问:
Gödel's Incompleness Theorems and the "Minds vs. Machines" conundrum
As a part of my self-learning about Gödel's Incompleness Theorems, I try to dive a little bit into their different interpretations and (possible) philosophical implications. One that immediately catches the eye relates to artifical intelligence. It seems like some people believe Gödel's theorems indicate that the mind is somehow transcendent to machines. As J. R. Lucas, puts it in his book (which I did not read) "Minds, Machines and Gödel": there is "some elusive and ineffable quality to human intelligence, which makes it unattainable by machines".

In short, Lucas' argument goes as follows:

However complicated a machine we construct, it will, if it is a machine, correspond to a formal system, which in turn will be liable to the Gödel procedure for finding a formula unprovable-in-that- system. This formula the machine will be unable to produce as true, although a mind can see that it is true. And so the machine will not be an adequate model of the mind.

Following this line of reasoning, it seems like the "elusive quality" Lucas mentions lies not within human consciousness, feelings or imagination but rather within the short textbook-proof that Gödel's sentence is true.

To me that sounds quite ridiculous (as you might have guessed). However, when I do try to point out exactly where Lucas' argument breaks down, I mainly get confused. So I'd appriciate your help with clearing things up.

It seems like the key here is to carefully follow the mathematical details of that textbook proof and try to see what's so "human" about it and why. Recall that given an arithmetical formal system $S$, its Godel sentence $G$ satisfies $S\vdash G\leftrightarrow \neg{\rm Prov}(\left\lceil G \right\rceil )$, where $${\rm Prov}(\left\lceil \varphi \right\rceil):=\exists x ({\rm Prf} (x, \left\lceil \varphi \right\rceil))$$ is the "provability" formula (and $\left\lceil \cdot \right\rceil$ means Gödel numbering).

Now, the textbook proof goes like this. In the proof of Gödel's first incompleteness theorem it is shown that if $S$ is consistent then it cannot prove $G$. Therefore, in that case, for every natural $n\in\mathbb{N}$, $\neg {\rm Prf} (n, \left\lceil G \right\rceil)$ is true (in the standard model $\mathbb{N}$). Thus $\forall x(\neg {\rm Prf} (x, \left\lceil G \right\rceil))$ is true (in $\mathbb N$) which is exactly $\neg{\rm Prov}(\left\lceil G \right\rceil )$. But it is known to be equivalent to $G$ by definition. Therefore $G$ is true.

How much of that does $S$ "know" (able to prove)? As shown in the proof of the second theorem, the system $S$ knows that if it is consistent it cannot prove $G$. It can also be shown that $S$ knows, for every $n\in \mathbb{N}$, that $\neg {\rm Prf} (n, \left\lceil G \right\rceil)$. So it seems like the only thing that $S$ is missing is the knowledge that if $\psi(n)$ is true for every $n\in\mathbb{N}$ then $\forall x (\psi(x))$ is true. This is where I get confused:

I was able to say "true in the standard model $\mathbb{N}$" but $S$ doesn't have that privilege; when it asserts that something is true, it has to be the case in all of its models (which might be non-standard). Wouldn't that be an "unfair advantage" that I have?

If "concluding that $\forall x (A(x))$ from $A(n)$ for every $n\in\mathbb{N}$ "is really the thing that seperates my mind from $S$ (regarding the truth of $G$), what is it about my human mind that causes that? What's so "human" here?

I know that $G$ is true only assuming that I know $S$ is consistent. So maybe Lucas' argument doesn't prove that the human mind transcends all machines, but only those it "knows" their consistence. In that case, the human mind might still reducable to a formal system, but simply one we will never be aware of its consistence. Perhaps that's where the Lucas' argument fails?

答:
First, I want to commend you on not dismissing Lucas' argument simply because the conclusion seems outlandish (this is what many people do: "Oh, so Lucas must be thinking that there is something about the human mind that cannot be captured by any formal system, but since by Lucas' own admission we can correspond a formal system with any physical system, that would mean that there is something about the human mind that is non-physical, and hence Lucas must be some weird dualist who believes that the human mind must have some 'soul-like' aspect to it that is non-physical. Well, Pffhbt to that!").  Sure, Lucas' argument may have an (to many people) unpalatable conclusion to it, but you are doing the right thing in demanding to know where the argument goes wrong if indeed it is wrong.

OK, so on that, a couple of comments:

Lucas writes:

However complicated a machine we construct, it will, if it is a machine, correspond to a formal system which in turn will be liable to the Gödel procedure for finding a formula unprovable-in-that- system.

Well, I would say Lucas needs to be very careful here, as there are a number of issues to be resolved:

First, I would say that Lucas is really glossing over a few things here. There is a big difference between something being a machine and something being a formal system for proving theorems. I can see how the human body and brain can be regarded as a 'machine' (a 'meat machine'), and how any machine can be an implementation of some formal system, i.e. how it realizes some computation.  However, any physical system can be implementing many kinds of computations all at the same time, simply depending on what level of abstraction you take. Moreover, computations are not the same thing as formal-systems-for-proving-theorems: even if I had some artificial neural network, whose mechanisms can be fully described computationally, it is still far from clear how that network would actually be used to prove theorems. In sum, even if we had a full description of how our brain works in terms of physical/chemical processes, it is far from clear how that would tell us what the proving abilities are as implemented by that physical system: that is a huge gap.

Second, it isn't even clear that a description of just the workings of our brain alone is sufficient to tell me what my formal abilities are. For example, I use pen and paper to solve any complex mathematical problems; do you think we can solve those just in our heads with our naked brain?!  In other words, the physical 'machine' that implements my abilities to prove theorems is likely to be a highly complex system that involves not just our brains, but also our hands, our eyeballs, and our environment that contains pens, paper, and, most importantly, handy dandy linguistic symbol systems.  This makes the translation from such a physical system to what we can prove even more difficult.

Then again, maybe it doesn't make it more difficult at all! If our abilities to prove theorems relies inherently on the very external formal/symbolic/linguistic systems that as a community of mathematicians and logicians we have created over time, then our abilities to prove mathematical theorems might be likewise limited by those very formal systems. So, whatever incompleteness results hold true for those formal systems would therefore automatically be true of our own abilities as well. Indeed, many mathematicians believe that we are 'working within ZFC', i.e. that all of human proving abilities can, in theory, be formalied by the ZFC system. But if ZFC is consistent, then that means that there is a Godel sentence for it that it cannot prove ... but that we would therefore not be able to prove either.

OK, but let's suppose that we can somehow got a full description of the formal system through which we prove theorems.  Lucas claims that we can then use Godel's procedure to point to a claim that we know is true, but that is not provable in that system ... and hence we cannot be that system:

Again, Lucas glosses over a number of serious potential pitfalls here:

As you yourself point out in point 3 of your post: Such a Godel sentence only exists for consistent systems, and so we would first need to know that the formal system that we are looking at is consistent. Well, that is again far from easy.  I already was pointing out that this could well be a super complicated system if it is indeed one that is implemented by the neural networks in our brain. And, from computability theory, we know that there are some very simple machines whose halting behavior is unknown (e.g. see the holdouts in the Busy Beaver problem), and if the system's ability to prove theorems is dependent on the halting behavior of its subsystems, I don't think there is a snowball's chance in hell for us to tell whether that system is consistent. And without that knowledge, we cannot say that there is something that that system cannot prove. And again, let's point to ZFC here here: compared to a neural networks involving trillions of connections, the consistency properties of ZFC should be child's play to analyze ... but we don't know if ZFC is in fact consistent ... and again, if that's the system that can capture all our proving abilities, then we will never be able to prove its consistency, by Godel's second Incompleteness theorem.

And finally, all of this is assuming that the system we are looking at is in fact consistent ... it may well be inconsistent, in which case Lucas' argument fails immediately.  And frankly, I don't believe humans are very consistent ... at least not individual ones. As a community, we can at least try to make sure that everything we do will still conform to the formal systems as laid out by something like ZFC (or at least the 'laws' of logic and math), so it isn't like we suddenly start proving that 1+1=3 either :)


问:
Let A and B be Boolean algebras and f : A → B is an isomorphism,and A is atomic.
Let A and B be Boolean algebras and f: A → B is an isomorphism, and A is atomic.

(a) B is atomic, (b) B is complete (c) A is atomless (d) B is atomless

I think the answer is (A) by the representation theorem

edit:

also, I found similar questions online

Let A and B be Boolean algebras and f : A → B is an isomorphism,
and A is complete.

(a) B is atomic, (b) B is complete (c) A is atomless (d) B is atomless

Let A and B be Boolean algebras and f : A → B is an isomorphism,
and A complete and atomic

(a) B is complete and atomic, (b) B is finite (c) A is atomless (B) is
atomless

答: The answer is $A$ because isomorphisms preserve the BA structure, so images of atoms are atoms etc. Not because of a representation theorem.

ContentLicense: CC BY-SA 4.0


问:
Question about subgroups on the group of homeomorphisms of a topological space
Let Y be path-connected, locally path-connected, and simply connected. Let $G_1$ and $G_2$ be subgroups of Homeo(Y) defining covering space actions of Y. (this means that each point y has a neighborhood U such that for $g_1, g_2 \in G_1, g_1(U) \cap g_2(U) \neq \emptyset\ \text {implies}\ g_1 = g_2$) Show that the orbit spaces $Y/G_1$ and $Y/G_2$ are homeomorphic iff $G_1$ and $G_2$ are conjugate subgroups of Homeo(Y).

Left to right is easy since we have that $G_1$ is isomorphic to $\pi _1(Y/G_1)$ because Y is simply connected. Same with $G_2$. (proposition 1.40 in Allen Hatcher's algebraic topology) It's right to left that I'm having trouble with. I was thinking of just building a map from $Y/G_1$ to $Y/G_2$ by directly sending one orbit to another, but I wasn't sure how to show it was continuous. Is there a way to use the quotient topology to do this? Or should I be looking at this problem another way entirely?

答:
Say $G_2 = h G_1 h^{-1}$.  Then the map $f: Y \to Y/G_2$ that is given by the action of $h$ followed by the quotient map, i.e., $y \mapsto \overline{h \cdot y}$

is continuous, since $h$ acts via a homeomorphism and the quotient map is continuous, and

collapses orbits of $G_1$: if $y' = g_1 \cdot y$ for some $g_1 \in G_1$, then $$f(y') = \overline{hg_1 \cdot y} = \overline{(h g_1^{-1} h^{-1}) \cdot hg_1 \cdot y} = \overline{h \cdot y} = f(y).$$

Therefore, by the universal property of quotient maps, there exists a continuous map $\bar{f}: Y/G_1 \to Y/G_2$ extending $f$.

By exchanging the roles of $G_1$ and $G_2$, you can construct another continuous map $Y/G_2 \to Y/G_1$ which you can check is inverse to $\bar{f}$.  Hence $Y/G_1 \cong Y/G_2$.


问:
Derivate of matrix conjucation
Consider the map from $GL(R,2)$ which sends a matrix $X$ to the map ($Y\mapsto XYX^{-1}$). This is a map smooth map $f:GL(R,2)\rightarrow End(GL(R,2)$ so we should be able to calculate its derivative.

Calculating the derivative at the identity applied to some $Y\in T_eGL(R,2)$ we want to calculate $d_I(f)(Y)$. We can pick a curve $\gamma$ with $\gamma(0)=I$ and $\dot{\gamma}(0)=Y$. Then $d_I(f)(Y)=\frac{d}{dt}f(\gamma)$.

I am not sure how to calculate this derivative. Because $f(\gamma)$ is a map sending $Z\mapsto \gamma(t)Z\gamma(t)^{-1}$. To calculate this do we need to take another curve in $Z$?

答:
First, recall the following result:

Proposition. Let $H$ be a square matrix of operator norm strictly less than $1$, then $I+H$ is invertible and
$$(I+H)^{-1}=\sum_{n=0}^{+\infty}(-1)^nH^n$$
which is called Neumann's series.

Proof. Since $\|H\|<1$, the series in the statement is absolutely convergent, hence convergent and a direct computation shows that:
$$(I+H)\sum_{n=0}^{+\infty}(-1)^nH^n=I,$$
whence the result claimed.

Remark. This result shows that $\operatorname{GL}(E)$ is open whenever $E$ is a Banach space.

Now, carry the following computations:
\begin{align*}
f(I+H)(Y)&=(I+H)Y(I+H)^{-1},\\
&=Y(I+H)^{-1}+HY(I+H)^{-1},\\
&=Y+HY-YH+O(H),\\
&=f(I)+\ell(H)(Y)+O(H),
\end{align*}
where $\ell(H)\colon Y\mapsto HY-YH$ and notice that $\ell$ is linear, whence, $d_If=\ell$.

Title: Derivate of matrix conjucation


问:
Two closed subsets $A$ and $B$ in $\mathbb{R}$ with $d(A,B)=0$
I am looking for two closed subsets A and B (with $A\cap B = \emptyset$) of $\mathbb{R}$ with $d(A,B)=0$. I found a solution in $\mathbb{R}^2$, namely $A=\{(x,\frac{1}{x})\mid x>0\}$ and $B=\{(x,0)\mid x>0\}$. I know that those subsets have to be unbounded because there is a theorem that says: the distance between two closed subsets of which one is bounded, is greater than 0.</p>

答: I used one of your sets to give another example: $$A=\{(x,y)\mid xy\ge -1,x<0\},~~B=\{(x,y)\mid xy\ge1,x>0\}$$

ContentLicense: CC BY-SA 3.0


问:
Minimum distance in a right-angled trapezoid

Let's define $ABCD$ a trapezoid, $AB \ \| \ CD$ and $AD \perp AB$. Let's consider $M$ a point on the altitude $(AD)$. Given $AB = 6$, $CD = 3$, $AD = 2$, where should be placed $M$ to have $CM + MB$ minimal? (Bulgarian Regional Mathematical Olympiad 1987, 7th grade)

Geometrical approach: Let's draw the circle $\mathcal{C}(C, r = MC)$ with the center $M$, through $C$ and $\mathcal{C} \cap BM = \{B, N\}$. Therefore $BN$ should be minimal to satisfy the condition. However, no clue here.

Algebraic approach: Let's observe that, denoting $AM = x$, we have the expression
$$CM + MB = \sqrt{9 + x^2}  + \sqrt{36 + (2 - x)^2}$$

答:
Reflect point $C$ over line $AD$ to end up in a new position $C'$ on the line $CD$ extended. Draw $C'M$ and $C'B$.

