问:
How to integrate $(x^2 - y^2) / (x^2 + y^2)^2$
How do I integrate
$$\int \int \frac{(x^2 - y^2)}{(x^2 + y^2)^2} dx dy?$$
The WolframAlpha page gives
$$
c_1 + c_2 + \tan^{-1}(x/y).
$$
And I kind of specifically need
$$
\int_{0}^{x} \frac{(x^2 - y^2)}{(x^2 + y^2)^2} dy.
$$

Note

I want to know integration technique to solve this without using $F' = f$.

For the double integral above, what I'm interested is Lebesgue integral, but I guess what Wolfram gave is in the Riemann sense.

答:
The main integral

$$I = \int\frac{x^2-y^2}{(x^2+y^2)^2}\,dx$$

can be solved using a tan substitution.

\begin{align*}
x &= y\tan\theta \\
dx &= y\sec^2\theta\,d\theta
\end{align*}

So that

\begin{align*}
I &= \int \frac{y^2(\tan^2\theta -1)}{y^4\sec^4\theta}\,y\sec^2\theta\,d\theta \\
&= \frac{1}{y}\int \frac{\tan^2\theta-1}{\sec^2\theta}\,d\theta \\
&= \frac{1}{y}\int \sin^2\theta - \cos^2\theta\,d\theta \\
&= - \frac{1}{y}\int \cos 2\theta\,d\theta \\
&= - \frac{1}{y}\frac{1}{2}\sin 2\theta + h(y) \\
&= - \frac{1}{2y} \frac{2\tan\theta}{1+\tan^2\theta}  + h(y) \\
&= - \frac{1}{2y} \frac{2 (y/x)}{1+(y/x)^2} + h(y) \\
&= - \frac{x}{x^2+y^2} + h(y).
\end{align*}


问:
Show that the sequence defined as $x_{n+1}=\frac{x_n^3+70}{39}, x_0=4$ converge
Show that the sequence $(x_n)$ defined as $x_{n+1}=\frac{x_n^3+70}{39}, x_0=4$ converge. I would like to have a feedback on my proof, please (not sure about 1 detail). My attempt is to prove the convergence by induction.

To show that the sequence $(x_n)$ converges, we must show that it is bounded and monotonic.

First of all, we show that $(x_n)$ is a decreasing sequence. Suppose $x_{n+1}<x_n$. The inequality is true for $n=1$ so we can suppose that it works up to a certain $n$. We show now that it holds for $n+1$, i.e $x_{n+2}<x_{n+1}$. Thus, by induction hypothesis we have that:

$x_{n+1}<x_n \iff x_{n+1}^3<x_n^3 \iff \frac{x_{n+1}^3+70}{39}<\frac{x_n^3+70}{39} \iff x_{n+2}<x_{n+1}$

so $x_{n+1}<x_n \forall n\ge0$ and the sequence $(x_n)$ is deacrising.

We show now that $(x_n)$ is bounded below by $2$. Suppose $2<x_n$. The inequality holds for $n=0$ so we can suppose that it works up to a certain $n$. the next step is to show that the inequality holds for $n+1$. By induction hypothesis we have

$2<x_n \iff 8<x_n^3 \iff 2=78/39<\frac{x_n^3+70}{39} \iff 2<x_{n+1}$ and so we proved that $2<x_n \ \forall n\ge0$.

Therefore $x_n$ is convergent and it's limit value is $2$ (We obtain it by solving an equation $l^3-39l+70=0$, where $l$ is $(x_n)$ limit as it goes to infinity)

答: Your proof is right. But you need not have proved that the sequence is bounded below by $2$, since it is trivial by induction that this is a positive sequence, so $0$ is a lower bound.

ContentLicense: CC BY-SA 4.0


问:
Compute the value of the surface integral .
Let $S$ denote the hemisphere $x^2+y^2+z^2=1$ , $z\ge0$, and let $F(x,y,z)=xi+yj$. Let $n$ be the unit outward normal of $S$. Compute the value of the surface integral $\iint_S{F.n} dS$ , using
the explicit representation $z=\sqrt{1-x^2-y^2}$

My attempt : I know the formula  $\iint_S{F.n} dS= \iint_{T} F.n |\frac{\partial{r}}{\partial u} \times \frac{\partial{r}}{\partial v}|dudv$

But i don't know how to apply  the formula in the given question

答:
Using divergence theorem, one can easily see that the result is volume of the unit sphere which is $\frac{4\pi}{3}$. But as the question specifically seeks to use representation $z=\sqrt{1-x^2-y^2}$ and do surface integral,

use the fact that radially outward normal vector for a sphere is $(x,y,z)$ which can be rewritten as $(\frac{x}{\sqrt{1-x^2-y^2}}, \frac{y}{\sqrt{1-x^2-y^2}}, 1)$ using the representation $z = \sqrt{1-x^2-y^2}$.

Or you can take partial derivative as below,

$-\displaystyle \frac{\partial z}{\partial x} = \frac{x}{\sqrt{1-x^2-y^2}}$

$-\displaystyle \frac{\partial z}{\partial y} = \frac{y}{\sqrt{1-x^2-y^2}}$

So the normal vector, $\vec n = (\frac{x}{\sqrt{1-x^2-y^2}}, \frac{y}{\sqrt{1-x^2-y^2}}, 1)$

$\vec F = (x,y,0)$

$\vec F \cdot \vec n = \frac{x^2 + y^2}{\sqrt{1-x^2-y^2}}$

Surface integral should be

$\displaystyle \int_{-1}^{1} \int_{-\sqrt{1-y^2}}^{\sqrt{1-y^2}} \frac{x^2 + y^2}{\sqrt{1-x^2-y^2}} \ dx \ dy$


问:
Which one is more likely to happen:$10$ tosses with $5$ head up and $100$ tosses with $50$ head up?
If we have a fair coin is tossed multiple times, which one is more likely to happen:
$10$ tosses with $5$ head up or $100$ tosses with $50$ head up.

So the question translates mathematically into comparing the following two numbers $${10\choose 5} \cdot \frac{1}{2^{10}}\quad\text{and}\quad{100\choose 50} \cdot \frac{1}{2^{100}}.$$

I used computer and knew the former one is (much) larger. So I am wondering if there is a mathematical way which we can show the result by hand? Thanks!

答:
For some $n$, there's

$${2n\choose n} \cdot 2^{-2n}=\frac{(2n)!}{n!n!2^{2n}}$$ chance of getting $n$ heads out of $2n$ flips.

For $n+1$, we have

$${2n+2\choose n+1} \cdot 2^{-(2n+2)}=\frac{(2n+2)!}{(n+1)!(n+1)!2^{2n+2}}$$

Divide the second result by the first:

$$=\frac{(2n+2)!n!n!2^{2n}}{(2n)!(n+1)!(n+1)!2^{2n+2}}=\frac{(2n+1)(2(n+1))}{4(n+1)(n+1)}=\frac{2n+1}{2n+2}$$

is the ratio of the chance of flipping $n+1$ heads out of $2n+2$ coins to the chance of flipping $n$ heads out of $2n$ coins, i.e. the amount by which your probability of half heads will change if you add another two flips to the mix. Since it's under $1$, by induction, it will always shrink, and so the probability of half heads is always larger with fewer coins.

Check by plugging in $n=1$ for one heads out of two flips compared to $n=2$ for two heads out of four flips, and we have $\frac{3}{4}=\frac{0.375}{0.5}$, which looks right to me for those first two cases.

Now that I look at it again, I guess you can use that result to simply say

$$P(n\text{ heads in }2n\text{ flips})=\frac{(2n-1)!!}{(2n)!!}=\frac{(2n-1)(2n-3)(2n-5)\cdots (5)(3)(1)}{(2n)(2n-2)(2n-4)\cdots (4)(2)}.$$

So yeah, that's another way to conclude $P$ must go down as $n$ increases.


问:
Evaluating limit without L'Hospital's
I would like to evaluate the limit $$\lim_{x\to0}\frac{2x\sin(x)}{1-\cos(x)}$$ It can be done with L'Hospital, but I would like to know if it is possible to do so without L'Hospital. I tried multiplying through by the conjugate of the denominator, which leads me to $$\lim_{x\to0}\frac{2x(1+\cos(x))}{\sin(x)}$$ which is not very helpful, because it is still $0/0$. Any help would be much appreciated, thanks!

Title: Evaluating limit without L'Hospital's

Body:
Hint

$$1-\cos(x)=2\sin^2\left(\frac{x}{2}\right),$$
and $$\lim_{x\to 0}\frac{\sin(x)}{x}=1.$$


问:
Why don't we study algebraic objects with more than two operations?
Undergraduates learn about algebraic objects with one operation, namely groups, and we learn about algebraic objects with two "compatible" operations, namely rings and fields. It seems natural to then look at algebraic objects with three or more operations that are compatible, but we don't learn about them. I asked one of my professors why this is so and she posed this question in response: we have left-modules and right-modules, but there are no top-modules or bottom-modules, or any other ways of combining two elements to produce a third. I can't think of any satisfactory answer to either of these questions. Can anyone shed any light on them?

Edit: Now that I know these objects are studied, what I meant by "we" is essentially "Why are these objects not introduced to undergraduates (at least in a standard curriculum) given how natural they seem?"

答:
This might be a bit artificial, but from the point of view of Universal Algebra, vector spaces are structures equipped with possibly infinite amount of unary operations - if $V$ is a vector space over a field $F$, we define $x \mapsto f\cdot x$ for each $x\in V$ and $f\in F$ as a unary operation in $V$. This way we need to add a unary operation for each element of $F$. In particular, if $F$ is infinite, this gives us infinite amount of unary operations.

The same is true for modules, group actions, semigroup actions.

This is so that the class of vector spaces over $F$ forms a variety, or equational class.

ContentLicense: CC BY-SA 3.0


答: From the later form you can use $\lim_{x\to 0} \frac{\sin x}{x} = 1$ directly with the product rule to show that the limit is $4$. This identity can be proven without L'Hopital so it's valid to use it. A lot of limits involving trigonometric functions can be solved this way

Title: Evaluating limit without L'Hospital's

Body: I would like to evaluate the limit $$\lim_{x\to0}\frac{2x\sin(x)}{1-\cos(x)}$$ It can be done with L'Hospital, but I would like to know if it is possible to do so without L'Hospital. I tried multiplying through by the conjugate of the denominator, which leads me to $$\lim_{x\to0}\frac{2x(1+\cos(x))}{\sin(x)}$$ which is not very helpful, because it is still $0/0$. Any help would be much appreciated, thanks!


问:
Proving that commensurability is transitive
We have that two groups $\Gamma$ and $\Gamma'$ are commensurable if there exist finite index subgroups $G \leq \Gamma$ and $G' \leq \Gamma'$ such that $G \cong G'$. We denote this $\Gamma \approx \Gamma'$.

I am trying to prove that this gives a transitive relation, but I just don't see why it needs to be. Clearly if $\Gamma \approx \Gamma'$ and $\Gamma' \approx \Gamma''$ we have finite index subgroups $G \leq \Gamma$, $G' \leq \Gamma'$, $H' \leq \Gamma'$ and $H'' \leq \Gamma''$ with $G \cong G'$ and $H' \cong H''$ but this doesn't mean that $G \cong H''$...

I feel that I must be missing something obvious, but I can't see why we have to have finite index subgroups of $\Gamma$ and $\Gamma''$ which are isomorphic to each other.

答:
An idea:

$$G\cong G'\implies \,\forall\,H'\le \Gamma'\;\exists\, H\le \Gamma\;\;s.t.\;\;G\cap H\cong G'\cap H'$$

because isomorphic groups have isomorphic subgroups. But then in fact

$$G'\cap H'\cong K'\cap H''\;,\;\;\text{for some finite index}\;\ K'\le\Gamma'\;\text{(same argument as above)} $$

and now remember that $\,[G:H]\,,\,[G:K]<\infty\implies [G:H\cap K]<\infty$


问:
How did they use Fermat's Little Theorem here?
This is the question:

In the solution,

How did they use FLT to get $6x+6$?

From what I can see here is that in mod7, all the terms between $x^{7}$ and 62 become $0$ in mod7 and the 62 is clearly 6 mod7 but I am not quite sure how the $x^{7}$ is $6x$ mod7.

答:
Fermat Little Theorem:  ${x^{p} \equiv x\bmod p}$ , where p is prime, i.e., ${x^p-x} $ is always divisible by p

All term are divisible by 7 except:

$x^7\equiv x \bmod 7$

$96\equiv 5\bmod 7\Rightarrow 96x\equiv 5x\bmod 7$

$62\equiv 6\bmod 7$

So:

$f(x)\equiv (x+5x+6)=(6x+6)\bmod 7$


问:
Problem from Bryant's book on combinatorics
I haven't found this question to be asked before.
Here is a problem that might seem easy at first glance:

"There are $n$ people in a queue for the cinema (and, being in England, the order of people in the queue never changes).
They are let into the cinema in $k$ batches, each batch consisting of one or more persons.
In how many ways can the k batches be chosen?"

From my own calculations the solution should be: $\dbinom{n-1}{k-1}$ where $n$ is the number of people in the queue.

Could someone help me find out the general pattern arising in this problem,
and/or correct my own solution(which may be incorrect)?

答:
André Nicolas gave the answer in a comment.

If you create $k$ batches by inserting $k-1$ separators into the queue  by giving the last person in a batch a separator, you should not give one to the last person in the queue.  So you have $n-1$ people to give $k-1$ separators to, and so have ${n-1 \choose k-1}$ possible ways of doing this.


答:
Use equivalents: it is  standard that  $\:\sin x\sim_0 x\:$ and $\:1-\cos x\sim_0\frac{x^2}2$. Therefore
$$\frac{2x\sin x}{1-\cos x}\sim_0\cdots$$

Title: Evaluating limit without L'Hospital's


问:
Show that the sequence defined by $x_{n+1}=\frac{1}{3}(x_n^2+x_n+1), x_0=\sqrt 2$ diverge
I would like to show that the sequence $(x_n)$ defined by $x_{n+1}=\frac{1}{3}(x_n^2+x_n+1), x_0=\sqrt{2}$ is divergent. I would like to know if my proof is correct and to have a critical feedback.

