问:
QR decomposition in Square Root Kalman Filters - How is root(Q) simply given?
Looking at this in the context of Square Root Kalman Filters (SRKF).

The covariance matrices of the estimate P and the measurement P(zz) must be reduced to their square root form S and S(zz), respectively.
This can be achieved through QR decomposition; however, when I see the QR decomposition, the formulas for S and S(zz) are given with the square root of the process noise Q and measurement noise R.

QR decomposition with root(Q) and root(R) given

I believed the point of QR decomposition was to avoid roots of matrices; how would I then go and find the root(Q) and root(R)? Another QR decomposition?

答:
A problem with the Kalman filter (KF) is numerical stability, which is due machine precision. For example, the covariance matrix is always positive semi-definite ignoring machine precision; however, due to  round off error, the covariance matrix may be indefinite. This is the point of the square root square root Kalman filter (SRKF).

The SRKF maintains the covariance matrix in the factored form $\mathbf{P} = \mathbf{S} \mathbf{S}^T$ where $\mathbf{P}$ is the covariance matrix. This form improves the stability of the Kalman filter because the square root form never allows the covariance matrix to have negative diagonal element or become asymmetric. Mathematically, this is simply because $\mathbf{P} = \mathbf{S}\mathbf{S}^T$.

To compute the matrix $\mathbf{S}$ from $\mathbf{P}$, you may use Cholesky decomposition, but the idea is that you propagate $\mathbf{S}$ instead of $\mathbf{P}$. The procedure of performing Cholesky decomposition does not improve the condition number of the matrix $\mathbf{P}$, but instead, the condition number is improved by maintaining $\mathbf{S}$ instead of $\mathbf{P}$. Therefore, you do not need to perform the decomposition at each time step, and furthermore, a true SRKF maintains $\mathbf{S}$ not $\mathbf{P}$.

Similarly, you can factor the matrices $\mathbf{Q}$ and $\mathbf{R}$ as well, and you do not need to perform the operation at each step unless these matrices are changing at each time step.


问:
Area of hemisphere without Integration?
I tried to find the curved surface area of a hemisphere without integration.
I found a glitch over there as I didn't found any mistake in my approach.
Anyone who has found please do answer.

I got a handle of bucket and I formed a hemisphere by rotating it along the circumference of a imaginary semicircle of same radius 'R'(just for an idea for the figure). I found the curved surface area to be $\pi R \cdot \pi R$=$\pi ^2R^2$.

But it is actually $2\pi R^2$.Figure for help...

答:
It can be a mistake in calculating the circumference of individual lines on the hemisphere.
Radius changes as we move outward. So, that can be avoided using Integration.</p>

ContentLicense: CC BY-SA 4.0

Title: Area of hemisphere without Integration?


问:
Poisson equation inside a ball $B(0, 1)$
Find the solution of the Poisson equation inside a ball $B(0, 1)$, i.e.:
$$\begin{cases} \Delta u=x^{2} \ \ \ in \ B(0,1)\\u=3 \ \ \ on \ \partial B(0,1)\end{cases}$$

($x^2=x\cdot x$, $x\in \mathbb{R}^{2}$)

I have the following formula for the solution of Poisson equation:
$$u(x)=-\int_{\partial\Omega}\frac{\partial G}{\partial n}(x,y) u(y) \ dS(y)-\int_{\Omega}G(x,y)\Delta u(y) \ dy$$

I know the formula for the Green function $G$ in this case.

I tried to evaluate these integrals with polar coordinates, but they turned out quite difficult...

答:
HINT: Try solving the equation in polar coordinates instead of that Green function integration. The heuristic is that the right hand side is unchanged with respect to the polar angle, i.e., spherically symmetric, the solution should inherit this property (This can be proved by the Green function integration formula you gave). Now in polar coordinates:
$$
\Delta u = \frac{1}{r} \frac{\partial u}{\partial r}
+ \frac{\partial^2 u}{\partial r^2}
+ \frac{1}{r^2} \frac{\partial^2 u}{\partial \theta^2},
$$
where $r = \sqrt{x^2+y^2}$. Now
$$
\frac{\partial^2 u}{\partial \theta^2} = 0
$$
because $u$ is constant on any circle $B(0,r)$, let $w(r) = w(\sqrt{x^2+y^2}) = u(x,y)$ then equation becomes:
$$
w'' + \frac{1}{r}w' -r^2 = 0,
$$
the boundary condition is:
$$
w(1) = 3.
$$
Let $v= w'$, and then try solving the first order ODE.  </p>

ContentLicense: CC BY-SA 3.0


问:
Fill blanks $\int_0^1 dx\int_{2x^2}^{x+1}f(y)dy=\int_0^1dy(...)+\int_{1}^2dy(...)$
How to fill the dots in $$\int_0^1 dx\int_{2x^2}^{x+1}f(y)dy=\int_0^1dy(...)+\int_{1}^2dy(...)?$$

Seems like the area of integration is the following, but I'm able to separate them, the hint says that the first should be $\sqrt{y/2}f(y)$

答:
The second integral when doing $x$ first has $\int_{y-1}^{\sqrt{y/2}}dx$ so the dots are $f(y)(\sqrt{y/2}-y+1)$

The first has a similar form with $0$ as the lower limit and the same upper limit, leading to the hinted value.


问:
ODE Non - Homogeneous
The equation is:
$$y''-2y'+1=0$$

I am trying to find the general solution and have tried a few different tactics. I can't use the char equation because it won't work. Does anyone have any ideas?

答:
If you do as Matthew Pilling suggests, you can see that there are only derivatives there.  So this means that you can set $w = y'$ and therefore $w' = y''$ and this becomes a first order equation.

$$ w' - 2w + 1 = 0 \\
w' = 2w - 1 \\
\frac{w'}{2w - 1} = 1 \\
\frac{dw}{2w - 1} = dx \\
\int \frac{dw}{2w - 1} = \int dx \\
\frac{1}{2}\ln(2w - 1) = x + C \\
\ln(2w - 1) = 2x + C \\
2w - 1 = Ce^{2x} \\
2y' - 1 = Ce^{2x} \\
2\,dy - dx = Ce^{2x}\,dx \\
2\int dy - \int dx = \int Ce^{2x}\,dx \\
2y - x = Ce^{2x} + D \\
y = Ce^{2x} + \frac{1}{2}x + D
$$
Reminder, C & D don't change with constant transformations, as they remain unknown constants.

Title: ODE Non - Homogeneous


问:
Why are chordal graphs always perfect graphs?
Chordal graphs are always perfect graphs, but I am not sure exactly why. How does the chord make sure that the graph cant be imperfect?</p>

答:
This is due to the Strong perfect graph theorem.

A hole is an induced cycle of length at least four. Its complement is called antihole. We call an hole odd if it has an odd number of vertices, and even otherwise.

Now, the Strong perfect graph theorem states, that a graph is perfect if and only if it contains no odd hole and no odd antihole.

Therefore, it is sufficient to show that a chordal graph $G$ contains no odd hole or odd antihole as an induced subgraph. $G$ clearly contains no odd hole, since the largest cycle in $G$ is a triangle. Suppose that $G$ contains an odd anti-hole. First observe, that the complement of $C_5$ (the odd hole of length $5$) is isomorphic to $C_5$. If the anti-hole in $G$ is of order $5$, then $C_5$ is an induced subgraph in the complement of $G$, but since $C_5=C_5^C$, $C_5$ is an induced subgraph in $G$. A contradiction.

Suppose, the odd anti-hole has order $2k+5,\ k \in \mathbb{N}$. It follows that the path $P_5$ of order $5$ is an induced subgraph in $G^C$, but $P_5^C$ is not chordal (it contains a $C_4$). A contradiction.


问:
Discrete metric space proof
Suppose that $A \neq \emptyset$ is a set. Let $d: A \times A \rightarrow \mathbb{R}$ be a metric defined as follows:

For $a, b \in A$,  $d(a, b) = \left\{
  \begin{array}{lr}
      1 & \mbox{if } a \neq b \\
      0 & \mbox{if } a = b
      \end{array}
\right.$

Determine the closed and open subsets of this metric space $(A, d)$.

I am new to metric spaces and have been trying for hours but I cannot figure out this problem. Any assistance is much appreciated.

答:
Take any $x \in A$ and consider the open ball of center $x$ and radius $1/2$, i.e.,  $B(x, 1/2)$. By the definition of the metric $d$, $B(0, 1/2)=\{x\}$. Therefore, every point is an open ball, and every subset of $A$ is open (why?).

From this fact, is easy to find the closed sets.

Title: Discrete metric space proof


问:
Please prove that $n\cos n^2 x$ doesn't converge as $n\to\infty$ for any $x\in\mathbb{R}$.
I am reading "An Introduction to Analysis vol. 2" (in Japanese) by Kazuo Matsuzaka.

There is the following example in this book:

Let $f_n(x) := \frac{\sin n^2 x}{n}$ for $n = 1,2,\dots$.

Then $f(x):=\lim_{n\to\infty} f_n(x)=0$, so $f'(x) = 0$.

On the other hand, $f_n^{'}(x) = n \cos n^2 x$ and $f_n^{'}(x)$ doesn't converge as $n\to\infty$.

So, $f_n\to f$ and, $f_n$ and $f$ are differentiable, but $f_n^{'}\to f^{'}$ doesn't hold.

The author wrote $n\cos n^2 x$ doesn't converge as $n\to\infty$ for any $x\in\mathbb{R}$.

But I cannot prove that $n\cos n^2x$ doesn't converge as $n\to\infty$ even for $x=1$.

Please prove that $n\cos n^2 x$ doesn't converge as $n\to\infty$ for any $x\in\mathbb{R}$.

答:
Hint

For any large positive integer $N$, you should be able to find integers $r,s$ such that $r > N$ and $s > N$ and such that $xr^2$ is reasonably close to $\pi \pmod{2\pi}$, while $xs^2$ is reasonably close to $0 \pmod{2\pi}.$


问:
How do you find $E(X^3)$ of a Poisson Distribution?
I know the proof to find the variance of a Poisson Distribution, and I tried to use that to find $E(X^3)$, but I can't get it to work. Any help would be great!!

答:
$E(X^3)=e^{-\lambda}\sum_0^\infty \frac{k^3 \lambda^k}{k!}$.

$k^3=k(k-1)(k-2)+3k(k-1)+k$

so the sum can be changed to three sums

$\lambda^3\sum_3^\infty \frac{\lambda^{k-3}}{(k-3)!}+ 3\lambda^2\sum_2^\infty \frac{\lambda^{k-2}}{(k-2)!}+\lambda\sum_1^\infty \frac{\lambda^{k-1}}{(k-1)!}$

Multiply by $e^{-\lambda}$ and sum to get $\lambda^3+3\lambda^2+\lambda$.


问:
On the asymptotic bound for $\arg\zeta(s)$ on the critical line
I am currently trying to prove

$$
N(T)={T\over2\pi}\log{T\over2\pi}-{T\over2\pi}+\mathcal O(\log T)
$$

in which $N(T)$ denotes the number of $\zeta$'s nontrivial zeros with imaginary part between $(0,T]$. Currently, using symmetric properties of $\xi(s)$, I am able to obtain

$$
N(T)={T\over2\pi}\log{T\over2\pi}-{T\over2\pi}+\frac78+\frac1\pi\arg\zeta\left(\frac12+iT\right)+\mathcal O\left(\frac1T\right)
$$

Apparently, the remaining job is to show that the argument of $\zeta$ on the critical line is of logarithmic growth, and I become stuck on interpreting the meaning of $\arg\zeta$. According to H. M. Edwards' Riemann's zeta function, this argument is bounded by the number of zeros of $\Re\zeta(s)$ on a certain curve (section 6.7 of his book), and I wonder if anybody could provide a more intuitive and clear explanation on that. Thank you!

答:
Contour integral over three line segments

It can be shown that

$$
\frac1\pi\arg\zeta\left(\frac12+iT\right)={1\over2\pi i}\int_\mathcal L{\zeta'\over\zeta}(s)\mathrm ds
$$

in which $\mathcal L$ represents line segments $1/2-iT\to2-iT\to2+iT\to1/2+iT$. Since $|\log\zeta(2+it)|\le\log\zeta(2)$ for all $t\in\mathbb R$, we see that the integral on the line segment $2-iT\to2+iT$ is bounded. Consequently, all we need is to estimate the upper bound for integrals on lines segments $1/2-iT\to2-iT$ and $2+iT\to1/2+iT$. Because the procedures are so similar, we here only present the proof for line segment $2+iT\to1/2+iT$.

Meromorphic expansion of $\zeta'/\zeta$

Since we know little information about the location of $\zeta$'s nontrivial zeros in the critical strip, we consider meromorphically expanding $\zeta'/\zeta$ near $2+iT$. Let $S=\{\rho\in\mathbb C:\zeta(\rho)=0\wedge|\rho-2-iT|\le2\}$, then define

$$
g(z)={\zeta(z+2+iT)\over\zeta(2+iT)}\prod_{\rho\in S}\left(1-{z\over\rho-2-iT}\right)^{-1}
$$

Because $|1-z/(\rho-2-iT)|\ge1$ on $|z|=4$, it follows from maximum modulus principle that if we set

$$
M=\max_{|z|\le4}\left|\zeta(z+2+iT)\over\zeta(2+iT)\right|
$$

then $\log|g(z)|\le\log M$. By Borel-Caretheodory lemma, we see that

$$
{g'\over g}(z)\ll\log M
$$

for all $|z|\le3/2$. Now, all we need is to find an explicit estimate of $M$:

Growth of $\zeta$ and $\zeta'/\zeta$

According to The theory of Riemann zeta-function by E.C. Titchmarsh, we know that there exists a fixed positive constant such that $\zeta(\sigma+it)\ll|t|^A$ for large $|t|$ for all $\sigma\in\mathbb R$. This indicates

$$\log|\zeta(\sigma+it)|\ll\log|t|$$

Moreover because

$$
|\zeta^3(\sigma)\zeta^4(\sigma+it)\zeta(\sigma+2it)|\ge1
$$

we see that

$$
-\log|\zeta(\sigma+it)|\ll\log|t|
$$

for all $\sigma>1$. Consequently we have $\log M\ll\log T$. This, combining with the definition of $g$, indicates that

$$
{\zeta'\over\zeta}=\sum_{\rho\in S}{1\over s-\rho}+\mathcal O(\log T)
$$

when $s$ is on the line segment $[1/2+iT,2+iT]$.

Final shot: the integral is $\mathcal O(\log T)$

$$
\begin{aligned}
\int_{2+iT}^{1/2+iT}{\zeta'\over\zeta}(s)\mathrm ds
&=\sum_{\rho\in S}\int_{2+iT}^{1/2+iT}{\mathrm ds\over s-\rho}+\mathcal O(\log T) \\
&\ll\sum_{\rho\in S}1+\mathcal O(\log T)
\end{aligned}
$$

To estimate the sum, we apply Jensen's inequality:

$$
\sum_{\rho\in S}1\le{\log M\over4/2}\ll\log T
$$

Therefore, the expansion of Riemann-von Mangoldt formula becomes

$$
N(T)={T\over2\pi}\log{T\over2\pi}-{T\over2\pi}+\mathcal O(\log T)
$$


问:
What does the notation $\alpha = \inf \{i > 0 : X_i > X_0 \}$ mean, where the $\{X_i\}_0^{\infty}$ are IID continuous?
I don't think I've seen this notation before $\alpha = \inf \{i > 0 : X_i > X_0 \}$. What does the colon here mean? What does this expression mean in words?

