问:
In the following procedure, how is the function (implicitly) chosen?
Given $x^3+y^3-3xy=0$ we differentiate with respect to $x$ on both sides, yielding $3x^2+3y^2\frac{dy}{dx} -3x\frac{dy}{dx} - 3y = 0$, and rearrange to get
$$
\frac{dy}{dx} = \frac{y-x^2}{y^2-x},\quad y^2\neq x
$$

The original relation $x^3+y^3-3xy=0$ implicitly defines any number of functions $y(x)$, since for $0 \leq x\leq 2^{2/3}$ it doesn't uniquely determine a value. Nevertheless this process somehow "selects" one of these (It seems to be the lower one). How does this happen? Does it always happen with implicit functions such as these? Why was this specific function selected, and by what process? Another valid function could be one that in the interval $0 \leq x\leq 2^{2/3}$ that varied between the upper and lower ones at rational and irrational points, and any other number of configurations.

答:
When you differentiate, you're assuming the function is differentiable, so this method can only produce differentiable solutions (and in particular, only continuous solutions, since differentiability implies continuity). The condition $y^2 \neq x$ means that $x \neq 0$ and $x \neq 2^{2/3}$, so the solutions can be considered separately on the intervals $(-\infty, 0)$, $(0, 2^{2/3})$, and $(2^{2/3}, \infty)$. On $(-\infty, 0)$ and $(2^{2/3}, \infty)$, the solution is unique, so there's no choice to be made.

For $0 < x < 2^{2/3}$, there are three choices of implicit function that are continuous. (That there are at most three follows from the fact that the curve is defined by a cubic equation. That there are at least three can be observed by graphing the curve.) All three of these functions satisfy the differential equation
$$\frac{dy}{dx} = \frac{y - x^2}{y^2 - x}.$$
When we consider only the interval $(0, 2^{2/3})$, there is nothing privileging any one of these three choices.

What makes one choice special in this case is how it behaves at the boundary points. All three branches intersect at $(0, 0)$, so the function extends to a continuous function on $(-\infty, 2^{2/3})$ regardless of which of the three branches we choose. However, only the middle branch gives differentiability at $x = 0$. At the right endpoint, only the lower branch allows us to extend to a continuous function on $(0, \infty)$.

So, if you just care about continuity, the lower branch is the "nicest" choice of the three, because it gives us a continuous function on the whole real line. But if you also want the implicit function to extend to something differentiable at the boundary points, there's no perfect choice: the middle branch gives differentiability at $x = 0$ but not $x = 2^{2/3}$, while the lower branch gives differentiability at $x = 2^{2/3}$ but not $x = 0$. And if you don't care about the boundary points, all three are equally reasonable.


问:
Number of solutions of $x^m-y^m=a \pmod p$
I'm interested in the total number of solutions of $x^m-y^m\equiv a \pmod p$ where $p$ is prime.

Is there any way to estimate the total number of solutions ?

I found https://www.degruyter.com/view/journals/crll/1935/172/article-p151.xml?language=en is related, but I cannot read German.

答:
I am assuming $m$ can be equal to anything and that $m$ can also be equal to $p$.
An easier solution is $a = 0$ when $x = y$ (and I hope you know how we get it).

Case 1 : $m = p$

$$\implies x^m \equiv x (\mod{p})$$
and $$y^m \equiv y (\mod{p})$$
(Using Fermat's little theorem)

$\implies x^m - y^m \equiv x - y (\mod{p})$
$\implies x - y = a$

So you have one solution here.

Case 2 : $m = np$ for some $n$

$$\implies x^m = (x^n)^p \equiv x^n (\mod{p})$$
and
$$y^m = (y^n)^p \equiv y^n (\mod{p})$$
$\implies x^m - y^m = (x^n)^p - (y^n)^p \equiv x^n - y^n \equiv a(\mod{p})$( Using Fermat's little theorem)

Another solution.

Case 3: $m$ and $p$ are relatively prime

If $x, y > p$ :

if $x \equiv u (\mod{p}), x^m \equiv u^m (\mod{p})$

if $y \equiv v (\mod{p}), y^m \equiv v^m (\mod{p})$

Here $a \equiv u^m - v^m (\mod{p})$

And $a$ should vary depending on $u$ and $v$, and for $u > v$, you can have $p$ values for $a$ ($0$ through $p - 1$).

For sure, there should be three solutions, plus some more from the third case.

Note : You can have only one solution for $a$ for a constant $u$ and $v$.

ContentLicense: CC BY-SA 4.0


问:
Where is $e^z(z-\bar{z})^2$ differentiable?
Question: Where is $f(z) = e^z(z-\bar{z})^2$ differentiable?

My attempts: By letting $z = x+iy$ and by using the Cauchy Riemann Equations, I was able to get to a point where $\sin y = 0$ and $\cos y = 0$. I've tried to question three times and I get the same result each time. For the real part of the function I got $u(x,y) = -4y^2\cos(y)e^x$ and for the imaginary part I got $v(x,y) = -4y^2\sin(y)e^x$. From there, using $\frac{\partial u}{\partial x} = \frac{\partial v}{\partial y}$ and $\frac{\partial u}{\partial y} = -\frac{\partial v}{\partial x}$. The equations I arrived at were:

$$ -4y^2\cos(y)e^x = -8y\sin(y)e^x - 4y^2\cos(y)e^x$$
and
$$-8y\cos(y)e^x+4y^2\sin(y)e^x=4y^2\sin(y)e^x.$$

It was from these questions that I arrived that $\sin(y) = 0$ and $\cos(y) = 0$. Have I made a mistake? Any help or guidance would be greatly appreciated!

答: Hint: If $e^{z}(z-\overline z)^{2}$ is differentiable at some point then $(z-\overline z)^{2} =e^{-z} (e^{z}(z-\overline z)^{2})$ is also differentiable at the point. That should make the computations much simpler.


问:
Silliness: $\exists~X~\text{s.t.}~AX=B \iff B\in R(L_A)$
So, I am asked to prove that the system of linear equations $AX=B$ has $\color{black}{a~solution}$ if and only if $B\in R(L_A)$. $R$ denotes the "range of" and $L_A$ is left multiplication by $A$. If the question says "a solution" does that mean that I can rely on the fact that it's invertible?

Friedberg's 1989 edition of Linear Algebra says the above are equivalent, namely that $\exists~X~\text{s.t.}~AX=B \iff B\in R(L_A)$:

How am I to "prove" this?

答:
I think it's best if we state the definition of range first.

Definition: Let $A$ be an $m\times n$ matrix and consider the linear mapping $L_A:\mathbb{F}^n\rightarrow \mathbb{F}^m$ induced by $A$ as $L_A(\mathbf{x}) = A\mathbf{x}$. We denote the range or image of $L_A$ as the set of all vectors $\mathbf{b}\in\mathbb{F}^m$ such that $\mathbf{b}=L_A(\mathbf{x})$ for some $\mathbf{x}\in \mathbb{F}^n$. We denote the range as
$$\mathrm{R}(L_A) = \{\mathbf{b}\in\mathbb{F}^m\mid \exists \mathbf{x}\in\mathbb{F}^n\ L_A(\mathbf{x})=\mathbf{b}\}$$
$$$$

Now, if there exists $\mathbf{x}$ such that $A\mathbf{x} = \mathbf{b}$ then by definition of $L_A$ we have $A\mathbf{x} = L_A(\mathbf{x}) = \mathbf{b}$. This means that $\mathbf{b}$ is a member of the range since there exists some $\mathbf{x}$ such that $L_A(\mathbf{x})=\mathbf{b}$.

Conversely, suppose that $\mathbf{b}$ is in the range of $L_A$. By definition of range, there exists some $\mathbf{x}$ such that $L_A(\mathbf{x}) = \mathbf{b}$. By definition of $L_A$ we have $L_A(\mathbf{x}) = A\mathbf{x} = \mathbf{b}$ so $\mathbf{x}$ satisfies the system.

ContentLicense: CC BY-SA 3.0


问:
Conjecture inequality $\frac{\cosh(a)}{b+c}+\frac{\sinh(b)}{a+c}+\frac{\tanh(c)}{a+b}>e$
Let , $a\geq b \geq c >0$ such that $a+b+c=\frac{10}{11}$ then prove (or disprove) :

$$\frac{\cosh(a)}{b+c}+\frac{\sinh(b)}{a+c}+\frac{\tanh(c)}{a+b}>e$$

We cannot use Jensen's inequality as proof for this inequality since the functions $\cosh,\sinh,\tanh$ are differents .

The use of Am-Gm is inefficient . Moreover I have tried something like :

$$\frac{\cosh(a)}{b+c}+\frac{\sinh(b)}{a+c}+\frac{\tanh(c)}{a+b}>\frac{2e}{3}\Big(\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}\Big)\geq e$$

Without success .

As a last resort I have tried power series but it's very ugly so not nice .

Update : 03/11/2020

I simplify a little bit the proof of user RiverLi  wich is a nice proof .

Due to convexity ($\cosh$ and $\sinh$) or concavity ($\tanh$) we have the following inequalities (always with the constraints above) :

$$\frac{\cosh(a)}{\frac{10}{11}-a}\geq f(a)=\frac{\cosh(\frac{10}{33})+(a-\frac{10}{33})\sinh(\frac{10}{33})}{\frac{10}{11}-a}$$

$$\frac{\sinh(b)}{\frac{10}{11}-b}\geq g(b)=\frac{\sinh(\frac{10}{33})+(b-\frac{10}{33})\cosh(\frac{10}{33})}{\frac{10}{11}-b}$$

$$\frac{\tanh(c)}{\frac{10}{11}-c}\geq h(c)=\frac{c\tanh(\frac{10}{33})\frac{33}{10}}{\frac{10}{11}-c}$$

Now it's less harder to show that the function $f,g,h$ are convex and continue the proof of RiverLi .

Any idea is welcome!

Thanks in advance

Body:
Hint.

Let $f(a) = \frac{\cosh a}{10/11 - a}$,
$g(b) = \frac{\sinh b}{10/11 - b}$ and $h(c) = \frac{\tanh c}{10/11 - c}$.

All $f, g, h$ are convex functions.

Then we have $f(a) \ge f(10/33) + f'(10/33)(a - 10/33)$,
$g(b) \ge g(10/33) + g'(10/33)(b-10/33)$
and $h(c) \ge h(10/33) + h'(10/33)(c-10/33)$.

It suffices to prove that
$$f(10/33) + f'(10/33)(a - 10/33)
+ g(10/33) + g'(10/33)(b-10/33) + h(10/33) + h'(10/33)(c-10/33) > \mathrm{e}.$$
Since $a\ge b\ge c$ and $f'(10/33) > g'(10/33) > h'(10/33)$, we have
$$\mathrm{LHS} \ge f(10/33) + g(10/33) + h(10/33) > \mathrm{e}.$$


答:
Partial answer: (if $x,y,m,p,a$ are all taken as variables)

The number of integers in the range $[1, n]$ which can be expressed in the form $a^x−b^y$ is asymptotically $$(\log n)^2 \over {2(\log a)(\log b)}$$ as $n \rightarrow \infty$. This is from a result of Pillai. See: Waldschmidt M., Perfect Powers: Pillai’s works and their developments

Also, the equation $p^x − b^y = c$ has been considered by R. Scott in this paper.

The equation $p^x − b^y = c$, where $p$ is prime, and $b > 1$ and $c$
are positive integers, has at most one solution $(x, y)$ when $y$ is
odd, except for five specific cases, and at most one solution when $y$
is even.

Answer: (if $a,p$ are given and $x,y.m$ are allowed to vary)

Let $g$ be a primitive element in the field $GF(p)$. So we have $x = g^r, y = g^s, a = g^t$ for some $r,s,t \in GF(p)$ and $x,y,a$ non-zero elements of $GF(p)$.

$$g^r - g^s \equiv g^t \mod p$$

$$g^{r-t} - g^{s-t} \equiv 1 \mod p$$

$$g^{r-t} \equiv g^{s-t} + 1 \mod p$$

The LHS is non-zero elements in $GF(p)$. The LHS cannot also be $1$ because then $g^{s-t}$ must be 0, which is not possible.

Since every element other than $0$ and $1$ in $GF(p)$ can be represented as a sum of its predecessor + 1 modulo $p$ and the predecessor can also be represented as the power of a primitive element $g$, there are solutions $\forall x,y,a \in GF(p)$ for any given $m$ where the LHS is not $0$ or $1$.

Since there are $p$ values in $GF(p)$ and $0$, $1$ are not representable in the form, we have $p-2$ solutions modulo $p$ for given $a, p, a \ne 0$.

If $a = 0$, we have $p$ trivial solutions $x^m \equiv y^m \mod p$ given by $x \equiv y \mod p$.

Title: Number of solutions of $x^m-y^m=a \pmod p$

Body:
I'm interested in the total number of solutions of $x^m-y^m\equiv a \pmod p$ where $p$ is prime.

Is there any way to estimate the total number of solutions ?

I found https://www.degruyter.com/view/journals/crll/1935/172/article-p151.xml?language=en is related, but I cannot read German.


问:
Evaluating $ \lim_{n\to\infty}\sum_{i=0}^n\frac{n}{n^2+i^2} $
A few days ago I confronted this limit question, and all my attempts to solve it failed somehow.
$$
\lim_{n\to\infty}\sum_{i=0}^n\frac{1}{n^2+i^2}
$$
I feel I should relate it to $\arctan(n)$. I tried integration and derivation to reach my point, it went well at first but eventually nowhere.

Any kind of help and enlightenment would be appreciated.

