问:
Disproving the metric of two nonempty closed sets
I am a novice in Analysis. I'd like to unleash my mental gymnastic on this exercise but I am not certain about what I did, If I am wrong, could anyone explain to me why my solution is not valid? Then giving an acceptable solution(s) are welcomed.

Let $(X, d)$ be a metric space.
Prove or disprove that $D(A, B) > 0$ for any nonempty disjoint closed $A,B \subset X.$

Where, $d(x, A) = \inf\limits_{a\in A} d(x, a)$ and $D(A, B) = \inf\limits_{a\in A, b \in B} d(a, b)$

Since $A$ is a closed set, it has a closed ball i.e., $x\in X, \ \epsilon >0,  \ \overline{B}_\epsilon(x): = \{y \in X \ |\ d(x,y) \le \epsilon\}$, the closed ball of $B$ is similarly defined $\overline{B}_\epsilon(z): = \{y \in X \ | \ d(z,y) \le \epsilon\}$. Observe that their discrete are $d(\cdot) \le \epsilon$. Then $D(A, B) \le \epsilon$ for any $\epsilon$ which contradicts $D(A, B) >0.$  We've disproved.

答:
It's not true in general. It is true if one of the sets is compact (then the infimal distance function to the other set is continuous on a compact set, and hence has a minimum), but it's not true if both are simply assumed to be closed.

Your proof is difficult for me to follow, but as best I can tell, it makes a couple of errors. First, closed, non-empty sets may not contain non-trivial balls. For example, singleton sets are closed, but when the point is not isolated, they will contain no ball of positive radius.

Secondly, and more seriously, you $\varepsilon > 0$ seems to depend on where you put the centre of your particular ball. You might get a large $\varepsilon > 0$ for one point $x \in A$, but a much smaller one for a different point $x' \in A$. You can't then fix a particular $\varepsilon > 0$ that applies for all the points $x \in A$, and therefore conclude that $d(A, B) \ge \varepsilon$.

As a specific counterexample, consider the metric space $\Bbb{R}^2$, and the sets
\begin{align*}
A &= \left\{(x, y) \in \Bbb{R}^2 : x > 0 \text{ and } y \ge \frac{1}{x}\right\} \\
B &= \left\{(x, y) \in \Bbb{R}^2 : x < 0 \text{ and } y \ge -\frac{1}{x}\right\}.
\end{align*}
Note that $A \cap B = \emptyset$ simply because they contain points with $x$ coordinates of different signs. Further, $(1/n, n) \in A$ while $(-1/n, n) \in B$, and
$$d\left(\left(\frac{1}{n}, n\right), \left(-\frac{1}{n}, n\right)\right) = \sqrt{\left(\frac{1}{n} + \frac{1}{n}\right)^2 + (n - n)^2} = \frac{2}{n} \to 0,$$
as $n \to \infty$. This implies the infimal distance is $0$.


问:
Automorphism group of $\mathbb{Z}_p\times \mathbb{Z}_p$

How to determine the automorphism group of $\mathbb{Z}_p\times \mathbb{Z}_p$ where $p$ is a prime? Or more specifically, how to determine the element of order $2$ in this group?

I got stuck here, since I only know that if two finite groups $H$ and $K$, where $(|H|,|K|)=1$, then $\text{Aut}(H\times K) = \text{Aut}(H) \times \text{Aut}(K)$. But $p$ and $p$ are not relatively prime.

Any hints or solutions are welcomed, thanks!

答:
An automorphism of a vector space is defined by its action on a basis.  So, take
the unique automorphism determined by </p>

<p>$(1,0)\rightarrow (0,1)$ and $(0,1)\rightarrow (1,0)$.</p>

ContentLicense: CC BY-SA 4.0

Body:
An automorphism of a vector space is defined by its action on a basis.  So, take
the unique automorphism determined by

$(1,0)\rightarrow (0,1)$ and $(0,1)\rightarrow (1,0)$.

ContentLicense: CC BY-SA 3.0


问:
How do you calculate a fee percentage to handle a fee being charged?
Problem is that we get charged a 3% fee.  We add this 3% fee to the invoice.  When we get the amount back they charge 3% on the invoice plus on the fee we added.  What formula can I use to figure out what percentage to charge on this so that we cover their 3% on both pieces?

Example 10.00 charge.  We add 0.30 to it to cover the fee, we send in 10.30.  They charge 3% on 10 and on the 0.30 so they send us back 9.99 when we expected 10.00.  How do we cover this?  Also, how do we cover these discrepancies when they charge fees and do rounding per line item?

-Edit-

For clarity we are sending an invoice in where we know they will charge 3% on it, we need to add a fee to our invoice that will negate this 3%.  It has to take into account that they will also charge 3% on the amount we add to negate the 3%.  Need to know what % to add to our invoice or what formula to calculate this.  Basically some sort of break even formula.

答:
The simplest I could make it.

You want to make say $30 and they are going to charge you 3% fee

$30/.97 <—(.03 or 3% less than 1 or 100%)

$30.93 (rounded up)

That’s it


问:
viewing module as a vector space

Corollary 3.10. Let $A$ be a local ring with maximal ideal $\mathfrak{m}$ and residue field $k:=A/\mathfrak{m}$, and let $M$ be a finitely generated module over $A$. The action of $A$ on $M/\mathfrak{m}M$ factors through $k$, and elements $a_1,...,a_n$ of $M$ generate it as an $A$-module if and only if the elements $$a_1+\mathfrak{m}M,...,a_n+\mathfrak{m}M$$
generate $M/\mathfrak{m}M$ as $k$-vector space.

In the corollary 3.10 of this book, 'The action of $A$ on $M/\mathfrak{m}M$ factors through $k$' so they view $M/\mathfrak{m}M$ as a vector space over $k$. Then how does the operation defined? Is it $(a+\mathfrak{m})(x+\mathfrak{m}M) = ax+\mathfrak{m}M$? And in general, if we have such form (some given action of a ring of given module factors through some field obtained by quotient), then can we always view such module as a vector space over that field? And is the operation defined as I wrote above?

答:
The action of $A$ on $M/\mathfrak mM$ is the ring homomorphism $\mu : A \to \operatorname{End}_{\textsf{Ab}}(M/\mathfrak mM)$ such that for each $a \in A$, $\mu(a) \in \operatorname{End}_{\textsf{Ab}}(M/\mathfrak mM)$ is left-multiplication by $a$.

Note that if $a \in \mathfrak{m}$, then $am \in \mathfrak mM$ for every $m \in M$, and so
$$
(\forall m \in M) \quad \mu(a)(m+\mathfrak mM) = a(m+\mathfrak mM) = am+\mathfrak mM = \mathfrak mM,
$$
that is, $\mu(a)$ is the trivial map in $\operatorname{End}_{\textsf{Ab}}(M/\mathfrak mM)$, which means that $a \in \ker \mu$.

Thus $\mathfrak m \subseteq \ker \mu$, and then the ring homomorphism $\tilde\mu : A/\mathfrak m \to \operatorname{End}_{\textsf{Ab}}(M/\mathfrak mM)$ given by $\tilde\mu(a+\mathfrak m) = \mu(a)$ is well-defined. Moreover, it is precisely the action that you proposed.

Title: viewing module as a vector space


问:
proving that $89 \mid 2^{44}-1$
i tried to prove that $2^{44} \equiv 1\pmod{89}$.

I noticed that by Fermat's little theorem $2^{88} \equiv 2^{44}\cdot 2^{44} \equiv 1\pmod{89}$
which means that $2^{44}$ is the inverse of itself $\rightarrow 2^{44} \equiv 1 \pmod{89}$ or $2^{44}\equiv 88 \pmod{89}$.

how can I rule out the second option?

Title: proving that $89 \mid 2^{44}-1$

Body: $2^{44} - 1 = (2^{22})^2 - 1 = (2^{22}+1)(2^{22} - 1) = (2^{22}+1)(2^{11} + 1)(2^{11} - 1) = (2^{22}+1)(2^{11} + 1)\times 2047 = (2^{22}+1)(2^{11} + 1)\times 23 \times 89$


问:
Continuous injection of $[a,b]$ is strictly monotonic from Zorich's book
I am reading Zorich's book "Analysis I" and ran into the following statement:

Let $f:[a,b]\to \mathbb{R}$ and $f$ is continuous on $[a,b]$. Then $f$ is injective if and only if $f$ is strictly monotonic.

Proof: $\Leftarrow$ is quite obvious.

The most interesting part is $\Rightarrow$. Suppose that $f$ is not strictly monotonic. Then we can find three points $x_1<x_2<x_3$ from $[a,b]$ such that $f(x_2)$ does not lie in between $f(x_1)$ and $f(x_3)$. In this case $f(x_3)$ lies in between $f(x_1)$ and $f(x_2)$ or $f(x_1)$ lies in between $f(x_2)$ and $f(x_3)$.

Question: Suppose $f$ is not strictly monotonic, i.e. $f$ is not strictly increasing and is not strictly decreasing. Then we can find points $x_1,x_2,x_3,x_4\in [a,b]$ such that $x_1<x_2$ with $f(x_1)\geq f(x_2)$ and $x_3<x_4$ with $f(x_3)\leq f(x_4)$.

I have spent smth like couple of hours trying to derive that exists three points with the above properties as in Zorich's book but I failed.

I'd be thankful if you can help me, please!

答:
Note that I am using $(a,b)_* = (\min(a,b), \max(a,b))$ in the following.

You are trying to prove that $f:[a,b] \to \mathbb{R}$ is strictly monotonic iff for all $x_1,x_2,x_3$ such that $a \le x_1<x_2<x_3\le b$ we have $f(x_2) \in (f(x_1),f(x_3))_*$.

If $f$ is strictly monotonic then we can presume that $f(a) < f(b)$ (otherwise take $-f$) and if $x_1<x_2<x_3$ then since $f(x_1) < f(x_2)< f(x_3)$ we see that $f(x_2) \in (f(x_1),f(x_3))_*$.

For the other direction suppose for all $x_1<x_2<x_3$ we have $f(x_2) \in (f(x_1),f(x_3))_*$. We would like to show that $f$ is strictly monotonic.

Since $f({1 \over 2}(a+b)) \in (f(a),f(b))_* $ we see that $f(a) \ne f(b)$. We can presume that $f(a)<f(b)$ (otherwise take $-f$). Note that for $x \in (a,b)$ we have  $f(x) \in (f(a),f(b))$.

Now suppose $a \le x < y \le b$.
If $x=a$ (or $y=b$) then the previous statement shows that $f(a)< f(x)$ (or
$f(y)< f(b)$), so we can assume $a<x<y<b$. In this case we have
$f(x) \in (f(a),f(y)$, in particular $f(x)<f(y)$.


答: For an intuitive counterexample that disproves the statement, let $A$ be the $x$-axis in $\Bbb R^2$ and let $B$ be the graph of $y= \frac 1x$.  Both sets are closed and they asymptotically approach each other, but they are disjoint.

Title: Disproving the metric of two nonempty closed sets

Body:
I am a novice in Analysis. I'd like to unleash my mental gymnastic on this exercise but I am not certain about what I did, If I am wrong, could anyone explain to me why my solution is not valid? Then giving an acceptable solution(s) are welcomed.

Let $(X, d)$ be a metric space.
Prove or disprove that $D(A, B) > 0$ for any nonempty disjoint closed $A,B \subset X.$

Where, $d(x, A) = \inf\limits_{a\in A} d(x, a)$ and $D(A, B) = \inf\limits_{a\in A, b \in B} d(a, b)$

Since $A$ is a closed set, it has a closed ball i.e., $x\in X, \ \epsilon >0,  \ \overline{B}_\epsilon(x): = \{y \in X \ |\ d(x,y) \le \epsilon\}$, the closed ball of $B$ is similarly defined $\overline{B}_\epsilon(z): = \{y \in X \ | \ d(z,y) \le \epsilon\}$. Observe that their discrete are $d(\cdot) \le \epsilon$. Then $D(A, B) \le \epsilon$ for any $\epsilon$ which contradicts $D(A, B) >0.$  We've disproved.


问:
What are the interesting applications of hyperbolic geometry?
I am aware that, historically, hyperbolic geometry was useful in showing that there can be consistent geometries that satisfy the first 4 axioms of Euclid's elements but not the fifth, the infamous parallel lines postulate, putting an end to centuries of unsuccesfull attempts to deduce the last axiom from the first ones.

It seems to be, apart from this fact, of genuine interest since it was part of the usual curriculum of all mathematicians at the begining of the century and also because there are so many books on the subject.

However, I have not found mention of applications of hyperbolic geometry to other branches of mathematics in the few books I have sampled. Do you know any or where I could find them?

答: It has been used in games, of which two of the most prominent examples are HyperRouge and Hyperbolica, both of which exist on Steam (Hyperbolica is still only in pre-release). Another example is Sokyokuban, a hyperbolic version of Sokoban that you can play in the web browser.</p>


问:
Having trouble finding this limit
The original question is:

$\lim _{x\to 0}\left(\frac{\sqrt{x+Δx}-\sqrt{x}}{Δx}\right)$

I've been stuck on this problem for a while now. I multiplied by the conjugate, simplified the numerator, canceled out Δx, and am left with $sqrt$(x+Δx) + $sqrt$(x), do I plug in "0" for the x's or did I do something wrong? I'm trying to understand this so steps would be very helpful.

答:
I would calculate it in the following way.
\begin{align*}
 \lim_{h\to 0}\frac{\sqrt{x+h}-\sqrt{x}}{h} &= \lim_{h\to 0}\frac{\sqrt{x+h}-\sqrt{x}}{h}\cdot\frac{\sqrt{x+h}+\sqrt{x}}{\sqrt{x+h}+\sqrt{x}}\\
&= \lim_{h\to 0}\frac{x+h-x}{h(\sqrt{x+h}+\sqrt{x})}\\
&= \lim_{h\to 0}\frac{h}{h(\sqrt{x+h}+\sqrt{x})}\\
&= \lim_{h\to 0}\frac{1}{\sqrt{x+h}+\sqrt{x}}=\frac{1}{2\sqrt{x}}\\
\end{align*}

Title: Having trouble finding this limit


问:
Give syntax proof of $\forall x (P(x) \lor Q(x))$ , $\forall y. \neg P(y)$ $\vdash$ $\forall x.Q(x)$
So the question is in the title already. I have to prove the following using first-order logic but I am failing to do so. Please someone help me understand how do I do it?

$\forall x (P(x) \lor Q(x))$ , $\forall y. \neg P(y)$ $\vdash$ $\forall x.Q(x)$

答:
$$\forall x (P(x) \lor Q(x)) , \forall y. \neg P(y) \vdash \forall x.Q(x)$$

Start at the end. Keeping an eye on where you want to go will help you get there.

