问:
Limit of a squared function over a cyclometric function
I am starting with limits, and this is a problem I stumbled upon:
$$ \lim_{x\to 0} \frac{\sqrt{e^{x^2}-1}}{arctan(x)}$$

This is how I proceeded:
$$ \lim_{x\to 0} \frac{\sqrt{e^{x^2}-1}}{arctan(x)} = \lim_{x\to 0} \frac{\sqrt{e^{x^2}-1}}{arctan(x)} \frac{1}{\frac{x}{x}} = \lim_{x\to 0} \frac   {\frac{\sqrt{e^{x^2}-1}}{x}}  {\frac{arctan(x)}{x}} = \lim_{x\to 0}\frac{\sqrt{e^{x^2}-1}}{x} $$
And this is where I got stuck. I would appreciate if someone could point me the right way as I am unsure about how to proceed. EDIT: The problem states that using derivatives is prohibited.

答:
$\lim_{x\to 0^-}\frac{\sqrt{e^{x^2}-1}}{x}=-\lim_{x\to 0^-} \sqrt{\frac{e^{x^2}-1}{x^2}}=-1$. Why? let's remember that in general $\displaystyle\lim_{f(x)\to 0}\frac{e^{f(x)}-1}{f(x)}=1$.

Now $\lim_{x\to 0^+}\frac{\sqrt{e^{x^2}-1}}{x}=\lim_{x\to 0^+} \sqrt{\frac{e^{x^2}-1}{x^2}}=1$. So the limit does not exist since the two limit sides are not equal.

EDIT: be careful that the difference on sign for the two limits it is due by the fact that $\sqrt{x^2}=|x|$, so in case $x\rightarrow 0^-$ comes out a minus sign!


问:
Determining the closed form of a particular inhomogeneous recursive relation
In my research I have come to the point where I would like to know the expression for $a_n$ which satisfies
\begin{equation}
a_n=2a_{n+1}-a_{n+2}+4, \hspace{4mm}a_1=3, \hspace{2mm}a_2=9
\end{equation}

Let me tell you where this come from. I have this sequence of numbers:
\begin{equation}
3,9,19,33,51,...
\end{equation}
You can see that the difference between two numbers is equal to the difference between the previous two numbers, increased by 4, e.g., $9-3=6$, $19-9=10=6+4$, ...
Hence I can write
\begin{equation}
a_{n+2}-a_{n+1}=a_{n+1}-a_{n}+4
\end{equation}
from which the above equation follows.

I am aware of the general procedure for solving these recursive equations. I started to look for a solution of the homogeneous equation, by setting $a_n=Ac^n$. This gives the equation $c^2-2c+1=0$, solved by $c=1$.
Then I looked for a particular solution, say $a_n=b$. However, this gives $b=2b-b+4$, from which $4=0$. Clearly this method does not work here.

Do you have an idea on how to find $a_n$? Does this equation even have a closed solution?

答:
$c=1$ is a double root of your characteristic equation $c^2-2c+1=0$, so that gives not just one but two linearly independent solutions of the homogeneous equation: $a_n = 1$ and $a_n = n$.  Since polynomials of degree $0$ and $1$ are solution of the homogeneous equation, your particular solution should be of the form $a_n = b n^2$.

You should end up with $a_n = 2 n^2 + 1$.

BTW, you could also try looking up your sequence in the OEIS.

ContentLicense: CC BY-SA 4.0


问:
Evaluating $\sum_{k=1}^{\infty} \frac{\cos^2(k)}{4k^2-1}$
What options do I have for this series? No idea how to do it.

$$\sum_{k=1}^{\infty} \frac{\cos^2(k)}{4k^2-1}$$

答:
Express the sum as follows:

$$\sum_{k=1}^{\infty} \frac{\cos^2{k}}{4 k^2-1} = \frac12 \sum_{k=1}^{\infty} \frac{1}{4 k^2-1} +  \frac12 \sum_{k=1}^{\infty} \frac{\cos{2 k}}{4 k^2-1}$$

Now, the first sum on the RHS is equal to $1/2$.  This may be shown using residue theory very easily:

$$\sum_{k=-\infty}^{\infty} \frac{1}{4 k^2-1} = -\text{Res}_{z=\pm 1/2} \frac{\pi \cot{\pi z}}{4 z^2-1} = 0$$

which means that

$$2 \sum_{k=1}^{\infty} \frac{1}{4 k^2-1} - 1 = 0$$

For the other sum, break into partial fractions and reorganize.  You end up with

$$\begin{align}\sum_{k=1}^{\infty} \frac{\cos{2 k}}{4 k^2-1} &= \cos{2} +  \sum_{k=1}^{\infty} \frac{\cos{2 (k+1)}-\cos{2 k}}{2 k+1}\\ &= \cos{2} - \sin{1} \sum_{k=1}^{\infty} \frac{\sin{(2 k+1)}}{2 k+1}\end{align}$$

That last sum may be derived from the well-known sum

$$\sum_{k=0}^{\infty} \frac{\sin{(2 k+1)}}{2 k+1} = \frac{\pi}{4}$$

Therefore, I get as the sum

$$\begin{align}\sum_{k=1}^{\infty} \frac{\cos^2{k}}{4 k^2-1} &= \frac14 + \frac14 \left (\cos{2} - \frac{\pi}{4}\sin{1} + 2 \sin^2{1} \right )\\ &= \frac{1}{2} - \frac{\pi}{8} \sin{1}\end{align}$$

You may verify that this checks out numerically in Mathematica or WA.

ADDENDUM

The "well-known" sum may be evaluated by considering the series expansion of the function $\text{arctanh}{z}$:

$$\text{arctanh}{z} = \sum_{k=0}^{\infty} \frac{z^{2 k+1}}{2 k+1}$$

so that

$$
\begin{align}\sum_{k=0}^{\infty} \frac{\sin{(2 k+1)}}{2 k+1} &= \Im{\left [\sum_{k=0}^{\infty} \frac{e^{i(2 k+1)}}{2 k+1}\right]}\\ &= \frac12 \Im{\left[\text{arctanh}{e^i}\right]}\\ &= \frac12 \Im{\left[\log{\left (\frac{1+e^i}{1-e^i} \right )}\right]}\\ &= \frac12 \Im{\left[\log{\left (i \frac{\cos{1/2}}{\sin{1/2}} \right )}\right]}\\ &= \frac{\pi}{4} \end{align}$$

ContentLicense: CC BY-SA 3.0


问:
Prove that $\lim_{n\to\infty}\sin(nx)/ \pi x$ is a delta function
I need to prove that $\delta_n(x)=\sin (nx)/\pi x$ is a delta function. That is, to prove that: $$\underset{n\rightarrow \infty}{\lim}\int^\infty_{-\infty}f(x)\frac{\sin( nx)}{\pi x}dx=f(0).$$ For that I made $y=x/n$ and took the limit under the integral since $f(x)$ is supposed to be analytic in $\mathbb{R}$. I have a fundamental limit for the $\sin$ leaving me with $$\int^\infty_{-\infty}f(0)\cdot1 \;dy.$$ But this is equal to $f(0)$ $\iff$ $f(x)$ have contribuition only at $x=0$, but here the only assumption is that $f(x)=0$ when $x\rightarrow \pm \infty$.

What is wrong here?

答:
There are indeed many ways to understand the conclusion, and they are all worthwhile. But some may be less reassuring or persuasive than others... of course depending on taste and prior experience.

For me, at this moment, ...  let $\langle,\rangle$ be the pairing of Schwartz functions and tempered distributions. By an extended form of Plancherel,
$$
\langle f,u\rangle \;=\; \langle \widehat{f},\widehat{u}\rangle
$$
for Schwartz $f$ and tempered $u$. If we get things normalized properly, for a suitable version $u_n(x)={\sin nx\over x}$, $\widehat{u}_n$ is the characteristic function of $[-n,n]$. This computation is most easily done by invoking Fourier inversion after computing the Fourier transform of that characteristic function. These converge as tempered distributions to $1$, which is the Fourier transform of $\delta$...


问:
What is a matrix of tuples?
Consider the prisoner's dilemma, in this game, you have a matrix,

$$A = \begin{bmatrix} (2,2) & (0,5) \\ (5,0) & (1,1) \end{bmatrix}$$

(or something like this)

But what is this object exactly?

Is $$A \in \mathbb{R}^{2^\mathbf{N} \times 2^\mathbb{N}}$$
I can't seem to know how to think about this object concretely.

答: I would say it’s a member of $(\Bbb R^2)^{2\times 2}$. If $A$ is some ring, $A^{2\times 2}$ is the set of two by two matrices with elements in $A$.

Title: What is a matrix of tuples?


问:
Show set of vectors with element-wise product is convex
Let C = $ \{x \in \mathbb{R}^n | \Pi x_i \geq 1\} $. The goal is to show that this is convex.

I have tried to use the definition with looking at the line segment formed by two vectors in C, but I was unable to conclude that the segment was in C. Instead, I am trying to find a convex set for which C is a sub-level set, but I am having no luck there either because I can't find an example of a convex set for which C would "fit in"

答:
This is the same as

$$
\left\{ x \in \mathbb{R}^n \mid \sum_{i=1}^n \log(x_i) \geq 0 \right\}
= \left\{ x \in \mathbb{R}^n \mid - \sum_{i=1}^n \log(x_i) \leq 0 \right\},
$$

the sublevel set of the convex function $-\sum_{i=1}^n \log(x_i)$ (since $\log$ is concave).


问:
Proving that $A\oplus\mathbb{C}\ni(a,\lambda)\mapsto L_{a}+\lambda I\in B(A)$ is injective
Let $A$ be a non-unital $C^{\star}$-algebra and let $B(A)$ be the Banach algebra of bounded linear operators from $A$ to itself. For $a\in A$ we consider the left-multiplier $L_{a}\colon A\to A$ defined by $L_{a}(b):=ab$. I am trying to prove that $$\phi\colon A\oplus\mathbb{C}\to B(A),\qquad\phi(a,\lambda):=L_{a}+\lambda I$$ is an injective (linear) map.

If $L_{a}+\lambda I=0$, then $ab+\lambda b=0$ for all $b\in A$. So I was looking for $b$'s that yield $a=0$ and $\lambda=0$. If I can first prove that $\lambda=0$, then it is easy to see that $a=0$ by considering $b:=a^{*}$ ($C^{\star}$-identity).

Thus I need to find a $b$ that guarantees $\lambda=0$. I can't figure out this seemingly easy looking problem. Any help would be greatly appreciated!

EDIT: It seems that I haven't yet used the fact that $A$ is non-unital.

答:
Suppose $\lambda\ne 0$. Then $ab+\lambda b=0$ implies $b=(-\frac{a}{\lambda})b$ for all $b\in A$. So if we let $e=-\frac{a}{\lambda}$ then $e$ is a left identity of $A$. By taking involution on the equality $b^*=eb^*$ we also obtain $b=be^*$ for all $b\in A$, which means $e^*$ is a right identity of $A$. Now, using the standard trick:

$e^*=ee^*=e$

And so $e$ is a two sided identity of $A$, a contradiction.


问:
Base of vector space $C([a,b], f)$
Let $ C[a, b] = F:[a, b] \rightarrow \mathbb {R}$ continuous functions.

Let $ \beta = \lbrace 1, x, x ^ {2}, x ^ {3}, ... \rbrace $ (with infinite elements), prove that $ \beta $ is not the basis of the space $ C [a, b ] $.

To do this, it must be shown that one of the following is not true:

$ \beta $ is a generator set.

$ \beta $ is linearly independent.

In section 1. take the function $ f(x) = nx $, $ \forall n \in \mathbb{N} $ such that:

$ f (x) = \lambda_{1} + \lambda_{2} x + \lambda_{3} x^{2}+... $

$ \Rightarrow nx= \lambda_{1} + \lambda_{2} x + \lambda_{3} x^{2}+... $

$\Rightarrow \lambda_{1} =0, \lambda_{2}=n, \lambda_{3}=0, \lambda_{4}=0,...$

But, $ \lambda_{2} = n $ is not unique, therefore it does not generate $f$.

Is this test correct for 1.?

How can I test section 2.?

答:
I do not really know what you consider a basis. But clearly, $nx \in \mathrm{span}\lbrace \beta \rbrace$. Just look again.

If you mean basis in an approximation sense, then consider Stone-Weierstraß-theorem.

If you mean it in a finite-dimensional sense, then your question boils down to whether all functions in $C[a, b]$ are polynomials. But they clearly are not: Note that for any polynomial $p$, if $k := \mathrm{deg}(p)$, then its $k+1$-st derivative vanishes. But for $f: [a, b] \rightarrow \mathbb{R}$, $f(x) := \exp(x)$ then $f^{(n)}(x) = \exp(x) \neq 0$ for all $n \in \mathbb{N}$.

Title: Base of vector space $C([a,b], f)$


问:
How to divide a Sine wave into n equal power/area sections?
I'm working on an electronics project which involves using a small micro-controller to dim an AC Mains light bulb. I want to keep it simple and instead of having a continuous dimming adjustment I'll limit it to 'n' settings. So Off, 1/10th brightness, 2/10th brightness... on up to 9/10ths and 10/10 full brightness, always on. I can't do that by simply dividing the sine wave in 10 equal durations, as in 1/10th the period, 2/10ths the period etc. Brightness is related to power which is related to the area under the curve.

I only have to consider one positive sine half period section. The same time values would be used for the negative half of the wave's period.

Area under the curve involves integrating the sine wave, if I remember my school calculus correctly, which is a cosine function, but the dividing it into equal areas has me stumped. I think I have 0%, 50% and 100% brightness figured out, but that's about it ;)

This question has probably been answered already, but I'm probably asking it wrong. There's probably a maths way of expressing the question in a way a search engine could find.

答:
Almost there. The area under the first half-period is $2$. You're therefore looking for the constants $a_i$ such that

$$\int_{a_i}^{a_{i+1}} \sin x dx = \frac2n$$

The integral is indeed (negative) cosine: the left-hand side is
$$-\cos x\big|_{a_i}^{a_{i+1}} = -\cos(a_{i+1}) + \cos(a_i)$$

Now, the first segment starts at $a_0 = 0$, so we have
$$-\cos(a_1) + \cos(0) = \frac2n$$
so that $a_1 = \cos^{-1}\left(1 - \frac2n\right)$.

Can you take it from here?


Title: Determining the closed form of a particular inhomogeneous recursive relation

Body:
In my research I have come to the point where I would like to know the expression for $a_n$ which satisfies
\begin{equation}
a_n=2a_{n+1}-a_{n+2}+4, \hspace{4mm}a_1=3, \hspace{2mm}a_2=9
\end{equation}

Let me tell you where this come from. I have this sequence of numbers:
\begin{equation}
3,9,19,33,51,...
\end{equation}
You can see that the difference between two numbers is equal to the difference between the previous two numbers, increased by 4, e.g., $9-3=6$, $19-9=10=6+4$, ...
Hence I can write
\begin{equation}
a_{n+2}-a_{n+1}=a_{n+1}-a_{n}+4
\end{equation}
from which the above equation follows.

