问:
what is the definition of restricted direct product ?
in an exercise , i'm given a definition of  restricted direct product , but it doesn't make sense for me !

they say , the  restricted direct product of groups $G_i$ is the set of elements of direct product which are the identity in all but finitely many compontents

i really don't understand what it means to be all but finitely !
i googled and found the same defintion with No clearification !

so , any help by example or explanation ?

答:
Note first that when we are taking the product of a finite number of groups, then the restricted and normal direct product are the same.

For convenience, let us assume that we have a family of groups indexed by the natural numbers, so we have a $G_i$ for each $i\in\mathbb{N}$.

The usual direct product of the groups $G_i$ can be seen as the set of sequences such that the $i$'th entry of the sequence is an element of $G_i$ (we can also write these as infinite tuples $(g_1,g_2,\dots)$ with each $g_i\in G_i$).

The restricted direct product is then the subgroup of the above direct product where we only take those elements such that for some $N$, the $i$'th entry is $e_i$ whenever $i\geq N$ (I use $e_i$ to denote the identity element of $G_i$). This means that we can see the restricted direct product as the set of finite tuples $(g_1,g_2,\dots,g_n)$ (but where $n$ can be as large as we want) and where we multiply such tuples by first extending the shortest of them by adding $e_i$'s to the end, such that they get the same length, and then multiply them as we usually would do with tuples (and we identify two such tuples if the only difference between them is how many $e_i$'s they end in).


问:
Solve the recurrence relation $T(n)=2T(n/2)+\log(n)$ given that $T(1) = 1$
Question:

Assume that a recurrence relation is given as below:

$T(n)=2T(n/2)+\log(n)$

and we know that $T(1) = 1$

We want to solve the relation (find an explicit definition of $T(n)$
which does not rely on itself).

My solving:

I wanted to solve it using substitution method but when i tried I get the series
$$\log(n)+2\log(n/2)+4\log(n/4)+8\log(n/8) ... 2^k\log(n/2^k)$$
which I am not able to solve.

答: The recurrence relation is not well defined. You can't calculate, for example, $T(3)$.

ContentLicense: CC BY-SA 4.0


问:
Solving ODE $y''+y'+2y=\tan(t) $ by laplace transform
If $y''+y'+2y=\tan(t) $
how can I solve this problem by Laplace?

There are some functions have not Laplace transform.
Is the tan function on of them?
And if it is one of them, how I can solve it?

答:
Wolfram Alpha finds a hair-raising solution to the ODE that involves the hypergeometric function with complex arguments.

I doubt that a human being can achieve that by hand in less than a day, and even less using the Laplace transform.


问:
Difficulty with uniqueness part of Cantor's Intersection theorem
Theorem: ( Cantor's intersection theorem ) Let $ (I_n) $ be a sequence of closed intervals on the real plane with the property $ I_{n+1} \subseteq I_n $ for all $ n \in \mathbb{N} $ and suppose $ \lim_{n \rightarrow \infty} |I_n| = 0 $ ( The limit is not on cardinality but on the length of the interval ). Then there exists a unique number $ c $ that belongs to all the intervals , meaning  $\bigcap\limits_{n=1}^\infty I_n = \{ c \} $.

Proof from lecture:

Denote $ I_n = [ a_n,b_n]$. The condition $ I_{n+1} \subseteq I_n $ implies that the sequence $ (a_n) $ is monotonically increasing and that the sequence $ (b_n) $ is monotonically decreasing. In addition, the given $ \lim_{n \rightarrow \infty} |I_n| = 0 $ is equivalent to $ | b_n - a_n | \rightarrow 0 $.

We need to show that there exists a unique point $ x $ that belongs to all the intervals, or equivalently that there exists a unique $ x $ such that $ a_n \leq x \leq b_n $ for all $ n \in \mathbb{N} $.

Uniqueness: If x,y are real numbers such that $ a_n \leq x \leq b_n $ and $ a_n \leq y \leq b_n $ then
$ 0 \leq | x - y | \leq | b_n - a_n | $ for all $ n \in \mathbb{N} $. The rightmost and leftmost sides in the last inequality both tend to zero so by sandwich theorem we get $ | x - y | = 0 $, meaning $ x=y $.

Existence: [ proof of existence ]

I haven't added the existence part because I understood it, I don't understand the uniqueness part though. How did they arrive to the inequality $ 0 \leq | x - y | \leq | b_n - a_n | $ ?

答: For each $n\in\Bbb N$, since $x,y\in[a_n,b_n]$, $|x-y|\leqslant b_n-a_n=|b_n-a_n|$.


答:
I assume you know what the direct product $\times_{i\in I} G_i$ (where $I$ is the indexing set) is. An arbitrary element of it looks like $(g_i)_{i\in I}$.

Now formally what they mean is that such a $(g_i)_{i\in I}$ belongs the restricted direct product iff there exists a finite subset $S\subset I$ such that $g_i=0$ for $i\notin S$ (where $0$ denotes the neutral element in each group).

ContentLicense: CC BY-SA 3.0

Title: what is the definition of restricted direct product ?

Body:
in an exercise , i'm given a definition of  restricted direct product , but it doesn't make sense for me !

they say , the  restricted direct product of groups $G_i$ is the set of elements of direct product which are the identity in all but finitely many compontents

i really don't understand what it means to be all but finitely !
i googled and found the same defintion with No clearification !

so , any help by example or explanation ?


答:
We have

$(1) \quad a_n \le x \le b_n$

and

$(2) \quad -b_n \le -y \le -a_n$

We add (1) and (2) and get

$$-(b_n-a_n) \le x-y \le b_n-a_n$$

This gives the desired inequality.

Title: Difficulty with uniqueness part of Cantor's Intersection theorem

Body:
Theorem: ( Cantor's intersection theorem ) Let $ (I_n) $ be a sequence of closed intervals on the real plane with the property $ I_{n+1} \subseteq I_n $ for all $ n \in \mathbb{N} $ and suppose $ \lim_{n \rightarrow \infty} |I_n| = 0 $ ( The limit is not on cardinality but on the length of the interval ). Then there exists a unique number $ c $ that belongs to all the intervals , meaning  $\bigcap\limits_{n=1}^\infty I_n = \{ c \} $.

Proof from lecture:

Denote $ I_n = [ a_n,b_n]$. The condition $ I_{n+1} \subseteq I_n $ implies that the sequence $ (a_n) $ is monotonically increasing and that the sequence $ (b_n) $ is monotonically decreasing. In addition, the given $ \lim_{n \rightarrow \infty} |I_n| = 0 $ is equivalent to $ | b_n - a_n | \rightarrow 0 $.

We need to show that there exists a unique point $ x $ that belongs to all the intervals, or equivalently that there exists a unique $ x $ such that $ a_n \leq x \leq b_n $ for all $ n \in \mathbb{N} $.

Uniqueness: If x,y are real numbers such that $ a_n \leq x \leq b_n $ and $ a_n \leq y \leq b_n $ then
$ 0 \leq | x - y | \leq | b_n - a_n | $ for all $ n \in \mathbb{N} $. The rightmost and leftmost sides in the last inequality both tend to zero so by sandwich theorem we get $ | x - y | = 0 $, meaning $ x=y $.

Existence: [ proof of existence ]

I haven't added the existence part because I understood it, I don't understand the uniqueness part though. How did they arrive to the inequality $ 0 \leq | x - y | \leq | b_n - a_n | $ ?


问:
Bidual of Banach Space
I am trying to prove that given a reflective Banach space $X$, i.e $X^{**}=X$, and a complet subspace $E \subseteq X$ then $E$ is reflective to. I had som problem proving it so i looked at a solution from a functional analysis class. Here the lecture states that it is obvious that $E \subseteq E^{**}$. I do not see why this should be true and have tryed to prove this without sucess. I would be very happy if someone could give an explenation or a proof of why this is true.

答: Any normed linear space $X$ is regarded as a subset of $X^{**}$ by identifying $x \in X$ with $F_x \in X^{**}$ defined by $F_x(f)=f(x)$ for all $f \in X^{*}$. In this sense $E \subseteq E^{**}$. The identification $x \to F_x$ is a nice map: it is linear, injective and preserves distances.

Title: Bidual of Banach Space


问:
Equivalence of categories preserves balancedness
I'm sorry for the silly question I'm going to ask.

Well, if $C$ and $D$ are two equivalent categories and $C$ is balanced, is it true that also $D$ is balanced? Intuitively, it seems to be true because equivalences preserve monomorphisms, epimorphisms and isomorphisms.

If not, can you provide a counterexample? Moreover, in this case, if $C$ and $D$ are isomorphic and $C$ is balanced, then also $D$ is?

答:
As you wrote, a category equivalence preserves monomorphisms and epimorphisms, and, as any functor, also isomorphisms.

Hence, in this case $C$ is balanced if and only if $D$ is balanced.

Since isomorphism of categories is a special case of equivalence, of course it also holds for isomorphic categories.


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

Problem 3.5:

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

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

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

The solution in the manual goes as following:

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

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

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

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

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

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

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

Why this is true?

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

答: If  a sequence converges (to a finite or infinite limit) then every subsequence of it also conveges to the same limit. This is immediate from the definition of limit.</p>


答:
We think of every element $x\in E$ as a functional $x:E^*\to\mathbb{C}$ defined by $x(f)=f(x)$. You can easily check that $x$ is indeed a bounded linear functional, i.e an element of $E^{**}$. So that way we have a map $E\to E^{**}$. It is linear, injective and preserves the norm. (the last two parts are not absolutely trivial, they follow from Hahn-Banach theorem).

So the "inclusion" $E\subseteq E^{**}$ is not a real inclusion as defined in set theory, it is just a very natural embedding.

Title: Bidual of Banach Space

Body: I am trying to prove that given a reflective Banach space $X$, i.e $X^{**}=X$, and a complet subspace $E \subseteq X$ then $E$ is reflective to. I had som problem proving it so i looked at a solution from a functional analysis class. Here the lecture states that it is obvious that $E \subseteq E^{**}$. I do not see why this should be true and have tryed to prove this without sucess. I would be very happy if someone could give an explenation or a proof of why this is true.


问:
Boundary of Convex set under Linear transformation
In $\mathbb{R}^n$, any line goes to line and any convex set goes to convex set under linear trasformation. But will boundary of convex set also will go to boundary of range of convex set?</p>

答:
Since in convex analysis, we often talk about relative interior points, if the OP also considers relative boundary of a convex set $C$, i.e. the boundary of $C$ within its affine hull, then even when the linear mapping $A$ is from $\mathbb{R}^n\rightarrow\mathbb{R}^n$, the relative boundary of $C$ may not be mapped to the relative boundary of the image $A(C)$. This happens when $A$ is not injective.

For example, consider a linear mapping $A: \mathbb{R}^2 \rightarrow \mathbb{R}^2$ defined as $(x,y)\mapsto (x,0)$ (the ''same'' map as in the previous answer, except its image is now in $\mathbb{R}^2$), then again we have $A(\partial [0,1]^2)=[0,1]\ne \{0,1\}=\partial A([0,1]^2)$, where $\partial$ means the relative boundary.


问:
Bibliography about the Axiom of Limitation of Size
I'm trying to study Von Neumann's Axiom of Limitation of Size, but I'm unable to find good bibliography about it. In fact, the only place I've found in which it is stated formally is Wikipedia. Does anyone know of good bibliography? I'm aware that there is already a similar question in Mathstackexchange, but couldn't found any references there either:
Axiom of Limitation of Size Reference Request</p>

答: In case it is useful, I've found the original source of the axiom: Von Neuman's 'An Axiomatization of Set Theory' (1925), that can be found on 'From Frege to Gödel' (page 400, axiom IV.2). In fact, in page 397 it is said "(the function) must be single valued, but it need not be one-to-one". Many textbooks (and the Stanford Encyclopedia) introduce the axiom as a bijection between a proper class and $V$, but that's not actually correct: it just needs to be surjective.


问:
How to denote a map $\kappa:M\rightrightarrows m_i$ where $m_i\in M$ is the input of $\kappa$?
Let $M=\{m_1,\dots\}$ be a set that contains several sets $m_i\in M$. I need to define a correspondence $\kappa$ that whenever inputed a set $m_i\in M$, returns me a subset of that particular set $m_i$. I do not know how to formalise this notion correctly.

My best attempt so far is this one: let $\kappa:M\rightrightarrows m_i$, where $m_i$ is the input of $\kappa$. This way of expressing such a simple idea seems unsatisfactory and imprecise: I need a better (more formal way) of specifying that "where $m_i$ is the input of $\kappa$". Could you please help me formalise it correctly?

I'm sorry for my rather unclear title, I hope somebody can understand my problem and provide some guidance.

答:
You will not find any standard notation for this I believe. One thing you could do is define $\kappa$ as a function

$\kappa: M \rightarrow \Pi_i \mathcal{P}(m_i)$

where $\kappa(m_i) \in \emptyset \times ... \times \mathcal{P}(m_i) \times \emptyset \times ...$ and by $\mathcal{P}$ we mean the powerset. This way, such a function would uniquely to a set $m_i$ assign a subset of $m_i$ by checking its value of the $i$'th coordinate.


问:
Curious Binomial identity that came up in derivative pricing
How can I show that, for a given integer $n$, that the following identity is true?
$$\left[(-1)^n\binom{n-1}{m}-\sum_{k=m}^{n-2}(-1)^k\binom{n}{k+1}\binom{k}{m}\right]=-(-1)^m$$

This came up in a derivation of arbitrary derivatives of the Black-Scholes function.  It is part of a larger equation for the $n^{th}$ order derivative
$$\sum_{m=0}^{n-2}\frac{H_m(d)}{(\sigma\sqrt{\tau})^m}\left[(-1)^n\binom{n-1}{m}-\sum_{k=m}^{n-2}(-1)^k\binom{n}{k+1}\binom{k}{m}\right]$$
which, in order to make contact with a known result (Carr, Peter. (2000). Deriving Derivatives of Derivative Securities. Journal of Computational Finance. 4. 101 - 128.
https://ieeexplore.ieee.org/document/844609) this identity must hold.  I've also checked that it's true in sagemath.

