问:
Unicity of solution of pde
Let the pde $$\dfrac{\partial^2 u}{\partial t^2} - \dfrac{\partial^2 u}{\partial x^2}=f(x)$$
The question is:

Find the limit condition such that this pde admit a unique solution in $[a,b] \times [0,T].$

For this, I suppose the existence of to sulutions $u_1$ and $u_2$ and we put $v = u_1 - u_2.$ Then, $$\dfrac{\partial^2 v}{\partial t^2} - \dfrac{\partial^2 v}{\partial x^2} = 0$$
myltiply this last equation by $\dfrac{\partial v}{\partial t}$ and we integrate over $[a,b]$ and we obtained $$\displaystyle\int_a^b \dfrac{\partial^2 v}{\partial t^2} . \dfrac{\partial v}{\partial t} dx - \int_a^b\dfrac{\partial^2 v} {\partial t^2}. \dfrac{\partial^2 v}{\partial x^2} dx = 0$$

I tried integration by parts, but didn't find the limit conditions to obtain the unicity of this pde. Thanks for any help.

答:
For the wave equation there is a conserved energy
$$
E(t)=\int_{-\infty}^\infty (u_t)^2+(u_x)^2 dy.
$$
This is obtained multiplying by $u_t$ and integrating by parts (as you did). For $v$ this quantity is zero initially, thus it vanishes for all times. Right?</p>

Title: Unicity of solution of pde


问:
How do I find the limit of $\lim_{x \to 0} \frac{x-\ln(x+1)}{x(\sin(2x))}$?
How do I find the limit of

$$\lim_{x \to 0} \frac{x-\ln(x+1)}{x(\sin(2x))}$$

I know it is equal to $\frac{1}{4}$, but how did we get that? Without using L'Hopital's rule.

I tried canceling $x$ with $\sin(2x)$ and $\ln(x+1)$, but still got the wrong  result...

答: $$\lim_{x \to 0} \frac{x-\ln(x+1)}{x(\sin(2x))} = \frac{0 }{ 0 },$$ you would make L'Hôpital's rule twice to have finally: $$\lim_{x \to 0} \frac{1/(x+1)}{ -4x \sin (2x) + 4 \cos (2x)}  = 1/4.$$ Hope my answer helps you!

ContentLicense: CC BY-SA 4.0


问:
How do I evaluate $\int_{0}^{\sqrt{2}}\int_{x}^{\sqrt{4-x^2}}{\sqrt{x^2+y^2}} \, dy \, dx$?
I'm having troubles evaluating this double integral. Can somebody help me? I've gone to the part that I need to use trigonometric substitution, but performing the said sub, I think I'm kind of unsure of the next steps.

$$\int \limits_{0}^{\sqrt{2}}\int \limits_{x}^{\sqrt{4-x^2}}{\sqrt{x^2+y^2}} \, dy \, dx $$

答:
Have you sketched the region of integration? It is a sector of a circle of radius $2$, one-eighth of the circle.

The function $z=\sqrt{z^2+y^2}$ forms a cone with its tip at the origin. Out to radius $2$, the volume of that cone would be $$V=\frac13\pi(2)^2(2)=\frac83\pi$$ using basic geometry formulas. The integral with the full circle as the region of integration would be the complement of that cone within a cylinder of radius $2$ and height $2$: $$\pi(2)^2(2)-\frac83\pi=\frac{16}{3}\pi$$

And since the region of integration is really a one-eight sector of that circle, the given integral is $$\frac{1}{8}\left(\frac{16}{3}\pi\right)=\frac23\pi$$


问:
Detecting weakly redundant constraints of a Convex Polytope
I am currently dealing with a problem where I need to detect all redundant constraints of a convex polytope defined by $\mathcal{P} = \{x\in\mathcal{R}^n: Ax\leq b \wedge Cx = d \}$, basically your standard convex polytope with equality constraints.

I already know the solutions for removing strongly redundant constraints. To check the strong redundancy of the $i^{th}$ constraint, we need to check the following linear program's feasibility. If it is infeasible, then it is strongly redundant.

$$
\min_{x} 0\\
\begin{align*}
\text{s.t. } Ax &\leq b\\
A_ix&=b_i\\
Cx &= d
\end{align*}$$

However, I am unsure how to identify weakly redundant constraints correctly. My first thought was to check the dimensionality of the polytope defined by $\mathcal{P_i} = \{x\in\mathcal{R}^n: Ax\leq b\wedge C x= d\wedge A_i x = b_i\}$ for each $i^{th}$ constraint.

I am not sure how to proceed. I know for polytopes defined by inequalities only, there is a procedure for eliminating weakly redundant constraints via a Chebyshev ball approach. One method I have thought of was enumerating all vertices of the polytope and checking each facet ( relating to a constraint) intersects at least $n$ vertices. Still, this method suffers from the curse of dimensionality and becomes computationally intractable quickly.

I suspect there is a linear programming formulation that will be much faster.

Any help would be greatly appreciated!

答:
I have figured it out. I am using a Chebychev ball approach. This is fairly performant.

Given a convex polytope defined as above, and assuming that we have already culled all strongly redundant constraints via the method above. we can use the following LP to check for weak degeneracy in the inequality constraints.

$$
\begin{align*}
R_i = \max_{x\in\mathcal{R}^n,r\in\mathcal{R}_+}& \quad r\\
\text{s.t. } A_jx & + ||A_j||_2r\leq b_j, \forall j \neq i\\
Cx &= d\\
A_i x &= b_i
\end{align*}
$$

For each inequality constraint $i$, if $R_i> 0$ then it is not weakly degenerate. This optimization problem is equivalent to the question "What is the largest euclidean ball I can fit inside of my feasible space?". The feasible space will be lower dimensional with a weakly redundant constraint active (a vertex, edge, ect), so the above optimization problem should identify weak redundancy.

I suspect that this can also cull strongly degenerate inequality constraints as it will just be infeasible.


问:
Bounds for Balls of $d_1,d_2,d_\infty$

Recall that if $(S,d_s)$ and $(T,d_T)$ are metric spaces, then the functions

$$\begin{align*}
d_1(\mathbf x,\mathbf y)&=d_s(x_1,y_2)+d_T(x_2,y_2)\\[5pt]
d_2(\mathbf x,\mathbf y)&=\sqrt{d_s(x_1,y_2)^2+d_T(x_2\color{red}{,}y_2)^2},\text{and}\\[5pt]
d_\infty(\mathbf x,\mathbf y)&=\text{max}(d_s(x_1,y_2)+d_T(x_2,y_2))\\[5pt]
\end{align*}$$

are all metrics on $\color{red}{S \times T}.$

Let $r \gt 0$ and $\mathbf x \in \color{red}{S \times T}.$

Prove that there is some $s \gt 0$ so that $B^{d_1}_s(\mathbf x) \subseteq B^{d_2}_r(\mathbf x)$.

I'm not sure how to approach this. I tried framing it in the form $d_1(x,x+s) \Leftarrow d_2(x,x+r)$ but am getting nowhere with it.

答:
Given $r>0$, pick $s > 0$ so that $s^2 = \frac{1}{2}r^2$, explicitly we define $s=\sqrt{\frac{r^2}{2}}$.

If then $\textbf{y} \in B^{d_1}_s(\textbf{x})$ we know that

$$d_S(x_1, y_1) \le d_1(\textbf{x},\textbf{y}) < s$$

and also

$$d_T(x_2, y_2) \le d_1(\textbf{x},\textbf{y}) < s$$

So that

$$d_S(x_1, y_1)^2 + d_T(x_2, y_2)^2 < 2s^2 = 2\cdot \frac{1}{2}r^2 = r^2$$

and it follows that

$$ d_2(\textbf{x}, \textbf{y}) < r $$

and so $\textbf{y} \in B^{d_2}_r(\textbf{x})$, as required.

Title: Bounds for Balls of $d_1,d_2,d_\infty$


问:
Recurrence Equation and sequences
$a_n $ is a sequence defined this way:

and we define:

need to prove that this happens:

i got exam soon and a lot of simmilar questions but i have no idea where to start

Title: Recurrence Equation and sequences

Body:
From the concept of GP, note that $T_{n+1}=r T_n \implies T_n=r^n K.$
Here we have

$$3X_{n+1}=X_n+1.$$ Let $X_n=F_n+C$, then $$3F_{n+1}+3C=F_N+C+1/3, \implies 3F_{n+1}=F_n, C=1/6 \implies F_n =3^{-n} D$$
So we have $$X_n=3^{-n}D+1/6,$$ $X_1=3$ gives $D=17/2$, so we have
$$X_n=\frac{17}{2} 3^{-n}+\frac{1}{6}.$$

Next, $$Y_{n+1}=Y_n/3+1/9$$ LEt $Y_n=G_N+P$, then
$$ G_{n+1}+P=\frac{1}{3}G_n+\frac{P}{3}+\frac{1}{9} \implies G_{n+1}=\frac{1}{3} G_n, P=1/4$$
$$\implies Y_n= 3^{-n} Q+\frac{1}{4}$$
$$Y_1=a_2=1 \implies Q=
9/4 \implies Y_n=\frac{3^{2-n}+1}{4}$$


问:
Hamiltonian Paths and Next Neighbor Algorithm in Complete Graphs
In a complete graph, can the next neighbor algorithm (NNA) ever produce the most optimal Hamiltonian path? The NNA is close enough to the most optimal path to be used in real-life applications, but is there ever an instance in which it happens to also be the most optimal path?</p>

答:
Consider $K_4$ with edge weights as given in the following figure.

The optimal Hamiltonian path is given by the following subgraph

Then the nearest neighbor algorithm can indeed find this optimal Hamiltonian path, but since the choice of the start vertex depends on random choice, we have no guarantee. If we start at $3$, then the algorithm will visit $2$, then $4$ and then $1$. However, if we start at $1$, it will proceed to $2$, then to $4$ and then to $3$.


问:
Solve in the set of integers $2^x +5^x = 3^x + 4^x$

Find the number of integer solutions(both positive and negative) of the equation:
$$2^x +5^x = 3^x +  4^x$$

With induction we see that for $x\geq 2$ we have $$5^x\geq 3^x+4^x$$

It is trivially true for $x=2$ and $x=3$. Now say it is true for $x$ and prove it for $x+2$: $$5^{x+2} \geq 5^2(3^x+4^x) = (3^2+4^2)(3^x+4^x) > 3^{x+2}+4^{x+2}$$

So equation has no solution for $x\geq 2$. Clearly $x=0$ and $x= 1$ are solutions.

If $x<0$ you can write $x=-n$ where $n$ is positive integer and we get $$3^n\cdot 4^n(2^n+5^n) = (3^n+4^n)2^n\cdot 5^n$$
so $ 3^n\mid 40^n$ which is clearly nonsense and thus no solution (for negative $x$).

Any easier solution?

答:
You can show $5^n\ge 3^n+4^n$ for $n\ge2$ faster: $5^2\ge 3^2+4^2$ and in the induction step
$$ 5^{n+1}=5\cdot 5^n\ge 5\cdot3^n+5\cdot4^n>3^{n+1}+4^{n+1}.$$
An alternative idea for negative $x$ s also by size comparison with $2^{-n}$ being quite big


问:
There is a metric on N that is not equivalent to the discrete metric
Show that there is a metric on N that is not equivalent to the discrete metric. Any hint or answer would be appreciated.

Thank you in advance.

答: Let $f: \Bbb N \to \Bbb Q$ be a bijection. Define $d(n,m)=| f(n)-f(m)|$.


问:
I am trying to prove gcd$(8,20) = 4$
I am trying to prove $\gcd(8,20) = 4$.

I am not quite sure how to go about it, but I think I need to prove that $\gcd(8,20)$ is not equal to $1$.
I have also set up $S = \{k \in \mathbb{N} : k = 8x + 10y \text{ for some } x, y \in \mathbb{Z}\}$. I am trying to somehow show that $4 \in S$.

Any suggestions on how I can prove my approaches? This was a personal challenge exercise, so I am just trying to further my learning.
I did take a look at other relevant posts on here, but they are too complex or advanced in comparison to where I am at. I just learnt how to do basic proofs with axioms and proofs for natural numbers. This exercise is given near the end of practice exercises after introduction on the Well Ordering Principle and [GCD].

Thanks for any help or solutions posted!

答: If $d$ divides both $a$ and $b$, then $d=\gcd(a.b)$ iff $\gcd ({a\over d}, {b\over d})=1.$ Therefore, you only need to find integers $x$ and $y$ so that $2x+5y=1.$

Title: I am trying to prove gcd$(8,20) = 4$


问:
If every point of $S$ is a limit point of $S$ and $S$ is closed, then we say that $S$ is.....?
I have a set $S \subset \mathbb{C}$ such that

$x \in S \implies x$ is a limit point of $S$

and

$S$ is closed

Can I conclude that $S$ is connected?

答:
your set is a perfect set. But not necessarily connected.

– amWhy Apr 23 at 4:27

For example, the union of two disjoint closed intervals has the properties in your question, but is not connected.

Connectedness stands apart from the properties such as bounded, closed, open, compact, having no isolated points... you will not get connectedness from any combination of those.

ContentLicense: CC BY-SA 3.0


问:
Sum of preimages of subspaces
Let $E$, $F$ be vector spaces over a field $K$.

Let $u$ be a linear map between $E$ and $F$ and $F_1,F_2,\dots, F_n$ subspaces of $F$.

It's quite easy to show that $u^{-1}(F_1)+\dots+u^{-1}(F_n) \subset u^{-1}(F_1+F_2+\dots +F_n)$

But do we have $u^{-1}(F_1+F_2+\dots +F_n) \subset u^{-1}(F_1)+\dots+u^{-1}(F_n)$ ?

I'm quite sure that the answer is no but I can't find a counterexample.

Comments :

Of course if $u$ is surjective, then the equality follows. So we need to find some subspaces such that $F_i \not \subset \text{im} u$...
I found no counterexample between $K^n$ to $K^p$...

Any ideas ?

答:
Here  is a counterexample.

Let $E:=\mathbb{R}$, $F:=\mathbb{R}^2$, and $u:E\to F$ be the linear map $t\mapsto (t,t)$.

Let $F_1=\{(x,0)\}$, $F_1=\{(0,y)\}$.

Then $u^{-1}(F_1)=\{0\}$, $u^{-1}(F_2)=\{0\}$ but $u^{-1}(F_1+F_2)=u^{-1}(F)=E$.

[This would work for any field, not just $\mathbb{R}$.]

Title: Sum of preimages of subspaces


问:
What is $(\mathbb R^3)^*$
Can anyone please tell me what is $(\mathbb R^3)^*$  in the following question ?

答:
$(\Bbb R^3)^\ast$ is the dual of $\Bbb R^3$: the vector space of all linear maps $\Bbb R^3 \to R$. So a "vector" (generic term for an element of a vector space) in this dual is a linear map from $\Bbb R^3$ to $\Bbb R$.

To show this fact, extend $\{u,v\}$ to a basis $\{u,v,w\}$ of $\Bbb R^3$ and define
$f: \Bbb R^3 \to \Bbb R$ by
$f(x_1 u + x_2 v + x_3 w) = x_3$ and check it is linear. What is its null space?

