问:
Intuition behind structural weakening
According to the Fitch deductive system I'm using in my logic class, the following is a valid inference:

Premise: Q

Step 1: Assume P

Step 2: Q by reiteration

Step 3: P → Q by conditional intro.

Basically it seems that, given the truth of any arbitrary atomic sentence, you can deduce that any other arbitrary atomic sentence, if also true, implies the first sentence. And it is this result that I'm having trouble wrapping my head around. Couldn't it be the case that two sentences can be true without any relation of implication holding between them? This certainly seems to be the case in informal reasoning and with facts about the world, so what's the rationale behind this being a valid step in propositional logic?

Edit: the initial title ("motivation for entailment in propositional logic") was misleading - this question is more about structural weakening than it is the material conditional or entailment per se. I made this mistake beacuse I myself wasn't clear on weakening until the question was answered.

答:
The key point is this:

Couldn't it be the case that two sentences can be true without any relation of implication holding between them?

That's quite true - and weakening does not assert a relationship, at least a meaningful one, between $P$ and $Q$!

Think of classical logic as describing the world as it happens to be. Classical logic doesn't "see" anything more complicated than what happens to be true and what happens to be false; things like causal relationships between facts are simply not treated by it. In particular, "$P\rightarrow Q$" just means "If it happens to be the case that $P$ is true then it happens to be the case that $Q$ is true." If we already know that $Q$ happens to be true, then of course there's no information gained by asserting $P\rightarrow Q$. But that doesn't mean that it would be improper for us to assert $P\rightarrow Q$, it only means that it would be a bit pointless to do so.

(On that note, the deduction rules of classical logic are created purely to characterize what operations on (sets of) sentences preserve "happens-to-be-tru"th. This is why we admit "From $P$ and $\neg P$, deduce $Q$" as a rule despite the absurdity of its premise: it is in fact exactly the absurdity of its premise that motivates its acceptance, since the fact that its premise can't occur ensures that that rule can never be used to accidentally conclude a falsehood!)

Now an editorial:

To my mind this is actually a strength of classical logic: working with it reveals the subtleties of notions like cause and effect and forces us, if we want to treat them rigorously, to really make clear what we mean by them. That is, its "starkness" enables us to use it as a jumping-off point for treating more nuanced notions without smuggling in preconceptions about them accidentally.

To see what I mean, consider modal logic (using the Kripke frame semantics for simplicity). Roughly speaking, in modal logic we consider not just a single "actual world" but a whole collection of "possible worlds." We can then (try to) frame causal relationships as facts across worlds: "$A$ causes $B$" means "In all possible worlds we have $A\rightarrow B$." This is quite different from the proposition "$A\rightarrow B$" holding at the actual world we're in, but perhaps failing at other "nearby" worlds. But while modal logic is at first a step away from classical logic, it can in fact be folded back into classical logic by broadening the notion of "world." Kripke frames are themselves classical mathematical objects, and so we can put the whole perspective above inside an overarching classical framework. Classical logic, that is, can serve as a context for studying non-classical logics! Of course it's not unique in this respect - any logic worth its salt can do the same - but the starkness of classical logic, in my opinion, keeps it "maximally neutral" in this role. I've said a bit more about this here.

I'm actually ignoring a huge issue here, namely the leap from propositional to predicate logics, but that's a whole separate thing which ultimately it doesn't affect the general picture I'm trying to paint here, so for now ignore it.

Title: Intuition behind structural weakening


问:
Find vector which minimises the sum.
Find the vector r which minimises the sum S = (r - a)^2 + (r - b)^2 + (r - c)^2, where a, b and c are constant vectors.

I'm unsure as to how to go about this. I was thinking it's some sort of constrained optimization, however I'm not well equipped with these tools.
enter image description here

答:
let $r = (r_x,r_y)$ and $a=(a_x,a_y),b=(b_x,b_y),c=(c_x,c_y)$.

Write out the expression $||r-a||^2+ ||r-b||^2 + ||r-c||^2$.

You should get a polynomial with the two variables $r_x$ and $r_y$ that doesn't have mixed terms of the form $r_x^ur_y^v$. Since it doesn't have mixed terms like this you just need to minimize the parts with $r_x$ and the parts with $r_y$ which is easy as these are just degree $2$ polynomials in one variable.

We now solve the subproblem of minimizing $(r-a)^2 + (r-b)^2 + (r-c)^2 = 3r^2-2(a+b+c)r + a^2+b^2+c^2$.

Taking derivative we have $6r-(a+b+c) = 0$ and so we take $r=(a+b+c)/3$

So the point we want is the incenter of the points $a,b,c$ which is given by the expression $(\frac{a_x+b_x+c_x}{3},\frac{a_y+b_y+c_y}{3})$

ContentLicense: CC BY-SA 4.0

Title: Find vector which minimises the sum.


问:
Let E be algebraic extension of F, an embedding maps a root of minimal polynomial to a root of minimal polynomial
Let E be algebraic extension of F, and let σ:E⟶E be an embedding of E into itself over F.

Let a ∈ E and let p(x) be its minimal polynomial over F.

My book says, without any comments, that σ must map a root of p(x) onto a root of p(x), but I don't understand why this should be the case.

Please help! Thanks in advance :)

答:
Recall that $\sigma$ is a ring homomorphism (that means that $\sigma(1)=1$ and $\sigma(x+y)=\sigma(x)+\sigma(y)$ and $\sigma(xy) = \sigma(x)\sigma(y)$ for any $x$ and $y$ in $E$) such that $\sigma(f)=f$ if $f \in F$.

Now, if you write $p(x) \in F[x]$ as $\sum_{k\geq0} f_kx^k$, apply $\sigma$ on both sides to the equation $\sum_{k\geq0} f_ka^k=0$ and see what happens.


问:
Prove that the sequence $(x_n)$ is Cauchy.
Can you find errors in my prove?

Problem:

Let $(x_n)$ be a sequence of real numbers such that $(y_n)$ is a limited sequence with general term $y_n = \sum_{k=1}^{n}|x_{k+1} - x_k|$. Prove that $(x_n)$ is Cauchy.

My attempt:

$y_n = \sum_{k=1}^{n}|x_{k+1} - x_k| = |x_{n+1} - x_1| \geq |x_{n+1}| - |x_1| \Rightarrow |x_1| + y_n \geq |x_{n+1}|$, where I use the telescopic property in the second equality. Since $y_n$ is convergent (because it is monotonic and limited), $|x_{n+1}|$ is also convergent by the comparision criterion, therefore $x_n$ converges absolutely, therefore it is Cauchy.

sorry if this is duplicated.

答:
The sum $\sum_{k=1}^n |x_{k+1}-x_k|$ is not telescopic, because of the absolute value ($\sum_{k=1}^n x_{k+1}-x_k$ would be). What is actually true is that $(y_n)_{n\in\mathbb{N}}$ is bounded and monotonic, thus convergent. Then $(y_n)_{n\in\mathbb{N}}$ is Cauchy, and for every $\epsilon\in\mathbb{R}^+$ there exists $N(\epsilon)\in\mathbb{N}$ such that $|y_{n+p}-y_n|<\epsilon$ for all $n\geq N(\epsilon)$, $p\in\mathbb{N}$. But using triangular inequality: $$|x_{n+1+p}-x_{n+1}|\leq \sum_{k=1}^{p} |x_{n+k+1}-x_{n+k}|=\sum_{k=1}^{n+p}|x_{k+1}-x_{k}|-\sum_{k=1}^{n}|x_{k+1}-x_{k}|=y_{n+p}-y_{n}<\epsilon$$
So $(x_n)_{n\in\mathbb{N}}$ is Cauchy sequence.


问:
Showing $\left(\frac{z+i}{z-i}\right)^n = -1$ implies $z$ is real
I have shown the following identity: $$ \frac{1+e^{i \theta}}{1-e^{i \theta}} = \frac{1}{2}\cot\left(\frac{\theta}{2}\right) $$ And I now need to use this to show that the the equation: $$ \left(\frac{z+i}{z-i}\right)^n = -1$$ has $\Im(z) = 0$.

I'm not sure where to get started on this, I have found a way to do it by conjugating the 2nd equation on both sides and showing that z is equal to its own conjugate, however this does not use the original identity, and I'd really like to see how it follows from that.

Any help would be much appreciated.

答:
$$
\frac {1+e^{i \theta}}{1-e^{i \theta}} = -\frac {e^{i\theta/2}+e^{-i\theta/2}}{e^{i\theta/2}-e^{-i\theta/2}} = -\frac {2\cos \frac \theta 2}{2i \sin \frac \theta 2} = -\frac 1i \cot \frac \theta 2 = i \cot \frac \theta 2
$$
You can recheck it here.

As for the second part
$$
\left ( \frac {z+i}{z-i}\right )^n = -1 \\
\frac {z+i}{z-i} = e^{\frac {(2k+1)\pi i}n} \\
z + i = (z-i) e^{\frac {(2k+1)\pi i}n} \\
z \left ( 1-e^{\frac {(2k+1)\pi i}n}\right) = -i \left ( 1+e^{\frac {(2k+1)\pi i}n}\right) \\
z = -i\ \left ( \frac {1+e^{\frac {(2k+1)\pi i}n}}{1-e^{\frac {(2k+1)\pi i}n}} \right )
$$
Based on first result
$$
\frac {1+e^{\frac {(2k+1)\pi i}n}}{1-e^{\frac {(2k+1)\pi i}n}} = i \cot \frac {(2k+1)\pi}{2n}
$$
so
$$
z = -i^2 \cot \frac {(2k+1)\pi}{2n} = \cot \frac {(2k+1)\pi}{2n}
$$
which is real, i.e. $\Im(z) = 0$.

ContentLicense: CC BY-SA 3.0


问:
Characterizing direct sum of modules
Suppose $M$ is an $R$-module (for a commutative ring $R$ with $1$) and that $(M_i)_{i\in I}$ is a collection of $R$-submodules of $M$. The inclusion maps $j_i:M_i\rightarrow M$ determine an $R$ module homomorphism $\operatorname{Hom}(M,N)\rightarrow\prod_{i\in I}\operatorname{Hom}(M_i,N)$ given by $\varphi\mapsto (\varphi\circ j_i)_{i\in I}$, where $N$ is any arbitrary $R$-module.

Suppose that this map is an isomorphism for any $N$. I'm trying to understand why this implies that $M$ is a direct sum of the $M_i$. I tried plugging in different values for $N$, such as $M$ and then mapping over the identity, but this doesn't seem to get me anywhere. Any suggestions? What am I missing?

Edit: I am looking for a reason that does not appeal to the universal property.

答: Supposing you meant the map is an isomorphism for all $N$, you have that $\operatorname{Hom} (\bigoplus M_i, -) \cong \operatorname{Hom} (M, -)$, where the isomorphism is a natural transformation. You can now use the Yoneda embedding and you are done (since this embedding is fully faithful).

Title: Characterizing direct sum of modules


问:
Simplify $2\log 4 + 3\log 8 - \log 2$ using logarithmic laws, then evaluate.
Firs time I've encountered this type of question where there is a number in front of the log, and it's throwing me a bit. I'm fine with using the logarithmic laws to simplify, but not sure if I need to get rid of those numbers first, and if so, how? thought perhaps the power law would apply but obviously there's no powers so I'm really not sure how to approach this one.</p>

Body:
Considering $\log(x)=\log_{10}(x)$

I have just made this consideration because the expression is nicely solved for $\log_{2}(x)$ if it is perhaps a typo. For $\log_{2}(x)$ we would have

$$2\log_{2}(4)+3\log_{2}(8)-\log_{2}(2) = 2\cdot 2 + 3 \cdot 3 - 1 = 12$$

\begin{align}
2\log(4)+3\log(8)-\log(2) & = \log(4^2)+\log(8^3)-\log(2)\\
& = \log(4^2)+\log(8^3)-\log(2) \\
& = \log(16)+\log(512)-\log(2)\\
& = \log(8192)-\log(2) \\
& = \log \left(\frac{8192}{2}\right) \\
& = \log(4096)\\
& = \log(2^{12})\\
& = \boxed{12\log(2)\approx 3.6}\\
\end{align}


问:
Find the m.g.f., the mean and the variance of the aggregate claims?
Assume that the number of losses $N$ has a geometric distribution with parameter $p$, while the claim amounts,${\{X_i}\}_{i≥1}$, is a sequence of i.i.d. Gamma distributed r.v. with parameters $a$ and $b$. Find the  m.g.f.,  the  mean and  the variance  of the aggregate claims.

[Note that if $X\sim Gamma(a,b)$ then its m.g.f. is given by $m_X(z)=(\frac{b}{b-z})^a$, while if $X\sim Geometric (p)$, then $P_N(z)=\Bbb{E}(z^N)=\frac{p}{1-qz}$.]

For this question there was no mark scheme so I wanted to check whether I was right or if there was a more efficient way of solving the question.

My solution:

$$
\begin{align}
m_S(z)&= \frac{p}{1-qm_X(z)} \\
&=\frac{p}{1-q(\frac{b}{b-z})^a} \\
&=\frac{p(b-z)^a}{(b-z)^a-qb^a}
\end{align}
$$

To find the expectation and variance we need to workout $E(S)=m'_s(0)=E(X)E(N)$ and $E(S^2)=m_s''(0)$.

*I have skipped the simplification steps to save space and time

$$
\begin{align}
m'_s(z)&=\frac{apqb^a(b-z)^{a-1}}{((b-z)^a-qb^a)^2} \\
m'_s(0)&=\frac{apqb^{2a}b^{-1}}{(b^a-qb^a)^2} \\
&=\frac{a}{b}\frac{q}{p}=E(X)E(N)=E(S)
\end{align}
$$

Which gives us the expectation.

$$
\begin{align}
m''_S(z)&=\frac{apqb^a(aqb^a-qb^a+a(b-z)^a+(b-z)^a)(b-z)^{a-2}}{((b-z)^a-qb^a)^3} \\
m''_S(0)&=\frac{apqb^a(aqb^a-qb^a+ab^a+b^a)b^{a-2}}{(b^a-qb^a)^3} \\
&=\frac{aq(a-ap+p)}{b^2p^2}
\end{align}
$$

Therefore $Var(S)=E(S^2)-E^2(S)$ so we get:

$$
\begin{align}
Var(S)&=\frac{aq(a-ap+p)}{b^2p^2}-\frac{a^2q^2}{b^2p^2} \\
&= \frac{a}{b^2}\frac{q}{p}
\end{align}
$$

The differentiation to do this took quite some time so I was wondering whether there was a more efficient way to solve this to avoid lots of time spent simplifying and differentiating (mainly focussed on the double prime differential to work out the variance).