答:
$$
\begin{aligned}
\sum_{0\le i\le n}{1\over n^2+i^2}
&={1\over n^2}+\sum_{1\le i\le n}{1\over n^2+i^2} \\
&={1\over n^2}+\int_{1-\varepsilon}^n \,{\mathrm d\lfloor t\rfloor\over n^2+t^2} \\
&={1\over n^2}+\left.{\lfloor t\rfloor\over n^2+t^2}\right|_{1-\varepsilon}^n-\int_1^n[t-\{t\}] \, \mathrm d\left(1\over n^2+t^2\right) \\
&={1\over n^2}+{1\over2n}-\int_1^nt \, \mathrm d\left(1\over n^2+t^2\right)+\int_1^n\{t\} \, \mathrm d\left(1\over n^2+t^2\right) \\
\end{aligned}
$$

wherein the integrals are evaluated as follows

$$
\begin{aligned}
\int_1^nt \, \mathrm d\left(1\over n^2+t^2\right)
&=\left.{t\over n^2+t^2}\right|_1^n-\int_1^n \, {\mathrm dt\over n^2+t^2} \\
&={1\over2n}-{1\over n^2+1}-\left.{1\over n}\arctan\left(\frac tn\right)\right|_1^n \\
&={1\over2n}-{1\over n^2+1}-{\pi\over4n}+\frac1n\arctan\left(\frac1n\right) \\
&=\mathcal O\left(\frac1n\right)
\end{aligned}
$$

$$
\int_1^n\{t\} \, \mathrm d\left(1\over n^2+t^2\right)=\mathcal O\left(\frac1n\right)
$$

As a result, we have

$$
\sum_{0\le i\le n}{1\over n^2+i^2}=\mathcal O \left(\frac1n\right)
$$


问:
Relationship between the number of atomic formulas in a given formula and the number of right arrows
Here's the problem I'm doing;

Let $L_P$ be the language of propositional logic (consisting of parentheses, $\lnot$, $\rightarrow$ and a countable set of atomic formulae). Let $\alpha$ be any formula of $L_p$. Let $s(\alpha)$ be the number of atomic formulas in $\alpha$ (counting repetitions) and let $c(\alpha)$ be the number of occurences of $\rightarrow$ in $\alpha$. Show that $s(\alpha) = c(\alpha)+1$.

Proof Attempt:

We proceed by induction on the number of left-parentheses of a given formula. Suppose that a given formula has $0$ left-parentheses. Then, it has $0$ right-parentheses. Hence, it can only be an atomic formula. So, $s(\alpha) = 1 = 1+0 = 1+c(\alpha)$, where $c(\alpha) = 0$ because $\rightarrow$ does not appear in an atomic formula. So, the assertion holds in the base case.

Now, suppose that the assertion holds for any formula with $n$ left-parentheses. Let $\alpha$ be such a formula. Then, there are only 3 ways to construct a new formula with $n+1$ left-parentheses:

$(\lnot \alpha)$

$(\alpha \rightarrow A_n)$

$(A_n \rightarrow \alpha)$

where $A_n$ is an arbitrary but fixed atomic formula. In the first case, the number of occurences of $\rightarrow$ does not change and the number of atomic formulae does not change. So, the assertion holds for this formula. In the last two cases, the number of occurrences of $\rightarrow$ and the number of atomic formulae in the given formulas each increase by $1$. So, the assertion will hold for them as well.

Hence, the assertion holds for any formula with $n+1$ left parentheses. This proves that the assertion holds for all formulas of $L_P$. $\Box$

Does the proof above work? If it doesn't, then why? How can I fix it?

答:
The induction must be on the rank of $\alpha$, i.e. on the number of connectives in the formula.

You have already verified the base case.

The case for $\lnot$ is trivial: no new atoms and no new occurrences of $\to$.

The only case left is $\to$.

We have that $\alpha$ is $(\alpha_1 \to \alpha_2)$ and we assume that the property holds for $\alpha_i, i=1,2$, i.e.

$s(\alpha_i)=c(\alpha_i)+1$.

Now compute: $s(\alpha)=s(\alpha_1)+s(\alpha_2)$; no new atoms added to those already present.

$c(\alpha)=c(\alpha_1)+c(\alpha_2)+1$, because we add a new occurrence of $\to$.

Substituting:

$$s(\alpha)=s(\alpha_1)+s(\alpha_2)=[c(\alpha_1)+1]+[c(\alpha_2)+1]=[c(\alpha_1)+c(\alpha_2)+1]+1=c(\alpha)+1.$$


问:
Prove that $g(z)$ has an antiderivative on $D_f$
I have some trouble with the following problem :

Let $ g(z) = \frac{1}{\sin^8(z)} $ , prove if $g$ has an antiderivative on its domain

So my first idea was to find out if  : $ k(z) = \frac{1}{\sin(z)} $ has an antiderivative on its domain : $D_f = \mathbb C$ \ {$k\pi , k\in\mathbb Z$} but it turns out that using $\gamma$ a circle of radius $1$ with $0$ as center I find : $\int_{\gamma} \frac{dz}{\sin z} = 2\pi i $ ,

And we know that $k$ has an antiderivative so ,$\displaystyle\oint_\gamma k(z)~\mathrm dz=0 $ on every closed curve $\gamma$  (%)

So  $ k(z) = \frac{1}{\sin(z)} $ doesn't have an antiderivative , but for $g(z)$ can I use what I found for $k(z)$ s.t $k^8(z)= g(z)$, or should I use the argument (%)

Thanks in advance for your help.

答:
$g$ is an even function, so that the Taylor series of $g$ at $z=0$ has only even coefficients. It follows that the residue of $g$ at $z=0$ is zero.

Then use the periodicity of $g$ to show that the residue at every $z=k$, $k\in \Bbb Z$, is zero.

Finally use the residue theorem to show that $\oint_\gamma g(z) \, dz =0$ for every closed curve $\gamma$ in $D = \Bbb C \setminus \Bbb Z$, which implies that $g$ has an antiderivative in $D$.


问:
Suppose $x \in X$ is a unit vector such that $||A|| = ||Ax||$; show that $A^TAx = ||A||^2 x$.
Let $X$ be a linear space with real Euclidean structure. Let $A : X \to X$ be a linear
map. Suppose $x \in X$ is a unit vector such that $||A|| = ||Ax||$; show that $A^TAx = ||A||^2 x$.

I tried in this way, considering $g(t)=||\frac{A(x+ty)}{||x+ty||}||$. It has a maximum at $t=0$ because of the definition of the norm of the matrix. Hence we get $g'(0)=0$. Now can I use this data to work with transpose?

Body: $$\|A^{T} Ax-\|A||^{2}x\|^{2}$$ $$=\|A^{T} Ax\|^{2}-2\|A||^{2} \langle A^{T}Ax, x \rangle+\|A\|^{4}$$ $$=\|A^{T} Ax\|^{2}-\|A\|^{4}.$$ Now  $\|A^{T} Ax\| \leq \|A^{T}\|\|A\|=\|A\|^{2}$ so we get $0 \leq \|A^{T} Ax-\|A||^{2}x\|^{2} \leq 0$


问:
Projection of space onto unit sphere - differentiating it!

I am trying to differentiate the function that projects each $x\in
 \mathbb{R}^3$ onto the unit sphere, namely:

$$f(x)=\frac{x}{||x||},\quad f(0)=0$$

I know that the derivative is $$Df\big|_{x}(h)=\frac{x\times (h\times x)}{||x||^3}=\frac{h}{||x||}-\frac{x(x\cdot h)}{||x||^3}$$
But I am having a lot of trouble proving it. I can reduce it to:
$$f(x+h)-f(x)-Df\big|_x(h)=x\left(\frac{1}{||x+h||}-\frac{1}{||x||}+\frac{(x\cdot h)}{||x||^3}\right)+o(h)$$
I am struggling to show that the expression in brackets is $o(h)$. It looks like the $||x+h||^{-1}$ is begging for expansion but I'm not sure how to deal with it because they are vectors. Help?

答:
I assume you mean computing the gradient restriction of the vector field $x/\|x\|$ defined over all of $\mathbb R^n \{0\}$, on the $(n-1)$-dimensional unit-sphere

The following fact is proved here https://math.stackexchange.com/a/1936635/168758.

Fact. Let $M$ be a smooth manifold embedded in $\mathbb R^n$ and let $f:\mathbb R^n \to \mathbb R^m$ be a differentiable function. Then $f|_M$ is differentiable and $df|_M(x;h) = \langle \mbox{proj}_{T_x M}(\nabla f(x)), h\rangle$, the projection of $\nabla f(x)$ onto $T_xM$, the tangent space of $M$ at $x$

In your case, $M = \mathbb S_{n-1} := \{x \in \mathbb R^n \mid \|x\| = 1\}$, the $(n-1)$-dimensional unit-sphere in $\mathbb R^n$.

One easily computes $T_x M = x^\perp := \{y \in \mathbb R^n \mid y^\top x = 0\}$, and the projection onto $T_x M$ is given by $\mbox{proj}_{T_x M}(v) = \mbox{proj}_{x^\perp}(v) = (I_n - xx^T)v$, for all $x \in M$.

In your specific problem, you have $f(x) := g(x/\|x\|)$ for $x \in \mathbb R^n\setminus\{0\}$ (with arbitrary continuation at $O$). Thus, $\nabla f(x) = g'(x/\|x\|)\cdot \dfrac{I_n-xx^\top/\|x\|^2}{\|x\|} \in \mathbb R^{n \times m}$, which is already a vector field on $T_x M$ (i.e no need to project).


答:
Take $g(t) = \frac12\frac{ \|A(x+ty)\|^2 } { \|x+ty\|^2 }$, i.e., take the square of your original function.
Then $g'(t)=0$, where
$$
g'(t) = \frac{ y^TA^TA(x+ty) } { \|x+ty\|^2 }
-\frac{ \|A(x+ty)\|^2  } { \|x+ty\|^4 } y^T(x+ty).
$$
Hence
$$
g'(0) =  y^TA^TAx
- \|Ax\|^2  y^Tx.
$$
This is zero for all $y$, hence
$$
A^TAx = \|Ax\|^2  x.
$$

Body:
Let $X$ be a linear space with real Euclidean structure. Let $A : X \to X$ be a linear
map. Suppose $x \in X$ is a unit vector such that $||A|| = ||Ax||$; show that $A^TAx = ||A||^2 x$.

I tried in this way, considering $g(t)=||\frac{A(x+ty)}{||x+ty||}||$. It has a maximum at $t=0$ because of the definition of the norm of the matrix. Hence we get $g'(0)=0$. Now can I use this data to work with transpose?


问:
Proving that the boundary set of $A=\{(x,y,z)\in\Bbb R^3:x+e^y<z^2\}$ is $\partial  = \{ (x, y, z) \in \Bbb R^3 ∶ x + e^y = z^2 \}$.

Consider metric space $\Bbb R^3$ with usual distance $d$. Let $A=\{(x,y,z)\in\Bbb R^3:x+e^y<z^2\}$. Prove that the boundary set $\partial A=\{(x,y,z)\in\Bbb R^3:x+e^y=z^2\}$

First of all, I denoted $B=\{ (, , ) \in \Bbb R^3 ∶  + ^ = ^2 \}$.
Then I am able to show that $\partial \subseteq B$ by showing there exists an element in the boundary set which is a solution of the equation.

However, I got stuck on showing the reverse direction.
Can I say the distance between every element in $B$ with $A$ is $0$ so that there exist $2$ convergence sequences (one is a subset of $A$ and the other is the subset of the complement of $A$) both converge to an arbitrary point in $B$. Thus, $A$ is a subset of $B$. I am confused about the construction of such two sequences.

答:
Let $b\in B$. Note that $B\cap A=\emptyset$ so the sequence $c_n\equiv b$ is a sequence converging to $b$ in the complement.
To show there exists a sequence in $A$ converging to $b$, denote $b=(x_0,y_0,z_0)$ and set $d_n=(x_0-1/n,y_0,z_0)$ then clearly $\forall n:d_n\in A$ while $d_n\to b$.


问:
Difficulty in understanding predicate symbols in FOL
I'm a beginner in First-Order Logic, and I have the following difficulty in understanding predicate symbols:

From what I know, $n$-ary predicate symbols, are represented by $p_i^n$. If $x$ is a variable, what does $p^1x$ really mean? The $1$ in the superscript simply denotes that $p$ is unary. I know that functions $f_i^n$ applied to terms return a value, i.e. another term $f^nt_1t_2...t_n$. What does $p^1x$ return? True or False, possibly?

I have an example in mind! Consider the binary predicate symbol $<$, and the wff $2 < 1$. This returns False, but is it always the case that predicate symbols return True/False? I'm not quite sure about the definition.

In addition, I was told that given an interpretation $\mathcal{I}$ and a domain of discourse $\mathcal{D}$, each predicate symbol $p_i^n$ is mapped to an $n$-ary relation, i.e. $\mathcal{I}(p_i^n) \subseteq D^n$. Could someone please give an example for this? In the example that I took above for the binary predicate $p^2$ = $<$, $\mathcal{D} = \mathbb{N}$, $<$ seems to be the following map - $<:\mathbb{N}\times \mathbb{N} \to \{T,F\}$ which doesn't seem like $\mathcal{I}(p_i^n) \subseteq D^n$. Where am I going wrong? Could someone please help me better understand these concepts? Thank you!

答:
If $x$ is a variable, what does $p^1x$ really mean?

As you say, $p^1$ is a unary predicate, i.e. a predicate with one argument-place.

Predicate symbols are syntactical objects, i.e. part of the formal language.

They are the formal counterpart of natural language predicates, like e.g. "... is a Philosopher".

Thus, the natural way to read $p(x)$ is: "$x$ is a $p$".

When we apply a unary predicate to a closed term (a "name") we get a meaningful statement, like: "Socrates is a Philosopher" (that has the form $\text {Phil}(\text {Socrates})$).

You example with $<$ as $p^2$ is correct.

If we use an interpretation $\mathfrak I$ with domain $\mathbb N$ and we interpret the binary predicate $p^2$ as $<$ we will have:

$(p^2)^{\mathfrak I} = \{ (n,m) \mid n,m \in \mathbb N \text {  and  } n < m \}$.

Thus, we have that $(p^2)^{\mathfrak I} \subseteq \mathbb N \times \mathbb N$ and we have that:

$p^2(x,y)[x/n,y/m]$ is True in $\mathfrak I$ iff  $n < m$.


