问:
Let $G$ be a group, and $A,B$ normal commutative subgroups of $G$. If the $A \cap B = \{e\}$, does this isomorphism hold?
Let $G$ be a group, and $A,B$ normal commutative subgroups of $G$. True or False

a) $\forall a \in A: \forall b \in B: ab = ba$

b) $\forall a \in A: \forall b \in B: aba^{-1}b^{-1} \in A \cap B$

c) If $A \cap B = \{e\}$, then $G \cong \frac{G}{A} \oplus \frac{G}{B}$

d) If $A \cap B = \{e\}$, then $AB$ is a commutative subgroup of $G$ and is isomorphic to $A+B$.

My answers:

a) False: take the group $D_4$ and let $A = \{e, a, a^2, a^3\}, B = \{e, b\}$. Then it holds that $ab = ba^2 \neq ba$.

b) True, since $A,B$ are normal subgroups, $aba^{-1} \in B$, so that $aba^{-1} \cdot b^{-1} \in B$, and $ba^{-1}b^{-1} \in A$, so that $a\cdot ba^{-1}b^{-1} \in A$,  which means that $aba^{-1}b^{-1} \in A \cap B$.

c) I am not sure how to prove this one.

d) False, take the example from a). Then $A \cap B = \{e\}$, but $AB$ is not commutative since $ab \neq ba$ and $ae \in AB, eb \in AB$.

I need help solving c, and are my answers for the other questions correct?

答:
Edit: okay, so, without requiring $AB = G$ we don't need to complicate ourselves that much. An isomorphism such as in $(c)$ would five $|G| = |G|^2/|A||B|$. So $|A||B| = |G|$. This need not be the case, for example, pick $A$ and $B$ of coprime order in a larger abelian group.

E.g. $G= \Bbb Z_2 \oplus \Bbb Z_3 \oplus \Bbb Z_5$, $A = \Bbb Z_2, B = \Bbb Z_3$.

If we insist on having $AB =G$, then the result is true.

Indeed, since $A \cap B  = 1$ we have $G = A \times B$, via $(a,b) \mapsto ab$. The map sends $A \times 1$ to $A$ and $1 \times B$ to $B$, hence $G/A = A \times B / A \times 1 = B, G/B = A \times B / 1 \times B = A$ and
thus

$$
G = A \times B = G/A \times G/B.
$$


问:
Find what a Infinite product number approaches
Find what the following number approaches:
$$\frac{6}{5}\times\frac{26}{25}\times\frac{626}{625}\times\frac{390626}{390625}\times...$$

What I found is that each fraction can be simplified as:$$1+\frac{1}{5^n}$$
Where $n$ is $1,2,4,8$ and so on (instead of $1,2,3,4,\ldots$)

What should I do next?

Note: Please use junior high school math if possible

ContentLicense: CC BY-SA 4.0

Body:
You have

$$\left(1+\frac{1}{5^1}\right)\left(1+\frac{1}{5^2}\right)\left(1+\frac{1}{5^4}\right)\left(1+\frac{1}{5^8}\right)\cdots$$

$$=\left(1+\frac{1}{5^1}+\frac{1}{5^2}+\frac{1}{5^3}\right)\left(1+\frac{1}{5^4}\right)\left(1+\frac{1}{5^8}\right)\cdots$$

$$=\left(1+\frac{1}{5^1}+\frac{1}{5^2}+\frac{1}{5^3}+\frac{1}{5^4}+\frac{1}{5^5}+\frac{1}{5^6}+\frac{1}{5^7}\right)\left(1+\frac{1}{5^8}\right)\cdots$$

$$=1+\frac{1}{5^1}+\frac{1}{5^2}+\frac{1}{5^3}+\frac{1}{5^4}+\frac{1}{5^5}+\frac{1}{5^6}+\frac{1}{5^7}+\frac{1}{5^8}+\cdots$$

$$= \frac{5}{5-1} \\= 1+\frac{1}{4}$$


问:
Is there a chain rule for integration?
I know the chain rule for derivatives. The way as I apply it, is to get rid of specific 'bits' of a complex equation in stages, i.e I will derive the $5$th root first in the equation $(2x+3)^5$ and continue with the rest.

I wonder if there is something similar with integration. I tried to integrate that way $(2x+3)^5$ but it doesn't seem to work. Well, it works in the first stage, i.e it's fine to raise in the power of $6$ and divide with $6$ to get rid of the power $5$, but afterwards, if we would apply the chain rule, we should multiply by the integral of $2x+3$!, But it doesn't work like that, we just need to multiply by $1/2$ and that's it.

So my question is, is there chain rule for integrals? I want to be able to calculate integrals of complex equations as easy as I do with chain rule for derivatives.

答:
If $g$ is invertible, then the integral
$$
\int f(g(x)) \, dx
$$
can be attacked using integration by substitution. Set $u=g(x)$, meaning that $x=g^{-1}(u)$ and $dx=(g^{-1})'(u)du$. The integral becomes
$$
\int f(u)(g^{-1})'(u) \, du \, .
$$
There are instances where the transformed integral is easier to solve than the original one, but there is no guarantee of this, and obviously we still don't have a 'rule' that tells us that
$$
\int f(g(x)) \, dx = F(x) + C
$$
for some function $F$.

For an example of when this technique might be useful, consider
$$
\int \sqrt{1-x^2} \, dx \, .
$$
This integral looks impossibly hard to solve until you make the substitution $\theta=\arcsin x$. Note that the inner function $g$ is in disguise:
$$
\int \sqrt{1-x^2} \, dx = \int \sqrt{1-\sin^2(\arcsin x)} \, dx \, .
$$
(In practice, we tend to say 'let $x=\sin \theta$' and proceed from there, but this is not quite correct because $x=\sin \theta$ does not imply $\arcsin x =\theta$.)

ContentLicense: CC BY-SA 3.0

Title: Is there a chain rule for integration?


答: Put $x=\frac{1}{5}$ then we have $$\lim_{n \to \infty}(1+x)(1+x^2)(1+x^4)(1+x^8)..(1+x^{2^n})$$ now multiply $1-x$ in numerator and denominator and use $(1-x^{2^k})(1+x^{2^k})=1-x^{2^{k+1}}$ repeatedly and see the magic!

Title: Find what a Infinite product number approaches

Body:
Find what the following number approaches:
$$\frac{6}{5}\times\frac{26}{25}\times\frac{626}{625}\times\frac{390626}{390625}\times...$$

What I found is that each fraction can be simplified as:$$1+\frac{1}{5^n}$$
Where $n$ is $1,2,4,8$ and so on (instead of $1,2,3,4,\ldots$)

What should I do next?

Note: Please use junior high school math if possible


问:
If $p \in U\subseteq M$ is an open subset, then $C^\infty_p(U) = C^\infty_p(M)$.
Let $M$ be a (smooth) manifold and $p \in M$. We define $C^\infty_p(M)$ to be the set of equivalence classes of smooth functions $M \to \mathbb{R}$ that are identified when they agree on some  neighborhood of $p$.

In Tu's book "Introduction to manifolds", p87, I read that if $U$ is an open subset of $M$ containing $p$, then
$$C_p^\infty(M) = C_p^\infty(U)$$

Questions: Evidently, this is not a strict equality because on the right we have (classes of) functions $U \to \mathbb{R}$ and on the left we have functions $M \to \mathbb{R}$. Hence, I guess that they mean that there is a canonical isomorphism (of $\mathbb{R}$-algebras)
$$C_p^\infty(M) \to C_p^\infty(U): f \mapsto f\vert_U$$

Clearly this is injective, because if $f$ and $g$ agree on $U$, then they are in the same class so $f= g$ in $C_p^\infty(M)$. Why is this mapping surjective? Don't we need that a smooth function $U \to \mathbb{R}$ can be smoothly extended to a smooth function $M \to \mathbb{R}$ for this?

答: You are right. However, germs are local, so pick a chart inside $U$ containing $p$, so that via a diffeomorphism with euclidean space you obtain a bump function $b : M \to \Bbb R$. Multiplying $f$ with $b$ may not be equal to $b$, but $f$ and $bf$ coincide near $p$ and thus they are equal as germs.


问:
Strategy of a game
Alice and Bob are playing a game. They write down the integers between $1$ and $100$ (inclusive). First, Alice chooses a number she likes and deletes it, and then Bob chooses to delete another number among the divisors and multiples of the one Alice deleted. Then Alice repeats what Bob has done, choosing another number among the divisors and multiples of the one Bob deleted. The game continues until one of them has no choices and the other wins.

For example, Alice chooses $94$. Bob chooses $47$. The only choice left for Alice now is $1$. And then Bob chooses $97$ to win the game.

To make the question interesting, we add the restriction on Alice that she can only begin with an even number.

The question is: Does Alice have a winning strategy?

It can be interpreted as a graph. The graph has $100$ vertices and has edges between every divisor-multiple pair. When a point is deleted, if every point with an edge to the deleted point has no winning strategy for Alice, then the deleted point has a winning strategy for Bob.   But I can't run the program on my PC. Maybe there is a wiser strategy?

答:
EDIT : Oops, misread the question. This would be too easy.

Alice chooses a prime number bigger than 50 (Grothendick would say 57)
; then Bob's only choice is 1, and then Alice can choose any number
again. She picks a new prime number bigger than 50, and Bob is stuck.

For your program, take care : the strategy depends on which numbers have already been played, so you can't decide whether a starting point is winning by analysing the other starting points' results.

Title: Strategy of a game


问:
is this symmetric
A is the set of all functions $\mathbb{R}$ $\to$ $\mathbb{R}$

f is related to g if and only if f(x) $\le$ g(x) for all x $\in$ $\mathbb{R}$

I said its reflexive since it is less than OR equal, so f(x)=f(x)

However would it be symmetric and transitive?

I said it would not be symmetric (counter-example) because take x,y $\in$ $\mathbb{R}$ if x=3, y=10 then it would not be symmetric but i'm not too sure...

Also since it is a partial order, is it a total order?

答: It is surely not symmetric, as you pointed out in your question. It is actually anti-symmetric, hence your relation (being also reflexive and transitive) turns out to be a (partial) order, not an equivalence.</p>

Title: is this symmetric


问:
$\# V(I) \le \text{dim}_K(K[X_1, \ldots, X_n]/I)$
Let $K$ be an algebraically closed field and $I \subseteq K[X_1, \ldots, X_n]$ be a zero-dimensional ideal, i.e. the set $V(I)$ is finite. This is equivalent to $\text{dim}_K (K[X_1, \ldots, X_n]/I)<\infty$. But the proof I know for this equivalence does not highlight any relation between $V(I)$ and $\text{dim}_K (K[X_1, \ldots, X_n]/I)$. I found recently that, in fact, one has the following inequality:

$$\# V(I) \le \text{dim}_K (K[X_1, \ldots, X_n]/I)$$

Here $\# V(I)$ denotes the number of points in $V(I)$. I am trying to prove this result. I know that the dimension of $K[X_1, \ldots, X_n]/I$ over $K$ is precisely the number of monomials from $K[X_1, \ldots, X_n]$ which do not belong to the initial ideal of $I$ (the classes of these monomials modulo $I$ form a $K$-basis of the vector space $K[X_1, \ldots, X_n]/I$). How can I relate these monomials with the number of solutions $(x_1, \ldots, x_n)$ of the set of equations $\{f(x_1, \ldots, x_n) = 0 \ | \ f\in I \}$?

答:
At @KReiser's suggestion, I'll post the solution of this problem. Let me denote the ring $K[X_1, \ldots, X_n]$ with $R$.

The key observation is the following: Since $K$ is algebraically closed, all the maximal ideals of $R$ are exactly ones of the form $(X_1-c_1, \ldots. X_n-c_n)$, with $c_i \in K$, so there is a bijection between the elements of $V(I)$ and the maximal ideals of $R$ which contain $I$, namely $(a_1, \ldots, a_n) \rightarrow (X_1-a_1, \ldots. X_n-a_n)$.

Now $R/I$ is a finite dimensional $K$-algebra, and as is shown here using CRT, the number of maximal ideals of $R/I$ (which is the number of maximal ideals of $R$ which contain $I$) is less than or equal to $\text{dim}_K(R/I)$.


问:
What exactly does "Uniqueness of inverse matrix" mean?
I have been trouble understanding the description: “Uniqueness of the inverse matrix”.

Or, more specifically, what exactly the term "Unique" here means? An explanation of the proof of this claim would also be appreciated and help with my understanding.

答:
Let $C$ and $B$ both be inverses of $A$:

$$B = BI = B(AC) =(BA)C = IC = C$$

Where $I$ is the identity matrix.

Unique means that there is one and only one, in other words take $C,B$ inverses of $A, C$ and $B$ must be equal they cannot be distinct, as shown in the proof.


问:
I am not able to find mistake in my solution of the finding $\displaystyle\lim _{h\to 0} \frac{q(1-\cos h )}{4 h^{2}}$, where $q\not=0 $.
I have to find the value of $$\lim _{h \rightarrow 0} \frac{q(1-\cos h )}{4 h^{2}}$$where $q$ is a non-zero constant.

Now in the book the answer is $\dfrac{q}{8}$. You can see the steps at this site.

This answer is different from my answer, but I am not able to find where I have done the mistake(s).

My steps are:
$$\lim _{h \rightarrow 0} \frac{q(1-\cos h)}{4 h^{2}}=\lim _{h \rightarrow 0}\left(\frac{q}{4 h} \times \frac{1-\cos h}{h}\right)$$
$$=\lim _{h \rightarrow 0} \frac{q}{4 h}  \times \lim _{h \rightarrow 0}\left(\frac{1-\cos h}{h}\right)$$
$$=\lim _{h \rightarrow 0} \frac{q}{4 h} \times 0
=0$$

答:
If one function tends to $0$ and another one is tending to $\infty$ (or is unbounded) you cannot say that the prodct tends to $0$. It is like saying $1=\lim_{x \to 0}1=_{x \to 0} [x (\frac 1 x)]= \lim_{x \to 0} \frac   1 x \lim_{x \to 0} x=(\lim \frac 1 x) (0) =0$. Do you see why  this argument doesn't work?