Any help would be massively appreciated!

答:
You can do it faster by noting that

$$m_S(z)=pm_G(q(\frac{b}{b-z})^\alpha),$$

where $m_G$ is the MGF for gamma with parameters 1,1.

Then (note that there may be mistakes)
$$m_S'(z)=pm_G'(q(\frac{b}{b-z})^\alpha) \alpha q(\frac{b}{b-z})^{\alpha}\frac{1}{b-z}$$
$$m_S''(z)=pm_G''(q(\frac{b}{b-z})^\alpha) (\alpha q(\frac{b}{b-z})^{\alpha}\frac{1}{b-z})^2+ pm_G'(q(\frac{b}{b-z})^\alpha) (\alpha^2+\alpha) q(\frac{b}{b-z})^{\alpha}\frac{1}{(b-z)^2}$$

$$=pm_G''(q(\frac{b}{b-z})^\alpha) (\alpha q(\frac{b}{b-z})^{\alpha}\frac{1}{b-z})^2+ m'_S(z)\frac{1+\alpha}{(b-z)}$$
And now
$$E(S)=m_S'(0)=m_G'(q) p\frac{\alpha q}{b}=\frac{1}{(1-q)^2} p\frac{\alpha q}{b}=\frac{\alpha q}{bp}$$
$$Var(S)=m_S''(0)-E(S)^2=E(S)(\frac{1+\alpha}{b}+\frac{2 \alpha q}{bp}-E(S))$$

$$=\frac{E(S)}{bp}(\alpha+p)$$


问:
Determine Greens function inside a semi circle
I need to determine Greens function $G(x,x_0), x \mbox{ and }x_0 \in \mathbb{R}^2$ inside a semi-circle $(0<r<a, 0<\theta<\pi)$ with $\nabla G = \delta(x - x_0)$ and $G = 0$ on the boundary.

I know that the formula of the Green function in a circle is :
$$\frac{1}{4\pi} \ln(a^2 \frac{r^2 + r_0^2 - 2rr_0cos\phi}{r^2r_0^2 + a^4 - 2rr_0a^2 cos\phi})$$. $\phi$ is the angle between $x$ and $x_0, r = |x|, r_0 = |x_0|$

I also know that I need to make a (positive or negative?) image such that $\nabla G  = \delta(x-x_0) \pm \delta(x - x^*_0)$, but that are the only ideas I have, I have no clue how to find that image and proceed.

答:
Presumably, $x_0$ lies inside of the upper semi-circle. Let $\bar{x}_0$ be its reflection in the horizontal axis. That is, if $ {x_0}$ has polar coordinates $(r_0, \theta_0)$, then $\bar{x}_0$ has polar coordinates $(r_0,-\theta_0)$.

Green's function for the upper semi-circle with pole at $x_0$ is $G(x,x_0)-G(x,\bar x_0)$. Why? Because it has the right kindof singularity at $x_0$, because it solves the Laplace's equation elsewhere in the domain, and because it is equal to $0$ on the boundary.

By the way, calculations in $\mathbb R^2$ are often made  easier by complex notation. For example,
$$\frac{1}{4\pi} \ln\left(a^2 \frac{r^2 + r_0^2 - 2rr_0\cos\phi}{r^2r_0^2 + a^4 - 2rr_0a^2 \cos\phi}\right) = \frac{1}{2\pi} \ln \frac{a\,|z-z_0|}{|a^2-\bar z_0 z|}$$
where $z=re^{i\theta}$ and $z_0=r_0e^{i\theta_0}$.


问:
Similar formula for difference between partial double sums similar to that of single partial sum?
Is there a double sum formula for
$$\left|\sum_{m=1}^{m_1}\sum_{n=1}^{n_1}x_{mn}-\sum_{m=1}^{m_2}\sum_{n=1}^{n_2}x_{mn}\right|$$
similar to how
$$\left|\sum_{n=1}^{n_1}x_n-\sum_{n=1}^{n_2}x_n\right|=\left|\sum_{n=n_1}^{n_2}x_n\right|?$$

If $m_2,n_2>m_1,n_1$, then you get basically the same formula as the single sum case. But if $m_1>m_2$ and $n_2>n_1$, or vise versa, then it does not appear to work out to a single partial double sum.

For example
$$\sum_{m=1}^2\sum_{n=1}^3x_{mn}-\sum_{m=1}^3\sum_{n=1}^2x_{mn}=\sum_{m=1}^2x_{m3}-\sum_{n=1}^2x_{3n}.$$
I suppose this maybe answers my question for me, but I don't know if anyone out there has any tricks up their sleeve?

If I can show that $$\left|\sum_{m=1}^{m_1}\sum_{n=1}^{n_1}x_{mn}-\sum_{m=1}^{m_2}\sum_{n=1}^{n_2}x_{mn}\right|\leq\left|\sum\sum x_{mn}\right|$$ for some particular sum $\sum\sum x_{mn}$, then I will be on my way to proving a theorem I've been stuck on.

答: It might be helpful to think of $x_{m,n}$ as a function of $\mathbb{N}\times \mathbb{N}\to \mathbb{R}$. The two sums correspond to "integrating" this function on rectangles of size $m_1\times n_1$ and $m_2\times n_2$. In the case of $m_1>m_2$, $n_2>n_1$ the two rectangles are not contained within each other, so their difference will have two parts. Drawing out a picture should allow you to come up with a formula.


答:
Cancelling the gradient,

$$\nabla S=2(r-a)+2(r-b)+2(r-c)=0$$

we get

$$r=\frac{a+b+c}3.$$

Title: Find vector which minimises the sum.

Body:
Find the vector r which minimises the sum S = (r - a)^2 + (r - b)^2 + (r - c)^2, where a, b and c are constant vectors.

I'm unsure as to how to go about this. I was thinking it's some sort of constrained optimization, however I'm not well equipped with these tools.
enter image description here


问:
Lemma about composite fields in Weintraub's "Galois Theory"
I'm working through Weintraub's Galois theory book and I'm having trouble understanding the following lemma (which as I understand will be used later on). So, given a base field $F$, its extension $E$, and subfields $B,D$ that are themselves extensions of $F$, i.e. $F\subset B,D\subset E$, Weintraub defines the composite field $BD$ as the smallest field containing both subfields and makes a passing comment about how it consists of all expressions of the form $\big(\sum_i b_id_i\big)\big(\sum_j b_jd_j\big)^{-1}$ (I suppose this is plausible). He goes on to state and prove the following lemma, about the case where the denominators $\big(\sum_j b_jd_j\big)^{-1}$ may be omitted:

[With $F,E,B,D$ defined as above] suppose $|D:F|$ is finite. Then $|BD:B|\leq|D:F|$ and $BD=\{\sum_i b_id_i\}$.

Proof: Let $K=\{\sum_i b_id_i\}$ and $\{\delta_j\}$ be a basis of $D$ over $F$, then $\{\delta_i\}$ spans $K$ over $B$ (I suppose this is true, just express the $d_i$s in $\sum_i b_id_i$ terms of the basis and you get a $B$-linear combination of the basis vectors), therefore $|K:B|\leq|D:F|$ and [by a previous lemma, about how a finite-dimensional integral domain is a field] $K$ is field. Therefore $BD=K$.

I do not understand the last step, why from the fact that $K$ is a field does it follow that it is equal to the composite $BD$? Aside from a technical explanation of why this works, some intuition as to why we can dispense with denominators in this case would be helpful as well. Thank you for reading.

答:
The fact that $K$ equals $BD$ follows because both $B$ and $D$ are definitely contained in $K$ (note that $K$ was defined as the set of all sums of products of elements of $B$ with elements of $D$; it contains $B$ by taking a sum with a single summand and expressing $b\in B$ as $b\cdot 1$, and symmetrically for $D$). Since $BD$ is, by definition, the smallest field that contains both $B$ and $D$, if $K$ is indeed a field then this shows that $BD\subseteq K$. But since you already noted that $BD$ is the set of all products of elements of the set $K$ and inverses of elements of the set $K$, then we have $K\subseteq BD$, giving the desired equality.

The intuition is essentially that in a finite field extension, the inverse of an element $\alpha$ can be written as a polynomial in $\alpha$: for if $F\leq E$ is an extension, and $\alpha\in E$, then $|F(\alpha):F|$ is finite, so there is an $n\gt 0$ such that $1$, $\alpha,\ldots,\alpha^n$ is linearly independent. Picking the smallest such $n$ we get an expression of the form
$$a_0 + a_1\alpha + \cdots +a_n\alpha^n=0$$
with $a_0\neq 0$, and from this we get $\alpha(a_1+a_2\alpha+\cdots+a_n\alpha^{n-1}) = -a_0$, so
$$\frac{1}{\alpha} = -\frac{1}{a_0}\left(a_1+a_2\alpha_\cdots+a_n\alpha^{n-1}\right).$$
Similarly, any element $\alpha=\sum_i b_id_i$ has an inverse, which can be expressed as a polynomial expression in $\alpha$, so you don't have to explicitly throw in the multiplicative inverse of $\alpha$, it comes for free once you throw  in $\alpha$ and close under ring operations.

Thus, the field $\mathbb{Q}(\sqrt{2},\sqrt{3})$ which includes $\sqrt{2}+\sqrt{3}$, does not require you to also throw in $1/(\sqrt{2}+\sqrt{3})$ explicitly, since it can be expressed as
$$10\left(\sqrt{2}+\sqrt{3}\right) - \left(\sqrt{2}+\sqrt{3}\right)^3$$
(you can verify this by multiplying this expression by $(\sqrt{2}+\sqrt{3})$ and seeing that you get $1$).


问:
Does $\partial_{\varepsilon_k} f(x^k)$ approximate $ \partial f (\bar{x})$ for $ x^k \to \bar{x} $ and $\varepsilon_{k} \to 0 $?
Suppose that $f: \mathbb{R}^n \to \mathbb{R}$ is convex and globally Lipschitz continuous.

Let $(x^k) \subset \mathbb{R}^n$ be a sequence such that $ x^k \to \bar{x} $ for some $\bar{x} \in \mathbb{R}^n$.

Assume $(\varepsilon_k) \subset \mathbb{R}_{+}$ a sequence of positive scalars such that $\varepsilon_{k} \to 0 \, .$

Does the following hold:

$$ \operatorname{dist} \left( \, \partial_{\varepsilon_k} f(x^k) \, , \, \partial f (\bar{x}) \, \right):= \sup_{g \in \partial_{\varepsilon_k} f(x^k)} \, \, \inf_{h \in  \partial f (\bar{x})} \, \| g - h \|_2  \to 0 $$

where

$\partial_{\varepsilon_k} f(\cdot)$ denotes the $\varepsilon_k$-subdifferential of $f$

$\partial f(\cdot)$ denotes the regular convex subdifferential of $f$

$\| \cdot \|$ denotes the Euclidean distance.

The following is all from Convex Analysis and Minimization Algorithms II and might be of use:

Definition 1.1.1 Given $x \in \mathbb{R}^n$ the vector $s \in \mathbb{R}^{n}$ is called an $\varepsilon$ -subgradient of $f$ at $x$ when the following property holds:
$$
f(y) \geqslant f(x)+\langle s, y-x\rangle-\varepsilon \quad \text { for all } y \in \mathbb{R}^{n}
$$

It follows immediately from the definition that

$\partial_{\varepsilon} f(x) \subset \partial_{\varepsilon^{\prime}} f(x)$ whenever $\varepsilon \leqslant \varepsilon^{\prime}$

$\partial f(x)=\partial_{0} f(x)=\cap\left\{\partial_{\varepsilon} f(x): \varepsilon>0\right\}\left[=\lim _{\varepsilon \downarrow 0} \partial_{\varepsilon} f(x)\right] $

Theorem 1.1.4 For $\varepsilon \geqslant 0, \partial_{\varepsilon} f(x)$ is a closed convex set, which is nonempty and bounded.

Proposition 4.1.1 Let $\left\{\left(\varepsilon_{k}, x_{k}, s_{k}\right)\right\}$ be a sequence converging to $(\varepsilon, x, s),$ with $s_{k} \in$ $\partial_{\varepsilon_{k}} f\left(x_{k}\right)$ for all $k .$ Then $s \in \partial_{\varepsilon} f(x)$.

Proposition 4.1.2 Let $\delta>0$ and $L$ be such that $f$ is Lipschitzian with constant $L$ on some ball $B(x, \delta),$ where $x \in \mathbb{R}^n$. Then, for all $\delta^{\prime}<\delta$
$$
\|s\| \leqslant L+\frac{\varepsilon}{\delta-\delta^{\prime}}
$$
whenever $s \in \partial_{\varepsilon} f(y),$ with $y \in B\left(x, \delta^{\prime}\right)$.

As a result, the multifunction $\partial_{\varepsilon} f$ is outer semi-continuous, just as is the exact subdifferential.

Theorem 4.1.3 Let $f: \mathbb{R}^{n} \rightarrow \mathbb{R}$ be a convex Lipschitzian function on $\mathbb{R}^{n} .$ Then there exists $K>0$ such that, for all $x, x^{\prime}$ in $\mathbb{R}^{n}$ and $\varepsilon, \varepsilon^{\prime}$ positive:
$$
\Delta_{H}\left(\partial_{\varepsilon} f(x), \partial_{\varepsilon^{\prime}} f\left(x^{\prime}\right)\right) \leqslant \frac{K}{\min \left\{\varepsilon, \varepsilon^{\prime}\right\}}\left(\left\|x-x^{\prime}\right\|+\left|\varepsilon-\varepsilon^{\prime}\right|\right)
$$

This result  implies the inner semi-continuity of $(x, \varepsilon) \longmapsto \partial_{\varepsilon} f(x)$ for a Lipschitz-continuous $f$.

