le \sin \theta+\cos \theta \le \sqrt{2}$$

How I got the formula

Consider $f(\theta)=a\sin \theta+b\cos \theta$

Now I will differentiate this function to get min/max.

$$f^{'}(\theta)=a\cos \theta-b\sin \theta=0$$(for min/max)

$$a\cos \theta=b\sin \theta$$

$$\frac{a}{b}=\tan \theta$$

When $\tan \theta=\frac{a}{b}$ then there is maxima and minima.

Now you can find that $\sin \theta=\frac{a}{\sqrt{a^2+b^2}}$ and $\cos \theta=\frac{b}{\sqrt{a^2+b^2}}$

Plug in these values in the formula and you'll get it.


问:
Prove that if a vector $v\in V$ is such that $f(v)=0$ for any linear function $f$ from $V$ to $W$ then $v=0$
Let be $V$ and $W$ vector spaces. So I ask to prove that if $\vec v\in V$ is such that $f(\vec v)=0$ for any linear function $f$ from $V$ to $W$ then $\vec v$ is the zero vector. Clearly my condition is equivalent to claim that $\vec v\in\ker f$ for each $f\in\hom(V,W)$ but ufortunately I do not be able to argue anything from this. So could someone help me, please?

答:
It's only true if $W$ is not the trivial vector space $\{0\}$.

If $v\ne 0$ then you can extend it to a basis $v_1=v,\,v_2,\,v_3,\,\dots\,$ of $V$.

Let $w\in W$ nonzero and define $f:V\to W$ on the basis by letting $f(v_1)=w$ and, say, $f(v_i)=0$ if $i>1$.


问:
proving the given random variable a stopping time
Let $T>0$ and $\{\mathcal{F_s}\}_{s\in[0,T]}$ be a filtration on the space $\Omega$. Define,$$T' = \sum_{n=0}^k\,t_n\,1_{X_n}\,,$$ where $0=t_0<t_1<...<t_k=T$ and $\{X_n\}$ be a partition of $\Omega$ and $X_n\in \mathcal{F_{t_n}}$ for all $n$.

I want to prove that $T'$ is $\{\mathcal{F_s}\}$ stopping time. i.e: $\{T'=s\}\in \mathcal{F_s} $ for all $s\in[0,T]$.

Let $s$ be fixed and define $T_n = t_n1_{X_n}$ for each $n$. I think it suffices to prove that $T_n$ is $\mathcal{F_s}$ m'ble but not sure how to proceed. Any hint on how to proceed.

答: Your definition of stopping time applies to discrete case but here you have the continuous case. The definition is $\{T'\leq s\} \in \mathcal F_s$ for all $s$. Verify that $\{T'\leq s\}=\bigcup_I X_n$ where $I=\{n: t_n \leq s\}$. Note that$X_n \in \mathcal F_{t_n} \subseteq \mathcal F_s$ for any $n$ with $t_n \leq s$. Hence  $\{T'\leq s\} \in \mathcal F_s$.


问:
Is there a way to do complex long division?
So I know that $z^2+z+1-i$ can be factored out as $(z-i)(z+i+1)$. Now if you were only given one of its factors, is there a way to figure out the other factor using complex long division? I tried it and it doesn't seem possible like polynomial long division.

答:
It works completly fine. For example:

$(z^2+z+1-i)\div (z-i)=\color{red}{z}+\color{blue}{i+1}$

First step:
$z^2\div z=\color{red}{z}$

and $z(z-i)=z^2-iz$.
Then $z^2+z+1-i-(z^2-iz)=(i+1)z+(1-i)$

Second step:
Now $(i+1)z\div z=\color{blue}{i+1}$

and $(z-i)(i+1)=(i+1)z+1-i$.
Then $(i+1)z+(1-i)-((i+1)z+1-i)=0$.


答:
By long division,

$$z^2+az+b=(z+a-c)(z+c)+(b-ac+c^2).$$

This identity is true in any commutative field.

(Obviously if $z+c$ is known to be a factor, then $z^2+az+b=(z+a-c)(z+c)$.)

Title: Is there a way to do complex long division?

Body: So I know that $z^2+z+1-i$ can be factored out as $(z-i)(z+i+1)$. Now if you were only given one of its factors, is there a way to figure out the other factor using complex long division? I tried it and it doesn't seem possible like polynomial long division.


答:
$24\mid 72\mid 9^{\large n}\!-9\,$ is easy: factor out $\,\color{#0a0}{c=9}\,$ via mod distributive law

$$\begin{align}
ca\:\!\bmod cm \,&=\, \color{#0a0}c\,(a\bmod m)\\[.2em]
\Rightarrow\ \ 9^n-9\,\bmod\:\!\ 72 \,&=\,  \color{#0a0}9\,(\color{#c00}9^{n-1}\!-1\bmod 8)\\[.2em]
      &=\,  9(\,\color{#c00}1^{n-1}\!-1\bmod 8)\ \ {\rm by}\ \ \color{#c00}{9\equiv 1}\!\!\!\!\pmod{\!8}\\[.2em]
      &=\,  9\,( 0)
\end{align}\quad\ $$

Remark  $ $ Numbers like $\,9\,$ above with $\,a^2=a\,$ are called idempotents. This implies $\,a^{\large n} = a\,$ for all $\,n\ge 2\,$ either as above or by a simple induction $\,a^{\large n+1}\! = a\,a^{\large n} = a\cdot a = a.\,$ More conceptually: note $a$ is a fixed point of $\,f(x) = ax\,$ i.e. $\,f(a) = a,\,$ and fixed points always stay fixed on iteration by a simple induction: if $\,\color{#c00}{f^n(a) = a}\,$ then  $\,f^{\large n+1}(a) = f(\color{#c00}{f^{\large n}(a)}) = f(\color{#c00}a) = a$.

The $\!\bmod\!\!$ Distributive Law allows us to do this induction purely arithmetically, by cancelling out a factor to reduce it to the trivial induction $\,\color{#c00}1^n\equiv 1,\,$ exactly as in the special case above, i.e.

$$\begin{align}
 a^n-a\,\bmod\, a(a\!-\!1) \,&=\,  \color{#0a0}a\,(\color{#c00}a^{n-1}\!-1\bmod a\!-\!1)\\[.2em]
      &=\,  a(\,\color{#c00}1^{n-1}\!-1\bmod a\!-\!1)\ \ {\rm by}\ \ \color{#c00}{a\equiv 1}\!\!\!\!\pmod{\!a\!-\!1}\\[.2em]
      &=\,  a\,( 0)
\end{align}\!\!\!\!\!\!\!\! $$

therefore: $\ m\mid a(a\!-\!1)\mid a^{\large n}\!-a\ \Rightarrow\ a^n\equiv a\pmod{\!m}$

Or $\ \color{#c00}{a^{\large n}\equiv a}\pmod{\!(a\!-\!1)a},\,$ by it is true mod $\,a\!-\!1\!:$   $\,1^{\large n}\!\equiv 1\,$ & $\bmod a\!:\,0^{\large n}\!\equiv 0,\,$  thus it's also true $\!\bmod {\rm lcm}(a\!-\!1,a) = (a\!-\!1)a\,$ by CCRT, so the congruence persists $\!\bmod m\mid (a\!-\!1)a$.

Or $ $ by the Factor Theorem: $\, a\!-\!1\mid a^{\large n-1}\!-1\,\Rightarrow\, m\mid a(a\!-\!1)\mid a(a^{\large n-1}\!-1) = a^n-a$.

Idempotents $\!\bmod m\,$ play a fundament role in factorization of integers (and rings) since they correspond to splittings of $n$ into coprime factors, cf.  last paragraph here.

The $\!\bmod\!$ Distributive Law is essentially an operational form of CRT = Chinese Remainder Theorem. Unlike CRT it is formulated operationally which often makes it more convenient to apply in concrete calculations like those above.


答:
Assuming $V$ and $W$ are finite dimensional:

Let $\mathcal{B}$ and $\mathcal{C}$ be ordered bases for $V$ and $W$ respectively, and let $x \mapsto [x]_{\mathcal{B}}$ and  $y \mapsto [y]_{\mathcal{C}}$ be the corresponding coordinate mappings. For every linear map $T:V \rightarrow W$ there exists a matrix $M_T$ such that for all $x \in V$

$$[T(x)]_{\mathcal{C}} = M_T [x]_{\mathcal{B}}, $$

and vice versa (i.e. each matrix like that defines a linear map). Our asssumption is that for each $M_T \in \mathbb{R}^{\dim W \times \dim V}$, we have

$$M_T [v]_{\mathcal{B}} = \mathbf{0},$$

which means that $[v]_{\mathcal{B}} = \mathbf{0} \implies v = \mathbb{0}_V$, where $\mathbf{0}$ is the zero vector in $\mathbb{R}^{\dim W}$ and $\mathbb{0}_V$ is the zero vector in $V$.

Title: Prove that if a vector $v\in V$ is such that $f(v)=0$ for any linear function $f$ from $V$ to $W$ then $v=0$

Body: Let be $V$ and $W$ vector spaces. So I ask to prove that if $\vec v\in V$ is such that $f(\vec v)=0$ for any linear function $f$ from $V$ to $W$ then $\vec v$ is the zero vector. Clearly my condition is equivalent to claim that $\vec v\in\ker f$ for each $f\in\hom(V,W)$ but ufortunately I do not be able to argue anything from this. So could someone help me, please?


问:
A problem on polynomial
Let $p(x)$ be polynomial of degree $10$. It is given that $p(0)=1$ and $p(x)=x$ for $x=1$ to $10$. What is the value of $p(11)$. </p>

<p>I am able to show that it is certainly greater than or equal to $10$, but unable to find the exact value.  </p>

答:
Hint: given the degree, these values specify uniquely the polynomial $p$. If you don't see why (and want to use "heavy machinery"), look at Lagrange interpolating polynomials; otherwise, just consider the fact that any degree-$d$ polynomial has at most $d$ roots, and that you can easily build an explicit degree-$10$ polynomial $q$ such that $\Delta=p-q$ (polynomial with degree at most $10$) is zero on $0,1,\dots,10$. Then prove with the above remark that $\Delta\equiv 0$, and conclude that $p=q$.