$\dfrac{A}{\forall x~.A[x\backslash t]}{\forall\mathsf{intro}}$ means that when given a predicate you may generalise to a universal by replacing all instances of the arbitrary free variable with a bound variable that does not occur free in the original predicate.

Your consequent is such a universal statement, so you will want this on the bottom of your proof (ie: the root of the tree).

$$\dfrac{~~\vdots\\Q(t)}{\forall x~Q(x)}{\forall\mathsf{intro}}$$

Now we know where to go, let us look at where we start.

$\dfrac{\forall x~.A}{A[t\backslash x]}{\forall\textsf{elim}}$ means that "when given a universal statement, we may infer the predicate with all instances of the bound variable replaced with a free variable."

In your premise of $\forall x~.(P(x)\lor Q(x)$ the predicate is $P(x)\lor Q(x)$ and bound variable is $x$, so the replacement of that with free variable $t$ is: $P(t)\lor Q(t)$.

In your other premise of $\forall y~.\lnot P(x)$ the predicate is $\lnot P(y)$ and bound variable is $y$, so the replacement of that with (the same) free variable $t$ is: $\lnot P(t)$.

So you will be using both of these.

$$\dfrac{~\forall x~.(P(x)\lor Q(x))~}{P(t)\vee Q(t)}{\forall\mathsf{elim}}\qquad\dfrac{\forall y~.\lnot P(y)}{\lnot P(t)}{\forall\mathsf{elim}}$$

These will occur in the undischarged leaves of your tree.

There will possibly be other leaves--which must be discharged assumptions--and the branches will need to be built between these and the root.

One of the things we may derive is a disjunction.  So let us seek the rules for eliminating this.

$\dfrac{\lower{1.5ex}{A\vee B}~~\dfrac{[A]\\~~\vdots}{C}~~\dfrac{[B]\\~~\vdots}{C}}{C}{\vee\mathsf{elim}}$ means that when given a derivation of $A\vee B$, a derivation of $C$ from assumed $A$ and a derivation of $C$ from assumed $B$, then we may infer $C$ (and discharge those assumptions).

You have a derivation for $P(t)\vee Q(t)$, you wish to derive $Q(t)$, so you shall need to derive it from assumptions.  Well, $Q(t)$ is trivially derived when you assume it, so that leaves deriving it from $P(t)$.

$$\dfrac{\dfrac{~\vdots~}{P(t)\vee Q(t)}\quad\dfrac{[P(t)]^1\\~~\vdots}{Q(t)}\quad\lower{1.5ex}{[Q(t)]^2}}{Q(t)}{\vee\mathsf{elim}^{1,2}}$$

But how?   Wait, there is also the other premise, from which we may derive $\lnot P(t)$.   So, let us take a look at the rules using negation.

$\dfrac{A~~\lnot A}{\bot}{\lnot\mathsf{elim}}$ means that when given derivations of a predicate and its negation we may infer the contradiction constant (also called falsum or bottom).

Okay, what can we do about that $\bot$ symbol?

$\dfrac{\bot}{B}{\bot\mathsf{elim}}$ means that whenever the contradiction constant is derived, then we may infer any predicate we may desire.   This is also known as the principle of explosion, or ex falso quodlibet (EFQ).

We look good to go!   You just need to stack it all it together.

$$\phantom{\tiny\dfrac{\dfrac{\dfrac{~\forall x~.(P(x)\lor Q(x))~}{P(t)\vee Q(t)}{\forall\mathsf{elim}}\quad\dfrac{\dfrac{\lower{1.5ex}{[P(t)]^1}\quad\dfrac{\forall y~.\lnot P(y)}{\lnot P(t)}{\forall\mathsf{elim}}}{\bot}{\lnot\mathsf{elim}}}{Q(t)}{\bot\mathsf{elim}}\quad\lower{1.5ex}{[Q(t)]^2}}{Q(t)}{\vee\mathsf{elim}^{1,2}}}{\forall x~.Q(x)}{\forall\mathsf{intro}^t}}$$


问:
Write $(1, 2) + (1, 3) + (2, 1) + (2, 3) + (3, 1) + (3, 2)$ in Summation Form
I understand that I can write
$$
(1, 2) + (1, 3) + (2, 1) + (2, 3) + (3, 1) + (3, 2)
$$
as
$$
\sum_{i = 1, j = 1, i \ne j}^3 (i, j).
$$
Now, I add the condition that I would treat terms like $(1, 2)$ and $(2, 1)$  or $(3, 1)$ and $(1, 3)$ as the same element. Meaning, the sum now looks like:
$$
2(1, 2) + 2(1, 3) + 2(2, 3)
$$
How should I write the summation now?

Body: $$2*\sum_{j>i}\sum_{i=1}^2 (i,j)$$


问:
With or without squeeze theorem
I am doing a proof involving finding the sum of a series $\sum_{x\geq1} \frac{x}{2^x} - \frac{x+1}{2^{x+1}}$ and I need to show that $\lim_{x\to\infty} \frac{x+1}{2^x}$ converges to $0$ without using L'Hopital's theorem.

My attempt:

$\lim_{x\to\infty} \frac{x+1}{2^x} = \lim_{x\to\infty} \frac{x}{2^x} + \lim_{x\to\infty} \frac{1}{2^x}$

We then get that $2^x > x^2 \implies \frac{x}{2^x} < \frac{x}{x^2} = \frac{1}{x}$ and also, $2^x > x \implies \frac{1}{2^x} < \frac{1}{x}$ for sufficiently large $x$. We have that

$$0 \leq \lim_{x\to\infty} \frac{x}{2^x} \leq \lim_{x\to\infty} \frac{1}{x} = 0$$

$$0 \leq \lim_{x\to\infty} \frac{1}{2^x} \leq \lim_{x\to\infty} \frac{1}{x} = 0$$

(Notice the change from $<$ to $\leq$ above. Is that allowed? Also, the series starts at $1$, not $0$.)

Using squeeze theorem, $\lim_{x\to\infty} \frac{x+1}{2^x} = 0$.

Is this correct? Notice my bracketed comment. Any assistance much appreciated.

答:
Since you are in the context of a series, $x$ is natural.

So, you can apply the binomial formula to $2^x = (1+1)^x=\sum_{n=0}^x\binom xn$:

$$0\leq \frac{x}{2^x} = \frac{x}{(1+1)^x} \stackrel{x\geq 2}{\leq} \frac{x}{\binom x2} = \frac 2{x-1}\stackrel{x\to \infty}{\longrightarrow}0$$

Similarly for the case $\frac 1{2^x}$.

Title: With or without squeeze theorem


问:
Reordering rows in an array for minimum total euclidean distance
I am looking for an algorithm that can find the shortest path between a set of vertices in 5 dimensions given a single source.

For example if I have a 10 x 5 array, where each row corresponds to a point in 5 dimensional space. How do I reorder the rows such that travelling from top to bottom of the array results in the minimum total Euclidean distance travelled.

My end goal is to perform this operation in python.

I feel this must be a well defined problem but my research has not led me anywhere conclusive. It seems to me that a solution would incorporate graph theory - a minimum spanning tree. I'd be very grateful for any suggestions.

答:
This is the Euclidean version of the Traveling salesman problem. It's an NP-hard problem, so finding the optimal solution is hard when your array is very large. (For $10$ points in five dimensions, even trying all $10!$ orders is not entirely out of the question.) Techniques based on integer programming can solve larger problems.

There are various approximation algorithms. Wikipedia mentioned a method based on a minimum spanning tree that gets within a factor of $2$ of the shortest tour. There is an improvement that gets a $\frac32$-approximation, and is still not too bad to implement (the Christofides algorithm).

For the Euclidean version in a fixed number of dimensions, there is an algorithm that can guarantee arbitrarily good approximations efficiently, but I suspect that this is too complicated to bother.


问:
Understanding a step in a property of $Alt(T)$ - Calculus on Manifolds, M. Spivak
In Calculus on Manifolds - Michael Spivak is proven that If $S \in \mathfrak{J}^k(V)$ and $T \in\mathfrak{J}^k(V)$ and $A l t(S)=0,$ then
$$Alt(S \otimes T)=Alt(T \otimes S)=0$$

where $\mathfrak{J}^m(V)$ is the set of all m-tensors and $Alt(T)$ is a tensor defined by $$Alt(T)(v_1,\dots,v_k)=\dfrac{1}{k!}\sum_{\sigma\in S_k}\operatorname{sgn}\sigma \cdot T\left(v_{\sigma(1)}, \ldots, v_{\sigma(k)}\right)$$. On the book the property is proven as follows

$$(k+l)!  Alt(S \otimes T)\left(v_{1}, \ldots, v_{k+l}\right)=\sum_{\sigma \in S_{k+l}} \operatorname{sgn} \sigma \cdot S\left(v_{\sigma(1)}, \ldots, v_{\sigma(k)}\right) \cdot T\left(v_{\sigma(k+1)}, \ldots, v_{\sigma(k+l)}\right)$$

If $G \subset S_{k+l}$ consists of all $\sigma$ which leave $k+1, \ldots$, $k+l$ fixed, then

\begin{align*}
&\sum_{\sigma \in G} \operatorname{sgn} \sigma \cdot S\left(v_{\sigma(1)}, \ldots, v_{\sigma(k)}\right) \cdot T\left(v_{\sigma(k+1)}, \ldots, v_{\sigma(k+l)}\right)\\
&=\left[\sum_{\sigma^{\prime} \in S_{k}} \operatorname{sgn} \sigma^{\prime} \cdot S\left(v_{\sigma^{\prime}(1)}, \ldots, v_{\sigma^{\prime}(k)}\right)\right] \cdot T\left(v_{k+1}, \ldots, v_{k+l}\right)\\
&=0
\end{align*}

(*) Suppose now that $\sigma_{0} \notin G .$ Let $G \cdot \sigma_{0}=\left\{\sigma \cdot \sigma_{0}: \sigma \in G\right\}$
and let $v_{\sigma_{0}(1)}, \ldots, v_{\sigma_{0}(k+l)}=w_{1}, \ldots, w_{k+l}$, Then

\begin{align*}
&\sum_{\sigma \in G \cdot \sigma_{0}} \operatorname{sgn} \sigma \cdot S\left(v_{\sigma(1)}, \ldots, v_{\sigma(k)}\right) \cdot T\left(v_{\sigma(k+1)}, \ldots, v_{\sigma(k+l)}\right)\\
&=\left[\operatorname{sgn} \sigma_{0} \cdot \sum_{\sigma^{\prime} \in G} \operatorname{sgn} \sigma^{\prime} \cdot S\left(w_{\sigma^{\prime}(1)}, \ldots, w_{\sigma^{\prime}(k)}\right)\right]\cdot T\left(w_{k+1}, \ldots, w_{k+l}\right)\\
&=0
\end{align*}

Notice that $G \cap G \cdot \sigma_{0}=\varnothing .$ In fact, if $\sigma \in G \cap G \cdot \sigma_{0}$, then $\sigma=\sigma^{\prime} \cdot \sigma_{0}$ for some $\sigma^{\prime} \in G$ and $\sigma_{0}=\sigma \cdot\left(\sigma^{\prime}\right)^{-1} \in G,$
a contradiction. We can then continue in this way, breaking $S_{k+l}$ up into disjoint subsets; the sum over each subset is $0$ , so that the sum over $S_{k+l}$ is $0 .$ The relation $\operatorname{Alt}(T \otimes S)=0$ is proved similarly. $\blacksquare$

I undertood all the proof but from (*). Is there another way to understand these logical steps?

答:
Note that $G$ is a subgroup of $S_{k+l}$ (with $|G| = k!$), and the right cosets $\{G \sigma| \sigma \in S_{k+l} \}$ form a partition of $S_{k+l}$ (hence there are $n={(k+l)! \over k!}$ of them).

Let $\sigma_1,..,\sigma_{n}$ be such that the $\{ G \sigma_k\}$ form the partition, we can assume that $\sigma_1 = e$, the identity.

Then we have
$\sum_{\sigma \in S_{k+l}} = \sum_{\sigma \in G} + \sum_{\sigma \in G\sigma_2} + \cdots + \sum_{\sigma \in G\sigma_{n}}$


问:
$G$ is bipartite iff $\chi(G)=2$
If $G$ is a loop-free undirected graph with at least one edge, prove that $G$ is bipartite if and only if $\chi(G)=2.$

Definition 11.18 A graph $G=(V,E)$ is called bipartite if $V=V_1 \cup V_2$ with $V_1 \cap V_2 = \emptyset$, and every edge of $G$ is of the form $\{a,b\}$ with $a\in V_1$ and $b\in V_2$.

I start by supposing $G$ is bipartite, coloring the vertices in $V_1$ $\color{blue}{blue}$ and the vertices in $V_2$ $\color{red}{red}$, then by definition of bipartite, every edge $\{a,b\}$ has vertices of different colors. $\chi(G)=2$.

Next, if $\chi(G)=2$, then if $\{a,b\}\in G$, then $a$ and $b$ are colored with different colors, and since $\chi(G)=2$ there are only two colors. There is no edge where both vertices are of the same color. You can write $V=V_1 \cup V_2$ where $V_1$ is all the vertices of color $1$, $V_2$ of color $2$. $V_1 \cap V_2 = \emptyset$ because no vertex is colored twice, and every edge is of the form $\{a,b\}$ where $a \in V_1$ and $b \in V_2$. Thus $G$ is bipartite.

Apparently I was supposed to make use of the fact that there was at least one edge?

答:
If there are no edges, then we have the empty graph on $n$ vertices (which only requires one color) or the null graph (the graph on no vertices, which has chromatic number $0$).

A graph $G$ with at least one edge is bipartite iff $\chi(G) = 2$. In general, a graph $G$ is bipartite iff $\chi(G) \leq 2$.

Note that in the definition of a bipartite graph, there is no stipulation that $V_{1}, V_{2}$ are non-empty.

Title: $G$ is bipartite iff $\chi(G)=2$


问:
Real Analysis. Convergence of sequences
I've found that the sequence is bounded between the values 1.5 and 2 which is option (a), but since this is a recursive limit and bounded, solving the equation should give us option (d) as an answer as well. Why is that not the case? A quick help would be appreciated. Thank you very much.

https://i.stack.imgur.com/QkPBc.jpg

答: Note that in fact we have that $a_n > \frac{1 + \sqrt{5}}{2}$ for all $n \geq 1.$ Indeed, this is clear for $n = 1$ and assuming it is true for a certain $n,$ we have that $a_{n + 1} = 2 - \frac{1}{1 + a_n} > \frac{4 + 2\sqrt{5}}{3 + \sqrt{5}} = \frac{1 + \sqrt{5}}{2}.$ Another way to see the same thing is assuming that there is $n \geq 1$ such that $a_n = \frac{1 + \sqrt{5}}{2}.$ Then we have that $a_{n - 1} = \frac{a_n - 1}{2 - a_n} = \frac{\sqrt{5} - 1}{3 - \sqrt{5}} = \frac{1 + \sqrt{5}}{2},$ so $a_1$ must have also been $\frac{1 + \sqrt{5}}{2},$ which is of course false.

