问:
Adding a real destroys strategic closure
The following should have an elementary argument, but I haven’t been able to think of it.

Suppose $P$ is a nontrivial partial order in $V$.  Let $r$ be a real not in $V$.  Show that in $V[r]$, $P$ is not $\sigma$-strategically-closed.

This question is a natural generalization of a question that came up in my research.  I can prove it if $r$ is added by a forcing $Q$ such that $Q^2$ is c.c.c.  The argument uses Unger's Theorem that for such $Q$, if $\dot R$ is a name for a $\sigma$-strategically-closed poset, then $Q*\dot R$ has the $\omega_1$-approximation property.

答:
Edit: The result here implies that $P$ is not $\sigma$-strategically closed in $V[r]$ if $r$ is added by a ccc forcing $Q$ (thus slightly weakening the condition "$Q^2$ is ccc" in your question), see below.

This is just a partial answer, showing that winning strategies witnessing $\sigma$-strategically-closedness cannot be approximated in $V$. To be precise:

If $V\subseteq W$ are models of $\mathrm{ZFC}$ with $\mathbb R^V\subsetneq\mathbb R^W$ then a winning strategy $\tau$ witnessing some $\mathbb P\in V$ to be $\sigma$-stategically closed cannot be approximated in $V$ in the following sense:

There is no function $T\in V$ mapping descending sequences $\vec p=\langle p_i\mid i< 2n+1\rangle$ in $\mathbb P$ of odd length to an antichain below $p_{2n}$ so that $\tau(\vec p)\geq a$ for some element of $T(\vec p)$.

This just means that "$\tau$ cannot be bounded in $V$ by antichains'' (the reason antichains come up is that $\tau$ is trivially bound by $T(\vec p)=\{q\mid q\leq p_{2n}\}$). Intuitively, any such $\tau$, if it exists, must make very large jumps.

So suppose that $T\in V$ were such an approximation. To ease notation, we will assume that $T$ depends only on the last element, so we can understand $T$ as a function on $\mathbb P$ (however slight adjustments of the following argument give the more general statement). Also, we can assume that $T(p)$ is always a maximal antichain below $p$ with at least two elements. Now, in $V$, given some $p\in \mathbb P$ we will label the elements of $T(p)$ with either $l_p$ or $r_p$ in some way so that both labels appear (think of this as a binary left or right decision on how to proceed after hitting $p$).

If $\vec p=\langle p_i\mid i<N\rangle$ ($N\leq\omega$) is a descending sequence through $\mathbb P$ then we decode an $l-r$ sequence $\mathrm{dec}(\vec p)$ from this: Start with $p_0$. If $\vec p$ is eventually below some (unique!) element $a_0$ of $T(p_0)$ then $\mathrm{dec}(\vec p)$ starts with $l$ or $r$ depending on whether $a$ has label $l_{p_0}$ or $r_{p_0}$. Now $\vec p$ may or may not eventually be below some element $a_1$ of $T(a_0)$ and if it is, $\mathrm{dec}(\vec p)$ continues depending on the $a_0$-label of $a_1$, etc. If the sequence reaches length $\omega$ we stop. Now suppose $\mathrm{dec}(\vec p)$ did indeed reach length $\omega$. Then $\vec p$ can only have a lower bound in $\mathbb P$ if $\mathrm{dec}(\vec p)\in V$: $V$ can define $\mathrm{dec}(\vec p)$ from $p_0$ and any lower bound $q$ of $\vec p$ by induction by using $q$ as a guide on whether $\vec p$ went left or right at each step.

We can use this to turn $\tau$ into a winning strategy $\tau^\ast$ for player $II$ in the Banach-Mazur game on $2^\omega$ in $W$ with payoff set $(2^\omega)^W\setminus(2^\omega)^V$: Remember that players alternate playing finite $0-1$ sequence which we interprete as $l-r$-sequences. We will describe $\tau^\ast$ now: Suppose $\delta$ is an opponent and suppose $\delta$ opens with $s_0$. By starting with $p_0=1_\mathbb P$ we can construct a descending sequence
$\vec p^0=\langle p_i\mid i\leq m_0\rangle$ in $\mathbb P$ with

$\forall i<m_0\ p_{i+1}\in T(p_i)$

$\mathrm{dec}(\vec p^0)=s_0$

Now $\tau$ has a reply $q_0$ to $p_{m_0}$ (considered as a first move by player $I$) and now we can further strengthen it to some $p_{m_0+1}\in T(p_{m_0})$. The reply $s_1$ of $\tau^\ast$ is simply $l$ or $r$ depending on the $p_{m_0}$ label of $p_{m_0+1}$.  We continue doing this: If $s_2$ is the next move of $\delta$ then we lengthen $\langle p_i\mid i\leq m_0+1\rangle$ to a sequence $\vec p^1=\langle p_i\mid i\leq m_1\rangle$ still satisfying 1. and 2. (with all 0's replaced by 1's) and choose our reply simiarly by looking at $q_1=\tau(p_{m_0}{}^\frown q_0{}^\frown p_{m_1})$.

This produces two infinite sequences
$$s=s_0^\frown s_1^\frown s_2^\frown \dots$$
and
$$\vec p^\ast = p_0\geq p_1\geq p_2\geq\dots$$
with $\mathrm{dec}(\vec p^\ast) = s$. Now $\vec p^\ast$ is itself not necessarily according to $\tau$, but from the construction above it is clear that the subsequence $\langle p_{m_i}\mid i<\omega\rangle$ constitute plays of player $I$ in a game that is according to $\tau$. Hence $\vec p^\ast$ has a lower bound, $s\in V$ and $\tau^\ast$ is indeed winning.

The upshot of this is that $\mathbb R^V$ is comeager in $W$, but this can only happen if $\mathbb R^V=\mathbb R^W$, contradiction.

Now let us see that $\mathbb P$ is not $\sigma$-strategically closed in a ccc extension $W$ with new reals. So suppose otherwise and let $\tau$ witness this. By the above, it is enough to show that, in $V$, $\tau$ is bounded by antichains to get a contradiction. Since $\tau$ lives in a ccc extension, $\tau(\vec p)$ can be covered in $V$ by countable subsets of $\mathbb P$ (uniformly in $\vec p)$. Thus it is enough to turn countable subsets $\{p_n\mid n<\omega\}\subseteq\mathbb P$ into antichains $A\subseteq\mathbb P$ so that any $p_n$ is above some $a\in A$.

First we will solve the problem in $W$, where $\mathbb P$ is $\sigma$-strategically closed. It is easy to see that
$$D=\{p\in\mathbb P\mid \forall n\ p_n\not < p\}$$
is dense open (start with some $q$ and strengthen it repeadetly making sure it is not strictly above $p_n$ for one $n$ at a time and follow $\tau$ so that there is a lower bound in the end). Still in $W$ we can build the desired antichain one step at a time inductively: Suppose we are at step $n$ and have constructed a finite antichain $A_n\subseteq D$ so far so that for all $m<n$ there is $a\in A_n$ with $a_i\leq p_m$. Now let $a_n\leq p_n$ be in $D$. If $A_n\cup\{a_n\}$ is an antichain this is our $A_{n+1}$. Otherwise, find some $a\in A_n$ compatible with $a_n$ and replace $a$ in $A_n$ by some $b\leq a, a_n$. We may have to strenghten some conditions infinitely many times in this construction. If we are careful we follow $\tau$ along the way to make sure that there is a lower bound in the end.

Thus we have an antichain $A\in W$ bounding all $p_n$'s. But the existence of such an antichain is absolute between $V$ and $W$: Consider the tree $S$ consisting of finite sequences $\langle a_i\mid i<n\rangle$ so that

All $p_m$, $m<n$ are above some $a_i$, $i<n$.

$\{a_i\mid i<n\}$ is an antichain

Ordered by endextension, this is indeed a tree of height $\omega$ in $V$ that has a branch in $W$. By absoluteness of wellfoundedness, $S$ has a branch in $V$ as well and the set of conditions appearing in this branch forms the antichain we were looking for in $V$.


问:
asymptotic infinity
I am very very new to math as a whole, so please excuse my n00biness.

I read: An asymptote of a curve is a line that continually approaches the curve but never meets it at any finite distance. The distance between the line and the curve approaches zero as they tend to infinity. When we say variable n tends to infinity it means as n gets very very large. If we look at 1/n as n tend to infinity, then n gets very large and 1/n goes to zero

So my question is: Based on the above can 1/n be analogous to the distance between a line and a curve, and n thought of as the increase in the size of the two?

答:
In math, $n$ is normally used for an integer number, so let me switch to the variable $x$.

Consider the $(x,y)$-plane. Write $L$ for the set of points whose $y$ coordinate is zero and $C$ for the graph of the function $x \mapsto 1/x$. In other words:
\begin{align*}
L &= \{ (x,0) \quad | \quad x \in \mathbb{R} \}\\
C &= \{ (x,1/x) \quad | \quad x \in \mathbb{R} \}
\end{align*}
Note that $L$ is also a graph but a really stupid one, it corresponds to the function $x \mapsto 0$.

We run into trouble at $x=0$, so let us just consider the part of the plane where $x>0$.

If you sketch both $L$ and $C$ you will see that at some fixed $x$, the distance between them is $1/x$. Since $1/x$ gets smaller and smaller as $x$ increases, indeed their distance tends to zero. So yes, your $1/n$ can be thought of in this way (and $L$ is an asymptote for $C$).

ContentLicense: CC BY-SA 3.0

Title: asymptotic infinity


问:
$\lim\limits_{n\to \infty}\frac{3^n+6^n-1}{6^n-1}$
$\lim\limits_{n\to \infty}\dfrac{3^n+6^n-1}{6^n-1} = 1$

I proved the statement above but now the exercise says that I have to find $n_0$ so that $\left\vert x_n-1\right\vert < \frac{1}{50}$ with $(x_n)$ as the given seqeunce above.

I wrote the sequence like this: $(x_n)=\frac{3^n}{6^n-1}+1$ and then I looked at $\left\vert \frac{3^n}{6^n-1}+1-1\right\vert < \frac{1}{50}$ which is equivalent to $\frac{\frac{1}{2^n}}{1-\frac{1}{6^n}} < \frac{1}{50}$. But now I don't know how to go on...

Can someone help me?

答: Since $\frac{3^n}{6^n-1}$ is clearly decreasing for all $n\in \mathbb{N}$ we can just test numbers. If $n=5$ the inequality does not hold but with $n=6$ we get under $1/50$ so $n_0=6$ works.

ContentLicense: CC BY-SA 4.0


问:
isomorphism between $\frac{\mathbb{F}_5[x]}{(x^2+x+1)} $ and $ \frac{\mathbb{F}_5[x]}{(x^2 -2)} $
I know $\frac{\mathbb{F}_5[x]}{(x^2+x+1)} $ and $ \frac{\mathbb{F}_5[x]}{(x^2 -2)} $ is isomorphic because they are both 2-degree extension of $ \mathbb{F}_5 $ .

But I cannot contract explicit isomorphism between them.

Could you show me isomorphism map between $\frac{\mathbb{F}_5[x]}{(x^2+x+1)}$  and $ \frac{\mathbb{F}_5[x]}{(x^2 - 2)} $?

I tried to show $$
f\colon\mathbb{F}_5[x]\to\mathbb{F}_5[x]/(x^2+x＋1)$$ is isomorphism,  which sends $x$ to $g(x)＋c＋(x^2＋x-1)$. I couldn't find $g(x)$, and show $f$ is ring hom and bijection（surjection is clear）. I couldn't find $g(x)$ because constant term doesn't vanish. Thank you for your help.

答:
As requested, I outline the isomorphism. If we fix $y:=2x+1\in\mathbb F_5[x]/\left(x^2+x+1\right),$ then we see that
$$y^2=4x^2+4x+1=2.$$
In particular, we can construct a map
$$\mathbb F_5[y]\to\frac{\mathbb F_5[x]}{(x^2+x+1)}$$
by sending $y\mapsto2x+1.$ It's not difficult to check that this is surjective, for $3y+2\mapsto x.$ It remains to show that our kernel is $\left(y^2-2\right).$ The kernel requires $p(y)\in\mathbb F_5[y]$ to have
$$p(2x+1)\equiv0\pmod{x^2+x+1}.$$
However, this is equivalent to
$$p(2x+1)\equiv0\pmod{(2x+1)^2-2},$$
so indeed $p(y)\in\left(y^2-2\right).$ This finishes the construction of our isomorphism.

At a high level, we were looking for an automorphism of $\mathbb F_{5^2},$ so it makes sense to look for linear maps to do the trick. This is what motivates the $y:=2x+1$ substitution.


问:
Absolute Galois group of Laurent power seris $k((t))$
Let $k$ be a field of characteristic $0$ and denote by $G$ its absolute Galois group. I heard several times that the absolute Galois group of the Laurent power series $k((t))$ is isomorphic to a semi-direct product $G\rtimes \hat{\mathbb{Z}}$, but I can't find a reference for this result. Qould you know where i can find this? Also, what makes it not hold if $k$ is of characteristic $p$?

答:
We write $\operatorname {Gal}(F)$ for the absolute Galois group of a field $F$. Let $k$ be a field of characteristic $0$ and $k^s$ a separable closure.

There is an exact sequence $$1\longrightarrow \operatorname{Gal}(k^s((t)))\longrightarrow \operatorname{Gal}(k((t)))\longrightarrow \operatorname{Gal}(k)\longrightarrow 1$$

It is true since $k$ has characteristic 0 that $\operatorname{Gal}(k^s((t)))\cong \hat{ \mathbb{Z}}$. See, for example, the Wikipedia page for "quasi-finite field." One sees this by proving that $k^s((t^{1/n}))$ is the unique extension of degree $n$ of $k^s((t))$. One problem in characteristic $p$ is that $k^s((t^{1/p}))$ isn't separable, so it doesn't contribute to the Galois group (and there are several other extensions of degree $p$ which make this Galois group more complicated).

There is a section of the map $\operatorname{Gal}(k((t))) \to \operatorname{Gal}(k)$ (an automorphism of $k^s$ over $k$ gives an automorphism of $k((t))^s/k((t))$ just by fixing elements of the form $t^{1/n}$). This is probably the easiest way to see the map is surjective in the first place. So this sequence splits and expresses $\operatorname{Gal}(k((t)))$ as a semidirect product $\operatorname{Gal}(k^s((t)))\rtimes \operatorname{Gal}(k)$.

Edits: a previous version of this post contained some false facts about $k^s((t))$ for positive characteristic.


问:
Does gradient descent for linear regression select the minimal norm solution?
I was told that Gradient Descent finds the weights of smallest norm.