It seems we're assigning the variable $\alpha$ to the RHS, where the RHS is the infimum of a set of random variables $X_1, \ldots, X_{\infty}$ are IID continuous?

答:
For each $\omega\in\Omega$, $\alpha(\omega)$ is the minimum index $i$ such that $X_i(\omega)>X_0(\omega)$.

For example: If $X_0(\omega_0)=2$ and $X_1(\omega_0)=0.4$, $X_2(\omega_0)=1.3$, $X_3(\omega_0)=2.5$,... Then $\alpha(\omega_0)=3$.

Note: That is wrong defined. $X_i>X_0$ must be changed to $X_i\geq X_0$ (because can be the empty set with $>$)


问:
Integration with unbounded spectral weight
I want to consider the solution of an equation in the space $L^2(e^{-2\alpha x} \mathbb{R}. )$ I am just wondering how to implement the weight when computing the norm.

So far, I use Hermite spectral method and I work out my weights by quadrature Q. I compute norm of a variable, q,in $L^2 (\mathbb{R}) $ space by simply doing $q^TQq$, and I’ve tried adjusting this by letting $q’ = e{-\alpha x$’ and then doing the quadrature but so far nothing is really working...

Kindest regards,

Catherine

答: With a change of variable $y = 2 \alpha x$, and depending upon the details of what you’re doing, the Gauss-Laguerre quadrature rules (see for example Stroud and Secrest) might prove fruitful.


问:
Finding the integral of $\frac{1}{x(x^n+a^n)}$
I was working on the problems in Mathematical Methods for Physics and Engineering by Riley,Hobson & Bence.
In Problem 2.34 (d) I'm supposed to find this integral: $$J=\int\frac{dx}{x(x^n+a^n)}.$$
I used partial fractions and arrived at the form
$$J=\frac{1}{a^n}\left[\log x-\int \frac{dx}{x^n+a^n}\right]$$
and now I'm stuck, I don't know how to integrate $1/(x^n+a^n)$.

答:
Alternatively, perform a u-sub. Let $x=1/t$. What will happen is that
(after little algebraic simplification) you get a monomial numerator which is one degree lower than a binomial denominator consisting of an $x$ term and a constant. Integration: A basic $ln$ term. It's very easy. Try it out...


问:
Finding Pi variables from matrix. From PageRank Algorithm.
$$\pmatrix{\pi_1 & \pi_2 & \pi_3} = \pmatrix{\pi_1 & \pi_2 & \pi_3}\pmatrix{\frac{1}{6} & \frac{4}{6} & \frac{1}{6} \\ \frac{5}{12} & \frac{2}{12} & \frac{5}{12} \\ \frac{1}{6} & \frac{4}{6} & \frac{1}{6} \\}$$

Answer:  $$\pi_1=\frac{5}{18}$$ $$\pi_2=\frac{8}{18}$$ $$\pi_3=\frac{5}{18}$$

Could someone please explain how to find these variables.

答:
The key terms are the Perron-Frobenius theorem and the stable state of a Markov chain. In practice, this vector $\pi$ is found as the eigenvector for the eigenvalue $1$, and is normalized so that the sum of its components is $1$.

Subtract $1$ along the diagonal to get
$$B=\pmatrix{-\frac{5}{6} & \frac{4}{6} & \frac{1}{6} \\ \frac{5}{12} & -\frac{10}{12} & \frac{5}{12} \\ \frac{1}{6} & \frac{4}{6} & -\frac{5}{6} }$$
Since we multiply by the matrix on the right, the desired vector is in the kernel of $B^T$. The Scilab command v=kernel(B')' outputs

```<code>v  =    0.4682929    0.7492686    0.4682929
</code>```

This is normalized to have length $1$, but we want the sum of components to be $1$. The command v/sum(v) does the job:

```<code>0.2777778    0.4444444    0.2777778
</code>```

which is your vector. Of course, with sufficient patience one can do it by hand as well.

Body:
The key terms are the Perron-Frobenius theorem and the stable state of a Markov chain. In practice, this vector $\pi$ is found as the eigenvector for the eigenvalue $1$, and is normalized so that the sum of its components is $1$.

Subtract $1$ along the diagonal to get
$$B=\pmatrix{-\frac{5}{6} & \frac{4}{6} & \frac{1}{6} \\ \frac{5}{12} & -\frac{10}{12} & \frac{5}{12} \\ \frac{1}{6} & \frac{4}{6} & -\frac{5}{6} }$$
Since we multiply by the matrix on the right, the desired vector is in the kernel of $B^T$. The Scilab command v=kernel(B')' outputs

v  =    0.4682929    0.7492686    0.4682929

This is normalized to have length $1$, but we want the sum of components to be $1$. The command v/sum(v) does the job:

0.2777778    0.4444444    0.2777778

which is your vector. Of course, with sufficient patience one can do it by hand as well.


问:
Is molarity a mathematical measure?
Molarity can simply be defined as

$$\text{c} \triangleq \frac{n}{V}$$

where $c$ is the molarity, $n$ is the number of moles of a substance of interest (theoretically a counting measure because a given volume contains a finite number of atoms, but concentrations are rarely estimated by 'counting' per se) divided by the volume of the system under consideration (possibly a Lebesgue measure for our purposes).

Holding the volume to be a non-zero constant that we might define as $\alpha \triangleq\frac{1}{V}$ gives us that $c = \alpha n$. Amir Dembo's Stanford University notes on probability theory tell me that a linear combination of measures whose coefficients $\alpha_i \geq 0 \forall i \cdots n$ is also a measure. The issue is that $V$ is not necessarily held to be a constant.

So, after accounting for $V$ being a variable, is the molarity a measure?

答:
I guess you are proposing that molarity as a set function would act on subsets of "physical space".

Moles are a direct count of the number of the chemical species, and would be the closest to a measure. But a molarity which is a measure(in the physics sense) of concentration is not additive;  two beakers with 1 litre each of identical fluids, e.g. at $1\ \mathrm c$ when combined (unioned?) is still $1\ \mathrm c$, instead of the $2\ \mathrm c$ demanded by additivity.

Since a measure (in the sense of measure theory) needs to be additive (in fact countably additive), this isn't a measure.

Title: Is molarity a mathematical measure?


答:
I'm not sure how rigorous you need to be, but it's a necessary condition for this sequence to converge that $\cos(n^2 x) \longrightarrow 0$. This is impossible.

To see why, consider that if $k^2x \approx (2m +1) \pi$ and
$(k+1)^2x \approx (2n +1) \pi$, then $(2k+1)x \approx 2j \pi$ for some integer $j$. It follows from this that $2jk^2 \approx (2k+1)(2n+1)$, but the left hand side is even, and the right hand side is odd. So, it's impossible.

Body:
I am reading "An Introduction to Analysis vol. 2" (in Japanese) by Kazuo Matsuzaka.

There is the following example in this book:

Let $f_n(x) := \frac{\sin n^2 x}{n}$ for $n = 1,2,\dots$.

Then $f(x):=\lim_{n\to\infty} f_n(x)=0$, so $f'(x) = 0$.

On the other hand, $f_n^{'}(x) = n \cos n^2 x$ and $f_n^{'}(x)$ doesn't converge as $n\to\infty$.

So, $f_n\to f$ and, $f_n$ and $f$ are differentiable, but $f_n^{'}\to f^{'}$ doesn't hold.

The author wrote $n\cos n^2 x$ doesn't converge as $n\to\infty$ for any $x\in\mathbb{R}$.

But I cannot prove that $n\cos n^2x$ doesn't converge as $n\to\infty$ even for $x=1$.

Please prove that $n\cos n^2 x$ doesn't converge as $n\to\infty$ for any $x\in\mathbb{R}$.


Title: Please prove that $n\cos n^2 x$ doesn't converge as $n\to\infty$ for any $x\in\mathbb{R}$.

Body: The sequence $\{n \cos n^2 \pi\}_{n=0}^\infty$ equals $\{0, 1, -4, 9, -16, 25\dots\}$


问:
Given $\int_{\frac13}^{\frac23}f(x)dx=0$, how to prove $4860(\int_0^1f(x)dx)^2\le 11\int_0^1|f''(x)|^2dx$?

Suppose $f\in C^2[0,1]$, and
  $\int_{\frac13}^{\frac23}f(x)dx=0$.
  Prove  that
  $$\left(\int_0^1f(x)dx\right)^2\le \frac{11}{4860}\int_0^1|f''(x)|^2dx.$$

This problem is quite similar to Prove the following integral inequality: $\int_{0}^{1}(f''(x))^2dx\ge 1920\left(\int_{0}^{1}f(x)dx\right)^2$.
I have tried to write
$$\int_0^1f(x)dx=\int_0^{\frac13}f(x)dx+\lambda \int_{\frac13}^{\frac23}f(x)dx+\int_{\frac23}^{1}f(x)dx$$ for any $\lambda \in \mathbb{R}$, and pick a suitable $g$ such that
$$\int_0^1f(x)dx=\int_0^1g(x)f''(x)dx$$
then we can use Cauchy-Schwarz inequality to get what we want. How can I get the function $g$?

Body:
By integration by parts, we have
$$
\begin{aligned}
\int_0^{\frac{1}{3}}g_1(x)f''(x)\mathrm{d}x
&=g_1(x)f'(x)\bigg|_0^{\frac{1}{3}}-\int_0^{\frac{1}{3}}g_1'(x)f'(x)\mathrm{d}x\\
&=g_1(x)f'(x)\bigg|_0^{\frac{1}{3}}-g_1'(x)f(x)\bigg|_0^{\frac{1}{3}}+\int_0^{\frac{1}{2}}g_1''(x)f(x)\mathrm{d}x
\end{aligned}\tag{1}
$$
$$
\begin{aligned}
\int_{\frac{1}{3}}^{\frac{2}{3}}g_2(x)f''(x)\mathrm{d}x
&=g_2(x)f'(x)\bigg|_{\frac{1}{3}}^{\frac{2}{3}}-\int_{\frac{1}{3}}^{\frac{2}{3}}g_2'(x)f'(x)\mathrm{d}x\\
&=g_2(x)f'(x)\bigg|_{\frac{1}{3}}^{\frac{2}{3}}-g_2'(x)f(x)\bigg|_{\frac{1}{3}}^{\frac{2}{3}}+\int_{\frac{1}{3}}^{\frac{2}{3}}g_2''(x)f(x)\mathrm{d}x
\end{aligned}\tag{2}
$$
$$
\begin{aligned}
\int_{\frac{2}{3}}^1g_3(x)f''(x)\mathrm{d}x
&=g_3(x)f'(x)\bigg|_{\frac{2}{3}}^1-\int_{\frac{2}{3}}^1g_3'(x)f'(x)\mathrm{d}x\\
&=g_3(x)f'(x)\bigg|_{\frac{2}{3}}^1-g_3'(x)f(x)\bigg|_{\frac{2}{3}}^1+\int_{\frac{2}{3}}^1g_3''(x)f(x)\mathrm{d}x
\end{aligned}\tag{3}
$$
Let $g(x)$ be the piecewise differentiable function, $g_1(x)$, $g_2(x)$ and $g_3(x)$ is polynomial.
$$g(x)=\begin{cases}
g_1(x),\quad x\in[0,\tfrac{1}{3})\\
g_2(x),\quad x\in[\tfrac{1}{3},\tfrac{2}{3})\\
g_3(x),\quad x\in[\tfrac{2}{3},1]
\end{cases}$$
Since $\textstyle\int_{\frac{1}{3}}^{\frac{2}{3}}f(x)\mathrm{d}x=0$, Let $g_1''(x)=g_3''(x)=c$, $g_2''(x)=k$, So $g_1(x),g_2(x),g_3(x)$ is quadratic polynomial.
$$\begin{cases}
g_1(x)=cx^2+a_1x+a_0,& x\in[0,\tfrac{1}{3})\\
g_2(x)=kx^2+b_1x+b_0,& x\in[\tfrac{1}{3},\tfrac{2}{3})\\
g_3(x)=cx^2+c_1x+c_0,& x\in[\tfrac{2}{3},1]
\end{cases}$$
Let,
$$g_1(x)f'(x)\bigg|_0^{\frac{1}{3}}+g_3(x)f'(x)\bigg|_{\frac{2}{3}}^1+g_2(x)f'(x)\bigg|_{\frac{1}{3}}^{\frac{2}{3}}=0$$
we get,
$$\tag{4}
\begin{cases}
g(0)=g(1)=0\\
g_1(\tfrac{1}{3})-g_2(\tfrac{1}{3})=0\\
g_2(\tfrac{2}{3})-g_3(\tfrac{2}{3})=0
\end{cases}\Rightarrow\quad \begin{cases}
a_0=c+c_1+c_0=0\\
\tfrac{1}{9}c+\tfrac{1}{3}a_1+a_0=\tfrac{1}{9}k+\tfrac{1}{3}b_1+b_0\\
\tfrac{4}{9}k+\tfrac{2}{3}b_1+b_0=\tfrac{4}{9}c+\tfrac{2}{3}c_1+c_0
\end{cases}
$$
Let,
$$g_1'(x)f(x)\bigg|_0^{\frac{1}{3}}+g_3'(x)f(x)\bigg|_{\frac{2}{3}}^1+g_2'(x)f(x)\bigg|_{\frac{1}{3}}^{\frac{2}{3}}=0$$
we get,
$$\tag{5}
\begin{cases}
g'(0)=g'(1)=0\\
g_1'(\tfrac{1}{3})-g_2'(\tfrac{1}{3})=0\\
g_2'(\tfrac{2}{3})-g_3'(\tfrac{2}{3})=0
\end{cases}\Rightarrow\quad
\begin{cases}
a_1=2c+c_1\\
\tfrac{2}{3}c+a_1=\tfrac{2}{3}k+b_1\\
\tfrac{4}{3}k+b_1=\tfrac{4}{3}c+c_1
\end{cases}
$$
by (4),(5),  it follows that
$$a_0=0,a_1=0,k=-2c,b_1=2c,b_0=-\frac{1}{3}c,c_0=c,c_1=-2c$$
Hence,
$$g(x)=\begin{cases}
g_1(x)=cx^2,& x\in[0,\tfrac{1}{3})\\
g_2(x)=-2cx^2+2cx-\frac{1}{3}c,& x\in[\tfrac{1}{3},\tfrac{2}{3})\\
g_3(x)=cx^2-2cx+c,& x\in[\tfrac{2}{3},1]
\end{cases}$$
by $(1)+(2)+(3)$, we have
\begin{align*}
\int_0^1g(x)f''(x)\mathrm{d}x
&=\int_0^{\frac{1}{3}}g_1(x)f''(x)\mathrm{d}x+\int_{\frac{1}{3}}^{\frac{2}{3}}g_2(x)f''(x)\mathrm{d}x+\int_{\frac{2}{3}}^1g_3(x)f''(x)\mathrm{d}x\\
&=\int_0^{\frac{1}{2}}g_1''(x)f(x)\mathrm{d}x+\underbrace{\int_{\frac{1}{3}}^{\frac{2}{3}}g''_2(x)f(x)\mathrm{d}x}_{=0}+\int_{\frac{2}{3}}^1g''_3(x)f(x)\mathrm{d}x\\
&=2c\int_0^1f(x)\mathrm{d}x
\end{align*}
By Cauchy-Schwarz inequality,
\begin{align*}
\left(2c\int_0^1f(x)\mathrm{d}x\right)^2&=\int_0^1g(x)f''(x)\mathrm{d}x\\
&\leqslant \int_0^1(g(x))^2\mathrm{d}x\int_0^1(f''(x))^2\mathrm{d}x
\end{align*}
After calculation, We finally get
$$\int_0^1(g(x))^2\mathrm{d}x=\frac{c^2}{1215}+\frac{c^2}{135}+\frac{c^2}{1215}=\frac{11c^2}{1215}$$
Therefore,
$$4860\left(\int_0^1f(x)\mathrm{d}x \right)^2\leqslant 11\int_0^1\big(f''(x)\big)^2\mathrm{d}x $$


问:
Slight generalisation of the Baire category theorem?
By the Baire category theorem, one cannot write a complete metric space $X$ as a countable union of closed nowhere dense subsets of $X$. Can this be generalised to say that there is no injection $f: X \hookrightarrow X$, $f(X) \subseteq \cup_{n \ge 1} X_n$, where the $X_n$ are closed and nowhere dense? It seems like not much of a jump, and that it should be true, but I can't quite get it out. Thanks for any insight.</p>

答: Since you don’t require continuity, there is an injection of $\Bbb R$ into the middle-thirds Cantor set $C$, which is closed and nowhere dense, because $|\Bbb R|=|C|=2^\omega=\mathfrak c$. If you insist on a union of countably infinitely many closed, nowhere dense sets, just inject each $[n,n+1)$ into a Cantor set in $\left[n,n+\frac12\right]$.


