问:
Reduce polynomial $x^5-x+2$
The polynomial $p(x) = x^5-x + 2$ is not reducible in $\mathbb {Z}$ (i.e mod $0$). Since it does not have a linear factor, so there is only the case where $p (x) = q (x) r (x)$, where $\deg (q) = 3$ and $\deg (r) = 2$. I have tried to show it by taking the polynomials $p (x) = (x^3+ax^2+bx+c)\cdot (x^2+dx+e)$ to get to the system of equations $$\left\{\begin{matrix}
d+a=0\\
e+ad+b=0\\
ae+bd+c=0\\
be+dc=-1\\
ce=2
\end{matrix}\right.$$

but I cannot come to a contradiction, I would appreciate your help

答:
HINT.-A way is to show, as you want to do, that $p (x) = (x^3+ax^2+bx+c)\cdot (x^2+dx+e)$ is not possible in $\mathbb Z[x]$ but avoiding to go to your system of five equations. For this note that $p(0)=p(1)=p(-1)=2$ so you have the system
$$\begin{cases}ce=2\\(1+a+b+c)(1+d+e)=2\\(-1+a+b+c)(1-d+e)=2\end{cases}$$ From this you have $(c,e)$ is equal to $(1,2)$ or $(2,1)$ then you get $d=0,\space\space a-b=1$ and $a+b=-2$.

It follows that $a=\dfrac{-1}{2}$ and $b=\dfrac{-3}{2}$.

Thus the coefficients $a$ and $b$ are not rational integers and your factors $q(x)$ and $r(x)$ are not in $\mathbb Z[x]$.

Title: Reduce polynomial $x^5-x+2$


问:
How to simplify this formula
I have this equation: $\frac{\sqrt{(a+b+\sqrt{a^2+b^2-2ab\cos\gamma})(-a+b+\sqrt{a^2+b^2-2ab\cos\gamma})(a-b+\sqrt{a^2+b^2-2ab\cos\gamma})(a+b-\sqrt{a^2+b^2-2ab\cos\gamma})}}{4}$

It calculates the area of a triangle using only 2 sides of a triangle $a, b$, and the angle of those sides, $\gamma$. It simply combines the law of cosines and Heron's Formula.
Is there any way to achieve the same function, and is it possible to simplify this equation?

答:
IMHO if you started with

$A = \sqrt {s(s-a)(s-b)(s-c)}$ with $s = \frac{a+b+c}{2}$ and used $\cos \gamma = \frac{a²+b²-c²}{2ab}$ like definied e.g. here: https://en.wikipedia.org/wiki/Heron%27s_formula

then rewrote the maths to avoid using $c$ through writing it as a function of $a$, $b$ and $ \gamma $ it should be fine and correct. Did you?

ContentLicense: CC BY-SA 4.0

Title: How to simplify this formula


问:
Conceptual Question about the formal definition of $\bigcap \mathcal F$
In Kenneth Kunen's "The Foundations of Mathematics, the following definition is provided:

When $\mathcal F \neq \emptyset$, $$\bigcap \mathcal F = \bigcap_{Y\in \mathcal F}Y=\{x:\forall Y \in \mathcal F(x\in Y)\}$$

Firstly, I will mention that I recognize why we want $\mathcal F \neq \emptyset$ if the formula $\varphi$ we choose to go with is $\varphi:=\forall Y \in \mathcal F (x \in Y)$. Namely, if $\mathcal F = \emptyset$, this formula would be vacuously true so any $x$ would satisfy this formula...and thus all $x$ in our universe would be admitted into the set. The universal set is a contradiction because it contains itself (and by Axiom of Regularity and Pairing Axiom, this cannot happen).

My question is: How should I encode in First Order Logic the fact that I do not want $\mathcal F$ to equal $\emptyset$?

At first I thought that the correct approach was to replace "$\forall Y \in \mathcal F(x\in Y)$" (which is equivalent to "$\forall Y [Y \in \mathcal F \rightarrow x \in Y]$) with:

$$\forall Y ([\mathcal F = \emptyset \lor Y \in F] \rightarrow x \in Y)$$

This seemed to make sense because if $\mathcal F$ does equal $\emptyset$, the universally quantified statement will be interpreted as a false statement (because, for whatever $x$ one choses, $x \notin \emptyset$)...in which case $\bigcap \emptyset = \emptyset$. There is no contradiction here.

However, after looking around on some posts and sites, it seems as though we want to retain the fact that $\bigcap \emptyset:= \text{Universal Set}$.

Given this, what is the formal way of encoding Kunen's definition (i.e. stipulating that $\mathcal F \neq \emptyset$)?

Thank you!

答:
What you're observing here is a tension between two approaches to "$\bigcap$:"

as a total operation, which takes in a set and spits out either a set or the proper class $V$ (I'm assuming we're working in $\mathsf{ZF}$ or similar, where there is no universal set), the latter iff the input is $\emptyset$; or

as a partial operation, which takes in a set and spits out a set or is undefined, the latter iff the input is $\emptyset$.

Kunen takes approach $(2)$, while some texts/sites take approach $(1)$ (or work in a framework where a universal set is accepted, in which case you get the best of both worlds but have to leave the $\mathsf{ZF}$-context). Certainly it's important for someone starting the subject to understand why the "naive" definition of $\bigcap$ leads to interpretation $(1)$, but approach $(2)$ is more convenient if we're working in first-order logic modified to accommodate partial functions (since at that point we can treat "$\bigcap$" as a genuine function), the issue with $(1)$ being that a proper class is not an "actual object" in the relevant sense. Ultimately there isn't a huge advantage one way or the other.

What you've written is a third alternative:

Interpret "$\bigcap$" as a total operation, which takes in a set and outputs a set, agreeing with $(1)$ and $(2)$ when the input is nonempty and using a "default value" of $\emptyset$ when the input is $\emptyset$.

In my opinion this is slightly less good: I don't think totality is worth paying the price of actively disagreeing with $(1)$. But this approach will also work fine, as long as one is careful to handle the emptyset (which one should be anyways).


答:
Let $c = \sqrt{a^2+b^2-2ab\cos\gamma}$ and $P_1,P_2,P_3,P_4$ be the 4 factors inside the square root. Notice

$P_1 P_4 = (a+b+c)(a+b-c) = (a+b)^2 - c^2 = 2ab(1+\cos\gamma)$,

$P_2 P_3 = (-a+b+c)(a-b+c) = c^2-(a-b)^2 = 2ab(1-\cos\gamma)$

We have
$$\frac{\sqrt{P_1P_2P_3P_4}}{4}
= \frac{\sqrt{(2ab)^2(1-\cos^2\gamma)}}{4}
= \frac12 ab\sin\gamma$$

Title: How to simplify this formula

Body:
I have this equation: $\frac{\sqrt{(a+b+\sqrt{a^2+b^2-2ab\cos\gamma})(-a+b+\sqrt{a^2+b^2-2ab\cos\gamma})(a-b+\sqrt{a^2+b^2-2ab\cos\gamma})(a+b-\sqrt{a^2+b^2-2ab\cos\gamma})}}{4}$

It calculates the area of a triangle using only 2 sides of a triangle $a, b$, and the angle of those sides, $\gamma$. It simply combines the law of cosines and Heron's Formula.
Is there any way to achieve the same function, and is it possible to simplify this equation?


答:
A much more efficient and practical way to prove $x^5 - x + 2$ is irreducible over $\mathbf Z$ is to use the reduction mod $p$ test: if there is a prime $p$ such that $f(x) \bmod p$ is irreducible in $(\mathbf Z/p\mathbf Z)[x]$ then $f(x)$ is irreducible in $\mathbf Z[x]$.  While $f(x) \bmod 2$ is reducible, $f(x) \bmod 3$ is irreducible: if $f(x)$ were reducible mod $3$ then it would have a linear or quadratic factor in $(\mathbf Z/3\mathbf Z)[x]$.  Since $f(a) \not\equiv 0 \bmod 3$ for $a = 0, 1, 2$, there is no linear factor.  For quadratic factors, the only quadratic irreducible polynomials in $(\mathbf Z/3\mathbf Z)[x]$ are $x^2 + x + 2$ and $x^2 + 2x + 2$.  You can check in $(\mathbf Z/3\mathbf Z)[x]$ that
$$
x^5 - x + 2 \equiv x+2 \bmod x^2 + x + 2, \ \ \ x^5 - x + 2 \equiv x+2 \bmod x^2 + 2x + 2
$$
so $x^5 - x + 2$ has no quadratic irreducible factor in $(\mathbf Z/3\mathbf Z)[x]$.  Thus $x^5 - x + 2$ is irreducible in $(\mathbf Z/3\mathbf Z)[x]$, so it is irreducible in $\mathbf Z[x]$.

The nice feature of working in $(\mathbf Z/p\mathbf Z)[x]$ is that there are finitely many polynomials of each degree, so at worst there are only finitely polynomial divisions to make to exhaustively check a polynomial in this ring has no irreducible factor of a specified degree without having to worry about trying to restrict the coefficients in some way first, as you may need to do when you work in $\mathbf Z[x]$, which has infinitely many (irreducible) polynomials of each degree.

Title: Reduce polynomial $x^5-x+2$

Body:
The polynomial $p(x) = x^5-x + 2$ is not reducible in $\mathbb {Z}$ (i.e mod $0$). Since it does not have a linear factor, so there is only the case where $p (x) = q (x) r (x)$, where $\deg (q) = 3$ and $\deg (r) = 2$. I have tried to show it by taking the polynomials $p (x) = (x^3+ax^2+bx+c)\cdot (x^2+dx+e)$ to get to the system of equations $$\left\{\begin{matrix}
d+a=0\\
e+ad+b=0\\
ae+bd+c=0\\
be+dc=-1\\
ce=2
\end{matrix}\right.$$

but I cannot come to a contradiction, I would appreciate your help


问:
When is the hexagon formed by the feet of perpendiculars and the midpoints of sides regular?
In a $\Delta ABC$, consider the hexagon $m_{a}f_{a}m_{b}f_{b}m_{c}f_{c}$ where $m$ and $f$ stand for the midpoint and foot of perpendicular of the respective sides. Now it can be shown quite easily, that $$m_{a}f_{a}= |\frac{b^2-c^2}{2a}|$$ With a little more effort, one can show also that, $$m_{c}f_{a}= \frac{R}{2}\frac{\sin(2B-\alpha)}{\cos(\alpha /2)}$$ Where $\alpha$ is the angle subtended by the segment $m_{a}f_{a}$ at the center of the hexagon, which is nothing but the nine-point center. Similarly, we can find the side-lengths of the other sides of this hexagon.

Under what conditions is this hexagon regular?

P.S. It can also be shown that $$\sin(\alpha /2) = \frac{\sin^2(B) - \sin^2(C)}{\sin^2(A)}$$

答: Never. A regular hexagon whose alternate sides lay on the sides of a triangle would force the triangle to be equiangular, and hence the midpoints would coincide with the feet of the perpendiculars.</p>

ContentLicense: CC BY-SA 3.0


问:
Solve the recurrence relation: $na_n = (n-4)a_{n-1} + 12n H_n$
I want to solve
$$ na_n = (n-4)a_{n-1} + 12n H_n,\quad n\geq 5,\quad a_0=a_1=a_2=a_3=a_4=0. $$

Does anyone have an idea, what could be substituted for $a_n$ to get an expression, which one could just sum up? We should use
$$ \sum_{k=0}^n \binom{k}{m} H_k = \binom{n+1}{m+1} H_{n+1} - \frac{1}{m+1} \binom{n+1}{m+1} $$
to simplify the result.

答:
Here are more details for @Phicar's suggested approach.
Let $A(z)=\sum_{n \ge 0} a_n z^n$ be the ordinary generating function of $(a_n)$.
Then $z A'(z)=\sum_{n \ge 0} n a_n z^n$, and the recurrence relation implies that
\begin{align}
z A'(z)
&= \sum_{n \ge 5} \left((n-4)a_{n-1} + 12n H_n\right) z^n \\
&= z \sum_{n \ge 5} (n-1) a_{n-1} z^{n-1} - 3 z \sum_{n \ge 5} a_{n-1} z^{n-1} + 12z \sum_{n \ge 5} n H_n z^{n-1} \\
&= z \cdot z A'(z) - 3z A(z) + 12z\left(\frac{d}{dz}\left(\frac{-\log(1-z)}{1-z}\right) -\sum_{n=1}^4 n H_n z^{n-1}\right) \\
&= z^2 A'(z) - 3z A(z) + 12z\left(\frac{1-\log(1-z)}{(1-z)^2} -1-3z-\frac{11}{2}z^2-\frac{25}{3}z^4\right)
\end{align}
The resulting solution for $A(z)$ is messy, so there must be a better way.


问:
Evaluate this limit of integration in Complex analysis
I am trying assignments of Complex analysis and I couldn't evaluate this particular integral.

Find the limit $\int_{|z|=R} |\sin(z)/z | |dz| $ as $R\to\infty$.

I am unable to see which result I should use and also there is confusion due to dz being in modulus and also $\sin(z) /z$.

Can you please tell significance of modulus here and how to compute the limit.

Thanks!!

答:
Hint : it is $\infty$.

$              $

Body:
Hint : it is $\infty$.

$              $


问:
If $r$ (mod $p$) is a primitive root, then $r+p$ is also a primitive root mod p
I've struggled with this problem for a while...I tried to figure it out using a reduced residue system's theorem but I honestly have no idea how to work with the given hypothesis.</p>

答:
$r$ (mod $p$) is a primitive root.

This means that there is a $k$ with $r^k = a$ mod $p$ for all $a$ coprime to $p$.

$(r+p)^k = r^k$ mod $p$

which means that l exists such as $l = k$ with $(r+p)^l = a$ mod $p$ for all $a$ coprime to $p$.

Therefore $(r+p)$ is a promitive root as well.


问:
Zero divisor or unit

Let $R$ be a finite ring with unity. Prove that every $x \neq 0$ is a zero divisor or a unit.

I made an attempt, but got stuck. I took a non-zero $x$ and defined the mapping $f: R \to R, y \mapsto xy$. This mapping may not be injective, but if it is, since it maps from $R$ to $R$ and is finite, $f$ is also surjective, and since $1$ is an element of $R$, there exists $y$ such that $xy = 1$. The ring is not necessarily commutative, however, so I also need to find a $y$ such that $yx = 1$. My thought was to define an analagous map $g: R \to R, y \mapsto yx$, which may or may not be injective. But then I run into problems where the first map is injective, but the second isn't.