Limit of  a product is the product of the limits when both limits are finite. Do not use this when one of the factors does not have a finite limit.


问:
Discontinuous function prove
I want to find an answer to this case
If function f is discontinuous then √f is discontinuous.
I am trying to disprove it and I put counter examples but it seems correct.
Is this true? Can you help me please</p>

答: Look at it this way: if $\sqrt f$ is continuous, then $f$ is continuous. This is clearly true, as the product of continuous functions is continuous (of course, you should start by saying that you are working over real numbers, that $f$ is positive, etc ...)

Title: Discontinuous function prove


Title: I am not able to find mistake in my solution of the finding $\displaystyle\lim _{h\to 0} \frac{q(1-\cos h )}{4 h^{2}}$, where $q\not=0 $.

Body:
I have to find the value of $$\lim _{h \rightarrow 0} \frac{q(1-\cos h )}{4 h^{2}}$$where $q$ is a non-zero constant.

Now in the book the answer is $\dfrac{q}{8}$. You can see the steps at this site.

This answer is different from my answer, but I am not able to find where I have done the mistake(s).

My steps are:
$$\lim _{h \rightarrow 0} \frac{q(1-\cos h)}{4 h^{2}}=\lim _{h \rightarrow 0}\left(\frac{q}{4 h} \times \frac{1-\cos h}{h}\right)$$
$$=\lim _{h \rightarrow 0} \frac{q}{4 h}  \times \lim _{h \rightarrow 0}\left(\frac{1-\cos h}{h}\right)$$
$$=\lim _{h \rightarrow 0} \frac{q}{4 h} \times 0
=0$$

Body:
Hint :

$q\dfrac{1-\cos ^2 h}{(1+\cos h)(4h^2)}=$

$\dfrac {\sin^2 h}{h^2}\dfrac {q}{4(1+\cos h)}.$


问:
Simple meaning to Center of a group
Recently I was learning Center of groups and on referencing the group table, I observed is that all the rows that are also present as columns are the centers of any group.
So, I made a small program to check it for various groups and found somewhat consistent answers. I wrote a small blog on it but I have a fear in my heart if it is only true or not and my post might send some wrong information to other person.

The way I made program was to find similar rows and columns instead of calculating $xg=gx$ as per definition. Is it perfectly fine to let it be so?

答: Yes it is. For some element $g \in G$, the $g$th row (let's write this short for the row corresponding to $g$) consists of the elements $gx$ for $x \in G$ in some particular order, the $g$th column of the elements $xg$, $x \in G$ (same order on the $x$s). If row and column are equal elementwise, this means $gx = xg$ for all $x \in G$, that is exactly true if $g \in Z(G)$.</p>

Title: Simple meaning to Center of a group


问:
Show $T$ is compact operator if $\langle Te_n,e_n \rangle$ tend to zero.
Suppose $\mathcal{H}$ is a Hilbert space, and $T\in B(\mathcal{H})$. If for each orthonormal (norm 1) basis $\{e_n\}\subseteq \mathcal{H}$, we have $\langle Te_n, e_n \rangle \rightarrow 0$. Can we deduce $T$ is compact?

I guess it may use spectra decomposition. Take adjoin, use the fact that $\langle Ae_n,e_n\rangle \rightarrow 0$ iff $\langle (A+A^*)e_n,e_n \rangle \rightarrow 0$ and $\langle (A-A^*)e_n,e_n\rangle \rightarrow 0$ , we can assume $T$ is self-adjoint. So $T$ can be viewed as a multiple operator(multiple by a $L^{\infty}(X,d\mu)$ function) on $L^2(X,d\mu)$, via unitary equivalence. Here $d\mu$ is an abstract $\sigma$-finite Borel measure on X. But I don’t know how to use the condition.. And, another way, if we use the spectra decomposition via $T=\int z dE$, E is the corresponding spectra measure. To show $T$ is compact, it is sufficient to prove the projection $E(-\infty,-\epsilon)$ and $E(\epsilon, +\infty)$ are all finite rank, for every $\epsilon>0$. And I feel the condition may can be use together with something like dominate converge theorem? But I fail.
Any help or hint? Thanks.

答: Perhaps, the following link provides an affirmative answer for the complex Hilbert space: $T$ is compact if and only if for any sequence $x_n$ weakly converging to $0$, $\langle Tx_n,x_n\rangle\to 0$.


答:
The trick here is that you have to use I'Hospital's Rule twice. Here is my attempt at solving the question.

We know that $\lim_{h \to 0} c f{\left(h \right)} = c \lim_{h \to 0} f{\left(h \right)}$ with $c= \frac{q}{4}$ and $f{\left(h \right)} = \frac{1 - \cos{\left(h \right)}}{h^{2}}$. Therefore, we can take out the constant so that we can focus on the variable $h$. Since $q$ is treated as a constant, we can take it and $4$ out. This step looks like this:

\begin{align}
{\lim_{h \to 0} \frac{q \left(1 - \cos{\left(h \right)}\right)}{4 h^{2}}} &= {\left(\frac{q \lim_{h \to 0} \frac{1 - \cos{\left(h \right)}}{h^{2}}}{4}\right)}  \\
&= \frac{q}{4} \cdot \lim_{h \to 0} \frac{1-\cos(h)}{h^2}\end{align}

Here we have to use l'Hopital's Rule because we have the indeterminate form $\frac{0}{0}$. Therefore:

\begin{align}
\frac{q \cdot {\lim_{h \to 0} \frac{1 - \cos{\left(h \right)}}{h^{2}}}}{4} &= \frac{q  \cdot {\lim_{h \to 0} \frac{\frac{d}{dh}\left(1 - \cos{\left(h \right)}\right)}{\frac{d}{dh}\left(h^{2}\right)}}}{4} \\
&= \frac{q \cdot {\lim_{h \to 0} \frac{\sin{\left(h \right)}}{2 h}}}{4}
\end{align}

By using the rule mentioned in the third line, you can take out the constants and focus in on the variable $h$. We have to once again use I'Hospital's Rule because of the indeterminate form $\frac{0}{0}$:

\begin{align}
\frac{q \cdot {\lim_{h \to 0} \frac{\sin{\left(h \right)}}{2 h}}}{4} &= \frac{q \cdot{\left(\frac{\lim_{h \to 0} \frac{\sin{\left(h \right)}}{h}}{2}\right)}}{4} \\
&\stackrel{\text{I'Hopital}}= \frac{q \cdot{\lim_{h \to 0} \frac{\frac{d}{dh}\left(\sin{\left(h \right)}\right)}{\frac{d}{dh}\left(h\right)}}}{8} \\
&= \frac{q \cdot{\lim_{h \to 0} \cos{\left(h \right)}}}{8} \\
&= \frac{q \cdot 1}{8} \\
&= \frac{q}{8}
\end{align}


答: You're making the mistake of equating $\infty\times0$ to $0$. You have to work the standard way of handling the indeterminate forms, usually converting into $0/0$ and using L'Hôpital's Rule.


答:
I think you've made a mistake by concluding that the product tends to zero, when one function goes to infinity and the other to zero. Why don't you try it this way:

$$\lim_{h \to 0}\frac{q(1-\cos h)}{4h^2}$$
We can write $\cos h= 1-2\sin^2\frac{h}{2}$ so $1-\cos h =2\sin^2\frac{h}{2}$

So, the limit becomes:
$$\lim_{h \to 0}\frac{2q\sin^2\frac{h}{2}}{4h^2}=\lim_{h \to 0}\frac{2q\sin^2\frac{h}{2}}{4\frac{h^2}{4}.4}$$

W.K.T. $\lim_{x \to 0}\frac{\sin x}{x}=1$, which we can also extend as $\lim_{x \to 0}\frac{\sin^2 x}{x^2}=1$
Thus,
$$\lim_{h \to 0}\frac{2q\sin^2 \frac{h}{2}}{4\frac{h^2}{4}.4}=\frac{2q}{16}\lim_{h \to 0} \frac{\sin^2\frac{h}{2}}{\frac{h^2}{4}}=\frac{2q}{16} =  \frac{q}{8}$$


问:
A little's law question to find average of the customer in some time
If shoppers enter a store at an average rate of $r$ shoppers per minute and each stays in the store for average time of $T$ minutes, the average number of shoppers in the store, $N$, at any one time is given by the formula $N=rT$. This relationship is known as Little's law.

The owner of the Good Deals Store estimates that during business hours, an average of 3 shoppers per minute enter the store and that each of them stays an average of 15 minutes. The store owner uses Little's law to estimate that there are 45 shoppers in the store at any time.

Little's law can be applied to any part of the store, such as a particular department or the checkout lines. The store owner determines that, during business hours, approximately 84 shoppers per hour make a purchase and each of these shoppers spend an average of 5 minutes in the checkout line. At any time during business hours, about how many shoppers, on average, are waiting in the checkout line to make a purchase at the Good Deals Store?

Please someone help me find an answer to this question. My steps:

$$ \frac{84}{5} = 16.8 $$
$$ N = rt \Rightarrow (16.8)(5) $$
$$ N = 84 $$

For me answer was coming 84 but the answer given was 7.

答:
You need to make sure that all the variables use the same unit of time.

If 84 shoppers make a purchase per hour, then you have $ \frac{84}{60} = 1.4 $ shoppers making a purchase per minute, so you have $r = 1.4$ and $t = 5$. Then, plugging into Little's law, you have $ N = 1.4*5 = 7$.

Alternatively, you could transform $t = 5 \ \text{min} \rightarrow t = \frac{1}{12} \ \text{hour} $, in which case you'd have $ N = 84*\frac{1}{12} = 7$.


问:
System of linear equations.
Let A be a matrix of order $3\times 3$ having entries from $Z_3$. If rank of $A$ is 1 how to find the number of solutions to the homogeneous system $Ax = 0$?
My attempt:- By rank  nullity theorem, $\dim(\ker(A)) = 2$. So trivial solution is not the only solution.
Can anyone please help me further?

答: The kernel is two dimensional by rank-nullity as you said, so there kernel is a two dimensional $\Bbb Z_3$ vector space, and has order $3^2=9$.

Title: System of linear equations.


问:
Rolle theorem for the Number of roots of a real function
I know I can apply the theorem for the polyniomials to know the number of roots. In fact polynomials are continous whatever $[a,b]$ considered and derivable whatever $(a,b)$.

If the function considered is not a polynomial can I apply this? I thought that I have to verify if the function is continous whatever $[a,b]$ in the domain and derivable whatever $(a,b)$ in the domain? It is right?

答:
Hint: actually there are other functions that are differentiable, and the theorem is general and doesn't restrict to polynomials.

Ex.: $e^x+e^{-x}$ satisfies this problem, as there are two points $a, b$ in some interval such that $f(a) = f(b)$, for some $c$, such that $a<c<b$, and as the derivative changes its sign, then Rolle's Theorem will be satisfied.

Antoher Hint: one thing interesting to note is that if the derivative doesn't change the sign, then it means there aren't two equal values for $f(a)$ and $f(b)$, which means, for example, that you can only have a real root, for the $f$
being zero in some interval.


答:
If at any time Bob chooses $1$ then Alice chooses $97$ and wins. So in the following I will ignore this choice for Bob.

Alice wins by playing $62$. There are two possibilities.

Bob chooses $31$

Alice now forces the following sequence where Alice's choices are asterisked. She always chooses a product of two primes greater than $50$ where one of the primes has already been used and so Bob has no choice (other than $1$).

$*93*,3,*51*,17,*85*,5,*95*,19,*57*$ and now $1$ is forced.

Bob chooses $2$

Alice now forces the following sequence.

$*58*,29,*87*,3,*51*$ and now as in the above sequence.

Title: Strategy of a game

Body:
Alice and Bob are playing a game. They write down the integers between $1$ and $100$ (inclusive). First, Alice chooses a number she likes and deletes it, and then Bob chooses to delete another number among the divisors and multiples of the one Alice deleted. Then Alice repeats what Bob has done, choosing another number among the divisors and multiples of the one Bob deleted. The game continues until one of them has no choices and the other wins.

For example, Alice chooses $94$. Bob chooses $47$. The only choice left for Alice now is $1$. And then Bob chooses $97$ to win the game.

To make the question interesting, we add the restriction on Alice that she can only begin with an even number.

The question is: Does Alice have a winning strategy?

It can be interpreted as a graph. The graph has $100$ vertices and has edges between every divisor-multiple pair. When a point is deleted, if every point with an edge to the deleted point has no winning strategy for Alice, then the deleted point has a winning strategy for Bob.   But I can't run the program on my PC. Maybe there is a wiser strategy?


问:
How to show that if f^N is a contraction mapping, then f has a unique fixed point.

If f is a mapping of a complete metric space $(X,d)$ into itself and $f^N$ is a contraction mapping for some positive integer $N$, then $f$ has precisely one fixed point.

The Banach fixed point theorem is already proven so we can show the following:

$f^N$ has a fixed point $\alpha$, then $f^N(\alpha)=\alpha$ and $f^{N+1}(\alpha)=f(\alpha)$ so $f^N(f(\alpha))=f(\alpha)$, hence $f(\alpha)$ is also a fixed point. By uniqueness we have $f(\alpha)=\alpha$

However, this only shows that $f^N$ has a unique fixed point because the Banach theorem states that the contraction mapping has a fixed point. $f^N$ is a contraction mapping but f is assumed to be not a contraction mapping. How can we show that $f$ also has a unique fixed point?

答: If $f(x_i) = x_i$ for $i = 1,2$, then $f^N(x_i) = x_i$ for $i = 1,2$. Use uniqueness on $f^N$.


问:
Limit as $x$ goes to the right and at the left of a point of not definition
The function $f(x)=\log(x^4-2x^2+1)+ \frac{-1}{2(x-1)}$ at the right of -1 goes to $-\infty$ and also at the left. But if I plot $f$ in geogebra it does not seem that that $f$ goes to $-\infty$ at the right and the left of -1. Why?