问:
Find the intersection point
For a Parabola   AB and CD are any two parallel chords having slope 1 .
C1 is a circle passing through O A B and C2 is a circle passing through O C D where O is the origin . C1 and C2 intersect at ?

We need to find the points where C1 and C2 intersect. I tried assuming a circle passing through OAB and using parametric from to generate a fourth degree equation. I was unable to proceed further . Can anyone give a hint ? I am in high school

答:
I found a quicker solution after getting the idea from intelligent pauca.
I also assumed a  = 1.

A $(t^2,2t)$ , B$((2-t)^2,2(2-t))$ are points of intersection of the lines with the parabola. (By equating slopes of two parametric points)

We assume a circle $x^2 + y^2 + cx + dy + c = 0$ passing through A and B . It intersects with the parabola $y^2=4x$ . Eliminating x we get the equation $y^4 + y^2(4c + 16) + 16dy = 0 $.
We observe that the sum of roots of this eqn = 0 . We aldready know three solutions to this equation.
$$ (0) + (2t) + (2(2-t)) + Y = 0$$ The fourth point is independent of t and has y - coordinate -4. So the point is (4,-4).
All circles passing through (0,0) and points of intersection of lines having slope 1 with parabola $y^2 = 4x$ must pass through point (4,-4) . Hence the two circles C1 and C2 must intersect at point (4,-4).

Title: Find the intersection point


问:
Solving functional equations including $f\circ f(x)$ functions
Find all $a\in\mathbb{R} $ for which there exists a function $f : \mathbb{R}\to\mathbb{R} $ , such that

(i) $f(f(x))=f(x)+x$, for all $x\in\mathbb{R} $,

(ii) $f(f(x)–x)=f(x)+ax$, for all $x\in\mathbb{R} $

Normally in such functional equations, I'd put different values of $x$ (like $x+1$, $x+2$ etc) and try to get an equation for $f(x)$. But here I'm not getting anywhere by that method. I can't use the coefficient comparison method either, because $f(x)$ may not be a polynomial.

答:
NOTE: $$f^{-1}f(x) = f(f^{-1}(x)) =x$$
GIVEN:

(i) $f(f(x))=f(x)+x$, for all $x$ $\in$ $ℝ$,

(ii) $f(f(x)–x)=f(x)+ax$, for all $x$ $\in$ $ℝ$

Substitute f(x) as x in equation (i). you can verify the equation below is equivalent to equation (i)  if you put $x = f(x)$ in the equation below.
$$f(x)= x+ f^{-1}(x)$$

Then in RHS of equation 2, substitute what we have above: $x = f(x)- f^{-1}(x)$:

$f(f(x)-(f(x)-f^{-1}(x)))=f(x)+ax \implies$
$x=f(x)+ax \implies f(x)=(1-a)x$

Substitute and solve... I got a = golden ratio and its conjugate...
Ask doubt if any below


问:
Does a group of order 24 have a sylow 11 subgroup
I am going through Algebra In Action by Shahriari and on page 147 it says

Definition 7.8 (Sylow subgroups). Let G be a finite group with |G| = $p^a$m, where
$p \nmid m$, and $a$ is a non-negative integer. A subgroup $P \le G$ with $|P| = p^a$ is called
a Sylow $p$-subgroup of $G$.
The set of all Sylow $p$-subgroups of $G$ is denoted by $\operatorname{Syl}_p(G)$.

Example 7.9. If $G$ is a group of order $24$, then a subgroup of order $3$ would be
one of its Sylow $3$-subgroups and a subgroup of order $8$ would be one of its Sylow
$2$-subgroups. The identity subgroup would be its Sylow $5$-subgroup as well as its
Sylow $11$-subgroup. While we know that $G$ has the identity subgroup, we need
Sylow’s theorem to know that $G$ actually must have a subgroup of order $3$ and a
subgroup of order $8$.

Why is the identity subgroup a $5$-subgroup and an $11$-subgroup??!!

EDIT $5^0 = \{$identity of $G\}$ is considered a sylow 5 group by this definition.

答:
$5$ and $11$ do not divide $24$, i.e. their highest powers which divide $24$ are $5^0$ and $11^0$. So the corresponding Sylow sub-groups have $5^0 = 11^0 = 1$ element. What's a subgroup with just one element? The identity subgroup!

Note that in the definition, $\alpha \ge 0$ and not just $\alpha > 0$. So, $\alpha = 0$ is perfectly okay to consider.


问:
How does a Kalman Filter with Constant Velocity estimate the velocity?
I'm gonna describe the constant velocity example (without acceleration) used in many textbooks.

My state is defined as 2D vector s = [x, v_x], where x describes position and v_x the velocity in x direction. Measurements are only the position. State transition matrix is:

```<code>H = 1 1
    0 1
</code>```

Assume fixed process noise covariance Q = diag(1) and measurement covariance R = diag(1). Now let's say my assumed initial state is s_0 = [0, 0] with covariance P_0 = diag(1), i.e. both position and velocity are 0.

Given the above setup, which part of the Kalman Filter algorithm updates the velocity? How is it possible that the velocity changes between iterations (i.e. after a prediction + update step)? I know for sure that in the prediction step, the state transition keeps constant velocity, so it cannot change there. Our measurements do not include velocity information, just position, so it shouldn't (?) happen in the update step either. So where then?

Sorry, this might be a stupid question. But I implemented it in python using the frame work filterpy (see below) and I cannot explain the behavior.

```<code>#!/usr/bin/python
from filterpy.kalman import KalmanFilter as KF
import numpy as np
import matplotlib.pyplot as plt

kf = KF(2,1)

kf.F = np.array([[1,1],    # State Transition Model
                 [0,1]])

kf.H = np.array([[1,0]])   # Measurement Model
kf.x = np.array([[0],[0]]) # Initial state
kf.P *= 1                  # State Uncertainty (diag)
kf.R *= 1                  # Measurement Noise (diag)
kf.Q *= 1                  # Process Noise     (diag)

T = range(10)
M = []
X = []

for t in T:
    kf.predict()

    m = t + np.random.randn()
    M.append(m)

    kf.update(m)
    X.append(kf.x[0,0])