I am aware of the general procedure for solving these recursive equations. I started to look for a solution of the homogeneous equation, by setting $a_n=Ac^n$. This gives the equation $c^2-2c+1=0$, solved by $c=1$.
Then I looked for a particular solution, say $a_n=b$. However, this gives $b=2b-b+4$, from which $4=0$. Clearly this method does not work here.

Do you have an idea on how to find $a_n$? Does this equation even have a closed solution?

Body:
Note that
$$
\eqalign{
  & a_{\,n + 2}  - 2a_{\,n + 1}  + a_{\,n}  = \Delta ^{\,2} a_{\,n}  = 4\quad  \Rightarrow   \cr
  &  \Rightarrow \quad a_{\,n}  = {4 \over {2!}}n^{\,\underline {\,2\,} }
  + c_{\,1} n^{\,\underline {\,1\,} }  + c_{\,0} n^{\,\underline {\,0\,} }  =   \cr
  &  = 2n\left( {n - 1} \right) + c_{\,1} n + c_{\,0}  \cr}
$$
where $n^{\,\underline {\,m\,} }$ indicates the Falling Factorial.


问:
Source for closed form of sum $\sum_{n=0}^\infty\zeta(2+2n)x^{2n}$
I recently came across the following summation:
$$\sum_{n=0}^\infty\zeta(2+2n)x^{2n}=\frac{1-\pi x\cot(\pi x)}{2x^2},$$
where $\zeta$ is the Riemann zeta function. Wolfram MathWorld cites Bailey et. al. 2006, but I can't seem to find that source in the references. Does anybody know where this reference is located (or any other place to find the evaluation of this sum)?

答:
In this MSE question the OP derives to a very similar expression
$$x\cot x = 1-2\sum_{n=1}^{\infty}\zeta(2n)\frac{x^{2n}}{\pi^{2n}}.$$
To get to this result he starts with the Weiertrass product of the sine.
So, how can we get from this expression to yours? First of all we can bring the $1$ as well as the $-2$ to the left hand side.
$$\frac{1-x\cot x}{2} = \sum_{n=1}^{\infty}\zeta(2n)\frac{x^{2n}}{\pi^{2n}}.$$
Now substitute $x\to\pi x$. This yields
$$\frac{1-\pi x\cot \pi x}{2} =\sum_{n=1}^{\infty}\zeta(2n)\frac{(\pi x)^{2n}}{\pi^{2n}}=\sum_{n=1}^{\infty}\zeta(2n)x^{2n}.$$

Start from $n=0$ instead from $n=1$, by writing every $n$ as $n+1$.
$$\sum_{n=1}^{\infty}\zeta(2n)x^{2n}=\sum_{n=0}^{\infty}\zeta(2(n+1))x^{2(n+1)} =x^2\sum_{n=0}^{\infty}\zeta(2n+2)x^{2n}.$$
Remember what the sum was equal to and divide both sides by $x^2$
$${\frac{1-\pi x\cot \pi x}{2x^2} = \sum_{n=0}^{\infty}\zeta(2n+2)x^{2n}}.$$


问:
Showing uniform continuity
I've been trying to show that a function $f$ on the real interval $[a,b]$ which satisfies

$$
f(x)=f(a)+\int_a^xf'(s)\,ds\qquad\text{($f'$ defined almost everywhere)}
$$

must be uniformly continuous on $[a,b]$.

Since the condition above is equivalent to absolute continuity I know that I could show what I need by means of the proof that absolute continuity - from its fundamental definition - implies uniform continuity: I have seen a proof of that. However, I would like to show the above without involving another form of continuity in the process.

I know that - since what I have stated is also equivalent to there existing any integrable function in place of $f'$ -the proof should not involve the properties of the derivative. However, in establishing a bound I get only as far as

$$
\left|f(y)-f(x)\right|=\left|\int_x^yf'(s)ds\right|\leq\int_x^y\left|f'(s)\right|\,ds
$$

Is it possible to show that an integrable first derivative (or indeed any integrable function) is bounded in sup norm?

Thank you.

Marko

答:
We assume that $\|f'\|_{L^1} = \int_{a}^{b} |f'|\,dt \lt \infty$. Let $A_{n} = \{x \,:\,|f'(x)| \leq n\}$. Put $g_{n} = [A_{n}] f'$, where $[A_n]$ denotes the characteristic function of $A_n$. Then we have $g_{n} \to f'$ almost everywhere, and, as Nate points out in his comment, dominated convergence implies that $\int_{a}^{b} |g_n - f'|\,dt \to 0$ as $n \to \infty$ (the integrand is bounded by the integrable function $2|f'|$).

Now, given $\varepsilon \gt 0$, choose $n$ so large that $\int_{a}^{b} |g_n - f'|\,dt \lt \varepsilon /2$. As $|g_{n}|$ is bounded by $n$, we have that $|\int_{x}^{y} g_{n}(t)\,dt| \leq n|y-x|$. Thus,
$$\left\vert \int_{x}^{y} f'(t)\,dt\right\vert \leq \left\vert\int_{x}^{y} |f'-g_n|\,dt\right\vert + \left\vert \int_{y}^{x} |g_n(t)|\,dt\right\vert \leq \varepsilon/2 + n \cdot |y-x|$$
and for $\delta = \frac{\varepsilon}{2n}$ we get for all $x,y$ with $|y-x| \lt \delta$
that
$$|f(y) - f(x)| = \left\vert \int_{x}^{y} f'(t)\,dt \right\vert \leq \varepsilon/2 + \delta n \lt \varepsilon$$
which is the very definition of uniform continuity of $f$.

In fact, we get the even more general estimate that for $\mu(E) \lt \delta$ we have $\int_{E} |f'|\,dt \lt \varepsilon$. But that's exactly absolute continuity of $f$.

It is of course not true that an integrable first derivative is bounded in the sup-norm. For instance, for $f(x) = \sqrt{x}$ we have $f'(x) = \frac{1}{2\sqrt{x}}$ which is not bounded  but integrable on $[0,1]$.

Title: Showing uniform continuity


问:
Proof that two sets are equal
$X$ is finite, if and only if, the affirmation is true,

$$Y\subseteq X \text{ and } f:Y\to X\text{ surjective } \Rightarrow Y=X$$

It is easy to see that this result is true, but I am not able to properly the solution, mainly the direction $(\Leftarrow)$. What I managed to write for the direction $(\Rightarrow)$ is the following,

We have that: $X$ is finite and $Y \subseteq X \text{ and } f:Y \to X \text{ surjective}$. Let's prove that $X=Y$.

Note that: if $X \neq Y$, then there is some element $x \in X$ such that $ x \notin Y$. By hypothesis, $Y$ is a subset of $X$, that is finite, so we have that $Y$ is finite and yet $|Y| \leq |X|$. Thus, there are bijections $\phi_{n}:I_n \to X$ and $\psi_{m}:I_m \to Y $, where $ m \leq n $. But, since $f:Y \to X$ is surjective, that is, $\forall x \in X, \exists \, y \in Y$ such that $ f(y)=x$, and also $|Y|\leq |X|$, there cannot be $x\in X $ such that $x\notin Y$. Thus, we conclude that $X =Y$.

Is this solution correct? Could someone help me in the other direction?

Edit:

A set $A$ is said to be finite, if it is empty or if there is a bijection $\varphi: I_n \to A$, where $A=\{1,\cdots,n\}$.

答:
⇒

Suppose X finite, $ Y \subset X $ then $ |Y| \leq | X |$, $ f: Y \to X $ surjective then $ |X| \leq |Y |$. We deduce that $ |X| = |Y| $. This together with $ X $
finite and $ Y \subset Y $ implies $ X = Y $.

⇐

Suppose by contradiction that $ X $ is infinite. We show that the implication is false. There exists $ g: \mathbb{Z} \to X $ injective. Denote by $ \bar{X} = X \setminus g(\mathbb{Z}) $. Define $ Y = X \setminus g(2\mathbb{Z}+1)$, and $ f : Y \to X $ such that $ f(x) = x $ for $ x \in \bar{X} $ and $ f(x) = g\left(\frac{g^{-1}(x)}{2}\right)$ for $ x \in g(2 \mathbb{Z})$.  It is straight-forward to show that $ f $ is surjective but $ X \neq Y $ which implies that the implication is false.

Title: Proof that two sets are equal


答:
Hint & caution:

If the load is constant (and just resistive), the instantaneous power is proportional to the square of the sine wave, and the power average over a period will be proportional to the integral of the $\sin^2$.

Additional note:

If you intend to use a modern led lamp, then you have to radically change your approach!

Title: How to divide a Sine wave into n equal power/area sections?

Body:
I'm working on an electronics project which involves using a small micro-controller to dim an AC Mains light bulb. I want to keep it simple and instead of having a continuous dimming adjustment I'll limit it to 'n' settings. So Off, 1/10th brightness, 2/10th brightness... on up to 9/10ths and 10/10 full brightness, always on. I can't do that by simply dividing the sine wave in 10 equal durations, as in 1/10th the period, 2/10ths the period etc. Brightness is related to power which is related to the area under the curve.

I only have to consider one positive sine half period section. The same time values would be used for the negative half of the wave's period.

Area under the curve involves integrating the sine wave, if I remember my school calculus correctly, which is a cosine function, but the dividing it into equal areas has me stumped. I think I have 0%, 50% and 100% brightness figured out, but that's about it ;)

This question has probably been answered already, but I'm probably asking it wrong. There's probably a maths way of expressing the question in a way a search engine could find.


问:
Cut-off function with first derivative property
I am looking for an explicit cut-off function $\xi(x)\in C^\infty_c(\mathbb{R}^d)$ such that $|\nabla_x\xi(x)|\leq C\xi(x)$ for some $C>0$. What I have found is the following one. Consider the two balls $B_m$ and $B_{m-1}$ of radius $m$ and $m-1$ both respectively centred in $0$. We can define $\xi_m(x)$ as $1$ in $B_{m-1}$, $0$ outside $B_m$, and $$\xi_m(x)=\frac{\exp\left\{\frac{1}{|x|_d-m}\right\}}{\exp\left\{\frac{1}{|x|_d-m}\right\}+\exp\left\{-\frac{1}{|x|_d-(m-1)}\right\}}\quad\text{ for }x\in B_m \setminus B_{m-1}. $$

Now, if I consider the gradient I am able to prove only that
$$|\nabla_x\xi_m(x)|^2 \leq C \xi(x)$$
and it fails the property that I am looking for ($|\nabla_x\xi_m(x)| \leq C \xi(x)$).

Do you have any suggestions?

答:
No, this is not possible. Let's go to $d=1$ for a counterexample. Assume $f\in C^1(\mathbb R ),$ with $f> 0$ on $(0,1)$ and $f=0$ elsewhere. Suppose, to reach a contradiction, that

$$|f'(x)|\le C|f(x)|$$

for $x\in (0,1).$ For such $x$ we then have

$$\tag 1 |f(x)|=|\int_0^x f'| \le \int_0^x |f'|\le C\int_0^x |f|.$$

For each such $x,$ define $M_x=\max_{[0,x]} |f|.$ By continuity, there exists $y(x)\in [0,x]$ such that $M_x= |f(y(x))|.$ So from $(1)$ we see

$$\tag 2 M_x = |f(y(x))| \le C\int_0^{y(x)} |f| \le C\int_0^{x} |f|\le CxM_x.$$

Now $M_x>0,$ so we can divide by $M_x$ in $(2)$ to get $1\le Cx.$ This is true for all $x\in (0,1).$ Clearly this is impossible.


问:
What is the total number of spanning trees on all subsets of vertices in a complete graph?
On a complete graph with $n$ vertices, I need to find the total number of spanning trees on every subset of vertices. The number of subsets in the graph is $2^n$. Cayley's formula $k^{k-2}$ then applies to each subset but I suppose I would need to know the number of vertices $v$ for each subset such that $k = n-v$.

答:
Since you want to count sum of the number of spanning trees for each subgraph of $K_n$, you need to:

1-  list all possible subgraphs

2- compute the number of spanning tree for each of these.

As you said, the number of spanning trees of the subgraphs of the complete tree depends only on the number of vertices chosen, since the subgraph are themselves complete, and is $k^{k-2}$ (where $k$ is the number of vertices of the subgraphs), except for $k=1$ where this number is $1$.

Then if you put $c(k)$ the number of way to chose $k$ vertices among $n$, then the answer to your question should be $\sum_{k=1}^{n}c(k)k^{k-2}$.

Now $c(k)$ is a well-known function, $c(k)={n\choose k}={n! \over k!(n-k)!}$ (see https://en.wikipedia.org/wiki/Binomial_coefficient). So the answer is $ {n\choose 1} + \sum_{k=2}^{n}{k\choose n}k^{k-2}$.

See also https://oeis.org/A270593 for the first terms of this sequence and  context.


问:
Determinant of Vector
Is posible obtain the determinant of any vector?.How I will be able to obtain the determinant of any vector $v=[v_1,v_2,\cdots,v_n]\in \mathbb{R}^n$?</p>

答:
Well, the determinant is defined not as a function of a vector, but rather as a multilinear function of $n$ vectors in $n$-space, in other words $\det : \prod_{i=1}^n \mathbb{R}^n\to\mathbb{R}$ and is multilinear. In this setup it doesn't make much sense talking about $\det(v)$ for some $v\in \mathbb{R}^n$, because $\det$ was built to receive as it's input $n$ vectors instead of just one.

If you insist in defining the determinant of a vector however, one possibility would be to talk about the new function obtained defining $f_v: \prod_{i=1}^{n-1} \mathbb{R}^n \to \mathbb{R}$ by setting:

$$f_v(w_1, \dots, w_{n-1})=\det(v, w_1\dots, w_{n-1})$$

So that we would me mapping $v \mapsto \det(v, \cdot, \dots, \cdot)$, that's probably a common approach when we deal with tensors in general: if we understand tensors as multilinear maps, then if we give just one argument to the tensor it gives us back a multilinear map that can receive one less input, because one of the inputs was fixed, in that framework $\det$ is a tensor and giving just one vector to it gives back a new function of $n-1$ vectors.

Title: Determinant of Vector


问:
What are the irreducible elements of $ \mathbb {Z}_2 $
We consider the following sub-ring of $ \mathbb{Q} $:

$$ \mathbb {Z} _2: = \left\{\frac {a} {2 ^k}; a \in \mathbb{Z}, k \in \mathbb {N} \right\} $$
the set of units of $\mathbb {Z}_2^× = \{ ±2^k, k \in \mathbb {Z}\}$

Problem

What are the irreducible elements of $ \mathbb {Z}_2 $

My effort:

We have $2$ and $−2 $ were units of $ \mathbb {Z}_2 $. Let $ p $ be a prime number other than 2. (The case $-p $ is treated in the same way). If $ p = \frac {a}{2 ^ l} \frac {b} {2 ^ k} = \frac {ab} {2 ^{k + l}} $ then $ ab = 2 ^{k + l } p $. Hence, $ a = p2 ^m $ and $ b = 2 ^{k + l − m} $ (or the other way around). In this case, $ \frac {b} {2 ^ k} $ is a unit and $ p $ remains irreducible in $ \mathbb {Z} _2 $.

Question:

can we say that: The irreducible elements of $ \mathbb {Z} _2 $ are therefore the $ p $ and $ −p $ for $ p$ prime other than 2.

