问:
A robust convex optimization problem
Consider a function $ f: \mathbb{R}^n \times \mathbb{R}^m \rightarrow \mathbb{R}  $ such that $\forall x \in \mathbb{R}^n$ the map $f(x,\cdot)$ is convex, and $\forall y\in \mathbb{R}^m$ the map $f(\cdot,y)$ is convex as well. We assume that $m<n$.

Given $y^1, ..., y^N \in \mathbb{R}^m$, and $c \in \mathbb{R}^n$, consider the following convex optimization problem.

$$ \min_{ x \in \mathbb{R}^n } c^\top x \ \ \ \text{ sub. to: } \ f(x,y^i) \leq 0 \ \ \forall i \in [1,N] $$

Let $x^\star$ be its optimizer, supposed to be unique.

Notice that the above optimization problem is equivalent to the following one.

$$ \min_{ x \in \mathbb{R}^n } c^\top x \ \ \ \text{ sub. to: } \ f(x,y) \leq 0 \ \ \forall y \in \text{conv}( \{y^1, ..., y^N\} ) $$

We now say that $f(x,y^k) \leq 0$ is a "true" constraint if it "really matters", i.e.
any optimizer of the problem

$$ \min_{ x \in \mathbb{R}^n } c^\top x \ \ \ \text{ sub. to: } \ f(x,y^i) \leq 0 \ \ \forall i \in ([1,N]\setminus\{k\}), $$

which we denote by $x_{ \neg k }^\star$, is such that $c^\top x_{ \neg k }^\star < c^\top x^\star$.

What is the number of true constraints?

答:
I guess that by "true" constraints you are meaning "active" or "blocking" constraints, that is constraints such that $f(x^*,y^i)=0$.

Because, by convexity, if a constraint does not block -- i.e., $f(x^*,y^i)<0$ -- then removing it from the set of constraints does not change the optimum.

To count the blocking constraints is quite easy a posteriori, that is once you know the optimum. In facts, most software provides you the "Lagrange multipier" $\frac {\partial x^*} {\partial y^i}$ which are null exactly when they do block.

To count the blocking constraints a priori, before to compute the optimum is a $NP$ problem. Even simplifying the convex hull of constraints $\{x|\forall i, f(x,y^i)\le0\}$ is $NP$.

Maybe you can find something starting there.

Title: A robust convex optimization problem


问:
Transformation of the integrand of an elliptic integral to the general form of an elliptic curve
First, some background.

An ellipse is defined by the following equation:

$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\implies y=b\sqrt{1-\frac{x^2}{a^2}}.$

To get the arclength of the ellipse, we use calculus to obtain:

$4\int_{0}^{a}\sqrt{1+\frac{b^2x^2}{a^2(a^2-x^2)}}dx.$

We introduce a change of variables $x=at$ to get:

$4a\int_{0}^{1}\sqrt{1+\frac{a^2b^2t^2}{a^2(a^2-a^2t^2)}}dt=4a\int_{0}^{1}\sqrt{\frac{1-t^2(1-b^2/a^2)}{1-t^2}}dt.$

Let $e^2=1-\frac{b^2}{a^2}$ be the square of the eccentricity of the ellipse, then the integral becomes:

$\int_{0}^{1}\sqrt{\frac{1-e^2t^2}{1-t^2}}dt.$

This is an elliptic integral.

Let $u=u(t)=\sqrt{\frac{1-e^2t^2}{1-t^2}}$ be the integrand, then:

$u^2(1-t^2)=1-e^2t^2.$

I believe that this equation defines an elliptic curve of the form:

$y^2=x^3+ax+b$

Such that:

$4a^3+27b^2\neq0.$

How does one transform the equation in terms of $u$ into an elliptic curve?

ContentLicense: CC BY-SA 4.0

Body:
$y^2=(1+e^2t^2)/(1+t^2)$

If $2\ne 0$,

$y=z/(1+t^2)$ gives $z^2=(1+e^2t^2)(1+t^2)=e^2(t^2+a)^2+b$

$z=w+ie(t^2+a)$ gives $w^2 +2wie(t^2+a)=b$

$t=u/w$ gives $w^2 +2ieu^2/w+2wiea=b$

or $$w^3+2w^2iea-bw=-2ieu^2$$

If $e\ne 0,3\ne 0$  it can be transformed into $U^2= W^3+AW+B$


问:
Matrix Derivative Product Rule $\frac{d}{dx} \langle x,y\rangle Ax $
Let $A\in\mathbb{R^{n\times n}}$,$x,y\in\mathbb{R^{n}}$. What is the derivative of the following expression?

$$
\frac{d}{dx} \langle x,y\rangle Ax
$$

What I tried:

Applying the product rule doesn't work because the first term (see next equation) is inconsistent dimensionally, or I am applying it in the wrong way. Let $f(x) = \langle x,y \rangle, g(x) = Ax$, then

$$ \frac{d}{dx} f(x)g(x) = \frac{df(x)}{dx} g(x) +  f(x)\frac{dg(x)}{dx} = xAx + \langle x,y \rangle A$$.

Here the first term in the last equation doesn't make sense dimensionally, any help or suggestion would be appreciated.

答: In Einstein notation and the standard basis for which $\langle x,\,y\rangle=x_ky_k$, you're differentiating a vector of $i$th component $x_ky_kA_{il}x_l$ with respect to a vector of $j$th component $x_j$, giving a matrix of $ij$ entry$$\delta_{jk}y_kA_{il}x_l+x_ky_kA_{il}\delta_{jl}=y_jA_{il}x_l+x_ky_kA_{ij}=(Axy^T+\langle x,\,y\rangle A)_{ij}.$$In other words, the derivative is $A(xy^T+\langle x,\,y\rangle I_n)$, with $I_n$ the $n\times n$ identity matrix.


问:
Representing an optimization program as a quadratic program with linear constraints.
I am trying to convert this minimization problem:

$$
\text{minimize}_{x \in \mathbb{R}^N, z \in \mathbb{R}}. \sum_{m=1}^M \text{max} (\mathbf{a}_m^T x, z) + \tau \Vert x \Vert_2
$$

to a quadratic program with linear constraints. How can I go about this problem?
Thanks.

答:
Introduce a new variable $y_m$ to represent the $\max$ and impose linear constraints
\begin{align}
y_m &\ge \mathbf{a}_m^T x \\
y_m &\ge z \\
\end{align}


问:
Linear programming (optimisation) problem in multiple dimensions
I am trying to solve the linear programming problem

\begin{align*}
(\text{P}):\,\text{Minimize }f({\bf x},{\bf y})&={\bf a}\cdot{\bf x}+{\bf b}\cdot{\bf y},\\
\text{subject to }A{\bf x}+B{\bf y}&={\bf c},\\
y_i\geq0&\forall1\leq i\leq n,
\end{align*}

where ${\bf a}$ and ${\bf x}$ are $m$-dimensional vectors,  ${\bf b}$ and ${\bf y}$ are $n$-dimensional vectors, and $c$ is a $k$-dimensional vector. I am considering the Lagrangian

$$\mathcal{L}={\bf a}\cdot{\bf x}+{\bf b}\cdot{\bf y}-\bf{\lambda}\cdot(A{\bf x}+B{\bf y}-{\bf c})-\bf{\mu}\cdot{\bf y}$$

and want to therefore show that the optimisation $\max_{{\bf{\lambda}},{\bf{\mu}}}\min_{{\bf x},{\bf y}}\mathcal{L}$ is equivalent to

\begin{align*}
(\text{P}):\,\text{Maximise } g(\bf{\lambda})&={\bf c}\cdot\bf{\lambda},\\
\text{subject to }A^\textrm{T}\bf{\lambda}&={\bf a},\\
B^\textrm{T}\bf{\lambda}&={\bf b}.
\end{align*}

I am very new to convex optimisation, so the optimisation knowledge I have is quite limited. Firstly, to solve for $\min_{{\bf x},{\bf y}}\mathcal{L}$, how would one write it formally? The hand-wavy physicist is tempted to write

$${\bf a}^\mathrm{T}{\bf x}+{\bf b}^\mathrm{T}{\bf y}-\bf{\lambda} ^\mathrm{T}(A{\bf x}+B{\bf y}-{\bf c})-\bf{\mu} ^\mathrm{T}{\bf y},$$

and "differentiate" with respect to ${\bf x}$ and ${\bf y}$ to get

$${\bf a}^\mathrm{T}-\bf{\lambda}^\mathrm{T}A={\bf 0}\,\textrm{and}\,{\bf b}^\mathrm{T}-\bf{\lambda}^\mathrm{T}B-\bf{\mu}^\mathrm{T}={\bf 0}.$$

I know this has absolutely zero rigour, so what is the formal way to write this when dealing with all the vectors and matrices, especially with the inequality? Furthermore for the second part, how exactly do I go about it? I have no idea at all. All help is appreciated!

答:
The primal problem is $\min_{x,y} \max_{\lambda \in \mathbb{R}^k, \mu \geq 0} L(x,y,\lambda,\mu)$ (note that the value of $\max_{\mu \geq 0} L(x,y,\lambda,\mu)$ is $\infty$ when $y_i<0$ for some $y$, so the feasible region of the primal is $y \geq 0$; similarly you can get the equality constraint from $\max_{\lambda}$).

You can derive the Lagrange dual problem by swapping min and max and simplyfing $\min_{x,y} L(x,y,\lambda,\mu)$. Taking the derivative is a valid approach but it does not seem like you fully understand why. Since $L$ is linear in $x$, if the derivative is not $0$, $\min_{x,y} L(x,y,\lambda) = -\infty$. So:
$$\max_{\lambda \in \mathbb{R}^k, \mu \geq 0}\min_{x,y} L(x,y,\lambda) = \max_{\lambda \in \mathbb{R}^k, \mu \geq 0} \begin{cases}c^T\lambda & \text{if } A^T\lambda=a, \; B^T\lambda +\mu = b  \\
-\infty & \text{otherwise.}
\end{cases}$$
It is not clear why $\mu$ is missing from your dual. You can also write $B^T\lambda \leq b$.

What is not part of your question but what I think you still want is an argument why strong duality holds, i.e., why the dual problem has the same optimal value as the primal problem. You could use Slater's condition.


问:
Natural density extended to real density?
The natural density of the set $A \subseteq \mathbb{N}$ is defined as
$$\delta(A)=\lim_{n \rightarrow \infty}\frac{ \# \{k \in A | k \leq n \} }{ \# \{k \in \mathbb{N} | k \leq n \} }.$$
My idea was to generalize this to sets with greater cardinality:

For example take the set $B = \{x \in \mathbb{C} | \ \lvert x \rvert \leq \frac{1}{2} \}$ instead of $A$ and the unit disk $U = \{x \in \mathbb{C} | \ \lvert x \rvert \leq 1 \}$ instead of $\mathbb{N}$. Now we know by some easy geometrical calculations that the ratio is $\frac{1}{4}$. How do we get this by a concept similar to the natural density or other densities with countable sets?

Is there a density concept (with limits) for general sets, including uncountable ones like $\mathbb{R}$?

Thanks for your ideas, comments and answers!

答:
A density always requires a superset of the considered set. Let's denote the superset of the considered set $A$ in the following by $B$. To transfer your concept to any other countable set should be easy, since for each countable set $B$ exists a injective function $f:B\to\mathbb N$. In this case just consider $f(A)$ resp. $f(B)$ instead of $A$ resp. $\mathbb N$ and proccede as you already did
$$\delta_f(A):=\lim_{n \rightarrow \infty}\frac{ \# \{k \in f(A) | k \leq n \} }{ \# \{k \in f(B) | k \leq n \} }.
$$
I think that it is obvious that it's important to take the same function $f$ (in the numerator and the denominator) for both sets $A,B$ although a lot of such functions $f$ exit. This density is not independent of the choice of $f$. I will give an example for this at the end of this answer.

To transfer your concept to an uncountable set one has to analyze what you have done so far: You use the counting measure when you were dealing whith natural numbers. This means you have to choose a new measure on the new superset $B$ you are now dealing with. As you mentioned $\mathbb C\cong\mathbb R^2$ the Lebesgue measure $\lambda$ would be a 'natrual' choice, but you can choose of course different. I think it is obvious that the density depends on the measure. What you were doing with the limit is, you intersected both sets with an sequence $\mathcal C:=(C_n)_{n\in\mathbb N}$ of increasing sets $C_n$ with the properties
$$\forall n\in\mathbb N:C_n\subseteq C_{n+1}\tag{1}\\
B\subseteq\bigcup_{n\in\mathbb N}C_n.
$$ These properties are just important if $\lambda(B)=\infty$, as measures satisfy a continuity property, you wouldn't need any intersection in the case of $\lambda(B)<\infty$. Otherwise you don't need this sequence. In your special case $A=\{x \in \mathbb{C} | \ \lvert x \rvert \leq \frac{1}{2} \}$ and $B= \{x \in \mathbb{C} | \ \lvert x \rvert \leq 1 \}$ you can directly use (with the Lebesgue measure)
$$\delta_{B,\lambda}(A)=\frac{\lambda(A)}{\lambda(B)}=\frac{\pi/4}{\pi}=\frac{1}{4}
.$$
In the general case this means
$$\delta_{B,\lambda,\mathcal C}(A):=\lim_{n \rightarrow \infty}\frac{\lambda(A\cap C_n)}{\lambda(B\cap C_n)}
.$$
In the space $\mathbb R^d$ you could use for example the balls with radius $n$ as $C_n=\{x\in\mathbb R^d: ||x||_2\leq n\}$. This density is in general not independent of the choice of the sequence $\mathcal C=(C_n)_{n\in\mathbb N}$ although measures satisfy a continuity property. I will give an example for this at the end of this answer.