Now,  observe that,  $AD$ is the perpendicular bisector of $CC'$ and hence $CM=C'M$. Thus, $CM+BM$ is, by the triangle inequality, least when $M$ is the intersection point of $C'B$ and $AD$. Therefore,  the minimum value of $CM+BM$ is $C'B$.  Calculate this length by pythagorean theorem now.


问:
Derive $\int_0^1 \frac{\ln(\sqrt2-1)-(\sqrt2-x)\ln x}{(\sqrt2-x)^2-1}\,dx=\frac{\pi^2}6+\frac14\ln^2(\sqrt2-1) $
I obtained the integral

$$\int_0^1 \frac{\ln(\sqrt2-1)-\ln(x)(\sqrt2-x)}{(\sqrt2-x)^2-1}\,dx=\frac{\pi^2}6+\frac14\ln^2(\sqrt2-1)
$$

as a by-product while carrying out some complex analysis on an engineering problem. In short, given the real nature of the physical outcome, the imaginary part of the final expression has to vanish, which implies the result above and is numerically verifiable.

However, the complex method employed is not intended to solve integrals such as this and produces it merely by accident. So, the question is: How to derive this elementary close-form directly, with real methods?

Note that the integrand may not look as innocent as it appears. There is a hole as the denominator becomes zero at $x=\sqrt2-1$ and the term $\ln(\sqrt2-1)$ is present in the numerator to remove the singularity, thus ensuring convergence.

答:
This will be a bit complicated, but I will try to explain clearly as I go.  We will first find a general antiderivative, and then use it to compute the desired result.  To make things a little easier to TeX, I'm going to write $a = \sqrt{2}$, so we're trying to find $$\int_0^1 \frac{\ln(a-1)-\ln(x)(a-x)}{(a-x)^2-1}\,dx.$$  I'm going to do this by computing the antiderivative
\begin{equation}
\int \frac{\ln(a-1)-\ln(x)(a-x)}{(a-x)^2-1}\,dx = \int\frac{\ln(a-1)}{(a-x)^2-1}\,dx - \int\frac{\ln{x}(a-x)}{(a-x)^2-1}\,dx.\tag{1}
\end{equation}
The first integral on the right of $(1)$ is relatively straightforward via partial fractions, so I will omit the computation: $$\int\frac{\ln(a-1)}{(a-x)^2-1}\,dx = \frac{\ln(a-1)}{2}\ln\left(\left|\frac{a-x+1}{x-a+1}\right|\right) + C.$$  For the second integral in $(1)$, we will do some manipulation first.  Since there is a $\ln(x)$ in the numerator, and the denominator is not obviously the derivative of the numerator, we're aiming for a polylogarithm.  We will also use a partial fraction decomposition, but I will not show the actual steps for that part.
\begin{align*}
\frac{\ln{x}(a-x)}{(a-x)^2-1} &= \ln(x)\frac{a-x}{a^{2} - 2ax + x^2 - 1}\\
&=\ln(x)\frac{a-x}{(a-x-1)(a-x+1)}\\
&=-\frac{\ln(x)}{2}\left(\frac{1}{a-x-1} + \frac{1}{a-x+1}\right).\tag{2}
\end{align*}
At this point, we're going to do something a bit non-obvious.  Note that $$\frac{1}{a-x-1} = \frac{a-1}{a-x-1}(a-1) = \frac{d}{dx}\ln\left(\left|\frac{a-x-1}{a-1}\right|\right).$$  We're going to do something akin to using an integrating factor, in the sense that we're going to extract a product rule derivative.  Continuing from $(2)$:
\begin{align*}
\frac{\ln{x}(a-x)}{(a-x)^2-1} &= -\frac{\ln(x)}{2}\left(\frac{1}{a-x-1} + \frac{1}{a-x+1}\right)\\[10pt]
&\qquad-\frac{1}{2x}\left(\ln\left(\left|\frac{a-x-1}{a-1}\right|\right) + \ln\left(\left|\frac{a-x+1}{a+1}\right|\right)\right)\\[10pt]
&\qquad+\frac{1}{2x}\left(\ln\left(\left|\frac{a-x-1}{a-1}\right|\right) + \ln\left(\left|\frac{a-x+1}{a+1}\right|\right)\right)\\[10pt]
&=-\frac{1}{2}\frac{d}{dx}\left\lbrace\ln(x)\left(\ln\left(\left|\frac{a-x-1}{a-1}\right|\right) + \ln\left(\left|\frac{a-x+1}{a+1}\right|\right)\right)\right\rbrace\\[10pt]
&\qquad+\frac{1}{2}\left(\frac{\ln\left(\left|1-\frac{x}{a-1}\right|\right)}{x} + \frac{\ln\left(\left|1-\frac{x}{a+1}\right|\right)}{x}\right)\\
&=-\frac{1}{2}\frac{d}{dx}\left\lbrace\ln(x)\left(\ln\left(\left|\frac{a-x-1}{a-1}\right|\right) + \ln\left(\left|\frac{a-x+1}{a+1}\right|\right)\right)\right\rbrace\\[10pt]
&\qquad-\frac{1}{2}\frac{d}{dx}\left\lbrace\Phi\left(\frac{x}{a-1}\right) + \Phi\left(\frac{x}{a+1}\right)\right\rbrace.
\end{align*}
Note that here $\Phi$ is a version of Spence's Function given by $$\Phi(x) = -\int_0^x \frac{\ln|1-u|}{u} \, du = \begin{cases}
\operatorname{Li}_2(x), & x \leq 1; \\ \frac{\pi^2}{3} - \frac{1}{2} \ln^2(x) - \operatorname{Li}_2(\frac{1}{x}), & x > 1.
\end{cases}$$
and $\operatorname{Li_2}$ is the  polylogarithm $$\operatorname{Li_2}(x) = -\int_0^x \frac{\ln(1-u)}{u} \, du.$$ Therefore, we have that \begin{align*}
\int\frac{\ln{x}(a-x)}{(a-x)^2-1}\,dx &= -\frac{1}{2}\left\lbrace\ln(x)\left(\ln\left(\left|\frac{a-x-1}{a-1}\right|\right) + \ln\left(\left|\frac{a-x+1}{a+1}\right|\right)\right)\right.\\[10pt]
&\qquad\quad + \left.\Phi\left(\frac{x}{a-1}\right) + \Phi\left(\frac{x}{a+1}\right)\right\rbrace + C.
\end{align*}
All in all, then, going back to $(1)$, we have that
\begin{equation*}
\int \frac{\ln(a-1)-\ln(x)(a-x)}{(a-x)^2-1}\,dx
\end{equation*}
is given by
\begin{align*}
\frac{\ln(a-1)}{2}&\ln\left(\left|\frac{a-x+1}{x-a+1}\right|\right)+\frac{1}{2}\left\lbrace\ln(x)\left(\ln\left(\left|\frac{a-x-1}{a-1}\right|\right)\right.\right.\\[10pt]
&+ \left.\ln\left(\left|\frac{a-x+1}{a+1}\right|\right)\right) + \left.\Phi\left(\frac{x}{a-1}\right) + \Phi\left(\frac{x}{a+1}\right)\right\rbrace + C.
\end{align*}
Replacing $a$ with $\sqrt{2}$, we get your original integral:
\begin{equation*}
\int \frac{\ln(\sqrt2-1)-\ln(x)(\sqrt2-x)}{(\sqrt2-x)^2-1}\,dx
\end{equation*}
is given by
\begin{align*}
\frac{\ln(\sqrt2-1)}{2}&\ln\left(\left|\frac{\sqrt2-x+1}{x-\sqrt2+1}\right|\right)+\frac{1}{2}\left\lbrace\ln(x)\left(\ln\left(\left|\frac{\sqrt2-x-1}{\sqrt2-1}\right|\right)\right.\right.\\[10pt]
&+ \left.\ln\left(\left|\frac{\sqrt2-x+1}{\sqrt2+1}\right|\right)\right) + \left.\Phi\left(\frac{x}{\sqrt2-1}\right) + \Phi\left(\frac{x}{\sqrt2+1}\right)\right\rbrace + C.
\end{align*}

Now, this function does not have any singularities on $(0,1)$, so we can use the Fundamental Theorem of Calculus to evaluate the original integral
\begin{equation*}
\int_0^1 \frac{\ln(\sqrt2-1)-\ln(x)(\sqrt2-x)}{(\sqrt2-x)^2-1}\,dx.
\end{equation*}
We have the following:
\begin{equation*}
\frac{\ln(\sqrt2-1)}{2}\ln\left(\left|\frac{\sqrt2-x+1}{x-\sqrt2+1}\right|\right)\Bigg|_{0}^{1}=\frac{1}{2}\ln(\sqrt{2}-1)\left( \ln\left(\frac{\sqrt2}{2-\sqrt2}\right)-\ln\left(\frac{\sqrt{2}+1}{\sqrt{2}-1}\right)\right)\tag{3}
\end{equation*}
\begin{equation*}
\frac{1}{2}\left\lbrace\ln(x)\left(\ln\left(\left|\frac{\sqrt2-x-1}{\sqrt2-1}\right|\right)+\ln\left(\left|\frac{\sqrt2-x+1}{\sqrt2+1}\right|\right)\right)\right\rbrace\Bigg|_{0}^{1} = 0\tag{4}
\end{equation*}
\begin{align*}
    \frac{1}{2}\left\lbrace\Phi\left(\frac{x}{\sqrt2-1}\right) + \Phi\left(\frac{x}{\sqrt2+1}\right)\right\rbrace\Bigg|_{0}^{1} &= \frac{1}{2}\left\lbrace\Phi\left(\frac{1}{\sqrt2-1}\right) + \Phi\left(\frac{1}{\sqrt2+1}\right)\right\rbrace\\
    &=\frac{1}{2}\left\lbrace\frac{\pi^{2}}{3} - \frac{1}{2}\ln^{2}\left(\frac{1}{\sqrt2-1}\right)\right.\\ &\qquad\left.- \operatorname{Li_{2}}\left(\sqrt2-1\right) + \operatorname{Li_{2}}\left(\frac{1}{\sqrt2+1}\right)\right\rbrace\\
    &=\frac{\pi^{2}}{6}-\frac{1}{4}\ln^{2}(\sqrt{2} - 1).\tag{5}
\end{align*}

Adding $(1)$, $(2)$, and $(3)$, we get
\begin{align*}
\int_0^1 \frac{\ln(\sqrt2-1)-\ln(x)(\sqrt2-x)}{(\sqrt2-x)^2-1}\,dx &=\frac{1}{2}\ln(\sqrt{2}-1)\left( \ln\left(\frac{\sqrt2}{2-\sqrt2}\right)-\ln\left(\frac{\sqrt{2}+1}{\sqrt{2}-1}\right)\right)\\ &\qquad+ \frac{\pi^{2}}{6}-\frac{1}{4}\ln^{2}(\sqrt{2} - 1)\\
&=\frac{1}{2}\ln(\sqrt2-1)\left(\ln(\sqrt2)-\ln(\sqrt2(\sqrt2-1))\right.\\&\qquad\left.-\ln(\sqrt2+1)+\ln(\sqrt2-1)\right)\\
&\qquad+\frac{\pi^{2}}{6}-\frac{1}{4}\ln^{2}(\sqrt{2} - 1)\\
&=-\frac{1}{2}\ln(\sqrt2-1)\ln(\sqrt2 +1)+\frac{\pi^{2}}{6}-\frac{1}{4}\ln^{2}(\sqrt{2} - 1)\\
&=\frac{\pi^{2}}{6} + \frac{1}{2}\ln^{2}(\sqrt2 - 1) - \frac{1}{4}\ln^{2}(\sqrt2 - 1)\\
&=\frac{\pi^{2}}{6} + \frac{1}{4}\ln^{2}(\sqrt2 - 1).
\end{align*}
Note that we used the fact that
\begin{align*}
\ln(\sqrt2 + 1) + \ln(\sqrt2 - 1) &= \ln((\sqrt2 + 1)(\sqrt2 - 1)) = \ln(1) =0\\ &\implies -\ln(\sqrt2 - 1)\ln(\sqrt2 + 1) = \ln^{2}(\sqrt2 - 1).
\end{align*}
Therefore, we have arrived at the desired result:
$$\boxed{\int_0^1 \frac{\ln(\sqrt2-1)-\ln(x)(\sqrt2-x)}{(\sqrt2-x)^2-1}\,dx = \frac{\pi^{2}}{6} + \frac{1}{4}\ln^{2}(\sqrt2 - 1)}$$


问:
What we need to show, $ab=1\bmod k$ or $(ab)\bmod n=1\bmod k$? For proving $U_k(n)≤U(n)$
I need to show that,

For each divisor $k$ of $n$, $U_k(n)$ is subgroup of $U(n)$ where, $U_k(n)=\{x\in U(n) : x=1\bmod k\}$

My attempt: as $U(n)$ is finite group for each $n\in\mathbb{Z}^+$.
Hence we may use, "finite subgroup test" to show that, $U_k(n)$  is subgroup of $U(n)$.

Clearly, $1\in U_k(n)$ since $1\equiv 1\bmod k$.

Let $a,b\in U_k(n)$ then $a=1\bmod k$ and $b=1\bmod k$, then clearly, $ab=1\bmod k$ so that $ab\in U_k(n)$ and so by finite subgroup test, $U_k(n)$ is subgroup of $U(n)$.

But, where we used $k$ divides $n$? Further, is we computed $ab$ under the operation multiplication modulo $n$? since by finite subgroup test, we need to show $U_k(n)$ is **closed under the operation of $U(n)$. Hence,  I think we need to show $(ab)\bmod n=1\bmod k$ Am I correct?

(If i am correct, then how to show $(ab)\bmod n=1\bmod k$)

Please help me......

答:
I assume by $U(n)$ you mean $\mathbb{Z}/n\mathbb{Z}$?

We consider the example n=16 and k=3, than for $a \in \mathbb{Z}/n\mathbb{Z}$ we can not say what $a \mod k$ is, since we can choose another representant. In our example let $a=[4]$ and clearly $4 \mod 3 = 1 \mod 3$, but if we use another representant, say $[4]=[20]$, than $20 \mod 3 \neq 1 \mod 3$. This is where you use that $k$ divides $n$.


