答:
The function is only defined for positive $x$. For $0<x\lt 1$, $\ln x$ is negative, large negative near $x=0$. Dividing by $x^2$ makes things much larger negative.

For $x$ large positive, the $x^2$ at the bottom crushes puny $\ln x$. So for $x$ large, our function is positive but close to $0$.

We might note that $\frac{\ln x}{x^2}$ is $0$ at $x=1$, and positive for $x\gt 1$.

For more detail, we compute the derivative.  This is $\frac{1-2\ln x}{x^3}$. Thus the derivative is $0$ at $x=e^{1/2}$, positive for $x\lt e^{1/2}$, and negative for $x\gt e^{1/2}$.

So our function reaches a maximum at $x=e^{1/2}$. The maximum value is $\frac{1}{2e}$.

Now we have enough information to draw a pretty good sketch. For fine detail about concavity, one can look at the second derivative. If you do that calculation, you will find that the curve is facing down up to $x=e^{5/6}$, and then facing up. This does not add much in the way of useful information in this case. However, an exam question might ask about concavity, since it is not too hard to deal with in this example.


问:
Evaluate $\lim_{n\to \infty} (n+1)^{\frac 23} -(n-1)^{\frac 23}$
Clearly, $\infty -\infty \not =0$

I have a feeling the squeeze theorem can be applied here, but I am not sure how to write the required terms separately. Can I get a hint?

答:
Using Taylor expansion

$$(1+\epsilon)^{2\over 3}= 1+{2\over 3}\epsilon +o(\epsilon)$$

We can write with $\epsilon ={1\over n}$

$$(n+1)^{2\over 3}-(n-1)^{2\over 3}=n^{2\over 3}\left(1+{2\over 3}\epsilon-1+{2\over 3}\epsilon+o(\epsilon)\right)={4\over 3}n^{-{1\over 3}}+o\left(n^{-{1\over 3}}\right)$$

And this quantity at $+\infty$ has limit $0$

ContentLicense: CC BY-SA 4.0


问:
Perimeter change when moving one point of a triangle along a circle
I am interested in the perimeter $P(\Delta')$ of a modified version of a triangle $\Delta = (A,B,C)$. I chose point $B$ to be the point I want to shift towards the inside (or on the boundary) of $\Delta$. I chose a distance $r<h$ where $h$ is the height of the triangle. $B$ can be moved to an arbitrary point $E$ which is of distance $r$ to $B$ and lays inside or on the boundary of $\Delta$. This means that $E$ must also lay on the circle of radius $r$ with the center point $B$. I denote $\Delta'=(A,E,C)$. What is the perimeter $P(\Delta')$ of the new triangle given the original side lengths, $r$ (and probably some representation of the arc length to E)? In what range does $P(\Delta)-P(\Delta')$ lie in?

I attached a small figure which visualizes both triangles.

Edit

Based on the comments, I added some notation:

Assuming $B=(0,0)$, then we have $E=(x,y)$ with $x=\lambda_A x_A + \lambda_C x_C$ and $y=\lambda_A y_A + \lambda_C y_C$ ($\lambda_A+\lambda_C = 1$).

This gives us the following set of equations:

$$r=\sqrt{(0-x)^2 + (0-y)^2} = \sqrt{(\lambda_A x_A + \lambda_C x_C)^2 + (\lambda_A y_A + \lambda_C y_C)^2}$$
$$|AE| = \sqrt{(x-x_A)^2+(y-y_A)^2}$$
$$|CE| = \sqrt{(x-x_C)^2+(y-y_C)^2}$$

Moreover, we have:
$$|AB| = \sqrt{x_A^2 + y_A^2}$$
$$|CB| = \sqrt{x_C^2 + y_C^2}$$

The perimeter change is in fact the difference $|AB|+ |CB| - |AE| - |CE|$. I would like to show that $r \leq |AB|+ |CB| - |AE| - |CE|$. I struggle with the proof because I still have six variables $x_A,x_C,y_A,y_C,\lambda_A$ and $\lambda_C$.

Background

This question arises from a computer science problem called lawn mowing/milling, where you are given a cutter (in my case a circular shape) and a Polygon to cover. I noticed that given a tour of the circular cutter one can change the cutter from a circular shape with a radius $s$ to a square with side length $2s$. Afterward, the tour can be shortened a bit. I have already proven a specific length $r$ that I can move the points of the tour "inwards" to shorten the tour. My goal is to find out how much shorter the final tour gets when I modify it. In my example, $A-B-C$ are part of the tour and $B$ is a point which can be moved inward by $r$ to a point $E$. My tour will later be $A-E-C$. I want to find out how much the tour is shortened after my transformation and I hope that it will be shortened by at least $r$.

答:
We can find perimeter change as a function of parameter $ \theta$

Let the circle be centered at (0,0) and radius r.

Let two points on a vertical  line be $ ( h,-p),(h,q) $

slant line lengths sum for $\theta_1,$ $ L1=$

$$ = \sqrt { r\cos \theta_1 -h)^2} + \sqrt{( r \sin \theta_1 +p)^2}$$
$$ + \sqrt { r\cos \theta_1 -h)^2} + \sqrt{( r \sin \theta_1-q)^2}$$
For $\theta_2$, $L_2=$
$$ = \sqrt { (r\cos \theta_2 -h)^2} + \sqrt{( r \sin \theta_2 +p)^2}$$
$$ + \sqrt { (r\cos \theta_2 -h)^2} + \sqrt{( r \sin \theta_2-q)^2}$$

The perimeter change is  difference  $ L_2- L_1$ where $L_2$ pair is marked in red. It depends on two $\theta_ s$ , the constants are  $( h,p,q,r).$


问:
Is $\sin(x^2)$ uniformly continuous for all $x \in (0, +∞)$? Where is my error?
The answer is that is not uniformly continuous. But I get the contrary. Here is my solution.
Let $f(x) = \sin(x^2)$, if $x ≠ 0$ and 0 if $x = 0$. In this case $f \in \Bbb{R}_{≥0}$ and $f(x)$ is continuous on $\Bbb{R}_{≥0}$.
According to Heine–Cantor's theorem, if a function is continuous on $[a, b]$ then it is uniformly continuous. Let $a = 0$, and $b_n \to +\infty$ is the sequence of all numbers that belong to $\Bbb{R}_{≥0}$. Then $f(x)$ is uniformly continuous on $[0, b_n]$. And so it is uniformly continuous on $\Bbb{R}_{≥0}$. Now, I think that we can claim that $\sin(x^2)$ is uniformly continuous for any real $x$.
Where is my error?

答:
The formal error in your proof is extrapolating from compact intervals to all of the real line. Other answers and comments have noted this.

The informal/intuitive reason is that uniform continuity is essentially a bound on how steep the function is, since the slope at any point determines (in a way) how the $\delta$ in the definition of uniform continuity depends on the $\epsilon$.

In this case it's easy to see that the derivative $2x\sin(x)$ is unbounded on the real line. (It is bounded on the open unit interval, as in the title of the question.)


问:
Randomly take $51$ numbers from the set 1, 2, ..., 159. Find variance of their sum.
We randomly take $51$ numbers from 159 natural numbers $1,...,159$ without replacement. Let $\alpha$ be a random variable equal to the sum of the selected numbers. Find the variance of $\alpha$.

Firstly I need to understand something about $\alpha$ destribution. There are totally $$C^{51}_{159} = \frac{159!}{51!108!}$$ kinds of sums. A lot of them are equal, because $$\sum_{i=1}^{51}i = 1326\leq\alpha\leq\sum_{i=109}^{159}i=6834$$
Consequenlty, I want to know how many subsets of $51$ numbers have the sum equal to $N$, where  $1362\leq N\leq6834$. I'm stuck here because I don't know how to do it.

答:
Replace 51 and 159 with $n, M$ respectively. We have a vector $\mathbf{x}_{n\times 1}$ which follows a multivariate distribution, and $\alpha = \sum_{i=1}^n x_i$ where $x_i$ is the $i^{th}$ component of $\mathbf x$.

Then, by symmetry,
$E(\alpha)=E(\sum x_i)=\sum_i E(x_i) =nE(x_1)= \frac{n(M+1)}{2}$.

$$E(\alpha^2)=E\left(\sum_i x_i\right)^2 = E\left(\sum_i x_i^2\right)+E\left(\sum_{i\neq j} x_i x_j \right)$$

Again by symmetry
$$
E\left(\sum_i x_i^2\right)=nE(x_1^2)=\frac 16 n(M+1)(2M+1)
$$

$$
E\left(\sum_{i\neq j} x_i x_j \right)=(n^2-n)E(x_1 x_2)=\frac{n^2-n}{M^2-M}\sum_{i\ne j}ij = \frac{n^2-n}{M^2-M}\left(\left(\frac{M(M+1)}{2}\right)^2 - \frac{M(M+1)(2M+1)}{6}\right) \\= \frac{1}{12} (n^2-n)(M+1)(3M+2)
$$

Therefore $$\text{var } \alpha = E(\alpha^2) - (E(\alpha))^2 = \cdots = 73440$$


问:
Show that the decimal representation map is ergodic using Fourier series.
The decimal expansion $T:([0,1],\mathcal{B}_{[0,1]},\lambda)\longrightarrow ([0,1],\mathcal{B}_{[0,1]},\lambda)$ is defined to be $$T(x)=10x\ \ \text{mod}\ \ 1,$$ where $\lambda$ is Lebesgue measure.

I know how to prove that $T$ preserves $\lambda$. And I am now trying to prove that $T$ is ergodic. To avoid complicated set manipulation, I want to do this using Fourier series.

Suppose there exists $f\in L^{2}([0,1],\mathcal{B}_{[0,1]},\lambda)$ such that $f\circ T=f$ $\lambda-$almost surely, I want to show $f$ is constant $\lambda-$almost surely.

Note that if $f\circ T=f$ $\lambda-$a.s. then $f\circ T^{p}=f$ $\lambda-a.s.$ for all $p\in\mathbb{Z}_{>0}$. Now, consider the Fourier series of $f$: $$f\sim \sum_{n=-\infty}^{\infty}c_{n}e^{2\pi nx}.$$ Then, $f\circ T^{p}$ has the Fourier series $$f\circ T^{p}\sim \sum_{n=\infty}^{\infty}c_{n}e^{2\pi n 10^{p}x}.$$ This two Fourier series should be the same, so comparing coefficients we see $$c_{10^{p}n}=c_{n}.$$ Suppose $n\neq 0$, then $10^{p}n\longrightarrow\infty$, as $n\rightarrow\infty$. It then follows from Riemann-Lebesgue lemma that $$c_{n}=c_{10^{p}n}\longrightarrow0,\ \text{as}\ n\rightarrow\infty.$$

Then I got stuck. How could I conclude from here that $c_{n}=0$ for all $n\neq 0$? Thank you so much!

答:
Use the Plancherel identity: since $f \in L^2$, we must have
$$
\infty > \int_{0}^{1} |f(x)|^2 \,dx = \sum_{n} |c_n|^2
$$
But you've shown that $c_n = c_{10n} = c_{100n} = \dots$, so if any $c_n$ is nonzero, then the right hand side is infinite, contradiction.


问:
Tricky logarithmic problem?
It is given that $\log_9 p = \log_{12} q = \log_{16} (p+q) $. Find the value of $q/p$. I can see that the bases have common factors, but I don't exactly know how to exploit that. I tried many approaches to this, but I couldn't get it. The farthest I got was probably $q/p=4/\sqrt{p}$ (if I remember right)  Thanks!</p>

ContentLicense: CC BY-SA 3.0

Title: Tricky logarithmic problem?

Body:
Hint: We have that $9^{t}=p$ and  $12^{t}=q$ where $t=\log_9(p)=\log_{12}(q)$, so
$$\frac{q}{p}=\frac{12^t}{9^t}=\frac{4^t}{3^t}=\frac{(16^t)^{1/2}}{(9^t)^{1/2}}=\frac{\sqrt{p+q}}{\sqrt{p}}.$$


问:
Example of a basis which is not a Riesz basis?
I'm looking for an example of a countable "basis" $B=(\phi_i)_{i\in I}$ in a real Hilbert space $\mathcal{X}$ which is not a Riesz basis. So, we only require that the closure of the span of $B$ is $\mathcal{X}$ and that, for every $N\in\mathbb{N}$ and every cardinality-$N$ subset $\{i_1,\ldots,i_N\}$ of $I$,
\begin{equation}
(\forall (\alpha_{i_1},\ldots,\alpha_{i_N})\in\mathbb{R}^N)\quad
\alpha_{i_1}\phi_{i_1} + \alpha_{i_2}\phi_{i_2}+\cdots+\alpha_{i_N}\phi_{i_N}=0\Rightarrow (\alpha_{i_j})_{1\leq j\leq N}\equiv 0.
\end{equation}
(note that, by "span" I mean finite linear combinations.)

A few observations / questions

This must occur in infinite-dimensional $\mathcal{X}$.

I think this would mean that the basis decomposition operator $L\colon\mathcal{X}\to\ell_2\colon x\mapsto(\langle x\,|\,\phi_i\rangle)_{i\in I}$ is unbounded?

答: Let $\{v_1,v_2,\dots\}$ be a complete orthonormal basis for $\mathcal{X}$. Let $w_1=v_1$, $w_2=\frac{1}{\sqrt 2}(v_1+v_2)$ and more generally for $k=3,4,\dots$ let $w_k=\frac{1}{\sqrt k}(v_1+v_2+\cdots v_k)$. Note that $\|w_k\|=1$. Since $v_k=\sqrt k w_k-\sqrt{k-1} w_{k-1}$, the closure of the span of $\{w_1,w_2,\dots\}$ is $\mathcal{X}$. Moreover, since $L(v_k)= (0,0,\dots,-\sqrt{k-1},\sqrt{k},0,\dots)$, $\|L(v_k)\|= \sqrt{k^2+(k-1)^2}$ so that the mapping into $\ell_2$ is unbounded.


问:
Euler circuits for the graph
How many different Euler circuits are there in the graph that begins at $A$ and then visits $D$ as the immediate next vertex?

My answer is $16$, and I suppose I have counted all possible such Euler circuits. I wonder whether I have missed additionally possible paths, and would be welcomed to know that.

答:
You are correct.

Notice that such an Euler circuit must use $e_2$. So you can reduce the problem to finding the number of Euler trails starting at $C$ and ending at $A$ in $G - D$.