Title: What is $(\mathbb R^3)^*$


问:
Can $ \int \sin(x)+\sum^{\infty}_{n=1} (-1)^n\frac{\sin^{(2n+1)}(x)}{(2n+1)!} dx$ be evaluated in terms of elementary functions?
How do you go about integrating this and can it even be done with elementary functions?
$$
\int \sin(x)+\sum^{\infty}_{n=1} (-1)^n\frac{\sin^{(2n+1)}(x)}{(2n+1)!} dx.
$$

I understand the concept of $$ \int \sum^{\infty}_{n=1}\sin^n (x) dx =\int \frac{dx}{1-\sin x}$$ but I have no clue if that will even help but its the first idea I had to try and clean it up or even how to use it to clean it up.

Additionally this is very similar to integrating $ \int \sin (\sin (x)) $ since the sum is so close to the Taylor series summation of $\sin (x)$ the only difference being $\sin (x)$ is just $x$, not sure if that would help but I'm trying to add as much as I know.

答:
$$\sum^{\infty}_{n=1} (-1)^n\frac{t^{(2n+1)}}{(2n+1)!}= \sin (t)-t$$

So, as you noticed, the problem is "simply" to compute
$$I=\int \sin(\sin(x))\,dx$$ which cannot be obtained aven using special functions.

If it was for definite integrals, the problem would be different since
$$\int_0^{\frac \pi 2} \sin(\sin(x))\,dx=\frac{\pi  }{2}\pmb{H}_0(1)$$ where appears  the Struve function.

But, from the answer to this question, we have, using Fourier series,
$$\color{blue}{\sin(\sin(x)) = 2\sum_{k=0}^\infty J_{2k+1}(1)\sin \left((2k+1)x\right)}$$ which makes
$$I=-2\sum_{k=0}^\infty \frac{ J_{2 k+1}(1) }{2 k+1}\cos ((2 k+1) x)$$
and the coefficients vary so fast that very few terms would be sufficient for a definite integral.

For example, using only three terms and integrating from $0$ to $\frac \pi 2$ would give $0.8932437398$ while the exact value given above is $0.8932437410$.


问:
What does Van Dalen mean when he writes "Show by 'algebraic' means"?
In Exercise 1.3.1, Van Dalen asks us to verify certain properties like contraposition, Peirce's Law, and several others by 'algebraic' means. What does he mean by this? My first instinct is to use a truth table.</p>

答:
He very likely means: derive them as identities in a Boolean algebra using the defining laws of such structures. Note he mentions such laws before and in Thm 1.3.1 as well.

E,g contraposition: $\phi \to \psi$ is (definitonally) equivalent to $\lnot \phi \lor \psi$ while $\lnot \psi \to \lnot \phi$ is equivalent to $\lnot \lnot \psi \lor \lnot \phi$ and these Boolean formulas are again equal to each other by commutativity of $\lor$ and double negation.


问:
Is $\sup_{\{x \in S : |f(x)|\leq |\inf_S f|\}} |f(x)| = |\inf_S f|$?
Let $f: \Bbb R \to \Bbb R$ be a bounded function and let $S \subset \Bbb R$. Suppose that the set
$$A=\{x \in S : |f(x)|\leq |\inf_S f|\}$$
is non empty.

Question: Is $\sup_A |f(x)| = |\inf_S f|$?

I was just able to prove that $\sup_A |f(x)|\leq |\inf_S f|$, since $|f(x)|\leq |\inf_S f|$ for all $x \in A$, but I can't get the reverse inequality.

答:
Yes.

If $\inf_S f = \min_S f$ then you also have $\sup_A |f| = \max_A |f| = |\min_S f|$, just by considering a point $x^*\in S$ such that $f(x^*)=\min_S f$.

Otherwise for all $x$ in $S$, you have $f(x) > \inf_S f$. Then for $A$ to be non-empty, you must have $\inf_S f < 0$.

Therefore there exists a negative decreasing sequence $f(x_n)$ converging to $\inf_S f$. It follows that $x_n \in A$ and $|f(x_n)|$ converges to $|\inf_S f|$.

Thus $\sup_A |f| = |\inf_A f|$.


问:
Show that the following f is integrable on every Jordan bounded set? (proof verifying)
Q: let $f:\mathbb{R}^n\to \mathbb{R}$ satisfy that for each open ball $U$ then $f$ is integrable (such that $\int|f|<\infty$)) over $U$ and $\int_{U}f=v(U)$ where v is the volume of $U$. Show that for every Jordan bounded set $E$ it is that $f$ is integrable over $E$ and $\int_{E}f=v(E)$

Jordan bounded sets are sets with negligible boundary.

Hey, I would really like to know if the following proof holds:

Let $E$ be a bounded Jordan set. for each $x\in E$ denote $B_x=B(x,r_x)$ where $r_x$ will be stated later.
then $\bigcup_{x\in E}B_x\supseteq E$.

Now, I claim that $ \{x\in E: \forall r. B(x,r)\cap E^\circ \neq \emptyset \land B(x,r)\cap E^{exterior}\neq \emptyset\}$ in negligible because each x that satisfies that belongs to $\partial E$ and hence this group is subset of negligible group and hence negligible itself.

Hence, for almost every x (except negligible set) we can choose $r_x$ such that $B_x\subseteq E$ and hence $\bigcup_{x\in E\\x\notin \partial E}B_x\subseteq E$ and therfore $$\int_{\bigcup_{x\in E\\x\notin \partial E}B_x}f = \int_Ef =(*)$$ and because $\int|f|<\infty$ over every open ball, then we got integrablity of $f$ over $E$

Second part: I claim that for each open set $U$ then $\int_Uf=v(U)$ because we can take an open ball $B$ such that $U\subseteq B$ and define $\tilde{f} = \cases{f\;\;x\in U\\0\;\;x \in B/U}$ and hence it is that $\int_B\tilde{f}=\int_Uf=v(U)$

Now we say that $(*)=v(\bigcup_{x\in E\\x\notin \partial E}B_x) = v(E)$ because the difference between the sets is negiligible.

Is my proof right? I am especially wondering if the claim in the second part is actually true and well explained?

Appreciate any help!

答:
Formal solution of what I have pointed in the comments:

Proof of the integrability of $f$:

claim: $\mathcal{B}_f$ (denotes the set of points where $f$ is not continous) is negligible.

proof: Indeed, $\mathcal{B}_f = \bigcup_{i=1}^{\infty}\mathcal{B}_{f|_{B(0,i)}}$ is a negligible set as $f$ is integrable over each open ball (corollary from Lebesgue sentence).

remark: The above proof does not assume that countable union of negligible sets is negligible (it isn't necessarily), but that for each $i$ the set $\mathcal{B}_{f|_{B(0,i)}}$ is negligible, and each Jordan set $E$ is contained in such ball

corollary: Again, from Lebesgue $f$ is integrable over every bounded Jordan set.

Proof of the property of $f$:

claim: $f|_{\mathbb{R}^n\setminus\mathcal{B}_f}=1$

proof: Otherwise, there is continuity point ,$x$ of $f$ which satisfies (WLOG) that $f(x)>1$. Hence, there is an open neighborhood, $U_x$ of $x$ where $f(U_x)>1$ and as there can be found a ball, $B(x,\delta_{U_x})$ contained in $U_x$ by the linearity of Integrals: $\int_{B(x,\delta_{U_x})}f>\int_{B(x,\delta_{U_x})}1=vol(B(x,\delta_{U_x}))$ which contradicts the given.

corollary: $\int_E{f} = \int_{E\setminus\mathcal{B}f|_E}1+\int_{\mathcal{B}f|_E}f = vol(E)$


问:
Conditional probability of the sum of r.v.
I have $n$ independent random variables $X_i$ with known PDF and CDF (say, Normal, but not necessarily with the same parameters).

Given $U_1, U_2 \subseteq \{1,...,n\} $ such that $U_1 \cup U_2 = \{1,...,n\}$ and $U_1 \cap U_2 = \emptyset $ I need to evaluate the probability $\Gamma$ defined as:

$\begin{equation}
\Gamma = P(\sum_{i=1}^n X_i \geq A \;\mid \; X_j\leq \alpha \;\forall j \in U_1, X_k\geq \beta \;\forall k \in U_2 )
\end{equation}$

The only practical approach I can think of is based on Monte Carlo simulation.

My questions are:

How can I compute $\Gamma$ analytically, knowing the PDFs/CDFs, for any $n$?

Are there other ways (besides Monte Carlo) to approximate $\Gamma$?

(Later edit after the first answer) What if $\alpha$ and $\beta$ can be different for each r.v. (i.e. $X_j\leq \alpha_j, X_k \geq \beta_k$)?

Thanks

答:
Assuming  $X_i$ are iid, with PDF $f(x)$ and CDF $F(x)$, the conditioning leaves us with $n_1$ and $n_2$ variables with respective PDF:

$$f_1(x) = \frac{1}{F(\alpha)}f(x) \, [x \le \alpha]$$
$$f_2(x) = \frac{1}{1-F(\beta)}f(x) \, [x > \beta]$$

Let $Y=\sum X_i$, the PDF $f_Y$ is formally given by the $n_1$-convolution of $f_1(x)$ with $n_2$ convolution of $f_2(x)$. This can be expressed more compactly with characteristic functions $\psi_Y(w)= \psi_1(w)^{n_1} \psi_2(w)^{n_2}$. But in any case, the calculation can be difficult.

A trivial approximation would be, if $n$ is large, to apply the CLT; this would only require to compute the mean and variance of the truncated PDF $f_1(x)$, $f_2(x)$.


答:
Note that $\inf_S f$ is a real number.

Case 1: $\inf_S f<0$.

Let $x_n\in S$ be such that $f(x_n)$ decreases to $\inf_S f$. This exists by definition of infimum. As $f(x_n)<0$ eventually, $|f(x_n)|$ increases to $|\inf_S f|$. Thus $x_n\in A$, and $$\sup_A |f|\ge \lim |f(x_n)| = |\inf_S f|.$$

Case 2: $\inf_S f\ge 0$.

Then $x\in S$ implies $f(x)\ge 0$ so the definition of $A$ simplifies slightly
$$ A = \{ x\in S : f(x) \le \inf_S f\}.$$
Now $x\in A$ implies $x\in S$ and $f(x)\le \inf_S f$, therefore $f(x)=\inf_S f$. Thus $$\sup_A |f| = \sup_A f = \inf_S f = \inf_S |f|.$$

Title: Is $\sup_{\{x \in S : |f(x)|\leq |\inf_S f|\}} |f(x)| = |\inf_S f|$?

Body:
Let $f: \Bbb R \to \Bbb R$ be a bounded function and let $S \subset \Bbb R$. Suppose that the set
$$A=\{x \in S : |f(x)|\leq |\inf_S f|\}$$
is non empty.

Question: Is $\sup_A |f(x)| = |\inf_S f|$?

I was just able to prove that $\sup_A |f(x)|\leq |\inf_S f|$, since $|f(x)|\leq |\inf_S f|$ for all $x \in A$, but I can't get the reverse inequality.


问:
Prove that $[0,1]$ is a connected space.

Background

I know that there are so many proofs on this that can be found easily on the internet, but I am not familiar with limit points and so on, so I tried to prove it (getting help from a book for the main ideas) using only the basic definitions of open sets, with respect to the topology with open balls.

Below you can find my attempt to prove, and as indicated in the paper. I have 3 questions:

What if $x\gt z-\epsilon$?

What if $z+\epsilon/2 \gt1$?

What if $z-\epsilon/2 \lt 0$?

Of course, I know that the idea is that $\epsilon$ is a small number, but how can I formally show that the inequalities that I listed as $1,2,3$ do not hold? Lastly, what is the contradiction in Case 2? I do not really get it.

My Attempt

答:
More simply, show that if $A, B$ are non-empty disjoint open subsets of the space $[0,1],$ with $0\in A,$ then $A\cup B\ne [0,1],$ as follows:

There exists $r\in (0,1]$ such that $[0,r)\subseteq A,$ so $[0,r/2]\subseteq A.$ So the set $$C=\{s\in (0,1]: [0,s]\subseteq A\}$$ is a non-empty subset of $(0,1].$ So $$d=\sup C$$ exists. And $d\in (0,1].$ We claim that $d\not\in A\cup B.$ Proof of claim:

(i).$\; [0,d)\subseteq A.$ Because by def'n of $d=\sup C,$ we have $$x\in [0,d)\implies \exists s\in (x,d)\cap C\implies$$  $$ \implies (\exists s\in (x,d)\cap C \land x\in [0,s]\subseteq A)$$ $$\implies x\in A.$$ (ii). $\;d<1.$ Otherwise by (i), $B=\emptyset$ or  $B=\{1\},$ contrary to $B$ being open and non-empty.

(iii). $\;d\not\in A.$ Otherwise there exists $t>0$ such that $A\supseteq (d-t,d+t)\cap [0,1]$ but then by (i) and (ii) we would have $A\supseteq [0,d)\cup (\, (d-t,d+t)\cap [0,1]\,)\supseteq [\,0, \min (1, d+(t/2)\,)\,],$ implying (by def'n of $C$) that $d=\sup C\ge \min (1, d+(t/2)\,)>d,$ which is absurd.

(iv). $\;d\not \in B.$ Otherwise there exists $u>0$ with $(d-u,d+u)\cap [0,1]\subseteq B.$ But $d>0$ so if $d\in B$ then by (i) we would have $A\cap B\supseteq [0,d)\cap (d-u,d+u)\cap [0,1]\supseteq (\max (0,d-u), d) \ne\emptyset,$ contrary to the hypothesis that $A,B$ are disjoint.


问:
inequality of $(ax+by+cz)^n$
The Cauchy Schwarz inequality says
$$
(ax+by+cz)^2 = (a^2+b^2+c^2)(x^2+y^2+z^2).
$$

I found that there is a kind of analogous inequality for $(ax+by+cz)^n$
$$
(x+y+z)^n \leq 3^n(x^n+y^n +z^n).
$$
if I remember it correctly.

How to prove this inequality?

Title: inequality of $(ax+by+cz)^n$

Body:
We have, for $x, y, z \ge 0$,
$$(x + y + z)^n \le 3^n \max\{x, y, z\}^n \le 3^n(x^n + y^n + z^n).$$


问:
Accessible Proof of Denjoy-Riesz Theorem?
The Denjoy-Riesz Theorem states that any compact zero-dimensional subset of the plane is contained in the image of an embedded copy of the arc $[0,1]$.  The only place I've seen this proved is in Kuratowski's Topology II, and it's proved as a trivial consequence of a huge body of theory on unicoherence, Janiszewski spaces, characterizations of the sphere etc.

The "original proof" by Riesz is known to have flaws.  Actually I've also seen it proved in a paper by Moore, but there he proves a generalization and it's atrocious to read as well.

Is there a more self-contained, modern proof accessible anywhere?

答:
Ok, I think the reason that this oft-cited result isn't proven in many books is because a stronger theorem was later proven using means that weren't really any more sophisticated, but the proof is still pretty long.  The only place besides Kuratowski that I found this is Moise's "Geometric Topology in Dimensions 2 and 3" where he proves the following, stronger theorem instead:

Let $C_1$ and $C_2$ be Cantor sets in the plane, and let $f: C_1 \rightarrow C_2$ be a homeomorphism.  Then there is a homeomorphism $\tilde{f}: \mathbb{R}^2 \rightarrow \mathbb{R}^2$ extending $f$.