In particular, for fixed $\varepsilon>0$

$$\partial_{\varepsilon} f(y) \subset \partial_{\varepsilon} f(x)+\|y-x\| B(0, \frac{K}{\varepsilon}) \quad \text{for all } x \text{ and } y \, .$$

Corollary 4.1.5 Let $f: \mathbb{R}^{n} \rightarrow \mathbb{R}$ be convex. For any $\delta \geqslant 0,$ there is $K_{\delta}>0$ such that
$$\Delta_{H}\left(\partial_{\varepsilon} f(x), \partial_{\varepsilon} f\left(x^{\prime}\right)\right) \leqslant \frac{K_{\delta}}{\varepsilon}\left\|x-x^{\prime}\right\| \quad \text{ for all } x \text{ and } x^{\prime} \in B(0, \delta) \, .$$

Theorem 4.2.1  Let be given $f: \mathbb{R}^n \to \mathbb{R}$, $x \in \mathbb{R}^n$ and $\varepsilon \geqslant 0 .$ For any $\eta>0$ and $s \in \partial_{\varepsilon} f(x),$ there exist $x_{\eta} \in B(x, \eta)$ and
$s_{\eta} \in \partial f\left(x_{\eta}\right)$ such that $\left\|s_{\eta}-s\right\| \leqslant \varepsilon / \eta$.

This result can be written in a set formulation:
$$
\partial_{\varepsilon} f(x) \subset \bigcap_{\eta>0} \bigcup_{\|y-x\| \leqslant \eta}\left\{\partial f(y)+B(0, \frac{\varepsilon}{\eta})\right\} .
$$

The following one is not from the above mentioned source but easy to prove:

Theorem Assume $f: \mathbb{R}^n \to \mathbb{R}$ is convex and let $x \in \mathbb{R}^n$. Then for every $\varepsilon > 0$ there is a $\delta > 0$ such that

$$
\bigcup_{y \in B_{\delta}(x)} \partial f(y) \subset \partial_{\varepsilon} f(x) \,.
$$

答:
The Clarke subdifferential is upper semi-continuous. By definition, for each $\epsilon>0$ and $k$ big enough we must have that $\partial f(x^k) \subset \partial f(x) +  B(0, \epsilon).$ Hence, for such $k$ big enough,

$$\sup_{g_k \in\partial f(x^k)} \inf_{g \in\partial f(x)} \| g_k - g \| \leq \epsilon. $$ Consequently,

$$\lim \sup_{g_k \in\partial f(x^k)} \inf_{g \in\partial f(x)} \| g_k - g \| = 0$$


问:
Distribution of a Markov Chain
I'm having a very hard time understanding a theorem in my notes stating the Markov property. It says that for a Markov-chain $(X_n)_n$ with initial distribution $\mu$ and transition matrix $p$ we have:

$\mathbb{P}((X_{m+n})_{n\geq0} \in C||\mathcal{F}_m)=\mathbb{P}((X_{m+n})_{n\geq0} \in C||X_m)=P_{X_m(.)}(C)$

where $P_{x}(C)$ denotes the distribution of a Markov chain with initial distribution $\delta_{x}$ and transition matrix $p$. I have a really hard time understanding why the second equality holds, and I think that's probably because I don't fully understand how the distribution of a Markov proces is supposed to work.

答:
The notation is a bit confusing, but here's what I'm assuming is going on.

The first equation is the actual Markov property. It says that the distribution of "future" states $X_{m+1}, X_{m+2}, X_{m+3}, \dots$ given the "present" state $X_m$ and the "past" states $X_0, X_1, \dots, X_{m-1}$ is the same as the distribution of the future states given only the present state.

The second equation is a time-homogeneity property - usually part of the definition, but not always. It says that the distribution of the future states $X_{m+1}, X_{m+2}, \dots$ given the present state $X_m$ is the same as the distribution we'd get for the entire Markov chain, if it had started at $X_m$.

An example of a Markov chain that does not satisfy the second property is a random walk on the integers that "slows down": on the $n^{\text{th}}$ step, it goes left with probability $\frac1{2n}$, goes right with probability $\frac1{2n}$, and stays put with probability $1 - \frac1n$. This decision does not depend on how it got to the current state, so the Markov property holds. But this Markov chain is not time-homogeneous. If you're in state $0$ at the beginning of time, you'll always end up in state $1$ or $-1$ next. If you're in state $0$ before the $100^{\text{th}}$ step, there's a $99\%$ chance you'll stay at $0$.

Title: Distribution of a Markov Chain


问:
Follow-up to question: Aut(G) for G = Klein 4-group is isomorphic to $S_3$
This is most likely a lack of understanding of wording on my part. I was considerind the Klein 4-group as the set of four permutations: the identity permutation, and three other permutations of four elements, where each of those is made up of two transposes, (i.e., 1 $\rightarrow$ 2, 2 $\rightarrow$ 1 and 3 $\rightarrow$ 4, 4 $\rightarrow$ 3) taken over the three possible such combinations of four elements.

Here, then, is my question. I am assuming (?) that Aut(G) in this case is the set of permutations of the four elements of the Klein 4-group - or the three non-identity ones for the purpose of showing isomorphic to $S_3$. If this is the case, what does it mean to have a permutation of these three permutations that I mentioned above? As always, thanks for your help.

答:
Let $G=\{1,a_1(=a_2a_3),a_2(=a_1a_3),a_3(=a_1a_2)\}$ be Klein's $4$-group. Consider the map:

\begin{alignat}{2}
  \varphi: S_3&\longrightarrow& \operatorname{Aut}(G) \\
  \sigma&\longmapsto& \varphi_\sigma:G &\longrightarrow G \\
  &&1&\longmapsto 1 \\
  &&a_i&\longmapsto a_{\sigma(i)} \\
\end{alignat}

Good definition of $\varphi$ (codomain). Firstly, for every $\sigma\in S_3$, $\varphi_\sigma\in \operatorname{Sym}(G)$; moreover, $\forall\sigma\in S_3$:

$$\varphi_\sigma(1a_i)=\varphi_\sigma(a_i)=1\varphi_\sigma(a_i)=\varphi_\sigma(1)\varphi_\sigma(a_i)$$

and ($i=1,2,3$; $\{i,j,k\}=\{1,2,3\}$):

\begin{alignat}{1}
&\varphi_\sigma(a_i^2)=(\varphi_\sigma(a_i))^2&&\iff 1=a_{\sigma(i)}^2 &&&:\text{true}\space\forall\sigma\in S_3 \\
&\varphi_\sigma(a_ia_j)=\varphi_\sigma(a_i)\varphi_\sigma(a_j)&&\iff a_{\sigma(k)}=a_{\sigma(i)}a_{\sigma(j)} &&&:\text{true}\space\forall\sigma\in S_3 \\
\end{alignat}

Therefore, $\varphi_\sigma\in\operatorname{Aut}(G)$ for every $\sigma\in S_3$.

Injectivity of $\varphi$.

\begin{alignat}{1}
\varphi_\sigma=\varphi_\tau &\Longrightarrow a_{\sigma(i)}=a_{\tau(i)}, \forall i \\
&\Longrightarrow \sigma(i)=\tau(i), \forall i \\
&\Longrightarrow \sigma=\tau \\
\end{alignat}

Surjectivity of $\varphi$. Every automorphism of $G$, say $\lambda$, must send $1$ to $1$, $a_1$ to some element of $\{a_1,a_2,a_3\}$ (say $a_{\bar i}$), $a_2$ to some element of $\{a_1,a_2,a_3\}\setminus\{a_{\bar i}\}$ (say $a_{\bar j})$, and $a_3$ to the lonely element of $\{a_1,a_2,a_3\}\setminus\{a_{\bar i},a_{\bar j}\}$ (say $a_{\bar k}$). Thus, $\lambda=\varphi_\sigma$, where $\sigma$ is the permutation defined by $\sigma(1)=\bar i$, $\sigma(2)=\bar j$, $\sigma(3)=\bar k$.

$\varphi$ is a group homomorphism. In fact:

\begin{alignat}{1}
\varphi_{\sigma\tau}(1) = 1 =\varphi_\sigma(1)=\varphi_\sigma(\varphi_\tau(1))=(\varphi_\sigma\varphi_\tau)(1)
\end{alignat}

and

\begin{alignat}{1}
\varphi_{\sigma\tau}(a_i) &= a_{(\sigma\tau)(i)} \\
&= a_{\sigma(\tau(i))} \\
&= \varphi_\sigma(a_{\tau(i)}) \\
&= \varphi_\sigma(\varphi_\tau(a_i)) \\
&= (\varphi_\sigma\varphi_\tau)(a_i) \\
\end{alignat}

whence $\varphi_{\sigma\tau}=\varphi_\sigma\varphi_\tau$.

By this all:

$$\operatorname{Aut}(G)\stackrel{\varphi}{\cong}S_3$$


问:
Partitions of $n$ into even number of parts versus into odd number of parts
I'm under the impression that if $n$ is even then the number of partitions of $n$ into an even number of parts exceeds the number of partitions of $n$ into an odd number of parts. And the opposite if $n$ is odd.

I feel like this is clear by looking at small examples, but I can't give a rigorous proof. I think it follows from how you make the partitions out of $n$ ones.

Would anyone be able to help me gain some more intuition about this or provide me with a proof? Thanks!

答:
Your conjecture is correct for $n>2$.

If $n>2$, $n$ has at least one self-conjugate partition; see the answer to this question. If $c(n)$ is the number of self-conjugate partitions of $n$, $$c(n)=(-1)^n(p_E(n)-p_O(n))\,,$$ where $p_E(n)$ and $p_O(n)$ are the numbers of partitions of $n$ with an even and an odd number of parts, respectively; this is proved in the answers to this question. Thus, $c(n)$ is positive for $n>2$, so $p_E(n)-p_O(n)$ must be positive for even $n>2$ and negative for odd $n>2$.


Title: Simplify $2\log 4 + 3\log 8 - \log 2$ using logarithmic laws, then evaluate.

Body: Firs time I've encountered this type of question where there is a number in front of the log, and it's throwing me a bit. I'm fine with using the logarithmic laws to simplify, but not sure if I need to get rid of those numbers first, and if so, how? thought perhaps the power law would apply but obviously there's no powers so I'm really not sure how to approach this one.</p>

Body:
$2log4=2log(2^2)=2\cdot 2log2=4log2$, using $log(2^2)=2log2$

$3log8=3log(2^3)=3\cdot 3log2=9log2$

$2log4+3log8-log2=4log2+9log2-log2=12log2$


问:
$ \int \frac{1+\sin 4x}{(\sin x -\cos x) \cdot \cos x}\, dx$ by substitution
I have to solve this integral $$ \int \frac{1+\sin 4x}{(\sin x -\cos x) \cdot \cos x}\, dx$$
It seems that the most convenient way to operate is doing the substitution $ \tan x = t$.
Then after some passages the integral becomes:
$$  \int \frac{t^4-4t^3+2t^2+4t+1}{(t-1)(t^2+1)^2}\, dx  $$

Now I'm trying the coefficient A,B,C,D,E and F such as:

$$\frac{t^4-4t^3+2t^2+4t+1}{(t-1)(t^2+1)^2} = \frac{A}{t-1}+\frac{2B(t-1)+C}{t^2+1}+ \frac {d}{dx} \frac{D(t-1)^2+E(t-1)+F}{(t-1)(t^2+1)}$$

Perhaps I made mistakes in this last passage because I find some coefficients that don't lead to the result suggested by the book.

答:
Substitute $t=2x+\frac\pi4$ as shown below

\begin{align}
&\int \frac{1+\sin 4x}{(\sin x -\cos x) \cos x}\, dx\\
=& \int \frac{2(\sin2x +\cos 2x )^2}{\sin2x-\cos2x-1}dx
=-\int \frac{2\sin^2t}{1+\sqrt2\cos t}dx\\
=& \int \left(- 1+\sqrt2\cos t -\frac1{1+\sqrt2\cos t}\right)dt\\
= &-t + \sqrt2\sin t - \ln| \frac{\sqrt2+1+\tan\frac t2}{\sqrt2+1-\tan\frac t2}|  +C\\
\end{align}


问:
Unsolved Problems due to Lack of Computational Power
I was recently reading up about computational power and its uses in maths particularly to find counterexamples to conjectures. I was wondering are there any current mathematical problems which we are unable to solve due to our lack of computational power or inaccessibility to it.

What exactly am I looking for?

Problems of which we know that they can be solved with a finite (but very long) computation?

(e. g. NOT the Riemann hypothesis or twin prime conjecture)

I am looking for specific examples.

答:
The Great Internet Mersenne Prime Search (GIMPS) comes to mind.  It's all about using computational muscle to find larger Mersenne primes.

Mersenne primes are primes of the form $2^n-1$.  There are $51$ which are known,  including the largest known prime number.  But there are conjectured to be infinitely many.


问:
Find integral limits for step functions
I have an equation of the form

$$f(t) + \int_0^t H(t') dt' = c$$.

For now, assume that $f(t)$ is linear, and $H(t)$ is some arbitrary step function (not necessarily just a standard heaviside step function. Is there any easy/ logical way of solving this to find the integral limit t. It feels like it should be easy, as the integral of the step function is 'just' a piecewise linear function, but I'm really struggling to do this.

While I don't have any specific values in mind, I am most interested in the following:

The values of the n components of H(t) form a non-negative decreasing sequence.

$f(t) = kt$, where $k > 0$.

$c > 0$.

Under these conditions, I think there should be a unique solution (though would be interested to hear if I am wrong in assuming this!).

