问:
Definition of quotient algebra in Bell & Slomson's Models and Ultraproducts
For a filter $F$ in a boolean algebra $B$, they define

$$ x \sim_F y \text{ iff for some } f\in F, x \wedge f = y \wedge f$$

They go on to prove that this is a equivalence relation and then define the quotient algebra $B/F$ where $\wedge, \vee, *$ applied to the equivalence classes are just the same operators applied to any two representative elements.

My question is, why is this the definition for $B/F$? Clearly any equivalence relation will allow a forming a quotient algebra, and there many equivalences relations on $B$; and though I can't think of any non-trivial one, there are probably some other equivalence relations on $B$ that "depend" somehow on $F$.

Right after the definition, they present some kind of analogue to the first homomorphism theorem for groups, so if $f: B_1 \to B_2$, then $f[B_1] \cong B_1/F$, where $F= \{ x \in B_1 \mid f(x)=1 \}$. I think this is a property that should be satisfied for any definition of quotient algebra, but again, maybe there are some other definitions of quotient algebra that satisfy it too.

Sure this definition has other nice properties like $B/F \cong 2$ whenever $F$ is an ultrafilter, etc. But all this seems like a coincidence for me right now. So my question is to justify the naturality of this construction.

答:
Although you mentioned the theorem, you seem to have missed the significance of the fact that if $f:B_1\to B_2$ is a homomorphism, then $\{x\in B_1:f(x)=1_{B_2}\}$ is a filter in $B_1$, and $f[B_1]\cong B_1/F$. The theorem basically says that if we want quotient algebras of $B_1$ to be homomorphic images of $B_1$, this is the only way to define them: the homomorphism determines the filter, and vice versa.

The equivalence relation is in any case a very natural one once you realize that one can think of a filter $F$ on a Boolean algebra $B$ as singling out a class of ‘large’ elements of $B$. This is perhaps clearest in the case of power set algebras. A simple example is the cofinite filter on $\wp(\Bbb N)$ consisting of all subsets of $\Bbb N$ with finite complements: being cofinite is clearly a notion of largeness. With respect to this notion the finite sets are small, and the infinite sets that are not cofinite are neither large nor small.

Defining $x\sim_Fy$ iff $x\land f=y\land f$ for some $f\in F$ is in a sense saying that $x$ and $y$ agree on a large element of $B$. Here again the power set algebras provide the intuition. For instance, if $\mathscr{F}$ is the cofinite filter on $\wp(\Bbb N)$, and $X,Y\subseteq\Bbb N$, then $X\sim_{\mathscr{F}}Y$ iff there is an $F\in\mathscr{F}$ such that $X\cap F=Y\cap F$, i.e., iff there is a finite $A\subseteq\Bbb N$ such that $X\setminus A=Y\setminus A$. This is sometimes expressed by saying that $X$ and $Y$ agree almost everywhere modulo finite sets: apart from a finite subset of $\Bbb N$, they have the same elements. Indeed, $X\sim_{\mathscr{F}}Y$ iff $X\mathrel{\triangle}Y$ is finite.


问:
On some doubts on tangent space of immersed submanifold
In Lee's book "Introduction to Smooth manifolds", he following lemma can be found.
Lemma 8.26 Let $M$ be a smooth manifold, let $S\subseteq M$ be an embedded submanifold, and let $Y$ be a smooth vector field on $M$. Then $Y$ is tangent to $S$ iff $Yf$ vanishes on $S$ for all $f\in C^\infty(M)$ such that $f|_S=0.$

My question is what if $S$ just an immersed submanifold? Is the same conclusion true? It should be because earlier Lee proves that any immersed submanifold is locally embedded submanifold. But why does Lee state the theorem only for embedded submanifod?

答:
Another example of an immersed but not embedded submanifold. Take $F: \mathbb{R}\to \mathbb{R^2}$ defined by $$F(t)=\left(2cos\left(t-\frac{\pi}{2}\right), 2sin\left(t-\frac{\pi}{2}\right)\right)$$ Then $(F, \mathbb{R})$ is an immersed submanifold of $\mathbb{R^2}$ but not an embedded submanifold of $\mathbb{R^2}$.

It looks like this and goes through the origin twice as the leminiscate loops around itself:

ContentLicense: CC BY-SA 4.0


问:
Existence of singular solution for linear and Riccati's differential equation
My textbook states that, under the assumption of its coefficients being continuous, first order linear differential equation (the same is said for Riccati's equation) does not have a singular solution.
Can someone help me understand this?

Thanks in advance!

答:
Because in these cases you easily can prove the local Lipschitz condition, providing uniqueness and thus prohibiting singular solutions or any tangency of different solutions.

$y'(x)=a(x)y(x)+b(x)$ has $y$-derivative $a(x)$ and thus the maximum over $a(x)$ over some fixed interval in its domain as global $y$-Lipschitz constant.

$y'(x)=a(x)y(x)^2+b(x)y(x)+c(x)$ has $y$-derivative $2a(x)y+b(x)$  which in any bounded set in the $x$-$y$ plane is bounded and thus gives a local $y$-Lipschitz constant.


问:
Serge Lang's projection
This question is a follow-up to Identity up to isomorphism treated as identity in proof. I thought that with all the kind help given there, now I would be able to work out the sketch of a proof given by Lang for the corollary dual to the one in the above thread, and to eliminate his assumptions of identity based on an identity up to isomorphism there, too. But I can't. Here is the problem:

In "Fundamentals of Differential Geometry", 1999, pp.18-19, Serge Lang gives the following definition:

And then this corollary to the inverse mapping theorem:

First of all, some clarifications: Morphism means $ C^p$ map, local isomorphism means local $ C^p$ diffeomorphism, toplinear isomorphism can be considered to be a linear isomorphism here. Furthermore, I understand to be $ V_1 \subseteq E_1 $ and $ V_2 \subseteq E_2 $, and the local inverse h, which Lang refers to, to be $ \varphi^{-1} $, and not the inverse of the derivative, as Lang's wording implies.

Again, what I don't see is how $ \varphi^{-1} $ satisfies the requirement of the corollary.

In order to eliminate the identification $ E_2=F $ in the proof, let instead be

$ \varphi: E_1 \times E_2 \rightarrow E_1 \times F $.

Then introduce the $ C^p $ diffeomorphism

$ g: E_1 \times E_2 \rightarrow E_1 \times F: \quad (x_1,x_2) \mapsto (id_1, D_2f(a_1,a_2))[x_1,x_2] $

and replace $ h:=\varphi^{-1} $ by the $ C^p $ diffeomorphism $ h \circ g: E_1 \times E_2 \rightarrow E_1 \times E_2 $. But with this, how does the resulting map $ f \circ h \circ g: E_1 \times E_2 \rightarrow F $ factor into an ordinary projection $ V_1 \times V_2 \rightarrow V_2 $ and a linear isomorphism $ V_2 \rightarrow W(0) \subseteq F $ with an open neighborhood W?

Can we state the local map $ \varphi^{-1} $ explicitly? Is it $ \varphi^{-1}(x_1,y) = (x_1, pr_2 \circ f^{-1}(y)) $ for $ y \in F $?

Clearly $ \varphi^{-1}(\varphi(x_1,x_2))= \varphi^{-1}(x_1,f(x_1,x_2)) = (x_1,x_2) $. But the other way around does not resolve properly:

$ \varphi(\varphi^{-1}(x_1,y))= \varphi(x_1, pr_2 \circ f^{-1}(y)) =(x_1,f(x_1,pr_2 \circ f^{-1}(y)) $.

And by the way, can we take f to be locally invertible, too? Evaluating the composition $ f \circ h \circ g $ seems to lead nowhere

$ f(h(g(x_1,x_2))) = f(h(x_1,D_2f(a_1,a_2)[x_2])) = f(x_1,pr_2 \circ f^{-1}(D_2f(a_1,a_2)[x_2])) $.

So, how to proceed? Where is the error, or what is the necessary idea? I thought about explicitly introducing the projection $ pr_2: E_1 \times E_2 \rightarrow E_2 \equiv (\{0\} \times E_2) \subseteq (E_1 \times E_2) $ at the beginning of the composition: $ f \circ h \circ g \circ pr_2 $, but unfortunately the projection is no $ C^p $-diffeomorphism.

答:
In this case it is much easier to get lost.

If we go through the proof lets redefine
$$\varphi:U\to E_1\times F, \quad (x,y)\mapsto (x,f(x,y))$$
this is also a bit different from what Lang is doing in that $\varphi$ is not defined on the entire space $E_1\times E_2$, since $f$ itself is only defined on the neighbourhood $U$. This remark is far from serious, however.

The derivative of this is:
$$D\varphi(x,y)\ [w_1,w_2]= \bigg[w_1, D_1f(x,y)\ [w_1] + D_2f(x,y)\ [w_2]\bigg]$$

This is invertible at $(a_1,a_2)$. You may use matrix notation, as Lang does, to simplify this - note that for $A, C$ invertible you have that
$$\begin{pmatrix}A&0\\ B& C\end{pmatrix}^{-1}= \begin{pmatrix}A^{-1}&0\\ -C^{-1}BA^{-1}& C^{-1}\end{pmatrix}$$

From the inverse function theorem it follows that there is some local inverse
$$h: V_1\times V_2\to E_1\times E_2$$
with $V_1\subseteq E_1, V_2\subseteq F$ open so that $\varphi(a_1,a_2)\in V_1\times V_2$ (and $h(V_1\times V_2)\subseteq U$).

Since its a local inverse you have $\varphi \circ h=\mathrm{id}_{V_1\times V_2}$. Write this composition out:
$$(\varphi\circ h)(x,y)=(h_1(x,y), f(h(x,y)) ) \overset!= (x,y)$$
hence $f(h(x,y)) = y$, which was the desired result.

What I did here was go through the proof and adapt it to be a proof of the statement without assuming that $E_2=F$. From reading your thoughts I think you wanted to do the same thing, but as an adaption you want to plug in the isomorphism $D_2f(a_1,a_2)$ at every stage where the identification takes place. This is also possible, and maybe its more systematic, but its easier to get lost.

A third way to do it would be to use the actual statement derived by Lang, meaning the case $E_2=F$, and work with this statement alone to derive the case $E_2\neq F$. Here we need to first use the identifications to get the situation $E_2=F$, then apply the theorem and after that use identifications to get back to the situation $E_2\neq F$.

In this vain let $T:F\to E_2$ be any isomorphism, for example $T=D_2f(a_1,a_2)^{-1}$. Then if
$$f:U\to E_1\times F$$
is a map with $D_2f(a_1,a_2)$ being invertible consider $\tilde f:=f\circ (\mathrm{id}_{E_1}, T): E_1\times F\to E_1\times F$. Here we have modified $f$ to be a map of the required form, note that
$$D_2\tilde f = D_2f(a_1,a_2)\circ T$$
which is invertible - you are thus in the situation of the lemma where $E_2=F$.

Apply the theorem: There exists a $\tilde h:V_1\times V_2\to E_1\times F$ so that $\tilde f \circ \tilde h$ is a projection to the second component. But:
$$\tilde f\circ \tilde h = f\circ ( (\mathrm{id}_{E_1},T)\circ \tilde h)$$
Defining $h:= (\mathrm{id}_{E_1},T)\circ \tilde h$ then allows you to recover the lemma where you just have $E_2\cong F$, rather than the full $E_2=F$.

Title: Serge Lang's projection


问:
Is any immersed manifold also embedded?
Let $M$ be a smooth manifold. To establish my criteria let me define immersed and embedded submanifolds:

A subset $N\subset M$ is an immersed submanifold when $N$ is itself a manifold and $\iota:N\rightarrow M$ is an (injective) immersion.

In the same way, a subset $N\subset M$ is an embedded submanifold if the inclusion is an embedding.

By theorem 5.8 of John M. Lee's book Introduction to smooth manifolds, any subset $N$ of a smooth manifold $M$ such that each point $p\in N$ is contained in the domain of a chart $(U,\varphi)$ of $M$ verifying

$$
\varphi(U\cap N)= V \times \{ c \},
$$

for  $V\subset\mathbb R^k$ open and $c\in \mathbb R^{n-k}$ constant, is a topological submanifold of $M$ and it admits a smooth structure making it into a $k$-dimensional embedded submanifold. Conversely, any embedded submanifold exhibits such a property.

On the other hand, Proposition 5.22 of the same book states that if $N\subset M$ is an immersed submanifold then, for each point $p\in N$ there is an open neighbourhood (with respect to $N$) $U'$ of $p$ such that $U$ is an embedded submanifold. However, if $U'$ is an embedded submanifold, by the previous result, there exists a chart $(U,\varphi)$ of $M$ such that $p\in U$ and

$$
\varphi(U'\cap U)=V \times \{ c\} ,
$$

for $V$ and $c$ as before. On the other hand, $U'=N\cap U''$, for some open neighbourhood (with respect to $M$) $U''$ of $p$. Then, if we set $\tilde U=U\cap U''$ and $\tilde\varphi = \varphi|_{\tilde U}$, we have found a chart of $M$ with $p\in \tilde U$ and such that

$$
\tilde\varphi(N\cap \tilde U)= V\times \{c \} .
$$

Then, again using theorem 5.8, we deduced that $N$ is an embedded submanifold. Maybe, warned by Lee's remark after proposition 5.22, the embedded structure of $N$ may or may not agree with the immersed structure. What I am saying is that for a subset $N$ being an immersed submanifold is such a nice property that the same set can be endowed with an embedded structure too.

Question. Am I right? Does any immersed submanifold admit a (possibly different) structure of embedded submanifold? If not, what is wrong with my previous reasoning?

Remark. Notice that being an immersed submanifold is not a trivial condition, i.e. not every subset can be endowed with such a strcture. For instance, the boundary of a square in $\mathbb R^2$ (Problem 5-9 of Lee's book). Then, not every set can be realised as an embedded submanifold.

答:
Here's a counterexample I just posted somewhere else. Take $F: \mathbb{R}\to \mathbb{R^2}$ defined by $$F(t)=\left(2\cos\left(t-\frac{\pi}{2}\right), 2\sin\left(t-\frac{\pi}{2}\right)\right)$$ Then $F(\mathbb{R})$ is an immersed submanifold of $\mathbb{R^2}$ but not an embedded submanifold of $\mathbb{R^2}$.

It looks like this and goes through the origin twice as the lemniscate loops around itself:

Title: Is any immersed manifold also embedded?


