问:
Infinity as supremum for a sequence of numbers
Let $X_1, X_2, · · ·$ be sequence of independent nonnegative random variables having a common distribution function $F$ such that $P(X_1=0)<1, E[X_1]=a<\infty$ and $\mathrm{Var}[X_1]=\sigma^2<\infty$. Let's denote $$S_0 = 0, S_n = X_1 + \cdots + X_n$$ for $n = 1, 2, \ldots $ and $$N_t =\sup\left\{n:S_n ≤t\right\}$$ for any $t≥0$.

I want to show that $N_t\rightarrow\infty$ as $t\rightarrow\infty$.

By common sense, this is obvious but I do not know how to write it down in a mathematical way. Any help would be appreciated!

答:
First observe that the condition $P(X_1=0)<1$ implies that there exists $k_0$ such that $P(0\leqslant X_1\leqslant 1/k_0)<1$. Indeed, let $A_k$ be the event $\{0\leqslant X_1\leqslant 1/k\}$; the sequence $(A_k)$ is non-increasing and its intersection has probability smaller than $1$, hence so should have one of the $A_k$.

Consider now the event $B_j:=\{  X_j\gt 1/k_0\}$. These events are independent and $\sum_{j\geqslant 1}P(B_j)$ diverges, hence the Borel-Cantelli lemma show that $B_j$ happens infinetely often, which means that for (almost every) $\omega$, the set of $j$ such that $\omega$ belongs to $B_j$ is infinite. This shows that $S_n\to \infty$ almost surely as $n$ goes to infinity.

Now suppose that we do not have $N_t\to\infty$ almost surely. This means that there is a set $E$ of positive probability such that for each $\omega\in E$, $N_t(\omega)$ doesn't go to infinity as $t$ goes to infinity. Since $t\mapsto N_t(\omega)$ is non-decreasing, this means that $N_t(\omega)$ is bounded for each $\omega\in E$, i.e. $N_t(\omega)\leqslant C(\omega)$ for each $t\geqslant 0$. Therefore, by definition of $N_t$, we would have for each $t$ that $S_{C(\omega)+1}\gt t$, which is a contradiction.


问:
Detecting graph topology.
I have a set of graphs and I need to classify them with respect to their topology. Is there an algorithm which can detect the topology (random, regular, scale-free, etc.) of a given undirected graph?</p>

答:
The things you are calling "topology" are a hodgepodge of different notions (especially given that it's unclear what you mean by "etc.") and there's no natural algorithm that handles all of them at once. However, many things of this flavor are easy to test for.

A graph is regular if the degree of every vertex is the same, and we can easily compute the degree sequence of a graph and check if this holds.

A graph is scale-free if its degree sequence approximately follows a power law. So we can kind of test for this property by finding the degree sequence and doing a regression. Say that $n_k$ is the number of vertices with degree $k$. Then we want $n_k \approx C k^{-\gamma}$ or $\log n_k \approx \log C - \gamma \log k$, so the $R^2$ value of a linear regression on the points $(\log n_k, \log k)$ will tell us how accurate a power law model is. We probably want to stick to a range of $k$ for which $n_k$ is decently large.

Of course, this will not give us a yes-or-no answer any longer: it will give us a value saying how close the network is to being scale-free. In that respect, this property is closer to what we have to deal with when we consider random graphs...

...where is meaningless to say that we "test a graph for being random". A specific graph can't be "random" or "not random", but we have different random graph models that can generate graphs. What we can ask is whether it's plausible that a graph we're looking at was generated by a certain model.

There are many random graph models. Off-hand, we have:

$G(n,\frac12)$, which will be equally likely to generate any labeled graph on $n$ vertices.

More generally, the Erdos-Renyi model $G(n,p)$, which independently decides the presence of each edge with probability $p$.

We can generate a uniformly random $d$-regular graph.

Many random graphs such as the Barabasi-Albert model generate a random graph with a scale-free degree distribution.

As a simple test, we can start by asking "what is the probability that random graph model $M$ generated the graph we see?" This is good at deciding between random graph models. It is less good at dealing with the alternative hypothesis "none of these models are a good fit".

We can also look at parameters of interest such as degree distribution, clustering coefficient, path lengths, and so forth, and compare them to what we'd expect from a random graph model. This is a more useful way to tell if any random graph model is a good model for our data.

ContentLicense: CC BY-SA 4.0

Title: Detecting graph topology.


问:
We have that $\left|\cos\alpha\right|=\frac{3}{5}$. What is $\sin\alpha-\cos\alpha$?

We have that $\left|\cos\alpha\right|=\dfrac{3}{5}$ and $\alpha \in(90^\circ;180^\circ).$ Find $\sin\alpha-\cos\alpha.$

For every angle  $\alpha \in(90^\circ;180^\circ)$ we have:
$\sin\alpha\in(0;1), \cos\alpha\in(-1;0),\tan\alpha<0$ and $\cot\alpha<0.$ Can you give me a hint how to find $\sin\alpha-\cos\alpha$ without using the basic trig identity $\sin^2\alpha+\cos^2\alpha=1$ (we have proven it only for acute angles) or some other well-known identities (we have proven such only for acute angles). Thank you!

答: Since $\sin\alpha>0$, $\sin\alpha=\frac45$ so $\sin\alpha-\cos\alpha=\frac75$.


问:
Finding the Taylor Series for a complicated function
Is there some trick to find the Taylor Series for this function or is the only to do it is just to differentiate .. ?

$\biggr(\sqrt{1+x^2} \cdot \cos(\sin x)\biggr)^{\frac{1}{\sin^6x}-\frac{1}{x^6}}$

答:
Yes, the Taylor expansion can help, and I think that's always superior to l'Hospital, especially since people rarely think about when that's applicable at all. So we have $$\sin x=x-x^3/6+O(x^5),$$ and thus
$$\cos(\sin x)=1-x^2/2+5x^4/24+O(x^6).$$ Since $$\sqrt{1+x^2}=1+x^2/2-x^4/8+O(x^6),$$ we see that $$\sqrt{1+x^2}\,\cos(\sin x)=1-x^4/6+O(x^6).$$ Concerning the exponent: $$\frac{x}{\sin x}=1+x^2/6+O(x^4),$$ so
$$\left(\frac{x}{\sin x}\right)^6=1+x^2+O(x^4)$$ and $$\frac{1}{\sin^6x} - \frac{1}{x^6}=x^{-4}(1+O(x^2)).$$ That means that the limit of your expression as $x\to0$ is the same as $$\lim_{x\to0}(1-x^4/6)^{x^{-4}}=e^{-1/6}.$$ In a deleted comment I wrote $e^{-1/4}$, but that was an error, sorry. :)


问:
Find $\lim _{n \rightarrow \infty} \int_{0}^{1} n x^{n} e^{x^{2}} d x$
Evaluate $$\lim _{n \rightarrow \infty} \int_{0}^{1} n x^{n} e^{x^{2}} d x$$
My attempt:
Consider
$$I=\int_{0}^{1}x^ne^{x^2}$$
using Taylor's series expansion for $e^{x^2}$ we get
$$I=\int_{0}^{1}x^n\left(1+x^2+\frac{x^4}{2!}+\frac{x^6}{3!}+\cdots\right)$$
So we get
$$I=\int_{0}^{1}x^n+\int_{0}^{1}\frac{x^{n+2}}{1!}+\int_{0}^{1}\frac{x^{n+4}}{2!}+\cdots$$
Integrating we get
$$I=\frac{1}{0!(n+1)}+\frac{1}{1!(n+3)}+\frac{1}{2!(n+5)}+\cdots$$
Now we have,
$$\lim _{n \rightarrow \infty} \int_{0}^{1} n x^{n} e^{x^{2}} d x=\lim_{n \to \infty}nI$$
Thus we obtain
$$\lim _{n \rightarrow \infty} \int_{0}^{1} n x^{n} e^{x^{2}} d x=\lim_{n \to \infty}\frac{n}{0!(n+1)}+\frac{n}{1!(n+3)}+\frac{n}{2!(n+5)}+\cdots=\frac{1}{0!}+\frac{1}{1!}+\frac{1}{2!}+..$$
So we get
$$\lim _{n \rightarrow \infty} \int_{0}^{1} n x^{n} e^{x^{2}} d x=e$$
is this a valid approach?

答:
Yes. You still have to justify the last step (i.e., the interchange of $\lim\limits_{n\to\infty}$ and the infinite summation), but that's easy, and can be viewed as an application of (a discrete form of) DCT.

Here are two approaches that look easier for me:

Integrate by parts, and then apply DCT: $$\int_0^1 nx^n e^{x^2}\,dx=n\frac{x^{n+1}}{n+1}e^{x^2}\Big|_0^1-\frac{2n}{n+1}\int_0^1 x^{n+2}e^{x^2}\,dx\underset{n\to\infty}{\longrightarrow}e-0=e.$$

Substitute $x^{n+1}=y$ and (again) apply DCT: $$\int_0^1 nx^n e^{x^2}\,dx=\frac{n}{n+1}\int_0^1 e^{y^{2/(n+1)}}\,dy\underset{n\to\infty}{\longrightarrow}\int_0^1 e^1\,dy=e.$$


问:
Orthonormal sets and inner products
How do I prove that a given basis $B$ is an orthonormal set relative to a given inner product $\langle x,y\rangle$?

I am unsure if I should use the Gram-Schmidt process.

答: Gram-Schmidt is for taking a set of vectors (usually a basis) and orthonormalizing them. If you want to prove that a given basis is orthonormal, you just need to show that the pairwise inner products of the elements of the basis are zero, and that they each magnitude 1.</p>

Title: Orthonormal sets and inner products


问:
Exercise in FOL
Section 4.1, ex. $3.(2)$ of Introduction to mathematical Logic by Richard Hodel.

Express the following statements:

(2) For every natural number $a$ greater than 1, there is a prime $p$ that divides $a$.

Is the following, in language of FOL, correct:

$\forall a\in \Bbb N \left[a>1\Rightarrow\left(\exists p\in \Bbb N(\forall b\in \Bbb N \lnot(b \vert p)\; \land\; p \vert a\right) \right].$

Edit

Reply to @Bram28:

Writing instead of $(\forall b\in \Bbb N \lnot(b \vert p))$ this:

$(\forall b\in \Bbb N(b>1\; \land b\neq p))$

would work?

答:
No.

You always have that $p|p$, as well as $1|p$, whether $p$ is prime or not.

Hence, the $\forall b\in \Bbb N \lnot(b \vert p)$ part will always be false.

It'll take a bit more work to express that $p$ is a prime ... but take into consideration what I just said, and try again!

Also, the rest of the expression has the right format, good job!

Title: Exercise in FOL


答:
To verify that a given basis $B$ is orthonormal, you need to first verify that, for any distinct $x,y\in B$, we have $\langle x,y\rangle=0$, so $B$ is orthogonal. Next you need to verify that, for any $x\in B$, we have $\langle x,x\rangle=1$, so each vector in $B$ is a unit vector.

The Gram-Schmidt process is to convert a given basis into an orthogonal (or orthonormal if you want to) basis which generates the same space.

Title: Orthonormal sets and inner products

Body:
How do I prove that a given basis $B$ is an orthonormal set relative to a given inner product $\langle x,y\rangle$?

I am unsure if I should use the Gram-Schmidt process.


问:
How do I bound $\int_{\gamma_{R}}\frac{\sin x}{x(1+x^2)}dx$ where $\gamma_R$ is a half circle on the upper half plane, so that it vanishes?
I get $\int_{\gamma_{R}}\frac{\sin x}{x(1+x^2)}dx$=$\int^\pi_0 \frac{sin(Re^{i\theta})}{Re^{i\theta}(1+R^2e^{2i\theta})}iRe^{i\theta}d\theta$

so
$\mid \int^\pi_0 \frac{sin(Re^{i\theta})}{Re^{i\theta}(1+R^2e^{2i\theta})}iRe^{i\theta}d\theta\mid\leq \int^\pi_0 \mid \frac{sin(Re^{i\theta})}{Re^{i\theta}(1+R^2e^{2i\theta})}iRe^{i\theta}\mid d\theta
$

Then I tried to simplify the integrand. I get

$\mid \frac{sin(Re^{i\theta})}{Re^{i\theta}(1+R^2e^{2i\theta})}iRe^{i\theta}\mid=\mid\frac{e^{iRcos\theta-Rsin\theta}-e^{-iRcos\theta+Rsin\theta}}{2(1+R^2e^{2i\theta})}\mid$, and when I send $R\to \infty$, I can't bound this term to make it zero. Can you please explain where I did wrong? Or, is there any simpler way to do this so that this integral vanishes as $R$ goes to infinity? I've seen an argument using big $O$ but I really don't understand.

答:
Hint: It is better if you consider
$
\dfrac{\sin x}{x(1+x^2)} = \dfrac{\Im e^{ix} }{x(1+x^2)}
$
and work with $f(z) = \dfrac{e^{iz}}{z(1+z^2)}$.  You just need to show that

$$
\left|\dfrac{1}{z (1+z^2)}\right| \leq \frac{M}{R^k}, k > 1.
$$


问:
Quotient of two geometric Brownian motions
Given $X=X_t$, $Y=Y_t$ and
\begin{align}
      dX_t &= μX_t\,dt +  σX_t\,dB_t,
\\
      dY_t &= σY_t\,dt + μY_t\,dB_t,
\\
      V_t &= \frac{X_t}{Y_t}
\end{align}
How to calculate $dV_t$?

I'm new to SDE and calculus as well, so I'm not sure if I solved it wrong or not...

Please help, thank you so much!

答:
It is often more simple to compute the derivative in a product formula instead of in a quotient formula. In the product formula for $X=YV$ you get, in infinitesimal increments,
$$
dX_t=d(V_tY_t)=(V_t+dV_t)(Y_t+dY_t)-V_tY_t=V_t\,dY_t+Y_t\,dV_t+dV_t\,dY_t
$$
While the last term as infinitesimal increment is random, this randomness is vanishingly small around a trend $d\langle V,Y\rangle_t$, which is the quadratic variation or correlation of the two processes. If $dV_t=a(V)dt+b(V)dB_t$, then $d⟨V,Y⟩_t=b(V)μY_t\,dt$. So let's insert all known equations
$$
X_t(μ\,dt+σ\,dB_t)=V_tY_t(σ\,dt+μ\,dBt)+Y_t\,dV_t+b(V)μY_t\,dt\\
V_t(μ-σ)(dt-dB_t)=a(V_t)\,dt+b(V_t)\,dB_t+μb(V)\,dt
$$
Comparing coefficients gives
\begin{align}
a(V_t)+μb(V_t)&=(μ-σ)V_t
\\
b(V_t)&=-(μ-σ)V_t
\\
\implies
a(V_t)&=(μ-σ)(1+μ)V_t
\end{align}

For a formal derivation of these results use the Ito theorem on transformations of stochastic processes which gives the same formula without mucking around with infinitesimals.