答:
Yes, almost – since $\mathbf Z_2$ is the localisation of $\mathbf Z$ w.r.t. the multiplicative set $S=\{2^k\mid k\in \mathbf N\}$, therefore it is a P.I.D. and its irreducible elements generate prime ideals, which correspond  bijectively to the prime ideals of $\mathbf Z$ which do not meet $S$, i.e.  to the irreducible elements of $\mathbf Z$ different from $2$.

As irreducible elements are defined up to a unit factor, the associated elements of an irreducible element $\frac p1\in\mathbf Z_2$ are all $\pm \frac p{2^k}$, where  $p$ is an odd prime of $\mathbf Z$ and $k$ is an integer.


问:
Multiplication operator $M_a:\ell^p\rightarrow \ell^q$

Let $1\leq q<p<\infty$. Let $(a_n)$ be a sequence such that for all $(x_n)\in\ell^p$ we have $(a_n x_n)\in\ell^q$. Show that $(a_n)\in\ell^r$ with $\frac{1}{p}+\frac{1}{r}=\frac{1}{q}$ and $\Vert M_a\Vert=\Vert a\Vert_{\ell^r}$, with $M_a(x_n)=(a_n x_n)\in \ell^q$

I can't show that $(a_n)\in\ell^r$, I just know that

$$\Vert M_a(x_n)\Vert =\left(\sum |a_n x_n|^q\right)^{\frac{1}{q}}\leq\left(\sum |a_n |^r\right)^{\frac{1}{r}}\left(\sum | x_n|^p\right)^{\frac{1}{p}}=\Vert a \Vert_{\ell^r}\Vert x\Vert_{\ell^p}$$

can someone help me show the rest?

答:
Firstly, we prove the Generalized Holder Inequality.

Generalized Holder Inequality: Let $(X,\mathcal{F},\mu)$ be a measure
space. Let $1\leq r<p<\infty$. Define $q=\frac{pr}{p-r},$ then $q\in(1,\infty)$
and $\frac{1}{p}+\frac{1}{q}=\frac{1}{r}.$ For any measurable functions
$f,g:X\rightarrow\mathbb{R}$, we have that $||fg||_{r}\leq||f||_{p}||g||_{q}$.
In particular, if $f\in L^{p}$ and $g\in L^{q}$, then $fg\in L^{r}$.

Proof: $\lambda=\frac{r}{p}$, then $\lambda\in(0,1)$ and $1-\lambda=\frac{r}{q}$.
Recall that $\ln:(0,\infty)\rightarrow\mathbb{R}$ is a concave function.
Let $x,y>0$ be arbitrary, we have that
\begin{eqnarray*}
\lambda\ln x^{p}+(1-\lambda)\ln y^{q} & \leq & \ln(\lambda x^{p}+(1-\lambda)y^{q})\\
x^{r}y^{r} & \leq & \lambda x^{p}+(1-\lambda)y^{q}.
\end{eqnarray*}
Note that the above inequality continues to hold if $x=0$ or $y=0$.
Now, let $f,g:X\rightarrow\mathbb{R}$ be measurable functions. If
$||f||_{p}=0$ or $||g||_{q}=0$, then $f=0$ or $g=0$ ($\mu$-a.e.)
and hence $fg=0$ ($\mu$-a.e.). In this case, we have $||fg||_{r}=0$
and hence $||fg||_{r}\leq||f||_{p}||g||_{q}.$ Suppose that $||f||_{p}\neq0$
and $||g||_{q}\neq0$. If $||f||_{p}=\infty$ or $||g||_{q}=\infty$,
we clearly have $||fg||_{r}\leq||f||_{p}||g||_{q}.$ Consider the
case that $||f||_{p},||g||_{q}\in(0,\infty)$. Define $F=|f|/||f||_{p}$
and $G=|g|/||g||_{q}$. Note that $||F||_{p}=||G||_{q}=1$

For each $t\in X$, we have that
$$
F^{r}(t)G^{r}(t)\leq\lambda F^{p}(t)+(1-\lambda)G^{q}(t).
$$
Integrating both sides with respect to $\mu$ yields
\begin{eqnarray*}
\int F^{r}(t)G^{r}(t)d\mu(t) & \leq & \lambda\int F^{p}(t)d\mu(t)+(1-\lambda)\int G^{q}(t)d\mu(t)\\
 & = & \lambda\cdot1+(1-\lambda)\cdot1\\
 & = & 1.
\end{eqnarray*}
On the other hand,
\begin{eqnarray*}
\int F^{r}(t)G^{r}(t)d\mu(t) & = & \frac{1}{||f||_{p}^{r}||g||_{q}^{r}}\int|f|^{r}|g|^{r}d\mu\\
 & = & \frac{||fg||_{r}^{r}}{||f||_{p}^{r}||g||_{q}^{r}}.
\end{eqnarray*}
Re-arranging terms, we obtain $||fg||_{r}\leq||f||_{p}||g||_{q}.$

We go back to your problem: (Note that the notations are different) Let
$1\leq r<p<\infty$. Define $q=\frac{pr}{p-r}$, then $q\in(1,\infty)$
and $\frac{1}{p}+\frac{1}{q}=\frac{1}{r}$.

(1) Let $a=(a_{n})$ be a sequence such that for any $x=(x_{n})\in l^{p},$
$(a_{n}x_{n})\in l^{r}$. Prove that $a\in l^{q}$.

Solution of (1): For each $n$, define $T_{n}:l^{p}\rightarrow l^{r}$
by $T_{n}x=(a_{1}x_{1},a_{2}x_{2},\ldots,a_{n}x_{n},0,0,\ldots),$
where $x=(x_{n})\in l^p$. Clearly $T_{n}$ is linear. Moreover, by the generalized
Holder inequality, we have that
\begin{eqnarray*}
||T_{n}x||_{r} & \leq & \left\{ \sum_{k=1}^{n}|a_{k}|^{q}\right\} ^{\frac{1}{q}}\left\{ \sum_{k=1}^{n}|x_{k}|^{p}\right\} ^{\frac{1}{p}}\\
 & \leq & \left\{ \sum_{k=1}^{n}|a_{k}|^{q}\right\} ^{\frac{1}{q}}\left\{ \sum_{k=1}^{\infty}|x_{k}|^{p}\right\} ^{\frac{1}{p}}\\
 &  & \left\{ \sum_{k=1}^{n}|a_{k}|^{q}\right\} ^{\frac{1}{q}}\cdot||x||_{p}.
\end{eqnarray*}
This shows that $T_{n}$ is a bounded linear map. Consider the family
of bounded linear maps $\{T_{n}\mid n\in\mathbb{N}\}$. Observe that
for each $x\in l^{p}$, $\{T_{n}x\mid n\in\mathbb{N}\}$ is a bounded
subset of $l^{r}$ (For, let $n\in\mathbb{N}$ be arbitrary, then
$||T_{n}x||_{r}^{r}=\sum_{k=1}^{n}|a_{k}x_{k}|^{r}\leq\sum_{k=1}^{\infty}|a_{k}x_{k}|^{k}<\infty$
because it is given that $(a_{n}x_{n})\in l^{r}$). By the Uniform
Boundedness Principle, $\sup_{n}||T_{n}||<\infty$.

We go to show that $||T_{n}||=\left\{ \sum_{k=1}^{n}|a_{k}|^{q}\right\} ^{\frac{1}{q}}$.
From the above discussion, we clearly have $||T_{n}||\leq\left\{ \sum_{k=1}^{n}|a_{k}|^{q}\right\} ^{\frac{1}{q}}$.
If $\left\{ \sum_{k=1}^{n}|a_{k}|^{q}\right\} ^{\frac{1}{q}}=0$,
we are done. Suppose that $\left\{ \sum_{k=1}^{n}|a_{k}|^{q}\right\} ^{\frac{1}{q}}>0$.
Let $x=\eta(|a_{1}|^{\frac{q}{p}},|a_{2}|^{\frac{q}{p}},\ldots,|a_{n}|^{\frac{q}{p}},0,0,\ldots)\in l^{p}$,
where $\eta=\left(\sum_{k=1}^{n}|a_{k}|^{q}\right)^{-\frac{1}{p}}$,
then $||x||_{p}=1$. Note that

\begin{eqnarray*}
||Tx||_{r}^{r} & = & \eta^{r}\sum_{k=1}^{n}\left|a_{k}\cdot|a_{k}|^{\frac{q}{p}}\right|^{r}\\
 & = & \eta^{r}\sum_{k=1}^{n}|a_{k}|^{(\frac{p+q}{p})r}\\
 & = & \eta^{r}\sum_{k=1}^{n}|a_{k}|^{q}.
\end{eqnarray*}
Hence, $||T_{n}||\geq||T_{n}x||_{r}\geq\eta\left\{ \sum_{k=1}^{n}|a_{k}|^{q}\right\} ^{\frac{1}{r}}=\left\{ \sum_{k=1}^{n}|a_{k}|^{q}\right\} ^{\frac{1}{q}}$.
Now, it follows that
\begin{eqnarray*}
||a||_{q} & = & \left\{ \sum_{k=1}^{\infty}|a_{k}|^{q}\right\} ^{\frac{1}{q}}\\
 & = & \sup_{n}\left\{ \sum_{k=1}^{n}|a_{k}|^{q}\right\} ^{\frac{1}{q}}\\
 & = & \sup_{n}||T_{n}||\\
 & < & \infty.
\end{eqnarray*}

(2) Define $T:l^{p}\rightarrow l^{r}$ by $Tx=(a_{n}x_{n})$, where
$x=(x_{n})\in l^{p}$. Prove that $T$ is a bounded linear map and
$||T||=||a||_{q}$.

Proof: Clearly, $T$ is linear. By the Generalized Holder Inequality, for
any $x\in l^{p}$, we have that $||Tx||_{r}\leq||a||_{q}||x||_{p}.$
This shows that $T$ is bounded and $||T||\leq||a||_{q}$. If $||a||_{q}=0$,
we are done. Suppose that $||a||_{p}\neq0$. Define $x=\eta(|a_{n}|^{\frac{q}{p}})$,
where $\eta=\left(\sum_{k=1}^{\infty}|a_{k}|^{q}\right)^{-\frac{1}{p}}$.
By direct verification, $||x||_{p}=1$. Therefore, $||T||\geq||Tx||_{r}=\left\{ \sum_{k=1}^{\infty}|a_{k}|^{q}\right\} ^{\frac{1}{q}}=||a||_{q}$.
This shows that $||T||=||a||_{q}$.


问:
The reason for the superiority of Lebesgue measure / Lebesgue integration
I'm learning about Lebesgue measure and Lebesgue integration.

I know that Lebesgue integration is superior to Riemann integration in the sense that we can integrate a much larger class of function with Lebesgue integral.

I would like to ask which is the main reason that creates the superiority of Lebesgue integral?

EDIT

As for me, basically, Lebesgue integration is another way to calculate the integration (or geometrically, the area under the curve), so why just changing the way we compute can make such a huge difference between those 2 kind of integration ?

答:
This is not a real answer to this question, I am not aiming at this since I see there are other nice answers. What I want to highlight is that Lebesgue is better than Riemann, as long as you are talking about definite integrals over, say, an interval $[a,b]$.

As long as you have improper integrals, the situation is reversed and the Lebesgue integral is no more a generalization of the Riemann (improper) Integral, since we may have functions that are Riemann integrable in the improper sense, but they are not Lebesgue integrable. As an example pick $f(x)= \frac{\sin(x)}{x}$, then $$\int_0^{\infty}\frac{\sin(x)}{x}dx$$ is a well defined real number, while $$\int_0^{\infty}\left|\frac{\sin(x)}{x}\right|dx$$ is not since the integral diverges.

Of course I agree that the Lebsegue integral is more general if we restrict to the definite notion of Riemann integral, but this answer was just to remind we have always to keep in mind there may be instances where the two notions do not agree in any sense.


问:
A $*$-homomorphism from the CAR algebra to $\mathfrak B(\mathcal H)$
Could a $*$-homomorphism $\pi:\text{CAR}\to\mathfrak B(\mathcal H)$ exist (with $\mathcal H$ separable) such that there is a compact and positive element $h\in\mathcal K$ commuting with the image of $\pi$? Some naive considerations have led me to conclude that this shouldn't be the case, but I hope I'm wrong.</p>

答:
No. The CAR algebra $A$ is simple, so any representation $\pi$ is necessarily injective. Assume $\pi$ is unital. If there is a compact $h$ in the commutant $\pi(A)'$, then the commutant contains a finite dimensional projection $p$ and $p \pi(\cdot) p$ would be a representation of $A$ on a finite dimensional Hilbert space, which is impossible.

Hope this helps.


问:
Given tan θ and cos θ (below), solved for sinθ but got multiple answers
$\tan \theta = \frac{\sqrt15}{3}$

$\cos \theta=\frac{\sqrt10}{4}$

We're supposed to solve for $\sin \theta$ using trig identities, so:

Pythagorean

$\sin^2{\theta}+\cos^2{\theta}=1$

$\sin^2{\theta}+\left(\frac{\sqrt10}{4}\right)^{2}=1$

$\sin^2{\theta}+\frac{10}{16}=1$

$\sin^2{\theta}=1-\frac{10}{16}$

$\sin^2{\theta}=\frac{6}{16}=\frac{3}{8}$

$\sin{\theta}=\sqrt{\frac{3}{8}}=\frac{\sqrt{24}}{8}=\frac{\sqrt{6}}{4}$

sin/cos=tan

$\frac{\sin\theta}{\cos\theta}=\tan\theta$

$\frac{\sin\theta}{\frac{\sqrt10}{4}}=\frac{\sqrt15}{3}$

$\sin{\theta}=\frac{\sqrt10}{4}*\frac{\sqrt15}{3}$

$\sin{\theta}=\frac{\sqrt150}{12}=\frac{5\sqrt6}{12}$

Using this method, $\sin{\theta}$ is somehow greater than one? The above answer simplifies to about $1.0206$.

1 + cot$^2\theta$ = csc$^2\theta$

$1+\cot^2{\theta}=\csc^2{\theta}$

$1+\left(\frac{3}{\sqrt15}\right)^2=\csc^2{\theta}$

$1+\frac{9}{15}=\csc^2{\theta}$

$\frac{24}{15}$ = csc$^2\theta$

$\frac{15}{24}=\sin^2{\theta}$

$\sqrt{\frac{15}{24}}=\sin{\theta}$

$\sqrt{\frac{5}{8}}=\sin{\theta}$

$\frac{\sqrt40}{8}=\sin{\theta}$

$\frac{2\sqrt10}{8}$ = sin $\theta$

$\frac{\sqrt10}{4}=\sin{\theta}$

This is the $\cos$ value (maybe I made a mistake?)

I have asked my math  teacher about this. She is also very confused as to what the answer is: the worksheet she printed out wasn't hers, and the answer listed is the second one, which is greater than one and is probably wrong somehow. I realized I could use the third identity only after I met with her, but she couldn't find any problem with the first two methods, and yet somehow they yielded different values.

Could someone please shed some light as to why I ended up with multiple answers, with one greater than one? Thanks for any answers.

