问:
Abelian Varieties and Neron Models over local fields: Quotient of points of k mod valuation ring is finite
I am working on a paper and I am not sure why the following fact is true. Suppose A/k is an abelian variety over a nn-archimedean local field k and $N/\mathcal{O}$ should be the Neron model. Then let $N^0$ be the connected component of the Neron Model, ie the normal open subgroup scheme, which is fibrewise the connected component of the identity. We are in the situation $N^0 \neq N$. Now they state that the quotient
$N^0(k)/N^0(\mathcal{O})$ is finite.
I am wondering, why this is true.
Maybe this has something to do with the reduction map, ie. the map $N(\mathcal{O}) \rightarrow N(\kappa)$, $\kappa$ the residue field, which is surjective?

I would be happy for an answer :)

答:
In case someone needs this fact too, here my suggestion:
For the definition and all the properties I use, see SGA 3, $VI_B$, §3.
First remark that $N^0$ is compatible with base change, hence $(N^0)_k= (N_k)^0=A^0=A$ since $A/k$ is ab abelian variety and hence connected. Using this we obtain $N^0(k)=A(k)=N(\mathcal{O})$ via the Neron mapping property.
So we are reduced to show $N(\mathcal{O})/N^0(\mathcal{O})$ is finite. Now go back to SGA, where they defined $N^0$ and we see that

$N^0(\mathcal{O})=Ker(N(\mathcal{O}) \rightarrow N(\kappa)/N_\kappa^0(\kappa))$.

ie you have to look at the fibers for $s \in Spec(\mathcal{O})$. But being a non-archimedean local field there were just two points, 0 and the unique prime ideal $\mathfrak{p}$. For $s=\mathfrak{p}$ we obtain the reduction map. This quotient is finite, since $\kappa$ is a finite field and we obtain the desired property.

If you have any comments, please feel free to write, I would be happy, if there were some things, which are not clear and we could discuss.


问:
Suppose $f$ is bounded on $[a,c]$ and $a<b<c$ prove $\overline{\int_{a}^{c}}f = \overline{\int_{a}^{b}}f + \overline{\int_{b}^{c}}f$
Suppose $f$ to be a bounded function on the interval $[a,c]$ and where $a<b<c$. Prove that

$$
\overline{\int_{a}^{c}}f = \overline{\int_{a}^{b}}f + \overline{\int_{b}^{c}}f
$$

I've been trying to directly prove it by assigning two partitions on each interval and making one which refines both - got further than what I was asked to prove.
Also tried on the contrary but the only  idea I had was to say suppose:
$$
\overline{\int_{a}^{c}}f \not= \overline{\int_{a}^{b}}f + \overline{\int_{b}^{c}}f
$$
then if $f$ was integrable the above was absurd but couldn't expain why should assuming something that isn't given could work.

much appreciate for any help

答:
Take arbitrary partitions $P_{ab} : a = x_0 < x_1 < \ldots < x_n = b$ and  $P_{bc} : b = y_0 < y_1 < \ldots < y_m = c$ of the intervals $[a,b]$ and $[b,c]$, respectively.
Together they form a partition of  the interval $[a,c]$: $$P_{ac}: a = x_0 < \ldots< x_{n-1} < b < y_1 < \ldots < y_m = c$$

The upper Darboux sums of $f$ over these intervals satisfy

$$U(f,P_{ab}) + U(f, P_{bc}) =U(f, P_{ac}) \geqslant \inf \,\{U(f,P)\,|\, P \text{ is a partition of } [a,c]\} := \overline{\int_{a}^{c}}f $$

Since the partitions $P_{ab}$ and $P_{bc}$ are independent, we can take infima succcesively to argue
$$ U(f, P_{bc})\geqslant - U(f,P_{ab}) + \overline{\int_{a}^{c}}f  \implies \overline{\int_{b}^{c}}f \geqslant - U(f,P_{ab}) + \overline{\int_{a}^{c}}f\\ \implies U(f,P_{ab})  \geqslant -\overline{\int_{b}^{c}}f + \overline{\int_{a}^{c}}f  \implies  \overline{\int_{a}^{b}}f \geqslant -\overline{\int_{b}^{c}}f + \overline{\int_{a}^{c}}f  \implies\\   \boxed{\overline{\int_{a}^{b}}f +\overline{\int_{b}^{c}}f  \geqslant \overline{\int_{a}^{c}}f}   $$

To prove the inequality in the other direction take any partition $P_{ac}$ of $[a,c]$ and insert the point $b$ to obtain a refined partition $P'_{ac}$ which induces partitions $P_{ab}$ and $P_{bc}$ of $[a,b]$ and $[b,c]$, respectively. Since upper sums decrease under refinement we have

$$U(f, P_{ac})\geqslant U(f,P'_{ac}) = U(f,P_{ab}) + U(f, P_{bc}) \geqslant \overline{\int_{a}^{b}}f +\overline{\int_{b}^{c}}f$$

Taking the infimum over all partitions of $[a,c]$ we get

$$ \boxed{\overline{\int_{a}^{c}}f\geqslant \overline{\int_{a}^{b}}f +\overline{\int_{b}^{c}}f}   $$

Altogether, this proves

$$\overline{\int_{a}^{c}}f = \overline{\int_{a}^{b}}f +\overline{\int_{b}^{c}}f$$

ContentLicense: CC BY-SA 4.0


问:
Prove $x^ny(x - y) + y^nz(y - z) + z^nx(z - x) \ge 0$

Prove the inequality with $x, y,z$ is the sides of a triangle and
  $n\in \mathbb Z \land n\ge2$ $${x^n}y(x - y) + {y^n}z(y - z) + {z^n}x(z - x) \ge 0 \tag 1$$

I can prove the inequality with $n=2$:
$$(1)\iff x(y-z)^2(x+z-x) + y(x-y)(x-z)(x+y-z) \ge 0$$
or apply Ravi Substitution: $$(1)\iff \frac {a^2}b+\frac {b^2}c + \frac {c^2}a \ge a+b+c\  $$
(alway true, by AM-GM)

But I can't prove the general problem above

Please forgive my grammar

答:
Notice that $$x^ny(x - y) + y^nz(y - z) + z^nx(z - x)=\left|\begin{matrix}x^ny & y^nz & z^nx\\1 & 1 & 1\\z & y & x\end{matrix}\right|=(x^ny,y^nz,z^nx)\cdot\left((1,1,1)\times(z,y,x)\right)$$
Without loss of generality, let $x\geqslant y\geqslant z>1$, for otherwise we can put a scalar on $x,y,z$. Then by drawing the vectors $(x^ny,y^nz,z^nx),(1,1,1),(z,y,x)$ you can see they are in right-hand order and hence the determinant is nonnegative.</p>

ContentLicense: CC BY-SA 3.0


问:
Closed form for $1-2+3-4+\cdots(-1)^{n-1}n$
How do I find a closed formula for the following?

$$\sum_{i=1}^n(-1)^{i-1}i$$

If $n$ is odd number, I can express it as $\frac{n+1}{2}$; if $n$ is even, then the expression if $\frac{-n}{2}$, but how do you find a expression that can express it in one formula?

答:
You can use the ceiling function: $(-1)^{n+1} \left\lceil \dfrac n2\right\rceil$.

For odd $n$, $\left\lceil \dfrac n2\right\rceil = \dfrac {n+1}2$ and $(-1)^{n+1} = 1$.

For even $n$, $\left\lceil \dfrac n2\right\rceil = \dfrac {n}2$ and $(-1)^{n+1} = -1$.


问:
Suppose $R$ be a ring and $I$ an ideal in R. prove that $rad(I)=R$ if and only if $R=I$.
Suppose $R$ be a commutative ring and $I$ an ideal in R. prove that $rad(I)=R$ if and only if $R=I$.

If $R=I$ then $\forall x \in R : x^1 =x \in I$ so $rad(I)=R.$

Can you help me to prove the other side? If it is incorrect please say a counterexample. Thanks.

$$rad(I)= \{ x\in R | \exists n ; x^n\in I \}.$$

Footnote : $R$ has unity.

答: If $R$ is a ring with unity and $Rad(I)=R$ then $1 \in Rad(I)$ and hence there exists some $n$ such that $1^n \in I$. This immediatelly implies $I=R$.


问:
Are Lie algebra complexifications $\mathfrak g_{\mathbb C}$ equivalent to Lie algebra structures on $\mathfrak g\oplus \mathfrak g$?
Given a real Lie algebra $\newcommand{\frakg}{\mathfrak{g}}\frakg$, its complexification $\frakg_{\mathbb C}$ is usually defined (e.g. on Wikipedia) as $\frakg_{\mathbb C}\equiv\frakg\otimes_{\mathbb R}\mathbb C$. As far as I understand it, this means that the elements of $\frakg_{\mathbb C}$ have the form
$$\sum_i \alpha_i (t_i\otimes e_1) + \sum_i \beta_i (t_i\otimes e_2),$$
where $\alpha_i,\beta_i\in\mathbb R$, $t_i\in\frakg$, and $e_1,e_2$ are a basis for $\mathbb C\simeq\mathbb R^2$ (e.g. $e_1\simeq 1$ and $e_2\simeq i$).
In other words, the elements of $\frakg_{\mathbb C}$ are pairs $(g,g')$ with $g,g'\in\frakg$, that is, elements of $\frakg\oplus\frakg$.

The Lie algebra structure on this space would then be derived from that of $\frakg$ as:
$$[(g,g'),(h,h')] \equiv ([g,h]-[g',h'], [g,h'] + [g',h]), \qquad g,g',h,h'.$$
So can we write that $\frakg_{\mathbb C}\simeq \frakg\oplus\frak g$? What about the case of $\frakg$ being a Lie algebra on some more generic field $\mathbb F$? Does this still hold in that case?

答:
Let $E \vert K$ be a field extension; to play it safe, let's assume the characteristic is $0$, and all dimensions are finite. It is useful for what follows to introduce an explicit notation for scalar restriction: If $V$ is any vector space (or later $\mathfrak g$ a Lie algebra) over $E$, let's call $R_{E \vert K} V$ (or later $R_{E\vert K} \mathfrak g$) the same additive group (Lie algebra) but viewed as a vector apace (Lie algebra) over $K$.

On the other hand one has scalar extension: If $V$ is a vector space (or $\mathfrak g$ a Lie algebra) over $K$, then $V_E := E \otimes_K V$ (or $\mathfrak g_E := E \otimes_K \mathfrak g$) is a vector space (Lie algebra) over $E$. (The Lie bracket on $\mathfrak g_E$ is defined as the unique bilinear map satisfying $[e_1 \otimes g_1, e_2 \otimes g_2] = e_1e_2 \otimes [g_1, g_2]$.)

Note that in spite of what one might think when first hearing the names, scalar restriction and scalar extension are not at all inverse to each other.

If we first look at vector spaces for a moment, everybody has learned in linear algebra that for an $E$-vector space $W$, we have that $R_{E\vert K} W$ is a $K$-vector space of dimension $[E:K] \cdot \mathrm{dim}_E(W)$, and for a $K$-vector space $V$, we have that $V_E$ is an $E$-vector space of dimension $\mathrm{dim}_E(V_E) = \mathrm{dim}_K(V)$. Consequently

$$R_{E\vert K}(V_E) \simeq V^{[E:K]} \qquad \text{ (iso of $K$-vector spaces)}$$

$$(R_{E\vert K}(W))_E \simeq W^{[E:K]} \qquad \text{ (iso of $E$-vector spaces)}$$

So now one might think that whilst not being the identity, the composition of scalar restriction with scalar extension would always come out as the $[E:K]$-fold direct sum in the respective categories. As the above shows this is the case for vector spaces. It is also true for abelian Lie algebras because they are basically the same as vector spaces. But as soon as we look at more intricate structures, it will in general not be the case. And, one runs into different problems depending on whether the scalars get "first restricted, then extended" or the other way around.

Extension followed by restriction:

You seem to think mostly about composition of restriction and extension this way around: Starting with a Lie algebra $\mathfrak g$ over the "small" field $K$ and relating its extension $\mathfrak g_E$ to some direct sum of it. Well here the first problem is that they are incomparable insofar as $\mathfrak g_E$ is an $E$-Lie algebra but $\mathfrak g$ (and any direct sum thereof) is a $K$-Lie algebra. So the only object that has a chance to be isomorphic to $\mathfrak g^{[E:K]}$ is not $\mathfrak g_E$ itself, but $R_{E\vert K} (\mathfrak g_E)$. However, in general it is not:

Example 1: Let $\mathfrak g$ be either $:= \mathfrak{sl}_2(\mathbb R)$ or the simple $\mathbb R$-Lie algebra $\mathfrak g := \mathfrak{su}_2$. Then in either case $\mathfrak g_{\mathbb C} \simeq \mathfrak{sl}_2(\mathbb C)$ and hence $R_{\mathbb C \vert \mathbb R}(\mathfrak g_{\mathbb C})$ is a simple Lie algebra, also known as $\mathfrak{so}(3,1)$, which is $\not \simeq \mathfrak g \oplus \mathfrak g$.

Example 1*: Actually, let $\mathfrak g$ be any absolutely simple Lie algebra over $K$ (that is, one such that any scalar extension of it stays simple; for example, all split simple Lie algebras are of that kind). Then by definition $\mathfrak g_E$ is simple, and one can show that this makes $R_{E\vert K}(\mathfrak g_E)$ simple (as a $K$-Lie algebra) as well, so it cannot be isomorphic to $\mathfrak g^{[E:K]}$ as soon as $[E:K] \ge 2$.

Of course if one chooses a $K$-basis $e_1, ..., e_r$ for $E$, one can identify the underlying $K$-vector space of $\mathfrak g_E$ with an $r$-fold direct sum of $\mathfrak g$, however, this direct sum is not a direct sum of Lie algebras, as discussed in the comments. This means that in general $R_{E\vert K} ( \mathfrak g_E) \not \simeq \mathfrak g^{[E:K]}$ if the RHS is understood as a direct sum of Lie algebras.

Restriction followed by extension:

In more advanced algebra, one learns that this can go wrong for various reasons when, instead of vector spaces or Lie algebras, one looks at fields.

Example 2: Let $K:=\mathbb Q, E:=\mathbb Q(\sqrt[3]{2})$. Then $$(R_{E\vert K}(E))_E = E\otimes_K E \simeq E \times L$$ where $L= E(\zeta_3) = E(\sqrt{-3})$ is the normal closure of $E$. Actually, that is an isomorphism of rings, which when viewed as isomorphism of $E$-algebras we should write
$$ (R_{E\vert K}(E))_E  \simeq E \times R_{L \vert E} (L).$$

In particular $(R_{E\vert K}(E))_E \not \simeq E \times E \times E$ as one would have naively hoped for in analogy with the vector space case. Compare https://en.wikipedia.org/wiki/Tensor_product_of_fields.

