问:
Is shrinking a square to a point reversible?
From Wikipedia, "Geometric points do not have any length, area, volume or any other dimensional attribute. A common interpretation is that the concept of a point is meant to capture the notion of a unique location in Euclidean space."

Suppose it shrinks by $1/2$ scale factor, so $\{{1/2^{n}}\}_{n=0}^\infty$ converges to point $0$. Does it have its Euclidean "square information" replaced with "location information"? At that point (pun intended), is unshrinking it back to its original shape irreversible? Thanks.

This idea came up when I was looking at the inscribed square problem and thought about shrinking a square to a point, so every point on the curve contains four vertices of information. That's "cheating". However, as noted in the answer below: "shrinking your square into a single point will make it lose its square information".

答:
Consider your square to be the set of points $X\subset\mathbb{R}^2$

Consider the function $\varphi:X\to\mathbb{R}^2$ that turns each point of your square into the same single point $(x,y)\in\mathbb{R}^2$. That means, for any point $(a,b)$ from your square, then
$$\varphi((a,b))=(x,y).$$
So it's clear that this $\varphi$ isn't injective (multiple points give the same image, all points in fact). Also, it isn't surjective, since only $(x,y)$ has a preimage in $\varphi$. Since $\varphi$ is not a bijection, you can't find any inverse function $\varphi^{-1}$ to undo what you did to your square using $\varphi$.

So yes, shrinking your square into a single point will make it lose its "square information".


问:
How to solve this quartic equation?
For the quartic equation:

$$x^4 - x^3 + 4x^2 + 3x + 5 = 0$$

I tried Ferrari so far and a few others but I just can't get its complex solutions. I know it has no real solutions.

ContentLicense: CC BY-SA 3.0

Title: How to solve this quartic equation?

Body:
Use the magic quartic formula!

$$a = -1, b = 4, c = 3, d = 5$$

$$u = -\frac{29}{12}, \Delta_0 = 85, \Delta_1 = -826, Q = \sqrt[3]{666i-541}$$
The discriminant is 65712, meaning that there are either four real or four complex roots.

$$v = \frac{(\sqrt[3]{666i-541})^2 + 85}{3\sqrt[3]{666i-541}} = \frac{(666i-541)\sqrt[3]{666i-541} + 85(\sqrt[3]{666i-541})^2}{3(666i-541)}$$

$$=\frac{736237\sqrt[3]{666i-541} - 85(666i+541)(\sqrt[3]{666i-541})^2}{2208711}$$

And you should be able to take it from there.


问:
Multivariate normal with diagonal variance implies its components are normally distributed.
I have a question that seems intuitive but I am not sure how to prove it. Let $X$ be a random vector with componants $X=[X_1,\ldots,X_n]$. Suppose $X\sim N(\mu,\Sigma)$ where $\Sigma = diag(\sigma_1^2,\ldots,\sigma_n^2)$ and $\mu = [\mu_1,\ldots,\mu_n]$. How can I properly show that $X_i\sim N(\mu_i,\sigma_i^2)$  (and that they are independent)?

答:
Let $X=(X_1,...,X_n)$ be a Gaussian vector in $\mathbb{R}^n$. If $X_1,...,X_n$ are independent then obviously $\Sigma$ is diagonal. Conversely, if $\Sigma$ is diagonal, then for all $u\in\mathbb{R}^n$ we can compute the characteristic function
$$\begin{align} \mathbb{E}\left[\exp\left(i\langle u, X\rangle\right)\right] &=\exp\left(i\langle u, \mu\rangle - \frac{u^\top \Sigma u}{2}\right)\\&=\exp\left(i\sum_{k=1}^nu_k\mu_k - \frac{\sum_{k=1}^nu_k^2\sigma_k^2}{2}\right) \\ &=\prod_{k=1}^n\exp\left(iu_k\mu_k -\frac{\sigma_k^2u_k^2}{2}\right) \\ &=\prod_{k=1}^n\mathbb{E}\left[\exp\left(u_kX_k\right)\right]\end{align}$$
hence independence.

Taking $u$ to be the vector with $0$ everywhere except at the $k$-th position where there's a $1$ shows that for all $k=1,...,n$ the component $X_k$ has the characteristic function of a $\mathcal{N}(\mu_k, \sigma_k^2)$ random variable.

ContentLicense: CC BY-SA 4.0


问:
Numerical Integration - Trapezium and Simpson's rule
I was reading through my professor's notes on Trapezoid and Simpson's rule in Numerical Integration and I was wondering if I get the intuition right.

Given a function $f$ and an interval $[a,b]$ these rules are guidelines on how you will split the initial interval $[a,b]$ into smaller intervals say $S_1,S_2...S_N$ where $S_1 =[a,x_i]$ and $S_N =[x_j,b]$.

Depending on the rule you know that each of these intervals contains a standard amount of points.

Let's take as an example a Simpson's interval which will contain 3 points.

Given the fact that we have 3 points we can :

Approximate $f$ using a 2nd degree polynomial $P_2$ on $S_i$

By integrating $P_2$ on $S_i$ we get $A_i$ which is an approximation of integrating $f$ on $S_i$

Repeat steps (1) and (2) for each $S_j$ and calculate each $A_j$

Finally add all $A_j$ and you get an approximation of integrating $f$ on $[a,b]$

I emphasize standard as in reality each $S_i$ contains infinitely many points and I understand that step (1) does not have to be repeated from scratch. The only thing you actually have to do is calculate the forward differences.

Is my process of thinking here right?

答:
Yes, you are right. The trapezium rule approximates $f(x)$ by a linear function (straight line) between $(x_i, f(x_i))$ and $(x_{i+1}, f(x_{i+1}))$. Simpson's rule (also known as Simpson's 1/3 rule) approximates $f(x)$ by a quadratic function (parabola) between $(x_i, f(x_i))$ and $(x_{i+1}, f(x_{i+1}))$; to fit the quadratic it uses one extra evaluation of $f$ in each interval, at its midpoint $x=\frac{x_i+x_{i+1}} 2$.

The trapezium rule and Simpson's rule are both examples of a family of numerical integration methods called closed Newton-Cotes formulas. The next method in the family is Simpson's 3/8 rule which approximates $f$ by a cubic function in each interval.

Newton-Cotes formulas are especially easy to apply since the function being integrated is evaluated at regular intervals across the range of integration. Other numerical integration methods, such as Gaussian quadrature, use interval sizes that vary.


问:
How to numerically plot the function $f(x)=\cos(x)\int^x_0\frac{\sin t}{\cos ^2(t) \sqrt{2+\sin^{2}\left(t\right)}}dt $
I'd like to numerically plot the function

$$f(x)=\cos(x)\int^x_0\frac{\sin t}{\cos ^2(t) \sqrt{2+\sin^{2}\left(t\right)}}dt \tag{1}$$

I can plot this analytically by evaluating the integral in $(1)$;

$$f(x)=\frac{\sqrt{2+\sin^{2}\left(x\right)}}{3}-\frac{\sqrt{2}\cos\left(x\right)}{3} \tag{2}$$

which exists on the range $(0,\pi)$. The problem is when I want to numerically plot this function from equation $(1)$ without explicitly integrating the integral in $(1)$.

I ask this because I would like to plot similar functions for powers other than $2$ and $1/2$, for example I would like to plot

$$f(x)=\cos(x)\int^x_0\frac{\sin t}{\cos ^2(t) \sqrt[m]{2+\sin^{m}\left(t\right)}}dt \tag{3}$$

but for larger powers the integral doesn't exist in closed form and so a nice formula like eq $(2)$ cannot be used and the plotting needs to be done numerically.

When I try and plot this numerically, for example by plotting $(1)$ with Desmos, the function is only defined up to $x=\frac{\pi}{2}$.

See the image below where I compare the numerical plot of $(1)$ in blue with the analytic plot of $(2)$ in red.

The integral in $(1)$ diverges as $x\to\frac{\pi}{2}$ however because $\cos x \to 0$ in the limit $f(\frac{\pi}{2})$ is finite,  I think because of the diverging integral my plotter registers the value of $(1)$ as undefined and stops plotting once the plotter reaches this point. Is there a way to plot this function in its entirety numerically?

Any suggestions would be welcome.

答:
Let $$J(x)=\int \frac{\sin x ~dx}{\cos ^2 x \sqrt{2+\sin^2 x}}=-\int \frac{dt}{t^2 \sqrt{3-t^2}}=\int\frac{udu}{\sqrt{3u^2-1}}$$
Here we have used \cos x=t$, t=1/u$, next we use $3u^2-1=v$, to get
$$J=\frac{1}{6}\int \frac{dv}{\sqrt{v}}=\frac{1}{3} \sqrt{v}=\frac{\sqrt{2+\sin^2 x}}{3\cos x}.$$
Hence, $$f(x)=\frac{1}{3}[\sqrt{2+\sin^2x}-\sqrt{2} \cos x]$$
This is a periodic function with the period as $2\pi$. $f(0)=0, f(\pi)=\frac{2\sqrt{2}}{3}, f(2\pi)=0$
$$f'(x)=\frac{\sin x \cos x}{3\sqrt{2+\sin^2x}}+\frac{\sqrt{2}}{3}\sin x$$
So $f'(0)=f'(\pi)=f'(2\pi)=0$, so $f_{min}=0$ and $f_{max}=f(\pi)=2\sqrt{2}/3$.
Now you can plot $f(x)$ for $[0, \pi]$ which is bell shaped touching  $x$-axis at $x=0,\pi$. See below:


答:
A rectangle with sides only under consideration, can not be contracted to a point.

There is no point $x_0$ on the sides of the rectangle such that for all x on the rectangle, the line segment from $x_0$ to x completely lies on the rectangle.

From Contractible space and star domain

Title: Is shrinking a square to a point reversible?

Body:
From Wikipedia, "Geometric points do not have any length, area, volume or any other dimensional attribute. A common interpretation is that the concept of a point is meant to capture the notion of a unique location in Euclidean space."

Suppose it shrinks by $1/2$ scale factor, so $\{{1/2^{n}}\}_{n=0}^\infty$ converges to point $0$. Does it have its Euclidean "square information" replaced with "location information"? At that point (pun intended), is unshrinking it back to its original shape irreversible? Thanks.

This idea came up when I was looking at the inscribed square problem and thought about shrinking a square to a point, so every point on the curve contains four vertices of information. That's "cheating". However, as noted in the answer below: "shrinking your square into a single point will make it lose its square information".


问:
$d$-dimensional integral with delta function
Consider the following integral in n-dimensions:

\begin{align}
g_n(\varepsilon)
 &= \frac{1}{(2\pi)^n} \int d^n  \mathbf k \; \delta(\epsilon +2t\cos (k_xa) + 2t\cos(k_y a)+...)
\end{align}
The boundaries of this integral is the hypercube with the boundaries $[0,\frac{2\pi}{a})$ along each axis. This integral has a physical meaning but that is not important now. Is it possible to calculate this in $n$-dimensions? Is it even possible to calculate it in two dimension?

I can handle this integral in the 1D case:

$$
 g_1(\varepsilon)=
\frac{1}{\pi} \int_0^{\pi/a} \delta(\epsilon +2t\cos (ka)) =
\frac{1}{2ta\pi} \frac{1}{\sqrt{1-\frac{\varepsilon^2}{4t^2}}}
$$

答:
Preliminaries

For $n \in \mathbb{N}$ define the tight-binding energy $E_n \colon \left[0, \frac{2 \pi}{a}\right)^n \to \mathbb{R}  , \, E_n (k) = - 2t \sum_{j=1}^n \cos(k_j a) \, .$ Due to the symmetries of the cosine function we may write the density of states as
$$ g_n \colon \mathbb{R} \to \overline{\mathbb{R}} \, , \, g_n (\varepsilon) = \int \limits_{\left[0,\frac{2 \pi}{a}\right)^n} \frac{\mathrm{d}^n k}{(2\pi)^n} \, \delta \left(\varepsilon - E_n (k)\right) \stackrel{p = a k}{=} \int \limits_{\left[0,\pi\right)^n} \frac{\mathrm{d}^n p}{2 t (\pi a)^n} \, \delta \left(\frac{\lvert\varepsilon \rvert}{2t} + \sum \limits_{j=1}^n \cos(p_j)\right) \, .$$
Note that $g_n (\varepsilon) = 0$ holds if $\lvert \varepsilon \rvert > 2 t n$, so we will only consider the case $\lvert \varepsilon \rvert \leq 2 t n$ from now on.

The density of states in one dimension

For $n = 1$ a short calculation yields $g_1 (\varepsilon) = \frac{1}{2 \pi a t \sqrt{1 - \left(\frac{\varepsilon}{2t}\right)^2}} $ as mentioned in the question.

The density of states in two dimensions

For $n = 2$ the integral is slightly more complicated but still manageable. We have
\begin{align}
g_2 (\varepsilon) &= \frac{1}{2 \pi^2 a^2 t} \int \limits_0^\pi \mathrm{d} p_1 \int \limits_0^\pi \mathrm{d} p_2 \, \delta \left(\frac{\lvert \varepsilon \rvert}{2t} + \cos(p_1) + \cos(p_2)\right) \\
&\!\!\!\!\!\!\!\!\!\!\!\!\!\stackrel{u_{1/2} = \pm \cos(p_{1/2})}{=} \frac{1}{2 \pi^2 a^2 t} \int \limits_{-1}^1 \mathrm{d} u_1 \int \limits_{-1}^1 \mathrm{d} u_2 \, \frac{\delta \left(\frac{\lvert \varepsilon \rvert}{2t} + u_1 - u_2\right)}{\sqrt{1-u_1^2} \sqrt{1-u_2^2}} \\
&= \frac{1}{2 \pi^2 a^2 t} \int \limits_{-1}^{1 - \frac{\lvert \varepsilon \rvert}{2t}} \frac{\mathrm{d} u_1}{\sqrt{1-u_1^2} \sqrt{1-\left(u_1 + \frac{\lvert \varepsilon \rvert}{2t}\right)^2}} \\
&\!\!\!\!\!\!\!\stackrel{u_1 + \frac{\lvert \varepsilon \rvert}{4t} = v}{=} \frac{1}{2 \pi^2 a^2 t} \int \limits_{-\left(1 - \frac{\lvert \varepsilon \rvert}{4t}\right)}^{1 - \frac{\lvert \varepsilon \rvert}{4t}} \frac{\mathrm{d} v}{\sqrt{1-\left(v - \frac{\lvert \varepsilon \rvert}{4t}\right)^2} \sqrt{1-\left(v + \frac{\lvert \varepsilon \rvert}{4t}\right)^2}} \\
&= \frac{1}{\pi^2 a^2 t} \int \limits_0^{1 - \frac{\lvert \varepsilon \rvert}{4t}} \frac{\mathrm{d} v}{\sqrt{\left[\left(1 - \frac{\lvert \varepsilon \rvert}{4t}\right)^2 - v^2\right] \left[\left(1 + \frac{\lvert \varepsilon \rvert}{4t}\right)^2 - v^2\right]}} \\
&\!\!\!\!\!\!\!\!\!\!\!\!\!\!\stackrel{v = \left(1 - \frac{\lvert \varepsilon \rvert}{4t}\right) \sin(\phi)}{=} \frac{1}{\pi^2 a^2 t \left(1 + \frac{\lvert \varepsilon \rvert}{4t}\right)} \int \limits_0^{\pi/2} \frac{\mathrm{d} \phi}{\sqrt{1 - \left(\frac{1 -\frac{\lvert \varepsilon \rvert}{4t}}{1 + \frac{\lvert \varepsilon \rvert}{4t}}\right)^2 \sin^2 (\phi)}} \\
&= \frac{\operatorname{K} \left(\frac{1 -\frac{\lvert \varepsilon \rvert}{4t}}{1 + \frac{\lvert \varepsilon \rvert}{4t}}\right)}{\pi^2 a^2 t \left(1 + \frac{\lvert \varepsilon \rvert}{4t}\right)} = \frac{\operatorname{K} \left(\sqrt{1 - \left(\frac{\varepsilon}{4 t}\right)^2}\right)}{2 \pi^2 a^2 t} \, .
\end{align}
Here, $\operatorname{K}$ is the complete elliptic integral of the first kind and the last step follows from an identity discussed in this question.