问:
How to find the number of positive devisors of $50,000$
How to find the number of positive devisors of $50,000$, I would like to know that what mathematical formulae  I need to use here as it is a big number to calculate mentally, I am sorry to ask if this is too silly question here to ask. Thank you. </p>

答: hint. break it into its prime factorisation. then consider the powers of the primes...

ContentLicense: CC BY-SA 3.0


问:
Seeking an intuitive understanding of formal definition of tangent
I am learning numerical optimization and I recently learned the formal definition of tangents. The definition is stated as follows:

Suppose $\Omega \subseteq \mathbb{R}^{n}$ is closed. If for all vector sequences $\left\{x_{i}\right\}$ satisfying $x_{i} \in \Omega$ and $\lim_{i\to\infty}x_{i} = x \in \Omega$, and for all positive scalar sequences $\left\{t_{i}\right\}$ satisfying $\lim_{i\to\infty}t_{i} = 0$, there exists a sequence $\left\{w_{i}\right\}$ satisfying $\lim_{i\to\infty}w_{i} = w$ and sequences $\left\{x_i + t_i w_i \right\}$ satisfy $x_i + t_i w_i \in \Omega$ for all $i$, then we say $w$ is tangent to $\Omega$ at $x \in \Omega$.

I can't associate this formal definition with an intuitive understanding, as it is too abstract. Could anyone provide an intuitive explanation for this definition?

答:
For each given $p \in \Omega$ and $t > 0$, consider the set $\tilde{T}_{p,t}\Omega$ defined by

$$ \tilde{T}_{p,t}\Omega = \left\{ \frac{p' - p}{t} : p' \in \Omega \right\}. $$

Intuitively, this is the set obtained by zooming $\Omega$ about the point $p$ at the magnification $t^{-1}$. The following figure demonstrates $\tilde{T}_{p,t}\Omega$ for some different values of $p$ and $t$ (represented by the circle of radius $t$ centered at $p$):

Then the definition of the tangent tries to capture the "limit" of the picture of $\tilde{T}_{y,t}\Omega$ as $y \to x$ and $t \to 0$. Indeed, the definition is now rephrased as:

A vector $w \in \mathbb{R}^n$ is called a tangent to $\Omega$ at $x$ if, for any $x_i \to x$ with $x_i \in \Omega$ and $t_i \to 0$ with $t_i > 0$, there exists a sequence $w_i \in \tilde{T}_{x_i,t_i}\Omega$ such that $w_i \to w$.

The definition becomes much more concise if we introduce the notion of distance between a point and a set. Let $\operatorname{dist}(x,A)=\inf\{\|x - a\| : a \in A\}$ denote the distance between the point $x\in\mathbb{R}^n$ and the set $A\subseteq\mathbb{R}^n$. Then

The set $T_x\Omega$ of all tangent vectors to $\Omega$ at $x$ is given by
$$ T_x\Omega = \{ w \in \mathbb{R}^n : \operatorname{dist}(w, \tilde{T}_{y,t}\Omega) \to 0 \text{ as $ (y, t) \to (x, 0)$ in $\Omega\times(0,\infty)$} \}. $$

It might be interesting to note that the right-hand side is nothing but the Kuratowski limit inferior of the sets $\tilde{T}_{y,t}\Omega$ as $(y,t) \to (x, 0)$ along $(y,t) \in \Omega\times(0,\infty)$.


Body: How to find the number of positive devisors of $50,000$, I would like to know that what mathematical formulae  I need to use here as it is a big number to calculate mentally, I am sorry to ask if this is too silly question here to ask. Thank you. </p>

Body: $$d(50,000)=d(2^45^5)=d(2^4)d(5^5)=(1+4)(1+5)=30$$


问:
Is $\pi^{2n} + a\pi^n + b$ transcendental, with $a,b$ algebraic?
I got this problem and I've tried something: Suppose $\pi^{2n} + a\pi^n + b = \beta$, $\beta$ algebraic, then

$$\pi^{2n} + a\pi^n + b = \beta \Rightarrow \beta - b = \pi^n(\pi^n + a).$$

I can't proceed, because the product of two transcendental isn't necessarily transcendental. How can I proceed? Thank you!

答: Complete the square: $\beta - b + a^2/4 = \pi^{2n} + a\pi^n + a^2/4 = (\pi^n + a/2)^2.$ Thus, $\pi^n = \sqrt{\beta - b + a^2/4} - a/2$ and we have our desired contradiction.


问:
$Q = \{18x^2 − 12x + 18, −36x^2 + sx + 9, 2sx + s\}$Find the two values of $s$ for which the set $Q$ is linearly dependent.
Consider the subset of P3

$$Q  =  \{18x^2 − 12x + 18, −36x^2 + sx + 9, 2sx + s\}$$

Find the two values of s for which the set $Q$ is linearly dependent.

Does it mean that whenever we plug a value $s$ into $Q$, we will get a vector that is part of set $Q$?
And if that so, I can't seem to find two value $s$ that will make them linearly dependent, the first element $18x^2-12x+18$ is not affected by any value $s$, and to find two different value $s$ that make it linear dependent, I don't think it is possible

答:
hint

The third polynom does not contain the term with $ x^2 $. The only way to eliminate it from the first and the second, is multiplying the first by $ 2 $ and add it to the second. it gives

$$(36x^2-24x+36)+(-36x^2+sx+9)=$$
$$(s-24)x+45$$

this must be equal to
$$\lambda(2sx+s)$$

Now, solve the system

$$s-24=2\lambda.s$$
$$\lambda.s=45$$


问:
How to find k'th integer not divisible by n?
Although this was a programming question I want the mathematical intuition behind it. So we were given two numbers $n$ and $k$. We were asked to find out $k{-th}$ number **not divisible** by $n$.
 For example, $n=3$ and
    $k=7$.
 If we see numbers $1$ to $12$. $$1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12$$ Then $3,6,9$ are divisible by $3$ so we remove it. So we are left with $1, 2, 4, 5, 7, 8, 10, 11$. Clearly the $7th$ number not divisible by $3$ is $10$. So answer is $10$. It turns out the answer is:
$$k+⌊\tfrac{k-1}{n-1}⌋$$

I want the mathematical intuition behind this formula.

答:
Firstly, we can think about how multiples of $n$ appear in a sequence. For example, when $n=4$:
$$1,2,3,[4],5,6,7,[8],9,10, ...$$
Secondly, we are asked to determine the $k^{th}$ position in the sequence above, "skipping" multiples of $n$. For example when $k=5$, using the same sequence above:
$$1,2,3,[4],5, \langle 6 \rangle ,7,[8],9,10, ...$$
The number $ \langle 6 \rangle $ comes in this natural $6th$ position, but because there's $1$ multiple of $4$ before it, its $k^{th}$ position is deducted by $1$, thus $k^{th}$ of the position $5$ equals $6$.

We can now generalise this counting of multiples by organising them into blocks. This is the spatial reasoning involved in the intuition for the solution:
$$\underline{[1,2,3]},4,\underline{[5,6,7]},8,\underline{[9,10,} ...$$
From the above, clearly, we're counting how many blocks of $3$ (that is $n-1$) precede the numbers that are multiples of $n$, until the $k^{th}$ provided:
$$\lfloor{\frac{k}{n-1}}\rfloor$$
The above will give us how many blocks of numbers, therefore how many multiples, we can "offset" our desired ${k^{th}}$ number, so:
$$k+\lfloor{\frac{k}{n-1}}\rfloor$$
This sort of works, let's see:
$$ \begin{aligned}
k=5, n=4\\
5+\lfloor{\frac{5}{4-1}}\rfloor\\
= 5+\lfloor{\frac{5}{3}}\rfloor\\
=  \langle 6 \rangle\\
\end{aligned}
$$
The problem seems to be if $k$ is a multiple of $n-1$ because we will be counting an extra multiple, for example $k=6$, with $n-1 = 3$, we'll have:
$$
k=6, n=4\\
6+\lfloor{\frac{6}{4-1}}\rfloor\\
= 6+\lfloor{\frac{6}{3}}\rfloor\\
= 8 \;\; \text{ WRONG! Correct will be $7$}
$$
Ok, when we hit a block size that divides $k$, we'll have to deduct $1$, since we don't want to count that extra block, so we have:
$$
answer\bigl(k,n \bigr) =
\begin{cases}
k+\lfloor{\cfrac{k}{n-1}}\rfloor - 1 & \text{, if $n-1 \mid k$}\\
k+\lfloor{\cfrac{k}{n-1}}\rfloor & \text{, if $n-1 \nmid k$ }
\end{cases}
$$
This could be the final answer to the problem. It's also $O(1)$ and easy to deconstruct. But we could go further, due to the identities:
$$
\begin{matrix}
\lfloor{\cfrac{m+1}{n}}\rfloor = \lfloor{\cfrac{m}{n}}\rfloor + 1 & \text{, if $n \mid m+1$} & (\alpha)\\
\lfloor{\cfrac{m+1}{n}}\rfloor = \lfloor{\cfrac{m}{n}}\rfloor & \text{, if $n \nmid m+1$} & (\beta)
\end{matrix}
$$
The identities above are easier to reason about, if we prove our intuition using same numbers from our cases above, like so:
$$
\lfloor{\cfrac{5+1}{4-1}}\rfloor = \lfloor{\cfrac{5}{3}}\rfloor + 1 \text{, because $4-1 \mid 5+1$}\\
\lfloor{\cfrac{6+1}{4-1}}\rfloor = \lfloor{\cfrac{6}{3}}\rfloor \text{, because $4-1 \nmid 6+1$}
$$
Or:
$$
\lfloor{\cfrac{6}{3}}\rfloor = \lfloor{\cfrac{5}{3}}\rfloor + 1 \text{, because 3 | 6}\\
\lfloor{\cfrac{7}{3}}\rfloor = \lfloor{\cfrac{6}{3}}\rfloor \text{, because $3 \nmid 7$}
$$
Or:
$$
\begin{matrix}
\lfloor{\cfrac{k}{n-1}}\rfloor = \lfloor{\cfrac{k-1}{n-1}}\rfloor + 1 & \text{, if $n-1 \mid k$} & (\alpha)\\
\lfloor{\cfrac{k}{n-1}}\rfloor = \lfloor{\cfrac{k-1}{n-1}}\rfloor & \text{, if $n-1 \nmid k$} & (\beta)
\end{matrix}
$$
That's where our $k-1$ comes from. So:
$$
k+\lfloor{\frac{k}{n-1}}\rfloor - 1 \text{, if $n-1 \mid k$}\\
\stackrel{\text{from $(\alpha)$}}{=} k+\lfloor{\frac{k-1}{n-1}}\rfloor + 1 - 1\\
= k+\lfloor{\frac{k-1}{n-1}}\rfloor
$$
And:
$$
k+\lfloor{\frac{k}{n-1}}\rfloor \text{, if $n-1 \nmid k$}\\
\stackrel{\text{from $(\beta)$}}{=} k+\lfloor{\frac{k-1}{n-1}}\rfloor
$$
We can combine the two answers now, since both will apply regardless of whether $n-1$ divides $k$ or not:
$$
k+\biggl\lfloor{\frac{k-1}{n-1}}\biggr\rfloor
$$
Q.E.D.


答:
Suppose $\beta$ is algebraic. Then
$$\pi^{2n} + a \pi^n + b - \beta =0$$
So that $\pi$ is a root of the polynomial $X^{2n} + a X^n + b - \beta$, and is algebraic over the field of algebraic numbers.

But the field of algebraic numbers is algebraically closed. This implies that $\pi$ is algebraic, which is a contradiction.

Title: Is $\pi^{2n} + a\pi^n + b$ transcendental, with $a,b$ algebraic?

Body:
I got this problem and I've tried something: Suppose $\pi^{2n} + a\pi^n + b = \beta$, $\beta$ algebraic, then

$$\pi^{2n} + a\pi^n + b = \beta \Rightarrow \beta - b = \pi^n(\pi^n + a).$$

I can't proceed, because the product of two transcendental isn't necessarily transcendental. How can I proceed? Thank you!


问:
Solve $f'(x)f''(x)=\frac12$ without using differential equations
The task seems to be trivial:

Twice differentiable gunction $f(x)$ maps segment of real numbers $[0,\infty)$ onto the same set of real numbers $[0,\infty)$ (each non-negative real value of $f(x)$ has a corresponding non-negative argument $x$). Find all functions $f(x)$ that satisfy the following criteria:

$$f'(x)f''(x)=\frac12$$

This differential equation is trivial and you don't need more than a few rows to prove that

$$f(x)=\frac23x^{\frac32}$$

However, the ask is to solve the problem without using approach based on differential equation! Any ideas?

答:
One possible solution is as follows, although I don't know if this satisfies your requirement that the approach shouldn't be based on differential equations.

Let $g = f'$. Then $(g^2)' = 2gg' = 1$, by your hypothesis. Thus $g^2 = x + c_0$. As $g$ is defined on $[0, \infty)$, we must have $c_0 \geq 0$. Then as $g$ is continuous and $g^2 \neq 0$ on $(0, \infty)$, we either have $g = +\sqrt{x + c_0}$ or $g = -\sqrt{x+c_0}$. In either case we may integrate and obtain

$$ f(x) = \pm\frac{2}{3}(x+c_0)^{\frac{3}{2}} + c_1,$$

for some $c_1 \in \mathbb{R}$. Now as $f$ is surjective, it is not hard to see that we need $f(0) = 0$, so solving for $c_1$ gives all such functions are of the form

$$f(x) = \pm \frac{2}{3}(x+c)^{\frac32} \mp \frac{2}{3}c^{\frac{3}{2}}$$

for $c \geq 0$.


答:
Hint:

If the prime factorization of a natural number $\,n\,$ is

$$n=\prod_{i=1}^k p_i^{a_i}\;,\;\;p_i\;\,\text{primes}\;,\;\;a_i\in\Bbb N$$

Then the number of different positive divisors of $\,n\,$ is

$$\prod_{i=1}^k (a_i+1)$$

Title: How to find the number of positive devisors of $50,000$


问:
Find a $T_pM$ and $N_pM$
I have the following Submanifold:
$$M = \{(x,e^{x})\in \mathbb{R^{2}:x \in \mathbb{R}}\}$$
to which I have to find a tangent and normal space $T_{p}M$ and $N_{p}M$ at the point $p \in M$.