One could also use the solution formulas for geometric Brownian motions
$$
X_t=X_0e^{(μ+σ^2/2)t+σB_t}\\
Y_t=Y_0e^{(σ+μ^2/2)t+μB_t}\\
V_t=V_0e^{((μ-σ)-(μ^2-σ^2)/2)t-(μ-σ)B_t}
$$
which directly implies, as again a geometric Brownian motion,
$$
dV_t=[(μ-σ)(1-(μ+σ)/2)-(μ-σ)^2/2]V_t\,dt-(μ-σ)V_t\,dB_t.
$$


问:
How to obtain the complete factorization of $P_n(X)=\frac{(X+i)^{2n+1} - (X-i)^{2n+1}}{2i} = 0$
I need to give the complete factorization of $P_n(X)=\frac{(X+i)^{2n+1} - (X-i)^{2n+1}}{2i} = 0$ and show that $P_n(X)=(2n+1)\prod_{k=1}^{n}(X^2-\frac{1}{t_{n,k}^{2}})$ with $t_{n,k}^{2}=\tan(\frac{k\pi}{2n+1})$.

Here is what I have done so far:
$$P_n(X)=\sum_{k=0}^{n}(-1)^{n-k}\binom{2n+1}{2k}X^{2k}=(2n+1)\left(X^{2n}+(-1)\frac{(1-2n)}{3}X^{2n-2}+...+(-1)^{n-1}nX^2+\frac{(-1)^n}{2n+1}\right)$$

But now I don't know how to progress further to get the asked result. I know that i can factorize a polynomial using the dominant coefficient $(2n+1)$ followed by the products of $(X-t_1)(X-t_2)...$ where $t_i$ are the roots, but how can i get the factorization with $X^2$ ?

答:
You already found out that $P_n$ has degree $2n$ and got the dominant coefficient. It remains to find the roots.

First, notice that $i$ is not a root of $P_n$. If $z$ is a root of $P_n$, then we necessarily have that $\left(\frac{z+i}{z-i}\right)^{2n+1}=1$ hence $\frac{z+i}{z-i}$ is a $(2n+1)$-th root of unity which is not one. For each of such roots of unity $\omega_{n,k}$, $1\leqslant k\leqslant 2n$, we can solve $\frac{z+i}{z-i}=\omega_{n,k}$.


问:
Proof in Propositional Logic of Peirce's Law
How can I proove in Propositional Logic (using only the basic axioms of P.L. and not a valuation function like it's used in Propositional Calculus) that :

$\vdash$ ((($\phi$ $\to$ $\psi$ )$\to$ $\phi$)$\to$ $\phi$)

where $\phi$,$\psi$ are given formulas ?

答: Using  form of natural deduction in which implication is left associative:

ContentLicense: CC BY-SA 3.0


答:
No Taylor, no Hospital.
Take the log to get
$$\left(\frac{1}{\sin^6 x}-\frac{1}{x^6}\right)\ln (\sqrt{1+x^2}\cos(\sin(x)))$$

Now
$$\left(\frac{1}{\sin^6 x}-\frac{1}{x^6}\right)=\frac{x^6-\sin^6x}{x^6\sin^6 x}=\frac{x^6}{\sin^6x}\frac{x^6-\sin^6x}{x^{12}}$$
And as $\frac{x^6}{\sin^6x}\to 1$ we can forget this factor.

Further

$$\frac{x^6-\sin^6x}{x^8}=\frac{x-\sin x}{x^3}\frac{x^2-x\sin x +x^2}{x^2}\frac{x^2+x\sin x +x^2}{x^2}\frac{x+\sin x}{x}$$
$$\to \frac{1}{6}\cdot 1\cdot 3 \cdot 2=1$$ So we can forget this also.

As $\frac{\ln(1+a)}{a}\to 1$ the problem reduces to
$$\lim\limits_{x\to 0} \frac{1}{x^4}\left(\sqrt{1+x^2}\cos(\sin x)-1\right)$$
This is where it gets tricky.
Consider now
$$\lim\limits_{x\to 0} \frac{1}{x^2}\left(\frac{\sqrt{1+x^2}}{x^2}\cos(\sin x)-\frac{1}{x^2}\right)$$
$$=\lim\limits_{x\to 0} \frac{1}{x^2}\left(\frac{\sqrt{1+x^2}-1}{x^2}-\frac{1}{2}\right)\cos(\sin x)+\frac{1}{x^2}\left(\left(\frac{1}{x^2}+\frac{1}{2}\right)\cos(\sin x)-\frac{1}{x^2}\right)$$

As $\cos(\sin x)=1$ when $x=0$ we ignore it also and evaluate first

$$\lim\limits_{x\to 0} \frac{1}{x^2}\left(\frac{\sqrt{1+x^2}-1}{x^2}-\frac{1}{2}\right)=-\frac{1}{8}$$

Leaving us with
$$\lim\limits_{x\to 0} \frac{1}{x^2}\left(\left(\frac{1}{x^2}+\frac{1}{2}\right)\cos(\sin x)-\frac{1}{x^2}\right)$$
to evaluate. Which can be done similarly, but no one really cares.

Title: Finding the Taylor Series for a complicated function

Body:
Is there some trick to find the Taylor Series for this function or is the only to do it is just to differentiate .. ?

$\biggr(\sqrt{1+x^2} \cdot \cos(\sin x)\biggr)^{\frac{1}{\sin^6x}-\frac{1}{x^6}}$


问:
$\sigma$-algebra of Borel in $\mathbb{R}$ is the power set of $\mathbb{R}$
We define the $\sigma$-algebra of Borel of $\mathbb{R}$ as the $\sigma$-algebra generated by the open intervals $(a,b)$ with $a,b \in \mathbb{R}$, $a<b$.

Let $x\in \mathbb{R}$ and construct an open interval $I=(a,b)$ such that $x\in I$. Then $x \in (a+\frac{n}{n+1}(x-a),b-\frac{n}{n+1}(b-x))$, $\forall n \in \mathbb{N}$.

We have $\{x\} = \cap(a+\frac{n}{n+1}(x-a),b-\frac{n}{n+1}(b-x))$. As any $\sigma$-algebra is closed under countable intersections, $\{x\}$ is contained in our $\sigma$-algebra.

We conclude that this $\sigma$-algebra is the power set of $\mathbb{R}$. However, I haven't seen this fact written in any text, so I wonder if my reasoning is wrong. Can anyone confirm or disprove my statement?

答:
How do you conclude that "this $\sigma$-algebra is the power set of $\mathbb{R}$" ?

You cannot write any element of the power set of $\mathbb{R}$ as a countable union of singletons.


问:
Homomorphisms between $\mathbb{Z} \star \mathbb{Z} $ and $\mathbb{Z} _2 $
What does a homomorphism between the free product $\mathbb{Z} \star \mathbb{Z} $ and $\mathbb{Z}_2 $ look like? I'm having trouble trying to do anything with this.

答:
Think about the epimorphism $\Bbb Z\rightarrow \Bbb Z_2: x\mapsto x\mod 2$.

For $\Bbb Z^2\rightarrow\Bbb Z_2$ there are two obvious possibilities

$(x,y)\mapsto x+y \mod 2$ and $(x,y)\mapsto x\cdot y \mod 2$

since $\mod n$, $n\geq 2$, is a congruence relation, i.e. compatible with the operations of addition and multiplication.


问:
The union of two simple planar graph have chromatic number $\leq 12$
Consider of graph $G$ being the union of two simple planar graph both on the same set of vertices. I want to show $\chi(G) \leq 12$.

Four colour theorem indicates that the chromatic number of a planar graph is less than 4, and for general graphs $\chi(G1 \cup G2) \leq \chi(G1)*\chi(G2)$.  But this would only yield an upper bound of 16. Is there anything particular to planar graph that helps to reduce the bound?

答:
Planar graphs satisfy $|E(G)| \le 3|V(G)|-6$, so the union of two planar graphs satisfies $|E(G)| \le 6|V(G)|-12$, which we can use to show that there is a vertex of degree less than $12$.

This leads us to a proof by induction: remove that vertex, color the rest, and then color that vertex. The inductive hypothesis applies because any subgraph of $G$ is also a union of two planar graphs: the corresponding subgraphs of $G_1$ and $G_2$.


问:
Intuitive understanding of the derivatives of $\sin x$ and $\cos x$
One of the first things ever taught in a differential calculus class:

The derivative of $\sin x$ is $\cos x$.

The derivative of $\cos x$ is $-\sin x$.

This leads to a rather neat (and convenient?) chain of derivatives:

```
sin(x)
cos(x)
-sin(x)
-cos(x)
sin(x)
...
```

An analysis of the shape of their graphs confirms some points; for example, when $\sin x$ is at a maximum, $\cos x$ is zero and moving downwards; when $\cos x$  is at a maximum, $\sin x$ is zero and moving upwards.  But these "matching points" only work for multiples of $\pi/4$.

Let us move back towards the original definition(s) of sine and cosine:

At the most basic level, $\sin x$ is defined as -- for a right triangle with internal angle $x$ -- the length of the side opposite of the angle divided by the hypotenuse of the triangle.

To generalize this to the domain of all real numbers, $\sin x$ was then defined as the Y-coordinate of a point on the unit circle that is an angle $x$ from the positive X-axis.

The definition of $\cos x$ was then made the same way, but with adj/hyp and the X-coordinate, as we all know.

Is there anything about this basic definition that allows someone to look at these definitions, alone, and think, "Hey, the derivative of the sine function with respect to angle is the cosine function!"

That is, from the unit circle definition alone.  Or, even more amazingly, the right triangle definition alone.  Ignoring graphical analysis of their plot.

In essence, I am asking, essentially, "Intuitively why is the derivative of the sine the cosine?"

答:
As a Physics Major, I would like to propose an answer that comes from my understanding of seeing sine and cosine in the real world.

In doing this, I will examine uniform circular motion.

Because of the point-on-a-unit-circle definition of sine and cosine, we can say that:

```<code>r(t) = < cos(t), sin(t) >
</code>```

Is a proper parametric function to describe a point moving along the unit circle.

Let us consider what the first derivate, in a physical context, should be.  The first derivative of position should represent, ideally, the velocity of the point.

In a physical context, we would expect the velocity to be the line tangent to the direction of motion at a given time t.  Following from this, it would be tangent to the circle at angle t.  Also, because the angular velocity is constant, the magnitude of the velocity should be constant as well.

```<code>r'(t)     = < -sin(t), cos(t) >
|r'(t)|^2 = (-sin(t))^2 + cos(t)^2
|r'(t)|^2 = sin(t)^2 + cos(t)^2
|r'(t)|^2 = 1
|r'(t)|   = 1
</code>```

As expected, the velocity is constant, so the derivatives of sine and cosine are behaving as they should.

We can also think about what the direction of the velocity would be, as well, compared to the position vector.

I'm not sure if this is "cheating" by the bounds of the question, but by visualizing the graph we can see that the velocity, by nature of being tangent to the circle, must be perpendicular to the position vector.

If this is true, then position * velocity = 0 (dot product).

```<code>                              r(t) * r'(t) = 0
  < cos(t), sin(t) > * < -sin(t), cos(t) > = 0
( cos(t) * -sin(t) ) + ( sin(t) * cos(t) ) = 0
              -sin(t)cos(t) + sin(t)cos(t) = 0
                                         0 = 0
</code>```

Life is good.  If we assume that the definition of cos(t) is -sin(t) and that the definition of sin(t) is cos(t), we find physical behavior exactly like expected: a constant velocity that is always perpendicular to the position vector.

We can take this further and look at the acceleration.  In Physics, we would call this the restoring force.  In a circle, what acceleration would have to exist in order to keep a point moving in a circle?

More specifically, in what direction would this acceleration have to be?

It takes little thought to arrive at the idea that acceleration would have to be center-seeking, and pointing towards the center.  So, if we can find that acceleration is in the opposite direction as the position vector, the we can be almost certain about the derivatives of sine and cosine.  That is, their internal angle should be pi.

```<code>                            r(t) * r''(t) = |r(t)| * |r''(t)| * cos(pi)
                            r(t) * r''(t) = |r(t)| * |r''(t)| * -1
< cos(t), sin(t) > * < -cos(t), -sin(t) > = |<cos(t),sin(t)>| * |<-cos(t),-sin(t)>| * -1
                    -cos(t)^2 + -sin(t)^2 = 1 * 1 * -1
               -1 * (cos(t)^2 + sin(t)^2) = -1
                                   -1 * 1 = -1
                                       -1 = -1
</code>```

Body:
As a Physics Major, I would like to propose an answer that comes from my understanding of seeing sine and cosine in the real world.

In doing this, I will examine uniform circular motion.

Because of the point-on-a-unit-circle definition of sine and cosine, we can say that:

r(t) = < cos(t), sin(t) >

Is a proper parametric function to describe a point moving along the unit circle.

Let us consider what the first derivate, in a physical context, should be.  The first derivative of position should represent, ideally, the velocity of the point.

In a physical context, we would expect the velocity to be the line tangent to the direction of motion at a given time t.  Following from this, it would be tangent to the circle at angle t.  Also, because the angular velocity is constant, the magnitude of the velocity should be constant as well.

r'(t)     = < -sin(t), cos(t) >
|r'(t)|^2 = (-sin(t))^2 + cos(t)^2
|r'(t)|^2 = sin(t)^2 + cos(t)^2
|r'(t)|^2 = 1
|r'(t)|   = 1

As expected, the velocity is constant, so the derivatives of sine and cosine are behaving as they should.

We can also think about what the direction of the velocity would be, as well, compared to the position vector.

I'm not sure if this is "cheating" by the bounds of the question, but by visualizing the graph we can see that the velocity, by nature of being tangent to the circle, must be perpendicular to the position vector.

If this is true, then position * velocity = 0 (dot product).

                              r(t) * r'(t) = 0
  < cos(t), sin(t) > * < -sin(t), cos(t) > = 0
( cos(t) * -sin(t) ) + ( sin(t) * cos(t) ) = 0
              -sin(t)cos(t) + sin(t)cos(t) = 0
                                         0 = 0

Life is good.  If we assume that the definition of cos(t) is -sin(t) and that the definition of sin(t) is cos(t), we find physical behavior exactly like expected: a constant velocity that is always perpendicular to the position vector.