Results in higher dimensions

In three or more dimensions this direct approach does not seem very helpful, as the resulting integrals will be too difficult. We can, however, easily compute the Fourier transform of the density of states in arbitrary dimensions:
$$ \hat{g}_n (s) \equiv \int \limits_{\mathbb{R}} \mathrm{d} \varepsilon \,  g_n(\varepsilon) \mathrm{e}^{\mathrm{i} s \varepsilon} = \int \limits_{\left[0,\frac{2 \pi}{a}\right)^n} \frac{\mathrm{d}^n k}{(2\pi)^n} \, \mathrm{e}^{\mathrm{i} s E_n (k)} = \left[\int \limits_0^{\frac{2\pi}{a}} \frac{\mathrm{d} k_1 }{2 \pi} \, \mathrm{e}^{- 2 \mathrm{i} t s \cos (k_1 a)} \right]^n = \left(\frac{\operatorname{J}_0 (2ts)}{a}\right)^n \, .$$
Here, $\operatorname{J}_0$ is a Bessel function of the first kind. The inverse transformations for $n \in \{1,2\}$ require some work (they are given in this question), but the results of the direct approach can be reproduced. The case $n = 3$ is the subject of said question and the result is very complicated, so numerical methods are probably better suited to finding the density of states in dimensions $n \geq 3$.


问:
Matrix derivative: $\frac{\partial}{\partial A_{ij}}(\mathbb{A}(\mathbb{A}^T\Sigma\mathbb{A})^{-1}\mathbb{A}^T)$
Let $\mathbb{A} = (A_{ij})$ be a rectangular, $m\times n$ real matrix, full-rank and with $m \le n$. Let $\mathbb{C}$ be a real $n\times n$ symmetric and pos-def matrix. With these assumptions, $\mathbb{A}^T\mathbb{C}\mathbb{A}$ is invertible. Finally, let $\mathbf{u}$ be an $m$-vector.

I need to compute the following derivative:

$$\frac{\partial}{\partial \mathbb{A}}(\mathbf{u}^T\mathbb{A}(\mathbb{A}^T\mathbb{C}\mathbb{A})^{-1}\mathbb{A}^T\mathbf{u})$$

where $\Sigma$ is considered a constant.

But I keep struggling with the algebra which gets really messy.

Any tricks I can use?

答:
$\def\R{\mathbb R}\def\e{\varepsilon}\def\p#1#2{\frac{\partial #1}{\partial #2}}$Consider the matrix-valued function $F$ defined by
$$\eqalign{
S &= \Sigma &\implies\quad S=S^T \\
B &= A^TSA &\implies\quad B=B^T \\
F &= AB^{-1}A^T &\implies\quad F=F^T \\
}$$
and calculate its differential
$$\eqalign{
dF &= dA\;B^{-1}A^T + AB^{-1}dA^T - AB^{-1}dB\,B^{-1}A^T \\
 &= dA\;B^{-1}A^T + AB^{-1}dA^T \\
 &\quad- AB^{-1}\;dA^TSAB^{-1}A^T - AB^{-1}A^TS\,dA\,B^{-1}A^T \\
 &= dA\;B^{-1}A^T + AB^{-1}dA^T - AB^{-1}\;dA^TSF - FS\,dA\,B^{-1}A^T \\
 &= (I-FS)\;dA\;B^{-1}A^T + AB^{-1}\;dA^T(I-SF) \\
}$$
In order to continue, we'll need the following two gradients
$$\eqalign{
\p{A}{A_{ij}} &= e_i\e_j^T \qquad\quad \p{A^T}{A_{ij}} &= \e_je_i^T \\
}$$
where $\Big\{e_i\in\R^{m},\;\e_j\in\R^{n}\Big\}$ denote the standard basis vectors for the respective dimensions.

Now the gradient can be calculated as
$$\eqalign{
\p{F}{A_{ij}}
 &= (I-FS)(e_i\e_j^TB^{-1}A^T) + AB^{-1}\e_je_i^T(I-SF) \\
}$$
So that is the answer to the question in the title. However the question in the body of the post concerns the gradient of a scalar-valued function $\phi$ with respect to the matrix $A$.
$$\eqalign{
\phi &= u^TFu \\
d\phi
 &= u^TdF\,u \\
 &= {\rm Trace}(u^TdF\,u) \\
 &= {\rm Trace}(uu^TdF) \\
 &= {\rm Trace}\Big(uu^T(I-FS)\;dA\;B^{-1}A^T\Big)
  + {\rm Trace}\Big(uu^TAB^{-1}\;dA^T(I-SF)\Big) \\
 &= {\rm Trace}\Big(B^{-1}A^Tuu^T(I-FS)\;dA\Big)
  \;+ {\rm Trace}\Big((I-SF)\,uu^TAB^{-1}\;dA^T\Big) \\
 &= {\rm Trace}\Big(2\,(I-SF)\,uu^TAB^{-1}\;dA^T\Big) \\
\p{\phi}{A}
 &= 2\,(I-SF)\,uu^TAB^{-1} \\
}$$


问:
Calculate $\lim_{k\to\infty}\int_{(1,\infty)}\frac{k\sin(x/k)}{x^3}d\mathcal{L}^1(x)$
I'm trying to solve the following exercise: Show that $\lim_{k\to\infty}\int_{(1,\infty)}\frac{k\sin(x/k)}{x^3}d\mathcal{L}^1(x)$ exists and compute its value.

I have switched to the improper Riemann integral for $k$ fixed using dominated convergence since $|\frac{k\sin(x/k)}{x^3}| \le k\cdot 1/x^3\in\mathcal{L}^1((1,\infty])$ but now I don't know how to continue since (again for $k$ fixed) I have been unable to compute the value of the improper integral above.

Is there a different way to approaching this problem?

答:
Note that the function $f: \mathbb{R} \to \mathbb{R}$ defined by $f(t)= \frac{\sin(t)}{t}$ is a uniformly bounded (by $1$), thus

$$
 \left| \frac{k \sin(x/k)}{x^3} \right| = \left| \frac{\sin(x/k)}{x/k} \right| \frac{1}{x^2} \leq \frac{1}{x^2}, \qquad  k\geq 1.
$$
Moreover, the standard limit $\lim_{t\to 0} \frac{\sin(t)}{t}=1$ implies that

$$ \lim_{k \to \infty} \frac{k \sin(x/k)}{x^3} = \frac{1}{x^2}.
$$

Since $x \mapsto 1/x^2$ is integrable on $[1,\infty)$, we may invoke the dominated convergence Theorem, which gives

$$
\lim_{k \to \infty} \int_{1}^{\infty} \frac{k \sin(x/k)}{x^3} dx = \int_{1}^{\infty} \frac{1}{x^2} dx = 1.
$$


问:
Negative elements in pivot column when solving LP Simplex method?
what do I do if all the elements in the pivot column are negative? Details are as follow:

$$
\begin{aligned}
\max &-60000y_1 + 4800y_2 + 900y_3\\
\text{s.t.:}&\begin{cases}-50y_1 +6y_2 +y_3 \leq3\\-75y_1+ 6.75y_2 + y_3 \leq 1.8\end{cases}
\end{aligned}
$$

The most negative elements in the bottom row is $-60000$ which defines pivot column as the first non-basic var vector $(-50, -75)$.
What do we do in this situation since pivot element is not allowed to be negative?

Thank you

Edit:

@callculus: This is what I have come up with. Similar problem.

答:
If you put the max objective function into the table the signs are flipping: $-z=60000y_1-4800y_2-900y_3$ The table is

$$\begin{array}{|m{cm}|m{1cm}|}
\hline y_1   & \color{blue}{y_2} & y_3  & s_1 & s_2  & \textrm{RHS}\\  \hline
\hline  60000& -4800&-900 & 0 &0&0\\ \hline -50& 6&1&1&0&3  \\ \hline -75&\color{green}{6.75} &1&0&1&\color{orange}{1.8} \\ \hline  \end{array}$$

Here $s_1$ and $s_2$ denote the slack variables. The most negative coefficient of the objective function is $-4800$. Thus $y_2$ is the pivot column. And $\min \{\frac{3}{6},\frac{1.8}{6.75}\}=\frac{1.8}{6.75}$. Thus the last row is the pivot row. I think you can go on. The optimal solution is $(y_1^*,y_2^*,y_3^*) = (0.048, 0,5.4)$.


问:
Determine the number of six-digit odd positive integers with exactly one zero and where all digits in the number are different
Determine the number of six-digit
a)  odd positive integers with exactly one zero and where all digits in the number are different
b) the number of positive six-digit integers with exactly one one or two zeros.

I have got that there are 9 * 8 * 7 * 6 * 5 * 4 sex-digit odd numbers, and there are 9 * 7 * 8 * 6 * 5 * 5 odd sex digit numbers but I don't know the rest.

答:
For the case of exactly one zero.

We have available six positions, each of which must be filled.

$$\underline{~~~}~~\underline{~~~}~~\underline{~~~}~~\underline{~~~}~~\underline{~~~}~~\underline{~~~}$$

Choose the final digit.  It must be odd.  There are $5$ choices (1,3,5,7,9)

For example, we might have chosen the digit $3$ to be the final digit, which would make our current choices make the number look like this:

$$\underline{~~~}~~\underline{~~~}~~\underline{~~~}~~\underline{~~~}~~\underline{~~~}~~\underline{~3~}$$

Next, choose the position of the $0$.  It must not be the final digit or the leading digit.  There are $4$ choices where it may go (second, third, fourth, or fifth position)

For example, we might have chosen the digit $0$ to take the third position from the left.  That would make our current choices make the number look like this:

$$\underline{~~~}~~\underline{~~~}~~\underline{~0~}~~\underline{~~~}~~\underline{~~~}~~\underline{~3~}$$

From left-to-right, choose the digit occupying each remaining unused position noting that it may not be any of the previously selected digits.  There are $8,7,6,$ and $5$ choices for these respectively.  (The first of which we have available each of the digits except $0$ and except whatever was chosen as the final digit regardless which it happened to be, leaving $8$ choices.  The next of which we have available each of the digits except $0$, except whatever was chosen as the final digit, and except whatever was chosen as the leading digit, regardless what those happened to be leaving $7$ choices and so on...)

For example, at this point we have available the eight digits $\{1,2,4,5,6,7,8,9\}$ t use.  If we were to choose the first remaining position to be the digit $1$ that would make our number look like this:

$$\underline{~1~}~~\underline{~~~}~~\underline{~0~}~~\underline{~~~}~~\underline{~~~}~~\underline{~3~}$$

We can continue filling in the spaces with the remaining unused digits as need be to arrive at a final example of a six-digit odd number with exactly one zero.

By noting that these steps adequately follow the requirements to use the multiplication principle/rule of product, we can multiply the number of choices for each step together to get the total count of possible six-digit odd numbers with exactly one zero as being:

$$5\cdot 4\cdot 8\cdot 7\cdot 6\cdot 5$$

The case of exactly two zeroes can be performed with similar logic, noting that when choosing the positions of the two zeroes the positions are chosen simultaneously and neither can be the leading or final digit.

Note, here we used the technique that we made choices for the most restrictive positions/digits first and then returned to fill out the rest.  We do not have to always perform tasks from left to right.


问:
$ \prod_{k=0}^\infty \prod_{j=1}^q ( 1- \chi(j) z/(kq+j) ) = (1-z) \prod_{k=2}^\infty ( 1 - \chi(k) z/k) $
Let $\chi$ be a primitive non-principal Dirichlet character with conductor $q>1$.
The following equation

$$ \prod_{k=0}^\infty\prod_{j=1}^q \biggl( 1- \chi(j) \frac{z}{kq+j} \biggr) = (1-z) \prod_{k=2}^\infty \biggl( 1 - \chi(k) \frac{z}{k} \biggr) $$

appears as (3.6) in https://doi.org/10.1016/j.aam.2018.05.003. How does one go about proving this identity?

答:
If $f$ is $q$-periodic then $$\prod_{k=0}^\infty\prod_{j=1}^q \biggl( 1- f(j) \frac{z}{kq+j} \biggr)=\prod_{k=0}^\infty\prod_{j=1}^q \biggl( 1- f(kq+j) \frac{z}{kq+j} \biggr)=\prod_{n=1}^\infty \biggl( 1- f(n) \frac{z}{n} \biggr)$$
It remains to show it converges (locally uniformly, thus to an entire function) iff $f$ has zero mean, by taking the $\log$ and using that for $|z|<N$, $$\sum_{n>2 N}  \log(1-f(n) \frac{z}n)=-\sum_{n>2N} (f(n) \frac{z}{n}+O(\frac{z^2}{n^2}))$$ converges uniformly.


问:
Represent the common term of the sequence as minus one in polynomial degree
Two days ago I've found my old notebook in which I had been writing some problems from journals/books/olympiads. And also I've found that many of these problems are still unsolved, but still call crazy interest.
I've tried some of them by  myself and in some cases got something. But there are lot of problems that still a sort of mystery for me.
So, I'll start a series of posts with those problems. Where I can I'll write from where I've tooked the problem.

There is the first one:
Is it possible to represent the common term of the sequence $1,1,-1,1,1,-1,...$ in the from of
$x_n=(-1)^{P(n)}$, where $P(n)$ - is the polynomial?
With which $r,s$ where $r$ - is the number of consecutive $1$, $s$ - number of consecutive $-1$?

In my thoughts to the fisrt part of question we need to somehow construct the polynomial that will have such properties $P(3n-2)=P(3n-1)=0$ $mod$ $2$ and $P(3n)=1$ $mod$ $2$, or prove that such polynomial does'nt exist.

