答:
This is one-dimensional motion. You're solving the problem as if it's two-dimensional motion. No vector addition needed!

Solve $6t+5=6$ and you're done!

Title: Find the time where their velocity is equivalent

Body:
The question is:

The equations of movement of 2 particles are:
$$x_1=3t^2+5t+6 \quad m$$ $$x_2=6t+8 \quad m$$
Find the exact time where both will have the same velocity.

So i've tried this: $$\vec{v}=\frac{dx}{dt}$$ basically differentiated both to get $$\vec{v_1}=6t+5\quad m/s$$ $$\vec{v_2}=6\quad m/s$$ and then i got their magnitude $|\vec{v_1}|$ and $|\vec{v_2}|$ as this $$|\vec{v_1}=\sqrt{36t^2+5^2}$$ $$|\vec{v_2}|=6$$ and then equated both to get the instant time where both velocities are equivalent, $$36t^2+5^2=6^2$$ $$t^2=\frac{11}{36}=\sqrt{\frac{11}{36}}$$ is this correct? i'm afraid i should have equated both velocity vectors and not their magnitude


问:
Finite-dimensionality of Ext¹ of simple modules
For simplicity let $\Bbbk$ be a field of characteristic $0$ and let $A$ be a finitely generated unital associative $\Bbbk$-algebra.

Is it true that for any two simple $A$-modules $S_1, S_2$, we have that $\operatorname{Ext}^1_A (S_1, S_2)$ is finite-dimensional?

If not, what would be a simple counterexample, and what sort of conditions do we need to ensure this?

(As far as I understand, the Weyl algebra $\Bbbk \langle x, y \rangle / (xy - yx - 1)$ has only infinite-dimensional simple modules, but their first extension groups are still finite-dimensional.)

答: Actually, your claim about the Weyl algebra is only true for simple holonomic modules. It turns out that there’s a simple non-holonomic module $M$ over the 2nd Weyl algebra $A=A_2(\Bbb C)$ for which $\operatorname{Ext}_A^1(M,M)$ is infinite dimensional. This is Corollary 1.3 of “Non-holonomic modules over Weyl algebras and enveloping algebras” by Stafford.

ContentLicense: CC BY-SA 4.0


问:
If the positive series $\sum a_n$ diverges and $s_n=\sum\limits_{k\leqslant n}a_k$ then $\sum \frac{a_n}{s_n}$ diverges as well
So I've been trying to figure out how to prove the following.

Let $(a_n)$ be a sequence of positive numbers such that $\sum\limits_{n=1}^\infty a_n =\infty$, and define $s_n=\sum\limits_{i=1}^n a_i$. Then $\sum\limits_{n=1}^\infty\frac{a_n}{s_n} =\infty$ as well.

I can prove it by comparing it to $\int_1^\infty \frac{1}{x} \, dx $ if the sequence $a_n$ is bounded by some $M$, but that's as far as I've been able to get.

答: Since $s_n\to+\infty$ when $n\to\infty$, for each $n\geqslant1$ there exists some finite $m\gt n$ such that $s_m\geqslant2s_n$. In particular, $\sum\limits_{k=n+1}^m\frac{a_k}{s_k}\geqslant\sum\limits_{k=n+1}^m\frac{a_k}{s_m}=1-\frac{s_n}{s_m}\geqslant1/2$. Thus, the rests of the series $\sum\limits_k\frac{a_k}{s_k}$ do not converge to zero, QED.</p>

ContentLicense: CC BY-SA 3.0


问:
Direct/Lagrange to show max/min
The domain Is defined by the half unit circle: $\left \{ \left ( x,y \right )\in \mathbb{R}^2: x^2+y^2\leq 1, x\geq 0 \right \}$

for the function $f:D\rightarrow \mathbb{R}$. The function is $f(x,y)=2xy^2-2x^2$. I have shown that a min/max must exists since the function is continuous and the domain is both bounded and closed. I do not know how to proceed to find these points where the function is equal to its min/max. I tried with the Lagrange multiplier but it is just too messy for me to do. How would one do it directly? I found that the gradient vanishes at $(0,0)$

答:
Since $x\geqslant 0, f(x,y) \leqslant 2x(1-x^2)-2x^2$. Now find $x$ between 0 and 1 to maximize this.

Finding the minimum is a lot easier since $2xy^2\geqslant 0$ so we must have $y=0$. Can you do the rest?

===============

We can also use polar coordinates. Let $x=R\cos \alpha, y=R\sin \alpha$, with $0\leqslant R \leqslant 1, -\frac{\pi}{2} \leqslant \alpha \leqslant \frac{\pi}{2}$.

(1) On boundary $S_1 = \left\{ x^2+y^2=1, x\geqslant 0 \right\}$ we have $R^2=1$ so $f(x,y) = 2\cos \alpha \sin^2 \alpha - 2 \cos^2 \alpha = 2\cos \alpha - 2\cos^3 \alpha - 2 \cos^2 \alpha$.

Now $\frac{\partial f}{\partial \alpha}= -2 \sin \alpha-6\cos^2\alpha (-\sin \alpha)-4 \cos \alpha (-\sin \alpha) \\
= 2 \sin \alpha (-1+3 \cos^2\alpha +2\cos \alpha)\\
= 2\sin\alpha (3\cos \alpha -1) (\cos \alpha+1)
$

$\frac{\partial{f}}{\alpha}=0 \Rightarrow \sin \alpha =0 $ or $\cos \alpha = \frac{1}{3}$

When $R=1, \sin \alpha =0, x=1, y=0, f(x,y) = -1$.

When $R=1, \cos \alpha = \frac{1}{3}, x=\frac{1}{3}, y=\pm \frac{2\sqrt{2}}{3}, f(x,y)=\frac{10}{27}.$

(2) On boundary $S_2 = \left\{ x=0, -1\leqslant y \leqslant 1\right\}$ we have $f(x,y)=0$.

Therefore the minimum is $-1$ at $(1,0)$ and the maximum is $\frac{10}{27}$ at $(\frac{1}{3}, \pm \frac{2\sqrt{2}}{3})$

Title: Direct/Lagrange to show max/min


问:
For which values $x,\alpha$ does $\sum_{n=1}^{\infty}\frac{n^{nx}}{(n!)^ \alpha}$ converge?
I want to know for which values of $x$ this series converges:
$$\sum_{n=1}^\infty \frac{n^{nx}}{(n!)^ \alpha};$$ here $\alpha \in\mathbb R$ is a constant

This series is defined $ \forall x \in \mathbb R$.

$$a_n=\frac{n^{nx}}{(n!)^ \alpha} \sim \frac{n^{nx}}{(\sqrt {2 \pi n}* (\frac{n}{e})^n)^ \alpha}= \frac{1}{(2 \pi n)^{\frac{\alpha}{2}}}n^{nx-n \alpha}e^{n \alpha} \sim 0 \Leftrightarrow  \alpha <0 \land x- \alpha <0.$$

Applying the root test:

$$ \sqrt[n]{\frac{1}{(2 \pi n)^{\frac{\alpha}{2}}}*n^{nx-n \alpha}*e^{n \alpha} }=\frac{1}{(2 \pi n)^{\frac{\alpha}{2n}}}n^{x- \alpha}e^{\alpha } <1 \iff x- \alpha<0   \iff x<\alpha $$

Is it right?

My doubt is regarding the necessary condition for the convergence in which I find
$\alpha <0 \land x- \alpha <0$ and the generality of $\alpha$ that doesn't make me say if $\alpha<$ or $>0$.

答:
For these kind of sums,
I use $n! \sim (n/e)^n$ so
$\sum_{n=1}^\infty \frac{n^{nx}}{(n!)^ a}
\sim \sum_{n=1}^\infty \frac{n^{nx}}{(n/e)^{an}}\\
=\sum_{n=1}^\infty (\frac{e^an^{x}}{n^a})^n\\
=\sum_{n=1}^\infty (e^an^{x-a})^n\\
$

By the n-th root test, we want
$|e^an^{x-a}| < 1$
and this requires
$x < a$.


问:
Infinity as a limit in an indeterminate form
There is this limit
$$\lim_{x\to1}\frac{\ln x}{\left(x-1\right)^{2}}$$
for which I built a graph and I know the answer, so the question is not about how to compute it, but about my observation that when $x$ approaches $1$ from the right, the $\ln x =  0^+ $, but when $x$ approaches $1$ from the left, the $\ln x =  0^-$.

Based on that fact, even though it's an an indeterminate form $0 : 0$, can I already say that the limit does not exist or is that not enough and I need to turn it into a determinate form? And if I have to turn it into a determinate form, how can I do that?

答:
You don't "turn" an expression into an indeterminate form. It is an indeterminate form until you have computed its limit.

You did recognize an indeterminate form of the type $\dfrac00$, now your task is to evaluate it. L'Hospital is a way.


问:
Is character values on $G$-modules additive?
The following definition is from Atiyah & Macdonald. Let $C$ be a class of $A$-modules, where $A$ is a commutative ring with identity. Let $\lambda$ be a function on $C$ with values in an abelian group $G$. The function $\lambda$ is $\textit{additive}$ if for each short exact sequence $$0\to M' \to M \to M'' \to 0,$$ whose terms are in $C$, we have $\lambda(M')-\lambda(M)+\lambda(M'')=0$. For example, a simple example is let $C$ be the class of all finite-dimensional $k$-vector spaces $V$, then the map $V\mapsto \dim{V}$ is an additive function on $C$.

Now consider $C$ to be the space of $G$-modules/$G$-representations (by definition this is not an $A$-modules in the sense of Atiyah and MacDonald since $G$ is not a ring), if we have a short exact sequence of $G$-modules $$0\to M'\to M\to M'' \to 0,$$
is the character value $\chi(g)$ of an element $g\in G$ addtive on $C$, that is if $\chi_{M'}(g)-\chi_M(g)+\chi_{M''}(g)=0$?

答:
Edit: Fun, but irrelevant fact: a $G$-representation actually is a module in the sense of Atiyah-MacDonald, over a ring called the group algebra (or group ring). It's fun to show that a $G$-representation really is a module and in the ordinary sense, so I'd recommend giving it a go! https://en.wikipedia.org/wiki/Group_ring

We will make extensive use of the fact that representations are $k$-vector spaces, and basically translate this into a linear algebra problem.  By basic properties of short exact sequences, we may assume that $M' \subseteq M$ and $M'' = M/M'$ is the quotient representation.

Let $\{v_1, \ldots, v_m\}$ be a basis for $M'$ and extend it to a basis $\{v_1,\ldots, v_n\}$ for $M$.

Let $T:M \to M$ be the linear transformation corresponding to the action of $g$. Then $g$ acts on $M'$ by the restriction $T\mid_{M'}$ and $g$ acts on $M'' = M/M'$ by the induced map $\bar{T}$. Let $T$ have matrix $A = (a_{ij})_{1\leq i, j \leq n}$ with respect to the basis $\{v_1,\ldots, v_n\}$.

The value $\chi_{M'}(g)$ is the trace of the linear transformation $T\mid_{M'}:M' \to M'$. Since $\{v_1,\ldots, v_m\}$ is a basis for $M'$, it is easy to see that $\chi_{M'}(g) = \operatorname{tr}(\bar{T}) = \sum_{i=1}^m a_{ii}$.

We have that $\{v_{m+1} + M', \ldots, v_{n} + M'\}$ is a basis for $M'' = M/M'$, and $\bar{T}$ acts on this basis by
$$
T(v_j + M') = \sum_{i=1}^n a_{ij}(v_i + M') = \sum_{i=m+1}^na_{ij}(v_i + M'),
$$
which means that the diagonal entries of the matrix of $\bar{T}$ are $a_{ii}$ for $m + 1 \leq i \leq n$, and hence the trace of this map is $\sum_{i=m+1}^n a_{ii}$, so $\chi_{M''}(g) = \sum_{i=m+1}^n a_{ii}$.

Clearly $\chi_{M}(g) = \sum_{i=1}^n a_{ii}$, so we have
$$
\chi_{M}(g) = \sum_{i=1}^n a_{ii} = \sum_{i=1}^m a_{ii} + \sum_{i=m+1}^n a_{ii} = \chi_{M'}(g) + \chi_{M''}(g).
$$


问:
Which of the following are subspaces?
Determine whether the given set $S$ is a subspace of the vector space $V$.

A. $V=\mathbb{R}^{nxn}$, and $S$ is the subset of all $n×n$ matrices with $\det(A)=0$.

B. $V$ is the space of three-times differentiable functions $\mathbb{R}→\mathbb{R}$, and $S$ is the subset of $V$ consisting of those functions satisfying the differential equation $y'''+2y=x^2$.

C. $V=P^3$, and $S$ is the subset of $P^3$ consisting of all polynomials of the form $p(x)=ax^3+bx$.

D. $V=\mathbb{R}^n$, and $S$ is the set of solutions to the homogeneous linear system $Ax=0$ where $A$ is a fixed $m×n$ matrix.

E. $V$ is the vector space of all real-valued functions defined on the interval $[a,b]$, and $S$ is the subset of $V$ consisting of those functions satisfying $f(a)=3$.

F. $V=\mathbb{R}^{nxn}$, and $S$ is the subset of all upper triangular matrices.

G. $V$ is the vector space of all real-valued functions defined on the interval $(−∞,∞)$, and $S$ is the subset of $V$ consisting of those functions satisfying $f(0)=0$.

I have attempted to solve this by using the fact that the zero vector/polynomial has to be in the set so the set is non-empty, check that it is closed under addition and under scalar multiplication. I have found that A,C,D,F and G  satisfy these properties, however I am not sure if I am correct and I do not understand how to work with B. I know that E is not a subspace because the zero element is not in the set. Can someone help me? Thank you.