Now this dampens our hope for Lie algebras (and other kinds of algebras), as indeed:

Example 3: Let $E \vert K$ as in example 2, and let $\mathfrak g := \mathfrak{sl}_n(E)$. Then

$$(R_{E\vert K}(\mathfrak{g}))_E \simeq \mathfrak g \oplus R_{L\vert E}(\mathfrak{g}_L)$$

Now funnily, that second factor throws us back to composition the other way around, first extension and then restriction, but we saw in example 1* that (although $[L:E]=2$) that factor is not $\simeq \mathfrak g \oplus \mathfrak g$ but rather a simple (six-dimensional) $E$-Lie algebra. So we also have
$$(R_{E\vert K}(\mathfrak{g}))_E \not \simeq \mathfrak g^{[E:K]}.$$

One good case for restriction followed by extension: If $E\vert K$ is Galois and $\mathfrak g$ a split simple $E$-Lie algebra, $$(R_{E\vert K}\mathfrak g)_E \simeq \mathfrak g^{[E:K]}.$$

(Example 3 showed that the condition of $E\vert K$ being Galois is necessary. I am not sure if the condition that $\mathfrak g$ is split is necessary, i.e. without it it I neither have a proof nor a counterexample to the statement.)

In any case,  both conditions are automatically satisfied for $E\vert K = \mathbb C \vert \mathbb R$. That is:

If $\mathfrak g$ is a complex semisimple Lie algebra, then $(R_{\mathbb C \vert \mathbb R}(\mathfrak g))_\mathbb C \simeq \mathfrak g \oplus \mathfrak g$.

Two final notes:

a) The examples above deal with (semi)simple Lie algebras and should give an idea what to expect and not to expect there. As noted earlier, the other extreme of abelian Lie algebras is trivially well-behaved like vector spaces. I have almost no idea what happens for Lie algebras between these extremes, i.e. solvable ones.

b) The answer in Precise connection between complexification of $\mathfrak{su}(2)$, $\mathfrak{so}(1,3)$ and $\mathfrak{sl}(2, \mathbb{C})$ which you link to in the comments would greatly benefit from using a notation for scalar restriction as above. Check for yourself that interpreted correctly, nothing in there contradicts anything in here, but one really needs to be careful to interpret it correctly. E.g. the first iso in there,

$$\mathfrak{so}(1,3) \simeq \mathfrak{su}(2)_\mathbb{C}$$

with my notation should be written

$$\mathfrak{so}(1,3) \simeq R_{\mathbb C \vert \mathbb R}(\mathfrak{su}(2)_\mathbb{C})$$

to make clear it's an iso of $\mathbb R$-Lie algebras, and actually came up in my example 1. And in point 2 of that answer, they're looking at an extension of a restriction of an extension, which would make any notation awkward, but if one in a first step abbreviated $\mathfrak{su}(2)_{\mathbb C} \simeq \mathfrak{sl}_2 (\mathbb C)$ it basically says the same as my last line in the "good case" applied to $\mathfrak g = \mathfrak{sl}_2(\mathbb C)$.

Added in response to comments: Indeed I stress that if $V = \mathbb C^2$, then $Res_{\mathbb C\vert \mathbb R}V \simeq \mathbb R^4$.

While the nomenclature might be a little misleading at first, this is indeed the standard meaning of scalar restriction. But of course the thing you have in mind is also studied. What you want is:

Given a complex Lie algebra $\mathfrak G$, is there a real Lie algebra $\mathfrak g$ such that $\mathfrak g_{\mathbb C} \simeq \mathfrak G$?

Such a $\mathfrak g$ is called a real form of $\mathfrak G$, and finding such a real form would be an actual "inverse" of scalar extension (compare my answer here, especially beginning of second paragraph, where I also note that we want this, but it's not what is called scalar restriction).

Now again, if we were just interested in vector spaces, this is trivial: If $V$ is an $n$-dimensional vector space over $E$, then $K^n$ is a $K$-form of $V$, and it is the unique one up to iso of $K$-vector spaces. But again, as soon as we are interested in more intricate structures, we face dire problems:

Problem A: Not every complex Lie algebra has a real form.

See example 1.36 here.

Widely celebrated result avoiding problem A for a large class of objects: Every semisimple complex Lie algebra has a real form. Actually, it even has a form over $\mathbb Q$; or even over $\mathbb Z$. This is widely credited to Chevalley and more or less the content of volume 8 of Bourbaki's treatise on Lie Groups and Algebras.

Sounds good. But:

Problem B: Every non-zero complex semisimple Lie algebra has more than one real form. I.e. there are non-isomorphic real Lie algebras which have isomorphic complexifications. E.g. $\mathfrak{su}_2 \not \simeq \mathfrak{sl}_2(\mathbb R)$, but both have complexification $\simeq \mathfrak{sl}_2(\mathbb C)$.

Actually, it is a celebrated result of E. Cartan's to classify all real semisimple Lie algebras, i.e. real forms of complex Lie algebras. I wrote my thesis partly about general approaches of this classification, partly about the exact classification over $p$-adic fields. Results over general characteristic $0$ fields are due largely to Tits in the 1960s and 1970s; I think that to this day some questions are open. Compare https://math.stackexchange.com/a/3677910/96384, Classification of real semisimple lie algebras or this recent question just about possible forms of $\mathfrak{sl}_2$: Lie algebra $\mathfrak{sl}_2 \mathbb{C}$ has only these two real forms $\mathfrak{sl}_2 \mathbb{R}$ and $\mathfrak{su}_2$?


问:
Show If a sequence converges, then each subsucession converges to the same limit in Euclidean space
Show if a sequence converges, then each subsucession converges to the same limit in Euclidean space

Then let $(x_{n}) \in \mathbb{R^n}$ a convergent sequence and $(x_{r}) \in \mathbb{R^n}$ a sub-succession of $(x_{n})$, it should be shown that $(x_{r})$ is convergent and its limit is equal to that of the $(x_{n})$ sequence.

proof:

let $(x_{r}) \in \mathbb{R^n}$ a sub-sequence of $(x_ {n}) \in \mathbb{R^n}$; given an $\epsilon>0$, exists $k \in \mathbb{N}$ such that $(x_{n}) \in B(a,\epsilon)$ for all $k>K$ where $a \in \mathbb{R^n}$ and is the limit of the sequence, that is to say, $\lim x_{k}=a$. now let's see what $\lim x_{r}=a$

I'm stuck, could someone help me?

答:
Given $\epsilon>0$, since $(x_n)$ converges to $a$, there is some $k\in\Bbb N$ such that $x_n\in B(a,\epsilon)$ for every $n\ge k$.

Since $(x_r)$ is a subsequence of $(x_n)$, each $r$ in $x_r$ actually is some $n$ in $x_n$ (meaning each $x_r$ is some $x_n$ from the original sequence). Then there is some $l\in\Bbb N$ such that $x_r\in(x_n)_{n\ge k}$, meaning there is a moment when the subsequence has run out of terms $x_n$ with $n<k$. For such terms of $(x_r)$ we have $x_r\in B(a,\epsilon)$, since each $x_r$ with $r\ge l$ represents some $x_n$ with $n\ge k$.

We conclude $(x_r)$ converges to $a$ as well.

Note: This argument is valid for any topological space associated to a metric (or even to a pseudo-metric).


问:
Being integral is preserved by quotient
Let $\varphi:B\rightarrow A$ be a homomorphism of commutative rings. We say that $\varphi$ is integral if every element of $A$ satisfies a polynomial equation with coefficients in $\varphi(B)$. I am trying to understand why the induced map $\varphi':B/J\rightarrow A/\varphi(J)A$ is also integral.

So, to start the obvious way, we know that there exists an $n\geq 1$ and some $b_i\in B$ such that $a^n+\varphi(b_{n-1})a^{n-1}+\cdots +\varphi(b_1)a+\varphi(b_0)=0$. But then simply reducing this equation mod $\varphi(J)A$ doesn't seem to work.

In particular, I don't see how reducing mod $\varphi(J)A$ tells us that the coefficients are in $\varphi'(B/J)$. What am I missing?

答:
Let $[a] \in A/\varphi(J)A$. Since $\varphi$ is integral and $a \in A$, we have that
$$ a^n + \varphi(b_1) a^{n-1} + \dots + \varphi(b_n) = 0,$$
for some $n$ and $b_i \in B$. Thus, the following holds in $A/\varphi(J)A$
$$ [a]^n + [\varphi(b_1)] [a]^{n-1} + \dots + [\varphi(b_n)] = 0.$$
Recall that $\varphi'$ is defined as follows
\begin{align*}
\varphi' \colon B/J & \to A/ \varphi(J) A \\
[b] & \mapsto [\varphi(b)].
\end{align*}
Hence, $\varphi'([b_i]) = [\varphi(b_i)]$, which implies that
$$ [a]^n + \varphi'([b_1]) [a]^{n-1} + \dots + \varphi'([b_n]) = 0.$$
Since $[a]$ was arbitrary, we conclude that $\varphi'$ is integral.

Title: Being integral is preserved by quotient


问:
How to find the solutions of $x^3-x^2-x+1=0$
I have to find the solutions of $x^3-x^2-x+1=0$. I know $x = 1$ but how to solve this? I'm lost on the step-by-step resolution.</p>

答:
HINT:

$$x^2(x-1)-1(x-1)=0\implies (x-1)(x^2-1)=0\implies (x-1)^2(x+1)=0$$

As the product is $0,$ at least one of the factors must be $0$


问:
Prove $(\forall x\ Q(x))\to P ⊢ \exists x (Q(x)\to P)$ using Natural Deduction
I have stumbled upon this predicate logic sequent which I have trouble proving using natural deduction:

$(\forall x\ Q(x))\to P ⊢ \exists x (Q(x)\to P)$

I started with ∀xQ(x) → P as the premise. I wanted to start using universal elimination but since the ∀x doesn't cover Q(x) → P completely, I refrained from doing so. I had to assume ∀xQ(x) but when I did, I couldn't reach ∃x(Q(x) → P) out of the assumption's scope.

I have also tried proof by contradiction but I couldn't progress. I tried to prove p → q ⊢ (¬p V q) with natural deduction to be able to incorporate it into the solution, but I couldn't prove it either.

So, I think I am completely stuck this one. Could you help me prove this (or tell me if this cannot be proved by natural deduction for any reason)?

If it helps, I am using the inference rules listed under
https://en.wikipedia.org/wiki/List_of_rules_of_inference

答: Hint: you will need to use the law of the excluded middle (LEM) in whatever form it is presented in your system of natural deduction. By LEM, it suffices to consider two cases: (i) when $P$ is true and (2) when $P$ is false. In case (i) any implication of the form $A \to P$ is true and hence so is $\exists x(A \to P)$ for any formula $A$. In case (ii), we must have that $\forall xQ(x)$ is false, given that $\forall xQ(x) \to P$, so $\exists x \lnot Q(x)$ and an $x$ that witnesses that will also witness $\exists x (Q(x) \to P)$. I leave it to you to translate this into the system of natural deduction you are using.


问:
I wonder what is really a reducible matrix?
Let A = \begin{bmatrix}
 0&  1&  0& 0\\
 0&  0&  1& 0\\
 0&  0&  0& 1\\
 1&  0&  0& 0
\end{bmatrix}
According to Matrix Analysis 2nd edition Definition 6.2.21, this is a reducible matrix if I exchange column 1 and 3.

However, obviously $A$ is irreducible as $(A+I)^{n-1}>0$, and $A$ represents a strongly connected graph too.

I read this question: Reducible matrices and strongly connected graphs. In the answer @thanasissdr mentioned that $B$ and $D$ must be square blocks. But my $A$ above is a $4\times4$ matrix and the result $B$ and $D$ are exactly square.

So this is what I'm confused: whether my $A$ is reducible or not?

答:
A matrix $M$ is reducible if there is a permutation matrix $P$ so that

$$PMP^T=\begin{bmatrix}B&C\\0&D\end{bmatrix}$$

where $B$ and $D$ are square.

On the other hand, $M$ is partly decomposable if there are permutation matrices $P$ and $Q$ so that

$$PMQ=\begin{bmatrix}B&C\\0&D\end{bmatrix}$$

where $B$ and $D$ are square.

The matrix

$$A = \begin{bmatrix}
 0&  1&  0& 0\\
 0&  0&  1& 0\\
 0&  0&  0& 1\\
 1&  0&  0& 0
\end{bmatrix}$$

happens to be partly decomposable, but not reducible.

You give a permutation $P^T$ of the columns so that $AP^T$ is of the form $\begin{bmatrix}B&C\\0&D\end{bmatrix}$, but for reducibility we need a simultaneous permutation of the rows and columns; $PAP^T$ must be of this form.


问:
Fourier series for $f(x) = \sqrt{1 - \cos x}$
I want to calculate the Fourier series for the $2\pi$ periodic function $f(x) = \sqrt{1 - \cos x}$ for $-\pi \leq x \leq \pi$

The Fourier series is given by:
$$f(x) = \frac{a_{0}}{2} + \sum\limits_{n = 1}^{\infty}a_{n}\cos nx + \sum\limits_{n = 1}^{\infty}b_{n}\sin nx$$

Where:

$a_{0} = \frac{1}{\pi}\int\limits_{-\pi}^{\pi}f(x)dx$

$a_{n} = \frac{1}{\pi}\int\limits_{-\pi}^{\pi}f(x) \cos nxdx$

$b_{n} = \frac{1}{\pi}\int\limits_{-\pi}^{\pi}f(x) \sin nxdx$

I know how to go about these questions and have no problem with the integration.

My question is do we need to split up the integrals into $0 \leq x \leq \frac{\pi}{2}$ and $\frac{\pi}{2} \leq x \leq \pi$ to account for the graphs on each side of the origin? Or can we just compute directly from the equations I wrote above with $-\pi \leq x \leq \pi$? Also I don't think we would, but would you consider odd/even functions?

Any help on understanding what to do would be great thank you

答:
$\cos x \le 1$, so $f(x)$ is well defined everywhere.  You don't need to split it.

