问:
Properties of a sequence generated by a function
Suppose $\mathcal{B} = \{a_1, a_2, a_3, \ldots\}$ is a sequence of integers generated by some function $\mathcal{D}$ such that we have

$$\prod_{i = 1}^{k}a_i \quad \text{always divides}\quad\prod_{i = n - k + 1}^{n}a_i$$

for every positive integer $k$. Note that here $n$ is greater than or equal to $k$. Let us call this property of the sequence as product property. What can be function $\mathcal{D}$ in general which will generate the sequence satisfying the product property? Are there any conditions on sequence $\mathcal{B} = \{a_1, a_2, a_3, \ldots\}$ which implies that the product property is satisfied? For example if we say function $\mathcal{D}$ that generates the sequence is $n^2$ that is we have $\mathcal{B} = \{1,4,9,16,25 \ldots\}$ then it's easy to see that it satisfies the property that is

$$\prod_{i = 1}^{k}i^2 \quad \text{always divides}\quad\prod_{i = n - k + 1}^{n}i^2$$

for every positive integer $k$. Similarly if we say function $\mathcal{D}$ that generates the sequence is $n(n + 1)$ then also it satisfies the product property that is we have
$$\prod_{i = 1}^{k}i(i + 1) \quad \text{always divides}\quad\prod_{i = n - k + 1}^{n}i(i + 1)$$
My main question is Are there any conditions on sequence $\mathcal{B} = \{a_1, a_2, a_3, \ldots\}$ which implies that the product property is satisfied? Are there any conditions on sequence $\mathcal{B} = \{a_1, a_2, a_3, \ldots\}$ which implies that the product property is not satisfied? What are some more example of the sequence $\mathcal{B}$ which satisfy the product property?

I have worked on this problem for a very long time but didn't succeed to get any strong conditions. However I have found many example of the sequence $\mathcal{B}$ which satisfy the product property. I have generated these sequences from some trial and error method, however I am struggling with the conditions part which would imply if the the sequence satisfies the product property or not. Any help or hint would be highly appreciated, Thanks.

答:
For any $a$, let $p_a$ denote the $a$th prime.  Define $e_a(n)$ to be the largest power of $p_a$ that divides $\mathcal{D}(n)$; that is, $$\mathcal{B}=\left\{\prod_{a=1}^{\infty}{p_a^{e_a(n)}}\right\}_n$$  (The product is finite because only finitely-many $e_a$ are nonzero.)

I do not know a simple necessary condition (but see equation (1) below).  A sufficient condition is that $e_a$ is nondecreasing, which holds for all your examples:

if $\mathcal{D}(i)=i^2$, then $e_a(i)=\left\lfloor2\log_{p_a}(i)\right\rfloor$;

if $\mathcal{D}(i)=i(i+1)$, then $e_a(i)=\left\lfloor\log_{p_a}(i(i+1))\right\rfloor$.

It is also satisfied by more exotic examples, like

$\mathcal{B}=\{2,4,8,16,\dots\}$ and

$\mathcal{B}=\{1,1,1,1,\dots\}$.

It is not satisfied by the following examples:

$\mathcal{B}=\{1,2,1,2,1,2,1,2,\dots\}$

$\mathcal{B}=\{1,2,1,2,3,4,5,6,\dots\}$

These results were not obtained from any deep insight, just some slight notational convenience, described below.

Your condition is equivalent to the following: for any $a\geq1$ and $n\geq k\geq1$, $$\sum_{j=n-k+1}^n{e_a(j)}\geq\sum_{j=1}^k{e_a(j)}$$  To see this, just note that if $x$ divides $y$, then any prime power dividing $x$ must also divide $y$.  The left-hand term is the exponent of $p_a$ dividing $\prod_{j=n-k+1}^n{\mathcal{D}(j)}$; the right-hand term is the exponent of $p_a$ dividing $\prod_{j=1}^k{\mathcal{D}(j)}$.

Now change variables to $x=n-k$; then we want, for any $a\geq1$ and $x,k\geq1$, $$\sum_{j=1}^k{e_a(j+x)}\geq\sum_{j=1}^k{e_a(j)}\quad\quad(1)$$  It is now immediate why $e_a$ increasing suffices: then $e_a(j+x)\geq e_a(j)$, and we can bound the sums termwise.


问:
Proof of a solution of a differential equation
Proof that  $$ x=a(\theta - \sin(\theta))$$ $$y=a(1-\cos(\theta))$$ where $a$ is a constant different from zero is a solution of the differential equation $$1+(y')^2+2yy'=0$$.

$\textit{What I've done:}$
I found $$y'= \frac{dx}{dy}=\frac{\frac{da-a\cos(\theta)}{d\theta}}{\frac{da-a\cos(\theta)}{d\theta}}=\frac{a\sin(\theta)}{a(1-\cos(\theta))}=\frac{\sin(\theta)}{1-\cos(\theta)}$$

So substituting this on our differential equation we have: $$1+(\frac{\sin(\theta)}{1-\cos(\theta)})^2+2[(a-a\cos(\theta))(\frac{\sin(\theta)}{1-\cos(\theta)})]=1+(\frac{\sin(\theta)}{1-\cos(\theta)})^2+2a\sin(\theta)$$ There's where I got stuck. I don't know what else to do. Help please

ContentLicense: CC BY-SA 4.0

Body:
$\require{cancel}$

$$\cancel{1}+\cancel{(\frac{\sin(\theta)}{1-\cos(\theta)})^2}+2[(a-a\cos(\theta))(\frac{\sin(\theta)}{1-\cos(\theta)})]=\cancel{1}+\cancel{(\frac{\sin(\theta)}{1-\cos(\theta)})^2}+2a\sin(\theta)$$

$$\cancel{2}[(a-a\cos(\theta))(\frac{\sin(\theta)}{1-\cos(\theta)})]=\cancel{2}a\sin(\theta)$$

$$(a-a\cos(\theta))\cdot\sin(\theta)=a\sin(\theta)-a\sin(\theta)\cos(\theta)$$

$$a\sin(\theta)-a\cos(\theta)\sin(\theta)=a\sin(\theta)-a\cos(\theta)\sin(\theta)$$

$$0=0$$
$\square$


问:
What is the fastest way to multiply and divide polynomials?
I want to get faster at multiplying and dividing polynomials. Is there a way I can do it faster? For multiplying I just do it normally. But for division I have to use the box method.</p>

答: You’re asking about getting faster. The only trick I know is to practice a lot. What’s always used to slow me down was making mistakes and having to redo my work.</p>


问:
Prove that a relation is antisymmetric
I am a bit confused about proving that a relation is antisymmetric. $x,y \in R$ and $x \sim y$ if $x=2y$ is the given relation. Antisymmetric holds true I think. Could I make this conclusion? Thanks!

答: In order to prove that $\sim$ is antisymmetric, you must show that for all $x,y\in R$ it is the case that if $x\sim y$ and $y\sim x$, then $x=y$; looking at one pair of $x$ and $y$ proves nothing. For this relation you can argue that if $x\sim y$, then $x=2y$, and if $y\sim x$, then $y=2x$, so if $x\sim y$ and $y\sim x$, then $x=2y$ and $y=2x$. From that you can infer that $x=4x$, which is true if and only if $x=0$. Thus, $\sim$ is antisymmetric if and only if it is a relation on $\{0\}$. You didn’t tell us what the underlying set is, but I expect that it contains more than just $0$, so … ?

Title: Prove that a relation is antisymmetric


问:
Measurability and Open Sets
I have a homework question, as follows: Can there be a Lebesgue measurable set $A$ such that

$$\lambda(A\cap U)=\frac{\lambda(U)}{2}$$
for all open sets $U$?

I’ve made two observations:

If such an $A$ exists, it’s measure must be infinity, since $$\lambda(A\cap \mathbb{R})=\lambda(A)=\frac{\lambda(\mathbb{R})}{2}$$

If such an $A$ exists, it cannot contain an open subset $U$, since $$\lambda(A\cap U)=\lambda(U)=\frac{\lambda(U)}{2} $$ which implies that an open set has nonpositive measure, a contradiction.

Past these observations I’m stuck, and have no real intuition on whether this thing should exist or not. My gut says no, as it seems too convenient to let us pick among every open set and this nice equality holds, but I can’t seem to pin a contradiction. I was hoping there would be some implication that the measure of $A$ was also finite, which would imply that it couldn’t exist.

答: By the regularity of the Lebesgue measure, there exists an open set $U\supset A$ such that $\lambda(U\setminus A) < \varepsilon$. Can you see how this, together with your observations thus far about $A$, leads to a contradiction?

Title: Measurability and Open Sets


问:
The operator norm of a linear functional bounded by the Minkowski functional defined by the Cesaro average
I am studying the existence of Banach limit, and I have figured out most of the things, except for one point. The following describes my question:

Let $V:=\ell_{\infty}(\mathbb{R})$, and for $\mathbf{x}\in V$, we define $$p(\mathbf{x}):=\limsup_{n\rightarrow\infty}\dfrac{1}{n}\sum_{k=1}^{n}x(k).$$ I have proved that $p(\mathbf{x}+\mathbf{y})\leq p(\mathbf{x})+p(\mathbf{y})$ and $p(\lambda\mathbf{x})=\lambda p(\mathbf{x})$ for any $\lambda\in [0,\infty)$, i.e. $p$ is a Minkowski functional on $V$.

Now, I need to prove that

Let $L:\ell_{\infty}(\mathbb{R})\longrightarrow \mathbb{R}$ be a linear functional such that $L(\mathbf{x})\leq p(\mathbf{x})$ for all $\mathbf{x}\in V$. Then, $L$ is continuous and $\|L\|_{V^{*}}\leq 1.$

I know that, if you work backwards from the Banach limit, the dual norm here is exactly $1$. But I am working in the converse direction, and the only thing I know is what was given in the statement.

I have proved the continuity, but I am not sure if I am correct:

Let $\mathbf{x}, \mathbf{y}\in \ell_{\infty}(\mathbb{R})$ be such that $$\|\mathbf{x}-\mathbf{y}\|_{V}=\sup_{k\in\mathbb{N}}|x(k)-y(k)|<\delta\ \ \text{for some}\ \ \delta>0.$$ Then,
\begin{align*}
|L(\mathbf{x})-L(\mathbf{y})|=|L(\mathbf{x}-\mathbf{y})|\leq |p(\mathbf{x}-\mathbf{y})|&=\limsup_{n\rightarrow\infty}\dfrac{1}{n}\sum_{k=1}^{n}|x(k)-y(k)|\\
&\leq \limsup_{n\rightarrow\infty}\dfrac{1}{n}\cdot n\cdot \sup_{k=1}^{n}|x(k)-y(k)|\\
&\leq \limsup_{n\rightarrow\infty}\dfrac{1}{n}\cdot n\cdot \sup_{k\in\mathbb{N}}|x(k)-y(k)|\\
&<\delta.
\end{align*}

This means that for any $\epsilon>0$, we can set $\delta:=\epsilon$ so that $$\mathbf{x},\mathbf{y}\in\ell_{\infty}(\mathbb{R}),\ \|\mathbf{x}-\mathbf{y}\|_{V}<\delta\implies |L(\mathbf{x})-L(\mathbf{y})|<\epsilon,$$ and hence $L$ is continuous.

Is this proof okay? I am not sure of this because I am not sure if $|L(\mathbf{x})|\leq |p(\mathbf{x})|$ is true.

Then, I do not know how to prove $\|L\|_{V^{*}}\leq 1$. The only thing I can think of is that as $V\neq \{0\}$, we can write $$\|L\|_{V^{*}}=\sup_{\|\mathbf{x}\|_{V}=1}|L(\mathbf{x})|=\sup_{\substack{\mathbf{x}\in V \\ \sup_{k\in\mathbb{N}}|x(k)|=1}}|L(\mathbf{x})|,$$ but I don't know what to do next. Any idea?

I am not sure if these are needed, but using only linearity of $L$ and $L\leq p$, I have proved that

For all $\mathbf{x}\in \ell_{\infty}(\mathbb{R})$, $$\liminf_{k\rightarrow\infty}x(k)\leq L(\mathbf{x})\leq \limsup_{k\rightarrow\infty}x(k).$$

$L$ is shift-invariant, i.e. $L\circ S=L$ where $S$ is the shift operator defined as $$S((x(1), x(2),\cdots))=(x(2), x(3),\cdots).$$

Thanks in advance for any help!!

答:
I'd say you are overthinking it.  Notice that for every $x$ in $\ell ^\infty ({\mathbb R})$, and every $n\in {\mathbb N}$,
one has that
$$
  \frac 1n \sum_{k=1}^n x(k) \leq    \frac 1n \sum_{k=1}^n \|x\|_\infty   = \|x\|_\infty .
  $$
Therefore $p(x)\leq \|x\|_\infty $, and hence
$$
  L(x) \leq  p(x) \leq  \|x\|_\infty .
  $$

Replacing $x$ with $-x$,  we also get
$$
  -L(x)=  L(-x) \leq  \|-x\|_\infty  =  \|x\|_\infty ,
  $$
so actually
$$
  |L(x)|\leq  \|x\|_\infty ,
  $$
whence  $\|L\|\leq 1$.


问:
Intersection of Closed and Compact Set is Closed
Fact

The intersection of a closed set $A$ and a compact set $B$ is closed.

Background

I wanted to prove that the intersection of closed and compact sets is compact. There might be other ways, but I am not interested in them for now and am looking to prove the above statement.

Relevant Facts

I don't have much idea what to use. $A^c$ is open. Every cover $\cup_{i\ge1}U_i$ of $B$ contains a subcover $\cup_{i=1}^kU_i$. I might want to prove $A\cap B^c$ is open. Limit points may be another points to consider but I don't get any idea in that direction.

I also don't have that $B$ is closed otherwise it would be trivial.

Some Clarification

I am aware that compact sets are not closed in arbitrary topological spaces. The reason why I was motivated to believe tha intersection of closed and compact set is compact is true is: See Matt E.'s Answer Here.

So if $K$ is compact in an arbitrary topological space (which is just to say that
  it is a compact topological space when given its induced topology) and
   $L$ is closed then $K \cap L$ is a closed subset of $K$ in its induced topology

Is it true if $A$ and $B$ are not whole spaces?

I would appreciate any hints.

答:
It is not always true. The following example witnesses it.

Example 1: Let $X= \mathbb R$ with finite complement topology. Note that $X$ is $T_1$, not $T_2$. It is not difficult to prove that $X$ is compact. Now let $A=X$; $B=\{0,1,2,\dots,n,\dots\}$. Note that $A$ of course is closed and $B$ is compact, however it is not closed. Therefore $A \cap B=B$ is not closed.

However it is true if $A$ and $B$ belong to a Hausdorff space.

Theorem 2: Every compact subspace of a Hausdorff space $X$ is a closed subspace of $X$.

Proof: Let $A$ be a compact subspace of $X$.  For every $x\in X\setminus A$ there exists an open set $V\subset X$ such that $x \in V$ and $A \cap V=\emptyset$, so that $X \setminus A$ is an open subset of $X$.