plt.scatter(T,M,c='red',s=10)
plt.plot(T,X)
plt.show()
</code>```

Below are 3 iterations. I can see that the velocity DID change after the update step.

```<code>--------1. Iteration--------
predict
[[0]
 [0]]
update
[[0.49755657]
 [0.16585219]]                <--- velocity changed, how?
--------2. Iteration--------
predict
[[0.66340876]
 [0.16585219]]
update
[[-0.09533335]
 [-0.21351886]]
--------3. Iteration--------
predict
[[-0.30885221]
 [-0.21351886]]
update
[[1.27699875]
 [0.60554702]]
</code>```

答:
I understand your confusion, and I see this occur often when students are introduced to the Kalman filter (KF). If I understand your confusion, you do not understand why the velocity is changing because you only have position measurements. The simple answer is the position and velocity are correlated, so the velocity is updated indirectly from the position measurements.

I encourage you to work out the equations symbolically for the update step, and you will understand why this occurs. However, I will provide a more general example for linear systems, and I hope you will be able to try working this out for your problem.

Illustration:

Recall, the Kalman gain is given by
$$
\begin{align}
  K_t &= P_t^- H_t^T(H_t P_t^- H_t^T + R_t)^{-1}
\end{align}
$$
where $K_t$ is the Kalman gain, $P_t^-$ is the covariance matrix before the measurement, and $H_t$ is the measurement model, and the updated state estimate is given by
$$
\begin{align}
  x_t^+ &= x^-_t + K_t(z_t - H_tx_t^-)
\end{align}
$$
where $x_t^-$ is the state estimate before the measurement, $x_t^+$ is the state estimate after the measurement, and $z_t$ is the measurement.
Note, the $x_t \in \mathbb{R}^n$ is an a vector, $z_t \in \mathbb{R}^m$ is a vector, and $P_t \in \mathbb{R}^{n \times n}$ is a matrix with rows and columns equal to the number of states. The observation model $H_t \in \mathbb{R}^{m \times n}$ is dependent on the length of the measurement vector.

Let us look at the Kalman gain in more detail. First, let $S_t = (H_t P_t^- H_t^T + R_t)$ where $S_t$ is the covariance of the innovation. In other words, $S_t$ is the covariance of $z_t - H_t x_t^-$, so rewriting $K_t$, we have

$$K_t = P_t^- H_t^TS_t^{-1}$$

where $S_t \in \mathbb{R}^{m \times m}$ is a square matrix with rows and columns each equal to the length of the measurement vector.  Based on this, we know that $K_t \in \mathbb{R}^{n \times m}$ without working out the equations.

Let $p_i$ be the $i$th row of $P$, and let $h_i$ be the $i$th column of $H^T$. Now,

$$
P_t^- H_t^T =
\begin{bmatrix}
p_0 h_0  & \cdots & p_0 h_m\\
\vdots & \ddots & \vdots \\
\vdots & \ddots & \vdots \\
p_n h_0 & \cdots & p_n h_m
\end{bmatrix}
$$

Now, if we let $s_{ij}$ represent the element of $S_t$ at the $i$th row and $j$th column, then the Kalman gain is given by
$$
K_t =
\begin{bmatrix}
p_0 h_0  & \cdots & p_0 h_m\\
\vdots & \ddots & \vdots \\
\vdots & \ddots & \vdots \\
p_n h_0 & \cdots & p_n h_m
\end{bmatrix}
\begin{bmatrix}
s_{00}  & \cdots & s_{0m} \\
\vdots & \ddots & \vdots \\
s_{m0} & \cdots & s_{mm}
\end{bmatrix}
$$

Now, if we go back and look at $x_t^+$, we can see that potentially, states are updated even if we don't directly measure those states because $P_t^- H_t^T$ projects the innovation covariance on the state space (i.e., the correlations are being considered in the Kalman gain). This may not always be the case, but if variable are correlated, then this will likely occur. As I mentioned, I encourage you to work this out for your system to gain a deeper understanding.


问:
How to proof that $H^{1/2}XH^{1/2}=\left(\frac{2\lambda_i^{1/2}\lambda_j^{1/2}}{\lambda_i +\lambda_j} \right)\circ\left( \frac{1}{2}(HX+XH) \right)$

I've got this from Hiai and Kosaki paper, could someone explain to me how to derive $H^{1/2}XH^{1/2}=\left(\frac{2\lambda_i^{1/2}\lambda_j^{1/2}}{\lambda_i +\lambda_j}  \right)\circ\left( \frac{1}{2}(HX+XH) \right)$ (red-boxed)

Also, what the meaning of "thanks to the standard $2\times 2$ matrix trick, one can reduce a proof to the case $H=K.$" What trick is this?

Any help would be appreciated.

Thanks so much

答:
The trick is probably
$$
\begin{bmatrix} H&0\\0&K\end{bmatrix}^{1/2}
\begin{bmatrix} 0&X\\0&0\end{bmatrix}
\begin{bmatrix} H&0\\0&K\end{bmatrix}^{1/2},
$$
since its norm is $\|H^{1/2}XK^{1/2}\|$ and
$$
\begin{bmatrix} H&0\\0&K\end{bmatrix}
\begin{bmatrix} 0&X\\0&0\end{bmatrix}+
\begin{bmatrix} 0&X\\0&0\end{bmatrix}\begin{bmatrix} H&0\\0&K\end{bmatrix}
=\begin{bmatrix}0&HX+XK\\0&0\end{bmatrix}.
$$
As for the fomula, doing the $2\times 2$ case to avoid dots,
\begin{align}
H^{1/2}XH^{1/2}
&=\begin{bmatrix} \lambda_1^{1/2}&0\\0&\lambda_2^{1/2}\end{bmatrix}
\begin{bmatrix} x&y\\ z&w\end{bmatrix}
\begin{bmatrix} \lambda_1^{1/2}&0\\0&\lambda_2^{1/2}\end{bmatrix}
=\begin{bmatrix} \lambda_1 x&\lambda_1^{1/2}\lambda_2^{1/2}y\\\lambda_1^{1/2}\lambda_2^{1/2} z&\lambda_2 w\end{bmatrix}\\[0.3cm]
&=\begin{bmatrix} \lambda_1& \lambda_1^{1/2}\lambda_2^{1/2}\\ \lambda_1^{1/2}\lambda_2^{1/2}&\lambda_2\end{bmatrix} \circ X,
\end{align}

and
\begin{align}
XH+HX&=\begin{bmatrix} \lambda_1 x&\lambda_1 y\\\lambda_2 z&\lambda_2w\end{bmatrix}
+\begin{bmatrix} \lambda_1 x&\lambda_2 y\\\lambda_1 z&\lambda_2w\end{bmatrix}
=\begin{bmatrix} 2\lambda_1 x&(\lambda_1+\lambda_2) y\\(\lambda_1+\lambda_2 )z&2\lambda_2w\end{bmatrix}\\[0.3cm]
&=\begin{bmatrix} 2\lambda_1&\lambda_1+\lambda_2\\\lambda_1+\lambda_2&2\lambda_2\end{bmatrix}\circ X.
\end{align}
So, denoting the Hadamard inverse of $A$ by $A^\#$,
\begin{align}
H^{1/2}XH^{1/2}
&=\begin{bmatrix} \lambda_1& \lambda_1^{1/2}\lambda_2^{1/2}\\ \lambda_1^{1/2}\lambda_2^{1/2}&\lambda_2\end{bmatrix}\circ \begin{bmatrix}2\lambda_1&\lambda_1+\lambda_2\\\lambda_1+\lambda_2&2\lambda_2\end{bmatrix}^\#\circ (HX+XH)\\[0.3cm]
&=\begin{bmatrix} \lambda_1^{1/2}\lambda_1^{1/2}& \lambda_1^{1/2}\lambda_2^{1/2}\\ \lambda_1^{1/2}\lambda_2^{1/2}&\lambda_2^{1/2}\lambda_2^{1/2}\end{bmatrix}\circ \begin{bmatrix}\lambda_1+\lambda_1&\lambda_1+\lambda_2\\\lambda_1+\lambda_2&\lambda_2+\lambda_2\end{bmatrix}^\#\circ (HX+XH)\\[0.3cm]
&=\begin{bmatrix} \frac{\lambda_1^{1/2}\lambda_1^{1/2}}{\lambda_1+\lambda_1}& \frac{\lambda_1^{1/2}\lambda_2^{1/2}}{\lambda_1+\lambda_2}\\ \frac{\lambda_1^{1/2}\lambda_2^{1/2}}{\lambda_1+\lambda_2}&\frac{\lambda_2^{1/2}\lambda_2^{1/2}}{\lambda_2+\lambda_2}\end{bmatrix}\circ (HX+XH)
\end{align}


问:
$2^{q-1}\equiv 1\pmod{q}.$

The question is asking to show that $q$ must be prime given
$$ 2^{q-1}\equiv 1\pmod{q}. $$

答:
If the order of the prime (ie its period) is $p-1$ or half of that, it means that power of 2 up to $2^{p-1}$ is a different value, mod $p$, and hence there are $p-1$ numbers between $0$ and $p$ that share no common factor with $p$.  Thence $p$ is a prime.

The case for numbers like $31$ and $73$ is that the order divides $p-1$, rather than being it.  This means that the powers of $2$ do not cycle through all of the roots of $p$, and hence one can not demonstrate that $p$ is coprime to all $0<n<p$.

In the case of $341$, the indicated period is $10$, which means that it can at most visit $10$ of the available co-primes to $341$.  In the main, the use of testing primes in this way is usually efficient, but there are classes of number, $1299936601 = 601 \cdot 1201 \cdot 1801$ has a period of $3600$ in any base under 601, which makes it appears prime.  It is in fact what is called a pseudoprime.  Any given base can have more pseudoprimes than the general populus.  For example, $2^{11}-1$ is a pseudoprime to base 2, since both the primes that divide this ($23, 89$), both have 11-digit periods in base 2.

Title: $2^{q-1}\equiv 1\pmod{q}.$


问:
Converts into $\frac{\partial^2 W}{\partial u^2}+\frac{\partial^2 W}{\partial v^2}=0$

Show that the substitution $u=x^2-y^2$, $v=2xy$ converts the equation $\frac{\partial^2 W}{\partial x^2}+\frac{\partial^2 W}{\partial y^2}=0$ into $\frac{\partial^2 W}{\partial u^2}+\frac{\partial^2 W}{\partial v^2}=0$

I have $$\left\lbrace u=x^2-y^2 \atop v=2xy \right. $$
So $$\frac{\partial W}{\partial x} = \frac{\partial W}{\partial u}\frac{\partial u}{\partial x}+\frac{\partial W}{\partial v}\frac{\partial v}{\partial y} $$
and $$\frac{\partial W}{\partial y} = \frac{\partial W}{\partial u}\frac{\partial u}{\partial y}+\frac{\partial W}{\partial v}\frac{\partial v}{\partial y} $$ I can compute for example $\frac{\partial u}{\partial x}=2x$, $\frac{\partial u}{\partial y}=-2y $,  $\frac{\partial v}{\partial x}=2y$ and $\frac{\partial v}{\partial y}=2x$  but how to find $\frac{\partial W}{\partial u}$ or $\frac{\partial W}{\partial v}$?

答:
Simply put, you don't!

You've correctly calculated that
\begin{align*}
W_{x} &= W_{u}\partial_{x}u + W_{v}\partial_{x}v = 2xW_{u} + 2yW_{v} \\
W_{y} &= W_{u}\partial_{y}u + W_{v}\partial_{y}v = -2yW_{u} + 2xW_{v}.
\end{align*}

Next, (note: $\partial_{x}u = \frac{\partial u}{\partial x}$ and $\partial_{xx}^{2}u = \frac{\partial^{2}u}{\partial x^{2}}$), applying product/chain rule again,
\begin{align*}
W_{xx} &= W_{uu}(\partial_{x}u)^{2} + W_{uv}(\partial_{x}u)(\partial_{x}v) + W_{u}(\partial_{xx}^{2}u) \\
&\qquad + W_{vu}(\partial_{x}u)(\partial_{x}v) + W_{vv}(\partial_{x}v)^{2} + W_{v}(\partial_{xx}^{2}v) \\
&= 4x^{2}W_{uu} + 8xyW_{uv} + 4y^{2}W_{vv} + 2W_{u}
\end{align*}
as well as
\begin{align*}
W_{yy} &= W_{uu}(\partial_{y}u)^{2} + W_{uv}(\partial_{y}u)(\partial_{y}v) + W_{u}(\partial_{yy}^{2}u) \\
&\qquad + W_{vu}(\partial_{y}u)(\partial_{y}v) + W_{vv}(\partial_{y}v)^{2} + W_{v}(\partial_{yy}^{2}v) \\
&= 4y^{2}W_{uu} - 8xyW_{uv} + 4x^{2}W_{vv} - 2W_{u}.
\end{align*}
Now what can we deduce?


问:
Positive elements in Hermitian * Banach Algebras
Let $A$ be a commutative Hermitian * Banach algebra, that is, a commutative Banach algebra with involution such that every self-adjoint element has real spectrum. It is known that if $x\in A$, then $y := x^\ast x \geq 0$, that is, $y^\ast = y$ and $\sigma(y)\subset[0,\infty)$, see "Bonsall and Duncan, Complete normed algebras, 41.Th 5.

My question is the following: let $y\in A$ be a positive element as before. Can we assume the existence of $x\in A$ such that $y = x^\ast x$? I'm particularly interested in the case $A = L^1 (\mathbb{R})$, the set of integrable functions over the real line, with convolution as multiplication. Note that in this case, $A$ is a $A^\star$-Banach algebra, that is, it is continuously embedded in a $C^\star$-algebra.

It is well known that the answer to this question is affirmative if $A$ is a $C^\star$-algebra, but I haven't been unable to find its answer to this more general case.

Edit: It seems that the answer to this question is positive if we add the condition that $0 \notin \sigma(y)$, i.e. $\sigma(y) \subset (0,\infty)$, see Theorem 11.20 in "W. Rudin, Functional analysis". However, I am still interested in the case $0 \in \sigma(y)$, so any help will be well received.

答:
I think this is an interesting question! I am not sure about the case $A=L^1(\mathbb{R})$, but I think I can give an example with $A=C^1(S^2)$, the $*$-algebra of continuously-differentiable, complex-valued functions on the standard 2-sphere embedded in $\mathbb{R}^3$. The multiplication is pointwise multiplication and the adjoint is pointwise conjugation.

I couldn't find a reference, so you should confirm that $C^1(S^2)$ is a Banach algebra. I am pretty sure that, if $M$ is any compact Riemann manifold, then $C^1(M)$ is a Banach algebra for the norm $\|f\|_1 := \|f\|_\mathrm{sup} + \|df\|_\mathrm{sup}$. The metric is needed to assign a norm to each $\mathbb{R}$-linear functional $df(x) : T_xM \to \mathbb{C}$ so that $\|df\|_\mathrm{sup} := \sup_{x \in M} \|df(x)\|$ is defined.

Here is the main claim:

Proposition: Suppose that $f \in C^1(S^2)$ vanishes at a single point $p \in S^2$ and is strictly positive everywhere else. Further suppose that, in some local coordinate system $(x,y)$ around $p$, we have $f(x,y) = x^2+y^2$. Then, there does not exist any $g \in S^2$ such that $|g|^2=f$.

Why do I think this proposition holds? This basically comes down to the following two claims.

Claim 1: Suppose $g : \mathbb{R}^2 \to \mathbb{C}$ is an $\mathbb{R}$-linear map satisfying $|g(x,y)|^2 = x^2+y^2$ for all $(x,y) \in \mathbb{R}^2$. Then, either $g(x,y) = x+iy$, $g(x,y) = x-iy$ or else it is a multiple of one of those by a complex phase of modulus one.

Claim 2: Suppose $g: \mathbb{R}^2 \to \mathbb{C}$ is a $C^1$ function with $
|g(x,y)|^2 = x^2+y^2$ for all $(x,y) \in \mathbb{R}^2$. Then, if $\gamma:[0,1] \to \mathbb{R}^2 \setminus \{(0,0\}$ is a small loop with winding number $1$, then $g \circ \gamma: [0,1] \to \mathbb{C} \setminus \{0\}$ has winding number $\pm 1$.

Claim 1 is elementary and I guess that the second claim should follow from the first using a linear approximation near $(0,0)$.

Now, suppose $f$ is as in the proposition above and $g \in C^1(S^2)$ has $|g|^2=f$. In particular, $g:S^2 \setminus \{p\} \to \mathbb{C} \setminus \{0\}$. By the second claim, if $\gamma:[0,1] \to S^2 \setminus \{p\}$ is a small loop traveling around the point $p$, then $g \circ \gamma:[0,1] \to \mathbb{C} \setminus \{0\}$ has winding number $\pm 1$. However, this is not possible because $S^2 \setminus \{p\}$ is contractible, so it should be possible to deform $g \circ \gamma$ to a constant loop.


问:
Tips for getting past a step in proving the Bessel function in summation form satisfies the Bessel differential equation
I want to prove that the Bessel function:

$$J_\nu(z)=\sum_{k=0}^{\infty}\frac{(-1)^k(z/2)^{2k+\nu}}{k!\Gamma(\nu+k+1)}$$

satisfies the Bessel differential equation:

$$\frac{d^2y}{dz^2}+\frac{1}{z}\frac{dy}{dz}+\left(1-\frac{\nu^2}{z^2}\right)y=0$$

The first thing that I did was take the derivative and second derivative of $J_\nu(z)$ to get:

$$J_\nu(z)=\sum_{k=0}^{\infty}\frac{(-1)^k(z/2)^{2k+\nu}}{k!(\nu+k)!}$$

$$J_\nu^{'}(x)=\sum_{k=0}^{\infty}\frac{(-1)^k(2k+\nu)(z/2)^{2k+\nu-1}}{2k!(\nu+k)!}$$

$$J_\nu^{''}(x)=\sum_{k=0}^{\infty}\frac{(-1)^k(2k+\nu)(2k+\nu-1)(z/2)^{2k+\nu-2}}{4k!(\nu+k)!}$$

After that I simply let $y=J_\nu(x)$ and plugged in the respective derivatives. After simplifying, I get:

$$\sum_{k=0}^{\infty}\frac{(-1)^k(z/2)^{2k+\nu}}{k!(\nu+k)!}\left[\frac{1}{4}(2k+\nu)(k+\nu-1)+\frac{1}{2z}(2k+\nu)+\left(1-\frac{\nu^2}{z^2}\right)\right]=0$$

This is where I am now stuck. I'm unsure how to prove that the left side of the equation equals zero. I've tried expanding the summation, where the first two terms are:

$$\left(\frac{(z/2)^\nu}{\nu!}\right)\left[\frac{1}{4}\nu(\nu-1)+\frac{1}{2z}\nu+\left(1-\frac{\nu^2}{z^2}\right)\right]-\left(\frac{(z/2)^{2+\nu}}{1!(\nu+1)!}\right)\left[\frac{1}{4}(2+\nu)(2+\nu-1)+\frac{1}{2z}(2+\nu)+\left(1-\frac{\nu^2}{z^2}\right)\right]+...=0$$

I still can't figure out how to get this expanded left side to equal zero. Should I try to prove that the summation I have in equation 6 converges to zero? Or am I not seeing terms that I am supposed to cancel out in equation 6 or 7? Or am I completely off the mark? Any advice would be greatly appreciated.

答: I doubt you’ll be happy with this, but you’ll want to do a bit more multiplying out: You want to expose the powers of $z$ explicitly. Assuming your work above to be correct, you’ll end up seeing that in order for the lowest order coefficients to cancel — and the final solution to be nontrivial — certain restrictions on $\nu$ will be required.


