问:
What is the minimum order of a 3-regular graph H containing G as an induced subgraph?
What is the minimum order of a 3-regular graph H containing the graph G(in diagram) as an induced subgraph?

答:
Since the question asked for the given graph to be an induced subgraph, the degree-$3$ vertex of $G$ cannot be in a triangle. All vertices of the triangular prism lie in triangles, so $G$ cannot be induced in there; it is also easy to show that the other non-isomorphic graph on $6$ vertices, $K_{3,3}$, does not have $G$ induced.

The minimum number of vertices in $H$ is therefore $8$:


问:
Unable to form an equation for the question
You own a hamburger stand. Your specialty is delicious cheeseburger that you sell for USD $7.00$ each. The cost of ingredients for each cheeseburger, such as meat, cheese, buns, tomato etc., is USD $1.65$. You hire workers at USD $12$ an hour. The production function for cheeseburgers is $Q = 40 \sqrt{L}$ , where L is the number of labor hours employed and $Q$ the number of cheeseburgers produced. There is a fixed cost of USD $172$.

There is no way to produce delicious cheeseburgers without also producing lots of grease, which needs to be hauled away. In fact, every cheeseburger that you produce, you also produce 1 ounce of grease. Lately, cooking grease has become popular as an ingredient to produce alternative fuel. So, a company begins to buy grease from you every day at the rate of USD $0.80$ per pound, thereby eliminating your cost of hauling away grease and providing you another revenue source.

Calculate the daily profit maximizing number of hamburgers, the amount of grease produced, the number of labor hours used and the consequent level of profit.

I have actually come up with an equation and differentiated it to get an answer. Just want to be sure it is right. My equation : $(-3x^2 + 2140x - 68800) / 400x$.

Number of burgers for maximum profit are coming out to be $151/152$. Does this sound right or am I going wrong somewhere?

答:
Math Lover and Henry have given nice answers to the OP's cheeseburger problem; this answer addresses the OP's italicized question, Does this sound right or am I going wrong somewhere?

One mistake is in ignoring the contribution of the sale of grease, so the $2140$ in the expression $(-3x^2+2140x-68800)/400x$ should really be $2160$.

But the real problem is that the expression $(-3x^2+2140x-68800)/400x$ represents the profit per cheeseburger, which is not what it is you want to maximize. Instead, you want to maximize the total profit, which is simply $(-3x^2+2140x-68800)/400$ (or, rather, $(-3x^2+2160x-68800)/400$, correcting for the omission of the grease contribution).  It may, superficially, sound like these are the same two things to maximize, but they aren't, and it'll be time well spent to think about why they're not the same.

Finally, two hopefully helpful suggestions. First, it's OK to use $x$ as your variable, but it would be easier to stick with $Q$; at the very least you should somewhere or other explicitly say what $x$ represents. And second, $151/152$ looks like a fraction, slightly less than $1$, which seems to suggest you've concluded the profit maximization occurs by making (and selling) something short of a single cheeseburger, which makes little sense. It took me a while to realize that you meant the profit (per burger) is maximized at a value of $x$ between $151$ and $152$ (i.e., $x=\sqrt{68800/3}\approx151.43755$). In short, please don't make us guess at the meaning of your expressions!

ContentLicense: CC BY-SA 4.0


问:
If a bounded $\mathcal{C}^n (\mathbb{R})$ function has a bounded n-th derivative, then all of its derivatives are bounded.
Let $f$ be bounded $\mathcal{C}^n (\mathbb{R})$ function such as that $f^{(n)}$ is bounded.

Prove that all of its derivatives $f^{(k)}$ for $k \in$ $ \{ 1,..,n-1 \}$ are bounded.

So far I've been able to prove it for the case : n=2 using Taylor's inequality, however I've not been able to generalise this using induction, as I must prove that one of its derivatives other than
$f^{(n)}$ is bounded.

答:
For $f \in \mathcal{C}^n (\mathbb{R})$ and $0 < k < n$, Taylor's theorem can be used to show that there are coefficients $c_1, c_2, \ldots, c_{n-1}$ such that
$$
\frac{f^{(k)}(x)}{k!} = \sum_{j=1}^{n-1} c_j \left(
 f(x+j) - f(x) - \int_x^{x + j} \frac{f^{(n)}(t)}{(n-1)!} (x+j - t)^{n-1} \, dt
 \right)
$$
for  all $x \in \Bbb R$, see for example If $f$ is Lebesgue integrable and $f''$ exists and is Lebesgue integrable, what can we say about the integrability of $f'$?.

In particular, the boundedness of $f$ and $f^{(n)}$ implies the boundedness of $f^{(k)}$ for $0 < k < n$.

See also Landau–Kolmogorov inequality for more estimates of this type.


问:
Embedding of finite groups in small $\operatorname{GL}(n,F)$
How does we obtain a small (not necessarily smallest) value of $n$ such that a $p$-group of order $p^m$ can be embedded in $\operatorname{GL}(n,\mathbb{F}_p)$?

[For embedding of a finite group $G$ in $S_n$, we can compute a small value of $n$ by looking at the minimum index of a subgroup $H$ containing no proper normal subgroup of $G$. Here, this value of $n$ need be necessarily smallest.]

答: For p > 2, check http://ysharifi.wordpress.com/2011/02/17/gln-z-has-only-finitely-many-finite-subgroups/, it is not hard to show that a group of order n can be embedded in $GL(n,\mathbb{F}_q)$ with q an arbitrary odd prime power.

ContentLicense: CC BY-SA 3.0


问:
What is the derivative of $\ln(4^x)$?
What is the derivative of $\ln(4^x)$  (which I believe is also equal to $x\ln4$)?

Is it $\dfrac{1}{x\ln4}$?

答:
Hint : $\ln 4^x=y \implies x \ln 4=y$

Since $\ln4$ is a constant. The derivative is simply the constant, i.e $\ln 4$

Title: What is the derivative of $\ln(4^x)$?


问:
Let $K$ be a Sylow subgroup of a finite group $G$. Prove that if $x \in N(K)$ and the order of $x$ is a power of $p$, then $ x \in K$.
Let $K$ be a Sylow subgroup of a finite group $G$. Prove that if $x \in N(K)$ and the order of $x$ is a power of $p$, then $ x \in K$.

This is how I tried...

Since $K$ is normal in $N(K)$, the factor group $$\frac{N(K)}{K}=\{K,xK,....\}
$$ is justified.

Then I took $$(xK)^{P^t}=x^{P^t}K=K$$.

This is where I am stuck.

答: Clearly, $K$ is normal in $N(K)$ so we can talk about the factor group $\frac{N(K)}{K}$. Since $G$ is finite and K is a Sylow p-subgroup of $G$ so $\frac{|N(K)|}{|K|}$ doesn't have any power of $p$ and it is given that $|x|$ is a power of $p$ so $\gcd(|x|,\frac{|N(K)|}{|K|}) = 1$ $\implies \exists \ m,n\in \mathbb{Z}$ such that $|x|m + n\frac{|N(K)|}{|K|} = 1$. and since  $x\in N(K)$ we know that $x^{\frac{|N(K)|}{|K|}} \in K$. Now $x = x^{|x|m + n\frac{|N(K)|}{|K|}} = (x^{|x|})^m (x^{\frac{|N(K)|}{|K|}})^n = (x^{\frac{|N(K)|}{|K|}})^n \in K$ as $x^{\frac{|N(K)|}{|K|}} \in K$


问:
Let $a,b$ be real numbers. Suppose that inequality $|(ax+by)(ay+bx)|\leqslant x^2+y^2$ holds for any real $x,y$. Prove that $a^2+b^2\leqslant2$
I know that this question has already been asked, but date of sending solutions passed (30th Sep), so this part of contest is inactive.

$$|(ax+by)(ay+bx)|\le x^2+y^2$$
$$|ab(x^2+y^2)+xy(a^2+b^2)| \le x^2+y^2$$
$$|k(a^2+b^2)+ab| \le 1$$
where  $k=\frac{xy}{x^2+y^2}$

$-\frac12\le k \le \frac12$  since $x^2+y^2\ge \pm 2xy$

Then I tried to treat left hand side as quadratic, but it did not lead to anything useful.
On the other hand from https://www.desmos.com/calculator/ljmcqbpobn it follows that there are $a,b$ s.t $a^2+b^2>2$, so I guess there are some mistakes in my approach.

答:
I will write a separate answer because I think you misunderstood the problem and therefore cannot benefit from the solutions.

You don't have to prove anything for all $x,y$. Instead you are told in advance that the inequality holds for all $x,y$ and you have to deduce something about $a,b$. If you can deduce all you need by only looking at two pairs of $x,y$ then, so to speak, you already win.

To give a hugely oversimplified example, consider the following problem:

Suppose that for all $x$ we have $ax^2\geq 0$. Prove that $a\geq 0$.

Of course you can make some argument that takes into account all values of $x$, but it is easier to say: if it holds for all $x$ then it holds in particular for $x=1$, and therefore $a\cdot 1\geq 0$, so we are done. Someone else could come and ask you also to plug in $x=2$ and $x=3$ but all you would get is $4a\geq 0$ and $9a\geq 0$, which you already know.

It is the same with the OM problem. If you can prove all you need by testing only two pairs $(x,y)$, good for you.


问:
We have $AB = BC$, $AC = CD$, $\angle ACD = 90^\circ$. If the radius of the circle is '$r$' , find $BC$ in terms of $r$ .

We have $AB = BC$, $AC = CD$, $\angle ACD = 90^\circ$. If the radius of the circle is '$r$' , find $BC$ in terms of $r$ .

What I Tried: Here is a picture :-

Let $AC = CD = x$. As $\angle ACD = 90^\circ$, we have $AD$ the diameter of the circle, so $AD = 2r$.
From here, using Pythagorean Theorem :- $$2x^2 = 4r^2$$
$$\rightarrow x = r\sqrt{2}$$

We have the green angles to be $45^\circ$, now as $ADCB$ is cyclic, $\angle ABC = 135^\circ$ and each of the brown angles are $22.5^\circ$. So we can use :-
$$\frac{a}{\sin A} = \frac{b}{\sin B}$$
$$\rightarrow a \div \frac{\sqrt{2 - \sqrt{2}}}{2} = r\sqrt{2} * \frac{2}{\sqrt{2}}$$
$$\rightarrow \frac{2a}{\sqrt{2 - \sqrt{2}}} = 2r$$
$$BC = a = r\sqrt{2 - \sqrt{2}}$$

However, the answer to my question is given as $r\sqrt{\sqrt{2}}$ , so where did I go wrong?

答: The given answer is wrong and you are correct. $ABC$ happens to be two consecutive sides of a regular octagon.


答:
Angle $ABC$ is $135°$ because it is half the concave angle $COA=270°$.

Thus angle $BAC=22.5°$ and $$BC=2r \sin 22.5°=2r\cdot \frac{\sqrt{2-\sqrt{2}}}{2}=r\sqrt{2-\sqrt{2}}$$

Body:
We have $AB = BC$, $AC = CD$, $\angle ACD = 90^\circ$. If the radius of the circle is '$r$' , find $BC$ in terms of $r$ .

What I Tried: Here is a picture :-

Let $AC = CD = x$. As $\angle ACD = 90^\circ$, we have $AD$ the diameter of the circle, so $AD = 2r$.
From here, using Pythagorean Theorem :- $$2x^2 = 4r^2$$
$$\rightarrow x = r\sqrt{2}$$

We have the green angles to be $45^\circ$, now as $ADCB$ is cyclic, $\angle ABC = 135^\circ$ and each of the brown angles are $22.5^\circ$. So we can use :-
$$\frac{a}{\sin A} = \frac{b}{\sin B}$$
$$\rightarrow a \div \frac{\sqrt{2 - \sqrt{2}}}{2} = r\sqrt{2} * \frac{2}{\sqrt{2}}$$
$$\rightarrow \frac{2a}{\sqrt{2 - \sqrt{2}}} = 2r$$
$$BC = a = r\sqrt{2 - \sqrt{2}}$$

However, the answer to my question is given as $r\sqrt{\sqrt{2}}$ , so where did I go wrong?


答: The given answer is larger than $r$, which clearly isn't true, as mirroring $B$ across $AC$ shows $|AB|<|AO|$, where $O$ is the center of the circle.

Title: We have $AB = BC$, $AC = CD$, $\angle ACD = 90^\circ$. If the radius of the circle is '$r$' , find $BC$ in terms of $r$ .


问:
On Canonical Forms for Basic Jordan Matrices
I am trying to express the matrix $A=\left(\begin{array}{ccc}
  0&-1\\
  1&0
\end{array}\right)$ in Jordan normal form. Decomposing it into its characteristic polynomial, it has eigenvalues $\lambda_{1,2}=\pm i$. By definition, it would have the canonical form $\Lambda_{\alpha\pm i\beta}=\left(\begin{array}{ccc}
  \alpha&-\beta\\
  \beta&\alpha
\end{array}\right)=\left(\begin{array}{ccc}
  0&-1\\
  1&0
\end{array}\right)$. However, I have worked through this problem by taking the canonical form as $\Lambda=\left(\begin{array}{ccc}
  i&0\\
  0&-i
\end{array}\right)$, with $A=T\left(\begin{array}{ccc}
  i&0\\
  0&-i
\end{array}\right)T^{-1}$ where $T=\left(\begin{array}{ccc}
  i&i\\
  1&-1
\end{array}\right)$, and surprisingly it works out. The form that I used above is of real eigenvalues $\lambda,\mu: \left(\begin{array}{ccc}
  \lambda&0\\
  0&\mu
\end{array}\right)_{\alpha,\beta}$. Can't really figure out why it works when you take $i$ as a real eigenvalue, and instead if you do it the correct way by taking $\Lambda_{\alpha+i\beta}$, the expression you get for $A$ is different – appreciate any pointers I can get, thanks.