Title: Real Analysis. Convergence of sequences


问:
Determine Whether $\int_1^\infty \frac{x^\frac{2}{3}+x^\frac{5}{3}}{x^2}dx$ diverges or converges
Yo, dealing with a particularly tricky integral right now. $\int_1^\infty \frac{x^\frac{2}{3}+x^\frac{5}{3}}{x^2}dx$. I want to know if it converges/diverges using the comparison theorem. Which function should I base off the comparison from? I can do something like $\frac{1}{x^\frac{4}{3}}$ for convergence and $\frac{1}{\sqrt[3]{x}}$ for divergence. That's all I have figured out for now.

Body:
Hint:
$$\frac{x^\frac{2}{3}+x^\frac{5}{3}}{x^2}\sim \frac{1}{\sqrt[3]{x}}$$


问:
Given two sets, finding two non trivial homomorphisms that are not isomorphisms
Is it possible to have two non trivial homomorphisms that are not isomorphisms for given two Groups?

I am specially interested in additive/remainder Group of Integers and multiplicative (arithmetic multiplication as group operator) group of complex roots of unity.

Thank You.

答: If the groups are isomorphic to start with and you’re asking about isomorphisms between them, you are in essence just asking about the set (group, actually) of automorphisms of a single group. This automorphism group will ordinarily be nontrivial, as is the case with the group of $n$-th roots of unity for a fixed $n>2$. If you were asking about the (countable) group of all roots of unity (in the complex numbers, for example), then the automorphism group is not even countable. So the quantitative answer to your question is, Yes, Many, Many.</p>


问:
How to get the confidence of interval?
Let $X$$_{1}$,$X$$_{2}$,...,$X$$_{n}$ be a random sample from
$f$($x$;$\theta$) $=$ $\theta$e$^{-\theta x}$ $I$$_{(0,\infty)}$($x$). Find $(1-\alpha)100$ percent confidence interval for the mean of the population.

答:
Known that $\Sigma_i X_i\sim Gamma (n;\theta)$ it is evident that $2\theta\Sigma_i X_i\sim \chi_{(2n)}^2$ thus it is a pivotal qty so you can calculate

$$a<2\theta\Sigma_i X_i<b$$

That is

$$\frac{a}{2\Sigma_i X_i}<\theta<  \frac{b}{2\Sigma_i X_i}  $$

This is the exact CI where $a,b$ are the 2 quantiles such that

$$1-\alpha=\mathbb{P}( a<2\theta\Sigma_i X_i<b)$$

If you want an approximate CI you can use gaussian approx via CLT

Example:

Suppose you have this random sample:

$$x_1=2.4;x_2=2.6;x_3=2.5$$

It is known that

$$2\theta\sum_{i=1}^3 X_i\sim \chi_{(6)}^2$$

thus,at $\alpha=5\%$, setting

$$P\left(a<2\theta\sum_{i=1}^3 X_i<b\right)=0.95$$

assuming you want an equiprobable tail interval you get

$$\frac{1.24}{15}<\theta<\frac{14.45}{15}$$

Title: How to get the confidence of interval?


问:
Dimensionality of datasets in multiple regression
As an example, let's say that a linear regression is performed of the form

$$Y = \beta_0 + \beta_1X_1 + \beta_2X_2 + \cdots + \beta_nX_n +\varepsilon$$

where $Y$ is a vector of $10,000$ measurements of peak acceleration of different car models, and the regressors correspond to different technical features of the cars.

From a linear algebra standpoint $Y$ lives in $\mathbb R^{10000}$, and the coefficients are found by minimizing the sum of the square distances of this vector on a hyperplane.

Now, from the point of view of dimension being the number of linearly independent vectors that span a space, this vector $Y$ is just $1$ dimension.

If it is truly $1$-dimension of a $\mathbb R^{10000}$ ambient space, the Euclidean projection on the hyperplane that underpins the process of finding the coefficients does not have any dimensionality issues (collinearity between the regressors being a separate topic). Otherwise, $L^2$ norms in high dimensions do pose problems.

So is $Y$ (the vector of $10,000$ observations) $1$-mimensional or high dimensional?

答:
Consider the function $$f(X_1,X_2) = \beta_0 + \beta_1 X_1 +\beta_2 X_2. $$ This is a plane in 3 dimensions no matter how many times you evaluate the function. Thus, your problem "lives" in a $2$-dimensional space.

As for using the $L^2$ norm, you are correct. This link might be useful in that regard Link


问:
Prove $\cos(2x)-\cos(4x)=2\sin(3x)\sin(x)$

Use the expansion of $\cos(3x-x)$ and $\cos(3x+x)$  to show that $\cos(2x)-\cos(4x)=2\sin(3x)\sin(x)$

I am so confused. I get that $\cos(2x)-\cos(4x)=\cos(3x-x)-\cos(3x+x)$

This part I don't get; the expression above equals to:

$(\cos(3x)\cos(x)+\sin(3x)\sin(x))-(\cos(3x)\cos(x)-\sin(3x)\sin(x))$ but I don't understand how.

and when simplified this is equal to $2\sin(3x)\sin(x)$

答:
$\cos (2x)-\cos (4x)=\cos (3x-x)-\cos (3x+x)=[\cos (3x) \cos x +\sin (3x) \sin x] -[\cos (3x) \cos x -\sin (3x) \sin x]=2\sin (3x) \sin x$.

I have used the formulas $\cos (A+B)=\cos A \cos B-\sin A \sin B$ and $\cos (A-B)=\cos A \cos B+\sin A \sin B$.


问:
Reflection matrix about a plane
How do you construct a matrix in $\mathbb{R}^3$ that reflects about the plane $y=z$? And is there a way to construct a reflection matrix about any plane in general?

答:
To answer the second part of your question, I offer a visual way to show that such a reflection matrix exists for any plane.

Suppose there exists a linear transformation $L$ that reflects all (column, let's use column for consistency) vectors about a certain plane.

Pick a vector $\vec{v}_1$ that lies on the plane. Next, pick another vector $\vec{v}_2$ that lies on the plane, such that there does not exist $k$ is real, $k\vec{v}_1 = \vec{v}_2$. (If you're a bit more hardworking, you can find a $\vec{v}_2$ that's perpendicular to $\vec{v}_1$.) It should be pretty easy for you to find $\vec{v}_1$ and $\vec{v}_2$ because the plane is usually defined as $\vec{r} = a + b\vec{v}_1 + c\vec{v}_2$, where $a$, $b$, and $c$ are real. We know that $L\vec{v}_1 = \vec{v}_1$ and $L\vec{v}_2 = \vec{v}_2$ because they lie on the plane, and vectors that lie on the plane remain the same after being transformed by $L$.

Next, we find $\vec{v}_3$ that's perpendicular to the plane. Since we already have $\vec{v}_1$ and $\vec{v}_2$ on the plane, we can easily define $\vec{v}_3$ as the cross product of $\vec{v}_1$ and $\vec{v}_2$, as the cross product must be perpendicular to both $\vec{v}_1$ and $\vec{v}_2$. It will be tough* to find $L\vec{v}_3$, but once you get it, you're almost done!

As $\vec{v}_1$, $\vec{v}_2$ and $\vec{v}_3$ are linearly independent, $\begin{pmatrix}\vec{v}_1 & \vec{v}_2 & \vec{v}_3\end{pmatrix}$ is invertible.

$$L\begin{pmatrix}\vec{v}_1 & \vec{v}_2 & \vec{v}_3\end{pmatrix} = \begin{pmatrix}\vec{v}_1 & \vec{v}_2 & L\vec{v}_3\end{pmatrix}$$

$$L = \begin{pmatrix}\vec{v}_1 & \vec{v}_2 & L\vec{v}_3\end{pmatrix} \begin{pmatrix}\vec{v}_1 & \vec{v}_2 & \vec{v}_3\end{pmatrix}^{-1}$$

*Simply find the distance from $\vec{v}_3$ to the plane, and then subtract $2$ times of that vector.

I'm not very good in linear algebra, but I picture that it could be solved in this manner. Do correct me if I'm wrong.

Title: Reflection matrix about a plane


问:
maximizing expectation of exponential of a function
is $\arg\max_{\theta} \mathrm{E}_{\tau} [e^{-F(\theta + \tau)}]$ = $\arg\min_{\theta} \mathrm{E}_{\tau}[F(\theta + \tau)]$ ?

答:
Note that $e^{-x}$ is convex. So we can apply Jensen's inequality, e.g. $f(E(X))\le E(f(X))$ with $f(x)=e^{-x}$

$$\begin{split}\arg\max_\theta E_\tau \left(e^{-F(\theta+\tau)}\right)&\ge\arg\max_\theta e^{-E_\tau(F(\theta+\tau))}\\
&=\arg\min_\theta E_\tau(F(\theta+\tau))\end{split}$$


问:
Curve defined by a vector
http://i.stack.imgur.com/tD4Bn.png

I'm studying line integrals with a curve as a vector, but I couldn't understand the 'dr' part.

 First of all: the curve isn't really a curve, it's like some points where a vector points from the origin. So this 'curve' does not really exists, rigth?

 And then, when we take the integral with respect to 'dr', shouldn't we take it do ||dr|| (the magnitude of dr)? I really didn't understand, so I need some better and intuitive explanation...

Thank you so much!

答:
Well, intuitively $dr$ is an infinitesimal tangent vector to the curve. If you imagine that the curve is the trajectory of some particle, then the derivative of $r$ is the velocity. If a small amount of time $dt$ passes, then you've moved through the vector $dr =r '(t)dt$.

I could stop there and pretend that this is all fine, but I won't. Elementary books insists on making people learn to think with those infinitesimals and when one needs to move to the modern approach get's really confused with questions like "but hey, the other book told it was fine to think like that! What's the problem?".

So I'll tell you the true idea. A curve in $n$-dimensional space is a map $\alpha : I \subset \mathbb{R} \to \mathbb{R}^n$ and it's tangent vector at the point $\alpha(t)$ is the derivative $\alpha'(t)$. Now, $\alpha'(t)$ already defines the tangent direction of the curve, so you don't really need that $dt$ multiplying. Now, the line integral is a measure of how much some vector field $F:\mathbb{R}^n \to \mathbb{R}^n$ projects on the direction of the curve.

If you recall a little about the dot product you'll soon be convinced that if we fix some point $t \in I$ and consider $\alpha(t) \in \mathbb{R}^n$, then $F(\alpha(t))\cdot \alpha'(t)$ is a good measure of how $F$ projects onto $\alpha$. We are considering the "local" direction of $\alpha$, also it'll be zero if $F$ is orthogonal to $\alpha$ and it'll be maximum if $F$ is in the direction of $\alpha$. Since this is local,  we can compute at each point and sum everything up, this will be a measure of how $F$ projects onto $\alpha$ in general (try constructing the partitions and so on, it'll be a good exercise). So we define the line integral of $F$ over $\alpha$ to be:

$$\int_\alpha F=\int_a^b F(\alpha(t))\cdot \alpha'(t)dt$$

Then, the people that love that thing of infinitesimals set $\alpha'(t)dt = dr$ and just write this as:

$$\int_a^b F\cdot dr$$

Personally I prefer the first notation because  it makes clear that you have to calculate $F$ at $\alpha(t)$, dot it with $\alpha'(t)$ and integrate this thing that will simply be a function from $\mathbb{R}$ to $\mathbb{R}$.

Title: Curve defined by a vector


问:
Localizing C[x,y]/(xy) to get a direct product of Laurent polynomials.
I am trying to show $$\left(\mathbb C[x,y]/(xy)\right)_{x+y}\cong \mathbb C[x^{\pm 1}] \times \mathbb C[y^{\pm 1}].$$ In class, my professor did the example $$\left(\mathbb C[x,y]/(xy)\right)_{x} \cong \mathbb C[x^{\pm 1}]$$ by using the fact that $A_f \cong A[t]/(tf - 1)$ for a commutative ring $A$. I am trying to mimic this approach. I get that $$\left(\mathbb C[x,y]/(xy)\right)_{x+y} \cong \left(\mathbb C[x,y]/(xy)\right)[t]/(t(x+y)-1)$$ $$=\mathbb C[x,y,t]/(xy, t(x+y) - 1).$$ I am not really sure how to simplify this quotient to get what I need. I would really appreciate any help! Thanks.

答:
Since $\Bbb C[x^{\pm 1}]\times \Bbb C[y^{\pm 1}]$ has two orthogonal idempotents, $(1,0)$ and $(0,1)$, we need to find two orthogonal idempotents in $\Bbb C[x,y,t]/(xy,t(x+y)-1)$. I claim that $xt$ and $yt$ suffice: it's not hard to see that $xt+yt=t(x+y)=1$, while seeing that $(xt)^2=xt(1-ty)=xt$ (and similarly for $(yt)^2=yt$) takes a touch more insight. This means that $R=\Bbb C[x,y,t]/(xy,t(x+y)-1)$ is isomorphic to $Rxt\times Ryt$.

Now we can check that $Rxt\cong \Bbb C[x^{\pm 1}]$: writing a general element in $R$ as $p(x,y,t)$, we can use the relation $xy=0$ to see that $p(x,y,t)xt=p(x,0,t)xt$, and then  our calcluation that $xt$ is idempotent means that we can write this uniquely as $xt(c+\sum c_ix^i + \sum d_jt^j)$ for $c,c_i,d_j\in \Bbb C$. Since $x^2t\cdot xt^2=x^3t^3=xt$, we can define a surjective map sending $x^2t\mapsto x$ and $xt^2\mapsto x^{-1}$, which is injective by the claim that every element of $Rxt$ can be uniquely written as $xt(c+\sum c_ix^i + \sum d_jt^j)$.

Intuitively, what's going on here is that you're taking the variety $V(xy)$, the union of two coordinate axes, and throwing away all the points lying on $V(x+y)$, the antidiagonal. This gives you two disjoint copies of a punctured line, which both have coordinate algebra $\Bbb C[x^{\pm 1}]$. By disjointness, the coordinate algebra of the pair is just the product.


答:
$$\cos(A-B)-\cos(A+B)=2\sin A\sin B$$
Let, ${A+B}=C$ and ${A-B}=D$

Thus, $$\cos D-\cos C=2\sin\left(\frac{C+D}{2}\right)\sin\left(\frac{C-D}{2}\right)$$
Now, using this, put the values of $C$ and $D$, then find the answer.