$f(x)$ is even, so $b_n=0$ for all $n$.  You can thus also write:

$$a_0 = \frac{2}{\pi} \int_0^\pi f(x) \, dx$$

$$a_n = \frac{2}{\pi} \int_0^\pi f(x) \cos n x \, dx$$


问:
Showing that an ideal is equal to $\mathbb{Z}[i]$
Let $a$, $b$, and $c$ be three elements of $\mathbb{Z}[i]$, the ring of Gaussian integers. We define the following to be  subset of $\mathbb{Z}[i]$:

$$(a,b,c)=\{ax+by+cz:\ x,y,z \in\mathbb{Z}[i]\}.$$

We know that $(a,b,c)$ is an ideal of $\mathbb{Z}[i]$. Show that the ideal $(-1+3i, -1+5i,-4+7i$) is equal to $\mathbb{Z}[i]$ .

What I have done so far:

I have factorised the three elements into a product of irreducible elments in $\mathbb{Z}[i]$:

$(-1+3i)=(1+i)(1+2i)$, $(-1+5i)=(1+i)(2+3i)$,$(-4+7i)=(2+3i)(1+2i)$

However I don't know how to show that the ideal is equal to  $\mathbb{Z}[i]$. I think it might have something to do with the norms of the three elements in the ideal...

答:
Easy way: the ideal $I$ contains the norms $\,w\bar w\,$ of every generator $\,w\,$ so by Bezout it also contains the gcd of the norms = $(10,2\color{#c00}6,5\color{#0a0}5)=(10,\color{#c00}6,\color{#0a0}5)=(10,1,5)\!=\!1\,$ by Euclid. So $\,1\!\in\! I\Rightarrow I\! =\! \Bbb Z[i]$.

Remark $ $ As explained here this is an instance of the method of simpler multiples (or, equivalently, the correspondence or Third Isomorphism Theorem,  from a more structural viewpoint).

Because the norm map is multiplicative it preserves many properties related to factorization. For example, in many favorable contexts (e.g. Galois) a number ring enjoys unique factorization iff its monoid of norms does. For further interesting examples see the papers of Bumby, Dade, Lettl, Coykendall cited in this answer.

Alternatively working $\!\bmod I\,$ (or, equivalently, in $\,\Bbb Z[i]/I)\,$ simplifies ideal arithmetic (as in Lubin's answer) to more intuitive arithmetic of numbers (ring elements), e.g. here we find by subtracting neighboring stacked congruences that

$\qquad\quad\qquad\begin{align} \color{#c00}{3i\equiv 1}\\ 5i\equiv 1\\ 7i\equiv 4\end{align}\,$
$\Rightarrow\ \begin{align}2i\equiv\color{#0a0}0\\[.4em] 2i\equiv \color{#0a0}3\end{align}\,$ $\Rightarrow\ \color{#0a0}{0\equiv 3}\,\Rightarrow\, 0\equiv \color{#c00}{3i\equiv 1}$

While it seems ad hoc at first glance, as long as we eventually keep finding smaller (norm) elements this will eventually yield the gcd, due to the fact that $\,\Bbb Z[i]\:\!$ is (norm) Euclidean. Thus, instead of mechanically applying the Euclidean algorithm, we can use our intuition to step outside the algorithm to optimize it when the occasion arises (one way is as above - using "simpler multiples", where "simpler" need not always imply smaller norm).


问:
How would you solve this equation by graphing? Or any other method?

A particle is moving so that its position at time $t$ is given by the parametric equations
\begin{align*}
 x &= 5\sin(-2t) \\
 y &= 5\cos(2t).
\end{align*}What is the speed of the particle?

A particle's position in the plane $t$ seconds after it starts moving is\begin{align*}
x &= 6\cos(kt),\\
y &= 6\sin(kt),
\end{align*}for some constant $k$. If the particle moves a distance of $120\pi$ every second, then what is $k$?

答:
HINT

The position of the particle is given by $s(t) = (5\sin(-2t),5\cos(2t))$. Consequently, the speed of the particle is given by $\|s'(t)\|$. Can you take it from here?

At this case, the distance is given by
\begin{align*}
d = \int_{t_{0}}^{t_{0} +1}\|s'(t)\|\mathrm{d}t
\end{align*}

where $d = 120\pi$ and $s(t) = (6\cos(kt),6\sin(kt))$. Can you take it from here?

EDIT

In the first problem, notice that $s'(t) = (-10\cos(-2t),-10\sin(2t))$. Thus we have that
\begin{align*}
\|s'(t)\| = \sqrt{10^{2}\cos^{2}(-2t) + 10^{2}\sin^{2}(2t)} = 10
\end{align*}
In the second problem, notice that $s'(t) = (-6k\sin(kt),6k\cos(kt))$. Hence we get that
\begin{align*}
120\pi = \int_{t_{0}}^{t_{0}+1}\|s'(t)\|\mathrm{d}t = \int_{t_{0}}^{t_{0}+1}\sqrt{36k^{2}\sin^{2}(kt) + 36k^{2}\cos^{2}(kt)}\mathrm{d}t = 6|k|
\end{align*}

Therefore $k = \pm 20\pi$.

Hopefully this helps.


问:
MLE estimation for exponentially distributed
A postal clerk has an operating time t that is exponentially distributed with density

$f\left(x\right)\:=\:\lambda e−^{\lambda x},\:x\:\ge \:0$

where $\lambda $ is an unknown parameter

Given $n$ observations $t_1,.\:.\:.\:,\:t_n$, find the probability maximization estimate $\overline{\lambda } $ for the unknown parameter $\lambda $. Find a numeric value for $\overline{\lambda } $ when you have the following 10 observed operating times: ten: $2, 1.4, 2.0, 0.5, 0.7, 2.0, 1.3, 1.1, 1.8, 0.2,$ where the unit of time is one minute.

To find the $\overline{\lambda } $ I just took the derivative of $n$ observations $t_1,.\:.\:.\:,\:t_n$ and found out that it was

$0=n\frac{1}{\lambda }-\left(t_1+t_2+......+t_n\right)$

$\lambda \:\:=\left(t_1+t_2+......+t_n\right)$

$\overline{\lambda } $ = 13

Would this be correct?

And to find a numeric value for $\overline{\lambda } $ what do I do then?

答:
$0=n\frac{1}{\hat \lambda }-\left(t_1+t_2+\cdots+t_n\right)$

does not suggest $\hat \lambda =\left(t_1+t_2+\cdots+t_n\right)$

but $\hat \lambda = \dfrac{n}{\left(t_1+t_2+\cdots+t_n\right)} $


问:
How to create a 3-form from 1-forms?
I am trying to understand how to create a 3-form from 1 triple wedge product of 1-forms.

I understand that $\alpha \wedge \beta = \alpha \otimes \beta - \beta\otimes\alpha$ where the outer product, $\otimes$, is dropped for brevity later.

In creating the triple wedge product we get:

$\alpha\wedge\beta\wedge\gamma = \alpha(\beta\gamma-\gamma\beta)+\beta(\gamma\alpha-\alpha\gamma)+\gamma(\alpha\beta-\beta\alpha)$

How is this triple wedge product made? I'm trying to prove it but quickly run into the following:

$\alpha\wedge\beta\wedge\gamma = (\alpha\wedge\beta)\gamma-\gamma(\alpha\wedge\beta)= (\alpha\beta-\beta\alpha)\gamma-\gamma(\alpha\beta-\beta\alpha)$

Am I missing an important step? Where can I go from here?

答:
The formula $\alpha\wedge\beta=\alpha\otimes\beta-\beta\otimes\alpha$ is only true when $\alpha$ and $\beta$ are $1$-forms. The more general formula for a $k$-form $\mu$ and a $l$-form $\nu$ is
$$
\mu\wedge\nu=\frac{(k+l)!}{k!l!}\operatorname{Alt}(\mu\otimes\nu)\ \ \ \ \ \mu\in\Lambda^kV^*,\nu\in\Lambda^lV^*
$$
Here $\operatorname{Alt}$ denotes the antisymmetrization operator.
$$
\operatorname{Alt}(\omega)(v_1,\cdots,v_m)=\frac{1}{m!}\sum_{\sigma\in S_m}\operatorname{sgn}(\sigma)\omega(v_{\sigma(1)},\cdots,v_{\sigma(m)})\ \ \ \ \ \omega\in T_mV
$$
Where $S_m$ is the set of permutations of $(1,\cdots,m)$ and $\operatorname{sgn}$ is the sign of a permutation.

It might be worth noting that some authors choose to define the wedge product of alternating tensors without the prefactor $\frac{(k+l)!}{k!l!}$. This is just a matter of convention, but one to be mindful of. In this alternate convenntion, the product of $1$-forms would be $\alpha\wedge\beta=(\alpha\otimes\beta-\beta\otimes\alpha)/2$.

Title: How to create a 3-form from 1-forms?


问:
Every element of $\mathbb{Q}[\alpha]$ is algebraic over $\mathbb{Q}$ for an $\alpha$ algebraic
So I am stumped by the following problem:

Let $\alpha \in \mathbb{R}$ be an algebraic number of degree $d$. We denote by $\mathbb{Q}[\alpha]$ the subspace of $\mathbb{R}$ consisting of real numbers that can be written as: $$a_0 + a_1\alpha +\dots+ a_{d-1}\alpha^{d-1}$$
for some rational numbers $a_0, ... ,a_{d-1} \in \mathbb{Q}$.

Show that every element $\gamma \in \mathbb{Q[\alpha]}$ is algebraic.

Does anyone have any pointers? I know the definition of an algebraic number.

答:
Hint $\mathbb Q[\gamma]$ is a subspace of $\mathbb Q[\alpha]$. What can you conclude about
$$\mbox{dim}_{\mathbb Q} \mathbb Q[\gamma] \,?$$


问:
Does a sum of measurable functions converge outside a set of measure zero?
I am trying to solve the following problem:

Let $\{f_n\}$ be a sequence of measurable functions defined on a probability space, such that
$$
P\left(f_n=\frac1n\right)=1-P(f_n=0)=\frac1n.
$$
Does $\sum_{n=1}^\infty f_n$ converge outside a set of measure zero?

A similar problem was solved here: Does the sum of sequence of measurable functions converge outside a set of measure zero?. The only difference  is that in my problem $P(f_n=1/n)$ and $P(f_n=0)$ are not summable, so the Borel-Cantelli lemma does not apply. Is there another way to solve this?

答:
Let $(\Omega,\mu)$ be the probability space.

$$
\int \sum_{n=1}^\infty f_n(\omega)\,d\mu(\omega)
= \sum_{n=1}^\infty\int f_n(\omega)\,d\mu(\omega)
= \sum_{n=1}^\infty \frac{1}{n^2} < \infty
$$
Therefore $\sum_{n=1}^\infty f_n(\omega)$ is integrable, so
$$
\mu\left\{\omega \in \Omega : \sum_{n=1}^\infty f_n(\omega) < \infty\right\} = 1 .
$$


问:
How many Assumptions can I make in Proof by Induction?
Let's say I want to prove by mathematical induction that a statement involving $n$ is true for all $n\in\mathbb N$. A classic example is proving that
$$\sum_{r=1}^nr=\frac{1}{2}n(n+1)$$
The base step is trivial. This requires only $1$ assumption to prove by induction; namely that for a given $n=k$ the statement holds and then we go on to prove that this implies the statement is also true for $n=k+1$. But what if one assumption is not enough to prove the general case? Let's say to prove the general case I needed to assume the statement held for $2$ values: $n=k-1$ and $n=k$ and then I used this to prove that the result follows for $n=k+1$. Would that be ok?

More generally, for a proof by induction must I only make one assumption or am I permitted to use many? Please explain why, whatever the answer is.

Thank you for your help.

答:
Just to complement the above answer, there is a form of induction called "course-of-values" or "complete" or "strong" induction that requires no base cases. It says that to prove $P(n)$ for all $n$, it is sufficient to prove that for all $n$, $P(n)$ holds, if $P(m)$ holds for all $m < n$. In symbols:
$$
(\forall n ((\forall m < n P(m)) \Rightarrow P(n))) \Rightarrow (\forall n P(n))
$$
You can use this, for example, to prove that any natural number $n > 1$ is a product of primes. (This form of induction sneaks the base case in secretly because "$\forall m < n P(m)$" is automatically true when $n = 0$).


问:
Prove using properties of divisors and or induction
How would I go about proving the statement that for all positive integers $n$, $$(n+1)\Big|\binom{2n}{n}$$

Induction in this case is possible but I wanted to use a different method. If I compute $$\frac{2n+1}{n+1}\binom{2n}{n}$$ I would be able to get $$\frac{(2n+1)!}{(n+1)!(n)!},$$ where $n$ is non negative. Would I be able to use a theorem with the greatest common divisor? If I were to use induction, how should I do it?

答:
As you observed
$$\frac{2n+1}{n+1}\binom{2n}{n}=\binom{2n+1}{n}$$

Therefore
$$n+1 | (2n+1) \binom{2n}{n}$$

Now, show that $n+1$ is relatively prime to $2n+1$.


答: Hint: $$x^3-x^2-x+1=x^2(x-1)-(x-1)$$ Keep the difference of squares formula in mind.

Title: How to find the solutions of $x^3-x^2-x+1=0$

Body: I have to find the solutions of $x^3-x^2-x+1=0$. I know $x = 1$ but how to solve this? I'm lost on the step-by-step resolution.</p>


答: There's no upper limit on how many base cases you are allowed to use, although too many can be unnecessary and more than you need. The way to tell how many cases you need is to formulate your problem as a recurrence relation. For example, your given series can be represented as $$a_{n+1}=a_n+(n+1)$$ with $a_1=1$. The number of $a_k$ terms on the RHS of the recurrence equation represents the number of base cases you need to solve any such recurrence relation.

Title: How many Assumptions can I make in Proof by Induction?

Body:
Let's say I want to prove by mathematical induction that a statement involving $n$ is true for all $n\in\mathbb N$. A classic example is proving that
$$\sum_{r=1}^nr=\frac{1}{2}n(n+1)$$
The base step is trivial. This requires only $1$ assumption to prove by induction; namely that for a given $n=k$ the statement holds and then we go on to prove that this implies the statement is also true for $n=k+1$. But what if one assumption is not enough to prove the general case? Let's say to prove the general case I needed to assume the statement held for $2$ values: $n=k-1$ and $n=k$ and then I used this to prove that the result follows for $n=k+1$. Would that be ok?

More generally, for a proof by induction must I only make one assumption or am I permitted to use many? Please explain why, whatever the answer is.

Thank you for your help.


