问:
Maximizing the probability of winning the prize in a game show

In a game show, there are three locked chests A, B and C. One of them holds a prize and the other
two are empty. In a turn, the contestant chooses one of the chests and flips a fair coin. If the result is heads, he gets the contents of the chest else not. Given that chests A, B and C may contain the
prize with probabilities of $\dfrac{8}{13}, \dfrac{4}{13}, \dfrac{1}{13}$ respectively and the contestant is given 2021 turns, then the number of times he must choose chest C to maximize his chances of obtaining the prize is:

(A) 0

(B) 155

(C) 672

(D) none

My attempt:

Let the number of times the contestant chooses chests A, B, C be $x$, $y$, $z$ respectively. Then, we have
$$\tag{1} x+y+z=2021$$
Now, if the contestant wants to open chest A, he must get at least one head out of $x$ tosses, in other words, we have to exclude the case where he obtains all tails. The probability that this happens is $1-\dfrac{1}{2^x}$. After opening the chest, the probability that he wins the prize is $\dfrac{8}{13}$. Thus, the probability that he obtains the prize from chest A is $\dfrac{8}{13} \left(1-\dfrac{1}{2^x} \right)$.

Performing a similar analysis for the other chests, the probability that he wins the prize is given by:
$$\tag{2} P = \dfrac{8}{13} \left(1-\dfrac{1}{2^x} \right) + \dfrac{4}{13} \left(1-\dfrac{1}{2^y} \right) + \dfrac{1}{13} \left(1-\dfrac{1}{2^z} \right)$$
The only thing left is to maximize $(2)$ under the condition $(1)$ and find the value of $z$ for optimum condition. I, however, do not seem to have any idea of how maximize the given function. Or perhaps, is there a simpler way to solve this?

答:
The problem description is somewhat ambiguous. If the contestant learns the result of the coin flip immediately, the correct strategy is to repeatedly choose chest A until the coin flips heads (if ever), then do the same with B and then with C.

If the contestant doesn't learn the result of the flip or has to pre-commit to her actions (which I assume is the intended version), one way to solve the problem is to compute the probability of obtaining the chest on a specific turn. If the contestant chooses A, this probability is $P_A = (\frac12)^x \cdot \frac8{13}$, where $x$ is the number of times the contestant has previously chosen A. Similarly, the probability for B is $P_B = (\frac12)^y \cdot \frac4{13}$. Since a greedy algorithm performs optimally in this problem, it suffices to choose options so that $P_A$ and $P_B$ and $P_C$ are all kept equally large. After choosing A once, $P_A$ shrinks to $P_B$. After choosing both A and B two more times, $P_A$ and $P_B$ both shrink to $P_C$. From here, they are  kept equally large by choosing them equally often. Thus, optimal play is such that $x = y + 1 = z + 3$.

Since $x + y + z = 2021$, this means that $3z + 5 = 2021$ so that $z = 672$.


问:
Manipulation of sums (in order to understand proof regarding Gauss sums)
I'm looking over the proof of $\left| \tau(\chi) \right|=\sqrt{q}$, when $\chi$ is a primitive character mod $q$, and $\tau$ the Gauss sum. (H. L. Montgomery and R. C. Vaughan. Multiplicative number theory. I. Theorem. 9.7) Anyhow, this is a universal exercise in manipulation of sums.

I need to simplify

$\sum_{a=1}^q\sum_{b=1}^q \chi (a) \overline{\mathbb{\chi(b)}}\mathbb{1}_{a=b} \cdot q$

I'm thinking something like this:

$\sum_{a=1}^q\sum_{b=1}^q \chi (a) \overline{\mathbb{\chi(b)}}\mathbb{1}_{a=b}\cdot q
= \sum_{a=1}^q\sum_{a=1}^q\left| \chi(a) \right| ^2 \cdot q
= \sum_{a=1}^q\sum_{a=1}^q \mathbb{1}_{gcd\left( a,q \right) =1} \cdot q = \phi(q) \cdot q$

where $\phi$ is the Euler phi-function. But this computation seems "illegal" to me. How would one do this?

(I need to end up with the expression $\phi(q) \cdot q$)

答:
Here is a derivation with an intermediate step which might be helpful.

We obtain
\begin{align*}
\color{blue}{\sum_{a=1}^q}&\color{blue}{\sum_{b=1}^q \chi (a) \overline{\chi(b)}\mathbb{1}_{a=b}\cdot q}\\
&=q\sum_{a=1}^q\left(\chi (a) \overline{\chi(a)}+\sum_{{b=1}\atop{b\ne a}}^q \chi (a) \overline{\chi(b)}\right)\mathbb{1}_{a=b}\\
&=q\sum_{a=1}^q\chi (a) \overline{\chi(a)}\cdot 1
+q\sum_{a=1}^q\sum_{{b=1}\atop{b\ne a}}^q \chi (a) \overline{\chi(b)}\cdot 0\\
&=q\sum_{a=1}^q\left|\chi (a)\right|^2=q\sum_{a=1}^q \mathbb{1}_{\gcd\left( a,q \right) =1}\\
&\,\,\color{blue}{=q\phi(q)}
\end{align*}

ContentLicense: CC BY-SA 4.0


问:
Finding F(10) and a relation between double derivative and function is given
Problem :

If $f''(x)=-f(x),$ where f(x) is a continuous double differentiable function and $g(x) =f'(x) $ If F(x) =(f$(\frac{x}{2}))^2+(g(\frac{x}{2}))^2$ and F(5)=5, then find F(10)

Attempt :

If $g(x) =f'(x) \Rightarrow g'(x) = f''(x) = -f(x) $

I am getting no clue how to proceed further.... please make this concept clear how to solve such problems, Thanks...

答: Hint: Evaluate $F'(x) =$ using the chain rule..

ContentLicense: CC BY-SA 3.0


问:
How to show the associativity in this group?
The group law is given by $a \oplus b = \frac{a+b}{1+a\,b}$ for $a,b \in G = (-1,1)$

For associativity I guess I have to show $(a \oplus b) \oplus c = a \oplus (b \oplus c)$, hence:

$$\begin{equation}\frac{\frac{a+b}{1+a\,b}+c}{1+\frac{a+b}{1+a\,b}\,c}= \frac{\frac{c+b}{1+c\,b}+a}{1+\frac{c+b}{1+c\,b}\,a}\end{equation}$$

but I got a hard time doing so. Is this actually the task?

答:
We have

$$\begin{align}
(a\oplus b)\oplus c&=\left(\frac{a+b}{1+ab}\right)\oplus c\\
&=\frac{\left(\frac{a+b}{1+ab}\right)+c}{1+\left(\frac{a+b}{1+ab}\right)c}\\
&=\frac{(a+b)+c(1+ab)}{(1+ab)+(a+b)c}\\
&=\frac{(a+b)+c+abc}{1+ab+ac+bc}\\
&=\frac{a(1+bc)+(b+c)}{(1+bc)+(b+c)a}\\
&=\frac{a+\left(\frac{b+c}{1+bc}\right)}{1+a\left(\frac{b+c}{1+bc}\right)}\\
&=a\oplus\left(\frac{b+c}{1+bc}\right)\\
&=a\oplus(b\oplus c).
\end{align}$$

But there's more: since $a,b\in (-1,1)$, we have

$$-1<a\oplus b=\frac{a+b}{1+ab}<1$$

by looking at extreme cases, so $a\oplus b\in(-1,1)$.


问:
perfect normal Lindelöf implies Souslin property.
I think this is quite awkward, but I decided to ask for help anyway.

This is the problem:

Show that every perfectly normal Lindelöf space has the Souslin property.

I've tried to slove this problem by using the fact that every open subset of a perfectly normal space is also Lindelöf, so the two open neighborhoods which seperates two closed subsets from each other (normality) have countable open cover. Now in worst case all of the open sets which these two countable open cover are obtained from, are disjoint, which is still countable and the wanted property holds. But I know there's something wrong with this and I don't know what to do.

Any help would be greatly appreciated.

答: Note that if $X$ is perfectly normal and Lindelöf, every open subspace $Y$ of $X$ is also Lindelöf, because these are $F_\sigma$ in $X$ and an $F_\sigma$ in a Lindelöf space is still Lindelöf. This is enough for our purposes: if $\mathcal{U}$ then is a pairwise disjoint family of non-empty open sets in $X$, the union $O=\bigcup \mathcal U$ is open and Lindelöf, and the $\mathcal{U}$ form a disjoint open cover of $O$, so in fact must be its own countable subcover (we cannot omit any member). So $\mathcal{U}$ is at most countable and $X$ is ccc.


问:
Find $\min(\max(x+y,2x-y+2))$ under $x+2y\ge 4, x\ge 0, y\ge 0$
Find $\min(\max(x+y,2x-y+2))$ under $x+2y\ge 4, x\ge 0, y\ge 0$.

My attempt: In the region defined by the constraints, we have that $f(x,y)=\max(x+y,2x-y+2)=2x-y+2$. So, I believe that it is enough to minimize $2x-y+2$ under the given constraints. There is no solution, because there is no level set that shares a unique point or side with the defined region.

However Wolfram Alpha tells that there is a minimum, which is the point $(0,2)$. How can I come to this conclusion? This point is intersection of $2x-y+2=0$ with $x+2y=4$.

答:
You can minimize the maximum via linear programming by introducing an additional variable $z$ and minimizing $z$ subject to the original constraints and
\begin{align}
z &\ge x+y\\
z &\ge 2x-y+2\\
\end{align}


问:
Finding the locus of $b^2-2x^2=2xy+y^2$
I was asked to determine the locus of the equation
$$b^2-2x^2=2xy+y^2$$

This is my work:

Add $x^2$ to both sides:
$$\begin{align}
b^2-x^2 &=2xy+y^2+x^2\\
 b^2-x^2 &=\left(x+y\right)^2
\end{align}$$

I see that this is similar to the equation of a circle. How can I find the locus of this expression?

答:
With variable changes
$$x= \sqrt{ \frac{5+\sqrt5}2}u + \sqrt{ \frac{5-\sqrt5}2}v,\>\>\>\>\>
 y= \sqrt{ \frac{5-\sqrt5}2}u - \sqrt{ \frac{5+\sqrt5}2}v
$$
the curve equation  $b^2-2x^2=2xy+y^2$  can be recast as

$$ \frac{u^2}{\left(\frac{\sqrt5-1}2b\right)^2} +\frac{v^2}{\left(\frac{\sqrt5+1}2b\right)^2} =1
$$
which reveals an ellipse of axis lengths $\frac{\sqrt5-1}2b$ and $\frac{\sqrt5+1}2b$.

Title: Finding the locus of $b^2-2x^2=2xy+y^2$


问:
How to express $2 \cos X = \sin X$ in terms of $\sin X$?
The Question was:

Express $2\cos{X} = \sin{X}$ in terms of $\sin{X}$ only.

I have had dealings with similar problems but for some reason, due to I believe a minor oversight, I am terribly vexed.

答:
Note: too long for a comment...

Squaring to force the use of $\cos^2x+\sin^2x=1$ results in an equation which is not equivalent to the original equation. This actually creates another countable set of solutions.

First note that, $\cos \theta=0$ does not occur when the equation is fulfilled, so:
$$(E): 2\cos x=\sin x \iff \tan x=2
$$
has the solution set $\arctan 2+\pi\mathbb{Z}$.

Now if you square $(E)$:
$$
(E)^2: 4\cos^2x=\sin^2x\iff (2\cos x-\sin x)(2\cos x+\sin x)=0\iff\tan x=\pm2
$$
has the solution set $(\arctan 2 +\pi\mathbb{Z})\sqcup(-\arctan 2+\pi\mathbb{Z})$. So that's not equivalent to $(E)$.

Replacing $\cos x$ by $\pm\sqrt{1-\sin^2x}$ does not really work either. The best we can say is that
$$
(E)\iff \cos x\geq 0 \wedge 2\sqrt{1-\sin^2 x}=\sin x\quad\mbox{or}\quad \cos x\leq 0 \wedge -2\sqrt{1-\sin^2 x}=\sin x.
$$


问:
The distribution of areas of a random triangle on the sphere - what are the second, third, etc. moments?
Suppose that we choose three points independently and uniformly at random on the surface of a unit sphere as the vertices of a triangle, and consider the area of this triangle. Call this random variable $X$.

The area of such a triangle is the sum of its angles minus $\pi$, so by linearity of expectation the expected value of $X$ is just $3q-\pi$, where $q$ is the expected value of one of the angles. But by symmetry we can fix one point to be at the north pole and the other to lie on the Prime Meridian, from which it is obvious that the angle distribution is uniform on $[0,\pi]$. Thus $\mathbb{E}[X]=\pi/2$, or one-eighth of the sphere's area.

However, because the angles are not independent (they cannot sum to less than $\pi$, for instance), we cannot use this sort of logic to easily infer the values of the second and third moments of this distribution.

After gathering some numerical data, it appears that $\mathbb{E}[X^2]=\frac{\pi^2}2$, but I am not sure how to prove this. (Note that this is equivalent to the statement that the standard deviation of $X$ is $\pi/2$, which may be easier to show?)

I also have $\mathbb{E}[X^3]\approx 20.36$, or that the third moment of $X$ is approximately $4.86$ (either can be inferred from the other, given lower-order moments - they should differ by $\pi^3/2$). I haven't found any particularly nice formulas that match either of these values, though I'm not sure about the last digit in either of these estimates.

In general, what is $\mathbb{E}[X^n]$ or $\mathbb{E}[(X-\frac{\pi}2)^n]$? If any of the values are open, has it been discussed in the literature? Is there a nice geometric argument for $\mathbb{E}[X^2]$?

Edit: Here is a histogram of the area distribution from $1000000$ samples. Interestingly, it seems not to decay to $0$ at the upper bound of $2\pi$.

答:
I found a cute geometric argument that $\mathbb{E}[X^2]=\frac{\pi^2}2$, and it seemed worth presenting here. (Many statements like "almost surely" and "if points are in general position" and "except for a set of measure $0$" have been omitted for ease of reading.)

Claim: If five points are randomly chosen on the sphere, their convex hull will be a triangle with probability $\frac5{16}$.

Proof: We start with a lemma.