问:
Is it a mistake of Proofs from The Book?
I saw  the crossing lemma  in the forty-five chapter titled Probability makes counting
(sometimes) easy.

Theorem 4. Let $G$ be a simple graph with n vertices and m edges, where $m \ge 4n$. Then  $${\rm{cr}}(G) \ge \frac{1}{64}\frac{m^3}{n^2}$$
What I am confused about is this sentence, marked in purple:

We know if $n=1$ , then $cr(K_1)-m+3n=3<6$.
It does not satisfy (4).

Similarly, $K_2$ is one and the last unsatisfied. Is this a mistake? Is it possible to use other methods to get this result:
$$E(X_p-m_p+3n_p)\ge 0?$$
The link to the electronic version is below.
Proofs from The Book

答:
Thank you all for your enthusiastic help. I seem to understand the source of the problem. Indeed, applying (4) alone is problematic. (4) only holds for $n \geq 3$.

But $E(X_p-m_p+3n_p)\ge 0$ is still correct .  Since when  $n$ is less than or equal to 2  the formula still holds.

To be more specific.

If $n=1$, then $cr(K_1)-m+3n=3>0$.

If $n=2$, then $cr(K_2)-m+3n=0-2+6=4>0$  and $cr(2K_1)-m+3n=0-0+6>0$


问:
How to prove that $\left\{\gamma \in \alpha^{+}: \beta + \gamma \leq \alpha\right\}$ is not a limit ordinal given that $\alpha \geq \beta$.
I am proving the following theorem and somewhat got stuck:

Suppose that $\alpha$ and $\beta$ are ordinals with $\alpha \geq \beta$. Let $X = \left\{\gamma \in \alpha^{+}: \beta + \gamma \leq \alpha\right\}$. Then $X$ is a successor ordinal.

I have proved that $X$ is an ordinal and that $X \neq \boldsymbol{0}$. But I am stuck at proving $X$ is not a limit ordinal. I tried to use transfinite induction on $\alpha$, but the process is very verbose, and I was stuck in the middle. Can anyone provide hints?

答:
To show that $X$ is (read: has the order type of) a successor ordinal is to show that $X$ has a maximum.

$X$ is a subset of $\alpha^+$, so $\gamma \le \alpha$ for all $\gamma \in X$, so $\sup\limits_{\gamma \in X} \gamma \le \alpha$, so $\sup\limits_{\gamma \in X} \gamma \in \alpha^+$.

Also, $\beta + \gamma \le \alpha$ for all $\gamma \in X$, so $\beta + \sup\limits_{\gamma \in X} \gamma = \sup\limits_{\gamma \in X} \,(\beta + \gamma) \le \alpha$.

Therefore, $\sup\limits_{\gamma \in X} \gamma \in X$, quod erat demonstrandum.


问:
Given $V= \pi \int_{1}^{c}-y \sqrt{1-y^2}\,\mathrm{d}y$, Find $f^\prime\left(x\right)$
Suppose that I have a function, $f\left(x\right)$ s.t. $f\left(x\right) > 0$ on $\left[0,a\right]$ and $f\left(0\right)=1$ and $f\left(a\right) = c$. The volume $V$ of the function is $$V= \pi \int_{1}^{c}-y \sqrt{1-y^2}\,\mathrm{d}y$$ (and $y=f\left(x\right)$). How do I obtain $f^\prime\left(x\right)$? On the surface, this looks like a question where I should integrate the volume. However, I have no way of finding out $f^\prime\left(x\right)$ using the fundamental theorem since the bound $c$ is unknown. Help would be appreciated.

答:
Disclaimer: It could be possible that I am misunderstanding the question because of the lack of context. So maybe take this with a grain of salt?

There is definitely some issue with this question. Primarily because, the volume of revolution about the $y-$axis for any given curve $y = f(x)$ $(x \in [a,b])$ can be obtained by first writing $x$ in terms of $y$ (if the inverse image of $f$ splits into branches; we take the branch that corresponds to our problem); and then computing the integral
$$V = \pi\int_{f(a)}^{f(b)} x^2 dy$$

In the light of this, when we consider your case, by looking at your question, it is reasonable to assume $c$ is arbitrary. So, we consider the case $c>1$, and here we see that we have $$x^2 = -y\sqrt{1-y^2}$$
Clearly, if we are doing real calculus; we want $y<0$ for any meaningful results; however it is clearly stated in your question that $y = f(x) > 0$. In other words, a strictly positive function that satisfies the given volume integral does not exist (for $c>1$). Here's a desmos plot of the two branches that may perhaps give you more intuition on why this problem has something wrong with it.

This is more speculation: Here's maybe a more nuanced problem. Let us assume $0<c<1$; this means we can rewrite the volume integral as
$$V = \pi\int_c^1 y\sqrt{1-y^2} dy$$
and in this case,
$$x^2 = y\sqrt{1-y^2} $$
and everything might look well on the onset; but even in this very nice case; $y$ is not a function of $x$; so we have to resort to simply looking at different branches of $y$.

$$x^2 = y\sqrt{1-y^2} \implies x^4 = y^2(1-y^2)$$
Letting $y^2 = t$, we have
$$t^2 - t + x^4 = 0$$
we use the quadratic formula to see
$$y^2 = t = \frac{1 \pm \sqrt{1-4x^4}}{2}$$
So, since $y$ is always positive, we can only consider the positive square root
$$y = \sqrt{\frac{1 \pm \sqrt{1-4x^4}}{2}}$$
Now the problem becomes more clear; the two branches of solutions correspond to different parts of the closed curve; (you can view this plot to gather more intuition) and will necessarily have two different derivatives depending on which branch you look at! (in other words, $f'(x)$ is not well defined at a given $x$). So, really; we might have to parameterize the problem; but that gets out of hand real fast.

I will perhaps edit this answer later; once I am able find a proper parameterization.

Edit: A possible parameterization for this problem could be
when $x\geq 0$, $\left(\frac{1}{\sqrt{2}}\sqrt{\sin(2t)}, \sin(t)\right)$
and when $x<0$, $\left(-\frac{1}{\sqrt{2}}\sqrt{\sin(2t)}, \sin(t)\right)$
for $t \in [0,\pi]$.

So, when $x\geq 0$ $$f'(x) = \frac{dy}{dx} = \frac{dy}{dt}\frac{dt}{dx} = \frac{\sqrt{2}\cos(t)\sqrt{\sin(2t)}}{\cos(2t)}$$
and when $x<0$
$$f'(x) = \frac{dy}{dx} = \frac{dy}{dt}\frac{dt}{dx} = -\frac{\sqrt{2}\cos(t)\sqrt{\sin(2t)}}{\cos(2t)}$$


问:
The function $w(z)=-(\frac1z+bz),-1<b<1 ,$ maps $|z|<1$ onto ...
I am stuck on the following problem that says:

The function $w(z)=-(\frac1z+bz),-1<b<1 ,$ maps $|z|<1$ onto

  (A) A half plane

  (B) Exterior of the circle

  (C) Exterior of an Ellipse

  (D) Interior of an ellipse

I tried by replacing $z=re^{i \theta}$ but I must admit that I have not gotten very far .Can someone point me in the right direction? Thanks in advance for your time.

答:
Ma Ming has provided an answer which works well even without the aid of multiple choices.  With the use of the four choices, we can find our answer with even less work:

Since $w$ maps $0 \mapsto \infty$, (d) cannot hold.

Since $w(z)$ can be continously extended to the boundary $\vert z \vert =1$, the image of this boundary under $w$ must be a (compact) closed curve.  This eliminates (a).

This leaves only (b)  and (c).  Technically, (b) implies (c), so we'd be done (in a multiple-choice testing environment).


问:
The Mean time of Moran Model
I am currently reading an article on Wikipedia on how to derive the mean time before fixation $k_i$ of the Moran Model.

But the derivation of this result rests on this equation:

$$ k_i^j = \delta_{ij} + k_{i-1}^{j}p_{i,i-1} + k_{i}^{j}p_{i,i} + k_{i+1}^{j}p_{i,i+1} $$

where $k_i^j$ is the mean time spent in state $j$ when starting in state $i$.

The problem I have is with the equation above, why does this equation hold? Or is it simply a definition?

I would appreciate if I get references for further reading.

Thank you.

答:
the Moran Model is a version of the gambler's ruin problem. The difference here is that the probabilities to go up or down depend on the position. However, the solution strategy is the same.

The computation uses something that is called "first step analysis". Let $T$ be the fixation time, then

$$k_i^j:=E\left[\sum_{n=1}^{T-1}1_{X_n=j}|X_0=i\right]$$

It should also be intuitive that
$$E(T|X_0=i)=1+\sum_{j=0}^{N-1}k_i^j$$

I think it is more intuitive/instructive to work with the latter quantity. For $i\in\{1,\ldots,N-1\}$

$$E[T|X_0=i]=\frac{1}{P(X_0=i)}E[T\cdot 1_{X_0=i}]
=\sum_{j=-1}^1\frac{1}{P(X_0=i)}E[T\cdot 1_{X_0=i}\cdot 1_{X_1=i+j}]\\
=\sum_{j=-1}^1\frac{P(X_1=i+j,X_0=i)}{P(X_0=i)}\cdot\frac{1}{P(X_1=i+j,X_0=i)}E[T\cdot 1_{X_0=i}\cdot 1_{X_1=i+j}]\\
=\sum_{j=-1}^1p_{i,i+j}E[T|X_0=i,X_1=i+j]=\sum_{j=-1}^1p_{i,i+j}E[T|X_0=i+j]
$$
The last step uses time homogeneity and Markov property, and $1_A$ is the indicator function of event $A$.

Now, abbreviate $h(i):=E[T|X_0=i]$ and you get the system of linear equations

$$h(i)=0, \text{if $i\in\{0,N\}$} \\
 h(i)=p_{i,i-1}h(i-1)+p_{i,i}h(i)+p_{i,i+1}h(i+1)
$$

You can do the same for $k_i^j$, but it's unnecessary information if you're only interested in the mean time to fixation.

You can find a more general description of the whole method in Nicolas Privault's book "understanding Markov Chains". He also has a free version of his script on his website.

Title: The Mean time of Moran Model


问:
Cup product formula.
Here is the section of a paper I was reading named "Note on Cup-Products" by I.M. James:

A formula for cup-products. The cohomology theory in what follows has coefficients in the ring of integers. Consider a $CW-$complex $X,$ and a pair of elements $a \in H^{p}(X), b \in H^{{q}}(X),$ such that $a \smile b=0,$ where $p, q \geqq 2 .$ Suppose that $H^{p+q-1}(X)$ is finite. Then I define an integer-valued function, $h,$ on $\pi_{p+q-1}(X)$ as follows. Let $\lambda \in \pi_{p+q-1}(X) .$ Let $X^{*}$ denote the complex which is obtained by attaching an oriented $(p+q)$ -cell to $X$ by a map of homotopy class $\lambda$. Let $c$ denote the cohomology class which is carried by the cell. There are unique elements $a^{\prime} \in H^{p}\left(X^{*}\right), b^{\prime} \in H^{q}\left(X^{*}\right),$ which map into $a, b$ respectively, under the injection. Since $a \cup b=0$ in $X,$ there is an integer $m$ such that $a^{\prime} \cup b^{\prime}=m c .$ I define $h(\lambda)=m .$ We prove:

THEOREM (4.1). The function h constitutes a homomorphism of $\pi_{p+q-1}(X)$ into the group of integers.

We prove (4.1) in the following form:

LEMMA (4.2). Let $\lambda_{1}, \lambda_{2}, \lambda_{3} \in \pi_{p+q-1}(X)$ be elements such that $\lambda_{1}+\lambda_{2}$
$+\lambda_{3}=0 .$ Then
$$
h\left(\lambda_{1}\right)+h\left(\lambda_{2}\right)+h\left(\lambda_{3}\right)=0
$$

In what follows, let $t$ be an indexing integer which takes values $1,2,3 .$ Let $X_{t}$ denote the complex which is obtained by attaching a $(p+q)$ -cell $e_{t}$ to $X$ by a map of homotopy class $\lambda_{t},$ so that
$$
X_{2} \cap X_{3}=X_{1} \cap X_{3}=X_{1} \cap X_{2}=X.
$$
Let $X^{\prime}=X_{1} \cup X_{2} \cup X_{3} .$ Consider the injections:
$$
H^{r}\left(X^{\prime}\right) \xrightarrow{j_{t}} H^{r}\left(X_{t}\right) \xrightarrow{i_{t}}{{H}}^{r}(X).
$$
Let $a^{\prime}, b^{\prime}$ denote the cohomology classes of $X^{\prime}$ such that $i_{t} j_{t}\left(a^{\prime}\right)=a,$ $i_{t} j_{t}\left(b^{\prime}\right)=b,$ and let $c_{t}$ denote the class which is carried by $e_{t} .$ Since $a \cup b=0$ in $X,$ there are integers $m_{t}$ such that
$$
a^{\prime} \cup b^{\prime}=m_{1} c_{1}+m_{2} c_{2}+m_{3} c_{3}
$$
However, by the naturality of the cup-product,

$$
\begin{aligned}
j_{t}\left(a^{\prime}\right) \cup j_{t}\left(b^{\prime}\right) &=j_{t}\left(a^{\prime} \cup b^{\prime}\right) \\
&=m_{1} j_{t}\left(c_{1}\right)+m_{2} j_{t}\left(c_{2}\right)+m_{3} j_{t}\left(c_{3}\right) \\
&=m_{t} j_{t}\left(c_{t}\right) .
\end{aligned}
$$
Hence $m_{t}=h\left(\lambda_{t}\right),$ by the definition of $h .$ Since $\lambda_{1}+\lambda_{2}+\lambda_{3}=0,$ by hypothesis, there is a map $f: S^{p+q} \rightarrow X^{\prime}$ which maps $S^{p+q}$ onto each of the three cells $e_{t}$ with degree $1 .$ Let $f^{*}: H^{r}\left(X^{\prime}\right) \rightarrow H^{r}\left(S^{p+q}\right)$ denote the homomorphism induced by $f,$ which is trivial unless $r=p+q .$ Then $f^{*}\left(c_{t}\right)=c,$ the cohomology class of $S^{p+q} .$ Hence
$$
\begin{aligned}
\left(m_{1}+m_{2}+m_{3}\right) c &=f^{*}\left(m_{1} c_{1}+m_{2} c_{2}+m_{3} c_{3}\right) \\
&=f^{*}\left(a^{\prime} \cup b^{\prime}\right) \\
&=f^{*}\left(a^{\prime}\right) \cup f^{*}\left(b^{\prime}\right),
\end{aligned}
$$
which is zero. Therefore $m_{1}+m_{2}+m_{3}=0,$ since $H^{p+q}\left(S^{p+q}\right)$ is freely generated by $c .$ Since $m_{t}=h\left(\lambda_{t}\right),$ this proves $(4.2) .$ The passage from (4.2) to (4.1) is elementary, and will be omitted.