答:
For any function $h$ (not necessarily step),

$$f(t)+\int_0^t h(t)\,dt=c$$

is

$$f(t)+H(t)-H(0)=c$$ where $H$ is an antiderivative of $h$. To obtain $t$, you solve this ordinary, univariate equation.

If $h$ is a step function, its antiderivative is piecewise linear and continuous. You can solve the equation in every piece independently (and check that the solution(s) belong to the definition interval of the piece).

If $f$ is $kt$, then $t=0$ is always a solution. There can be a second solution if the values of $h$ are larger, then smaller than $k$.

Title: Find integral limits for step functions


问:
Why does $p$ divide $\binom{p}{k}$ where $p$ is prime?
On the Wikipedia page about Binomial coefficients, it states that $p$ divides $\binom{p}{k}$ where $p$ is prime. It offers a proof that I don't understand from the very first line. It says that it is true since $\binom{p}{k}$ is  a natural number and $p$ divides the numerator but not the denominator. I don't get how that shows us anything really.

Please could you explain why the above statement is true?

Many thanks.

答: We know that $\binom pk = n = p \frac ab$ where $a, b$ are integers and $p$ is not a factor of $b$.  Therefore, $pa=bn$ so $p \mid bn$.  We also know, because $p$ does not divide $b$, that $p$ and $b$ are relatively prime.  Therefore, it must be the case that $p$ divides the other factor of the product, which is $n = \binom pk$.


问:
Is this a correct $CDF$
For a uniformly distributed r. v. $X$ on $[0, 3]$ find $CDF$ of $Y = X^2$.

Here's what I do:

$$p\left(x\right)=\begin{cases}
\dfrac{1}{3},x\in\left[0;3\right]\\
0,x\notin\left[0;3\right]
\end{cases}$$

If $x\in [0, 3]$, then $y\in [0, 9]$.

$$F_{Y}\left(Y\le y\right)=F_{Y}\left(X^{2}\le y\right)=F_{Y}\left(X^{2}\le y\right)=F_{Y}\left(-\sqrt{y}\le X\le\sqrt{y}\right)$$

$$F_{Y}\left(Y\le y\right)=\frac{1}{3}x\bigg|_{-\sqrt{y}}^{\sqrt{y}}=\frac{2\sqrt{y}}{3}$$

Thus, $$F_{Y}\left(y\right)=\begin{cases}
0,y<0\\
\frac{2\sqrt{y}}{3},y\in\left[0;9\right]\\
1,y>9
\end{cases}$$

Is this a correct solution? I'm a bit confused that at $y=9$ the $F_y$ is greater than $1$.

答: Note that $p$ is $0$ on the interval $[- \sqrt{y}, 0),$ so actually $\mathbb{P}(Y \leq y) = \frac{1}{3} x \vert_0^{\sqrt{y}} = \frac{\sqrt{y}}{3}.$ Then it is immediate that this is indeed a cdf. I hope this helps. :)

Title: Is this a correct $CDF$


答:
$F_Y(y)$ certainly cannot be $\frac23\sqrt y$ for $y$ close to $9$, because $\lim_{y\to 9^-}\frac23\sqrt y=2>1$.

If you want to calculate $P(-\sqrt{y}\le X\le \sqrt y)$ using the pdf of $X$, which you can do, then you must integrate the pdf of $X$. The pdf of $X$ is the function $p_X(t)=\begin{cases}0&\text{if }t\le 0\lor x\ge 3\\ \frac13&\text{if }0<t<3\end{cases}$ and therefore, for $y\ge 0$, $$P(-\sqrt y\le X\le y)=\int_{-\sqrt y}^{\sqrt y}p_X(x)\,dx=\int_0^{\min\{\sqrt y,\,3\}} \frac13\,dx=\frac13\min\{\sqrt y,\,3\}=\\=\begin{cases}1&\text{if }y\ge9\\ \frac13\sqrt y&\text{if }0\le y<9\end{cases}$$

Title: Is this a correct $CDF$

Body:
For a uniformly distributed r. v. $X$ on $[0, 3]$ find $CDF$ of $Y = X^2$.

Here's what I do:

$$p\left(x\right)=\begin{cases}
\dfrac{1}{3},x\in\left[0;3\right]\\
0,x\notin\left[0;3\right]
\end{cases}$$

If $x\in [0, 3]$, then $y\in [0, 9]$.

$$F_{Y}\left(Y\le y\right)=F_{Y}\left(X^{2}\le y\right)=F_{Y}\left(X^{2}\le y\right)=F_{Y}\left(-\sqrt{y}\le X\le\sqrt{y}\right)$$

$$F_{Y}\left(Y\le y\right)=\frac{1}{3}x\bigg|_{-\sqrt{y}}^{\sqrt{y}}=\frac{2\sqrt{y}}{3}$$

Thus, $$F_{Y}\left(y\right)=\begin{cases}
0,y<0\\
\frac{2\sqrt{y}}{3},y\in\left[0;9\right]\\
1,y>9
\end{cases}$$

Is this a correct solution? I'm a bit confused that at $y=9$ the $F_y$ is greater than $1$.


问:
Are there any interesting non standard models of $Q$?
I've encountered a few of the models of Robinson arithmetic here a quick list:

$\mathbb{N}\cup {\infty}$ (used to show Robinson arithmetic has a non standard model)

$\mathbb{N}\cup \{a,b\}$ where the operations are defined so $a$ and $b$ do not commute.

$\mathbb{Z}[x]^+$ which is the set of all integer polynomials with positive leading coefficient. (In this model you can show Robinson Arithmetic cannot prove all elements are odd or even.)

All these models seem to be used only to show how weak Robinson Arithmetic is or rather what things it's unable to prove. Besides the natural numbers are there some models that have some importance beyond being used as a counter example? I am aware that any model of $Q$ must contain an isomorphic copy of the natural numbers. But for weak theories like the group axioms there are plenty of important "models" for example the symmetric group is studied for other reasons besides proving that the group axioms can't prove commutativity.

答:
While $\mathbb{N}\cup\{\infty\}$ is a bit boring, it's an initial segment of a much more interesting model of $\mathsf{Q}$: namely, cardinal arithmetic (up to whatever cutoff point you want). (Note that we take $S\kappa=\kappa+1=\kappa$, not $\kappa^+$, when $\kappa$ is infinite.) Of course this is a very boring arithmetic system, but it's definitely relevant outside of $\mathsf{Q}$-flavored topics.

Ordinal arithmetic is more interesting, but it doesn't quite get $\mathsf{Q}$: successor and multiplication play a bit messily with each other. If we want $\omega$ to be a successor we need to interpret $S\alpha$ as $1+\alpha$ rather than $\alpha+1$, and that breaks the multiplication axioms. Still, ordinal arithmetic does form a very natural model of an "alternative arithmetic," just not $\mathsf{Q}$ specifically.

And here's another one coming from set theory which is much less trivial (if less useful as well):

Work in $\mathsf{ZF}$ (= set theory without choice). A Dedekind-finite set is one which does not have a non-surjective self-injection; every finite set is Dedekind-finite, but the converse is not provable in $\mathsf{ZF}$ alone. It's easy to prove that the Dedekind-finite sets are closed under finite unions, finite Cartesian products, and adding or removing (if the set is nonempty) single elements.

Ignoring set/class issues for simplicity, this leads to the following:

The collection of Dedekind-finite cardinalities consistently forms a nonstandard model of $\mathsf{Q}$.

Specifically, the collection of Dedekind-finite cardinalities is always a model of $\mathsf{Q}$, and it is nonstandard iff there is an infinite Dedekind-finite set.

Interestingly, there is a lot of flexibility re: how good the arithmetic of Dedekind-finite sets can be. In every "easily-constructible" model of $\mathsf{ZF}$ with Dedekind-finite sets, there are incomparable Dedekind-finite cardinalities and so we get a model of $\mathsf{Q}$ with no good ordering (in contrast to e.g. $\mathbb{Z}[x]^+$). On the other hand, it turns out that $(i)$ we can also (nontrivially) get all Dedekind-finite cardinalities comparable and $(ii)$ whenever we do so, the resulting model of $\mathsf{Q}$ in fact satisfies all of true first-order arithmetic! See here; note that there is a weak large cardinal assumption used, but its necessity is open.

As a coda, let me push back against the claim that $\mathbb{Z}[x]^+$ is unnatural. In my opinion, it's extremely natural; it's equivalent to $\mathbb{Z}[x]$ in a very strong sense, namely bi-interpretability, in the same way that we can build the integers from the natural numbers.

The issue is that negative elements are convenient in algebra but somewhat inconvenient in logic (since "induction for $\mathbb{Z}$" is messier than "induction for $\mathbb{N}$"). But everything we do algebraically with $\mathbb{Z}[x]$ could be appropriately (if tediously) rephrased to live entirely in $\mathbb{Z}[x]^+$, and conversely we could have whipped up "$\mathsf{Q}$-for-$\mathbb{Z}$" instead of $\mathsf{Q}$ (and similarly "$\mathsf{PA}$-for-$\mathbb{Z}$" and so on) with the same effect. So I think this is a case where what we're really seeing is a less convenient presentation of a common and extremely useful notion.


问:
Proof $f(n)=f(n-1)+2f(n-2)$ by induction.
I am having a hard time trying to prove the following recurrence relation
$$f(n)=f(n-1)+2f(n-2),\quad n\geq 2, \quad f(1)=1 \quad \& \quad f(2)=3.$$

$f(n)$ is the number of ways to tile a $(2\times n)$-rectangle with tiles of the form $2 \times 1$, $1 \times 2$, or $2\times 2$.

I attempt this by induction but I am confused in the base case since I get $f(2)=f(1)+2f(0)$ but we do not know $f(0)$.

Can someone help?

答:
Hint

Suppose you start by using the tile $2\times 1$. So, you get exactly the same problem but facing a $(2\times (n-1))-$rectangle. So, you get $f(n-1)$ ways of to tile a $(2\times (n-1))-$rectangle.

However if first use $2\times 1$ tile, you find yourself forced to use another $2\times 1$ tile in the upper part. So, you get exactly the same problem but facing a $(2\times (n-2))-$rectangle. So, you get $f(n-2)$ ways of to tile a $(2\times (n-2))-$rectangle.

Can you understand what happens if you first use the $2\times 2$ tile?


问:
Evaluating $\iint_s \vec F \cdot \hat n ds $ around the curved surface of cylinder cut by plane at $45^\circ $
I need to calculate the surface integral of $F(x,y,z) = \hat i x +\hat j y + \hat k z$ on the curved part of surface $x^2+z^2 = 1, x+y=2, $ and $y$ goes from $1$ to $3$ as shown in following figure. How do I evaluate $\displaystyle \iint_S \vec F .\hat n ds$
 this surface?

EDIT::I couldn't do it via parametrization, I got the above figure which is incorrect. Using this formula $\iint_s \vec F \cdot \frac{\nabla \phi }{|\nabla \phi|}\sqrt{1 + (z_x)^2 + (z_y)^2} dx dy$ I got the following. Not sure if it's correct.
$$\int_1^3 \;dy \int_{-1}^{2-y} \vec F(x, y , \sqrt{1-x^2})\cdot \frac{x \hat i + \sqrt{1-x^2}\hat k}{\sqrt{1-x^2}} dx  \\ + \int_1^3 \;dy \int_{-1}^{2-y} \vec F(x, y , -\sqrt{1-x^2})\cdot \frac{x \hat i - \sqrt{1-x^2}\hat k}{\sqrt{1-x^2}} dx$$

答: Instead of calculating the integral over this surface, you can calculate it over the two surfaces that would close the shape (the base circle and the slanted circle from the cutting plane) and then negate the result. Over these surfaces, which are flat, recall also that the integral is just the flux of the field passing through the surface.</p>


问:
Proof verification of absolutely convergent series .
I didn’t quite understand the Absolute Convergence of a series, so  I thought that looking at the proof would help me understand it, but   Unfortunately it didn’t, from my personal understanding to the theorem, i think that it says :

If the series: $$\sum_{n=1}^{\infty}\mid{a_n}\mid$$
Is convergent then the series $$\sum_{n=1}^{\infty}{a_n}$$ is also convergent.

So if this is what really says, then i have a proof of it  and not only a regular proof it has an equivalence relation $\iff$.

Proof:

$\Leftarrow$

Assume that The series $\sum_{n=1}^{\infty}{a_n}$ is convergent so $-\sum_{n=1}^{\infty}{a_n}$ is also convergent and $$-\sum_{n=1}^{\infty}{a_n}= \sum_{n=1}^{\infty}{-a_n}$$
which means that $\sum_{n=1}^{\infty}\mid{a_n}\mid$ is convergent.

$\Rightarrow$

Assume that $\sum_{n=1}^{\infty}\mid{a_n}\mid$ is convergent, so by definition :
$$\sum_{n=1}^{\infty}\mid{a_n}\mid=\cases{\sum_{n=1}^{\infty}{-a_n} \\ \sum_{n=1}^{\infty}{a_n}}$$
So one of these series is convergent so if the first series is convergent we can just multiply by $-$ and we get that the second onr is also convergent, the same thing if the second one is convergent.

Over all I have two questions:

Is my understanding about the absolute convergent Theorem right?

If so, is my proof valid?

答: No, what you proved is not correct and it seems you don't exactly understand what $\sum_{n \geq 0} \vert a_n \vert$ means. It is just the series of the absolute values of the initial terms. As pointed before, the two are not equivalent (a series which is not absolutely convergent is either outright divergent or it is convergent, in which case it is called a conditionally convergent series). Riemann has a pretty nice theorem on conditionally convergent series, I encourage you to check it out. To prove the statement you wrote, it is enough to show that the tail of $\sum_{n \geq 0} a_n$ is Cauchy. But this is immediate since the tail of $\sum_{n \geq 0} \vert a_n \vert$ is Cauchy and $\vert a_n + a_{n + 1} + \dots + a_m \vert \leq \vert a_n \vert + \vert a_{n + 1} \vert + \dots + \vert a_m \vert.$ I hope this helps. :)


问:
when can't methods of undetermined coefficient be used in this case
I have solved the equation $y'' +3y'+2y= \frac{1}{1+e^{x}}$ and I have solved through varation of parameters.