Think of an Euler trail in this case as a sequence of five edges $e_{i_1}e_{i_2}\cdots e_{i_5}$. Perhaps one of the easier ways to count is to single out an edge, say $e_3$, and count the number of valid sequences with $e_3$ in the first position, then the second, and so on. Since we are starting at $C$, you may notice that a sequence representing an Euler trail can only have $e_3$ in the first, third, and fifth position. You obtain

First: $4$ trails. Traverse $e_3$. There are $4$ ways to go from $A$ to $C$, back to $A$, that is two choices from $A$ to $B$, two choices from $B$ to $C$, and the way back is determined.

Third: $8$ trails. You can go $CBCABA$ of which there are four ways, or $CBACBA$, another four ways.

Fifth: $4$ trails, similar to the first case.

While this answer is long winded, at least in thinking this way you can be confident that you've exhausted all possibilities in your count. I believe in general it is hard ($NP$-hard?) to count the number of Euler circuits in a given undirected graph.

Title: Euler circuits for the graph


答:
Let all of them equal to $x$. We then have
$$p=9^x; q = 12^x; p+q = 16^x$$
Hence, we need solutions for
$$9^x + 12^x = 16^x \implies 1 + \left(\dfrac43\right)^x = \left(\dfrac{16}9\right)^x$$
Now let $\left(\dfrac43 \right)^x = y$. We then have
$$y^2 = y+1 \implies y = \dfrac{1+\sqrt5}2$$
Hence,
$$\dfrac{p}q = \dfrac{12^x}{9^x} = \left(\dfrac43\right)^x = y = \dfrac{1+\sqrt5}2$$</p>

Title: Tricky logarithmic problem?

Body: It is given that $\log_9 p = \log_{12} q = \log_{16} (p+q) $. Find the value of $q/p$. I can see that the bases have common factors, but I don't exactly know how to exploit that. I tried many approaches to this, but I couldn't get it. The farthest I got was probably $q/p=4/\sqrt{p}$ (if I remember right)  Thanks!</p>


问:
A question involving complex numbers and polynomials
Let $\iota$ be a root of the equation $x^2+1=0$ and let $\omega$ be a root of the equation $x^2+x+1=0$. Construct a polynomial

$$f(x)=a_0+a_1x+\dots+a_nx^n$$

where $a_0,a_1,\dots,a_n$ are all integers such that $f(\iota+\omega)=0$.

My approach was to substitute $(\iota + \omega)$ in place of $x$ and try some expansion but that resulted in nothing.
This question was asked in a 10 marks question of a very popular entrance examination at the 10+2 level.
A rigorous answer is appreciable.
Thank you.

答:
Hint:

suppose we have:

$p(x)=x^2-1=(x-1)(x+1)$; $x_p=1, -1$

$q(x)=x^2-5x+6=(x-3)(x-2)$; $x_q=2, 3$

And we want a polynomial its root are $x_p+x_q$, we must have:

$x_1=1+2=3$, $x_2=1+3=4$, $x_3=-1+2=1$ and $x_4=-1+3=2$

So our polynomial is:

$P(x)=(x-3)(x-4)(x-1)(x-2)$

Now consider:

$P(x)=(x-a)(x-b)(x-c)(x-d)=x^4-(a+b+c+d)x^3+(ab+cd)x^2+abcd$

Therefore to avoid confusion and mess it is better you first find the coefficients and then construct the polynomial.In your question:

$x_1=i+\omega$, $x_2=i+\overline{\omega}$, $x_3=-i+\omega$, $x_4=-i+\overline{\omega}$


问:
Is the open ball compact?
The open ball $B(f,2)$ in $(C[0,1],\mathbb R)$, $d_{\infty}$), where $f:[0,1] \to \mathbb R$ denotes the constant function $f(t)=5$ and $d_{\infty}(f,g)=max_{t\in[0,1]} |f(t)-g(t)|$

So I think the answer is that its not compact since the $B(f,2)$ is not closed and therefore according the the Heine-Borel theorem, the subset is not compact.

Is this right or have I messed up?

答:
The Heine-Borel theorem in general does not applies on subsets of functional spaces (such as $C([0,1])$) but is stated for $\mathbb R^n$. This is because in infinite dimensional spaces it can be that a closed and bounded set is not compact, and even that a compact set is not closed.

However, it is true that compact sets are closed in a Hausdorff space. And $C([0,1])$ is a Hausdorff space, since it is a metric space.

To see directly that the open ball $B(2)$ can't be compact, just take a sequence $\{B_n\}$ of balls, such that $B_n$ has radius $2-1/n$. Its union is the open ball $B(2)$, but you can't take any finite family of it which covers the open ball $B(2)$.

Title: Is the open ball compact?


问:
if $f(x)\geq0$ for all $x$, then $f+f'+f''+f'''+ \cdots + f^n \geq0$
Suppose $f$ is a polynomial of degree $n$, and $f\geq0$ for all $x$, prove that $f+f'+f''+f'''+\cdots +f^n\geq0$. (so just functions of the form $ax^n+bx^{n-1}+.....c$. We are not going to include irrational functions, exponential, logs, etc.)

So for this problem, I looked at a few functions and graphed them to see why this statement would be true. The first thing I found out is that the polynomial's highest degree must be even, otherwise there would be some $x$ where $f(x)$ is negative. Now this implies the odd numbered derivatives will have an odd number for the term with the highest degree.

Having figured that out, I was wondering whether the sum of every successive would change the function so that $f+f'$ has negative values, and $f+f'+f''$ would have positive values and so on, and this does turn out to be true for the ones I've tried. If this is true in general, that would mean that $f+f'$ has some negative values, $f+f'+f''$ has not negative values, and $f+f'+f''+f'''+....f^{n-1}$ has negative values until we get to $f+f'+f''+f'''+....f^{n-1}+f^n$ which has no negative values.

Ok but having figured this out, I'm not sure as to how to proceed with the proofs because the above assumptions I've made are not lemmas or proofs that I was given.

答: Since $g(x)$ has positive leading coefficient, $\exists$ $g(a) = \min g(x)$: $$g(x) \geqslant g(a)=g'(a)+f(a)=0+f(a) \geqslant 0$$


问:
The ideal $(0)$ is maximal in $\mathbb{Q}$ but not $\mathbb{Z}$?
I'm watching Richard Borcherds talk on Commutative algebra: https://www.youtube.com/watch?v=GTz9laU7c30&list=PL8yHsr3EFj53rSexSz7vsYt-3rpHPR3HB&index=11

And 8 minutes he says that The ideal $(0)$ is maximal in $\mathbb{Q}$ but not $\mathbb{Z}$.

So if $(0)$ is a maximal ideal then $\mathbb{Q} \backslash (0) \cong \mathbb{Q}^*$ is a field, right? How does this field relate to the field $\mathbb{Q}$.

Also, can somebody explain this claim to me without resorting to showing that the quotient rings are field (or not a field), and just show me directly $(0)$ is not a maximal ideal in $\mathbb{Z}$?

Thank you

答:
So if $(0)$ is a maximal ideal then $\mathbb{Q} \backslash (0) \cong \mathbb{Q}^*$ is a field, right? How does this field relate to the field $\mathbb{Q}$.

No, you would write $\mathbb{Q}/ (0) \cong \mathbb{Q}$ is a field. It looks like you might be confusing the quotient ring with the set difference. The quotient ring is the forward slash, and the backwards slash is typically used for set difference. In the context of group theory, I think perhaps sometimes the two slashes are used to indicate left/right cosets, but this isn't the case in ring theory.

Perhaps you should read this previous post with regards to gaining an understanding of the quotient argument.

just show me directly (0) is not a maximal ideal in ℤ?

Ok, pick an ideal of $\mathbb Z$ that isn't $\mathbb Z$ or $(0)$. Got it? You're done. That shows $(0)$ isn't maximal.


答:
$\mathbb{Q}\setminus\{0\}$ is not a field, it doesn't even have a zero element. Don't get confused, $I$ being a maximal ideal in $R$ means that the quotient ring $R/I$ is a field, not the set $R\setminus I$.

For example, if $I=\{0\}$ then there is a natural isomorphism $\pi:R\to R/(0)$ defined by $\pi(r)=r+(0)$. So this is a not very interesting quotient, this is just the ring $R$ itself. But in particular it means that $(0)$ is a maximal ideal if and only if $R$ is a field. So $(0)$ is indeed maximal in $\mathbb{Q}$.

On the other hand, it is not maximal in $\mathbb{Z}$. For example, $I=2\mathbb{Z}$ is a proper ideal which properly contains $(0)$.

Title: The ideal $(0)$ is maximal in $\mathbb{Q}$ but not $\mathbb{Z}$?

Body:
I'm watching Richard Borcherds talk on Commutative algebra: https://www.youtube.com/watch?v=GTz9laU7c30&list=PL8yHsr3EFj53rSexSz7vsYt-3rpHPR3HB&index=11

And 8 minutes he says that The ideal $(0)$ is maximal in $\mathbb{Q}$ but not $\mathbb{Z}$.

So if $(0)$ is a maximal ideal then $\mathbb{Q} \backslash (0) \cong \mathbb{Q}^*$ is a field, right? How does this field relate to the field $\mathbb{Q}$.

Also, can somebody explain this claim to me without resorting to showing that the quotient rings are field (or not a field), and just show me directly $(0)$ is not a maximal ideal in $\mathbb{Z}$?

Thank you


问:
is my answer about probability correct?
Probability that a shooter with a shot to hit the mark is 0,78 what is probability if from 5 shots the shooter hit the mark 3 times. My answer is this: Let B be the event that the mark is hit 3 times. Then :
P(B)=(0,78)(0,78)(0,78)(1-0,78)(1-0,78). I'm supposing that three first times is hitting the mark, and the two lasts not. Is this answer correct and it would be same idea if it asks prob.to hit the mark 5 times P(B)=(0,78)^5

答:
No it is not correct.

The 3 hits can be in any position in the 5 shots and not necessarily in the first 3 positions as you stated. Thus you have to multiply your probability by $\binom{5}{3}=10$ say all the possible combinations of the 3 hits in the 5 shots

Thus the correct answer is

$$P(X=3)=\binom{5}{3}\cdot 0.78^3 \cdot 0.22^2$$

$P(X=5)=0.78^5$ is right because this is the unique possible combination

Title: is my answer about probability correct?


问:
Is a ring automorphism restricted to the invertibles the identity automorphism?
In the solution of an exercise in my algebra book, the author says:

Given $\varphi \in \text{Aut}(\mathbb{Q}[x])$, we have that $\varphi(1)=1 \Rightarrow \varphi(a)=a$ for every $a \in \mathbb{Q}$

I really can't figure out why there is that implication, and this also made me think if is it possible to generalise the question the following way:

Given $A$ commutative ring with $1$ and $\varphi \in \text{Aut}(A)$, can we say that $\varphi_{|A^*}$ is the identity automorphism of $A^*$ (where $A^*$ is the group of the invertible elements of $A$)?

答:
As stated in the comments, the last statement is wrong, a counterexample being complex conjugation on $\mathbb C$.

As for the implication, observe that for any $n > 0$:

$$\varphi(n) = n \cdot \varphi(1) = n.$$

From the definition of a ring homomorphism, it is easy to deduce that $\varphi(0) = 0$. Hence

$$0 = \varphi(0) = \varphi(1+(-1)) = \varphi(1) + \varphi(-1)$$

and so $\varphi(-1) = -1$. It follows that $\varphi(-n) = \varphi(n)\varphi(-1) = -n$ for any $n \geq 0$. Hence $\varphi$ is the identity on $\mathbb Z$.

Let now $\frac{a}{b} \in \mathbb Q$, where $a,b \in \mathbb Z$, $b \neq 0$. Then
$$a = \varphi(a) = \varphi(b \cdot \frac{a}{b}) = b \cdot \varphi(\frac{a}{b})$$

and so finally $\varphi(\frac{a}{b}) = \frac{a}{b}$ for every rational number $\frac{a}{b}$.


问:
Are rotations of $(0,1)$ by $n \arccos(\frac{1}{3})$ dense in the unit circle?
Under which conditions will successive rotations of $(0, 1)$ by an angle $\theta$ guarantee that given $\delta > 0$ and some point $p$ on the unit circle, there exists some $n$ such that rotating $(0, 1)$ by $n\theta$ yields a point within $\delta$ radians of $p$?  That is, are there certain angles that will guarantee that the set generated by rotating $(0, 1)$ by integral multiples of $\theta$ is dense in the unit circle?

More specifically, I am working with the angle $\cos^{-1}\left(\frac{1}{3}\right)$ and wish to show that this property is true, but I'm not sure how to go about proving this.  It seems likely I'm missing something simple here...  Does it hold for any angle that is an irrational multiple of $\pi$?  If so, why?

答:
Yes, a necessary and sufficient condition is that the angle not be a rational multiple of $\pi$.

The problem is equivalent to considering the subgroup generated by $\theta$ in the additive group $\mathbb{R}/\{2k\pi\}$, viewed as the interval $[0,2\pi)$. Under the map $x\mapsto \frac{x}{2\pi}$, we can consider $\mathbb{R}/\mathbb{Z}$, and the subgroup generated by $r+\mathbb{Z}$. The assertion is that the subgroup is dense if and only if $r$ is irrational.

If $r$ is rational, then $\langle r+\mathbb{Z}\rangle$ is finite, so it is not dense.

That the subgroup is dense when $r$ is irrational is a consequence of the Equidistribution theorem. Added. Or more simply, a consequence of Dirichlet's approximation theorem (as noted by Gerry Myerson).

Here is Dirichlet's Theorem:

Theorem. (Dirichlet) For any real number $\alpha$ and positive integer $N$, there exist integers $p$ and $q$, $1\leq q\leq N$, such that
$$\left|q\alpha - p\right|\lt\frac{1}{N+1}.$$

To use it to derive the desired result, let $r$ be an irrational number in $[0,1]$ (corresponding to an angle which is not a rational multiple of $\pi$), and let $\beta\in[0,1]$ be any real number We want to show that for every $\epsilon\gt0$ there exists an integer $n$ such that $nr - \lfloor nr\rfloor$ is within $\epsilon$ of $\beta$ (that is, there is a multiple of the rotation which will land $(0,1)$ within $\epsilon$, along the circle, of $\beta$). Find an integer $M$ such that $\frac{1}{M+1}\lt\epsilon$. By Dirichlet's Theorem, you can find integers $p$ and $q$ such that $|q r - p|\lt \frac{1}{M+1} \lt \epsilon$. That means that if you take $qr$ modulo $1$, you lie within $\frac{1}{M+1}$ of either $0$ or $1$. Either way, by taking multiples of $qr$ you divide the circle into segments of length at most $\frac{1}{M+1}$, and by the Archimedean property some multiple of $qr\bmod 1$ will be strictly larger than $\beta$ (Note that $qr-p$ cannot be equal to $0$, because we are assuming that $r$ is irrational). The distance from $\beta$ to the smallest multiple which is strictly larger is less than $\frac{1}{M+1}$, which shows that there is some multiple of $r\bmod 1$ which is within $\epsilon$ if $\beta$, as desired.