Lemma: Given any three random points on the sphere, adding a random fourth point has a $50\%$ chance of producing a triangular convex hull.

Proof of lemma: Draw the three great circles connecting each pair of the three points, which subdivides the sphere into eight regions. Four of these regions have the property that a fourth point inside them will yield a triangular convex hull, and each such region is opposite a congruent region inside which a fourth point would not yield a triangular convex hull. So for any location $P$ of the fourth point, exactly one of $P$ and $-P$ would work. This completes the proof.

Now, back to the main theorem.

Given five points, consider ways of labeling one point $A$ and another $B$. Obviously, every five points have $20$ such labelings. Say that a collection of labeled points is good if, starting with the three unnamed points, we can add $A$ to get a triangular convex hull, and add $B$ onto the previous four to again produce a triangular convex hull.

For instance, in the following diagram, the left arrangement is good, but the right two are not (the second one has a quadrilateral convex hull when we add $A$, the third when we add $B$).

Now, fixing a labeling, what are the odds that starting with the unnamed points, adding $A$ gives a triangular hull and adding $B$ to those four points does as well? By our lemma, $\frac12\cdot\frac12=\frac14$. So every choice of five points produces an expected $\frac{20}4=5$ good labelings.

Conversely, given $5$ points whose convex hull is a triangle, how many good labelings does it have? Well, any choice of three points gives a valid starting triangle, and by the time we add the fifth point, we'll have a triangle once again, so the only way it could fail is if the first four points form a quadrilateral. How many ways can that happen?

It shouldn't be too hard to convince yourself that every $5$-point arrangement with a triangular convex hull looks like this, in terms of incidences and intersections (I've drawn lines as straight for convenience):

Inspecting this a bit, we can see that only one of the five points, when removed, leaves a quadrilateral convex hull (the top one, in the above diagram). So that has to be our choice of $B$ for things to fail, and then any of the other four could be $A$. So there are $20-4=16$ good labelings that produce a given configuration.

So, a random five-point arrangement produces $5$ good labelings on average, and a working arrangement produces $16$ good labelings. (Obviously, if a five-point arrangement doesn't have a triangular hull, it won't produce any good labelings.) So the fraction of five-point arrangements that work must be $\frac{5}{16}$.

Okay, proof complete. Now what?

Observe that the probability five points have a triangular convex hull is $10$ times the probability that they have a triangular convex hull and the three points on said triangle were the first three chosen (since any of the ten possible three-subsets could have just as easily been that triangle). So there is a $\frac1{32}$ chance that, if we pick three random points, the next two random points we choose will be inside that triangle.

But given a triangle, the odds that two random points lie inside it is just the square of the fraction of the area of the sphere that it takes up! So what we have shown is that $\mathbb{E}[\left(\frac{X}{4\pi}\right)^2]=\frac1{32}$, and hence $\mathbb{E}[X^2]=\frac{\pi^2}{2}$.

Note that this sort of argument can't extend to $6$ points, because the resulting probability of a triangular convex hull will be ${6\choose 3}\cdot \mathbb{E}[\left(\frac{X}{4\pi}\right)^3] = \frac{15}{32} - \frac{15\log(2)}{4 \pi^2}$, which is almost certainly irrational.

What breaks with $6$ points? I think the fundamental problem is that there is more than one point/line arrangement with positive probability, so the number of good labelings isn't constant. Consider the below two configurations:

In the left case, removing any of the outer triangle's vertices will yield a quadrilateral convex hull, but in the right case, any vertex can be removed while preserving triangularity.


问:
Is $\mathbb{Z}[{ \sqrt 8 } ] $ a Euclidean domain?

Is  $\mathbb{Z}[{ \sqrt 8 } ] $ a Euclidean domain ?

I have some  confusion  that is   what is difference between  euclidean  domain and euclidean Norms ?

My attempt : I thinks  yes

i know  that  $d( a+b \sqrt 8) = |a^2 - 8b^2 | $ as i can show it is  euclidean domain  by same pattern $\mathbb{Z}[{ \sqrt 2 } ]$ is  euclidean  domain

答:
It is more straight forward to give a counter example. Since $4=2 \cdot 2 = (\sqrt{8}+2)(\sqrt{8}-2)$, $\mathbb{Z}[\sqrt{8}]$ is not a unique factorisation domain (UFD), hence not a Euclidean domain. Note that those factors are irreducible.

Suppose to the contrary that $2$ is not irreducible. There exists $a,b \in \mathbb{Z}[\sqrt{8}]$ with $N(2)=4=2 \cdot 2=N(a)N(b)$. It requires that $a,b \notin U(\mathbb{Z}[\sqrt{8}])$, the set of units. Therefore $N(a)=N(b)=2$.

Let $a=u+v\sqrt{8}$ with $u,v \in \mathbb{Z}$ and $u,v$ not both zero. It follows that $N(a)=|u^2-8v^2|=2$ $\implies$ $8v^2-u^2= \pm 2$. Therefore,

\begin{equation}
(2v)^2 = \frac{u^2}{2} \pm 1
.
\end{equation}

LHS is even. If $u$ is even, RHS is odd; a contradiction. If $u$ is odd, $ \frac{u^2}{2}\notin \mathbb{Z}$; another contradiction. Therefore $2$ is irreducible.

Note that $2 \nmid \sqrt{8} \pm 2$. Hence $\mathbb{Z}[\sqrt{8}]$ is not a UFD.

Link to a bigger image:
Hasse diagram from rng to ED https://i.stack.imgur.com/jUcJX.png


问:
Complex integral $\int_{-\infty}^{\infty} e^{-zt^2+2wt}dt$
The goal is to show this integral
$\int_{-\infty}^{\infty} e^{-zt^2+2wt}dt = \sqrt{\frac{\pi}{z}} e^\frac{w^2}{z}$ for $z,w \in \mathbb{C}$ with $\Re(z)>0$ using the identity:

$
\int_{-\infty}^{\infty}e^{-\pi x^2}e^{-2\pi ix \psi}dx = e^{-\pi \psi^2}
$ for $\psi \in \mathbb{R}$

I have proven in a previous far that the function is holomorphic as a function of $z$ on all of $\mathbb{C}$ but I'm not sure how to do this next part. I looked for a substitution that would help but I only know the substitution formula for real number so I cannot just replace the $x$ in the identity with some complex number.

Any help welcome :)

Body:
$$\int_{-∞}^∞ e^{-zt^2+2wt}dt= \int_{-∞}^∞ e^{-z(t-\frac{w}{z})^2+\frac{w^2}{z}}dt$$
$$= e^{\frac{w^2}{z}} \int_{-∞}^∞ e^{-z(t-\frac{w}{z})^2}dt$$
Take $(t-\frac{w}{z})=u$ the integral becomes
$$e^{\frac{w^2}{z}}\int_{-∞}^∞ e^{-zu^2}du=\sqrt{ \frac{π}{z}}e^{\frac{w^2}{z}}$$


答:
Let $X$ be perfectly normal and Lindelöf, and suppose that $\mathscr{U}$ is an uncountable family of pairwise disjoint, non-empty open subsets of $X$. Let $F=X\setminus\bigcup\mathscr{U}$, and let $G$ be any open set containing $F$. Then $\{G\}\cup\mathscr{U}$ is an open cover of $X$, so there is a countable $\mathscr{U}(G)\subseteq\mathscr{U}$ such that $\{G\}\cup\mathscr{U}(G)$ covers $X$. It follows that $U\subseteq G$ for each $U\in\mathscr{U}\setminus\mathscr{U}(G)$.

Now $F$ is closed, so there are open sets $G_n$ for $n\in\Bbb N$ such that $F=\bigcap_{n\in\Bbb N}G_n$. Each family $\mathscr{U}(G_n)$ is countable, so there is a $U\in\mathscr{U}\setminus\bigcup_{n\in\Bbb N}\mathscr{U}(G_n)$. Then $U\subseteq G_n$ for each $n\in\Bbb N$, so $U\subseteq F$, which is impossible. Thus, $X$ has no uncountable family of pairwise disjoint, non-empty open sets and is therefore ccc (i.e., has the Suslin property).

Title: perfect normal Lindelöf implies Souslin property.

Body:
I think this is quite awkward, but I decided to ask for help anyway.

This is the problem:

Show that every perfectly normal Lindelöf space has the Souslin property.

I've tried to slove this problem by using the fact that every open subset of a perfectly normal space is also Lindelöf, so the two open neighborhoods which seperates two closed subsets from each other (normality) have countable open cover. Now in worst case all of the open sets which these two countable open cover are obtained from, are disjoint, which is still countable and the wanted property holds. But I know there's something wrong with this and I don't know what to do.

Any help would be greatly appreciated.


问:
Question involving Moore-Penrose pseudo-inverse
The problem is this:
Suppose $A,B \in \mathcal{M}_{n \times n}(\mathbb{R})$ satisfying:

$ABA = A$ and $BAB =B$

$A^2 = A$

Is it necessarily true that $B^2=B$?

I have tried playing around a little bit with the relations to see if I can get the result to pop out but that hasn't worked. It is a part of a section on pseudo-inverses but I don't see an obvious way to apply that. Any hints would be appreciated.

答:
No. Let $X$ and $Y$ be any two matrices such that the products $XY$ and $YX$ are well defined. Let
$$
A=\pmatrix{I&0\\ 0&0},\ B=\pmatrix{I&X\\ Y&YX},\ B^2=\pmatrix{I+XY&X+XYX\\ Y+YXY&YX+YXYX}.
$$
Then $A$ is idempotent, $ABA=A$ and $BAB=B$, but $B$ is idempotent if and only if $XY=0$. Surely this is not always the case.


问:
representation on the tensor algebra over a Lie algebra
I'm now reading <Differential Geometry, Lie Groups, and Symmetric Spaces> by Helgason.

In Proposition 1.1 of Chapter 2, he says:

If $\rho$ is a representation $\mathfrak{a} \to \mathfrak{gl}(V)$, then there is a unique representation $\tilde{\rho} : T(\mathfrak{a}) \to \mathfrak{gl}(V)$

such that $\tilde{\rho}(X) = \rho (X)$ for all $X \in \mathfrak{a}$, where $\mathfrak{a}$ is a Lie algebra and $T(\mathfrak{a})$ is the tensor algebra over $\mathfrak{a}$.

But there is no explanation about the existence or the uniqueness of the representation $\tilde{\rho}$.

Since the tensor algebra $T(\mathfrak{a})$ is the direct sum
$T(\mathfrak{a}) = \bigoplus_{k=0}^{\infty} T^k \mathfrak{a}$, I think we can consider $\tilde{\rho}$ as an extension of $\rho$.

To define such extension on $T(\mathfrak{a})$, we need to define an algebra homomorphism on each $k$-th tensor of $\mathfrak{a}$.

And then I got lost in the middle, since it should be a representation on $V$.

I know I can do tensor products of representations, but then it would result a representation on  $T^k V$.

How should I understand this?

答:
The elements of, say, $T^2\mathfrak{a}$ are linear combinations of elementary tensors $a\otimes a'$, $a,a'\in\mathfrak{a}$. Such a tensor will act on a vector $v\in V$ by the rule
$$(a\otimes a')\cdot v=a\cdot (a'\cdot v)$$
as it must for otherwise the action would not match with the multiplication of the tensor algebra. The same rule is applied to longer elementary tensors:
$$(a_1\otimes a_2\otimes\cdots\otimes a_m)\cdot v=a_1\cdot(a_2\cdot(\cdots (a_m\cdot v)\cdots).$$

This leads to a representation of $T(\mathfrak{a})$ by the universal property of the tensor algebra. Any linear transformation $\phi$ from a vector space $W$ to an associative algebra $A$ (both over the same field) gives rise to a unique homomorphism of algebras $T(W)\to A$ extending $\phi$. Here $W$ takes the role of $\mathfrak{a}$ and $\mathfrak{gl}(V)$ plays the part of $A$.

Observe that we don't need $W$ to be a Lie algebra for this to work.

Lie algebra people use the so called universal enveloping algebra instead of the tensor algebra. It is a quotient algebra of $T(\mathfrak{a})$ gotten by imposing the relations $a\otimes a'-a'\otimes a=[a,a']$, so identifying certain elements (="commutators") of $T^2(\mathfrak{a})$ with elements of $\mathfrak{a}$. We also impose all consequences of those relations so that the associative law holds. The way to do that is to mod out the ideal $I$ generated by those relations, and we end up with $U(\mathfrak{a})=T(\mathfrak{a})/I$.

Then, assuming that $\rho:\mathfrak{a}\to\mathfrak{gl}(V)$ is a representation of a Lie algebra, the ideal $I$ is automatically mapped to zero (because all its generators do), and the homomorphism $T(\mathfrak{a})\to \mathfrak{gl}(V)$ we got from the universal property induces a homomorphism of associative algebras $U(\mathfrak{a})\to \mathfrak{gl}(V)$ extending the original representation.


问:
$x^{2}+y^{2}=6xy$ find the value of: $\frac{x+y}{x-y}$
$x$ and $y$ are both positive real numbers :

My attempt:

$$x^{2}+y^{2}=6xy \Leftrightarrow x^{2}+y^{2}-2xy=4xy \Leftrightarrow (x-y)^{2}=4xy \Leftrightarrow (x-y)=\frac{4xy}{x-y} $$
A little note here: $x$ doesn’t equal to $y$.

And now let’s plug this result to $\frac{x+y}{x-y} $:
$$\frac{x+y}{x-y} = (x+y) \cdot \frac{x-y}{4xy}=\frac{x^{2}-y^{2}}{4xy}=\frac{x}{4y}-\frac{y}{4x}$$
And the problem here is to figure out the value of:
$$\frac{x}{y}-\frac{y}{x} $$
Thank you for your help

答:
Hint: Treat the original equation as a quadratic in $x/y$, so you can find that value (2 possible values, corresponding to $\pm$).

Then, express the desired expression in term of $x/y$, and hence find the expression. (Again, there are 2 possible values).


问:
Distribution of the sum of random variables (are those dependent or independent?)
The weight of the citizens in a city is distributed normally with expected value of 70kg and Standard deviation of 6kg. How is the weight of 9 people is distributed?