This is what I understood in the linear regression setting:

$f_w(x)=w^\top x$ are the linear functions $ \mathbb{R}^n \rightarrow \mathbb{R} $.

$ \mathcal{D}=\{(x_i,y_i) \in \mathbb{R}^d\times\mathbb{R} \mid i=1,...,n\} $ is a dataset

Then I choose some loss function $L$ to get a cost function
$ C_0(w)=\frac{1}{n} \sum_i L(f_w(x_i),y_i) $

Finally, I define a regularized cost $ C_\lambda(w)=C_0(w)+\frac{\lambda}{2n}||w||^2 $

Then there must be some sort of equivalence between (and it looks it should be some classical result):

minimizing $C_\lambda(w)$ for an opportune $ \lambda \ge 0 $

minimizing $C_0(w)$ with Gradient Descent

My questions:

Can somebody clarify a bit this fact? By providing some hypothesis (e.g. on $L$), some insight, some reference...

Does this apply in a similar way to the linear classification setting ($y_i\in\{-1,+1\}$)? In that case, I would expect some equivalence bewteen:

maximizing $M(w)= \min_{i=1...n} y_i \langle w, x_i \rangle $ over $||w||=1$

minimizing $||w||$ over $ y_i \langle w, x_i \rangle \ge 1 $

maybe the previous 1. and 2. points above (those regarding regression)

答:
I will prove that when we initialize the weights with zero and apply Gradient Descent in the limit of infinitesimal stepsize, then the solution is guaranteed to converge against the minimal norm solution.

Some notation

Let the loss function be given as

$$(1) \qquad L_\lambda(w) = \tfrac{1}{2N}\|Xw-y\|_2^2 + \tfrac{1}{2}\lambda\|w\|_2^2$$

With gradient $\nabla_w L_\lambda = (\lambda I + \tfrac{1}{N}X^TX)w - \tfrac{1}{N}X^Ty$ and Hessian $\nabla^2_w L_\lambda = \lambda I + \tfrac{1}{N}X^TX$

Note that $\nabla^2_w L_\lambda$ is positive semi-definite for all $\lambda \ge 0$ and even positive definite if $\lambda>0$. Consequently, $L_{\lambda>0}$ is strongly convex and has a unique global minimum, whereas $L_0$ is only guaranteed to be convex and not strongly convex, and so can potentially have multiple global minima.

This does indeed happen when the columns of $X$ are not linearly independent: To any optimal $w$, we can add an element of $\ker(X)$ and get another optimal $w'$.

Claim 1: For $\lambda>0$, let $\hat w(\lambda) = \arg\min_w L_\lambda(w)$, which is unique by strong convexity. Then

$$(2)\qquad \arg\min_w L_0(w) = \hat w_0 + \ker(X) \qquad\text{where}\qquad \hat w_0 =\lim\limits_{\lambda\searrow  0}\hat w(\lambda)=X^+y $$

Here $X^+$ is the pseudoinverse. We will refer to $\hat w_0$ as the unique minimal norm solution among solutions of the unregularized problem $L_0$. (One can also show $\hat w_0 = \underset{w\in\arg\min(L_0)}{\arg\min}\|w\|_2$)

Proof: Exercise.

Connection of Gradient Descent to Ordinary Differential Equations

Now, a hypothesis we could make is that "under normal circumstances" Gradient Descent (GD) should converge against the minimal norm solution. The question is what do we mean with "normal circumstances"? Well, if we start GD at a point that is already a global minimum other than the minimal norm minimum, then of course GD will never move away from that point. Another thing that can happen of course is non-convergence due to too high step size. To avoid these issue, we will consider GD with infinitesimal stepsize and fixed starting location at the origin. In this case, the iterative scheme is equivalent to the solution of the ODE

Claim 2: GD on $L_\lambda$ with starting point $w_0$ and infinitesimal step size is equivalent to the ODE

$$(3) \qquad
\begin{aligned}&\dot w = -\nabla L_{\lambda}(w)
\\ &w(0) = w_0\end{aligned}
\qquad \nabla L_{\lambda}(w)= \big(\lambda I+\tfrac{1}{N}X^TX\big)w+X^Ty \qquad $$

Proof: Compare GD applied to (1) with the explicit Euler method applied to (3)

This is actually a general fact for any differentiable loss function / model.

Claim 3: The solution of the ODE (3) with initial condition $w_0=0$ converges against $\hat w(\lambda)$ for all $\lambda\ge 0$ as $t\to\infty$. (We will only consider the interesting case $\lambda=0$)

Proof: Recall that a linear ODE $\dot x = Ax+b, x(t_0)=x_0$ with paramters $\theta=(t_0, x_0, A, b)$ has the analytic solution

$$x^*(t, \theta) = e^{A(t-t_0)}x_0 + \frac{e^{A(t-t_0)}- I}{A}b$$

Note that $\frac{e^{A(t-t_0)}- I}{A}$ is only shorthand for the power series $\sum_{n=0}^{\infty} \frac{A^{n}(t-t_0)^{n+1}}{(n+1)!}=\int_{0}^{t} e^{A(s-t)}ds$ which exists and converges even when $A$ is singular. Then, with $t_0=0$, $w_0=0$, $A=-\tfrac{1}{N}X^TX$, $b=\tfrac{1}{N}X^Ty$:

$$ w^*(t) = \frac{I - e^{-\frac{1}{N}X^TX t}}{X^TX}X^Ty $$

So, we are left with showing that $\lim_{t\to\infty} \frac{I-e^{-\frac{1}{N}X^TX}}{X^TX} X^T = X^+$; which we can do by showing the limit satisfies the four  Moore-Penrose conditions

$$\begin{aligned}
&1.&AA^+A&=A  & &3.&(AA^+)^T &= AA^+
\\ &2.&A^+AA^+&=A^+ & &4.&(A^+A)^T &=A^+A
\end{aligned}$$

Indeed, we actually already have (I think?) $\lim_{t\to\infty} \frac{I-e^{-At}}{A}=A^+$ whenever all eigenvalues of $A$ have non-negative real part, which we shall show exemplary for the first MP-condition. Note that the matrix exponential satisfies $e^{U^{-1} A U} = U^{-1} e^A U$ and $e^{\operatorname{diag}(v)} = \operatorname{diag}(\exp(v))$ element-wise, hence:

$$\begin{aligned}
A  \frac{I-e^{-At}}{A} A
&= A (I - e^{-At})
= U^{-1} \big(\underbrace{\Lambda  (I - e^{-\Lambda t} )}_{\to \Lambda}\big) U \to A
\end{aligned}$$

And obviously all eigenvalues of $\tfrac{1}{N}X^TX$ are non-negative reals, so

$$\begin{aligned}
\lim_{t\to\infty} w^*(t) = \lim_{t\to\infty}\frac{I - e^{-\frac{1}{N}X^TX t}}{X^TX}X^Ty = (\tfrac{1}{N}X^TX)^+ \tfrac{1}{N}X^Ty = X^+y = \hat w_0
\end{aligned}$$

Q.E.D.

Corollary: If we start at a general point $w_0= v + v^\perp$ where $v\in\ker(X)$ and $v^\perp\in\ker(X)^\perp$, then GD converges against $\hat w_0 + v$. In particular, if $X$ has full column rank then we still have convergence against the minimal norm solution.

Proof: $e^{-X^TXt} = U e^{-\Lambda t} U^T$, thus for any eigenvalue, eigenvector pair $(\lambda_i, u_i)$ of $X^TX$ we have

$$\lim_{t\to\infty}e^{-X^TXt} u_i = \lim_{t\to\infty}e^{-\lambda_i t}u_i=\begin{cases}0:&\lambda_i>0 \\ 1:& \lambda_i=0\end{cases}$$

Consequently, $\lim_{t\to\infty}e^{-X^TXt}=\operatorname{proj}_{\ker(X)}$.

Some final remarks: The connection between gradient based optimizers and ODEs is actually pretty deep, see for example A Differential Equation for Modeling Nesterov’s Accelerated
Gradient Method: Theory and Insights by Sue, Boyd and Candès


问:
What is the least nonnegative number $a$ congruent to $3^{340}\pmod{341}$?
Find the least nonnegative number $a$ congruent to $3^{340} \pmod{341}$.

What steps should I take to get to the answer?

Body:
As $341=11\cdot31,$

Using Fermat's Little Theorem, $ 3^{10}\equiv1 \pmod {11}$

$\implies 3^{340}=(3^{10})^{34}\equiv1^{34}\equiv1\pmod{11}\ \ \ \ (1)$

Again applying  Fermat's Little Theorem, $3^{30}\equiv1\pmod {31}$

$\implies 3^{340}=(3^{30})^{11}\cdot 3^{10}\equiv 1\cdot3^{10}\equiv 3^{10} \pmod {31}$

Now, $3^3\equiv-4\pmod {31}\implies 3^{10}=3\cdot (3^3)^3\equiv3\cdot (-4)^3\equiv 3(-2)\equiv-6$

$\implies 3^{340}\equiv-6\ \ \ \ (2)$

Now, use CRT in $(1),(2)$ to find $x\equiv56\pmod{341}$


问:
How can one prove that the set of reflexive relation on $\mathbb{N}$ is uncountable
I'm seeking for a proof that the set $S$ of a reflexive relations over $\mathbb{N}$ is uncountable. It would be also helpful if someone could give me some general method for finding an injective or surjective function that proves this.

答:
For each subset $A$ of $\Bbb N$ of size greater than $1$, consider the relation $\mathcal{R}_A = \{(x,y)\in \Bbb N\times \Bbb N~:~x=y\vee (x\in A\wedge y\in A)\}$

That is to say, given a subset $A$, let everything in $A$ be related to everything else inside of $A$ and everything outside of $A$ be related only to themselves.

It is clear that each subset of $A$ defines a unique such relation and that each relation defined this way is reflexive.

It is clear then there are at least as many such relations as there are subsets of $\Bbb N$ which is known to be uncountable.


问:
Orthogonality in normed spaces
Let $M$ and $N$ be nonzero closed vector subspaces of a normed vector space $E$. Prove that if $M \neq N$, then
$M^\perp \neq N^\perp$.

答:
Although the question was answered in the post I linked, I thought I add some details.

Take w.l.o.g. $x\in M \setminus N$. Since $N$ is closed, $X/N$ is also a normed vector space and $q\colon X\to X/N, y \mapsto y+N$ is linear and continuous.

By Hahn-Banach, we can find a $\lambda \in (X/N)'$ such that $\lambda(x+N) = \|x+N\|_{X/N} \neq 0$, since $x\notin N$ (and therefore $x+N\neq N$).

Now $f=\lambda \circ q \in X'$ and $f|_N = 0$ but $f(x) \neq 0$, so $f\in N^\perp$ but $f\notin M^\perp$.

Title: Orthogonality in normed spaces


问:
How to sketch/graph the conic section $2x^2−4xy+5y^2=36$ in the (x, y)-plane by making use of the coordinate transformations $2(x−y)^2+3(y)^2 =36$
How do I sketch/graph the conic section $$2x^2−4xy+5y^2=36$$ in the (x, y)-plane by making use of the coordinate transformations $$2(x−y)^2+3(y)^2 =36$$
With $x'=x-y$ and $y'=y$, so
$$2(x')^2+3(y')^2 =36$$

答:
First draw the ellipse in $(x',y')$. It will be aligned with the axes. Then, using $(x,y) = (x'+y', y')$, transform every point. This will be a skew transformation, since the second co-ordinate is left unchanged.

The transformation from $(x',y')$ to $(x,y)$, including a box enclosing the ellipse for reference, will look like so:

$\hskip{3cm}$


问:
How to do these Geometry proofs using vectors?
I'm new to vectors and need to answer the following two things:

Show by using vectors that the mid segment of a trapezoid is parallel to the bases and $1/2$ as long as the sum of them.

That's what I got so far.

Show by using vectors that the line segment m in a quadrilateral is equal to the sum of both not divided side-vectors divided by two.

The work sheet only provides the solutions but I can't figure out how to get there.

Thanks!

答:
We do not work with free vectors. We work with position vectors, which are fixed on the Euclidean plane wrt some origin. (See figure.) The underlying coordinate plane provides coordinates to any points $P,Q$. This facilitates writing $\vec{OP}=P-O$ and $\vec{PQ}=Q-P$. For example if $P=(3,4)$ on coordinate plane, then $\vec{OP}=(3-0,4-0)=3\hat{i}+4\hat{j}$.

So let position vectors of the vertices of trapezium be $A,B,C,D$. Since $E$ is midpoint of $AD$, its position vector will be (think coordinates)
$$E=\frac{A+D}{2} \, , \quad F=\frac{B+C}{2}$$

Now lengths are given by magnitudes of vectors
$$|AB|=|\vec{AB}|=B-A \, , \quad |DC|=|\vec{DC}|=C-D$$

Since $AB // DC$, $\,\vec{DC}=k\vec{AB}$.

Now $$|EF|=F-E=\frac{(B-A)+(C-D)}{2}=\frac{1}{2}(|AB|+|DC|)$$

You can also conclude about parallelism.

This was Ques $1$. Now you can do Ques $2$.


问:
$\begin{align}\color{#c00}a&\mid b\\\color{#c00}c&\mid d\end{align}\Rightarrow\ \color{#c00}{ac}\mid bd$ $\ \ \bf\small [Divisibility\ Product\ Rule]$
I just need to check the reasoning in my proof is correct, I think it is valid although I'm not totally convinced because I can't follow the logic; does proving that $x$ is an integer prove that $ac|bd$?

Theorem: Let $a$, $b$, $c$, $d$ be integers. If $a|b$ and $c|d$, then $ac|bd$.

Proof: $aj=b$ and $ck=d$ for integers $j$, $k$. Then, $ac|bd$ implies $acx=ajck$ and thus $x=jk$ for some $x$. Since the product $jk$ is an integer, $x$ is an integer and thus $acx=ajck$ and thus $ac|bd$.

答:
Hint $\  $ Noting that $\displaystyle\rm\ \ \ a\mid b\,\iff\, \frac{b}a\in \mathbb Z$

we observe that $\rm\ \ a\mid b,\ c\mid d\ \ \Rightarrow\ \ ac\mid bd $

is equivalent to $\,\rm\displaystyle\,\ \ \frac{b}a,\,\frac{d}c\in\mathbb Z\ \ \Rightarrow\ \frac{b\,d}{a\,c}\in\, \mathbb Z$

in your notation $\rm\ \ \ j,\ k\ \in\ \mathbb Z\ \ \Rightarrow\ \  j\, k\ \in \, \mathbb Z$

That's true: the product of integers is an integer, i.e. integers are closed under product. So this divisibility product law is equivalent to this product closure law (except if our definition of divisibility includes $\,0\mid n\iff n= 0\,$ then we need to handle such cases separately - which is easy).