This  is one of the examples that I have but in the same  excercise I have like 10 different submanifolds to which i'll have to find $T_{p}M$ and $N_{p}M$. I am clear with their idea visually and we also had a Theorem about the Basis of $T_{p}M$ and $N_{p}M$.

if somebody could help me out with the practical solutions in these cases (since the excercises has like 10 different cases with different functions i assume that there might be a practical way to see them)
i would be very thankful

答:
Let $$M=\{(x,e^x): x\in \mathbb{R}\}$$ be a submanifold of $\mathbb{R^2}$. Let $\phi: \mathbb{R^2}\to \mathbb{R}$ defined by $\phi(x,y)= e^x-y$ on some open subset $U$ of $\mathbb{R^2}$. Notice $M$ is the level set $\phi^{-1}(0)$. Then since $T_p(M) = (\nabla f\vert p)^{\perp}$, then we can find the tangent space this way: $(\nabla f)=(e^x, -1)$. Therefore the complement is generated by $(1,e^x)$.

Notice that all tangent lines to $M$ in $\mathbb{R}^2$ are precisely $$span\{(1,e^x):x\in \mathbb{R}\}$$

If you have 50 similar exercises, I would follow this procedure.

Title: Find a $T_pM$ and $N_pM$


答:
We need to see can there exist scalars $\alpha, \beta, \gamma$ not all zero such that
\begin{align}
0 &= \alpha(18x^2-12x+18) + \beta(-36x^2+sx+9)+\gamma(2sx+s)\\
&= x^2(18\alpha - 36\beta) + x(-12\alpha + s\beta + 2s\gamma) + (18\alpha+9\beta+s\gamma)
\end{align}
or
$$\begin{cases}
18\alpha - 36\beta = 0 \\ -12\alpha+s\beta+2s\gamma = 0 \\ 18\alpha+9\beta+s\gamma = 0\end{cases}$$
The first equation implies $\alpha = 2\beta$ which we can plug into the second and third equation to obtain
$$\begin{cases}
(-24+s)\beta +2s\gamma = 0 \\ 45\beta + s\gamma = 0\end{cases}$$
Now multiplying the second equation by $-2$ and adding it to the first gives
$$(-114+s)\beta = 0.$$
If $s = 114$, we cannot conclude $\beta = 0$. In fact $\alpha = -57, \beta = -114, \gamma = 45$ is a nontrivial solution.

Moving on, if $s \ne 114$ we get $\beta = 0$ and then $s\gamma = 0$. If $s = 0$, we cannot conclude $\gamma = 0$. In fact $\alpha = \beta = 0, \gamma = 1$ is a nontrivial solution.

Moving on, if $s \ne 114, 0$ we get $\gamma = 0$ and then $\alpha = 0$ as well so there are no nontrivial solutions.

We conclude that $Q$ is linearly dependent if and only if $s \in \{0,114\}$.

Title: $Q = \{18x^2 − 12x + 18, −36x^2 + sx + 9, 2sx + s\}$Find the two values of $s$ for which the set $Q$ is linearly dependent.

Body:
Consider the subset of P3

$$Q  =  \{18x^2 − 12x + 18, −36x^2 + sx + 9, 2sx + s\}$$

Find the two values of s for which the set $Q$ is linearly dependent.

Does it mean that whenever we plug a value $s$ into $Q$, we will get a vector that is part of set $Q$?
And if that so, I can't seem to find two value $s$ that will make them linearly dependent, the first element $18x^2-12x+18$ is not affected by any value $s$, and to find two different value $s$ that make it linear dependent, I don't think it is possible


问:
Number of solutions of the quaternion
I have the quaternions $q = x_0 + ix_i + jx_2 + kx_3$ where $x_i \in \mathbb{R}$ and $i,j,k$ satisfy the relations $$ i^2=j^2=k^2 = -1, \hspace{0.3cm} ij=-ji=k,\hspace{0.3cm}  jk=-kj=i,\hspace{0.3cm}  ki=-ik=j$$
I also have that $f(q) =q^2$ defines a smooth map from $\mathbb{R}^4 \cup \{\infty\} \cong S^4$ to itself.

I have to find how many solutions are there to the equation $q^2 = 1$.

Can someone help me ?

答:
Write the quaternion $q$  as $x_0 + v$ where $v = i x_1 + j x_2 + k x_3$, and note
that $q^2 = x_0^2 + 2 x_0 v - (x_1^2 + x_2^2 + x_3^2)$.  Thus $q^2 = 1$ implies $x_0 v = 0$, i.e. either $x_0 = 0$ or $v=0$.  With $x_0 = 0$ we would have $x_1^2 + x_2^2 + x_3^2 = -1$, which has no real solutions.  With $v=0$ we have $x_0 = \pm 1$.

Title: Number of solutions of the quaternion


问:
Remainder when $20^{15} + 16^{18}$ is divided by 17
What is the reminder, when $20^{15} + 16^{18}$  is divided by 17.
I'm asking the similar question because I have little confusions in MOD.
If you use mod then please elaborate that for beginner.
Thanks in advance.</p>

答:
The key thing to remember about the operation "mod" is that it behaves "well" with respect to product (hence powers), and of course addition. This means that if you can simplify your life a lot distributing the calculation into many steps and taking "mod" at each stage.

To compute $20^{15}$, you can first notice that  $20 \pmod{17} = 3$. Then $20^{15} \pmod{17} = 3^{15} \pmod{17}$. The power $15$ is quite large, but you can for instance take: $3^3 \pmod{17} = 27 \pmod{17} = 10$, hence $  3^{15} \pmod{17} = 10^5 \pmod{17} = 10 \cdot 100^2 \pmod{17} = 10 \cdot (-2)^2 \pmod{17} = 40 \pmod{17} = 6 \pmod{17}$.

The term $16^{18}$ is much easier: $16^{18} \pmod{17} = (-1)^{18} \pmod{17} = 1$.

Hence, the answer is $6 + 1 = 7$.


问:
Calculate third derivative of multivariate function
I really need to understand the algorithm of solving these kind of problems.

$$f(x,y) = x^3 + y^3 +xy(x - y)$$
I need to calculate:
$$f'''_{xxy}(1,1) + f'''_{xxx}(1,1) + f'''_{yyy}(1,1)$$

I would really appreciate your help!

Body:
$$ f_x = 3x^2 +2xy - y^2;
f_{xx} = 6x + 2y;
f_{xxx} = 6; f_{xxy} = 2;
f_y = 3y^2 + x^2 - 2xy;
f_{yy} = 6y - 2x;
f_{yyy} =6
$$
So the answer to your question is $14$.


问:
Generate sine wave signal with time-varying frequency
I try to generate a sine wave signal with a frequency changing over time. The frequency is randomly defined by generating some random values in the range [0.5, 2] and interpolating the points between them.

The expected output signal is a sine wave with a non-changing amplitude and a changing frequency.
But there are some smaller bumps and the signal is not a 'smooth' sine wave. E.g. the period at x = 200 should be larger than the period at x = 10, but the opposite is the case.