Back to my first example above, this concept means you define a counting measure on the set $B$ by $\lambda(B):=\#f(B)$ (or directly $\lambda(B):=\# B$, as it is the counting measure where all elements have the same weight and $f$ is injektive, it makes no difference) and $\mathcal C_f=(C_n)_{n\in\mathbb N}$ with $C_n:=\{b\in B: f(b)\le n\}$:
$$\delta_{B,\lambda,\mathcal C_{f}}(A):=\lim_{n \rightarrow \infty}\frac{\lambda(A\cap C_n)}{\lambda(B\cap C_n)}=\lim_{n \rightarrow \infty}\frac{ \# \{k \in f(A) | k \leq n \} }{ \# \{k \in f(B) | k \leq n \} }=\delta_f(A).
$$
From this you see that sequence of the sets $C_n$ is affected by the choice of $f$. This is the reason why the density $\delta_f=\delta_{B,\lambda,\mathcal C_{f}}$ depends on $f$. To give a simple example consider the set $B=\mathbb N$ and the two different choices $f_1=\operatorname{Id}_{\mathbb N}$ and $f_2$, whereas $f_2$ takes always the lowest two odd numbers mapping to the next two natural numbers then take the next lowest even number mapping to the next natural number and so on, i.e. $$f_2:B\to\mathbb N:(1,3,\color{green}{2},5,7,\color{green}{4},9,11,\color{green}{6},\dots)\mapsto(1,2,\color{green}{3},4,5,\color{green}{6},7,8,\color{green}{9},\dots).$$
Now consider the set $A$ consisting of all even natural numbers. I think it's obvious to see that $$\delta_{f_1}(A)=\delta_{B,\lambda,\mathcal C_{f_1}}(A)=\frac{1}{2}\\\text{and }\delta_{f_2}(A)=\delta_{B,\lambda,\mathcal C_{f_2}}(A)=\frac{1}{3}.$$ So this was the example of which you see that the density $\delta_f$ of the beginnig is dependent on the function $f$ and the general density $\delta_{B,\lambda,\mathcal C}$ is depnendent on the sequence $\mathcal C$, which is not obvisous in the first view. The other dependencies $B,\lambda$ are easier to see.


问:
Why do people use "K" to represent a complete graph?
Why do people use the letter "K", rather than "C", to represent a complete graph? Does it come from German "komplett"?</p>

答:
A search in Wikipedia gives:

The complete graph on $n$ vertices is denoted by $K_n$.  Some sources claim that the letter K in this notation stands for the German word komplett, but the German name for a complete graph, vollständiger Graph, does not contain the letter K, and other sources state that the notation honors the contributions of Kazimierz Kuratowski to graph theory.

ContentLicense: CC BY-SA 3.0


问:
Is $\det(X^T A X)/\det(X^T B X) = \det(A)/\det(B)$?
Let's assume $A,B$ are symmetric, positive definite real $n \times n$ matrices. Does it hold that
$$\frac{\det(X^T A X)}{\det(X^T B X)} = \frac{\det(A)}{\det(B)}$$ for all $X$ real $n \times p$ matrices for which the denominator on the left is nonzero?

答: Not true. Take $A=I_{2\times 2}, B=2I_{2\times 2}$ and $X=(1,2)^T$.


问:
Is Bayes' Theorem really that interesting?
I have trouble understanding the massive importance that is afforded to Bayes' theorem in undergraduate courses in probability and popular science.

From the purely mathematical point of view, I think it would be uncontroversial to say that Bayes' theorem does not amount to a particularly sophisticated result. Indeed, the relation
$$P(A|B)=\frac{P(A\cap B)}{P(B)}=\frac{P(B\cap A)P(A)}{P(B)P(A)}=\frac{P(B|A)P(A)}{P(B)}$$
is a one line proof that follows from expanding both sides directly from the definition of conditional probability. Thus, I expect that what people find interesting about Bayes' theorem has to do with its practical applications or implications. However, even in those cases I find the typical examples being used as a justification of this to be a bit artificial.

To illustrate this, the classical application of Bayes' theorem usually goes something like this: Suppose that

1% of women have breast cancer;

80% of mammograms are positive when breast cancer is present; and

10% of mammograms are positive when breast cancer is not present.

If a woman has a positive mammogram, then what is the probability that she has breast cancer?

I understand that Bayes' theorem allows to compute the desired probability with the given information, and that this probability is counterintuitively low. However, I can't help but feel that the premise of this question is wholly artificial. The only reason why we need to use Bayes' theorem here is that the full information with which the other probabilities (i.e., 1% have cancer, 80% true positive, etc.) have been computed is not provided to us. If we have access to the sample data with which these probabilities were computed, then we can directly find
$$P(\text{cancer}|\text{positive test})=\frac{\text{number of women with cancer and positive test}}{\text{number of women with positive test}}.$$
In mathematical terms, if you know how to compute $P(B|A)$, $P(A)$, and $P(B)$, then this means that you know how to compute $P(A\cap B)$ and $P(B)$, in which case you already have your answer.

From the above arguments, it seems to me that Bayes' theorem is essentially only useful for the following reasons:

In an adversarial context, i.e., someone who has access to the data only tells you about $P(B|A)$ when $P(A|B)$ is actually the quantity that is relevant to your interests, hoping that you will get confused and will not notice.

An opportunity to dispel the confusion between $P(A|B)$ and $P(B|A)$ with concrete examples, and to explain that these are very different when the ratio between $P(A)$ and $P(B)$ deviates significantly from one.

Am I missing something big about the usefulness of Bayes' theorem? In light of point 2., especially, I don't understand why Bayes' theorem stands out so much compared to, say, the Borel-Kolmogorov paradox, or the "paradox" that $P[X=x]=0$ when $X$ is a continuous random variable, etc.

答:
There are two main issues here. One is that on a Bayesian interpretation of probability (this term doesn't reference the theorem, but they're both named for Bayes), probability quantifies how well we know individual events, not detailed available frequency statistics. The best-of-both-worlds hope, if you combine Bayesian and frequentist perspectives, is that past data give us the mammogram values you cited, and an individual woman can be diagnosed based on Bayes's theorem.

The second issue is that $P(A|B)$ need not be remotely close to $P(B|A)$. To wit:

A test that's usually right may still have most of its positives be false, which warrants some scepticism, as well as further testing.

Conflating $P(A|B)$ with $P(B|A)$ is a danger in the legal system. Will we arrest people based on accuracy, precision etc., even if their guilt is unlikely? Will "this evidence is unlikely if they're innocent" get them convicted, even though it may not mean their innocence is unlikely? And yes, this has had real-world fallout in both policing and court decisions.

Statistics tests what probability assumes (e.g. "if this is Gaussian then..."). Statistical tests often boil down to, "we can't measure the probability the null hypothesis is true, but we'll assess it based on the probbaility on the null hypothesis that data at least this surprising would occur". Indeed, which statement gets to be the null hypothesis is more about its facilitating such calculations than its being a "default" or "reasonable" assumption.


问:
Show $\cos\frac\gamma2=\sqrt{\frac{p(p-c)}{ab}}$

Show that the equality $$\cos\dfrac{\gamma}{2}=\sqrt{\dfrac{p(p-c)}{ab}}$$ holds for a $\triangle ABC$ with sides $AB=c,BC=a, AC=b$, semi-perimeter $p$ and $\measuredangle ACB=\gamma$.

I have just proved that $$l_c=\dfrac{2ab\cos\dfrac{\gamma}{2}}{a+b}$$

Is this a well known formula for angle bisectors?

Can we use it somehow?

Thank you in advance!

答:
Your formula for length of angle bisectors is correct. To derive for sine/cosine of half-angles, a straightforward way is using the formulas : $$1+\cos 2\theta = 2\cos^2 \theta \, , \quad 1-\cos 2\theta = 2\sin^2 \theta$$

By cosine-rule in $\triangle ABC$,
$$\cos \gamma = \frac{a^2+b^2-c^2}{2ab}$$

$$\Rightarrow \cos \frac{\gamma}{2}=\sqrt{\frac{1+\cos \gamma}{2}}$$
which simplifies to

$$ \cos \frac{\gamma}{2}=\sqrt{\frac{p(p-c)}{ab}}$$

Similarly you can also find $\sin (\gamma/2)$, $\tan (\gamma/2)$ etc.


问:
Combinatorial proof of a Stirling number identity.
Consider the identity
$$\sum_{k=0}^n (-1)^kk!{n \brace k} = (-1)^n$$
where ${n\brace k}$ is a Stirling number of the second kind. This is slightly reminiscent of the binomial identity
$$\sum_{k=0}^n(-1)^k\binom{n}{k} = 0$$
which essentially states that the number of even subsets of a set is equal to the number of odd subsets.

Now there is an easy proof of the binomial identity using symmetric differences to biject between even and odd subsets. I am wondering if there is an analogous combinatorial interpretation for the Stirling numbers. The term $k!{n\brace k}$ counts the number of set partitions of an $n$ element set into $k$ ordered parts. Perhaps there is something relating odd ordered partitions with even ordered partitions?

As an added note, there is a similar identity
$$\sum_{k=1}^n(-1)^k(k-1)!{n\brace k}=0$$
for $n \geq 2$.
A combinatorial interpretation of this one would also be appreciated.

答:
These are not combinatorial interpretations, but they are simple.

The defining equation for Stirling numbers of the second kind is
$$
\sum_{k=0}^n\begin{Bmatrix}n\\k\end{Bmatrix}\binom{x}{k}k!=x^n\tag{1}
$$
That is, Stirling numbers of the second kind tell how to write monomials as a combination of falling factorials (or combinatorial polynomials).

Noting that $\displaystyle\binom{-1}{k}=(-1)^k$ and setting $x=-1$ yields
$$
\begin{align}
\sum_{k=0}^n\begin{Bmatrix}n\\k\end{Bmatrix}(-1)^kk!=
\sum_{k=0}^n\begin{Bmatrix}n\\k\end{Bmatrix}\binom{-1}{k}k!=(-1)^n
\end{align}
$$

Since $\displaystyle\binom{x}{k}=\binom{x-1}{k-1}\frac{x}{k}$ and $\begin{Bmatrix}n\\0\end{Bmatrix}=0$ for $n\ge1$, we can rewrite $(1)$ as
$$
\sum_{k=1}^n\begin{Bmatrix}n\\k\end{Bmatrix}\binom{x-1}{k-1}(k-1)!=x^{n-1}\tag{2}
$$
Setting $x=0$ yields
$$
\sum_{k=1}^n\begin{Bmatrix}n\\k\end{Bmatrix}(-1)^{k-1}(k-1)!=0^{n-1}
$$
where $0^0=1$ for the case $n=1$.


答:
Form half-angle trig-identity

$$\cos\gamma=2\cos^2\frac{\gamma}{2}-1\iff \cos \frac{\gamma}{2}=\sqrt{\frac{1+\cos\gamma}{2}}$$

Substituting the value from cosine formula: $\cos \gamma=\frac{a^2+b^2-c^2}{2ab}$
$$\cos \frac{\gamma}{2}=\sqrt{\frac{1+\frac{a^2+b^2-c^2}{2ab}}{2}}$$
$$=\sqrt{\frac{(a+b)^2-c^2}{4ab}}$$

$$=\sqrt{\frac{(a+b+c)(a+b-c)}{4ab}}$$
$$=\sqrt{\frac{\left(\frac{a+b+c}{2}\right)\left(\frac{a+b+c}{2}-c\right)}{ab}}$$
$$=\sqrt{\frac{p(p-c)}{ab}}$$

Title: Show $\cos\frac\gamma2=\sqrt{\frac{p(p-c)}{ab}}$

Body:
Show that the equality $$\cos\dfrac{\gamma}{2}=\sqrt{\dfrac{p(p-c)}{ab}}$$ holds for a $\triangle ABC$ with sides $AB=c,BC=a, AC=b$, semi-perimeter $p$ and $\measuredangle ACB=\gamma$.

I have just proved that $$l_c=\dfrac{2ab\cos\dfrac{\gamma}{2}}{a+b}$$

Is this a well known formula for angle bisectors?

Can we use it somehow?

Thank you in advance!


答:
We have standard formula for $R = {abc\over 4S}$ and $S = rp$, where $S$ is area and $p$ perimeter of triangle.

$\cot {\gamma \over 2} = {p-c\over r}$

$\sin \gamma  = {c\over 2R}$ (the law of sine)

If we multiply these two we get $$\cos ^2 {\gamma\over 2} = {S(p-c)\over rab} = {p(p-c)\over ab}$$
(Remember that $\sin \gamma = 2\sin{\gamma\over 2}\cos {\gamma \over 2}$)


问:
What can we conclude about the convergence and uniform convergence of an infinite series, if it's radius of convergence is infinite?

Question: What can we conclude about the convergence and uniform convergence of an infinite series, if it's radius of convergence is infinite ?

For example: Let us consider an infinite series $~\sum_{n=1}^\infty (x/n)^n~$.
We have to conclude the decision for $x\in(-\pi,\pi)$.

Radius of convergence of the series is $$\dfrac 1R=\lim_{n\to\infty}\dfrac{n^n}{(n+1)^{n+1}}=\lim_{n\to\infty}\dfrac{1}{(1+1/n)^n}\cdot\dfrac{1}{n+1}=0\implies R=\infty$$
So here radius of convergence is infinite.

Now wikipedia says, "The radius of convergence is infinite if the series converges for all complex numbers $z$." So following this statement we can say that the series is convergent.
Is it okay ? What about the uniform convergence ?

答: For a power series, let $I$ be the interior of the domain of convergence. Then the series converges uniformly on compact subsets of $I$. This is true even when the domain of convergence is the whole space.


问:
Integration of $\int_0^{\infty}\frac{e^{(\lambda +is)t}-e^{-(\lambda +is)t}}{e^{\pi t}-e^{-\pi t}}dt$
Integration of $$\int_0^{\infty}\frac{e^{(\lambda +is)t}-e^{-(\lambda +is)t}}{e^{\pi t}-e^{-\pi t}}dt$$

I am thinking about this integral from last half and hour but still can't figure it out how to solve this. what i was think is to make this in form

$$\int_{0}^{\infty}\frac{\sinh(\lambda+is)t}{\sinh (\pi t)} dt$$ but still this will get me nowhere. please just give me a hint how to proceed this.

Thank you.

答:
Considering $\displaystyle \mathcal{I}(\alpha, \beta) = \int_{0}^{\infty} \frac{\sinh\alpha z}{\sinh \beta z}\,\mathrm{d}z$, we show that $\displaystyle \mathcal{I}(\alpha, \beta) = \frac{\pi}{2\beta} \tan\left(\frac{\alpha \pi}{2\beta}\right)$.