答: The problem has no solutions because there is no $\theta$ such that $\tan\theta=\frac{\sqrt{15}}3$ and that $\cos\theta=\frac{\sqrt{10}}4$. In fact, if $\cos\theta=\frac{\sqrt{10}}4$, then$$\tan^2\theta=\frac{\sin^2\theta}{\cos^2\theta}=\frac{1-\cos^2\theta}{\cos^2\theta}=\frac{\frac6{16}}{\frac{10}{16}}=\frac35,$$and therefore $\tan\theta=\sqrt{\frac35}$.


答:
Note that $\|M_a\|=\|a\|_{\ell^r}$.

Proof: Set $s=p/q$ and $t=r/q$. Note that $1<s,t$ and that $1/s+1/t=1$. We can apply Holder's inequality:
If $x=(x_k)\in\ell^p$ we have that
$$\|M_ax\|_{\ell^q}^q=\sum_k|a_kx_k|^q\leq\sum_k\bigg(\big(|a_k|^q\big)^t\bigg)^{1/t}\cdot\bigg(\sum_k(|x_k|^q)^s\bigg)^{1/s}$$
But $qt=r$ and $qs=p$ so raising both sides to $1/q$ we get that
$$\|M_ax\|_q\leq\|a\|_r\cdot\|x\|_p$$
and this shows that $\|M_a\|\leq\|a\|_r$. For the other inequality, set $\delta=\frac{r}{q}-2\in\mathbb{R}$ and for $n\in\mathbb{N}$ set
$$x_n=(\bar{a_1}\cdot|a_1|^{\delta},\bar{a_2}\cdot|a_2|^\delta,\dots,\bar{a_n}\cdot|a_n|^\delta,0,0,0,0\dots)\in\ell^p$$

Now $$\|M_a\|=\sup_{x\neq0}\frac{\|M_ax\|_q}{\|x\|_p}\geq\frac{\|M_ax_n\|_q}{\|x_n\|_p}$$

But $\|M_ax_n\|_q=\bigg(\sum_{k=1}^n|a_k|^{(\delta+2)q}\bigg)^{1/q}=\bigg(\sum_{k=1}^n|a_k|^r\bigg)^{1/q}$ and $\|x_n\|_p=\bigg(\sum_{k=1}^n|a_k|^{(\delta+1)p}\bigg)^{1/p}=\bigg(\sum_{k=1}^n|a_k|^r\bigg)^{1/p}$, so
$$\|M_a\|\geq\bigg(\sum_{k=1}^n|a_k|^r\bigg)^{1/r}$$
and this is true for all $n$, so letting $n\to\infty$ yields $\|a\|_r\leq\|M_a\|$.

Since $(a_nx_n)\in\ell^q$ for all $(x_n)\in\ell^p$, the linear map $M_a$ acts as $M_a:\ell^p\to\ell^q$. Now $\ell^p$ and $\ell^q$ are Banach spaces, so in order to show that $M_a$ is bounded we can use the closed graph theorem.

Let $(x_n)\subset\ell^p$ be a sequence (of sequences, i.e. $x_n=(x_k^{(n)})_{k=1}^\infty\in\ell^p$ for all $n$) such that $x_n\to0$ in $\ell^p$. We use the closed graph theorem: assuming that $M_ax_n\to y$, where $y\in\ell^q$, we will show that $y=0$.

Proof: Indeed, if $y=(y_k)$, then for any $k$ we have that
$$|y_k-a_kx_k^{(n)}|\leq\|y-M_ax_n\|_q\to0.$$
Therefore $y_k=\lim_{n\to\infty}a_kx_k^{(n)}$. But
$$|x_k^{(n)}|\leq\bigg(\sum_{j}|x_j^{(n)}|^p\bigg)^{1/p}=\|x_n\|_p\to0$$
so $x_k^{(n)}\to0$, and therefore $a_kx_k^{(n)}\to0$, i.e. $y_k=0$. Since this was true for any $k$, $y=0$ as we wanted.

The closed graph theorem now yields that $M_a$ is a bounded operator, so $\|M_a\|<\infty$. Since $\|M_a\|=\|a\|_{\ell^r}$, we conclude that $\|a\|_{\ell^r}<\infty$, i.e. $a\in\ell^r$.

Title: Multiplication operator $M_a:\ell^p\rightarrow \ell^q$

Body:
Let $1\leq q<p<\infty$. Let $(a_n)$ be a sequence such that for all $(x_n)\in\ell^p$ we have $(a_n x_n)\in\ell^q$. Show that $(a_n)\in\ell^r$ with $\frac{1}{p}+\frac{1}{r}=\frac{1}{q}$ and $\Vert M_a\Vert=\Vert a\Vert_{\ell^r}$, with $M_a(x_n)=(a_n x_n)\in \ell^q$

I can't show that $(a_n)\in\ell^r$, I just know that

$$\Vert M_a(x_n)\Vert =\left(\sum |a_n x_n|^q\right)^{\frac{1}{q}}\leq\left(\sum |a_n |^r\right)^{\frac{1}{r}}\left(\sum | x_n|^p\right)^{\frac{1}{p}}=\Vert a \Vert_{\ell^r}\Vert x\Vert_{\ell^p}$$

can someone help me show the rest?


问:
Using definition of limits

Let $c∈\mathbb{R}$ and let $f:\mathbb{R}\setminus{c}\rightarrow\mathbb{R}$ be a function such that $f(x)>0$ for all $x∈\mathbb{R}$. Use the definition of limits to prove that
$$
\lim_{x\to c}f(x)=\infty  \space\space\space\space\space\space\space\space\space\space\space   \text{iff}  \space\space\space\space\space\space\space\space\space\space\space    \lim_{x\to c}\frac{1}{f(x)}=0.
$$

Proving the "$\Rightarrow$": Here is the definition: $\lim_{x\to c}f(x)=\infty$ if $\forall M∈\mathbb{R},\exists\delta>0$ such that $\forall x∈\mathbb{R}, 0<|x-c|<\delta\Rightarrow f(x)>M$. Here is my proof:

Let $\epsilon >0$ and set $M=\frac{1}{\epsilon}$. Since $\lim_{x\to c}f(x)=\infty$, we can find a $\epsilon >0$ such that $f(x)>M$ whenever $0<|x-c|<\delta$.

Thus $0<\frac{1}{f(x)}<\frac{1}{\epsilon}$ whenever $0<|x-c|<\delta$. This implies that it is possible to find a $\delta>0$ such that $|\frac{1}{f(x)}|<\epsilon$ whenever $0<|x-c|<\delta$. Since $\epsilon$ is arbitrary, we have proved that $\lim_{x\to c}\frac{1}{f(x)}=0$.

Proving the "$\Leftarrow$": This proof I am unsure of. I know that by the definition of a limit, $\lim_{x\to c} f(x) = 0$ if $\forall\epsilon>0, \exists\delta>0$ such that $\forall x∈\mathbb{R}\setminus{c}, 0<|x-c|<\delta \Rightarrow |f(x)-0|<\epsilon$. I am unsure of how to define $\lim_{x\to c}\frac{1}{f(x)}=0$ in a similar way. Any advice would be greatly appreciated.

答: For the $\leftarrow$ direction. Let $M > 0$ be given, $\exists \delta > 0$ such that: $0 < |x-c| < \delta\implies \dfrac{1}{f(x)} < \dfrac{1}{M}\implies f(x) > M$. this shows that the limit is $\infty$ as claimed.

Title: Using definition of limits


问:
Understanding a problem on the binomial distribution
I've been trying to understand how to go around the third requirement in this problem:

This is how its solved on Slader:

How do we get info about whether the $30\%$ figure applies by assuming that that's the case and then finding the probability of getting exactly five failures for our new observation (plant), it's already clear that in the new observation $25\%$ of the failures are due to operator error. I can't seem to draw the link between doing this helps know whether the $30\%$ figure applies or not.
Thank you.

答:
The explanation given in the image is a little hand-wavey, but the overall idea is sound. To explain why, suppose instead that $20$ out of $20$ of the failures were caused by operator errors. If the theoretical probability of an operator error really is $30 \%$, then the chance of this occuring is
$$
0.3^{20}=\text{very small number...}
$$
From a practical perspective, what seems more likely: that (a) this was simply a freak event, and that the probability of an operator error really is $30\%$; or (b) the $30\%$ figure is not accurate. I think you will undoubtedly agree that (b) is the more sensible option.

Now let's consider your example: if the theoretical probability of an operator error  is $30\%$, then getting $5$ out of $20$ failures stemming from operator errors does not seem out of the ordinary. We have no reason to suspect that the $30\%$ is inaccurate.

To do this more rigorously, you can conduct a hypothesis test. We'll consider a batch of $20$ errors, and agree beforehand that if, say, $10$ or more failures are caused by operator errors, we will conclude that there is sufficient evidence to suggest that the $30\%$ figure is inaccurate. This seems reasonable: after all, if the theoretical probability of an operator error is $30\%$, then the chance of getting $10$ or more failures being caused by operator errors is only about $4.7\%$. Again, it's a question of which interpretation of the events is more reasonable: that the $30\%$ is accurate, and we just experienced an unlikely event; or that the events witnessed are enough evidence to suggest that the $30\%$ figure is not right.

Note that context is really important here. If we were very confident that the chance of an operator error being the cause of failure really is $30\%$, then we will need a lot of evidence to overturn our beliefs. This might mean, for example, that we will only conclude that the $30\%$ figure is innaccurate if $12$ or more failures are caused by an operator error.


问:
Probability of at least $2$ balls in random bin when distributing $K$ identical balls to $N$ distinct bins?
If we distribute $k$ identical balls to $n$ distinct bins, what is the probability a randomly selected bin will contain at least 2 balls, assuming all balls have a uniform probability of being placed in any bin and $k < n$.

Since it's probability, not "number of ways to arrange," and all distributions have equal probability, first I found the total number of distributions by assuming $k$ distinct balls, $n^k$, which means there are $n^{k+1}$ bins across all distributions.

I can figure out that if there are $k < n$ balls, the number of distributions where every bin has at most $1$ ball is $k \choose n$ (we have $k$ balls and each chooses 1 distinct bin), so guaranteed ${k \choose n} n$ bins across all distributions have fewer than $2$ balls. But I'm not sure how to figure out how many of the $n$ bins contain fewer than $2$ balls in the cases where some bins have $2$ or more balls.

I could easily be going about this from the wrong angle completely though.

答:
Total number of ways by which $k$ identical balls can be put into $N$ boxes is ${N-1+k \choose k}$, by partitioning argument.

Number of ways a particular bin will contain 0 balls = $N-2+k \choose k$, by leaving the bin out (event $A_1$)

Number of ways a particular bin will contain exactly 1 ball = $N-2+k-1 \choose k-1$, by keeping a ball into the bin and leaving the bin out (event $A_2$)

By classical definition of probability, with $P(A^c)=1-p(A)$ and the above two events being mutually exclusive, with $P(A)=P(A_1 \bigcup A_2) = P(A_1) + P(A_2)$, the required probability will be = 1 -  $\frac{{N-2+k \choose k} + {N-2+k - 1 \choose k-1}}{N-1+k \choose k}$


问:
Expressing tricky sets as intervals precisely.
Fix $\delta \in \mathbb{R}$. Consider $f(x)$ is a real-valued function which is defined whenever $x \in [\delta, \infty) \cup$ or $x\in (-\infty, -\delta]\setminus \left\{-\frac{1}{n^2}:n \in \mathbb{Z}^+\right\}$. Is it possible to express the domain of $f(x)$ using proper (rigorous) notation as a union of intervals? I think $\left\{-\frac{1}{n^2}:n \in \mathbb{Z}^+\right\}$ is a set while $(-\infty, -\delta]$ and $[\delta, \infty)$ are intervals, so I won't be correct to say that the domain of $f(x)$ is $(-\infty, -\delta] \setminus \left\{-\frac{1}{n^2}:n \in \mathbb{Z}^+\right\} \cup [\delta, \infty)$.

答:
I don't know if I really understood your question, is this ok?
$$\operatorname{Dom}f=[\delta,\infty)\cup\left(\left(\bigcup_{n\in\mathbb{Z}^+} \left(-\frac{1}{n^2},-\frac{1}{(n+1)^2}\right)\right) \cap (-\infty,\delta]\right)$$


答: The probability no balls are in a random bin is $P_0=(\frac{n-1}{n})^k$ and the probability that exactly one ball is in the bin is $P_1=\frac{k}{n}(\frac{n-1}{n})^{k-1}$  What you want is $P_{2+}=1-P_0-P_1$.

Title: Probability of at least $2$ balls in random bin when distributing $K$ identical balls to $N$ distinct bins?

Body:
If we distribute $k$ identical balls to $n$ distinct bins, what is the probability a randomly selected bin will contain at least 2 balls, assuming all balls have a uniform probability of being placed in any bin and $k < n$.

Since it's probability, not "number of ways to arrange," and all distributions have equal probability, first I found the total number of distributions by assuming $k$ distinct balls, $n^k$, which means there are $n^{k+1}$ bins across all distributions.

I can figure out that if there are $k < n$ balls, the number of distributions where every bin has at most $1$ ball is $k \choose n$ (we have $k$ balls and each chooses 1 distinct bin), so guaranteed ${k \choose n} n$ bins across all distributions have fewer than $2$ balls. But I'm not sure how to figure out how many of the $n$ bins contain fewer than $2$ balls in the cases where some bins have $2$ or more balls.

I could easily be going about this from the wrong angle completely though.


问:
Why is the genus (in terms of Euler characteristic) the maximum number of disj. closed curves that can be embedded in a surface without separating it?
Let $S$ be a compact surface and $g$ its genus defined in terms of Euler characteristic.
Why is $g$ the maximum number of disjoint simple closed curves that can be embedded in the surface without disconnecting it?
I would be grateful if anyone could tell me how to prove it or give me a reference. Thanks.

答:
Welcome to MSE. I want to assume some familiarity about pants decompositions (see the references linked here). For simplicity, I'll assume that $S$ is orientable without boundary, but I think you can modify the argument to adapt to boundary components.

A pants decomposition for a connected genus $g$ surface $S$ consists of $2g - 2$ pairs of pants. We'll need to glue at least $(2g - 2) - 1 = 2g - 3$ of these pairs of pants together in order to form a connected surface.

At the outset, there are $3(2g - 2) = 6g - 6$ boundary cuffs which are all pairwise-glued to form $S$. In other words, we performed $(6g-6)/2 = 3g - 3$ gluings total. But we were forced to do at least $(2g - 3)$ of them, meaning that we did
$$(3g - 3) - (2g - 3) = g$$
gluings "just for fun." In other words, there were $g$ gluings that we didn't have to make in order to keep $S$ connected. These curves are the maximal set of disjoint scc's in this alternate characterization of genus.


问:
On counting and generating all $k$-permutations of a multiset
Let $A$ be a finite set, and $\mu:A \to \mathbb{N}_{>0}$.  Let $M$ be the multiset having $A$ as its "underlying set of elements" and $\mu$ as its "multiplicity function".  (Hence $M$ is finite.)

Let $0 \leq k \leq \left|M\right|$ be an integer.

This question is about the set $\mathbf{P}(M, k)$ of all $k$-permutations of elements from the multiset $M$.