The proof of this comprises chapters 12 and 13 of his book.  Let $I = [0,1]$.  I'm lazy, but sometime in the next few days I'll post my simpler proof of the following, ostensibly stronger theorem as another answer:

Let $C_1$ and $C_2$ be Cantor sets in the plane, and let $f: C_1 \rightarrow C_2$ be a homeomorphism.  Then there is an ambient isotopy $F: \mathbb{R}^2 \times I \rightarrow \mathbb{R}^2$ sending $C_1$ to $C_2$ which equals $f$ on $C_1$ at time $t=1$.

To be fair to Moise this is implied by other theorems he proves elsewhere in the book, namely that any stable homeomorphism of the plane (i.e. one that leaves some open set in $\mathbb{R}^2$ fixed) is isotopic to the identity.  The construction he uses produces a stable homeomorphism - but then you also need the theorem on stable homeomorphisms.

Either of these theorems immediately implies the Denjoy-Riesz Theorem, since we can cover the usual middle-thirds Cantor set with an arc and then isotope the arc onto any Cantor set.

Either theorem can be restated for compact, totally disconnected sets (equiv. zero-dimensional or totally separated) since each resides in a Cantor set.

No such set separates the plane, since no subset of an arc does.

For those who were curious, a different generalization by Moore and Kline goes as follows:

A compact set $X$ in $\mathbb{R}^2$ can be covered by an arc iff each component of $X$ is a point or a closed arc $A$ such that if $a \in A$ is a limit point of $X \setminus A$ then $a$ is an endpoint of $A$.

The proof is nigh-unreadable and makes references to many technical results in other papers.  I will try to simplify this one as well, but my technique seems ill-suited to it without something smart.  Maybe you can help once you see my method for the isotopy extension version.


问:
For what $k$ are the numbers $4n+1, kn+1$ coprime?
Already knowing the answer gives me sort of a hint ($k=\pm2^m+4 : m \in  \mathbb{N} \cup \{0\}$), but I can't really find a way through it.

Bonus question: how about a generalisation of this problem? When we are given any natural number m instead of 4?

答:
The two facts that you need to know:

$\gcd(a, b) = \gcd(a, b + ac)$.

If $\gcd(a, b) = 1$, then $\gcd(a, bc) = \gcd(a, c)$.

Combining these, we have:
$$\gcd(4n + 1, kn + 1) = \gcd(4n + 1, (k - 4)n) = \gcd(4n + 1, k - 4).$$
The first equality comes from 1, the second from 2 and the fact that $\gcd(n, 4n + 1) = \gcd(n, 1) = 1$, which again comes from 1.

The rest should be quite clear. You may also work out your generalized problem along the same lines.


问:
Random Variable Multiplied with Independent Random Vector - Derivation of Product Distribution
Let $x$ be a random variable and $\mathbf{Y}$ be a random vector. Is there a general formula to derive the probability density function of
$$
\mathbf{Z} = \frac{1}{\sqrt{x}} \mathbf{Y} \; ?
$$

I could only find the rule for the case where $\mathbf{Y}$ is also a random variable (Wikipedia).

FYI:
In the end I would like to use it for the case where $x \sim \Gamma(k = \nu/2, \theta = 2 / \nu)$ and $\mathbf{Y} \sim N(\boldsymbol{\mu}, \mathbf{\Sigma})$ to arrive at the multivariate t-distribution. And I know about the easier derivation in this case by utilizing $\int_0^{\infty} f(\mathbf{Z} | x) f(x) dx$, but I just want to do it another way just for fun.

答:
The formula should be
$$
\begin{align}
 f(\mathbf{Z})
 &= \int_{-\infty}^{\infty} f_x(x) f_\mathbf{Y}(\mathbf{Z} \sqrt{x}) x^{p/2} dx,
\end{align}
$$
where the $x^{p/2}$ part is the Jacobian of the transformation from $\mathbf{Y}$ to $\mathbf{Z}$, $J(\mathbf{Y} \rightarrow \mathbf{Z})$.

Using this formula to derive the multivariate-$t$ distribution from its stochastic representation given in the question above we get
$$
\begin{align}
 f(\mathbf{Z})
 &= \int_{-\infty}^{\infty} f_x(x) f_\mathbf{Y}(\mathbf{Z} \sqrt{x}) x^{p/2} dx, \\
   &= \int_{0}^{\infty} \frac{1}{\Gamma(\nu/2) (2/\nu)^{\nu/2}} x^{\nu/2-1} \exp(- x \nu/2) \frac{\exp\left(- \frac{\mathbf{Z}' \sqrt{x} \boldsymbol{\Sigma}^{-1} \mathbf{Z} \sqrt{x}}{2}\right)}{(2 \pi)^{p/2} |\boldsymbol{\Sigma}|^{1/2}} x^{p/2} dx \\
   &= \frac{1}{\Gamma(\nu/2) (2/\nu)^{\nu/2} (2 \pi)^{p/2}} |\boldsymbol{\Sigma}|^{-1/2} \int_{0}^{\infty}  x^{\frac{\nu + p}{2} - 1} \exp\left[- x\frac{\nu}{2} \left(1 + \frac{\mathbf{Z}' \boldsymbol{\Sigma}^{-1} \mathbf{Z}}{\nu}\right)\right] dx \\
   &= \frac{1}{\Gamma(\nu/2) (2/\nu)^{\nu/2} (2 \pi)^{p/2}} |\boldsymbol{\Sigma}|^{-1/2} \frac{\Gamma\left((\nu + p)/2\right)}{\left(\frac{\nu}{2}\right)^{(\nu + p)/2} \left(1 + \frac{\mathbf{Z}' \boldsymbol{\Sigma}^{-1} \mathbf{Z}}{\nu}\right)^{(\nu + p)/2}} \\
   &= \frac{\Gamma\left((\nu + p)/2\right)}{\Gamma(\nu/2) \nu^{p/2} \pi^{p/2}} |\boldsymbol{\Sigma}|^{-1/2} \frac{\nu^{\nu/2} 2^{(\nu + p)/2}}{2^{\nu/2} 2^{p/2} \nu^{\nu/2}}  \left(1 + \frac{\mathbf{Z}' \boldsymbol{\Sigma}^{-1} \mathbf{Z}}{\nu}\right)^{-(\nu + p)/2} \\
   &= \frac{\Gamma\left((\nu + p)/2\right)}{\Gamma(\nu/2) \nu^{p/2} \pi^{p/2}} |\boldsymbol{\Sigma}|^{-1/2} \left(1 + \frac{\mathbf{Z}' \boldsymbol{\Sigma}^{-1} \mathbf{Z}}{\nu}\right)^{-(\nu + p)/2},
\end{align}
$$
where in the third step can be checked here.


问:
Is integral of greater function still greater?
If I have $f(x) \geq g(x) $ for all $x$ over $[a;b]$, does it imply that $\int_a^b f(x)dx \geq \int_a^b g(x)dx$?
Or should any other requirements be added? Such as $f(x)$ should be convex or positive, etc...

答: Define $h(x):=f(x)-g(x)$ so $h\ge0$, and your desired conclusion is $\int_a^bhdx\ge0$. Indeed, this integral is an increasing function of $b$ that's $0$ if $b=a$, and so is $\ge0$ for all $b\ge a$.


答:
The definite integral operator is linear, so

$$\int_a^b f(x)\,dx\ge\int_a^b g(x)\,dx\iff\int_a^b (f(x)-g(x))\,dx\ge0.$$

Then

$$f(x)\ge g(x)\implies f(x)-g(x)\ge 0$$ and the integral of a non-negative function cannot yield a negative result.

Title: Is integral of greater function still greater?

Body:
If I have $f(x) \geq g(x) $ for all $x$ over $[a;b]$, does it imply that $\int_a^b f(x)dx \geq \int_a^b g(x)dx$?
Or should any other requirements be added? Such as $f(x)$ should be convex or positive, etc...


问:
dimension of kernel of linear transformation on vector space of matrix
Let $M_n(\mathbb{C})$ be the set of $n \times n$ matrices thought of as vector space of $n^2$ dimension over $\mathbb{C}$. Let $F_A: M_n(\mathbb{C}) \rightarrow M_n(\mathbb{C})$ be defined by $B \mapsto AB$.  Show that $\ker(F_A)$ for any $A$ has dimension $kn$ for some $k \in \mathbb{N}$

What I have tried that is that since $\ker(F_A)$ is finite dimensional, it is isomorphic to a subspace spanned by some of the canonical basis of $M_n(\Bbb C)$, which I call $B_{ij}$, then I showed that $AB_{ij}=0$ iff $i^{th}$ column of $B$ is 0, which tells me that the $i^{th}$ row of any matrix in the kernel must be free, but then I cannot finish the argument.

答:
For a matrix $B$, let $B_i$ denote the $i^{th}$ column of $B$. We can think of a matrix as just a tuple of its column vectors, ie $B \in M_n (\mathbb C)$ is thought of as $(B_1, \dots, B_n) \in (\mathbb C^n)^n$. We do this because the columns of $AB$ are found by applying $A$ to each column of $B$, so $F_A(B_1, \dots, B_n) = (AB_1, \dots AB_n)$. Then for $F_A(B) = 0$, we must have $A B_i = 0$ for all $i$. That is, each $B_i$ must be in $ker(A)$ where $A$ is thought of as a linear map on $\mathbb C^n$.

Since $B \in ker(F_A)$ if and only if each column $B_i \in ker(A)$, we can identity $ker(F_A)$ as $(ker(A))^n$. Again, we are viewing a matrix in $M_n(\mathbb C)$ as its tuple of columns in $(\mathbb C^n)^n$. Let $k = dim(ker(A))$. Then in conclusion, $dim(ker(F_A)) = dim(ker(A)^n) = kn$.


问:
First moment inequality implies tail distribution inequality?
Let $U,V$ be two continuous random variables, both with continuous CDF. Suppose that $\mathbb E V \geq \mathbb E U$. Can one conclude that $\mathbb P(V> x) \geq \mathbb P(U>x)$ for all $x\geq 0$? If not, what additional conditions are needed?</p>

答:
Take for example $U \sim N(0,2)$, $V \sim N(1,1)$, i.e. both normally distributed but with different means and variances.

Then $\mathbb{E} V = 1 > 0 = \mathbb{E} U$, but since the variance of $U$ is larger than that of $V$, after some point $x_0$ you'll have that $\mathbb{P}(U > x) > \mathbb{P}(V > x)$.


问:
Proof by induction: Inductive step struggles
Using induction to prove that:
$$
1 - \frac{1}{2} + \frac{1}{4} - \frac{1}{8} +...+\left( \frac{1}{2}\right)^{n} = \frac{2^{n+1}+(-1)^{n}}{3\times2^{n}}
$$

where $ n $ is a nonnegative integer.

Preforming the basis step where $ n $ is equal to 0

$$
1 = \frac{2^{1}+(-1)^{0}}{3\times2^{0}} = \frac{3}{3} = 1
$$
Now the basis step is confirmed.

Then I started the inductive step where $ n = k $ is assumed true and I needed to prove $ n = k+1 $
$$
1 - \frac{1}{2} + \frac{1}{4} - \frac{1}{8} +...+\left(- \frac{1}{2}\right)^{k} + \left(- \frac{1}{2}\right)^{k+1} = \frac{2^{k+1+1}+(-1)^{k+1}}{3\times2^{k+1}}
$$
Using the inductive hypothesis
$$
\frac{2^{k+1}+(-1)^{k}}{3\times2^{k}} + \left(- \frac{1}{2}\right)^{k+1} = \frac{2^{k+1+1}+(-1)^{k+1}}{3\times 2^{k+1}}
$$
After this I am struggling here trying to get around to the end. I would appreciate any guidance.

Body:
Multplying numerator and denominator of the first term by $2$ we get $$
\frac{2^{k+1}+(-1)^{k}}{3*2^{k}} + (- \frac{1}{2})^{k+1} =\frac{2^{k+2}+2(-1)^{k}}{3*2^{k+1}} + (- \frac{1}{2})^{k+1} =\frac  {2^{k+2}+2(-1)^{k}+3(-1)^{k+1}} {3*2^{k+1}}
$$ Now use the fact that $2(-1)^{k}+3(-1)^{k+1}=(-1)^{k+1}$ (which can be checked by considering the case $k$ even and $k$ odd separtely).


Title: Proof by induction: Inductive step struggles

Body:
Using induction to prove that:
$$
1 - \frac{1}{2} + \frac{1}{4} - \frac{1}{8} +...+\left( \frac{1}{2}\right)^{n} = \frac{2^{n+1}+(-1)^{n}}{3\times2^{n}}
$$

where $ n $ is a nonnegative integer.

Preforming the basis step where $ n $ is equal to 0

$$
1 = \frac{2^{1}+(-1)^{0}}{3\times2^{0}} = \frac{3}{3} = 1
$$
Now the basis step is confirmed.

Then I started the inductive step where $ n = k $ is assumed true and I needed to prove $ n = k+1 $
$$
1 - \frac{1}{2} + \frac{1}{4} - \frac{1}{8} +...+\left(- \frac{1}{2}\right)^{k} + \left(- \frac{1}{2}\right)^{k+1} = \frac{2^{k+1+1}+(-1)^{k+1}}{3\times2^{k+1}}
$$
Using the inductive hypothesis
$$
\frac{2^{k+1}+(-1)^{k}}{3\times2^{k}} + \left(- \frac{1}{2}\right)^{k+1} = \frac{2^{k+1+1}+(-1)^{k+1}}{3\times 2^{k+1}}
$$
After this I am struggling here trying to get around to the end. I would appreciate any guidance.

Body:
\begin{align}
\frac{2^{k+1}+(-1)^k}{3 \times 2^k}+\left(-\frac{1}{2}\right)^{k+1}
&=\frac{2 \times 2^{k+1}+2 \times(-1)^k}{3 \times 2^{k+1}}+\frac{3 \times (-1)^{k+1}}{3 \times 2^{k+1}} \\
&=\frac{2^{k+2}+2 \times (-1)^k+3 \times (-1)^{k+1}}{3 \times 2^{k+1}} \\
&=\frac{2^{k+2}-2 \times (-1)^{k+1}+3 \times (-1)^{k+1}}{3 \times 2^{k+1}} \\
&=\frac{2^{k+2}+(-1)^{k+1}}{3 \times 2^{k+1}}
\end{align}


问:
Show that $\lim_{x\to \alpha_{i}^{\pm}}\frac{p_n(x)}{p_{n+1}(x)}=\pm \infty$ where $p_n\in \mathbb{R}[x]$ of degree $n$
Let $p_n\in \mathbb{R}[x]$ of degree $n$ with positive leading coefficient, having $n$ simple real roots. Let $\alpha_1<\dots <\alpha_{n+1}$ denote the real roots of $p_{n+1}$. If $p_n/p_{n+1}$ is strictly decreasing, why is it true that
$$
\lim_{x\to \alpha_{i}^{\pm}}\frac{p_n(x)}{p_{n+1}(x)}=\pm \infty
$$
for $i=1,\dots, n+1$?

Edit: It is assumed that $p_n$ and $p_{n+1}$ have no common zeroes.

答:
Let $a$ be a root of $p_{n+1}$. Then $p_n(a) \ne 0 = p_{n+1}(a)$, so that $\lim_{x\to a}\left | \frac{p_n(x)}{p_{n+1}(x)} \right |= \infty$.

Therefore it suffices to prove the following:

Let $f:(a, b) \to \Bbb R$ be a decreasing function.

If $\lim_{x \to a^+} |f(x)| = \infty $ then $\lim_{x \to a^+} f(x) = +\infty $.