Unfortunately, I did'nt mark from where I've took this problem.

Any help will be appreciated!

答:
There is no such polynomial. As you have observed, as $(-1)^2 = 1$, we only care about the value of the polynomial modulo $2$.

Observe that for any $n>0$, $x^n$ has the same parity as $x$. Hence for any polynomial $P(x) = a_nx^n+a_{n-1}x^{n-1}+\dots +a_1x+a_0$, we can reduce it modulo $2$ to:

$$P(x) \equiv (a_n+a_{n-1}+\dots + a_1)x+a_0\equiv b_1x+b_0 \pmod 2$$

where $b_0, b_1$ are either $0$ or $1$. Now it is obvious that $P(x)$ either alternates in parity (when $b_1=1$) or is of constant parity (when $b_1=0$). Therefore there is no polynomial satisfying the condition.


问:
Finding the volume bounded to the area of $z = x + y+ 2, z=0$ and cylinder $x^2 + y^2 = 16$
I want to find the volume bounded to the area of $z = x  + y+ 2, z=0$ and cylinder $x^2 + y^2 = 16$ with triple Integral.

I appreciate someone to give me the approach or a hint on what to do?

Thank you.

答:
Hint:

$$V=4\int^4_0dx\int^{\sqrt {16-x^2}}_0dy\int^{x+y+2}_0dz$$

More explanation:

Let's start with simple form We want to find volume of a cylinder bounded by $x^2+y^2=16$, $z=0$ and $z=a$.We find a quarter of volume and multiply it by 4. The upper bound of x is 4, y is $y= \sqrt {16-x^2}$ and z is a and we can write:

$$V=4\int^4_0dx\int^{\sqrt {16-x^2}}_0dy\int^a_0dz$$

Now if a is a function of x and y it would mean the upper bound of z is a function of x and y(as mentioned in question).For y we considered $y=\sqrt{16-x^2}$ which is a circle, similarly we put $z=x+y+2$ , which is a plane,as the upper bound of z.If a is constant the integration will be easy, if it is a function of x and y , then integration becomes a bit complicated and messy, that's why other answers to this question suggest simpler method with easier calculations.


问:
Prove the identity $ \sum_{n=0}^{N} \frac{1}{(n+1)(n+2)} = \frac{N+1}{N+2} $
I have the identity $$ \sum_{n=0}^{N} \frac{1}{(n+1)(n+2)} = \frac{N+1}{N+2} $$ that I am supposed to prove by using the partial fraction decomposition $$ \frac{1}{n+1}- \frac{1}{n+2} $$ but I'm not sure how to proceed

Body:
Notice that this is a telescopic sum, meaning that:
$$ 1 - \frac{1}{2} + \frac{1}{2} - \frac{1}{3} + ... + \frac{1}{n+1} - \frac{1}{n+2} = 1 - \frac{1}{n+2} = \frac{n+2-1}{n+2}= \frac{n+1}{n+2} $$


问:
In how many ways can 4 couples sit in a row if no 2 women sit next to each other?
The Numbers of ways?

I am so confussed- I have looked at it tones of different ways and its not working.
The previous question was the same thing but using 7 couples and the answer was 203,212,800 and no matter what I tried I couldn't get that answer or the one for this version of the question.

Please help me understand the method of doing this type of problem?

答:
Assuming a couple to be one man and one woman, each seat has two possibilities,
 man or woman. With 7 men and 7 women, and for there not to be any women sitting next to one another, each woman must be delimited by at least one man.

_ M _ M _ M _ M _ M _ M _ M _

There are 8 spaces that the 7 women can be placed into:

$$\frac{8!}{(8-7)!}=8!$$

In addition to that, there are 7 men that can be placed into $7!$ different permutations, so by the rule of product the final expression is:

$$8!\space 7!=203212800$$

Use the same process for 4 couples.


问:
For which positive integer $n$, is $\frac{2^n-1}{3}$ a factor of $4m^2+1$ for some integer $m$? (How to prove that one answer group is the only one?)
Edit:
Dietrich Burde kindly gave me a hint but I, in my inexperience, am not able to understand the logic. I've tried but failed. So, right now my problem is understanding him. So, if you can explain the logic to me, I would be grateful.

For which positive integer $n$, is $\frac{2^n-1}{3}$ a factor of $4m^2+1$ for some integer $m$?
So, I've gotten that $n=2^q,\ q\in\mathbb{Z}_+$ is one solution, and I'm pretty certain that it is the only solution, but I don't know how to prove this (correct or wrong).

If you're interested on how I got to this point, well here's briefly what I've done:
$n$ must be an even integer, as $3\mid 2^n-1$ as $2^{2k}-1=4^k-1\equiv 1^k-1\equiv 0\ (mod\ 3)$

$n=2$ is a quite trivial solution.

$\frac{2^n-1}{3}$ must be a factor of $4m^2+1$, which gives us this (Diophantine) equation: $$4m^2+1=\frac{d(2^n-1)}{3}$$
For $d=1$, we get $4m^2=\frac{2^n-4}{3}$ and because $n$ must be even, we get:
$$4m^2=\frac{4^k-4}{3}$$
$$m^2=\frac{4^{k-1}-1}{3}$$
$$3m^2=(2^{k-1})^2-1^2$$
$$3m^2=(2^{k-1}-1)(2^{k-1}+1)$$
From this you can quite easily get $k=1, m=\pm 1$ as a solution so $n=4$.

But what about for other values of $d$?

Well, we have $n=2,4=2^1, 2^2$. Maybe it works for all $n=2^q$.
I proved this with induction using the fact that $2^{(2^{l+1})}=(2^2)^{2^l}$.

But is there other solutions? I don't know. I feel like there is no other ones because I haven't been able to find anything for the integers in between.

答: Hint: Every divisor $d$ of $4m^2+1$ satisfies $d\equiv 1\bmod 4$. And $\frac{2^n-1}{3}$ is an integer if and only if $n$ is even. Suppose that $n=2k$ is not a power of $2$. Then $\frac{2^n-1}{3}$ has a prime divisor $p$ with $p\equiv 3\bmod 4$, a contradiction.


问:
How many paths of length $n$ on a cycle of length $k$ that terminate at the start vertex?
Given a cycle of even length $k$ with vertices $v_1,v_2,\dots,v_k$ ($k=4$ is a square) and a start vertex $v_1$, how many paths of length $n$ that terminate in $v_1$ exist?

I started by trying to model the counting argument with distance from $v_1$ as $\delta(v_i)$ so we have

$\delta(v_1) = 0$

$\delta(v_2) = \delta(v_k) = 1$

$\delta(v_3) = \delta(v_{k-1}) = 2$

$\dots$

$\delta(v_{\frac{k}{2}+1})= \delta(v_{\frac{k}{2}-1})=\frac{k}{2}-2$

$\delta(v_{\frac{k}{2}}) = \frac{k}{2} - 1$

Now, for every vertex except $v_0$ and $v_{\frac{k}{2}}$ we either increase the distance or decrease it.

From $v_0$ we can only increase, and from $v_{\frac{k}{2}}$ we can only decrease.

From here on I'm quite stuck, it might not be the way but couldn't see a better way. How can we calculate this? I do see from this argument that for odd $n$ there are no such paths since the distance increased can't be the same as the distance decreased, but couldn't go much further.

Just to make it clear, $v_1$ is connected to $v_2$ and $v_k$, $v_2$ is connected to $v_1$ and $v_3$ and so on... $v_{k-1}$ is connected to $v_{k-2}$ and $v_k$ and $v_k$ is connected to $v_{k-1}$ and $v_1$ i.e., the vertices are ordered clockwise (or anti clockwise - doesn't matter).

答:
HINT

A good way to analyse this problem is to consider a path of length $n$ as a sequence of clockwise and anti-clockwise moves:  CAACAC ...

Then you are looking for the number of strings of length $n$ for which the number of As and Cs differ by a multiple of $k$.

Example $n=10,k=4$

The number of As in the string of length 10 is either 1,3,5,7 or 9. So you need

$\begin{pmatrix}10\\1\\\end{pmatrix}+\begin{pmatrix}10\\3\\\end{pmatrix}+...+\begin{pmatrix}10\\9\\\end{pmatrix}$.


答: Note that the average height at the center of the cylindrical shape is $h=2$, which leads its volume $V=\pi r^2 h= 32\pi$.

Title: Finding the volume bounded to the area of $z = x + y+ 2, z=0$ and cylinder $x^2 + y^2 = 16$

Body:
I want to find the volume bounded to the area of $z = x  + y+ 2, z=0$ and cylinder $x^2 + y^2 = 16$ with triple Integral.

I appreciate someone to give me the approach or a hint on what to do?

Thank you.


问:
How to construct a bijective mapping from a 4D cuboid to a 4D sphere?
I thought about the parameterizing the 4D sphere using
$$x_1= r\cos(\theta_1), x_2= r\sin(\theta_1)\cos(\theta_2), x_3= r\sin(\theta_1)\sin(\theta_2)\cos(\theta_3),\\ x_4= r\sin(\theta_1)\sin(\theta_2)\sin(\theta_3),$$
where $x_1,x_2,x_3,x_4$ are my 4D space coordinates.
But what will be the range of values of $\theta_1, \theta_2, \theta_3$ to make the mapping from $(r,\theta_1, \theta_2, \theta_3)$ to $(x_1,x_2,x_3,x_4)$ bijective ?

答:
I think the other commenters were too quick to disregard this question. I personally think that this question is interesting, so I'm going to give a hint.

When dealing with higher dimensional geometry, it usually helps to start first on a low dimensional case and then work your way up. WLOG, I will assume the radius of the sphere we want to map to is $1$. The way we do this is to shoot out rays radially that intersect the box and the sphere. Here's a diagram of the one dimensional case:

Can you construct a bijective mapping that takes a point on the circle to a point on the sphere? Think about the line that goes through both points. How can we generalize this to more dimensions?


问:
Transition matrix of reversible process.
I am reading about time reversal for Markov Chain and I am not exactly sure I got it right. The chain is reversible means that $P(X_{n+1}=j|X_n=i)=P(X_{n} = j|X_{n+1}=i)$ is that correct? Now in general, if we don't know that the chain is reversible and if we don't know about any invariant distribution, how do we compute $P(X_{n} = j|X_{n+1}=i)$?

答:
Let $(X_n)$ be Markov$(\lambda, P)$, where $P$ is irreducible. Then $(X_n)$ is reversible if, for all $N\geq1$, the sequence $(X_{N-n})_{0\leq n\leq N}$ is also Markov$(\lambda,P)$. Implicitly this requires that our initial distribution $\lambda$ is invariant for $P$.

The main result is the following:

Theorem: Let $P$ be irreducible, and $\pi$ an invariant distribution. Suppose $(X_n)_{0\leq n\leq N}$ is Markov$(\pi,P)$. Then $(X_{N-n})_{0\leq n\leq N}$ is Markov$(\pi,\hat{P})$, where $\hat{P}=(\hat{p}_{ij})$ is the matrix satisfying
$$\pi_j\hat{p}_{ji}=\pi_ip_{ij},$$
for all $i,j$. Further, $\hat{P}$ is irreducible, and has $\pi$ as an invariant distribution.

The proof is fairly boring -- you can try it yourself if you wish. There are two things worth pointing out:

The theorem says that a Markov chain, in equilibrium, is still a Markov chain when run backwards. The transition matrix could be different though.

If we want time-symmety, we must begin in an equilibrium of the Markov chain. This is intuitively clear -- else convergence to an equilibrium gives us a clue as to which "direction" of time we're going in.

So if $X_n)$ is Markov$(\lambda,P)$, it is reversible iff $\hat{P}=P$, i.e. $$\lambda_jp_{ji}=\lambda_ip_{ij}$$
for all $i,j$. These are the detailed balance equations. To prove that a chain is reversible, you need to show that there is a distribution $\lambda$ satisfying these equations.

As an exercise, you might like to show that a simple symmetric random walk on a finite connected graph is reversible.


问:
How can I show that an entire function where the sum of real and imaginary part are less or equal to the absolute of the function is constant?
Suppose there is an entire function f with

$\Re(f(nz))+\Im(f(nz))\leq|f(z)|,\qquad \qquad \forall z\in\partial\mathbb{D}, \forall n\in\mathbb{N} $,

How could I show that $f$ is constant?

答:
Can't we just use $|f(nz)| \leq |f(z)|$ for each $n, z,$

and the Maximum Modulus Principle to say $f$ is bounded by $\max_{z \in \partial D} |f(z)| = M < \infty?$

If $\forall w \in \mathbb{C}$ we find a larger radius $n > |w|,$ MMP bounds $|f(w)| \leq |f(nz)| \leq M.$ Such $f$ should be a constant by Liouville.


问:
Why should we use the fundamental definition of derivative while checking differentiability?
I have seen in many examples that using the fundamental definition of derivative only gives the correct results while checking for differentiability at a point. But why do we get wrong results while we directly apply the rules of differentiation and taking the limit at the point. Here is an example

$$y = \begin{cases}
x^2, &-\infty < x < 1\\
2x, &1\leq x < \infty
\end{cases}
$$

If we check the differentiability at the point $x=1$ then when we apply the rules of differentiation left hand and right hand derivative becomes same. But by using fundamental definition of derivative it will be different.

But if I am right the rules of differentiation are in fact derived from the fundamental definition of derivative? So both of these should give correct result. Or it seems like I am missing some basic concept.

I have seen same question like this one here. The answer to this question only says it gives wrong result.But it doesn't answer the question of why?

答:
When considering limits, you always have to look at what happens as $h$ approaches $0$ both from the left hand side and right hand side. The limit
$$
\lim_{h \to 0}\frac{f(x+h)-f(x)}{h}
$$
exists if and only if the left-hand limit and the right-hand limit both exist and are equal.

If
$$
f(x) = \begin{cases}
x^2, &-\infty < x < 1\\
2x, &1\leq x < \infty
\end{cases}
$$
then
$$
\lim_{h \to 0^+}\frac{f(1+h)-f(1)}{h} = \lim_{h \to 0^+}\frac{2+2h-2}{h} = \lim_{h \to 0^+}\frac{2h}{h}=2
$$
but
\begin{align}
\lim_{h \to 0^-}\frac{f(1+h)-f(1)}{h} &= \lim_{h \to 0^-}\frac{(1+h)^2-2}{h} \\
&=\lim_{h \to 0^-}\frac{h^2+2h-1}{h} \\
&= \lim_{h \to 0^-}h+2-\frac{1}{h} \\
&= \infty \, .
\end{align}

You can't use the derivative rule
$$
\frac{d}{dx}(x^2)=2x
$$
since $f(x) \neq x^2$ when $x=1$.