答:
That is because
$$\log(x^2-2x^2+1)=\log\bigl((x^2-1)^2\bigr),$$
and the argument of the log tends to $0^+$ in both cases


问:
How many digits does $2^{25964951}-1$ have?
Mersenne prime numbers are prime numbers of the form $2^n - 1 \,(\mathbb{N}\ni n>1)$such as
$2^2 -1$ or $2^{25964951}-1$: How many digits does the latter have?

I found this here:

$$\log_{10}(2^{25964951}-1)+1 =  7 816 230$$

Is this the way how to proof this or do I have to do something other?

答:
Taking the 10-logarithm is basically the easiest way to do it. Remember to always round down, no matter what the decimal part is (for instance, $\log_{10}(99)+1\approx2.996$, and we want the answer to be $2$).

However, if all you have is a classic calculator that won't work with such large numbers, or you want to get an approximate answer with just mental calculations, some more manipulations must be done first.

Start with noting that $2^{25\,964\,951}-1$ and $2^{25\,964\,951}$ have the same number of digits. From there, we know that
$$
\log_{10}(2^{25\,964\,951})={25\,964\,951}\cdot\log_{10}(2)
$$
And then either let the calculator do the work, or if you go the mental route, know that $\log_{10}(2)\approx 0.3$ (from $2^{10}\approx 10^3$) and just get
$$
{25\,964\,951}\cdot\log_{10}(2)\approx 7\,800\,000
$$
For the mental calculation, adding 1 here (although it's technically what you're supposed to do) makes no sense in my opinion.


问:
Any set of $d$ points in a projective space is the zero locus of polynomials of degree $d-1$?
This is an exercise from Harris' "Algebraic geometry: A First Course":

Let $\Gamma\subset \mathbb{P}^n$ be a finite set of $d$ points, not all on the same line. Show that $\Gamma$ is the zero locus of polynomials of degree $d-1$ or lower.

Here's a proof I came up with: Let us consider the set of all distinct lines, passing through each pair of distinct points in $\Gamma$. Then for each point let us consider all lines in that set that pass through it, $V(L_i)$, and let's take the product of forms that define them, $\Pi_i L_i$. Then the common zeroes of such products are precisely the points in $\Gamma$.

This proof doesn't quite work however in the case of points in general position. For example, if we take 4 points on the plane in general position, then there is an extra point in which the diagonals intersect. An obvious way to avoid is to not consider the diagonals at all. But what will then be a procedure for choosing the lines $V(L_i)$ as above?

答:
As requested by the OP, I'm converting my comment in to an answer.

Hint: Lagrange interpolation.


答:
If squeezing of the domain is allowed then this is your counterexample:

$f(x)= 1$ for $x>0$ and $f(x)= -1$ for $x\le0$ which is discontinuous but the square root of it i.e. $\sqrt{f(x)}$ is continuous on its domain.

Title: Discontinuous function prove

Body:
I want to find an answer to this case
If function f is discontinuous then √f is discontinuous.
I am trying to disprove it and I put counter examples but it seems correct.
Is this true? Can you help me please</p>


问:
Prove non computability of a function defined on partial and total functions
I have to prove that the following function is not computable

$$ f(x) = if\ (x\ is \ even\ and\ f_{x/2}\ is \ total\ ) \ then\ 1\ else \ (x\ is \ even\ and\ f_{x/2}\ is \ partial\ )\ then\ 0\ else\ \bot$$

I was trying to count on the fact that I have no possibility  to show that $f_{x/2}$ is total by means of a TM (I think). But I have no clue on how to prove this. I was also trying to reduce the function to a set on TM indexes and try to apply the Rice Theorem, but I couldn't do it. Is someone able to give me a clue on how to proceed?

EDIT: I thought that maybe the proof hides in the meaning of total and partial, even though I do not know how to formally prove this. My guess is that for a given index $x$, even if I could prove that $f_{x/2}$ is not defined for a least 1 value, I cannot be sure that for such value the TM cannot stop (i.e. accept the input) in the future, and therefore I cannot conclude it is partial. Moreover, in order to define the function as total I would need to try every possible input, and even after a long time where all the input I tried caused the machine to stop I cannot conclude that $f_{x/2}$ is total.

答:
Your intuition is correct: if your function was computable, then you would be able to decide the TOTAL problem (deciding whether or not a Turing machine halts on all inputs) - which is undecidable.

We first fix an encoding of Turing Machines.

We use the following reduction: suppose that you have a machine $M$ that computes $f$. Now, we design a machine $N$, which accepts $x$ if the machine $M_x$ is total.

$N$ work as follows: on input $x$, it runs $M$ on the input $2x$, and gets the result $y = f(2x)$. Now, if $y = 1$, $N$ accepts; otherwise, it rejects.

It is easy to prove that $N$ accepts $x$ $\Leftrightarrow$ $M_x$ is total. Hence, $f$ is undecidable.

One point that I overlooked in the previous proof is the fact that TOTAL is undecidable. This can be done easily, by considering a machine which does the same thing regardless of its input (and so it hats on all inputs iff it halts on any input). With this idea in mind, the reduction to the Halting Problem is then easy to define.


问:
$x^2+y^n-1$ is absolutely irreducible
I have encountered this while going over the wikipedia page "irreducible polynomial".

A polynomial that is irreducible over any field containing the coefficients is absolutely irreducible. By the fundamental theorem of algebra, a univariate polynomial is absolutely irreducible if and only if its degree is one. On the other hand, with several indeterminates, there are absolutely irreducible polynomials of any degree, such as $x^{2}+y^{n}-1$, for any positive integer $n$.

How can I show that $x^2+y^n-1$ is absolutely irreducible for any positive integer $n$? I'm new so I don't have enough tools to tackle this. Any help is appreciated.

答:
a) Over a field $k$ of characteristic $2$ and for $n=2s$ even, it is not true that the polynomial is  irreducible.

Indeed, since $-1=1$ in characteristic $2$:$$x^2+y^{2s}-1=(x+y^s-1)^2$$
b) In all the other cases the polynomial $x^2+y^n-1$ is irreducible because $1-y^n$ is not a square in $k[y]$.

The reason is that   in the decomposition in linear factors  of $1-y^n$ over an algebraic closure $\bar k$ of $k$ some factor will appear with odd exponent:

(i) This is clear in characteristic $0$ since  $1-y^n$ is separable, i.e. all its linear factors appear with exponent $1$.

(ii) And in characteristic $p\gt2$ you can write $1-y^n=(1-y^r)^{p^e}$ with $1-y^r$ separable (equivalently $p$ does not divide $r$), so that all linear factors appear with exponent $p^e$, an odd number, so that $1-y^n$ is not a square either

Title: $x^2+y^n-1$ is absolutely irreducible


问:
Need Greatest common divisor explaination
In the book the problem says:
What is the GCD of two natural numbers m and n if after increasing the number m by 6 the GCD of two numbers increased 9 times.

And the solution it gives is 2,3 or 6.
Can someone explain how is it. And what values of m and n satisfy the equation. I can't find any. May be i couldn't understand the question or the book is miss printed. Idk. Please help me understand this with some examples. Thanks

答:
$\gcd(a;b)=d$ and $\gcd(a;b+6)=9d$ so $$d\mid b+6\mbox{ and }d\mid b\Rightarrow d\mid 6$$

Assume $d=1$. Then $a$ and $b$ are coprime and $a=9x$ and $b+6=9y$ with coprime $x,y$. This implies $3\mid a$ and $3\mid b$ so contradiction.

So $d\mid 6$ and $d\neq 1$, so $d$ can only be $2,3$ or $6$


问:
Motion on a slope - A Level
A particle P of mass 5kg and particle Q of mass 1kg are connected by a light inextensible string. P lies on a slope inclined at a 60 degree angle to the horizontal. The string passes from P parallel to the line of greatest slope, and runs over a light pulley that can freely rotate. Then descends vertically to Q. The coefficient of friction between P and the slope is 0.7.
Find the force of friction acting on P and it’s direction and find the acceleration of P when the system is released from rest.

I was able to find the frictional force 17.15N up the slope) but I can’t find the acceleration. I have tried to use the equation 5gsin(60)-17.15-9.8 = 5a with 9.8 being the weight of Q, but I get the answer of 3.1m/s/s for a, bit the mark scheme says that a (acceleration) is 2.58m/s/s but I don’t see how. Could you please give me some advice on how to get 2.58?

答:
Remember the mass is $5\,\text{kg}$ for $\text{P}$ and $1\,\text{kg}$ for $\text{Q}$, so $$5g\sin60°-17.15-9.8 = 6a$$

This will give you $a = 2.58\,\text{m}/\text{s}^2$.

Title: Motion on a slope - A Level


问:
Can Ker(T) = Im(T) of a linear transformation of R raised to a power?
So I know this can be true for when T:R -> R, but does this apply for when R is raised to any power? Would this also work if, for example, we had R^3?
Thanks in advance</p>

答:
If $T:\Bbb R^n\to\Bbb R^n$ is a linear transformation, then Rank-Nullity Theorem asserts that $$\dim \ker(T)+\dim\operatorname{Im}(T)=n.$$

So, if $\dim\ker T=\dim\operatorname{Im} T=k$, then $n=2k$.

So if $n=3$ or any other odd number, this is IMPOSSIBLE.

For EVEN $n$ it is possible:
Let $\{e_1,\dots,e_k,e_{k+1},\dots,e_{2k}\}$ be a basis for $\Bbb R^{2k}$.
Define $T(e_j)=e_{k+j}$ for $1\le j\le k$ and $T(e_j)=0$ for $k+1\le j\le 2k$.

Then $\ker t=\operatorname{Im} T=span\{e_{k+1},\dots,e_{2k}\}$.


答:
In $\mathbb R$, there are no such application. In $\mathbb R^2$, there are many... for example, if $T$ is defined s.t. $$T(1,0)=(0,1)\quad \text{and}\quad T(0,1)=(0,0),$$
extend over $\mathbb R^2$ by $\mathbb R-$linearity.

Title: Can Ker(T) = Im(T) of a linear transformation of R raised to a power?

Body:
So I know this can be true for when T:R -> R, but does this apply for when R is raised to any power? Would this also work if, for example, we had R^3?
Thanks in advance</p>


问:
Why is cartesian product of $A\times\varnothing=\varnothing$
I am studying some preliminaries of set theory.

I have a question why is Cartesian Product of $A\times\varnothing=\varnothing$ not the set $\{(a,\varnothing)~\mid~a\text{ belongs to }A \}$ using the definition of $A\times B=\{ (a, b)~\mid~a\text{ belongs to }A,~b\text{ belongs to }B\}$?

Can someone please explain!

答: The set $A×B$ is defined by $ A×B = \{(a,b)|a\in A \wedge b\in B \}$ but $a\in A \wedge b\in B$ is never satisfied if either $A$ or $B$ is $\emptyset$.


问:
$\# V(I) \le \text{dim}_K(K[X_1, \ldots, X_n]/I)$
Let $K$ be an algebraically closed field and $I \subseteq K[X_1, \ldots, X_n]$ be a zero-dimensional ideal, i.e. the set $V(I)$ is finite. This is equivalent to $\text{dim}_K (K[X_1, \ldots, X_n]/I)<\infty$. But the proof I know for this equivalence does not highlight any relation between $V(I)$ and $\text{dim}_K (K[X_1, \ldots, X_n]/I)$. I found recently that, in fact, one has the following inequality:

$$\# V(I) \le \text{dim}_K (K[X_1, \ldots, X_n]/I)$$

Here $\# V(I)$ denotes the number of points in $V(I)$. I am trying to prove this result. I know that the dimension of $K[X_1, \ldots, X_n]/I$ over $K$ is precisely the number of monomials from $K[X_1, \ldots, X_n]$ which do not belong to the initial ideal of $I$ (the classes of these monomials modulo $I$ form a $K$-basis of the vector space $K[X_1, \ldots, X_n]/I$). How can I relate these monomials with the number of solutions $(x_1, \ldots, x_n)$ of the set of equations $\{f(x_1, \ldots, x_n) = 0 \ | \ f\in I \}$?

答:
Here you get an approach to the problem over any field or Artinian ring:

Isomorphism of product of local ring of variety

I copy the post given in the above link here:

Let $B:=k[x_1,..,x_n]$ with $k$ any field and let $I \subseteq B$ be any ideal.
Let $X:=Spec(A)$ with $A:=B/I$. By the Noether normalization lemma it follows there is a finite set of elements $y_1,..,y_d \in A$ with $d=dim(X)$ and the property that the sub-ring $k[y_1,..,y_d] \subseteq A$ generated by the elements $y_i$ is a polynomial ring. The ring extension $k[y_1,..,y_d] \subseteq A$ is an integral extension of rings. If $d=0$ it follows from the same lemma the ring extension $k \subseteq A$ is integral and in particular it follows $A$ is an artinian ring. In particular it follows $A\cong R_1 \prod \cdots \prod R_l$ is a direct product of artin local rings $R_i$ with $\mathfrak{p_i} \subseteq R_i$ the maximal ideal (this is Atiyah-Macdonald's book Thm 8.7). Hence $Spec(A) \cong \prod Spec(R_i)$ is a disjoint union of the schemes $U_i:=Spec(R_i)$. We may write $X:=\{ \mathfrak{p_1}, \ldots , \mathfrak{p_l}\}$ as a disjoint union of $l$ distinct points, with $\mathfrak{p_i} \subseteq  R_i$ the unique maximal ideal in the local ring $R_i$. Since $U_i \subseteq X$ is an open subscheme it follows there are isomorphisms $\mathcal{O}_{X,\mathfrak{p_i}} \cong \mathcal{O}_{U_i,\mathfrak{p_i} } \cong (R_i)_{\mathfrak{p_i}} \cong R_i$ since $R_i$ is a local ring and every element in $R_i - \mathfrak{p_i}$ is a unit. Here $\mathcal{O}_{U_i}$ is the restriction of $\mathcal{O}_X$ to the open set $U_i$. Hence for any base field $k$ it follows from Atiyah-Macdonalds (AM) book "Commutative Algebra" that your ring $A:=B/I$ decompose as a finite direct sum of Artinian local rings $R_i$. At the level of sheaves you get the following formulas:

$\mathcal{O}_X \cong  \mathcal{O}_{U_1} \oplus \cdots \oplus \mathcal{O}_{U_l}$

and

$  \mathcal{O}_{X,\mathfrak{p_1}} \oplus \cdots \oplus \mathcal{O}_{X,\mathfrak{p_l}} \cong  \mathcal{O}_{U_1,\mathfrak{p_1}} \oplus \cdots \oplus \mathcal{O}_{U_l,\mathfrak{p_l}} \cong \mathcal{O}_{U_1}(U_1) \oplus \cdots \oplus \mathcal{O}_{U_l}(U_l) .$

These formulas answers the following questions:

Question 1: "But if k is not algebraically closed, is it still true that, if V(I) is finite, k[X1,…,Xn]/I is a direct product of local rings?"