First of all we can show by induction that the sequence is increasing.

Suppose that $x_{n+1}>x_n \ \forall n\ge0$. It is true for $n=0$, because $1+\frac{\sqrt{2}}{3}>\sqrt{2}$. Now we want to show that is holds for $n+1$. So, we want to show that $x_{n+2}>x_{n+1}$. But,

$x_{n+2}>x_{n+1} \iff x_{n+1}^2+x_{n+1}>x_n^2+x_n \iff \underbrace{x_{n+1}^2-x_n^2}_{> 0}+\underbrace{x_{n+1}+x_n}_{>0}>0$

which is true because we supposed by induction that $x_{n+1}>x_n$. So, $(x_n)$ is an increasing sequence.

We know as well that $\lim_{n \to \infty}x_n=\lim_{n \to \infty}x_{n+1}$, so we can resolve the following equation to find a limit $l$ of $(x_n)$ as it goes to infinity:

$3l=l^2+l+1 \Leftrightarrow l^2-2l+1=0 \Leftrightarrow (l-1)^2=0$, so the solution is $l=1$.

But, we have proved that $x_n$ is an increasing sequence and it is clearly bounded below by $\sqrt{2}$, so the solution $l=1$ doesn't hold. We conclude that $(x_n)$ diverge.

答:
Your proof is right, as others have mentioned in comments. Here is an alternative proof:
\begin{align}
x_{n+1}-x_{n} &= \dfrac13(x_n^2+x_n+1)-x_n\\
&=\dfrac13(x_n-1)^2
\end{align}
Note that, above shows in one go that the given sequence is non-decreasing and the limit, if exists, is $1$. So there is no convergence.


问:
Homotopy with rational coefficients of simply connected manifolds that are not-rational homology spheres
I have been reading a book on differential geometry/dynamical systems and this question comes from a fact used in a proof. The author says the following :

Note first that $W$ is simply connected and closed and is not a rational homology sphere, then $n:=\dim W\geq 4$ and there exists $i_0\neq 0,1,n-1,n$ such that $b:=\dim H_i(W,\mathbb{Q})>0$ and $\dim H_i(W,\mathbb{Q})=0$ for $1\leq i<i_0$.

Now why is this the case ? I guess the part about $\dim W\geq 4$ could follow from the Poincare Conjecture but this seems a bit overkill. Now I don't know why the rest is true .

Any help is appreciated. Thanks in advance.

答:
As $W$ is simply connected, we have $H_1(W; \mathbb{Q}) \cong \pi_1(W)^{\text{ab}} = 0$. Moreover, since $W$ is closed and orientable, we have $H_{n-1}(W; \mathbb{Q}) \cong H^1(W; \mathbb{Q}) \cong \operatorname{Hom}(H_1(W; \mathbb{Q}), \mathbb{Q}) = 0$ where the first isomorphism is given by Poincaré duality and the second follows from the universal coefficient theorem.

If $n = 3$, it would follow that $W$ is a rational homology sphere. Since $W$ is not a rational homology sphere, we see that $n \geq 4$ and $H_j(W; \mathbb{Q}) \neq 0$ for some $j \in \{2, \dots, n-2\}$ (we already know that $H_1(W; \mathbb{Q})$ and $H_{n-1}(W; \mathbb{Q})$ are zero). Let $i_0$ be the smallest such value of $j$. Then by definition, $H_{i_0}(W; \mathbb{Q}) \neq 0$ and $H_i(W; \mathbb{Q}) = 0$ for $1 \leq i < i_0$. As $H_{i_0}(W; \mathbb{Q})$ is a non-zero vector space over $\mathbb{Q}$, we see that $b = \dim H_{i_0}(W; \mathbb{Q}) > 0$.


答:
This is a good opportunity to use power series: near $0$, we have $$\sin x=x-\frac{x^3}{6}+\cdots\quad\text{and}\quad\cos x = 1-\frac{x^2}{2}+\frac{x^4}{4!}+\cdots$$ so $$2x\sin x=2x^2-\frac{x^4}{3}+\cdots\quad\text{and}\quad 1-\cos x = \frac{x^2}{2}-\frac{x^4}{4!}+\cdots$$

where the dots represent terms of higher degree. It follows that

$$\frac{2x\sin(x)}{1-\cos(x)}
=
\frac{2x^2-\frac{x^4}{3}+\cdots}{\frac{x^2}{2}-\frac{x^4}{4!}+\cdots}
=
\frac{2-\frac{x^2}{3}+\cdots}{\frac{1}{2}-\frac{x^2}{4!}+\cdots}
$$

By continuity of the sum of power series, we obtain

$$\lim_{x\to0}\frac{2x\sin(x)}{1-\cos(x)}=\frac{2}{1/2}=4$$

Note: because of the term-by-term differentiation formula for power series, the fact that the valuation of the power series is $2$ implies that we can get the limit by using L'Hospital's Rule twice.

Title: Evaluating limit without L'Hospital's


问:
What is the probability of choosing the bomb?
Five crackers are put in a bowl full of bran. In four of them, there is a check for $ 2 million, one contains a bomb, strong enough to kill the person who pulls the cracker.

Mr. Adam and Mr. Bill reach for the bowl together. Mr. Adam opens up the lid and picks up one of the crackers. What is the probability of choosing the bomb for Mr. Adam and Mr. Bill? (It should be 1/5 for Mr. Adam and 1/5 for Mr. Bill)

$$P(X_1) =1/5$$

$$P(X_2) =P(X_2|X_1)*P(X_1) + P(X_2|¬X_1)*P(¬X_1)$$

where $X_1$ means choosing the bomb on the first turn and $X_2$ means choosing the bomb on the second turn

$$P(X_2) =0*(1/5) + (1/4)*(4/5)$$
$$P(X_2) =1/5$$
2. Mr. Adam and Mr. Bill decide to draw for turns to pick from the bowl. What are the probabilities of choosing the bomb in this case?
Let's say they toss a coin and if it comes head then Mr. Adam goes and if it comes tails then Mr. Bill. So, they could choose a bomb if they get heads and also if they get tails and the other (first mover) does not choose the bomb.
$$P(X) =P(X|H)*P(H) + P(X|T,¬X_1)*P(T)*P(¬X_1)$$
$$P(X) =(1/5)*(1/2) + (1/4)*(1/2)*(4/5)$$
$$P(X) =(1/5)$$

So, there is no difference in the probabilities of choosing the bomb whether they draw for turns or not. Am I getting it right?

答:
Say we have two persons A and B. If person A starts first, the probability of A choosing a specific item should be higher than that of person B.

Probability of person A (first person) choosing a specific box $ = \frac{1}{5} + \frac{4}{5} \cdot \frac{3}{4} \cdot \frac{1}{3} + \frac{4}{5} \cdot \frac{3}{4} \cdot \frac{2}{3} \cdot \frac{1}{2} = \frac{3}{5}$

Then probability of person B choosing the specific item $ = 1 - \frac{3}{5} = \frac{2}{5}$

In the above working, person A has probability of choosing a specific item as follows -

i) in the first chance, it is $\frac{1}{5}$

ii) gets a second chance only if fails to choose the specific item in the first chance (probability of $\frac{4}{5}$) and person B also fails to choose the item next, probability of which is $\frac{3}{4}$.

iii) is certain to choose the item if both of them fail to choose the item in their first two chances.

Now if the first person can be either of them with probability $\frac{1}{2}$ based on toss of a coin, then it is easy to see that they have equal probability of picking a specific item: $\displaystyle \frac{1}{2}$.

$\big(\frac{1}{2} \cdot \frac{3}{5} + \frac{1}{2} \cdot \frac{2}{5} = \frac{1}{2} \big)$.


问:
Concentration for stochastic integral involving the Brownian motion itself
Let $\{W_t\}_{0 \le t \le h}$ be Brownian motion in $\mathbb{R}$. If $\{g_t\}_{0 \le t \le h} \subset \mathbb{R}$ is a fixed scalar field (independent of the Brownian motion) parameterized by time, then I know how to obtain concentration of the following integral:
$$\int_0^h g_t dW_t$$
because it is equivalent via time-change to a draw from the distribution
$$\mathcal{N}\left(0, \int_0^h g_t^2 dt \right). $$
My question is what to do in the case when we are dealing with a scalar field that depends on the Brownian motion. In particular, suppose $g: \mathbb{R} \rightarrow \mathbb{R}$ is some scalar field (say it is $1$-Lipschitz, so it can grow if the Brownian motion gets too large), and I want to understand concentration of
$$\int_0^h g(W_t)dW_t.$$
Is there a standard way of doing this? My apologies if this is a naive question.

Edit: I am trying to be competitive with the following strategy -- first, apply some tail bound on how large $W_t$ ever gets, and then using this tail bound obtain a bound on the variance in
$$\mathcal{N}\left(0, \int_0^h g(W_t)^2 dt\right). $$
Then, apply a standard Gaussian tail bound. However, I feel this doesn't make any sense since if I tell you the trajectory of the Brownian motion then the integral is no longer a random variable, so it doesn't seem right to treat the result as a draw from an appropriate Gaussian.

答:
Answer for linear g

By definition for partitions $\pi=\{0=t_0<...<t_n=h\}$, $|\pi| = \sup_{i} |t_{i+1} - t_i |$

$$\int_0^h g(W_t)dW_t = \lim_{|\pi|\to 0} \sum_{i=1}^{n-1} g(W_{t_i})(W_{t_{i+1}}-W_{t_i}) $$

So for $g(x) = ax +b$ we can do the following:

$$\begin{align}
\sum_{i=1}^{n-1} (aW_{t_i} + b)(W_{t_{i+1}}-W_{t_i})
&= a \sum_{i=1}^{n-1} W_{t_i} (W_{t_{i+1}}-W_{t_i}) + b\sum_{i=1}^{n-1} (W_{t_{i+1}}-W_{t_i}) \\
&= a \sum_{i=1}^{n-1} W_{t_i} (W_{t_{i+1}}-W_{t_i}) + b(W_h - W_0)
\end{align}$$

So we only really need $g(x) = x$, which we can get using the following argument:
$$x(y-x) = xy - x^2 =\frac{x^2 + y^2 -(x-y)^2}{2} - x^2 = \frac{1}{2}((y^2-x^2) -(y-x)^2)$$

Which means

$$\begin{align}
\sum_{i=1}^{n-1} W_{t_i} (W_{t_{i+1}}-W_{t_i})
&= \frac{1}{2}\left(\underbrace{\sum_{i=1}^{n-1} (W_{t_{i+1}}^2-W_{t_i}^2)}_{=W^2_h - W^2_0} + \underbrace{\sum_{i=1}^{n-1} (W_{t_{i+1}}-W_{t_i})^2}_{\to [W]_h-[W]_0 = h-0}\right)\\
&\to\frac{1}{2}(W_h^2 - h)
\end{align}$$

where $[W]_h=h$ is the quadratic variation of the brownian motion. So for linear $g(x)=ax+b$, we get:

$$\int_0^h g(W_t)dW_t = \frac{a}{2}(W_h^2 - h) + bW_h$$

Possible approach for non linear, differentiable g

By the Ito Formula

$$F(W_h) - F(W_0) = \int_0^h \underbrace{\frac{\partial  F}{\partial x}(W_t)}_{\stackrel{!}{=}g(W_t)} dW_t + \frac{1}{2}\int_0^h \underbrace{\frac{\partial^2 F}{\partial x^2}(W_t)}_{=g'(W_t)} d[W]_t$$

So

$$\int_0^h g(W_t)dW_t = \int_0^{W_h}g(x)dx - \frac{1}{2}\int_0^h g'(W_t) dt$$

The issue here of course is the calculation of the integral over $g'$. If the derivative is bounded this might provide you with some bounds though.


问:
On Humphreys’ proof of conjugacy of Borel subalgebras
I am reading Humphreys’ Introduction to Lie Algebras and Representation Theory and have trouble understanding the first step of the proof of conjugacy of Borel subalgebras (§16.4).

Hereinafter $ L $ is a finite-dimensional semisimple Lie algebra over an algebraically closed field of characteristic $ 0 $.

Some Definitions and Facts

A Borel subalgebra of $ L $ is a maximal solvable subalgebra of $ L $.

Let $ H $ be a Cartan subalgebra of $ L $, $ \Phi $ be the root system of $ (L, H) $, and
$$
L = H \oplus \bigoplus_{\alpha \in \Phi} L_\alpha
$$
be the root space decomposition of $ L $. Fix a basis $ \Delta $ of $ \Phi $ and let
\begin{align*}
B(\Delta) &= H \oplus \bigoplus_{\alpha \succ 0} L_\alpha, \\
N(\Delta) &= \bigoplus_{\alpha \succ 0} L_\alpha.
\end{align*}
$ B(\Delta) $ is called a standard Borel subalgebra of $ L $ (relative to $ H $).

Then, as stated in §16.3,

$ N(\Delta) = [B(\Delta), B(\Delta)] $.

$ B(\Delta) $ is a Borel subalgebra of $ L $, and $ N(\Delta) $ is a nilpotent subalgebra of $ L $. Moreover, we can see that $ \operatorname{ad}_L x $ is nilpotent for $ x \in N(\Delta) $.

Question

Fix a standard Borel subalgebra $ B $ of $ L $ and take an arbitrary Borel subalgebra $ B' $. Let
$$
N' = \{x \in B \cap B' \mid \text{$ \operatorname{ad}_L x $ is nilpotent}\},
$$
and consider the case that $ N' \neq 0 $. We can see that $ N' $ is an ideal in $ B \cap B' $ and that its normalizer $ K = N_L(N') $ is a proper subalgebra of $ L $ (p. 85, ll. 2–5). Next, Humphreys says that (p. 85, ll. 6–11. [[ ]] is by the quoter):

Next we show that $ B \cap B' $ is properly contained in both $ B \cap K $, $ B' \cap K $. For consider the action of $ N' $ on $ B/(B \cap B') $ induced by $ \operatorname{ad} $. Each $ x \in N' $ acts nilpotently on this vector space, so by Theorem 3.3 [[the main step of the proof of Engel’s Theorem]] there must exist nonzero $ y + (B \cap B') $ killed by all $ x \in N' $, i.e., such that $ [xy] \in B \cap B' $, $ y \notin B \cap B' $. But $ [xy] $ is also in $ [BB] $, so is nilpotent; this forces $ [xy] \in N' $, or $ y \in N_B(N') = B \cap K $, while $ y \notin B \cap B' $. Similarly, $ B \cap B' $ is properly contained in $ B' \cap K $.