Does anyone know, what happened here?

```<code>import numpy as np
from scipy.interpolate import interp1d
import matplotlib.pyplot as plt

x_samples = np.arange(-100, 3100, 50)
freq_samples = np.random.random(x_samples.shape) * 1.5 + 0.5

x = np.arange(0, 3000, 0.1)
interpolation = interp1d(x_samples, freq_samples, kind='quadratic')
freq = interpolation(x)
y = np.sin(freq * x)

plt.plot(x, y, label="sin(freq(x) * x)")
plt.plot(x, freq, label="freq(x)")
plt.legend()
plt.show()
</code>```

Plot

答:
This is known as Frequency Modulation in Electronics. As you observed the variable crowding and loosening along time axis is its principal feature and is quite unlike the amplitude modulated wave.

The instantaneous frequency of carrier wave is variable and occurs in the wave expression argument inside the parenthesis, depending upon how argument ( carrier frequency) is modified.


问:
The lower bound of all positive convergent series
Is there any lower bound on a convergent, non-negative series?

For example, if $\sum_{n=1}^\infty a_n$ is convergent, and $a_n \geq 0$ $\forall n$ in the bounds $\{1, 2, 3, 4, ...\}$, then can it be said that $a_n \leq \frac{1}{n^{1+\epsilon}}$, for arbitrarily small $\epsilon > 0$.

I am seeing if this holds generally, since $\sum_{n=1}^\infty \frac{1}{n}$ diverges (harmonic series), and $\sum_{n=1}^\infty \frac{1}{n^{1+\epsilon}}$ converges $\forall \epsilon > 0$.

Therefore, is it the case that $\frac{1}{n^{1+\epsilon}} \geq a_n$, for some $\epsilon > 0$, given that the sum $\sum_{n=1}^\infty a_n$ converges? I.e., in general, is it the case that for a non-negative series that converges, that some power greater than 1 for the denominator can be found such that $a_n \leq \frac{1}{n^{1+\epsilon}}$?

答:
The answer is no, the simplest counterexample is

$$\sum_{n=1}^\infty \frac{1}{n (1+\ln(n)^{1+\delta})}$$
for any $\delta >0$.

This series is convergent, but it is not bounded by any $\frac{1}{n^{1+\epsilon}}$.

And you can take this one step further $\sum_{n=1}^\infty \frac{1}{n (1+\ln(n))(1+\ln(\ln(n)))^{1+\delta})}$ and so on.


问:
Clarifications on definition of an indicator random variable
Let's say I have a probability space $(\Omega, \mathcal{F}, P)$ and a random variable defined on it, say $X$, which is equal to $0$ and $1$ with probabilities $p_0 > 0, p_1 > 0$ respectively, with $p_0 + p_1 = 1$.

I have two questions:

I know the notation for $X$ would be something like $X: (\Omega, \mathcal{B}) \to (\{0, 1\}, \mathcal{M})$  (i.e., a mapping over measurable spaces). Is $\mathcal{B}$ considered a sub-$\sigma$-algebra of $\mathcal{F}$?

What would the $\sigma$-algebra $\mathcal{M}$ be? Is it $\sigma(\{\{0\}, \{1\}\})$?

These are questions that I can't seem to find answered in any textbook I own on measure theory, so I apologize for what may seem like a lack of effort.

答:
Thanks to Balarka Sen for clarifications via chat.

When we say that $X$ is defined over the probability space $(\Omega, \mathcal{F}, P)$, what we mean is that $X: (\Omega, \mathcal{F}) \to (\{0, 1\}, \sigma(\{\{0\}, \{1\}\}))$. Keep also in mind that the universal set that would be in $\sigma(\{\{0\}, \{1\}\})$ would be $\{0, 1\}$, as opposed to $\Omega$ like it usually is.


问:
$\operatorname{Var}(X+Y\mid X+Y)=\operatorname{Var}(X\mid X+Y)+\operatorname{Var}(Y\mid X+Y)$?
Is this equality true for independent random variables $X$ and $Y$? Or maybe we should also assume the same distribution?

答:
The conditional distribution of a random variable $Z$ given $Z=z$ is just the unit mass at $z$.  This has variance $0$. Therefore $\text{Var}(X+Y \mid X+Y) = 0$.
On the other hand, $\text{Var}(X \mid X+Y) \ge 0$; equality means $X+Y$ determines $X$ almost surely.  That can happen, but usually is not the case.  So your equality is not true in general.


问:
Intended Audience for N Is a Number?
Is the film N is a Number appropriate (mathematically) for a student entering high school?

答: In my opinion, yes. It is focused mostly around anecdotes and mathematicians talking about their personal experiences with mathematics (Erdős related, of course). The explanations of mathematical theorems are oriented towards the general public, so they remain quite elementary. I really enjoyed this film, so I would highly recommend it to anyone. (I only wish more films like this were made. =))</p>

Title: Intended Audience for N Is a Number?


问:
How to show that If a is relatively prime to n, then multiplication by a permutes the elements 1,2,3,....,n-1?
Lemma#3: If a is relatively prime to n, then multiplication by a permutes the elements  1,2,3,....,n-1.

I didn't understand what permutes means here. If I consider $\mathbb{Z_n}$ then any element a relatively prime to n has the same order as 1 and also it generates $\mathbb{Z_n}$. Thats the maximum I know. Please I need help to understand this.Thank you.

答:
Hint

Suppose there exists $i,j \in \{1,2,\cdots,n-1\}$ such that

$(a \times i) \equiv (a \times j) \pmod{n}.$

What does this say about whether $n$ divides $a(i - j)$?

Of what relevance is the fact that $a$ relatively prime to $n$?

With $i \neq j$, is it possible for $n$ to divide $(i - j).$


答:
You can draw a picture and see what we expect:

Here is a rigorous calculation, using curves and derivations: Let $p=(x_0,e^{x_0})\in M$. Define $\gamma:[-\epsilon, \epsilon]\to M$ by $\gamma(t)=(t+x_0,e^{t+x_0}).$ Then, $\gamma(0)=p$ and $\gamma'(0)=a\frac{\partial}{\partial x}+b\frac{\partial }{\partial y}$ is an arbitrary tangent vector at $p$. Operating with $\gamma'(0)$ on the projections $r_1(x,y)=x$ and $r_2(x,y)=y,$ respectively, we get $\gamma'(0)r_1=a$  and $\gamma'(0)r_2=b$.

Now, by definition of $\gamma'(0)$, we have

$a=\gamma'(0)r_1=\frac{d (r_1\circ \gamma)}{dt}|_{t=0}=\frac{d (t+x_0)}{dt}|_{t=0}=1$ and $b=\gamma'(0)r_2=\frac{d (r_2\circ \gamma)}{dt}|_{t=0}=\frac{d (e^{t+x_0})}{dt}|_{t=0}=e^{x_0}$.

Therefore, in terms of components, $v\in T_pM$ is given by vectors in  $\operatorname {span}\{(1,e^{x_0})\}.$

And then $N_pM$ is just the orthogonal complement, so we need to solve $A\cdot 1+B\cdot e^{x_0}=0.$ We know that this subspace has dimension $1$ (because $\mathbb R^2$ has dimension $2$) so all we need is one vector satisfying this condition. Insepction gives the vector $(e^{x_0},-1)$ and so we may take $N_pM=\operatorname {span}\{(e^{x_0},-1)\}.$

Title: Find a $T_pM$ and $N_pM$

Body:
I have the following Submanifold:
$$M = \{(x,e^{x})\in \mathbb{R^{2}:x \in \mathbb{R}}\}$$
to which I have to find a tangent and normal space $T_{p}M$ and $N_{p}M$ at the point $p \in M$.

This  is one of the examples that I have but in the same  excercise I have like 10 different submanifolds to which i'll have to find $T_{p}M$ and $N_{p}M$. I am clear with their idea visually and we also had a Theorem about the Basis of $T_{p}M$ and $N_{p}M$.

if somebody could help me out with the practical solutions in these cases (since the excercises has like 10 different cases with different functions i assume that there might be a practical way to see them)
i would be very thankful


问:
A semisimple ring has only finitely many isomorphism classes of simple modules
Consider the ring
$$R= M_{n_1}(D_1)\oplus \dots \oplus M_{n_k}(D_k)$$
where $D_1,\dots , D_k$ are skew fields.

Is there an easy way to see that there are, up to isomorphism of left $R$-modules, only finitely many simple left $R$-modules?

Attempt: I managed to prove this for one factor, but my proof does not generalise and I don't think I can get an induction started.

答: You can prove quite generally that if $R \times S$ is a product of rings, every $R \times S$-module $M$ canonically decomposes as a product $M_R \times M_S$ of an $R$-module and an $S$-module, where $M_R = (1, 0) M$ and $M_S = (0, 1) M$. It follows that a simple $R \times S$-module is either a simple $R$-module or a simple $S$-module (and the same for indecomposable modules), hence that the number of simple modules over $R \times S$ is the number of simple modules over $R$ plus the number of simple modules over $S$. Then you're done by induction.


答:
When you use the modulo operator, multiplication is conserved, e.g:
$$20 \equiv 3 \pmod {17}$$
$$40 \equiv 6 \pmod {17}$$
So that:
$$40*20 \equiv 6*3 \pmod {17}\equiv 1 \pmod {17}$$

The same holds for powers (since this is just multiplication many times), so:
$$20^{15} \equiv 3^{15} \pmod {17}$$
$$16^{18} \equiv (-1)^{18} \pmod {17}$$
Now use: $$3^{15} = (3^5)^3 \equiv (243)^{3} \equiv (5)^{3} \equiv 6 \pmod {17}$$
To see that the remainder is $7$.

Title: Remainder when $20^{15} + 16^{18}$ is divided by 17

Body:
What is the reminder, when $20^{15} + 16^{18}$  is divided by 17.
I'm asking the similar question because I have little confusions in MOD.
If you use mod then please elaborate that for beginner.
Thanks in advance.</p>


问:
How to extend the tensor construction over $\mathbb{Z}$ to one over a commutative ring $C$
Let $(E_1,\ldots,E_n)$ be a family of $C$-modules for some commutative ring $C$. Write $E:=\prod_{i=1}^nE_i$. Let $\phi:E\rightarrow\mathbb{Z}^{(E)}$ be defined by the equation $\phi(x):=(\delta_{xy})_{y\in E}$ for all $x\in E$.

Let $H$ be the $\mathbb{Z}$-submodule of $\mathbb{Z}^{(E)}$ generated by elements of the form
$$\phi(x_j+y_j,z)-\phi(x_j,z)-\phi(y_j,z)$$
for all $1\leq j\leq n$, $x_j,y_j\in E_j$ and $z\in\prod_{i\ne j}E_i$. Write $\bigotimes_{i=1}^nE_i:=\mathbb{Z}^{(E)}/H$ and let $\pi:\mathbb{Z}^{(E)}\rightarrow\bigotimes_{i=1}^nE_i$ be the canonical $\mathbb{Z}$-linear surjection. Finally, for $x\in E$, we write
$$x_1\otimes x_2\otimes\ldots\otimes x_n:=\pi(\phi(x)).$$

What is the most natural way to quotient $\bigotimes_{i=1}^nE_i$ in order to get a $\mathbb{Z}$-module in which
$$(\gamma x_1)\otimes'x_2\otimes'\ldots\otimes' x_n=x_1\otimes'(\gamma x_2)\otimes'\ldots\otimes'x_n=\ldots=x_1\otimes'\ldots\otimes' (\gamma x_n)$$
for $\gamma\in C$ and $x\in E$? Note that $\otimes'$ is the new tensor to be obtained.

答:
Without bothering to find an explicit description of the kernel, we can use the universal property to move from one product to another. Specifically(letting the ring here be $R$ just because), consider the canonical $R$-multilinear map from $E_1 \times \cdots \times E_n \to E_1 \otimes_{R} \cdots \otimes_{R} E_n$ given by

\begin{equation*}
    (e_1, \ldots, e_n) \mapsto e_1 \otimes_{R} \cdots \otimes_{R} e_n.
\end{equation*}

Now, such a map is also $\mathbb{Z}$-multilinear, as in general the $\mathbb{Z}$-action on a module $M$ is given by a (unique) homomorphism $\mathbb{Z} \to \operatorname{End}(M)$, so it is equal to the composition $\mathbb{Z} \to R \to \operatorname{End}(M)$, which is to say that the action of any integer is given by an action of a ring element(this is more or less obvious). Thus by the universal property we obtain a (surjective) map $E_1 \otimes_{\mathbb{Z}} \cdots \otimes_{\mathbb{Z}} E_n \to E_1 \otimes_{R} \cdots \otimes_{R} E_n$.


问:
Extensions of the Turing jump
The Turing jump $0^{(\alpha)}$ is defined for ordinals $\alpha<\omega_1^{\mathit{CK}}$ with

$0^{(0)} = \varnothing$,

$0^{(\alpha+1)}$ is the diagonal halting problem using $0^{(\alpha)}$ as an oracle,

$0^{(\lambda)}$ for a limit $\lambda$ is the effective join of the $0^{(\lambda_n)}$ for a computable sequence $\lambda_n$ of ordinals with computable notations and $\lambda_n\rightarrow\lambda$.

The set of natural numbers corresponding to ordinal notations in this scheme is denoted by $\mathcal O$ and $\omega_1^\mathit{CK}$ is the supremum of the ordinals encoded in $\mathcal O$. The hyperjump $X\mapsto\mathcal O^X$ allows the sequence in the limit case of the above definition to be $X$-computable.

This suggests the following extension of the Turing jump beyond $\omega_1^\mathit{CK}$:

$0^{(\omega_1^\mathit{CK})}=\mathcal O$

$0^{(\omega_{\alpha+1}^\mathit{CK})} = \mathcal O^{0^{(\omega_{\alpha}^\mathit{CK})}}$

$0^{(\omega_\lambda^\mathit{CK})}$ for a limit $\lambda$ is the effective join of the $0^{(\omega_{\lambda_n}^\mathit{CK})}$ for a $0^{(\omega_{\alpha}^\mathit{CK})}$-computable sequence of ordinals with $0^{(\omega_{\alpha}^\mathit{CK})}$-computable notations for some $\alpha<\lambda$ and $\lambda_n\rightarrow\lambda$.

Is this construction well-defined? In particular, is it true that the hyperjump of each admissible $\omega_\alpha^\mathit{CK}$ allows us to denote ordinals up to $\omega_{\alpha+1}^\mathit{CK}$, and the definition for the limit case up to $\omega_{\lambda}^\mathit{CK}$?

What can we say about the ordinals that obtain notations this way? What is their supremum?

The second part of my question is less rigorous and assumes that there are meaningful answers to the above questions. It seems natural to further abstract from the construction described above: Whenever the process we have defined so far reaches a limit, we can take the set of natural numbers corresponding to notations in the scheme defined so far as an oracle for the next ordinal, define another jump operator and enumerate its applications using the already-defined ordinals. If we manage to formalize this creation of new jump operators and their limits in a systematic way, we will again have ordinal stages and this process will have a supremum. At this point we could devise a scheme for creating jump operator schemes, and so on. My question is: Can this abstraction process be formalized in a way that cannot be extended meaningfully beyond its supremum? What would that supremum look like? I've found this paper1 on mastercodes thanks to this site but I don't understand the notation in it and I'm not sure the idea is the same.