Answer 1: Yes, this is true by the above argument.

Question 2: "And if so, then some factors are most likely given by localizations at the points of V(I), whereas other factors are not; how to describe these latter factors?"

Answer 2: All factors are given as localizations at points by the above argument.

Consider your example with $A:=\mathbb{Q}[x,y]/(x^2+y^2-3)$ and let $C:=Spec(A)$. By Zorn's Lemma (if you accept this) there is a maximal ideal $\mathfrak{m} \subseteq A$, and the residue field $\kappa(\mathfrak{m})$ is a finite extension of $\mathbb{Q}$. Hence $C$ - which as a set is the set of prime ideals in $A$ - is non-empty. If $k$ is the algebraic closure of $\mathbb{Q}$ and you pass to the fiber product $C\otimes k$ you get a curve. By Hartshorne Prop I.1.13 it follows $dim(C\otimes k)=1$ since it is defined by a hypersurface in $k[x,y]$.

Let $k$ be any field and let $A:=k[x]$ be the polynomial ring in the variable $x$. Let $f(x) \in A$ be any polynomial. We may write $f(x)=\prod_{i=1}^l p_i(x)^{d_i}$ where $p_i(x)$ is an irreducible polynomial and  $d_i \geq 1$ is an integer for every $i$. Here $p_i(x)$ and $p_j(x)$ are distinct for $i\neq j$. Let $B_i:=k[x]/(p_i(x)^{d_i})$ and let $B:=A/(f(x))$.
There is a canonical isomorphism

$\phi: B \rightarrow B_1\oplus \cdots \oplus  B_l$

defined by

$\phi(x + (f(x))) :=(x+(p_1(x)^{d_1}),..,x+(p_l(x)^{d_l}))$.

This is Prop 1.10 in the AM-book. The ideals $(p_i(x)^{d_i})$ and $(p_j(x)^{d_j})$ are coprime for $i\neq j$. Since the ideals $(p_i(x))$ are maximal, it follows the rings $B_i$ are Artinian local rings, hence the decomposition given by $\phi$ decompose the Artinian ring $B$ into a direct sum of Artinian local rings $B_i$. Hence in this case Prop 1.10 in AM implies Thm 8.7 in the AM book (the "stucture theorem on Artin rings").

Question 3: "And is there an higher-level explanation why it (apparently, not quite sure) works in one variable, that is, k[X]/f is a direct product of local rings for any f∈k[X], even if k is not algebraically closed, and even if V(f) is empty?"

Here you must understand the fact that any polynomial $f(x) \in k[x]$ factors as a product of powers of irreducible polynomials, and that an irreducible polynomial defines a maximal ideal in $k[x]$. Moreover if $p(x)$ is an irreducible polynomial, it follows the ring $k[x]/((p(x)^d)$ for $d\geq 1$ is an Artinian ring. Then you must read AM, Chapter1 where it is proved that the map $\phi$ is an isomorphism of rings.

Here the copied post ends.

This is an answer to your question: In your case you may write

$$D1.\text{   }A \cong \oplus_{i=1}^l R_i$$

where $R_i$ is a local ring with maximal ideal $\mathfrak{p}_i$. It follows

$$dim_k(A) :=d = \sum_{i=1}^l dim_k(R_i)$$

hence $l \leq d:=dim_k(A)$$.

Question: "I found recently that, in fact, one has the following inequality:

$$\# V(I)≤dim_k(k[x_1,…,x_n]/I)$$

I am trying to prove this result."

Answer: Here I interpret $\# V(I)$ as the "number of points" in the scheme $V(I)$. By the above argument it follows the number of maximal ideals in $A$ is less than or equal to $dim_k(A)$. Here $k$ is any field. Hence the general case (over any field) can be proved using methods from Atiyah-Macdonalds book "Commutative Algebra". In fact:

It seems to me the methods introduced above proves the result for any Artinian ring $A$ over any field $k$. Let $V(I) \subseteq Spec(A[x_1,..,x_n]):=\mathbb{A}^n_A$ and let $B:=A[x_i]/I$.  It follows

$$ \# V(I) \leq length_A(B)$$

where $length_A(B)$ is the "length" of $B$ as left $A$-module.

Formula $D1$ gives

$$D1.\text{   }B\cong R_1 \oplus \cdots \oplus R_l$$

where $R_i$ is a local ring, and each $R_i$ is an $A$-algebra. We get

$$length_A(B) = length_A(\oplus_{i=1}^l R_i)=\sum_{i=1}^l length_A(R_i).$$

Hence $\# V(I) \leq length_A(B)$.


问:
PROBLEM Absolute Value Function and Coordinate Geometry
I have tried solving it, but I'm not sure if I did it right. If someone can help me with the complete solution. I'd highly appreciate

The equation I found was $y = 2x + 14$. I'm not sure if that was right.

A straight line passing through the points $(-5, 4)$ and $(2, 18)$, intersects an absolute value function with a vertex at $(-6, 22)$ that passes through the point $(5, -11)$ at points $A$ and $B$. Find the distance between points $A$ and $B$.

答:
How to find the absolute value function? You are given the vertex $(-6,22)$ and another point $(5,-11)$

Since vertex is given, the function is of the form

$$y = k(|x+6|) + 22$$

Here, the $k$ will change the slope of the absolute

Now, we know that the point $(5,-11)$ lies on it

$$\implies -11 = 11k + 22$$

$$\implies k = -3$$

Therefore the absolute value function is given as

$$y = -3|x+6| + 22$$

Now can you proceed?


答:
Firstly, we have to make sure that there are no positive integer value of $n$ such that:

$$2^{25 964 951} = 10^n$$

Cause if $2^{25 964 951}$ is equal $10...0$ then after minus $1$, number of digits will be different.

And that's true, cause:
$$log_{10}(2^{25 964 951}) = n$$
$$25964951\times log_{10}(2) = n$$

And $n$ will not be an integer.

So, $2^{25 964 951}$ and $2^{25 964 951} - 1$ have the same number of digits.

With the equation above, $n \approx 7816229$.

Then adding $1$, we have the results.

Title: How many digits does $2^{25964951}-1$ have?

Body:
Mersenne prime numbers are prime numbers of the form $2^n - 1 \,(\mathbb{N}\ni n>1)$such as
$2^2 -1$ or $2^{25964951}-1$: How many digits does the latter have?

I found this here:

$$\log_{10}(2^{25964951}-1)+1 =  7 816 230$$

Is this the way how to proof this or do I have to do something other?


问:
Can $RanA=KerA^T$ for a real matrix $A$? And for complex $A$?
My question is divided in two: $\DeclareMathOperator{\Ran}{Ran}\DeclareMathOperator{\Ker}{Ker}\DeclareMathOperator{\span}{span}$

$1$- Can $\Ran  A=\Ker A^T$ for real matrices?

$2$ - What happens with complex matrices?

As for the first part, I know there is a relation that states $\Ker A=(\Ker A^*)^\perp$ as a general case (for real and complex matrices), where $A^*$ is the hermitian adjoint and $\perp$ signifies the orthogonal complement. For real matrices, this translates to $\Ker A=(\Ker A^T)^\perp$, so I'm $99\%$ sure that the answer to this question is that $\Ran A=\Ker A^T$ is not possible for real matrices. But it would be nice to have this confirmed. Also, I have no clue how I would prove this without using the aforementioned relation. Is there another way to prove this?

This leads to the obvious question of what happens for complex matrices, which is the second part of my question. I have the following example:

$$A=\begin{pmatrix}1&i\\i&-1\end{pmatrix}$$

For this matrix, we can do row operations to reduce it to echelon form:

$$A_{e}=\begin{pmatrix}1&i\\0&0\end{pmatrix}$$

So, $\Ker A=\span\left\{\begin{pmatrix}-i\\1\end{pmatrix}\right\}$ and $\Ran A=\span\left\{\begin{pmatrix}1\\i\end{pmatrix}\right\}$

If we scale the basis of $\Ker A$ by $-i$, we can clearly see that $\Ker A=\Ran A$. As this matrix is symmetric, then $\Ker A=\Ker A^T=\Ran A=\Ran A^T$. So, all the fundamental subspaces of the matrix $A$ are the same. What is the condition for this to happen? Is it coincidence that the matrix is symmetric? I'm guessing it's not, but I'd be grateful if someone could shed some light into what the the reason for this is.

EDIT: I know there have been some similar questions asked in the past, but they still don't address the topic of complex matrices very much. I'm hoping this question can do so in a bit more detail.

答:
Suppose we have a vector $v$ with real components such that $v\in\operatorname{Ran}A$ and that $\operatorname{Ran}A\subseteq\operatorname{Ker}A^T$. Then $v=Aw$ for some $w$ and
$$
0=A^Tv=A^TAw
$$
implies $w^TA^TAw=0$, that is, $v^Tv=0$ and so $v=0$. Thus $\operatorname{Ran}A=\{0\}$ and thus $\operatorname{Ran}A=\operatorname{Ker}A^T$ is not possible (unless you allow $0\times0$ matrices).

Your idea is good for complex matrices: finding a $2\times2$ matrix having rank $1$ so that its column space coincides with the null space of the transpose. Thus the matrix has to be of the form
$$
A=\begin{pmatrix} a & ka \\ b & kb \end{pmatrix}
$$
and we need that
$$
\begin{pmatrix} a & b \\ ka & kb \end{pmatrix}\begin{pmatrix} a \\ b \end{pmatrix}=0
$$
that amounts to saying that $a^2+b^2=0$. Thus the choice of $a=1,b=i$ is the simplest one. Note that $k$ can be any complex number.

Since $\dim\operatorname{Ran}A=1$, the dimension of the null space of $A^T$ is $1$ as well. Hence they're equal.


问:
Permutation and combination #probability
I am currently studying Extension 1 Mathematics. I missed two classes and I figured out that tomorrow I will have a quiz. Can you help me to solve this permutation and combination question:

In how many ways 3 cards be selected from a pack of 52 playing cards
if:

(i) at least one of them is an ace;

(ii) not more than one is an ace.

In how many ways can 9 books be distributed amongst a man, a woman
and a child, if the man receives 4, the woman 3, and the child 2?

I know the solution (below), but I don't understand how to get there.

1 (i) 4804 (ii) 21808

2 1260

Thanks in advance!

答:
Hint: 1. (i) There are $\binom{52}{3}$ ways to draw $3$ cards. There are $\binom{48}{3}$ ways to draw $3$ cards without aces.

Hint: 1. (ii) There are $4$ aces and $\binom{48}{2}$ ways to choose $2$ cards without aces. Remember that no aces need be drawn.

Hint: 2. First think of the woman and child as one group and the man as another. How many ways are there to give $9$ books to them if the man gets $4$ and the group gets $5$? For each of those, how many ways can the $5$ books be given to the woman and child if the woman gets $3$ and the child gets $2$?

Bonus: Consider that $\binom{9}{5}\binom{5}{2}=\dfrac{9!}{4!\,3!\,2!}$ and how that relates to 2.

The answers in the book are correct.


问:
Continuous function on unit disc with $({\text{Re}}(z))^3$ on boundary?
I'm trying to answer the following question:

Let $D$ denote the open unit disc. Does there exist a continuous function $f:\overline{D}\rightarrow\mathbb{C}$ that is analytic on $D,$ and whose restriction to $\partial\overline{D}$ is the function $f|_{\partial\overline{D}}(z)=({\text{Re}}(z))^3$?

My instinct is no. Supposing there were, I would like to use the [maximum/minimum] modulus principle to state that $|f|$ does not attain its [maximum/minimum] value within $D.$ Because $|f|$ on the boundary has maximum value $1$ and minimum value $0,$ this implies that $0<|f(z)|<1$ for all $z\in D.$ Are we then able to find some $z\in D$ that has $f(z)=0,$ obtaining a contradiction?

I'm a bit stuck. Any hint/guidance is appreciated. Thanks!

答:
$|\operatorname{Im} f|$ attains its maximum on $\partial D$. But $|\operatorname{Im} f|=0$ on $\partial D$, hence $f$ takes on only real values and is therefore constant, a contradiction.

(To prove that $|\operatorname{Im} f|$ attains its maximum on $\partial D$ just copy the proof of the maximum modulus theorem using the open mapping theorem)


问:
Challenging Integrals for High School Students
I am now in my last year of high school. We have covered all the techniques useful for indefinite integration that are included in our Maths and Further Maths courses. This includes:

Integration by parts, inspection, substitution, partial fraction decomposition

Integration of regular and inverse trigonometric, regular and inverse hyperbolic, exponential, logarithmic, polynomial functions

I would like to have some challenging integrals to attack that are possible for me to solve at my current level of knowledge. By challenging, I mean integrals similar to the ones in this document. They were generally enjoyable and very satisfying to solve. If you have an integral that you think I could do that is more challenging than those in the aforementioned link, so much the better.