Edit $\ $ Since, based on comments, at least one reader seems to have misconstrued the intent of my answer, I elaborate below. When I posted this answer, there were already a few answers explaining the logical flaw in the proposed proof in the question. That done, my intent was instead to address another point, namely how one may exploit the the innate arithmetical structure in order to attain a simpler and more conceptual proof. Such simplification may make the proof more intuitive, which may help one to avoid making such errors. To elaborate, below I show how the "arithmetic" of the divisibility relation is intimately connected with the arithmetic of the subring $\rm\,\mathbb Z\subset\mathbb Q.\,$

First, notice how the above proof makes clear that this product rule for divisibility is essentially equivalent to the product rule for integrality, i.e. if $\rm\,j,k\,$ are integers then so too is their product $\rm\,jk.\,$ Indeed, as we saw above, this integrality product rule easily implies the divisibility product rule. Conversely, if the divisibility product rule is true, then specializing $\rm\ a = 1 = c\ $ we infer $\rm\, 1\mid b,\ 1\mid d\ \Rightarrow\ 1\mid bd,\, $ i.e. $\rm\, b,d\in\mathbb Z\ \Rightarrow\ bd\in \mathbb Z\ $ (note $\rm\, 1\mid n\iff n\in \mathbb Z,\,$ by definition).

Similarly, one deduces the equivalence between the divisibility difference rule, namely that $\rm\ a\mid b,\,c\ \Rightarrow\ a\mid b\!-\!c\ $ and the fact that $\rm\,\mathbb Z\,$ is closed under difference $\rm\,j,\,k\in \mathbb Z\ \Rightarrow\ j\!-\!k\in\mathbb Z$.

Combining these observations leads to the following equivalence between the arithmetic of divisibility relations and subrings $\rm\,Z\,$ of $\,\mathbb Q\,$ (or any field).

Theorem $\ $ Let $\rm\,Z\, $ be a subset of $\rm\,\mathbb Q\,$ with $\rm \,1\in Z.\,$ Let $\:\mid\: $ denote the divisibility relation  $\rm\,  a\mid b \iff b/a\in Z\,$ for $\rm\,a,b\in Z,\, a\ne 0,\ $ and $\rm\ 0\mid b \iff b = 0.\,$ Then the following two statements are equivalent.

$\rm(1)\ \ \ Z\ $ is a subring of $\rm\,\mathbb Q$

$\rm(2)\ \ $ The relation $\:\mid\: $ satsifies the following properties.

$\rm\qquad a\mid b,\,c\,\ \Rightarrow\,\ a\mid b-c\qquad $ for all $\rm\,a,b,c\in Z$

$\rm\qquad  a\mid b,\ c\mid d\,\ \Rightarrow\,\ a\,c\mid b\,d\ \ \ $ for all $\rm\,a,b,c,d\in Z$

Remark $ $ The test in $(2)$  is the Subring Test in divisibility language.


问:
Find the ordered pair $(a,b)$ of real numbers for which $x^2+ax+b$ has a non-real root whose cube is $343$.
Because $x^2 + ax + b$ has all real coefficients, we know the roots must be complex conjugates. So let $\lambda e^{i \theta}$ be a root, so we know that $(x - \lambda e^{i \theta})(x - \lambda e^{-i \theta}) = x^2 - \lambda x (e^{i \theta} + e^{-i\theta}) + \lambda^2$ is our polynomial.

We also know that $(\lambda e^{i \theta})^3 = 7^3 \implies \lambda e^{i \theta} = 7$, and since $|\lambda e^{i \theta}| = |\lambda| = 7,$ we know that $\lambda = 7$.

This means that $e^{i\theta} = 1$ and so $\theta = 0$. But clearly this doesn't give the right answer, since our polynomial isn't supposed to have real roots. Where is the flaw in my reasoning?

答: We have $x^3 = 343$ because we need to find an $x$ that when cubed is $343$. Now rearrange to form $x^3 - 343=0$. Use the difference of cubes factorization which is $a^3-b^3=(a-b)(a^2+ab+b^2)$ by letting $a=x$ and $b=7$. So, we get $(x-7)(x^2+7x+49) = 0$. This simplifies down to $(x^2+7x+49) = 0$, which is a quadratic with roots that fit our condition that $x^3$ = $343$, because we derived this equation from that fact. Use the quadratic formula on $x^2+7x+49$ to get $\frac{-7}2 \pm {7i\sqrt{3} \over 2}$ as our nonreal roots that when cubed make $343$.


答:
Suppose $M^\perp = N^\perp$, then by double annihilator theorem $\overline{M} = \overline{N}$. Since $M$ and $N$ are closed, it's a contradiction.

If $X$ is a normed vector space and $U\subseteq X$, how do we show $(U^\perp)_\perp=\overline U$?

Title: Orthogonality in normed spaces

Body:
Let $M$ and $N$ be nonzero closed vector subspaces of a normed vector space $E$. Prove that if $M \neq N$, then
$M^\perp \neq N^\perp$.


问:
Transform Karush–Kuhn–Tucker conditions
Consider the problem
\begin{align}
\min\limits_{x\in \mathbb{R}^n} J(x) \quad \text{s.t.} \quad e(x)=0, \ g(x)\leq0.
\end{align}
Reformulate this problem into a problem with equality- and box constraints.

By box constraints there are just simple bounds meant, i.e.
\begin{align}
x\in \{x\in \mathbb{R}^n: x_a\leq x\leq x_b \ \text{componentwise in } \mathbb{R}^n\}.
\end{align}

答:
You could add a slack variable $s$ and write

$$
\min_{x,s} J(x) \quad \text{s.t.} \quad e(x) = 0,\; g(x) + s = 0
$$

with the box constraint $s \geq 0$.

Title: Transform Karush–Kuhn–Tucker conditions


问:
Solving this system of trigonometric equations
How would I go about solving:
$$
\begin{cases}
\sin(x) + \cos(y) = \frac{1}{2}\\
\frac{\cos(x)}{\sin(y)} = -\frac{\sqrt{3}}{2}
\end{cases}
$$
Where
$$
x \in \left(\frac{-\pi}{6},\frac{7\pi}{6}\right) \, , \, \, y \in \left(\frac{-2\pi}{3},\frac{2\pi}{3}\right)
$$
I tried taking inverse trigonometric functions but it got super messy so I assume there must be a simpler way to solve this. Can I for example shift $\sin(x)$ by $\pi/2$ or some constant to obtain a cosine function?

答:
Hint: squaring both
$$
\begin{cases}
\sin^2 x=0.25-\cos y +\cos^2 y\\
\\cos^2 x=0.75 \sin^2 y
\end{cases}
$$
Adding both equations yields: $0.25\cos^2 y-\cos y=0$. Can you take it from here?


答:
Let's just square the second equation, we will check at the end if squaring increases any solution.

$\frac{\cos^2x}{\sin^2y} = \frac34 \implies \sin^2x = 1 - \frac34\sin^2y$

$\implies \sin x = \pm \sqrt{1 - \frac{3}{4}\sin^2y}$

Substitute this in equation 1.

$\pm \sqrt{1 - \frac{3}{4}\sin^2y} = \frac{1}{2} - \cos y$

Square both sides and change $\sin^2x$ to $\cos^2x$ to get:

$4\cos y = \cos^2y$

$\implies \cos y = 0\text{ or }4$
$\implies \cos y = 0 \implies y =(2n + 1)\frac{\pi}{2} $

Substitute this in equation 1, $\sin x = \frac{1}{2} \implies x = n\pi + (-1)^{n}\frac{\pi}{6}$

From the given range of y and x in question,

$$y \in \{-\frac\pi 2, \frac\pi 2\}$$

$$x \in \{\frac\pi 6, \frac{5\pi}{6}\}$$

But according to the second equation, we know that $\cos x$ and $\sin y$ can't be of the same signs, this will reduce some solutions.

So, the ordered pair $(x,y)$ satisfying the above two equations are

$$(\frac\pi 6, -\frac\pi 2)$$

$$(\frac{5\pi}{6}, \frac\pi 2)$$

Title: Solving this system of trigonometric equations

Body:
How would I go about solving:
$$
\begin{cases}
\sin(x) + \cos(y) = \frac{1}{2}\\
\frac{\cos(x)}{\sin(y)} = -\frac{\sqrt{3}}{2}
\end{cases}
$$
Where
$$
x \in \left(\frac{-\pi}{6},\frac{7\pi}{6}\right) \, , \, \, y \in \left(\frac{-2\pi}{3},\frac{2\pi}{3}\right)
$$
I tried taking inverse trigonometric functions but it got super messy so I assume there must be a simpler way to solve this. Can I for example shift $\sin(x)$ by $\pi/2$ or some constant to obtain a cosine function?


问:
Characterization of proper maps using filters
A continuous function $f \colon X \to Y$ is called proper if it is closed and if $f^{-1}(\{y\})$ is compact for every $y \in Y$. I want to prove that a continuous function $f \colon X \to Y$ is proper if and only if for every ultrafilter $\mathcal U$ on $X$ and every point $y \in Y$ such that $f[\mathcal U]$ converges to $y$ (here, $f[\mathcal U]$ is the filter generated by sets of the form $f(U)$, $U \in \mathcal U$), there is an $x \in X$ such that $\mathcal U$ converges to $x$ and $f(x) = y$.

Here is what I have been able to argue so far: Suppose $f$ is proper. Pick an ultrafilter $\mathcal U$ on $X$ and a $y \in Y$ such that $f[\mathcal U]$ converges to $y$. Since $f$ is proper, $A := f^{-1}(\{y\})$ is compact. Since $\mathcal U$ is an ultrafilter, it either contains $A$ or it doesn't. In the form case, $\mathcal U$ converges to some $x \in A$ and we're done. In the latter case, $\mathcal U$ contains some set $B$ such that $A \cap B$ is empty. I do not know how to proceed from here. Any tips? I have not attempted the reverse implication.

Edit: In the argument above, I meant to say that either $A \in \mathcal U$ or $A' := X \setminus A$ is in $\mathcal U$. In this latter case, $f(A')$ is in $f[\mathcal U]$. Since $f[\mathcal U]$ converges to $y$, the closure of every set in $f[\mathcal U]$ contains $y$, which means $y \in \overline{f(A')}$. Since $f$ is closed, $\overline{f(A')}=f(\overline{A'})$. Thus, there is an $x$ in the closure of $A'$ such that $f(x) = y$. Now I just need to argue that $\mathcal U$ converges to $x$. Suppose it does not. Then $\mathcal U$ contains the compliment $U'$ of some open set $U$ containing $x$, which means that $f(U')$ is contained in $f[\mathcal U]$, and since $f(U)$ is in $f[\mathcal U]$, $f(U) \cap f(U') \in f[\mathcal U]$. I'm not sure where to go from here...

答:
Suppose that $\mathscr{U}$ does not converge to any point of $A$. Then each $x\in A$ has an open nbhd $V_x$ such that $V_x\notin\mathscr{U}$. $A$ is compact, so there is a finite $F\subseteq A$ such that $\{V_x:x\in F\}$ covers $A$. Let $V=\bigcup_{x\in F}V_x$; then $V$ is an open nbhd of $A$, and $V\notin\mathscr{U}$.

Since $f$ is closed, $f[X\setminus V]$ is closed in $Y$; let $W=Y\setminus f[X\setminus V]$. Then $W$ is an open nbhd of $y$. And $f[\mathscr{U}]\to y$, so $W\in f[\mathscr{U}]$. Thus, there is some $U\in\mathscr{U}$ such that $f[U]\subseteq W$. And this leads immediately to a contradiction; how?

Added: You didn’t ask about it, but I’ve added a spoiler-protected proof of the converse.

 For the converse, let $f$ be a continuous map with the stated property. Suppose that there is a $y\in Y$ such that $A=f^{-1}[\{y\}]$ is not compact; then there is a non-convergent ultrafilter $\mathscr{U}$ on $X$ such that $A\in\mathscr{U}$. Clearly $y\in f[U]$ for each $U\in\mathscr{U}$, so $f[\mathscr{U}]\to y$ and by hypothesis there is an $x\in A$ such that $\mathscr{U}\to x$, contradicting the choice of $\mathscr{U}$. Thus, $f$ has compact fibres, and it only remains to show that $f$ is closed. $\hspace{10 in}$ Suppose that $F\subseteq X$ is closed, but $A=f[F]$ is not closed in $Y$. Fix $y\in\operatorname{cl}A\setminus A$, and let $\mathscr{A}$ be an ultrafilter on $A$ converging to $y$. Let $\mathscr{F}=\{f^{-1}[S]:S\in\mathscr{A}\}$; $\mathscr{F}$ is a filter base on $F$. Extend $\mathscr{F}$ to an ultrafilter $\mathscr{U}$ on $X$. Let $N$ be a nbhd of $y$ in $Y$. Then $N\cap A\in\mathscr{A}$, $f^{-1}[N]\in\mathscr{F}\subseteq\mathscr{U}$, and $N\supseteq f\big[f^{-1}[N]\big]\in f[\mathscr{U}]$, so $N\in f[\mathscr{U}]$. Thus, $f[\mathscr{U}]\to y$. But $F\in\mathscr{U}$, and $F$ is closed, so any limit of $\mathscr{U}$ must be in $F$ and therefore is not sent to $y$ by $f$. This contradiction shows that $A$ must be closed and hence that $f$ is a closed map.


问:
Find the locus of this point P

Find the locus of all points inside $\triangle ABC$ such that $PA^{2}+PB^{2}=PC^{2}$.

At first, i tried finding a right angled triangle and then tried to go on applying Pythagorean Theorem and finding other trivia but that didn't seem to work. I think the locus is most likely to be some line segment rather than an arc of some circle as that seems to be quite unrelated seen from an Euclidean perspective. Applying Apollonius' Theorem and Stewart's Theorem on some triangles might be the key to this problem although i could not find such triangles. I don't know whether this can be solved by trigonometry or not but since this is a problem from a chapter on euclidean geometry, i am sure that there is some clever way to look at this problem.

答:
Use vectors with $C$ at the origin. Then
$$|\mathbf{PA}|^2+|\mathbf{PB}|^2=|\mathbf{PC}|^2\implies|\mathbf P|^2-2(\mathbf A+\mathbf B)\cdot\mathbf P+|\mathbf A|^2+|\mathbf B|^2=\mathbf 0.$$
Then if $\mathbf D=\mathbf A+\mathbf B$, this rearranges to
$$|\mathbf P-\mathbf D|^2=2\mathbf A\cdot\mathbf B.$$
So as long as $\angle BCA<90^{\circ}$, the locus will be a circle with centre $D$ radius $\sqrt{2ab\cos C}$.