Another perspective on this hyper-abstraction process could be a computational one: I might very well be mistaken, but it seems to me that at each individual stage I would be able to find a canonical way to generalize the abstraction to an abstraction on an even higher level. I could write a Turing machine (using oracles) not only computing ordinal notations, but also one computing codes of Turing machines computing ordinal notations, and so on. But could there be a "final" such Turing machine able to carry out all the work I would be doing there?

1Harold T. Hodes: Jumping Through the Transfinite: The Master Code Hierarchy of Turing Degrees, The Journal of Symbolic Logic, Vol. 45, No. 2 (Jun., 1980), pp. 204-220; DOI: 10.2307/2273183, JSTOR

答:
The process you have in mind can be formalized as follows. We define a sequence of Turing degrees (not individual sets of natural numbers) as follows. I'm going to talk about (hyper)jump sequences along well-orderings rather than families of fundamental sequences of ordinals, since the former is ultimately much easier to work with.

I'll use the following definition (although there are many equivalent ones out there). Suppose I have a map $\mathfrak{m}:\mathcal{P}(\omega)\rightarrow\mathcal{P}(\omega)$, which below will be either the jump or hyperjump operator. Given a set $A$ and a well-ordering $R$ of $\omega$ (not just an ordinal, a literal relation on the natural numbers) we let the $\mathfrak{m}$ sequence along $R$ starting with $A$ be the unique set $U$ such that

$$U_k=\begin{cases}
A &\mbox{ if $k$ is the $R$-least element},\\
A\oplus \mathfrak{m}(U_l) & \mbox{ if $l$ is the immediate predecessor of $k$ in $R$, and}\\
A\oplus \{\langle i,j\rangle: j<_Rk, i\in U_j\} & \mbox{ if $k$ is an $R$-limit point.}\\
\end{cases}$$

where "$X_n$" is the $n$th column of $X$, namely $\{m: \langle m,n\rangle\in X\}$. Folding $A$ in at each level isn't something we ever have to do, but it cleans things up a bit in my opinion.

It's a good exercise to prove that a unique such $U$ indeed exists. (HINT: transfinite recursion/induction.)

We let $r_0=\bf 0$ be the degree of the computable sets.

Having defined $r_\alpha$, we let $r_{\alpha+1}={r_\alpha}'$.

Now suppose $\lambda$ is a limit and we've defined $r_\alpha$ for all $\alpha<\lambda$. There are now four cases:

Small $\lambda$. suppose there is some $\alpha<\lambda$ such that there is a well-ordering $R$ of $\omega$ which has ordertype $\lambda$ is computable from $r_\alpha$. Fixing some such $\alpha$, some representative $A\in r_\alpha$, and some $R$ as above, we let $r_{\lambda}$ be the degree of the jump sequence starting with $A$ along $R$.

Wait, is that even well-defined? Yes it is, but this is nontrivial. It's easiest to first consider a single appropriate choice of $\alpha$. It turns out that if $R,S$ are isomorphic well-orderings of $\omega$ which are each computable from some degree ${\bf a}$, then for any $A,B\in{\bf a}$ the jump sequence along $R$ starting with $A$ is Turing-equivalent to the jump sequence along $S$ starting with $B$. This isn't too hard to prove (see Sacks' book), but it should be surprising in light of various examples of the opposite phenomenon (e.g. with iterated consistency principles or with the Ershov hierarchy). Now it's a good exercise to generalize this to show that the choice of $\alpha$ doesn't matter either.

Medium $\lambda$. Suppose the case above doesn't hold, but there is some $\alpha<\lambda$ such that $\lambda=\omega_1^{CK}(\alpha)$. Fixing the least such $\alpha$, we let $r_\lambda=\mathcal{O}^{r_\alpha}$.

Large $\lambda$. Suppose $\lambda$ is neither small nor medium, but there is some $\alpha<\lambda$ and some $r_\alpha$-computable well-ordering $R$ of $\omega$ such that the set of admissible ordinals $<\lambda$ has ordertype $R$. In this case we follow the "small $\lambda$" approach above with hyperjump in place of jump; and as there, we have a crucial - and nontrivial - well-definedness result.

Ginormous $\lambda$. Suppose $\lambda$ is neither small, medium, or large. Then this process ends.

It turns out that the stopping point of this construction - that is, the first ginormous ordinal - is exactly the first recursively inaccessible ordinal, that is, the first ordinal which is both admissible and a limit of admissibles. The proof is basically unwinding the definitions and applying Sacks' connection between computability and admissibility.  It's worth noting that the first recursively inaccessible ordinal is galactically larger than the first limit of admissibles, but is in turn cosmically smaller than the first $2$-admissible; see this survey of Madore for some useful information.

More generally, if we replace ${\bf 0}$ with $deg(A)$ for some $A$ in the start of the construction, the stopping point is the least $\kappa$ such that there is an admissible set $\mathbb{M}$ of height $\kappa$ with $A\in \mathbb{M}$ and $\mathbb{M}\models$ "Every set is contained in an admissible set."

Finally, let me end by saying a bit about how "long jump sequences" can be created:

The approach you've taken is in some sense the most direct: we keep writing down concrete operators which get us a bit further each time, and take the sup of what we can get along them as above.

The "mastercodes" approach is more model-theoretic: rather than thinking of operators on reals we think of operators on structures, and the degrees we look at are the degrees of the theories of those structures (or similar objects). This has a surprising advantage of providing canonical representatives of those degrees, namely the actual reals coding the relevant theories themselves.

Finally, there's the "higher type oracle" approach. We can define "$\omega_1^{CK}(\mathfrak{F})$" for a higher type function $\mathfrak{F}$ as the least ordinal with no $\mathfrak{F}$-computable copy via Kleene's theory of computability in higher types. Even relatively concrete choices of $\mathfrak{F}$ get us far beyond the first recursively inaccessible ordinal. Madore's survey I linked to above refers to these frequently.

Title: Extensions of the Turing jump


问:
Vice versa: Model generated curve fitting to real experiment results
Phd student here. For my thesis I have a problem to solve. I have written an algorithm, which generates a curve for some model (alpha-hemolysin makes spores on cell membrane over time). That model uses like $20$ constants: $k_{01},\dotsc,k_{20}$. Now, what I need to do is make it the other way around: I have a curve (from real experiment data). Now need to find those constants $k_{xx}$, so that the generated curve would best fit the real experiment curve, marked by arrow between blue marks in the image below.

I will probably need to write some function which would return total difference between the generated curve and the real curve and try to minimize it by applying various $k_{xx}$ sets. Would machine learning be suitable for it? Is there any other approach? Any suggestions or hints are very welcome, as I had no experience before and this is very new to me. PS will use Orange as a tool. Thank you!

答:
We can compare the behavior/tendency to a particular case of a well known  tan  hyperbolic function  of given starting slope that goes asymptotically to zero.

Like for example

$$ y= y_{max} \tanh c \cdot x $$
of starting slope

$$ y'= y_{max} \cdot  c $$

where $y_{max}$ is the  constant steady state asymptotic value  it tends to and $c$ is a constant that represents the rate at change at start.

Another example is a trigonometric tangent related function

$$ y= 1-\frac{1}{x}$$

that has such a tendency towards either axis:


问:
Generic mathematical equation for deriving two numbers from the same pool with a dependency
Sorry for the title, I did not know how else to describe my question.  I'm a programmer and I'm looking for a general form to answer the following type of question:

It takes 1 hour to produce a widget.  It takes 5 hours and 1 widget to produce a gadget.  If I have 100 hours, how many gadgets can I produce assuming I have no pre-made widgets?

This one seems simple, since it takes 6 hours per gadget, I can make 100 / 6 gadgets in 100 hours.

If I have 10 widgets, how does the math change?

It'd be something like 100 / 5 up to 10 (so 50 hours burned); then 50 / 6 for the remainder because I have run out of pre-made widgets.

Is there a simpler or more compact way to write out this formula?

答:
I think this is easiest to just split into cases. Since you are a programmer, you shouldn't have any trouble implementing something like this. Suppose a widget takes time $t_{\text{w}}$ to make, a gadget takes one widget and another $t_{\text{g}}$ hours to make, and we have $H$ hours in total, starting initially with $W$ spare widgets. Let $G$ be the total number of gadgets we can make in the allotted time. If $W\geq H/t_{\text{g}}$ then simply
$$G=\lfloor H/t_{\text{g}}\rfloor$$
However, if this is not the case, then
$$G=\left\lfloor W+\frac{H-W\cdot t_{\text{g}}}{t_{\text{g}}+t_{\text{w}}}\right\rfloor$$
Example - if we have $H=100$ hours, a widget takes $t_{\text{w}}=1$ hour to make, and a gadget takes $t_{\text{g}}=5$ hours and one widget to make, and we start with say $13$ spare widgets, then
$$G=\left\lfloor13+\frac{100-13\times 5}{6}\right\rfloor=\lfloor 18.833...\rfloor=18 \text{ gadgets }$$


问:
Why is logarithm in BMO
I read that $\log|x|$ is supposedly a typical example of a BMO function. How do I see that it is in BMO actually?</p>

答:
$\log|x|$ is funny:

  Scaling $x$ adds a constant

    No matter for us.

Take a cube of size $1$.

  If it's placed near $0$,

    Then $\log|x|$ is not big.

On a far-away cube

  There's a gradient bound

    $\log|x|$ is Lipschitz.

Title: Why is logarithm in BMO


问:
Emedding of $\mathbb{ RP}^3$
Is there a simple formula for an embedding (homeomorphic onto its image) of $\mathbb{RP}^3$ in some Euclidean space? I have seen a simple formula for $\mathbb{RP}^2$ in $\mathbb R^4$, but I can't find much of anything on $\mathbb{RP}^3$. I am not asking for an immersion, but a true embedding.

答:
Thinking of $\mathbb{RP}^n$ as $S^n$ mod antipode, and thinking of $S^n$ as the unit $n$-sphere $\{ x_0 + \dots + x_n^2 = 1 \} \subsetneq \mathbb{R}^{n+1}$, we can consider the map

$$S^n \ni (x_0, \dots x_n) \mapsto (x_0^2, \dots x_{n-1}^2, x_0 x_1, x_1 x_2, \dots x_{n-1} x_n) \in \mathbb{R}^{2n}.$$

This is a slight variant of a Veronese embedding. It factors through $\mathbb{RP}^n$ and produces an embedding (at least topologically; I haven't checked whether it's an immersion). Note that $x_n^2$ is omitted since it's determined by $x_0^2, \dots x_{n-1}^2$. The first $n$ coordinates produce a map which identifies two points of $S^n$ if each of their coordinates differs by a possibly different sign, and the remaining $n$ coordinates are there to ensure that all the signs match, so we only identify points which differ by a single global sign.

Title: Emedding of $\mathbb{ RP}^3$


问:
Show that B is a Boolean subalgebra of $\mathbb{P(N)}$ which cannot be Boolean isomorphic to some $\mathbb{P(M)}$
Let $B = \{X \subseteq \mathbb{N} | X  $
is finite or its complement
$ \subseteq \mathbb{N}$ is finite$\}$. Show that
B is a Boolean subalgebra of $\mathbb{P(N)}$ which cannot be Boolean isomorphic to
some $\mathbb{P(M)}$.

答: You should show yourself that it is a Boolean algebra. $B$ cannot be Boolean isomorphic to the power set of some set because the former is countably infinite and the latter is either finite of uncountable.


问:
Reducing this non-linear system
I'm trying to solve the following non-linear system:

$(A_x - P_x)^2 + (A_y - P_y)^2 + (A_z - P_z)^2 = (v(t_a - t_0))^2$ $\{1\}$

$(B_x - P_x)^2 + (B_y - P_y)^2 + (B_z - P_z)^2 = (v(t_b - t_0))^2$ $\{2\}$

$(C_x - P_x)^2 + (C_y - P_y)^2 + (C_z - P_z)^2 = (v(t_c - t_0))^2$ $\{3\}$

$(D_x - P_x)^2 + (D_y - P_y)^2 + (D_z - P_z)^2 = (v(t_d - t_0))^2$ $\{4\}$

Knowns:
$A_x, A_y, B_x, B_y, C_x, C_y, D_x, D_y, v, t_a, t_b, t_c, t_d$

Unknowns: $P_x, P_y, t_0$

The end goal is to determine $P_x, P_y$

I know that a slight variation of this system can be solved in the following manner.

$(A_x - P_x)^2 + (A_y - P_y)^2 = (v(t_a - t_0))^2$ $\{1\}$

$(B_x - P_x)^2 + (B_y - P_y)^2 = (v(t_b - t_0))^2$ $\{2\}$

$(C_x - P_x)^2 + (C_y - P_y)^2 = (v(t_c - t_0))^2$ $\{3\}$

Now, open the brackets, and form two new equations, {2'} = {3}-{2}, and {3'} = {1} - {3}:

$C_x^2-B_x^2 -2*(C_x-B_x)*P_x + C_y^2-B_y^2 -2*(C_y-B_y)*P_y = v^2*(t_c^2-t_b^2 -2*(t_c-t_b)*t_0)$

$A_x^2-C_x^2 -2*(A_x-C_x)*P_x + A_y^2-C_y^2 -2*(A_y-C_y)*P_y = v^2*(t_a^2-t_c^2 -2*(t_a-t_c)*t_0)$

These equations have the form:

$HP_x + JP_y = Kt_0 + L$

$TP_x + UP_y = Vt_0 + W$

Calculating $T(2') - H(3')$ and $U(2') - J(3')$ gives:

$P_x = \frac{T(Kt_0 + L) - H(Vt_o + W)}{TJ-HU}$, $P_y = \cdots$

Substitute these into {1} to obtain a quadratic equation in $t_0$. We then simple solve for the root of this quadratic, and plug into the formulae for $P_x$, $P_y$.

I'd like to follow a similar process for this system. I'll set aside {1} (in the first system), and form the linear system {2'} = {2}-{1}, {3'} = {3} - {1}, and {4'} = {4} - {1}. With these, I should be able to do the same as what I'd done above, obtaining an expression for $P_x, P_y, P_z$ in terms of $t_0$.

I then substitute these into {1} to obtain a quadratic in terms of $t_0$. I solve for $t_0$, then plug into the formulae to find $P_x, P_y, P_z$

My problem, and what I've tried:

I'm not sure what precisely this would look like. I've spent some time attempting to perform the manipulations by hand to obtain the expressions for $P_x, P_y, P_z$ and the quadratic in $t_0$, however I've failed. I've gotten several different answers, none of which look right to me.

Moreover, I've tried running this through a computer algebra system, and it choked up (unfortunately, I do not have access to something "more popular" like Mathematica). I tried running it on my school's CAS software, and it too failed. Even my maths professor has come up with multiple answers.