Thank you for your suggestions.

答:
Since you asked for integrals, here is an integral! This one I found quite challenging, but perhaps you'll find a more elegant way of solving it than I did:
$$
\int \frac{dx}{(9-36x^2)^{3/2}} \, .
$$
NB this definitely can be solved using the methods outlined in your post.


问:
Application of Mean Value Theorem on 2 differentiable functions simultaneously
MCQ question

Let $f$ and $g$ be two differentiable functions on $(0,1)$ such that $f(0)=2$, $f(1)=6$, $g(0)=0$ and $g(1)=2$

Then there exists $\theta \in (0,1)$ such that $f'(\theta)$ equals

(a) $1/2 g'(\theta)$

(b)$2g'(\theta)$

(c) $6g'(\theta)$

(d) $1/6 g'(\theta)$

Now I know that we have to apply Mean Value Theorem here. But my question is wouldn't we get two different $\theta$ that is $\exists\, \theta_{1}$ such that $f'(\theta_{1})=4$ and $\exists\, \theta_{2}$ such that $g'(\theta_{2})=2$

So, how can option (b) be the right option unless $\theta_{1}=\theta_{2}$?

答:
Hint

Apply Rolle's theorem to the function $f-2g$.


问:
Given f(x,y), x(t,u) and y(t,u), how do I intuitively calculate the second partial derivative with respect to u?
I'm a bit confused when dealing with higher order derivatives. Suppose we have a function f(x,y) where x(t,u) and y(t,u).

Can someone help me understand how I should think about solving $\frac{\delta^2f}{\delta x^2}$ or $\frac{\delta^2f}{\delta u^2}$?

I've read my litterature but at some point I started getting confused by the indexing, and I feel now that I don't understand it fully anymore.

I'll walk you through my though process.
When doing the second partial derivative of $\frac{\delta^2f}{\delta u^2}$, my first though is to do $\frac{\delta f}{\delta u}(\frac{\delta f}{\delta u})$, where $\frac{\delta f}{\delta u}=\frac{\delta f}{\delta x}\frac{\delta x}{\delta u}+\frac{\delta f}{\delta y}\frac{\delta y}{\delta u}$ which can be written as: $\frac{\delta x}{\delta u}*f_1+\frac{\delta y}{\delta u}*f_2$, correct? And now to get $\frac{\delta^2f}{\delta u^2}$ one would simply need to compute the result of $\frac{\delta f}{\delta u}(\frac{\delta x}{\delta u}*f_1+\frac{\delta y}{\delta u}*f_2)$, correct?

It is here I start to get confused. When reviewing an example from the litterature they seem to do the opposite of an exercise that I later solved, and my head starts to spin because of all the indexing and so on... But this is what I think I should do...
$\frac{\delta x}{\delta u}(\frac{\delta x}{\delta u}*f_{11}+\frac{\delta x}{\delta t}*f_{12})+\frac{\delta y}{\delta u}(\frac{\delta y}{\delta u}*f_{21}+\frac{\delta y}{\delta t}*f_{22})$

And why is becuase when I review the indexing, $f_{11}$ tells me that I should derivate with respect to the function $f$s first variable, $x$, and again with respect to the first variable of $x$ which is $u$, but is my thinking correct?

So my question is, what is $\frac{\delta^2f}{\delta x^2}$ and $\frac{\delta^2f}{\delta u^2}$ and how can I memorize the steps requiered to compute these things in the future?

EDIT:

Can I perhaps use the fact that $\frac{\delta^2 f}{\delta u^2}=(\frac{\delta f}{\delta x}\frac{\delta x}{\delta u}+\frac{\delta f}{\delta y}\frac{\delta y}{\delta u})(\frac{\delta f}{\delta x}\frac{\delta x}{\delta u}+\frac{\delta f}{\delta y}\frac{\delta y}{\delta u})$  and then simply expand with taking the products of the terms?

If I expand upon this I get:
$\frac{\delta f}{\delta x}\frac{\delta x}{\delta u}(\frac{\delta f}{\delta x}\frac{\delta x}{\delta u}+\frac{\delta f}{\delta y}\frac{\delta y}{\delta u})+\frac{\delta f}{\delta y}\frac{\delta y}{\delta u}(\frac{\delta f}{\delta x}\frac{\delta x}{\delta u}+\frac{\delta f}{\delta y}\frac{\delta y}{\delta u})$.

I can evaluate $\frac{\delta f}{\delta x}\frac{\delta f}{\delta x}=f_{11}$ and the others also get evaluated to such terms, which simplifies it to:

$\frac{\delta x}{\delta u}(f_{11}\frac{\delta x}{\delta u}+f_{12}\frac{\delta y}{\delta u})+\frac{\delta y}{\delta u}(f_{21}\frac{\delta x}{\delta u}+f_{22}\frac{\delta y}{\delta u})$

This is however in contradiction with what I guessed before. Furthermore, if this is correct (which I percieve it to be), how do I then now proceed to derivate these terms now? Given that I have formulas for $x(t,u)$ and $y(t,u).$

答:
$$\frac{\partial f}{\partial u} = \frac{\partial f}{\partial x}\frac{\partial x}{\partial u} + \frac{\partial f}{\partial y}\frac{\partial y}{\partial u}$$

$$\implies \frac{\partial^2 f}{\partial u^2} = \frac{\partial}{\partial u}\left(\frac{\partial f}{\partial x}\right)\frac{\partial x}{\partial u} + \frac{\partial f}{\partial u}\frac{\partial ^2 x}{\partial u^2} + \frac{\partial}{\partial u}\left(\frac{\partial f}{\partial y}\right)\frac{\partial y}{\partial u} + \frac{\partial f}{\partial u}\frac{\partial ^2 y}{\partial u^2}$$

$$\implies \frac{\partial^2 f}{\partial u^2} =  \left(\frac{\partial ^2f}{\partial x^2}\frac{\partial x}{\partial u}+ \frac{\partial ^2f}{\partial x \partial y}\frac{\partial y}{\partial u}\right)\frac{\partial x}{\partial u} + \frac{\partial f}{\partial u}\frac{\partial ^2 x}{\partial u^2} + \left(\frac{\partial ^2f}{\partial x \partial y}\frac{\partial x}{\partial u}+ \frac{\partial ^2f}{\partial y^2}\frac{\partial y}{\partial u}\right)\frac{\partial y}{\partial u} + \frac{\partial f}{\partial u}\frac{\partial ^2 y}{\partial u^2}$$


问:
Calculate the limit of the multivariable function using the definition
I want to show that the function $e^{x+2y}$ is differentiable in the point $(2,2)$ by using the definition.

Definition:
A multivariable function is differentiable in a point a if there exist constants $A_{1} + \dotsb + A_{n}$ and a function $\varrho(h)$ such that
$$f(a+h)-f(a)=A_{1}h_{1}+ \dotsb+ A_{n}h_{n}+ |h|\varrho(h)$$
and $\varrho(h)\to 0$ as $h \to (0, 0)$.

I find the expression for $\varrho(h)$ to be the following:

$$\frac{e^{h+2k}-1-h-2k}{\sqrt{h^2+k^2}}$$

I write the function in terms of polar coordinates ($x= r \cos \theta, y =r \sin\theta$) and obtain:

$$\frac{e^{r \cos\theta +2r\sin\theta}-1-r \cos \theta-2r\sin\theta}{r}$$

Using the squeeze theorem I obtain:

\begin{align}
\left|\frac{e^{r \cos \theta +2r\sin\theta}-1-r\cos \theta-2r\sin\theta}{r}\right|&\leq \left|\frac{e^{r+2r}+r+2r}{r} \right|\\
&= \frac{e^{r+2r}}{r} + \frac{r}{r}+ \frac{2r}{r}\\
&= \frac{e^{r+2r}}{r} +3
\end{align}

The limit of this as $r \to 0$ is $3$, but the answer should be $0$. I don't know where I went wrong.

Any help would be much appreciated!

答:
Firstly your expression for $\rho(h)$ is not correct since it is missing a factor $e^6$. By your definition

$$ \rho(h) = \frac{f(a+h)-f(a)-A_1h_1-A_2h_2}{\vert h \vert } = \frac{e^{(2+h_1)+2(2+h_2)} - e^6- A_1h_1-A_2h_2}{\vert h \vert}$$

where the constants are $A_1 = \frac{\partial f }{\partial x}(2,2) = e^6$ and $A_2 = \frac{\partial f}{\partial y}f(2,2) = 2e^6.$
Hence

$$ \rho(h) = e^6\frac{e^{h_1+2h_2}-1-h_1-2h_2}{\sqrt{h_1^2+h_2^2}}.$$

Regardless we still need to compute the limit as $h \to 0 $ of

$$\frac{e^{h_1+2h_2}-1-h_1-2h_2}{\sqrt{h_1^2+h_2^2}}$$

as you did.

Secondly $\frac{e^{r+2r}}{r} + 3$ does not converge to $3$ since $e^{3r}/r$ diverges as $r \to 0$. Even if the limit was $3$ the squeeze theorem does not allow you to conclude that

$$ \lim_{(x,y) \to 0} \frac{e^{x+2y} -1 -x -2y}{\sqrt{x^2 + y^2}} = 3.$$

To use the squeeze theorem here you need a bound which converges to $0$.

Here's a solution to the problem.

Remember that $P(x) = 1 + x$ is the first order Taylor polynomial for $e^x$ at $0$. Hence by definition we have

$$ \lim_{x \to 0}\frac{e^x - P(x)}{x} = 0$$
i.e.

$$ \forall \epsilon , \exists \delta : \vert x \vert \leq \delta \implies \vert e^x - P(x)\vert \leq \epsilon \vert x \vert $$

Notice that

$$ \left\vert\frac{e^{x+2y} -1 - x -2y}{\sqrt{x^2+y^2}} \right\vert= \left\vert \frac{e^{x+2y} - P(x+2y)}{\sqrt{x^2+y^2}}\right\vert.$$

Let $\epsilon > 0$ and take $\vert (x,y) \vert < \tilde \delta$ with $\tilde \delta$ small enough to guaranty that $\vert x+ 2y\vert< \delta$ and we have

$$ \left\vert\frac{e^{x+2y} -1 - x -2y}{\sqrt{x^2+y^2}} \right\vert \leq \epsilon \frac{\vert x+2y\vert }{\sqrt{x^2+y^2}} \leq \epsilon \frac{3r}{r} = 3\epsilon$$

Therefore by definition we have that

$$ \lim_{(x,y) \to 0} \frac{e^{x+2y}-1 -x-2y}{\sqrt{x^2+y^2}} = 0$$

and $e^{x+2y}$ is differentiable at $(2,2)$


问:
Limit with modified Bessel function
I'm trying to determine the limit for $x\to \infty$ of the following expression:
$$
x-\frac{x-1}{x}2e^{-1/x}K_1(2/x)=x \left[1- \frac{x-1}{x^2}2e^{-1/x}K_1(2/x) \right],
$$
where $K_1(\cdot)$ denote the modified Bessel function (e.g., https://dlmf.nist.gov/10.25). My attempt:

Using the limiting form in 10.30.2 provided at https://dlmf.nist.gov/10.30, I would obtain
\begin{equation}
\begin{split}
2e^{-1/x}K_1(2/x) &\sim 2e^{-1/x} \frac{1}{2}\Gamma(1)\left(\frac{2/x}{2} \right)^{-1}\\
&=e^{-1/x} x\\
&\sim \left(1-\frac{1}{x} \right) x
\end{split}
\end{equation}
as $x \to \infty$. This would leave me with
\begin{equation}
\begin{split}
x \left[1- \frac{x-1}{x^2}2e^{-1/x}K_1(2/x) \right]&\sim x \left[1- \frac{x-1}{x^2}\left(1-\frac{1}{x} \right) x\right]\\
&=x\left[1- \frac{1}{x}\left(1-\frac{1}{x} \right)^2 x\right]\\
&= x \left[ \frac{2}{x}-\frac{1}{x^2}\right]\\
&\to 2.
\end{split}
\end{equation}

QUESTION: Is the above result correct? My intuition and some attempts to plot the function would have suggested me that the limit is $0$, so I'm wondering whether there's something I'm missing in the above reasoning. Any help is appreciated.

答:
Let $x=\frac 2y$ to make the expression
$$\frac{2}{y}+e^{-y/2} (y-2) K_1(y)$$ Expanding around $y=0$ gives
$$2-y \left(\log (y)+\gamma +\frac{1}{4}-\log (2)\right)+O\left(y^2\right)$$

Title: Limit with modified Bessel function


问:
Find χ(G) (Chromatic number of the graph)
Let G be a graph with vertex set V = {1, . . . , 20} Vertices u, v ∈ V are adjacent if and only if gcd(u, v) = 1. Find χ(G) and explain your answer.

I couldn't finish this answer all the way, so any help would be appreciated

答:
The answer is $9$.

$u, v \in V$ being adjacent if and only if $\gcd(u, v) = 1$ is equivalent to saying that $u$ and $v$ are not connected provided that they share a non-trivial common factor.

In particular, we can color all vertices that are multiples of $2$ with one common color since they will share a common factor (and so, they won't be directly connected). Among the remaining numbers, we can paint the multiples of $3$ a second color since they will share a common factor. Continuing in this fashion, we end up requiring exactly one color for each prime number. If you are familiar with the Sieve of Eratosthenes, this process is identical.

We require eight colors for each prime number in $V = \{1, 2, \ldots, 20\},$ and we require one final color for Vertex $1$ since $\gcd(1, x) = 1$ for all integers $x$.


问:
Is the sum of the first N Laguerre polynomials (with alternating signs) always positive?
I have noticed that the following simple sum of Laguerre polynomials (weighted with alternating signs) seems to be positive for any $N$ when $x>0$:

$$\sum_{k=0}^{N}\;(-1)^{k}\;L_{k}(x)$$

More precisely, when $N$ is even, the sum is positive $\forall\;x$ and when $N$  is uneven, the sum has only one zero in $x=0$.