问:
Is the action functional "locally well defined up to an additive constant"?
I'm looking at an arbitrary symplectic manifold $(P, \omega)$, and the action functional on the set of contractible loops $\Lambda P:$ $$A: \Lambda P \to \mathbb{R}, \hspace{5pt} A(z) = \int_{D^2} \bar{z}^*\omega,$$where $\bar{z}$ is some extension of $z: S^1 \to P$ to $D^2$, which exists by contractibility.

Of course, this function is not well defined, because the value of $A$ will almost certainly depend on the choice of extension $\bar{z}$ (unless for example $\pi_2(P) = 0$). However, according to Floer in his paper "The unregularized gradient flow of the symplectic action," this function is locally well-defined, up to an additive constant, so we can talk about its derivative $dA$.

For a fixed $z \in \Lambda P$, how does one choose the neighborhood $z \in U \subset \Lambda P$ on which $A$ can be well-defined, with a suitable, coherent choice of extensions of loops in $U$?

For what it's worth, here's my attempt, even though I don't think it needs to be that complicated.

Choose $z$ and a "small," contractible neighborhood $U$ of $z$ (small enough so that all loops in $U$ are homotopic to $z$). Choose also an arbitrary extension $\bar{z}$ of $z$ to $D^2$. For any other $z' \in U$, choose a homotopy $h: z \simeq z'$. Since $D^2 \times \{0\} \cup S^1 \times I$ is a retract of $D^2 \times I$, there exists a map $\tilde{h}: D^2 \times I \to P$ with $\tilde{h}(x, 0) = \bar{z}(x)$ and $\tilde{h}|_{S^1 \times I} = h$. Thus I can define a lift of  $\bar{z}' := \tilde{h}|_{D^2 \times \{1\}}$, and set $$A(z') := \int_{D^2} \bar{z}'^* \omega.$$However, this is no better than choosing an arbitrary extension, because the homotopy $h$ was arbitrary. So instead of looking at the extension problem

I should be looking at all of the homotopies simultaneously, something like

where ideally I'd be able to choose the homotopies to depend continuously (even smoothly) on $z' \in U$. Since $U \times D^2 \times I$ is contractible, there's definitely an extension up to homotopy, but that's not what I need. If $U \times D^2 \times I$ had a CW-complex structure, I could find a solution, but it doesn't in general because $U$ lives in an infinite-dimensional manifold, and they don't admit CW-structures, according to https://mathoverflow.net/questions/357477/cw-structure-on-infinite-dimensional-manifolds/357522#357522.

P.S: I know that people usually work with some quotient of the universal cover of $\Lambda P$ instead etc, but I'm just interested to see if this approach can work too.

答: Answering your edit: This follows from your choice of neighbourhood $U$, namely from its contractibility. Two continuous maps $f,g \in C(X,Y)$ lie in the same path component (using a suitable topology) if and only if they are connected. Since $U$ is contractible, so are $U_{[z]}$ and $U_{[\tilde{z}]}$, and hence $[\bar{z}]$ ($[\tilde{z}]$) and $[\bar{z}']$ ($[\tilde{z'}]$) are homotopic.


问:
A question on one-to-one holomorphic functions on the unit disk
Let $D = \{ z \in \mathbb{C}; |z| < 1 \}$ be the open unit disk, and let $f$ be a one-to-one holomorphic function such that $f(0) = 0$ and $\bar{D} \subseteq f(D)$, where $\bar{D}$ is the closure of $D$, i.e. the closed unit disk. This may turn out to be a trivial question but, is it necessarily the case that $|f'(0)| >= 1$? I am wondering if some kind of lower bound version of Schwarz's lemma may hold. There may be an easy counterexample, but I just don't know many examples of such functions...

Edit: can we just apply the usual Schwarz lemma to $f^{-1}$? I will have to think a bit...

答:
If $f,g : 0 \in U \to \mathbb C, f(0)=g(0), g$ univalent, where $U$ is some region (open connected), we call $f$ subordinate to $g$ if $f(U) \subset g(U)$
($0$ here is a convenience, any base point will do).

If in addition $U =\mathbb D$ the unit disc, $\omega =g^{-1} \circ f$ gives a map that satisfies the conditions of Schwarz Lemma, for which $f=g \circ \omega$ which is sometimes taken as the definition of subordination. This immediately implies that $|a_1(f)|=|f'(0)| \le |g'(0)| =|a_1(g)|$.

Less obvious results are the fact that $\int_0^{2\pi}|f(re^{i\theta})|^pd\theta \le \int_0^{2\pi}|g(re^{i\theta})|^pd\theta$ for all $0 <r<1, p>0$ and $\sum_{k=1}^n|a_k|^2 \le \sum_{k=1}^n|b_k|^2$ for all $n \ge 1$ where $f(z)=f(0)+\sum_{k \ge 1}a_kz^k, g(z)=f(0)+\sum_{k \ge 1}b_kz^k$ and in all cases equality happens only when $f$ is a rotation of $g$ so $f(z)=\alpha g(\beta z), |\alpha|=|\beta|=1$

In our case the fact that $\mathbb D \subset f(\mathbb D), f(0)=0$ implies that $z$ is subordinate to $f$ hence indeed $1 \le |f'(0)|$. Since by hypothesis we assume that the inclusion is strict we actually have strict inequality too.

(note also that for any $n \ge 2$, $z^n$ is subordinate to $z$, so in general, we cannot have simple inequalities of the type $|a_n| \le |b_n|$ for $n \ge 2$ and the ones above are close to the best we can do)


问:
Range of composite function method
Let $f(x)$ and $g(x) $be functions I've made a method to find range of $f(g(x))$

$$
f(x)=
\begin{cases}
2+x,&   0\leq x<2\\
x^2,& 2\leq x<4
\end{cases}
$$

$$
g(x)=
\begin{cases}
1+x,&   0\leq x<1\\
3-x,& 2\leq x<5
\end{cases}
$$

Now to find the range what I did was put $g(x)$ in $f(x)$ and find intersection of function and x
Like putting

2+(1+x)......0<=(1+x)<2  and 0<=x=<1

Taking the intersection I got this function will be satisfactory for 0=<x<1.
Apparently the answer isnt coming sme and I'm not able to figure out what is wrong in my method . Is there any simpler method

答:
Step 1: examine the range of $g$

if $x\in[0,1)$ then $g(x)\in A=$ ?

$A=[1,2)$

if $x\in[2,5)$ then $g(x)\in B=$ ?

$B=[-2,1)$

Step 2: intersect the range of $g$ with the domain of $f$

$C=(A\cup B)\cap[0,4)=$ ?

$C=[-2,2)\cap[0,4)=[0,2)$

Step 3: the final range is $R=f(C)=$ ?

$R=[2,4)$

Bonus: it is not asked but the domain of $f\circ g$ is $D=g^{-1}(C)=$ ?

this part is a bit more difficult
$C=[0,2)=[0,1)\cup[1,2)$

the first output interval $[0,1)$ is covered by $B$ so we search antecedents $x$ in $[2,5)$, this is the subset $(2,3]$

the second output interval $[1,2)$ is exactly $A$ so antecedents $x$ are the whole $[0,1)$

therefore $D=[0,1)\cup(2,3]$

Title: Range of composite function method


问:
Region D is bounded by below by $z=0$, and above by $x^2+y^2+z^2=4$, and on sides $x^2+y^2=1$ is required to be setup in spherical coordinate
I started Spherical coordinate literally yesterday so please bear with me, im still fairly new to the topic, so i came upon this question and graphically it should look like this to me

I already have the final answer given by :$\displaystyle\int _0^{2\pi }\int _0^2\int _0^{\frac{\pi }{6}}\rho \ ^2\sin \phi \ d\phi \ d\rho \ d\theta +\ \ \int _0^{2\pi }\int _0^1\int _{\frac{\pi }{6}}^{\frac{\pi }{2}}\rho \ ^2\sin \phi \ d\phi \ d\rho \ d\theta +\ \int _0^{2\pi }\int _1^2\int _{\frac{\pi }{6}}^{\sin ^{-1}\left(\frac{1}{\rho \ }\right)}\rho \ ^2\sin \phi \ d\phi \ d\rho \ d\theta \ $

i understood the limit for $\theta$ which is $0<\theta <2\pi $, I even understood why we took $0<\rho \ <1$ and $0<\rho \ <2$ and $1<\rho <2$

what i didn't understand at all are these $\phi $ angles limits ($\sin ^{-1}\left(\frac{1}{\rho }\ \right), \frac{\pi }{6},\frac{\pi }{2}$) I want to be able to form the limits of these $\phi$ on my own but I'm thoroughly confused. Please if anybody can help explain the limits of the $\phi$, it'd be greatly appreciated.

Also side note: this particular example has already been asked once but i didnt understand the answer that was given to that question which is why im asking again. So please kindly don't mark this as duplicate, thank you.

答:
As an aside, I find the integrating bounds easier to set up if you integrate in the order $d\rho d\phi d\theta$.

Use the formulas $z=\rho \cos(\phi)$, $r=\rho \sin(\phi)$, $x=r\cos(\theta)=\rho \sin(\phi)\cos(\theta)$, $y=r\sin(\theta)=\rho \sin(\phi)\sin(\theta)$.

The first integral is easy to understand, since $\rho$ is always integrating from $0$ to the sphere of radius $\rho=2$. The angle $\phi$ starts at $0$ on the positive $z$ axis and the integral stops just shooting $\rho=0$ to $\rho=2$ when the sphere intersects the cylinder. Note that by symmetry with respect to $\theta$, this intersection will hold for all $\theta$. Hence, when the cylinder $x^2+y^2=1$ intersects $x^2+y^2+z^2=4$ (so $\rho=2$), this implies $z=\sqrt{3}$ so $2\cos(\phi)=\sqrt{3} \implies \phi=\frac{\pi}{6}$
Thus, the setup for the first region is done.

For the other region, we consider as we keep increasing $\phi$ beyond $\frac{\pi}{6}$. For this region, the integration with respect to $\rho$ is always from $0$ to the cylinder, with $\frac{\pi}{6}\leq \phi \leq \frac{\pi}{2}$. In this region, $x^2+y^2=1$ so $r=1$. From the formula above, this means $1=\rho\sin(\phi)$. As we can see, we can either set up the integration to be $\rho=0$ to $\rho=\frac{1}{\sin(\phi)}$ with $\phi=\frac{\pi}{6}$ to  $\phi=\frac{\pi}{2}$, or set it up so that we first integrate we respect to $\phi$, then $\rho$ (which is what your answer has).

To see why this splits as two integrals with the order $d\phi d\rho$, make a $\phi$-$\rho$ plot with the $y$-axis being $\rho$ and the $x$-axis being $\phi$. Plot the function $\rho(\phi)=\frac{1}{\sin(\phi)}$ for $\frac{\pi}{6}\leq \phi \leq \frac{\pi}{2}$.

If you integrate with respect to $\phi$ first, you are integrating from $\phi=\frac{\pi}{6}$ to $\frac{\pi}{2}$ when $0\leq \rho \leq 1$. After that, $\phi$ goes from $\phi=\frac{\pi}{6}$ to $\phi=\arcsin(\frac{1}{\rho})$. This occurs from $\rho=1$ to $\rho=2$.

As I mentioned above, if you take this same picture, but integrate with respect to $\rho$ first, you are always integrating $\rho=0$ to $\rho=\frac{1}{\sin(\phi)}$. And this occurs from $\phi=\frac{\pi}{6}$ to $\phi=\frac{\pi}{2}$.


Title: Fourier series for $f(x) = \sqrt{1 - \cos x}$

Body:
I want to calculate the Fourier series for the $2\pi$ periodic function $f(x) = \sqrt{1 - \cos x}$ for $-\pi \leq x \leq \pi$

The Fourier series is given by:
$$f(x) = \frac{a_{0}}{2} + \sum\limits_{n = 1}^{\infty}a_{n}\cos nx + \sum\limits_{n = 1}^{\infty}b_{n}\sin nx$$

Where:

$a_{0} = \frac{1}{\pi}\int\limits_{-\pi}^{\pi}f(x)dx$

$a_{n} = \frac{1}{\pi}\int\limits_{-\pi}^{\pi}f(x) \cos nxdx$

$b_{n} = \frac{1}{\pi}\int\limits_{-\pi}^{\pi}f(x) \sin nxdx$

I know how to go about these questions and have no problem with the integration.

My question is do we need to split up the integrals into $0 \leq x \leq \frac{\pi}{2}$ and $\frac{\pi}{2} \leq x \leq \pi$ to account for the graphs on each side of the origin? Or can we just compute directly from the equations I wrote above with $-\pi \leq x \leq \pi$? Also I don't think we would, but would you consider odd/even functions?

Any help on understanding what to do would be great thank you

Body:
$$
f(x)=1-\cos x =2\sin^2 (x/2)
$$
and hence
$$
\sqrt{1-\cos x}=\sqrt{2}|\sin (x/2)|
$$
Hence, since is even, then $f(x)=\frac{a_0}{2}+\sum_{n=1}^\infty a_n\cos (nx)$, where
$$
a_n=\frac{2}{\pi}\int_0^\pi f(x)\cos (nx)\,dx=\frac{2\sqrt{2}}{\pi}\int_0^\pi
\sin(x/2)\cos (nx)\,dx \\ =\frac{\sqrt{2}}{\pi}\int_0^\pi
\big(\sin(nx+x/2)-\sin(nx-x/2)\big)\,dx  \\=\frac{\sqrt{2}}{\pi}
\left(-\frac{1}{n+½}\cos(nx+x/2)+\frac{1}{n-½}\cos(nx-x/2)\right)\Big|_{x=0}^{x=\pi}
\\=\frac{\sqrt{2}}{\pi}\left(\frac{1}{n-½}-\frac{1}{n+½}\right)=
\frac{\sqrt{2}}{\pi}\cdot\frac{4}{4n^2-1}.
$$
So finally
$$
\sqrt{1-\cos x}=\frac{2\sqrt{2}}{π}−\frac{4\sqrt{2}}{π}\sum_{n=1}^\infty\frac{1}{4n^2−1}\cos nx
$$


问:
Conditional Probability of $P(X|X^2)$ for uniform distribution.
Assume $X \sim U[-1,1]$ is a uniform distribution. We can tell intuitively that
$$P(X=k|X^2=x^2) =
  \begin{cases}
    1/2 \;\;\;\;\;\;\;\text{if} \;\; k=-x  \\
    1/2 \;\;\;\;\;\;\;\text{if} \;\; k=x
  \end{cases}$$

But I want to confirm this using regular conditional probability. However, when I try the definition as $P(X\in A|Y=y):=\lim\limits_{\{Y=y\}\in U}\frac{P(A\cap U)}{P(U)}$, the conditional probability is $0$ since for example if trying to compute $P(X=2|X^2=4):=\lim\limits_{\epsilon  \to 0}\frac{P(X=2 \;\cap\; X^2 \in [4-\epsilon,4+\epsilon])}{P(X^2 \in [4-\epsilon,4+\epsilon])}$, which is just $0$ because $X$ is continuous distribution. Maybe I got something wrong. Please help me compute this conditional probability formally.