So in a Hausdorff space, $B$ is always closed as it is compact, and hence $A \cap B$ is closed, as the intersection of two closed set is closed.

ContentLicense: CC BY-SA 3.0


问:
Why the substitution is not working even though its bijective?
The following integral i was trying to evaluate:
$$\int_0^\infty \left( \sin(1/x) - \frac{\sin(\pi/x)}{\pi} \right) \,dx.$$ ,what i did was substituting pi/x = 1/t in the second integral , which converts to first integral which tells integral is zero ,but answer is ln(pi) , whats the fault that i wanted to know .

答:
Integrate by parts to get $-x\dfrac{\sin\left(\frac{{\pi}}{x}\right)}{{\pi}}+x\sin\left(\dfrac{1}{x}\right)+\operatorname{Ci}\left(\dfrac{{\pi}}{x}\right)-\operatorname{Ci}\left(\dfrac{1}{x}\right)$

$(1).$ As $x\to \infty,$ the first two terms combine to give $-1+1=0.$ As $x\to 0,$ all four terms tend to $0$. Now,

${\displaystyle \operatorname {Ci} (x)=-\int _x^\infty \frac {\cos t}{t}}dt$ so ${\displaystyle \operatorname {Ci} (\pi/x)=-\int _{\pi/x}^\infty \frac {\cos t}{t}}dt$ and $\displaystyle\operatorname {Ci} (1/x)=-\int _{1/x}^\infty \frac{ \cos t}{t}dt$

and therefore

$(2).\ \operatorname{Ci}\left(\dfrac{{\pi}}{x}\right)-\operatorname{Ci}\left(\dfrac{1}{x}\right) = \displaystyle \int _{1/x}^{\pi/x} \frac {\cos t}{t}dt=\int _{1/x}^{\pi/x}\left(\frac{1}{t}-\frac{t}{2}+O(t³)\right)dt=$

$\ln(\pi/x)-\ln(1/x)+O(1/x^2)=\ln \pi+O(1/x^2)$

To finish, let $x\to \infty$ in $(2)$ and combine the result with $(1).$


问:
Pre-annihilator of the annihilator of a closed subspace $Y$ is $Y$
Let $Z$ be a linear subspace in $X^*$, the dual space of $X$,  where $X$ is a normed linear space. We define $$Z_ \bot = \{x \in X : \phi(x) = 0, \forall \phi \in Z     \}. \quad  (I)$$

Now, I have to prove that for any closed subspace $Y$ in $X$, $(Y^ \bot )_ \bot = Y.$ Where, $ Y ^ \bot$ is the annihilator or orthogonal complement of $Y$. That is, $ Y ^\bot  = \{ \phi \in X^*: \phi(y) =0, \forall y \in Y\}$.

Well, my idea is to show that the two sides are subsets of each other: $(Y^ \bot )_ \bot \subseteq Y$ and $Y \subseteq (Y^ \bot )_ \bot$.

We know that $Y^ \bot$ is indeed a subspace of $X^*$. Thus, we can plug in $Y^ \bot$ as $Z$ in the definition (I), and form $$(Y^ \bot )_ \bot = \{x \in X,  \phi(x) = 0, \forall \phi \in Y^\bot  \} \quad (II).$$ Now, let's pick an arbitrary $y \in Y$ and an arbitrary $\psi \in Y ^\bot$. Obviously, $y$ satisfies the conditions in (II). Thus, we have $ y  \in (Y^ \bot )_ \bot. $ Thus, $Y \subseteq (Y^ \bot )_ \bot$. Am I right so far? This whole conclusion, though obvious, seems a bit cyclic to me.

Also, I have difficulty in the proof of the inverse direction. That is, to show that if $ y \in (Y^ \bot )_ \bot $, then $y \in Y.$ Plus, and perhaps more importantly, where should I use the closedness of $Y$? Thanks for any help.

答: Because $Y$ is closed, if $Y$ is properly contained in $(Y^\perp)_\perp$, it's a closed subspace. Assume we have proper containment. By Hahn-Banach, we  can define a nonzero functional on $(Y^\perp)_\perp$ which is zero on $Y$ by choosing any nonzero linear functional on $(Y^\perp)_\perp/Y$, and extending it to a functional on all of $X$. Call such a functional $\phi$. Then $\phi\in Y^\perp$, so $\phi|_{(Y^\perp)_\perp}=0$, a contradiction.


问:
Proof of relation between divided difference and interpolation polynomial coefficient
Given data points $\{(x_i,f(x_i)\}_{i=0}^{m}$, if we define divided difference recursively as:

$$f[x_0,\cdots,x_{k+1}] = \frac{f[x_1,\cdots,x_{k+1}]-f[x_0,\cdots,x_k]}{x_{k+1}-x_0} \text{ with the definition that} f[x_0] = f(x_0)$$

then it is true that the unique interpolating polynomial $L_m(x)$ of degree $m$ such that $L_m(x_i) = f(x_i)$ for all $i = 0\cdots m$ is given by:
$$
L_m(x) = \sum_{k=0}^mf[x_0,\cdots,x_k]\omega_k(x), \text{ where } \omega_k(x) = \prod_{i=0}^{k-1}(x-x_i).
$$
The question asks to show this. That is, show the coefficients of the interpolating polynomial of degree $m$ in the basis $\{\omega_k\}_{k=0}^{m}$ is exactly $a_k = f[x_0,\cdots,x_k]$, the divided differences.

To tackle this problem, I've attempted to use induction. The base step is trivial as $a_0 = f(x_0)$, but I am stuck on the inductive step. I supposed the degree $m$ interpolating polynomial of some data $\{(x_i,f(x_i)\}_{i=0}^{m}$ is indeed given by:
$$
L_m(x) = \sum_{k=0}^mf[x_0,\cdots,x_k]\omega_k(x)
$$
and I want to argue that $L_{m+1}(x) = \sum_{k=0}^{m+1}f[x_0,\cdots,x_k]\omega_k(x)$ is the interpolating polynomial of degree $m+1$ interpolating $\{(x_i,f(x_i)\}_{i=0}^{m+1}$, with any additional point $(x_{m+1},f(x_{m+1}))$ added.

Clearly, $L_{m+1}(x)$ interpolates $\{(x_i,f(x_i)\}_{i=0}^{m}$ by definition of the $\omega_k$'s, but I am stuck at showing $L_{m+1}(x_{m+1}) = f(x_{m+1})$. Hints will be extremely appreciated!

答:
This proof was more delicate than I expected.
First observe that by an easy induction, for all $x_0,\ldots, x_m$ we have
$$f[x_0,\ldots, x_m]=f[x_m,\ldots,x_0]\,.   \quad \quad (1)$$
Recall $L_m$ is defined to be the degree $m$ polynomial interpolating $f$ at $x_0,\ldots,x_m$. Our goal is to verify by induction that
$$L_m(x)+f[x_0,\ldots,x_{m+1}] \, \omega_{m+1}(x)=L_{m+1}(x) \, \quad \quad (2) $$
for all choices of $x,x_0,\ldots, x_m.$

Since both sides of (2) are degree $m+1$ polynomials and (2) is clear for $x=x_0,\ldots,x_m$, it suffices to verify it also holds for $x=x_{m+1}$. To this end, define $Q_k$ as the degree $k$ polynomial interpolating $f$ at $x_1,\ldots,x_{k+1}$ and denote
$$\psi_m(x):=\prod_{j=1}^m (x-x_j)=\frac{\omega_{m+1}(x)}{x-x_0} \,. \quad \quad (3)$$

By the induction hypothesis, equation (2) holds for polynomials of lower degree, so
$$Q_m(x)=Q_{m-1}(x)+f[x_1,\ldots,x_{m+1}] \psi_{m}(x)  \, \quad \quad (4) $$
and by considering the points in the reverse order  $x_m,\ldots,x_1,x_0$
$$L_m(x)=Q_{m-1}(x)+f[x_m,\ldots,x_0] \psi_{m}(x)  \, \quad \quad (5) $$
Subtracting the last two equations and setting $x=x_{m+1}$  gives
$$Q_m(x_{m+1})-L_m(x_{m+1})=\Bigl(f[x_1,\ldots, x_{m+1}]- f[x_m,\ldots,x_0] \Bigr)\psi_{m}(x_{m+1})\quad \quad \quad \quad \quad \; \; \; \;$$
$$=f[x_0,\ldots, x_{m+1}]\, \omega_{m+1}(x_{m+1}) \,, \quad \quad (6) $$
using equations (1) and (3) in the last step.
Since $Q_{m}(x_{m+1})=f(x_{m+1})=L_{m+1}(x_{m+1})$, equation (6) gives (2) for $x=x_{m+1}$. This completes the induction step and the proof of (2).


问:
Is there a function that takes in a binary number and yields the corresponding decimal number, or vice versa?
For example,

f(10111010) = 186

f(101110101101001) = 23913

Correspondingly, is there an inverse function to this?

答: There is a standard algorithm to convert between the representations of a number in different bases.</p>


答: The function exists, because you just defined it.  I believe what you meant to ask is, "does a simple formula exist for this function?"  I don't think that there is a formula much simpler than performing the algorithm to do the conversion.  I expect it would be a mess of fractions and floor functions and modulo reductions.</p>

Title: Is there a function that takes in a binary number and yields the corresponding decimal number, or vice versa?

Body:
For example,

f(10111010) = 186

f(101110101101001) = 23913

Correspondingly, is there an inverse function to this?


问:
Invariant subspaces of a linear operator that commutes with a projection
I have an assignment problem, the statement is:

Let $V$ be a vector space and $P:V \to V$ be a projection. That is, a linear operator with $P^2=P.$ We set $U:= \operatorname{im} P$ and $W:= \ker P.$ Further suppose that $T:V\to V$ is a linear operator such that $TP = PT.$ Prove that $TU \subseteq U$ and $TW\subseteq W.$

Here is my attempt:

Suppose $u\in U:= \operatorname{im} P$ so $u= P(v)$ for some $v\in V.$ Then $Tu = TPv = PTv \in \operatorname{im} P := U$ so $TU\subseteq U.$

Suppose $w\in W:= \ker P$ so that $Pw=0.$ Then $P (Tw) = T(Pw) = T(0)=0$ so $Tw\in \ker P := W$ so $TW\subseteq W.$

It seems fine to me, but nowhere did I use that $P$ was a projection, I only used $TP=PT.$ Is my proof okay?

答:
Yes, that's all. $P$ doesn't have to be projection for this particular exercise.

However, we can just start out from the fact that $P$ is a projection in a solution: now it projects to subspace $U$, in the direction of $W$, and we also have $U\oplus W=V$, and $P|_U={\rm id}_U$.

Having these, an operator $T$ commutes with $P$ iff both $U$ and $W$ are $T$-invariant subspaces. Your proof can be reformulated for one direction, and the other direction goes as:

If $TU\subseteq U$ and $TW\subseteq W$, then $TP(u+w)=Tu$, as it is $\in U$, it $=PTu$, and as $Tw\in W$, we finally have
$$TP(u+w)=Tu=PTu=PTu+PTw=PT(u+w)\,.$$


问:
Probability of sum of iid variables being greater than $0$
Let $S_n=X_1+X_2+X_3+...+X_n$ where $X_i$ are i.i.d with $E[X_i]=0$ and finite variance.

In a course I'm taking, the following solution was given:

$$\lim_{n \to \infty} P(S_n\geq 0)=1-\Phi(0)=0.5$$

While this makes some intuitive sense, I'm wondering where the mistake in the following line of reasoning using the law of large numbers is :

$$\lim_{n \to \infty} P(S_n\geq 0)= \lim_{n \to \infty} P(\frac{S_n}{n}\geq 0) \geq \lim_{n \to \infty} P(\frac{S_n}{n}= 0)=1 $$

答: The law of large numbers says that $P\big( \lim_{n \to \infty} \frac{S_n}{n}= 0 \big) = 1$, but that's not the same assertion as $\lim_{n \to \infty} P\big( \frac{S_n}{n}= 0 \big) = 1$.


答:
For each prime, $p$ let
$$
v_p(n)=\# \text{times $p$ occurs in the prime decomposition of $a_1\times \dots \times a_n$}
$$
In order to for $a_1\dots a_k$ to divide $a_{n-k+1}\cdots a_n$, it must be the case that there are at least as many copies of $p$ in the former as in the latter. Since the number of copies of $p$ in the former is $v_p(n)-v_p(n-k)$. Therefore, it needs to be true that
$$
v_p(n)\ge v_p(k)+v_p(n-k)
$$
An equivalent way of saying this is
$$
v_p(n+m)\ge v_p(n)+v_p(m)\tag{$*$}
$$
A sequence $v_p(1),v_2(2),\dots$ which satisfies $(*)$ is called super-additive. We just argued that this condition is necessary, for each $p$. It is also sufficient; as long as $(*)$ holds for all primes, the ratio will be an integer. Furthermore, the sequences $v_p(n)$ uniquely determine the prime decompositions of each $a_n$ (up to a sign). The only constraint between the sequences is that for each $n$, there can only be finitely many $p$ for which $v_p(n)$ is nonzero.

We have the following complete characterization of sequences with the divisibility property:

The sequence $(a_1,a_2,\dots)$ satisfies the divisibility property if and only if for each prime $p$, $(v_p(1),v_p(2),\dots)$ is super-additive.

A collection of super-additive sequences $(v_p(1),v_p(n),\dots)$ determines a corresponding valid sequence $(a_1,a_2,\dots)$ if and only if for each natural number $n$, there are a finite number of primes $p$ for which $v_p(n)\neq 0$.

There is a lot of freedom here, and there is no real easy way to characterize all super-additive sequences. For some more examples, and a reference to a Knuth paper with more information, see this answer of mine: Properties of sequence

Title: Properties of a sequence generated by a function

Body:
Suppose $\mathcal{B} = \{a_1, a_2, a_3, \ldots\}$ is a sequence of integers generated by some function $\mathcal{D}$ such that we have

$$\prod_{i = 1}^{k}a_i \quad \text{always divides}\quad\prod_{i = n - k + 1}^{n}a_i$$

for every positive integer $k$. Note that here $n$ is greater than or equal to $k$. Let us call this property of the sequence as product property. What can be function $\mathcal{D}$ in general which will generate the sequence satisfying the product property? Are there any conditions on sequence $\mathcal{B} = \{a_1, a_2, a_3, \ldots\}$ which implies that the product property is satisfied? For example if we say function $\mathcal{D}$ that generates the sequence is $n^2$ that is we have $\mathcal{B} = \{1,4,9,16,25 \ldots\}$ then it's easy to see that it satisfies the property that is

$$\prod_{i = 1}^{k}i^2 \quad \text{always divides}\quad\prod_{i = n - k + 1}^{n}i^2$$

for every positive integer $k$. Similarly if we say function $\mathcal{D}$ that generates the sequence is $n(n + 1)$ then also it satisfies the product property that is we have
$$\prod_{i = 1}^{k}i(i + 1) \quad \text{always divides}\quad\prod_{i = n - k + 1}^{n}i(i + 1)$$
My main question is Are there any conditions on sequence $\mathcal{B} = \{a_1, a_2, a_3, \ldots\}$ which implies that the product property is satisfied? Are there any conditions on sequence $\mathcal{B} = \{a_1, a_2, a_3, \ldots\}$ which implies that the product property is not satisfied? What are some more example of the sequence $\mathcal{B}$ which satisfy the product property?

