问:
Coordinates reachable from $(x, y) = (2,5)$
If we are on coordinate $(m,n)$ then we can either go to coordinate $(m + n, n)$ or $(m, m + n)$. If we start at $(2,5)$ then what all coordinates we can reach in finitely many moves?

In our case we have: From coordinate $(2,5)$ we can go to $(7,5)$ or $(2,7)$, now if you choose to go to $(7,5)$ then from $(7, 5)$ you can go to $(12,5)$ or $(7,12)$ and so on we keep on iterating this process and we need to find all such reachable coordinates. Any help would be highly appreciated. Thanks.

答:
First, note that the map $(m, n) \mapsto (m+n, n)$ corresponds to the matrix $E_1 = \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}$ and the map $(m, n) \mapsto (m, m+n)$ corresponds to the matrix $E_2 = \begin{bmatrix} 1 & 0 \\ 1 & 1 \end{bmatrix}$.  Therefore, it will be useful to study what matrices can be formed as products of these two matrices.

Lemma: A matrix in $M_{2\times 2}(\mathbb{Z})$ can be expressed as a product of $E_1$ and $E_2$ (for example $E_1^5 E_2^2 E_1^7 E_2^3 E_1$) if and only if it has determinant 1 and every entry is a nonnegative integer.

Proof: The forward direction is easy to see.  For the reverse direction, we will prove that any such matrix $\begin{bmatrix} a & b \\ c & d \end{bmatrix}$ with $ad - bc = 1$ and $a,b,c,d \in \mathbb{N}$ is a product of $E_1$ and $E_2$ by strong induction on $a + c$.

First, if any of $a,b,c$ or $d$ is zero, it certainly cannot be either $a$ or $d$, and $a = d = 1$.  Then if $b = 0$, we have
$$\begin{bmatrix} 1 & 0 \\ c & 1 \end{bmatrix} = E_2^c.$$
Similarly, if $c = 0$, we have
$$\begin{bmatrix} 1 & b \\ 0 & 1 \end{bmatrix} = E_1^b.$$

It now remains to show the case where all of $a,b,c,d > 0$.  Then if $a = c$, we must have the common value is $a = c = 1$, and the matrix is of the form $$\begin{bmatrix} 1 & b \\ 1 & b+1 \end{bmatrix} = E_2 \begin{bmatrix} 1 & b \\ 0 & 1 \end{bmatrix} = E_2 E_1^b.$$

If $a > c$, then $ad - bc = 1$ implies that $\frac{a}{c} - \frac{b}{d} = \frac{1}{cd}$.  Therefore, $\frac{b}{d} = \frac{a}{c} - \frac{1}{cd} \ge \frac{a}{c} - \frac{1}{c} \ge \frac{c}{c} = 1$, so $b \ge d$.  We now see that
$$\begin{bmatrix} a & b \\ c & d \end{bmatrix} = E_1 \begin{bmatrix} a - c & b - d \\ c & d \end{bmatrix}.$$
However, by inductive hypothesis since $(a-c) + c < a+c$, we have $\begin{bmatrix} a-c & b-d \\ c & d \end{bmatrix}$ is a product of $E_1$ and $E_2$, so $\begin{bmatrix} a & b \\ c & d \end{bmatrix}$ is also a product of $E_1$ and $E_2$.

If $a < c$, then a similar argument shows that $b \le d$, and $\begin{bmatrix} a & b \\ c & d \end{bmatrix} = E_2 \begin{bmatrix} a & b \\ c - a & d - b \end{bmatrix}$ where the second factor can be expressed as a product of $E_1$ and $E_2$ by inductive hypothesis. $\square$

Therefore, the points reachable from $(2, 5)$ via the operations $E_1$ and $E_2$ are precisely the points which can be expressed as $(2a + 5b, 2c + 5d)$ where $a,b,c,d \in \mathbb{N}$ and $ad - bc = 1$.

(I'm unsure of exactly what kind of characterization the original problem is asking for.  This does provide a straightforward way of generating reachable points - for example by choosing random natural numbers $b,c$, factoring $bc + 1$ and assigning a random factorization to $a,d$.  On the other hand, if you're given a pair and you want to determine whether it's reachable, it's probably easier to use the procedure described by Hagen von Eitzen's answer rather than trying to solve for $a,b,c,d$ in this description.)


问:
The inequality of Submartingale and Stopping time
First there have theorem

\begin{aligned}
&\text { Theorem 5.4.1. If } X_{n} \text { is a submartingale and } N \text { is a stopping time with } \mathbb{P}(N \leq k)=1\\
&\text { then } \mathbb{E} X_{0} \leq \mathbb{E} X_{N} \leq \mathbb{E} X_{k}
\end{aligned}

but this theorem is related to bounded stopping time

Then there have an exercise

Let $ \left \{X_ {n}, \mathcal{F}_{n} \right \}_{n \geq 0} - $ martingale or nonnegative submartingale, and let $ \tau- $ stop time with respect to $ \left \{\mathcal{F}_{n} \right \}_{n \geq 0} $ such that $ \mathrm{P} (\tau <\infty) = 1. $

How to prove that $ \mathrm {E} \left | X_{\tau} \right | \leq \lim_{n \rightarrow \infty} \mathrm{E} \left | X_{n} \right | $

答:
$|X_n|$ is a non-negative submartingale in both cases.

$E|X_{\min \{\tau,n\}}|\leq E|X_n|$ by Theorem 5.4.1 and Fatou's Lemma gives $E|X_{\tau}| =E \lim |X_{\min \{\tau,n\}}|\leq \lim \inf E|X_{\min \{\tau,n\}}| \leq \lim \inf E|X_n| $. Since $(E|X_n|)$ is monotonic its limit exists so  $\lim \inf E|X_n| $ is same as $\lim E|X_n|$.

ContentLicense: CC BY-SA 4.0


问:
Formal notation related to a sequence or a set
My question is quite naive...

I just want to represent a finite sequence of Natural number, is it the best way to write it like this?:

$\langle a_0, \ldots, a_n \rangle $, where $\forall i \in [ 0, n ], a_i \in \mathbb{N}$

I have also seen somewhere $\{ a_i \}$, does it mean a set (or a sequence) of variables? is it finite or infinite? where could i add the constraint like $\forall i \in [ 0, n ], a_i \in \mathbb{N}$?

Also, is there any difference between "tuple" and "sequence"?

Hope my question is clear, I just want to make sure what I write matches the convention...

Thank you very much

答:
There are several usual notations:

Let $\bar a = \langle a_i\mid i<n\rangle \in \mathbb N^{<\mathbb N}$... (where $\mathbb N^{<\mathbb N}$ means the collection of all finite sequences of natural numbers, and $\mathbb N^\mathbb N$ is the collection of all infinite sequences)

Let $\bar a = \langle a_i\mid i<n\rangle$ be a finite sequence of natural numbers...

And you can always replace $\langle a_i\mid i<n\rangle$ by $\langle a_0,\ldots,a_{n-1}\rangle$.

Note that $[0,n]$ can be read as the real numbers between $0$ and $n$.

The important thing is to be clear in your intention, and consistent in your notation.

And lastly, the difference between a "tuple" and a "sequence" is the context, usually tuple is used for finite length, while sequence is for infinite ones. Also, tuples are usually for a constant length, while sequences are a variable length.

The notation $\{a_i\}$ means that this is a set whose elements are distinct. Naturally it says $i\in I$ for some index set $I$, which in turn might be that this is just a sequence (or a directed set). The meaning is usually clear from context, read more - a lot more - and things will slowly clarify.

ContentLicense: CC BY-SA 3.0


问:
Infinite series for arctan of x
this is a bit of a vague question so I won't be too surprised if I get vague responses.

$$\tan^{-1}(x) = x - (x^3 / 3) + (x^5 / 5) - (x^7 / 7) + \cdots $$ ad infinitum

I'm using this, where $x = (\sqrt{2} - 1)$ to calculate $\pi$, as $\pi = 8  \tan^{-1}(x)$

I have never really learnt about infinite series before, and obviously when I translate this into a python code, I can't use a range from $(0, \infty)$.

So, my question is this; How do I/can I represent this idea in the form of a geometric series or is that the wrong way of going about it?

Thanks.

答:
You will have to truncate at some point. For this kind of alternating series (terms alternate in sign, decrease in absolute value, and have limit $0$) the truncation error is $\lt$ the first "neglected" term.

Remark: Note that $\tan^{-1}\frac{1}{2}+\tan^{-1}\frac{1}{3}=\tan^{-1}1 =\frac{\pi}{4}$. This may be a more pleasant way to approximate $\pi$ than stuff involving $\sqrt{2}$. There are many such "Machin-like" formulas.

Title: Infinite series for arctan of x


问:
Open subset and metric space intersections
Suppose $(A, d)$ is a metric space. If $B \subseteq A$ is a subset and $O \subseteq A$ is an open subset, show that $O \cap \overline{B} \subseteq \overline{O \cap B}$ and $\overline{O \cap \overline{B}} = \overline{O \cap B}$. Deduce that if $O \cap B = \emptyset$ then $O \cap \overline{B} = \emptyset$.

I have an extremely long direct proof of this (2 pages) but I am looking for a shorter version. Does anyone have any short or alternative proofs of this? Any assistance much appreciated.

答:
Proof: Let $x∈O∩\overline B$. Then, in order to prove that $x∈ \overline{O∩B}$, take an open $U$ containing $x$, and keep in mind that we want to show that $U∩(O∩B) \neq \varnothing$. Now, since $U∩O$ is an open set containing $x$, and $x∈ \overline B$, it follows that $(U∩O)∩B≠\varnothing.$
Next, the equality $$\overline{O \cap \overline{B}} = \overline{O \cap B}$$ follows by applying $U \subseteq V \implies \overline U \subseteq \overline V$ to the inclusions $O \cap \overline B \subseteq \overline{O \cap B}$ (which we previously proved) and $O \cap B \subseteq O \cap \overline B$ (this is trivial).


问:
Two maximum values for $|iz+3-4i|$ what went wrong?
I need to find the maximum value of $|iz+3-4i|$ given that $|z|\leq4$

$|iz+3-4i|\leq |iz|+|3-4i|$

$|iz+3-4i|\leq 4+5$

But I could write an inequality for $|3-4i|$ whose maximum value would be 7, not 5 so I get the maximum value to be 11 when doing so rather than the original answer of 9.

What went wrong here? , thanks in advance.

答: The reason is $|3-4i|$ represents a real number associated to the Eulidean norm of the point which you call the distance from origin. There is no way you can maximize a real number $c$(say) unless it is a real variable.


问:
Polynomial such that $e^{2i\pi P(n)} \rightarrow 1$
here is a  problem i’ve been having quite a lot of trouble with .

Let $P$ be a polynomial such that the sequence $e^{2i\pi P(n)}$ converges to 1 $(i^2=-1).$

Show that $\forall n ,P(n)$ is an integer.

Taking the imaginary part we are left with some $P$ polynomial such that

$$\sin(2\pi P(n))\rightarrow0$$
But a half angle factorization also gives  that: $$\sin(\pi P(n))\rightarrow 0$$
And we wish to force P to , i guess even if it’s not equivalent, to have only integers coefficients.

May you help me please.

答:
(Some years ago, I saw a proof.)

We use Mathematical Induction to prove the following statement:

Let $P(x)$ is a polynomial with real coefficients. If $\lim_{n\to\infty} \mathrm{e}^{2\pi \mathrm{i}P(n)} = 1$, then $P(n)$ is an integer
for any $n\in \mathbb{N}_{> 0}$.

When $\mathrm{deg}(P) = 0$, the statement is true.

Assume that the statement is true for $\mathrm{deg}(P) = k$ ($k\ge 0$).

Let us prove that the statement is also true for $\mathrm{deg}(P) = k + 1$.

We have $\lim_{n\to \infty} \mathrm{e}^{2\pi \mathrm{i}(P(n) - P(n-1))} = 1$.

Let $Q(x) = P(x) - P(x-1)$. Then $\mathrm{deg}(Q) = k$. By the induction hypothesis, $Q(n)$ is an integer for any $n\in \mathbb{N}_{> 0}$.

Since $P(n) = Q(n) + Q(n-1) + \cdots + Q(2) + P(1)$, we have
$\lim_{n\to\infty} \mathrm{e}^{2\pi \mathrm{i}P(n)} = \lim_{n\to\infty} \mathrm{e}^{2\pi \mathrm{i}P(1)} = 1$ and thus $P(1)$ is an integer.
Thus, $P(n)$ is an integer for any $ n\in \mathbb{N}_{> 0}$.

We are done.


问:
Find general solution of differential equation $x\frac{dy}{dx} + y = a \sin^2x.$ and $a \in \mathbb{R}$
The real number $a$ and the differential equation are given $$x\frac{dy}{dx} + y = a \sin^2x.$$

a) Find a general solution of the equation.

b) Find all $a \in \mathbb{R}$ that solutions of $y (x)$ of the differential equation
$$\lim_{x \to \infty}y (x) = 3.$$
Which of these solutions have limited functions?

So in example a) I got function $$y(x)=\frac{a(2x-\sin2x)+4C}{4x}$$
When I calculate limit of this function I got $\frac{a}{2}$, so $a = 6$.
Is this the only solution or is some different?

答:
$$\begin{align}x\frac{dy}{dx} + y &= a\sin^2(x) \\  x\frac{dy}{dx} + \frac{dx}{dx}y &= a\sin^2(x) \\ \frac{d}{dx}(xy) &= a\sin^2(x)\\ \therefore xy = \int a\sin^2x\ dx &= \frac{ax}{2} - \frac{a\sin(2x)}{4} + c\\ \implies y &= \frac{a}{2} - \frac{a\sin(2x)}{2x}+ \frac{c}{x}\end{align} $$

Therefore, as $ \lim_{x \to \infty} y = 3$ we have $\frac{a}{2} = 3 \implies a =6$.


答:
You are only going to get an approximation good to some number of decimal places.  If you use $N$ terms in the series, then the error is approximately the magnitude of the $N+1$th term.  The question you must ask yourself is, if I want $M$ decimal places of accuracy, then how big must $N$ be?

Example: say you want $M=6$ places of accuracy.  Then

$$\frac{(\sqrt{2}-1)^{2 N+1}}{2 N+1} < 10^{-6}$$

By trial and error, I get $N=6$.  That means you only need $6$ terms in the series to get that level of accuracy.