It is worth mentioning that $h(\lambda)$ can equally well be defined to be the functional cup-product [12] of $a$ and $b$ with respect to $\lambda.$

first part

Second part

Third part

I have some questions about it:

1- Why " since $a \smile b = 0$ in $X,$ there is an integer $m$ such that $a' \smile b' = mc$" where $c$ is the cohomology class which is carried by the oriented $(p + q)$-cell. is there a proof for this?

2- Why the section is named " a formula for cup-products", where is the formula?

3- I need an example of spaces which has infinite $H^{p+q-1}(X)$?

4- why the author defined $h(\lambda) = m$ like this? is there any thing in the literature that say that in a situation similar to what we have we should define our function like that? (I got some help that this is a generalization of hopf invariant)

5- Is there a book that contains a proof of naturality of cup product?

6- when the author said "Hence $m_t = h(\lambda_t)$", I do not understand how the previous equality leads to this, what is the relation between $h$ and $j_t$?

7- How is $f^*(c_t) = c$?

8- Why $f^{*}\left(a^{\prime}\right) \cup f^{*}\left(b^{\prime}\right)$ is zero?

Maybe I am overthinking things, frustrated from not understanding and hence beginning to ask trivial questions. Forgive me please if I did that.

Any help will be greatly appreciated.

答:
By Mayer-Vietoris (or excision) applied to $X^* = X \cup_{S^{p+q-1}, \lambda} D^{p+q-1}$, there is an exact sequence $$\cdots \to \mathbb{Z} \{c\} \cong H^{p+q-1} S^{p+q-1} \to H^{p+q} X^* \to H^{p+q} X \to \cdots.$$  The element $a' \cup b' \in H^{p+q} X^*$ is sent to $a \cup b \in H^{p+q} X$, which is $0$ by assumption, so by exactness $a' \cup b'$ can be identified with an element coming from $H^{p+q-1} S^{p+q-1}$ which has the form $mc$ for some $m \in \mathbb{Z}$.

Perhaps $a' \cup b' = mc$ is the formula for cup products referred to.

Just take an infinite wedge of $(p+q-1)$-spheres.

Asking for a "reason" why an author decides to do something depends on the background context and their goals.  As you point out, this is a generalization of the Hopf invariant, which should be reason enough to study it.

Any good algebraic topology book will do this, for example Hatcher, Proposition 3.10.

This is the definition of $h$, applied to $j_t(a')$ and $j_t(b')$, which are lifts of $a$ and $b$ in the cohomology of $X$ to the cohomology of $X_t$.

By construction $f$ maps $S^{p+q-1}$ (which represents $c$) onto $e_t$ (which represents $c_t$) with degree $1$.  So $f^* c_t = 1 \cdot c = c$.

$f^*(a') \smile f^*(b') = 0$ because $f^*(a') = 0$ and $f^*(b') = 0$, which follows from the fact that $H^* S^{p+q}$ has no cohomology in those degrees.

Title: Cup product formula.


问:
Can Cauchy's integral formula be used on the boundary point?
Can Cauchy's integral formula be used if the point a is on the boundary of the circle ? The theorem is given below.

答:
No. The boundary of the circle is $\gamma$, and if $a\in\gamma$, then the integrand doesn't exist at $a$.

If you need to work with an $a$ on the boundary of $D$, then enlarge $D$ a little. You can do that because $\overline{D}\subset U$.


问:
Composition of linear orders on the same set
I need to show an example of finite linear orders $K_1$ and $K_2$ are defined on the same set, such that their composition $K_1\circ K_2$ is not a transitive relation.

Whatever orders I choose, their composition is a transitive relation.

Any ideas?
Thank you for your time.

答: Consider the 5-element set consisting of the five elements a, b, c, d, e. Let $K_1$ order these elements in the order e, c, d, a, b. Let $K_2$ order them in the order d, e, b, c, a. Under the composite relation $K_1\circ K_2$, transitivity fails for the triple a, c, e.</p>


问:
conv(A) equals to the intersection of all convex sets containing A
Let A be a finite set, and let $ conv(A)=\{ \sum _{i=1}^n\lambda_i v_i|\sum _{i=1}^n\lambda_i=1, \space \lambda_i\ge 0,v_i\in A\}$.

I wish to prove that, as written in the title, conv(A) equals to the intersection of all convex sets containing A.

It's very intuitive why it is true, but i have no clue how to prove it.

I'm trying to prove via containment in both directions, But i've encountered the following difficulties -

proving that a given x in the intersection is in conv(A): ok, so i have a x. If x is in A, we're done. If x is not in A, we have to show some how that it's imposible? How?

proving that a given x in conv(A) is in the intersection : ok, so i have a x as a convex combination of the members of A. How do i prove that x must be in the intersection?

答:
This is probably too late but let me answer it anyway.

Assume some set $S$ is a convex set containing all the $v_i$'s in $A$ and assume further the element $\sum_i t_i v_i$ is not in $S$. If $t_0=1$, we get a contradiction. If $t_0\neq 1$ we can write $\sum_{i=0}^n t_i v_i=t_0 v_0+(1-t_0)\sum_{i=1}^n\frac{t_i}{1-t_0}v_i$ which means $\sum_{i=0}^n t_i v_i$ is in the line segment connecting $v_0$ and $\sum_{i=1}^n\frac{t_i}{1-t_0}v_i$. If $\sum_{i=1}^n\frac{t_i}{1-t_0}v_i\in S$, by definition of a convex set,$\sum_{i=0}^n t_i v_i$ must also be in $S$ so our assumption indicates $\sum_{i=1}^n\frac{t_i}{1-t_0}v_i\notin S$. Proceed inductively, we will end up claiming one of the $v_i$'s is not in $S$, which is a contradiction.


问:
Is there a complex Geometric Hahn Banach theorem for imaginary part?
I know the complex version of the geometric Hahn Banach theorem, e.g. in here https://math.stackexchange.com/a/2497406/789584

Let $X$ be a (complex) locally convex space, and $A,B\subset X$ disjoint, nonempty, convex, with $A$ compact and $B$ closed. Then there exists a continuous linear functional $\varphi:X\to\mathbb C$  and $c,d\in\mathbb R$ with
$$
\text{Re}\,\varphi(v)<c<d<\text{Re}\,\varphi(w),\ \ \ v\in A,\ \ w\in B.
$$

I am wondering if the imaginary part of $\varphi$ above satisfy a similar relations, namely if there is $c', d'\in\mathbb R$ such that $$
\text{Im}\,\varphi(v)<c'<d'<\text{Im}\,\varphi(w),\ \ \ v\in A,\ \ w\in B.
$$

答:
It's clear that can't be any essential difference between taking the real and imaginary parts here. Lest there be confusion about the formal details:

The theorem says there exists $\psi\in X^*$ such that $\Re\psi$ separates $A$ and $B$ (meaning that that inequality holds). Let $\phi=i\psi$. Then $\phi\in X^*$ and $\Im\phi=\Re\psi$, so $\Im\phi$ separates $A$ and $B$, as desired.


问:
Towards a consistent notation for entropy and cross-entropy
I am learning about the cross entropy, defined by Wikipedia as
$$H(P,Q)=-\text{E}_P[\log Q]$$
for distributions $P,Q$.

I'm not happy with that notation, because it implies symmetry, $H(X,Y)$ is often used for the joint entropy and lastly, I want to use a notation which is consistent with the notation for entropy:
$$H(X)=-\text{E}_P[\log P(X)]$$

When dealing with multiple distributions, I like to write $H_P(X)$ so it's clear with respect to which distribution I'm taking the entropy. When dealing with multiple random variables, it thinks it's sensible to make precise the random variable with respect to which the expectation is taken by using the subscript $_{X\sim P}$. My notation for entropy thus becomes
$$H_{X\sim P}(X)=-\text{E}_{X\sim P}[\log P(X)]$$

Now comes the point I don't understand about the definition of cross entropy: Why doesn't it reference a random variable $X$? Applying analogous reasoning as above, I would assume that cross entropy has the form \begin{equation}H_{X\sim P}(Q(X))=-\text{E}_{X\sim P}[\log Q(X)]\tag{1}\end{equation}
however, Wikipedia makes no mention of any such random variable $X$ in the article on cross entropy. It speaks of

the cross-entropy between two probability distributions $p$ and $q$

which, like the notation $H(P,Q)$, implies a function whose argument is a pair of distributions, whereas entropy $H(X)$ is said to be a function of a random variable. In any case, to take an expected value I need a (function of) a random variable, which $P$ and $Q$ are not.

Comparing the definitions for the discrete case:
$$H(p,q)=-\sum_{x\in\mathcal{X}}p(x)\log q(x)$$
and
$$H(X)=-\sum_{i=1}^n P(x_i)\log P(x_i)$$

where $\mathcal{X}$ is the support of $P$ and $Q$, there would only be a qualitative difference if the events $x_i$ didn't cover the whole support (though I could just choose an $X$ which does).

My questions boil down to the following:

Where is the random variable necessary to take the expected value which is used to define the cross entropy $H(P,Q)=-\text{E}_{P}[\log Q]$

If I am correct in my assumption that one needs to choose a random variable $X$ to compute the cross entropy, is the notation I used for (1) free of ambiguities.

答:
Your notation $H(X)=-\text{E}_P[\log P(X)]$ is really redundant. In general, if $X$ is a random variable, and $g$ is any function, then $E[g(X)]$ is defined without ambiguity, it's not necessary (actually is makes no sense) to specify "with respect to which variable the expectation is taken".

The confusion might arise here because we are dealing with two things, random variables and distributions, and using different letters for them. But, in essence, a random variable is  a distribution.

If we understand that $X$ is a rv with distribution $p$ , and the same for $Y$ and $q$, then we can write, unambiguosly (let me use $\tilde H$ for the cross entropy to distinguish it from the joint entropy):

$$\tilde H(p,q) = - E[ \log q(X)] = - \sum_x p(x) \log q(x) $$

If the above causes some confusion, consider that $\log (q(\cdot))$ is just a function, like $\sin(\cdot)$ or $\sqrt{\cdot}$

It mightbe more clear and consistent (but also more verbose) to use $P_X()$ and $P_Y()$ to denote the distributions of random variables $X,Y$.

$$\tilde H(X,Y) = - E[ \log P_Y(X)] = - \sum_x P_X(x) \log P_Y(x) $$

Notice that the lowercase $x$, used in the sums, is a dummy variable (not a random variable!) and we could also use $u$ instead or any other letter.

Edit: An attempt to clarify.

First: if $X$ is a given random variable, then its probability distribution is given, and the expression $E[X]$ is perfectly well defined, with zero ambiguity: eg, in the discrete case, $E[X] = \sum_x x P(x) $ where $P(\cdot)$ is the pmf of $X$. Period. It would be nonsensical to specify "with respect with which variable or distribution" the expectation should be taken.

This is also true when  some  funcion is applied to the rv (which just produces another rv), as in $E[g(X)]$, or when the variable is multidimensional, as in $E[g(X,Y)]$.

Hence, the notations $E_Q[X]$, $E_Q[g(X)]$ ,  $E_Q[g(X,Y)]$ (where $Q$ is some distribution) are wrong, they make no sense.

Having said that: sometimes it's not feasible or practical to stick with that notation, where each random variable has a letter like $X$. For example,
suppose we want consider the family of zero mean gaussian distributions, parametrized by the deviation $\sigma$, and we want to denote by $r(\sigma)$ the respective differential entropy. Letting $\phi(x;\sigma)$ be the gaussian density, we'd write

$$r(\sigma) = - \int \phi(x;\sigma) \log \phi(x;\sigma) dx$$

Notice that here $x$ is not a random variable, is just a dummy integration variable, and we could replace it with any letter.
Now, surely, this is an expectation, and so might want to write something like

$$r(\sigma) = - E[\log \phi(x;\sigma)] \tag2$$

... but this is not right, because the argument of $E[\cdot]$ is not a random variable! We might instead write

$$r(\sigma) = - E[\log \phi(X_{\sigma};\sigma)] \tag3$$

adding the definition: "$X_{\sigma}$ is a rv that follows the distribution $N(0,\sigma^2)$", but this is rather ugly. Hence we often abuse the notation in the following way:

$$r(\sigma) = - E_{\phi(x;\sigma)}[\log \phi(x;\sigma)] \tag4$$

But, be careful, here the $E[]$ notation must be understood not in the probabilistic setting but mere as a functional operator; here random variables are not involved, only functions (some of them, densities or distributions). That is

$$E_{f(x)}[g(x)] \triangleq \int f(x) g(x) dx \tag 5 $$

In this setting, the original notation $H(P,Q)=-\text{E}_P[\log Q]$ is, indeed, totally correct.

And as for the plain entropy, it should either be

$$H(X) = -E [\log P(X)] \tag 6$$

(entropy of a random variable $X$ is a rv which has distribution $P$) or

$$H(P) = -E_P [\log P] \tag 7$$

(entropy of a distribution $P$, functional expection operator, no random variables appear). You are mixing both.


问:
An exact equation to calculate the extrema of $\dfrac{\sin(x)}{x}$
A couple of days ago I calculated that the $m^{th}$ extrema of $\dfrac{\sin(x)}{x}$ denoted by $y_m$ is given by this equation below:

$$ \sqrt{1-y_m^2} +y_m \sin^{-1}(y_m)-y_m(-1)^m \left(2m+1\right)\dfrac{\pi}{2}=0$$

which can be simplified to the following :

$$ \sin^{-1}(y_m)+\cot(\sin^{-1}(y_m))-(-1)^m \left(2m+1\right)\dfrac{\pi}{2}=0$$

I cannot find any reference to this expression I calculated anywhere online in any published work or in general. Could anyone tell me whether it is an already known equation?