I have worked on this problem for a very long time but didn't succeed to get any strong conditions. However I have found many example of the sequence $\mathcal{B}$ which satisfy the product property. I have generated these sequences from some trial and error method, however I am struggling with the conditions part which would imply if the the sequence satisfies the product property or not. Any help or hint would be highly appreciated, Thanks.


问:
Probability of rolling a $6$ between $90$ and $110$ times when rolling a die $600$ times

Show that if you roll a fair die $600$ times, then with probability at least $1/6$ you will get the roll $6$ between $90$ and $110$ times.

This is what I have so far

Let $E(x) = 100 \Rightarrow U(x) = 100 \cdot 1/6 = 600$

But, I don't know what more I need to show?

Thanks for your help.

答:
You have a binomial distribution so you need to demonstrate

$$\sum_{k=90}^{110}\binom{600}{k}\left(\frac{1}{6}\right)^k\left(\frac{5}{6}\right)^{n-k}\ge \frac{1}{6}$$

You can either demonstrate this directly or by arguing from the expected value and the variance using a normal distribution approximation.

Since the probability is in fact $\gt\frac{3}{4}$ you do not need to add up all of these terms. $\sum_{k=98}^{101}$ is $\gt\frac{1}{6}$


问:
Find a conformal mapping that maps $U$ to the unit disk.
Let $D = \{|z| < 1\}$ be the unit disk and define $U = D - \{(x,0): \text{$-1 < x \leq 0$}\}$. Find a conformal mapping from $U$ to $D$.

Here $U$ is essentially the unit disk minus the negative real axis. I think this is a relatively simple problem, but I am unable to solve it. I know that with Mobius transformations this can be done, we can probably construct a transformation to solve this. My idea was to map the interval $(-1,0]$ to the non-positive real line. From here I would need to keep track of where $U$ went to and adjust accordingly.

Any help is appreciated.

Edit: After some time looking for similar problems this is what I came up with. First, use the transformation $T(z) = i \frac{z+1}{1-z}$ to map our $U$ to the upper half-plane with a slit going from $(0,i]$. Next, we will extend this slit, so we use the transformation $S(z) = -1/z$. This maps our new domain to the upper half-plane where the slit now is from $[i,\infty)$ (imaginary axis). Then from here, we can follow the solution provided here: Conformal Mappings dealing with Slits.

Keep in mind that the $\sqrt{z}$ function provided in the link is not the typical one, this one is defined for when $z \notin [0,\infty)$.

Is this correct?

答: First, map your domain $U$ to a half-disk, by $z\to \sqrt{z}.$ Then use the method described here: Find a conformal map from semi-disc onto unit disc


问:
A compact subspace of a Banach space must have a finite dimension?
Let $X$ be a Banach space and $A \subset X$ a compact subspace. Doesn't this necessarily imply that $A$ is finite-dimensional? I've read that this is not true, but I can't figure out why my reasoning is incorrect.

<s>Wrong</s> proof. If $A$ is compact and $X$ is a Banach space (i.e complete metric space), then $A$ is closed in $X$ and totally bounded. Since $A$ is closed in $X$, it's complete with respect to the restriction of metric $d_{A \times A \to \mathbb{R}}$ induced by the norm of $X$. Since $(A, d)$ is complete and totally bounded, it's sequentially compact (i.e every sequence in $A$ has a convergent subsequence in $A$), which is more than what we need to show that unit ball is compact and $\dim A < \infty$.

Where did I go wrong? Thanks.

答:
Your assertion is true, and the proof can be shortened even more: the unit ball is closed in $A$, so if $A$ is compact, so is the unit ball, hence $A$ is finite dimensional.

Anyway, it doesn't really matter because only the zero subspace is compact.  Every nonzero subspace $E$ of $X$ admits at least one continuous, surjective linear map $E \rightarrow k$, where $k = \mathbb R$ or $\mathbb C$, so $E$ cannot be compact.


问:
Checking whether a relation is an equivalence relation or not
I am a bit confused whether the given relation is an equivalence relation or not.
$\sim$ is defined on the set of all integers $x \sim y$ if and only if $x - y$ is a multiple of $4$. I think this is an equivalence relation. I checked it for reflexivity, symmetry, and transitivity and according to my scratch work, all of these conditions seem to be met. Am I missing something or am I on the right track? Any input would be helpful. Thanks!

答:
Yes you are right, and for showing that, that is the demonstration

x-y is a multiple of 4 so $x-y=4k$ that mean $4/x-y$  implies that $x=y[4]$

And we know $x=y[4]$$\Leftrightarrow $$y=x[4]$ so this relation is symmetry

and also we know  $x=x[4]$ so it's reflexive

And we know if $x=y[4] $ and $y=z[4]$ that's implies$ x=z[4]$ So it's transitive

And after this(transitivity, reflexivity, symmetry) we can say this relation is  an equivalence relation


问:
Is this complex exponential conversion correct?
I am trying to convert $5\cos(100\pi t + 25)$ to a complex exponential.

I get that Euler's formula is such that $\cos(nt) = \frac{1}{2}(e^{nit} + e^{-nit})$.

I have my answer as $\frac{5}{2}(e^{(100\pi i(t+25)} + e^{-100(\pi i(t+25)})$. Is this correct?

答: It is not correct because the $100$ multiplies the $25$ in your solution, while it does not in the original.  You should use the angle sum formula to get the $25$ out of the argument of the cosine.


问:
Joint PDF for two independent random variables along $Y=X$
Suppose I have two independent random variables $X$ and  $Y$, both have support on $(0,\infty)$.

How do I find the p.d.f. for joint distribution along line $Y=X$?

For simplicity, we can assume both have exponential distribution. Can we just write $y=x$ in the joint pdf so that $f(x,y)=g(y)$?

I would love some insight and correction. Thank you very much!

答:
I wonder if intuitively you’re fishing for something like
$$g(x) = f(x, x)\, /\int_0^\infty\, f(t, t)\, dt.$$


问:
How do I solve $3=\frac{k}{5+\frac{k}{5+\frac{k}{5+\cdots}}}$ for $k$?

How do I solve this for $k$?

$$3=\frac{k}{5+\frac{k}{5+\frac{k}{5+\cdots}}}$$

I know $k$ can be estimated by graphing, but is there a way to solve using algebra? If so, how?

答:
Let: $x = 5+\dfrac{k}{5+\dfrac{k}{5+\dfrac{k}{...}}} \implies x = 5 +\dfrac{k}{x} \implies x^2-5x=k$

Substitute this in the original equation: $3= \dfrac{k}{x} \implies 3 = \dfrac{x^2-5x}{x} \implies x = 8$

$x =8 \implies k = 24.$


问:
To calculate an improper integral
Is $$\int_1^\infty \frac{\log x}{x^2}dx$$
finite? How to solve this?

答:
We integrate by parts

\begin{align*}
\int_{1}^{\infty}\frac{\ln(x)}{x^2}dx &= \lim\limits_{t\to\infty}\int_{1}^{t}\frac{\ln(x)}{x^2}dx\\
&= \lim\limits_{t\to\infty}\left(\left[-\frac{\ln(x)}{x}\right]_{1}^{t}-\int_{1}^{t}-\frac{1}{x}\cdot\frac{1}{x}\text{ }dx\right)\\
&= \lim\limits_{t\to\infty}\left(\frac{\ln(1)}{1}-\frac{\ln(t)}{t}-\int_{1}^{t}-\frac{1}{x^2}dx\right)\\
&= \lim\limits_{t\to\infty}\left(-\frac{\ln(t)}{t}-\left[\frac{1}{x}\right]_{1}^{t}\right)\\
&= \lim\limits_{t\to\infty}\left(-\frac{\ln(t)}{t}-\left(\frac{1}{t}-\frac{1}{1}\right)\right)\\
&=\lim\limits_{t\to\infty}\left(1-\frac{\ln(t)}{t}-\frac{1}{t}\right)
\end{align*}

As $t\to\infty$, $\frac{\ln(t)}{t}$ and $\frac{1}{t}$ both tend to $0$, so

$$\int_{1}^{\infty}\frac{\ln(x)}{x^2}dx=\lim\limits_{t\to\infty}\left(1-\frac{\ln(t)}{t}-\frac{1}{t}\right)=1$$

Title: To calculate an improper integral


问:
Calculating the inverse of a CDF which is a polynomial
I would like to generate random numbers from a CDF. I have looked at the inverse sampling method but I am struggling to find the inverse of my CDF given that it is a 6th order polynomial.

$F_X \left ({x} \right)= Ax^2-Bx^4 + Cx^6$

Where $A$, $B$ and $C$ are real positive and non-zero. Is there a different method to generate a random number based on this CDF? If so, I would really appreciate any guidance (especially for closed-form solutions).

My "tries": I have learnt about the Box-Mueller but it cannot be applied for this case (correct me if I am wrong). Secondly, I have had a look at rejection sampling but I don't quite understand how to apply it (working on MATLab by the way).

Edit: The region of interest is $0 < x < 80$

答:
You can use Wolfram Alpha as you did with the plot of the cdf to find the inverse function: solve (3.4675*10^-4)*x^2 -(3.5116*10^-8)*x^4+(8.361*10^-13)*x^6 ==p for x.

If you have Mathematica, you can generate samples in the following manner:

```<code>(* CDF and PDF *)
cdf = 3.4675 10^(-4) x^2 - 3.5116 10^(-8) x^4 + 8.361 10^(-13) x^6;
pdf = D[cdf, x];

(* Generate random samples *)
n = 1000;
data = x /. Solve[cdf == #][[4]] & /@ RandomReal[{0, 1}, n];

(* Fit a nonparametric density estimate and plot results *)
skd = SmoothKernelDistribution[data,
   Automatic, {"Bounded", {0, 80}, "Gaussian"}];
Plot[{pdf, PDF[skd, x]}, {x, 0, 80},
  PlotLegends -> {"True density", "Estimated density"}]
</code>```

The appropriate solution for $x$ given $p$ (a random sample from a Uniform[0,1] distribution) in the following equation

$$3.4675*10^{-4} x^2 - 3.5116*10^{-8} x^4 + 8.361*10^{-13} x^6 = p$$

is given by

```<code>x /. Solve[cdf == p, x][[4]]
</code>```

The imaginary parts cancel out (eventually).  I don't know if this completely qualifies as a case of casus irreducibilis.

Body:
You can use Wolfram Alpha as you did with the plot of the cdf to find the inverse function: solve (3.4675*10^-4)*x^2 -(3.5116*10^-8)*x^4+(8.361*10^-13)*x^6 ==p for x.

If you have Mathematica, you can generate samples in the following manner:

(* CDF and PDF *)
cdf = 3.4675 10^(-4) x^2 - 3.5116 10^(-8) x^4 + 8.361 10^(-13) x^6;
pdf = D[cdf, x];

(* Generate random samples *)
n = 1000;
data = x /. Solve[cdf == #][[4]] & /@ RandomReal[{0, 1}, n];

(* Fit a nonparametric density estimate and plot results *)
skd = SmoothKernelDistribution[data,
   Automatic, {"Bounded", {0, 80}, "Gaussian"}];
Plot[{pdf, PDF[skd, x]}, {x, 0, 80},
  PlotLegends -> {"True density", "Estimated density"}]

The appropriate solution for $x$ given $p$ (a random sample from a Uniform[0,1] distribution) in the following equation

$$3.4675*10^{-4} x^2 - 3.5116*10^{-8} x^4 + 8.361*10^{-13} x^6 = p$$

is given by

x /. Solve[cdf == p, x][[4]]

The imaginary parts cancel out (eventually).  I don't know if this completely qualifies as a case of casus irreducibilis.


问:
how for any $(a , e)$ we will get an ellipse on the double cone by the intersection of a Plane and the cone
I understand Dandelin spheres.

But I can not understand how for any $(a , e)$ we will get an ellipse on the double cone by the intersection of a Plane and the cone.

Here $a$ is the half of the major axis and $e$ is the eccentricity.

答:
Expanding-upon a comment ...

As mentioned in this answer the eccentricity of a conic is given by
$$e = \frac{\sin U}{\sin V} \tag1$$
where $U$ is the angle of inclination of the cutting plane, and $V$ is the angle of (the generators of) the cone.*

By varying $U$ from $O$ to $V$, the value of $e$ varies from $0$ (circle) to $1$ (parabola), so that every ellipse shape is achieved by simply tilting the plane.

To achieve every possible size, we vary the distance from the cutting plane to the cone's vertex. For specificity, consider this "side view" of the situation, with the cutting plane meeting the cone's axis at $P$, and the vertices of the resulting ellipse are $Q$ and $R$.

The Law of Sines tells us that, in $\triangle PVQ$,
$$\frac{q}{\sin\angle PVQ}=\frac{p}{\sin\angle PQV} \quad\to\quad
\frac{q}{\sin(90^\circ-V)}=\frac{p}{\sin(V+U)} \quad\to\quad
q = \frac{p\cos V}{\sin(V+U)} \tag2$$
and likewise, in $\triangle PVR$,
$$\frac{r}{\sin\angle PVR}=\frac{p}{\sin\angle PRV} \quad\to\quad
\frac{r}{\sin(90^\circ-V)}=\frac{p}{\sin(V-U)} \quad\to\quad
r = \frac{p\cos V}{\sin(V-U)} \tag3$$

Consequently, $|QR|$, the major axis of the ellipse, is given by
$$|QR| = q+r = \frac{p\cos V(\sin(V-U)+\sin(V+U))}{\sin(V-U)\sin(V+U)}
= \frac{p\cos U\sin 2V}{\sin(V-U)\sin(V+U)} \tag4$$

(Sanity check: When $U=0$, this reduces to $2p\cot V$, the width of the cone at the level of $P$, as we would expect with a circle. When $U=V$, $\sin(V-U)=0$ so that the length becomes infinite, as we would expect with a parabola.)

So, for a given pair of angles $U$ and $V$, we can choose $p$ so that $(4)$ yields whatever major axis length we desire.

*These angles are measured against "the horizontal". If instead one chooses to measure against, say, the "vertical" axis of the cone, then the ratio would be $\cos U/\cos V$.

I prefer sines, as they make zero-values correspond to zero-angles, so that in particular a circle ($e=0$) corresponds to a horizontal cutting plane with $0^\circ$ inclination.


答: If you don't like the $\log$, then you can turn it back to exponent by setting $u = \log x \implies x = e^u\implies dx = e^udu\implies I = \displaystyle \int_{0}^\infty ue^{-2u}(e^{u}du)= \displaystyle \int_{0}^\infty ue^{-u}du= -\displaystyle \int_{0}^\infty ud(e^{-u})=- \left(ue^{-u}|_{u=0}^\infty-\displaystyle \int_{0}^\infty e^{-u}du\right)=\displaystyle \int_{0}^\infty e^{-u}du = 1.$

Title: To calculate an improper integral

Body:
Is $$\int_1^\infty \frac{\log x}{x^2}dx$$
finite? How to solve this?