答:
More generally Dunford decomposition (https://fr.wikipedia.org/wiki/D%C3%A9composition_de_Dunford) $A=D+N$ with $D$ diagonalisable (or simply diagonal) and $N$ nilpotent is only possible if the minimal polynomial is split.

Here the characteristic polynomial $\lambda^2+1$ is irreducible on $\mathbb R$ (so it has no divisors, and minimal polynomial is the same, thus not split).

Jordan normal form is a specialized case of Dunford with an $N$ which has only one upper diagonal of $0$'s and $1$'s, and similarly $A$ cannot be put in this form for real domain.

On $\mathbb C$ the characteristic polynomial is always split $\lambda^2+1=(\lambda-i)(\lambda+1)$ so the matrix you found $D=\operatorname{diag}(i,-i)$ is effectively a Jordan form with $N=0$ (it is diagonalisable).


问:
Applications of information geometry to the natural sciences
I am contemplating undergraduate thesis topics, and am searching for a topic that combines my favorite areas of analysis, differential geometry, graph theory, and probability, and that also has (relatively) deep mathematical applications to biology or physics (I am not planning to pursue a Ph.D. in pure math).

To this extent, I recently stumbled across information geometry. By this I refer to the field of using data to generate a Riemannian manifold with the Fisher information metric. Could you tell me which applications the field has, particularly to (mathematical) biology or (statistical) physics?

Also, are there any good references, at the level of a Ph.D. student well-versed in probability, analysis, and geometry (but not as much so in statistical inference)?

(And, this goes a bit beyond the question, but if you have any other thoughts on what would be interesting subjects given my preferences above, please share!)

答:
Personally, I worked in this area producing three papers appeared in conference proceedings (see this paper, this other and this one). I can provide you some book titles:

S. Amari&al., Differential Geometry in Statistical Inference, Institute of Mathematical Statistics (1987). This is currently free online here.

Shun-ichi Amari; Hiroshi Nagaoka, Methods of Information Geometry, AMS (2000).

Khadiga A. Arwini, Christopher T.J. Dodson, Information Geometry, Springer (2008).

You can also read the Wikipedia entry about this matter.

I think these hints should provide you a good starting point.


问:
Open subsets of Jet Bundles
I am currently reading through 'Stable mappings and their singularities' by Golubitsky and Guillemin. There they define
$$
J^k(X, Y)_{x, y}
$$
to be the set of equivalence classes of smooth maps $f: X\to Y$ with $f(x) = y$ under the equivalence $\sim_k$, where $f \sim_k g$ if $f, g$ have $k$-th order contact at $x$. Then they define the Jet Bundle as
$$
J^k(X, Y) := \sqcup_{\{x, y\} \in X\times Y} J^k(X, Y)_{x, y}
$$

Later they go on to define charts on $J^k(X, Y)$: given $(\phi, U), (\psi, V)$ charts on $X, Y$ respectively, they define a chart
$$
J^k(U, V)\to \phi(U) \times \psi(V) \times B_{n, m}^k
$$
where $B_{n, m}^k$ is a space consisting of $m$-tuples of polynomials in $n$ variables up to and including degree $k$, with zero constant term. What bothers me is related to the fact that they do not even define a topology on $J^k(X, Y)$. Implicit in their definition of the above charts is the assumption that $J^k(U, V)$ is even a subset of $J^k(X, Y)$, clearly it isn't strictly a subset with the above definitions, but what would a natural injection look like? I know that jets carry local information, so there should be a natural injection, but I can't see one with these definitions.

答: From what I can tell they are defining the topology on $J^k(X,Y)$ by the given charts. $J^k(U,V)$ is a subset of $J^k(X,Y)$, bearing in mind that in the jet setting we are viewing functions purely as the coordinates of the jet space. i.e. you don't have to worry about the difference between $f|_U$ and $f$, they both have the same coordinates (value of $f$ at its $k$ derivatives at the point).

Title: Open subsets of Jet Bundles


问:
Can you say that $Sp\{v\} \cup Sp\{w\} = Sp\{v,w\} - Sp\{v\cap w\}$
Is it correct to write the union of two spans as the span of the vectors from the two spans minus the intersection between the two spans?

i.e.

$Sp\{v\} \cup Sp\{w\} = Sp\{v,w\} - Sp\{v\cap w\}$

答: No, this is not true. Cansider the case $v=w.$


问:
Difficulty in solving differentiation of Acosx/3.
For the differentiation of $a\frac{ \cos x}{3}$ , I get $\frac{1}{3}\frac {\sin x}{3}$

But in answer they showed a as well in my textbook.Why is $a$ not differentiated?

答: I presume that it has been differentiated: $$\frac{d}{dx}\left[a\frac{\cos x}3\right]=\frac{\cos x}{3}\frac{d}{dx}[a]+a\frac{d}{dx}\left[\frac{\cos x}{3}\right]=0\cdot \frac{\cos x}{3}+a\cdot\left(-\frac{\sin x}3\right)=-\frac{ a\sin x}{3}$$


问:
Every measurable set in $X \times Y$ is contained in a measurable rectangle
If $X$ and $Y$ are any two sets (not necessarily subsets of the
same space), the Cartesian product $X \times Y$ is the set of all ordered
pairs $(x,y)$, where $x \in X$ and $y \in Y$. Thus, for instance, if $A \subset X$ and $B \subset Y$, we shall call the set $E = A \times B$ (a subset of $X \times Y$) a rectangle
and we shall refer to the component sets $A$ and $B$ as its sides.

In our context the word "class" should be understood as a set of sets. So the class of measurable rectangles refers to the set of measurable rectangles. So I want to use the fact that a rectangle $A \times B$ in the Cartesian product of two measurable spaces $(X, S)$ and $(Y, T)$, (where $S$ and $T$ are $\sigma$-rings of subsets of $X$ and $Y$ respectively), is measurable if $ A \in S $ and $ B \in T $.

I am reading the Measure Theory book by Paul Halmos, I have found this.
Exercise (6) Section 33.

If $(X,S)$ and $(Y,T)$ are measurable spaces, then every  measurable set in $X \times Y$ is contained in a measurable rectangle.

My idea is the following:

the class of all those sets which way be covered by a measurable rectangle is a $\sigma$-Ring.

I think this class contains every measurable set of $ X\times Y$, on the other hand if I test that it is a $ \sigma $-Ring I have finished

答:
Halmos develops the Measure theory based on $\sigma$-rings. Measurable set of $X\times Y$ means sets in product  $\sigma$-ring, that is generated by $A \times B$, $A \in S$, $B \in T$ where $S$ and $T$ are $\sigma$-rings.  In general, $X\times Y$ may not be a measurable rectangle.

Your idea on how to solve exercise 6 (section 33) is correct. It is exactly the way to go.

Here is a detailed solution to exercise 6 in section 33:

Let $$C=\{ E \subseteq X\times Y : \textrm{ there are } A\in S, B \in T \textrm{ such that } E \subseteq A\times B\}$$

Just checking the definition of $\sigma$-ring, it is immediate that $C$ is $\sigma$-ring.

It is also immediate that for all $A\in S, B \in T$, $A\times B \in C$.

So $C$ is a $\sigma$-ring and all measurable rectangles are in $C$.

Since $S \times T$ is the smallest $\sigma$-ring  having all  measurable rectangles,
we have $S \times T\subseteq C$. It means that every measurable sets of $X \times Y$  is contained in a measurable rectangle.


问:
Explicit Matrix Form of the Time-Evolution Operator
I have been given a time-dependent Hamiltonian $H = \eta$ cos $\omega t$ $\begin{pmatrix} 0 & 1 \\ 1 & 0 \\ \end{pmatrix}$
and asked to calculate explicitly in matrix form the time-evolution operator $U(0, t)$ associated to $H$.

I am completely stuck on how to do this. Do I use $U(0,t) = e^\frac{-iHt}{\hbar}$   ? Although I believe this only holds if H is time-independent.

And if so, how do I write this explicitly in matrix form?

Thanks for any help!

答:
Use $U(t,0) = exp [ -\frac{i}{\hbar} \int_{0}^{t} H(t') dt' ] = exp [ - \frac{i}{\hbar} \begin{pmatrix}0 & \frac{\eta}{\omega}sin\omega t\\ \frac{\eta}{\omega}sin \omega t & 0 \end{pmatrix}]$

Then let $A = - \frac{i}{\hbar} \begin{pmatrix}0 & \frac{\eta}{\omega}sin\omega t\\ \frac{\eta}{\omega}sin \omega t & 0 \end{pmatrix}$

Work out $e^A$ by diagonalizing A (writing $A=SDS^{-1}$) and using the power series expansion $e^{A} = \sum_{n=0}^{\infty} \frac{A^n}{n!}$

And you should get $U(t,o) = e^{A} = \begin{pmatrix}cos\alpha & -isin\alpha \\ -isin\alpha & cos\alpha \end{pmatrix}$ where $\alpha = \frac{\eta}{\hbar \omega} sin\omega t$


问:
Category of adjunctions inducing a particular monad
Every pair $F \dashv G$ of adjoint functors $F: \mathcal C \to \mathcal D$, $G: \mathcal D  \to \mathcal C$ induces a monad $\mathbb T = (T,\eta,\mu)$ on $\mathcal C$. Given a  monad $\mathbb T = (T,\eta,\mu)$ on $\mathcal C$, we define $\operatorname{Adj}(\mathbb T)$ to be the category of adjunctions inducing $\mathbb T$. Its objects are adjoint pairs of functors $F \dashv G$ between $\mathcal C$ and some category $\mathcal D$ such that the monad induced by the adjunction is $\mathbb T$, i.e. $GF = T$, the unit is $\eta$, and $G\varepsilon F = \mu$ where $\varepsilon$ is the counit. The morphisms between two such adjunctions $F \dashv G$, $F' \dashv G'$ are functors $H: \mathcal D \to \mathcal D'$ such that $HF=F'$ and $G' H = G$.

I am now worried whether this is really a category. Consider the example where $\mathcal C$ is a one-object category with only the identity morphism and $\mathbb T = (T,\eta,\mu)$ is the trivial monad. For every category $\mathcal D$ which has an initial object $I$, we get an adjunction $F \dashv G$, where $F: \mathcal C \to \mathcal D$ sends the one object of $\mathcal C$ to $I$, and $G: \mathcal D \to \mathcal C$ sends everything to the one object (it's easy to check that this is really an adjunction). Its induced monad on $\mathcal C$ is necessarily $\mathbb T$. So $\operatorname{Adj}(\mathbb T)$ contains at least one object for each category having an initial object. However, $\operatorname{Adj}(\mathbb T)$ itself has an initial object, namely the Kleisli adjunction, which in this case is just $1_{\mathcal C} \dashv 1_{\mathcal C}$. If $\operatorname{Adj}(\mathbb T)$ is a category, it seems that in a way it contains itself, in the form of the adjunction $\mathcal C \rightleftarrows \operatorname{Adj}(\mathbb T)$ defined by the Kleisli adjunction. (How) Is that possible?

答: Well, either one works in a framework where self-containing categories are permitted (but I do not know of any), or else one works more carefully. So I would say something like, fix a universe $\mathbf{U}$ such that $\mathcal{C}$ is a $\mathbf{U}$-small category, and let $\mathbf{Adj}(\mathcal{C})$ be the category of $\mathbf{U}$-small categories equipped with an adjunction etc.; then $\mathbf{Adj}(\mathcal{C})$ will not be $\mathbf{U}$-small, and so will not member of itself.</p>


答: Hope this helps you understand what the symbols mean:

Title: Can you say that $Sp\{v\} \cup Sp\{w\} = Sp\{v,w\} - Sp\{v\cap w\}$

Body:
Is it correct to write the union of two spans as the span of the vectors from the two spans minus the intersection between the two spans?

i.e.

$Sp\{v\} \cup Sp\{w\} = Sp\{v,w\} - Sp\{v\cap w\}$


问:
Find the radius of the circle inscribed in the overlapping region of the $2$ semi-circles, given that the circle touches line $P_1P_2$.

$C_1$ and $C_2$ are centres of $2$ identical semi-circles with radius $r$ units. We have $(C_1P_1):(PQ) = 1:7$. Find the radius of the circle inscribed in the overlapping region of the $2$ semi-circles, given that the circle touches line $P_1P_2$.

What I Tried: Here is a picture in Geogebra :-

What I did is assume $C_1P_1 = C_2P_2 = x$, and we have $PQ = 7x$. Now :-
$$PC_1 + C_1P_2 + P_2C_2 + C_2Q = PQ$$
We have :- $PC_1 = C_1P_2 = C_2Q = r$, and we get :-
$$3r + x = 7x$$
$$\rightarrow r = 2x$$
This implies that :- $$C_1P_1 = P_1P_2 = P_2C_2$$

I am not sure how to start finding the radius of the inscribed circle, one thing I did is drop a perpendicular from $E$ to $PQ$ and then try to use Pythagorean Theorem, but that did not help as I expected.

Can anyone help me? Thank You.

答:
Drop $ED \perp PQ$.

$\triangle EC_{1}D $ is right-angled with $C_1 D=3r/4$, $ED=R$, $C_1E=r-R$


问:
Evaluate $\int \frac{1}{\sqrt{x}} \, dx$ without using the power rule
I am trying to evaluate $\int \frac{1}{\sqrt{x}} \, dx$ without using the power rule. I tried using integration by parts but it seems out of luck.