Then compute $q(11)$.

Title: A problem on polynomial


答:
Since @Aqua has  resolved your query here ia  an alternate proof$$\frac{z_1+z_2}{1+z_1z_2}=\frac{\frac{1}{\bar {z_1}}+\frac{1}{\bar {z_2}}}{1+\frac{1}{\bar {z_1}\bar{z_2}}}=\frac{\bar z_1+\bar z_2}{1+\bar z_1\bar z_2}$$

As $X=\bar X \Rightarrow X$ is real  we complete the proof

Title: $\frac{z_1+z_2}{1+z_1z_2} \in \mathbb{R}$. Is my solution correct? Why?

Body:
I have to show that if numbers:

$z_1$ and $z_2 \in \mathbb{C} $

$ |z_1| = |z_2| = 1 $

$z_1*z_2 \neq -1$

then:
$$\frac{z_1+z_2}{1+z_1z_2} \in \mathbb{R}$$

So I thought I can transform that expression like that:
\begin{align*}
\frac{z_1+z_2}{1+z_1z_2}
&= \frac{(z_1+z_2)(1-z_1z_2)}{(1+z_1z_2)(1-z_1z_2)}
= \frac{z_1-z_1*z_1z_2+z_2-z_2*z_1z_2}{|1+z_1z_2|^2}
\\
&= \frac{z_1-|z_1|^2*z_2+z_2-|z_2|^2*z_1}{|1+z_1z_2|^2}
\\
&= \frac{z_1-z_2+z_2-z_1}{|1+z_1z_2|^2} = \frac{0}{|1+z_1z_2|^2}
\\
&= 0
\end{align*}

Is that correct? Could you tell me exactly why $(1+z_1z_2)(1-z_1z_2) = |1+z_1z_2|^2$ ?

I know that $z * \bar z = |z|^2$. But that equality gets me confused (if it is even right).


问:
Example of not locally finitely generated sheaf
Take the sheaf $\mathbf{H}_{\mathbb{C}}$ of holomorphic functions on $\mathbb{C}$. In a book I am reading they define the subsheaf $I$ by setting the stalks to be

$I_{z}:=\begin{cases}
0\quad \text{if } z=0\\
\mathbf{H}_{\mathbb{C},z} \quad \text{else}
\end{cases}$

and claim that this sheaf is not locally finitely generated as $\mathbf{H}_{\mathbb{C}}$-module.

Could anyone explain to me why this is true?

Let $U$ some neighbouhood of $0$. I will have to show that there is no surjective morphism of sheaves

$(\mathbf{H}_{\mathbb{C}}^{n})_{\vert U}\rightarrow I_{\vert U}$.

(Clearly, the only point around which finding a suitable surjective map could possibly fail is $0$.)

答: We can actually show that there is no non-zero morphism $f:\mathbf{H}_{\mathbb{C}}|_U\to I|_U$ if $U$  is connected. This will prove what you want. To show this, note that $s=f(1)$ (where 1 is the constant unit section of $\mathbf{H}_{\mathbb{C}}$) is a section of $I_z$ hence a section of $\mathbf{H}_{\mathbb{C}}$ since $I$ is a subsheaf of $\mathbf{H}_{\mathbb{C}}$. We have $s_0=0$ since $I_0=0$ hence $s|_V=0$ in some neighborhood $V$ of $0$. By the principle of isolated zeros, $s=0$. It follows that $f=0$.


问:
How are the top items calculated?
Imagine a list of items e.g. books. How do you determine which books are the most popular/best? So for example, you would put a book rated 4.7 on average with 100 ratings, higher than a book rated 4.9 on average with 10 ratings. What is the formula for that?</p>

答:
Use a weighted score mechanism, weighted against the amount of ratings. For example:

$$ \frac{average\space rating\space of\space book \times number\space of\space ratings\space per\space book}{number\space of\space total\space ratings} $$

Title: How are the top items calculated?


问:
Is there an operation over a finite set of integers (which size is $2^{128}$, or a power of two) that makes a non-abelian group?
I know that there is at least one operation on permutations that makes a non-abelian group: we can "multiply*" them, if we consider them as permutation matrices.
And naturals are directly mapped to permutations.
However, I need a non-abelian group which is not only finite, but also with a size which is a power of two due to the need of direct correspondence between them and hash digests (128bits).

Besides the non-abelian requirement, the results of the operation should be uniformly distributed along the $2^{128}$ numbers, just like the result of permutation matrix multiplication is.

P.S.: I suppose any power of two is welcome, since a pair of 64-bit  numbers, e.g., can be used to represent a 128-bit number; with the disadvantage of two separate operations - probably not ideal due to the uniformity requirement cited above.

-> "compose"?

答: There are non-abelian groups with as few as 8 elements, namely the dihedral group D4 and the quaternion group H. So there are plenty of nonabelian groups with 2^128 elements!</p>


问:
Determine whether $f(x)={\sin x \over x}$ is uniformly continuous in $\mathbb R$
Determine whether the function$$f(x)={\sin x \over x}$$is uniformly continuous in $\mathbb{R}$.

I am using the definition that for $\epsilon>0$ there exists $\delta>0$ such that $|f(x)-f(y)|<\epsilon$ whenever $|x-y|<\delta$.

答:
This query has been used as a reference to justify voting to close
a new query.  However, the new query specifically requires that $~\epsilon, \delta~$
be employed.  Therefore, I am filling the gap so that the new
query is answered here.

Define $f(x)$ so that $~f(0) = 1,~~$ and
for all $~\displaystyle x \neq 0, ~~f(x) = \frac{\sin ~x}{x}
~\implies$

[since $\lim_{x \to 0} ~f(x) ~= ~1$] $~f(x) ~$ is continuous everywhere.

I will employ an $\epsilon, \delta$ approach to demonstrate that $f$ is
uniformly continuous throughout $\mathbb{R}.$  This answer will then constitute
a stronger result than the OP in the new query is asking for.

To Prove

For any $~\epsilon > 0~$ there exists  $~ \delta > 0 ~$  such that

$|~f(x) ~-~ f(y) ~| ~<~ \epsilon~$ whenever $~|~x ~-~ y ~| ~<~ \delta$.

Let $~g(x) ~\equiv ~x (\cos x) - (\sin x) ~\implies ~g(0) ~= ~0.$

Let $~h(x) ~\equiv ~(\sin x) ~- ~x (\cos x) ~- ~x^2 ~\implies ~h(0) ~= ~0.$

For any fixed value of $~x, ~$ and any variable value of $~\delta,$

$~~~~~~~~~~~~$let $~J_x(\delta) ~\equiv ~f(x) ~- ~f(x + \delta) ~- ~\delta ~\implies ~~J_x(0) ~= ~0.$

Let $~k(x) ~\equiv ~f(0) ~- ~f(x) ~- ~x ~\implies ~k(0) ~= ~0.$

Let $~m(x) ~\equiv ~(\sin ~x) ~- ~x ~\implies ~m(0) ~= ~0.$

$\underline{\textbf{Lemma 1:}}$

$-\pi/2 ~< ~x ~< ~0 ~\implies ~g(x)  ~> ~0 ~~$ and

$0  ~< ~x ~< ~\pi/2 ~\implies ~g(x)  ~< ~0.$

$\underline{\text{Proof:}}$

$g'(x) ~= ~x (-\sin ~x) ~+ ~(\cos ~x) ~- ~(\cos ~x) ~= ~-x (\sin ~x).$

Therefore,

for $~-\pi/2 ~< ~x ~< ~0, ~g(x) ~$ is strictly decreasing $~~$and

for $~0 ~< ~x ~< ~\pi/2, ~g(x) ~$ is strictly decreasing.

$\underline{\textbf{Lemma 2:}}$

For $-\pi/2 ~< ~x ~< ~0, ~f(x) ~$ is strictly increasing $~~$ and

for $~0 ~< ~x ~< ~\pi/2, ~f(x) ~$ is strictly decreasing.

$\underline{\text{Proof:}}$

For $\displaystyle ~x ~\neq ~0, ~f'(x) ~=
~\frac{x (\cos ~x) ~- ~(\sin ~x)}{x^2} ~= ~\frac{g(x)}{x^2}.$

Invoke Lemma 1.

$\underline{\textbf{Lemma 3:}}$

For $0 ~< ~x, ~h(x) ~< ~0 ~~$ and

for $~x ~< ~0, ~h(x) ~< ~0.$

$\underline{\text{Proof:}}$

$h'(x) ~= (\cos ~x) ~+ ~x(\sin ~x) ~- ~(\cos ~x) ~- ~2x$

$= ~x(\sin ~x) ~- ~2x  ~= x(\sin ~x - 2).$

Noting that $~h(0) ~= ~0,$

for $~x ~< ~0, ~h(x) ~$ is strictly increasing $~~$ and

for $~x ~> ~0, ~h(x) ~$ is strictly decreasing.

$\underline{\textbf{Lemma 4:}}$

For $~0 ~< ~x, ~0 ~< ~\delta, ~J_x(\delta) < 0.$

$\underline{\text{Proof:}}$

$\displaystyle ~J_x(\delta) ~= ~\frac{\sin ~x}{x} ~- ~~\frac{\sin ~(x + \delta)}{x + \delta} ~- ~\delta ~\implies$

$\displaystyle
~\frac{d ~J_x(\delta)}{d ~\delta} ~=
~- ~\frac{(x + \delta) (\cos[x + \delta]) ~- ~(\sin[x + \delta])}
{(x + \delta)^2} ~- ~1$

$\displaystyle
~= ~\frac{(\sin[x + \delta]) ~- ~(x + \delta)(\cos[x + \delta]) ~- ~(x + \delta)^2}{(x + \delta)^2}$

$\displaystyle
~= ~\frac{h(x + \delta)}{(x + \delta)^2}.$

Invoke Lemma 3, noting that $~J_x(0) ~= ~0.$

$\underline{\textbf{Lemma 5:}}$

For $~0 ~< ~x ~< \pi/2, ~k(x) < 0.$

$\underline{\text{Proof:}}$

Since $~f(0) ~= ~1, ~k(x) ~= ~1 ~- ~f(x) ~- ~x.$

By the analysis in the proof to Lemma 2,

$\displaystyle k'(x) ~= ~- ~\frac{g(x)}{x^2} ~- ~1 ~=
 ~\frac{h(x)}{x^2}.$

By Lemma 3,
this implies that for $0 ~< ~x ~< \pi/2, ~k'(x) < 0.$

Noting that $~k(x) ~= ~0, ~$ the Lemma is proven.