问:
When finding the basis of a subspace and having a free variable, would the dimension be just one?

So I am not sure if this is the correct way and answer to this problem but here is my solution

The problem:
\begin{equation*}
{\bigg( x,y,z \bigg)| x+y=0 \space and \space x-2z=0}
\end{equation*}
Solution:
$$
x  \ -2z=0\\
    x+y   =0\\x=2z\\y=-2z\\So,\\ \bigg(x,y,z\bigg)=\bigg(2,-2,1)z\\ basis :\\<(2,-2,1)>\\dimension \space is \space 1
$$

So is my solution correct and would it be a dimension of 1?

答: Indeed your work is correct. Another way of looking at it would be geometrically. The set of points in question you're looking at is $$ \{(x,y,z): x+y=0 \text{ and } x-2z=0\}$$ can be thought of as the intersection of two planes. In this case since the planes are not parallel, the intersection is an object of dimension 1, a line, whose direction vector you've found in your solution.


答:
Your solution is Correct!

In terms of matrix language, we have to solve $$AX=0$$ where $A=\begin{pmatrix} 1 &1&0\\1&0&-2 \end{pmatrix}$ , $X=(x,y,z)^T$. So Treat $A$ as a linear map form $\Bbb R^3$ to $\Bbb R^2$. Since rank of $A$ is 2, by dimension theorem, null space of $A$ has dimension 1.

Title: When finding the basis of a subspace and having a free variable, would the dimension be just one?

Body:
So I am not sure if this is the correct way and answer to this problem but here is my solution

The problem:
\begin{equation*}
{\bigg( x,y,z \bigg)| x+y=0 \space and \space x-2z=0}
\end{equation*}
Solution:
$$
x  \ -2z=0\\
    x+y   =0\\x=2z\\y=-2z\\So,\\ \bigg(x,y,z\bigg)=\bigg(2,-2,1)z\\ basis :\\<(2,-2,1)>\\dimension \space is \space 1
$$

So is my solution correct and would it be a dimension of 1?


答:
It's a binomial distribution. The mean is $100$, and the standard deviation is calculated as follows:

$$\sigma^2=np(1-p)$$
$$\sigma^2=600*\frac{1}{6}\frac{5}{6}=83.3333$$
$$\sigma\approx9.12$$

So the range for this interval is found as:

$$[\mu-\sigma,\mu+\sigma]$$

Title: Probability of rolling a $6$ between $90$ and $110$ times when rolling a die $600$ times

Body:
Show that if you roll a fair die $600$ times, then with probability at least $1/6$ you will get the roll $6$ between $90$ and $110$ times.

This is what I have so far

Let $E(x) = 100 \Rightarrow U(x) = 100 \cdot 1/6 = 600$

But, I don't know what more I need to show?

Thanks for your help.


答:
Notice if it does equal anything (or if it even makes sense to talk of it) then

$3 = \color{green}{\frac{k}{5+\frac{k}{5+\frac{k}{5+\cdots}}}}$

$8 = 5 +\color{green}{\frac{k}{5+\frac{k}{5+\frac{k}{5+\cdots}}}}$

$\frac 1 8 = \frac {1}{5 +\color{green}{\frac{k}{5+\frac{k}{5+\frac{k}{5+\cdots}}}}}$

$\frac k8 =  \frac {k}{5 +\color{green}{\frac{k}{5+\frac{k}{5+\frac{k}{5+\cdots}}}}}=3$

so

$\frac k8 = 3$

....

or  $3 = \frac k{5 + \color{blue}{\frac{k}{5+\frac{k}{5+\frac{k}{5+\cdots}}}}}$

$3 = \frac k{5 + \color{blue}3}$

.....

The real question is does it equal anything and does it make sense to talk of it?

.....

Here's a case of something that we can't talk about equaling anything.

$2x = 2^{(2^{(2^{\dots)})})}$  Then $2x = 2^x$ so $x = 2$. Doesn't make sense and $2^{2^{2^{\dots}}}$ obviously gets bigger with each step and there's no way doing it "infinitely many times" would settle down to simply $4$.

Likewise $S = 1 + 2 + 4 + 8 + 16 + 32 + .....$

$2S = 2 + 4 + 8 +16+32 + 64 +..... $

$1 + 2S = 1 + 2 + 4 + 8 +16+32 + 64 +..... $

$1+2S = S$ so $S = -1$ ... uh, say what?

So we do have to ask does $3 = \color{green}{\frac{k}{5+\frac{k}{5+\frac{k}{5+\cdots}}}}$ make sense?

......

Which it will if the terms get close to and converge to somthing.

Let $k = 24$.

Then $\frac {24}{5 + 3} = 3$.

And $\frac {24}{5 + \frac {24}{5+3}} = \frac {24}{5+3} = 3$.

And $\frac {24}{5 + \frac {24}{5+\frac {24}{5+3}}}=\frac {24}{5+3}=3$

and so on... so in this case... it's okay.

Title: How do I solve $3=\frac{k}{5+\frac{k}{5+\frac{k}{5+\cdots}}}$ for $k$?

Body:
How do I solve this for $k$?

$$3=\frac{k}{5+\frac{k}{5+\frac{k}{5+\cdots}}}$$

I know $k$ can be estimated by graphing, but is there a way to solve using algebra? If so, how?


问:
Why is the linear Span of the idempotents in a Banach algebra a Lie ideal?
In https://arxiv.org/pdf/1505.04503.pdf it was mentioned that "the linear span of the idempotents is a lie ideal".

I wonder where I can find a proof for that.

Let $A$ be a Banach algebra and let $\chi$ be the linear span of the idempotents. We want to show that $ax-xa \in \chi$ for all $x\in \chi, a\in A$. From here, I'm not sure how to use the properties of idempotents to show that.

Any references or suggestions will be appreciated.

答:
The commutator $[a,x]$ is linear in $x$, so it suffices to consider the case where $x$ itself is idempotent.  Now note that $$[a,x]=[a,x^2]=x[a,x]+[a,x]x$$ so it suffices to show that $x[a,x]$ is in the span of the idempotents (since then so is $[a,x]x$ by symmetry).  But observe that $$x[a,x]x=xax-xax=0$$ so $$(x+x[a,x])^2=x+x[a,x]$$ and thus $x[a,x]$ is a linear combination of the idempotents $x$ and $x+x[a,x]$.

The following perspective may help illuminate what's going on here (and is in fact how I came up with the argument above).  Since $x$ is idempotent, we can decompose $A$ as a direct sum $$xAx\oplus xA(1-x)\oplus (1-x)Ax\oplus (1-x)A(1-x).$$  Moreover, if we represent elements of $A$ as $2\times 2$ matrices using this direct sum decomposition, we can multiply them using matrix multiplication, with $x$ corresponding to $\begin{pmatrix} 1 & 0 \\ 0 & 0\end{pmatrix}$ (where I write the top left entry as $1$ since $x$ is in fact the unit of the algebra $xAx$, and also acts as a unit on $xA(1-x)$ and $(1-x)Ax$).  We can then calculate that the commutator of $x$ with a matrix $\begin{pmatrix} a & b \\ c & d\end{pmatrix}$ is $\begin{pmatrix}0 & -b \\ c & 0\end{pmatrix}$.  But now notice that any matrix of the form $\begin{pmatrix} 1 & 0 \\ c & 0\end{pmatrix}$ is idempotent, and so by taking linear combinations we can get any matrix $\begin{pmatrix} 0 & 0 \\ c & 0\end{pmatrix}$.  Similarly, we can get any matrix $\begin{pmatrix} 0 & b \\ 0 & 0\end{pmatrix}$ as a linear combination of idempotents, and thus our commutator is a linear combination of idempotents.


问:
Cardinality of the set of continuous functions versus set of all functions

Let $F$ denote the set of all functions $f: \mathbb{R} \to \mathbb{R}$, and let $C \subset F$ denote the subset of all continuous functions. Prove that $|\mathbb{R}| = |C| < |F|$. (Hint: use the fact that a continuous function on $\mathbb{R}$ is determined by its values on the rational numbers $\mathbb{Q} \subset \mathbb{R}$.

I've done some searching on this website about this proof and almost every answer seems to involve arithmetic of infinite cardinals, but this is not something I am familiar with, so I am hoping there is a proof that doesn't require it, even if it's longer and has more moving parts.

Here is a sketch of what I have so far.

By the Cantor-Schroeder Bernstein theorem, we can prove $|\mathbb{R}| = |C|$ by finding injections $f: \mathbb{R} \to C$ and $g: C \to \mathbb{R}$. For $f$, we can take $c \mapsto f(x) = c$, the constant function, which is of course continuous (take $\delta  = \epsilon$). For $g$, we enumerate the rationals (which are countable), $q_1, \ldots, $ and map a function $f$ to the sequence $(f(q_1), f(q_2), \ldots)$. This map is injective because a continuous function from $\mathbb{R}$ to $\mathbb{R}$ is uniquely determined by how it acts on the rationals (by density). If I could prove that the reals have the same cardinality as the sequences of reals, I can compose injections to conclude that, by Schroeder Bernstein, $|\mathbb{R}| = |C|$.

I don't know how to prove that $|C| < |F|$. I believe the definition of a strictly larger cardinality is that there is an injection from $C$ to $F$ but not surjection. (Is this correct?) If $so$, I can definitely inject $C$ into $F$ by the identity function. I would need only show there is no surjection from $C$ to $F$, which I don't know how to do.

答:
The cardinality of the set of all functions $f:\mathbb{R} \to \mathbb{R}$ is simply $|\mathbb{R}|^{|\mathbb{R}|}$ which by cardinal arithmetic can be looked at as $2^{| \mathbb{R}|}$ or denoted as $2^{\mathfrak{c}}$, $\mathfrak{c}$ being the cardinality of the continuum.

Now concerning the cardinality of the set of all continuous functions from $\mathbb{R}$ to $\mathbb{R}$, we must be careful. let's use the fact that $\mathbb{Q}$ is dense in $\mathbb{R}$. Let's go ahead and apply the Cantor-Schroder-bernstein theorem as you tried to.

The cardinality of the set of all continuous functions is at least that of the continuum $\frak{c}$ since we can associate injectively to each constant function a real number.

The cardinality of the set of all continuous functions is at most that of the continuum $\mathfrak{c}$ since we can inject this set into the set $\mathbb{R}^{\mathbb{N}}$. Essentially one associates to each continuous function its values on all the rational points (infinite sequence of integers).

Since $\mathbb{Q}$ is dense in $\mathbb{R}$ this uniquely determines the continuous function.

Hence nay the Bernstein-Cantor-Schroder theorem, $$|\{f: f:\mathbb{R} \to \mathbb{R} \text{ is continuous}\}|=\mathfrak{c}$$


答:
To finish your argument for the first part I’m going to assume that you know that $|\Bbb R|=|\wp(\Bbb N)|=\left|\{0,1\}^{\Bbb N}\right|$, and that $|\Bbb N\times\Bbb N|=|\Bbb N|$. Thus,

$$\left|\Bbb R^{\Bbb N}\right|=\left|\left(\{0,1\}^{\Bbb N}\right)^{\Bbb N}\right|=\left|\{0,1\}^{\Bbb N\times\Bbb N}\right|=\left|\{0,1\}^{\Bbb N}\right|=|\Bbb R|\,.\tag{1}$$

To rephrase that in terms of mappings, let $f:\Bbb R\to\{0,1\}^{\Bbb N}$ be a bijection. Then

$$g:\Bbb R^{\Bbb N}\to\left(\{0,1\}^{\Bbb N}\right)^{\Bbb N}:\langle x_n:n\in\Bbb N\rangle\mapsto\langle f(x_n):n\in\Bbb N\rangle$$

is a bijection, establishing the first equality in $(1)$. The second follows from the general result that for sets $A,B$, and $C$ there is a bijection between $\left(A^B\right)^C$ and $A^{B\times C}$; it’s a little messy to write down the details, but it’s not too hard and is a good exercise. The third follows easily from the existence of a bijection between $\Bbb N\times\Bbb N$ and $\Bbb N$, and I’ll leave that one to you.

For the inequality note that each $A\subseteq\Bbb R$ has an indicator (characteristic) function

$$\chi_A:\Bbb R\to\Bbb R:x\mapsto\begin{cases}
1,&\text{if }x\in A\\
0,&\text{if }x\notin A\,.
\end{cases}
$$

The map from $\wp(\Bbb R)$ to $\Bbb R^{\Bbb R}$ that takes $A$ to $\chi_A$ is easily seen to be injective, so $|\wp(\Bbb R)|\le\left|\Bbb R^{\Bbb R}\right|$. (In fact they are equal, but you don’t need that.) Finally, Cantor’s theorem says that $|\Bbb R|<|\wp(\Bbb R)|$, so $|\Bbb R|<\left|\Bbb R^{\Bbb R}\right|$.

Title: Cardinality of the set of continuous functions versus set of all functions

Body:
Let $F$ denote the set of all functions $f: \mathbb{R} \to \mathbb{R}$, and let $C \subset F$ denote the subset of all continuous functions. Prove that $|\mathbb{R}| = |C| < |F|$. (Hint: use the fact that a continuous function on $\mathbb{R}$ is determined by its values on the rational numbers $\mathbb{Q} \subset \mathbb{R}$.

I've done some searching on this website about this proof and almost every answer seems to involve arithmetic of infinite cardinals, but this is not something I am familiar with, so I am hoping there is a proof that doesn't require it, even if it's longer and has more moving parts.

Here is a sketch of what I have so far.

By the Cantor-Schroeder Bernstein theorem, we can prove $|\mathbb{R}| = |C|$ by finding injections $f: \mathbb{R} \to C$ and $g: C \to \mathbb{R}$. For $f$, we can take $c \mapsto f(x) = c$, the constant function, which is of course continuous (take $\delta  = \epsilon$). For $g$, we enumerate the rationals (which are countable), $q_1, \ldots, $ and map a function $f$ to the sequence $(f(q_1), f(q_2), \ldots)$. This map is injective because a continuous function from $\mathbb{R}$ to $\mathbb{R}$ is uniquely determined by how it acts on the rationals (by density). If I could prove that the reals have the same cardinality as the sequences of reals, I can compose injections to conclude that, by Schroeder Bernstein, $|\mathbb{R}| = |C|$.

I don't know how to prove that $|C| < |F|$. I believe the definition of a strictly larger cardinality is that there is an injection from $C$ to $F$ but not surjection. (Is this correct?) If $so$, I can definitely inject $C$ into $F$ by the identity function. I would need only show there is no surjection from $C$ to $F$, which I don't know how to do.


问:
convolution of exponential distribution and uniform distribution
Given $X$ an exponentially distributed random variable with parameter $\lambda$ and $Y$ a uniformly distributed random variable between $-C$ and $C$. $X$ and $Y$ are independent. I'm supposed to calculate the distribution of $X + Y$ using convolution.