If $\lim_{x \to b^-} |f(x)| = \infty $ then $\lim_{x \to b^-} f(x) = -\infty $.

Here it suffices to prove the first part, the second part works in the same way. The idea is that $f$ cannot approach $-\infty$ for $x$ close to $a$ since it is decreasing, therefore it must approach $+\infty$. A formal proof could look like this:

Let $c = (a+b)/2$ and $K_0 = |f(c)|$. Then
$$ \tag{*}
 f(x) \ge f(c) \ge -|f(c)| = -K_0
$$
for $a < x \le c$.

We have $\lim_{x \to a^+} |f(x)| = \infty $, so for all $K > K_0$ there is  a $\delta \in (0, c-a)$ such that for all $x \in (a, a + \delta)$:
$$
\begin{align}
 |f(x)| > K &\iff \left( f(x) < -K \text{ or } f(x) > K\right) \\
&\implies f(x) > K
\end{align}
$$
where the last implication holds because $f(x) < -K$ is not possible, according to $(*)$.

This proves that $\lim_{x \to a^+} f(x) = +\infty$.


问:
Real analysis - Problem on Continuity
Is it possible to have a function on a metric space which is discontinuous at every point of the metric
space but the restriction of that function on a dense set is continuous?</p>

答:
Yes $f: \Bbb R \to  \Bbb R$ defined by: $$f(x) = \begin{cases} 1 & x \in \Bbb Q\\
                            0 & x \notin \Bbb Q\\
\end{cases}$$

is everywhere discontinuous but constant (so continuous) on both dense sets $\Bbb Q$ and $\Bbb R\setminus \Bbb Q$.

Title: Real analysis - Problem on Continuity


问:
Covariance, matrix inverse, and second order derivatives of log-likelihood
On page 35 of the book Analysis of survival data by D.R. Cox and D. Oakes, I see the following idea: the observed information matrix is defined as the matrix of minus the second derivatives of $l,$ the log likelihood of getting a certain value of parameters. (We are working with a parameter family of distributions, and want to perform hypothesis testing on these parameters $\phi$ (with is a vector) base on experiments.) If we write $v_{\omega \omega}$ for the leading submetrix of the inverse of the observed information matrix (here $\omega$ denotes the vector formed by components of $\phi$ of particular interest, and we extract submatrix at the rows/columns corresponding to these entries).

Then, the book says:  $v_{\omega \omega}$ can be regarde as the estimated covariance matrix of $\hat \omega,$ the maximum-likelihood estimate of $\omega.$

Why is this the case?

答:
The basic logic is first derive the asymptotic distribution of MLE estimator, then use the so called 'information equality'.

Suppose we have $n$ iid samples $\left\{z_{i}\right\}_{1\leq i\leq n}$ each following true distribution with a pdf given by $g\left(z_{i}\right)$. As mentioned, we are working with a parameter family of distributions, further assume $g\left(\cdot\right)=f\left(\cdot,\phi_{0}\right)$, where function $f$ is known to us, and $\phi$ takes value in a compact set with $\phi_{0}$ being an interior point.

We derive MLE estimator from a moment condition perspective. Since $f\left(\cdot,\phi_{0}\right)$ is a pdf, we have
$$\int f\left(z,\phi_{0}\right) dz=1,$$
differentiating both sides w.r.t. $\phi$ gives
$$\int\frac{\partial f\left(z,\phi_{0}\right)}{\partial \phi}dz=0,$$
multiply and divide both sides by $f\left(z,\phi_{0}\right)$ gives (here we can choose the domain of integral to avoid the denominator becoming $0$, this won't affect the identity)
$$\int\frac{\partial \log f\left(z,\phi_{0}\right)}{\partial \phi}f\left(z,\phi_{0}\right)dz=0, \tag{$\ast$}$$
which is (all expectations are taken w.r.t. the true probability measure $f\left(z,\phi_{0}\right)$)
$$\mathbb{E}\left[\frac{\partial\log f\left(z,\phi_{0}\right)}{\partial \phi}\right]=0.$$
Apply the traditional sample analogue idea, we want to choose $\hat{\phi}$ such that
$$\frac{1}{n}\sum_{i=1}^{n}\frac{\partial \log f\left(z_{i},\hat{\phi}\right)}{\partial \phi}=0,$$
clearly, this is same as FOC of fundamental MLE objective function. Assuming twice continuous differentiability of the log-likelihood, the (element-wise) mean-value theorem gives
$$\frac{1}{n}\sum_{i=1}^{n}\frac{\partial \log f\left(z_{i},\phi_{0}\right)}{\partial \phi}+\left(\frac{1}{n}\sum_{i=1}^{n}\frac{\partial^{2}\log f\left(z_{i},\bar{\phi}\right)}{\partial \phi\partial \phi^{\top}}\right)\left(\hat{\phi}-\phi_{0}\right)=0,$$
where $\bar{\phi}$ lies 'between' $\hat{\phi}$ and $\phi_{0}$. Multiplying by $\sqrt{n}$ and solving for $\sqrt{n}\left(\hat{\phi}-\phi_{0}\right)$ gives
$$\sqrt{n}\left(\hat{\phi}-\phi_{0}\right)=-\left(\frac{1}{n}\sum_{i=1}^{n}\frac{\partial^{2}\log f\left(z_{i},\bar{\phi}\right)}{\partial \phi\partial \phi^{\top}}\right)^{-1}\frac{1}{\sqrt{n}}\sum_{i=1}^{n}\frac{\partial \log f\left(z_{i},\phi_{0}\right)}{\partial \phi}.$$
Denote $H=\mathbb{E}\left[\frac{\partial^{2}\log f\left(z_{i},{\phi}_{0}\right)}{\partial \phi\partial \phi^{\top}}\right]$ be the negative information matrix, and $V=\mathbb{E}\left[\frac{\partial \log f\left(z_{i},\phi_{0}\right)}{\partial \phi}\frac{\partial \log f\left(z_{i},\phi_{0}\right)}{\partial \phi^{\top}}\right]$ be the variance matrix. A mixture use of LLN, continuous mapping theorem, and CLT gives the following results
$$\left(\frac{1}{n}\sum_{i=1}^{n}\frac{\partial^{2}\log f\left(z_{i},\bar{\phi}\right)}{\partial \phi\partial \phi^{\top}}\right)^{-1}\overset{\mathbb{P}}{\to}H^{-1} \quad \text{ and }\quad \frac{1}{\sqrt{n}}\sum_{i=1}^{n}\frac{\partial \log f\left(z_{i},\phi_{0}\right)}{\partial \phi}\overset{d}{\to}\mathcal{N}\left(0,V\right),$$
hence by Slutsky theorem,
$$\sqrt{n}\left(\hat{\phi}-\phi_{0}\right)\overset{d}{\to}\mathcal{N}\left(0,H^{-1}VH^{-1}\right).$$
What remains is to prove $H^{-1}VH^{-1}=-H^{-1}$. We prove this by showing $H=-V$, recall equation ($\ast$), differentiate both sides w.r.t. $\phi^{\top}$ gives
$$\int\left(\frac{\partial^{2}\log f\left(z,\phi_{0}\right)}{\partial \phi\partial\phi^{\top}}f\left(z,\phi_{0}\right)+\frac{\partial \log f\left(z,\phi_{0}\right)}{\partial \phi}\frac{\partial \log f\left(z,\phi_{0}\right)}{\partial \phi^{\top}}f\left(z,\phi_{0}\right)\right)dz=0,$$
which exactly implies $H+V=0$ as required.


问:
Rate of convergence to zero of an integrand given its integral is convergent?
Let $h(x)$ be real-valued, monotonically-decreasing function on the non-negative reals such that $\int_{0}^{\infty} x^{k+1} h(x) \, dx$ is convergent for some positive integer $k$.

Does it follow that there exists $\beta > 0$ such that $h(x) = O\!\left( \frac{1}{x^{k+2+\beta}} \right)$ as $x \to \infty$?

Remark: Naively, $h(x)$ needs to decrease to zero sufficiently quickly to override the growth of $x^{k+1}$ ... but I'm worried about whether I've accidentally asserted the existence of a "slowest function with a converging integral" (which doesn't exist - this question and this question).

Why I am asking: I have a function $f(x)$ and am seeking to establish whether $\int_{0}^{\infty} x^{k} f(x) \, dx$ is convergent.  I applied integration by parts to the finite integral
$$\begin{align*}
  \int_{0}^{t} x^{k} f(x) \, dx
  &=
  \left[
    \frac{x^{k+1} f(x)}{k+1}
    -
    \int \frac{x^{k+1}}{k+1} f'(x) \, dx
  \right]_{0}^{t}
  \\&=
    \frac{t^{k+1} f(t)}{k+1}
    -
    \int_{0}^{t} \frac{x^{k+1}}{k+1} f'(x) \, dx
\end{align*}$$
From other information, I know that $f'(x)$ is monotonically decreasing, $\int_{0}^{\infty} x^{k+1} f'(x) \, dx$ is convergent and $f(t) \to 0$ as $t \to \infty$. (In fact in my particular application, $f'(x)$ turns out to be the tail probability of a cumulative distribution function on the non-negative reals.)
But I appear to need to know how quickly $f(t)$ tends to zero if I'm to conclude that the first term on the right-hand side has a finite limit as $t \to \infty$.  I'm hoping that this information can be deduced from the convergence of $\int_{0}^{\infty} x^{k+1} f'(x) \, dx$.

(This question is a refinement of an earlier one.  I am asking a new question as per guidance.  The condition that $h(x)$ is monotonically decreasing was added after I realized I forgot it, following comments by @user10354138.  Many thanks in advance.)

Edit: See this question & answer for a revised question that got to an answer that I applied.

答:
(Writing out the answer as indicated by @[Greg Martin].)

Let
$$
  h(x) = \begin{cases}
    \displaystyle\frac{1}{x^{k+2}(\log x)^2} & \text{if $x \geq e$}
  \\
    0 & \text{otherwise}
  \end{cases}
$$
Then for $x \geq e$, $h(x)$ is monotonically decreasing and
$$
  \int x^{k+1} h(x) \, dx
  =
  \int \frac{1}{x(\log x)^2} \, dx
  \underset{y \, = \, \log x}{=}
  \int \frac{1}{y^2} \, dy
  =
  -\frac{1}{y}
  =
  -\frac{1}{\log x}
$$
so
$$
  \int_{e}^{\infty} x^{k+1} h(x) \, dx
  =
  \left[ -\frac{1}{\log x} \right]_{e}^{\infty}
  =
  1
$$
hence $ \int_{0}^{\infty} x^{k+1} h(x) \, dx $ is convergent.  But there is no $\beta > 0$ for which $\log(x) = O\!\left(\frac{1}{x^\beta}\right)$ as $x \to \infty$.


问:
How to solve for the function $f(x)$ such that the area underneath is always equal to the arclength?
First, I set up my equation. The area is obviously $\int f(x) \space dx$. The arclength is $\int \sqrt{1+(\frac{dy}{dx})^2}  \space dx$. Thus, I have the equation $\int f(x) \space dx = \int \sqrt{1+(\frac{dy}{dx})^2}  \space dx$. Now I differentiatie both sides to get $f(x) = \sqrt{1+(\frac{dy}{dx})^2} $. I will now make some substitutions to make it look simpler like this. $y = \sqrt{1 +y'^2}$. This becomes $y^2 = 1+y'^2$. Since we have a function that is similar to its own derivative, I will guess the solution looks like $y = Ae^{kx}$. $y' = Ake^{kx}$. Pugging these values into our equation, I get $(Ae^{kx})^2 = (Ake^{kx})^2 + 1$ or  $A^2e^{2kx} = A^2k^2e^{2kx} + 1$. I am now stuck here, please help.

答:
So you got to
$$
y=\sqrt{1+y'^2}\implies 1=y^2-y'^2
$$
This is the equation for a hyperbola, so use hyperbolic coordinates $y=\cosh(u(t))$, $y'=\sinh(u(t))$. The derivative of the first equation combined with the second then implies for $y'\ne0$ that $u'(t)=1$, so $y(t)=\cosh(t+C)$.

The other case $y'(t)=\sinh(u(t))=0$ gives $y=1$, which is also a solution.


问:
why $\tilde{A}/A$ is isomorphic to $\mathbb{C}$?
When I am reading the answer of this question, I found the statement $\tilde{A}/A\simeq \mathbb{C}$ suspicious to me. Here $A$ is a non-unital C* algebra and $\tilde{A}$ is it unitization. I don't understand why $A$ is a maximal ideal in $\tilde{A}$, I thought $A$ is only an essential ideal. Can anyone explain why?

答:
Recall how the unitization is created: we have that $\tilde{A}=A\oplus\mathbb{C}$ as vector spaces and then we set $$(a,\lambda)\cdot(b,\mu)=(ab+\mu a+\lambda b, \lambda\mu)$$
for the multiplication operation and
$$(a,\lambda)^*=(a^*,\bar{\lambda})$$
for involution.
It is a standard proof that $\tilde{A}$ with these operations admits a (unique) $C^*$-norm under which it it complete and also $\|(a,0)\|_{\tilde{A}}=\|a\|_A$. So we have a $*$-isomorphism $A\cong A\oplus0\subset\tilde{A}$, so we identify $A$ with the set $A\oplus0=\{(a,0):a\in A\}$. Now since $(a,0)\cdot(b,\mu)=(ab+\mu a,0)$ we have that $A$ is an ideal in $\tilde{A}$.

Also, when $A$ is non-unital, $A$ is an essential ideal in $\tilde{A}$: let $(b,\mu)\in\tilde{A}$ be such that $(b,\mu)\cdot(a,0)=0$ for all $a\in A$. We will show that $(b,\mu)=(0,0)$. We have that $ba+\mu a=0$ for all $a\in A$. For $a=b^*$ we have that $bb^*=-\mu b^*$, so $\|b\|^2=\|bb^*\|=|\mu|\cdot\|b\|$. Therefore, if $b\neq0$ we have that $|\mu|=\|b\|$. Now take an approximate unit $(e_\lambda)$ of $A$ and note that $be_\lambda+\mu e_\lambda=0$ for all $\lambda$, so $\lim_{\lambda}e_\lambda=\frac{-b}{\mu}$. But this shows that $\frac{-1}{\mu}b$ is a unit of $A$, a contradiction. So $b=0$ and $\mu=0$ follows immediately.

Also, the quotient map $\pi:\tilde{A}\to\mathbb{C}$ defined by $\pi(a,\lambda)=\lambda$ is a well-defined $*$-homomorphism. Note that this is onto and that the kernel of $\pi$ is precisely $A$. So $\tilde{A}/A\cong\mathbb{C}$ indeed.


问:
Question(s) regarding understanding of denseness
I think I have some problem understanding the definition and the implication of denseness in metric spaces.

Kolmogorov and Fomin in "Elements of the Theory of Functions and Functional Analysis" state the following (some parts omitted):

The space of bounded sequences ($l_\infty$ but called $M^\infty$ in the book) is not separable. In fact, let us consider all possible sequences consisting of zeros and ones. They form a set with cardinal number that of continuum. The distance between two such distinct points is 1. We surround each of these points with a sphere of radius $\frac 1 2$. These spheres do not intersect. If some set is everywhere dense in the space under consideration, then each of the indicated spheres should contain at least one point of this set and consequently it cannot be denumarable.

Here, I am wondering how we can conclude that the full space is not separable if we only consider the subset of sequences consisting of zeros and ones? Is there a more detailed explanation?