答:
The issue here is that the derivative $y'(1)$ is not defined.  I'll start with the definition and then discuss why the differentiation rules don't apply at $x = 1$.  First, for the derivative to exist at $x = 1$, then by definition,the following limit has to exist:
$$\lim_{h\to 0}\frac{y(1+h) - y(1)}{h} \tag{1}.$$
Now, in order for $(1)$ to exist the left and right limits have to exist and agree.  So, we can look at the left and right derivatives:

Right Derivative:
$$y'_{+}(1) = \lim_{h\to 0^{+}}\frac{f(1+h) - f(1)}{h} = \lim_{h\to 0^{+}}\frac{2(1+h) - 2}{h} = \lim_{h\to 0^{+}}\frac{2h}{h} = 2$$
There's no trouble there, however we run into problems with the left derivative:

Left Derivative:
\begin{align}
y'_{-}(1)&=\lim_{h\to 0^{-}}\frac{y(1+h) - y(1)}{h}\\
&= \lim_{h\to 0^{-}}\frac{(1+h)^2 - 2}{h}\\
&= \lim_{h\to 0^{-}}\frac{1 +2h+h^2-2}{h}\\
&= \lim_{h\to 0^{-}}\frac{h^2+2h-1}{h}\\
&= \lim_{h\to 0^{-}}\left(h + 2 - \frac{1}{h}\right)\\
&= \infty
\end{align}

So, since the left and right derivatives don't agree, we can say that $y$ is not differentiable at $y = 1$.  As has been pointed out in the comments we could also have concluded that from continuity, because if $y'(1)$ had existed, then $y$ would need to be continuous at $x = 1$, and because it is not we can immediately rule out differentiability there.

Now, to understand why it doesn't work to take the derivatives of $2x$ and $x^2$ and take a limit, we start by noting that we can certainly write
$$y'(x) = \begin{cases}
2x,&-\infty<x<1\\
2, &1<x<\infty
\end{cases}$$

However, if we were to write $$y'(1) = \lim_{x\to 1}y'(x)$$ we would be making two unfounded assumptions:

First, that $y$ is actually differentiable at $x = 1$, and so writing $y'(1)$ makes sense, and second that $y'$ is continuous.  We don't know a priori that either of those things are true, and in fact as previously mentioned, we know that $y'(1)$ can't exist because $y$ isn't continuous at $x = 1$.

To give a bit more of an intuitive explanation, remember that the derivative at a point relies on the value of the function at that point.  So, when we use $2x$ for the derivative on the left, instead of using $y(1) = 2$, which lies on the right piece, we're really using $\lim_{x\to 1^{-}}y(x) = 1$ on the left piece.

Title: Why should we use the fundamental definition of derivative while checking differentiability?

Body:
I have seen in many examples that using the fundamental definition of derivative only gives the correct results while checking for differentiability at a point. But why do we get wrong results while we directly apply the rules of differentiation and taking the limit at the point. Here is an example

$$y = \begin{cases}
x^2, &-\infty < x < 1\\
2x, &1\leq x < \infty
\end{cases}
$$

If we check the differentiability at the point $x=1$ then when we apply the rules of differentiation left hand and right hand derivative becomes same. But by using fundamental definition of derivative it will be different.

But if I am right the rules of differentiation are in fact derived from the fundamental definition of derivative? So both of these should give correct result. Or it seems like I am missing some basic concept.

I have seen same question like this one here. The answer to this question only says it gives wrong result.But it doesn't answer the question of why?


问:
Splitting field of a polynomial over $\mathbb{Z_3}$
Let $E$ be the splitting field of $f(x)=x^4-x-2 $ over $\mathbb{Z_3}$ determine $[E:\mathbb{Z_3}]$ and factor $f(x)$ in linear factors in $E[x]$

It's clear that $2$ is a root of $f(x)$ in $\mathbb{Z_3}$, hence we can write $f(x)=(x-2)(x^3+2x^2+x+1)$ in $E[x]$, I'm unsure how I could go on from here since the cubic we got it's not easily factorable, it should have a real root and two complex conjugate solutions.

How can I determine $[E:\mathbb{Z_3}]$? and how could I factor further? The solution presented on the notes where I found the exercise states the following factorization: $f(x)=(x-2)(x-\alpha)(x-\alpha^3)(x-\alpha^9)$ where $\alpha$ is a root of $x^3+2x^2+x+1$

答:
I assume that $\mathbb{Z}_3$ refers to the finite field with $3$ elements. First show that $g=x^3+2x^2+x+1$ is irreducible. Note that it is a degree $3$ polynomial, so it is sufficient to show that it has no roots (in $\mathbb{Z}_3$).

So you know there is a degree $3$ extension of $\mathbb{Z}_3$ in which $g$ has a root $\alpha$. Now show that this is already a splitting field of $g$ and therefore a splitting field of $f$ and determine the conjugates of $\alpha$ in order to factor $f$.
Hint for this part: Extension of finite fields are Galois-extensions. What is the Galois group?


问:
Probabillity problem using Central limit theorem
I have the following problem:

On average, a pearldiver finds a pearl in every 50th shell.

How many shells does he at least need to find to get at least one pearl with a probability of at least 95%?

A hint suggests using the central limit theorem.

Now, I tried using the CLT:

Let $S_n=X_1+X_2+...+X_n$ ,$\quad X_i$ i.i.d. random variables,

$\mathbb{E}(S_n)=n\mu$,$\quad$ Var(X)=$n\sigma^2$,$\quad$ where $\mathbb{E}(X_i)=\mu$ and Var($X_i)=\sigma^2$.

Let $Z_n= \frac{S_n-n\mu}{\sigma\sqrt{n}}$ .

The CLT states that:

$\lim_{n\rightarrow \infty} P(Z_n\leq z)=\phi(z)$$\qquad$$\phi(z):=$CDF of $\mathcal{N}(0,1) \quad \forall z\in \mathbb{R}$

I saw in the standard normal table that $\phi_{0;1}(1.65)=0.95953$ is the smallest number whose probability is $\geq0.95$.

But where do I go frome here?

Since I don't know $\sigma^2$ and only $\mu$, I really can't do anything with the CLT.

I would really appreciate some advice.

答: Note that you haven't used the first sentence of the question statement yet. In this particular problem, I think you are supposed to assume that $X_1,\ldots, X_n$ are i.i.d. $\text{Bernoulli}(1/50)$. From there you can compute the mean and variance and follow the rest of your work.


答:
Dominated Convergence works, but here is an alternate approach.
$$
\begin{align}
\int_1^\infty\frac{k\sin(x/k)}{x^3}\,\mathrm{d}x
&=\frac1k\int_{1/k}^\infty\frac{\sin(x)}{x^3}\,\mathrm{d}x\tag{1a}\\
&=\frac1k\int_{1/k}^\infty\frac{x}{x^3}\,\mathrm{d}x\\
&+\color{#C00}{\frac1k\int_{1/k}^1\frac{\sin(x)-x}{x^3}\,\mathrm{d}x+\frac1k\int_1^\infty\frac{\sin(x)-x}{x^3}\,\mathrm{d}x}\tag{1b}\\
&=\frac1k\cdot k+\color{#C00}{O\!\left(\frac1k\right)}\tag{1c}
\end{align}
$$
Explanation:

$\text{(1a)}$: substitute $x\mapsto kx$

$\text{(1b)}$: break integrand into pieces

$\text{(1c)}$: $\left|\frac{\sin(x)-x}{x^3}\right|\le\min\left(\frac16,\frac1{x^2}+\frac1{x^3}\right)$

Thus,
$$
\lim_{k\to\infty}\int_1^\infty\frac{k\sin(x/k)}{x^3}\,\mathrm{d}x=1\tag2
$$

Title: Calculate $\lim_{k\to\infty}\int_{(1,\infty)}\frac{k\sin(x/k)}{x^3}d\mathcal{L}^1(x)$

Body:
I'm trying to solve the following exercise: Show that $\lim_{k\to\infty}\int_{(1,\infty)}\frac{k\sin(x/k)}{x^3}d\mathcal{L}^1(x)$ exists and compute its value.

I have switched to the improper Riemann integral for $k$ fixed using dominated convergence since $|\frac{k\sin(x/k)}{x^3}| \le k\cdot 1/x^3\in\mathcal{L}^1((1,\infty])$ but now I don't know how to continue since (again for $k$ fixed) I have been unable to compute the value of the improper integral above.

Is there a different way to approaching this problem?


问:
Diagonalizable matrices in $M_{2\times 2}(\mathbb{F}_2)$
List all diagonalizable $2\times 2$ matrices over the a field $F$ consisting of two elements $0$ and $1$.

I want to try and do this using C++, but perhaps this isn't the place to ask. I have an idea as to how I'd do it.

答:
Consider similarity classes(equivalence class consisting of similar matrices)
There are only three classes having representatives:

$$
\pmatrix{0& 0\\0&0}\;\pmatrix{1& 0\\0&1}\;\pmatrix{1& 0\\0&0},$$
respectively.

Then consider conjugation with invertible matrices in $M_{2\times 2}(\mathbb{F}_2)$.

There are only 6 of them. (i.e. $|GL_2(\mathbb{F}_2)|=6$)

Edit:  If we allow finite extension of $\mathbb{F}_2$, then the following one is also a diagonalizable matrix.
$$
\pmatrix{1&1\\1&0}$$
which has eigenvalues $\lambda=\lambda_1$, $\lambda_2$ satisfying
$$\lambda^2+\lambda+1=0.$$

So if we consider diagonalizable matrices in $M_{2\times 2}(\mathbb{F}_2)$ over $\overline{\mathbb{F}_2}$ are not necessarily projection.


问:
Writing the following set using set builder notation
I'm learning DM and currently stuck on finding the correct set builder for:

B={−1, 1/2, −1/3, 1/4, −1/5, 1/6,···}

how do i find the steps? Thanks in advance

答: The series is of the form (-1)^n * 1/n</p>

Body: The series is of the form (-1)^n * 1/n


问:
Coin that lands in a plane tessellated by triangles..
A coin of radius 1 cm is tossed onto a plane surface that has been tessellated by right triangles whose sides are 8 cm, 15 cm, and 17 cm long. Find the probability that the coin lands within a triangle.

I know that this has to do with similarity because the inner triangle that is formed by the area where the coin can land is similar to the outer triangle. Therefore, I know the angles of this triangle, but I am not sure how to find one side of this triangle.

答:
First since $I$ is the incenter of $\triangle ABC$, you can find $$IH={2S_{ABC}\over P_{ABC}}={120\over 40}={3}$$ where $S$ denotes area and $P$ denotes perimeter.

Next you find $$IM=IH-1=2$$

Notice $I$ is also the incenter of the inner triangle. Now you know the side ratio between the two triangles is $2\over 3$.


问:
Find $\frac{a+b}{ab}$ such that $\int_{-1/2}^{1/2} \cos x\ln\frac{1+ax}{1+bx}dx=0$
Let $f(x) = \cos(x) \ln\left(\frac{1+ax}{1+bx}\right)$ be integrable on $\left[-\frac{1}{2} , \frac{1}{2}\right]$. Let
$$\displaystyle \int_{-1/2}^{1/2}f(x)\operatorname{dx}=0$$

where $a$ and $b$ are real numbers and not equal. Find the value of $\frac{a+b}{a\cdot b}$.

At first I applied integration by parts but then it got even more complicated and I couldn't solve it. How can I solve this problem?

答:
To be zero the integral, $h(x)=\log \left(\frac{1+a x}{1+b x}\right)$ must be odd, thus

$$\log \left(\frac{1+a x}{1+b x}\right)=-\log \left(\frac{1-a x}{1-b x}\right)$$
$$\log \left(\frac{1+a x}{1+b x}\right)+\log \left(\frac{1-a x}{1-b x}\right)=0$$
$$\log\frac{1-a^2x^2}{1-b^2x^2}=0$$
$$1-a^2x^2=1-b^2x^2\to b=-a$$
since the problem says $a\ne b$.

$h(x)=-h(-x)\to b=-a$

Must be added that  $h(x)$ exists on $[-1/2,1/2]$ only if
$$\frac{1+a x}{1-a x}>0\to -2<a<2$$


问:
Locally free modules are flat
The following theorem is well-known in commutative algebra:

Let $A$ be a ring. Suppose that, for all prime ideals $\frak p$ of $A$, the module $M_\frak p$ is flat over $A_\frak p$. Then $M$ is flat over $A$.

Up until now, I never used this theorem but always:

free modules are flat

if a module is torsion over a domain, then it is not flat

extension of scalars preserves flatness

etc. I would like to see some concrete, non-trivial examples of the above theorem in use, where using this theorem is favorable over using any three of the above.

Any help would be appreciated.

答:
Let $R=\mathbb Z$ and $M=\{a/b\in\mathbb Q\mid b\text{ is squarefree}\}$. Let $p$ be a prime. We claim that $M_{(p)}\cong\Bbb Z_{(p)}$. To this end, let $m\in M_{(p)}$. Then we can write $$m=\frac{a}{bp^{n}},$$ where $b\notin(p)$ and $n\in\{0,1\}$. Define a linear map $\phi:M_{(p)}\to\Bbb Z_{(p)}$ by $\phi(m)=pm$. Conversely, define $\psi:\Bbb Z_{(p)}\to M_{(p)}$ by $\psi(f)=f/p$. Notice that $$\frac{f}p=\frac{(a/p)}{b}$$ for some $b\notin (p)$ and $a/p\in M$. Verify that $\psi$ and $\phi$ are inverses of each other.

Thus $M_{(p)}$ is free and hence flat as a $\mathbb Z_{(p)}$-module. It follows that $M$ is a flat $\mathbb Z$ module.

Title: Locally free modules are flat


问:
Ratio of sides in a trapezoid
The problem :

Let the lengths of bases $AD$ and $BC$ of trapezoid $ABCD$ be $a$ and $b$ ($a>b$). Find the length of segment $MN$ whose endpoints divide $AB$ and $CD$ in the ratio $AM:MB$ = $DN:NC$ = $p:q$.

I tried constructing a segment $BF$ ($F$ lies on $AD$, $BF || CD$), But can't go further.

How do I go about it?

答:
Continue with your construction of BF || CD and observe that the triangles BAF and BME are similar (E: intersection of MN and BF), which establishes

$$\frac{MN- b}{a-b}= \frac q{p+q}$$
and yields

$$MN= \frac{aq +bp }{p+q}$$

Title: Ratio of sides in a trapezoid


问:
Asymptotic behavior of $x_{n+1}=x_n+\frac{1}{x_n}, \space\space\space x_0=1$
I have defined a sequence $x_n$ as follows:
$$x_{n+1}=x_n+\frac{1}{x_n}, \space\space\space x_0=1$$
After convincing myself that there is no nice closed-form formula for $x_n$, I decided to try and find an asymptotic formula for $x_n$ (unfortunately, I have very little experience with asymptotic formulae). I noticed that the recurrence is equivalent to
$$\Delta x_{n}=\frac{1}{x_n}$$
and so a solution can be approximated by solving the corresponding differential equation for $y(t)$:
$$y'=\frac{1}{y}$$
This differential equation (with initial value $y(0)=1$) yielded
$$y=\sqrt{2t+1}$$
which led me to believe that
$$x_n\approx \sqrt{2t+1}$$
When I plotted $x_n$ and $y(n)$ side by side, it did indeed appear that they were very close to each other. However, this isn't enough for me... I would like to prove that $y(n)$ is a good approximation for $x_n$, either by proving that
$$\lim_{n\to\infty}(x_n-\sqrt{2n+1})=0$$
...or, even better, by finding a zero-approaching function $f$ satisfying
$$x_n=\sqrt{2n+1}+O(f(x))$$
This is where I got stuck. How do can I prove (or disprove?) the statement in the first equality, and find $f$ satisfying the second?