(That is: use Dirichlet's Theorem to get a multiple of the rotation that is smaller in absolute value than $\frac{2\pi}{M+1}$, and note that since $\theta$ is not a rational multiple of $\pi$, this small angle cannot be $0$; then taking multiples of this rotation you get within $\frac{2\pi}{M+1}$ of any point in the circle; pick $M$ sufficiently large so that $\frac{2\pi}{M+1}\lt \epsilon$, and you are done).

So the answer to your question is "yes": the images of $(0,1)$ under successive rotations by an angle $\theta$ are dense in the unit circle if and only if $\theta$ is not a rational multiple of $\pi$.


问:
Existence Universal goto-programm (turing machine)
May you can help me out with my problems with source codes. Well first of all we proved that for recursive functions $N:\mathbb N^2\rightarrow \mathbb N$ and $A^k: \mathbb N^k\rightarrow \mathbb N$ and for every goto programm (I am not 100% sure, but I think this is identical to a turing machine) there exist a source code $\mathcal{P}=<p_0,...,p_{L-1}>$ where $L$ is the number of lines of the program and $p_i$ is the code of the i-th line.

Question: How to construct a simple universal program $U$? I.e a program which delivers the ouput for the input $(\mathcal{P},n)$, this is what the program $P$ would give for the input $n$

This is nothing more than a goto-interpreter, similar to a java-interpreter, but what about a formal argument, why this program exist?

EDIT: Next to my original question (the construction of $U$), two other questions arises:Let $f(n)$ is the output of the universal program $U$ on the input $(n,n)$. Is $f$ recursiv? Same with $g(n)=f(n)+1$. Is it recursiv?

答:
You have to be a bit careful with asking "is $f(n)$" recursive, where $f(n) = U(n,n)$ and $U$ is a universal GOTO program (or recursive function or turing machine or whatever).

$f$ isn't total, so if you distinguish between recursive and partially recursive functions, it's obviously not recursive. It's partially recursive though, since every GOTO program (or turing machine or recursive function or whatever) is.


问:
why $\frac{d^2y}{dx^2}=\frac{\frac{d}{dt}\left(\frac{dy}{dx}\right)}{\frac{dx}{dt}}$?
I have $x=\cos t$ and want to evaluate $\cfrac{d^2y}{dx^2}$. I already calculated $\cfrac{dy}{dx}$ to find second derivative the book I am reading from, used this:

$$\cfrac{d^2y}{dx^2}=\cfrac{\dfrac{d}{dt}\left(\dfrac{dy}{dx}\right)}{\dfrac{dx}{dt}}$$

My question is how we obtained this formula?

Body: $$\cfrac{d^2y}{dx^2}=\cfrac{d}{dx}\left(\cfrac{dy}{dx}\right)=\cfrac{\dfrac{d}{dt}\left(\dfrac{dy}{dx}\right)}{\dfrac{dx}{dt}}$$


问:
Is uniform convergence of a function sequence equivalent to uniform equicontinuity of the set of members of the sequence?
There is a plethora of questions on this topic on Math.stackexchnge.

Can the above question be answered for $C(\mathbb R ,\mathbb R)$ with an unambiguous "yes"?

答: No, but it can be answered with an unambiguous “no”. Take, for instance, $f_n(x)=x+n$, for each $n\in\Bbb N$.


问:
Suppose $f$ is a real-differentiable function on $[a,b]$ and suppose $f'(a)<c<f'(b)$. Prove then there is a point $x \in (a,b)$ such that $f'(x)=c$
This is what i have:

Put $g(t) = f(t) - ct$.

Then $g'(a)<0$ so that $g(t_{1}) < g(a)$ for some $t_{1} \in (a,b)$ and
$g'(b)>0$ so that $g(t_{2}) < g(b)$ for some $t_{2} \in (a,b)$.

Hence $g$ attains its minimum on $[a,b]$ at some point x such that $a<x<b$.

Then, $g'(x)=0$, hence $f'(x)=c$

答:
This is known as the Intermediate Value Theorem. You can find a proof worked out searching Google or Youtube.

Also check out Darboux's theorem, which states:

Let ${I}$ be an open interval, $f\colon I\to \mathbf{R}$ a real-valued differentiable function. Then $f'$ has the intermediate value property: If $a$ and $b$ are points in $I$ with $a\leq b$, then for every $y$ between $f'(a)$ and $f'(b)$, there exists an x in [a,b] such that $f'(x)=y$


问:
How do I solve this exercise of double integral $\iint_{D}\frac{1}{\sqrt{x^{2}+y^{2}}}dA$?
I have the double integral $\iint_{D}\frac{1}{\sqrt{x^{2}+y^{2}}}dA$ where $D$ is the domain given by:

$D = \left \{ (x, y ) \in \mathbb R^{2} : x^{2}+y^{2} \le 2y\right \} $

I tried this in polar coordinates: $$\left ( r cos \Theta \right )^{2} + \left ( r sin \Theta - 1 \right )^{2} = 1 $$

$$r = 2sin\Theta $$ $$\int_{-\Pi /4}^{\Pi /4}\int_{0}^{2sin\Theta } \frac{1}{r} rdrd\Theta$$

Am I the right way? How to resolve then?

答:
$$x^2+y^2\le 2y$$
$$\Rightarrow x^2+y^2-2y\le 0$$
$$x^2+(y-1)^2-1\le 0$$
$$x^2+(y-1)^2\le 1$$
so this is the circular region you are integrating over


问:
If $f$ is positive decreasing and continuous on $[0, \infty)$ I want to show that $\sum_{n=m+1}^{\infty} f(n) \leq \int_m^{\infty} f(t) \, dt$.
If $f$ is positive and decreasing and continuous on $[0, \infty)$ I want to show that

$$\sum_{n=m+1}^{\infty} f(n) \leq \int_m^{\infty} f(t) \, dt \leq \sum_{n=m}^{\infty} f(n)$$

for all nonnegative integers $m$.

Would we induct on $m$?

答:
Hint:

Induction is not necessary here. Show that for each $N>m+1$, $\:\sum_{n=m+1}^{N} f(n) $ is a lower Riemann sum and $\sum_{n=m}^{N} f(n)$ an upper Riemann sum of the decreasing function $f$.


答:
can somebody explain this claim without showing that the quotient rings are field

Hint: it's the special case ${\cal P} = 0\,$ of the following generalization of Bezout.

Lemma $\,{\cal P}\ \color{#c00}{\rm maximal}$ and $\,a\not\in {\cal P}\Rightarrow\, a\,$ is invertible $\!\bmod \cal P$

Proof $\,\ a\not\in{\cal P}\Rightarrow P\subsetneq {\cal P}+aR\color{#c00}{\Rightarrow} P+aR = (1)\Rightarrow \underbrace{p + ab = 1}_{\!\!\!\!\!\!\!\!\!\!\!\!\!\textstyle \bmod {\cal P}\!:\ ab\equiv 1}\,$ for $\,p\in {\cal P},\ b\in R.$

Remark $\ $ Note $\,r\equiv s\pmod{\! 0}\iff 0\mid r-s\iff r=s.\,$ Said structurally $\,R/0\cong R$. Therefore $\,ab\equiv 1\pmod{0}\iff ab=1\iff a\,$ is a unit (invertible).

As explained here, one reason we allow $0$ as a prime ideal is that we can often factor out prime ideals so reducing to the case ${\cal P}=0\,$ i.e. reducing to the nicer case of an integral domain.


问:
Evaluation of $\lim_{x\to 0^+}\frac{(1+x)^x - 1}{x^2}$
I'm interested in evaluating mentioned limit without using l'Hopital's rule, if possible.
What I'm thinking of is doing the following:

Let $$f(x) = \frac{(1+x)^x - 1}{x^2} = \frac{\left((1+x)^\frac{1}{x}\right)^{x^2} - 1}{x^2}$$

Now if we define:
$$g(x) = (1+x)^\frac{1}{x} \Rightarrow \lim_{x\to 0^+}g(x) = e > 0$$
and $$h(x) = x^2 \Rightarrow \lim_{x\to 0^+}h(x) = 0$$
then $f$ becomes a compound function $f(x) = \frac{g(x)^{h(x)} - 1}{h(x)}$

Now I'd like to argue that since $g(x) \xrightarrow{x\to 0} e > 0$, then $$\lim_{x\to 0^+}f(x) = \lim_{x\to 0^+}\frac{g(x)^{h(x)} - 1}{h(x)} \stackrel{?}{=} \lim_{x\to 0^+}\frac{\Big(\displaystyle\lim_{x\to 0^+}g(x)\Big)^{h(x)} - 1}{h(x)} = \lim_{x\to 0^+}\frac{e^{h(x)} - 1}{h(x)} = 1$$

However, the step with the question mark is troublesome for me and I struggled to prove it. Which got me thinking, is such statement even true or possible in this case? Or were such cases namely the very reason why L'Hopital's rule was discovered and is applicable?

Any clarification is greatly appreciated! Thanks a lot in advance!

Body:
Using
$$ \lim_{u\to0}\frac{e^u-1}{u}=1, \lim_{u\to0}\frac{\ln(1+u)}{u}=1 $$
one has
$$\lim_{x\to0}\frac{(1+x)^x-1}{x^2} = \lim_{x\to0}\frac{e^{x\ln (1+x)}-1}{x^2}=\lim_{x\to0} \frac{e^{x\ln (1+x)}-1}{x\ln (1+x)}\cdot\frac{x\ln (1+x)}{x^2}=\lim_{x\to0} \frac{e^{x\ln (1+x)}-1}{x\ln (1+x)}\cdot\frac{\ln (1+x)}{x}=1.$$


问:
Unbounded Integral
Given $\lim_{t \rightarrow \infty} f^\prime(t) = L$ and $c \leq L|b|,$ where $a,b$ are some constants, I want to show that
$$\lim_{t \rightarrow \infty}  \int_{a}^{a + t|b|} f^\prime(s)ds - ct = \infty$$

So I started with a lower bound,
$$\lim_{t \rightarrow \infty}  \int_{a}^{a + t|b|} (f^\prime(s) - L)ds \leq \lim_{t \rightarrow \infty}  \int_{a}^{a + t|b|} f^\prime(s)ds - ct$$

I also know that for all $\epsilon>0$, there exists $N > 0$ such that $|f^\prime(t)-L|< \epsilon$ for all $t>N$. I was hoping to show that $ 0< \epsilon_0 < |f^\prime(s)-L|$, so that I can show that the improper integral is unbounded. However, I am not sure how to proceed

答:
You should assume that $b \not= 0$ and $c < L|b|$.  Then

choose $\epsilon > 0$ small enough that $(L - \epsilon)|b| > c$, and

select $M > a$ with the property that $s \ge  M$ implies $f'(s) > L-\epsilon$.

If $a+t|b| > M$ then
$$\int_a^{a+t|b|} f'(s) \, ds = \int_a^M f'(s) \, ds + \int_M^{a + t|b|}f'(s) \, ds > f(M) - f(a) + (L-\epsilon)(a+t|b|-M)$$
which leads to
$$\int_a^{a+t|b|} f'(s) \, ds - ct > f(M) - f(a) + (L-\epsilon)(a-M) + t \left[(L - \epsilon)|b| - c\right].$$
The right-hand side of this inequality is unbounded above as $t \to \infty$.

Title: Unbounded Integral


答:
Let $n \le t \le n+1$, since $f$ is decreasing, we have
$$
f(n+1) \le f(t) \le f(n)
$$
so
$$
f(n+1) \le \int_n^{n+1} f(t) dt \le f(n)
$$
and
$$
\sum_{n=m+1}^{N+1} f(n) \le \int_m^{N+1} f(t) dt \le \sum_{n=m}^N f(n)
$$
thus by letting $N \to +\infty$
$$
\sum_{n=m+1}^{\infty} f(n) \leq \int_m^{\infty} f(t) \, dt \leq \sum_{n=m}^{\infty} f(n).
$$

Title: If $f$ is positive decreasing and continuous on $[0, \infty)$ I want to show that $\sum_{n=m+1}^{\infty} f(n) \leq \int_m^{\infty} f(t) \, dt$.

Body:
If $f$ is positive and decreasing and continuous on $[0, \infty)$ I want to show that

$$\sum_{n=m+1}^{\infty} f(n) \leq \int_m^{\infty} f(t) \, dt \leq \sum_{n=m}^{\infty} f(n)$$

for all nonnegative integers $m$.

Would we induct on $m$?


答:
If we have $ \dfrac{u}{v} = c$ constant temporarily, then by chain rule we also have  $ \dfrac{u'}{v'} = c$ when differentiated w.r.t. $t$

$$\cfrac{d^2y}{dx^2}=\dfrac{d[dy/dx]}{dx}$$

From independent variable $x$ now we newly consider switching to  $t$ so that

$$\cfrac{d^2y}{dx^2}=\cfrac{\dfrac{d}{dt}\left(\dfrac{dy}{dx}\right)}{\dfrac{dx}{dt}}.$$

Title: why $\frac{d^2y}{dx^2}=\frac{\frac{d}{dt}\left(\frac{dy}{dx}\right)}{\frac{dx}{dt}}$?

Body:
I have $x=\cos t$ and want to evaluate $\cfrac{d^2y}{dx^2}$. I already calculated $\cfrac{dy}{dx}$ to find second derivative the book I am reading from, used this:

$$\cfrac{d^2y}{dx^2}=\cfrac{\dfrac{d}{dt}\left(\dfrac{dy}{dx}\right)}{\dfrac{dx}{dt}}$$

My question is how we obtained this formula?


答:
$\mathbf Q^*$ is not a field, not even a ring, since it has not neutral element for addition. $\{0\}$ is a maximal ideal in $\mathbf Q$ just because it is the only proper ideal in $\mathbf Q$ (the same is true for all fields).

In $\mathbf Z$, which is  an integral domain, not a field, $\{0\}$ is a prime ideal, as in all integral domains, which is contained in all ideals, in particular in the maximal ideals of $\mathbf Z$, which are the ideals generated by prime elements.


问:
Inversion problem for four points

If $A$, $B$, $C$, $D$ are four non-coplanar points, then show, by inversion with respect to any one of those points, that the sum of any two of the products $BC\cdot AD$, $CA\cdot BD$, $AB\cdot CD$ is greater than the third.

I have shown that if $P'$ and $Q'$ are inverses of $P$ and $Q$ with respect to an origin of inversion $O$, then $\dfrac{P'Q}{PQ} = \dfrac{OP'}{OQ}$ or $\dfrac{OQ'}{OP}$, but cannot apply this to the proof.