Then your integral is $\displaystyle \mathcal I(\lambda+is, \pi)$. Starting with   $z \mapsto \frac{1}{\beta} \log(z)$, we have:

\begin{aligned} \displaystyle \mathcal{I}(\alpha, \beta) & = \frac{1}{\beta} \int_0^1 \frac{z^{\frac{\alpha}{\beta}} -z^{-\frac{\alpha}{\beta}}}{z^2-1}\;{dz} \\& = \frac{1}{\beta} \int_0^{1} \left( z^{-\frac{\alpha}{\beta}}-z^{\frac{\alpha}{\beta}}\right) \sum_{k \ge 0}z^{2k} \;{dz}\\& = \frac{1}{\beta} \sum_{k \ge 0}\int_0^1 z^{-\frac{\alpha}{\beta}+2k} -z^{\frac{\alpha}{\beta}+2k}\;{dz} \\& = \frac{1}{\beta} \sum_{k \ge 0} \frac{1}{1-\frac{\alpha}{\beta}+2k}-\frac{1}{1+\frac{\alpha}{\beta}+2k} \\& = \frac{1}{\beta} \sum_{k \ge 0} \frac{2\alpha /\beta }{(1+2 k)^2-(\alpha/\beta)^2}  \\& = \frac{1}{2\beta} \sum_{k ~ \text{odd} } \frac{4\alpha/\beta}{k^2-(\alpha/\beta)^2} \end{aligned}

Consider the series $\displaystyle \displaystyle \pi\cot(\pi z) = \frac{1}{z}+ \sum_{n \ge 1} \frac{2z}{z^2-n^2} $. Map $\displaystyle z \mapsto \frac{z}{2} $ and divide by two then we have:
$\displaystyle \displaystyle \frac{\pi}{2}\cot\left(\frac{\pi}{2} z\right) = \frac{1}{z}+\frac{1}{2}\sum_{n \ge 1} \frac{z}{(z/2)^2-n^2} $$\displaystyle \displaystyle =  \frac{1}{z}+ \frac{1}{2}\sum_{n \ge 1} \frac{4z}{z^2-(2n)^2} = \frac{1}{z}+\frac{1}{2}\sum_{n ~ \text{even}} \frac{4z}{z^2-n^2}$

Thus $\displaystyle \displaystyle \frac{1}{z}+\sum_{n \ge 1} \frac{2z}{z^2-n^2} -\frac{1}{z}-\frac{1}{2} \sum_{n ~ \text{even}} \frac{4z}{z^2-n^2}$ $\displaystyle \displaystyle  =\pi\cot(\pi z)-\frac{\pi}{2}\cot\left(\frac{\pi}{2} z\right) = -\frac{\pi}{2} \tan\left(\frac{z\pi}{2}\right)$

Hence  $\displaystyle \displaystyle \frac{1}{2} \sum_{n \textrm{ odd}} \frac{4z}{z^2-n^2} = -\frac{\pi}{2}\tan\left(\frac{\pi z}{2}\right)$ and so $\displaystyle  \displaystyle  \mathcal I(\alpha, \beta)=\frac{1}{2\beta} \sum_{k ~ \text{odd} } \frac{4\alpha/\beta}{(\alpha/\beta)^2-k^2} = \frac{\pi}{2\beta} \tan\left(\frac{\alpha \pi}{2\beta}\right).$


问:
Topological conjugacy between linear maps in dimension $1$
Let $A_{\alpha}: \mathbb{R} \to \mathbb{R}$ denote the linear map

\begin{align}
A_{\alpha}(x)=\alpha x.
\end{align}

By definition two homeomorphisms $f:X \to X$ and $g:X \to X $ are topologically conjugate to each other if there is a homeomorphism $h:X \to X $ such that $hf=gh$.

I wish to prove that if $ 0 < \alpha < 1$ and $ 0 < \beta < 1 $ then $A_{\alpha}$ and $A_{\beta}$ are topologically conjugate. Could anyone help me find this homeomorphism?

答:
Although it is "overkill" let us show how the more general result Any two orientation-preserving homeomorphisms of $[a,b]$ without fixed point in$(a,b)$ are topologically conjugate can be used.

Let us consider the maps $B_\alpha : [0,\infty) \to [0,\infty), B_\alpha(x) = \alpha x$. It suffices to show that all of them are conjugate:

In fact, each homeomorphism $h : [0,\infty) \to [0,\infty)$ must keep $0$ fixed because otherwise $h$ would map the connected set $(0,\infty)$ onto the non-connected set $[0,\infty) \setminus \{h(0)\}$. Thus $h$ has the odd extension $h' : \mathbb R \to \mathbb R$ given by $h'(x) = h(x)$ for $x \ge 0$ and $h'(x) = -h(-x)$ for $x \le 0$ which is a well-defined homeomorphism. Note that

$B'_\alpha = A_\alpha$.

$(h_1h_2)' = h'_1h'_2$.

Thus, given $h$ such that $h B_\alpha = B_\beta h$, then $h' A_\alpha = h' B'_\alpha = (h B_\alpha)' = (B_\beta h)' = B'_\beta h'  = A_\beta h'$.

Now define $\phi : [0,\infty) \to [0,1), \phi(t) = \dfrac{t}{1+t}$. This is a homeomorphism with inverse $\phi^{-1}(t)  = \dfrac{t}{1-t}$. This reduces our task to show that all $C_\alpha = \phi A_\alpha \phi^{-1} : [0,1) \to [0,1)$ are conjugate.

We have $C_\alpha(x) \to 1$ as $x \to 1$, thus $C_\alpha$ extends to a homeomorphism $D_\alpha : [0,1] \to [0,1]$ which keeps the boundary points $0,1$ fixed and therefore is orientation preserving. Moreover $D_\alpha$ has no fixed point in $(0,1)$: If $x \in (0,1)$ satisfies $D_\alpha(x) = C_\alpha(x) = x$, then $y = \phi^{-1}(x) \in (0,\infty)$ and $\alpha y = A_\alpha(\phi^{-1}(x)) = \phi^{-1}(x) =y$ which is impossible for $\alpha < 1$.

By the general result $D_\alpha$ and $D_\beta$ are conjugate and therefore also $C_\alpha$ and $C_\beta$ are conjugate.


问:
Proving the sums of three cubes conjecture by the Hasse principle
In his Cours d'arithmétique Serre applies the Hasse-Minkowski theorem to quadratic forms of the form:
$$
x^2 + y^2 + z^2 = n
$$
for $n \in \mathbb{N}$ to prove that a natural number $n$ is a square if and only if $-n$ is not a square in $\mathbb{Q}_2$, and hence that a natural number can be expressed as the sum of three squares provided that it is not of the form $4^a(8b-1)$ for some $a\in \mathbb{N}_0$, $b \in \mathbb{N}$.

My question is about if one might prove the sums of three cubes conjecture by analogous means.

Now we know that the Hasse principle does not hold for a general cubic forms, if we knew that it holds for cubic forms of the form:
$$
x^3 + y^3 + z^3 = n
$$
for some $n \in \mathbb{N}$, could we reduce the proof of the sum of three cubes conjecture to a local statement, analogous to the above.

Has any work been done to prove the sum of the cubes conjecture by these means?

Is the Hasse principle known to hold for cubic forms of the above form?

I suppose this would naturally entail the introduction of something like an analogy of the Hilbert symbol in the cubic case.

Many thanks.

Edit: This would naturally not work if the Hasse principle were known to fail for a cubic form:
$$
x^3 + y^3 + z^3 = n
$$
for some $n \in \mathbb{N}$, so a natural way of dismissing the above proof strategy would be to provide such a counter-example. - Is any such counter-example known to exist?

答:
I don't think that the Hasse principle is useful here. First, Hasse-Minkowski fails in general for cubic polynomials, e.g., for $3x^3+4y^3+5z^3$. For $x^3+y^3+z^3=n$ I don't see how to use it.

Secondly, Ramanujan has used generating functions to obtain parametrised solutions for $x^3+y^3+z^3=1$ and $x^3+y^3+z^3=2$, and this direction appears to be more interesting, see for example the article by Harper.

In $1992$, Roger Heath-Brown conjectured that every
$n$ unequal to $4$ or $5$ modulo $9$ has infinitely many representations as sums of three cubes.Conversely it easy to see that the other $n$ cannot be a sum of three cubes. There are many computational aspects, too, which made the problem popular for $n=33$ and $n=42$.


问:
How do you show that $(x-y,y^2)$ is primary in $k[x,y,z]$?
I know that $Rad((x-y,y^2))=(x,y)$ which is prime but I can’t conclude that the original ideal is primary. Any quick methods?

答:
Here's a nice criterion for being primary. (See Dummit and Foote, $\S15.2$ Proposition $19$ for a reference.)

Proposition. Let $R$ be a commutative ring and $Q \trianglelefteq R$ be an ideal. Then $Q$ is primary iff every zero divisor in $R/Q$ is nilpotent.

Applying this to your problem, first note that
\begin{align*}
\frac{R}{Q} = \frac{k[x,y,z]}{(x-y,y^2)} \cong \frac{k[y,z]}{(y^2)} \, .
\end{align*}
Suppose $0 \neq \overline{a} \in R/Q$ is a zero divisor, so $\overline{a}\overline{b} = 0$ for some $\overline{b} \neq 0$. Then $ab \in (y^2)$ but $a \notin (y^2)$ and $b \notin (y^2)$, so $y^2 \mid ab$ but $y^2 \nmid a$ and $y^2 \nmid b$. Thus we must have $y \mid a$ and $y \mid b$. Then $a = y a'$ for some $a' \in R$, so $\overline{a}^2 = (\overline{y a'})^2 = 0$.


问:
Numerical method to solve a non-linear ODE
I want to solve numerically the following non-linear ordinary differential equation:
$$f''(x)=A(1+f'(x)^2)^{3/2}-\frac{f'(x)}{x}(1+f'(x)^2)$$
where $A$ is a constant and $x\in[0,R]$. The ODE is also subjected to the following boundary conditions:
$$f'(0)=0$$
$$f(R)=0$$
My struggle is with what method should I adopt to approximate the solution due to its high non-linearity. Thanks in advance.

答:
So, I adopted a finite difference approximation with a Newton-Rapson scheme to solve the non-linear system and it worked just fine. The singularity at $x=0$ ended up not being a problem since my system of nonlinear equations was:
$$
\left\{\begin{array}{lcr}
f_2-f_1 = 0 \\
\frac{f_{i+1}-2f_{i}+f_{i-1}}{h^2} - \frac{\gamma}{C}\left(\frac{f_{i+1}^2 -2f_{i+1}f_{i-1}+f_{i-1}^2+4h^2}{4h^2}\right)^{3/2} + \frac{\left(f_{i+1}-f_{i-1}\right)\left(f_{i+1}^2 -2f_{i+1}f_{i-1}+f_{i-1}^2+4h^2\right)}{8h^4(i-1)}\quad\forall 2 \le i\le N-1\\
f_N=0
\end{array}\right.
$$
Thank you all for the replies!


问:
If sequences $x_n$ and $y_n$ in a metric space converge to the same limit, does $d(x_n, y_n)$ converge?
$(x_n)$ and $(y_n)$ are sequences in a metric space $M$ with $\lim_{n\to\infty} x_n = \lim_{n\to\infty} y_n = a$. Show that given $\varepsilon>0$ there exists $n_0$ such that $d(x_n, y_n)<\varepsilon$ $\forall n≥n_0$

Hello, this is from my course in general topology. I know I'm supposed to use the definition of convergence for $x_n$ and $y_n$ and the triangle inequality, but I can't seem to get it right. Any help?

答:
Let $\varepsilon > 0.$

Since we know that $x_n \to L$, we know that there exists $N_x$ such that when $n>N_x,$ then $d(x_n,L)<\frac{\varepsilon}{2}$.

Likewise, since $y_n \to L$, we know that there exists $N_y$ such that when $n>N_y$, $ $ $d(y_n,L)< \frac{\varepsilon}{2}.$

Set $N =$max$\{N_x,N_y\}$.

Then when $n >N,$ by the Triangle Inequality: $$d(x_n,y_n) \leq d(x_n,L) + d(y_n,L) = \frac{\varepsilon}{2} + \frac{\varepsilon}{2} = \varepsilon.$$

Thus, we have shown that if $x_n \to L$ and $y_n \to L,$ $d(x_n,y_n) \to 0.$ $\square$


问:
Combinatorial proof of ${2n \choose 3} = 2 \times {n \choose 3} + 2 \times {n \choose 1}{n\choose 2}$
I need to prove ${2n \choose 3} = 2 \times {n \choose 3} + 2 \times {n \choose 1}{n\choose 2}$ forming my
counting problem as such:

Suppose that we are trying to count the number of ways
to form a sports team of (insert the correct number here) students picked from (complete the rest of
this sentence, being very precise/specific).

I am entirely new to binomials and any help for where to begin or general guidance would be appreciated.

答:
Say you have total of $2n$ players of which $n$ are experienced and $n$ are new.

You are making a team of $3$ pplayers by choosing $3$ out of $2n$ players - which is $2n \choose 3$.

The selection of $3$ will contain either i) All $3$ experienced ii) All $3$ new iii) $2$ new, $1$ experienced and iv) $1$ new and $2$ experienced.

If we add selection for all of these cases, it should equal to $2n \choose 3$.

So ${2n \choose 3} = {n \choose 3} + {n \choose 3} + {n \choose 1} {n \choose 2} + {n \choose 2} {n \choose 1} = 2 {n \choose 3} + 2 {n \choose 1} {n \choose 2}$


问:
The limit of the product $\prod_{i=1}^n\frac{1 - (2i + 1)a/(2n)}{1 - ia/n}$ as $n\to\infty$
Given that $0 < a < 1$, what is the value of
$$ P = \lim_{n\to \infty} \prod_{i=1}^n \frac{1 - (2i + 1)a/(2n)}{1 - ia/n} $$
Thus, $P_n$, the $n$th term, is
$$ P_n = \frac{1-3a/2n}{1-a/n}\cdot \frac{1-5a/2n}{1-2a/n}\cdots \frac{1-(2n+1)a/2n}{1-a} $$

Motivation

The product describes a real-live physical problem. Put simply, if you have a river or stream in which you're sampling at an upstream site A and a downstream site B and you analyze the chemistry of the stream at A and B, if the concentration has changed, it is because of ground water which has seeped in between A and B. By making some simplifying assumptions, the concentration and amount of ground water which has entered are calculated using the evaluation of this infinite product.

答:
Take logs of both sides to get

$$\log{P_n} = \sum_{i=1}^n \left(\log{\left [ 1- a\frac{i}{n} - \frac{a}{2 n}\right]} -\log{\left [ 1- a\frac{i}{n}\right]} \right) $$

Noting that $n$ is large, so we can rewrite the summand as

$$\log{\left [ 1- a\frac{i}{n} - \frac{a}{2 n}\right]} -\log{\left [ 1- a\frac{i}{n}\right]} = \log{\left( 1-\frac{a/(2 n)}{1-a i/n}\right)} \sim -\frac{a}{2} \frac{1}{n} \frac{1}{1-a i/n} $$

In the limit as $n \to \infty$, this is a Riemann sum and becomes the integral

$$\log{P} = -\frac{a}{2} \int_0^1 \frac{dx}{1-a x} = \frac12 \log{(1-a)}$$

Therefore

$$P=\sqrt{1-a}$$

BONUS

Here are a few plots of log base $2$ of the relative error between the above result $P$ and the finite product $P_n$ vs $\log_2{n}$.  The values of $a$ from left to right are $1/4$, $1/3$, $2/3$, $9/10$, and $99/100$.