NOTE: There may be a closed-form that I wasn't able to find. In fact, the similar sequence
$$y_{n+1}=\frac{y_n}{2}-\frac{1}{2y_n}$$
has closed form formula
$$y_n=\frac{1}{\tan(2^n\arctan(y_0^{-1}))}$$
...but even if you do find a closed form, I would still like to know how to prove the above statements, since the techniques would be useful to know for future problems.

答:
I will borrow some notation from Will Fisher's answer. We can estimate your limit by looking at it this way: we have
$$
x_n - x_0 = \sum_{i=0}^{n-1} \Delta x_i = \sum_{i=0}^{n-1} \frac 1{x_i}.
$$
We also have
$$
b_n - b_0 = \sum_{i=0}^{n-1} \Delta b_i = \sum_{i=0}^{n-1} (\sqrt{2i+3} - \sqrt{2i+1}) = \sum_{i=0}^{n-1} \frac 2{\sqrt{2i+3} + \sqrt{2i+1}}.
$$
Since we have no closed form for $x_n$ but our approximation $b_n$ is very good, I will now look at the closed forms of $\Delta b_n$ and $\frac 1{b_n}$.

We have
$$
\frac 1{b_n} - \Delta b_n
= \frac 1{\sqrt{2n+1}} - \frac 2{\sqrt{2n+3}+\sqrt{2n+1}} \\
= \frac{\sqrt{2n+3} - \sqrt{2n+1}}{\sqrt{2n+1}(\sqrt{2n+3} + \sqrt{2n+1})} \\
= \frac{(2n+3) - (2n+1)}{\sqrt{2n+1}(\sqrt{2n+3} + \sqrt{2n+1})^2} \\
= \frac 2{\sqrt{2n+1}(\sqrt{2n+3} + \sqrt{2n+1})^2}
$$
By replacing $3$'s by $1$'s and vice-versa, we obtain the following bounds for all $n \ge 0$:
$$
\frac 1{2(2n+3)^{3/2}} \le \frac 1{b_n} - \Delta b_n \le \frac 1{2(2n+1)^{3/2}}.
$$
This suggests that if we add up the $\frac 1{b_i}$'s instead of $\Delta b_i$ in the expression $b_n - b_0 = \sum_{i=0}^{n-1} \Delta b_i$, we would have an error term which would look roughly like a sum of terms of the form $i^{-3/2}$, thus giving an error roughly like $n^{-1/2}$, which goes to $0$ with $n$. Therefore, we can use the $\frac 1{b_i}$ instead of $\Delta b_i$ to compare with $\frac 1{x_i}$ (we will do this more formally in a second).

Writing $\varepsilon_n \overset{def}= x_n - b_n$, we also have as in the other answer
$$
f(b+\varepsilon) - f(b)
 = \left( (b+\varepsilon) + \frac 1{b+\varepsilon} \right) - \left( b + \frac 1b \right) \\
 = \varepsilon - \frac {\varepsilon}{b(b+\varepsilon)}
$$
Therefore
$$
x_{n+1} - b_{n+1}
 = \left( x_n + \frac 1{x_n} \right) - (\Delta b_n + b_n) \\
 = \left( x_n + \frac 1{x_n} \right) - \left( b_n + \frac 1{b_n} \right) + \left( \frac 1{b_n} - \Delta b_n \right) \\
 = \left( f(x_n) - f(b_n) \right) + \left( \frac 1{b_n} - \Delta b_n \right) \\
 = (x_n - b_n)\left( 1 - \frac 1{x_nb_n} \right) + \left( \frac 1{b_n} - \Delta b_n \right).
$$
so that letting $c_n = 1 - \frac 1{x_nb_n}$ and $d_n = \frac 1{b_n} - \Delta b_n$, we have $x_{n+1} - b_{n+1} = (x_n - b_n) c_n + d_n$, hence by induction on $k$, we deduce
$$
x_{n+k} - b_{n+k} = (x_n - b_n) \left( \Pi_{i=0}^{k-1} c_{n+i} \right) + \sum_{i=0}^{k-1} \left( \Pi_{j=i+1}^{k-1} c_{n+j} \right) d_{n+i}.
$$
We now show that as $n+k$ go to infinity, both terms go to zero.

The second term goes to $0$ with $n$ since
$$
\sum_{i=0}^{k-1} \left( \Pi_{j=i+1}^{k-1} c_{n+j} \right) d_{n+i}
\le \sum_{i=0}^{k-1} d_{n+i} \\
\le \sum_{i=0}^{k-1} \frac 1{2(2(n+i)+1)^{3/2}} \\
\le \int_{n-1}^{\infty} \frac 1{2(2x+1)^{3/2}} dx \\
= \frac 12 \left( \left. \frac {-1}{\sqrt{2x+1}} \right|_{n-1}^{\infty} \right) \\
= \frac 12 \left( \frac 1{\sqrt{2n-1}} \right) \to 0.
$$
Note that this bound is uniform over $k$, i.e. only depends on $n$.

For the first term, note that $|x_n - b_n| \le C$ is uniformly bounded for all $n$ (see Will Fisher's answer for why $C \approx 0.56$), so it suffices to show that $\Pi_{i=0}^{k-1} c_{n+i}$ goes to $0$ with $k$ for $n$ fixed. In this way, we can fix $n$ so that the second term is arbitrarily small, and then take $k$ large enough to make the first term arbitrarily small; this implies the sum of both terms can be arbitrarily small, given that $n+k$ is large enough.

Since $x_n - b_n \ge 0$, we have $x_n \ge b_n$, hence $\frac 1{x_n} \le \frac 1{b_n}$, so that $c_n \le 1 - \frac 1{x_n^2} < 1$. Using $\log(1-x) \le -x$ for $x \ge 0$, it follows that
$$
\log \left( \Pi_{i=0}^{k-1} c_{n+i} \right)
= \sum_{i=0}^{k-1} \log c_{n+i}
\le \sum_{i=0}^{k-1} \log \left( 1 - \frac 1{x_{n+i}^2} \right)
\le - \sum_{i=0}^{k-1} \frac 1{x_{n+i}^2} \underset{k \to \infty}{\longrightarrow} -\infty
$$
where the limit over $k$ follows from the following argument: since $x_n \le b_n + C \le 2b_n$ where $C \approx 0.56 \le 1 = b_0 \le b_n$ (see Will Fisher's answer for why $C \approx 0.56$), we have $\frac 1{x_n} \ge \frac 1{2b_n}$, thus
$$
\sum_{i=0}^{k-1} \frac 1{x_{n+i}^2} = \sum_{m=n}^{n+k-1} \frac 1{x_m^2} \ge \frac 14 \sum_{m=n}^{n+k-1} \frac 1{2m+1} \underset{k \to \infty}{\longrightarrow} \infty.
$$
Hope that helps,


问:
Formalizing composition of sequences
The following first equality is intuitively evident:
$$
\lim_{n\to \infty} \left(1+\frac{x+y+\frac{xy}{n}}{n}\right)^n
= \lim_{n\to \infty} \left(1+\frac{x+y}{n}\right)^n
=e^{x+y}
$$
how can I give a formal argument?

Informally it is something like
$$
\lim_{n\to \infty} \left(1+\frac{x+y+\frac{xy}{n}}{n}\right)^n
=\lim_{n\to \infty} \left(1+\frac{\lim_{n\to \infty}\left(x+y+\frac{xy}{n}\right)}{n}\right)^n
=\lim_{n\to \infty} \left(1+\frac{x+y}{n}\right)^n
$$
Is there a "classical" property of sequences that backs this argument? Something like limit of composition, but here we speak of sequence only:
$$
(b_n),\quad b_n \to b\quad\text{ for }n\to\infty \implies
\lim_{n\to \infty} \left(1+\frac{b_n}{n}\right)^n
=\lim_{n\to \infty} \left(1+\frac{b}{n}\right)^n
$$

EDIT As Qiaochu Yuan is important to define what I have already defined.
This computation is at the beginning of "Real Analysis", lets say in the chapter of sequences. Later, in the notes of complex analysis, I develop formal series with derivation and everything is the standard way (I think).

By this question I tried another approach, very early (1) define
$$
e^x = \lim_{n\to\infty}\left(1+\frac{x}{n}\right)^n
$$
(2) derive the product as depicted in this question
(3) derive the derivation of the $e^x$.

Thus I have (a) no logarithm, (b) no derivation of $e^x$ yet.

答:
Another approach: Note that

$$\left(1+\frac{x+y+\frac{xy}{n}}{n}\right)^n
= \left(1+\frac x n\right)^n\left(1+\frac y n\right)^n.$$

Addendum

We have that
\begin{align*}
\left(\frac{1+\frac{x+y+\frac{xy}{n}}{n}}{1+\frac{x+y}{n}}\right)^n
&= \left(
\frac{\left(1+\frac x n\right)\left(1+\frac y n\right)}{1+\frac{x+y}{n}}\right)^n
\longrightarrow \frac{e^x e^y}{e^{x+y}} \quad(n\to\infty).
\end{align*}
It remains to show that this limit is unity.
See this answer, for example.

Title: Formalizing composition of sequences


问:
Negative Binomial Series
On this webpage, the expansion of the negative binomial series is given below.

$$
(x+a)^{-n} = \sum^\infty_{k=0} (-1)^k \begin{pmatrix}n + k - 1\\k\end{pmatrix} x^ka^{-n-k}
$$
when $|x| < a$. My questions are as follows:

Are we assuming that $a > 0$?

What happens when $|x| < a$ condition is not true?

答:
I'm not an expert, but I think this condition is not meant to be restrictive, but rather an indication of how to chose $x$ and $a$ between the two. If $x = a$, there is no need for the formula, because $(x + x)^{-n} = 2^{-n} x^{-n}$. If both $x$ and $a$ are negative, you can factor a $(-1)^{-n} = (-1)^n$ out of the expression and multiply it in the formula. If one of the terms is positive and the other is negative, check which term has a smaller absolute valua and you can factor $(-1)^n$ if the negative one does (Note that the terms can not be the opposite of each other). Then, you choose it to be $x$. Otherwise, if both terms are positive, chose the smaller one as $x$.

Examples:

Both negative: $(-1 - 2)^{-n} = (-1)^{n}(1+2)^n$, $x=1, a = 2$

Both positive: $(3 + 5)^{-n}$, $x=3, a=5$

One positive, one negative: $(-1 + 3)^{-n}, |-1| < |3|, x=-1, a=3$,
$(4 - 2)^{-n}, |4| > |-2|, x=-2, a=4$,
$(-4+2)^{-n} = (-1)^n (4-2)^{-n}$

Title: Negative Binomial Series


问:
Perturbation and real rank zero $C^*$algebra
Let $A$ be a unital $C^*$-algebra. $RR(A) = 0$ iff for any $\epsilon > 0$ and for any pair of orthogonal positive elements $a, b \in A_+$ there is a projection $p \in A$ such that $||pa - p || < \epsilon$ and $(1 - p) b = b$.

Is there any idea how to prove both sides? I'm stucked.

答:
$\textbf{(ii)}$The elements in $A_{sa}$ with finite spectra are dense in $A_{sa}$, then $\textbf{(iii)}$ every hereditary $C^*$-subalgebra of $A$ has an approximate unit consisting of projections.

$\textbf{proof}$:This is the implication proved in G. K, PEDERSEN, The linear span of projections in simple $C^*$-algebras.

By the hypothesis we can have $\textbf{(v)}$ there is a projection $p \in A$ such that $||(1-p)a||\leq \epsilon$ and $||pb|| \leq \epsilon$.

We have $\textbf{(v)} \Rightarrow \textbf{(i)} = RR(A) = 0$. Since, if $a \in A_{sa}$ and $\epsilon >0$, consider the orthogonal elements $a_+$ and $a_-$ in $A_+$. Choose a projection $p$ in $A$ such that $||(1-p)a_+|| \leq \epsilon$ and $||pa_- || \leq \epsilon$. We get $|| a - (pap + (1-p)a(1-p))|| 2\epsilon - \epsilon p \leq pap$. Thus the element $y = pap + 2\epsilon p + (1-p)a(1-p)-2\epsilon(1-p)$ is invertable in $A_{sa}$ and $||a - y|| \leq 4 \epsilon$.

Also, with a little effort we can show that $ \textbf{(v)} \Rightarrow \textbf{(ii)}$ and $ \textbf{(v)} \Rightarrow \textbf{(iii)}$

And of course it is easy to see that $ \textbf{(i)} \Rightarrow \textbf{(vi)}$ and $\textbf{(vi)}$: for each pair $a, b \in A_{+}$ and $\epsilon >0$ such that $||ab|| < \epsilon^{2}$, there is a projection $p$ in $A$, such that $||(1-p)a|| < \epsilon$ and $||pb||<\epsilon$


问:
Proving $6n^4>(n+2)^4$ for $n\geq 4$, by induction

I need to prove that:
$$6n^4>(n+2)^4$$
for $n\geq4$ is true using proof by induction.

I've started by showing that when $P(4)$ is true since $\text{LHS}=1536$ and $\text{RHS}=1296$.

I then let $k≥4$ and assumed $P(k)$ is true, which is $6k^4>(k+2)^4$.

So we need to show that P(k+1) is true, so,
$$6(k+1)^4>(k+3)^4$$

This is where I get stuck, so would be very grateful if anybody could help finish the proof.

Thanks
Kara :-)

Body: $$\sqrt[\leftroot{-3}\uproot{3}4]{6}(k+1)=\sqrt[\leftroot{-3}\uproot{3}4]{6}k+\sqrt[\leftroot{-3}\uproot{3}4]{6}>(k+2)+1=k+3$$


问:
Diagonalizable Operators: An Operational Extension
Let $T$ be a diagonalizable operator on a vector space $V$. Prove that the operator
$$a_nT^n + a_{n-1}T^{n-1}+\cdots+a_1T+a_0 Id_V$$
 on $V$ is also diagonalizable for any scalars $a_1, a_1,\dots,a_n$.

First off, what is $Id_V$? I've never seen this before.

答: I found this in Michael Artin's Algebra, and I feel it will be of use here:


问:
Why does replacing dx and dy with finite values yield this result?