References:

Locations and amplitudes of the extrema of the sinx/x function

Edit: Since this is confusing some people. I'll clarify that I do not want to solve for $x_k$ (the location of the extrema) but the extrema itself, there are a lot of papers which already have done that in different manners. I instead was looking for a function that directly gives you the extrema. The function which I mentioned above directly gives you the extrema $y(x_m)$ (and not its location $x_m$) when numerically solved for its root. I wanted to know whether this equation which I found was already known/used before. I cannot find any published work where they have given a equation that directly gives you the extremum values for $\dfrac{\sin(x)}{x}$.

Body:
Let's find the first positive one.
Where $(x_{1},y_{1}) \approx (4.5,-0.2)$

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

$$\frac{d}{dx} f(x) = \frac{x \cdot cos(x) - sin(x)}{x^2}$$

$$\frac{x_{1} \cdot cos(x_{1}) - sin(x_{1})}{x_{1}^{2}} = 0$$

$$x_{1} \cdot cos(x_{1}) - sin(x_{1}) = 0$$

$$x_{1} - tan(x_{1}) = 0$$

$$x_{1} = tan(x_{1})$$

It cannot to be solved exactly.


问:
Unique questions about pentagon
Consider the (non-regular) pentagon with consecutive vertices at (-1,-1), (-1,1), (0,2), (1,1), and (1,-1).
a) Prove that there is no circle that is tangent to all 5 sides of the pentagon
b) Is there an ellipse that is tangent to all five sides of the pentagon? If so, find its equation.

Here is what I have done so far:
I have drawn out the pentagon. For part a) I am very bad at proving. For part b) I have some idea. I drew the ellipse and tried to match the ellipse inside of the pentagon and tried to see its tangency. My best idea is to take the derivative. I may be completely off from the true solution, I am not sure. Please click the solution button below and tell me the correct solution. This is a question that is not for homework, so if you posted the solution there would be no issue.

答:
If a polygon is circumscribed to a circle, then the bisectors of its internal angles all meet at a same point. Hence, for part a) just show that this is not the case.

For b) it is a general theorem that there always exists a conic tangent to five given lines. In particular, if the lines form a convex pentagon then the conic is an ellipse.

There is a simple geometric construction for the ellipse inscribed in a given convex pentagon $ABCDE$. One can, first of all, find tangency points $PQRST$ (see figure below). Draw, for instance, diagonals $AC$ and $BD$, meeting at $F=(1/2,1/2)$. Line $EF$ meets then side $BC$ at tangency point $P=(1,1/3)$.

This construction depends on Brianchon's theorem:

The three opposite diagonals of every hexagon circumscribing a conic
are concurrent.

In fact, if $P$ is the tangency point on $BC$, we can view $ABPCDE$ as a limiting case of a hexagon circumscribed to the ellipse. Hence diagonals $AC$, $BD$ and $EP$ must intersect at the same point $F$.

For a generic pentagon one would go on finding the other tangency points and remembering then that the line, passing through the intersection point of two tangents to an ellipse and through the midpoint of their tangency points, also passes through the center of the ellipse. Once the center is found one can construct a pair of conjugate diameters and finish then the construction by standard techniques.

In this particular case the construction is simpler, as by symmetry we know that major axis of the ellipse lies on $y$-axis and $PR$, parallel to $x$-axis, is minor axis. Hence semi-minor axis $PO$ has a length $b=1$ and semi-major axis $QO$ has a length $a=4/3$. Focal distance is then $c=\sqrt{a^2-b^2}=\sqrt7/3$ and from that we can find foci $L$ and $M$.

Title: Unique questions about pentagon


问:
Formal notation related to a sequence or a set
My question is quite naive...

I just want to represent a finite sequence of Natural number, is it the best way to write it like this?:

$\langle a_0, \ldots, a_n \rangle $, where $\forall i \in [ 0, n ], a_i \in \mathbb{N}$

I have also seen somewhere $\{ a_i \}$, does it mean a set (or a sequence) of variables? is it finite or infinite? where could i add the constraint like $\forall i \in [ 0, n ], a_i \in \mathbb{N}$?

Also, is there any difference between "tuple" and "sequence"?

Hope my question is clear, I just want to make sure what I write matches the convention...

Thank you very much

答: Sometimes I see $a_1, a_2, ...$ without the brackets.</p>


问:
Does this set of vectors span the space of all polynomials of degree at most 3?
I'm having trouble understanding this homework problem.

Suppose four polynomials are defined by the following:

$
p_{1}(x) = x^3 - 2x^2 + x + 1 \\
p_{2}(x) = x^2 - x + 2 \\
p_{3}(x) = 2x^3 + 3x + 4 \\
p_{4}(x) = 3x^2 + 2x + 1 \\
$

Does the set $S = $ { ${p_{1}, p_{2}, p_{3}, p_{4}}$ } span $P_{3}$ (the space of all polynomials of degree at most 3)?

So if I start with the polynomial $y = ax^3 + bx^2 + cx + d$ I understand(I think) that in order for $S$ to span $P_{3}$ then $y$ must be a linear combination of $S$ but I'm not sure where to go from there.

EDIT:

$$
A =
\begin{bmatrix}
1 & -2 & 1 & 1\\
0 & 1 & -1 & 2\\
2 & 0 & 3 & 4\\
0 & 3 & 2 & 1
\end{bmatrix}
 \sim
\begin{bmatrix}
1 & 0 & 0 & 0\\
0 & 1 & 0 & 0\\
0 & 0 & 1 & 0\\
0 & 0 & 0 & 1
\end{bmatrix}
$$

So $\operatorname{rank}(A) = 4$ which means the vectors in the set are linearly independent because there are only 4 column vectors in the matrix (AND there is only the trivial solution to the matrix) and therefore we span $P_{3}$?

答: Hint: The answer can be shortened by determining the determinant of $A$ and see if it is not equal to $0$. Hence by Theorem if the determinant is not equal to zero the set is linearly independent else its linearly dependent.</p>


问:
How to relate Frobenius norm and trace of a matrix?
I came across an expression relating the Frobenius Norm with Trace as follows :

$$\|UU^T\hat X\|^2_F = tr((UU^T\hat X)^T(UU^T\hat X))$$

Could someone explain briefly the equivalence ?
THanks

答:
For any matrix $M$, we have
$$
\|M\|_F^2 = \operatorname{tr}(M^TM).
$$
In particular, note that the $j,j$ entry of $M^TM$ is equal to $\sum_{i=1}^n m_{ij}^2$, so that the sum of the diagonal entries (the trace) is indeed equal to the sum of the squares of all entries of $M$.


问:
Find the probability that a family has $k$ kids and two of them are boys (no more boys)
I came across this problem in an exercise book.

The actual question contains two parts. Let $P(k)$ denotes the probability that a family has $k$ kids. Given $P(0)=P(1)=a$ and

$P(k)=(1-2a)\cdot 2^{-k+1}$, $k\geq 2$

And the probability of having boys and girls are the same (i.e. $Q(B)=Q(G)=\frac{1}{2}$) $Q$ just the notation denotes the probability of having boys and girls.

Suppose a family has already has two boys (and just two boys),

Find the probability that the family only have 2 kids.

Find the probability that the family have another two girls (and just two girls)

I am guessing to use beysian theorem for the first question that
$P(k=2|BB) = \frac{P(BB|k=2)P(k=2)}{P(BB)}$ .
And for the numerator is
$P(BB|k=2)P(k=2) = \frac{1}{2}\cdot\frac{1}{2}\cdot(1-2a)\cdot\frac{1}{2}$
denominator is
$P(BB) = \sum^\inf_{k=2}\frac{1}{2}\cdot\frac{1}{2}\cdot\frac{1}{2}^{k-2}\cdot P^k_2\cdot(1-2a)\cdot 2^{-k+1}$, $P^k_2$ is the permutation with coefficient k and 2 i.e. $\frac{k!}{2!(k-2)!}$

How can one simplify the expression on the denominator?

答:
If I correctly understood your question you'd like to compute:
$$
(1-2a)\sum_{k=2}^\infty \binom k2 2^{-2k+1}=2(1-2a)\sum_{k=2}^\infty \binom k2\left(\frac14\right)^k
$$

For this you could use the easy to remember expression:
$$
S_n(x)=\sum_{k=0}^\infty \binom knx^k=\frac{x^n}{(1-x)^{n+1}}.\tag1
$$

To prove the expression observe that:
$$
S_{n}(x)=\frac{x^n}{n!}\frac{d^n}{dx^n}\sum_{k=0}^\infty x^k=\frac{x^n}{n!}\frac{d^n}{dx^n}\left(\frac1{1-x}\right)=\frac{x^n}{n!}\frac{n!}{(1-x)^{n+1}}.
$$


问:
Truth value of statements about empty set
I have a problem:

Let P be the statement " x $\in$ A and x $\in$ $\mathbb{Z}$ "

Determine the truth value of statement: ($\forall$x)P $\Longrightarrow$ ($\exists$x)P

Is there a set A for which the truth value of the above statement is false? Explain.

My approach is:

The statement is only false when the antecedent is true and the consequent is false.

This is not possible because the antecedent is ... (I have no clue why there is no set A for which the hypothesis ($\forall$x)P of the statement is true).

Can someone help, please?

答: Hint: " x ∈ A and x ∈ Z " is false, as $A$ is the empty set.


问:
The n-th derivative of $\frac{1}{(x-1)(x+1)}$
Let's say that we know that the (n-th derivative of) $\frac{1}{n}^{(n)} = (-1)^nn!x^{-(n+1)}$

Can we then use the formula and the product form of a derivative for $(\frac{1}{(x-1)(x+1)})^{(n)} = \frac{1}{x-1}^{(n)}*\frac{1}{x+1}+\frac{1}{x-1}*(\frac{1}{x+1})^{(n)}$

And then we use the formula.

Or is this not correct ?

答:
No, it is not correct.

On the other hand,$$\frac1{(x-1)(x+1)}=\frac{1/2}{x-1}-\frac{1/2}{x+1},$$and you can use this.


问:
Maclaurin Series Terms and Convergence or Divergence
This is one of the HW questions I'm trying to solve

Find the first four terms of this Maclaurin Series of $$f(x)=\cos\left(\frac{x}{x+1}\right)$$

I tried using $\sum_{n=0}^\infty (-1)^n  \frac{x^{2n}}{2n!} = \cos x $ and   $ \sum_{n=0}^\infty (-1)^n x^n= \frac{1}{1+x}$ but no luck! I can't properly substitute the derivative of the series in the Maclaurin Series.

Also related to this question is its second part$$ \sum_{} (n)^p (\cos (\frac{1}{n-1})-\cos(\frac{1}{n})$$
for what values of $p$ it converges or diverges.

I would really appreciate it if someone could type in a detailed solution so I can improve my understanding of this topic.

答:
HINTS

The route you want to go notes that if $g(x) = \frac{x}{x+1}$, then $f(x) = \cos(g(x))$ and expansion into series gives
$$
f(x)
 = \cos(g(x))
 = \sum_{n=0}^\infty(-1)^n \frac{g(x)^{2n}}{(2n)!}
 = \sum_{n=0}^\infty \frac{(-1)^n}{(2n)!} \left(\sum_{m=0}^\infty (-1)^mx^m \right)^{2n},
$$
which is very messy to deal with.

You are better off writing the first four terms using the definition, i.e. compute $f'(x), f''(x)$ and $f'''(x)$ and evaluate at $0$. You may find it helpful to write $g(x) = 1 - \frac{1}{x+1}$.

As for the second part, note that what governs the conergence of the sum is what happens when $n$ is large, in which case $\frac{1}{n}$ and $\frac{1}{n-1}$ become very small. Can you use the Maclaurin series to expand the cosine around zero and approximate the cosine using the expansion which includes the first meaningful term which depends on $n$?

Happy to guide you further if you post progress on these in the comments below.

UPDATE

Your first attempt is correct, you can see Wolfram Alpha computes the Maclaurin series to be
$$
\cos \left(\frac{x}{x+1}\right)
 = 1 - \frac{x^2}{2} + x^3 - \frac{35}{24} x^4 + \frac{11}{6} x^5 + \mathcal{O}\left(x^6\right)
$$

As for the second part, note that for small $x$ we have
$$
\cos x \approx 1 - \frac{x^2}{2}
$$
so $\cos(1/n) = 1 - n^2/2$ and you can find $\cos(\frac1{n-1})$ similarly and subtract, then finding the limit of the difference as $n \to \infty$.


问:
Rewriting two coupled linear matrix equations as a single linear matrix equation
The two coupled matrix equations I have are
$$Y_1 = X_1 + AX_1A^T + BX_2B^T$$
$$Y_2 = X_2 + CX_1C^T + DX_2D^T$$

I would like to write them as a single $2 \times 2$ matrix equation where each matrix element consists of the previous matrices.

For example a simpler version of this problem that I can solve is
$$Y_1 = X_1 + AX_1 + BX_2$$
$$Y_2 = X_2 + CX_1 + DX_2$$
which can be rewritten as
$$Y = (I+M)X$$
where
$$Y = \begin{bmatrix} Y_1 & 0 \\ 0 & Y_2 \end{bmatrix}, \qquad M = \begin{bmatrix} A & B \\ C & D \end{bmatrix}, \qquad X = \begin{bmatrix} X_1 & 0 \\ 0 & X_2 \end{bmatrix}$$
and $I$ is the identity matrix.

EDIT:
I don't think the solution to the simple example that I gave is correct. The $X$ and $Y$ in that case should be
$$X = \begin{bmatrix} X_1 \\ X_2 \end{bmatrix}$$
$$Y = \begin{bmatrix} Y_1 \\ Y_2 \end{bmatrix}$$

答:
$\def\v{\operatorname{vec}}\def\m#1{\left[\begin{array}{c}#1\end{array}\right]}$You could define the vectors
$$\eqalign{
x_k &= \v(X_k) \quad &y_k = \v(Y_k) \\
x &= \m{x_1\\x_2} \quad &y = \m{y_1\\y_2} \\
}$$
and vectorize the equations using Kronecker products to obtain
$$\eqalign{
{\cal K} &= \m{(I+A\otimes A)&B\otimes B\\C\otimes C&(I+D\otimes D)}\\
y &= {\cal K} x \\
}$$


问:
related rates waterskier rising off ramp
I feel this should be an easy one, but something is tripping me up. here is the question:

A waterskier skis over the ramp at a speed of 30ft/s. How fast is she rising as she leaves the ramp?