$\underline{\textbf{Lemma 6:}}$

For $~0 ~< ~x, ~m(x) < 0.$

$\underline{\text{Proof:}}$

$m'(x) ~= ~(\cos ~x) ~- ~1 ~\implies$

for $~x ~< ~\pi/2,  ~m'(x)~$ is strictly negative.

Thereafter, $~m'(x)~$ is never $~> ~0.$

Therefore, with $~m(0) = 0, ~m(x)$

is strictly decreasing on $~(0, ~\pi/2)~$ and

thereafter, is strictly non-increasing.

$\underline{\text{Identity 7:}}$

For $~0 ~< ~|a - b| ~< ~(\pi/4),$

$\displaystyle |(\sin ~a) ~- ~(\sin ~b)|
~= ~2\left|\left(\sin\frac{a - b}{2}\right) ~\left(\cos\frac{a + b}{2}\right)\right|$

$\displaystyle \leq ~2\left|\left(\sin\frac{a - b}{2}\right) \right|$

$\displaystyle = ~2 \times \left(\sin ~\left|\frac{a - b}{2}\right|\right)
~< ~\text{[by Lemma 6]} ~|a - b|.$

WLOG $~x ~< ~y.$

Set $\displaystyle ~\delta = \min\left[\left(\frac{\pi}{4} \times \epsilon/2\right), ~\left(\pi/4\right)\right]
~\implies \frac{\delta}{\pi/4} ~< ~(\epsilon/2).$

Note that $~\forall ~x, ~f(x) ~= ~f(-x). ~$ Assume that $~|x - y| ~< ~\delta.$

$\underline{\text{Case 1:} ~~x ~< ~0 ~\leq ~y}$

$|f(x) - f(y)| ~\leq ~|f(x) - f(0)|  ~+ |f(0) - f(y)|~$

$= ~|f(-x) - f(0)| ~+ ~|f(0) - f(y)|.$

By Lemma 5, each of the two terms in the line above is $~< ~\delta.$

Therefore, their sum is less than $~\epsilon.$

$\underline{\text{Case 2:} ~~x ~= ~0 ~< ~y}$

By Lemma 5, $~|f(x) - f(y)| ~< ~\delta ~< ~\epsilon.$

$\underline{\text{Case 3:} ~~0 ~< ~x ~< ~\pi/4 ~~~~
\text{and} ~~~~x ~< ~y}$

Let $~\alpha ~= ~(y - x) ~\implies ~\alpha ~< ~\delta.$

Then, by Lemma 4,

$~|f(x) - f(y)| ~= ~J_x(\alpha) ~+ ~\alpha ~< ~\alpha ~< ~\delta ~< ~\epsilon.$

$\underline{\text{Case 4:} ~~\pi/4 ~\leq ~x ~< ~y}$

Let $~\alpha ~= ~(y - x) ~\implies ~\alpha ~< ~\delta.$

$\displaystyle |f(x) - f(y)| ~= ~\left|\frac{\sin ~x}{x} ~- ~\frac{\sin ~y}{y}\right|$

$\displaystyle = ~\left|\frac{\sin ~x}{x} ~- \frac{\sin ~x}{y} ~+ \frac{\sin ~x}{y} ~- ~\frac{\sin ~y}{y}\right|$

$\displaystyle \leq ~\left|\frac{\sin ~x}{x} ~- \frac{\sin ~x}{y}\right| ~+ ~\left|\frac{\sin ~x}{y} ~- ~\frac{\sin ~y}{y}\right|.$

Using Lemma 6,

$\displaystyle ~ \left|\frac{\sin ~x}{x} ~- \frac{\sin ~x}{y}\right| ~= \left(\frac{1}{x} - \frac{1}{y}\right) ~|\sin ~x|$

$\displaystyle < ~\left(\frac{y - x}{xy}\right) \times (x) ~= ~\frac{\alpha}{y} ~< ~\frac{\delta}{\pi/4} ~< ~\frac{\epsilon}{2}.$

Using Identy 7,

$\displaystyle \left|\frac{\sin ~x}{y} ~- ~\frac{\sin ~y}{y}\right| ~= ~\left(\frac{1}{y}\right) \times \left|\sin ~x ~- ~\sin ~y\right|$

$\displaystyle < ~\left(\frac{1}{\pi/4}\right) \times (\alpha) ~< ~\frac{\delta}{\pi/4} < \frac{\epsilon}{2}.$

Thus, $~|f(x) - f(y)| ~< ~\epsilon.$

$\underline{\text{Case 5:} ~~x ~< ~y ~\leq ~0}$

Since $~f(x) = f(-x), ~|f(x) - f(y)| ~= ~|f(-x) - f(-y)| .$

Therefore, Case 5 is proven by Cases 2 through 4, collectively.


问:
Why does the value of $b$ cause parabola to shift down and left if positive and down and right if negative?
It makes perfect sense why $a$ and $c$ affect the parabola in the ways they do but I don't understand how $b$ has this effect on the graph.

答:
For some intuition, consider this: for the graph $y=ax^2+bx+c$, the coefficient $c$ denotes the height at which the it intersects the $y$-axis. The coefficient $b$, entirely analogously, denotes the slope at which the graph intersects the $y$-axis.

If $b$ is positive and $a$ is positive, then with the intuition from the above paragraph, it makes complete sense that this makes the parabola move down and to the left, compared to if $b=0$.


问:
Conditional density function of $Z_1\mid Z_1+Z_2$
Say that we have two random variables $Z_1,Z_2$ such that $Z_1$ has ${\mathcal{B}}_{1/2,1}$ and $Z_1+Z_2$ follows ${\mathcal{B}}_{1,1/2}$ where ${\mathcal{B}}_{\alpha,\beta}$ is the beta density function. Is it possible to find the conditional density function of $Z_1\mid Z_1+Z_2$?

答:
Once  you assume independence between $Z_1,Z_2$ the problem is not difficult. Otherwise you cannot solve it unless you have further information to find the joint density $f_{Z_1,Z_2}(x,y)$

So let's assume independence and, just to simplify the notation, let's indicate $X$ and $Y$ instead of $Z_1$, $Z_2$

Setting

$$\begin{cases}
u=x+y \\
v=x,
\end{cases}$$

That is

$$\begin{cases}
x=v \\
y=u-v,
\end{cases}$$

it is self evident that the Jacobian $|J|=1$ thus

$$f_{UV}(u,v)=\frac{1}{4\sqrt{v(1+v-u)}}\mathbb{1}_{(0;1)}(v)\mathbb{1}_{(v;v+1)}(u)$$

As you can see, the joint density of $(U,V)$ is defined on the following parallelogram

and thus it is easy to integrate it in $V$ to derive $f_U(u)$

The conditional density you are looking for is obviously

$$f_{X|X+Y}(u,v)=\frac{f_{UV}(u,v)}{f_U(u)}$$

I am trying to compute $Pr(v<1/3,u<2/3)$ with the joint distribution function

Answering to your question, take a look at the integration area you are interested in (yellow area)

$$\mathbb{P}\Bigg[V<\frac{1}{3};U<\frac{2}{3}\Bigg]=\int_0^{\frac{1}{3}}\frac{1}{2\sqrt{v}}\Bigg[\int_{v}^{\frac{2}{3}}\frac{1}{2\sqrt{1+v-u}} du  \Bigg]dv\approx0.19475$$


问:
How to show that $f(x) = o{(x^2)}$ as $x \to \infty$. when $f$ is differentiable on $(0, \infty)$ and $f'(x) = o(x)$ as $x\ \to \infty$
Suppose $f$ is differentiable on $(0, \infty)$ and $f'(x) = o(x)$ as
$x \to \infty$.

How to show  that $f(x) = o{(x^2)}$ as $x  \to \infty$?

答:
Apply De l'Hospital's theorem in the form $$\left[ \frac{\text{anything}}{\infty}\right].$$ Then
$$
\lim_{x \to \infty} \frac{f(x)}{x^2} = \lim_{x \to \infty} \frac{f'(x)}{2x}=0
$$
by assumption.</p>


答:
Since the divisor is of degree one, you could also use synthetic division:

$$
\begin{array}{c|ccc}
i & 1 & 1 & 1-i \\
& & i & i-1\\
\hline
& 1 & 1+i & 0 \\
\end{array}
$$


问:
Why has a diffeomorphism full rank?
Let $f:U \subset \mathbb{R}^n \to \mathbb{R}^n$, $U$ open, be a diffeomorphism, i. e. $f$ is bijective and $f,f^{-1}$ are continuously differentiable. Why does the Jacobian $f'(x)$ has full rank for all $x \in U$?

(It's clear that if $f$ is continuously differentiable and $f'(x) \neq 0$ then f is locally invertible (inverse function thm.))

答:
Suppose $f$ and $f^{-1}$ are smooth. Then $f^{-1}\circ f = \mathrm{id}$, and the chain rule shows that
\begin{align}
\forall x \in U~ \mathrm{d}f^{-1}(f(x))\circ \mathrm{d}f(x) = \mathrm{id}
\end{align}
thus, $\mathrm{d}f(x)$ is invertible.

Title: Why has a diffeomorphism full rank?


问:
Struggling to find solution: question 4 (Ordinary Differential equation) (Reducible exact equation)
Could someone please help with this problem?
$$(2y+\frac{y^2}{x})dx+(2y+x)dy=0$$
I have tried and found the integrating factor $F(x)= 1/x$ and then re-arranged the equation to be:
$$(2\frac{y}{x}+\frac{y^2}{x^2})dx+(2\frac{y}{x}+1)dy=0$$
After this point it seemed so similar to a reducible seperable equation so I did this:
$$y=ux$$ $$y'=u'x+u$$
With this transformation
$$\frac{du}{dx}x+u=-\frac{2u+u^2}{2u+1}$$
$$\frac{du}{dx}x=-u-(u+1-\frac{1}{2u+1})$$

答:
Your denominator is wrong: it should be $2u+1$ since the ODE is $\left(2\frac{y}{x}+\frac{y^2}{x^2}\right)dx+\left(\color{red}2\frac{y}{x}+1\right)dy=0$.