答: You're correct on all the ones you answered.  That leaves B.  But note that the function $y=0$ does not satisfy the differential equation, so you don't have the zero element in your subspace.

OwnerDisplayName: Thekla Lambri

Title: Which of the following are subspaces?


问:
How do you prove the property that you can move limits in and out of continuous functions?
Suppose $$\lim_{x\to c} f(x) = L,$$ and if $g$ is continuous, then $$g\left(\lim_{x \to c} f(x)\right) = g(L) \Rightarrow \lim_{x\to c} g(f(x)) = g(L).$$ A popular application of this would be to take the logarithm of both sides and then move the logarithm inside the limit. How do you prove this property?

答:
Let $\epsilon>0$. By continuity at $L$, there exists $\eta>0$ such that $|f(x)-L|<\eta \implies |g(f(x))-g(L)\rvert<\epsilon$. Since $\lim_{x\to c} f(x)=L$, there exists $\delta>0$ such that $0<|x-c|<\delta \implies |f(x)-L|<\eta$. Hence, $0<|x-c|<\delta \implies |g(f(x))-g(L)|<\epsilon$.

By definition,
$$
\lim_{x\to c} g(f(x))=g\left(\lim_{x\to c}f(x)\right)=g(L).
$$


问:
The set $\mathbb{C} \setminus \sigma(x)$ is connected.
Suppose that $A$ is a commutative Banach algebra with unit $e$ and the polynomials in $x \in A$ are dense in $A$. Show that the set $\mathbb{C} \setminus \sigma(x)$ is connected, where $\sigma(x)$ denotes the spectrum of $x$. I can show that $\lambda \notin \sigma(x)$ implies $P_n(x) \to (x-\lambda e)^{-1}$ for a sequence of polynomials $\{P_n(x)\}$ in $A$. However, I cannot prove that $P_n(z) \to (z-\lambda e)^{-1}$ uniformly for $z \in \sigma(x)$ without using the Gelfand transform. Any good suggestion?

答:
$
\newcommand{\C}{{\mathbb{C}}}
\newcommand{\N}{{\mathbb{N}}}
\newcommand{\inv}{^{-1}}
$

Suppose by contradiction that $\C\setminus \sigma (x)$ is disconnected,  so it admits at least one bounded connected
component,  say $\Omega $.

Picking any point $\lambda  \in \Omega $, we have that $x-\lambda $ is invertible and we may find  a sequence $\{p_n\}_n$ of polynomials
such that
$$
  \lim_{n\to \infty }\|p_n(x)-(x-\lambda )\inv\|=0.
  $$
Observing that the  Gelfand transform
$$
  A\to C(\sigma (x))
  $$
is contractive and
sends $x$ to the identity function on $\sigma (x)$,
we deduce that
$$
  \lim_{n\to \infty }\sup_{z\in \sigma (x)}|p_n(z)-(z-\lambda )\inv|=0.
  $$
Notice that
for every $n\in \N$ and every $z\in \sigma (x)$, one has that
$$
  |(z-\lambda )p_n(z)-1| = |z-\lambda ||p_n(z)-(z-\lambda )\inv|\leq  \alpha \beta _n,
  \tag{1}
  $$
where
$$
  \alpha =\sup_{z\in \sigma (x)}|z-\lambda |,
  $$
and
$$
  \beta _n=\sup_{z\in \sigma (x)}|p_n(z)-(z-\lambda )\inv|.
  $$
Observing that the boundary of $\Omega $ is contained in $\sigma (x)$, we deduce from the maximum principle that (1) also holds
for every $z\in  \Omega $, and hence
$$
  |p_n(z)-(z-\lambda )\inv|\leq  {\alpha \beta _n\over |z-\lambda |}, \quad\forall z\in \Omega \setminus\{\lambda \}.
  $$
Letting $\gamma (t)=\lambda +re^{it}$, for $t\in [0,2\pi ]$, with $r$ sufficiently small so that $\gamma $ lies in $\Omega $, we than have that
$$
  2\pi i=
  \left|\int_\gamma  p_n(z)-(z-\lambda )\inv \,dz\right| \leq
  {\alpha \beta _n\over r}\text{Length}(\gamma ) =2\pi \alpha \beta _n.
  $$
Since $\beta _n\to 0$, as $n\to \infty $, this is a contradiction.


问:
Rest with congruences
Find the remainder of the division of $619^{1367}$ by $15$.

I cannot see what the initial congruence associated with the problem would be. Any tips?

Title: Rest with congruences

Body: Hint:  Note $619 = _{15} 4$ and that $4^2 = _{15}1$.


问:
confusion about differential eq question
i was using sympy to solve Differential eqs  this was the code
enter image description here

what does that C mean in the output and also is the answer right?Im new to solving D.E and im actually clueless where to start with

答:
The C1 is an arbitrary constant. It appears during the integration. Let's try to solve the equation. ($f(x) = y$ for simplicity)

$-5y+y'=2$

$\frac{dy}{dx}=2+5y$

$\frac{1}{2+5y}dy=dx$

Then we integrate both sides. Notice that an arbitrary constant will appear since the integration is not definite.

$\frac{1}{5}ln(5y+2)=x+C$

$5y+2=e^{5x+C}=e^{5x}e^C$

Since $e^C$ is a constant, we can substitute it with an arbitrary constant which is $C1$. You can try to solve for $y$ and you will get the result.


问:
Series Expansion of Exponential Function is Cauchy
Define $p_n (x) = 1 + \frac{x}{1!} + \frac{x^2}{2!} + \dots + \frac{x^n}{n!}$.
Show that $\{ p_n (x) \}_{n=0}^{\infty}$ is a Cauchy sequence.

I'd just like to know the facts required to show this. We know that if a function $f: A \mapsto \mathbb{R}$, $A \subseteq \mathbb{R}$, is uniformly continuous on $A$ then, if $(x_n)$ is a Cauchy sequence in $A$, then $(f(x_n))$ is also a Cauchy sequence. Is this a relevant fact? Do I need to find some sequence, say $y_n$, that is Cauchy, and then find an explicit function that maps the terms of $y_n$ to the terms of $p_n(x)$?

Is there an easier way to do this?

Edit: I need to expand on this... the norm is $||p|| =$max$_{[0,1]} | p(x)|$. So for $p_n$ and $p_m$ the difference is $\sum _{ i = \min (n,m)} ^{\max (n,m)} \frac {x^i}{i!}$. The maximum term is the first term, because I've shown this to be a decreasing sequence. Now, is it still bounded by the geometric series with the factor 1/2?

答:
A sequence $f_n(x)$ is uniformly Cauchy if $\lim_{n\to \infty} ||f_{n+q}(x)-f_n(x)||_{\infty}=0,\quad \forall q\geq 1 $.

Now, applying this to our problem, we have

$$| p_{n+q} (x)-p_n(x) | = \Big| \sum_{k=0}^{n+q}\frac{x^k}{k!}-\sum_{k=0}^{n}\frac{x^k}{k!}\Big|=
\Big| \sum_{k=n+1}^{n+q}\frac{x^k}{k!} \Big|\leq \sum_{k=n+1}^{n+q}\frac{x^k}{k!}\leq \sum_{k=n+1}^{\infty}\frac{x^k}{k!}<\epsilon \rightarrow (1) $$

$$\implies \sup_{0\leq x\leq 1} |p_{n+q} (x)-p_n(x)|=||p_{n+q} (x)-p_n(x)||_{\infty}<\epsilon.$$

The last inequality in $(1)$ follows from the fact that $ \sum_{k=0}^{\infty}\frac{x^k}{k!} $ is a convergent series.


答:
We have $619^{1367}\equiv 4^{1367}\mod 15$.

Since $4^2=16$ and $16\equiv 1\mod 15$, we can use this here, by manipulationg the exponent.

$4^{1367}=4\cdot 4^{1366}=4\cdot 16^{683}$.

So $4^{1367}\equiv 4\cdot 16^{683}\equiv 4\cdot 1\equiv 4\mod 15$

Title: Rest with congruences

Body:
Find the remainder of the division of $619^{1367}$ by $15$.

I cannot see what the initial congruence associated with the problem would be. Any tips?


问:
Simple algebra, finding x
How and why did the sign + and - switch places in the solution?

Question:

$$\frac{x-6}x - \frac6x = \frac{x-6}{x+6}$$

Answer:

Step one

$$\frac{x-6-6}x = \frac{x+6}{x-6}$$

答:
The original solution is like this :

$$\dfrac{x - 6}{x} - \dfrac 6x = \dfrac{x - 6}{x + 6} \implies \dfrac{x - 12}{x} = \dfrac{x - 6}{x + 6}\\
\implies x(x -6) = (x - 12)(x +6)\\
\implies x^2 - 6x = x^2 +6x-12x-72\\
\implies -6x = -6x - 72\\
\implies 6x = 6x + 72$$

Now that is unsolvable, since $6x-6x = 0 \neq 72$.

Therefore, the actual question must be $\dfrac{x - 6}{x} - \dfrac 6x = \dfrac{x + 6}{x - 6} $.

If this is the question, the solution in the picture is the right solution.

Title: Simple algebra, finding x


问:
Dirichlet process Posterior notation
While reading through Yee Whye Teh's tutorial on Dirichlet Process, I came across the Posterior distribution for DP. After observing some draws $\theta_1,\theta_2, ... ,$, the posterior is defined as $$G | \theta_1,...,\theta_n \sim DP(\alpha + n, \frac{\alpha}{\alpha + n}H +\frac{n}{\alpha + n}\frac{\sum_{i=1}^n \delta_{\theta_i}}{n})$$
I understand the posterior distribution of the DP is thus generated in the following manner: Draw an infinity of $\rho's$: $\rho_1,\rho_2, ...$ following a $GEM(\alpha + n)$ distribution. And then draw an infinity of $\theta$'s from the mixture of base distribution $H$ and the empirical distribution.

Can someone explain what $\delta_{\theta_i}$ means in this case ?

答:
$\delta_{\theta_{i}}$ is the dirac measure on $\theta_{i}$:

$$\delta_{\theta_{i}}(A) = 1_{A}(\theta_{i})$$

It is present in the posterior of the DP because, for a measurable partition $B_1,...,B_k$ of $\Theta$, the marginal distribution becomes:

$$(G(B_1),...,G(B_k)) | \theta_{1},...,\theta_{n} \sim DP(\alpha H + \sum_{i=1}^n \delta_{\theta_i}(B_1), ..., \alpha H + \sum_{i=1}^n \delta_{\theta_i}(B_k)) \sim DP(\alpha H + n_1, ..., \alpha H + n_k)$$

since $n_j = \sum_{i=1}^n 1_{B_j}(\theta_i) = \sum_{i=1}^n \delta_{\theta_i}(B_j)$

Title: Dirichlet process Posterior notation


问:
Proving part of the Wedderburn Structure Theorem
I'm having trouble with an exercise from "Noncommutative Algebra," by Farb & Dennis proving part of the Wedderburn structure theorem for semisimple rings.

If $R$ is a semisimple ring and $\{M_1,\ldots,M_n\}$ is a set of representatives for the isomorphism classes of simple $R$-modules, let $D_i=End_R(M_i).$ We want to show that $R$ is isomorphic to $\prod_{i=1}^nEnd_{D_i}M_i,$ and to do this we would like to show that $M_i$ is finite-dimensional over $D_i$. (In particular, if the decomposition of $R$ as a left $R$-module is $\bigoplus M_i^{n_i},$ then it will turn out that $n_i=dim_{D_i}(M_i)$).

Hoever, I've been having difficulties proving why $M_i$ is finite-dimensional over $D_i$, although I see why it should be. What exactly are the details of the proof that $dim_{D_i}(M_i)$ is finite? Is there some trick involved that I'm missing, or does the proof just consist in following all the definitions of semisimple rings and simple modules?

答:
Here is an explanation of the relevant theory, including the fact you're looking for. The thing is that the way I see it, I learn the structure of the $M_i$ modules after I prove Wedderburn's result. So I wrote the entire thing down:

Let $R$ be a semisimple ring. That is, we have equality of $R$-modules $R=\oplus_{i\in I}M_i$.

First, we claim that $I$ is a finite set. This is true because:

$R$ is generated by $1_R$ over $R$.

By definition of a direct sum, the element $1_R$ (as any other element of $R$) has a finite number of nonzero components in the decomposition $R=\oplus_{i\in I}M_i$.

So, from now on we will write: $R=\oplus_{i=1}^r M_i^{n_i}$  with each $M_i$ being a simple $R$-module and with $M_i\not\cong M_j$ for $i\neq j$..

Next, we claim that all simple $R$-modules are isomorphic to one of the $M_i$.

Let $M$ be a simple $R$-module. Fix $0\neq m\in M$. The map $\varphi:R\rightarrow M$ defined by $\varphi(r)=rm$ is a nonzero homomorphism of $R$-modules (verify). Now look at $\varphi:\oplus_{i=1}^r M_i^{n_i}\rightarrow M$ again. By Schur's lemma, the simplicity of $M$ implies that the restriction of $\varphi$ to each $M_i$ is either zero or an isomorphism. But it can't be zero for all those restrictions because $\varphi$ is not the zero map. Thus, at least one of the $M_i$ is mapped isomorphically onto $M$.

We now prove Wedderburn's theorem (this is independent of the previous discussion) which gives a ring isomorphism $R\cong \prod_{i=1}^r M_{n_i}(\text{End}_R(M_i)^{op})
= \prod_{i=1}^r M_{n_i}(D_i)$:

$R^{op}\cong \text{End}_R(R)$ (because $\text{End}_R(R)$ is exactly multiplication on the right by elements of $R$).