Title: Prove $\cos(2x)-\cos(4x)=2\sin(3x)\sin(x)$

Body:
Use the expansion of $\cos(3x-x)$ and $\cos(3x+x)$  to show that $\cos(2x)-\cos(4x)=2\sin(3x)\sin(x)$

I am so confused. I get that $\cos(2x)-\cos(4x)=\cos(3x-x)-\cos(3x+x)$

This part I don't get; the expression above equals to:

$(\cos(3x)\cos(x)+\sin(3x)\sin(x))-(\cos(3x)\cos(x)-\sin(3x)\sin(x))$ but I don't understand how.

and when simplified this is equal to $2\sin(3x)\sin(x)$


问:
If an ODE has only periodic solutions and one equilibrium point, is that equilibrium point Lyapunov stable?
Consider the non-linear IVP
$$\dot x=f(x),$$
$$x(0)=x_0.$$
where $f$ is locally Lipschitz and
$$f(0)=0$$
and $$f(x)\ne 0\ \forall x\ne 0.$$
If all solutions to this IVP for various initial conditions are periodic does that mean that the origin is Lyapunov stable, i.e.,
$$\forall \varepsilon>0, \ \exists\delta>0, \ |x(0)|<\delta\implies|x(t)|<\varepsilon \ \forall t\ge 0.$$

My professor said yes in class and he "justified" this by drawing periodic orbits that are circular. I know that periodic solutions are bounded, but what if a solution close starts very close to the origin, moves far away from the origin, then returns to the same point? Is there an example of such a system or is my professor right?

I was thinking of a solution like
$$x(t)=\begin{cases}\frac{x(0)}{|x(0)|}\sin(t)+x(0)(1-\sin(t))&x\ne 0\\0&x=0\end{cases}.$$
$x(t)$ oscillates between $x(0)$ and a point with norm 1, so no matter how small $\delta$ is, the solution will never satisfy the definition of Lyapunov stability with $\varepsilon=\frac{1}{2}$. The origin is an equilibrium point and the only equilibrium point and $x(t)$ satisfies the initial condition. $x(t)$ is differentiable, $2\pi$ periodic and satisfies the time-invariance property. I know that $x(t)$ does not work because it is not continuous with respect to the initial condition and $f$ is locally Lipschitz but I'm wondering if there is a way to alter $x(t)$ to make it a solution to an autonomous ODE and still have the same properties.

答:
This is not a satisfactory proof of the OP's question.  It's a sketch of a proof of a slightly weaker claim (see assumptions below).  However, since this question hasn't gotten attention from an expert, as I hoped for in a comment above, then I'll write this down as the best that I can do at the moment.  There may be counterexamples to my argument, but I think they would have to be pathological.

One important point: I'm interpreting that the question requires that every point be in a strictly periodic orbit, in the sense that for every initial condition $x(0)$, there will be some $t > 0$ such that $x(t) = x(0)$.  I think that this is the sense in which the OP meant that "all solutions...are periodic."  (A physicist like me might informally say that all solutions are periodic if they asymptotically approach one or more periodic attractors, which is a completely different situation.)

I'll make the following assumptions that are stronger than the OP's question.  These can probably be relaxed, but I have not spent any time trying to do so.

$f$ is differentiable (not just Lipschitz).

The Jacobian of $f$ at $x=0$, which I'll call $J$, is not identically zero.

Proof: I will try to prove that your professor's picture is essentially correct, i.e., that initial conditions near the origin follow orbits that remain near the origin.  Sufficiently close to the origin,
$$ \dot{x} = f(0) + J x + \dots \approx J x $$
since $f(0)=0$.  This linearization has the solution
$$ x(t) = e^{Jt} x(0) $$
as long as $x(t)$ is sufficiently close to $0$.

The eigenvalues of $J$ are the local Lyapunov exponents.  Note that these can be complex, but since $J$ is real, they must come in complex conjugate pairs; linear combinations of these are used to create real solutions for $x(t)$.  (I will ignore this subtlety below but it doesn't change anything.)  In the next paragraph, I try to show that $\text{Re}(\lambda) = 0$ for all eigenvalues $\lambda$.

Suppose that there is some eigenvector $v$ whose associated eigenvalue $\lambda$ has a negative real component, i.e., $\text{Re}(\lambda) < 0$.  Suppose we let $x(0) = v$.  Then $x(t) = v e^{\lambda t}$, and $x(t)$ will asymptotically approach $0$, a fixed point.  Specifically, $x(t)$ will never return to $x(0)$, which we assumed that it would.  Note that the linearization of the original ODE is still valid within this entire neighborhood, so we can't say that somewhere along the path, it will get swept off in a different direction and complete a periodic orbit without going to $0$.

Dealing with the case that $\text{Re}(\lambda) > 0$ is slightly harder.  This is because, after some time, $x(t)$ will leave the neighborhood where the linearization is valid, and after that we cannot say what will happen.  But here I will employ a trick: Make time go backwards.  Note that a periodic orbit must remain periodic whether time runs forward or backward.  Now consider the original ODE,
$$ \dot{x} = f(x) $$
under the substitution $t \to -t$ (running time backwards). We get the ODE
$$ \dot{x} = -f(x) $$
which corresponds to taking $J \to -J$ and $\lambda \to -\lambda$ in the discussion above.    So if we have $\text{Re}(\lambda) > 0$ when time is running forward, we can run time backward and apply the same argument as we did for $\text{Re}(\lambda) < 0$.

Note that because we can choose the initial condition to be exactly on the eigenvector, the above discussion applies even if $J$ has both positive and negative eigenvalues, because we are singling out the effect of a single eigenvalue.

We have shown that the cases where $x(t)$ move rapidly toward or away from the origin are excluded, so $x(t)$ must remain at approximately the same distance from the origin.  In particular, the eigenvalues of $J$ must be pure imaginary, so $x(t)$ will be a sum of oscillating components.  In particular, if $x(0) = \sum_i c_i v_i$ for eigenvectors $v_i$, then $x(t) = \sum c_i e^{\lambda_i t} v_i$.  Obviously $|x(t)|$ can change, but it is bounded by the triangle inequality.

The main place where this argument needs more rigor is that one would need to do an $\epsilon$-$\delta$ analysis (or invoke relevant theorems) to validate that we really can ignore the effect of the higher-order terms in the Taylor series.  But I think the basic outline is sound.


问:
Find a $m$ divides the area under and above x-axis of $y=x (1 - x)$
My questions is, suppose that for specific $m$, $y=mx$ divides the area under and above $x$-axis of $y=x(1- x)$ into $2$ equal areas, so what is $m$? In an interval, I estimated fast with $m \in (\frac{1}{5}, \frac{2}{5}]$, is this right? I plot this equation on graphs, and only estimated it. I didn't calculate integral precisely.

Body:
First find the area of $x(1-x)$ from $0$ to $1$ (which is the area above the x axis) which comes out to be $\frac{1}{6}$, now the area between the line $y=mx$ and $y=x(1-x)$ would be $\frac{1}{12}$, so $$\int_0^{1-m}\left[x(1-x)-mx\right]dx=\frac{1}{12}$$
So, $$\frac{(1-m)^2}{2}-\frac{(1-m)^3}{3}-\frac{m(1-m)^2}{2}=\frac{1}{12}$$
$$\frac{m^2-2m+1}{2}+\frac{m^3-3m^2+3m-1}{3}-\frac{m^3-2m^2+m}{2}=\frac{1}{12}$$
$$3m^2-6m+3+2m^3-6m^2+6m-2-3m^3+6m^2-3m=\frac{1}{2}$$
$$-m^3+3m^2-3m+1=\frac{1}{2}$$
$$(1-m)^3=\frac{1}{2}$$
$$m=1-\frac{1}{\sqrt[3]{2}}$$


问:
Calculate $\int_0^{\infty} \frac{\cos 3x}{x^4+x^2+1} dx$
Calculate $$\int_0^{\infty} \frac{\cos 3x}{x^4+x^2+1} dx$$

I think that firstly I should use Taylor's theorem, so I have:$$\int_0^\infty \frac{1-\frac{x^2}{2!}+\frac{x^4}{4!}-\dots}{(x^2+1)^2}dx$$
However I don't know what I can do the next.

答:
Note that $I(a) = \int_0^\infty \frac{\sin at} {t(t^2+1)}dt
= \frac\pi2 (1-e^{-a}) $, which can be obtain by solving
$$I’’(a)-I(a) = -\int_0^\infty \frac{\sin at}t dt= -\frac\pi2$$

Then
\begin{align}
\int_{0}^{\infty}\frac{\cos 3x}{x^4+x^2+1} dx
= &\frac14\int_{-\infty}^{\infty}
 \overset{x=\frac{\sqrt3}2 t-\frac12}{\frac{(1+x)\cos 3x}{x^2+x+1}} +\overset{x= \frac{\sqrt3}2 t +\frac12}{\frac{(1-x)\cos 3x}{x^2-x+1}} d x\\
=& \frac1{\sqrt3}\cos\frac3{2} \int_{0}^{\infty} \frac{\cos \frac {3\sqrt3}{2}t}{t^2+1} dt + \sin\frac3{2} \int_{0}^{\infty} \frac{t\sin\frac {3\sqrt3}{2}t}{t^2+1} dt \\
=& \frac1{\sqrt3}\cos\frac3{2}\cdot I’(\frac {3\sqrt3}{2})
-\sin\frac3{2}\cdot I’’(\frac {3\sqrt3}{2})\\
=&\frac\pi2 e^{-\frac {3\sqrt3}{2} }\left( \frac1{\sqrt3}\cos\frac32+ \sin\frac32\right)
  =\frac\pi{\sqrt3}e^{-\frac {3\sqrt3}{2} }\sin\left(\frac\pi6+\frac32\right)
\end{align}


问:
Solving a special rational equation on a very small interval
I need to solve the following equation (for $x$):
\begin{equation}
\mathcal f(x):=\sum_{i=1}^n b_i \left( \frac{a_i}{1+b_i x}\right)^2-\phi=0, \quad \text{with} \quad -1/b_1< x \le 0.
\end{equation}
Assume that the equation has at least one solution in $(-1/b_1,0]$ and $b_1\ge b_2\dots\ge b_n>0$. In practice, $b_1$ can be very large, so that the interval of interest $(-1/b_1,0]$ can be very small. Therefore, numerical issues can happen (I think because $1/(1+b_1x)$ can blow up, at least when a numerical method captures a solution that is very close to $-1/b_1$). What is the best practical idea to find a solution of $f(x)=0$ that falls into $(-1/b_1,0]$?

By the way, I am currently using the bisection method and $b_1$ is around 1000 for example. The quantity $n$ could be up to 50 in my case. Further, such an equation must be solved iteratively in my algorithm for different values of $ a_i$'s for $i\in [n]$.
Thank you very much in advance for your time and help.

答:
As said in comments, let $x=\frac y{b_1}$ and $c_i=\frac {b_i}{b_1}$ to make
$$\sum_{i=1}^n b_i \left( \frac{a_i}{1+b_i x}\right)^2=b_1\Bigg[\frac{a_1^2}{(1+y)^2}+\sum_{i=2}^n \frac{ a_i^2\, c_i}{(1+ c_i\,y)^2}\Bigg]$$ Now, to get rid of the vertical asymptote at $y=-1$, we shall muliply everything by $(1+y)^2$ and consider the function

$$g(y)=b_1\Bigg[{a_1^2}+{(1+y)^2}\sum_{i=2}^n \frac{ a_i^2\, c_i}{(1+ c_i\,y)^2}\Bigg]-\phi(1+y)^2$$ for which
$$g(0)=-\phi+b_1 \sum_{i=1}^n a_i^2\qquad \qquad g(-1)=b_1\,a_1^2 >0 \qquad \qquad g'(-1)=0$$ $$g''(-1)=-2\phi+2b_1 \sum_{i=2}^n \frac { a_i^2\, c_i } {(1-c_i)^2 }$$

Assuming $g(0)<0$, using Taylor around $y=-1$, we  have
$$g(y)=g(-1)+\frac 12 g''(-1) (y+1)^2+ O\left((y+1)^3\right)$$ and then a first estimate
$$y_0=-1 +\sqrt {-\frac {b_1 a_1^2} {g''(-1)}}$$

Let us try using $n=5$, $b_i=10^{5-i}$, $a_i=p_{i+5}^2$ and $\phi=12345678987654321$. This gives as an estimate
$$y_0=-1+\sqrt{\frac{175914852251400405000}{15208068862859378242920969041} }\sim -0.999892$$
while the exact solution, obtained using Newton method is $-0.999848$

Starting with this guess, Newton iterates would be
$$\left(
\begin{array}{cc}
 n & y_n \\
 0 & \color{red}{-0.99989}244905795162394 \\
 1 & \color{red}{-0.9998}3867358692746307 \\
 2 & \color{red}{-0.999847}63616543148912 \\
 3 & \color{red}{-0.999847899}77068160754 \\
 4 & \color{red}{-0.999847899999109}03399 \\
 5 & \color{red}{-0.99984789999910920552}
\end{array}
\right)$$

Very immodestly, let me precise that this method bears my name.

Edit

If you are coding, for a complete safety,  combine bisection and Newton methods . If you go to this place, on page $359$, you will find subroutine RTSAFE (here and here) which does exactly that. It is very robust.

I apologize for giving you a reference to Fortran coding. If you can access the books of Numerical Recipes, you will find the equivalent in C and C++.


问:
Do I need to prove $\lim_{\|P\|\to 0} L(f,P)=\sup_P L(f,P)$, and $\lim_{\|P\|\rightarrow 0} U(f,P)=\inf_P U(f,P)$? Or is it trivial?
I'm trying to show that $f$ is bounded and Riemann integrable on $[a, b]$ implies that there exists a sequence $\{P_1, P_2, P_3, \ldots\}$ of partitions of $[a, b]$ s.t.
$$
\lim_{n\to \infty} [U_f(P_n)-L_f(P_n)] = 0
$$
in which case
$$
\int\limits_a^b f(x) dx = \lim_{n\to \infty} U_f(P_n) = \lim_{n\to \infty} L_f(P_n).
$$
(This is one part of an iff proof of the sequential characterization of the Riemann integral)

I wanted to do a proof by contradiction, and I found something I think would help finish my proof from this answer, but I'm not sure if I need to prove this or if it's some kind of definition or something. I've looked everywhere and can't find anything.

I'll lay out what I have so far:

Suppose $\lim_{n\to \infty} [U_f(P_n)-L_f(P_n)] \not= 0$. Then $\lim_{n\to \infty} U_f(P_n) - \lim_{n \to \infty}L_f(P_n) \not= 0$

$\Rightarrow \lim_{n\to \infty} U_f(P_n) \not= \lim_{n \to \infty}L_f(P_n)$.

Observe that for $\|P\|$ defined as the length of the largest subinterval $[x_{i-1}, x_i]$, $1 \leq i \leq n$, as $n \rightarrow \infty, \|P\| \rightarrow 0$ since the more partitions there are over the fixed interval, the smaller they must be in length in order to fit into the interval.