For example, if $k = 3$ and $M$ is the multiset $\{0, 0, 1, 2\}$, then $\mathbf{P}(M, k)$ comprises these 12 $k$-permutations:

$$
(0, 0, 1), (0, 0, 2), (0, 1, 0), (0, 1, 2), (0, 2, 0), (0, 2, 1), \\
(1, 0, 0), (1, 0, 2), (1, 2, 0), (2, 0, 0), (2, 0, 1), (2, 1, 0)
$$

I am interested in

an efficient (and preferably non-recursive) algorithm for generating all the elements of $\mathbf{P}(M, k)$ (in any order);

a way to determine the cardinality of the set $\mathbf{P}(M, k)$ without generating all its elements (IOW, without resorting to the solution of (1)).

(I had no trouble coming up with a recursive algorithm to generate all the elements of $\mathbf{P}(M, k)$, but it is impractical, for two reasons, both having to do with my algorithm's recursive structure: (1) its space requirements grow very rapidly with $\left|A\right|$, and (2) there is no simple way to implement it so that it can be "gracefully interrupted" (e.g. in an interactive setting)1.  These two shortcomings are not independent: the severity of (1) is what makes the interruptibility mentioned in (2) desirable.)

My guess is that a fair bit of work has been done on this problem, but all the searches I've tried are swamped by hits pointing to similar-sounding but substantially different problems.  (Many of these false hits turn out to be easier special cases of the problem described here.)  Similarly, I spent some time flipping through Stanley's Enumerative Combinatorics, v. 1, but I was not able to find what I was looking for (even though odds are that the problem is discussed somewhere in this work).

Therefore, in lieu of answers to 1 and 2 above, I could also use

search keywords that will lead to the prior work on this problem; (in particular, is there a name for the integers $\mathbf{P}(M, k)$?)

1By this I mean that the computation can be aborted (e.g. through some user-initiated signal or some pre-configured time-out) without having to bring down the entire process.

答:
I am not sure if this is still relevant, but I needed this myself so here is my (non-recursive) Python code in case anyone wants it:

```<code>def kpm(R, k):
    class P:
        def __init__(self, R):
            self.R = R
        def push(self, v):
            self.R[v] += 1
        # pops smallest value >= v if one exists
        # returns None otherwise
        def pop(self, v):
            while v < len(self.R):
                if self.R[v] > 0:
                    self.R[v] -= 1
                    return v
                v += 1
            return None

    pool = P(R)
    T = [None] * k
    t = 0

    while True:
        if T[t] is None:
            T[t] = pool.pop(0)
            if t < k - 1:
                t += 1
            else:
                print(T)
        else:
            pool.push(T[t])
            T[t] = pool.pop(T[t] + 1)
            if T[t] is None:
                if t > 0:
                    t -= 1
                else:
                    break
            else:
                if t < k - 1:
                    t += 1
                else:
                    print(T)

kpm([2,1,1], 3)
</code>```

This generates

```<code>[0, 0, 1]
[0, 0, 2]
[0, 1, 0]
[0, 1, 2]
[0, 2, 0]
[0, 2, 1]
[1, 0, 0]
[1, 0, 2]
[1, 2, 0]
[2, 0, 0]
[2, 0, 1]
[2, 1, 0]
</code>```

Edits: Minor optimizations

Body:
I am not sure if this is still relevant, but I needed this myself so here is my (non-recursive) Python code in case anyone wants it:

def kpm(R, k):
    class P:
        def __init__(self, R):
            self.R = R
        def push(self, v):
            self.R[v] += 1
        # pops smallest value >= v if one exists
        # returns None otherwise
        def pop(self, v):
            while v < len(self.R):
                if self.R[v] > 0:
                    self.R[v] -= 1
                    return v
                v += 1
            return None

    pool = P(R)
    T = [None] * k
    t = 0

    while True:
        if T[t] is None:
            T[t] = pool.pop(0)
            if t < k - 1:
                t += 1
            else:
                print(T)
        else:
            pool.push(T[t])
            T[t] = pool.pop(T[t] + 1)
            if T[t] is None:
                if t > 0:
                    t -= 1
                else:
                    break
            else:
                if t < k - 1:
                    t += 1
                else:
                    print(T)

kpm([2,1,1], 3)

This generates

[0, 0, 1]
[0, 0, 2]
[0, 1, 0]
[0, 1, 2]
[0, 2, 0]
[0, 2, 1]
[1, 0, 0]
[1, 0, 2]
[1, 2, 0]
[2, 0, 0]
[2, 0, 1]
[2, 1, 0]

Edits: Minor optimizations


问:
Is there $z \in \mathbb{C}$ with $\mathrm{Re}(z) > 0$ such that $e^{2z}+\frac{z+1}{z-1} = 0$?
Let $z$ be a complex number with positive real part. Is it possible that $e^{2z}+\frac{z+1}{z-1} = 0$?

答:
Near $z=1$, the second term dominates the first; far from $z=1$, the first dominates the second.  By numerical methods, the curve in the right half plane where $\left|e^{2z}\right| = \left|(z+1)/(z-1)\right|$ looks like this:

and nowhere on that curve (except the origin) is there a zero of your function.


问:
Why does the alternatization of a $k$-linear map indeed produce an alternating $k$-linear map?
The alternating operator $A$ produces for any $k$-linear map $f$ an alternating $k$-linear map $Af$ (the alternatization of $f$):

$$Af(v_1, \ldots, v_k) = \sum_{\sigma \in S_k} \text{sgn}(\sigma)f\left(v_{\sigma(1)}, \ldots, v_{\sigma(k)}\right)$$

I've tried to prove that the definition of an alternating $k$-linear map holds (and most of all, to find an intuitive explanation for this!), but haven't made much progress...

If I think about the definition where having two equal arguments must make the map evaluate to $0$, then having two equal arguments means transposing them doesn't change anything, so two terms cancel out in the sum, but what about the rest?

Proof idea for alternative definition

Not what I was looking for, but I figured out a proof outline for an alternative definition of an alternating map. An alternating map is also defined as a map where transposing two arguments changes the sign of the map (e.g. $f(y, x, z) = - f(x, y, z)$).

Composing any permutation with a transposition changes its sign, so if the sum gets decomposed into a sum of all even permutations and all odd permutations, all even permutations will get mapped to odd permutations and vice-versa, so the two sums change signs. As a consequence, the entire alternatized function also changes sign.

答:
Fix $\pi \in S_k$.

\begin{align*}
Af(v_{\pi(1)},\dots,v_{\pi(k)})&= \sum_{\sigma\in S_k}\operatorname{sgn}(\sigma)f(v_{\sigma(\pi(1))},\dots,v_{\sigma(\pi(n))}) \\
&=\operatorname{sgn}(\pi)\sum_{\sigma\in S_k}\operatorname{sgn}(\pi)\operatorname{sgn}(\sigma)f(v_{\sigma(\pi(1))},\dots,v_{\sigma(\pi(n))}) \\
&=\operatorname{sgn}(\pi)\sum_{\sigma\in S_k}\operatorname{sgn}(\sigma\pi)f(v_{\sigma\pi(1))},\dots,v_{\sigma\pi(n))}) \\
&=\operatorname{sgn}(\pi)Af(v_1,\dots,v_k)
\end{align*}
where the last equality holds since right multiplication $\sigma\mapsto\sigma\pi$ yields a bijection $S_k\to S_k$.


问:
Exponential With Step Function
I would like to know how to differentiate the following function:
$f(t)=e^{{-5}(t-2)}$. I think this function exists at $t=0$, but the derivative is valid only for $t>0$ or $t=0^+$.

$$f'(t)=-5e^{{-5}(t)}$$ Is this correct? How would I express this better? I'm a M.S.E.E., not an applied mathematician. This function describes capacitor current after a sudden change of voltage. Thanks. Fred

答:
To compute the derivative of any exponential function, use the following formula:

$$\dfrac{d}{dx}\left(a^x\right)=\left(a^x\right)\left(\ln{a}\right)\left(\dfrac{d}{dx}(x)\right)$$

Hence,

$$\dfrac{d}{dt}\left(e^{-5(t-2)}\right)=\left(e^{-5(t-2)}\right)\underbrace{\left(\ln{e}\right)}_{= \ 1}\underbrace{\left(\dfrac{d}{dt}(-5(t-2))\right)}_{= - 5}$$

$$=\boxed{-5e^{-5(t-2)}}$$

Title: Exponential With Step Function


问:
What is the meaning of $A =\prod_{j<k \hspace{2mm} j,k \neq c,d}^{}(i_j-i_k)$ in Hungerford's book?
I am trying to parse this proof in Hungerford's book:

When he gets into computing this:

$$A =\prod_{\substack{j<k\\ j,k \neq c,d}}^{}(i_j-i_k)$$

What does this means?

Does it means that $j<k$ and  $j\neq c$ and $k\neq d$; or

Does it means that $j<k$ and  $j\neq c$ and $k\neq d$ and $h\neq d$ and $k\neq c$?

I computed the indices I'd get in both cases in a simple example where $c=3$, $d=5$ and $j=9$ and it for the first case, I'd get the following indices:

$$\begin{array}{cccccccc}
\begin{array}{cc}
 1 & 2 \\
\end{array}
 & \text{} & \text{} & \text{} & \text{} & \text{} & \text{} & \text{} \\
\begin{array}{cc}
 1 & 3 \\
\end{array}
 &
\begin{array}{cc}
 2 & 3 \\
\end{array}
 & \text{} & \text{} & \text{} & \text{} & \text{} & \text{} \\
\begin{array}{cc}
 1 & 4 \\
\end{array}
 &
\begin{array}{cc}
 2 & 4 \\
\end{array}
 &
\begin{array}{c}
 \text{$\bullet$ $\bullet$} \\
\end{array}
 & \text{} & \text{} & \text{} & \text{} & \text{} \\
\begin{array}{c}
 \text{$\bullet$ $\bullet$} \\
\end{array}
 &
\begin{array}{c}
 \text{$\bullet$ $\bullet$} \\
\end{array}
 &
\begin{array}{c}
 \text{$\bullet$ $\bullet$} \\
\end{array}
 &
\begin{array}{c}
 \text{$\bullet$ $\bullet$} \\
\end{array}
 & \text{} & \text{} & \text{} & \text{} \\
\begin{array}{cc}
 1 & 6 \\
\end{array}
 &
\begin{array}{cc}
 2 & 6 \\
\end{array}
 &
\begin{array}{c}
 \text{$\bullet$ $\bullet$} \\
\end{array}
 &
\begin{array}{cc}
 4 & 6 \\
\end{array}
 &
\begin{array}{cc}
 5 & 6 \\
\end{array}
 & \text{} & \text{} & \text{} \\
\begin{array}{cc}
 1 & 7 \\
\end{array}
 &
\begin{array}{cc}
 2 & 7 \\
\end{array}
 &
\begin{array}{c}
 \text{$\bullet$ $\bullet$} \\
\end{array}
 &
\begin{array}{cc}
 4 & 7 \\
\end{array}
 &
\begin{array}{cc}
 5 & 7 \\
\end{array}
 &
\begin{array}{cc}
 6 & 7 \\
\end{array}
 & \text{} & \text{} \\
\begin{array}{cc}
 1 & 8 \\
\end{array}
 &
\begin{array}{cc}
 2 & 8 \\
\end{array}
 &
\begin{array}{c}
 \text{$\bullet$ $\bullet$} \\
\end{array}
 &
\begin{array}{cc}
 4 & 8 \\
\end{array}
 &
\begin{array}{cc}
 5 & 8 \\
\end{array}
 &
\begin{array}{cc}
 6 & 8 \\
\end{array}
 &
\begin{array}{cc}
 7 & 8 \\
\end{array}
 & \text{} \\
\begin{array}{cc}
 1 & 9 \\
\end{array}
 &
\begin{array}{cc}
 2 & 9 \\
\end{array}
 &
\begin{array}{c}
 \text{$\bullet$ $\bullet$} \\
\end{array}
 &
\begin{array}{cc}
 4 & 9 \\
\end{array}
 &
\begin{array}{cc}
 5 & 9 \\
\end{array}
 &
\begin{array}{cc}
 6 & 9 \\
\end{array}
 &
\begin{array}{cc}
 7 & 9 \\
\end{array}
 &
\begin{array}{cc}
 8 & 9 \\
\end{array}
 \\
\end{array}$$

For the second case, I'd get the following indices:

$$\begin{array}{cccccccc}
\begin{array}{cc}
 1 & 2 \\
\end{array}
 & \text{} & \text{} & \text{} & \text{} & \text{} & \text{} & \text{} \\
\begin{array}{c}
 \diamond \;\diamond  \\
\end{array}
 &
\begin{array}{c}
\diamond \;\diamond  \\
\end{array}
 & \text{} & \text{} & \text{} & \text{} & \text{} & \text{} \\
\begin{array}{cc}
 1 & 4 \\
\end{array}
 &
\begin{array}{cc}
 2 & 4 \\
\end{array}
 &
\begin{array}{c}
 \text{$\bullet$ $\bullet$} \\
\end{array}
 & \text{} & \text{} & \text{} & \text{} & \text{} \\
\begin{array}{c}
 \text{$\bullet$ $\bullet$} \\
\end{array}
 &
\begin{array}{c}
 \text{$\bullet$ $\bullet$} \\
\end{array}
 &
\begin{array}{c}
 \text{$\bullet$ $\bullet$} \\
\end{array}
 &
\begin{array}{c}
 \text{$\bullet$ $\bullet$} \\
\end{array}
 & \text{} & \text{} & \text{} & \text{} \\
\begin{array}{cc}
 1 & 6 \\
\end{array}
 &
\begin{array}{cc}
 2 & 6 \\
\end{array}
 &
\begin{array}{c}
 \text{$\bullet$ $\bullet$} \\
\end{array}
 &
\begin{array}{cc}
 4 & 6 \\
\end{array}
 &
\begin{array}{c}
 \diamond \;\diamond  \\
\end{array}
 & \text{} & \text{} & \text{} \\
\begin{array}{cc}
 1 & 7 \\
\end{array}
 &
\begin{array}{cc}
 2 & 7 \\
\end{array}
 &
\begin{array}{c}
 \text{$\bullet$ $\bullet$} \\
\end{array}
 &
\begin{array}{cc}
 4 & 7 \\
\end{array}
 &
\begin{array}{c}
 \diamond \;\diamond \\
\end{array}
 &
\begin{array}{cc}
 6 & 7 \\
\end{array}
 & \text{} & \text{} \\
\begin{array}{cc}
 1 & 8 \\
\end{array}
 &
\begin{array}{cc}
 2 & 8 \\
\end{array}
 &
\begin{array}{c}
 \text{$\bullet$ $\bullet$} \\
\end{array}
 &
\begin{array}{cc}
 4 & 8 \\
\end{array}
 &
\begin{array}{c}
 \diamond \;\diamond  \\
\end{array}
 &
\begin{array}{cc}
 6 & 8 \\
\end{array}
 &
\begin{array}{cc}
 7 & 8 \\
\end{array}
 & \text{} \\
\begin{array}{cc}
 1 & 9 \\
\end{array}
 &
\begin{array}{cc}
 2 & 9 \\
\end{array}
 &
\begin{array}{c}
 \text{$\bullet$ $\bullet$} \\
\end{array}
 &
\begin{array}{cc}
 4 & 9 \\
\end{array}
 &
\begin{array}{c}
 \diamond \;\diamond \\
\end{array}
 &
\begin{array}{cc}
 6 & 9 \\
\end{array}
 &
\begin{array}{cc}
 7 & 9 \\
\end{array}
 &
\begin{array}{cc}
 8 & 9 \\
\end{array}
 \\
\end{array}$$

Where $\bullet \;\bullet$ and $\diamond \;\diamond$ are the deleted pairs of indices. My guess is that it means the second case because it seems only this way would allow us to have $\sigma(A)=A$, since no $(i_j - i_k)$ would be changed. Is my reasoning correct?