  So, $R^{op}\cong \text{End}_R(\oplus_{i=1}^r M_i^{n_i}) \cong\prod_{i=1}^r\text{End}_R(M_i^{n_i})$ (this follows from Schur's lemma)

  So, $R^{op}\cong\prod_{i=1}^rM_{n_i}(\text{End}_R(M_i))$.

  Taking $-^{op}$ on both sides now yields the desired result.

Note that by Schur's lemma, each $D_i=\text{End}_R(M_i)^{op}$ is a division ring.

Now that we know that as a ring we have $R\cong \prod_{i=1}^r M_{n_i}(D_i)$, we deduce that as an $R$-module we have $R\cong \oplus_{i=1}^r(D_i^{n_i})^{n_i}$. It is easy to see that each $D_i^{n_i}$ is a simple $R$-module. Thus, these are the simple $R$-modules. That is - the $M_i$ modules are the $D_i^{n_i}$ modules. We also see that the dimension of $M_i$ over $D_i$ equals the number of times it appears in the decomposition of $R$ as a semisimple $R$-module.


问:
Find the time where their velocity is equivalent
The question is:

The equations of movement of 2 particles are:
$$x_1=3t^2+5t+6 \quad m$$ $$x_2=6t+8 \quad m$$
Find the exact time where both will have the same velocity.

So i've tried this: $$\vec{v}=\frac{dx}{dt}$$ basically differentiated both to get $$\vec{v_1}=6t+5\quad m/s$$ $$\vec{v_2}=6\quad m/s$$ and then i got their magnitude $|\vec{v_1}|$ and $|\vec{v_2}|$ as this $$|\vec{v_1}=\sqrt{36t^2+5^2}$$ $$|\vec{v_2}|=6$$ and then equated both to get the instant time where both velocities are equivalent, $$36t^2+5^2=6^2$$ $$t^2=\frac{11}{36}=\sqrt{\frac{11}{36}}$$ is this correct? i'm afraid i should have equated both velocity vectors and not their magnitude

答:
The question has already been answered by others. The answer is the solution of the equation $6t+5 = 6$ which means $t=\frac{1}{6}$

I just wanted to clarify some things for Xetrez that I think he might be misunderstanding.

'Velocity' is a vector by definition. The corresponding scalar (magnitude) is 'speed'.

Note that the problem doesn't give any vector notation or any direction information, so we can assume all velocities are along a single axis. You can presume there's a unit vector $\hat i$ that is left off for convenience.

So if we're dealing with a single axis, does that mean direction doesn't matter? No! You can still have forward (positive) or backwards (negative) along a single axis. So if a problem asks about velocity, that means the sign matters. So you wouldn't take absolute values when the question is about velocities. If it asks about 'speed' then only magnitude matters, hence the sign wouldn't matter in that case, and you'd take absolute values.

You didn't take the absolute value correctly. This was an unnecessary step, but I just wanted to point out the error anyway.

$|\vec{v_1}| = \sqrt{(6t+5)^2} = \sqrt{36t^2+60t+25}$.

If the problem restricted t to $t \ge 0$, then $|\vec{v_1}| = |6t+5| = 6t+5$

Since the whole thing is along a single component $\vec{v_1} = (6t+5)\hat i$. You just take the absolute value of that one component, just like taking the absolute value of any number.

I wanted to point this out because I thought maybe you got confused between taking the magnitude of a vector with two perpendicular components.

If you had this:

$\vec{v_1} = (6t)\hat i + 5\hat j$

Then the magnitude of the vector would be: $\sqrt{(6t)^2+5^2}$. This is because $\hat i$ and $\hat j$ are perpendicular unit vectors, one along the x-axis and one along the y-axis.

If you weren't confusing these two ideas, then I apologize.

One final thing... if the problem were asking about all times when the speeds were the same, then what we'd want is

$|\vec{v_1}| = |\vec{v_2}| \iff $

$|6t+5| = |6| \iff$

$|6t+5| = 6 \iff$

$6t+5 = 6$ OR $6t+5 = -6$

or you could solve $\sqrt{36t^2+60t+25} = 6$ and get the same solutions.

So if the problem were asking about times when speeds were the same we'd get $t=\frac{1}{6}$ or $t=\frac{-11}{6}$. If the problem had the additional restriction that $t \ge 0$ we'd have only $t=\frac{1}{6}$ for the speed case also.


问:
limit with two variables
The limit I need to calculate is $\lim_{(x,y)\rightarrow (0,0)}\frac{xy^{2}}{x^{4}+y^{2}}$. Using polar coordinates I get:
$lim_{r\rightarrow 0}\frac{r\cos(\theta)\sin^{2}(\theta)}{r^{2}\cos^{4}(\theta)+\sin^{2}(\theta)}$. Now if $\sin(\theta)\neq 0$ then the limit is $0$. How do I handle the case where $\theta=0$ or $\theta = \pi$?
And is there a better way to approach this limit?

答:
For any $ x\ne 0$ and $ y\ne 0$, we have

$$x^4+y^2>y^2>0$$

$$\frac{1}{x^4+y^2}<\frac{1}{y^2}$$

$$|\frac{xy^2}{x^4+y^2}|<|x|$$

Thus, the limit is zero.

By polar coordinates, you will find that
$$|F(r,\theta)|\le r|\cos(\theta)|\le r$$

Title: limit with two variables


问:
Vanishing particular solution for $yy'-ay=-bx$
Consider a differential equation of the form:
$$y\frac{dy}{dx}-ay=-bx$$
where both both $a$ and $b$ are known real numbers
$$a,b\neq0.$$
When I attempt to solve this with a particular solution of:
$$y_p=Cx+D$$
Results to
$$CD-aD=0$$
$$C^2-aC=-b$$
By inspection, we can say that:
$$C=a$$
But this does not make sense since:
$$0=-b$$
Which is impossible!

I even tried using quadratic but same thing. This looks very simple, but I am now a bit lost here, what is the best way to solve this differential equation?

答:
$$y(x)y'(x)-ay(x)+bx=0$$ is not an easy one. May be they are some tricks I miss but I give you my attemps
$$y=\frac 1 2x(z(x)+a) \implies(4b-a^2)+x (a+z(x)) z'(x)+z(x)^2=0$$ Switch variables
$$(4b-a^2)+\frac{x(z)}{x'(z)}(a+z)+z^2=0$$
$$x(z)=e^{u(z)}\implies (4b-a^2)+\frac{a+z}{u'(z)}+z^2=0$$ which is separable
$$u'(z)=\frac{a+z}{(a^2-4 b)-z^2}$$
$$u(z)+C=-\frac{1}{2} \log \left((4b-a^2)+z^2\right)-\frac{a }{\sqrt{4 b-a^2}}\tan ^{-1}\left(\frac{z}{\sqrt{4
   b-a^2}}\right)$$


问:
How to solve a integer part inequality?

$[x+2]>3$

I was thinking to just do $x + 2 >3$ and then it will give me $x > 1$ so I add the difference between $3$ and $2$ only to the right part, because not adding it will give me $x \in (1,+\infty)$ so if I plug in $1.99$ it will give me $3$ which is not good because $3 > 3$ is false.

答:
$$\lfloor x+2\rfloor=\lfloor x\rfloor+2$$ is always true and your equation reads

$$\lfloor x\rfloor>1$$ or

$$\lfloor x\rfloor\ge2$$ (as the LHS is an integer).

This occurs for all

$$x\ge2.$$

Title: How to solve a integer part inequality?


答:
To evaluate indetermination just plug in the limit value for numerator and denominator to see that it is in the form $\frac 0 0$.

For the solution we can avoid l'Hospital by $x-1=t \to 0$ to obtain

$$\frac{\ln x}{\left(x-1\right)^{2}}=\frac{\ln (1+t)}{t^{2}}=\frac1t\frac{\ln (1+t)}{t}$$

with $\frac{\ln (1+t)}{t}\to 1$ therefore we can conclude that limit doesn't exist.

Title: Infinity as a limit in an indeterminate form

Body:
There is this limit
$$\lim_{x\to1}\frac{\ln x}{\left(x-1\right)^{2}}$$
for which I built a graph and I know the answer, so the question is not about how to compute it, but about my observation that when $x$ approaches $1$ from the right, the $\ln x =  0^+ $, but when $x$ approaches $1$ from the left, the $\ln x =  0^-$.

Based on that fact, even though it's an an indeterminate form $0 : 0$, can I already say that the limit does not exist or is that not enough and I need to turn it into a determinate form? And if I have to turn it into a determinate form, how can I do that?


问:
Commutativity of a sheaf of groups from an epimorphism
Let $F$ and $G$ be sheaves of groups $\mathcal{S}^{op}\to Groups$ and $f:F\to G$ an epimorphism (of sheaves of sets).

If $F$ is a sheaf of commutative groups, is $G$ also a sheaf of
  commutative groups?

If $f$ would be a surjection on every section then for $x,y\in G(V)$ there would be $x', y'\in F(V)$ with $f(V)(x')=x$ and $f(V)(y')=y$ and
$$
xy=f(V)(x')f(V)(y')=f(V)(x'y')=f(V)(y'x')=f(V)(y')f(V)(x')=yx.
$$
but $f$ is only surjective on stalks.

答:
Yes. You may find the following fact helpful: a sheaf of groups on a topological space is a sheaf of abelian groups if and only if all its stalks are abelian groups.

The claim is also true for a general site, but then it makes no sense to talk about stalks. Instead the best approach is to treat the question internally in the topos of sheaves of sets and observe that being an abelian group is a condition that can be expressed in terms of the commutativity of certain diagrams of sheaves of sets; thus, the defining property of epimorphisms is enough to imply that quotients of sheaves of groups are sheaves of abelian groups. This amounts to internalising the usual proof that any group quotient of an abelian group is again an abelian group. (I reiterate my view that sheaves of algebraic structures are best thought of as internal models in a topos rather than sheaves taking values in some category.)


问:
Prove that IK, AC, BD are concurrent.

Given a circle (O, R) with diameter AB. Point M on (O), A, B are not coincident. Two lines through O and perpendicular to AM, BM intersects the tangent of (O) through M at C, D, respectively. OC intersects AM at I, OD intersects BM at K. Prove that IK, AD, BC are concurrent.

Attempts: I tried drawing an altitude through M of triangle ABC, intersecting IK at some point but still stuck on proving that it is the midpoint of that altitude. AC, BD are tangents of (O) and I, K are midpoints of AM, BM respectively has been proved.

答:
Let $IK$ intersects $AD$ at $S$. Then homothety at $S$ which takes $A$ to $D$ takes $I$ to $K$. If we prove that it takes also $C$ to $B$ then we are done.

Now $CA =CM$ so the line $CA$ is also tangent to (semi) cricle at $A$ and thus $CA\bot AB$. The same way we have $BD\bot AB$, so $CA||BD$. But this now means that $C$ goes to $B$ and we are done.

Title: Prove that IK, AC, BD are concurrent.


答:
It's straight from definition of continuous and limits.

Let $\lim_{x\to c}f(x) = L$.  $g$ is continuous so for any $\epsilon > 0$ there is a $\delta > 0$ so that $|y-L| < \delta \implies |g(y) - g(L)| < \epsilon$.

But $\lim_{x\to c}f(x) = L$ so for that $\delta$ then there is $\gamma > 0$ so that if $|x - c|< \gamma$ then $|f(x) - L| < \delta$ and therefore $|g(f(x)) - g(L)| = |g(f(x)) - g(\lim_{x\to c}f(x))| < \epsilon$.

So by definition.   $\lim_{x\to c} g(f(x)) = g(L)=g(\lim_{x\to c} f(x))$.

Title: How do you prove the property that you can move limits in and out of continuous functions?

Body: Suppose $$\lim_{x\to c} f(x) = L,$$ and if $g$ is continuous, then $$g\left(\lim_{x \to c} f(x)\right) = g(L) \Rightarrow \lim_{x\to c} g(f(x)) = g(L).$$ A popular application of this would be to take the logarithm of both sides and then move the logarithm inside the limit. How do you prove this property?


Title: Rest with congruences

Body:
hint

$$15=3×5$$
$$\phi(15)=15(1-\frac 13)(1-\frac 15)=8$$

By Fermat, since $gcd(619,15)=1$

$$619^8\equiv 1 \mod 15$$

now, observe that $ 1376=172×8$.


问:
Show that $(E|X|^{p_1})\leq (E|X|^{p_2})^\frac{p_1}{p_2}$, where $0<p_1<p_2<\infty$
Let $X$ be a random variable on $(\Omega,\mathcal{F},\mu)$. for $h\in L^1(\Omega,\mathcal{F},\mu)$, define $Eh(X)=\int h(X)d\mu$.

Show that $(E|X|^{p_1})\leq (E|X|^{p_2})^\frac{p_1}{p_2}$, where $0<p_1<p_2<\infty$

Further show that equality holds iff $|X|$ is constant a.e.

I honestly don't have a good handle on how to start this one.

I sort of want to use jensen's inequality, but $x^{p_1},x^{p_2}$ aren't convex if $p_1$ or $p_2$ are less than $1$. Even if this could be remedied by multiplying  by a negative, it's not clear how to proceed. The fact that $\frac{p_1}{p_2}<1$ seems relevant, but I don't see where exactly that could be relevant to this problem.

I considered trying to show that $(E|X|^{p_1})^{p_2}\leq (E|X|^{p_2})^{p_1}$, but that didn't really get me very far.

Any thoughts suggestions or hints would be appreciated.

Thanks in advance.

答:
Let $\phi(x) = |x|^{\frac{p_1}{p_2}}$. Note that $\phi$ is concave due to $p_1\leq p_2$. Then
$$E|X|^{p_1} = E\left(|X|^{p_2}\right)^{\frac{p_1}{p_2}} = E\phi\left(|X|^{p_2}\right) \leq \phi\left(E|X|^{p_2}\right) = \big(E|X|^{p_2}\big)^{\frac{p_1}{p_2}}$$
by Jensen's inequality and the fact that $-\phi$ is convex.