EDIT

@Did has pointed out that the step in the second line needs more justification, i.e., a quantification of the error term inherent in the $\sim$ symbol.  Along these lines, for sufficiently large $n$, we may write

$$\log{\left( 1-\frac{a/(2 n)}{1-a i/n}\right)} = -\frac{a}{2} \frac{1}{n} \frac{1}{1-a i/n} + \frac12 \left ( \frac{a}{2} \frac{1}{n} \frac{1}{1-a i/n} \right )^2 + \cdots$$

As we did for the first term on the RHS, we may also sum over $i$ for the subsequent terms.  For example, the first error term has behavior, for large $n$

$$\frac{a^2}{8 n}\; \frac1n \sum_{i=1}^n \frac1{(1-a i/n)^2} $$

which, in the limit as $n\to\infty$, is another Riemann sum; the term therefore behaves as

$$\frac{a^2}{8(1-a)} \frac1n$$

It should be clear that higher order terms in the log behave as higher orders of $1/n$.  Thus, the error terms arranged in this way do not contribute to the sum for sufficiently large $n$.


答:
Say we have red and blue line, each containig $n$ points. How many triples can we choose?

First answer is ${2n\choose 3}$

Second we can choose 3 points from one line so that is ${n\choose 3}$ triples times 2 and we can choose 2 points fome one line and 1 point from other, so that is ${n\choose 2}{n\choose 1}$ times 2.

Title: Combinatorial proof of ${2n \choose 3} = 2 \times {n \choose 3} + 2 \times {n \choose 1}{n\choose 2}$

Body:
I need to prove ${2n \choose 3} = 2 \times {n \choose 3} + 2 \times {n \choose 1}{n\choose 2}$ forming my
counting problem as such:

Suppose that we are trying to count the number of ways
to form a sports team of (insert the correct number here) students picked from (complete the rest of
this sentence, being very precise/specific).

I am entirely new to binomials and any help for where to begin or general guidance would be appreciated.


问:
solve non-convex quadratic constrained quadratic programming
$\min_{\beta}\beta^{T} A \beta$

$s.t. \ \beta^{T} C \beta=1\ and\ \beta\geqslant 0$

Here $A,C\in \mathbb{R}^{M\times M}$, $\beta \in  \mathbb{R}^{M}$

I saw in one paper saying that it could be solved via its semidefinite programming relaxation by adding an auxiliary variable $B \in \mathbb{R}^{M \times M}$ like this:

$\min_{\beta ,B}trace(AB)$

$s.t.trace(CB)=1$,

$\beta \geqslant 0$,

$\begin{bmatrix}
1 & \beta^{T}\\\\
 \beta& B
\end{bmatrix}\succeq 0$

where $\succeq 0$ means left matrix is positive semidefinite.

I don't get how this is done, and besides, how to solve such a problem using any possible C/C++ software?

Thanks. $;)$

答:
If you set $B = \beta \beta^T$,
the objective becomes
$$ \beta^T A \beta =
  \text{trace}( \beta^T A \beta ) =
  \text{trace}( A \beta \beta^T ) =
  \text{trace}( A B ). $$

Similarly, the constraint becomes $\text{trace}(CB) = 1$.

The problem is therefore equivalent to

\begin{array}{ll}
\text{Find}\ &\beta, B \\
\text{to minimize}\  &\text{trace}(AB) \\
\text{such that}\  &\text{trace}(CB) = 1 \\
& \beta \geqslant 0 \\
& B = \beta \beta^T
\end{array}

The last constraint implies that
$\begin{pmatrix} 1 & \beta^T \\ \beta & B \end{pmatrix} =
\begin{pmatrix} 1 \\ \beta \end{pmatrix} \begin{pmatrix} 1 \\ \beta \end{pmatrix}^T
\succcurlyeq 0$, hence the convex relaxation you mention.
(A relaxation is not an equivalent problem: you have dropped some of the constraints
to make the problem easier to solve.)

For a C/C++ implementation, you can check the coin-or project.


问:
Simplification of permutation $(1 2 3 4) (4 5 6 7)$
Basic problem,

I'm confused on how to simplify this permutation into disjoint permutations?

$(1 2 3 4) (4 5 6 7)$

I am struggling between $(1 2 3 4 5 6 7)$ and $(1 2 3 5 6 7 4)$?

Thanks!

答:
If you decompose these two permutations into product of transpositions, you will get:
$$(1234)(4567)=(12)(23)(34)(45)(56)(67),$$
so from here you can notice that this is the decomposition into transpositions of the permutation $(1234567)$, so we conclude that
$$(1234)(4567)=(1234567).$$

If you want to calculate it directly, you have to take a look to where does each number go (doing the permutations from right to left, so the first arrow points to the number after the right permutation and the second points to the number after the left permutation):
$$ 1 \to 1\to 2$$
$$ 2 \to 2\to 3$$
$$ 3 \to 3\to 4$$
$$ 4 \to 5\to 5$$
$$ 5 \to 6\to 6$$
$$ 6 \to 7\to 7$$
$$ 7 \to 4\to 1$$
And, as you see, this gives us the same permutation result: $(1234567)$.


问:
Is $x^3$ a possible solution for a "monic" second order ODE?

Is it possible to find two continuous function $p(x), q(x)$ on $\mathbb{R}$ such that $y=x^3$ is a solution to $$y'' + p(x)y' + q(x)y = 0 \quad ?$$

We want to find $p,q$ such that

$$
6x + 3x^2p(x)+ x^3q(x)= 0
$$
If $p$ is any polynomial then it is impossible, and if $p,q$ is continuously differentiable then by using power serieses it is also impossible. So $p,q$ must be merely continuous (on my suspection) if this is true. I strongly doubt it is true, but I cannot give a proof on this.

答:
More generally, if $p$, $q$ are continuous functions, the differential equation
$y'' + p(x) y' + q(x) y = 0$ satisfies the hypotheses of the existence and uniqueness theorem, so for any initial condition $y(x_0) = y_0$, $y'(x_0) = y'_0$ there is a
unique solution defined in a neighbourhood of $x_0$ (in fact it turns out that the solution is defined globally, but that's not relevant here).  So from the fact that $y = 0$ is a solution that satisfies the initial condition $y(0)=0$, $y'(0)=0$ we can conclude that $y = x^3$, which would satisfy the same initial condition, can't be a solution.


问:
Consider the sequence of functions and compute the limit of the integral
Consider the sequence of functions $f_n:[0,1] \to R$ given by

$$f_n(x) = \frac{e^{-(n + x^2)}}{n^2 + x}$$
for $x \in [0,1]$ and $n \geq 1$

Compute $$\lim_{n \to \infty} \int_{0}^{1} f_n(x) dx$$

Maybe I'm overthinking this, but it seems like all I have to do is compute the integral of $f_n(x)$ and then take the limit. However, I am wondering if there is more to it since I can't seem to find an easy way to compute the integral. Can anyone lend me a hand?

Thanks!

答:
I am going to give a solution avoiding any reference to Lebesgue.

First, observe that $f_n(x)\ge 0$ for all $n$ and $x\in[0,1]$, so $$\int_0^1f_n(x)dx\ge 0.$$

Secondly, $f_n(x)\le \frac{1}{n^2+x}$ (observe that $n+x^2>0$ so $e^{-(n+x^2)}\le e^0=1$), so
$$\int_0^1f_n(x)dx\le \int_0^1 \frac{1}{n^2+x}dx=\log|n^2+x|\bigg|_0^1=\log(n^2+1)-\log(n^2)=\log(1+1/n^2).$$

Putting al together, $$0\le\int_0^1f_n(x)dx\le\log(1+1/n^2).$$

But $\lim_{n}\log(1+1/n^2)=0$.
So the squeeze theorem assures that $$\exists\lim_n \int_0^1 f_n(x)dx=0.$$


问:
Prove the dimension of a vector space.
Suppose $V$ is a vector space over a field F, has dimension $d$. Then I think it is natural to consider $F$ is a vector space over its subfield $S\subset F$ because it satisfies the following properties,
$$
x + y \in F \\
c\cdot x\in F
$$
for $x, y\in F$ and $c\in S$. Now suppose we say that the vector space ($F$ over $S$) above has dimension n, then how to explain if the vector space V over this subfield $S\subset F$, inherited the addition and scalar multiplication from $V$ over $F$, has dimension $n\cdot d$?

答:
That's pretty simple: by hypothesis, $F\simeq_S S^n$ and $V\simeq_F F^d$, so
$$V\simeq_S\bigl(S^n\bigr)^d \simeq_S S^{nd}.$$

Title: Prove the dimension of a vector space.


问:
Theorem 11.10 Rudin Analysis
Following is from Rudin's Principles of Mathematical Analysis.

In the last step it supposes $\mu^*(A_n) < \infty$ for all $A_n \in M_F(\mu) $. But when introducing $M_F(\mu) $ or $\sigma$-ring it did not suppose such condition . Is it necessary to include  hypothesis  $\mu^*(A_n) < \infty$ or the theorem still holds for $\mu^*(A_n) = \infty$ and if so how  to prove that case (I tried but couldn't make any progress) .

答:
In definition 11.7 Rudin defines $\mu$ to be finite on elementary sets. Fix $A \in \mathfrak{M}_F(\mu)$. By definition this means that there exists elementary sets $A_n$ such that $d(A_n, A) \to 0$, where $d(A, B) := \mu^*((A-B) \cup (B-A))$.
Thus by subadditivity and by monotonicity one has
$$\mu^*(A) = \mu^*(A_n \cup (A-A_n)) \leq \mu^*(A_n) + \mu^*(A-A_n) \leq \mu^*(A_n) + d(A, A_n).$$
Since $d(A_n, A) \to 0$, we may choose $n$ large enough that $d(A_n, A)< \infty$. Then since $\mu^*(A_n)$ is assumed to by finite since $\mu$ is assumed to be finite on elementary sets, we have both terms on the right hand side are finite, and hence $\mu^*(A) < \infty$. Thus any $A \in \mathfrak{M}_F(\mu)$ has $\mu^*(A) < \infty$.

Rudin is known for leaving out these little details.

Title: Theorem 11.10 Rudin Analysis


问:
Closed set is $F_{\sigma}$ and open set is $G_{\delta}$

Let $(X,d)$ a metric space with $A \subseteq X$ closed and $B \subset X$.

Prove that:

a) $A$ is an $F_{\sigma}$ set;

b) $B$ is open if and only if $B$ is a $G_{\delta}$ set.

The approach I was thinking of involves writing $A$ as a countable reuniun of closed sets and $B$ as a countable intersection of open sets, but I'm not sure how to write the detailed proofs with explanations. Thanks in advance!

答: If $A_n=A$ for each $n\in\Bbb N$, then $A_n$ is closed and $A=\bigcup_{n\in\Bbb N}A_n$. A similar argument shows that, if $B$ is open, then $B$ is a countable intersection of open sets.


问:
Is the set $A = \{1,1/2,1/3,\ldots,1/n,\ldots\}$ open or closed
I don't think it is closed because $0$ is a limit point of the set (by the Archimedean principle) but $0$ is not in $A$ but $A$ is clearly not open, does than mean it's not open nor closed. Is that even possible?

答:
$A=\{1,\frac{1}{2},\frac{1}{3},...\}$

I will now prove to you that $A$ is neither open nor closed.

In order for a set $X$ to be open, for all $x \in X$, $x \in X^o$. In other words, we need each $x \in X$ to have an $\varepsilon$-neighborhood such that $(x-\varepsilon,x+\varepsilon) \in X.$

Counterexample: Consider $a \in A, a=1.$ Surely, you would agree that $(1-\varepsilon,1+\varepsilon) \not\in A.$ Thus, we have found $a \in A$ such that $(a-\varepsilon,a+\varepsilon) \not\in A$. Thus, $A$ is not open.

In order for a set to be closed, it must contain all of its accumulation points. In other words, $\bar A = A.$

Counterexample: Consider $a \in A, a = 0.$ Notice that $0 \in \bar A$ but $0 \not\in A.$ Thus, $\bar A = [0,1]$ and $A = (0,1].$ Since $[0,1] \neq (0,1],$ it is true that $\bar A \neq A.$ Thus, A is not closed.

Now, we have shown that $A = \{1,\frac{1}{2},\frac{1}{3},...\}$ is neither open nor closed.


问:
find the upper bound of $|f'(\alpha)|$
find  $\max_{f} |f'(\alpha)|$ where  $f$ range over the class  of  analytics functions bounded  by $1$ in the unit disc  and  $\alpha$ is  a fixed  point of $|z|< 1$

My attempt : By using the  schwarz pick lemma . we have

$$
\left| \frac{f(z_1)-f(z_2)}{1- \overline{f(z_1)}f(z_2)} \right| \le \left| \frac{z_1-z_2}{1 - \overline{z_1}z_2} \right|,
$$

Put $z_1= z$  and $z_2 =\alpha$,then
$$
\left| \frac{f(z)-f(\alpha)}{1- \overline{f(z)}f(\alpha)} \right| \le \left| \frac{z-\alpha}{1 - \overline{z}\alpha} \right|,
$$

$$
\left| \frac{f(z)-f(\alpha)}{z- \alpha} \right| \le \left| \frac{1}{1 - \overline{z}\alpha} \right|,
$$

taking limit  $z \to \alpha$ on  both side then

$$
\lim_{z\to\alpha}\left| \frac{f(z)-f(\alpha)}{z- \alpha} \right| \le \lim_{z\to \alpha} \left| \frac{1}{1 - \overline{z}\alpha} \right|,
\implies |f'(\alpha) |\le | \frac{1}{1-\alpha^2}|$$

Therefore the  $\max_{f} |f'(\alpha)|$ is $| \frac{1}{1-\alpha^2}|$

Is it's correct ? yes/no

答:
There is an error in your calculation because $\lim_{z \to \alpha} \overline z \alpha = |\alpha|^2$ and not $\alpha^2$.

You have
$$
\left| \frac{f(z)-f(\alpha)}{z-\alpha} \right|
\le \left| \frac{1- \overline{f(z)}f(\alpha)}{1 - \overline{z}\alpha} \right|
$$
and taking the limit $z \to \alpha$ gives
$$
 |f'(\alpha)| \le \frac{1-|f(\alpha)|^2}{1-|\alpha|^2} \le \frac{1}{1-|\alpha|^2}\, .
$$
(The left inequality is also known as part of the Schwarz-Pick theorem.)

Equality holds for $f(z) = \frac{z-\alpha}{1-\overline\alpha z}$, so that the bound is sharp.

Title: find the upper bound of $|f'(\alpha)|$


问:
Find the zeros of $f(x) = x^2 - 150 x\log(x)$
I have to find the zeros of $f(x) = x^2 - 150 x\log(x)$

It is is pretty straightforward to find $x = 0$ because $x(x-150 \log(x)) = 0$, but how can I find the zeros of $x - 150 \log(x)$?