However there is an addendum to the question I was asked, and that was, why can't methods of undetermined coefficient be used to solved this?

I am at a loss for how to start to answer this, or the reasons why variation breaks down

答:
Basically, undetermined coefficients can be used when the derivatives of the source term terminate in some useful way, either by going to zero (polynomials) or by going into some kind of loop. (exponentials, sine and cosine)

Here the derivatives will never fall into one of these nice patterns and we will never be able to make a form for $y_p$ which includes every derivative.


问:
Proof of square-free polynomial factorization over finite fields
Why should it be that the greatest common divisor of a polynomial and it's derivative produce the product of the repeated factors?

What about the case where the derivative is zero in $\mathbb{F}_p$? Why does the approach continue to work in this case?

This is considered a trivial step in most algorithms, and that may be the case, but I'm having trouble finding a proof of the correctness of the basic algorithm described here.

A proof that $f / \gcd(f,f')$ eliminates repeated factors would be appreciated.

答:
If $f = q^k g$ where $q \nmid g$ and $q$ is irreducible then

$$ f' = k q' q^{k-1} g+ q^k g'$$

so the highest power of $q$ dividing $f'$ is $k-1$ (it is not difficult to show that if $q$ is irreducible then $(q,q')= 1$).

So dividing by $(f,f')$ is dividing by every irreducible factor to one power less than what appears in $f$.

EDIT: proof of $(q,q') = 1$ in the finite field case

If $q' \neq 0$ we are done because $q$ is irreducible.

If $q' = 0$, $q = \sum a_i x^{pk}$ where $p$ is the characteristic of our finite field $k$.

Note that $q(x+y) = q(x) + q(y)$. If $q(x_0) = 0$ for some $x_0 \in k$, then $(x-x_0) | q$, contradicting irreducibility of $q$.

Then, by pigeonhole, $q(x_0) = q(y_0)$ for some $x_0, y_0 \in k$. However, then $q(x_0 - y_0) = 0$, reaching a contradiction again.


问:
Self-Paced Graduate Math Courses for Independent Study
Does anyone know of any graduate math courses that are self-paced, for independent study?

I am a high school math teacher at a charter school in Texas. While I am quite happy with where I am right now, but my goal is to earn at least 18 graduate credits on math subjects so that I can teach higher-level math, and be certified as dual-credit math teacher. (HS class where students earn high school as well as college credits at the same time.) I am aware of many online graduate classes offered by some respectable universities, but all of them are semester-based, which may not be very feasible since my full-time teaching is extremely demanding, not to mention that math is anything but a casual subject.

I welcome any suggestions even for programs from outside of US, as long as they are accredited and conducted in English. (For example, I was told that the college system in the Philippine is an exact "copy cat" of the US.) For your information, I am quite comfortable studying independently, in fact, I took lots of prerequisite math classes successfully under this study mode. By the way, last year I took GRE for this purpose, my verbal + quantitative score is a decent 1200 under old scoring scale.

Thank you very much for your time and help.

答:
This posting is in response to DFEUER who wrote: "Options for online math programs are very limited. I haven't personally heard of an online, accredited, graduate-level math program of any sort. Where have you seen such? – dfeuer 16 hours ago"

Hi dfeure, here is the list I gather so far up to early last year. I suspect the landscape is very different this year since more and more schools are jumping in. Note the first two in the list are top universities, at least in Texas. UTB has the lowest tuition; UH has the least credit hours (33) to graduate from its MA Teaching program.

(1) Texas AM - College Station: http://distance-ed.math.tamu.edu/FAQ.html

(2) University of Houston: http://www.mathematics.uh.edu/graduate/master-programs/master-of-arts/index.php

(3) University of Texas - Pan America: http://portal.utpa.edu/utpa_main/daa_home/ogs_home/ogs_imagesfiles/forms/domestic/grad/master_of_science_in_mathematics.pdf

(4) University of Texas - Brownsville: http://www.utb.edu/its/olt/Pages/pMSM.aspx

(5) Emporia State University, Kansas: http://www.emporia.edu/mathcsecon/programs/master-of-science-mathematics.html

Keep in mind those above are only a short list with the most affordable tuition, there may be some other schools that have higher recognition but also charge higher tuition. If you don't mind going online to foreign schools, you may also want to check out the University of London (the oldest, century-old independent study program) or the University of Philippines - Open University. (The UP is the flag-ship university in the Philippines, at the fraction of the cost of US schools.) They do have master programs but unfortunately not in math.

Hope these info will help and thanks for responding.


问:
Seeing how Independent Random Variables are Uncorrelated
In my textbook for my measure theory class (Marek Capinski and Ekkehard Kopp:
Measure, Integral and Probability), in the chapter on the independence of two random variables, the author(s) make the following claim after defining the covariance of two random variables and correlation:

As we can see $\operatorname{Cov}(X,Y)= \mathbb{E}(XY)-\mathbb{E}(X)\mathbb{E}(Y)$, so clearly independent random variables $X$ and $Y$ are uncorrelated; it is sufficient to take $f(x)=x-\mathbb{E}(X)$ and $g(x)=x-\mathbb{E}(Y)$ in theorem $5.18$.

Where theorem $5.18$ states:

The random variables $X$ and $Y$ are independent if and only if $\mathbb{E}(f(X)g(Y))=\mathbb{E}(f(X))\mathbb{E}(g(Y))$ holds for all Borel measurable bounded functions $f,g$.

My question:

Since the author did not provide a proof, i'm not understanding how using $f(x)=x-\mathbb{E}(X)$ and $g(x)=x-\mathbb{E}(Y)$ in theorem $5.18$ shows that independent random variables $X$ and $Y$ are uncorrelated. If any one could explain or provide insight, it would be a great help.

答:
Suppose two random variables are independent. Then (choose identity functions) $E(XY)=E(X)E(Y)$.

It remains to note that

$$Cov(X,Y)=E(XY)-E(X)E(Y).$$

EDIT: Using the given functions, we have

$$Cov(X,Y)=E((X-E(X))(Y-E(Y))=E(X-E(X))E(Y-E(Y))=0$$


问:
Beginner proof that there can't exist a sample space for given conditions
Given the following conditions ($A, B, C$ are events)$$
\operatorname{Pr}[A]=\operatorname{Pr}[B]=\operatorname{Pr}[C]=\frac{5}{6} \text { and } \operatorname{Pr}[A \cap B \cap C]=0
$$

I think that there isn't a sample space that can exhibit the above conditions, intuitively, but I don't know how to show it rigorously.

答:
HINT: use complementary sets, that is, observe that

$$
\Pr [A^\complement \cup B^\complement \cup C^\complement ]=1\,\land\, \Pr [A^\complement ]=\Pr [B^\complement ]=\Pr [C^\complement ]=\frac16
$$

cannot be true using the fact that any measure is sub-additive.


问:
Function for concatenated semicircles
I am looking for a closed-form formula for something like this:

Can anybody help - Thank you!

答:
a good approximation is
y = |sin x|^(sin x) + 11/7 sin x
here is my research. I took a sine wave and circles and plotted the difference in their y values then made an equation that roughly matched it, then added sin x to that.

Title: Function for concatenated semicircles


问:
How to find focus and directrix of $4y = x^2 -16$
How would I go about finding the focus and directrix of $4y = x^2 -16$ $(1)$

I did the $SP^2 = QP^2$ and got $(y+b)^2 = (x-a)^2 + (y-b)^2$.

This gave me: $4by = x^2 -2ax +a^2$. Comparing this to $(1)$. I get $a = 0$ but I don't know what to do to get $b$. Can someone advise?

Edit: my workings:

答:
Let $(P): 4y=x^2-16$ be the parabola of focus $F$ and directrix $(d)$. By definition, the distance from any point $M$ on $(P)$ to the directrix $(d)$ equals the distance from $M$ to the focus $F$. Moreover, since $(P)$ is vertical and directed upwards, the focus is above the vertex and the directrix is horizontal and below it. We will use this in two steps.

1- Note that $(P)$ has a vertex $V(0,-4)$. This means that $F$ has a zero abcissa, say $(d):y=a$ and $F(0,b)$. Since $(d)$ is below the vertex and $F$ is above it, $a<-4$ and $b>-4$.
As the vertex $V$ is a point on $(P)$, then:

$VF=distance(V$$\rightarrow$$(d))$. Meaning $|-4-b|=|-4-a|$. Which gives $b+4=-4-a$. This implies that $a=-b-8$.

This enables us to write $F(0,b)$ and $(d):y=-b-8$ where $b$ is to be determined.

2- Choose another point on $(P)$, say $M(4,0)$. Then:

$MF^2=distance(M$$\rightarrow$$(d))^2$. Meaning $(4-0)^2+(0-b)^2=(-b-8)^2$, which gives $b=-3$. This gives $a=-5$.

Hence the focus is $F(0,-3)$ and the directrix is$(d):y=-5$.


问:
Number of students in the school
An School has $1200$ students. Each pupil takes $5$ classes a day. Each teacher teaches $4$ classes. Each class has $30$ students and $1$ teacher. How many teachers are there at School?

$\text{(A)}\ 30 \qquad \text{(B)}\ 32 \qquad \text{(C)}\ 40 \qquad \text{(D)}\ 45 \qquad \text{(E)}\ 50$

Although this question seems very easy at first glance, It confused me a lot. my guess is $\frac{1200}{30}=40$. but it doesn't make sense because I didn't use many information in the problem. can you please explain how to solve this?

答:
I will take a different approach and use dimensional analysis to solve the problem.

Define the units of $[C]$ as classes , $[S]$ as students, $[T]$ as teachers and $[h]$ as hours.

We have $1200 [S]$, $30 [S]/[C]$, $1 [T]/[C]$ $5 [h]$ for each student and $4 [h] $ for each teacher.

Notice that $$\frac{[S]}{\frac{[S]/[C]}{[h]}}=[c][h]$$
and $$\frac{[T]}{\frac{[T]/[C]}{[h]}}=[c][h]$$

This gives us a relationship between all the units i.e.
$$\frac{[S]}{\frac{[S]/[C]}{[h]}}=\frac{[T]}{\frac{[T]/[C]}{[h]}}=[c][h]$$

This, together with the values we have, gives us $$\frac{1200 \cdot 5 }{30}=4 \cdot x$$ where $x$ is the number of teachers, solving for $x$ we get $$x=50$$.

Title: Number of students in the school


问:
$\sum_{i=1}^r\sum_{j=r+1-i}^{n-r}{r \choose i}{{n-r}\choose j}=2^n-2^{n-r}+\sum_{i=1}^r\left({{n-r}\choose i}-{n\choose i}\right)$
I am trying to prove the following

$$
\sum_{i=1}^r\sum_{j=r+1-i}^{n-r}{r \choose i}{{n-r}\choose j}=2^n-2^{n-r}+\sum_{i=1}^r\left({{n-r}\choose i}-{n\choose i}\right)
$$

I want to know the hint to the correct path.

Edit

I am trying to prove this and now I got the following:

$$
-\sum_{i=1}^r \sum_{j=i}^r {r\choose i}{{n-r}\choose {j-i}}
$$
that must be equal to
$$
\sum_{i=1}^r\left({{n-r}\choose i}-{n\choose i}\right)$$

答:
A straightforward combinatorial proof is also possible. Rewrite the identity as

$$\begin{align*}
&2^n-2^{n-r}=\\
&\sum_{i=1}^r\left(\binom{n}i-\binom{n-r}i\right)+\sum_{i=1}^r\sum_{j=r+1-i}^{n-r}\binom{r}i\binom{n-r}j\tag{1}
\end{align*}$$

The lefthand side is the number of subsets of $[n]$ that are not subsets of $[n]\setminus[r]$, i.e., the number of subsets of $[n]$ that intersect $[r]$.

Now fix $i\in[r]$. $\binom{n}i-\binom{n-r}i$ is the number of subsets of $[n]$ of cardinality $i$ that intersect $[r]$, and $\binom{r}i\binom{n-r}j$ is the number of subsets of $[n]$ that have $i$ members in $[r]$ and $j$ members in $[n]\setminus[r]$, where $r+1\le i+j\le n-r+i$.

Let $S$ be a subset of $[n]$ that intersects $[r]$. If $|S|\le r$, then $S$ is counted in the $i=|S|$ term of the first summation in $(1)$ and nowhere else. If $|S|>r$, let $i=|S\cap[r]|$ and $j=|S|-i$; then $S$ is counted in the $\binom{r}i\binom{n-r}j$ term of the second summation in $(1)$ and nowhere else. Thus, every subset of $[n]$ that intersects $[r]$ is counted exactly once in $(1)$, and $(1)$ is therefore equal to $2^n-2^{n-r}$.


问:
Expectations of mean squared error
I am currently studying the expected mean squared error and the derivations of this are as follows:

\begin{align}
&E[(y_i - \hat{y}_0 | x_0]\\
&=E [ (f(x) + \epsilon_i - \hat{f}(x_i))^2 | x_0]\\
&=E [(f(x) - \hat{f}(x_i))^2|x_0] + 2 E [(f(x_i)-\hat{f}(x_i))\epsilon_i | x_0] + E[\epsilon_0^2 |x_0]\\
\end{align}

From which we can further get the bias and variance. But I do not understand the following, what rule is used in order to rewrite it using the following part: $2 E [(f(x_i)-\hat{f}(x_i))\epsilon_i | x_0]$ ?

答:
It’s important to note that the expected value operator $\mathbb{E}[\cdot]$ is a linear operator. Therefore, it holds with two random variables $X, Y$ and constant scalar $\lambda$:

$\mathbb{E}[X+Y] = \mathbb{E}[X] + \mathbb{E}[Y]$, and

$\mathbb{E}[\lambda X] = \lambda \mathbb{E}[X]$.