(the ramp is a right triangle with adjacent 15ft and opposite 4ft.)

I want to set it up as a $x^2+y^2=z^2$
so I think I need to differentiate first before plugging values in.
$2x \frac {dx}{dt} + 2y \frac {dy}{dt} = 2z \frac {dz}{dt}$

divide both sides by 2 and plug the values I have in, that being x=15, y=4, dz/dt = 30ft/s, z = $sqrt241$ (because $15^2+4^2 = 241$)
so this leaves me with:

$15 \frac {dx}{dt} + 4 \frac {dy}{dt} = sqrt241*30$

how should I go about finding dx/dt? where did I go wrong? I get really stuck on related rates so any help in the right direction or approach is greatly appreciated.

答:
The previous answers are incorrect because they all assume that the water skier's speed along the ramp is 30 ft/s, but that is the horizontal speed of the boat pulling her and thus her horizontal speed as well (assuming the tow rope does not stretch). Her speed relative to the surface of the ramp will be greater than 30 ft/s. But what we are after is her vertical speed.

If we let $x$ and $y$ represent her horizontal and vertical position respectively and let $s$ represent her position along the ramp then we can relate these changing quantities via the Pythagorean theorem:
$$
s^2 = x^2 + y^2
$$
Implicitly differentiating with respect to time, $t$, yields:
$$
2s\cdot s' = 2x\cdot x' + 2y\cdot y' \qquad \Longrightarrow \qquad
s\cdot s' = x\cdot x' + y\cdot y'
$$
We don't know $s$ nor $s'$, but we can use similar triangles to express them in terms of $x$ and $x'$ respectively,
$$
\frac{s}{x} = \frac{\sqrt{241}}{15} \quad \Longrightarrow \quad
s = \frac{\sqrt{241}}{15} x \quad \Longrightarrow \quad s' = \frac{\sqrt{241}}{15} x'
$$
We can now replace $s$ and $s'$ in our previous equation above:
$$
\frac{\sqrt{241}}{15} x \cdot \frac{\sqrt{241}}{15} x' = x\cdot x' + y\cdot y'
\quad \Longrightarrow \quad
y' = \left( \frac{4}{15} \right)^{2} \frac{x\cdot x'}{y}
$$
Finally, substituting $x=15$ ft, $x'=30$ ft/s and $y=4$ ft into the formula for $y'$ above yields:
$$y'=8 \text{ ft/s.}$$


问:
Real world applications of Pythagoras' Theorem
I have a school assignment, and it requires me to list a few of the real world applications of Pythagoras Theorem. However, most of the ones I found are rather generic, and not special at all.

What are some of the real world applications of Pythagoras' Theorem?

答:
Here is a true life application of the Pythagorean theorem (the 3-dimensional version, which is a corollary of the 2-dimensional version).

My wife and I needed to have a long iron rod manufactured for us, to use as a curtain rod.

I measured the length $L$ of the rod we wanted.

But we forgot to take into account that we live on the 24th floor of an apartment building and therefore the only way the rod could get into our apartment was by coming up the elevator.

Would the rod fit in the elevator?

My wife measured the height $H$, the width $W$, and the depth $D$ of the elevator box. She then calculated the diagonal of the elevator box by applying the Pythagorean theorem: $\sqrt{H^2 + W^2 + D^2}$. She compared it to $L$, and thankfully, it was greater than $L$. The rod would fit!

I would like to say that we realized this problem BEFORE we asked them to manufacture the rod, but that would be a lie. However, at least my wife realized it before the manufacturers arrived at our apartment building with the completed curtain rod, and she quickly did the measurements, and the Pythagorean Theorem calculation, and the comparison. So PHEW, we were saved.


问:
Question about initial topology and dual vector space
Consider the following fragment from the book "Lectures on von Neumann algebras".

Why is the line
$\varphi$ is  $\sigma(\mathcal{E}, \mathcal{F})$-continuous $\implies$ there exist $\psi_1, \dots, \psi_n \in \mathcal{F}$ such that $|\varphi| \leq \sum_k p_{\psi_k}$ true? This is probably an easy consequence of the initial topology definition, but I can't see it.

答:
Hints:

There is a neighborhood $0\in \mathscr N$ such that $|\varphi(x)|<1$ whenever $x\in \mathscr N$ and there are seminorms $\psi_1,\cdots, \psi_n$ and an $ \epsilon>0$ such that $B_{\epsilon}(\psi_1,\cdots, \psi_n)\subseteq \mathscr N.$ Then $|\varphi(x)|<1$ if $|\psi_i(x)|<\epsilon$ for $1\le i\le n.$

It follows that, for any integer $n,\ |\varphi(nx)|<1$ whenever $x\in \ker \psi_i.$ But this means that $x\in \ker \varphi$ and so $\bigcap \ker \psi_i\subseteq \ker \varphi,$ which in turn implies that $\varphi=\sum c_i\psi_i.$


问:
Derive $\int_0^1 \frac{\ln(\sqrt2-1)-(\sqrt2-x)\ln x}{(\sqrt2-x)^2-1}\,dx=\frac{\pi^2}6+\frac14\ln^2(\sqrt2-1) $
I obtained the integral

$$\int_0^1 \frac{\ln(\sqrt2-1)-\ln(x)(\sqrt2-x)}{(\sqrt2-x)^2-1}\,dx=\frac{\pi^2}6+\frac14\ln^2(\sqrt2-1)
$$

as a by-product while carrying out some complex analysis on an engineering problem. In short, given the real nature of the physical outcome, the imaginary part of the final expression has to vanish, which implies the result above and is numerically verifiable.

However, the complex method employed is not intended to solve integrals such as this and produces it merely by accident. So, the question is: How to derive this elementary close-form directly, with real methods?

Note that the integrand may not look as innocent as it appears. There is a hole as the denominator becomes zero at $x=\sqrt2-1$ and the term $\ln(\sqrt2-1)$ is present in the numerator to remove the singularity, thus ensuring convergence.

Body:
We have
\begin{align}
I &\equiv \int \limits_0^1 \frac{\log(\sqrt{2}-1) - (\sqrt{2}-x)\log(x)}{(\sqrt{2}-x)^2-1} \, \mathrm{d} x \\
&= \int \limits_0^1 \frac{- (\sqrt{2} - 1 -x) \log(\sqrt{2}-1) - (\sqrt{2}-x)\log\left(\frac{x}{\sqrt{2}-1}\right)}{(\sqrt{2}-x)^2-1} \, \mathrm{d} x \\
&= -\log(\sqrt{2}-1) \int \limits_0^1 \frac{\mathrm{d} x}{1 + \sqrt{2} - x} + J = \log^2(1+\sqrt{2}) - \log(1 + \sqrt{2}) \log(\sqrt{2}) + J \, ,
\end{align}
where
\begin{align}
J &= \int \limits_0^1 \frac{-\log\left(\frac{x}{\sqrt{2}-1}\right) (\sqrt{2}-x) }{(\sqrt{2}-x)^2-1} \, \mathrm{d} x \\
&= \int \limits_0^{\sqrt{2}-1} \frac{-\log\left(\frac{x}{\sqrt{2}-1}\right) (\sqrt{2}-x) }{(\sqrt{2}-x)^2-1} \, \mathrm{d} x + \int \limits_{\sqrt{2}-1}^1 \frac{-\log\left(\frac{x}{\sqrt{2}-1}\right) (\sqrt{2}-x) }{(\sqrt{2}-x)^2-1} \, \mathrm{d} x \\
&\!\!\stackrel{\text{IBP}}{=}\frac{1}{2} \int \limits_0^{\sqrt{2}-1} \frac{-\log[(\sqrt{2} - x)^2 - 1]}{x} \, \mathrm{d} x - \frac{1}{2} \log(1+\sqrt{2}) \log \left(\frac{1+\sqrt{2}}{2}\right) \\
&\phantom{=} ~ + \frac{1}{2} \int \limits_{\sqrt{2}-1}^1 \frac{-\log[1 - (\sqrt{2} - x)^2]}{x} \, \mathrm{d} x \\
&= \frac{1}{2} \int \limits_0^1 \frac{-\log(\lvert (\sqrt{2}-x)^2 - 1\rvert)}{x} \, \mathrm{d} x - \frac{1}{2} \log(1+\sqrt{2}) \log \left(\frac{1+\sqrt{2}}{2}\right) \\
&= \frac{1}{2} \int \limits_0^1 \frac{-\log\left[\left\lvert 1 - \frac{x}{\sqrt{2}-1}\right\rvert \left(1 - \frac{x}{1 + \sqrt{2}}\right)\right]}{x} \, \mathrm{d} x - \frac{1}{2} \log(1+\sqrt{2}) \log \left(\frac{1+\sqrt{2}}{2}\right) \\
&= \frac{1}{2} \int \limits_0^1 \frac{-\log\left(1 - \frac{x}{1 + \sqrt{2}}\right)}{x} \, \mathrm{d} x + \frac{1}{2} \int \limits_{\sqrt{2}-1}^1 \frac{-\log\left(\frac{x}{\sqrt{2} - 1} - 1\right)}{x} \, \mathrm{d} x \\
&\phantom{=} ~ + \frac{1}{2} \int \limits_0^{\sqrt{2}-1} \frac{-\log\left(1 - \frac{x}{\sqrt{2} - 1}\right)}{x} \, \mathrm{d} x - \frac{1}{2} \log(1+\sqrt{2}) \log \left(\frac{1+\sqrt{2}}{2}\right) \, .
\end{align}
Now let $x = (1 + \sqrt{2}) t$ in the first, $x = \frac{\sqrt{2} - 1}{t}$ in the second and $x = (\sqrt{2} - 1) t$ in the third integral. This yields
\begin{align}
J &= \frac{1}{2} \int \limits_0^{\sqrt{2} - 1} \frac{-\log(1-t)}{t} \, \mathrm{d} t + \frac{1}{2} \int \limits_{\sqrt{2}-1}^1 \frac{-\log\left(\frac{1}{t} - 1\right)}{t} \, \mathrm{d} t + \frac{1}{2} \int \limits_0^1 \frac{-\log(1-t)}{t} \, \mathrm{d} t \\
&\phantom{=} ~ - \frac{1}{2} \log(1+\sqrt{2}) \log \left(\frac{1+\sqrt{2}}{2}\right) \\
&= \int \limits_0^1 \frac{-\log(1-t)}{t} \, \mathrm{d} t - \frac{1}{4} \log^2(1 + \sqrt{2}) - \frac{1}{2} \log(1+\sqrt{2}) \log \left(\frac{1+\sqrt{2}}{2}\right) \\
&= \frac{\pi^2}{6} - \frac{3}{4} \log^2(1+\sqrt{2}) + \log(1+\sqrt{2}) \log(\sqrt{2}) \, ,
\end{align}
so
$$ I = \frac{\pi^2}{6} + \frac{1}{4} \log^2 (1 + \sqrt{2}) \, . $$

A similar calculation yields the more general integral
\begin{align}
f(a) &\equiv \int \limits_0^1 \frac{\log(a) - \log(x) (1 + a -x)}{(1 + a - x)^2 - 1} \, \mathrm{d} x \\
&= \frac{\pi^2}{6} - \log(\sqrt{a}) \log \left(\frac{(1-a)(2+a)}{(1+a)\sqrt{a}}\right) + \frac{1}{2} \left[\operatorname{Li}_2 \left(\frac{1}{2+a}\right) - \operatorname{Li}_2(a)\right]
\end{align}
for $a \in (0,1]$. While special values like $f(1) = \frac{\pi^2}{12} + \frac{1}{2} \operatorname{Li}_2 \left(\frac{1}{3}\right)$ are reasonably nice, the polylogarithms only cancel for $a = \sqrt{2} - 1$, leading to the particularly simple expression for $I = f(\sqrt{2} - 1)$.


问:
Show that the maximum of $f(x)= \sqrt{x^{2}+a} - |x| $ on $[-1,1]$ for some real $a>0$ is $\sqrt{a}$.
Show that the maximum of $f(x)= \sqrt{x^{2}+a} - |x| $ on $[-1,1]$ for some real $a>0$ is $\sqrt{a}$.

Plotting the graph of both $\sqrt{x^{2}+a}$ and $|x|$, it is clear that the maximum difference is at $x=0$ and is equal to $\sqrt{a}$. However, as $f(x)= \sqrt{x^{2}+a} - |x| $ is not differentiable at $x=0$ I am unsure of how to prove this.

Any help would be greatly appreciated.

答:
If we write $|x| = \sqrt{x^2}$, we obtain:

$$f(x) = \sqrt{x^2+a} - \sqrt{x^2} = \frac {(\sqrt{x^2+a} - \sqrt{x^2})(\sqrt{x^2+a} + \sqrt{x^2})}{\sqrt{x^2+a} + \sqrt{x^2}} = \frac {a}{\sqrt{x^2+a}+\sqrt{x^2}}$$

The denominator is obviously strictly increasing as $|x|$ increases.

Hence the maximum of $f$ occurs at $x=0$, with $f(0) = \sqrt a$.


答:
If $A$ is empty, then every statement of the form

"For all $x \in A$, $x$ has Property $P$"

is vacuously true, even if Property $P$ is something impossible or logically inconsistent, for the simple reason that $A$ contains no counterexamples (because it's empty).

Meanwhile, every statement of the form

"There exists $x \in A$ such that $x$ has Property $P$"

is false, because there does not exist $x \in A$, period.

The conditional statement in the original post has the hypothesis

"If for all $x \in A$, $x$ has Property $P$..."

and the conclusion

"...then there exists $x \in A$, such that $x$ has Property $P$."

A conditional statement is true except when the hypothesis is true and the conclusion is false (i.e. its truth value is defined to be "true, unless we can give a counterexample"). If $A$ is empty, what truth values does that assign to the hypothesis, the conclusion, and the conditional statement as a whole?

Title: Truth value of statements about empty set

Body:
I have a problem:

Let P be the statement " x $\in$ A and x $\in$ $\mathbb{Z}$ "

Determine the truth value of statement: ($\forall$x)P $\Longrightarrow$ ($\exists$x)P

Is there a set A for which the truth value of the above statement is false? Explain.