Title: Find the locus of this point P


问:
What does "a function is minorised" mean?
I'm reading a paper about Lasry-Lions regularisation.
In Theorem 1 of the paper, considered function $f$ is assumed to be minorised.

What does it mean?

EDIT: Is it equivalent to "bounded below"?

答: It means that the range of $f$ has a lower bound. In other words, there is some $L\in\Bbb R$ such that$$(\forall x\in X):f(x)\geqslant L.$$


问:
distribution - exercises
Let $f \in \mathcal{C}(\mathbb{R})  \cap L^1(\mathbb{R}).$ We suppose that $f$ is bounded and even $(f(-x)=f(x))$. let $\lambda \in \mathbb{R}.$

We assume that:

$\int_{-1}^1 \dfrac{1 - f(\lambda x)}{x} dx = \lim_{\varepsilon \to 0} \int_{\varepsilon < |x| < 1} \dfrac{1 - f(\lambda x)}{x} dx= 0 $

and forall $\varphi \in \mathcal{D}(\mathbb{R}),$ we have:

$$\int_{-\infty}^{+ \infty} \dfrac{1 - f(\lambda x)}{x} \varphi(x) dx = \int_1^{+ \infty} \dfrac{1 - f(\lambda x)}{x} (\varphi(x) - \varphi(-x)) dx + \int_{-1}^1 \dfrac{1 - f(\lambda x)}{x} (\varphi(x) - \varphi(0)) dx$$

and

$< T_{\lambda} , \varphi > = \int_{\infty}^{+\infty} \dfrac{1 - f(\lambda x)}{x} \varphi(x) dx$ for all $\varphi \in \mathcal{D}(\mathbb{R}.$ We have

$\forall \varphi \in \mathcal{D}(\mathbb{R}): |\int_1^{+ \infty} \dfrac{1 - f(\lambda x)}{x} (\varphi(x) - \varphi(-x)) dx| \leq C N_1(\varphi).$

The question is:

1- prove that

$$\exists C > 0, \forall \varphi \in \mathcal{D}(\mathbb{R}): |<T_{\lambda} , \varphi >| \leq C N_1(\varphi)|.$$

My solution is:

$|<T_{\lambda} , \varphi > | = |\int_{-\infty}^{+ \infty} \dfrac{1 - f(\lambda x)}{x} \varphi(x) dx| = |\int_{1}^{+ \infty} \dfrac{1 - f(\lambda x)}{x} (\varphi(x) - \varphi(-x)) dx| + \int_{-1}^1 \dfrac{1 - f(\lambda x)}{x} (\varphi(x) - \varphi(0)) dx$

$\leq |\int_{1}^{+\infty} \dfrac{1 - f(\lambda x)}{x} (\varphi(x) - \varphi(-x)) dx| + \sup_{[-1,1]} |\varphi(x) - \varphi(0)| \int_{-1}^1 \dfrac{1 - f(\lambda x)}{x} dx|$

$\leq C N_1(\varphi)$

my solution is correct?

EDIT: i have an other question.

I have an other question please.

2- Prove that $$\forall \varphi \in \mathcal{D}(\mathbb{R}), \exists M > 0, \forall n \in \mathbb {N}: |<T_n - vp \dfrac{1}{x} , \varphi >| \leq M \displaystyle\int_{-\infty}^{+\infty} |f(nx)| dx$$ use to this proof that $< vp \dfrac{1}{x},\varphi> = \displaystyle\int_0^{+\infty} \dfrac{\varphi(x) - \varphi(-x)}{x} dx$

my solution is:$|<T_n - vp \dfrac{1}{x},\varphi>| = |\int_{-\infty}^{+\infty} \dfrac{1 - f(\lambda x)}{x} \varphi(x) dx - \int_0^{+\infty} \dfrac{\varphi(x) - \varphi(-x)}{x} dx| = |\int_{-\infty}^{+\infty} \dfrac{1 - f(\lambda x)}{x} \varphi(x) - \int_0^{+\infty} \dfrac{\varphi(x)}{x} dx - \int_{-\infty}^0 \dfrac{\varphi(y)}{y} dx| = |\int_{- \infty}^{+ \infty} - \dfrac{f(\lambda x)}{x} \varphi(x) dx| \leq \int_{-\infty}^{+ \infty} |f(nx)| dx / \sup_{x \in K} \dfrac{|\varphi(x)|}{x} dx$ where $K$ is support of $\varphi,$ then $ |<T_n - vp \dfrac{1}{x},\varphi>| \leq M \int_{-\infty}^{+\infty} |f(nx)|dx$ . My solution is correct? ?

QUESTION 3.

3- Deduce that $T_n \longrightarrow vp \dfrac{1}{x}$ in $S'(\mathbb{R})$ when $n \rightarrow + \infty.$

My solution is,
we have $\forall \varphi \in \mathcal{D}(\mathbb{R}), |< T_n - vp \dfrac{1}{x} , \varphi>| \leq M \int_{-\infty}{+\infty} |f(nx)|dx$

so $f(nx)$ converge almost every where to 0 when $n \rightarrow + \infty$ because $f(nx)$ is bounded and $f \in L^1,$ then by Lebesgue theorem (dominated convergence), $\int_{-\infty}^{+\infty} |f(nx)| dx \rightarrow 0$ when $n \rightarrow + \infty$

then, $<T_n , \varphi > \longrightarrow < vp \dfrac{1}{x},\varphi>$ so $T_n \longrightarrow vp \dfrac{1}{x}$ in $S'.$

QUESTION: (EDIT)

Let $\forall \varphi \in \mathcal{D}(\mathbb{R}): < T_{\lambda},\varphi> = \int_{-\infty}^{+\infty} \dfrac{1 - f(\lambda x)}{x} \varphi(x) dx$ where $\lambda \in \mathbb{R}.$ Prouve that $$\forall \varphi \in \mathcal{D}(\mathbb{R}): |\int_1^{+\infty} \dfrac{1 - f(\lambda x)}{x} (\varphi(x) - \varphi(-x)) dx|\leq C N_1(\varphi)$$

my question is, if we can write this solution; $|\int_1^{+\infty} \dfrac{1 - f(\lambda x)}{x} (\varphi(x) - \varphi(-x)) dx| = \int_K |\dfrac{1 - f(\lambda x)}{x}||\varphi(x) - \varphi(-x)| dx \leq \int_K |\dfrac{1 - f(\lambda x)}{x} 2x |\varphi'(\xi_x)| dx \leq C \sum_{x \in K} |\varphi'(x)| \leq C N_1 (\varphi).$

They are errors in this solution please?

答:
Question 1: You have
$$
|<T_\lambda, \varphi>| \leq CN_1(\varphi) + \int_{-1}^1 \Big|\frac{1-f(\lambda x)}{x}(\varphi(x)-\varphi(0))\Big| dx. $$
By Taylor, $\varphi(x)-\varphi(0) = x\varphi'(\xi_x)$, so the integral becomes
$$
\int_{-1}^1 \Big|1-f(\lambda x)\Big| |\varphi'(\xi_x)| dx \leq \int_{-1}^1 |1-f(\lambda x)| dx \sup_{\xi\in[-1,1]} |\varphi'(\xi)|  \leq C' N_1(\varphi).$$

Question 2: You have (see your proof) - treating $\int_{-\infty}^\infty \frac{\dots}{x} dx$ as $\lim_{\epsilon\to 0} \int_{\mathbb{R}\setminus [-\epsilon,\epsilon]} \frac{\dots}{x} dx$ -
$$ |<T_n - vp\frac{1}{x},\varphi>| \leq \int_{-\infty}^\infty \Big| \frac{f(nx)}{x} \varphi(x)\Big| dx \leq  \int_{-\infty}^\infty \Big|\frac{f(nx)}{x}\varphi(0)\Big| dx + \int_{-\infty}^\infty |f(nx)\varphi'(\xi_x)| dx $$
again using Taylor's $\varphi(x)=\varphi(0)+x\varphi'(\xi_x)$. The first summand is $0$ as
$f$ is even and so you conclude with $M:=N_1(\varphi)$ for all $n$
$$ |<T_n - vp\frac{1}{x},\varphi>| \leq M \int_{-\infty}^\infty |f(nx)| dx.$$

Question 3: Substituting $nx\mapsto t$ gives $$ \int_{-\infty}^\infty |f(nx)| dx = \frac{1}{n} \int_{-\infty}^\infty |f(t)| dt.$$
As $f$ is integrable, you can conclude.

Title: distribution - exercises


问:
Is there always a linear isometry from a normed vector space into its dual?
Let $\mathbb K=\mathbb R$ or $\mathbb K=\mathbb C$ and $E$ be a normed $\mathbb K$-vector space.

Can we show that there is a linear isometry from $E$ into $E'$ or is there a counterexample?

I think this should be true: My idea is that we should be able to consider the linear isometry $\iota_1$ from $E$ into its completion $\tilde E$ and the linear isometry $\iota_2$ from $\tilde E$ into its dual $\tilde E'$ ...

答: This is not true. Take $E=c_0$ and $E'=\ell _1$. Assume there is any linear isometry from $E$ to $E'$, i.e., $T : E\to E'$ linear and continuous, and such that $\|Tx\|_{\ell^1} = \|x\|_{c_0}$. Let $x_n \rightharpoonup x$ in $E$, then $Tx_n \rightharpoonup Tx$ in $\ell^1$. By the Schur property of $l^1$, $\|Tx_n -Tx\|_{\ell^1}\to 0 $. Due to the isometry property, $x_n \to x$ in $c_0$. Hence every weakly converging sequence in $c_0$ is strongly converging, which is absurd (take the sequence of unit vectors).


问:
Minimizing costs to make a cylinder
A cylinder must have a volume $V=4000$ and its made from just one rectangle sheet, so its top and bottom must be cut from this sheet.

I have to find the dimensions of this sheet such that the parts I lost (when I cutoff the top and the bottom) are the lowest.

I thought about minimize the difference between the area of the original sheet and the parts I'll use to make my cylinder:

(I imagine this is the best way to don't waste sheet - the original problem doesn't have any image).

So the area of the original sheet is
$$(h+2R)\cdot4R=4hR+8R^2$$
and the area I'll use to make the cylinder is
$$4hR+2\cdot\pi R^2$$
so the difference between them is
$$8R^2-2\pi R^2=0$$
and derivating it:
$$16R-4\pi R$$
which the only root is $R=0$. What is wrong?

答:
This is an equality constrained optimization problem, i.e.

\begin{equation}
\text{minimize} \quad 4hR + 8R^2 \quad \text{subject to} \quad \pi R^2 h = 4000,
\end{equation}

which in turn can be turned in to

\begin{equation}
\text{minimize} \quad f(R) = 4 \frac{4000}{\pi R^2} R + 8R^2 =  \frac{16000}{\pi R} + 8R^2 .
\end{equation}

This problem can be solved by setting $f'(R) = 0$ and solve for $R$, which yields $R^* = \left(\frac{1000}{\pi}\right)^{1/3}$. You can verify that it is indeed a minimum with $f''(R^*) > 0$.

Title: Minimizing costs to make a cylinder


问:
Linear Algebra - Find the two numbers
The sum of two numbers is $19,712$.

One of the two numbers is divisible by $10$.

If the units digit of the number is erased, the other number is obtained.

What is the product of the two numbers?

Title: Linear Algebra - Find the two numbers

Body:
Here we can see

$10x+x = 19712.$

$11x = 19712.$
$x = 1792.$

$1792 + 17920 = 19712.$

Ans: $1792 \times 17920 = 32112640.$


问:
Is the space of all constant functions closed in $C^1[0,1]$ with respect to $L^2$-norm?

Let $C$ be the space of all constant polynomials over $\Bbb R.$ Is $C$ closed in $C^1[0,1]$ with respect to $L^2$-norm?

Let $f \in C^1[0,1]$ be a limit point of $C.$ We need to show that $$\displaystyle\inf \left \{\int_{0}^{1} \left (f(x) - c \right )^2\ dx\ \bigg |\ c \in \Bbb R \right \} = 0 \implies f \in C.$$ Let us define a function $g : \Bbb R \longrightarrow \Bbb R$ by $$g(c) = \int_{0}^{1} \left (f(x) - c \right )^2\ dx,\ c \in \Bbb R.$$ Then $g$ is differentiable on $\Bbb R$ and by Leibnitz integral rule we have $$g'(c) = \int_{0}^{1} -2 \left (f(x) - c \right )\ dx.$$ Now if the infimum of $g$ is attained at $c_0 \in \Bbb R$ (the infimum has to be attained at some real number because if $c \to \pm \infty$ then $g(c)$ blows up to $+ \infty$) then $g'(c_0) = 0 \implies \displaystyle \int_{0}^{1} f(x)\ dx = c_0.$ Now if we compute $g(c_0)$ we find that $$g(c_0) = \int_{0}^{1} \left (f(x) \right )^2\ dx - \left ( \int_{0}^{1} f(x)\ dx \right )^2.$$ So by Cauchy-Schwarz inequality we have $$g(c_0) \geq 0\ \ \ \ (\because \left \lvert \left \langle f, 1 \right \rangle \right \rvert \leq \left \|f \right \|_2 \left \|1 \right \|_2 ).$$ So $g(c_0) = 0 \iff f$ is a scalar multiple of $1$ i.e. $f \in C.$ But this proves that $C$ is closed in $C^1[0,1].$

Is my above reasoning correct at all? Would anybody please verify it?

Thanks for your time.

答:
Seems ok to me now. However, it could have been much shorter: $g(c) \ge 0, \inf_{c \in \mathbb R}g(c) = 0$ implies that there is $c_0 \in \mathbb R$ such that

$$g(c_0) = 0 = \int_0^1 (f(x) - c_0)^2 \quad \Rightarrow \quad f(x) = c_0 ~~\text{a.e.}$$


问:
Derivative of multivariable function $f: \mathbb{R}^n\setminus\{0\}\to \mathbb{R}^n, f(x)=g(\|x\|)x$?

Derivative of multivariable function $f: \mathbb{R}^n\setminus\{0\}\to
 \mathbb{R}^n, f(x)=g(\|x\|)x$

My attempt: I used the product and chain rule.

$$J_f(x)=x\cdot g'(\|x\|)\cdot \frac{1}{\|x\|}\cdot x+g(r)I_n$$

The solution says it's $J_f(x)=g'(r)\frac{1}{\|x\|}xx^T+g(r)I_n$

why do we have $xx^T$? Why the transpose?

答:
You need to be careful when you write the dot where you really mean matrix multiplications. If $x$ is a column vector, $xx^T$ is an $n\times n$ matrix; but $x^Tx$ is not. In any case, $xx$ does not make sense and $x\cdot x$ would be a scalar if you do inner product.