答:
Since $U(-1,1)$ is a continuous distribution, $\Pr(X=k|X^2=x^2)$ is not well defined. Instead you first find the CDF given $u \leqslant X^2 \leqslant u+\Delta u$, for some small $\Delta u$ (you can assume it's positive), then take the derivative w.r.t. $k$, and let $\Delta u \to 0$ to get your conditional PDF.

So in this case your unconditional probability is
$$
\Pr(u \leqslant X^2 \leqslant u+\Delta u)=\sqrt{u+\Delta u} - u. \tag 1
$$

Then you need to calculate
$$
\Pr(X \leqslant k \text{ and } u \leqslant X^2 \leqslant u+\Delta u).\tag 2
$$

This is a piece-wise linear function of $k$. I will leave the details to you.

Finally you calculate the ratio of (2) to (1), take the derivative, let $\Delta u \to 0$, you will get your conditional probability mass function.

============
Update:

I found this interesting: https://www.probabilitycourse.com/chapter4/4_3_2_delta_function.php

It basically says that the "symbolic derivative" of the Heavyside step function $u(x)$ is the Dirac delta function $\delta(x)$, where

$$
u(x) = \begin{cases} 1, & x\geqslant 0 \\ 0, & x < 0 \end{cases}
$$

and
$$
\delta(x) = \begin{cases} \infty, & x=0 \\ 0, & \text{elsewhere} \end{cases}
$$

And any discrete distribution has a generalized PDF in terms of the delta function:
$$
f_X(x) = \sum \Pr(X=x_k) \delta (x-x_k).
$$

Now back to OP's question. We look at the joint distribution of $(X,Y)$ on $[-1,1] \times [0,1]$, where $X \sim U[-1,1], Y=X^2, a.s.$

On one hand,
$$
f_{X|Y}(x | y) =
\frac 12 \left( \delta(x+\sqrt y) + \delta(x-\sqrt y) \right)
$$

$$
f_Y(y) = \frac{1}{2\sqrt y}
$$

On the other hand,
$$
F_{X,Y} (x,y) = \Pr(X\leqslant x, X^2\leqslant y)=
\Pr\left(-\sqrt y \leqslant X\leqslant \min(x,\sqrt y)\right) \\
= \begin{cases}
\frac{x+\sqrt{y}}{2}, & x < \sqrt y\\
\sqrt y, & x>\sqrt y
\end{cases}\\
= \frac{x+\sqrt y}{2} - \frac{x-\sqrt y}{2} u(x-\sqrt y)
$$

We want to show that

$$
f_{X,Y} (x,y) = f_{X|Y} (x|y) f_Y(y)
$$

But the link I provided did not discuss the multivariate distribution case. Getting $f_{X,Y} (x,y)$ involves taking the derivative of $F_{X,Y}(x,y)$ w.r.t. to $x$ and $y$, which in term requires the derivative of the delta function. I only found this: https://en.wikipedia.org/wiki/Unit_doublet but there is little information.


问:
Lebesgue Integral of 1 over $\mathbb{R}$
We've been studying Lebesgue integrals in my analysis class and I'm unsure how to evaluate,
$$\int_{\mathbb{R}} 1 d\lambda(t). $$
I know it's a simple function, is $\lambda(\mathbb{R})=+\infty$? If so, would this integral just be infinity?

答: Yes. It is $+\infty$ because $\lambda(\mathbb{R}) = +\infty$. You may check this  without regarding $1$ (let's call it $f$) as a non-negative simple function and evaluating $1·\infty$. Simply regard $f$ as a non-negative function.  Consider the sequence of simple functions $\varphi_n = \mathbf{1}_{[-n,n]}$ (here $\mathbf{1}_{[-n,n]}$ stands for the indicator function of $[-n,n]$), all of which are $\leq f$ everywhere. The integral of $\varphi_n$ is $1\cdot 2n$ and so $\int f\geq 2n$ for all $n\in\mathbb{N}$, thus, $\int f = +\infty$.


问:
combining two partial order relations
Relation $R_1$ on set $A$ and Relation $R_2$ on set $B$ are defined.

$R_1$ and $R_2$ are both partial orders.

We define $R$ like below:
$$(a, b) R (x, y) \quad\text{ if }\quad aR1x \;\text{ and }\; bR2y$$
Prove that $R$ is also a partial order.

I know that a relation $R$ on a set $S$ is called a partial order if it is reflexive, anti-symmetric, and transitive.

答:
What the problem requires of you is to manually check if the three partial order properties are satisfied.

Reflexivity: $(a,b)R(a,b)$ is in $R$ since $aR_1a$ and $bR_2b$ are satisfied due to their being partial orders.

Anti-symmetry: suppose we have $(a,b)R(c,d)$ and $(c,d)R(a,b)$. Then $aR_1c$ and $cR_1a$, which implies $a=c$ by anti-symmetry of $R_1$, and similarly for $R_2$. Hence $(a,b)=(c,d)$, as required.

Transitivity: let $(a,b)R(c,d)$ and $(c,d)R(e,f)$. Then $aR_1c$ and $cR_1e$, which implies $aR_1e$ by transitivity of $R_1$, and similarly for $R_2$. Hence we have $(a,b)R(e,f)$, as required.

Title: combining two partial order relations


问:
Closed form for $1-2+3-4+\cdots(-1)^{n-1}n$
How do I find a closed formula for the following?

$$\sum_{i=1}^n(-1)^{i-1}i$$

If $n$ is odd number, I can express it as $\frac{n+1}{2}$; if $n$ is even, then the expression if $\frac{-n}{2}$, but how do you find a expression that can express it in one formula?

答:
You can also solve the linear induction relation $S_n-S_{n-1}=(-1)^{n-1}n$

Homogeneous equation $H_n-H_{n-1}=0$ has solution $H_n=H_0$

A particular solution can be found of the form $(an+b)(-1)^{n-1}$ (since $-1$ is not root of homogeneous characteristic equation $x-1=0$, we can find a polynomial of same degree).

$(-1)^{n-1}n=(an+b)(-1)^{n-1}-(a(n-1)+b)(-1)^{n-2}=(-1)^{n-1}(an+b+an-a+b)$

$\iff a=\frac 12,\ b=\frac 14$

Therefore we get $S_n=H_0+(-1)^{n-1}(\frac 12n+\frac 14)$

Solving for $S_0=0=H_0-\frac 14$ we finally find back the formula given by Arthur, but via another resolution method: $$S_n=\frac 14+(-1)^{n-1}\left(\frac 12n+\frac 14\right)$$

Of course in this situation the closed formula can be guessed, but this theoretical method is more general and can be applied to more complicated sums.
I think it's good to understand it on a simple problem too.


问:
Real projective space is Hausdorff
Define the projective quotient space as follows:

$\mathbb{R}P^n$ is the quotient of $\mathbb{R}^{n+1}-\{0\}$ by the quotient equivalence relation $(x_0,\cdots,x_n)\sim (\alpha x_0,\cdots,\alpha x_n)$ for all $\alpha\neq 0$. Denote the quotient map as $r$.

Consider the set $V_i=\{x\in\mathbb{R}^{n+1}:x_i=0\}$. Denote $U_i$ as $r(\mathbb{R}^{n+1}-V_i)$.

Question: Prove $\mathbb{R}P^n$ is Hausdorff.

My approach:

I have proved that $U_i$ is open and Hausdorff for all $i$. Thus, for two distinct points $x$, $y$ in $\mathbb{R}P^n$, if there are in the same $U_i$, we can find two distinct open sets to separete them. But there are also points that are not in the same $U_i$. How do we deal with this kind of points?

Update:

Consider the diagonal set $\Delta(\mathbb{R}P^n)=\{(x,x):x\in\mathbb{R}P^n\}=\cup_{i}\{(x,x):x\in U_i\}=\cup_{i}\Delta(U_i)$. (The second equality is because $\{U_i\}$ forms a cover for $\mathbb{R}P^n$.)

Because $U_i$ is Hausdorff, each $\Delta(U_i)$ is closed. This implies $\mathbb{R}P^n$ is Hausdorff.

Is my proof correct? Thank you!

答: Hint: (I think the sets $U_i$ and $V_i$ are a bit of a distraction). Each point in $\Bbb{RP}^n$ is represented by a unique pair of points $\pm x \in \Bbb{R}^{n+1} \setminus \{0\}$ with $\|x\| = 1$.Given two distinct pairs $\pm x$ and $\pm y$ with $\|x\| = \|y\| = 1$ you can find disjoint open cones $C_x$ and $C_y$ such that $\pm x \in C_x$ and $\pm y \in C_i$. The images of $C_x$ and $C_y$ in $\Bbb{RP}^n$ are open sets that separate $r(x)$ and $r(y)$.

Title: Real projective space is Hausdorff


问:
Tensor products, projectivity, and subrings
Let $R$ be a unital subring of a unital ring $S$. Let $M$ be an $S$-bimodule that is projective as a right and left module, and let $N$ be a $R$-submodule of $M$, that is again projective as a right and left module. Consider the canonial map
$$
N \otimes_R N \to M \otimes_S M, ~~~~~~~ n \otimes n' \mapsto n \otimes n'.
$$
Will this map be an embedding in general?

答: No. Consider the case $R = \mathbb{R}$, $S = \mathbb{C}$, $N = M = \mathbb{C}$. Then $N \otimes_R N = \mathbb{C} \otimes_{\mathbb{R}} \mathbb{C} \cong \mathbb{R}^4$ and $M \otimes_S M = \mathbb{C} \otimes_{\mathbb{C}} \mathbb{C} \cong \mathbb{C}$. The map $N \otimes_R N \to M \otimes_S M$ you described is always $R$-linear, and $\dim_{\mathbb{R}} (\mathbb{R}^4) = 4 > 2 = \dim_{\mathbb{R}}(\mathbb{C})$, so the map will not be injective in this case.


问:
Problem designing a control law
I am trying to build a posture regulation control which works with acceleration inputs.

Doing it with velocity inputs,I have well understood how it works. Moreover,for now, the controller only controls the position $x,y$ and sends to zero these two coordinates, so it brings $(x,y)=(0,0)$ and does not control the orientation.

I have built the acceleration level control for this case, and what I get is:

where we can see that the $x$ and $y$ coordinates go to zero, and $\theta$ is not controlled, as I expected. Even if the acceleration profile is a bit weird (and this makes me think that the error may be in the implementation, but I am not sure):

I think the acceleration is weird because I was expecting a profile of the type:

where on top there is the linear acceleration and the bottom one is the angular acceleration.

So I am not sure I am doing well. In this case, I have defined the polar coordinates:

$R = \sqrt{x^2+y^2}$

$\gamma=atan2(y.x)-\theta$

and the controller is:

```<code> x = state(1);
 y = state(2);
 theta = state(3);
 v_state = state(4);
 omega_state = state(5);
 theta_w = state(6);

 k_1 = gain(1);
 k_2 = gain(2);
 k_3 = gain(3);
 k_4 = gain(4);
 k_5 = gain(5);

 temp = x^2+y^2;

 R = sqrt(temp);
 gamma = atan2(y,x)-theta;
 delta = gamma + theta;

 sinc = sin(gamma)*sign(cos(gamma));

v=k_1*R*sign(cos(gamma));
omega = k_2*(k_1*sinc*(gamma+k_3*delta)/(2*gamma + k_3*delta));

a = k_4*(v-v_state);
alpha = k_4*(omega-omega_state);

inputs = [a;alpha];
</code>```

where $v$ and $\omega$ can be obtained with the same procedure that I have written below, or also from the feedback linearization as it is done in the paper.

Moreover, v_state and omega_state are respectively linear and angular velocities of the system, that now have become states, since I want acceleration inputs.

And the model, written in Matlab , is:

```<code>a = inputs(1);
alpha = inputs(2);

x = state(1);
y = state(2);
theta = state(3);
v_state = state(4);
omega_state = state(5);

x_dot = -v_state*cos(theta)+y*omega_state;
y_dot = -v_state*sin(theta)-x*omega_state;
theta_dot = omega_state;
theta_w_dot = -omega_state;
v_dot = a;
omega_dot = alpha;
</code>```

Note: the above codes are from Simulink blocks.

Now, I would like to control also the final orientation,and bring it to zero, so I define the polar coordinate(in the code above I have already written the control law that should control also the final orientation that I am going to derive below):

$\delta = \theta + \gamma$

and if I take the derivatives of all these polar coordinates, I get the following kinematic model:

$\dot{R}=-vcos\gamma$

$\dot{\gamma}=v\frac{sin\gamma}{R}-2\omega$

$\dot{\delta}=v\frac{sin\gamma}{R}-\omega$

and the objective is to have $R=0, \gamma=0,\delta=0$.

Now, to derive the control law that controls also the orientation I use a Lyapunov function:

$V = \frac{1}{2}(R^2+\gamma^2+\delta^2)>0$

$\dot{V}=R\dot{R}+\gamma\dot{\gamma}+\delta\dot{\delta}= R(-vcos\gamma)+\gamma(v\frac{vsin\gamma}{R}-2\omega)+\delta(\frac{vsin\gamma}{R}-\omega)= v(-Rcos\gamma + \gamma\frac{sin\gamma}{R}-\delta \frac{sin\gamma}{R})-\omega(2\gamma+\delta)$

and this has to be less than or equal to zero for the Lyapunov criterion.

From here, I derive that

$v = k_1\cdot Rcos\gamma$

$\omega = \frac{k_1\cdot sin\gamma\cdot cos\gamma(\gamma+k_2\delta)}{2\gamma+k_2\delta}$

but if i try these and use them as desired linear and angular velocities, I get :

so the orientation still does not go to zero.

I don't understand what I am doing wrong, and I have been on this for a while now and I am really stuck. The reasoning seems correct, at least to me, but apparently I am doing something wrong that I don't get.

Can somebody please help me understand what I am doing wrong? Do you think my approach is correct, or should I try another way for solving the problem?

For the implementation I have used Matlab and Simulink.

Could the problem be connected to the initial configuration? I really don't get what I am doing wrong, i am trying from a while now, but I am really stuck.