My approach is:

The statement is only false when the antecedent is true and the consequent is false.

This is not possible because the antecedent is ... (I have no clue why there is no set A for which the hypothesis ($\forall$x)P of the statement is true).

Can someone help, please?


问:
elementary substructure of $H(\kappa)$
$H(\kappa)=\{x:|TC(x)|<\kappa\}$, with $\kappa$ regular card. and $TC(*)$ the transitive closure.

Now we defined an elementary substructure:

$M\prec H(\kappa)$ if for every formula $\varphi$ and all $a_{1},\dots, a_{n}\in M$ holds: $\varphi^{M}(a_{1},\dots,a_{n})\leftrightarrow\varphi^{H(\kappa)}(a_{1}.\dots, a_{n})$.

My first question is, what does this exactly mean? It was the end of the lecture, and so we defined this quite hastily.

And afterwards the professor wrote:

Let $\kappa > \omega_{1}$ and $M\prec H(\kappa)$ be a countable elementary substructure.

Note: $M$ is defenitely not transitive.

How do I see that fast, that $M$ can not be transisive?

(Definition: set $T$ transitive, if $x\in T$ implies $x\subset T$.)

答:
This means that the same first order sentences are true, using one binary relation symbol in the language, interpreted as '$\in$'.

$M$ is a substructure of some $(T,\in)$ means that the relation symbol of $M$ is also interpreted as $\in$, and that it is an elementary substructure means that in addition all formulas have the same truth values when variables are evaluated in $M$.

$\kappa>\omega_1$ implies that  $H(\kappa)\models\,$"$ \exists x:\nexists f:x\hookrightarrow \omega$".

This latter one means that there is a set which cannot be embedded into $\omega$, and this sentence can be written as a correct first order formula, using only '$\in$'.

By 1., the same sentence also holds in $M$, though it doesn't hold in any countable transitive set.

Title: elementary substructure of $H(\kappa)$


问:
How to prove $(P-\lambda)^{-1} = \frac{1}{\lambda(1 - \lambda)}P - \frac{1}{\lambda}I$?
I have a question regarding Functional Analysis:

Assume that $X$ is a Banach space and that $P \in B(X)$ is a projection with $ran P \neq {[0]}$.

Then I have easily shown that $\lambda = 0$ and $\lambda = 1$ are eigenvalues of $P$.

I have troubles proving that if $\lambda \notin\{0,1\}$, then it follows that we have $(P-\lambda)^{-1} = \frac{1}{\lambda(1 - \lambda)}P - \frac{1}{\lambda}I$.

Any help would be grateful.

答:
There are a few ways to get this answer depending on how lazy you want to be about showing it. In order of decreasing laziness:

The first way is just to multiply your proposed inverse by $P-\lambda I$ and show you get the identity.

The second way is to note that since $P^2 = P$, it stands to reason that the inverse should be of the form $aI + bP$ and then solve $(aI + bP)(P-\lambda I) = I$ for $a$ and $b$. (This is informed by 3.)

The third way is to write $(P - \lambda I)^{-1}$ as a geometric series. You have some cases to consider here based on the nature of $\lambda$ (i.e. $|\lambda|>1$, $|\lambda| < 1$, and $|\lambda| = 1$). If $|\lambda| > 1$, you'll need to factor out $\lambda^{-1}$ in that term and then expand to ensure convergence. The conditional convergence for $|\lambda| = 1$ is a bit tricky. Note that in the series expansion for the inverse, $P^n = P$ for $n\ge 1$ which informs the ansatz in 2.


问:
Sufficient condition for convergence in mean
I want to prove that if $\xi_n$ is sequence of non-negative random variables, $\xi_n\xrightarrow {a.e.}\xi$ and $\mathbb{E}\xi_n^p \to \mathbb{E}\xi^p $ than $$\mathbb{E}|\xi_n-\xi|^p\to0 \text{ (i.e. }\xi_n\xrightarrow {L^p}\xi)$$
I don't know where to start. I would be very grateful for help!

答:
This is the Brezis-Lieb lemma and goes as follows: suppose that $p>1$.

Prove first that there is a constant $C$ depending on $p$ such that for all $a,b\in\mathbb{R}$ you have $$\left||a+b|^p-|a|^p-|b|^p \right|\leq C\left(|a|^{p-1}|b|+|a||b|^{p-1} \right).$$

Assume in addition that $\|\xi_n\|_{L^p}\leq M$ for some $M\in\mathbb{R}$ and prove that
$$\mathbb{E}\left[|\xi_n|^p-|\xi_n-\xi|^p\right]\to \mathbb{E}[|\xi|].$$

Now remove the above assumption, and conclude.

For the case $p=1$.

Prove that for all $a,b\in\mathbb{R}$ you have
$$||a+b|-|a|-|b||\leq 2|b|.$$

Assume in addition that $\|\xi_n\|_{L^1}\leq M$ for some $M\in\mathbb{R}$ and prove that
$$\mathbb{E}\left[|\xi_n|-|\xi_n-\xi|\right]\to \mathbb{E}[|\xi|].$$

Now remove the above assumption, and conclude.

Here is a way to prove the result without the Brezis-Lieb lemma. Indeed, for $p\geq 1$ we have the inequality
$$|a-b|^p\leq 2^{p-1}(|a|^p+|b|^p),\;\;\text{ for all }a,b\in\mathbb{R}. $$
Therefore
$$|\xi_n-\xi|^p\leq 2^{p-1}(|\xi_n|^p+|\xi|^p).$$
Also since $\xi_n \xrightarrow {a.e.} \xi$ then you clearly have that $$2^{p-1}(|\xi_n|^p+|\xi|^p)\xrightarrow {a.e.} 2^{p}|\xi|^p\in L^1\;\;\; (\text{since }\xi\in L^p),$$
and by the assumption $\mathbb{E}[|\xi_n|^p]\to \mathbb{E}[|\xi|^p]$  we have
$$ \mathbb{E}[2^{p-1}(|\xi_n|^p+|\xi|^p)]\to \mathbb{E}[2^p\,|\xi|^p]. $$
Therefore by (a slightly stronger version of) dominated convergence applied to
$$ f_n:=|\xi_n-\xi|^p, \;\;\; f:=0,\;\;\;g_n:=2^{p-1}(|\xi_n|^p+|\xi|^p),\;\;\;g:=2^p|\xi|^p,$$
you get that $\mathbb{E}[f_n]\to 0 $ as desired.


答:
$$\begin{aligned}
(P-\lambda)^{-1} = \frac{1}{\lambda(1 - \lambda)}P - \frac{1}{\lambda}I\quad
&\Leftrightarrow\quad I = (P-\lambda)\left(\frac{1}{\lambda(1 - \lambda)}P - \frac{1}{\lambda}I\right)\\
&\Leftrightarrow\quad 0 = \frac{1}{\lambda(1 - \lambda)}P^2 - \frac{1}{\lambda}P-\frac{1}{(1 - \lambda)}P \\
&\Leftrightarrow\quad 0 = \frac{1}{\lambda(1 - \lambda)}P^2 - \frac{1}{\lambda(1 - \lambda)}P \\
&\Leftrightarrow\quad 0 = \frac{1}{\lambda(1 - \lambda)}\left(P^2 - P\right)\\
&\Leftrightarrow\quad P^2=P \\
\end{aligned}$$

Title: How to prove $(P-\lambda)^{-1} = \frac{1}{\lambda(1 - \lambda)}P - \frac{1}{\lambda}I$?

Body:
I have a question regarding Functional Analysis:

Assume that $X$ is a Banach space and that $P \in B(X)$ is a projection with $ran P \neq {[0]}$.

Then I have easily shown that $\lambda = 0$ and $\lambda = 1$ are eigenvalues of $P$.

I have troubles proving that if $\lambda \notin\{0,1\}$, then it follows that we have $(P-\lambda)^{-1} = \frac{1}{\lambda(1 - \lambda)}P - \frac{1}{\lambda}I$.

Any help would be grateful.


问:
Can this be an onto function?

A function $f:R\to R$, where $R$ is set of real numbers, is defined by $$f(x)=\frac{\alpha x^2 + 6x - 8}{\alpha + 6x -8x^2}$$
Find the interval of values of $\alpha$ for which $f$ is onto. Justify your answer.

My attempt:

Since the domain is all real values, the denominator in $f(x)$ must have no real roots, so discriminant of the quadratic (denominator) is less than zero :

$$9 + 8\alpha <0 {\space}{\space}{\space}{\space}{\space}{\space}{\space}{\space}{\space}{\space}{\space}{\space}{\space}{\space}{\space}{\space}{\space}{\space}{\space}{\space}..(1)$$

But then I noticed that the expression for discriminant is the same for both numerator and denominator, if $(1)$ is true, numerator will also have no real roots which means there is no $x∈R$ such that $f(x)=0$   $\\\\\\$ and so $f(x)$ can never be onto for any value of $\alpha$

Textbook Solution:

The expression can be written as (Note: $f(x)=y$):
$$ (8y+\alpha)x^2 + 6(1-y)x-(8+\alpha y)=0{\space}{\space}{\space}{\space}{\space}{\space}{\space}{\space}{\space}{\space}{\space}{\space}{\space}{\space}..(1)$$

Since $x$ is real, $D\ge0$:

$$(8\alpha +9)y^2 + (46 + {\alpha}^2)y + (8\alpha + 9)\ge 0{\space}{\space}{\space}{\space}{\space}{\space}{\space}{\space}{\space}{\space}{\space}{\space}{\space}{\space}..(2)$$

For $y$ to be real, $9+8\alpha\gt0$ and $(46+{\alpha}^2)^2-4(9+8\alpha)^2\le0$

Solving these inequalities, we get $2\le\alpha\le14$

I've looked for solutions online but they all give the textbook answer which led me to believe that maybe I was doing something wrong. Please help me.

答:
The question is poorly worded. They do not want the domain to be the set of real numbers, they want it to be $\{x \in \mathbb{R} : \alpha + 6x - 8x^2 \neq 0\}$.

In calculus and analysis it's common for functions not to be defined everywhere and many times the phrase "real valued function" means the domain is an appropriate subset of the real numbers. The proper term for this is a "partial function" but since these are so common in this area of mathematics, it's common to just call these "functions."

Whether it's still appropriate to write "$f : \mathbb{R} \to \mathbb{R}$" I would argue no. I've always seen $f : D \subseteq \mathbb{R} \to \mathbb{R}$ where $D$ is an unspecified set but is assumed to be the largest subset of $\mathbb{R}$ for which the definition of $f$ makes sense. It would seem that whoever wrote this textbook have a different opinion than me.

I would add: while I think it's appropriate for a calculus book to be a bit sloppy about domains, if this were a book on elementary set theory/logic/basic proofs, then being this sloppy is wholly inappropriate.

Title: Can this be an onto function?


问:
Function contradiction
Let $f(x)$ and $g(x)$ be two functions and $f(x),g(x):\mathbb{R}\rightarrow\mathbb{R}$, such that $$f(f(...f(x)))=g(x)$$ thus $f(g(x))=g(x)\Rightarrow f(x)=x\Rightarrow f(f(...f(x)))=x$, what's wrong if $g(x)\neq x$ ?

答:
The iteration sequence $x, f(x), f(f(x)), ...$, if it converges, will always converge to a fixed (or stationary) point $x_0$ (that is, $f(x_0) = x_0$), as you proved.

That said, $g(x)$ will be identity only on those stationary points. Otherwise it will be equal to the stationary point you reach starting $x$. Thus, the range of $g$ will be the set of stationary points. As an example (like Conrad mentioned),  $f(x)=\frac{x}{2}$ and $g(x)=0$ works.

Title: Function contradiction


答:
The first derivative (where it is defined) is $$f'(x)=\frac x{\sqrt{x^2+a^2}}-\frac{x}{|x|}$$
Notice that there are no roots for this. But you have discontinuities. So you need to check these points separately. $f'>0$ so $f$ is increasing for $x<0$, and $f'<0$ for $x>0$. You need to calculate $$\lim_{x\to \pm 0}f(x)=\sqrt a$$
Notice that I've calculate the limits on both sides. It just happens that they are equal.

Body:
Show that the maximum of $f(x)= \sqrt{x^{2}+a} - |x| $ on $[-1,1]$ for some real $a>0$ is $\sqrt{a}$.

Plotting the graph of both $\sqrt{x^{2}+a}$ and $|x|$, it is clear that the maximum difference is at $x=0$ and is equal to $\sqrt{a}$. However, as $f(x)= \sqrt{x^{2}+a} - |x| $ is not differentiable at $x=0$ I am unsure of how to prove this.

Any help would be greatly appreciated.


问:
What is this derivative? Confusing derivatives
if $g(x)=y $ and  $y'= \frac{\partial y}{\partial x}$

then what is this $\frac{\partial y'}{\partial y}$ ? would it be zero since $y'$ is not dependent on $y$?
what if $y'$ is an implicit derivative, obtained from $f(x,y)=0$? what if $y'$ is defined in a differential equation such as $y'=h(x,y)$?

these kind of questions about partial and total derivatives make me very confused. I think I always get confused about the mathematical notation.

can you recommend me any books that could help me clear these notation problems I have?

答:
I do not know any books about that, but I can recommend a few online resources, along with how I would answer it.
Let's start with my explanation. So you start with the derivative chain rule, modified to fit in with partial derivatives;

$\frac{\partial y'}{\partial y}=\frac{\partial
 y''}{\partial y'}$

Using that, and the fact that $y=g(x)$, you just have to find the derivative of $g(x)$ (since the prime notation is the same as the $\frac{dx}{dy}$ notation, and yes, $y'$ is dependant upon $y$), and the equation from the beginning becomes $y'=\frac{\partial [g(x)]}{\partial x}$, which we've just simplified down to the partial derivative of $g(x)$.

Now for the resources:

https://www.khanacademy.org/math/in-in-grade-12-ncert/xd340c21e718214c5:continuity-differentiability/xd340c21e718214c5:chain-rule/a/chain-rule-overview

https://www.khanacademy.org/math/ap-calculus-ab/ab-differentiation-1-new/ab-2-1/a/derivative-notation-review

https://www.khanacademy.org/math/multivariable-calculus/multivariable-derivatives/partial-derivative-and-gradient-articles/a/introduction-to-partial-derivatives

Hopefully this was helpful in answering your questions about the confusing notations and other stuff that come along with learning differential calculus.

