问:
Show that $|A| = 0$.
Let $A \subset \mathbb{R}$ such that $\forall x \in \mathbb{R}$ and $r>0$ holds $|A \cap [x,x+r]| \leq rf(r)$ where $\lim_{r \rightarrow 0} f(r) = 0$. Show that $|A| = 0$.

I tried to prove the statement when $A$ is bounded, but i failed. Therefore any help is apperciated, but rather a hint than a solution. I guess the general case follows from the bounded case.

答:
Suppose that $A\subset [u,v]$, for every $n>0$, denote $A_{i,n}=[u+{i\over n}(v-u),u+{{i+1}\over n}(v-u)]\cap A$, $A\subset \cup_{i=0}^{i=n-1}A_{i,n}$, $\mu(A_{i,n})\leq {{u-v}\over n}f({{u-v}\over n})$.

This implies that $\mu(A)\leq\sum_{i=1}^{i=n-1}{{v-u}\over n}f({{v-u}\over n})$.

For every $c>0$, there exists $N$ such that for every $n>N$, $f({{v-u}\over n})<c$, this implies that $\mu(A)\leq \sum_{i=1}^{i=n-1}{{v-u}\over n}f({{v-u}\over n})\leq \sum_{i=1}^{i=n-1}c{{v-u}\over n}\leq c(v-u)$. This implies that $\mu(A)=0$/

Title: Show that $|A| = 0$.


问:
Calculating the plim of a posterior belief about a Bernoulli r.v. formed from observing an infinite sequence of other Bernoulli r.v.'s
Suppose $Y$ is a Bernoulli random variable s.t. Prob$\{Y=1\} = \alpha\in(0,1)$. Further suppose I observe a sequence $\{X_t\}_{t=1}^\infty$ of i.i.d. Bernoulli random variables s.t. $\text{Prob} \{X_t=1|Y = y\} = \beta_y\in(0,1)$ $\forall y\in\{0,1\}$.

How would I calculate $\underset{t\to\infty}{\text{plim}}\ \text{Prob}\Big\{Y=1|\{X_{\tau}\}_{\tau=1}^t\ \Big\}?$ Thank you very much in advance!

P.S. If there are conditions (on $\alpha,\beta_0,\text{ and }\beta_1$) under which the above plim may not exist, I have a followup question. (But I suppose we'll cross that bridge when/if we get there!!) Thanks again!

答:
Using Bayes' Theorem and the Law of Total Probability we have:
$$\textrm{Prob}\left\{Y = 1 \mid \left\{X_\tau\right\}_{\tau=1}^t\right\} = \frac{\textrm{Prob}\left\{\left\{X_\tau\right\}_{\tau=1}^t \mid Y = 1\right\}\textrm{Prob}\left\{Y = 1\right\}}{\textrm{Prob}\left\{\left\{X_\tau\right\}_{\tau=1}^t \mid Y = 1\right\}\textrm{Prob}\left\{Y = 1\right\} + \textrm{Prob}\left\{\left\{X_\tau\right\}_{\tau=1}^t \mid Y = 0\right\}\textrm{Prob}\left\{Y = 0\right\}}.$$
Let $S_t := \sum_{\tau=1}^t X_\tau$. From your description we have
$$\textrm{Prob}\left\{\left\{X_\tau\right\}_{\tau=1}^t \mid Y = y\right\} = \beta_y^{S_t} \left(1-\beta_y\right)^{t-S_t}.$$
Hence
$$\textrm{Prob}\left\{Y = 1 \mid \left\{X_\tau\right\}_{\tau=1}^t\right\} = \frac{\beta_1^{S_t} \left(1-\beta_1\right)^{t-S_t}\alpha}{\beta_1^{S_t} \left(1-\beta_1\right)^{t-S_t}\alpha + \beta_0^{S_t} \left(1-\beta_0\right)^{t-S_t}(1-\alpha)}.$$
If $\beta_0 = \beta_1$ then this probability will equal $\alpha$ no matter what data we observe. Otherwise, let
$$R_t = \frac{\beta_0^{S_t} \left(1-\beta_0\right)^{t-S_t}}{\beta_1^{S_t} \left(1-\beta_1\right)^{t-S_t}}.$$
Then
$$\textrm{Prob}\left\{Y = 1 \mid \left\{X_\tau\right\}_{\tau=1}^t\right\} = \frac{\alpha}{\alpha + R_t(1-\alpha)}.$$
Now we need to work out what $R_t$ converges to. We first note that $S_t = t\bar{X}_t$. Given $Y = y$, we know that $\underset{t\to\infty}{\text{plim}}\ \bar{X}_t = \beta_y$.

Considering the case $Y = 1$,
$$\underset{t\to\infty}{\text{plim}}\ R_t = \lim_{t\to\infty} \frac{\beta_0^{\beta_1t} \left(1-\beta_0\right)^{t\left(1-\beta_1\right)}}{\beta_1^{\beta_1t} \left(1-\beta_1\right)^{t\left(1-\beta_1\right)}} = \lim_{t\to\infty} \left[\frac{\beta_0^{\beta_1} \left(1-\beta_0\right)^{\left(1-\beta_1\right)}}{\beta_1^{\beta_1} \left(1-\beta_1\right)^{\left(1-\beta_1\right)}}\right]^t.$$
Using differentiation you can see that $x^{\beta_1}(1-x)^{1-\beta_1}$ is maximised when $x = \beta_1$. So
$$\beta_0^{\beta_1} \left(1-\beta_0\right)^{\left(1-\beta_1\right)} < \beta_1^{\beta_1} \left(1-\beta_1\right)^{\left(1-\beta_1\right)}$$
and hence
$$\underset{t\to\infty}{\text{plim}}\ R_t = \lim_{t\to\infty} \left[\frac{\beta_0^{\beta_1} \left(1-\beta_0\right)^{\left(1-\beta_1\right)}}{\beta_1^{\beta_1} \left(1-\beta_1\right)^{\left(1-\beta_1\right)}}\right]^t = 0,$$
and so
$$\underset{t\to\infty}{\text{plim}}\ \textrm{Prob}\left\{Y = 1 \mid \left\{X_\tau\right\}_{\tau=1}^t\right\} = \underset{t\to\infty}{\text{plim}}\ \frac{\alpha}{\alpha + R_t(1-\alpha)} = 1.$$

Considering the case $Y=0$, we can use similar reasoning to see that
$$\underset{t\to\infty}{\text{plim}}\ R_t = \infty$$
and so
$$\underset{t\to\infty}{\text{plim}}\ \textrm{Prob}\left\{Y = 1 \mid \left\{X_\tau\right\}_{\tau=1}^t\right\} = \underset{t\to\infty}{\text{plim}}\ \frac{\alpha}{\alpha + R_t(1-\alpha)} = 0.$$

In conclusion,

if $\beta_0 = \beta_1$ then $\underset{t\to\infty}{\text{plim}}\ \textrm{Prob}\left\{Y = 1 \mid \left\{X_\tau\right\}_{\tau=1}^t\right\} = \alpha$.

Otherwise, if $Y = 1$ then $X_{\tau} \sim \textrm{Bernoulli}(\beta_1)$ and $\underset{t\to\infty}{\text{plim}}\ \textrm{Prob}\left\{Y = 1 \mid \left\{X_\tau\right\}_{\tau=1}^t\right\} = 1$.

And if $Y = 0$ then $X_{\tau} \sim \textrm{Bernoulli}(\beta_0)$ and $\underset{t\to\infty}{\text{plim}}\ \textrm{Prob}\left\{Y = 1 \mid \left\{X_\tau\right\}_{\tau=1}^t\right\} = 0$.

ContentLicense: CC BY-SA 4.0


问:
Another proof of $|\vec{a}\times\vec{b}|^2 = |\vec{a}|^2\cdot|\vec{b}|^2 - (\vec{a}\cdot\vec{b})^2$.
I know how to prove the formula $|\vec{a}\times\vec{b}|^2 = |\vec{a}|^2\cdot|\vec{b}|^2 - (\vec{a}\cdot\vec{b})^2$, but my teacher asked me to prove it by using the formula $(\vec{a}\times\vec{b})\times\vec{c} = (\vec{a}\cdot\vec{c})\vec{b} - (\vec{b}\cdot\vec{c})\vec{a}$, how to prove it?

答:
Slightly differently:

Let $\vec c= \vec a \times \vec b$, and use $(\vec P \times \vec Q). \vec R=\vec P. (\vec Q \times \vec R)$, and $\vec P \times (\vec Q \times \vec R)=(\vec P. \vec R)\vec Q-(\vec P. \vec  Q) \vec R$, then
$$|\vec a \times \vec b|^2=
(\vec a \times \vec b).(\vec a \times \vec b)= (\vec a \times \vec b).\vec c=\vec a.(\vec b \times \vec c)= \vec a.[\vec b \times (\vec a \times \vec b)]$$ $$=\vec a.[ b^2 \vec a- (\vec b. \vec a) \vec b]= a^2b^2-(\vec a.\vec b)^2.$$


问:
Potential function of $F(x, y)=\left(\lambda x-x^3, y\right)$
I have the following ODE: $$x''=\lambda x-x^3$$
To work with it, I turned it into a system of equations:
$$\begin{matrix}
x' & = & y \\
y' & = & \lambda x -x^3
\end{matrix}$$
Now I want to find a first integral for this system, as well as a potential function for $F(x, y)=\left(\lambda x-x^3, y\right)$. To find the first integral I've done:
$$\frac{dy}{dx}=\frac{\lambda x-x^3}{y} \quad \Longleftrightarrow \quad \frac{y^2}{2}=\lambda \frac{x^2}{2}-\frac{x^4}{4} +C$$
Therefore the first integral is:
$$E(x,y)=y^2-\lambda x^2+\frac{x^4}{2}$$
I've tried calculating the potential function but I got the exact same result, and I don't think this is correct. Can someone help me with the potential? Thank you in advance

答:
There was two mistakes in the equation :
$$y^2=\lambda x^\color{red}{2}-\frac{x^4}{\color{red}{2}  }+C$$
( Corrected in red ).

Then :
$$y=x'=\pm \sqrt{\lambda x-\frac{x^4}{2}+C}$$
$$\frac{dx}{dt}=\pm \sqrt{\lambda x-\frac{x^4}{2  }+C}$$
$$t(x)=\pm \int \frac{dx}{\sqrt{\lambda x^2-\frac{x^4}{2  }+C}}$$
This is an elliptic integral.

The inverse function $x(t)$ involves the Jacobi elliptic functions.


问:
Is this definition of a function correct?
So im trying to teach myself some logic and how quantifier scope and bound works and I saw the following defintion of a function which left me scratching my head.

$$\forall a\in A \exists b\in B:((a,b)\in R\land \forall b'\in B:(a,b')\in R\implies b=b').$$

We can see from the defintion that the variables $a$ and $b$ are bound in the entire expression by the brackets however my confusion arises for the variable $b'$ clearly it is bounded in the set $B$ however for the implication no brackets are used to bound $b'$ in the implication so basically my question is should we instead write

$$\forall a\in A \exists b\in B:((a,b)\in R\land \forall b'\in B:((a,b')\in R\implies b=b')).$$

With the addition of the extra brackets we have bounded the $b'$ in the implication as it falls within the quantifiers scope is this correct unnecessary or wrong?.

Thanks in advance.

答:
Should we instead write

$$\forall a\in A \exists b\in B:((a,b)\in R\land \forall b'\in B:((a,b')\in R\implies b=b')).$$

Yes.  You would use this to prove that the set of ordered pairs $R$ is a function mapping each element of $A$ to a unique element of $B$.

If you have established or simply assumed that $R$ is a function mapping each element of $A$ to a unique element of $B$, you could write

$$\forall a\in A: (R(a)\in B)$$

where $R(a)$ is the image of $a$ under $R$.


问:
Support of a measure on Borel sets not well defined.
Let $(X, \tau$) be a topological space and let $B$ be the Borel $\sigma$-algebra generated by $\tau$. Let  $(X,B,\mu)$ be a measure space. Then the support of the distribution $\mu$ is defined as :
$$\operatorname{supp}({\mu})=\{x\in X : \mu(U)>0\text{ for all U open neighborhood of }x\}.$$ Now in the wikipedia article about  Radon measures, it is written

A common problem is to find a good notion of a measure on a
topological space that is compatible with the topology in some sense.
One way to do this is to define a measure on the Borel sets of the
topological space. In general there are several problems with this:
for example, such a measure may not have a well defined support.
Another approach to measure theory is to restrict to locally compact
Hausdorff spaces[...]

But I don't see why the support is not well defined when we use the above expression.

答:
If the topological space is not second countable, the definition
$$\operatorname{supp}({\mu})=\{x\in X : \mu(U)>0\text{ for all $U$ open neighborhood of }x\}$$
can still be applied, but it may not yeld what is typically expected of the support of measure.

So while the definition of support of a measure as you posted, or as we can find here, (they are the same definition with slightly different wording) is a definition that can be applied to the general case, it is only useful (or meaningful) if there are some condition on the topology (such as being second countable).

Let us see a concrete simple example:

Let $X=[0,1]$ and $\tau = \{ A \subseteq [0,1] \}=2^X$. Please, note that $(X, \tau)$ is not second countable.

Let $\mathcal{B}$ be the Borel $\sigma$-algebra corresponding to $\tau$. So  $\mathcal{B}= 2^X$.

Let $\mu$ be a measure defined by, for any $E \in \mathcal{B}$, $\mu(E)=0$ if $E$ is countable and  $\mu(E)=+\infty$ if $E$ is uncountable. It is easy to check that $\mu$ is a measure.

Now what is the support of $\mu$?  Let us simply apply the definition:

$$\operatorname{supp}({\mu})=\{x\in X : \mu(U)>0\text{ for all $U$ open neighborhood of }x\}$$

But, in our case, for all $x \in X$, $\{x\}$ is an open neighborhood of $x$ and $\mu(\{x\})=0$. So, for all $x \in X$, $x \notin \operatorname{supp}({\mu})$. So, $\operatorname{supp}({\mu}) = \emptyset$.

Why is it odd to have $\operatorname{supp}({\mu}) = \emptyset$? Because, then we have  some $E\subseteq \operatorname{supp}({\mu})^c$, such that $\mu(E) >0$ (just take any uncountable set $E \subseteq X$).