Thanks in advance!

答: Suppose there is a  countbale dense set $(x_n)$. Then each of the spheres mentioned above contains at least one of the points $x_n$. Pick one such for each sphere and map the sphere to the the integer $n$ you obtained. You get a one-to-one map from an uncountable set to the set of positive integers. This contradiction proves that $\ell^{\infty}$ is not separable.


问:
Is the third derivative of the absolute value at x=0 equal to 0?
I drew pictorially what each derivative looks like. For the second derivative, the derivative limits on both sides of x=0 approach 0, so we should have the third derivative equals 0 even though 1) the first derivative is undefined and 2) the second derivative is undefined.

答:
The definition of the third derivative is
$$
f'''(0) =
\lim_{h \to 0} \frac{f''(h) - f''(0)}{h}
$$
so if $f''(0)$ doesn't exist (as in this case) it's meaningless to even start talking about this limit, and therefore $f'''(0)$ doesn't exist either.


问:
Elementary proof of $e > (1+\frac{1}{n})^{n+\frac{1}{k}} $ for $k>2~$
I saw proofs of $e < (1+\frac{1}{n})^{n+0.5}$, but how to prove the inequality:

$
e > (1+\frac{1}{n})^{n+\frac{1}{k}}
$

for $k>2$ and sufficiently large $n$? I'm interested in elementary solutions (without using derivatives or Taylor expansion), suitable for early stages of a calculus course.

答:
Cauchy-Schwarz says
$$
\begin{align}
\log\left(1+\frac1n\right)
&=\int_n^{n+1}\frac1x\,\mathrm{d}x\tag{1a}\\
&\le\left(\int_n^{n+1}\frac1{x^2}\,\mathrm{d}x\right)^{1/2}\left(\int_n^{n+1}1\,\mathrm{d}x\right)^{1/2}\tag{1b}\\
&=\frac1{\sqrt{n(n+1)}}\tag{1c}
\end{align}
$$
That is,
$$
\left(1+\frac1n\right)^{\sqrt{n(n+1)}}\le e\tag2
$$
For any $\epsilon\gt0$, if $n\ge\frac1{8\epsilon}+\frac\epsilon2-\frac12$, then
$$
\begin{align}
\left(n+\frac12-\epsilon\right)^2
&=n(n+1)+\overbrace{\frac14+\epsilon^2-2\left(n+\frac12\right)\epsilon}^{\le0}\tag{3a}\\
&\le n(n+1)\tag{3b}
\end{align}
$$
Therefore, putting together $(2)$ and $(3)$, we get that for $n\ge\frac1{8\epsilon}+\frac\epsilon2-\frac12$
$$
\left(1+\frac1n\right)^{n+\frac12-\epsilon}\le e\tag4
$$


问:
Prove this identity using Jacobi's Triple Product Identity
Using Jacobi's Triple Product Identity prove that\begin{align*}
\prod_{n \geq 1}\left(1-x^{n}\right)^{6}=&\ \frac{1}{2}\left\{\prod_{n \geq 1}\left(1+x^{2 n-1}\right)^{2}\left(1-x^{2 n}\right)\times\left(1+4 x \frac{d}{d x}\right) 2 \prod_{n>1}\left(1+x^{2 n}\right)^{2}\left(1-x^{2 n}\right)\right.\\
& \left.-2 \prod_{n>1}\left(1+x^{2 n}\right)^{2}\left(1-x^{2 n}\right) \times 4 x \frac{d}{d x} \prod_{n \geqslant 1}\left(1+x^{2 n-1}\right)^{2}\left(1-x^{2 n}\right)\right\}
\end{align*}

I found this identity in the A Simple proof of Jacobi's Four Square Theorem. I have given a picture. See that after the step

\begin{align*}\prod_{n\geq 1}(1-x^n)^6\ & =\frac{1}{2}\Bigg\{ \sum_{s=-\infty}^{\infty}x^{s^2}  \sum_{r=-\infty}^{\infty} (2r+1)^2x^{r^2+r}-\sum_{r=-\infty}^{\infty}x^{r^2+r}\sum_{r=-\infty}^{\infty}(2s)^2x^{s^2}
  \Bigg\}\\  & =   \frac{1}{2}\Bigg\{ \sum_{s=-\infty}^{\infty}x^{s^2}  \times\bigg ( 1+4x\frac{d}{dx}\bigg)\sum_{r=-\infty}^{\infty} x^{r^2+r}-\sum_{r=-\infty}^{\infty}x^{r^2+r}\times 4x\frac{d}{dx}\sum_{r=-\infty}^{\infty}x^{s^2}
  \Bigg\}\end{align*}

How did they got the identity using Jacobi's Triple Product Identity

答:
The Jacobi Triple Product identity states that $$\sum_{n\in\mathbb{Z}} z^nx^{n^2}=\prod_{n=1}^{\infty} (1-x^{2n})(1+zx^{2n-1})(1+z^{-1}x^{2n-1})\tag{1}$$ for all complex numbers $x, z$ such that $z\neq 0,|x|<1$.

Putting $z=1$ we get $$\sum_{n\in\mathbb {Z}} x^{n^2}=\prod_{n=1}^{\infty} (1+x^{2n-1})^2(1-x^{2n})\tag{2}$$ Putting $z=x$ in $(1) $ we get $$\sum_{n\in\mathbb {Z}} x^{n^2+n}=\prod_{n=1}^{\infty} (1+x^{2n})(1+x^{2n-2})(1-x^{2n})$$ Note that the factor $(1+x^{2n-2})$ equals $2$ if $n=1$ and thus we can write above equation as $$\sum_{n\in\mathbb{Z}} x^{n^2+n}=2\prod_{n=1}^{\infty} (1+x^{2n})^2(1-x^{2n})\tag{3}$$ The paper of M D Hirschhorn uses the identities $(2),(3)$ in their derivation. Let me know if there is anything more which is troubling you in that proof.


问:
Does there exist a basis function where its inverse function is in the same basis?
I'm interested in finding a basis function
$\phi(x)$, which I can use to approximate some function

$y(x) \approx \hat{y}(x) = \sum\limits_i c_i \phi_i((x - d_i)/s_i)$,

where its inverse function, $\phi^{-1}(x)$, is in the same basis. I.e.:

$\phi^{-1}(x) = l\phi((x - m)/n)$.

Does such a function exist? Or is there some way to prove it doesn't exist?

答:
Consider $$\phi(x)=\begin{cases}-2x,\quad &x\le 0 \\ -x/2, \quad & x\ge 0\end{cases}$$
Clearly $\phi^{-1}=\phi$. And you can use the translations and dilations of $\phi$ to produce any piecewise linear function, up to an additive constant: indeed, $\phi'$ is a point mass at zero, which means that a linear combination of scaled and translated copies of $\phi$ can be any finite combination of point masses. </p>


问:
How to solve for theta in this equation
This is part of a large problem regarding double integrals with polar equations. I have successfully converted the needed equations to polar form, however I can't solve for theta to find where the two equations intersect.

The first equation is a circle centered at the origin, of radius $2$, so it's just $r = 2$.

The second equation is for a circle of radius $1$, centered at $(1, 1)$, which is

$$\cos(\theta) + \sin(\theta) + \sqrt {\sin(2\theta)}.$$

When I set them equal to each other, I have no idea what to do. I've tried a trig identity for $\sin(2 \theta)$, but it doesn't help much.

I've tried using basic online algebra calculators, and none can solve it. But I know there is a solution, since I can get the equation to be both above and below 2 (and the 2 equations intersect at 2 different points on a graph).

Any help would be greatly appreciated.

答:
If I read it properly, you want to find the zero's of
$$f(t)=\cos (t)+\sin (t)+\sqrt{\sin (2 t)}-2$$ There is a obvious symmetry around $t=\frac \pi 4$.

Using Taylor expansion, we have
$$f(t)=\left(\sqrt{2}-1\right)-\left(1+\frac{1}{\sqrt{2}}\right) \left(t-\frac{\pi
   }{4}\right)^2+\frac{1}{24} \left(\sqrt{2}-4\right) \left(t-\frac{\pi
   }{4}\right)^4+O\left(\left(t-\frac{\pi }{4}\right)^6\right)$$ So, approximate soltions are obtained solving the quadratic equation in
$\left(t-\frac{\pi }{4}\right)^2$ and the acceptable solutions are
$$t=\frac \pi 4 \pm \sqrt{\frac{2}{7} \left(-15-9 \sqrt{2}+\sqrt{345+333 \sqrt{2}}\right)}$$ that is to say $t_1\sim 0.29649$ and $t_2\sim 1.27431$ while the "exact" solutions given by Newton method are $0.29870$ and $1.27209$.

If you are ready to solve cubic equations, using one more term in the series expansion will give as estimates $0.29815$ and $1.27265$.

Title: How to solve for theta in this equation


问:
Sketch the image under $w=\log(z)$ of the set, $A=\{z\in\mathbb{C}:e^{-\frac{\pi}{4}}<|z|<e^{\frac{\pi}{4}},\Re(z)>0\}$
Sketch the image under $w=\log(z)$ of the set, $$A=\{z\in\mathbb{C}:e^{-\frac{\pi}{4}}<|z|<e^{\frac{\pi}{4}},\Re(z)>0\}$$

$\log(z)=\ln |z| +iArg(z)$ where $-\pi<Arg(z)\leq\pi$ is the principal argument of $z$.

$w=\log z$ or $e^{u+iv}=x+iy$ shows that
\begin{cases}
e^u\cos v=x>0 \quad(\Re(z)>0)\implies -\dfrac{\pi}{2}<v<\dfrac{\pi}{2},\\
e^u\sin v=y.
\end{cases}

But I have no idea how to use the condition $$e^{-\frac{\pi}{4}}<|z|<e^{\frac{\pi}{4}}\implies e^{-\frac{\pi}{4}}<\sqrt{x^2+y^2}<e^{\frac{\pi}{4}}$$ of the open half annulus. Is it give the bound of $u$? How?

Any solution/hint will be appreciated.

答: Just take logarithms in $e^{-\frac{\pi}{4}}<|z|<e^{\frac{\pi}{4}}$  and note that $\ln |z|$ is the real part of $log \, z$. The answer is $\{w: -\frac {\pi} 4 <\Re w <\frac {\pi} 4, -\frac {\pi} 2<\Im w <\frac {\pi} 2\}$


Body:
Use instead the perturbation method from Concrete Mathematics:
\begin{align}
S_{n} + (-1)^{n+1}\frac{1}{2^{n+1}}&= \sum_{k=0}^{n}(-1)^{k}\frac{1}{2^{k}} +(-1)^{n+1}\frac{1}{2^{n+1}}=1 -\frac{1}{2}\sum_{k=0}^{n}(-1)^{k}\frac{1}{2^{k}} \\
&= 1-\frac{1}{2}S_n
\end{align}
\begin{align}
\frac{3}{2}S_n&=1 - (-1)^{n+1}\frac{1}{2^{n+1}} \\
S_n &=\frac{2(1 - (-1)^{n+1}\frac{1}{2^{n+1}})}{3} = \frac{2^{n+2}-(-1)^{n+1}}{3\cdot2^{n+1}}
\end{align}


答:
The limit of the third derivative $f'''(x)$ as $x$ approaches zero exists and is indeed zero. However, for the derivative to be continuous in this point (or even to exist in this point), it is necessary for this limit to be equal to the function ($f'''(x)$) evaluated in that point. In this case, $f'''(x)$ is not defined in $x=0$, as $f''(x)$ is also not defined in $x=0$ $(f^{(n+1)}(x)$ can only be defined at most in the intervals where $f^{(n)}(x)$ is defined), so $f'''(0)$ does not exist.

In summary, it is clear that the existence of the limit and the existence of the function are two different things. A function can be undefined in a point and yet have a limit at that point, as is the case here for $f'''(x)$.

Title: Is the third derivative of the absolute value at x=0 equal to 0?

Body: I drew pictorially what each derivative looks like. For the second derivative, the derivative limits on both sides of x=0 approach 0, so we should have the third derivative equals 0 even though 1) the first derivative is undefined and 2) the second derivative is undefined.


问:
Solve in the set of integers $2^x +5^x = 3^x + 4^x$

Find the number of integer solutions(both positive and negative) of the equation:
$$2^x +5^x = 3^x +  4^x$$

With induction we see that for $x\geq 2$ we have $$5^x\geq 3^x+4^x$$

It is trivially true for $x=2$ and $x=3$. Now say it is true for $x$ and prove it for $x+2$: $$5^{x+2} \geq 5^2(3^x+4^x) = (3^2+4^2)(3^x+4^x) > 3^{x+2}+4^{x+2}$$

So equation has no solution for $x\geq 2$. Clearly $x=0$ and $x= 1$ are solutions.

If $x<0$ you can write $x=-n$ where $n$ is positive integer and we get $$3^n\cdot 4^n(2^n+5^n) = (3^n+4^n)2^n\cdot 5^n$$
so $ 3^n\mid 40^n$ which is clearly nonsense and thus no solution (for negative $x$).

Any easier solution?

Body:
Re write the Eq. as $$3^x-2^x=5^x-4^x....(1)$$
Use LMVT on $f(t)=t^x$ in $(2,3)$ and $(4,5)$
to write $$\frac{3^x-2^x}{3-2}=xt_1^{x-1}, t_1\in(2,3)~~~~(2)$$
and
$$\frac{5^x-4^x}{5-4}=xt_2^{x-1}, t_2\in (4,5)~~~~(3)$$
so $t_1 \ne t_2$, By (1), we can equate $xt_1^{x-1}=xt_2^{x-1} \implies x=0~or~ x=1.$

So (1) has only two solutions: $x=0,1$.


答:
A subset $A$ is dense in a metric space $X$ when every point $x\in X$ is in the closure of $A$. Thus every open ball $B_r(x)$ contains a point of $A$ distinct from $x$. So, the centers of disjoint open balls cannot be dense in $X$.

In the given example, the centers of the balls are the $0$-$1$ sequences, which are uncountable. Given any dense set $A$, however, each of these balls must contain at least one point of $A$, all distinct from each other. This means that any dense set $A$ must be uncountable.

Title: Question(s) regarding understanding of denseness

Body:
I think I have some problem understanding the definition and the implication of denseness in metric spaces.

Kolmogorov and Fomin in "Elements of the Theory of Functions and Functional Analysis" state the following (some parts omitted):

The space of bounded sequences ($l_\infty$ but called $M^\infty$ in the book) is not separable. In fact, let us consider all possible sequences consisting of zeros and ones. They form a set with cardinal number that of continuum. The distance between two such distinct points is 1. We surround each of these points with a sphere of radius $\frac 1 2$. These spheres do not intersect. If some set is everywhere dense in the space under consideration, then each of the indicated spheres should contain at least one point of this set and consequently it cannot be denumarable.

Here, I am wondering how we can conclude that the full space is not separable if we only consider the subset of sequences consisting of zeros and ones? Is there a more detailed explanation?

Thanks in advance!


问:
How to find the Taylor series of function $f(x)=\frac{a\left(1-x^2\right)+b\sqrt{x}\left(1+x\right)}{2\left(1-x\right)^2}$, at $x=0$
How to find the Taylor series of function $f(x)=\frac{a\left(1-x^2\right)+b\sqrt{x}\left(1+x\right)}{2\left(1-x\right)^2}$, at $x=0$. I know that the Taylor series of the functions:

$\frac{t}{1-t}=\sum_{n=0}^{+\infty}t^{n+1}, \frac{t}{(1-t)^2}=\sum_{n=0}^{+\infty} nt^{n}$.