Does anyone have any tips on how to do this?

I understand that the convolution is represented by
$$\int_{-\infty}^{+\infty} f_1(x) \cdot f_2(z-x)dx\tag {1}$$

and so given $$f_1(x) = \begin{cases}
\lambda e^{-\lambda x} &, x\ge 0\\ 0 &, x < 0 \end{cases}$$

and $$f_2(y) = \begin{cases} \frac{1}{2C} &, y \in [-C,C]\\ 0 & , \text{otherwise}\end{cases}$$

(1) becomes: $$\int_{-\infty}^{+\infty} \lambda e^{-\lambda x}\cdot \frac{1}{2C} dx$$

but I don't know how to procede from here. How do I choose which intervals of $z$ to integrate?

答:
Your final integral is incorrect; where is $z$ - it needs to be in your integral limits?

It is probably easier to calculate

$$\int_{-\infty}^{+\infty}f_1(z-x)\cdot f_2(x)dx=
\begin{cases}
\int_{-C}^{+C}\lambda e^{-\lambda (z-x)}\cdot \frac{1}{2C} &, z \ge x\\
0 &, z\lt x
\end{cases}$$

On second thoughts - your way is better

$$\begin{align}\int_{-\infty}^{+\infty}f_1(x)\cdot f_2(z-x)dx&=
\begin{cases}
\int_{z+C}^{z-C}\lambda e^{-\lambda x}\cdot \frac{1}{2C} dx &,z \ge C \\
\int_{z+C}^0\lambda e^{-\lambda x}\cdot \frac{1}{2C} dx &,C \ge z \ge -C \\
0 &,z\lt -C
\end{cases}
\end{align}
$$


答:
Let $t= \ln x.$ When we differentiate this we get $dt=\frac{1}{x}dx$. Also $e^{t} = x.\Rightarrow \color{red}A$

By $\color{red}A$ we get $e^{t}dt=dx\Rightarrow \color{red}B$.
So by substituting $\color{red}B$ we get,

$$\int_{0}^{\infty} te^{-t}dt=\int_{0 }^{\infty} t\frac{d\left [ -e^{-t} \right ]}{dt}dt$$

By integration by parts we get,
$$\int_{0 }^{\infty} te^{-t}dt=\left [ -te^{-t} \right ]_{0}^{\infty}-\int_{0}^{\infty}-e^{-t}dt=\left [ -te^{-t} \right ]_{0 }^{\infty}+\int_{0}^{\infty}e^{-t}dt$$

$$\int_{0 }^{\infty} te^{-t}dt=\left [ -te^{-t} \right ]_{1}^{\infty}+\left [ -e^{-t} \right ]_{0}^{\infty}=\left [ -te^{-t}-e^{-t} \right ]_{0}^{\infty}={\infty}=\left [ -e^{-t}(t+1) \right ]_{0}^{\infty}$$

So we have to find $\lim_{t\to\infty}e^{-t}(t+1)$ value,

$$\lim_{t\to\infty}e^{-t}(t+1)=\lim_{t\to\infty}\frac{t+1}{e^{t}}$$

So by L'hopital rule,

$$\lim_{t\to\infty}e^{-t}(t+1)=\lim_{t\to\infty}\frac{t+1}{e^{t}}=\lim_{t\to\infty}\frac{1}{e^{t}}=0$$

$$\lim_{t\to \infty} \int_{0 }^{t} te^{-t}dt=\lim_{t \to \infty} \left [ -e^{-t}(t+1) \right ]_{0}^{t}$$
So we get $$\lim_{t\to \infty} \int_{0 }^{t} te^{-t}dt=0+1=1$$

So as our final answer we get,

$$\int_1^\infty \frac{\log x}{x^2}dx=1$$

I got this as my answer. If there is any miscalculations. Please correct me.

Title: To calculate an improper integral


问:
Why am I wrong about this Gradient? Help
$f(x,y) = \langle 3x+y+1,2x-y-1 \rangle = A\vec{x}-\vec{b}$

Why are we unable to take the gradient of $f(x,y)$?

Is this a trick question. I can take the partial derivative of each function

$\frac{\partial f_x\:}{\partial \:x}\left(\left(3x+y\right)\right)=3$

$\frac{\partial f_y\:}{\partial \:y}\left(\left(2x-y\right)\right)=-1$

So why is the gradient not?

$=\begin{bmatrix}3\\ -1\end{bmatrix}$

答: The gradient of a real-valued function $f: \mathbb{R^2} \to \mathbb{R}$ is defined as $\nabla f = \langle f_x,f_y\rangle$. Your function is not a scalar-valued function hence gradient is not applied here.


问:
Show the measure is finite
Let $\{\mu_n\}$ be measures on $(X,M)$ with $\sup\{\mu_n\} \leq M < \infty$.

Let $\mu = \sum_{n \geq 1 } 2^{-n} \mu_n$

Show that $\mu$ is a finite measure.

Attempt:

It's easily verified that $\mu(\emptyset)=0$.

For countable additivity,

Let $(E_k) \subset M$ be disjoint. Then we have $$\mu (\cup_{k\geq 1} E_n)= \sum_{n \geq 1 } 2^{-n} \mu_n(\cup_{k \geq 1 }E_k)= \sum_{n \geq 1 } 2^{-n} (\sum_{k\geq 1}\mu_n(E_k))= \sum_{k\geq 1}(\sum_{n \geq 1} 2^{-n} \mu_n (E_k)) = \sum_{k\geq 1} \mu(E_k)$$

The part I'm having trouble is the finiteness of $\mu$. I know that $\sum_{n \geq 1} 2^{-n}=1$. But why do we have $\sum_{n\geq 1} \mu_n < \infty$ from $\sup\{\mu_n\} \leq M < \infty$?

Thank you.

答:
The idea is to use the Weierstrass M test: https://en.wikipedia.org/wiki/Weierstrass_M-test.

Define $\lambda_n : = 2^{-n} \mu_n$ and note that
\begin{equation*}
\lambda_n = 2^{-n} \mu_n \leq 2^{-n} M, \ \forall n \in \mathbb{N}\ .
\end{equation*}
Now, observe that
\begin{equation}
\sum_{n=0}^{\infty} 2^{-n}M = M \left( \sum_{n=0}^{\infty} 2^{-n} \right)\ .
\end{equation}
Therefore, the series the series $\displaystyle \sum_{n=0}^{\infty} 2^{-n}M$ converges, i. e.,
\begin{equation}
\sum_{n=0}^{\infty} 2^{-n}M < \infty\ .
\end{equation}
Then by the weierstrass M test the series $\displaystyle \sum_{n=0}^\infty \lambda_n$ converges, i. e.,
\begin{equation}
\mu = \sum_{n=0}^\infty 2^{-n}\mu_n < \infty.
\end{equation}

Title: Show the measure is finite


问:
Taylor's Theorem with integral form of remainder - clarification on requirements
In multiple places, the requirements for Taylor's Theorem with integral form of remainder state that the assumption is slightly stronger then the usual form of Taylor's theorem, since as opposed to assuming only that the (n+1)th derivative exists, we now assume that the (n+1)th derivative is continuous

(e.g. https://en.wikipedia.org/wiki/Taylor%27s_theorem#Taylor's_theorem_in_one_real_variable and
https://brilliant.org/wiki/taylors-theorem-with-lagrange-remainder/)

My question is, why is assuming the (n+1)th derivative is continuous necessary, and assuming that is is merely integrable not sufficient? I am familiar with the typical proof using induction coupled with integration by parts, and I don't see any point at which continuity seems necessary, but it was so consistently stated as a condition in online sources I worried I might have missed something important.

答:
The different requirements can already be seen in the case $n=0$: The mean-value theorem
$$
 f(x) = f(a) + (x-a) f'(\xi)
$$
holds whenever $f$ is differentiable on $[a, x]$, but the Fundamental theorem of calculus
$$
 f(x) = f(a) + \int_a^x f'(t) \, dt
$$
requires that $f$ is absolutely continuous. That is in particular the case if $f'$ is continuous.

The derivation for the integral form of the remainder uses the Fundamental theorem of calculus and then integration by parts on the terms
$$
 \int_a^x \frac{f^{(k)}(t)}{(k-1)!} (x-t)^{k-1} \, dt
$$
for $k=1, \ldots, n$. Again these steps are valid if the $k$-th derivative of $f$ is absolutely continuous (see for example Weakest hypothesis for integration by parts).

Therefore Taylor's theorem with the integral remainder is valid if $f^{(n)}$ is absolutely continuous.

If $f^{(n+1)}$ is continuous then $f^{(n)}$ is absolutely continuous, so that is a stronger (but perhaps easier to verify) requirement.


问:
Questions on ring fixed by automorphisms
Suppose $A$ is an integral domain and $F$ is its field of fractions. Let $G \leq Aut(F)$ be a group of automorphisms of $F$. Assume $g(a) \subseteq A$ for all $g \in G$ and let $A^G$ be the fixed ring
$$A^G=\{a \in A\ \mid g(a)=a, \forall g\in G\}.$$
I have managed to show that $A$ is integral over $A^G$ by considering the polynomial $\prod_{g\in G}(x-g(a))$ and showing it has coefficients in $A^G[x]$. Clearly it is monic, so the result follows. I have some more questions about this setup. I would also like to show that the fraction field of $A^G$ is $F^G$ and I am not sure how to do it. I think I might have to use the norm or trace maps $N,T:A \rightarrow A^G$
$$N(a)=\prod_{g\in G} g(a),$$
$$T(a)=\sum_{g\in G} g(a).$$
Also, I would like to show that $A$ Noetherian $\implies A^G$ Noetherian and $A$ Dedekind $\implies A^G$ Dedekind.

答:
It is true that the fraction field of $A^G$ is $F^G$, at least when $G$ is finite (which you seem to assume is your proof of integrality).

It is well-known that that the trace map of a separable extension is always surjective. Hence (the extension $F/F^G$ is Galois, so in particular separable) any element of $F^G$ is of the form $Tr(\alpha)=Tr(\frac{x}{y})$ for $\alpha\in F$, $x,y\in A$. This is equal to
$$\sum_{g\in G} \frac{gx}{gy}$$
$$\frac{\sum_{g\in G} \frac{N(y)}{gy}gx}{N(y)}$$
Both the numerator and denominator are in $A^G$, which shows that $F^G \subseteq \text{Frac}(A^G)$ (the reverse inclusion is trivial).

For Noetherianness, my idea is that Eakin-Nagata theorem might be helpful; we would need to show that $A$ is finitely generated as an $A^G$ module. I can't see right now how to do that, though.


问:
show $\sigma(X,Y)$ where $X= \ell_\infty, Y = \text{span}\{e_n \mid n \in \mathbb{N}\}$ is not equal to the weak star topology
The problem is to show that $\sigma(X,Y)$ is not equal to the weak star topology, although they coincide on bounded subsets of $\ell_\infty$, where  $X= \ell_\infty, Y = \text{span}\{e_n \mid n \in \mathbb{N}\}$. The definition of $\sigma(X,Y)$ is the coarsest topology that makes all $y \in Y$ continuous.

From these definitions, I think
$$\sigma(X,Y) = \left\{ \left\{x \in \ell_\infty : \sum_{i = 1}^\infty x_i y_i \in U \right\} : y \text{ has finitely many nonzero entries}, U \text{ open} \right\}$$
and the weak star topology is given by
$$\sigma(X^*,X) = \left\{ \left\{y \in \ell_\infty   : \sum_{i = 1}^\infty x_i y_i \in U \right\} : x \in \ell_\infty, U \text{ open} \right\}.$$

I also think that the closure of $Y$ is $c_0$, the set of all sequences that converge to $0$.

I spent a long time on this problem and am completely stuck. How do I proceed?

答:
Let $X=l^{1}$. Then $X^{\ast}=l^{\infty}.$ Let $e_{n}\in l^{1}$
be such that $e_{n}(k)=\delta_{nk}$. Let $Y=\mbox{span}\{e_{n}\mid n\in\mathbb{N}\}$.
Let $\mathcal{T}_{1}=\sigma(X^{\ast},X)$ be the weak*-topology
on $X^{\ast}$. Let $\mathcal{T}_{2}=\sigma(X,Y)$.

Claim 1. $\mathcal{T}_{1}\neq\mathcal{T}_{2}$.

Proof of Claim 1: We define a sequence $(x_{n})$ in $X^\ast$ such that
for each $y\in Y$, $\langle x_{n},y\rangle\rightarrow0$ as $n\rightarrow\infty$
but $\langle x_{n},z\rangle\not\rightarrow0$ for some $z\in X$.

Let $x_{n}\in X^{\ast}$ be defined as $x_{n}(k)=\begin{cases}
0, & \mbox{if }k<n\\
n^{2}, & \mbox{if }k\geq n
\end{cases}.$ That is, $x_{n}=(0,0,\ldots,0,n^{2},n^{2},\ldots)$. Let $y\in Y$,
then $y=\sum_{k=1}^{N}\alpha_{k}e_{k}$ for some $\alpha_{k}\in\mathbb{K}$
(=$\mathbb{R}$ or $\mathbb{C}$). Clearly, for any $n>N$, we have
$\langle x_{n},y\rangle=0$. This shows that $x_{n}\rightarrow0$
with respect to $\mathcal{T}_{2}$-topology.

Let $z=(1,\frac{1}{2^{2}},\frac{1}{3^{2}},\ldots)\in X$, then $\langle x_{n},z\rangle=\sum_{k=n}^{\infty}n^{2}\cdot\frac{1}{k^{2}}\geq1$.
Hence, $\langle x_{n},z\rangle\not\rightarrow0$ as $n\rightarrow0$.
Therefore, $x_{n}\not\rightarrow0$ with respect to $\mathcal{T}_{1}$-topology.

Since $Y\subseteq X$, we have $\mathcal{T}_{2}\subseteq\mathcal{T}_{1}$.
Combine the above observation, we conclude that $\mathcal{T}_{2}\subsetneq\mathcal{T}_{1}$.

Claim 2. $\mathcal{T}_{1}=\mathcal{T}_{2}$ when restricted on any
bounded subset of $X^{\ast}$.

Proof of Claim 2: Clearly $\mathcal{T}_{2}\subseteq\mathcal{T}_{1}$
when restricted on any bounded subset of $X^{\ast}$. We go to show
that $\mathcal{T}_{1}\subseteq\mathcal{T}_{2}$ when restricted on
any bounded subset of $X^{\ast}$.