Title: Infinite series for arctan of x

Body:
this is a bit of a vague question so I won't be too surprised if I get vague responses.

$$\tan^{-1}(x) = x - (x^3 / 3) + (x^5 / 5) - (x^7 / 7) + \cdots $$ ad infinitum

I'm using this, where $x = (\sqrt{2} - 1)$ to calculate $\pi$, as $\pi = 8  \tan^{-1}(x)$

I have never really learnt about infinite series before, and obviously when I translate this into a python code, I can't use a range from $(0, \infty)$.

So, my question is this; How do I/can I represent this idea in the form of a geometric series or is that the wrong way of going about it?

Thanks.


问:
Expressing an integral in terms of incomplete beta function
How do we prove the following identity :
$I(p,q)=\int_0^1 t^p\,(2-t)^q\,dt=2^{p+q+1} B_{\frac{1}{2}}(p+1,q+1)$ where $B$ stands  for incomplete beta integral?Any help or hint will be greatly appreciated.

答:
I just realised it can be easily shown.In the integral
$$ I(p,q)=\int_0^1 t^p\,(2-t)^q\,dt$$
substitute $$ t=2u$$  so that  $dt=2du$.In terms of the new variable the new variable beocomes:
$$ I(p,q)=\int_0^{1/2} (2u)^p\,(2-2u)^q\,2du$$ which simplifies to :
$$I(p,q)=2^{p+q+1}\int_0^{1/2} u^p\,(1-u)^q\,du=2^{p+q+1} B_{\frac{1}{2}}(p+1,q+1)$$


问:
Question about the proof of $d(\omega \wedge \mu) = d\omega\wedge \mu + (-1)^k\omega \wedge d\mu$
In John M. Lee's Introduction to smooth manifolds on p. 217 he proves the equality

$$d(\omega \wedge \mu) = d\omega\wedge \mu + (-1)^k\omega \wedge d\mu$$

I just don't understand a step in his proof and I can't get my head around it:

The crucial part is here:

$$(g\,df+f\,dg)\wedge dx^I\wedge dx^J = (df\wedge dx^I) \wedge (g\,dx^J) + (-1)^k(f\,dx^I)\wedge (dg\wedge dx^J)$$

I simply don't understand how we obtain the right hand side. What I did was:

$$(g\,df+f\,dg)\wedge dx^I\wedge dx^J = (g\,df)dx^I\wedge dx^J + (f\,dg)dx^I\wedge dx^J$$

How exactly do I get from my right hand side to the desired one? I know about the anticommutativity of the wedge product, but I just don't get why

$(g\,df)dx^I\wedge dx^J$ becomes (or is equal to) $(df\wedge dx^I) \wedge (g\,dx^J)$

whereas

$(f\,dg)dx^I\wedge dx^J$ all of a sudden takes care of the anticommutativity and turns into $(-1)^k(f\,dx^I)\wedge (dg\wedge dx^J)$

Any help is appreciated.

答:
As @Ted Shifrin mentioned, switching a 1-form and a $k$-form introduces $(-1)^k$. In general, if $\omega$ is a $k$-form and $\mu$ is a $l$-form, then $$\omega\wedge\mu=(-1)^{kl}\mu\wedge\omega.$$

We can show it for $\omega=dx_1\wedge\cdots\wedge dx_k$ and $\mu=dx_{k+1}\wedge\cdots\wedge dx_{k+l}$ and extend linearly.

To switch from $\omega\wedge\mu$ to $\mu\wedge\omega$, we have to pass $dx_k$ through all $l$ factors of $\mu$, so we get $$\omega\wedge\mu=(-1)^l dx_1\wedge\cdots\wedge dx_{k-1}\wedge \mu\wedge dx_k$$

We repeat this for $k$ many factors, so $\omega\wedge\mu=(-1)^{kl}\mu\wedge\omega.$


问:
What are the elements of metric space?Is it always an interval are it can be set of integers etc.
What are the elements of metric space? Is it always an interval or it can be set of integers, etc.?</p>

答:
An metric space can, as mentioned above, be any non-empty set. Hence, the elements of a metric space can be almost anything you can imagine, as long as you can equip that set with a metric. A metric is intuitively a way to measure the distance between elements. Here are some examples:

Example 1: $\mathbb{R}^n$ with the Euclidean metric. The elements are in this case vectors $\mathbf{x} = (x_1, \ldots , x_n)$ and the metric is

$$d(\mathbf{x}, \mathbf{y}) = \lVert \mathbf{x} - \mathbf{y} \rVert = \bigg( \sum_{j=1}^n (x_1-y_1)^2 \bigg)^{1/2}.$$

Example 2: The discrete metric. Let $X$ be a non-empty set. We can make it a metric space by endowing it with the discrete metric, given by

$$d(x,y) = \begin{cases}
0, \quad & x = y \\
1, \quad & x \neq y
\end{cases}.$$

Example 3: The vector space of continuous real-valued functions on $[a, b]$. Consider the vector space $C([a, b])$ of continuous functions $f : [a, b] \to \mathbb{R}$. We can give it with the following metric:

$$d(f, g) = \bigg( \int_a^b f(x)g(x) \, dx \bigg)^{1/2}.$$

Notice that in these three examples, the elements of the metric spaces are different objects. Mind you that a set can often be equipped different metrics and still be a metric set. For example are the metrics I have used in example 1 and 2 not the only metrics compatible with $\mathbb{R}^n$ and $C([a, b])$.


答: A metric space is an arbitrary non-empty set $X$, together with a metric, i.e., a map $d\colon X\times X\to\Bbb R$ with the appropriate properties.

Title: What are the elements of metric space?Is it always an interval are it can be set of integers etc.

Body: What are the elements of metric space? Is it always an interval or it can be set of integers, etc.?</p>


问:
On the notion of sheaf
Citing form Borceux, Handbook of categorical algebra, in the preface to volume 3:

The crucial idea behind the notion of a sheaf is to work not just with a "plain" set of elements, but with a whole system of elements at various levels. Of course, reasonable rules are imposed concerning the interactions between the various levels: an element at some level can be restricted to all lower levels and, if a compatible family of elements is given at various individual levels, it is possible to "glue" the family into an element defined at the global level covered by the individual ones. The various notions of sheafs depend on the way the words "level", "restriction" and "covering" are defined.

The easiest examples are borrowed from topology, where the various "levels" are the open subsets of a fixed space $X$: for example a continuous function on $X$ may very well be defined "at the level of the open subset $U\subseteq X$", without being the restriction of a continuous function defined on the whole of $X$.

Could someone explain me the bold part?

答: Rephrasing it, that means that you can have a continuous function $f:U \to Y$ that can't be extended continuously to the whole $X$. Moreover in that case the restriction morphism $p_{XU}$ is not surjective, of course.</p>

Title: On the notion of sheaf


问:
Convergence exercise related to improper integral
I'm doing the following exercise in mathematical analysis: (about improper integral)

Let $f\in C^1(0,1]$ be a continuously differentiable function on the half interval $(0,1]$. Suppose $f$　is nonnegative everywhere and $\int_0^1 x|f'(x)|dx<+\infty$. Prove:$\int_0^1 f(x)dx<+\infty$.

So of course I think the idea is to use integration by parts: $\int x f'(x) dx=xf(x)-\int f(x)dx$, so
$$\int_\epsilon ^1 x f'(x) dx=f(1)-\epsilon f(\epsilon)-\int_\epsilon ^1 f(x)dx.$$
Now it's time for $\epsilon \rightarrow 0+$, but I can't deal with $\lim_{\epsilon\rightarrow 0+}\epsilon f(\epsilon)$. Can anyone give me some help? Thanks in advance!

答: $\int_{\epsilon}^{1} f(x)dx \leq \int_{\epsilon}^{1}x f'(x)dx-f(1)$ since $\epsilon f(\epsilon) \geq 0$. Hence, $\int_{0}^{1} f(x)dx \leq \int_{0}^{1}x f'(x)dx-f(1)<\infty$.


答: Put $z = a + bi\implies |iz+3-4i|= |i(a+bi)+3-4i|=|3-b+(a-4)i|=\sqrt{(a-4)^2+(b-3)^2}$. The problem translates to: Maximize $f(a,b) = \sqrt{(a-4)^2+(b-3)^2}$ given that $a^2+b^2 \le 16$. First observe that if the max occurs at an interior point inside the disk $D: a^2+b^2 \le 16$ at $M$, then consider the point of intersection $B$ of the line joining $A = (4,3)$ to $M$ and the circle $C: a^2+b^2 = 16$, we have: $AM \le AB$ and $AB$ is the max value. This can be proven rigorously but is quite easy. Thus the constraint is reduced to the circle $a^2+b^2 = 16$, and $f(a,b) = \sqrt{a^2-8a+16+b^2-6b+9}=\sqrt{41-8a-6b}$. Again observe that $f(a,b)$ is maximized iff $g(a,b) = 41 - 8a-6b$ is maximized under the condition $a^2+b^2=16$. This can be done with Lagrange Multiplier from calculus. Indeed, $\nabla g(a,b) = \lambda\nabla h(a,b), h(a,b) = a^2+b^2-16\implies (-8,-6)=(2\lambda a, 2\lambda b)\implies 2\lambda a= -8, 2\lambda b = -6\implies \lambda a = -4, \lambda b = -3\implies \lambda^2(a^2+b^2)= (-4)^2+(-3)^2=25\implies 16\lambda^2=25\implies \lambda = \pm \dfrac{5}{4}$. We can see that only $\lambda = \dfrac{5}{4}$ yields the max value, and it corresponds to $a= -\dfrac{16}{5}, b = -\dfrac{12}{5}\implies g_{\text{max}} = 41-8\cdot \dfrac{-16}{5}-6\cdot \dfrac{-12}{5}=81\implies f_{\text{max}}= \sqrt{g_{\text{max}}}=\sqrt{81} = 9$, and this is the maximum value of $|iz+3-4i|$.

Title: Two maximum values for $|iz+3-4i|$ what went wrong?

Body:
I need to find the maximum value of $|iz+3-4i|$ given that $|z|\leq4$

$|iz+3-4i|\leq |iz|+|3-4i|$

$|iz+3-4i|\leq 4+5$

But I could write an inequality for $|3-4i|$ whose maximum value would be 7, not 5 so I get the maximum value to be 11 when doing so rather than the original answer of 9.

What went wrong here? , thanks in advance.


问:
The probability that a linear Brownian motion will hit a curve

Summary

I am trying to estimate the probability that a standard linear Brownian motion will hit some curve. To make things a bit simple, I can assume that the curve is a graph of a function, that is is positive at $t=0$, that it is bounded from left by $0$ and by right by some $T>0$, that it is continuous, or even differentiable, and many other nice curvish features that may help making this question more feasible.

Formalizing

Let $\{B(t)\mid t\ge 0\}$ be a standard linear Brownian motion, and let $f:[0,T]\to\mathbb{R}$ be infinitely-differentiable (in $(0,T)$) real function with $T>0$ and $f(0)>0$. Let $A_f$ be the event "$\exists t\in(0,T]):\ B(t)=f(t)$", that is, the Brownian motion "hits" the graph of the function $f$.

The question is as follows: given $f$, what is $\mathbb{P}\left(A_f\right)$?

Attempt

All I could do is solve this for $f\equiv c>0$; in that case, if we define $M(t)=\max\{B(s)\mid 0\le s\le t\}$ we have that
$$\mathbb{P}\left(A_f\right)=\mathbb{P}\left(M(T)\ge c\right)$$
and by reflection principle, the last probability equals
$$2\mathbb{P}\left(B(T)\ge c\right)$$
and that can be solved using straight-forward normal cdf.

However, even for non-constant linear $f$'s that trick won't do; and $f(x)=1/x$ (with something at $0$, bounded from the right by some $T$) seems much harder. This is where I stop and post a question.

答:
Denote $G(x,t,y,\tau)$ the Green function for the first BVP of the corresponding heat equation in the domain $\Omega=\{x<f(t)\,|\,0<t<T\}$. The first event of hitting the curve can be interpreted as a particle to stick to the curve or leave the domain.  The distribution of probability with the source of particles at the origin is described by $G(x,t,0,0)$.  Therefore, the required probability is equal to the flow of particles via the boundary:
$$
P=-\int_0^T\partial_xG(f(t),t,0,0)\,dt=1-\int_{-\infty}^{f(T)}G(x,T,0,0)\,dx.
$$


问:
Vector fields pointing from one fiber to another - Notation
Let $(P,M,\pi)$ be a Principal $G$ bundle.

I am quoting following lines from the book Modern Differential geometry for Physicists by C J Isham.

The basic idea of parallel transport/covariant differentiation is to compare points in neighbouring fibers ina way that is not dependent on any particular local bundle trivialisation. This suggests looking for vector fields that point from one fiber to another.

I do not understand why does idea of comparing fibers of two nearby points should suggest looking for some vector fields. It is not clear what does it mean to say vector field that points from one fiber to another.

I thought I might understand if I go further but still it is not clear.

In search of some vector fields (which is not clear at this point what kind of vector fields they are) he defines vector field $X^A$ on $P$ given an element $A\in T_eG$. I am not defining it because definition is not relavent here. He then says with out proof that $$\pi_*\left( X^A_p\right)=0$$
I have checked this and it turns out to be true. Before saying this, he says

How ever, the particular vector fields are not suitable for our purposes since they do not point from one fiber to another. On the contrary, the vectors $X^A_p$ are tangent to the fiber at $p\in P$ and hence they point along the fiber, rather than away from it. Technically, $X^A_p$ is said to be a vertical vector i.e., it belongs to the vertical subspace $V_pP$ of $T_pP$ defined by

$$V_pP= \left\{ \tau\in T_pP : \pi_*\tau=0\right\}$$
where $\pi$ is the projection map of the bundle.

I do not understand what does it mean to say vectors are tangent to the fibers or point along the fibers.