Multiplying both the numerator and denominator with $\sqrt{x}$ we have

$$\int \frac{1}{\sqrt{x}} \, dx = \int \frac{\sqrt{x}}{x} \, dx$$

Integrate by parts: $u = \frac{1}{x}, v' = \sqrt{x}$. We obtain the same integral as above with

$$\int \frac{\sqrt{x}}{x} \, dx = \frac{2}{3} x\sqrt{x} + \frac{2}{3} \int \frac{\sqrt{x}}{x} \, dx$$

答:
Evaluate $\int \frac{1}{\sqrt{x}} \; dx$ without using the power rule

Consider the $u$-substitution formula backwards by treating $u$ as the dummy variable in the integral on the right such that

$$\int \frac{1}{\sqrt{\phi(x)}} \; \phi'(x) \; dx = \int \frac{1}{\sqrt{u}} \; du$$

and if you find the antiderivative on the left with respect to $x$ then you have to undo the substitution with $x = \phi^{-1}(u)$ so that $u$ also becomes a dummy variable.

In this case, we clearly see that by choosing $\phi(u) = u^2$ then the integral on the right greatly simplifies so that we obtain

\begin{align}
  \int \frac{1}{\sqrt{\phi(x)}} \; \phi'(x) \; dx & = \int \frac{1}{\sqrt{x^2}}2x \; dx \\
                                                  & = 2 \int dx                         \\
                                                  & = 2x \Big\vert_{x = \sqrt{u}}       \\
                                                  & = 2\sqrt{u} + C
\end{align}

We eventually have the following result with $x$ as the dummy variable such that

$$\int \frac{1}{\sqrt{x}} \; dx = 2\sqrt{x} + C$$

This result is valid on $\mathbb{R}_{>0}$ because $\phi(u)$ is also continous and bijective on that restriction.

Note: The integral $\int dx$ is not evaluated with the power rule. If we consider $\int dx$ as the definite integral $\int^x_0 dt$ then we clearly see it corresponds to the summation of all differentials $dt$ from $0$ up to $x$, so that $\int^x_0 dt = x + C$.


问:
Show that the parametric graphs intersects at the point $P=(2,1,3)$
Show that the following graphs intersects at the point $P=(2,1,3)$:
\begin{align}
r_1(t)&=2e^{-t} \hat i+\cos t \hat j+(t^2+3) \hat k \\
r_2(t) &=(1-t) \hat i+t^2 \hat j+(t^3+4)\hat k
\end{align}
We have to do the following

$ 2e^{-t_1}=(1-t_2), \\ \cos t_1=t_2^2, \\ t_1^2+3=t_2^3+4$

If we show, $t_1=0  $ and $t_2=-1$, then we get

$r_1(0)=(2,1,3) \\ r_2(-1)=(2,1,3)$.

How to show that ?

Any help

答: That the problem gave you the putative point of intersection makes it very easy; all you had to do was to solve, say, $2e^{-t_1}=2=1-t_2$ to get a value for $t_1$ and $t_2$, then substitute in the other coordinates to see if they match. If the intersection had not been given it would have become much harder.


问:
How foundation of mathematics can be based on a first order logic system?
Let's assume that we have an axiomatic system that utilises the second order logic. Can this system be reformulated in terms of first order logic? I guess, it is not the case, since the second order logic is more expressive that the first order logic. If I am right, how than Zermelo-Fraenkel set theory can be used as foundation of mathematic (it uses only first order logic).</p>

答:
You need to ask yourself what you mean by a "foundation" for mathematics. This is important since claims about the nature of mathematics will carry unstated assumptions of an extra-mathematical character.

Several years ago controversy erupted on Internet sites because two advocates for homotopy type theory attempted to participate in the FOM mailing list started many years ago by Harvey Friedman. The details involve unfortunate personality traits exacerbated by strong differences of opinion. The HOTT advocates left the mailing list disgruntled by the lack of acceptance of their ideas.

Nothing about this story is meant to be critical.  Within the HOTT literature you can find the explanation of how a foundation for mathematics is merely an "encoding" of mathematical ideas. In making these claims, HOTT theorists are attempting to place HOTT on the same footing as first-order logic with set theory. But, HOTT differs significantly.

In the discussions which ensued, many interested in HOTT proclaimed how HOTT satisfied their personal intuition for what a foundation for mathematics ought to be,  whereas first-order logic with set theory did not.

One reason for this may be the historical account claiming that mathematics is arithmetized. A paper by Reyes sitting on my bookshelf asserts that one of the aims of categorical logic is to introduce a more "geometrical" account of logic.

It is the case that much of what is posted on the FOM mailing list is traditional with respect to arithmetization and Hilbert's arithmetical metamathematics. And, during the controversy Friedman discussed his differences with Lawvere. Of course, HOTT uses Lawvere's category-theoretic notion of set.

Disputes like this are not new. Brouwer's intuitionism declares mathematics to be the product of human creativity. On the other hand, Russell clearly explains that he is crafting an account of mathematics that will exclude the philosophical position called monism. Like Frege, he had been a critic of formalists. The idea that mathematics is extensional comes from the metaphysical motivations held by Frege and Russell.

The introduction of Heyting's "Intuitionism" contains an imaginative discussion among presumed advocates for different views. This kind of text is valuable for understanding that "foundation" means different things to different people.

As yet unmentioned is the rejection of completed infinities by the Russian constructivists. Markov's normal algorithms do not presume the "infinite tape" of Turing machines.

Positions like Brouwer's or the Russian constructivists do not try to portray a foundation as an " encoding" for all of mathematics in the same sense as set theory or HOTT. So, even that metamathematical view may be disputed.

With this said, your question falls in line with the received views concerning logic and mathematics. You should find the links provided by Mr. Allegranza helpful and informative.


问:
How many different values of weight of edges can I fit into this maximally?
Suppose I have $n$ vertices and the graph is complete. Now each edge is assigned with a weight. The requirement is that any triangle that can be found in the graph must have at least 2 sides having the same weight (similar to an isosceles triangle but the triangular inequality is not required.) The best situation would be that all edges have the same weight, but how many different values of edges can I fit into this graph maximally?

I am thinking about using some combinatorics like n choose 3. But it doesn't seem to work on the small examples that I tried on my own. Can someone help me out?

答:
We claim that the answer is $n-1$. Suppose instead that $G$ is a graph that satisfies the property but has greater than $n-1$ distinct weights.

Without loss of generality, assume that the vertex set of the graph is $[n]=\{1,2,\ldots,n\}$. For $1\leq k\leq n$, denote by $G_k$ the induced subgraph of $G$ on the vertex set $[k]$.

Let $k_0$ be the minimum $k$ such that $G_k$ has greater than $k-1$ distinct weights. $G_{k_0-1}$ has at most $k_0-2$ distinct weights. This implies that the vertex $k_0$ has (at least) two edges in $G_{k_0}$ incident on it with distinct weights that do not appear in $G_{k_0-1}$. However, in this case, the triangle formed by (the vertices corresponding to) these two edges has $3$ distinct weights!

This is a contradiction, and therefore, the answer is at most $n-1$.

This bound is also attained for the graph where an edge from $i$ to $j$ ($1\leq i<j\leq n$), is labelled $i$.


问:
What is a manifold on a Euclidean space?
In this semester I study differential geometry and in this chapter we want to define what is a surface. In order to do that we first define what a manifold is on a Euclidean space, not generally what is a manifold, and Euclidean space I mean $(\mathbb{R}^{n},\left \| \cdot  \right \|)$.

That been said, if someone who doesn't study math ask you "what this a manifold?" how would you answer, in simple terms (as Feynman says).

Can we just say it's a homomorphic function from open sets to open sets ?

Older posts had been made on this topic (but I don't think they answer well  my question, in my option) and have already been answered if there isn't  anything new to add  I will delete this.

答:
In layman's terms, it's a space that at least for a point and it's immediate neighbors behaves like a Euclidean space (which is a set of n-tuples $\mathbb{R}^n$ equipped with a dot product, or if you're more comfortable with Physics, at least for a 3 dimensional space, is like the 3d space you use for Newtonian mechanics).

More specifically, a Manifold is a Topological space (a set of points and their neighbors satisfying some axioms), such that each point has a neighborhood of points that can be mapped with a continuous and invertible function (Homeomorphism) to a Euclidean space.

An example of what this last paragraph looks like graphically:
Manifold mapping to Euclidean space example

To answer your question: no, it's not an Homeomorphic function. As I've written above, it is a space satisfying some particular properties.


问:
one example on game playing with some details, how this average step was calculated?
I have one example in my TA notes:

We have a two players games. two players will be agree on number $n$. game start with $x=2$. first player select by Coin flipping and start game. after that each of players arbitrary do one of the following: $x=x^2$ or $x=x^3$.

for example if "David" start game and choose $x^3$ then $x$ will be equals to $8$. after playing of each players that condition $x>n$ occurred, the game will be stopped and the last player announced as winner. at average this game ended after $O(\log \log n)$ steps.

This is very nice example, but how we can detect this average order can be achieved?

答:
$x$ is replaced with $x^2$ or $x^3$.
That is, $\log x$ is replaced with $2\log x$ or $3\log x$.
That is, $\log \log x$ is replaced with $\log 2+\log\log x$ or $\log 3+\log\log x$. So for $\log\log x$, we have a simple addition of random vars. In particular, the stopping time is between $\approx\frac{\log\log n-\log\log 2}{\log3}$ and $\approx\frac{\log\log n-\log\log 2}{\log2}$.


问:
Prove that a set $E$ is closed iff it's complement $E^{c}$ is open
I was wondering if this proof was right.

$\Leftarrow$ Suppose $E$ is closed. Then choose $x\in E^{c}$, then $x\notin E$, and so $x$ is not a limit point of $E$.

Hence there exists a neighborhood $N$ of $x$ such that $E \cap N$ is empty, such that $N \subset E^{c}$. Thus $x$ is an interior point of $E^{c}$ and $E^{c}$ is open

$\Rightarrow$ Then suppose $E^{c}$ is open. Let $x$ be a limit point of $E$, then every neighborhood of $x$ contains a point of $E$, so that $x$ is not an interior point of $E^{c}$. Since $E^{c}$ is open, this means that $ x \in E$, Therefore $E$ is closed.

答: Looks good. All true statements and they make logical sense. However, it kinda depends on where you start. Is this meant to be a proof for metric spaces? Then it's absolutely fine. Or is it a proof for topological spaces? For topological spaces, the definition of being a closed set is usually that the complement is open. You could use the same definition for metric spaces, but that's usually not done. So it depends on the definitions from which you start.</p>


答:
Notwithstanding the slightly broken English and all, the problem may be reformulated without players:

Start with $2$ and square or cube it arbitrarily. Show that the number of operations to exceed $n$ is $O(\log\log n)$.

We can take two logs, since $\log$ is strictly increasing, and get the following:

Start with $\log\log2$ and add $\log2$ or $\log3$ arbitrarily. Show that the number of operations to exceed $\log\log n$ is $O(\log\log n)$.

This is trivially true, since $\log2$ and $\log3$ are constants that are not accounted for explicitly in $O$.

Title: one example on game playing with some details, how this average step was calculated?

Body:
I have one example in my TA notes:

We have a two players games. two players will be agree on number $n$. game start with $x=2$. first player select by Coin flipping and start game. after that each of players arbitrary do one of the following: $x=x^2$ or $x=x^3$.

for example if "David" start game and choose $x^3$ then $x$ will be equals to $8$. after playing of each players that condition $x>n$ occurred, the game will be stopped and the last player announced as winner. at average this game ended after $O(\log \log n)$ steps.

This is very nice example, but how we can detect this average order can be achieved?


问:
Prove this function is continuous at all points to conclude a topological space is connected
I need to show whether the circle $x^2+y^2=4$ is connected in $\mathbb R^2$ with topology induced by $d_1$. I have changed it to polar coordinates and constructed the following: $$f:[0,2\pi ) \longrightarrow \mathbb R^2, \ \ \ f(\alpha ) = (2\cos (\alpha ), 2\sin (\alpha ))$$

I am trying to show that $f$ is continuous at every $\alpha$ which would show $f$ is continuous and then I may conclude that since $[0,2\pi )$ is connected and $f$ is a continuous map between topological spaces, the image is also connected, but I am unsure as to how to start this at all.

答:
You can use the metric to prove continuity.

\begin{align*}
\lim_{x\to \alpha} d(f(x),f(\alpha)) &= \lim_{x\to \alpha} 2(|\cos x - \cos \alpha|+|\sin x - \sin \alpha|) \newline
&=0
\end{align*}

This last limit you need to use $\cos$, $\sin$ and absolute value continuity.


问:
How can I prove $(n^2)!$ greater growth than $(n!)^n$
I tried take limit n goes to infinity.
Wolfram solved the limit and $(n^2)!$ has greater growth rate but there is not step by step solution.

I think I can simplify $(n!)^n$ to $(n^n)^n$
But how can I prove $(n^2)! > (n^n)^n$

Wolfram link

答: Hint: Both $(n^2)!$ and $(n!)^n$ can be written as a product of a bunch of smaller terms. In fact, they can both be written as a product of $n^2$ terms. How do these terms compare to each other?


问:
Find the number of ordered triples

Determine, with proof, the number of ordered triples (A_1,A_2,A_3) of
sets which have the property that $$(i) A_1 \cup A_2 \cup A_3 = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\},$$ and $$(ii) A_1 \cap A_2 \cap A_3 = \emptyset,$$ where $\emptyset$ denotes the empty set. Express the answer in the form $$2^{a}3^{b}5^{c}7^{d},$$
where a, b, c, and d are nonnegative integers.

Hello, first, I want to say that I know the problem is well known and there are questions here about the problem. But, I don't want to see how to solve it too early, so I am posting this question to show my work, which actually is wrong. So, it is not a duplicate and is not a homework check, because I already know it is wrong. I would appreciate knowing where it is wrong.

I am not sure if I am interpreting the problem right, but what I imagine they are asking is, for example the number of ways to do things like this:
(1234567,8,9 10).

So, following it, I thought that would be a nice way to attack the problem considering $12$ empty spaces. So we will need to fill the 12 empty spaces with $10$ numbers plus two commas.