P.S.

It's a really hard part of learning calculus that you are struggling on. A few months ago, I was EXTREMELY confused with that very part of it. I know what it's like to feel overwhelmed by the extremities of this subject. It's really tough, but I assure you, that by the end of this month, you'll be a master at this.


问:
Finding $r$ in quadratic integer equation
$$M=f(n)=\frac{n(n+1)}{2}$$
$$M=g(k)=40k+30+r$$
I have found that $r=2,6$ but $r$ can only be one. Since $n, k \in \mathbb{N}$, I wanted to see which vaue for r would lead to integer solutions for k and n.
$$n^2 +n -2(40k+30+r)=0$$
$$n=\frac{-1+\sqrt{1-4(1)(-1)(80k+60+2r)}}{2}$$
$$n= \frac{-1+\sqrt{320k +241+8r}}{2}$$
$$n=\frac{-1+\sqrt{320k +257}}{2},\frac{-1+\sqrt{320k+289}}{2}$$

now $n$ is only an integer if $320k +257$ or $320k + 289$ are perfect squares.

I don't want to add any more context.

答:
Sufficiency

So we must look at when $320k+257$ and $320k+289$ , separately, are perfect squares. Indeed, note that if these are perfect squares then they are odd hence perfect squares of odd numbers . In particular, we have that the numerator of the expression for $n$ is an even number so $n$ is an integer provided they are perfect squares.

$320k+257$

The idea behind showing that no number of the form $320k+257$ is a square, is to show that no number of a simpler form is a square. Let me show you.

Every number of the form $320k+257$ is of the form $5l+2$ for some $l$.

This is easy, since $320k+257 = 5(64k) + 55 + 2 = 5(64k+11)+2$ so we can take $l = 64k+11$.

No number of the form $5k+2$ can be a perfect square.

Suppose that $x^2 = 5k+2$. Let $5r$ be the smallest multiple of $5$ such that $5r<x$. Then $0<x-5r < 5$. We have $$(x-5r)^2 = x^2-10xr+25r^2 = 5(5r^2-2xr +k) + 2$$

So $(x-5r)$ has the same property as $x$ in terms of having a square $2$ more than a multiple of $5$, which means that some number out of $0,1,2,3,4$ must have a square of the form $5k+2$. But this never happens, as can be easily checked!

This shows that no number of the form $5k+2$ is a perfect square, and hence that no number of the form $320k+257$ is a perfect square.

$320k+289$

Using the logic we used previously works somewhat : if $x^2 = 320k+289$ then $(x-320r)$ also has the same property for any $r$. In particular, having all the solutions from $0,1,2,...,319$ whose squares leave a remainder of $289$, will give us the set of all $x$ for which $x^2$ leaves a remainder of $289$ upon division by $320$.

Why? For example, take $k=0$, so $320k+289 = 289 = 17^2$. Then $x = 17$. Now take any number of the form $17+320l$, and square it, it will also leave a remainder $289$ upon division by $320$. For example, $320+17, 17-320,17-10*320$ would all have this property.

Which means, all we need are the numbers from $0,1,...,319$ whose squares leave a remainder $289$ upon division by $320$. I won't explain how one can simplify this (which one can) but the solutions are $17,47,113,143,177$.

That is, let $S = \{17,47,113,143,177\}$ then every $k$ such that $320k+289$ is a perfect square, is of the form $\left\{\frac{(s+320l)^2 + 289}{320} : s \in S, l \in \mathbb Z\right\}$.

Further reading

You may look up the notion of quadratic congruences, which has a large theory dedicated to it. In particular, the Jacobi symbol may be used in some instances to decide if squares of the type $ak+b$ exist, and we can use prime factorizations of $a$ to obtain solutions to the problem or provide bounds on the number of solutions.

Quadratic congruences, incidentally, are motivated by the kind of problem you consider i.e. when is a certain kind of quadratic form the square of another number.


问:
How many chocolate chips are required to ensure that 90\% of the $k$ cookies has at least 1 chip?
There's a cookie dough, and you randomly cut this dough into $k$ equal sized cookies. You want to put $n$ chips so that 90% of the cookies have at least 1 chip.

Does this $90\%$ also represent the the probability that a single cookie has at least 1 chip? If it does, what justifies the equivalence?

答:
It implies that the expected number of cookies with chips would be $90\%$ of all the cookies.

It does not implies that the number of cookies with chips would always be at least $90\%$ of all the cookies. You will never be certain, no matter how many chips you use, and for example there is a small chance all the chips end up in the same cookie.

Nor does it imply that the probability would be at least $90\%$ for the number of cookies with chips being at least $90\%$ of all the cookies. Some batches will have strictly more than $90\%$ of the cookies having chips and to offset this, some batches will have strictly fewer than $90\%$ of the cookies having chips.

For example, suppose there are $k=10$ cookies and $n=23$ chocolate chips, with each chip placed independently uniformly at random in one of the cookies.  The probability that a particular cookies has no chips is $\left(1 -\frac{1}{k}\right)^n \approx 0.0886$ in this example.  So each cookie has a probability over $91\%$ of having a chip.

But the probability that at least $9$ of the $10$ cookies have chips is less than $80\%$.  Here is a simulation in R illustrating this, with each of the $10^5$ runs simulating how many cookies receive chips:

```<code>set.seed(2021)
cookieswithchips <- replicate(10^5, length(table(sample(10, 23, replace=TRUE))))
mean(cookieswithchips)
# 9.11145
table(cookieswithchips)
# cookieswithchips
# 5     6     7     8     9    10
# 1   164  2671 17613 44955 34596
</code>```

Body:
It implies that the expected number of cookies with chips would be $90\%$ of all the cookies.

It does not implies that the number of cookies with chips would always be at least $90\%$ of all the cookies. You will never be certain, no matter how many chips you use, and for example there is a small chance all the chips end up in the same cookie.

Nor does it imply that the probability would be at least $90\%$ for the number of cookies with chips being at least $90\%$ of all the cookies. Some batches will have strictly more than $90\%$ of the cookies having chips and to offset this, some batches will have strictly fewer than $90\%$ of the cookies having chips.

For example, suppose there are $k=10$ cookies and $n=23$ chocolate chips, with each chip placed independently uniformly at random in one of the cookies.  The probability that a particular cookies has no chips is $\left(1 -\frac{1}{k}\right)^n \approx 0.0886$ in this example.  So each cookie has a probability over $91\%$ of having a chip.

But the probability that at least $9$ of the $10$ cookies have chips is less than $80\%$.  Here is a simulation in R illustrating this, with each of the $10^5$ runs simulating how many cookies receive chips:

set.seed(2021)
cookieswithchips <- replicate(10^5, length(table(sample(10, 23, replace=TRUE))))
mean(cookieswithchips)
# 9.11145
table(cookieswithchips)
# cookieswithchips
# 5     6     7     8     9    10
# 1   164  2671 17613 44955 34596


问:
Every neighbourhood of $1$ is of form $U\setminus \mathbb{N} \cup \{1\}$?
From Topology without tears:

Let $X$ be a set $(\mathbb{R}\setminus\mathbb{N})\cup {1}$.

Define a function $f:\mathbb{R}\to X$:
\begin{equation}
f(x) =
    \begin{cases}
      x & \text{if $x \in \mathbb{R}\setminus\mathbb{N}$}\\
      1 & \text{if $x \in \mathbb{N}$}
     \end{cases}
\end{equation}

Further define topology $\tau $ on $X$ by

$$\tau=\{U:U\subseteq X \text { and } f^{-1}(U) \text{ is open  in the euclidean topology on } \mathbb{R} \}$$

Then prove:

1)$f$ is continuous.

2)Every open neighbourhood of $1$ in (X,\tau) is of form $(U\setminus \mathbb{N}) \cup \{1\}$,where $U$ is open in $\mathbb{R}$

I am not getting how every open neighbourhood of $1$ in $(X,\tau)$ of form $U\setminus \mathbb{N} \cup \{1\}$

Because this open neighbourhood will also be a open set in $(X,\tau)$.And According to me, if a $S\subseteq X$ contains $1$,then $f^{-1}(S)=\{x\in \mathbb{R}:f(x)\in S\}$ will also contain $\mathbb{N}$(since for $\forall x \in \mathbb{N},f(x)=1\in S)$.Then in that case how $f^{-1}(S)$ will be open in eucliedean topology??As $\mathbb{N}$ is itself not open in Euclidean topology in \mathbb{R}??

And hence according to me,none of the open set in $(X,\tau)$ must contain $1$

Any hint  where I am wrong is appreciable.Thanks in advance...

答:
You are correct about $f^{-1}(\{1\})$ is $\mathbb N$. To give an example of $S\subset X$ open, pick a neighborhood $U_n$ for each $n\in\mathbb N$. Let $S_0=\bigcup_{n\in\mathbb N}U_n$, and $S=(S_0\setminus\mathbb N)\cup\{1\}$. Then we have $f^{-1}(S)=S_0$, which is open in $\mathbb R$.

Now, can you prove that every open sets in $X$ containing $1$ looks like that?


问:
Convergence of $A^n v$ for a matrix $A$ and a vector $v$
In this question, Martin Sleziak shows that for a complex square matrix $A$, we have $$\lim A^n = 0 $$ if and only if the spectral radius of $A$ is less than $1$. I was wondering what happens if we are interested only in the convergence with respect to a given vector.

To be more specific, suppose now that all eigenvalues of $A$ have modulus $\ge 1$ and let v be some given vector such that $$\lim A^n v = 0. $$ Does it imply that $ v = 0 $?

I suspect the answer is yes, but I don't know how to deal with the case where two distinct eigenvalues of $A$ have the same modulus.

Edit: The comments encouraged me to add more details. Let $J$ be the Jordan form of $A$ and $P$ such that $A=PJP^{-1}$. Then
$$A^n =
P\begin{pmatrix}
J_1^n & \;     & \; \\
\;  & \ddots & \; \\
\;  & \;     & J_m^n
\end{pmatrix}P^{-1} $$
where $J_i$ is a Jordan block of size $k_i\times k_i$, so that

$$ J_i^n = \begin{pmatrix}
\lambda_i^n & \binom{n}{1}\lambda_i^{n-1} & \binom{n}{2}\lambda_i^{n-2} & \cdots & \cdots & \binom{n}{k_i-1}\lambda_i^{n-k_i+1} \\
 & \lambda_i^n & \binom{n}{1}\lambda_i^{n-1} & \cdots & \cdots & \binom{n}{k_i-2}\lambda_i^{n-k_i+2} \\
 &  & \ddots & \ddots & \vdots & \vdots\\
 &  & & \ddots & \ddots & \vdots\\
 &  & &  & \lambda_i^n & \binom{n}{1}\lambda_i^{n-1}\\
 &  &  &  &  & \lambda_i^n
\end{pmatrix}. $$
Thus every entry of $A^n v$ has the form
$$\tag{1}\label{sum}\sum_{i=1}^m p_i(n)\lambda_i^n$$
where $p_i$ is a complex polynomial such that $\deg p_i< k_i$ for $i=1,\dots,m$.

So essentially what I'm asking is this:

Suppose $|\lambda_i|\ge1$ for $i=1,\dots,m$ and that \eqref{sum} converges to $0$ (when $n\to\infty$). Does it imply that the sequence \eqref{sum} is identically zero?

答:
Yes: if $A$ has all eigenvalues of modulus at least one, and $\lim_{k\to\infty} A^kv = 0$, then $v=0$.

Let me first give a slick answer in the case that all eigenvalues of $A$ have modulus $r>1$. The spectral radius of $A^{-1}$ is $r^{-1} < 1$. For all $v$,
$$ |v| = |A^{-n} A^n v| \leq |A^{-n}| |A^n v|.$$
The spectral radius $\rho$ satisfies $\rho(B) = \lim_{k \to \infty} |B^k|^{1/k}$ for all matrix norms $|\cdot|$. Hence $\lim_{n\to\infty}|A^{-n}|^{1/n} = r^{-1}$.
Given $\epsilon>0$, there exists a natural number $N$ such that $|A^nv| \leq \epsilon$ and $|A^{-n}|^{1/n} < 1$ for $n \geq N$, hence $|v| \leq \epsilon$. This shows that $v = 0$.

The following works for general (upper-triangular) Jordan blocks of eigenvalue with modulus $\geq 1$. Let $J$ be a block of size $n$ with eigenvalue $\alpha$. We prove that if $\lim_{k\to\infty} J^k v =0$, then $v = 0$. The proof is by induction on $n$. For $n=0$, $v=0$ and we're done. In general, the $n$th entry of $J^k v$ is $\alpha^k v_n$, which tends to zero if and only if $v_n = 0$. Hence $v$ is in the domain of $J$ restricted to the upper $(n-1)\times(n-1)$-block, which is a Jordan block of size one smaller. Now induct to finish the proof.

Essentially, the polynomials $p_i$ you reference are not so complicated since one is always constant.


问:
Write $\sum\limits_{i=0}^n 2^i + 2^{n+1}$ as a single summation: any guidance?
I'm working on an assignment for Discrete Math, and sort of stuck on this question. I'm finding it a bit challenging to wrap my head around some of these concepts doing it remotely, and being removed from other students doing it.

Anyway: I'm kind of confused - $\sum\limits_{i=0}^n 2^i + 2^{n+1}$ seems to be a single summation already.

The direction I was going was that by Theorem 1 in my textbook for a geometric sum
$\sum\limits_{i=0}^n 2^i = 2^{n+1} - 1$

So I was thinking the entire statement would be equal to $2^{n+2} - 1$ (edited from a silly mistake: $4^{n+1} - 1$), but that doesn't really seem to line up with what the question is asking.

Thanks in advance for your patience and advice.

答: If the expression means $\sum_{i=0}^n(2^i+2^{n+1})$, since $i$ takes $n+1$ values the result is$$2^{n+1}-1+(n+1)2^{n+1}=(n+2)2^{n+1}-1.$$If the expression means $(\sum_{i=0}^n2^i)+2^{n+1}$, the result is$$2^{n+1}-1+2^{n+1}=2\cdot2^{n+1}-1=2^{n+2}-1.$$The latter meaning is more likely, as this calculation provides the inductive step in proving $\sum_{i=0}^n2^i=2^{n+1}-1$.