Note that, for this example to work, it was key that any $E$ uncountable could not be written as countable union of the open sets $\{x\}$.

Remark:  The notion of support of measure is a minimal set outside of which the measure is null (similar to the notion of support of a function). It means that,
for any measurable set $E$, if $E \subseteq \operatorname{supp}({\mu})^c$ then $\mu(E)=0$.  However, using the definition of support of a measure as you posted, or as we can find here, this does not need to be the case for all  topological space, as we have seen in the example above,

Let us now prove that, if the topological space is second countable, then for any measurable set $E$, if $E \subseteq \operatorname{supp}({\mu})^c$ then $\mu(E)=0$.

Proof: Note that

$$\operatorname{supp}({\mu})^c=\{x\in X : \mu(U)=0\text{ for some $U$ open neighborhood of }x\}$$

A base of the topological space is a collection $B$ of open sets such that any open set $U \subseteq X$ is the union of elements (open sets) in $B$. The open sets in $B$ are called basic open sets.

So, if there is an open set $U$ such that $x\in U$ and $ \mu(U)=0$ then there is a basic open set $U'$ such that    $x\in U'\subseteq U$ and $\mu(U') =0$.
On the other, if there is a basic open set $U'$ such that    $x\in U'$ and $\mu(U') =0$, then we can take $U'$ as $U$, and we have that there is an open set $U$ such that $x\in U$ and $ \mu(U)=0$ . So, we have that: there is an open set $U$ such that $x\in U$ and $ \mu(U)=0$ if and only if  there is a basic open set $U'$ such that    $x\in U'$ and $\mu(U') =0$.

So, we have that
$$\operatorname{supp}({\mu})^c=\{x\in X : \mu(U')=0\text{ for some $U'$ basic  open set such that }x \in U'\}$$

So, for every $x \in \operatorname{supp}({\mu})^c$, there is a basic open set $U'$ such that $x\in U'$ and  $\mu(U')=0$. So
$$ \operatorname{supp}({\mu})^c \subseteq \bigcup_{U' \in B \textrm{ and }\mu(U')=0} U'\tag{1}$$

On the other hand, if $U'$ is a basic open set and $\mu(U')=0$, then for all $x\in U'$, $x \in \operatorname{supp}({\mu})^c$. So $U' \subseteq \operatorname{supp}({\mu})^c$. So,
$$ \bigcup_{U' \in B  \textrm{ and }\mu(U')=0} U' \subseteq \operatorname{supp}({\mu})^c \tag{2}$$
From $(1)$ and $(2)$,
$$ \operatorname{supp}({\mu})^c = \bigcup_{U' \in B \textrm{ and }\mu(U')=0} U'$$

So, it is clear that $\operatorname{supp}({\mu})^c$ is an open set.

Now, suppose that the topological space is second countable, that means, it has a countable base $B$. If $B$ is countable, we have
$$ \mu(\operatorname{supp}({\mu})^c) = \mu \left (\bigcup_{U' \in B \textrm{ and }\mu(U')=0} U' \right ) \leqslant \sum_{U' \in B \textrm{ and }\mu(U')=0}\mu(U') = 0$$

So, for any measurable set $E$, if $E \subseteq \operatorname{supp}({\mu})^c$ then $\mu(E)=0$.


问:
If $x^{T}A^{T}Ax = x^Tx$ holds for every $x$, then $A^{T} A = I_n$

Given $A \in \mathbb R^{n \times n}$, if $$\left( \forall x \in\mathbb R^n \right) \left(x^{T} A^{T} A x = x^T x \right)$$ how to conclude that $A^{T}A = I_n$?

I appreciate any help!

答:
Slightly different approach (in the final stage of proof).

We have $$\left( \forall x \in\mathbb R^n \right) \left(x^{T} (A^{T} A -I) x =0 \right)$$

$S=A^{T} A -I$ is obviously symmetric hence it is diagonalizable.

Equation above must be hold also for any eigenvector $v$ of $S$.
i.e. $v^TSv=0 =v^T\lambda v$,

hence every eigenvalue $\lambda$ of $S$ is zero and it's diagonal form must be zero matrix what means $S$ is zero matrix itself.


问:
Proving a set is a manifold

Prove that the following set is a manifold and find it's dimension $$M=\{(x,y)\in\Bbb{R^3}\times\Bbb{R^3}|x\cdot y=0, \|x\|=\|y\|=1\}$$

So $M$ can be thought of as the intersection of the following equations: $$x_1y_1+x_2y_2+x_3y_3=0$$ $$x_1^2+x_2^2+x_3^2=1$$$$y_1^2+y_2^2+y_3^2=1$$

I tried defining an implicit function $F(x,y)$ s.t $F(x,y)=0$ iff $(x,y)\in M$ but didn't know how to take it from here. Any help would be appreciated.

答:
What if you consider the $\mathcal C^1$ function $\mathbf F : \mathbb R^6 \to \mathbb R^3$ defined as: $$\mathbf F(\mathbf x)= \begin{bmatrix}
    x_1y_1+x_2y_2+x_3y_3 \\
    x_1^2+x_2^2+x_3^2 -1 \\
    y_1^2+y_2^2+y_3^2-1 \\
    \end{bmatrix} $$

Can you check $\text{rank}(D \mathbf F(\mathbf x))$ now?

Title: Proving a set is a manifold


问:
Green's Theorem Double Integration Changing the Order
I've been trying to work out how to change the order of integration (my tutor requested both ways - this is not a homework assignment) for the following problem when calculating using Green's Theorem:

$\oint\limits_C (6y - 3y^2 + x)~ \delta x + yx^3~ \delta y$ where C is shown in the image below.

I believe the correct solution is:
$\oint\limits_{C} (6y - 3y^2 + x) \delta~ x + yx^3 \delta~ y = \int\int\limits_D \frac{\delta Q}{\delta x} - \frac{\delta P}{\delta y}~ \delta A
= \int\limits_0^1\int\limits_{3x}^{5x} 3yx^2 - 6 + 6y~ \delta y~ \delta x = 14.8$

I am trying to calculate the same integral but with the order of integration the other way round (dx first then dy but can't seem to find the right limits that yield the same solution.

I believe the limits for the outer integral is 0 to 5 and that the inner integral involves using y/5 and y/3 but I'm not sure how to do this.

Hope that makes sense and thanks in advance for your help!

答:
If you are changing the order and integrating wrt $dx$ first, then you need to split your integral into two.

i) First between $0 \leq y \leq 3$, notice that $x$ is bound between lines $y = 5x$ and $y = 3x$.

So $\frac{y}{5} \leq x \leq \frac{y}{3}$

ii) Now between $3 \leq y \leq 5$, note that the lower bound of $x$ continues to be defined by the line $y = 5x$ but the upper bound is defined by line $x = 1$.

So the second integral will have bounds $3 \leq y \leq 5, \frac{y}{5} \leq x \leq 1$.

Can you take it from here? Let me know if any questions.


问:
What is the attitude to this problem?
I am trying to show that for ever $z,w \in \mathbb{C}$ the following hold : $$
|1-z \bar{w}|^{2}+|z+w|^{2}=\left(1+|z|^{2}\right)\left(1+|w|^{2}\right)
$$

I am not sure if my approach here is true, I notice that I can mark $$z=a+bi$$ $$w=c+di$$ and than subtitude it in the above expression, but this seems too much algebra and I think this is not the point of the problem. (I am not even sure if that approach is true)

答: The approach you're suggesting is right. And yes, it would involve some algebra. However, there is an identity you can use to solve it and the exercise seems to be an application of that identity. I'll provide that much and leave the rest to you. Use the fact that for any $z \in \mathbb{C}$, $z\bar{z}=|z|^2$.

Title: What is the attitude to this problem?


问:
Finding covariance of two negative binomial distributions
Take a set of random variables $X_n$ which indicate number of tosses of a coin ($P(H)=p$) to achieve $n$ heads. What is $\operatorname{cov}(X_1,X_5)$?

I reasoned that the number of tosses till the first heads is not going to affect the number of tosses for the next four. So I can split $X_5$ into $X^{'}_1+X^{'}_4$ which are the number of tosses to get the first and latter 4 respectively of $X_5$. Since those two are independent, $\operatorname{cov}(X_1,X_5)$ becomes $\operatorname{cov}(X_1,X^{'}_1)$. Since $X_1$ and $X^{'}_1$ are the same, $\operatorname{cov}(X_1,X^{'}_1)=\operatorname{var}(X_1)=\frac{1-p}{p^2}=\operatorname{cov}(X_1,X_5)$

Is this reasoning sound?

答: Community wiki answer so the question can be marked as answered: Yes, the argument is sound.</p>

ContentLicense: CC BY-SA 3.0


问:
$(1+x+...+x^{2n})^{\frac{1}{2n}}$ is convex?
For natural number $n$, Let $f_n=(1+x+...+x^{2n})^{\frac{1}{2n}}$.  Then I think that $f_n$ is convex. Is it true? And how to I prove that? I tried to prove that $f_n''>0$. But i cannot to prove that.

答: Hint: Use the identity $1 + x + x^2 + \cdots + x^k = \dfrac {x^{k+1}-1}{x-1}$ to rewrite the function and take the second derivative. To prove $f''(x) \ge 0$, you will need to prove that $1 + x + x^2 + \cdots + x^k \ge 0$ for all $x$ if $k$ is even.


问:
Is it possible to calculate $card \{ (x, y) | xy = (x \& y) (x | y) \}$, where $x, y$ are integers>=0 in $[a, b]$ and $\&, |$ are bitwise AND and OR?
I saw a programming task of calculating $card \{ (x, y) | x * y= (x \& y) * (x | y) \}$ where $x, y \in [ 0, 31 ]$, then I came up with a thought: is this workable by Maths?

答:
With $c=x\mathop\&y$, we have $x=a+c$, $y=b+c$, $x\mathop| y=a+b+c$, where $a,b,c$ are "bit-disjoint". The condition then reads
$$(a+c)(b+c)=c(a+b+c)$$
and simplifies to
$$ab=0. $$
Thus at least one of $a,b$ must be $=0$, in other words, one of $x,y$ must be a "bit-subset" of the other. If the range in question is $[0,2^n-1]$, we find pairs with "$x\subseteq y$" by having three choices for each of the $n$ bits: is not in $y$, is in $y$ but not $x$, is in $x$ (and $y$). This gives us $3^n$ pairs. Likewise, we have $3^n$ pairs with "$y\subseteq x$". But we counted the $2^n$ cases with $x=y$ twice, hence the final answer is
$$ 2\cdot 3^n-2^n.$$


问:
Determining the normal functionals on the direct sum of von Neumann algebras
Consider a collection $\{M_i: i \in I\}$ of von Neumann algebras and consider their $\ell^\infty$-direct sum $M$. I'm trying to show that a normal functional $\omega$ on $M$ is of the form
$$\omega(m) = \sum_i \omega_i(m_i), m = (m_i)_i \in M$$
where $\omega_i \in (M_i)_*$ and $\sum_i \|\omega_i\|<\infty$.

In another post, user Ruy suggested to note that

$$
  (M_*)^* \cong
  M \cong
  \oplus_i^\infty M_i \cong
  \oplus_i^\infty ({M_i}_*)^* \cong
  (\oplus_i^1{M_i}_*)^*.
  $$
and then invoke uniqueness of the predual of a von Neumann algebra to obtain
$$M_* \cong \bigoplus_i^1 {M_i}_*.$$

However, to prove the above result I guess I need to know explicitely how this isomorphism looks like. How can I see this?

答:
Let me try another perhaps more pedestrian  approach.

So we're given that  $M=\ell ^\infty (M_i)$ (I guess I prefer this notation over ${\oplus}_i^\infty  M_i$), where the $M_i$ are von
Neumann algebras.   Denoting by $M_*$ the space of normal linear functionals, we thus need to prove that
$$
  M_*=\ell ^1({M_i}_*).
  $$

Given $\omega \in M_*$, denote by $\omega _i$ the restriction of $\omega $ to $M_i$, so that $\omega _i\in {M_i}_*$, and
let us prove that
$$
  \|\omega \| = \sum_i\Vert \omega _i\Vert.
  \tag 1
  $$

Fixing  $\rho $ in the interval $(0, 1)$,  and for each $i$,  choose some  $m_i$ in $M_i$ such that $\Vert m_i\Vert =1$, and
$\omega _i(m_i)\geq \rho \Vert \omega _i\Vert $.

If $F$ is a finite subset of the index set $I$, and $m_F\in M$ is defined to have coordinates $m_i$, for $i$ in $F$, and zero
elsewhere, then $\Vert m_F\Vert =1$, and
$$
  \Vert \omega \Vert  \geq  |\omega (m_F)| = \sum_{i\in F}\omega _i(m_i) \geq  \sum_{i\in F} \rho \Vert \omega _i\Vert .
  $$
Since both $F$ and $\rho $ are arbitrary,  it follows that
$\Vert \omega \Vert  \geq \sum_{i\in I} \Vert \omega _i\Vert $.

In order to prove the reverse inequality, let $N$ be the subalgebra of $M$ formed by the elements
with finitely many nonzero coordinates.  Then, for every $m$ in the unit ball of $N$, we have that
$$
  |\omega (m)| =
  \big |\sum_{i\in I}\omega _i(m_i)\big | \leq
  \sum_{i\in I}|\omega _i(m_i) | \leq  $$$$ \leq
  \sup_{i\in I}\|m_i\|\sum_{i\in I}\|\omega _i\| \leq
  \sum_{i\in I}\|\omega _i\|.
  \tag 2
  $$

Using von Neumann's double commutant Theorem, it is easy to see that $N$ is weakly dense in $M$.  Furthermore, by Kaplansky's
density theorem, we have that the unit ball of $N$ is dense in the unit ball of $M$ relative to $\sigma (M,M_*)$, aka the weak$^*$
topology.  Therefore the inequality in (2) is preserved if, instead,  $m$ is taken  in  the unit ball of $M$.  This concludes
the proof of claim (1)

Therefore this defines  an isometric  linear map
$$
  \omega \in M_*\mapsto (\omega _i)_i\in \ell ^1({M_i}_*),
  \tag 3
  $$
and we'll be done once we prove it to be  onto.