It simplifies to $u'x=-u-\frac{2u+u^2}{2u+1}=-3\left(\frac{u^2+u}{2u+1}\right)$, which is variable separable.

We get $\int\frac{2u+1}{u^2+u}~du=-3\int\frac{dx}x$.

To integrate, take $m=u^2+u$. Can you complete?

$$\begin{align*}&m=u^2+u\implies dm=(2u+1)du\\\implies&\int\frac{dm}m=-3\int\frac{dx}x\\\implies&mx^3=k\\\implies&(y^2+xy)x=k\end{align*}$$


答: well, you may easily factor $p(x)-x$, and from this you should not have many troubles to find the result. (Remember that the value in 0 is used to find the value of the multiplicative constant, after factorizing $p(x)-x$)</p>

Title: A problem on polynomial

Body:
Let $p(x)$ be polynomial of degree $10$. It is given that $p(0)=1$ and $p(x)=x$ for $x=1$ to $10$. What is the value of $p(11)$. </p>

<p>I am able to show that it is certainly greater than or equal to $10$, but unable to find the exact value.  </p>


问:
How badly are distances distorted by mapping a rectangle to a line?
Purely out of curiousity, I was thinking about the following question:

Let $[n]$ be the set $\{0,1,2,\ldots,n-1\}$ and consider the cartesian product $[n]\times [m]$. Given any bijection $f:[n]\times[m]\rightarrow [nm]$, let's define its distortion as the greatest difference $|f(x_1)-f(x_2)|$ in its evaluation between two adjacent elements $x_1$ and $x_2$ of $[n]\times[m]$ where "adjacency" is considered that they agree in one coordinate and differ by exactly $1$ in the other coordinate. What is the minimum possible distortion of any bijection?

One might consider this as something about minimizing the Lipschitz constant of a bijection between a discrete line segment and a lattice rectangle under the taxicab metric.

One can constrain the answer fairly well: for a lower bound to distortion, observe that some element $x_1$ of $[n]\times [m]$ must have $f(x_1)=0$ and some $x_2$ must have $f(x_2)=nm-1$. There has to be some path of at most $n+m-2$ points in $[n]\times [m]$, each adjacent to the prior, connecting $x_1$ and $x_2$. Thus, some difference along this path must be at least $\frac{nm-1}{n+m-2}$.

On the other side, one can consider the explicit bijection $f(a,b)=ma+b$ or $g(a,b)=a+nb$ to get that some bijection has distortion at most $\min(n,m)$. There are other maps that attain this bound (e.g. you can adapt Cantor's pairing function to this finite domain to get this bound - essentially by ordering the points of the rectangle by diagonals and the filling each diagonal in in a consistent order). My guess is that this bound is optimal, although I lack a good argument to prove it. There's roughly a ratio of $2$ between the two bounds I give here in the worst case.

As far as exact results, this the distortion is clearly at least $1$ for any map $[1]\times [n]\rightarrow [n]$ where $n\geq 1$ and at least $2$ for any map $[2]\times [n]\rightarrow [2n]$ where $n\geq 2$ simply due to the pigeonhole principle. One can work out that all maps $[3]\times [3]\rightarrow [9]$ (and by extension $[3]\times [n]\rightarrow [3n]$ for $n\geq 3$) have distortion at least $3$ by extending the lower bound argument a touch. I checked with computer that $[4]\times [4]\rightarrow [16]$ maps have distortion at least $4$ and $[5]\times [5]\rightarrow [25]$ maps have distortion at least $5$. I told my computer to try working on the $[6]\times[6]$ case but I'm not optimistic that it will survive the combinatorial explosion.

答: We claim that for any natural numbers $n,m\ge 2$ any bijection $f:[n]\times[m]\rightarrow [nm]$ has distortion at least $d=\min\{n,m\}$. Indeed, let $k$ be the smallest number such that a set $S=f^{-1}([k])$ contains a number from each row or a number from each column of a rectangular set $[n]\times[m]$. It is easy to check that there are at least $d$ numbers of $[n]\times[m]\setminus S$ adjacent to a number of $S$. So $f(x)\ge k+d$ for some of these $d$ numbers. Since $x$ is adjacent to some number $y\in S$, we have $|f(x)-f(y)|\ge k+d-k=d$.


问:
Definition of linear dependence
My linear algebra teacher stated the following definition:

Let $V$ be a vector space and $v_1,...,v_n \in V$ vectors. We say that $v_1,...,v_n$ are linearly dependent if $\exists \alpha_1,...,\alpha_n$ not all $0$ such that:
$$\sum_{k=1}^n \alpha_kv_k = 0$$

The part that is confusing me is: "$\exists \alpha_1,...,\alpha_n$ not all $0$". Does this mean that, if at least 2 vectors are linearly dependent, then the whole set of vectors is said to be linearly dependent?

答:
Yes, if a subset of vectors is linearly dependent, then the set itself is dependent. The proof is simply - take your equation with some non zero scalars, and add whatever vectors you want multiplied by $0$.

The converse is not true. If a set is linearly dependent, it is possible that all its subsets are independent (think of $\{(1,0),(0,1),(1,1)\}$).

Title: Definition of linear dependence


问:
Show that the addition function, $f: \mathbb{R}^2 \longrightarrow R$, given by $f(x,y)=x+y$ is a continuous function.

Show that the addition function, $f: \mathbb{R}^2 \longrightarrow R$, given by $f(x,y)=x+y$ is a continuous function

If someone can give me a most clean proof this was very nice.
Thanks for read my proof

答:
$$\lvert f(x,y) - f(x_0,y_0) \rvert = \lvert x + y  - x_0 - y_0 \rvert = \lvert x - x_0  + y - y_0 \rvert \le \lvert x - x_0  \rvert + \lvert y - y_0 \rvert  \\ = \lVert (x,y) - (x_0,y_0) \rVert_1 .$$
Here $\lVert (a,b) \rVert_1 = \lvert a  \rvert + \lvert b \rvert_1$ is a norm on $\mathbb R^2$ which induces the usual topology on $\mathbb  R^2$ (i.e. the product topology on $\mathbb  R \times \mathbb  R$). This norm is equivalent to the Euclidean norm $\lVert (a,b) \rVert_2 = \sqrt{a^2 + b^2}$ since $\lVert (a,b) \rVert_2 \le \lVert (a,b) \rVert_1$ and $\lVert (a,b) \rVert_1 \le 2\lVert (a,b) \rVert_2$.


问:
i.i.d. random variables' questions.

Hi I was wondering if anyone could advise me on this question.

So from previous assistance, I have learned that I should be focusing on the equation:

$P(Y>1)=1-P(Y\leq1)$ so that it suffices to compute $P(Y\leq1)$

and hence use:

$P(Y\leq1)=P(x_1\leq1,X_2\leq1,X_3\leq1)$ $=P(X_1\leq1)P(X_2\leq1)P(X_3\leq1)$

and I have been shown that $P(X_i\leq1)=F(1)$ for each $i$

So from this I was wondering if this means to substitute $x=1$ into the equation of $F(x)$ which gives me the value 3. And if I'm going off the advice I have been given this should mean that:

$P(Y\leq1)=3*3*3$ which I just don't think can be correct.

I think I may be misinterpreting the information. Can someone please advise me on where I'm going wrong and how to actually approach this.

Thanks in advance.

Title: i.i.d. random variables' questions.

Body: Minor mistake: $F(1)=1-2^{-1}=1-1/2=1/2$.


答:
Let $k>0$.

Then $f'(x) < \frac{k}{2} x$ for all $x >a$.

By the MVT, for all $x > a$ there exists some $c_x$ so that

$$\frac{f(x)-f(a)}{x-a}=f'(c_x) < \frac{k}{2}c_x <\frac{k}{2}x \,.$$

this implies that

$$f(x)-f(a)< \frac{k}{2}x (x-a) < \frac{k}{2}x^2 \,.$$

Since $a$ is fixed, it is easy to finish the proof.

Title: How to show that $f(x) = o{(x^2)}$ as $x \to \infty$. when $f$ is differentiable on $(0, \infty)$ and $f'(x) = o(x)$ as $x\ \to \infty$

Body:
Suppose $f$ is differentiable on $(0, \infty)$ and $f'(x) = o(x)$ as
$x \to \infty$.

How to show  that $f(x) = o{(x^2)}$ as $x  \to \infty$?


答:
Inputting $x=1$ in $F(x)=1-2^{-x}$ does not give you $3$. It gives you
$$
1-2^{-1}=1-\frac12=\frac12
$$
Apart from that, your answer seems perfectly fine.

Title: i.i.d. random variables' questions.

Body:
Hi I was wondering if anyone could advise me on this question.

So from previous assistance, I have learned that I should be focusing on the equation:

$P(Y>1)=1-P(Y\leq1)$ so that it suffices to compute $P(Y\leq1)$

and hence use:

$P(Y\leq1)=P(x_1\leq1,X_2\leq1,X_3\leq1)$ $=P(X_1\leq1)P(X_2\leq1)P(X_3\leq1)$

and I have been shown that $P(X_i\leq1)=F(1)$ for each $i$

So from this I was wondering if this means to substitute $x=1$ into the equation of $F(x)$ which gives me the value 3. And if I'm going off the advice I have been given this should mean that:

$P(Y\leq1)=3*3*3$ which I just don't think can be correct.

I think I may be misinterpreting the information. Can someone please advise me on where I'm going wrong and how to actually approach this.

Thanks in advance.


问:
Questions related to group theory
If $H$ and $K$ are two normal subgroups of a group $G$, such that $H \subseteq K$, then show that $K/H$ is a normal subgroup of $G/H$.

Here how do I prove that $hk=kh$ because in this question as we know $H$ is normal to $G$ so $hg=gh$ and $K$ is also normal to $G$ so $kg=gk$. And also $H$ is contained in $K$. Which means all $h$ belong to $K$.