Then $\lim_{\|P\|\to 0} U_f(P_n) \not= \lim_{\|P\| \to 0}L_f(P_n)$ by the above explanation.

This is where I'm stuck, because if I can just say that $\lim_{\|P\|\to 0} L(f,P)=\sup_P L(f,P)$, and $\lim_{\|P\|\rightarrow 0} U(f,P)=\inf_P U(f,P)$ then we know the upper and lower Darboux integrals aren't equal, and thus $f$ isn't Riemann integrable, which is a contradiction.

Any help is appreciated, thanks in advance.

答:
If $f$ is Riemann integrable, then upper and lower Darboux integrals are equal and

$$\sup_P L(P,f) = \underline{\int_a}^b f(x) \, dx = \underbrace{\int_a^b f(x) \, dx}_I= \overline{\int_a^b} f(x) \, dx =\inf_P U(P,f)$$

For any $n \in \mathbb{N}$, by basic properties of $\sup$ and $\inf$, there exist partitions $P_n'$ and $P_n''$ such that

$$I-\frac{1}{n} < L(P'_n,f) \leqslant I \leqslant U(P''_n,f)< I+\frac{1}{n}$$

Let $P_n = P_n'\cup P_n''$ be the common refinement.  Since lower sums increase and upper sums decrease as the partition is refined, we have

$$I-\frac{1}{n} <L(P'_n,f) \leqslant L(P_n,f) \leqslant U(P_n,f) \leqslant U(P_n'',f) < I + \frac{1}{n}$$

Thus, $ 0 \leqslant U(P_n,f)- L(P_n,f) < \frac{2}{n}$ for all $n$ which implies that

$$\lim_{n \to \infty}[U(P_n,f) - L(P_n,f)] = 0.$$

Proof that $\lim_{\|P\| \to 0}U(P,f) = \inf_P U(P,f)$ and $\lim_{\|P\| \to 0}L(P,f) = \sup_P L(P,f)$.

Suppose $f:[a,b] \to \mathbb{R}$ is bounded and $|f(x)|\leqslant B$ for all $x \in [a,b]$.

Let $\bar{I} = \inf_PU(P,f)$.  For any $\epsilon > 0$ there exists  a partition $P_\epsilon=(x_0,x_1,\ldots,x_m)$ such that $\bar{I}\leqslant U(P_\epsilon,f) < \bar{I} + \frac{\epsilon}{2}$.

Note that $P_\epsilon$ has $m$ subintervals. Let $\delta = \frac{\epsilon}{4mB}$ and let $P$ be any partition with $\|P \| < \delta$.   The common refinement  $Q = P \cup P_\epsilon$ must have at most $m-1$ more points than $P$. (The largest discrepancy in the number of points occurs when each of the points $x_1,\ldots, x_{m-1}$ of $P_\epsilon$ is in the interior of a subinterval of $P$.)

Since $Q$ is a refinement of $P_\epsilon$, we have

$$\tag{1}\bar{I}\leqslant U(Q,f) \leqslant U(P_\epsilon,f) < \bar{I} + \frac{\epsilon}{2}$$

We claim that the difference in the upper sums for $P$ and $Q$ is bounded as

$$\tag{2}U(P,f) - U(Q,f) < (m-1) \cdot 2B \cdot \delta  = (m-1) \cdot 2B \cdot \frac{\epsilon}{4mB} < \frac{\epsilon}{2}$$

To prove the claim suppose first that $Q$ has only one more point $y^*$ in the interior of the subinterval $(y_{j-1},y_j)$ of partition $P$.  It follows that

$$U(P,f) - U(Q,f) \\= \sup_{x \in [y_{j-1},y_j]} f(x)(y_j- y_{j-1}) - \sup_{x \in [y_{j-1},y^*]} f(x)(y^*- y_{j-1})- \sup_{x \in [y^*,y_j]} f(x)(y_j- y^*) \\ \leqslant 2B \cdot (y_j - y_{j-1}) \leqslant 2B \|P\| < 2B\delta =  2B \frac{\epsilon}{4mB} = \frac{\epsilon}{2m}$$

By induction as we add up to $m-1$ additional points to $Q$ we get the claimed result (2).

Thus, $U(P,f) < U(Q,f) + \frac{\epsilon}{2}$ and using (1), we have with $\|P\| < \delta$,

$$\bar{I} \leqslant U(P,f) < U(Q,f) + \frac{\epsilon}{2} < \bar{I} + \epsilon$$

This proves that $\lim_{\|P\| \to 0} U(P,f) = \bar{I}= \inf_PU(P,f)$.

In a similar way we can show that $\lim_{\|P\| \to 0} L(P,f) = \underline{I}= \sup_P L(P,f)$.


问:
If $T$ is invertible and $K$ is compact, is $0\ne\lambda\mapsto(\lambda T+K)^{-1}$ holomorphic?
Let $X$ be a Banach space, $T\in\mathfrak L(X)$ be bijective, $K\in\mathfrak L(X)$ be compact and $$B:\mathbb C\to\mathfrak L(X)\;,\;\;\;\lambda\mapsto\lambda T+K.$$

Can we show that $B(\lambda)$ is bijective for all $\lambda\in\mathbb C\setminus\{0\}$ and $$\mathbb C\setminus\{0\}\to\mathfrak L(X)\;,\;\;\;\lambda\mapsto B(\lambda)^{-1}$$ is holomorphic?

This is clearly, if correct, an application of the analytic Fredholm theorem.

The trick is clearly to write $$B(\lambda)=\lambda T\left(1+\frac1\lambda T^{-1}K\right)\tag1\;\;\;\text{for all }\lambda\in\mathbb C\setminus\{0\}.$$ $\mathbb C\ni\lambda\mapsto\lambda T$ and $\mathbb C\setminus\{0\}\ni\lambda\mapsto\frac1\lambda T^{-1}K$ are trivially holomorphic. Moreover, $T^{-1}K$ is compact.

So, the only missing piece is that we need to show that $B(\lambda)$ is bijective for at least one $\lambda\in\mathbb C$. But why is that the case?

答: You seem to assume invertibilty of $T$. In that case $T+\frac 1 n K \to T$ in operator norm. Since the set of all invertble operators is open it folows that $T+\frac 1 n K$ is invertible  for large enough $n$ and this implies that $nT+K$ is invertible for such $n$.


问:
Dynamical limitation and minimization
I come across the following question: Under what conditions (on the series of functions $f_n$ or perhaps the domain of minimization) the following holds
$$
\lim_{n\to\infty}\min_{x_1,\ldots,x_n}f_n(x_1,\ldots,x_n) = \min_{\left\{x_i\right\}_i}\lim_{n\to\infty}f_n(x_1,\ldots,x_n)
$$
I can show that the above holds true when the minimization is carried over a finite number of variables independently of $n$, namely,
$$
\lim_{n\to\infty}\min_{x_1,\ldots,x_M}f_n(x_1,\ldots,x_M) = \min_{x_1,\ldots,x_M}\lim_{n\to\infty}f_n(x_1,\ldots,x_M)$$
But, I'm not sure how to approach the first question.

Thanks

答:
Let's a functions $F_n:\mathbb{X}\subset\mathbb{R}^{\mathbb{N}}\to \mathbb{R}$ such that $F_n(x_1,\ldots, x_n,\ldots )=f_n(x_1,\ldots, x_n )$ for all $f_n:\Pi_n(\mathbb{X})\subset\mathbb{R}^n\to \mathbb{R}$ of sequence $\{f_n\}_{n\in\mathbb{N}}$.  Here $\Pi_n$ is the projection $(x_i)_{i\in\mathbb{N}}\mapsto (x_1,\ldots, x_n)$. Most of the times when the minimum of a function $F:\mathbb{X}\subset\mathbb{R}^{\mathbb{N}}\to\mathbb{R}$ is achieved can admit the following conditions.

Condition: supose there is $x^*_n=(x_1^*,\ldots, x_n^*,\ldots)$ such that  $$F_n(x^*_n)=\min_{x\in \mathbb{X} }F_n(x).$$

Condition: supose thare is a sequence $\mathbb{R}^{\mathbb{N}} \ni x_{nk}\to x_{n\infty}\in \mathbb{R}^{\mathbb{N}}$ such that $$\lim_{k\to \infty}F_n(x_{nk})=F_n(x_{n\infty})=\min_{x\in \mathbb{X} }F_n(x).$$

Under these conditions the problem can be stated as
$$
\lim_{k\to \infty}\lim_{n\to \infty}F_n(x_{nk}) \mathop{=}^{?}\lim_{n\to \infty}\lim_{k\to \infty}F_n(x_{nk})
$$

For exemple for $\mathbb{X}\subset\ell^p$ whit usal topology of norm $\|\,\cdot\,\|_p$, we could to use the classical theorem

Theorem. Let $\{ F_t ; t\in T\}$ be a family of functions $F_t : Y \rightarrow \mathbb{C}$ depending on a parameter t; let $\mathcal{B}_X$ be a base $Y$ and $\mathcal{B}_{T}$ a base in $T$. If the family converges uniformly on $Y$ over the base $\mathcal{B}_{T}$ to a function $F : Y \rightarrow \mathbb{C}$ and the limit $\lim_{\mathcal{B}_{T}} F_t(y)=A_t$ exists for each $t\in T$, the both repeated limits $\lim_{\mathcal{B}_{Y}}(\lim_{\mathcal{B}_{T}}F_t(y))$ and $\lim_{\mathcal{B}_{T}}(\lim_{\mathcal{B}_{Y}}F_t(y))$ exist and the equality and holds
  $$
\lim_{\mathcal{B}_{Y}}(\lim_{\mathcal{B}_{T}}F_t(y))=\lim_{\mathcal{B}_{T}}(\lim_{\mathcal{B}_{Y}}F_t(y)).
$$

This theorem can be found in books of Zorich (Mathematical Analysis II p. 381).

Title: Dynamical limitation and minimization


问:
Confusion About Theorem for Finitely Generated Abelian Groups
I was wondering about finitely generated abelian groups. Am I understanding this correctly? We may apply the structure theorem for finitely generated abelian groups in the following manner:

$\mathbb{Z}/12\mathbb{Z}\cong \mathbb{Z}/3\mathbb{Z}\oplus \mathbb{Z}/4\mathbb{Z}$

as well as

$\mathbb{Z}/10\mathbb{Z} \cong \mathbb{Z}/2\mathbb{Z}\oplus \mathbb{Z}/5\mathbb{Z}$?

On top of this, I was wondering if we can claim in general that $\mathbb{Z}/n\mathbb{Z}$ is abelian under addition? Thanks.

答:
In both examples, your decompositions are correct. However, the Fundamental Theorem of Finitely Generated Abelian Groups only provides that decompositions exist. It doesn't tell you how to decompose specific Abelian groups.

In the case of your second example, we use the fact that the prime factors of $10$ only appear with multiplicity one; i.e., $10 = 2 \cdot 5$. So the primary decomposition formulation of the Fundamental Theorem necessitates that $\mathbb{Z}/10\mathbb{Z} \cong \mathbb{Z}/2\mathbb{Z} \oplus \mathbb{Z}/5\mathbb{Z}$.

As $12 = 2^{2} \cdot 3$, the Fundamental Theorem doesn't tell us whether the direct factor of order $4$ for $\mathbb{Z}/12\mathbb{Z}$ is $\mathbb{Z}/4\mathbb{Z}$ or $\mathbb{V}_{4}$. One needs to look at the group more closely to make this determination.

On top of this, I was wondering if we can claim in general that Z/nZ is abelian under addition?

Yes- $\mathbb{Z}/n\mathbb{Z}$ is cyclic. Namely, $\mathbb{Z}/n\mathbb{Z} = \langle \overline{1} \rangle$.


问:
Finding $\frac{S_{\triangle BKC}}{S_{\triangle ABC}}$
Suppose circle $I$ is $\triangle ABC$’s incircle and $D, E,$ and $F$ are the tangent points of $BC, CA, AB$ and circle $I.$ Given that $DI$ intersects $EF$ at $K, BC = 10, AC = 8,$ and $AB = 7$ find $\frac{S_{\triangle BKC}}{S_{\triangle ABC}}.$

答:
Since you let me have the pleasure of solving the problem (by not sharing your solution or approach), it is only fair that I reciprocate and let you enjoy the exercise of at least some parts of the proof. Here we go:

Lemma 1 - If in $\triangle ABC$ shown below, $AX$ bisects $\widehat A$ then $\frac{BX}{CX}=\frac{AB}{AC}$ .

Proof is left as an exercise for the interested reader.

Lemma 2- As in the figure below, extend $IK$ to $J$, such that $\widehat{JEK}=\widehat{IEK}$ . Then $\triangle EJI$ and $\triangle ABC$ are similar and $\frac{JI}{BC}=\frac{EJ}{AB}=\frac{EI}{AC}$.

Proof is somewhat beautiful, therefore it is left as an exercise for the interested reader.

Now for calculations, let us refer to the radius of the in-circle as $r$. Let us also write the lengths of $BC$ , $CA$ and $AB$ as $a$ , $b$ and $c$ , respectively. We have:
$$\frac{S_{\triangle BKC}}{S_{\triangle ABC}} = \frac{S_{\triangle BIC} + S_{\triangle BIK} + S_{\triangle CIK}}{S_{\triangle ABC}} $$
$$= \frac{a}{a+b+c} + \frac{a.IK}{(a+b+c)r} \qquad (1)$$
To find $IK$ , note that in $\triangle EJI$ , $EK$ bisects $\widehat{JEI}$ . From Lemma 1 we have:
$$\frac{IK}{JK} = \frac{EI}{EJ} $$
$$\therefore \frac{IK}{IJ} = \frac{EI}{EI+EJ} \qquad (2)$$
From Lemma 2 we have:
$$\frac{EI}{EI+EJ} = \frac{b}{b+c} \qquad (3)$$
and
$$\frac{IJ}{EI} = \frac{a}{b} $$
$$\therefore IJ = \frac{a.r}{b} \qquad (4)$$
From (2) , (3) and (4) we can calculate $IK$ :
$$IK = \frac{a.r}{b+c} \qquad (5)$$
Now from (5) and (1) we can calculate the desired ratio of areas:
$$\frac{S_{\triangle BKC}}{S_{\triangle ABC}} = \frac{a}{a+b+c} (1 + \frac{a}{b+c}) = \frac{a}{b+c}$$
For the particular values of $a=10$ , $b=8$ and $c=7$ we have:
$$\frac{S_{\triangle BKC}}{S_{\triangle ABC}} = \frac{10}{8+7} = \frac23$$


问:
$S_{\mathfrak{p}}(0)$ is the smallest $\mathfrak{p}$-primary ideal if $\mathfrak{p}$ is a minimal prime ideal of $A$.
The question is in A&D exercise 4.11 and I asked this in the previous post but I changed the question.
I want to show

If $\mathfrak{p}$ is a minimal prime ideal of a ring $A$, show that $S_{\mathfrak{p}}(0)$ is the smallest $\mathfrak{p}$-primary ideal.