答: It is indeed incorrect to say a vector field along a fiber is pointing to another point on another fiber. Just like a a tangent vector at a point on a sphere isn't pointing to another point on the sphere; it isn't because it's in the tangent space. That said, one needs to create a a means to transport vectors, otherwise calculus on manifolds couldn't be done. That's why connection 1-forms exist (also why Lie derivatives exist).</p>


问:
Suppose that $x, y\in\mathbb{R}$ with $x^2+ y^2= 1$ and $y\geq 0$. Show that there exists $0\leq t\leq\pi$ such that $x= \cos t$ and $y= \sin t$
Suppose that $x, y\in\mathbb{R}$ with $x^2+ y^2= 1$ and $y\geq 0$.Show that there exists $0\leq t\leq\pi$ such that $x= \cos t$ and $y= \sin t$

This is an analysis question and I can only use the fundamental definition of $\sin$ and $\cos$ that is the infinite sum. I can also use the Taylor theorem, but I am stuck and clearly do not understand what I need to assume and what I need to prove. I can assume that $\sin$ and $\cos$ are infinitely differentiable. $\pi$ is defined as$:\quad\pi= 2\inf\{x\geq 0:\quad cos x= 0\}$.
Intuitively I should be using the Intermediate Value Theorem but I am not sure on which function.

答:
$y=+\sqrt{1-x^2}>0.$
Let $x=\cos t ,t\in[0,\pi] \implies y=\sin t >0$


问:
Remainder When Divided By 70 : Remainder Theorem Problem
If $x$ = $16^3$ + $17^3$ + $18^3$ + $19^3$ , then $x$ divided by $70$ leaves a remainder of?

I tried to solve this problem by using the remainder theorem which states that remainder of $$Rem[\frac{a+b+c+....}{x}] = Rem[\frac{a}{x}] + Rem[\frac{b}{x}] + Rem[\frac{c}{x}] + .... $$ where 'Rem' means 'Remainder Of'.

Using this same logic I can write $$Rem[\frac{16^3+17^3+18^3+19^3}{70}] = Rem[\frac{16^3}{70}] + Rem[\frac{17^3}{70}] + Rem[\frac{18^3}{70}] + Rem[\frac{19^3}{70}] = 18 +13+11+69 = 111 \Rightarrow Rem[\frac{111}{70}] = 41$$

So my answer comes up as 41. Can someone please explain to me what I might be doing wrong or what I have understood wrong?

答:
Another way, where there is no need to compute the numeric values of the four cubes.

Note that $16+19=17+18=35=70/2$, hence
\begin{align*}
16^3+17^3+18^3+19^3
&=16^3+(35-18)^3 +18^3+(35-16)^3\\
&\equiv 16^3+35^3-18^3+18^3+35^3-16^3\\
&=2\cdot 35^3\equiv 0\pmod{70}.
\end{align*}
In particular, when we expand the cube $(35-2n)^3$ we find
$$35^3-3\cdot 35^2\cdot 2n+3\cdot 35\cdot (2n)^2-(2n)^3\equiv 35^3-(2n)^3\pmod{70}$$
because $3\cdot 35^2\cdot 2n$ and $3\cdot 35\cdot (2n)^2$ are divisible by $70$.


问:
Proof of $\ln(p_k(x))=\delta_k(x)$
In "An introduction to Statistical Learning in R" by James, Witten, Hastie, and Tibshirani, on page 139-140, in the section concerning Linear Discriminant Analysis for p=1, assuming $f_k(x)\sim $ Gaussian

We are given

$p_k(x)=\frac{\pi_k\cdot\frac{1}{\sqrt{2\pi}\sigma_k}\exp\big(-\frac{1}{2\sigma_k^2}(x-\mu_k)^2\big)}{\sum_{l=1}^k\pi_l\cdot\frac{1}{\sqrt{2\pi}\sigma_k}\exp\big(-\frac{1}{2\sigma_l^2}(x-\mu_l)^2\big)} \qquad(A)$

and they say that "it is not hard to show that...taking the log of this and rearranging the terms" brings one to

$\delta_k(x)=x\cdot \frac{\mu_k}{\sigma^2}-\frac{\mu_k^2}{2\sigma^2}+\ln(\pi_k) \qquad (B)$

Just about the only part I understand about this is the variances are assumed to be the same, other than this I cannot figure out how taking the log goes from A to B

Might someone be able to demonstrate this or point to a resource?

答:
The authors are not claiming that expression (B) is equal to the logarithm of expression (A), it is not. What they are saying is that, under the given assumptions, an $x$ that is a maximum of $\delta_k$ will also be a maximum of $p_k$ (and vice-versa). To see this, you have to take the $\log$ as suggested :
$$\begin{align}\log(p_k(x) ) &=\log\left(\frac{\pi_k\cdot\frac{1}{\sqrt{2\pi}\sigma_k}\exp\big(-\frac{1}{2\sigma_k^2}(x-\mu_k)^2\big)}{\sum_{l=1}^K\pi_l\cdot\frac{1}{\sqrt{2\pi}\sigma_k}\exp\big(-\frac{1}{2\sigma_l^2}(x-\mu_l)^2\big)}\right) \\
&= \log(\pi_k) -\frac{1}{2\sigma^2}(x-\mu_k)^2 +\log\left(\frac{1}{\sqrt{2\pi}\sigma}\right) - \log\left(\sum_{l=1}^K\pi_l\cdot\frac{1}{\sqrt{2\pi}\sigma}\exp\big(-\frac{1}{2\sigma^2}(x-\mu_l)^2\big)\right) \\
&= \log(\pi_k) -\frac{x^2}{2\sigma^2} + x \cdot\frac{\mu_k}{\sigma^2}-\frac{\mu_k^2}{2\sigma^2} -\log\left(\sum_{l=1}^K\pi_l\cdot\frac{1}{\sqrt{2\pi}\sigma}\exp\big(-\frac{1}{2\sigma^2}(x-\mu_l)^2\big)\right) + C \\
&=\underbrace{\log(\pi_k) + x \cdot\frac{\mu_k}{\sigma^2}-\frac{\mu_k^2}{2\sigma^2}}_{\delta_k(x)} + F(x)  \end{align} $$
So we see where the expression of $\delta_k$ is coming from.

Now notice that all of the $K$ classes have a normal distribution with the same variance $\sigma^2$, so for any $k \in \{1,\ldots,K\}$, the term $F(x)$ as written above is the same. It is therefore enough to find the $k$ such that $\delta_k(x)$ is maximal to predict the class of $x$.

Title: Proof of $\ln(p_k(x))=\delta_k(x)$


问:
How to determine degenerate critical points of $f(x,y)$ (the 2nd Derivative Test has failed)

Find the critical points of the function
  $$f(x,y) = y^2\left(\sin x - \frac{x}{2}\right)$$
  and state their nature.

I have determined that this functions has an infinite number of critical points and they are: $$(0,0), \ (1.895,0), \ (-1.895,0), \ \left(\frac{5\pi}{3} + 2\pi n,0\right), \ \left(\frac{\pi}{3} + 2\pi n,0\right)$$

However, notice that the Hessian of this function is:

$H_f(x,y)$ = $\begin{bmatrix}-y^2\sin x & 2y\cos x - y\\2y\cos x - y & 2\sin x - x\end{bmatrix}$

and that

$\det(H_f(x,0)) = 0$

So the second derivative test fails in this example.

How do I go about classifying these critical points as either maxima, minima, saddle points or degenerate critical points?

Notice the Hessian matrix is:

$H_f(x,y) = \begin{bmatrix}-y^2\sin x & 2y\cos x - y\\2y\cos x - y & 2\sin x - x\end{bmatrix}$

and that

$H_f(a,0) = \begin{bmatrix} 0\ &0\\0  &2\sin a - a\end{bmatrix}$ for $a \neq 0$

then

$rank\ (H_f(a,0)) = 1 \le 2$ for all $a \neq 0$

and

$rank\ (H_f(0,0)) = 0 \le 2$

According to Folland's Advanced Calculaus these should be degenerate points right?

If $f: \mathbb{R}^n \to \mathbb{R}$ is $C^2$ and $c$ is a critical point of $f$, then $c$ is a degenerate critical point of $f$ if $rank\ (H_f(c)) \le n$.

Is this correct? Are all the critical points here "degenerate" critical points? And if so, are degenerate critical points analogous to "inflection points" for functions $\mathbb{R} \to \mathbb{R}$.

Please help.

Thank you.

答:
One of the ways in which a second-partial-derivative  test  can fail us is where the Hessian determinant is zero not at an isolated point, but on a line passing through that point.  A "slice" through the surface for the function $ \ f(x,y) \ = \ y^2· \left(\sin x - \frac{x}{2}\right) \ $ at, say, $ \ y \ = \ \pm 1 \ $ , looks like this:

Except for $ \ y \ = \ 0 \ $ , where $ \ f(x,0) \ = \ 0 \ $ , all parallel slices are a "vertically-stretched" version of this slice.  The consequence of this is that each slice at a specific value of $ \ x \ $ is a parabola $ \ z \ = \ ky^2 \ $ , except for those values of $ \ x \ $ at which $ \ \left(\sin x - \frac{x}{2}\right) \ = \ 0 \ \ . $

So we  have "flat" lines in the surface of the function for $ \ f(x,0) \ = \ 0 \ $  , $ \ f(0,y) \ = \ 0 \ $ , $ \ f(\approx 1.895 , y) \ = \ 0 \ $ , and $ \ f(\approx -1.895 , y) \ = \ 0 \ $ .  If instead of evaluating the Hessian determinant at each individual critical point, we look at the determinant function

$$ \mathbf{H_f} \ \ = \ \ -2y^2 \ · \ \left[ \ \sin x · \left( \sin x \ - \ \frac{x}{2} \right) \ + \ 2 · \left( \cos x \ - \ \frac{1}{2} \right)^2 \ \right] \ \ , $$

we see that every point along the $ \ x-$ axis on the surface gives a value of zero, so  those will be the points that Folland calls "degenerate" (I happen to have a copy of that text and I believe this is the sort of situation he is referring to).

So the short answer to your main question is that all of your critical points are degenerate, but that is not the same as saying that they are all inflection points.  It is still possible to characterize them by looking at the behavior of the function in their neighborhoods.

As for recognizing the types of the critical points you've identified, things are a bit  complicated then.  The graph will be helpful for this, since the expression in brackets in the determinant function is not straightforward to analyze.

Since the slices parallel to the $ \ y-$ axis are parabolic, they "open upward" where $ \left(\sin x - \frac{x}{2}\right)  >  0 \ $ and "downward" when this expression is $  <  0 \ $ .

For the critical points $ \  \left(\frac{5\pi}{3} + 2\pi n \ , \ 0 \right)  \ , \ n \ge 0 \ , $ the surface is "concave up" in the $ \ x-$ direction and "down" in the $ \ y-$ direction, making these saddle points (negative curvature); for $ \  \left(\frac{\pi}{3} + 2\pi n \ , \ 0 \right)  \ , \ n \ge 1 \ , $  the surface is "concave down" in both directions (positive curvature), so these are local maxima.

Since our function has odd symmetry about the $ \ yz-$ plane, the direction of local extrema is reversed, telling us that $ \  \left(-\frac{5\pi}{3} + 2\pi n \ , \ 0 \right)  \ , \ n \le 0 \ , $ are saddle points and $ \  \left(-\frac{\pi}{3} + 2\pi n \ , \ 0 \right)  \ , \ n \le -1 \ , $   are local minima.

The points $ \  \left(  \frac{\pi}{3}  \ , \ 0 \right)  \ $ has $ \left(\sin x - \frac{x}{2}\right)  >  0 \ $ , so it is "concave upward" in the $ \ y-$ direction, but it lies at a local maximum of that function, so it is "concave downward" in the $ \ x-$ direction; hence it is a saddle point.  The directions of concavity are both reversed for $ \  \left(  -\frac{\pi}{3}  \ , \ 0 \right)  \ , $ so it is likewise a saddle point.

The two points $ \ (\approx 1.895 \ , \ 0) \ $ and $ \ (\approx -1.895 \ , \ 0) \ $ are zeroes of $ \ f(x,y) \ $ and are sitting on "declining slopes" in the $ \ x-$ direction but on a "flat line" in the $ \ y-$ direction; I don't know what the correct term for such a critical point is.

Finally, near the origin, the points on $ \ (x \ , \ 0) \ $ in the neighborhood have opposed concavities in the perpendicular directions, so those are all saddle points.  The orientation of these saddle points reverses at the origin, so perhaps the origin is also characterized as a sort of saddle point.  (The term seems a bit inadequate to describe the behavior of the surface there.)


问:
Prove positive definite matrix and determinant inequality
$A$ and $B$ are two real symmetric matrices and $A \succeq B$ (that means $A-B$ is a psd matrix), does that hold $|A|\ge|B|$? and why?</p>

答:
Of course not. Consider, e.g. $A=I_2$ and $B=-2I_2$ with $\det A=1<4=\det B$.

Edit: However, the statement is true if $B$ is positive semidefinite. That $A\succeq B$ implies that $A$ is positive semidefinite too. If $\det B=0$, the statement obviously holds. If $\det B>0$, then both $A$ and $B$ are positive definite. Hence $A\succeq B$ implies that $B^{-1/2}AB^{-1/2}\succeq I$ and in turn $\det(B^{-1/2}AB^{-1/2})\ge 1$, i.e. $\det A\ge\det B$. (In fact, if $A\succeq B\succ0$, we have $\lambda_k(A)\ge\lambda_k(B)$ for every $k$. This is a property of positive definite ordering, but we do not need this fact here.)


答:
I made a mistake in my solution with the remainders. Below is the correct approach and answer :-

Using this same logic I can write $$Rem[\frac{16^3+17^3+18^3+19^3}{70}] = Rem[\frac{16^3}{70}] + Rem[\frac{17^3}{70}] + Rem[\frac{18^3}{70}] + Rem[\frac{19^3}{70}] = 36 +13+22+69 = 140 \Rightarrow Rem[\frac{140}{70}] = 0$$

Title: Remainder When Divided By 70 : Remainder Theorem Problem

Body:
If $x$ = $16^3$ + $17^3$ + $18^3$ + $19^3$ , then $x$ divided by $70$ leaves a remainder of?

I tried to solve this problem by using the remainder theorem which states that remainder of $$Rem[\frac{a+b+c+....}{x}] = Rem[\frac{a}{x}] + Rem[\frac{b}{x}] + Rem[\frac{c}{x}] + .... $$ where 'Rem' means 'Remainder Of'.

Using this same logic I can write $$Rem[\frac{16^3+17^3+18^3+19^3}{70}] = Rem[\frac{16^3}{70}] + Rem[\frac{17^3}{70}] + Rem[\frac{18^3}{70}] + Rem[\frac{19^3}{70}] = 18 +13+11+69 = 111 \Rightarrow Rem[\frac{111}{70}] = 41$$

So my answer comes up as 41. Can someone please explain to me what I might be doing wrong or what I have understood wrong?