Any help would be appreciated.

EDIT: The linked question does not answer this because it assumes that the ring is commutative. This ring is not commutative.

答:
What you have shown is:

If $x \neq 0$ is not a left (resp. right) zero divisor, then there exists $y$ such that $xy = 1$ (resp. $yx = 1$).

If $x$ is a left zero divisor but not a right zero divisor, then we will have $xz = 0$ for some $z \neq 0$ and $yx = 1$ for some $y$.

We then get $$0 = y\cdot 0 = yxz = 1 \cdot z = z,$$ which contradicts $z \neq 0$.

Therefore this situation cannot happen. Every left zero divisor is automatically a right zero divisor, and vice versa, so we may say unambiguously "zero divisor".

Thus when $x$ is not a zero divisor, we will have $xy = 1$ and $zx = 1$ for some $y, z$. This then gives $$y = 1 \cdot y = zxy = z \cdot 1 = z,$$ and we see that $y = z$ is the multiplicative inverse of $x$, hence $x$ is a unit.

Title: Zero divisor or unit


问:
How to show that for $\vert x \vert \lt 1$, $\lim \limits_{n \to \infty} x^n = 0$?
I have tried to use the monotone convergence theorem, but that still runs into the same problem as trying to prove it with a normal $N$ and $\epsilon$ argument, which I somehow cannot figure out how to do, especially when we have $x = 0.999$ or something like that.

答:
The definitios goes like :

$$\forall \epsilon >0 \exists N : n \geq N \implies |x^n| = |x|^n < \epsilon $$

So for any arbitrary $\epsilon$ you have

$$|x|^n < \epsilon \iff n > \frac{\ln \epsilon}{\ln |x|}$$

then you can take $N = \newcommand{\ceil}[1]{\left\lceil #1 \right\rceil} \ceil{\frac{\ln \epsilon}{\ln |x|}}$.


问:
$\int_\Omega \phi d\mu_n\to\int_\Omega\phi d\mu,\forall\ \phi\in C_0(\Omega) $ and $\mu_n\geq 0$ implies $\mu\geq 0$?
Let $\Omega\subset\mathbb{R}^N$ be a bounded domain and suppose that $\mu_n$ is a sequence of non-negative Radon measures, that converges to a Radon measure $\mu$ in the weak star sense: $$\int_\Omega \phi d\mu_n\to\int_\Omega\phi d\mu,\forall\ \phi\in C_0(\Omega) $$

where $C_0(\Omega)$ denotes the set of continuous compacted support functions. Can I conclude that $\mu\geq 0$?

Thank you.

答:
Yes. For every nonnegative $\phi\in C_o(\Omega)$ you have
$$\int\phi \, d\mu\ge 0,$$
so $d\mu$ is a positive linear functional. By the Riesz Representation Theorem you can conclude that $d\mu\ge 0$.</p>


答:
Notice the sequence $|x|^{n}$ is decreasing when $|x| < 1$. Indeed, one has that
\begin{align*}
|x| < 1 \Rightarrow |x|^{2} < |x| < 1 \Rightarrow |x|^{3} < |x|^{2} < |x| < 1 \Rightarrow \ldots
\end{align*}

Moreover, $|x|^{n}$ is bounded below by zero. Hence it converges.

Once $|x|\xrightarrow{n\to\infty}|x|$ (because it is a constant) and every subsequence converges to the same limit, one has
\begin{align*}
L = \lim_{n\to\infty}|x|^{n+1} = \lim_{n\to\infty}|x|\times|x|^{n} = \lim_{n\to\infty}|x|\times\lim_{n\to\infty}|x|^{n} = |x|\times L \Rightarrow L(1 - |x|) = 0 \Rightarrow L = 0
\end{align*}

At last but not least, one has that
\begin{align*}
\lim_{n\to\infty}|x|^{n} = \lim_{n\to\infty}|x^{n}| = 0 \Rightarrow \lim_{n\to\infty} x^{n} = 0
\end{align*}
and we are done

Title: How to show that for $\vert x \vert \lt 1$, $\lim \limits_{n \to \infty} x^n = 0$?

Body: I have tried to use the monotone convergence theorem, but that still runs into the same problem as trying to prove it with a normal $N$ and $\epsilon$ argument, which I somehow cannot figure out how to do, especially when we have $x = 0.999$ or something like that.


问:
Find a winning strategy in a stone game.

Alice and Bob play the following game: There are piles of stones and in each turn the player can do one of the following: Remove one stone from a pile, or take two piles with $x$ and $y$ stones in them and replace them with $1$ pile of $xy$ stones. The player who has no move loses. Who has the winning strategy?

The answer can depend on the number of piles and the number of stones in each pile.  I think I got an extremely ugly inductive solution. What I got is that the first player wins if and only if there is an odd number of stones, or there is an even number of stones and a positive even amount of piles with even amount of stones in them. I might be mistaken somewhere though. Does anyone have something elegant?

答:
A situation consists of $e$ even-sized and $o$ odd-sized non-empty heaps. I claim that winning or losing depends only on $(e,o)$.
Let $W$ be the set of positions $(e,o)$ that are winning and $L$ the set of $(e,o)$ that are losing positions.

Claim. We have
$$W=\{\,(e,o)\mid o\text{ odd}\lor(e\text{ even}\land e\ne 0)\,\}$$
and
$$L=\{\,(e,o)\mid o\text{ even}\land (e\text{ odd}\lor e=0)\,\}.$$

Proof.
Since the game must end after finitely many moves, it ssuffices to show that every valid move from a situation $\in L$ leads to a situation $\in W$, and at for every situation $\in W$, there exists a valid move to a situation $\in L$.

Let us start with $(e,o)\in L$:

First case: $o$ is even and $e=0$.
Removing a stone from any (necessarily odd) heap decreases $o$ to an odd number, hence takes us to $W$.
Combining two (necessarily odd) heaps also decreases $o$ by one, hence takes us to $W$. We conclude that $(o,0)\in L$ for odd $o$.

Second case: $o$ is even and $e$ odd.
Removing a stone from an odd heap or combining two odd heaps or combining an odd and an even heap, decreases $o$ to odd, hence takes us to $W$
Removing a stone from an even heap increases $o$ to odd, hence takes us to $W$.
Finally, combining two even heaps (which is possible only if $e\ge 3$) takes us to $(e',o')=(e-1,o')$ with $e'$ even and positive, so again to $W$.

So indeed every valid move from a situation $\in L$ takes us to a situation $\in W$.

Next consider $(e,o)\in W$:

First case: $e$ is even and positive. If $o$ is even, we can combine two even heaps  to arrive at $(e',o')=(e-1,o)\in L$. If $o$ is odd, we can remove a stone from one of the even heaps and arrive at $(e',o')=(e-1,o+1)\in L$.

Second case: $o$ is odd and $e=0$. By removing a stone from an odd heap, we arrive at either $(e',o')=(1,o-1)\in L$ or  (if we emptied a heap) $(e',o')=(0,o-1)\in L$.

Third case: $o$ is odd and $e$ is odd. Combine an odd and an even heap to arrive at $(e',o')=(e,o-1)\in L$.

These cases logically cover all of $W$. So indeed, from every situation in $W$, there exists a valid move to $L$. $\square$


答:
Suppose $0 < x < 1$ and set $x_1 := \frac{1}{x} > 1$. Let $\varepsilon > 0$ be arbitrary and set $K := \frac{1}{\varepsilon}$. Then you can find a $n \in \mathbb{N}$ such that
$$
x_1^n > \frac{1}{\varepsilon} \iff x^n < \varepsilon,
$$
meaning that $\lim\limits_{n \to \infty} x^n = 0$. The case $-1 < x <0$ is proved analogously,


问:
2-adic valuation of $5^{2^{k-2}}-1$
I have the following exercice :

Consider that the 2-adic valuation of $x$ is $k \geq 2$. Prove that the 2-adic valuation of $(1+x)^2- 1$ is $k+1$.

Using the previous question, conclude that for $k\geq 2$, the 2-adic valuation of $5^{2^{k-2}}-1$ is $k$.

For the first l question, we know that $(1+x)^2 - 1= x(x+2) $. So, $$v_2((1+x)^2-1) = v_2(x)+\min(v_2(x),v_2(2))= k+1.$$

For the last question, I tried several methods but didn't work. Can someone give me a hint please?

Thanks!

Title: 2-adic valuation of $5^{2^{k-2}}-1$

Body:
If $v_2(x)=k\ge 2$ then $v_2(x+2)=1$ and $v_2( (1+x)^2-1)=v_2 (x(x+2)) =v_2 (x)+v_2(x+2) =k+1$.

Let $x_{k+2}=5^{2^k}-1=(1+x_{k+1})^2-1, v_2(x_2)=2$, by induction if $v_2(x_k)=k$ then $v_2(x_{k+1})=k+1$.


答:
To do this at the very beginning of the course, you can use Bernoulli's inequality.  If $y>0$, then $(1+y)^n > 1+ny$.  Prove it by induction.  Then use this to show:
If $z>1$, then $z^n \to \infty$.  Then use this to show:
If $0<x<1$, then $x^n \to 0$.  Finally, $|-x| = |x|$ to get the case $-1<x<0$.


问:
Can the intersection of open or closed balls be empty, if their radii are bounded from below?
I am wondering about the following question: </p>

<p>Given a (countable) sequence of nested open balls: </p>

<p>$$ B_1 \supseteq B_2 \supseteq \cdots $$
Not necessarily having the same same center. All having radius bounded from below, say by $r > 0$. Then can we say that $$\bigcap_{i=1}^{\infty} B_i \neq \varnothing$$
it is certainly true in the reals, as one can simply go to the point where the radii are close to $r$ then take the center of that ball. However, I'm having trouble seeing whether or not it is true in the general case. So thanks in advance for proof or counterexample. </p>

答:
The below argument is incorrect as pointed out by TCL. I'll leave the answer here anyway, just incase.

Note that $\bigcap_i^n B_i=B_n$ because each $B_i$ is a subset of $B_{i+1}$. Let $d_i=\mbox{diam}(B_i)$. It should be clear that if $\bigcap_i^{\infty} B_i=\emptyset$, then $\lim_{i\rightarrow\infty}d_i=0$ and so there exists some $i\in\mathbb{N}$ such that $d_i< 2r$, but then $B_i$ is an open ball of radius less than $r$ which contradicts the assumption that each $B_i$ has radius bounded below by some non-zero $r$.

I should add that i'm using the definition of the diameter of a subset $A$ of a metric space $X$ with metric $d$ to be $\mbox{diam}(A)=\sup\{d(x,y)|x,y\in A\}$.


问:
Finding the matrix representation of $T$
Suppose that $T \in \mathscr{L}(\mathbb{R^n};\mathbb{R^m})$ for some $n,m \in \mathbb{N}$.

If $T(1,1) = (3,\pi,0)$ and $T(0,1) = (4,0,1)$. Find the matrix representation of $T$.

So I know exactly what this is going to look like in matrix form, but I'm not sure if my reasoning is correct here. So since we have $T(1,1) = (3,\pi,0)$, we get that the column vector from this is given as
\begin{bmatrix}
           3 \\
           \pi \\
           0
           \\
         \end{bmatrix}
A similar reasoning shows what the other column vector is, however, what's the logic behind this and how is it determined? I was thinking that we just obtain the column vector by looking at what $T$ is equivalent to.

答:
When we are interested in the matrix representation of a linear transformation, we have to choose a basis for the domain and the counter domain. At your case, it has been given the value of $T$ at the vectors $(1,1)$ and $(0,1)$, whose images are described in terms of the standard basis $\mathcal{B}_{2} = \{(1,0,0),(0,1,0),(0,0,1)\}$ of $\mathbb{R}^{3}$.

If your are interested in the representation of $T$ in terms of $\mathcal{B}_{1} = \{(1,0),(0,1)\}$, notice that
\begin{align*}
T(1,0) = T((1,1) - (0,1)) = T(1,1) - T(0,1) = (-1,\pi,-1)
\end{align*}

Consequently, the sought matricial representation is given by:
\begin{align*}
[T]_{\mathcal{B}_{1}}^{\mathcal{B}_{2}} =
\begin{bmatrix}
-1 & 4\\
\pi & 0\\
-1 & 1
\end{bmatrix}
\end{align*}

Hopefully this helps.


问:
Cannot find a mixed-strategy Nash equilibrium in a matrix
I have the follow matrix, but whenever I try to calculate the probabilities, they are all.. equal? I can't deduce the formulas any more and I am incredibly confused since I don't think this is how it is done. Can someone point me out what I am doing wrong?

player 2:
3p+1(1-p) = 3p+1(1-p)
3p+1-1p = 3p+1-1p

player 1:
1q+3(1-q) = 1q+3(1-q)
1q+3-3q=1q+3-3q

答: This game has only a pure strategy Nash equilibrium of $(T, R)$. There is no reason to mix strategies because changing strategies can not improve either players reward. Note that this would technically be considered a trivial mixed strategy where for example player one plays $T$ with probability 1 and plays $B$ with probability 0.


问:
Bound on approximation error of Bernstein polynomial
Consider a continuous function $f: [0,1] \to [0,1]$. Let $B_n$ be its $n$-th order Bernstein polynomial,
$$
B_n(x) = \sum_{k=0}^n f\left(\frac{k}{n}\right) \binom{n}{k}x^k (1-x)^{n-k}.
$$
As is well known, $B_n(x) \rightarrow f(x)$ uniformly on $[0,1]$ as $n \rightarrow \infty$. I am interested in bounding the approximation error $B_n(x)-f(x)$.

This reference, section 4, contains one such bound:
$$
|B_n(x)-f(x)| \leq \left( 1 + \frac{1}{4n^2} \right) \omega(n^{-1/2})
$$
where $\omega$ is the modulus of continuity of $f$, that is, $\omega(\delta) = \sup_{|x-x'|<\delta} |f(x)-f(x')|$.

My questions are

Is here any reference or proof to that result?

Are there any similar results that provide a bound on $|B_n(x)-f(x)|$?