The first and the last space can not have comma (it would represent $A_1$ or $A_3$ equal the empty space), so we can have in the first space $10$ numbers, in the last space $9$ numbers. Having done that, we yet have $10$ characters to distribute between the spaces. Since the order matters, it means that we have $10!$ possible ways to do it.

Summarizing, we have $$10 \cdot 9 \cdot 10!$$ way.

(What I am doing here is considering the space before the first comma as $A_1$, between commas $A_2$, and after the second comma $A_3$)

What is the problem??

答: You are not counting partitions like $A_1=\{4\},A_2=\{8\},A_3=\{1,2,3,5,6,7,9,10\}$ that have their numbers out of order. The sets can be empty too since the question did not rule it out.

Title: Find the number of ordered triples


答: Yeah your proof looks fine, but you should give us the definition of closed and open. Because it is not unusual to define a set is closed when its complement is open. You should maybe defend what happens when $E=\varnothing$ or $E^C = \varnothing$. </p>

Title: Prove that a set $E$ is closed iff it's complement $E^{c}$ is open

Body:
I was wondering if this proof was right.

$\Leftarrow$ Suppose $E$ is closed. Then choose $x\in E^{c}$, then $x\notin E$, and so $x$ is not a limit point of $E$.

Hence there exists a neighborhood $N$ of $x$ such that $E \cap N$ is empty, such that $N \subset E^{c}$. Thus $x$ is an interior point of $E^{c}$ and $E^{c}$ is open

$\Rightarrow$ Then suppose $E^{c}$ is open. Let $x$ be a limit point of $E$, then every neighborhood of $x$ contains a point of $E$, so that $x$ is not an interior point of $E^{c}$. Since $E^{c}$ is open, this means that $ x \in E$, Therefore $E$ is closed.


问:
Series convergence $\sum_{n=1}^\infty \frac{1}{1+2^{(-1)^nn}} $
$$\sum_{n=1}^\infty \frac{1}{1+2^{(-1)^nn}} $$

How can I show whether this series converges or not

答:
Let $a_n=\frac{1}{1+2^{(-1)^nn}}.$ Then $a_n=\frac{1}{1+2^{-n}}$ if $n$ is odd.
Therfore $(a_n)$ contains a subsequence , which   converges to $1$.

This shows that $(a_n)$ does not converge to $0$.


问:
Must then A ∩ B also have a maximum?
Let $\sup(A) = \max\{A\}$ and $\sup(B) = \max\{B\}$ hold,

that is, that the suprema are contained in the respective amount and thus maxima.

Must then $A \cap B$ also have a maximum?

If so, prove it. If not, give a counterexample.

I literally have no clue what to do. Please help, I'd appreciate it.

答:
Assuming further that $A\cap B \neq \emptyset$ otherwise this could already be a good case for a counterexample.

Take $A= (0,1)\cup \{2\}$ and $B=(0,1)\cup\{3\}$ then $A\cap B = (0,1)$ has no maximum.

Title: Must then A ∩ B also have a maximum?


问:
Prove the existence of the derivatives of $f(x)$
Problem:
Consider the function $f(x)=\exp(-1/x^2)$ when $x\neq0$ and $f(x)=0$ when $x=0$.

I have proved $f'$ exist by Newton Quotient, and the problem asks to prove the existence of further derivatives fo $f(x)$.

I am not sure where to start. I tried to prove this by induction with Newton Quotient, but I cannot find a general form of $f^n(x)$.

A solution was given but I don't really understand:

We can see that $f^{(n)} (x)$ will contain a term proportional to $x^{−3n}e^{−1/x^2}$ along with smaller inverse powers of $x$ (the $x^{−3n}$ is the most singular as $x \to 0$). Since
$$\lim_{x\to0}\dfrac{e^{−1/x^2}}{x^{3n}}= 0,$$
we also define $f^{(n)}(0) = 0$ and hence all higher derivatives exist.

I don't really understand what "most singular" mean in here. Also, how is the limit of $f'$  and $f^n$ equals to 0 related with the existence of derivative?

答:
Consider functions of the form
$$\tag1g(x)=\begin{cases}p(x^{-1})e^{-\frac1{x^2}}&\text{if }x\ne 0\\0&\text{if }x=0\end{cases} $$
where $p$ is a polynomial.

Prove that $g$ is

continuous also at $x=0$,

differentable at $x\ne 0$,

also differentiable at $x=0$ with $g'(0)=0$

and conclude that $g'$ is also of the form $(1)$.

Now as $f=f^{(0)}$ is of form $(1)$ with $p$ constant, conclude by induction that $f^{(n)}$ is of that form for all $n$.


问:
Finite Extensions and Roots of Unity
Two questions; the hint I've been provided is that they are, in fact, related.

Prove that a finite extension of $\mathbb{Q}$ contains finitely many roots of unity.

What is the largest (finite) order an element of $GL_{10}(\mathbb{Q})$ can have?

I am not sure how to approach the first problem. Does the second have something to do with the relationship between field extensions and vector spaces over a base field?

答:
Suppose $K$ is a number field of degree $d$ which contains infinitely many roots of unity. It then contains roots of unity of arbitrary high order (beecause there are finitely many of each order), and therefore it contains roots of cyclotomic polynomials $\Phi_n$ for infinitely many $n$. Let $S$ be the set of such $n$'s

Now the degree of $\Phi_n$ is $\phi(n)$ and $\phi(n)\geq\sqrt n$ for $n>6$. Since $S$ is infinite, there is an element $m\in S$ such that $\phi(m)>d$.

But then $K$ contains the field generated by a root of the polynomial $\Phi_m$, which is of degree $\phi(m)$. This is absurd.

For the second problem, notice that if a matrix in $GL_{10}(\mathbb Q)$ has finite order, it generates a subring of the ring of matrices $M_{10}(\mathbb Q)$ isomorphic to an extension of $\mathbb Q$ by a root of unity. This extension has degree at most 100, because it is contained in $M_{10}(\mathbb Q)$ (of course you can do much better in bounding this...)

Title: Finite Extensions and Roots of Unity


Title: How can I prove $(n^2)!$ greater growth than $(n!)^n$

Body:
I tried take limit n goes to infinity.
Wolfram solved the limit and $(n^2)!$ has greater growth rate but there is not step by step solution.

I think I can simplify $(n!)^n$ to $(n^n)^n$
But how can I prove $(n^2)! > (n^n)^n$

Wolfram link

Body:
Use Stirling's approximation
$$(n^2)!\sim \sqrt{2 \pi } e^{-n^2} n^{2 n^2+1}$$
and
$$(n!)^n\sim (2 \pi n)^{n/2} e^{-n^2} n^{n^2}>e^{-n^2} n^{n^2+n}$$
Therefore $$(n!)^n<(n^2)!;\;\forall n\in\mathbb{N},n>1$$


问:
Ackermann Function: Proof that n < A(m, n) for all m, n in N
I've been stuck at this problem for a while now and can't get it solved. The prof wants me to proof n < A(m,n) for all m and n being positive Integers.

More details on the Ackermann Function:
$$(1) A(0,n) = n + 1$$
$$(2) A(m + 1, 0) = A(m, 1)$$
$$(3) A(m + 1, n + 1) = A(m, A(m + 1, n))$$

My approach was the following:

Inductionstart_m: $m=0 ;  A(0, n) = n+1$

A(0, n) = n+1 is defined and can be assumed to be true

Inductionassumption_m: $n < A(m, n)$ for one m and all n

Inductionproof_m (What we want to proof): $n < A(m+1, n)$ for all n

InductionEnd_m -->

Inductionstart_n: $n=0 ; A(m+1, 0) = A(m, 1)$ --> $1 < A(m, 1)$

Inductionassumption_n: $n < A(m+1, n)$ for one n and all m

Inductionproof_n: $n+1 < A(m+1, n+1)$ for all m

InductionEnd_n: $n+1 < A(m+1, n+1) = A(m, A(m+1, n))$
we already proofed: $n < A(m+1, n)$
Now Im stuck. I dont know how to proof $n+1 < A(m+1, n)$

If that can be proven, then the proof would be done since we know that:
$A(m+1,n) < A(m, A(m+1,n)) $
is true from the Inductionassumption_m because it has the form: $n < A(m, n)$ for one m and all n

答:
Notice that the Ackermann function is discrete. This means we may use $a>b\iff a\ge b+1$, which may be used on the step you found troublesome:

$$A(m+1,n+1)=A(m,A(m+1,n))\ge A(m+1,n)+1\ge(n+1)+1>n+1\tag{$\star$}$$

and the whole proof would be as follows:

Base case $m=0$: For all $n$, $A(0,n)=n+1\ge n$ follows.

Induction hypothesis $m\implies m+1$: Assume for some $m$ that for all $n$, $A(m,n)\ge n$. Then:

Base case $n=0$: $A(m+1,0)=A(m,1)>1>0$ follows.

Induction hypothesis $n\implies n+1$: See $(\star)$.

Therefore $A(m+1,n)>n$ for all $n$.

Therefore ($A(m,n)>n$ for all $n$) for all $m$.

Notice the indented steps all use a fixed $m$ i.e. the same value of $m$. If you write (for all $m$) on those lines, then you are not proving ($A(m+1,n)>n$ for all $n$), which has a fixed value of $m$.


问:
space of essentially bounded measurable functions on the unit circle
Why the set $A =\{ \phi \in L^{\infty}(S):\phi\text{ is invertible in }L^{\infty}(S)\}$ is dense in $L^{\infty}(S)$? Here $S$ is the unit circle in the complex plane. $L^{\infty}(S)$ is the space of all essentially bounded complex valued measurable functions on $S$.
I know that this is true but I am looking for its proof.

答:
Let $f\in L^\infty$. For simplicity of the argument we fix some measurable and bounded representant of this class and denote it also with $f$. Define:
$$f_n(x) = \begin{cases}f(x) & |f(x)|>\frac1n \\ \frac1n & |f(x)|\leq\frac1n \end{cases}$$
This construction remains measurable, as you can check and $\|f_n\|_\infty \leq \|f\|_\infty + \frac1n$, ie $f_n\in L^\infty$.

By definition $|f_n(x)| \ge \frac1n$ for all $x$ so $f_n$ is invertible in $L^\infty$ with $\|f_n^{-1}\|_\infty \leq n$. Further it is obvious again from construction that $\|f_n -f \|_\infty \leq \frac1n$, so the sequence converges to $f$ in $L^\infty$.


答:
for $x>0 \ $  with integration by parts
$$u=\frac{1}{x} , \  \ u'=\frac{-1}{x^2} \\ v=\frac{2}{3}x^\frac{3}{2}  \ \ \ \ v'=\sqrt{x}$$
$$ \begin{align} \int \frac{1}{\sqrt{x}}dx =\int \frac{\sqrt{x}}{x} dx & = \frac{2}{3}x^\frac{3}{2}.\frac{1}{x}-\int  \frac{-1}{x^2}.\frac{2}{3}x^\frac{3}{2} dx +c\\ & = \frac{2}{3}\sqrt{x}+ \int \frac{2}{3}  \frac{\sqrt{x}}{x}  dx +c \end{align} $$
so $$ \Big(1-\frac{2}{3}\Big) \int \frac{\sqrt{x}}{x} dx  = \frac{2}{3}\sqrt{x}+c  $$
then $$  \int \frac{\sqrt{x}}{x} dx = 2 \sqrt{x}+c$$
with power rule
$$ \begin{align} \int \frac{1}{\sqrt{x}}dx = \int x^\frac{-1}{2}dx &= \frac{1}{1-\frac{1}{2}} x^{1-\frac{1}{2}}+c \\ & =2 \sqrt{x}+c \end{align}$$

Title: Evaluate $\int \frac{1}{\sqrt{x}} \, dx$ without using the power rule

Body:
I am trying to evaluate $\int \frac{1}{\sqrt{x}} \, dx$ without using the power rule. I tried using integration by parts but it seems out of luck.

Multiplying both the numerator and denominator with $\sqrt{x}$ we have

$$\int \frac{1}{\sqrt{x}} \, dx = \int \frac{\sqrt{x}}{x} \, dx$$

Integrate by parts: $u = \frac{1}{x}, v' = \sqrt{x}$. We obtain the same integral as above with

$$\int \frac{\sqrt{x}}{x} \, dx = \frac{2}{3} x\sqrt{x} + \frac{2}{3} \int \frac{\sqrt{x}}{x} \, dx$$


问:
Probability that a particle travels to each point in a random walk
A particle is placed at point A in the figure below. It travels to each of its adjacent points with equal probability. What is the probability that it has traveled to all the points after infinite time?

I am not sure how to go about this problem. First of all, most of the random walk questions I have encountered till now are about returning to the starting point. I am not able to understand how to tackle the problem of traveling to all the points. Secondly, this seems a pretty random graph. I could try using Markov chains, which is something I am not that comfortable with. I am looking for a direct probabilistic approach. Any hints would be appreciated.

Thanks.

答: The graph is connected. Since the particle has a nonzero probability of moving to any of the neighbours of its current position, it also has a nonzero probability of reaching any position from any current position. Thus the probability that the particle has hit all the points in the infinite limit is $1$ (almost sure).


答:
You can write the terms as
$$(n!)^n = \underbrace{1 \cdot \dotso \cdot 1}_{n\text{ times}} \cdot \underbrace{2 \cdot \dotso \cdot 2}_{n\text{ times}} \cdot \dotso \cdot \underbrace{n \cdot \dotso \cdot n}_{n\text{ times}}$$
and
$$ (n^2)! = \underbrace{(1 \cdot \dotso \cdot n)}_{n \text{ terms}} \cdot \underbrace{((n+1) \cdot \dotso \cdot 2n)}_{n \text{ terms}} \cdot \dotso \cdot \underbrace{((n(n-1)+1) \cdot \dotso \cdot n^2)}_{n \text{ terms}}$$
You can look at the numbers as groups of $n$ numbers. You can see that the second term is always larger.