Notation: $S_{\mathfrak{p}}(0)=\text{Ker}(A\to A_{\mathfrak{p}})$.

I found some proof about this problem: Since $\mathfrak{p}$ is minimal, $r(S_{\mathfrak{p}}(0)) = \mathfrak{p}$. Next, if $xy\in S_{\mathfrak{p}}(0)$ then there is $n>0$ such that $x^ny^n\in\mathfrak{p}$. If $x\notin S_{\mathfrak{p}}(0)$, then $x^n\notin\mathfrak{p}$ for all $n$ so that $y^n\in\mathfrak{p}$. Hence, there is $m>0$ such that $y^{nm}\in S_{\mathfrak{p}}(0)$.

In the proof, they said 'If $x\notin S_{\mathfrak{p}}(0)$ then $x^n\notin \mathfrak{p}$ for all $n$'. I don't understand why this is true. Could you explain this?

答: Here's a straightforward solution: Suppose that $xy\in S_{\mathfrak p}(0)$. We need to show that either $x\in S_{\mathfrak p}(0)$ or $y^n\in S_{\mathfrak p}(0)$ for some $n>0$. Assume that $y\notin \mathfrak p=r(S_{\mathfrak p}(0))$, that is, $y^n\notin S_{\mathfrak p}(0)$ for all $n>0$. Since $xy$ is in $S_{\mathfrak p}(0)=\ker(A\to A_{\mathfrak p})$, we can find some $s\notin \mathfrak p$ such that $$sxy=0=(sy)x.$$ As $y\notin \mathfrak p$, we see that $sy\notin\mathfrak p$, hence we obtain $x\in \ker(A\to A_{\mathfrak p})=S_{\mathfrak p}(0)$.


问:
Finding the number of integer solutions, why is this wrong?
The question is to find the number of solutions such that $(x, y)$ are integers: $(x-8)(x-10)=2^y$. Here's what I did: $u(u-2)=2^y$. From the quadratic formula, $u=1+\sqrt{1+2^y}$. This is where I think this question was meant to be solved by modular arithmetic here, but I don't know it so please try not to use it in your answers. I now constructed two right angle triangles. Triangle 1 has sides $1$, $2^{y/2}$,a hypotenuse of $\sqrt{1+2^y}$, and an angle $t$ such that $\sin(t)=\dfrac {1}{\sqrt{1+2^y}}$. Triangle 2 has sides of  $1$, $w$, a hypotenuse of $u$, and an angle $t$ such that $\cos t=\dfrac{1}{u}$. Now we want $\cot(t)+\dfrac1u=1$, or $\cot(t)+\cos(t)=1$. The problem now just reduces to finding the number solutions to that equation. However, this gives me an infinite number of solutions. Why is this wrong? Thanks!

答:
Let us continue your calculation at the point you reached $u=1+\sqrt{1+2^y}$. The calculation after that was not a good idea: introducing transcendental functions is unlikely to help in a problem of elementary number theory.

The solutions are $u=1\pm\sqrt{1+2^y}$. We want $\sqrt{1+2^y}$ to be an integer. So we want $1+2^y$ to be a perfect square, say $t^2$. We have arrived at the equation $1+2^y=t^2$ or equivalently
$$2^y=t^2-1=(t-1)(t+1).$$
Now we can argue precisely as in the other solutions: $t-1$ and $t+1$ are each a power of $2$. That forces $t=\pm 3$. Since we are using $1\pm\sqrt{1+2^y}$, the choice $t=3$ is enough.

To finish, we have $u=1\pm 3$. So $u=-2$ or $u=4$. That gives $x=u+8=6$ or $x=12$. Each has $y=3$.

Remark: The OP mentions modular arithmetic. That cannot quite work. The reason is that the equation does have a solution. A pure congruential argument would show there are no solutions at all. We can use congruences to rule out certain types of solutions, but certainly not all.


问:
Cauchy's Integral test for Convergence/Divergence
I don't understand how we are able to relate the convergence of a series seeing its Improper integral like

$\int_1 ^{\infty} f(n)dn < a_1 + a_2 +a_3+...... a_\infty$

If the integral diverges we can say that the series also diverges but if it converges how are we able to say that the series converges too. I am new to this. Any help is appreciated.

答:
That's because we have$$\sum_{n=1}^\infty f(n)\geqslant\int_1^\infty f(x)\,\mathrm dx
\geqslant\sum_{n=2}^\infty f(n).$$It follows from the second inequality the the convergence of the integral implies the convergence of the series and from the first inequality that the convergence of the series implies the convergence of the integral.


问:
Prove: $(L_0)^0 = L$
Let $W \subseteq V$ and $L \subseteq V^{*}$.

Let $W^0 = \{ f\in V^{*} \mid f(w)=0 ~~ \forall w \in W \}$ be the annihilator of $W$.

Let $L_0 = \{ v\in V \mid f(v)=0 ~~ \forall f \in L \}$.

How can I show that $(L_0)^0=L$ ?

What is the name of $L_0$ ? (Like $W^0$ is the annihilator of $W$)

Thank you.

答:
If $f \in L$, then $f(v) = 0$ for every $v \in L_0$ (by definition of $L_0$), that is, $f$ annihilates every element of $L_0$; and this means precisely that $f \in (L_0)^0$. Thus $L \subseteq (L_0)^0$.

To see that it is not necessarily true the equality $L = (L_0)^0$, take $V$ to be the real vector space of all continuous functions from $[0,1]$ to $\mathbb R$, and for each $n \in \mathbb N$ (the zero is included) consider $\phi_n \in V^*$ such that $$(\forall f \in V) \quad \phi_n(f) = \int_0^1 x^nf(x)dx.$$

Now, in virtue of the Weierstrass's Theorem we have that if $L := \{\phi_n : n \in \mathbb N\}$, then $L_0 = \{0\}$ and so $(L_0)^0 = \{0\}^0 = V^* \supsetneq L$.

However, the equality $L = (L_0)^0$ is true when $V$ is finite-dimensional, as @Mathlover points in his answer.

Title: Prove: $(L_0)^0 = L$


答:
Both results you quote are true by CRT.  In general, if $(m,n)=1$, then $\Bbb Z_{mn}\cong\Bbb Z_m\times\Bbb Z_n$ (and conversely).  This also follows from the fundamental theorem of finite abelian groups (FTFAG), which the theorem you mentioned generalizes.

Next $\Bbb Z_n:=\Bbb Z/n\Bbb Z$ is cyclic, hence abelian.

Title: Confusion About Theorem for Finitely Generated Abelian Groups

Body:
I was wondering about finitely generated abelian groups. Am I understanding this correctly? We may apply the structure theorem for finitely generated abelian groups in the following manner:

$\mathbb{Z}/12\mathbb{Z}\cong \mathbb{Z}/3\mathbb{Z}\oplus \mathbb{Z}/4\mathbb{Z}$

as well as

$\mathbb{Z}/10\mathbb{Z} \cong \mathbb{Z}/2\mathbb{Z}\oplus \mathbb{Z}/5\mathbb{Z}$?

On top of this, I was wondering if we can claim in general that $\mathbb{Z}/n\mathbb{Z}$ is abelian under addition? Thanks.


问:
Is mathematics one big tautology?
Is mathematics one big tautology? Let me put the question in clearer terms:

Mathematics is a deductive system; it works by starting with arbitrary axioms, and deriving therefrom "new" properties through the process of deduction. As such, it would seem that we are simply creating a string of equivalences; each property can be traced back logically to the axioms. It must be that way, that's how deductive systems work!

If that be the case, then in what sense are we introducing novel or new ideas? It would seem that everything is simply equivalent to the fundamental set of axioms that we choose to start with. Is there a precise step in mathematical derivation that we can isolate as going beyond pure logic? If so, then how does this fit in with the fact that mathematics is deductive? Must we change our view of mathematics as purely deductive? And if not, then is there a way to reconcile the feeling of creativity in mathematics with the fact that it boils down to pure logic?

I'm trying to figure out the true nature of what's going on here.

答:
Edit: expanded

The purely formal process of combining axioms with laws of deduction to move from one "truth" to another "truth" is tautological.

But the meaning which we assign to mathematical statements exists outside of this formal system.

To illustrate the difference, consider the following thought experiment. Suppose there existed an oracle that told us which statements are true (say, in ZFC). Then our job as mathematicians would be made easier in some sense, but still it would by no means be rendered obsolete.

First of all, mathematics is not just the quest for all truths. Rather, mathematicians work within a dynamic field of meaning, a constantly evolving language consisting of concepts that they find useful. An oracle, on the other hand, would produce statements in a fixed language, and without regard for their "meaning".

Secondly, mathematicians want to know why things are true. And they are not content with just any proof -- they want, if possible, a proof that illuminates, that has a potential for shedding insight, or for being generalizable or for embodying a technique that can be used in other problems. If we liken mathematics to a novel, then we are not content with just a list of facts or of minute details of cause-and-effect -- we are interested in a larger insight into the relationships between the characters.

Finally, as has been commented upon, a large part of mathematics is in finding the "right" questions to ask -- the ones which we find interesting and which could lead to further insight.

I hope this illustrates some of the differences between the "meaningless" nature of a truth-producing oracle and the meaningful, or meaning-seeking, nature of mathematics

Title: Is mathematics one big tautology?


问:
Automorphism group of $\mathbb{Z}_p\times \mathbb{Z}_p$

How to determine the automorphism group of $\mathbb{Z}_p\times \mathbb{Z}_p$ where $p$ is a prime? Or more specifically, how to determine the element of order $2$ in this group?

I got stuck here, since I only know that if two finite groups $H$ and $K$, where $(|H|,|K|)=1$, then $\text{Aut}(H\times K) = \text{Aut}(H) \times \text{Aut}(K)$. But $p$ and $p$ are not relatively prime.

Any hints or solutions are welcomed, thanks!

答:
As suggested by vadim123, the automorphism group is isomorphic to $\operatorname{GL}(2, p)$, the group of $2 \times 2$ invertible matrices over $\Bbb{Z}$.

If $p = 2$, the group $\operatorname{GL}(2, 2)$ is isomorphic to $S_{3}$, so you should be able to determine the involutions, which are
$$
\begin{bmatrix}0&1\\1&0\end{bmatrix}, \quad
\begin{bmatrix}1&1\\0&1\end{bmatrix}, \quad
\begin{bmatrix}1&0\\1&1\end{bmatrix}.
$$

If $p > 2$, you will have
$$
c = \begin{bmatrix}-1&0\\0&-1\end{bmatrix},
$$
a central element, and then the conjugacy class of
$$
b = \begin{bmatrix}1&0\\0&-1\end{bmatrix},
$$
which has order
$$
\frac{\lvert \operatorname{GL}(2, p) \rvert}{\lvert C_{\operatorname{GL}(2, p)}(b) \rvert}
=
\frac{(p^{2} - 1)(p^{2} - p)}{(p-1)^{2}}
=
(p+1)p.
$$


问:
A Friedrichs' Inequality with p = 2
Consider a bounded domain $\Omega \subset R^d$ with boundary $\Gamma$. I am trying to prove the Friedrichs' inequality which says that for some constant $C$, for all $v \in C^1$:

$||v||_{L_2(\Omega)} \leq C(||\nabla v||^2_{L_2(\Omega)} + ||v||^2_{L_2(\Gamma)})^{1/2}$

The hint says to use integration by parts on the identity:

$\int_{\Omega} v^2 dx = \int_{\Omega} v^2 \Delta \phi dx$, where $\phi (x) = \frac{1}{2d}|x|^2$

So:

$\displaystyle ||v||^2_{L_2(\Omega)}
= \int_{\Omega}v^2 \Delta \phi dx
\\\\
= \displaystyle \int_{\Gamma} (\nabla \phi) \cdot \eta v^2 ds - \int_{\Omega} (\nabla \phi) \cdot \nabla (v^2) dx
\\\\
\displaystyle = \int_{\Gamma} (\nabla \phi) \cdot \eta v^2 ds - \int_{\Omega} (\nabla \phi) \cdot \nabla (v^2) dx
\\\\
\displaystyle = \int_{\Gamma} \frac{x}{d} \cdot \eta v^2 ds - \int_{\Omega} \frac{x}{d} \cdot (2v \nabla v) dx
$

I can see the first norm arising, but not the second? My vector calculus is currently rusty... So any guidance towards a solution would greatly be appreciated.

答:
We have :
$$
\|v\|^2_{L^2(\Omega)} = \left|\int_{\Gamma} \frac{x}{d}\cdot \eta v^2 ds - \int_{\Omega} \frac{|x|}{d} (2v \nabla v) ds \right| \\ \leq \max_{\Omega_1}\frac{|x|}{d}\left|\int_{\Gamma} v^2 ds\right| + \max_{\Omega_1} \frac{2|x|}{d} \left|\int_{\Omega} v \nabla v ds\right|  \\ \leq \max_{\Omega_1} \frac{|x|}{d}\|v\|^2_{L^2(\Gamma)} + \max_{\Omega_1} \frac{2|x|}{d} \left|\int_{\Omega} v^2 ds\right|^{\frac 12} \left|\int_{\Gamma} |\nabla v|^2ds\right|^{\frac 12}
$$

by Cauchy-Schwarz. We trudge on :
$$
\max_{\Omega_1} \frac{|x|}{d}\|v\|^2_{L^2(\Gamma)} + \max_{\Omega_1} \frac{2|x|}{d} \left|\int_{\Omega} v^2 ds\right|^{\frac 12} \left|\int_{\Gamma} |\nabla v|^2ds\right|^{\frac 12}\\ = \max_{\Omega_1} \frac{|x|}{d}\|v\|^2_{L^2(\Gamma)} + \max_{\Omega_1} \frac{2|x|}{d} \|v\|_{L^2(\Omega)} \|\nabla v\|_{L^2(\Omega)}
$$

whence we are at :
$$
\|v\|^2_{L^2(\Omega)}\leq \max_{\Omega_1} \frac{|x|}{d}\|v\|^2_{L^2(\Gamma)} + \max_{\Omega_1} \frac{2|x|}{d} \|v\|_{L^2(\Omega)} \|\nabla v\|_{L^2(\Omega)}
 \\ \leq C'(\|v\|^2_{L^2(\Gamma)} + \|v\|_{L^2(\Omega)} \|\nabla v\|_{L^2(\Omega)})$$

Now, desperate times call for beautiful inequalities. Infact, the entirety of PDE theory is littered with inequalities that will blow anyone's mind, from the sublime to the ridiculous.