答:
So all you have is the power series for $\sin x$ and $\cos x$, and you know that $\frac \pi2$ is the smallest positive zero of $\cos x$.

Lemma. Let $f\colon\Bbb R\to\Bbb R$ be a twice differentiable function with
$$ f''(t)+f(t)=0\text{ for all }t,\quad f(0)=0,\quad f'(0)=0. $$
Then $f$ is identically $0$.

Proof. Indeed,
$$ \frac{\mathrm d}{\mathrm dx}(f(x)^2+f'(x)^2)
=2f'(x)f(x)+2f''(x)f'(x)=2f'(x)(f(x)+f''(x))=0$$
so that $f(x)^2+f'(x)^2$ is constant, and by evaluating at $0$, is constant zero.  But as squares of reals are non-negative, this implies that also $f$ is identically zero. $\square$

From the power series, you readily find that $$\sin't=\cos t,\qquad\cos't=-\sin t,$$ hence $$\sin''t=-\sin t,\qquad \cos''t=-\cos t.$$

As $\cos0=1$ and $\cos\frac\pi2=0$, the Intermediate Value Theorem gives us some $a\in[0,\frac\pi2]$ where $\cos a=|x|$ and then necessarily $|\sin a|=y$.
Note that $\sin t>0$ for  $t\in(0,\frac\pi2]$ (and in particular, $\sin\frac\pi2=+1$) as otherwise Rolle's theorem would give us a critical point of $\sin t$ and hence a zero of $\cos t$  in $(0,\frac\pi2)$, contradicting the definition of $\pi$. We conclude that in fact $\sin a=y$. Thus

Result 1. If $x^2+y^2=1$, $x\ge 0$, $y\ge 0$, then there exists $a\in[0,\frac\pi2]$ with $\sin a=y$, $\cos a=x$.

By switching some roles, this also gives us

Result 2. If $x^2+y^2=1$, $x\le 0$, $y\ge 0$, then there exists $b\in[0,\frac\pi2]$ with $\sin b=-x$, $\cos b=y$.

If the $x$ in our original problem is $\ge0$, result 1 gives us the desired answer. So assume $x<0$ and let $b$ be as in result 2.

Define $f(t)=\sin t+\cos(t+\frac\pi2)$. Then

$f(0)=\sin 0+\cos\frac\pi2=0+0=0$

$f'(0)=\cos 0-\sin\frac\pi2=1-1=0$

$f''(t)=-\sin t-\cos(t+\frac\pi2)=-f(t)$

Hence by the lemma, $f(t)=0$ for all $t$, i.e., $$\cos(t+\tfrac\pi2)=-\sin t$$
and similarly
$$\sin(t+\tfrac\pi2)=\cos t$$ for all $t\in\Bbb R$. In particular, with $c:=b+\frac\pi 2\in [0,\pi]$, we have

$$ \sin c = \cos b = y,\qquad \cos c = -\sin b = x$$
as desired.

Title: Suppose that $x, y\in\mathbb{R}$ with $x^2+ y^2= 1$ and $y\geq 0$. Show that there exists $0\leq t\leq\pi$ such that $x= \cos t$ and $y= \sin t$

Body:
Suppose that $x, y\in\mathbb{R}$ with $x^2+ y^2= 1$ and $y\geq 0$.Show that there exists $0\leq t\leq\pi$ such that $x= \cos t$ and $y= \sin t$

This is an analysis question and I can only use the fundamental definition of $\sin$ and $\cos$ that is the infinite sum. I can also use the Taylor theorem, but I am stuck and clearly do not understand what I need to assume and what I need to prove. I can assume that $\sin$ and $\cos$ are infinitely differentiable. $\pi$ is defined as$:\quad\pi= 2\inf\{x\geq 0:\quad cos x= 0\}$.
Intuitively I should be using the Intermediate Value Theorem but I am not sure on which function.


问:
The Image of Hyperbola Under Inversion
Consider a hyperbola given by the equation $x^2-y^2=1$. I want to find its image under the inversion $f(z)=\frac{1}{z}$. Can I use this as the following argument?

First, I can observe if $z=re^{i\theta}$, then $f(z)=\frac{1}{r}e^{-i\theta}$. Then, I can observe that $x^2-y^2=1$ can be written in terms of polar coordinates as $r^2\cos(2\theta)=1$. So, as every point $(r,\theta)$ maps to $(\frac{1}{r},-\theta)$, then the equation $r^2\cos(2\theta)=1$ will become
$$(\frac{1}{r})^2\cos(2(-\theta))=1$$
which gives us
$$r^2=cos(2\theta)$$
i.e. the graph of lemniscate.

Title: The Image of Hyperbola Under Inversion

Body:
Let $w=u+iv=\frac{1}{z}$, $z=x+iy \implies u+iv=\frac{\bar z}{z \bar z} \implies u=x/(x^2+y^2), v=-y/(x^2+y^2)~~~~(1)$
$$x^2-y^2=1=Re[z^2]=Re[\frac{1}{w^2}]=1 \implies Re[\frac{\bar w^2}{(\bar w w)^2}]=\frac{u^2-v^2}{(u^2+v^2)^2}=1~~~~(2)$$

So the image of $x^2-y^2=1$ under inversion $(w=1/z)$ is the curve $u^2-v^2=(u^2+v^2)^2$, which can be finally written as
$$x^2-y^2=(x^2+y^2)^2.$$


问:
Calculate combined speed
I have a robotic arm that has to move at a defined speed. it has a speed for the X axis, and it has a speed for the Y axis. So if it moves horizontally it uses the X-axis speed, and if it moves Vertically it uses the Y-axis speed. If it moves diagonally it should combine the two speeds. In the image you can see some samples. the vertical and horizontal speeds are given, the angle between the X axis and the diagonal Line is given too.

What I would need to know is how I can calculate the length of the diagonal line (composite speed) with the given informations.

EDIT: the combined speed should follow the ellypse like in the image.

答: If you have $v_x$ and $v_y$, use the Pythagorean Theorem to get the total velocity $\sqrt{v_x^2+v_y^2}$.

Title: Calculate combined speed


问:
How prove $a_{n}>0$ with Komal 661 problem
Let $K$ be a fixed positive integer,Let $(a_{0},a_{1},\cdots )$ be the sequence of real numbers that satisfies $a_{0}=-1$ and
$$\sum_{i_{0},i_{1},\cdots,i_{K}\ge 0,i_{0}+i_{1}+\cdots+i_{K}=n}\dfrac{a_{i_{1}}a_{i_{2}}\cdots a_{i_{K}}}{i_{0}+1}=0$$
for every postive integer $n$,show that $a_{n}>0$ for $n\ge 1$

This problem is from Komal problem 661. https://www.komal.hu/feladat?a=feladat&f=A661&l=en

答:
This only a partial answer but here is a way to get an expression for the sequence. Define the generating function of the sequence $\displaystyle f(X)=\sum_{i\geq 0} a_i X^i$ and let $\displaystyle g(X)=\sum_{i\geq 0} \cfrac{X^{i}}{i+1}$. Then we can rewrite the recurrence as $(f(X))^Kg(X)=-1$. Thus $f(X)=-(g(X))^{-1/K}=-(1+h(X))^{-1/K}$. We now use the binomial theorem to get $\displaystyle f(X)=-1-\sum_{n\geq 1} \dbinom{-1/K}{n}(h(X))^n$. Equating the coefficients of $X^i$ on both sides we get $$a_i=-\sum_{\substack{s\geq 1,m_1,\ldots, m_s\geq 1 \\ m_1+\ldots+m_s=i}}  \dbinom{-1/K}{i}\cdot \cfrac{1}{(m_1+1)\ldots(m_s+1)}$$
for $i\geq 1$. After this you need to do some bounding to show that the first term $\cfrac{1}{Ki}$ dominates the negative terms.


问:
Continuity of $U:\Omega \rightarrow \mathbb{R}$
Let $\Omega \in \mathbb{R^n}$, suppose we have a function $ U \in \mathcal{C}({\Omega})$ i.e $U:\Omega \rightarrow \mathbb{R}$ is continuous.

Let $x_1 \in \Omega$
and suppose $\exists M \gt 0$ s.t $U(x_1) \lt M$. Then Why by continuity of $U$ at $x_1$, there exists $\delta \gt 0$ s.t $U \lt M$ on $B(x_1,\delta)$?

Can I please have a proof or explanation?

答:
$O:=U^{-1}[(-\infty, M)]$ is an open subset of $\Omega$ (by continity of $U$) with $x_1 \in O$ by definition. As $O$ is open, $x_1$ is an interior point of $O$ so there is some $\delta>0$ so that

$$B(x_1, \delta) \cap \Omega \subseteq O= U^{-1}[(-\infty, M)] \text{ so: } \forall x \in \Omega: d(x,x_1) < \delta \to U(x) < M$$

That last condition is what's meant by "$U < M$ on $B(x_1, \delta)$".

If $\Omega$ is open, we can also ensure that $B(x_1, \delta) \subseteq \Omega$ as well, because then $x_1$ is an interior point of $\Omega$ as well.


问:
Why is $5$ unique when it comes to terminating decimals?
When $n$ limited to odd numbers, for $\frac{1}{n}$ to be a terminal decimal$,\quad n$ must be in the form of $5^k$ ($k$ is positive integer). Why is $5$ so unique in this case?

Guess it is related to decimal system? I checked https://en.m.wikipedia.org/wiki/Repeating_decimal, where there is a statement “A fraction in lowest terms with a prime denominator other than $2$ or $5$ (i.e. coprime to $10$) always produces a repeating decimal” without further expectations. Just curious if there is a simple proof for this observation out there. Thanks!

答: If $\frac1n$ terminates after $m$ decimals, then $10^m\cdot\frac 1n$ is an integer, so $n$ is a divisor of $10^m$. All these divisors are of the form $2^a5^b$, but as you wanted $n$ odd, we must have $a=0$.


问:
The Gaussian and Mean Curvatures of a Parallel Surface
This is a homework problem from do Carmo. Given a regular parametrized surface $X(u,v)$ we define the parallel surface $Y(u,v)$ by $$Y(u,v)=X(u,v) + aN(u,v)$$ where $N(u,v)$ is the unit normal on $X$ and $a$ is a constant. I have been asked to compute the Gaussian and mean curvatures $\overline{K}$ and $\overline{H}$ of $Y(u,v)$ in terms of those of X, $K$ and $H$. Now, I know how to do this by brute force: calculate the coefficients of the first and second fundamental forms of $Y$ in terms of those of $X$. However, this is a lengthy and messy calculation. do Carmo says that $$\overline{K}=\frac{K}{1-2Ha+Ka^2}$$ and $$\overline{H}=\frac{H-Ka}{1-2Ha+Ka^2}.$$ The denominator of these fractions is actually something that arose earlier in the problem; I calculated $$Y_u\times Y_v=(1-2Ha+Ka^2)(X_u\times X_v).$$ So, it seems like I should be able to calcuate $\overline{K}$ and $\overline{H}$ from this initial step. Is there something I'm missing? Or, is it actually just a brute force calculation?

Thanks.

答:
This is already implicitly included in the previously posted answers, but if you just want to understand where the formulas for $\overline K$ and $\overline H$ come from, without caring much for rigor, that is actually very easy. Once expressed in terms of the principal radii of curvature, the expressions for $\overline K$ and $\overline H$ are equivalent to
\begin{align*}
\overline R_1+\overline R_2&=R_1+R_2-2a,\\
\overline R_1\overline R_2&=(R_1-a)(R_2-a).
\end{align*}
These relations then follow from the intuitively obvious fact that upon translating by $a$ along the normal vector, the principal radii of the surface simply shift by $a$.


问:
How many times the solution will intersect the line
Hello everyone I have a problem I'm struggling with :

I have the following systems :

$$
\begin{cases}
& {x}'=x+y
 \\
& {y}'= y-x
\end{cases}
$$

I found the solutions using the Method of Matrix Exponential,

For the first system I found :

$X_1(t) = e^{t} \begin{pmatrix}
\cos(t)&&\sin(t)\\
-\sin(t)&&\cos(t)
\end{pmatrix}$ $\begin{pmatrix}
C_1\\
C_2
\end{pmatrix}$

Best wishes

Edit: ANSWERED

答:
For each solution $X_i(t)$ you got, you should equal the first component (that is to say, the $x-$coordinate) to $0$ and see how many solutions for $t$ you get.

In the first case,
$$\frac{1}{2}e^{5t} + \frac{1}{2}e^{-t} = 0,$$
which cannot have a solution due to both terms are positive.

In the second case,
$$e^{2t}\sin(3t) = 0$$
which has infinite solutions since it is zero when $3t = k\pi$ for $k\in\mathbb{Z}$.


问:
secant method in maple

secant method in maple. Find a root of the statement $x^3-3x^2+4x-1=0$ with the initial value $x_0=0$ and $x_1=1$ with 5 digits point approximation.

答:
If this is homework and you're supposed to program it yourself then you could show what you've accomplished on your own already.