问:
Finite/Infinite groups
Which of the following statements is/are true?

a) There are infinitely many finite groups in which every non-identity element have order $2$.

b)There exists an infinite group in which every non-identity element have order $2$.

c)There exists an infinite group in which there are elements of order $n$ for all $n\in \Bbb N$.

d)There are infinitely many infinite groups in which each non-identity element have finite order.

My attempt: Option 2 seems correct as I have an example of such a group( Power set of Natural numbers under the binary operation of symmetric difference is one such example) but I am unable to conclude anything about other options.

About option 1, it seems that it is true keeping in mind $\Bbb Z_2\times \Bbb Z_2\times\cdots$ but still I am confused. For the last two options, I don't have any conclusive idea which I can apply here. Please guide.

答:
If it's easier for you to conceptualize, you can take each element to be finite, even though the group as a whole is infinite.

For instance, for (2), we can take each element of $G$ to be a finite sequence consisting of $1$ and $-1$ and ending in $-1$ (the identity is the null sequence). To find $ab$, we multiply the values of $a$ and $b$ pairwise, then terminate the sequence at the last $-1$. If they are of different lengths, we append $1$ to the shorter one enough times to make them the same length, then multiply. $G$ is infinite because although each individual element has a finite length, there's an infinite number of those finite lengths.

A similar strategy works for (3) and (4). For (3), we take finite sequences in which the $n$th value is an element of $\mathbb Z_n$, unless the sequence is length 1, the last value isn't $1$. For (4), we simply take variations on (3). For instance, we can start the sequence at $\mathbb Z_k$. Or take $k$ copies of (3), etc.

Title: Finite/Infinite groups


问:
GCD and divisibility
Update: I think there was a typo in the text. Please don't waste your time with this problem. :)

If gcd$(a,b) = p$, a prime, then $p|am$ and $p|an$ such that  gcd$(m,n) = 1$

Why does gcd$(m,n)$ have to be equal to 1? And does this statement hold true if gcd$(a,b)$ is not equal to a prime?

答: Your original statement seems like it is miscopied from somewhere, or missing additional context.  If $\gcd(a,b)=p$, prime or not, then $a=pc$ and $b=pd$, for some integers $c,d$ such that $\gcd(c,d)=1$.</p>

Title: GCD and divisibility


问:
linear approximation in point $(0,\ldots,0)$ for $f(x_1,\ldots,x_n)= e^{-x_1}+\cdots +e^{-x_n}$
hello i dont really know how to do this linear approximation for the function $f(x_1,\ldots,x_n)= e^{-x_1}+\cdots +e^{-x_n}$ in point $(0,\ldots,0)$
please help me
i started with  i know how to do it for endless functions but in this case i dont know how to do it maybe with the hesse matrix?

答:
In multivariate calculus we have the standard formula
$$f({\bf x})=f({\bf 0})+\nabla f({\bf 0})\cdot{\bf x}+o\bigl(|{\bf x}|\bigr)\qquad({\bf x}\to{\bf 0})\ .$$
For your function $f$ one computes
$$f({\bf 0})=n,\qquad\nabla f({\bf 0})=(-1,-1,\ldots,-1)\ .$$
It follows that the linear approximation of $f$ near ${\bf 0}$ is
$$f_{\rm lin}({\bf x})=n-\sum_{i=1}^n x_i\ .$$


问:
Finding best players in a tournament with a probabilistic comparison function
I am currently facing the following problem in my research and I have no clue how to tackle this kind of question.

The problem

Imagine you have a tournament with $n$ players $P=\{p_1,...,p_n\}$. My goal is to determine one of the best players in my tournament.

I do have a comparison function $f: P x P\to \{0,1\}$ that can tell me which of two given players is better, i.e. $f(p_1,p_2)=1$ iff player two is better than player one and $f(p_1,p_2)=0$ iff player one is better than player two. You can think of $f$ as the $<$ relation.

The kicker is that my comparison function $f$ has an error, meaning that it will give me the correct result of my comparison with a probability $p>0.5$. Calculating $f$ will take some time and thus I want to find a good player for my tournament with the least amount of queries. My current approach is to compare all players with each other which gives me a total amount of $b \in O(n^2)$ comparison calls. I then chose the player $p_i$, which "won" the most comparisons.

Edit:

Please be aware that my comparison function will give me the same result for a call $f(p_i,p_j)$ no matter how often I call it. So the probability that the result is correct is $p$, but the function itself is deterministic. My example below is a bit misleading. However, each comparison call is only done once so this won't be a problem.

Key questions

What is the probability that the chosen player is the best player?

What is the probability that the chosen player is in the top k percent?

My thoughts

I think that question one might be easier to calculate as my best player will win all comparisons if $p=1$ and I can deduce the probability that $k$ comparisons were correct. However, I am stuck at the point at which I have to calculate the probability that it in fact is the player that "won" the most comparisons as others might be evaluated incorrectly.

My dream is to get a formula that allows me to calculate the desired probabilities for different $p,n$, and budget $b$.