We can take this further and look at the acceleration.  In Physics, we would call this the restoring force.  In a circle, what acceleration would have to exist in order to keep a point moving in a circle?

More specifically, in what direction would this acceleration have to be?

It takes little thought to arrive at the idea that acceleration would have to be center-seeking, and pointing towards the center.  So, if we can find that acceleration is in the opposite direction as the position vector, the we can be almost certain about the derivatives of sine and cosine.  That is, their internal angle should be pi.

                            r(t) * r''(t) = |r(t)| * |r''(t)| * cos(pi)
                            r(t) * r''(t) = |r(t)| * |r''(t)| * -1
< cos(t), sin(t) > * < -cos(t), -sin(t) > = |<cos(t),sin(t)>| * |<-cos(t),-sin(t)>| * -1
                    -cos(t)^2 + -sin(t)^2 = 1 * 1 * -1
               -1 * (cos(t)^2 + sin(t)^2) = -1
                                   -1 * 1 = -1
                                       -1 = -1

ContentLicense: CC BY-SA 2.5


问:
If $AB = I$ then $BA = I$

If $A$ and $B$ are square matrices such that $AB = I$, where $I$ is the identity matrix, show that $BA = I$.

I do not understand anything more than the following.

Elementary row operations.

Linear dependence.

Row reduced forms and their relations with the original matrix.

If the entries of the matrix are not from a mathematical structure which supports commutativity, what can we say about this problem?

P.S.: Please avoid using the transpose and/or inverse of a matrix.

答:
Coincidentally, a totally unrelated MathSciNet search turned up this article, which gives a result along the lines of (but slightly stronger than) the one in Martin Brandenburg's answer.

In particular:

Theorem (Hill, 1967): Let $R$ be a ring satisfying the ascending chain condition on right ideals.  Then:

a) For any $n \in \mathbb{Z}^+$, the matrix ring $M_n(R)$ also satisfies ACC on right ideals.

b) If $a,b \in R$ are such that $ab = 1$, then also $ba = 1$.

Title: If $AB = I$ then $BA = I$


问:
Excercise: Ordering functions using the BigO notation
This was one of the previous year's exam questions. I have to order the following functions according to their growth rates using the $\mathcal O(n)$ notation.

$f_1(n) = 2010 * \log_3(n^n)$

$f_2(n) = n^{1+2+...+ \log_2 \log_2 n }$

$f_3(n) = 4^{100+\log_2 n}$

How do I deal with these kind of problems?

答:
First you forget about all multiplicative constants. We are left with
$$ \log (n^n) \quad n^{\sum_{i=1}^{\log_2\log_2 n} i} \quad 4^{\log_2 n} $$
Now you simplify all the terms:
$$ n\log n \quad n^{\Theta(\log_2\log_2 n)^2} \quad n^2 $$
In principle you can first calculate the exact sum in $f_2$, but here it really doesn't matter. Now let's look at all the exponents of $n$:
$$ 1+o(1) \quad \Theta(\log_2\log_2 n)^2 \quad 2 $$
The $o(1)$ term is all that remains of the logarithm. Some other exercise might require comparison of such terms, but here we need not be so exact. The ordering is now clear: $$1+o(1) < 2 < \Theta(\log_2\log_2 n)^2.$$ So $f_1 \ll f_3 \ll f_2$.</p>


问:
$X_t$ and $Y_t$ are id for each $t$. $T_1$, $T_2$ are iid. Are $X_{T_1}$, $Y_{T_2}$ id?
I am working on a problem and as an intermediary step I think I need to use the following:

$X_t$ and $Y_t$ are identically distributed for each $t\in[0, \infty)$. $T_1$, $T_2$ are iid, continuous RVs. Then, $X_{T_1}$, $Y_{T_2}$ are identically distributed.

Intuitively I think that should be true but I don't know how to prove it.

If I am not mistaken, to show that property for $T_1$, $T_2$ discrete I could go through the characteristic function + the law of total expectation. What would be the analogue of this for $T_1, T_2$ continuous?

答: I don't think this needs to be true.  If we take $X_t$ and $Y_t$ to be independent Brownian motions, $T_1 := \inf \{t : X_t = 1 \}$, and $T_2 := \inf\{t : Y_t = -1 \}$ then $T_1$ and $T_2$ are i.i.d. by the symmetry properties of Brownian motion, but $X_{T_1} = 1$ and $Y_{T_2} = -1$ so $X_{T_1}$ and $Y_{T_2}$ are not identically distributed.


问:
Is the cardinality of a topology always of the form $2^n$?
Just to recap: Let $M$ be any set. A set $\mathcal{O}_M \subseteq P(M)$ is a topology on $M$ if it satisfies the three things that it satisfies (I won't write them down since it's not relevant to my question).

My question is directed toward the fact that $\mathcal{O}_M \subseteq P(M)$.

We know that for any set $M, |P(M)| = 2^{|M|}.$

Does this imply that all topologies always have a cardinality of $2^n$ for some $n \in \mathbb{N}$?

答: No. Consider the topology $\{\emptyset,M,\{x\}\}$ for $x\in M\ne\{x\}$.


答:
Since $G_1$ and $G_2$ are planar, we have $|E(G_1)| \le 3|V(G_1)|-6$ and $|E(G_2)| \le 3|V(G_2)|-6$. Since $V(G_1) = V(G_2)$ given, let $V = V(G_1) = V(G_2)$ for convenience. Then we have
$$|E(G_1)|+|E(G_2)| = |E(G_1 \cup G_2)| \le 6|V|-12$$
Now, by Handshaking Lemma, we also know that
$$\sum_{v \in V}d_{G_1 \cup G_2}(v) = 2|E(G_1 \cup G_2)| \le 12|V| - 24$$
So, one can easily see that there exists a vertex $v \in V$ such that $d_{G_1 \cup G_2}(v) < 12\ (*)$. Now, we will use induction on $|V|$.

Now, if $|V| = 1$, the result is clear (actually it is clear for $|V| \le 12$). Now, suppose inductively it holds for planar graphs with $|V| = n$. Then, for a vertex set with $|V| = n+1$, consider a graph with vertex set $V-v$ (Note that removing a vertex cannot make a planar graph non-planar). Then, by induction hypothesis, we can color this graph with at most $12$ colors. Now add $v$ back with the removed edges. By $(*)$, $d_{G_1 \cup G_2}(v) < 12$ so we can color this vertex with at least one of the colors used in the graph with vertex set $V-v$ and we are done.

Title: The union of two simple planar graph have chromatic number $\leq 12$

Body:
Consider of graph $G$ being the union of two simple planar graph both on the same set of vertices. I want to show $\chi(G) \leq 12$.

Four colour theorem indicates that the chromatic number of a planar graph is less than 4, and for general graphs $\chi(G1 \cup G2) \leq \chi(G1)*\chi(G2)$.  But this would only yield an upper bound of 16. Is there anything particular to planar graph that helps to reduce the bound?


问:
Exponential integral question
How would I solve the following problem?
$$f(x)=\int\!\frac{4}{\sqrt{e^x}}\,dx$$

Using $u$ substitution I have set $u=e^x$ andd $du=e^x dx$

so would I have $$4\int\!\frac{du}{\sqrt{u}}$$

What would I do from here?

Title: Exponential integral question

Body:
$$\sqrt{e^x}= e^{x/2},$$
so you have
$$
\int e^{-x/2} \, dx = \int e^u \,\Big( -2\,du \Big) = -2e^u + C \ -2e^{-x/2} + C.
$$


答: Let $X =\{ a,b,c \} $ and note that $ T = \{ \varnothing, X, \{a\}, \{a,b\}, \{a,c\} \}$ is a topology on $X$.  Moreover cardinality of $T$ is $5$.

Title: Is the cardinality of a topology always of the form $2^n$?

Body:
Just to recap: Let $M$ be any set. A set $\mathcal{O}_M \subseteq P(M)$ is a topology on $M$ if it satisfies the three things that it satisfies (I won't write them down since it's not relevant to my question).

My question is directed toward the fact that $\mathcal{O}_M \subseteq P(M)$.

We know that for any set $M, |P(M)| = 2^{|M|}.$

Does this imply that all topologies always have a cardinality of $2^n$ for some $n \in \mathbb{N}$?


问:
Prove that $\sum_{k=0}^nn\binom{n-1}{k-1} = n2^{n-1} $ ( on basis of $\sum_{k=0}^n\binom{n}{k}=2^n$). What is the part I don't understand?
I know that on the basis of that equality: $\displaystyle \sum_{k=0}^n\binom{n}{k}=2^n$

I can prove that those both are true.

$$\sum_{k=0}^nn\binom{n-1}{k-1} = n2^{n-1} $$

$$\sum_{k=0}^nn(n-1)\binom{n-2}{k-2} n(n-1)2^{n-2}$$

But I am not sure how. Let's take the first one:

$$\sum_{k=0}^nn\binom{n-1}{k-1} = n\sum_{k=0}^n\binom{n-1}{k-1} = n\sum_{k=-1}^{n-1}\binom{n-1}{k} = (?) = n2^{n-1} $$

I don't know hat to put in the place of that question mark. I see that the last form is not the one from my basic equality (in the sum, on bottom i got $k=-1$ and that is not equal to $k$).

答: We assume by convention that $\large n\choose x$ $=0$ whenever $x<0$


答:
Note\begin{align}
dV &= \frac{dV}{dX}dX+ \frac{dV}{dY}dY
+\frac12 \frac{d^2V}{dX^2}(dX)^2
 +\frac12 \frac{d^2V}{dY^2}(dY)^2
+\frac{d^2V}{dXdY}(dXdY)\\
&= \frac{1}{Y}dX- \frac{X}{Y^2}dY
+\frac12 \cdot0\cdot(dX)^2
 +\frac{X}{Y^3}(dY)^2
- \frac{1}{Y^2}(dXdY)\\
\end{align}
Substitute $dX/X= μdt+  σdB$ and $dY/Y =σdt+ μdB$ into above to obtain
$$dV = V[\mu dt+\sigma dB - \sigma dt-\mu dB
+\mu^2(dB)^2- \mu\sigma (dB)(dB)] \\
= (\mu-\sigma+\mu^2-\mu\sigma)Vdt + (\sigma-\mu)VdB
$$

Title: Quotient of two geometric Brownian motions

Body:
Given $X=X_t$, $Y=Y_t$ and
\begin{align}
      dX_t &= μX_t\,dt +  σX_t\,dB_t,
\\
      dY_t &= σY_t\,dt + μY_t\,dB_t,
\\
      V_t &= \frac{X_t}{Y_t}
\end{align}
How to calculate $dV_t$?

I'm new to SDE and calculus as well, so I'm not sure if I solved it wrong or not...

Please help, thank you so much!


问:
Find point farthest on $(x-5)^2+(y+1)^2+(z-7)^2=69$ from point $(2, 3, -6)$
So, I have the equation of a sphere $$(x-5)^2+(y+1)^2+(z-7)^2=69$$ and a point, $P(2, 3, -6)$. I am supposed to find the point on the sphere farthest away from $P$.

I tried to figure out how to use parametric equations to represent the sphere and maximize that using the distance formula. However, they are multiple variables in this equation, and this led me to nowhere.

Body:
The center of the sphere is $C(5,-1,7)$. The line passing through $C$ and $P(2,3,-6)$ has parametric equations
$$
\begin{cases}
x=5+(2-5)t&x=5-3t\\
y=-1+(3-(-1))t&y=-1+4t\\
z=7+(-6-7)t&z=7-13t\\
\end{cases}
$$
Substitute in the equation of the sphere
$$(x-5)^2+(y+1)^2+(z-7)^2=69$$
$$(5-3t-5)^2+(-1+4t+1)^2+(7-13t-7)^2=69$$
$$194 t^2 = 69\to t=\pm\sqrt{\frac{69}{194}}$$
the point on the sphere farthest away from $P$ is:
$$\left(3 \sqrt{\frac{69}{194}}+5,-2 \sqrt{\frac{138}{97}}-1,13 \sqrt{\frac{69}{194}}+7 \right)$$


问:
A system of quadratic diophantine equations
Find $(p,x,y)$, where $p$ is prime and $x,y$ natural numbers, such that:

$p+1=2x^2$

$p^2+1=2y^2$

A solution to this problem is $(7,2,5)$ and I tried to show that it is the only one, but I didn't get far. Can anyone help?

答:
From $p+1=2x^2, p^2+1=2y^2$ we have that $$p(p-1)=2(y-x)(y+x).$$ Since $p$ is prime it is $y-x=1$ or $y+x=p.$

If $y-x=1$ then $p=3$ and there is no solution.

If $y+x=p$ then $p+1=4x=2x^2\implies x=2.$ And thus $y=5,p=7.$


问:
How can I evaluate $\sum_{n=1}^{\infty} \frac{1+\sin^2 n}{n+n^{1.5}}$?
How can I evaluate this sum? My teacher asked but I can't get it.

$$\sum_{n=1}^{\infty} \frac{1+\sin^2 n}{n+n^{1.5}}$$

答:
You cannot expect a nice closed form answer to this. However, it is easy to show that it is bounded and thereby it converges. We have $\sin^2(n) \in (0,1)$. Hence, $1+\sin^2(n) \in (1,2)$.

We also have $n+n^{1.5} \in (n^{1.5}, 2n^{1.5})$. Hence, we have
$$\dfrac{1+\sin^2(n)}{n+n^{1.5}} \in \left(\dfrac1{2n^{1.5}}, \dfrac2{n^{1.5}} \right)$$
We hence have
$$\sum_{n=1}^{\infty}\dfrac{1+\sin^2(n)}{n+n^{1.5}} \in \left(\sum_{n=1}^{\infty}\dfrac1{2n^{1.5}}, \sum_{n=1}^{\infty}\dfrac2{n^{1.5}} \right) = \left(\dfrac{\zeta(3/2)}2, 2 \zeta(3/2)\right)$$


问:
Power series sum
Find the radius of convergence and the sum of the series:
$\sum \limits_{k=0}^\infty (2k^2+8k+5)x^k$.

I know how to find the radius of convergence but I have no idea how to calculate the sum of the series.
Where do I begin? Thanks.