答:
$$
\begin{align}
&(-1)^n\binom{n-1}{m}-\sum_{k=m}^{n-2}(-1)^k\binom{n}{k+1}\binom{k}{m}\\
&=\sum_{k=m}^{n-1}(-1)^{k+1}\binom{n}{n-k-1}\binom{k}{k-m}\tag1\\
&=(-1)^{m+1}\sum_{k=m}^{n-1}\binom{n}{n-k-1}\binom{-m-1}{k-m}\tag2\\
&=(-1)^{m+1}\binom{n-m-1}{n-m-1}\tag3\\[9pt]
&=(-1)^{m+1}[n\ge m+1]\tag4
\end{align}
$$
Explanation:

$(1)$: $(-1)^n\binom{n-1}{m}$ becomes the $m=n-1$ term of the sum

$\phantom{\text{(1):}}$ furthermore, $\binom{k}{m}=\binom{k}{k-m}$

$(2)$: $\binom{k}{k-m}=(-1)^{k-m}\binom{-m-1}{k-m}$ (negative binomial coefficients)

$(3)$: Vandermonde's Identity

$(4)$: Iverson Brackets


问:
Sums factor as products in internal hom functor
In case this is helpful, I am working in the category of condensed abelian groups, but I think this can be phrased in a more general categorical way.

Suppose we have a closed symmetric monoidal category $\mathcal{C}$, with internal hom ${\mathscr{H}\kern -.5pt om}$. By the adjunction we have that

$$ {\mathscr{H}\kern -.5pt om} (\lim_{\to} A_i , B) = \lim_{\leftarrow} {\mathscr{H}\kern -.5pt om} (A_i, B)$$

Any direct sum can be written as a filtered colimit (consider finite direct sums). This would mean that

$$ {\mathscr{H}\kern -.5pt om} (\bigoplus A_i, B) = \lim_{\leftarrow} {\mathscr{H}\kern -.5pt om}(A_i, B)$$

One would expect this to be a product but due to my example (more on this below), this does not seem to always be the case. Could it be that the resulting limit just isn't a product?

MY COUNTEREXAMPLE (although I don't think this is strictly necessary to answer the question):

It is not difficult to show that for any set $I$,

$$ \underline{\bigoplus_I \mathbb{Z}} = \bigoplus_I \underline{\mathbb{Z}}$$

where the underline is the standard condensation functor.

Also, for Hausdorff topological groups $A$ and $B$ with $A$ compactly generated, where $\underline{\operatorname{Hom}}(-,-)$ is the internal hom,

$$\underline{\operatorname{Hom}} (\underline{A}, \underline{B}) = \underline{ \operatorname{Hom}( A, B)}$$

where $\operatorname{Hom}(A,B)$ carries the compact-open topology. If sums factored as products,

$$ \underline{\operatorname{Hom}}(\underline{\bigoplus_I \mathbb{Z}}, \underline{\mathbb{Z}}) = \underline{\operatorname{Hom}}(\bigoplus_I \underline{\mathbb{Z}}, \underline{\mathbb{Z}}) = \prod_I \underline{\mathbb{Z}} = \underline{\prod_I \mathbb{Z}} $$

where $\prod_I \mathbb{Z}$ has the discrete topology.

However, we also have

$$\underline{\operatorname{Hom}} (\underline{\bigoplus_I \mathbb{Z}}, \underline{\mathbb{Z}}) = \underline{\operatorname{Hom}(\bigoplus_I \mathbb{Z}, \mathbb{Z})} = \underline{\prod_I \mathbb{Z}}$$

but it is not difficult to see that with the compact-open topology $\prod_I \mathbb{Z}$ does not have the discrete topology.

Of course, there could be some mistake in the above but I can't spot it.

答:
Just found my mistake. It is not true that

$$\prod_I \underline{\mathbb{Z}} = \underline{\prod_I \mathbb{Z}}$$

The condensation functor commutes with limits of compact Hausdorff spaces, but this is not necessarily the case for discrete groups.


问:
Determine Butcher Table from a method
If I have a method, for example $y_{n+1} = y_n + \frac{h}{2}[f(t_n,y_n)+f(t_{n+1},y_n+hf(t_n,y_n)) ]$, how would I go about determining its Butcher Table to write it as a Runge-Kutta method?

答:
Usually we can observe that we have a linear combination of terms of $f$, plus a $y_n$ that is always there.
What I personally observe first are the coefficients in the linear combination. The coefficients of these terms are the $b$'s in the final RK form (excluding $h$ from coefficients).
Here you have 1/2 for both terms.

Secondly I find it simplest to identify the $c$'s. They reveal themselves from the time points which the $f$'s are evaluated at, with respect to $t_n$.
Here you have $t_n+0$ and $t_{n+1} = t_n + h\cdot 1$, so the c's are $0, 1$.
Each $c_i$ corresponds to a $k_i$.

Each $k_i$ is defined through a set of $a$'s. Then I find it simplest to start with the lowest $c$, usually zero (could be negative though), call this $c_1$.
The $k$ could be implicitly defined, where this would be clear if any term of $f$ was to be evaluated in $y_{n+i}, i>0$ on the form you specified above.
Your method is clearly explicit with $k_1 = f(t_n, y_n + h \sum_j a_{1, j}k_j) = f(t_n, y_n) $, so $a_{1, j} = 0$ for all $j$. This gves both $a$ and $k$ for this $c$.

Then move on to $k_2$ and $c_2$. Here we find that $a_{2, 1} = 1$, and the rest zero.
Then we are done.

Title: Determine Butcher Table from a method


问:
Is $A=\left\{x:x \text{ belongs to } (1,2) \cup [2,3)\right\}$ connected?
My question is
whether the given set
$$A=\left\{x:x \text{ belongs to } (1,2) \cup [2,3)\right\}$$is connected?
If so then how?
I know $x$ belongs to $(1,2)$ implies that $1<x<2$ and $x$ belongs to $[2,3)$ implies that $2\leq x<3$.
So these two are "not" disjoint sets since they have at least one element in common, i.e., $2$.

答:
A set $A$ is connected if there are no Two non-empty separated subsets $B$ and $C$ such that $A = B \cup C$
And according to Rudin's "Principles of Mathematical Analysis" Theorem 2.47:

A subset E of the real line $R^1$ is connected if and only if it has the following property: If $x \in E, y \in E,\space and \space x < z < y, then \space z \in E$

Which clearly hold here.


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

答: let $p$ divide $x$, where $x = p_1p_2\cdots p_n$ so we have $\frac {x}{p} = \frac {p_1p_2\cdots p_n}{p}=k , k \in \mathbb N$ but if $p \neq p_1 , p \neq p_2 \cdots p \neq p_n  $ that means one of the primes $p_1p_2\cdots p_n$ was divisble by p ! contradicting the fact that it was a prime. So no it is not possible.</p>


答:
Let's subistute $y(t) =u(t)e^{\lambda t}$, where $\lambda$ i some yet unspecified constant. We get
$$ y' = (u'+\lambda u)e^{\lambda t}$$
$$ y'' = (u''+2\lambda u' + \lambda^2 u)e^{\lambda t}$$
So the equation takes form
$$ \big(u''+(2\lambda +1)u' + (\lambda^2 + \lambda +2) u\big)e^{\lambda t} = \tan t$$
If we now choose $\lambda$ such that $\lambda^2+\lambda+2=0$ (that is $\lambda=\frac{-1\pm i\sqrt{7}}{2}$) the equation becomes
$$ \big(u''+(2\lambda +1)u' \big)e^{\lambda t} = \tan t$$
$$ u''+(2\lambda +1)u'  = e^{-\lambda t}\tan t$$
$$ e^{-(2\lambda +1)t}\big(e^{(2\lambda +1)t}u'\big)'  = e^{-\lambda t}\tan t$$
$$ \big(e^{(2\lambda +1)t}u'\big)'  = e^{(\lambda +1) t}\tan t$$
$$ e^{(2\lambda +1)t}u'  = \int e^{(\lambda +1) t}\tan t\, dt$$
$$ u' =e^{-(2\lambda +1)t}  \int e^{(\lambda +1) t}\tan t\, dt$$
$$ u = \int e^{-(2\lambda +1)t}  \left(\int e^{(\lambda +1) t}\tan t\, dt\right) dt$$
$$ y = e^{\lambda t}\int e^{-(2\lambda +1)t}  \left(\int e^{(\lambda +1) t}\tan t\, dt\right) dt$$

Title: Solving ODE $y''+y'+2y=\tan(t) $ by laplace transform

Body:
If $y''+y'+2y=\tan(t) $
how can I solve this problem by Laplace?

There are some functions have not Laplace transform.
Is the tan function on of them?
And if it is one of them, how I can solve it?


问:
On which open set in $\mathbb{R}^2$ is the $1$-form $(xdy-ydx)/(x^2+y^2)$ exact?
I think on an open set in the first quadrant not containing zero, there exists a function $f$ such that $df(x,y) = (xdy-ydx)/(x^2+y^2)$.

I think its like solving a differential equation with $x=rcos \theta$ and $y=rsin \theta$.

But I am not being able to do anything rigorous. Kindly help.

答:
Yes, it is exact differential, see
$$\frac{d \tan^{-1} (y/x)}{dx}=\frac{1}{1+y^2/x^2} \frac{xdy-ydx}{x^2}=\frac{xdy-ydx}{x^2+y^2}, (x,y) \ne (0, 0)$$


问:
Help with a Vardi Integral
I ran into some trouble trying to evaluate this integral:

$$\mathcal{I}=\int_{0}^{1}\frac{x}{x^{2}+1}\log\log\left(\frac{1}{x}\right)\mathrm dx$$

Basically, what I have at the moment is a generalized expansion

$$\int_{0}^{1}\frac{x^{p-1}}{\delta + x^{w}}\log\log\left(\frac{1}{x}\right)\mathrm dx = -\gamma\int_{0}^{1}\frac{x^{p-1}}{\delta+x^{w}}\mathrm dx-\frac{1}{\delta}\sum\limits_{k\geq 0}(-1)^{k}\cdot\frac{\log(wk+p)}{\delta^{k}(wk+p)}$$

Hence,

$$\mathcal{I} = -\gamma\int_{0}^{1}\frac{x}{1+x^{2}}\mathrm dx - \sum\limits_{k\geq 0}(-1)^{k}\cdot\frac{\log(2k+2)}{(2k+2)}$$

I can't figure out how to solve the sum. I'm positively certain there's some trick with digamma functions or Hurwitz-Zeta, but I'm too sleep-deprived to make sense of it. Maybe I'm going about this all wrong and there exists a more straightforward method to solve $\mathcal{I}$? Please let me know your opinion!

Title: Help with a Vardi Integral

Body:
In fact your formula is very close to the answer (provided by @Claude Leibovici).

The sum in your formula

$- \sum\limits_{k\geq 0}(-1)^{k}\cdot\frac{\log(2k+2)}{(2k+2)}=-\frac{\log2}{2} \sum\limits_{k\geq 0}(-1)^{k}\cdot\frac{1}{(k+1)}-\frac{1}{2}\sum\limits_{k\geq 0}(-1)^{k}\cdot\frac{\log(k+1)}{(k+1)}$

a) $-\frac{\log2}{2} \sum\limits_{k\geq 0}(-1)^{k}\cdot\frac{1}{(k+1)}=\frac{\log2}{2} \sum\limits_{k\geq 1}(-1)^{k}\cdot\frac{1}{k}=-\frac{\log^22}{2} $

b)$ -\frac{1}{2}\sum\limits_{k\geq 0}(-1)^{k}\cdot\frac{\log(k+1)}{(k+1)}=-\frac{1}{2}\sum\limits_{k\geq 1}(-1)^{k+1}\cdot\frac{\log(k)}{k}=-\frac{1}{2}\Bigl(\frac{\log(1)}{1}-\frac{\log(2)}{2}+\frac{\log(3)}{3}-+...\Bigr)$

After regrouping the terms in the parentheses

$b) =\frac{1}{2}\lim_{s\to+0}\frac{\partial}{\partial{s}}\Bigl(\frac{1}{1^{s+1}}-\frac{1}{2^{s+1}}+\frac{1}{3^{s+1}}-\frac{1}{4^{s+1}}+-...\Bigr)=$$=\frac{1}{2}\lim_{s\to+0}\frac{\partial}{\partial{s}}\Bigl((\frac{1}{1^{s+1}}+\frac{1}{3^{s+1}}+...)-(\frac{1}{2^{s+1}}+\frac{1}{4^{s+1}}+...)\Bigr)=$$=\frac{1}{2}\lim_{s\to+0}\frac{\partial}{\partial{s}}\Bigl((\frac{1}{1^{s+1}}+\frac{1}{2^{s+1}}+\frac{1}{3^{s+1}}+...)-2(\frac{1}{2^{s+1}}+\frac{1}{4^{s+1}}+...)\Bigr)=$$=\frac{1}{2}\lim_{s\to +0}\frac{\partial}{\partial{s}}\Bigl(\zeta(1+s)-\frac{2}{2^{s+1}}\zeta(1+s)\Bigr)$

But at $s\to0$ $\zeta(1+s)=\frac{1}{s}+\gamma+O(s)$

Therefore b) $=-\frac{\log2}{2}(\frac{\log2}{2}-\gamma)$


问:
When a $A$ commute with $e^B$, does it also commute $B$?
Let $A$ and $B$ be $n\times n$ matrices, and let $[A,B]=AB-BA$ be the commutator of $A$ and $B$.

Then, if we consider the matrix exponential $e^B=\sum_n \frac{B^n}{n!} $, we have
$$ [A,e^B]=\sum_n \frac{1}{n!}[A,B^n] $$

Now, if $[A,B]=0$, then $[A,B^n]=0$ $\forall n$ and therefore $[A,e^B]=0$. But what about the converse? Is it true as well?

In other words, does $[A,e^B]=0$ imply $[A,B]=0$ ?

答:
No. E.g. let $B=\pmatrix{0&0\\ 0&2\pi i}$. Then $e^B=I$. Now, all matrices commute with $e^B$, but some of them don't commute with $B$.