Simulation

I wrote a small simulation in Python which revealed some interesting facts about the influence of $p$. In my example, the tournament players are represented as numbers $0,...,63$. The function $f$ is the standard $<$ relation with a given probability. In the plot below I have plotted the mean position (y-axis) that was selected as the best individual for different $p$ (x-axis). You can find the source code below.

```<code>import random
import numpy as np
from itertools import combinations
from tqdm import tqdm
import matplotlib.pyplot as plt

x, y = [], []

n = 64 # How many players
nums = np.arange(n).tolist() # Player strengths
count = 1000 # The amount of tests (O(n^2)) combinations that should be made

for p in tqdm(np.arange(0, 1, 0.01)):
    x.append(p)

    def compare(a, b):
        r = random.random()
        if r <= p:
            return a < b
        else:
            return a >= b

    def tournament():
        scores = [0] * n
        for a, b in combinations(nums, 2):
            result = compare(a, b)
            if result:
                scores[b] += 1
            else:
                scores[a] += 1

        best = max(nums, key=lambda x: scores[x])
        return best

    vals = []

    for _ in range(count):
        vals.append(tournament())

    y.append(np.mean(vals))

plt.plot(x, y)

plt.show()
</code>```

答:
The score of player $p_i$ is roughly normal with mean $p(i-1)+(1-p)(n-1-i)$ and constant variance $(n-1)p(1-p)$.  Integrate the pdf of the top one multiplied by the cdfs of all the others.  That won't have a nice formula but might have a nice approximation.  With a gap of $2p-1$ between players' means and standard deviation of $\sqrt{np(1-p)}$, the odds of the top player winning could be $O((2p-1)/\sqrt{np(1-p)})$.  Probabilities $p=1-k/ n$ for $k=O(1)$ might be the transition region where the top player's chances falls from 1.  Conversely, $p=(1/2)+k/\sqrt n$ might be the region where the weakest players' chances drop from $1/n$ to much less.

There is a small error in that cross-correlation between any two comes from a single comparison.

With a budget $b$, the means and variances are both scaled down in proportion to $b$, so the spreads overlap more.

Simulations using the normal distribution approximation suggest the chance of number-one winning is
$$\frac1{1+\left(\frac{\sqrt{np(1-p)}}{4p-2}\right)}$$
or, if you do $c{n\choose2}$ of the comparisons with $0\lt c\lt 1$,

$$\frac1{1+\left(\frac{\sqrt{np(1-p)/c}}{4p-2}\right)}$$


问:
Hyperbolic Geometry Theorem

Prove the perpendicular bisector to AB is perpendicular lo the line IJ and IJ and AB are parallel divergent.

Prove $IJ=\dfrac{1}{2}ED$ prove for angle $A$ acute and obtuse to deduce that in hyperbolic geometry you have $IJ<\dfrac{1}{2} AB$.

So if the angle $A$ is acute or obtuse then I am almost sure there most be a saccheri cuadrilateral $BAED$ but i cant prove the cuadrilateral meets axiom of $BE$  $AD$ congruent to $BE$.

So i have to prove that a segment C perpendicular to segment $AB$ is also perpendicular to other segments parallel to $A$, $ED$ but the triangle doesnt make sense to me any ideas would be great thanks !!!

By angle side angle the bisector divides angle C into equal angles according to bisector definition, so the triangle by $ASA$ into congruent triangles according to me but i cant see a solution-

答:
Using SAA criterion, prove that $\triangle BIE\equiv \triangle CIF$ and $\triangle AJD\equiv \triangle CJF$.

From these congruences we get $BE\equiv CF\equiv AD$, hence $BEDA$ is a Saccheri quadriteral with base $ED$.

The line passing through the midpoints of the base and the summit of a Saccheri quadrilateral is perpendicular both to the base and the summit.

Lines which have common perpendicular line are divergent parallel.

Summit of a Saccheri quadrilateral is always greater than the base, hence $AB>ED$.

Finally you have $EI=IF$ and $DJ=JF$ and to prove $IJ=\frac{1}{2}ED$ you have to consider few cases:

Angles $\angle A$ and $\angle B$ are acute.
Then $ED=EI+IF+FJ+JD=2IF+2FJ=2IJ$

Angle $\angle A$ is right. Then $F=J=D$.

Angle $\angle A$ is obtuse. Then $ED=EF-FD=2IF-2JF=2IJ$.

Angle $\angle B$ is right. Case symmetric to 2.

Angle $\angle B$ is obtuse. Case symmetric to 3.

Title: Hyperbolic Geometry Theorem


问:
The union of a locally finite family of functionally closed- RYSZARD ENGELKING
In the book RYSZARD ENGELKING General Topology he states

Show that the union of a locally finite family of functionally closed sets is not necessarily functionally closed. Note that in a perfectly normal space the union of a locally finite family of functionally closed sets is functionally closed.

A topological space $X$ is called a perfectly normal space if $X$ is a normal space and every closed subset of $X$ is $G_{\delta}$-set.

A subset $A$ of a topological space $X$ is called functionally closed if $A=f^{-1}(0)$ for some continous function $f:X \to [0,1]$

Every closed subset of $X$ is $G_{\delta}$-set means that it countable intersections of open sets

On the other hand,

observation 1: I have proven that a subset $A$ of a normal space is closed $G_{\delta}$ if and only if there exists a continuous function $f:X \to [0,1]$ such that $A=f^{-1}(0)$. i.e, in a normal space functionally closed sets coincide with closed $G_{\delta}$.

A family $\{A_s\}_{s \in S}$ of subsets of a topological space $X$ is locally finite if for every point $x \in X$ there exists a neighbourhood $U$ such that the set $\{s \in S : U \cap A_s \neq \emptyset \}$ is finite.

Any ideas:
Suposse that $X$ is perfectly normal espace.
Let $A=\{A_s\}_{s \in S}$ a locally finite family of functionally closed sets, for observation 1, the $A_s$ are $G_{\delta}$-sets.  We have to find a continuous function $f:X \to [0,1]$ such that $\cup_{s \in S} A_s= f^{-1}(0)$. For each $A_s$ there is a continuous function $f_s: X \to [0,1]$ such that $f_s^{-1}(0)=A_s$.I don't know how to use my hypotheses to build my continuous function f? some help?
On the other hand, I can't think of a counterexample for the first part.

答:
As you discovered, the set $\Bbb Q$ in the space that I constructed in my answer to this question is the union of a closed, discrete (hence locally finite) family of functionally closed sets but is not itself functionally closed.

For the rest of the question, the union of a locally finite family of closed sets is closed, so if the space is perfectly normal, the union is a $G_\delta$-set and therefore functionally closed.


问:
Min cost max flow variant
Let $$ be the set of customers and let $_⊆$ be the set of items from which customer $∈$ wants to buy only one, where $$ are the items. Denote $(,)≥0$ as the price customer $$ is willing to pay for item $$. Assume $||=||=.$

The question is to find a polynomial time algorithm which maximizes the total revenue independent of the number of customers that actually get an item.

To do this, I created a bipartite graph $I$ and $J$ with source $s$ and sink $t$.
$$
(s,i), i \in I \\
(j,t), j \in J
$$
with capacity $1$ and cost $0$. The cost function we make
$$
w^*(i, j) = \begin{cases}\frac{1}{w(i,j)} & w(i,j) \neq 0 \\ 0 &otherwise\end{cases}.
$$

where edge $(i,j)$ has capacity 1 if $j \in J_i$.
Now, we solve this problem using a min cost max flow algorithm. However, there can be scenario's where a higher flow is selected with a lower cost (and hence lower revenue) where a lower flow with a higher cost should have been selected.

How do I use the min-cost max flow algorithm to select the highest cost unrelated to the amount of flow?

答:
Make it so that everyone is willing to buy every item (but possibly only willing to pay a price of $0$ for it) and you ensure that the maximum flow is $n$.

Your cost function doesn't order $w(i,j)=0$ correctly, so instead use $w^*(i,j) = -w(i,j)$.

At this point, a min-cost max flow will maximize the total revenue. (Ignore edges with $w(i,j)=0$ in the solution, if you like; it won't change the answer.)

This can also be solved as an assignment problem, of course, but this is the min-cost max-flow approach.

Title: Min cost max flow variant


问:
What is value of following integral ( Solve by complex analysis)
I am trying exercises of textbook Ponnusamy and Silvermann and I am struck on this problemof Section 9.37.

Evaluate the following integral along different simple closed curves not passing through 0, +1, -1.

$\int_{C} \frac{e^{z-1} -1 } {z^2 (z^2 -1) }  dz $.

I am weak in these type of problems ( ie Evaluating of integrals). I shall be really thankful if you can tell how to attempt it. I have read examples from textbook and could understand them.

Kindly tell me 2 curves around which I should calculate the integral and why you are choosing that curve. I will first try to calculate it myself.

Thanks!!

答:
Let $f(z):=\frac{e^{z-1} -1 } {z^2 (z^2 -1) }$

Here are 3 possible circular closed loops.

let $R_k$ be the residue at pole $P_k$.

Do you agree that for example, if $C$ is the red circle:

$$\int_C f(z)dz=2i \pi(R_2+R_3)$$

(by residue theorem) ?

Do you know how to compute residues ? (beware, $0$ is a double pole).

In order you are able to check them, here is what I have found:

$$R_1=\tfrac12(e^{-2}-1), \ \ R_2=0, \ \ R_3=-e^{-1}.$$

A full answer would consider each of the $2^3$ subsets of $\{P_1,P_2,P_3\}$ and conclude that there are eventually 4 possible different results:

$$\int_C f(z)dz \ \ = \ \ 0, \ \ 2i \pi R_1, \ \  2i \pi R_3, \ \ 2i \pi(R_1+R_3)$$


问:
Problem ordinary differential equation
Consider the ordinary differential cquation
$y^{\prime}=y(y-1)(y-2)$

Which of the following statements is true ?

(a) If $y(0)=0.5$ then $y$ is decreasing

(b) If $y(0)=1.2$ then $y$ is increasing

(c) If $y(0)=2.5$ then $y$ is unbounded

(d) If $y(0)<0$ then $y$ is bounded below

I solve using separation  of variable and get implicit relation as

$\frac{ y(y-1)}{(y-2)^2}= A e^ {2x} $

Now this becomes very difficult to apply condition and find the value of $A$  and check option

is there easy method  to solve such questions ?

答:
There is no need to actually solve the differential equation—you can extrapolate the behavior of the solution by looking at the behavior of $y’$ for the given initial conditions.

For example, consider the first two statements regarding whether $y’$ is increasing or decreasing for given values of $y$. Simply insert the stated $y$ value into the expression for $y’$ and determine whether it’s positive or negative, and that tells you whether or not $y$ is increasing or decreasing.

For the latter two cases, the process is a little more subtle—you need to be able to note the regions of initial $y$’s where $y$ will continue to monotonically increase or decrease, and use that information to determine what behavior is true or false.

Title: Problem ordinary differential equation


问:
Set theory: Terminology question
I have some numbers $(a, b)$ which can take any values subject to certain inequalities

$$a + b \geq s_1$$
$$a \geq s_2$$
$$b \geq s_3$$

What is the correct way to describe the set of all valid $(a, b)$? Is it the convex hull of $(a, b)$ satisfying the inequalities or is there some other way to describe this set?

答:
It may differ from style to style. I would write something like this:
$$D_f = \{(a, b) \in \mathbb{R} : a+b \geq s_1 \wedge a \geq s_2 \wedge b \geq s_3\}$$

where $s_1, s_2, s_3 \in \mathbb{R}.$

Title: Set theory: Terminology question


问:
A question about normal space and uniform space
Let $X$ be a compact and Hausdorff space. Hence there is a uniformity $\mathcal{U}$ on $X$, also  $X$ is normal space i.e. for every point $x\in X$ and every closed set $A\subseteq X$ there exist open sets $U, V$ such that $x\in U$, $A\subseteq V$ and $U\cap V=\emptyset$.

Can I say that there is an entourage $U\in \mathcal{U}$ such that $U[x]\cap U[A]=\emptyset$, where $U[x]=\{y: (x, y)\in U\}$?

Please help me to know it.

答: Start with an entourage $W$ such that $W[x]$ is disjoint from $A$. Since $\mathcal U$ is a uniformity, there is an entourage $U$ such that $U=U^{-1}$ and $U\circ U\subseteq W$. Then $U[x]\cap U[A]\varnothing$.


问:
exact homology sequence associated to a fibration
let $p:E\rightarrow B$ be a fibration with fiber $F$, ($E$ and $B$ are cw complexes). $B^k$ denotes the $k$-skeleton of $B$.

1) what does this sentence mean :"we denote the restriction of $E$ to $B^k$ by $E^k$".

2) why there is an exact sequence
$$\cdots \rightarrow H_{i+1}(E^{k+1},E^k)\rightarrow H_i(E^k,F)\rightarrow H_i(E^{k+1},F)\rightarrow \cdots$$

3)What is a trivialization of $E$ over the $(k+1)$-cells $D_\alpha^{k+1}$ of $B^{k+1}$

from which we conclude that
$$H_{i+1}(E^{k+1},E^k)=H_i(\sqcup_\alpha D_\alpha^{k+1}\times F \, , \, \sqcup_\alpha S_\alpha^{k}\times F)$$

答:
To answer (1), the restriction is defined by $E^k=p^{-1}(B^k)$, which forms a fibration over $B^k$.

To answer (2), this is the long exact sequence for the triple  $F\subset E^k\subset E^{k+1}$.

Finally, to answer (3), a trivialization of a fibre bundle means that it is isomorphic to a "trivial" fibre bundle, which is just a product of the base with the fibre. (I'm not sure why we switched to a bundle rather than a fibration.) In any event, the fibre bundle is a product when restricted to the interior of each $k+1$-cell, so that it looks like $D^{k+1}\times F$ on that cell. When we take the relative homology $H_{i+1}(E^{k+1},E^k)$, we are modding out by the boundary of each $k+1$-cell so that the result is the same as looking at each cell separately, noticing that $E^{k+1}$ is a product there, and then modding out by the boundary, which gives the right-hand side of the formula you stated.


答:
It doesn't. I think something has been misread.

Let $a=7$ and $b=14$, then $\gcd(a,b) = 7$ which is prime. Obviously $7|7m$ and $7|7n$ for any positive integers $m$ and $n$. Let $m = 2$ and $n = 4$ then $7|7\times 2$ and $7|7\times 4$, yet $\gcd(2,4) = 2 \neq 1$.

Title: GCD and divisibility

Body:
Update: I think there was a typo in the text. Please don't waste your time with this problem. :)

If gcd$(a,b) = p$, a prime, then $p|am$ and $p|an$ such that  gcd$(m,n) = 1$

Why does gcd$(m,n)$ have to be equal to 1? And does this statement hold true if gcd$(a,b)$ is not equal to a prime?


问:
Relationship between the length of the tangent line through a point on sphere and great-circle distance
As an aviator I'm familiar with the concept of great-circle navigation because when we fly a route between 2 points on the globe we know the shortest distance between these two points is the great circle distance.

I'm developing a navigation app in Google Earth and I need to calculate the shortest distance from the surface of the "spherical" Earth to any point on the tangent line through A (origin) when flying the great circle path.

Also, I'm using a mean earth radius of 6,371.009 km for WGS84 ellipsoid.

Just to be clear, I'd like to refer to the diagram in the following link:

http://www.alaricstephen.com/main-featured/2017/5/22/the-haversine-formula

I use the Haversine formula to calculate the distance, d, between the points A and D (see diagram). What I'd like to calculate is the distance D to E as a function of d.

In the diagram this is referred to as the external secant (exsec) which is the portion DE of the secant exterior to the circle.

答:
$$r=OE\cos\theta$$

and

$$DE=r\sec\frac{\stackrel\frown{AD}}r-r.$$


问:
How to solve this optimisation problem for $x_1^*$ and $x_2^*$ using Lagrangian
I am attempting the following optimization problem, to yield $x_1^*$ and $x_2^*$ in terms of $p$ and $u$. Although I cannot seem to get to the solution outlined below, what I believe should be a simple algebraic exercise has left me with pages of workings and no result. I would appreciate some guidance on the logic and order of operations here.

My approach so far has been to find $x_1$ and $x_2$ from the first two equations and then substitute into the final equation, although I am constantly left with an expression of $x_1$ in terms of $x_2$.

答:
From the first tw0 partial derivatives we have the following two equations:

$$p_1=\lambda\cdot \alpha k\cdot \left(\frac{x_2}{x_1}\right)^{1-\alpha}\Rightarrow p_1=\lambda\cdot \alpha\cdot  k\cdot x_2^{1-\alpha}\cdot x_1^{\alpha-1}$$

$$p_2=\lambda\cdot (1-\alpha) k\cdot \left(\frac{x_1}{x_2}\right)^{\alpha}\Rightarrow p_2=\lambda\cdot (1-\alpha)\cdot  k\cdot x_1^{\alpha}\cdot x_2^{-\alpha}$$

Dividing the first equation by the second equation. $\lambda$ and $k$ are cancelled out directly.

$$\frac{p_1}{p_2}=\frac{\alpha\cdot  x_2^{1-\alpha}\cdot x_1^{\alpha-1}}{(1-\alpha)\cdot   x_1^{\alpha}\cdot x_2^{-\alpha}}$$

$$\frac{p_1}{p_2}=\frac{\alpha}{1-\alpha}\cdot \frac{  x_1^{\alpha-1}}{  x_1^{\alpha}}\cdot \frac{ x_2^{1-\alpha}}{  x_2^{-\alpha}}$$

Using exponent-rules

$$\frac{p_1}{p_2}=\frac{\alpha}{1-\alpha}\cdot \frac{  x_2}{  x_1}\Rightarrow x_2=\frac{p_1}{p_2}\cdot \frac{1-\alpha}{\alpha}\cdot x_1$$

Input the expression for $x_2$ into the third FOC (first order condition).

$$\overline u=k\cdot x_1^{\alpha}\cdot \left(\frac{p_1}{p_2}\cdot \frac{1-\alpha}{\alpha}\cdot x_1\right)^{1-\alpha}$$

$$\overline u=k\cdot x_1^{\alpha}\cdot \left(\frac{p_1}{p_2}\cdot \frac{1-\alpha}{\alpha}\right)^{1-\alpha}\cdot x_1^{1-\alpha}$$

$x_1^{\alpha}\cdot x_1^{1-\alpha}=x_1$

$$\overline u=k\cdot \left(\frac{p_1}{p_2}\cdot \frac{1-\alpha}{\alpha}\right)^{1-\alpha}\cdot x_1$$

We can write the term in the brackets at the LHS by swapping the numerators and the denominators.

$$\overline u\cdot \left(\frac{p_2}{p_1}\cdot \frac{\alpha}{1-\alpha}\right)^{1-\alpha} =k\cdot x_1$$

$$x_1^*=\frac{\overline u}k\cdot \left(\frac{p_2}{p_1}\cdot \frac{\alpha}{1-\alpha}\right)^{1-\alpha} $$


问:
Smallest graph with non-disjointly generated automorphism group
Consider the automorphism group $\text{Aut}(H)$ of a graph $H$. I say that $\text{Aut}(H)$ is disjointly generated if there exists a set of generators $A$ of $\text{Aut}(H)$ such that for each vertex $v$ of $H$, there exists at most one generator $g \in A$ with $g(v) \neq v$.

I'm interested in the size $n_{\text{min}}$ of the smallest graph $H$ such that $\text{Aut}(H)$ is not disjointly generated. Are there any lower bounds (either by mathematical proof or reference to literature) on $n_{\text{min}}$? I'm particularly interested in whether $n_{\text{min}} > 8$.

It is easy to see that $n_{\text{min}} > 4$ by brute-forcing all graphs with $\leq 4$ vertices. However, this approach quickly becomes infeasible to do by hand. One can of course design an algorithm to compute this automatically, but any approach I've come up with is both hard to implement and becomes infeasible even to a very fast computer before $n=8$. I would also accept an answer that discusses a reasonably quick algorithm to verify that $n_{\text{min}} > 8$, but prefer mathematical proofs or references to literature.

答: If $\operatorname{Aut}(H)$ is disjointly generated, it should be abelian. (Any two generators commute, because any two generators permute disjoint sets of vertices.) So the smallest graph with a non-disjointly-generated automorphism group is $K_3$, because $\operatorname{Aut}(K_3) = S_3$, and $n_{\min} =3$.


问:
Does Tarjan's algorithm actually find all SCCs?
So I learned about Tarjan's algorithm the other day and I'm not quite sure if I understand it 100% correctly. I read a lot about the statement that Tarjan's algorithm is able to decompose a directed graph (digraph) into its strongly connected components (SCCs). An SCC being defined as a partition of the digraph with the condition that all of the component's nodes can be reached from every other node in that component.

Suppose I have the following digraph:

Example Digraph

To my understanding, the graph has two SCCs: one formed by the three nodes to the left, the second by the three nodes to the right. If I understand Tarjan's algorithm correctly however, it will only detect the "left" SCC, because the "right" one does not contain a cycle, is that correct? In this case, the statement that Tarjan's algorithm finds SCCs is not entirely correct?

答:
I did a quick and dirty python implementation of the algorithm as described in the Wikipedia article, and tested it on your digraph.