To see that this is indeed the case, pick $(\omega _i)_i\in \ell ^1({M_i}_*)$, and
define for each subset $J\subseteq I$, the linear functional $\omega _J$ on $M$ by
$$
  \omega _J(m) =\sum_{i\in F} \omega _i(m_i),\quad\forall m\in M.
  $$

For $J$ finite it is clear that $\omega _J$ is normal,  and hence so is $\omega :=\omega _I$,  since the latter is the norm limit of the
$\omega _F$,  for $F$ finite.

Finally notice that the map in (3) sends $\omega $ to $(\omega _i)_i$, concluding the proof.


问:
Compute $\sum_{n \in Z} \exp \left(ia*(n^2-2fn)\right) $ with $a>0\ $
I want to Compute $$\sum_{n \in Z} \exp \left(ia*(n^2-2fn)\right)$$
with $a>0\ $ being a real number. My final result should be in the form of
$$
\sum_{n=-\infty}^{\infty} \boldsymbol{c}(\boldsymbol{n}) \boldsymbol{\delta}(\boldsymbol{f}-\boldsymbol{\beta} \boldsymbol{n})
$$
where I have to express c and $\beta$ coefficients. I tried to break the sum into even and off and make use of the DTFT table but with no result.
Any suggestions??

答:
$$g(x)=\sum_{n \in Z} \exp \left(ia(n^2-2xn)\right), \qquad h(x)=\sum_{n=-\infty}^{\infty}c(n)\delta(x-bn)$$ both series converge in the sense of distributions.

$g$ is $\pi/a$ periodic. If $g=h$ then $h$ is $\pi/a$ periodic ie. $b=\frac{\pi}{ak}$ and $c$ is $k$-periodic.

Whence its Fourier coefficients $e^{ian^2}$ are $k$ periodic.

If $a \not \in 2\pi\Bbb{Q}$ then this is impossible.

If $a=2\pi l/k $ then $b=\frac{\pi}{ak}$ and $c(n)/b$ is the inverse FFT of $e^{ian^2},n\in 0,\ldots k-1$.


问:
What are a , b and c?
$$y = ax^2 + bx + c$$

which is tangent at the origin with the line $y=x$, It is also tangential with the line $y=2x + 3$. Determine the function! Draw a figure!

My main question is this solvable? I am doubtful?

答:
Note: The method below is very similar to my answer to the question "Find equation of quadratic when given tangents?".

Since the derivative of $y=f(x)=ax^{2}+bx+c$ is $f^{\prime }(x)=2ax+b$, the
equations of the tangents to the graph of $f(x)$ at points $(x_{i},f(x_{i}))$, with $i=1,2$ are

$$\begin{eqnarray*}
y &=&f^{\prime }(x_{i})x-f^{\prime }(x_{i})x_{i}+f(x_{i}) \\
&=&\left( 2ax_{i}+b\right) x-\left( 2ax_{i}+b\right)
x_{i}+ax_{i}^{2}+bx_{i}+c.
\end{eqnarray*}$$

One of the points is $(x_{1},f(x_{1}))=(0,0)$. As the equation of the
tangent at $(0,0)$ is $y=x$ we must have

$$bx+c\equiv x.$$

Comparing coefficients we get $b=1,c=0$. Hence $f(x)=ax^{2}+x$. Similarly
for the tangent at $(x_{2},f(x_{2}))$ we must also have

$$\left( 2ax_{2}+1\right) x-\left( 2ax_{2}+1\right) x_{2}+ax_{2}^{2}+x_{2}\equiv 2x+3.$$

Comparing again coefficients, we get the following system in $a$ and $x_2$, which enables us to find $a$:

$$\left\{
\begin{array}{c}
2ax_{2}+1=2\qquad\qquad\qquad \\
-\left( 2ax_{2}+1\right) x_{2}+ax_{2}^{2}+x_{2}=3.%
\end{array}%
\right. $$

From the first equation we get $x_{2}=1/(2a)$, which by substitution in the
second equation gives $a=-1/12$.

Therefore the quadratic equation $y=f(x)$ is

$$y=-\frac{1}{12}x^{2}+x.$$

Below is the graph of $y=f(x)$ together with its two tangents at points $(0,0)$ and $(-6,-9)$.

Title: What are a , b and c?


问:
Math behind RSA
Now I very recently wrote an investigation paper on RSA Encryption and how it works. After completing it, I thought of looking up real life key generators. To my surprise, they all have characters within them, and not just numbers... Everything I've read till now has been based on the fact that RSA works solely on prime numbers + Euler's Totient function + Modular Arithmetic. I know how to encrypt when the public and private keys are solely numbers, but how is it done with characters?

Just to make it clear, I do know how to use RSA encryption and know the math behind it.

PS: Here are a few links :

https://travistidwell.com/jsencrypt/demo/

https://www.devglan.com/online-tools/rsa-encryption-decryption

答:
Here is another RSA article for your list. One of the problems with some of the programming languages is that their primitive data types have limited sizes. Say the longest integer possible

In  Java it's long and it's 64 bits

In C++ it's long long and is also 64 bits.

Obviously, that's not enough to hold a key of $1024$ bits or $2048$ bits. Java has a "any size" BigInteger, but it's different from the C++ equivalent (or Python). Some sort of a standard is required, for keeping the keys, so that all the languages could follow it. And there are quite a few such standards.

Essentially, big numbers are converted into chunks or arrays of bytes (there are systems with different order of bits, thus I use the word "converted" and thus another reason for standards). However, persisting those bits in binary form, again, isn't always portable (Java arrays and C++ arrays, persisted in binary files, look different). One way to convert binary to "readable" ASCII is base64 encoding. It is used to encode email attachments, because email protocols like POP3, SMTP, IMAP are text based, binary "gibberish" would make the life of the protocol (and the engineers developing and maintaining it) very difficult.

An example, because we are on a maths forum, say a $2048$ bits key would require $2048/8=256$ bytes. But base64 "converts" chunks of 6 bits to "readable" ASCII characters ($2^6=64$, this is the reason for base64 name), thus $2048/6 \approx 342$ characters.

Title: Math behind RSA


问:
Why an interval is lost here when squaring this inequality
$$\sqrt{x-2}\geq -1$$

When squaring the inequality I lose the interval [-2,3) why so ?

答: Because $a\geqslant b$ and $a^2\geqslant b^2$ are not equivalent assertions. Actually, none of them implies the other one.


问:
What is the motivation behind the steps in this 'simple' proof that $\pi$ is irrational?
In $1947$, Ivan Niven published A Simple Proof that $\pi$ is irrational, which only requires knowledge of elementary calculus to understand. Since then, many variations of this proof have been published, one of which I will briefly outline here:

Assume, for the sake of contradiction, that $\pi$ can be expressed in the form $a/b$, where $a$ and $b$ are integers.

Let $f$ be the polynomial function defined by
$$
f(x)=\frac{x^n(a-bx)^n}{n!} \, ,
$$
where $n$ is a positive integer.

Consider the integral $I=\int_{0}^{\pi} f(x)\sin x \, dx$. Since $f$ and $\sin$ are positive on the interval $(0,\pi)$, we have $0<I$. Moreover, if $n$ is large enough, then $I<1$. This is because
$$
f(x)\sin x =\frac{x^n(a-bx)^n}{n!}\sin x<\frac{x^n\pi^n}{n!}
$$
and $$\int_{0}^{\pi}\frac{x^n\pi^n}{n!} \, dx <1$$ for sufficiently large $n$.

On the other hand, integration by parts tells us that
\begin{align}
\int f(x) \sin x \, dx = &-f(x)\cos x +f'(x)\sin x + f''(x) \cos x - f'''(x)\sin x \\ &-f''''(x)\cos x +\ldots\pm f^{(2n)}(x)\cos x
\end{align}
Every function $f^{(k)}$ (with $0 \leq k\leq 2n$) takes integer values when $x=0$ and $x=\pi$. The same is true for $\sin$ and $\cos$. Hence, $I$ must be an integer. But earlier we showed that if $n$ is sufficiently large, $0<I<1$, thereby reaching a contradiction. $\blacksquare$

Although the steps in the proof are not too difficult to follow, it still leaves me scratching my head thinking why on earth someone would try to prove $\pi$ is irrational by defining a bizzarre-looking function and then proceeding to integrate it. The proof seems entirely unmotivated. Is there any insight into how someone could have thought of such a proof? Is there something about the function $f$ and the trigonometric functions that provide us with information about the irrationality of $\pi$?

答:
We can reason as follows.

If we want to force a contradiction from $\pi=a/b$, we need to exploit the one thing we know about $\pi$: that certain functions have a period proportional to $\pi$, and nice values at certain rational multiples of it. Let's try working with the sine function for example, which is $0$ at multiples of $\pi$, even though its period is $2\pi$.

Its half-period from $0$ to $\pi$ tells the whole story, thanks to $\sin x$ being odd. And on this half-period, $\sin x$ looks a bit like a parabola, so one obvious approximation of it is a multiple of$$x(\pi-x)=x(a/b-x)=x(a-bx)/b.$$Of course, $x(a-bx)$ is upper-bounded by e.g. completing the square.

Now it's only natural to see which functions of the form $f(x(a-bx))\sin x$ we can integrate from $0$ to $\pi$, presumably using integration by parts if $f$ is a polynomial. If the integrand is non-negative in our choice of integration range, the integral is at least $0$. And if $f(y)$ is proportional to some power of $y$, we even get an upper bound on the integral.

You can almost tell without calculation that $\int_0^\pi x^n(a-bx)^n\sin xdx$ will be an integer. A special integer? I hope so. Integers are funny things, you see. Two different real numbers' difference can be arbitrarily small, but when integers differ, they differ by at least $1$. So $\int_0^\pi x^n(a-bx)^n\sin xdx\ge1$, which is a much more interesting lower bound. If we could just show the LHS gets arbitrarily small with large $n$, we'd be done. The rest is just trying IBP to see what happens.

But now, a bit of bad news: our upper bound on $x^n(a-bx)^n$ is $(\pi/2)^{2n}$, which grows exponentially. Now, luckily $\sin x$ is bound, but that's not much consolation. It's looking less likely the integral will shrink enough as $n$ grows to finish the argument.

Then again, polynomials getting differentiated many times for repeated IBP introduces potentially rather large combinatorial factors. If we want to get rid of all the polynomial factors so we know how to finish the integration with e.g. $\int\sin xdx=-\cos x+C$, we might even get superexponential factors, like $n!$. That's cool, because if some positive integer $k(n)$ divides $\int_0^\pi x^n(a-bx)^n\sin xdx$, that integral will have lower bound $k(n)$. So if $k$ grows superexponentially, our original strategy is basically saved.

And then Niven or Bourbaki (not a real person, by the way) or anyone tries it out, and polishes the finding into a proof, rather than the above thinking-out-loud.


问:
The true nature of division in light of Euclid's understanding of multiplication
I'm not a math major, but a philosophy major that likes to know that he knows what he's talking about. This may seem like a super stupid question, but here I go.

So Euclid made a lot of sense when he gave the example of the nature of multiplication. For example.
"2 x 3" is really 2 added to itself 3 times, and vice versa, and this is different from 2 + 3, which is five, two units added to three.

But, what about division?

It seems to be the opposite of multiplication like a mirror image from 3rd grade, but I'm starting to doubt that after digging deeper.

For example, this site gave the following diagram for division:

Based on this image, for 6/3, division seems to be the counting of groups, not the inverse of multiplication according to Euclid.

Can someone explain the true nature of division in light of the euclid example I gave?

答:
E.g., $6:3 = 2$ is the number of blocks of length 3 that fit into a block of 6.

Similarly, $7:3$ is the number of blocks of length 3, namely 2, that fit into a block of length 7, with a residue block of 1:

$7 = 2\cdot 3 + 1$ (division with remainder).


问:
On a proof about Jacobson radical
I have encountered the following proof that shows if an element $a$ of a ring $A$ is in the Jacobson radical $J(A)$, intersection of all maximal ideals, then $1-ab$ must be invertible $\forall b \in A$. Here is the proof (taken from "Commutative algebra" by Andrea Ferretti).

I don't quite get the first line of the proof:

"if $a \in J(A)$ then $1-ab$ is not contained in any maximal ideal, so $<1-ab> = A$".

What is the reasoning behind this?

答:
Suppose for contradiction  that $1-ab$ is contained in a maximal ideal $M$. Since $a$ is in the Jacobson radical, $a\in M$, so $ab\in M$, and thus $1 = (1-ab) + ab\in M$, contradicting the fact that $M$ is proper.

Now we know that $1-ab$ is not contained in any maximal ideal. Let $I = (1-ab)$ be the principal ideal generated by $1-ab$. Since every proper ideal is contained in a maximal  ideal, $I$ must be improper, so $(1-ab) = A$.

Title: On a proof about Jacobson radical


问:
Show the converse of this theorem about local maxima is false
Theorem: Let $f$ be continuous on the interval $[a, b]$ and let $c\in (a, b)$. Assume that $f$ is differentiable on $(a, c)$ and $(c, b)$.
If there exists a neighbourhood $(c − \delta, c + \delta) \subset [a, b]$ such that $$f'(x)\begin{cases} \geq 0, &\text{for } x \in (c − \delta, c)\\
\leq 0, &\text{for } x \in (c, c + \delta)\end{cases},$$
then $f$ has a local maximum at $c$.

To prove the converse doesn't hold, I am told to use the function $f(x) = 2x^4 + x^4\sin(1/x)$ for $x\neq 0$ and $f(x) = 0$ for $x = 0$.

I guess I am supposed to show that $f$ has a local maximum at $0$ another way and then show there is no neighbourhood such that $f'(x)$ is as stated in the theorem. I am really struggling with finding a starting place for this as when I look at the neighbourhood around $0$, I cannot determine whether terms are positive or negative as $\sin(1/x)$ and $\cos(1/x)$ in the differentiated form could be positive or negative.
Any help in where to start would be much appreciated thank you!