答:
If you want to circumvent the problems underlined by @Gary and have an expansion using only integer powers, let $x=t^2$; using $t>0$, you need to work with
$$f(t)=\frac{a(1- t^4)+b \left(1+t^2\right) t}{2 \left(1-t^2\right)^2}=a\frac {(1+t^2)}{2(1-t^2) }+b\frac{ \left(1+t^2\right) t}{2 \left(1-t^2\right)^2}$$ which is not difficult using long division if you just need a few tems. If you do it and look at the patterns, you could conjecture that
$$f(t)=\frac a 2+\frac 14\sum_{n=1}^\infty \Big[(2 a \left(1+(-1)^n\right)+b \left(1-(-1)^n\right) n\Big]\,t^n$$


问:
Prove that $f(iz)=\overline{f(\bar z)}$ for analytic $f$ satisfies some condition
Let $f$ is entire function in $\mathbb{C}$ satisfying $$ e^{-a|x-y|}\leq |e^{if(x+iy)}|\leq e^{b|x-y|}$$ for some $a,b>0$ . Then to show that $$f(iz)=\overline{f(\bar z)}$$ for all $z\in\mathbb{C}$

I think I need to use Schwarz's reflection principle but I am not getting how to use the given condition.

答:
As mentioned in the comments, the given condition implies that $f(z)$ is real-valued on the line $y=x$. It follows that
$$
 f(r(z)) = \overline{f(z)}
$$
where
$$
 z = x+iy \mapsto r(z) = y+ix = i \bar z
$$
is the reflection at the line $y=x$.


问:
Continuum between arithmetic mean, AGM, and geometric mean
NOTE: I am aware of this possible duplicate, but my question is slightly different as it also involves arithmetic-geometric mean.

The arithmetic mean  of two numbers is defined as:
$$\text{am}(a,b) = \frac{a}{2}+\frac{b}{2}$$
For example, $\text{am}(3,12) = 7.5$. Similarly, geometric mean of two numbers is defined as:
$$\text{gm}(a,b) = \sqrt{ab}$$

So $\text{gm}(3,12)=6$. The arithmetic-geometric mean is a mean "between" these two, calculated as follows:
$$
a_{0}=a,g_{0}=b\\
a_{n+1}=\text{am}(a_{n},g_{n})\\
g_{n+1}=\text{gm}(a_{n},g_{n})\\
\text{agm}(a,b)=\text{lim}_{n\rightarrow\infty}\;\;(a_n)
$$

For instance, $\text{agm}(3,12) \approx 8.7407$.
Note that $\text{gm}(a,b)\leq \text{agm}(a,b)\leq \text{am}(a,b)$.

My question: I need to define a general mean $M(a,b,c)$ for $0\leq c \leq 1$ that satifies these properties:
$$
M(a,b,0) = \text{am}(a,b)\\
M(a,b,0.5) = \text{agm}(a,b)\\
M(a,b,1) = \text{gm}(a,b)
$$
Power mean is not what I'm looking for, as $PM_{0.5}\;(a,b) \neq \text{agm}(a,b)$ (e.g. $PM_{0.5}\;(3,12) = 8.7464... \neq 8.7407...$).

答:
Standard quadratic interpolation, as suggested by  @Martin-R

\begin{align}
M(a,b,t)&=
(2t-1)(t-1)\operatorname{AM}(a,b)
\\
&-4t(t-1)\operatorname{AGM}(a,b)
\\
&+t(2t-1)\operatorname{GM}(a,b)
.
\end{align}

Or in expanded form,

\begin{align}
M(a,b,t)&=
 (\sqrt a+\sqrt b)^2\,t^2
 -(\tfrac32\,a+\tfrac32\,b+\sqrt{ab})\,t
 +\tfrac12\,a+\tfrac12\,b
 \\
 &+\frac{\pi t(1-t)}
 {\displaystyle\int_0^1 \left(\sqrt{1-x^2}\sqrt{(a+b)^2-(a-b)^2\,x^2}\right)^{-1} \, dx }
.
\end{align}

Edit

At the second thought, I must admit that
such interpolation is a bad choice for general $a,b\ge0$.
It's ok for just switching between the three values $t=0,\tfrac12,1$ and works
relatively reasonable for some ranges of $a,b$, but it could
as bad as returning negative output, for example,
$M(1000000,1,0.9)\approx-2081$, so it needs a serious treatment.


问:
Which one result in mathematics has surprised you the most?
A large part of my fascination in mathematics is because of some very surprising results that I have seen there.

I remember one I found very hard to swallow when I first encountered it, was what is known as the Banach Tarski Paradox. It states that you can separate a ball $x^2+y^2+z^2 \le 1$ into finitely many disjoint parts, rotate and translate them and rejoin (by taking disjoint union), and you end up with exactly two complete balls of the same radius!

So I ask you which are your most surprising moments in maths?

Chances are you will have more than one. May I request post multiple answers in that case, so the voting system will bring the ones most people think as surprising up. Thanks!

答: The fact the every set can be well-ordered (given the Axiom of Choice, of course).</p>

LastEditorDisplayName: anon


问:
What is the intuitive meaning of the covariant derivative on a sphere?
I get that it is used to construct covariant equations in physics. But otherwise, what does it mean?</p>

答:
It is the most natural directional derivative when your space might be curved and "direction" therefore becomes a bit iffy.

More specifically, the standard directional derivative along a vector $v$ at a point $p$ is
$$
\partial_v F(p)=\lim_{h\to 0}\frac{F(p+hv)-F(p)}{h}
$$
This doesn't make sense on a general manifold, as the space might not be a vector space and $p+hv$ doesn't work. Secondly, $F$ could be, say, a vector field and comparing two tangent vectors at different points is not easily done.

Covariant derivative fixes this to work with tangent vectors as direction in a consistent way, as long as we have either curvature information (through a connection) or a metric (through a Riemannian inner product structure) on our manifold.


问:
Independence of functions of independent random variables.
Suppose $X_1,..,X_n$ are independent random variables and $X_i'$ is an independent copy of $X_i$, then how does one show that $$E[f(X_1,..,X_n)|X_1,..,X_{i-1},X_{i+1},..,X_n]$$ and  $$E[f(X_1,..,X_i',..,X_n)|X_1,..,X_{i-1},X_{i+1},..,X_n]$$ are equal almost everywhere and why is $f(X_1,..,X_n)$ and $f(X_1,..,X_i',..,X_n)$ conditionally independent over $G:=\sigma(X_1,..,X_{i-1},X_{i+1},..,X_n)$?

I am trying to argue using measure theoretic arguments but I can't seem to be able to show this. I am stuck trying to understand the $\sigma$-algebras generated by both functions which are measurable on $\sigma(X_1,..,X_{i-1},X_{i+1},..,X_n)$. Any ideas?

i.e. Possible proof of conditional independence:

Note: $\sigma(f(X_1,..,X_n))\subset \sigma(X_1,..,X_n)$ and $\sigma(f(X_1,..,X_i',..,X_n))\subset \sigma(X_1,..,X_i',..,X_n)$. Let $A\in  \sigma(X_1)\cup...\cup\sigma(X_n) $ and $B\in \sigma(X_1)\cup..\cup\sigma(X_i')..\cup\sigma(X_n)$ then $$E[1_A1_B|G]=E[1_A|G]E[1_B|G]$$
This implies that it holds for all $A\in \sigma(X_1,..,X_n)$ and $B\in \sigma(X_1,..,X_i',..,X_n)$.

答:
Let $Y=[X_1,\ldots,X_{n-1}]$. Then
$$
\mathsf{E}[f(X_1,\ldots,X_{n-1},X_n)\mid Y]=\varphi(Y) \quad\text{a.s.}
$$
and
$$
\mathsf{E}[f(X_1,\ldots, X_{n-1},X_n')\mid Y]=\varphi'(Y) \quad\text{a.s.},
$$
where $\varphi(y)=\mathsf{E}f(y_1,\ldots, y_{n-1}, X_n)$ and $\varphi'(y)=\mathsf{E}f(y_1,\ldots, y_{n-1}, X_n')$. (See, e.g., Lemma 6.2.1 on page 236 here.) But $\varphi(y)=\varphi'(y)$ because $X_n\overset{d}{=}X_n'$.


问:
Sum of squares of normal distribution
I have a problem in which I need to apply weak law of large numbers to sum of squares of normal distribution with mean $1$ and variance $1$.

$X_1, X_2,\dots \sim \mathcal{N}(1,1).$  They are iid.

$V(X_1^2+X_2^2+\dots\X_n^2)/n$ what will be the limit value of this when $n$ tends to infinite.

My views:

I know sum of squares of SNV is chi square.

Variance of this chi square will be $2n$.

But I don’t know if it follows central chi square in case when mean is not $0$.

In this particular case, will my variance be $2n$?

答:
My views:
I know sum of squares of SNV is chi square.
Variance of this chi square will be 2n.
But I don’t know if it follows central chi square in case when mean is not 0.

sum of square of SNV is a chi-squared but your Gaussian are not centered thus the sum of your iid reduced gaussian is a Noncentral chi-squared distribution with variance $2(k+2\lambda)$ where $\lambda$ is the noncentrality parameter.

As @Kavi Rama Murthy already asked, we have to know what is there in the numerator:

$\frac{1}{n}\mathbb{V}\left[\sum_{k=1}^{\infty}X_k^2\right]$

$\frac{1}{n}\mathbb{V}\left[\sum_{k=1}^{n}X_k^2\right]$

Something else?

EDIT

I have a finite sum up to n

Thus the result is simply

$$\frac{1}{n}\mathbb{V}\left[\sum_{k=1}^{n}X_k^2\right]=\frac{1}{n}\cdot n\mathbb{V}[X_1^2]=2(1+2)=6$$

Title: Sum of squares of normal distribution


答:
The geometric mean can be turned to an arithmetic one by a nonlinear mapping,

$$\sqrt{ab}=f^{-1}\left(\frac{f(a)+f(b)}2\right)$$ where $f(x):=\log(x)$ is the logarithm.

Hence a way to construct "intermediate" means is by using a function intermediate between the identity and the logarithm. You could obtain yours by solving the functional equation

$$2f(\text{agm}(a,b))=f(a)+f(b).$$

By homogeneity, it should be enough to solve

$$2f(\text{agm}(a,1))=f(a)+f(1).$$

Of course solving this equation seems arduous. I guess that you can solve it numerically and tabulate $f$. Then you'll have a generalization of the arithmetico-geometric mean for $n$ arguments.

Whether you can find a natural "interpolation" between these three means seems even less tractable.

Title: Continuum between arithmetic mean, AGM, and geometric mean

Body:
NOTE: I am aware of this possible duplicate, but my question is slightly different as it also involves arithmetic-geometric mean.

The arithmetic mean  of two numbers is defined as:
$$\text{am}(a,b) = \frac{a}{2}+\frac{b}{2}$$
For example, $\text{am}(3,12) = 7.5$. Similarly, geometric mean of two numbers is defined as:
$$\text{gm}(a,b) = \sqrt{ab}$$

So $\text{gm}(3,12)=6$. The arithmetic-geometric mean is a mean "between" these two, calculated as follows:
$$
a_{0}=a,g_{0}=b\\
a_{n+1}=\text{am}(a_{n},g_{n})\\
g_{n+1}=\text{gm}(a_{n},g_{n})\\
\text{agm}(a,b)=\text{lim}_{n\rightarrow\infty}\;\;(a_n)
$$

For instance, $\text{agm}(3,12) \approx 8.7407$.
Note that $\text{gm}(a,b)\leq \text{agm}(a,b)\leq \text{am}(a,b)$.

My question: I need to define a general mean $M(a,b,c)$ for $0\leq c \leq 1$ that satifies these properties:
$$
M(a,b,0) = \text{am}(a,b)\\
M(a,b,0.5) = \text{agm}(a,b)\\
M(a,b,1) = \text{gm}(a,b)
$$
Power mean is not what I'm looking for, as $PM_{0.5}\;(a,b) \neq \text{agm}(a,b)$ (e.g. $PM_{0.5}\;(3,12) = 8.7464... \neq 8.7407...$).


问:
Grade 8 Random Number Question - Which concept to use
I found a question in Math Challengers 2012 exam (Provencial, part2, Q8). It states

Betty and Ben each select independently and at random an integer between 0 and 5 (inclusive). What is the average non-negative difference between their numbers? Express the answer as a common fraction.

You get about 2 minutes to do this without a calculator. Which concept do you use? Does this belong to the Combinatorics/Probability Generating Functions?

答:
you have to calculate

$$\mathbb{E}|X-Y|=1\times \frac{10}{36}+2\times \frac{8}{36}+3\times \frac{6}{36}+4\times \frac{4}{36}+5\times \frac{2}{36}=\frac{70}{36}$$

the fastest manual way is to write down a $6\times6$ table, remembering that any outcome is equiprobable


问:
Exponential map and connection
Suppose you have a Riemannian manifold $(M,g)$ and a point $p\in M$ fixed. Let $v: s\mapsto v(s)$ be a curve in $T_pM$. Now consider the map $f(s):=\exp_p(v(s))$.

Can one get an explicit formula for $f'(s)$? Maybe that is confusing but I thought of something like
$f'(s)=d(\exp_p)_{v(s)}(\nabla_sv(s))$, where $\nabla_s$ denotes the covariant derivative.

Formally, $d(\exp_p)_{v(s)}$ is a map from $T_{v(s)}T_pM$ to $T_{\exp_p(v(s))}M$ and one can identify $T_{v(s)}T_pM\cong T_pM$. I would like to know if the formula above is true and how this identification is related to the Levi-Civita-connection on $M$. Thanks for any explanation.

答: Since $T_pM$ is a vector space, the derivative of $v(s)$ is defined without need to resort to "covariant derivative", so in your formula $\nabla_s v(s)$ can be replaced simply by the derivative $v'(s)$.  The formula you wrote down is true by chain rule.  In fact, it would be true for any map and not just the exponential map.  To define the exponential map, one certainly needs the Levi-Civita connection.</p>

Title: Exponential map and connection


问:
Why is $\sup_{n \in \mathbb{N}} E[|X_n|^p] \leq B < \infty$ different from $\sup_{n \in \mathbb{N}} E[|X_n|^p] < \infty$ as a requirement?
This is part of a convergence theorem for $L_p$ martingales.

Theorem. We have $p > 1$ and $B>0$. Let further $(X_n)_n$ be a martingale with $\sup_{n \in \mathbb{N}} E[|X_n|^p] \leq B < \infty$. Then there exists a random variable $X_{\infty}$ such that $E[|X_{\infty}|^p] \leq B$ and s.t.

$X_n \longrightarrow X_{\infty}$ a.s. and in $L_p$ norm.

A powerful theorem but with a strong requirement condition. My question concerns this requirement. Why do we demand that

$\sup_{n \in \mathbb{N}} E[|X_n|^p] \leq B < \infty$ $(*)$

instead of just

$\sup_{n \in \mathbb{N}} E[|X_n|^p] < \infty$ $(**)$

I know that, in general, these are different concepts. For example, only demanding that for all $n \in \mathbb{N}$

$|X_n| < \infty$ does not imply that $\sup_{n \in \mathbb{N}} |X_n| < \infty$

But in the case of $(*)$, we already have a supremum. And in my understanding, if $(**)$ holds, we must have some $B>0$ such that $(*)$ must also hold. Since if the supremum does not go to infinity, it must be smaller than some value.