Consider the change in the function $f(x)=x^2$ at $x=1$  when $x$ is changed by some value $\Delta x$.
$$
\Delta y = f(x+\Delta x)-f(x)\\
\to \Delta y = (1+\Delta x)^2-1^2
$$
Let's set this value $\Delta x$ to $0.1$. Then
$$
\Delta y = (1+0.1)^2-1^2=0.21
$$

Now let's consider this change in the function utilizing its derivative.
$$
\frac{dy}{dx}=2x\\
dy=2x(dx)
$$
Let's replace the infinitesimal differences $dx$ and $dy$ with a finite difference (is this valid?).
$$
\Delta y = 2x(\Delta x)
$$
Again, let's set $\Delta x$ to 0.1.
$$
\Delta y = 2x(0.1)\\
\to\Delta y=2(1)(0.1) = 0.2
$$
Why does $\Delta y$ end up being different than it was previously? I would think that since we're replacing $dx$ with a finite difference, $dy$ would be the same as  $\Delta y$ in #1. There must be a flaw in my thinking somewhere.

Edit:
Thanks for the answers. Here's the justification I worked out.
$$
\frac{\Delta y}{\Delta x} = \frac{(x + \Delta x)^2 - x^2}{\Delta x}\\
\frac{\Delta y}{\Delta x}=\frac{x^2+2x\Delta x + (\Delta x)^2 - x^2}{\Delta x}\\
\frac{\Delta y}{\Delta x}= 2x+\Delta x\\
\to\Delta y = \Delta x(2x + \Delta x)\\
$$
As $\frac{\Delta y}{\Delta x}$ approximates $\frac{dy}{dx}$ with smaller and smaller values of $\Delta x$, a remainder term is always present. Since $x^2$ is exponential, $\Delta y$ must be greater than multiplying the current growth rate, $f'(x)$ by $\Delta x$.

Using 0.1 for $\Delta x$ gives us
$$
\Delta y = 0.1(2(1) + 0.1) = 0.21
$$
the same value calculated in #1.

答:
Note that $\displaystyle \frac {dy} {dx} = \lim_{\Delta x \to 0} \frac{\Delta y} {\Delta x} $.

That limit is very important. Taking $\frac {dy} {dx} \approx \frac{\Delta y} {\Delta x} $ is an approximation that holds only for small values of $\frac{\Delta x} {x} $, and even then, only approximately (although the approximation gets better and better as $\frac{\Delta x} {x} \to 0 $).

So your first algebraically calculated value was exact, while the second value using calculus was an estimate, albeit a reasonably good one, since $\frac{\Delta x} {x} = \frac{0.1}{1} = 0.1 $, a fairly small value.


答:
By definition of derivative of a function as $f'(a)=\lim\limits_{h\to 0}\frac{f(a+h)-f(a)}{h}$, it follows that we can write
\begin{align}
f(a+h)-f(a)&= f'(a)h + R(h)
\end{align}
whereby the "remainder term" $R(h)$ is "small" in the sense that it satisfies $\lim\limits_{h\to 0}\frac{R(h)}{h}=0$.

If you want to modify the notation, we can write this as
\begin{align}
\Delta f_a(h) &= df_a(h) + R(h)
\end{align}
As you can see, this equation shows in general, the actual difference $\Delta f_a(h)$ is NOT equal to the linear approximation $df_a(h)$ (afterall that's why we call it an approximation), because of the extra term $R(h)$.

Or if you use the classical Leibniz notation which suppresses all the arguments, you can write this as $\Delta y = dy + R$, for some "small" remainder term, $R$. So as you can see, it is NOT true that $\Delta y = dy$. The correct way of writing things is that
\begin{align}
\Delta y &= \frac{dy}{dx}\cdot \Delta x + R
\end{align}
or simply $\Delta y \approx \frac{dy}{dx}\cdot \Delta x$. This is only an approximation, NOT always an equality. It is only an equality for affine functions (i.e of the form $y=mx+b$).

Title: Why does replacing dx and dy with finite values yield this result?

Body:
Consider the change in the function $f(x)=x^2$ at $x=1$  when $x$ is changed by some value $\Delta x$.
$$
\Delta y = f(x+\Delta x)-f(x)\\
\to \Delta y = (1+\Delta x)^2-1^2
$$
Let's set this value $\Delta x$ to $0.1$. Then
$$
\Delta y = (1+0.1)^2-1^2=0.21
$$

Now let's consider this change in the function utilizing its derivative.
$$
\frac{dy}{dx}=2x\\
dy=2x(dx)
$$
Let's replace the infinitesimal differences $dx$ and $dy$ with a finite difference (is this valid?).
$$
\Delta y = 2x(\Delta x)
$$
Again, let's set $\Delta x$ to 0.1.
$$
\Delta y = 2x(0.1)\\
\to\Delta y=2(1)(0.1) = 0.2
$$
Why does $\Delta y$ end up being different than it was previously? I would think that since we're replacing $dx$ with a finite difference, $dy$ would be the same as  $\Delta y$ in #1. There must be a flaw in my thinking somewhere.

Edit:
Thanks for the answers. Here's the justification I worked out.
$$
\frac{\Delta y}{\Delta x} = \frac{(x + \Delta x)^2 - x^2}{\Delta x}\\
\frac{\Delta y}{\Delta x}=\frac{x^2+2x\Delta x + (\Delta x)^2 - x^2}{\Delta x}\\
\frac{\Delta y}{\Delta x}= 2x+\Delta x\\
\to\Delta y = \Delta x(2x + \Delta x)\\
$$
As $\frac{\Delta y}{\Delta x}$ approximates $\frac{dy}{dx}$ with smaller and smaller values of $\Delta x$, a remainder term is always present. Since $x^2$ is exponential, $\Delta y$ must be greater than multiplying the current growth rate, $f'(x)$ by $\Delta x$.

Using 0.1 for $\Delta x$ gives us
$$
\Delta y = 0.1(2(1) + 0.1) = 0.21
$$
the same value calculated in #1.


答:
The derivative of a function is not the value of
$$
\frac{\Delta y}{\Delta x}
$$
for any finite value of $\Delta x$, but rather the limit of the above ratio as $\Delta x \to 0$. More formally, if $y=f(x)$, then
$$
f'(x)=\lim_{\Delta x \to 0}\frac{\Delta y}{\Delta x}=\lim_{\Delta x \to 0}\frac{f(x+\Delta x)-f(x)}{\Delta x} \, .
$$
If you pick a very small value of $\Delta x$, such as $\Delta x=0.1$ as you did in your question, then you will find that
$$
\frac{\Delta y}{\Delta x} \approx \frac{dy}{dx} \, .
$$
Thus, the derivative can be informally considered as the change in $y$ divided by the change in $x$ for small changes in $x$, even though, as we have discovered, this is not strictly the case.


答:
Having $a<0$ is fine.

The sum on the RHS, as you've written it, doesn't actually converge unless $\lvert\frac{x}{a}\rvert<1$. This is why assuming $|x|<|a|$ is necessary for using the series expansion.

Title: Negative Binomial Series

Body:
On this webpage, the expansion of the negative binomial series is given below.

$$
(x+a)^{-n} = \sum^\infty_{k=0} (-1)^k \begin{pmatrix}n + k - 1\\k\end{pmatrix} x^ka^{-n-k}
$$
when $|x| < a$. My questions are as follows:

Are we assuming that $a > 0$?

What happens when $|x| < a$ condition is not true?


问:
Help with modular arithmetic
If$r_1,r_2,r_3,r_4,\ldots,r_{ϕ(a)}$ are the distinct positive integers less than $a$ and coprime to $a$, is there some way to easily calculate, $$\prod_{k=1}^{\phi(a)}ord_{a}(r_k)$$</p>

答:
The  claim is true, with the stronger condition that there is some $i$ with $e_i=1$ and all other exponents are zero.  The set of $r_i$'s is called a reduced residue system.

The second (now deleted) claim is false.  Let $a=7$.  Then $2^13^1=6^1$, two different representations.

Title: Help with modular arithmetic


问:
Implication for a Sobolev function
Suppose $f\in H^2(\mathbb{R})$ with $\lVert f\rVert_{H^1}<\varepsilon$.

Does this imply that $f\in C^1(\mathbb{R})$?

I know that for open and bounded $\Omega\subset\mathbb{R}$, one can embed $H^2(\Omega)$ into $C^1(\Omega)$.

But I guess this is not true for $\Omega=\mathbb{R}$.

答:
The Sobolev spaces on $\mathbb R$ are actually easier to study than the ones on domains, because you can use the Fourier transform. The $H^2$ norm can be defined to be

$$
\lVert f\rVert_{H^2}=\sqrt{\int_{-\infty}^{\infty} \lvert \hat{f}(k)\rvert^2(1+k^2)^2\, dk}, $$
and the Fourier inversion formula, which reads
$$
f(x)=\int_{-\infty}^\infty \hat f(k)e^{ixk}\,dk, $$
implies, via Cauchy-Schwarz, the inequality
$$\tag{1}
\lvert f(x)\rvert \le C\lVert f\rVert_{H^2},\quad \forall x\in \mathbb R,$$
with $C=\sqrt{ \int_{-\infty}^{\infty} (1+k^2)^{-2}\,dk}$. This is the embedding $H^2(\mathbb R)\subset C(\mathbb R)$.

I now read your question more carefully and realised you wanted the embedding $H^2(\mathbb R)\subset C^1(\mathbb R)$. This is obtained the same way, starting with
$$f’(x)=i\int_{-\infty}^{\infty} k\hat f(k) e^{ikx}\, dk,$$
yielding the inequality
$$\tag{2}
\lvert f’(x)\rvert \le C_1 \lVert f\rVert_{H^2}, $$
with $C_1=\int_{-\infty}^{\infty} k^2(1+k^2)^{-2}\, dk$. The inequalities (1) and (2) are the desired embedding $H^2(\mathbb R)\subset C^1(\mathbb R)$.

Title: Implication for a Sobolev function


问:
Calculate $\int_{0}^{\infty} e^{(\frac{-x^2}{2a})}dx$
Given that the derivative of $e^{(\frac{-x^2}{2a})}$  is $e^{(\frac{-x^2}{2a})}$ $ * \frac{-x}{a}$,

and that   $\int_{0}^{\infty}x*e^{\frac{-x^2}{2a}} = 1$

I need to calculate  $\int_{0}^{\infty} e^{(\frac{-x^2}{2a})}dx$.

I've been trying to do this using integration in parts, but I've failed in every attempt.

$\int_{0}^{\infty} \frac{x}{a}*e^{(\frac{-x^2}{2a})}dx = \frac{x}{a}|_{0}^{\infty}*\int_{0}^{\infty}e^{(\frac{-x^2}{2a})} - \frac{1}{a}\int_{0}^{\infty}e^{(\frac{-x^2}{2a})} $

$1 = \int_{0}^{\infty}e^{(\frac{-x^2}{2a})}(\frac{x}{a}|_{0}^{\infty}-\frac{1}{a})$

$\frac{1}{\frac{x}{a}|_{0}^{\infty}-\frac{1}{a}} = \int_{0}^{\infty}e^{(\frac{-x^2}{2a})}$

$0 = \int_{0}^{\infty}e^{(\frac{-x^2}{2a})}$

It can't be 0 since it's the gaussian integral.

What am I doing wrong?

答: It follows from the property of normal distribution that $\int_{-\infty}^{\infty} e^{-\frac{x^2}{2a}}dx =\sqrt{2\pi a}$. At the same time, since the integrand is an even function $$\int_{0}^{+\infty} e^{-\frac{x^2}{2a}}dx = \frac{1}{2} \int_{-\infty}^{\infty} e^{-\frac{x^2}{2a}}dx = \sqrt{\frac{\pi a}{2}}$$


问:
Inverse of function of projective space $\mathbb{R}P^1$ in $\mathbb{S}^1$?
Is it possible to explicitly define the inverse function of this function of the projective space $\mathbb{R}P^1$ in $\mathbb{S}^1$?

$$f([x,y]) = \left( \dfrac{2xy}{x^2+y^2}, \dfrac{x^2-y^2}{x^2+y^2}\right).$$

答:
EDIT: The map with a correct geometrical intuition should be
$$
f([x,y]) = \left( \frac{2xy}{x^2+y^2}, \frac{y^2-x^2}{x^2+y^2} \right).
$$
Note that this map differs by the map in the question by an automorphism of $\mathbb RP^1$, namely the map $[x,y] \mapsto [y,x]$. To avoid confusing the geometrical intuition behind the question, I will use my map $f$, not the one from the OP. Keep that in mind if you need the map in coordinates.

Yes, it is very easy. You just have to take
$$
g : \mathbb{S}^1 \to \mathbb{R}P^1, \qquad (x,y) \mapsto [1-y,x]
$$
when seeing $\mathbb{S}^1$ as the unit circle in $\mathbb R^2$ centered at the origin. Interpret this over affine space for the moment, i.e. take $x=t$ and $y=1$. The map $f$ becomes the map
$$
f(t) = \left( \frac {2t}{t^2+1} , \frac {1-t^2}{t^2+1} \right).
$$
This is exactly the map which takes a real number $t$ (seen on the image as $(t,0)$) and associates the corresponding point on the unit circle as in the image below (and this map is also used to characterize all Pythagorean triples when $t = a/b \in \mathbb Q$, so that $[t,1] = [a,b] \in \mathbb QP^1$ is a $\mathbb Z$-point):

You need to interpret $f$ geometrically to understand this. The map $f$ basically sends the point on the real line to the intersection of that other line through $(0,1)$ with the unit circle. Inverting it just means running this bijection backwards, i.e. take the point $q$ on the unit circle, trace the line $L$ through $p = (0,1)$ and $q$, and the inversion map sends $q$ to the intersection of $L$ with the embedding of $\mathbb R^1$ in $\mathbb R^2$ depicted above ($t \mapsto (t,0)$).

In geometric terms, consider the unit circle centered at $(0,0)$. Given $(x,y)$ such that $x^2 + y^2 = 1$, the line $L$ passing through $(0,1)$ and $(x,y)$ intersects this circle in precisely $2$ points; the point $(x,y)$ and the point $(0,1)$. Its slope is $\Delta y / \Delta x = (y-1)/(x-0) = (y-1)/x$, so the map $g$ sends the point $(x,y)$ to $[1-y,x]$ if $(x,y) \neq (0,1)$ (the slope argument stops working for $(x,y) = (0,-1)$ because $\Delta x = 0$, but the result in $\mathbb RP^1$ is still correct). The vector $(1-y,x)$ is the one orthogonal to $L$. If $(x,y) = (0,1)$, the tangent line to the circle is the one with two points of intersection at $(0,1)$ (counting the point of tangency as a double point), so we associate the line $[1,0]$ to the point $(0,1)$.

(A hint of why we use the vector orthogonal to $L$ to represent $L$: it was a fundamental observation of Grothendieck that the duality between lines and points in the plane generalizes better when we see it as the duality of hyperplanes and points in a vector space, where a hyperplane is seen as the locus of an affine map. Therefore, in the plane, a line should be interpreted as the affine map whose set of solutions corresponds to the $1$-dimensional affine subspace we are familiar with, and that line is represented by a point in its locus and a vector orthogonal to that line. This representation works for lines as well as for hyperplanes, and when we compute things this way, the coordinate systems fit nicely with the abstract theory.)