I have checked this numerically up to $N=10{}^{2}$.

I cannot find any reference to this sum in the literature. Are my observations correct?   If this is a known result, could someone point me in the right direction to find a reference or some documentation for this?

答:
Maybe some help on the way but not a full answer.

According to Wikipedia, Laguerre polynomials build a Sheffer sequence

$$\frac{\partial }{\partial x} L_n = \left(\frac{\partial }{\partial x}-1\right)L_{n-1}$$

Which can be rewritten as :

$$L_{n-1} = \frac{\partial }{\partial x}(L_{n-1}-L_n)$$

So by the fundamental theorem of calculus we can say

$$\int_0^xL_{n-1}(t)dt = L_{n-1}(x)-L_n(x)$$

So your sum is a pairwise sum of such integrals.

Remains to be proven that sum of such integrals for even $n-1$ are positive.


问:
Can Linear Constant coefficient Difference Equations always represent an LTI system?
Can we say that a Linear Constant Coefficient Difference Equation can always represent a Linear Shift Invarient system ? Are there any conditions which need to be satisfied additionally by these kind of equations to be able to do that? </p>

答:
WHat about final rest condition? (i.e if x(t) = 0, for t > t0 then y(t) is also 0 for t > t0 )
whether now the differential equation satisfies LSI or not?</p>


问:
Compute $\int_{-\infty}^{\infty}\frac{x\sin(\pi x)}{(x-3)(x-2)}dx$ using residue thoerem
I am trying to compute $\int_{-\infty}^{\infty}\frac{x\sin(\pi x)}{(x-3)(x-2)}dx$ using the residue theorem. To do so, I am integrating the function $f(z)=\frac{ze^{i\pi z}}{(z-3)(z-2)}$ over a the frontier of $\{z:|z|<R,Im(z)>0\}-(\{z:|z-2|<r\}\cup\{z:|z-3|<r\})$ (a semicircle that "avoids" the singularities of $f$).

I have used Jordan's lemma to prove that the integral over the big semicircle is null when $R\longrightarrow \infty$. My problem is I do not know how to compute the integral over the small semicircles when $r\longrightarrow 0$. How could I approach this?

Thanks in advance.

答:
For a pole $z=\alpha_i$ lying on $X-$axis in the contour you described, the value of the integral:

$\int_{C_{r_i}}f(z)dz=-\pi i\times Res[f(z), z=\alpha_i]$ where $C_{r_i}:|z-z_i|<r_i$ traced clockwise in upper half plane.

The negative sign accounts for the clockwise direction you trace on the $X-$axis and $\pi i$ rather than $2\pi i$ is due to semicircle.

ADDED- $f(z)$ has simple pole at $z=\alpha$. From Laurent's series

$f(z)=\frac{a_{-1}}{z-\alpha}+g(z)$

Integrate both sides and parametrize by circle $C_r:z-\alpha=re^{i\theta},0<\theta<\pi$ to get

$\int_{C_r}f(z)dz={a_{-1}}\int_0^\pi id\theta +\int_0^\pi g(\alpha+re^{i\theta})re^{i\theta}d\theta$

As the second integral goes to zero as $r\rightarrow 0$ using  ML-inequality

($\because |g(\alpha+re^{i\theta})|\le M\implies |I_2|\le \pi rM$)

$\int_{C_r}f(z)dz=\pi i\times Res[f(z), z=\alpha]$


问:
show that the set $ A'=\left\{ x:x\in\partial(A\setminus\{x\}\right\} $is closed
I'm studying Calculus 3, and our teacher tends to give us extremely difficult questions.
one of those questions was this:
"let $A \subseteq \Bbb R^k $. show that the set $ A' = \{ x \in \Bbb R^k\mid x \in \partial(A \setminus \{ x \})\}$ is closed".
I've tried to show using sequence characterization or showing that $(A')^c$ is open, but I don't have a clue of how to start, or even if I'm in the right direction.
I could really use some help or a guiding hand with this.

答:
Since you are looking at the boundary of $B = A-{x}$, the point $x$ is the point not in $B$ for any ball around $x$. So if $x$ is in the boundary of $B$ then $x$ is a limit point of $A$ since in the definition of limit point, every neighbourhood of the point $x$ in question must contain a point of the set $A$ other than $x$ itself. The set of all limit points of a set $S$ is called the derived set of $S$.

So based on how you have defined the boundary in your class the set you are analyzing $\{x\in \mathbb{R}^{k}| x\in A\setminus{\{x\}}$ is called the derived set of $A$.

So if you can use the theorem:

If $X$ is a $T1$ space, the derived set of every subset of $X$ is closed in $X$.

Then the derived set of $A$ is closed. If not we can elaborate that proof a bit more.

Let's call the derived set of $A$, $A'$.

Let q be a limit point of A′. If it is proved that $q \in A′$, then the proof is done.

Let $G_{q}$ be the open set containing q. Since q is a limit point of A′,$G_{q}$ contains at least one point $r \in A′$ different from q. But $G_{q}$ is an open set containing $r \in A′$; hence $G_{q}$ contains infinitely many points of A.

So there exist $a \ in A$ such that $a\neq q,a\neq r$ and $a\in G_{q}$. That is,each open set containing q contains infinitely many points of A. Hence $q\in A′$.


问:
Are we able to compute the probability of the given function to be positive?
I have the following function
$$   \cos \frac{n x}{2}\;\cos \frac{(m-n)x}{2} \;\bigg( 2\cos \frac{m x}{2} +\cos \frac{(2n-m)x}{2}\bigg)       $$
with three parameters $m,n,x>0$ where $m>n$. These parameters are continuous and distributed uniformly.

Then, is it possible to compute the probability of which this function is negative or positive, at least for fixed values of $m$ and $n$, say $n=1$ and $m=3$?

In other words, we know that $\sin mx$ for a random value of $x$ is $50$ percent positive and $50$ percent negative, so the probability of being positive and negative is $\frac{1}{2}$. I mean can we say something like this for the given function?

答:
You cannot have a uniform distribution over an infinite interval, so you seem to be looking for some limit of the proportion. It seems $m$ and $n$ (or at least their ratio) do make a difference: if $m=n$ then the expression is never negative, while otherwise it seems to be negative for a relatively small proportion of the possible $x$.

If $\frac mn$ is rational the expression is periodic and you can do calculations over a cycle.  You might start by looking at when the expression is $0$, which are in effect double roots, and the gaps between the roots, noting that the expression is $3$ when $x=0$ and at the end of the cycle.

So for example if $m=3$ and $n=1$ then the cycle is of length $2\pi$, and the roots in $[0, \frac2\pi)$ are:

$x=\cos^{-1}\left(\frac14\right)$

$x=\frac12 \pi$

$x=\pi$ (a double root)

$x=\frac32 \pi$

$x=2\pi-\cos^{-1}\left(\frac14\right)$

so the proportion of the cycle for which this expression is positive is $\frac12 + \frac1{\pi}\cos^{-1}\left(\frac14\right) \approx 0.91957$ and is negative is $\frac12 - \frac1{\pi}\cos^{-1}\left(\frac14\right) \approx 0.08043$


问:
Holomorphic functions with same restrictions on $\mathbb{R}$ - Paley-Wiener's Theorem
The Theorem has two statements; the first one says that (proved):

If $f:\mathbb{R}\to\mathbb{C}$ is such that $f\in C(\mathbb{R})\cap L^2(\mathbb{R})$, and $Support(\hat{f})\subset[-\sigma,\sigma],$ then $\tilde f(x+iy):=\frac{1}{\sqrt{2\pi}}\int_{-\sigma}^{\sigma}\hat{f}(\lambda)e^{i\lambda(x+iy)}d\lambda$ $\forall x,y\in \mathbb{R}$ is an holomorpic extension of $f$ and $|\tilde f(x+iy)|\leq C e^{\sigma |y|}$ with $C$ and $\sigma>0$.

The second one says that ( to prove):

If $f:\mathbb{C}\to\mathbb{C}$ is such that it is holomorphic, $|f(x+iy)|\leq C e^{\sigma |y|}$ with $C$, $\sigma>0$ and $f|_{\mathbb{R}}\in L^2(\mathbb{R})$ $\Rightarrow$

$f(x+iy)=\frac{1}{\sqrt{2\pi}}\int_{-\sigma}^{\sigma}\hat{f|_{\mathbb{R}}}(\lambda)e^{i\lambda(x+iy)}d\lambda$ a.e.$\forall x,y\in \mathbb{R}.$

I know that $f(x)=\frac{1}{\sqrt{2\pi}}\int_{-\sigma}^{\sigma}\hat{f|_{\mathbb{R}}}(\lambda)e^{i\lambda(x)}d\lambda$ a.e. $\forall x\in \mathbb{R}$ by using the fact that $Support(\hat{f|_{\mathbb{R}}})\subset[-\sigma,\sigma]$ and the inverse Fourier transform for real variable function.

I think to prove the equality in $\mathbb{C}$ by showing that

$|f(x+iy)-\frac{1}{\sqrt{2\pi}}\int_{-\sigma}^{\sigma}\hat{f|_{\mathbb{R}}}(\lambda)e^{i\lambda(x+iy)}d\lambda|\leq  2\sigma C||f||_{2,{\mathbb{R}}}e^{\sigma|\rho|},$ $\forall x+iy\in B_{\mathbb{C}}(0,\rho)$

and using the Liouville's Theorem to obtain that by the boundedness, the holomorphic property and the fact that their restrictions to real numbers are the same function, then the two functions are equal in $B_{\mathbb{C}}(0,\rho)$, $\forall \rho>0$.
I'm not sure if that works; my skills in complex analysis are very poor.

答: I think that I found the answer: the identity principle holomorphic functions says that except for the null function, a holomorphic function always has isolated zeroes.</p>


问:
REVISITED $^2$: Does a solution in $\mathbb{R}^n$ imply a solution in $\mathbb{Q}^n$?
Let $A ∈ M_{m\times n}(\mathbb{Q})$ and $B ∈ \mathbb{Q}^m$. Suppose that the system of linear
equations $AX = B$ has a solution in $\mathbb{R}^n$. Does it necessarily have a solution
in $\mathbb{Q}^n$?

Where do I start?

I feel I can use this to help:

$$x_i=\frac{1}{a'_{ii}}\left(b'_i - \sum_{j=i+1}^{k} a'_{ij} x_j \right).$$

Here is a thought that just crossed my mind, but I'm not sure if this is legal:

Suppose $Y\in \{\mathbb{R}\setminus \mathbb{Q}\}^n$ and $Y'\in\mathbb{Q}^n$ are solutions to $AX=B$. Then $AY=AY'=B$, or $A(Y-Y')=A(Y+(-Y'))=0$, but  addition between $\{\mathbb{R}\setminus \mathbb{Q}\}^n$ and $\mathbb{Q}^n$ is . . .

答:
Of course! $\mathbb{Q}$ is a field, therefore if there is solution to linear system, the solutions will be rational.

You can check it by writing the explicit formulas for the solution -- for example, the ones you provided as help or Cramer's rule -- and notice that the formulas involves only additions and multiplications of rationals.

So if your system has a solution in $\mathbb{R}^n$, they are in facts in $\mathbb{Q}^n$. And if there is no solution in $\mathbb{R}^n$, there cannot be one in $\mathbb{Q}^n$.

It is impossible that $Y \in (\mathbb{R}∖\mathbb{Q})^n$ and $Y′∈\mathbb{Q}^n$ are solutions of the same linear system.

The crucial point is that the system is linear. The answer is totally different for systems of polynomials equations.


问:
Motion on a slope - A Level
A particle P of mass 5kg and particle Q of mass 1kg are connected by a light inextensible string. P lies on a slope inclined at a 60 degree angle to the horizontal. The string passes from P parallel to the line of greatest slope, and runs over a light pulley that can freely rotate. Then descends vertically to Q. The coefficient of friction between P and the slope is 0.7.
Find the force of friction acting on P and it’s direction and find the acceleration of P when the system is released from rest.

I was able to find the frictional force 17.15N up the slope) but I can’t find the acceleration. I have tried to use the equation 5gsin(60)-17.15-9.8 = 5a with 9.8 being the weight of Q, but I get the answer of 3.1m/s/s for a, bit the mark scheme says that a (acceleration) is 2.58m/s/s but I don’t see how. Could you please give me some advice on how to get 2.58?

答:
You have done all correct, you just messed up in the calculation. The equation
$$5g\sin60° -17.15-9.8=6a$$
has the solution $\boxed{a=2.58\,\text{m}\,\text{s}^{-2}}$ as required.

Hope this helps. Ask anything if not clear :)

Title: Motion on a slope - A Level


问:
A question about splitting fields.
Let $E_{1}$ and $E_{2}$ be two splitting fields of polynomial $p(x)\in \mathbb{F[x]}$ over $\mathbb{F}$.

My textbook has a long proof for proving that $E_{1}$ and $E_{2}$ are isomorphic.

But isn't this obvious? Aren't $E_{1}$ and $E_{2}$ equal, as both of them are the smallest field extensions of field $\mathbb{F}$ such that all the roots of $p(x)$ are contained within them?

EDIT: Obviously $E_{1}$ and $E_{2}$ can be constructed differently. But when they're fully constructed, they have to be the very same field! Hence, even if we considered different basis sets for them, $[E_{1}:F]=[E_{2}:F]$.

答:
Trying to shed some light to this by proffering several splitting fields of the polynomial $x^2+1$ over the rationals.

One splitting field is the field $\mathbb{Q}[i]$, i.e. the usual subfield of complex numbers.

Another splitting field is the quotient field $\mathbb{Q}[x]/\langle x^2+1\rangle$ of the ring of polynomials with rational coefficients by the maximal ideal generated by the polynomial $x^2+1$.