答:
The kernel of the natural projection $\varphi : G \rightarrow G/K$ contains $H$ because $H \subset K$, therefore it induces a map $\tilde{\varphi} : G/H \rightarrow G/K$ whose kernel is $\mathrm{Ker}{(\varphi)}/H=K/H$. As a kernel, $K/H$ is normal in $G/H$.

However, as said in comments, you should work a little bit more about normal subgroups before considering this proof.

Title: Questions related to group theory


问:
Trace of permutation matrix
If we have the symmetric group $S_k$ I know we can represent this as a permutation matrix and that this representation is a group homomorphism defined by say $f$. Where $f$ sends the symmetric group to an $n \times n$ matrix.

I was wondering how to show that the trace of $f$ was equal to the number of $x={1,... k}$ such that $h(x)=x$, where $h$ is our permutation?

So I can see for example that $(1,4,2,5,3)$ there are no elements on the diagonal so the trace is zero as no elements sends itself to itself. However I was wondering if there was a better way to show this more generally?

答:
If $\sigma \in S_n$, the corresponding permutation matrix is
$$
P_\sigma = \begin{pmatrix} e_{\sigma(1)}\\ \vdots\\  e_{\sigma(n)}  \end{pmatrix}
$$ where the $e_k$'s are standard basis row vectors: $e_k (j) = \delta_{k,j} = \begin{cases} 1 \ k = j \\ 0 \ k \neq j \end{cases}$. So
$$
\text{tr}(P_\sigma) = \sum_{i = 1}^nP_\sigma(i, i) = \sum_{i = 1}^n e_{\sigma(i)}(i) = \sum_{i = 1}^n
\delta(\sigma(i), i) = \sum_{i = 1 : \sigma(i) \neq i}^n 0 + \sum_{i = 1 : \sigma(i) = i}^n 1
$$ which is just $\#\{1 \leq i \in \mathbb N \leq n : \sigma(i) = i\}$.

Title: Trace of permutation matrix


问:
$0\leq a \leq b $ implies $ b^{-1} \leq a^{-1}$ in C*-algebra?
I'm studying C*-algebra in Conway's Functional Analysis.

Problem is this.

$0\leq a \leq b $, then $ b^{-1} \leq a^{-1}$ in C*-algebra?

I tried this problem via this way.

1st trial) Since $  a^{-1} - b^{-1} = a^{-1}(b-a)b^{-1}$ so product of three positive element is positive.

But this is false because product of positive element in C*-algebra is not positive.

2nd trial) How about using below Lemma?

$a \geq 0$  is equivalent to $a= a^*, ||t-a||\leq t$ for some $t \geq ||a|| $

Then, since $||a^{-1}(b-a)b^{-1}||  \leq ||a^{-1}|| \cdot ||b-a|| \cdot ||b^{-1}||$ so set $t = ||a^{-1}|| \cdot ||b-a|| \cdot ||b^{-1}||$ and try to solve this problem. What I stuck is is $||a||^{-1} = ||a^{-1}||$? It is just kind of passing thought, but in generally $a$ is hermitian, then $||a^{2^n}|| = ||a||^{2^n}$ so i think this might be true.

Can you help me?

答:
One way to prove the result is to first realise the $C^*$-algebra as sub-algebra of $\mathcal{B}(H)$ for a Hilbert space $H$ (GNS-construction) and then use the Lemmata below.

Lemma 1 Suppose $H$ is a Hilbert space and $S,T\in \mathcal{B}(H)$ are invertible, self-adjoint operators. Then $\Vert Sx\Vert \le \Vert T x \Vert$ for all $x\in H$ implies $\Vert T^{-1}x\Vert\le \Vert S^{-1} x\Vert$ for all $x\in H$.

Proof. Let $x,y\in H$, then
$$
\vert \langle T^{-1}x,y\rangle \vert = \vert \langle S^{-1}x,ST^{-1}y\rangle\vert\le \Vert S^{-1}x\Vert \cdot \Vert ST^{-1}y\Vert \le \Vert S^{-1}x\Vert \cdot\Vert y \Vert.
$$
Here we first used the Cauchy-Schwarz inequality and then the assumed estimate $\Vert Sz\Vert \le \Vert Tz \Vert$ for $z=T^{-1}y.$  The Lemma follows from taking the supremum over all $y\in H$ with $\Vert y \Vert = 1$.

Lemma 2 Suppose $H$ is a Hilbert space and $A,B\in \mathcal{B}(H)$ are invertible, self-adjoint operators. Then $A\le B$ implies $B^{-1}\le A^{-1}$.

Proof. Apply the previous Lemma for $S=A^{1/2}$ and $T=B^{1/2}$.

I've stolen the trick above from
a paper by Kato, where he shows a similar result for unbounded operators. I.e. when one defines $A\le B$ correctly (paying attention to domains), then $0<A\le B$ implies $0<B^{-1}\le A^{-1}$. This can be used to derive operator monotonicity of the maps $f(z)=z^\nu$ ($0\le \nu \le 1$), i.e. $0\le A \le B$ implies $f(A)\le f(B)$.

One more remark: It is not true in general that $\Vert a^{-1} \Vert = \Vert a \Vert^{-1}$. To see this, assume that $a_n$ is a sequence of invertible elements with limit $a_*\neq 0$. Then the sequence $\Vert a_n^{-1} \Vert = \Vert a_n \Vert^{-1}$ would be bounded by some $C<\infty$. Hence $\Vert a_n^{-1} - a_m^{-1}\Vert \le C^2 \Vert a_n - a_m \Vert \rightarrow 0$, which means that $a_n^{-1}$ must have a limit and it's easy to see that this must be an inverse of $a_*$. But this would imply that the set of invertible elements is open and closed in the set of nonzero elements and thus every nonzero $a$ was invertible, which is absurd unless the $C^*$-algebra is $\mathbb{C}$.


问:
Is every commutative simple ring a field, even if we don't assume it to be unital a priori?
I'm aware that every commutative simple ring with unity is a field. Is this true even for rings for which we don't know if they have a unity?

答:
Note that the in the setting of nonunital rings, a ring $R$ is called simple if its multiplication is not trivial (i.e., $R^2\neq0$) and it has no ideals other than $0$ and $R$. Hence we have two possible answers here:

We allow for trivial multiplication. If $R$ has trivial multiplication then the ideals are exactly the subgroups of $(R,+)$, so $R$ is simple iff $(R,+)$ has no nontrivial proper subgroups, hence it is $0$ or a cyclic group of prime order. Therefore $R$ is $0$ or $(\mathbb Z_p,+)$ with trivial multiplication for some prime $p$.

We restrict to the usual definition of simplicity given above. Then (as I show below) $R$ must have identity element, so it is a simple commutative unital ring, hence a field by the usual proof.

Proposition. A simple commutative ring is unital.

Proof: Since $R^2\neq0$ there is $0\neq a\in R$ such that $Ra=R$, as $Ra$ is a nonzero ideal. That is, $R$ is generated by a single element. Let us see that we can pick an idempotent generator. Since $R=Ra$, there is $e\in R$ such that $a=ea$, whence $ea=e^2a$, i.e., $(e-e^2)a=0$. Suppose $e$ is not idempotent; then $e^2-e\neq0$, so $R=R(e^2-e)$ and there is $b\in R$ such that $e=b(e^2-e)$, so $a=ea=b(e^2-e)a=0$, a contradiction. Hence $e^2=e$. Now let us see that $e$ is actually the identity element of $R$. Let $x\in R$ and suppose $ex-x\neq0$. Then again $R=(ex-x)R$, so there is $y$ (depending on $x$) such that $e=(ex-x)y$, so $e=e^2=e(ex-x)y=(ex-ex)y=0$, a contradiction. Therefore $ex=x$ for all $x\in R$.

Remark: A generalization of the previous proposition is: Let $R$ be a (not necessarily unital nor commutative) semiprime ring and $I$ be a minimal left ideal of $R$. Then there is an idempotent $e\in I$ such that $I=Ie=Re$.


问:
Solve $(x^2-y^2)dx+2xy\ dy=0$
I'm at the beggining of a differential equations course, and I'm stuck solving this equation:
$$(x^2-y^2)dx+2xy\ dy=0$$
I'm asked to solve it using 2 different methods. I proved I can find integrating factors of type $\mu_1(x)$ and $\mu_2(y/x)$. If I'm not wrong, these two integrating factors are:
$$\mu_1(x)=x^{-2} \ \ , \ \ \mu_2(y/x)=\left(1+\frac{y^2}{x^2}\right)^{-2}.$$
Then, I've used $\mu_1(x)$ to find the function $\psi(x,y)$ (I think it's called "potential function") so that
$$\frac{\partial \psi}{\partial x}=x^{-2}(x^2-y^2) \ \ , \ \ \frac{\partial \psi}{\partial y}=2xy.$$
Then, a solution to my equation is $\psi(x,y)=C$, being $C$ a constant.
I've got as solution:
$$\boxed{\psi(x,y)=x+\frac{y^2}{x}=C.}$$
I want to know first if this solution is correct. Then, I know that given two different integrating factors of different type, in my case $\mu_1(x)$ and $\mu_2(y/x)$, then
$$\frac{\mu_1(x)}{\mu_2(y/x)=K} \ \ , \ \ \text{$K$ constant}$$
defines also the solution to the equation. But what I get is
$$\boxed{\frac{x^{-2}}{\left(1+\frac{y^2}{x^2}\right)^{-2}}=K.}$$
I've graphed both solutions and they are not the same (correct me if I'm wrong). What of these is the right solution? Did I did something wrong along the way? Are both solutions the same? Any help will be appreciated.

PD: I know it can also be solved as an homogeneus one, but I'm interested in using these two methods and finding what I did wrong along the way.

EDIT: I realised I got wrong the integrating factor $\mu_2(y/x)$. Instead it it
$$\mu_2(y/x)=\left(1+\frac{y^2}{x^2}\right)^2.$$
Now, I can state as solution
$$\frac{\mu_2}{\mu_1}=K \ \implies \ \boxed{\left(x+\frac{y}{x}\right)^2=K}$$
which is equivalent to my first solution.