问:
Using Farkas' Lemma to Prove Strong Duality
I've been trying to prove the Strong Duality Theorem following these slides. In those slides, we start with proving Farkas' lemma, then a corollary followed by the proof of the Strong Duality Theorem. To prove Farkas' lemma, I first proved that $\{A(x)|x \in X_n\}$ where $X_j = \{x=(x_1,x_2 \dots x_j) \in \mathbb{R}^j | x_i \geq 0 \text{ for all } 1\leq i \leq j\}$ and $A \in M(m,n)$, is a closed, convex set. This was deceptively hard and has been discussed in these answers. Using this I managed to prove the following:
Let $b\in \mathbb{R}^m$. Exactly one of these two linear programs have a solution:
\begin{align}
        Ax&=b\\
        x &\in X_n \nonumber
\end{align}
\begin{align}
        (A^Ty) &\in X_n\\
        \langle b,y \rangle &< 0 \nonumber\\
        y &\in \mathbb{R}^m
\end{align}
This is Farkas' lemma. I also managed to prove this "corollary". I put it in quotes because I could not use the Farkas' lemma directly to get the result. The corollary is as follows: Let $b\in \mathbb{R}^m$. Exactly one of these two linear programs have a solution:
\begin{align}
        Ax+s&=b\\
        x &\in X_n \nonumber\\
        s &\in X_m \nonumber
 \end{align}
\begin{align}
        A^Ty &\in X_n \nonumber\\
        \langle b,y \rangle &< 0 \\
        y &\in X_m \nonumber
 \end{align}

The final step of this puzzle, which directly proves the Strong Duality Theorem is what I am trying to solve. This is what I am trying to prove now: For any $\alpha \in \mathbb{R}$, $b\in \mathbb{R}^m$, and $c\in \mathbb{R}^n$, prove that exactly one of these two linear programs have a solution:
\begin{align}
        Ax+s&=b\\
        \langle c,x \rangle &\leq \alpha \nonumber\\
        x &\in X_n \nonumber\\
        s &\in X_m \nonumber
 \end{align}
\begin{align}
         \langle b,y \rangle + \alpha z&< 0 \\
        A^Ty + cz&\in X_n \nonumber\\
        y &\in X_m \nonumber\\
        z &\in \mathbb{R}_+ \nonumber
 \end{align}
Here are my questions:

Is there a way to prove Farkas' corollary using Farkas' lemma?

How do I prove the last step from Farkas' corollary?

答:
The first system of the last alternative is
$$
Ax+s = b, \ \langle c,x\rangle \le \alpha, \ x\ge 0, \ s \ge0.
$$
Introducting a slack variable $t\ge 0$, this can be written as
$$
\pmatrix{A& I & 0 \\ c^T&0 & 1} \pmatrix{ x \\ s \\ t} = \pmatrix {b \\ \alpha }
$$
with $x\ge0,s\ge0,t\ge0$. By Farkas Lemma, either this system or the following one is solvable:
$$
\pmatrix{A^T & c\\  I & 0 \\ 0 & 1} \pmatrix{y \\ z} \ge 0
$$
$$
b^Ty + \alpha z < 0,
$$
which is the second alternative in the last system.


问:
Show that cartesian product of a measurable set with $\mathbb{R}$ is measurable
Let $S\subset \mathbb{R}$ be measurable. Show that $S$ is measurable if and onlt if $S\times \mathbb{R}$ is measurable.

My attempt:
If $S$ is measurable, there exist open $U\subset \mathbb{R}$, closed $V\subset \mathbb{R}$ s.t.

$$U\supset S \supset V$$

s.t. $\lambda(U\setminus V)<\epsilon$. Also $U\times \mathbb{R} \supset S\times \mathbb{R} \supset V\times \mathbb{R}$ but I'm not sure how to show the difference in Lebesgue measure can be controlled part. Do I need to take any subset of $S\times \mathbb{R}$ and look at top and bottom approximations and then estimate the difference in Lebesgue measure?

I think this will help me figure out how to prove the other direction as well.

答:
Let's show first that if $A$ is measurable then $A\times \mathbb{R}$ is measurable.  It is enough to consider the case $A$ bounded.  Then it is enough to show that $A \times I$ is measurable for a bounded open interval. Now, for every $\epsilon > 0$ there exists  $K\subset A \subset U$, $K$ compact, $U$ open, such that $\mu(U \backslash K) < \epsilon$. Take $J$ compact, $J\subset I$. We get $J\times K \subset A\times I \subset U\times I$, $\ \ J\times K$ compact, and $U\times I$ open $\ldots$.

Assume now that $A$ is not measurable. Let us show that $A\times I$ is not measurable.  May assume that $A$ is bounded.  Take $M\subset A$, $M$ measurable, of largest possible measure, and $N\supset A$, $N$, measurable, of smallest possible measure. Since $A$ is not measurable, we have $\mu(M) < \mu (N)$.

Let us show that $M\times I \subset A\times I$ is a measurable subset of $A\times I$ of largest possible measure. For this, it is enough to show that every measurable subset $K$ of $(A\backslash M) \times I $ has measure zero. We may assume $K$ compact. The projection of $K$ onto the first component is a compact subset $L$ of $A\backslash M$, and so of measure $0$. Since $K \subset L\times I$, we get $\mu(K) \le \mu(L\times I) = 0\cdot |I| = 0$. Similarly we show that $N\times I$ is a measurable set of smallest measure containing $A\times I$.

From the above it follows that $A\times I$ is not measurable.


问:
Rotating the hyperboloid of two sheets $z^2-y^2-x^2=1$
Revolving $f(x)=\frac{1}{x}$ about $f(x)=x$ yields the two-sheeted hyperboloid $2xy-2=z^2$ in $\Bbb R^3.$

I'm trying to rotate the surface so that the forward sheet lies in the first octant $(+,+,+)$ and the backward sheet lies in $(-,-,-).$

What's the equation?

Here's what I have so far. This is the graph of $2xy-2=z^2:$

It would also work to rotate the two-sheeted hyperboloid $z^2-y^2-x^2=1.$

My attempt was to use a $3$d rotation matrix that rotates points about the origin. I was able to find the right rotation matrices, but I could not reason what to multiply the rotation matrix with to get the desired result. I could probably get the final equation, with a few hints.

I believe I have to change the entries in the matrix $Q$ (as shown in the hint below by @NinadMunshi). I'm not sure what to change them to, but the new entries should rotate the hyperboloid.

答:
I'm trying to rotate the surface so that the forward sheet lies in the first octant $(+,+,+)$ and the backward sheet lies in $(-,-,-).$

Let $O=(0,0,0)$ be the origin of the space $\Bbb R^3$.

By the construction, the first hyperboloid has a rotation axis $\ell_1=\Bbb R (1,1,0)$ so it is contained in its the two-sided asymptotic cone $C_1$ based at $O$, with an axis $\ell_1$ and an angle $\pi/2$ at $O$. The cone $C$ covers $\tfrac {2-\sqrt{2}}{2}$ of an area of a unit sphere $S$ centered at $O$. On the other hand,  two octancts cover only $1/4$ of an area of $S$, so a required rotation with a base at $O$ is impossible.

It is easy to see that the second hyperboloid has a rotation axis $\ell_2=\Bbb R (0,0,1)$ and it is contained in its the two-sided asymptotic cone $C_2$ based at $O$, with an axis $\ell_2$ and an angle $\pi/2$ at $O$. Similarly to the first case we see that a required rotation with a base at $O$ is impossible.


问:
Need a Intuitive understanding of solving double integral where integrand has max function
I'll try my best to keep the question brief. So I came upon this question:

$$\int \int _R^{ }\cos \left(\max \left\{x^3,\ y^{\frac{3}{2}}\right\}\right)dA$$ where R region is $[0,1]\times \ [0,1]$. This particular example has already been asked like 3 times. But none of them answered me completely. The way how its solved is like this:

you have max function = $max({x^3,y^{\frac{3}{2}})}$, which means either $x^3$ is biggest or $y^{\frac{3}{2}}$ is the biggest, which can be represented as $x^3>y^{\frac{3}{2}}$  $\Rightarrow $$x^2>y$ OR

$y^{\frac{3}{2}}>x^3$ $\Rightarrow $ $\sqrt{\ y}>x\ $

from this we can say that $y=x^2$ ; Now our region is already given by $[0,1]\times \ [0,1]$ so we divide the region R like this:

My problem is that our limit is clearly given as $y=0, y=1, x=0, x=1$ first of all why is there a need to divide our region R into 2? and 2nd, the Max function is inside our integrand, how are we taking that from integrand and using it in the limits? I mean the integrand $f(x,y) =z$ usually means its the top surface of the volume, so how is that hindering the limits on x-y plane? I want to understand it intuitively and pictorially/graphically instead of just memorizing this method of solving. Any help is greatly appreciated.

答: The $\max$ function can't be directly integrated. The integrand has different definitions above and below the line $y=x^2$ so you will have to split the integral into two integrals, one over the region under $y=x^2$ and the other over the region above it. You have$$z(x,y)=\max\{x^3,y^{3/2}\}=\begin{cases}x^3,&y<x^2\\y^{3/2},&y\ge x^2\end{cases}$$The required integral is$$\begin{align*}I&=\int_0^1\left[\int_0^{x^2} x^3~dy+\int_{y^2}^1 y^{3/2}~dy\right]~dx\\&=\underbrace{\int_0^1\int_0^{x^2} x^3~dy~dx}_{\text{below }y=x^2 }+\underbrace{\int_0^1\int_{y^2}^1 y^{3/2}~dy~dx}_{\text{above }y=x^2}\end{align*}$$


答:
Re: the "set vs. function" issue: these are really two different ways to phrase the same thing.

There is a natural way to identify a map $f$ from a set $X$ to $\{True,False\}$ with a subset $Set_f$ of $X$: namely, we use $$Set_f=\{x: f(x)=True\}.$$ Note that this is "dual" to the Boolean version of the characteristic function of a set, that is, the assignment to a set $A\subseteq X$ of the function $$Func_A: x\mapsto\begin{cases}
True & \mbox{ if }x\in A,\\
False & \mbox{ if }x\not\in A.\\
\end{cases}$$ In fact, you should check that in fact we always have $$Set_{Func_A}=A\quad\mbox{and}\quad Func_{Set_f}=f,$$ so these are literally inverse constructions.

Similarly, we can equivocate between $n$-ary relations on $X$ (= subsets of $X^n$) and $n$-ary Boolean functions on $X$ (= maps from $X^n$ to $\{True,False\}$). This should explain how e.g. "$<$" can be thought of both as a set of ordered pairs and as a function which takes in a pair of inputs and spits out either "True" or "False." The former approach tends to be used more for whatever reason, but per the above they're really equivalent in an explicit way.

This takes us to the question of how to interpret expressions with free variables. Let's think about terms first. If I have a unary function symbol $f^1$ and a variable $x$, "$f^1x$" is a term. But this term isn't "determined" yet in a sense: even after I specify a structure (and so in particular an interpretation of $f$), I haven't given a value for the variable $x$. So this shouldn't be thought of as referring to a definite object. Rather:

The term "$f^1x$" describes a way of taking a structure $(D,\mathcal{I})$ and a variable assignment $\nu$ for that structure and outputting an element of the domain $D$ of that structure.

It may help to consider a slightly less trivial example, say the term "$g^2xx$" where $g^2$ is a binary function symbol. This (unlike the above example) isn't just a function symbol "repackaged," I'm doing something interesting to the inputs.

Of course an individual term doesn't actually "use" the entirety of any variable assignment; e.g. "$f^1x$" only cares about what gets assigned to $x$. So really we should be a bit more parsimonious:

A term with some free variables corresponds, given a structure $(D,\mathcal{I})$, to a function $D^n\rightarrow D$ where $n$ is the number of free variables occurring in that term.

(There's some nuance here, but I'd ignore that at first.)

Predicates - or more generally, formulas (possibly with free variables) - will behave the same way:

Given a structure $(D,\mathcal{I})$, a formula $\varphi$ describes a subset of $D^n$ where $n$ is the number of free variables occurring in $\varphi$ - or, per the start of this answer, it describes a map $D^n\rightarrow\{True,False\}$.

Title: Difficulty in understanding predicate symbols in FOL

Body:
I'm a beginner in First-Order Logic, and I have the following difficulty in understanding predicate symbols:

From what I know, $n$-ary predicate symbols, are represented by $p_i^n$. If $x$ is a variable, what does $p^1x$ really mean? The $1$ in the superscript simply denotes that $p$ is unary. I know that functions $f_i^n$ applied to terms return a value, i.e. another term $f^nt_1t_2...t_n$. What does $p^1x$ return? True or False, possibly?

I have an example in mind! Consider the binary predicate symbol $<$, and the wff $2 < 1$. This returns False, but is it always the case that predicate symbols return True/False? I'm not quite sure about the definition.

In addition, I was told that given an interpretation $\mathcal{I}$ and a domain of discourse $\mathcal{D}$, each predicate symbol $p_i^n$ is mapped to an $n$-ary relation, i.e. $\mathcal{I}(p_i^n) \subseteq D^n$. Could someone please give an example for this? In the example that I took above for the binary predicate $p^2$ = $<$, $\mathcal{D} = \mathbb{N}$, $<$ seems to be the following map - $<:\mathbb{N}\times \mathbb{N} \to \{T,F\}$ which doesn't seem like $\mathcal{I}(p_i^n) \subseteq D^n$. Where am I going wrong? Could someone please help me better understand these concepts? Thank you!


问:
How calculate: $ P(Z \geq 2\mid X\leq 1) $
I am trying to understand an example from my textbook.
I know that it is conditional probability and joint density, but I don't know on which theory to base myself to solve it.

Let $Z$, be a random variable with distribution $\Gamma \left( z \right) = (2,1)$ and $X$ another random variable such that its distribution given conditional $Z=z$ is $U[0,z]$.
Calculate:
$$ P(Z \geq 2\mid X\leq 1) $$

Other question, how do I identify if they are continuous or discrete variables?
I appreciate your help.

答:
First observe that the joint distribution is

$$f_{XZ}(x,z)=\frac{1}{z}\cdot z e^{-z}=e^{-z}\cdot\mathbb{1}_{[0\leq x\leq z<\infty)}$$

Thus $X\sim Exp(1)$. In fact, integrating you get

$$f_X(x)=\int_x^{\infty}f(x,z)dz$$

that is

$$f_X(x)=e^{-x}\cdot \mathbb{1}_{[0;\infty)}(x)$$

Concluding

$$\mathbb{P}[Z\geq2|X \leq 1]=\frac{\int_0^1dx\int_2^{\infty}e^{-z}dz}{1-e^{-1}}=\dots=\frac{1}{e(e-1)}$$


答:
Why do you denote the gamma distribution as $\Gamma(x)$? Simply write $Z\sim \Gamma(2,1)$.