答:
I'm hoping you got:
\begin{align*}
f'(x) &= 8x^3 + 4x^3\sin\left(\frac{1}{x}\right) - x^2\cos\left(\frac{1}{x}\right) \\
&= x^2\left(4x\left(2 + \sin\left(\frac{1}{x}\right)\right) - \cos\left(\frac{1}{x}\right)\right).
\end{align*}
Note that $2 + \sin\left(\frac{1}{x}\right) \in [1, 3]$, so for $x \in (0, 1/24)$, we have
$$4x\left(2 + \sin\left(\frac{1}{x}\right)\right) \in \left(0, \frac{1}{2}\right).$$
Now, consider the sequences $a_n = \frac{1}{2n\pi}$ and $b_n = \frac{1}{(2n + 1)\pi}$. Both sequences converge to $0$ from above, so both sequences lie eventually in $(0, 1/24)$. Note that $\cos(1/a_n) = 1$ and $\cos(1/b_n) = -1$. So, when $a_n \in (0, 1/24)$,
\begin{align*}
f'(a_n) &= a_n^2\left(4a_n\left(2 + \sin\left(\frac{1}{a_n}\right)\right) - \cos\left(\frac{1}{a_n}\right)\right) \\
&< a_n^2\left(\frac{1}{2} - \cos\left(\frac{1}{a_n}\right)\right) \\
&= a_n^2\left(\frac{1}{2} - 1\right) < 0.
\end{align*}
When $b_n \in (0, 1/24)$, then
\begin{align*}
f'(b_n) &= b_n^2\left(4b_n\left(2 + \sin\left(\frac{1}{b_n}\right)\right) - \cos\left(\frac{1}{b_n}\right)\right) \\
&> b_n^2\left(0 - \cos\left(\frac{1}{b_n}\right)\right) \\
&= b_n^2\left(0 + 1\right) > 0.
\end{align*}
That is, $f'(x)$ changes sign infinitely often as $x \to 0^+$.


答:
Suppose $1-ab$ is contained in maximal ideal $I$.

Then since $a$ is in each maximal ideal (and so is the multiple $ab$), $(1-ab) + ab = 1$ would be in maximal ideal $I$.

Title: On a proof about Jacobson radical

Body:
I have encountered the following proof that shows if an element $a$ of a ring $A$ is in the Jacobson radical $J(A)$, intersection of all maximal ideals, then $1-ab$ must be invertible $\forall b \in A$. Here is the proof (taken from "Commutative algebra" by Andrea Ferretti).

I don't quite get the first line of the proof:

"if $a \in J(A)$ then $1-ab$ is not contained in any maximal ideal, so $<1-ab> = A$".

What is the reasoning behind this?


问:
This asymptotic representation of $\ln(1+1/x)$?
I found this asymptotic approximation in an explanation for Bertrand's test, and I'm afraid I don't get it. Can someone explain what is meant by:

Using little o notation as an asymptotic approximation when $x$ tends toward infinity:

$$\ln\left(\frac{1+x}{x}\right)=\frac{1}{x}+o\left(\frac{1}{x}\right)$$

$$\left(1+\frac{1}{x}\right)^p=1+\frac{p}{x}+o\left(\frac{1}{x}\right)$$

I think I understand that in the second one, all the terms are $c/x^n$, where $n\ge 2$, so are dominated by $1/x$ but I don't understand the first one.

答:
For the first one:
$$\ln\left(\frac{1+x}{x}\right) = \ln\left(1+\frac{1}{x}\right).$$  As $x \to\infty$, we can consider the transformation $x\to \frac{1}{y}$ and consider the Taylor expansion as $y\to 0$.  We get $$\ln\left(\frac{1+x}{x}\right) = \ln(1+y) \sim y + O(y^2),\quad\text{as }y\to 0.$$  Going back to $x$, we get $$\ln\left(\frac{1+x}{x}\right) \sim \frac{1}{x} + O\left(\frac{1}{x^2}\right)\sim\frac{1}{x} + o\left(\frac{1}{x}\right)\quad\text{as }x\to \infty.$$  Note that the change from $O$ to $o$ comes from the fact that $$f\in O\left(\frac{1}{x^2}\right)\implies \lim_{x\to\infty}\frac{f(x)}{1/x^2}<\infty \implies\lim_{x\to\infty}\frac{f(x)}{1/x} = 0\implies f\in o\left(\frac{1}{x}\right).$$


问:
Equality of two elliptic integrals
by wolfram 1   and wolfram 2

It's true that $\displaystyle \int_0^{\frac{\pi}{2}}\int_0^{\frac{\pi}{2}} \frac{2}{\sqrt{1-\sin^2\theta\sin^2\phi}} d\phi d\theta =\int_0^{\frac{\pi}{2}}\int_0^{\frac{\pi}{2}} \frac{1}{\sqrt{\sin\theta\sin\phi}} d\phi d\theta$ ?

If yes,  I need  a way to prove the equality of the following two integrals

I tried everything  but I am unable to convert  into a standard form so How do I solve this problem.

Addition 1: For the second integral
$\int_0^{\frac{\pi}{2}}\int_0^{\frac{\pi}{2}} \frac{1}{\sqrt{\sin\theta\sin\phi}} d\phi d\theta =\bigg(\int_0^{\frac{\pi}{2}}\frac{1}{\sqrt{\sin\theta}}d\theta\bigg)^2 $ and we can use that $2 \int^{\frac{\pi}{2}}_{0}\frac{1}{\sqrt{\sin x}} \mathrm dx = \frac{\Gamma(1/2)\Gamma(1/4)}{\Gamma(3/4)} = \frac{\Gamma \left( \frac{1}{4}\right)^2}{\sqrt{2\pi}}$

Addition 2: For the first integral

Let $\displaystyle K(k)=\int_0^{\frac{\pi}{2}}\frac{1}{\sqrt{1-k^2\sin^2 t}}dt$  ( Complete Elliptic Integral of the First Kind). we know that $ \displaystyle K(k)=\frac{\pi}{2}\sum_{n=0}^\infty \left(\frac{(2n)!}{2^{2n}(n!)^2}\right)^2k^{2n}$. Then it's not difficult de show that $\int_0^{\frac{\pi}{2}}\int_0^{\frac{\pi}{2}} \frac{1}{\sqrt{1-\sin^2\theta\sin^2\phi}} d\phi d\theta=(\pi /2 )^2 \sum _{n=0}^{\infty }(\frac{(2n)!}{4^n(n!)^2})^3$.

The egality of the two integrals hold if we can calculte $$\sum _{n=0}^{\infty }(\frac{(2n)!}{4^n(n!)^2})^3$$
Wolfram gives

Addition 3:

This link gives

答:
The equivalence of the elliptic integrals can be shown in the following way:
$$\begin{align}
\int\limits_0^{\frac{\pi}{2}}\int\limits_0^{\frac{\pi}{2}} \frac{d\phi\, d\theta}{\sqrt{1-\sin^2\theta\sin^2\phi}}
& =\int\limits_0^{\frac{\pi}{2}}K(\sin\theta)d\theta\tag1\\
&=\int\limits_0^{1}\frac{K(r)dr}{\sqrt{1-r^2}}\tag2\\
&=\int\limits_0^{1}\frac{K\left(\frac{2\sqrt k}{1+k}\right)dk}{\sqrt{k}(1+k)}\tag3\\
&=\int\limits_0^{1}\frac{K\left(k\right)dk}{\sqrt{k}}\tag4\\
&=\int\limits_0^{1}\frac{dk}{\sqrt{k}}
\int\limits_0^{\frac{\pi}{2}} \frac{d\phi}{\sqrt{1-k^2\sin^2\phi}}\tag5\\
&=\int\limits_0^{\frac{\pi}{2}}d\phi\int\limits_0^{1}
 \frac{dk}{\sqrt{k}\sqrt{1-k^2\sin^2\phi}}\tag6\\
&=\int\limits_0^{\frac{\pi}{2}}d\phi\int\limits_0^{\phi}
 \frac{d\theta}{\sqrt{\sin\theta\sin\phi}}\tag7\\
&=\frac12\int\limits_0^{\frac{\pi}{2}}\int\limits_0^{\frac{\pi}{2}}
 \frac{d\phi\, d\theta}{\sqrt{\sin\theta\sin\phi}}.\tag8\\
\end{align}$$

Explanation:

$(1)$: Definition of the complete elliptic integral of the first kind
$K(k)=\int\limits_0^{\frac{\pi}{2}}\frac{d\phi}{\sqrt{1-k^2\sin^2\phi}}$.

$(2)$: $\sin\theta\mapsto r$.

$(3)$: $r\mapsto\frac{2\sqrt k}{1+k}$.

$(4)$:  Landen's transformation $K(k)=\frac1{1+k}K\left(\frac{2\sqrt k}{1+k}\right)$.

$(5)$: Definition of the complete elliptic integral of the first kind.

$(6)$: Interchange of integration order.

$(7)$: $k\mapsto\frac{\sin\theta}{\sin\phi}$.

$(8)$: Use of the integrand symmetry.

Title: Equality of two elliptic integrals


问:
Relationship between Arithmetic Mean(AM), Geometric Mean(GM) and Harmonic mean(HM)
Does the relationship $GM^2 = AM * HM$ hold when the number of elements is greater than $2$ ($3$ or more elements). It doesn't seem to work for some examples that I have chosen randomly, like the numbers $2$, $5$ and $9$, which gives $57.7 = 59$, a gap of $1.3$. Too large to be attributed to errors related to using floating point numbers.

Does it only work when having two elements?

答: No, the case of $n=2$ values' AM, GM and HM is special. (The value $n=1$ is even more special!) In general, you're comparing $(\prod_{i=1}^nx_i)^{2/n}$ to $\frac{\sum_ix_i}{\sum_j1/x_j}=\frac{\sum_ix_i}{\sum_j\prod_{k\ne j}x_k}\prod_kx_k$. When $n=2$, the LHS's exponent is $1$, and the RHS has denominator $\sum_jx_j$. But for $n\ge3$, the denominator is of degree $n-1\ge2$, so you won't get the identity you want.


问:
Proving an inequality without an integral: $\frac {1}{x+1}\leq \ln (1+x)- \ln (x) \leq \frac {1}{x}$
I would like to prove the following inequality without integration; could you help?

$$\frac {1}{x+1}\leq \ln (1+x)- \ln (x) \leq \frac {1}{x}, \quad x > 0. $$

I can however differentiate this.

Thanks in advance.

答:
This is trivial using the definition of $\ln(x)$ as the area below the hyperbola $1/x$ between 1 and $x$.

$\ln(x+1)-\ln(x)$ is the hyperbola area between $x$ and $x+1$, bounded by the rectangles with area $1/(x+1)$ and $1/x$.


问:
Finding contrapositive of a universal statement

For all $x$, if $x^2$ is even, then $x$ is even.

The contrapositive to this statement is:

For all $x$, if $x$ is odd, then $x^2$ is odd.

Why do we ignore the "For all $x$" and not say "For some $x$..."?

答:
Symbolically

$~~~~\forall x \in N: [x^2 \in E \implies x \in E]$$

$~~~\equiv ~~~\forall x \in N: [x \notin E \implies x^2 \notin E]$

$~~~\equiv ~~~\forall x \in N: [x \in O \implies x^2 \in O]$

where $N$ is the set of natural numbers, $E$ is the set of even numbers, and $O$ is the set of odd numbers.

We use the contrapositive of the original implication. The quantifier is unaffected.


问:
GNS construction for a normal state on a von Neumann algebra
Let $M$ be a von Neumann algebra. For any faithful normal state $\omega$ on $M$, according to the GNS construction, we have $\omega(x)=\langle x \Omega,\Omega\rangle$, where $\Omega$ is a cyclic vector.

For any two different faithful normal states, are the cyclic vectors in the above construction the same?

答:
Yes, they both coincide with the unit of $M$, so in a sense they are indeed the same.  However the two GNS
representations take place in DIFFERENT Hilbert spaces,  namely the completion of $M$ relative to two different inner
products.

If the states are $\omega _1$ and $\omega _2$,  and the corresponding GNS representations are $(\pi _1,H_1,\xi _1)$ and $(\pi _2,H_2,\xi _2)$,
then both  $H_1$ and $H_2$
contain a dense copy of $M$ and, for every $m$ on $M$,  the diagram
$\require{AMScd}$
\begin{CD}
  H_1 @<\iota_1<<M @>\iota_2>> H_2\\
  @V\pi_1(m)VV @VL_mVV @VV\pi_2(m)V\\
  H_1 @<\iota_1<<M @>\iota_2>> H_2\\
  \end{CD}
commutes, where $L_m$ is the operator of left-multiplication by $m$, and $\iota_1$ and $\iota_2$ are the inclusion maps.

This might give an impression that $π_1$ and $π_2$ are equivalent representations, and indeed they are (!) when
restricted to the dense subspaces $\iota_1(M)$ and $\iota_2(M)$.

However, the extra vectors added in the completion process strongly depend  on the states $\omega_1$ and $\omega_2$, so the representations $\pi_1$ and $\pi_2$ might very well be inequivalent!


问:
Number of functions with a fixed number of elements in the range
I have a question that goes:

Find the number of functions $f:A \longrightarrow B$, such that the range contains exactly $3$ elements. Given that $n(A)= 4$ and $n(B)= 5$

Here's what I tried. Range has to have $3$ elements. So the way to choose $3$ elements out of $5$ is ${5 \choose 3} = 10$

Now, I faced a problem while trying to order the choices. What I tried to do was, use the stars and bars method to find the solutions to $$x_1+x_2+x_3=4$$ where,

$x_1=$ Number of elements in the domain mapped to 1st element in range

$x_2=$ Number of elements in the domain mapped to 2nd element in range

$x_3=$ Number of elements in the domain mapped to 3rd element in range

And the Answer would have been $10\times $(The number of solutions obtained for the above equation)

But the answer is said to be $360$ and my method doesn't work out, can someone point out where I went wrong?

答:
Your approach is correct; you forgot that the three elements in the range can be arranged in $3!$ ways as first, second, third.

Hence answer should be
$$\binom{5}{3} \times 3! \times \binom{4}{2}=360$$

Note that you're not looking for number of solutions to $x_1+x_2+x_3=4$. Rather the positive integral solutions of this equation : $(2,1,1)$ and its permutations; indicate the type of mapping. $(2,1,1)$ means exactly one element in the range maps to $2$ elements of the domain. Hence $\binom{4}{2}$ (onto) functions on any range.


答:
Because the contrapositive refers to an equivalent form of the implication, it is thus a tautological equivalence. It refers to the "realm" of propositional logic and not first order logic.