What should $P_x, P_y, P_z$ and the quadratic look like?

Note: I'm aware that I'll end up with two roots from the quadratic. In my case, this is ok, I want both.

答:
You should not compute anything directly with components. Instead, you should express the relation in terms of matrices. There are enough structure among your variables and finding the quadratic equation is more or less equivalent to inverting some $4 \times 4$ matrix.

Instead of points in $3$-d, think in terms of events.

For each event, say point $A = (A_x,A_y,A_z)$ at time $t_A$, we will associate a $4\times 1$ column vector $(A_x,A_y,A_z,v t_a)^T$ to it. Let

$u_1, u_2, u_3, u_4$ be the column vectors corresponding to $A, B, C, D$.

$p$ be the column vector $(P_x,P_y,P_z,vt)^T$.

$\eta$ be the $4 \times 4$ diagonal matrix with entries $(1,1,1,-1)$ along
its diagonal.

In terms of these, you equations becomes
$$(p - u_i)^T \eta (p - u_i) = 0
\iff 2 u_i^T \eta p = p^T\eta p + u_i^T\eta u_i
\quad\text{ for } 1 \le i \le 4
$$
Let $s = p^T\eta p$ and

$\Omega$ be the $4 \times 4$ matrix whose $i^{th}$ row is $2u_i^T$.

$U$ be the $4 \times 1$ column vector whose $i^{th}$ entry is $u_i^T\eta u_i$.

$\theta$ be the $4 \times 1$ column vector with all entries $1$.

In terms of these, your set of equations become

$$\Omega \eta p = s \theta + U
\implies p = \eta \Omega^{-1}(s \theta + U)\tag{*1}
$$
Substitute this back into definition of $s$, one obtain a quadratic equation in $s$.

$$s = p^T \eta p = (s\theta + U)^T\Omega^{-T}\eta \Omega^{-1}(s \theta + U)$$

Solving this give you two possible values of $s$. Plug that into $(*1)$ will give you the $p$ and hence $(P_x,P_y,P_z,t)$.

As one can see, if one doesn't insists writing down everything in terms of components, setting up the column vectors $u_i, U$ and matrices $\Omega$
is pretty straight forward. The only thing that is slightly more complicated is taking the inverse of the $4\times 4$ matrix $\Omega$.

Title: Reducing this non-linear system


问:
Kernel and image of a linear operator
Let $T:V \rightarrow V$ be a linear operator then $ker(T) \subseteq Im(T)$ if and only if $\forall v \in V$, if $ \ T(v) \in Ker(T)$ then $v \in Ker(T)$

I was trying this for the first part, let $v \in V$ such that $T(v) \in Ker(T) \subseteq Im(T)$ then $\exists w \in V$ such that $T(w) = T(v)$ as T is linear $T(w-v) =T(w)-T(v)=0$ then $w-v \in Ker(T)$ from here I don't know how to conclude that $v \in Ker(T)$

答:
Let $V=\mathbb{R}^2$.  Let $T$ be the operator represented by the matrix: $$\left(\begin{array}{cc}0&0\\ 1&0\end{array}\right)$$

Then $\ker(T)=\left<\left(\begin{array}{c}0\\ 1\end{array}\right)\right>={\rm Im}(T)$.

However $T\left(\begin{array}{c}1\\ 0\end{array}\right)=\left(\begin{array}{c}0\\ 1\end{array}\right)\in \ker(T)$, but $\left(\begin{array}{c}1\\ 0\end{array}\right)\notin \ker(T)$.

Now let $T$ be the operator represented by the matrix: $$\left(\begin{array}{cc}0&0\\ 0&1\end{array}\right)$$

Then $T^2=T$, so $T^2v=0 \implies Tv=0$.  That is $Tv\in\ker(T)\implies v\in \ker(T)$.

However $\left(\begin{array}{c}1\\ 0\end{array}\right)\in \ker(T)$, but $\left(\begin{array}{c}1\\ 0\end{array}\right)\notin {\rm Im}(T)$.

Thus neither implication is true.

Title: Kernel and image of a linear operator


问:
mathematical symbol, what does the open circle mean?
What does the circle in d) and e) means?

In this exercise, I have to find inverse matrix.
$h:\mathbb{R}^3 \to \mathbb{R}^3$ is given by $$h(x,y,z) = (4x+y,x+z,-2x+3y+z)$$

(d) $g \circ h$

(e) $h \circ g$

答:
It means composition,  for Instance $g(x)=x^2$ and $f(x)=-1/x$.
So $g(f(x)) = (-1/x)^2 = \frac{1}{x^2}$ and $f(g(x))= \frac{-1}{x^2}$


问:
What is the cokernel of this map $\bar{1} \mapsto \bar{13}$?
I am calculating the cokernel of this map: $f: \mathbb Z_9 \rightarrow \mathbb Z_{39}$ defined by $\bar{1} \mapsto \bar{13},$ I know that its image is $\{\bar{0}, \bar{13}, \bar{26}\} = 13\mathbb Z_{39} .$ Now, I know that the cokernel is defined by $\mathbb Z_{39}/\operatorname{Img} f = \mathbb Z_{39}/13\mathbb Z_{39}$ then by Lagrange theorem the number of cosets are 13. Is this correct?

I am not sure if my application of Lagrange theorem is correct: I considered my $G = \mathbb Z_{39}$ and so its order is $39$ and my $H = 13\mathbb Z_{39}$ and its order is $39/13,$ then the cokernel is isomorphic to $\mathbb Z_{13}$ is that correct?

答: It's correct.  Because $\Bbb Z_{39}/\langle 13\rangle\cong\Bbb Z_{13}$, not so much by Lagrange, but by, say, the first isomorphism theorem.  That is, it is the homomorphic image of the cyclic $\Bbb Z_{39}$ (so it's cyclic) and its order is $13$.


问:
Continuous function on the circle and real line
If $\varphi$ is a continuous function defined on the unit circle $T =\{z \in \mathbb{C} : |z| = 1 \}$, then $f (t) = \varphi(e^{it})$ is a continuous periodic function with period $2\pi$. Conversely, if $f$ is a continuous function on the real line of period $2\pi$, there exists a unique continuous function $\varphi$ defined on the unit circle $T$ such that $f (t) = \varphi(e^{it})$. Is my guess correct? Can anyone lead me a proof for this?</p>

答: Let $X$ be a topological space and $f\colon \mathbb R \to X$ be continuous and $2\pi$-periodic. Since $f$ is $2\pi$-periodic $\phi\colon T \to X$ is well-defined by $$\phi(e^{it}) = f(t).$$ On the other hand with the condition $$\forall t\in \mathbb R\colon f(t) = \phi(e^{it}),$$ there is only this possible definition, showing the uniqueness. Finally let us prove the continuity of $\phi$. Let $\tau\colon T \to [0,2\pi)$ be defined by $$\tau(e^{it}) = t$$ for an adequate $t\in [0,2\pi)$. Then $\tau$ is certainly continuous (but not homeomorphic). Continuity now follows from $\phi = f \circ \tau$.</p>


问:
How to find the number of positive devisors of $50,000$
How to find the number of positive devisors of $50,000$, I would like to know that what mathematical formulae  I need to use here as it is a big number to calculate mentally, I am sorry to ask if this is too silly question here to ask. Thank you. </p>

答:
Write the prime factorization
$$
n = p_1^{e_1} p_2^{e_2} \cdots p_r^{e_r}.
$$
Notice that there is a one-to-one correspondence between positive integer factors of $n$ and $r$-tuples of integers $(d_1, d_2, \dots, d_r)$ such that $0 \le d_i \le e_i$ for each $i$.  Explicitly,
$$
k = p_1^{d_1} p_2^{d_2} \cdots p_r^{d_r} \longmapsto (d_1, d_2, \dots, d_r).
$$
There are $e_i + 1$ possible powers of $p_i$ for each $i$.  Thus, the total number of divisors, $\sigma_0$ is given by
$$
\sigma_0(n) = (e_1 + 1)(e_2 + 1)\cdots(e_r + 1).
$$

For your example,
$$
50\,000 = 2^4 \cdot 5^5,
$$
so
$$
\sigma_0(50\,000) = (4 + 1)(5 + 1) = 30.
$$


问:
Using Cauchy criterion to prove that sequence $x_n = 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} +$...$+ (-1)^{n-1}\frac{1}{n}$ is convergent.
I'm stuck at $$|a_{n+p} - a_n| \leq \frac{1}{n+1} + \frac{1}{n+2} + ...+\frac{1}{n+p}$$,where $\forall p \in\mathbb{N}$. How can I find a N where $\forall n > N, |a_{n+p} - a_n| < \epsilon$ ?

答:
You need to modify the series expression with $b_n=a_{2n}-a_{2n+1}=\frac{1}{2n}-\frac{1}{2n+1}=\frac{1}{2n(2n+1)}$.

Use Cauchy criterion on $\sum b_n$.


问:
Total and partial derivatives
I have a function $Y(x,t)$ and another function $\eta(x,t)$.

I have the total derivative $\dfrac{dY}{d\eta}$.

So, what is the correct?

$$\dfrac{dY}{d\eta}\dfrac{\partial \eta}{\partial x}=\dfrac{\partial Y}{\partial x}\dfrac{\partial x}{\partial \eta}\dfrac{\partial \eta}{\partial x}+\dfrac{\partial Y}{\partial t}\dfrac{\partial t}{\partial \eta}\dfrac{\partial \eta}{\partial x}=\dfrac{2\partial Y}{\partial x}$$

or

$$\dfrac{dY}{d\eta}\dfrac{\partial \eta}{\partial x}=\dfrac{\partial Y}{\partial x}\dfrac{\partial x}{\partial \eta}\dfrac{\partial \eta}{\partial x}+\dfrac{\partial Y}{\partial t}\dfrac{\partial t}{\partial \eta}\dfrac{\partial \eta}{\partial x}=\dfrac{\partial Y}{\partial x}+\dfrac{\partial Y}{\partial t}\underbrace{\dfrac{\partial t}{\partial x}}_{0}=\dfrac{\partial Y}{\partial x}$$

?

Thank you so much.

答:
$$\frac{\mathrm{d}Y}{\mathrm{d}\eta}=\frac{\partial Y}{\partial x}\frac{\mathrm{d}x}{\mathrm{d}\eta}+\frac{\partial Y}{\partial t}\frac{\mathrm{d}t}{\mathrm{d}\eta}$$
Now,
$$\frac{\mathrm{d}\eta}{\mathrm{d}x}=\frac{\partial \eta}{\partial x}+\frac{\partial \eta}{\partial t}\frac{\mathrm{d}t}{\mathrm{d}x}$$
$$\frac{\mathrm{d}\eta}{\mathrm{d}t}=\frac{\partial \eta}{\partial x}\frac{\mathrm{d}x}{\mathrm{d}t}+\frac{\partial \eta}{\partial t}$$
Invert these and plug them into the above. This all tends to be much easier when we have some information about $Y,\eta$.

Title: Total and partial derivatives


问:
Showing $E(X + a) = E(X) + a$ for Non-Negative Random Variables
I have been working through the lecture notes given here. Specifically I have been looking at page 4

Theorem 1.2. part a) and part b)

Let $X$ and $Y$ be a non-negative random variable.

a) $X$ $\leq$ $Y$, then $E(X)$ $\leq$ $E(Y)$

b) Then for $a\geq 0$ $E(X+a) = E(X) +a$.

The proof given in the text says given $E(Z + a) = E(Z) + a$ for simple random variables $Z$.
We need to simply follow the proof of part a). However, I cannot make the jump from the proof in part a) to the proof in part b). The proof in part a) is as follows:

Let $\epsilon>0$. Since $X \geq 0$, there exists a simple random variable $Z$ s.t $Z \leq X$ and $E(Z) \geq E(X) - \epsilon$. Since $X \leq Y$, we have that $Z \leq Y$. Hence, $E(X) - \epsilon \leq E(Z) \leq E(Y)$. Taking $\epsilon$ to zero gives us the desired result.

Thank you.

答:
The notes do not say "Then".  As you say, the proof given of (b) starts with $E(a+Z) =a+EZ$ for simple $Z$ and then suggests repeating the steps for $X$ in the proof of part (a)

So possibly: given $\epsilon_1 > 0$ and since $X \geq 0$ and $X+a\ge 0$, there exists a simple random variable $Z_1$ such that  $Z_1 \leq X$ and $E(Z_1) \geq E(X) - \epsilon$ and a simple random variable equal to $(Z_2+a)$ satisfing $(Z_2+a) \leq (X+a)$ and $E(Z_2+a) \ge E(X+a) - \epsilon$.  Putting these together gives

and $$E(X+a) \geq E(Z_1+a)=E(Z_1)+a \geq E(X)+a - \epsilon$$ and
$$E(X)+a \geq E(Z_2)+a=E(Z_2+a) \geq E(X+a) - \epsilon$$

Sending $\epsilon$  towards zero gives $E(X)+a = E(X+a)$.


问:
Homomorphisms of a Finite Group & Perfect Squares
I'm working on a textbook problem:

Come up with a finite group $G$ and homomorphism: $f: G \rightarrow G$ such that $3$% of the elements of in $G$ satisfy $f(x) = x^2$ (where $x$ is an element in $G$).

So far, I've tried a lot of different ideas. When I made the binary operator for $G$ additive, I had a tough time since the homomorphism must follow a lot of conditions such as $f(2) = 2f(1)$, not leaving a lot of space for me to try and come up with a way to get $f(x) = x^2.$

I tried making the group a power group and multiplicative. In both scenarios, I didn't really get anything too useful – every idea that I tried ultimately failed.

Does anyone know how to do this?

答:
Here is a solution where I explain all the "false starts" I tried along the way. Let's start by looking at a very simple example, namely the multiplication-by-$m$ map $f(x) = mx$ on the cyclic group $C_n \cong \mathbb{Z}/n$ (written additively). The subset of elements satisfying $f(x) = mx = 2x$ is the subset of elements satisfying $(m-2) x = 0$, which has size $\gcd(m-2, n)$ (exercise). So the density of this subset is $\frac{\gcd(m-2, n)}{n}$, which unfortunately as a fraction in lowest terms always has numerator $1$ and so cannot be equal to $\frac{3}{100}$. So no example of this form can work.

The next group I tried was $\mathbb{Z}/n \times \mathbb{Z}/n$, but I ran into the following problem generalizing the previous observation: if $A$ is any abelian group and $f : A \to A$ is a homomorphism then the subset of elements satisfying $f(a) = 2a$ is $\text{ker}(f - 2)$ which is a subgroup, so its density always has numerator $1$ when written as a fraction in lowest terms (with denominator the index). This means an example must be nonabelian.

However, abelian examples do help us for the following reason: if $f_1 : G_1 \to G_1$ is an endomorphism of a group and $f_2 : G_2 \to G_2$ is an endomorphism of another group, we can take their direct product to get an endomorphism $f_1 \times f_2 : G_1 \times G_2 \to G_1 \times G_2$. If we write $d(G, f)$ for the density of elements satisfying $f(g) = g^2$, then

$$d(G_1 \times G_2, f_1 \times f_2) = d(G_1, f_1) d(G_2, f_2).$$

So when we're looking for nonabelian examples it suffices to find a nonabelian group with density of the form $\frac{3}{n}$ where $n \mid 100$, and then we can take the direct product with a suitable abelian group to get density $\frac{3}{100}$.

If $f : G \to G$ is a complicated endomorphism of a complicated nonabelian group then it could be quite hard to count the number of elements satisfying $f(g) = g^2$, so for starters let's see what happens if we just set $f(g) = e$ to be the zero endomorphism. Then we're counting the elements satisfying $g^2 = e$, which when $G$ is nonabelian need not be a subgroup. For example, $G = S_3$ has $4$ such elements, with density $\frac{4}{6} = \frac{2}{3}$, which does not have numerator $1$, so this is promising.

Let's see if we can get density $\frac{3}{4}$. The smallest possible example with this property would be a group of order $8$, so let's consider the dihedral group $D_4$. The dihedral group consists of a copy of $C_4$, of which $2$ elements square to the identity, plus $4$ reflections, all of which square to the identity, so the density is $\frac{6}{8} = \frac{3}{4}$ as desired.