Attempt:

Let X be the weight of a person in that city. So $X\sim\mathrm{N}\left(70,36\right)$.

Now Let Y=9X, so as Y is the sum of normally distributed variables, it is also distributed normally. then:

$E\left[Y\right]=E\left[9X\right]=9E\left[X\right]=9\cdot70=630$

$Var\left(Y\right)=Var\left(9X\right)=9^{2}Var\left(X\right)=9^{2}\cdot36=2916$

So we conclude that $Y\sim\mathrm{N}\left(630,2916\right)$

On the other side:

My teacher told me that $Var\left(Y\right)=9\cdot36=324$ as we can conclude it from the Central Limit Theorem. I don't understand how is the CLT related here, as it is a theorem about a limit and also its premise is that the variables are independent no?

He added me the following theorem:

Let $X_1,\ldots,X_n$ be a sample out of a distribution with expected value $\mu$ and variance $\sigma^2$. For large $n$ we have that the sum $\sum_{i=1}^n X_i$ is closely distributed as $N(n\mu,n\sigma^2)$

Please help me figure it out

Thanks

答:
No, the random variable $Y$ describing the sum of weights of 9 people is:

$$Y = X_1+X_2+\dots +X_9 \neq 9X_1$$

For $9X_1$, we assume perfect correlation between the 9 random variables. In reality, they are independent. That is the reason your approach produces a much higher variance than your teachers approach. It's all the covariance terms.

Also, the argument in terms of the central limit theorem, it isn't needed in this case since the individual random variables are normally distributed. So, the distribution of the sum of 9 people is exactly $N(9 \mu, 9 \sigma^2)$


答:
Another approach:

What is the numerator and denominator of $ \left( \frac{x+y}{x-y} \right) ^ 2$ in terms of $xy$?

Now, because $x,y$ can be swapped, explain why the expression can take on both positive and negative values.

Title: $x^{2}+y^{2}=6xy$ find the value of: $\frac{x+y}{x-y}$

Body:
$x$ and $y$ are both positive real numbers :

My attempt:

$$x^{2}+y^{2}=6xy \Leftrightarrow x^{2}+y^{2}-2xy=4xy \Leftrightarrow (x-y)^{2}=4xy \Leftrightarrow (x-y)=\frac{4xy}{x-y} $$
A little note here: $x$ doesn’t equal to $y$.

And now let’s plug this result to $\frac{x+y}{x-y} $:
$$\frac{x+y}{x-y} = (x+y) \cdot \frac{x-y}{4xy}=\frac{x^{2}-y^{2}}{4xy}=\frac{x}{4y}-\frac{y}{4x}$$
And the problem here is to figure out the value of:
$$\frac{x}{y}-\frac{y}{x} $$
Thank you for your help


问:
Proof of that the graph of subdifferential is closed when $f$ is closed proper convex function on $\mathbb{R}^n$.
In the Theorem 24.4 of Convex Analysis written by R. Tyrrell Rockafellar, if $f$ is a closed proper convex function on $\mathbb{R}^n$, then the graph of $\partial f$ is a closed subset of $\mathbb{R}^n \times \mathbb{R}^n$.

In the proof of the theorem, the author takes "lim inf" and uses the fact that $f$ and $f^*$ are closed.

I don't fully understand the proof because I'm not familiar with detailed convex analysis. Can anybody explain it in detail for me?

For example, why do we have to take "lim inf" instead of "lim"? Further, how is the closedness of $f$ and $f^*$ used there?

答:
Since $f$ is a closed proper function we know that $f^*$ is a closed proper function.

Then both $f,f^*$ are lsc. and so we have $\lim_i f(x_i) \ge f(x)$,  $\lim_i f^*(x_i^*) \ge f^*(x^*)$.

Applying Theorem 23.5 we get
$\langle x, x^* \rangle = \lim_i\langle x_i, x_i^* \rangle = \liminf_i\langle x_i, x_i^* \rangle =  \liminf_i f(x_i)+f^*(x_i^*) \ge f(x)+f^*(x^*)$
and applying Theorem 23.5 again we get $x^* \in \partial f(x)$.


答:
Proceed as you did:
\begin{equation}
x^2+y^2=6xy\Longrightarrow\left\{\begin{array}{l}
x^2+y^2+2xy=8xy\Longrightarrow(x+y)^2=8xy,\\
x^2+y^2-2xy=4xy\Longrightarrow(x-y)^2=4xy.
\end{array}\right.
\end{equation}
Then divide both right-side equations and you have it.
$$\left(\frac{x+y}{x-y}\right)^2=\frac{8xy}{4xy}=2\Longrightarrow\frac{x+y}{x-y}=\pm\sqrt{2}.$$
Then,
$$x>y\Longrightarrow\frac{x+y}{x-y}=\sqrt{2},\quad x<y\Longrightarrow\frac{x+y}{x-y}=-\sqrt{2}.$$


Body:
hint
$$x^2+y^2+2xy=6xy+2xy$$

$$\implies (x+y)^2=8xy$$
$$\implies (x+y)=\pm 2\sqrt{2xy}$$
and
$$x^2+y^2-2xy=6xy-2xy$$
$$\implies (x-y)^2=4xy$$


答:
Note that $$(x+y)^2-(x-y)^2=4xy$$

Use this in your equation
$$(x-y)^2=4xy$$
$$(x-y)^2=(x+y)^2-(x-y)^2$$
$$\Rightarrow \dfrac{(x+y)^2}{(x-y)^2}=2$$
$$\Rightarrow \dfrac{x+y}{x-y}=\pm \sqrt{2}$$

Here both positive and negative values are possible depending on whether $x$ is larger or smaller than $y$.


答:
The equation
$$Ax^2 + Bxy+Cy^2 +Dx + Ey + F = 0$$
is:

an ellipse if $B^2 - 4AC < 0$ (could be circle or a point if $A = C$, $F > 0$ and $B = 0$)

a parabola if $B^2 -4AC = 0$

a hyperbola if $B^2 - 4AC > 0$

In this case,  $B^2-4AC = -4 <0$. Hence it is an ellipse.

Title: Finding the locus of $b^2-2x^2=2xy+y^2$

Body:
I was asked to determine the locus of the equation
$$b^2-2x^2=2xy+y^2$$

This is my work:

Add $x^2$ to both sides:
$$\begin{align}
b^2-x^2 &=2xy+y^2+x^2\\
 b^2-x^2 &=\left(x+y\right)^2
\end{align}$$

I see that this is similar to the equation of a circle. How can I find the locus of this expression?


问:
Introductory text on voting theory
Some years ago I read about a famous theorem by K. Arrow on rank-order voting systems. My interest in voting theory topic has been spurred by the use of MeekSTV in our moderator elections.

As such, I'd like to know if there is some nice, introductory text on voting theory where I could learn more about this subject.

答: Here is an elementary text: Mathematics and Politics: Strategy, Voting, Power, and Proof. </p>

Title: Introductory text on voting theory


问:
$\mathbb Z[\sqrt{13}]$ is not a UFD.
I have seen complicated arguments to establish whether $\mathbb Z$ extended by this or that radical is a UFD.  They usually suppose that $u$ is a unit and then compute the reciprocal and then make arguments about each component being an integer.  Some of them go by defining a norm and proving things about that.  I'm a little confused about why these arguments are so elaborate when it seems to me that there is a very straight-forward argument in many cases.  My guess is that I'm overlooking some important caveat in one of my steps but I'm not sure where.

Take for example $\mathbb Z[\sqrt{7}]$.  To prove it's not a UFD I look at two factorizations of 6:

$$ 6=3\cdot 2 $$

$$ 6 = (\sqrt 7 - 1)(\sqrt 7 + 1) $$

I then argue that there is no unit (in fact no element at all) such that $3u=\sqrt 7-1$.  Since $u=a+b\sqrt 7$ then we would need

$$ 3a = -1 $$

but this is not true for any $a\in\mathbb Z$.  Of course the same argument could be used on any of $\pm 3,\pm 2$ on the left and $\pm (\sqrt 7-1),\pm (\sqrt 7+1)$ on the right.

Is there some logical error in saying that if $w+x\sqrt 7 = y+z\sqrt 7$ for integers $w,x,y,z$ then we must have $w=y$ and $x=z$?

答:
There is a counter example. Since $4=2 \cdot 2 = (\sqrt{13}+3)(\sqrt{13}-3)$, it suffices to show those factors are irreducible. Suppose to the contrary that $2$ is not irreducible. It follows that there exists $a,b \in \mathbb{Z}[\sqrt{13}]$ such that $2=ab$, $N(a) \ne 1$ and $N(b) \ne 1$. It boils down to show that there are no solutions to $u^2=13v^2 \pm 2$ in $\mathbb{Z}$. Observe that $2 \nmid \sqrt{13} \pm 3$ and we are done.

Someone asked the same about $\mathbb{Z}[\sqrt{8}]$. You might find Is $\mathbb{Z}[{ \sqrt 8 } ] $ a Euclidean domain? relevant.

Title: $\mathbb Z[\sqrt{13}]$ is not a UFD.


问:
Solve integral $\int\frac{1}{(x^2-1)\sqrt{x^2+1}}dx$
$$\int\frac{1}{(x^2-1)\sqrt{x^2+1}}dx$$
I'm trying to solve this integral. First I substituted : $x=\tan(t)$; $t=\arctan(x)$

Then $$ dx=\frac{1}{\cos^2(t)}\,dt$$

Now by simplifying I'm to this step $$ \int\frac{\cos(t)}{\sin^2(t)-\cos^2(t)}\,dt$$

What can i do now ..

Thank you in advance :)

答:
Substitute  $t= \frac{\sqrt2 x}{\sqrt{x^2+1}}$

\begin{align}
I=& \int\frac{1}{(x^2-1)\sqrt{x^2+1}}dx\\
= &\frac1{\sqrt2} \int \frac{1}{t^2-1} dt=\frac1{2\sqrt2} \ln|\frac{t-1}{t+1}|
= \frac1{2\sqrt2} \ln|\frac{\sqrt2 x-\sqrt{x^2+1}}{\sqrt2 x+\sqrt{x^2+1}}|
\end{align}
Alternatively
$I=-\frac1{2\sqrt2}\coth^{-1}\frac{3x+\frac1x}{2\sqrt2 \sqrt{x^2+1}}
$


问:
Different Representations of Dunce Cap
I am having a bit of confusion concerning the Dunce Cap while studying simplicial homology, hope someone can help!

Given a solid triangle with vertices $a, b, c$ I usually see the Dunce Cap defined as the quotient on the edges as follows: $[a,b], [a,c]$ and $[b,c]$ are identified, using that orientation on the edges.  However, the "n-fold Dunce Cap" is usually defined where all the edges are identified like a cycle:

My understanding is that the $n$-fold Dunce Cap has $\mathbb{Z}/n\mathbb{Z} = H_1$.  But that doesn't make sense to me in the usual case, since the classical Dunce Cap is known to be contractible so should have trivial $H_1$ - which means it doesn't make sense in the case $n = 3$.

Can someone tell me what I'm messing up?  Are those two identification schemes on the triangle the same space or not?  And is it accurate that $\mathbb{Z}/n\mathbb{Z} = H_1$ for the $n$-fold Dunce Cap?

In general, does it matter what order we apply to the edges at the start for any $n$?  Or as long as we're identifying all edges/vertices of some $n$-gon will we always get the same thing?

答:
Everything you say looks right to me. In both of the spaces you've described, a cell structure exists with one 2-cell, one 1-cell, and one 0-cell.

In the classical dunce cap, the boundary of the 2-cell is $1+1-1=1$ times the generator of $C_1$, so the first homology vanishes.

In the 3-fold dunce cap, the boundary of the 2-cell is $1+1+1=3$ times the generator of $C_1$, so the first homology is $\mathbb{Z}/3\mathbb{Z}$.

Evidently these are different spaces.