答:
Here is another way where $x=0$ is not a problem, you have $$x^2-y^2+2xyy'= 0$$

Let set $\ u=x^2+y^2\ $ then $\ u'=2x+2yy'$

Giving  $\require{cancel}\cancel{x^2}-(u-\cancel{x^2})+x(u'-\cancel{2x})=-u+xu'=0$

ODE $\ u=xu'\ $ has solutions $u=Cx$ therefore $$x^2+y^2=Cx$$

We can also solve it using polar coordinates:

$\begin{cases}
x^2-y^2=r^2\cos(2\theta)\\
2xy = r^2\sin(2\theta)\\
dx=\cos(\theta)dr-r\sin(\theta)d\theta\\
dy=\sin(\theta)dr+r\cos(\theta)d\theta\\
\end{cases}$

When you report in the ODE, you can cancel the factor $r^2$, also collecting $dr$ and $d\theta$ you'll find addition formulas for angle $(2\theta-\theta)$ and you will eventually get:

$$\cos(\theta)dr+r\sin(\theta)d\theta=0$$

This can also be writen $\frac {r'}r=-\tan(\theta)\ $ which solves to $$r=C\cos(\theta)$$

Which you can verify is a circle of diameter $C$ passing through origin and the same as above solution.

Title: Solve $(x^2-y^2)dx+2xy\ dy=0$


问:
Can $ y(t)=\left((t-1)^2+1\right)\cdot \cos(t) $ be a solution for this ODE
I've been asked if
$$
y(t)=\left((t-1)^2+1\right)\cdot \cos(t)
$$
can be a solution for the ODE:
$$
y''+y+(y-\cos(t))P(t,y,y') = 0
$$
when $P$ is smooth.

So I concluded that there exists a unique solution to the ODE, by Peano's theorem if:
$$
y_{1}=y,\quad y_{2}=y'.
$$
Then the ODE is of the form:
$$
y_{2}'=-y_{1}-P(t,y_{1},y_{2})(y_{1}-\cos(t)).
$$
$P$ is smooth and a function of $y_{2}$ and $y_{1}$, therefore the ODE has a unique solution, but I don't know how to continue any further.  Any help/hints would be appreciated!

答:
The function $\bar{y}(t)=\cos(t)$ is the $\textit{unique}$ solution of the considered IVP
$$
\left\{
\begin{array}{l}
y''+y+(y-\cos(t))P(t,y,y') = 0,\\
y(1)=\cos(1),\quad y'(1)=-\sin(1).
\end{array}
\right.
$$
We remark that the function $y(t)=\left((t-1)^2+1\right)\cdot \cos(t)$ satisfies the same initiale conditions, so it can't be a solution of the EDO.


答:
You may also note that if $a,b$ are invertible and
$$b-a ≥0$$
then:
$$a^{-1/2}(b-a)a^{-1/2}= a^{-1/2}ba^{-1/2}- 1≥0$$
and $a^{-1/2}ba^{-1/2}≥1$. Since $1$ commutes with that element you may take the inverse of both sides, turning the inequality around to get:
$$1-a^{1/2}b^{-1}a^{1/2}≥0$$
now conjugate with $a^{-1/2}$ again to get
$$a^{-1}-b^{-1}≥0$$

Title: $0\leq a \leq b $ implies $ b^{-1} \leq a^{-1}$ in C*-algebra?

Body:
I'm studying C*-algebra in Conway's Functional Analysis.

Problem is this.

$0\leq a \leq b $, then $ b^{-1} \leq a^{-1}$ in C*-algebra?

I tried this problem via this way.

1st trial) Since $  a^{-1} - b^{-1} = a^{-1}(b-a)b^{-1}$ so product of three positive element is positive.

But this is false because product of positive element in C*-algebra is not positive.

2nd trial) How about using below Lemma?

$a \geq 0$  is equivalent to $a= a^*, ||t-a||\leq t$ for some $t \geq ||a|| $

Then, since $||a^{-1}(b-a)b^{-1}||  \leq ||a^{-1}|| \cdot ||b-a|| \cdot ||b^{-1}||$ so set $t = ||a^{-1}|| \cdot ||b-a|| \cdot ||b^{-1}||$ and try to solve this problem. What I stuck is is $||a||^{-1} = ||a^{-1}||$? It is just kind of passing thought, but in generally $a$ is hermitian, then $||a^{2^n}|| = ||a||^{2^n}$ so i think this might be true.

Can you help me?


问:
Seeking a geometric characterisation of a certain property of a circle, a tangent ellipse within it, and a point strictly within the ellipse.
Consider a circle $C$, an ellipse $E$ strictly within $C$ except for a tangent point $p$, and a point $b$ strictly within $E$. Consider a point $q$ on $E$, and a point $q'$ given by the intersection of the extension of $qb$ with $E$, such that both $q$ and $q'$ are distinct from $p$. Define the points $r$ and $r'$ as the intersection of $C$ with the extensions of $pq$ and $pq'$ respectively.

We are interested in whether the point $j$, the intersection of the extension of $pb$ with $rr'$, lies inside the ellipse.

We can show (algebraically using Mathematica page 1,page 2) that the point $j$ is a property only of $C$, $E$, and $b$, independent of $q$, as illustrated in the figure:

<img src="https://i.stack.imgur.com/VTl7m.png" alt="illustration of the independence of $j$ on $q$" />

So we would like some way to characterize this property (whether $j$ is interior to $E$) of $C$, $E$, and $b$, geometrically without introducing $q$.

答:
Consider an ellipse in standard position with equation
$$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1 \tag1$$
and define $c^2 := a^2 - b^2$. At point $T=(a\cos t,b\sin t)$, with $0\leq t\leq \pi$ (to avoid sign complications), the ellipse has unit inward normal $\vec{u}$ and unit clockwise tangent vector $\vec{v}$ given by
$$
\vec{u} := \frac1s\left(-b\cos t, -a \sin t\right) \quad
\vec{v} := \frac1s\left(a\sin t, -b\cos t\right) \quad
s := \sqrt{a^2\sin^2t+b^2\cos^2t} \tag2$$
We can parameterize points on the ellipse thusly:
$$E(\theta) = T + \frac{2 a b s^3 \cos\theta }{\left(
\begin{array}{c}
(c^2 (a^2 \sin^2\theta - b^2 \cos^2\theta)+a^2b^2) \cos^2\theta \\
+ a^2 b^2 \sin^2\theta
+ a b c^2 \sin2t \cos\theta \sin\theta
\end{array}\right)}\left( \vec{u} \cos\theta +  \vec{v} \sin\theta \right)
 \tag3$$
where $\theta$ is the angle that $\overrightarrow{TE}$ makes with normal vector $\vec{u}$. (We've effectively converted to a $uv$-coordinate system centered at $T$.) Note that $|TE|$ is the big multiplied factor, which amounts to a "polar" representation of the ellipse in the new coordinate system; the rectangular representation in that system (with $u$ and $v$ giving distances in the $\vec{u}$ and $\vec{v}$ directions from $T$) can be read immediately from the coefficients of $\cos^2\theta$, $\sin^2\theta$, $\sin\theta\cos\theta$ in the denominator, and $\cos\theta$ (with a sign change) in the numerator:

$$\begin{align}
u^2 (c^2 (a^2 \sin^2t- b^2 \cos^2t)+a^2b^2) + v^2 a^2 b^2 + u v a b c^2 \sin2t - 2 u a b s^3 = 0 \tag4
\end{align}$$

From this we find that (ignoring special cases) the axes of the ellipse are rotated relative to the $uv$-frame by an angle $\phi$ satisfying
$$\tan2\phi = \frac{\text{coeff of}\; uv}{(\text{coeff of}\; u^2) - (\text{coeff of}\; v^2)} = \frac{a b\sin2t}{a^2\sin^2t-b^2\cos^2t} \tag5$$

Now, we'll get at OP's construction by considering points $P$ and $Q$ on the ellipse for angles $\theta=p$ and $\theta=q$, and taking $R$ to be the intersection of $\overline{PQ}$ with the line through $T$ making angle  $r$ with vector $u$. (This $R$ plays the role of OP's point $b$, with $P$ and $Q$ corresponding to $q$ and $q'$.) We find that
$$|TR| = \frac{2 a b s^3}{
\left(\begin{array}{c}
   (a^4\sin^2t + b^4 \cos^2t) \cos r + a b c^2 \sin2t \sin r \\
 + a^2 b^2 ((\tan p + \tan q) \sin r - \tan p \tan q \cos r)
\end{array}\right)
} \tag6$$
We can manipulate this into
$$\begin{align}
\frac1{a^2b^2}&\left(
\frac{2 a b s^3}{|TR|}
- (a^4 \sin^2t + b^4 \cos^2t) \cos r
- a b c^2 \sin2t \sin r
\right) \\[4pt]
&= (\tan p + \tan q ) \sin r - \tan p \tan q \cos r
\end{align} \tag7$$

The significance of $(7)$ is that its right-hand side is devoid of ellipse-specific parameters, while its left-hand side is devoid of $P$- and $Q$-specific parameters.

To see why this is helpful, consider another ellipse (generalizing OP's circle) with parameters, say, $\hat{a}$, $\hat{b}$ (and $\hat{c}$), and a point $\hat{T}:=(\hat{a}\cos\hat{t}, \hat{b}\sin\hat{t})$ on it. We can parameterize that ellipse relative to unit normal and tangent vectors $\hat{u}$ and $\hat{v}$ (with $\hat{s}$) at $\hat{T}$. Let's move this new ellipse to be tangent to the first, aligning $\hat{T}$, $\hat{u}$, and $\hat{v}$ with with $T$, $\vec{u}$, and $\vec{v}$. Then we can use the lines through $T=\hat{T}$ with the earlier direction angles $p$, $q$, $r$ (no hats!) determine points $\hat{P}$ (collinear with $T$ and $P$), $\hat{Q}$ (likewise), $\hat{R}$ (likewise) on this new ellipse. The last of these points satisfies a suitably-hatted counterpart to $(6)$, and thus also a counterpart to $(7)$; since the right-hand sides of $(7)$ and this new relation match, their left-hand sides do, too. That is, we have
$$\begin{align}
\frac1{a^2b^2}&\left(
\frac{2 a b s^3}{|TR|}
- (a^4 \sin^2t + b^4 \cos^2t) \cos r
- a b c^2 \sin2t \sin r
\right) \\[4pt]
= \frac1{\hat{a}^2\hat{b}^2}&\left(
\frac{2 \hat{a} \hat{b} \hat{s}^3}{|T\hat{R}|}
- (\hat{a}^4 \sin^2\hat{t} + \hat{b}^4 \cos^2\hat{t}) \cos r
- \hat{a} \hat{b} \hat{c}^2 \sin2\hat{t} \sin r
\right)
\end{align} \tag8$$
In this way, we have confirmed OP's assertion (generalized to not-necessarily-circular ellipses) that the relative positions of $R$ and $\hat{R}$ along the $r$-line are independent of the points along the $p$- and $q$-lines.