It suffices that for any bounded net $(x_{\alpha})$ in $X^{\ast}$,
if $x_{\alpha}\rightarrow0$ with respect to $\mathcal{T}_{2}$, then
$x_{\alpha}\rightarrow0$ with respect to $\mathcal{T}_{1}$. Let
$(x_{\alpha})$ be a bounded net in $X^{\ast}$. Choose $M>0$ such
that $||x_{\alpha}||\leq M$. Suppose that $x_{\alpha}\rightarrow0$
with respect to $\mathcal{T}_{2}$. Let $y\in X$ be fixed. Let $\varepsilon>0$
be arbitrary. Choose $N$ such that $\sum_{n=N+1}^{\infty}|y(n)|<\frac{\varepsilon}{2M}.$
Define $z=\sum_{n=1}^{N}y(n)e_{n}\in Y$. By assumption, there exists
$\alpha_{0}$ such that $|\langle x_{\alpha},z\rangle|<\frac{\varepsilon}{2}$
whenever $\alpha\succeq\alpha_{0}$. Let $\alpha\succeq\alpha_{0}$
be arbitrary, then
\begin{eqnarray*}
 &  & \left|\langle x_{\alpha},y\rangle\right|\\
 & = & \left|\sum_{n=1}^{\infty}x_{\alpha}(n)y(n)\right|\\
 & \leq & \left|\sum_{n=1}^{N}x_{\alpha}(n)y(n)\right|+\left|\sum_{n=N+1}^{\infty}x_{\alpha}(n)y(n)\right|\\
 & \leq & \left|\langle x_{\alpha},z\rangle\right|+\sum_{n=N+1}^{\infty}\left|x_{\alpha}(n)y(n)\right|\\
 & \leq & \left|\langle x_{\alpha},z\rangle\right|+\sum_{n=N+1}^{\infty}M|y(n)|\\
 & < & \varepsilon/2+M\cdot\frac{\varepsilon}{2M}\\
 & = & \varepsilon.
\end{eqnarray*}
This shows that $x_{\alpha}\rightarrow0$ with respect to $\mathcal{T}_{1}$.


问:
Characterizing continuity in terms of closed sets
So I know that $f: X \to Y$ is continuous (that is, preimages of open sets are open) is equivalent to either of the following: preimages of closed sets are closed; $f$ is continuous at each point of $X$.

I am wondering what happens if we replace the word "open" in the definition of continuity at a point $x$. Then, is it true that we still have a characterization of continuity? To be precise, what I mean is:

Is it true that $f: X \to Y$ is continuous if and only if for all $x \in X$, and all closed $C \subseteq Y$ containing $f(x)$, there exists a closed $B \subseteq X$ containing $x$ such that $f(B) \subseteq C$?

My attempt:

$\implies$: $f$ is continuous so preimages of closed sets are closed. So put $B := f^{-1}(C)$ which is closed and contains $x$. And $f(B) = f(f^{-1}(C)) \subseteq C$.

$\impliedby$: This is where I got stuck. In the proof that I am aware of that continuity at each point implies continuity of the function, we seem to use the result that arbitrary unions of open sets are open. But this isn't necessarily true for closed sets, so I can't just blindly replace the word "open" with "closed". Any suggestions is appreciated.

答:
The second implication is not true in general.

If $X$ is a $T_1$ space, $f:X\to Y$ is any function at all, $x\in X$, and $C$ is a closed subset of $Y$ containing $f(x)$, then $\{x\}$ is a closed subset of $X$ containing $x$, and certainly $f[\{x\}]=\{f(x)\}\subseteq C$.


答: Your condition is satsified whether or not $f$ is continuous (as long as singleton sets are closed, which is true in metric spaces or just a $T_1$ space). Given any closed set $C$ containing $f(x)$ just take $B=\{x\}$.

Title: Characterizing continuity in terms of closed sets

Body:
So I know that $f: X \to Y$ is continuous (that is, preimages of open sets are open) is equivalent to either of the following: preimages of closed sets are closed; $f$ is continuous at each point of $X$.

I am wondering what happens if we replace the word "open" in the definition of continuity at a point $x$. Then, is it true that we still have a characterization of continuity? To be precise, what I mean is:

Is it true that $f: X \to Y$ is continuous if and only if for all $x \in X$, and all closed $C \subseteq Y$ containing $f(x)$, there exists a closed $B \subseteq X$ containing $x$ such that $f(B) \subseteq C$?

My attempt:

$\implies$: $f$ is continuous so preimages of closed sets are closed. So put $B := f^{-1}(C)$ which is closed and contains $x$. And $f(B) = f(f^{-1}(C)) \subseteq C$.

$\impliedby$: This is where I got stuck. In the proof that I am aware of that continuity at each point implies continuity of the function, we seem to use the result that arbitrary unions of open sets are open. But this isn't necessarily true for closed sets, so I can't just blindly replace the word "open" with "closed". Any suggestions is appreciated.


问:
What's the structure that's a combination of "Time" and "Time differences" called?
Let's say we call 2021/03/11 12:34:56 a Time