So you only need to apply binomial expansion to obtain:
$$\mathbb{E}[((f(x) - \hat{f}(x_i)) + \epsilon_i)^2] =\\
\mathbb{E}[(f(x) - \hat{f}(x_i))^2 + 2 (f(x_i)-\hat{f}(x_i))\ \epsilon_i + \epsilon_i^2] =\\
\mathbb{E}[(f(x) - \hat{f}(x_i))^2] + 2 \mathbb{E}[(f(x_i)-\hat{f}(x_i))\ \epsilon_i] + \mathbb{E}[\epsilon_i^2]$$.

Title: Expectations of mean squared error


问:
2021 Mathcounts practice competition target #4 question #2
What is the greatest number of positive consecutive numbers that sum to 400?

My approach:

We have terms from : $(x-n)..... x.....(x+n)$

This would mean the sum of these terms is

$(2n+1) * 2x * 1/2  = x(2n+1) = 400$

$2n+1$ MUST be an odd number, so $x$ is even

$400 = 20^2  = 2^4 * 5^2$

The odd multiple is either $5$, with $x$ being 80, or we have $25*16$
&25*16& would have more terms. Since x is 16, we know we have the range (x-12) = 4 to (x+12) = 28

Therefore, we have 25 terms as the maximum.

The correct answer is 27, and I don't know how to get that. I considered the possibility of an even amount of terms , but 27 isn't even so that wouldn't matter.  Where did I go wrong / Is the real answer wrong?

PROOF THIS IS NOT AN ONGOING TEST:

I have the real answer.

答:
Let the sum be $\sum_{k=1}^n (x+k)$, where $x\geqslant 0$.
This sum has $n$ positive consecutive terms and equals

$$nx + (1+2+\dots+n) = nx + n(n+1)/2 = \frac n2(2x+n+1)$$

In other words, we need to find the maximal integer $n$ such that for some integral $x\geqslant 0$ we have

$$\frac n2(2x+n+1) = 400 \iff n\big(2x+(n+1)\big) = 800$$

From this, we have that $n$ divides $800$, so $27$ cannot be the answer.

Of course, $n\big(2x+(n+1)\big) > n^2$ so $n$ must be less than $\sqrt{800} \simeq 28.28$, which implies $n\leqslant 28$.

Now, $26$, $27$ and $28$ do not divide $800$, so our first candidate is $25$.
We get

$$25(2x+26) = 800 \implies 2x + 26 = 32 \implies x = 3,$$

which confirms $n=25$ as being the correct answer.


答: Let $400=(x+1)+(x+2)+\cdots+(y-1)+y$, where $y\ge x>0$. Now, we have $400=\frac{x(x+1)}2-\frac{y(y+1)}2$, or equivalently, $800=(x-y)(x+y+1)$. Thus, we wish to find a factorization $800=n\cdot m$ where $m$ and $n$ have different parities, $m>n\ge0$, and which maximizes $n$ (the length of the sum). After some experimentation we see that $n=25$ and $m=32$ is the best possible.

Title: 2021 Mathcounts practice competition target #4 question #2

Body:
What is the greatest number of positive consecutive numbers that sum to 400?

My approach:

We have terms from : $(x-n)..... x.....(x+n)$

This would mean the sum of these terms is

$(2n+1) * 2x * 1/2  = x(2n+1) = 400$

$2n+1$ MUST be an odd number, so $x$ is even

$400 = 20^2  = 2^4 * 5^2$

The odd multiple is either $5$, with $x$ being 80, or we have $25*16$
&25*16& would have more terms. Since x is 16, we know we have the range (x-12) = 4 to (x+12) = 28

Therefore, we have 25 terms as the maximum.

The correct answer is 27, and I don't know how to get that. I considered the possibility of an even amount of terms , but 27 isn't even so that wouldn't matter.  Where did I go wrong / Is the real answer wrong?

PROOF THIS IS NOT AN ONGOING TEST:

I have the real answer.


答:
In characteristic $p$ there is a subtlety with things like $f=x^p(x-1)$ which gives $f' = x^p=\gcd(f,f')$.

In general factoring in irreducibles $f=\prod_j h_j^{e_j}$ we'll have $\gcd(f,f')=(\prod_{j,p\ \nmid\ e_j} h_j^{e_j-1})(\prod_{j,p\ |\ e_j} h_j^{e_j})$, this follows from $f' = \sum_j e_j h_j^{e_j-1} \prod_{i\ne j} h_i^{e_i}$ and $\gcd(f,f')=\prod_j h_j^{d_j}$ for some $d_j$ (as $\gcd(f',f)$ divides $f$).

Title: Proof of square-free polynomial factorization over finite fields

Body:
Why should it be that the greatest common divisor of a polynomial and it's derivative produce the product of the repeated factors?

What about the case where the derivative is zero in $\mathbb{F}_p$? Why does the approach continue to work in this case?

This is considered a trivial step in most algorithms, and that may be the case, but I'm having trouble finding a proof of the correctness of the basic algorithm described here.

A proof that $f / \gcd(f,f')$ eliminates repeated factors would be appreciated.


问:
Prove that $G$ is an equivalence relation, and describe $ f ^ * (G) $.

Let $G$ be the relation in $\Bbb {R}^2$ given by $$ G = \{((a, b), (c, d)) \in \Bbb {R}^2 \times \Bbb {R}^2: a ^ 2 + b ^ 2 = c ^ 2 + d ^ 2 \}. $$ Prove that $ G $ is an equivalence relation. Also describe
$$ f ^ * (G) = \{(x, y) \in \mathbb R^2: (f (x), f (y)) \in G \} $$ where $ f: \Bbb {R} \to \Bbb  {R} ^ 2 $ is given by $ f (t) = (\sin (t), \cos (t)) $.

I already proved that $ G $ is an equivalence relation, it is quite clear and easy to do, that.  But how do I describe $ f ^ * (G) $.

答:
For $x,y\in\Bbb R$, we have
$$\begin{align} (x,y)\in f^*(G)&\iff (f(x),f(y))\in G\\&\iff \bigl((\sin x,\cos x),(\sin y,\cos y)\bigr)\in G\\&\iff \sin^2x+\cos^2x=\sin^2y+\cos^2y\end{align}$$
but the latter is true for all $x,y$ (because both sides are $=1$), hence $f^*(G)=\Bbb R\times \Bbb R$.


答: What you need is the set of pairs $(x,y)$ such that the pairs $(\sin x,\cos x)$ and $(\sin y,\cos y)$ are related. According to the definition of the relation, this happens if $\sin^2x+\cos^2x=\sin^2y+\cos^2y$. But this is always the case, since both are equal to $1$. Therefore, $f^*(G)=\mathbb{R}^2$.

Title: Prove that $G$ is an equivalence relation, and describe $ f ^ * (G) $.

Body:
Let $G$ be the relation in $\Bbb {R}^2$ given by $$ G = \{((a, b), (c, d)) \in \Bbb {R}^2 \times \Bbb {R}^2: a ^ 2 + b ^ 2 = c ^ 2 + d ^ 2 \}. $$ Prove that $ G $ is an equivalence relation. Also describe
$$ f ^ * (G) = \{(x, y) \in \mathbb R^2: (f (x), f (y)) \in G \} $$ where $ f: \Bbb {R} \to \Bbb  {R} ^ 2 $ is given by $ f (t) = (\sin (t), \cos (t)) $.

I already proved that $ G $ is an equivalence relation, it is quite clear and easy to do, that.  But how do I describe $ f ^ * (G) $.


问:
Evaluating $\int_0^{\pi/2} \log(1 - x \cot x) \, dx\;$ (Is there a closed form?)
I am interested in knowing if it is possible to find a closed-form solution to the following challenging log-cotangent integral

$$\int_0^{\frac{\pi}{2}} \log(1 - x \cot x) \, dx$$

I very much doubt a closed form in terms of known mathematical constants can be found and have no reason to suspect one exists.

In the absence of a simple closed-form value being found, maybe the integral can be evaluated in terms of an infinite series. In this direction, since $0 < x \cot x < 1$ for all $x \in (0,\frac{\pi}{2})$ the log term appearing in the integrand can be expanded. Doing so produces
$$\int_0^{\frac{\pi}{2}} \log(1 - x \cot x) \, dx = -\sum_{n = 1}^\infty \frac{1}{n} \int_0^{\frac{\pi}{2}} (x \cot x)^n \, dx = -\sum_{n = 1}^\infty \frac{a_n}{n},$$
where
$$a_n = \int_0^{\frac{\pi}{2}} (x \cot x)^n \, dx. \tag1$$
Noting that
\begin{align*}
a_1 &= \int_0^{\frac{\pi}{2}} x \cot x \, dx = \frac{\pi}{2} \log (2)\\
a_2 &= \int_0^{\frac{\pi}{2}} (x \cot x)^2 \, dx = \pi \log (2) - \frac{\pi^3}{24}\\
a_3 &= \int_0^{\frac{\pi}{2}} (x \cot x)^3 \, dx = \frac{9\pi}{16} \zeta (3) - \frac{\pi^3}{16} + \frac{3\pi}{2} \log (2) - \frac{\pi^3}{8} \log (2),
\end{align*}
perhaps it is possible to find a general expression for $a_n$. Indeed, an attempt at finding (1) can be found here.

Any other approaches or suggestions to this challenging integral would be welcome.

答:
Edited to make the answer self-contained

We introduce first a representation of the integral which makes use of Bessel functions:
\begin{align}
 I&=\int_0^{\frac{\pi}{2}} \log(1 - x \cot x) \, dx\\
 &=\int_0^{\frac{\pi}{2}} \log\left( \frac{\sin x-x\cos x}{\sin x} \right) \, dx\\
 &=\int_0^{\frac{\pi}{2}} \log\left( \sqrt{\frac{\pi}{2}}\frac{x^{3/2}J_{3/2}(x)}{\sin x} \right) \, dx\\
 &=-\frac{3\pi}{4}+\pi\ln \pi-\frac{\pi}{4}\ln 2-\frac{\pi}{2}\ln 2 -\int_0^{\frac{\pi}{2}}\log\left( J_{3/2}(x) \right)\,dx
\end{align}

In a paper by Dickinson a series expansion is given for the derivative of the logarithm of a Bessel function:
\begin{equation}
 \frac{d}{dx}\log J_\nu(x)=\frac{\nu}{x}-\sum_{n=1}^\infty\sum_{k=0}^\infty \frac{2x^{2k+1}}{j_{\nu,n}^{2k+2}}
\end{equation}
$j_{\nu,n}$ is the $n$-th zero of the Bessel function of order $\nu$. This result holds provided $0<x<j_{\nu,1}$ (which is the case here). By integrating this result,
\begin{equation}
 \log J_\nu(x)=\nu\log(x)-\sum_{n=1}^\infty\sum_{k=0}^\infty \frac{1}{k+1}\frac{x^{2k+2}}{j_{\nu,n}^{2k+2}}+C
\end{equation}
The constant is determined by adjusting the result for $x\to0$. As $\log J_\nu(x)= \nu\log x-\nu\log 2-\log\Gamma(\nu+1)+O(x^2)$
\begin{equation}
 C=-\nu\log 2-\log\Gamma(\nu+1)
\end{equation}
Then,
\begin{equation}
 \log\left(J_{3/2}(x) \right)=\frac{3}{2}\log x-\frac{3}{2}\log 2-\log\left(\frac{3\sqrt{\pi}}{4}\right)-\sum_{k=0}^\infty \frac{x^{2k+2}}{k+1}\sum_{n=1}^\infty\frac{1}{j_{3/2,n}^{2k+2}}
\end{equation}
The integral reads thus
\begin{equation}
 I=\frac{\pi}{2}\left( 3\ln\pi-3-\ln(12) \right)-\sum_{k=0}^\infty\frac{(\pi/2)^{2k+3}}{(k+1)(2k+3)}\sum_{n=1}^\infty \frac{1}{j_{3/2,n}^{2k+2}}
\end{equation}
The sum of inverse even powers of zeros of Bessel functions appears in problems involving the diffusion equation and a broad literature exists on the topic. The quantity
\begin{equation}
 \sigma(p,\nu)=\sum_{n=1}^\infty\frac{1}{j_{\nu,n}^{2p}}
\end{equation}
is discussed here. A recent analysis can be found in a paper by Jorge L. deLyra.
The first values for $\nu=3/2$ are
\begin{align}
 &\sigma(1,3/2)=\frac{1}{10}\;;\;\sigma(2,3/2)=\frac{1}{350}\;;\;\sigma(3,3/2)=\frac{1}{7875}\\
 &\sigma(4,3/2)=\frac{37}{6063750}\;;\;\sigma(5,3/2)=\frac{59}{197071875}
\end{align}
we find
\begin{align}
 I&=\frac{\pi}{2}\left( 3\ln\pi-3-\ln(12) \right)-\sum_{k=0}^\infty\frac{\sigma(k+1,3/2)}{(k+1)(2k+3)}\left(\frac{\pi}{2}\right)^{2k+3}\\
 &=\frac{\pi}{2}\left(3\ln\frac{\pi}{e}-\ln(12) \right)-\\
&\hspace{1cm}-\frac{\pi^3}{8}\left(\frac{1}{30}+\frac{(\pi/2)^2}{3500}+ \frac{(\pi/2)^4}{165375}+\frac{37(\pi/2)^6}{218295000}+\frac{59(\pi/2)^8}{10838953125}+\cdots \right)
\end{align}
which is identical to the series proposed by @ClaudeLeibovici.

As remarked by @JamesArathoon in a comment, it is also possible to perform first IBP on the integral. Then, the problem boils down to integrate $J_{1/2}(x)/J_{3/2}(x)$. By using the recurrence formula of the Bessel function, it requires to integrate $J_{5/2}(x)/J_{3/2}(x)$. A series expansion for the ratio of consecutive Bessel functions is also given in the above cited paper. It also makes use of the $\sigma(k,3/2)$, but the convergence is slower in this case.


问:
neutral Tannakian categories
I'm trying to understand the following Corollary in Deligne and Milne's Tannakian Categories. I don't know which obvious thing I'm missing, but why is the last sentence "Obviously $F \mapsto F^*$ and $f \mapsto \omega^f$ are inverse maps" true? Maybe someone who has read the book could help me out.