For example, I'm sure you know that an equivalent form of $\phi \to \psi$ is $\lnot \phi \lor \psi$. Would you not then substitute in $\forall x (Fx \to Qx)$ for $\forall x (\lnot Fx \lor Qx)$?

Title: Finding contrapositive of a universal statement

Body:
For all $x$, if $x^2$ is even, then $x$ is even.

The contrapositive to this statement is:

For all $x$, if $x$ is odd, then $x^2$ is odd.

Why do we ignore the "For all $x$" and not say "For some $x$..."?


答:
As you said, we first choose any three elements, which can be done in $\binom{5}{3}$ ways.

Now, you need to apply inclusion-exclusion principle, for mapping a set of $4$ elements to a set of $3$ elements. Basically, you only need to count the number of onto functions from a set of $4$ elements to a set of $3$ elements, which can be given by $$\sum_{k=0}^3 (-1)^k \binom{3}{k}(3-k)^4$$

Which evaluates to $36$, now multiplying this by $10$ gives $360$.

For calculating the number of onto functions from a set of $m$ elements to a set of $n$ elements, you may see this nice post.

Title: Number of functions with a fixed number of elements in the range

Body:
I have a question that goes:

Find the number of functions $f:A \longrightarrow B$, such that the range contains exactly $3$ elements. Given that $n(A)= 4$ and $n(B)= 5$

Here's what I tried. Range has to have $3$ elements. So the way to choose $3$ elements out of $5$ is ${5 \choose 3} = 10$

Now, I faced a problem while trying to order the choices. What I tried to do was, use the stars and bars method to find the solutions to $$x_1+x_2+x_3=4$$ where,

$x_1=$ Number of elements in the domain mapped to 1st element in range

$x_2=$ Number of elements in the domain mapped to 2nd element in range

$x_3=$ Number of elements in the domain mapped to 3rd element in range

And the Answer would have been $10\times $(The number of solutions obtained for the above equation)

But the answer is said to be $360$ and my method doesn't work out, can someone point out where I went wrong?


问:
The completness of ring and its power series ring

Let $R$ be a ring and $I$ an ideal of $R$. If $R$ is $I$-adically complete, why then $R[[x]]$ is $(IR[[x]]+(x)R[[x]])$-adically complete? (Matsumura, Commutative Ring Theory, Exercise 8.6.)

Take a Cauchy sequence $(a_n)$. By definition for any $k$ there exists $N$ s.t. $m,n\ge N$ then $a_m-a_n\in J^k\dots$

答:
Here is a short calculation (whose steps can be justified by the reader):

$\lim_n R[[x]]/(I,x)^n = \lim_n R[[x]]/(I^n,x^n) = \lim_m \lim_n R[[x]]/(I^n,x^m)$
$= \lim_m \lim_n R/I^n[[x]]/(x^m) =\lim_m R[[x]]/(x^m) = R[[x]].$


问:
Prove that $E (|xi -a|)$ is minimized when $a = md$
This was given to us as an assignment, but I have no idea how I should do it.
Can someone give me some hints or give me a name or the topic?

$\min E(|X-a|) = E|X-md|$.

$X$ is an arbitrary continuous random variable.

答:
$$\mathbb{E}|X-a|=\xi(a)=\int_a^{+\infty}(x-a)f_X(x)dx+\int_{-\infty}^a(a-x)f_X(x)dx$$

In order to minimize $\xi$ let's derivate it w.r.t. $a$

$$\xi'(a)=-\int_a^{+\infty}f_X(x)dx+\int_{-\infty}^af_X(x)dx=-[1-F_X(a)]+F_X(a)$$

Set it $=0$:

$$-[1-F_X(a)]+F_X(a)=0$$

$$F_X\left(a\right)=\frac{1}{2}$$

$$a=F_X^{-1}\left(\frac{1}{2}\right)$$

which is exaclty the definition of Median


答:
The solutions to the equation (with $x_1,x_2,x_3\geq 1$) are $(2,1,1)$, $(1,2,1)$, $(1,1,2)$, i.e. exactly one range element has two preimages.

choose range: $\binom{5}{3}$ choices

choose range element that has $2$ preimages: $3$ choices

choose two preimages for that element: $\binom{4}{2}$ choices

choose one preimage for second range element: $\binom{2}{1}$ choices

$$\binom{5}{3} \cdot 3 \cdot \binom{4}{2} \cdot \binom{2}{1} = 360$$


答:
You cannot use stars and bars here, because the objects to be placed are assumed there to be indistinguishable. You should use the Stirling number of the second kind instead:
$$
N=\binom53{4 \brace 3}3!=360,
$$
where the factor $3!$ accounts for the fact that our "boxes" are also distinguishable.


问:
Does $f(\epsilon)=o(\epsilon\ln(\epsilon))$ imply $\frac{f(\epsilon)}{\epsilon}=o(1)$?
I have the following homework question:

Does $f(\epsilon)=o(\epsilon\ln(\epsilon))$ imply
  $\frac{f(\epsilon)}{\epsilon}=o(1)$ ?

It doesn't seem correct to me, using the definition I could only get
$$\frac{f(\epsilon)}{\epsilon}=o(\ln(\epsilon))$$ and it doesn't seem
that if $g(\epsilon)=o(\ln(\epsilon))$ then $g(\epsilon)=o(1)$.

To contradict this I wish to find a function $c(\epsilon)$ s.t $$\lim_{\epsilon\to0}c(\epsilon)=0$$
but $$\lim_{\epsilon\to0}|c(\epsilon)\ln(\epsilon)|>0$$

I am having difficulty finding such a function, I tried a few and
checked them using WA, they all satisfied $$\lim_{\epsilon\to0}c(\epsilon)=0$$
but they all also satisfied that $$\lim_{\epsilon\to0}|c(\epsilon)\ln(\epsilon)|=0$$

Can someone please help me find such $c(\epsilon)$ ? (or surprise
me and show that this statement is in fact true)

答:
The notation
$$f(x)=_a o(g(x))$$
means $$\lim_{x\to a}\frac{f(x)}{g(x)}=0\iff \frac{f(x)}{g(x)}=_a o(1) $$
and to contradict your implication just take $f(\epsilon)=\epsilon$</p>


问:
Laplace's equation on the plane: how much boundary data must be specified to guarantee existence and uniqueness?
My question stemmed from a specific problem, so let's jump right in. I want to solve Laplace's equation in plane polars $(r, \theta)$ on the domain $(r,\theta) \in [1,\infty)\times[0,2\pi)$ subject to boundary conditions $$u(1,\theta) = f(\theta), \quad \lim_{r\to\infty} u(r,\theta) = 0,$$
where $f$ is a known function on $[0,2\pi)$. It can be shown (e.g. via separation of variables) that the general solution for $u$ that is bounded at infinity is given by
$$
u(r,\theta) = a_0 + \sum_{n=1}^\infty r^{-n} \left(a_n\cos n\theta + b_n \sin n\theta\right),
$$
which we can match against the Fourier coefficients of $f$ to find $a_n$ and $b_n$. However we see that the solution only strictly decays at infinity if $a_0 = 0$. This is a restriction on the allowed boundary data $f(\theta)$ in order for the solution to exist.

So my question comes in two parts:

Is there a way to get around this? I've been reading a solution in which they write the constant $a_0$ as a Fourier series expansion of a square wave defined on the domain $[0,4\pi)$. However surely this throws up half-angles that cannot be matched with u.

Assuming the answer to the above question is no, can someone provide an argument as to why the constant seen at $r=1$ has to be the same as the constant at infinity? It seems to me that in the above formulation we have specified too much boundary data and therefore the solution doesn't exist. But why??

Thanks in advance!

答:
Let $u$ a (real) harmonic function on the exterior of a disc of radius $k$ centered at the origin and continuous on the boundary circle.

By Green's theorem, one gets that there is a constant $B$ (not depending on $R$) given by $2\pi B=\int_{0}^{2\pi}R\frac{\partial u}{\partial r}(R,\theta)d\theta, R>k$, which immediately implies that the harmonic function $u(r, \theta)-B \log r$ has a harmonic conjugate, so it is the real part of an analytic function on the disc exterior.

This implies that $\int_{0}^{2\pi}u(R,\theta)d\theta=A+2\pi B \log R, R >k$

(if $2u=2\Re f= f+\bar f$ the Laurent series of $f$ gives that $\int_{0}^{2\pi}u(R,\theta)d\theta$ is constant in $R$ since only the free term in the Laurent series has non zero integral on a circle and then apply the above to $u-B \log r$).

All the above is general stuff about harmonic function on an annulus or disc exterior; applied to the OP problem, we get that $A=B=0$ by taking $R \to \infty$ and using that $\lim_{r\to\infty} u(r,\theta) = 0$; but now taking $R \to 1$ we get the neceesary condition that $\int_{0}^{2\pi}f(\theta)d\theta =0$. The condition is also sufficient as noted in the OP.


问:
Induced metric on $S^2$ from pullback of metric on $\mathbb{R}^3$
I'm going over some GR from more of a differential geometry perspective and had a quick question about a simple calculation - my differential geometry background isn't too strong so I apologise if any of the terminology is incorrect, but I'd be grateful for any clarification.

I'm following an example in Appendix A of Sean Carroll's Introduction to GR, where there is a map $\phi:M \to N$, given by
$$ \phi(\theta,\phi) = (\sin\theta \cos \phi , \sin \theta \sin \phi, \cos \phi) ,
$$
and $M=S^2$ is a submanifold of $N =\mathbb{R}^3$ (i.e. the two-sphere embedded in $\mathbb{R}^3$). The coordinates on the manifolds are $x^{\mu} = (\theta,\phi)$ on $M$, and $y^{\alpha} = (x,y,z)$ on $N$. The induced metric on $M$ is just the pullback of the flat-space metric $\phi^*g$, which is given by the formula

$$
(\phi^*g)_{\mu \nu} = \frac{\partial y^{\alpha} }{\partial x^{\mu}} \frac{\partial y^{\beta} }{\partial x^{\nu}} g_{\alpha \beta}.
$$

I understand how to calculate the individual Jacobian matrices of partial derivatives $\frac{\partial y^{\alpha} }{\partial x^{\mu}}$ (e.g. just using $y^1 = \sin\theta \cos \phi $, $y^2 = \sin \theta \sin \phi$ and $y^3 = \cos \phi$ as defined by $\phi$), however I was confused as to how to treat the full expression above.

The Jacobian matrices are $2 \times 3$ matrices, so the second has to be transposed to be a $3 \times 2$ matrix in order to give the required $2 \times 2$ metric $g_{\mu \nu}$. My question is, in the pullback equation above, how does the metric $g_{\alpha \beta}$ act on the Jacobian $\frac{\partial y^{\beta} }{\partial x^{\mu}}$, and how should I be writing this down? I can see that $y^{\beta}$ should be replaced with $y^{\alpha}$, but should any indices be lowered, and how should I interpret the metric tensor transposing the Jacobian matrix?

Edit - Ted Shriffin's comments are correct, the last component of the map should be $\cos \theta$ not $\cos \phi$, and all my matrices should be transposed.

答:
Seeing as I'm getting new comments on this old question I thought I'd post the answer myself. Thanks to Ted Shifrin's help in the comments and also pointing out the typo in the mapping: the map $\Phi:M \to N$ should of course be
$$ \Phi(\stackrel{x^{\mu} \ \rm{coords}}{\theta \, , \, \phi})^T = (\stackrel{y^\alpha \ \rm{coords}}{\sin\theta \cos \phi , \sin \theta \sin \phi, \cos \theta})^T \ ,
$$
where the transpose makes clear these are vectors not rows. I've also made clear the $x^{\mu}$ and $y^{\alpha}$ coordinates on $S^2$ and $\mathbb{R}^3$ respectively.
My mistake came from being unfamiliar with using matrix notation with tensors (it's been a while since I've done linear algebra). The final matrix of course must've been $2 \times 2$ because the indices $\alpha \beta$ were summed over whilst $\mu = \{0,1\}$. Similarly, all the expressions I had should have been transposed, but in the textbook they weren't written this way, probably to save space. e.g. the Jacobian was written as a $2 \times 3$ matrix but should have been $3 \times 2$,
$$
\frac{\partial y^\alpha}{\partial x^{\mu}} =
\begin{pmatrix}
\frac{\partial y^1}{\partial x^1} & \frac{\partial y^1}{\partial x^2}  \\
\frac{\partial y^2}{\partial x^1} & \frac{\partial y^2}{\partial x^2}   \\
\frac{\partial y^3}{\partial x^1} & \frac{\partial y^3}{\partial x^2}
\end{pmatrix}
=
\begin{pmatrix}
\cos \theta \cos \phi & -\sin \theta \sin \phi \\
\cos \theta \sin \phi & \sin \theta \cos \phi  \\
-\sin \theta & 0
\end{pmatrix} \ .
$$

Then the induced metric can be written as
$$ \begin{align}
(\Phi^* g)_{\mu \nu} &= \frac{\partial y^{\alpha}}{\partial x^{\mu}}  \frac{\partial y^{\beta}}{\partial x^{\nu}} g_{\alpha \beta} \\
&= \frac{\partial y^{\alpha}}{\partial x^{\mu}}  \frac{\partial y^{\beta}}{\partial x^{\nu}}
\delta_{\alpha \beta} \ ,
\end{align}
$$
with the Kronecker delta from the flat Euclidean metric on $\mathbb{R}^3$. Here we don't just use matrix multiplication as I originally thought (this wouldn't make sense anyway). We can find the components of the induced metric by summing over the repeated index,
$$
(\Phi^* g)_{\mu \nu} = \sum_{\alpha = 1}^3 \frac{\partial y^{\alpha}}{\partial x^{\mu}}  \frac{\partial y^{\alpha}}{\partial x^{\nu}} .
$$
E.g. the $\mu \nu = 0 0$ component is just $$
(\cos \theta \cos \phi)^2 + (\cos \theta \sin \phi)^2 + (- \sin \theta)^2 = 1 \ ,
$$
and repeat for the others.
One then thankfully finds the induced metric to be
$$
g_{\mu \nu}(x) = \begin{pmatrix}
1 & 0 \\
0 & \sin^2 \theta \\
\end{pmatrix} \ .
$$


答: Be careful what you wish for -- setting $c(x)=\frac{42}{\ln x}$ satisfies your conditions, but doesn't help much with the original question... or does it? :-)</p>

Title: Does $f(\epsilon)=o(\epsilon\ln(\epsilon))$ imply $\frac{f(\epsilon)}{\epsilon}=o(1)$?