The inequality we use is this one. Recall that for any real $a,b$ we have $a^2+b^2 \geq 2ab$. We use this to write for any $C>0$ :
$$
2ab = 2\left(\frac aC\right) (bC) \leq \frac{a^2}{C^2} + C^2b^2 \implies ab \leq \frac{a^2}{2C^2} + 2C^2b^2
$$

for any real numbers $a$ and $b$. Let $a = \|v\|_{L^2(\Omega)} , b = \|\nabla v\|_{L^2(\Omega)}$. Then we have :
$$
\|v\|_{L^2(\Omega)} \|\nabla v\|_{L^2(\Omega)} \leq \frac{\|v\|^2_{L^2(\Omega)}}{2C^2} + 2 \|\nabla v\|^2_{L^2(\Omega)} C^2
$$

Put this back into the expression, and we get :
$$
\|v\|^2_{L^2(\Omega)} \leq C'(\|v\|^2_{L^2(\Gamma)} + \|v\|_{L^2(\Omega)} \|\nabla v\|_{L^2(\Omega)}) \\ \leq C'\left(\|v\|^2_{L^2(\Gamma)} + \frac{\|v\|^2_{L^2(\Omega)}}{2C^2} + 2 \|\nabla v\|^2_{L^2(\Omega)} C^2\right) \\
\leq C'\|v\|^2_{L^2(\Gamma)} + 2C'C^2 \|\nabla v\|^2_{L^2(\Omega)} + \frac{C'}{2C^2}\|v\|^2_{L^2(\Omega)}
$$

since we can choose $C$ as anything we want, we ensure that it is large enough so that $2C^2 > C'$. Then we can take $\frac{C'}{2C^2}\|v\|^2_{L^2(\Omega)}$ to the other side of the inequality, factorize and bring the $1 - \frac{C'}{2C^2}$ to the other side to get :
$$
\|v\|^2_{L^2(\Omega)} \leq \frac{C'}{1 - \frac{C'}{2C^2}} \|v\|^2_{L^2(\Gamma)} + \frac{2C'C^2}{1 - \frac{C'}{2C^2}} \|\nabla v\|^2_{L^2(\Omega)} \\ \leq D\left( \|\nabla v\|^2_{L^2(\Omega)} + \|v\|^2_{L^2(\Gamma)}\right)
$$

where $D$ is the larger of the coefficients. Hence, we are done!

The inequality I used is often seen in interpolation-type inequalities for Sobolev spaces. You will do well to remember it.

In general, to see the tricks and traps that PDE theory uses, it is recommended that you hike your way through the proofs of difficult results in the field, such as the Sobolev inequality , the embedding theorems, the Rellich-Kondrasov inequality, and the Poincare inequality (of course). Each of the proofs contain key algebraic observations like the one I make above, that allow you to go from one end to the other smoothly. These may be found in Kesavan's book titled "Topics in functional analysis and applications".

Title: A Friedrichs' Inequality with p = 2


问:
Finding the Normal line to a plane
I feel like I am missing something here, but maybe someone can help me out :)

I was asked to find the normal line to the plane $Ax+By+Cz+D=0$ that passes through the point $(x_0,y_0,z_0)$.

The equation I am seeing for the normal line is dependent on one of the variables. I think I am just forgetting how this works and then misreading what is in front of me.

Perhaps someone would be able to clear things up here?

答:
The normal vector to $Ax+By+Cz+D=0$ is $\left(\begin{matrix}A\\B\\C\end{matrix}\right)$

So your line is defined by $\vec{x}=\left(\begin{matrix}x_0\\y_0\\z_0\end{matrix}\right) +
\lambda \left(\begin{matrix}A\\B\\C\end{matrix}\right),\ \lambda \in \Bbb R$

Title: Finding the Normal line to a plane


问:
Find the value: $\int_{0}^{π/6}\ 4\sin^{2}xdx$
Solving: $\int_{0}^{\frac{\pi}{6}} 4\sin^2(x)dx$

This is my work

but I can not seem to get to the answer $\frac{1}{6}(2\pi-3\sqrt{3})$. Any help is much appreciated. Thank you!

答:
We have the following identity: $\operatorname{cos}(2x) = 1 - 2\operatorname{sin}^2(x)$. Therefore, $4\operatorname{sin}^2(x) = 2(1 - \operatorname{cos}(2x))$.

$$\begin{aligned}\int_0^{\frac{\pi}{6}}4\operatorname{sin}^2(x) dx &= \int_0^{\frac{\pi}{6}}2(1 - \operatorname{cos}(2x)) dx\\ &=\int_0^{\frac{\pi}{6}}2 dx - \int_0^{\frac{\pi}{6}}2\operatorname{cos}(2x)dx\\ &= 2x|^{\frac{\pi}{6}}_0 - \frac{2\operatorname{sin}(2x)}{2}\vert^{\frac{\pi}{6}}_0\\&=2\cdot\left(\frac{\pi}{6} - 0  \right) - \left[ \operatorname{sin}\left(2\cdot \frac{\pi}{6}\right)  - \operatorname{sin}\left(2\cdot 0\right)\right] \\ &=\frac{\pi}{3} - \operatorname{sin}\left( \frac{\pi}{3}\right)=\frac{\pi}{3}-\frac{\sqrt3}{2}\end{aligned}$$


问:
Determine Whether $\int_1^\infty \frac{x^\frac{2}{3}+x^\frac{5}{3}}{x^2}dx$ diverges or converges
Yo, dealing with a particularly tricky integral right now. $\int_1^\infty \frac{x^\frac{2}{3}+x^\frac{5}{3}}{x^2}dx$. I want to know if it converges/diverges using the comparison theorem. Which function should I base off the comparison from? I can do something like $\frac{1}{x^\frac{4}{3}}$ for convergence and $\frac{1}{\sqrt[3]{x}}$ for divergence. That's all I have figured out for now.

答:
$$ \frac{x^\frac{2}{3}+x^\frac{5}{3}}{x^2} = \frac {1}{x^\frac 43} + \frac {1}{x^\frac 13}$$
$$\frac {1}{x^\frac 43} + \frac {1}{x^\frac 13} > \frac {1}{x^\frac 13}$$

$$\int_1^{\infty} \frac {1}{x^\frac 13}dx = [\frac 32 x^{2/3}]_1^\infty{\to \infty} $$
Since, $\int_1^{\infty} \frac {1}{x^\frac 13}dx$ diverges, the given integral also diverges.


问:
$y=x$ and $y^n$ In the First Quadrant
I wanted to ask a question about an equation, $A(n)$, where $n \in \mathbb{N}$ be the area in the first quadrant bound by $y = x$ and $y = x^n$. Would the following be true:

$A(n)=A(\frac{1}{n})$

$\displaystyle \lim_{n \to \infty} = 1$?

$A(n)=A(\frac{1}{n})$?

In general I visualize $y=x$ and $y=x^n$ as taking up most of the first quadrant, but takes up less space as the $x$ value increases. Does this come into play in any way?

答:
It is possible to answer both of these questions without actually working out the integrals, albeit unrigorously.

For the first, think of the shape of the curve $y=x^n$ as $n$ increases. You'll notice the curve becomes more "square". Flatter close to $x=0$ then steeper as $x$ gets close to $1$ and beyond. Taken to the limit, the bounded area will tend toward forming an isosceles right triangle with both catheti (perpendicular sides) equal to $1$ and a hypotenuse formed by a segment of the line of identity $y=x$ with length $\sqrt 2$. That has area $\frac 12$, and it is the same value that you'd calculate using calculus (you can refer to the formula in Kavi's comment and take the limit quite easily).

For the second, note that if $f(x) = x^n$, then $f^{-1}(x) = x^{\frac 1n}$. The graph of the inverse function is found by reflecting the graph of the original function about the line of identity, and the line of identity itself remains invariant. Which means the area does not change when you replace $n$ with $\frac 1n$. You will, of course, arrive at the same result using the precise formula you get after integration and imposing bounds (again, refer to Kavi's formula and replace $n$ by its reciprocal).

So, first one false, second one true.

Title: $y=x$ and $y^n$ In the First Quadrant


问:
Why are upper bounds only defined on subsets of preordered sets?
So I recently was reading through spivaks calculus and just learned the least upper bound property.
I noticed that in the definition the sets in consideration are always subsets of $\mathbb{R}$
This made me curious as to why only subsets of $\mathbb{R}$ were considered and why they have a notion of upper bounds.

Having looked up the definition of what an upper bound of a set is on Wikipedia I came across this rather confusing definition.

an upper bound or majorant
of a subset S of some preordered set (K, ≤) is an element of K which is
greater than or equal to every element of S.

I do not know what a preordered set is but im willing to guess that $\mathbb{R}$ is an example of one .

So basically my question is what is a preordered set and what about them makes them able to have upper and lower bounds?

Thanks in advance.

答:
A pre-ordered set is a set $X$ with a relation, often denoted by $\le$, that obeys two simple axioms: $x \le x$ for all $x \in X$ and transitivity: If $x \le y$ and $y \le z$ both hold, then so does $x \le z$. If you think of $x \le y$ as "$y$ is at least as big as $x$", these properties make intuitive sense. A common example is set inclusion, so $X$ is all subsets of $\Bbb N$ say and $x \le y$ is defined to mean $x$ is a subset of $y$. Then $x=\{0,1\}$ and $y=\{2,3\}$ are not "comparable", neither $x \le y$ nor $y \le x$ holds, but the set $\{x,y\}$ has a least upper bound, namely $z=\{0,1,2,3\}$: both sets are smaller: $x \le z$ and $y \le z$ and there can be no smaller common upper bound. One can easily show that this $X$ with this (non-linear, unlike $\Bbb R$) pre-order also has the least upper bound property.

It's an important defining feature of $\Bbb R$ that it has this lub property, it's what distinguishes it from $\Bbb Q$; it "fills up" the rational number line with relevant real numbers like $\sqrt{2}, \pi$ and $e$.. That's why it's used a lot in analysis. But the phenomenon of preordered sets is everywhere in maths and having the lub property can be very relevant there as well.


问:
Convergence of series $\frac{x^{n}}{\sqrt{n}}$ when $0<x<1$
I was stuying 1-D random walk example in Borel-Cantelli Lemma

Event $A_n$ = We return back to origin in 2n steps

Probability of taking right step at each point = p

Probability of taking left step at each point = q

$\sum_{n=1}^{\infty}(P(An)) =$ $2n\choose n$ $(p^n)(q^n)$

After using sterling's approximation it comes to the form: $k\cdot \sum_{n=1}^{\infty} \frac{x^{n}}{\sqrt{n}}$

So I need to find if series $\sum_{n=1}^{\infty} \frac{x^{n}}{\sqrt{n}}$ when $0 < x < 1$ converges or not to determine which Borel-Cantelli Lemma would be applicable.

答: $\sum \frac {x^{n}} {\sqrt n} \leq \sum x^{n} =\frac x {1-x} <\infty$.


答:
The series converges for $-1<x<1$. This can be established with the Ratio Test.

\begin{align*}
\lim_{n\to\infty}\left|\frac{\frac{x^{n+1}}{\sqrt{n+1}}}{\frac{x^n}{\sqrt{n}}}\right| &= \lim_{n\to\infty}\left|\frac{\sqrt{n}}{\sqrt{n+1}}x\right|\\
&= |x|\lim_{n\to\infty}\frac{\sqrt{n}}{\sqrt{n+1}}\\
&= |x|<1\text{ if } -1<x<1
\end{align*}

Title: Convergence of series $\frac{x^{n}}{\sqrt{n}}$ when $0<x<1$

Body:
I was stuying 1-D random walk example in Borel-Cantelli Lemma

Event $A_n$ = We return back to origin in 2n steps

Probability of taking right step at each point = p

Probability of taking left step at each point = q

$\sum_{n=1}^{\infty}(P(An)) =$ $2n\choose n$ $(p^n)(q^n)$

After using sterling's approximation it comes to the form: $k\cdot \sum_{n=1}^{\infty} \frac{x^{n}}{\sqrt{n}}$

So I need to find if series $\sum_{n=1}^{\infty} \frac{x^{n}}{\sqrt{n}}$ when $0 < x < 1$ converges or not to determine which Borel-Cantelli Lemma would be applicable.


问:
Category of sheaves on a topological space is not $\mathbf{AB4^{*}}$
I want to show that the category of sheaves on a topological space ($\mathbf
{Sh}(X)$) is not $\mathbf{AB4^{*}}$. Recall that $\mathbf{AB4^{*}}$ means that the product of epimorphisms is an epimorphism.

What I have tried:

Let $X = \mathbb{C}$ and consider the sheaf of holomorphic functions: that is, for every open set $U$, we have the abelian group
$$\mathcal{O}(U) := \left\{f\colon U \subseteq \mathbb{C} \longrightarrow \mathbb{C} : f \text{ is holomorphic in } U\right\}$$
Now consider the sheaf morphism that, given in components, is the complex differentiation map:
$$
\begin{array}{rcl}
     d(U) \colon \mathcal{O}(U)& \longrightarrow &\mathcal{O}(U)\\
     f & \longmapsto & d(U)(f) := \dfrac{df}{dz}
\end{array}
$$
I've seen that $d$ is an epimorphism of sheaves (not of presheaves: the function $f(z) = 1/z$ in $\mathbb{C}\setminus{\left\{0\right\}}$ is a classical counterexample), but I suspect that the product morphism is not epic. Any ideas? I accept other counterexamples, my goal is to show that $\mathbf{AB4^{*}}$ doesn't hold. $\mathbf{EDIT:}$

Given $\left\{f_{i} \colon A_{i} \longrightarrow B_{i}\right\}_{i \in \mathcal{I}}$, the product morphism is the unique morphism $$\varphi \colon \prod_{i \in \mathcal{I}} A_{i} \longrightarrow \prod_{i \in \mathcal{I}} B_{i}$$
so in my example, by product morphism I mean $d = f_{i}$ for $i \in \mathcal{I}$, where $\mathcal{I}$ is an infinite set.

答:
Another example: take $X=\{\infty\}\cup \mathbb{N}^+$ and give it the topology generated by opens $U_k=\{\infty,k,k+1,\dots\}$ and $V_k=U_k-\{\infty\}$ for $k=1,2,\dots$.

For a sheaf $F$, the stalk $F_k=\Gamma(V_k,F)$, $F_{\infty}=colim_kF(U_k)$.

1.the constant presheaf $\underline{Z}$ is a sheaf. For $j\ge 1$, define a morphism $f_j:\underline{Z}\to \underline{Z}$: on $V_k$ is always identity,  on $U_k$ is $0$ when $k<j$ and is identity when $k\ge j$. Then $f_j$ is surjective.

2.Suppose that $\prod_{j\ge 1}f_j:A\to B$ is surjective. Let $t=(1,1,\dots)\in\Gamma(X,B)$. There exists $k\ge 1$ and $s=(g_1,g_2,\dots)\in\Gamma(U_k,A)=\prod\mathbb{Z}$, such that $f_j(g_j)=1\in\Gamma(U_k,B^{(j)})$ for any $j$. However, $f_j|_{U_k}=0$ when   $j>k$.