```<code># Implement Trajan's algorithm for strongly connected components

SCC = []
indices = { }
lowlinks = { }
stack = []
index = 0

G = {1:[3], 2:[1,4],3:[2],4:[5],5:[4,6],6:[5]}

def strongConnect(v):
    global index

    indices[v] = index
    lowlinks[v] = index
    index += 1
    stack.append(v)
    for w in G[v]:
        if w not in indices:
            strongConnect(w)
            lowlinks[v] = min(lowlinks[v], lowlinks[w])
        elif w in stack:
            lowlinks[v] = min(lowlinks[v], indices[w])
    if lowlinks[v] == indices[v]:
        scc = []
        while True:
            w = stack.pop(-1)
            scc.append(w)
            if w == v:
                break
        SCC.append(scc)

for v in G:
    if v not in indices:
        strongConnect(v)

for scc in SCC:
    print(scc)
</code>```

This outputs

```<code>[6, 5, 4]
[2, 3, 1]
</code>```

so the algorithm does indeed work in this case.  The algorithm is very well known, and the chance that it is incorrect is negligible.

Body:
I did a quick and dirty python implementation of the algorithm as described in the Wikipedia article, and tested it on your digraph.

# Implement Trajan's algorithm for strongly connected components

SCC = []
indices = { }
lowlinks = { }
stack = []
index = 0

G = {1:[3], 2:[1,4],3:[2],4:[5],5:[4,6],6:[5]}

def strongConnect(v):
    global index

    indices[v] = index
    lowlinks[v] = index
    index += 1
    stack.append(v)
    for w in G[v]:
        if w not in indices:
            strongConnect(w)
            lowlinks[v] = min(lowlinks[v], lowlinks[w])
        elif w in stack:
            lowlinks[v] = min(lowlinks[v], indices[w])
    if lowlinks[v] == indices[v]:
        scc = []
        while True:
            w = stack.pop(-1)
            scc.append(w)
            if w == v:
                break
        SCC.append(scc)

for v in G:
    if v not in indices:
        strongConnect(v)

for scc in SCC:
    print(scc)

This outputs

[6, 5, 4]
[2, 3, 1]

so the algorithm does indeed work in this case.  The algorithm is very well known, and the chance that it is incorrect is negligible.


问:
Clarify about primary ideals in a commutative ring
In a commutative ring, if one takes a primary ideal $I$, then $\sqrt I$ is prime. It is not true in general that an ideal with such property is primary. For example, given a prime ideal $\mathfrak p$, one has that the radical of $\mathfrak p^n $ is $\mathfrak p$, clearly, but a $\mathfrak p^n $ is not always primary. The notes from which I'm studying prove that, in $\mathbb Z$, every ideal $I$ such that $\mathfrak q:=\sqrt I$ is prime is a power of $\mathfrak q$. Then they conclude noticing that every power of a prime ideal is primary; however, it seems to me that they only proved that in $\mathbb Z $, for an ideal, being a power of a prime is equivalent to have the property that its radical is prime. This proves that every primary is a power of a prime (in $\mathbb Z$) but not the converse, which is what we need. What am I missing? Thanks

答:
Take $n=p_1^{a_1}p_2^{e_2}\dotsm p_k^{e_k}$. Then it's not difficult to show that $\sqrt{\smash[b]{(n)}}=(p_1p_2\dots p_n)$, so the radical is prime if and only if $n$ is a prime power.

Conversely, $(p)^e=(p^e)$ being a $(p)$-primary ideal is easy to show.

It is generally false that the power of a prime ideal is primary, but it can hold for particular prime ideals. Indeed, if $p$ is a prime element of a domain, then $(p^e)=(p)^e$ is $(p)$-primary.


问:
Matrix Concentration for Products- Orthogonality Property
The following comes from page 3 of the paper "Matrix Concentration for Products" found here https://arxiv.org/pdf/2003.05437.pdf.

Given two independent, complex, square matrices Y and Z:
$$ YZ=\mathbb{E}(Y)Z+(Y-\mathbb{E}(Y))Z.$$
I understand that $$ \mathbb{E}((Y-\mathbb{E}(Y))Z|Z)=0. $$
Now, this is called an orthogonality property (Why?) and apparently used to show:
$$ \mathbb{E}(\left\| YZ\right\|_2^2) =\mathbb{E}(\left\|(\mathbb{E}(Y))Z\right\|_2^2) + \mathbb{E}(\left\| (Y-\mathbb{E}(Y))Z\right\|_2^2) $$

where $\left\| .\right\|_2$ refers to the Schatten 2-norm or Frobenius-norm.

Could someone clarify how exactly this property is used here? I assume the tower property comes into play here but I'm not sure how exactly.

Thank you.

答:
this problem suffers from a proliferation of symbols and parenthesis

$A:= \mathbb{E}(Y)$

$\mathbb{E}\Big(\left\| YZ\right\|_F^2\Big) $

$=\mathbb{E}\Big(\text{trace}\big((YZ)^* YZ\big)\Big) $

$=\text{trace}\Big(\mathbb{E}\big((YZ)^* YZ\big)\Big) $

$=\text{trace}\Big(\mathbb{E}\big((AZ+(Y-A)Z)^* (AZ+(Y-A)Z)\big)\Big) $

$=\text{trace}\Big(\mathbb{E}\big(((AZ)^*AZ+ (Y-A)Z)^*(Y-A)Z)\big)\Big) + \text{trace}\Big(\mathbb{E}\big(C\big)\Big)$

$=\mathbb{E}(\left\|(A)Z\right\|_F^2) + \mathbb{E}(\left\| (Y-A)Z\right\|_F^2)+ \text{trace}\Big(\mathbb{E}\big(C\big)\Big)$

$=\mathbb{E}(\left\|(A)Z\right\|_F^2) + \mathbb{E}(\left\| (Y-A)Z\right\|_F^2)+0$

where $C$ is a matrix containing the cross terms.  In particular $C$ is given by

$C= \big(AZ\big)^*\big(Y-A\big)Z + \big((Y-A)Z\big)^*\big(AZ\big)$

using linearity of both expectations and trace, and that trace and expectations commute we can focus on the first term

$\text{trace}\Big(\mathbb{E}\Big(\big(AZ\big)^*\big(Y-A\big)Z\Big)\Big) $

$=\text{trace}\Big(\mathbb{E}\Big(Z^*A^*\big(Y-A\big)Z\Big)\Big)$

$=\mathbb{E}\Big(\text{trace}\Big(Z^*A^*\big(Y-A\big)Z\Big)\Big)$

$=\mathbb{E}\Big(\text{trace}\Big(\big(Y-A\big)ZZ^*A^*\Big)\Big)$

$=\text{trace}\Big(\mathbb{E}\Big(\big(Y-A\big)ZZ^*A^*\Big)\Big)=0$

because applying the Tower Property and your orthogonality relation

$\mathbb{E}\Big(\big(Y-A\big)ZZ^*A^*\Big)$

$\mathbb{E}\Big(\big(Y-A\big)ZZ^*\Big)A^*$

$=\mathbb{E}\Big[\mathbb{E}\Big(\big(Y-A\big)ZZ^*\Big\vert Z\Big)\Big]A^*$

$=\mathbb{E}\Big[\mathbb{E}\Big(\big(Y-A\big)\Big\vert Z\Big)ZZ^*\Big]A^*$

$=\mathbb{E}\Big[\mathbf 0ZZ^*\Big]A^*$

$=\mathbf 0$

so the trace is zero.  The second term comprising $C$ is the conjugate transpose of the first one, so it has trace zero as well.


答: We can prove that any power of a prime in $\mathbb{Z}$ is primary by hand. Let $I=(p^n)$ for a prime $p\in\mathbb{Z}$. Suppose $xy\in I$, but $x\not\in I$. Since $xy\in I$, $p^n m = xy$ for some $m$. $p^n$ divides the LHS, so $p^n$ must divide the RHS. Since $x\not\in I$, we have $p^j\mid x$ and $p^k\mid y$ for some $j$ strictly less than $n$ and some $k$ strictly greater than 0. Thus $y^n\in (p^{kn})\subset (p^n)$.

Title: Clarify about primary ideals in a commutative ring

Body: In a commutative ring, if one takes a primary ideal $I$, then $\sqrt I$ is prime. It is not true in general that an ideal with such property is primary. For example, given a prime ideal $\mathfrak p$, one has that the radical of $\mathfrak p^n $ is $\mathfrak p$, clearly, but a $\mathfrak p^n $ is not always primary. The notes from which I'm studying prove that, in $\mathbb Z$, every ideal $I$ such that $\mathfrak q:=\sqrt I$ is prime is a power of $\mathfrak q$. Then they conclude noticing that every power of a prime ideal is primary; however, it seems to me that they only proved that in $\mathbb Z $, for an ideal, being a power of a prime is equivalent to have the property that its radical is prime. This proves that every primary is a power of a prime (in $\mathbb Z$) but not the converse, which is what we need. What am I missing? Thanks


问:
Find the least integral value of $n$ for $f(n)={\left(\frac{9}{10}\right)}^n<\frac{1}{2}$
I was doing a probability question where it was required to find out the least integral value of n with $$f(n)=\left(\frac 9 {10}\right)^n<\frac{1}{2}$$

A little calculation(by calculator of course) yields $\min(n)=7$.

as:

$f(5)=0.59$

$f(6)=0.53$

$f(7)=0.43$

My question is there a good way to approximately tell which integer $n$ can be without computational software and log tables (even $\log 2,\log 3$ etc.)vCalculus seems to be an unlikely option

答:
hint

For small $ X $, We can write

$$(1-X)^n\approx 1-nX$$

or, for more accuracy
$$(1-X)^n\approx 1-nX+\frac{n(n-1)}{2}X^2$$
and

$$\frac{9}{10}=1-\frac{1}{10}$$


问:
studying the topology of a real algebraic set
Let $f_1,\ldots,f_n \in \mathbb{R}[x_1,\ldots,x_m]$ be polynomials with real coefficients and let $I$ be the ideal that they generate. Denote by $V_{\mathbb{R}}(I)$ the corresponding real variety, i.e. $V_{\mathbb{R}}(I)=V_{\mathbb{C}}(I) \cap \mathbb{R}^m$.

Main question: What are the tools that i can use in order to study the topology of $V_{\mathbb{R}}(I)$ induced by e.g. the Euclidean topology of $\mathbb{R}^m$? More specifically, i need to perform computations in order to determine the number of connected components and possibly find equations that describe them. Note that i already have my hands on the book "topology of real algebraic sets" by Akbulut and King, however, after skimming through it, i still have no ideal how to proceed.

Side question: is it true that an algebraic set is the union of connected components in the metric topology?

答:
It turns out that instead of studying real algebraic sets, it is more natural to study real semialgebraic sets, i.e. those defined by systems of polynomial equalities and inequalities. You can find some tools for studying the topology of real semialgebraic sets described here. The most useful, as far as I recall, are CAD (Cylindrical Algebraic Decomposition) and a theorem that any real semialgebraic set $X$ is homotopy-equivalent to $X\cap B(0,r)$, where $B(0,r)$ is the open ball around $0$ of radius $r$, for sufficiently large $r$ (IIRC there is an effective upper bound on $r$).

To answer your side question, all subsets of a metric space are unions of their connected components. For real semialgebraic sets we can do better: they have only finitely many connected components.


问:
Proving a limit property
I want to prove the following property:

Let $(a_n),(b_n)$ be Sequences in $\mathbb{R}$, $a_n\to a $ and $b_n\to b$

$(\exists\ n_0 \in\mathbb{N}\ \forall n \geq n_0: a_n \leq b_n)\Rightarrow a\leq b $,
suppose $a>b\ , \epsilon:=(a-b)/2 $

At some point my script mentions that $-(a_n-a)<\epsilon $ and $b_n - b<\epsilon$

Could someone explain to me the intution behind the step above.

Thanks in advance.

答:
You want to prove that, provided the first sequence converges to $a$ and the second one to $b$, it holds $a\le b$.

Suppose not, so $a>b$. Therefore $\varepsilon=(a-b)/2>0$ and we can find $n_1\ge n_0$ such that, for $n\ge n_1$,
$$
|a_n-a|<\varepsilon,\qquad |b_n-b|<\varepsilon \tag{*}
$$
This translates to
$$
-\varepsilon<a_n-a<\varepsilon,\qquad -\varepsilon<b_n-b<\varepsilon
$$
The first pair of inequalities implies $-(a_n-a)<\varepsilon$. You also can isolate $b_n-b<\varepsilon$ and derive a contradiction.

On the other hand, I'd more simply rewrite (*) as
$$
a-\varepsilon<a_n<a+\varepsilon,\qquad b-\varepsilon<b_n<b+\varepsilon
$$
and observe that $b+\varepsilon=(a+b)/2=a-\varepsilon$, so this becomes
$$
\frac{3b-a}{2}<b_n<\frac{a+b}{2}<a_n<\frac{3a-b}{2}
$$
and the central inequalities provide the contradiction.

Title: Proving a limit property


答:
Convex hull language could be used, but not the way you are using it.

First, the set of all valid $(a,b)$ is simply the SET of all ordered pairs $(a,b)$ satisfying the inequalities. The set builder notation for this set is described in the other answer.

If you want to formulate this set in geometric language, then you want to think of $(a,b)$ as an ordered pair representing a point in the Cartesian coordinate plane. You can then graph the line $a+b=s_1$ and then use one color to shade in the portion of the plane to the upper-right of that line to represent the inequality $a+b \ge s_1$. Then graph the line $a=s_2$ and use a second color to shade in the portion of the plane to the right of that line to represent the inequality $a \ge s_2$. Finally graph the line $b=s_3$ and use a third color to shade in the portion of the plane above that line to represent the inequality $b \ge s_3$. The portion of the plane with all three colors shaded is then a picture of the SET of all points $(a,b)$ satisfying those inequalities.