Bayes' rule tells you that
$$P(Z\geq 2 | X\leq 1)=\frac{P(Z\geq 2,X\leq 1)}{P(X\leq 1)}.$$

The joint probability of $(X,Z)$ is given by
$$f_{(X,Z)}(x,z) = f_{X|Z}(x|z)f_Z(z),$$
where $f_{X|Z}(x|z) = \frac{1}{z} {\bf 1}_{[0,z]}(x)$ and $f_Z(z)=ze^{-z}{\bf 1}_{[0,\infty)}(z)$. Hence,
$$P(Z\geq 2, X\leq 1) = \int_{2}^\infty \int_{0}^1 \frac{1}{z} {\bf 1}_{[0,z]}(x)ze^{-z}{\bf 1}_{[0,\infty)}(z) dxdz=\int_{2}^\infty \int_{0}^1 e^{-z}dxdz =e^{-2}.$$
In the last part we used that
$$\int_0^1 {\bf 1}_{[0,z]}(x) dx = 1 \quad \mbox{since } z\geq 2.$$

On the other hand,
$$P(X\leq 1)= \int_0^\infty \int_0^1  \frac{1}{z} {\bf 1}_{[0,z]}(x)ze^{-z}{\bf 1}_{[0,\infty)}(z) dxdz =  \int_0^\infty \int_0^1  {\bf 1}_{[0,z]}(x)e^{-z}{\bf 1}_{[0,\infty)}(z) dxdz  =$$
$$=\int_0^1 \int_0^1  {\bf 1}_{[0,z]}(x)e^{-z}{\bf 1}_{[0,\infty)}(z) dxdz +  \int_1^\infty \int_0^1  {\bf 1}_{[0,z]}(x)e^{-z} dxdz =$$
$$=\int_0^1 ze^{-z} dxdz +  \int_1^\infty e^{-z} dxdz =1-2e^{-1}+e^{-1} = 1-e^{-1}.$$

Altogether,

$$P(Z\geq 2 | X\leq 1)=\frac{P(Z\geq 2|X\leq 1)}{P(X\leq 1)}=\frac{e^{-2}}{1-e^{-1}} = \frac{1}{e(e-1)}.$$

The idea and procedure is clear: use Bayes' rule, find the joint and marginal distributions and then use the definition of probability. The only tricky part is to be careful with the values of $X$ and $Z$ and the integration regions. I hope this helps, good luck!

Title: How calculate: $ P(Z \geq 2\mid X\leq 1) $

Body:
I am trying to understand an example from my textbook.
I know that it is conditional probability and joint density, but I don't know on which theory to base myself to solve it.

Let $Z$, be a random variable with distribution $\Gamma \left( z \right) = (2,1)$ and $X$ another random variable such that its distribution given conditional $Z=z$ is $U[0,z]$.
Calculate:
$$ P(Z \geq 2\mid X\leq 1) $$

Other question, how do I identify if they are continuous or discrete variables?
I appreciate your help.


问:
How to solve the integral $\int \frac{1}{x^{8}\left(1+x^{2}\right)} \ \mathrm{d} x$?
I encountered a very difficult problem, to calculate the answer of this formula:
$$
\int \frac{1}{x^{8}\left(1+x^{2}\right)} \ \mathrm{d} x
$$
Can you help me to find out how it solved?

Body:
$$\frac{1}{x^8(1+x^2)}=\frac{1+x^2-x^2}{x^8(1+x^2)}=\frac{1}{x^8}-\frac{1}{x^6(1+x^2)}=\cdots$$
and so on, finally you will get:
$$\frac{1}{x^8(x^2 + 1)} = \frac{1}{x^8} - \frac{1}{x^6} + \frac{1}{x^4} - \frac{1}{x^2} + \frac{1}{x^2 + 1}.$$


答: Using partial fraction decomposition (left as an exercise for the reader), notice that $$\frac{1}{x^8(x^2 + 1)} = \frac{1}{x^8} - \frac{1}{x^6} + \frac{1}{x^4} - \frac{1}{x^2} + \frac{1}{x^2 + 1}.$$ Therefore, $$\int \frac{1}{x^8(x^2 + 1)} ~ dx = \int \left[ \frac{1}{x^8} - \frac{1}{x^6} + \frac{1}{x^4} - \frac{1}{x^2} + \frac{1}{x^2 + 1} \right] ~ dx.$$ From here, one can notice that the first $4$ terms have simple anti-derivatives, and the anti-derivative of the last term is $\arctan x.$ One can do a sanity check by computing the derivative of $\arctan x$ to see that it is indeed the case. Therefore, we have that $$\int \left[ \frac{1}{x^8} - \frac{1}{x^6} + \frac{1}{x^4} - \frac{1}{x^2} + \frac{1}{x^2 + 1} \right] ~ dx = \frac{1}{x} - \frac{1}{3x^3} + \frac{1}{5x^5} - \frac{1}{7x^7} + \arctan x + C.$$

Title: How to solve the integral $\int \frac{1}{x^{8}\left(1+x^{2}\right)} \ \mathrm{d} x$?

Body:
I encountered a very difficult problem, to calculate the answer of this formula:
$$
\int \frac{1}{x^{8}\left(1+x^{2}\right)} \ \mathrm{d} x
$$
Can you help me to find out how it solved?


问:
Notation: Address elements of a vector
I am dealing with sets of vectors $\big\{x_1, x_2, x_3, \dotsc, x_m \big\}$ from some abstract vector space $\mathcal{V}$. Occasionally, $\mathcal{V}=\mathbb{R}^n$ an I need to address the elements these vectors e.g. sum over all elements of the vector $x_j$. However, the subindex is already used to denote a specific vector.

Is there a common notation address the elements of a vector? E.g. $x_j[i]$ or $x_j^i$ or $x_j^{(i)}$?

答:
Well, there is no common notation here. The policy is to keep the notation as simple as possible.

In your case, I would write $x_{ij}$ for the $j$th component of vector $x_i$.

Title: Notation: Address elements of a vector


答:
We need to integrate:
$$
I = \int \frac{1}{x^{8}\left(1+x^{2}\right)} \ \mathrm{d} x
$$
Continuing from my comment, substitute $x = tan\theta$. Then, $dx = sec^2\theta. d\theta$. We also know that $1+ tan^2\theta = sec^2\theta$. Substituting back into the integral, we get:

$$I =\int \frac{sec^2\theta}{tan^{8}\theta \left(1+tan^{2}\theta\right)} \ \mathrm{d}\theta$$
$$ \implies I = \int \frac{1}{tan^{8}\theta} \ \mathrm{d}\theta$$
$$ \implies I =\int \cot^8\theta \ \mathrm{d}\theta$$

We can now handle this integral quite easily by repeatedly using the formula, $cot^2\theta = cosec^2\theta - 1$.

We finally get,
$$ I = -\frac{cot^7\theta}{7}+\frac{cot^5\theta}{5}-\frac{cot^3\theta}{3}+cot\theta\ + \theta + C$$

Now substitute back $\theta = tan^{-1}x$ for your final answer!


问:
Wellposedness of differentiation.
A common example of an illposed problem is the following differentiation example:

Let $y(t)\in C^1([0,1])$ with $y(0)=0$. We consider the problem of determining $x(t) = y'(t)$ and show why it is not wellposed due to stability issues. For example, let $y(t) = t$. Then $x(t)=1$. Then let us add a small perturbation $\tilde y(t) = t + \epsilon \sin(kt)$ for $\epsilon >0$ arbitrarily small and $k>0$ arbitrarily large. The perturbation in the input data $y(t)$ is small. The perturbation in the output data $x(t)$ is not because, if for example we use the supremum norm,
$$\max_{t\in [0,1]} | \tilde x - x | =\max_{t\in [0,1]} | 1+k\epsilon \cos (kt)-1 | = k\epsilon$$
which is arbitrarily large.
Now here is where I got confused. Why is the problem wellposed when using the $C^1$ norm, instead of the $C^0$ norm?
The error is then $k^2\epsilon$ which is still arbitrarily large?!

答: The problem is well-posed if the pre-image space is $C^1$. Then the $C^0$-error is still $k\epsilon$, but this is equal to $\|y-\tilde y\|_{C^1}$, hence bounded against the error in the data.

Title: Wellposedness of differentiation.


问:
ODE direction field doesn’t match solution
For the ODE $y’=\sqrt y$ and the initial condition $y(0)=0$ the solution is $y=\frac{1}{4}x^2$. This doesn’t match the direction field for the ODE. I wondered why this was happening here? I feel I am missing/overlooking something very basic

答: On solving the differential equation by variable separable method, we get$$\int\frac{y'}{\sqrt y}=\int dx\implies x-2\sqrt y=C$$The initial condition gives $C=0$. Note that in the solution $x=2\sqrt y$, both $x,y\ge0$. But when you square it to obtain $y=x^2/4$, it becomes defined for negative $x$. The curve over $x<0$ does not actually satisfy the ODE:$$y=x^2/4,x<0\implies \sqrt y=|x/2|=-x/2$$ giving $y'=-\sqrt y$ which is not the original ODE. You are probably seeing the mismatch between direction field and the curve tangents only over $x<0$ (look at the graph).


问:
Satisfying an equation in natural numbers
I am playing a game where I can gain xp from crafting certain things using some material. The xp is given by$$E=250a+375b+500c$$
$a$ represents how many sets of $50$ of the material I will put in, $b$ of $75$ units, and $c$ of $100$ units. Thus, the total quantity I am putting is
$$Q=50a+75b+100c$$
This is divisible by $25$, so I know that if the quantity I have is $Q'=25q+r$, then I will work with $Q=Q'-r$.

It can be seen that $E=5Q$ for any choice of $a$, $b$, and $c$ that satisfies the $Q$ equation.

Given $Q$, how can I find $(a,b,c)$ that satisfy $Q=50a+75b+100c$?

答:
Presumably you are asking for solutions in nonnegative integers. As pointed out by Morgan Rodgers in their comment, you can set $Q=25S$ and $(a,b,c)=(25u,25v,25w)$ and solve $S=2u+3v+4w$.

Note that $S$ and $v$ are either both even or both odd. If $S$ is even, then write $S=2T$ and $v=2x$ and solve $T = u+3x+2w$; if $S$ is odd, then write $S=2T+3$ and $v=2x+1$ and solve $T=u+3x+2w$.

In either case, you can choose any values for $x$ and $w$ such that $3x+2y \le T$, and then $u=T-3x-2y$ is valid and uniquely determined.


问:
Abstract Algebra Automorphism - Is my answer Correct?
Hope you are well.

Im posting the following as a favor on behalf of a friend who is not a native speaker, she just wants to confirm her answer is indeed correct.

Q: show that the set of all automorphism functions of G is a group under composition.

image of full answer

A: (h ∘ g)∘ f = h ∘ (g∘ f)

(h ∘ g)(a)∘ f(a) = h( g(a))∘
f(a) =h [ f(a)g(a) ]

h(a)∘(g ∘ f)(a) =h(a) ∘
[ g(f(a))]= h [f(a)g(a)]

Thanks in advance!

答:
Note that $\operatorname{Aut}(G)\subseteq \operatorname{Sym}(G)$, and then $\operatorname{Aut}(G)$ is a group if and only if $\operatorname{Aut}(G)\le\operatorname{Sym}(G)$. Therefore we have to prove that $\forall \sigma,\tau\in\operatorname{Aut}(G)$, both $\sigma\tau\in\operatorname{Aut}(G)$ and $\sigma^{-1}\in\operatorname{Aut}(G)$, namely that both $\sigma\tau$ and $\sigma^{-1}$ are operation-preserving:

\begin{alignat}{1}
(\sigma\tau)(gh) &= \sigma(\tau(gh)) \\
&= \sigma(\tau(g)\tau(h)) \\
&= \sigma(\tau(g))\sigma(\tau(h)) \\
&= (\sigma\tau)(g)(\sigma\tau)(h) \\
\tag 1
\end{alignat}

and:

\begin{alignat}{1}
gh=gh &\Rightarrow \sigma(\sigma^{-1}(gh))=\sigma(\sigma^{-1}(g))\sigma(\sigma^{-1}(h)) \\
&\Rightarrow \sigma(\sigma^{-1}(gh))=\sigma(\sigma^{-1}(g)\sigma^{-1}(h)) \\
&\Rightarrow \sigma^{-1}(gh)=\sigma^{-1}(g)\sigma^{-1}(h) \\
\tag 2
\end{alignat}


问:
Hirsch's Differential Topology vs Rudin Functional analysis definition of weak and strong topology.
I'm currently reading through chapter 2 - Function spaces of Hirsch's Differential topology I don't however get a feeling of what exactly it means.

From the book:

A weak topology on a compact-open $C^r$ topology on $C^r(M,N)$ is generated by the sets defined as follows. Let $f \in C^r(M,N)$. Let $(\varphi,U),(\psi,V)$ be charts on $M, N$; Let $K \subset U$ be a compact set such that $f(K) \subset V$; let $0 < \epsilon \leq \infty$. Define a weak subbasic neighborhood
$$
\mathcal{N}(f;(\varphi,U),(\psi,V),K,\epsilon)
$$
to be the set $C^r$ maps $g : M \to N$ such that $g(K) \subset V$ and
$$
\left\lVert D^k(\psi f \varphi^{-1})(x) - D^k(\psi g \varphi^{-1})(x)\right\rVert < \epsilon
$$
for all $x \in \varphi(K), k = 0,\ldots,r$. This means the local representation of $f$ and $g$, together with their first $k$ derivatives, are within  $\epsilon$ at each point of $K$.
The weak topology on $C^r(M,N)$ is generated by these sets; It defines a topological space $C_W^r(M,N)$. A neighborhood of $f$ is thus any set containing the intersection of a finite number of sets of this type.

I was comparing this definition with the one I'm more used to from functional analysis (see Rudin's Functional analysis Chapter 3, section 3.8

Suppose next that $X$ is a set and $\mathcal{F}$ is a nonempty family of mappings $f : X \to Y_f$, where each $Y_f$ is a topological space. Let $\tau$ be the collection of all unions of finite intersections of sets $f^{-1}(V)$, with $f \in \mathcal{F}$ and $V$ open in $Y_f$. Then $\tau$ is a topology on $X$, and it is in fact the weakest topology on $X$ that makes every $f \in \mathcal{F}$ continuous: if $\tau'$ is another topology with that property, then $\tau \subset \tau'$. This $\tau$ is called the weak topology on $X$ induced by $\mathcal{F}$, or more succintly, the $\mathcal{F}$-topology of $X$.

I might be wrong here but the two definitions look completely different to me, there's no resemblance. However while in fucntional analysis is the weak topology is the one that makes certain family of mappings all continuous I struggle to understand what weak topology means in the context of differential topology. To me the two have nothing in common.

Can you clarify and maybe explain what Hirsch is trying to capture with his definition?

答:
First of all, charts are there because calculus on manifolds is (typically) done by way of focusing on patches small enough that coordinates are well-defined. So let's focus on the flat case.

Let $U\subseteq \mathbb{R}^n$ and $V\subseteq\mathbb{R}^m$ be open subsets, fix $r,s\in \mathbb{Z}_{\geq0}$ with $r\geq s$ and consider the set $C^r(U,V)$ of all $r$-times continuously differentiable functions from $U$ to $V$. We would like to endow $C^r(U,V)$ with a $C^s$-topology of sorts, which should incorporate the first $s$ derivatives when it comes to proximity, the zeroth derivative of a function being the function itself. In other words for two functions $f,g\in C^r(U,V)$, we want

$$f\approx_{C^s} g \iff f \approx g, f' \approx g', f'' \approx g'', ... , f^{(s)}\approx g^{(s)},$$

where $\approx$ means "approximately", and derivatives of functions are certain multilinear functions (e.g. as in Lang's Fundamentals of  Differential Geometry).