答:
I found a few references:

"Iterated Bernstein polynomial approximations", by Zhong Guan: theorem 1 mentions that if $f$ is $C^r$ for $r=0$ or $1$,

$$|B_n(x)-f(x)| \leq C_rn^{-r/2}\omega_r(n^{-1/2})$$

where $\omega_r$ is the modulus of continuity of the $r$-th derivative; and $C_0=5/4$, $C_1=3/4$. If $f$ is $C^r$ with $r>1$ the rate $1/n$ cannot be improved.

"On the Rate of Approximation of Functions
by the Bernstein Polynomials", by Telyakovskii
, contains the above results and some refinemets for the error at a specific $x$.

"Rate of convergence of Bernstein polynomials for functions with derivatives of bounded variation", by Bojanic and Cheng, improves $C_0$ from the above $5/4$ to

$$\frac{4306 + 837\sqrt 6}{5832}$$.

"Approximation of Hölder Continuous Functions by Bernstein Polynomials", by Mathé, establishes that

$$|B_n(x)-f(x)| \leq L \left( \frac{x(1-x)}{n} \right)^{\alpha/2}$$

when $f$ is Hölder with exponent $\alpha$ for some $0<\alpha\leq 1$ and constant $L$.

"The Weierstrass Approximation Theorem and Large Deviations", by Gzyl and Palacios, mentions that if $f$ is $C^2$

$$|B_n(x)-f(x)| \leq \frac{\sup_x|f''(x)|}{8n}.$$


问:
how to prove the transfer function of the decomposed form is the following?
We know that for an uncontrollable state space model of the following form:

$\begin{array}{l}
\dot x\left( t \right) = Ax\left( t \right) + Bu\left( t \right)\\
y\left( t \right) = Cx\left( t \right) + Du\left( t \right)
\end{array}$

it has a transfer function $G\left( s \right) = C{\left( {SI - A} \right)^{ - 1}}B + D$.

Furthermore, we can associate a controllable Kalman decomposition:

$\begin{array}{l}
\left[ {\begin{array}{*{20}{c}}
{\frac{{d\left( {{{\bar x}_c}} \right)}}{{dt}}}\\
{\frac{{d\left( {{{\bar x}_{\bar c}}} \right)}}{{dt}}}
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
{{{\bar A}_c}}&{{{\bar A}_{12}}}\\
0&{{{\bar A}_{\bar c}}}
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
{{{\bar x}_c}}\\
{{{\bar x}_{\bar c}}}
\end{array}} \right] + \left[ {\begin{array}{*{20}{c}}
{{{\bar B}_c}}\\
0
\end{array}} \right]u\\
y = \left[ {\begin{array}{*{20}{c}}
{{{\bar C}_c}}&{{{\bar C}_{\bar c}}}
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
{{{\bar x}_c}}\\
{{{\bar x}_{\bar c}}}
\end{array}} \right] + Du
\end{array}$

my question is: how to prove that the new transfer function is: $G\left( s \right) = {{\bar C}_c}{\left( {SI - {{\bar A}_c}} \right)^{ - 1}}{{\bar B}_c} + D$

答:
The controllable Kalman decomposition is a linear transformation, i.e. $\bar{x} = T\,x$ with $T$ non-singular. So the state space model in the controllable Kalman decomposition $(\bar{A}, \bar{B}, \bar{C}, \bar{D})$ is related to the initial state space model $(A,B,C,D)$ according to

\begin{align}
\bar{A} &= T\,A\,T^{-1}, \\
\bar{B} &= T\,B, \\
\bar{C} &= C\,T^{-1}, \\
\bar{D} &= D.
\end{align}

For any such non-singular $T$ is follows that

$$
G(s) = \bar{C}\,(s\,I - \bar{A})^{-1} \bar{B} + \bar{D} = C\,T^{-1}\,(s\,I - T\,A\,T^{-1})^{-1} T\,B + D, \tag{1}
$$

which simplifies to the following when using the push-through identity

$$
G(s) = C\,(s\,I - A)^{-1} B + D. \tag{2}
$$

So any non-singular $T$ does not change the transfer function $G(s)$ associated with the state space model.

If there exists such a $T$ such that

$$
\bar{A} =
\begin{bmatrix}
\bar{A}_c & \bar{A}_{12} \\
0 & \bar{A}_\bar{c}
\end{bmatrix}, \quad
\bar{B} =
\begin{bmatrix}
\bar{B}_c \\ 0
\end{bmatrix}, \quad
\bar{C} =
\begin{bmatrix}
\bar{C}_c & \bar{C}_\bar{c}
\end{bmatrix},
$$

it follows that the transfer function $G(s)$ can be obtained using $(1)$. It can be noted that due to upper block triangular structure of $\bar{A}$ it holds that

\begin{align}
(s\,I - \bar{A})^{-1} &=
\begin{bmatrix}
s\,I-\bar{A}_c & -\bar{A}_{12} \\
0 & s\,I-\bar{A}_\bar{c}
\end{bmatrix}^{-1}, \tag{3a} \\
&=
\begin{bmatrix}
(s\,I-\bar{A}_c)^{-1} & (s\,I-\bar{A}_c)^{-1} \bar{A}_{12} (s\,I-\bar{A}_\bar{c})^{-1} \\
0 & (s\,I-\bar{A}_\bar{c})^{-1}
\end{bmatrix}. \tag{3b}
\end{align}

From this it follows that multiplying $(3)$ by $\bar{B}$ yields

$$
(s\,I - \bar{A})^{-1} \bar{B} =
\begin{bmatrix}
(s\,I-\bar{A}_c)^{-1} \bar{B}_c \\ 0
\end{bmatrix},
$$

after which I assume it is clear that

$$
G(s) = \bar{C}_c(s\,I-\bar{A}_c)^{-1} \bar{B}_c + D.
$$


问:
Diagonalisation of a linear operator
Given the following linear operator
$$T(x,y,z)=(2x+z,-3y+z,-3z)$$
with $T\colon\mathbb{R}^{3}\to\mathbb{R}^{3}$, I found that its matrix in the canonical basis is
$$[T]=\begin{bmatrix} 2 & 0 & 1 \\ 0 & -3 & 1 \\ 0 & 0 & -3 \end{bmatrix}$$
such that its eigenvalues and eigenvectors are:
$$\lambda_{1}=2,\vec{v_{1}}=(1,0,0),\quad\lambda_{2}=-3,\vec{v_{2}}=(0,1,0)$$
How do I conclude that $T$ is not diagonalisable?

答: I am sorry for my late reply. Ben Grossman already answered your question. Now, you know the definition of geometric multiplicity. I just want to add something with his answer. In general, the geometric multiplicity/dimension of eigenspace of an eigenvalue is less than or equal to the algebraic multiplicity of that eigenvalue. If the equality holds for every eigenvalue, then the operator is diagonalizable. Converse is also true. The proof is elementary. You should find it in any standard Linear Algebra text.</p>

Title: Diagonalisation of a linear operator


答:
Hint: By the binomial theorem, all the terms of $(r+p)^k$ except the first have a factor of $p$.  Thus each of them is zero $\bmod p$.

Or even more elegantly, by the congruence power rule $(r+p)^k=r^k$.  See @BillDubuque's comment below.

Title: If $r$ (mod $p$) is a primitive root, then $r+p$ is also a primitive root mod p

Body: I've struggled with this problem for a while...I tried to figure it out using a reduced residue system's theorem but I honestly have no idea how to work with the given hypothesis.</p>


问:
About convexity of a set in $\mathbb{R}^n$
Can we say that a set $E\subset \mathbb{R}^n$ is convex iff $E\cap L$ is contractible for every real affine line $L$ in $\mathbb{R}^n$?

I know that $E\subset \mathbb{R}^n$ is convex iff $E\cap L$ is convex for every real affine line $L$ in $\mathbb{R}^n$. And a convex set is contractible.

Also if we consider $E\subset \mathbb{R}^n\subset \mathbb{C}^n$ , then is $E$ convex if and only if $E\cap L$ is contractible for every complex affine line $L$ in $\mathbb{C}^n$?

答:
Strictly speaking this is not true since $\emptyset$ is not contractible. So let us interpret the condition in the sense that for every real affine line $L$ in $\mathbb{R}^n$, the set $E\cap L$ is either empty or contractible.

You know that $E$ is convex iff $E\cap L$ is convex for every real affine line $L$ in $\mathbb{R}^n$. Thus your first question is equivalent to showing that a subset $A \subset \mathbb R$ is convex iff it is contractible.

Clearly nonempty convex sets are contractible. Now assume that $A$ is not convex. This means that there exist $x,y \in A$ such that $z_t = tx + (1-t)y \notin A$ for some $t \in [0,1]$. W.l.o.g. we may assume that $x \le y$. It is impossible that $x = y$ because then $z_t = x$ for all $t \in [0,1]$. Thus $x < y$. Hence $A$ is not connected because $A \cap (-\infty, z_t)$ and $A \cap (z_t,\infty)$ form a partition of $A$ in disjoint nonempty open sets. But a non-connected set is not contractible.


问:
Simple Math Problem
A bat and ball cost \$1.10.

The bat costs a dollar more then the ball.

How much does the ball cost?

If this is not the correct place to ask a question like this please tell me and I will remove it :)

The ball costs \$0.05, not \$0.10

答:
Many people come to the quick conclusion that the bat costs a dollar or x = 1.00 so when asked what the ball costs or y = ? they form the equation x + y = 1.10 and plugin 1 for x and solve for y.  Giving them that the ball is $0.10.

This is incorrect the bat cost a dolar more so x-1.00 = y, y being the cost of the ball.  And to find what the ball or bat costs on its own you have to set up a system of equations like so:

```<code>x-1.00 = y
x+y = 1.10

-x+1.00+x+y = 1.10-y
1.00+2y = 1.10
2y = 0.10
y = 0.05
</code>```

Wether or not the person who is asked this question gets it correct or not depends on if they think you are trying to trick them.  If they do think that you are tricking them they will really think about the problem and find the flaw in their original assumption.  If not they will quickly answer with the incorrect answer to this simple fourth grade style math problem.

Title: Simple Math Problem

Body:
Many people come to the quick conclusion that the bat costs a dollar or x = 1.00 so when asked what the ball costs or y = ? they form the equation x + y = 1.10 and plugin 1 for x and solve for y.  Giving them that the ball is $0.10.

This is incorrect the bat cost a dolar more so x-1.00 = y, y being the cost of the ball.  And to find what the ball or bat costs on its own you have to set up a system of equations like so:

x-1.00 = y
x+y = 1.10

-x+1.00+x+y = 1.10-y
1.00+2y = 1.10
2y = 0.10
y = 0.05

Wether or not the person who is asked this question gets it correct or not depends on if they think you are trying to trick them.  If they do think that you are tricking them they will really think about the problem and find the flaw in their original assumption.  If not they will quickly answer with the incorrect answer to this simple fourth grade style math problem.


问:
Find the probability of getting a heart on the first draw and a club on the second draw while drawing 2 cards successively and without replacement
I am currently having some difficulty with a card probability question (I guess combinatorics will haunt me for the whole life...).
The question is to find the probability of getting a heart on the first draw and a club on the second draw while drawing 2 cards successively and without replacement from the standard poker deck. There are 2 approaches that I came with but they seem to produce different results.

Logically, I believe the answer to be
$$
P(C|H)=\frac{\binom{13}{1}}{\binom{52}{1}}*\frac{\binom{13}{1}}{\binom{39}{1}}=\frac{1}{12}=0.083
$$
However, I can't formally prove that it is the right approach, so I tried using the conditional probability formula.

$$
P(C|H)=\frac{P(C\cap H)}{P(H)}=\frac{\binom{13}{1}*\binom{13}{1}}{\binom{52}{2}}/\frac{\binom{13}{1}}{\binom{52}{1}}=\frac{2}{51}=0.0392
$$

One(or even 2) of my answers are clearly wrong, so I would be glad if someone could tell me where my mistake is and how to fix one(or both) approaches.

答:
The straightforward way to answer this question is to see that the probability of a heart on the first draw is $13/52$, or, equivalently, $1/4$.

Then there are $51$ cards left, so the probability of a spade the next time is $13/51$.

Multiply.

(I haven't addressed your question about what's wrong with your proposed approaches. You don't say why you think either might be right.)


问:
Hensel's lemma question
Find a root$\bmod{3^4}$of the polynomial $f (x) = x^4 + 10x^2 + x + 3$.

I tried to successively derive the polynomial and study the congruences at $f (x) = 0,$ but I got nowhere. Is it the right way?

答:
First solve $x^4+10x^2+x+3\equiv0\bmod3$.

$x^4+x^2+x\equiv0\bmod3$

$x(x^3+x+1)\equiv0\bmod3$

$x(x-1)(x^2+x+2)\equiv0\bmod3$.

Solutions are $x\equiv0,1\bmod3$.

Now solve $x^4+10x^2+x+3\equiv0\bmod 9$.  $x=3k+0$ or $1$.

With $x=3k,$ it's $(3k)^4+(3k)^2+3k+3\equiv0\bmod9$

or $3k+3\equiv0\bmod9$ or $k+1\equiv0\bmod3$ or $k\equiv2\bmod3.$

With $x=3k+1$, it's $(3k+1)^4+(3k+1)^2+3k+1+3\equiv0\bmod9$

or $12k+1+6k+1+3k+3\equiv0\bmod9$ or $21k+6\equiv0\bmod9$ or

$7k+2\equiv0\bmod3$ or $k\equiv1\bmod3$.

Solutions are $x\equiv6,4\bmod9$.

Now solve $x^4+10x^2+x+3\equiv0\bmod27$.  $x=9n+4$ or $6$.

With $x=9n+4$, it's $(9n+4)^4+10(9n+4)^2+9n+4+3\equiv0\bmod27$

or $4\cdot9n\cdot4^3+4^4+10\cdot72n+160+9n+7\equiv0\mod27$ or $3033n+423\equiv0\bmod27$

or $337n+47\equiv0\bmod3$ or $n+2\equiv0\bmod3$ or $n\equiv1\bmod3$.

With $x=9n+6$, it's $(9n+6)^4+10(9n+6)^2+9n+6+3\equiv0\bmod27$

or $4\cdot9n\cdot6^3+6^4+1080n+360+9n+9\equiv0\bmod27$ or $8865n+1665\equiv0\bmod27$

or $985n+185\equiv0\bmod3$ or $n+2\equiv0\bmod3$ or $n\equiv1\bmod3$.

Solutions are $x\equiv13, 15\bmod 27$.

The solutions mod $27$ can then be lifted to solutions mod $81$,

and that is left as an exercise for the reader.

Title: Hensel's lemma question


问:
Probability question dealing with standard normal variables and De Moivre Laplace theorem
To examine the accuracy of an algorithm that selects random numbers from the set $\{1, 2, …, 40\},$ $100,000$ numbers are selected and there are $3500$ ones. Given that the expected number of ones is $2500,$ is it fair to say that this algorithm is not accurate?

The picture here is the steps I have so far

I uploaded this question on chegg, but I am still very confused with how the last three lines are determined. How do we know that $Z$ can be between $-3.8$ and $3.8.$ Any hints would be incredibly appreciated!!

答:
Let the number of 1's in $n=100,000$ draws with probability $p=1/40$ of 1 at each draw be $X \sim \mathsf{Binom}(n,p).$ In R, $P(X \ge 3000) = 1 - P(X \le 2999) \approx 0.$

```<code>1 - pbinom(2999, 100000, 1/40)
[1] 0
</code>```

As mentioned in the question, $\mu_X=E(X) = np = 2500.$ Also, $\sigma_X = SD(X) = \sqrt{np(1-p)} = 49.3710.$

```<code>sqrt(100000*(1/40)*(39/40))
[1] 49.37104
</code>```

Thus $3000$ is $\frac{3000-2500}{49.27} = 8.7152$ standard deviations above the mean. Binomial distributions with such large $n$ are approximately normal, and the Empirical Rule
states that there is almost no probability of having an
observation more than 3 standard deviations above the mean.