问:
Why is this a proof for Brouwer's fixed point theorem
our teacher said that we can prove the Brouwer's fixed point theorem using the index $n(f\circ \gamma_r,0)$ where $\gamma_r (t)=re ^{it}$ , but I don't understand how come any help would be a lot appreciated.Here's what he's done

Let $D$ the closed unit disk. $f : D\rightarrow D$ a continuous function.$f$ has a fixed point in $D$.

So we have : $n(f\circ \gamma_0,0) = 0 $ if  $f$ has no fixed point in $D$, we can show that :

The image of $\partial D$ by $1+\frac{f}{g}$ is in the open half plane {$z=x+iy|x>0$}, where $g:D\rightarrow D$ s.t $z \mapsto -z.\;\;$   (1)

We can conclude that : $n(f\circ \gamma_1,0) = n(g \circ \gamma_1,0) = 1.\;\;$ (2)

Hence we have a contradiction.

I don't see exactly why (1) & (2) are true and what gives us the contradiction.Thanks in advance for your help.

答:
One has to be careful here as $n(f\circ \gamma_1,0)$ makes sense only if $f$ doesn't have zeroes on the unit circle, while Brower theorem holds in general for any continuous self-map of the closed disc (which may have zeroes on the boundary), but the argument is correct if presented properly.

Note that on the unit circle $|f/g|=|f| \le 1$, hence $\Re (f/g)(w) \ge |(f/g)(w)| \ge -1$, and in particular $\Re (f/g)(w)=-1$ if and only if $(f/g)(w)=-1$ or $f(w)=w$ which means that $f$ has a fixed point on the unit circle and we are done!.

So we can move to the case where $\Re (f/g)(z)>-1, |z|=1$, which gives precisely the first point above.

But now this means $f+g$ doesn't have a zero on the unit circle (and obviously $g$ doesn't either), nor does $1+f/g$ while $n((1+ f/g)\circ \gamma_1,0)=0$ since its image is contained in the right half plane hence the argument can vary by at most $\pi$.

But then $n((f+g)\circ \gamma_1,0)=n(g\circ \gamma_1,0)+n((1+ f/g)\circ \gamma_1,0)=n(g\circ \gamma_1,0)=1$. This means that $f+g$ has at least a zero inside the unit disc, so $f$ has a fixed point and we are done!


问:
Number of iterations to find the root of $x^3+2x-54$ using Newton's Method
I am asked to calculate the (theoretical) minimum number of iterations needed to find the root $\alpha$ of $x^3+2x-54$ using Newton's Method, guaranteeing an absolute error less than $10^{-8}$, and starting from an interval $I$ and $x_0$ of my election.

I have searched the root in $I=[3,4]$, with $x_0=3.5$ (which in fact is very close to the root). I tried to find the number of iterations by two ways:

1st option. Here we need to know the value of $\alpha$. As the requested analysis is theoretical, I think this is not a sin. Using Wolfram, $\alpha\approx3.60$. Searching in Wikipedia I found that $|e_{n+1}|\leq M|e_n|^2$, where $M=\sup_{x\in I}\frac{1}{2}|\frac{f''(x)}{f'(x)}|$ and $|e_k|=|x_k-\alpha|$.

In this case, $M=\frac{1}{2}|\frac{6\cdot3}{3\cdot3^2+2}|=0.310$

$$|e_n|\leq M^{\sum_{i=0}^{n-1}2^i}|e_o|^{2^n}=0.31^{2^n-1}|3.5-\alpha|^{2^n}\approx0.31^{2^n-1}\cdot0.1^{2^n}$$

If we want $|e_n|<10^{-8}$, then
$$(0.31\cdot0.1)^{2^n}<10^{-8}\cdot0.31\to2^n>\frac{\log(10^{-8}\cdot0.31)}{\log(0.031)}\to n>\frac{\log(\frac{\log(10^{-8}\cdot0.31)}{\log(0.031)})}{\log(2)}\approx2.5$$

So we would need a minimum of $3$ iterations.

2nd option. Using the method shown here. $N(x)=x-\frac{f(x)}{f'(x)}\implies f(N(x))=\frac12f''(\tilde x)\frac{f(x)^2}{f'(x)^2}$

As $\max_{x\in I}|f''(x)|=24$, $\min_{x\in I}|f'(x)|=29$, then $$|f(N(x))|\leq\frac{12}{29^2}|f(x)|^2\to|f(x_n)|\leq(\frac{12}{29^2})^{\sum_{i=0}^{n-1}2^i}|f(x_0)|^{2^n}$$

$|f(x_0)|=|f(3.5)|\approx3.70$, and as $|x-\alpha|\leq0.31|f(x)|$, and we want $|x_n-\alpha|<10^{-8}$:

$$0.31(\frac{12}{29^2})^{2^n-1}\cdot3.7^{2^n}<10^{-8}\to(\frac{12\cdot3.7}{29^2})^{2^n}<\frac{10^{-8}\cdot12}{0.31\cdot29^2}\to0.0528^{2^n}<0.046\cdot10^{-8}\to$$
$$\to n>\frac{\log(\frac{\log(0.046\cdot10^{-8})}{\log(0.0528)})}{\log(2)}\approx2.87$$

So we would need a minimum of $3$ iterations.

If my procedure is not wrong, both methods give the same number of iterations (once have been rounded). The first one is tighter, probably due to the fact that we use the value of $\alpha$. Am I right? From a theoretical point of view, is it better to use the first or the second approach?

答:
The Rheinboldt-Ortega strategy for the proof of the Newton-Kantorovich theorem tells you that the convergence of the Newton method for $f(x)=0$ is majored by the convergence of the Newton method for the quadratic polynomial $$0=p(t)=\frac{L}2t^2-|f'(x_0)|t+|f(x_0)|=\frac{L}2(t-t^*)(t-t^{**}),$$ starting at $t_0=0$ towards its smaller root $t^*$. $L$ is a bound on the second derivative for a large enough neighborhood around $x_0$, $|f'(x_0)|$ gets replaced with $\|f'(x_0)^{-1}\|^{-1}$ for systems and their Jacobi matrix. This means foremost that for the iterates one gets the relation
$$
|x_{k+1}-x_k|\le t_{k+1}-t_k
$$

Here you get $L=24$, $f(x_0)=-4.125$, $f'(3.5)=38.75$, so that the roots are $t^{**}=3.11895341$, $t^{*}=0.11021325$. For the convergence of the $t_k$ sequence one gets the nice relation
$$
\theta_{k+1}=\frac{t_{k+1}-t^*}{t_{k+1}-t^{**}}
=\frac{p'(t_k)(t_k-t^*)-p(t_k)}{p'(t_k)(t_k-t^{**})-p(t_k)}
=\frac{t_k-t^*}{t_k-t^{**}}\frac{2t_k-(t^*+t^{**})-(t_k-t^{**})}{2t_k-(t^*+t^{**})-(t_k-t^{*})}=\theta_k^2
$$
so that
$$
t_k=\frac{t^*-θ_0^{2^k}t^{**}}{1-θ_0^{2^k}},~~θ_0=\frac{t^*}{t^{**}},
~~t^*-t_k=\frac{θ_0^{2^k}(t^{**}-t^*)}{1-θ_0^{2^k}}
$$
To get $|x_k-x^*|\le t^*-t_k\le 10^{-8}$ one would need approximately
$$
2^k\ge \frac{\ln(10^{-8})-\ln(t^{**}-t^*)}{\ln(θ_0)}=5.840012=2^{2.55}
$$
which again gives $k=3$ as the answer.


问:
Evaluate $\int \frac{1}{\sqrt{x}} \, dx$ without using the power rule
I am trying to evaluate $\int \frac{1}{\sqrt{x}} \, dx$ without using the power rule. I tried using integration by parts but it seems out of luck.

Multiplying both the numerator and denominator with $\sqrt{x}$ we have

$$\int \frac{1}{\sqrt{x}} \, dx = \int \frac{\sqrt{x}}{x} \, dx$$

Integrate by parts: $u = \frac{1}{x}, v' = \sqrt{x}$. We obtain the same integral as above with

$$\int \frac{\sqrt{x}}{x} \, dx = \frac{2}{3} x\sqrt{x} + \frac{2}{3} \int \frac{\sqrt{x}}{x} \, dx$$

答:
Let $x= \sin^2 t$ so that $dx = 2\sin t \cos t \; dt.$  Then

$$\int \frac{1}{\sqrt{x}} \; dx = \int \frac{1}{\sin t} 2\sin t \cos t \; dt = 2\int \cos t \; dt$$

$$ =2\sin t + C = 2\sqrt{x} + C.$$


问:
Embedding of finite groups in small $\operatorname{GL}(n,F)$
How does we obtain a small (not necessarily smallest) value of $n$ such that a $p$-group of order $p^m$ can be embedded in $\operatorname{GL}(n,\mathbb{F}_p)$?

[For embedding of a finite group $G$ in $S_n$, we can compute a small value of $n$ by looking at the minimum index of a subgroup $H$ containing no proper normal subgroup of $G$. Here, this value of $n$ need be necessarily smallest.]

答: Embed your group in a permutation group $S_n$ using Cayley's theorem, and then embed $S_n$ into a $GL_n$ using permutation matrices.</p>


问:
Matrix Determinant Identity
I have come across an observation about the determinant of a matrix, but I don't know how to prove it in general. Let me demonstrate it through an example.

$$
\begin{align}
\left|
\begin{matrix}
 1 & 2 & 3 & 4 \\
 5 & 6 & 7 & 8 \\
 9 & 10 & 11 & 12 \\
 13 & 14 & 15 & 16 \\
\end{matrix}
\right|
\cdot
\left|
\begin{matrix}
 6 & 7 \\
 10 & 11 \\
\end{matrix}
\right|
&=
\left|
\begin{matrix}
 1 & 2 & 3 \\
 5 & 6 & 7 \\
 9 & 10 & 11 \\
\end{matrix}
\right|
\cdot\left|
\begin{matrix}
 6 & 7 & 8 \\
 10 & 11 & 12 \\
 14 & 15 & 16 \\
\end{matrix}
\right|\\
&-
\left|
\begin{matrix}
 2 & 3 & 4 \\
 6 & 7 & 8 \\
 10 & 11 & 12 \\
\end{matrix}
\right|
\cdot
\left|
\begin{matrix}
 5 & 6 & 7 \\
 9 & 10 & 11 \\
 13 & 14 & 15 \\
\end{matrix}
\right|
\end{align}
$$

What this essentially says is that you can find the determinant of a larger matrix by breaking it down into this "ad-bc"-style of smaller determinants, however I can't work out why this should be true. I've expanded out the determinant for a 3x3 example in full generality and confirmed that it works, but I'd like to know if there's a more straightforward reason for this observation.

答: This is a determinant identity for Schur complement of a matrix. It's also closely related to http://en.wikipedia.org/wiki/Dodgson_condensation. "It is named for its inventor Charles Dodgson (better known as Lewis Carroll)." </p>

Title: Matrix Determinant Identity


问:
Rotate a shape 360 degrees on one axis and not be the same as the starting shape?
Are there shapes in nth dimensions (in euclidian geometry), that can be rotated 360 degrees on one axis and have more than one point not in the same place?

Alternatively, in non-euclidian geometry, are there any objects or shapes that can be rotated the full rotation of a closed circle in the relative geometry that fulfil the same properties? (For example, in a non-euclidian plane where 360 degrees is not enough for a full rotation, it would be rotated a full rotation instead.)

答:
A not very serious answer here:

A USB plug! When you first stick it into the receptacle it doesn't fit. Then you turn it over, a rotation of 180 degrees, and it still doesn't fit. Then you turn it over again, another rotation of 180 degrees, and not it fits. Thus it must have changed by the total 360 degrees rotation.


问:
What is a manifold on a Euclidean space?
In this semester I study differential geometry and in this chapter we want to define what is a surface. In order to do that we first define what a manifold is on a Euclidean space, not generally what is a manifold, and Euclidean space I mean $(\mathbb{R}^{n},\left \| \cdot  \right \|)$.

That been said, if someone who doesn't study math ask you "what this a manifold?" how would you answer, in simple terms (as Feynman says).

Can we just say it's a homomorphic function from open sets to open sets ?

Older posts had been made on this topic (but I don't think they answer well  my question, in my option) and have already been answered if there isn't  anything new to add  I will delete this.

答:
We are all familiar with a line or a plane in $\mathbb R^n$.

Perhaps we are also familiar with the generalization to arbitrary dimension $k$ with $1 \le k \le n-1$ (the cases $k=0$ and $k=n$ could also be included, but for intuition's sake I'll leave them out of this discussion). Let's refer to this as a "flat $k$-dimensional subspace" in $\mathbb R^n$ (a more mathematical term is an "affine $k$-space").

At the most intuitive level one can now say:

A $k$-dimensional manifold in $\mathbb R^n$ is a subset of $\mathbb R^n$ which looks, locally, like a flat $k$-dimensional subspace in $\mathbb R^n$.

And I might go on to give examples to shore up the intuition, for example a sphere in $\mathbb R^3$.

To formalize this one further step but still keep some intuition, one can use the idea of graphs. A 1-manifold in $\mathbb R^2$ is a subset which looks, locally, like a line. But, we might also want to capture the intuition that the line can be curved. To do this, we add one more layer of formality, which might not be understandable to someone who studies no math, but would at least be understandable to an undergraduate who has had some multivariable calculus.  Here's one special case:

A 1-manifold in $\mathbb R^2$ is a subset which looks, locally, like the graph of a smooth function, having one of two forms: $y=f(x)$; or $x=g(y)$.