If you just want to see answers from such an algorithm then you could use the Student:-NumericalAnalysis command. Eg,

```<code>[Student:-NumericalAnalysis:-Secant(x^3-3*x^2+4*x-1,x=[0,1],
                                    tolerance=1e-5,
                                    maxiterations=50)]:
evalf[5](%)[];

                                0.31767

[Student:-NumericalAnalysis:-Secant(x^3-3*x^2+4*x-1,x=[0,1],
                                    tolerance=1e-5,
                                    output=sequence,
                                    maxiterations=50)]:
evalf[5](%)[];

 0., 1., 0.50000, 0.20000, 0.33624, 0.31947, 0.31764, 0.31767, 0.31767
</code>```

Programming questions about Maple are better asked on stackoverflow, unless it is the mathematics behind the algorithm that is your central concern.

Title: secant method in maple

Body:
If this is homework and you're supposed to program it yourself then you could show what you've accomplished on your own already.

If you just want to see answers from such an algorithm then you could use the Student:-NumericalAnalysis command. Eg,

[Student:-NumericalAnalysis:-Secant(x^3-3*x^2+4*x-1,x=[0,1],
                                    tolerance=1e-5,
                                    maxiterations=50)]:
evalf[5](%)[];

                                0.31767

[Student:-NumericalAnalysis:-Secant(x^3-3*x^2+4*x-1,x=[0,1],
                                    tolerance=1e-5,
                                    output=sequence,
                                    maxiterations=50)]:
evalf[5](%)[];

 0., 1., 0.50000, 0.20000, 0.33624, 0.31947, 0.31764, 0.31767, 0.31767

Programming questions about Maple are better asked on stackoverflow, unless it is the mathematics behind the algorithm that is your central concern.


问:
Need two sequences be positive for the limit comparison test to hold?
According to the wikipedia, the limit comparison test requires that $\lim _{n \to \infty} \frac{a_n}{b_n} = c, 0<c<\infty \textrm{ where } a_n, b_n > 0 \textrm { for all n}$. My question is can we get rid of the requirement $a_n, b_n > 0$ and loosen the limit requirement to $\lim _{n \to \infty} |\frac{a_n}{b_n}| = c$. Then we prove that $|a_n| \textrm{converges} \Rightarrow |b_n| \textrm{converges}$ and vice versa, and use absolute convergence to say $a_n$ converges.

答: In terms of convergence you can conclude. But divergence of  $\sum_{n}|a_n|$ does not imply that of $\sum_{n}a_n$. Take $a_n=\frac{(-1)^n}{n},~b_n=\frac{1}{n+1}$, then the modulus limit is 1, although $\sum_n |a_n|,~\sum_n |b_n|$ are both divergent but $\sum_n a_n $ is convergent by Leibniz test.


问:
Basic question about finding flow given eigenvalues
I am reading the ODE/ dynamical systems book by Hirsh, Devaney, and Smale, Differential Equations, Dynamical Systems, and an Introduction to Chaos (3e).

How do I go about finding the flow of a $4\times4$ matrix with eigenvalues with $4$ complex eigenvalues, all of which are purely imaginary?

For example, let's say the eigenvalues are: $i\sqrt{2},-i\sqrt{2},i\sqrt{3},-i\sqrt{3}$.

My guess would be a spiral? Is this correct? If so how I would I decide which way the spiral is "spiraling"?

Edit: According to the book (pp.114-115):

This is pretty much the same thing so I have my solution. But how do I describe the "flow"? Is finding a solution enough?

答:
A flow is a function that given a time and a point in your phase space maps it forwards or backwards in time. So to convert that general solution to a flow you turn it into a function such that if I insert time and a starting value I get as output the value of the solution at that time for that starting value.

More formally, a flow corresponding to an ODE is a function $\phi : T \times M \rightarrow M$  with $T$ the time set and $M$ the phase space such that $\phi(t,x_0)$ is a solution of the ODE satisfying $\phi(0,x_0) =x_0$. (There is a more general definition of flows for example see here http://www.scholarpedia.org/article/Dynamical_systems#Flows, the one I give here follows from that definition if you assume you are dealing with an ODE)


问:
Finding if a function is one-one and onto
Consider the function $f:\mathbb R \to \mathbb R$ given by $f(x)= |x| (x - \sin x)$. How do I prove that this function is one-one. Is there any method apart from checking if the function is increasing / decreasing?

答:
Study the function for when $x \ge 0$ and for when $x<0$. You immediately see your function is odd (so it’s symmetric with respect to the origin) so you can study its injectivity only for the $x \ge 0$ part. Then suppose $f(x)=f(y)$ for some $x,y$ and try to conclude that your equation has solutions only if $x=y$.

Beside the fact that here the monotonicity test is way easier.


问:
PDF of $Y=\frac{8}{X^3}$
Consider the random variable X with probability density function $$f(x) = \begin{cases}
 \frac{24}{x^4}; & \text{ if, } x>2 \\
 0; & \text{ otherwise }
\end{cases}$$ Find the probability density function of $Y=\frac{8}{X^3}$.

I tried using the formula $$f_Y(y)=f_X(g^{-1}(y)) \vert\frac{\partial}{\partial y}g^{-1}(y)\vert.$$

where

$Y=g(x)=\frac{8}{x^3}$,

$g^{-1}(Y)=\frac{2}{y^\frac{1}{3}}$

but it ended up as $f_Y(y)=$$\frac{y^\frac{4}{3}}{|y\frac{4}{3}|}$

Can anyone tell me where it's gone wrong?

2nd attempt:
$P(Y \leq y) = P(\frac{8}{X^3} \leq y) = P(X \geq \frac{2}{y^\frac{1}{3}})$ = $1$- $P(X \leq \frac{2}{y^\frac{1}{3}})$ = $\frac{8}{y}$ using the integral on $f(x)$ over $2$ to $\frac{8}{y\frac{1}{3}}$

答:
What you have done is correct for $0<y<1$. There is no need for absolute value so the density is $1$. Note that $X>2$ implies that $\frac 8 {X^{3}} \in (0,1)$ so $Y$ is uniformly distributed on $(0,1)$.

In your second method $P(Y \leq y)=P(X \geq  2 y^{-1/3})=\int_{ 2 y^{-1/3}}^{\infty} \frac {24}{x^{4}}dx=y$ if $y$ is between $0$ and $1$.

Title: PDF of $Y=\frac{8}{X^3}$


答:
If the absolute value of velocity has to follow the ellipse its square can be computed as
$$v^2=\frac {v_x^2v_y^2}{v_x^2\sin^2\theta+v_y^2\cos^2\theta},$$
where $\theta $ is angle with respect to the $x $-axis.

Title: Calculate combined speed

Body:
I have a robotic arm that has to move at a defined speed. it has a speed for the X axis, and it has a speed for the Y axis. So if it moves horizontally it uses the X-axis speed, and if it moves Vertically it uses the Y-axis speed. If it moves diagonally it should combine the two speeds. In the image you can see some samples. the vertical and horizontal speeds are given, the angle between the X axis and the diagonal Line is given too.

What I would need to know is how I can calculate the length of the diagonal line (composite speed) with the given informations.

EDIT: the combined speed should follow the ellypse like in the image.


问:
What is the best way to indexing a component of a indexed vector?
I'm writing some academic stuff. I'd like to make it clear and precise especially in the mathematical expression; but, I got an ambiguity.

When I index something, I often use the subscript. So, I indexed the e-th vector as $\mathbf{v}_e$. Also, I use the subscript indexing the component of a vector:

$\mathbf{u}=\left[u_1,u_2,\cdots,u_n\right]^\mathsf{T}$

Then I'd like to specify the component of an (already) indexed vector like:

$\mathbf{v}_e=\left[\cdots,\rm{Here},\cdots\right]^\mathsf{T}$

I've tried:

$v_e^i$ but it can be confused with exponent...

$v_{e,i}$ but it can be confused with differential (Einstein notation).

$v_e^{(i)}$ I think it is a bit verbose and can be confused with $v_i^{(e)}$.

Can anyone recommend anything? In the professional mathematical world, which is the mainstream?

答:
I'd suggest matrix notation. Let $\mathbf{v} = [\mathbf{v}_1 \ \cdots \ \mathbf{v}_n]^\top$ be the matrix whose rows are your vectors. (Maybe you'd want to use $\mathbf{v} = \begin{bmatrix}\mathbf{v}_1^\top \\ \ldots \\ \mathbf{v}_n^\top\end{bmatrix}$.) Then $\mathbf{v}_{ej}$ is the $j$th component of $\mathbf{v}_e$.

Another option would be $\mathbf{v}^{(1)}, \ldots, \mathbf{v}^{(n)}$ for the vectors with $\mathbf{v}^{(e)}_j$ for the $j$th component. I've seen similar things in papers and used them myself.


答:
Your proposition is quite correct. In fact it belongs to the set  of Mclaurin? curves.

$$ r^n= a ^n  \cos( n\cdot \theta ) $$
Obtained by exponentiation of $ z = a e^{i \theta}$ to power of $n$ and then taking real part.

The subsets are also the inversions got by
$$ r\to \dfrac{a^2}{r}$$

$ n=2, -2 $ are the Lemniscate and equi- hyperbola $ x^2-y^2= a^2$

$ n=1, -1 $ are the straight lines and circles through origin

$$r= a \sec \theta,r= a\cos \theta $$

The above and all the Rosettes formed this way may be termed Mclaurin Rosettes. Iirc it was mentioned in the second chapter among exercise examples in Heinrich Guggenheimer's of Differential Geometry text book... with reference to both aspects: Inversions and their generation from real part of exponentiated unit complex number.

Title: The Image of Hyperbola Under Inversion

Body:
Consider a hyperbola given by the equation $x^2-y^2=1$. I want to find its image under the inversion $f(z)=\frac{1}{z}$. Can I use this as the following argument?

First, I can observe if $z=re^{i\theta}$, then $f(z)=\frac{1}{r}e^{-i\theta}$. Then, I can observe that $x^2-y^2=1$ can be written in terms of polar coordinates as $r^2\cos(2\theta)=1$. So, as every point $(r,\theta)$ maps to $(\frac{1}{r},-\theta)$, then the equation $r^2\cos(2\theta)=1$ will become
$$(\frac{1}{r})^2\cos(2(-\theta))=1$$
which gives us
$$r^2=cos(2\theta)$$
i.e. the graph of lemniscate.


问:
Confusion regarding the definition and generators of Borel $\sigma$-algebra
We have defined the Borel $\sigma$-algebra as following:

Let $\Omega$ be a topological space. The $\sigma$-algebra generated by open sets (i.e. the smallest $\sigma$-algebra contains all open sets) is called the Borel $\sigma$-algebra, and noted $\mathcal{B}(\Omega)$. An element of $\mathcal{B}(\Omega)$ is called a Borel(measurable) set.

And then we were given the following example:

On $\mathbb{R}$, the Borel $\sigma$-algebra is also generated by intervals $]a,b[,-\infty \leq a < b \leq \infty$. It is also generated by intervals $[a,b], ]a,b]$ or $[a,b[$.

What does it mean here, when we say that Borel $\sigma$-algebra is generated by the interval $]a,b[$? In my understanding a $\sigma$-algebra generated by for ex. $\varepsilon$ is the intersection of all $\sigma$-algebras containing  $\varepsilon$.

If I apply this to our example doesn't that mean that on $\mathbb{R}$ the Borel $\sigma$-algebra can be generated by any set, or any number. Fox example:

The Borel $\sigma$-algebra generated by $\{1\}$ is the intersection of all $\sigma$-algebras containing $\{1\}$. This can be done for any number, for any set. Is that right?

答:
If $\mathcal{A}$ is a family of subsets of $X$ then $\sigma(\mathcal{A})$ is defined as the intersection of all $\sigma$-algebras on $X$ that contain $\mathcal{A}$. It is the same as the smallest $\sigma$-algebra containing $\mathcal{A}$ and could also be constructed conceptually by starting with $\mathcal{A}$ and then adding only those subsets that we have to add because of the properties of a $\sigma$-algebra, iteratively. So $\emptyset,X$ have to be in it and we need to add all countable unions and intersections, and complements of sets that we already added, until we get nothing new anymore. This can be formalised as a transfinite recursion over ordinals $< \omega_1$, but this is usually not what's done in measure theory. The definition I gave at the beginning is (with some cleverness ) enough to work with the $\sigma$-algebra with generators $\mathcal{A}$.

Read the remark correctly: the Borel $\sigma$-algebra is $\sigma(\mathcal{A})$ where $\mathcal{A} = \{]a,b[: a < b, a, b \in \Bbb R\}$, so not a single interval, but the set of all open intervals. This $\sigma$ algebra contains all open sets, because open sets are all countable unions of such intervals. So it will also contain all Borel sets.

The same applies to all other types of intervals: we need all (or lots, really) of them to generate all open sets and hence all Borel sets.
If a $\sigma$-algebra contains all sets of the form $[a,b)$ it will also contain all open intervals as $(a,b) = \bigcup_n [a+\frac{1}{n}, b)$ and hence all open sets etc.


问:
Order of differentiation. Example
Let $y=x^2$
and $y=t^4$
$$\frac{\mathrm{d}y}{\mathrm{d}x}=2x$$
$$\frac{\mathrm{d}y}{\mathrm{d}t}=4t^3$$

Now, $\frac{\mathrm{d}}{\mathrm{d}t}\left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)=\frac{\mathrm{d}}{\mathrm{d}t}\left(2x\right)=\frac{\mathrm{d}}{\mathrm{d}t}\left(2t^2\right)=4t$

$\frac{\mathrm{d}}{\mathrm{d}x}\left(\frac{\mathrm{d}y}{\mathrm{d}t}\right)=\frac{\mathrm{d}}{\mathrm{d}x}\left(4t^3\right)=\frac{\mathrm{d}}{\mathrm{d}x}\left(4x^{\frac{3}{2}}\right)=6t$

Why am I getting $\frac{\mathrm{d}}{\mathrm{d}t}\left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)$ is not equal to $\frac{\mathrm{d}}{\mathrm{d}x}\left(\frac{\mathrm{d}y}{\mathrm{d}t}\right)$?
Isn't it a law of calculus?

答:
$$\frac{d}{dt}\frac{dy}{dx}=\frac{d}{dt}\frac{dy}{dx}\frac{dx}{dx}= \frac{d}{dx} \left(\frac{dy}{dx}\right)\frac{dx}{dt}=\frac{d^2y}{dx^2}\frac{dx}{dt}~~~(1)$$
Next
$$\frac{d}{dx} \left(\frac{dy}{dt}\right)=\frac{d}{dx} \left(\frac{dy}{dt} \frac{dx}{dx}\right)=\frac{d}{dx} \left(\frac{dy}{dx} \frac{dx}{dt}\right)$$
$$=\frac{d^2y}{dx^2}\frac{dx}{dt}+\frac{dy}{dx}\frac{d}{dx} \left(\frac{dx}{dt}\right)=\frac{d^2y}{dx^2}\frac{dx}{dt}+\frac{dy}{dx}\frac{d}{dx} \left(\frac{dx}{dt}\right)\frac{dt}{dt}$$
$$=\frac{d^2y}{dx^2}\frac{dx}{dt}+\frac{dy}{dx}\frac{d}{dt} \left(\frac{dx}{dt}\right)\frac{dt}{dx}=\frac{d^2y}{dx^2}\frac{dx}{dt}+\frac{dy}{dx}\frac{d^2x}{dt^2} \frac{dt}{dx} .~~~~(2)$$

One may check the correctnes of (1) and (2) and their inequality by taking $x=t^2, y=t^2, y^2=x^3$, etc.

Title: Order of differentiation. Example


问:
Can a number have a prime factor that isn't a part of the number's prime factorization?
Is it possible to have a number, $x$, divisible by some prime, such that that prime does not appear in the unique prime factorization of $x$?</p>

答:
If we are working in a unique factorisation domain, then this is impossible: if $p$ divides $x$, then $x = p y$ for some $y$, and so $p$ (or an associate) must appear in the prime factorisation of $x$, which is unique by hypothesis.

In fact, if we go by the strict definition of prime, then if $p$ is prime and divides $x$, it (or an associate) must appear in any factorisation of $x$ into irreducibles, regardless of whether the factorisation is unique.

However, there are rings which are not unique factorisation domains. For example, in the ring $\mathbb{Z}[\sqrt{-5}]$, $(1 - \sqrt{-5})(1 + \sqrt{-5}) = 6 = 2 \cdot 3$, but $2$, $3$, and $1 \pm \sqrt{-5}$ are all irreducible and non-associate. (Of course, this also shows that none of them are prime.)


问:
Finding biconnected components (cut vertices) with a special property of a directed graph
I'd like to find vertices of a directed graph such that it divides the graph into (at least) two in a special way. The successors of a vertex must be only reachable from its predecessors through the specified vertex.

Let me illustrate. Here's a directed graph with all edges directed towards from left to right.