问:
How to prove that the following statements are equivalent: $\lim_{n\to \infty} a_n = a$ and $\lim_{x\to 0} \bar{\alpha}(x) = a$
In an assignment for the course Real-Analysis I need to proof that that the following statements are equivalent:

$\lim_{n\to \infty} a_n = a$

$\lim_{x\to 0} \bar{\alpha}(x) = a$

Given that $(W, d_w)$ is a metric space and $\alpha: \mathbb{N} \longrightarrow W$ and $\alpha(n) = a_n$ a row in $W$. We can use the function $\bar{\alpha}:\mathbb{R} \supset \longrightarrow W$ with:

$$Dom(\bar{\alpha}) = \{\frac{1}{n+1} : n \in \mathbb{N}\} \subset \mathbb{R}$$

and

$$\bar{\alpha}(x) = \alpha (\frac{1}{x} -1), x \in Dom(\bar{\alpha})$$

Let $a \in W$ and prove that the following statements are equivalent:

$\lim_{n\to \infty} a_n = a$

$\lim_{x\to 0} \bar{\alpha}(x) = a$

However, me and my fellow students have been trying to figure it out for some time now, but we can't...

答:
Let's write $x=\frac{1}{n+1}~~~n \in \mathbb{N}$.   $\bar{\alpha}(\frac{1}{n+1})=\alpha(n)=a_n~~~\forall n$.

$1. \Rightarrow 2.$  For given $\epsilon >0~~~\exists N \in \mathbb{N}$ such that $d_w(a_n,a)<\epsilon$ for all $n >N$.
Then for $\frac{1}{n+1}< \frac{1}{N+1}$ ($|x|<\delta$),

$d_w(\bar\alpha(\frac{1}{n+1}),a)\leq d_w(\bar\alpha(\frac{1}{n+1}),a_n)+d_w(a_n,a)<\epsilon$

$2. \Rightarrow 1.$   Given $\epsilon >0~~~\exists\delta>0$ such that $d_w(\bar\alpha(\frac{1}{n+1}),a)<\epsilon$ whenever $\frac{1}{n+1}<\frac{1}{M+1}<\delta$ (for some large $M \in \mathbb{N}$). Then for $n >M$,
$d_w(a_n,a)\leq d_w(\bar\alpha(\frac{1}{n+1}),a_n)+d_w(\bar\alpha(\frac{1}{n+1}),a)<\epsilon$


问:
How do you find $E(X^3)$ of a Poisson Distribution?
I know the proof to find the variance of a Poisson Distribution, and I tried to use that to find $E(X^3)$, but I can't get it to work. Any help would be great!!

Body:
Observe that $$E[e^{Xt}]=\sum_{k=0}^{\infty} e^{kt}P(X=k) = \sum_{k=0}^{\infty} e^{kt}e^{-\lambda } \frac{\lambda ^k}{k!} = e^{\lambda(e^t-1)} \quad (*).
$$

Also $$\frac{d^3}{d^3t}E[e^{Xt}]|_{t=0} = \frac{d^3}{d^3t}(1+E[X]+\frac{1}{2!}t^2E[X^2]+\frac{1}{3!}t^3E[X^3]+ O(t^4X^4))|_{t=0}=  E[X^3] \quad (**).$$
Combining  (*) and (**) results

$$\frac{d^3}{d^3t}E[e^{Xt}]|_{t=0} = \frac{d^3}{d^3t}e^{\lambda(e^t-1)}|_{t=0}=  \lambda e^{\lambda (e^t-1) + t} (1 + 3 \lambda e^t + \lambda^2 e^{2 t})|_{t=0}=\lambda+3\lambda^2+\lambda^3.$$


问:
Solving the problems by matrix in linear algebra
I learned from the class that if you want to solve the problems by matrix in

linear algebra, the problem has to satisfy linearity(homogeneity,superposition).

However, linear systems equation like (x + 2y = 3, 4x + 5y = 6) do not satisfy

linearity because they do not satisfy homogeneity and superposition. But we can

solve this linear system by constructing matrix $$
    \begin{pmatrix}
    1 & 2 \\
    4 & 5 \\
    \end{pmatrix}\begin{pmatrix}
    x\\
y\\
    \end{pmatrix} = \begin{pmatrix}
    3\\
6\\
    \end{pmatrix}
$$and getting inverse of the matrix \begin{pmatrix}
    1 & 2 \\
    4 & 5 \\
    \end{pmatrix}.
In short, if I want to solve the problem using matrices, I learned that the problem needs to satisfy linearity but linear equations do not satisfy the linearity(superposition, homogeneity). So isn't it wrong to use matrix to solve the linear system problem?

答:
The proposed system of linear equations can be written as
\begin{align*}
\begin{bmatrix}
1 & 2\\
4 & 5
\end{bmatrix}
\begin{bmatrix}
x\\
y
\end{bmatrix} =
\begin{bmatrix}
3\\
6
\end{bmatrix}
\end{align*}

Since the coefficient matrix is invertible, we have the following solution
\begin{align*}
\begin{bmatrix}
x\\
y
\end{bmatrix} =
\begin{bmatrix}
1 & 2\\
4 & 5
\end{bmatrix}^{-1}
\begin{bmatrix}
3\\
6
\end{bmatrix} =
-\frac{1}{3}
\begin{bmatrix}
5 & -2\\
-4 & 1
\end{bmatrix}
\begin{bmatrix}
3\\
6
\end{bmatrix} =
\begin{bmatrix}
-1\\
2
\end{bmatrix}
\end{align*}
That is to say, the solution set is given by $S = \{(-1,2)\}$.

In general, given the system of linear equations
\begin{align*}
\begin{bmatrix}
a & b\\
c & d
\end{bmatrix}
\begin{bmatrix}
x\\
y
\end{bmatrix} =
\begin{bmatrix}
e\\
f
\end{bmatrix}
\end{align*}
where $ad - bc \neq 0$, the solution can be written as
\begin{align*}
\begin{bmatrix}
x\\
y
\end{bmatrix} =
\frac{1}{ad-bc}
\begin{bmatrix}
d & -b\\
-c & a
\end{bmatrix}
\begin{bmatrix}
e\\
f
\end{bmatrix}
\end{align*}

Another way to interpret such problem is the following.

Consider the basis $\mathcal{B} = \{(1,0),(0,1)\}$ such that $T:\mathbb{R}^{2}\to\mathbb{R}^{2}$ satisfies
\begin{align*}
[T]_{\mathcal{B}} =
\begin{bmatrix}
a & b\\
c & d
\end{bmatrix}
\end{align*}

Thus the proposed problem consists in verifying if $(e,f)\in T(\mathbb{R}^{2})$.


问:
Indiscrete topology regular and normal?
I came across some tables in "counterexamples for topology" that contradict some of the properties I have determined in the following table I have created.It says that none of the properties on the chart I created are satisfied by the indiscrete topology. I thought the indiscrete topology was regular, normal, and separable, the chart says it is not? What is going on here? It says $4$ is for the indiscrete.

$$\begin{array}{cc|c}
 & \mbox{T1} & \mbox{Hausdorff}&\mbox{Regular} & \mbox{Normal}&\mbox{Separable}\\
\hline
\mathbb{R}&Y & Y & Y&Y&Y\\
\mathbb{R}^n&Y & Y & Y&Y&Y\\
\mbox{indiscrete}&N & N & Y&Y&Y\\
\mbox{discrete}&Y & Y & Y&Y&Y\\
\mbox{Cofinite}& Y& N&N &N&Y \\
\mbox{Cocountable}&Y&N&N&N&Y\\
\mathbb{R}_l&Y &Y &Y &Y&Y\\
\mbox{line w  2 origins} &Y &N &N &N&Y\\
\mbox{ordered  square} & Y&Y &Y &Y&N \\
\mathbb{R}_k &Y &Y &N &N&Y\\
\{0,1\}^A &Y &Y &Y &Y&Y
\end{array}$$

答: Some people, unfortunately, including Steen and Seebach, fail to recognize that the $T_k$ designations are supposed to form a hierarchy, so that $T_k$ implies $T_\ell$ whenever $k\ge\ell$. For you and me $T_3$ means regular plus $T_1$, and $T_4$ means normal plus $T_1$. For people like Steen and Seebach who follow the other convention, regular and normal are our $T_3$ and $T_4$, respectively, and $T_3$ and $T_4$ are our regular and normal, respectively. Thus, for them a regular or normal space must be $T_1$, but a $T_3$ or $T_4$ space need not be $T_1$. They also switch $T_{3½}$ and completely regular.

Title: Indiscrete topology regular and normal?


问:
Degree of transitive constituents is odd implies $|G|$ is odd
I want to prove that: the order of a permutation group $G \le S^\Omega$ is odd if and only if the degrees of all transitive constituents of $G$ and the degrees of all transitive constituents of each $G_\alpha (\alpha \in \Omega)$ are odd.  This is Exercise 3.13 in Wielandt's text.

(Let me recall that if $\Delta \subseteq \Omega$ is a fixed block of $G$, then $G$ restricted to $\Delta$, denoted $G^\Delta$, is called a constituent.  It is transitive iff $\Delta$ is a minimal fixed block (i.e. an orbit).  So the assertion asks to show that $|G|$ is odd iff each orbit of $G$ and each orbit of each $G_\alpha$ has odd length.)

The hint given is to use these three facts: (i) $|G_\alpha|~|\alpha^G|=|G|$. (ii) $|G:G_{\alpha \beta}| = |\alpha^G| |\beta^{G_\alpha}| = |\beta^G| |\alpha^{G_\beta}|$. (iii) If prime $p$ divides $|G|$, then $G$ contains an element whose cycle decomposition contains a $p$-cycle.

The implication is proved using (ii): If $|\alpha^G|$ or $|\beta^{G_\alpha}|$ is even for any $\alpha,\beta$, then by (ii) $|G|$ is even.  How do I prove the converse?

答:
The question is essentially answered in my comment and Derek Holt's comment above.  This answer elaborates on these two comments.

Let $G \le S^\Omega$. We need to show that if $|G|$ is even, then some orbit $\alpha^G$ of $G$ or some orbit $\beta^{G_\alpha}$ of some $G_\alpha$ $(\alpha \in \Omega)$ has even length.

Since $|G|$ is even, by (iii) above there exist $g,\alpha,\beta$ such that $g=(\alpha,\beta) \cdots \in G$. It suffices to show that for this $\alpha,\beta$, $|G:G_{\alpha \beta}|$ is even; by (ii) it would then follow that one of $|\alpha^G|$ or $\beta^{G_{\alpha}}|$ is also even.  The permutation $g$ contains the 2-cycle $(\alpha,\beta)$.  Hence $g$ fixes the set $\{\alpha,\beta\}$ setwise but not pointwise, i.e. $g \in G_{\{\alpha,\beta\}} - G_{\alpha \beta}$.  Also, every element in $G_{\{\alpha,\beta\}}$ is either in the subgroup $G_{\alpha \beta}$ or in the coset $g G_{\alpha \beta}$, whence the index $|G_{\{\alpha,\beta\}}:G_{\alpha \beta}|$ equals 2. Thus, the index $|G:G_{\alpha \beta}| = |G:G_{\{\alpha,\beta\}}| |G_{\{\alpha,\beta\}}:G_{\alpha \beta}|$ is even.


问:
Orthogonal projection onto a subspace

Let $v_1=\begin{pmatrix}1\\ 1\\ 1\\ 0\end{pmatrix}$ and $v_2 = \begin{pmatrix}0\\ 1\\ 1\\ 1\end{pmatrix}$. Find the orthogonal projection of $v = \begin{pmatrix}3\\ 2\\ 1\\ 0\end{pmatrix}$ onto $V= \operatorname{span}
(v_1,v_2)$.

If I have $A= \begin{pmatrix}1&0\\ \:1&1\\ \:1&1\\ \:0&1\end{pmatrix}$ and compute the projection matrix $P=A(A^TA)^{-1}A^T = \begin{pmatrix}\frac{3}{5}&\frac{1}{5}&\frac{1}{5}&-\frac{2}{5}\\ \frac{1}{5}&\frac{2}{5}&\frac{2}{5}&\frac{1}{5}\\ \frac{1}{5}&\frac{2}{5}&\frac{2}{5}&\frac{1}{5}\\ -\frac{2}{5}&\frac{1}{5}&\frac{1}{5}&\frac{3}{5}\end{pmatrix}$ then how does this help? I'm not sure I've entirely understood the idea of orthogonal projection?

答: What you had was the projection matrix for projection onto the span of the two vectors. Just multiply it by your given vector! (You might find some of my YouTube lectures, linked in my profile, helpful.)</p>

Title: Orthogonal projection onto a subspace


问:
Conjunction/disjunction of types
Let $T$ be a theory. A type $p(x)$ is a set of formulas in variable $x$ such that $T\cup p(x)$  is finitely satisfiable.

Question. What does disjunction (or conjunction) of types mean? More precisely, if $p(x)$ and  $q(x)$ are types, then what do $p(x)\wedge q(x)$ and $p(x)\vee q(x)$ mean?

答:
Well, there is an obvious intended meaning of $p(x)\land q(x)$ and of $p(x)\lor q(x)$ in terms of satisfaction:

$a\in M$ satisfies $p(x)\land q(x)$ if $a$ satisfies $p(x)$ and $a$ satisfies $q(x)$.

$a\in M$ satisfies $p(x)\lor q(x)$ if $a$ satisfies $p(x)$ or $a$ satisfies $q(x)$.

Now the interesting thing is that these satisfaction conditions on $a$ are themselves definable by types.

$a$ satisfies $p(x)$ and $a$ satisfies $q(x)$ if and only if $a$ satisfies $p(x)\cup q(x)$. More generally, for a family of types $\{p_i(x)\mid i\in I\}$, we can use the notation $\bigwedge_{i\in I} p_i(x)$ for the type $\bigcup_{i\in I} p_i(x)$.

$a$ satisfies $p(x)$ or $a$ satisfies $q(x)$ if and only if $a$ satisfies $$\{\varphi(x)\lor \psi(x)\mid \varphi(x)\in p(x) \text{ and }\psi(x)\in q(x)\}.$$ More generally, for a finite family of types $\{p_1(x),\dots,p_n(x)\}$, we can use the notation $\bigvee_{i=1}^n p_i(x)$ for the type $$\{\bigvee_{i=1}^n \varphi_i(x)\mid \varphi_i(x)\in p_i(x)\text{ for all }1\leq i \leq n\}.$$

On the other hand, an infinite disjunction of types is typically not definable by a type.

The fact that types are closed under arbitrary conjunctions and finite disjunctions corresponds to the fact that closed sets are closed under arbitrary intersections and finite unions, since a (partial) type corresponds to a closed set in the Stone space of complete types.

Title: Conjunction/disjunction of types


问:
Calculate a series
Let $y_n = 1/3y_{n-1} + 2/3y_{n-2}$ for $n>2$

Is the below correct? It is from my book but I don't understand how.

$\sum_{n=0}^{\infty}y_n = y_1(1 + 2/3 + 2/3^2 +...) + y_2(1 + 1/3 + 1/3^2 +...)$

I try to calculate $y_3$,$y_4$,$y_5$,... but their sum is not clean like the above and there are a bunch of extra terms that don't seem to cancel out. Can someone guide me on this? Thanks!

What I have done:
$y_3 = 1/3y_2 + 2/3y_1$ This one is fine
but starting from $y_4$,
$y_4 = 1/3^2y_2+2/3^2y_1+2/3y_2$,

$y_5 = 1/3^3y_2 + 2/3^3y_1 + 2/3^2y_2 + 2/3^3y_2 + 2/3^2y_1$

For the $y_2$ I still can get $1/3,(1/3)^2,...$, but going forward there seem to be a lot of extra terms that I don't get how they can add up to the above sum.

答:
If uou properly solved $$y_n = \frac 13 y_{n-1} + \frac 23 y_{n-2}$$ you are supposed to have obtained
$$y_n=c_1+c_2\left(-\frac{2}{3}\right)^n$$ since
$$r^2=\frac 13 r+\frac 23=0 \implies (r-1)\left(r+\frac{2}{3}\right)=0$$ Now, applying the consitions $y_1=a$ and $y_2=b$ gives
$$c_1=\frac{2 a+3 b}{5}  \qquad \text{and} \qquad c_2=\frac{9(b-a)}{10}$$

From here, it starts to be simpler.

Notice that you have a sign error and that $y_0$ is still undefined.

Title: Calculate a series


问:
Proof of Lemma 6.3 in Carothers' Real Analysis

Lemma 6.3 (Chapter 6 - Connectedness): Let $E$ be a subset of a metric space $(M,d)$. If $U$ and $V$ are disjoint, open
sets in $E$, then there are disjoint open sets $A$ and $B$ in $M$ such that $U= A \cap E$
and $V = B \cap E$.

The proof roughly goes as follows (I haven't typed it out exactly, but I have captured all the important details) -

Since $U$ is open in $E$, for each $x\in U$, $\exists\epsilon_x > 0$ such that $E\cap B(x,\epsilon_x) \subset U$. Since $V$ is open in $E$, for each $y\in V$, $\exists\delta_y > 0$ such that $E\cap B(y,\delta_y) \subset V$. Since $U\cap V = \varnothing$, we also have $E\cap B(x,\epsilon_x) \cap B(y,\delta_y) = \varnothing$.

Now, the author claims that:

For every $x\in U$ and $y\in V$, $$B\left(x,\frac{\epsilon_x}{2}\right) \cap B\left(y,\frac{\delta_y}{2}\right) = \varnothing$$

Why is this true? I'm trying to show this in the following way. Suppose $z \in B\left(x,\frac{\epsilon_x}{2}\right) \cap B\left(y,\frac{\delta_y}{2}\right)$. So, $d(x,z) < \epsilon_x/2$ and $d(y,z) = \delta_y/2$. To show that the intersection is non-empty would amount to showing $d(x,z) < \epsilon_x/2$ and $d(y,z) = \delta_y/2$ cannot hold together. How do I do this?

Thus, $$A = \bigcup_{x\in U} B\left(x,\frac{\epsilon_x}{2}\right) \text{ and } B = \bigcup_{y\in V} B\left(y,\frac{\delta_y}{2}\right)$$ work.

Why does this choice of $A, B$ work? All I need to show for this, is that $U = A\cap E$ and $V= B\cap E$ right? Is anything else needed to be done?

Thank you.

Picture for Reference:

Attempting a Proof by Contradiction:

As mentioned in an answer, $B(x,r/2)\subset B(x,r)$. Suppose $\exists x \in U, \exists y\in V$ for which $B\left(x,\frac{\epsilon_x}{2}\right) \cap B\left(y,\frac{\delta_y}{2}\right) \ne \varnothing$. Also, $$E\cap B(x,\epsilon_x) \subset U \implies E\cap B(x,\epsilon_x/2) \subset U$$ and
$$E\cap B(y,\delta_y) \subset V \implies E\cap B(y,\delta_y/2) \subset V$$
What's next? I'm not able to find a contradiction. $U\cap V = \varnothing \implies E\cap B(x,\epsilon_x) \cap B(y,\delta_y) = \varnothing$, and I believe this is what we want to contradict.

答:
Using your notations, suppose that:$\exists z\in E$ such that $z \in B\left(x,\frac{\epsilon_x}{2}\right) \cap B\left(y,\frac{\delta_y}{2}\right)$

It follows that: $d(x,z)\lt \epsilon_x/2$ and $d(y,z)\lt \delta_y/2\implies d(x,z)\lt \epsilon_x$ and $d(y,z)\lt \delta_y$

Therefore, $z\in B(x,\epsilon_x) $ and $z\in B(y, \delta_y)$ whence it follows that $z\in B(x,\epsilon_x) \cap B(y, \delta_y) $, which is a contradiction as $E\cap B(x,\epsilon_x) \cap B(y, \delta_y)=\phi$