答:
As Joshua said, this is not solvable by elementary algebra.  But it can be solved using the Lambert W function.  Recall the definition: $$xe^x = y \;\Longleftrightarrow\; x = W(y).$$

Start with $x^2 - 150 x \log x = 0$.  Factor, $x(x-150 \log x) = 0$.  So either $x=0$ or $x-150 \log x = 0$.  For the first case, $x=0$, we should note that $\log 0$ is not defined, but $\lim_{x \to 0^+}(x^2 - 150 x \log x) = 0$.  So $x=0$ is a solution in that sense.

Here is the second case.
$$
x = 150\log x\\
e^x = x^{150}\\
e^{x/150} = x\\
1=xe^{-x/150}\\
\frac{-1}{150}=\frac{-x}{150}e^{-x/150}\\
W\left(\frac{-1}{150}\right) = \frac{-x}{150}\\
-150W\left(\frac{-1}{150}\right)= x
$$
Actually, there are two real branches of the $W$ function, so we get two real solutions
$$
-150W_0\left(\frac{-1}{150}\right) \approx 1.006734134,\qquad
-150W_{-1}\left(\frac{-1}{150}\right) \approx 1042.390835.
$$
Here are two graphs of $x^2 - 150 x\log(x)$.

Here is the classic paper on Lambert W. Perhaps few papers can be labeled "classic" after only 25 years, but this is one!

Corless, R. M.; Gonnet, G. H.; Hare, D. E. G.; Jeffrey, D. J.; Knuth, D. E., On the Lambert (w) function, Adv. Comput. Math. 5, No. 4, 329-359 (1996). ZBL0863.65008.


问:
Prove that a map is Injective
How can I prove that $f: (0, \infty) \times (0,\pi) \to \mathbb{R}^2$ where $f(x,y) = (\sinh(x)\sin(y),\cosh(x)\cos(y))$ is injective? </p>

答:
If $f$ is not injective, then there exist $a\not=c$, $b\not=d$ such that

$\cosh(a) \cos(b)=\cosh(c) \cos(d)\\
\sinh(a) \sin(b)=\sinh(c) \sin(d)$

W.L.O.G let $a>c$. Then $\sinh(a)>\sinh(c)$ and $\cosh(a)>\cosh(c)$ since both functions are increasing in $(0,\infty)$. Therefore $\sin(b)<\sin(d)$ and $\cos(b)<\cos(d)$, which can only occur in $(0,\pi)$ if $\cos(b)<0$ and $\cos(d)>0$ (see graph), but then the above could not be true because $\cosh(a)$ is positive.

Title: Prove that a map is Injective


问:
Find expression for $c^2\in \mathbb{C}$ given certain parameters

Let $a,b,\theta,\phi \in \mathbb{R}^+$. Let $z,w,c \in \mathbb{C} / z=ae^{i(\theta+\phi)},w=be^{i\phi},c=|z-w|$.

Then:

(A) $c^2=a^2+b^2$

(B) $c^2=a^2+b^2 -2absin(\theta)$

(C) $c^2=a^2+b^2 -2abcos(\theta)$

(D) $c^2=a^2+b^2 +2abcos(\theta)$

答:
Let $z\in \mathbb{C}$ and we have $|z|$ and $\phi$. Then if $z=a+bi$ we know that $a=|z|cos(\phi)$ and $b=|z|sin(\phi)$.

So $z=a.e^{i(\theta+\phi)}=a.cos(\theta+\phi)+a.sin(\theta+\phi)i$.

Similarly, $w=b.e^{i\phi}=b.cos(\phi)+b.sin(\phi)i$.

So now

$z-w=a.cos(\theta+\phi)+a.sin(\theta+\phi)i-b.cos(\phi)-b.sin(\phi)i$

Then

$c=|z-w|=\sqrt{[a.cos(\theta+\phi)-b.cos(\phi)]^2+[a.sin(\theta+\phi)-b.sin(\phi)]^2}$

Some math later

$c=\sqrt{a^2cos^2(\phi+\theta)+b^2cos^2(\phi)-2ab.cos(\theta+\phi)cos(\phi)+a^2sin^2(\phi+\theta)+b^2sin^2(\phi)-2ab.sin(\theta+\phi)sin(\phi)}$

Now we use trigonometric properties (sum of squares and product)

$c=\sqrt{b^2+a^2-2ab[\frac{1}{2}(cos(\theta+2\phi)+cos(\theta))]-2ab[\frac{1}{2}(cos(\theta)-cos(\theta+2\phi))]}$

Some math later again

$c=\sqrt{a^2+b^2-2ab.cos(\theta)}$

Finally

$c^2 = a^2+b^2-2ab.cos(\theta)$


问:
Characterization of the denominator of a polynomial $P \in \mathbb Q[X]$ by looking at roots?
Let $P(X) \in \mathbb Q[X]$ be a monic polynomial of degree $n$ and $a \in \mathbb Z$. We have the implications
$$a P(X) \in \mathbb Z[X] \implies a^n P(X/a) \in \mathbb Z[X] \iff \text{the roots of $P$ lie in } \overline{\mathbb Z}/a \,.$$

I would like to add extra conditions to the statements on the right to make this an equivalence.

This question is a bit vague, naturally, because I don't know what conditions to add. Of course, I don't want anything trivial that amounts to saying that $aP \in \mathbb Z[X]$. My motivation below can help to get a feeling for what type of condition I'd like to see. Something like "$aP$ sends integers to integers" would be acceptable.

Motivation. I started thinking about this when answering (or trying to) the following questions:

If $f(x)\in\mathbb Q[x]$ and $f(f(x)),f(f(f(x)))\in\mathbb Z[x]$ prove that $f(x)\in\mathbb Z[x]$

Can composition of integer polynomial and rational polynomial with a non-integer coefficient result in integer polynomial?

Ideas. Requiring that $aP(\mathbb Z) \subset \mathbb Z$ is not sufficient: take $a = 2$, $P = X^3 + \frac{X^2+X}4 = \frac{(2X)^3 + (2X)^2 + 2 (2X)}{8}$, which has roots in $\overline{\mathbb Z}/2$.

Even imposing the same on all derivatives of $aP$ and imposing that $c-d \mid aP(c) - aP(d)$ is not enough: $a = 2$, $P = X^5 + \frac{X^4+X^2}4$.

答:
Maybe you're looking for something like this:

Let $P$ be a primitive polynomial in $\mathbb{Z}[X]$. Then $P$ is monic if and only if all roots of $P$ are in $\overline{\mathbb{Z}}$.

Proof: For $\Rightarrow$, this is true by definition of $\overline{\mathbb{Z}}$. For $\Leftarrow$, let $a \in \mathbb{Z}$ be the leading coefficient of $P$; the coefficients of $\frac{1}{a}P$ are sums of products of elements of $\overline{\mathbb{Z}}$, hence are themselves contained in $\overline{\mathbb{Z}}$; but $\overline{\mathbb{Z}} \cap \mathbb{Q} = \mathbb{Z}$ (i.e. $\mathbb{Z}$ is integrally closed).


问:
Confused About Fields: Two Proofs for a0 = 0
For any $a \in F$, $a0 = 0$.

I know this proof works:
$a0 = a(0+0) = a0 + a0$; taking the additive inverse of $a0$, we get $a0=0$.

But how about this? $a0 = a(1+(-1)) = a + (-a) = 0$

The reason I'm skeptical is because not all fields have -1 as an element. I tried to work around this issue by saying that "-1" exists in all fields in the sense that all fields have 0,1 and "-1" stands for the additive inverse of "1." For example, {0,1} is a field with 1 + 1 = 0: here, 1 is "-1" (it's the additive inverse). To make sense of this idea, I redid the proof without thinking of numbers; instead I used variables to represent an element of a field and its additive inverse. For example, let j be an element of a field and w its additive inverse:

$a0 = a(j+w) = aj + aw$

Where do I go from here? How do I show that $aj$ and $aw$ are additive inverses of each other?

答: In your notation, let $j$ be the multiplicative identity and $w$ be the additive inverse of the mulitplicative identity. Then it works out as $$a\cdot 0 = a (j+w)= aj + aw = aj-aj=0.$$ Here we have used the fact that $aw=-aj$ which needs the fact that $$0=a(w+j)=a\cdot 0.$$ Do you see the cirularity? Better to stick to the standard proof that you stated.


问:
Calculate this limit without using L'Hôpital rule ..Any suggestions( I got to the solution with the suggestions, why do you close the questions?)
I can't think of any change of variable to solve this limit, I can't use derivatives or series. Can you please guide me on how to do it?
Thank you

$$\underset {x\to 6} {\text {lim}}\frac {x -\sqrt[3] {x + 2} - \sqrt {3 x - 2}} {x -\sqrt[3] {4 x + 3} - \sqrt {x + 3}}$$

Edit
taking the suggestion of @Paramanand Singh
making x=6+h , x->6 ,h->0  I arrive at the following  ??

$$\underset {h\to
     0} {\text {lim}}\frac {(h +
     6) - \sqrt[3] {h + 6 + 2} - \sqrt {3 (h + 6) - 2}} {(h +
     6) - \sqrt[3] {4 (h + 6) + 3} - \sqrt {h + 6 + 3}}$$

$$\underset {h\to 0
     } {\text {lim}}\frac {(h +
     6) - \sqrt[3] {h + 8} - \sqrt {3 h + 16}} {(h +
     6) - \sqrt[3] {4 h + 27} - \sqrt {h + 9}}$$

$$\displaystyle \huge \lim_{h->0}\frac{\frac{(h+6)-\sqrt[3]{8+h}-\sqrt{3h+16}}{h}}{\frac{(h+6)-\sqrt[3]{4h+27}-\sqrt{h+9}}{h}}$$

Edit 2  : by manipulating the numerator and denominator, I get 4 limits and when I develop them individually I reach the desired result.  Thanks to all

答:
Well one way(which may not be satisfactory to you) is to expand the numerator and denominator about $x=6$. We get the following result for the numerator
$$x-\sqrt{3x-2}-\sqrt[3]{x+2} = \frac{13}{24}(x-6)+O((x-6)^2)$$
and for the denominator
$$x - \sqrt{x+3} - \sqrt[3]{4x+3} = \frac{37}{54}(x-6)+O((x-6)^2)$$
And therefore the result is after dividing and neglecting the higher order terms $\frac{117}{148}$.

For what it's worth, Wolframalpha gives the same result.


问:
Question about an almost split sequence.
On page 124 of the book Elements of representation theory of associative algebras, volume 1, Example 3.10, I computed the modules in this example.

$$
S(3)=0\leftarrow 0 \rightarrow K \leftarrow 0, \\
P(2) =  K\overset{1}{\leftarrow} K \overset{1}{\rightarrow} K \leftarrow 0,\\
P(4) =  0\overset{}{\leftarrow} 0 \overset{}{\rightarrow} K \overset{1}{\leftarrow} K,\\
P(2) \oplus P(4) =  K\overset{1}{\leftarrow} K \overset{\left(\begin{matrix} 0 \\ 1 \end{matrix}\right)}{\rightarrow} K^{2} \overset{\left(\begin{matrix} 1 \\ 0 \end{matrix}\right)}{\leftarrow} K.
$$

Let $f_1: S(3) \to P(2)$ be the embedding and $f_2: S(3) \to P(4)$ be the embedding. Let $f=\left(\begin{matrix} f_1 \\ f_2 \end{matrix}\right): S(3) \to P(2) \oplus P(4)$. Then $f$ is injective. Is $f: S(3) \to P(2) \oplus P(4)$ the map in the sequence
$$
0 \to S(3) \overset{f}{\to} P(2) \oplus P(4) \to (P(2) \oplus P(4))/S(3) \to 0 \quad (1)
$$
in Example 3.10?

I think that
$$
(P(2) \oplus P(4))/(Im(S(3))) = (P(2) \oplus P(4))/(Im f) = K\overset{1}{\leftarrow} K \overset{0}{\rightarrow} 0 \overset{0}{\leftarrow} K, (2) \\
(P(2) \oplus P(4))/S(3) = K\overset{1}{\leftarrow} K \overset{1}{\rightarrow} K \overset{0}{\leftarrow} K. (3)
$$
I am not sure about the maps in (3). Should $(P(2) \oplus P(4))/S(3)$ the sequence (1) in Example 3.10 be $(P(2) \oplus P(4))/(Im(S(3)))$? I think that the direct sum of $S(3)$ and (3) is $P(2) \oplus P(4)$ which contradicts the fact the (1) is non-split. Thank you very much.

答:
Their notation in the example is sloppy, but understandable if you keep the preceeding lemma in mind.  Note that the end terms of an almost split sequence are both indecompoable modules, so both of your guess are incorrect.  What they meant by $M=(P(2)\oplus P(4))/S(3)$ is the indecomposable module given by the non-split extension:  $0 \to P_4 \to M \to P(2)/S(3) \to 0$.

To be slightly more precise yet intuitive, the quiver is: $1 \xleftarrow{\alpha} 2 \xrightarrow{\beta} 3 \xleftarrow{\gamma} 4$.
The maps from $S(3) \to P(2)\oplus P(4)$ is $a \overline{e_3} \mapsto  a(\beta+\gamma)$ for any $a\in kQ$.  By using the philosophy as mentioned in Specific projective dimension of a module over bound quiver , the cokernel $M$ of the injection is by "tearing half of the $S(3)$ from $P(2)$ and half from $P(4)$, and then glue the remaining together", hence it is an indecompoable module with sincere support.  You can verify this by first writing the basis of $M$, and then consider the action of $\beta$ and $\gamma$ on it.


答:
First, all fields DO have $-1$ as an element: every field has $1$ as an element; every field has an additive inverse for every element in the field; and therefore every field has $-1$ as an element.

But as you seem to have intuited, your new proof does require knowledge that $(-1) \cdot a = -a$. And you ask how to prove this, so here is a chain of equivalences each of which only uses one of the field axioms:
\begin{align*}
(-1) \cdot a = -a &\iff (-1) \cdot a + a = 0 \\ &\iff (-1) \cdot a + 1 \cdot a = 0 \\ &\iff ((-1) + 1) \cdot a = 0 \\ &\iff 0 \cdot a = 0
\end{align*}
As you can see, you can deduce either of these two statements from the other, and so you'll need to prove one of them first before you can deduce the other.

Your question has a pretty simple proof that $0 \cdot a = 0$, so that seems like the one to prove first.

Title: Confused About Fields: Two Proofs for a0 = 0

Body:
For any $a \in F$, $a0 = 0$.