Any help would be appreciated.

答: Take any point,say D, and invert with radius so that A,B,C are inside the sphere. Then,say, B'C'= r^2 BC/DB.DC.The inverse points are not collinear, so, say, A'B' + B'C' is greater than A'C'. Use the foregoing expression for each term,and an easy simplification gives the required proof.</p>

Title: Inversion problem for four points


问:
Are the continuous functions on $G$ dense in $L^{1}(G)$?
If $G$ is a locally compact group, is the set $C_{c}(G)$ of all continuous functions on $G$ with compact support dense in $L^{1}(G)$?</p>

答:
Yes. A more general result is the following:

Proposition. Let $X$ be a locally compact Hausdorff space, and let $\mu$ be a Radon measure on $X$. Then for any $p \in (0, \infty)$, $C_c(X)$ is a dense subset of $L^p(X, \mu)$.

See Proposition 3 in Terry Tao's blog post for a proof (this post also appears as a chapter in Tao's book An Epsilon of Room). Another reference is Proposition 7.9 in Folland's Real Analysis.


问:
Cardinality of Hamel basis of a separable incomplete inner product space
I know that cardinality of Hamel basis of an infinite dimensional separable Hilbert space is always equal to the cardinality of the continuum. But if an infinite dimensional separable inner product space is incomplete then is it true that the cardinality of Hamel basis may or may not be countable. Do we have examples for both the situation and how to determine the cardinality of Hamel basis under such conditions.</p>

答:
Let $c_{00}$ be the space of all sequences $(a_n)_{n\in\Bbb N}$ of real numbers such that $a_n=0$ if $n\gg1$ and consider the inner product$$\bigl\langle(a_n)_{n\in\Bbb N},(b_n)_{n\in\Bbb N}\bigr\rangle=\sum_{n=1}^\infty a_nb_n.$$If, for each $n\in\Bbb N$, $e(n)$ is the sequence $0,0,\ldots,0,1,0,0,\ldots$, whith the $1$ as the $n$th element, then $\{e(n)\mid n\in\Bbb N\}$ is a Hamel basis of $c_{00}$.

And if $H$ is a non-closed hyperplane of $\ell^2$, then $H$ is a non-complete Hilbert space and any Hamel basis of $H$ will have the same cardinal as a Hamel basis of $\ell^2$, which is the cardinality of the continuum.


问:
How do I solve this equation analytically?
I have attempted to solve the equation many times analytically but unfortunately I got to a point where I cannot further proceed. The equation:

$$T(t)e^{0.269} =\int6.187e^{0.269}dt + \int-0.807 \sin\left(\frac \pi{12}t\right)e^{0.269}dt$$

The point I got to:
$$(0.296)T(t) =(23)(0.296t) + \ln\left(0.807e^{0.296}+\cos\left(\frac \pi{12}t\right)-\frac \pi{12}\right)+ \ln\left(C-\frac \pi{12}\right)$$

Thanks a lot for your help!!!

答:
We can divide by $e^{0.269}$, and we'd get

$$ T(t) = \int 6.187 \mathrm{d}t + \int -0.807 \sin \Biggl ( \dfrac{\pi}{12} t \Biggr ) \mathrm{d}t $$

Integrating this is pretty simple,

$$ T(t) = 6.1867t + 0.807 \cos \Biggl (\dfrac{\pi}{12} t \Biggr ) \cdot \dfrac{12}{\pi} + C $$
I'm not sure where the $\ln$ came from, since the integral of $6.187$ and $\sin(\ldots)$ have nothing to do with logarithms.


Body: $$\cfrac{\cfrac{d}{dt}\left(\cfrac{dy}{dx}\right)}{dx/dt}=\cfrac{dt}{dx}\left(\cfrac{d}{dt}\right)\left(\cfrac{dy}{dx}\right)=\cfrac{d}{dx}\left(\cfrac{dy}{dx}\right)=\cfrac{d^2y}{dx^2}$$


问:
Boundedness in probability of Stochastic Process
This question is related to some other similar ones I did in the recent past.

Let $X_t$ be a Stochastic Process defined through the equation $$\text{d}X_t=f(X_t,t)\text{d}t+\text{d}W_t,$$
where $f$ is a twice differentiable function such that $f(x)<a<0$ for all $x>k$, and $X_0<k$.

What I am wishing to show is:

For all $\epsilon>0$ there is $M>0$ (which we can WLOG assume to be bigger than $k$) such that for all $t\ge0$ it holds that $$\mathbb{P}[X_t>M]\le\epsilon.$$

What I tried to do is define $\tau_t$ as $\sup\{0<s<t: X_s=k\}$, i.e. the last hitting time of the level $k$ before time $t$, where the sup is set equal to $0$ if the set $\{0<s<t: X_s=k\}$ is empty. This should be a stopping time, even if last hitting times in general are not.

Then, we can manipulate the given probability in this way:
\begin{align*}\mathbb{P}[X_t>M]&=\underbrace{\mathbb{P}[X_t>M,\tau_t=0]}_{=0}+\underbrace{\mathbb{P}[X_t>M,\tau_t\ne0, X_t<k]}_{=0}+\mathbb{P}[X_t>M,\tau_t\ne0, X_t>k]\\
&\le\mathbb{P}[X_t-X_{\tau_t}>M-k,\tau_t\ne0]\\
&\le \mathbb{P}[a(t-\tau_t)+W_t-W_{\tau_t}>M-k, \tau_t\ne0]\\
&\le \mathbb{P}[a(t-\tau_t)+W_t-W_{\tau_t}>M-k].\end{align*}

Since $a<0$ we can bound this probability uniformly over $t$, and the claim should be proven. However, I think I've been too slick with some steps, and I want to make sure they are correct (in particular, the $\tau_t$ seems to be not a stopping time). Can someone help me find any potential errors? Also, is there a more straightforward way to prove this? Maybe I just didn't see an easier solution.

EDIT: the $f$ in my specific problem satisfies the hypothesis $|f(x,t)-k_0|<\theta\cdot |x|+ \mu,$ for some constants $k_0, \theta$ and $\mu$: is there some kind of comparison principle with the absolute value bound? I can't find it online, but if it exists, then I would be done, as the Ornstein-Uhlenbeck process is bounded in the sense I am looking for.

答:
Small discussion before presenting the main result

Clearly, from the above work, we can have the desired conclusion if we can prove the following estimation:
$$\limsup \mathbb{E}(t- \sigma_{N(t)-1}) <+\infty$$

Also, our question is now converted from a question on the behaviour of $X_t$ (for a specific $t$) to some question on the behaviour related to sequences of stopping times.

IMHO, the main advantages of this conversion are three things in follows: 1) the nice form of the expression we have to control now 2) If things go fine, we can provide an $O(1/M^2)$ upper bound for our $\sup \mathbb{E}(X_t>M)$ 3) The Markov property.

After all the lengthy work above, I think it is nice to have an affirmative answer in some cases.

Here I'll show those "some cases" and this also the central fruit of my work.

An uniform bound on $\mathbb{E} (X_t)_+^2 $

1.Main result

In this section, I'll take the advantage of some probabilistic (also analytic) tools in their purest form by assuming the following.

Condition C2 $f$ is time-homogeneous, i.e $f(t,x)=f(x) \square$

Here is the main result:

Theorem 8

Under the assumption A1 and the conditions C1 and C2, there is a constant $C$, which does not depend on $t$, such that:

$$ \mathbb{E}( (X_t-m_2)_+^2) < A \quad \forall t$$
$\square$

Remark 9: Yes, it is $m_2$.

2.Premilinaries

Theorem 9 (A different bound for $(X_t)_+$ )
Under the condition, $C_1$ and the assumption A1, we have:
$$ \mathbb{E}( X_t-m_2)_+^2 \le \mathbb{E}( t\vee \tau_{N(t)-1}-\tau_{N(t)-1})$$
where $N(t)$ stays the same as above.
$\square$

Remark 10: I have to replace the upper bound in theorem 6 by this bound because it is not straightforward to obtain a bound for $t-\sigma_{N(t)-1}$. The underlying reason is the fact that we have little no information about $f(x)$ when $x<k$. Thankfully for all $N$, $X$ is bigger than $m_1$(hence bigger than $k$) on $[\tau_N,\sigma_{N+1}]$

Demonstration:(summarized) Just add more technical manipulation in the proof of theorem 6.

Let:

$A(t) := \mathbb{E}( t\vee \tau_{N(t)-1}-\tau_{N(t)-1})$

$\gamma_1 \sim \tau_1-\sigma_1$ i.e the random time that $X$ has to spend to reach $m_2$ if it starts from $m_1$ ( under the assumption of homogeneity of $f$)

$\gamma_2 \sim \sigma_{2}-\tau_1$ i.e the random time that $X$ has to spend to reach $m_1$ if it starts from $m_2$ ( under the assumption of homogeneity of $f$)

$F$ and $G$ are the distribution of $\gamma_1$ and $\gamma_2$ respectively.

Remark 11 $F*G(t) = P(\gamma_1+\gamma_2 \le t)$

Theorem 12 Under condition C1, C2 and the assumption A1, let $X_0=m_1$ (i.e $\sigma_1=0$), we have:
$$A(t)=\int_{0}^t A(t-s)dF*G(s)+\underbrace{\int_{t}^{\infty} B(t,s)dF*G(s)}_{=:h(t)} $$
where $B(t,s)= \mathbb{E}( t\vee \gamma_1-\gamma_1 | \gamma_1+\gamma_2=s)$

Demonstration(Omitted)

Message I get a little bit tired (of all technical details), please feel free to ask me to fill in details at any point in my work that is obscure in your opinion. By all means, I can always be wrong and making blunders.

Remark 13 In short, $A$ satisfies the renewal equation :
$$ A=h+ A*K$$
where $K=F*G$

Remark 14: To be honest, $h(t)= \mathbb{E}( t\vee \gamma_1-\gamma_1 ,\gamma_1+\gamma_2 \ge t)$

Theorem 15 (Regularity $\gamma_2$)
$\mathbb{E}(\gamma_1^2)<+\infty$
Demonstration (Summarized) As noted, as we know, during $[0,\gamma_2]$, $X_t \ge m_1> k$, thus our SDE behaves nicely as we have control over its drift. What's left is some technical argument.
$\square$

3.Demonstration

Clearly, $$h(t) = \mathbb{E}( t\vee \gamma_1-\gamma_1 ,\gamma_1+\gamma_2 \ge t) \le  \underbrace{\mathbb{E}( \gamma_2, \gamma_1+\gamma_2 \ge t )}_{=:h^*(t)} \le \mathbb{E}(\gamma_2)  $$
Hence $h \in L^{\infty}$, which implies that (by renewal theory):

$$H(t) = \int_{0}^t h(t-s)dU(s)$$
where $U(s) = \mathbb{E}(N_s)$

Thus,
$$H(t) \le \int_{0}^t h^*(t-s)dU(s)$$
Besides, $h^*$ is clearly decreasing and $\int_{0}^{\infty} h(t)dt<+\infty$,
Thus by the key renewal theorem, we have:
$$\lim \int_{0}^t h^*(t-s)dU(s) = \frac{1}{\mathbb{E}(\gamma_1+\gamma_2)} \int_{0}^{\infty} h(t)dt<+\infty, $$
Thus our conclusion.
$\square$

Remark 16 $\mathbb{E}(\gamma_1)$ can be infinite and to be honest, most of the work in this section to get around that uncertainty.

Remark 16.1 The solution above is only applicable when $\mathbb{E}(\gamma_1) <+\infty$, if $\mathbb{E}(\gamma_1) =+\infty$ some technical manipulations are required

Remark 17 We have assumed $X_0= m_1$ while in fact this condition can be relaxed easily, however, I decided to omit that part.


问:
Tricky logarithmic problem?
It is given that $\log_9 p = \log_{12} q = \log_{16} (p+q) $. Find the value of $q/p$. I can see that the bases have common factors, but I don't exactly know how to exploit that. I tried many approaches to this, but I couldn't get it. The farthest I got was probably $q/p=4/\sqrt{p}$ (if I remember right)  Thanks!</p>

答:
For some $a$, we have $9^a=p$, $12^a=q$, and $16^a=p+q$. Note that $12^2=9\cdot 16$. It follows that $q^2=p(p+q)$. </p>

<p>Divide through by $p^2$, and let $x=\frac{q}{p}$. We find that $x^2-x-1=0$. </p>

Title: Tricky logarithmic problem?

Body:
For some $a$, we have $9^a=p$, $12^a=q$, and $16^a=p+q$. Note that $12^2=9\cdot 16$. It follows that $q^2=p(p+q)$.

Divide through by $p^2$, and let $x=\frac{q}{p}$. We find that $x^2-x-1=0$.


问:
Prove ${M}\choose{N_1}$ ${M}\choose{N_2}$ $\leq$ ${2M}\choose{N_1+N_2}$ for $M, N_1, N_2 \in \mathbb{N},\,M\geq N_1,N_2$.
I need to prove the inequality ${M}\choose{N_1}$ ${M}\choose{N_2}$ $\leq$ ${2M}\choose{N_1+N_2}$ for $M, N_1, N_2 \in \mathbb{N},\,M\geq N_1,N_2$.

I wanted to prove it by induction. But since we are given 3 variables running, I was confused if the natural induction, as I know it, can be used in this case. Can you tell me if proving by induction is possible and necessary in this example ?

We get $N_1, N_2, N_1+N_2$ fractions from ${M}\choose{N_1}$,${M}\choose{N_2}$, and ${2M}\choose{M_1+M_2}$, respectively. Without loss of generality, we assume $N_1\geq N_2.\,$An appropriate approximation of all fractions from left to right leads to: $$ \frac{M}{N_1}\frac{M-1}{N_1 -1}\frac{M-2}{N_1-2}\cdots \frac{M-(N_1-1)}{1} \cdot\frac{M}{N_2}\frac{M-1}{N_2 -1}\frac{M-2}{N_2-2}\cdots \frac{M-(N_2-1)}{1}\leq \frac{2M}{N_1+N_2}\cdots $$
For the second fraction from the left I get: $\frac{M-1}{N_1 -1}=\frac{2M-2}{2N_1-2}\geq\frac{2M-1}{2N_1-1}\leq\frac{2M-1}{N_1+N_2-1}.$ The last term in the last inequality is not the convenient approximation for $\frac{M-1}{N_1 -1}$ unless we got $\leq$ instead of $\geq$ sign in the last inequality.

Can somebody provide some hint how to go further ?
Thanks.

答:
There is simple combinatorial argument.