So we're done. We can take our final group $G$ to be $\boxed{ D_4 \times C_{25} }$ and our final endomorphism to be $\boxed{ f(g, h) = (e, h) }$.


问:
Prove if $I$ is an ideal of $R$ ring, then $(I \cup \{x\})/(x) \cong I$.
I am trying to prove the statement above where $(x)$ is the principal ideal generated by the polynom $x$ in $R[x]$, and similarly with $(I \cup \{x\})$. Additionally, I have to prove that $R[x]/(I \cup \{x\}) \cong R/I$ with the help of this statement. For this part, I know that $R[x]/(x) \cong R$ and I should use the 3rd (or maybe 2nd, I don't know which one it is in English) isomorphism theorem, but I'm clueless with the initial statement. I appreciate any sort of help.

答: Hint: Use the first isomorphism theorem for the natural map $R[x]\to R/I$ sending $x\mapsto 0$.


问:
CDF of $\frac{T_1}{T_1+T_2}$ where $T_1,T_2$ are the first two delays in a Poisson process?
$P$ is a Poisson Process with rate $\lambda$. Let $T_1$ be the time of the first event and let $T_2$ be the time of the from the first to the second event. Let $Y = \frac{T_1}{T_1+T_2}$. Find the density of $Y$.

I think I should find the CDF first and then take the derivative, but I do not know how to find $P(Y\leq t)$. Do I need to find the joint density?

答:
Hints:

$$P(Y \le y) = P(T_1 \le y(T_1 + T_2)) = P(\frac{1-y}{y}T_1 \le  T_2).$$

Show that $T_1$ and $T_2$ are independent $\text{Exponential}(\lambda)$ random variables using properties of Poisson processes.

Then, for $y \in (0,1)$,
$$P(Y \le y) = \int_0^\infty \int_{\frac{1-y}{y} t_1}^\infty \lambda^2 e^{-\lambda t_1} e^{-\lambda t_2} \, dt_2 \, dt_1 = \int_0^\infty \lambda e^{-\lambda t_1} e^{-\lambda\frac{1-y}{y} t_1} = y$$


问:
Show using induction (coupled linear recurrences)
Some homework help would be greatly appreciated, took a screenshot and made an image to make it easier to show and get help with.

(2) Consider the numbers defined recursively by $a_1=3$, $c_1=5$, $$a_{n+1}=3a_n+2c_n+1\text{ and }c_{n+1}=4a_n+3c_n+2\;.$$ Show using induction that for all $n\in\Bbb Z^+$, $$a_n^2+(a_n+1)^2=c_n^2\;.$$

答:
Let $P(n)$ be the statement that $a_n^2+(a_n+1)^2=c_n^2$; you want to prove that $P(n)$ is true for all $n\ge 1$. You can easily check that $P(1)$ is true: that’s just arithmetic. For the induction step you need to let $n\ge 1$ be arbitrary, assume that $P(n)$ is true, and somehow use that assumption to prove that $P(n+1)$ must then also be true. In other words, your induction hypothesis will be that

$$a_n^2+(a_n+1)^2=c_n^2\;,\tag{1}$$

where $n$ is some unspecified positive integer, and from this you’ll try to prove that

$$a_{n+1}^2+(a_{n+1}+1)^2=c_{n+1}^2\;.$$

You’ll do this by using the recurrences to expand $a_{n+1}$ and $c_{n+1}$ in terms of $a_n$ and $c_n$ and using the relation ship between $a_n$ and $c_n$ assumed in $(1)$.


答:
HINT For convenience, let $T=T_1$ and $S=T_2$. Note they are iid, so if $f_T(t)$ is the pdf of $T$ then $f_T(s) = f_S(s)$ and $f_{T,S}(t,s) = f_T(t)f_S(s) = f_T(t)f_T(s)$.

Once you know the joint pdf,
$$
F_Y(y) = \mathbb{P}[Y \le y]
$$
reduces to a regular 2D integral.

Title: CDF of $\frac{T_1}{T_1+T_2}$ where $T_1,T_2$ are the first two delays in a Poisson process?

Body:
$P$ is a Poisson Process with rate $\lambda$. Let $T_1$ be the time of the first event and let $T_2$ be the time of the from the first to the second event. Let $Y = \frac{T_1}{T_1+T_2}$. Find the density of $Y$.

I think I should find the CDF first and then take the derivative, but I do not know how to find $P(Y\leq t)$. Do I need to find the joint density?


问:
Two independent Poisson Processes
$X$ and $Y$ are two independent Poisson Processes with rates $\lambda_1, \lambda_2$ respectively.  What is the probability that at least two events from $X$ occur before a total of two events from $Y$? A valid example would be $X_1, Y_1, X_2, Y_2...$ since two events from $X$ occurred before a total of 2 events from $Y$.

I think I use complimentary counting to solve this problem. The invalid cases I've noted are $X_1, Y_1, Y_2...$; $Y_2, Y_2...$; $Y_1, X_1, Y_2...$ Am I on the right track or is there a more efficient/cleaner method?

答:
HINT

Derive the distribution of $T_2$, the second event occurrence from a Poisson process and you have to find $\mathbb{P}[T_2<S_2]$ where $T_2$ is 2nd event from the first and $S_2$ is 2nd event from the 2nd process.

Title: Two independent Poisson Processes


问:
Covariance of sum random process
Sum random process is defined as:
$$
X[n] = \sum_{i=0}^n U[i] \text{ for } n \ge 0
$$
$$
E[U[i]]=0 \text{, } \text{var}(U[i]) = \sigma^2_U \text{ for } i \ge 0 \text{ and } U[i] \text{ are IID}
$$
This tells us that $E[X_n]=0$.

Show that $c_x[n_1, n_2] = \sigma^2\text{min}(n_1, n_2)$

答:
The covariance is $E(X[n_1]E[n_2])=\sum_{i=0}^{n_1}\sum_{j=0}^{n_2}E(U(i)U(j))$.

For $j\ne i)$, $E(U(i)U(j))=0$.

The only terms remaining are $E(U(i)^2)=\sigma^2$ for $i\le min(n_1,n_2)$.

The answer is slightly incorrect- should be $\sigma^2 min(n_1+1,n_2+1)$.

Title: Covariance of sum random process


问:
Construction of type III von Neumann algebras
I saw a conclusion: type III von Neumann algebras can be obtained by means of generalized Hilbert algebras and weights.

Can anyone show me the concrete conatructions or recommending some reference books on type III von Neumann algebras.

答: See chapter XII in Takesaki, Masamichi, Theory of operator algebras. II, Encyclopaedia of Mathematical Sciences 125. Operator Algebras and Non-commutative Geometry 6. Berlin: Springer (ISBN 3-540-42914-X/hbk). xxii, 518 p. (2003). ZBL1059.46031.


答:
Hints:

Calculate the probabilities that the first event is from $X$, and that the first event is from $Y$

You only care about the first two or three events

Use the memoryless property of Poisson processes to find the probabilities of

An event from $X$ followed by another event from $X$ without any events from $Y$

An event from $X$ followed by an event from $Y$ followed by another event from $X$ without more events from $Y$

An event from $Y$ followed by an event from $X$ followed by another event from $X$ without more events from $Y$

Add up those probabilities

Title: Two independent Poisson Processes

Body:
$X$ and $Y$ are two independent Poisson Processes with rates $\lambda_1, \lambda_2$ respectively.  What is the probability that at least two events from $X$ occur before a total of two events from $Y$? A valid example would be $X_1, Y_1, X_2, Y_2...$ since two events from $X$ occurred before a total of 2 events from $Y$.

I think I use complimentary counting to solve this problem. The invalid cases I've noted are $X_1, Y_1, Y_2...$; $Y_2, Y_2...$; $Y_1, X_1, Y_2...$ Am I on the right track or is there a more efficient/cleaner method?


问:
Prove that $\rho$ is irreducible iff $Mρ(x)=ρ(x)M$ implies $M = \lambda E$

Let $\rho \colon G \rightarrow \operatorname{GL}(n; \mathbb{C})$ be a representation of a finite group $G$. Prove that $\rho$ is irreducible exactly when each matrix $M \in \operatorname{Mat}(n;\mathbb{C})$ with $Mρ(x)=ρ(x)M$, for each $x \in G$, is of the shape $λE$ for a scalar $λ \in \mathbb{C}$.

Can anyone give me some help?

答:
Let $\lambda$ be an eigenvalue of $M$.  Then $M-\lambda E$ has non-trivial kernel $N\subseteq \mathbb{C}^n$.

Then $N$ will be a nontrivial subrepresentation of $\rho$.  As $\rho$ irreducible, $N=\mathbb{C}^n$.  That is $M-\lambda E=0$, so $M=\lambda E$


答:
You are on the right track. Considering the combined process as you are doing, you only need to look at the first three arrivals. There are eight possibilities (XXX, XXY, XYX, XYY, YXX, YXY, YYX, YYY), and you can figure out which ones to count. (I think counting the complement is no easier/harder than counting what you actually want, but you have the right idea already.)

Now comes the question of how to compute the probabilities associated with the above eight outcomes. It turns out that the outcomes have the same distribution as three independent coin flips: the probability that an arrival is from $X$ is $\frac{\lambda_1}{\lambda_1+\lambda_2}$, and the probability that an arrival is from $Y$ is $\frac{\lambda_2}{\lambda_1+\lambda_2}$. Summing up the probabilities of the "good" outcomes will give you the answer.

See this page for an explanation of why the arrivals in the combined process behave like coin flips with the above probabilities.

Title: Two independent Poisson Processes


问:
Subspace Topology and Limit Points
Trying to show that for every positive integers $i,j$ there is a set $A_{i,j}\subseteq[0,1]$ ($[0,1]$ has the Subspace Topology from $\mathbb{R}_\text{std.}$) such that $|A^{(i)}|=j$ and $|A^{(i+1)}|=\emptyset$, where $A^{(m)}$ denotes the $m$th order limit points. I would show my attempted solution(s), however, I really have no idea how to start such a problem. The $m$th order limit points means: $A^{(1)} = A'$, $A^{(2)} = (A')'$, and so on.

答:
SKETCH: Let $A_{1,1}=\{0\}\cup\left\{\frac1{2^n}:n\in\Bbb Z^+\right\}$. To get $A_{2,1}$, we add a sequence converging downward to each $\frac1{2^n}$. Show that this can be done by adding the points $\frac1{2^n}+\frac1{2^{n+m}}$ for $n,m\in\Bbb Z^+$. For $A_{3,1}$  add points $\frac1{2^n}+\frac1{2^{n+m}}+\frac1{2^{n+m+k}}$ for $n,m,k\in\Bbb Z^+$ and show that this works. At this point the general construction should be clear, as well as how to see that it works.

It may be helpful to think in binary. Apart from $0$, the points of $A_{1,1}$ are $0.1_2$, $0.01_2$, $0.001_2$, and in general all of the numbers in $[0,1)$ whose binary expansions contain a single $1$. The new points in $A_{2,1}$ are those numbers in $[0,1)$ whose binary expansions contain exactly two $1$s, and in general the points of $A_{n+1,1}\setminus A_{n,1}$ are those numbers in $[0,1)$ whose binary expansions contain exactly $n+1$ $1$s.

Once you have all of the sets $A_{n,1}$, you can get $A_{n,m}$ by starting with $$\bigcup_{k=0}^{m-1}(A_{n,1}+k)\,,$$ where $A+k=\{a+k:a\in A\}$, and multiplying it by $\frac1m$.

Title: Subspace Topology and Limit Points


答:
nice question! Here is an alternative approach if you have studied transformations of random variables:

Let $X = T_1 + T_2 $ and $Y = \frac{T_1}{T_1 + T_2} $

We have that $(X,Y) =  (T_1+T_2 ,  \frac{T_1}{T_1 + T_2}) := g(T_1,T_2)$

And define $h(X,Y):=g^{-1}(X,Y) = (XY,(1-Y)X)$

The Jacobian of $h$ is the determinant of \begin{pmatrix}
x & y \\
-x & 1-y
\end{pmatrix}
Which is $x$

And hence you can easily find the joint density of $(X,Y)$ and then integrate out $x$ to find the pdf of $Y$


问:
Find area of right triangle △ with circumcircle $O$ and tangent circle $K$.
The triangle $ABC$, with $\angle A = 90^\circ$,
is inscribed in the circle $\mathcal{O}$.
The circle $\mathcal{K}$ is tangent to the circle $\mathcal{O}$,
the sides $AB$ and $AC$.
Suppose that the radius of $\mathcal{O}$ is $3$
and the radius of $\mathcal{K}$ is $2$. Find the area of $\triangle ABC$.

I drew this problem out by making circle $\mathcal{K}$
internally tangent to circle $\mathcal{O}$.
I was able to make it tangent to $AB$ and $AC$.
However, I have gotten nowhere using trigonometry to solve this.

I also tried to name the sides and angles
to find the area, which did not work, either

答:
One possible solution is to assume a special case for which the second circle information is superfluous.

For example, inscribe a square into circle O with a corner at point A. Then, per a source, to quote:

We can show this using a symmetry argument – the square is symmetrical across its diagonal, so the diagonal must pass through the center of the circle.

The diagonal line also creates two right triangles, one at Point A. So, with a hypotenuse of 6 (as the radius is given as 3), the sides of the triangle (and square) are ${3\sqrt{2}}$.

This implies an area of the square of 18 sq units and for the triangle, as one possible answer for the question, 9 sq units.


问:
Can all linear operators be represented as matrices? (In infinite dimension.)
Old title: Components of unbounded endomorphism wrt Schauder basis? Changed for discoverarbility.

In finite-dimensional linear algebra, given an endomorphism $T$ and a basis $\boldsymbol{e}_i$, we can easily find a matrix representation of $T$:
$$
 T(v)
 = T(\sum_i v^i \, \boldsymbol{e}_i)
 \overset{*}{=} \sum_i T(v^i \, \boldsymbol{e}_i)
 = \sum_i v^i T(\boldsymbol{e}_i)
 =: \sum_{i,j} v^i \, T_{ij} \, \boldsymbol{e}_j
$$

Let us now assume that $X$ is a separable Banach space with a Schauder basis $\{\boldsymbol{e_i}\}_{i \in \mathbb{N}}$ and $T: X \to X$ is an unbounded linear operator. If we try to replicate the previous equation, the sum will become infinite, and since for unbounded operators $T(\lim_n v_n) = \lim_n T(v_n)$ doesn't hold, the equality $(\overset{*}{=})$ above doesn't hold.

This would seem to indicate that unbounded operators don't have a coordinate representation in a Schauder basis, at least not in the same sense as bounded operators do. Or is there a simple generalization that I just don't see?

If we now consider $X$ to be a separable Hilbert space and $\boldsymbol{e}_i$ an orthonormal basis, we know that it's isomorphic to $\ell_2$ with basis:
$$
  \begin{pmatrix}1\\0\\0\\\vdots\end{pmatrix},
  \begin{pmatrix}0\\1\\0\\\vdots\end{pmatrix},
  \begin{pmatrix}0\\0\\1\\\vdots\end{pmatrix},
  \;\dots
$$
This means that either the endomorphisms on the space of „infinite column vectors“ aren't always „infinite matrices“, or that my previous conclusion about the „non-representability“ of unbounded operators was wrong. Which of these is true?

In summary, these are my questions:

Is it possible to represent an unbounded operator using its coordinates wrt. a Schauder basis of a separable Banach/Hilbert space?

If yes, how? If no, are there other practical ways to represent it? (Eg. using Hamel basis? Or as an integral kernel on $L^p$?)