We could consider the product topology, which would give the topology of pointwise convergence of each derivative. Or we may use the supremum (i.e. $C^0$-) norm for each one of the derivatives, which would give the topology of uniform convergence. In fact since all the functions on the RHS are continuous if we restrict our attention to compact subsets $K\in\mathcal{K}(U)$ of $U$ the supremum (semi-)norm becomes nicer to deal with, and we get a topology more rigid (= stronger = finer = larger = more expensive = higher resolution = harder convergence) than the product topology and more flexible (= weaker = coarser = smaller = cheaper = lower resolution = easier convergence) than the uniform topology. See the discussion here for further details on this comparison.

To that end define a one parameter family $\mathfrak{g}_\bullet^s$ of seminorms on $C^r(U,V)$ by

$$\mathfrak{g}_\bullet^s:\mathcal{K}(U)\to F(C^r(U,V), \mathbb{R}_{\geq0}),\quad
K\mapsto \left[f\mapsto \max_{k\in\{0,1,...,s\}}\max_{x\in K}\left|f^{(k)}(x)\right|\right].$$

(Here the target is the set of all functions from $C^r(U,V)$ to $\mathbb{R}_{\geq0}$.)

Using this family of seminorms parametrized by compact subsets instead of just one norm (e.g. the supremum norm on the whole open set $U$) gives uniform convergence on compacta (or compact convergence), and the topology $\mathcal{N}^s$ generates is precisely the topology of uniform convergence of the first $s$ derivatives on compacta.

Observe that this is not the unique family of seminorms that does the job that we set forth above (though taking the maximum over the order of derivatives has the benefit of playing nicely with intersection). Then the weak subbasic $C^s$-neighborhoods $\mathcal{N}^s$ of Hirsch would precisely be

$$\mathcal{N}^s: C^r(U,V)\times \mathcal{K}(U)\times ]0,\infty]\to \mathcal{P}(C^r(U,V)), \quad (f,K,\varepsilon)\mapsto \{g\in C^r(U,V)\mid \mathfrak{g}_K^s(f-g)<\varepsilon\}.$$

(Here the target is the set of all subsets of $C^r(U,V)$.)

To generalize this to the manifold case we need to replace $C^r(U,V)$ by $C^r(M,N)$ for $M,N$ two manifolds. But the condition for a function to be in $C^r(M,N)$ is it being in $C^r(U,V)$ where $U\in\operatorname{Chart}(M)$ and $V\in\operatorname{Chart}(N)$, which is why the neighborhood-valued map $\mathcal{N}^s$ now gets two additional parameters $\operatorname{Chart}(M)$ and $\operatorname{Chart}(N)$.

Now let's focus on Rudin's definition of weak topology. Let me rephrase it in a more categorical way (seeing spaces parametrized by functions bothers me; the targets and functions come at the same time, the way I see it).

Consider the forgetful functor $\operatorname{Forget}: \operatorname{Top}\to \operatorname{Set} , (S,\mathcal{T}(S))\mapsto S$, let $X\in \operatorname{Set}$ and consider the comma category $(X\searrow \operatorname{Forget})$. An object in $(X\searrow \operatorname{Forget})$ is a pair $(f,Y)$ where $Y$ is a topological space and $f:X\to\operatorname{Forget}(Y)$ is a function.

For any $(f,Y)\in(X\searrow \operatorname{Forget})$, we could pull back the topology of $Y$ via $f$ (i.e. we could look at preimages; which are denoted by $^\ast$ throughout) and we would get a subset $f^\ast(\mathcal{T}(Y))$ of $\mathcal{P}(X)$. Likewise for any collection $\mathfrak{F}\subseteq(X\searrow \operatorname{Forget})$ we can put all the pullbacks together and consider

$$\bigcup_{(f,Y)\in\mathfrak{F}}f^\ast(\mathcal{T}(Y))\subseteq \mathcal{P}(X).$$

Let us denote by $\mathcal{T}(X,\mathfrak{F})$ the smallest topology of $X$ that contains $\bigcup_{(f,Y)\in\mathfrak{F}}f^\ast(\mathcal{T}(Y))$, i.e.

$$\mathcal{T}(X,\mathfrak{F})=\bigcap\left\{\mathcal{T}\subseteq\mathcal{P}(X)\left\vert \mathcal{T}\supseteq \bigcup_{(f,Y)\in\mathfrak{F}}f^\ast(\mathcal{T}(Y))\mbox{ and }\mathcal{T}\mbox{ is a topology}\right\}\right..$$

There are certain universal properties associated to $\mathcal{T}(X,\mathfrak{F})$ as expected. In particular $\mathcal{T}(X,\mathfrak{F})$ is the smallest topology that makes all maps $f:(X,\mathcal{T}(X,\mathfrak{F}))\to (Y, \mathcal{T}(Y))$, $(f,Y)\in\mathfrak{F}$, continuous. See wikipedia for further categorical musings. Also see the discussion here.

Finally let's put these two things together. We are trying to topologize $C^r(U,V)$, so set $X:=C^r(U,V)$. Here are some families in $(C^r(U,V)\searrow \operatorname{Forget})$ that are worthy of mention in our discussion:

$\mathfrak{F}_1^s:=\{(\mathfrak{g}_{\{x\}}^s,\mathbb{R}_{\geq0})\mid x\in U\}$ (singletons are compact subsets)

$\mathfrak{F}_2^s:=\{(\mathfrak{g}_K^s,\mathbb{R}_{\geq0})\mid K\in\mathcal{K}(U)\}$ (seminorms we defined above)

$\mathfrak{F}_3^s:=\{(\mathfrak{h}_{f,K}^s:=\mathfrak{g}_K^s(\bullet-f),\mathbb{R}_{\geq0})\mid K\in\mathcal{K}(U), f\in C^r(U,V)\}$ (translates of the seminorms in $\mathfrak{F}_2^s$)

$\mathfrak{F}_4^s:=\left\{\left(g\mapsto \max_{k\in\{0,1,...,s\}}\sup_{x\in U}\left|g^{(k)}(x)-f^{(k)}(x)\right|, \mathbb{R}_{\geq0}\right)\mid f\in C^r(U,V)\right\}$

$\mathfrak{F}_5^s:=\left\{\left(D^k,C^0\left(U,\operatorname{Lin}\left({(\mathbb{R}^n)}^{\otimes k}, \mathbb{R}^m\right)\right)\right)\mid k\in \{0,1,...,s\}\right\}$ ($D^k: f\mapsto f^{(k)}$ is the $k$th derivative operator)

In the first four options the targets are topologized in the expected manner. They are somewhat more general in that their analogs apply to more general situations. The final option is making use of differentiation specifically. Here the set $\operatorname{Lin}\left({(\mathbb{R}^n)}^{\otimes k}, \mathbb{R}^m\right)$ of $k$-linear functions from $\mathbb{R}^n$ to $\mathbb{R}^m$ are topologized by the operator norm (say), and the target set is endowed with the compact-open topology (= $\mathcal{N}^0$ topology = topology of uniform convergence on compacta). I have suppressed some topolinear isomorphisms and embeddings here because I think I'm already pushing your patience.

It is straightforward that $\mathcal{T}(C^r(U,V),\mathfrak{F}_1^s)$ is the topology of pointwise convergence of the first $s$ derivatives and that $\mathcal{T}(C^r(U,V),\mathfrak{F}_4^s)$ is the topology of uniform convergence of the first $s$ derivatives.

Here are some observations to compare the remainder (I'll suppress the quantifiers for the sake of sanity):

Observation 1: $\mathcal{N}^s(f,K,\varepsilon) = (\mathfrak{h}_{f,K}^s)^\ast([0,\varepsilon[).$

Observation 2: $|\mathfrak{h}_{f,K}^s(g)-\mathfrak{h}_{f,K}^s(h)|\leq \mathfrak{g}_K^s(g-h).$

Observation 3: $|\mathfrak{g}_K^s(f)-\mathfrak{g}_K^s(g)|\leq \mathfrak{g}_K^s(f-g).$

Observation 4: $\mathfrak{h}_{0,K}^s=\mathfrak{g}_K^s$, provided that $0\in V$.

Observation 5: $\mathcal{N}^s(f,K,\varepsilon)=\bigcap_{k\in\{0,1,...,s\}}(D^k)^\ast(\mathcal{N}^0(f^{(k)},K,\varepsilon))$.

All these observations are straightforward to verify.

Observation 1 shows that the $\mathcal{N}^s$-topology is contained in any topology w/r/t which the $\mathfrak{h}_{\bullet,\bullet}^s$ are continuous, hence

$$\mathcal{N}^s\mbox{-topology } \subseteq \mathcal{T}(C^r(U,V),\mathfrak{F}_3^s).$$

Observation 2 shows that the $\mathfrak{h}_{\bullet,\bullet}^s$ are ($1$-Lipschitz) continuous when the domain is endowed with the $\mathcal{N}^s$-topology, so

$$\mathcal{N}^s\mbox{-topology } \supseteq \mathcal{T}(C^r(U,V),\mathfrak{F}_3^s).$$

Likewise, by Observation 3 the $\mathfrak{g}_\bullet^s$ are ($1$-Lipschitz) continuous when the domain is endowed with the $\mathcal{N}^s$-topology, so

$$\mathcal{N}^s\mbox{-topology } \supseteq \mathcal{T}(C^r(U,V),\mathfrak{F}_2^s).$$

By Observation 4, if $0\in V$, so that the function that is constantly $0$ is in $C^r(U,V)$, then

$$\mathcal{T}(C^r(U,V),\mathfrak{F}_2^s)\subseteq \mathcal{T}(C^r(U,V),\mathfrak{F}_3^s).$$

I believe the assumption that $V$ contains the origin is mild. Indeed, a general $V$ need not contain $0$, but since $V$ is supposed to be a chart of a manifold we could arrange it to contain $0$.

This final inclusion could be strict, provided that $V$ contains a nonzero $y$ and its negative $-y$. The point is that antipodes are not distinguishable by open sets coming from $\mathfrak{g}_\bullet^s$; see the discussion here.

Finally by Observation 5 we have that

$$\mathcal{N}^s\mbox{-topology }= \mathcal{T}(C^r(U,V),\mathfrak{F}_5^s),$$

which I think is the most interesting claim. In words, it says that the $\mathcal{N}^s$-topology on $C^r(U,V)$ is the smallest topology that makes the first $s$ derivatives continuous operators. An alternative reformulation by way of multiindex formalism is in Kelley & Namioka's Linear Topological Spaces, p. 82:

Putting all the inclusions and equalities together we have:

\begin{align*}
\mathcal{T}(C^r(U,V),\mathfrak{F}_1^s)
&\subseteq \mathcal{T}(C^r(U,V),\mathfrak{F}_2^s)\\
&\subseteq \mathcal{N}^s\mbox{-topology }
= \mathcal{T}(C^r(U,V),\mathfrak{F}_3^s)
= \mathcal{T}(C^r(U,V),\mathfrak{F}_5^s) \\
&\subseteq \mathcal{T}(C^r(U,V),\mathfrak{F}_4^s).
\end{align*}

(Of course on top of this we also have a grading w/r/t the order $s$ of derivatives; the higher the order of derivatives the smaller the open sets become. This would give refinements.)


答:
Below I have a plot of the slope field and of $y(x) = \frac14x^2$.

We see that the right half fits the slope field perfectly, but the left half doesn't. So clearly something is wrong with the left half.

Not really. Or, it depends. Note that $\frac14x^2$ perfectly fulfills $(y')^2 = y$. So one could say that the non-fit is a result of choosing the "wrong sign" of the square root to the left of the $x$-intersection.

But also note that $y(x) = 0$ is a perfectly valid solution to your differential equation. And for any $a\in \Bbb R$, setting $y(x) = 0$ for $x\leq a$ and $y(x) = \frac14(x-a)^2$ for $x>a$, is also a valid solution.

So with your initial condition, and with the original differential equation $y' = \sqrt y$, all we can really say is that $a\geq 0$, possibly "$a = \infty$" (i.e. $y(x) = 0$ everywhere). With an initial condition like $y(1) = \frac14$, on the other hand, we fix $a = 0$.

Title: ODE direction field doesn’t match solution

Body: For the ODE $y’=\sqrt y$ and the initial condition $y(0)=0$ the solution is $y=\frac{1}{4}x^2$. This doesn’t match the direction field for the ODE. I wondered why this was happening here? I feel I am missing/overlooking something very basic


问:
If $X$ and $Y$ are Lévy processes with $X_t\sim Y_t$ for all $t$, can we infer that $X\sim Y$?
Let $(\Omega,\mathcal A,\operatorname P)$ be a probability space, $(\mathcal F_t)_{t\ge0}$ be a filtration on $(\Omega,\mathcal A)$, $E$ be a $\mathbb R$-Banach space and $(X_t)_{t\ge0}$ be an $E$-valued process on $(\Omega,\mathcal A,\operatorname P)$. Remember that $X$ is called $\mathcal F$-Lévy if

$X$ is $\mathcal F$-adapted;

$X_0=0$;

$X_{s+t}-X_s$ and $\mathcal F_s$ are independent for all $s,t\ge0$;

$X_{s+t}-X_s\sim X_t$ for all $s,t\ge0$.

Assume $X$ is $\mathcal F$-Lévy and let $(Y_t)_{t\ge0}$ be another $E$-valued $\mathcal F$-Lévy process on $(\Omega,\mathcal A,\operatorname P)$ (or possibly on another probability space) with $X_t\sim Y_t$ for all $t\ge0$.

Are we able to show that $X$ and $Y$ have the same (finite-dimensional) distribution(s)?

EDIT: To be precise, the question is whether we can show that $$\operatorname P\left[X_{t_1}\in B_1,\ldots,X_{t_k}\in B_k\right]=\operatorname P\left[Y_{t_1}\in B_1,\ldots,Y_{t_k}\in B_k\right]\tag1$$ for all $B_1,\ldots,B_k\in\mathcal B(E)$, $k\in\mathbb N$ and $0\le t_1<\cdots<t_k$ or even $$\operatorname P\left[X\in B\right]=\operatorname P\left[Y\in B\right]\;\;\;\text{for all }B\in\mathcal B\left(E^{[0,\:\infty)}\right)\tag2.$$

答:
Not sure what the question is, but in general, $X_t \sim Y_t$ for all $t$ does not imply that $X\sim Y$ as random variables in the (inifnite dimensional) space of functions $[0,T]\to E$. For example, $X_t\sim Y_t$ for all $t$ does not necessarily imply that $\int_0^1 X_tdt \sim \int_0^1 Y_tdt$, for the latter to hold you need the stronger condition that $X\sim Y$ as random variables.