You may work on the case when $n=2$ and write everything in terms of components to see what happens.


问:
Prove that the composition is homeomorphic to a disjoint union
How to prove that X × S is homeomorphic to the disjoint union of |S| copies of the space X, if X - some topological space, and let S - be a space with a discrete topology?

P.S I know that a subset of a disjoint union is called open if its intersection with each of the components is open. And also I actually know the definition of disjoint union. But still have no idea how to prove it based on this.

答:
Recap: The disjoint sum $\oplus_{s \in S} X_s$ of $|S|$ many copies of $X$, is the set of all pairs $\{(x,s): x \in X, s \in S\}$, where the $s$ coordinate serves to make the copies of $X$, $X \times \{s\}$, pairwise disjoint.

Note that as a set this is just $X \times S$. Denote its two projections as $\pi_S$ and $\pi_X$ resp.

The disjoint sum topology $\mathcal{T}_s$ is defined as

$$\{O \subseteq X \times S\mid \forall s \in S: \pi_X[(X \times \{s\}) \cap O] \text{ open in } X\}$$

while the product topology $\mathcal{T}_p$ is defined as the smallest topology such that $\pi_S$ and $\pi_X$ are continuous and which has as a base $$\mathcal{B}=\{O \times O'\mid O \text{ open in } X, O' \text{ open in } S\}$$

I'll now show that these topologies on $X \times S$ are actually identical.

Note that if $O$ is open in $X$, then $\pi^{-1}[O] = O \times S$ is open in $\mathcal{T}_s$, as for all $s$, $$\pi_X[(X \times \{s\}) \cap (O \times S)] = \pi_X[O \times \{s\}]= O$$

and if $O'$ is open in $S$, then $\pi_S^{-1}[O'] = X \times O'$ is open in $\mathcal{T}_s$ as

$$\forall s \in S: \pi_X[(X \times \{s\}) \cap (X \times O')]=\begin{cases}
X & s \in O'\\
\emptyset & s \notin O'
\end{cases}$$

and both sets are open in $X$. So both projections are open and hence $\mathcal{T}_p \subseteq \mathcal{T}_s$ by minimality.

Conversely, if $O \in \mathcal{T}_s$ and $(x,s) \in O$ then we know by definition
$O' := \pi_X[(X \times \{s\}) \cap O]$ is open in $X$ and $$(x,s) \in O' \times \{s\} \subseteq O$$

and $O' \times \{s\} \in \mathcal{B}$ (using that $\{s\}$ is open in $S$) and so $(x,s)$ is an interior point of $O$ for $\mathcal{T}_p$, and as $(x,s)$ was arbitrary, $O \in \mathcal{T}_p$ and so $\mathcal{T}_p \subseteq \mathcal{T}_s$ and we have equality of topologies.

Hence $X \times S \simeq \oplus_{s \in x} X_s$.


问:
Inverse gamma sum convergence
I have been working on the following problem:

Assume $X_k$ is a random variable that follows a $\Gamma(3,k)$ distribution. Then

$$\sum_{i=1}^n \frac{1}{X_i}-\frac{1}{2} \log n\overset{p}{\to}0$$

Well, I know that the random variable $Y_k=\frac{1}{X_k}$ follows an inverse gamma distribution,and clearly $-\frac{1}{2}\log (n)=\log(n^{-1/2})$ so I am guessing  $\sum_{i=1}^n \frac{1}{X_i}$ can be written in terms of logarithm in some way....

I am wondering if anyone has any other ideas.

答:
This does not converge to 0 in probability, it converges to $\gamma/2$, where $\gamma$ is the Euler-Mascheroni constant. To see this, note that the expectation of $X_i$ is $1/2i$ for each $i$, and that convergence in expectation implies convergence in probability. We have
$$ \mathbb{E}\left[\sum_{i = 1}^n \frac{1}{X_i} - \frac{1}{2}\log(n)\right] = \mathbb{E}\left[ \sum_{i = 1}^n \frac{1}{X_i} - \sum_{i = 1}^n \frac{1}{2i} + \sum_{i = 1}^n \frac{1}{2i} - \frac{1}{2}\log(n)\right] = \frac{1}{2}\left(\sum_{i = 1}^n \frac{1}{i} - \log(n)\right) \to \frac{\gamma}{2}. $$
The last convergence is the definition of the Euler-Mascheroni constant. From this we get that
$$ \sum_{i = 1}^n \frac{1}{X_i} - \frac{1}{2}\log(n) \to \frac{\gamma}{2} \text{ in }\mathbb{P}. $$

Title: Inverse gamma sum convergence


问:
If $f(z)$ is differentiable, show that $|f'(z)|^2=(\frac{\partial u}{\partial x})^2+(\frac{\partial v}{\partial x})^2$.
Let $f(z)$ be a complex valued function which can be represented by $$f(z)=u(x,y) + iv(x,y).$$

If $f(z)$ is differentiable, show that $$\left|f'(z)\right|^2=\left(\frac{\partial u}{\partial x}\right)^2+\left(\frac{\partial v}{\partial x}\right)^2.$$

I am a beginner in complex analysis, so bear with me. In my solution, it is given that
$$f'(z)=u_x + iv_x=\frac{\partial u}{\partial x} + i \frac{\partial v}{\partial x}$$
and things follow from here.
My doubt is, how does the above came into picture. As far as i know, $f(z)$ is a complex valued function with real part as $u(x,y)$ and imaginary part being $v(x,y)$ which are both functions of $x$ and $y$.
How does one calculate the derivative of $f(z)$ by taking partial derivative with respect to $x$ only?

Any help would be appreciated from the community.

答: Because, if $z=a+bi$,\begin{align}f'(z)&=\lim_{h\to0}\frac{f(z+h)-f(z)}h\\&=\lim_{h\to0,\ h\in\Bbb R}\frac{f(z+h)-f(z)}h\\&=\lim_{h\to0,\ h\in\Bbb R}\frac{u(a+h,b)-u(a,b)+v(a+h,b)i-v(a,b)i}h\\&=\lim_{h\to0,\ h\in\Bbb R}\frac{u(a+h,b)-u(a,b)}h+\lim_{h\to0,\ h\in\Bbb R}\frac{v(a+h,b)-v(a,b)}hi\\&=u_x(a,b)+v_x(a,b)i.\end{align}


问:
Using norms to prove no zero-divisors exist
I was reading a post on a specific problem and there was an equation $(a+b\sqrt{k})(c+d\sqrt{k})\equiv0 \mod p$ where $a,b,c,d,k$ are integers in the integers $\mod p$, $p$ is prime. Supposedly this equation is equivalent to $(a+b\sqrt{k})(a-b\sqrt{k})(c+d\sqrt{k})(c-d\sqrt{k})\equiv 0 \mod p$, why is this true, why can we multiply the first two factors by their conjugate in the first equation, and have the same equation?

答:
It's closely connected to your prior question. If $\,k\,$ is not a square $\bmod p\,$ then a null norm yields

$$ 0\equiv \bar w w\equiv (a-b\sqrt k)(a+b\sqrt k ) \equiv a^2-kb^2\Rightarrow k\equiv (a/b)^2,\ \text{contra hypothesis}\qquad$$

when $\,b\not\equiv 0\ $ (and if $\,b\equiv 0\,$ then $\,w\bar w\equiv a^2\equiv 0\iff 0\equiv a\equiv w,\,$ by $\,p\,$ prime $\Rightarrow \Bbb Z_p\,$ a field.

Thus norms $\bar w w\,$ of nonzero $w$ are nonzero so invertible (being in a field $\,\Bbb Z_p)\,$ so its factor $w$ is also invertible (explicitly $\, \dfrac{1}{w} \equiv \dfrac{\bar w}{w\bar w}$ by rationalizing the denominator  (well-defined by $\bar w w\not\equiv 0)$

Hence $\,w\equiv 0\iff N(w) = \bar w w\equiv 0.\,$ Applied to a product using norm multiplicativity we get

$$wv\equiv 0 \iff 0\equiv N(wv) \equiv N(w)N(v) \iff N(w)\equiv 0\ \, {\rm or}\ \, N(v)\equiv 0\qquad$$

which is the sought equivalence.

Remark $ $ The idea generalizes to higher degree algebraic extensions. by using norms, minimal polynomials or the Euclidean algorithm to compute the inverse in a similar way (the linked answer also explains how to view this as a special case of the method of simpler multiples).

The standard way to construct this ring is as a quotient $\,R = \Bbb Z_p[\sqrt k] := \Bbb Z[x]/(x^2-k)$ or, equivalently, by congruences, i.e. $\Bbb Z[x]\bmod (p,\,x^2-k)$.  By here or here the quotient ring is a field $\!\iff\! f(x) = x^2-k\,$ is irreducible or $\,\Bbb Z_p\!\iff\! f\,$ has no root in $\,\Bbb Z_p\!\iff\! k\,$ is nomsquare in $\,\Bbb Z_p.\,$

On the other hand, if $\,k\,$ is a square, say  $\,0\not\equiv k\equiv j^2,\,$ and $\,p\neq 2\,$ then by CRT  we have$\,R = \Bbb Z_p[x]/(x-j)(x+j)\cong \Bbb Z_p[x]/(x-j)\times \Bbb Z_p[x]/(x+j)\cong \Bbb Z_p^2,\,$ and such a  product of rings is never a field since it has idempotent zero-divisors $(1,0)(0,1)=(0,0);\,$ equivalently $\,(x-j)(x+j)\equiv 0\,$ yields zero-divisors in $R.\,$ Each component of the product may be viewed as a "possible world" where the sqrt is the negative of the value in the other world, e.g. if $k=1$ then in effect we create a ring by adjoining $\,x = \pm1$. Thus the pairs in the product ring keep track of all the possible values arising from both possible "signs" (branches) of the sqrt, i.e. the arithmetic in the product ring does the computation in parallel in both possible worlds - one world where $\,x=1\,$ and another where $\,x = -1$.

And, of course, if $\,k\equiv 0\,$ then $\,R = \Bbb Z_p[x]/(x^2)$ is also not a field since here $\,x\,$ is zero-divisor - being nilpotent $\,x^2\equiv 0.\,$ This is known as the ring of dual numbers (it proves handy for constructing algebraic models of derivatives and tangent spaces).


问:
Show every open set in standard topology is open in lower limit topology.
Let $\mathbb{R}_{std}$ be the standard topology on $\mathbb{R}$. Let $\mathbb{R}_{LL}$ be the lower limit topology, whose basis elements consists of all intervals of the form $[a,b)$. Prove that every open set in $\mathbb{R}_{std}$ is open in $\mathbb{R}_{LL}$, but not every open set in $\mathbb{R}_{LL}$ is open in $\mathbb{R}_{std}$.

Attempt: For any open set $U$ in $\mathbb{R}_{std}$ for each $x  \in U$, there is an open interval $(a,b)$ containing $x$. Then $[x,b)$ is in $\mathbb{R}_{LL}$ and $[x,b) \subset (a,b) \subset U$, so $U$ is open in $\mathbb{R}_{LL}$.

In $\mathbb{R}_{LL}$, however for any  $[x,b)$ there is no open interval $(c,d)$ containing $x$ with $(c,d) \subset [x,b)$ thus not every open set in $\mathbb{R}_{LL}$ is open in $\mathbb{R}_{std}$.

答:
Yes, this is the right way. The way you saw is also correct: if basic Euclidean sets (like open intervals) are all LL-open, then so are all Euclidean open sets. This is perfectly rigorous.

But the final argument why $[x,b)$ is not Euclidean open can be slightly expanded: suppose $x \in (c,d) \subseteq [x,b)$ then we can pick $r$ with $c < r < x$ (we know that $c < x < d$ after all, and $\Bbb R$ is a dense order) but then the inclusion tells us that $r \in [x,b)$ so $r \ge x$ while $r < x$, contradiction.

It's clear from the "picture", but a proof shouldn't rely on that only.


问:
How to find the integers $a$ and $b$?
The question says:

Find a and b integers where:

$671a-654b = 18$ and $\gcd(a,b) = 18$

My attempt was:

$a = cd$ where $d$ is $\gcd$ and $c$ is just an integer.

$b = kd$ where $k$ is just an integer too.

$k$ and $c$ are co-prime, btw (This is all according to a law in math, I don't know its name though)

By putting them in the equation we find:

$d(671c - 654k) = 18$

which is

$671c - 654k = 1$

By using the euclidean algorithm, I only found this which is not correct:

$76(671) - 78(654) = -16$

-$16$ should be $1$ in its place but it isn't, so it's not right, either.

So how do I continue from here?

EDIT: I redid my algorithm and I found that $c = 77$ and $k = -79$ which after putting it in a and b, I find their values. Thank you all!

答:
Hint:

Use first the extended Euclidean algorithm to find $u$ and $v$ such that
$$671u-654v=1$$
since $671$ and $654$ are coprime, then multiply $u$ and $v$ by $18$ to obtain $a$ and $b$.

Added. The algorithm:

\begin{array}{rrrl}
r_k & u_k & v_k & q_k \\ \hline
671 & 1 & 0 \\
654 & 0 & 1 & 1 \\ \hline
17 & 1 & -1 & 38 \\
8 & -38 & 39 & 2 \\
1 & \color{red}{77} & \color{red}{-79} \\ \hline
\end{array}

Title: How to find the integers $a$ and $b$?


问:
Prove that $ta+(1-t)b \in I$ for all $t\in[0,1]$ without using the fact that an interval is convex?
Let $a,b \in \mathbb{R}$ and $I=[\min\{a,b\},\max\{a,b\}]$.

Is it possible to give a simple proof that $ta+(1-t)b \in I$ for all $t\in[0,1]$ without using the fact that an interval is convex?

Body:
For $t \in (0,1)$ and $a<b$,
$$a  = (1 + t-t) a = ta + (1 - t)a < ta + (1 - t)b < tb + (1 - t)b  = b$$
and we conclude.


答: Pitt's theorem asserts that every bounded linear operator from $\ell_p$  into $\ell_q$ is compact whenever $1 ≤ q<p< ∞$.  Thus the answer is also  negative for $\ell_p$, for every finite $p>2$.

Title: Is there always a linear isometry from a normed vector space into its dual?

Body:
Let $\mathbb K=\mathbb R$ or $\mathbb K=\mathbb C$ and $E$ be a normed $\mathbb K$-vector space.

Can we show that there is a linear isometry from $E$ into $E'$ or is there a counterexample?

I think this should be true: My idea is that we should be able to consider the linear isometry $\iota_1$ from $E$ into its completion $\tilde E$ and the linear isometry $\iota_2$ from $\tilde E$ into its dual $\tilde E'$ ...