```<code>(3000 - 2500)/57.3710
[1] 8.715205
</code>```

A normal approximation to $P(X \ge 3000) = P(X \ge 2999.5)\approx 0$
can be obtained directly in R by using a normal PDF:

```<code>1 - pnorm(2999.5, 2500, 57.3719)
[1] 0
</code>```

Ordinarily, you could also come close to a normal approximation by standardizing and using printed normal tables, as follows:

$$P(X \ge 3000) = P(X > 2999.5)\\
= P\left(\frac{X-\mu_x}{\sigma_x} >
\frac{2999.5-2500}{57.3790}\right)\\
\approx P(Z > 8.7043) \approx 0.$$
where the final result is far beyond tabled values,
and so must be taken as $0.$

```<code>(2999.5-2500)/57.3790
[1] 8.705275
</code>```

Your posted handwritten work seems mainly in the
right direction, except that I can't read the
last few lines.

Body:
Let the number of 1's in $n=100,000$ draws with probability $p=1/40$ of 1 at each draw be $X \sim \mathsf{Binom}(n,p).$ In R, $P(X \ge 3000) = 1 - P(X \le 2999) \approx 0.$

1 - pbinom(2999, 100000, 1/40)
[1] 0

As mentioned in the question, $\mu_X=E(X) = np = 2500.$ Also, $\sigma_X = SD(X) = \sqrt{np(1-p)} = 49.3710.$

sqrt(100000*(1/40)*(39/40))
[1] 49.37104

Thus $3000$ is $\frac{3000-2500}{49.27} = 8.7152$ standard deviations above the mean. Binomial distributions with such large $n$ are approximately normal, and the Empirical Rule
states that there is almost no probability of having an
observation more than 3 standard deviations above the mean.

(3000 - 2500)/57.3710
[1] 8.715205

A normal approximation to $P(X \ge 3000) = P(X \ge 2999.5)\approx 0$
can be obtained directly in R by using a normal PDF:

1 - pnorm(2999.5, 2500, 57.3719)
[1] 0

Ordinarily, you could also come close to a normal approximation by standardizing and using printed normal tables, as follows:

$$P(X \ge 3000) = P(X > 2999.5)\\
= P\left(\frac{X-\mu_x}{\sigma_x} >
\frac{2999.5-2500}{57.3790}\right)\\
\approx P(Z > 8.7043) \approx 0.$$
where the final result is far beyond tabled values,
and so must be taken as $0.$

(2999.5-2500)/57.3790
[1] 8.705275

Your posted handwritten work seems mainly in the
right direction, except that I can't read the
last few lines.


问:
A matrix game without a pure strategy equilibrium
I would like to see a matrix $2\times 2$ game without a pure strategy Nash equilibrium.

答:
\begin{array}{|c|c|}
\hline
& H & T \\
\hline
H & (1, 0) & (0, 1) \\ \hline
T & (0, 1) & (1, 0) \\ \hline
\end{array}

There is no Nash equilibrium in pure strategies, but there exists a Nash equilibrium in mixed strategies, both players choose H with probability $\frac{1}{2}$ and T with probability $\frac{1}{2}$.


问:
Quotient of $k$-algebra by a maximal ideal
If $k$ is a field, and $A$ a $k$-algebra which is finite dimensional as a k-vector space.  Are we allowed to say that for a maximal ideal $m$ the field $A/m$ is isomorphic to $k$ or not necessarily? If not, does $A/m$ have some special relationship with $\bar{k}$?

答: If $K/k$ is a finite field extension, then $K$ is naturally a $k$-algebra and, of course, finite-dimensional as a $k$-vector space. However, since $K$ is a field, the unique maximal ideal of $K$ is $(0)$ and the quotient field is $K$ itself. Converseley, in general, $A/\mathfrak{m}$ is a finite field extension of $k$ and as such embeds into $\overline{k}$.


问:
Can a triangle be divided into four similar triangles such that not all four of the triangles are congruent to each other?
I understand that you can divide a triangle into four congruent triangles by connecting the midpoints of each side. Can ANY NON-EQUILATERAL triangle be divided into four similar triangles with the restriction that not all four triangles can be congruent to each other?
As I explore this question, I keep running into dead ends, and I ask if any of you can help.

EDIT: You guys revealed that there are multiple ways to do this with right triangles. I've been experimenting with a general case and right triangles, but the closest I've gotten is splitting the triangle three times (on triangle ABC, drawing a line from Angle BAC that is perpendicular to Side BC, calling the point of intersection on Line BC Point D, then drawing lines from Angles ADB and ADC to be perpendicular with Lines AB and AC, respectively) yet I cannot prove that the triangles within ACD are similar to the triangles within ABD unless they are all within a right triangle. How to proceed?

答: For a right isosceles triangle, bisect its right angle.  Select one of the smaller triangles thus formed and bisect that right angle.  Do the latter step again.  QEF (Latin, which was to be done).</p>


答:
Let $N$ be the set of positive integers. Define a metric on $N$ as follows:
$$d(m,n)=\left\{\begin{array}{ll} 1+ \frac{1}{mn}& \mbox{if }m \neq n\\ 0 &\mbox{if }m=n \end{array}\right.$$
It is straightforward to check that this is a complete metric on $N$ (complete because this metric is discrete).

The closed balls  $$B(n, 1+1/n^2)=\{m: m\ge n\}, n=1,2,\cdots$$ are decreasing and have
empty intersection. Of course, the corresponding open balls also have empty intersection.

Title: Can the intersection of open or closed balls be empty, if their radii are bounded from below?

Body:
I am wondering about the following question: </p>

<p>Given a (countable) sequence of nested open balls: </p>

<p>$$ B_1 \supseteq B_2 \supseteq \cdots $$
Not necessarily having the same same center. All having radius bounded from below, say by $r > 0$. Then can we say that $$\bigcap_{i=1}^{\infty} B_i \neq \varnothing$$
it is certainly true in the reals, as one can simply go to the point where the radii are close to $r$ then take the center of that ball. However, I'm having trouble seeing whether or not it is true in the general case. So thanks in advance for proof or counterexample. </p>


问:
Under what conditions does $ \lim_{n \to \infty} \frac{1}{n} \sum_{i=1}^n a_{i,n}= \lim_{n \to \infty} \frac{1}{n} \sum_{i=1}^n a_{i,\infty}$
Suppose we have  sequence $\{a_{i,n} \}$ of non-negative numbers.

Consider the following two limits
\begin{align}
 L_1&=\lim_{n \to \infty} \frac{1}{n} \sum_{i=1}^n a_{i,n}\\
 L_2&= \lim_{n \to \infty}   \frac{1}{n}  \sum_{i=1}^n a_{i,\infty}
\end{align}
where $a_{i,\infty} =\lim_{n \to \infty} a_{i,n}$.

My question: under what conditions does $L_1=L_2$?  Are there any good references for this or keywords?

答:
A sufficient condition is that $a_{i,n} \to a_{i,\infty}$ as $n \to \infty$ uniformly for all $i \in \mathbb{N}$.

Note that

$$|L_1- L_2| = \left|L_1 - \frac{1}{n}\sum_{i=1}^n a_{i,n} +\frac{1}{n}\sum_{i=1}^n a_{i,n} - \frac{1}{n}\sum_{i=1}^n a_{i,\infty}+\frac{1}{n}\sum_{i=1}^n a_{i,\infty}- L_2\right|\\ \leqslant  \left|L_1 - \frac{1}{n}\sum_{i=1}^n a_{i,n}\right| +  \left|L_2 - \frac{1}{n}\sum_{i=1}^n a_{i,\infty}\right|+ \left|\frac{1}{n}\sum_{i=1}^n a_{i,n} - \frac{1}{n}\sum_{i=1}^n a_{i,\infty} \right| \\ \leqslant \left|L_1 - \frac{1}{n}\sum_{i=1}^n a_{i,n}\right| +  \left|L_2 - \frac{1}{n}\sum_{i=1}^n a_{i,\infty}\right|+ \frac{1}{n}\sum_{i=1}^n |a_{i,n} - a_{i,\infty}|$$

Since $a_{i,n} \to a_{i,\infty}$ uniformly, there exists $N \in \mathbb{N}$  such that $|a_{i,n} - a_{i, \infty}|< \epsilon$ for all $i \in \mathbb{N}$ when $n > N$, and, hence,

$$ \frac{1}{n}\sum_{i=1}^n |a_{i,n} -  a_{i,\infty}| < \frac{1}{n}\sum_{i=1}^n\epsilon =\epsilon$$

Thus for all $n > N$, we have

$$|L_1- L_2|< \left|L_1 - \frac{1}{n}\sum_{i=1}^n a_{i,n}\right| +  \left|L_2 - \frac{1}{n}\sum_{i=1}^n a_{i,\infty}\right| + \epsilon$$

and

$$|L_1- L_2| = \limsup_{n \to \infty} |L_1 - L_2| \\< \limsup_{n \to \infty}\left|L_1 - \frac{1}{n}\sum_{i=1}^n a_{i,n}\right| +  \limsup_{n \to \infty}\left|L_2 - \frac{1}{n}\sum_{i=1}^n a_{i,\infty}\right| + \epsilon = \epsilon$$

Since $\epsilon > 0$ can be arbitrarily small it follows that $L_1 = L_2.$


问:
What's the difference between a fingerprint and a hash?
My understanding is that both a fingerprint and a hash are functions that take as input some arbitrarily long bitstring, and output a bitstring of a fixed size.  The Wikipedia page for Hash Functions says:

Hash functions are related to (and often confused with) ... fingerprints

But try as I may, I can't find any sources that mention the difference between the two.  Do you know what the difference between a hash function and a fingerprinting function  is?

答:
I had the same question, it looks like the difference is the same as the difference between $P \Rightarrow Q$ and $P \iff Q$. Strictly speaking if $h$ is a hash function then it must be a fingerprinting function however the converse is not true, i. e., if $h$ is a fingerprinting function then we don't know if it is a hash function or not.

Notice that here by a hash function we mean what is colloquially referred to as a cryptographic hash function. If the context differentiates between a cryptographic hash function and just a hash function, that function may be a fingerprinting function and not a hash function. The Wikipedia article: Fingerprint (computing) says:

Mainstream cryptographic grade hash functions generally can serve as high-quality fingerprint functions, are subject to intense scrutiny from cryptanalysts, and have the advantage that they are believed to be safe against malicious attacks.

Therefore it feels like a set of all hash functions $\mathcal{H}$ is a smaller subset of the set of all fingerprinting functions $\mathcal{F}$. Therefore, we may write:

$$
f \in \mathcal{H} \Rightarrow f \in \mathcal{F}
$$


问:
Probability of getting the flu
Suppose that the probability of exposure to the flu during an epidemic is .6. Experience has
shown that a serum is 80% successful in preventing an inoculated person from acquiring the
flu, if exposed to it. A person not inoculated faces a probability of .90 of acquiring the flu if
exposed to it. Two persons, one inoculated and one not, perform a highly specialized task in a
business. Assume that they are not at the same location, are not in contact with the same people,
and cannot expose each other to the flu. What is the probability that at least one will get the flu?

The correct answer is 0.5952, I can't find my mistake.

A: vaccinated person catching the flu

B: unvaccinated person catching the flu

F: exposed to flu

thus, I must find P(A ∪ B)

P(F)=0.60

P(A | F)=0.20

P(B | F)=0.90

A, B ⊆ F

$P(A│F)=(P( A∩F))/(P(F))→P(A│F)=P( A)/(P(F))→ 0.20=P( A)/0.60→P(A)=0.12$

$P(B│F)=(P( B∩F))/(P(F))→P(B│F)=P( B)/(P(F))→ 0.90=P( B)/0.60→P(B)=0.54$

$P(A ∪ B)=P(A)+ P(B) -P(A∩B)=0.12+0.54-0=0.66$

Ans: 0.66

答:
Response already given addressing OP's specific question - therefore open season.

Alternative approach:

Chance of inoculated person getting flu is

$$p_1 = .6 \times .2 = 0.12.$$

Therefore, chance of inoculated person not getting flu is

$$q_1 = 1 - p_1 = 0.88.$$

Chance of non-inoculated person getting flu is

$$p_2 = .6 \times .9 = 0.54.$$

Therefore, chance of non-inoculated person not getting flu is

$$q_2 = 1 - p_2 = 0.46.$$

Chance of both not getting flu is

$$q_1 \times q_2 = (0.88) \times (0.46) = 0.4048.$$

Therefore, chance of at least one getting flue is

$$1 - (q_1 \times q_2) = 0.5952.$$

Title: Probability of getting the flu


问:
Some equation over algebraically closed field
Let $m\in\mathbb{N}$ and $k$ be an algebraically closed field. Consider $$X_1:=(x^1_1,\cdots,x^1_n), X_2:=(x^2_1,\cdots,x^2_n),\cdots,X_m:=(x^m_1,\cdots,x^m_n)$$ which are nonzero and whose components are in $k$. (nonzero means $X_i\neq(0,0,\cdots,0)$)

My question is:

Is there a $\Lambda:=(\lambda_1,\cdots,\lambda_n)\in k^n$ such that $X_i \circ \Lambda\neq 0$ (inner product) for every $i$?.