答:
Hints:

For example,

$$f(x)=\sum_{k=0}^\infty x^k\,,\,\,|x|<1\implies f'(x)=\sum_{k=1}^\infty kx^{k-1}\implies f''(x)=\sum_{k=2}^\infty k(k-1)x^{k-2}\implies$$

$$f''(x)=\sum_{k=2}^\infty k^2x^{k-2}-\sum_{k=1}^\infty kx^{k-1}$$

but we know

$$f(x)=\frac1{1-x}\implies f'(x)=\frac1{(1-x)^2}\;\;\ldots\text{ and etc.}$$

Title: Power series sum


问:
How do we solve the ODE $(3-x)y' = y$
How do you solve

$$(3-x)y' = y$$

What I did was
\begin{align*}
(3-x)\mathrm{d}y = y\mathrm{d}x & \Longleftrightarrow
\frac{\mathrm{d}y}{y} = \frac{\mathrm{d}x}{3-x}\\\\
& \Longleftrightarrow \ln(y) = \ln(3-x) + C\\\\
& \Longleftrightarrow y = (3-x)c
\end{align*}

But the answer is $\dfrac{c}{3-x}$. How do you get it?

答: You have forgotten to put $-$ with $\ln(3-x)$, that will come due to $$\int \frac{dx}{3-x}=-\ln(3-x)$$

Title: How do we solve the ODE $(3-x)y' = y$


问:
Expressing k-cycle as a product of transpositions
Suppose $\pi = (a_1a_2a_3...a_k)$ is a $k$-cycle representing a permutation in $S_n$ for some $n \geq k$

I'm aware that $\pi$ can be expressed as a product of $k-1$ transpositions in the following way: $\pi = (a_1a_k)(a_1a_{k-1})...(a_1a_3)(a_1a_2)$

However I think that $\pi$ can also be expressed as the following product of $k-1$ transpositions:

$(a_1a_2)(a_2a_k)(a_2a_{k-1})...(a_2a_4)(a_2a_3)$

So I don't think the representation of $\pi$ as a product of transpositions is unique. If that's the case, does any representation of $\pi$ as a product of transpositions always contain $k-1$ transpositions or do different products of transpositions representing $\pi$ contain different numbers of transpositions? Is it a general rule that any $k$-cycle can be expressed only as a product of $k-1$ transpositions?

答:
So I don't think the representation of π as a product of transpositions is unique.

This is correct, and you found an example indeed.

Is it a general rule that any $k$-cycle can be expressed only as a product of $k−1$ transpositions?

Well, this is not true since we can write $(a_1a_2) = (a_1a_2)(a_1a_2)(a_1a_2)$ for instance. What does not change here is the parity of the number of transpositions. But it is true that any $k$-cycle can be written as product of $k-1$ transpositions and we cannot write a $k$-cycle as product of less than $k-1$ transpositions (see here).


问:
Bayesian networks problem
Attempting to understand Exercise 20 (pdf page 44) in this book:

The party animal problem corresponds to the network in g(3.14). The boss is angry and the worker has a headache - what is the probability  the worker has been to a party? All variables are binary and

$p(U =tr|P =tr,D=tr)=0.999$, $p(U =tr|P =fa,D=tr)=0.9$, $p(U =tr|P =tr,D=fa)=0.9$, $p(U =tr|P =fa,D=fa)=0.01$.

$\begin{matrix} & & D \\ & & \downarrow \\ P & \rightarrow & U \\ \downarrow & & \downarrow \\ (H) & & (A)\end{matrix} $

I'm unsure of how to solve this problem given only the above information, and help greatly appreciated. Can this even be solved with the information given in the question?

Title: Bayesian networks problem

Body:
Using the values provided in the book and by Maxim:

$p(U=tr|P=tr,D=tr)=0.999$

$p(U=tr|P=fa,D=tr)=0.9$

$p(U=tr|P=tr,D=fa)=0.9$

$p(U=tr|P=fa,D=fa)=0.01$

$p(H=tr|P=tr)=0.9$

$p(H=tr|P=fa)=0.2$

$p(A=tr|U=tr)=0.95$

$p(A=tr|U=fa)=0.5$

$p(P=tr)=0.2$

$p(D=tr)=0.4$

As the possibility of having headache only depends on the possibility of the worker going to a party (not directed path from any other component), $P(H = T|P = T)$ and $P(H = T|P = F)$ do not depend on any other factor. Hence, the given condition's probability is the weighted sum of each of the cases:

Equation 1:

$P(H = T, A = T) = P(H = T|P = T) P(H = T, A = T, P = T) + P(H = T|P = F) P(H = T, A = T, P = F)$

Equation 2: Aim is to find:

$P(P = T| A = T, H = T) = (P(H = T|P = T) P(P = T, H = T, A = T))/ P(H = T, A = T)$

Finding $P(H = T, A = T, P = T)$ required in Equation 1:

$P(P = T, H = T , A = T) =$

$P(P=T) P(D=T) P(U=T|D=T,P=T) P(A=T|U=T)$
$+ P(P=T) P(D=T) P(U=F|D=T,P=T) P(A=T|U=F)$
$+ P(P=T) P(D=F) P(U=T|D=F,P=T) P(A=T|U=T)$
$+ P(P=T) P(D=F) P(U=F|D=F,P=T) P(A=T|U=F) $

$= (0.2)(0.4)(0.999)(0.95) + (0.2)(0.4)(0.0001)(0.5) + (0.2)(0.6)(0.9)(0.95) + (0.2)(0.6)(0.1)(0.5)$

$ = 0.075924 + 0.00004 + 0.1026 + 0.006 $
$ = 0.184564$

Finding $P(H = T, A = T, P = F)$ required in Equation 1:

$P(P = F, H = T , A = T) = $

$P(P=F) P(D=T) P(U=T|D=T,P=F) P(A=T|U=T)$
$+ P(P=F) P(D=T) P(U=F|D=T,P=F) P(A=T|U=F)$
$+ P(P=F) P(D=F) P(U=T|D=F,P=F) P(A=T|U=T)$
$+ P(P=F) P(D=F) P(U=F|D=F,P=F) P(A=T|U=F)$

= $(0.8)(0.4)(0.9)(0.95) + (0.8)(0.4)(0.1)(0.5) + (0.8)(0.6)(0.01)(0.95) + (0.8)(0.6)(0.99)(0.5)$

$= 0.2736 + 0.016 + 0.00456 + 0.2376$

$= 0.53176$

Substituting obtained values into Equation 1:

$P(H = T, A = T) = P(H = T|P = T) P(H = T, A = T, P = T) + P(H = T|P = F) P(H = T, A = T, P = F)$

$P(H = T, A = T) = P(H = T|P = T)(0.184564) +  P(H = T|P = F)(0.53176)$

$P(H = T, A = T) = (0.9)(0.184564) + (0.2)(0.53176)$

$P(H = T, A = T) = 0.2724596$

Reiterating Equation 2:

$P(P = T| A = T, H = T) = (P(H = T|P = T) P(P = T, H = T, A = T))/ P(H = T, A = T)$

$P(P = T| A = T, H = T) = P(H = T|P = T)(0.184564)/P(H = T, A = T)$

$P(P = T| A = T, H = T) = (0.9)(0.184564) / 0.2724596$

$P(P = T| A = T, H = T) = 0.6096595$


问:
Memoryless source of an alphabet
I am doing an exercise where I have to find an upper bound Shannon's entropy for a memoryless source of an alphabet with $A$ letters. I don't understand what memoryless source of an alphabet means. Does that mean that the probability $p_i$ are all equal to $1/A$? How should I think about this?

答: The phrase "memoryless source" typically refers to the situation where you are drawing a long string of letters from the alphabet $A$, and each choice of letter is independent of all the others and each choice follows the same probability distribution.  It does not mean that that distribution has to be uniform; it can be any probability distribution on $A$.

Title: Memoryless source of an alphabet


答:
Hint: you series is a positive one and

$$\frac{1+\sin^2n}{n+n^{1.5}}\le\frac2{n^{3/2}}$$

Thus your series converges.

Title: How can I evaluate $\sum_{n=1}^{\infty} \frac{1+\sin^2 n}{n+n^{1.5}}$?

Body:
How can I evaluate this sum? My teacher asked but I can't get it.

$$\sum_{n=1}^{\infty} \frac{1+\sin^2 n}{n+n^{1.5}}$$


问:
Einstein manifold varying the $\lambda$ value
Given an Einstein manifold with constant negative Ricci curvature, so $Ric = \lambda g$ with $\lambda <0$, if I change the value of $\lambda$, the solution will only be the one given by its rescaled metric, or it is not always so simple, because there can be more Einstein manifolds corresponding to that value of $\lambda$ and not only the one to which it is sufficient to rescaling the metric?

答:
First of all, if $g$ is Einstein with Einstein constant $\lambda$, then $\mu\cdot g$ is Einstein with Einstein constant $\frac{\lambda}{\mu}$, then you are right while saying that an Einstein metric gives solution up to rescaling to any Einstein equation.

But there is no need for this metric to be unique. For example, on a compact surface of genus $\gamma \geqslant 2$, there exist many non-conformal hypebolic metrics. They are all Einstein as they are all of constant sectionnal curvature $-1$.

More generally, if there exist two different (say, non-conformal) metric $g_1$ and $g_2$ that are Einstein on $M$ with negative constant, that is $Ric_{g_i} = \lambda_ig_i$and $\lambda_i<0$, the equation
$$Ric_g = \Lambda g,~~~\Lambda <0$$
has at least two solution, which are are $\frac{\lambda_i}{\Lambda}g_i$.


问:
finitely generated projective module over local ring is free
I have a question about an argument from a proof in Hideyuki Matsumura's "Commutative Ring Theory" on page 9, Theorem 2.5:

Let $(A,\mathfrak{m})$ be a local ring; then a finitely generated projective module $M$ over $A$ is free.

The proof works as follows:

Choose a minimal $A$-basis $\omega_1,...,\omega_n$ of
$M$. Take into account that "minimal" means that there cannot exist another system of generators $b_1,..., b_m \in M$ with $m < n$ and $M = \sum_{i=1} ^m A b_i$.

Define a surjective map $\varphi:F \to M$ from the free module $F = Ae_1 \oplus \cdots \oplus Ae_n$ to $M$ by $\varphi(\sum a_i e_i) = \sum a_i\omega_i$. If we set $K = \operatorname{Ker}(\varphi)$ then, from the minimal basis property

$$\sum a_i \omega_i =0 \Rightarrow a_i \in \mathfrak{m} \text{ for all } i. $$

Thus $K \subset \mathfrak{m}F$.  Because $M$ is projective, there exists $\psi: M \to F$ such that
$F = \psi(M)\oplus K$, and it follows that $K = \mathfrak{m}K$. (???)

On the other hand, $K$ is a quotient of $F$, therefore finite over $A$, so that $K = 0$ by Nakayama and $F = M$.

Question: why the fact that $F = \psi(M)\oplus K$ implies that $K = \mathfrak{m}K$?

$K \subset \mathfrak{m}F$,
then we can deduce  $K＝K∩mF$＝$K∩（m \psi(M)\oplus mK）$
But why the last is equal to $mK$?

Thank you for your kind help.

答: You have $K=K\cap \mathfrak mF=K\cap(\mathfrak m\psi(M)\oplus \mathfrak mK)$. Since $K\cap \psi(M)=0$ ($\because F=\psi(M)\oplus K$), you get $K\cap\mathfrak m\psi(M)=0$. Thus $K=\mathfrak mK$ as required.


问:
Is $X - Y(Y^\top Y)^{-1}Y^\top XY(Y^\top Y)^{-1}Y^\top$ positive semi-definite or negative definite?
Let $X$ is a positive definite matrix and $Y$ is a matrix of full column rank. Is the following matrix positive semi-definite or indefinite:

$$
X - Y(Y^\top Y)^{-1}Y^\top XY(Y^\top Y)^{-1}Y^\top
$$

I tried to decompose it into a product of $(\cdot)$ and $(\cdot)^\top$, but couldn't.

答:
Suppose $X-PXP\succeq0$ where $P=Y(Y^YY)^{-1}Y^T$ is the orthogonal projection onto $V=\operatorname{range}(Y)$. Then $0\le(v+w)^T(X-PXP)(v+w)=2v^TXw+w^TXw$ for any $v\in V$ and $w\in V^\perp$. Therefore $v^TXw=0$, or else we can scale $w$ down by a (possibly negative) small number $\epsilon$ to make the RHS of the previous inequality negative. It follows that $V$ and $V^\perp$ are invariant subspaces of $X$. Since $P|_V=\operatorname{id}$ and $P|_{V^\perp}=0$, we see that $XP=PX$.

Conversely, if $XP=PX$, then $X-PXP=(I-P)X(I-P)\succeq0$. Therefore $X-PXP\succeq0$ if and only if $XP=PX$.

In case $X-PXP$ is not positive semidefinite, we must have $P\ne I$. Hence there exists some nonzero vector $w\in V^\perp$ and $w^T(X-PXP)w=w^TXw>0$. Therefore $X-PXP$ cannot be negative semidefinite and it must be indefinite.


问:
Hamiltonian - Classical Mechanics
I'm studying classical mechanics reading Mathematical Methods of classical mechanics, by Arnold and doing some exercises lists. As I'm studyng by myself, I got stuck on this exercise.

Consider the system $$\dot q=F_1(q,p), \dot p= F_2(q,p)$$ with $q,p$ in $\mathbb{R}$. Assuming that the vector $(F_1(q1,q2),F_2(q1,q2))$ is perpendicular to $(q1,q2)$ for all $(q,p)\in \mathbb{R}^2$, and that the functions $F_1,F_2$ are class $C^2$. Show that there is a Hamiltonian, such that the ODEs above are equal to Hamilton equations if, and only if, the function $G(q,p)=G_1(q,p)^2 + G_2(q,p)^2$ depends only on the norm of $(q,p)$.

Thoughts:
So, if  $G$ only depends on the norm, then the vector field $(F1,F2)$ is tangent to circles around the origin. Is this correct? To be honest I don't know how to continue, I'm a little lost.

答:
Given $qF_1+pF_2=0$, we want to prove some $H$ satisfies $F_1=H_p,\,F_2=-H_q$ iff some $f$ satisfies $F_1^2+F_2^2=f(q^2+p^2)$. Given such an $H$, can we get such an $f$; given such an $f$, can we get such an $H$?

The constraint on $H$ is equivalent to $H_{q^2}=H_{p^2}$, i.e. $H$ is a function of $q^2+p^2$, say $H=g(q^2+p^2)$. Then $F_1=2pg^\prime,\,F_2=-2qg^\prime,\,f=4(p^2+q^2)g^{\prime2}$.

So the real question is, can we get from $f$ to $g$ or vice versa (not necessarily uniquely)? Yes, viz. $f(x)=4xg^{\prime2}(x)$ with $x:=q^2+p^2$.