问:
Symplifying Inequality involving Binomials
Is there any easy way of simplifying the following inequality

$$ (a+\frac{\Delta}{2})^{\gamma} -  (b-\frac{\Delta}{2})^{\gamma} > a^{\gamma} - b^{\gamma} $$

if $\gamma \in \mathbb{R}^{+}$ and $a>b>\Delta$.

Another option might be trying to find a bound for how large this difference can be.

Thanks!

答:
Write the question as follows, and obtain the consequent equivalent formulations:
$$
(a+\frac{\Delta}{2})^{\gamma} -  a^{\gamma} >  (b-\frac{\Delta}{2})^{\gamma}- b^{\gamma}\\
 \int_0^\Delta \frac{\gamma}{2}(a+\frac{t}{2})^{\gamma-1} \, dt>  - \int_0^\Delta \frac{\gamma}{2}(b-\frac{t}{2})^{\gamma-1} \, dt\\
\frac{\gamma}{2} \int_0^\Delta (a+\frac{t}{2})^{\gamma-1} + (b-\frac{t}{2})^{\gamma-1}  \, dt>  0
$$
Now since the integrand is positive, this proves the claim.$\qquad \Box$


问:
The reflective property of ellipses
I have following question to proof:

An ellipse is revolved about its major axis to generate an ellipsoid. The inner surface of the ellipsoid is silvered to make a mirror. Show that a ray of light emanating from one focus will be reflected to the other focus. Sound waves also follow such paths, and this property is used in constructing "whispering galleries." (Hint: Place the ellipse in the standard position in the xy-plane and show that the lines from a point P on the ellipse to the two foci make congruent angles with the tangent to the ellipse at P.)

And I don't know how to prove the hint. How to prove the hint? Thanks.

答:
Consider the following ellipse with focus $ S_1,S_2.$

Let P($a $ $cos θ $ , $b $ $sin θ $) be any point on the ellipse.

We know that $$PS_1=a(1+ecos θ )$$  $$  PS_2=a(1-ecos θ )$$.

Let the normal through $P$ meet the x axis at $Q$.

Then the  equation of the normal $PQ$ is $$\frac{ax}{cos θ}-\frac{by}{sin θ}=a^2-b^2$$

Hence the coordinate of Q is $(\frac{(a^2-b^2)cos θ}{a},0)$.

Findingthe lenth of $S_1Q$ and $S_2Q$we get $$S_1Q=ae+ae^2cos θ $$   $$S_2Q=ae-ae^2cos θ $$.

Now we get  $$\frac{S_1Q}{S_2Q}=\frac{ae+ae^2cos θ}{ae-ae^2cos θ}=\frac{a(1+ecos θ )}{a(1-ecos θ )}=\frac{PS_1}{PS_2}$$
$$⇒\frac{S_1Q}{S_2Q}=\frac{PS_1}{PS_2}$$
Hence by converse of angle bisector theorem the normal $PQ$ is the angle bisector of $∠P$.

So a ray of light emanating from one focus will be reflected to the other focus.

Title: The reflective property of ellipses


问:
Need help evaluating this infinite sum involving a sine function and squared denominator
I need major help in evaluating the following infinite sum:
$$\sum_{k=1}^{\infty} \frac{k\sin(kx)}{(k^2+a^2)^2}\tag{1}\label{whatiwant}$$

where a is a constant. I know that (from Gradshteĭn et al, 2014, page 47)
$$\sum_{k=1}^{\infty} \frac{k\sin(kx)}{(k^2+a^2)}=(\pi/2)\frac{\sinh(a(\pi-x))}{\sinh(a\pi)}\tag{2}\label{whatihave}$$
so that \eqref{whatiwant} is simply the left-hand side of \eqref{whatihave} with its denominator squared, but I am extremely confused as to how to proceed.

答:
I'm afraid there is no too easy way to evaluate this sum. The regular approach to evaluate sums like $\sum_{k=-\infty}^{k=+\infty}f(k)$ is to apply integration in the complex plane with a special factor, which residuals give the required sum.

In our case $\sum_{k=1}^{\infty} \frac{k\sin(kx)}{(k^2+a^2)^2}=-\frac{1}{2a}\frac{\partial}{\partial{a}}\sum_{k=1}^{\infty} \frac{k\sin(kx)}{(k^2+a^2)}$, so we just have to evaluate $S(a,x)=2\sum_{k=1}^{k=\infty} \frac{k\sin(kx)}{(k^2+a^2)}=\sum_{k=-\infty}^{k=\infty}\frac{k\sin(kx)}{(k^2+a^2)}=\Im(\sum_{k=-\infty}^{k=\infty}\frac{k\exp(ikx)}{(k^2+a^2)})=\Im(\hat{S})$

Let's now consider $F(t)=\frac{t\exp(itx)}{t^2+a^2}\pi\frac{\exp(-i\pi{t})}{\sin(\pi{t})}$,
where the factor  $\pi\frac{\exp(-i\pi{t})}{\sin(\pi{t})}=2\pi{i}\frac{1}{\exp(2\pi{i}t)-1}$ has simple poles at integers $t$ ($t=0$, $-1$, $+1$, ...) with the residuals equal to $1$. The factor $\frac{t\exp(itx)}{t^2+a^2}$ in turn has two poles at $t=-ia$ and $t=ia$.

Let's consider the integral of $F(t)$ along a big circle of the radius $R$: $$I(a,x)=\oint\frac{t\exp(itx)}{t^2+a^2}\frac{2\pi{i}}{\exp(2\pi{i}t)-1}dt$$

According to the residual theorem  $I(a,x)=2\pi{i}\sum_{\text{all poles}}{Res}(\frac{t\exp(itx)}{t^2+a^2}\frac{2\pi{i}}{\exp(2\pi{i}t)-1})=2\pi{i}\left(\sum_{k=-\infty}^{k=\infty}\frac{k\exp(ikx)}{k^2+a^2}+\sum_{t=+-ia}{Res}\frac{t\exp(itx)}{t^2+a^2}\frac{2\pi{i}}{\exp(2\pi{i}t)-1}\right)=2\pi{i}\left(\hat{S}(a,x)+2\pi{i}[\frac{ia\exp(-ax)}{2ia(\exp(-2\pi{a})-1)}+\frac{-ia\exp(ax)}{-2ia(\exp(2\pi{a})-1)}]\right)$  $$I(a,x)=2\pi{i}\left(\hat{S}(a,x)+\pi{i}\frac{\sinh(a(x-\pi))}{\sinh(a\pi)}\right)$$

On the other hand, if $0<x<2\pi$ at $R\to\infty$ in the lower half-plane the integrand has a factor $\sim\frac{e^{itx}}{e^{2\pi{i}t}}\sim\exp(it(x-2\pi))$, and in the upper half-plane the factor becomes $\sim\exp(itx)$. Also, $\frac{t}{t^2+a^2}\to0$ at $|t|\to\infty$. So, at $0<x<2\pi$ Jordan's lemma is applicable in the whole complex plane, and the integrale along a big circle $\to0$ as soon as $R\to0$

$I(a,x)=0$ $\Rightarrow$  $\hat{S}(a,x)=-\pi{i}\frac{\sinh(a(x-\pi))}{\sinh(a\pi)}$  $\Rightarrow$ $$\sum_{k=1}^{\infty} \frac{k\sin(kx)}{k^2+a^2}= \frac{\pi}{2}\frac{\sinh(a(\pi-x))}{\sinh(a\pi)}, x\in(0, 2\pi)$$
$$\sum_{k=1}^{\infty} \frac{k\sin(kx)}{(k^2+a^2)^2}= -\frac{\pi}{2a}\frac{\partial}{\partial{a}}\left(\frac{\sinh(a(\pi-x))}{\sinh(a\pi)}\right)$$


问:
Proof of the equality of the difference of two sets iff sets are equal (direct vs. indirect)
I have a problem with the following (really) basic result:

$$A\backslash B=B\backslash A \Longleftrightarrow A=B$$

More specifically, I am able to prove it only by contradiction (in particular in the necessary condition - for the sufficient condition I get a contradiction following the steps of a direct proof).

So my problem is, can we actually prove it without using contradiction?

Moreover (and this is probably a very dumb question, so I apologize), is there a general way to know if a given result that we know is true can be proven only by contradiction or there exist a direct proof of it?

Looking forward to any feedback!

答:
Here is a direct proof:

Take any element $a\in A$.  We know that $a\notin B\setminus A=A\setminus B$, so $a\in B$.  It follows that $A\subset B$.  The reverse containment is similar.


问:
Generalized spectral radius and matrix norm
I am reading a paper that discusses the design of an approximate matrix in the context of numerical methods for PDEs. However, I do not understand the following step, which I believe should be independent of all the context:

We are trying to design a matrix $A$ such that its norm (when seen as an operator $A:V\rightarrow W$, where V and W are ordinary column vector spaces) is equal to that of another matrix $B$ that belongs to the same space. The inner product to be considered in $V$ is defined with the help of a positive-definite matrix $M$ as

$$
(v, w) = v^T M^{-1} w  \tag{1}
$$

where $v,w \in V$ and where its corresponding norm is denoted as $\left | \cdot \right |_{M^{-1}}$. The norm of any operator $C:V\rightarrow W$ is then expressed as

$$
\left \| C \right \| = \sup_{\left | x \right |_{M^{-1}} = 1}{x^T C x}  \label{norm} \tag{2}
$$

Closely paraphrasing the paper, a practical way to impose $\left \| A \right \| = \left \| B \right \|$ is to compute the spectrum of matrices $A$ and $B$, with respect to matrix $M^{-1}$ and impose that at least the spectral radius be the same. According to the explanations given in the same paper, the referred spectral radius (of an operator $C$) is to be taken as the largest of the absolute values of the solutions for $\lambda$ of the generalized eigenvalue problem $CU = \lambda M^{-1} U$.

Could someone help me break down the logic behind this equivalence? That is, why does the equality of the $M^{-1}$-spectral radii of $A$ and $B$ imply $\left \| A \right \| = \left \| B \right \|$?

EDIT:
I have realized that the matrices that the paper is applying this equivalence to are symmetric, so I am not sure (although I suspect so) if the equivalence is intended to be true only in this case, since it is not explicitly mentioned in the abstract discussion. For this case, I have found an explanation which I will post as an answer.

答:
I have found a way to explain this equivalence for the case of both $A$ and $B$ symmetric, based on this material.

Note that the norm above can be expressed as

$$
\left \| C \right \| = \sup_{x \neq 0}{R(x)}  \tag{2}
$$

where the generalized Rayleigh coefficient, $R(x)$, is defined as
$$
R(x) := \frac{x^T C x}{x^T M^{-1} x}
$$
The maximum of $R$ is attained at a stationary point of R (sum is assumed over repeated indices):
$$
\frac{\partial }{\partial x_m} \left ( \frac{x_i C_{ij} x_j}{x_k M^{-1}_{kl} x_l} \right ) = 0 \Rightarrow \left ( \delta_{im} C_{ij} x_j +  x_i C_{ij} \delta_{jm}\right ) x_k M^{-1}_{kl} x_l - x_i C_{ij} x_j \left ( \delta_{km}  M^{-1}_{kl} x_l - x_k M^{-1}_{kl} \delta_{lm}\right ) \\
= 2C_{jm} x_k M^{-1}_{kl} x_l - 2 x_i C_{ij} x_j M^{-1}_{ml} x_l = 0 \Rightarrow (x^T M^{-1} x) C x = (x^T C x) M^{-1}x
$$
where $\delta_{ij}$ is the Kronecker delta. Note that we have used the symmetry of $A$ and $B$. That is,
$$
C x = R(x) M^{-1}x
$$
So that the maximum is attained at an eigenvalue of the generalized problem. In particular, the maximum eigenvalue will yield the greatest value of $R$. Thus, imposing that the (generalized) spectral radius of the two matrices coincides will force their norms to be equal too.


问:
For every $p$-subgroup $Q$ there exists a $p$-Sylow $P$, such that $Q \leq P$
I am trying to prove a consequence of Sylow's theorems.
Let $G$ be a group with $|G|=p^km$, where $p$ is a prime and $k, m$ are integers such that $p$ does not divide $m$. Let $Q$ be a $p$-subgroup of $G$, that is $|Q|=p^i$, for some $i \leq k$.

Then there esists a $p$-Sylow $P$ such that $Q \leq P$.

Here's my attempt:

Let $\Omega$ be set of $p$-Sylow of G. Let us consider the action $\Omega \times Q \to \Omega$, which sends the pair $(K,g)$ to $K^g=\{g^{-1}kg|k\in K\}$.

Since $Q$ is a $p$-group, we have that $|Fix(Q)|\equiv |\Omega|\pmod p$. Thanks to Sylow's theorems, we know that $|\Omega|\equiv 1\pmod p$, then $|Fix(Q)|\equiv 1\pmod p$.

This means that $Fix(Q) \neq \emptyset$. Let $P$ be a fixed point of $Q$.

We have that $P^x=P$, for every $x \in Q$.

Which means that $Q \leq N_G(P)$, where $N_G(P)$ is the normalizer of $P$ in $G$. It is true then that the set $PQ$ is a subgroup of $G$.

We must have that $|PQ|=\cfrac{|P||Q|}{|P \cap Q|}=\cfrac{p^kp^i}{|P \cap Q|}$.

Here, $|P|=p^k$, because $P$ is in $\Omega$.

To conclude the proof I would need to show that $|PQ|=p^k$, so that $|P \cap Q|=p^i$ and, since $P \cap Q$ is a subgroup of $Q$, we would have $Q=P \cap Q$, that is $Q \leq P$, but that is where I am struggling, what do you think?

答:
$P$ is normal and Sylow in $N_G(P)$. If $Q \subseteq N_G(P)$, then the subgroup $PQ \subseteq N_G(P)$, and as you showed, $PQ$ is a $p$-group containing $P$.
But $P$ is Sylow, so a maximal $p$-subgroup, implying $PQ=P$, whence $Q \subseteq P$.


问:
How many meals to choose if I can skip some or all courses?
This question was asked in class before our professor disconnected online, trying to take a crack at it myself and would like a push in the right direction.

"A restaurant has $5$ appetizers, $6$ entrees, $3$ salads, and $7$ deserts. How many ways to choose a complete meal? How many ways to choose if I can skip some or all courses?"

Initially, I used the product rule to answer the first part: |$A$|= $5 * 6 * 3 * 7 = 630$

My thought process going into the second part lies along that since I can skip some or all courses, I can take the powerset of the courses, multiply the courses, then add the sums for my final answer.

For example (zero for the empty set): $P(A)$ = $0 + 5 + 6 + 3 + 7 + (5 * 6) + (5 * 3) + (5 * 7) ... $

We just started combinatorics today, so I am not sure where to head. Any advice would be appreciated.