Title: Different Representations of Dunce Cap


问:
What is this function called?
What is this function called?
Here is a graph:

```<code>          _______
         /
        /
_______/
</code>```

Sort of like:

$f(x, a, b) = 0$, if $x < a; \quad \quad \displaystyle \frac {(x - a)} {(b - a)}$ if $a < x  b$.

Closest name is Sigmoid function, but they all are smooth. Also I remember I saw similar formula for saturation, but not sure about that, googling didn't help.

Edit: Ramp function is closer:

But it is general one, while my is more specific one, used for normalization.

答: I call it a ramp function.  At least some others do.</p>

Title: What is this function called?


问:
A special curve $\left( \frac { x^{ 2 }+1 }{ 1-x^{ 2 } } \right)$.
Let $a,x\in \mathbb{Q}$ and $n\in\mathbb{N}$

For $0<x<1$, $a>0$ and $n>1$, show that

$$\ \frac { x^{ 2 }+1 }{ 1-x^{ 2 } }   \neq { a }^{ n }$$.

For example, for $x = 0.3$, then
$\frac {109 }{ 91 }$ can not be shown as $a^2$ or $a^3$ etc.

答:
Assume on the contrary that $\dfrac{1+x^2}{1-x^2}=a^n$ for some $x, a \in \mathbb{Q}, n \in \mathbb{N}, 0<x<1, a>0, n>1$. Write $x=\frac{s}{t}, s<t, s, t \in \mathbb{Z}^+, \gcd(s, t)=1$, so $\dfrac{t^2+s^2}{t^2-s^2}=a^n$

We consider 2 cases:

Case 1: $s, t$ have different parities.

Then $\gcd(t^2+s^2, t^2-s^2)=1$, so we have $t^2+s^2=b^n, t^2-s^2=c^n, a=\frac{b}{c}$ for some $b, c \in \mathbb{Z}^+$. Thus $c^n=(t-s)(t+s)$, so since $\gcd(t-s, t+s)=1$, we have $t-s=d^n, t+s=e^n, c=de$ for some $d, e \in \mathbb{Z}^+$. Thus $d^{2n}+e^{2n}=(t-s)^2+(t+s)^2=2(t^2+s^2)=2b^n$.

If $n \geq 3$, by results of Darmon and Merel, we have no non-trivial primitive solutions to the equation $x^n+y^n=2z^n$ (where trivial means $|xyz| \leq 1$) Clearly $\gcd(d, e)=1$, so $(x, y, z)=(d^2, e^2, b)$ is a primitive solution to the Fermat-like equation $x^n+y^n=2z^n$. It must thus be trivial, so $|d^2e^2b| \leq 1$, so $d=e=b=1$ since $d, e, b$ are positive integers. This implies that $2s=e^n-d^n=0$, a contradiction.

Therefore $n=2$. Now $t^2+s^2=b^2, t^2-s^2=c^2$, so $t^4-s^4=(t^2-s^2)(t^2+s^2)=(bc)^2$. We proceed via infinite descent (by comparing the size of $b$) to show that there are no positive integer solutions $(t, s, b, c)$ satisfying both equations simultaneously, where $t, s$ are of different parities and $\gcd(t, s)=1$.

Note that if $t^2=s^2+c^2 \not \equiv 0 \pmod{4}$, so $t$ cannot be even, so $s$ is even. Using the parameterization of primitive Pythagorean triplets we get $t^2=k^2+l^2, s=2kl, c^2=k^2-l^2$, where $k, l$ are of different parities and $\gcd(k, l)=1$. Then $(k, l, t, c)$ is also a solution of the 2 equations, and we have $t<b$. We have thus obtained a smaller positive integer solution, and this leads (via infinite descent) to a contradiction.

Case 2: $s, t$ are of the same parity.

Since $\gcd(s, t)=1$, $s, t$ are both odd. We have $\gcd(t^2+s^2, t^2-s^2)=2$, so $t^2+s^2=2b^n, t^2-s^2=2c^n$ for some $b, c \in \mathbb{Z}^+$. If $n$ is even, then we may take $n$ to be $2$, since we can replace $b, c$ with $b^{\frac{n}{2}}, c^{\frac{n}{2}}$ respectively. We thus have $t^2+s^2=2b^2, t^2-s^2=2c^2$, so $t^2=b^2+c^2, s^2=b^2-c^2$. Note that $\gcd(b, c)=1$ and $b, c$ are of different parities (since $s, t$ are odd), so applying the result from Case 1 (With a variable change $(t, s, b, c) \to (b, c, t, s)$), we get a contradiction.

Therefore $n$ is odd. Thus $2b^n=(t^2-s^2)+2s^2=2c^n+2s^2$, so $b^n+(-c)^n=s^2$. If $n \geq 5$, then again by results of Darmon and Merel, we have no non-trivial primitive solutions to the equation $x^n+y^n=z^2$. Clearly $\gcd(b, c)=1$, so $(x, y, z)=(b, -c, s)$ is a primitive solution to the equation $x^n+y^n=z^2$. It must thus be trivial, so $|bcs| \leq 1$, so $b=c=s=1$ since $b, c, s$ are positive integers. However this implies that $t^2=s^2+2c^n=3$, a contradiction.

Therefore $n=3$, and we have $t^2+s^2=2b^3, t^2-s^2=2c^3$. We have $2c^3=(t-s)(t+s)$, so we have $\{(t-s), (t+s)\}=\{2d^3, e^3\}$ for some $d, e \in \mathbb{Z}^+$. Notice that it does not matter which of $t+s, t-s$ is $2d^3$ and which is $e^3$, since we will only consider $4b^3=2(s^2+t^2)=(t-s)^2+(t+s)^2=4d^6+e^6$. Note that $\gcd(t-s, t+s)=2$, so $e$ is even and $d$ is odd. This also implies that $\gcd(d, e)=1$.

Write $e=2f, f \in \mathbb{Z}^+, \gcd(d, f)=1$. Then $4b^3=4d^6+64f^6$, so $b^3=d^6+16f^6$, so $b^3+(-d^2)^3=2(2f^2)^3$. Since $\gcd(-d^2, 2f^2)=1$ and $|2bd^2f^2|>1$, $(x, y, z)=(b, -d^2, 2f^2)$ is a non-trivial primitive solution to the equation $x^3+y^3=2z^2$, which again by results of Darmon and Merel (as in Case 1) leads to a contradiction.

Conclusion: We have obtained a contradiction in both cases, so the equation $\dfrac{1+x^2}{1-x^2}=a^n$ has no solutions for $x, a \in \mathbb{Q}, n \in \mathbb{N}, 0<x<1, a>0, n>1$.


答:
$$\int\frac{\cos t}{\sin^2 t-\cos^2 t}dt$$

Use $\cos^2 t = 1-\sin^2 t$ to modify the denominator
$$\int\frac{\cos t}{2\sin^2 t-1}dt$$
Substitute $u=\sin t$, $du =\cos t \, dt$
$$\int\frac{1}{2u^2-1}du$$
Now integrate using partial fractions or otherwise.

Title: Solve integral $\int\frac{1}{(x^2-1)\sqrt{x^2+1}}dx$

Body:
$$\int\frac{1}{(x^2-1)\sqrt{x^2+1}}dx$$
I'm trying to solve this integral. First I substituted : $x=\tan(t)$; $t=\arctan(x)$

Then $$ dx=\frac{1}{\cos^2(t)}\,dt$$

Now by simplifying I'm to this step $$ \int\frac{\cos(t)}{\sin^2(t)-\cos^2(t)}\,dt$$

What can i do now ..

Thank you in advance :)


问:
Are all morphisms in category theory are actually some kind of mappings? What is the importance of associativity rule in category theory?
I always found the identity and the composition rule trivial while thinking about categories. But I think the associativity rule is NOT trivial and it is hard to create a composition table that obeys the associativity rule. Do you think there is something about the associativity rule that makes it very special and useful in mathematics, or am I just confused? I know we can create an associative composition table using mappings between sets and using permutations. So can we say all morphisms in category theory are actually some kind of mappings?

For example, this is a poset with 3 objects. And the morphisms are not mappings, they represent some relation between these 3 objects.

But I could turn the objects into specific sets, and turn the morphisms into specific mappings between those sets. So that the composition table stays the same and arrows are still associative. (I don't have to add every possible mapping as arrows, I just added one of them.)

So is it possible to turn every category into a combination of some sets and mappings between those sets?

Note: I am very new to category theory and abstract algebra, sorry if this question is confusing.

答:
Thanks to @Berci, I found out that every small category is isomorphic to some subcategory of Set. This is explained by Cayley's Theorem and Yoneda Lemma, which I now understand are very fundamental subjects of category theory. This means every object in a small category can be represented as a set, and every morphism can be represented as a mapping.

Cayley's Theorem: https://proofwiki.org/wiki/Cayley%27s_Theorem_(Category_Theory)#Theorem

Yoneda Lemma:
https://en.wikipedia.org/wiki/Yoneda_lemma

Thanks to everybody who answered my question. I can edit my answer if I got something wrong. I will try to learn more about Cayley's Theorem, Yoneda Lemma, large categories, and why we need them.


问:
Well ordered set that isn't topologically complete
Does there exist an well ordered set $(A,<)$ that isn't topologically complete, i.e. that for every $$B\subseteq A,~ B\not = \emptyset$$
such that $B$ is bounded above with some $c\in A$, $B$ has supremum in $A$.

$\mathbb N$ is obviously complete since it is locally finite. And ideas such as $\mathbb N\times \{0,1\}$ or $\left(\mathbb N\times\{0\}\cup  \mathbb Z\times\{1\}\right)$ dont work either.

答: No, there is no such well-ordered set. Suppose $(A, \le)$ is well-ordered, and $\emptyset \ne B \subseteq A$ with $B$ bounded above by some element of $A$. Then the set of upper bounds of $B$ is a nonempty subset of $A$, and so has a least element $b_0$. It's easy to see that $b_0 = sup(A)$.


答: Actually i think i proved it. Given $B \subseteq A$ defined as above, we have some $c \in A$ such that $c \ge b, \forall b \in B$. Now we know that set of all upper limits of $B$ is non-empty, therefore, it contains the minimum $m$. It is easily seen that $m$ is the supremum of $B$ since it is the smallest upper limit of $B$.

Title: Well ordered set that isn't topologically complete

Body:
Does there exist an well ordered set $(A,<)$ that isn't topologically complete, i.e. that for every $$B\subseteq A,~ B\not = \emptyset$$
such that $B$ is bounded above with some $c\in A$, $B$ has supremum in $A$.

$\mathbb N$ is obviously complete since it is locally finite. And ideas such as $\mathbb N\times \{0,1\}$ or $\left(\mathbb N\times\{0\}\cup  \mathbb Z\times\{1\}\right)$ dont work either.


答:
Rephrasing your definition of topological completeness to note that it's exactly the same as the least upper bound property, we note that any well-order has that property.

This has the standard proof: let $C$ be the set of upper bounds of $B$. It's non-empty by definition, and so by the well-order property has a least element.

This least element is the supremum of $B$.


问:
$A=\{a_{\lambda_1}+a_{\lambda_2}+...+a_{\lambda_k}| k\in\mathbb N, \lambda_1,...,\lambda_k\in\mathbb R \;\text{are all distinct}\}$ is not bounded
For any real number $\lambda\in\mathbb R$ choose a positive real number $a_\lambda>0$.
Show that the subset $A$ below is unbounded.
$$A=\{a_{\lambda_1}+a_{\lambda_2}+...+a_{\lambda_k}| k\in\mathbb N, \lambda_1,...,\lambda_k\in\mathbb R \;\text{are all distinct}\}$$

And consider the case when $\lambda$s are from integers.

Here is an answer A boundedness question.

By hypothesis,

$$\begin{array}{l|rcl}
b : & \mathbb R & \longrightarrow & \mathbb R \\
    & \lambda & \longmapsto & b_\lambda \end{array}$$ is supposed to be a one-to-one map. Without loss of generality by changing $b$ into $-b$, we can suppose that $b(\mathbb R) \cap (0,\infty)$ is uncountable. Therefore, one of the sets $B_n = b(\mathbb R) \cap (1/n, \infty)$ is uncountable  (a union of countably many countable sets is countable). Taking $k$ distinct elements of $B_n$ we have $\sum_{i=1}^k b_{\lambda_i} \ge k/n$ proving that $B$ in unbounded.

The result doesn't hold if the $\lambda$s belong to $\mathbb N$. Take for example $b_n = 1/n^2$. As $\sum 1/n^2$ converges, every partial sum is also finite.

but I am not satisfied.

Why we write -b into b? What is the point?

It is obvious that $b(\mathbb R)\cap (1/n,\infty)$ is uncountable but $a_{\lambda_i}s$ ($a_{\lambda_1}+a_{\lambda_2}+...+a_{\lambda_k}$) are not supposed to be inside $(1/n,\infty)$
I understand why we take $(1/n,\infty)$ because we want an lower bound that can be go to the infinity. But it is not good with me.

And for the second consideration with $\lambda$s from $\mathbb Z$ I didnot quite understand why the $1/n^2,1/n^3$ sums are working?

答:
There are not any new ideas in my answer, but I thought the below might help. I will write $a_\lambda$ as $a(\lambda)$ and use standard function notation.

There is nothing in the hypotheses that implies the map $a:\mathbb R \to (0,\infty)$ is $1-1.$ For example, you could simply define $a(\lambda) = 1$ for all $\lambda.$ (It's the $\lambda$'s that are distinct, not the $a(\lambda)$'s.) So the "proof" in 1. is a nonproof as far as I can see.

All we need to assume is that $a:\mathbb R \to (0,\infty).$ The desired result will follow.

Proof: For $n=1,2,\dots$ let $E_n=a(\mathbb R)\cap (1/n,\infty).$ Then $\mathbb R = \cup_n a^{-1}(E_n).$ Since $\mathbb R$ is uncountable, $a^{-1}(E_n)$ must be infinite for some $n,$ say $n_0.$ Now every infinite set contains a countably infinite set, so there exist countably many distinct $y_1,y_2,\dots \in a^{-1}(E_{n_0}).$ We then have, for each $k,$

$$\tag 1 \sum_{j=1}^{k}a(y_j) > k\cdot \frac{1}{n_0}$$

As $k\to \infty,$ the right side of $(1)\to \infty,$ showing the set $A$ is unbounded.

To show the result need not hold for $a:\mathbb Z\to (0,\infty), $  define $a(n) =2^{-n}$ for $n\in \mathbb N,$ $a(n)=0$ for $n<1.$ Suppose $n_1,\dots, n_k\in \mathbb Z$ are distict. Then

$$\sum_{j=1}^{k}a(n_j) \le \sum_{n=1}^{\infty}2^{-n} =1.$$

Thus the set $A$ in this case is bounded above by $1.$


Body:
Other approach

Put $ x=\sinh(t) $ with

$$dx=\cosh(t)dt$$

$$\sqrt{x^2+1}=\sqrt{\sinh^2(t)+1}=\cosh(t)$$

$$x^2-1=(\sinh^2(t)-1)$$

It becomes
$$\int \frac{dt}{(\sinh(t)+1)(\sinh(t)-1)}$$

$$=\frac 12\int (\frac{1}{\sinh(t)-1}-\frac{1}{\sinh(t)+1})dt$$


问:
Differential equation where forcing function depends on variable
The force acting on a body depends on the displacement $x(t)$ of the body from the origin.
This dependence is given by

$$ F(x) =
\begin{cases}
F_0 - mg, &-L < &x < 0,\\
-mg,& &x > 0.
\end{cases}
$$

where $L$ is a positive number, $F_0 > 0$ is a constant resting force and $g$ is the gravitational acceleration of Earth.