More generally, if $B$ is diagonalisable over $\mathbb C$ and it possesses two eigenvalues $\lambda_1$ and $\lambda_2$ such that $\lambda_1-\lambda_2$ is a nonzero integer multiple of $2\pi i$ (i.e. $B$ is a diagonalisable matrix whose eigenvalues are not $2\pi i$ congruence-free), then the family of matrices that commute with $B$ is a proper subset of the family of matrices that commute with $e^B$. Hence this $B$ serves as a counterexample to your hypothesis. In particular, any such counterexample must be non-Hermitian.


问:
Help regarding the limit of an integral
The  function $g(t)$  givenby
$$ g(t)=\frac{n(n-1)}{2}\left[\frac{t^{n-2}(2-t)^{n+1}-t^{2n-1}}{n+1}-\frac{t^{n}(2-t)^{n-1}-t^{2n-1}}{n-1}\right]$$  integrates to 1 form $t=0$ to $t=1$.From numerical simulations one can see  that limit of $\int_0^1 tg(t) dt$ as n goes to infinity approaches one i.e $ \lim_{n \to \infty}\int_0^1 tg(t) dt$ is expected to be 1.However, i am not able to prove it analytically.Any help/hint in this regard is greatly appreciated

答:
The only thing to use is
$$I(p,q)=\int_0^1 t^p\,(2-t)^q\,dt=2^{p+q+1} B_{\frac{1}{2}}(p+1,q+1)$$ where appears the incomplete Beta function.

Multiplying by $t$ we then have

$$\int_0^1\Big[t^{n-1}(2-t)^{n+1}-t^{2n}\Big]\,dt=2^{2 n+1} B_{\frac{1}{2}}(n,n+2)-\frac{1}{2 n+1}$$
$$\int_0^1\Big[t^{n+1}(2-t)^{n-1}-t^{2n}\Big]\,dt=2^{2 n+1} B_{\frac{1}{2}}(n+2,n)-\frac{1}{2 n+1}$$ All of this gives
$$\int_0^1 t\,g(t)\, dt=$$ $$\frac{n \left(4^n \left(2 n (2 n+1) B_{\frac{1}{2}}(n,n+1)-(2 n (n+1)+1)
   B_{\frac{1}{2}}(n,n)\right)+1\right)}{(n+1) (2 n+1)}$$ But, hoping that I am not mistaken, this tends to $1$ and not $0$.

Could you check if there is or not a typo in the definition of $g(t)$ ?

Title: Help regarding the limit of an integral


问:
Express $(\theta, \phi)$ of a torus using $(x,y,z)$
I know how to write the $(x,y,z)$ coordinates of a torus using $(\theta, \phi)$:
$$
\begin{align}
x(\theta, \phi) &= (R+r\cos(\theta))\cos(\phi) \\
y(\theta, \phi) &= (R+r\cos(\theta))\sin(\phi) \\
z(\theta, \phi) &= r\sin(\theta)
\end{align}
$$

For $\theta, \phi\in [0, 2\pi]$. How can I find the inverse relation? I would like to find expressions
$$
\theta(x,y,z) = \ldots \\
\phi(x,y,z)=\ldots
$$

Given that I have a point on the torus in cartesian coordinates $(x, y,z)$. How can I find its expression in terms of angles $(\theta, \phi)$?

From the last expression I can get
$$
\theta = \arcsin\left(\frac{z}{r}\right)
$$
and then I guess from the first equation one could find
$$
\phi = \arccos\left(\frac{x}{R + r\cos(\theta)}\right)
$$
which works when the denominator is not zero. To avoid having the denominator being zero we need
$$
R+r\cos(\theta) \neq 0 \implies \theta \neq \arccos\left(-\frac{R}{r}\right)
$$

答:
Your torus $T \subset \mathbb R^3$ is parameterized by
$$f : [0,2\pi] \times [0,2\pi] \to \mathbb R^3, f(\theta,\phi) = ((R+r\cos(\theta))\cos(\phi), (R+r\cos(\theta))\sin(\phi), r\sin(\theta)) .$$
Note that this function is not injective. We have $f(\theta,\phi) = f(\theta',\phi')$ iff both $\theta - \theta', \phi - \phi' \in \{0, 2\pi\}$. This reflects the "usual" construction of $T$ by identifying opposite edges of the square $[0,2\pi] \times [0,2\pi]$: First identify $[0,\phi]$ with $[2\pi,\phi]$ which yields a tube. In this tube identify $[\theta,0]$ with $[\theta,2\pi]$ which yields a torus.

You can consider the restriction $g = f\mid_{[0,2\pi) \times [0,2\pi)}$ to get a bijective parameterization of $T$. You want to determine $g^{-1}$.

In his answer Adam Latosiński has given four equations which uniquely determine $\theta,\phi \in [0,2\pi)$ in terms of $x, y, z$. Unfortunately $\sin, \cos : [0,2\pi] \to [-1,1]$ are not injective, thus we cannot naively take the functions $\arcsin, \arccos$ to compute  $\theta,\phi$. In fact we have bijections $\arcsin : [-1,1] \to [-\frac{\pi}{2},\frac{\pi}{2}]$ and $\arccos : [-1,1] \to [0,\pi]$ (principal values). Using them correctly gives ugly formulae involving many cases. For example we get
$$\phi  = \begin{cases} \phantom{-}\arccos(\frac{x}{\sqrt{x^2+y^2}}) & y \ge 0 \\ 2\pi -\arccos(\frac{x}{\sqrt{x^2+y^2}}) &  y \le 0 \end{cases}$$
It is much better to work with complex numbers. We know that $\exp : [0,2\pi) \to S^1 =  \{ w \in \mathbb C \mid \lvert w  \rvert = 1\}$ is a bijection. Since $\exp(t) = \cos(t) + i \sin(t)$ we get from Adam Latosiński equations (with $w = x + iy$)
$$\exp(\phi) = \frac{w}{\lvert w  \rvert} ,$$
$$\exp(\theta) = \frac{\lvert w  \rvert - R +iz}{r} .$$

Update:

An alternative approach is this.

We can identify $\mathbb C$ and $\mathbb R^2$. Doing so, $S^1 = \{ (u,v) \in \mathbb R^2 \mid u^2 + v^2 = 1\}$.

The function $\exp : [0,2\pi] \to S^1, \exp(t) =(\cos(t), \sin(t))$, is a surjective closed map between compact Hausdorff spaces, hence a quotient map. This implies that $f$ induces a homeomorphism $\bar f : S^1 \times S^1 \to T$ given by
$$\bar f((a,b),(c,d)) = ((R+rc)a, (R+rc)b, rd).$$
Its inverse $\bar f^{-1} : T \to S^1 \times S^1$ is given by
$$a(x,y,z) = \frac{x}{\sqrt{x^2 + y^2}} ,$$
$$b(x,y,z) = \frac{y}{\sqrt{x^2 + y^2}} ,$$
$$c(x,y,z) = \begin{cases} \phantom{-}\sqrt{1 - (\frac{z}{r})^2} & x^2 +y ^2 \ge R^2 \\ -\sqrt{1 - (\frac{z}{r})^2} &  x^2 +y ^2 \le R^2 \end{cases}$$
$$d(x,y,z) = \frac{z}{r} .$$
Note that  $x^2 +y ^2 = R^2$ means $z = \pm r$ for $(x,y,z) \in T$.

This gives you the "$S^1$-coordinates" of $\bar f^{-1}(x,y,z)$. But assocating to a point $(u,v) \in S^1$ an angle $\psi$ such that $(cos(\psi), \sin(\psi)) = (u,v)$ is easy. Either take
$$\psi  = \begin{cases} \phantom{-}\arccos(u) & v \ge 0 \\ 2\pi -\arccos(u) &  v \le 0 \end{cases}$$
or
$$\psi  = \begin{cases} \phantom{-}\arcsin(v) & u \ge 0 \\ \pi - \arcsin(v) &  u \le 0 \end{cases}$$
Note that the first formula gives you $\psi \in [0,2\pi]$ and the second $\psi \in [-\frac{\pi}{2},\frac{3\pi}{2}]$.


问:
Mapping cylinder is Hausdorff

If $f:X\to Y$ is a map between Hausdorff spaces, then $M_f$ and $C_f$ are Hausdorff. Here, $M_f$ and $C_f$ are mapping cyliner and mapping cone i.e. $M_f = X\times I\sqcup Y/(x,0)\sim f(x)$ for $x\in X$ and $C_f = M_f/(X\times\{1\})$

I first tried to prove $M_f$ is $T_2$. Before, as $X$ and $Y$ are $T_2$, $X\times I\sqcup Y$ is $T_2$. Now let $[x],[y]\in M_f$ such that $[x]\neq[y]$. Then $x\neq y$ so we can find open nbds $U,V$ of $x,y$ in $X\times I\sqcup Y$ respectively such that $U\cap V = \emptyset$. I have no idea how to get further. Could you give any hint?

答:
Hint: distinguish cases.
The classes in $M_f$ can be singletons, for most points this is the case, i.e. all $[(x,t)]$ where $t \neq 0$ are their own class.

If we have the class of $(x,0)$: this is a non-trivial class as $(x,0) \sim f(x)$ by definition, and the class becomes $(f^{-1}[\{f(x)\}] \times \{0\}) \cup \{f(x)\}$: it will also
contain all $(x',0)$ where $f(x')=f(x)$, as these are forced to be in the same class via transivity and the connecting $f(x)$.

How would you separate the points $\{(x,t)\}$ and $(f^{-1}[\{f(p)\}] \times \{0\}) \cup \{f(p)\}$ in $M_f$? The former has a saturared neighbourhood $X \times (a,1)$ in $X \times I \sqcup Y$ for some $0 < t$, while for the latter we can use the saturated neighbourhood $Y \sqcup X \times [0,a)$ in that same space. The quotient images of these disjoint saturated open sets are still disjoint (general fact) and as required.

A little more thougth is required to separate $(f^{-1}[\{f(x)\}] \times \{0\}) \cup \{f(x)\}$ and $(f^{-1}[\{f(x')\}] \times \{0\}) \cup \{f(x')\}$ for distinct $x,x'$ in $X$ with $f(x) \neq f(x')$; here you must use the Hausdorffness of $X$ and $Y$ really.

The class of some $y \in Y$ is either of the same form as the previous case, namely $\{y\} \cup (f^{-1}[\{y\}])\times \{0\}$ if $y\in f[X]$, or else just $\{y\}$.

I'll leave you to puzzle out the details, but you do need to go in such case by case detail and realise what the classes actually are as subsets of the sum.

Title: Mapping cylinder is Hausdorff

Body:
Hint: distinguish cases.
The classes in $M_f$ can be singletons, for most points this is the case, i.e. all $[(x,t)]$ where $t \neq 0$ are their own class.

If we have the class of $(x,0)$: this is a non-trivial class as $(x,0) \sim f(x)$ by definition, and the class becomes $(f^{-1}[\{f(x)\}] \times \{0\}) \cup \{f(x)\}$: it will also
contain all $(x',0)$ where $f(x')=f(x)$, as these are forced to be in the same class via transivity and the connecting $f(x)$.

How would you separate the points $\{(x,t)\}$ and $(f^{-1}[\{f(p)\}] \times \{0\}) \cup \{f(p)\}$ in $M_f$? The former has a saturared neighbourhood $X \times (a,1)$ in $X \times I \sqcup Y$ for some $0 <a < t$, while for the latter we can use the saturated neighbourhood $Y \sqcup X \times [0,a)$ in that same space. The quotient images of these disjoint saturated open sets are still disjoint (general fact) and as required.

A little more thougth is required to separate $(f^{-1}[\{f(x)\}] \times \{0\}) \cup \{f(x)\}$ and $(f^{-1}[\{f(x')\}] \times \{0\}) \cup \{f(x')\}$ for distinct $x,x'$ in $X$ with $f(x) \neq f(x')$; here you must use the Hausdorffness of $X$ and $Y$ really.

The class of some $y \in Y$ is either of the same form as the previous case, namely $\{y\} \cup (f^{-1}[\{y\}])\times \{0\}$ if $y\in f[X]$, or else just $\{y\}$.

I'll leave you to puzzle out the details, but you do need to go in such case by case detail and realise what the classes actually are as subsets of the sum.


问:
If $x_ny_n$ converges, then $\lim_{n\to\infty} x_ny_n = \lim_{n\to\infty} x_n \lim_{n\to\infty}y_n$

If $x_ny_n$ converges, then $\lim_{n\to\infty} x_ny_n =  \lim_{n\to\infty} x_n  \lim_{n\to\infty}y_n$

I need to prove this given statement or give a counterexample.

I started like this:

If $x_ny_n$ converges there will be a $N_1\in\mathbb N$ such that $$|x_ny_n-L|<\epsilon.$$
I am confused about how to continue further. Should I also conside $|x_n-L_1|<\epsilon$
and $|y_n-L_2|<\epsilon$?

答: Hint: can you find sequences $x_n$ and $y_n$ which do not converge, such that $x_ny_n$ does converge?


问:
If $\lim_{x\to \infty} f(x) = c\neq 0$ and $\lim_{x\to \infty} g(x) = 0$, then $\lim_{x\to \infty} \frac{f(x)}{g(x)}$ does not exist.
This is just a minor part of another problem and intuitively clear, but I think that it is necessary to have a proof of this assertion.

If $\lim_{x\to \infty} f(x) = c\neq 0$ and $\lim_{x\to \infty} g(x) = 0$, then $\lim_{x\to \infty} \frac{f(x)}{g(x)}$ does not exist.