Consider two disjoint partitions of  set $A=\{ 1,2,3,\ldots, 2M\}$ : $B=\{ 1,2,3,\ldots, M\}$ and $C=\{ M+1,M+2,M+3,\ldots, 2M\}$

LHS denotes choosing $N_1$ members from $B$ and $N_2$ members from $C$.

RHS denotes choosing $N_1 + N_2$ members from $A$ where there are more possibilities. The $N_1$ choices can be made from both halves (same for $N_2$), for which there are clearly more number of ways. In fact, RHS includes LHS case. The inequality follows.

Also equality occurs for $N_1=N_2=M$, when number of choices (each binomial coefficient) is $1$.


问:
Perimeter change when moving one point of a triangle along a circle
I am interested in the perimeter $P(\Delta')$ of a modified version of a triangle $\Delta = (A,B,C)$. I chose point $B$ to be the point I want to shift towards the inside (or on the boundary) of $\Delta$. I chose a distance $r<h$ where $h$ is the height of the triangle. $B$ can be moved to an arbitrary point $E$ which is of distance $r$ to $B$ and lays inside or on the boundary of $\Delta$. This means that $E$ must also lay on the circle of radius $r$ with the center point $B$. I denote $\Delta'=(A,E,C)$. What is the perimeter $P(\Delta')$ of the new triangle given the original side lengths, $r$ (and probably some representation of the arc length to E)? In what range does $P(\Delta)-P(\Delta')$ lie in?

I attached a small figure which visualizes both triangles.

Edit

Based on the comments, I added some notation:

Assuming $B=(0,0)$, then we have $E=(x,y)$ with $x=\lambda_A x_A + \lambda_C x_C$ and $y=\lambda_A y_A + \lambda_C y_C$ ($\lambda_A+\lambda_C = 1$).

This gives us the following set of equations:

$$r=\sqrt{(0-x)^2 + (0-y)^2} = \sqrt{(\lambda_A x_A + \lambda_C x_C)^2 + (\lambda_A y_A + \lambda_C y_C)^2}$$
$$|AE| = \sqrt{(x-x_A)^2+(y-y_A)^2}$$
$$|CE| = \sqrt{(x-x_C)^2+(y-y_C)^2}$$

Moreover, we have:
$$|AB| = \sqrt{x_A^2 + y_A^2}$$
$$|CB| = \sqrt{x_C^2 + y_C^2}$$

The perimeter change is in fact the difference $|AB|+ |CB| - |AE| - |CE|$. I would like to show that $r \leq |AB|+ |CB| - |AE| - |CE|$. I struggle with the proof because I still have six variables $x_A,x_C,y_A,y_C,\lambda_A$ and $\lambda_C$.

Background

This question arises from a computer science problem called lawn mowing/milling, where you are given a cutter (in my case a circular shape) and a Polygon to cover. I noticed that given a tour of the circular cutter one can change the cutter from a circular shape with a radius $s$ to a square with side length $2s$. Afterward, the tour can be shortened a bit. I have already proven a specific length $r$ that I can move the points of the tour "inwards" to shorten the tour. My goal is to find out how much shorter the final tour gets when I modify it. In my example, $A-B-C$ are part of the tour and $B$ is a point which can be moved inward by $r$ to a point $E$. My tour will later be $A-E-C$. I want to find out how much the tour is shortened after my transformation and I hope that it will be shortened by at least $r$.

答:
(Expanding a comment.)

I would like to show that
$$r \leq |AB|+|CB|−|AE|−|CE| \tag1$$

Without additional conditions, $(1)$ may not be true. Writing the inequality as
$$|AE|+|CE|\leq |AB|+|CB|-r \tag{1'}$$ leads us to consider the locus of points $P$ satisfying the corresponding equality:
$$|AP|+|CP|=|AB|+|CB|-r \tag2$$
This is an ellipse with foci $A$ and $C$, with major radius $\frac12(|AB|+|CB|-r)$. This ellipse separates the points with a larger sum from those with a smaller sum. The described points $E$, then, must lie on the arc where $\bigcirc B$, with radius $r$, overlaps this ellipse; depending upon $r$, this arc may-or-may-not include all (or even any!) points of the arc overlapping the triangle itself.

For instance, this first figure shows a situation with points $E_-$, $E_0$, $E_+$ on $\bigcirc B$ and inside $\triangle ABC$, but respectively inside, on, and outside the ellipse:

That is, we have
$$\begin{align}
|AE_-|+|CE_-| < |AB|+|BC|-r \quad\to\quad r < |AB|+|BC|-|AE_-|-|CE_-|\\
|AE_0\;|+|CE_0\,| = |AB|+|BC|-r \quad\to\quad r = |AB|+|BC|-|AE_0\,|-|CE_0\;|\\
|AE_+|+|CE_+| > |AB|+|BC|-r \quad\to\quad r > |AB|+|BC|-|AE_+|-|CE_+|
\end{align} \tag3$$
so that only $E_-$ and $E_0$ satisfy $(1)$.

With appropriate conditions, we can have a circumstance where the circle-triangle overlap is contained within the ellipse (which seems to be the case OP envisions):

On the other hand, we can have a case where the circle doesn't meet the ellipse at all, so all points on the circle are in the $E_+$ family that don't satisfy $(1)$:

For a given triangle, the threshold $r$-values that keep the circular arc within the ellipse can be determined by intersecting the circle with the ellipse. Unfortunately, conic-conic intersections like this lead to quartic equations, which are exceedingly messy to solve in general.

Unless/until OP provides additional constraints or context, this is about as far as this investigation will go.


问:
Derivative of arcsin, question on provided proof
The following is proof of the derivative of $\arcsin$.  Does anyone know why and with what axiom the $y'$ is introduced into the equation in the following step?

$\sin y = x$

$y'\cos y = 1$

答:
The step involves applying the chain rule:
\begin{align}
\frac{d \sin (y)}{dx} &= \frac{d \sin (y)}{dy} \cdot \frac{dy}{dx} \\
&= \cos(y) \cdot y' \, .
\end{align}
Applying the chain rule in this way, where you have an expression in terms of $y$ and want to differentiate with respect to $x$, is known as implicit differentiation. In general,
$$
\frac{df(y)}{dx} = \frac{df(y)}{dy} \cdot \frac{dy}{dx} \, .
$$
In practice, implicit differentiation is easy because you just differentiate as you would normally, and then multiply by $dy/dx$. Examples:
\begin{align}
\frac{d(y^2)}{dx}&=2y\frac{dy}{dx} \\[4pt]
\frac{d(\log(y))}{dx} &= \frac{1}{y}\frac{dy}{dx} \\[4pt]
\frac{d\left((2y^3+5)^2\right)}{dx} &= 12y^2(2y^3+5)\frac{dy}{dx} \, .
\end{align}
Let me know if you have any questions.


答:
We prove by induction on $M$. Base case is easy. If it's true for $M=k$, we now prove it's true for $M=k+1$.

First note that if $N_1=0, N_1=k+1, N_2=0$ or $N_2=k+1$ it's trivial.

If not, then $1 \le N_1, N_2 \le k$.

$$
\binom{k+1}{N_1} \binom{k+1}{N_2}=\left(\binom{k}{N_1}+\binom{k}{N_1-1}\right) \left(\binom{k}{N_2}+\binom{k}{N_2-1}\right) $$

Can you start from here?

 $$=\binom{k}{N_1}\binom{k}{N_2}+\binom{k}{N_1}\binom{k}{N_2-1}+\binom{k}{N_2}\binom{k}{N_1-1}+\binom{k}{N_1-1}\binom{k}{N_2-1}\\ \le \binom{2k}{N_1+N_2}+\binom{2k}{N_1+N_2-1}+\binom{2k}{N_1+N_2-1}+\binom{2k}{N_1+N_2-2}\\ =\binom{2k+1}{N_1+N_2} + \binom{2k+1}{N_1+N_2-1} = \binom{2k+2}{N_1+N_2}$$

Title: Prove ${M}\choose{N_1}$ ${M}\choose{N_2}$ $\leq$ ${2M}\choose{N_1+N_2}$ for $M, N_1, N_2 \in \mathbb{N},\,M\geq N_1,N_2$.

Body:
I need to prove the inequality ${M}\choose{N_1}$ ${M}\choose{N_2}$ $\leq$ ${2M}\choose{N_1+N_2}$ for $M, N_1, N_2 \in \mathbb{N},\,M\geq N_1,N_2$.

I wanted to prove it by induction. But since we are given 3 variables running, I was confused if the natural induction, as I know it, can be used in this case. Can you tell me if proving by induction is possible and necessary in this example ?

We get $N_1, N_2, N_1+N_2$ fractions from ${M}\choose{N_1}$,${M}\choose{N_2}$, and ${2M}\choose{M_1+M_2}$, respectively. Without loss of generality, we assume $N_1\geq N_2.\,$An appropriate approximation of all fractions from left to right leads to: $$ \frac{M}{N_1}\frac{M-1}{N_1 -1}\frac{M-2}{N_1-2}\cdots \frac{M-(N_1-1)}{1} \cdot\frac{M}{N_2}\frac{M-1}{N_2 -1}\frac{M-2}{N_2-2}\cdots \frac{M-(N_2-1)}{1}\leq \frac{2M}{N_1+N_2}\cdots $$
For the second fraction from the left I get: $\frac{M-1}{N_1 -1}=\frac{2M-2}{2N_1-2}\geq\frac{2M-1}{2N_1-1}\leq\frac{2M-1}{N_1+N_2-1}.$ The last term in the last inequality is not the convenient approximation for $\frac{M-1}{N_1 -1}$ unless we got $\leq$ instead of $\geq$ sign in the last inequality.

Can somebody provide some hint how to go further ?
Thanks.


问:
What does $X \not \models \beta$ really mean?
I have a question.

Let X be some set. Let $v$ be valuation.

I saw definition which says that $X \models \beta$  means that for any valuation $v$: if $v(X)\subseteq\{1\}$, then $v(\beta)=1$.

Question 1. Does $()\subseteq \{1\}$ mean at the same time
that "all premisses in X are true or X is empty set"?

Question 2. If so, what does really $X\not\models\beta$ says?

Option (1): "there exists $v$ such that $v(X)\subseteq \{1\}$ and $v(\beta)=0$". If so, does it mean that at the same time: "all premisses are true and $\beta$ is true or $X=\emptyset$ and $\beta$ is true"?

or

Option (2): "there exists $v$ such that $v(X)=1$ and $v(\beta)=0$". If so, should I read it  "all premisses are true and $\beta$ is false?

Which option is proper? Or maybe both options are wrong? The question basically is: what does $X\not \models \beta$ mean and how to write it by using valuation?

Please, help me understand.

答:
$X$ is a set of formulas. Thus, to say that $v(X) \subseteq \{ 1 \}$ means that all formulas in $X$ a satisfied (evaluated to $1$, i.e. True) by $v$.

More customarily: $v \vDash X$.

When $\beta$ is a formula, to say that $X \nvDash \beta$ means that $\beta$ is not logical consequence of the set $X$ of premises, i.e. that there is some valuation $v$ such that $v \vDash X$ and $v \nvDash \beta$, i.e. such that $v$ satisfies all formulas in $X$ and falsifies $\beta$.

Thus, option (2) is the correct one.

When $X$ is the empty set. we have that $\emptyset \vDash \beta$ means that $\beta$ is a tautology, i.e. a formula that is true in every interpretation. We usually write it: $\vDash \beta$.

Thus, to write $\nvDash \beta$ means: it is not true that $\vDash \beta$, i.e. it means that $\beta$ is not a tautology.


问:
Infinite sum equation
I have some doubts about the following question:

$$Find\ the\ value\ a\ for\ which\ \sum _{n=0}^{\infty } {e^{na}}= 2$$

What I tried to do to solve it:

I've noted that I can rewrite this as the following:

$$1 + \sum _{n=1}^{\infty } {e^{na}}= 2$$

Hence, I can derive that:

$${e^{a}} + {e^{2a}} + {e^{3a}} + .... + {e^{na}} = 1$$

But from here I don't really know what to do. I know this isn't a hard problem, maybe I need to study something I don't understand yet. Could you point out a way to solve this problem? I just need a direction / general tip of where to go.

Title: Infinite sum equation

Body:
Let $b=e^a$.

$$1+b+b^2+b^3+\cdots=\frac1{1-b}=2.$$

Hence

$$b=\frac12,\\a=\log\frac12.$$


Title: Infinite sum equation

Body:
I have some doubts about the following question:

$$Find\ the\ value\ a\ for\ which\ \sum _{n=0}^{\infty } {e^{na}}= 2$$

What I tried to do to solve it:

I've noted that I can rewrite this as the following:

$$1 + \sum _{n=1}^{\infty } {e^{na}}= 2$$

Hence, I can derive that:

$${e^{a}} + {e^{2a}} + {e^{3a}} + .... + {e^{na}} = 1$$

But from here I don't really know what to do. I know this isn't a hard problem, maybe I need to study something I don't understand yet. Could you point out a way to solve this problem? I just need a direction / general tip of where to go.

Body:
$$1+{e^{a}} + {e^{2a}} + {e^{3a}} + .... + {e^{na}} +\cdots= 2\\\times e^{a}\\
{e^{a}} + {e^{2a}} + {e^{3a}} + .... + {e^{na}}+e^{(n+1)a} +\cdots= 2e^a\\\underbrace{{e^{a}} + {e^{2a}} + {e^{3a}} + .... + {e^{na}}+e^{(n+1)a}+\cdots}_{1}=2e^a\\1=2e^a\\$$can you take over from here ?


问:
Show that $(D^\bot)^\bot=D$
Let $D$ be a non-empty subset of Hilbert space $H$. I need to find $(D^\bot)^\bot$. I guess $(D^\bot)^\bot=D$. We have that $D \subseteq (D^\bot)^\bot$ and that $D^\bot$ is a closed linear subspace of . But for the other inclusion I have problems. Many thanks for your help!

Edit: $D$ is not assumed to be closed!

答:
Denote by $F$ the closed linear span, $\overline{\text{span}(D)}$ of $D$. We claim that $(D^\perp)^\perp=F$. Write $H$ as $F\oplus F^\perp$. Then, $D\subseteq (D^\perp)^\perp$ implies $F\subseteq (D^\perp)^\perp$. Hence, we are only left with the other inclusion. Pick $h\in (D^\perp)^\perp$ decompose $h$ as
$$\tag{1}h=h_1+h_2\, \ \ ,\ h_1\in F,\, h_2\in F^\perp$$
Now, $h_2\in F^\perp=(\overline{\text{span}(D)})^\perp$. Also, $D\subseteq \overline{\text{span}(D)}$ implies $(D^\perp)^\perp \subseteq \bigl((\overline{\text{span}(D)})^\perp\bigr)^\perp$. Hence, since $h\in (D^\perp)^\perp$ implies $h\in \bigl((\overline{\text{span}(D)})^\perp\bigr)^\perp$. Which means that $\langle h,h_2\rangle =0$. Taking inner products in $(1)$ with $h_2$ and using $\langle h_1,h_2\rangle =0$ we obtain
$$||h_2||^2=\langle h,h_2\rangle -\langle h_1,h_2\rangle =0$$
Hence, $h_2=0$ and $h=h_1 \in F$. Which means that $(D^\perp)^\perp \subseteq F$.