Newton's second law gives $m\ddot{x} = F(x)$ and we impose the initial conditions $\dot{x}(0) = 0, x(0) = -L.$

How does one solve an equation like this? My main problem is that if we restrict to the branch $x > 0$ we get $x = -g\frac{t^2}{2} + At + B$. I don't know if I am allowed to plug in the boundary conditions here, but if I were to, I would get $x = -g\frac{t^2}{2} - L$. But this function is clearly always negative, so then how could this satisfy the branch $x>0$?

Any tips or clarifications?

答: You just have to solve two ODE's. First solve the ODE for $x\in[-L,0)$ using your initial conditions $\dot{x}(0)=0$, $x(0)=-L$. That equation is valid until $x=0$. Then you can get values of $\dot{x}$ for when $x=0$. That is your new initial condition to use when solving the second ODE.


问:
Distance to a vector subspace
Let E be the set of real sequences $(x_n)_{n\in N}$ such that the series $\sum u_{n}^2$ converges.

For $x=(x_n)_{n \in N} $and $y=(y_n)_{n \in N}$

Inner product :

(x|y) = $\sum_{n=0}^{+\infty} x_ny_n$

Let F be the vector subspace of E formed from zero sequences starting from a certain rank.

If $ z=(\frac{1}{n+1})_{n\in N} $, show that $z \in E $and calculate $d(z,F)$

Showing that $z\in E$ is easy with Riemann. I tried to find an orthogonal base of F to calculate $d(z,F)$ but I couldn’t find one. Is there another way to do the calculus of a distance without using  an orthogonal projection? Here the spaces considered have infinite dimension so there has to be another way !

答:
Hint

The given space $E$ is the Hilbert space $\ell^2(\mathbb R)$. The space $F$ of eventually vanishing sequences is usually denoted by $c_{00}$.

$c_{00}$ is dense in $\ell^2(\mathbb R)$. This is quite easy to prove. Hence $d(x,F)=0$ for any $x \in E$.

Title: Distance to a vector subspace


问:
Compute the integral $\int_{\delta_{B_1(-2)}} \frac{1}{z(z+2)^3}dz$
Could you please help me to solve this integral. I have no idea on how to proceed. Thank you.

I think, i should rewrite $\frac{1}{z(z+2)^3}$ as  a sum from a geometric series and then use the Cauchy integral formula, but I don´t see a trick..

$\int_{\delta_{B_1(-2)}} \frac{1}{z(z+2)^3}dz=\oint\frac{\frac{1}{z}}{(z+2)^3}dz$

Then

$\oint\frac{\frac{1}{z}}{(z+2)^3}dz=\left(\frac{2\pi i }{2!}\right)\left[\frac{d^2f(z)}{dz^2}\right]_{z=-2}$

$=\left(\frac{2\pi i }{2!}\right)\left[\frac{d^2}{dz^2}(\frac{1}{z})\right]_{z=-2}$

$=\left({\pi i}\right)\left[{\frac{2}{(z)^3}}\right]_{z=-2}$

$=\color{blue}{-\frac{1\pi i}{4}}$

Is that correct?

Solution (geometric progression)

$\int_{\delta_{B_1(-2)}} \frac{1}{z(z+2)^3}dz$

$\frac{1}{z(z+2)^3} = 2*(1-(1+\frac{z}{2}))^{-1}*\frac{1}{z(z+2)^3}$

$=\frac{2}{z(z+2)^3}*\frac{1}{1-(1+\frac{z}{2})}$

We know, that

$\frac{1}{1-(1+\frac{z}{2})}=\sum_{n=0}^∞ {(1+\frac{z}{2})^n}=1+(1+\frac{z}{2})^1+(1+\frac{z}{2})^2+(1+\frac{z}{2})^3+ ...)$

So we get

$\frac{2}{z(z+2)^3}*\frac{1}{1-(1+\frac{z}{2})}=\frac{2}{z(z+2)^3}*(1+(1+\frac{z}{2})^1+(1+\frac{z}{2})^2+(1+\frac{z}{2})^3+ ...)=$

$=\frac{4}{(z+2)^3}+\frac{z}{(z+2)^3}+\frac{1/2}{(z+2)}+\frac{1}{4}+\frac{2+z}{8}+...$

=>
$f(z)=1/2, z=-2
=>
2\pi if(z)=2\pi i*1/2=\pi i$

In this solution I have $\color{blue}{\pi i}$

And in the top solution I have another answer $\color{blue}{-\frac{\pi i}{4}}$.

What's my mistake?

答: The only enclosed pole is a third-order one at $z=-2$, so the residue is $\tfrac12\lim_{z\to-2}\tfrac{d^2}{dz^2}[(z-2)^3f(z)]$. I leave you to evaluate this, then multiply by $2i\pi$.


问:
Why combining two quadratic equations of a circle and a parabola creates extra solutions for $x$
We have a parabola and a circle with the following equations and their graph placed at the end of my question.

Parabola: $y^2 = 4x -4$

Circle: $(x-2)^2 + y^2 = 9$

My goal was to calculate their intersection points so I substituted $y^2$ from the parabola equation into the circle equation and I got

$(x-2)^2 + (4x-4)=9 \implies x^2 - 4x + 4 + (4x - 4) = 9 \implies x^2 = 9 \implies x = \pm3$

$x=3$ is the only correct solution but why is $x=-3$ produced as an extra invalid solution?

What is the exact mathematical explanation behind this? Why substituting one equation into the other has produced extra answers?

update

When I calculate $x$ from the parabola equation and substitute it in the circle equation, I don't get any extra answers for $y$:

$y^2=4x-4 \implies y^2 +4 = 4x \implies x = \frac{y^2}{4} + 1$

$(x-2)^2 +(4x-4)=9 \implies ((\frac{y^2}{4} + 1) - 2)^2 + (4x - 4)=9 \implies y^4 +8y - 128 = 0 \implies y^2=8,-16$

$y^2 = -16$ cannot be true so $y^2 = 8 \implies y=\pm 2\sqrt{2}$ and these are correct answers for $y$.

2nd update

I made a mistake in the calculation in the previous update although the final solutions where correct. I write the correct calculation:

$(x-2)^2 +y^2=9 \implies
((\frac{y^2}{4} + 1) - 2)^2 + y^2=9 \implies
(\frac{y^2}{4} - 1)^2 + y^2=9 \implies
(\frac{y^4}{16} - \frac{y^2}{2} + 1) + y^2=9 \implies
\frac{y^4}{16} + \frac{y^2}{2} + 1=9 \implies
(\frac{y^2}{4} +  1)^2=9 \implies
(\frac{y^2}{4} +  1)=\pm3 \implies
\frac{y^2}{4} =2,-4 \implies
y^2=8,-16$

答:
Total there are 4 roots, two real roots you have given are ok.
The remaining two with $x=-3$ substitution give imaginary $ y= \pm 4i$ and they should be discarded as extraneous, spurious etc. when looking for real solutions.

The above  $x,y$ parts of a point should both be real for a full real solution.

$(x,y)= (-3,4i)$ are admissible as complex roots.

Had you computed $y$ values instead of $x,$ you would have noticed different imaginary values , with different signs as above.


问:
Find all homomorphisms from $\mathbb{Q}[\sqrt[4]{2},i]$ to the algebraic closure of $\mathbb{Q}$ and its images
I know, that $\mathbb{Q}[\sqrt[4]{2},i]$ has dimension $8$ as a vector space, that means the image, under any homomorphism, should have dimension $\leq8$.
And since $\mathbb{Q}[\sqrt[4]{2},i]\subset$ algebraic closure$(\mathbb{Q})$, I believe that the possible images are all the subfields of $\mathbb{Q}[\sqrt[4]{2},i]$. Is my idea correct?

答:
$\mathbb{Q}[\sqrt[4]{2},i]$ is the splitting field of $X^4-2$ over $\mathbb{Q}$. It is easy to check that it factorises as
$$
(X-\sqrt[4]{2})(X-i\sqrt[4]{2})(X+\sqrt[4]{2})(X+i\sqrt[4]{2})
$$

Over $\mathbb{Q}$, splitting fields are Galois extensions, which mean that they have some nice properties: the Galois group has the same order of the extension degree, which is $8$ in this case. So you are looking for $8$ automorphisms of $\mathbb{Q}[\sqrt[4]{2},i]$. These are generated by

$$
\begin{array}{crcl}
\sigma:&i&\to&-i \\
\tau:&\sqrt[4]{2}&\to&i\sqrt[4]{2}\\
\end{array}
$$

So the automorphisms you are looking for are
$$
\{\text{id},\sigma,\tau,\tau^2,\tau^3,\sigma\tau,\sigma\tau^2,\sigma\tau^3\}
$$


问:
About compact operator
When seeing a proof of Fredholm's alternative I don't get the following:

Let $T$ be compact from a Banach space $X$ to itself, and $\lambda \neq 0$. Define $S=I-T$, $S^k$ its $k$-th power and $N_k=$ker$(S^k)$.

Is it true that $S(N_{k-1})=N_{k}$? One of the inclusions is clear, but the other one isn't, because $S$ is not necessarily onto (is it?). And in the proof it seems to me that we need that to happen.

Thank you.

答:
Even in the finite-dimensional case and for $k=2$ it is not necessarily true that $S(\ker S^k) = \ker S^{k-1}$: $\text{ran}(S) \cap \ker S^K$ might be $\{0\}$, e.g. for
$$ S = \pmatrix{1 & 0\cr
                0 & 0\cr}$$</p>

Title: About compact operator


问:
Proof of Gallai Theorem for factor critical graphs
I am trying to find the proof of the following theorem:

Definition 1.2. A vertex $v$ is essential if every maximum matching of $G$ covers $v$ (or $\nu(G-v) = \nu(G)-1$). It is avoidable if some maximum matching of $G$ exposes $v$ (or $\nu(G-v) = \nu(G)$). A graph $G$ is factor-critical if $G-v$ has a perfect matching for any $v \in V(G)$.

Theorem 1.3. Let $G$ be a connected graph. Then $G$ is factor-critical if and only if every vertex is avoidable.

I don't think this is something I can prove on my own. However, I can't seem to find the proof of it online either.

It makes sense intuitively. If for every vertex $v$, in graph $G$, there exists a perfect matching that doesn't include $v$, then removing $v$ and its incident edges would result in a perfect matching because one existed that didn't include $v$ in the first place.

Could someone provide a more formal proof for this theorem, or provide a link to where I might find the proof?

Thank You

答:
This can be proved quickly using the Tutte-Berge formula. For a subset of vertices $U \subseteq V$, let $q(U)$ denote the number of odd components of $G - U$. The Tutte-Berge formula states, in effect, that there is some $U \subseteq V$ such that a maximum size matching of $G$ avoids exactly $q(U) - |U|$ vertices. Now it is easy to see that any matching of $G$ avoids at least $q(U) - |U|$ vertices of $G-U$: from each of the $q(U)$ odd components at least one vertex must be either unmatched or matched to a vertex in $U$, and the latter can happen for at most $|U|$ vertices.
It follows that the vertices in $U$ are not avoidable, so by assumption $U$ must be empty. Then, since $G$ is connected, $q(U) - |U| = q(U)$ is either $0$ or $1$. The former would mean that $G$ has a perfect matching, so that none of the vertices are avoidable, a contradiction. So every maximum matching avoids exactly one vertex, which, together with the assumption that every vertex is avoidable, implies that $G$ is factor-critical.

In general, the behaviour of the set of avoidable vertices is described by the so-called Edmonds-Gallai decomposition of the graph, see e.g. Theorem 2 of these lecture notes. Gallai's Theorem immediately follows from this decomposition.


问:
On the Gaussian Poincare inequality
Let $X$ be a standard normal random variable. Then, for any differentiable $f:\mathbb{R}\to\mathbb{R}$ such that $\mathbb{E}f(X)^2<\infty,$ the Gaussian Poincare inequality states that

$$\mathrm{Var}(f(X))\leq \mathbb{E}[f^\prime(X)^2].$$

Suppose this inequality is proved for all functions that are twice continuously differentiable with compact support. Can you please tell me the precise argument that allows one to extend this to all differentiable functions $f$ with $\mathbb{E}f(X)^2<\infty$?

答:
Let $f$ be absolutely continuous with $E[f'(X)^2] < +\infty$.

Assume that $f_n$ is a sequence of functions, infinitely differentiable with compact support, such that
$$E[(f_n(X) - f(x))^2] + E[(f'(X) - f_n(X))^2] \to 0.$$
Then by the triangle inequality for $E[(\cdot)^2]^{1/2}$, we have $E[f_n^2(x)]\to E[f^2(X)]$ and $E[f_n'(X)^2]\to E[f'(X)^2]$, as well as $E[f_n(X)]\to E[f(X)]$ by Jensen's inequality by a similar argument. This implies $Var[f_n(X)]\to Var[f(X)]$, so all we have to do is find a sequence $f_n$ such that the above display holds.

The Sobolev space $W^{1,2}(\gamma_p)$ of weakly differentiable functions $f$ with $E[f(g)^2 + \|\nabla f(g)\|^2]<+\infty$ for $g\sim N(0,I_p)$ is actually well studied, see for instance section 1.5 in the book Gaussian Measures by Bogachev.
In particular this space is equal to the completion of the set of infinitely differentiable functions with compact support with respect to the norm
$\|f\|_{2,1} = E[f(g)^2]^{1/2} + E[\|\nabla f(g)\|^2]^{1/2}$. With this definition, set of infinitely differentiable with compact support is dense and for any $f\in W^{1,2}(\gamma_p)$ there exists a sequence of infinitely differentiable with compact support $f_n$ such that $\|f_n - f\|_{2,1} \to 0$. The case $p=1$ is of interest for the question, but the same strategy would work verbatim for $p>1$ and the Gaussian Poincare inequality $Var[f(g)] \le E[ \|\nabla f(g)\|^2]$.

Title: On the Gaussian Poincare inequality


问:
Cusps on quadratic Bezier curve
A quadratic Bezier curve is given in parametric form by:
$$C(t) = (1-t)^2P_0 + 2(1-t)tP_1 + t^2P_2.$$

My points are: $(1,1)$, $(2,2)$ and $(3,3)$. How do I show that this curve has cusps?

Best regards,
Sergey

答:
Compute curvature with standard parametric form. Zero radius of curvature gives a cusp.

Primed with respect to $t$,

$$\kappa= \dfrac{(x'y''-y'x'')}{ (x'^2+y'^2)^{\frac32} }\to \infty $$

At a cusp slope $\dfrac{y'}{ x'} $ remains stationary or maximum/minimum.

Title: Cusps on quadratic Bezier curve


问:
Number of solutions of the equation $a-z-e^{-z}=0$.
I am trying to calculate the number of solutions of the equation $a-z-e^{-z}=0$, with $a\in \mathbb{R}, a>1,$ that are in the semiplane $\{z\in\mathbb{C}:Re(z)>0\}$.

I have tried using Rouché's Theorem and the Argument Principle but the fact that the function in the equation is not a polynomial is making it difficult for me to apply these results.

It would be very helpful if anyone could give me a hint as to how to apply these theorems in this case or point out a different approach in case using the results I have thought is not viable.

Thanks in advance.

答:
Fix $R>a+1$ and let $\Gamma$ be the contour consisting of the semicircle in the half-plane of radius $R$ and the straight line $[-iR, iR].$ We now apply the Rouche's theorem with $f(z)=a-z$ and $g(z)=e^{-z}.$