答:
It can be easy to confuse functions and equations at first.
A function can be linear, such as $f(x,y) = 2x + 3y$.
An equation can be homogeneous, such as $f(x,y) = 2x + 3y = 0$, or inhomogeneous, such as $f(x,y) = 2x + 3y = 7$, but both are called linear equations.
Systems of linear equations can be solved with matrices.  If the equations are homogeneous, then if you find a solution, any constant multiple of that solution is also a solution.  For example, $(-3, 2)$ is a solution to $2x + 3y = 0$, so $(-30, 20)$ is also a solution.  If you find two such solutions, the sum of those solutions is also a solution.

In more advanced math, some equations are called linear if they don't have constant terms and affine if they do.  Those handle slightly more complicated functions.

Some equations such as differential equations have functions as the unknown variables in them.  These equations are called linear and homogeneous if the sum of two solutions is another solution, and any constant multiple of a solution is a solution. Such a combination is called a linear superposition of solutions.  For example (don't worry about all the details, just the result): $y" + y = 0$ has $y = sin x$ as a solution and $y = cos x$ as another.  So $5 sin (x)$ is also a solution, as is $sin (x) + cos (x)$, as is $3 sin (x) + 5 cos (x)$.

Title: Solving the problems by matrix in linear algebra

Body:
I learned from the class that if you want to solve the problems by matrix in

linear algebra, the problem has to satisfy linearity(homogeneity,superposition).

However, linear systems equation like (x + 2y = 3, 4x + 5y = 6) do not satisfy

linearity because they do not satisfy homogeneity and superposition. But we can

solve this linear system by constructing matrix $$
    \begin{pmatrix}
    1 & 2 \\
    4 & 5 \\
    \end{pmatrix}\begin{pmatrix}
    x\\
y\\
    \end{pmatrix} = \begin{pmatrix}
    3\\
6\\
    \end{pmatrix}
$$and getting inverse of the matrix \begin{pmatrix}
    1 & 2 \\
    4 & 5 \\
    \end{pmatrix}.
In short, if I want to solve the problem using matrices, I learned that the problem needs to satisfy linearity but linear equations do not satisfy the linearity(superposition, homogeneity). So isn't it wrong to use matrix to solve the linear system problem?


问:
Is the radical of a Lie algebra equal to the radical of its Killing form?
A Lie algebra $\mathfrak{g}$ is semi-simple if the maximal solvable ideal ${\rm rad}(\mathfrak{g})$ is trivial (let's take this as the definition for this question). Cartan's Second Criterion says that $\mathfrak{g}$ is semi-simple if and only if the Killing form $\kappa_{\mathfrak{g}}$ is non-degenerate, which is equivalent to saying that the radical ${\rm rad}(\kappa_{\mathfrak{g}})=\mathfrak{g}^\perp$ (as in the radical of a symmetric bilinear form) is trivial.

So I am asking myself if we have that ${\rm rad}(\mathfrak{g})={\rm rad}(\kappa_{\mathfrak{g}})$ in general.

We always have that ${\rm rad}(\kappa_{\mathfrak{g}})$ is solvable by Cartan's First Criterion, for if $X \in {\rm rad}(\kappa_{\mathfrak{g}})$ and $Y \in {\rm rad}(\kappa_{\mathfrak{g}})'$, then in particular $Y$ is in ${\rm rad}(\kappa_{\mathfrak{g}})$ and so $\kappa_{\mathfrak{g}^\perp}(X,Y) = \kappa_{\mathfrak{g}}(X,Y)=0$. This implies that ${\rm rad}(\kappa_{\mathfrak{g}}) \subseteq {\rm rad}(\mathfrak{g})$.

I am struggling with the reverse inclusion. Namely, I would like to show that if $\mathfrak{h}$ is an arbitrary solvable ideal of $\mathfrak{g}$, then $\mathfrak{h} \subseteq{\rm rad}(\kappa_{\mathfrak{g}})$. I could not immediately see how to use Cartan's First Criterion again here (in other words, I couldn't see how $\kappa_{\mathfrak{h}}(\mathfrak{h},\mathfrak{h}')=0$ helps). Is this only true if $\mathfrak{g}$ is assumed to be semi-simple so one can use $\mathfrak{g} = \mathfrak{h}\oplus \mathfrak{h}^\perp$ somehow?

答: For the two-dimensional solvable but non-nilpotent Lie algebra $\mathfrak g$ with basis vectors $x,y$ and bracket $[x,y]=y$ we have $\kappa(x,x)= 1$ so $\mathrm{rad}(\kappa) \neq \mathfrak g = \mathrm{rad}(\mathfrak g)$.


问:
Show that for the two sided left shift operator in $\ell_2(\mathbb{Z})$ has $\{|\lambda | =1\} $ in its spectrum
Let $S:\ell_2(\mathbb{Z}) \to \ell_2(\mathbb{Z})$ be the left shift operator $S(\ldots, x_{-2}, x_{-1}, x_0, x_1, x_2, \ldots) = (\ldots, x_{-1}, x_0, x_1, x_2, x_3 \ldots)$

I want to show that $\{|\lambda | =1\} \subset \sigma(S)$

The hint that was given is to show that if $ |\lambda | =1$, then it's an almost eigenvalue of $S$.
Namely, for every $\epsilon > 0$ there is $x \in \ell_2(\mathbb{Z})$ such that $||x||=1$ and $||Sx - \lambda x|| \le \epsilon$

If could find such $x$ for a given $\epsilon$, then I can easily show that $ (S - \lambda I) x$ is not invertible.

However, I am struggling to find such an $x$.

Help would be appreciated.

答:
The suggestion by amsmath is the canonical one. We have
$$
\sum_{k=-N}^N|\lambda^k|^2=2N+1.
$$
So we can define
$$
x=\frac1{2N+1}\sum_{k=-N}^N\lambda^ke_k,
$$
with $N$ to be chosen later. So $\|x\|=1$ and

\begin{align}
Sx-\lambda x&=\frac1{2N+1}\Big(\sum_{k=-N}^N\lambda^ke_{k-1}-\sum_{k=-N}^N\lambda^{k+1}e_k\Big)
=\frac1{2N+1}\Big(\sum_{k=-N-1}^{N-1}\lambda^{k+1}e_{k}-\sum_{k=-N}^N\lambda^{k+1}e_k\Big)\\[0.3cm]
&=\frac1{2N+1}\Big(\lambda^{-N}e_{-N-1}-\lambda^{N+1}e_N\Big).
\end{align}
Then
$$
\|Sx-\lambda\|=\frac{\sqrt2}{2N+1}.
$$
Now can choose $N$ such that $2N+1>\sqrt2/\varepsilon$.


问:
Is $\{\{1,2,3\},\emptyset\}$ a partition of $\{1,2,3\}$?
Title says it all. Is $\{\{1,2,3\},\emptyset\}$ a partition of $\{1,2,3\}$? I'm guessing it isn't, because the definition in Naive Set Theory states the following:

A partition of $X$ is a disjoint collection $\mathbb{C}$ of non-empty subsets of $X$ whose union is $X$.

答: Thus violates the “non-empty” part of the definition, so no, it’s not a partition according to this definition.</p>


问:
Ring of order $n$ is isomorphic to $\Bbb Z/n\Bbb Z$, with $n$ square-free

Let $R$ be a ring of order $n$ and suppose $n$ has no square in its prime decomposition. How do I see that $R$ is isomorphic to $\Bbb Z/n\Bbb Z$?

I bet that the map $\Bbb Z  \to R, \, 1\mapsto 1_R$ descends to an iso $\Bbb Z/n\Bbb Z \to R$ but I don't see how $n$ having no squares implies the desired descent.

答:
We can smuggle the core idea of Cayley's theorem from group theory into ring theory and show that $R$ embeds in the ring of endomorphisms of $R$ as an additive group:

$$ \phi : R \to \text{End}(R,+) \quad \text{as} \quad \phi(r) = [s \mapsto rs] \text.$$

Now, $(R, +)$ is cyclic because $n = |R|$ is squarefree, so $\text{End}(R, +) \cong \text{End}(C_n)$, where $C_n$ is the cyclic group of order $n$.

And $\text{End}(C_n) \cong \mathbb{Z}/n\mathbb{Z}$. (If $n \in \mathbb{Z}$, then $g \mapsto n \cdot g = \sum_{i = 1}^n g$ is an endomorphism of any abelian group; in the case of $C_n$, the image of the generator determines the endomorphism.)

So $R$ embeds in $\mathbb{Z}/n\mathbb{Z}$. The embedding must be an isomorphism since the two rings in question are finite.


问:
How to know the price of a given day assuming continuous growth on log scale
First i want to say that my math level is too basic and I dont know how to ask my problem in the right way, but i'll try to explain it in my words:

I have a chart with two axes: Day and Price

Both axes are on arithmetic scale

I know two things:

On day $1$, price is $\$1$

On day $30$, price is $\$2$

From that information, I know the price has changed $100\%$ from day $1$ to day $30$

Assuming continuous growth (a straight line) from day $1$ to day $30$ I want to know the price (or percentage  change) for any day in between $1$ and $30$.

So I do this, $\frac{100}{30} = 3,33$ (my logic: $100\%$ change / $30$ days)

I draw a line from day $1$ to day $3$ in the chart and verify it, for example on day $11$, the change is $33,33\%$ this is fine, this is the expected number from my calculation ($\frac{100}{30}$ * ndays from day $1$)... So I can know the change or price for any day I want

Arithmetic scale line

Now, I want to do that same thing (I want know any day price change) but with the price axe on logarithmic scale, having only information about day 1 price and day 30 price

When I change the scale of the price, from arithmetic to logscale, my growth line doesnt match that $33,33\%$ anymore

semilog scale line

As I understand it, if the scale is logaritmic, maybe the growth should be logaritmic too or something like that... not lineal so my calculation ($\frac{100}{30}$) is not valid any more

How can I do this? What is the calculation I should do to get the right %change for any day using log scale graphic?

Summary of my question:

I want to know how to calculate any day price (or its % change) of a growth line in log scale that was constructed connecting two prices (from day $1$ to day $30$).

Body:
If linear growth, simply p(t)=$29t-28$

If exponential groth, let p(t)=$Ae^{at}$ , with conditions $p(1)= 1, p(30)=2$

1=$Ae^{(a)}$ , 2=$Ae^{30(a)}$

2= $e^{29a}$ , $\ln 2 =29a$, $a=\frac{\ln2}{29}$,

1=$Ae^{\frac{\ln2}{29}}$ , $A=e^{-\frac{\ln2}{29}}$

So, the answer $\to$

$p(t)=e^{(t-1)\frac{ln2}{29}}$


问:
Giving a 1-hour talk to highschool math club: any topic suggestion?
I've been invited (by my kid) to give a one hour talk to her highschool math club.  Last year (right before the pandemic hit) I did two such talks on probability, and they loved it.  I'm looking for topic suggestions.

Logistics: There will be about $10-15$ kids total, from grades $9-12$.  These kids are mostly "A" students, and some of them have skipped a math grade, but they're nowhere near Olympiad standard.  The talk will be on March $1$st, $2021$, and will be on zoom.  I will be making slides this weekend.

I don't want this to become an "educational" lecture, so I'm looking for fun examples to motivate certain math areas.  E.g. last year my talks on probability culminated with Polya Urn and Buffon's Noodle.  Also, I'd like everything (well, $95\%$) to be fully explainable within the one hour, i.e. I don't want to just spew facts and then tell them "just trust me".

My own math skill tends toward discrete: comp-sci, combinatorics, etc.  However I'm open to any idea - esp. if it has worked for you before! - as long as I can learn it during one weekend.  :)

Some ideas:

modulo arithmetic esp. finite fields: Perfect shuffle is fun.  Would cryptography (e.g. RSA) be too difficult?  What about codes e.g. Hamming?  Any other fun examples of modulo arithmetic?

non-Euclidean geometry: especially how a POINT in elliptic geometry is actually $2$ "ordinary points".  Can the alternates to the parallel postulate (and some consequence e.g. angle sum in a triangle) be explained sufficiently at highschool level in an hour?

finite geometry: Personally I find Finite projective planes very beautiful.  Do they (or finite affine planes) have any application?  (Besides the kiddie game Spot It?)

algorithm: This is actually my work area, but I don't know what I can cover in an hour.  Maybe the $O(N \log N)$ lower bound for sorting?  Some of these kids don't even have coding experience... :(

graph theory: Edge coloring of a complete graph is one of my favorite pictorial proofs.  Eulerian tour.  Euler characteristic (of a planar graph) is related.  What else?  Shortest path would require discussion of algorithm.  Hall's marriage theorem is surprising and neat but I don't think I can prove it from scratch in an hour.

combinatorics: Start with ${N \choose m}$ and then stars-and-bars for sure.  What are your favorite (elementary) examples?  Can Burnside's lemma be proved from scratch (at highschool level, without group theory) in an hour?

Comments on above and/or any other suggestion most welcome!

(Apologies if I'm tagging too widely...)

答:
Again, thanks all of you for answering and commenting.  I thought some of you might be interested in what actually happened.

I decided to go with a theme of "finite mathematics" and talk mostly about modulo arithmetic, using it to illustrate $3$ different topics:

(A) I started by performing a crowd-pleasing "magic trick" related to the perfect shuffle.  I actually cannot do a perfect shuffle at all, so instead I did a sort of reverse: I dealt a deck of cards into 4 piles, re-assembled them into a deck, and repeated the process $3$ more times.  As expected, the kids were very surprised to see that the original deck order was restored!  Which I then explained via $4^4 = 256 = 1 \pmod{51}$ as in the wikipedia article.  (BTW if some of you are not aware of this trick, it's actually quite neat and also easy to do in practice.) This took about $15$ minutes.

(B) Then I talked about error correction codes, starting with the background: why you need it, message, encoding, codeword, error, decoding, etc.  For motivation, I found out that the ASCII for lower-case "l" and lower-case "d" differ by just $1$ bit, so I made up a story how, with a $1$-bit error, "Trump lied" would become "Trump died". :D

Anyway, I discussed the Hamming $(7,4)$ code in detail, with actual examples of encoding and decoding, doing parity checks (which I called mod $2$ addition) by hand, and introducing the concept of Hamming distance.  I did not use linear algebra, as most kids haven't learned that.

Then I discussed the crucial fact that, when $N$ is prime, mod $N$ arithmetic supports division (I did not use the term finite field), which in turn means you can solve systems of equations.  This segued into polynomials and Reed Solomon codes, which I covered briefly using the approach of this excellent blogpost by @JyrkiLahtonen - thanks!

The whole error correction code segment took about $30$ minutes.

(C) Lastly, I talked about the kiddie game Spot-It - I have asked my kid to poll her friends before hand and many are familiar with it.  While the game itself is uninteresting, the math behind the card design is fascinating: it's based on the finite projective plane of order $7$.  I spent many slides using modulo $5$ arithmetic to construct the finite affine plane of order $5$, then showing how it satisfies some of Euclid's axioms and therefore deserves the fancy name of "finite plane".  Then, I showed how it can almost be turned into the cards of Spot-It, discussing point-line duality briefly.  Finally I added the points and line at infinity to turn it into a finite projective plane for the full solution.  This segment took about $25$ minutes.

The whole thing took about $70-75$ minutes and was quite well received, both during the event, and when my kid told me the others' reactions afterwards.  She said the older kids mostly got it, while the younger kids had some trouble but still got the flavor.

I spent way, waaaaay too much time over the weekend making the slides, as I was having such a great time!  I just couldn't bear to leave out any of the $3$ topics since they're all fun (and even "educational" in the case of error correction codes) and also show a great variety between them.  I ended up with about $60$ slides!  Had to talk really fast to go through them in $70$ minutes.  :)


问:
$|G:H|$ and $|H|$ are coprime