I have trouble understanding the last sentence “Similarly, $ B \cap B' $ is properly contained in $ B' \cap K $”; it seems to me that we use the fact that $ B $ is standard in “$ [xy] $ is also in $ [BB] $, so is nilpotent” and that this argument is no longer valid when $ B $ is replaced by $ B' $.

答:
I also thought that I need the condition that B is standard Borel for that part. However, it turned out that the only condition I need is that B is solvable.

Lie's Theorem (4.1 in Humphreys' book) says that a solvable Lie subalgebra of $\mathfrak{gl}(V)$ for a finite dimensional vector space $V$ must be contained in the Lie algebra of upper triangular matrices with respect to some basis of $V$. We can apply this on the solvable Lie algebra $\text{ad}_L B$. Then $[\text{ad}_L B, \text{ad}_L B]=\text{ad}_L [B,B]$ must be strictly upper triangular, so the elements of $[B,B]$ are nilpotent in $L$.


问:
Prove the Recursion Theorem
Let $g$ be a function on a subset $A\times\Bbb N$ into $A$, and $a \in A$

Prove there is a unique sequence, $f$ of elements of $A$ such that

a) $f_{0} = a$

b) $f_{n+1} = g(f_{n}, n)$ for all $n \in\Bbb N$ such that $(n+1) \in\operatorname{dom}f$

c) $f$ is either an infinite series or $f$ is a finite sequence of length $k+1$ and $g(f_{k}, k)$ is undefined

I have no idea how to approach this :(

答:
Uniqueness is very often easier to prove than existence, so I’d start there. Suppose that $f$ and $h$ both satisfy (a)-(c):

$f_0=h_0=a$;

$f_{n+1}=g(f_n,n)$ for all $n\in\Bbb N$ such that $n+1\in\operatorname{dom}f$;

$h_{n+1}=g(h_n,n)$ for all $n\in\Bbb N$ such that $n+1\in\operatorname{dom}h$;

$f$ is either an infinite sequence or a finite sequence of length $k+1$ such that $g(f_k,k)$ is undefined; and

$h$ is either an infinite sequence or a finite sequence of length $k+1$ such that $g(h_k,k)$ is undefined.

If $f\ne h$, one of two things must be true: either there is a least $n\in\Bbb N$ such that $f_n$ and $h_n$ are both defined, but $f_n\ne h_n$, or $f$ and $h$ agree wherever both are defined, but one has a larger domain than the other. In the second case we may without loss of generality assume that $f$ is a finite sequence of length $k+1$ such that $g(f_k,k)$ is undefined, that $f_n=h_n$ for $n\le k$, and that $k+1\in\operatorname{dom}h$. In either case it’s not hard to derive a contradiction, thereby showing that in fact $f=h$.

You can prove existence by looking at approximations to $f$. Say that finite sequence $\sigma$ of elements of $A$ of length $k+1$ is a k-approximation to $f$ if $\sigma_0=a$ and $\sigma_{n+1}=g(\sigma_n,n)$ for $n<k$. Clearly $f$ has a $1$-approximation. Now show that either

there is some $k\in\Bbb N$ such that $f$ has a $k$-approximation $\sigma$ such that $g(\sigma_k,k)$ is undefined,

or

$f$ has a $k$-approximation for each $k\in\Bbb N$.

Show that in the first case the $k$-approximation $\sigma$ is the desired $f$. Show in the second case that if $k,\ell\in\Bbb N$ with $k\le\ell$, $\sigma$ is a $k$-approximation to $f$, and $\tau$ is an $\ell$-approximation to $f$, then $\sigma=\tau\upharpoonright(k+1)$ (i.e., $\tau_i=\sigma_i$ for all $i\le k$) and that the union of these $k$-approximations over $k\in\Bbb N$ is the desired $f$.

Title: Prove the Recursion Theorem


答:
It is easy to demonstrate that $$p_n=\frac{(2n)!}{n!n!}\frac1{2^{2n}}$$
is decreasing function of $n$. In this answer a practical way to estimate the value of the coefficient will be shown.

For large $n$ we can compute the probability using the Stirling approximation as follows:
$$
p_n\approx
\frac{\sqrt{4\pi n}\left(\frac{2n}e\right)^{2n}}{2\pi n\left(\frac{n}e\right)^{2n}}\frac1{4^n}=\frac1{\sqrt{\pi n}}
$$

This means that for large $n$ and $k$, the ratio of probabilities is:
$$
\frac{p_n}{p_k}\approx\sqrt{\frac kn}.
$$

Somewhat surprisingly it works well already in your example:
$$
\left(\frac{p_{5}}{p_{50}}\right)^2\approx 9.56076.
$$

Title: Which one is more likely to happen:$10$ tosses with $5$ head up and $100$ tosses with $50$ head up?

Body:
If we have a fair coin is tossed multiple times, which one is more likely to happen:
$10$ tosses with $5$ head up or $100$ tosses with $50$ head up.

So the question translates mathematically into comparing the following two numbers $${10\choose 5} \cdot \frac{1}{2^{10}}\quad\text{and}\quad{100\choose 50} \cdot \frac{1}{2^{100}}.$$

I used computer and knew the former one is (much) larger. So I am wondering if there is a mathematical way which we can show the result by hand? Thanks!


问:
$k[X_1,...,X_{m}]\otimes_kk[X_{m+1},...,X_{n+m}]\simeq k[X_1,...,X_{n+m}]$

The ring $k[X_1,...,X_m,X_{m+1},...,X_{n+m}]$, together with the obvious inclusions $$k[X_1,...,X_{m}]\hookrightarrow k[X_1,...,X_{n+m}]\hookleftarrow k[X_{m+1},...,X_{n+m}]$$
is the tensor product of the $k$-algebras $k[X_1,...,X_m]$ and $k[X_{m+1},...,X_{n+m}]$ where $k$ is a commutative ring with unity.

First I want to show this by showing the above two inclusions have a universal property but I don't know how to define the unique $k$-algebra homomorphism $\Phi$ in the following diagram:

where $\phi$ and $\psi$ are the given $k$-algebra homomorphism. How can I define it?

Second, the proof of this statement in Milne's textbook is the following: To verify this we only have to check that, for every $k$-algebra $R$, the map $$\text{Hom}(k[X_1,...,X_{m+n}],R)\to \text{Hom}(k[X_1,...],R)\times\text{Hom}(k[X_{m+1},...],R)\tag{1}$$
induced by the inclusion is a bijection. But this map can be identified with the bijection $$R^{n+m}\to R^m\times R^n.\tag{2}$$
I understand that showing $(1)$ is same as showing universal property but how can $(1)$ be identified with $(2)$? Could you explain this to me?

Milne's book and the statement is in p.42 Example 10.4

Edit: Universal property of polynomial ring in Computational commutative algebra 1 by Martin & Lorenzo

答:
I believe that Milne's book wants you to verify that $k[X_1, \dots, X_{n+m}]$ satisfies the universal property of the coproduct of $k[X_1, \dots, X_n]$ and $k[X_{n+1}, \dots, X_{n+m}]$ in the category of commutative unital rings (which is the tensor product of rings).

Note that $\Phi$ is the unique ring morphism satisfying $\Phi i = \phi $ and $\Phi j = \psi$. In particular, $\Phi(X_i) = \phi(X_i)$ for $i=1,\dots, n$ and $\Phi(X_i) = \psi(X_i)$ for $i = n+1, \dots, n+m$. By the universal property of the polynomial algebra, there is a unique morphism $\Phi: k[X_1, \dots, X_{n+m}]\to R$ satisfying this and you are done.

This universal property (of the polynomial algebra) asserts that if $R$ is a commutative algebra, then we have a bijection
$$\text{Hom}_{Alg}(k[X_1, \dots, X_n], R) \cong R^n: f \mapsto (f(X_1), \dots, f(X_n))$$  I.e., an algebra morphism $k[X_1, \dots, X_n]\to R$ is uniquely determined by the choice of $n$ points of the ring $R$. This explains why $(1)$ and $(2)$ are equivalent.


问:
Infinite convex sum
On page 524 of paper https://www.jstor.org/stable/2039405?seq=1#metadata_info_tab_contents the following result mentioned is:

If $X$ is a convex subset of $\mathbb C$, $(z_n)$ be sequence of elements in $X$ and $(\alpha_n)$ be sequence of positive numbers such that $\sum\limits_{n=1}^{\infty}\alpha_n =1$, then $\sum\limits_{n=1}^{\infty}\alpha_n z_n$ is in $X$.

How can we prove it? Is it also true in any normed space or atleast finite dimensional normed space?

答:
Let $X$ be a convex subset of a finite-dimensional affine space $E$.  Let $x_k \in X$, $\alpha_k > 0$, such that $\sum_{k=1}^\infty \alpha_k = 1$.
Assume $\sum_{k=1}^\infty \alpha_k x_k$ converges.  Then $\sum_{k=1}^\infty \alpha_k x_k \in X$.

(If we like, we may subtract $x_1$ from everything to reduce to the case where $0 \in X$, so $E$ is a vector space, not merely an affine space.  But I have not done that.)

The proof is by induction on the affine dimension $n$ of $E$.

First consider the case $n=0$.  Then $X$ is a single point $X = \{x_0\}$.  So $x_k =x_0$ for all $k$ and $\sum_{k=1}^\infty \alpha_k x_0 = x_0 \in X$.

Now, for the induction step, let $n \ge 1$ be the affine dimension of $E$ and assume our result is true for any  dimension ${} < n$.

We have a convex set $X \subseteq E$,  points $x_k \in X$ ($k=1,2,\cdots$), scalars $\alpha_k > 0$ ($k=1,2,\cdots$),
such that $\sum_{k=1}^\infty \alpha_k =1$ and
$\sum_{k=1}^\infty \alpha_k x_k$ converges to a point $x_0 \in E$.  We want
to show $x_0 \in X$.  Assume (for purposes of contradiction) that $x_0 \notin X$.
Apply the Hyperplane Separation Theorem to the disjoint nonempty convex sets $X$ and $\{x_0\}$.  There is a nonconstant linear functional $L$ on $E$ and a scalar $\lambda \in \mathbb R$ such that $L(x) \ge \lambda$ for all $x \in X$, and $L(x_0) \le \lambda$.  I claim that $L(x_k) = \lambda$ for all $k$.  Indeed,
$L(x_k) \ge \lambda$ since $x_k$ in $X$; if $x_k > \lambda$ for some $k$, then
$$
\lambda = L(x_0) = L\left(\sum_{k=1}^\infty\alpha_k x_k\right) = \sum_{k=1}^\infty\alpha_k L(x_k)
> \sum_{k=1}^\infty \alpha_k \lambda = \lambda,
$$
a contradiction.  [A linear functional on a finite-dimensional space is automatically continuous.] So $L(x_k) = \lambda$ for all $k$. Now consider the affine space $E_\lambda = \{x \in E : L(x) = \lambda\}$ of dimension $n-1$, and the convex set
$X_\lambda = \{x \in X : L(x) = \lambda\}$ contined in $E_\lambda$.  We have $x_k \in X_\lambda$ for all $k$ and $x_0 \in E_\lambda$.  By the induction hypothesis, $x_0 \in X_\lambda$ so $x_0 \in X$.

This completes the induction.

Title: Infinite convex sum


问:
Singular power cardinals?
I am recently exploring set-theoretical postulates that contradict GCH. One particularly interesting one is the proposition "$2^\kappa$ is singular for each infinite cardinal $\kappa$". However I could not prove that this proposition is at all consistent with ZFC.

It seems that a slightly weakened form "$2^\kappa$ is singular for each infinite regular cardinal $\kappa$" is indeed consistent with ZFC, since the function $\kappa\mapsto\aleph_{\kappa^+}$ satisfies the conditions of Easton's theorem and there therefore exists a model where $2^\kappa=\aleph_{\kappa^+}$ for each regular $\kappa$. Can this result also be extended to all singular cardinals?

答: As it turns out, my question has already been answered in this post, where other related topics have also been discussed.</p>

Title: Singular power cardinals?


问:
Does half-life mean something can never completely decay?
Caffeine has a half-life of approximately six hours. I understand this to mean that every six hours, the amount of caffeine in the body is half of what it was six hours prior. Does that mean that caffeine never completely leaves the body? It just keeps reducing to half after a fixed time ad infintum.

I guess a broader statement of my question is: When calculating the half life of a thing, is it true that it will never reach zero? And how is this used/reconciled in a more practical setting like when calculating the amount of caffeine in the body?

答:
For half-life to make sense, the assumption that the metabolism that breaks down caffeine at a speed proportional the total amount in the body, and not on concentration or other things. Then exponential decay holds. This assumption doesn't exactly hold in practice, and the exponential decay is only approximate.

And of course, the caffeine is not a continuous quantity, therefore at some point we are talking about fraction of a molecule. Continuous dynamics is a good approximation for the number of caffeine molecules, but eventually they will break down completely.

See also http://en.wikipedia.org/wiki/Half-life for more information.


问:
similar matrices, similar polynomials?
Let $\mathbb{K}$ be a field and $p \in \mathbb{K}[X]$ a polynomial.

If $A,B \in \mathbb{K}^{n \times n}$ are similar, $p(A)$ and $p(B)$ are also similar

my attempt:

$A=S^{-1}BS$

$S,T\in \mathbb{K}^{n \times n}$

$p(B) = \sum^{n}_{i=0} a_i B^i$

$p(A)= \sum^{n}_{i=0} a_i(S^{-1}BS)^i$

$p(A) = T^{-1}p(B)T$

$\sum^{n}_{i=0} a_i(S^{-1}BS)^i = T^{-1} (\sum^{n}_{i=0} a_i B^i )T$

Im not sure how to solve this or if Im even on the right track

答:
Hint

This is a consequence of the identity

$$p(A) = p(S^{-1} B S) = S^{-1} p(B) S$$ which is easy to verify due to linearity and the equality

$$(S^{-1} B S)^n = \underbrace{(S^{-1} B S) \dots (S^{-1} B S)}_{n \text{ times}} = S^{-1} B^n S$$ valid for all $n \in \mathbb N$.