答: Given a nonzero vector $X$ in $k^n$, the set of all $\Lambda$ in $k^n$ such that $X\cdot\Lambda=0$ is a proper subspace of $k^n$. $k$, being algebraically closed, is an infinite field, and no (nontrivial) vector space over an infinite field can be expressed as a finite union of proper subspaces (there's a question about proving this, either here or on MathOverflow, maybe you can search for it if you're interested). So there must be $\Lambda$ in $k^n$ such that there is no $i$ for which $X_i\cdot\Lambda=0$.


答: java’s two-variable solution is perfectly fine, but you can also go straight to a single equation in one unknown. If $x$ is the cost of the ball in dollars, then the bat costs $x+1$ dollars, and the two together cost $x+(x+1)=2x+1$ dollars. You know that the total cost is $\$1.10$, so $2x+1=\$1.10$, $2x=\$0.10$, and $x=\$0.05$. It’s entirely a matter of personal preference which approach you take.</p>

Title: Simple Math Problem

Body:
A bat and ball cost \$1.10.

The bat costs a dollar more then the ball.

How much does the ball cost?

If this is not the correct place to ask a question like this please tell me and I will remove it :)

The ball costs \$0.05, not \$0.10


问:
How to find area covered by a car windshield wiper when it swaps a certain angle?
The problem is as follows:

A buggy is set to cross over a snowy terrain. The driver seat has a
rectangular window featured in the diagram from below. When the driver
activates the mechanism for cleaning the window from the snow, the
wiper spins $120^{\circ}$ using the point $M$ as a pivot. Assuming the
rotation is clockwise. Find the area covered by the windshield wiper
in the buggy. Assume the wiper has a T shape and the rectangular
window is $ABCD$ where $BC=2AB=2AE=2AM=1\,m$. Also assume AE is
perpendicular to $MN$ and $N$ is midpoint between $AE$.

The alternatives given in my book are as follows:

$\begin{array}{ll}
1.&\left(\frac{\pi}{6}-\frac{\sqrt{3}}{16}\right)\,m^{2}\\
2.&\left(\frac{\pi}{16}-\frac{\sqrt{3}}{6}\right)\,m^{2}\\
3.&\left(\frac{\pi}{4}-\frac{\sqrt{3}}{8}\right)\,m^{2}\\
4.&\left(\frac{\pi}{16}-\frac{\sqrt{3}}{16}\right)\,m^{2}\\
\end{array}$

So far the only thing which I could come up with was the diagram from below:

However I don't know how to find such area. I remember how to calculate the area of a half circle sector by means of the equation:

$A=\frac{\alpha}{360}\pi r^2$

Assuming $\alpha$ is an angle given in sexagesimal units. But in this case the figure doesn't help much. How exactly should I find that weird surface.

Can someone help me with a method relying euclidean geometry or something along precalculus?. I think integrals can be used but I am not looking for such sort of answer.

Please include a drawing in the answer because for me it is not easy to spot how to find such problematic area. Does it require some sort of construction?.

答:
The area we want to find out is the one surrounded by red in the figure above. This is the worst windshield wiper ever.

To verify this, we know that since $\overline{AE}\perp\overline{MN}$, the inner circle — with center $M$ and radius $\overline{MN}$ — will never be reached. Same with the two blue areas at the left and right bottom corner.

I assume that it was the inaccurate figure you have that misled you. Anyway, here's my solution. The length and area unit below are  $m$ and $m^2$, respectively.

Let's find out the length of $\overline{MN}$ first by noting that $\triangle ANM$ is a right triangle with $\angle AMN=30^{\circ}$. Therefore $$\overline{AM}=0.5\Longrightarrow \overline{MN}=\frac{\sqrt3}4$$

Now let's compute the red area. Let it be $S$.

$$\begin{align}S&=\frac{1}2\cdot (0.5)^2\pi-\frac{120}{360}\cdot \left(\frac{\sqrt3}4\right)^2\pi-2\triangle ANM\\
&=\frac{\pi}8-\frac{\pi}{16}-2\cdot \frac{1}2\cdot \frac{1}4\cdot\frac{\sqrt3}4\\&=\color{red}{\frac{\pi}{16}-\frac{\sqrt3}{16}}
\end{align}$$

which is the final answer.


问:
i'm 16 years old, what will I have to do from now on in the next few years to become a great mathematician?
I have 5 months to go to 17 years and I want to finish calculus 1 and in the other 5 months maybe finish lang linear algebra. I don't think I have talent or creativity (I'm so scared) but I want to commit myself to the maximum so that with hard work I can compete with guys maybe born with more creativity than me (is it possible?). Before I go to university I want to finish what you do in the first three years, and my question is how can I make the most of my time? Is it worth spending all day on a problem? (when I can read the solution), how can I improve my creativity in math? (I'm afraid I don't have it), I have so much difficulty with the problems of the mathematics Olympics but I am also working on those in order to have a medal at the national level (if only I had committed myself a few years earlier. ..). How should I approach the proofs (they also scare me), in my calculus book there were some problems that asked to prove simple things like 0b = 0 but I couldn't do it, is it serious? Could you give me a method to follow in order to become an excellent mathematician and solve important open problems? Thanks</p>

答:
Don't put pressure on yourself to be a great mathematician. It's no way to start out. It's also, I think, the wrong way to approach math.

Math is a massively collaborative discipline with many, many overlapping and intersecting subfields that run the gamut from applied to theoretical. Mathematical discoveries are generally not the product of a lone genius making a sudden breakthrough, but the fruit of a community of mathematicians making incremental progress on a problem.

The point being: you should think less about how you'll become a great mathematician and more about how you'll eventually fit into the mathematical community.

If you really want to be ahead of your peers, see if you can find a topic or a result that interests you. Dive into the literature and textbooks related to that topic or result and see what you can learn. Don't understand what you're reading? Try to figure out what you need to learn in order to learn that thing. Maybe even pick another topic. Have a university in mind? What are the professors at the university researching? Can you understand their work? What would you need to know to understand their work? Is it interesting? These sorts of investigations are how you get involved with the mathematical community and start to become a productive mathematician.

If you were destined to be "great" (whatever that means), you will be, but only after you've figured out how to work with other mathematicians and build on the results of others. If you can get a jump on these sorts of things, I promise you'll be well ahead of your peers, even if you don't know as much math as them at the start of your degree.

I hope this isn't too scattershot. These are just some of the things I wish someone would have told me before I started my undergraduate math degree. Great mathematicians didn't just start great (some of them may have even started out quite mediocre), they also knew how to keep getting better. A big part of that continual growth is knowing how to learn from your peers and how to research on your own.


问:
Proof of the Cayley Transform. Showing that $H=i(I-U)(I+U)^{-1}$ where $U=(I+iH)(I-iH)^{-1}$ for a Hermitian matrix $H$.
I am working on showing that  a Hermitian matrix $H=i(I-U)(I+U)^{-1}$ where $U=(I+iH)(I-iH)^{-1}$.

I have already shown that $I-iH$ is invertible and that $U$ is unitary, but am stuck on showing that $U$ can be written in this form. Any help would be greatly appreciated !

Thank you so much!

答:
If you start with $U=(I+iH)(I-iH)^{-1}$, and you want to find $H$ in terms of $U$, start by multiplying by sides on the right by $I-iH$:
$$
               U(I-iH)=(I+iH) \\
               U-iUH = I+iH \\
               U-I = i(U+I)H \\
               -i(U-I)(U+I)^{-1}=H\\
               H=i(I-U)(I+U)^{-1}.
$$


问:
Let $G$ be a finite abelian group, and let $n$ divide $|G|$. Let $m$ be the number of solutions of $x^n=1$. Prove that $n\mid m$.

Let $G$ be a finite abelian group, and let $n$ divide $|G|$.  Let $m$ be the number of solutions of $x^n=1$.  Prove that $n\mid m$.

My attempt

It's tempting to find a way to use Lagrange's theorem.  Maybe something here is a subgroup of something else?  We can fix $n$ and take the subgroup of $G$ of all elements which solve $x^n=1$.  Proof that this is a subgroup:  Inverses of solutions are always solutions.  Because the group is abelian, products of solutions are solutions.  QED.

Great, so it's a subgroup, so $m$ divides the order of $G$.  So does $n$.  I'm not sure that this really got me anywhere.  It'd be nice if there were some relevant subgroup of order $n$.

Being finite and abelian then it has a representation as $G\cong C_{p_1^{n_1}}\times\dots\times C_{p_k^{n_k}}$, a product of cyclic groups of prime power order.  The solutions are exactly the product of solutions "in each factor", i.e. solutions of the form $\langle e, \dots, e, x, e, \dots, e\rangle$ where $x\in C_{p_i^{k_i}}$ for some $i$.  So perhaps something comes from thinking about the number of solutions to $x^n=1$ where $x$ is taken from $C_{p_i^{k_i}}$.

Again this is a subgroup so the number of solutions divides $p_i^{k_i}$, and $p_i^{k_i}$ divides $|G|$. And $n$ divides the order of $G$.  But at this point I'm not sure whether I'm on a productive path, since these facts don't seem to be enough to show that $n|m$.

In fact the more that I think about how $n$ is so-to-speak missing factors from $|G|$ the more I think that finding numbers which divide $|G|$ just isn't a productive path.

答: By the fundamental theorem of finite abelian groups we may choose a $G$ subgroup $G_n$ of size $n$. Lagrange's theorem gaurantees $G_n\leq\ker(\varphi_n)$ where $\varphi_n$ denotes the $G$ endomorphism $\varphi_n:x\mapsto x^n$ and $\leq$ denotes subgroup inclusion. Finally, by Lagrange's theorem once again, $$n=|G_n|\;\Big\vert\;|\ker(\varphi_n)|=m$$


问:
Analytic function with $f(z)=z$ for all $z$
I'm working on the following problem:

Suppose $f:U\to U$ is analytic on a bounded connected domain $U\subset\mathbb C$. Show that if $0\in U$, $f(0)=0$ and $f'(0)=1$, then $f=\text{id}$ on all of $U$.

A hint I am given is to compute the first nonzero coefficient in the Taylor expansion of the function $f_n(z)-z$ at $0$, where $f_n=\underbrace{f\circ\cdots\circ f}_{n}$, but I'm not quite sure how to use this hint. Does anybody have any hints on how to solve this problem?

答:
The crucial result one uses here is that if $0 \in U$ bounded region and $g:U \to U$ an analytic self-map of $U$ with Taylor series at the origin $\sum a_k(g)z^k$ then $|a_k(g)| \le C_{k,U}$ for some constant independent of $g$ (depending only on $k,U$)

Assume this and use the hint considering $f(z)=z+a_2z^2+...a_nz^n+..$ as in the hypothesis and let's prove that $a_n=0, n \ge 2$ by induction (this proves that $f(z)=z$ on an open set containing zero where the expansion is valid, hence by the identity theorem, $f(z)=z$ in the whole region $U$).

Notice that for $f_n=\underbrace{f\circ\cdots\circ f}_{n}$ we have $f_2(z)=f(z)+a_2f(z)^2+..=z+2a_2z^2+z^3g_2$ so immediately $f_n(z)=z+na_2+z^3g_n$, while if $U$ is contained in a ball centered at $0$ and of radius $M$, we have $|f_n(z)| \le M, z \in U$ as all $f_n$ are self-maps of $U$ hence $n|a_2|=|a_2(f_n)| \le C_2$ which forces $a_2=0$

Now assuming $a_2=..=a_{k-1}=0, k \ge 3$. let's show that $a_k=0$ and the method should be clear as now $f(z)=z+a_kz^k+z^{k+1}h_1$ implies $f_2(z)=f(z)+a_kf(z)^k+...=z+2a_kz^k+z^{k+1}h_2$ and hence $f_n(z)=z+na_kz^k+z^{k+1}h_n$, so again by the result above $n|a_k| \le C_k$ for all $n$ hence $a_k=0$ and we are done modulo said result

(which is an easy consequence of Cauchy theorem applied to a closed disc centered at zero of fixed radius $r$ contained in $U$ with boundary circle $C$, for which $|a_k(g)|=|g^{(k)}(0)/k!|=|\frac{1}{2\pi i}\oint_C \frac{g(w)dw}{(z-w)^{k+1}}| \le \frac{M}{2\pi r^k}=C_{k,U}$ as required, where as before the disc centered at zero and of radius $M$ conatins $U$, so any self-map of $U$, satisfies $|g(z)| \le M$))


问:
Using $\gcd(a,b)$ to find gcd of other values $\gcd(a^2,b)$ and $\gcd(a^3,b)$
If $\gcd(a,b) = p$, a prime, what are the possible values of $\gcd(a^2,b)$ and $\gcd(a^3,b)$?

Through examples I've been able to find the answer, but I don't know how to come up with a proof.

Update: I think gcd of either is $p^2$, but again this is just by working examples.

I'm not familiar with much number theory. I've just started the second section of my textbook, so all I know is just divisibility and primes.

答: It is easy to prove that is $\gcd(a, b) = m$ then $\gcd(a / m, b / m) = 1$, $\gcd(k a, k b) = k \gcd(a, b)$, and if $\gcd(a, b) = \gcd(a, c) = 1$ then $\gcd(a, b c) = 1$. Splice those together.</p>


问:
Annulus of convergence of $e^{z+1/z}$
I'm working on the following problem:

Determine the Laurent series $f(z)=e^{z+1/z}$ and find its annulus of convergence about the origin.

What I've done so far: Using $e^w=\sum_0^\infty w^n/n!$ I can find that
$$
\begin{aligned}e^ze^{1/z}&=\Big(\sum_{k=0}^\infty \frac{z^k}{k!}\Big)\Big(\sum_{m=0}^\infty\frac1{m!z^m}\Big)\\&=\sum_{n=-\infty}^\infty\Big(\sum_{m=0}^\infty\frac1{(m+n)!m!}\Big)z^n,\end{aligned}
$$
where $1/k!=0$ for $k<0$.

Now, I'm not sure how to compute the annulus of convergence... any help would be appreciated!