I know this proof works:
$a0 = a(0+0) = a0 + a0$; taking the additive inverse of $a0$, we get $a0=0$.

But how about this? $a0 = a(1+(-1)) = a + (-a) = 0$

The reason I'm skeptical is because not all fields have -1 as an element. I tried to work around this issue by saying that "-1" exists in all fields in the sense that all fields have 0,1 and "-1" stands for the additive inverse of "1." For example, {0,1} is a field with 1 + 1 = 0: here, 1 is "-1" (it's the additive inverse). To make sense of this idea, I redid the proof without thinking of numbers; instead I used variables to represent an element of a field and its additive inverse. For example, let j be an element of a field and w its additive inverse:

$a0 = a(j+w) = aj + aw$

Where do I go from here? How do I show that $aj$ and $aw$ are additive inverses of each other?


问:
Elementary proof of Power Rule For differentiation $f'(x)=rx^{r-1}$ for $f(x)=x^r$
I could  not find any elementary proof of the Power Rule for differentiation:

Given $x,r \in R$ ,$x>0$ and a function $f(x)=x^r$, then its derivative is $f'(x)=rx^{r-1}$

Defintion:Let a sequence of rational numbers {$r_n$} tend to $r$ then $x^r=\lim_{r_n \to r}x^{r_n}$

All the proofs I had seen utilized the derivative of $logx$ but I dont think this is  elementary because one has to show that for $e=\text{Euler's number}$ , $t \in R$and $n \in N$ $\lim_{n \to \infty}(1+\frac{1}{n})^n=\lim_{t \to 0}(1+t)^\frac{1}{t}=e$ which can be proved using the Power Rule

Ofcourse the proof for the Power Rule is straight forward if one is familiar with the result of Power Rule if $r$ is a rational number and theorem  regarding  derivative of a sequence of derivatives that are uniformly convergent.

答:
Possibly the most "elementary" way to do it is to define logarithms first, and only later define general exponentials (and not to do it the way you do, except to show the new definition agrees with the standard one for integers and rationals).

For $x\gt 0$, define the function $\log(x)$ be
$$\log(x) = \int_1^x \frac{1}{t}\,dt.$$

The following properties are now straightforward:

$\log(x)$ is strictly increasing and differentiable.

$\frac{d}{dx}\log(x) = \frac{d}{dx}\int_1^x\frac{1}{t}\,dt = \frac{1}{x}$.

$\lim_{x\to0^+}\log(x) = -\infty$ and $\lim_{x\to \infty}\log(x)=\infty$.

$\log(1)=0$.

Then we also have, by letting $u=\frac{1}{t}$, that
$$\begin{align*}
\log\left(\frac{1}{x}\right) &= \int_1^{1/x}\frac{1}{t}\,dt
&= \int_1^x u\left(-\frac{u}{u^2}\right)\,du\\
&= -\int_1^x\frac{1}{u}\,du\\
&= -\log(x).
\end{align*}$$

Also: fix $a\gt 0$. Then letting $au=t$, we have
$$\begin{align*}
\log(ax) &= \int_1^{ax}\frac{1}{t}\,dt\\
&= \int_{1/a}^x\frac{1}{au}a\,du\\
&= \int_{1/a}^x\frac{1}{u}\,du  = -\int_1^{1/a}\frac{1}{u}\,du + \int_1^x\frac{1}{u}\,du\\
&= -\log\left(\frac{1}{a}\right) + \log(x)\\
&= \log(a)+\log(x).
\end{align*}$$
So now we also have the following further properties: for $a,b\gt 0$,

$\log(ab)=\log(a)+\log(b)$.

$\log(a/b)=\log(a)-\log(b)$.

By induction, it follows that for positive integers $n\gt 0$, $\log(a^n) = n\log(a)$; and that $\log(a^{-n}) = -n\log(a)$.

If $r=\frac{p}{q}$ is a rational, with $q\gt 0$, then we have that
$$q\log(a^{p/q}) = \log(a^{p}) = p\log(a),$$
hence $\log(a^{p/q}) = \frac{p}{q}\log(a)$.

Now, since $\log(x)$ is strictly increasing, it has an inverse. Let us denote the inverse by $E$. So $E(x)=y$ if and only if $\log(y)=x$. From the properties of $\log(x)$, we get:

$E(x)$ is strictly increasing and differentiable.

Since $x=\log(E(x))$, differentiating both sides and using the Chain Rule we have
$$\begin{align*}
1 &= \frac{1}{E(x)}\left(\frac{d}{dx}E(x)\right)\\
E(x) &= \frac{d}{dx}E(x).
\end{align*}$$

$\lim_{x\to-\infty}E(x) = 0^+$ (that is, $E(x)$ is positive for all $x$, and approaches $0$ as $x\to-\infty$), and $\lim_{x\to\infty}E(x)=\infty$.

$E(0)=1$.

Now, say $E(a)=r$ and $E(b)=s$. Then $\log(r)=a$ and $\log(s)=b$. Therefore,$a+b=\log(r)+\log(s) = \log(rs)$ and $a-b=\log(r)-\log(s)=\log(r/s)$, hence
$$E(a+b) = rs = E(a)E(b)\quad\text{and}\quad E(a-b) = \frac{E(a)}{E(b)}.$$
giving
5. $E(a)E(b) = E(a+b)$.
6. $\frac{E(a)}{E(b)} = E(a-b)$.

If $E(a)=r$ and $n\gt 0$ is an integer, then by induction we have that $E(na) = E(a)^n = r^n = E(na)$. This immediately extends to negative integers, and finally to rational exponents, so that $E(a)^{p/q} = E(ap/q)$ for all integers $p,q$, with $q\gt 0$.

Definition. If $a\gt 0$ and $b$ is any real number we define $\exp(a,b) = E(b\log(a))$.

Proposition. If $b$ is a rational number, then $\exp(a,b)=a^b$.

Proof. First, assume $b$ is a positive integer. By induction: $\exp(a,1) = E(1\log(a)) = E(\log(a)) = a = a^1$. If $\exp(a,k) = a^k$, then
$$\exp(a,k+1) = E((k+1)\log(a)) = E(k\log(a)+\log(a)) = E(k\log(a))E(\log(a)) = a^ka = a^{k+1}.$$
Thus, the proposition holds for $b$ a positive integer.

If $b$ is a negative integer, then
$$\exp(a,b) = E(b\log(a)) = E(-(-b)\log(a)) = \frac{1}{E((-b)\log(a))} = \frac{1}{a^{-b}} = a^b.$$

Finally, if $b=\frac{p}{q}$ with $p,q$ integers and $q\gt 0$, then
$$(\exp(a,b))^q = (E(b\log(a)))^q = E(qb\log(a)) = E(p\log(a)) = a^p,$$
so taking $q$th roots on both sides we get $\exp(a,b) = a^{p/q}=a^b$. $\Box$

In light of the proposition, we simply write $a^b$ instead of $\exp(a,b)$; we know it agrees with the "old" definition when $b$ is a rational number.

Thus, if $\{q_n\}_{n\in\mathbb{N}}$ is a sequence of rational numbers and $q_n\to r$, then
$$\lim_{n\to\infty} a^{q_n} = \lim_{n\to\infty}E^{q_n\log(a)} = E\left(\lim_{n\to\infty}q_n\log(a)\right) = E(r\log(a)) = a^r,$$
since $E$ is continuous. Thus, this definition agrees with the definition you propose for $a^r$.

So, finally, using the Chain Rule, we get
$$\begin{align*}
\frac{d}{dx}x^r &= \frac{d}{dx} E(r\log(x)) \\
&= E(r\log(x))\left(\frac{d}{dx}r\log(x)\right) \\
&= \frac{r}{x}E^(r\log(x))\\
&= rx^{-1}E(r\log(x))\\
&= rE(-\log(x))E(r\log(x))\\
&= rE(-\log(x)+r\log(x)) = rE((r-1)\log(x))\\
&= rx^{r-1},
\end{align*}$$
giving the desired differentiation formula.

Of course, we let $e$ be the solution to $\log(x)=1$, and then we have that
$$E(x) = E(x\log(e)) = e^x,$$
which gives us the "usual" notation. Then Taylor's Theorem will give the equivalence of this definition with the power series definition.


问:
Question about the action of the operator $(-\Delta+i)^{-1}$
It is well-known that minus the Laplace operator $-\Delta:\mathcal{D}(-\Delta)\to L^{2}(\mathbb{R}^{3})$ is self-sdjoint on the domain $\mathcal{D}(-\Delta)=H^{2}(\mathbb{R}^{3})$ (Sobolev space). Therefore, the spectrum of $-\Delta$ is contained in $\mathbb{R}$, which means that $i$ is in the resolvent and hence the inverse of $(-\Delta+i):\mathcal{D}(-\Delta)\to L^{2}(\mathbb{R}^{3})$ exists and

$$(-\Delta+i)^{-1}\in\mathcal{B}(L^{2}(\mathbb{R}^{3}))$$

where $\mathcal{B}(\mathcal{H})$ denotes the set of bounded, linear operators on the Hilbert space $\mathcal{H}$.

Now, in a lecture I have seen the following formula:

$$\frac{1}{-\Delta+i}f(x)=\int_{\mathbb{R}^{3}}\frac{\hat{f}(k)}{4\pi^{2}k^{2}+i}e^{2\pi ik\cdot x}\,\mathrm{d}^{3}k$$

for some $f\in L^{2}(\mathbb{R}^{3})$ and where $\hat{f}$ denotes the Fourier transform $f$. I have two questions about this:

(1) Is $\frac{1}{-\Delta+i}$ here just a notation for the inverse operator $(-\Delta+i)^{-1}$? I guess that this is the case, but just to get sure, because when talking about real functions for example, the symbols $1/f$ and $f^{-1}$ mean completely different things.

(2) How is the above formula derived?

Thanks a lot!

答:
(1) Yes, $(-\Delta +i)^{-1}$ refers to the inverse operator relative to the operation of composition.
There is an issue regarding what is the inverse of a
function, since one may either be talking about the  pointwise multiplication or the composition of functions,  hence
the two somewhat well established notations $1/f$ and $f^{-1}$.  For operators, the pointwise multiplication makes no
sense so people often use these notations indistinctly

(2) By a simple integration by parts argument,  one can show that for every $f$ in the Schwartz space,
$$
  \widehat{\frac {\partial f}{\partial \xi _i}}(\xi ) = 2\pi  i \xi _i \widehat f(\xi ), \quad \forall \xi \in  \widehat {\mathbb R^3} = \mathbb R^3.
  $$

Thus,  denoting by $\mathfrak F$ the Fourier transform,  we have that
$$
  \mathfrak F \circ  \frac{\partial} {\partial \xi _i} = 2\pi  i M_{i} \circ  \mathfrak F,
  \tag 1
  $$
where $M_i$ is the multiplication operator
$$
  M_i(f)|_\xi  = \xi _if(\xi ).
  $$

Another way to state (1) is to say that
$$
  \mathfrak F  \frac{\partial} {\partial \xi _i}  \mathfrak F^{-1}= 2\pi  i M_{i},
  $$
whence
$$
  \mathfrak F  \Delta    \mathfrak F^{-1}=
  \sum_{i=1}^3   \mathfrak F  \frac{\partial^2} {\partial \xi _i^2}  \mathfrak F^{-1} =
  -4\pi ^2  \sum_{i=1}^3   M_i^2 =   -4\pi ^2 M,
  $$
where $M$ is now the multiplication operator
$$
  M(f)|_\xi  = |\xi |^2f(\xi ).
  $$
Since an operator ($\Delta $) and its conjugate ($-4\pi ^2 M$) are essentially a mirror image of each other, all you need to do
is compute $(4\pi ^2 M+i)^{-1}$, and this is now much easier since you only have to deal with multiplication operators!


问:
Show $\frac{(b+c)^2}{bc}l_a^2+\frac{(c+a)^2}{ca}l_b^2+\frac{(a+b)^2}{ab}l_c^2=4p^2$
Show that the equality $$\dfrac{(b+c)^2}{bc}l_a^2+\dfrac{(c+a)^2}{ca}l_b^2+\dfrac{(a+b)^2}{ab}l_c^2=4p^2$$ holds for a $\triangle ABC$ with sides $AB=c,BC=a, AC=b$, semi-perimeter $p$ and angle bisectors $l_a,l_b$ and $l_c$.

I am a little confused because I have recently seen so many angle bisector formulas. Here we can probably use that $$l_a=\dfrac{2}{b+c}\sqrt{bcp(p-a)}$$ and $$l_b=\dfrac{2}{a+c}\sqrt{acp(p-b)}$$ and $$l_c=\dfrac{2}{a+b}\sqrt{abp(p-c)}.$$ Can you show me? I don't think I can remember these formulas. What can I do if I forget them and can't check because I am not allowed to? What if we don't have them?

答:
The standard angle bisector formula is $$l_a = \frac{2bc \cos\left(\dfrac{A}{2}\right)}{b + c}$$ which, using $\cos\left(\dfrac{A}{2}\right) = \sqrt{\dfrac{p(p - a)}{bc}}$, transforms to $$l_a = \frac{2\sqrt{bcp(p - a)}}{b + c}$$

Now putting these formulas into the LHS, $$ \sum \frac{(b+c)^2}{bc} \left( \frac{2\sqrt{bcp(p - a)}}{b + c} \right)^2 = \sum 4 p(p-a) = 4p \left( \sum (p - a) \right) = 4p^2$$

Hence proved :)


问:
torsion points on complex torus
I’m not sure how to convince myself that "set of torsion points is dense on complex torus".

Is this obvious? I could not find a reference or a proof.

Could anyone help me clarify or verify this?

答:
Let $E = \mathbb{C}/\Lambda$ be an elliptic curve over $\mathbb{C}$ (i.e., a complex torus), where $\Lambda = \mathbb{Z}\omega_1 + \mathbb{Z}\omega_2$ is a lattice (and the $\omega$'s are linearly independent over $\mathbb{R}$). We want to show that for any nonempty open ball $U$ contains a torsion point $p$.

We have the following claim: the (image under the quotient map of) $p = (m_1\omega_1 + m_2 \omega_2) / n$ is torsion for each $n, m_1, m_2 \in \mathbb{Z}$. But this is easy to see since $np \in \Lambda$.

Well, now let $U' \subset \mathbb{C}$ be the inverse image of the ball $U$ under the quotient map $\mathbb{C} \to \mathbb{C}/\Lambda$. Note that $U'$ is nonempty since $U$ is. The result follows from the claim that $U'$ contains a $p$ as above.