Title: similar matrices, similar polynomials?


问:
Algorithm to solve a puzzle
Firstly, note that this is from an exercise sheet, and we were asked to try and solve it, for a discussion.

We are given $n$ chips that can test each other in the following way: if two of the chips are connected,
then both determine whether the other one is faulty or correct. A correct chip decides properly about
the other, while a faulty chip can give an arbitrary answer. Assume, that more than half of the chips
are correct. Give an algorithm that uses less than $n$ tests of the above type to find a correct chip.

My attempt:

First, we categorize the possible outcome of different combinations. Chips will be in uppercase, their outcome will be lowercase.C/c = correct, F/f = fault. X will denote an unknown chip.

CF = {fc,ff}, CC = {cc}, FF = {fc,ff,cf,cc}, FC = {cf,ff}.

First set of observations:

if the outcome is fc, we know the chips were XF

if the outcome is cf, we know the chips were FX

My current idea is by first combining any 2 random chips together, then remove every outcome but cc, which can only be produced by FF or CC. Combine a chip from these 'cc'-pair with another chip from another 'cc'-pair. From here,if the possible combinations are

CC-FF = {fc,ff}, CC-CC = {cc}, FF-CC = {cf,ff}, FF-FF = {fc,ff,cf,cc}.

After this, I do not know how to proceed. Should I keep on focusing on just 'cc'- outcomes? Should I start eliminating faulty ones(by filtering out fc/cf). Of course, your own solution would also be very helpful to me. Thank You

答:
Use three piles 1, 2, 3, where originally all $n$ chips are in pile 1.
Repeatedly pick two chips from pile 1 and test. If the result is "cc" put one of the to pile 2 and the other to pile 3 (knowing that both chips are of the same type). In all other cases discard both. When discarding, we will always discard at least as many F as C, i.e., this method keeps the C chips in majority.
We repeat this until pile 1 is empty or has at most one chip X. We also know that the C/F statistics of pile 2 exactly equals that of pile 3.

If no X was left ($n$ was even), move pile 2 to pile 1 and discard pile 3, and start over again.

If an X was left, add it to either of the two piles, making one pile even and the other odd. If C is in minority in the odd pile, then there is at most a tie in the even pile, hence overall a minority of C, which is absurd. Therefore C is in majority in the odd pile. Move this pile to pile 1 and discard the other pile, and start over again.

Note that each test between two chips leads to removal of at least one of them: Either they are both discarded immediately, or one of them is discarded together with a whole pile. Hence after at most $n-1$ tests, we will have only one chip left, which must be C.

Title: Algorithm to solve a puzzle


问:
Find all $n \in \mathbb{N}$ such that $(n+2) \mid (n^2+5)$
I'm trying to find all $n \in \mathbb{N}$ such that

$(n+2) \mid (n^2+5)$

as the title says, I've tried numbers up to $20$ and found that $1, 7$ are solutions and I suspect that those are the only $2$ solutions, however I have no idea how to show that.

I've done nothing but basic transformations:

$(n+2) \mid (n^2+5)$

$\iff n^2+5 = k(n+2)$

$\iff n^2+5 \mod(n+2) = 0$

$\iff (n^2 \mod(n+2) + 5 \mod(n+2)) \mod(n+2) = 0$

Now I suspect the next step is to find all possible solutions for

$n^2 \mod(n+2)$, which I have no idea how to do.

答:
$(n+2) \mid (n^2+5)\implies \exists m \in \mathbb N$ such that :
$$\frac{n^2+5}{n+2}= m \iff n^2-mn+(5-2m)=0$$

By Solving the quadratic equation in $n$, We get:
$$n = \frac{m\pm \sqrt{m^2-20+12m}}{2}$$
Now since $n\in \mathbb N \implies m^2-20+12m$, Must be a perfect square:
$$m^2+12m-20 = x^2 \iff (m+6)^2-x^2=56$$
$$(m+6-x)(m+6+x)=56$$
You can look at the Factors of $56$ and try to solve the systems of equations, but you will end up with :
$$(m,x) \in \{(9,13), (3,5)\}$$
By Plugging the values of $m$ to the quadratic equation you will get :
$$n  \in \{1,7\}$$

I think that solving the quadratic equation is extremely useful in questions about perfect squares or divisibility.


问:
Interesting and unexpected applications of $\pi$
$\text{What are some interesting cases of $\pi$ appearing in situations that do not seem geometric?}$

Ever since I saw the identity $$\displaystyle \sum_{n = 1}^{\infty} \frac{1}{n^2} = \frac{\pi^2}{6}$$

and the generalization of $\zeta (2k)$, my perception of $\pi$ has changed. I used to think of it as rather obscure and purely geometric (applying to circles and such), but it seems that is not the case since it pops up in things like this which have no known geometric connection as far as I know. What are some other cases of $\pi$ popping up in unexpected places, and is there an underlying geometric explanation for its appearance?

In other words, what are some examples of $\pi$ popping up in places we wouldn't expect?

Body:
I find several generalized continued fraction expansions of $\pi$ also quite interesting. It is really astonishing to see how fractions arranged in regular pattern on the RHS gives rise to $\pi$! (which is not only irrational, but transcendental)
\begin{align}
\displaystyle \pi &={\cfrac {4}{1+{\cfrac {1^{2}}{2+{\cfrac {3^{2}}{2+{\cfrac {5^{2}}{2+{\cfrac {7^{2}}{2+{\cfrac {9^{2}}{2+\ddots }}}}}}}}}}}}\\
&={\cfrac {4}{1+{\cfrac {1^{2}}{3+{\cfrac {2^{2}}{5+{\cfrac {3^{2}}{7+{\cfrac {4^{2}}{9+\ddots }}}}}}}}}}\\
&=3+{\cfrac {1^{2}}{6+{\cfrac {3^{2}}{6+{\cfrac {5^{2}}{6+{\cfrac {7^{2}}{6+{\cfrac {9^{2}}{6+\ddots }}}}}}}}}}\\
&=2+{\cfrac {2}{1+{\cfrac {1}{1/2+{\cfrac {1}{1/3+{\cfrac {1}{1/4+\ddots }}}}}}}}\\
&=2+{\cfrac {2}{1+{\cfrac {1\cdot 2}{1+{\cfrac {2\cdot 3}{1+{\cfrac {3\cdot 4}{1+\ddots }}}}}}}}\\
&=2+{\cfrac {4}{3+{\cfrac {1\cdot 3}{4+{\cfrac {3\cdot 5}{4+{\cfrac {5\cdot 7}{4+\ddots }}}}}}}}
\end{align}
A particularly interesting that I find is the following:
\begin{equation}
\pi=3+{\cfrac {1^{3}}{6+{\cfrac {1^{3}+2^{3}}{6+{\cfrac {1^{3}+2^{3}+3^{3}+4^{3}}{6+{\cfrac {1^{3}+2^{3}+3^{3}+4^{3}+5^{3}+6^{3}}{6+{\cfrac {1^{3}+2^{3}+3^{3}+4^{3}+5^{3}+6^{3}+7^{3}+8^{3}}{6+\ddots }}}}}}}}}}
\end{equation}
Tell me if someone finds a geometrical intuition for these:)


问:
Reshape operator
I am reshaping matrices into a single vector. This is the function that I am using. Is there a conventional symbol for this operation (latex command)?</p>

答:
This is typically denoted by the operator $\text{vec}(\cdot)$.

In MATLAB, you could also just use the built-in colon operator (e.g. b = A(:)), which will stack entries of a matrix in column-major order.

Title: Reshape operator


问:
Boundedness of gradient
Let $f:\mathbb{R}^N\to\mathbb{R}^N$ is such that $|\langle f(x),y\rangle|\leq C|x||y|$ holds for all $x,y\in\mathbb{R}^N$.

Then, there exists a fixed positive constant $C$ such that
$$
|f(x)|\leq C|x|,\,\forall x\in\mathbb{R}^N.
$$

I tried to prove this way: We can write
$$
|f(x)|=\sup_{y\in\mathbb{R}^N\setminus\{0\}}\frac{|\langle f(x),y\rangle|}{|y|}.
$$
Hence, using the given condition $|\langle f(x),y\rangle|\leq C|x||y|$, we get $|f(x)|\leq\frac{C|x||y|}{|y|}=C|x|$. Therefore, the result follows.

Please inform me if I am missing something. Thanks.

答:
Since The hypothesis is true for all $\mathbf x, \mathbf y \in \mathbb R^n$ then it's true for $\mathbf y = \mathbf 0$.

So let $\mathbf y = \mathbf 0$.

$$\Vert ( \mathbf f( \mathbf x),\mathbf y)\Vert= \Vert ( \mathbf f( \mathbf x),\mathbf 0)\Vert = \Vert\mathbf f( \mathbf x)\Vert \le C \,\Vert \mathbf x\Vert \, \Vert\mathbf y\Vert= C \,\Vert \mathbf x\Vert \, \Vert\mathbf 0\Vert=0.$$

Hence $\Vert\mathbf f( \mathbf x)\Vert \le 0$, which implies that $\Vert\mathbf f( \mathbf x)\Vert = 0$.

Therefore, for all $\mathbf x \in \mathbb R^n$, $\Vert\mathbf f( \mathbf x)\Vert \le C$ for any $C \ge 0$.

Edit:

I don't know whether it was supposed to be the case that $\mathbf y \neq \mathbf 0$; in that case take $\mathbf y = \mathbf e_i$, so:
$$\Vert \mathbf f(\mathbf x)\Vert \le \Vert (\mathbf f (\mathbf x),\mathbf e_i)\Vert \le C \, \Vert \mathbf x\Vert \, \Vert \mathbf e_i\Vert=C \, \Vert \mathbf x\Vert$$

Title: Boundedness of gradient


问:
Probability of a biased coin
If we toss a biased coin that is twice as likely to come up heads as tails, $5$ times,

what is

a. the probability of observing $2$ heads out of the $5$ tosses

b. the probability of observing at least $2$ heads out of the $5$ tosses

c. the probability of observing at most $2$ heads out of the $5$ tosses

I think it is a basic problem but I am finding a hard time on figuring it out.

Is it binomial distribution?

答:
The probability of observing $k$ head out of $n$ coin tosses is $P(n,k)=\binom{n}{k} \cdot p^k\cdot (1-p)^{n-k} $, where $p$ is the probability to observing head, and as consequence $1-p$ is probability of observing cross. In this problem $p$ is twice as $1-p$, so $p=2\cdot(1-p) \to 3p=2 \to p=\frac{2}{3}$.
For example, point a. gives:

$P(2,5)=\binom{5}{2} \cdot \bigl( \frac{2}{3}\bigr)^2\cdot \bigl(\frac{1}{3}\bigr)^{3} \approx 0.16 $

Title: Probability of a biased coin


问:
Prove that $V = \text{null }\varphi \oplus \{au:a \in \mathbb{F}\}$
As I am self learning Linear Algebra Done Right, I would like to make sure I learn it correctly. Please help on the following the proof.

Suppose $\varphi \in \mathcal{L}(V, \mathbb{F})$. Suupose $u \in V$ is not in null $\varphi$. Prove that $V = \text{null }\varphi \oplus \{au:a \in \mathbb{F}\}$

My approach:

Suppose $\varphi \in \mathcal{L}(V, \mathbb{F})$, and  $u \in V$ is not in null $\varphi$. Let $U = \{au:a \in \mathbb{F}\}$ .

Let $w_1,...,w_m$ be the basis of null $\varphi$. Extend $w_1,...,w_m$ to be a basis of $V$ as $w_1,...,w_m, v_1,...,v_n$. Then $v_1,...,v_n$ is the basis of $\{au:a \in \mathbb{F}\}$ [I cannot convince myself here. Seems right.Not sure]

For any $v \in V$, $v = a_1m_1 + ... + a_mw_m + b_1v_1+...+b_nv_n = \text{null }\varphi + U$.

Suppose $v \in \text{null }\varphi \cap U$, then $v = a_1m_1 + ... + a_mw_m = b_1v_1+...+b_nv_n $. $a_1m_1 + ... + a_mw_m - b_1v_1 -...-b_nv_n = 0$. Since $w_1,...,w_m, v_1,...,v_n$ is a basis of $V$, $a_1 = ... = a_m = b_1 = ... = b_n = 0$ which  implies $v = 0$.

Therefore $V = \text{null }\varphi \oplus \{au:a \in \mathbb{F}\}$.

答:
As the author-@Sheldon Axler-has mentioned, there is no assumption that $V$ is finite-dimensional. Thus you could not use bases.

I'll give you a complete answer now. Based on the following theorem:

Direct sum of two subspaces: Suppose $U$ and $W$ are subspaces of $V$. Then $U+W$ is a direct sum if and only if $U \cap W = {0}$.

It is obvious that $\text{null} \varphi \cap \{au: a \in F\} = 0$. (You can verify it yourself.)

What is left to do is to prove that for any vector $v \in V$, $v$ can be written as $v=k+au$ where $k\in \text{null} \varphi$. That is, we wanna prove $V = \text{null} \varphi + \{au: a \in F\}$.

Because $u$ is not in $\text{null} \varphi$, let $T(u)=f_1\neq 0$. For an arbitrary $v$, let $T(v)=f_0$ ($f_0$ is any number in $F$). If $T(v) = f_0=0$, then $v$ is in $\text{null} \varphi$. If $T(v) = f_0 \neq 0$, then $T(v)= mT(u)= T(mu)$, where $m=\frac{f_0}{f_1}$. So $T(v-mu) = 0$ and $v=k+mu$ where $k\in \text{null} \varphi$. In both cases, $v$ can be written as $v=k+au$.

Proof completed.

Be careful! You can't use the following theorems in the book, which use basis in their proofs or require space is finite dimensional.

Every subspace of V is part of a direct sum equal to V: Suppose $V$ is finite-dimensional and $U$ is a subspace of $V$. Then there is a subspace $W$ of $V$ such that $V = U \oplus W$.

or

Suppose $V$ is finite-dimensional and $T \in \mathcal{L}(V,W)$. Then range $T$ is finite-dimensional and $\text{dim} V = \text{dim} T + \text{dim range} T$.