As a famous example, take $g(x)=x/2,\,f(x)=x$.

Title: Hamiltonian - Classical Mechanics


问:
Counting Problem - N unique balls in K unique buckets w/o duplication $\mid$ at least one bucket remains empty and all balls are used
I am trying to figure out how many ways one can distribute $N$ unique balls in $K$ unique buckets without duplication such that all of the balls are used and at least one bucket remains empty in each distribution?

Easy, I thought. I'll just hold a bucket in reserve, distribute the balls, and place the empty bucket. I get:

$ K\cdot N! / (N-K-1)! $

Even were I sure this handles the no duplicates condition, what if $K \geq N$?

Then I get a negative factorial in the denominator. Is the solution correct and/or is there a more general solution?

Thanks!

答:
If the number of buckets is greater than the number of balls, then all distributions qualify, so there are $K^N$ ways to do the job.

If $K\le N$, we can use Inclusion/Exclusion. There are $(K-1)^N$ ways to distribute the balls so that bucket $i$ is empty. So our first estimate is $K (K-1)^N$.  But this double counts, for each $i$ and $j$, the $(K-2)^N$ distributions that have $i$ and $j$ empty, So from $K(K-1)^N$ we must subtract $\binom{K}{2}(K-2)^N$. But we have subtracted once too many times the $(K-3)^N$ distributions in which $i$, $j$, and $k$ remain empty. And so on. We end up with
$$K(K-1)^N -\binom{K}{2}(K-2)^N+\binom{K}{3}(K-3)^N -c\dots.$$


问:
Why does integrating a test function allow us to probe badly behaved functions?
As a follow up to: Motivation behind weak form and the test function?

Why does integrating against a test function as a "probe" for functions $f$ and $g$? So when one integrates the equation against a test function:

$\Delta f = g \longrightarrow \int \Delta f \phi dxdy = \int g\phi dxdy$

What is really happening? I understand the test function is a sort of "distribution" of functions in some sense. How can we use this to "probe" $f$ and $g$ if these are not well behaved functions? Is the test function just a "trick" of sorts to shift the higher order derivatives onto $\phi$ and turn the PDE into a linear set of equations?

答:
Since in Motivation behind weak form and the test function?
 you are given a really good "physicist" point of view, I will try to give a "mathematician" point of view.

Let's say you are given a problem
$$\tag{1} -\Delta f = g \, ,$$
with $g \in C^{\infty}(\Omega)$ and $\Omega \subset \mathbb{R}^n$ is a smooth bounded domain. We want to find if a (or maybe more than one) solution $f$ exists such that $f|_{\partial \Omega} = 0$. I will show now two attempts to "solve" this problem and you will see where weak formulation and test functions naturally arises.

Attempt 1: we search for $f \in C^2(\overline{\Omega})$ and work with the PDE (1) in the point wise sense.

This may seem the most natural approach, but it turns out that this really is not the case. Since we are asking $f$ to be $C^2$, even only proving that one solution exists basically reduces in finding one. You have few weapons of attacking the problem different from building an explicit $f$ that solves the problem point wise and is twice differentiable. This turns out to be really hard in practice, even for simple domains, for example for $\Omega = B_{R}(0) \setminus B_{r}(0)$, constructing an explicit solution relies on finding the so called Green's function of that domain, and it's far from being easy even for very simple domains. Here we have come to the clue of the problem: building an explicit solution is really domain-dependant. This is really bad, mathematically speaking, because it means that even if you find an explicit formula for an annulus let's say, just bending the annulus a little bit makes you restart from scratch on the whole problem, since different domain means different Green's function, and different Green's function means different solution construction. I don't want to go into the details of Green's functions but believe me that it's hard (sometimes close to impossible) to find for most of domains. But let's say you have a magic box that computes Green's functions for all smooth domains, then you are done right? Speaking about existence theory of PDEs, it turns out you are farm from being done! Just let me change the problem a little bit. We work in some smooth $\Omega \subset \mathbb{R}^2$ and try to solve the problem
$$ \tag{2} (1+\sin(x+y)^2)f_{xx}+ (1+y^2x^4)f_{yy}=g$$
for some $f$ satisfying $f|_{\partial \Omega} = 0$. Guess what? For this problem your magic box that computes Green's functions is completely useless! You haven't build a general and strong theory of existence, every problem is drastically different if you look at it point wise, and solving two similar problems could mean using completely different techniques. As we will see, this problem is completely equivalent to the first one for the second attempt, here we go:

Attempt 2: we search for $f$ in some space of functions much bigger than $C^2(\Omega)$ and at first we don't ask $f$ to be a solution point wise, but just that it is solution in some "weak sense". Then, in a second moment we will build a theory that discovers if and when weak solutions are strong solutions.

This approach may let you think that us (mathematicians) always try to find the most difficult solutions to simple problems, but the reality is exactly the opposite. We start stating the problem weakly, that is defining $f$ to be a weak solution of (1) if: for every $\phi \in C^{\infty}_c (\Omega)$ we have
$$ \int_{\Omega} \nabla f \cdot \nabla \phi \, dx = \int_{\Omega} g\phi \, dx \, .$$
This formulation has already endowed the equation we want to solve (the PDE) with it's boundary data, in some sense now the boundary data and the PDE are coupled in the same formulation. Already at this point, one could guess that solving this problem probably will not depend on the domain much, at least not drastically. From a mathematical point of view, this problem is super well posed, on the LHS we have a scalar product in some Hilbert space $(H, \langle \cdot, \cdot \rangle)$ and on the right hand side we have a linear functional $l : H \to \mathbb{R}$. So this problem is really equivalent in finding a vector $f$ (let me call it a vector, as we imagine it as living in the vector space $H$ that I won't specify much, but it's some Sobolev space) such that
$$ \langle f, \phi \rangle = l(\phi) \,,$$
for all $\phi \in H$ (there's some density result I have used since now I allow the test functions in the whole Hilbert space $H$ and not in the subspace $C^{\infty}_c(\Omega)$, but let me skip these details). The existence of such an $f$ is guaranteed by simple and really general theorems as Rietsz's Representation Theorem, even for rough data $g$, for example in $L^2(\Omega)$. This procedure is, in contrast with the previous one, highly domain independant. It turns out that, by exactly the same procedure, one can solve problem (2) and similar problems in arbitrary dimension. This is because shifting one derivative to the text function $\phi$ completely changed the symmetry of the problem, making the LHS somewhat symmetric in $(f, \phi)$, and in mathematics when you find symmetries you are on the right track.

Ok but if you arrived at this point maybe you see that there's something that I have swept under the carpet carefully: is this "weak solution" a solution in the point wise sense? If the answer is "yes" then we are all happy and we have solved a large class of problems by a very general theory, if the answer is "no" then I am an impostor and this post makes little to no sense. Fortunately, it turns out that for regular domains and some regularity on the data the answer is yes! This gives rise to the beautiful Schauder theory, but I don't want to go off topic.

To summarize, weak formulation of a PDE is just a "trick" to enlighten symmetries where they are hidden. If you try to solve a problem in the point wise sense you realize that you are really asking too much, and you have little room to change the problem in order to fit some general result in mathematics. Thus the track becomes: first you relax your definition of solution so that the problem may be "symmetrized" to fit some strong/abstract theorem, and when you have existence of a weak solution by this method, then you have something to work with to prove that it is indeed smooth. This track turns out to be very effective in a lot of areas in mathematics.

EDIT: let me make one last comment. The whole point of the story is that derivatives are not point operators, they are local operators, meaning that the derivative of $f$ at $x_0 \in \Omega$ does not depend only on the value $f(x_0)$, it does depends on the values of $f$ around $x_0$. So if you think about it, taking a PDE and treating it point wise does not make much sense, since $-\Delta f = g $ is not a point wise equation, is a local one. If I say $-\Delta f(x_0)=g(x_0)$ to someone, I am implicitly prescribing the behavior of $f$ also in a neighborhood of $x_0$, I'n not only saying what happens at $x_0$. Then it turns out to be really natural to make the formulation of the PDE not point wise but local, multiplying by some arbitrary weight and then integrating, because in this way you capture the entire information the PDE can give you. You are not pretending anymore to be dealing with point operators.


答:
First of all, rule out $p\in\mathbb \{2,3\}$. Hence, $2\mid p-1$. Then, observe that your equations yield $$p\cdot \frac{p-1}2=(y-x)(y+x)$$ We could have $p\mid y-x$, but since $p>\frac{p-1}2$, this would imply that $y-x>y+x$. Contradiction. Thus, we have $p\mid x+y\implies p\leqslant x+y\implies 2p\leqslant 2(x+y)$. At the same time, we infer that $p-1\geqslant 2(y-x)$.

Add these inequalities to obtain $4x\geqslant p+1\iff 4x\geqslant 2x^2\iff x\in\mathbb \{0,1,2\}$. It is easy to see that $(x,y,p)\equiv (2,5,7)$ is the only solution.

Title: A system of quadratic diophantine equations

Body:
Find $(p,x,y)$, where $p$ is prime and $x,y$ natural numbers, such that:

$p+1=2x^2$

$p^2+1=2y^2$

A solution to this problem is $(7,2,5)$ and I tried to show that it is the only one, but I didn't get far. Can anyone help?


答:
The formula $\;\sum_{k=0}^n\binom{n}{k}=2^n$ stems from the binomial formula
$$\sum_{k=0}^n\binom{n}{k}x^k=(1+x)^n, \tag{1}$$
in which we set $x=1$.
Differentiating both sides of this formula, we get
$$n(1+x)^{n-1}=\sum_{k=0}^nk\binom{n}{k}x^{k-1},$$
and we can observe that
$$k\binom nk=k\frac{n!}{k!\,(n-k)!}= \frac{n!}{(k-1)!\,(n-k)!}=n\binom{n-1}{k-1}.$$
Can you proceed?

Title: Prove that $\sum_{k=0}^nn\binom{n-1}{k-1} = n2^{n-1} $ ( on basis of $\sum_{k=0}^n\binom{n}{k}=2^n$). What is the part I don't understand?

Body:
I know that on the basis of that equality: $\displaystyle \sum_{k=0}^n\binom{n}{k}=2^n$

I can prove that those both are true.

$$\sum_{k=0}^nn\binom{n-1}{k-1} = n2^{n-1} $$

$$\sum_{k=0}^nn(n-1)\binom{n-2}{k-2} n(n-1)2^{n-2}$$

But I am not sure how. Let's take the first one:

$$\sum_{k=0}^nn\binom{n-1}{k-1} = n\sum_{k=0}^n\binom{n-1}{k-1} = n\sum_{k=-1}^{n-1}\binom{n-1}{k} = (?) = n2^{n-1} $$

I don't know hat to put in the place of that question mark. I see that the last form is not the one from my basic equality (in the sum, on bottom i got $k=-1$ and that is not equal to $k$).


答:
$$(3-x)y' = y$$
Substitute $u=3-x \implies du=-dx$:
$$u\dfrac {dy}{dx}=y$$
$$u\dfrac {dy}{du}\dfrac {du}{dx}=y$$
$$-u\dfrac {dy}{du}=y$$
$$-\dfrac {dy}{y}=\dfrac {du}u$$
Integrate.

Title: How do we solve the ODE $(3-x)y' = y$

Body:
How do you solve

$$(3-x)y' = y$$

What I did was
\begin{align*}
(3-x)\mathrm{d}y = y\mathrm{d}x & \Longleftrightarrow
\frac{\mathrm{d}y}{y} = \frac{\mathrm{d}x}{3-x}\\\\
& \Longleftrightarrow \ln(y) = \ln(3-x) + C\\\\
& \Longleftrightarrow y = (3-x)c
\end{align*}

But the answer is $\dfrac{c}{3-x}$. How do you get it?


问:
Is the cardinality of a topology always of the form $2^n$?
Just to recap: Let $M$ be any set. A set $\mathcal{O}_M \subseteq P(M)$ is a topology on $M$ if it satisfies the three things that it satisfies (I won't write them down since it's not relevant to my question).

My question is directed toward the fact that $\mathcal{O}_M \subseteq P(M)$.

We know that for any set $M, |P(M)| = 2^{|M|}.$

Does this imply that all topologies always have a cardinality of $2^n$ for some $n \in \mathbb{N}$?

答: Since finite examples were given, let me give an infinite example. Consider the cofinite topology on $\Bbb N$. Since there are countably many finite subsets of $\Bbb N$, the cofinite topology is countable. However, there is no set $A$ such that $|2^A|=\aleph_0$.


问:
Proving the parity of subsets
We've been told that a set $T$ is called even if it has even number of elements. Letting $n$ be a positive even integer, and letting $S_1, S_2, \dots, S_n$ be even subsets of the set $S = ${$1,2,\dots,n$}, I am trying to find a way to prove that there exist $i$ and $j$, $1 \leq i < j \leq n$, such that $S_i \cap S_j$ is even. If anyone might be able to help with general guidance on how to conceptualize this problem to get started on a proof, I would greatly appreciate it! I am thinking there might be a way to prove it using matrices, but I'm not positive... Thanks!</p>

答:
Suppose by way of contradiction that all intersections are odd.

Let $A$ be the $n× n$ matrix whose rows are the indicator vectors for the subsets; that is, $A_{i,j}=1$ when $j∈ S_i$, zero otherwise. Think of the entries as integers modulo $2$.

Let $I$ be the $n×n$ identify matrix, and let $J$ be the $n× n$ matrix with all ones. The condition on even sizes and assumption on odd intersections implies
$$
AA^T = J-I
$$
Furthermore, $J-I$ is invertible mod $2$, since $(J-I)^2\equiv J^2 + I^2\equiv 0+I\pmod 2$. Therefore, we have
$$
n\ge\def\r{\text{rank }}\r A \ge \r AA^T = \r J-I = n,
$$
so $\r A=n$. However, letting $\def\one{{\bf 1}}\one$ be the $n× 1$ column vector of all ones, then $A\one={\bf 0}$, since each row has an even number of ones. Having a nonzero vector in the kernel of $A$ contradicts $\r A=n$.

Title: Proving the parity of subsets


问:
Topological extension property

Let $X$ and $Y$ be topological spaces. We say that the extension property holds if, whenver $S$ is a closed subset of $X$ and $f:S\rightarrow Y$ is continuous, $f$ can be extended to a continuous map on $X$.

A question asks whether the extension property holds for $X$ being any topological space and $Y=[-1,1]$.

I have no idea, but I am thinking whether there is some $X$ on which continuous functions are either constant or unbounded.

Can anyone offer any ideas? Thanks!