答:
I feel like Hungerford may have painted himself into a corner with that notation, so let me offer you a different way of doing exactly what the proof is doing.

For a fixed $n$, consider commuting variables $x_1,\ldots,x_n$ and form the Vandermonde matrix $V=V(x_1,\ldots,x_n)$ with $V(x)_{ij} = x_i^{j-1}$. It is a pleasant exercise to verify that $
\det V$ is equal to $\prod_{i<j}(x_i-x_j)$ and, in particular, it evaluates to a non-zero value whenever we pick distinct values for the variables, as Hungerford stated. Let us call this polynomial $\Delta$.

The symmetric group $S_n$ acts on the variables $x_i$ by its action on the indices, so in particular it acts on each polynomial in the variables $x_1,\ldots,x_n$ and hence on $\Delta $.

The first observation we make is that for a transposition the determinant changes by $-1$,  since in the way we have written it, it is clear two rows are swapped. This means that if $\sigma$ is written as a product of $N$ transpositions, then $\sigma \Delta  = (-1)^N\Delta$. Assume that $\sigma$ is also written as a product of $M$ traspositions.

To conclude that $N=M$ modulo $2$, it suffices we evaluate $\Delta$ at a the point say $p=(1,2,\ldots,n)$ as suggested in your post, in which case $\Delta(p)\neq 0$ and we obtain that $(-1)^N = (-1)^M$.


问:
Is the composite of an uniformly continuous sequence of functions with a bounded continuous function again uniformly continuous?
Let $\{f_n\}$ be a sequence of functions $f_n: J\to \mathbb{R}$ that converges uniformly to $f:J\to \mathbb{R}$ where $J\subseteq \mathbb{R}$ is an interval.

It is clear that for a uniformly continuous function $g:\mathbb{R}\to\mathbb{R}$, the sequence $\{g\circ f_n\}$ converges uniformly to $g\circ f:J\to \mathbb{R}$. There is a counterexample, if $g$ is only continuous.

If $J$ is compact, there is no such counterexample because then every continuous function $g$ is uniformly continuous. If $J$ is not compact, bounded and continuous for $g$ does not imply uniformly continuous.

Let $g:\mathbb{R}\to\mathbb{R}$ be bounded and continuous and $\{f_n\}$ a sequence of functions $f_n: J\to \mathbb{R}$ that converges uniformly to $f:J\to \mathbb{R}$. Does the sequence $\{g\circ f_n\}$ converges uniformly to $g\circ f:J\to \mathbb{R}$? If not, what is a counterexample?

答: Hint.  $g(x) = \sin(x^2)$, $f(x)=x$, and find constants $a_n \to 0$ so that $f_n(x) = x+a_n$ does what you want.  So that for each $n$ there is $x$ with $g(x)=\sin(2\pi n) = 0$ and $g(f_n(x))=\sin(2\pi(n+1/2)) = 1$.</p>


问:
Is a continuous map between two topological groups homotopic to a homomorphism between them?
Let $G$ and $H$ be two topological groups and $f:G\to H$ be a continuous map.

Is there a continuous homomorphism $g:G\to H$ homotopic to $f$?

答:
Neat question!

With suitable hypotheses, this question has an affirmative answer. Here is one such answer:

Let $G$ be a compact connected topological group, and let $H$ be a
locally compact abelian topological group. Then every continuous
function $f \colon G \to H$ sending the identity of $G$ to the
identity of $H$ is homotopic to exactly one continuous
homomorphism. Moreover, the homotopy can be chosen to preserve the
identity.

The above statement is Corollary 2 of the following paper of Wladimiro Scheffer:

Maps between topological groups that are homotopic to homomorphisms. Proc. Amer. Math. Soc. 33 (1972), 562-567.
https://doi.org/10.1090/S0002-9939-1972-0301130-8

I'm not sure what other answers to this question might look like (i.e. with different hypotheses on $G$ and $H$), but it's an interesting question.

For example, one cannot drop the assumption that $H$ is abelian, for the homomorphism
$$e^{i\theta} \mapsto \begin{pmatrix}e^{i\theta}&0\\0&e^{-i\theta}\end{pmatrix}$$
from $S^1$ to $SU(2)$ is at once non-trivial and homotopic (relative to the identity) to the trivial homomorphism since $SU(2) \cong S^3$ is simply connected.

See also:

Bogatyi, Semeon Antonovich, and Olga Dmitrievna Frolkina. "On commutativity of connected groups." Sbornik: Mathematics 197.1 (2006).

Rezk, Charles. "Classifying spaces for 1–truncated compact Lie groups." Algebraic & geometric topology 18.1 (2018): 525-546.

https://mathoverflow.net/questions/334473/when-does-bg-to-ba-loop-to-a-homomorphism


答:
For you $\Longrightarrow$ direction, I have some critiques.

Let $\epsilon >0$ and set $M=\frac{1}{\epsilon}$. Since $\lim_{x\to c}f(x)=\infty$, we can find a $\epsilon >0$ such that $f(x)>M$ whenever $0<|x-c|<\delta$.

I think that this is a typo, but I believe that the third $\epsilon$ should be a $\delta$. Otherwise, this is wrong and doesn't seem to make sense.

Thus $0<\frac{1}{f(x)}<\frac{1}{\epsilon}$ whenever $0<|x-c|<\delta$.

This is incorrect. If you have $$f(x)>M=\frac{1}{\epsilon}$$ then you should end up with $$\epsilon>\frac{1}{f(x)}>0 $$whenever $0<|x-c|<\delta$

After that then your proof is fine.

For the $\Longleftarrow$ direction, I have some suggestions.

You say you have some confusion with writing out the definition of the limit for $1/f(x)$. Hopefully the following helps:

We have a function $f:\mathbb{R}\setminus\{c\}\to\mathbb{R}$. Define the function $g:\mathbb{R}\setminus\{c\}\to\mathbb{R}$ as $g(x)=1/f(x)$. This is well-defined since $f(x)>0$ for all $x\in\mathbb{R}\setminus\{c\}$. Now $$\lim_{x\to c}g(x)=\lim_{x\to c}\frac{1}{f(x)}$$ so, writing out the definition in terms of $g$, we have $$\forall\epsilon>0\,\exists\delta>0\,\forall x\in\mathbb{R}\setminus\{c\}\left(0<|x-c|<\delta\Longrightarrow |g(x)-0|<\epsilon\right)$$

Replace $g(x)$ with $1/f(x)$ to see how that definition works.

Title: Using definition of limits

Body:
Let $c∈\mathbb{R}$ and let $f:\mathbb{R}\setminus{c}\rightarrow\mathbb{R}$ be a function such that $f(x)>0$ for all $x∈\mathbb{R}$. Use the definition of limits to prove that
$$
\lim_{x\to c}f(x)=\infty  \space\space\space\space\space\space\space\space\space\space\space   \text{iff}  \space\space\space\space\space\space\space\space\space\space\space    \lim_{x\to c}\frac{1}{f(x)}=0.
$$

Proving the "$\Rightarrow$": Here is the definition: $\lim_{x\to c}f(x)=\infty$ if $\forall M∈\mathbb{R},\exists\delta>0$ such that $\forall x∈\mathbb{R}, 0<|x-c|<\delta\Rightarrow f(x)>M$. Here is my proof:

Let $\epsilon >0$ and set $M=\frac{1}{\epsilon}$. Since $\lim_{x\to c}f(x)=\infty$, we can find a $\epsilon >0$ such that $f(x)>M$ whenever $0<|x-c|<\delta$.

Thus $0<\frac{1}{f(x)}<\frac{1}{\epsilon}$ whenever $0<|x-c|<\delta$. This implies that it is possible to find a $\delta>0$ such that $|\frac{1}{f(x)}|<\epsilon$ whenever $0<|x-c|<\delta$. Since $\epsilon$ is arbitrary, we have proved that $\lim_{x\to c}\frac{1}{f(x)}=0$.

Proving the "$\Leftarrow$": This proof I am unsure of. I know that by the definition of a limit, $\lim_{x\to c} f(x) = 0$ if $\forall\epsilon>0, \exists\delta>0$ such that $\forall x∈\mathbb{R}\setminus{c}, 0<|x-c|<\delta \Rightarrow |f(x)-0|<\epsilon$. I am unsure of how to define $\lim_{x\to c}\frac{1}{f(x)}=0$ in a similar way. Any advice would be greatly appreciated.


问:
How to convert a matrix of decision variables in a vector for solving a linear programming problem?
I want to solve this problem :

find the minimum of the function F = sum(sum(A - Q*E)),

were A is an i x j matrix, Q is an i x k matrix and E is an k x j matrix. This is basically a least absolute deviation optimisation, where A are the data, Q are know coefficients and I'm searching for the matrix E.

Apparently it is possible to solve with linear programming, I read that I need to vectorize the matrix E, but I don't know what to do at this point... Can anyone give me some help please ? It will be great !

答:
I am going to assume that you actually want sum(sum(abs(A-Q*E))), since you said you want the least absolute deviation. I'm also going to offer some Matlab code in here since your expression looks like it is expressed in it.

It is common in optimization literature to define a linear operator $\mathop{\textbf{vec}}:\mathbb{R}^{m\times n}\rightarrow\mathbb{R}^{mn}$ which maps a matrix to a vector. Typically this is done "Fortran-style": that is, by stacking the columns one on top of another. For instance, suppose $A\in\mathbb{R}^{m\times n}$ is subdivided by its columns as follows:
$$A=\begin{bmatrix} a_1 & a_2 & \dots & a_n \end{bmatrix}, \quad a_i\in\mathbb{R}^m,~i=1,2,\dots,n$$
Then
$$\mathop{\textbf{vec}}(A)\triangleq\begin{bmatrix} a_1 \\ a_2 \\ \vdots \\ a_n \end{bmatrix}\in\mathbb{R}^{mn}.$$
So that's really all you are doing here, on one hand. Not surprisingly, this operator is linear, so $\mathop{\textbf{vec}}(A\pm B)=\mathop{\textbf{vec}}(A)\pm\mathop{\textbf{vec}}(B)$ and $\mathop{\textbf{vec}}(\alpha A)=\alpha\mathop{\textbf{vec}}(A)$. In MATLAB, the colon notation gives you the $\mathop{\textbf{vec}}$ operator: A(:)

Now, you do have one challenge: what do do about $\mathop{\textbf{vec}}(QE)$? For this, you need to know about Kronecker products. In particular, you need to represent the matrix-matrix product in terms of the vectorized version of $E$, $\mathop{\textbf{vec}}(E)\in\mathbb{R}^{kj}$. For that you need the formula offered in the "Matrix equations" section of the Wikipedia page. Translated to your problem, you have:
$$\mathop{\textbf{vec}}(QE)=(I_{jj}\otimes Q)\mathop{\textbf{vec}}(E).$$

In MATLAB, this computation is simply kron(eye(j),Q)*E(:)

In summary, create $\bar{A}=\mathop{\textbf{vec}}(A)$ (barA), $\bar{Q}=I\otimes Q$ (barQ), and define the variable $\bar{E}\triangleq \mathop{\textbf{vec}}(E)$ (barE), and your expression simply becomes sum(abs(barA-barQ*barE)). If you're really using MATLAB, this is just sum(abs(A(:)-kron(eye(j),Q)*E(:))).

EDIT: since you have confirmed that you intend to use MATLAB, may I recommend my software CVX. It is a modeling framework for convex optimization that does all of these contortions for you. Here, for instance, is the CVX model for this problem, including your condition that $E\geq 0$:

```<code>cvx_begin
    variable E(k,j)
    minimize(sum(sum(abs(A-Q*E))))
    subject to
        E >= 0
cvx_end
</code>```

See that? No Kronecker products! Here's an alternate approach:

```<code>cvx_begin
    variable E(k,j)
    minimize(norm(vec(A-Q*E),"inf"))
    subject to
        E >= 0
cvx_end
</code>```

Body:
I am going to assume that you actually want sum(sum(abs(A-Q*E))), since you said you want the least absolute deviation. I'm also going to offer some Matlab code in here since your expression looks like it is expressed in it.

It is common in optimization literature to define a linear operator $\mathop{\textbf{vec}}:\mathbb{R}^{m\times n}\rightarrow\mathbb{R}^{mn}$ which maps a matrix to a vector. Typically this is done "Fortran-style": that is, by stacking the columns one on top of another. For instance, suppose $A\in\mathbb{R}^{m\times n}$ is subdivided by its columns as follows:
$$A=\begin{bmatrix} a_1 & a_2 & \dots & a_n \end{bmatrix}, \quad a_i\in\mathbb{R}^m,~i=1,2,\dots,n$$
Then
$$\mathop{\textbf{vec}}(A)\triangleq\begin{bmatrix} a_1 \\ a_2 \\ \vdots \\ a_n \end{bmatrix}\in\mathbb{R}^{mn}.$$
So that's really all you are doing here, on one hand. Not surprisingly, this operator is linear, so $\mathop{\textbf{vec}}(A\pm B)=\mathop{\textbf{vec}}(A)\pm\mathop{\textbf{vec}}(B)$ and $\mathop{\textbf{vec}}(\alpha A)=\alpha\mathop{\textbf{vec}}(A)$. In MATLAB, the colon notation gives you the $\mathop{\textbf{vec}}$ operator: A(:)

Now, you do have one challenge: what do do about $\mathop{\textbf{vec}}(QE)$? For this, you need to know about Kronecker products. In particular, you need to represent the matrix-matrix product in terms of the vectorized version of $E$, $\mathop{\textbf{vec}}(E)\in\mathbb{R}^{kj}$. For that you need the formula offered in the "Matrix equations" section of the Wikipedia page. Translated to your problem, you have:
$$\mathop{\textbf{vec}}(QE)=(I_{jj}\otimes Q)\mathop{\textbf{vec}}(E).$$

In MATLAB, this computation is simply kron(eye(j),Q)*E(:)

In summary, create $\bar{A}=\mathop{\textbf{vec}}(A)$ (barA), $\bar{Q}=I\otimes Q$ (barQ), and define the variable $\bar{E}\triangleq \mathop{\textbf{vec}}(E)$ (barE), and your expression simply becomes sum(abs(barA-barQ*barE)). If you're really using MATLAB, this is just sum(abs(A(:)-kron(eye(j),Q)*E(:))).