问:
Is the empty set empty in all models of set theory?
In Timothy Bays' excellent article on Skolem's paradox, I have been stumped by the following lines:

Let's grant that the specific element which serves as “the empty set”
will not remain constant as we move from one model of set theory to
another—with the empty set in the first model becoming, perhaps, a
singleton in the second. Nonetheless, we can still use a formula in
the language of set theory to capture the notion “x is the empty
set” in an essentially absolute way. In any model of our axioms, an
element m' ∈ M will satisfy the open formula “∀y y ∉ x” if
and only if the set {m | M ⊨ m ∈ m' } is really empty.
Hence, there's at least a sense in which we can still capture the
notion “x is the empty set” from within the algebraic framework. And
this point extends more widely—a similar argument would apply to
notions like “x is a singleton” or “x has seventeen members.” Even
on the algebraic conception of axiomatization, therefore, there are
some set-theoretic notions which we can still pin down pretty
precisely.

At first he says that the element for "the empty set" will not remain constant from one model to another, but then he goes on to show that the empty set is really the self-same empty set in all models. What am I missing here?

答:
This is the difference between intensional and extensional definitions. We like to think of sets as classes, i.e. every set is definable (remember that we allow parameters, so $A$ is given by the class $x\in y$, where $A$ is taken as the parameter $y$).

But in reality that can easily fail when the model's $\in$ relationship is not the "real $\in$".

The internal empty set is simply the object $\varnothing^M$ (in $M$) such that $M\models\forall x(x\notin\varnothing^M)$. We can show that it is unique in $M$, using the axiom of extensionality. But we can also talk about the empty class, which is the subset of $M$ which corresponds to the elements of $\varnothing^M$ inside $M$. So the empty class of $M$ is always empty, since $M$ thinks that there are no elements inside $\varnothing^M$.

What we do, normally, is identify a set with its class, so $\{x\in M\mid M\models x\in\varnothing^M\}$ is identified with $\varnothing^M$, even if it's not equal to it.

However, since the empty set defines the empty class, and that is externally empty, it means that the empty class of all models is really just the empty set.

Finally, let me now point out that every model of $\sf ZF$ has an initial segment which is truly well-founded, so by Mostowski's collapse lemma we can collapse that initial segment. It is easy to show that this initial segment must always include $V_\omega$, so it includes all the truly hereditarily finite sets, which of course include $\varnothing$.

So in a way, $\varnothing$ is the same in all models, up to a small re-labeling issue.


问:
About the definition of directional derivative in the clousure of an open set
Let be $X$ a normed space and thus we prove that if $x_0\in\overset{\circ}Y\subseteq X$ then for any $\vec v\in X$ such that $\lVert\vec v\lVert=1$ the set
$$
V:=\{h\in\Bbb R:(x_0+h\vec v)\in\overset{\circ}Y\}
$$
is a not empty open set that has $0$ as a cluster point. So first of all we observe that if $x_0\in\overset{\circ}Y$ then $0\in V$ and thus it is not empty. Anyway if $x_0\in\overset{\circ}Y$ then  there exists $\delta>0$ such that $B(x_0,\delta)\subseteq\overset{\circ} Y$ and thus $(x_0+h\vec v)\in \overset{\circ}Y$ for any $h\in(-\delta,\delta)$ so that $0$ is a cluster point of $V$. Then if $(x_0+h\vec v)\in\overset{\circ}Y$ there exists $\delta>0$ such that $B(x_0+h\vec v,\delta)\subseteq\overset{\circ}Y$ and thus $(x_0+t\vec v)\in\overset{\circ}Y$ for any $t\in(h-\delta,h+\delta)$ so that $(h-\delta,h+\delta)\subseteq V$ and thus this it is open. Now if $\overset{\circ}Y$ is not empty and so that for any $x_0\in\overset{\circ}Y$ there exists $\delta>0$ such that $B(x_0,\delta)\subseteq\overset{\circ}Y$ then $V\neq\emptyset$ for any $\vec v\in V$ such that $\lVert\vec v\lVert=1$ (indeed an open ball is a convex set) and so there exist at least a net in $V$ converging to $0$ so that if $f:Y\rightarrow\Bbb R^n$ is a function then the limit
$$
\lim_{h\rightarrow 0}\frac{f(x_0+h\vec v)-f(x)}h
$$
is well defined and we call it the deriviative of $f$ in the direction $\vec v$. Anyway if $x_0\notin\overset{\circ}Y$ it could happen that one of the limits
$$
\lim_{h\rightarrow 0^+}\frac{f(x_0+h\vec v)-f(x)}h\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\lim_{h\rightarrow 0^-}\frac{f(x_0+h\vec v)-f(x)}h
$$
could be defined and we can think to exted the notion of directional derivative: e.g. the upper half space $H^k:=\{x\in\Bbb R^k:x_k\ge0\}$ is closed and if $x_0\in\partial H^k=\{x\in\Bbb R^k:x_k=0\}$ the limit
$$
\lim_{h\rightarrow 0^+}\frac{f(x_0+h\vec e_k)-f(x)}h
$$
is well defined for any function $f$ whose domain is the upper half space. So I ask if $x_0$ is a cluster point of $\overset{\circ}Y$ $\Biggl($and thus $x_0\in\overline{\overset{\circ\,}Y}\Biggl)$ then at least for one $\vec v\in X$ such that $\lVert\vec v\lVert=1$ the set $V$ as above defined is a not empty open set that has $0$ as cluster point so that it is possible to exted the notion of directional derivative in the clousure of an open set. So could someone help me, please?

答:
It is not true that $V$ will always be non-empty.

We should note that the vector $\vec{v}$ was chosen first, then $V$ is a subset of the real numbers such that $x_0+\vec{v}V \subset \mathring{Y}$. It is a linear crossection of a particular direction of $Y$, particularly in the direction of $\vec{v}$.

To construct a counterexample for any $\vec{v}$ consider the set of points in $\mathbb{R}^2$ satisfying $$(x, y) \in \mathring{Y} \iff x>0 \wedge 0<y<x^2.$$ (Convince yourself that this is an open set.)

The limit point we will consider is $(0, 0)$. Take any vector $\vec{v}$. The set $V$ is either empty or a half infinite open interval that does not contain $0$. It is clear that $0$ is not a cluster point of any $V$.


问:
Totally ordered abelian group
Let $\Gamma$ be a totally ordered abelian group (written additively), and let $K$ be a field. A valuation of $K$ with values in $\Gamma$ is a mapping $v:K^* \to \Gamma$ such that

$1)$ $v(xy)=v(x)+v(y)$

$2)$ $v(x+y) \ge \min (v(x),v(y))$,

for all $x,y \in K^*$.

Show that the set $A=\{x\in K^*:v(x) \ge 0\} \cup \{0\}$    is a valuation ring of $K$.

I want to show that $-1 \in A$. From $v(1)=v(1)+v(1)$, $v(1)=0$ and $v(1)=v(-1)+v(-1)$ so that $2v(-1)=0$. Is it possible to conclude that $v(-1)=0$?

In general for any totally ordered abelian group, does $2x=0$ imply $x=0$?

答:
It does follow that $\nu(-1)=0$.

Here is an elementary argument. Suppose firstly that $\nu(-1)>0$. Then $2\nu(-1)=\nu(-1)+\nu(-1)>0$; now $2\nu(-1)=0$ implies $0>0$, contradiction.

The same argument, \textit{mutatis mutandis}, shows that you cannot have $\nu(-1)<0$.

Title: Totally ordered abelian group


答:
There is always a whole number of molecules, and at some point in time there will be zero molecules left. With time, it becomes increasingly likely that all the caffeine molecules have completely disappeared.

However, in chemistry the "concentration" is the mean, or expected, amount of molecules. This is not a whole number. This is much like the way the expected roll of a dice is $3\frac{1}{2}$ even though there is no such number on the dice - the average of your dice rolls over many rolls will be $3\frac{1}{2}$.

This expected number gets smaller and smaller but never reaches zero.

Perhaps this simulation works on you computer.

Title: Does half-life mean something can never completely decay?

Body:
Caffeine has a half-life of approximately six hours. I understand this to mean that every six hours, the amount of caffeine in the body is half of what it was six hours prior. Does that mean that caffeine never completely leaves the body? It just keeps reducing to half after a fixed time ad infintum.

I guess a broader statement of my question is: When calculating the half life of a thing, is it true that it will never reach zero? And how is this used/reconciled in a more practical setting like when calculating the amount of caffeine in the body?


问:
A few questions about cyclic groups
I cannot understand how is $\langle 1 \rangle$ the creator of $(\mathbb{Z},+)$. I know that this element can create all-natural numbers, simply by adding-1 a certain amount of times, but the negative numbers cannot be created. So what I thought is that each element has a negative number, but still, I can't see how $\langle 1 \rangle$ can create by itself all elements of $\mathbb{Z}$, only both -1 and 1 can create together all elements.

In addition, I have seen a couple of times that for example - the group $Z^{*}_8$ if defined as $Z^{*}_8=\{[1],[3],[5],[7]\}$. In order to check the elements that a certain element create, which means the order, we check the power of the element until we start returning on the elements created, let say $\langle 5 \rangle=\{[5]^{1},[5]^{2},[5]^{3}\}=\{[5],[1],[5]\}$ so the order is 2. Now, I know that power represents the amount of time we run the method $*$ of the group, but if so, why is here with $Z^{*}_8$ we see (*) as ($\cdot $), and in the group - $(\mathbb{Z},+)$ we check as $\langle 2 \rangle=\{2,2+2,2+2+2,...\}$ and so on?

I hope you can help me understand this. Thanks!

答:
Notice that $\langle 1\rangle$ contains $-1$, since without it it isn't a group ($-1$ is the inverse element of $1$, because $1+(-1)=0$ in $(\mathbb{Z},+)$). Hence, you should understand that $\langle 1\rangle$ being a group implies $-1\in \langle 1\rangle$ so then you can easily see how it generates all integers.

In the second question, in $\mathbb{Z}_8^\times$ you're referring to the multiplicative group of $\mathbb{Z}_8$, to be said, the $\mathbb{Z}_8$ elements that have inverse for multiplication (Notice this is NOT the same as inverse for addition, the operation you use in integer groups). So in $\mathbb{Z}_8^\times$ the operation you're using is multiplication, while in $\mathbb{Z}$ or $\mathbb{Z}_8$ you're using addition, and powers in addition and multiplication are different though both can share the same $a^n$ notation:

In multiplication, $a^n=a\cdot a\cdots a$

In addition, $a^n=a+a+\cdots+a$

Title: A few questions about cyclic groups


问:
How do I solve this limit $\lim_{x\to \infty}\left(x-\sqrt{\frac{4x^3+3x^2}{4x-3}}\right)$?
I need to find the limit of $$\lim_{x\to \infty}\left(x-\sqrt{\frac{4x^3+3x^2}{4x-3}}\right)$$
I did rooting like this:
$$\frac{\left(x-\sqrt{\frac{4x^3+3x^2}{4x-3}}\right)*\left(x+\sqrt{\frac{4x^3+3x^2}{4x-3}}\right)}{x+\sqrt{\frac{4x^3+3x^2}{4x-3}}} = \frac{x^2-\frac{4x^3+3x^2}{4x-3}}{x+\sqrt{\frac{4x^3+3x^2}{4x-3}}}$$
but I got stuck. How can I continue from here? Thanks in advance.

答:
Putting the terms in the numerator over the common denominator one obtains:
$$\begin{align}\frac{x^2-\frac{4x^3+3x^2}{4x-3}}{x+\sqrt{\frac{4x^3+3x^2}{4x-3}}}&=\frac{-6x^2}{(4x-3)x\left(1+\sqrt{\frac{4x+3}{4x-3}}\right)}\\
&=\frac{-6}{(4-\frac3{x})\left(1+\sqrt{\frac{4x+3}{4x-3}}\right)}.
\end{align}$$

Can you get it from here?


问:
Can we have x-intercept instead of y-intercept in equation of a line
As I have started learning co-ordinate geometry for ML, so got one question. Line equation

Let say in x,y dimension if a line cross over the x-axis instead of y (as shown in image above, unlike y=mx+c ) then how equation will change or will the equation has any impact apart from the x-intercept in this case? Is this a valid case?
Please help me clearing this doubt.

答:
I think if the line is parallel to y - axis (as shown by the image you have attached). Then the slope of that line  will not be defined  because tan90 is not define .

If only slope is not define then the question of repsenting that in slope -intercept form doesn't make any sense.


答:
A group $G$, written multiplicatively, is called cyclic if there is an elemet $a \in G$ with $G = \langle a \rangle =\{a^n \mid n \in \mathbb{Z}\}$.  Note that $n$ spans all the integers.  If the group is written additively, this becomes
$$
G = \langle a \rangle =\{na \mid n \in \mathbb{Z}\}.
$$
So in exponential notation we get terms like
$$
\ldots, a^{-2}, a^{-1}, a^0=e, a, a^2, \ldots
$$
In additive notation this would read
$$
\ldots, -2a, -a, 0, a, 2a, \ldots
$$
With $a=1$ this explains why $1$ generates all of $\mathbb{Z}$ (as does $-1$).

Title: A few questions about cyclic groups

Body:
I cannot understand how is $\langle 1 \rangle$ the creator of $(\mathbb{Z},+)$. I know that this element can create all-natural numbers, simply by adding-1 a certain amount of times, but the negative numbers cannot be created. So what I thought is that each element has a negative number, but still, I can't see how $\langle 1 \rangle$ can create by itself all elements of $\mathbb{Z}$, only both -1 and 1 can create together all elements.

In addition, I have seen a couple of times that for example - the group $Z^{*}_8$ if defined as $Z^{*}_8=\{[1],[3],[5],[7]\}$. In order to check the elements that a certain element create, which means the order, we check the power of the element until we start returning on the elements created, let say $\langle 5 \rangle=\{[5]^{1},[5]^{2},[5]^{3}\}=\{[5],[1],[5]\}$ so the order is 2. Now, I know that power represents the amount of time we run the method $*$ of the group, but if so, why is here with $Z^{*}_8$ we see (*) as ($\cdot $), and in the group - $(\mathbb{Z},+)$ we check as $\langle 2 \rangle=\{2,2+2,2+2+2,...\}$ and so on?