Let us consider the skewfield of Hamiltonian quaternions
$$
\mathbb{H}=\{a+bi+cj+dj\mid a,b,c,d\in\mathbb{R}\},
$$
where the usual relations are satisfied: $i^2=j^2=k^2=ijk=-1$. Let us pick any triple $(b,c,d)$ of real numbers such that $b^2+c^2+d^2=1$. Then the quaternion
$$
u=bi+cj+dk
$$
satisfies the equation $u^2=-1$. It is easy to see that the subset
$$
\mathbb{Q}[u]=\{a+fu\in\mathbb{H}\mid a,f\in\mathbb{Q}\}
$$
is then a splitting field of $x^2+1$ over the rationals.

The set of matrices of the form
$$
\left(\begin{array}{rr}a&b\\-b&a\end{array}\right),
$$
with $a,b$ arbitrary rational numbers is yet another splitting field of $x^2+1$.

No two of these splitting fields are equal in an obvious sense. They are all isomorphic to each other, though.

Another example. You often think of the splitting field of $x^2-2$ over the rationals as $\mathbb{Q}[\sqrt2]$ - a subset of the real numbers. So then $\sqrt2$ may be a certain equivalence class of Cauch sequences. Why is that particular equivalence class of Cauchy sequence equal to the coset of $x$ in the quotient ring $\mathbb{Q}[x]/\langle x^2-2\rangle$?

IOW. This questions is philosophical in the sense that we really are talking about the identity of the roots. Not just up to isomorphism!

Title: A question about splitting fields.


问:
probability problem for three events
there are three newspapers A, B and C in a certain city. below are the percentages how the people read the newspapers:

A : 50

B: 25

C: 35

A & B: 10

A & C: 8

B & C: 5

A & B & C: 3

a) probability to read at least on of the newspapers?

$$P(A \cup B\ \cup C)=P(A)+P(B)+P(C)-P(A \cap B)-P(A \cap C)-P(B \cap C)+P(A \cap B \cap C)$$
$$P(A \cup B\ \cup C)=.5+.25+.35-.1-.08-.05+.03=.9$$
b)probability that a person reads B or C but not A.
$$P(B \cap C \cap A^c ) = ?$$
I draw a Venn diagram and I got $0.02$ as result, but I do not khow to show this using set properties.

c)probability that the person reads B OR C given he reads A.

$$P(B \cup C | A )=\frac{P(B \cup C \cap A)}{P(A)}$$
again, using Venn diagran I got $\frac{15}{50}$ but I do not know how to get it using set properties.

I would appreciate help with this.

答:
Probability for two sets is given by following rule

$$ P(A \cup B) = P(A) + P(B) - P(B \cap C)$$

Now since, we know that sets $$ (B \cap C) \cap A^C $$ and $$ (B \cap C) \cap A $$ are mutually disjoint and further their union is equal to $$ (B \cap C) $$ i.e. $$ [(B \cap C) \cap A^C]
 \cup  [(B \cap C) \cap A^C] = (B \cap C) $$ we can get

$$ P(B \cap C) = P([(B \cap C) \cap A^C]) + P([(B \cap C) \cap A])$$

which can be solved to
$$ P(B \cap C \cap A^C) = P(B \cup C) - P(B \cap C \cap A) $$ and thus solving your first problem.

Now your second problem can be solved by distributive law, which states that

$$ (B \cup C) \cap A  = (B \cap A) \cup (C \cap A)$$
but since
$$ (B \cap A) \cap (C \cap A) = A \cap B \cap C $$, we can get
$$ P[(B \cap A) \cup (C \cap A)] = P((B \cap A) + P(C \cap A) - P[(B \cap A) \cap (C \cap A)]$$
which can solve your second problem.

Title: probability problem for three events


问:
How do i calculate this multivariable limit?
I am really uncertain about how to approach this but here's my attempt:

Let $$x=(x_1,x_2,x_3), \; \mathopen|x\mathclose|=(x_1^{2}+x_2^{2}+x_3^{2})^{\frac{1}{2}}$$

If the following limits exist, find their value:

$$\lim_{x \to 0} \frac{ln(1+\mathopen|x\mathclose|^2)}{\mathopen|x\mathclose|^2+sin(x_1x_2x_3)}$$

My attempt is simply to estimate an upper bound, and then using the standard limit $\lim_{x \to 0}\frac{ln(1+x)}{x}=1$,

$$\frac{\mathopen|ln(1+\mathopen|x\mathclose|^2)\mathclose|}{\mathopen|\mathopen|x\mathclose|^2+sin(x_1x_2x_3)\mathclose|}\leq\frac{ln(1+\mathopen|x\mathclose|^2)}{\mathopen|x\mathclose|^2}$$
Because the $ln$ function with that input is positive, and the $sin$ function takes on a max value of $1$. But, $$\mathopen|x\mathclose|^2\rightarrow0 \;\text{when}\;x\rightarrow0$$ so $$\frac{ln(1+\mathopen|x\mathclose|^2)}{\mathopen|x\mathclose|^2}\rightarrow1$$

Body:
Hint: Use AM-GM inequality and $|\sin(t)|\leq|t|$ to bound $|\sin(x_{1}x_{2}x_{3})|$ in terms of $|x|$.

Here's a full solution: factor the equation into
$$
\frac{\ln(1+|x|^{2})}{|x|^{2}} \times \frac{|x|^{2}}{|x|^{2} + \sin(x_{1}x_{2}x_{3})}
$$
where the first part goes to 1 as $|x| \to 0$. For the second part, AM-GM inequality gives
$$
|x|^{2} = x_{1}^{2} + x_{2}^{2} + x_{3}^{2} \geq 3 (x_{1}x_{2}x_{3})^{2/3}
$$
so we have
$$
|\sin(x_{1}x_{2}x_{3})| \leq |x_{1}x_{2}x_{3}| \leq \left(\frac{|x|^{2}}{3}\right)^{2/3} = \frac{|x|^{3}}{3^{2/3}} = O(|x|^{3}).
$$
Hence the second part also converges to 1 since
$$
\Bigg| \frac{|x|^{2} + \sin(x_{1}x_{2}x_{3})}{|x|^{2}} - 1\Bigg| = \frac{|\sin(x_{1}x_{2}x_{3})|}{|x|^{2}} \leq \frac{|x|}{3^{2/3}} \to 0.
$$
So the answer is 1.


问:
Show that the error of quadratic interpolation $\xi$ is given by $\xi\le|E|+\frac{5}{4}\epsilon$
Show that the error of quadratic interpolation $\xi$ in equidistant points, the values of which are obtained by rounding and the error is lesser or equal to $\epsilon$, is given by:

$$\xi\le|E|+\frac{5}{4}\epsilon$$

I got as far as
$$\prod^n_{i=0} (x-x_i)\frac{f^5}{5!} \le |E|+\frac{5}{4}\epsilon$$

I don't know what |E| is. Help?

答:
You know that $|\hat y_i-f(x_i)|\le ϵ$, where $\hat y_i$ are the rounded function values.

Now
\begin{alignat}{1}
\left|f(x)-\sum_{i=0}^n\hat y_iL_i(x)\right|
&\le \left|f(x)-\sum_{i=0}^n  f(x_i)L_i(x)\right|&+\sum_{i=0}^n |\hat y_i-f(x_i)|·|L_i(x)|\\
&\le |E(x)|&+ϵ\sum_{i=0}^n|L_i(x)|
\end{alignat}
where $L_i(x)$ is the Lagrange interpolation kernel for position $x_i$.
The graph of the sum in the last term for $x_i=i$, $i=0,1,2=n$ is


答:
Let $F(t,u)=f(x(t,u),y(t,u))$.

Then
$F_u=f_x x_u  + f_y y_u$ by the Chain Rule.

Note that this is a sum of two products, so we use the Sum, Product rules to get
$$
F_{uu}=
(f_x)_u x_u +f_x x_{uu}
+ (f_y)_u y_u + f_y y_{uu}.
$$
Now use the Chain Rule twice more to get
$$
F_{uu}=
(f_{xx} x_u +f_{xy} y_u ) x_u +f_x x_{uu}
+ (f_{yx} x_u +f_{yy}y_u   ) y_u + f_y y_{uu}.
$$

Collect the terms and we  have

$$
F_{uu}= f_{xx}x_u^2 +2 f_{xy} x_u y_u + f_{yy} y_u^2 +f_x x_{uu} + f_y y_{uu}.
$$

Title: Given f(x,y), x(t,u) and y(t,u), how do I intuitively calculate the second partial derivative with respect to u?

Body:
I'm a bit confused when dealing with higher order derivatives. Suppose we have a function f(x,y) where x(t,u) and y(t,u).

Can someone help me understand how I should think about solving $\frac{\delta^2f}{\delta x^2}$ or $\frac{\delta^2f}{\delta u^2}$?

I've read my litterature but at some point I started getting confused by the indexing, and I feel now that I don't understand it fully anymore.

I'll walk you through my though process.
When doing the second partial derivative of $\frac{\delta^2f}{\delta u^2}$, my first though is to do $\frac{\delta f}{\delta u}(\frac{\delta f}{\delta u})$, where $\frac{\delta f}{\delta u}=\frac{\delta f}{\delta x}\frac{\delta x}{\delta u}+\frac{\delta f}{\delta y}\frac{\delta y}{\delta u}$ which can be written as: $\frac{\delta x}{\delta u}*f_1+\frac{\delta y}{\delta u}*f_2$, correct? And now to get $\frac{\delta^2f}{\delta u^2}$ one would simply need to compute the result of $\frac{\delta f}{\delta u}(\frac{\delta x}{\delta u}*f_1+\frac{\delta y}{\delta u}*f_2)$, correct?

It is here I start to get confused. When reviewing an example from the litterature they seem to do the opposite of an exercise that I later solved, and my head starts to spin because of all the indexing and so on... But this is what I think I should do...
$\frac{\delta x}{\delta u}(\frac{\delta x}{\delta u}*f_{11}+\frac{\delta x}{\delta t}*f_{12})+\frac{\delta y}{\delta u}(\frac{\delta y}{\delta u}*f_{21}+\frac{\delta y}{\delta t}*f_{22})$

And why is becuase when I review the indexing, $f_{11}$ tells me that I should derivate with respect to the function $f$s first variable, $x$, and again with respect to the first variable of $x$ which is $u$, but is my thinking correct?

So my question is, what is $\frac{\delta^2f}{\delta x^2}$ and $\frac{\delta^2f}{\delta u^2}$ and how can I memorize the steps requiered to compute these things in the future?

EDIT:

Can I perhaps use the fact that $\frac{\delta^2 f}{\delta u^2}=(\frac{\delta f}{\delta x}\frac{\delta x}{\delta u}+\frac{\delta f}{\delta y}\frac{\delta y}{\delta u})(\frac{\delta f}{\delta x}\frac{\delta x}{\delta u}+\frac{\delta f}{\delta y}\frac{\delta y}{\delta u})$  and then simply expand with taking the products of the terms?

If I expand upon this I get:
$\frac{\delta f}{\delta x}\frac{\delta x}{\delta u}(\frac{\delta f}{\delta x}\frac{\delta x}{\delta u}+\frac{\delta f}{\delta y}\frac{\delta y}{\delta u})+\frac{\delta f}{\delta y}\frac{\delta y}{\delta u}(\frac{\delta f}{\delta x}\frac{\delta x}{\delta u}+\frac{\delta f}{\delta y}\frac{\delta y}{\delta u})$.

I can evaluate $\frac{\delta f}{\delta x}\frac{\delta f}{\delta x}=f_{11}$ and the others also get evaluated to such terms, which simplifies it to:

$\frac{\delta x}{\delta u}(f_{11}\frac{\delta x}{\delta u}+f_{12}\frac{\delta y}{\delta u})+\frac{\delta y}{\delta u}(f_{21}\frac{\delta x}{\delta u}+f_{22}\frac{\delta y}{\delta u})$

This is however in contradiction with what I guessed before. Furthermore, if this is correct (which I percieve it to be), how do I then now proceed to derivate these terms now? Given that I have formulas for $x(t,u)$ and $y(t,u).$


答:
Divide all the terms by $|x|^{2}$ and use the fact that $|\sin (x_1x_2x_3)|
\leq |x_1x_2x_3| \leq \frac  1 2 |x_3| |x|^{2}$ (since $|x_1x_2| \leq \frac1 2 (x_1^{2}+x_2^{2}))$. Final answer : The limit is $\frac 1{1+0}=1$.

Title: How do i calculate this multivariable limit?

Body:
I am really uncertain about how to approach this but here's my attempt:

Let $$x=(x_1,x_2,x_3), \; \mathopen|x\mathclose|=(x_1^{2}+x_2^{2}+x_3^{2})^{\frac{1}{2}}$$

If the following limits exist, find their value:

$$\lim_{x \to 0} \frac{ln(1+\mathopen|x\mathclose|^2)}{\mathopen|x\mathclose|^2+sin(x_1x_2x_3)}$$

My attempt is simply to estimate an upper bound, and then using the standard limit $\lim_{x \to 0}\frac{ln(1+x)}{x}=1$,

$$\frac{\mathopen|ln(1+\mathopen|x\mathclose|^2)\mathclose|}{\mathopen|\mathopen|x\mathclose|^2+sin(x_1x_2x_3)\mathclose|}\leq\frac{ln(1+\mathopen|x\mathclose|^2)}{\mathopen|x\mathclose|^2}$$
Because the $ln$ function with that input is positive, and the $sin$ function takes on a max value of $1$. But, $$\mathopen|x\mathclose|^2\rightarrow0 \;\text{when}\;x\rightarrow0$$ so $$\frac{ln(1+\mathopen|x\mathclose|^2)}{\mathopen|x\mathclose|^2}\rightarrow1$$


问:
Bowyer-Watson algorithm for Delaunay triangulation fails, when three vertices approach a line.
Bowyer-Watson algorithm as described on Wikipedia (https://en.wikipedia.org/wiki/Bowyer%E2%80%93Watson_algorithm) doesn't generate any triangles for this valid point list:

```<code>p0 = (50, 50), p1 = (150, 50), p2 = (250, 67)
</code>```

super-triangle is chosen to be:

```<code>p3 = (0,0), p4 = (1600,0), p5 = (0, 1200)
</code>```

When point p2 is inserted, it is exactly outside triangle (p0,p1,p5), so it is not a bad triangle. Therefor it is not removed, and edge (p0,p1) is not part of the polygonal hole and triangle (p0,p1,p2) is thus never created.