I have also tried to not make the values of $\gamma$ and $\delta$ don't go to large, so I have added this lines of code:

```<code>gamma_t = atan2(y,x)-theta;
delta_t = gamma_t + theta;

gamma = mod(gamma_t+pi,2*pi)-pi;
delta = mod(delta_t+pi,2*pi)-pi;
</code>```

but this didn't solve the problem.

About the plots, if it can add something, I have done in the following way:

```<code>x=configuration.signals.values(:,1);
y=configuration.signals.values(:,2);
theta=configuration.signals.values(:,3);
t=configuration.time;
a=inputs.signals.values(:,1);
alpha=inputs.signals.values(:,2);

figure(1);
plot(t,x,t,y,t,theta);grid
legend('x','y','theta');

figure(2);
plot(t,a,t,alpha);grid
legend('a','alpha');
</code>```

if I have to add something to the question to make it clearer please tell me. I really don't know how to move on.

I am trying to read several papers and references, such as:

Stabilized Feedback Control of Unicycle Mobile Robots

Posture regulation for unicycle-like robots with prescribed performance guarantees

and they seem to use the same approch I have used, so there has to be some mistake in some step I have done, which I do not understand.

Can someone please help me?

答:
Something is wrong with your Lyapunov analysis.

\begin{align}
\dot{V}&=R\dot{R}+\gamma\dot{\gamma}+\delta\dot{\delta} \\
&= R(-vcos\gamma)+\gamma(v\frac{sin\gamma}{R}-2\omega)+\delta(v\frac{sin\gamma}{R}-\omega) \\
&= v(-Rcos\gamma + \gamma\frac{sin\gamma}{R}+\delta \frac{sin\gamma}{R})-\omega(2\gamma+\delta)
\end{align}
You suggested using,
\begin{align}
v &= k_1\cdot Rcos\gamma \\
\omega &= \frac{k_1\cdot sin\gamma\cdot cos\gamma(\gamma+k_2\delta)}{2\gamma+k_2\delta}
\end{align}
Substituiting this in we get that,
\begin{align}
\dot{V} &= k_1\cdot Rcos\gamma(-Rcos\gamma + \gamma\frac{sin\gamma}{R}+\delta \frac{sin\gamma}{R})-\frac{k_1\cdot sin\gamma\cdot cos\gamma(\gamma+k_2\delta)}{2\gamma+k_2\delta}(2\gamma+\delta) \\
&= -k_{1}R^{2}cos^{2}\gamma + k_{1}(\gamma + \delta)cos\gamma sin\gamma -\frac{k_1\cdot sin\gamma\cdot cos\gamma(\gamma+k_2\delta)}{2\gamma+k_2\delta}(2\gamma+\delta)
\end{align}
If we assume that $k_{2} =1$ then,
\begin{align}
\dot{V} &= -k_{1}R^{2}cos^{2}\gamma + k_{1}(\gamma + \delta)cos\gamma sin\gamma -\frac{k_1\cdot sin\gamma\cdot cos\gamma(\gamma+\delta)}{2\gamma+\delta}(2\gamma+\delta) \\
&= -k_{1}R^{2}cos^{2}\gamma + k_{1}(\gamma + \delta)cos\gamma sin\gamma - k_1\cdot sin\gamma\cdot cos\gamma(\gamma+\delta) \\
&= -k_{1}R^{2}cos^{2}\gamma
\end{align}
and $\dot{V}\leq 0$ if $k_{1}>0$. Therefore, $\dot{V}=0$ when $\gamma = \frac{\pi}{2} + \pi k$ or $R = 0$. Though your dynamics when $R=0$ are not well defined.

If $k_{1}\neq 1$ then
\begin{align}
\dot{V} &= -k_{1}R^{2}cos^{2}\gamma + \left(k_{1}(\gamma + \delta) -\frac{k_1(\gamma+k_2\delta)}{2\gamma+k_2\delta}(2\gamma+\delta)\right)cos\gamma sin\gamma \\
&= -k_{1}R^{2}cos^{2}\gamma + k_{1}\left((\gamma + \delta) -\frac{(\gamma+k_2\delta)}{2\gamma+k_2\delta}(2\gamma+\delta)\right)\frac{1}{2}sin(2\gamma)
\end{align}
As you increase $\gamma$ the second term will oscilate from being positive and negative and therefore $\dot{V}\not\leq 0$.

Title: Problem designing a control law


问:
Find the distributional derivative of a piecewise continuous function $f(x)=x^2$, when $x <3$ and $f(x)=x^3,$ when $x\geq 3$.
Find the distributional derivative of a piecewise continuous function $f(x)=x^2$, when $x <3$ and $f(x)=x^3,$ when $x\geq 3$.

I know that I want to set this up as $$f(\phi)=-\int_{-\infty}^\infty f(x)\phi'(x)=-\int_{-\infty}^3x^2\phi'(x)dx - \int_{3}^\infty x^3\phi'(x)dx.$$

Using integration by parts, I can express the integrals as $$\int_{-\infty}^3x^2\phi'(x)dx = x^2\phi(x)\bigg|_{-\infty}^3-2\int_{-\infty}^3x\phi(x)dx = 9\phi(3)-2\int_{3}^\infty x\phi(x)dx$$ $$\int_{3}^\infty x^3\phi'(x)dx = x^3\phi(x)\bigg|_{3}^\infty-3\int_{3}^\infty x^2\phi(x)dx=27\phi(3)-3\int_{3}^\infty x^2\phi(x)dx.$$

but I'm unsure how to proceed from here. There isn't an obvious way for me to evaluate the remaining integrals, or to reduce them into the form of something else known.

Any thoughts would be greatly appreciated.

Thanks in advance.

答:
You wrote $9$ for $9\phi (3)$ and $27$ for $-27\phi (3)$ So you get $-9\phi (3)+2 \begin {align} \int\limits_3^{\infty} x\phi(x)dx \end {align}+27\phi (3)$ $+3\begin {align} \int\limits_3^{\infty} x^{2}\phi (x)dx \end {align}$. Hence the derivative is $=18\delta_3+2x\chi_{(-\infty, 3)}+3x^{2}\chi_{(3,\
\infty)}$.

Notation: $\delta_3(\phi)=\phi(3)$ for any test function $\phi$.


问:
Dimension(s) concept - is there a difference in meanings?
In physics, a vector (say p) is in 3D when it has 3 components.

In linear algebra, a matrix (say Q) made of 2 vectors (say v and w) is

2D when v and w are independent.

1D when two vectors are dependent.

0D when both are zero vectors.

It does not matter how many components (say 75) they have.

why is there such a difference in meaning? is anyone wrong.

答:
You are mixing 3 things which speaking loosely can be called dimensions. And you do seem to be speaking loosely (not too rigorously in math sense).

dimension of a vector - the number of its components/elements

dimension(s) of a matrix - it is just the pair of two natural numbers $(n, m)$ which says how many rows/columns that matrix has

rank of a matrix - the dimension of the vector space generated by its rows/columns

That's where your confusion comes from.


问:
An algebra with no von Neumann unitization
What is an example of a non unital $C^*$ algebra which can not be an essential ideal of any von Neumann algebra?

答:
I believe $C_0(\mathbb R)$ is  an example.

If $C_0(\mathbb R)$ is an essential ideal in some unital algebra $A$,  then
$A$ embedds in the multiplier algebra of $C_0(\mathbb R)$,  so $A$ must be commutative.

The spectrum of $A$ will then
be a compact space containing an open dense copy of $\mathbb R$.   Since the closure of a connected set is connected,  we see that the spectrum of $A$ is connected.

This implies that $A$ has no nontrivial projections
so it cannot be a von Neumann algebra.


答:
You have made errors: $\left. x^2 \phi(x) \right|_{-\infty}^{3} = 9 \phi(3)$, not just $9$. Likewise for the $x^3\phi(x)$ part. And how did you go from $\int_{-\infty}^{3}$ to $\int_3^\infty$?

What you should get is
$$
\langle f', \phi \rangle
= -\langle f, \phi' \rangle
= -\int_{-\infty}^3x^2\phi'(x)\,dx - \int_{3}^\infty x^3\phi'(x)\,dx
\\= -9\,\phi(3) + 27\,\phi(3) + \int_{-\infty}^3 2x\,\phi(x)\,dx + \int_3^\infty 3x^2\,\phi(x)\,dx
\\= \langle 18\,\delta(x-3),\phi(x)\rangle + \langle g(x), \phi(x) \rangle,
$$
where
$$
g(x) = \begin{cases}
2x & (x<3) \\
3x^2 & (x>3)
\end{cases}
$$

Thus,
$$
f'(x) = 18\,\delta(x-3) + g(x).
$$

Body:
Find the distributional derivative of a piecewise continuous function $f(x)=x^2$, when $x <3$ and $f(x)=x^3,$ when $x\geq 3$.

I know that I want to set this up as $$f(\phi)=-\int_{-\infty}^\infty f(x)\phi'(x)=-\int_{-\infty}^3x^2\phi'(x)dx - \int_{3}^\infty x^3\phi'(x)dx.$$

Using integration by parts, I can express the integrals as $$\int_{-\infty}^3x^2\phi'(x)dx = x^2\phi(x)\bigg|_{-\infty}^3-2\int_{-\infty}^3x\phi(x)dx = 9\phi(3)-2\int_{3}^\infty x\phi(x)dx$$ $$\int_{3}^\infty x^3\phi'(x)dx = x^3\phi(x)\bigg|_{3}^\infty-3\int_{3}^\infty x^2\phi(x)dx=27\phi(3)-3\int_{3}^\infty x^2\phi(x)dx.$$

but I'm unsure how to proceed from here. There isn't an obvious way for me to evaluate the remaining integrals, or to reduce them into the form of something else known.

Any thoughts would be greatly appreciated.

Thanks in advance.


问:
Is ${}^{*}\mathbb{Q}={}^{*}\mathbb{R}$?
This is a follow-up question of a previous one.

Is it true ${}^{*}\mathbb{Q}={}^{*}\mathbb{R}$ ?

It seems to me it's true because for any $x \in{}^{*}\mathbb{R}$, there is a $y \in {}^{*}\mathbb{Q}$ such that $y < x$ and vice versa, and because for any $x' \in \mathbb{R}$, there is a $y' \in {}^{*}\mathbb{Q}$ such that $y' = x'$.

答: No, ${}^*\Bbb Q\ne{}^*\Bbb R$. It’s not true that $\Bbb R\subseteq{}^*\Bbb Q$. In the ultrapower construction this is especially easy to see: in ${}^*\Bbb R$, $\Bbb R$ is identified with the set of equivalence classes of constant sequences of reals, and the class of a constant sequence with irrational value is easily seen to be different from every element of ${}^*\Bbb Q$, though it’s also easy to see that there are elements of ${}^*\Bbb Q$ infinitely close to each such class.</p>

Title: Is ${}^{*}\mathbb{Q}={}^{*}\mathbb{R}$?


问:
Find the limit of the function without L'Hôpital's rule
I have a problem, such as:

$$\lim_{x \to 0} \left(\frac{\cos(ax)}{\cos(bx)}\right)^\frac{1}{x^2}$$

How do I solve this problem without using L'Hôpital's rule or small-o?

Thanks!

Body:
Basically a variant of user's answer: Using familiar limits $(1-u)^{1/u}\to e^{-1}$ and ${\sin u\over u}\to1$ as $u\to0$, we have

$$\begin{align}
(\cos ax)^{1/x^2}
&=(\cos^2ax)^{1/(2x^2)}\\
&=(1-\sin^2ax)^{1/(2x^2)}\\
&=\left((1-\sin^2ax)^{1/\sin^2ax}\right)^{(a^2/2)(\sin ax/(ax))^2}\\
&\to (e^{-1})^{(a^2/2)\cdot1^2}\\
&=e^{-a^2/2}
\end{align}$$

So in general we have

$$\left(\cos ax\over\cos bx\right)^{1/x^2}={(\cos ax)^{1/x^2}\over(\cos bx)^{1/x^2}}\to{e^{-a^2/2}\over e^{-b^2/2}}=e^{(b^2-a^2)/2}$$


答:
I think what you're trying to do is to find conditional law for $X$ and this one question is closely related to the problem of probability kernel.

Theorem

Let $(E_1, \mathcal{E}_1 )$ and $(E_2, \mathcal{E}_2 )$ be two measurable function. Suppose further that $(E_2, \mathcal{E}_2 )$  is regular  ( Remark: $\mathbb{R}$ is regular).

Let $m$ be a finite positive measure on $E_1 \otimes E_2$, with $\mu$ as its first marginal law.

Then there is a unique (in almost sure sense) probability kernel from $(E_1, \mathcal{E}_1 )$ to $(E_2, \mathcal{E}_2 )$  such that $m= \mu \otimes p$, that is:

$\int_{E_1 \times E_2} gdm= \int_{E_1} \mu(dx) \int_{E_2} p(x, dy)g(x,y)$

for all measureable function $g:E_1\times E_2 \longrightarrow \mathbb{R_+}$

In particular, we have:

Corollary:

$\mathbb{E}(f(Y) | X) = \int_{\mathbb{R}} f(y) p(X,dy)   $ for all positive measureable function $f$, a.s

where $p$ is the respective kernel for $ m= \mathcal{L}(X,Y)$

Remark 1 In fact, there are a lot of technical questions in these statements.(measurability, regularity, almost sureness, spaces of measures, etc.) So if you may, please don't be concerned too much of those because it'll be a heavy burden on you.

Remark 2 Roughly speaking, $p$ represents conditional law.

Remark 3 $p(X=2|X^2=2)$ doesn't mean anything because $P(X^2=2)=0$.We can replace it by anything we want, so "calculating it" also doesn't mean anything. But $\mathcal{L}(X| X^2)$ is a different thing. That's why we introduce the notion of probability kernel.

Back to your question, what you're trying to find is the probability kernel for $(X^2,X)$
.  You can find it by your intuition or anything and check it with the unicity in the above theorem ( yeah, this is the right way of *calculating it* in my opinion) then use it in form of the above corollary when you want.

**Appendix **

$\mathcal{L}(X)$: law of $X$.

Disclaim any errors in these statements are mine any maybe due to my reluctance to check the question of almost sureness.

Title: Conditional Probability of $P(X|X^2)$ for uniform distribution.