You could then describe this set as the convex hull of the union of the two rays and the line segment that form the boundary of the triply shaded region.

Title: Set theory: Terminology question

Body:
I have some numbers $(a, b)$ which can take any values subject to certain inequalities

$$a + b \geq s_1$$
$$a \geq s_2$$
$$b \geq s_3$$

What is the correct way to describe the set of all valid $(a, b)$? Is it the convex hull of $(a, b)$ satisfying the inequalities or is there some other way to describe this set?


问:
using Lagrange multipliers to determine shortest distance between a point and straight line

Find the shortest distance between $(0,2,0)$ and the straight line
$x=y=z$ .

Let the required point on the straight line be $(x,y,z)$ . Now we set up the Lagrangian as $$L=x^2+(y-2)^2+z^2+\lambda(x-y)+\mu(y-z)+\nu(z-x)$$ where $\lambda,\mu,\nu$ are Lagrange multipliers. The extremizing equations are $$\frac{\partial L}{\partial x}=\frac{\partial L}{\partial y}=\frac{\partial L}{\partial z}=0$$ which gives $$2x+\lambda-\nu=2(y-2)+\mu-\lambda=2z-\mu+\nu=0$$ Solving for $x,y,z$ and substituting in $x=y=z$ we get $$\begin{pmatrix}2&-1&-1 \\ 1&-2&1 \\ 1&1&-2\end{pmatrix}\begin{pmatrix}\lambda\\ \mu \\ \nu\end{pmatrix}=\begin{pmatrix}-4\\ -4 \\ 0\end{pmatrix}$$ But I keep getting the coefficient matrix as singular matrix, so it doesn't give us the unique value of Lagrange multipliers which help to minimize the distance required.

What is possibly going wrong there? Any help is appreciated.

答:
The line is a $1$-dimensional manifold in ${\mathbb R}^3$. In order to define it you need $3-1=2$ constraints; say $x-y=0$ and $x-z=0$. Try the problem again with just these equations, and $2$ Lagrange multipliers.

As noted in a comment there are simpler ways to solve this problem, e.g., by parametrizing the line as $t\mapsto(t,t,t)$.


问:
Given some arbitrary function $y = f(x)$, if you only know $y$ when given the associated $x$, what is the fastest way of finding $x$ s.t. $f(x) = 0$?
For full disclosure, this was a problem I saw on glassdoor that was supposedly asked in an interview. I hope it's OK to ask here; I am not being asked to solve this, I just thought it was a very interesting problem.

I am not sure if the problem was exactly posed as stated in the title, so I make some clarifications: (i) Assume the function definitely takes $0$ for some value in the domain. (ii) When we say "we only know $y$ when given $x$", I think we can treat the function as if it is some sort of oracle, telling you its value at that point in the domain. But we don't have an explicit function.

I can't even see how this is necessarily solvable as stated. I just don't see how finite (or even countable) sampling of a continuous function will guarantee our ability to find the zero. Especially since the function is arbitrary, we could end up with pathological functions that mean some clever approach would fail. The usual computational methods I am aware of such as Newton's method, bijection, Secant etc. all don't work.

I am quite baffled, can anyone please give me some hint or mathematical way of approaching this?

答:
You can't find a unique function under such conditions without a second point or without the graph. There could be hundreds of functions that might yield the same $y$ at the same $x$ as in your question.

However, if some conditions are added as hints as in here, you can do at least something algebraic to get hold of a possible function.


问:
What does it mean to find a linear combination?
I am going through Gilbert Strang's Intro To Linear Algebra class. I believe I am misunderstanding how the questions in the textbook are written, however.

The first question of section 1.3 is:

Find the linear combination $3s_1 + 4s_2 +5s_3=b$

My understanding of a linear combination was that it was the multiplication of matrices itself that creates a linear combination of the columns.

So, is the answer here:

$\begin{bmatrix}s_1 & s_2 & s_3\end{bmatrix}\begin{bmatrix}3\\4\\5\end{bmatrix}$ = b?

I am confused on the rest of the question as well. It states to:

Write b as a matrix-vector multiplication multiplication $Sx$ with 3,4,5 in $x$. Compute the three dot products (row of S)$\cdot x$  such that the following go into columns of S:
$\\ s_1 = \begin{bmatrix}1\\1\\1\end{bmatrix}, s_2 = \begin{bmatrix}0\\1\\1\end{bmatrix}, s_3 = \begin{bmatrix}0\\0\\1\end{bmatrix} $

Then, is this asking me to set the question up in the following form:

$\begin{bmatrix}1 & 0 & 0 \\
                1 & 1 & 0 \\
                1 & 1 & 1 \end{bmatrix}\begin{bmatrix}3\\4\\5\end{bmatrix}$

And then solve to get

$b = \begin{bmatrix}3 & 0 & 0 \\
                3 & 4 & 0 \\
                3 & 4 & 5 \end{bmatrix}$?

I know this question is long winded and basic, but understanding what's being asked would be really helpful for me

答:
We have
$$3 \underbrace{\begin{bmatrix}1 \\ 1\\ 1\end{bmatrix}}_{s_1} + 4\underbrace{\begin{bmatrix}0 \\ 1\\ 1\end{bmatrix}}_{s_2} + 5\underbrace{\begin{bmatrix}0 \\ 0\\ 1\end{bmatrix}}_{s_3}=\begin{bmatrix}1 & 0 & 0 \\
                1 & 1 & 0 \\
                1 & 1 & 1 \end{bmatrix}\begin{bmatrix}3\\4\\5\end{bmatrix} = \underbrace{\begin{bmatrix} 3 \\ 7 \\ 12\end{bmatrix}}_{b}$$


问:
How to accurately calculate the error function $\operatorname{erf}(x)$ with a computer?
I am looking for an accurate algorithm to calculate the error function

$$\operatorname{erf}(x)=\frac{2}{\sqrt{\pi}}\int_0^x e^{-t^2}\ dt$$

I have tried using the following formula,

(Handbook of Mathematical Functions, formula 7.1.26), but the results are not accurate enough for the application.

Body:
You could get good theoretical approximations of the error function using
$$\text{erf}\left(x\right)\sim \text{sgn}(x)\sqrt{1-\exp\Big(-\frac 4 {\pi}\,\frac{1+P_n(x^2)}{1+Q_n(x^2)}\,x^2 \Big)}$$
For example
$$P_1(x^2)=\frac{10-\pi ^2}{5 (\pi -3) \pi }x^2\qquad Q_1(x^2)=\frac{120-60 \pi +7 \pi ^2}{15 (\pi -3) \pi }$$ while
$$P_2(x^2)=\frac{\left(105840-110880 \pi +37800 \pi ^2-4260 \pi ^3+69 \pi ^4-17 \pi ^5\right)
   }{3 \pi  \left(-12600+12600 \pi -3360 \pi ^2-30 \pi ^3+73 \pi ^4\right)}x^2+$$ $$\frac{-2116800+1270080 \pi ^2-504000 \pi ^3+48510 \pi ^4+503 \pi ^6}{315 \pi ^2
   \left(-12600+12600 \pi -3360 \pi ^2-30 \pi ^3+73 \pi ^4\right)}x^4$$

$$Q_2(x^2)=\frac{60480-70560 \pi +27720 \pi ^2-3600 \pi ^3-143 \pi ^4+43 \pi ^5}{\pi
   \left(-12600+12600 \pi -3360 \pi ^2-30 \pi ^3+73 \pi ^4\right)}x^2+$$
$$\frac{-6350400+8467200 \pi -4515840 \pi ^2+1192800 \pi ^3-145320 \pi ^4+2380 \pi
   ^5+793 \pi ^6}{105 \pi ^2 \left(-12600+12600 \pi -3360 \pi ^2-30 \pi ^3+73 \pi
   ^4\right)}x^4$$ and so on.

The first one gives a maximum absolute error of $0.000150$ while the second  gives a maximum absolute error of $0.000012$.

With regard to the infinite norms, they are respectively $3.04\times 10^{-8}$ and  $1.20\times 10^{-10}$.


问:
Which of the following are subspaces?
Determine whether the given set $S$ is a subspace of the vector space $V$.

A. $V=\mathbb{R}^{nxn}$, and $S$ is the subset of all $n×n$ matrices with $\det(A)=0$.

B. $V$ is the space of three-times differentiable functions $\mathbb{R}→\mathbb{R}$, and $S$ is the subset of $V$ consisting of those functions satisfying the differential equation $y'''+2y=x^2$.

C. $V=P^3$, and $S$ is the subset of $P^3$ consisting of all polynomials of the form $p(x)=ax^3+bx$.

D. $V=\mathbb{R}^n$, and $S$ is the set of solutions to the homogeneous linear system $Ax=0$ where $A$ is a fixed $m×n$ matrix.

E. $V$ is the vector space of all real-valued functions defined on the interval $[a,b]$, and $S$ is the subset of $V$ consisting of those functions satisfying $f(a)=3$.

F. $V=\mathbb{R}^{nxn}$, and $S$ is the subset of all upper triangular matrices.

G. $V$ is the vector space of all real-valued functions defined on the interval $(−∞,∞)$, and $S$ is the subset of $V$ consisting of those functions satisfying $f(0)=0$.

I have attempted to solve this by using the fact that the zero vector/polynomial has to be in the set so the set is non-empty, check that it is closed under addition and under scalar multiplication. I have found that A,C,D,F and G  satisfy these properties, however I am not sure if I am correct and I do not understand how to work with B. I know that E is not a subspace because the zero element is not in the set. Can someone help me? Thank you.

答:
I'm afraid you got one wrong.

A. False (not closed for sums)

B. False (zero not in the set) Here zero is the zero constant function.

C. True

D. True

E. False (zero not in the set)

F. True

G. True

Why is A false? In the case of $n=2$, consider
$$
\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}=
\begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix}+
\begin{bmatrix} 0 & 0 \\ 0 & 1 \end{bmatrix}
$$
and generalize for all $n>1$. On the other hand, A is true for $n=1$, because the only $1\times1$ matrix having zero determinant is the null matrix.

Title: Which of the following are subspaces?


问:
Prove that the image of a set is closed and bounded
Let $f:A \subseteq \mathbb{R}^n \rightarrow \mathbb{R}^m$ where $A$ is closed and $f$ continuous in $A$. Prove that $D \subseteq A$ is bounded then $\overline{f(D)}$ is closed and bounded

We know that $\overline{f(D)}$ always is closed but I think maybe using Heine-Borel Theorem we want to show that $\overline{f(D)}$ is compact I woul appreciate any advice.

答: Note that $\overline D\subset\overline A=A$. Besides, since $D$ is bounded, $\overline D$ is bounded too. So, $\overline D$ is a compact subset of $A$. Therefore, $f\left(\overline D\right)$ is compact too, and so it is bounded. In particular, $f(D)$ is bounded, and so, $\overline{f(D)}$ is bounded too. Sinse it is a closure, it is also a closed set.


答:
As
$$
\left(\frac{9}{10}\right)^n \lt\frac{1}{2} \Rightarrow \left(\frac{10}{9}\right)^n \gt 2
$$
$$
\Rightarrow \left(1 +\frac{1}{9}\right)^n\gt 2
$$
Bernoulli's Inequality shows $n=9$ can be the solution.

Now we start descending from $n=9$ to $n=8$ and $n=7$ and eventually see $n=6$ does not suffice, hence the solution is $n=7$.

Title: Find the least integral value of $n$ for $f(n)={\left(\frac{9}{10}\right)}^n<\frac{1}{2}$

Body:
I was doing a probability question where it was required to find out the least integral value of n with $$f(n)=\left(\frac 9 {10}\right)^n<\frac{1}{2}$$

A little calculation(by calculator of course) yields $\min(n)=7$.

as:

$f(5)=0.59$

$f(6)=0.53$

$f(7)=0.43$

My question is there a good way to approximately tell which integer $n$ can be without computational software and log tables (even $\log 2,\log 3$ etc.)vCalculus seems to be an unlikely option


问:
prove $\sum_\text{cyc}\frac{a+2}{b+2}\le \sum_\text{cyc}\frac{a}{b}$

Prove if $a,b,c$ are positive $$\sum_\text{cyc}\frac{a+2}{b+2}\le \sum_\text{cyc}\frac{a}{b}$$

My proof:After rearranging we have to prove $$\sum_\text{cyc} \frac{b}{b^2+2b} \le \sum_\text{cyc} \frac{a}{b^2+2b}$$

As inequality is cyclic:

let $a\ge b\ge c$ then $$\frac{1}{a^2+2a}\le \frac{1}{b^2+2b}\le \frac{1}{c^2+2c}$$.The rest follows by rearrangement inequality.

The case $a\ge c\ge b$ is analogous.

Thus Proved!

Is it correct?...And any other alternative ways possible?

答:
I apologize for the first non-obvious proof and I give you a refinement

Hint :$a\geq b \geq c$

First prove that :

$$\frac{2(x-y)}{x+y+y^2}\leq \frac{x}{y}-\frac{x+2}{y+2}\quad (1)$$

For that multiply by $y(y+2)(x+y+y^2)$ , put in factor and it becomes :

$$2(x-y)^2\geq 0$$

Apply $(1)$ for $(a,b)$,$(b,c)$,$(c,a)$

Now we need to show :

$$\frac{2(a-b)}{a+b+b^2}+\frac{2(b-c)}{b+c+c^2}+\frac{2(c-a)}{a+c+a^2}\geq 0$$

Now introducing $f(c)$

$$\frac{2(b-c)}{b+c+c^2}+\frac{2(c-a)}{a+c+a^2}=f(c)$$

Using derivatives prove that $f(c)$ is decreasing when $c$ increases .

Now we put $b=c$ and the inequality becomes :

$$\frac{2(a-b)}{a+b+b^2}+\frac{2(b-a)}{a+b+a^2}\geq 0$$

Wich is obvious with the condition $a\geq b \geq c $


问:
A question with infinity
I am a sophomore in high school and my math teacher did a very short lesson on infinity, here's how it went:  (Try to solve each part yourself the first two are easy)

Part 1

You have an inf. number of boxes each labeled like so