Where am I going wrong?

答:
$(*)$ and $(**)$ are both equivalent in that they tell you the sup is finite. $(*) $ gives a name to the sup ($B$), which can be used later.

So for a sequence $(a_n)$ both conditions: $\sup_n a_n<+\infty$ and $\sup_n a_n\le B<+\infty$ are equivalent. The latter obviously is equivalent to: $a_n\le B$ for all $n$, which is of course strictly stronger than: $a_n<+\infty$ for all $n$.


问:
Prove that $\liminf_{x \to p} f(x) \ge f(p)$
Let $f : A \to \mathbb{R} $ and let $ p \in A$ be  a limit point of  $A$ .Then prove that  $f$ is lower semicontinious   at $p$   if and only if $\liminf_{x \to p} f(x) \ge f(p)$

My attempt: since  $p$ is a limit point of $A \implies  B(p, \epsilon) \cap A \neq \emptyset.$
we know that for  every $\epsilon >0$ there exist  $\delta >0$

Therefore we can say that  $B(p, \delta) \cap A \neq \emptyset.$

Now  assume that  $f$ is lower semicontinious at $p$ .Then $f(p) -\epsilon  <f(x)$ for all  $x \in B(p, \delta) \cap A \neq \emptyset.$

After that im not able to proceed further

答:
The following two definitions are equivalent:

$f$ is lower semicontinuous at $p$ if for every $\epsilon > 0$ there is a $\delta > 0$ such that
$$
 f(x) > f(p) - \epsilon
$$
for all $x \in B_\delta(p) \cap  A$.

$f$ is lower semicontinuous at $p$ if $\liminf_{x \to p} f(x) \ge f(p)$.

$(1) \Rightarrow (2):$  For every $\epsilon > 0$ there is a $\delta > 0$ such that
$
 f(x) > f(p) - \epsilon
$
for all $x \in A$ with $|x-p| < \delta$. It follows that
$$
 \liminf_{x \to p} f(x) \ge f(p) - \epsilon \, ,
$$
and since this holds for all $\epsilon > 0$,
$$
 \liminf_{x \to p} f(x) \ge f(p) \, .
$$

$\neg (1) \Rightarrow \neg(2):$ If $(1)$ does not hold then there is a $\epsilon > 0$ such that for all $\delta > 0$
$$
 f(x) \le f(p) - \epsilon
$$
for some $x \in A$ with $0 < |x-p| < \delta$. It follows that
$$
\liminf_{x \to p} f(x) = \lim_{\delta \to 0} \, \bigl( \inf \{ f(x) \mid x \in B_\delta(p) \setminus \{p \} \} \bigr) \\
\le f(p) - \epsilon < f(p) \, ,
$$
i.e. $(2)$ does not hold.


问:
Is the Cardinality of all Lines in the Extended Euclidean Plane $\mathfrak{c}$?
Consider the surjection $[0,2\pi)*\left(\mathbb{R}^2\cup\ell_\infty\right)\rightarrow L$ (such that $L$ is the set of all lines determined by the binary operation $*$ between a point and an angle). Therefore, $\left|L\right|\leq\left|[0,2\pi)*\left(\mathbb{R}^2\cup\ell_\infty\right)\right|=\mathfrak{c}*(\mathfrak{c}^2+\mathfrak{c})=\mathfrak{c}*\mathfrak{c}$. Is $\mathfrak{c}*\mathfrak{c}=\mathfrak{c}$? How would I show this?

Similar Questions:

Cardinality of the set of all straight lines in $\mathbb R^2$

Is the set of points or the set of lines on a plane "larger"?

What is the Cardinality of the Extended Euclidean Plane?

答:
Yes it is. There is a general proof of the statement $|A\times A|=|A|$ for every infinite set $A$ that relies on Axiom of Choice. But for this particular case you can do a much easier constructive proof of $|\mathbb{R}\times\mathbb{R}|=|\mathbb{R}|$:

Obviously $|\mathbb{R}|\leq |\mathbb{R}\times\mathbb{R}|$.

Let's see that there is a surjective map from $[0,1]$ to $[0,1]\times[0,1]$, and thus $|\mathbb{R}\times\mathbb{R}|=|[0,1]\times[0,1]|\leq |[0,1]|=|\mathbb{R}|$. Let $x:=0.x_1x_2x_3\ldots$ denote the decimal expansion of $x$. In case of $x$ having more than one, consider the shorter one. Consider the map: $h:[0,1]\longrightarrow[0,1]\times[0,1]$ $$x=0.x_1x_2x_3\ldots\longmapsto h(x):=(0.x_1x_3x_5\ldots,0.x_2x_4x_6\ldots)$$
This map is surjective. If we take $(0.x_1x_2x_3\ldots,0.y_1y_2y_3\ldots)\in[0,1]\times[0,1]$, we have that: $$h(0.x_1y_1x_2y_2x_3y_3\ldots)=(0.x_1x_2x_3\ldots,0.y_1y_2y_3\ldots)$$
Proving the claim.


问:
Are there non-continuable functions that become continuable when raised to some power?
Let $X$ be a complex algebraic variety (integral, separated scheme of finite type over $\mathbb C$) and $U\hookrightarrow X$ an open subvariety. I will say that $f\in\mathscr O_X(U)$ is continuable if there exists some $g\in \mathscr O_X(X)$ with $g|_U=f$. My question is simply: Can it happen that $f$ is not continuable, but there exists a $k\in\mathbb N^+$ with $f^k$ continuable?

I was debating with a friend who thinks it is possible but could not give an example, while I thought that it is impossible but wasn't able to give a satisfying proof. So, I'd be really glad if you could provide one of the two (that is, a simple example of such a phenomenon or a proof that it can not happen).

答:
What about the following:
$X = Spec \mathbb{C}[x_1,x_2]/(x_1^3+x_2^2)$, and $U=D(x_1)$, i.e $Spec \mathbb{C}[x_1^{\pm 1},x_2]/(x_1+(x_2/x_1)^2)$. Take $f =(x_2/x_1)$ on it. It is not a global section, but surely $(x_2/x_1)^2 = -x_1$ is.</p>


问:
Prove $\lim\limits_{x \to \infty} x \sin^2 x$ does not exist.
I need to prove $\lim\limits_{x \to \infty} x \sin^2 x$ does not exist, and request your help verifying my argument is correct. I've got somewhat decent  at making $\epsilon-\delta$ arguments, but proving a limit does not exist has been a bit challenging actually.

We need to prove $\lim\limits_{x \to \infty} x \sin^2 x \neq \ell$, for every $\ell \in \mathbb{R}$, so take $\ell \in \mathbb{R}$. The result follows if we can find some $\epsilon > 0$ such that for every $N \in \mathbb{R}$, there is some real $x$ such that $x > N$ but $\lvert x \sin^2 x - \ell \rvert \geq \epsilon$.

If $\epsilon =\lvert \ell \rvert + 1$, then for every $N \in \mathbb{R}$ we can find $x_0 > \max(N, \quad 2\lvert \ell \rvert + 1)$ such that $x_0 = \cfrac{\pi}{2} + 2\pi k$, for some big enough $k \in \mathbb{Z}$. For this $x_0$ we have that
$$x_0 \sin^2 x_0 = x_0 \cdot 1 = x_0 > 2\lvert \ell \rvert + 1 \geq 1 + \lvert \ell \rvert + \ell$$.
This implies $x_0 \sin^2 x_0 - \ell = x_0 - \ell > 1 + \lvert \ell \rvert = \epsilon$ and the proof is complete.

Any comments or suggestions for improvement are very appreciated.

答:
This is just a variation of Martin Argerami's answer. He constructed a sequence $(x_k)$ such that $\lim_{k\to\infty}x_k\sin^2x_k=\infty$. Let us show that for each $a \in [0,\infty)$ there exists a sequence $(y_k)$ such that
$$\lim_{k\to\infty}y_k\sin^2y_k=a .$$

We have $\sin^2 x = 0$ for $x = m\pi$ and $\sin^2 x = 1$ for $x = \pi/2 + m\pi$ for all $m \in \mathbb Z$. Thus by the IVT the function $f(x) = x \sin^2x$ maps the interval $[m\pi, \pi/2 + m\pi]$ onto the interval $[0,\pi/2 + m\pi]$. Hence for $m \ge m_0 = \max(0,(a-2)/\pi)$ we find a point $y_m \in [m\pi, \pi/2 + m\pi]$ such that $f(y_m) = a$. Let $y_m = m\pi$ for $m < m_0$. Then clearly $y_m \to \infty$ as $m \to  \infty$ and $\lim_{m\to\infty}y_m\sin^2y_m=a$.


问:
How can I evaluate $\lim_{n\to\infty}\Big[ \cot(\pi\sqrt{100n^2+n+1}\Big]$
While scrolling through questions in Brilliant I saw the following question on limit

$$\lim_{n\to\infty}\Big[ \cot(\pi\sqrt{100n^2+n+1}\Big]$$

If we take $n^2$ outside square root we are left with ${100+\frac {1} {n} +\frac {1}{n^2}}$ inside square root and we can't use expansion of $\sqrt{(1+\frac{1}{n})}$ because of three terms inside square root.

If we use
$$\pi\cot(\pi x)=\frac{1}{x}+\sum_{k=1}^{\infty}\Big(\frac{1}{x-k}+\frac{1}{x+k}\Big)$$

with $x=\sqrt{100n^2+n+1}$,then our limit becomes

$$\lim_{n\to\infty}\frac{1}{\pi}\Big[\frac{1}{ \sqrt{100n^2+n+1} }+\sum_{k=1}^{\infty}\Big(\frac{1}{\sqrt{100n^2+n+1} -k}+\frac{1}{\sqrt{100n^2+n+1} +k}\Big)\Big]$$

Now I got stuck on summation with this method.

How can we show that this limit equals those messy square roots?

答:
Hint:

$$\sqrt{100n^2+n+1}=10n\sqrt{1+\dfrac{n+1}{100n^2}}\sim10n+\dfrac{n+1}{20n}\to10n+\frac1{20}.$$

So the question amounts to finding the cotangent of $\dfrac\pi{20}=9°$.

The following table can be very useful: https://math.la.asu.edu/~surgent/mat170/Exact_Trig_Values.pdf.


问:
Recurrence Equation and sequences
$a_n $ is a sequence defined this way:

and we define:

need to prove that this happens:

i got exam soon and a lot of simmilar questions but i have no idea where to start

答:
From the given information, $y_{n+1}=a_{2n+2}=\dfrac13a_{2n+1}=\dfrac13\left(a_{2n}+\dfrac13\right)=\dfrac13y_n+\dfrac19$.

Now can you show that $x_{n+1}=\dfrac13x_n+\dfrac13$?

Title: Recurrence Equation and sequences


问:
Sequence of $C^2$ function with compact support generating borel-sigma-algebra.
I'm looking for a sequence of functions $(f_n)_{n\geq 0}$ all $\mathbb{R} \mapsto\mathbb{R}$, twice continuously differentiable and with compact support such that the sigma-algebra generated by them is the borel sigma-algebra on $\mathbb{R}$.
Does anyone know of a such?</p>

答:
For rationals $a,b$ and a positive integer $n$, let $f_{a,b,n}$ be a $C^2$ function which is $1$ on the interval $[a,b]$ and $0$ outside $(a-\frac{1}{n}, b+\frac{1}{n})$.  As $n \to \infty$, $f_{a,b,n} \to 1_{[a,b]}$ pointwise, so any $\sigma$-algebra that makes all the $f_{a,b,n}$ must contain all the rational intervals $[a,b]$, and hence all the Borel sets.  So the countable set $\{f_{a,b,n} : a,b \in \mathbb{Q}, n \in \mathbb{N}\}$ generates the Borel $\sigma$-algebra.  You can re-index to make it into a sequence if you really want.

There are lots of other possibilities.  It's worth noting that you could replace $C^2$ by $C^\infty$ by a convolution argument.


问:
Moment generating function for a standardized sum of random variables
I ran into a homework problem to derive the mgf for a standardized sum of random variables. Having already solved for when $\mu$ = 0, I am now asked to generalize for when $\mu$ =/= 0. Furthermore, it is required to use L'Hopital rule to solve the question. My issue is with the highlighted portion, where I do not know how to proceed.

$S_n^* = \frac{S_n - n\mu}{\sqrt{n} \sigma} = \frac{X_1 + ... + X_n - n\mu}{\sqrt{n}\sigma}$

$M_{S_n^*}(t) = E[e^t(\frac{X_1 + ... + X_n - n\mu}{\sqrt{n}\sigma})] = M_x(\frac{t}{\sqrt{n}\sigma})^n.e^{t(\frac{-\sqrt{n}\mu}{\sigma})}$

$log[M_{S_n^*}(t)] = n log[M_x(\frac{t}{\sqrt{n}\sigma})] - \frac{t\mu}{\sigma(1/n)}$

Let $\theta$ = $\frac{1}{\sqrt{n}}$. When n tends to $\infty$, $\theta$ tends to 0.

$\lim_{n \to \infty} log[M_{S_n^*}(t)]$ = $\lim_{\theta \to 0} \frac{log[M_x(\frac{t}{\sigma}\theta)]}{\theta^2}$ - $\lim_{\theta \to 0} \frac{t\mu}{\sigma \theta}$.

I do not know how to manipulate the second term. Ideally, I want to make the term disappear (i.e. limit tends to 0), but taking limits makes the highlighted term $\infty$.

Image of workings thus far

The answer should be the mgf of a standard normal variable i.e. $e^{t^2/2}$.

答:
You can't separate the limit into a difference if it is of the form $\infty-\infty$. If you want to use L'Hopital's rule, you ought to combine the terms into a single fraction:

$$ \lim_{\theta\to0}\frac{\ln M_X(\frac{t}{\sigma}\theta)-\frac{t\mu}{\sigma}\theta}{\theta^2}. $$

This limit is of the form $0/0$, and you can use the rule. If you differentiate the top and bottom, it will still be of the form $0/0$, but then you can use the rule again and you're good!

Keep in mind $M_X(0)=1$, $M_X'(0)=\mu_X$, $M_X''(0)=\mu_{X^2}$.


答:
Firstly, take n out common from the squareroot

$\lim _{n\rightarrow \infty }\cot \left(\pi \sqrt{100n^2+n+1}\right)=\lim _{n\rightarrow \infty }\cot \left(\frac{\pi \sqrt{100+\frac{1}{n}+\frac{1}{n^2}}}{\frac{1}{n}}\right)$

now let $t=\frac{1}{n}$, and we have

$\lim _{n\rightarrow \infty }\cot \left(\frac{\pi \sqrt{100+t+t^2}}{t}\right)$

As it is a 0/0 form, we can now use the L's hospital rule:

$\lim _{n\rightarrow \infty }\cot \left(\frac{\pi \sqrt{100+t+t^2}}{t}\right)=\lim _{n\rightarrow \infty }\cot \left(\frac{\frac{\pi \left(1+2t\right)}{2\sqrt{100+t+t^2}}}{1}\right)=\cot \left(\frac{\pi }{20}\right)$

Can you proceed from here?