And, more generally,

A $k$-manifold in $\mathbb R^n$ is a subset which looks, locally, like the graph of a smooth function, having one of several possible forms: for example, $(x_{k+1},...,x_n) = f(x_1,...,x_k)$; or more generally of the form $(x_{i(k+1)},...,x_{i(n)}) = f(x_{i(1)},...,x_{i(k)})$, where $i : \{1,...,n\} \to \{1,...,n\}$ is a permutation.

Again, to shore up the intuition I might write down some formulas for the sphere in $\mathbb R^3$, such as $z = \sqrt{1-x^2-y^2}$ for the upper half-hemisphere and five other similar formulas to cover the lower, left, right, front, and back hemispheres.

Only in the next step of formality would I bring in open subsets in order to formalize the meaning of the phrase "looks, locally". I wouldn't expect someone who studies no math to follow at this point, nor even an undergrad calc student, but at least this could make sense to an undergraduate math major. And at this stage the definition would no longer be just intuitive, instead it becomes the complete definition:

A subset $M \subset \mathbb R^n$ is a $k$-dimensional manifold if for each $p \in M$ there exists smooth function of the form $(x_{i(k+1)},...,x_{i(n)}) = f(x_{i(1)},...,x_{i(k)})$, where $i : \{1,...,n\} \to \{1,...,n\}$ is a permutation, and there exist open subsets $U \subset \mathbb R^n$ and $V$ of $\{(x_{i(1)},...,x_{i(k)})\}$, such that $U \cap M$ is equal to the graph of $f$.


问:
What is a function independent of a variable?
I am having trouble to interpret if a given functions if independent of a parameter or not.
Suppose $f(x) = x^2$ and $x = cy$. I ask you, is this function $y$ independent?
I don't know if it is a misinterpretation of the literature or not, but i am not sure how to answer.

You can tell me yes, since $y$ varies, $x$ varies and $f$ varies. But i could reply that no, $f$ varies because $x$ varies, not because $y$ varies.

I am a little confused

答:
This $y$ is not a function. It may be the independent variable of a certain function $$g:\quad y\mapsto x=g(y):=c\,y\ ,$$ which maps the $y$-space into the $x$-space. The $\>c\>$ here is a parameter, i.e., a  constant whose numerical value is immaterial  for the algebraic calculations.

From the two functions $$f: \quad x\mapsto z=f(x):=x^2$$ and $g$ one may form the composed function
$$h:=f\circ g:\qquad y\mapsto z=f\bigl(g(y)\bigr)=(cy)^2=c^2y^2\ .$$
Depending on the circumstances, the function $g$ is not viewed as an actual map, but as a change of coordinate, say due to a change of the measuring unit. In such a case people sometimes call the $h$ again $f$, so that now $f$ appears as "function  $z=f(y)$". This can lead to mistakes when applying the chain rule.


问:
find the vector a {1, 1, 1} length of the projection in the vector b
I need to find the vector a {1, 1, 1} length of the projection in the vector b

Given:
vector b with the OX axis forms a 30 degree angle

vector b with the OZ axis forms a 60 degree angle

What formulas do I need to use?

答:
Hint:

$\vec{a} \cdot \vec{b} = ||a|| \,||b|| \,\cos \theta$

Say $\vec {b} = (b_1, b_2, b_3)$

Unit vector in the direction of $x-$axis $= (1,0,0)$

So $(b_1, b_2, b_3) \cdot (1,0,0) = \sqrt{b_1^2 + b_2^2 + b_3^2} \, \cos 30^0$

Or the direction angle of a vector with $x$ and $z$ axis is simply given by

$\cos \alpha = \frac{b_1}{\sqrt{b_1^2 + b_2^2 + b_3^2}}, \cos \gamma = \frac{b_3}{\sqrt{b_1^2 + b_2^2 + b_3^2}} \,$ respectively.

That gives you two equations.

Solve for $b_1, b_2, b_3$ in terms of one of them and divide by magnitude to get the unit vector in the direction of $\vec{b}$.

Now dot product of $(1,1,1)$ with unit vector in the direction of $\vec{b}$ will give the projection of $(1,1,1)$ in vector $\vec{b}$.


答:
So the OP asked this:

if someone who doesn't study math ask you "what this a manifold?" how would you answer, in simple terms (as Feynman says).

My emphasis. So clearly the interest is about informing people who have no practice with math and requires analogies and explanations that do not include jargon. Which is great, because I am not a mathematician so I will provide the answer that was provided to me when I had this same questions hovering in my head.

You can start by saying that, the simplest example of a manifold is surface that is, in general, globally curved. As an example you can give a sphere. The surface of a sphere. Then explain that despite the fact that the sphere is curved, you can divide it in several little squares or triangles and if your division is fine enough and each little triangle is small enough, you can tile the entire surface of the sphere very tightly even though its surface is curved and the small polygonal tiles are flat. So it is locally flat, despite being globally curved. This, incidentally, is why the surface of the Earth seems so flat when seen from our vantage point.

Here you make your point that if you can tile a surface with such minuscule flat polygons, then it is a manifold because even though it is generally curved, it is locally flat. As a general example you can give a waving flag, relief terrain maps etc. 2D manifolds for which global curvature isn't constant, the most general examples you can think of.

The next point in the instruction is harder because then you need to show your interlocutor that the notion is general. That is, there is no reason why manifolds should be 2D, tiled by flat polygons. They may very well be 3D, tiled by flat polygons or 4D, tiled by flat polytopes.

Once you do that you can say that an entire manifold can, in fact, be flat. In which case it will be both globally and locally flat. When you study a manifold embedded inside a flat Euclidean space it is just a fancy way of saying that you, as the observer, is watching your curved study manifold from an external flat manifold. For instance, when you study the curvature of the surface of a sphere, a 2D manifold, inserted in a 3D flat space like the room where you are studying it, which is a 3D flat manifold. You can contrast this type of study with a study of the surface of the sphere where the observer isn't in an external room, but constrained to walk on the surface of the sphere itself.

Again, the notion of observing a sphere inside a room can be extended to the general case of $n-1$ dimensional manifolds embedded in $n$ dimensional flat manifolds.

I will leave it as homework how to explain what you mean by "Euclidean space".

I hope this helps, good luck.

Title: What is a manifold on a Euclidean space?

Body:
In this semester I study differential geometry and in this chapter we want to define what is a surface. In order to do that we first define what a manifold is on a Euclidean space, not generally what is a manifold, and Euclidean space I mean $(\mathbb{R}^{n},\left \| \cdot  \right \|)$.

That been said, if someone who doesn't study math ask you "what this a manifold?" how would you answer, in simple terms (as Feynman says).

Can we just say it's a homomorphic function from open sets to open sets ?

Older posts had been made on this topic (but I don't think they answer well  my question, in my option) and have already been answered if there isn't  anything new to add  I will delete this.


问:
Sum in a discrete moment generating function
I am reviewing my class tutorials and we were working in a MGF in the case of a discrete random variable.
We arrive at this stage :
$$M_{X}(t)=E(e^{tx})=\sum_{x=0}^{n}{e^{tx}\left(\frac{1}{2}\right)^{x+1}}=\frac{1}{2^{n+1}}(2^{n-1}e^t+2^{n-2}e^{2t}+...+2e^{(n-1)t}+e^{nt}+2^n)$$
Could you explain to me why the sum does that ? I think I am missing this concept that is very usefull.
(Would you mind doing the same for the product from i=0 to n)

Edit - How did we get the expansion from the sum?

Body:
To prove:

$$\sum_{x=0}^{n}{e^{tx}\left(\frac{1}{2}\right)^{x+1}}=\frac{1}{2^{n+1}}(2^{n-1}e^t+2^{n-2}e^{2t}+...+2e^{(n-1)t}+e^{nt}+2^n)$$

Let's see how $\sum_{x=0}^{n}{e^{tx}2^{n-x}}$ expandes

At $x=0$ $\Rightarrow$ $2^n$

At $x=1$ $\Rightarrow$ $e^t2^{n-1}$

At $x=2$ $\Rightarrow$ $e^{2t}2^{n-2}$

I think you can see the pattern so the expansion will be
$$\sum_{x=0}^{n}{e^{tx}2^{n-x}}=(2^{n-1}e^t+2^{n-2}e^{2t}+...+2e^{(n-1)t}+e^{nt}+2^n)$$
Now divide both sides by $2^{n+1}$ and you'll get the original expansion.
$$
\begin{align}
\sum_{x=0}^{n}{e^{tx}2^{n-x-n-1}} & = \sum_{x=0}^{n}{e^{tx}\left(\frac{1}{2}\right)^{x+1}} \\
& = \frac{1}{2^{n+1}}(2^{n-1}e^t+2^{n-2}e^{2t}+...+2e^{(n-1)t}+e^{nt}+2^n)
\end{align}
$$


问:
Set of accumulation points of a sequence and minimal or maximal element of a convergent sequence
Let $(a_n)_{n \in \mathbb{N}}$ be a sequence in $\mathbb{R}$. A number $ \alpha \in \mathbb{R}$ is called an accumulation point (or cluster point) of $(a_n)$ if for every $\varepsilon> 0$ the set $\{n \in \mathbb{N} : a_n \in U_{\varepsilon}(\alpha)\}$ contains infinitely many elements. Here, $U_{\varepsilon}(\alpha)$ denotes the open ball with centre $\alpha$ and radius $\varepsilon > 0$, i.e.,
$$
U_{\varepsilon}(\alpha) = \{ a \in \mathbb{R} : |a - \alpha| < \varepsilon\}.
$$
Denote by $H(a_n) = \{ \alpha \in \mathbb{R}: \alpha \text{ is accumulation point of } (a_n)\}$ the set of all accumulation points of the sequence $(a_n)$.

I have to show the following:

For every bounded sequence $(a_n)_{n \in \mathbb{N}}$ we have $\sup \{a_n : n \in \mathbb{N}\}, \inf\{a_n : n \in \mathbb{N}\} \in \{a_n : n \in \mathbb{N}\} \cup H(a_n) $.

If $(a_n)_{n \in \mathbb{N}}$ is convergent, then the set $\{a_n : n \in \mathbb{N}\}$ has a minimal or maximal element.

Hope, someone can help me

答:
For part $1$, if $\sup \{a_n : n \in \mathbb{N}\}$ and $\inf \{a_n : n \in \mathbb{N}\}\in \{a_n : n \in \mathbb{N}\},$ then we are done. So, we may assume that $\sup \{a_n : n \in \mathbb{N}\}\notin \{a_n : n \in \mathbb{N}\}.$ Then by the definition of supremum, we can find a subsequence whose limit is $\sup \{a_n : n \in \mathbb{N}\}$ (Think of this). Similar argument applies to the infimum case.

For part $2$, if $(a_n)_{n \in \mathbb{N}}$ is convergent, then $H(a_n)$ contains only one point. By part $1$, at least one of $\sup \{a_n : n \in \mathbb{N}\}$ and $\inf \{a_n : n \in \mathbb{N}\}$ will be in $\{a_n : n \in \mathbb{N}\}.$


问:
Find the number of ordered triples

Determine, with proof, the number of ordered triples (A_1,A_2,A_3) of
sets which have the property that $$(i) A_1 \cup A_2 \cup A_3 = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\},$$ and $$(ii) A_1 \cap A_2 \cap A_3 = \emptyset,$$ where $\emptyset$ denotes the empty set. Express the answer in the form $$2^{a}3^{b}5^{c}7^{d},$$
where a, b, c, and d are nonnegative integers.

Hello, first, I want to say that I know the problem is well known and there are questions here about the problem. But, I don't want to see how to solve it too early, so I am posting this question to show my work, which actually is wrong. So, it is not a duplicate and is not a homework check, because I already know it is wrong. I would appreciate knowing where it is wrong.

I am not sure if I am interpreting the problem right, but what I imagine they are asking is, for example the number of ways to do things like this:
(1234567,8,9 10).

So, following it, I thought that would be a nice way to attack the problem considering $12$ empty spaces. So we will need to fill the 12 empty spaces with $10$ numbers plus two commas.

The first and the last space can not have comma (it would represent $A_1$ or $A_3$ equal the empty space), so we can have in the first space $10$ numbers, in the last space $9$ numbers. Having done that, we yet have $10$ characters to distribute between the spaces. Since the order matters, it means that we have $10!$ possible ways to do it.

Summarizing, we have $$10 \cdot 9 \cdot 10!$$ way.

(What I am doing here is considering the space before the first comma as $A_1$, between commas $A_2$, and after the second comma $A_3$)

What is the problem??

答:
Your main problem is that the order within a set $A_i$ does not matter, so for example the list $12345678,9,10$ is the same as $87654321,9,10$, but you are counting those as different.

Another more minor problem is that you are treating the two commas as distinct, so are double counting every string. Lastly, you are not allowing $A_1$ and $A_3$ to be empty but are allowing $A_2$ to be empty, which is inconsistent. As Parcly points out, there is nothing in the question that says they are not allowed to be empty.

Edit:

There is another problem with your approach that I also missed until I saw JMoravitz's solution. A number can be an element of two of the three sets $A_i$. The requirement that $A_1\cap A_2\cap A_3 = \emptyset$ only rules out that a number is in all three sets simultaneously.

Title: Find the number of ordered triples


问:
Line under curve on the interval (proof)
What is the easiest way how to prove that $3x+y=1$ is under $y=(1-x)^{3}$ without drawing graph? I mean by using some matematical technique. It is enough for interval $[-1,1]$. Thank you for advice.

答:
HINT

You have to prove that the graph of $y=(1-x)^3$ is over the graph of $y=1-3x$ over $x \in [-1,1]$. It suffices to prove that
$$
\begin{split}
(1-x)^3 &\ge 1-3x \\
1 - 3x + 3x^2 - 3x^3 &\ge 1 - 3x \\
3x^2 - 3x^3 &\ge 0
\end{split}
$$
Can you finish?