I hope you can help me understand this. Thanks!


问:
Property of open neighborhood in Euclidean space.
Let $a\in \mathbb{R}^{2n}$, $z \in \mathbb{R}^n$, and $\epsilon >0$. There exists or not $x=(x_1,x_2) \in B_{\epsilon}(a)=\{ y \in \mathbb{R}^{2n} \mid ||y-a||<\epsilon \}$ and $\lambda \neq 0$ such that

$$x_1+x_2=\lambda z$$

My idea is using the fact that the open neighborhood $B_{\epsilon}(a)$ is diffeomorphic to the whole space $\mathbb{R}^{2n}$. However, I did not success yet.

Any other answers or ideas are very welcome.

Thanks in advance.

答:
Hint

If $\|a\|<\varepsilon $, the function $g:B_\varepsilon (a)\times (\mathbb R\setminus \{0\})\to \mathbb R^n$ defined by $$g(x,y,\lambda ):=\frac{x+y}{\lambda },$$
is surjective. If $\|a\|\geq \varepsilon $, then $z=0$ won't be reached.


问:
'Cofinality' and base cardinality of convergent net less than cardinality of index set, implies constant from some index onwards?
I have seen the following statement:

Let $\{ x_i\}_{i\in I}\subseteq \mathbb{R} $ be a net converging to an element $x_0\in \mathbb{R}$, where $\vert I\vert>\aleph_0$, $I$ is totally ordered and every lower set of $I$ is countable. Then there exists $i_0\in I$ such that $x_i=x_0$ for all $i\geq i_0$, i.e., $x_i$ is constant from some point onwards.

The solution\proof of this statement revolves around the fact that there exists a countable local base of $x_0$ and that a countable union of countable sets is countable.

It seems to me that using this argument, I can claim the following more general statement:

Let $X$ be a Hausdorff space and $x_0\in X$ such that $x_0$ has a local base $\mathcal{B}_{x_0}$ of cardinality less than $\kappa>\aleph_0$. Then any convergent net $\{ x_i\}_{i\in I} \subseteq X $ satisfying that $x_i\to x_0$, $\vert I\vert=\kappa$, and every lower set of $I$ has at most $\alpha<\kappa$ elements, must satisfy that $x_i=x_0$ from some point onwards.

I think this is proven by noticing that

$$ \bigcup_{U\in \mathcal{B}_{x_0}}\big( U^c\cap \{x_i \}_{i\in I}\big) = \{ x_i\}_{i\in I}\setminus \{x_0\}, $$

and hence

$$ \big\vert \{ x_i\}_{i\in I}\setminus \{x_0\}\big\vert\leq \vert \mathcal{B}_{x_0}\vert \cdot \alpha \leq \max\{ \alpha, \vert\mathcal{B}_{x_0}\vert \} <\kappa.$$

I would appreciate if anyone can verify whether this argument is generally sound. Secondly, I think this statement can be rephrased using cofinality.
I think that $\{ x_i\} $ is not constant from some index onwards if and only $\{ j\in I: x_i\neq x_0\} $ is cofinal and that we can also say that every cofinal set in $I$ would have to be of cardinality greater than $\vert \mathcal{B}_{x_0}\vert$ in the previous case. So I was wondering whether the following seems true:

Let $X$ be a Hausdorff space and $x_0\in X$ such that $x_0$ has a local base $\mathcal{B}_{x_0}$ of cardinality less than $\kappa>\aleph_0$. Then any convergent net $x_i\to x_0$ satisfying $\kappa>\text{cf}(I)>\vert \mathcal{B}_{x_0}\vert$ and $\vert I\vert\geq \kappa$, must be constant from some index onwards.

答:
The proof is correct but the `noticing' needs a bit more explaining and will reveal more about the ordered set $I$, without assuming the order is linear.

For every $i\in I$ the complement, $L_i$ say, of $\{j:j\ge i\}$ is a lower set and hence countable. This helps in showing that $U^c\cap\{x_i:i\in I\}$ is of cardinality at most $\alpha$ for every neighbourhood $U$.

If $A$ is a countable subset (in the original case), or of cardinality less than $\kappa$ (in the second case) of $I$ then $\bigcup_{i\in A}L_i$ is countable or of cardinality less than $\kappa$, respectively. If you take $j$ in its complement then $j$ is an upper bound for $A$. So if you have a small neighbourhood base $\mathcal{B}$ and you have $i_U\in I$ for $U\in\mathcal{B}$ then you can take an $i$ above all $i_U$ and the net will be constant above that $i$.

Your general statement is false: let $I$ be the ordinal $\omega_\omega$. Then the cardinality of $I$ is $\aleph_\omega$ and its cofinality is $\aleph_0$.
Define $x_i=2^{-n}$ if $\omega_{n-1}\le i<\omega_n$ (where $\omega_{-1}=-1$ for the nonce). This net converges to $0$ in $\mathbb{R}$ but it is not constant on a tail.

Part 2 The above reflects an earlier version of the question. The new final statement is correct but there is no need to introduce a cardinal $\kappa$. If $\mathrm{cf}(I)>|\mathcal{B}|$ then every net that converges to the point in question is constant on a tail: given $i_U\in I$ for $U\in\mathcal{B}$ we know that $\{i_U:U\in\mathcal{B}\}$ is not cofinal, so, as the order is linear, the set has an upper bound $j$ and then $\{x_i:i\ge j\}\subseteq\bigcap\mathcal{B}=\{x_0\}$.

Also, in the first statement $|I|=\aleph_1$, and in the second one $\kappa$ is the cardinal successor of $\alpha$. This follows from the general statement that a linear order in which all initial segments have cardinailty at most $\lambda$ has cardinality at most $\lambda^+$.


问:
No extension to complex numbers?
Complex numbers are 2D. It is a commonly cited result that there is no 3D or 4D analogue of the complex numbers.

I just want to be clear on exactly what this result says:

It is impossible to construct a 4D field.

It is impossible to construct a 4D field which the complex numbers are a subset of.

It is impossible to construct a 4D field which the real numbers are a subset of.

Something else?

(I'd also like to know why it's true - but given that I didn't understand the answer the first $2^7$ times, I doubt I'll understand it this time either. :-} Oh well...)

答:
A question is, why one would need exactly a field. The bicomplex numbers aka tessarines are a commutative associative ring.

Yes, there are some elements that have no inverse, but that is not bad at all: dual numbers and split-complex numbers also have such elements.

Bicomplex numbers have subrings, isomorphic to complex numbers and hyperbolic (split-complex) numbers. They can be represented as commutative real matrices.

They are a really great thing, generalizing the various notions of complex number.

Title: No extension to complex numbers?


问:
How do I solve $\min_x \max(c_1^Tx, c_2^Tx, \dots, c_k^Tx)$ for $\lVert x \rVert_2 = 1$.
Let $f(x) = \max(c_1^Tx, c_2^Tx, \dots, c_k^Tx)$.
where $x, c_1, c_2, \dots, c_k \in \mathbb R^n$.
What fast iterative methods are available for finding the (approximate) min of $f$ with the constraint $\lVert x \rVert_2 = 1$?

Notes:

$f$ is convex and and non negative, that is, $c_i$ positively span $\mathbb R^n$

For my use case $k \gg n$ and (rougly) $3 \le n \le 50$ and $100 \le k \le 10000$.
I tried projected subgradients (projected to the sphere) but it can be slow to converge and improving the initial guess doesn't seem to accelerate the method. From the literature, this seems to be equivalent to Riemannian manifold subgradient methods which use the exponential map. I haven't tried bundle methods yet but I am investigating them now. They seem like they might be too slow.

I've also tried approximating $f$ with a smooth maximum LogSumExp and doing gradient descent with projection. I also tried unconstrained gradient descent with a quadratic penalty. The approximation is too inaccurate and the evaluation of exp is too slow for my use case. I'm not interested in SQP because I suspect whatever method used to smooth (e.g. LogSumExp, hyperbolic) will be too costly/inaccurate to evaluate.

Edit:
I believe this problem is equivalent to a linear programming problem with quadratic equality constraints. I haven't been able to to find much literature on this type of problem. Perhaps I can use a SDP method but I'm not sure.

答:
Convert it to a quadratically constrained linear program by the following trick: Add an auxiliary variable $t$ such that: $\max(c_1^Tx, c_2^Tx, \dots, c_k^Tx)\leq t$
Then, this will imply $c_i^Tx\leq t~~ \forall i$.

Now your optimization program can be rewritten as:
$$
\text{minimize}_{x, t} ~~t\\
\quad\text{subject to}\\
\quad\|x\|_2= 1\\
\quad \quad \quad \quad \quad \quad \quad c_i^Tx\leq t~~ \forall i\in\{1, \ldots, k\} (*)
$$

Up to this point the optimization program is exactly the same as the original one. From a convex optimization perspective relaxing the norm constraint results in an efficiently solvable problem.

If you relax  $\|x\|_2= 1$ to $\|x\|_2\leq 1$, then the program is convex and can be efficiently solved using any convex optimization technic such as interior point method. The solution might satisfy the original constraint $\|x\|_2 = 1$ by itself. But there is no guarantee for that as that program is not convex unfortunately. However, you can perform Step two, and project back $x$ to $\|x\|_2 = 1$ by setting $x = x / \|x\|_2$

If the solution turns out to be zero, you should solve Problem (*) using iterative projection method, where in each iteration you project back to  $\|x\|_2 = 1$. This will give you an approximate solution to the non-convex problem you have.


问:
Quotient of $ \operatorname{Proj} A$ by the action of a finite group
Let $X$ be $ \operatorname{Proj}(A)$ for some graded ring A, and let $G$ be a finite group acting on $A$ with morphisms of graded rings; consequently $G$ acts on $X$.

I know I can write $X = \bigcup_{f\in A^+}X_f$ where $X_f=\{p \in X \mid f \notin p\}$.

Is it true that the quotient $X/G$, as a topological space, can be written as $X/G = \bigcup_{f\in A^G}p(X_f)$ ?
With $p:X \to X/G$ I intend the projection, while $A^G $ is the subring of $G$ invariants element of $A$.

I tried to prove it, but no idea on how to proceed. Any help/hint would be appreciated!

答: Yes; in fact, $X$ itself can already be written as $\bigcup_{f\in (A^G)^+}X_f$.  Suppose $p\in X$ and $p$ is not in $X_f$ for any $f\in (A^G)^+$, i.e. $p$ contains all of $(A^G)^+$.  Let $f$ be any element of $A^+$.  Enumerate the elements of $G$ as $g_1,\dots,g_n$ and observe that every elementary symmetric polynomial in $g_1f,\dots,g_nf$ is in $(A^G)^+$, and thus in $p$.  Thus, the polynomial $q(t)=(t-g_1f)\cdots(t-g_nf)$ has all coefficients except the leading coefficient in $p$.  Taking this polynomial mod $p$, then, it reduces to just $t^n$.  But $f$ is a root of $q(t)$, so this means $f^n$ is $0$ mod $p$.  Since $p$ is prime, this means $f\in p$.  Since $f\in A^+$ was arbitrary, this means $p$ is all of $A^+$, which is a contradiction.


问:
$\lim_{n \to \infty} (u_{n+1}-u_n)$ does not converge then $\lim_{n \to \infty} u_n$ does not converge
Prove that if $\lim_{n \to \infty} (u_{n+1}-u_n)$ does not converge then $\lim_{n \to \infty} u_n$ does not converge.

My try-
Contrapositive statement - If $\lim_{n \to \infty} u_n$ converges then $\lim_{n \to \infty} (u_{n+1}-u_n)$ converges.

Let , $\lim_{n \to \infty} u_n=L$

Then $\lim_{n \to \infty} u_{n+1}=L$

$\lim_{n \to \infty} (u_{n+1}-u_n)=L-L=0$

Is the proof correct? Any other proof is also desirable.

答:
First of all, a bit of terminology: a limit does not converge. Sequences converge, while limits (may) exist and equal something. Another thing: please be sure to define the environment where you are working. What is $u_n$? A sequence of real numbers? A sequence of functions? A sequence with values in which space?

Suppose we are talking about sequences of real numbers. If the sequence converges, then it is a cauchy sequence, meaning that for every $\varepsilon > 0$ there exist a $N \in \mathbb{N}$ such that for all $n,m > N$ we have $ |u_{n+1} - u_n| < \varepsilon$.

Now define the following: $a_n := u_{n+1} - u_n$. You easily see that $a_n$ converges by definition of cauchy sequence.


问:
Prove that $S^3\setminus S^1$ is connected
As the titles says, I have to prove that $S^3\setminus S^1$ is connected.

I'm having a hard time solving this problem. The best idea I've come up with is to show somehow that, if $S^3\setminus S^1$ wasn't connected, then $S^1$ would be a non-trivial clopen in $S^3$, and that's a contradiction; to do this I've tried to use the fact that in these hypotheses $S^3\setminus S^1=A\cup B$ disjoint open sets and $S^3\setminus S^1=C \cup D$ disjoint closed sets, but I haven't been able to conclude anything (and I'm not sure this is the right way to proceed). Maybe I'm missing something trivial.

Any hints?

答:
Using the stereographic projection, you can identify $S^3-\{point\}$ (the point not in $S^1$) to $\mathbb{R}^3$. The image of the circle by the stereographic projection is a circle if the center of the stereographic projection is not in the circle. If you remove a circle to $\mathbb{R}^3$ it is still connected.

This implies that $S^3-\{point\}\cup S^1$ is connected and $S^3-S^1$ is connected.


问:
Implicit differentiation: $x^2y - 2x^3 - y^3 + 1 = 0$
Hi I'm stuck on one problem in my study guide. It's the only one with an = 0 at the end of it.