Title: How can I evaluate $\lim_{n\to\infty}\Big[ \cot(\pi\sqrt{100n^2+n+1}\Big]$

Body:
While scrolling through questions in Brilliant I saw the following question on limit

$$\lim_{n\to\infty}\Big[ \cot(\pi\sqrt{100n^2+n+1}\Big]$$

If we take $n^2$ outside square root we are left with ${100+\frac {1} {n} +\frac {1}{n^2}}$ inside square root and we can't use expansion of $\sqrt{(1+\frac{1}{n})}$ because of three terms inside square root.

If we use
$$\pi\cot(\pi x)=\frac{1}{x}+\sum_{k=1}^{\infty}\Big(\frac{1}{x-k}+\frac{1}{x+k}\Big)$$

with $x=\sqrt{100n^2+n+1}$,then our limit becomes

$$\lim_{n\to\infty}\frac{1}{\pi}\Big[\frac{1}{ \sqrt{100n^2+n+1} }+\sum_{k=1}^{\infty}\Big(\frac{1}{\sqrt{100n^2+n+1} -k}+\frac{1}{\sqrt{100n^2+n+1} +k}\Big)\Big]$$

Now I got stuck on summation with this method.

How can we show that this limit equals those messy square roots?


问:
Time complexity of a simple recurrence relation
My professor told me that the following recurrence relation has linear complexity, but I'm not sure.

$T(n) = 3 T(n-1)+5$:
\begin{equation}
\begin{split}
T(n) & = 3 T(n-1)+5 \Rightarrow \\
T(n-1)     & = 3T(n-2) +5 \Rightarrow \\
T(n)-T(n-1)     & = 3T(n-1)-3T(n-2) \Rightarrow\\
T(n)     & = 4T(n-1)-3T(n-2) \Rightarrow \\
q^2 & = 4q - 3 \Rightarrow \\
q_{1,2} &= 3, 1 \Rightarrow\\
T(n) &= \alpha 1^n + \beta 3^n
\end{split}
\end{equation}

From these $T(0)=0, T(1)=5$ $\Rightarrow$ $\alpha=-5/2$ and $\beta=5/2$ $\Rightarrow$ $T(n) = \frac{-5}{2} +\frac{5}{2}3^n = O(3^n)$

How can be this linear? What am I missing?

答:
what your professor meant: the amount of calculations required to calculate $T(n)$ is linear in $n$ (most recurrences are).

what you did is calculating a closed form (or it seems to be an approximation) of $T(n)$ for a specific $n$.


问:
Rotation composition in $R^3$ and linear mappings
I'm trying to understand why the result of a composition of linear transformations (two rotations) it is different from mapping the base vectors from $R^3$ to $R^3$ twice.
I'll explain.
I have a mapping: ϕ that maps vectors in $R^3$ to $R^3$ as a linear combinations of the canonical base vectors (e1,e2,e3).
The transformation matrix associated with ϕ is the following:

\begin{bmatrix}\cos(a)&0&\sin(a)\\0&1&0\\-\sin(a)&0&\cos(a)\end{bmatrix}
That is the classic counterclockwise rotation with respect to the e2 axis.

Now I want to map the canonical base vectors (e1,e2,e3) using ϕ.

$ϕ(\mathbf{e1}) = \cos(a) \bigl( \begin{smallmatrix} 1 \\ 0 \\ 0 \end{smallmatrix} \bigr) + 0 \bigl( \begin{smallmatrix} 0 \\ 1 \\ 0 \end{smallmatrix} \bigr) - \sin(a) \bigl( \begin{smallmatrix} 0 \\ 0 \\ 1 \end{smallmatrix} \bigr)$ = $\bigl( \begin{smallmatrix} \cos(a) \\ 0 \\ -\sin(a) \end{smallmatrix} \bigr)$

$ϕ(\mathbf{e2}) = 0 \bigl( \begin{smallmatrix} 1 \\ 0 \\ 0 \end{smallmatrix} \bigr) + 1 \bigl( \begin{smallmatrix} 0 \\ 1 \\ 0 \end{smallmatrix} \bigr) + 0 \bigl( \begin{smallmatrix} 0 \\ 0 \\ 1 \end{smallmatrix} \bigr) = \bigl( \begin{smallmatrix} 0 \\ 1 \\ 0 \end{smallmatrix} \bigr)$

$ϕ(\mathbf{e3}) = \sin(a) \bigl( \begin{smallmatrix} 1 \\ 0 \\ 0 \end{smallmatrix} \bigr) + 0 \bigl( \begin{smallmatrix} 0 \\ 1 \\ 0 \end{smallmatrix} \bigr) + \cos(a) \bigl( \begin{smallmatrix} 0 \\ 0 \\ 1 \end{smallmatrix} \bigr) = \bigl( \begin{smallmatrix} \sin(a) \\ 0 \\ \cos(a) \end{smallmatrix} \bigr)$

After that, I want to perform a counterclockwise rotation around the rotated base vector e3 (e.g. ϕ(e3)).

To do this, i'm using this transformation matrix:

\begin{bmatrix}\cos(b)&-\sin(b)&0\\\sin(b)&\cos(b)&0\\0&0&1\end{bmatrix}

And I map the "old" rotated basis onto the new basis using Ψ:

$Ψ(ϕ(\mathbf{e1})) = \cos(b) \bigl( \begin{smallmatrix} \cos(a) \\ 0 \\ -\sin(a) \end{smallmatrix} \bigr) + \sin(b) \bigl( \begin{smallmatrix} 0 \\ 1 \\ 0 \end{smallmatrix} \bigr) + 0 \bigl( \begin{smallmatrix} \sin(a) \\ 0 \\ \cos(a) \end{smallmatrix} \bigr) = \bigl( \begin{smallmatrix} \cos(a)\cos(b) \\ \sin(b) \\ -\sin(a)\cos(b) \end{smallmatrix} \bigr)$

$Ψ(ϕ(\mathbf{e2})) = -\sin(b) \bigl( \begin{smallmatrix} \cos(a) \\ 0 \\ -\sin(a) \end{smallmatrix} \bigr) + \cos(b) \bigl( \begin{smallmatrix} 0 \\ 1 \\ 0 \end{smallmatrix} \bigr) + 0 \bigl( \begin{smallmatrix} \sin(a) \\ 0 \\ \cos(a) \end{smallmatrix} \bigr) = \bigl( \begin{smallmatrix} -\cos(a)\sin(b) \\ \cos(b) \\ \sin(a)\sin(b) \end{smallmatrix} \bigr)$

$Ψ(ϕ(\mathbf{e3})) = 0 \bigl( \begin{smallmatrix} \cos(a) \\ 0 \\ -\sin(a) \end{smallmatrix} \bigr) + 0 \bigl( \begin{smallmatrix} 0 \\ 1 \\ 0 \end{smallmatrix} \bigr) + 1 \bigl( \begin{smallmatrix} \sin(a) \\ 0 \\ \cos(a) \end{smallmatrix} \bigr) = \bigl( \begin{smallmatrix} \sin(a) \\ 0 \\ \cos(a) \end{smallmatrix} \bigr)$

Assuming that what is written is true, now I have tried to do the verification, checking if the composition of the two rotations actually gives me the same result.

Therefore:

\begin{equation}
\begin{bmatrix}\cos(b)&-\sin(b)&0\\\sin(b)&\cos(b)&0\\0&0&1\end{bmatrix}*\begin{bmatrix}\cos(a)&0&\sin(a)\\0&1&0\\-\sin(a)&0&\cos(a)\end{bmatrix}=\begin{bmatrix}\cos(a)\cos(b)&-\sin(b)&\sin(a)\cos(b)\\\sin(b)\cos(a)&\cos(b)&\sin(a)\sin(b)\\-\sin(a)&0&\cos(a)\end{bmatrix}
\end{equation}

If now I tried to make a linear mapping transforming the canonical basic vectors with this composition of rotations, I would find a different result from the one found before.

Therefore (calling τ this linear mapping):

τ(e1) = \cos(a)\cos(b) $\bigl( \begin{smallmatrix} 1 \\ 0 \\ 0 \end{smallmatrix} \bigr)$ + \sin(b)\cos(a) $\bigl( \begin{smallmatrix} 0 \\ 1 \\ 0 \end{smallmatrix} \bigr)$ - \sin(a) $\bigl( \begin{smallmatrix} 0 \\ 0 \\ 1 \end{smallmatrix} \bigr)$ = $\bigl( \begin{smallmatrix} \cos(a)\cos(b) \\ \sin(b)\cos(a) \\ -\sin(a) \end{smallmatrix} \bigr)$

τ(e2) = -\sin(b) $\bigl( \begin{smallmatrix} 1 \\ 0 \\ 0 \end{smallmatrix} \bigr)$ + \cos(b) $\bigl( \begin{smallmatrix} 0 \\ 1 \\ 0 \end{smallmatrix} \bigr)$ + 0 $\bigl( \begin{smallmatrix} 0 \\ 0 \\ 1 \end{smallmatrix} \bigr)$ = $\bigl( \begin{smallmatrix} -\sin(b) \\ \cos(b) \\ 0 \end{smallmatrix} \bigr)$

τ(e3) = \sin(a)\cos(b) $\bigl( \begin{smallmatrix} 1 \\ 0 \\ 0 \end{smallmatrix} \bigr)$ + \sin(a)\sin(b) $\bigl( \begin{smallmatrix} 0 \\ 1 \\ 0 \end{smallmatrix} \bigr)$ + \cos(a) $\bigl( \begin{smallmatrix} 0 \\ 0 \\ 1 \end{smallmatrix} \bigr)$ = $\bigl( \begin{smallmatrix} \sin(a)\cos(b) \\ \sin(a)\sin(b) \\ \cos(a) \end{smallmatrix} \bigr)$

Clearly:
Ψ(ϕ(e1)) ≠ τ(e1), Ψ(ϕ(e2)) ≠ τ(e2), Ψ(ϕ(e3)) ≠ τ(e3)

When instead I should have expected the same result...

I noticed that by reversing the order of the matrix product, I get the same result instead... I can't understand why.

答: Note that$$\phi(e_1)=(\cos(a),0,-\sin(a))=\cos(a)e_1-\sin(a)e_3.$$Therefore\begin{align}\psi\bigl(\phi(e_1)\bigr)&=\psi\bigl(\cos(a)e_1-\sin(a)e_3\bigr)\\&=\cos(a)\psi(e_1)-\sin(a)\psi(e_3).\end{align}There is no reason for you to assume that this is equal to$$\cos(b)\phi(e_1)+\sin(b)\phi(e_2)+0\times\phi(e_3),$$which is what you did.


问:
In how many ways can the letters of the word ELEMENTARY be permutated such that no two E's are next to each other?
What I did was to separate out the 3 E's (EEE) from the word ELEMENTARY, which leaves L M N T A R Y. For no two E's to be next to each other, we can place the E's in 8 positions (between the alphabets
L M N T A R Y and also before L and after Y). Firstly, since all the letters L M N T A R Y are distinct the number of ways to permute them is 7! = 5040. Then to place the 3E's in the 8 locations; firstly the first E have 8 choices to choose from, then second E has 7 choices to choose from and 3rd E has 6 choices to choose from. Then using product rule, 5040 x 8 x 7 x 6 = 1693440.

However, my answer is wrong, the method above is correct till the "7! = 5040". In the solution that was given to me, there are indeed 8 positions to place the 3 E's. However, instead of using 8 x 7 x 6, the solution used C(8,3) = 56. Then, using product rule, 5040 x 56 = 282240 ways of permutating ELEMENTARY such that no two E's are next to each other.

Why is my answer wrong? Thank you! :")

答:
As you state, the difference between your answer is that you multiply by $8*6*5$ whereas the correct answer multiplies by $C(8,3)$. By definition, $C(8, 3) = \frac{8!}{3! 5!} = \frac{8*6*5}{3!}$, so your approach missed thus division by $3!$.

Essentially, your approach counted the same configuration multiple times. Let's label these 3 E's as $E_1, E_2, E_3$. Then your argument correctly shows that there are $7! * 8 * 7 * 6$ ways to arrange the letters of $E_1LE_2ME_3NTARY$ so that no $E_i$'s are adjacent. But of course, we aren't working with $E_i$, we're just working with $E$! Your counting would consider $E_1LE_2ME_3NTARY$ and $E_2LE_1ME_3NTARY$ to be different but all we did was shuffle the same letter around. That is, permuting the 3 E's doesn't lead to a new configuration. And how many permutations of 3 elements are there? $3!$ (And I'm not just excited!)

To summarize, for every correct configuration of $ELEMENTARY$, your method counts each $E_{\sigma(1)}LE_{\sigma(2)}ME_{\sigma(3)}NTARY$ as distinct while they should be the same, for $\sigma$ a permutation of $\{1,2,3\}$. There are $3!$ such permutations so you have to divide your result by $3!$ to get the correct one.


问:
How to find the Taylor series of function $f(x)=\frac{a\left(1-x^2\right)+b\sqrt{x}\left(1+x\right)}{2\left(1-x\right)^2}$, at $x=0$
How to find the Taylor series of function $f(x)=\frac{a\left(1-x^2\right)+b\sqrt{x}\left(1+x\right)}{2\left(1-x\right)^2}$, at $x=0$. I know that the Taylor series of the functions:

$\frac{t}{1-t}=\sum_{n=0}^{+\infty}t^{n+1}, \frac{t}{(1-t)^2}=\sum_{n=0}^{+\infty} nt^{n}$.

Body:
Setting $\sqrt x=t$ leads to
$$\frac{a \left(1-t^4\right)+b t \left(t^2+1\right)}{2 \left(1-t^2\right)^2}=-\frac{a}{2}+\frac{1}{4}\left[(b+2a)\frac{1}{1+t}-(b-2a)\frac{1}{1-t}+b\frac{1}{ (t-1)^2}-b\frac{1}{ (t+1)^2}\right]=$$
$$-\frac{a}{2}+\frac{1}{4}\left[(b+2a)\sum _{n=0}^{\infty } (-1)^n t^n-(b-2a)\sum _{n=0}^{\infty } t^n+b\sum _{n=0}^{\infty } (n+1) t^n-b\sum _{n=0}^{\infty } (-1)^n (n+1) t^n\right]=$$
$$-\frac{a}{2}+\frac{1}{4} \sum _{n=0}^{\infty } \left(2 a \left((-1)^n+1\right)-b \left((-1)^n-1\right) n\right)t^n=$$
$$=-\frac{a}{2}+\frac{1}{4} \sum _{n=0}^{\infty } \left(2 a \left((-1)^n+1\right)-b \left((-1)^n-1\right) n\right)x^{n/2}$$


问:
Why is there a point on the unit circle that is not represented by these parametric equations?
$$x=\frac{2t+1}{2t^{2}+2t+1}$$ $$y=\frac{2t^{2}+2t}{2t^{2}+2t+1}$$By squaring $x$ and $y$ and adding them up, I obtained $x^2 + y^2 = 1$ after some algebraic manipulation. But the question asks which point is not represented by these parametric equations. The first thing that came to my mind was to set the denominator does not equal to zero but obviously, it was the wrong this to do. What am I supposed to do in this situation?

Part 2 of the question asks whether the mapping from parameters to points is one to one or many to one. I have identifies that $x$ is odd, so it is definitely one-to-one. However, $y$ is represented by an even function so it cannot by one to one. The answer says overall it is indeed one-to-one - how can this be if $y$ is even? Unless the mapping is only defined by the $x$ function?

答: It can never be the point $(0,1)$. Note that$$\frac{2t^2+2t}{2t^2+2t+1}=1\iff2t^2+2t=2t^2+2t+1,$$which is impossible.