答:
For example, the term in the product that is a constant times $z^6$ should be
$$
\frac 1 {6!0!} z^6\cdot z^0 + \frac 1 {7!1!} z^7\cdot z^{-1} + \frac 1 {8!2!} z^8\cdot z^{-2} + \frac 1{9!3!} z^9\cdot z^{-3} + \cdots
$$
And the term that is a constant times $z^{-6}$ is the same except that, for example, instead of $\dfrac 1 {9!3!} z^9\cdot z^{-3}$ one has $\dfrac 1 {9!3!} z^{-9}\cdot z^3.$ Thus both of those terms have the same coefficient.

So that coefficient is
$$
\sum_{k=0}^\infty \frac 1 {(|n|+k)!k!}
$$
So we can write
$$
e^{z + 1/z} = 1 + \sum_{n=1}^\infty \sum_{k=0}^\infty \frac 1 {(n+k)!k!} \left( z^n + \frac 1 {z^n} \right).
$$

Title: Annulus of convergence of $e^{z+1/z}$


答: We wish to use induction to show $f(z)=z+a_nz^n+O(z^{n+1})$ for all $n>1$. Given some $n$ for which this holds, use induction to show that $f_k(z)=z+ka_nz^n+O(z^{n+1})$, where $f_k$ represents $k$ compositions of $f$ with itself. Now use the Cauchy inequalities to show that $|f_k^{(n)}(0)|=n!k|a_n|$ is bounded by something in terms of $n$ but not in terms of $k$. Then we may take $k\to\infty$ to show that $a_n=0$. This holds for all $n\in\mathbb N$, which completes the argument.

Title: Analytic function with $f(z)=z$ for all $z$

Body:
I'm working on the following problem:

Suppose $f:U\to U$ is analytic on a bounded connected domain $U\subset\mathbb C$. Show that if $0\in U$, $f(0)=0$ and $f'(0)=1$, then $f=\text{id}$ on all of $U$.

A hint I am given is to compute the first nonzero coefficient in the Taylor expansion of the function $f_n(z)-z$ at $0$, where $f_n=\underbrace{f\circ\cdots\circ f}_{n}$, but I'm not quite sure how to use this hint. Does anybody have any hints on how to solve this problem?


问:
$\frac{\partial u}{\partial t} = \frac{\partial^2 u}{\partial x^2}$ with $u(x,0)=\frac{e^{2x}-1}{e^{2x}+1}$. Then $\lim_{t \to \infty}u(1,t)$ equals
Let $u(x,t)$ be a bounded solution of $\frac{\partial u}{\partial t} = \frac{\partial^2 u}{\partial x^2}$ with $u(x,0)=\frac{e^{2x}-1}{e^{2x}+1}$. Then $\lim_{t\to \infty}u(1,t)$ equals:

In the above problem, the boundary is not given but let's assume that the boundary conditions are as follows: $u(0,t)=u(\pi,t)=0$. Then how to solve for the bounded equation and how to find the above limit?

I will show my working below: with the given (assumed) initial conditions,
$$u(x,t)=\sum_{n=1}^{\infty}a_n e^{-\frac{n^2\pi^2t}{\pi^2}}\sin(\frac{n\pi x}{\pi})=\sum_{n=1}^{\infty}a_n e^{-{n^2t}}\sin(nx)$$
Now,  $u(x,0)=f(x)=\frac{e^{2x}-1}{e^{2x}+1}=\sum_{n=1}^{\infty}a_n \sin(nx)$, which is a fourier sine series, thus the constants $a_n$ are given by:
$$a_n=\frac{2}{\pi}\int_{0}^{\pi}f(x)\sin(nx)dx$$

But I am unable to determine $a_n$ or solve the problem. Can someone please help me solve this problem?
Thanks in advance.
The ans is :

$ -\frac{1}{2}$

答:
Case $x\in \mathbb{R}$

Here I suppose that the domain is $(x,t)\in \mathbb{R}\times \mathbb{R}^+$ and compute the solution as convolution of the initial datum $u(x,0)=\frac{e^{2x}-1}{e^{2x}+1}=\tanh(x)$ with the fundamental solution:
$$
u(1,t)=\frac{1}{\sqrt{4\pi t}}\int_\mathbb{R} e^{-\frac{|y-1|^2}{4t}} \tanh(y) \,dy.
$$
With the change of variable $w=\frac{y-1}{2\sqrt{t}}$ the integral becomes
$$
2\sqrt{t}\int_\mathbb{R} e^{-w^2} \underbrace{\frac{e^{4\sqrt{t}w+2}-1}{e^{4\sqrt{t}w+2}+1}}_{=:f(w,t)} \,dw.
$$
Now we split the domain in $(-\infty,0)$ and $[0,\infty)$. The integrand satisfies $|f(w,t)|\leq e^{-w^2}\in L^1(\mathbb{R})$ hence we can apply the Dominate convergence theorem and get
$$
\int_{-\infty}^0 e^{-w^2} \frac{e^{4\sqrt{t}w+2}-1}{e^{4\sqrt{t}w+2}+1} \,dw\to \int_{-\infty}^0 e^{-w^2}(-1) \,dw=-\frac{\pi}{2},\qquad
\int_0^{+\infty} e^{-w^2} \frac{e^{4\sqrt{t}w+2}-1}{e^{4\sqrt{t}w+2}+1} \,dw\to \int_{0}^{+\infty} e^{-w^2} \,dw=\frac{\pi}{2}, \quad \text{as} \quad t \to +\infty
$$
Hence we obtain $u(1,t) \to 0$ as $t\to +\infty$.

$\lim\limits_{t\to \infty}u(1,t)=-1/2$ if $u(x,0)\equiv 0$ for $x> 0$.

The only way I got $\lim_{t\to \infty}u(1,t)=-1/2$ as answer is by using the initial condition $u(x,0)=\tanh(x)$ in $(-\infty,0)$ and defining it $u(x,0)\equiv 0$ for $x\geq 0$. Use the computation above only for $x<0$.

Case $x\in \mathbb{R}^+$

Some computations using the domain $(0,\infty)$ with the additional boundary condition $u(0,t)=0$, which is consistent with $u(x,0)=\tanh(x)$, gives the same result since $u(x,0)$ is odd.


答:
The drawing explains how you do it.

There are at least three different solutions.
You did not ask for a proof so I just demonstrated a potential solution

Title: Can a triangle be divided into four similar triangles such that not all four of the triangles are congruent to each other?

Body:
I understand that you can divide a triangle into four congruent triangles by connecting the midpoints of each side. Can ANY NON-EQUILATERAL triangle be divided into four similar triangles with the restriction that not all four triangles can be congruent to each other?
As I explore this question, I keep running into dead ends, and I ask if any of you can help.

EDIT: You guys revealed that there are multiple ways to do this with right triangles. I've been experimenting with a general case and right triangles, but the closest I've gotten is splitting the triangle three times (on triangle ABC, drawing a line from Angle BAC that is perpendicular to Side BC, calling the point of intersection on Line BC Point D, then drawing lines from Angles ADB and ADC to be perpendicular with Lines AB and AC, respectively) yet I cannot prove that the triangles within ACD are similar to the triangles within ABD unless they are all within a right triangle. How to proceed?


问:
prove $\sum_\text{cyc}\frac{a+2}{b+2}\le \sum_\text{cyc}\frac{a}{b}$

Prove if $a,b,c$ are positive $$\sum_\text{cyc}\frac{a+2}{b+2}\le \sum_\text{cyc}\frac{a}{b}$$

My proof:After rearranging we have to prove $$\sum_\text{cyc} \frac{b}{b^2+2b} \le \sum_\text{cyc} \frac{a}{b^2+2b}$$

As inequality is cyclic:

let $a\ge b\ge c$ then $$\frac{1}{a^2+2a}\le \frac{1}{b^2+2b}\le \frac{1}{c^2+2c}$$.The rest follows by rearrangement inequality.

The case $a\ge c\ge b$ is analogous.

Thus Proved!

Is it correct?...And any other alternative ways possible?

答:
I have an alternative proof.

We need to prove that
$$
\frac{a}{b} +
\frac{b}{c} +
\frac{c}{a} \geqslant
\frac{a + 2}{b + 2} +
\frac{b + 2}{c + 2} +
\frac{c + 2}{a + 2}
$$
Here we can write this inequality in $2$ forms:
$$
\frac{c}{a} -
\frac{c + 2}{a + 2} \geqslant
\frac{a + 2}{b + 2} -
\frac{a}{b} +
\frac{b + 2}{c + 2} -
\frac{b}{c}
$$
$$
\frac{c - a}{a^2 + 2a} =
\frac{b - a}{a^2 + 2a} + \frac{c - b}{a^2 + 2a}
\geqslant
\frac{b - a}{b^2 + 2b} +
\frac{c - b}{c^2 + 2c}
$$
And
$$
\frac{b}{c} - \frac{b + 2}{c + 2} +
\frac{c}{a} - \frac{c + 2}{a + 2} \geqslant
\frac{a + 2}{b + 2} - \frac{a}{b}
$$
$$
\frac{b - c}{c^2 + 2c} +
\frac{c - a}{a^2 + 2a} \geqslant
\frac{b - a}{b^2 - 2b}
$$
Let $\min{(a,b,c)} = a$.

Case I: $c\geqslant b\geqslant a$: Write inequality in the first form.

Case II: $b\geqslant c\geqslant a$: Write inequality in the second form.


问:
To use the two-point formula to find the linear equation relating $C$ and $F$:
I've tried to solve a problem which I'm going to give below. What I don't understand is that which variable is dependent and which is independent among $C$ and $F$. I think we can relate $C$ and $F$ using two equations. In one of the equations $C$ can be the dependent variable and $F$ the independent and in the other equation $C$ can be independent and $F$ can be dependent. However, the problem is as follows:

Let $C$ denote Celsius temperature and $F$ Fahrenheit temperature. Thus, $C=0$ and $F=32$  at the freezing point of water, while $C=100$ and $F=212$ at the boiling point of water. Use the two point formula to find the linear equation relating $C$ and $F$.

My solution:

Let $C_1 = 0$ and $C_2 = 100$. Likewise, $F_1=32$ and $F_2=212$. If $P(C_1,F_1)$ and $Q(C_2,F_2)$ are the two points on the line $L$ then we have the following equation for $L$:

$F-F_1= \bigg( \frac{F_2-F_1}{C_2-C_1}\bigg)(C-C_1)$

OR

$F-32= \bigg( \frac{212-32}{100-0}\bigg)(C-0)$

OR

$F = \frac{180}{100}C + 32$

OR

$F = \frac{9}{5}C+32$

Do I need to find the other relation between $C$ and $F$?

答:
Your steps in using the two point formula to find a relation between $F$ and $C$ is correct, and done well.

The resulting relation (equation) is your solution: $$F = \frac{9}{5}C+32$$

Note that given this equation, you can also represent $C$ as a function of $F$:

$$F = \frac{9}{5}C+32 \iff F - 32 = \frac 95 C \iff \frac 59(F - 32) = C$$

So $$F = \frac{9}{5}C+32 \quad \text{and} \quad C = \frac 59(F - 32)$$ represent equivalent equations, where the first expresses $F$ as a function of $C$, and the second expresses $C$ as a function of $F$.


问:
How do I reason with Skolem functions?
Given the first-order sentence.

$\forall x \forall y : (Overlap(x,y) \iff  (\exists z Part(z,x) \land Part(z,y)))~~~~~~~~~~~~~~~~$  (1)

Here is an outline (partial) proof that $Overlap$ is symmetric:

Proof 1
\begin{align}
& \text{1}~~\forall x \forall y : (Overlap(x,y) \iff  (\exists z Part(z,x) \land Part(z,y))) & \tag{Axiom}\\
&  \text{2}~~~~~~~Overlap(a,b) \tag{Assume} \\
&  \text{3}~~~~~~~\exists z (Part(z,a) \land Part(z,b)) \tag{1,2}\\
&  \text{4}~~~~~~~\exists z (Part(z,b) \land Part(z,a)) \tag{3}\\
&  \text{5}~~~~~~~Overlap(b,a)  \tag{1,2,4}\\
&  \text{6}~~\forall x \forall y : (Overlap(x,y) \implies Overlap(y,x)) \tag{$\forall$-Intro 2-5} \\
\end{align}
Note in line 4 the same existential $z$ is used.

Now I wish write the  axiom (1) in an equisatisfiable form that can be proved using equational logic and term rewriting. This usually requires Skolemization.  Sentence (1) can be Skolemized as follows:

$\forall x \forall y : (Overlap(x,y) \iff   (Part(Sk(x,y),x) \land Part(Sk(x,y),y))~~~~~~~~~~~$ (2)

Now I try to prove symmetry using (2) with a similar approach as in Proof 1:

Proof Attempt 2
\begin{align}
& \text{1}~~\forall x \forall y : (Overlap(x,y) \iff   (Part(Sk(x,y),x) \land Part(Sk(x,y),y))& \tag{Axiom}\\
&  \text{2}~~~~~~~~Overlap(a,b) \tag{Assume} \\
&  \text{3}~~~~~~~~Part(Sk(a,b),a) \land Part(Sk(a,b),b)  \tag{1,2}\\
&  \text{4}~~~~~~~~Part(Sk(a,b),b) \land Part(Sk(a,b),a)  \tag{3}\\
&  \text{5}~~~~~~~~Overlap(b,a)\tag{XX}\\
&  \text{6}~~\forall x \forall y : (Overlap(x,y) \implies Overlap(y,x)) \tag{XX} \\
\end{align}

I am stuck at lines 4-5 in Proof Attempt 2.By following the approach of Proof 1 there is an issue with the order of arguments to Skolem functions.

答:
This is a sketch of the main idea. The axiom is the same if the  universal quantifiers are interchanged.

Skolemization of this gives
$$\exists Sk \forall y \forall x \; Overlap(x, y) \Leftrightarrow Part(Sk(y, x), x) \land Part(Sk(y, x), y) \tag{a}\label{eqa}$$

which is equivalent to

$$\exists Sk \forall x \forall y \; Overlap(y, x) \Leftrightarrow Part(Sk(x, y), y) \land Part(Sk(x, y), x) \tag{b}\label{eqb}$$

by changing the names of the variables.