```<code>Z      C - D - E
      /   /
A - B - F - G
 \
  H - I
</code>```

Disregarding root and leaf vertices and the isolated vertex $Z$ that trivially fit the criteria, the vertices that I'm looking for are: $B$, $D$ and $H$.

For a graph containing cycles:

```<code>      3 - 4 - 6
     /     \
1 - 2 <---- 5
</code>```

The vertices I'm looking for are $2$ and $4$. $2$ is the waypoint for access from $1$, and $4$ restricts access to $6$. When it comes to other vertices in the cycle, each could fit the criteria, but as in the case for leaves and roots above, I'm not really interested in them. I would prefer if they were left out, but having them fit is not the worst thing in the world.

Here's my attempt at understanding the properties of such vertices. Biconnected components or cut vertices of a graph determine the vertices whose removal increases the number of connected components. This is too broad of a property, since vertices like $F$ that leave one unconnected component are included but its successor $D$ can be reached through $C$. Conversely, requiring a vertex to only contain bridges as outgoing edges is too strict, since in $B$ the later vertices form a diamond shape.

In my experience often it's simply a matter of finding a name for the thing I'm looking for. So if one exists, what is it? If not, could there be an elegant way of formulating the property I've described, without requiring an implementation to loop through the graph a bunch.

答:
Here's my attempt. It was quite self-evident, almost said in the question, but here it is.

On a similarly connected undirected graph, we remove each vertex one at a time. For that vertex to fit the criteria, each direct predecessor must be unconnected with each direct successor.

This criteria also leaves bare cycles out, which was what I wanted. By also requiring that a vertex has predecessors and successors, roots and leaves are left out as well.

Implementing such criteria seems to be possible quite elegantly by determining the connected components of the undirected graph after deleting the vertex. Each component must never simultaneously contain a vertex from the set of predecessor vertices and a vertex from the set of successor vertices.


问:
The inverse type of Bernhard Leeb's solution for IMO‐1983–inequality

Given three side-lengths $a, b, c$ of a triangle. Prove that
$$a^{2}b\left ( a- b \right )+ b^{2}c\left ( b- c \right )+ c^{2}a\left ( c- a \right )\geq 3\left ( a+ b- c \right )c\left ( a- b \right )\left ( b- c \right )$$
Source: StackMath/@haidangel ft.@tthnew

I used discriminant to create this inequality, also $constant\!:\!=\!\!3$ is the best here. See_ on.StackMath, that's what I'm doing research on, of course this inequality is a result. Not an answer. I think we use Bernhard Leeb's result to help a real lot
$$a^{2}b\left ( a- b \right )+ b^{2}c\left ( b- c \right )+ c^{2}a\left ( c- a \right ):= \left ( c+ a- b \right )\left ( c- a \right )^{2}b- \left ( b+ c- a \right )\left ( a- b \right )\left ( b- c \right )c$$

答:
Just a comment but it's too long for a comment.

Given $a, b, c$ be three side-lengths of a triangle. Prove that
$$a^2b(a-b)+b^2c(b-c)+c^2a(c-a)\geqslant 3(a+b-c)(a-b)(b-c)c$$
Source: By MSE/@tthnew

Some try to proof:  Since $a,b,c$ be three sid-lengths of a triangle, if $(a-b)(b-c)<0$ it's obvious.

So, we only need to consider this inequality when $(a-b)(b-c)\geq 0\to b\equiv \text{mid}\{a,b,c\}.$

But I can't prove this.

By the way if we change $3\to \frac{12}{5}=2.4,$ it's much easy to prove (for the weaker version).

Indeed, let $a=x+y,b=y+z,c=z+x$ and multiply our inequality with $x+y+z.$ The inequality becomes:
$$\frac{1}{20} \left( x-y \right)^2z^2 \left( 15x+31y \right) +\dfrac{1}{20}
 \left( 2x-y-z \right)^2y \left( 20x^2+19xy+46zx+37
z^2 \right) $$ $$+{\frac {21}{20}}
 \left( zx+y^2-2yz \right)^2x+\frac {17}{10} \left( x-z
 \right)^2y^2z+\frac {3}{20} \left( z+x-2y \right) ^2z
^2y$$
$$+\frac{2}{5} \left( x^2-yz \right)^2 \left( 2y+5
z \right) +\frac{1}{5} \left( y-z \right) ^{2}x^3\ge 0.$$

According to Mr. RiverLi's idea,
After let $a=x+y,b=y+z,c=z+x$ then $f(a,b,c)=f(x+y,y+z,z+x)\equiv f(x),$ and we only need to prove
$$\Delta_x=-16 y^2 (y-z)^4 (y+z) (3 y+z) \left(y^2-4 y z+27 z^2\right)\le 0,$$
which is clearly true. I'm still try to find another way.


问:
Number of ways to arrange 7 distinct objects in 10 distinct spots with constraints

There are 7 distinct objects and 10 distinct spots numbered 1-10. Two of spots 7, 8, and 9 must be open. How many ways are there to arrange these objects?

I was helping a friend with this problem and was struggling to understand why our solutions differ numerically. Here are our approaches.

Solution 1:

There are ${3 \choose 2}$ = 3 ways to select the blocked spots, leaving 8 spots open.
There are ${8 \choose 7}$ = 8 ways to select available spots. Multiply by 7! for the total = 24 * 7! = 120,960

Solution 2:

Consider the total number of possibilities - all 3 filled - only 2 filled.

Total: ${10 \choose 7}*7!$ = $\frac{10!}{3!}$

Choose the 7 spots and order the objects.

3 filled: $7 * 6* 5 * {7 \choose 4} * 4! = 7 * 6 * 5 * \frac{7!}{3!}$

Fill the 3 spots (7 choices for spot 1, 6 for 2, 5 for 3), leaving 7 spots available. Choose 4 from these 7 and order the remaining objects.

Only 2 filled: ${3 \choose 2}*7*6*{7 \choose 5}*5! = 3 * 7 * 6 * \frac{7!}{2!}$

Choose the 2 spots to fill, and then fill them with the objects (7 choices for spot 1, 6 for 2). Finally, choose the 5 remaining spots from the 7 available and order the objects. There are only 7 available since we assume the third spot must be open.

Result: $\frac{10!}{3!} - (7 * 6 * 5 * \frac{7!}{3!}) - (3 * 7 * 6 * \frac{7!}{2!})$ = 110,880

I believe Solution 1 is right, but I'm not sure what specifically is wrong with Solution 2. Shouldn't they produce the same number? If not, is Solution 1 wrong?

答: Contrary to your intuition the first solution is wrong. You count 3 times any combination with no object present in positions 7,8,9.</p>


问:
What are the roots of the polynomial?
Suppose $A\in \mathbb{R}^{n\times n}$ is a PSD matrix. $diag(a_1,\ldots,a_n)-A$ is positve definite.
Define    $p(x_1,\cdots,x_n) = \det[diag(x_1,\cdots,x_n)-A] .$
Define
$q(x) = p(x,\cdots,x,a_{k+1},\cdots,a_n),$
which is a degree-$k$ univariate polynomial.

My question: Are all the roots of polynomial $q$ positive?

答:
It is true. Let $D=\operatorname{diag}(a_{k+1},\ldots,a_n)$ and $A=\pmatrix{P&R\\ R^T&S}$ where $S$ has the same size as $D$. Since both $A$ and $\operatorname{diag}(a_1,\ldots,a_n)-A$ are positive definite, $P$ and $D-S$ are positive definite. Now, by using Schur complement, we obtain
\begin{aligned}
q(x)
&=\det\left((xI_k\oplus D)-A\right)\\
&=\det\pmatrix{xI-P&-R\\ -R^T&D-S}\\
&=\det(D-S)\det\left(xI-P-R(D-S)^{-1}R^T\right).
\end{aligned}
Therefore $q(x)=0$ if and only if $x$ is an eigenvalue of the positive definite matrix $P+R(D-S)^{-1}R^T$. It follows that all roots of $q(x)=0$ are positive.

Title: What are the roots of the polynomial?


答:
Here is another way to do it -

First, we choose spots such that exactly two spots out of number $7, 8$ and $9$ are free and one of them has an object. That can be done in

$\displaystyle {3 \cdot {7 \choose 6}} = 21$ ways.

Then we choose spots such that all three spots $7, 8, 9$ are free. There is only one way to do that.

Now we permute objects in $7$ spots.

So total number of arrangements $\displaystyle = 7! \cdot 22 = 110880$

As your first solution triple counts arrangements where $7, 8, 9$ are empty, it can be fixed by subtracting $2 \cdot 7!$.

Title: Number of ways to arrange 7 distinct objects in 10 distinct spots with constraints

Body:
There are 7 distinct objects and 10 distinct spots numbered 1-10. Two of spots 7, 8, and 9 must be open. How many ways are there to arrange these objects?

I was helping a friend with this problem and was struggling to understand why our solutions differ numerically. Here are our approaches.

Solution 1:

There are ${3 \choose 2}$ = 3 ways to select the blocked spots, leaving 8 spots open.
There are ${8 \choose 7}$ = 8 ways to select available spots. Multiply by 7! for the total = 24 * 7! = 120,960

Solution 2:

Consider the total number of possibilities - all 3 filled - only 2 filled.

Total: ${10 \choose 7}*7!$ = $\frac{10!}{3!}$

Choose the 7 spots and order the objects.

3 filled: $7 * 6* 5 * {7 \choose 4} * 4! = 7 * 6 * 5 * \frac{7!}{3!}$

Fill the 3 spots (7 choices for spot 1, 6 for 2, 5 for 3), leaving 7 spots available. Choose 4 from these 7 and order the remaining objects.

Only 2 filled: ${3 \choose 2}*7*6*{7 \choose 5}*5! = 3 * 7 * 6 * \frac{7!}{2!}$

Choose the 2 spots to fill, and then fill them with the objects (7 choices for spot 1, 6 for 2). Finally, choose the 5 remaining spots from the 7 available and order the objects. There are only 7 available since we assume the third spot must be open.

Result: $\frac{10!}{3!} - (7 * 6 * 5 * \frac{7!}{3!}) - (3 * 7 * 6 * \frac{7!}{2!})$ = 110,880

I believe Solution 1 is right, but I'm not sure what specifically is wrong with Solution 2. Shouldn't they produce the same number? If not, is Solution 1 wrong?


问:
Strong Law of Large Numbers for Stationary Process
In the text, "Function Analysis" by Stein and Shakarchi volume 4, there is a proof for the strong law for a sequence of I.I.D. random variables $\{f_n\}$ that involves showing that a shift $\tau$ defined on a product space $Y$ is both measure-preserving and ergodic.

Specifically, $Y = R^\infty = \prod_{j = 0}^\infty R_j$ where each $R_j = \mathbb{R}$ with $m^*$ the corresponding product measure of $\mu$, the common distribution function of $\{f_n\}$. Also, the shift $\tau:Y \to Y$ is given by $\tau(y) = (y_{n + 1})_{n = 0}^\infty$, if $y = (y_n)_{n = 0}^\infty$.

The theorem was proven by defining $\{g_n\}$ to be the coordinate functions on $Y$.

I'm having trouble with the case where instead of $\{f_n\}$ being I.I.D., it is stationary. That is, for every $N$, the joint probability distribution of $f_r, f_{r + 1}, \dots, f_{N + r}$ is independent of $r$. The text says consider the probability space $Y$ as constructed from above, and then to show that whenever ${f_n}$ is a stationary process, then it has the same joint distribution as the sequence ${g_0(\tau^n(y))}$, where $g_0$ is a suitable function on $Y$. However, I am unsure about how to go about defining such a $g_0$.

This is a problem that a professor and I have been puzzled with for a while during our independent study together. Any help would be much appreciated.

答: $g_0$ is simply the first projection: $g_0(y_0,y_1,y_2,...)=y_0$. Use statinarity to show that $(f_n)$ and $(g_0(\tau^{n}(y))$ have the same finite dimensional distribution. That also implies that they induce the same probability measure on $Y$.


问:
How to take integral of a second order partial derivative of a function?
My textbook shows that for a function $u(x,y)$ satisfies $\frac{\partial^2 u}{\partial x\partial y}= 0$, we can integrate this relation twice and get $u(x,y) = F(x) + G(y)$, which makes me confused.

I think integrating $\frac{\partial^2 u}{\partial x\partial y}= 0$ will lead to $\frac{\partial u}{\partial y}= C_1(y)$ because when integrating $x$, $y$ could be seen as a constant, and continue integrating it we will get $u(x,y)=C_2(y)+C$ where $\frac{dC_2(y)}{dy} =C_1(y)$. What's wrong with that?

答:
You can see that something is wrong with your solution from symmetry: $x$ and $y$ should be interchangable. Your solution doesn't reflect that.
When you integrate $\frac{\partial u}{\partial y} = C_1(y)$ you left out the $x$-dependency of $C$:

$$
u(x,y) = \int \frac{\partial u}{\partial y}(y) \text{d} y + C(x)
$$


问:
Let $E=C([-1,1])$ and $f,g \in E$ such that $f(x) = x^2-1$ and $g(x)= 1-x^2$. Determine if $B(f,2) \cap B(g,2) = \emptyset?$

Let $E=C([-1,1])$ the set of continuous functions with the sup norm and $f,g \in E$ such that $f(x) = x^2-1$ and $g(x)= 1-x^2$. Determine if $B(f,2) \cap B(g,2) = \emptyset?$