To see this notice that the $\mathbb{Q}$-span of $\{\omega_1, \omega_2\}$ is dense in $\mathbb{C}$, so has nonempty intersection with $U'$ - say containing $p = (m_1/n_1) \omega_1 + (m_2/n_2)\omega_2$. Thus (the image of) $p = (m_1n_2\omega_1 + m_2n_1 \omega_2)/(n_1n_2)$ is a torsion point, contained in $U$.

Title: torsion points on complex torus


问:
Solving a bivariate recursion $f(k, n) = n^m f(k-n, m)$
I have encountered an unknown object that seems to obey the following recursion, for any integers $k \geq n, m \geq 1$,

$$
f(k, n) = n^m f(k-n, m)
$$

I think that $f(k, n) = 0$ is a solution to this (or is it?), but I am wondering if there are any other non-trivial solutions to this recursion, given the value $f(k, 1)$. (Or if it is possible to prove this has no non-trivial solution).

Thanks.

答: OK, so now with $m=1$, $f(k,n) = n f(k-n,1)$.  In particular for $n=1$, $f(k,1) = f(k-1,1)$, i.e. $f(\cdot,1)$ is constant.  If $c = f(1,1)$, $f(k,n) = n c$, and your equation then says $ n c = n^m m c$.  Taking e.g. $m=2$ and $n > 0$, we see that the only possibility is $c=0$.


问:
Legendre's three-square theorem
Problem:

If $ n \in \mathbb{N} $ can be represented as $ n = n_1^2 + n_2^2 + n_3^2 ,\quad n_1, n_2, n_3 \in \mathbb{Z}, $ show that then $ n \neq 4^a(8m+7)$ for $a,m \in \mathbb{Z}.$

My solution:

Since this is the beginning of a math course in Discrete Mathematics, I don't think we're supposed to use the quadratic reciprocity law, Dirichlet's theorem on arithmetic progressions nor the equivalence class of the trivial ternary quadratic form. as listed on Wikipedia.

It also states on Wikipedia (Why?) that every square is congruent to either $0,1,4$ in $\mathbb{Z}_8$. So the sum of three squares $n \in \{0,1,2,3,4,5, 6\} \subset \mathbb{Z}_8.$ My initial plan was to show that $ 4^a(8m+7) \in \{6,7\}$. This not true though, as $ 4^a \equiv 4 $ or $ \equiv 0 $ for $ a \geq 1 $. Thus $ 4^a(8m+7) \equiv 0 $ or $4$.

Any hints?

Muchos gracias!

答:
Well as the equation $$n=n^2_1+n^2_2+n^2_3$$ has no integral solutions if $n$ is of the form $n=8m+7$ for some integer $m$--established in the comments, we can prove that the equation
$$n=n^2_1+n^2_2+n^2_3$$ has no integral solutions if $n$ is of the form $n=4^a(8m+7)$ for some integers $m, a \ge 1$, by induction on $a$.

[To elaborate, as shown, there are no solutions to  $$n=n^2_1+n^2_2+n^2_3$$ for $n$ of the form $n=8m+7$ i.e., $a=0$. So fix an $a >0$. We show that if there is a solution to $$n=n^2_1+n^2_2+n^2_3$$ for any $n$ of the form $n=4^a(8m+7)$, then there must be a solution to $$\tilde{n}=\tilde{n}^2_1+\tilde{n}^2_2+\tilde{n}^2_3$$ for some $\tilde{n}$ of the form $\tilde{n}=4^{a-1}(8m+7)$. As it has already been established in the comments that there is no solution to $$\tilde{n}=\tilde{n}^2_1+\tilde{n}^2_2+\tilde{n}^2_3$$ for some $\tilde{n}$ of the form $\tilde{n}=(8m+7) =4^0(8m+7)$ this establishes the desired result.]

The crux of the proof is this: Suppose $n = 4^a(8m+7)$ for $a \ge 1$. Then $n \mod 4 = 0$ and as every even square is $0 \mod 4$ and every odd square is $1 \mod 4$ [check this for yourself] it follows that $n_1,n_2,n_3$ must all be even. So then this implies
$$\frac{n}{4} = \left(\frac{n_1}{2}\right)^2 +\left(\frac{n_2}{2}\right)^2 +\left(\frac{n_3}{2}\right)^2$$ must hold, with the $\left(\frac{n_i}{2}\right)$s; $i=1,2,3$ all integers and with $\frac{n}{4} = 4^{a-1}(8m+7)$.

Title: Legendre's three-square theorem


问:
Is there any approach using real methods to evaluate $\int_{0}^{\infty} \frac{x \sin (x)}{x^{4}+1} d x$
What is the approach to solve the integral

$$\int_{0}^{\infty} \frac{x \sin (x)}{x^{4}+1} d x$$

Is there any real methods to solve it?

答:
Let  $I(a) = \int_0^\infty \frac{\sin at} {t(t^2+1)}dt$. Then
$I’(a) = \int_0^\infty \frac{\cos at} {t^2+1}dt,\>
I’’(a)= \int_0^\infty \frac{t\sin at} {t^2+1}dt
$ and
$$I’’(a)-I(a) = -\int_0^\infty \frac{\sin at}t dt= -\frac\pi2$$
which leads to $I(a) = \frac\pi2 (1-e^{-a})$ and $I’(a) = \frac\pi2 e^{-a}$. Then,
\begin{align}
\int_{0}^{\infty} \frac{x \sin x}{x^{4}+1} d x
= &\frac1{4\sqrt2}\int_{-\infty}^{\infty}
( \underset{x-\frac1{\sqrt2} =t}{\frac{\sin x}{x^2-\sqrt2 x+1}} - \underset{x+\frac1{\sqrt2} =t}{\frac{\sin x}{x^2+\sqrt2 x+1}} ) d x\\
= &\frac1{4\sqrt2}\int_{-\infty}^{\infty}
\left( \frac{\sin(t+\frac1{\sqrt2})}{t^2+\frac12} - \frac{\sin(t-\frac1{\sqrt2})}{t^2+\frac12} \right) d t \>\>\>\>\>(\sqrt2t\to t)\\
=& \sin\frac1{\sqrt2} \int_{0}^{\infty} \frac{\cos \frac t{\sqrt2}}{t^2+1} dt = \sin\frac1{\sqrt2}\cdot I’(\frac1{\sqrt2})\\
=&\frac\pi2 e^{-\frac1{\sqrt2} } \sin\frac1{\sqrt2}
\end{align}


答:
(Unlike most posts: the indented details in this post are extra details you don't have to read.  They just go into more detail.  The main points are in the parts that are not indented.)

Your argument that it is not closed is exactly correct.

[$0$ is limit point as for any $\epsilon > 0$ and $n > \frac 1\epsilon$ we have $\frac 1n \in N_{\epsilon}(0)$. And $0\not \in A$.]

Your statement that it is "clearly not open" is true but not good enough.  You will have to give a more thorough proof/argument.  Buy yes it is not open.

[Easy enough to prove $1\in A$ and for any $\epsilon > 0$ then let $w$ be such that $\max(1-\epsilon, \frac 12) < w < 1$ then $w \in N_{\epsilon}(1)$ but $w\not \in A$ so $1$ is not an interior point so $A$ is not open.]

So, yes, it is neither open nor closed.  Not only is that possible it is very common.  The definition of open and the definition of closed are not mutually exhaustive so it is theoretically
possible to have both i) not all its limit points contained in it and ii) not all of its points being limit points.

A much simpler counter example is $[1,2)$.  It's not closed because $2$ is a limit point in it.  And it's not open because $1$ is not an interior point.  If you think about it, a limit point not being in it, in no way will force every point (including those that have nothing to do with the missing limit point) to be an interior point.

"open" and "closed" despite the english meaning are not an "either/or" situation.

Nor are they mutually exclusive. A set can be both open and closed.  $\emptyset$ is both, and in $\mathbb R$ with the Euclidean/"usually" metric $\mathbb R$ is too.  Those are the only examples with $\mathbb R$ and the Euclidean metric but there are other metrics where open and closed sets are more common.

(For example the discrete metric where $d(x,y) =\begin{cases} 0& x=y\\1& x\ne y\end{cases}$ is a metric where EVERY set is both open and closed.  [A set $A$ is open because if $x\in A$ then $N_{1}(x)=\{w| d(w,x)< 1\} = \{w|d(w,x)=0\} = \{w|w=x\} = \{x\}\subset A$.  So all points of $A$ are interior points.  A set $A$ is closed because if $w$ is a limit point of $A$ then $N_1(w) = \{x|d(x,y)< 1\}=\{x|d(w,x)=0\} =\{x|w=x\} =\{w\}$ must contain a element of $A$.  But $N_1(w)$ contains only one point at all.  So that point $w\in A$.  So the limit point is in $A$.])

so "open" does not mean "not closed" and vice versa, and both open and closed and neither open nor closed are possible.

....

But, however, (as if this post isn't long enough), it is a theorem that one connection between a set being open and a set being close is that:  $A$ is open $\iff$ $A^c$ is closed.

But as for $A$ being open if or only if $A$ is or is not closed... there's nothing that can be said about that. All cases are possible:  A set $A$ can be opened; closed; neither; or both.

Title: Is the set $A = \{1,1/2,1/3,\ldots,1/n,\ldots\}$ open or closed

Body: I don't think it is closed because $0$ is a limit point of the set (by the Archimedean principle) but $0$ is not in $A$ but $A$ is clearly not open, does than mean it's not open nor closed. Is that even possible?


问:
Finding $(a,b)$ in some given subset of $\mathbb{Z}^2$ with a specific gcd.
Given the subset of $\mathbb{Z}\times\mathbb{Z}\supset S=\{(9(k+3),4k), k\in \mathbb{Z}\}$ I’m trying to find $(x,y)\in S$ such that $\mathrm{gcd(x,y)}=18$. The set $S$ arises as the set of solutions to the equation $4x-9y=108,\; x,y\in\mathbb{Z}$. The equation isn’t particularly hard to solve, but I can’t quite figure out how to go about finding the couples $(x,y)$ with gcd 18.

Playing around a little I found that since I want $18$ to be the gcd of $9(k+3)$ and $4k$, 18 has to divide both, which implies $k$ is an odd number and $2k=9q$ with $q\in\mathbb{Z}$ but this still allows for the possibility of $\mathrm{gcd}(x,y)>18$, that is, those are necessary conditions but certainly not sufficient ones. And I’m not entirely sure where to go from there. This feels like it’s very easy and I’m just not seeing something obvious. Anyway, if anyone has a suggestion, a hint or something, I’d appreciate, thanks.

答: Write $x=18a$ and $y=18b$ where $a,b$ are coprime, so we have $$4a-9b = 6$$ Clearly $2\mid 9b$ so $2\mid b$ and thus $b=2d$. Similary we see that $3\mid a$ so $a=3c$. Now we have $c-3d=1$ and thus $c= 3d+1$ where $d$ is arbitrary. Now we get $$x= 54(3d+1)$$ and $$y= 36d$$


问:
Why is the exponential distribution specified with parameter $X$ instead of $T$
One thing that always confused me about the exponential distribution was why the probability distribution function or even the expected value is modeled as a function of $X$ instead of a variable like $T$ for time--note capitalization means a random variable. This is not meant to be a nit-picky notation question, but really just trying to understand if I am missing something about the distribution and why $X$ is used.

The exponential is used for waiting time or inter-arrival times in general.

If you look at the Wikipedia article for example, the pdf is specified as:
$$
f(x;\lambda) = \lambda e^{-\lambda x}
$$

The exponential distribution models the probability that an event happens within some continuous interval $[0, t]$, given an intensity parameter $\lambda$. Now other distributions such as the poisson distribution, explicitly reference time as the variable of interest: $P(k, t; \lambda) = \frac{(\lambda t)^k e^{-\lambda t}}{k!}$

Using $x$, the notation for the pdf of the exponential becomes something awkward like $ P(X \leq T; \lambda)$ or equivalently $P(x \in [0, T])$. Wouldn't it seem more intuitive to specify the pdf as $P(T \leq t; \lambda)$, where the random variable is specified as $T$?

答:
Wouldn't it seem more intuitive to specify the pdf as $P(T≤t;λ)$, where
the random variable is specified as T?

The simple answer is no. "Intuitive" is context dependent term. For me, $X$ for exponential r.v. is much more intuitive than $T$, as I'm used to the standard notations of capital $X$ and $Y$ for random variables. Whatever they designate of applied to. You are concentrated much on the application, however, both exponential and Poisson distributions exist independently of their application and the very notion of time. E.g., Poisson is frequently used to count something, however you don't have to involve the time in this. You can count events per square meter, or per any other time independent measure-unit, or per subject, etc.

Regarding the exponential r.v., note that exponential distribution is a special case of Gamma distribution with $\alpha = 1$, and pace $\lambda$. Where Gamma is not associated that much with time-to-event modelling like the exponential distribution, so it is natural to denote Gamma by $X$, and thus it will be weird switch to $T$ for a special case of $\alpha=1$.

However, note that notations are generally context dependent. Such that, in the context of survival analysis, time-to-event is usually denoted by $T$, where $T$ can be distributed exponentially. The same holds for queuing theory, the waiting times are also usually denoted by $T$. That is, where the context is time, the $T$ notation is adopted, however for general introduction, there is no special reason to use $T$ and not the common notation of $X$.


问:
Proving $\sum_{n=0}^\infty\frac{x^n}{n!}>0$ for all $x\in\Bbb R$ without invoking any Taylor series knowledge base.
While studying basic Taylor expansions, I came across this expansion of $$e^x=\sum_{n=0}^\infty\frac{x^n}{n!}.$$We already know this function is positive for all $x$. But let us assume we don't know about the Taylor expansion , and try to prove this in the expanded form only, how to will we prove it ?

It's already clearly  visible for $ x > 0$ as well as for $ -1<x<0 $ . But how to prove it for $x<-1$ ? I have tried much but not able to build up a strong convincing proof.

Any help would be highly appreciated. thanks!

P.S : prove it 'WITHOUT' using $e^{x}$ function.

答: I'll use the Kronecker delta, so the $0^0=1$ we use for an $x^0$ term in Taylor series gives $0^n=\delta_{n0}$ for integers $n\ge0$. Let $[x^n]f(x)$ denote the $x^n$ coefficient in $f(x)$. As @leoli hints,$$[x^n]\sum_{k,\,l\ge0}\frac{x^k(-x)^l}{k!l!}=\sum_{k=0}^n\frac{1}{k!(n-k)!}=\frac{1}{n!}\sum_{k=0}^n\binom{n}{k}(-1)^k=\frac{0^n}{n!}=\delta_{n0},$$so $\sum_{k,\,l\ge0}\frac{x^k(-x)^l}{k!l!}=\sum_n\delta_{n0}x^n=1$ and, for $x<0$,$$\sum_l\frac{(-x)^l}{l!}>0\implies\sum_k\frac{x^k}{k!}=\frac{1}{\sum_l\frac{(-x)^l}{l!}}>0.$$


问:
Find map $f : \mathbb{R}\to {\mathbb{R}}$ such that inverse image $f^{-1}(${$x$}$)$ has two elements for any $x\in \mathbb{R}$.
Find map $f : \mathbb{R} \rightarrow \mathbb{R}$ such that inverse image $f^{-1}(${$x$}$)$ has two elements for any $x\in \mathbb{R}$.