Proof:

$(\lim_{x\to \infty} f(x) = c\neq 0)\Rightarrow (\lim_{x\to \infty} \frac{1}{f(x)}=\frac{1}{c})$; this is already proven. $(\lim_{x\to \infty} \frac{1}{f(x)}=\frac{1}{c})\land(\lim_{x\to \infty} g(x) = 0)\Rightarrow (\lim_{x\to \infty} \frac{g(x)}{f(x)}=\frac{1}{c}\cdot0=0)$; this is also already proven. Now, $\lim_{x\to \infty} \frac{g(x)}{f(x)}=0$ means that for every $\varepsilon> 0$ there is a $N\in\mathbb{R}_{>0}$ such that, for all x, $$\textrm{if  } x>N,\textrm{then}\quad\left|\frac{g(x)}{f(x)}-0\right|=\left|\frac{g(x)}{f(x)}\right|=\frac{|g(x)|}{|f(x)|}<\epsilon, $$$$\textrm{so  } \frac{1}{\epsilon}< \frac{|f(x)|}{|g(x)|}=\left|\frac{f(x)}{g(x)}\right|. $$
Thus, $\lim_{x\to \infty} \left|\frac{f(x)}{g(x)}\right|=\infty$.

Since $\lim_{x\to \infty} h(x) = 0$ if and only if $\lim_{x\to \infty} |h(x)| = 0$ (this can be easily proved), $\lim_{x\to \infty} \frac{f(x)}{g(x)}\neq0$. Suppose $\lim_{x\to \infty} \frac{f(x)}{g(x)}=l\neq0$, then $$\lim_{x\to \infty}f(x)=\lim_{x\to \infty} \frac{f(x)}{g(x)} \cdot\lim_{x\to \infty}g(x)=l\cdot0=0,$$ a contradiction! $\square$

First of all is this proof correct? It seems a bit long for this relatively easy assertion, so are there any superfluous parts in it? Would it be sufficient to end with the statement $\lim_{x\to \infty} \left|\frac{f(x)}{g(x)}\right|=\infty$?

答: An easier proof: Prove by contradiction. (If $g(x)=0$ for $x$ arbitrarily large then $\frac {f(x)} { g(x)}$ won't even exist, so assume that  $g(x) \neq 0 $ for large enough $x$).  Now $f(x)=\frac {f(x)} {g(x)} g(x)$ So if $l=\lim \frac {f(x)} {g(x)}$ exists and is finite and  $g(x) \to 0$ then $f(x) \to (l)(0)=0$ a contradiction.


问:
Coupon Collector's problem with unequal probabilities
Given 4 different items, each with different chance of being selected, select an item, replace it and select another.

What is the expected mean number of selections that is required to select at least one of each item.

Through some research, this appears to be a variation of the coupon collector's problem where each of my "coupons" doesn't have the same probability.

From the final formula this post, I deduced that the average number of draws is given by:

$$
n\sum\limits_{j=1}^n {1\over j}
$$

How can I amend this for items of unequal probability of selection please?

From the partial answer from @John:

$p_1 = 1$

$p_2 = p_a (p_b + p_c + p_d) + p_b (p_c + p_d + p_a) + p_c (p_d + p_a + p_b) + p_d (p_a + p_b + p_c)$

Am I then right in thinking $p_3$ would be
$
p_ap_b(p_c + p_d) +
p_ap_c(p_b + p_d) +
p_ap_d(p_b + p_c) +
p_bp_c(p_a + p_d) +
p_bp_d(p_a + p_c) +
p_cp_d(p_a + p_b)
$

and $p_4$ would be
$
p_ap_bp_c(p_d) +
p_ap_bp_d(p_c) +
p_ap_cp_d(p_b) +
p_bp_cp_d(p_c)
$ simplified to $4(p_ap_bp_cp_d)$

I put these values into a spreadsheet, but there's something I've done wrong. For the purpose of testing, I gave my coupons equal probability where I know the result should be 8.33. However, I'm getting the result 71.67 and can't see what is wrong. Any suggestions please?

答:
$\def\m{\boldsymbol\mu}\def\p{\mathbf p}$

The expected value in question reads:
$$
\mathbb E=\sum_i\frac1{p_i}-\sum_{(i,j)}\frac1{p_i+p_j}+\sum_{(i,j,k)}\frac1{p_i+p_j+p_k}-\cdots-\frac{(-1)^n}{\underbrace{p_1+p_2+\cdots+p_n}_{=1}}.\tag1
$$

It can be written in a compact notation as:
$$
\mathbb E=\sum_{\m\ne0}\frac{(-1)^{|\m|-1}}{\m\cdot\p},\tag2
$$
where $\m$ are $2^n$ binary vectors of the length $n$ with $|\m|=\sum_{i=1}^n\mu_i$, and $\p=(p_1,p_2,\dots, p_n)$. The summation in (2) runs over all $\m$ except for $(0,0,\dots,0)$.


问:
Sum of two sinusoidal function
I want to calculate the sum of this two functions

$$ y_1(t) = A_1 \cdot \sin(\omega_1 \cdot t)  $$
$$ y_2(t) = A_2 \cdot \sin(\omega_2 \cdot t + \frac{\pi}{2}) $$
What would be ?? $$ y = y_1 +y_2 $$

答: The sum is just $$y_1+y_2=A_1\cdot \sin( \omega_1\cdot t) + A_2\cdot \sin( \omega_2\cdot t+\dfrac{\pi}{2}).$$ You could also use $$\sin(t+\dfrac{\pi}{2})=\cos(t)$$ to get that $$y_1+y_2=A_1\cdot \sin( \omega_1\cdot t) + A_2\cdot \cos( \omega_2\cdot t).$$

Title: Sum of two sinusoidal function


问:
Example of non-filtered colimit not commuting with a finite limit
I know that the canonical morphism $\mathrm{Colim}_i\mathrm{Lim}_jD(i,j)\to\mathrm{Lim}_j\mathrm{Colim}_iD(i,j)$ for a diagram $D:\textbf{I}\times\textbf{J}\to \textbf{Set}$ is not in general an isomorphism, but it is in the case in which the categories $\textbf{I}$ and $\textbf{J}$ are respectively filtered and finite.

I'm able to think about counterexamples for $\textbf{I}$ filtered (and of course $\textbf{J}$ infinite), and I would like to find another counterexample in the case of $\textbf{J}$ finite (and $\textbf{I}$ non-filtered).

Thanks in advance for any suggestion!

答: You can take $I=J$ be discrete categories with just two objects each. Then nearly any diagram $I\times J \to \mathbf {Set}$ will produce a morphism that is not an isomorphism.


问:
$dx/dt=rx-\ln(1+x)$
I have the equation $\frac{dx}{dt}=rx-\ln(1+x)$
I know the bifurcation point is at $r=1$ but I saw this through the geometric approach. How would I get this from a purely algebraic method?

答:
The solution of $$rx-\log(1+x)=0$$ is
$$x=-1-\frac{1}{r}W\left(-e^{-r} r\right)$$ where $W(.)$ is Lambert function.

In the real domain, $W(t)$ exists if $t \geq -\frac 1e$ and then the result.

Title: $dx/dt=rx-\ln(1+x)$


问:
Keno Game: What percentage of the sum of 20 balls has a value of 210 to 695?
In the Keno game, a player starts by selecting 20 numbers from the numbers 1 to 80. What percentage of the sum of 20 balls has a value of 210 to 695?. I don't have an idea to solve this. May anyone gives me a hint to solve this problem? Thanks</p>

答:
A related question: Lottery "Sum" forecasting

The smallest possible draw is $1+2+\cdots +20 =210$, so this lower constraint does not really affect the calculation, but it does no harm.

The probability of getting a sum $s$ is the number of partitions of $s$ into $20$ distinct positive parts each not greater than $80$, divided by the number of ways of choosing $20$ numbers from $80$.  This partition number can be found with a recursion, where if $n$ is the number of parts and $k$ is the largest part:

$$r(s,n,k) = \sum\limits_{i=1}^{s-k} r(s-i,n-1,i)$$ starting at $r(0,0,0)=1$.  Here we want $\sum\limits_{s=210}^{695} \sum\limits_{k=1}^{80} r(s,20,k)$.  The following R code does the recursion and finds the probability of $0.1025153$ (exact up to rounding)

```<code>library(matrixStats)
balls <- 80
draws <- 20      # from 2 to balls-1

maxsum <- draws*(balls + (balls-draws+1))/2
partitions <- matrix(rep(0, balls*maxsum), ncol=balls)
partitions[cbind(1:(balls-draws+1), 1:(balls-draws+1))] <- 1
cumpartitions <- rowCumsums(partitions)

for (nthdraw in 2:draws){
  for (thisdraw in nthdraw:(balls-draws+nthdraw )){
    partitions[, thisdraw] <-
      c(rep(0, thisdraw), cumpartitions[1:(maxsum-thisdraw), thisdraw-1])
    }
  partitions[, 1:(nthdraw-1)] <- 0
  cumpartitions <- rowCumsums(partitions)
  }

results <- cumpartitions[, balls]
c(sum(results), choose(balls, draws))  # check should be equal
# 3.535316e+18 3.535316e+18
sum(results[210:695]) / sum(results) # answer to the question
# 0.1025153
</code>```

Other approaches include simulation, such as

```<code>set.seed(2021)
kenosum <- function(balls, drawn){sum(sample(balls, drawn))}
simdata <- replicate(10^6, kenosum(80, 20))
mean(simdata >= 210 & simdata  <= 695)
# 0.102791
</code>```

which is quite close.