And let's call 30 minutes a TimeDiff

Then we have the following operations.

```<code>Time - Time -> TimeDiff
Time + TimeDiff -> Time
Time - TimeDiff -> Time
TimeDiff + TimeDiff -> TimeDiff
TimeDiff - TimeDiff -> TimeDiff
TimeDiff * Scalar -> TimeDiff
</code>```

Note, we for example, can't add Times. Nor can we multiply TimeDiffs with each other but we can multiply TimeDiffs by scalars.

Is there a general name for this sort of structure?

TimeDiff on it's own I think would just be called a Vector, as you can add, subtract and multiply by scalars.

But what would the combination of Time+TimeDiff be called? I feel like this should have a name, as it's a combination that might come up in other places (although I am finding it hard to find as good an example as "time").

答:
More generally, this kind of structure is called a torsor. See Torsors Made Easy by John Baez.

A torsor is like a group that has forgotten its identity.

An affine space is a torsor for a vector space.

An affine space is like a vector space that has forgotten its origin.


问:
Given that $|\vec u| = 8$ and $|\vec v| = 10$ and the angle between them is $60^\circ $ determine $|5 \vec u + 2 \vec v|$
I tried to use the cosine law but I got the wrong answer. I got $36.331$. The correct answer is supposed to be $52.915$.

Is there a mistake here? $$|5 \vec u + 2 \vec v|^2 = 5(8)^2 +2(10)^2 -2(40)(20)\cos(120^\circ)$$

答:
Let $\theta$ be the angle between $\vec{u}$ and $\vec{v}$. We compute $|5\vec{u}+2\vec{v}|^2$ using the dot product:

$$
|5\vec{u}+2\vec{v}|^2 = (5\vec{u}+2\vec{v})\cdot(5\vec{u}+2\vec{v})
=
25\vec{u}\cdot\vec{u}+20\vec{u}\cdot\vec{v}+4\vec{v}\cdot\vec{v}
=25|\vec{u}|^2 + 20|\vec{u}||\vec{v}|\cos\theta + 4|\vec{v}|^2
$$

so $|5\vec{u}+2\vec{v}|^2 = 1600+1600\cos\theta+400=2800$ (since $\cos 60^\circ=\frac{1}{2}$)

Therefore, $|5\vec{u}+2\vec{v}|=\sqrt{2800}=20\sqrt{7}$.


问:
Is there a non-abelian group $G$ such that $\operatorname{Aut}(G)$ is abelian?
Is there a non-abelian group $G$ such that $\operatorname{Aut}(G)$ is abelian?  I have not been able to find an example.</p>

答:
A non-abelian group can have an abelian group of isomorphisms (automorphisms).
An affirmative answer to this question was given by G. A. Miller  in 1913 .
Title of the paper is "A non abelian group whose group of isomorphism is abelian"
He constructed a non-abelian group $G$ of order $64$ such that $Aut(G)$ is abelian and
has order $128$.

Added In this paper Author has defined  $p + 1$ non-isomorphic groups of order $p^8$ (for each $p$) whose automorphism groups are elementary abelian of order $p^{16}$.

This link may also help you.


问:
Radon-Nikodym with a smaller $\sigma$-algebra
Let $(X,M, \mu)$ be a $\sigma$-finite measure. Let $M' \subset M$ be a smaller $\sigma$-algebra. Let $\mu' = \mu|_M$.

Show that there exists a unique $g \in L^1(\mu')$ such that $\int_E f d\mu = \int_E g d\mu'$ for all $E\in M'$ with $\|g\|_1 \leq \|f\|_1$.

My attempt:

I want to use Radon-Nikodym theorem here. But I'm not sure how to deal with the $\mu'$, since the version I learned requires both measures to be $\sigma$-finite.

Thank you for your help.

答:
I suppose $f$ is a given element of $L^{1}(\mu)$.

Consider $(X,M)$ with the measures $\mu'$ and $v$ wher $\nu (E)=\int_E f d\mu$. $\nu$ is  real (or comlex)  measure and $\nu << \mu'$. Hence. there is a unique $g \in L^{1}(\mu')$ such that $\nu(E)=\int_E gd\mu'$ for al $E \in M'$.

Using a simple function approximation you can see that $\int ghd\mu' =\int fh d\mu$ for any bounded measurable function $h$. Taking $h=\frac {|g|}g$ when $g \neq 0$ and $1$ when $g=0$ we se that $\int |g|d\mu' =\int fhd\mu\leq \int|f|d\mu$ since $|h|=1$


问:
Combination Calculations with Sub Groups
Given 16 Trials equally distributed to 4 individuals ( 4 Trials per person).

How many combinations are there where these conditions are met:

Condition A: 5 Total Successes

Condition B: 2 or Fewer Successes per Individual.

I understand Condition A in exclusivity would be:

$$\binom{16}{5} = 4368$$

I believe Condition B in exclusivity would be the following but please correct me if I am wrong:

$$\left(\binom42 + \binom41 + \binom40\right)^4 = 14641$$

My problem is I do not know how to solve for the number of combinations when both conditions are applied.

答: There are only two ways to write $5$ as a sum of $4$ nonnegative integers $\leq2$:$$1+1+1+2\\0+1+2+2$$  For the first of these we have $$4\binom41^3\binom42$$possibilities, and for the second we have $$12\binom40\binom41\binom42^2$$ possibilities.


问:
See if $\sum_{n=2}^{\infty}(\ln n)^{-20}$ converge
I want some help here, use comparison test to see if it converges or diverge. $\sum_{n=2}^{\infty}(\ln n)^{-20}$. This is $\sum \frac{1}{(\ln n)^{20}}$. So it's $\geq 0$, I wonder if it's using $\sum \frac{1}{n^{20}}$ to compare it to, because then $\sum \frac{1}{n^{20}}$ converge and if you compare it then $\sum \frac{1}{(\ln n)^{20}}$ does too

答:
Let's see if we can use the comparison test with $\sum_{n=2}^{\infty}(\ln n)^{-20}$ and $\sum_{n=2}^{\infty}n^{-20}$.

We know that $$\sum_{n=2}^{\infty}n^{-20}=\sum_{n=2}^{\infty}\frac{1}{n^{20}}$$ converges,
so to show that another series converges by comparison we need to show that the value is less than this value.

Note that:
$$n\geq \ln n$$
So then:
$$\frac{1}{n}\leq\frac{1}{\ln n}\implies \frac{1}{n^{20}}\leq\frac{1}{(\ln n)^{20}}$$

Which implies that our sum is bigger than the convergent value. Could be $\infty$, or an actual number, but this application doesn't show it.

As @Stinking Bishop mentioned, if you compare with $\sum\frac{1}{n}$ via $\lim_{n\to\infty}\frac{(\ln n)^{20}}{n}=0$ you see that:

$$\infty<\sum\frac{1}{n}\leq\sum(\ln n)^{-20}$$
Thus, the sum diverges.


问:
Why is $\int_0^\infty \frac{2 \cos(x) + x^2 - 2}{x^4}dx = \frac{\pi}{6}$?
Wolfram alpha computes that $\int_0^\infty \frac{2 \cos(x) + x^2 - 2}{x^4}dx = \frac{\pi}{6}$.

Why is this true? I thought this looked like something which could be justified with contour integrals, but I wasn't able to find a contour (or at least an easy contour) which gives the result.

What's a technique that can be used to compute this integral?

答:
While there is surely a contour integral, one can compute the indefinite integral as

$$\int\frac{2\cos(x) + x^2 - 2}{x^4}dx = \frac{\text{Si}(x)}{3}+\frac{2}{3 x^3}+\frac{\sin (x)}{3 x^2}+\frac{\left(x^2-2\right)
   \cos (x)}{3 x^3}-\frac{1}{x},$$
wherefrom the result follows.


问:
does uniform convergence imply convergence in distribution?
Suppose that $(X,\mathcal{M},\mu)$ is a measure space, and $f_n$ is a sequence of complex valued measurable functions converging uniformly to $f$.
Is it then true for every almost every $\lambda > 0$, that $\mu{(\{x: |f_n(x)| > \lambda\})}$ converges to $ \mu(\{x:|f(x)| > \lambda\}) $?

If it's not true, are there conditions that can be given so that it becomes true?

Edit: Someone has pointed out a counter example for non infinite measure spaces, is it true for finite measure spaces?

答:
In a finite measure space even point-wise converegence implies convergence in measure.

A simple counter-example in the case of an  infinite measure is $f_n(x)=1+\frac  1n$ on $[0,\infty)$ and $f_n(x)=1-\frac  1n$ on $(-\infty, 0)$. In this case $f_n \to 1$ uniformly, $\mu (|f_n| >1)=\infty$ for all $n$ but $\mu (|f| >1)=0$.


答:
You have two possibilities for number of trials across $4$ individuals that meet both conditions. Those are $ \{ 2 \ 2 \ 1 \ 0\}$ with permutations $\displaystyle \frac{4!}{2!}$ and $ \{ 2 \ 1 \ 1 \ 1\}$ with permutations $\displaystyle \frac{4!}{3!}$.

So the total number of combinations of trials (considering trials are distinct) meeting both conditions

$ = \displaystyle \frac{4!}{2!} \cdot {4 \choose 2}^2 \cdot {4 \choose 1} + \frac{4!}{3!} \cdot {4 \choose 2} \cdot 4^3 = 3264$

Title: Combination Calculations with Sub Groups

Body:
Given 16 Trials equally distributed to 4 individuals ( 4 Trials per person).

How many combinations are there where these conditions are met:

Condition A: 5 Total Successes

Condition B: 2 or Fewer Successes per Individual.

I understand Condition A in exclusivity would be:

$$\binom{16}{5} = 4368$$

I believe Condition B in exclusivity would be the following but please correct me if I am wrong:

$$\left(\binom42 + \binom41 + \binom40\right)^4 = 14641$$

My problem is I do not know how to solve for the number of combinations when both conditions are applied.


答: Another thing to point out is: $\ln n < n^{\alpha}$ for any $\alpha > 0$ and for sufficiently large $n$. This means $(\ln n)^{20} < n^{20\alpha}\implies \displaystyle \sum \dfrac{1}{(\ln n)^{20}} \ge \displaystyle \sum \dfrac{1}{n^{20\alpha}}=\infty$ if $ 20\alpha \le 1$. Thus to see the series diverges to $\infty$, you can take $\alpha  = \dfrac{1}{25}$.

Title: See if $\sum_{n=2}^{\infty}(\ln n)^{-20}$ converge

Body: I want some help here, use comparison test to see if it converges or diverge. $\sum_{n=2}^{\infty}(\ln n)^{-20}$. This is $\sum \frac{1}{(\ln n)^{20}}$. So it's $\geq 0$, I wonder if it's using $\sum \frac{1}{n^{20}}$ to compare it to, because then $\sum \frac{1}{n^{20}}$ converge and if you compare it then $\sum \frac{1}{(\ln n)^{20}}$ does too


问:
A compact subspace of a Banach space must have a finite dimension?
Let $X$ be a Banach space and $A \subset X$ a compact subspace. Doesn't this necessarily imply that $A$ is finite-dimensional? I've read that this is not true, but I can't figure out why my reasoning is incorrect.

<s>Wrong</s> proof. If $A$ is compact and $X$ is a Banach space (i.e complete metric space), then $A$ is closed in $X$ and totally bounded. Since $A$ is closed in $X$, it's complete with respect to the restriction of metric $d_{A \times A \to \mathbb{R}}$ induced by the norm of $X$. Since $(A, d)$ is complete and totally bounded, it's sequentially compact (i.e every sequence in $A$ has a convergent subsequence in $A$), which is more than what we need to show that unit ball is compact and $\dim A < \infty$.

Where did I go wrong? Thanks.

答: It is not true that every compact subset of a Banach space is finite dimensional. Let $H$ be  a Hilber psace with orthnormal basis $(e_n)$. Then $\{0,\frac {e_1} 1,\frac {e_2}2,\frac {e_3} 3...\}$ is a compact set which is not contained in any finite dimensional subsapce of $H$.


问:
Solving trigonometric equations: do I need to consider different cases in this example?
I'm trying to solve this set of trigonometric equations:
\begin{align}
\begin{aligned}
\tan\left(\frac{\delta+\beta}{2}\right) &= c\cdot\tan\left(\frac{\theta}{2}\right)\\
b\cdot\tan\left(\frac{\delta-\beta}{2}\right) &= a\\
\cos\left(\frac{\delta+\beta}{2}\right)\cos\left(\frac{\gamma}{2}\right) &= \cos\left(\frac{\theta}{2}\right)
\end{aligned}
\end{align}
Where a,b,c and $\theta$ are some given (but random) real numbers (they could be $0$ and $\pi$). The goal is to solve for the real values of $\gamma,\beta$ and $\delta$. I'm trying to solve these equations by combining some variables. Let
\begin{align}
M = \frac{\delta+\beta}{2}, N = \frac{\delta-\beta}{2}, P = \frac{\gamma}{2}
\end{align}
then it's easy to solve for $M$ from equation $1$. However, to solve for $\delta$ and $\beta$, we also need to figure out $N$. I'm worried that if I need to discuss the solutions by cases, like if $\theta = \pi$ (so that $\tan\frac{\theta}{2}$ is undefined) or if $a$ or $b$ is equal to $0$. Any suggestions on solving for this question? Is it possible to obtain a generalized solution? Thanks!!

答:
Look at:
$$M+N=\delta$$
$$M-N=\beta$$

You can find $M,N$ from your first two equations, and then calculate $\delta,\beta$ from those.

For $\gamma$, note that $$\tan^2(\alpha)+1=\frac{1}{\cos^2\alpha}$$
From this, you can write $$\cos\alpha=\pm\frac{1}{\sqrt{\tan^2\alpha+1}}$$
(depending on the angle).

Then this:
$$
\cos\left(\frac{\delta+\beta}{2}\right)\cos\left(\frac{\gamma}{2}\right) = \cos\left(\frac{\theta}{2}\right)
$$
Becomes this:

$$\cos\left(\frac{\gamma}{2}\right)=\cos\left(\frac{\theta}{2}\right)\cdot\sqrt{\tan^2(\frac{\delta+\beta}{2})+1}$$


问:
$\sum_{n=1}^{\infty} \frac{n^2}{(1+n^3)^p}$, find values of p
I have a sum $\sum_{n=1}^{\infty} \frac{n^2}{(1+n^3)^p}$ and $p>1$. Use integral tests and determine which values of $p$ can the partial sum, $s_5$ estimates true sum of $s$ to within $10^{-6}$ error. So this is a question about errors. It says using $s_5$, so error formula is $error \leq \int_{5}^{\infty} \frac{n^2}{(1+n^3)^p}$. So I calculate the sum and I can get p?  Here I need $10^{-6}$ maximum error, so I equal
$10^{-6} \leq \int_{5}^{\infty} \frac{n^2}{(1+n^3)^p}$. And finally it's get $p = 3.246$.

答:
You have
$$S_p=\sum_{n=1}^{\infty} \frac{n^2}{(1+n^3)^p}\qquad \text{and} \qquad I_p=\int_{5}^{\infty} \frac{n^2}{(1+n^3)^p}\,dn$$The integral is easy to compute since
$$J=\int \frac{n^2}{(1+n^3)^p}\,dn=\frac 13\int \frac{3n^2}{(1+n^3)^p}\,dn=\frac 13\int \frac{dx}{(1+x)^p}\,dx=\frac 13\int \frac{dy}{y^p}$$that is to say
$$I_p=\frac{3^{1-2 p} \,14^{1-p}}{p-1}$$ So, making the problem more general,you want to know $p$ such that
$$10^{-k} \leq \frac{3^{1-2 p} \,14^{1-p}}{p-1}$$ or
$$(p-1)\,126^{p-1} \geq 3 \times 10^k$$ Rewrite it as
$$\big[(p-1)\log(126)\big] e^{\big[(p-1)\log(126)\big]} \geq 3 \log(126)\times 10^k$$ and the solution of the equation is given is terms of Lambert function
$$(p-1)\log(126)=W\big[ 3 \log(126)\times 10^k \big]\implies $$ $$\color{blue}{p=1+\frac{W\big[ 3 \log(126)\times 10^k \big] }{\log(126)}}$$ Since that argument is large, you can evaluate it using
$$W(t)\approx L_1-L_2+\frac{L_2}{L_1}+\frac{L_2(L_2-2)}{2L_1^2}+\frac{L_2(2L_2^2-9L_2+6)}{6L_1^3}+\cdots$$ where $L_1=\log(t)$ and $L_2=\log(L_1)$.

For $k=6$, this will give $p\sim 3.866$.


问:
Proof verification: a planar graph on $n$ vertices has at most $10^n$ spanning trees.
As stated in the title, we want to show

If $G$ is a planar graph on $n$ vertices, then $G$ has at most $10^n$ spanning trees.

Proof attempt. By a corollary of Euler's formula, we have $e\leq 3n-6$. A spanning tree either contains a particular edge or does not, so there are at most $2^{3n-6}$ spanning trees of $G$. We claim $2^{3n-6}\leq 10^n$ for all $n\geq 3$. The base case is obvious, so suppose there is some $n>3$ satisfying $2^{3n-6}\leq 10^n$. Then $2^{3(n+1)-6}=2^{3n-6}\cdot 2^3 \leq 10^n\cdot 2^3 \leq 10^{n+1}$, where the first inequality holds by the inductive hypothesis. $\blacksquare$

答: The proof is correct, as pointed out in the comments, but in fact one can get a sharper bound. A spanning tree has $n-1$ edges, which gives a bound of $\binom{3n-6}{n-1}$ spanning trees (still far from sharp, since most subgraphs with $n-1$ edges are neither spanning nor trees), which gives growth like $(\frac{27}{4})^n.$ Note that $\frac{27}4 = 6\frac34.$ It is apparently not known whether we can get this below $6$ - see Maximum number of spanning trees of a planar graph with a fixed number of edges


问:
Show that $\frac{3\> + \>\cos x}{\sin x}$ cannot have any value between $-2\sqrt2$ and $2\sqrt2$
Show that

$$\dfrac{3+\cos x}{\sin x}\quad \forall \quad x\in R $$

cannot have any value between $-2\sqrt{2}$ and $2\sqrt{2}$.

My attempt is as follows:

There can be four cases, either $x$ lies in the first quadrant, second, third or fourth:-

First quadrant: $\cos x$ will decrease sharply and sinx will increase sharply, so $y_{min}=3$ at $x=\dfrac{\pi}{2}$.

$y_{max}$ would tend to $\infty$ near to $x=0$

Second quadrant: $\cos x$ will increase in magnitude and sinx will decrease sharply, so $y_{min}=3$ at $x=\dfrac{\pi}{2}$.

$y_{max}$ would tend to $\infty$ near to $x=\pi$

Third quadrant: $\cos x$ will decrease in magnitude and sinx will increase in magnitude but negative, so $y_{min}$ would tend to $-\infty$ near to $x=\pi$

$y_{max}$ would be $-3$ at $x=\dfrac{3\pi}{2}$

Fourth quadrant: $\cos x$ will increase sharply and sinx will decrease in magnitude, so $y_{min}$ would tend to $-\infty$ near to $x=2\pi$