Body:
I have the following homework question:

Does $f(\epsilon)=o(\epsilon\ln(\epsilon))$ imply
  $\frac{f(\epsilon)}{\epsilon}=o(1)$ ?

It doesn't seem correct to me, using the definition I could only get
$$\frac{f(\epsilon)}{\epsilon}=o(\ln(\epsilon))$$ and it doesn't seem
that if $g(\epsilon)=o(\ln(\epsilon))$ then $g(\epsilon)=o(1)$.

To contradict this I wish to find a function $c(\epsilon)$ s.t $$\lim_{\epsilon\to0}c(\epsilon)=0$$
but $$\lim_{\epsilon\to0}|c(\epsilon)\ln(\epsilon)|>0$$

I am having difficulty finding such a function, I tried a few and
checked them using WA, they all satisfied $$\lim_{\epsilon\to0}c(\epsilon)=0$$
but they all also satisfied that $$\lim_{\epsilon\to0}|c(\epsilon)\ln(\epsilon)|=0$$

Can someone please help me find such $c(\epsilon)$ ? (or surprise
me and show that this statement is in fact true)


问:
Smooth convergent sequence
Suppose that you have a convergent sequence of real numbers $x_n$ and assume that you want to impose conditions so that this sequence converges to some limit $x^*$ for $n\rightarrow \infty$ in a smooth way that is without jumps. In other terms I want to impose that it does not converge too fast and that it doesn't converge to $x^*$ abruptly at some point in time. In other terms, I want the function $|x_n - x^*|$ to be continuous and with the x-axis as a horizontal asymptote.

If I just imposed the sequence to be a Cauchy sequence, I'd not get the result since a constant sequence is Cauchy. I'd modify the Cauchy property this way:

for every positive real number $\epsilon$ there is a positive integer $N$ such that for all natural numbers $n,m>N$ we have $|x_n - x_m|<\epsilon$ and $|x_n - x_m|=\epsilon$ for at least one pair $m,n$

Is this correct? Is there a more elegant way to get the result?

答: The problem with this idea is that sequences are discrete objects, so it doesn't make much sense to discuss their "continuity". In fact, any sequence viewed as a map $\mathbb{N} \to \mathbb{R}$ is continuous. The best you can probably say is to impose the restriction, $$|x_1 - x^*| > |x_2 - x^*| > |x_3 - x^*| > \cdots > 0$$ and this sequence $|x_n - x^*| \searrow 0$.

Title: Smooth convergent sequence


问:
Calculate $a+2b^2+3c^3+4d^4$
Hello I wanted to get some help in how to follow.

I have an excercise that it's like this:

It is known that

$$\int_{2}^{3} \frac{x^2-3x+2}{x^2-x+1} \,\mathrm dx = a\ln(7)+b\ln(3)+c\ln(2)+d$$
where a,b,c,d are integers.

Calculate : $$a+2b^2+3c^3+4d^4$$
And I know the answer is $5$, however I don't know how to arrive to it.
what I did was calculate this integral and it gave me $1+\ln 12-\ln 28$ so I know that $d=1$
however I am lost on how to follow.
I tried comparing it like this:
$$\ln(7^a.3^b.2^c) +1 =1+\ln\left(\frac{12}{28}\right) $$
But this is as far as I got, I imagine there must be something I am missing here so I would appreciate if you could give me a small tip so I can arrive at the solution

答:
The integral results to me

$$1+\log 3-\log 7$$

thus

$$\begin{cases}
a=-1 \\
b=1 \\
c=0 \\
d=1
\end{cases}$$

that is

$$-1+2+0+4=5$$

Title: Calculate $a+2b^2+3c^3+4d^4$


问:
Time and Work : Alternating Workers
Two taps $A$ and $B$ can fill a tank in $20$ minutes and $30$ minutes respectively. An outlet pipe $C$ can empty the full tank in $15$ minutes. $A$,$B$ and $C$ are opened alternatively each for $1$ minute. How long will the tank take to be filled $?$

Below is my approach :-

For 1$^{st}$ set of $3$ minutes ; net work done by the opening of two inlet taps and one outlet tap = $\frac{1}{20} + \frac{1}{30} - \frac{1}{15} = \frac{1}{60}$

For the $59$ sets of 3 minutes ; work done = $\frac{59}{60}$ $\Rightarrow$ Total time till now = $59\times3=177$ minutes

Work left = $\frac{1}{60}$

Now after $177$ minutes, again tap $A$ will be opened and as we can see that tap $A$ can do $\frac{1}{20}$ units of work in $1$ minute and the units of work left is pretty much smaller quantity i.e. $\frac{1}{60}$ units, then the whole tank will be filled under a minute only by the tap $A$ which means the answer to this problem should lie between $177$ and $178$ minutes but to my bad luck the correct answer has been given as $167$ minutes. Can someone please explain me where I am going wrong?

答:
Your thought that instead of being full at the end of the minute C discharges, which would give $180$ minutes, you should look at shorter times when the tank gets full is a good one but you didn't push it far enough.  After some number $n$ of $3$ minute intervals the tank is $\frac n{60}$ full.  In the next $2$ minutes the $A$ tap fills $\frac 1{20}$ and then $B$ fills $\frac 1{30}$.  We want the tank full at this moment.  We have
$$\frac n{60}+\frac 1{20}+\frac 1{30}=1\\
\frac{n+5}{60}=1\\
n=55$$
The three minute intervals take $3 \cdot 55=165$ minutes, then there are the two minutes for the $A$ and $B$ taps making $167$ in total.

Title: Time and Work : Alternating Workers


问:
Lagrange Multiplier equal zero?
I have the following analytical cost function: $$ \textit{J}(\bf{w}) = \textit{k} +\bf{q}^{T}\bf{Bw}+\bf{w}^{T}\bf{Cw}-\lambda\bf{w}^{T}\bf{Dw} $$
 where $\textit{k}$ is a constant, bold lowercase variables are vectors (dimension Nx1) and bold upercase are matrices (dimension NxN). The only variable is $\bf{w}$ (others are constant). The constraint is  $\bf{w}^{T}\bf{Dw}\geq0$ and $\lambda$ is the lagrange multiplier. The derivatives are:

\begin{array}{l}
\frac{{\partial {J(\bf{w})}}}{{\partial {\bf{w}}}} = {{\bf{B}}^T}{\bf{q}} + \left( {{\bf{C}} + {{\bf{C}}^T}} \right){\bf{w}} - \lambda \left( {{\bf{D}} + {{\bf{D}}^T}} \right){\bf{w}} = {\bf{0}}\\
\frac{{\partial {J(\bf{w})}}}{{\partial \lambda }} =  - {{{\bf{w}}}^T}{\bf{D}\bf{w}} = 0\\
\lambda  \ge 0
\end{array}

Pre-multiplying the first derivative by $\textbf{w}^{T}$ results in

${{\bf{\hat w}}^T}\left[ {{{\bf{B}}^T}{\bf{q}} + \left( {{\bf{C}} + {{\bf{C}}^T}} \right){\bf{\hat w}}} \right] = 0$

As a result

${\bf{\hat w}} = {\left( { - {\bf{C}} - {{\bf{C}}^T}} \right)^{ - 1}}{{\bf{B}}^T}{\bf{q}}$

The constraint disapeared! What happened with it? Was it intrinsically satisfied? For any $\bf{D}$?

答: Your last equation doesn't follow from anything you'd written above; certainly not from the equation above it, which is a scalar equation and thus can only determine at most one degree of freedom of $\hat{\mathbf w}$. If you want to get a vector equation such as the last equation, you have to solve the equation $\partial J(\mathbf w)/\partial\mathbf w=\mathbf 0$ without projecting it along a single direction, and then $\mathbf D+\mathbf D^T$ does appear in the matrix whose inverse you need to take.</p>

Title: Lagrange Multiplier equal zero?


问:
Area enclosed by a curve and the coordinate axes
While finding out the area enclosed by curves and the $2$-D coordinate axes, I found something strange. Let's assume we have a parabola $y^2 = 16x$. Now the y coordinate would be given by $4x^{1/2}$. Integrating this expression with limits from, say, a to 0, we get a certain value we call the area. But, why do we always obtain the area in the first quadrant? The $x$ coordinate is positive in the fourth quadrant too, so shouldn't the area we obtain, include that in the fourth quadrant as well? Mathematically, I don't see a particular reason for this. Any help in understanding this would be great.Thanks in advance.

答: Yes, "the area enclosed by" these things is ambiguous. If you're asked to find it, the questioner should be sure to be clear and specific, like "the finite region above the $x$-axis and bounded by $y^2=16x$ and $x=a$ ($a>0$)". Or if a question doesn't mention the axes and asks for "the area of the region enclosed by $y^2=16x$ and $x=a$ ($a>0$)", then this region would be in both the first and fourth quadrants.


问:
Set defined for proving existence of $ r > 0 $ s.t. $ r^2 = 2 $ is problematic
Theorem: There exists a positive real number $ r $ s.t. $ r^2 = 2 $

The proof in my book begins with defining a set $ L = \{ x \in \mathbb{R} | x \geq 0 , x^2 \leq 2 \} $  and then proceeds to show that $ r^2 = 2 $ is the maximum of this set. However, I don't understand the following:

Since the set L contains real numbers $ x $ such that $x \geq 0, x^2 \leq 2 $, then specifically, it contains a number $ x>0 $ s.t. $ x^2 = 2 $ . How am I supposed to show the existence of such a number ( $ r^2 = 2 $ ) if it already exists  in the set I've defined? ( In which case, the set $ L $ is problematic since it already contains the number which I need to prove exists )

答:
The point is that you know there exists some $x$ such that $x \ge 0$ and $x^2 \le 2$, (e.g., $1$) and you know that for all large enough $x$, $x^2 > 2$ (e.g., that is true for $x \ge 2$). Hence (without knowing whether there is an $x$ with $x^2=2$), you can define $L = \{x \in \Bbb{R} \mid x \ge 0, x^2 \le 2\}$ and you know that it is non-empty ($1 \in L$) and bounded above (if $x \in L$, then $x < 2$). But you also know that every non-empty bounded-above set of real numbers has a supremum $s$, say. And then you can prove that $s^2 = 2$. There is no circularity here, even though $s \in L$.

In this case (as pointed out by fleablood in a comment), if you were to change $\le$ in the definition of $L$ to $<$, you would get the same result, but the supremum $s$ would not be a member of $L$. However, there are proofs where you need to reason about a set that must contain a particular number of interest in order to establish what that number is. For example, to prove that any family $F$ of open intervals that covers the closed interval $I = [0, 1]$ has a finite subfamily that covers $I$, you look at the set $M$ of $x \in [0, 1]$ such that a finite subfamily of $F$ covers $[0, x]$ and then prove that $1 \in M$. It would be hard to recast that without allowing $1 \in M$.


问:
Simple convex problem, reference needed
I am writing a paper, where I need to apply a quite simple convexity argument. But it's lengthy, and in order to save space in the paper, I would like to just cite a reference. The claim is as follows:

Let $A\in\mathbb R^n$ be convex such that $\overline B_\varepsilon(0)\subset A$. If $x_0\in\mathbb R^n$ such that $\operatorname{span}\{x_0\}\subset\overline A$, then $\operatorname{span}\{x_0\}+ B_\varepsilon(0)\subset A$.

Here,

$\overline A$ denotes the closure of $A$

$\overline B_\varepsilon(0)$ denotes the closure of the $\varepsilon$-ball around zero

$\operatorname{span}\{x_0\} = \{tx_0 : t\in\mathbb R\}$.

Sketch of proof: Let $x\in\operatorname{span}\{x_0\}$ and connect every point from $\overline B_\varepsilon(0)$ with $x_\delta$, where $x_\delta\in A$ is close to $x$. The union of all these lines is in $A$ since $A$ is convex. Also, the closer $x_\delta$ is to $x$, the more of the segment $[0,x]$ is contained in this union. Letting $\delta\to 0$, it follows that $[0,x)\subset A$. So, $\operatorname{span}\{x_0\}\subset A$. Now, we can even connect every point in $\operatorname{span}\{x_0\}$ with every point in $\overline B_\varepsilon(0)$. The union of all these lines is in $A$ and should contain $B_\varepsilon(x)$ for every $x\in\operatorname{span}\{x_0\}$.

答:
A two-sentence proof:

For any $y \in B_{\varepsilon}(0)\backslash\{0\}$ and any $t \in \Bbb R$, choose $z \in A$ such that $|z - \frac {\varepsilon + |y|}{\varepsilon - |y|}tx_0| < |y|$. We then have $$\left|\frac{\varepsilon + |y|}{2|y|}(y + tx_0 - \frac{\varepsilon - |y|}{\varepsilon + |y|}z)\right| \leq \frac{\varepsilon + |y|}{2|y|}\cdot (|y| + |tx_0 - \frac{\varepsilon - |y|}{\varepsilon + |y|}z|) < \varepsilon$$ and hence $$y + tx_0 = \frac{2|y|}{\varepsilon + |y|}\cdot \left(\frac{\varepsilon + |y|}{2|y|}(y + tx_0 - \frac{\varepsilon - |y|}{\varepsilon + |y|}z)\right) + \frac{\varepsilon - |y|}{\varepsilon + |y|} \cdot z\in A.$$

It's now up to the referee to check that everything is correct.

Title: Simple convex problem, reference needed


问:
Finding the amplitude of the difference between two sinusoids
I have the following function:

$$y(t)=a\sin(wt+p)-b\sin(wt+p+\frac{2\pi}{3})\tag1$$

Question: what is the amplitude of the function $y(t)$ in terms of $a,b,w$ and $p$?

I thought I can find:

$$\frac{\partial y(t)}{\partial t}=0\space\Longleftrightarrow\space t=\dots\tag2$$

And than use that $t$ to find $\hat{y}(t)$. But I can not solve $(2)$.