问:
Why homeomorphism between $R$ and $(0,1)$ makes $(0,1)$ completely metrizable space?
I know that we can have few homeomorphisms between $R$ and $(0,1)$ such as $f(t)=\tan(-\frac{\pi}{2}+t\pi)$ but I can not understand why this metric $d(x,y)=|f(x)-f(y)|$ makes $(0,1)$ a complete metric space? I can not understand why now the Cauchy sequence such this one $\{\frac{1}{n}\}_n$ has limit in $(0,1)$. How the measure affects limit points?

答: With this new metric, $\left\{1\over n\right\}$ is not a Cauchy sequence anymore.


答: Because now the distance from $\frac1m$ to $\frac1n$ is $\left|f\left(\frac1m\right)-f\left(\frac1n\right)\right|$, which is not the original distance. And now $\left(\frac1n\right)_{n\in\Bbb N}$ is not a Cauchy sequence anymore, and therefore the fact that it doesn't converge is not an obstacle to the fact that $(0,1)$ is complete.

Title: Why homeomorphism between $R$ and $(0,1)$ makes $(0,1)$ completely metrizable space?

Body: I know that we can have few homeomorphisms between $R$ and $(0,1)$ such as $f(t)=\tan(-\frac{\pi}{2}+t\pi)$ but I can not understand why this metric $d(x,y)=|f(x)-f(y)|$ makes $(0,1)$ a complete metric space? I can not understand why now the Cauchy sequence such this one $\{\frac{1}{n}\}_n$ has limit in $(0,1)$. How the measure affects limit points?


问:
Topological and algebraic groups
I learnt that an algebraic group $G$ is not a topological group because Zariski topology on $G$ x $G$ is not the product topology. Are there any other topologies, other than Zariski, that do not generate a topological group? If so, is it for the same reason (topology on the product is not the product topology)?

答: $\mathbb R$ with the Sorgenfrey topology is not a topological group; not only with respect to the usual sum and $x\mapsto -x$, but with respect to every group operation whatsoever; to wit: there is no group operation on $\mathbb R$ making the multiplication and inversion Sorgenfrey-continuous.

Title: Topological and algebraic groups


问:
Can anybody help with this matrices problem?
I am home schooling my child at the moment due to the covid-19. He's doing well in his Maths lessons and his tutor has asked him to self-study some matrices, I have no background in mathematics past A-levels, however I have been interested in his work and have been attempting some of the questions and topics he's set myself as a pass time in the evening.

He presented this question to me this morning and we're both unsure how to solve it. I've tried to look online but can't seem to find a question similar (probably lack of awareness / knowledge of what to actually search for). If anybody could show me a step by step on how to solve this it would be great.

"Let $a = 5, b = 9.$

Consider the following system of equations, where λ is some scalar:

$ax − 3az = −3a$

$−2x − λy + z = 2$

$bx + 2by + λbz = b$

By forming the augmented matrix corresponding this system of equations and reducing to row echelon form, find the values of λ that imply the system:
(a) is inconsistent;
(b) has infinitely many solutions;
(c) has a unique solution.
In this case find the unique solution in terms of λ."

After sorting into row echelon form. I'm unsure how to proceed.

Many thanks

Chahid

答:
Begin by scaling rows 1 and 3 to obtain the following:

\begin{bmatrix}
1 & 0 & -3 & -3\\
-2 &-\lambda & 1 & 2\\
1 & 2 & \lambda & 1
\end{bmatrix}

Perform the following elementary operations: $2R_1+R_2\mapsto R_2$ and $-R_1+R_3\mapsto R_3$ to obtain
\begin{bmatrix}
1 & 0 & -5 & -4\\
0 &-\lambda & 1 & 2\\
0 & 2 & 3+\lambda & 4
\end{bmatrix}

Next, perform $2R_2+\lambda R_3\mapsto R_3$
\begin{bmatrix}
1 & 0 & -5 & -4\\
0 &-\lambda & 1 & 2\\
0 & 0 & -10+\lambda(3+\lambda) & -8+4\lambda
\end{bmatrix}

We'll focus our attention on the last row. If this row is all 0, i.e. if there's a value for $\lambda$ such that $-10+\lambda(3+\lambda)=-8+4\lambda=0$, then the system has a free variable and infinitely many solutions. $-10+\lambda(3+\lambda)$ is 0 when $\lambda=-5,2$, and $-8+4\lambda=0$ when $\lambda=2$, so the system has a infinitely many solutions when $\lambda=2$.

If $\lambda=-5$, we return $0=-28$, a false statement, leading to the system of equations having no solutions.

If $\lambda\neq -5,2$, then $-10+\lambda(3+\lambda)\neq 0$ and we can divide by the expression to get a value for $z$, which we can substitute back up the chain to get
$$z=\frac{4}{\lambda+5}$$
$$y=\frac{4}{\lambda+5}$$
$$x=-\frac{3(1+\lambda)}{\lambda+5}$$


问:
For a compact object $H$, is true that $H\simeq H_{\alpha_0}$ for some $\alpha_0$?
Let $H$ be a compact object, i.e., the functor $[H,-]$ preserves filtered colimits.

Suppose $(H,\pi)=\text{colim}(H_{\alpha})$ with $(H_{\alpha})$ filtered. Then $(H\xrightarrow{\text{id}_H} H)=(H\xrightarrow{f}H_{\alpha_0}\xrightarrow{\pi_{\alpha_0}} H)$ for some $f$ and $\alpha_0$. Hence, $\pi_{\alpha_0}\circ f = \text{id}_H$.

I want to see $f \circ \pi_{\alpha_0} = \text{id}_{H_{\alpha_0}}$ but maybe this is not necessarily true.

Any help or counterexample?

答: No, what you show is all that’s true: $H$ is a retract of some object in the diagram. Indeed if $e:H’\to H’$ is any idempotent which $H$ splits, then $H$ is the filtered colimit of a couple of different filtered diagrams made up of copies of $H’$, most obviously $H’\to H’\to\cdots$, where all the maps are $e$.


问:
Probability problem: what is the probability one card from each suit will be represented when 6 cards are dealt?
What is the probability one card from each suit will be represented when 6 cards are dealt?</p>

答:
Let $N_s , N_h, N_d $ and $N_c$ be the events that we get no Spades, Hearts, Diamonds  or Clubs respectively when dealing 6 cards. Thus the event $X := N_s  \cup N_h\cup N_d \cup N_c$ Is such that at least $1$ suit is unrepresented.  By the inclusion exclusion principle we have that $\mathbb{P}(X) = \mathbb{P}(N_s) + \mathbb{P}(N_h) + \mathbb{P}(N_d)+ \mathbb{P}(N_c) - \mathbb{P}(N_s \cap N_h) - \mathbb{P}(N_s \cap N_d) - \mathbb{P}(N_s \cap N_c) - \mathbb{P}(N_h \cap N_d) - \mathbb{P}(N_h \cap N_c) - \mathbb{P}(N_d \cap N_c) + \mathbb{P}(N_s \cap N_h \cap N_d) + \mathbb{P}(N_s \cap N_h \cap N_c) + \mathbb{P}(N_s \cap N_d \cap N_c) + \mathbb{P}(N_h \cap N_d \cap N_c) - \mathbb{P}(N_s \cap N_h \cap N_d \cap N_c)$

This might look horribly complicated but its actually very simple. Here is what it is doing: Add all the $\textbf{singletons}$, Subtract all the $\textbf{doubletons}$, Add all the $\textbf{triplets}$ and Subtract all the $\textbf{quadruples}$. And from symmetry we know that all the singletons are equal so are all the doubletons so are all the triplets and so is the quadruple.  Let us solve for them all now:

$\textbf{Singletons:}$

The singletons, $\mathbb{P}(N_i)$, are simply the probability there are none of a particular suit. Which occur simply with probability: $\frac{39}{52} \cdot \frac{38}{51} \cdot \frac{37}{50} \cdot \frac{36}{49} \cdot \frac{35}{48} \cdot \frac{34}{47} = 0.160258358663.  $ This is the same as 39 chose 6 over 52 chose 6

$\textbf{Doubletons:}$

The doubletons are the event that there are neither of 2 suits appearing in 6 cards. $\mathbb{P}(N_i \cap N_j) = $ having 26 cards available to make up 6 over the total number of combinations = $\frac{C^{26}_6}{C^{52}_6}$

$\textbf{Triplets:}$

The event $\mathbb{P}(N_i \cap N_j \cap N_k)$ is the event that none of 3 suits appear. i.e only 1 suit and occurs with probability $\frac{C^{13}_6}{C^{52}_6}$

$\textbf{Quadruple:}$

$\mathbb{P}(N_s \cap N_h \cap N_d \cap N_c)$ is the only quadruple which is clearly 0 as it is stating no spades, no hearts, no clubs and no diamonds.

Hence $\mathbb{P}(X) = 4\cdot \frac{C^{39}_6}{C^{52}_6} -6\cdot\frac{C^{26}_6}{C^{52}_6} +4\cdot\frac{C^{13}_6}{C^{52}_6}= 0.573517917805 $ This is the value that at least 1 suit is

$\textbf{Un}$represented and hence the answer to your question is $1-0.573517917805 = 0.426482082195 \approx 0.426 $


问:
How to prove this inequality? $ab+ac+ad+bc+bd+cd\le a+b+c+d+2abcd$
let $a,b,c,d\ge 0$,and $a^2+b^2+c^2+d^2=3$,prove that

$ab+ac+ad+bc+bd+cd\le a+b+c+d+2abcd$

I find this inequality are same as Crux 3059 Problem.

Body:
Lagrange multipliers is OK and not very complex:
$f= a+b+c+d+2abcd−ab−ac−ad−bc−bd−cd,g=a^2+b^2+c^2+d^2-3,F=f+\lambda g$

$F_{a}=1+2bcd-(b+c+d)+2a\lambda=0$......<1>

$F_{b}=1+2acd-(a+c+d)+2b\lambda=0$......<2>

$F_{c}=1+2abd-(a+b+d)+2c\lambda=0$......<3>

$F_{d}=1+2abc-(a+b+c)+2d\lambda=0$......<4>

$F_{\lambda}=a^2+b^2+c^2+d^2-3=0$......<5>

<1>-<2>,<2>-3>,<3>-<4>:

$(a-b)(1-2cd+2\lambda)=0$......<6>

$(b-c)(1-2ad+2\lambda)=0$......<7>

$(c-d)(1-2ab+2\lambda)=0$......<8>

WLOG, we only consider 4 cases:

case I: $a-b=0,b-c=0.c-d=0 \to a=b=c=d=\dfrac{\sqrt{3}}{2}$

case II: $a-b=0,b-c=0,1-2ab+2\lambda=0 \to$$ \lambda=\dfrac{2a^2-1}{2}$ if $a^2\not= 0.5$ put into <4>,$d=\dfrac{1+2a^3-3a}{1-2a^2}=\dfrac{(a-1)(2a^2+2a-1)}{1-2a^2}$ put into <5>

$(a-1) \left[ 3(a+1)+(a-1) \left (\dfrac{(2a^2+2a-1)}{1-2a^2} \right)^2 \right]=0 $

$a=1,\to a=b=c=1,d=0$

when $3(a+1)+(a-1) \left (\dfrac{(2a^2+2a-1)}{2a^2-1} \right)^2 =0 \to a<1$

now I will proof for any root $a<1,f>0$, in this case :

$f=3a+d+2a^3d-3a^2-3ad=3a(1-a)+d(2a^3-3a+1)=(1-a)[3a-d(2a^2+2a-1)]$

if  $2a^2+2a-1<0$, then $f>0$

if  $2a^2+2a-1>0$,$f>0 \iff 3a>d(2a^2+2a-1) \iff$$ 3a^2>(3-3a^2)(2a^2+2a-1)^2 \iff 2a^4+2a^3-2a+1>0 \iff a^2(2a^2+2a-1)+(a-1)^2>0$

which is true!

for $a^2=0.5$, we will have same result.

case III: $a=b,1-2ad+2\lambda=0,1-2ab+2\lambda=0 \to a(b-d)=0$

when $a=b=0,\to c+d=1,c^2+d^2=3 \to d=\dfrac{1+\sqrt{5}}{2} \to c=\dfrac{1-\sqrt{5}}{2} <0$, so there is no solution in this case.

when $b-d=0 \to a=b=d,$ which is same as case II.

case IV: $1-2cd+2\lambda=0,1-2ad+2\lambda=0,1-2ab+2\lambda=0 \to d(a-c)=0 $and $ a(b-d)=0$ again there are 4 sub cases:

A:$a=0,d=0 \to b+c=1$ which is case III.

B:$d=0,b=d $ which is case A

C:$a=c,a=0$,which is case A

D:$a=c,b=d \to 2\lambda=2ab-1 \to 2a^2+2b^2=3,1+2a^2b+2ab^2-2a-2b=0$ WLOG, let $b \ge a \to 1+b-2b^3+2a(b^2-1)=0 \to (1-b)((1+2b+2b^2-2a(1+b))=0 \to b=1,a=\dfrac{\sqrt{2}}{2} $

verify all solutions, when $a=b=c=1,d=0, f=0$, all other cases $ f>0 \to f_{min}=0$.QED


问:
angle and plane of rotation in 3D geometric algebra
Suppose that we perform a rotation of an orthonormal basis $\textbf{e}_1$, $\textbf{e}_2$ and $\textbf{e}_3$ in an Euclidean vector space. We want to rotate  a certain angle $\varphi$ about a certain plane defined by the bivector B. As far as I know, individual rotation of single vector can be accomplished by sandwich operation of rotors, e.g. $R=e^{\frac{\theta}{2}\hat{B}}$. Now suppose we know the final position of the rotated basis, i.e., $\textbf{e}^{\prime}_1$, $\textbf{e}^{\prime}_2$ and $\textbf{e}^{\prime}_3$. How can we recover $\varphi$ and B if we know the initial and the rotated basis?