This said, if your question is whether $(X_{t_1},\dots,X_{t_n})\sim (Y_{t_1},\dots,Y_{t_n})$ for arbitrary $t_1,\dots,t_n$, $n\in \mathbb{N}$ (equality of finite dimensional distributions) the answer is yes. Let us just look at $n=2$ for the sake of simpler notation: Take $0\leq s\leq t<\infty$ then,

$$\varphi_{(X_t,X_s)}(u,v) = \mathbb{E}[e^{iuX_t + iv X_s }] = \mathbb{E}[e^{iu(X_t-X_s+X_s) + iv X_s }] = \mathbb{E}[e^{iu(X_t-X_s)}] \mathbb{E}[e^{i(u+v)(X_s)}] = \mathbb{E}[e^{iu(X_{t-s})}] \mathbb{E}[e^{i(u+v)(X_s)}] =\varphi_{X_{t-s}}(u) \varphi_{X_{s}}(u+v),$$
where we used first that $X_t-X_s$ is independent of $X_s$ and then that $X_t-X_s\sim X_s$. Now, you can finally use that $X_t\sim Y_t$ and hence $\varphi_{X_t}(u)=\varphi_{Y_t}(u)$ and go backwards. As a result,
$$(X_{t_1},\dots,X_{t_n})\sim (Y_{t_1},\dots,Y_{t_n}),$$
which, as said, is a consequence of the independent and stationary increments.


问:
How to calculate Risk Function given that we have loss function and pdf
Given $f(x|\theta)= \frac{\mathrm{e}^{-\theta}\theta^x}{x!} , x = 0,1,2,...,\theta$ and $\mathcal{L}(a-\theta)^2$

a) Calculate $\mathcal{R}(\theta,d)$ for $d(x)=x$

b) Given the gamma density $\pi(\theta) = \frac{\lambda^\alpha\theta^{\alpha-1}\mathrm{e}^{-\lambda\theta}}{\Gamma(\alpha)}$. $\mathcal{L}>0, \lambda>0, \theta>0$.
Calculate $\mathcal{r}(\pi,d)$ for $d(x)=x$.

From what I know I guess:
$\mathcal{R}$ is Risk Function and $\mathcal{L}$ is Loss function.

This is an exercise problem in Introduction to Statistical Theory by Charles J. Stone, Paul Gerhard Hoel, and Sidney Charles Port. (Chapter 1 Exercise 7). Someone please help me understand and solve this problem.

答:
a)

$\mathcal{R}(\theta, d) = \sum_{x=0}^{\infty} \mathcal{L}(\theta, a)f(\theta|x)$  - This is the formula to calculate Risk Function for discrete distributions.

Given, $\mathcal{L}(a-\theta)^2$ , we can substitute this in the formula

$ = \sum_{x=0}^{\infty} (x-\theta)^2 f(\theta|x)$

$= \sum_{x=0}^{\infty} (x^2+\theta^2-2x\theta) f(\theta|x)$

$= \sum_{x=0}^{\infty} x^2f(\theta|x) + \sum_{x=0}^{\infty} \theta^2f(\theta|x) - \sum_{x=0}^{\infty} 2x\theta(\theta|x) $

Actually if we look closely we can tell that $f(\theta|x)$ is Poisson Distribution. Therefore,

$= E[X^2] + \theta^2(1) - 2\theta E[X] $

$= \theta + \theta^2 + \theta^2 - 2\theta^2$

$= \theta$

$\mathcal{R}(\theta, d) = \theta$

b)

$r(\pi, d) = E[\mathcal{R}|\theta, d]$ - This is the formula to calculate mean risk

We calculated $\mathcal{R}(\theta, d) = \theta$, in section (a). Using that

$= \int_{0}^{\infty} \theta \pi(\theta) \, d\theta $

$= \int_{0}^{\infty} \frac{\lambda^{\alpha}\theta^{(\alpha +1)-1}\mathrm{e}^{-\lambda\theta}}{\Gamma(\alpha)} \, d\theta $

$= \frac{\alpha}{\lambda}\int_{0}^{\infty} \frac{\lambda^{\alpha + 1}\theta^{(\alpha +1)-1}\mathrm{e}^{-\lambda\theta}}{\Gamma(\alpha + 1)} \, d\theta $

$= \frac{\alpha}{\lambda} (1)$         -- (Because of Gamma Distribution - pdf)

$= \frac{\alpha}{\lambda}$

$r(\pi, d) = \frac{\alpha}{\lambda}$


问:
Vectors : finding scalar $\mu$ in $\overrightarrow a = \mu \overrightarrow b + 4\overrightarrow c $

Non-zero vectors $\overrightarrow a ,\overrightarrow b \& \overrightarrow c $ satisfy $\overrightarrow a .\overrightarrow b  = 0$, $\left( {\overrightarrow b  - \overrightarrow a } \right).\left( {\overrightarrow b  + \overrightarrow c } \right) = 0$ and $2\left| {\overrightarrow b  + \overrightarrow c } \right| = \left| {\overrightarrow b  - \overrightarrow a } \right|$. If $\overrightarrow a  = \mu \overrightarrow b  + 4\overrightarrow c $, then find all possible values of $\mu $.

My approach is as follows

Given $\left( {\overrightarrow b  - \overrightarrow a } \right).\left( {\overrightarrow b  + \overrightarrow c } \right) = 0\& 2\left| {\overrightarrow b  + \overrightarrow c } \right| = \left| {\overrightarrow b  - \overrightarrow a } \right|$

$\overrightarrow u  = \left( {\overrightarrow b  - \overrightarrow a } \right);\overrightarrow v  = \left( {\overrightarrow b  + \overrightarrow c } \right)$

$2\left| {\overrightarrow v } \right| = \left| {\overrightarrow u } \right| \Rightarrow 4\left( {{{\left| {\overrightarrow b } \right|}^2} + {{\left| {\overrightarrow c } \right|}^2} + 2\overrightarrow b .\overrightarrow c } \right) = \left( {{{\left| {\overrightarrow b } \right|}^2} + {{\left| {\overrightarrow a } \right|}^2} - 2\overrightarrow a .\overrightarrow c } \right)$

$\overrightarrow a  = \mu \overrightarrow b  + 4\overrightarrow c  \Rightarrow \overrightarrow b  - \overrightarrow u  = \mu \overrightarrow b  + 4\left( {\overrightarrow v  - \overrightarrow b } \right)$

$ \Rightarrow \overrightarrow b .\overrightarrow v  - \overrightarrow u .\overrightarrow v  = \mu \overrightarrow b .\overrightarrow v  + 4\left( {\overrightarrow v .\overrightarrow v  - \overrightarrow b .\overrightarrow v } \right) \Rightarrow 5\overrightarrow b .\overrightarrow v  = \mu \overrightarrow b .\overrightarrow v  + 4{\left| {\overrightarrow v } \right|^2}$\

$ \Rightarrow 5\overrightarrow b .\left( {\overrightarrow b  + \overrightarrow c } \right) = \mu \overrightarrow b .\left( {\overrightarrow b  + \overrightarrow c } \right) + 4{\left| {\overrightarrow v } \right|^2} \Rightarrow 5\overrightarrow b .\overrightarrow b  + 5\overrightarrow b .\overrightarrow c  = \mu \overrightarrow b .\overrightarrow b  + \mu \overrightarrow b .\overrightarrow c  + 4{\left| {\overrightarrow b } \right|^2} + 4{\left| {\overrightarrow c } \right|^2} + 8\overrightarrow b .\overrightarrow c $

Not able to proceed from here onward

答:
As $\vec a \perp \vec b$ like $\hat i \perp \hat j$, it's strategic to represent

$$ \vec c = \frac{1}{4}\vec a - \frac{\mu}{4}\vec b$$

Remaining two conditions are,

$$ (\vec b - \vec a)\cdot(\frac{1}{4}\vec a + \frac{4-\mu}{4}\vec b)=0$$

and

$$ (\vec b - \vec a)\cdot(\vec b - \vec a) = 2^2(\frac{1}{4}\vec a + \frac{4-\mu}{4}\vec b)\cdot(\frac{1}{4}\vec a + \frac{4-\mu}{4}\vec b)$$

which should give you a quadratic in $\mu$.


问:
Give an $\epsilon$-$\delta$ proof of the following: For all $a>0$, $f(x) = 1/x$ is continuous at $a$
I'm fairly new to epsilon-delta proofs as we've just started learning them. I'm very confused about choosing a delta for this, do we have to choose an arbitrary delta with a range? Why so?
Thank you very much!</p>

答:
Consider $|x-a|0;$

Then $a/2 <x <(3/2)a;$

Let $\epsilon >0$ be given.

Choose $\delta =\min (a/2, \epsilon(a^2/2))$;

Then $|x-a|<\delta$ implies

$|1/x-1/a|=$

$\dfrac{|x-a|}{|xa|} \le (2/a^2)|x-a| <$

$(2/a^2)\delta < \epsilon$.

Body:
Consider $|x-a|<a/2; a>0;$

Then $a/2 <x <(3/2)a;$

Let $\epsilon >0$ be given.

Choose $\delta =\min (a/2, \epsilon(a^2/2))$;

Then $|x-a|<\delta$ implies

$|1/x-1/a|=$

$\dfrac{|x-a|}{|xa|} \le (2/a^2)|x-a| <$

$(2/a^2)\delta < \epsilon$.


问:
Find the max. and min. points of $g(x,y,z) = x^2 + y^2 + z^2$
The problem

Let $X$ a submanifold of $\mathbb{R}^3$ defined by $$ X = \left\{ (x,y,z) \mid (x^2 + y^2 + z^2 - 5)^2 = 16(1-z^2) \right\} $$

Let $\Phi(x,y,z) = (x^2 + y^2 + z^2 - 5)^2 - 16(1-z^2) $. Let $f$ a function defined by
$$ f : \mathbb{R}^3 \rightarrow \mathbb{R} \mbox{, } (x,y,z) \mapsto x^2 + y^2 + z^2 $$

We define $g$ by the restriction of $f$ on $X$ : $ g = \left. f \right|_X $

Give the points in which the differential of $g$ is zero and deduce the max. and min. points.

My works

$X$ is compact and $g$ is continuous on $X$, then $g$ has a global max. and min. point. Thus, $g$ is $\mathscr{C}^{\infty}$ on $X$. Then, we can apply the method of Lagrange multipliers. We have

$$ \frac{ \partial g }{ \partial x } = 2x \mbox{, } \frac{ \partial g }{ \partial y } = 2y \mbox{, } \frac{ \partial g }{ \partial z } = 2z $$

and

$$ \frac{ \partial \Phi }{ \partial x } = 4x(x^2 + y^2 + z^2 - 5) \mbox{, } \frac{ \partial \Phi }{ \partial y } = 4y(x^2 + y^2 + z^2 - 5) \mbox{, } \frac{ \partial \Phi }{ \partial z } = 4z(x^2 + y^2 + z^2 + 3) $$

We get the following system

$$ \left( \begin{array}{c}
2x\\ 2y\\ 2z\\
\end{array} \right) = \lambda \left( \begin{array}{c}
4x(x^2 + y^2 + z^2 - 5) \\
4y(x^2 + y^2 + z^2 - 5) \\
4z(x^2 + y^2 + z^2 + 3) \\
\end{array} \right) $$
with $\lambda$ a real scalar. And I don't know how to resolve this system... Actually, I am not sure if this is the system we have to get. I think I made a mistake somewhere but I don't know where.

Thank you for your help.

答:
Your working is correct -

$\left( \begin{array}{c}
2x\\ 2y\\ 2z
\end{array} \right) = \lambda \left( \begin{array}{c}
4x(x^2 + y^2 + z^2 - 5) \\
4y(x^2 + y^2 + z^2 - 5) \\
4z(x^2 + y^2 + z^2 + 3) \\
\end{array} \right)$

Also, $(x^2 + y^2 + z^2 - 5)^2 = 16(1-z^2)$.

From above system of equations, please note the possible points to check for max and min.

One is $(x = 0, y = 0, z = 0)$ but it does not meet the fourth equation constraint.

If $x \ne 0$ then from first equation $\lambda = \frac{1}{2(x^2+y^2+z^2-5)}$ but that will lead to $z = 0$ so you have another set of points $(x, y, 0)$.

Similarly if $z \ne 0$ then from the third equation $\lambda = \frac{1}{2(x^2+y^2+z^2+3)}$ but that leads to $x = y = 0$. So the next set of possible points are $(0, 0, z)$.

Taking points $(x, y, 0)$ and plugging into the fourth equation, you have -

$x^2+y^2 - 5 = \pm4 \implies x^2+y^2 = 9$ or $x^2+y^2 = 1$.

Substituting $(0, 0, z)$ in fourth equation shows no real solution.

So max and min of $x^2 + y^2 + z^2$ for the given equality constraint seem to be $9$ and $1$ resp, and for all points on $x^2 + y^2 = 9, z = 0$ and $x^2 + y^2 = 1, z = 0$ resp.


答:
You want to rotate the surface so that the direction $(1,0,0)$ becomes $\dfrac1{\sqrt3}(1,1,1)$. A possible rotation matrix is

$$\begin{pmatrix}\frac1{\sqrt3}&0&-\frac2{\sqrt6}
\\\frac1{\sqrt3}&\frac1{\sqrt2}&\frac1{\sqrt6}
\\\frac1{\sqrt3}&-\frac1{\sqrt2}&\frac1{\sqrt6}\end{pmatrix}$$

Now your equation is

$$\frac{(u+v+w)^2}3-\frac{(v-w)^2}2-\frac{(-2u+v+w)^2}6=1$$

or

$$u^2+v^2+w^2-4uv-4vw-4wu+3=0.$$

Caution: sign and permutations errors are guaranteed, I didn't check. But the final result must not be far from correct.

https://www.wolframalpha.com/input/?i=plot+x%5E2%2By%5E2%2Bz%5E2-4xy-4yz-4zx%2B3%3D0

Title: Rotating the hyperboloid of two sheets $z^2-y^2-x^2=1$

Body:
Revolving $f(x)=\frac{1}{x}$ about $f(x)=x$ yields the two-sheeted hyperboloid $2xy-2=z^2$ in $\Bbb R^3.$

I'm trying to rotate the surface so that the forward sheet lies in the first octant $(+,+,+)$ and the backward sheet lies in $(-,-,-).$

What's the equation?