答:
$$\frac{\partial y(t)}{\partial t}=w \left(a \cos (p+t w)+b \sin \left(p+t w+\frac{\pi }{6}\right)\right)$$ Let $p+t w=x$ and you need to solve
$$a \cos (x)+b \sin \left(x+\frac{\pi }{6}\right)=0$$ Expand the sine
$$(2 a+b) \cos (x)+b\sqrt{3}  \sin (x)=0$$ which looks to be simple.


问:
Combination of two sets of basis
Given two sets of basis $V = \{v_1,...,v_n\}$ and $U = \{u_1,...,u_n\}$, each spanning $\mathbb{R}^n$, is it possible to select a subset of the two sets $T$ such that T = $\{v_1\, ...,v_m\} \cup \{u_1\, ...,v_o\}$ with $m+o = n$ such that $T$ spans $\mathbb{R}^n$?

答:
First, we need to impose the condition that $m\neq 0$ and $o\neq 0$, or the question becomes trivial because we can always choose $T=V$ or $T=U$.

Now suppose $m,o\neq 0$. We want to know if we can swap any $u_i$ with any $v_j$. We claim that this is possible. Consider $u_1$. If we can swap $u_1$ with $v_n$, that is, $\{u_1,v_1,\cdots,\cdots,v_{n-1}\}$ is linearly independent, then we are done. If not, $u_1$ is a linear combination of $v_1,\cdots,v_{n-1}$. Now do the same thing for $u_2,\cdots,u_n$. If we can swap any of these $u_i$ with $v_n$, then we are done. If not, each of $u_2,\cdots,u_n$ is a linear combination of $v_1,\cdots,v_{n-1}$. Since $\{u_1,\cdots,u_n\}$ spans $\mathbb{R}^n$, this implies $\{v_1,\cdots,v_{n-1}\}$ spans $\mathbb{R}^n$, which is a contradiction. Therefore, we must be able to swap $v_n$ with some $u_i$ and create a new basis.

Title: Combination of two sets of basis


问:
Prove $\frac{a^2}{b^3}+\frac{b^2}{c^3}+\frac{c^2}{a^3}≥\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$
Question : Prove $$\frac{a^2}{b^3}+\frac{b^2}{c^3}+\frac{c^2}{a^3}\geq \frac{1}{a}+\frac{1}{b}+\frac{1}{c}$$
$(a, b, c \in \mathbb{R}^+)$

I tried to solve it like this :
$$\frac{a^2}{b^3}+\frac{b^2}{c^3}+\frac{c^2}{a^3}+\frac{1}{a}+\frac{1}{b}+\frac{1}{c} \geq 2 \; (\frac{a}{b^2}+\frac{b}{c^2}+\frac{c}{a^2})$$

Am I doing this right? How can I finish this problem?

答:
Problems of this type can often be solved by showing, using AM-GM, that each term on the right hand side is not greater than some linear combination of the terms on the left hand side.

In this particular situation,
$$
4\cdot \frac{a^2}{b^3} + 6\cdot \frac{b^2}{c^3} + 9\cdot\frac{c^2}{a^3} \ge 19 \cdot \sqrt[19]{\left(\frac{a^2}{b^3}\right)^4 \left(\frac{b^2}{c^3}\right)^6 \left(\frac{c^2}{a^3}\right)^9} = 19 \cdot \sqrt[19]{\frac{a^8}{a^{27}}} = 19 \cdot \frac{1}{a}.
$$
Adding this inequality to its cyclic variants yields the result.


答:
There are million ways to prove this but the most immediate one would be the Rearrangement Inequality.
No matter the order of $a,b,c,$ the triplets $\left(\dfrac{1}{a^3}, \dfrac{1}{b^3}, \dfrac{1}{c^3}\right)$ and $\left(a^2, b^2, c^2\right)$ have the reverse arrangement.

Title: Prove $\frac{a^2}{b^3}+\frac{b^2}{c^3}+\frac{c^2}{a^3}≥\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$

Body:
Question : Prove $$\frac{a^2}{b^3}+\frac{b^2}{c^3}+\frac{c^2}{a^3}\geq \frac{1}{a}+\frac{1}{b}+\frac{1}{c}$$
$(a, b, c \in \mathbb{R}^+)$

I tried to solve it like this :
$$\frac{a^2}{b^3}+\frac{b^2}{c^3}+\frac{c^2}{a^3}+\frac{1}{a}+\frac{1}{b}+\frac{1}{c} \geq 2 \; (\frac{a}{b^2}+\frac{b}{c^2}+\frac{c}{a^2})$$

Am I doing this right? How can I finish this problem?


问:
which of these is a Hausdorff Space?
I have found the topologies on the set $\{a, b\}$:

$\{\emptyset, \{a, b\}\}$,

$\{\emptyset, \{a\}, \{a, b\}\}$,

$\{\emptyset, \{b\}, \{a, b\}\}$

and $\{\emptyset, \{a\}, \{b\}, \{a, b\}\}$,

and now want to determine which of these is Hausdorff/separated.

I have the definition that in order of for this each distinct point of the space must have a disjoint neighbourhood.

My thinking is that the only separated space in the list is $\{\emptyset, \{a, b\}\}$, because it is the only one that has no two elements with both $a$ and $b$?

答:
Actually, that one is not separated: there are no neighborhoods of $a$ and $b$ with empty intersection. And the same thing occurs with $\bigl\{\emptyset,\{a\},\{a,b\}\bigr\}$ and with $\bigl\{\emptyset,\{b\},\{a,b\}\bigr\}$.

On the other hand, in the case of $\bigl\{\emptyset,\{a\},\{b\},\{a,b\}\bigr\}$ you do get a Hausdorff space: $\{a\}$ is a neighborhood of $a$, $\{b\}$ is a neighborhood of $b$, and $\{a\}\cap\{b\}=\emptyset$.

Title: which of these is a Hausdorff Space?


问:
If $a,b,c,d,e$ be real number other than $-2≤a≤b≤c≤d≤e≤2$, prove the following inequation $\frac{1}{b-a}+\frac{1}{c-b}+\frac{1}{d-c}+\frac{1}{e-d}≥4$
Question
If $a,b,c,d,e$ be real numbers $-2≤a≤b≤c≤d≤e≤2$, prove the following inequation
$$\frac{1}{b-a}+\frac{1}{c-b}+\frac{1}{d-c}+\frac{1}{e-d}≥4$$

I tried to use Cauchy-Schwarz like
$$\left(\frac{1}{b-a}+\frac{1}{c-b}+\frac{1}{d-c}+\frac{1}{e-d}\right)(b-a+c-b+d-c+e-d)≥16$$
I think I should prove it using $-2≤a≤b≤c≤d≤e≤2$, but I'm confused about how to do it. Can anyone help me?

答:
Cauchy-Schwarz : $\left(\frac{1}{b-a}+\frac{1}{c-b}+\frac{1}{d-c}+\frac{1}{e-d}\right)(e-a)≥16$

From $−2≤a≤b≤c≤d≤e≤2$,  $e-a≤4$

∴$\frac{1}{b-a}+\frac{1}{c-b}+\frac{1}{d-c}+\frac{1}{e-d}≥4$


答:
You got it the other way around: to be Hausdorff, each couple of points of the space must have two disjoint open neighborhoods (one neighborhood for each point!). This is what gives a Hausdorff space its "separatedness", two points can always be separated by two disjoint neighborhoods.

In your example, the only topology which is Hausdorff is the discrete topology $\{\varnothing, \{a\},\{b\},\{a,b\}\}$, because $a$ and $b$ need to be separated, and you can do this only if both $\{a\}$ and $\{b\}$ (the only "plausible" neighborhoods of the two points) are open sets.

$\{a,b\}$ would also be a neighborhood for both $a$ and $b$, yes, but it doesn't separate them.
Similarly, if $\{a\}$ is open and $\{b\}$ isn't, you can't separate $a$ and $b$ because $\{a\}$ and $\{a,b\}$ aren't disjoint neighborhoods.

Title: which of these is a Hausdorff Space?

Body:
I have found the topologies on the set $\{a, b\}$:

$\{\emptyset, \{a, b\}\}$,

$\{\emptyset, \{a\}, \{a, b\}\}$,

$\{\emptyset, \{b\}, \{a, b\}\}$

and $\{\emptyset, \{a\}, \{b\}, \{a, b\}\}$,

and now want to determine which of these is Hausdorff/separated.

I have the definition that in order of for this each distinct point of the space must have a disjoint neighbourhood.

My thinking is that the only separated space in the list is $\{\emptyset, \{a, b\}\}$, because it is the only one that has no two elements with both $a$ and $b$?


问:
M/GI/1 service time distribution
I want to compute the distribution of the waiting time and the number of jobs for M/GI/1 where the service time is Heavy-Tailed or more specifically Pareto. I found this paper http://dl.acm.org/citation.cfm?id=1340307. However, I cannot understand at the end of the day how I can compute these distributions. Can someone explain it to me simple?

答:
You can use the Pollaczek–Khinchine formula to compute the transform of the response time W as
$$W^\ast(s) = \frac{(1-\rho)s g(s)}{s-\lambda(1-g(s))}$$
(where g(s) is the transform of service time pdf). See Diagle's book for more details

Daigle, N. The Basic M/G/1 Queueing System. Queueing Theory with Applications to Packet Telecommunication (2005): 159-223 (pages 165-66 available on Google Books).

These may also be of interest, in particular the explicit results for your case (heavy tailed distributions).

Takacs, L. Delay distributions for one line with Poisson input, general holding times, and various orders of service. Bell System Technical Journal 42 (1963): 487-504.

Abate, Joseph, and Ward Whitt. Explicit M/G/1 waiting-time distributions for a class of long-tail service-time distributions. Operations Research Letters 25.1 (1999): 25-31.

Title: M/GI/1 service time distribution


问:
Differentiation with respect to non-independent variable
If a function is defined as $F(x(t), x'(t), t) = \int(E+V)dt$ would $dF/dE$ equal zero? We know that $E$ and $V$ are made up of $x, x', t$ but $F$ is a function of $x, x', t$ not $E$. So the confusion is that it seems like the function $F$ contains $E$, but the way the function is defined does not explicitly say so.

答:
The answer is no: $\mathrm{d}F/\mathrm{d}E$ is never equal to zero. However, explaining the concept need some clarifications, so let's start from the very concept of functional dependence.

First of all, the writing
$$
F\big(x(t), x'(t), t\big) = \int(E+V)\,\mathrm{d}t
$$
is too imprecise to help in the understanding of what's going on. It is not safe to use the same $t$ symbol for the integration variable and the independent (time?) variable on the left side of the equality, thus in the following I'll use $s$ for the former and reserve $t$ for the latter.

Now let's try to analyze what can be thought the most general form of a integral function-functional as given in the original post:
$$
F\big(x(t), x'(t), t\big) = \int\limits_{[a(t),b(t)]}\Big[E\big(t, x(t), x'(t), s, x(s), x'(s)\big)+ V\big(t, x(t), x'(t), s, x(s), x'(s)\big)\Big]\,\mathrm{d}s\label{1}\tag{1}
$$
where $[a(t),b(t)]\subseteq\Bbb R$, meaning with this that $a=-\infty$ and $b=+\infty$ are acceptable choices. Then, calculating the functional derivative of \eqref{1} (as described, for example, in this long Q&A) we have
$$
\begin{split}
\frac{\delta F}{\delta E}& =\frac{\delta F}{\delta E}[E_o]\big(x(t), x'(t), t\big)\\
&=\frac{\mathrm{d}}{\mathrm{d} \epsilon}\int\limits_{[a(t),b(t)]}\Big[E\big(t, x(t), x'(t), s, x(s), x'(s)\big)+ \epsilon E_o\big(t, x(t), x'(t), s, x(s), x'(s)\big)\\
& \qquad \qquad\qquad \qquad\qquad+V\big(t, x(t), x'(t), s, x(s), x'(s)\big)\Big]\,\mathrm{d}s\Bigg|_{\epsilon=0}\\
&= \int\limits_{[a(t),b(t)]} E_o\big(t, x(t), x'(t), s, x(s), x'(s)\big)\,\mathrm{d}s
\end{split}\label{2}\tag{2}
$$
where $E_o=E_o\big(t, x(t), x'(t), s, x(s), x'(s)\big)$ is an admissible variation of the quantity (energy?) $E$.

The meaning of \eqref{2} is simply that the functional derivative of the functional $F$ is the integration on the given interval $[a(t), b(t)]$, i.e.
$$
\frac{\delta F}{\delta E}[\cdot] = \int\limits_{[a(t),b(t)]}[\cdot]\,\mathrm{d}s
$$

Notes

In my opinion, \eqref{1} is the "most general" form of the functionals described in the OP since its value depend both directly on the pointwise values of $t, x(t)$ and $x'(t)$ than on their "global" shape as a whole.

The calculations show that, if the structure of the functional is the one given in the OP, its functional derivative cannot be zero: in this case, it is the operator of integration over a given (fixed or variable, finite or infinite) interval. And the Euler-Lagrange equations loose their significance in this setting.

Note also that, assuming even a slightly simpler structure i.e. choosing
$$
E=E\big(s, x(s), x'(s)\big)
$$
i.e. allowing $E$ to be independent on the actual values $t, x(t), x'(t)$,  does not change the structure of the functional derivative, which is again the integration on the interval $[a(t), b(t)]$.


问:
Application of Poisson summation formula
I am currently self reading " Spectral theory of Riemann zeta function" by Yoichi Motohashi. The example is in first chapter and of poincare series. I want to know how the Poisson sum formula on the last sum led to the final equation. I dont need any exact answer. Any help/hint would be appreciated.

Edit 1:
Confusion in proof for writing Poincares series in terms of Kloosterman sum

This question I asked earlier today related to this topic.

答:
Please type your questions in mathjax, more readable and easier for us to copy/paste. $$f(x+iy)=\sum_n |x+iy+h/l+n|^{-2s}\exp(\frac{-2i\pi m}{ l^2(x+iy+h/l+n)})$$ is smooth $1$-periodic, it is equal to its Fourier series $$\sum_k \exp(2i\pi k x) c_k(y)$$ where $$c_k(y)=\int_0^1 f(x+iy)\exp(-2i\pi k x) dx$$ $$=\int_{-\infty}^\infty |x+iy+h/l|^{-2s}\exp(\frac{-2i\pi m}{ l^2(x+iy+h/l)})\exp(-2i\pi k x) dx$$

What do you get then?