问:
$\Omega=\{1,2,3,4\}$. Consider the events $E = \{1,2\}, F = \{1,3\}$ and $G = \{3,4\}$,Which option correct regarding independence of events?
Let $P \ $ be a probability function that assigns the same weight to each of the points of the sample
space $\Omega = \{1,2,3,4\}$. Consider the events $E = \{1,2\}, F = \{1,3\}$ and $G = \{3,4\}$. Which of the following statement(s) is (are) true?

$(1)$ $E$ and $F$ are independent

$(2)$ $E$ and $G$ are independent

$(3)$ $F$ and $G$ are independent

$(4)$ $E$, $F$ and $G$ are independent

I am not sure what they are asking. I know for independence we need to show $P(A \cap B)=P(A)P(B)$
but I am not able to understand events here. Does that mean event $E$ is selecting $1$ and $2$?

$P(\text{selecting $1$} )=\dfrac{1}{4}=P(\text{selecting $2$} )$

Therefore $P(E)=\dfrac{1}{16}=P(F)$

But $E\cap F = \{1\}$ therefore $P(E\cap F)=\dfrac{1}{4}\ne P(E)P(F)$. Is it true that they are not independent?

答: No, the event $E$ contains the outcomes $1$ and $2$ of the sample space, so $P(E) = P(\{1\}) + P(\{2\}) = \frac{1}{4} + \frac{1}{4}$. Otherwise you are correct that you need to check $P(E \cap F) \overset{?}{=} P(E) P(F)$, etc.


问:
Does $|\Bbb R^I| = |\Bbb R|$ imply that $I$ is (at most) countable?
$\newcommand{\R}{\Bbb R}$
This question is inspired from this question where the OP was inquiring about the cardinality of $\Bbb R^I$ for different $I$.

As in my answer, for $I$ finite or countable, we have that $|\R^I| = |\R|$. For $I = \R$, we have $|\R^I| = |2^\R| > |\R|.$

However, for $I$ with cardinality strictly between $|\Bbb N|$ and $|\R|$, can we say anything?

I am aware that $\sf{CH}$ states that there is no such set. For now, let us assume $\sf{AC}$ or at least the hypothesis that any two cardinalities are comparable. Does $|\R^I| = |\R|$ then imply that $|I| \le |\Bbb N|$?

My work so far has shown that $|\R^I| = |R| \implies |I| < |\R|$.

答:
Suppose $|I| = \aleph_1$ where $\aleph_0 < \aleph_1 < \mathfrak{c}=2^{\aleph_0}$, which can hold in some models of set theory.

Then $2^{|I|}$ is at least $2^{\aleph_0}$, but could be more or equal, depending on what model of set theory you are (e.g. MA (Martin's axiom) will imply they're equal). See Easton's theorem for other options.

So in short: you cannot even be sure there will be intermediate cardinalities, but if there are we cannot say more than that it's at least $\mathfrak{c}$. So the short answer to the title question is no it need not be.


问:
Duality between universal enveloping algebra and algebras of functions
There is a well known duality (of Hopf algebras)  between universal enveloping algebra $U(\mathfrak{g})$ of a complex Lie algebra  $\mathfrak{g}$ of a compact group $G$ and the algebra of continuous functions $C(G)$.

My question is, is there in the literature any place where this is presented in some detail? Bonus: also some references pointing to the generalization of this fact to quantized universal enveloping algebras?

Wikipedia says it is related to Tannaka-Krein theory, but I don`t know much about it and from a preliminary search I found nothing about this duality in the texts.

答:
I would maybe recommend some classical literature on quantum groups. I have in particular two books in mind:

Christian Kassel: Quantum groups

Anatoli Klimyk, Konrad Schmüdgen: Quantum Groups and Their Representations

Kassel brings quite detailed overview over both the Lie group $SL(2)$ as well as the Lie algebra ${\frak sl}(2)$ and then describes in detail the duality between $U({\frak sl}(2))$ and $SL(2)$ in section V.7.

Klimyk–Schmüdgen describes this duality for a general Lie group $G$ and its Lie algebra $\frak g$ (although quite briefly) in Sections 1.2.5–1.2.6.

In order to bring yet another source, it is maybe worth reading also the book of Timmermann, where he describes duality of Hopf algebras in Section 1.4 and in particular discusses Lie groups and Lie algebras in Example 1.4.7.

Now as for the quantum group case, both Kassel and Klimyk–Schmüdgen study in detail the quantum groups $SL_q(2)$ and $U_q({\frak sl}(2))$ and then describe their duality (Kassel in §§VII.4–VII.5, Kl–Sch in §4.4).

Applications of Tannaka–Krein duality for quantum groups is the main focus of my research, but to be honest, I am not sure how is this related to this kind of duality. (But maybe it is. Could you maybe provide a link for the wiki article?)


问:
If $A$ has the same cardinality as $B$, then $A \setminus B$ has the same cardinality as $B \setminus A$.
If $A$ has the same cardinality as $B$, then $A \setminus (A \cap B)$ has the same cardinality as $B \setminus (A \cap B)$.

If the sets are finite, then this is easy to see. But how to prove this in general?

I'm thinking along the following lines. Define $C = A \cap B$, $D = A \setminus C$, $E = B \setminus C$. So $B = C \cup D$ and $B = C \cup E$.
Suppose $f$ is the bijection from $A$ to $B$. Write
$$
B = (C \cap f(C)) \cap (C \cap f(E)) \cap (E \cap f(C)) \cap (E \cap f(E)).
$$
I think I need to redefine the map based on this decomposition. But I don't see how. And I'm not sure if this is fruitful.

答: You cannot prove it in general, take $A=\Bbb Q$, $B=\Bbb N$. Both are countable, but $A\setminus B$ is infinite, $B \setminus A$ is empty.


问:
Maximization of minimum difference
Suppose we have a function of the form: $(x_1 - x_2) + (x_3 - x_4) + (x_5 - x_6)$ and we have maximized this summation using linprog (using some constraints which are not important for this matter). This provides us with a value for the different x variables.

The problem I now want to solve is the maximization of the minimum $(x_i - x_j)$ and with the constraint that the solution $(x_1, x_2, ...)$ filled in in the original summation should have a higher or the same value (constraint).

This would distribute the difference between the variables $x_1,x_2; x_3,x_4, ...$

How can this problem be solved in Matlab (maximization of the minima)?

答:
In order to maximize a minimum using linear programming techniques, you can introduce an additional variable $\delta$ and add constraints of the form

$$x_i-x_j\ge \delta$$

The variable $\delta$ must then be maximized. For your problem to be well-defined there should of course be additional constraints on the variables $x_i$, probably like the ones you used in the first problem.

Title: Maximization of minimum difference


答:
Even under $\mathrm{GCH}$, if $\lambda$ is a cardinal with $\text{cf } \mathfrak c \le \lambda \lt \mathfrak c$, you have $\mathfrak c^\lambda = \mathfrak c^+$.

See Jech - Set Theory, theorem 5.15.

Title: Does $|\Bbb R^I| = |\Bbb R|$ imply that $I$ is (at most) countable?

Body:
$\newcommand{\R}{\Bbb R}$
This question is inspired from this question where the OP was inquiring about the cardinality of $\Bbb R^I$ for different $I$.

As in my answer, for $I$ finite or countable, we have that $|\R^I| = |\R|$. For $I = \R$, we have $|\R^I| = |2^\R| > |\R|.$

However, for $I$ with cardinality strictly between $|\Bbb N|$ and $|\R|$, can we say anything?

I am aware that $\sf{CH}$ states that there is no such set. For now, let us assume $\sf{AC}$ or at least the hypothesis that any two cardinalities are comparable. Does $|\R^I| = |\R|$ then imply that $|I| \le |\Bbb N|$?

My work so far has shown that $|\R^I| = |R| \implies |I| < |\R|$.


问:
Finding $\frac{S_{\triangle BKC}}{S_{\triangle ABC}}$
Suppose circle $I$ is $\triangle ABC$’s incircle and $D, E,$ and $F$ are the tangent points of $BC, CA, AB$ and circle $I.$ Given that $DI$ intersects $EF$ at $K, BC = 10, AC = 8,$ and $AB = 7$ find $\frac{S_{\triangle BKC}}{S_{\triangle ABC}}.$

答:
We use the identity $A = r \cdot s, $ where $s$ is sub-perimeter, $r$ is the inradius and $A$ is the area of the triangle.

So area of $\displaystyle \triangle ABC  = \frac{10+8+7}{2} \cdot r = \frac{25r}{2}$

Area of $\triangle BIC = \frac{1}{2} BC \cdot DI = 5r$

Area of $\triangle BIK = \frac{1}{2} BD \cdot KI$, Area of $\triangle CIK = \frac{1}{2} CD \cdot KI$

So all we are left with is to find $KI$. For that, we draw internal angle bisector of $\angle A$ which meets $BC$ at $M$.

As $AF = AE, FI = IE, \angle AFI = \angle AEI = 90^0$, therefore $AI$ is perpendicular bisector of $FE$.

So $\angle IFK = \frac{\angle A}{2}$. Also, $\angle FIK = 180^0 - \angle FID = \angle B$.

That leads to $\triangle FIK \sim \triangle ABM$ (by A-A-A)

$\frac{KI}{FI} = \frac{BM}{AB} = \frac{14/3}{AB} \implies KI = \frac{2r}{3}$

(to calculate $BM$, we used the fact that $AM$ is the angle bisector and hence $BM:MC = AB:AC = 7:8$ and $BM + MC = 10$)

Getting to concluding steps, area of $\triangle BKC = 5r + \frac{1}{2} (BD+CD) \frac{2r}{3} = \frac{25r}{3}$

$\therefore \frac{S_{\triangle BKC}}{S_{\triangle ABC}} = \frac{2}{3}$.


问:
Substituition of the integral limit in the first fundemantal theorem of calculus
As I was reading G. Strang's Calculus book, I came across an example problem I couldn't understand. The topic is the First Fundamental Theorem of Calculus. The form proposed for the theorem is (Picture): $$F( x) \ =\ \int ^{x}_{a} f( t) dt\\$$ After proving this, there is an example problem. $$Let\ F( x) \ =\ \int ^{\sqrt{x}}_{1} \sin(t)dt.
Find\ F'( t)$$Book solves the question by (Picture) $$u(x) = \sqrt{x}\ then\ F(x) = \int ^{u(x)}_{1}\sin(t)dt\\F'( x) \ = \sin( u( x))\frac{du}{dx}\\
=\sin( u( x)) \cdot \left(\frac{1}{2} x^{-\frac{1}{2}}\right)\\
=\frac{\sin\sqrt{x}}{2\sqrt{x}} \cdot \\$$My question is that, in it's original form, there is an $x$ in the upper limit of the integral. When we substitute it with $u(x)$, shouldn't we change $F(x)$ with $F(u(x))$ also? How does it stay unchanged?

答: Since $F(x)=\int_1^{\sqrt x}\sin t\,\mathrm dt$, $F\bigl(u(x)\bigr)=\int_1^{\sqrt{\sqrt x}}\sin t\,\mathrm dt$. But that's not the function that you want to differentiate. Instead, it's $F(x)$.


问:
How much classical algorithmics and probability theory is necessary for quantum algorithmics?
I would like to learn more about quantum algorithmics in next months/years.

I wonder - do I need to be very very good in "classical" algorithmics to be good in quantum algorithmics, or are they almost "independent" and I don't need more than basics or things I will learn "ad-hoc" when I will be needing them?

The same about probability - does quantum algorithmic uses very advanced probability theory, like more than 1 semester course of probability on Math studies? If Yes, does it use only discrete probability or also continuous probability?

Thanks in advance for answers.
Stay cool!

答:
Quantum Computing brings in a lot of background information. Linear Algebra, some familiarity with complex variables, and trig are necessary. Everything else can really be picked up in the process of taking the course. A lot of Quantum Computing courses also introduce a lot of the linear algebra (e.g., Hilbert Spaces, inner products, tensor products).

With that said, there are a lot of subjects that will almost certainly contribute to a better understanding. In a one-qubit system, the qubit states correspond precisely to the unit vectors in $\mathbb{C}^{2}$. These unit vectors sit on the surface of the three dimensional unit sphere, which I'll call $S^{3}$. In Quantum Computing, $S^{3}$ is referred to as the Bloch Sphere. Now $S^{3}$ has nice algebraic structure; namely, the points on the unit sphere correspond precisely to the quaternions of norm $1$.

Recall, the Quaternions (or Hamiltonians) are the elements of:
$$\mathbb{H} = \{ a + b\textbf{i} + c \textbf{j} + d\textbf{k} : a, b, c, d \in \mathbb{R}, \textbf{i}^{2} = \textbf{j}^{2} = \textbf{k}^{2} = -1, \textbf{ij} = -\textbf{ji}, \textbf{ik} = -\textbf{ki}, \textbf{jk} = -\textbf{kj} \}$$

That is, $\textbf{i}, \textbf{j}, \textbf{k}$ are square roots of $-1$ and anti-commute.

We can get our hands around the Hamiltonians by representing them as $2 \times 2$ unitary matrices, where the norm of a Hamiltonian element corresponds to the determinant of the corresponding matrix. So the Hamiltonians of norm $1$ are represented as special unitary matrices. Note that $U(2)$ and $SU(2)$ (the $2 \times 2$ unitary and special unitary matrices) form groups under matrix multiplication. Precisely, we are looking at the representation theory of a Lie group. Pollatsek's Lie Groups text is an accessible introduction to these ideas for undergrads, which only assumes Multivariable Calculus and Linear Algebra (https://www.amazon.com/Lie-Groups-Introduction-Mathematical-Association/dp/0883857596).

Quantum Computers are also very good at solving number theoretic and algebraic problems (e.g., factoring integers, Hidden Subgroup for Abelian groups, etc.). So having some number theory and group theory will be useful.

The key (and perhaps, only) algorithmic technique in Quantum Computing is the Fourier transform. In particular, it is nice to think of the Fourier transform over finite Abelian groups as a change of basis into the character basis. For an Abelian group $G$, the characters form an orthonormal basis of the group algebra $\mathbb{C}[G]$ (which is isomorphic to the vector space of functions from $G \to \mathbb{C}$). Having some Fourier Analysis and (again) Group Theory will be particularly helpful. Even if you don't have Group Theory, just having some Fourier Analysis will be very useful for picking up on the ideas.

From a Complexity standpoint, it would be useful to look over some randomized complexity classes like $\textsf{BPP}$. Note that $\textsf{BQP}$ generalizes $\textsf{BPP}$, rather than $\textsf{NP}$. Quantum Computing is inherently random, so having some perspective on error and randomness in classical computing will be useful.