An alternative is a normal approximation.  The mean score is $\frac{20 \times 81}{2} = 810$ and the standard deviation is $\sqrt{\frac{20 \times (80-20) \times 81}{12}}=90$ so we could try about $\Phi\left(\frac{695+ \frac12-810}{90}\right) - \Phi\left(\frac{210 -\frac12-810}{90}\right)$ or in R:

```<code>pnorm(695.5, 810, 90) - pnorm(209.5, 810, 90)
# 0.1016471
</code>```

which is not too far away from the exact answer.

Body:
A related question: Lottery "Sum" forecasting

The smallest possible draw is $1+2+\cdots +20 =210$, so this lower constraint does not really affect the calculation, but it does no harm.

The probability of getting a sum $s$ is the number of partitions of $s$ into $20$ distinct positive parts each not greater than $80$, divided by the number of ways of choosing $20$ numbers from $80$.  This partition number can be found with a recursion, where if $n$ is the number of parts and $k$ is the largest part:

$$r(s,n,k) = \sum\limits_{i=1}^{s-k} r(s-i,n-1,i)$$ starting at $r(0,0,0)=1$.  Here we want $\sum\limits_{s=210}^{695} \sum\limits_{k=1}^{80} r(s,20,k)$.  The following R code does the recursion and finds the probability of $0.1025153$ (exact up to rounding)

library(matrixStats)
balls <- 80
draws <- 20      # from 2 to balls-1

maxsum <- draws*(balls + (balls-draws+1))/2
partitions <- matrix(rep(0, balls*maxsum), ncol=balls)
partitions[cbind(1:(balls-draws+1), 1:(balls-draws+1))] <- 1
cumpartitions <- rowCumsums(partitions)

for (nthdraw in 2:draws){
  for (thisdraw in nthdraw:(balls-draws+nthdraw )){
    partitions[, thisdraw] <-
      c(rep(0, thisdraw), cumpartitions[1:(maxsum-thisdraw), thisdraw-1])
    }
  partitions[, 1:(nthdraw-1)] <- 0
  cumpartitions <- rowCumsums(partitions)
  }

results <- cumpartitions[, balls]
c(sum(results), choose(balls, draws))  # check should be equal
# 3.535316e+18 3.535316e+18
sum(results[210:695]) / sum(results) # answer to the question
# 0.1025153

Other approaches include simulation, such as

set.seed(2021)
kenosum <- function(balls, drawn){sum(sample(balls, drawn))}
simdata <- replicate(10^6, kenosum(80, 20))
mean(simdata >= 210 & simdata  <= 695)
# 0.102791

which is quite close.

An alternative is a normal approximation.  The mean score is $\frac{20 \times 81}{2} = 810$ and the standard deviation is $\sqrt{\frac{20 \times (80-20) \times 81}{12}}=90$ so we could try about $\Phi\left(\frac{695+ \frac12-810}{90}\right) - \Phi\left(\frac{210 -\frac12-810}{90}\right)$ or in R:

pnorm(695.5, 810, 90) - pnorm(209.5, 810, 90)
# 0.1016471

which is not too far away from the exact answer.


问:
what is the definition of restricted direct product ?
in an exercise , i'm given a definition of  restricted direct product , but it doesn't make sense for me !

they say , the  restricted direct product of groups $G_i$ is the set of elements of direct product which are the identity in all but finitely many compontents

i really don't understand what it means to be all but finitely !
i googled and found the same defintion with No clearification !

so , any help by example or explanation ?

答:
For each $i\geq 0$, let $G_i$ be a group. You have probably already met the infinite direct product $$\prod_{i\geq 0}G_i=\{(g_0,g_1,\ldots)|g_i\in G_i\}$$ where we can choose any $g_i\in G_i$ for the $i$-th position in an element of the product. This set forms a group under pairwise multiplication of each component of two elements.

It turns out that this group has a subgroup called the restricted direct product (or direct sum if the $G_i$ are all abelian) which is the following set
$$\tilde{\prod_{i\geq 0}}G_i=\{(g_0,g_1,\ldots)|g_i\in G_i,\exists N\in\mathbb{N},(n>N\Rightarrow g_n=e_n)\}$$ where $e_n\in G_n$ is the identity of the group $G_n$. This is just the group of infinite sequences of elements in the product which are eventually always the identity. You should easily be able to show that this set forms a subgroup of $\prod G_i$.


问:
Asumptions required for Inverse Limit on Infinite Galois Group
I am currently trying to get some knowledge on Infinite Galois Theory.

During the construction of the Galois Group of an infinite Galois extension through an inverse limit of Galois group of intermediary finite Galois extensions, there is one point that puzzles me.

Some authors require the inverse system of the inverse limit to be based on a directed set (for example Milne), while some authors only require the weaker asumption of a partially ordered set.

So for example in Milne's approach we have to check that the compositum EF of two finite galois extensions E and F is still a finite galois extension before we are able to talk about the inverse limit.

Do we gain anything from this extra property of the inverse limit in this context, or is it just a difference of convention used by different authors ?

答:
For an infinite normal extensions $E/F$, for each $a\in E$ take a finite normal extension $K_a/F$ such that $a\in K_a$. For simplicity take $K_a$ the normal closure of $F(a)/F$.

Then $Aut(E/F)$ is the group $G$ whose elements are maps $f:E\to \bigcup_a Aut(K_a/F)$ such that $f(a)\in Aut(K_a/F),f(a)|_{K_a\cap K_b}=f(b)|_{K_a\cap K_b}$.

It is immediate that $Aut(E/F)$ injects naturally into $G$.

It remains to check that every element of $G$ defines an automorphism on the whole of $E$. For this you need to show that $f(a),f(b)$ gives an automorphism of $K_aK_b$.

As you see there is nothing abstract here.

If $E/F$ is separable then the primitive element theorem gives $K_aK_b\subset K_c$ and $f(c)|_{K_aK_b}$  is our automorphism. When $E/F$ is not separable you can replace $E$ by $E\cap F^{sep}$ as any automorphism of $E\cap F^{sep}/F$ extends easily to $E/F$.


问:
Prove that event will happen an infinite number of times
I want to prove that if $X_n$ is sequence of collectively independent exponential random variables and $\mathbb{E}(X_n) = 1$ than $$\text{event }\{X_n\le\ln{n}+\ln{\ln{n}}+\ln^2{\ln{\ln{n}}}\}\text{ will happen an infinite number of times.}$$ May somebody can help me? I would be grateful.

答: This is  simple application of Borel Cantelli Lemma. If $\sum P(X_n \leq a_n)=\infty$ then $X_n \leq a_n$ holds infinitely often with probability $1$. So you only have to check that $\sum [1-e^{-(\ln n+ \ln \ln n +\ln^{2} \ln \ln n)}]=\infty$. Can you check this? [The general term of this series does not tend to $0$].


问:
Cauchy sequence is convergent if it has a convergent subsequence
I know that there are several proofs of this on this site and on the internet, however, my doubt is about something very specific. I will write my question first and my proof below.

My doubt is if I can always choose $ X_m = (X_n)_k$ on the step 3 and 4 on the proof below, I am not sure why this makes sense.

Proof

I am doing the proof on R for simplicity, but I know that it can be easily generalized for any metric space, feel free to make any comment in terms of metrics/metric spaces.

Step 1:  The sequence $\{x_n\}$ is Cauchy so for any $l$ > 0 exists P $\leq$ m,n such that  $\mid x_n - x_m \mid < l/2 $

Step 2:  The subsequence $\{(x_n)_k\}$ is convergent so for any $l$ > 0 exists N $\leq$ n_k such that  $\mid (x_n)_k - x \mid < l/2 $

Step 3:  Chose M = max(m,n,n_k) and x_m = $\{(x_n)_k\}$, such that $\mid x_n - (x_n)_k  \mid + \mid (x_n)_k - x \mid$ < l/2 + l/2

Step 4:  By the triangle inequality, $\mid x_n - x  \mid$ < l

答:
Write it down in a more structured way. You have the idea, I think, but it needs a better write-up:

So suppose $(x_n)_n$ is Cauchy in $(X,d)$ and we have a subsequence $(x_{n_k})_k$ that converges to some limit $p \in X$.

The claim is that $x_n \to p$ as well. So we start by letting $\varepsilon>0$ be arbitrary.

We apply the Cauchy condition to $\frac{\varepsilon}{2}$ and conclude that there exists some $N \in \Bbb N$ such that

$$\forall n,m \ge N: d(x_n, x_m) < \frac{\varepsilon}{2}$$

Now using that $x_{n_k} \to p$ as $k \to \infty$, pick $M \in \Bbb N$ so that

$$\forall k \ge M: d(x_{n_k}, p) < \frac{\varepsilon}{2}$$

Now there exists some $k' \ge M$ so that $n_{k'} > N$ as the subsequence indices are increasing without bound in $\Bbb N$. Now apply the triangle inequality:

$$\forall n \ge N: d(x_n, p) \le d(x_n, x_{n_{k'}}) + d(x_{n_{k'}}, p) < \frac{\varepsilon}{2} + \frac{\varepsilon}{2} = \varepsilon$$

where the first $<$ is because of the Cauchy condition (we take $m= n_{k'}$ there) and the second because of the convergence condition.

As $\varepsilon >0$ was arbitrary, $x_n \to x$ as required.


问:
Is $(p \to q) \to (\neg q \to \neg p)$ a theorem in (Johansson's) minimal logic?
From a related question we know that $(P\to Q)\to (\neg Q \to \neg P)$ is a theorem in intuitionistic logic. I'm asking if that's also true for the positive fragment of intuitionistic logic aka minimal logic. Note that minimal logic does allow a notion of negation, because it does allow one to define negation relative to a (an arbitrarily chosen) false element $\bot$, i.e. ($\neg P =_{\mathrm{def}} P \to \bot $), it's just that the axiom $\bot \to X$ is not included in minimal logic.

Interestingly, as Wikipedia says

The propositional form of modus ponens, $(A \land (A\to B))\to B$ is clearly valid also in minimal logic.

The proof of this latter claim (not included in Wikipedia) is actually slightly involved, at least in the Hilbert calculus (of minimal logic). Recall (only the axioms I'm going to use) in this proof:

THEN-2: $(\phi \to (\chi \to \psi )) \to ((\phi \to \chi ) \to (\phi \to \psi ))$

AND-1: $\phi \land \chi \to \phi$

AND-2: $\phi \land \chi \to \chi$

Let $P = A \land (A\to B)$. By AND-1 and AND-2 we have that both $P \to A$ and $P \to (A\to B)$, but we really want to show that $P \to B$. However by MP application to $P \to (A\to B)$ and THEN-2 we get $(P \to A) \to (P \to B)$. And since (we had by AND-1 that) $P \to A$ it follows by another MP application to these last two formulas that $P \to B$.

As Wikipedia (also) says, from the fact that the propositional form of MP holds for minimal calculus, i.e. $(A \land (A\to B))\to B$ is a theorem, it also follows by simple substitution of $B$ with $\bot$ (and applying the definition of negation) that the propositional form of the law of non-contradiction

$\neg (A \land \neg A)$ can still be proven in minimal logic.

However, I don't quite see how/if one ca pull off a similar trick for $(P\to Q)\to (\neg Q \to \neg P)$. So, is the latter a theorem in minimal logic (too, somehow), or is there a proof that it isn't? (And if the later is the case, does minimal logic plus this propositional form of modus tollens result is some interesting logic?)

(I actually have not idea how to attempt a proof that something isn't a theorem in minimal logic. For intuitionistic logic one can easily use the 3-element linear lattice with the appropriate implication table [making it a Heyting algebra] to show e.g. that $(\neg Q \to \neg P) \to (P\to Q)$ is not a theorem in intuitionistic logic, by taking $P = 1$ and $Q = \frac{1}{2}$; that yields $(0 \to 0) \to (1 \to \frac{1}{2}) = (1) \to (\frac{1}{2}) = \frac{1}{2}$, proving that this formula is not a theorem in intuitionistic (or minimal) logic, but this is the "converse" of the formal I'm asking about. I have no idea what can serve as a similar device for minimal logic that can show that something is a non-theorem just in minimal logic.)

答:
Yes it is.

The usual ND rules for managing $\to$ are available also in ML.

$(p \to q)$ --- premise

$(q \to \bot)$ --- assumed [a]

$p$ --- assumed [b]

$q$ --- from 3) and 1) by $\to$-elim

$\bot$ --- from 4) and 2) by $\to$-elim

$(p \to \bot)$ --- from 3) and 5) by $\to$-intro, discharging [b]

$(q \to \bot) \to (p \to \bot)$ --- from 2) and 6) by $\to$-intro, discharging [a]

In an Hilbert-style proof system, you need the two "THEN-axioms" to prove the Deduction Theorem.

With this and Modus Ponens you have the necessary resources to manage $\to$.


问:
Derivative of the expected value of the truncated normal distribution wrt upper limit
I am looking for the derivative of the expected value of the truncated normal distribution with respect to one of the upper limits. It looks like this:
$$f=\frac{1}{\sqrt{\vert T \vert} (2\pi)^{3/2}} \int_{-\infty}^{x_3=a_3} dx_3 \int_{-\infty}^{x_2=a_2} dx_2 \int_{-\infty}^{x_1=a_1} dx_1 \begin{bmatrix}{x_1\\x_2\\x_3}\end{bmatrix}
   exp\bigg(-0.5\begin{bmatrix}{x_1  x_2  x_3}\end{bmatrix} T^{-1} \begin{bmatrix}{x_1\\x_2\\x_3}\end{bmatrix} \bigg)$$\
$$\frac{\partial f}{\partial a_2}?$$

I have tried using the Fundamental Theorem of Calculus but I am a bit stuck what to do with the other integrals and elements of the vectors.

答:
The answer is
\begin{equation}
\frac{\partial f}{\partial a_2}=\frac{1}{\sqrt{\vert T \vert} (2\pi)^{3/2}} \int_{-\infty}^{x_3=a_3} dx_3 \int_{-\infty}^{x_1=a_1} dx_1 \begin{bmatrix}{x_1\\a_2\\x_3}\end{bmatrix}
   exp\bigg(-0.5\begin{bmatrix}{x_1  a_2  x_3}\end{bmatrix} T^{-1} \begin{bmatrix}{x_1\\a_2\\x_3}\end{bmatrix} \bigg).
\end{equation}

Consider the antiderivative $F(x)$ of the function
\begin{equation}
F(x) = \int f(x) dx
\end{equation}
or
\begin{equation}
\frac{d F(x)}{dx} = f(x),
\end{equation}
so definite integral is the difference of antiderivatives
\begin{equation}
\int\limits_a^b f(x)dx = F(b) - F(a),
\end{equation}
i.e. the derivative is
\begin{equation}
\frac{\partial}{\partial b} \int\limits_a^b f(x)dx = F^{\prime}(b) = f(b).
\end{equation}


问:
How can I manipulate the solution of my differential equation into a nice form?
$$\frac{dP}{dt}=\frac{1}{2}\left(2P^{2}-P\right)\cos t$$
I can simply integrate both sides to obtain $$\frac{\left(2P-1\right)^{2}}{P}=ke^{\frac{1}{2}\sin t}$$ I have troubles from here on. I need to separate it to be in terms of a single $P$, and I've tried completing the square, quadratic formula (after timesing up the $P$), and squaring both sides. What should I do? A hint would be nice