答: This should have been a comment, but I'm not permitted to comment. I think all that is meant is that when you form the two composites, you "obviously" get the identity map.</p>

Title: neutral Tannakian categories


问:
Characterizing direct sum of modules
Suppose $M$ is an $R$-module (for a commutative ring $R$ with $1$) and that $(M_i)_{i\in I}$ is a collection of $R$-submodules of $M$. The inclusion maps $j_i:M_i\rightarrow M$ determine an $R$ module homomorphism $\operatorname{Hom}(M,N)\rightarrow\prod_{i\in I}\operatorname{Hom}(M_i,N)$ given by $\varphi\mapsto (\varphi\circ j_i)_{i\in I}$, where $N$ is any arbitrary $R$-module.

Suppose that this map is an isomorphism for any $N$. I'm trying to understand why this implies that $M$ is a direct sum of the $M_i$. I tried plugging in different values for $N$, such as $M$ and then mapping over the identity, but this doesn't seem to get me anywhere. Any suggestions? What am I missing?

Edit: I am looking for a reason that does not appeal to the universal property.

答:
Recall that $\bigoplus_{i \in I} M_i = \{(m_i)_{i \in I} \in \prod_{i \in I} M_i : \{i \in I: m_i\neq0\} \textrm{ is finite}\}$ comes together with a family $\{\iota_i : M_i \to \bigoplus_{i \in I} M_i\}_{i \in I}$ of canonical "inclusions" (what are they?).

So, take $N = \bigoplus_{i \in I} M_i$ and $\varphi \in \operatorname{Hom}(M,N)$ the preimage of $(\iota_i)_{i \in I} \in \prod_{i \in I} \operatorname{Hom}(M_i,N)$ under your isomorphism. Prove that $\varphi$ is the inverse of the $R$-module homomorphism $\psi : N \to M$ given by $(m_i)_{i \in I} \mapsto \sum_{i \in I} m_i$.

Title: Characterizing direct sum of modules


问:
Isomorphic quotient groups
Suppose that $G$ is a finite group, and that $H$ and $K$ are normal subgroups of $G$ with trivial intersection, and suppose that $H$ and $K$ are isomorphic. Is it true that the quotient groups $G/H$ and $G/K$ are isomorphic? </p>

答: Let $G=\mathbb Z_2\times\mathbb Z_4$. Find two subgroups in $G$ isomorphic to $\mathbb Z_2$ and intersecting trivially such that the quotients of $G$ by them are not isomorphic.</p>

Title: Isomorphic quotient groups


问:
If $\gcd(a,b)=1$ and $a$ and $b$ divide $c$, then so does $ab$

Using divisibility theorems, prove that if $\gcd(a,b)=1$ and $a|c$ and $b|c$, then $ab|c$.

This is pretty clear by UPF, but I'm having some trouble proving it using divisibility theorems. I was looking at this Math.SE problem, and if this could be proven, then the solution would follow immediately.

答:
Applying $\rm\color{#c00}{Euclid's}$ Lemma: $\,\ b\mid an\overset{\color{#c00}{(a,\,b)\ {\bf =}\ 1}}\Longrightarrow\,b\mid n\,\overset{\large \times\ a\ }\Longrightarrow\,ab\mid an\quad  \small{\bf QED}$

Remark $ $ They're equivalent, i.e. conversely, this lcm property $\color{#0a0}{\rm L\ implies}$ Euclid's Lemma

$$ (a,b)=1,\ a\mid bc\ \Rightarrow\ a,b\mid bc\,\color{#0a0}{\stackrel{\large \rm L}\Rightarrow}\, ab\mid bc\,\Rightarrow\,a\mid c$$

Unlike proofs by Bezout, this proof by Euclid's Lemma works in UFDs which have no Bezout identity, e.g. polynomial rings like $\,\Bbb Z[x]\,$ and $\,\Bbb Q[x,y].\,$ The Bezout-based proofs essentially substitute inline into the above proof a Bezout-based proof of Euclid's Lemma, yielding essentially the following proof $\ a,b\mid c\,\Rightarrow\,ab\mid ac,bc\,\Rightarrow\,ab\mid (ac,bc) = (a,b)c = c\ $ in gcd (or ideal) language (compare various forms of Euclid's Lemma).

There are many useful properties known  equivalent to Euclid's Lemma over arbitrary domains, e.g. see the end of this post for a handful, and some closely related gcd properties.


问:
Linear Algebra: Change of Basis question on Wiki definitions
I'm reading about change of basis and am confused with definitions on Wiki and less so on this site.  I present the definitions which are more understandable first, then follow with Wiki.  Here are the definitions which I understand:

$$
B = \{\mathbf b_1, \mathbf b_2, ..., \mathbf b_n\}
$$
where $B$ is a basis of a vector space $V$.   Then the coordinates $(\alpha_z)_{z \in I}$ of a vector $\mathbf x \in V$ using basis $B$ and field $F$ are:
$$
\phi_B : V \rightarrow F^n
$$
$$
\phi_B(\mathbf x) = [\mathbf x]_B
$$
$$
[\mathbf x]_B = \begin{bmatrix} \alpha_1 \cr \alpha_2 \cr ... \cr \alpha_n \end{bmatrix}_B = (\alpha_1, ..., \alpha_n)
$$
$$
\mathbf x = [\mathbf x]_S = \mathbf b_1 \alpha_1 + ... + \mathbf b_n \alpha_n = \sum_{i=1}^n \mathbf b_i \alpha_i
$$
The basis transformation matrix or vector is denoted $P_B$ and is from basis $B$ to $S$, where $S$ is the standard basis:
$$
P_B = [\mathbf b_1 \mathbf b_2 ... \mathbf b_n] = (\mathbf b_1, ..., \mathbf b_n)
$$
$$
P_B = [P]_S^B = [P]^B = [[\mathbf b_1]_S [\mathbf b_2]_S ... [\mathbf b_n]_S] = ( [\mathbf b_1]_S, ..., [\mathbf b_n]_S)
$$
$$
\mathbf x = P_B [\mathbf x]_B
$$
$$
\mathbf x = [\mathbf x]_S = [P]_S^B [\mathbf x]_B
$$
$$
[\mathbf x]_B = P_B^{-1} \mathbf x = ([P]_S^B)^{-1} \mathbf x
$$
where $\det(P_B) \ne 0$ and the change of basis from basis $B$ to $B'$ is as follows.

Convert from standard coordinates to coordinates in $B'$:
$$
[\mathbf x]_{B'} = P_{B'}^{-1} \mathbf x
$$

Question : does above appear correct?

I'm the most confused by the Wiki definitions and explanation.  The forms do not explicitly agree with above:

$$
\mathbf x_{old} = A \mathbf x_{new}
$$
is this form reversed?

Wiki then defines the change of basis formula as:
$$
B_{old} = (\mathbf v_1, ..., \mathbf v_n), B_{new} = (\mathbf w_1, ..., \mathbf w_j)
$$
For $j = 1, ... n$ and $w_j$'s coordinates are $a_{i,j}$ over $B_{old}$
$$
\mathbf w_j = \sum_{i=1}^n a_{i,j} \mathbf v_i
$$

Question: Why are the coordinates $a_{i,j}$ two dimensional when they are one dimensional in earlier examples.

Question: Can the above form be rewritten in notation used earlier?  For instance:

$$
P_{B_{old}} = [\mathbf v_0 ... \mathbf v_n] = (\mathbf v_0, ..., \mathbf v_n)
$$
$$
\mathbf w_j = P_{B_{old}} [\mathbf w_j]_{B_{old}}
$$
$$
[\mathbf w_j]_{B_{old}} = (a_{1,j}, ..., a_{n,j})
$$

答:
1) does above appear correct? Yes. Combining the equations, we have $$[\mathbf x]_B = P_B^{-1} \mathbf x= P_B^{-1} P_{B'}[\mathbf x]_{B'}=R[\mathbf x]_{B'}$$

2) No. We have
$$
T(\mathbf x) = P_B \mathbf x=P_B P_B[\mathbf x]_B,
$$
so the correct map is $S(\mathbf x)=P_B^{-1} \mathbf x$.

3) is this form reversed? It is not reversed, if "old" is the standard basis and "new" is $B$, then in the other notation, the wiki equation reads
$$
\mathbf x= P_B[x]_B,
$$
which agrees with the equation derived in the first part of your question. But i think it would be less confusing if written $\mathbf x_{new}=A^{-1}\mathbf x_{old}$.

4) Yes, the equations are equivalent, the way you wrote is just the compact matricial form of the equation. If you are not convinced of that, i recommend doing with a 2x2 or 3x3 example.

Answer to EDIT:
Why are the coordinates $a_{i,j}$ two dimensional when they are one dimensional in earlier examples? Can the above form be rewritten in notation used earlier?

$a_{i,j}$ are just the coefficients of the linear combination of the vectors that produces $\mathbf w_j$, they are also the coordinates of $\mathbf w_j$ in the old basis, as you noticed. This equation is equivalent to
$$
\mathbf w_j=P_{B_{old}}\mathbf [\mathbf w_j]_{B_{old}},
$$
where $[\mathbf w_j]_{B_{old}}=(a_{1,j},\dots,a_{n,j})$, note that this equation is the same as the one in display in 3).     i'm not sure what you mean with "two dimensional" though.


问:
Determining whether a language $L_{a}$ is recursively enumerable.
I'm trying to determine whether a language $L_{a}$ is recursively enumerable, but first I'm having trouble deciphering the definition $L_{a}$ given the following question:

Given an recursively enumerable language $L$, we can define $L_{a} = \{ x : \exists\langle M\rangle \in L \text{ such that } x \in L(M)\}$. Prove $L_{a}$ is recursively enumerable.

I understand it as the language $L_{a}$ is searching for a Turing Machine encoding $\langle M \rangle$ that exists in $L$, but I'm not sure this is correct.

Furthermore, when trying to prove a language is recursively enumerable is generally easier to prove this through some sort of contradiction or by using the property of decidability?

答:
There are two equivalent criteria for a set $A$ to be r.e.: (1) the existence of an algorithm $M$ that implements a total function whose domain is $\Bbb{N}_+$ and whose range is $A$. (2) the existence of an algorithm $M$ that implement a partial function whose domain is $A$ (i.e., $M$ terminates iff its input belongs to $A$). Here "algorithm" means a Turing machine or any equivalent notion.

In your example, it is convenient to use both of these characterizations (quite a common occurrence). We are given an r.e. set $L$ that we view as comprising codes defining Turing machines. I will write $\{e\}$ for the function computed by the Turing machine with code $e$. By characterization (1) there is an algorithm $M$ that implements a total function $f$ say $\Bbb{N}_+ \to A$ whose range is $L$. We want to show that the set $L_a$ of $x$ such that $x \in \mathrm{dom}(\{e\})$, for some $e \in L$, is r.e.. To do this we use characterization (2) and design an algorithm $N$ that terminates on input $x$ iff $x \in L_a$. Given an input $x$, $N$ works in stages. At stage $n$, $N$ has a list of states $s_1, \ldots, s_n$ resulting from partial executions of the Turing machines coded by $f(1), \ldots, f(n)$. At stage $0$, we have an empty list of states. To get from stage $n$ to stage $n+1$, $N$ simulates one extra step for the execution of the machines coded by $f(1), \ldots, f(n)$ in the states $s_1 \ldots, s_n$ giving a new list of states $s_1', \ldots, s_n'$ and sets $s_{n+1}'$ to be  the initial state of the machine coded by $f(n+1)$ on input $x$. $N$ terminates the computation as soon as one of the $s_i$ is a terminating state for the machine coded by $f(i)$ (meaning that $x \in \mathrm{dom}(\{f(i)\})$. It should be clear that $N$ terminates on input $x$ iff $x$ is in the $\mathrm{dom}\{e\}$ for some $e \in L$, i.e., iff $x \in L_a$.


问:
Can the gradient operator $\nabla_\mathbf{x}$ be treated as a standalone vector?
Let $f:\mathbb{R}^n\rightarrow\mathbb{R}$ be a scalar function. Then, the gradient of $f(\mathbf{x})$ is defined as:

$$
\nabla_\mathbf{x} f(\mathbf{x}) =
\begin{bmatrix} \frac{\partial f(\mathbf{x})}{\partial x_1} \\
\frac{\partial f(\mathbf{x})}{\partial x_2} \\
\vdots \\
\frac{\partial f(\mathbf{x})}{\partial x_n}
\end{bmatrix}
$$

However, if $f(\mathbf{x})$ is a scalar, then wouldn't the following also be valid?

$$
(\nabla_\mathbf{x}) (f(\mathbf{x})) =
\begin{bmatrix} \frac{\partial}{\partial x_1} \\
\frac{\partial}{\partial x_2} \\
\vdots \\
\frac{\partial}{\partial x_n}
\end{bmatrix} f(\mathbf{x}) =
\begin{bmatrix} \frac{\partial f(\mathbf{x})}{\partial x_1} \\
\frac{\partial f(\mathbf{x})}{\partial x_2} \\
\vdots \\
\frac{\partial f(\mathbf{x})}{\partial x_n}
\end{bmatrix}
$$

In other words, the gradient $\nabla_\mathbf{x}$ is treated as a standalone vector, and then it is multiplied by the scalar $f(\mathbf{x})$. Interestingly, if this is indeed true, then I can build the Hessian matrix of $f(\mathbf{x})$ using the following outer product:

$$
(\nabla_x \nabla_x^T) (f(\mathbf{x}))
$$

In other words, I left-multiply the row vector $\nabla_x^T$ by the column vector $\nabla_x$ to get an $n \times n$ matrix of second-order partial derivative operators. I then multiply this matrix by the scalar $f(\mathbf{x})$ to get the Hessian matrix. However, I am wondering if this is just a coincidence.

答:
This notation is often formally used, notably in the definition of the Laplace operator. We can make it explicit though.

For convenience, lets only work with functions that are $C^\infty$, such that all their partial derivatives exist and can again be partially differentiated and so on (think about the issues that happen when we don't do this).