We can also show similarly from the Axiom that

$$\exists S \forall x \forall y \; Overlap(x, y) \Leftrightarrow Part(S(x, y), y) \land Part(S(x, y), x) \tag{c}\label{eqc}$$

From (\ref{eqb}) and (\ref{eqc}) and the assumption on the Skolem functions that $\exists Sk \exists S \forall x \forall y \; Sk(x, y) = S(x, y)$, it follows

$$\forall x \forall y \;  Overlap(x,y) \Leftrightarrow Overlap(y, x) $$

Title: How do I reason with Skolem functions?


问:
Linear Operator on $\mathcal{H} = l^2(\mathbb{N}\cup \{0\})$
This is taken from Conway's a course in functional analysis section 1.3 problem 3. It's a 3 part question I wanted to see if what I had below was correct and to ask about how I would approach part (c).

Let $\mathcal{H} = l^2(\mathbb{N}\cup \{0\})$.

(a) Show that if $\{\alpha_n\}\in \mathcal{H}$, then the power series $\sum_{n=0}^\infty\alpha_nz^n$ has radius of convergence $\geq 1$.

Since $\{\alpha_n\} \in \mathcal{H}$, we have $\sum_{n=0}^\infty |\alpha_n|^2 < \infty$. We hope to show that $\left\lvert\sum_{n=0}^\infty \alpha_nz^n\right\rvert< \infty$ for any $z<1$. For $z = 0$ the series is $0$ so let us show the series converges where $z \neq 0$. By Cauchy Schwarz inequality:
\begin{align}
    \left(\sum_{n=0}^\infty \alpha_nz^n\right)^2 &\leq \left(\sum_{n=0}^\infty |\alpha_n|^2\right)\left(\sum_{n=0}^\infty z^{2n}\right)\\
    &=\left(\sum_{n=0}^\infty |\alpha_n|^2\right)\left(\frac{1}{1 - z^2}\right)\\
    &<\infty
\end{align}

where the last inequality comes from $\{\alpha_n\} \in \mathcal{H}$, and the result follows.

(b) If $|\lambda|<1$ and $L:\mathcal{H}\rightarrow \mathbb{F}$ is defined by $L(\{\alpha_n\}) = \sum_{n=0}^\infty \alpha_n\lambda^n$ find th vector $h_0$ in $\mathcal{H}$ such that $L(h) = \langle h, h_0\rangle$ for every $h$ in $\mathcal{H}$.

For this we can let $h_0(n) = \overline{\lambda^n}$.

(c) What is the norm of the linear functional $L$ defined in (b)

We have $||L|| = \sup\{|L(h)|: ||h||= 1\}$, and here I"m thinking that I should use the Cauchy Schwarz inequality from part (a), but it doesn't seem to suggest a good candidate for $||L||$.

答:
Parts (a) and (b) are good. For part (c), the norm of a linear functional $h \mapsto \langle h, h_0\rangle$ is precisely $\|h_0\|$. And yes, this is proven by Cauchy-Schwarz: $|\langle h, h_0\rangle| \le \|h_0\|\|h\|$, with equality occurring when taking $h = h_0$.

So, we just need to compute $\left\|\left(\overline{\lambda^n}\right)_n\right\|$, which is a straightforward geometric series:
$$\left\|\left(\overline{\lambda^n}\right)_n\right\|^2 = \sum_{n=0}^\infty|\lambda|^{2n} = \frac{1}{1 - |\lambda|^2} \implies\|L\| = \frac{1}{\sqrt{1 - |\lambda|^2}}.$$


问:
Simple Math Problem
A bat and ball cost \$1.10.

The bat costs a dollar more then the ball.

How much does the ball cost?

If this is not the correct place to ask a question like this please tell me and I will remove it :)

The ball costs \$0.05, not \$0.10

答: Since a ball costs one dollar less than the bat, two ball cost one dollar less than a bat and a ball. Thus two balls cost 10 cents....</p>

Title: Simple Math Problem


问:
Evaluate the integral $\int_0^{\infty} \frac{b\ln{(1+ax)}-a\ln{(1+bx)}}{x^2} \,dx$
Evaluate the following integral

$$I=\int_0^{\infty} \frac{b\ln(1+ax)-a\ln(1+bx)}{x^2} \,dx$$ with $\ a,b\in\mathbb{R},\ 0<a<b$.

My first attempt was to write $b\ln(1+ax)-a\ln(1+bx)$ as another integral, so I could substitute in the initial integral and then, by reversing the order of integration, try to calculate $I$.

I've tried writing $$b\ln(1+ax)-a\ln(1+bx)=\frac{ab}{t}\cdot\ln(1+tx)\Biggr|_{t=b}^{t=a}=ab\int_b^a\left(\frac{x}{t(1+tx)}-\frac{\ln(1+tx)}{t^2}\right)\,dt$$ but it doesn't seem to help me evaluating $I$.

What else could I try?

答:
Consider first the antiderivative
$$f(c)=\int \frac{\log (c x+1)}{x^2}\,dx$$
A first integration by parts gives
$$f(c)=-\frac{\log (c x+1)}{x}+\int\frac{c}{x (c x+1)}\,dx$$ Partial fraction decomposition give
$$\int\frac{c}{x (c x+1)}\,dx=\int \left(\frac{c}{x}-\frac{c^2}{c x+1} \right)\,dx=c \log (x)-c\log (c x+1)$$ As a total
$$f(c)=c \log (x)-c \log (c x+1)-\frac{\log (c x+1)}{x}$$
Now, you consider
$$b f(a)- a f(b)=\frac{a (b x+1) \log (b x+1)-b (a x+1) \log (a x+1)}{x}$$ Taking the limits at the bounds
$$\int_0^\infty \left(b f(a)- a f(b) \right) dx=a b \log \left(\frac{b}{a}\right)$$


问:
Why is the derivative of $y = f(x^3)$ equal to $f'(x^3) * 3x^2$
Why is the derivative of this function:

$ y = f(x^3)$

equal to

$y'= f'(x^3)* 3x^2$

instead of just:

$y' = f'(x^3)$

I know it has something to do with one of the rules (chain?) but I am not sure how it would apply in this situation.

答: This is because of a particular rule that we call the chain rule. It can be derived as follows: let $f(g(x))$ be a composition of two differentiable functions. Then, $$ f(g(x))' = \lim_{x \to x_0} \frac{f(g(x))-f(g(x_0))}{x-x_0},\quad x=x_0 $$ Which is precisely $$\lim_{x\to x_0} \frac {f(g(x))-f(g(x_0))}{g(x)-g(x_0)}\cdot\frac{g(x)-g(x_0)}{x-x_0}$$ since $$\lim_{y\to y_0} \frac {f(y)-f(y_0)}{y-y_0} =  f'(y_0)$$ and $$\lim_{x\to x_0} \frac {g(x)-g(x_0)}{x-x_0}= g'(x_0)$$ we have, setting $g(x)=y,\quad g(x_0)=y_0$ $$ f(g(x_0))' = f'(g(x_0)) g'(x_0)$$ which is precisley the chain rule. Plugging in $g(x)=x^3$, we get $$f(g(x_0))'=f'(x_0^3)\cdot 3x_0^2$$ which is just the derivative of $f(x^3)$ at $x=x_0$.


问:
How to find the maximum distance from the front of a house when a certain amount of fence can be used?
The problem is as follows:

Alice and Willy had bought a new house. The figure from below is a
sketch of the house. The couple wants the home to be surrounded
completely by a wooden fence, which must occupy the maximum area as
posible in front of the house. This is is indicated by a lighter shade
in the drawing where $ABCD$ is a rectangle. The couple however, only
got the materials including the wood, nails and paint from the
hardware store to complete $98$ meters of the fence. Assuming the
maximum area the couple wants to enclose starts in front of the house
as indicated in the arrow up to the entrance of the house, How many
meters in a straight line from the house will the entrance to the
fence be located?

The alternatives in my book are as follows:

$\begin{array}{ll}
1.&18\,m\\
2.&14\,m\\
3.&16\,m\\
4.&25\,m\\
\end{array}$

I'm assuming that to solve this problem it is required the use of derivatives, since it will be generated a quadratic equation. But I don't know how to establish such equation. Can someone help me with this part?.

Since what they require is to get the maximum area, I believe it will be given by:

$a(b+c+2)=A$

Assuming the small segments b and c are between the entrance door in the fence.

Since $a+b+c=98$

Then this means:

$a(98-a+2)=A(a)$

Then this would mean:

$A(a)=a(100-a)=100a-a^2$

Since the function will be as follows:

$A(a)=100a-a^2$

Thus the way to attain the maximum will be the derivate equal to zero.

$A'(a)=100-2a=0$

$a=50$

Therefore this would be the maximum.

But this answer does not appear in the alternatives. Which part did I made a mistake?. Can someone help me here?. Please an answer must include a drawing since for me it is difficult to spot where to establish the right equation in order to maximize it.

答:
Your formula for the area is wrong.

Suppose that the side lengths AB and CD are each $x$ units.

Then, the width of the rectangle will be

$$98 - (2 \times x) + 2 = 100 - 2x.$$

This means that the area will be

$$f(x) = (100 - 2x) \times x = 100x - 2x^2.$$

Taking the derivative,

$$f'(x) = 100 - 4x.$$

This means that $f(x)$ will have a maximum when

$$100 - 4x = 0.$$

Addendum

Per OP's request:

Reactions to his recent questions.

First, see my comment that immediately follows your query.

I adopted the interpretation that the width of the fence is variable,
only because that was the only way that the problem could be attacked.

Let's take it one step at a time.

There are two possibilities:

either the width of the fence is some fixed distance a.

or the width of the fence is variable.

If the width of the fence is fixed, at the value $a$, then the problem makes
no sense!  Naturally, you would then use all of the fence, so the length
of each of the sides would be

$$\left(\frac{1}{2}\right) \times \left[98 - (a-2)\right] ~=~
\left[50 - \left(\frac{a}{2}\right)\right].$$

Since the assumption of a fixed-width fence leads to the conclusion that the
dimensions of the fence are fixed, the assumption is untenable.

Assuming that the dimensions of the width of the fence equals the variable $a$,
and that the dimensions of each side of the fence is $x$, then you have that

$$(a-2) + 2x = 98 \implies a = (100 - 2x).$$

Therefore, the problem is reduced to only dealing with one variable, $x$.

You have raised a separate question - why couldn't the problem be attacked
using two variables, $a$ and $x$.  Although this is feasible, as long as you
remain aware that $a$ must equal $(100 - 2x)$ it adds an unnessary complication.

The main reason that you had trouble relates to the first sentence that I posted
in my original answer: your formula for the area is wrong.

You had

$$a(b + c + 2) = A. \tag1$$

I soon as I saw that, I stopped reading your analysis, because that formula for
the area makes no sense.  If AB and CD have lengths $b$ and $c$ respectively,
where $b = c$, then the area is

$$A = (a \times b).$$

If (for some bizarre reason), $b \neq c$, then you have a trapezoid and the area
is

$$A = \left(a \times \frac{b+c}{2}\right).$$

Therefore, there was no way that your formula  in line (1) above, was correct.

You raised another point:

The other thing about BC it is just as taking it as a reference point
but not as a changing parameter.

I always assumed that points B and C were fixed.  The variable $x$ was intended
to represent the length of line BA, which I presumed to be equal to the length
of line CD.

Another question

As it stands can this problem be solved?

My original answer stands; I have already solved the problem.  The key points
in my solution are

The width of the fence is variable, rather than being the fixed length
of line BC.

The problem can be reduced to a single variable problem (which is clearly best)
by understanding that the width $a$ must equal $(100 - 2x).$


问:
solving linear recurrence - general solution confusion
I've been trying to get my head around this for days. I understand what is going on with the calculation of a linear recurrence and I also understand how the characteristic is obtained.

What is confusion me is the general solution.

The general solution to the recurrence $at(n) + bt(n-1) ct(n-2) = 0$

For a unique root $r$, the general solution is: $t(n) = (A + Bn)  r^n$. For two distinct roots $r_1$ and $r_2$, the general solution is: $t(n) = Ar_1^n + Br_2^n$

What are $A$ and $B$? What do they represent in the linear recurrence and in the general solution

I just can't see it and I need to get it so I can start my discrete math assignment ...

Thanks heaps in advance :)

答: $A$ and $B$ are related to the initial conditions of the linear recurrence.  To completely solve a linear recurrence of the form: $$at_n+bt_{n-1}+ct_{n-2}=0$$you need $a,b,c$, and also $t_0,t_1$, where the sequence starts. Given initial conditions, you can find a unique $A$ and $B$ that will give an exact solution.</p>


问:
Prove the equation using a combinatorial argument
Prove using a combinatorial argument that both sides count the same thing:

$${{n \choose 2} + 1 \choose 2} = {n \choose 2} + 3{n \choose 4} + 3{n \choose 3}$$

答:
Claim. Both sides represent the number of ways to choose one or two edges from a complete graph on n vertices.

Proof. The left hand side is obvious, while the three summands on the right hand side are respectively equal to the number of ways in which the chosen edges are incident to precisely 2,4,3 vertices.


问:
Subgroup generated by a subset of the group G
I'm trying to prove the set $A = \{a_1^{m_1}a_2^{m_2}\cdots a_n^{m_n}\mid m_i \in \mathbb{Z} \textrm{ and } 1 \leq i \leq n \}$ is equal to the set $\displaystyle \bigcap_{H \in \mathbb{\mathscr{L}}} H$
where $\mathscr{L}$ is the collection of all subgroups $H$ of $G$ that contains the set $S = \{a_1, a_2, \cdots, a_n\}$

For the first inclusion it was okay, it's easy to show that

$A \subset \displaystyle \bigcap_{H \in \mathbb{\mathscr{L}}} H$ , since $S \subset H$ and $H$ is a subgroup, so we can operate the elements $a_1, \cdots, a_n$ and find another element in $H$.

But I'm having some issues in the second inclusion.

We consider some $k \in \displaystyle \bigcap_{H \in \mathbb{\mathscr{L}}} H$, if $k \in A$, it means that $A$ is a subgroup of $G$ that contains $S.$

My attempt was the One-step test for subgroups,

$x, y \in A \implies xy^{-1} \in A$, so $A \leq G$

We consider $x,y \in A$, it follows that

$x = a_1^{m_1}a_2^{m_2}\cdots a_n^{m_n}$

$y = a_1^{l_1}a_2^{l_2}\cdots a_n^{l_n}$, with $m_i, l_i \in \mathbb{Z}$

Because of the sock-shoes property we have that

$y^{-1} = a_n^{-l_n}a_{n-1}^{-l_{n-1}}\cdots a_1^{-l_1}$

Then we have to show that

$xy^{-1} \in A$, I got stuck here, because for me it seems that I need some commutative property here.