Also $k = 1$ ( from the initial value problem in the original question ), although I dont think that matters too much.

答:
You get
$$\int \frac{dp}{2p^2-p}=\frac{1}{2} \int \cos t~dt \implies \ln \frac{2p-1}{p}=\frac{1}{2} \sin t+ c$$
$$\implies 2-\frac{1}{p}=ke^{\frac{1}{2}\ln \sin t}\implies p= \frac{1}{2-ke^{\frac{1}{2}\sin t}}.$$


答:
Yes, this is provable. Unwinding the definition, your formula is

$$(P\to Q)\to((Q\to\bot)\to(P\to\bot)).$$

This is an instance of the more general tautology

$$(P\to Q)\to((Q\to R)\to(P\to R)),$$

which is easily provable in minimal logic: for example, recalling that the deduction theorem holds for minimal logic, it suffices to show

$$P\to Q,Q\to R,P\vdash R,$$

which follows by two applications of modus ponens.

Title: Is $(p \to q) \to (\neg q \to \neg p)$ a theorem in (Johansson's) minimal logic?

Body:
From a related question we know that $(P\to Q)\to (\neg Q \to \neg P)$ is a theorem in intuitionistic logic. I'm asking if that's also true for the positive fragment of intuitionistic logic aka minimal logic. Note that minimal logic does allow a notion of negation, because it does allow one to define negation relative to a (an arbitrarily chosen) false element $\bot$, i.e. ($\neg P =_{\mathrm{def}} P \to \bot $), it's just that the axiom $\bot \to X$ is not included in minimal logic.

Interestingly, as Wikipedia says

The propositional form of modus ponens, $(A \land (A\to B))\to B$ is clearly valid also in minimal logic.

The proof of this latter claim (not included in Wikipedia) is actually slightly involved, at least in the Hilbert calculus (of minimal logic). Recall (only the axioms I'm going to use) in this proof:

THEN-2: $(\phi \to (\chi \to \psi )) \to ((\phi \to \chi ) \to (\phi \to \psi ))$

AND-1: $\phi \land \chi \to \phi$

AND-2: $\phi \land \chi \to \chi$

Let $P = A \land (A\to B)$. By AND-1 and AND-2 we have that both $P \to A$ and $P \to (A\to B)$, but we really want to show that $P \to B$. However by MP application to $P \to (A\to B)$ and THEN-2 we get $(P \to A) \to (P \to B)$. And since (we had by AND-1 that) $P \to A$ it follows by another MP application to these last two formulas that $P \to B$.

As Wikipedia (also) says, from the fact that the propositional form of MP holds for minimal calculus, i.e. $(A \land (A\to B))\to B$ is a theorem, it also follows by simple substitution of $B$ with $\bot$ (and applying the definition of negation) that the propositional form of the law of non-contradiction

$\neg (A \land \neg A)$ can still be proven in minimal logic.

However, I don't quite see how/if one ca pull off a similar trick for $(P\to Q)\to (\neg Q \to \neg P)$. So, is the latter a theorem in minimal logic (too, somehow), or is there a proof that it isn't? (And if the later is the case, does minimal logic plus this propositional form of modus tollens result is some interesting logic?)

(I actually have not idea how to attempt a proof that something isn't a theorem in minimal logic. For intuitionistic logic one can easily use the 3-element linear lattice with the appropriate implication table [making it a Heyting algebra] to show e.g. that $(\neg Q \to \neg P) \to (P\to Q)$ is not a theorem in intuitionistic logic, by taking $P = 1$ and $Q = \frac{1}{2}$; that yields $(0 \to 0) \to (1 \to \frac{1}{2}) = (1) \to (\frac{1}{2}) = \frac{1}{2}$, proving that this formula is not a theorem in intuitionistic (or minimal) logic, but this is the "converse" of the formal I'm asking about. I have no idea what can serve as a similar device for minimal logic that can show that something is a non-theorem just in minimal logic.)


答:
Assuming your differential equation is correct, then $$\frac 1 {P(2P - 1)}\mathrm{d}P = \frac 1 2 \cos t\ \mathrm{d}t$$

and we further decompose the LHS using partial fractions into $$\left(\frac 2 {2P - 1} - \frac 1 P\right)\mathrm{d}P = \frac 1 2 \cos t\ \mathrm{d}t.$$

Integrating both sides, we have $$\ln (2P - 1) - \ln P = \frac 1 2 \sin t + C, C \in \mathbb {R}$$

Now, we perform some magic (i.e. algebraic manipulation).

$$\ln \frac {(2P - 1)} P = \frac 1 2 \sin t + C$$

$$\frac {(2P - 1)} P = ke^{\frac 1 2 \sin t}, k \in \mathbb {R}$$

$$2 - \frac 1 P = ke^{\frac 1 2 \sin t}$$

$$\frac 1 P = 2 - ke^{\frac 1 2 \sin t}$$

Finally, $$P = \frac 1 {2 - ke^{\frac 1 2 \sin t}}$$ which is as nice as it gets.

Comments

You probably did not get this right as your integration was incorrect to begin with. However, you are right that the value of $k$ is irrelevant if all you are concerned with is making $P$ the subject of the formula.

Title: How can I manipulate the solution of my differential equation into a nice form?

Body:
$$\frac{dP}{dt}=\frac{1}{2}\left(2P^{2}-P\right)\cos t$$
I can simply integrate both sides to obtain $$\frac{\left(2P-1\right)^{2}}{P}=ke^{\frac{1}{2}\sin t}$$ I have troubles from here on. I need to separate it to be in terms of a single $P$, and I've tried completing the square, quadratic formula (after timesing up the $P$), and squaring both sides. What should I do? A hint would be nice

Also $k = 1$ ( from the initial value problem in the original question ), although I dont think that matters too much.


问:
Finite union of compact Hausdorff spaces
Is it true that the union of two compact Hausdorff subspaces of an arbitrary space is Hausdorff? What if the ambient space is weakly Hausdorff, that is, the image of any continuous map from a compact Hausdorff space to the given space is closed? (Then it is in particular, a $T_1$-space.)

答:
No, I think the following is a counterexample: let $X = \Bbb N \cup \{-\infty, +\infty\}$ and define the topology on $X$ by declaring all subsets of $\Bbb N$ open, and a subset $A \cup \{-\infty\}$ of $X$ is an open neighbourhood of $-\infty$ iff $\Bbb N\setminus A$ is finite, and similarly for $+\infty$. Then $X$ is not Hausdorff, (as witnessed by $-\infty, +\infty$) despite being the union of the compact Hausdorff subspaces $\Bbb N \cup \{+\infty\}$ and $\Bbb N \cup \{-\infty\}$ (which also have a non-compact intersection BTW).

This $X$ is not weakly Hausdorff as the embedding $N \cup \{\infty\}$ into $X$ does not have a closed image in $X$.

The other answer (by user10354138) shows why that is. The argument only holds for finite unions of compact Hausdorff subsets.


问:
Two sport shooters shoot : Calculate probabilities
Two sport shooters shoot one after the other independently of one another at a target with 10 rings.

Given that all possible outcomes are equally probable, calculate the probability that

(a) both score at least 7 points

(b) at least one scores 9 points or more

(c) the two points differ at most by 1

$$$$

I have done the following:

(a) Since the two shoot independently from the other the probability is equal to $\frac{4}{10}\cdot \frac{4}{10}$, since the possible outcome is $7,8,9,10$.

Is that correct?

(b) Do we take the sum of probabilities : propability that S1 scores at least 9 + propability that S2 scores at least 9 ?

(c) How do we calculate that probability?

答:
With the assumption that all outcomes are equally probable as the question states (though it is difficult to see how scoring one point has same probability as scoring ten points), for $(b)$, apply $P(A \cup B) = P(A) + P(B) - P(A\cap B)$ where $A$ and $B$ are events of player $1$ and $2$ scoring at least $9$ points.

For $(c)$, consider player one scoring $1$ point. In that case, player two must score either $1$ point or $2$ points to have at most one point difference with player one. The same is true when player one scores $10$ points. But for score of $2$ points to $9$ points by player one, player two has $3$ choices of score - either one less, equal or one more.

So desired probability $ = 2 \cdot\frac{1}{10} \cdot \frac{2}{10} + 8 \cdot\frac{1}{10} \cdot \frac{3}{10} = \frac{28}{100} = \frac{7}{25}$

Or you can look at all permissible outcomes -

i) Difference of one point in their score - $(1,2), (2,1), (2, 3), (3,2),...(9,10), (10,9)$ which are $9\cdot 2 = 18$ outcomes.

ii) Both have same score - $10$ possible outcomes.

That is $28$ favorable outcomes out of $100$.


答:
The answer to the first question is no, as seen by, e.g., the indiscrete two-point space having the two single point as trivially compact Hausdorff.

The second answer is yes.

Claim If $K$ is compact Hausdorff, $X$ is weakly Hausdorff, $u\colon K\to X$ is continuous, then $u(K)$ is compact Hausdorff in the subspace topology inherited from $X$.

Proof (gist): Let $L=u(K)$ and so $u\colon K\to L$ is closed since any closed subset $E$ of $K$ is compact Hausdorff so $u(E)$ is closed in $X$, hence in $L$ too.  If $a,b\in L$ with $a\neq b$ then $u^{-1}(a),u^{-1}(b)$ are disjoint closed subsets of $K$.  Since compact Hausdorff implies $T_4$, you can find disjoint $U,V$ in $K$ containing $u^{-1}(a)$ and $u^{-1}(b)$ respectively.  Then $U'=\{x\in X\mid u^{-1}(x)\subseteq U\}$ and similarly $V'$ shows $L$ is Hausdorff.$\quad\square$

Apply this to $i_1\cup i_2\colon K_1\amalg K_2\to X$ and you have the union.

Title: Finite union of compact Hausdorff spaces

Body: Is it true that the union of two compact Hausdorff subspaces of an arbitrary space is Hausdorff? What if the ambient space is weakly Hausdorff, that is, the image of any continuous map from a compact Hausdorff space to the given space is closed? (Then it is in particular, a $T_1$-space.)


问:
Is there a standard compact notation for $k!{n\brace k}$?
Is there a standard compact notation for the numbers
$$
?(n,k)=\sum_{i=0}^k(-1)^i\binom ki(k-i)^n\equiv k!{n\brace k}
$$
where ${n\brace k}$ is the Stirling number of the second kind? In probabilistic applications the combination (which counts the number of ways to partition a set of $n$ objects into $k$ non-empty distinguishable subsets) seems to appear even more often than the Stirling numbers themselves. And if $k$ is a very long expression I would prefer to avoid to write it once more as $(\dots)!$

答:
The numbers you are using are the triangular arrray
OEIS sequence A019538.
As mentioned in an OEIS comment, they were known as the
"differences of zero" denoted by $\,\Delta^k0^n.$
They are the numbers in the leading diagonal of the
forward difference table of $\,x^n\,$ at $\,x=0.\,$

For example, when $\,n=4\!:$
$$ \Delta^k x^4|_{x=0} =:
\Delta^k 0^4 = 0,\;1,\;14,\;36,\;24 $$
from the forward difference table:
$$ \begin{matrix}
\Delta^4x^4 &|  &&&&& 24 \\
\Delta^3x^4 &|   &&&& 36 && 60 \\
\Delta^2x^4 &|   &&& 14 && 50 && 110 \\
\Delta^1x^4 &|   && 1 && 15 && 65 && 175  \\
x^4 &|   & 0 && 1 && 16 && 81 && 256 \\
x &|     & 0 && 1 && 2 && 3 && 4
\end{matrix} $$

This notation is used in older books such as George Boole's
A Treatise on The Calculus of Finite Differences. The Wikipedia article
Finite difference explains some of this.

In the matter of how standard, compact or convenient
this notation is in practice, there may be some differences
of  opinion.


问:
Change the order of a double integration
How should I change the order of the double integral of : $\int _0^1\:\int _{-x}^x 1\,dy\,dx$

答:
Draw a picture ! Then you should see:

$$\int _0^1\:\int _{-x}^x\:1\:dydx=\int_{-1}^1\:\int _{|y|}^1\:1\:dxdy.$$


答:
Look at the picture here:

Since in the first integral you fixed $x$ and let $y$ vary from $-x$ to $x$ then the region of integration must be the triangle between vertical line $x = 1$ and the lines $y = x$ (blue one) and $y = -x$ (the red one).

Here a picture that shows how you integrate in the first integral:

Then you can now fix $y$ and you must see that $x$ vary from the absolute value of y to 1.

Here the picture:

Title: Change the order of a double integration

Body: How should I change the order of the double integral of : $\int _0^1\:\int _{-x}^x 1\,dy\,dx$


问:
Norm characterization on Sobolev Spaces $W^{k,p}(\mathbb{R}^n)$.
I need to prove that $u \in W^{k,p}(\mathbb{R}^n)$ if and only if $u \in L^p(\mathbb{R}^n)$ with $D^l u \in L^p(\mathbb{R}^n)$ for every multi-index $l$ with $|l| = k$. This is related to the fact that
$$\lVert u \rVert_p +\sum_{|l| = k}\lVert D^lu \rVert_p$$
is an equivalent norm for the space. Any advice?

答:
Added later: Upon revisiting this question I realised the proof I originally wrote was incorrect; crucially one cannot just iteratively apply the case $\ell=1, k=2$ to deduce the general case. My answer has been amended to more closely follow the proof in the cited text.

This is true, and is a consequence of the Gagliardo-Nirenberg interpolation inequality which can be found in many sources. We will follow Chapter 5 of the following reference:

Adams, Robert A.; Fournier, John J. F., Sobolev spaces, Pure and Applied Mathematics 140. New York, NY: Academic Press (ISBN 0-12-044143-8/hbk). xiii, 305 p. (2003). ZBL1098.46001.

The key ingredient is the one-dimensional inequality
$$ |f'(0)|^p \leq \frac{2^{p-1}9^p}{\delta} \left(\delta^{-p}\int_0^1 |f(t)|^p \,\mathrm{d}t+ \delta^p\int_0^1 |f''(t)|^p \,\mathrm{d}t \right ), $$
for all $f \in C^2([0,\delta])$ with $\delta>0,$ and $1 \leq p < \infty.$

By rescaling we can assume $\delta=1.$ We use the fundamental theorem of calculus
$$ f'(0) = f'(z) - \int_0^z f''(t) \,\mathrm{d}t, $$
and for $0 \leq x<y \leq 1$ the mean-value theorem gives $z \in (x,y)$ such that
$$ f'(z) = \frac{u(y)-u(x)}{y-x}. $$
Chaining these we deduce that
$$ |f'(0)| \leq |x-y|(|f(x)|+|f(y)|) \int_0^1 |f''(t)| \,\mathrm{d}t.$$
From here we can integrate along $x \in (0,1/3),$ $y \in (1/3,1)$ and apply Hölder to conclude.