问:
Uniform convergence of $f_n(x)= nx/(n^2x^2+1)$
I am trying to find whether $f_n(x)= nx/(n^2x^2+1)$ is uniformly convergent or not.

I found that the pointwise limit was $0$ except for $x=1/n$. Can I then say that $f$ is uniformly convergent on $(-\infty,-a) \cup (a,\infty)$ for all $a>0$ so that, for $n$ large enough, I cannot take $1/n$ because it would be less than $a$, and then I always have that $|f_n(x)-f(x)| < \epsilon$ ?

Thank you!

答:
Asserting that the pointwise limit of a sequence $(f_n)_{n\in\Bbb N}$ of functions from $\Bbb R$ into $\Bbb R$ is some $f(x)$ unless $x=\frac1n$ makes no sense, since $\frac1n$ is not a fixed number. Besides, for every real number $x$ you do have $\lim_{n\to\infty}\frac{nx}{n^2x^2+1}=0$. So, your sequence converges pointwise to the null function.

However, since $(\forall n\in\Bbb N):f_n\left(\frac1n\right)=\frac12$, you know that $\sup_{x\in\Bbb R}\bigl|f_n(x)-0\bigr|\geqslant\frac12$, and therefore the convergence is not uniform. Otherwise, you would have $\lim_{n\to\infty}\sup_{x\in\Bbb R}\bigl|f_n(x)-0\bigr|=0$.


问:
A special case of amalgamation in $\mathrm{Ind}(\mathcal{C})$
Let $\newcommand{\C}{\mathcal{C}}\C$ be a category with amalgamation, i.e. such that every span has a cocone. In this MO question, it is said that one can prove by induction that if $B ← A → C$ is a span in $\DeclareMathOperator{\Ind}{Ind}\Ind(\C)$ with $A ∈ \C$, then it has a cocone. I can only show this if $B$ and $C$ are colimits of chains of objects in $\C$ (to continue the induction, I need to drop the hypothesis that $A$ is in $\C$), or if amalgamation in $\C$ is functorial without the condition on $A$. How does one proceed? If it is false, is there something else similar which is true? The case of interest to me is when both $A$ and $B$ are in $\C$.

答:
Lemma: any object $X$ in $Ind(\mathcal C)$ which is not finitely presentable is a retract of the colimit of a smooth chain of objects $X = \varinjlim_{\alpha < \lambda} X_\alpha$ of lower presentability rank. Chain means the colimit is indexed by and ordinal $\lambda$ and smooth means that the functor $\alpha \mapsto X_\alpha$ preserves colimits (i.e. $X_\beta = \varinjlim_{\alpha<\beta} X_\alpha$ when $\beta < \lambda$ is limit).

Now to prove the claim. Let $B \leftarrow A \to C$ be an amalgamation problem in $Ind(\mathcal C)$ with $A \in \mathcal C$ and assume by induction that $Ind(\mathcal C)$ has amalgamation for spans $B'\leftarrow A \to C'$ where $rk(B'),rk(C') \leq rk(B),rk(C)$ and moreover that the amalgam $D'$ may be chosen to have $rk(D') \leq max(rk(B'),rk(C'))$ where $rk$ means presentability rank. Write $A = \varinjlim A_\alpha$, $C = \varinjlim C_\alpha$ as colimits of smooth chains of objects of lower presentability rank. We may assume without loss of generality that the maps $B \leftarrow A \to C$ factor through $B_0 \leftarrow A \to C_0$. We build a chain of amalgamations $D_\alpha$ of $B_\alpha \leftarrow A \to C_\alpha$, functorial in $\alpha$ [even though amalgamation in the whole category is not assumed to be functorial], by induction on $\alpha$. When $\alpha = 0$, let $D_0$ be any amalgamation. When $\alpha$ is limit, take a colimit of over the amalgams already constructed.

To amalgamate $B_{\alpha+1} \leftarrow A \to C_{\alpha+1}$, first choose an arbitrary amalgam $D_{\alpha+1}^0$. Then we have an amalgamation problem $D_\alpha \leftarrow A \to D_{\alpha+1}^0$, which we amalgamate to get $D_{\alpha+1}$. One must check that all the requiste diagrams commute (since we need this amalgam to be functorial in $\alpha$ in order to take a colimit at the limit stages).


问:
Does the sign of curvature remain constant under the sum of their metrics?
Does the sign of curvature remain constant under the sum of their metrics over a fixed manifold? e.g. If $(M,g_1)$ and $(M,g_2)$ are closed Riemannian manifolds of positive curvature then $(M,g_1+g_2)$ is also of positive curvature?

Update: Q2: What if we weaken the problem by allowing manifolds having boundary?

答:
Edit  As C.F.G. noted, I had originally answered different but related question.  So, let me answer the actual question asked with a No: the sum of two positively curved metrics can have negative curvature.  I'll construct an example below.

To do so, consider $S^3$ with a metric $g_\lambda$ making it a Berger Sphere.  Recall that this is obtained by beginning with the usual round metric $g_0$ and scaling it in the direction of the Hopf fibers.  More precisely, given any vectors $v,w\in T_p S^3$, we define $g_\lambda(v,w) = g_0(v,w) + \lambda g_0(v',w')$ where $v'$ is the $g_0$-projection of $v$ to the tangent space of the Hopf circle through $p$, and likewise for $w'$.

It is known that the resulting metric $g_\lambda$ is positive curved iff $-1<\lambda < 1/3$.  (If $\lambda \leq -1$, then $g_\lambda(v,v) \leq 0$ for all $v$ tangent to a Hopf circle, so we don't even get a metric in this case).

Pick any $\lambda$ strictly between $0$ and $1/3$.

Now, consider the metric $g:=g_{\lambda} + t g_0$ on $S^3$.  For any $t > 0$, this is a sum of positively curved metrics.  However, I claim that for any $t$ with $0< t < 3 \lambda -1$, that $g$ has negative curvature planes.

To see this, note that \begin{align*} g(v,w) &= g_\lambda(v,w) + t g_0(v,w)\\ &= (1+t) g_0(v,w) + \lambda g_0(v',w'),\end{align*} so $$\frac{1}{1+t} g(v,w) = g_0(v,w) + \frac{\lambda}{1+t} g_0(v',w')$$ is a Berger metric $g_{\lambda/(1+t)}$.  Since scaling doesn't change the sign of curvature, it follows that if we can show that $\lambda/(1+t) > \frac{1}{3}$, we are done.  Since $t< 3 \lambda - 1$, we find $\lambda/(1+t) > \lambda/(1 + 3\lambda - 1) = \frac{1}{3}.$

End edit - the old answer is left below

Proposition:  There is a positive curved metric $g_b$

No, the sum of metrics of fixed signed curvature is, in general, not of the same sign.

Proposition:  If the sum of any two positively curved metrics is positively curved, then every closed manifold of dimension $2$ or more admits a metric of positive curvature.

Proof:  First, if $g$ is a metric with positive curvature, then $\lambda g$ is for any positive $\lambda$.  Also, an easy induction argument shows that the hypothesis imply that any finite sum of positively curved metrics is positively curved.

Now, given any closed manifold $M^n$, compactness allows us to cover $M$ by a finite collection of open sets $U_1,..., U_k$, each diffeomorphic to $\mathbb{R}^n$ with the property that their closures $\overline{U_i}$ are contained in open sets $V_i$ which are also diffeomorphic to $\mathbb{R}^n$.  Let $\lambda_i$ be a partition of unity supported on $\overline{U}_i$.

Since $n\geq 2$, each $V_i$ admits a metric $g_i$ which is positively curved.  For example view $\mathbb{R}^n$ as the complement of a point in $S^n$ then pull the induced metric back to $g_i$.

Now the metic $g = \sum \lambda_i g_i|_{U_i}$ is positively curved by the first paragraph.  $\square$

Of course, there are obstructions to admitting a positively curved metric.  So, a consequence of this is that the sum of positively curved metrics need not be positively curved.


问:
Line-grid intersection in 2D
Say I have a line segment with known length L and a grid of pixels with dimension dX and dY along the X and Y axes. The line can begin from any arbitrary point within the grid and can have any arbitrary angle w.r.t the X and Y axes but its length is always the same (L). I would like to know what the maximum number of cells in the grid is that can be intersected with the line?

I was thinking something like this but am not sure:

ceil(L/dX) + ceil(L/dY) + 1

Numerical Simulations

In order to actually find what the correct answer is, I did a numerical simulation experiment where basically hundreds of millions of randomly generated lines with length L were started from a random point and fired randomly along various directions and the exact number of intersected grid pixels was counted to see what the max value is (since this was numerical simulations, I cannot say for sure that these values are correct):

L = 200, dX = 5, dY = 10 => Max no of intersected voxels = 47

L = 200, dX = 15, dY = 2 => Max no of intersected voxels = 103

L = 200, dX = 10, dY = 10 => Max no of intersected voxels = 31

答:
If a line segment from one corner of a pixel to another corner has length less than $L$, we can move the endpoints a tiny bit to avoid ever intersecting a corner between pixels, and still have length less than $L$.

In this case, if the original corners were at horizontal distance $x \cdot dX$ and vertical distance $y \cdot dY$, the new segment will cross $x+1$ vertical cell boundaries and $y+1$ horizontal cell boundaries; since it will cross $x+y+2$ cell boundaries total, it will pass through $x+y+3$ cells.

So we want to maximize $x+y+3$ subject to $(x \cdot dX)^2 + (y \cdot dY)^2 < L^2$; the $<$ condition will usually not bother us, but it is there to make sure that a line segment beginning at a corner of a cell does not count as intersecting the cell.

If we forget about the integer constraint (that $x,y \in \mathbb Z$), then the method of Lagrange multipliers tells us that the maximum occurs close to the point where the ellipse $(x \cdot dX)^2 + (y \cdot dY)^2 = L^2$ has a tangent line with slope $-1$ (equal to the slope of $x+y+3=k$). This solution is
$$
   (x,y) = \left(\frac{dY}{dX} \cdot \frac{L}{\sqrt{dX^2 + dY^2}}, \frac{dX}{dY} \cdot \frac{L}{\sqrt{dX^2 + dY^2}}\right)
$$
with $x+y+3 = L \sqrt{\frac1{dX^2} + \frac1{dY^2}} + 3$.

If we replace $x,y$ by $\lceil x\rceil - 1$ and $\lceil y \rceil-1$, we get an integer point where $(x \cdot dX)^2 + (y \cdot dY)^2 < L^2$, and the value $x+y+3$ decreases by at most $2$. (If it decreased by exactly $2$, then the original value $x+y+3$ was already unachievable, since it required having $(x \cdot dX)^2 + (y \cdot dY)^2 = L^2$ exactly.) Therefore this gives us the correct answer up to an error of $1$.

Unfortunately, fixing this error would require a search of integer points in some elliptical sliver of the form $(x \cdot dX)^2 + (y \cdot dY)^2 < L^2$ and $x+y \ge k$. Essentially, we are looking for fractional solutions that are worse than the optimal one but "round better". This is something that has to be done case-by-case.

But if you are okay with being off by $1$ sometimes, then we do have a closed-form solution: we get $$\left\lceil \frac{dY}{dX} \cdot \frac{L}{ \sqrt{dX^2+dY^2}}\right\rceil +\left\lceil \frac{dX}{dY} \cdot \frac{L}{ \sqrt{dX^2+dY^2}}\right\rceil +1$$ (or possibly $+2$) intersections.

Title: Line-grid intersection in 2D


问:
How to calculate probabilties of specific dice showing up while having a minimum total sum when rolling multiple dice.
I am trying to design a dice game (up to 5 dice) and here is the problem I am looking to solve:

I am rolling $N$ dice, and I want to know the probability of rolling exactly $M$ ones (1's) & the total being $\geq K$.

Example:

I roll 4 dice, and I want to know what the probability of rolling exactly 1 one while the total sum of the dice is $\geq 10$.

答:
There are $4$ possibilities for the die that shows the $1$.  The other three dice cannot sum to $6,7,\text{ or }8.$  Since the $1$ doesn't appear on these dice the numbers must be anything but$$\begin{align}
&2,2,2,\\&2,2,3,\\&2,3,3,\text{ or }\\&2,2,4\end{align}$$ which can occur in $1,3,3,\text{ or }3$ ways, respectively, for a total of $10$ ways.  There are $5^3=125$ ways to roll three numbers greater than $1$, so we have $4\cdot115=460$ acceptable rolls.  The probability of success is $$\frac{460}{6^4}=\frac{460}{1296}$$

The best way to approach the general problem is through the use of generating functions, but that's probably overkill for your application, since you won't, I imagine, be using more than a few dice.  If that's true, the approach exemplified above should suffice.


问:
Optimal interest compounding frequency
I am trying to figure out what is the optimal compounding frequency for annual interest $\mathbf r$ and compounding count $\mathbf n$.

I calculated that for $\mathbf n \rightarrow \infty$ the compounded interest approaches $\mathbf e^r$, but compounding infinitely many times is not very practical.

Especially when the compounding must be done manually or there might be a small fixed transaction fee charged for every compounding (in cryptocurrency yield farming for example).

To start I am trying to figure out how many times do I need to compound to reach 90% of the theoretical maximum $\mathbf e^r$ so I figured I need to solve the following equation for $\mathbf n$:

$$\left( 1+\frac{r}{n}\right)^n = 0.9e^r$$

How do I do that? Using natural log does not seem to help much. Also how would I modify the equation to account for a fixed transaction fee (every time compounding is done)?

答:
The only explict solution of
$$\left( 1+\frac{r}{n}\right)^n = k(r)$$ is given by
$$n=-\frac{r \log (k(r))}{r W_{-1}\left(-\frac{k(r)^{-1/r} \log (k(r))}{r}\right)+\log(k(r))}$$ where $W_{-1}(.)$ is the secondary branch of Lambert function

Make $k(r)=\frac 9{10} e^r$ and obtain

$$n=-\frac{r \left(r-\log \left(\frac{10}{9}\right)\right)}{r
   W_{-1}\left(-\frac{\left(\frac{10}{9}\right)^{\frac{1}{r}} \left(r-\log
   \left(\frac{10}{9}\right)\right)}{e r}\right)+r-\log
   \left(\frac{10}{9}\right)}$$
use the series expansion given in the linked page.

Title: Optimal interest compounding frequency