Pick the coordinates of $[x,y]$ in such a way so that $x^2 + y^2 = 1$. It follows that $f([x,y]) = (2xy, y^2-x^2)$. This point lies on the unit circle since
$$
(y^2-x^2)^2 + (2xy)^2
= x^4 + y^4 - 2x^2 y^2 + 4x^2y^2 \\
= x^4 + y^4 + 2x^2y^2 \\
= (x^2+y^2)^2 \\
= 1 \\
$$
Now consider a point $(t,0)$ on the real line (see it as in the above image). The line passing through $(0,1)$ and $(t,0)$ intersects the unit circle in a second point $(0,1) + \lambda((t,0)-(0,1)) = (\lambda t,1-\lambda)$ precisely when
$$
(\lambda t)^2 + (1-\lambda)^2 = 1 \\
\lambda^2(t^2+1) - 2\lambda + 1 = 1 \\
\lambda = 0 \qquad \text{ or } \qquad \lambda(t^2+1) - 2 = 0
$$
so that the $\lambda$ of interest is $\frac 2{t^2+1}$. This gives the coordinates of the point on the unit circle:
$$
(x', y') = \left( \frac{2t}{t^2+1}, 1 - \frac 2{t^2+1} \right) = \left( \frac{2t}{t^2+1}, \frac{t^2 - 1}{t^2+1} \right).
$$
Another way to interpret the real number $t$ is as follows: consider the point $p=(0,1)$ and the line $L_p$ passing through $(0,1)$ and $(t,0)$. A vector orthogonal to $(t,0) - (0,1) = (t,-1)$ is $(1,t)$, so $[1,t]$ is the correct representation for $L_p$. Replacing $[1,t]$ by $[1,x/y] = [y,x]$ and re-writing our expression, we recover the original $f$ (note that the flip $[x,y] \mapsto [y,x]$ needs to occur here, so this is probably why the OP formulated $f$ the way it was). So now, to invert $f$, given a point $(x,y)$ on the unit circle, we need to consider the line through $(x,y)$ and $(0,1)$. A vector orthogonal to $(x,y)-(0,1) = (x,y-1)$ is $(1-y,x)$, so the point $(x,y)$ maps to $[1-y,x]$. This gives
$$
g(x,y) = \left\{ \begin{matrix} [1-y,x] & \text{ if } (x,y) \neq (0,1) \\ [0,1] & \text{ if } (x,y) = (0,1) \end{matrix} \right..
$$

Just as a confirmation, we evaluate
$$
g(f([x,y]))
= g\left( \frac{2xy}{x^2+y^2}, \frac{y^2-x^2}{x^2+y^2} \right) \\
= \left[1-\frac{y^2-x^2}{x^2+y^2}, \frac{2xy}{x^2+y^2} \right] \\
= \left[(x^2+y^2) - (y^2-x^2), 2xy \right] \\
= \left[2x^2, 2xy\right] \\
= [x,y]
$$
This solution assumes $x\neq 0$, so one also computes $g(f([0,1])) = g(0,1) = [0,1]$. Conversely, when $x \neq 0$ and $x^2+y^2 = 1$, we have
$$
x^2+(1-y)^2 = x^2 + (y^2 -2y+1) = 2-2y = 2(1-y), \\
x^2 - (1-y)^2 = (1-y^2)-(1-y)^2 = (1-y^2)-(1-2y+y^2) = 2y(1-y),
$$
therefore
$$
f(g(x,y)) = f([1-y,x]) = \left( \frac{2(1-y)x}{(1-y)^2 + x^2}, \frac{ x^2 - (1-y)^2}{(1-y)^2 + x^2} \right) = (x,y).
$$

A third interpretation of $f$ goes as follows: look at the unit circle $S^1$ and consider its blow-up at the point $p=(0,1)$. Given two distinct points $p,q \in \mathbb R^2$, there is a unique affine map $L : \mathbb R^2 \to \mathbb R$ satisfying $L(p) = L(q) = 0$. We can write it by letting $x = (x_1,x_2)$, $p = (p_1, p_2)$, $q = (q_1,q_2)$ and
$$
L(x) = L(x) - L(p) + L(p) \overset{(!)}= \langle (w,z), x-p \rangle = w(x_1-p_1) + z(x_2-p_2).
$$
because if $L(x) = Ax + b$ for some matrix $A$ and vector $b$, then $L(x) - L(y) = A(x-y)$, so $L(x)-L(y)$ is linear in $x-y$, hence (!). The condition $L(q) = 0$ gives the equation
$$
w(q_1-p_1) + z(q_2-p_2) = 0,
$$
which determines $(w,z)$ up to a scalar multiple; when $L(p) = 0$ (meaning that the line $L$ passes through the point $p$), we write $L_p = [w,z]$ and we write $q \in L_p$ when $L(q) = 0$. When $q = (x,y)$, we also write $L_p(x,y)$ instead of $L_p((x,y))$. In this interpretation, the vector $(w,z)$ is orthogonal to $q-p$, and the geometric line $L_p$ corresponds to the set of points $q$ such that $q-p$ is orthogonal to $(w,z)$.

We wish to have a parametrization of this blow-up (especially for the lines), which in general looks like
$$
\mathrm{Bl}_p(\mathbb RP^2) = \{ ((x,y),L_p) \in \mathbb RP^2 \times \mathbb RP^1 \, | \, L_p(x,y) = 0 \}.
$$

Restricting our attention to $S^1$, we have
$$
\mathrm{Bl}_p(S^1) = \{ ((x,y),[w,z]) \in \mathbb RP^2 \times \mathbb RP^1 \, | \, x^2 + y^2 = 1, wx + z(y-1) = 0 \}.
$$
Now since $S^1$ is smooth and we blew up at a smooth point of a curve, the correspondence
$$
S^1 \leftarrow \mathrm{Bl}_p(S^1) \rightarrow \mathbb RP^1, \quad (x,y) \leftarrow ((x,y),[w,z]) \mapsto [w,z]
$$
actually consists of two isomorphisms of varieties. Going from right to left gives the map $f$, and from left to right gives the map $g$.

To see this more in detail, the line $[w,z] \in \mathbb RP^1$ maps to $((x,y), [w,z]) \in \mathrm{Bl}_p(S^1)$, then $wx + z(y-1) = 0$, so $wx = z(1-y)$. Solving the system of equations
$$
wx + z(y-1) = 0, \qquad x^2 + y^2 = 1
$$
by assuming $w \neq 0$ gives
$$
z^2(y^2-2y+1) = (z(1-y))^2 = (wx)^2 = w^2x^2 = w^2(1-y^2) \quad \Rightarrow \quad (w^2+z^2)y^2 + (-2z^2)y + (z^2-w^2) = 0
$$
which means
$$
y = \frac{2z^2 \pm \sqrt{4z^4 - 4(z^2-w^2)(z^2+w^2)}}{2(w^2+z^2)} = \frac{z^2 \pm w^2}{w^2+z^2}.
$$
One solution is not of interest (it corresponds to $(x,y) = p$) and the other solution is the point $\left(\frac{2zw}{w^2+z^2}, \frac{z^2-w^2}{w^2+z^2}\right)$. Note that $f([w,z]) = (x,y)$ corresponds to our map of interest. When $w=0$, a similar process occurs except the quadratic will have a double root, which will lead to $f([0,1]) = (0,1)$.

In the other direction, suppose $x^2 + y^2 = 1$. The line $L_p$ through $p = (0,1)$ and $(x,y)$ is given by coordinates $[w,z]$ satisfying $w(x-0) + z(y-1) = 0$. An easy choice of $w$ and $z$ is $[w,z] = [-(y-1),x] = [1-y,x]$. Note that $g(x,y) = [1-y,x]$.

Hope that helps,


问:
What is the use of homotopy pushouts in Lurie's HTT 2.1.4.10?
I have trouble understanding the usage of homotopy pushouts in the proof of proposition 2.1.4.10 in Lurie's Higher Topos Theory. My trouble is the way he shows that for a map $j\colon S\to S'$, the induced functor $j_!\colon (\mathcal{Set}_\Delta)_{/S}\to(\mathcal{Set}_\Delta)_{/S'}$ preserves weak equivalences in the covariant model structure.

Given a covariant equivalence $f\colon X\to Y$ (over $S$), I want to show that $j_!f$ is a covariant equivalence as well. This seems to be the relevant diagram. The top horizontal arrow is induced by $f$ and hence by assumption an equivalence of simplicial categories. The bottom horizontal arrow is $j_!f$. From what I understand, the diagram Lurie draws (one of the backsides of my diagram) is a pushout. This follows from the pasting laws for pushouts and the facts that $\mathfrak{C}[-]$ preserves colimits. He then states that $\mathfrak{C}[-]$ preserves monomorphisms and hence (I believe by A.2.4.4) he deduces that it must be a homotopy pushout as well. According to Lurie we are done.

Here are my questions:

Why does this imply that the map $j_!f$ (the bottom horizontal map in my diagram) is an equivalence?

Is my assertion correct that he uses A.2.4.4 to deduce that the square is a homotopy pushout? I don't really see the relevant cofibration.

In a similar vein he mentions in the proof of 2.1.4.7 (2) that a square is a homotopy pushout to conclude that the bottom arrow in a sqaure is an equivalence. This is equally confusing to me: Do homotopy pushouts preserve weak equivalences in general?

Any hint would be helpful, so thanks in advance.

答:
Suppose $f: X \to Y$ is a morphism in $(\mathcal{S}et_{\Delta})_{/S}$ and $j: S \to S'$.   If $f$ is a covariant equivalence, then $\mathfrak{C}[X^\triangleleft \sqcup_X S] \to \mathfrak{C}[Y^\triangleleft \sqcup_Y S]$ is a categorical equivalence.  Applying the functor $\mathfrak{C}[S'] \sqcup_{\mathfrak{C}[S]} -$ to this equivalence, we see that $$\mathfrak{C}[S'] \sqcup_{\mathfrak{C}[S]} \mathfrak{C}[X^\triangleleft \sqcup_X S] \to \mathfrak{C}[S'] \sqcup_{\mathfrak{C}[S]} \mathfrak{C}[Y^\triangleleft \sqcup_Y S]$$ is also a categorical equivalence.  But we can identify these pushouts using Lurie's claim: this is just the morphism $\mathfrak{C}[X^\triangleleft \sqcup_X S'] \to \mathfrak{C}[Y^\triangleleft \sqcup_Y S']$.  So $j_! f$ is also a covariant equivalence.

Sure.  The morphism $S \to X^\triangleleft \sqcup_X S$ is a monomorphism of simplicial sets (since it is a pushout of the monomorphism $X \to X^\triangleleft$).  As will be explained by Lurie, $\mathfrak{C}[-]$ converts monomorphisms to cofibrations, so $\mathfrak{C}[S] \to \mathfrak{C}[X^\triangleleft \sqcup_X S]$ is a cofibration of simplicial categories, and you are in a setting to apply proposition A.2.4.4.

Yes - the idea is that in good cases you compute homotopy pushouts by replacing one of the morphisms with a cofibration and computing the ordinary pushout, and weak equivalences are preserved by pushouts, again in good cases.  (Here, "good cases" should mean something like left proper, but don't quote me on this.)  Anyway, if $i: X \to X'$ is a cofibration, then you can show that $X^\triangleleft \sqcup_X S \to (X')^\triangleleft \sqcup_{X'} S$ is a monomorphism, so that again by Jacob's claim $\mathfrak{C}[X^\triangleleft \sqcup_X S] \to \mathfrak{C}[(X')^\triangleleft \sqcup_{X'} S]$ is a cofibration.  So, as a pushout of a weak equivalence along a cofibration, the morphism $\mathfrak{C}[(X')^\triangleleft \sqcup_{X'} S] \to \mathfrak{C}[(Y')^\triangleleft \sqcup_{Y'} S]$ is also a weak equivalence.


问:
Elementary Linear Algebra
What does it mean when someone says "find a fundamental set of solutions for the system y' $=A$ y"?

That is, the system

$$ {\bf{y'}} =A {\bf{y}}. $$

答:
Assuming you know the meaning of linear independence of $n$ functions on an interval $I$, we have:

Any set $y_1,y_2,...,y_n$ of $n$ linearly independent solutions of the homogeneous $n$th-order equation on an interval $I$ is said to be a fundamental set of solutions on the interval.

Edit: As @Gerry remarked me, note that you are asking about the first order homogenous OE.

Title: Elementary Linear Algebra


问:
If $ab^p = b^qa$ then show that $b=e$ for a group $G$ with order $pq, p$ and $q$ primes.
Let $p$ and $q$ be primes, and the order of a group $G$ be $pq$. If there is $a,b$ in the group $G$ such that

$$ab^p = b^qa,$$

then show that $b=e.$

My attempt:

Group $G$ is cyclic because $p$ and $q$ are both primes and therefore the group is abelian.

$a^{-1}ab^p= a^{-1}b^qa$ and I get $b^p=b^q$ because $a^{-1}b^qa = b^q$ do to that the group is abelian. Does this imply that $b=e$?

I don't understand why

答:
I am assuming $p$ and $q$ are distinct. If not then $p=q$ and a cyclic group $G$ with $pq=p^2$ elements gives a counterexample; indeed let $b$ be any generator of $G$.

Let $r$ be the smallest positive integer such that $b^r =e$. Then $r\mid pq$. [Indeed, the set $\{b,b^2,\ldots, b^{r-1}, e\}$ of elements generated by $b$ has cardinality precisely $r$ and forms a subgroup of $G$.] So $r$ must be either $1,p,q,$ or $pq$. Now if $r$ is 1, then we are done: then $b^r=b^1=b=e$. So we show that $r$ cannot be $p,q$, nor $pq$.

So suppose that $r = p$. Then $b^r=b^p=e$ so plugging this into $ab^p=b^qa$, we'd get $ae =b^qa$ which would imply $a=b^qa$ which would imply $b^q=e$. So then both $b^p=e$ and $b^q=e$ would hold for prime distinct $p,q$. This and Claim 1 [below] however, together imply that then there would be a positive integer $c<\min\{p,q\}$ so that $b^c=e$. So $r$ as defined cannot be $p$ after all.

Likewise $r$ cannot be $q$ after all either.

So now suppose that $r=pq$. But then this would imply $G$ is a cyclic group generated by $b$ and so $G$ is abelian. But then this gives $ab^p = ab^q$ which in turn gives [assuming $q> p$ WLOG] $a=ab^{q-b}$ which in turn gives $e=b^{q-p}$ which implies $r \le q-p < pq$ a contradiction.  So $r$ cannot be $pq$ either.

So the only possibility left now is $r$ must be 1 and so $b^1=b^r=b=e$ it must be.