A partial derivative is just a function (operator) that sends functions to functions:
$$\frac{d}{dx_i}: C^\infty(\mathbb{R}^n, \mathbb{R}) \to C^\infty(\mathbb{R}^n, \mathbb{R})$$
We can regard the gradient as a $1 \times n$ matrix with such operators as coefficients, so the gradient is then an element of the set $C^\infty(\mathbb{R}^n, \mathbb{R})^{1\times n}$. This gradient, and any other matrix of operators, is then itself an operator that returns a (real) matrix, by applying the operators pointwise to a function.

To be able to multiply matrices with coefficients in $C^\infty(\mathbb{R}^n, \mathbb{R})$, we need to be able to multiply the coefficients with each other. The multiplication of operators can be defined as composition. In this setting we indeed have
$$ \nabla^T \nabla = L$$
$$ \nabla \nabla^T = H$$
where $L$ is the Laplace operator and $H$ is the Hessian operator.


问:
What does LSQR stand for
One of the most popular and efficient iterative methods to solve large sparse systems of equations in the least squares sense is LSQR. It is related to CGLS (Conjugate Gradient Least Squares) in that it has the same iterates (mathematically, not numerically).

But what does its name mean? is it

Least Squares with QR factorisation? If so, why is it called that, does it provide Q and R? I suppose not because Q would not be sparse.

Least SQuaRes?

Least SQuares Regression?

Something else?

I have looked at the three references over at the SciPy documentation, but none of them seem to explain the naming. In the MATLAB documentation, the call it "the Least Squares Method", even stating it as "... the least squares (LSQR) algorithm...", suggesting that my second option is correct, but with no source.

答:
You’re right, the name is not explained anywhere, but there is some logic behind it. When Chris Paige visited Stanford in 1972, we started several methods that needed names. LSQR means that it’s for least-squares problems and uses a QR factorization at each iteration k (updated from the previous iteration).  The QR factorization is used to solve a (k+1) by k least-squares subproblem involving Bk, the lower bidiagonal matrix from the Golub-Kahan bidiagonalization process. This explains the strange name SYMMLQ (for symmetric systems using an LQ factorization of the tridiagonal matrix from the Lanczos process). By this plan, MINRES should have been called SYMMQR, but luckily Chris’s more pronounceable and decipherable name prevailed.

In 1981 I used the unsymmetric orthogonal tridiagonalization for two methods named USYMLQ and USYMQR by the same rule, ultimately described by Saunders, Simon, and Yip in SINUM 1988 (http://dx.doi.org/10.1137/0725052). I couldn’t think of a 6-character name that meant an unsymmetric form of MINRES. (Fortran 66 allowed at most 6 characters for variable and subroutine names.)

LSQR is equivalent to CG on the normal equations, so it could have been called LSCG.  When the MINRES companion came along (David Fong's PhD thesis), it took many months to think of the name LSMR
(SISC 2011 https://doi.org/10.1137/10079687X).

Analogous to LSQR, in SIMAX 2019 there's a paper about LSLQ by Ron Estrin, Dominique Orban, and me (http://www.siam.org/journals/simax/40-1/M111355.html). We have another paper in SIMAX 2019 about LNLQ for the least-norm problem min ||x|| st Ax=b
(https://doi.org/10.1137/18M1194948).

It's great to know that you use LSQR every day on large problems.  Keep LSMR in mind -- it will always stop sooner because it minimizes ||A'r||, which is most of ||A'r||/(||A|| ||r||),the quantity that both LSQR and LSMR look at to decide when to stop.  The difference can be significant
(http://stanford.edu/group/SOL/software/lsmr/).

The same is true for MINRES compared to CG on SPD systems!
See Fong and Saunders (2012), "CG versus MINRES: An empirical comparison" (http://stanford.edu/group/SOL/software/minres/).

Best wishes for your use of iterative solvers!
Michael

Title: What does LSQR stand for


答:
$\nabla \nabla^T$ is not a common notation.

Although sometimes the notation $(\nabla \cdot \nabla) f = \nabla^2 f$ is used to denote $\Delta f$, where $\Delta$ is the Laplace operator, it's important to note that
$$(\nabla \cdot \nabla) f = \Delta f = \operatorname{tr} H_f$$
results in the trace of the Hessian $H_f$ of $f$. Some consider this notation strongly misleading: The square of first order derivatives, as $(\nabla \cdot \nabla) f$ might suggest, does not equate to the second order derivative. After all, with $f = x^2$, the second order derivative is $f'' = 2$ and not $\left(f'\right)^2 = (2x)^2 = 4x^2$.

Your observation however has a common application in optimization. Second-order optimization methods use the Hessian to improve an initial guess $x_0$ of a minimum of $f$. Instead of calculating the exact Hessian at $x_0$, the Gauss–Newton algorithm approximates the Hessian using
$$H_f(x_0) \approx J^T(x_0)\,J(x_0),$$
where $J$ is the Jacobian. This question discusses this approximation.

Title: Can the gradient operator $\nabla_\mathbf{x}$ be treated as a standalone vector?

Body:
Let $f:\mathbb{R}^n\rightarrow\mathbb{R}$ be a scalar function. Then, the gradient of $f(\mathbf{x})$ is defined as:

$$
\nabla_\mathbf{x} f(\mathbf{x}) =
\begin{bmatrix} \frac{\partial f(\mathbf{x})}{\partial x_1} \\
\frac{\partial f(\mathbf{x})}{\partial x_2} \\
\vdots \\
\frac{\partial f(\mathbf{x})}{\partial x_n}
\end{bmatrix}
$$

However, if $f(\mathbf{x})$ is a scalar, then wouldn't the following also be valid?

$$
(\nabla_\mathbf{x}) (f(\mathbf{x})) =
\begin{bmatrix} \frac{\partial}{\partial x_1} \\
\frac{\partial}{\partial x_2} \\
\vdots \\
\frac{\partial}{\partial x_n}
\end{bmatrix} f(\mathbf{x}) =
\begin{bmatrix} \frac{\partial f(\mathbf{x})}{\partial x_1} \\
\frac{\partial f(\mathbf{x})}{\partial x_2} \\
\vdots \\
\frac{\partial f(\mathbf{x})}{\partial x_n}
\end{bmatrix}
$$

In other words, the gradient $\nabla_\mathbf{x}$ is treated as a standalone vector, and then it is multiplied by the scalar $f(\mathbf{x})$. Interestingly, if this is indeed true, then I can build the Hessian matrix of $f(\mathbf{x})$ using the following outer product:

$$
(\nabla_x \nabla_x^T) (f(\mathbf{x}))
$$

In other words, I left-multiply the row vector $\nabla_x^T$ by the column vector $\nabla_x$ to get an $n \times n$ matrix of second-order partial derivative operators. I then multiply this matrix by the scalar $f(\mathbf{x})$ to get the Hessian matrix. However, I am wondering if this is just a coincidence.


问:
Show that $\sin6\alpha\equiv \sin2\alpha(16\cos^4\alpha-16\cos^2\alpha+3)$
$$\sin6\alpha\equiv \sin2\alpha(16\cos^4\alpha-16\cos^2\alpha+3)$$

Can you help me with De Moivre's theorem and how I would go about tackling this question.

I understand that De Moivre's theorem states that $(\cos\alpha+i\sin\alpha)^n \equiv \cos n\alpha+i\sin n\alpha$.

But I don't see how this would come to use in this question.

答:
HINT:

$$\sin 6\alpha=\sin (3\cdot 2\alpha)=3\sin2\alpha-4\sin^32\alpha=\sin2\alpha(3-4\sin^22\alpha)$$

Now, $$3-4\sin^22\alpha=3-4(2\sin\alpha\cos\alpha)^2=3-16\cos^2\alpha(1-\cos^2\alpha)$$


问:
Orthogonality Relations Exercise 2.19, Brezis' Book Functional Analysis
I am studying some topics of Brezis's book and I am trying to solve this excercises:

Let $E$ a Banach space and let $A\,\colon\, D(A)\subset E \rightarrow E^*$ be a densely defined unbounded operator.

Assume that there exists a constant $C$ such that
$$
\langle Au, u \rangle \geq -C \|Au\|^2\quad \forall u\in D(A).
$$
Prove that $N(A)\subset N(A^*)$.

I was trying some solutions but, I coudn't. In fact, I search the hint in the back of the book and say:

Recall that $N(A^*)=R(A)^{\perp}$. Let $u\in N(A)$ and $v\in D(A)$; we have
$$
\langle A(u+tv),u+tv\rangle \geq -C \|A(u+tv)\|^2\quad \forall t\in\mathbb{R},
$$
which implies that $\langle A v,u\rangle=0$. Thus $N(A)\subset R(A)^\perp$.

Could anyone explain me why this inequality implies the result?.

答: $R(A)^{\perp} \subset N(A^{*})$: If $ \langle y, Ax \rangle=0$ for all $x \in D(A)$ then $ \langle A^{*}y, x \rangle=0$ for all $x \in D(A)$. Since $D(A)$ is dense it follows that $A^{*}y=0$ so $y \in N(A^{*})$.


问:
Plot $2|z|=|z-2|$ over the complex plain
I was given a home work to Sketch $2|z|=|z-2|$ on the complex plane.I tried to simplify and found out that the last expression was $3x^2+3y^2+4x-4$. However I don't know how to sketch this on the complex plane and how I can describe the set of points

答:
Hint

$$2|z|=|z-2| \Leftrightarrow \\
4|z|^2 = |z-2|^2 \Leftrightarrow \\
4z \bar{z}= (z-2)(\bar{z}-2) = z \bar{z}-2z -2 \bar{z}+4 \Leftrightarrow \\
3z \bar{z} +2z +2 \bar{z}-4 =0 \Leftrightarrow \\
9z \bar{z} +6z +6 \bar{z}-12 =0\Leftrightarrow \\
(3z+2)(3 \bar{z}+2)=8\Leftrightarrow \\
|3z+2|^2=8
$$


问:
Computing the gradient of a two-layer neural network w.r.t. the data
I want to verify that I am computing the following gradient correctly. I am looking at this conjecture, and the gradient is paramount. Using the definitions from the the paper, consider the data $\vec{x} = \{ x_1, x_2, ..., x_d \} \in \mathbb{R}^d$, 2 vectors $\vec{a} = \{ a_1, a_2, ..., a_k \} \in \mathbb{R}^k$, and $\vec{b} = \{ b_1, b_2, ..., b_k \} \in \mathbb{R}^k$ and a matrix which we represent as a vector of vectors $\vec{w_l} = \{ w_{l,1}, w_{l,2}, ..., w_{l,d}\} \in \mathbb{R}^d$ for $1 \le l \le k$. We also have a Lipschitz function $\varphi$. Define a function
$$f: x \mapsto \sum_{l=1}^k a_l \varphi( \langle \vec{w}_l, \vec{x} \rangle + b_l)$$

where$\langle \vec{w}_l, \vec{x} \rangle =\sum_{i=1}^d w_{l,i}x_i$. I want to properly write out the following gradient vector $\nabla f$ as a function of $x$.
$$\nabla f(x)=\begin{bmatrix} \frac{\partial}{\partial x_1} f\\
\frac{\partial}{\partial x_2} f \\
... \\
\frac{\partial}{\partial x_d} f
\end{bmatrix}$$

Now, for each $x_i$ I compute $\frac{\partial}{\partial x_i} f$:

$$\frac{\partial}{\partial x_i} f = \sum_{l=1}^k a_l \varphi'(\langle \vec{w}_l, \vec{x} \rangle + b_l)w_{l,i}$$

where $\varphi'$ is the derivative of $\varphi$. The conclusion here is that we see each element of $\nabla f$ is itself a sum of a function of all of the data. This is my main doubt. Do we in fact have the entire dot product $\langle \vec{w_l}, \vec{x} \rangle$ inside each element of the sum inside each element of the gradient $\nabla f$? Basically I would just like confirmation this computation is correct. Thanks.

答:
Yes, the computation is correct.

Recall the chain-rule: if $h(t) = f(g(t))$ then $$h'(t) = f'(g(t))g'(t).$$ Note that where $f$ and $f'$ are evaluated does not change. Both are evaluated at $g(t)$.

That is what is going on in your calculation. Before differentiation $\phi$ is evaluated at $\langle \vec w_l, \vec x \rangle + \vec b_l$, and after differentiation $\phi'$ is still evaluated at the same point.


问:
Finding the corresponding probabilities for the position of a random walk
Suppose I have the following probability statement for the event $B$:

$$P(B) = 1 - \frac{P(S_{a+b} = b - a - 2)}{P(S_{a+b} = a - b)}$$

where $S_{n}$ is a simple random walk with the corresponding probabilities:

$$P(\text{moving up 1}) = \frac{1}{2}\text{ and }P(\text{moving down 1}) = \frac{1}{2}$$

I cannot figure out what the probabilities are in the probability statement, unfortunately.

I have considered a general case where we want to find, for example, $P(S_{n} = x)$ for a simple random walk, and I concluded this:

If $S_{n} = x$, then let $a$ be the number of positive steps and $b$ the number of negative steps. Then, $a + b = n$ and $a - b = x$.

Well, that would mean $a = \frac{n+x}{2}$ and $b = \frac{n-x}{2}$. So, $P(S_{n} = x) = {n\choose\frac{n+x}{2}} 2^{-n}$.

答:
One way to go about it, note that $X=2(Bern(p))-1$

Therefore
$$S_k=\sum_{n=1}^k X_k =\sum_{n=1}^k 2(Bern(p))-1=-k+2\sum_{n=1}^k (Bern(p))$$
$$=-k+2Binom(k,p)$$

Then
$$P(S_k=a)=P(-k+2Binom(k,p)=a)=P(Binom(k,p)=\frac{a+k}{2})$$

Therefore,
$$P(S_{a+b}=b-a-2)=P(Binom(a+b,p)=b-1)$$
$$P(S_{a+b}=a-b)=P(Binom(a+b,p)=a)$$

And you should be able to finish from here...