Suppose $G$ is a finite group with a normal subgroup, $K$, and another subgroup, $H$, such that $[G:H]$ and $|H|$ are coprime.

Is it true that $[HK:H]$ will divide both $[G:H]$ and $|H|$? Why is this?

答:
It is true that $[HK:H]$ divides $[G:H]$: because $K$ is normal, $HK$ is certainly a subgroup that contains $H$, so $H\leq HK\leq G$, hence
$$[G:H]=[G:HK][HK:H],$$
proving that $[HK:H]$ divides $[G:H]$.

However, it need not be the case that $[HK:H]$ divides $|H|$. For a counterexample, take $C_6$, the cyclic group of order $6$, $K$ the subgroup of order $3$, and $H$ the subgroup of order $2$. We have $[G:H]=3$, and $|H|=2$, so $[G:H]$ and $|H|$ are coprime. But $HK=G$, so $[HK:H]=[G:H] = 3$, which certainly does not divide $|H|=2$.

In comments you state that what you actually want to do is prove that $[K:K\cap H]$ and $|K\cap H|$ are coprime.

Note that
$$[HK:H] = \frac{|HK|}{|H|} = \frac{|H||K|}{|H||H\cap K|}=\frac{|K|}{|H\cap K]} = [K:K\cap H].$$
So $[K:K\cap H]$ divides $[HK:H]$, which divides $[G:H]$.
On the other hand,
$K\cap H\leq H$. So $|K\cap H|$ divides $|H|$. Therefore, $[K:K\cap H]$ and $|K\cap H|$ are coprime.

Title: $|G:H|$ and $|H|$ are coprime


答:
$HK\le G$, thus $|HK|\mid |G|$ by Lagrange.  Thus we have that $[HK:H]\mid[G:H]$ pretty easily by Lagrange.

But it is not true that  $[HK:H]\mid|H|$, in general.  For instance, in $A_4$, let $K=V_4$ and $H=\langle(123)\rangle$.  Then $HK=A_4$.  So $[HK:H]=4\not\mid3=|H|$.

(In fact, if $[G:H]$ and $|H|$ are coprime, then when $[HK:H]$ divides both,  $[HK:H]=1$, or $HK=H$.  This will probably help us find counterexamples.)

Title: $|G:H|$ and $|H|$ are coprime

Body:
Suppose $G$ is a finite group with a normal subgroup, $K$, and another subgroup, $H$, such that $[G:H]$ and $|H|$ are coprime.

Is it true that $[HK:H]$ will divide both $[G:H]$ and $|H|$? Why is this?


问:
How to obtain the equality between two integrals with different integrands?
I am looking into the proof of Ramanujan's approximation formula for the partition function $p(n)$ by Stein and Shakrachi. I am confused about one step towards the end of the proof just before we apply the method of steepest descent. Stein obtains

$$-\mu^{\frac{3}{2}}\int_{0}^{\pi}e^{2s\sin{(\theta)}}e^{\frac{3i\theta}{2}}\sqrt{i} \, d\theta = \mu^{\frac{3}{2}}\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}e^{2s\cos{(\theta)}}\left(\cos{\left(\frac{3\theta}{2}\right)}+i\sin{\left(\frac{3\theta}{2}\right)}\right) \, d\theta.$$

How we go from one integral to the other is what is confusing me at the moment. Is there a transformation identity? Or am I missing something elementary?

答: First we can replace $\theta $ with $\pi-\theta$ and get the integrand as $$\exp\left(2s\sin \theta+\frac{i\pi} {4}+\frac{3i\pi}{2}-\frac{3i\theta}{2}\right)$$ Next we put $\theta=\pi/2-t$ to get the integrand as $$\exp\left(2s\cos t+i\pi+\frac{3it}{2}\right)$$ The $i\pi$ leads to a negative sign, another negative comes from $-dt$ and finally one more negative sign comes from the reversal of integral limits $[\pi/2,-\pi/2]$ to $[-\pi/2,\pi/2]$. So all in all we have the change seen as a "-" sign.


问:
Convergence of a sequence to a differentiable function

Let $\{f_n\}$ be a sequence of real-valued differentiable functions on
$[0,1]$ whose derivatives are continuous. Further, suppose $\{f_n\}$
converges pointwise to a function $f$ defined on $[0,1]$ and the
sequence of the derivatives $\{f'_n\}$ is uniformly bounded and
equicontinuous. Show that $f$ is differntiable on $[0,1]$ and
$\{f_n\}$ converges uniformaly to $f$.

I am aware of the similar statement that assumes the uniform convergence of $f'_n$ to some function $g$ and gives the same results here. But this statement here is quite different. In particular, I cannot apply the fundamental theorem of calculus since $f'_n$ is not assumed to be convergent. I also tried applying Arzela-Ascoli on $\{f'_n\}$ but that proved the uniform convergence of a subsequence. Any suggestions?

答:
By Arzela-Ascoli Theorem there is a subsequence $(f_{n_k}')$ converging uniformly to some function. By the arrgument you are familiar with it follows that $f$ is differentiable and $(f_{n_k}') \to f'$ uniformly.

Suppose $f_n$ does not converge uniformly to $f$. There exist $\epsilon >0$, points $x_i \in [0,1]$ and a subsequence $(f_{m_i})$ such that $|f_{m_i}(x_i)-f(x_i)| >\epsilon$ for all $i$. Now repeat the previous argument with $(f_n)$ replaced by $(f_{m_i})$. You will get a subsequence along which $f_{m_i}' $ tends to $f'$ uniformly. But then $f_{m_i}(x_i)-f(x_i)=f_{m_i}(0)-f(0)+\int_0^{x} [f_{m_i}'(t)-f'(t)] dt \to 0$  along that subsequence leading to a contradiction.

[ For differentiability of $f$ here is the argument: $f_{n_k}(x)-f_{n_k}(0)=\int_0^{x} f_{n_k}'(t)dt \to \int_0^{x} h(t)dt $ where $h$ is the uniform limit of $f_{n_k}'$. Since $h$ is continuous and $f(x)-f(0)=\int_0^{x} h(t)dt$ it follows that $f$ is differentiable].


问:
Is matrix transpose a linear transformation?
This was the question posed to me. Does there exist a matrix $A$ for which $AM$ = $M^T$ for every $M$. The answer to this is obviously no as I can vary the dimension of $M$. But now this lead me to think , if I take , lets say only $2\times2$ matrix  into consideration. Now for a matrix $M$, $A=M^TM^{-1}$ so $A$ is not fixed and depends on $M$, but the operation follows all conditions of a linear transformation and I had read that any linear transformation can be represented as a matrix. So is the last statement wrong or my argument flawed?

答: Notice that the answer from Troy Woo can be written simply as $(M^\top)_{ij}=\sum_{k,l=1}^{n}\delta^k_j\delta^l_im_{kl}$. Writing $I=(e₁,\ldots,e_n)$, then $M^\top=\sum_{i,j=1}^{n}e_ie_j^\top Me_ie_j^\top$, which is another form of the answer from Kevin Bumgartner.


问:
Groups in which $[a, b][b^{-1}, a^{-1}] = 1$
Let $G$ be a group, I'm in a situation where this holds for all $a,b \in G$: $[a, b][b^{-1}, a^{-1}] = 1$, I believe this implies that $[a, b] = 1$ for all $a$, $b$, however, I cannot seem to prove this. Is this assertion correct? If so why?

答:
This is not true. In any group, $[x,y]^{-1}=[y,x]$. And in any nilpotent group of class at most $2$, the commutator bracket is bilinear, so $[x^a,y^b]=[x,y]^{ab}$ for all $x$ and $y$ and all integers $a$ and $b$. Therefore, if $G$ is nilpotent of class at most $2$, then $[a,b][b^{-1},a^{-1}]=[a,b][b,a]=1$ for all $a,b$, but if $G$ is not abelian, then you do not get $[a,b]=1$ for all $a$ and $b$.

(The above holds regardless of whether your convention is that $[x,y]=x^{-1}y^{-1}xy$ or that $[x,y]=xyx^{-1}y^{-1}$).

For a specific example, take the Dihedral group of order $8$,
$$D_8 = \langle r,s\mid r^4=s^2=1, sr=r^3s\rangle.$$
For any two elements $a,b$ of $D_8$, if they commute then $[a,b]=1$, and if $a$ and $b$ do not commute then $[a,b]=r^2$. Therefore, $[a,b][b^{-1},a^{-1}]=1$ in all cases (since $a$ and $b$ commute if and only if $b^{-1}$ and $a^{-1}$ commute). But $D_8$ is not abelian.

To characterize these groups: if $[a,b][b^{-1},a^{-1}]=1$, then $[a,b]=[b^{-1},a^{-1}]^{-1} = [a^{-1},b^{-1}]$.  Thus, the given condition is equivalent to

$$[x,y]=[x^{-1},y^{-1}]\quad\text{for all }x,y\in G.$$

Your groups are precisely the $2$-Engel groups. Here's a proof.

I will use the convention that $[x,y]=x^{-1}y^{-1}xy$, and the following identities, which can be verified by direct calculation:
$$[xz,y] = [x,y]^z[z,y]\quad\text{and}\quad [x,zy]=[x,y][x,z]^y$$

Recall that a group $G$ is said to be $n$-Bell if $[x^n,y]=[x,y^n]$ for all $x,y\in G$, and $n$-Levi if $[x^n,y]=[x,y]^n$ for all $x,y\in G$.

Proposition. A group $G$ satisfies $[a,b]=[a^{-1},b^{-1}]$ for all $a,b\in G$ if and only if $G$ is $(-1)$-Bell.

Proof. If $G$ is $(-1)$-Bell, then $[a^{-1},b^{-1}] = [a,(b^{-1})^{-1}] = [a,b]$. Conversely, if $G$ satisfies the identity, then
$$[x^{-1},y] = [x^{-1},(y^{-1})^{-1}] = [x,y^{-1}],$$
proving that $G$ is $(-1)$-Bell. $\Box$

The following can be deduced from the description of the Bell semigroup of a group given in

Kappe , L.C. On n-Levi groups. Arch. Math. (Basel) 47 (1986), no. 3, 198–210. MR0861866 (88a:20048)

but here is a direct argument.

Lemma. If $G$ is $n$-Bell (that is, $[x^n,y]=[x,y^n]$ for all $x,y\in G$) then it is also $(1-n)$-Bell (that is, $[x^{1-n},y]=[x,y^{1-n}]$ for all $x,y\in G$). If $G$ is both $n$-Bell and $m$-Bell, then it is also $nm$-Bell.

Proof. Suppose that $[x^n,y]=[x,y^n]$. Expanding we have
$$x^{-n}y^{-1}x^ny = x^{-1}y^{-n}xy^n.$$
Left multiplying by $yx$ and right multiplying by $y^{-1}x^{-1}$ we get
$$yx^{1-n}y^{-1}x^{n-1} = y^{1-n}xy^{n-1}x^{-1}$$
or equivalently, that $[y^{-1},x^{n-1}] = [y^{n-1},x^{-1}]$. Setting $a=y^{-1}$, $b=x^{-1}$, we get that
$$[a,b^{1-n}] = [a^{1-n},b]$$
holds for all $a,b\in G$.

If $G$ is both $n$-Bell and $m$-Bell, then
$$[x^{nm},y] = [(x^m)^n,y] = [x^m,y^n] = [x,(y^n)^m] = [x,y^{nm}]$$
for all $x,y\in G$, so $G$ is also $nm$-Bell, as claimed. $\Box$

Note that this means that a group is $2$-Bell if and only if it is $-1$-Bell, since $-1 = 1-2$ and $2=1-(-1)$.

Proposition. Let $G$ be a group. Then $[x^{-1},y]=[x,y^{-1}]$ for all $x,y\in G$, if and only if $G$ is $n$-Bell for all integers $n$.

Proof. It is clear that if $G$ is $n$-Bell for all $n$, then the result holds (take $n=-1$). Conversely, if $G$ is $-1$-Bell, then by the Lemma it is also $2$-Bell, since $1-(-1)=2$. If $G$ is $k$-Bell, then it is also $-k$-Bell (since it is both $k$-Bell and $-1$-Bell), and therefore it is also $(k+1)$-Bell (since $1-(-k) = 1+k$). This proves $G$ is $n$-Bell for all positive $n$; and therefore, since it is also $-1$-Bell, it is $n$-Bell for all integers $n$, as claimed. (Note that every group is $0$-Bell and $1$-Bell). $\Box$.

Corollary. Let $G$ be a group. The following are equivalent:

$G$ satisfies $[x,y]=[x^{-1},y^{-1}]$ for all $x,y\in G$.

$G$ is $-1$-Bell.

$G$ is $2$-Bell.

$G$ is $n$-Bell for every integer $n$.

Proof. The equivalence of 2, 3, and 4 is given above. If $G$ is $n$-Bell for all integers $n$, then $[x^{-1},y^{-1}] = [x,(y^{-1})^{-1}] = [x,y]$. Conversely, if $[x,y]=[x^{-1},y^{-1}]$ for all $x,y\in G$, then
$$[x^{-1},y] = [x^{-1},(y^{-1})^{-1}] = [x,y^{-1}]$$
so $G$ is $-1$-Bell, and therefore by the Proposition above it is $n$-Bell for all $n$. $\Box$

Recall that a group $G$ is $2$-Engel if $[y,x,x]=1$ for all $x$ and $y$ (here, $[a,b,c] = [[a,b],c]$). This is equivalent to $[x,y,x]=1$ for all $x,y\in G$, since $x$ commutes with $[y,x]$ if and only if it commutes with $[x,y]=[y,x]^{-1}$. A group is $2$-Engel if and only if every $2$-generated subgroup is nilpotent of class at most $2$. We have:

Proposition. Let $G$ be a group. The following are equivalent:

$G$ is $2$-Engel.

$G$ is $2$-Levi.

$G$ is $2$-Bell.

Proof. If $G$ is $2$-Engel, then
$$\begin{align*}
[x^2,y] &= [xx,y] = [x,y]^x[x,y]\\
&= x^{-1}[x,y]x[x,y]\\
&= [x,y][x,y]^{-1}x^{-1}[x,y]x[x,y]\\
&= [x,y][[x,y],x][x,y]\\
&= [x,y][x,y,x][x,y]\\
&= [x,y]^2,
\end{align*}$$
so $G$ is $2$-Levi.

If $G$ is $2$-Levi, then it is also $2$-Bell: if $[x^2,y]=[x,y]^2$ for all $x,y\in G$, then
$$[x,y^2] = ([y^2,x])^{-1} = ([y,x]^2)^{-1} = ([y,x]^{-1})^2 = [x,y]^2,$$ so $[x^2,y]=[x,y]^2=[x,y^2]$.

Finally, if $G$ is $2$-Bell, it is $n$-Bell for all $n$, hence $-1$-Bell. We have
$$1 = [1,y] = [x^{-1}x,y] = [x^{-1},y]^x[x,y] = [x,y^{-1}]^x[x,y]$$
and
$$1 = [x,1] = [x,y^{-1}y] = [x,y][x,y^{-1}]^y = [x,y][x^{-1},y]^y.$$
Thus, $[x,y^{-1}]^x[x,y] = [x,y][x^{-1},y]^y$, so
$$([x,y^{-1}]^x)^{[x,y]} = [x,y]^{-1}[x,y^{-1}]^x[x,y] = [x^{-1},y]^y,$$
which gives, using the $-1$-Bell property
$$[x^{-1},y]^{x[x,y]} = [x^{-1},y]^y.$$
Therefore, $[x^{-1},y]$ commutes with $x[x,y]y^{-1} = x(x^{-1}y^{-1}xy)y^{-1} = y^{-1}x$. Thus,
$$[x^{-1},y,y^{-1}x]=1$$
holds for all $x,y$. In particular, substituting $yx$ for $x$,
$$\begin{align*}
1&=[(yx)^{-1},y,y^{-1}(yx)]\\
&= [(yx)^{-1},y,x]
&= [yx,y^{-1},x]\\
&= \Bigl[[yx,y^{-1}],x\Bigr]\\
&= \Bigl[ [y,y^{-1}]^x[x,y^{-1}],x\Bigr]\\
&= [x,y^{-1},x].
\end{align*}$$
Thus, for all $x,y\in G$, $[x,y^{-1},x]=1$. Therefore, $[x,y,x]=1$ holds, and thus $[y,x,x]=1$ holds for all $x,y\in G$, proving that $G$ is $2$-Engel. $\Box$

Thus we conclude:

Theorem.  A group $G$ satisfies the identity $[a,b][b^{-1},a^{-1}]=1$  for all $a,b\in G$ if and only if $G$ is $2$-Engel.


问:
Sharpest upper bound on $|f(x)|$ using multivariable Taylor approximation, Hessian, or any other method
(Based on a question in University of Pittsburgh's Preliminary examination: Aug 2016, question 5.)

Let $f\colon \mathbb{R}^2 \to \mathbb{R}$ be a $C^2$ function. Let $B=\{x \in \mathbb{R}^2\colon |x|\leq 1\}$ be the closed unit ball. Suppose $f(o)=Df(o)=o$ and that for an $M\geq 0$, we have
$$
\left \vert \frac{\partial^2 f}{\partial_{x_1}^2}(x)\right \vert^2 + 2\left \vert \frac{\partial^2 f}{\partial_{x_1}\partial_{x_2}}(x)\right \vert^2 +\left \vert \frac{\partial^2 f}{\partial_{x_2}^2}(x)\right \vert^2 \leq M^2 , \quad \text{for any $x=(x_1,x_2) \in B$}.
$$

Give the sharpest bound on $|f(x)|$ for $x \in B$.

By that, I mean something in terms of $M$ and (powers of) $|x|$, or maybe a different norm, like, $|x_1|+|x_2|$.

Please include the details of your argument with clear reference to any lemma/theorem you use.
Your time is appreciated!

答:
Im assuming "$o$" means $0$. By Taylor's theorem with Lagrange remainder, for $x \in B$,
$$f(x) = \frac{1}{2}x \cdot H(\theta x)x$$
where $\theta \in (0, 1)$ depends on $x$, and $H$ is the hessian of $f$. Note that the bound given in the problem statement is equivalent (by Clairaut's theorem since $f$ is $C^2$) to saying that for $x \in B$,
$$||H(x)||_2^2 \leq M^2.$$
So $||H(x)||_2 \leq M$ for all $x \in B$. Let $x \in B$. Then
\begin{align}
|f(x)| &= \frac{1}{2}|x \cdot H(\theta x)x| \\
&\leq \frac{1}{2}||x|| \cdot ||H(\theta x)x|| \\
&\leq \frac{1}{2}||x|| \cdot ||H(\theta x)||_2 \cdot ||x|| \\
&\leq \frac{1}{2}M||x||^2.
\end{align}
The first inequality is by Cauchy Schwarz. The second inequality uses the result that $||AB||_2 \leq ||A||_2 \cdot ||B||_2$, which can be proved using Cauchy Schwarz. I believe the inequalities are sharp in the sense that they are achieved when $H(x) = MI$ on $B$.


问:
Solve for martingale progression in CRAPS
In craps the house has a 1.45% edge (51.45% chance to lose).  If you martingale+1 (double up + 1 unit, every time you lose) and do this for 15 outcomes, then double up the initial bet for 30 more outcomes you can almost 4x your money

The odds of 4 losses in a row would be .5145^4 = about 7%

But what are the odds I will make it 40 rolls(outcomes) without losing more then 4 in a row?

答:
We can treat this as a Markov chain.  We have $6$ states.  For $k=0,1,2,3,4$ state $k$ means that we currently have a string of $k$ losses.  State $5$ means that at some time, a string of $5$ losses has occurred, so state $5$ is absorbing.  If the probability of a loss is $p$ then the transition matrix is
$$\begin{bmatrix}
1-p&p&0&0&0&0\\
1-p&0&p&0&0&0\\
1-p&0&0&p&0&0\\
1-p&0&0&0&p&0\\
1-p&0&0&0&0&p\\
0&0&0&0&0&1
\end{bmatrix}$$  and the initial distribution is $$X=\begin{bmatrix}1&0&0&0&0&0\end{bmatrix}$$
We simply have to compute $XP^{45}$ and read off the answers.

I wrote a python script to do this:

```<code>import numpy as np

p = (1+.0145)/2
P = np.zeros((6,6))
for i in range(5):
    P[i,0] = 1-p
    P[i,i+1] = p
P[5,5] = 1
X = np.array([1,0,0,0,0,0])
Y = X@np.linalg.matrix_power(P, 45)
print(Y)
</code>```

This produced the output

```<code>[0.2337404  0.12075448 0.06238393 0.03222866 0.0166499  0.53424264]
</code>```

so the probability of $5$ losses in a row is $$.53424264$$

Body:
We can treat this as a Markov chain.  We have $6$ states.  For $k=0,1,2,3,4$ state $k$ means that we currently have a string of $k$ losses.  State $5$ means that at some time, a string of $5$ losses has occurred, so state $5$ is absorbing.  If the probability of a loss is $p$ then the transition matrix is
$$\begin{bmatrix}
1-p&p&0&0&0&0\\
1-p&0&p&0&0&0\\
1-p&0&0&p&0&0\\
1-p&0&0&0&p&0\\
1-p&0&0&0&0&p\\
0&0&0&0&0&1
\end{bmatrix}$$  and the initial distribution is $$X=\begin{bmatrix}1&0&0&0&0&0\end{bmatrix}$$
We simply have to compute $XP^{45}$ and read off the answers.

I wrote a python script to do this:

import numpy as np

p = (1+.0145)/2
P = np.zeros((6,6))
for i in range(5):
    P[i,0] = 1-p
    P[i,i+1] = p
P[5,5] = 1
X = np.array([1,0,0,0,0,0])
Y = X@np.linalg.matrix_power(P, 45)
print(Y)

This produced the output

[0.2337404  0.12075448 0.06238393 0.03222866 0.0166499  0.53424264]

so the probability of $5$ losses in a row is $$.53424264$$


问:
Proving geometrical inequalities
Let $u$ and $v$ be the distances of an arbitrary point on the side $AB$ of an acute-angled triangle $ABC$ to its sides $BC$ and $AC$. Let $h_a$ and $h_b$ be the lengths of the altitudes from its vertices $A$ and $B$ respectively. Prove that
$$$$ $min$ {$h_a,h_b$} $\leq$ $u+v$ $\leq$ $max$ {$h_a,h_b$}
$$$$

I was able to prove the first part of the question like this: Let the arbitrary point be X.
WLOG, let $h_a\geq h_b$. Then$$\frac{AX}{XB}={h_a\over v}={u\over h_b}$$
$$h_ah_b=uv$$
$$u+v\geq2\sqrt{uv}=2\sqrt{h_ah_b}\geq2h_b$$
$$Or, u+v\geq min\{h_a,h_b\}$$
But now I'm stuck in the second part. $$$$
Edit: The above proof is INCORRECT; the correct proof is given below in @player3236 's answer.

答:
Your idea (of using similar triangles) is correct but the formula is wrong. We should have:

$$\frac {AX}{AB} = \frac {v}{h_b}, \quad \frac {BX}{AB} = \frac {u}{h_a}$$

Hence

$$u+v = h_a\frac {BX}{AB} + h_b\frac {AX}{AB}$$

Therefore

$$\min\{h_a,h_b\}\frac {BX}{AB} + \min\{h_a,h_b\}\frac {AX}{AB} \le h_a\frac {BX}{AB} + h_b\frac {AX}{AB} \le \max\{h_a,h_b\}\frac {BX}{AB} + \max\{h_a,h_b\}\frac {AX}{AB}$$

which simplifies to

$$\min\{h_a,h_b\} \le u+v \le \max\{h_a,h_b\}$$

Title: Proving geometrical inequalities


问:
How do I find the coordinates of this point, and then the area?
I'm struggling w the wording of this question (been stuck for 15+mins). </p>

<p>'The line $3x -4y = 8$ meets the $y$-axis at $A$. The point $C$ has coordinates $(-2,9)$. The line through $C$ perpendicular to $3x - 4y = 8$ meets it at $B$. Calculate the area of the triangle.'</p>

答:
Here is a picture of the situation.  I was looking for another way to find the area, but I don't think you can avoid the work of calculating the coordinates of point $ \ B  \ $ using what you've learned before calculus.

What you are given which is important is that side $ \ BC \ $ is perpendicular to side $ \ AB \ $ .  Refaat M. Sayed and sango have obtained the coordinates of $ \ B \ $ as $ \ (4 \ , \ 1 ) \ . $  What is "cute" in this problem (and was probably intended to save you a bit of work) is that the difference in the $ \ x-$ coordinates of points $ \ A \ $ and $ \ B \ $ is $ \ 4 \ $ and the difference in their $ \ y-$ coordinates is $ \ 3 \ $ .  So $ \ \overline{AB} \ $ is the hypotenuse of a  $ \ 3-4-5 \ $ triangle.  Similarly, the differences in the $ \ x-$ and $ \ y-$ coordinates of points $ \ B \ $ and $ \ C \ $ are $ \ 6 \ $ and $ \ 8 \ , $ respectively.  Thus $ \ \overline{BC} \ $ is the hypotenuse of a $ \ 6-8-10 \ $ triangle.  The right triangle $ \ \Delta ABC \ $ consequently has area $ \ \frac{1}{2} · 5 · 10 \ = \ 25 \ \ . $

Further along in your courses, when you learn about vectors, there are slick ways to get the area of $ \ \Delta ABC \ $ with and without finding the position of $ \ B \ $ .


问:
How many digits of $\pi$ are needed to approximate $\pi^n$ correctly to the nearest whole number?
In his piece on $\pi^{\pi^{\pi^\pi}}$, Matt Parker plots a graph showing that if $\pi$ is approximated to $n$ digits, then whenever $m \leq 2 n$ (roughly), $\pi^m$ is correct to the nearest whole number.

Is there any theoretical underpinning for this?

答:
Starting from @Winther's answer, the only explicit solution of
$$m\,\pi^m=10^{n}$$ is given by
$$m=\left\lceil \frac{W\left(10^n \log (\pi )\right)}{\log (\pi )}\right\rceil$$ where $W(.)$ is Lambert function. In practice, $m$ is close to $2n$ (as already mentioned) but not sufficiently accurate.

For $n=50$, this would give $m=97$. Checking

$$97\, \pi^{97}-10^{50}=6.23 \times 10^{49} <10^{50}$$ while
$$98\, \pi^{98}-10^{50}=4.15 \times 10^{50} >10^{50}$$ For large value of $x$, you can approximate the value of Lambert function by
$$W(x)\approx L_1-L_2+\frac{L_2}{L_1}+\frac{L_2(L_2-2)}{2L_1^2}+\frac{L_2(2L_2^2-9L_2+6)}{6L_1^3}+\cdots$$ where $L_1=\log(x)$ and $L_2=\log(L_1)$.

Using this approximation for $n=50$ would give
$$W\left(10^{50} \log (\pi )\right)\sim 110.55887520$$ while its exact value is                               $110.55887517$


问:
Poisson equation inside a ball $B(0, 1)$
Find the solution of the Poisson equation inside a ball $B(0, 1)$, i.e.:
$$\begin{cases} \Delta u=x^{2} \ \ \ in \ B(0,1)\\u=3 \ \ \ on \ \partial B(0,1)\end{cases}$$

($x^2=x\cdot x$, $x\in \mathbb{R}^{2}$)

I have the following formula for the solution of Poisson equation:
$$u(x)=-\int_{\partial\Omega}\frac{\partial G}{\partial n}(x,y) u(y) \ dS(y)-\int_{\Omega}G(x,y)\Delta u(y) \ dy$$

I know the formula for the Green function $G$ in this case.

I tried to evaluate these integrals with polar coordinates, but they turned out quite difficult...

答:
Some suggestions along the same lines as Shuhao's, but perhaps on the lazier side.

For any rotation $T:\mathbb R^2\to \mathbb R^2$ the function $u\circ T$ solves the same boundary problem, because $\Delta(u\circ T)=(\Delta u)\circ T$.

By an appropriate uniqueness theorem (maximum principle for harmonic functions, really), $u=u\circ T$. That is, $u$ is a function of $r$ only.

Since the Laplacian of $u$ is homogeneous of degree $2$, and is the second derivative of $u$, the function $u$ itself should be homogeneous of degree $4$. Up to a harmonic (constant) term, that is. All that remains is to find two real constants.