EDIT: since you have confirmed that you intend to use MATLAB, may I recommend my software CVX. It is a modeling framework for convex optimization that does all of these contortions for you. Here, for instance, is the CVX model for this problem, including your condition that $E\geq 0$:

cvx_begin
    variable E(k,j)
    minimize(sum(sum(abs(A-Q*E))))
    subject to
        E >= 0
cvx_end

See that? No Kronecker products! Here's an alternate approach:

cvx_begin
    variable E(k,j)
    minimize(norm(vec(A-Q*E),"inf"))
    subject to
        E >= 0
cvx_end


问:
$ x_n = 1 + 2a + 3a ^ 2 + \cdots + na ^ {n-1}$ has any three consecutive terms coprime
Let $ a \in \mathbb N $. For $ n \in \mathbb N $, let $$ x_n = 1 + 2a + 3a ^ 2 + \cdots + na ^ {n-1}. $$

$ x_1 = 1, x_2 = 1 + 2a, \ldots $.

For $ n \in \mathbb N $, show that $ \gcd (x_n, x_ {n + 1}, x_ {n + 2}) = 1 $.

Any help, suggestion on how to start solving that exercise.

答: Notice that $x_{n+2}-(a+1)x_{n+1}+ax_{n}=a^{n+1}$ so $gcd(x_{n},x_{n+1},x_{n+2})$ divides $a^{n+1}$, so if it is not 1 , say $m$ , $1<m<a^{n+1}+1, m \in \mathbb N,$ then since $x_{n+2}$ is equivalent to 1 modulo a , we have that $x_{n+2}$ is coprime to all prime divisors of $a$, so $m$ is coprime to $x_{n+2}$ since the prime divisors of $m$ are prime divisors of $a$ , so $m$ is not a divisor of $x_{n+2}$ so it is not the $gcd(x_{n},x_{n+1},x_{n+2})$, a contradiction.


问:
Can you please advise me how to simplify these terms?
e^(iπ/c) −sgrt(−1) + e^(−iπ/c) -1/i*sin(π/2)− cos(π/c),

e^(iπ/4) + e^(−iπ/2) −(1 + i)/sgrt(2)+ 2sgrt(−1).

答:
Expression 1

$$e^{iπ/c} −\sqrt{-1} + e^{−iπ/c} -\frac{1}{i\cdot\sin(π/2)}− \cos\left(\frac{\pi}{c}\right).$$
We can put the two exponential terms together. By the complex cosine representation we have $e^{iπ/c}+e^{-iπ/c}=2\cos(\pi/c).$ Next, the square root of $-1$ is defined as $i$. Furthermore $\sin(\pi/2)=1$ and $1/i=-i$. Putting all of this together gives us
$$-i+i+2\cos\left(\frac{\pi}{c}\right)-\cos\left(\frac{\pi}{c}\right)=\cos\left(\frac{\pi}{c}\right).$$

Expression 2

With the help of expression $1$ you should be able to also solve the second expression. If you still have questions just ask in the comments.


问:
$\forall\varepsilon > 0\exists\delta > 0:(|f(x) - \ell| < \varepsilon)\implies(0<|x-a| < \delta )$. $f(x)=x$ satisfies this for all $a$ and $\ell$?
This is to be found in the answer to problem 26 (b) of chapter 5 of Michael Spivak's Calculus 3rd edition. The problem reads as follows:

26 Give examples to show that the following definitions of $\lim_{x\to a}f(x)=\ell$ are not correct.

(b) For all $\varepsilon > 0$ there is a $\delta > 0$ such that if $|f(x) - \ell| < \varepsilon$, then $0<|x-a| < \delta$.

I thought of constant functions because they do not satisfy this condition, but approach a limit "everywhere". This means, let $f(x)=c, c\in \mathbb R$ and let $a\in \mathbb R$ and let $\varepsilon > 0$, then $|f(x) - c|=0 < \varepsilon$ for all $x$ satisfying $0<|x-a| < \delta$ for any $\delta > 0$ and not just some $\delta > 0$. (It is already shown in the text of this chapter that $\lim_{x\to a}c=c$.)

In his answer Spivak also mentions constant functions but then adds the following:

Moreover, the function $f(x)=x$, for example, satisfies this condition no matter what $a$ and $\ell$ are.

What does he mean by that? It seems to me that if $|f(x) - \ell| < \varepsilon$ holds, then $|x - \ell| < \varepsilon$ so $a=\ell$ does hold, which means it does matter what $a$ and $\ell$ are. Unfortunately, I really don't understand what he is getting at.

答:
If $|f(x)- l| < \varepsilon$ holds, then $|x-l|<\varepsilon$

This is true.

so $a = l$ does hold

This is also true, but not necessary. You have assumed that $\delta$ is small (equal to $\varepsilon$), but this need not be the case. We are free to chose $\delta$ as we please, so $|a-l|$ can actually be as big as we like (using this incorrect definition).

In the true definition, we are asking "Can we find $x$ close enough to $l$ such that $|f(x) - l| < \varepsilon$". In the bad definition above, we are asking "Can we find a $\delta$ big enough such that if $|f(x) - l| < \varepsilon$, then $|x - a| < \delta$".

The important thing to take note of is the order which each section appears. Let's break down the 'definition' given in the question step by step to see where we go wrong.

Step 1 - What is given?

We are trying to show that $\lim_{x\to a} f(x) = l$, so $a$ and $l$ are fixed.

Step 2 - Choose an epsilon

We have been told that what follows must hold for all $\varepsilon > 0$, so we let $\varepsilon > 0$ be arbitrary but, from this point onwards, fixed.

Step 3 - Find a delta

We want to find a $\delta$ that 'works'. By this we mean, can we find a $\delta$ such that, if |$f(x) - l| < \varepsilon$ then $|x-a| < \delta$?

Suppose that $|f(x) - l| < \varepsilon$. Then, by definition of $f$, $|x-l| < \varepsilon$, so we have $x$ being 'close' to $l$. Can we find a delta such that $|x-a| < \delta$? Sure we can! We just pick delta to be as big as we need. So, using this definition, as long as we pick $\delta$ large enough, it works for all choices of $a$ and $l$.

Example

Let's consider the following example:

Using this (incorrect) definition, show that $\lim_{x\to 5}f(x) = 27$, where $f(x) = x$

Let $\varepsilon > 0$ be arbitrary. We seek $\delta > 0$ such that, if $|x-27|<\varepsilon$, then $|x - 5| < \delta$.

If $|x - 27| < \varepsilon$, and $\delta > 22 + \varepsilon$, then
\begin{align*}
|x - 5| &= |x - 5 + 27 - 27| \\
 &\leq |x - 27| + |-5 + 27| \\
 &= |x - 27| + |22| \\
&< \varepsilon + 22 \\
&< \delta
\end{align*}

Hence $\lim_{x\to 5}f(x) = 27$.

Clearly, this is nonsense, but it is a good exercise in demonstrating why the order of the statements in the definition is important.


问:
Compactness and Totally Bounded Sets
Suppose that $(K,d)$ is a compact metric space, and $\varepsilon>0$. Prove that there is a finite subset $S⊆K$ so that for every $y∈K$ there is $x∈S$ with $d(x,y)<\varepsilon$, and moreover so that for any distinct $x,w∈S$ we have $d(x,w)>\varepsilon/2$.

I dont know how to approach $d(x,w)>\varepsilon/2$.

So far I have, $d(x_i,w) - d(w,y) \le d(x,w)\ge d(x_i,w) - \varepsilon/2 \le d(x,w)$ but it seems wrong.

答:
I assume you have proved that there is a finite set $S$ such that every point of $X$ is within distance $\varepsilon$ of some point of $S$ (cover every point with $\varepsilon$-balls, extract a finite subcover).

For the second part, try induction on the size of the covering. A valid inductive hypothesis might be, "suppose that for any compact metric space admitting a covering of $\leq n$ points, such a configuration with pairs of points distance $\varepsilon/2$ apart exists". The base case of $n = 1$ (or $n=0$ with the empty space) is trivial. If you have a minimal covering with $n+1$ points (i.e. if you remove some $\varepsilon$-ball, you no longer have a cover), then apply the inductive hypothesis to the closure of the subspace $Y$ covered by $n$ points, and judiciously choose the spot for the $(n+1)$-st (e.g. on the boundary of $Y$).

Title: Compactness and Totally Bounded Sets


问:
Show that $\lambda \in \sigma (T)$.
Let $H$ be a Hilbert space, let $T \in B(H)$, and let $\lambda \in \mathbb{C}$. Suppose there is a sequence $\{x_n\}^{\infty}_{n=1}$ in $H$ such that $\|x_n\| = 1$ for all $n$ and $\|(T - \lambda I)x_n\| \to 0$. Show that $\lambda \in \sigma(T)$.

I am new to Operator Algebra and stuck with this problem. Here is what I've tried.
$$
\|(T - \lambda I)x_n\| \leq |T - \lambda I| \|x_n\| = |T - \lambda I| \to 0
$$
It follows that $\lambda I - T = 0$. Namely, $\lambda I - T$ is not invertible. Hence, $\lambda \in \sigma(T)$.

Please tell me is it alright? Any help would be appreciated!

答:
$|T-\lambda I| \to 0$ makes no sense.

Suppose, if possible, $\lambda \notin \sigma(T)$. Then $(T-\lambda I)$ has a bounded inverse. Now $1=\|x_n\|=\|(T-\lambda I)^{-1}(T-\lambda I)x_n\| \leq \|(T-\lambda I)^{-1}\|\|(T-\lambda I)x_n\| \to 0$ which is a  contradiction.

Title: Show that $\lambda \in \sigma (T)$.


问:
Derivative of conditional expectation of integral of stochastic process
Let $T>0$. Let $(\Omega, \mathcal{F}, \mathbb{P})$ be a probability space and let $\mathbb{F}=(\mathcal{F}_u)_{u\in[0;T]}$ be a filtration such that $\mathcal{F}_T=\mathcal{F}$. Let $(\alpha_u)_{u\geq 0}$ be a strictly positive, $\mathbb{F}$-adapted stochastic process.

I would like to show in a rigorous way that
$$
\frac{d\mathbb{E}_{\mathbb{P}}\left[\int_t^\tau\alpha_u du\biggr|\mathcal{F}_t\right]}{d\tau}\Big|_{\tau=\tau'}=\mathbb{E}_{\mathbb{P}}\left[\alpha_{\tau'}\Big|\mathcal{F}_t\right],
$$
where $t,\tau'\in(0;T)$ and $\tau'>t$.

I could only come up with a intuitive argument where $\alpha_u$ is constant over a small interval $[\tau;\tau+\Delta\tau]$, but I suspect there might be a more rigorous way to prove this statement. What are some theorems that one could use to construct such a rigorous argument? If so, which are those theorems? Is there maybe a general method for proving such statements?

EDIT: Imposing conditions on $(\alpha_u)_{u\geq 0}$ may be required to give a definitive answer to this question.

答:
Added note: If $H$ is bounded and $\mathcal F_t$-measurable, then
$$
\eqalign{
\Bbb E_{\Bbb P}\left[\int_t^\tau\alpha_u du\cdot H\right]
&=\int_t^\tau\Bbb E_{\Bbb P}\left[\alpha_u \cdot H\right]du\cr
&=\int_t^\tau\Bbb E_{\Bbb P}\left[\Bbb E_{\Bbb P}[\alpha_u\mid\mathcal F_t]\cdot H\right]du\cr
&=\int_t^\tau\Bbb E_{\Bbb P}\left[A_u\cdot H\right]du\cr
&=\Bbb E_{\Bbb P}\left[\int_t^\tau A_u du\cdot H\right].\cr
}
$$
As $\int_t^\tau A_u du$ is $\mathcal F_t$-measurable, this shows that $\int_t^\tau A_u du$ is a version of the conditional expectation $\Bbb E_{\Bbb P}\left[\int_t^\tau \alpha_u du\mid \mathcal F_t\right]$.


问:
How to show that the following set $\mathbb{C}/\mathbb{T}$ is compact?
Let $w_1, w_2$ be two complex numbers that are linearly independent over $\mathbb{R}$, i.e., $\frac{w_1}{w_2}$ does not belong to $\mathbb{R}$.

Let $\mathbb{T}= \mathbb{Z} w_1+ \mathbb{Z}w_2$.

We define an equivalence relation on $\mathbb{C}$ by saying that two complex numbers $z_1, z_2$ are equivalent if and only if $z_1-z_2 \in \mathbb{T}$.

Let $\mathbb{C}/\mathbb{T}$ be the set of all equivalence classes of the equivalence relation defined above.

Let $\pi:\mathbb{C}\rightarrow \mathbb{C}/\mathbb{T}$ be the canonical projection.

We define a topology $\tau$ on $\mathbb{C}/\mathbb{T}$ by saying that a subset $U$ of $\mathbb{C}/\mathbb{T}$ is open if and only if $\pi^{-1}(U)$ is an open subset of $\mathbb{C}$.

How to prove that $\mathbb{C}/\mathbb{T}$ with the topology $\tau$ defined above is compact?

答:
Consider the subset
$$D = \{a w_1 + b w_2 \mid (a,b) \in [0,1] \times [0,1]\} \subset \mathbb C
$$
Clearly $D$ is homeomorphic to $[0,1] \times [0,1]$ and so is compact. Also, the restriction of $\pi : \mathbb C \to \mathbb C / \mathbb T$ to the subset $D$ is surjective, because every equivalence class contains an element of $D$. To see why this is true, start with an arbitrary element of $\mathbb C$, written in the form $x w_1 + y w_2 \in \mathbb C$ (where $x,y \in \mathbb R$). Take $a = x - \lfloor x \rfloor$ and $b = y - \lfloor y \rfloor$, where $\lfloor t \rfloor$ is the largest integer $\le t$. It follows that $a w_1 + b w_2 \in D$, and that $a w_1 + b w_2$ is equivalent to $x w_1 + y w_2$, because
$$(x w_1 + y w_2) - (a w_1 + b w_2) = \lfloor x \rfloor w_1 + \lfloor y \rfloor w_2 \in \mathbb Z w_1 + \mathbb Z w_2
$$
Therefore, $\mathbb C / \mathbb T$ is the continuous image of the compact set $D$, hence is compact.


问:
Finding the support of a distribution
Definition. Let
 $u :C^{\infty}_{c}(\Omega) \rightarrow \mathbb{C}$ a linear continuous functional, that is, $u$ is a distribution. We define the support of $u$ as $$\Omega- \{x \in \Omega; \exists \; \mbox{an open}\; V_x \; \mbox{such that}\; \; u|_{V_x}=0\}.$$ Here $u|_{V_x}$ means $$u|_{V_x}(\varphi)=u(\varphi)\quad \forall \varphi \in C_{c}^{\infty}(V_x),$$ where $\Omega \subset \mathbb{R}^{n}$ is open.

How can I find the support of the distribution \begin{equation} \label{1}u(\varphi)=\sum_{n=1}^{\infty}[\varphi (\frac{1}{n})-\varphi(0)-\frac{1}{n}\varphi'(0)] \quad \forall \varphi \in C^{\infty}_{c}(\mathbb{R})? \end{equation}

I would like an answer or a tip on how to solve this question.

I observe this serie is convergent, because using the Taylor Formula $$\varphi(\frac{1}{n})=\varphi(0)+\frac{1}{n}\varphi'(0)+\frac{1}{n^2}\alpha(\frac{1}{n})$$ we conclude $u(\varphi)$ is convergent.