```<code>[1][2][3][4].....
</code>```

and that there is a person in each one of these boxes.  But you have another set of five boxes like so

```<code>[1][2][3][4][5]
</code>```

You want to only have one set of boxes what do you do?  You must be able to tell each person what box they should goto next.

Spoiler

Each person is told to move down five boxes.  Or n+5 is you new box, and the five people from the other boxes are able to move into the first five boxes or just n+0.

Part 2

Well you actually have two of these inf. box sets, but you really only want one.  What now? (same rules apply)

Each person from set one is told to go to box 2n and each person from set two is told to box 2n-1

Part 3

Now you have an inf. number of these inf. box sets arranged like so

```<code>1: [1][2][3][4][5]...
2: [1][2][3][4][5]...
3: [1][2][3][4][5]...
4: [1][2][3][4][5]...
5: [1][2][3][4][5]...
</code>```

and you want to fit them all into one inf. box set, how do you do it.  She refused to tell us the answer and no one in my class figured it out, but I think I just did (correct me if I'm wrong, I'm also looking for alternate solutions).

How I went about solving this problem

Step 1

I assumed that there had to be some sort of pattern so I tried a few

Step 2

I picked the following patter.

1: [1][3][6][10][15]...

2: [2][5][9][14][]...

3: [4][8][13][][]...

4: [7][12][][][]...

5: [11][][][][]...

The numbers in the box represent the new box of each person

Step 3

I started to create a formula for the people to find there new room based of there current room number and box set number.  I found the following sequence in for people in the first row, 1 3 6 10 15.  Or +2 +3 +4 +5.  Which told me that it was quadratic so the equation had to be ax^2 + bx + c = new room number

Step 4

I found three equation by plugging in room numbers and their correct new room numbers to find these three equations:

(1^2)a + 1b + c = 1 or a + b + c = 1

(2^2)a + 2b + c = 3 or 4a + 2b + c = 3

(3^2)a + 3b + c = 6 or 9a + 3b + c = 6

Step 5

After solving this system I got the following equation 1/2x^2 + 1/2x + 0 = new room number where x = current room number.  This of course only tell the people in the boxes on the bottom where to go. Now I needed to find how to place the rest of the people.  The first thing I noticed was that people on the same diagonal as seen bellow were very much related to each other when you look at the rooms they will end up in.

[x-4][][][][]

[][x-3][][][]

[][][x-2][][]

[][][][x-1][]

[][][][][x]

The solution

Working this knowledge into my equation was fairly easy (I was able to do it while going for a run so I won't explain how I got my equation).  The equation is:

 (1/2(r+c-1)^2 + 1/2(r+c-1) + 0)-r+1 = new room number | r = row | c = column

Now my question to all of you is

Is my equation correct?

Are there any alternate solutions to this problem?

How could you fit a 3D array of boxes (inf boxes of people in all direction) into a 1D line of boxes?

Can you find a way to fit a n dimensional array of boxes into a 1D line of boxes?

For any of you into JAVA here's a a short snippet of code I made to prove my answer

```<code>    package main;

public class Main {

    /**
     * @param args
     */
    public static void main(String[] args) {
        // TODO Auto-generated method stub
        int f = 2;
        int r = 1;
        int c = 1;
        while(true){
            double x = (.5*(r+c-1)*(r+c-1) + .5*(r+c-1) + 0)-r+1;
            System.out.print(x +", ");
            c++;
            r--;
            if(r < 1){
                r = f;
                f++;
                c = 1;
            }
        }
    }
}
</code>```

答:
There are various ways to do this.  Here's one:
$$
\begin{array}{rrrrrrrrrrrrrrrr}
1 & \rightarrow & 2 & & 6 & \rightarrow & 7 & & 15 & \rightarrow & 16 \\
  & \swarrow & & \nearrow & & \swarrow & & \nearrow & & \swarrow \\
3 & & 5 & & 8 & & 14 \\
\downarrow & \nearrow & & \swarrow & & \nearrow \\
4 & & 9 & & 13 \\
  & \swarrow & & \nearrow \\
10 & & 12 \\
\downarrow & \nearrow \\
11
\end{array}
$$

Here's another:
$$
\begin{array}{rrrrrrrrrrrrrrr}
1 & \rightarrow & 2 & & 9 & \rightarrow & 10 & & 25 & \rightarrow \\
  & & \downarrow & & \uparrow & & \downarrow & & \uparrow \\
4  & \leftarrow & 3 & & 8 & & 11 & & 24 \\
\downarrow & & & & \uparrow & & \downarrow & & \uparrow \\
5 & \rightarrow & 6 & \rightarrow & 7 & & 12 & & 23 \\
 & & & & & & \downarrow & & \uparrow \\
16 & \leftarrow & 15 & \leftarrow & 14 & \leftarrow & 13 & & 22 \\
\downarrow &  & & & & & & & \uparrow \\
17 & \rightarrow & 18 & \rightarrow & 19 & \rightarrow & 20 & \rightarrow & 21
\end{array}
$$

Title: A question with infinity


问:
proof of misplacement in array
suppose we have an array $A$ and we have a misplacement such if $i < j$ then $A[i] > A[j] $

prove that if we have a misplacement in our array then we have at least $ j - i$ misplacement in our array

答:
We have a misplacement with: If $i < j$ then $A[i] > A[j] $

$i < j$ means that there is a $k$ such as $i<k<j$ and $j-i=k$

Let's prove that we have at least $k$ misplacements in our array $A$.

For all $k$ a defined above $A[k] = A[j-i]$

Let's now use a reductio ad absurdum argument. Suppose that there is only one misplacement and for all $k$ we have $A[i] < A[k] < A[j]$

This contradictory to $A[i] > A[j]$

This means that $A[j] < A[k] < A[i]$ for all $k = j-i$

Thus we have at least k misplacements such as  $A[j] < A[k]$

Title: proof of misplacement in array


问:
Cross Product Question (Simple Physics)
A positively charged particle with velocity $v\hat k$ moves
through an unknown (but uniform) magnetic field. The force on the particle is observed to be $\vec F = F_0 (3 \hat{i} + 4\hat j)$. What is the $x$-component of the magnetic field?

The equation we use to solve these problems: $\vec F = q\vec v \times \vec B$

So I know $\vec F$, so I plug that in: $F_0 (3 \hat{i} + 4\hat j) = v\hat k \times \vec B$.

How do I proceed to get the $x$-component?

Note: The answer to this problem is $\frac{4F_0}{qv}$

答:
You have that
$$
F_0\,\left(3\hat{i}+4\hat{j}\right) = q\, v\,\hat{k}\times\vec{B},
$$
which means that
$$
F_0\,\left(3\hat{i}+4\hat{j}\right) = q\, v\,\hat{k}\times\left(B_x\hat{i}+B_y\hat{j}+B_z\hat{k}\right),
$$
therefore, recalling that $\hat{k}\times\hat{i}=\hat{j}$ and that the corresponding components of the vectors shall be equal, you get:
$$
4\,F_0\hat{j}=q\,v\,B_x\,\left(\hat{k}\times\hat{i}\right),
$$
i.e.
$$
4\,F_0\hat{j}=q\,v\,B_x\,\hat{j},
$$
hence
$$
4\,F_0=q\,v\,B_x\,
$$
which results precisely in
$$
B_x={4\,F_0\over q\,v}.
$$

Title: Cross Product Question (Simple Physics)


问:
Proving $\mu$ is a measure on $A$ iff for every decreasing sequence in $A$ $\mu(\cap_{n=1}^{\infty} E_n)= \lim_{n\to \infty} \mu(E_n)$
Let $(X,A)$ a measurable space.
$\mu: A\to [0,\infty]$ satisfying:

$\mu(X)<\infty$

$\mu(\emptyset)=0$

$\forall{E,F}\in{A}$ disjoint sets,
$\mu(E\cup F)=\mu(E)+\mu(F)$

Prove:
$\mu$ is a measure on $A$ iff
for every deceasing sequence $E_n$ in $A$ $\mu(\cap_{n=1}^{\infty} E_n)= \lim_{n\to \infty} \mu(E_n)$.

First: $\mu (\emptyset)=0$ (given).
The forward case is obvious since $\mu$ is a measure by assuming, and $\mu(X) < \infty$ implies that $\mu(E) <\infty$ for all $E\in{A}$.
So using a theorem we finish it.

The second condition is to show that for a sequence of disjoint sets $F_n \in{A}$ , $\mu(\cup_{n=1}^{\infty} F_n) =\sum_{n=1}^{\infty} \mu(F_n)$.

Let's define a new decreasing sequence $E_n$.

$E_n=X\setminus (\cup_{i=1}^{n} F_i)=\cap_{i=1}^{n} F_i
^{c}$.

$\mu(X)=\mu(E_n\cup E_n^{c})=(by 2)=\mu(E_n)+\mu(E_n^{c})$.

$\mu(E_n^{c})=\mu(\cup_{i=1}^{n} F_i)=\mu(X)-\mu(\cap_{i=1}^{n} F_i^{c})$

Now let n approaches infinity so:

$\mu(\cup_{i=1}^{\infty} F_i)=\mu(X)-lim_{n\to \infty} \mu(\cap_{i=1}^{n} F_i^{c})=\mu(X)-lim_{n\to \infty} \mu(X\setminus \cup_{i=1}^{n} F_i)$

Then can I use that $\mu(lim)=lim(\mu)$ ?

(Then I get

$\mu(\cup_{i=1}^{\infty} F_i)=\mu(X)-\mu(lim_{n\to \infty} X\setminus {\cap_{i=1}^{n} F_i)}=\mu(X)-\mu(lim_{n\to \infty} \cup_{i=1}^{n}  F_i)$ .

So I can reduce $\mu(X)$ with its limit since $\mu(X)$ is finite (by 2), and use 3 after it.

Can someone help in this.

答:
Hint

Let $\{F_i\}_{ i\in\mathbb N}$ a collection of disjoint sets.
\begin{align*}
\mu\left(\bigcup_{i=1}^\infty F_i\right)&=\mu(X)-\mu\left(\bigcap_{i=1}^\infty F_i^c\right)\\
&=\mu(X)-\lim_{n\to \infty }\mu\left(\bigcap_{i=1}^n F_i^c\right)\\
&=\mu(X)-\lim_{n\to \infty }\left(\mu(X)-\mu\left(\bigcup_{i=1}^n F_i\right)\right).
\end{align*}
I let you justify all steps and conclude.


问:
$\int_0^1\arctan\,_4F_3\left(\frac{1}{5},\frac{2}{5},\frac{3}{5},\frac{4}{5};\frac{1}{2},\frac{3}{4},\frac{5}{4};\frac{x}{64}\right)\,\mathrm dx$
I need help with calculating this integral:
$$\int_0^1\arctan\,_4F_3\left(\frac{1}{5},\frac{2}{5},\frac{3}{5},\frac{4}{5};\frac{1}{2},\frac{3}{4},\frac{5}{4};\frac{x}{64}\right)\,\mathrm dx,$$
Where $_pF_q$ is a generalized hypergeometric function.

I was told it has a closed-form representation in terms of elementary functions and integers.

答:
This hypergeometric function is not an elementary function, but its inverse is - see Bring radical.

\begin{align}
I&= \int_0^1\arctan{_4F_3}\left(\frac15,\frac25,\frac35,\frac45;\frac12,\frac34,\frac54;\frac{x}{64}\right)\,dx \\
&=\frac{3125}{48}\left(5+3\pi+6\ln2-3\alpha^4+4\alpha^3+6\alpha^2-12\alpha\\-12\left(\alpha^5-\alpha^4+1\right)\arctan\frac1\alpha-6\ln\left(1+\alpha^2\right)\right)\\
&=0.7857194\dots
\end{align}
where $\alpha$ is the positive root of the polynomial $625\alpha^4-500\alpha^3-100\alpha^2-20\alpha-4$. It can be expressed in radicals as follows:

$$\alpha=\frac15+\sqrt\beta+\sqrt{\frac15-\beta +\frac1{25\sqrt\beta}},$$
where
$$\beta=\frac1{30}\left(\frac\gamma5-\frac4\gamma+2\right),$$
where
$$\gamma=\sqrt[3]{15\sqrt{105}-125}.$$


问:
Simple algebra, finding x
How and why did the sign + and - switch places in the solution?

Question:

$$\frac{x-6}x - \frac6x = \frac{x-6}{x+6}$$

Answer:

Step one

$$\frac{x-6-6}x = \frac{x+6}{x-6}$$

答:
Here's another way to solve the equation
$$\frac{x - 12}{x} = \frac{x - 6}{x + 6}$$
by corrensponding addition: subtract the numerator from the denominator on both fractions to get
$$\frac{x - 12}{x-(x-12)} = \frac{x - 6}{x + 6-(x-6)}\iff\frac{x - 12}{12} = \frac{x - 6}{12}$$
which is impossible.

NB: The given solution solves
$$\frac{x - 12}{x} = \frac{x + 6}{x - 6}.$$
Using the method above gives
$$\frac{x - 12}{12} = \frac{x + 6}{-12}
\iff x-12=-6-x$$
that is $x=3$.

Title: Simple algebra, finding x


问:
How many squares fit in a circle?
I am reading a book on computer architecture. The author discusses how processor chips are made from wafers. Dies (the heart of the chip, that does calculations) are cut from circular wafer plates.

Because dies are square (or rectangular I should say), and the wafer is a circle, not all of the area of the wafer is used for dies.

To calculate how many dies can go into a single wafer, the author uses this equation:
 $$\text{Dies per wafer}=\frac{\pi\times(\text{Wafer diameter}/2)^2}{\text{Die area}}-\frac{\pi\times\text{Wafer diameter}}{\sqrt{2\times\text{Die area}}}.$$

The second part of the equation compensates for the problem of squares in a circle. I am interested in the derivation of the second part. Can you help me out?

答:
The first part of the equation gives us the number of squares (including all the squares we had to chop at the circles edge). The second part is an estimate of square dies on the edge of the circle. This is estimated by counting number of squares that can be layed along the edge of circle so that diagonal of square overlaps with edge of circle.

This esitmate should work alright only if the square is significantly smaller than the circle. If you were for example fitting a square which was of edges $1$ unit long on a circle $\sqrt{2}$ unit in diameter, The equation gives us $- \frac{1}{2} \pi$ as the number of squares.

Title: How many squares fit in a circle?