答:
Let $X$ be any space that is not completely regular, and let $p\in X$ and $F\subseteq X$ be a point and a closed set such that $p\notin F$, and there is no continuous function $f:X\to[0,1]$ such that $f(p)=0$ and $f(x)=1$ for each $x\in F$. Then it’s easy to check that there is no continuous function $f:X\to[0,1]$ such that $f(p)=-1$ and $f(x)=1$ for each $x\in F$.

Let $S=\{p\}\cup F$, and let

$$f:S\to[-1,1]:x\mapsto\begin{cases}-1,&\text{if }x=p\\1,&\text{if }x\in F\;;\end{cases}$$

then $f$ is continuous, but it cannot be extended continuously to $X$. And as long as $X$ is $T_1$, $S$ is closed in $X$.

This answer has a nice example, due to John Thomas, of a $T_3$-space $X$ with distinct points $p$ and $q$ such that every continuous $f:X\to[0,1]$ satisfies $f(p)=f(q)$, which is enough: you don’t actually need each continuous $f:X\to[0,1]$ to be constant, though Eric van Douwen showed some years ago that any space like the Thomas example can be used to build a $T_3$-space on which every continuous real-valued function is constant.

Title: Topological extension property


问:
Why is the cone measurable?
Problem

Denote by $\lambda_n$ the Lebesgue measure on $\mathbb {R}^n$.
Let $h \in \mathbb {R}_{>0}$. Let $A \subset \mathbb {R}^{n-1}$ be Lebesgue-measurable with finite measure.
Define the cone $C(A,h)$ as the union of all straight lines connecting a point in $A \times\{0\}$ with the point $(0, h) \in \mathbb {R}^{n-1} \times \mathbb{R}$. Prove that $\lambda_n (C(A,h))=\frac{h}{n}\cdot \lambda_{n-1}(A)$.

Where I struggle

Assuming that $C(A,h)$ is (Lebesgue-)measurable, this is a simple application of Fubini. How Fubini is used is all clear to me.

However, why is $C(A,h)$ measurable?
It seems one could roughly argue as follows: As $A$ and $[0,h]$  are measurable so is $A \times [0,h]$. As $A \times [0,h]$ is measurable it is the union of a null set and a Borel set. $C(A,h)$ is the image of $A \times [0,h]$ under a certain diffeomorphism (c.f. Deducing a formula for the volume of a cone over a $J$-measurable set.). Diffeomorphisms map null sets to null sets, and Borel sets to Borel sets. Hence, $C(A,h)$ is (as a finite union of measurable sets) measurable.

Unfortunately, I am not allowed to use the above properties of diffeomorphisms. What are other ways to prove the claim? Please only give a hint to set me thinking.

答:
By regarding $h$ as fixed, I will suppress $h$ from notation and simply write $C(A) = C(A, h)$. Now let $U = \mathbb{R}^{n-1} \times [0, h) $. Then $C(A)$ is Lebesgue-measurable if and only if

$$ C'(A) := C(A)\cap U = C(A) \setminus \{(0, h)\} $$

is Lebesgue-measurable. This modification provides a better control over the cone, as it preserves various set operations:

$ C'(\cup_{i\in I} A_i) = \cup_{i\in I} C'(A_i) $ for an arbitrary collection of subsets $\{A_i\}_{i\in I}$ of $\mathbb{R}^{n-1}$.

$ C'(\cap_{i\in I} A_i) = \cap_{i\in I} C'(A_i) $ for an arbitrary collection of subsets $\{A_i\}_{i\in I}$ of $\mathbb{R}^{n-1}$.

$C'(\mathbb{R}^{n-1}\setminus A) = U\setminus C'(A)$ for any subset $A$ of $\mathbb{R}^{n-1}$.

$C'(\varnothing) = \varnothing$ and $C'(\mathbb{R}^{n-1}) = U$.

All of these can be easily proved by noting that, if $f : U \to \mathbb{R}^{n-1}$ is defined by $f(x, x_n) = \frac{x}{1-(x_n/h)}$, then

$$ C'(A) = f^{-1}(A). $$

So, if $\mathcal{F}$ denote the set of all Lebesgue-measurable subsets of $\mathbb{R}^{n-1}$ and

$$ \mathcal{G} = \{ A \in \mathcal{F} : \text{$C'(A)$ is Lebesgue-measurable}\}, $$

then we observe the following:

$\mathcal{G}$ is a $\sigma$-algebra.

$\mathcal{G}$ contains all the open subsets of $\mathbb{R}^{n-1}$. If applicable, this is an immediate consequence of the fact that $f$ is continuous. Of course, it is possible to directly show that all the points in $C'(A)$ are interior points of $C'(A)$ in $U$.

These two properties already show that $\mathcal{G}$ contains all the Borel subsets of $\mathbb{R}^{n-1}$ by the Dynkin's $\pi$-$\lambda$ theorem.

$\mathcal{G}$ contains all the Lebesgue-null subsets $N$ of $\mathbb{R}^{n-1}$.

Indeed, for each open set $G$ containing $N$ and $\lambda_{n-1}(G) < \infty$, the Fubini–Tonelli theorem tells that $\lambda_n(C'(G)) = \frac{h}{n}\lambda_{n-1}(G)$. So by letting $\lambda_{n-1}(G) \downarrow 0$ along the open sets $G$ containing $N$, we find that $C'(N)$ is also a Lebesgue-null subset of $U$.

This and the previous observations altogether shows that $\mathcal{G} = \mathcal{F}$.

Title: Why is the cone measurable?


问:
The use of combination in statistics
Why can't we use 0 in combination?
Like for example, count the probability of a person wins at least one prize if he buys 10 out of 1000 tickets in which 5 tickets will be drawn to be the winner. The solution shows that the P is 1- 5C990/5C1000.
Why cant I use 1 - 0C10/5C1000 instead? Thank you

答:
Actually, the correct calculation is $$P=1-\frac{\binom{10}{0}\binom{990}{5}}{\binom{1000}{5}}$$

Representing that out of the $5$ cards picked from $1000$, $0$ winners are picked from the $10$ tickets that he bought, and  $5$ winners are picked from the $990$ he didn't

But $\binom{10}{0}=1$, so it is just left out of the solution.

For a calculation of the form $$\frac{\prod_{j\in\Bbb N}\binom{x_j}{y_j}}{\binom xy}$$ to be accurate in representing a probability we require that $\sum_{j\in \Bbb N} y_j= y$

Title: The use of combination in statistics


问:
Applications of the wreath product?
We recently went through the wreath product in my group theory class, but the definition still seems a bit unmotivated to me. The two reasons I can see for it are 1) it allows us to construct new groups, and 2) we can use it to reconstruct imprimitive group actions. Are there any applications of the wreath product outside of pure group theory?</p>

答:
Here are two places the wreath product is used in "real" applications.

The representation theory of $S_m\wr S_n$ can be used to prove results about voting for committees of a certain type.  (Published at https://doi.org/10.1016/j.aam.2020.102077)

The wreath product has been used in analyzing a number of different music theoretic concerns; see Peck's article for a survey.

In both cases, basically the object of study and its symmetries forced consideration of the wreath product rather than a larger symmetric group.  I think that's pretty cool.

Title: Applications of the wreath product?


问:
Convolution limits of positive function
Why are the limits of the convolution of two positive functions $0$ and $t$?

$$(f \star g)(t) = \int_0^t f(x)g(t-x)\,dx \text{ for } f,g : [0,\infty) \rightarrow \mathbb{R}$$

I understand the lower limit of 0 because both are non-negative functions, but why $t$?

Thank you!

答: Consider the integrand term $f(x)g(t-x)$. If $x<0$, then $f(x)=0$. If $x>t$, $g(t-x)=0$. Thus for a fixed $t$, $f(x)g(t-x)$ is 0 outside the interval $x\in [0,t]$.

Title: Convolution limits of positive function


问:
Stereographic Projection Mapping of Points on Z-plane
In a stereographic projection why is that points on the z-plane inside the sphere mapped from southern hemisphere and points on the z-plane outside the sphere mapped from the northern hemisphere?

Note: The question was edited.

答:
Consider this picture demonstrating the stereographic projection:

If the ray starting at the north pole intersects the northern hemisphere (pictured: the ray through $A$), it leaves the sphere before intersecting the complex plane. Therefore, it can only intersect the complex plane outside the sphere.

Conversely, if the ray intersects the southern hemisphere (pictured: the ray through $B$), it must first intersect the plane to do so, and the ray only leaves the sphere afterwards. Therefore, it intersects the complex plane inside the sphere.


问:
A linear operator $F : \mathbb{R}^3 \rightarrow \mathbb{R}^3$ projects vectors along (1,2,1) on a subspace
My teacher uploaded this question:

And uploaded this answer:

And I have spent way too long trying to understand what is going on. It seem like he is pulling numbers from thin air... :/

I think I got this figured out:

He renamed the components of the plane equation from $x_1, x_2, x_3 $ to a,b,c

The components of an arbitary vector $u$ is now $u=(x_1,x_2,x_3)$

I understand what is going on visually but I am having some trouble understanding the calculations. $F(u)=u+t_u(1,2,1)$ is this some kind of formula? What does $t_u$ mean? In the next step, the components of $F(u)$ are multiplied and added to the scalars of the plane equation. Why does this work? Is it because it is scaling $F(u)$ the same amount as the plane is scaling itself?

I am overall really close to figuring this out but I need some help.

Unfortunately due to covid all the teachers get bombarded with questions so trying to get an answer before the end of this year is not possible, that's why I am asking here instead of him.

答:
We have that $F(u) = u + t_u(1,2,1) = \cdots$ is the projection of $u$ along $(1,2,1)$ on $M$, and therefore satisfies...

This is how your teacher makes sense of "the projection along $(1,2,1)$ onto $M$". For an input point $u = (x_1,x_2,x_3)$, the output $F(u)$ is a specific point on the line through $u$ parallel to $(1,2,1)$. In particular, it is the point on the line where the line intersects the plane $M$.

With that in mind, note that the line through $u$ parallel to $(1,2,1)$ can be parameterized with
$$
r(t) = u + t(1,2,1) = (x_1 + t,x_2 + 2t, x_3 + t).
$$
The output $F(u)$ should be equal to $r(t)$ for some value of $t$ that we will call $t_u$. In other words, we have $F(u) = r(t_u)$. What makes $t_u$ special is that $r(t_u)$ is an element of $M$. Since $r(t_u) = (x_1 + t_u,x_2 + 2t_u, x_3 + t_u)$, $r(t_u)$ is an element of $M$ if
$$
(x_1 + t_u) + 2(x_2 + 2t_u) - (x_3 + t_u) = 0.
$$
Solving this equation for $t_u$ gives us
$$
t_u = -\frac 14 (x_1 + 2x_2 - x_3).
$$
From there, I think the rest should make sense.


问:
Does $n n^{1/n} =O(n)$?
I was asked does $n n^{1/n} =O(n)$ ?

I can see that the left hand side is always bigger than $n$ but how would you prove the equality is false?

Title: Does $n n^{1/n} =O(n)$?

Body: $$\frac{n\sqrt[n]n}n=\sqrt[n]n\xrightarrow[n\to\infty]{}1\implies n\sqrt[n]n=\mathcal O(n)$$


问:
Differential equation - Solve for y
The question is
$$\frac{dy}{dx} = 4xy^\frac12$$

I did

$$y^{-\frac12}\ dy = 4x\ dx$$
thus root $$y = 4x^2 + 2c$$
but the answer is $(x^2+c)^2$.

Help?

答:
Hint:

You correctly separated this equation, but you need to add integral signs to both sides and then integrate them

$$\frac{dy}{dx}=4xy^{\frac 12} \implies \int y^{-\frac 12}\ dy =\int 4x\ dx$$

Can you do this?

Title: Differential equation - Solve for y


问:
Power series sum
Find the radius of convergence and the sum of the series:
$\sum \limits_{k=0}^\infty (2k^2+8k+5)x^k$.

I know how to find the radius of convergence but I have no idea how to calculate the sum of the series.
Where do I begin? Thanks.

答:
Hint:

You can rewrite the sums as
\begin{align}
\sum_{k=0}^\infty 8k x^{k}&=8x\sum_{k=0}^\infty k x^{k-1},\\[1ex]
\sum_{k=0}^\infty 2k^2x^k&=2x^2\sum_{k=0}^\infty k(k-1)x^{k-2}+2x\sum_{k=0}^\infty kx^{k-1},
\end{align} whence
$$\sum \limits_{k=0}^\infty (2k^2+8k+5)x^k=2x^2\sum_{k=0}^\infty k(k-1)x^{k-2}+10x\sum_{k=0}^\infty kx^{k-1}+5\sum_{k=0}^\infty x^{k}.$$

Title: Power series sum


问:
Sufficient conditions for $f$ to be a rotation of $g$
Assume that $f$ and $g$ are analytic and nonvanishing on $B_{2}(0)=$ {$z \in \mathbb{C} :|z|<2$} and that $|f(z)|=|g(z)|$ when $|z|=1$. Can we show that there is an $|a|=1$ such that $f(z)=ag(z)$ for any $z$ in $B_{2}(0)$?

I believe we should look at $f/g$ which would also be analytic. I feel like the $a$ will come out of Schwarz Lemma. Thanks for the help.

答: You're right, you want to look at $f/g$ (call this $h$).  However, Schwarz Lemma would only help if $f(0) = 0$.  Instead, note that $1/\overline{h(1/\overline{z})}$ is analytic in $\{z: |z| > 1/2\}$ and coincides with $h(z)$ for $|z|=1$.  What can you conclude from this?</p>


答:
The equality is true. Recall that $\mathcal{O}(n)$ means the function is eventually $\leq C n$, where $C$ is a positive constant.

In your case, show that $n^{1/n} < e^{1/e}$ by considering the function $x^{1/x}$. We hence have for all $n \in \mathbb{N}$, $$n \cdot n^{1/n} < e^{1/e} \cdot n$$ Since, we have $n \cdot n^{1/n} < \underbrace{e^{1/e}}_{\text{const}} \cdot n$, $\forall n \in \mathbb{N}$, we have that $n \cdot n^{1/n} \in \mathcal{O}(n)$.

Title: Does $n n^{1/n} =O(n)$?

Body:
I was asked does $n n^{1/n} =O(n)$ ?

I can see that the left hand side is always bigger than $n$ but how would you prove the equality is false?


问:
Proving that a function $f: \mathbb{C} \to \mathbb{C}$ is bijective and find the inverse

Prove that the function $f : \mathbb{C} \rightarrow \mathbb{C}$, $f(x+iy) = x + (-1)^{\lfloor y \rfloor} + i(y + (-1)^{ \lfloor x \rfloor} ) $ for all $x,y \in \mathbb{R}$ is bijective and find its inverse.