With the sup norm I have that \begin{align*}
d(f,g) &= \|f-g\|_{\infty}\\
&= \sup_{x\in [-1,1]}|f(x)-g(x)|  = \sup_{x\in [-1,1]}|x^2-1-(1-x^2)| \\
&= \sup_{x\in [-1,1]}|2x^2-2|=2
\end{align*}

So this would seem that the intersection is empty, but I'm not entirely sure this concludes it?

答:
I don't think your result implies that the open balls do not intersect. Try to visualize it, if you take an open ball with the center $g$ then they will intersect obviously, as $g$ lies in the boundary of $B(f,2)$

In fact, $h(x)=x^2$ belongs to both of these open balls


问:
Solving the system of linear homogeneneous differential equations.
I have stumbled upon quite a strange issue, while solving the following system of differential equations:
$$ \left\{\begin{matrix}
\; \dot{x} &=&2x+6y-15z \\
\; \dot{y} &=&x+y-5z \\
\; \dot{z} &=&x+2y-6z
\end{matrix}\right. \left |A=\begin{pmatrix}
2 & 6 & -15 \\
1 & 1 & -5 \\
1 & 2 & -6 \\
\end{pmatrix}\right. \left\|\dot{f} = Af, f = \begin{pmatrix}
x\\
y\\
z\\
\end{pmatrix}.\right.$$
Getting eigenvalues for the matrix $A$, I got that $|A-\lambda I| = 0 \Leftrightarrow (\lambda + 1)^{3} = 0.$ That is the Characteristic equation has only one solution (the matrix has only one eigenvalue), s.t. $mult(\lambda_{1} = -1) = 3.$

Thus, I have had to find the eigenvectors, gotten this eigenvalue I got that:

I. $\lambda = -1:$

$(A-\lambda I) = \left(\begin{array}{ccc|c}
3 & 6 & -15 & 0\\
1 & 2 & -5 & 0\\
1 & 2 & -5 & 0\\
\end{array}\right) \xrightarrow[]{\text{RREF}} \left(\begin{array}{ccc|c}
1 & 2 & -5 & 0\\
0 & 0 & 0 & 0\\
0 & 0 & 0 & 0\\
\end{array}\right),$ thus $v_{\lambda} =
\begin{pmatrix}
5c-2b\\
b\\
c\\
\end{pmatrix}.$ For some $a, b\in\mathbb{C}$, i.e. the eigenvectors of A are:
$$
\tilde{e}_{1} = \begin{pmatrix}
5\\
0\\
1\\
\end{pmatrix}, \tilde{e}_{2} = \begin{pmatrix}
-2\\
1\\
0\\
\end{pmatrix},\mathbb{E_{\lambda}} = \left\{\tilde{e}_{1}, \tilde{e}_{2}\right\}.$$
Now, as $|\mathbb{E_{\lambda}}| = 2$, we have to create a joint vector: $\tilde{e}_3 = v^{(1)}_{\lambda}$, but I don't know, what should I do there, in order to obtain $\tilde{e}_3$, as we have TWO eigenvectors, corresponding to one eigenvalue. If I pick $v_{\lambda}$ with $b = c = 1$ , then the joint vector $v^{(1)}_{\lambda}$, s.t. $(A-\lambda I)\cdot v^{(1)}_{\lambda} = v_{\lambda}$, is:
$$v^{(1)}_{\lambda} =  \begin{pmatrix}
4\\
1\\
1\\
\end{pmatrix}, v^{m}_{\lambda} = \left(v_{\lambda} \cdot t + v^{(1)}_{\lambda}\right)$$
However, when I derive that:
$$ f = e^{-t}\left(\tilde{C}_1\cdot\tilde{e}_{1}+ \tilde{C}_2\cdot\tilde{e}_{2}+ \tilde{C}_3\cdot  v^{m}_{\lambda}\right)$$
And then check the result, assuming that $\tilde{C}_1 = 1, \tilde{C}_2 = 2, \tilde{C}_1 = 3$, I got that $\dot{x} = (-6e^{-t}-9e^{-t}t)$, however, by $f$ it should be $(-4e^{-t}-9e^{-t}t).$

I just really want to understand, what am I mistaken in? I have tried to reach my course teacher, but he doesn't seem to respond, so I decided to post my question there.

Thank you in advance for the help!

答:
It seems like I have made a mistake in calculations, when checking my answer, as after recalculating it once again, I got the following: (Assuming $\tilde{C}_k = k$)

$ \frac{\partial x}{\partial t} = \frac{\partial}{\partial t}\left(e^{-t}\left(5\tilde{C}_1 -2\tilde{C}_2 +\tilde{C}_3(3t+4) \right)\right) = \frac{\partial}{\partial t}\left(e^{-t}\left(5 -2\cdot2 +3(3t+4) \right)\right) = -e^{-t}(4+9t) [LHS] = -e^{-t}(4+9t)[RHS] = e^{-t}\left(2\cdot(5-2\cdot2+3(3t+4))+6\cdot(2+3(t+1))-15\cdot(1+3(t+1))\right)$

That means, that the solution that I have provided there works correctly.


问:
Use substitution $u=\cos(x^2/4)$ to find $\int_0^\sqrt{\pi} 2x\cos(x^2/4) dx$
I found that $du=-x\sin(x^2/4)dx/2$ but this is way more complicated then the $u$ substitutions I have done. I am stuck and from here don't know what to do. Any help is much appreciated. Thank you.

答: Use $t=x^2/4$ to write the integral as $\int_0^{\pi/4}4\cos tdt$. As @user10354138 notes, $u=\sin t$ would help much more than $u=\cos t$: the integral is clearly $[4\sin t]_0^{\pi/4}=2\sqrt{2}$. If we instead use $u=\cos t$, we get$$\int_{1/\sqrt{2}}^1\frac{4u}{\sqrt{1-u^2}}du=[-4\sqrt{1-u^2}]_{1/\sqrt{2}}^1=2\sqrt{2}.$$


问:
area of intersecting part of triangles
Let ABCD be a rectangle and AC one of its diagonals.Length of AD is 4m and of AB is 1 m.A line is drawn from point D towards the midpoint of BC which we will label E.Let DE intersect AC at F.How do we figure out the area of the triangle EFC?Any hint or link to previous solutions will be appreciated.</p>

答:
HINT: For $\text{h}$, you must know the length  of $\text{EF}$ or $\text{FC}$
. Do you know in what ratio $\text{DE}$ cuts $\text{AC}$? Since $\triangle \text{AFD}$ and $\triangle \text{EFC}$ are similar, $\text{AF}=2\times\text{FC}$.

Title: area of intersecting part of triangles


问:
Method of characteristics hyperbolic PDE
Consider the hyperbolic PDE :

$$2u_{xx} + 8u_{xy} + 6u_{yy} = 0.$$

It can be shown using the method of characteristics that the above PDE has the following general solution:

$$u(x,y) = F(y-x) + G(3x-y).  $$

I want to find a solution to the equation with the following general boundary conditions,

$u(x,0) = g_0(x) $ and $u_y(x,0) = g_1(x)$ for $x \in \mathbb{R}.$

My working thus far:

$u_y(x,0) = F'(-x) - G'(3x) = g_1(x)\,\,\,\,\,\, (1)$

$u(x,0) = F(-x) + G(3x) = g_0(x) \,\,\,\, (2)$

Differentiating (2), we get

$-F'(-x) + 3G'(3x) = g_0'(x) \,\,\,\, (3)$

Adding $(1)$ and $(3),$ we get

$2G'(3x) = g_0'(x) + g_1(x) \,\,\,\, (4)$

I want to integrate $(4)$ to get expression for $G(3x) $ but I don't know how to do it. Is this simply:

$$G(3x) = \frac{3}{2}\int g_1(x) dx + \frac{3}{2}g_0(x) + C.  $$

Also, once $G(3x)$ is computed, how do I obtain expression for $G(3x-y)?$ I'm stuck here and need help.

答:
Let we consider a system:
$$u_y(x,0) = F'(-x) - G'(3x) = g_1(x)\,\,\,\,\,\, (1)$$
$$u(x,0) = F(-x) + G(3x) = g_0(x)\,\,\,\,\,\, (2)$$

From (2) we get $F(x) = g_0(-x) - G(-3x)$. Now we have a new form for solution of this problem $u(x, y) = g_0(x-y)-G(3 x-3 y)+G(3 x-y)$.
Using condition $u_y(x,0) = g_1(x)$ we can define the function $G$.

$$
2 G'(3 x)-g_0'(x)=g_1(x),\text{ or } 2G'(x)=g_1\left(\frac{x}3\right)+g_0'\left(\frac{x}3\right)
$$

By integrating we obtain $G(x) =C+ \frac{1}{2}\int\limits_0^x \left(g_1\left(\frac{z}3\right)+g_0'\left(\frac{z}3\right) dz\right) $, where $C$ is some constant. And we can write $F(x) = -C-\frac{1}{2} \int_0^{-3 x}
   \left(g_0'\left(\frac{z}{3}\right)+g_1\left(\frac{z}{3
   }\right)\right) \, dz+g_0(-x)$.

So, we have found the solution
$$u(x, y) = -\frac{1}{2} \int_0^{3 x-3 y}
   \left(g_0'\left(\frac{z}{3}\right)+g_1\left(\frac{z}{3
   }\right)\right) \, dz+\frac{1}{2} \int_0^{3 x-y}
   \left(g_0'\left(\frac{z}{3}\right)+g_1\left(\frac{z}{3
   }\right)\right) \, dz+g_0(x-y).$$


问:
Criteria of convergence for gaussian integral with complex parameters
Consider the following Gaussian integral:

\begin{equation}
I=\int_{-\infty}^{+\infty} e^{\mathrm{i}ax^2+bx}\mathrm{d}x
\end{equation}

Where $a\in \mathbb{R}$.

If $b=0$ then:
$$I=\int_{0}^{\infty} e^{i a x^{2}} d x=e^{i \pi \operatorname{sgn}(a) / 4} \sqrt{\frac{\pi}{4|a|}}$$
If $b$ is purely imaginary then (from wikipedia):
$$I=\left(\frac{ \pi i}{a}\right)^{\frac{1}{2}} e^{\frac{-i \mathrm{Im}[b]^{2}}{4a}}$$

My questions:

If $b\in \mathbb{C}$ where $\mathrm{Re}[b]\neq 0$ what are the constraints needed on $a$ so that the integral still converges?

For general $a,b\in\mathbb{C}$ How must $a$ relate to $b$ so that the integral converges? More generally, what are the general conditions for $I$ to converge? (Setting $\mathrm{Re}[a]<0$ is not the most general condition: indeed if $b=0$ and $a$ purely imaginary then $I$ still converges).

Any reference or link to a specific book/article is always appreciated. Thanks for the help.

答:
For real $c $ and $d$ $e^{cx^{2}+dx}$ is integrable iff $c <0$. Since $|e^{ia x^{2}+bx}|=e^{\Re (ia x^{2}+bx)}=e^{-\Im a x^{2}+\Re b x}$ it folows that the required condition  is $\Im a >0$.

The integral converges for all complex $b$ if the imaginary part of $a$ is positive and diverges for all complex $b$ if the imaginary part of $a$ is non-negative.


答:
Using the hinted substitution:

$$2\int x \cos\frac{x^2}4 dx=-4\int \frac{xu\,du}{x\sin\dfrac{x^2}4}=-4\int\frac{u\,du}{\sqrt{1-u^2}}=4\sqrt{1-u^2}=4\sin\frac{x^2}4.$$

Anyway, the substitution $u=\dfrac{x^2}4$ is preferred.

Title: Use substitution $u=\cos(x^2/4)$ to find $\int_0^\sqrt{\pi} 2x\cos(x^2/4) dx$

Body: I found that $du=-x\sin(x^2/4)dx/2$ but this is way more complicated then the $u$ substitutions I have done. I am stuck and from here don't know what to do. Any help is much appreciated. Thank you.


问:
Remainder When Divided By 70 : Remainder Theorem Problem
If $x$ = $16^3$ + $17^3$ + $18^3$ + $19^3$ , then $x$ divided by $70$ leaves a remainder of?

I tried to solve this problem by using the remainder theorem which states that remainder of $$Rem[\frac{a+b+c+....}{x}] = Rem[\frac{a}{x}] + Rem[\frac{b}{x}] + Rem[\frac{c}{x}] + .... $$ where 'Rem' means 'Remainder Of'.

Using this same logic I can write $$Rem[\frac{16^3+17^3+18^3+19^3}{70}] = Rem[\frac{16^3}{70}] + Rem[\frac{17^3}{70}] + Rem[\frac{18^3}{70}] + Rem[\frac{19^3}{70}] = 18 +13+11+69 = 111 \Rightarrow Rem[\frac{111}{70}] = 41$$

So my answer comes up as 41. Can someone please explain to me what I might be doing wrong or what I have understood wrong?

答:
To determine the remainder of some computation, $C, \pmod{70}$, simply determine the remainder of $C, \pmod{2}, \pmod{5}$, and $\pmod{7}.$  Then employ Number Theory methods to solve the resultant congruencies.

Let $C = 16^3 + 17^3 + 18^3 + 19^3.$

Since the second and fourth terms of $C$ (only) are odd, it is immediate that $C \equiv 0\pmod{2}.$

You can prove, by binomial expansion, that if $r,s,k,n$ are all positive integers, then $(r+ns)^k \equiv r^k \pmod{n}.$

Therefore, the problem immediately reduces to the twin problems of
computing

$$D = 1^3 + 2^3 + 3^3 + 4^3 \pmod{5}$$

and computing

$$E = 2^3 + 3^3 + 4^3 + 5^3 \pmod{7}.$$

The problem permits shortcuts, by recognizing that if

$a \equiv (-b) \pmod{n},$ then $a^3 \equiv (-b)^3 = (-1)^3b^3 = -(b^3) \pmod{n}.$

Using this approach, note that when applying $\pmod{5}$ congruencies against $D$,

$4 = (-1),$ and $3 = (-2).$

Similarly, when applying $\pmod{7}$ congruencies against $E$

$5 = (-2),$ and $4 = (-3).$

Therefore, it is immediate that $D \equiv 0\pmod{5}$ and $E\equiv 0\pmod{7}$.

Therefore, it is immediate that $C$ is a multiple of $2$, $5$, and $7$.


问:
99% of all readings are within WHAT temperature range?
I am a bit stumped.

"The temperature reading from a thermocouple placed in a constant-temperature medium is
normally distributed with mean of 82.5C and standard deviation .1C. 99% of all readings
are within what temperature range?"

The question states that multiple readings are taken - no sample size is given. The mean is 82.5 with standard deviation of 0.1.
I am not sure whether the question is badly phrased or some other error in it.

I attempted is as a Confidence Interval, however a N would be necessary.

I would appreciate at least a hint because I feel like I am going mad.

答:
If $X \sim \mathsf{NORM}(\mu=82.5,\sigma=0.1),$
then you seek $L$ and $U$ for which
$P(L < Y \le U) = 0.99.$ There are many solutions,
but it seems easiest to cut probability $0.005$ from
each tail of the distribution, as @tommik (+1)
has suggested.

In R, where pnorm is a binomial CDF and pnorm is
a quantile function (inverse CDF), we find
$L=82.24242, U= 82.75758.$

```<code>LU=qnorm(c(.005,.995), 82.5, .1); LU
[1] 82.24242 82.75758
diff(pnorm(LU, 82.5, .1))  # check
[1] 0.99

curve(dnorm(x, 82.5, .1), 82, 83, lwd= 2, ylab="PDF",
      xlab="x", main="Central 99% of NORM(82.1, 0.1)")
 abline(h=0, col="green2")
 abline(v=LU, lwd=2, col="red", lty="dotted")
</code>```

Note: This can be done by standardizing and using printed
standard normal CDF tables, but in such tables
z-scores are usually given to only two places.