答: HINT Treat skipping a course as another dish.</p>


问:
$div(t_{\nabla u})=\nabla div u$? In Navier Stoke Equation
How to, in general prove, that:

div($T_{\nabla u})=\nabla$div$u$?

$T$:the transpose.

$u$: is a function s.t $(x,t) \rightarrow u(x,t):=x'(t)=(x_1'(t),...,x_n'(t))$

is the velocity field of the fluid.

$\nabla$ :nabla operator.

div: divergence.

In a reference, I have seen that:

1)Tr(Du)=Tr∇u = divu ,
2)div(t∇u)=∇(divu)
3)div(2Du) = div(∇u +t∇u)=Δu +∇(divu)
4)div((div u)Id) = ∇(divu). ,Tr:trace,t:transpose.

And point 2 is my question(while divergenve gives a scalar, how might we apply a gradient on it! I am new to Navier Stokes equation lesson). And I need some help pls, or clarification.

答:
If $A = (a_{ij})_{ij}$ then $\operatorname{div}(A)_i$ (that is, the $i$-th component of the vector $\operatorname{div}(A)$) is given by $$\sum_{j}\partial_j a_{ij}.$$

If $A = (\nabla u)^t$, that is, if $A$ is the transpose of $\nabla u$, then $a_{ij} = \partial_i u_j$ and therefore $$\operatorname{div}((\nabla u)^t)_i = \sum_{j}\partial_j\partial_iu_j.$$
On the other hand, since $\operatorname{div}u = \sum_j \partial_ju_j$, the $i$-th component of the vector $\nabla \operatorname{div}(u)$ is precisely $$\partial_i \operatorname{div}(u) = \partial_i \sum_j \partial_j u_j = \sum_j \partial_j\partial_i u_j.$$
Notice that in the very last step we have used Schwartz's theorem to swap the other of the derivatives. Hence these computations are rigorous if $u$ is sufficiently regular. Since you mention Navier-Stokes, chances are that you are working with different classes of functions. However, I hope this clarifies things a little bit!


答:
No, for the second part, let $0$ designate that you are skipping that course.  So, instead of $a \times b \times c \times ...$ you have $(a+1) \times (b+1) \times (c+1) \times ...$

Actually unclear if this is equivalent to what you came up with.  I didn't read your query that closely.

Title: How many meals to choose if I can skip some or all courses?

Body:
This question was asked in class before our professor disconnected online, trying to take a crack at it myself and would like a push in the right direction.

"A restaurant has $5$ appetizers, $6$ entrees, $3$ salads, and $7$ deserts. How many ways to choose a complete meal? How many ways to choose if I can skip some or all courses?"

Initially, I used the product rule to answer the first part: |$A$|= $5 * 6 * 3 * 7 = 630$

My thought process going into the second part lies along that since I can skip some or all courses, I can take the powerset of the courses, multiply the courses, then add the sums for my final answer.

For example (zero for the empty set): $P(A)$ = $0 + 5 + 6 + 3 + 7 + (5 * 6) + (5 * 3) + (5 * 7) ... $

We just started combinatorics today, so I am not sure where to head. Any advice would be appreciated.


问:
Prove that for every $a \in ℝ$ and for every $x>0$ the inequality $x^{-a}+a · e · \ln(x) \geq 0$ holds
I have tried some techniques to solve the following problem but I was unable to solve it:

"Prove that for every $a \in ℝ$ and for every $x>0$ the inequality $x^{-a}+a · e · \ln(x) \geq 0$ holds".

I tried to take the derivative and see if it is always positive, I tried a direct approach manipulating the expression, I also tried to divide it in cases, but I always find troubles.
I would appreciate any help or hint you could give me. Thanks.

答:
Note that we can transform the inequality:
\begin{align}
x^{-a}+ae\ln(x) &\geq 0\\
\implies \ln\left(e^{x^{-a}}\right) + \ln\left(x^{ae}\right) &\geq 0\\
\implies \ln\left(e^{x^{-a}}x^{ae}\right) \geq 0\\
\implies e^{x^{-a}}x^{ae} \geq 1\\
\implies e^{x^{-a}} \geq x^{-ae}.
\end{align}

Now, setting $y = x^{-a}$, we note that $x>0, a\in \mathbb{R} \implies y > 0$, so we wish to prove that for all $y>0$ we have$$e^{y} \geq y^{e},$$ which is discussed here.


答:
Suppose that $A\setminus B=B\setminus A$. Since $B\setminus A\subseteq B$, this implies that $A\setminus B\subseteq B$. But $(A\setminus B)\cap B=\varnothing$, so $A\setminus B=\varnothing$, and it follows immediately that $A\subseteq B$. Similarly, $B\subseteq A$, so $A=B$.

I suspect that you didn’t really use a proof by contradiction, however: more likely you found a proof of the contrapositive and dressed it up as a proof by contradiction. The contrapositive of this implication is that if $A\ne B$, then $A\setminus B\ne B\setminus A$. If $A\ne B$, then without loss of generality there is some $x\in A\setminus B$ that is not in $B\setminus A$. But then $x\in A$ and $x\notin B$, so $A\ne B$. If you began by assuming that $A=B$, this is a proof by contradiction, but that assumption is unnecessary: without it you have a direct proof of the contrapositive of the original implication.

Title: Proof of the equality of the difference of two sets iff sets are equal (direct vs. indirect)

Body:
I have a problem with the following (really) basic result:

$$A\backslash B=B\backslash A \Longleftrightarrow A=B$$

More specifically, I am able to prove it only by contradiction (in particular in the necessary condition - for the sufficient condition I get a contradiction following the steps of a direct proof).

So my problem is, can we actually prove it without using contradiction?

Moreover (and this is probably a very dumb question, so I apologize), is there a general way to know if a given result that we know is true can be proven only by contradiction or there exist a direct proof of it?

Looking forward to any feedback!


问:
Find the value of the parameter $L$ via Cardano-Vieta's Formula knowing a condition on two roots
Given $L\in \mathbb{R}$, the equation $f(x)=x^4-Lx^3-2x^2+6x-4=0$ has two roots such that their sum is equal to $-1$. ¿What is the value of $L$? Here is my attempt. Asumming the four roots are $a,b,c,d$ and $a+b=-1$. The Cardano-Vieta's formulas read as
$$ \begin{cases} abcd = -4 \\ abc +abd+bcd+acd = -6 \\ ab+ac+ad+bc+bd+cd=-2 \\ a+b+c+d=L\end{cases}$$
Using $a+b=-1$, the last equations arises $c+d=1+L$. So we have 5 equations and 5 incognites. The problem is that I'm stuck in solving this. I could reduce the system to three equations and three variables $(L,d,a)$, but I still  can't solve. Which is the best way to attack this kind of system? Maybe Cardano-Vieta's formulas don't fix here very well... Any comment or help is welcome. Thanks in advance!

答:
Suppose one solution is $x_1=a$ the other one is $x_2=-1-a$.

Thus the polynomial $f(x)=x^4-L x^3-2 x^2+6 x-4$ has a factor
$$(x-a)(x+1+a)=x^2+x-a^2-a$$
and another factor $x^2+bx+c$. Thus their product is
$$(x^2+x-a^2-a)(x^2+bx+c)=$$
$$=x^4+(b+1) x^3+x^2 \left(-a^2-a+b+c\right)+x \left(-a^2b-a b+c\right)-a^2 c-a c$$
To be equal to $f(x)$ must be
$$\begin{cases}
b+1=-L\\
-a^2-a+b+c=-2\\
-a^2b-a b+c=6\\
-a^2 c-a c=-4\\
\end{cases}
$$
subtract $3$nd less $2$rd
$$a + a^2 - b - a b - a^2 b=8\to b=\frac{-a^2-a+8}{-a^2-a-1}$$
solve the $4$th wrt $c$
$$c=\frac{4}{a^2+a}$$
and plug these results in the second equation
$$-a^2+\frac{-a^2-a+8}{-a^2-a-1}+\frac{4}{a^2+a}-a=-2\to a=-2;\;a=1$$
substitute $a=1$ to get $b=-2$ and finally substitute in the first equation to have the result $L=1$.


答:
For i):

Domain: Because of the $\ln$ you need that $1+\exp(-(x+1)(y-1))>0$ but this is always the case.

Range: For $x=-y$ you get $h(-y,y)=\ln(1+\exp((y-1)^2))>\ln(\exp((y-1)^2))=(y-1)^2$, hence the function is unbounded. Furthermore you can show that $$\lim_{x\rightarrow \infty} h(x,x)=\ln(1+0).$$
The range is $(0,\infty)$.

For ii)
Domain: $\mathbb{R}^2$ as well.

Range: You can easily check that $h(x,x)$ is unbounded. Furthermore $h(2,2)=1$, hence the range is $\left[1,\infty\right)$.

Title: What is the domain and the range of $h(x,y)=\ln(1 + \exp(-(x+1)(y-1))$ and $h(x,y)=\exp((y-x)^2+2(2-x)^2)$?

Body:
i. $h(x,y)=\ln(1 + \exp(-(x+1)(y-1))$

idea: due to the $\ln$

$(1 + \exp(-(x+1)(y-1))> 0$

due to the $\exp$

$-(x+1)(y-1)>0$ at this point I don't know how to continue

ii. $h(x,y)=\exp((y-x)^2+2(2-x)^2)$

due to the $\exp$
$((y-x)^2 +2(2-x)^2) >0$


答:
Hint:

Define the function $f:(0,+\infty)\to\Bbb R$ by $$f(x)=x^{-a}+ae\ln(x)$$
Show that $f(x)\to+\infty$ as $x\to0^+$ and $x\to +\infty$. Conclude that $f$ attains its global minimum. Use its derivative to find where it is. You will see that $f(x)=0$ at that point, so we get $f(x)\geq0$ for all $x\in (0,+\infty)$

Body:
I have tried some techniques to solve the following problem but I was unable to solve it:

"Prove that for every $a \in ℝ$ and for every $x>0$ the inequality $x^{-a}+a · e · \ln(x) \geq 0$ holds".

I tried to take the derivative and see if it is always positive, I tried a direct approach manipulating the expression, I also tried to divide it in cases, but I always find troubles.
I would appreciate any help or hint you could give me. Thanks.


问:
Determine the truth value of the following statements.

$(\forall x \in \Bbb R)( \sqrt{x^2}= x)$ .

$(\exists x \in \Bbb R)(\exists y \in \Bbb R):(x^2 + y^2=9 )$.

$(\exists x \in \Bbb R)(\forall y \in \Bbb R)(x + y =7 )$.

答: Statement ‘3.’ is false because there does not exist any real number $x$ such that $x+y=7$ is true for any real number $y$, indeed $x+y=7$ is true only for $y=7-x$.


答:
Let $$f_x(a):=e^{-a\ln x}+ae\ln x.$$
For $x=1$, it is clear that $f_1(a)=1\geq0$. Now let $x\neq0$, then
$$f'_x(a)=-e^{-a\ln x}\ln x  +e\ln x\overset{!}{=}0\Leftrightarrow a=-1/\ln x.$$
Now you just have to think about, why this is a minimum point and thus the minimum of the function is $$f_x(-1/\ln x)=0.$$

Title: Prove that for every $a \in ℝ$ and for every $x>0$ the inequality $x^{-a}+a · e · \ln(x) \geq 0$ holds


答:
Since you can "skip any or all courses" there are $6$ appetizers- the $5$ given and "no appetizer".  There are $7$ entrees- the $6$ given and "no entree".  There are $4$ salads, the three given and "no salad".  There are $8$ deserts, the $7$ given and "no desert".

So there are $6\times7\times4\times8$ possible means.  That would include not eating anything, eating only a desert or only an appetizer, etc,.


问:
How to find standard deviation error bound from variance error bound
I'm working on a robotics problem where I'm keeping state in 3D. The state has an error bound described by a covariance matrix I'm keeping internally. A cool property of matrices is that you can take their dot product with (in 2D)

$$[cos(t) \ sin(t)]^T, \ t \in [0, 2\pi)$$

or (in 3D)

$$ [sin(\phi)cos(\theta) \ \ sin(\phi)sin(\theta) \ \ cos(\phi)]^T, \ \phi \in [0, \pi), \ \theta \in [0, 2\pi)$$

to plot an ellipse (or ellipsoid) which neatly shows how they warp a circle (or sphere). For covariance matrices this corresponds to a notion about how uncertain your state estimate is in various directions.

Right now my ellipses are showing variance, because they're from the covariance matrix, but I actually would like to plot ellipses to show standard deviation. My question is: How can I calculate the "co-standard-deviations" matrix from the covariance matrix?

I've tried just taking a matrix square root (and also an element-wise square root), but the ellipse ends up skewed.

I know this can't be right, because the spatial directions along which my sensors have maximum uncertainty should be the same, whether I'm showing standard deviation or variance.

答:
The ellipsoid I want can be expressed as

$$ \{x | x^TP^{-1}x = 1 \} $$

where $P$ is a covariance matrix and $x$ is a vector/point. This is all the points on the boundary an ellipsoid with principal axes along the eigenvectors of $P$, with length equal to the square roots of the corresponding eigenvalues. source

But what I'm really looking for to do my visualization is a matrix $A$, such that $A \cdot spheroid$ = this ellipsoid, whereas before I was doing $P \cdot spheroid$ = variance ellipsoid.

The answer turns out to be that you want the matrix $A \ s.t. AA^T = P$. I was missing the transpose!

Here are a few different proofs to solidify the intuition:

If $P$ is positive semidefinite and symmetric, as it must be if it's a covariance matrix, then it is diagonalizable as $P = VDV^{-1}$, where $V$ holds the eigenvectors and $D$ is a diagonal matrix holding eigenvalues.

If we choose unit length eigenvectors, noting that eigenvectors are orthogonal for a symmetric matrix, then $V$ is an orthonormal matrix, and $V^{-1} = V^T$.

Next, we note that because $D$ is diagonal, we can split it apart really easily as $D = D_{\sqrt{}}D_{\sqrt{}}$, where $D_{\sqrt{}}$ holds the square root of the eigenvalues on its diagonal. Once again, this will work because $P$ doesn't have any negative eigenvalues by property of being symmetric positive semidefinite.

Putting it all together we get

$$ P = VD_{\sqrt{}}D_{\sqrt{}}V^T = (VD_{\sqrt{}}) (D_{\sqrt{}}V^T) = (VD_{\sqrt{}})(VD_{\sqrt{}}^T)^T = (VD_{\sqrt{}})(VD_{\sqrt{}})^T = AA^T$$

So I can find $A$ with a simple eigenvalue-and-vector decomposition of $P$. Note $A$ is no longer symmetric, so I couldn't repeat this process recursively.