A mini example which shows that $D \subsetneq (D^\perp)^\perp$, is to take as $D$ the space $c_{00}$ as a subspace of $\ell_2$. Then, its easy to check that $D$ is dence in $\ell_2$. Hence, $(D^\perp)^\perp= \overline{D}=\ell_2$ and of course $D=c_{00}\subsetneq \ell_2$.

Title: Show that $(D^\bot)^\bot=D$


问:
Solution to a definite integral as $\int_0^{1}(x^{m}\left(1+x)^{n}dx\right)$
While solving a question based on Binomial series, I encountered a definite integral as : $$\int_0^{1}(x^{m}\left(1+x)^{n}dx\right)$$ where $m,n$ are natural numbers.
My approach : wrote the above expression in two ways (1)- $$\int_0^1\sum_{r=0}^n \binom{n}{r} \left( {x}\right)^{m+r}dx $$ and also as (2)- $$\int_0^1\sum_{r=0}^m \binom{m}{r} \left(-1)^{r}( {1+x}\right)^{m+n-r}dx$$ and tried to solve using both the above mentioned expressions. Did some other simplification of expression (1) , But was unable to proceed further. Can anyone guide me to a "general answer based on any m,n" ? Any help would be highly appreciated. Thanks in advance.!

答:
I found an answer to this problem :
$$ \fbox{$\begin{array}{rcl}\displaystyle\int_{0}^{1}{x^{m}\left(1+x\right)^{n}\,\mathrm{d}x}=\sum_{r=0}^{n}{\left(-1\right)^{r}\binom{m}{r}\frac{[2^{m+n+1-r}-1]}{m+n+1-r}} \end{array}$}$$
I did it by writing $ x^{m} $ as $ (1+x-1)^{m} $ and finally expanding binomial and integrating
Can anyone confirm this answer please .!


问:
Seeking a specific book with graphs of interesting functions, but can't remember the name
I am looking for a specific book that was recommended on youtube in either a Numberphile or Matt Parker video. it was a visual dictionary of pictures of interesting plots of math functions. the cover was blue and black. I think the name of the book was similar to "A List of Interesting Functions and their Graphs" IIRC. What is the name or ISBN of the book?

EDIT: I already have the NIST book. The book I am looking for would likely be in the category of recreational mathematics or mathematics as art. The video was definitely more than 6 months ago. maybe even a year ago. it was a new video at that time.

答:
I am not %100 sure but I think I found the book. I still can't find the video where Matt Parker suggested it.

Curves for the Mathematically Curious: An Anthology of the Unpredictable, Historical, Beautiful, and Romantic

ISBN-10 : 0691180059

ISBN-13 : 978-0691180052

https://www.amazon.com/Curves-Mathematically-Curious-Unpredictable-Historical/dp/0691180059/


问:
Suppose $f$ is a real-differentiable function on $[a,b]$ and suppose $f'(a)<c<f'(b)$. Prove then there is a point $x \in (a,b)$ such that $f'(x)=c$
This is what i have:

Put $g(t) = f(t) - ct$.

Then $g'(a)<0$ so that $g(t_{1}) < g(a)$ for some $t_{1} \in (a,b)$ and
$g'(b)>0$ so that $g(t_{2}) < g(b)$ for some $t_{2} \in (a,b)$.

Hence $g$ attains its minimum on $[a,b]$ at some point x such that $a<x<b$.

Then, $g'(x)=0$, hence $f'(x)=c$

答: This is known as Darboux's theorem. You need to do a bit more work because it could happen that you have to pick $x=a$ or $x=b$.</p>


问:
homogenous equation substitution mistake
My teacher gave me this equation to solve which is a homogeneous equation. In this type of equation, we usually use t substitutions. But in this one I cannot find what should I substitute t with? Any help?

$$ 2xdy+(x^2y^4+1)ydx=0$$

Any help would be appreciated)

答:
This equation is not homogeneous. Hint: Write it as
$$ y'+\frac{1}{2x}y=-\frac12xy^5 \tag1$$
which is a Bernoulli equation. Then let $u=y^{1-5}=y^{-4}$ to change (1) into a linear equation.


问:
For a given group to be cyclic, does the generator have to produce exactly all of its elements?
Take for instance $\Bbb Z_5^*$ under addition.

Take the generator of $\langle1\rangle=\{1,2,3,4,0,\ldots\}$, yes it generates the entire group but it has an additional $0$. Is this still a generator?

答: You are mixing two things here: the group $(\Bbb Z_5,+)$ and the group $(\Bbb Z_5\setminus\{0\},\times)$. Yes, the first group is generated by $1$. However, $1$ is the identity element of the second one, and therefore it does not generate it (the subgroup generated by $1$ is $\{1\}$). It generated by $2$ and $3$, but not by $1$ (nor $4$).


问:
22 students were in the class. 16 took math, 14 took English and 18 took history.
22 students were in the class. 16  took math, 14 took English and 18 took history. At maximum how many took math, English, and history? All took at least one subject.

I was thinking to solve it with this equation as sets.n(A ∪ B ∪ C) = n(A) + n(B) + n(C) - n(A ∩ B) - n(B ∩ C) - n(C ∩ A) + n(A ∩ B ∩ C) trying to find n(A ∩ B ∩ C). But unfortunately there is no information regarding n(A ∩ B) , n(B ∩ C) , n(C ∩ A). So I am confused which approach should I take to solve it

答:
If you were asked exactly how many students took all three subjects, you would need more information than what is given, but the question asks for the maximum number of students who took all three subjects.

There can't be more than $14$, since only $14$ took English.  Can there be as many as $14$?  That would leave $2$ who took math, but not history, and $4$ who took history, but not math, for a total of $$14+2+4=20$$ but there are $22$ students.  Therefore, fewer than $14$ took all three subjects.

The point is that if $14$ students took all three courses, and the other data are as given in the problem, then there are $20$ students in all, but we are told in the problem that there are $22$ students.  Therefore, it is not the case that $14$ students took all three subjects.

Can you extend this line of reasoning to answer the question?


问:
Prove $\left(\frac{a+1}{a+b} \right)^{\frac25}+\left(\frac{b+1}{b+c} \right)^{\frac25}+\left(\frac{c+1}{c+a} \right)^{\frac25} \geqslant 3$
$a,b,c >0$ and $a+b+c=3$, prove
$$\left(\frac{a+1}{a+b} \right)^{\frac25}+\left(\frac{b+1}{b+c} \right)^{\frac25}+\left(\frac{c+1}{c+a} \right)^{\frac25} \geqslant 3$$

I try to apply AM-GM

$$\left(\frac{a+1}{a+b} \right)^{\frac25}+\left(\frac{b+1}{b+c} \right)^{\frac25}+\left(\frac{c+1}{c+a} \right)^{\frac25} \geqslant 3\cdot \sqrt[3]{\left(\frac{a+1}{a+b} \right)^{\frac25}\left(\frac{b+1}{b+c} \right)^{\frac25}\left(\frac{c+1}{c+a} \right)^{\frac25}}$$
Thus it remains to prove
$$\left(\frac{a+1}{a+b} \right)\left(\frac{b+1}{b+c}\right)\left(\frac{c+1}{c+a} \right) \geqslant 1   $$ with the condition $a+b+c=3.$

 But I found the counter example for $$\left(\frac{a+1}{a+b} \right)\left(\frac{b+1}{b+c}\right)\left(\frac{c+1}{c+a} \right) \geqslant 1   $$ :(

答:
Partial Hint :

With your work one can show that :

$$\left(\frac{a+1}{3-x-a+a} \right)\left(\frac{3-x-a+1}{3-x-a+x}\right)\left(\frac{x+1}{x+a} \right) \geqslant 1   $$

With $1\leq a\leq 2$ and $x\in[2-a,1]$

I cannot prove it but using Bernoulli's inequality we have :

$$ \left(\frac{a+1}{x+a} \right)^{\frac25}+\frac{1}{1+\frac{2}{5}\left(\frac{3-x-a+a}{3-x-a+1}-1\right)}+\left(\frac{x+1}{3-x-a+x} \right)^{\frac25}\geq 3$$

With $x\in[0,1]$ and $1\leq a\leq 1.5$


问:
Finding the rank and dimker of a matrix.
Hi so im working on a question about finding the $rank(A)$ and the $dim(Ker(A))$ of a 7x5 Matrix. Without being given an actual matrix to work from.

I have been told that that the homogeneous equation $A\vec x=\vec0$ has general solution $\vec x=\lambda \vec v$ for some non zero $\vec v$ in $R^{5}$.

So my thinking so far is that I know for an $m*n$ matrix we know that:

$rk(A)+dimker(A)=n$
which must mean that $rk(A)+dimker(A)=5$

but this is where I get stuck and dont know how to proceed.

Any help is greately appreciated.

.

This is the exact question for the person who asked.

答: Hint: If you understand the definition of $\ker(A)$ and the definition of dimension, then you can deduce $\dim \ker A$ from the description of the solution to the homogeneous equation.


问:
can we say that $\lim\limits_{x\to0} \sqrt{-x^2}=0$ even if the function domain is a singleton $\{0\}$
In Real analysis, the strict definition of a limit, we take a sequence of point that converging to the point a "limit point". which is not verified here in this case we have only one element in the function domain :

$$\lim_{x\to 0} \sqrt{-x^2}$$

but by replacing $0$ in the function we get $0$

so my question is : in a formal way can we say that

$$\lim_{x\to 0} \sqrt{-x^2}=0$$

答:
In complex analysis, the assertion $\lim_{z\to0}\sqrt{-z^2}=0$ is defensibly true even if you don't bother to specify the sign of the square root for nonzero numbers. But as a real-valued function of a real variable, the expression $\sqrt{-x^2}$ makes no sense except at $x=0$, so it really doesn't make sense to talk about its limit as $x$ tends to $0$. Most formal definitions of limit require the point being tended to to be a "limit point" for the domain of the function, meaning every (punctured) neighborhood of the point has nonempty intersection with the domain. The natural domain for the expression $\sqrt{-x^2}$ as a real-valued function of a real variable is the singleton set $\{0\}$, which has no limit points.

Added later: To put things another way, asking for the meaning of $\lim_{x\to c}f(x)$ when $c$ is not a limit point for the domain of $f$ is a little like asking for the meaning of $\gcd(\sqrt2,\sqrt3)$.  That is, the greatest common divisor is well defined for pairs of integers (not both $0$), but not for any old pair of numbers; likewise limits are only defined at limit points.  Expressions like $\gcd(\sqrt2,\sqrt3)$ and $\lim_{x\to0}\sqrt{-x^2}$ can, perhaps, be usefully likened to Noam Chomsky's grammatically correct but semantically nonsensical sentence "Colorless green ideas sleep furiously."


答:
The best way to interpret this limit as a complex number. Which can be written as: $$\underset{x\to0}{lim}\hspace{0.3cm}x^2i+0$$
As always, you can separate both parts of the limit to get a more clear view of why the result is a real 0:
$$=\underset{x\to0}{lim}\hspace{0.3cm}x^2i+\underset{x\to0}{lim}\hspace{0.3cm}0 $$
And when you finally evaluate both parts of the limits, you can actually be sure that the real part and the imaginary part are both reaching 0 the closer x is to 0 from left or right (Even if the real part domain is $\{0\}$ since the imaginary part domain is $[0,\infty)$ ).

Title: can we say that $\lim\limits_{x\to0} \sqrt{-x^2}=0$ even if the function domain is a singleton $\{0\}$

Body:
In Real analysis, the strict definition of a limit, we take a sequence of point that converging to the point a "limit point". which is not verified here in this case we have only one element in the function domain :

$$\lim_{x\to 0} \sqrt{-x^2}$$

but by replacing $0$ in the function we get $0$

so my question is : in a formal way can we say that

$$\lim_{x\to 0} \sqrt{-x^2}=0$$


答:
The function under consideration is $f:x \to \sqrt{-x^2} $ with domain {$0$}. In other words, your function is,  $f:$ {$0$} $\to ${$0$} with $f(0)=0.$ Your question is, "what is $\lim_{x \to 0}f(x)$?"

Now, let's talk about the domain. $0$ is not a limit point of {$0$} because every neighbourhood of $0$ in {$0$} is {$0$}, so there are no neighbourhoods of $0$ that contains a point $q \neq 0$ such that $q \in ${$0$}. Note that this is stronger than we needed: we only needed to show that there exists a neighbourhood of $0$ that does not contain a point $q \neq 0$ such that $q \in ${$0$}.

Now, in order for us to consider the limit $\ \lim_{x \to 0}f(x)$ (e.g. going by Rudin's definition), $\ 0$ must be a limit point of the domain which we have just seen it is not.

Therefore using the conventional definition of limit of a function at a point in the domain does not tell us anything about $\ \lim_{x \to 0}f(x),\ $ and in fact, I am not sure what the definition or meaning of "$\lim_{x \to 0}f(x)\ $" is for the above function, so let's stop talking about things that haven't been defined.


答:
The answer to your question is already written explicitly in the proof: "We next take the derivative of both sides of the equation ...".

Apply the chain rule to the left-hand side of the equation $\sin(y)=x$.

$\ $

Your $y'=\frac{1}{\cos(y)}$ comes also from the inverse rule of differentiation $[f^{-1}]'(x)=\frac{1}{f'(f^{-1}(x)}$, from the Inverse function theorem:

Set $f=\sin$, $f^{-1}=\text{arcscin}$, $y=f^{-1}(x)$.

Set $f^{-1}(x)=\text{arcsin(x)}$ on the left-hand side of the inverse rule and $f^{-1}(x)=y$ on the right-hand side and procced as in the proof above.

Title: Derivative of arcsin, question on provided proof

Body:
The following is proof of the derivative of $\arcsin$.  Does anyone know why and with what axiom the $y'$ is introduced into the equation in the following step?