Here's what I have so far. This is the graph of $2xy-2=z^2:$

It would also work to rotate the two-sheeted hyperboloid $z^2-y^2-x^2=1.$

My attempt was to use a $3$d rotation matrix that rotates points about the origin. I was able to find the right rotation matrices, but I could not reason what to multiply the rotation matrix with to get the desired result. I could probably get the final equation, with a few hints.

I believe I have to change the entries in the matrix $Q$ (as shown in the hint below by @NinadMunshi). I'm not sure what to change them to, but the new entries should rotate the hyperboloid.


问:
What are the prerequisites to Jech's Set theory text?
I'm looking for a book to self-study axiomatic set theory, and heard this was a classic.  What are the main prerequisites for this text? My knowledge of set theory isn't too great. Probably the only time I came across nontrivial set theory was when I read the proof that every nonzero ring has a maximal ideal (Zorn's Lemma).</p>

答:
Jech's tome (at least the Third Millennium Edition with which I am familiar) does start out at the very basics, going through each of the axioms of $\sf{ZFC}$ is quite a bit of detail.  To a certain extent it is quite self-contained.

However, at the very least it would be beneficial to have seen some basic first-order logic (including Gödel's Completeness and Incompleteness Theorems).

Additionally, it is quite brisk, eventually leading to topics that are at the forefront of current research (for example, some exercises have as references articles written this century).  I would not recommend this as a first serious set-theory text, especially for self-study.  In my opinion, its main purpose is as more of a reference than a text book.

I find that the two volume Discovering Modern Set Theory by W. Just and M. Weese is a very apt introduction to set-theory.  The first volume is fittingly entitled "The Basics" and the second "Set-Theoretic Tools for Every Mathematician".  They do not cover forcing, but do cover the related concept of Martin's Axiom.


问:
Difficulty in understanding predicate symbols in FOL
I'm a beginner in First-Order Logic, and I have the following difficulty in understanding predicate symbols:

From what I know, $n$-ary predicate symbols, are represented by $p_i^n$. If $x$ is a variable, what does $p^1x$ really mean? The $1$ in the superscript simply denotes that $p$ is unary. I know that functions $f_i^n$ applied to terms return a value, i.e. another term $f^nt_1t_2...t_n$. What does $p^1x$ return? True or False, possibly?

I have an example in mind! Consider the binary predicate symbol $<$, and the wff $2 < 1$. This returns False, but is it always the case that predicate symbols return True/False? I'm not quite sure about the definition.

In addition, I was told that given an interpretation $\mathcal{I}$ and a domain of discourse $\mathcal{D}$, each predicate symbol $p_i^n$ is mapped to an $n$-ary relation, i.e. $\mathcal{I}(p_i^n) \subseteq D^n$. Could someone please give an example for this? In the example that I took above for the binary predicate $p^2$ = $<$, $\mathcal{D} = \mathbb{N}$, $<$ seems to be the following map - $<:\mathbb{N}\times \mathbb{N} \to \{T,F\}$ which doesn't seem like $\mathcal{I}(p_i^n) \subseteq D^n$. Where am I going wrong? Could someone please help me better understand these concepts? Thank you!

答:
$\newcommand{\sem}[1]{[\![#1]\!]^{(\mathcal{D}, \mathcal{I}), v}  }
\newcommand{\semI}[1]{[\![#1]\!]^{(\mathcal{D}, \mathcal{I}), v'}  }
\newcommand{\tpl}[1]{\langle #1 \rangle}$

We can write "$\sem{\cdot}$" to mean "the semantic value of the string of symbols $\cdot$ in the model with domain $\mathcal{D}$, interpretation $\mathcal{I}$ and variable assignment $\mathcal{v}$".

The semantic value of a term (= a variable like $x$, an individual constant like $0$, or a function symbol applied to a matching number of arguments like $5 + 4$) will be an object from the domain.

The semantic value of a formula (= a predicate applied to a matching number of arguments like $2 < 1$, or a complex formula involving connectives and quantifiers like $\forall x (x > 0 \to even(x))$) will be a truth value.

So your guess is correct: An $n$-ary predicate symbol applied to an appropriate number of arguments returns a truth value. This follows from the definition of the semantics of FOL:

If $P$ is an $n$-ary predicate symbol and $t_1, \ldots, t_n$ are terms, then
$\sem{P(t_1, \ldots t_n)}=
\begin{cases}
\text{True} & \text{iff} \langle \sem{t_1}, \ldots, \sem{t_n} \rangle \in \sem{P}\\
\text{False} & \text{otherwise}
\end{cases}$

The semantic value of non-logical symbols (= individual constants, function symbols and predicate symbols) is determined by the interpretation function; and indeed, the interpretation of an $n$-ary predicate symbol is an $n$-ary relation on $\mathcal{D}$:

If $c$ is a constant symbol, then $\sem{c} = \mathcal{I}(c) \in \mathcal{D}$

If $f$ is an $n$-ary function symbol, then $\sem{f} = \mathcal{I}(f) : \mathcal{D}^n \to \mathcal{D}$

If $P$ is an $n$-ary predicate symbol, then $\sem{P} = \mathcal{I}(P) \subseteq \mathcal{D}^n$

A reasonable interpretation for the languge in your example would be

$\mathcal{D} = \mathbb{N}\\
\mathcal{I}(0) = 0, \mathcal{I}(1) = 1, \ldots\\
\mathcal{I}(<) = \{\tpl{x, y} \in \mathbb{N} \times \mathbb{N}: x < y\} = \{\tpl{0, 1}, \tpl{0, 2}, \ldots, \tpl{1, 2}, \tpl {1, 3}, \ldots\}$

We then have

$\sem{2 < 1} = \text{True}\\
\Longleftrightarrow \tpl{\sem{2}, \sem{1}} \in \sem{P}\\
\Longleftrightarrow \tpl{\mathcal{I}(2), \mathcal{I}(1)} \in \mathcal{I}(P)\\
\Longleftrightarrow\ \tpl{2, 1} \in \{\tpl{x,y}: x < y\}\\
\text{Since} \tpl{2, 1} \not \in \{\tpl{x,y}: x < y\},\\
\sem{2 < 1}= \text{False}$

One could say that interpretation of an $n$-ary predicate symbol itself is an $n$-ary relation on the domain, and the process of evaluating an atomic formula, i.e. a predicate occurring together with an argument vector, consists of implicitly forming its characteristic function (see Noah Schweber's post) to combine the interpretation of the predicate symbol as a set and the interpretation of its arguments as objects into a truth value; this is what happens in the definition of $\sem{P(t_1, \ldots t_n)}$ above.

If $x$ is a variable, what does $p^1x$ really mean?

Variables are like pronouns: $p^1x$ means "It is $P$", or "This is $P$". To make sense of variables, we need an assignment function, which maps variables to objects in the domain:

$v: VAR \to \mathcal{D}$

If $x$ is a variable, then $\sem{x} = v(x) \in \mathcal{D}$

A variable assignment is a way of pointing at objects: By fixing e.g. $v: x \mapsto 1$, I am pointing with my finger at the object $x$, thereby giving a meaning to the pronoun "it". Without specifying an assignment, i.e. pointing at an object, we can not make sense of the pronoun.

We often have to deal with more than one variable, but luckily I also have more than one finger: The assignment $v : x \mapsto 1, y \mapsto 0$ could mean that with my left index finger which I'm using when saying "this" I'm pointing at 1 and with my right hand which I use for "that" I'm pointing at 0. Under this particular assignment, the formula $x < y$, "This is smaller than that" means that 1 is smaller than 0, so $\sem{x < y}$ comes out as false; under another assignment where I point $v': x \mapsto 3, y \mapsto 5$, it means that 3 is smaller than 5, and $\semI{x < y}$ = True.

Usually variables occur in the scope of quantifiers. $\forall x (x > 0 \to even(x))$ means "Every way of pointing at an object makes the proposition "If this is larger than 0, then it is even"  true". Likewise, $\exists x (x > 0 \land even(x))$ means "I can find a way of pointing at an object so that the proposition "This object is larger than 0 and even" true." The semantics of the quantifers $\forall, \exists$ is to loop over variable assignments, i.e. ways of pointing at things.

$\sem{P(x)}$ then means "$P$ holds of the object $x$", whatever it is that $v$ tells us $x$ refers to: $\sem{P(x)} = \text{True iff } v(x) \in \mathcal{I}(P)$.


问:
Find the inverse of the matrix
Find the inverse of the matrix

```<code>B=    1 0 2
      0 -1 1
      1 1 0

Thats easy -1  2  2
            1 -2 -1
            1 -1 -1
</code>```

But how do i show this

```<code>Hence solve BX = -3   where X = x1
                  4             x2
                  5             x3
</code>```

答:
Let

$$
C = \begin{bmatrix}
-3 \\
4 \\
5
\end{bmatrix}
$$

Now,
$$
\begin{eqnarray}
BX&=&C\\
B^{-1}BX &=& B^{-1}C\\
X = B^{-1}C &=& \begin{bmatrix}
-1 & 2 & 2 \\
1 & -2 & -1 \\
1 & -1 & -1
\end{bmatrix} \cdot
\begin{bmatrix}
-3 \\
4 \\
5
\end{bmatrix} = \begin{bmatrix}
3+8+10 \\
-3-8-5\\
-3-4-5
\end{bmatrix} = \begin{bmatrix}
21 \\
-16 \\
-12
\end{bmatrix}\\
X=\begin{bmatrix}
x_1 \\
x_2 \\
x_3
\end{bmatrix} &=& \begin{bmatrix}
21 \\
-16 \\
-12
\end{bmatrix}
\end{eqnarray}
$$

Title: Find the inverse of the matrix


问:
Finding out all questions in test bank
Let’s Say that a school has a test bank for an online multiple choice math exam which contains 100 questions. When a student starts his exam the system randomly selects 15 questions for him.
If my friends and I decide to find out all the 100-questions, how many trials do we need if that is possible??

Edit:(as suggested in comments)

How many trials are needed, on average, to see all the questions??

答:
The answer was intended to be a comment but appears to be too long for this.

Let the full number of questions be $q$ and the number of questions on a ticket be $t$. All questions on a ticket are assumed to be distinct, so that $1\le t\le q$. We are looking for the expected value of the number of tickets required to reveal all $q$ questions. This is an extended version of the classical coupon collector problem (where $t=1$).

As shown by Peter Franek the probability that not all questions are covered after $k$ drawn tickets is
$$
Q_k=\frac{-\sum_{i=1}^{q-t} (-1)^i\binom qi\binom{q-i}t^k}{\binom qt^k}.
$$

From this (as already mentioned in a comment) the expected value in question can be computed as:
$$
E(T)=1+\sum_{k=1}^\infty Q_k=1-\sum_{i=1}^{q-t}
\frac{(-1)^i\binom qi}{\frac{\binom qt}{\binom{q-i}t}-1}.
$$

It can be checked that the last expression reduces to the correct value $qH_q$ for $t=1$.

Particularly for $q=100, t=15$ one obtains: $E(T)\approx 32.5588$.

Title: Finding out all questions in test bank


答:
This IVP
$$
y'=\sqrt{y}, \quad y(0)=0
$$
does not enjoy uniqueness. (This is because the flux $\sqrt{y}$ is not sufficiently smooth.)

It possesses infinitely many solutions:
$$
y_c(x)=\max\left\{0,\frac{1}{4}(x-c)^2\right\}, \quad c \ge 0.
$$


问:
Sequential compactness of unit sphere implies sequential compactness of closed balls
Let $(V,\left\lVert\cdot\right\rVert)$ be a normed vector space whose unit sphere $\left\{v \in V: \left\lVert v \right\rVert = 1\right\}$ is sequentially compact. Show that any closed ball $\left\{v \in V: \left\lVert v \right\rVert \leq R\right\}$ must be sequentially compact. Show that $V$ is complete.

My only idea was to note that if $(v_n)$ is a sequence in the closed ball, then $\left(\frac{v_n}{\left\lVert v_n \right\rVert}\right)$ is a sequence in the unit sphere, so has a convergent subsequence, but that doesn't really seem to help.

答:
You have started the proof correctly. Now consider two cases:

Case 1) $0$ is  a limit point of $(\|v_n\|)$. In this case show that $(v_n)$ has a subsequence converging to $0$.

Case 2. There is a subsequence of $(\|v_n\|)$ converging to some positive number $r$. In this case use $v_n=\frac {v_n} {\|v_n\|} \|v_n\|$ to get a subsequence of $(v_n)$ which converges.


答:
In Banach spaces, where we often consider elements of sequence spaces (e.g. $x\in c_0$ so that $x=(x_1, x_2, \ldots)$) the problem of indexing comes up a lot.  Typically when we need to index into a sequence of sequences we use $x_j(n)$ to indicate the $n^{\mathrm{th}}$ component of the $j^{\mathrm{th}}$ element of the master sequence.  I would disagree with Wuestenfux here and say that this is reasonably standard in Banach Space Theory.

Other suggestions, such as raised indices ($x^i_j$) run into problems when you need to consider powers of the series, and multiple sub-indices ($x_{i,j}$) can get confusing and hard to read (there's a Banach space called Schreier space where some of the proofs require considering indices $j$ such that $p_{n_k +1} \leq j \leq p_{n_{k+1}+1}$ which is not only hard to read, but hard to think about!).  Provided you are clear that $x_j$ refers to a sequence and not a function you shouldn't have much difficulty with people understanding $x_j(n)$.

That said, whatever your choice, state it clearly upfront :)

Title: Notation: Address elements of a vector

Body:
I am dealing with sets of vectors $\big\{x_1, x_2, x_3, \dotsc, x_m \big\}$ from some abstract vector space $\mathcal{V}$. Occasionally, $\mathcal{V}=\mathbb{R}^n$ an I need to address the elements these vectors e.g. sum over all elements of the vector $x_j$. However, the subindex is already used to denote a specific vector.

Is there a common notation address the elements of a vector? E.g. $x_j[i]$ or $x_j^i$ or $x_j^{(i)}$?


问:
How does the rewriting of the following two equations work?
I am failing to understand the proof of coming to the steady-state formula in queueing theory. This is probably due to the fact that I may have forgotten (and cannot find it back) some of the algebra from my school days. The step I am missing is the transition from the first to the second statement below:

$$(\lambda + \mu)\frac{\lambda}{\mu}P_0 = \lambda P_0 + \mu P2$$

$$\frac{\lambda^2}{\mu}P0 + \lambda P_0 = \lambda P_0 + \mu P2$$

My main question is how does $$(\lambda + \mu)\frac{\lambda}{\mu}P_0$$ become
$$\frac{\lambda^2}{\mu}P_0 + \lambda P_0$$
and what did happen to: $$\mu$$

Can anybody help me out?