You might get better accuracy using the bottom
row of a t table (perhaps labeled 'Inf', '$\infty$'. or
'Norm'). There you can find that $2.576$ cuts
probability $0.005$ from the upper tail of a
standard normal distribution.

$$P(L < X < U) =
P\left(\frac{L-\mu}{\sigma} < Z =\frac{X-\mu}{\sigma}
<\frac{U-\mu}{\sigma}\right)=0.99,$$
where is $Z$ is a standard normal random variable, and $\mu$ and $\sigma$ are known.

Set $(U-\mu)/\sigma = (U-82.5)/.1 = 2.576$ and solve for $U.$ Similarly for $L$ and $-2.576.$

Body:
If $X \sim \mathsf{NORM}(\mu=82.5,\sigma=0.1),$
then you seek $L$ and $U$ for which
$P(L < Y \le U) = 0.99.$ There are many solutions,
but it seems easiest to cut probability $0.005$ from
each tail of the distribution, as @tommik (+1)
has suggested.

In R, where pnorm is a binomial CDF and pnorm is
a quantile function (inverse CDF), we find
$L=82.24242, U= 82.75758.$

LU=qnorm(c(.005,.995), 82.5, .1); LU
[1] 82.24242 82.75758
diff(pnorm(LU, 82.5, .1))  # check
[1] 0.99

curve(dnorm(x, 82.5, .1), 82, 83, lwd= 2, ylab="PDF",
      xlab="x", main="Central 99% of NORM(82.1, 0.1)")
 abline(h=0, col="green2")
 abline(v=LU, lwd=2, col="red", lty="dotted")

Note: This can be done by standardizing and using printed
standard normal CDF tables, but in such tables
z-scores are usually given to only two places.

You might get better accuracy using the bottom
row of a t table (perhaps labeled 'Inf', '$\infty$'. or
'Norm'). There you can find that $2.576$ cuts
probability $0.005$ from the upper tail of a
standard normal distribution.

$$P(L < X < U) =
P\left(\frac{L-\mu}{\sigma} < Z =\frac{X-\mu}{\sigma}
<\frac{U-\mu}{\sigma}\right)=0.99,$$
where is $Z$ is a standard normal random variable, and $\mu$ and $\sigma$ are known.

Set $(U-\mu)/\sigma = (U-82.5)/.1 = 2.576$ and solve for $U.$ Similarly for $L$ and $-2.576.$


问:
What does expectation ( E(x) ) practically mean in Statistics?
I know the mathematical definition of Expectation- Sum of the product of the probability of an event occurring - P(x), and the value corresponding with the actual observed occurrence of the event. ( in the case of a discrete random variable )

But I don't get the practical aspect of it. I don't understand what the expectation of an event actually is trying to tell and how it can be useful, I just understand the mathematical aspect of it

答:
The expectation is simply a probability-weighted average of all events that may occur. For example, if you roll a fair six-sided die, the probability of getting each outcome $1,2,3,4,5$ or $6$ is $1/6$. Then the expected value is $$\sum_{i=1}^6\frac16\cdot i=3.5$$
which is the weighted average of each outcome. This means that if you roll a dice for many many times, the average of all the rolls would approach $3.5$.


Body:
$$I=\int_{0}^{\sqrt{\pi}} 2x \cos (x^2/4) dx$$
Let $x^2/4=t \implies xdx=2dt$
Then
$$I=\int_{0}^{\pi/4} 4 \cos t dt =4 \sin t|_{0}^{\pi/4}=2 \sqrt{2}$$

Now in OP's way: Let $\cos(x^2/4)=u \implies x dx=-\frac{2du}{\sin(x^2/4)}=\frac{-2udu}{\sqrt{1-u^2}}$, then
$$I=-4\int_{1}^{1/\sqrt{2}}\frac{u}{\sqrt{1-u^2}} du$$
Next, let $u^2=v$, then
$$I=-2\int_{1}^{1/2} \frac{dv}{\sqrt{1-v}} =4\sqrt{1-v}|_{1}^{1/2}=2\sqrt{2}. $$


问:
Constructing a "limit" function. But is it measurable?
Let $\alpha_n:\,(0,T)\rightarrow A$, where $A\subset\mathbb{R}^d$ compact, a sequence of measurable functions. Since $A$ is compact, for almost every $t\in (0,T)$, there exists a subsequence $\alpha_{n_k}(t)$ and $\alpha_t\in A$ such that $\alpha_{n_k}(t)\rightarrow \alpha_t$ as $k\rightarrow\infty$.
Now, we define $\alpha:(0,T)\rightarrow A$ via:
\begin{align*}
\alpha(t):=\alpha_t.
\end{align*}
Is $\alpha$ also measurable?

Thank You in advance!

答:
The limit superior of a sequence of measurable functions is measurable (e.g. see Rudin's Real and Complex Analysis Thm 1.14.).
In your case, $a$ is the limit (and hence the limit superior) of the sequence of measurable functions $a_{n_k}$, hence measurable.


问:
prove that: $\limsup_{n \rightarrow \infty} (a_n+b_n) \le \limsup_{n\rightarrow\infty}(a_n) + \limsup_{n\rightarrow \infty} (b_n)$
I am reading Rudin's "Principles of Mathematical Analysis", and the following problem appears in it:

Problem 3.5:

For any two real sequences $\{a_n\}$, $\{b_n\}$,

prove that: $\limsup_{n \rightarrow \infty} (a_n+b_n) \le \limsup_{n\rightarrow\infty}(a_n) + \limsup_{n\rightarrow \infty} (b_n)$

provided the sum on the right is not of the form $\infty - \infty$

The solution in the manual goes as following:

Since the case when $\limsup_{n\rightarrow\infty}(a_n) = +\infty$ and $\limsup_{n\rightarrow\infty}(b_n) = -\infty$ has been excluded from consideration , we note that the inequality is obvious if $\limsup_{n\rightarrow\infty}(a_n) = +\infty$. Hence we shall assume that $\{a_n\}$ is bounded above.

Let $\{n_k\}$ be a subsequence of the positive integers such that $\lim_{k\rightarrow\infty}(a_{n_k} + b_{n_k}) = \limsup_{n\rightarrow\infty}(a_n + b_n)$.

Then choose a subsequence of the positive integers $\{k_m\}$ such that

$$\lim_{m\rightarrow\infty}(a_{n_{k_m}}) = \limsup_{k\rightarrow\infty}(a_{n_k})$$

The subsequence $a_{n_{k_m}} + b_{n_{k_m}}$ still converges to the same limit as $a_{n_k} + b_{n_k}$.

And proof continue.
I understand all the proof except for the last sentence, precisely:

The subsequence $a_{n_{k_m}} + b_{n_{k_m}}$ still converges to the same limit as $a_{n_k} + b_{n_k}$.

Why this is true?

I was thinking that we can also assume that $\{b_n\}$ also bounded so that it contains a series that converges and so $\limsup_{n\rightarrow \infty} (b_n)$ is finite but I don't know if this is useful and I couldn't go further.

答:
After the answer from Kavi, and some extra effort I see that we can solve the problem as follows:

If $\limsup_{n\rightarrow\infty}(b_n) = +\infty$ then the inequality is also obvious so we can also (as we did with $\{a_n\}$) assume that $\{b_n\}$ is bounded.

Since both sequences are bounded, $\{a_n+b_n\}$ also bounded, and there is a subsequence of $\{a_n+b_n\}$ that converges so $\limsup_{n\rightarrow\infty}(a_n+b_n)$ is finite and since:
$$\lim_{k\rightarrow\infty}(a_{n_k} + b_{n_k}) = \limsup_{n\rightarrow\infty}(a_n + b_n)$$

Then $\lim_{k\rightarrow\infty}(a_{n_k} + b_{n_k})$ is also finite and therefore converges.

Then every subsequence of it also converges to the same limit.

And this answers my question, but please correct me if there is a flaw somewhere.


问:
Showing a group is abelian knowing there's only one isomorphism
Suppose $G$ is a group and let $S$ be the set of all isomorphisms $f: G \rightarrow G$

Show, if $|S|=1$, that the group G is abelian with elements of order 1 or 2

Now so far I know that since $|S|=1$ there is a canonical map or canonical isomorphism but how do I use this fact concretely?

Thank you in advance

答:
If $G$ is not abelian, then there are two elements $a,b\in G$ such that $ab\neq ba$. Therefore $x\mapsto bxb^{-1}$ is a non-trivial group isomorphism, contradicting $|S|=1$.

On the other hand $x\mapsto x^{-1}$ is an automorphism as well, now we know $G$ is Abelian. So if $|S|=1$ then this automorphism has to be the identity $x\mapsto x$. Meaning for any $x\in G$ we have $x=x^{-1}$ which means that the order is either $1$ or $2$.


问:
Self Learning Stochastic Process By Sheldon Ross
I have just started self-learning Stochastic Processes by Sheldon Ross (2nd Edition). I am finding the exercises really tough and time-consuming. I even tried searching for a solution manual but couldn't find it anywhere.

I have gone through the book - A first course in Probability by the same author before. But, Stochastic Processes is on a completely different level for me. I wanted some advice from people who have learned from this book on how did they go about it and verify their solutions. If someone knows about the availability of a solution manual, could someone direct me towards the same?

答:
What specifically are you having trouble with in Ross's Stochastic Processes? I am familiar with this text and I would have to say it has its shortcomings. Although the preface states

This text is a nonmeasure theoretic introduction to stochastic processes, and
as such assumes a knowledge of calculus and elementary probability.

The first chapter begins with the formal measure-theoretic definition of a probability space, and proceeds to introduce and prove the Borel-Cantelli lemmas, which are statements about the $\limsup$ of a sequence of sets. It is unlikely the notion of limit superior would have been introduced in a typical undergraduate calculus and introductory probability courses; and it is not mentioned at all in First Course in Probability - so I could see how this maybe be confusing.

The concept of expectation is defined in terms of Riemann-Stieltjes integrals, as opposed to Lebesgue integrals, however, and indeed this is treated in 7.9 of the 10th edition of First Course in Probability. Also, conditional expectation is not defined rigorously in terms of the Radon-Nikodym theorem, so the "nonmeasure theoretic" claim is justified, I suppose.

Going chapter-by-chapter in Stochastic Processes:

Most of this material is treated in First Course in Probability. Characteristic functions and Laplace transforms are not, but the general concept is similar enough to that of the moment-generating function that it shouldn't be too hard to pick up on. The last section which formally introduces stochastic processes is of course fundamental to the rest of the text, and the definitions and examples seem reasonable to me.

First Course in Probability introduces the Poisson process in much the same manner as the beginning of this chapter. The remainder rigorously proves important and often-used properties of the Poisson process, although section 2.6 "Conditional Poisson Processes" is a bit more advanced (such processes are also known as doubly stochastic Poisson processes or Cox processes), and could probably be skipped on a first reading.

Renewal processes are a generalization of Poisson processes and are extremely important in the study of stochastic processes. A rigorous proof of the strong law of large numbers is given in First Course in Probability, and the techniques used there are  important for being able to follow the proofs of the results in this chapter. But for example, stopping times are not defined formally in the measure-theoretic sense, nor is Wald's identity stated in full generality - so this should be understandable, with a bit of effort. The proof of the elementary renewal theorem is a bit more involved, though not beyond the level of an introductory real analysis course. The Key renewal theorem doesn't seem to be explained well, but the coverage on alternating renewal processes, the inspection paradox, delayed and equilibrium renewal processes, renewal reward processes, and regenerative processes all seem solid to me.

The chapter on Markov chains seems a bit lacking, only covering some of the more fundamental results and focusing heavily on examples which to me do not seem well-motivated. In particular semi-Markov processes seem a bit out of place in a first introduction to Markov Chains.

The chapter on continuous-time Markov chains is very thorough and does a good job of covering the essential topics, as well as providing important applications of queueing theory.

The topic of martingales really should be left to a second course in stochastic processes which assumes prior knowledge of measure theory, in my opinion. I would not recommend this text as an introduction to the topic.

The general random walk is another topic for which "a knowledge of calculus and elementary probability" is insufficient, in my opinion.

Brownian motion is an extremely important topic in the study of stochastic processes...but I really don't think it is treated well in this text or given the proper motivation. The construction of the process is not compelling, the reflection principle does not seem to be mentioned, and the pathological properties of sample paths (i.e. continuous with probability one but differentiable nowhere) not discussed either.
9-10. These chapters are a bit advanced for an introductory course, in my opinion.

Ross's book does have good exercises and many examples, but I find the theory and motivation a bit lacking. I might suggest Resnick's Adventures in Stochastic Processes instead.

If you have any specific questions about the material feel free to elaborate, of course.