That's the right way to solve this problem? if it's not, how can I show it? Don't give the answer please, just some tips.

答:
Please check again the question.
As stated by @NL1992, $a_2a_1\notin A$ but $a_2a_1\in \bigcap_{H \in \mathbb{\mathscr{L}}} H$. Unless $a_ia_j=a_ja_i$ for all $a_i,a_j\in S$,  otherwise these two sets may not be equal.

The correct version of $A$ should be
$$A=\{g_1^{m_1}\dots g_k^{m_k}:k\geq 1,m_1,\dots,m_k\in \Bbb{Z},g_1,\dots g_k\in S\}.$$

You can first show that $A$ is a subgroup of $G$ containing $S$. Since $\bigcap_{H \in \mathbb{\mathscr{L}}}H $ is the intersection of all subgroups containing $S$, we must have $A\subseteq \bigcap_{H \in \mathbb{\mathscr{L}}}H $.


问:
Simple COVID Probability Question
This is a relatively simple question that I have been struggling to answer for some reason. When conducting a thought experiment, I was wondering what the approximate probability of catching COVID would be if one were to host a 15 person house gathering. Just as a reference, if approximately 3% of the U.S has either come into contact with the virus or has it, and we were to account for 15 people in a confined space (approximately 4 independent families), how would one calculate the probability of one person catching the virus in a 15 person get-together?

Also, more broadly, what is the general structure of a statistical problem that outlines this kind of an issue? Is there a method or formulaic approach used to calculate questions like these? (I'm new to statistics and learning more each day!)

答:
You need to know the probability $P_v$ of catching the virus if you are in contact with someone who has.  The probability that no one has the virus among 15 people  (assuming independence - which is questionable, since they are in family groups) is $ P_0=0.97^{15}$, so your infection probability is $P_v(1-P_0)$.

An additional complication to the calculation is that $P_v$ would be somewhat dependent on the number of people present who are infected.  However in this scenario, the effect would be small.

Title: Simple COVID Probability Question


问:
Suppose that the function $f:(-1,1) \longrightarrow \mathbb{R}$ has n derivatives.
Suppose that the function $f:(-1,1) \longrightarrow \mathbb{R}$ has n derivatives. Assume that there is an $M>0$ such that $|f(x)| \leq M|x|^n$ for all $x\in (-1,1)$. Prove that $ f(0) = f'(0) = ... = f^{(n-1)}(0) = 0$ where n is a natural number

My attempt at the proof follows.

We are given that $|f(x)| \leq M|x|^n$ for all x on the interval $(-1,1)$. Since f is differentiable by defintion, it follows that $f(x)$ is continuous on (-1,1). Therefore if we let $ x = 0$, it follows that $|f(0)| \leq M|0|^n = 0$ and therefore it follows that $f(0) = 0$. Since $f$ has n derivatives it follows that $f'(0) = \lim_{x \to 0} \frac{f(x)- f(0)}{x}$ and therefore $|f'(0)| = \lim_{x \to 0} |\frac{f(x)- f(0)}{x}| = \lim_{x \to 0} |\frac{f(x)}{x}| \leq \lim_{x \to 0} \frac{M|x|^n}{|x|} = \lim_{x \to 0} M|x|^{n-1} = 0$.

Similarly, $f''(0) = \lim_{x \to 0} \frac{f'(x) - f'(0)}{x}$. This is where I get stuck. If we take the absolute value of $f''$ it would follow that $|f''(0)| = \lim_{x \to 0} |\frac{f'(x)}{x}|$. Ideally, I want to be able to get f(x) in the numerator so we can use the given inequality, but I am not sure how to do this. I was wondering if this proof is on the right track, and any help would be greatly appreciated.

答:
UPDATE: My answer is correct now. Sorry for the inconvenience.

We have that $|f(x)|\le M|x|^n$. We want to prove $f^{(k)}(0)=0$ for every $0\le k\le n-1$.

The case $k=0$ is trivial. Now we'll use strong induction, so suppose we have the result up to some $k$ such that $0\le k\le n-2$. Then $f^{(l)}(0)=0$ for every $0\le l\le k$. Since $f\in C^n\big((-1,1)\big)$ and $k\le n-2$, for every $x\in(-1,1)$ we can consider the Maclaurin series (Taylor series at $x=0$) of degree $k+1$ and express the remainder as $\dfrac{f^{(k+2)}(\xi_x)}{(k+2)!}x^{k+2}$ with $\xi_x$ strictly between $0$ and $x$.

Then, for $x\in(-1,1)$,

$f(x)=f(0)+f'(0)x+\dfrac{f''(0)}{2!}x^2+\ldots+\dfrac{f^{(k+1)}(0)}{(k+1)!}x^{k+1}+\dfrac{f^{(k+2)}(\xi_x)}{(k+2)!}x^{k+2}$.

Remember we have $f^{(l)}(0)=0$ for every $0\le l\le k$ by induction hypothesis, so

$f(x)=\dfrac{f^{(k+1)}(0)}{(k+1)!}x^{k+1}+\dfrac{f^{(k+2)}(\xi_x)}{(k+2)!}x^{k+2}$, from what we get

$f^{(k+1)}(0)=(k+1)!\dfrac{f(x)}{x^{k+1}}-\dfrac{f^{(k+2)}(\xi_x)}{k+2}x$

Now, since $f\in C^n\big((-1,1)\big)$ and $k+2\le n$, and taking $x\to0$ provoques $\xi_x\to0$, we have that

$f^{(k+2)}(\xi_x)\xrightarrow{x\to0}f^{(k+2)}(0)$.

Also remember that we had $|f(x)|\le M|x|^n\Rightarrow \dfrac{|f(x)|}{|x|^{k+1}}\le M|x|^{n-k-1}$ (and $n-k-1\ge1$).

Finally, $\left|f^{(k+1)}(0)\right|\le(k+1)!\dfrac{|f(x)|}{|x|^{k+1}}+\dfrac{\left|f^{(k+2)}(\xi_x)\right|}{k+2}|x|\le (k+1)!M|x|^{n-k-1}+\dfrac{\left|f^{(k+2)}(\xi_x)\right|}{k+2}|x|\xrightarrow{x\to0}0$

We get that $f^{(k+1)}(0)=0$, so we conclude it is true for every $0\le k\le n-1$ (note we can't extend this to $k=n$, since we needed the remainder of the Taylor series, which would be with $f^{(n+1)}$ in that case).


问:
Map a 3D point into its corresponding 2D coordinates on a plane
My problem is the following:

I have three 3D points, having coordinates $P_0 = (x_{p0}, y_{p0}, z_{p0})$, $P_1 = (x_{p1}, y_{p1}, z_{p1})$, $P_2 = (x_{p2}, y_{p2}, z_{p2})$. I know the 2D coordinates of (the image of) each of these 3D points onto a 2D plane, namely $T_0 = (x_{t0}, y_{t0})$, $T1_0 = (x_{t1}, y_{t1})$, $T_2 = (x_{t2}, y_{t2})$

Given a chosen point $Q = (x_{q0}, y_{q0}, z_{q0})$ lying onto the plane defined by the 3D points, I need to find its corresponding 2D coordinates.

Possible solutions

I found suggestions regarding the use of the barycenter of the 3D triangle, but I still cannot figure out how to do it.

I also thought of calculating transformation matrix, but that seems like an overcomplicated solution.

答: I'm not so sure what you mean by defining 2D coordinates from a 3D plane, but it seems like you could just remove the Z axis, so $Q=(x_{q0},y_{q0})$.


问:
Can this problem have solutions for $N>3$:Find three $(N=3)$ positive integers a, b and c such that: $a+b+c=k^2$, $a+b=t^2$, $b+c=m^2$ and $a+c=n^2$
Can this question have solutions for N>3:

Find three ($N =3$) positive integers a, b and c such that: $a+b+c=k^2$, $a+b=x^2$, $b+c=m^2$ and $a+c=n^2$; $x, m, n∈\mathbb   N$

Solution:

Let $a+b+c=(x+1)^2=x^2+2x+1$

If $a+b=x^2$ then third number $c=2x+1$.

If $b+c=(x-1)^2=x^2-2x+1=m^2$ then :

$a=4x$, so $b=x^2-4x$

Now due to statement we must have $a+c=6x+1=n^2$

$n^2=6x+1$ can have infinite solutions such as:

$(x, n^2)=(20, 121=11^2), (60, 361=19^2), (140, 841=29^2)\cdot\cdot\cdot$

Which give:

$(a, b, c, k)=(80, 320, 41, 21), (240, 3360, 121, 61), (560, 19040, 281, 141),\cdot\cdot\cdot$

I tried to solve this problem for $N=4$, four numbers but no success. Now I have two questions:

1-Does this problem have solutions for N>3?

2-Any idea for better method?

Body:
We have:

$a + b = u^2$, $c + d = v^2$

$a + c = w^2, b + d = x^2$

$a + d = y^2, b + c = z^2$

$a + b + c + d = k^2$

Let:

$u=m^2-n^2$ & $v =2mn$

$w=p^2-q^2$ & $x=2pq$

$y=r^2-s^2$ & $z=2rs$

We apply the condition:

$m^2+n^2=p^2+q^2=r^2+s^2=k^2$

Hence, $u^2+v^2=w^2+x^2=y^2+z^2=k^2$

Since:

$2a=u^2+w^2-z^2$

$2b=x^2+z^2-v^2$

$2c=v^2+w^2-y^2$

$2d=v^2+x^2-z^2$

Substituting value of $(u,v,w,x,y,z)$ from above we get:

$a=1/2(m^4+n^4+p^4+q^4)-(m^2n^2+p^2q^2+2r^2s^2)$

$b=2(p^2q^2+r^2s^2-m^2n^2)$

$c=1/2(p^4+q^4-r^4-s^4)+(2m^2n^2-p^2q^2+r^2s^2)$

$d=2(m^2n^2+p^2q^2-r^2s^2)$

For: $(m,n,p,q,r,s,k)=(15,10,17,6,18,1,325)$ we get:

$(a,b,c,d)=(39169,-23544,24840,65160)$


问:
Find the coefficient of $x^n$ in the generating functions: $g(x) = \frac{x^3}{(1+x)^5 (1−x)^6}$
One of the problems in my Discrete Math course states that we need to find the coefficient of $x^n$ in generating function $g(x) = \frac{x^3} {(1+x)^5 (1−x)^6}$

I separated $g(x) = \frac{x^3} {(1+x)^5 (1−x)^6}$ into $x^3$ * $\frac{1}{(1+x)^5}$ * $\frac{1}{(1-x)^6}$

Then got $f(x) = x^3 * \sum_{k=0}^\infty \binom{n+4}{4}(-1)^kx^k * \sum_{k=0}^\infty \binom{n+5}{5}x^k$

I dont know what to do from here, can you tell me what is the next step?

答:
The key is the generalized binomial theorem
in the handy special case
$(1-x)^{-s}
=\sum_{k=0}^{\infty} \binom{s+k-1}{k}x^k
=\sum_{k=0}^{\infty} \binom{s+k-1}{s-1}x^k
$.

This implies that

$(1+x)^{-s}
=\sum_{k=0}^{\infty} \binom{s+k-1}{k}(-1)^kx^k
=\sum_{k=0}^{\infty} \binom{s+k-1}{s-1}(-1)^kx^k
$

and,
for any $m$,

$(1-x^m)^{-s}
=\sum_{k=0}^{\infty} \binom{s+k-1}{k}x^{km}
=\sum_{k=0}^{\infty} \binom{s+k-1}{s-1}x^{km}
$.

For $m=2$ this is

$(1-x^2)^{-s}
=\sum_{k=0}^{\infty} \binom{s+k-1}{k}x^{2k}
=\sum_{k=0}^{\infty} \binom{s+k-1}{s-1}x^{2k}
$.

Your problem is

$\begin{array}\\
g(x)
&= \dfrac{x^3} {(1+x)^5 (1−x)^6}\\
&= \dfrac{x^3(1+x)} {(1+x)^6 (1−x)^6}\\
&= \dfrac{x^3(1+x)} {(1−x^2)^6}\\
&= \dfrac{x^3} {(1−x^2)^6}+\dfrac{x^4} {(1−x^2)^6}\\
\end{array}
$

The results above
should make this
straightforward.

Actually,
it's not that simple.
Here's the rest.

$\begin{array}\\
\dfrac{x^3} {(1−x^2)^6}
&=x^3\sum_{k=0}^{\infty} \binom{k+5}{5}x^{2k}\\
&=\sum_{k=0}^{\infty} \binom{k+5}{5}x^{2k+3}\\
&=\sum_{k=0}^{\infty} \binom{k+5}{5}x^{2(k+1)+1}\\
&=\sum_{k=1}^{\infty} \binom{k+4}{5}x^{2k+1}\\
\\
\dfrac{x^4} {(1−x^2)^6}
&=x^4\sum_{k=0}^{\infty} \binom{k+5}{5}x^{2k}\\
&=\sum_{k=0}^{\infty} \binom{k+5}{5}x^{2k+4}\\
&=\sum_{k=0}^{\infty} \binom{k+5}{5}x^{2(k+2)}\\
&=\sum_{k=2}^{\infty} \binom{k+3}{5}x^{2k}\\
\end{array}
$

This gives the coefficients
of the even and odd terms.

To map these cases
into a single one,
use
$h(k)
=k-2\lfloor k/2 \rfloor
$
where
$h(k) = 0$ for even $k$
and
$h(k) = 1$ for odd $k$.

Since we want
even $k \to \lfloor k/2 \rfloor+3$
and
odd $k \to \lfloor k/2 \rfloor+4$,
we want

$\begin{array}\\
k
&\to \lfloor k/2 \rfloor+3+h(k)\\
&\to \lfloor k/2 \rfloor+3+k-2\lfloor k/2 \rfloor\\
&\to k+3-\lfloor k/2 \rfloor\\
\end{array}
$

Therefore
$g(x)
=\sum_{k=3}^{\infty} x^k\binom{k+3-\lfloor k/2 \rfloor}{5}
$.