答:
Recall the distributive law: If $a,b,c$ are terms, we have $(a+b)c=ac+bc$. In this case,

$$(\lambda+\mu)\frac{\lambda}{\mu}P_0 = \lambda\frac{\lambda}{\mu}P_0 + \mu\frac{\lambda}{\mu}P_0= \frac{\lambda \cdot \lambda}{\mu}P_0 + \frac{\mu \cdot \lambda}{\mu}P_0 = \frac{\lambda^2}{\mu}P_0 + \lambda P_0.$$

I hope this helps. Let me know if you need more detail.


问:
proof $S_n$ has no subgroup of order $ \frac {n!}{4}$

proof $S_n$ has no subgroup of order $ \frac {n!}{4}$for $n>5$.

We know only subgroup of index $2$ in $S_n$ is $A_n$ and $A_n$ is simple group  And we know every  non trivial subgroup of $S_n$ has atleast one even permutation now let $K$ be subgroup of $S_n$ with order $ \frac {n!}{4}$ then $K \cap A_n$ is not trivial and $K \cap A_n$ is subgroup of $A_n$ but how we can show $K \cap A_n$ is not normal in $A_n$ ?

答:
This is a standard trick I learned from my supervisor Jef Laga.

$n\geq 5$. Suppose $K\leq S_n$ with index $4$. From $A_n$ simple some easy manipulations would tell you the only normal subgroups of $S_n$ are $1$, $S_n$ or $A_n$.

Consider the action of $S_n$ on the cosets of $K$ by left multiplication. This gives a homomorphism $\varphi: S_n\rightarrow S_4$, and $\ker \varphi \trianglelefteq S_n$. One could easily see $\ker \varphi \neq S_n$ as the action is transitive by definition; and $\ker \varphi \neq 1$ as $\varphi$ cannot be injective ($n>4$). Therefore $\ker \varphi =A_n$. This would clearly cause a contradiction, as $\ker \varphi \leq K$.

Similarly, one could show that the index $t$ of a subgroup in $S_n$ cannot have $2<t<n$.


问:
Show that the $G$-action on $R$ induces a $G$-action on $R/I$ defined by ${}^g \pi(r) = \pi({}^gr).$
Here is the question that I want to answer part(d) in it:

Let $R$ be a ring and $I \subset R$ a two-sided ideal, with quotient homomorphism $\pi : R \rightarrow R/I.$ Let $\operatorname{End_{I}(R)}$ be the set of $\varphi \in \operatorname{End(R)}$ such that $\varphi(I) \subset I,$ and let $\operatorname{Aut_{I}(R)} = \operatorname{End_{I}(R) \cap Aut(R)}.$

$(a)$ Given $\varphi \in \operatorname{End_{I}(R)},$ show that there exists $\bar{\varphi} \in \operatorname{End(R/I)}$ such that $\bar{\varphi} \pi = \pi \varphi.$

$(b)$ Given $\varphi, \psi \in \operatorname{End_{I}(R)},$ show that $\overline{\varphi \psi} = \bar{\varphi} \bar{\psi}.$

$(c)$ Given $\varphi \in \operatorname{Aut_{I}(R)},$ show that $\varphi(I) = I$ and $\bar{\varphi} \in \operatorname{Aut(R/I)}.$\ Hint: Use part $(b).$

$(d)$ Let $G$ be a group which acts on $R$ by ring automorphisms and restricts to $I.$ Show that the $G-$action on $R$ induces a  $G-$action on $R/I$ defined by ${}^g \pi(r) = \pi({}^gr).$

My question is:

If I managed to prove parts (a), (b) and (c), could anyone give me a hint on how to prove (d)?

答:
It seems to me that after a) b) and c) have been checked part d) is just an obvious rephrasing of them.

Just convince yourself that to have a group $G$ acting on $R$ by automorphisms amounts to have a group homomorphism $G\rightarrow{\rm Aut}(R)$ and then that given that the image lands in ${\rm Aut}_I(R)$ the induced map
$$
G\rightarrow{\rm Aut}(R/I),\qquad g\mapsto\bar\phi_g
$$
is again an homomorphism.


问:
Every finite-dimensional vector space over $\mathbb{F}$ has a unique Hausdorff topology?
When I was reading my notes, I encountered a problem that I don't know how to start with.

Prove that if $V$ is a finite-dimensional vector space over $\mathbb{F}$ there is a unique topology on $V$ making it a Hausdorff topological vector space.

My guess is that we can construct a topology on $V$ and show that it is Hausdorff and unique but what kind of topology should I start with? I was thinking to start with a topology that every points are closed s.t. it is Hausdorff...

答:
There is an isomorphism $T: V \to F^{n}$ for some $n$ and we can  declare $U$ to be open in $V$ if $T(U)$ is open in $F^{n}$. This makes $V$ a Hausdorff TVS.

It is a well known fact that all linear maps on finite dimensional TVS's are continuous.  Hence, if $\tau_1$ and $\tau_2$ are two Hausdorff vector topologies on $V$ then the identity map $: (V,\tau_1) \to (V,\tau_2)$ is a homeomorphism which proves that $\tau_1=\tau_2$.


答:
Let $\vec a=t \vec i$, $\vec b= s \vec j$ then $\vec c=(t\vec i-\mu s \vec j)/4$.
$$2|\vec b+ \vec c|=|\vec a- \vec b| \implies |s \vec j+(t/4) \vec i-(s\mu/4)\vec j|=|t \vec i- s\vec j|$$
$$\implies (4-\mu)^2s^2+t^2=t^2+s^2\implies (\mu^2-8\mu+12)s^2=3t^2~~~(1)$$
Next $$(\vec b-\vec c).(\vec a- \vec b)=0 \implies  t^2=(4-\mu) s^2~~~(2)$$
From (1) and (2) equate $t^2/s^2$, to get
$$3(4-\mu)=\mu^2-8\mu+12 \implies \mu^2-5\mu=0 \implies \mu =0, 5$$

Title: Vectors : finding scalar $\mu$ in $\overrightarrow a = \mu \overrightarrow b + 4\overrightarrow c $

Body:
Non-zero vectors $\overrightarrow a ,\overrightarrow b \& \overrightarrow c $ satisfy $\overrightarrow a .\overrightarrow b  = 0$, $\left( {\overrightarrow b  - \overrightarrow a } \right).\left( {\overrightarrow b  + \overrightarrow c } \right) = 0$ and $2\left| {\overrightarrow b  + \overrightarrow c } \right| = \left| {\overrightarrow b  - \overrightarrow a } \right|$. If $\overrightarrow a  = \mu \overrightarrow b  + 4\overrightarrow c $, then find all possible values of $\mu $.

My approach is as follows

Given $\left( {\overrightarrow b  - \overrightarrow a } \right).\left( {\overrightarrow b  + \overrightarrow c } \right) = 0\& 2\left| {\overrightarrow b  + \overrightarrow c } \right| = \left| {\overrightarrow b  - \overrightarrow a } \right|$

$\overrightarrow u  = \left( {\overrightarrow b  - \overrightarrow a } \right);\overrightarrow v  = \left( {\overrightarrow b  + \overrightarrow c } \right)$

$2\left| {\overrightarrow v } \right| = \left| {\overrightarrow u } \right| \Rightarrow 4\left( {{{\left| {\overrightarrow b } \right|}^2} + {{\left| {\overrightarrow c } \right|}^2} + 2\overrightarrow b .\overrightarrow c } \right) = \left( {{{\left| {\overrightarrow b } \right|}^2} + {{\left| {\overrightarrow a } \right|}^2} - 2\overrightarrow a .\overrightarrow c } \right)$

$\overrightarrow a  = \mu \overrightarrow b  + 4\overrightarrow c  \Rightarrow \overrightarrow b  - \overrightarrow u  = \mu \overrightarrow b  + 4\left( {\overrightarrow v  - \overrightarrow b } \right)$

$ \Rightarrow \overrightarrow b .\overrightarrow v  - \overrightarrow u .\overrightarrow v  = \mu \overrightarrow b .\overrightarrow v  + 4\left( {\overrightarrow v .\overrightarrow v  - \overrightarrow b .\overrightarrow v } \right) \Rightarrow 5\overrightarrow b .\overrightarrow v  = \mu \overrightarrow b .\overrightarrow v  + 4{\left| {\overrightarrow v } \right|^2}$\

$ \Rightarrow 5\overrightarrow b .\left( {\overrightarrow b  + \overrightarrow c } \right) = \mu \overrightarrow b .\left( {\overrightarrow b  + \overrightarrow c } \right) + 4{\left| {\overrightarrow v } \right|^2} \Rightarrow 5\overrightarrow b .\overrightarrow b  + 5\overrightarrow b .\overrightarrow c  = \mu \overrightarrow b .\overrightarrow b  + \mu \overrightarrow b .\overrightarrow c  + 4{\left| {\overrightarrow b } \right|^2} + 4{\left| {\overrightarrow c } \right|^2} + 8\overrightarrow b .\overrightarrow c $

Not able to proceed from here onward


问:
Quantum mechanics for mathematicians
I'm looking for books about quantum mechanics (or related fields) that are written for mathematicians or are more mathematically inclined.

Of course, the field is very big so I'm in particular looking for books that explain how C*-algebras and von Neumann-algebras are used, I want to see "practical" applications of those.

What I know is not much more than Griffiths' book (I don't like that book!) contains. I have in my possession the books by Claude Cohen-Tannoudji but they don't seem to contain this.

答:
It may not have the specific topics you want, but I like Leon Takhtajan's book entitled, coincidentally, Quantum Mechanics for Mathematicians.

Books like Dirac tended to frustrate me with drawn-out developments of "this is a ket, this is a bra, they have such-and-such properties and we combine them in the following ways."  Takhtajan is kind enough to come out and say "take a Hilbert space and a self-adjoint operator."

Title: Quantum mechanics for mathematicians


问:
Is there a continuous Taylor/MacLaurin transform (like the Fourier transform)?
My consideration might be total nonsense (as a high school student, I lack the mathematical knowledge to really check my idea), but I was just wondering whether one could find a continuous generalization of the Taylor series comparable to the continuous Fourier transform.

Just like the Fourier transform decomposes a function $f(t)$ into a sum of trigonometric functions with a result $\hat{f}(\omega)=\mathcal{F}(f)(\omega)$ with the domain of frequencies, could one define a Taylor transform $\mathcal{T}(f)$ that would operate on $n$th derivatives?

Instead of the trigonometric functions in Fourier transform, $f$ would be decomposed into a sum of polynomials of form $\frac{1}{n!}x^n$.

So is it a valid operation to generalize the Taylor series for continuous values of $n$ (i.e. using integrals instead of sums)? My naive approach would be

$$f(t) = \intop_0^{\infty} \mathcal{T}(f)(n)\cdot \frac{t^n}{\Gamma(n+1)} \mathrm{d}n$$

So in fact, the transform would define (or at least require) non-integer derivatives of a function. Do these exist (like a non-integer iterates of functions can be defined too)?

Example:
Assuming that $\frac{\mathrm{d}^ne^x}{\mathrm{d}x^n}(0) = 1$ for any $x\in \mathbb{R}$, I'd come up with

$$e^t = \intop_0^{\infty} \frac{t^n}{\Gamma(n+1)} \mathrm{d}n$$ as a continuous version of

$$e^t = \sum_{n=0}^{\infty} \frac{t^n}{n!}$$

Thus $$\mathcal{T}(e^t)(n) = 1,\,n\in\mathbb{R}$$

Is a transform like I explained possible or - in case it is - even somehow useful?

答:
as David Z says, the Mellin transform is basically what you found.
if you look at Ramanujans's master theorem on wikipedia and change the notation a bit you will find the desired relation that for an analytic function $f$ which has a (almost Taylor) expansion

$$
  f(x) = \sum_{n=0}^\infty \frac{(-x)^n}{n!} \mathcal{T}[f](n)
$$

with

$$
  \mathcal{T}[f](-s) = \int_{0}^\infty \frac{x^{s-1}}{\Gamma(s)} f(x) dx
$$

(I am not sure, however, why they use this minus-sign convention)


问:
Putnam 1990: Problem A-4
I was told this lovely puzzle recently which I thought people here might enjoy.

Consider a paper punch that can be centered at any point
of the plane and that, when operated, removes from the
plane precisely those points whose distance from the
center is irrational. How many punches are needed to
remove every point?

答:
Three punches suffice if they are made at collinear points $A,B,C$ such that $B$ is the midpoint of $AC$ and the square of the length of $AB$ is irrational. For any point $P$, Apollonius's theorem ensures that
$$
PA^2+PC^2-2\cdot PB^2=2\cdot AB^2.
$$
As the right hand side is irrational, it follows that one of the distances $PA,PB,PC$ is irrational. Therefore the point $P$ has been removed.

Note that Apollonius's theorem applies even when the triangle $\Delta PAC$ is degenerate.

Title: Putnam 1990: Problem A-4


问:
Could someone please explain sin, cos, and tan in a simple way?
I haven't taken this in school yet but I love math and physics and join lots of competitions. I have a problem with sine, cosine, and tangent, that I really need a SIMPLE explanation along with an example. I know for a first that they are related to angles and are used in physics, so how are they calculated and how are they used?

Thanks in advance

答:
It is all about triangles and relations between the side lengths and angle amplitudes.

In a right triangle, by definition,

$$\cos(\theta)=\frac AC$$

$$\sin(\theta)=\frac BC$$

$$\tan(\theta)=\frac BA$$

Using these three interrelated functions, you can solve a real lot of geometric problems, such as relations between angles and sides of a general triangle, or even a triangle drawn on a sphere, and do topography, mechanics, geography, GPS, optics, astronomy... and more.

It turns out that these functions also have a deep meaning in maths and they are about as essential as the four basic operations and exponentiation. But this is an advanced topic.

To compute them, calculators have built-in algorithms, mostly based on approximation polynomials. E.g., for an angle expressed in radians,

$$\cos(\theta)\approx1-\frac{\theta^2}2+\frac{\theta^4}{4!}-\frac{\theta^6}{6!}+\cdots$$

and the more terms you add, the closer you get to the exact value.

Amazingly, to solve the apparently innocuous equation

$$4x^3-3x=\frac13,$$
you need the trigonometric functions.