Body:
Assume $X \sim U[-1,1]$ is a uniform distribution. We can tell intuitively that
$$P(X=k|X^2=x^2) =
  \begin{cases}
    1/2 \;\;\;\;\;\;\;\text{if} \;\; k=-x  \\
    1/2 \;\;\;\;\;\;\;\text{if} \;\; k=x
  \end{cases}$$

But I want to confirm this using regular conditional probability. However, when I try the definition as $P(X\in A|Y=y):=\lim\limits_{\{Y=y\}\in U}\frac{P(A\cap U)}{P(U)}$, the conditional probability is $0$ since for example if trying to compute $P(X=2|X^2=4):=\lim\limits_{\epsilon  \to 0}\frac{P(X=2 \;\cap\; X^2 \in [4-\epsilon,4+\epsilon])}{P(X^2 \in [4-\epsilon,4+\epsilon])}$, which is just $0$ because $X$ is continuous distribution. Maybe I got something wrong. Please help me compute this conditional probability formally.


问:
Obstruction to a Spin structure on a bundle ξ, and ξ ⊕ $n$ det ξ
In Ref, it says that:

The obstruction to putting a Spin structure on a bundle $ξ (= Rn → E → B)$ is $w_2(ξ) \in H^2(B;Z/2Z)$.

Pin± structures is that

Pin− structures on ξ correspond to Spin structures on ξ ⊕ det ξ.

Pin+ to Spin structures on ξ ⊕ 3 det ξ where det ξ is the determinant line bundle.

Question

Why do we have det ξ and 3 det ξ instead of other numbers? I suppose the 2 det ξ is related to the $w_1(ξ)^2$?

How to show: a variety of “descent” theorems of the type: a Pin± structure on ξ⊕η descends to a Pin+, Pin−, or Spin structure on ξ when dim η=1 or 2 and various conditions on η are satisfied.

答:
Your questions are addressed in Lemma 1.7 of the paper you cite, but I shall give an explanation of this from the perspective of classifying spaces.

Let me answer your questions in reverse order.  First, a general setting for the "descent theorems" you are looking for can be encapsulated as the existence of pullback squares of the form
\begin{equation}
\tag{1}
\begin{array}{ccc}
BG_1 & \rightarrow & BG_2 \\
\downarrow & & \downarrow \\
BO(n) & \xrightarrow{\oplus \eta} & BO(n+r),
\end{array}
\end{equation}
which should be read as: "a $G_1$-structure on a $n$-plane bundle $\xi$ corresponds to a $G_2$-structure on $\xi \oplus \eta$".  Any time you can produce such a square will give you a descent theorem.

Where might these pullback squares come from?  For the rest of this answer, we focus primarily on the case $G_1, G_2 \in \{\mathrm{Spin}(n), \mathrm{Pin}^\pm(n)\}$ and $\eta = r \det \xi$ for some integer $r$.

Suppose we had a diagram
\begin{equation}
\tag{2}
\begin{array}{ccc}
BO(n) & \xrightarrow{\oplus r \det} & BO(n+r) \\
\downarrow & & \downarrow \\
K(\mathbb{Z}/2, 2) & \xrightarrow{\sim} & K(\mathbb{Z}/2, 2).
\end{array}
\end{equation}

From the cohomology of $BO(n)$, we know that the vertical maps are spanned by $w_2$ and $w_1^2$.  Suppose they are given by $w_2 + a_1 w_1^2$ and $w_2 + a_2 w_1^2$ respectively, where $a_1, a_2 \in \mathbb{Z}/2$.  Then, we can determine when this square commutes up to homotopy based on $r$, $a_1$, and $a_2$ by solving the equation $$(w_2 + a_1 w_1^2)(\xi) = (w_2 + a_2 w_1^2)(\xi \oplus r \det \xi).$$

We summarize the results in the following table:
$$
\begin{array}{|ccc|}
\hline a_1 & a_2 & r \\ \hline
0 & 0 & 0, 3 \pmod{4} \\
0 & 1 & 2, 3 \pmod{4} \\
1 & 0 & 1, 2 \pmod{4} \\
1 & 1 & 0, 1 \pmod{4} \\ \hline
\end{array}
$$

Once we have a commuting square (2), we can take the vertical fibers of the diagram to get a diagram that looks like (1), where $G_i = \mathrm{Pin}^+$ (resp. $\mathrm{Pin}^-$) if $a_i = 0$ (resp. $a_i = 1$).

It remains to show that this resulting diagram (1) is cartesian.  To see why, note that we may back up the fibration one more time to get a map $K(\mathbb{Z}/2, 1) \to K(\mathbb{Z}/2, 1)$ which is an equivalence iff (1) is a pullback.  But this map is the loop of the map $K(\mathbb{Z}/2, 2) \xrightarrow{\sim} K(\mathbb{Z}/2, 2)$, which is an equivalence by construction.

There is one more observation to make.  If $r$ is odd, then the map $BO(n) \xrightarrow{\oplus r \det} BO(n+r)$ factors through $BSO(n+r)$, and we may replace $G_2$ with $\mathrm{Spin}(n+r)$.  Combining this with the table above, we obtain:

$r \equiv 0 \pmod{4}$: a $\mathrm{Spin}$- or $\mathrm{Pin}^+$- or $\mathrm{Pin}^-$-structure on $\xi$ corresponds to the same on $\xi \oplus r \det \xi$.

$r \equiv 1 \pmod{4}$: a $\mathrm{Pin}^-$-structure on $\xi$ corresponds to a $\mathrm{Spin}$-structure on $\xi \oplus r \det \xi$.

$r \equiv 2 \pmod{4}$: a $\mathrm{Pin}^+$-structure on $\xi$ corresponds to a $\mathrm{Pin}^-$-structure on $\xi \oplus r \det \xi$, and vice-versa.

$r \equiv 3 \pmod{4}$: a $\mathrm{Pin}^+$-structure on $\xi$ corresponds to a $\mathrm{Spin}$-structure on $\xi \oplus r \det \xi$.

This proves Lemma 1.7 in your reference.


问:
If $m|n$ and $a$ is a primitive root of $n$, show that $a$ is a primitive root of $m$ (understanding a tip)
I saw the answer to this question and it is the same problem, but i didn't get how to use the tip S.C.B gave.

This was the tip:

$"$First, assume that it is not $a$ primitive root $(\text{mod m})$. Then we have that there exists such $r<\phi(m)$ such that $$a^r\equiv 1(\text{mod m})$$
Now use that, if $n=mk$
$$ϕ(mk)=ϕ(m)ϕ(k)\frac{d}{ϕ(d)}≥ϕ(m)ϕ(k)>rϕ(k)$$
where $d=gcd(m,k)"$

and i saw the answer to this question using group theory, but i want an answer using elementary number theory, if you have a different answer, or tip, it would be good as well.

What i tried was this:

$n=mk$, then
$$a^{\phi(n)}\equiv 1(\text{mod n})\Rightarrow a^{\phi(mk)}\equiv1(\text{mod mk}) \Rightarrow a^{\phi(mk)}\equiv1(\text{mod m})$$

But i don't know how to follow from here

答:
Here's another approach.

Let $\phi_n$ denote the set of elements which are coprime to $n$.

First, show that $a$ is a primitive root modulo $n$ iff $ \{ a^r \pmod{n} \} = \phi_n $.

Second, show that for $m \mid n$,  $ \{ k \pmod{m} | k \in \phi(n) \} = \phi_m$

 (Proof of the hard part) Suppose we have $ \gcd(a, m) = 1$, and we want to "lift" it to an element in $\phi(n)$.

 Use CRT to solve the following system of congruences:

 - $ A \equiv a \pmod{p^i}$, for every prime $p$ that divides $m$ and $n$, and $ p^i \mid\mid n$

 - $ A \equiv 1 \pmod {q^i}$, for every prime $q$ that doesn't divide $m$ but divides $n$, and $q^i \mid \mid n$

 CRT guarantees us a solution since the congruences are coprime, and we have $ A \equiv a \pmod{m} $ as well as $ \gcd(A, n) = 1$.

Hence, conclude that $ a$ is a primitive root mod $m$ in the problem.


问:
Splitting in F8 Confusion
I am so confused about the following question:

Consider the field F8 = F2[θ]/($0^3$ + θ + 1). Check that the
polynomial $x^3$ + x + 1 splits in this field as
$x^3$ + x + 1 = (x + θ)(x + $θ^2$
)(x + $θ^2$ + θ).
Now take the other degree three irreducible polynomial x
3 + x
2 + 1 over F2.
Show that it also splits in F8 and find its roots in F8.

First, what is the θ? Why are θ and x different variables? So is θ then a constant in F8?

For $x^3+x^2+1$, I think we would do $x^3$ = -x - 1 = x + 1, so $x^3 + x^2 + 1$ = $x^2$ + x = x(x+1) with roots 0 and 1. But I'm not sure if this is right since it doesn't use $\theta$.

I'm really just confused about what the $\theta$ represents.

答:
$\Bbb F_8$ has six elements in addition to $0$ and $1$. They each need a name. Here they have chosen to use $\theta$ to construct these names. We get that $\theta$ is an element of $\Bbb F_8$. Also, the two expressions $\theta^3+\theta + 1$ and $0$ denote the exact same element of $\Bbb F_8$.

Then, on top of this, we are making polynomials with elements of $\Bbb F_8$ as coefficients. These polynomials need a variable, here they have chosen $x$ as the name of that variable. Such a polynomial could look like
$$
\theta^2x^5 + (\theta + 1)x^4 + (\theta^2+1)x^3 + x^2 + (\theta^2 + \theta).
$$
This is a fifth degree polynomial, with a variety of coefficients from $\Bbb F_8$.

We have that $x^3+x+1$ is a third-degree polynomial with coefficients from $\Bbb F_8$ (the coefficients all happen to be $0$ and $1$). And $\theta$ is one of its roots. As such, the linear polynomial $x+\theta$ (with constant term $\theta$) is a factor of $x^3+x+1$.

Yes, the concrete construction of $\Bbb F_8$ is as a quotient of the polynomial ring $\Bbb F_2[\theta]$, so technically the elements of $\Bbb F_8$ are equivalence classes of polynomials over $\Bbb F_2$ in the variable $\theta$. And these equivalence classes are then used as the coefficients in a new polynomial ring in the variable $x$. Following along here does take a little getting used to.

There is no big reason to use $\theta$ to make the names of elements in $\Bbb F_8$. We could just as well have used $x$, and said that $\Bbb F_8 = \Bbb F_2[x]/(x^3+x+1)$. However, we would still need a variable name for the polynomials with $\Bbb F_8$-coefficients, and you can't use $x$ for both; confusion would be unavoidable, even for seasoned algebraists. The authors of your problem wanted the polynomial variable to be called $x$, and they wanted to use $\theta$ to name the elements of $\Bbb F_8$.

Title: Splitting in F8 Confusion


问:
Does a sum of measurable functions converge outside a set of measure zero?
I am trying to solve the following problem:

Let $\{f_n\}$ be a sequence of measurable functions defined on a probability space, such that
$$
P\left(f_n=\frac1n\right)=1-P(f_n=0)=\frac1n.
$$
Does $\sum_{n=1}^\infty f_n$ converge outside a set of measure zero?

A similar problem was solved here: Does the sum of sequence of measurable functions converge outside a set of measure zero?. The only difference  is that in my problem $P(f_n=1/n)$ and $P(f_n=0)$ are not summable, so the Borel-Cantelli lemma does not apply. Is there another way to solve this?

答: $Ef_n=\frac  1n P(f_n=\frac  1n)+0P(f_n=0)=\frac  1{n^{2}}$. Hence $\sum Ef_n=\sum \frac 1 {n^{2}} <\infty$. By Tonelli's Theorem this implies $E \sum f_n <\infty$. In turn this implies $\sum f_n <\infty$ almost surely.


问:
Why are "~B" and "$\{x : x\notin B\}$" not the same thing? Trying to understand basic set theory.
Why are they different: "~B" vs. $\{x:x \notin B\}$?  They both mean elements not in B. As long as that's satisfied, it should work, right?

How we use them in class:

$A\setminus B$ = {$x: x \in A \text{ and } x\notin B $}

$x \in A\setminus B$  iff $(x \in A) \land (x \notin B)$

Example:

Let B = {2, 4, 6, 8}.

the integers 1, 3, 5, 7, 9

the real numbers greater than 25

the function $f(x)=x^2+3$

the circle of radius 1 centered at the origin of the plane
the Empire State Building
my uncle Wilbur

Solution explains they are different:

$\{x:x \notin B\}$ contains all of the above.

1, 3, 5, 7 should be included in "~B", and perhaps real numbers greater than 25. But, most likely not the others. And certainly, knowing that my uncle Wilbur is not a member of the set B would contribute little to any discussion of B.

Question:
What's wrong with including the others? They are clearly not in B. I'm so confused right now.

答: Although it's true that the notion of "complement of a set" presupposes some notion of universal set, I would expect this universal set to include all the entities under discussion, even Wilbur, unless the contrary was explicitly stated in the question. And I don't consider "would contribute little to any discussion" relevant at all. Mathematical entities like sets don't exist solely to contribute to our discussions. As far as I'm concerned, that "solution", considered as mathematics, is hopelessly confused.</p>


问:
Positive semi-definite Matrices
Assume that we have a positive semi-definite matrix A. Can we decompose it as follows,

$$A= U^T U \quad ?$$

Or is it only feasible for a positive definite matrix?

答:
Every positive semidefinite matrix $A\in\mathbb{R}^{n\times n}$ have a unique square root, ie there exists unique $C\in\mathbb{R}^{n\times n}$ such that $A=C^2$. The square root $C$ is symmetric, thus $U=C$ will do the job.
Conversely, if the matrix $A$ is not symmetric, there can not exist such a decomposition: because matrices of the form $U^TU$ are symmetric.

Title: Positive semi-definite Matrices


问:
Prove $(\forall x\ Q(x))\to P ⊢ \exists x (Q(x)\to P)$ using Natural Deduction
I have stumbled upon this predicate logic sequent which I have trouble proving using natural deduction:

$(\forall x\ Q(x))\to P ⊢ \exists x (Q(x)\to P)$

I started with ∀xQ(x) → P as the premise. I wanted to start using universal elimination but since the ∀x doesn't cover Q(x) → P completely, I refrained from doing so. I had to assume ∀xQ(x) but when I did, I couldn't reach ∃x(Q(x) → P) out of the assumption's scope.

I have also tried proof by contradiction but I couldn't progress. I tried to prove p → q ⊢ (¬p V q) with natural deduction to be able to incorporate it into the solution, but I couldn't prove it either.

So, I think I am completely stuck this one. Could you help me prove this (or tell me if this cannot be proved by natural deduction for any reason)?