For the question-proper, we seek the locus of $R$ when $\hat{R}$ traces the $ab$-ellipse.

That is, when $\hat{R}=E(r)$, so that $|T\hat{R}|$ is the big factor in $(3)$, with $\theta=r$. Substituting this into $(6)$ and solving for $|TR|$ gives
$$\begin{align}
|TR| &=
\frac{-D \cos r}{A \cos^2r + B \cos r\sin r + C \sin^2r} \\[8pt]
A &=
  \hat{a}\hat{b} c^2 (a b \hat{s}^3 + \hat{a} \hat{b} s^3 ) (a^2 \sin^2t - b^2 \cos^2t) \\
&\phantom{=}- a^2 b^2 \hat{c}^2 s^3 (\hat{a}^2 \sin^2\hat{t} - \hat{b}^2 \cos^2\hat{t}) \\
&\phantom{=}+ a^3 b^3 \hat{a} \hat{b} \hat{s}^3 \\[4pt]
B &= a b \hat{a}\hat{b} (c^2 ( a b \hat{s}^3 + \hat{a} \hat{b} s^3) \sin2t - a b \hat{c}^2 s^3 \sin2\hat{t}) \\[4pt]
C &= a^3b^3 \hat{a}\hat{b}\hat{s}^3 \\[4pt]
D &= -2 a b s^6 \hat{a}^2 \hat{b}^2
\end{align}
\tag9$$

This is the "polar" representation (in our $uv$ coordinate frame) of yet another ellipse tangent at $T=T'$, as suggested by Intelligenti pauca's sketch. Its rectangular representation (with $u$ and $v$ representing the distances from $T$ in the $\vec{u}$ and $\vec{v}$ directions) is
$$A u^2 + B u v + C v^2 + D u + E v + F = 0 \quad (E=F=0)$$

Gleaning the exact geometry of this ellipse a bit of a chore in general. Let us simplify to OP's specific case, where the $\hat{a}\hat{b}$-ellipse is a circle; we have $\hat{a}=\hat{b}=\hat{s}$ and $\hat{c}=0$, so that

$$\begin{align}
A &= c^2 (a b \hat{a} + s^3 ) (a^2 \sin^2t - b^2 \cos^2t) + a^3 b^3 \hat{a} \\[4pt]
B &= a b c^2 ( a b \hat{a} + s^3) \sin2t \\[4pt]
C &= a^3b^3 \hat{a} \\[4pt]
D &= -2 a b s^6
\end{align}
\tag9$$

Then we readily find that this ellipse is rotated relative to the $uv$-frame by $\phi$ satisfying
$$\tan2\phi=\frac{B}{A-C} = \frac{ab\sin2t}{a^2\sin^2t-b^2\cos^2t} \tag{10}$$
which matches $(5)$, confirming Intelligenti pauca's suspicion that the axes of this ellipse are parallel to those of the $ab$-ellipse. The center of this  ellipse is
$$(u,v) = \frac{D}{B^2-4AC}(2C,-B) = \frac{s^2}{2mn} (2a^2 b^2 \hat{a}, -c^2 p\sin2t) \\[16pt]
m := a b \hat{a} - c^2 s \cos^2t \qquad
n := a b \hat{a} + c^2 s \sin^2t \qquad
p := a b \hat{a} + s^3
\tag{11}$$

Rotating the ellipse by $\phi$ to align its axes with $\vec{u}$ and $\vec{v}$ gives

$$u^2 b^2 m
 + v^2 a^2  n
 + 2 u a b^2 s^3 \cos t
 + 2 v a^2 b s^3 \sin t = 0 \tag{12}$$

That is,
$$\left(u + \frac{a s^3 \cos t}{m}\right)^2 \frac{m^2 n}{a^3 b \hat{a} s^6}
 + \left(v + \frac{b s^3 \sin t}{n}\right)^2 \frac{mn^2}{a b^3 \hat{a} s^6}
 = 1 \tag{12'}$$
From this, we read that the major and minor radii (say, $a'$ and $b'$) of this third ellipse are
$$a' = \sqrt{\frac{a^3 b \hat{a} s^6}{m^2 n}} \qquad b' = \sqrt{\frac{a b^3 \hat{a} s^6}{mn^2}} \quad\to\quad (c')^2 := (a')^2-(b')^2 = \frac{a b c^2 \hat{a} p s^6}{m^2n^2} \tag{13}$$
and its eccentricity, $e'$, is
$$e' := \frac{c'}{a'} = \frac{c}{a}\; \sqrt{\frac{p}{n}} = e\;\sqrt{\frac{p}{n}} \tag{14}$$
where $e$ is the eccentricity of the $ab$-ellipse.

Determining other properties of these figures is left to the reader. $\square$


问:
Given $\angle A = 60°, \angle B = 80°, \angle C = 40°$, find $\alpha$

I tried making an equilateral triangle with side $b$ but it didn't seem to be of any use. Please help.

答:
Hint : you need a calculator and the law of sines.

You can express both b and c in terms of a, via the law of sines.

then, you can use the law of sines again on the left hand triangle re
100 degrees -> (a + b + c) and $\alpha$ -> a.


问:
take Fourier transform of equation
I was trying to solve this problem:

By taking the Fourier transform of the equation $$\frac{d^2\phi}{dx^2}-K^2\phi=f(x)$$ show that its solution, $\phi(x)$, can be written as $$\phi(x)=\frac{-1}{\sqrt{2\pi}}\int^{+\infty}_{-\infty} \frac{e^{ikx}\tilde{f(k)}}{k^2+K^2}dk$$ where $\tilde{f(k)}$ is the Fourier transform of $f(x)$.

I tried this but I'm stuck..

Could someone show me how to take this Fourier transform?

$$\frac{1}{\sqrt{2\pi}}\int^{+\infty}_{-\infty} \frac{d^2\phi}{dx^2}-K^2\phi e^{-iwx} dx = \frac{1}{\sqrt{2\pi}}\int^{+\infty}_{-\infty}f(x)e^{-iwx}dx $$

$$\frac{1}{\sqrt{2\pi}}\int^{+\infty}_{-\infty} \frac{d^2\phi}{dx^2} e^{-iwx}dx- \frac{1}{\sqrt{2\pi}}\int^{+\infty}_{-\infty} K^2\phi e^{-iwx} dx = \frac{1}{\sqrt{2\pi}}\int^{+\infty}_{-\infty}f(x)e^{-iwx}dx $$

答:
Hint:

Take the Fourier transform of both sides (you do not need to write down the integrals involved).

The left-hand side can be simplified by using:
$$ \widehat{af+bg}=a\hat{f}+b\hat{g}$$
together with the rule for $\hat{g’}$.

Now you should have an expression like
$$\hat{\phi}(t)(at^2+b)= \hat{f}(t)$$

Solve for $\hat{\phi}$ and take the inverse Fourier transform.

Title: take Fourier transform of equation


问:
Solving a 2nd order partial differential equation
The equation at hand is $\Big[\frac{-\hbar}{2m}\frac{\partial^2}{\partial x^2} + V(x) \Big] \psi(x) = E\psi(x)$. Given are $V(x) = V_0 $ for $x>0$ and $V(x) = 0$ for $x \le 0$ and $\psi(0)=-k/\kappa$. How can I solve this equation and find $\psi(x)$ ?

EDIT:

I proceeded to solve for $\psi(x)$ when $V(x)=0$ by setting $\psi(x)=Ae^{zx}$ and solving for $A,z$. Then I use $\psi(0) = -k/\kappa$ to arrive at $\psi(x)$. Is this correct?

答:
Hint:

This is a second order linear ODE with piecewise constant coefficients. You can solve for the two pieces using the standard technique of the characteristic equation. Expect to discuss the sign of $V-E$.

Then join the two partial solutions, enforcing the known "boundary" condition, and possibly other continuity constraints.


问:
How to solve the Frobenius-norm minimization?
I am looking for a way to determine a matrix $W \in \mathbb{R}^{n\times m}$, such that

$$||W^TW-\lambda^2\mathbf{I}||_{\rm F}^2$$

gets minimized, where $\mathbf{I}$ is the $m$-dimensional identity matrix, $||\cdot ||_{\rm F}$ is the Frobenius-norm. In fact, I'm looking for a way in which the matrix $W$ gets as close as possible to the identity matrix. I'd be grateful for all helpful advices or even solutions!

答:
The singular value decomposition (SVD) of the real matrix $\mathbf W$ is a factorization of the form $$\mathbf W=\mathbf U\mathbf \Lambda \mathbf V^T$$ where $\mathbf U$ and $\mathbf V$ are unitary matrices and the diagonal of $\mathbf \Lambda\triangleq\operatorname{diag}\left\{w_1,w_2,\cdots,w_m\right\}, w_k\geq 0\forall k$, contains all singular of $\mathbf W$.