答:
I am simply going to quote the result from Doran & Lasenby (2003) Geometric Algebra for Physicists p. 103 (with minor adaptations of symbols to your problem statement).  Note use of the summation convention on indices.

$$R = \frac{1+e'_k e_k}{|1+e'_k e_k|} = \frac{\psi}{\sqrt{\psi\tilde{\psi}}}$$

where $\psi = 1+e'_k e_k$.  The case of $\psi=0$ (180-degree rotation) must be handled separately, essentially by inspection.

$\frac{\mathbf{B}}{2}$ must then be recovered as the logarithm of $R$.  If we wish, $\mathbf{B}=\varphi\hat{B}$ can be resolved into $\varphi = |\mathbf{B}|$ and $\hat{B}$, with $|\hat{B}|=1$.


问:
Estimate of $\frac{\cosh^2(x)}{\sinh^2(x)}-\frac{\cosh^2(y)}{\sinh^2(y)}$ depending on $x$ and $y$
Can we get any estimate of $\frac{\cosh^2(x)}{\sinh^2(x)}-\frac{\cosh^2(y)}{\sinh^2(y)}$ depending on $x$ and $y$, where $x, y$ real with $x>y.$

I have tried using increasing property of hyperbolics, get nothing.

Please help. Thanks.

答:
Not sure if this helps but, if $x,y \ge a > 0$, Lagrange's theorem gives you
$$
f(x)-f(y) = f'(\xi) (x-y), \quad \xi \in (x,y)
$$

which would yield
$$
\left|\frac{\cosh^2 x}{\sinh^2 x}-\frac{\cosh^2 y}{\sinh^2 y} \right| \leq 2 |\coth a | \textrm{csch}^2 a |x - y|
$$


问:
zero divisors in polynomial Ring $R[x]$
How can I find all zero-divisors in the ring $GF(3)[x]_{x^2 + 2x}$?
I know for sure that is a ring and not a field because $x^2+2x$ is reducible.
What I tried is to take: if the  $gcd(m,n)$ not equal $1$ then you get a zero-divisors.
I'm really lost and I don't know how to approce the problem. Is someone that can help me?

答:
Well, the nonzero elements in the quotient ring $\Bbb Z_n$ are either units or zero divisors. The elements $m$ with $gcd(m,n)=1$ are units and the remaining nonzero elements are zero divisors.

The same holds for the quotient ring $GF(3)[x]/\langle x(x+2)\rangle$. Those elements which have gcd 1 with $x(x+1)$ are units, the others are zero divisors - bar $0$ which is neither nor.

You can list all elements of $GF(3)[x]/\langle x(x+2)\rangle$ and start from there.

Title: zero divisors in polynomial Ring $R[x]$


答:
Hint:

Each of the numbers $1,2,3,4,\dots,10$ will fall into exactly one of the following six cases:

In $A_1$ only and not $A_2$ or $A_3$

In $A_2$ only and not $A_1$ or $A_3$

In $A_3$ only and not $A_1$ or $A_2$

In both $A_1$ and $A_2$ but not in $A_3$

In both $A_1$ and $A_3$ but not in $A_2$

In both $A_2$ and $A_3$ but not in $A_1$

The condition that $A_1\cup A_2\cup A_3 = \{1,2,3,\dots,10\}$ implies that it is never the case that there are values who are in none of $A_1,A_2,A_3$.  The condition that $A_1\cap A_2\cap A_3=\emptyset$ implies that it is never the case that there are values who are in all of $A_1,A_2,A_3$ simultaneously.

 Apply rule of product (multiplication principle).  Pick which case the element $1$ falls under.  Pick which case the element $2$ falls under.  Continue in this fashion deciding which case each element falls under.

$~$

 $6^{10}=2^{10}\cdot 3^{10}$

Title: Find the number of ordered triples

Body:
Determine, with proof, the number of ordered triples (A_1,A_2,A_3) of
sets which have the property that $$(i) A_1 \cup A_2 \cup A_3 = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\},$$ and $$(ii) A_1 \cap A_2 \cap A_3 = \emptyset,$$ where $\emptyset$ denotes the empty set. Express the answer in the form $$2^{a}3^{b}5^{c}7^{d},$$
where a, b, c, and d are nonnegative integers.

Hello, first, I want to say that I know the problem is well known and there are questions here about the problem. But, I don't want to see how to solve it too early, so I am posting this question to show my work, which actually is wrong. So, it is not a duplicate and is not a homework check, because I already know it is wrong. I would appreciate knowing where it is wrong.

I am not sure if I am interpreting the problem right, but what I imagine they are asking is, for example the number of ways to do things like this:
(1234567,8,9 10).

So, following it, I thought that would be a nice way to attack the problem considering $12$ empty spaces. So we will need to fill the 12 empty spaces with $10$ numbers plus two commas.

The first and the last space can not have comma (it would represent $A_1$ or $A_3$ equal the empty space), so we can have in the first space $10$ numbers, in the last space $9$ numbers. Having done that, we yet have $10$ characters to distribute between the spaces. Since the order matters, it means that we have $10!$ possible ways to do it.

Summarizing, we have $$10 \cdot 9 \cdot 10!$$ way.

(What I am doing here is considering the space before the first comma as $A_1$, between commas $A_2$, and after the second comma $A_3$)

What is the problem??


问:
Why does $7^{2\ln x}\cdot \ln(7) \cdot (2/x)$ equal to $7^{2\ln x}\cdot \ln(49) /x$?
While reviewing, I came upon this problem which has the derivative

$7^{2\ln x}\cdot \ln(7) \cdot (2/x)$

simplified to

$7^{2\ln x}\cdot \ln(49) /x$

How/why is it simplified like that?

答:
You can use the rules of exponents and logarithms to come up with a few equivalent forms.  For example

$$7^{2\ln x}=(7^2)^{\ln x}=49^{\ln x}$$

If you take the derivative of that, you'll get the form you seek.  It's also possible to rewrite the original as

$$7^{2\ln x}=49^{\ln x}=(e^{\ln 49})^{\ln x}=(e^{\ln x})^{\ln 49}=x^{\ln 49}$$

Take the derivative of that to get

$$x^{\ln49-1}\ln49=x^{\ln49-\ln e}\ln 49=x^{\ln\frac{49}e}\ln49$$


问:
Is my proof to this number theory question valid?
I had the following question:

Does there exist a nonzero polynomial $P(x)$ with integer coefficients satisfying both of the following conditions?

$P(x)$ has no rational root;

For every positive integer $n$, there exist an integer $m$ such that $n$ divides $P(m)$.

I created a proof showing that there was no polynomial satisfying both of these conditions:

Suppose that we have a nonzero polynomial with integer coefficients $P(x)=\sum_i c_i x^i$ without a rational root, and for all positive integer $n$, we have an integer $m_n$ such that $n|P(m_n)$. This would imply $P(m_n)\equiv0\pmod n\Rightarrow \sum_i c_i m_n^i\equiv0$. By Freshman's Dream we have $P(m_n+an)=\sum_i c_i(m_n+an)^i\equiv\sum_i c_im_n^i+c_ia^in^i\equiv\sum_ic_im_n^i=P(m_n)\equiv0\pmod n$ for some integer $a$. Therefore if $b\equiv m_n\pmod n$ then $P(b)\equiv P(m_n)\equiv0\pmod n$.

Now the above conditions and findings imply for all prime $p$, we have a number $m_p$ such that $p|P(m_p)$, and that if $b\equiv m_p\pmod p$ then $P(b)\equiv 0\pmod p$. Consider the set of the smallest $n$ primes $\{p_1,p_2,p_3,\cdots,p_n\}$. By Chinese Reamainder Theorem there exists an integer $b$ such that $b\equiv m_{p_1}\pmod{p_1},b\equiv m_{p_2}\pmod{p_2},b\equiv m_{p_3}\pmod{p_3},\cdots,b\equiv m_{p_n}\pmod{p_n}$ by Chinese Remainder Theorem. Then $p_1,p_2,p_3\cdots,p_n|P(b)\Rightarrow p_1p_2p_3\cdots p_n|P(b)$. As $n$ approaches infinity (as there are infinitely many primes), $p_1,p_2,p_3\cdots,p_n$ approaches infinity. Therefore either $P(b)=\infty$ or $P(b)=0$. Since for finite $b$ and integer coefficients $P(b)$ must be finite, then $P(b)=0$. However as $a$ is an integer, this implies $P$ has a rational root, a contradiction.

I'm not sure if my proof is correct, and my main concern is that I am incorrectly using Chinese Remainder Theorem since I am not sure if I can apply it to infinitely many divisors.

Is this proof correct, and if not, how do I solve this question?

EDIT: It appears not only is my proof incorrect (as$b$ does not converge) as Paul Sinclair has shown, but that according to Jaap Scherphuis there are examples of polynomials that satisfy the conditions. Therefore, my question now is how one can prove the existence of these polynomials while using elementary methods (as this is an IMO selection test problem).

答:
Your definition of $b$ depends on $n$. So it is not a constant $b$, but rather each $n$ has its own $b_n$. And it is not the case that $P(b) = \infty$ or $P(b) = 0$, but instead
$$\lim_{n\to\infty} P(b_n) = \infty\text{ or }\lim_{n\to\infty} P(b_n) = 0$$
But this completely spoils your conclusion. Since you do not have a finite fixed $b$, you do not have a root $P(b) = 0$.


问:
definition of "edge disjoint union" in graph theory
Could someone give a precise definition of what 'edge disjoint union' means in graph theory?
i.e. what does it mean to say a graph $G$ can be written as the edge disjoint union of graphs $H$, $K$?

I've been trying to look for a precise definition and annoyingly cannot find one.

Is it just a matter of $e(H)\sqcup e(K)=e(G)$?
Or do we require the graph $H\cup K$ include all the vertices of $G$ as well (e.g. in the case that $G$ has isolated vertices with no edges which may not appear in $H,K$)

答:
If we say $G$ can be written as the edge disjoint union of graphs $H$ and $K$, first of all, this means $V(G) = V(H) \cup V(K)$ because otherwise we couldn't have the graph $G$ right? And as you suggested, we must have $E(H)\sqcup E(K)=E(G)$ as disjoint union.

In this way, we also cover the isolated vertices case you mentioned as there are no edges incident to those vertices at all. So, we can put those vertices to both $V(H)$ and $V(K)$ as well without having any trouble, as long as they are contained in $V(H) \cup V(K)$.


答:
If $y$ is close to $x$, you can use Taylor series and get
$$\coth ^2(x)-\coth ^2(y)=\sum_{n=1}^p a_n (y-x)^n+O\left((y-x)^{p+1}\right)$$ and the first coefficients will be
$$a_1=-2 \coth ^3(x)+2 \coth (x)$$
$$a_2=3 \coth ^4(x)-4 \coth ^2(x)+1$$
$$a_3=-4 \coth ^5(x)+\frac{20}{3} \coth ^3(x)-\frac{8 }{3}\coth (x)$$ and so on.

Title: Estimate of $\frac{\cosh^2(x)}{\sinh^2(x)}-\frac{\cosh^2(y)}{\sinh^2(y)}$ depending on $x$ and $y$

Body:
Can we get any estimate of $\frac{\cosh^2(x)}{\sinh^2(x)}-\frac{\cosh^2(y)}{\sinh^2(y)}$ depending on $x$ and $y$, where $x, y$ real with $x>y.$

I have tried using increasing property of hyperbolics, get nothing.

Please help. Thanks.


答:
$3x+y=1$ can be written as $y=1-3x$

We have to solve
$$1-3x<(1-x)^3$$
Expand and rearrange
$$1-3x<1-3x+3x^2-x^3$$
$$x^3-3x^2<0$$
$$x^2(x-3)<0$$
$$x<3$$
The line is under the curve for $x<3$

Title: Line under curve on the interval (proof)

Body: What is the easiest way how to prove that $3x+y=1$ is under $y=(1-x)^{3}$ without drawing graph? I mean by using some matematical technique. It is enough for interval $[-1,1]$. Thank you for advice.


问:
Sign of dual variable in dual lagrange problems
When I consider a general minimization problem
\begin{align}
&\min f_0(\textbf{x})\\
&s.t \quad  g_i(\textbf{x}) \le 0 \qquad i=1,\dots ,p\\
& \qquad h_i(\textbf{x}) = 0 \qquad i=1,\dots ,m.
\end{align}

and consider the primal function $L(\textbf{x},\lambda,\nu)$, I would expect that the dual problem becomes something like this

\begin{align}
&\max g(\lambda,\nu)\\
&s.t \quad \lambda \le 0
\end{align}

(Since $\lambda$ is a vector, the inequality is intended componentwise), and $g$ is the dual function $g(\lambda,\nu) := \inf_{\textbf{x}} (L(\textbf{x},\lambda,\nu))$. I would expect this because I have a minimization problem with a $\le$ inequality constraints.

Why on Convex Optimization by Stephen Boyd, even if he clearly refers to a general problem of the form I put at the beginning, he says that in the dual problem we have $\lambda \ge 0$??

答:
Good for you that you do not just accept the theory, but try to link it to your previous knowledge on linear optimization.

In linear optimization, if you have $\min\{c^Tx : Ax \leq b\}$, the dual variable associated with $Ax \leq b$ is negative. So, the objective is $b^Ty$ with $y\leq 0$.

To cast that linear problem in convex optimization form, you get $g_i(x) = Ax - b \leq 0$. The dual objective has a term $(Ax-b)^T \lambda$ with $\lambda \geq 0$. Now $x$ will be such that the term with $(Ax)^T\lambda$ cancels with another term, so you will end up with $-b^T\lambda$ with $\lambda \geq 0$.

As you can see, both problems are identical with $y = -\lambda$.