If it helps, I am using the inference rules listed under
https://en.wikipedia.org/wiki/List_of_rules_of_inference

答:
Here's a skeleton of a proof by reduction to absurdity.  Prove that it is absurd to assume that such a conditional does not exist.

$$\def\fitch#1#2{~~~~\begin{array}{|l}#1\\\hline#2\end{array}}
\fitch{(\forall x~Qx)\to P}{\fitch{\lnot\exists x~(Qx\to P)}{\fitch{\boxed a}{~~\vdots\\Qa}\\\forall x~Qx\\P\\\exists x~P\\\fitch{\boxed c~P}{~~\vdots\\\bot}\\\bot}\\\lnot\lnot\exists x~(Qx\to P)\\\exists x~(Qx\to P)}$$


问:
Can the function has singularities at some points but still entire?
This is from Stein's Complex Analysis textbook. From page 87, it reads,

We contend that the function $H=f-f_{\infty}-\sum_{k=1}^{n}f_k$ is
entire and bounded. Indeed, near the pole $z_k$ we subtracted the
principal part of $f$ so that the function $H$ has a removable
singularity there. Also, $H(1/z)$ is bounded for $z$ near 0 since we
subtracted the principal part of the pole at $\infty$. This proves our
contention, and by Liouville's theorem we conclude that $H$ is
constant.

It said that $H$ has a removable singularity at some points, but then it concludes that $H$ is also entire. Singularity is defined as some point that is not defined but is defined in the neighborhood of that point, but then how do we conclude that it is entire if the function is not even defined at these singularity points?

I somehow understand that if we subtract principal part from the function, we can differentiate at that point. But I don't know how to rigorously formalize this in the context of the function $H$, because it seems that we are subtracting a lot of $f_k$'s from only one function $f$.

Sorry if this is not an organized question, but can somebody give an insight? Thanks.

答: To prove that  a function is entire you only have to prove its differentiability at each point. The points where there is a singularity have no limit points and these are removable singularities. Hence, be redefining the function at these points we get an entire function. By an abuse of language we say that the function is entire. The fact that there may be infinitely many of these singularities does not matter.


答:
When a function has a removable singularity you can define the value at the singularity to make the function analytic there. It's an acceptable abuse of terminology to call the function $H$ both before and after patching it at the removable singularity.

You subtract one $f_k$ for each pole $z_k$. Those $f_k$ are analytic everywhere except at $z_k$.

Title: Can the function has singularities at some points but still entire?

Body:
This is from Stein's Complex Analysis textbook. From page 87, it reads,

We contend that the function $H=f-f_{\infty}-\sum_{k=1}^{n}f_k$ is
entire and bounded. Indeed, near the pole $z_k$ we subtracted the
principal part of $f$ so that the function $H$ has a removable
singularity there. Also, $H(1/z)$ is bounded for $z$ near 0 since we
subtracted the principal part of the pole at $\infty$. This proves our
contention, and by Liouville's theorem we conclude that $H$ is
constant.

It said that $H$ has a removable singularity at some points, but then it concludes that $H$ is also entire. Singularity is defined as some point that is not defined but is defined in the neighborhood of that point, but then how do we conclude that it is entire if the function is not even defined at these singularity points?

I somehow understand that if we subtract principal part from the function, we can differentiate at that point. But I don't know how to rigorously formalize this in the context of the function $H$, because it seems that we are subtracting a lot of $f_k$'s from only one function $f$.

Sorry if this is not an organized question, but can somebody give an insight? Thanks.


问:
Functional equation $ f(x) + f\left(1-\frac{1}{x}\right) = \tan^{-1}(x) $ and definite integral

Let $f(x)$ be a function $f :\mathbb{R}\to \mathbb{R}$ such that
$$ f(x) + f\left(1-\frac{1}{x}\right) = \tan^{-1}(x) $$
for all real $x$ except $0$.

Find $\int_0^1f(x)\ \mathrm dx$.

My approach till now:

Put $x = \frac{1}{x}$ in the functional equation and consider the domain of integration $(0,1)$ such that $\tan^{-1}\frac{1}{x} = \cot^{-1}(x)$ and add the original functional equation and the resulting equation after the substitution to get:

$$f(x) + f(1-x) = \frac{\pi}{2} - f\left(\frac{1}{x}\right) - f\left(1-\frac{1}{x}\right)$$
and integrate both sides from $0$ to $1$.

Let $I = \int_0^1f(x)\ \mathrm dx $, then LHS of above functional equation becomes $2I$. Now I am not able to evaluate the RHS, some $\frac{\ln(2)}{2}$ term always creeps up and doesn't gets cancelled and its not even in the answer.

Please help me with this problem.

答:
Let $g(x)=1-\frac1x$. Then $g(g(x))=\frac1{1-x}$ and $g(g(g(x)))=x$.

Thus,
$$
f(x)+f(g(x))=\tan^{-1}(x)\tag1
$$
$$
f(g(x))+f(g(g(x)))=\tan^{-1}(g(x))\tag2
$$
$$
f(g(g(x)))+f(x)=\tan^{-1}(g(g(x)))\tag3
$$
Since $2f(x)=(1)-(2)+(3)$, we get
$$
f(x)=\frac12\left(\tan^{-1}(x)-\tan^{-1}(g(x))+\tan^{-1}(g(g(x)))\right)\tag4
$$
As mentioned in comments, this function is not continuous:

What is interesting is that the derivative of this function is continuous (if we subtract $\frac\pi2$ from $f$ in $[0,1]$ to counter the jump discontinuities):

In any case,
$$
\begin{align}
&\int_0^1f(x)\,\mathrm{d}x\\
&=\frac12\left(\int_0^1\tan^{-1}(x)\,\mathrm{d}x-\int_0^1\tan^{-1}(g(x))\,\mathrm{d}x+\int_0^1\tan^{-1}(g(g(x)))\,\mathrm{d}x\right)\tag5\\
&=\frac12\left(\int_0^1\tan^{-1}(x)\,\mathrm{d}x-\int_{-\infty}^0\tan^{-1}(x)\,\mathrm{d}g(g(x))+\int_1^\infty\tan^{-1}(x)\,\mathrm{d}g(x)\right)\tag6\\
&=\scriptsize\frac12\left(\frac\pi4-\color{#C00}{\int_0^1\frac{x}{1+x^2}\,\mathrm{d}x}+\color{#090}{\int_{-\infty}^0\frac1{(1-x)\left(1+x^2\right)}\,\mathrm{d}x}+\frac\pi2-\color{#00F}{\int_1^\infty\frac{x-1}{x\left(1+x^2\right)}\,\mathrm{d}x}\right)\tag7\\
&=\frac12\left(\frac\pi4-\color{#C00}{\frac{\log(2)}2}
+\color{#090}{\frac\pi4}
+\frac\pi2-\color{#00F}{\frac{\pi-2\log(2)}4}\right)\tag8\\[3pt]
&=\frac{3\pi}8\tag9
\end{align}
$$
Explanation:

$(5)$: apply $(4)$

$(6)$: apply $g$ and $g\circ g$ to get $\tan^{-1}(x)$ in each integral

$(7)$: integrate by parts

$(8)$: evaluate the integrals by partial fractions

$(9)$: simplify


问:
Moving 'eye' along a vector while focused on a target
I am struggling a little bit with my trigonometry. I am writing a little 3D simulation in which I have a character (the 'eye') which is always fixed on a target.

I want to move along a vector from any given position left or right along a vector. This would be the equivalent of walking left or right in the real world, but while always facing the target.

I have drawn a picture that I hope will help.

I think what I need to do is figure out the angle my eye to the target with respect to the x and y axis (in 2D), but am stumbling to figure out how that fits into moving along the 'blue line' in my diagram.

Thank you for any hints or guidance!

答:
If you can assume that the line going from your eye to the target, $L$, is always on the XY plane, i.e. the same plane containing "left" and "right," you can calculate the perpendicular line $P$ by finding the dot product (and expanding into components):

$$ L \cdot P = 0 \\
\implies x_L \times x_P + y_L \times y_P = 0$$

We then have an underconstrained problem because this is the equation of the entire line passing through the eye perpendicular to the target, meaning we have a free parameter left over - which makes it awkward to write code for. To find a unique solution, I'm going to arbitarily assert that I also want my answer to be a unit vector, which lets me substitute:

$$ x_L \times x_P + y_L \times \sqrt{1-x_P^2} = 0$$

Wolfram Alpha solves this to say that the x-component is given by,

$$ x_p = \frac{y_L}{\sqrt{x_L^2 + y^2_L}}$$

You can then find the corresponding $y_p = \sqrt{1-x_p^2}$ for the two components of the perpendicular vector, i.e. the direction of left or (depending on sign) right relative to the target.

Note that if you just calculate this once and step multiple times along it, you won't get a circle, because each step also moves the perpendicular. You'll also need to rotate the eye to point back towards the target - the angle to rotate by is given by $\arctan(\frac{|s|}{|L|})$ where $|s|$ is your step size. (The original position, the new position, and the target form a right triangle with the former at the right angle.)


答:
No. The theory of $\Bbb R$ is not the same first-order theory of $\Bbb Q$. If you define $^*\Bbb Q$ and $^*\Bbb R$ as ultrapowers then they have the same first-order theories as their standard counterparts.

In particular, $^*\Bbb Q$ satisfies $\forall x(x\cdot x\neq(1+1))$; whereas $^*\Bbb R$ satisfies $\exists x(x\cdot x=1+1)$, and that $x$ is the equivalence class of $f(n)=\sqrt2$ in the ultrapower.

You can do a bit better, e.g. since in $\Bbb R$ it is true that every odd-degree polynomial has a root, so we can find many algebraic numbers in $^\ast\Bbb R$ which do not exist in $^*\Bbb Q$.

One more thing on the use of the transfer principle. The transfer principle is really just Los theorem, that the ultrapower of a structure is elementary equivalent to it, i.e. a sentence is true in the ultrapower if and only if it is true in the structure.

Structures, and models come with a language. If your language is the language of ordered-fields then $\Bbb Q$ is not definable. This is because definable sets of real numbers are finite unions of intervals (possibly one-point intervals). The rationals are clearly not such set.

If, on the other hand, you are free to add whatever predicate that you choose to add and whatever function, constants or otherwise, then clearly you can add an unary predicate symbol $Q$ which is interpreted as true for $x$ if and only if $x\in\Bbb Q$. Since $\Bbb Q\models\forall x.Q(x)$ we have that $^*\Bbb Q$ must satisfy the same sentence, and by a similar reasoning we have that $Q^{^*\Bbb R}=\, ^*\Bbb Q$. So it's quite obvious now that $\exists x.\lnot Q(x)$ is true in $\Bbb R$ and therefore in $^*\Bbb R$, and so we have $^*\Bbb R\neq\, ^*\Bbb Q$.

But usually when we talk about $^*\Bbb R$ we talk about it as an ordered field, in which case there is no $Q$ predicate like that, and the rationals are not definable there. So we have that there is no formula which states $\Bbb Q\neq\Bbb R$.

Title: Is ${}^{*}\mathbb{Q}={}^{*}\mathbb{R}$?

Body:
This is a follow-up question of a previous one.

Is it true ${}^{*}\mathbb{Q}={}^{*}\mathbb{R}$ ?

It seems to me it's true because for any $x \in{}^{*}\mathbb{R}$, there is a $y \in {}^{*}\mathbb{Q}$ such that $y < x$ and vice versa, and because for any $x' \in \mathbb{R}$, there is a $y' \in {}^{*}\mathbb{Q}$ such that $y' = x'$.


问:
How to Evaluate $\sum_{n=1}^{\infty} \frac{\ln(n+2)}{n^2}$
How can one evaluate the following sum :

$$ S=\sum_{n=1}^{\infty} \frac{\ln(n+2)}{n^2} \approx 2.33444 $$

The similar Sum can be given by  :

$$ \sum_{n=1}^{\infty} \frac{\ln(n)}{n^2} = -\frac{\pi^2}{6}\left(\,-12\ln(A)+\gamma+ \ln(2)+\ln(\pi) \,\right)$$

I have been unsuccessful with using definitions of the Glaisher Constant $(A)$ to evaluate $S$ , or with integration and directly evaluating the sum $S$.

Q = Is there a closed form for the above Sum $S$ ?

Thank you kindly for your help and time.

EDIT

The Glaisher - Kinkelin constant A is given by the following limit

$$ A= \lim_{n \to \infty } \frac {K(n+1)}{n^{n^2/2+n/2+1/{12}}\exp(\frac{-n^2}{4})} $$

$$ K(n) = \prod_{t=1}^{n-1} t^t $$

Furthermore the following product was found as

$$\prod_{n=1}^{\infty} n^{1/{n^2}} = \left(\frac{A^{12}}{2 \pi e^{\gamma}}\right)^{\pi^2/6}$$

https://en.wikipedia.org/wiki/Glaisher–Kinkelin_constant

You may find the limit also as :

closed form of $\prod_{n=1}^{\infty}\left(\frac{n}{n+1}\right)^{(-1)^{n-1}n}$

Similarity :

$$ S = \ln \left(\prod_{k=1}^{\infty} (k+2)^{1/k^2}\right) = ? $$

答:
This is a long comment.

This is a "Feynman Trick" inspired method that appears to me illegitimate and which I certainly can't defend, but that nevertheless leads to what appears to be an asymptotic expansion for $S$

Starting with
$$S(m)=\sum _{n=1}^{\infty } \frac{\log (n+m)}{n^2}$$
and differentiating under the summation sign
$$\sum _{n=1}^{\infty } \left( \frac{\partial } {\partial m}\frac{\log (m+n)}{n^2}\right)=\frac{\pi ^2}{6 m}-\frac{H_m}{ m^2}$$

Using the asymptotic expansion for the Harmonic Number, $H_n$ involving the Bernoulli Numbers, $B_{2k}$

$$H_m \approx \ln m+\gamma+\frac{1}{2m}-\sum _{k=1}^{r} \frac{B_{2 k}}{2 k m^{2 k}}$$

and integrating indefinitely with respect to $m$ and then setting $m=2$ only in the resulting function (avoiding setting a lower limit for the integral, say $m=0$ which is undefined for this function), we obtain the asymptotic approximation given in (1)

$$\sum _{n=1}^{\infty } \frac{\log (n+2)}{n^2} \approx \left(\frac{1}{6} \pi ^2 \log (2)+\frac{\log (2)}{2}+\frac{\gamma }{2}+\frac{1}{2}+\frac{1}{4\ 2^2}-\sum _{k=1}^r \frac{2^{-2 k-1} B_{2 k}}{2 k (2 k+1)}\right)\tag{1}$$