答:
We will prove that $$ \mbox{supp}\, u=\{0,1,\frac{1}{2}, \ldots, \frac{1}{n}, \ldots\}:=S.$$

Fisrtly we notice that $S$ is closed in $\mathbb{R}$, because $\frac{1}{n}\rightarrow 0$. Now, if $\varphi \in C^{\infty}_{c}(\mathbb{R}-S)$ then $$ \langle u, \varphi\rangle=\langle u|_{\mathbb{R}-S}, \varphi \rangle=\sum_{n=1}^{\infty}\left(\varphi(\frac{1}{n})-\varphi(0)-\frac{1}{n}\varphi'(0))\right)=0,$$ because $\mbox{supp}\, \varphi' \subset \mbox{supp}\, \varphi$ and $S \cap \mbox{supp}\, \varphi=\varnothing$. Thus, if $x \in \mbox{supp}\, u$ then $x \in S$, because $\mathbb{R}-S$ is open and $u|_{\mathbb{R}-S}=0$. So $$\mbox{supp}\, u \subset S. $$

Now, let $x=\frac{1}{p} \in S$ and $V$ an open subset of $\mathbb{R}$ with $x \in V$. Define \begin{equation}\Omega=
\begin{cases}
V \cap ]\frac{1}{p+1}, \frac{1}{p+1}[ \quad p >1\\
V \cap ]\frac{1}{2},2[, \quad p=1.
\end{cases}
\end{equation} Thus there exists $\varphi \in C^{\infty}_{c}(\Omega)$, $0 \leq \varphi \leq 1$ and $\varphi(x)=1$. Therefore $\varphi(0)=\varphi(\frac{1}{j})=0$ if $j \neq p$, and we obtain \begin{equation}
\langle u|_{V},\varphi \rangle=\langle u, \varphi \rangle=\sum_{n=1}^{\infty}\left(\varphi(\frac{1}{n})-\varphi(0)-\frac{1}{n}\varphi'(0)\right)=\varphi(\frac{1}{p})=1.
\end{equation}
Therefore $x \in \mbox{supp}\, u$ and we have what we wish for, that is,  $$\{1, \frac{1}{2}, \ldots, \frac{1}{n}, \ldots\}\subset\mbox{supp} u .$$

Besides $0 \in \mbox{supp}\, u$, because $\frac{1}{n}\rightarrow 0$ and $\mbox{supp}\, u$ is closed in $\mathbb{R}$.

Title: Finding the support of a distribution


问:
finding the shape of the higher order equations
Is there any way to find the shape of the higher-order equation like this one? I am more interested in making an initial guess about the shape (fast prediction) and then the process to find it. Any process or tool would be helpful.

$f(x) = x^2z^3 + 9/80 * y^2z^3  - (x^2 + 9/4 * y^2 + z^2 - 1)^3 $

I tried to plot it using wolframalpha but I am not succeeded.

答:
You have a function
$$f(x, y, z) = x^2 z^3 + \frac{9}{80} y^2 z^3 - (x^2 + \frac{9}{4} y^2 + z^2 - 1)^3$$
This describes an implicit surface $f(x, y, z) = 0$.  There are a number of applications that can plot implicit surfaces (or more generally, isosurfaces $f(x, y, z) = c$).

In this particular case,
$$\begin{aligned}
f(x,y,z) & = 0 \\
x^2 z^3 + \frac{9}{80} y^2 z^3 - (x^2 + \frac{9}{4} y^2 + z^2 - 1)^3 & = 0 \\
z^3 ( x^2 + \frac{9}{80} y^2 ) & = (x^2 + \frac{9}{4} y^2 + z^2 - 1)^3 \\
z ( x^2 + \frac{9}{80} y^2 )^{1/3} & = x^2 + \frac{9}{4} y^2 + z^2 - 1 \\
z^2 - z ( x^2 + \frac{9}{80} y^2 )^{1/3} + 1 - x^2 - \frac{9}{4} y^2 & = 0 \\
\end{aligned}$$
which is a quadratic function in $z$, whose solution is
$$\begin{aligned}
z & = \left(\frac{9 y^2 + 80 x^2}{640}\right)^\frac{1}{3} \pm \frac{1}{640^\frac{1}{3}}\sqrt{ (9 y^2 + 80 x^2)^\frac{2}{3} - 2160^\frac{2}{3} y^2 - 640^\frac{2}{3} x^2 + 640^\frac{2}{3} } \\
~ & = \frac{1}{640^\frac{1}{3}} \left( (9 y^2 + 80 x^2)^\frac{1}{3} \pm \sqrt{ (9 y^2 + 80 x^2)^\frac{2}{3} - 2160^\frac{2}{3} y^2 - 640^\frac{2}{3} x^2 + 640^\frac{2}{3} } \right) \\
\end{aligned}$$
The shape is a 3D heart. The $+$ side contains the two bumps on top, and the $-$ side the conical bottom part.

Personally, I would use GNU Octave for this.  It is free, and available for Windows, macOS, Linux, and BSDs.

```<code>function value = f (x, y, z);
    value = x**2 * z**3 + y**2 * z**3 * 9/80 - ( x**2 + y**2 * 9/4 + z**2 - 1)**3;
endfunction;
xrange = linspace(-1.6, +1.6, 64);
yrange = linspace(-1.6, +1.6, 64);
zrange = linspace(-1.6, +1.6, 64);
[x, y, z] = meshgrid(xrange, yrange, zrange);
v = arrayfun(@f, x, y, z); isosurface(v, 0); pbaspect([1, 1, 1]);
</code>```

Body:
You have a function
$$f(x, y, z) = x^2 z^3 + \frac{9}{80} y^2 z^3 - (x^2 + \frac{9}{4} y^2 + z^2 - 1)^3$$
This describes an implicit surface $f(x, y, z) = 0$.  There are a number of applications that can plot implicit surfaces (or more generally, isosurfaces $f(x, y, z) = c$).

In this particular case,
$$\begin{aligned}
f(x,y,z) & = 0 \\
x^2 z^3 + \frac{9}{80} y^2 z^3 - (x^2 + \frac{9}{4} y^2 + z^2 - 1)^3 & = 0 \\
z^3 ( x^2 + \frac{9}{80} y^2 ) & = (x^2 + \frac{9}{4} y^2 + z^2 - 1)^3 \\
z ( x^2 + \frac{9}{80} y^2 )^{1/3} & = x^2 + \frac{9}{4} y^2 + z^2 - 1 \\
z^2 - z ( x^2 + \frac{9}{80} y^2 )^{1/3} + 1 - x^2 - \frac{9}{4} y^2 & = 0 \\
\end{aligned}$$
which is a quadratic function in $z$, whose solution is
$$\begin{aligned}
z & = \left(\frac{9 y^2 + 80 x^2}{640}\right)^\frac{1}{3} \pm \frac{1}{640^\frac{1}{3}}\sqrt{ (9 y^2 + 80 x^2)^\frac{2}{3} - 2160^\frac{2}{3} y^2 - 640^\frac{2}{3} x^2 + 640^\frac{2}{3} } \\
~ & = \frac{1}{640^\frac{1}{3}} \left( (9 y^2 + 80 x^2)^\frac{1}{3} \pm \sqrt{ (9 y^2 + 80 x^2)^\frac{2}{3} - 2160^\frac{2}{3} y^2 - 640^\frac{2}{3} x^2 + 640^\frac{2}{3} } \right) \\
\end{aligned}$$
The shape is a 3D heart. The $+$ side contains the two bumps on top, and the $-$ side the conical bottom part.

Personally, I would use GNU Octave for this.  It is free, and available for Windows, macOS, Linux, and BSDs.

function value = f (x, y, z);
    value = x**2 * z**3 + y**2 * z**3 * 9/80 - ( x**2 + y**2 * 9/4 + z**2 - 1)**3;
endfunction;
xrange = linspace(-1.6, +1.6, 64);
yrange = linspace(-1.6, +1.6, 64);
zrange = linspace(-1.6, +1.6, 64);
[x, y, z] = meshgrid(xrange, yrange, zrange);
v = arrayfun(@f, x, y, z); isosurface(v, 0); pbaspect([1, 1, 1]);


问:
Convex set. Proof.
Prove that if A is convex set and $\alpha, \beta ≥ 0$ then $(\alpha + \beta)A = \alpha A + \beta A$

What came first on my mind is that I have to show that $(\alpha + \beta)A\subset \alpha A + \beta A$ and $\alpha A + \beta A\subset (\alpha + \beta)A$. But I don't know how to use convexity of A to show this.

答: For the open half-space (I suppose in an n-vector space over the real numbers), consider that up to an isomorphism they're $\{ (x_1,...,x_n): x_n >0\} $ so that each convex combination remains into that set.</p>

Title: Convex set. Proof.


问:
Why is this map from $\mathbb{R}^n$ to $\mathbb{R}^{n+1}$ continuous?
Let $f_i: \mathbb{R}^n \to \mathbb{R}^{n + 1}$ be defined by
$$
f_i(x_1, \ldots, x_n) = (x_1, \ldots, x_{i - 1}, 1,x_i, \dots, x_n).
$$
I'm trying to show as explicitly as possible why this is continuous. The image has the subspace topology in $\mathbb{R}^{n + 1}$. Our open sets are intersections of the hyperplane with unions of $n + 1$-dimensional open balls. They sure look like they should be open, but I don't know how to say it other than by saying "the map sends $\mathbb{R}^n$ to an embedding of $\mathbb{R}^n$ in $\mathbb{R}^{n + 1}$ and doesn't 'change' anything, i.e. acts like the identity", which seems hand-wavy.

答:
The universal property of product spaces applied to $\Bbb R^{n+1}$, where $X$ can be any space:

A map $f:X \to \Bbb R^{n+1}$ is continuous iff $\pi_j \circ f: X \to \Bbb R$ is continuous for all $n+1$ projections $\pi_1, \ldots , \pi_{n+1}$ on $\Bbb R^{n+1}$.

Then note that $\pi_j \circ f_i = \pi'_j$ for $j \in \{1, 2, \ldots, i-1\}$ (where $\pi'_j$ are the continuous projections on $X=\Bbb R^n$), and $\pi_i \circ f_i \equiv t$ and a constant map is continuous, and $\pi_j \circ f_i = \pi'_{j-1}$ for $j \in \{i+1, \ldots, n+1\}$, so it immediately follows from the universal property (which holds for all maps into products) that $f_i$ as defined is continuous.


问:
Solution of a problem of size
I know that the set of all groups does not exist. It's the same in the case of the topological spaces, and the sets themselves. My question is: How the Tarki's axiom solve this problem?

I can't see how the universes of Grothendieck gives sense to the staments "set of all groups", and the same.

答:
It doesn't, it gets around it.

There is (in $\mathsf{ZF}$ at least) no "set of all groups," and since that's a theorem of $\mathsf{ZF}$ adding additional axioms isn't going to change that. However, we can look at the particular ways in which we're tempted to use a hypothetical "set of all groups" in practice, and ask whether there might be some actually-consistent-with-$\mathsf{ZF}$ (+etc.) proxy for that.

Now $\mathsf{ZF}$ has a very nice framework for developing such ideas - namely, the cumulative hierarchy. For each (limit, for niceness) ordinal $\alpha$ we can talk about the set of all groups of set-theoretic rank (this is unrelated to the group-theoretic notion of rank) $<\alpha$; this will be a genuine set, and in fact an element of $V_{\alpha+1}$. The idea now is that if $V_\alpha$ is "nice enough," this approximation to the naive thing we want will actually do all the specific things we want it to.

This is where Tarski comes in. Things are very nice indeed if $V_\alpha$ is a Tarski universe (= if $\alpha$ is a strongly inaccessible cardinal), and we wind up with the following basic theme:

Suppose there is a proper class of strongly inaccessible ordinals $\alpha$. Then for any particular application we care about, we can find a set $\mathfrak{S}$ which is "sufficiently like" the class of all groups for our purposes - specifically, we'll take $\mathfrak{S}$ to be the set of groups in $V_\alpha$ for $\alpha$ a "sufficiently large" strongly inaccessible cardinal.

Title: Solution of a problem of size


问:
Using definition of limits

Let $c∈\mathbb{R}$ and let $f:\mathbb{R}\setminus{c}\rightarrow\mathbb{R}$ be a function such that $f(x)>0$ for all $x∈\mathbb{R}$. Use the definition of limits to prove that
$$
\lim_{x\to c}f(x)=\infty  \space\space\space\space\space\space\space\space\space\space\space   \text{iff}  \space\space\space\space\space\space\space\space\space\space\space    \lim_{x\to c}\frac{1}{f(x)}=0.
$$

Proving the "$\Rightarrow$": Here is the definition: $\lim_{x\to c}f(x)=\infty$ if $\forall M∈\mathbb{R},\exists\delta>0$ such that $\forall x∈\mathbb{R}, 0<|x-c|<\delta\Rightarrow f(x)>M$. Here is my proof:

Let $\epsilon >0$ and set $M=\frac{1}{\epsilon}$. Since $\lim_{x\to c}f(x)=\infty$, we can find a $\epsilon >0$ such that $f(x)>M$ whenever $0<|x-c|<\delta$.

Thus $0<\frac{1}{f(x)}<\frac{1}{\epsilon}$ whenever $0<|x-c|<\delta$. This implies that it is possible to find a $\delta>0$ such that $|\frac{1}{f(x)}|<\epsilon$ whenever $0<|x-c|<\delta$. Since $\epsilon$ is arbitrary, we have proved that $\lim_{x\to c}\frac{1}{f(x)}=0$.

Proving the "$\Leftarrow$": This proof I am unsure of. I know that by the definition of a limit, $\lim_{x\to c} f(x) = 0$ if $\forall\epsilon>0, \exists\delta>0$ such that $\forall x∈\mathbb{R}\setminus{c}, 0<|x-c|<\delta \Rightarrow |f(x)-0|<\epsilon$. I am unsure of how to define $\lim_{x\to c}\frac{1}{f(x)}=0$ in a similar way. Any advice would be greatly appreciated.

答:
Both directions use the same concept.

$\implies$

$\lim\limits_{x\to c} f(x) = \infty$ so for any $\epsilon> 0$ and $M =\frac 1\epsilon > 0$ there is a $\delta$ so that $|x-c|< \delta \implies f(x) > M =\frac 1{\epsilon}$ so $|x-c| < \delta \implies 0 < \frac 1{f(x)} = |\frac 1{f(x)} -0| < \frac 1M =\epsilon$.

So $\lim\limits_{x\to c} \frac 1{f(x)} = 0$.

$\Leftarrow$

$\lim\limits_{x\to c}\frac 1{f(x)} =0$ means for every $M > 0$ and $\epsilon = \frac 1M > 0$ there is a $\delta > 0$ so that $ |x-c|< \delta\implies |\frac 1{f(x)} - 0|=\frac 1{f(x)} < \epsilon = \frac 1M$.

And as $f(x) > 0$ then $|x-c| < \delta \implies f(x) > \frac 1{\epsilon} = M$.

So $\lim\limits_{x\to c}\frac 1{f(x)} =\infty$.

.......

For the desired $\epsilon$ or $M$ you just use $\frac 1\epsilon, \frac 1M$ to show there is a $\delta$ so that $|x -c| \delta$ implies the limit you know for the reciprical of the value you want, and then by take the reciprical you get the limit of the reciprical fits the value you want.

The exact same procedure either way.

Title: Using definition of limits