I'm really struggling to find an example so any hints or examples would be much appreciated.

答:
For every $n\in\Bbb Z$, map the interval $[n,n+1)$ to itself twice over, by:

$$f(x) =
\begin{cases}
n+2(x-n)  & \text{if } x<n+\frac12 \\
n+2(x-n-\frac12) & \text{if }x\ge n+\frac12
\end{cases}
$$


问:
The limit of the product $\prod_{i=1}^n\frac{1 - (2i + 1)a/(2n)}{1 - ia/n}$ as $n\to\infty$
Given that $0 < a < 1$, what is the value of
$$ P = \lim_{n\to \infty} \prod_{i=1}^n \frac{1 - (2i + 1)a/(2n)}{1 - ia/n} $$
Thus, $P_n$, the $n$th term, is
$$ P_n = \frac{1-3a/2n}{1-a/n}\cdot \frac{1-5a/2n}{1-2a/n}\cdots \frac{1-(2n+1)a/2n}{1-a} $$

Motivation

The product describes a real-live physical problem. Put simply, if you have a river or stream in which you're sampling at an upstream site A and a downstream site B and you analyze the chemistry of the stream at A and B, if the concentration has changed, it is because of ground water which has seeped in between A and B. By making some simplifying assumptions, the concentration and amount of ground water which has entered are calculated using the evaluation of this infinite product.

答:
The product can be rewritten as
$$\mathcal{P}_n=\prod_{i=1}^{n}\frac{1 - (2i + 1)a/(2n)}{1 - ia/n}=\prod_{i=1}^{n}\frac{\frac{n}{a}-\frac12-i}{\frac{n}{a}-i}=\frac{\Gamma\left(\frac{n}{a}-n\right)}{\Gamma\left(\frac{n}{a}\right)}\cdot\frac{\Gamma\left(\frac{n}{a}-\frac12\right)}{\Gamma\left(\frac{n}{a}-n-\frac12\right)}$$
Now taking the logarithm and using Stirling formula
$$\ln\Gamma(x)=x(\ln x-1)+\frac12(\ln2\pi-\ln x)+O\left(x^{-1}\right)\qquad \text{as}\;
x\rightarrow\infty, $$
we arrive at
$$\ln\mathcal{P}_{\infty}=\frac12\ln(1-a),$$
which reproduces Ron Gordon's answer.</p>


问:
Be $H$ Sylow p-subgroup of $S_{2p}$ with $p > 2$ prime number. Be $N(H)$ the normalizer of H. Find the cardinality of $N(H)$
I've got some trouble doing this exercise

Be $H$ Sylow p-subgroup of $S_{2p}$ with $p > 2$ prime number. Be $N(H)$ the normalizer of H. Find the cardinality of $N(H)$

I know that H is generated by two disjoint p-cycle, so H is isomorphic to $\mathbb{Z}_{p}\times \mathbb{Z}_{p}$ but I'm stuck.

答:
We note that $N(H) = \text{Stab}(H)$ under the conjugation action of $S_{2p}$ on $H$. By the Orbit-Stabilizer Theorem, we have that $n_{p} \cdot N(H) = |S_{2p}|$, where $n_{p}$ is the number of Sylow $p$-subgroups.

Start by determining $n_{p}$. We have two disjoint $p$-cycles. We select the elements for the first $p$-cycle in $2p \cdot (2p-1) \cdot (2p-2) \cdots (p+1)$ ways. Two $p$-cycles are equivalent if one is obtained from the other by cyclic rotation. So we divide out by $p$, the number of cyclic rotations, to obtain:

$$2p(2p-1) \cdots (p+1)/p$$

such ways to choose the first $p$-cycle. Similarly, the second $p$-cycle is chosen in $p!/p = (p-1)!$ ways. The selections are independent. So by the rule of product, we have $n_{p} = (2p)!/p^{2}$. So $|N(H)| = p^{2}$.


问:
Is Bayes' Theorem really that interesting?
I have trouble understanding the massive importance that is afforded to Bayes' theorem in undergraduate courses in probability and popular science.

From the purely mathematical point of view, I think it would be uncontroversial to say that Bayes' theorem does not amount to a particularly sophisticated result. Indeed, the relation
$$P(A|B)=\frac{P(A\cap B)}{P(B)}=\frac{P(B\cap A)P(A)}{P(B)P(A)}=\frac{P(B|A)P(A)}{P(B)}$$
is a one line proof that follows from expanding both sides directly from the definition of conditional probability. Thus, I expect that what people find interesting about Bayes' theorem has to do with its practical applications or implications. However, even in those cases I find the typical examples being used as a justification of this to be a bit artificial.

To illustrate this, the classical application of Bayes' theorem usually goes something like this: Suppose that

1% of women have breast cancer;

80% of mammograms are positive when breast cancer is present; and

10% of mammograms are positive when breast cancer is not present.

If a woman has a positive mammogram, then what is the probability that she has breast cancer?

I understand that Bayes' theorem allows to compute the desired probability with the given information, and that this probability is counterintuitively low. However, I can't help but feel that the premise of this question is wholly artificial. The only reason why we need to use Bayes' theorem here is that the full information with which the other probabilities (i.e., 1% have cancer, 80% true positive, etc.) have been computed is not provided to us. If we have access to the sample data with which these probabilities were computed, then we can directly find
$$P(\text{cancer}|\text{positive test})=\frac{\text{number of women with cancer and positive test}}{\text{number of women with positive test}}.$$
In mathematical terms, if you know how to compute $P(B|A)$, $P(A)$, and $P(B)$, then this means that you know how to compute $P(A\cap B)$ and $P(B)$, in which case you already have your answer.

From the above arguments, it seems to me that Bayes' theorem is essentially only useful for the following reasons:

In an adversarial context, i.e., someone who has access to the data only tells you about $P(B|A)$ when $P(A|B)$ is actually the quantity that is relevant to your interests, hoping that you will get confused and will not notice.

An opportunity to dispel the confusion between $P(A|B)$ and $P(B|A)$ with concrete examples, and to explain that these are very different when the ratio between $P(A)$ and $P(B)$ deviates significantly from one.

Am I missing something big about the usefulness of Bayes' theorem? In light of point 2., especially, I don't understand why Bayes' theorem stands out so much compared to, say, the Borel-Kolmogorov paradox, or the "paradox" that $P[X=x]=0$ when $X$ is a continuous random variable, etc.

答:
You are mistaken in thinking that what you perceive as "the massive importance that is afforded to Bayes' theorem in undergraduate courses in probability and popular science" is really "the massive importance that is afforded to Bayes' theorem in undergraduate courses in probability and popular science." But it's probably not your fault: This usually doesn't get explained very well.

What is the probability of a Caucasian American having brown eyes? What does that question mean? By one interpretation, commonly called the frequentist interpretation of probability, it asks merely for the proportion persons having brown eyes among Caucasian Americans.

What is the probability that there was life on Mars two billion years ago? What does that question mean? It has no answer according to the frequentist interpretation. "The probability of life on Mars two billion years ago is $0.54$" is taken to be meaningless because one cannot say it happened in $54\%$ of all instances. But the Bayesian, as opposed to frequentist, interpretation of probability works with this sort of thing.

The Bayesian interpretation applied to statistical inference is immune to various pathologies afflicting that field.

Possibly you have seen that some people attach massive importance to the Bayesian interpretation of probability and mistakenly thought it was merely massive importance attached to Bayes's theorem. People who do consider Bayesianism important seldom explain this very clearly, primarily because that sort of exposition is not what they care about.


问:
Limit of a martingale sequence
I see this question on Stochastic Processes by Ross.

Let $X_n$ be a Markov process for which $X_0$ is uniform on (0, 1) and,

$$
    X_{n+1}=
\begin{cases}
    \alpha X_n + 1 - \alpha,& \text{w.p. } X_n,\\
    \alpha X_n,              & \text{w.p. }1-X_n.
\end{cases}
$$

where $0<\alpha<1$. Discuss the limiting properties of the sequence $\lim_{n\to\infty} X_n$.

I was able to show $E[X_{n+1}\mid X_n]=X_n$, so the sequence is a martingale. Also since it is bounded between 0 and 1, a limit exists by the martingale convergence theorem. Also the mean of this limit is 0.5, as $E[X_0]=0.5$ and the mean is constant for a martingale. Can we precisely say what the limit is? Is there anything else we could say about the limit?

答:
Denote $X_\infty$ the limit of the process. Let $x\in(0,1)$ and define $A_x$ as
$$A_x\equiv\left\{\omega\in\Omega:\lim_{n\rightarrow\infty}X_n(\omega)=x\right\}$$
Now, fix $\omega\in A_x$ and $\varepsilon>0$. By definition
$$\exists N\in\mathbb{N}, \forall n\geq N:\quad X_n(\omega)\in (x-\varepsilon, x+\varepsilon)\equiv I$$
In particular, for all $n\geq N$, $X_{n+1}(\omega)$ must lie in the set
$$(\alpha(x-\varepsilon),\alpha(x+\varepsilon))\cup(\alpha(x-\varepsilon)+1-\alpha,\alpha(x+\varepsilon)+1-\alpha)\equiv J$$
Notice that if we take $\varepsilon$ small enough then $I\cap J=\emptyset$, which is impossible since $X_{n+1}(\omega)\in I\cap J$. This implies that for all $x\in(0,1)$, the set $A_x$ is empty. Therefore, $X_\infty$ a.s. takes values in $\left\{0,1\right\}$, and its distribution can be inferred from the equality $\mathbb{E}X_\infty=1/2$.

Title: Limit of a martingale sequence


问:
Connection between vector space isomorphisms and dimensions
Assuming the Axiom of Choice, and therefore assuming that every vector space has a Hamel basis, it isn't too hard to show that if $U$ and $V$ are vector spaces with the same dimension (i.e. their bases have the same cardinality), then $V\cong U$. However, I'm not so sure how to deal the converse implication (assuming it's true in general in the first place):

Let $\mathcal B_1$ be a basis for $V$ and $\mathcal B_2$ a basis for $U$. If $V\cong U$ then there exists a linear map $\phi: V\to U$ that is bijective, and so particularly, $\text{Im}(\phi) = U = \langle \mathcal B_2\rangle$.

I'm not sure how to conclude that $\phi(\mathcal B_1)=\mathcal B_2$. I thought of using the First Isomorphism Theorem ($V/\text{ker}(\phi)\cong \text{Im}(\phi)$) to mimic the Rank-Nullity Theorem being used for the finite-dimensional case, but it doesn't really help in this scenario.

答:
Suppose that $\mathcal B_1 = (v_\alpha : \alpha \in I)$ is a basis of $V$ and that $\phi: V \to U$ is a bijective linear map. We now consider $\mathcal B_2 = \phi(\mathcal B_1) = (\phi(v_{\alpha}): \alpha \in I)$.

Claim: $\phi(\mathcal B_1)$ spans $U$.

Proof: Consider an arbitrary $u \in U$. $\phi$ is surjective, so there exists a $v \in V$ with $\phi(v) = u$. Because $\mathcal B_1$ is a (Hamel) basis, there exist elements $v_{\alpha_1}, \dots, v_{\alpha_n}$ and coefficients $c_1,\dots,c_n$ such that
$$
v = \sum_{i=1}^n c_i v_{\alpha_i} \implies u = \phi(v) = \sum_{i=1}^n c_i \phi(v_{\alpha_i}).
$$

Claim: $\phi(\mathcal B_1)$ is linearly independent.

Proof: Suppose that $v_{\alpha_1},\dots,v_{\alpha_n}$ and coefficients $c_1,\dots,c_n$ are such that $\sum_{i=1}^n c_i \phi(v_{\alpha_i}) = 0$. It follows that
$\phi \left(\sum_{i=1}^n c_i v_{\alpha_i} \right) = 0$. Because $\phi$ is injective, it must holds that $\sum_{i=1}^n c_i v_{\alpha_i} = 0$. Because $\mathcal B_1$ is a basis, this can only hold if $c_{\alpha_1} = \cdots = c_{\alpha_n} = 0$.

So, $\phi(\mathcal B_1)$ is indeed linearly independent.

Because $\phi(\mathcal B_1)$ is a linearly independent spanning set, it is a basis.


问:
Restriction of symplectic form on submanifold
Consider a symplectic manifold $(M,\omega)$, for a 2 dimensional submanifold $N \subset M$, do we always have $\omega|_{N}$ gives a volume form on $N$?If not, is there any condition we can put on $N$ to make this happen?

答:
Consider $M = T^4$ with the symplectic form $\omega = dx^1\wedge dx^2 + dx^3\wedge dx^4$ and $N = T^2$ embedded in $M$ via the map $i : N \to M$ given by $i(y^1, y^2) = (y^1, 1, y^2, 1)$. Then $\omega|_N = i^*\omega = 0$ which is not a symplectic form on $N$.

A situation where $\omega|_N$ is necessarily a symplectic form is the following: $M$ is a Kähler manifold and $N$ is a complex submanifold (and hence Kähler).


问:
Cardinality of set of functions
Let $\mathbb{N}=\{0,1,2,\ldots\}$ and we consider the following set:

$$\{f:\mathbb{N}\longrightarrow\mathbb{N} \mid\forall n\in\mathbb{N} \ , f(n)=f(n+1)+f(n+2)\}$$

I can tell that this set has at least one element $f:\mathbb{N}\to\mathbb{N}$ such that $f(n)=0$ but can we say more about its cardinal? and what would happen in the case of decreasing functions

答:
Take $a,b\in\Bbb N$. Then consider the function $f\colon\Bbb N\longrightarrow\Bbb N$ such that:

$f(0)=a$;

$f(1)=b$;

$f(2)=-f(1)+f(0)=-b+a$;

$f(3)=-f(2)+f(1)=2b-a$

and so on. So, $f$ belongs to your set and, clearly, every element form your set can be obtained by this process. But then$$n>0\implies f(n)=(-1)^n\left(F_{n-1}a-F_nb\right),$$where $F_n$ is the $n$th Fibonacci number (this can be proved by induction), with $F_0=0$ and $F_1=1$. But then, unless $a=b=0$, you will always have $f(n)<0$ for some $n\in\Bbb N$. So, your set consists only of the null function.

Title: Cardinality of set of functions