If I choose a larger super-triangle it works, but then when p2 moves closer to the line defined by p0 and p1, the same happens.

Is Bowyer-Watson broken by design, or am I doing something wrong?

答:
I think I know what you are after. The pseudo code on Wikipedia just specifies that the initial super triangle should contain all points in its interior. I could not pin down a corresponding statement in Bowyer's original work, but at least Watson's article states explicitly

Starting with $n+1$ arbitrary points, known to form an $n$-simplex whose convex hull contains the whole data set, [...]

And while both authors discuss degeneracies in general, it seems that one should rather demand that the vertices of the super triangle (or more generally: simplex) have to be outside all circumcircles of any three given points to begin with (which is hard when any three points are almost collinear). Then again, such a constellation is problematic only if it survives until all points have been treated.

How to solve? Personally, I would use symbolic vertices $(-\infty,-\infty)$, $(0,\infty)$, $(\infty,0)$ as super triangle. One then has to contemplate how to define the circumscribed circle of three vertices when some are infinite (spoiler: the disk becomes a half plane).


问:
Taylor series of $e^z$ at point $z_0 = 1+i$
I solved this over a year ago but fail to do so now. I remember that for problems like this one can use $e^z = e^{z_0}e^{z-(z_0)}$.
I can see the answer right away if I had $z_0 = \pi i$. But in my case, with $z_0 = 1+i$, I get stuck or at least I am feeling unsure if this is the best way to write the answer.

$$ \begin{align} e^z & = e^{1+i}e^{z-(1+i)} \\
 & = \sum_{k=0}^{\infty} \frac{e^{1+i}}{k!} (z-(1+i))^k
\end {align}$$

This should be what I was looking for (solved it now that I wrote it down carefully here).
Since I already wrote this, any remarks?

答:
As mentioned in the original post, the main idea here is
$$e^z = e^{1+i}e^{z-(1+i)}.$$

Nothing else to do here.


问:
Probability exercise, dependent events.
7 persons take a tram, boarding at random in 3 wagons. What is the probability that 4 persons will get in the first wagon?

I tried to solve it like this:

P(1 person will get in the first wagon) = 1/3

P(2 persons) = 1/3 * P(1 person), since the events depend on each other.
So P(4 persons) = ( 1/3 )^ 4 = 1/81

Is this correct?

答:
Hint:

The answer is

$$\frac{N\text{(umerator)}}{D\text{(enominator)}}.$$

Here,

$D = 3^7$ and $N = $ the number of different ways that the 7 people can board so that (presumably) exactly $4$ of the $7$ go in the first wagon.

Questions:

How many ways are there of selecting the 4 people out of 7 who will go into the first wagon?

For a given selection of 4 people out of 7 to go into the first wagon, how many ways are there of distributing the remaining 3 people among the remaining 2 wagons?

Title: Probability exercise, dependent events.


问:
Estimate a value with square roots squared without calculating
If $n$ is a natural number and $n<(6+\sqrt{29})^2<n+1$, find the value of $n$.

This problem is, of course, very easy to solve with a calculator. But I am looking for a solution that does NOT need to find the value of $(6+\sqrt{29})^2$. Is there a way to do that?

答: Expand the expression to get $36+12\sqrt{29}+29=65+\sqrt{4176}$. Now you only need to find the integer $m$ such that $m<\sqrt{4176}<m+1$, i.e. $m^2<4176<(m+1)^2$.


问:
Area of triangle formed by points where medians are intersecting the sides
I am looking for a way to find the area of triangle that is formed by points where medians are intersecting the sides in order to solve another question related to medians and triangle.

What will be the area of triangle DEF if the area of the given triangle is 'A'?

Thanks!

答:
Let the position vectors of $B,C$ are $\vec b, \vec c$, then the area of the triangle ABC is $|\vec b \times \vec c|/2$. The position vectors of points F, E, D are  $\vec b/2, \vec c/2, (\vec b+ \vec c)/2$. Area of the small triangle DEF is $|\vec{DE} \times \vec{DF}|=\frac{1}{8}|
\vec b \times \vec c|$. So the required fraction is $1/4$.


问:
How to Construct Graphs with High Chromatic Number but no Complete Subgraphs
How does one go about constructing graphs with high chromatic number, but which don't contain any copies of $K_4$ (and thus any larger complete graphs) as subgraphs?  In particular, how do you start with a graph $G$ and glue together some copies of $G$ to get a graph with higher chromatic number?

My impression is that there are various procedures for doing this in a sort of ad hoc way, but I couldn't find a collection or discussion of the topic in particular, and I'm not active in graph theory.  Does anyone know a reference for such things, or are willing to show a few of the tricks of the trade?

Thanks!

Btw, this is a more general version of a question I asked over here: https://mathoverflow.net/questions/382311/ways-to-increase-chromatic-number-of-graph if you want to see a much more complicated and specific formulation.

答:
A simple, undirected graph is perfect if for each induced subgraph $H$ of $G$, $\omega(H)=\chi(H)$. A nice generalization due to Gyárfás is called chi-boudedness.

A graph $G$ is chi-bounded if there is a function $f$ such that $\chi(H) \leq f(\omega(H))$ for each induced subgraph $H$ of $G$.

The question whether a graph is chi-bounded or not is far from trivial, since there are graphs which are triangle-free while having arbitrarily large chromatic number and hence they are obviously not chi-bounded.

There is a nice collection of procedures for generating triangle-free graphs with arbitrary large chromatic number in "A survey of $\chi$-boundedness" by Alex Scott and Paul Seymour. You can find a preprint here.

Here is a list with references:

Tutte's construction, nicely described in this book

Mycielski’s construction

Kneser graph

Shift graphs

Zykov’s construction

Ramsey graphs

The Burling graph

There might be more...


问:
Orbit of a point under the action of the stabiliser of another point
I'm wondering, for a group $G$ acting on a set $X$, what can be said about the orbit of a point under the action of a stabiliser subgroup of $G$. Let $x\in X$ and let $H = {\rm Stab}_G(x)$. Obviously $|H\cdot x|$ is going to be $1$, but what can we say about $|H\cdot y|$ for some other $y\in X$, perhaps in relation to $|G\cdot y|$?  If that not much can be said in general, feel free to put interesting restrictions on $y$, e.g., $y$ in $G\cdot x$.

Edit. I guess the original question was far too vague. I am still interested in the general question, but in the specific examples I am thinking of, $X$ is the set of vertices of a finite graph and $G$ acts by graph automorphisms. Does this extra information tell us more about $H\cdot y$?

答:
As of little/no interest as it may be, I'd warm up by noting that, if $\operatorname{Stab}_G(x)\unlhd G$, then $|\operatorname{Stab}_G(x)\cdot y|=1$ for every $y\in O_G(x)$ (the orbit of $x$ under the $G$-action). In fact:

\begin{alignat}{1}
\operatorname{Stab}_G(x)\cdot y &= \{h\cdot y\mid h\in \operatorname{Stab}_G(x)\}
\end{alignat}

If $y\in O_G(x)$, then $\exists g\in G\mid y=g\cdot x$, and hence:

\begin{alignat}{1}
\operatorname{Stab}_G(x)\cdot y &= \{h\cdot (g\cdot x)\mid h\in \operatorname{Stab}_G(x)\} \\
&= \{(hg)\cdot x\mid h\in \operatorname{Stab}_G(x)\} \\
\end{alignat}

If, further, $\operatorname{Stab}_G(x)\unlhd G$, then:

\begin{alignat}{1}
\operatorname{Stab}_G(x)\cdot y &= \{(hg)\cdot x\mid h\in \operatorname{Stab}_G(x)\} \\
&= \{(gh')\cdot x\mid h'\in \operatorname{Stab}_G(x)\} \\
&= \{g\cdot (h'\cdot x)\mid h'\in \operatorname{Stab}_G(x)\} \\
&= \{g\cdot x\} \\
&= \{y\} \\
\end{alignat}

and $|\operatorname{Stab}_G(x)\cdot y|=1$.


问:
On perfect matches of $K_{m,m}$

Alice has a collection of $2m-1$ different perfect matchings of
$K_{m,m}$ ($\mathcal{M}_1,\mathcal{M}_2...\mathcal{M}_{2m-1}$). Prove
that Alice can pick exactly one edge from each of
$\mathcal{M}_{i_1},\mathcal{M}_{i_2},...\mathcal{M}_{i_m}$ for some
$\{i_1,i_2,...i_m\}\subset\{1,2,...,2m-1\}$ such that the picked edges
form themselves a perfect matching of $K_{m,m}$.

Note: $K_{m,m}$ is the complete bipartite set with $m$ vertices on each side and a perfect matching is a set of edges such that each vertex has degree $1$.

I thought myself and My main idea was to look at subgraphs of $K_{m,m}$, $K_{n,n}$ such that for each $i$, $\mathcal{M}_i$ forms a perfect matching for $K_{n,n}$ and induction  would finish the job.

I also thought that interpreting edges as lattice points, but this didn't work.

I would appreciate any help!

答:
What you're looking for is named a "rainbow matching", and its existence is proved in https://www.sciencedirect.com/science/article/pii/S0097316598928941

EDIT : the exact link between these results may not be clear, so check https://en.wikipedia.org/wiki/Rainbow_matching for a more direct explanation :)

Title: On perfect matches of $K_{m,m}$


问:
Solve equation $\sqrt{s+13} - \sqrt{7-s} = 2$
Solve the equation

$$\sqrt{s+13}-\sqrt{7-s} = 2$$

I moved the $-\sqrt{7-s}$ to the right side

Thus, I had
$$\sqrt{s+ 13} = 2 +\sqrt{7-s}$$

I then squared both sides
$$\sqrt{s+ 13}^2 = \left(2 +\sqrt{7-s}\right)^2$$

Using the product formula $(x + y)^2 = x^2 + 2xy + y^2$

I got  $$s + 13 = 4 + 4\sqrt{7-s}+  7 – s$$

I then combined like terms
$$2s + 2= 4 \sqrt{7-s}$$

I’m stuck at this point. Does anyone have an idea how to solve this equation?

答:
Wolframalpha helps you solving equations and can even show you
a step-by-step solution.</p>


问:
graph points distance problem.
there are $n \geq 3$ points in plane where each two points have distance at least $1$.
I know there are at most $ 3n - 6 $ pairs with the distance exactly $1$.

I want to prove this fact.

答:
(Answer to the unedited question which asked to prove there are at most 2n-6 pairs with distance 1)
The hexagon shown in the figure has side 2. Consider a similar hexagon of side $n$.

Then it´s easy to show it will have $3n^2+3n+1$ points. Moreover, each point not in the edge ($3n(n-1)+1$ points) will have 6 points at distance 1. The non vertex points of the edges ($6(n-1)$ points) will have 4 adjacent points, and the vertices ($6$ points) will have 3 adjacent points. Taking into account that we´ve counted each pair 2 times, we have a total of $\frac{1}{2}\cdot(6(3n(n-1)+1)+4(6(n-1))+6\cdot3)=9n^2+3n$ pairs of points at distance one. For $n\geq2$, that is bigger than $2\cdot(3n^2+3n+1)-6$, so your conjecture is false.

A bound for the number of pairs is: if you have $k$ points at distance $\geq1$, then there can´t be more than $3k$ pairs of distance $1$. This is obvious since each point corresponds to at most other six points at distance 1, that is, there are at most $6k$ ordered pairs of points at distance 1, and as you count the pairs 2 times, there are at most $3k$ pairs of points at distance 1. This bound can probably be lowered a bit with a more refined argument looking at the points "in the edge of the set".

Title: graph points distance problem.


问:
Not enough digits of 6 in Pi?
I have downloaded the first 1 million digits of pi and was checking which number appears most often. All numbers except the 6 are in the lead at some point in time. But after 1 million digits the 6 falls more than 800 occurrences behind the most frequent digit, which is 5 at that position. So I guess it must take quite a long time until the 6 will catch up and I wonder when the number 6 will be the most frequent one (should happen one day, no?).

Could someone either tell me when the 6 will be in the lead or point me to a place where I can download let's say 10 million digits of pi?

答: Kester Habermann just sent me the answer: After 990,213,633 digits the number 6 takes for the first time the lead of the most frequent numbers in the race of digits of $\pi$. It took a while, but finally the 6 made it to the top!

Title: Not enough digits of 6 in Pi?


问:
Prove that there is a set S of 10 students containing exactly 1 person from every algebra team and exactly 1 person from every calculus team.
60 students attend the one or both of the following courses: algebra and calculus. On algebra they are
divided into 10 teams of equal size. On calculus they are divided into different 10 teams of equal size. Prove that there is a set S of 10 students containing exactly 1 person from every algebra team and exactly 1 person from every calculus team.

I have to prove this applying concepts of graph theory, I'd appreciate if any of you could prove it.

答:
You can, indeed, see the problem as a graph problem.

Consider a bipartite graph, with vertices are $a_1, ..., a_{10}, c_1, ..., c_{10}$.
Each vertex represents a team (either a algebra or a calculus team).

Now, you put an between $a_i$ and $c_j$ if and only if there is a student who belongs to the $i$-th algebra team, and to the $j$-th calculus one.

Your graph has $60$ edges, one per student, representing the student's teams.

The problem can now be reformulated as "Does this graph admit a perfect matching ?" Indeed, you would like to pick students (=edges), such that you "cover" all the teams both in algebra and calculus (="perfect"), and you want a single student per team (=matching).

If you still need help to solve the problem, I can try to give some hints, but at least you can use some graph theory now !