First observe that $|g(z)|\le 1$ on $\Gamma.$ Check that $|f(z)|>1$ on $\Gamma.$ It follows that $a-z-e^{-z}$ has the same number of zeroes in the Half-plane as $a-z.$ (In fact it also tells us that the zero is contained in $\{|z|\le a+1\}\cap \{Re(z)>0\}.$) Note that $z-a$ has exactly one zero in the half-plane.

Also as an aside note that $h(z)=a-z-e^{-z}$ is real on the real line and $h(0)>0$ and $h(a)<0.$ This tells us that the zero of $h$ is actually a real number in $(0, a).$


问:
Contradiction in two ways to derive Fourier transformation of error function
In the question, Fourier transformation of error function is derived as follows:
$$
\int_{-\infty}^\infty \operatorname{erf}(x)e^{-ixk}dx=\frac{2}{ik} \exp \left(\frac{-k^{2}}{4}\right).
$$
However, in the paper, following equation holds for any $a,b\in\mathbb{C}$:
$$
    \int \operatorname{erf}(a z) e^{b z} d z=\frac{1}{b} e^{b z} \operatorname{erf}(a z)-\frac{1}{b} \exp \left(\frac{b^{2}}{4 a^{2}}\right) \operatorname{erf}\left(a z-\frac{b}{2 a}\right).
$$
According this formula, Fourier transformation of error function is not well-defined because
$$
\int_{-\infty}^\infty \operatorname{erf}(x)e^{-ixk}dx=-\frac{1}{ik} e^{-ik \infty}-\frac{1}{ik} e^{ik \infty}+\frac{2}{ik} \exp \left(\frac{-k^{2}}{4}\right).
$$
Which part of the above discussion is wrong?

答:
The first equation gives a correct formula for the Fourier transform of the error function. As metamorphy writes, the integral on the left hand side is not defined, but the Fourier transform in a distributional sense is defined and equals
$$
\frac{2}{i} \exp\left(-\frac{k^2}{4}\right) \operatorname{pv}\frac{1}{k}
$$
where $\operatorname{pv}\frac{1}{k}$ is the principal value distribution.

The second equation gives a primitive function of the error function, but since the integral over all the reals is not defined, this formula cannot be used to determine the Fourier transform.


问:
Show that $4mn-m-n$ can never be a square
Let $m$ and $n$ be positive integers. Show that $$4mn-m-n$$ can never be a square.

In my attempt I started by assuming for the sake of contradiction that $$4mn-m-n=k^2$$ for some $k \in \mathbb{N_0^+}$. Then I considered $k^2 \pmod3$  (I couldn't find a way for $\pmod4$ or $\pmod8$ to help)
$$k^2 \equiv 0,1 \pmod3$$

This yield three possibilities for $m$ and $n$ $\pmod3$ either $$m,n \equiv 0 \pmod3$$ $$m \equiv 0,\;\;\;n \equiv 2 \pmod3 \; \; \;(WLOG)$$ $$m,n \equiv 2 \pmod3$$  Case $1$:   $m,n \equiv 0 \pmod3$

Let $m=3m'$ and $n=3n'$ then $k^2=36m'n'-3m'-3n'$ Hence $3|k^2 \implies 3|k$ Let $k=3k'$ then $$9k'^2=36m'n'-3m'-3n'$$
$$\implies3k'^2=12m'n'-m'-n'$$ Here I'm stuck and don't know if anything I've done is helpful and don't feel inclined to pursue the other cases given what happened here. I was hoping to reduce this to some sort of infinite descent argument - that didn't happen. I now think that my approach was probably completely wrong. Any help would be greatly appreciated.

答: Assume on the contrary that $4mn-m-n=k^2$. Then $(4m-1)(4n-1)=(2k)^2+1$. Since $4m-1 \equiv 3 \pmod{4}$ and $4m-1>0$, $4m-1$ has a prime factor $p \equiv 3 \pmod{4}$. Then $p \mid (2k)^2+1$, which gives a contradiction since $(\frac{-1}{p})=(-1)^{\frac{p-1}{2}}=-1$.</p>


问:
Find number X, that if you increase it by 10% its sum of digits decrease by 11%
I am stuck with this task:
Find the smalest positive number X, for which apply: If you increase this number by 10%, digit sum will decrease by 11%.

Mathematicaly I formulated this problem as:
find number X, for which is true
DS(X) = 0,89DS(X1,1)
and
X = abs(X)

I have tried brute force, without success.
Can anybody help me?
Any help will be welcome...

BR
KJ

答:
This is a good construction / structure-finder-and-avoider problem, so I would encourage you to work this out yourself one step at a time. If you're stuck in any of these steps, explain what you've tried and what you're stuck on.

The number is a multiple of 10. Let the number be $10X$.

Since $DS(10X)$ decreases by 11%, hence $DS(10X)$ is a multiple of 100.

$0.89 DS(10X) = DS(10X + X ) \equiv 2DS(X) \pmod{9}$.

Hence $DS(10X)$ is a multiple of 3, and thus 300.

Assume that / In the event that $DS(10X) = 300$.

Note that in $10X + X$, any carry over is either a 1 or a 0.

In $10X + X$, each carry over of 1 reduces the digit sum by 9. A carry over of 0 doesn't change the digit sum.

Hence, there are exactly $ \frac{ 600 - 300 \times 0.89 } { 9 }  = 37$ carry overs of 1.

Since the units digit doesn't allow for a carry over (it is 0), so $10X$ has at least 38 digits.

Assume that / In the event that $10X$ has exactly 38 digits

In order to get a carry over, remember that we're adding at most 1 each time, so the sum of each pair of neighboring digits is at least 9.

For the leading digit to have a carry over, it must have a digit of $\underline{ \quad } $ (fill in the blank).

Recall that we don't need the units digit to have a carry over, and that the units digit is actually 0.

Now, greedily assign digits to get the smallest possible 38 digit number, subject to all of the above constraints:

 $90909093999999999999999999999999999990$.

Verify that this number works (which it should because of the above arguments).

Note: If we couldn't find such a 38 digit number, then we could try for a 39+ digit number (or subsequently $DS(10X) = 600$), while relaxing some of the above constraints.


问:
Question : 3 Of Section : Solvability by Radicals Hungerford' s Algebra

Let $K$ be a field, $f \in  K[x] $ an irreducible polynomial of degree $n \geq 5$  and $F$ a splitting
field of $f$ over $K$. Assume that $\operatorname{Aut}_K{F}$ is isomorphic to $S_n $. Let $u$ be a root of $f$ in $F$. Then

(a) Show that $\operatorname{Aut}_K {K(u)}  =1$ and $[K(u):K]=n$.

(b) There is no radical extension field $E$ of $K$ such that $K \subset  K(u) \subset  E$.

(c) Every normal closure over $K$ that contains u also contains an isomorphic
copy of $F$.

I have proved that $[K(u): K] =n$ , but I don't know how to show that for all $\sigma \in \operatorname{Aut}_K {K(u)} $ and for all $v \in K(u)$  we have $\sigma(v)= v$ and (b) although I have done (c).

I have thought a lot about this question and I will not be able to do it by myself.

So, Please guide me..

答:
For (a): Note that the $K$-automorphisms of $K(u)$ correspond bijectively to the conjugates of $u$ over $K$ contained in $K(u)$, so we want to show that $u$ is the only root of $f$ in $K(u)$. Assume that there is some other root $v$ of $f$ contained in $K(u)$. Then we may divide $f$ by $(x-u)(x-v)$ over $K(u)$ to get a polynomial of degree $n-2$. This will have a splitting field of degree at most $(n-2)!$ over $K(u)$. Hence $f$ has a splitting field of degree at most $n\cdot(n-2)!$ but this is a contradiction as we know that the splitting field has $K$-automorphism group $S_n$ and hence degree at least $n!$.

For (b): Since the splitting field of a polynomial of degree $n$ always has degree at most $n!$ and $[F:K]\geq|\operatorname{Aut}_KF|=n!$ we see that $[F:K]=n!$. So in particular $F/K$ is separable and we may apply the following theorem:

A finite separable extension $E/K$ is contained in a radical extension iff the Galois-group of its normal hull is solvable.

In this case $E=K(u)$. Since $F$ is the normal closure of $E$ and $G(F/K)=S_n$ is not solvable (as $n\geq5$) we see that $K(u)$ cannot be contained in some radical extension.

Edit: The 'separable' property is actually only needed for the other direction of the above theorem (i.e. if the group is solvable then the extension is solvable by radicals), so we technically didn't need to show that.


问:
A not compact operator defined in $l^2$
I'm trying to solve the following problem presented in Kreyszig's Introductory Functional Analysis with Applications. Consider the operator $T:l^2\to l^2$ defined by
$$Tx = (\alpha_k\xi_k), \quad x=(\xi_k)\in l^2,$$
where the sequence $(\alpha_k)$ is dense in $[0,1]$. I have to prove that $T$ is not compact, so I consider the bounded sequence $(e_n)$ given by
$$e_n = (0,0,\ldots,1_n,0,0,\ldots).$$
We then obtain the sequence $(Te_n)$ given by
$$Te_n = (0,0,\ldots,\alpha_n,0,0,\ldots).$$

I don't know how to justify that $(Te_n)$ doesn't have a convergent subsequence, or even worse if this sequence is the appropiate to proceed. In the solution given, the author says "every $\alpha_k$ is an eigenvalue of $T$" (so you can apply a result that every sequence of eigenvalues of a compact operators must converge to zero and obtain a contradiction) but that clearly is not correct.

I would appreciate if someone can help me with this.

答:
You could also proceed with the same idea in your original proof without referring to the spectral theory of compact operators:

Since $\alpha_n$ is bounded and not tending to zero, there is a subsequence $n_k$ with $\alpha_{n_k}\to c\ne0$. Now the sequence $e_{n_k}$ is bounded, and its image sequence $Te_{n_k}$ has no convergent subsequence. Indeed, if it would have a convergent subsequence $Te_{n_{k_\ell}}\to y$ then its limit $y$ would have to be $0$ (why?), but $\alpha_{n_{k_\ell}}\to c\ne0$, a contradiction (why?).

Title: A not compact operator defined in $l^2$


问:
Estimating volume of the solid using MidPoint Rule

If the region shown in the figure is rotated about the $x$-axis to form a solid, use Midpoint rule with $n=4$ to estimate the volume of the solid.

I am not sure what this question is asking me to do. I believe at $x=3,\,5,\,7$, and $9$ the figure will be rotated around the $x$-axis. Is it the average of the volume of these points that will be the answer?

答: Are you familiar with the disk method for determining the volume of solids of revolution?  You are being asked to approximate that integral with four disks, each one having a radius of the midpoint of its thickness, which you have correctly determined are $3,5,7,9$.  So compute the volume of the four disks and add them up.


问:
Notation help for matrices - Linear Algebra
I am first told that there is a matrix $A.$

What does this then mean when
$$\left[A \mid\, \vec{0}\right]$$ is the augmented matrix of a linear system?

What does it mean for $A$ to have $\mid\; \vec 0$ next to it?

答:
$[A\ |\ \vec 0]$ simply means that a column of zeros is added to the matrix $A$. For example, if
$$
A = \pmatrix{1&2&3\\2&1&0\\-1&2&-3},
$$
then
$$
[A\ \ |\ \ \vec 0]  = \pmatrix{1&2&3&0\\2&1&0&0\\-1&2&-3&0}.
$$
Often, this matrix is written with a vertical line. That is, we might write
$$
[A \ \ | \ \ \vec 0] = \left(
\begin{array}{ccc|c}
1&2&3&0\\2&1&0&0\\-1&2&-3&0
\end{array}
\right).
$$
Mathematically, it makes no difference.


问:
Probability that $n$ random walks (1D) intersect a single point
I am struggling finding a relation to determine the probability $n$ random walks (1D) intersect in a single point at step $s$. In the method below my attempts. My method is somewhat intuitive based. I am looking for more rigorous proof.

note:
This question arises from someone who claims that a matching cumulative digit sum of: $\pi$, $e$ and $\varphi$ (golden ratio) is unique and "cosmological" [1]. I tend to disprove it. This digit sum can be seen as a random walk if the constants are normal (every digit occurs with same frequency).

Method:

For every step $s$ on the random walk we can determine the probability density function if we know the standard deviation on every step $s$.

The standard deviation of a single step can be calculated it's a discrete uniform distribution "equally likely outcomes", where $q$ is the number of outcomes e.g. the number of digits $[0,1,2,3,4,5,6,7,8,9]$, $q=10$:

$$\sigma=\sqrt{\frac{q^{2}-1}{12}} $$

All the (1D) random walks start in the origin for this example. The standard deviation will grow with every step $s$, the variance is proportional to the number of steps [2].

$$Var(s)=s \cdot \sigma^{2}$$

$$\sigma(s)=\sqrt{s} \cdot \sigma$$

While the bins grow rapid I assume a normal approximation of the Binomial distribution.

$$f(x)=\int_{-\infty}^{\infty} {\frac {1}{\sigma {\sqrt {2\pi }}}}e^{-{\frac {1}{2}}\left({\frac {x}{\sigma }}\right)^{2}}dx $$

The probability that $n$ random walks intercept in a single point is (not sure):

$$p(s)=\int_{-\infty}^{\infty} \left[ {\frac {1}{\sigma {\sqrt {2\pi }}}}e^{-{\frac {1}{2}}\left({\frac {x}{\sigma }}\right)^{2}} \right]^n dx $$

With help of Wolfram Alpha [3] the solution is found for $n=3$ meaning the probability of $3$ point intersecting random walks.

$$p(s)=\frac{1}{2 \sqrt{3} \ \pi \ \sigma^{2}} \cdot \frac{1}{s}$$

The total probability $p(n)$ is proportional to the reciprocal sum of $s$. So the total probability is proportional to the harmonic series:

$$p(n)=\frac{1}{2 \sqrt{3} \ \pi \ \sigma^{2}} \cdot \sum_{s=1}^{\infty}\frac{1}{s}$$

This series diverges, meaning there are infinate point intersections of $n=3$ random walks. So a matching cumulative digit sum of $\pi$, $e$ and $\varphi$ is not unique, probability $\sim 8 \%$ for the first 1200 digits (see graph).

Question

Does anyone know the general formula for the probability $p(n)$ that $n$ random walks (1D) intersect a single point?