$\sin y = x$

$y'\cos y = 1$


问:
Probability for Poisson question
Can anyone help me with this question?
[1]: https://i.stack.imgur.com/1LJJH.png

this is my solution, is this in the right track?
[2]: https://i.stack.imgur.com/wodoa.jpg

答:
I simply would use the Neyman Pearson's lemma

$$\frac{L(\mathbf{x}|\lambda_0)}{L(\mathbf{x}|\lambda_1)}\leq k$$

that is

$$\frac{e^{-14}2^{\Sigma_i X_i}}{e^{-7}}\leq k$$

or equivalently

$$\Sigma_i X_i\leq k^*$$

leading to the following Critical region

$$P(\Sigma_i X_i\leq k^*|\lambda=2)=0.01$$

Where $\Sigma_i X_i\sim Po(14)$

Concluding, you will reject $H_0$ iff

$$\sum_{i=1}^7 X_i\leq5$$

Title: Probability for Poisson question


问:
Show that D is a ring of sets
Let $\nu$ a pre-measure on a ring $\mathcal{R}$.

show: $\mathcal{D}=\{A\in\mathcal{R}|\nu(A)=0\}$ is a ring

closed under union: $\nu(A\cup B)=\nu(A)+\nu(B)-\nu(A\cap B)=-\nu(A\cap B)$

closed under relative complement: $\nu(A\setminus B)=\nu(A)-\nu(B)=0$  but only if $B\subset A$

how can I show 1) and what if $B$ is not a subset of $A$ in 2)?

答:
Like Bernard Pan said, $\nu(S)=0$ for any $S\subset A$.

And since $(A\cap B) \subset A$ and $(A\setminus B) \subset A \qquad \nu(A\cap B)=0\quad$ and $\quad\nu(A\setminus B)=0$

Title: Show that D is a ring of sets


问:
Proof that the adjoint operator of $\Phi$ is $\Phi(\sigma)=\sum_{i}A_{i}^{*}\sigma B_{i}$ in terms of Kraus operators.
I am studying basics of quantum information theory and I am trying to prove the fact that the adjoint operator of $\Phi\in\mathcal{L}(\mathcal{L(\mathcal{H})},\mathcal{L(\mathcal{K})})$ is $\Phi^{*}(\sigma)=\sum_{i}A_{i}^{*}\sigma B_{i}$ where $\mathcal{H}$ and $\mathcal{K}$ are finite-dimensional Hilbert spaces. I know that $\Phi(\rho)=\sum_{i}A_{i}\rho B_{i}^{*}-(1)$ for finite set of Kraus operators $\{A_{i},B_{i}\in\mathcal{L}(\mathcal{H},\mathcal{K})\}$ for any $\rho\in\mathcal{H}$.

Now, I am trying to use the definition of adjoint operators that $\rho\cdot\Phi^{*}(\sigma)=\Phi(\rho)\cdot\sigma$ to prove it. I think I should plug in the formula $(1)$ into $\Phi(\rho)$ but after that I am stucked about how to calculate the inner product "$\cdot$". Which fact should I use to proceed the calculation?

答:
In the context of QIT, the inner-product over $\mathcal L(\mathcal H)$ is defined by
$$
\langle \rho_1,\rho_2 \rangle_{\mathcal L(\mathcal H)} = \operatorname{Tr}(\rho_1 \rho_2^*).
$$
With that in mind, we want to show that for any $\rho \in \mathcal L(\mathcal H)$ and $\sigma \in \mathcal {L(K)},$ your formulas for $\Phi$ and $\Phi^*$ are such that
$$
\langle \Phi(\rho),\sigma \rangle_{\mathcal {L(K)}} = \langle \rho, \Phi^*(\sigma)\rangle_{\mathcal {L(H)}}.
$$
To that end, we note that
\begin{align}
\langle \Phi(\rho),\sigma \rangle_{\mathcal {L(K)}} &=
\operatorname{Tr}\left[ \left(\sum_i A_i \rho B_i^*\right) \sigma^* \right]
\\ & = \sum_i\operatorname{Tr}\left[ A_i \rho B_i^* \sigma^* \right]
\\ & = \sum_i\operatorname{Tr}\left[ \rho B_i^* \sigma^* A_i \right]
\\ & = \sum_i\operatorname{Tr}\left[ \rho \left(A_i^*\sigma B_i\right)^* \right]
\\ & = \operatorname{Tr}\left[ \rho \left(\sum_i A_i^*\sigma B_i\right)^* \right]
= \langle \rho, \Phi^*(\sigma) \rangle_{\mathcal{L(H)}}.
\end{align}


答:
From the definition of a limit in a metric space:
$$
\forall\epsilon\gt0,\exists\delta\gt0:0\lt|x-0|\le\delta\implies|f(x)-0|\le\epsilon\tag1
$$
Since there are no $x$ in the domain of $f(x)=\sqrt{-x^2}$ that satisfy $0\lt|x-0|\le\delta$, the implication is vacuously true.

However...

Thanks to Adam Rubinson for pointing out my error. As described in Wikipedia, to take the limit of a function at a point $p\in S\subset\mathbb{R}$, it is required that $p\in\overline{S\setminus\{p\}}$. Unfortunately, in this case, $S=\{p\}$, so $p\not\in\overline{S\setminus\{p\}}=\emptyset$. So, although condition $(1)$ is satisfied, we cannot say that
$$\require{cancel}
\cancel{\lim_{x\to0}\sqrt{-x^2}=0}\tag2
$$
When $S=\mathbb{R}$, $p\in\overline{S\setminus\{p\}}$ for all $p\in S$, so $(1)$ is sufficient.


问:
How many different ways can you distribute 5 apples and 8 oranges among six children?
How many different ways can you distribute 5 apples and 8 oranges among six children if every child must receive at least one piece of fruit?
If there was a way to solve this using Pólya-Redfield that would be great, but I cannot figure out the group elements.</p>

答:
I am too lazy to calculate the numbers of elements with $k$ cycles in $S_8$ but if you do that yourself a solution could work as follows. (I will use this version of Redfield-Polya and $[n]$ shall denote $\{1,\dots,n\}$.)

Let us take $X = [13]$ the set of fruits, where $G= S_5 \times S_8$ acts on $X$ such that the first five apples and the later eight oranges are indistinguishable. Then
$$K_n = |[n]^X/G|$$
is the number of ways to distribute these apples and oranges among six distinguishable children. And
$$ N_n = K_n -n\cdot K_{n-1}$$
the of ways to distribute these apples and oranges among six distinguishable children such that every child must recieve at least one piece of fruit.

Now by the Theorem
$$K_n = \frac{1}{|G|} \sum_{g\in G} n^{c(g)} = \frac{1}{5!\cdot 8!} \left(\sum_{g\in S_5} n^{c(g)}\right)\left(\sum_{g\in S_8} n^{c(g)}\right) = \frac{1}{5!\cdot 8!} \left(\sum_{i\in [2]} d_i n^{i}\right) \left(\sum_{i\in [4]} e_i n^{i}\right),$$
where $c(g)$ is the number of cycles of $g$, $d_i$ the number of permutations of $S_5$ with exactly $i$ cycles and $e_i$ the number of permutations of $S_8$ with exactly $i$ cycles.

The number that we are looking for in the end is $N_6$.


问:
Determining the average area as the point $X$ varies
We consider a fixed triangle $ABC$ with side lengths $a = BC$, $b = CA$, $c = AB$, and
a variable point $X$ in the interior. The lines through $X$ parallel to $AB$ and $AC$,
together with line $BC$, determine a triangle $T_a$. The triangles $T_b$ and $T_c$ are defined
in a similarly way .

Let $S$ and $p$ denote the average area and perimeter, respectively, of the three triangles $T_a, T_b, T_c$.

(a) Determine all possible values of $S$ as $X$ varies, in terms of $a, b, c.$

(b) Determine all possible values of $p$ as $X$ varies, in terms of $a, b, c$.

Can someone give me hints for this problem ? I don't know how to start (a).

I got that (b) is $\frac {1}{3}(a+b+c)$, because of the parallel property , sum of perimeters of $T_a, T_b, T_c$  is $a+b+c$.

答:
Let $[ABC]=S_0$ , its perimeter $= P_0$

Let bases (parallel to $a$) of ${T_a, T_b, T_c}$ be ${x,y,z}$ respectively.

Note  each $T_i \sim \triangle ABC$ with similarity ratios  ${x/a, y/a, z/a}$ respectively.

These three triangles leave three parallelograms out of $\triangle ABC$. Equating base lengths, we get $x+y+z=a$

Thus $$P=\frac{1}{3} \cdot \big( \dfrac{x}{a}+\dfrac{y}{a}+\dfrac{z}{a} \big)P_0$$

$$S=\frac{1}{3} \cdot \big( \dfrac{x^2}{a^2}+\dfrac{y^2}{a^2}+\dfrac{z^2}{a^2} \big)S_0$$

Now find the range of $$\dfrac{x^2+y^2+z^2}{a^2}$$


问:
Statistic Mode on continuous data
I have a discussion with a co-worker, he insist on try to get the mode of a dataset with continuous data, with numbers that go from 0 to 3000, I say that the result are irrelevant because the repeated values will be too lower compared to the total of the data.

Can you point me to some literature that proves or disproves my point?
and what percentage of the total data can be acepted as a valid mode?

Thanks.

答:
Suppose you have 1000 observations from a gamma distribution with shape 5 and rate 0.01. The following
random sample is chosen using R.

```<code>set.seed(2020)
x = rgamma(1000, 5, .01)
summary(x)
   Min. 1st Qu.  Median    Mean 3rd Qu.    Max.
  66.37  339.38  463.70  499.28  619.42 1636.80
</code>```

Because this is an (unrounded) sample from a continuous
distribution all 1000 observations are different, and
the sample can have no mode.

However, the continuous distribution of the population
has a mode. Sometimes the population mode is approximated by looking at a histogram of the data, and using some
kind of interpolation to say where in the tallest bar
the population mode may lie. (The assessment of the mode
depends on the choice of histogram bins used. Sometimes the tallest bar of a histogram, if there is one, is called its modal bar.)

As @heropup suggests, a better way of finding the shape (hence mode) of the population distribution is to use a kernel density estimator (KDE).

The figure below shows a histogram of the 1000 observations. The actual population density for
$\mathsf{Gamma}(5, 0.01)$ is shown as a black curve.
(We are able to plot this curve because this is a
simulation in which we know the exact population
distribution.)

If the population distribution is unknown, then a KDE
often does a better job of estimating the mode of the
population. A KDE of the sample is shown as a dotted red line in the figure.

```<code>hist(x, prob=T, br=20, col="skyblue2",
     ylim=c(0,.002), main="1000 Obs. from GAMMA(5, .01)")
 curve(dgamma(x, 5, .01), add=T)
 lines(density(x), col="red", lwd=2, lty="dotted")
</code>```

Notes: (1) By differentiating this gamma density function, one can find that the mode of its distribution (unique maximum of its density function) is at $400.$