Differentiate:

$x^2y - 2x^3 - y^3 + 1 = 0$

答:
Implicit differentiation:

$$2xy\,dx+x^2\,dy-6x^2\,dx-3y^2\,dy=0\implies (x^2-3y^2)dy=(6x^2-2xy)dx\implies$$

$$y'=\frac{dy}{dx}=\frac{6x^2-2xy}{x^2-3y^2}\;,\;\;x\neq\pm\sqrt 3\, y$$


问:
Show that the mapping $w (z) = z^2 $ takes the upper-half complex plane $\text{Im} (z) \ge 0$ into the whole plane.
I am supposed to Show that the mapping $w (z) = z^2 $ takes the upper-half complex plane $\text{Im} (z) \ge  0$ into the whole plane.
$w = u+iv$ and $z=x+iy$

I tried doing it as follows:

$$w = u+iv = (x+iy)(x+iy) = x^2-y^2 + i2yx  $$

This implies that  $$u = x^2-y^2  \space\space\space\space\space  \text{and} \space\space\space\space\space v=2yx$$

The condition $\text{Im}(z) \geq 0$ would mean that $v \geq0$ and that $x^2 \geq u$.
Since x can take any value from the condition we see that u can be any positive value.

This would mean that we have transferred the initial complex number into the upper right quadrant but not the entire quadrant. I probably did something wrong but am very confused.

Any help would be appreciated.

答: You can find a parametric equation of the upper half plane, namely $$\text{upper-half plane}= \{ \alpha +i\beta:-\infty\lt\alpha\lt+\infty ,0\leq\beta\lt+\infty\}$$ Then the image of the upper half plane under $f(z)={z^2}$ is ${(\alpha+i\beta)^2}={\alpha^2}-{\beta^2}+i\cdot2\alpha\beta $. Since $-\infty\lt\alpha\lt+\infty$ and $0\leq\beta\lt+\infty$ , the negative sign in the real part guarantees: $${-\infty\lt\alpha^2}-{\beta^2\lt+\infty}$$ and because $\alpha$ can be negative: $$-\infty\lt2\alpha\beta\lt+\infty$$ Since both ${\alpha^2}-{\beta^2}$ and $2\alpha\beta$ are real numbers, positive or negative, ${\alpha^2}-{\beta^2}+i\cdot2\alpha\beta$ is a parametric equation for the whole complex plane. I am not sure of the rigour of this approach, I just know that parametric equations work in most of the cases. Any feedback is appreciated!


问:
Prove the the given matrix is totally unimodular
I need to prove following matrix is totally unimodular:

I know that I can delete the second, sixth and ninth columns, and then the sixth row, since each of them contains only one non-zero number, and I got to this matrix:

$\hskip 1.7 in$

I don't know how to continue from here: The matrix is too big tin order to use the definition, and in each row and column there are more then two non-zero numbers, so I can't use the row/column partition method.

How can I show it?

答:
The following result is mentioned in section 8 of Fulkerson and Gross and also in Wikipedia:

The consecutive-ones property: if $A$ is (or can be permuted into) a 0-1 matrix in which for every row, the 1s appear consecutively, then $A$ is TU. (The same holds for columns since the transpose of a TU matrix is also TU.)


问:
Sylow 2-subgroup and Sylow 3-subgroup of $D_{24}$ (of order $48$).

Find Sylow 2-subgroup and Sylow 3-subgroup of $D_{24}$

I have found
$n_2 = 1$ or $3$ and $n_3 = 1$ or $4$ or $16$, where $n_2$ & $n_3$ are the number of Sylow $2$-subgroups and Sylow $3$-subgroups respectively, but I am unable to pinpoint the exact solution. I am having difficulty in understanding $D_{24}$ also. Can someone help with this?

答: Hint:  $D_{n}\cong\Bbb Z_n\rtimes\Bbb Z_2$.


问:
Meaning of "non-trivial three-term arithmetic progressions"
In this paper "ON ROTH’S THEOREM ON PROGRESSIONS" by Tom Sanders, the author gives a bound related to "non-trivial three-term arithmetic progressions", but what exactly means "non-trivial" in this context? Or what is an example of a trivial three-term arithmetic progression?</p>

答: "Non-trivial" just means "non-constant".  That is, a non-trivial three-term arithmetic progression is a triple of elements of the form $(a,a+b,a+2b)$ where $b\neq 0$ (the $b\neq 0$ condition being what makes it "non-trivial").


问:
Yet another Collatz generalization and conjecture
Observe that Collatz (the $\frac{3n+1}{2}$ version) can be written as: $$n_{i+1}:= \begin{cases}n_i+\left\lceil\frac{n_i}{2}\right\rceil\quad\text{ if odd }n_i \\
n_i-\left\lceil\frac{n_i}{2}\right\rceil\quad\text{ if even }n_i
\end{cases}$$

Looking at it this way gives the following pleasingly simple table for the difference between between consecutive terms, where $\Delta n := n_{i+1}-n_i$:

$$
\begin{array}{|l|l|} \hline
n_i&\Delta n\\ \hline
1&+1 \\ \hline
2&-1 \\ \hline
3&+2 \\ \hline
4&-2 \\ \hline
5&+3 \\ \hline
\ldots&\ldots\\ \hline
\end{array}
$$

I apologize for not having worked out a nice short-form version of this, but consider a generalization of the above to give a construction like:

$$
\begin{array}{|l|l|} \hline
n_i&\Delta n\\ \hline
1&+1 \\ \hline
2&+2 \\ \hline
3&-1 \\ \hline
4&-2 \\ \hline
5&+3 \\ \hline
6&+4 \\ \hline
7&-3 \\ \hline
8&-4 \\ \hline
9&+5 \\ \hline
\ldots&\ldots\\ \hline
\end{array}
$$

In fact, let $k$ be the number determining that cyclic count, so that $k=1$ for the standard Collatz in the first table, and $k=2$ in the second table, and $k=3$ would give $\{1,2,3,-1,-2,-3,4,5,6,-4\ldots\}$. And by that definition, let $f(n,k)$ return the corresponding $\Delta n$ using $k$ as described.

Then the next term for regular Collatz could be found as $C(n):=n+f(n,1)$. More generally, let's define $$C_k(n):=n+f(n,k),$$ so e.g. by examining that second table, we see that $C_2(8)=8-4=4$ and $C_2(9)=14$.

This construction seems like it might have some interesting properties. In particular, I conjecture that

$\forall k,n\in\mathbb N : C_k(n)$ will eventually reach $k$, assuming Collatz itself holds.

Furthermore, if you take any such sequence and number it starting with $S_0=n$, it seems we have

$$\gcd(S_i,k) \leq \gcd(S_{i+1},k).$$

This means that once you have $k|S_j$ for some $j$, all future terms can be divided through by $k$ to give a standard Collatz sequence, but this is definitely not the case before reaching $S_j$.

I'm interested to see if anyone can find a counterexample or simple reason why this should or should not hold. Any reasonably constructive feedback would be acceptable as an answer, and of course if this has been investigated before, by all means let me know.