```<code>import numpy as np

#Elements of digits [0,1,2,3,4,5,6,7,8,9] rescaled to fit random walk
array=[-9,-7,-5,-3,-1,1,3,5,7,9]

#steps, in single random walk, x walks to intercept, number of trial to find intercept
steps=2500
xwalks=3
trials=1500

#Set output array to zero
count=np.full([steps],0)

for n in range(trials):

    #Identify initial array, set total array to zero
    w0=np.random.choice(array,steps)
    w0=np.cumsum(w0)
    total=np.full([steps],0)

    #Select x random walk check for intercept
    for m in range(xwalks-1):

        #Next current random walk
        w=np.random.choice(array,steps)
        w=np.cumsum(w)

        #Compare previous and current random walk
        eq=np.equal(w0,w)
        eq=eq.astype(int)

        #Count intercepts
        total=total+eq

        #Set current walk to previous
        w0=w

    #Sum all interceptions for all trials
    count=count+np.where(total==(xwalks-1),1,0)

#Print output
print(count)
print(np.sum(count))
print(np.sum(count)/trials)
</code>```

答:
I am looking at this problem without any prior assumed knowledge of random walks, so I am not sure whether or not my analysis is useful - it might offer some fresh perspectives or just cover ground which is already known.

Firstly, it might make sense to consider the binary representation of the numbers in question to simplify matters. Then the probability that n random walks intersect at a single point after m steps/digits is p[m,n]=Sum[(nCi/2^m)^n,{i,1,m}] using pseudo-Mathematica syntax.

The probability that n random walks intersect at a single point at any step in the future is Sum[p[m,n],{m,1,infinity}].

Note that p[m,n] vanishes to zero as m increases towards infinity. This means that the probability of n random walks intersecting also vanishes to zero.

In fact, it is known that if we have two possible binary digits 0 and 1, then the expected value of the sum of the binary digits is 1/2 multiplied by the number of digits, and the standard deviation is proportional to the square root of the number of digits. (Consider the absolute value of the deviation from the expected value, for example, which makes more intuitive sense.) Therefore, for n different random or pseudorandom irrational numbers such as pi, we would expect the digit sums to diverge (even though the expected digit sums are the same), and therefore I actually agree with the person who suggested that each random or pseudorandom number has an unique digit sum footprint. This concurs with the fact that the probabilities vanish to zero.

Alternatively, instead of considering an one dimensional random walk, we can consider a n dimensional random walk.

To demonstrate this, first consider a 2D random walk for the case n=2. Instead of consider the binary digits 0 or 1, consider the digits to be either -1 or 1, which is the standard form of the random walk usually considered. To convert between binary digits 0 or 1 and -1 or 1, a translation and scaling factor can be applied.

With each step, we can go either in the positive or negative direction for each of the two dimensions, so we are essentially travelling diagonally in the plane with each step. For the two random walks to intersect after m steps, we look at the points (x,x) along the diagonal of the plane i.e. (0,0), (1,1), (-1,-1), (2,2), (-2,-2) etc.

By applying a 45 degree (or pi/4 radian) rotation followed by scaling by a factor of 1/Sqrt[2], the problem can be restated so that at each step, we are moving either left, right, up or down, and for the two random walks to intersect after m steps, we are now looking at the points (x,0) along the negative/positive x-axis.

This can now be extended beyond two dimensions to n dimensions.


问:
Prove that a property holds for the roots of an equation, given some constraints.
For example, in this question:

The cubic expression $$ ax^3 + bx^2 + cx + d $$ has a pair of roots which are reciprocals of each other. Prove that $$a^2-d^2=ac-bd$$.

I understand that the correct approach is probably to use the formulae relating coefficients and roots, but I'm not exactly sure where to go for there.

Is there are a more generic framework for solving these types of problems?

答:
Let the roots be $ y, z, \frac{1}{z}$.

As you said, "use the formulae relating coefficients and roots" to show that

$y+z + \frac{1}{z} = -b/a$

$yz + y/z + 1 = c/a$

$y = -d/a$

Finally, verify that

$$1 - (d/a)^2 = 1-y^2 = (c/a) - (b/a)(d/a).$$

Hence, $ a^2 - d^2 = ac - bd$.


问:
Why was Apery's constant believed to be rational?
Apéry's theorem states that Apéry's constant $\zeta(3) := \sum_{n=0}^\infty n^{-3}$ is irrational. The Wikipedia article claims that this result was "wholly unexpected" to the point that the mathematical community initially did not believe it.

But for me personally, Apéry's theorem states exactly what I expected. Broadly speaking, I find that almost all "interesting" or "natural" series have an irrational sum (with pretty much the only exception being geometric series with a rational initial term and common coefficient). Indeed, as this question points out, it's arguably more natural to expect that Apéry's constant is a rational multiple of $\pi^3$ (and therefore irrational).

Why was the mathematical community so convinced that Apéry's constant was rational before Apéry published his proof to the contrary?

(Note: I don't really care exactly who believed what at the time. What I'm wondering is which mathematical heuristic(s) the community was using to conjecture that Apéry's constant was rational.)

Edit: the consensus view of the comments and answer is that the proof was unexpected, but the theorem itself was not. Since the standard usage of the word "result" refers to a theorem but not to a particular proof, I have edited the Wikipedia page to change the phrase "unexpected nature of the result" to "unexpected nature of the proof" in order to clarify the sentence's meaning.

答:
We can read in the  Wikipedia article

... However, in June 1978, Roger Apéry gave a talk titled Sur l'irrationalité de $\zeta(3)$. During the course of the talk he outlined proofs that $\zeta(3)$ and $\zeta(2)$ were irrational, the latter using methods simplified from those used to tackle the former rather than relying on the expression in terms of $\pi$. Due to the wholly unexpected nature of the result and Apéry's blasé and very sketchy approach to the subject, many of the mathematicians in the audience dismissed the proof as flawed. ...

Here the phrase unexpected nature of the result addresses the unexpected technique to tackle the problem using methods known from showing the irrationality of $\zeta(2)$, while the experts rather expected methods similar to those which were used in trying to show that the quantities
\begin{align*}
\frac{\zeta(2n+1)}{\pi^{2n+1}}
\end{align*}
are transcendental.

At the end of this  paragraph there is also a nice reference to the paper A proof that Euler missed... by A. v. d. Poorten. The author tells us:

... I heard with some incredulity that, for one, Henri Cohen (Bordeaux, now Grenoble) believed that these claims might well be valid. Very much intrigued, I joined Hendrik Lenstra (Amsterdam) and Cohen in an evening's discussion in which Cohen explained and demonstrated most of the details of the proof. We came away convinced that Professeur Apéry had indeed found a quite miraculous and magnificient demonstration of the irrationality of $\zeta(3)$. ...

Conclusion: What we can read here addresses the interesting techniques to prove the irrationality of $\zeta(3)$. Irrationality was not seriously questioned.


问:
Confusing moments about the Law of exponents for natural powers
Suppose that $a\in \mathbb{R}$ and $a>1$ and we define in the recursive way the following operation: $$a^1:=a,$$ $$a^{n+1}:=a^n\cdot a \quad\text{for each} \quad n\in \mathbb{N}.$$

I want to prove that $a^{n+m}=a^na^m$ for any $n,m\in \mathbb{N}$.

Proof: Let $E=\{m\in \mathbb{N}: a^{n+m}=a^na^m \ \text{for any} \ n\in \mathbb{N}\}$

For any $n\in \mathbb{N}$ we have $a^{n+1}=a^na=a^na^1$ which shows us that $1\in E$.

Suppose that $m\in E$ and we'll try to show that $m+1\in E$.

Indeed, for any $n\in \mathbb{N}$ we have $a^na^{m+1}=a^n(a^ma)=(a^na^m)a=a^{n+m}a=a^{(n+m)+1}$.

As you see I am almost done but one probably stupid moment bothers me a lot. Probably I am too pedantic.

We know that $(n+m)+1=n+m+1$ by associativity of natural numbers. How it follows that $$a^{(n+m)+1}=a^{n+m+1} \ ?$$

I do not think the we can just substitute $(n+m)+1$ by $n+m+1$.

May be I cannot formulate my question quite explicitly but I'd be very thankful for your help!

答:
Another way of formulating it is that we have a function $exp: \mathbb R \times \mathbb N \rightarrow \mathbb R$ such that $exp(a,n+1) = n* exp(a,n)$ (in other words, I am rewriting $a^n$ as $exp(a,n)$). You are asking whether we can take $exp(a,(n+m)+1)$ to be the same as $exp(a,n+m+1)$. A fundamental property of a well-defined function is that if $x$ and $y$ are different expressions representing the same number, then $f(x)=f(y)$. So, for instance, $f_1(\frac a b) = \frac {2a}b$ is a well-defined function, because applying it to two different representation of the same fraction gives the same result (e.g. $f_1(\frac 1 2)=f_1(\frac 2 4)$). $f_2(\frac a b) = a+b$, on the other hand, is not a well-defined function, because $f_2(\frac 1 2)=3$, while $f_2(\frac 2 4)=6$.

You've said that you've defined $exp$ recursively, and implicit in saying that this is a definition is saying that the function is well-defined.


问:
Solutions to a second-degree DE

$y_1,y_2$ are solutions to a differential equation $y''-ty'+3y=t$.
Which of the following functions are also the solutions?

a) $y_1+y_2$

b) $5y_1$

c) both of them

d) none of them

I think the solutions is none of them, but I am not sure and not sure how to verify it in general. Can you point me to the right direction?

答:
Well, for b), you can check:

$$5y_{1}'' - 5ty_{1}' + 15y_{1} = 5(y_{1}'' - ty_{1}' + 3y_{1}) = 5t \ne t,$$ so we know that $5y_{1}$ is not a solution.  You can use a similar procedure to check a).

Title: Solutions to a second-degree DE


问:
Improper integral convergence problem comparison
Does the following improper integral converge ?

$$\int _1^{\infty}\:\cfrac{e^{1/x}-1}{x}dx$$

I have tried to compare it to some known improper integrals but with no luck.

Thanks for helping.

答:
HINT:

Note that as $x\to \infty$, $e^{1/x}=1+\frac1x+O\left(\frac1{x^2}\right)$.

Can you finish now?


问:
Prove that dimension of solution of linear equation $a_1x_1 + a_2x_2 + \cdots + a_nx_n = 0$. ($\exists a_i \not= 0$) is $n - 1$
I have already seen question like this here. But I don't know the theory that was applied in there in order to show that.

I was given a hint that I could consider such linear map $A$ that does this: $$A(X) = a_1x_1 + a_2x_2 + a_3x_3 + \cdots + a_nx_n$$

$X$ is basically column with these $x$s. After that I should somehow apply the fact that $\dim \ker A + \dim \operatorname{im} A = \dim V$ where $V$ is vector space which vectors $A$ takes as a parameters.

Could you, please, explain how can I use it in order to prove that dimension of solution is $n - 1$?

答:
Hint.

The $1$-by-$n$ matrix $A=[a_1 \cdots a_n]$ is a linear map from $\mathbb{R}^n$ to $\mathbb{R}$.

The image of $A$ is of dimension $1$ since there exists $a_i\ne 0$. Also $\mathbb{R}^n$ is of dimension $n$.

Now you can apply the rank-nullity theorem.


问:
Formula of intersection of 2 points with the x axis
I'm trying to figure out how to get the point x = 3 :
What's given here are the points S and G .
(Assuming the 2 angles are equal)

Apparently, we can assume that the ball does not bounce off the x-axis. Instead, we can change the target with respect to the symmetrical point on the x-axis so that we can now look at the intersection of the x-axis and the line connecting the current and target point.

The Formula to get x=3 is below :

$$\frac{S_x.G_y + G_x.S_y}{S_y+G_y}$$

Is there any explanation for this formula ?

答:
Slightly different notation for co-ordinates.

To arrive at that equation equate slopes of straight lines  $(SI, IG2)$ because after reflection the three points now are in a straight line.

( Reflection law requires that incidence and reflected angles be equal).

It can be also derived by total time minimization using Fermat's Law for given light speed.


问:
Differential equation where forcing function depends on variable
The force acting on a body depends on the displacement $x(t)$ of the body from the origin.
This dependence is given by

$$ F(x) =
\begin{cases}
F_0 - mg, &-L < &x < 0,\\
-mg,& &x > 0.
\end{cases}
$$

where $L$ is a positive number, $F_0 > 0$ is a constant resting force and $g$ is the gravitational acceleration of Earth.

Newton's second law gives $m\ddot{x} = F(x)$ and we impose the initial conditions $\dot{x}(0) = 0, x(0) = -L.$

How does one solve an equation like this? My main problem is that if we restrict to the branch $x > 0$ we get $x = -g\frac{t^2}{2} + At + B$. I don't know if I am allowed to plug in the boundary conditions here, but if I were to, I would get $x = -g\frac{t^2}{2} - L$. But this function is clearly always negative, so then how could this satisfy the branch $x>0$?

Any tips or clarifications?