Put the above decomposition into the initial problem
\begin{equation}
\begin{aligned}
\Vert\mathbf W^T\mathbf W-\lambda^2 \mathbf I\Vert_F^2&=\Vert\mathbf V\mathbf \Lambda^2 \mathbf U^T-\lambda^2 \mathbf I\Vert_F^2\\
&=\Vert\mathbf\Lambda^2-\lambda^2 \mathbf I\Vert_F^2\\
&=\left(w_1^2-\lambda^2\right)^2+\left(w_2^2-\lambda^2\right)^2+\cdots+\left(w_m^2-\lambda^2\right)^2
\end{aligned}
\end{equation}
Then, we need to minimize all subproblems as follows
\begin{equation}
\begin{array}{ll}
\min\limits_{w_k} & \vert w_k^2-\lambda^2\vert
\end{array}
\end{equation}
So, $w_k=\vert\lambda\vert$ if $w_k\neq0$. Furthermore, each singular value of $\mathbf W$ is $\lambda$ or $0$, i.e., $$\mathbf W=\mathbf U\operatorname{diag}\{\underbrace{\lambda,\lambda,\cdots,\lambda}_{\min\{m,n\}\lambda},0,0,\cdots,0\}\mathbf V^T$$for any unitary matrices $\mathbf U$ and $\mathbf V$.


问:
Given $\mathcal{F}\subseteq \mathcal{G}$, show $\bigcup\mathcal{F} \subseteq \bigcup\mathcal{G}$
Just proof verification. Given $\mathcal{F}\subseteq\mathcal{G}$, show $\bigcup\mathcal{F} \subseteq \bigcup\mathcal{G}$. Here is my attempt:

Let $A$ be an arbitrary set and $A\in\mathcal{F}$. This implies $A\in\mathcal{G}$. Let $x$ be arbitrary. Suppose $x\in\bigcup\mathcal{F}$. Then $x\in\bigcup_{i=0}^{n} S_i \, \forall S_i \in \mathcal{F}\,\forall i,n\in\mathbb{N}$. Then $\exists i\big|S_i=A$.

It follows that $x\in A$, $A\in\mathcal{G}$ and $x\in \mathcal{G}$. By the definition of $\bigcup\mathcal{G}$, $A\in\bigcup\mathcal{G} \therefore x\in \bigcup\mathcal{G}$. Now, $x\in\bigcup\mathcal{F} \Longrightarrow x\in\bigcup\mathcal{G}$. Since $x$ and $A$ were arbitrary, we can conclude that $\bigcup\mathcal{F} \subseteq\bigcup\mathcal{G}$.

If you notice any places where I can be more concise, if my notation/wording is off, or if my argument/reasoning is wrong, I would appreciate it if you let me know. Thanks.

答:
Suppose $\mathcal{F}\subseteq\mathcal{G}$. If $A$ is a set and $A\in\mathcal{F}$, then $A\in\mathcal{G}$.

Fix $x\in\bigcup\mathcal{F}$. Then $x\in A$ for some $A\in\mathcal{F}$. By the above, we have $x\in\bigcup\mathcal{G}$ since $A\in G$. Therefore $\bigcup\mathcal{F}\subseteq\bigcup\mathcal{G}$.


问:
Las Vergnas theorem - do vertices need to be ordered by their degree?
I'm learning graph theory and I'm not sure one thing about the Las Vergnas theorem.  Do vertices need to be ordered by their degree?

Las Vergnas theorem: $G$ is a graph and $V(G)=(v_1,...,v_n)$. There is no such $i,j$ such that $i<j, i+j \geq n, v_iv_j \notin E(G),
deg(v_i) \leq i, deg(v_j) \leq j-1, deg(v_i) + deg(v_j) \leq n-1$. Then $G$ is hamiltonian

Thanks

答:
Las Vergnas theorem states that if there exists an arrangement of vertices $x_1, \dots, x_n$ of the graph $G$ such that for all $i, j$:

if
$$1 \leq i < j \leq n, \hspace{1cm} i+j \geq n, \hspace{1cm} x_ix_j \notin E(G),$$
$$\text{d}(x_i) \leq i,\hspace{0.5cm} \text{and} \hspace{0.5cm} \text{d}(x_j) \leq j-1$$
then
$$\text{d}(x_i)+\text{d}(x_j) \geq n $$

then the graph $G$ is Hamiltonian. To answer the question, that means that the vertices do not need to be ordered by their degree - there must only exist at least one arrangement of the vertices for which the condition holds.


问:
What justifies the existence of two independent rvs $\overline{X}$ and $\overline{Y}$ that have the same distribution as some rvs $X$ and $Y$
What justifies the existence of two independent rvs $\overline{X}$ and $\overline{Y}$ that have the same distribution as some rvs $X$ and $Y$

In the proof:

$X,Y$ are independent random variables $\iff$ $E[\exp(i\langle (t,s),(X,Y)\rangle)]=E[\exp(i\langle t,X\rangle)]E[\exp(i\langle s,Y\rangle)](*)$

for the $\Leftarrow$ part, I have seen a proof that uses the fact there exist independent random variables $\overline{X}$ and $\overline{Y}$ with $X$ ~ $\overline{X}$ and $Y$ ~ $\overline{Y}$.

My question pertains to why it is always possible to find such independent random variables $\overline{X}$ and $\overline{Y}$ that follow such a distribution. To me this assertion seems very far-reaching and I am having trouble to believe that we are always able to construct such an $\overline{X}$ and $\overline{Y}$. Any intuition or proofs?

答: Consider the probability space $(\Omega,\mathcal{F},P)$, where $\Omega=\mathbb{R}^2$, $\mathcal{F}$ is Borel sets of $\Omega$ and $P=P_X\otimes P_Y$. Then let $\bar X=(s,t)=s$ and $\bar Y(s,t)=t$ for $(s,t)\in\Omega$.


问:
Is there a natural way to model events from two different probability spaces?
My question is motivated by the question here: How to calculate probabilities for football match?. In that question, we have a situation where we have calculated (based on historical data), the probabilities of football teams winning in particular situations. So for example, Chelsea wins at home 50% of the time, and Liverpool wins an away game 20% of the time. So a fundamental question in that post is "Given these two probabilities, what is the probability that Chelsea win when playing Liverpool at thome?"

This is also my question, but I am looking for an answer in terms of rigorous probability theory. Both the answers to the other question don't explain things rigorously. In particular, my understanding is that we effectively have two different probability spaces, let's call them $(C, \mathscr{A_1},P_1)$ and $(L, \mathscr{A_2},P_2)$, representing the probability measures on Chelsea and Liverpool respectively. Here the space on $C$ represents Chelsea playing games at home, and the second space represents Liverpool playing away.

It seems reasonable to me that we may define $C = L = \mathbb{N}^2$, where we would interpret the elementary event $\{(i,j)\}$ as meaning that Chelsea (assuming we are in the first probability space) has scored $i$ goals, while the opponent has scored $j$ goals. Here the 'opponent' is not necessarily Liverpool. I think this means that the probability space models what happens 'on average' when the team plays against any opponent, but I'd like to know if this is actually a reasonable interpretation? I also think it seems reasonable to define the sigma algebra as the Borel sigma algebra on $\mathbb{R}^2$.

Then the only difference between the two spaces is the probability functions $P_1$ and $P_2$. Is there any meaningful way to assign a probability to the event that Chelsea beat Liverpool (in a home game for Chelsea)?

The problem I see s that $P_1$ and $P_2$ inherently encode information that we want to model. For example, the function $P_1$ models Chelsea playing an arbitrary opponent at home, while $P_2$ is a model for Liverpool playing an arbitrary opponent away. So while the underlying state-space and sigma algebras are equal mathematically, the probability function 'encodes' information to mean that the event $\{(1,2)\}$ in $C$ is not the same as $\{(1,2)\}$ in $L$, since in the former, it's the event that Chelsea scores one goal and another team score 2, while in the latter, it's the event that Liverpool score one goal and another team score 2. So while mathematically they are the same objects, the interpretation is different.

How do we reconcile this difference of interpretations? It seems to me that due to this issue of interpretation, the state spaces fundamentally look different, so I am not sure how to define the event that is "Chelsea beats Liverpool"?

答:
I extend the comment that I wrote here. I will not give an answer to your question, but this will help you to find an answer yourself (that maybe an answer to the original question does not even exist).

So to extend the comment that I wrote, if in some sense $X$ is the random variable that corresponds to the score of Chelsea and $Y$ the score of Liverpool, then you are basically after
$$\mathbb P(X>Y),\ \ \ .... \text{right?!}$$
Since $\{X>Y\}$ is nothing else but
$$\{\omega\in \Omega\ : \ X(\omega)>Y(\omega)\} $$
these two random variables must live in the same probability space.

So assume you got that right, you defined your probability space and all.
If you know joint law and marginal law, then I hope you see that the statement
$$\text{"Liverpool wins 20% at home"}$$
does not say much about the joint law. While the one that you are interested in requires the joint law. Do we have the joint law? As long as the answer is no, then you cannot answer the question. Why? See it as follows: consider a case where the Premier League is so unpopular that there are only three teams playing, Chelsea, Liverpool and Arsenal. Since you want to practice how to do this rigorously, I leave this exercise for you:

Exercise. Construct a probability space where Liverpool has probability $\frac 1 5$ to win in an arbitrary match, Chelsea has probability $\frac 4 5 $ to win an arbitrary match, and Chelsea wins from Liverpool with probability $x$ (you choose $x$ yourself from the infinite amount of choices).

Sorry, not every story has a happy ending.

PS: oh, also if you ever encounter occasions where the state space is countable, then taking the power set as the $\sigma$-algebra may not be a bad idea.


问:
A sequence of continuous functions and integration.
Let $a, b \in \mathbb{R}$ such that $a<b$. For each $n \in \mathbb{N}$, let $f_n : [a,b] \subset \mathbb{R} \longrightarrow \mathbb{R}_+$ be a continuous functions. Since, for each $n \in \mathbb{N}$, the function $f_n$ is continuous, then $f_n$ is integrable. Hence, there exists a constante $M_n >0$ (depending of $n \in \mathbb{N}$) such that
$$\int_{a}^{b} f_n(x) \; dx < M_n.\tag{1}$$

Question. There exists $M>0$ (not depending of $n \in \mathbb{N})$ so that
$$\int_{a}^{b} f_n(x) \; dx < M, \; \forall \;  n \in \mathbb{N}? \tag{2}$$

It seems to me unlikely that happen, since we do not have more properties of functions $ f_n $, with $ n \in \mathbb{N} $, for instante, $ f_n $ is not necessarily bounded.

答:
What about $f_n(x)=n$ for every $x \in [a,b]$ ?

(Note also that your last sentence is not correct : as continuous functions over $[a,b]$, all the $f_n$'s are bounded)