My approach: For to prove that $f$ is injective, we need to show that if $z_{1},z_{2} \in \mathbb{C}$, $$f(z_{2})=f(z_{1}) \implies z_{1}=z_{2}$$
Suppose that $a+bi,c+di \in \mathbb{C}$ and that $f(a+bi)=f(c+di )$, so we have $$a+(-1)^{\lfloor b\rfloor}+i(b+(-1)^{\lfloor a \rfloor})=c+(-1)^{\lfloor d\rfloor}+i(d+(-1)^{\lfloor c \rfloor}) \implies \left\{ \begin{aligned} a+(-1)^{\lfloor b\rfloor}=c+(-1)^{\lfloor d\rfloor} \\ b+(-1)^{\lfloor a \rfloor}=d+(-1)^{\lfloor c \rfloor} \end{aligned}\right. \implies \left\{\begin{aligned} a=c \\ b=d \end{aligned}  \right.$$ so $f$ is injective.

Now, for to prove that $f$ is surjective we need to prove that if $c+di \in \mathbb{C}$, so there exists a $a+bi \in \mathbb{C}$ such that $f(a+bi)=c+di$, but it's it immediate? Isn't it just, for example, taking $a+bi=c+di$ and Q.E.D?

How can I prove that $f$ is surjective and how can I find the inverse of the funcion?

答:
To show surjective you have to show that given any $a$ and $b$  in $a+ib$ they can be written as
$$x + (-1)^{\lfloor y \rfloor} + i(y + (-1)^{ \lfloor x \rfloor} )$$
for some $x$ and $y$

Case 1: Both $a$ and $b$ are even integers.

Then take $x = a+1$, $y=b+1$. (Then $x$ and $y$ are both odd).

This is a solution because
$$x + (-1)^{\lfloor y \rfloor} + i(y + (-1)^{ \lfloor x \rfloor} )$$
$$=(x - 1) + i(y - 1)$$
$$=(a)+i(b)$$

Case 2: $a$ is even integer and $b$ is odd integer.

Then take $x = a-1$, $y=b+1$. (Then $x$ is odd and $y$ is even).

This is a solution because
$$x + (-1)^{\lfloor y \rfloor} + i(y + (-1)^{ \lfloor x \rfloor} )$$
$$=(x + 1) + i(y - 1)$$
$$=(a)+i(b)$$

Case 3: $a$ is odd integer and $b$ is even integer.

Then take $x = a+1$, $y=b-1$. (Then $x$ is even and $y$ is odd).

This is a solution because
$$x + (-1)^{\lfloor y \rfloor} + i(y + (-1)^{ \lfloor x \rfloor} )$$
$$=(x - 1) + i(y + 1)$$
$$=(a)+i(b)$$

Case 4: Both $a$ and $b$ are odd integers.

Then take $x = a-1$, $y=b-1$. (Then $x$ and $y$ are both even).

This is a solution because
$$x + (-1)^{\lfloor y \rfloor} + i(y + (-1)^{ \lfloor x \rfloor} )$$
$$=(x + 1) + i(y + 1)$$
$$=(a)+i(b)$$

Hence, we conclude:

For any $a+ib$ from the range of $f$ we can find $x+iy$ from the domain of $f$ such that $a+ib = f(x+iy)$


问:
Is it possible to evaluate $\lim_{n\to\infty}n(\sqrt[n]{4}-1)$ without applying L'Hospital's rule?
I can evaluate the limit with L'Hospital's rule:

$\lim_{n\to\infty}n(\sqrt[n]{4}-1)=\lim_{n\to\infty}\cfrac{(4^{\frac1n}-1)}{\dfrac1n}=\lim_{n\to\infty}\cfrac{\dfrac{-1}{n^2}\times 4^{\frac1n}\times\ln4}{\dfrac{-1}{n^2}}=\ln4$

But is there any way to do it without using L'Hospital's rule?

Body:
You could try this.  As $n \to \infty$,
$$
4^{1/n} = \exp\left(\frac{\log 4}{n}\right)
= 1 + \frac{\log 4}{n} + O(1/n^2)
\\
4^{1/n}-1 = \frac{\log 4}{n} + O(1/n^2)
\\
n\left(4^{1/n}-1\right) = \log 4 + O(1/n)
\\
\lim_{n\to\infty} n\left(4^{1/n}-1\right) = \log 4
$$


答: If $f(x)=4^x$, then $f'(x)=\log(4)4^x$ and, in particular, $f'(0)=\log(4)$. In other words,$$\lim_{h\to0}\frac{4^h-1}h=\log(4)$$and therefore$$\lim_{n\to\infty}\frac{4^{1/n}-1}{1/n}=\log(4),$$which is the same thing as asserting that$$\lim_{n\to\infty}n\left(\sqrt[n]4-1\right)=\log(4).$$Note that all that I used was the definition of derivative together with the knowledge of $(4^x)'$.

Title: Is it possible to evaluate $\lim_{n\to\infty}n(\sqrt[n]{4}-1)$ without applying L'Hospital's rule?

Body:
I can evaluate the limit with L'Hospital's rule:

$\lim_{n\to\infty}n(\sqrt[n]{4}-1)=\lim_{n\to\infty}\cfrac{(4^{\frac1n}-1)}{\dfrac1n}=\lim_{n\to\infty}\cfrac{\dfrac{-1}{n^2}\times 4^{\frac1n}\times\ln4}{\dfrac{-1}{n^2}}=\ln4$

But is there any way to do it without using L'Hospital's rule?


答:
You could use a Maclaurin expansion for $e^x$:

$$\begin{align}
n\left(\sqrt[n]{4}-1\right)
&=n\left(e^{\ln(4)/n}-1\right)\\
&=n\left(\frac{\ln(4)}{n}+\frac{1}{2}\left(\frac{\ln(4)}{n}\right)^2+\cdots\right)\\
&=\ln(4)+\frac{1}{2}\frac{\ln(4)^2}{n}+\frac{1}{6}\frac{\ln(4)^3}{n^2}+\cdots\\
\end{align}$$

As $n\to\infty$, convince yourself that the trailing terms collectively converge to $0$, leaving $\ln(4)$.


答:
Another approach using known limit

$$ \lim_{n \to +\infty} \frac{\mathrm{e}^{a_n}-1}{a_n} = 1 $$
where $a_n$ is a sequence such that  $ \lim_{n \to +\infty} a_n = 0$.

$$
\begin{split}
\lim_{n \to +\infty} \frac{4^{\frac{1}{n}}-1}{\frac{1}{n}}&=\lim_{n \to +\infty} \frac{\mathrm{e}^{\frac{\ln 4}{n}}-1}{\frac{1}{n}}\\
&=\lim_{n \to +\infty} \frac{\mathrm{e}^{\frac{\ln 4}{n}}-1}{\frac{\ln 4}{n}}\cdot \frac{\frac{\ln 4}{n}}{\frac{1}{n}}\\
&=\lim_{n \to +\infty} \frac{\mathrm{e}^{\frac{\ln 4}{n}}-1}{\frac{\ln 4}{n}}\cdot \lim_{n \to +\infty}\frac{\frac{\ln 4}{n}}{\frac{1}{n}}\\
&=  1 \cdot \ln 4 = \ln 4
\end{split}
$$


问:
Dice tracker and roll probability percentage
I have a dice tracker that graphs the combined roll of 2 fair six sided dice (numbers 2 through 12). I want to display the variance percentage that the current number is over/under as a positive/negative percentage. How would I do something like this?</p>

答:
It really depends on what you want to show:

One way to do is to compute what the expected value would be of such a roll and then contrast the current draw with that expected value as a percentage deviation.

Another approach is to compute a rolling average and show how the rolling average approaches the expected value as the throws increase.


问:
A Variation of the even-town odd-town problem

Let assume facebook has $n$ users. Mark Zuckerberg decided that people are allowed to open groups under the following restrictions:

1) No two different groups can exactly the same participants.

2) The number of users in each group is even.

3) Every two group has an odd number of common users.

Denote by $m$ the maximal number of groups that can be formed in facebook under the restrictions mentioned above. Prove $m \leq n$.

I tried solving this by defined a matrix, multiplying it by its transpose and using some rank considerations I thought I could derive the bound, but I didn't succeed. Could you please give me a hint?

The question had anther subsection : prove that in the case that $n$ is even than we have stron inequality

答:
Let $A$ be the $m\times n$ whose rows correspond to groups, whose columns correspond to people, with a $1$ in an entry if that person is in that group, and $0$ otherwise. From now on, we will think of these entries as elements of $\mathbb Z/2\mathbb Z$. The conditions on even sizes/odd intersections then imply
$$
AA^T\equiv J_m-I_m,
$$
where $J_m$ is the $m\times m$ matrix filled with ones, and $I_m$ is the $m\times m$ identity matrix. Therefore,
$$
\def\r{\text{rank }}\r A\ge \r AA^T=\r (J_m-I_m).
$$
Furthermore, the matrix $A$ has a nontrivial kernel; letting $v$ be the $n\times 1$ column vector with all ones, then $Av\equiv{\bf 0}\pmod 2$, since each row has an even number of ones. This means $\r A\le n-1$, so that
$$
\r (J_m-I_m)\le n-1 \tag{*}
$$
Now, we need the following:

Lemma: $$\r (J_m-I_m)=\begin{cases}m &  m \text{ is even}\\ m-1 & m \text{ is odd}.\end{cases}$$

Proof: When $m$ is even, we have $(J_m-I_m)^2\equiv (J_m)^2+(I_m)^2\equiv 0+I_m$, so that $J_m-I_m$ is invertible. When $m$ is odd, then $(J_m-I_m)v=0$, where $w$ is the $m\times 1$ vector of all ones. This proves $\r(J_m-I_m)\le m-1$; since the upper $(m-1)\times (m-1)$ submatrix is $J_{m-1}-I_{m-1}$, which was just shown to be invertible, the rank is indeed $m-1$. $\square$

In either case, we have $m-1\le \r(J_m-I_m)$, which combined with $(*)$ implies $m\le n$. This answers your first question. Furthermore, when $n$ is even, we cannot have $m=n$, otherwise we would have $\r(J_m-I_m)=\r(J_n-I_n)=n\not \le n-1$, contradicting $(*)$. Therefore, when $n$ is even, we can further conclude $m\le n-1$.


答: Let $t=\frac 1n \to 0$: $$\lim_{t\to 0} \frac{4^t-1}{t}  $$ which is of the well-known form $\lim_{x\to 0} \frac{a^x-1}{x} =\ln a $.


问:
Show the sum of a sequence of trace is convergent. $\operatorname{trace}( \sum_{t=0}^{\infty} (A')^t C' \Sigma_S C A^t S)$
I encountered this problem when doing research.

Matrices are all real-valued. $\Sigma_S$ is a covariance matrix. One can assume $\Sigma_S$ is positive definite. $A$ is a square matrix with dimension $n\times n$. $C$ is a $n\times m$ matrix  where $m\leq n$. $C$ is a bit redundant here. One can assume that $C'\Sigma_S C$ is positive definite.

Now I want to find the conditions on $A$ and $C'\Sigma_S C$ such that
$$
\operatorname{trace}( \sum_{t=0}^{\infty} (A')^t C' \Sigma_S C A^t S) < \infty
$$
where $S$ is positive definite. Here, prime $'$ means transpose and $A^t$ refers to the $t$th power of matrix $A.

Ideas: We can write $C\Sigma_SC = LL'$ and assume that $(A,L)$ is controllable. We can see $\sum_{t=0}^\infty (A')^t LL'A^t$ is actually the infinite-horizon controllability Gramian. If all eigenvalues of $A$ have magnitudes less than $1$, then there is a unique positive definite solution of
$$
W_\infty  -A'W_\infty A = LL'.
$$
And the controllability Gramian can be expressed as $W_\infty = \sum_{t=0}^\infty (A')^t LL'A^t.$ Then, we have $trace( \sum_{t=0}^{T} (A')^t C' \Sigma_S C A^t S)\rightarrow \operatorname{trace}(W_\infty S)$ as $T\rightarrow \infty$.

But this convergence requires that $A$ is table and $(A,L)$ is controllable. Can we conclude $$
\operatorname{trace}( \sum_{t=0}^{\infty} (A')^t C' \Sigma_S C A^t S) < \infty
$$ using only the assumption that $(A,L)$ is controllable?

More Background:

Consider a linear-quadratic-gaussian optimal control problem with a discrete-time linear system
$$
x_{t+1} = Ax_t + Bu_t + Cw_t,
$$
where $x_t$ is the state, $u_t$ is the control and $w_t$ is Gaussian noise with zero mean and variance $\mathbf{E}[w_t' w_\tau ]=\Sigma_S\delta(t-\tau)$. Here, $\delta(\cdot)$ is the Kronecker delta function. Given $x_0 =x$, let $\mathcal{F}_t = \{x_0,u_0,\dotsc,u_{t-1}\}$. And the problem is associated with quadratic stage cost $x_t'Qx_t + u_t' Ru_t$ with $Q\geq 0$ and $R\geq 0$.

We have $\hat{x}_t = \mathbf{E}\left[ x_t \middle \vert   \mathcal{F}_t \right] = A^{t-1}x +\sum_{\tau=0}^{t-1}A^{t-1-\tau}B u_\tau$. And the variance of estimation error $P_t(\mathcal{F}_t) := \mathbf{E}\left[(x_t -\hat{x}_t)'(x_t - \hat{x}_t)\middle\vert \mathcal{F}_t\right]$ is propagated as $P_{t+1} = A' P_t A + C'\Sigma_SC$ with $P_0 =0$. Thus, $P_t(\mathcal{F}_t)$ can be expressed as
$$
P_t(\mathcal{F}_t) = \sum_{\tau=0}^{t-1} (A')^\tau C' \Sigma_S C A^\tau.
$$

Let $S = A' PB (R+ B' P B)^{-1}B' PA$ where $P$ is a positive definite solution of an algebraic Riccati equation. Then the control performance degradation at time $t$ by not having observations from times $0$ to $t$ is
$$
\mathbf{E}\left[ (x_t - \hat{x})'S(x_t - \hat{x}) \middle\vert \mathcal{F}_t\right] = \operatorname{trace}(P_{t}(\mathcal{F}_t) S) = \operatorname{trace}(\sum_{\tau=0}^{t-1} (A')^\tau C' \Sigma_S C A^\tau S).
$$

Wants to find the loosest conditions such that the degradation is bounded as $t\rightarrow \infty$.