Below is Claim 1. In fact, Claims 2 and 3 are also true and are included for completeness, but they aren't really needed here.

Claim 1: Let $b$ be a group element, and let $p$ and $q$ be relatively prime positive integers satisfying $b^p=b^q=e$. There is a positive integer $c<\min\{p,q\}$ such that $b^c=e$.

Claim 2: Let $b,p$, and $q$ be as in Claim 1. Then in fact $b^1=b=e$.

Proofs of Claims 1 and 2:  Let us assume WLOG that $p<q$. Divide $q$ by $p$ to get $q=mp+c$ where $m$ is an integer, and where $c$ is the remainder $c=p \mod q$.

We now show Claim 1. That $p$ and $q$ are relatively prime implies that when $q$ is divided by $p$ there has to be a nonzero remainder and so $c$ is positive and less than $\min\{p,q\}$. So we now claim that $b^c=e$. Indeed, $b^q=e$ and $q=mp+c$ together give $b^{mp+c}=b^{mp}b^c=e$. This and that $b^p=e$ gives $e=b^q=b^{mp}b^c =(b^{p})^mb^c = e^mb^c = eb^c$. So this string of equations imply $e=eb^c$, which gives $b^c =e$, and so Claim 1 follows.

We now show Claim 2. We use induction on $p+q$. Repeat the reasoning as in the proof of Claim 1 to get $b^c=e$. In fact, because $p$ and $q$ are relatively prime, $c$ as above also satisfies the following: $c$ and $p$ are relatively prime as well. Thus
$b^c=e$ and $b^p=e$ for two relatively prime integers $p$ and $c$ satisfying $p+c < p+q$. So by induction it follows that $b^1=b=e$ and so Claim 2 follows as well. $\surd$

In fact the following is also true:

Claim 3: Let $b$ be a group element, and let $p$ and $q$ be [not necessarily relatively prime] positive integers satisfying $b^p=b^q=e$ then $b^{r'}=e$
for $r' =\gcd(p,q)$.


问:
Differentiable function such that $\left| f(x) \right| \leq C x^{-k}$ and $\left| f'(x) \right| \leq C' x^{-k}$ has limit at zero.
The problem is:

Let $f$ be differentiable real valued function, such that for $x>0$ we have

$$
\left|f(x) \right| \leq C x^{-k}
$$

for some positive constants $k$ and $C$, and the rate of growth is optimal, i.e there is no smaller $k$ that works.

If

$$
\left|f'(x) \right| \leq C' x^{-k}
$$

then $\lim\limits_{x \to 0+} f(x)$ exists.

I don't really know what to do here, I assume we have to use the mean value theorem to show that if $x_n \to 0$ then $f(x_n)$ is a Cauchy sequence but I can't see how.

答: I think $f(x) = ln(x)$ is a contadiction, maybe I didn't understand the question.


问:
Find the value of :- $2010 - \sum_{k=1}^{2010}\bigg\lceil\frac{2010}{k} - \bigg\lfloor \frac{2010}{k}\bigg \rfloor\bigg\rceil $

Find the value of :- $$2010 - \sum_{k=1}^{2010}\bigg\lceil\frac{2010}{k} - \bigg\lfloor \frac{2010}{k}\bigg \rfloor\bigg\rceil $$

What I Tried: Quite stuck at the problem. I need to find some short cut to evaluate the summation but not been able to do so till now.

I know that the factors of $2010$ are $1,2,3,5,6,10,15,30,67,134,201,335,402,670,1005,2010$ . So for all these values in $k$ , the sum of the terms will be $0$ . So we can actually cancel these terms out from the list of $\{1,2,3, ... ,2010\}$ .

The problem is, what about the other numbers? For example, $\frac{2010}{4} = 402.5$ , which can be actually considered $402$ like $5$, but for $\frac{2010}{9} = 223.33$ , it will be considered $223$ , which is actually one number not like $10$ though.

So can anyone help me?

答:
Note that if $k$ is not a divisor of 2010, then

$\bigg\lceil\frac{2010}{k} - \bigg\lfloor \frac{2010}{k}\bigg \rfloor\bigg\rceil=1$

and if $k$ is a divisor, $\bigg\lceil\frac{2010}{k} - \bigg\lfloor \frac{2010}{k}\bigg \rfloor\bigg\rceil=0$

(That's actually because $\frac{2010}{k} - \bigg\lfloor \frac{2010}{k}\bigg \rfloor=${$\frac{2010}{k}$}  ({n} is the fractional part of n) and {$\frac{2010}{k}$} is between $0,1$ and is equal to $0$ whenever $\frac{2010}{k}$ is an integer and is equal to $1$ otherwise.
So actually $\sum_{k=1}^{2010}\bigg\lceil\frac{2010}{k} - \bigg\lfloor \frac{2010}{k}\bigg \rfloor\bigg\rceil=2010-$(number of positive divisors of 2010)

So the final value you're asking for is equal to the number of positive divisors of 2010 which is 16.


问:
How do you find a plane that is both parallel and a distance of X units from a given plane
I've been asked to come up with equations for two parallel planes that are X units apart. I've come up with the plane $x+z-1=0$ and a parallel plane $x+z+4=0$. Would it be correct to say these planes are 5 units apart since the $d$ constant is increased by 5?

I calculated the distance between the point $(1,0,0)$ on the 1st plane and the 2nd plane and got $3.54$ which I guess proves that the the planes are not 5 units apart. But if that's the case, I'm not sure how else to go about it.

答:
Find the normal vector to your first plane, which is $\vec{n} = (1,0,1)$. The unit vector in this direction is $\vec{u} = \frac{\vec{n}}{|\vec{n}|} = \frac{1}{\sqrt{2}}(1,0,1)$. Now pick a point on the first plane, for example, you already mentioned $(1,0,0)$. Not just add $5$ times the unit normal vector to this point to get a point which is 5 units away from the plane:

$$ (1,0,0) + \frac{5}{\sqrt{2}}(1,0,1) = \left( 1 + \frac{5}{\sqrt{2}}, \; 0, \; \frac{5}{\sqrt{2}} \right)$$

Now just find the plane which is parallel and goes through this point. In other words, plug this point in to the equation $x + z + d = 0$ and solve for $d$.


问:
What is nedeed to prove that the tangent space on a manifold is a vector space?
I am currently working with the definition of tangent vectors being equivalence classes of curves. So $v =[\gamma]$ and $w=[\sigma]$ where $v,w$ are the vectors. I want to prove that the sum of this two equivalence class is also a equivalence class to give it the vector space structure.  We define the sum as:

$v+w= [\phi^{-1} \circ (\phi \circ \gamma + \phi \circ \sigma)]$

where $\phi$ is a chart. I understand that the curves can not be added when they belong to the manifold, that's why we first map then into the reals where that can be performed. So now that the sum is defined, how can I prove that is another class at the point $p$ on $M$. Is it enough to make sure that the map passes by $p$? The same questions goes for the multiplication by the scalar. Thanks

答:
Having read your post more carefully, here's a one-sentence summary of your mistake: you're trying to add (and scalar multiply) the curves in $\Bbb{R}^n$, rather than their velocities. As you observed, adding the curves messes up things with the base points.

As a set, we have $T_pM$ is the set of equivalence classes of smooth curves, $[\gamma]$, where $\gamma$ is defined on some open interval containing $0$ such that $\gamma(0)=p$. Now, for any chart $(U,\phi)$ about the point $p$, consider the function $F_{\phi,p}:T_pM \to \Bbb{R}^n$ defined as
\begin{align}
F_{\phi,p}([\gamma]):= (\phi\circ \gamma)'(0).
\end{align}
This function is well-defined because of how the equivalence relation is defined. Notice the intuitive meaning: $\gamma$ is a curve with values in the manifold $M$, so if we use a chart, we can get a corresponding curve $\phi\circ \gamma$ with values in the Banach space (i.e a normed vector space) $\Bbb{R}^n$, and we know how calculus works in the setting of vector spaces. So, all this map $F_{\phi,p}$ does is it takes a curve $[\gamma]$ and maps it to the "velocity vector" $(\phi\circ \gamma)'(0)$. I hope this is intuitive (otherwise, just draw a few pictures to see where each object is).

Now, it is also easy to verify that $F_{\phi,p}$ is a bijective function; I leave it to you to verify that $G_{\phi,p}:\Bbb{R}^n\to T_pM$ defined as
\begin{align}
G_{\phi,p}(v):= [t\mapsto \phi^{-1}(\phi(p)+tv)]
\end{align}
is the inverse function. In words, what we're doing is we're taking a vector $v\in\Bbb{R}^n$, and considering the straight line $t\mapsto \phi(p)+tv$. This is a curve based at the point $\phi(p)$, in the direction $v$. Since $\phi$ is a homeomorphism, it follows that for small enough values of $t$, we have $\phi(p)+tv\in \phi(U)=\text{domain}(\phi^{-1})$, hence we can consider the equivalence class of the curve $t\mapsto \phi^{-1}(\phi(p)+tv)$.

So, what has all this extra notation yielded? Well, we have a bijective function $F_{\phi,p}:T_pM\to \Bbb{R}^n$, and of course, $\Bbb{R}^n$ is a vector space, so by basic linear algebra, we can "pull back" the vector space structure of $\Bbb{R}^n$ so as to make $F_{\phi,p}$ a linear isomorphism. Explicitly, what I mean is that we can define addition and scalar multiplication $+_{\phi}$ and $\cdot_{\phi}$ (I put the subscript because everything is chart-dependent so far) as follows:
\begin{align}
\begin{cases}
[\gamma_1]+_{\phi} [\gamma_2]&:= F_{\phi,p}^{-1}\bigg(F_{\phi,p}([\gamma_1])+
F_{\phi,p}([\gamma_2])\bigg)\\
c\cdot_{\phi}[\gamma]&:= F_{\phi,p}^{-1}\bigg(c\cdot F_{\phi,p}([\gamma])\bigg)
\end{cases}
\end{align}

If you unwind all the definitions, then
\begin{align}
c\cdot_{\phi}[\gamma_1]+_{\cdot}[\gamma_2]= [t\mapsto \phi^{-1}\left(\phi(p) + t(c\cdot (\phi\circ \gamma_1)'(0)+(\phi\circ \gamma_2)'(0))\right)]
\end{align}
Hopefully, the idea is clear enough: you have a bijection, so you just push everything forward, do the computations in $\Bbb{R}^n$, then bring everything back to $T_pM$, and that's how addition and scalar multiplication is defined. I leave it to you that all the vector space axioms are satisfied and that $F_{\phi,p}$ is a linear isomorphism etc.

One final thing to note is that so far the addition and scalar multiplication have been defined using a particular chart $(U,\phi)$, but actually, it is a simple chain rule exercise to verify that if you have a different chart $(V,\psi)$, then $+_{\phi}=+_{\psi}$ and $\cdot_{\phi}=\cdot_{\psi}$, so the vector space structure on $T_pM$ is actually chart-independent, hence we just denote it as $+$ and $\cdot$ as usual. I leave it to you to unwind definitions, use chain rule etc to verify this. If you have trouble, let me know, maybe I can elaborate more.


问:
Bijective association between the linear space of symmetric tensors and linear space of homogeneous polynomials example
edit:
Given a symmetric tensor G there is a bijective mapping between the $d$ indexes and a vector of $K$ numbers representing the time each variable $x_t$ appears in the associated monomial. For example, imagine $G_{112}$ being a 3 dimensional tensor with every index spanning from 1 to 3 (so $G_{333}$ is the last element)

The element is associated with $G_{112}x_1^2x_2^1x_3^0 = G_{112}x^{f([1,1,2])}$ where $f([1,1,2]) = [2,1,0]$.
More precisely
\begin{equation}
p(x) = \sum G_ix^{f(i)}
\end{equation}

So i want to "see" this relation using an example so i took two 2-dimensional symmetric tensors like \begin{pmatrix}
1 & 0 \\
0 & 1
\end{pmatrix}
which is associated with the homogeneous polynomial
\begin{equation}
p(x_1,x_2) =x^2_1+x^2_2
\end{equation}
and
\begin{pmatrix}
0 & 1 \\
1 & 0
\end{pmatrix}
which is associated with the homogeneous polynomial
\begin{equation}
q(x_1,x_2) = 2x_1x_2
\end{equation}
Now if i do the product
\begin{equation}
p(x_1,x_2)q(x_1,x_2) = 2x_1^3x_2 + 2x_1x_2^3
\end{equation}
i should obtain a homogeneous polynomial which is associated to a symmetric tensor, and that symmetric tensor is
\begin{equation}
\begin{pmatrix}
0 & \frac{1}{2} \\
\frac{1}{2} & 0
\end{pmatrix}
\begin{pmatrix}
\frac{1}{2} & 0 \\
0 & \frac{1}{2}
\end{pmatrix}
\begin{pmatrix}
\frac{1}{2} & 0 \\
0 & \frac{1}{2}
\end{pmatrix}
\begin{pmatrix}
0 & \frac{1}{2} \\
\frac{1}{2} & 0
\end{pmatrix}
\end{equation}

But i can also calculate the tensor product of the two original tensors which is 4-dimensional and is
\begin{equation}
\begin{pmatrix}
0 & 1 \\
1 & 0
\end{pmatrix}
\begin{pmatrix}
0 & 0 \\
0 & 0
\end{pmatrix}
\begin{pmatrix}
0 & 0 \\
0 & 0
\end{pmatrix}
\begin{pmatrix}
0 & 1 \\
1 & 0
\end{pmatrix}
\end{equation}
They should be the same... What am i doing wrong? Can you give an example of this link between tensors and polynomials?

答:
This is because the tensor product of symmetric tensors is not necessarily a symmetric tensor. In your example, the rank-$4$ tensor
$$
T = \begin{pmatrix}
\begin{pmatrix}
0 & 1 \\
1 & 0
\end{pmatrix} &
\begin{pmatrix}
0 & 0 \\
0 & 0
\end{pmatrix}\\
\begin{pmatrix}
0 & 0 \\
0 & 0
\end{pmatrix} &
\begin{pmatrix}
0 & 1 \\
1 & 0
\end{pmatrix}\\
\end{pmatrix}
$$
is not symmetric. For example, the entries $T_{1112} = 1$ and $T_{1211} = 0$ are not equal.

There's another indication that simply taking the tensor product doesn't work, i.e. tensor products of symmetric tensors are not commutative, but multiplying polynomials certainly is.

I'm not familiar with this mapping between homogeneous polynomials and symmetric tensors that you have described, but I think what you want is probably the symmetric product, which means that after taking the tensor product you need to average all entries with indices in the same orbit w.r.t. permutation. Note that the symmetric product is also commutative, which is a good sign. Again taking the $1112$ set of indices as an example, we have $T_{1112} = 1$, $T_{1121} = 1$, $T_{1211} = 0$, $T_{2111} = 0$, so the symmetrized version should have $T_{(1112)} = \frac{1}{2}$. You can check that this gives you the same matrix as the one obtained from the product of the polynomials.