Now given $u \in L^p(\Bbb R^n)$ whose $k$th order derivatives are in $L^p,$ by mollifying we can assume $u$ is smooth. We apply the above estimate to $f(r) = u(x+r\omega)$ for $\omega \in S^{n-1},$ and integrate over all $\omega \in S^{n-1}$ and $x \in B_R$ to obtain
\begin{equation*}\begin{split}
 &\int_{B_R} |\nabla u|^p \,\mathrm{d}x \\
 &\leq \frac{C(n,p)}{\delta} \int_{B_R}  \int_{S^{n-1}} \left( \delta^{-p}\int_0^\delta |u(x+t\omega)|^p \,\mathrm{d} t  +\delta^p\int_0^{\delta} |\nabla^2u(x+t\omega)|^p \,\mathrm{d}t \right) \,\mathrm{d} \omega\,\mathrm{d} x \\
 &\leq C(n,p) \left(\delta^{-p} \int_{B_{R}} |u|^p \,\mathrm{d}x + \delta^p\int_{B_{R}} |\nabla^2u|^p \,\mathrm{d} x \right).
\end{split}\end{equation*}
I've been a bit brief here, but the last line uses an application of Fubini by (temporarily) assuming $u$ and its derivatives vanish outside of $B_R.$ We refer the reader to Lemma 5.4 of the cited text for the full details.

From here the idea is to apply similar estimates to $\nabla^j u$ and chaining the inequalities. This requires an induction argument, together with a carefully choosing the parameters $\delta$ so the relevant terms can be absorbed, as is done in Lemma 5.6. This will eventually give
$$ \int_{B_R} |\nabla^j u|^p \,\mathrm{d}x \leq C(n,p,k) \left( \int_{B_{R}} |u|^p \,\mathrm{d}x + \int_{B_{R}} |\nabla^ku|^p \,\mathrm{d} x \right) $$
for each $1 < j < k.$ Note one could have proved this in the case $R=1,$ and deduced the general case by scaling. To conclude, we send $R \to \infty.$


问:
On the requirement of the word "identical" in the phrase "Independent but not Identically distributed"?
Assuming that I have 3 independent exponential random variable $X_1,X_2,X_3$ with parameters ${\lambda _1},{\lambda _2},{\lambda _3}$, it is trivial that when ${\lambda _1} = {\lambda _2} = {\lambda _3}$ I can immediately call them Independent Identically Distributed.

Question 1, in the case ${\lambda _1} \ne {\lambda _2} = {\lambda _3}$, can I say that $X_1,X_2,X_3$ are independent but not identically distributed ?

Question 2, again if I have three independent random variable $X_1,X_2,X_3$ that is not exponential anymore but for example Binomial, Gaussian, Gamma, Can I said they are independent non-identically distributed ?

Question 3, does independent but not identically distributed require the independent random variable to have the same distribution but difference in parameters ?

Edit: added the word independent to avoid confusion

答: The phrase "independent but not identically distributed" simply means that the random variables are independent, but they do not follow the same distribution. Here "the same distribution" refers to the same type of distribution with the same parameters. For example, if $X_1,X_2,X_3$ are independent and $X_1,X_2\sim B(5,0.5),\:X_3\sim B(5,0.6)$ then they are independent but not identically distributed. If $Y_1,Y_2,Y_3$ are independent and $Y_1,Y_2\sim \text{Po}(5),\:Y_3\sim N(0,1)$ then they are also independent but not identically distributed.


问:
Pigeonhole principle to prove that given $4$ numbers the difference between two numbers is divisble by $3$

"Use the Pigeonhole Principle to show that among any four numbers one can find two numbers so that their difference is divisible by $3$."

I am struggling with this supposedly basic question on one of our past paper as its only worth $3$ marks.

答:
There are 3 possible remainders when we divide a num by 3.thus by pigeon hole principal we have 4 num.two of them must have same remainder when we divisible by 3.so we can write these n1 And n2 as

$N1=3q1+r,$

$N2=3q2+r,$

$N1-N2=3q1+r-3q2-r =3(q1-q2)$

Which is divisible by 3


问:
Does the weak closure of a set S always contain the convex hull of S (infinite dimension)?
Let $X$ be an infinite dimensional normed vector space. I learned that the weak closure of $S^1=\{x\in X:\|x\|=1\}$ is the ball $\{x\in X:\|x\|\le 1\}$. I was wondering if the weak closure always contains the convex hull of the set we consider if we are in the infinite dimension vector space.

答: Let ${e_n}$ be an orthonormal basis of a Hilbert space. Together with $0$ this gives a weakly closed set. Its weak closure is itself and does not contain convex combinations of $e_n$'s.


问:
Unless in a Logic Problem
This is a problem from Kenneth H. Rosen's Discrete mathematics and its applications:

Translate the given statement into propositional logic using the propositions provided.

You cannot edit a protected Wikipedia entry unless you are an administrator.

Express your answer in terms of $e$:“You can edit a protected Wikipedia entry” and $a$: “You are an administrator.”

Based on the book, "$q$ unless $\neg p$" is logically equal to $p \rightarrow q$, which was hard to understand, but by accepting this, the solution would be $\neg a \rightarrow \neg e$ or equally $e \rightarrow a$. This means that "if you can edit a Wikipedia entry, then you are (necessarily) an administrator".

Comparing the two statements in bold, this is indeed true, but from my point of view, the converse statement is also true, which is:

"if you are an administrator, then you can (definitely) edit a Wikipedia entry."

In fact, the first answer I came up with was $a \rightarrow e$, and this was my logic; there may be others rather than administrators who can edit Wikipedia entries, but if the person is an administrator, there is no way he/she cannot edit them!

Now I understand that this is only what I imagined and is not based on the statement, but my question is why the biconditional statement is not acceptable? Is this a common natural language mistake which leads me to believe $e \leftrightarrow a$ should be the answer? And if so, then what is the point of solving these problems and using them in Logical Puzzles, while the natural language can make everything confusing?

答:
Confusion often arises in tackling problems like this one. The issue is that in everyday speech language is often (usefully) used to say things that are not literally present in the words if they are interpreted very strictly.

So in problems like this it can help to remember it can be a slightly artificial game. It can help to remove the question from the real world by altering what it is talking about so that your commonsense knowledge about Wikipedia, editing, and administrators does not get in the way.

The form of the problem is:

You cannot do [onething] unless you are [otherthing]

Which means

The only people who can do [onething] are those who are [otherthing]

or, to make it closer to propositional logic:

If you can do [onething] then you are [otherthing]

The tools you are allowed are:

$e$: You can do [onething]
and $a$: you are [otherthing]

As stated in the question, the translation into propostional logic is
$e \to a$. The problem says nothing about whether $a \to e$ or not, but it looks like it might because of the familiar context.

Title: Unless in a Logic Problem


问:
Integral proof of logarithm of a product property
In one of my textbooks, the expansion of a logarithm product is proved using integrals.

$$\ln xy = \ln x + \ln y\iff \int_1^\left(xy\right)dt/t$$

$$\ = \int_1^xdt/t + \int_x^\left(xy\right)dt/t$$

Then, let $ u = t/x $ and substitute in the second integral:

$$= \int_1^x dt/t + \int_1^y du/u = ln x + ln y $$

While the expansion is rather quaint, I find the u-substitution difficult to follow:

If $u = t/x$ then $du/dt = 1/x$ and $du = dt/x $ ... I am unsure how $ du/u $ is obtained for substitution.

Any help or hints are appreciated!

答: Because $ u = \frac{t}{x} $, we have that $ t = ux \implies dt = x \cdot du $. Hence, $ \frac{1}{t} \ dt = \frac{1}{ux} \cdot x \ du = \frac{1}{u} \ du $. The bounds follow from the transformation and calculated independently of the integrand in this case. </p>


问:
Give an example of a set that is closed under a norm and not closed under another
Q: Give an example of a vector space $E$ and a subset $F$ of $E$ such that $F$ is closed under a norm topology and not closed under another norm topology.

I tried to find an example but i was not able to come up with one. I only now that $E$ can not be a vector space with finite dimension, because then any two norms are equivalent.

I imagine that a good example can be found in $C([a,b], \mathbb{R^n})$, but i wasn't able to find it. Can somebody help me?

答: On that space, the set of functions $f$ such that $f(a)=0$ is closed with respect to $\|f\|_{\text{sup}}=\sup|f|$, but it is not closed with respect to the norm $\|f\|_{\text{int}}=\int_a^b|f|$. Can you justify it?


问:
How to approach the following problem
Let $f:[0,1]\to\mathbb{R}$ be a continuous function then the maximum value of $\int_0^1 f(x)x^2 \,\text{d}x -\int_0^1x(f(x))^2\,\text{d}x$ for all such function(s) is?
I first thought differentiating it but the limits are not variable so no use then I spent a lot of time thinking over it but literally have no idea how to begin.
Please help.

答:
Hint:

$$f(x)x^2 -xf(x)^2 = xf(x) ( x-f(x)) $$

Think of $f(x-f)$ as a quadratic in $f$. Complete the square to get $$\frac{x^2}{4} - \left(f-\frac x2\right)^2 $$

So, your integral is $$\int_0^1 \frac{x^3}{4} - x\left(f(x) -\frac x2\right)^2 dx $$

What function $f$ maximizes this?

Title: How to approach the following problem


答:
Seems like functioncal calculus to me:

We want to optmize:

$$ g[f] =  \int_0^1  \left[ f(x)x^2 - x f^2 (x) \right]dx$$

Add a peturbation to $g$:
$$g[f+ \epsilon] = \int_0^1 \left[ (f + \epsilon)x^2 - x (f+\epsilon)^2 \right] dx$$

It is easy to find that terms of order $\epsilon$ are:

$$ \epsilon \int_0^1 \left[x^2 - 2 xf \right] dx$$

So, for the minimum condition, this term in bracket must be zero, which makes our function as:

$$  f(x) = \frac{x}{2}$$

Now you can integrate easily

All of this is based on this article that I read

Meta commentary:

This is completely analogous to how if we are a maxima point in single variable calculus , then first order variation is zero. I.e:

$$ f(x+h) = f(x) + h f'(x) + \text{junk}$$

Now, if you are at maxima then $f'(x)=0$ and hence if you increase $h$ by some teeny amount, the function doesn't change much. Similar notion here though more abstract.

Another notion is that, similar to how we neglect term of $dx^2$, we neglect terms of $\epsilon^2$ saying they are tinnier than tiny petrubations.

Title: How to approach the following problem

Body:
Let $f:[0,1]\to\mathbb{R}$ be a continuous function then the maximum value of $\int_0^1 f(x)x^2 \,\text{d}x -\int_0^1x(f(x))^2\,\text{d}x$ for all such function(s) is?
I first thought differentiating it but the limits are not variable so no use then I spent a lot of time thinking over it but literally have no idea how to begin.
Please help.


问:
Search for simpler (other) idea to show $(1.001)^{1000}>2$
The below question asked me to solve: (without using logarithm) which one is bigger $$2 \ , \ (1.001)^{1000}$$ I use two Idea to show $(1.001)^{1000}$ is bigger.

First: $$\lim_{n \to \infty}\left(1+\frac 1n\right)^n=e=2.71...\\$$and $1000$ is big enough to say $$(1.001)^{1000} \sim e>2$$

Second: Using binomial expansion $$(1.001)^{1000}=(1+0.001)^{1000}=\\1+1000\times 1^{999} \times (0.001) +\frac {1000\times 999}{2}1^{998}(.001)^2+...=\\1+1+\frac{999}{2000}+...>2$$   Now ,my question is there any simpler or other Idea to show $(1.001)^{1000}>2$

Thanks in advance for any hint.

答:
According to Bernoullli's inequality, we have $(1+x)^r\ge1+xr$. for this problem consider $x=0.001$ and $r=1000$.

Note: Here $(1+x)^r=1+xr$ is not the case because $r\neq0$


问:
Permuting a linearly independent set of vectors to obtain a spanning set.
Consider a set of linearly independent vectors $\{v_1,v_2,\ldots ,v_n \} \subseteq \mathbb{C}^{2n}$ such that for each $v_i$, the $2n^{\text{th}}$ entry is zero and every other entry is non-zero. If we define the permutation $P:\mathbb{C}^{2n} \rightarrow \mathbb{C}^{2n}$ by its action on the standard basis as follows: $P(e_i)=e_{i+1}$ (where the index addition is modulo $2n$), then is $$\text{span}\{v_1,\ldots, v_n,Pv_1,\ldots ,Pv_n \} = \mathbb{C}^{2n} ?$$

Body:
What about
$$v_1 = (1, 2, 3, 0)$$
$$v_2 = (1, 3\sqrt{2}i-1, -6, 0)$$
Then
$$-\frac{1+\sqrt{2}i}{3}v_1 + \frac{1+\sqrt{2}i}{3}v_2+2P(v_1) + P(v_2) = 0$$
As they are linearly dependent, they can't span $\mathbb{C}^4$.


Body:
Set $U=2-P^{-1}$, $U(0)=2-P(0)^{-1}=1$, then you get
$$
U'=P^{-2}P'=\frac12(2-P^{-1})\cos t=\frac{U}2\cos t
$$
and consequently
$$
U(t)=U(0)e^{\sin(t)/2}
$$
Now solve backward
$$
2-P(t)^{-1}=e^{\sin(t)/2}.
$$


问:
Prove that $\mathbb{Q}(\sqrt{2},\sqrt[3]{3})=\mathbb{Q}(\sqrt{2}\sqrt[3]{3})=\mathbb{Q}(\sqrt{2}+\sqrt[3]{3})$
Showing that $\mathbb{Q}(\sqrt{2},\sqrt[3]{3})$ has degree 6 over $\mathbb{Q}$ is straighforward:
It contains $\mathbb{Q}(\sqrt{2})$ and $\mathbb{Q}(\sqrt[3]{3})$ which are degree 2 and 3 over $\mathbb{Q}$ and since 2 and 3 are coprime it follows that $[\mathbb{Q}(\sqrt{2},\sqrt[3]{3}):\mathbb{Q}]=6$.

Then since $\mathbb{Q}(\sqrt{2}\sqrt[3]{3})\subset \mathbb{Q}(\sqrt{2},\sqrt[3]{3})$ and $x^6-72$ is the minimal polynomial of $\sqrt{2}\sqrt[3]{3}$ it follows that  $[\mathbb{Q}(\sqrt{2},\sqrt[3]{3}):\mathbb{Q}(\sqrt{2}\sqrt[3]{3})]=1$ and so $\mathbb{Q}(\sqrt{2},\sqrt[3]{3})=\mathbb{Q}(\sqrt{2}\sqrt[3]{3})$.

But I can't figure out how to prove the last equality. Any help is appreciated!

答:
Show that $x^3-3$ is irreducible over $\Bbb{Q}$ and $x^2-2$ is irreducible over $\Bbb{Q}(\sqrt[3]3)$,

which implies that $[\Bbb{Q}(\sqrt2,\sqrt[3]3):\Bbb{Q}]=6$.

Next show that $\sqrt2\sqrt[3]3$ has $6$ distinct $\Bbb{Q}$-conjugates, which implies that $[\Bbb{Q}(\sqrt2\sqrt[3]3):\Bbb{Q}]=6$.

Same for $\sqrt2+\sqrt[3]3$.


Title: Search for simpler (other) idea to show $(1.001)^{1000}>2$

Body:
The below question asked me to solve: (without using logarithm) which one is bigger $$2 \ , \ (1.001)^{1000}$$ I use two Idea to show $(1.001)^{1000}$ is bigger.