Added Questions: Under what conditions will the accumulated degradation with a discounted factor $\beta<1$
$$
\sum_{t=0}^\infty \beta^t \operatorname{trace}(P_{t}(\mathcal{F}_t) S)
$$
be bounded?

答:
Because $C'\Sigma_S C$ is positive definite, the function
$$
\|M\| = \sqrt{\operatorname{trace}(M' S^{1/2}[C'\Sigma_S C]S^{1/2}M)}
$$
defines a matrix norm. With Gelfand's formula, we have
$$
\rho(M) = \lim_{k \to \infty} \|M^k\|^{1/k}.
$$
Now, note that
$$
\operatorname{trace}( (A')^t C' \Sigma_S C A^t S) = \\
\operatorname{trace}( S^{1/2}(A')^t C' \Sigma_S C A^t S^{1/2}) = \\
\operatorname{trace}(([S^{-1/2}AS^{1/2}]')^t S^{1/2}C' \Sigma_S CS^{1/2} [S^{-1/2}AS^{1/2}]^t ) = \\
\|(S^{-1/2}AS^{1/2})^t\|^2.
$$
Note that $S^{-1/2}AS^{1/2}$ has the same eigenvalues as $A$. With Gelfand's formula, we can conclude by the root test that the sum will necessarily diverge when $A$ has any eigenvalue of magnitude strictly greater than $1$.

We can also see that if $A$ has an eigenvalue of maximal magnitude (exactly) $1$, then the sum diverges. Indeed, if $v$ is a (possibly complex) unit eigenvector of $S^{-1/2}AS^{1/2}$ associated with the eigenvalue $\lambda$ with $|\lambda| = 1$, then we have
$$
\|(S^{-1/2}AS^{1/2})^t\|^2 \geq
v'[(S^{-1/2}AS^{1/2})^t]' S^{1/2}[C'\Sigma_S C]S^{1/2}[(S^{-1/2}AS^{1/2})^t]v =\\
|\lambda|^{2t} v'S^{1/2}[C'\Sigma_S C]S^{1/2}v = \\
v'S^{1/2}[C'\Sigma_S C]S^{1/2}v > 0.
$$
So, the sequence being added has a positive lower bound. So, the sum necessarily diverges.

In other words, the sum will converge if and only if all eigenvalues of $A$ have magnitude smaller than $1$.


问:
dy/dx treated as fraction?
My question might seem very dumb but here goes.
Say we have y = f(x).

If we have g = dy/dx, we can also say that gdx = dy, and this works.
Why does it work? I know that dy/dx isn't a fraction but what underlying properties are being implicitly used when we change g = dy/dx into gdx = dy?

An example, to derive one of the laws of motion we can start with a = dv/dt, then say that adt = dv and by integration of both sides we conclude that at = v-u <=> v = u + at

答:
My question might seem very dumb but here goes.

Your question is one of the most commonly discussed topics in undergraduate mathematics.

It is true that $\frac {dy}{dx}$ is not a quotient. However, it is the limit of a quotient. And for most things we do, limits are actually quite well-behaved. In particular, if $\lim_x f(x)$ and $\lim_x g(x)$ exist, then $\lim_x f(x)g(x) = \lim_x f(x)\lim_x g(x)$.

And this is why the chain-rule works: $\frac{\Delta y}{\Delta x}\frac{\Delta x}{\Delta t} = \frac{\Delta y}{\Delta t}$. Now if $\lim_{\Delta t\to 0}\frac{\Delta x}{\Delta t}$ exists, then it must be that as $\Delta t\to 0, \Delta x\to 0$ as well. Otherwise the fraction would blow up, not converge. So, assuming the two right-side derivatives exist,

$$\begin{align}\frac {dy}{dt} &= \lim_{\Delta t\to 0} \frac{\Delta y}{\Delta t}\\
&= \lim_{\Delta t\to 0}\frac{\Delta y}{\Delta x}\frac{\Delta x}{\Delta t}\\
&= \left(\lim_{\Delta t\to 0}\frac{\Delta y}{\Delta x}\right)\left(\lim_{\Delta t\to 0}\frac{\Delta x}{\Delta t}\right)\\
&=\left(\lim_{\Delta x\to 0}\frac{\Delta y}{\Delta x}\right)\frac{dx}{dt}\\
&=\frac{dy}{dx}\frac{dx}{dt}
\end{align}$$

Now, hopefully you recognize that I've taken a few liberties in that demonstration, skipping over some fiddly details. But that is the point. Because derivatives are limits of quotients, they tend to behave like quotients. It is just the fiddly details that make it wrong to actually think of them as quotients.

If we have $g = \frac{dy}{dx}$, we can also say that $g\,dx = dy$, and this works.
Why does it work?

As stated (and barring the use of some sophisticated higher mathematical concepts), it doesn't work, because it doesn't even exist. There is no such thing as $dx$ or $dy$ by themselves. Instead we can only think of this as a short hand for either $$g\frac{dx}{dt} = \frac{dy}{dt}$$ (i.e., the chain-rule, which I've already explained) or for $$\int \frac{dy}{dx}\,dx = \int dy$$

But what are $\int \frac{dy}{dx}\,dx$ and $\int dy$? They are once again limits:

$$\begin{align}\int \frac{dy}{dx}\,dx &= \lim_{\Delta x \to 0} \sum\frac{\Delta y}{\Delta x}\Delta x\\
&= \lim_{\Delta x \to 0} \sum \Delta y\\
&= \lim_{\Delta y \to 0} \sum \Delta y\\
&= \int dy\end{align}$$

and again we see that if we sweep those fiddly details under the rug and just look at the overall form of the calculation, it is normal quotient behavior that isn't messed up by taking limits.

Title: dy/dx treated as fraction?


问:
Finding the solution of the nonhomogenous linar system of differential equations
I have the system $$\dot{x_1}=x_1+2x_2-6, \enspace \enspace x_1(0)=1 \\ \dot{x_2}=6x_1-3x_2+24 \enspace\enspace x_2(0)=-1$$
and want to find the solution of this IVP. I wrote the system as $$\dot{x}=Ax+b(t)$$ where $$ A = \left( \begin{matrix} 1 & 2 \\ 6 & -3 \end{matrix} \right) \text{ and }  b(t) = \left( \begin{matrix} -6 \\ 24 \end{matrix} \right). $$

I am studying from Perko's book and it says the following. If $e^{At}$ is any fundamental matrix solution of the system $\dot{x}=Ax$ then the solution of the nonhomogeneous linear system has the form $$ x(t)=e^{At}x_0 + e^{At} \int_{0}^{t} e^{-A\tau} b(\tau) d\tau. $$

To find $e^{At}$, I diagonalize $A$ by finding its eigenvalues $\lambda_1,\lambda_2$ and then corresponding eigenvectors $v_1,v_2$. So that I construct the matrix $P$ which has columns $v_1$ and $v_2$. Thus, I could be able to write $e^{At}=Pe^{Bt}P^{-1}$ where $B=diag(\lambda_1,\lambda_2)$. However, I think that there must be some easier way to find the solution as $b(t)$ does not depend on $t$ actually.

To sum up,

How can I find the solution of these type of systems where the nonhomogeneous part does not depend on the parameter?

How can I convert the system to a (probably) second order differential equation?

答:
$$\dot{x_1}=x_1+2x_2-6, \enspace \enspace x_1(0)=1 \\ \dot{x_2}=6x_1-3x_2+24 \enspace\enspace x_2(0)=-1$$
Substract both differential equations:
$${x'_1}-x'_2=-5x_1+5x_2-30$$
$$({x_1}-x_2)'=-5(x_1-x_2)-30$$
$$(({x_1}-x_2)e^{5t})'=-30e^{5t}$$
It's easy to integrate:
$$({x_1}-x_2)e^{5t}=-6e^{5t}+c_1$$
$${x_1}=x_2-6+c_1e^{-5t}$$
Note that since we have :
$$x_1(0)=1, x_2(0)=-1 \implies c_1=8$$
Substitute this in the second differential equation and solve to get $x_2$. Here substitute $x_1$:
$$\dot{x_2}=6\color {red}{x_1}-3x_2+24 $$
$$\dot{x_2}=6\color {red}{(x_2-6+8e^{-5t} )}-3x_2+24 $$
Then solve the equation, it's a first order DE:
$$\dot{x_2}-3x_2=48e^{-5t}-12 $$
$$({x_2}e^{-3t})'=48e^{-8t}-12e^{-3t} $$

2. How can I convert the system to a (probably) second order differential equation?

Just differentiate one of the two differential equation and substitute in the second one. Then you have a second order linear differential equation that you can easily solve.


问:
How to show $\frac{\sigma(m!)}{m!}>ln(m)$?
I am trying to show that $\frac{\sigma(m!)}{m!}>ln(m)$, where $m$ is a positive integer, and I can't seem to get anywhere in my proof. Where $\sigma(n)$ is the sum of the divisors of $n$.

I've tried to write $\sigma(m!)$ after taking an arbitrary prime power decomposition for m! and can't seem to get anywhere.

I'm personally thinking this has something to do with Merten's function since: $log(x)=\sum_{n\leq x} \frac{\Lambda(n)}{n}-O(1)$ with $\Lambda(n)=log(p)$ if $n$ is a prime power and 0 otherwise. I've tried expanding this function and playing around with it but still can't make headway relating this back to $\sigma$.

Any suggestions or hints would be greatly appreciated!

答:
If you are interested in the harder question of the asymptotics for $\sigma(n!)/n!$ then one can prove that it essentially behaves as $$ a_n = \mathrm e ^\gamma  \log n$$ where $\gamma $ is the Euler--Mascheroni constant. By this I mean that $$\lim_{n\to \infty}  \frac{\sigma(n!)}{n! a_n} =1 .$$ For an integer $m $ let $\alpha_p(m)$ be the largest power of a prime $p$ that divides $m$. Then we have
$$ \frac{\sigma (m)}{m} = \sum_{d\mid m} \frac{d}{m}=\sum_{d\mid m } \frac{1}{d}=\prod_{p\mid m } \left( 1+ \frac{1}{p}+\ldots + \frac{1}{p^{\alpha_p(m) }} \right) .$$ For an upper bound we have $$ \frac{\sigma (n!)}{n!} \leq \prod_{p\leq n } \frac{1}{1-1/p} ,$$ which, when divided by $a_n $ tends to $1$ by Mertens' third theorem. The lower bound is a bit harder. First,
$$ 1+ \frac{1}{p}+\ldots + \frac{1}{p^\alpha}
=\frac{1}{1-1/p} \left(1 -\frac{1}{  p^{\alpha +1 }   } \right) $$ so that $$ \frac{\sigma(n!)}{n! a_n } \geq \frac{\prod_{p\leq n } (1-1/p)}{a_n}  \prod_{p\leq n }
\left(1 -\frac{1}{  p^{\alpha_p(n) +1 }   } \right).$$ The first fraction in the right side tends to $1$ again by Mertens' third theorem. Hence, we only have to prove that $$ \lim_{n\to \infty } \prod_{p\leq n }
\left(1 -\frac{1}{  p^{\alpha_p(n) +1 }   } \right)  =1 .$$  Using the inequality $ (1-x) \geq e^{-2x}$ which is valid for all $0<x\leq 1/2$ we obtain $$  \prod_{p\leq n } \left(1 -\frac{1}{  p^{\alpha_p(n) +1 }   } \right) \geq \exp\left(-2 \sum_{p\leq n }   \frac{1}{  p^{\alpha_p(n) +1 }   } \right) $$ and hence it suffices to prove $$ \sum_{p\leq n }   \frac{1}{  p^{\alpha_p(n) +1 }   } \to 0 .$$ For this, one has to use that $\alpha_p(n!)$ is always kind of large for every small $p$. By Legendre's lemma   $$ \alpha_p(n!) = \sum_{k=1}^\infty
\left[ \frac{n}{p^k} \right]   $$   we get that $a_p(n)\geq 1$ for $ n^{1/2 }<p \leq n $. Hence, $$ \sum_{\sqrt{n}   < p\leq n }   \frac{1}{  p^{\alpha_p(n) +1 }   }
=\sum_{\sqrt{n}  < p\leq n }   \frac{1}{  p^2   }  \ll \frac{1}{\sqrt{n}    }.$$
For the remaining primes $p$ we have $p\leq n^{1/2 } \leq  n/2$ hence $[n/p] \geq n/(2p) $ hence $$ \alpha_p(n) \geq \frac{n}{2 p} \geq \frac{n}{2 n^{1/2} } =\frac{ \sqrt{n} }{2}  \Rightarrow \frac{1}{p^{1+\alpha_p(n) } } \leq  \frac{1}{2^{1+\alpha_p(n) } } \leq  \frac{1}{2^{ \sqrt{n }  /2  } }\Rightarrow \sum_{p\leq \sqrt{n} } p^{-1-\alpha_p(n)} \leq \frac{\sqrt{n}  }{2^{ \sqrt{n }  /2  } } \to 0  .$$ This concludes the proof. With a bit more details the same argument gives an explicit rate of convergence, namely $$ \frac{\sigma (n!)}{n!} =\mathrm e ^\gamma (\log n) +O\left(\frac{1}{n }\right).$$


问:
I am asking if it is possible to choose the following case: $λ_{2}<λ_{5}$ and $λ_{2}λ_{6}>0$
Let $A,B,C$ three coprime positive integers, i.e., there exist six integers $λ_{1},λ_{2},λ_{3},λ_{4},λ_{5},λ_{6}$ such that

$$λ_{1}A+λ_{2}B=1$$

$$λ_{3}A+λ_{4}C=1$$

$$λ_{5}B+λ_{6}C=1$$

I know that by the Bézout theorem (https://en.wikipedia.org/wiki/B%C3%A9zout%27s_identity#Structure_of_solutions) that there exists an infinity of them.

I am asking if it is possible to choose the following case: $λ_{2}<λ_{5}$ and $λ_{2}λ_{6}>0$ and $$λ_{3}>(-λ_{1}λ_{5})/(λ_{2}-λ_{5})$$ and $$λ_{3}>(-λ_{1}λ_{4}λ_{5})/(λ_{2}λ_{6})$$

答:
I'm paraphrasing my comment.

If, $A,B,C > 0$, yes: add $pB$ to $\lambda_1$, substract $pB$ from $\lambda_2$, add $pC$ to $\lambda_5$ and substract $pB$ from $\lambda_6$, with $p$ large enough.

For the edited version (with the first added condition), yes, you add $pC$ to $\lambda_3$ and take $pA$ from $\lambda_4$.