$y_{max}$ would be $-3$ at $x=\dfrac{3\pi}{2}$

So in this way I have proved that $\dfrac{3+\cos x}{\sin x}$ cannot lie between $-2\sqrt{2}$ and $2\sqrt{2}$, but is their any smart solution so that we can calculate quickly.

答:
If $\frac{3 + \cos x}{\sin x} = k, k \in \mathbb R$, then $3 + \cos x = k \sin x \Rightarrow 9 + 6 \cos x + \cos^2 x = k^2 - k^2 \cos^2 x$, thus $(k^2+1)\cos^2 x + 6 \cos x + (9 - k^2) = 0$. Let $u = k^2$. Now for no value of $\cos x$ to exist, the discriminant must be less than $0$:

$$6^2 - 4(u+1)(9-u) < 0 \Rightarrow 4(u+1)(9-u) > 36$$
$$\Rightarrow -u^2 + 8u + 9 > 9 \Rightarrow u(-u+8) > 0 \Rightarrow 0 < u < 8$$

where in the last step, a sketch of the quadratic shows that it is concave up, hence the direction of the inequality.

Thus $0 < k^2 < 8$. This implies there are no values of $x$ such that $-2 \sqrt{2} < \frac{3 + \cos x}{\sin x} < 2 \sqrt{2}$.


问:
Relation of the kernels of one bounded operator and its extension
Sorry for this long and formal post. The application in PDEs is mentioned just at the end.

Let
 $$V \hookrightarrow H \text{  and } Q_H' \hookrightarrow Q',$$
 where $V$ and $Q$ are Banach and $H$ and $Q_H$ are Hilbert spaces. The hooked arrow $\hookrightarrow$ denotes the continuous embedding, which is basically $V \subset H$ and $\|v\|_V \geq \|v\|_H$ for all $v\in V$. The same with $Q_H'\hookrightarrow Q'$.

Also, assume that the embeddings are dense, i.e. $\overline V = H$ and $\overline {Q_H'} = Q'$, where the overline denotes the closure of the space with respect to the norm of its superspace.

Consider the linear bounded operator $$J\colon V \to Q_H'.$$ Then $V_0:=\ker(J)$ is a closed subspace of $V$. From an inf-sup condition, I have that
$$\|Jv\|_{Q_H'} \geq \gamma \|v\|_V\quad (1)$$
 for all $v \in V_1$, where $V_1$ is complementary to $V_0$, i.e. $V=V_0 \oplus V_1$.

Furthermore I have that $J\colon V\subset H\to Q'$ is bounded, so that one can define the natural extension $\bar J\colon H \to Q'$, using that $V$ is dense in $H$, that is bounded as well. Also for $\bar J$ I assume this boundedness from below, see $(1)$, for functions that are not in the kernel of $\bar J$.

Now my question is: Is the kernel of $J$ dense in the kernel of $\bar J$?

Or, equivalently, is $\overline V_0 = H_0$, where $H_0$ is the kernel of $\bar J$?

What I have tried so far:

I have shown that $\overline{V_0} \subset H_0$. To show the converse direction, I thought of taking $h \in H_0$ and show that there is a sequence $\{v_{0,n}\} \subset V_0$ that goes to $h_0$ (in the norm of $H$).

Since $\overline V = H$, there is $\{v_n\} \subset V$ that goes to $h_0$ (in the norm of $H$).

Because of $(1)$ there is a bounded projector $P_V\colon V \to V$, with $P(V)=V_0$. Then one can split up every $v_n$ into $v_{0,n}:=Pv_n$ and the remainder $v_{1,n}$ that is in $V_1$.

Now I want to show, that $\{v_{1,n}\}$ goes to $0$ (in $H$) what would make $\{v_{0,n}\} \subset V_0$ approaching $h_0$.

......

In terms of PDEs, this would answer the questions, whether the (sub)space of divergence free elements of $H_0^1(\Omega)^3$ is dense in the (sub)space of these functions in $L^2(\Omega)^3$. In this case:

$J:=div$

$V:= H_0^1(\Omega)^3$ and $H:=L^2(\Omega)^3$

$Q_H := L^2(\Omega)/\mathbb R$ and $Q' = (H^1(\Omega)/\mathbb R)'$

And the question becomes: Is
$$ \{v\in H_0^1(\Omega)^3:\text{div } v = 0 \in L^2(\Omega)/\mathbb R \} \text{ dense in } \{v \in L^2(\Omega)^3:\text{div } v = 0 \in (H^1(\Omega)/\mathbb R)' \} $$

答:
Some updates: Somehow I think the argument doesn't need to be so long. Actually the divergence free vector field in $H(\mathrm{div})$ can be defined as

$L^2$-divergence free vector fields are the closure of $C^{\infty}$-divergence free vector fields in $L^2$-norm.

This is in Luc Tartar's book page 35 here. The the density argument of $C^{\infty}\subset H^1 \subset L^2$ would imply the density of the $H^1$-divergence free vector fields in $L^2$-divergence free vector fields. The proof I presented was just try to replicate what Tartar did for the density of $C^{\infty}$-divergence free vector fields in
$H^1$-divergence free vector fields.

Tool we use to prove density: Suppose subspace $\mathscr{X} \subset X$, they are both Banach. Define $$X^{\perp} = \{l\in X': \langle l,v\rangle =0 \;\forall v\in X\},$$
and
$$\mathscr{X}^{\perp} = \{l\in X': \langle l,v\rangle =0 \;\forall v\in \mathscr{X}\},$$ where $X'$ is the set of all bounded linear functional on $X$. Then we have

Claim: $\mathscr{X}$ is dense in $X$. $\Longleftrightarrow$ $\mathscr{X}^{\perp} =X^{\perp} $.

Sketch of the proof: First $\mathscr{X}^{\perp} \supset X^{\perp} $ holds always. "$\Rightarrow$" is like standard exercise.  For "$\Leftarrow$", we want to prove $\mathscr{X}^{\perp}  \subset X^{\perp}$ implying the left: suppose the density does not hold, then we could find an open subset $Z\subset X$ so that $\overline{\mathscr{X}} \cap Z =\varnothing $. Choose $z\in Z$, we can find a non-zero bounded linear functional $g\in X'$ such that $\langle g,z\rangle \neq 0$. Consider a functional $L$ on $\overline{\mathscr{X}} + \{z\} $:
$$
\langle L,x+tz\rangle = \langle l,x\rangle + t\langle g,z\rangle, \quad \text{ for } x\in \mathscr{X}, t\in \mathbb{R}, l\in \mathscr{X}^{\perp},
$$
then we can extend $L$ to whole $X$. It can be checked that $L\in \mathscr{X}^{\perp}$, but $\langle L,z \rangle = \langle g,z \rangle\neq 0$ implies  $L\notin X^{\perp}$. Thus $\mathscr{X}^{\perp}  \not\subset X^{\perp}$ and the claim.

Now we move on to prove

Divergence free vector fields in $H^1$ is dense in divergence free vector fields in $L^2$.

Denote
$$
V :=H_0^1(\Omega)^3,\quad  V_0 :=  \{v\in H_0^1(\Omega)^3:\mathrm{div}\, v = 0\},
$$
and
$$
H := L^2(\Omega)^3,\quad H_0 :=  \{v \in L^2(\Omega)^3:\mathrm{div}\, v = 0 \},
$$
then what you wanted to show is:

(A) $V_0$ is dense in $H_0$.

We can prove this using above claim. Define
$$
H(\mathrm{div}) =  \{v \in L^2(\Omega)^3,\mathrm{div}\, v \in L^2(\Omega) \},
$$
and we can check this is a Hilbert space under the norm:
$$
\|\cdot\|_{H(\mathrm{div})}^2 = \|\cdot \|_{L^2(\Omega)^3}^2
+ \|\mathrm{div}(\cdot)\|_{ L^2(\Omega)}^2.
$$
Now all the relevant spaces are Hilbert now and we can associate the bounded linear functional with a specific inner product.

First Let $l\in H(\mathrm{div})'$, representation theorem in Hilbert space says there is some $u_l \in H(\mathrm{div}) \subset L^2(\Omega)^3$ :
$$
\langle l,v\rangle = \int_{\Omega} u_l \,v + \int_{\Omega}(\mathrm{div} \,u_l )\,(\mathrm{div}\, v).
$$

Consider some $l$ vanishes on $V_0$:
$$V_0^{\perp} = \{l\in H(\mathrm{div})': \langle l,v\rangle =0 \;\forall v\in V_0\}\subset \{l\in (H_0^1(\Omega)^3)': \langle l,v\rangle =0 \;\forall v\in V_0\}. $$
We also know that
$$
\mathrm{div}: H_0^1(\Omega)^3 \to L^2(\Omega),\quad \text{ and }\quad \mathrm{div}^* = -\nabla : (L^2(\Omega))'\simeq L^2(\Omega) \to  ( H_0^1(\Omega)^3)'.
$$
Closed range theorem reads:
$$
R(-\nabla ) = (\mathrm{ker}(\mathrm{div}))^{\perp} = \{l\in  ( H_0^1(\Omega)^3)': \langle l,v\rangle =0 \;\forall v\in \mathrm{ker}(\mathrm{div}) = V_0 \} \supset V_0^{\perp},
$$
and this means $\langle l,v\rangle =0 $ for any $v\in V_0$, then $u_l = \nabla \phi$ for some $\phi\in L^2(\Omega)/\mathbb{R}$ in the sense of isomorphism:
$$
\langle l,v\rangle =  \int_{\Omega} u_l \,v =  \int_{\Omega} \nabla \phi \,v,
$$
for $v\in H^1_0(\Omega)^3$ and divergence free.

Now we want to show
$$
V_0^{\perp}\subset \{l\in H(\mathrm{div})': \langle l,v\rangle =0 \;\forall v\in H_0\} = H_0^{\perp}.
$$
For the above $l$ that vanishes on $V_0$, $u_l = \nabla \phi$, for $u_l \in H(\mathrm{div})\subset L^2(\Omega)^3$, we can pin down this $\phi\in H^1_0(\Omega)$ by solving:
$$
\int_{\Omega} \nabla \phi \cdot \nabla v = \int_{\Omega} u_l \cdot \nabla v,\quad \text{ for } \forall v\in H^1_0(\Omega).
$$
We can use Green's identity which is valid for $u\in H(\mathrm{div})$ and $\phi \in H^1$, this result can be found in Tartar's book: for $u\in H_0\subset  H(\mathrm{div})$
$$
\langle l,u\rangle = \int_{\Omega} \nabla \phi \cdot u = -\int_{\Omega} \phi\,\mathrm{div}\,u + \int_{\partial \Omega} (u\cdot n)\phi \,dS,
$$
and the boundary term vanish for $\phi \in H^1_0(\Omega)$. Therefore $\langle l,u\rangle = 0$ for $u\in H_0$, and we have:

(B) $V_0^{\perp}\subset H_0^{\perp}$.

By the claim, we have (A).


问:
Is a pseudometric $d(\cdot,x)$ continuous for fixed $x$?
Let $(A,d)$ be a pseudometric space with semimetric $d$ defined by :

Def. $d$ is a pseudometric on some space $A$ as soon as:

$\forall x\in A, d(x,x)=0$;

$\forall (x,y,z)\in A\times A\times A, d(x,y)\leq d(x,z)+d(z,y)$.

$d(x,y)=d(y,x)\geq 0, \forall x,y\in A (Scott's appointment)$.

Question. If $x\in A$ is fixed, is $d(\cdot,x): A\to\mathbb R$ a continuous function?

My Attempt

The topology can be determined by sequences in $(A,d)$.

One can prove that

$d(\cdot,x):A\to\mathbb R$ is continuous at $y\in A$ iff
$\lvert d(y,x)- d(y_n,x)\rvert \to 0$ for any sequence such that $d(y_n,y)\to 0$ as $n\to\infty$.

So take any sequence $y_n$ so that $y_n\to y$.
Let $\epsilon>0$. Then there will be $n_0:\forall n>n_0: d(y,y_n)<\epsilon$. Hence

(1) $ d(y_n,x)\leq d(y_n,y)+d(y,x)=d(x,y)+\epsilon$.

On the other hand,

(2) $ d(x,y)\leq d(x,y_n)+d(y_n,y)\leq d(x,y_n)+\epsilon$.

Combining $(1)-(2)$, we have $\epsilon-d(y_n,x)\leq d(x,y)\leq d(y_n,x)+\epsilon$. So $\lvert d(y,x)-d(y_n,x)\rvert\to 0$, which shows that $d(\cdot,x)$ is continuous at $y$.

I still have some doubts: (Q1) Is there any other way to show the result more directly? (Q2) Implicitly in this proof, did I use the topology on $(A,d)$ induced by $d$ and the topology on $(\mathbb R, \lvert\cdot,\cdot\rvert)$ induced by $\lvert\cdot,\cdot,\rvert$?

答:
The topology on $A$ is the one generated by $d$ itself, say $\tau_d$. You are basically trying to show that the function $f_x:A \rightarrow \mathbb{R}$ defined by $f_x(y)=d(y,x)$ is continuous on $A$ with respect to $\tau_d$.

Suppose $(X,d)$ and $(Y,s)$ are pseudometric spaces, then $f:X \rightarrow Y$ is continuous at $a \in X$ iff $f(x_n) \xrightarrow{s} f(a)$ for every sequence $x_n \xrightarrow{d} a$.

So I think your approach is quite good. Just in (1) the $=$ should be $\leq$ and in the second last paragraph it should be $d(y_n,x)-\epsilon \leq \cdots$.

You are using the topology on $(A,d)$ when you are talking about the convergence of $\{y_n\}_n$ and using the usual topology of $\mathbb{R}$ while talking about the convergence of the sequence $\{d(y_n,x)\}_n$. Here note that every metric is a pseudometric, so $|\cdot|$ on $\mathbb{R}$ is kind of treated as a pseudometric here.


问:
Solving system of ODE with elimination method for $x'-2x+y'=0$ and $x'-x+y'-y=0$
Suppose that a system of ODE is comprised of
$$x'-2x+y'=0$$
$$x'-x+y'-y=0$$

Naturally, I'll reformat this into the following:

$$(1) \ \ \ (D-2)x+(D)y=0$$
$$(2) \ \ \ (D-1)x+(D-1)y=0$$

And when I eliminate the y, I get $(-2D+2)x=0$ and solving for $D$ gets $D=1$ so one of my general solution is $x(t)=c_1e^x$ but this is where the problem happens. Say I plug that solution into $(1)$, then we'd get $-c_1e^x+(D)y=0$. How am I supposed to find the solution on the Y side? Or is there a better way of solving for this ODE?

答: Hint: Apply $D-1$ to the top and $D-2$ to the bottom equations respectively and subtract the new top equation to the new bottom equation to get a ODE equation only $y$ and solve for $y$, then come back to solve for $x$ using linear first order ODE standard.


问:
Convergence using the ratio test
I am trying to determine the smallest value for natural number $k$ such that $$\sum\limits_{n=1}^\infty\frac{(n!)^{k-1}(27n)^n}{(kn)!e^n}$$ converges. I attempted to solve using the ratio test:

$$\text{If} \lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|=L<1, \text{then the series $\sum\limits_{n=1}^\infty a_n$ is absolutely convergent.}$$

Using $a_n = \frac{(n!)^{k-1}(27n)^n}{(kn)!e^n}$ and $a_{n+1} = \frac{((n+1)!)^{k-1}(27(n+1))^{n+1}}{(k(n+1))!e^{n+1}}$, I tried to simplify the following ratio like this:

$$\lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|= \lim_{n\to\infty}\left|\frac{\frac{((n+1)!)^{k-1}(27(n+1))^{n+1}}{(k(n+1))!e^{n+1}}}{\frac{(n!)^{k-1}(27n)^n}{(kn)!e^n}}\right|$$

$$= \lim_{n\to\infty}\left|\frac{\frac{(n+1)^{k-1}(n!)^{k-1}(27)^{n+1}(n+1)^{n+1}}{(kn+k)(kn+(k-1))\cdots(kn+1)(kn)!e^{n+1}}}{\frac{(n!)^{k-1}(27^n)n^n}{(kn)!e^n}}\right|$$

$$= \frac{27^{n+1}e^n}{27^ne^{n+1}}\lim_{n\to\infty}\left|\frac{\frac{(n+1)^{k-1}(n+1)^{n+1}}{(kn+k)(kn+(k-1))\cdots(kn+1)}}{\frac{n^n}{1}}\right|$$

$$ = \frac{27}{e}\lim_{n\to\infty}\left|\frac{(n+1)^{k-1}(n+1)^{n+1}}{(kn+k)(kn+(k-1))\cdots(kn+1)n^n}\right|$$

From here, I consider the only the highest order term in the numerator and denominator, with there only being $k$ terms remaining from the factorial, yielding leading term $(kn)^k = k^kn^k$,

$$ = \frac{27}{e}\lim_{n\to\infty}\left|\frac{n^{k+n} + O(n^{k+n-1})}{k^kn^{k+n}+O(n^{k+n-1})}\right|$$

Solving the limit by dividing top and bottom by ${n^{k+n}}$ gives

$$ = \frac{27}{e}\frac{1}{k^k}$$

Which would be first less than 1 and therefore convergent at $k=3$, i.e. $\frac{27}{e}\frac{1}{3^3} = \frac{1}{e} < 1$. Then, by the ratio test, $k=3$ would be the first instance that this is true, because at $k=1$ we get $\frac{27}{e} > 1$ and at $k=2$ we get $\frac{27}{4e} > 1$, which diverge by the ratio test.

However, when I punch $\sum\limits_{n=1}^\infty\frac{(n!)^{3-1}(27n)^n}{(3n)!e^n}$ into Wolfram Alpha, it states that the series diverges, when the ratio test above shows that it converges for $k=3$. I'm not sure where my work is wrong and any pointers to the right answer of $k=4$ would be much appreciated. Thanks in advance!

答:
We can write $$\begin{align}
\frac{a_{n+1}}{a_n}&=
\left(\frac{(n+1)!}{n!}\right)^{k-1}\frac{(27(n+1))^{n+1}}{(27n)^n}\frac{(kn+k)!}{(kn)!}\frac{e^n}{e^{n+1}}\\
&=(n+1)^{k-1}27(n+1)\left(1+\frac1n\right)^n\frac1e\frac1{(kn+k)\cdots(kn+1)}\\
&=27\left(1+\frac1n\right)^n\frac1e\frac{(n+1)^k}{(kn+k)\cdots(kn+1)}\\
&=27\left(1+\frac1n\right)^n\frac1e\frac1{k\left(k-\frac1n\right)\left(k-\frac2n\right)\cdots\left(k-\frac{k-1}{n}\right)}\\
&\to\frac{27}{k^k} \text{ as }n\to\infty
\end{align}$$
Thus the series certainly converges when $k>3$ and diverges when $k<3$ and the case $k=3$ is doubtful.

For the case $k=3$, we have $$a_n=\frac{(n!)^2(27n)^n}{(2n)!e^n}\sim\frac{(27n)^n\sqrt{n\pi}}{4^ne^n}$$ by Stirling's approximation.  Thus $a_n\to\infty$ as $n\to\infty$, and the series diverges for $k=3$.

Title: Convergence using the ratio test


答:
Geometric method: Refer to the figure:

Note:
$$S_{\Delta AFO}=\frac12\cdot AO\cdot FO\cdot \sin 60^\circ=\frac12\cdot 20\cdot 16\cdot \frac{\sqrt3}{2}=80\sqrt3$$
From the triangle $AFO$
$$AD=\frac{2S_{\Delta AFO}}{FO}=\frac{2\cdot 80\sqrt3}{16}=10\sqrt3=BE$$
From the triangle $ADO$:
$$DO=\sqrt{AO^2-AD^2}=\sqrt{20^2-(10\sqrt3)^2}=10=CE$$
Finally, from the triangle $BEO$:
$$|5\vec{u}+2\vec{v}|=BO=\sqrt{EO^2+BE^2}=\sqrt{50^2+(10\sqrt3)^2}=\sqrt{2800}\approx 52.915.$$

Title: Given that $|\vec u| = 8$ and $|\vec v| = 10$ and the angle between them is $60^\circ $ determine $|5 \vec u + 2 \vec v|$

Body:
I tried to use the cosine law but I got the wrong answer. I got $36.331$. The correct answer is supposed to be $52.915$.

Is there a mistake here? $$|5 \vec u + 2 \vec v|^2 = 5(8)^2 +2(10)^2 -2(40)(20)\cos(120^\circ)$$


问:
An infinite dimensional limit
I am trying to show
$$\lim_{n\rightarrow \infty} \int_{[0,1]^n} \frac{|x|}{\sqrt{n}}dx=\frac{1}{\sqrt{3}}$$
where $[0,1]^n$ is the unit cube in $\mathbb{R}^n$. I would appreciate any hint.

答: If $X_1,X_2,...$ is an i.i.d. sequence of random variables with uniform distribution on $[0,1]$ then the given limit is $\lim \frac 1 {\sqrt n} E(X_1^{2}+X_2^{2}+\cdots +X_n^{2})^{1/2}$. By SLLN we have $(X_1^{2}+X_2^{2}+\cdots +X_n^{2}) /n \to \int_0^{1}x^{2}dx=\frac 1 3$ a.s. and in $L^{1}$ and the result follows from this.

Title: An infinite dimensional limit