Next, find $P^{-1}$:

$$ P = VDV^T \rightarrow P^{-1}P = P^{-1}VDV^T \rightarrow I = P^{-1}VDV^T$$
and now start right multiplying by inverses
$$ \rightarrow V = P^{-1}VD \rightarrow VD^{-1}V^T = P^{-1} $$

Next, use this $A$ in the constraint of the ellipse we want:

We're going to take a whole bunch of $VD_{\sqrt{}} \cdot spheroid\ point$ to get our $x$s that satisfy the constraint. Let's call an spheroid point $s$. The constraint now becomes:

$$ (VD_{\sqrt{}}s)^T P^{-1} (VD_{\sqrt{}}s) = 1 $$

Is this true?

$$ \rightarrow (VD_{\sqrt{}}s)^T VD^{-1}V^T (VD_{\sqrt{}}s) $$
$$ = s^T(VD_{\sqrt{}})^T VD^{-1}V^T (VD_{\sqrt{}}s) $$
$$ = s^TD_{\sqrt{}}^T V^T VD^{-1}V^T VD_{\sqrt{}}s $$
$$ = s^TD_{\sqrt{}} V^{-1} VD^{-1}V^{-1} VD_{\sqrt{}}s $$
$$ = s^TD_{\sqrt{}} D^{-1} D_{\sqrt{}}s $$
$$ = s^Ts = 1$$

the last because any point on the spheroid has magnitude 1, so its dot product with itself will be 1.

Plotting all $x = VD_{\sqrt{}}s$, I see ellipses now align!

One last proof to really solidify why this is:

$V$ is orthonormal, so it's a rotation matrix. So for the variance case we first rotate our spheroid with $V^T$ to recover the same spheroid, then scale by $D$, then rotate that ellipsoid by $V$. And in the standard deviation case, we scale the spheroid by $D_{\sqrt{}}$ and the rotate that sqrts-scaled ellipsoid by $V$.

And now we see the answer to this puzzle to make $A$ symmetric again: Use $Q = V^T$, so $A = VD_{\sqrt{}}V^T$. This extra rotation won't affect the rotation of the final ellipsoid, because, as in the variance case, it's applied to a spheroid and just returns the same spheroid.


答:
$\Rightarrow$

Let $a \in A$. We want to prove $a\in B$

If $a \in B$ we're done.

If $a\not\in B$ then $a \in A \setminus B$ so $a\in B\setminus A \subseteq B$ so $a\in B$

So $A\subseteq B$

Similarly $B\subseteq A$

So $A=B$

$\Leftarrow$

Suppose $A=B$

$A\setminus B = \emptyset = B\setminus A$


问:
Proof of Heron's Formula for the area of a triangle

Let $a,b,c$ be the lengths of the sides of a triangle. The area is given by Heron's formula:

$$A = \sqrt{p(p-a)(p-b)(p-c)},$$

where $p$ is half the perimeter, or $p=\frac{a+b+c}{2}$.

Could you please provide the proof of this formula?

Thank you in advance.

答:
I noticed that many proofs for Herons formula use the properties of angles, which is why I wanted to show a proof that doesn't involve any, and seems to be simpler than the usual methods that are applied.

A triangle with side lengths $a, b, c,$ where $c$ is the greatest side length and an altitude($h$) that intercepts $c$ such that $c$ is the  sum of two side lengths ,
$c = m + n$, using Pythagorean theorem and the area formula of a triangle,  we can prove Herons formula.

First we find $a$ and $b$ in terms of $m$ and $n$ using Pythagorean theorem:

$a^2 - m^2 = h^2$

$b^2 - n^2 = h^2$

$a^2 - m^2 = b^2 - n^2$

$a^2 - b^2 = m^2 - n^2$

Then using $c = m + n$, eliminate $m$:

$a^2 - b^2 = (c-n)^2 - n^2$

$a^2 - b^2 = c^2 - 2cn$

$n = \frac{1}{2c}(c^2 + b^2 - a^2)$

We can eliminate $n$ from $b^2 - n^2 = h^2$ to find $h$:

$h = \sqrt{b^2 - (\frac{1}{2c}(c^2 + b^2 - a^2))^2}$

Plugging this into the area formula ($A = \frac{1}{2}ch$) gives:

$A = \frac{1}{2}c\sqrt{ b^2 - (\frac{1}{2c}(c^2 + b^2 - a^2))^2} $

$A = \frac{1}{2}c\sqrt{ \frac{1}{4c^2}(2a^2b^2+2a^2c^2-a^4-b^4+2b^2c^2-c^4)} $

$A = \sqrt{\frac{1}{16}(c^2 - (a - b)^2)(( a + b)^2 - c^2)} $

$A = \sqrt{\frac{1}{16}(a + b - c)( a + b + c)( b + c - a)(a + c - b)} $

$A = \sqrt{s(s - a)(s- b)(s- c)}$

Q.E.D.


问:
Is homotopy of smooth maps $M \to \mathbb{S}^n$ equivalent to smooth homotopy?
I'm trying to solve the following exercise from Milnor's Topology from the Differentiable Viewpoint

Let $M$ be a compact smooth manifold, show that every continuous map $M \to \mathbb{S}^n$ can be uniformly approximated by a smooth map. If  two smooth maps are continuously homotopic, show that they are smoothly homotopic.

I solved the first part approximating every coordinate thanks to Stone-Weierstrass' Theorem and normalizing. I think it should work. Then I tried to use the first part to solve the second one (about smooth homotopy), but I'm not able to understand how I should use approximation to change a continuous map to a smooth one. How could I solve this problem?

答:
I think I found the solution in the case $\partial M = \emptyset$. If someone could check it I'll be grateful.

Let $f,g : M \times \to \mathbb{S}^n$ be two smooth maps and let $H : M \times [0,1] \to \mathbb{S}^n$ be a continuous homotopy between $f,g$. Since $\partial M = \emptyset$, we can say that $ M \times [0,1]$ is a compact smooth manifold. Hence exists  $\widetilde{H} : M \times [0,1] \to \mathbb{S}^n$ such that $\lVert H-\widetilde{H}\rVert < 2$. Now using $$\frac{t\widetilde{H}(x,0) - (1-t)f(x)}{\lVert t\widetilde{H}(x,0) - (1-t)f(x) \rVert}$$ we conclude that $f$ is smoothly homotopic to $\widetilde{H}(\cdot,0)$ and in the same way we can show that $g$ is smoothly homotopic to $\widetilde{H}(\cdot,1)$. Since being smoothly homotopic is an equivalence relation, $f$ and $g$ are smoothly homotopic.


问:
$M$ is torsion-free iff $M \longrightarrow \operatorname{Hom} _ R\bigl(\operatorname{Hom} _ R(M, R),R\bigr)$ is injective

Let $R$ be a domain (ring without zero divisors) and $M$ be an $R$-module. I want to show that $M$ is torsion free iff the natural map $f: M \longrightarrow\operatorname{Hom}_ R\bigl(\operatorname{Hom}_R(M, R), R\bigr);
m \mapsto (\phi \mapsto \phi(m))$ is injective but I don't have any idea how to manage it.

I can probably prove "<=": $M$ is torsion, therefore there exist $r_0 \neq 0, m_0 \in M$ with
$r_0 m_0=0$. How can I show that then $f$ is not injective? A natural guess
is to show that $f(m_0) =0$.
$m_0$ is mapped by $f$ to $r$-module map $m_0^*:\phi \mapsto \phi(m_0)$.
Since $m_0$ is torsion and every $\phi\in M^*$ respects $R$-module
structure we have $r_0 \phi(m_0)= \phi(r_0 m_0)=0$. This imply that
$\phi(m_0)=0$ since $R$ domain. This argument seems to work.

As alternative way we kno that every submodule of a torsion free module
is torsion free. But is
$\operatorname{Hom}_ R\bigl(\operatorname{Hom}_R(M, R), R\bigr)$
torsion free?

"=>": is not clear. Assume $f$ is not injective, why is $M$ torsion?

答:
The modules for which the natural map $f\colon M\rightarrow M^{\ast\ast}$ is injective are called torsion-less. This is not equivalent to being torsion-free. Over a domain, torsion-free implies torsion-less, as your argument correctly shows. To see the converse fails, note that $\mathbb{Q}$ is a torsion-free $\mathbb{Z}$-module, but $\mathbb{Q}^{\ast}=0$.

As an alternative argument, if $f$ is injective, then $M$ embeds as a submodule of $M^{\ast\ast}$, so if the latter is torsion-free, we are done. I will prove the following slightly more general assertion: if $N$ is a torsion-free $R$-module and $M$ is any $R$-module, then $\operatorname{Hom}_R(M,N)$ is torsion-free. To this end, assume that $f\in\operatorname{Hom}_R(M,N)$ is torsion, so there is a $0\neq r\in R$ such that $rf=0$. By definition, this means $rf(m)=(rf)(m)=0(m)=0$ for every $m\in M$. Since $r\neq0$ and $N$ is torsion-free, this implies $f(m)=0$ for all $m\in M$, but that just means $f=0$.


答:
$$
x^{-a}+ae \ln(x) = e^{-a \ln(x)} + ae \ln(x)
$$
so that, with the substitution $u = -a \ln (x)$, this is equivalent to showing that
$$
 e^u \ge eu
$$
for all $u \in \Bbb R$, and that is true because $f(u) = e^u$ is convex:
$$
 e^u = f(u) \ge f(1) + (u-1)f'(1) = e + (u-1)e = eu \, .
$$


问:
Regarding a space X as a CW complex
(My notations and definitions are similar to those in Hatcher's book.)

In Hatcher's AT, we construct a CW complex in an inductive way. This creates a new topological space.

Why don't we do a converse thing? Let $X$ be an already-existing topological space. Now I'm trying to give a CW complex structure on $X$ as follows, with some assumptions on $X$.

Assume there are index sets $\Lambda_1$, $\Lambda_2$, $\cdots$. For each $n=1$, $2$, $3$, $\cdots$ and each $\alpha \in \Lambda_n$, there is an $n$-cell $e^n_\alpha$ in $X$. Let $X^0$ be a discrete subspace in $X$. Let $X^n = \left( \bigcup_{\alpha \in \Lambda_n} e^n_\alpha \right) \cup X^{n-1}$.  At the same time we have a continuous map $\phi^n_\alpha : D^n_\alpha \to X$ such that $\phi^n_\alpha (\partial D^n_\alpha) \subset X^{n-1}$ and $\phi^n_\alpha (\textbf{int}D^n_\alpha ) = e^n_\alpha$. Finally, assume the topology on $X$ is the coherent topology of its subspaces $X^n$. These are the assumptions.

Now is it enough to regard $X$ as a CW complex? Or, more assumption needed?

答:
Not sure about your description since you said you wanted to start from a space, not to build one. The skeleta $X^0, X^1,..$ should be part of the data. You should look at proposition A.2 in Hatcher's to see the correct statement of what you are trying to set up. Essentially you start with a collection of maps $\Phi_\alpha: D_\alpha^n \to X$ and it gives (sufficient and necessary) conditions for these maps to be the characteristic maps of a CW-structure on $X$.

An alternative point of view (at least for the finite case) is the following: a finite CW-complex $X$ is a Hausdorff space together with a finite sequence of closed subsets $X^0 \subset X^1  \subset ... \subset X^n=X$ such that:

$X^0$ is a finite set of points.

$X^i - X^{i-1} = \amalg Z_{i_k}$ where for every index $i_k$ there is an homeomorphism  $\phi_{i_k}: \mathring{D^i} \to Z_{i_k}$.

For every index $i_k$ the previous homeomorphism  $\phi_{i_k}$ extends to a continuous map $D^i \to \bar{ Z}_{i_k}$.

Title: Regarding a space X as a CW complex


问:
Examples of nonsymmetric spaces
In learning about symmetric spaces, I have come across many examples ($\mathbb{R}^n, S^n$ and $\mathbb{P}^n$ to name a few), but I haven't come across many counterexamples.

I am aware that symmetric spaces are complete, so that would imply that any example of a noncomplete space, such as $\{\mathbb{R}-\{0\}\}$ should work.

I am also aware that simply connected symmetric spaces are uniquely determined by there curvature at a point, which leads me to believe that the "egg" surface, shown below, which is a $C^2$ surface given by the union of a sphere and the surface of revolution of a polynomial cannot be a symmetric space either.

Are there any other examples of (preferably visualisable) nonsymmetric spaces?

答:
You have to start out with a homogeneous space $G/H$, so the group of Riemannian isometries has to act transitively, first of all. Yes, the curvature tensor has to be parallel, so in the case of a surface, the manifold must have constant curvature. So any surface in $\Bbb R^3$ with non-constant curvature cannot be a symmetric space.

Grassmannians are the first "interesting" example of symmetric spaces. These are not spaces of constant (sectional) curvature. $G(k,n)$ is the space of $k$-dimensional subspaces of $\Bbb R^n$. (You can also do oriented subspaces, and you can do complex subspaces of $\Bbb C^n$, as well.)

The easiest example I know of a homogeneous space that fails to be (locally) symmetric is a flag space like
$$X = \{(\ell,H)\in G(1,3)\times G(2,3): \ell\subset H\},$$
the collection of ordered pairs of lines through the origin in $\Bbb R^3$  and planes through the origin with the property that the line is contained in the plane. It is a circle bundle over $\Bbb RP^2$. This example generalizes to all dimensions, considering subspaces contained in larger-dimensional subspaces.

Title: Examples of nonsymmetric spaces


问:
What does Poincare's Lemma tell me about the existence of a vector potential and what restrictions have to be considered for the domain?
I am studying Vector Calculus and Poincare's lemma was mentioned in the class notes.

Poincaré Lemma: If $U \subset \Bbb{R}^n$ is an open star-shaped set, then for every $k=\{0,...,n\}$ every CLOSED k-form in $U$ is EXACT.

I want to know if I understand the concept or the significance of the result applied to vector fields: basically, if we have a vector field defined in a star-shaped domain and its divergence is zero, then the vector field has a vector potential?
Now to my second question. I've seen some people use contractible domain in the definition instead of star-shaped. I know that a star-shaped domain is a contractible domain but a contractible domain isn't necessarily a star-shaped domain. Is it necessary for that the domain is star-shaped or it holds if the domain is just contractible? (a lot of sources I have read/asked always mention that the domain needs to be able to be continuously contracted to a point).

答: Many texts assume a star-shaped domain so that one can write down an explicit integral along rays from the "star point" to give a primitive of the closed form. But all one needs is a contractible set in order for the Poincaré lemma to be valid, and there is, in fact, a general formula obtained by integrating over the fibers of the null-homotopy.</p>