First: $$\lim_{n \to \infty}\left(1+\frac 1n\right)^n=e=2.71...\\$$and $1000$ is big enough to say $$(1.001)^{1000} \sim e>2$$

Second: Using binomial expansion $$(1.001)^{1000}=(1+0.001)^{1000}=\\1+1000\times 1^{999} \times (0.001) +\frac {1000\times 999}{2}1^{998}(.001)^2+...=\\1+1+\frac{999}{2000}+...>2$$   Now ,my question is there any simpler or other Idea to show $(1.001)^{1000}>2$

Thanks in advance for any hint.

Body:
By elementary arithmetic:

$$1.001^2=1.002001\ge1.002$$

$$1.001^4=1.004006004001\ge1.004$$
$$\cdots$$

$$1.001^{128}>1.128.$$
$$1.001^{256}>1.256.$$
$$1.001^{512}>1.512.$$

Finally,

$$1.001^{1000}>1.001^{512+256+128}>1.512\cdot1.256\cdot1.128>2.$$


问:
Proof of relation between divided difference and interpolation polynomial coefficient
Given data points $\{(x_i,f(x_i)\}_{i=0}^{m}$, if we define divided difference recursively as:

$$f[x_0,\cdots,x_{k+1}] = \frac{f[x_1,\cdots,x_{k+1}]-f[x_0,\cdots,x_k]}{x_{k+1}-x_0} \text{ with the definition that} f[x_0] = f(x_0)$$

then it is true that the unique interpolating polynomial $L_m(x)$ of degree $m$ such that $L_m(x_i) = f(x_i)$ for all $i = 0\cdots m$ is given by:
$$
L_m(x) = \sum_{k=0}^mf[x_0,\cdots,x_k]\omega_k(x), \text{ where } \omega_k(x) = \prod_{i=0}^{k-1}(x-x_i).
$$
The question asks to show this. That is, show the coefficients of the interpolating polynomial of degree $m$ in the basis $\{\omega_k\}_{k=0}^{m}$ is exactly $a_k = f[x_0,\cdots,x_k]$, the divided differences.

To tackle this problem, I've attempted to use induction. The base step is trivial as $a_0 = f(x_0)$, but I am stuck on the inductive step. I supposed the degree $m$ interpolating polynomial of some data $\{(x_i,f(x_i)\}_{i=0}^{m}$ is indeed given by:
$$
L_m(x) = \sum_{k=0}^mf[x_0,\cdots,x_k]\omega_k(x)
$$
and I want to argue that $L_{m+1}(x) = \sum_{k=0}^{m+1}f[x_0,\cdots,x_k]\omega_k(x)$ is the interpolating polynomial of degree $m+1$ interpolating $\{(x_i,f(x_i)\}_{i=0}^{m+1}$, with any additional point $(x_{m+1},f(x_{m+1}))$ added.

Clearly, $L_{m+1}(x)$ interpolates $\{(x_i,f(x_i)\}_{i=0}^{m}$ by definition of the $\omega_k$'s, but I am stuck at showing $L_{m+1}(x_{m+1}) = f(x_{m+1})$. Hints will be extremely appreciated!

答:
Consider adding the table entries one by one...

$p_0(x) = f_0$ interpolates the data on $x_0$.

The interpolant on $x_0, x_1$ is given by
$$
p_1(x) = p_0(x) + c_1 (x-x_0).
$$
We can compute $c_1$ simply by prescribing   $p_1(x_1)= f_1$, which leads to
$$
p_0(x_1) + c_1(x_1-x_0)= f_1 \Leftrightarrow c_1 =\frac{f_1-f_0}{x_1-x_0}
$$
hence,
$$
p_1(x) = f_0 + \frac{f_1-f_0}{x_1-x_0} (x-x_0)
$$

The interpolant on $x_0, x_1, x_2$ is given by
$$
p_2(x)=p_1(x) + c_2 (x-x_0)(x-x_1)
$$
again, $c_2$ is computed by demanding $p_2(x_2)=f_2$:
$$
f_0+\frac{f_1-f_0}{x_1-x_0}(x_2-x_0) + c_2(x_2-x_0)(x_2-x_1) \Leftrightarrow c_2 = \cdots
$$

This way, you recover the recursive definition of the divided differences, with the correspondence $c_k = f[x_0, \cdots, x_k]$. I think you can now reconstruct the inductive step.


问:
Is there a use for a wave made of alternating parabolas?
Messing around in Desmos, I constructed (with a lot of copy+pasting) a wave-like object made of the heads of a parabola I'd shifted, flipped, and connected together. The result closely resembles a cosine wave with equation $$y = \cos(\frac{\pi x}{2})$$

Original Desmos graph: https://www.desmos.com/calculator/cmhoghswxo

Picture of the graph: https://i.stack.imgur.com/PEZTa.png

Is there any application for this type of wave? The parabola is a shockingly useful shape so I'd think there might be some hyperspecific use for a wave like this, but I have no idea what that might be when the cosine and sine functions are that much better for modeling wave mechanics.

Entirely possible this thing's pretty useless, but thanks for any information that can be provided.

答:
I arrived at this exact function myself just now while trying to imagine what a "one-dimensional orbit" might look like. I googled "alternating parabola wave" to see if anyone has talked about this, and found this question lol.

In a 3D world, gravity is governed by the inverse-square law. Here's Newton's equation for gravity:

$$|F|=\frac{Gm_1m_2}{r^2}$$

Notice this contains an inverse $r^2$. This is an effect of the inverse-square law, which states that in three dimensions, effects emanating from a point decrease in magnitude proportional to the inverse-square of the distance from the point. The reason for this is that the effect spreads out proportionally to the surface area of a sphere.

This means that in two dimensions, instead of gravity spreading out proportional to the surface area of a sphere, it would spread out proportional to the circumference of a circle, i.e. proportional to the inverse of $|r|$. One can imagine Newton's law of two-dimensional gravitation being the following:

$$|F|=\frac{km_1m_2}{|r|}$$

Likewise, in a one-dimensional world, the effect would not decrease at all and would remain constant. Thus:

$$|F|=km_1m_2$$

This would mean the equation governing the acceleration of an object orbiting a point would be:

$$a=-k*\text{sgn}(x)$$

And, of course, $a=x''$. The solution to this equation is precisely this alternating parabola wave (let's call it $P$), ignoring any constants:

$$x=P(t)$$

Using some trig hacking, I was able to create a single formula for this wave equation:

$$x=\left(1-\frac{2}{5}\left(\arcsin\left(\cos\left(\frac{\pi}{2}t\right)\right)\right)^2\right)*\text{sgn}\left(\sin\left(\frac{\pi}{2}t\right)\right)$$

Here it is in Desmos: https://www.desmos.com/calculator/egn9orerco

I got lost in the math and am still confused about where exactly the $\frac{2}{5}$ came from. But oh well.

PS: It might alternatively be reasonable to assume that the force exerted by one-dimensional gravity would actually not change direction, and would therefore be completely independent of the relative position of the object. This would make the "orbit" a regular parabola. But that's much less fun.


问:
Word for curvature change along a path?
In natural coordinates, the curvature $k$ is defined as the along path rate of change of the unit tangent vector, or equivalently the along path change of angle between the x-axis and the velocity vector. This quantity is also the inverse of the local radius of curvature $R$.
$$ k = \frac{\partial \phi}{\partial s} = \frac{1}{R}$$

(For details see the natural coordinate diagram from Wenegrat, J. O., & Thomas, L. N. (2017). Ekman transport in balanced currents with curvature. Journal of Physical Oceanography, 47, 1189– 1203.)

I would like know if a there is a word/definition for curvature changing along a path. It feels like there should be a word for this... would it just be inflection? I am basing that simply off of the definition of inflection point: the location on a function where its curvature changes sign, seemingly implying that "inflection" describes curvature change in general.

$$\frac{\partial k}{\partial s} = ?$$

答:
I´m not sure if there exists a generally accepted word for the change of curvature, but I usually use some probabilistic terms that are closely related.

When given a function $f(x)$, its first derivative $f'(x)$ is called 'slope' (or 'velocity' in some contexts), its second derivative $f''(x)$ is called 'curvature' (or 'acceleration'), but there are generally accepted names for higher derivatives.

In probabilistic terms, when given a random variable $X$, its first (raw) moment $\mu=\mathbb{E}[X]$ is the so-called 'mean', and the second (central) moment $\sigma^2= \mathbb{E}[X-\mu]$ is the variance. In this case, higher-order (standardized) moments also have a name:

Third moment: Skewness

Fourth moment: Kurtosis

At this point, since the moments of the random variable $X$ and the derivatives of a function are closely related by the notion of the characteristic function of $X$ (more precisely, its moment-generating function), I usually refer to the third derivative of a function $f'''(x)$ as the 'skewness of $f$' and to its fourth derivative $f''''(x)$ as the 'kurtosis of $f$'.

I know that this reasoning is not $100\%$ accurate since the idea of 'moments' is applied to random variables and not to functions, and that is not generally accepted, but it looks pretty and sounds nice to me.

Title: Word for curvature change along a path?


答:
$$f(x):=\left(1+\frac1x\right)^x$$ is a strictly growing function of $x$, and $f(1)=2$.

Indeed for $x>0$,

$$(\log(f(x))'=\log\left(1+\frac1x\right)-\frac1{x+1}>0$$

because

$$(\log(f(x))''=-\frac1{x(x+1)^2}<0$$ and the first derivative decreases to $0$.


问:
Distribution of normalized sum of exponential order statistics
Let $X_1, \dots, X_n$ be i.i.d. $\exp(\lambda)$ and denote the order statistics with $Y_1 \leq \dots \leq Y_n$.

I want to show that the distribution of $ \frac{ \sum_{i = 1}^j Y_i } {\sum_{i=1}^n Y_i}$ is independent of $\lambda$.

I know that $Y_i = \sum_{j=1}^i \frac{Z_j}{n-j+1}$, where the $Z_j's$ are i.i.d. $\exp(\lambda)$.

答:
I am supposing the probability density function of $X_1$ is
$$
f_{X_1}(x) = \lambda \exp(-\lambda x)1(x>0).
$$
Therefore the distribution of $\lambda X_1$ is an exponential distribution with parameter $1$. Now you can write your quantity of interest as

$$
\frac{ \sum_{i = 1}^j\lambda Y_i } {\sum_{i=1}^n \lambda Y_i} = \frac{ \sum_{i = 1}^j Y_i' } {\sum_{i=1}^n Y_i'},
$$,
where $Y_i'$s are the order statistics of $\lambda X_1,\ldots,\lambda X_n$. We already observed that $\lambda X_i$'s are exponential distribution with parameter $1$. (of course it is independent of $\lambda$)


问:
Differential equation with signum function
I'm trying to solve differential equation to get EOM for a dynamical system. Firstly, I reduced order for my equation by substitution
$$
\dot{v}_x(t) = \frac{\mathrm{d}x}{\mathrm{d}t}.
$$

I solved it and got this equation:
$$
\frac{\mathrm{sgn}(v_x)}{v_x} = At + B,
$$
where $A,B$ are  const.

I know, that I can now look at two cases with $v_x>0$ or $v_x<0$, but I wonder if I can calculate it with signum function. I habe no idea what to do, when I come back and put
$$
v_x = \frac{\mathrm{d}x}{\mathrm{d}t}
$$

How to calculate integral of derivative inside signum function?

答:
$$\frac{\text{sgn} (v_x)}{v_x} $$ is always positive, so $At+B\gt 0$. There are two solutions to the equation you got, owing to the fact that $v_x \mapsto -v_x$ doesn’t change the equation:

$$v_x = \pm \frac{1}{At+B} =\frac{dx}{dt} \\ x(t) = \pm \frac 1A \ln(At+B)+C $$


问:
Hint about proving $\lim_{n \to \infty}( \sqrt{\log{(x_n + a)}} - \sqrt{\log{x_n}} ) = 0$
I need to prove that for any $a \in \mathbb{R}$ and while $\lim x_n = \infty$, $\lim_{n \to \infty}( \sqrt{\log{(x_n + a)}} - \log{\sqrt{x_n}} ) = 0$. While a full proof would be nice (it may avoid future questions), I am only requesting a simple hint about how to simplify $\sqrt{\log{(x_n + a)}} - \log{\sqrt{x_n}}$. This is an exercise from "Analise Real. Funções de Uma Variável - Volume 1" by Elon Lima. Thanks in advance.

Edit: Yes, this is typo in the translation I was reading. My apologies to everyone. $\sqrt{\log{(x_n + a)}} - \sqrt{\log{x_n}}$ is the correct expression. I modified the title accordingly.

答:
Composing Taylor series gives the result
$$\log(x+a)=\log (x)+\frac{a}{x}-\frac{a^2}{2 x^2}+\frac{a^3}{3
   x^3}+O\left(\frac{1}{x^4}\right)$$
$$\sqrt{\log(x+a)}=\sqrt{\log (x)}+\frac{a}{2 x \sqrt{\log (x)}}-\frac{a^2 (2 \log (x)+1)}{8 x^2
   \log (x)\sqrt{\log (x)}}+\cdots$$
$$\sqrt{\log(x+a)}-\sqrt{\log(x)}\sim \frac{a}{2 x \sqrt{\log (x)}}$$

Try for $x=10^6$ and $a=12345$. The "exact" value is $0.001650$ and the approximation gives $0.001661$.


问:
Flux of $v=(y_1e^{y_2}-e^{y_1},y_2e^{y_3}-e^{y_2},y_3e^{y_1}-e^{y_3})$ through a sphere

$v=(y_1e^{y_2}-e^{y_1},y_2e^{y_3}-e^{y_2},y_3e^{y_1}-e^{y_3})$

$Y = (y_1^2+y_2^2+y_3^2=r^2, y_2>=0, y_3>=0)$

Calculate flux of v through Y.

My first intuition was to use spherical coordinates so I set $\varphi(x)=(r\sin x_2\cos x_1, r\sin x_2 \sin x_1, r\cos x_2)$, with $x_1$ between $0$ and $\pi$ and $x_2$ between $0$ and $\pi/2$.

Then I wanted to use the Stokes-Ampère theorem, but I don't really know how to use it. I saw that div v = $0$ and found $\omega_v^2$. Then I found $\omega_u^1$ with $u=(0,y_3e^{y_1}-y_1e^{y_3},y_1y_2e^{y_3}-y_1e^{y_2})$.

But what now? How do I set the integral? In class we saw an example with polar coordinates and a simple u and it seemed easy with just one variable. But here I have spherical coordinates and the vector u is very confusing.

Could someone please explain how to use this theorem correctly and how do I integrate from here?

答:
$\vec F =(xe^y - e^x, ye^z - e^y, z e^x - e^z)$

Given surface is $S: x^2+y^2+z^2 = r^2; \ y, z \geq 0$

We can apply divergence theorem by closing the surface with semi-disks in plane $z = 0$ and $y = 0$.

$\nabla \cdot \vec F = e^y - e^x + e^z - e^y + e^x - e^z = 0$

As the divergence of the vector field is zero, its surface integral over closed surface is zero. Now we find the flux through semi-disks at $y = 0$ and at $z=0$ and subtract from the total flux which is zero. That would give us the flux through surface $S$.

For disk at $y = 0$, outward normal vector is $(0, -1, 0)$ and we can parametrize the surface as $(\rho \cos \theta, 0, \rho\sin \theta), 0 \leq \rho \leq r, 0 \leq \theta \leq \pi$.

For disk at $z = 0$, outward normal vector is $(0, 0, -1)$ and we can parametrize the surface as $(\rho \cos \theta, \rho\sin \theta, 0), 0 \leq \rho \leq r, 0 \leq \theta \leq \pi$.

It is easy to see that the total outward flux through both semi-disks is equal to the area of a circle of radius $r$.

Can you take it from here, fill in all the details and complete the working?