Are all operators on $\ell_2$ „infinite matrices“?

答:
Claim: Given a Schauder basis $\{ \boldsymbol{e}_j \} \subset X$ and a linear operator $T: \mathrm{D}(T) \subset X \to X$, the vectors $T(\boldsymbol{e}_j)$ aren't always sufficient to describe the action of the operator on a general vector of $X$.

Or in other words: There exists a non-zero linear operator, whose matrix is zero.

Proof: Let $X = L^2([0, 2\pi])$  be the space of all square-integrable functions equipped with the Fourier basis $\{ \boldsymbol{e}_j \} = (1,\, \sin x,\, \cos x,\, \sin 2x,\, ...)$. We now use the distribution theory to construct an unbounded linear operator $T$.

Let us take $\mathrm{D}(T)$ to be the space of bounded piece-wise continuous functions. Since $\mathrm{D}(T) \subset L^1_{\mathrm{loc.}}(\mathbb{R})$, all vectors $f \in X$ can be considered regular distributions on $\mathcal{D}(\mathbb{R})$ and assigned a distributional derivative $Df$. The result will be a sum of a regular function and (singular) delta distributions in discontinuities: $Df = (Df)_{\mathrm{reg.}} + (Df)_{\mathrm{sing.}}$. Finally we chose a test function $\varphi \in \mathcal{D}(\mathbb{R})$ that is nonzero on $[0, 2\pi]$ (ie. $\varphi(x)=1$ on $[0, 2\pi]$). We define:
$$
  T(f) = (Df)_{\mathrm{sing.}}[\varphi] \; \boldsymbol{e}_1
$$
This operator is zero for all continuous functions, including the entire basis $\{ \boldsymbol{e}_j \}$, but it is generally nonzero for discontinuous functions.

Corollary: Since, given a Schauder basis, there's a natural isomorphism $\iota: L^2 \to \ell_2$:
$$
  \underbrace{f}_{L^2} = \sum_j \underbrace{c_j}_{\ell_2} \boldsymbol{e}_j
  \: , \quad
  \iota(f) = \{ c_j \}
$$
we can construct a non-zero operator on $\ell_2$ whose matrix is only zeros:
$$
  S(a_j) = \iota \, T \, \iota^{-1} \; a_j
  \quad \neq \quad
  \sum_{j,k} S_{jk} \, a_j \, \boldsymbol{e}_k = 0
$$

Claim: Let $X$ be a separable Banach space and $T: \mathrm{D}(T) \subset X \to X$ a densely defined closed operator. Given a Schauder basis $\{ \boldsymbol{e}_j \} \subset \mathrm{D}(T)$ (such a basis always exists), we construct the coordinates $T_{jk}$ of the operator $T$:
$$
 T(\boldsymbol{e}_j) =: \sum_k T_{jk} \, \boldsymbol{e}_k
$$
Now we define an operator $T_0: \mathrm{D}(T_0) \to X$ using those coordinates:
$$
 T_0(v) = T_0( \, \sum_j v^j \boldsymbol{e}_j \, ) = \sum_{j,k} T_{jk} \, v^j \, \boldsymbol{e}_k
 \: , \qquad
 \mathrm{D}(T_0) = \{ v \in X \,|\, \text{the sum converges}\}
$$
Then $T_0$ is also closed and $T_0 \subset T$.

Proof: The Schauder basis in question exists, because $\mathrm{D}(T)$ is dense in $X$ – therefore it is a generating set in the sense of countable linear combinations. We can throw away its linearly dependent vectors until we're left with a basis.

The defining feature of closed operators is:
$$
  T \text{ is closed} \;\wedge\;
  x_n \to x  \;\wedge\;
  T(x_n) \to y
  \quad \implies \quad
  x \in \mathrm{D}(A) \;\wedge\;
  T(x) = y
$$
To show that $T_0 \subset T$, we chose an arbitrary $v \in \mathrm{D}(T)$ and define:
$$
  x_n = \sum_{j=1}^n v^j \, \boldsymbol{e}_j
  \: , \quad
  x_n \to v
$$
Then:
$$
  T(x_n)
  = \sum_{j=1}^n v^j \, T(\boldsymbol{e}_j)
  = \sum_{j=1}^n \sum_k T_{jk} \, v^j \, \boldsymbol{e}_k
$$
If that sum converges for $n\to\infty$, by the definition of $T_0$, it holds that $v \in \mathrm{D}(T_0)$ and the result is equal to $T_0(v)$. However, from the closedness criterion, we also know that $v\in\mathrm{D}(T)$ and the result is equal to $T(v)$. If the sum doesn't converge, by definition $v\notin\mathrm{D}(T_0)$, but it doesn't tell us anything about $T$. Thus, $T_0 \subset T$.

It remains to prove that $T_0$ is also closed. We assume $① \Leftrightarrow x_n \to x$ and $② \Leftrightarrow T_0(x_n) \to y$ and want to prove $T_0(x) = y$. If we expand the expressions, we get:
$$
  ① \; \implies \;
  \sum_{j=1}^m (x_n)^j \, \boldsymbol{e}_j \xrightarrow{n}
  \sum_{j=1}^m x^j \, \boldsymbol{e}_j
  \\[8pt]
  ② \; \Longleftrightarrow \;
  \sum_{j,k} T_{jk} \, (x_n)^j \, \boldsymbol{e}_k \xrightarrow{n} y
$$
To simplify the matters, we introduce a double-sequence $\Phi_{mn}$:
$$
  \Phi_{mn} = \sum_{j=1}^m \sum_k T_{jk} \, (x_n)^j \, \boldsymbol{e}_k
$$
The assumption $②$ is equivalent to $\lim_{n\to\infty} \lim_{m\to\infty} \Phi_{mn} = y$. The assumption $①$ implies that $\lim_{n \to \infty} \Phi_{mn}$ exists – as it turns out, the convergence is even uniform wrt. $m$ (is it really tho?). Then it follows from the Moore-Osgood Theorem, that $\lim_{m\to\infty} \lim_{n\to\infty} \Phi_{mn}$ exists and is equal to $y$, too.


问:
Show that the group ordering in the direct product is doesn't matter to the group structure.
Show that the group ordering in the direct product is doesn't matter to the group structure.

For instance, for $A,B,C \ne \emptyset$, $A×B×C$ and $A×C×B$ is same (That is, the group structure of $A×B×C$ is same with $A×C×B$).

Any idea?

答:
You could use group presentations.

If $X\cong\langle G_X\mid R_X\rangle$, then, by this result, we have, assuming without loss of generality the presentations are given by disjoint symbols, that

$$\begin{align}
A\times B\times C&\cong\langle G_A\cup G_B\cup G_C\mid R_A\cup R_B\cup R_C\cup\{ xy=yx, xz=zx, yz=zy\mid x\in G_A, y\in G_B, z\in G_C\}\rangle\\
& \cong\langle G_A\cup G_C\cup G_B\mid R_A\cup R_C\cup R_B\cup\{ xy=yx, xz=zx, yz=zy\mid x\in G_A, z\in G_C, y\in G_B\}\rangle\\
&\cong A\times C\times B.
\end{align}$$


Body:
\begin{align}
& \Pr(Y\le t) = \Pr( T_1 \le t(T_1+T_2)) = \Pr((1-t)T_1\le tT_2) \\[8pt]
= {} & \iint\limits_{(u,v)\,:\,(1-t)u\,\le\,tv} e^{-u} e^{-v} \, d(u,v) \\[8pt]
= {} & \int_0^\infty \left( \int_0^{tv/(1-t)} e^{-u} e^{-v} \, du \right) \, dv \\[8pt]
= {} & \int_0^\infty \big( 1 - e^{-tv/(1-t)} \big) e^{-v} \, dv \\[8pt]
= {} & 1 - (1-t) = t \quad \text{provided $0<t<1$}. \\
& \text{So the distribution is uniform on $[0,1]$.}
\end{align}


问:
Principal value of integral
How to find the Cauchy principal value of the integral $$\int_0^\infty \left(\frac{1}{x^2}-\frac{\cot(x)}{x} \right) dx?$$</p>

答:
A few preliminary notes

$\displaystyle f(z)=\frac1{z^2}-\frac{\cot(z)}{z}$ has a removable singuarity at $0$.

the other singularities of $f$ have residue $-\dfrac1{n\pi}$ at $z=n\pi$.

$\left\{\dfrac x\pi\right\}=\dfrac12\implies\cot(z)=i\tanh(y)$

$|\cot(z)+i\,|\sim2e^{-2|y|}$ as $y\to\infty$.

We will integrate along the following contours:

$\hspace{7mm}$

Where the blue segments are along the lines $x=\pm\left(k+\frac12\right)\pi$ and the green segment is along $y=\sqrt{k}$ as $k\to\infty$.

Since the contour contains no singularities, we have
$$
\begin{align}
0=
&\mathrm{PV}\int_{-\infty}^\infty\left(\frac1{x^2}-\frac{\cot(x)}{x}\right)\,\mathrm{d}x
+\color{#C00000}{\int_{\large\gamma_s}\left(\frac1{z^2}-\frac{\cot(z)}{z}\right)\,\mathrm{d}z}\\
&+\color{#0000FF}{\int_{\large\gamma_v}\left(\frac1{z^2}-\frac{\cot(z)}{z}\right)\,\mathrm{d}z}
+\color{#00A000}{\int_{\large\gamma_h}\left(\frac1{z^2}-\frac{\cot(z)}{z}\right)\,\mathrm{d}z}
\end{align}
$$
Due to 2., the integral along $\color{#C00000}{\gamma_s}$ tends to $0$; that is, the residue at $k\pi$ cancels the residue at $-k\pi$.

Due to 3., the integral along $\color{#0000FF}{\gamma_v}$ tends to $0$; that is, $|\cot(z)|\le1$ on those segments.

Due to 4., the integral along $\color{#00A000}{\gamma_h}$ tends to $-\pi$; that is, $\cot(z)\to-i$.

Putting this all together, we get
$$
0=
\mathrm{PV}\int_{-\infty}^\infty\left(\frac1{x^2}-\frac{\cot(x)}{x}\right)\,\mathrm{d}x
+\color{#C00000}{0}+\color{#0000FF}{0}+\color{#00A000}{-\pi}
$$
Therefore,
$$
\mathrm{PV}\int_0^\infty\left(\frac1{x^2}-\frac{\cot(x)}{x}\right)\,\mathrm{d}x=\frac\pi2
$$

Title: Principal value of integral


问:
1-manifold not locally homeomorphic to a 2-manifold

Prove that a $1$-manifold ($1$ dimensional manifold) cannot be locally
homeomorphic to a $2$-manifold.

A $k$-manifold is a topological space $(X,\tau)$ which is Hausdorff, second countable and for all $x\in X$ exists a neighbourhood $N$ of $x$ which is homeomorphic to $\mathbb{R}^k$.

My attempt:

Let $(X,\tau_X)$ be a $1$-manifold, and $(Y,\tau_Y)$ a $2$-manifold. Suppose that $X$ is locally homeomorphic to $Y$. Then there is a function $f\colon X \to Y$ such that
$$ \forall x\in X \ \ \ \exists x\in U_x \in \tau_X \colon \ \ f(U_x)\in \tau_Y \text{ and } f|_{U_x}\colon U_x \to f(U_x) \text{ is a homeomorphism.} $$
Take an arbitrary point $x\in X$. Then for the above $U_x \cong f(U_x)$.
But since $(X,\tau_X)$ is a $1$-manifold, $U_x \cong \mathbb{R}$. And since $(Y,\tau_Y)$ is a $2$-manifold, $f(U_x) \cong \mathbb{R}^2$, which leads to $\mathbb{R} \cong \mathbb{R}^2$, a contradiction.

Can you verify my proof? Can you correct me if I'm wrong? Thank you in advance!

答:
Let $A$ be a $1$-manifold and $B$ be a $2$-manifold. To show that $A$ cannot be locally homeomorphic to $B$, you can show this:

if $p \in A$, and if $U$ is a little enough connected neighbourhood of $p$, then $U\setminus \{p\}$ has at least two connected components

if $q \in B$, and if $V$ is a connected neighbourhood of $q$, then $V\setminus\{q\}$ is connected

Now, if $f : A \to B$ is a local homeomorphism around $p \in A$, and if $U$ is a connected neighbourhood of $p$, then $f(U)$ is a connected neighbourhood of $f(p)$. Moreover, $U \setminus \{p\}$ is homeomorphic to $f(U\setminus \{p\}) = f(U) \setminus\{f(p)\}$ and $U\setminus \{p\}$ has to be connected because of $2.$, which can be easily contradicted thanks to $1.$


问:
Prove that if $G$ is finite and $G$ has composition series $1=N_0\lhd...\lhd N_r=G$ and $1=M_0\lhd M_1\lhd M_2=G$ then $r=2$.

Prove that if $G$ is finite and $G$ has composition series $1=N_0\lhd...\lhd N_r=G$ and $1=M_0\lhd M_1\lhd M_2=G$ then $r=2.$

So if $r=1$ we would have $G$ is simple since $G/1$ is simple, which contradicts that $M_1\lhd G$.

So assume $r>2$ then there must exists some $N_j$ in the first composition series such that $\vert N_j\vert<\vert M_1\vert$ or $\vert N_j\vert>\vert M_1\vert$

Then by second isomorphism theorem we know that $N_j\cap M_1$ is normal in both $M_1$ and $N_j$. Which by correspondence theorem gives me $N_j\cap M_1/M_0\lhd M_1/M_0$. And $N_j\cap M_1/N_{j-1}\lhd N_j/N_{j-1}$

What I need to know is that $N_j\cap M_1$ isn't trivial so that one of these composition factors isn't simple. Which I believe if $N_j\cap M_1/M_0$ is a trivial normal subgroup then either $N_j\cap M_1=M_1$ or $N_j\cap M_1=M_0=1$, but then $N_j\cap M_1/N_{j-1}$ will only be trivial if $N_j\cap M_1=N_{j-1}$ or $N_j$, so this is a contradiction that the $N_i$ were a composition series.

答:
You are almost there. Let’s be a bit more precise in choosing $i$.

(I will assume that you have proper inclusion in your series.)

Note that $1=N_0\cap M_1\unlhd N_1\cap M_1\unlhd N_2\cap M_1\unlhd\cdots \unlhd N_r\cap M_1 = M_1$. Because $M_1$ is simple (since $1=M_0\triangleleft M_1 \triangleleft M_2=G$ is a composition series), in this subnormal series we have that there exists $i$, $0\leq i\lt r$ such that $N_i\cap M_1=\{e\}$, and $N_{i+1}\cap M_1 = M_1$. So $N_i$ is the “last” normal subgroup that has trivial intersection with $N_1$; and $M_1$ is therefore contained in $N_{i+1}$.

Then $\frac{N_{i+1}}{M_1}\triangleleft \frac{M_2}{M_1}=\frac{G}{M_1}$. But since $M_1/M_1$ is simple, either $N_{i+1}=M_1$, or else $N_{i+1}=G$.

I’ll leave it to you to complete the argument that if $N_{i+1}=M_1$ then we must have $i=0$ and $r=2$.

If $N_{i+1}=G$, then $i=r-1$. Now, because $M_1\cap N_j=\{e\}$ for $j=0,\ldots,r-1$, then $M_1N_j$ properly contains $M_1$ for $j=1,\ldots,r-1$. Also, $M_1N_j\unlhd M_1N_{j+1}$, since $N_j\triangleleft N_{j+1}$ and $M_1$ is normal in $G$. So we have
$$M_1=M_1N_0 \lhd M_1N_1\unlhd\cdots\unlhd M_1N_{r-1}\triangleleft M_1N_r=G.$$
But $G/M_1$ is simple. So...

By the way, some of these ideas are behind Schreier’s Refinement Theorem, and the Jordan-Holder theorem; no doubt D&F are noddding in that direction.


问:
Proving x and y is divisible by p (prime).
If p is a prime number and x and y are integers, how do I prove "if xy and x+y are both divisible by p, then x and y is divisible by p"?

I started like this..

1) p divides xy, so p divides x or p divides y, since p is a prime number.

and then I'm already stuck :/

Help me to get to the next step please.

答:
Hints:

$$(1)\;\;\;p\,\mid\,xy\implies\;p\,\mid\,x\;\;\vee\;\;p\,\mid\,y$$

Now suppose WLOG that $\,p\mid x\,$, and also we have $\,x+y=tp\,$ , so

$$(2)\;\;\;x=kp\implies tp=x+y=kp+y\implies\ldots\ldots$$


问:
Stuck within Induction
I am taking a course in Analysis and have to prove that
$$
\sum_{j=0}^n j^3 = \frac{n^4}{4} + \frac{n^3}{2} + \frac{n^2}{4}
$$

I think until I had $n^4/4 + n^3/2 + n^2/4 + (n + 1)^3$ I did not make a mistake.

From there I seem to always make the same mistake and end up with
$$
\frac{n^4 + 6n^3 + 13n^2 + 12n + 4}{4}
$$

Any suggestions, hints, solutions?

Title: Stuck within Induction

Body:
It's easier if you notice that
$$
\frac{n^4}{4}+\frac{n^3}{2}+\frac{n^2}{4}+(n+1)^3=\frac{n^2}{4}(n^2+2n+1)+(n+1)^3
$$
and that $n^2+2n+1=(n+1)^2$. Collect and find
$$
\frac{1}{4}(n+1)^2(n^2+4n+4)
$$
Can you finish now?

 $$\frac{1}{4}(n+1)^2(n+2)^2=\frac{1}{4}\bigl((n+1)^2((n+1)+1)^2\bigr)=\frac{1}{4}\bigl((n+1)^4+2(n+1)^3+(n+1)^2\bigr)$$


答: Hint: for $f(n) = \frac{n^4}{4}+\frac{n^3}{2}+\frac{n^2}{4}$, evaluate what $f(n+1)$ is

Title: Stuck within Induction

Body:
I am taking a course in Analysis and have to prove that
$$
\sum_{j=0}^n j^3 = \frac{n^4}{4} + \frac{n^3}{2} + \frac{n^2}{4}
$$

I think until I had $n^4/4 + n^3/2 + n^2/4 + (n + 1)^3$ I did not make a mistake.

From there I seem to always make the same mistake and end up with
$$
\frac{n^4 + 6n^3 + 13n^2 + 12n + 4}{4}
$$

Any suggestions, hints, solutions?