(2) The following R code shows that the maximum of the KDE is at $374.5.$

```<code>mean(density(x)<span class="math-container">$x[density(x)$</span>y==max(density(x)$y)])
[1] 374.5053
</code>```

Title: Statistic Mode on continuous data

Body:
Suppose you have 1000 observations from a gamma distribution with shape 5 and rate 0.01. The following
random sample is chosen using R.

set.seed(2020)
x = rgamma(1000, 5, .01)
summary(x)
   Min. 1st Qu.  Median    Mean 3rd Qu.    Max.
  66.37  339.38  463.70  499.28  619.42 1636.80

Because this is an (unrounded) sample from a continuous
distribution all 1000 observations are different, and
the sample can have no mode.

However, the continuous distribution of the population
has a mode. Sometimes the population mode is approximated by looking at a histogram of the data, and using some
kind of interpolation to say where in the tallest bar
the population mode may lie. (The assessment of the mode
depends on the choice of histogram bins used. Sometimes the tallest bar of a histogram, if there is one, is called its modal bar.)

As @heropup suggests, a better way of finding the shape (hence mode) of the population distribution is to use a kernel density estimator (KDE).

The figure below shows a histogram of the 1000 observations. The actual population density for
$\mathsf{Gamma}(5, 0.01)$ is shown as a black curve.
(We are able to plot this curve because this is a
simulation in which we know the exact population
distribution.)

If the population distribution is unknown, then a KDE
often does a better job of estimating the mode of the
population. A KDE of the sample is shown as a dotted red line in the figure.

hist(x, prob=T, br=20, col="skyblue2",
     ylim=c(0,.002), main="1000 Obs. from GAMMA(5, .01)")
 curve(dgamma(x, 5, .01), add=T)
 lines(density(x), col="red", lwd=2, lty="dotted")

Notes: (1) By differentiating this gamma density function, one can find that the mode of its distribution (unique maximum of its density function) is at $400.$

(2) The following R code shows that the maximum of the KDE is at $374.5.$

mean(density(x)$x[density(x)$y==max(density(x)$y)])
[1] 374.5053


问:
sequence proof that $b_{n+1}<b_n$ and $a_{n+1}>a_n$
Let $ \{a_n\}_{n \geq 0},\{b_n\}_{n \geq 0}  $ be two sequences. The sequences are given as $ b_n:= \dfrac{a_{n-1}+b_{n-1}}2 $ $a_n = \frac{c}{b_{n}} $ for $n \geq 1$ with $b_0 = b $ and $a_0 = \dfrac{c}{b}$.

I know that: $$ \frac{a_{0}+b_{0}}{2} > \sqrt{a_{0}b_{0}} $$

And I already proofed that: $$ a_n < \sqrt{c} < b_n  $$

I want to show that for all integers $n$: $b_{n+1}<b_n$ and $a_{n+1}>a_n$ and that: $b_n-a_n \le \frac{1}{2^{n-1}}\cdot (b_1-a_1) $

I started with:
$b_{n} > \sqrt{c} = \frac{c}{\sqrt{a_{n+1}b_{n+1}}} >  \frac{c}{0.5({a_{n+1}b_{n+1})}} = \frac{c}{b_{n+2}}$

But I don't know how to go on :/
Can someone help? :)

Body:
Update: Forgot to prove $b_n>b_{n+1}$. This is because $a_n < c < b_n$, so $b_n=\frac{a_{n-1}+b_{n-1}}{2}<\frac{b_{n-1}+b_{n-1}}{2}=b_{n-1}$.

$$2(b_n-a_n)=a_{n-1}+b_{n-1}-2a_n \le a_{n-1}+b_{n-1} - 2a_{n-1}=b_{n-1}-a_{n-1}$$
The rest is straightforward.


问:
3D polyhedron with no axis of symmetry
In learning to work with inertia tensors I came to wonder if there were any polyhedra with no axis of symmetry; and if so, are there any regular polyhedra ? My workings so far : the only shape I can think of in 3D that has no axis of symmetry is of course an ellipsoid; so I was wondering if this is due to the fact that it has no angles... Thanks and sorry if this is too much of an open question; I'll delete it if needed.</p>

答:
The following polyhedra all have some symmetry – either a rotation axis or a mirror plane (which has an associated normal axis):

Platonic solids and Kepler–Poinsot solids

Prisms and antiprisms, including star forms

Archimedean solids and uniform star polyhedra

Johnson solids

So no regular polyhedron, by any reasonable definition, is wholly asymmetric. However, it is very easy to make such a polyhedron. Just take six distinct edges that differ by any sufficiently small amount – $7,8,9,10,11,12$ will do – and build a tetrahedron with them.

Title: 3D polyhedron with no axis of symmetry


问:
Modularity by Newman (Networks)
I need help with modularity of a network.
i want to understand the modularity defined by Newman : https://arxiv.org/pdf/cond-mat/0308217.pdf
He says that M is between 0 and 1 if the modularity M is determined as
\begin{equation}
M = \sum_i (e_{ii} - a_i^2)
\end{equation}
while $e_{ij}$  "is the fraction of all edges in the network that link vertices in community i to vertices in community j" and \begin{equation}
a_i = \sum_j e_{ij},
\end{equation}
You get a $kxk$ Matrix $e$ ( k communities) with $e_{ij}$ as elements.
Now, i constructed a Graph with such a clustering : selfmade graph

For this i get the matrix:

e =

(1/7  --  2/7 )

(2/7 --  4/7 )   (sorry my latex matrix code is not woriking here)

Now if i put these into the modularity formula i get M = -0,204. With optimal clustering into two communites i get M = 0,2.
Can someone explain? Somewhere else it is stated taht the modularity is between [-1,1), but im not sure if the other (spektral formula with Adjacency Matrix) is meant.
Thanks!

答:
Modularity involves a few ingredients. First off there is the graph/network, then there is the partition of the vertices, then there are the details of the definition of modularity. Embedded in the definition is a null model which is used to specify what we mean exactly by "random". Typically the choice of null model is the configuration model which is a random graph model that can be read about here.

Once the definition of modularity is fixed, we essentially have a function that maps a graph/network and a partition of the vertices to a value. This value is between $-1$ and $1$ (assuming a certain normalization---sometimes it can be between $-1/2$ and $1$). Given a graph/network that seems to have community structure (such as the example you linked to), you can still get a negative modularity score if the partition used to calculate modularity doesn't correspond to the community structure. However, computing the modularity score corresponding to your example graph and a 'good' partition of the vertices results in a positive modularity score.

A lot of community detection focuses on finding this partition in a process called modularity maximization, which are typically approximative algorithms for finding the partition of vertices that results in the largest modularity score (the number of partitions of vertices grows very, very fast). For a graph/network with a large number of vertices, it's typical some partitions have positive scores and some will have negative scores. Finding a partition with a positive score is not proof that the graph/network has community structure. For an excellent paper on this topic, check out: Cristopher Moore's
The Computer Science and Physics of Community Detection: Landscapes, Phase Transitions, and Hardness.

EDIT/UPDATE
We can find the modularity score for each graph and each partition of its vertices. As we said, this can be a negative value. We can also assign a maximum modularity score (i.e., a single value) to a graph by taking the maximum score over all of the partitions of its vertices. I believe this is what Newman & Girvan are referring to in the paper that you linked to in your question. The trivial partition, where all vertices are in the same community, leads to a modularity score of $0$ and there are graphs where the trivial partition is the maximally modular partition, meaning that all other partitions have non-positive modularity scores. Thus $0$ is the minimum that the maximum modularity can take.  I believe that for most graphs there will be both partitions with negative modularity scores and partitions with positive modularity scores.

Title: Modularity by Newman (Networks)


答:
Let $X = $ then number of students who took all three.

Let $ME =$ then number of students who took english and math only.

Let $MH= $ then number of students who took math and history only.

Let $HE= $ then number of students who took history and english only.

Then $14-ME-HE - X=$ then number of students who only took english.

$16-ME - MH-X=$ the number of students who only took math.

$18-MH - HE - X= $ then number of students who only took history.

$X + ME + MH + HE + (14-ME-HE - X) + (16-ME-MH-X) + (18-MH-HE -X) = 22$

$-2X-ME-MH-HE + 48 =22$

$26 -ME -MH -HE = 2X$

we want to maximize $X$ so we can do that by minimizing $ME,MH,HE$.

Can they all be $0$?  That would mean $13$ students took all three.  Then of the $14$ students who took english $14-13=1$ of them didn't take all three. That person couldn't take two classes so that person just took english. So $1$ person took only English.

Of the $16$ who took math, $16-13=3$ of then didn't take all three so $3$ took just math. And of the $18$ who took history $18 -13= 5$ took just history.

And that means $1+3+5=9$ people took just one class, $0$ people took two, and $13$ took all three and $13 + 0 + 9 =22$ so... that's possible.

The maximum number that took all three is $13$.

Body:
22 students were in the class. 16  took math, 14 took English and 18 took history. At maximum how many took math, English, and history? All took at least one subject.

I was thinking to solve it with this equation as sets.n(A ∪ B ∪ C) = n(A) + n(B) + n(C) - n(A ∩ B) - n(B ∩ C) - n(C ∩ A) + n(A ∩ B ∩ C) trying to find n(A ∩ B ∩ C). But unfortunately there is no information regarding n(A ∩ B) , n(B ∩ C) , n(C ∩ A). So I am confused which approach should I take to solve it


答:
First, note that you have to suppose that $b^2 \neq c$ if you want to write strict inequalities.
To show that $b_{n+1} \leq b_n$, write
$$b_{n+1} - b_n = \frac{a_n - b_n}{2}$$
and this is negative since $a_n \leq b_n$. The fact that $a_{n+1} \geq a_n$ follows. To show that $b_n - a_n \leq \frac{1}{2^{n-1}} (b_1 - a_1)$, write for $n > 1$,
\begin{align*}
b_n - a_n & = \frac{a_{n - 1} + b_{n - 1}}{2} - a_n \\
& \leq \frac{a_{n - 1} + b_{n - 1}}{2} - a_{n - 1} \\
& = \frac{b_{n - 1} - a_{n - 1}}{2}
\end{align*}
and conclude by induction.

Title: sequence proof that $b_{n+1}<b_n$ and $a_{n+1}>a_n$

Body:
Let $ \{a_n\}_{n \geq 0},\{b_n\}_{n \geq 0}  $ be two sequences. The sequences are given as $ b_n:= \dfrac{a_{n-1}+b_{n-1}}2 $ $a_n = \frac{c}{b_{n}} $ for $n \geq 1$ with $b_0 = b $ and $a_0 = \dfrac{c}{b}$.

I know that: $$ \frac{a_{0}+b_{0}}{2} > \sqrt{a_{0}b_{0}} $$

And I already proofed that: $$ a_n < \sqrt{c} < b_n  $$

I want to show that for all integers $n$: $b_{n+1}<b_n$ and $a_{n+1}>a_n$ and that: $b_n-a_n \le \frac{1}{2^{n-1}}\cdot (b_1-a_1) $

I started with:
$b_{n} > \sqrt{c} = \frac{c}{\sqrt{a_{n+1}b_{n+1}}} >  \frac{c}{0.5({a_{n+1}b_{n+1})}} = \frac{c}{b_{n+2}}$

But I don't know how to go on :/
Can someone help? :)


问:
If $f$ is a polynomial, $a_1=f(0)$ and $a_n$ = $f(a_{n-1})$ , conclude the integer values of $a_1$ or $a_2$

Let $f$ be a polynomial with integer coefficients.

We define $a_1= f(0)$

$a_2 = f (a_1)= f(f(0))$

Similarly $a_n$ = $f(a_{n-1})\ \  \forall n \geq3$

If there exists a natural number $k$ greater than or equal to $3$ such that $a_k = 0$ prove that either $a_1 =0$ or $a_2 = 0$

Source: ISI UGB 2019 paper

What I thought was that if for some $k$, $a_k$ is zero then intuitively $a_1$, $a_2$ all are zero $\dots$ but this is not correct. What I need is a rigorous proof for the above claim.

答:
EDITED (to avoid zero denominators):
I'll work through the case $k=3$, where $a_3 = 0$ and hopefully you can generalize this. Apply the Mean Value Theorem ($f$ is a polynomial, so infinitely differentiable) on $(a_2,a_3)$ (or $(a_3,a_2)$ if $a_2 > 0$) to assert that there exists $\xi$ in this interval for which
$$
f(a_3)-f(a_2) = (a_3-a_2)f'(\xi),
$$
i.e.
$$
a_1 = -a_2f'(\xi).
$$
Similarly, MVT on $(a_1,a_2)$ (or $(a_2,a_1)$) gives the existence of $\zeta$ in this interval such that
$$
f(a_2)-f(a_1)=(a_2-a_1)f'(\zeta),
$$
i.e.
$$
-a_2 = (a_2-a_1)f'(\zeta).
$$
Putting this all together, $$
a_1 = -a_2f'(\xi) = (a_2 - a_1)f'(\zeta),
$$
so either both $a_1$ and $a_2$ are zero, or you can divide through by whichever is non-zero to form an equation just in the other, and hence conclude that that must be zero.

For the general case, observe that we may always reduce an index $j \in [2,\dots,k-1]$ via
$$
f(a_{j-1}) - f(a_{k-1}) = (a_{j-1} - a_{k-1})f'(\xi_j),
$$
i.e.
$$
a_j = (a_{j-1} - a_{k-1})f'(\xi_j).
$$
Also,
$$
f(a_k) - f(a_{k-1}) = (a_k - a_{k-1})f'(\zeta),
$$
i.e.
$$
a_1 = - a_{k-1}f'(\zeta).
$$
This allows us to reduce to an equation in $a_1$: use the base case as a guide in how to conclude. (You may want to try $k=4, 5$ to see if you have the idea.)


答:
This is a partial answer, I can only reduce the sum to a single integral.

For any $d \ge 1$, let $\Delta_d$ be the simplex
$$\Delta_d = \{ (x_1,\ldots,x_d) \in [0,\infty)^d : x_1 + \cdots x_d \le 1 \}$$
The summands at hand can be represented as integrals over $\Delta_3$.

$$\frac{m!n!p!}{(m+n+p+3)!} = \int_{\Delta_3} x^m y^n z^p dxdydz$$

This leads to
$$\begin{align}
\mathcal{S} \stackrel{def}{=} & \sum_{m,n,p \ge 0}\frac{m!n!p!}{(m+n+p+3)!}
= \int_{\Delta_3}\frac{dxdydz}{(1-x)(1-y)(1-z)}\\
= &\int_{\Delta_2}  \left(\int_0^{1-y-z}\frac{dx}{1-x}\right) \frac{dydz}{(1-y)(1-z)}
\\
= & -\int_{\Delta_2} \frac{\log(y+z)}{(1-y)(1-z)} dydz\\
= & -\int_{\Delta_2} \left(\frac{1}{1-y} + \frac{1}{1-z}\right)\frac{\log(y+z) dydz}{2 - y - z}\\
= & -2\int_{\Delta_2} \frac{\log(y+z)dydz}{(1-y)(2-y-z)}
\end{align}$$
Change variable to $(y,\lambda) = (y,y+z)$, we obtain

$$\begin{align}\mathcal{S}
&= -2\int_0^1 \int_0^\lambda \frac{\log \lambda}{(1-y)(2-\lambda)} dy d\lambda
= 2\int_0^1 \frac{\log\lambda\log(1-\lambda)}{2-\lambda} d\lambda\\
&\approx 0.4861407
\end{align}
$$
This is as far as I can go.

Title: $\sum_{m\ge 0}\sum_{n\ge 0}\sum_{p\ge 0}\frac{m!n!p!}{(m+n+p+3)!}$

Body:
Evaluate: $\sum_{m\ge 0}\sum_{n\ge 0}\sum_{p\ge 0}\frac {m!n!p!}{(m+n+p+3)!}$

my working :
Claim:$\sum_{k\ge0}\frac{k!}{(k+l)!}=\frac{1}{(l-1)((l-1)!)}$

proof:$\frac{k!}{(k+l)!}=\frac{1}{l-1}\left(\frac{k!}{(l+k-1)!}-\frac{(k+1)!}{(l+k)!}\right)$

$\implies\sum_{k\ge0}\frac{k!}{(k+l)!}=\frac{1}{(l-1)((l-1)!)}$

now $\sum_{m\ge 0}\sum_{n\ge 0}\sum_{p\ge 0}\frac {m!n!p!}{(m+n+p+3)!}=\sum_{m\ge 0}\sum_{n\ge 0}m!n!\sum_{p\ge 0}\frac {p!}{(m+n+p+3)!}$
$=\sum_{m\ge 0}\sum_{n\ge 0}m!n!\frac {1}{(m+n+2)((m+n+2)!)}$


问:
Existence of a sequence
Let $p \in [1, + \infty)$ consider the Banach space $l^p=\{f=(f_n)_{n \in \mathbb{N}} : \sum_{n=1}^{\infty}|f_n|^p<+\infty\}$. Let $(a_n)_{n \in \mathbb{N}} \in l^2$ a sequence such that $a_n \neq0$ for every $n \geq1$. Show that exist a sequence $(b_n)_{n \in  \mathbb{N}}$ such that $(\frac{b_n}{a_n})_{n \geq 1} \notin l^2$. I try with the following:

I consider variuos cases:

If $(a_n)_{n \in \mathbb{N}} \in l^1$ then $(a_n)_{n \in \mathbb{N}}=(b_n)_{n \in \mathbb{N}}$, i.e $(\frac{b_n}{a_n})_{n \geq 1} \notin l^2$

If  $(a_n)_{n \in \mathbb{N}} \notin l^1$ we can take $(a_n)^{\frac{3}{2}}_{n \in \mathbb{N}}=(b_n)_{n \in \mathbb{N}}$ where we have two subcases:

1.1 If  $(a_n)^{\frac{3}{2}}_{n \in \mathbb{N}} \in l^1$ then $(a_n)^{\frac{3}{2}}_{n \in \mathbb{N}}=(b_n)_{n \in \mathbb{N}}$, which implies that $(\frac{b_n}{a_n})_{n \geq 1}=(a_n)^\frac{1}{2} \notin l^2$

1.2 If  $(a_n)^{\frac{3}{2}}_{n \in \mathbb{N}} \notin l^1$ we can take $(b_n)=(a_n)^{\frac{5}{2}}$ then $(\frac{b_n}{a_n})_{n \geq 1}=(a_n)^\frac{3}{2} \notin l^2$.

I don't know if my argument is correct, I don't know if I can continue in that way or are there other procedures to solve the exercise.

答: If we take $b_n=a_n$, we get $\frac{b_n}{a_n}=1 \not \in l^2$.

Title: Existence of a sequence