Sample Mathematica code if anyone wants it:

```<code>Clear[f, c];
SetAttributes[{f, c}, Listable];
f[n_, k_] := With[{m = Mod[n, 2 k, 1]}, If[m <= k, 1, -1] (Floor[(n - 1)/(2 k)] (k) + Mod[n, k, 1])]
c[n_, k_] := NestWhileList[# + f[#, k] &, n, # != k &, 1, 1000]

c[27, Range[6]] // Column
</code>```

will give you sequences for $C_k(27)$ for $k \leq 6$.

答:
This is an interesting idea but the behaviour of the new sequences do follow from the behaviour of the standard series.

For given $k$, if $x_{n+1}<x_n$, then $x_{n+1}$ is divisible by $k$ and thereafter, as you say, the sequence is just $k$ times the standard series.

So, all that needs proving is that one always reaches a multiple of $k$. If $x_{n+1}>x_n$ then, modulo $2k$, $1\le x_n\le k$ and $x_{n+1}\equiv {2x_n}\pmod {2k}$. So, modulo $2k$, the terms of the sequence double each time until a residue greater than $k$ is reached.


答: Look at $n^3 - (n-1)^3 = 3n^2 - 3n + 1$. Now substitute.</p>

Title: Telescoping sum of powers

Body:
$$
\begin{array}{rclll}
n^3-(n-1)^3 &= &3n^2 &-3n &+1\\
(n-1)^3-(n-2)^3 &= &3(n-1)^2 &-3(n-1) &+1\\
(n-2)^3-(n-3)^3 &= &3(n-2)^2 &-3(n-2) &+1\\
 \vdots &=& &\vdots & \\
 3^3-2^3 &= &3(3^2) &-3(3) &+1\\
 2^3-1^3 &= &3(2^2) &-3(2) &+1\\
 1^3-0^3 &= &3(1^2) &-3(1) &+1\\
\underline{\hphantom{(n-2)^3-(n-3)^3}} & &\underline{\hphantom{3(n-2)^2}} &\underline{\hphantom{-3(n-2)}} &\underline{\hphantom{+1}}\\
n^3-0^3 &= & 3f_2(n) &-3f_1(n) &+n
\end{array}
$$

Can somebody explain me how these results are disposed intuitively? I didn't understand why $$(n-1)^3 -(n-2)^3$$ became equals to $$3(n-1)^2 - 3(n-1) + 1$$

 How do this transformation was done?

Thanks!


问:
Solving $L=\sqrt{(D-h\tan\theta)^2+(H-h\cos\theta)^2 } $ for $\theta$
I'm finding it herculean to solve for $\theta$ in the following equation. Kindly help please:
$$ L=\sqrt{(D-h\tan\theta)^2+(H-h\cos\theta)^2 } $$

答:
Using the tangent half-angle substitution $\theta=2 \tan ^{-1}(x)$, we end with
$$\sum_{n=0}^8 a_n\,x^n=0$$ the coefficients being
$$\left(
\begin{array}{cc}
 n & a_n \\
 0 & D^2+(h-H)^2-L^2 \\
 1 & -4 D h \\
 2 & 4 h H \\
 3 & -4 D h \\
 4 & -2 \left(D^2-7 h^2+H^2-L^2\right) \\
 5 & 4 D h \\
 6 & -4 h H \\
 7 & 4 D h \\
 8 & D^2+(h+H-L) (h+H+L)
\end{array}
\right)$$ and again an octic polynomial as @Blue already commented.

Solving for all the roots is not a problem since there are plenty of subroutines available for all coding environments.

Now, what we can try is to consider that this expression as a Taylor series to $O(x^9)$ and use series reversion. This would give
$$x=t+a_1\sum_{n=2}^8 b_n\,u^n \qquad \text{with}\qquad t=-\frac {a_0}{a_1} \qquad \text{and}\qquad u=\frac {t}{a_1}$$

$$\left(
\begin{array}{cc}
 n & b_n \\
 2 & -a_2 \\
 3 & 2 a_2^2-a_1 a_3 \\
 4 & -5 a_2^3+5 a_1 a_3 a_2-a_1^2 a_4 \\
 5 & 14 a_2^4-21 a_1 a_3 a_2^2+6 a_1^2 a_4 a_2+a_1^2 \left(3 a_3^2-a_1
   a_5\right) \\
 6 & -42 a_2^5+84 a_1 a_3 a_2^3-28 a_1^2 a_4 a_2^2+7 a_1^2 \left(a_1
   a_5-4 a_3^2\right) a_2+a_1^3 (7 a_3 a_4-a_1 a_6)
\end{array}
\right)$$

Coefficients $b_7$ and $b_8$ are really too long for being printed here but they are available upon request.

For a single test, using $D=100$, $h=20$, $H=50$, $L=95$, using the above approach, we  end with
$$x=\frac{9233947126838127045}{36893488147419103232}$$ which leads to
$\color{blue}{\theta=0.4904968}$ while the exact solution is
$\color{red}{\theta=0.4904992}$.


问:
Methodical way of finding the number of $n$ such that $\lfloor\sqrt n\rfloor\mid n$ and $\lfloor \sqrt{n+1}\rfloor\mid n+1$
I can't seem to find a methodical/systematic way of solving:

How many integers $n$ are there with $4\le n\le 2019$ such that $\lfloor\sqrt n\rfloor\mid n$ and $\lfloor \sqrt {n+1}\rfloor\mid n+1$?

I found this question in the Philippine Math Olympiad 2020, Qualifying Rounds and I am lacking skills to find a method in answering this question.

答: First of all assume that $\lfloor\sqrt{n}\rfloor=\lfloor\sqrt{n+1}\rfloor=a$. Then $a|n$ and $a|n+1$ $\implies a=1$ which is only true if $n=1$. Otherwise $\lfloor\sqrt{n}\rfloor\not=\lfloor\sqrt{n+1}\rfloor$ and thus $\lfloor\sqrt{n}\rfloor+1=\lfloor\sqrt{n+1}\rfloor \iff n+1$ is a perfect square $\iff n=k^2-1$ for some positive integer $k$. Then $$k-1=\lfloor\sqrt{k^2-1}\rfloor|k^2-1 \hspace{50 pt}k=\lfloor\sqrt{k^2}\rfloor|k^2$$ which are both true. So all positive integers satisfying the two conditions are $1$ and the numbers of form $k^2-1$.


答:
$\lfloor \sqrt{n+1} \rfloor$ is either same as $\lfloor \sqrt{n} \rfloor$ or exceeds it by $1$ when $n+1$ is a perfect square.

So $n+1=m^2$, $\lfloor \sqrt{n+1} \rfloor=m$ and $n=m^2-1$, $\lfloor \sqrt{n} \rfloor=m-1$.

Since $m \mid m^2$ and $m-1 \mid m^2-1$, we're looking for all $n$'s for which $n+1$ is a perfect square.

Title: Methodical way of finding the number of $n$ such that $\lfloor\sqrt n\rfloor\mid n$ and $\lfloor \sqrt{n+1}\rfloor\mid n+1$

Body:
I can't seem to find a methodical/systematic way of solving:

How many integers $n$ are there with $4\le n\le 2019$ such that $\lfloor\sqrt n\rfloor\mid n$ and $\lfloor \sqrt {n+1}\rfloor\mid n+1$?

I found this question in the Philippine Math Olympiad 2020, Qualifying Rounds and I am lacking skills to find a method in answering this question.


问:
Show that a sequence is convergent - Strategy to divide through by dominant term
We need to show that the sequence $a_n = \frac{(2n+1)(n+2)}{3n^2 + 3n}$ for $n=1, 2, ...$ is convergent and find its limit.

Now in my book they suggest:

Rearrange the expressions for $a_n$ by dividing both numerator and denominator by the dominant term $n^2$, which gives the following result:

$$a_n = \frac{(2n+1)(n+2)}{3n^2 + 3n} = \frac{(2+1/n)(1+2/n)}{3+3/n}$$

I understand that if we divide $3n^2 + 3n$ by $n^2$ we will get $3+3/n$ in the denominator.

But I can't follow how they get to this numerator. I tried $\frac{(2n+1)(n+2)}{n^2} = \frac{2}{n^2} + \frac{5}{n} + 2$ but that doesn't resemble the numerator they have in this example. Am I overlooking something? I would like to have some help to understand the example given and what algebraic manipulation leads to this result.

Body: $$(2n+1)(n+2)\frac {1}{n^2}= \left(\frac{2n+1}{n} \right) \left( \frac{n+2}{n} \right) = (2+ \frac 1n)(1+\frac 2n)$$


问:
How to find if a function with multiple variables has a limit
As the title says - I need to find a limit (if it exists) in (0, 0) for this function:

$$f(x, y) = (x^2 + y^2)\arctan(1/(x^2+y^2)),$$ where $(x, y) = (0, 0)$.

I've just started this chapter - functions with multiple variables - so I don't really know how to resolve this exercise.

Can anybody help me with this one and also explain their logic? This is something new for me and I'd like to learn it :)

Thanks.

答:
If you are dealing with this type of multivariable limits, one good thing to do is to go in polar coordinates.

Briefly, if you look at the $(xy)$ plane, any point $(x,y)$ can be written like $\left ( r\cos \left( \alpha \right) , r\sin \left( \alpha \right)  \right )$, where $r$ and $\alpha$ are functions of $x$ and $y$.

Now you see that $r \xrightarrow[]{(x,y) \to (0,0)} 0$, but $\alpha$ is still free to vary whatever the way it wants, so that $(x,y)$ can approach $(0,0)$ following any path imaginable (and therefore all paths).

The interest here is that $x^2+y^2 = r^2$, and this particular case enable us to deal with a single variable limit !

In our case : $$\lim_{(x,y) \to (0,0)} \, (x^2 + y^2) \, \arctan \left ( \frac{1}{x^2+y^2} \right ) = \lim_{r \to 0} \, r^2 \, \arctan \left ( \frac{1}{r^2} \right ) = 0$$

I hope this will help


Title: Telescoping sum of powers

Body: $(x-1)^3=x^3-3x^2+3x-1$; now set $x=n-1$.


问:
Optimizing a Conical Frustum Using Partial Differentiation
I am working on minimizing the surface area of a frustum with a known volume, however, the equations for the volume and surface area (listed below) have three variables. I am trying to use partial differentiation to optimize surface area with volume as a constraint, yet, differentiating with respect to any variable results in very messy work that can't be solved for a variable. Is there a better way to do it, or can anyone please show me how to do it using partial differentiation?

$
V=1/3(πh)(r^2+rR+R^2)
$

$
S.A=π(r+R)(\sqrt{(r-R)^2+h^2}+πr^2+πR^2
$

Thank you.

答:
Define the Lagrangian function

$$\Lambda(r,R,h;\lambda):=\pi(r+R) \sqrt{(r-R)^2+h^2}+\pi r^2+\pi R^2 -\lambda(V_0-(\pi h) (r^2+rR+R^2)/3), $$

with $V_0$ denoting the allocated volume.

The partial derivatives are

$$\begin{align}\Lambda_r&=\pi  \sqrt{h^2+(r-R)^2}+\frac{\pi  (r-R) (r+R)}{\sqrt{h^2+(r-R)^2}}+\frac{1}{3} \pi  h \lambda  (2 r+R)+2 \pi  r, \\ \Lambda_R&=\frac{\pi  \left(R^2-r^2\right)}{\sqrt{h^2+(r-R)^2}}+\pi  \sqrt{h^2+(r-R)^2}+\frac{1}{3} \pi  h \lambda  (r+2 R)+2 \pi  R, \\ \Lambda_h &=\frac{\pi  h (r+R)}{\sqrt{h^2+(r-R)^2}}+\frac{1}{3} \pi  \lambda  \left(r^2+r R+R^2\right), \\ \Lambda_\lambda&=\frac{1}{3} \pi  h \left(r^2+r R+R^2\right)-V_0. \end{align}$$

Proceed with a change of variables, motivated by the square root in the above
$$
\begin{align}
h &= \rho \cos \theta, \\
r &= \frac{s+\rho \sin \theta}{2}, \\
R &= \frac{s-\rho \sin \theta}{2}.
\end{align}
$$

Setting the partial derivates above equal to zero, and applying the substitution gives

$$
\begin{align}
0&=\frac{1}{6} \pi  \lambda  \rho  \cos (\theta ) (\rho  \sin (\theta )+3 s)+\pi  (\sin (\theta )+1) (\rho +s),\\
0&=-\frac{1}{6} \pi  \lambda  \rho  \cos (\theta ) (\rho  \sin (\theta )-3 s)-\pi  (\sin (\theta )-1) (\rho +s), \\
0&=\frac{1}{24} \pi  \lambda  \left(-\rho ^2 \cos (2 \theta )+\rho ^2+6 s^2\right)+\pi  s \cos (\theta ),\\
0 &=\frac{1}{12} \pi  \rho  \cos (\theta ) \left(\rho ^2 \sin ^2(\theta )+3 s^2\right)-V_0.
\end{align}
$$

The first three equations can be written in linear form as

$$
\left(
\begin{array}{ccc}
 \pi  (\sin (\theta )+1) & \pi  (\sin (\theta )+1) & \frac{1}{6} \pi  \rho  \cos (\theta ) (\rho  \sin (\theta )+3 s) \\
 \pi  (1-\sin (\theta )) & \pi  (1-\sin (\theta )) & \frac{1}{6} \pi  \rho  \cos (\theta ) (3 s-\rho  \sin (\theta )) \\
 0 & \pi  \cos (\theta ) & \frac{1}{24} \pi  \left(-\rho ^2 \cos (2 \theta )+\rho ^2+6 s^2\right) \\
\end{array}
\right) \begin{pmatrix} \rho \\ s \\ \lambda \end{pmatrix} = \begin{pmatrix} 0\\ 0 \\ 0 \end{pmatrix}.
$$

If the matrix in the above is invertible, we must have $\rho=0$, which implies a planar frustum ($h=0$). Hence, we can assume that the matrix is singular, meaning its determinant
$$\frac{1}{3} \pi ^3 \rho  \sin (\theta ) \cos ^2(\theta ) (\rho -3 s)=0. $$

This vanishing determinant condition reduces the search to two seemingly admissible cases:

$\rho=3s \implies h =2 \sqrt{(2 r+R) (r+2 R)}$

$\rho \sin \theta =0 \implies r=R$

Case 1:

The first three equations, upon divisions allowed by $s>0, \cos \theta \neq 0$ become
$$\frac{1}{2} \pi  (3 \lambda  s \cos (\theta )+8) =0, \\\frac{1}{2} \pi  (3 \lambda  s \cos (\theta )+8)=0, \\\pi  \cos (\theta )+\frac{1}{8} \pi  \lambda  s (5-3 \cos (2 \theta ))=0. $$
It takes a little work to deduce from this that $\sin^2 \theta = \frac{1}{9}$, which gives $r=0$ or $R=0$. This renders Case 1 inadmissible.

Case 2:

Let us return to the Cartesian form of the system, assuming that $r=R$:

$$\pi  (h \lambda  r+h+2 r)=0, \\ \pi  (h \lambda  r+h+2 r)=0, \\\pi  r (\lambda  r+2)=0, \\\pi  h r^2-V_0 .$$

Since $r>0$ the third equation immediately gives $\lambda = -2/r$. It follows from there that $h=2r$ and finally that $r=\sqrt[3]{\frac{V_0}{2 \pi }}.$

Since there is a unique critical point, it has to be the minimum. That is, the frustum of volume $V_0$, of least surface area is a cylinder with radii $R=r=\sqrt[3]{\frac{V_0}{2 \pi }}$ and height $h=2\sqrt[3]{\frac{V_0}{2 \pi }}$.

Remark: This is somewhat expected based on the iso-perimetric inequality.


问:
Question about prime factorization of ideals in ring of integers
Suppose that $p$ is a prime number, and $\mathcal{O}_K$ is the ring of integers of a number field $K$. Suppose we have $p\mathcal{O}_K=Q_1^{r_1}\cdots Q_n^{r_n}$, where the $Q_i$ are distinct nonzero prime ideals of $\mathcal{O}_K$ and $r_i\geq 1$. Suppose $P$ is a prime ideal of $\mathcal{O}_K$ such that $P\cap\mathbb{Z}=p\mathbb{Z}$. Then I'd like to show that $P$ is one of the $Q_i$.

My idea is that if I can show that $P$ contains $Q_1^{r_1}\cdots Q_n^{r_n}$, then it contains one of the $Q_i$ and so must be equal to that $Q_i$ since prime ideals are maximal in a Dedekind domain.

Now, $p\mathbb{Z}=p\mathcal{O}_K\cap\mathbb{Z}=Q_1^{r_1}\cdots Q_n^{r_n}\cap\mathbb{Z}=P\cap\mathbb{Z}$, but I'm not sure where to go from here to get what I want. What can I do?

答: You know that $p \in P \cap \mathbb Z$, so $p \in P$, so $p \mathcal O_K \subset P \mathcal O_K = P$.


问:
$D_n = \{x\in\mathbb R: F(x) - F(x^-) \ge 1/n\}$ is finite for $n\in\mathbb N$ if $F$ is a distribution function

Let $D$ be the set of discontinuity points of a distribution function $F$. $D_n = \{x\in\mathbb R: F(x) - F(x^-) \ge 1/n\}$ for $n\in\mathbb N$. Show that each $D_n$ is finite. Hence, show that a distribution function cannot have an uncountable number of discontinuities.

So, I know that a distribution function $F$ must be such that:

$F$ is an increasing function on $\mathbb R$.

$\lim_{t\to\infty} F(t) = 1$ and $\lim_{t\to -\infty} F(t) = 0$

$F$ is right continuous.

If $F$ has no jump discontinuities in $[a,b]$, then $F(a) = F(b)$.

How do I show the required statement now? I am trying to construct a proof by contradiction. Suppose $D_m$ is not finite for some $m\in\mathbb N$. What now?

I was able to do the second part of the proof anyway, i.e. if the $D_n$'s are all finite then $D$ is countable. Here's how: $$D = \{a\in\mathbb R: F(a) - F(a^-) > 0\} = \bigcup_{n=1}^\infty D_n$$

Assuming that all $D_n$'s are finite, hence countable - we see that $D$ is countable (a countable union of countable sets is countable).

Thanks!

答:
Note that you can actually bound the cardinality of $D_n$ by $n$ itself. Indeed, suppose that $x_1, \ldots, x_{n+1}$ are distinct elements in $D_n$ listed in an increasing order.

Then, $f(x_{i+1}) - f(x_i) \ge 1/n$ for $i = 1, \ldots, n$. (We are only using monotonicity here to note that $f(x_{i+1}^-) \ge f(x_i)$.)

Adding the equations for all $i$ together gives us that $$f(x_{n+1}) - f(x_1) \ge \dfrac{n + 1}{n} > 1.$$
However, this contradicts that the image of $f$ is contained in $[0, 1]$.


答:
Notice that
$$(n-2)^3=((n-1)-1)^3=(n-1)^3-3(n-1)^2+3(n-1)-1$$

Anyway, related to the image you posted (which I didn't notice at first sight), the result was really obtained by substitution from the first row ($n\to n-1$), as other people also observed.

Title: Telescoping sum of powers


问:
Different definitions of cographs
The inductive definition of cographs states, that:

$K_1$ is a cograph

if $G_1$ and $G_2$ are cographs, then $G_1 \cup G_2$ is a cograph

if $G$ is a cograph, then $\overline{G}$ is a cograph

Another definition states that:
$$G \text{ is cograph } \iff G \text{ or } \overline{G} \text{ is not connected}$$

The question is, how these two definitions are equivalent, especially how the second definiton implies the first one.

答:
These are not equivalent. For example, take the disjoint union of two copies of $P_4$ (the path on $4$ vertices). This satisfies the second definition, since it's not connected.

However, it does not satisfy the first definition - the only way to build $2P_4$ out of smaller cographs would be if $P_4$ were a cograph, and $P_4$ does not satisfy either definition.

The first definition is the standard definition of a cograph. As far as I know, the second definition does not describe an interesting class of graphs.

Title: Different definitions of cographs


问:
Can we have x-intercept instead of y-intercept in equation of a line
As I have started learning co-ordinate geometry for ML, so got one question. Line equation

Let say in x,y dimension if a line cross over the x-axis instead of y (as shown in image above, unlike y=mx+c ) then how equation will change or will the equation has any impact apart from the x-intercept in this case? Is this a valid case?
Please help me clearing this doubt.

答:
As the other poster said, the formula basically doesn't work for vertical slopes.  However, there are two ways you can save it.

The first is by rewriting it so that there is an $x$-intercept and a slope.  So, if the basic formula is:

$$y = mx + b$$

We can make a similar formula for $x$:

$$x = qy + c$$

Where $q$ is the $\frac{\text{run}}{\text{rise}}$ and $c$ is the x-intercept.  Here, the slope would be $q = 0$, and the $x$-intercept $c$.

Alternatively, If you wanted to "save" the original formula, you could use the hyperreals and utilize infinities.  Here, you could say that the line isn't really vertical, but that its slope is indeed infinite, i.e., $\omega$ (in the hyperreals, there are more than one infinite value - you still can't "divide by zero", but you can divide by infinitely small values).

So, with hyperreals, the formula would be:

$$y = \omega x + b$$

If the $x$-intercept is $c$, then we can make the formula:

$$0 = \omega c + b \\
\omega c = -b \\
b = -\omega c
$$

So, for an $x$-intercept of $c$, the new equation is

$$ y = \omega x + -\omega c$$

It's probably easier to just say to use a different formula for this situation, but I wanted to point it out because many times the places where formulas stop working because of a "divide by zero" you can often get a valid formula by extending into the hyperreals.