答:
From

$$
m\ddot x = F(x) = \left(\theta(x+L)-\theta(x)\right)(F_0-m g)- m g\theta(x)
$$
where $\theta(x)$ is the Heaviside step function, we have

$$
m\ddot x\dot x = F(x)\dot x
$$

or equivalently with $v = \dot x$

$$
m v dv = F(x)dx
$$

and then

$$
\frac 12 m v^2 = (x+L)(F_0-m g)\theta(x+L)-x F_0\theta(x)+C_0
$$

now assuming that $v(-L) =  0\Rightarrow C_0=0$ we have

$$
\frac{dx}{\sqrt{(x+L)(F_0-m g)\theta(x+L)-x F_0\theta(x)}} = \sqrt{\frac 2m}dt
$$

now assuming $-L \le x \le 0$

$$
\int_{-L}^x\frac{du}{\sqrt{(u+L)(F_0- m g)}}= \sqrt{\frac 2m}t
$$

giving

$$
x(t) = \frac{t^2 (F_0-g m)}{2 m}-L
$$

and for $x > 0$

$$
\int_{-L}^0\frac{du}{\sqrt{(u+L)(F_0- m g)}}+\int_0^x\frac{du}{\sqrt{(u+L)(F_0- m g)-u F_0}}= \sqrt{\frac 2m}t
$$

giving

$$
\frac{2L}{\sqrt{L(F_0-m g)}}+\frac{2 \left(\sqrt{L (F_0-m g)}-\sqrt{F_0 L-m g(L+x)}\right)}{m g} = \sqrt{\frac 2m}t
$$

and now inverting we can find $x(t)$

$$
x(t) = \frac{F_0}{g m-F_0}-1+\frac{ F_0 L \sqrt{\frac{2}{m}}}{\sqrt{L (F_0-g m)}}t-\frac{g}{2}t^2
$$


问:
Estimating $1 + H_k$ by $H_{k'}$
The harmonic numbers are $H_k = 1 + 1/2 + \cdots + 1/k$.

For real $x \geq 1$ let $\kappa(x)$ be the
largest integer $k$ for which $H_k \leq x$. For instance $\kappa(2) = 3$ since $H_3 \leq 2$ and $H_4 > 2$. In a sense $H_{\kappa(x)}$ approximates $x$.

Show that $\kappa(1 + H_k) \sim e k$ as $k \rightarrow \infty$.

答:
Working out Steven Stadnicki's hint, thank you Steven!

For an $n$ where $x \approx H_n$ we have $x \approx \log n + \gamma$, which shows
$\kappa(x) \approx e^{x - \gamma}$. So
\begin{align}
\kappa(1 + H_m) &\approx \exp(1 + H_m - \gamma) \\
&\approx \exp(1 + \log m + \gamma - \gamma) \\
&= e m.
\end{align}
Same sort of argument shows $\kappa(H_m + H_n) \sim e^\gamma mn$ as $m, n \rightarrow \infty$. Nice!

Title: Estimating $1 + H_k$ by $H_{k'}$


问:
Block-matrix inverse of $\left(\begin{array}{ccc} I_N & - A & - B\\ - A^{\top} & NI_M & 0\\ - B^{\top} & 0 & NI_K \end{array}\right)$
Consider a matrix $M$ written in block-form as follows:

$$M = \left(\begin{array}{ccc}
  I_N & - A & - B\\
  - A^{\top} & NI_M & 0\\
  - B^{\top} & 0 & NI_K
\end{array}\right)$$

where $A\in\mathbb{R}^{N\times M}$, $B\in\mathbb{R}^{N\times K}$ are rectangular matrices, and $I_N$ is the $N\times N$ identity matrix.

I want to find the inverse matrix $M^{-1}$. I suspect that I should be able to write down some expression for $M^{-1}$ exploiting the block structure of $M$.

Can we invert $M$?

答:
I will assume for the purpose of this answer that $N < M + K$, but even without this assumption the same approach can be taken with appropriate adjustments.

As the comment on your question notes, we can write $M$ in the form
$$
M = \pmatrix{I_N&-F\\ -F^T&NI_{M+K}},
$$
and from there it is indeed possible to use the Schur complement to obtain a formula for the inverse. However, the fact that the diagonal entries are both multiples of the identity matrix leads to a nice alternative approach using the SVD. Suppose that $F$ has singular value decomposition $F = U\Sigma V^T$ where $U$ is orthogonal and $N \times N$, $V$ is orthogonal and $(M+K)\times(M+K)$ and
$$
\Sigma = \pmatrix{\Sigma_0&0}
$$
where $\Sigma$ has the same shape as $F$ and $\Sigma_0$ is square and diagonal with non-negative entries. We can write $M$ in the form $M = WPW^T$, where
$$
W = \pmatrix{U & 0\\0 & V}, \quad P = \pmatrix{I_N & \Sigma \\ \Sigma^T & N I_{M + K}}.
$$
We can write
$$
P =
\left( \begin{array}{cc|c}
I_N & \Sigma_0 & 0\\
\Sigma_0 & N \cdot I_N & 0\\
\hline
0 & 0 & N \cdot I_{M + K - N}
\end{array}\right).
$$
It now suffices to take the inverse of each block. It is easy to verify that
$$
\pmatrix{I & \Sigma_0\\ \Sigma_0 & N\cdot I}\pmatrix{N \cdot I & -\Sigma_0\\ -\Sigma_0 & I} = \pmatrix{N\cdot I - \Sigma_0^2 & 0\\0 & N \cdot I - \Sigma_0^2} \implies\\
\pmatrix{I & \Sigma_0\\ \Sigma_0 & N\cdot I}^{-1} =
\pmatrix{N \cdot I & -\Sigma_0\\ -\Sigma_0 & I}
\pmatrix{N\cdot I - \Sigma_0^2 & 0\\0 & N \cdot I - \Sigma_0^2}^{-1} = \\
\pmatrix{N(N\cdot I - \Sigma_0^2)^{-1} & -(N\cdot I - \Sigma_0^2)^{-1}\Sigma_0\\
(N\cdot I - \Sigma_0^2)^{-1} \Sigma_0 & (N\cdot I - \Sigma_0^2)^{-1}}.
$$
From there, we get
$$
P^{-1} = \pmatrix{N(N\cdot I - \Sigma_0^2)^{-1} & -(N\cdot I - \Sigma_0^2)^{-1}\Sigma_0 & 0\\
(N\cdot I - \Sigma_0^2)^{-1} \Sigma_0 & (N\cdot I - \Sigma_0^2)^{-1} & 0\\
0 & 0 & \frac 1{N} I_{M + K - N}}.
$$
Finally,
$$
M^{-1} = [WPW^T]^{-1} = W P^{-1}W^T.
$$


问:
Checking whether a code is cyclic by using the generator matrix
I was wondering whether it is possible to conclude whether a code is cyclic or not from the generator matrix of a code. If so, how could one do this?
Thank you in advance!</p>

答:
Not very directly.

Recall that for a $(n,k)$ cyclic code any codeword is of the form $c(x) = g(x) u(x)$ where $g(x)$, the GP, is a factor of $x^n+1$ and has degree $n-k$.

First factor $x^n+1$ to find all the possible factors of degree $n-k$ (candidate GPs).

Then represent each row of $G$ as a polynomial $c(x)$, factor them, and check if all are multiple of one of the candidate GPs. If so, the code is cyclic.


问:
Maximizing an unusual function (Putnam 1996)
“Fish," he said, "I love you and respect you very much. But I will kill you dead before this day ends.”

-- Ernest Hemingway, The Old Man and the Sea

I have, with varying degrees of concentration, been trying to resolve this problem for most of this evening:

If we require $\{a_1, a_2, ... a_n \} = \{1, 2, ... n\}$, determine the maximum value of $a_1 a_2 + a_2 a_3 + \dots + a_{n-1}a_n + a_n a_1$ as a function of $n$.

(That is, how can we - by permuting the first $n$ positive integers to form a sequence $\{a_n\}$ - maximize the expression above for arbitrary $n$?)

My thinking was that the fundamental difficulty with the problem was the lack of symmetry. So, my approach below was motivated by trying to introduce some.

Notice that $\sum_{k = 1}^n a_k^2 = \sum_{k = 1}^n k^2$, regardless of what the sequence $\{a_k\}$ is. Now, we do a neat rearrangement by completing the squares. (In the first and second lines, we are just adjusting the sigma notation.) If $S$ is our original expression,

\begin{eqnarray} 2 \sum_{k = 1}^n a_k^2 - 2S = 2 \sum_{k = 1}^n a_k^2 - 2 (a_1 a_2 + a_2 a_3 + \dots + a_{n-1}a_n + a_n a_1) \\
= \left( \sum_{k = 1}^{n-1} a_k^2 + a_{k+1}^2 \right) + (a_n^2 + a_1^2) - 2 (a_1 a_2 + a_2 a_3 + \dots + a_{n-1}a_n + a_n a_1) \\
= \left( \sum_{k = 1}^{n-1} a_k^2 + a_{k+1}^2 \right) + (a_n^2 + a_1^2) + \left( \sum_{k = 1}^{n-1} - 2 a_k a_{k+1} \right) - 2a_n a_1
\end{eqnarray}

$$= \sum_{k = 1}^{n-1} (a_k - a_{k+1})^2 + (a_n - a_1)^2 \qquad (*)$$

Now, if $2\left(\sum_{k = 1}^n k^2 - S\right) \ge m$ with equality for some arrangement $\{a_k\}$, then $\sum_{k = 1}^n k^2 - \frac{m}{2} \ge S$; thus, $$\frac{(n)(n+1)(2n+1)}{6} - \frac{m}{2} \qquad(**)$$

would be our maximum value. If we can minimize $(*)$ then, since $m$ will be this minimum, we're done.

What this has boiled down to is trying to show that $(*)$ is a friendly enough expression to minimize. Although it's difficult for me to display in notation, I've come up with a heuristic which suggests that, for a fixed $n$,

$$\min (*) = m = (n-2)(2)^2 + 2(1)^2 = 4n - 6$$

This does, in fact, lead to the correct solution (checking the numerical answer).

Can anyone help me prove that this assertion is correct?

答:
Hint: Prove by induction for $n\geq 3$. Show that adding $n$ into the sequence (in a special place) would increase the sum by at least 4.

This solution follows naturally from considering the equality case.

Note: The $n=1, 2$ case will have to be dealt with separately, since they give values that don't fit into your sequence.


问:
What does this ODE mean?
I'm interested in understanding the following system of linear ODE's. Consider $\mu \leq 0$ and the system
\begin{cases}
\frac{d}{dt}u(t) = \mu u, \quad \quad \quad \quad \quad \quad \quad u(0)=1,\\
\frac{\partial}{\partial t} v(s,t) = \delta(t-s)u + \mu v \quad v(s,0)=0 \: \forall s
\end{cases}
I know the solutions to be $u(t) = e^{\mu t}$ and $$v(s,t) = \begin{cases}
e^{\mu t} \quad \text{if } t\geq s \geq 0 \\
0 \quad \quad \text{otherwise}
\end{cases}$$

I'm a bit confused as to what it means to be a solution to the second ode in our system above. The Dirac delta distribution is throwing me off. What I expect the solution of the second system be is the following; $v$ is a solution to the second ODE if for all $\phi \in C^\infty_c(\mathbb{R}^2)$ that
$$\int_\mathbb{R} v (s,t)\frac{\partial}{\partial t} \phi(s,t) = u(s)\phi(s) +  \mu \int_\mathbb{R} \phi (s,t)v(s,t).$$

This is merely an educated guess from my limited knowledge of distribution theory. When I go to verify this I get a minus sign on the left hand side, but also I'm not even sure if this is even correct.

Many Thanks

答:
You have made an error and missed a minus sign. The distributional derivative of $f$ is defined by
$$
\int_{-\infty}^{\infty} f'(x) \, \phi(x) \, dx = - \int_{-\infty}^{\infty} f(x) \, \phi'(x) \, dx
$$
so it should be
$$
-\int_\mathbb{R} v (s,t)\frac{\partial}{\partial t} \phi(s,t) = u(s)\phi(s) +  \mu \int_\mathbb{R} \phi (s,t)v(s,t).
$$

Practically, if $f$ is piecewise smooth, $f'(x) = \delta(x-a)$ means that $f$ has a step discontinuity at $x=a$ with $f(a+) - f(a-) = 1,$ where $f(a\pm) = \lim_{x\to a\pm} f(x).$

Title: What does this ODE mean?


问:
Name of the solid built on a sphere's surface
I don't know the name of the solid built on a sphere's surface. I mean, given a sphere of radius r1, create another sphere with the same origin and radius $r_2$, with $r_2 > r_1$. I'm interested in the solid that results when you "subtract" the first sphere from the second one.

If we were in 2D the object would be a ring.

答: It is called a spherical shell.</p>

Body: It is called a spherical shell.


答:
Okay, so if $r$ and $\frac 1r$ are two roots of $ax^3 + bx^2 + cx + d$ then $ax^3 + bx^2 + cx + d = a(x-r)(x-\frac 1r)(x+k)$ for some $k$.

So $ka = d$

And $ak -ar -a\frac 1r = d-ar -a\frac 1r = b$.

$-akr-ak\frac 1r+a = -dr-d\frac 1r + a=c$

If we assume $a\ne 0$ then $r+\frac 1r = \frac {d-b}a$ and $a = c+\frac {d-b}a\cdot d$.

So $a^2 =ca +d(d-b) = ca + d^2 -bd$ and $a^2 -d^2 =ca -db$.

And if $a = 0$ then $d=b$. and the result follows.

(If $a = 0; b\ne 0$ then $bx^2 + cx +d = b(x-r)(x-\frac 1r)= bx^2 -b(r+\frac 1r)x + b$ so $b=d$ )

Title: Prove that a property holds for the roots of an equation, given some constraints.

Body:
For example, in this question:

The cubic expression $$ ax^3 + bx^2 + cx + d $$ has a pair of roots which are reciprocals of each other. Prove that $$a^2-d^2=ac-bd$$.

I understand that the correct approach is probably to use the formulae relating coefficients and roots, but I'm not exactly sure where to go for there.

Is there are a more generic framework for solving these types of problems?


问:
How to derive order of magnitude when given any base raised to some exponent?
How does one quickly derive order of magnitude without needing to compute the value and count digits?

Say you’re given $26^{25}$ which has order of magnitude $10^{35}$.

How does one get from an exponent of $25$ to $35$ (in a different base)?

答: You can use logarithms:$$26^{25} = 10^x \implies 25\log_{10} 26 = x\implies x \approx 35.37.$$  This tells us that $26^{25}$ has order of magnitude $10^{35}$.