问:
Elementary connection between linear algebra and differential equations
While studying linear algebra I encountered the following result regarding the solutions of a non homogeneous system of linear equations: if $Sol(A,\pmb{b})\neq \emptyset$

$$ Sol(A,\pmb{b})= \pmb{x} + Sol(A,\pmb{0})$$

Where x is a particular solution.

Then, while studying differential equations, I found that the solutions for

$$y'=a(x)y+b(x)$$

are all of the form

$$y=y_p+Ce^{A(x)}$$

Where the last term refers to the solutions of the associate homogeneous equation and $y_p$ is a particular solution.

These two concepts seem related to me, but I've not been able to find/think about a satisfying reason, so I'd like to know whether there is actually a relation or not.

I should specify that I've not been able to study properly systems of differential equations of the first order or equations of the n-th order yet, I suspect that the answer is there but I can't be sure. However I'd like to get the complete answer, so if it involves the latter topics it's not a problem.

答:
This can be useful to visualize the correlation between linear algebra and systems of  differential equations:

If $A$ and $B$ are two linear spaces, let $\Gamma:A\to B$ be a linear operator with $b\in B$, then the level set of $b$ of $\Gamma$ can be obtained translating $Ker(\Gamma)$ of a factor $\Gamma^{-1}(b)$, hence
$$\forall \gamma\in\Gamma^{-1}(b):\Gamma^{-1}(b)=\gamma+Ker(\Gamma).$$
Let's now consider $y\in\mathcal C^{(n)}(I)$ and define
$$\Gamma_{a_0,\dots,a_{n-1}}(y):=y^{(n)}+\overset{n-1}{\underset{i=0}{\sum}}a_i(x)y^{(i)}.$$
The derivative defines a linear operation, so $\Gamma$ is a linear operator such that $\Gamma:\mathcal C^{(n)}(I)\to\mathcal C^{(0)}(I)$.

Now solving the system $(*)\begin{cases}y(x_0)=y_0\\\vdots\\y^{(n-1)}(x_0)=y_0^{(n-1)} \end{cases}$ is equivalent to $\Gamma(y)=b$, and solving the homogeneous equation associated to $(*)$ is equivalent to $\Gamma(y)=0$.

We have showed that the the general integral of $(*)$ is a linear variety and for this reason it can be expressed by the sum of a particular solution of $(*)$ with a solution of the associated homogeneous equation.


问:
Probability Game Optimal Strategy
You're playing a game where you have a .45 probability of winning and .55 probability of losing. You start out with 2000 chips and the winning condition is that if you bet a cumulative total of 10000 chips, then you win. Otherwise if you run out of chips, you lose.

For example, if you bet 20 chips, no matter the outcome, that will contribute 20 to the total. And you'll be left with either 1980 or 2020 current chips.

The min bet size is 1$ and max is as much as you currently have. What is the best strategy to maximize your probability of winning?

What I said was, you'd want to use a strategy akin to betting as much as possible at each round. Since otherwise, from LLN, you are losing in general and just prolonging with small bets will make you lose more in the long run. I said you'd be able to maybe calculate the exact strategy using dynamic programming, with f(s,t) representing what you should bet at each combination of (current money, total cumulative so far). However I do not know if this is correct nor did I have time to fully solve it.

Thank you!

答:
Your optimal, risk-minimizing strategy is to bet a single chip each round

Other users have provided strong evidence of this via simulations. However, it can be mathematically proven that the one-chip strategy is superior to any other betting method, with an absurdly high probability of success.

OP believes the law of large numbers argues in favor of making big bets in the hope of reaching $10,000$ total chips bet very quickly:

What I said was, you'd want to use a strategy akin to betting as much as possible at each round. Since otherwise, from LLN, you are losing in general and just prolonging with small bets will make you lose more in the long run. I said you'd be able to maybe calculate the exact strategy using dynamic programming, with f(s,t) representing what you should bet at each combination of (current money, total cumulative so far). However I do not know if this is correct nor did I have time to fully solve it.

But it's the other way around.

Suppose in each round $k \geq 1$, I stake $C_k$ chips, where $C_k$ is a positive integer.

Since I win with probability $0.45$, but lose with probability $0.55$, my expected change in wealth at round $k$ is $$0.45 C_k - 0.55 C_k = -0.1 C_k,$$ with a variance of $$1.1^2 C_k^2 (0.45) + 0.9^2 C_k^2 (0.55) = 0.99 C_k^2.$$ This implies that the more I stake on round $k$, the more I expect to lose, and the higher the volatility/risk I assume. My loss- and volatility-minimizing strategy for each individual round is then to bet $C_k = 1$ each time, so that I lose an average of $0.1$ chip per round. Since variance of independent variables is additive, the variance per round is about $0.99$ "chips squared".

This may not sound promising in the long run, as my wealth $W_k$ at the start of round $k$ is expected to trend inexorably downwards: $$\Bbb{E}[W_k] = W_0 - 0.1 k,$$ where $W_0$ is my starting bankroll.

But remember--we're not playing to amass chips, we're playing to stay in the game as long as possible. And placing any bet $C_k > 1$ actually decreases $\Bbb{E}[W_{k+1}]$ relative to the one-chip strategy. There is no betting strategy for $C_k$ that can improve on the one-chip expectation of $\Bbb{E}[W_{k+1}] = W_0 - 0.1 (k+1)$.

Furthermore, the one-chip strategy also achieves the lowest possible volatility: placing any bet $C_k > 1$ increases $\operatorname{Var}(W_{k+1})$, relative to the one-chip strategy, which has the following variance and standard deviation of bankroll:
\begin{align*}
\operatorname{Var}(W_k) &= 0.99 k \\
\sigma_{W_k} = \sqrt{\operatorname{Var}(W_k)} &= \sqrt{0.99k}, \
\end{align*} where $\sigma_{W_k}$ is the standard deviation of my bankroll $W_k$ at the start of round $k$.

So if we only bet a single chip each round, our starting bankroll of $W_0 = 2,000$ chips puts us in an excellent position to place $10,000$ chips worth of bets before we go bust, albeit at a snail's pace.

If we bet a single chip per round, it would take us $k = 10,000$ rounds to achieve the stated winning condition. But since we started with $2,000$ chips, we have $$\Bbb{E}[W_{10,000}] = W_0 - 0.1 (10,000) = 2,000 - 1,000 = 1,000.$$ That is, we expect to amble up to round $10,000$, and win, with a very comfortable cushion of $1,000$ chips to spare!

Not only that, but we would deviate from this amount, on average, by only $$\sigma_{W_{10,000}} = \sqrt{0.99 \times 10,000} = 99.5 \text{ chips.}$$ Notice how teeny that is compared to our expected bankroll? That implies we have a strong chance of success--in fact, a very strong chance of success.

My winning probability if I use the one-chip strategy is insanely high

My actual bankroll $W_{10,000}$ after $10,000$ rounds, using the one-chip strategy, is given by $$W_{10,000} = 2,000 + \sum_{i = 1}^{10,000} X_i,$$ where $X_i$ are iid $\{ \pm 1 \}$-valued random variables with $\Bbb{P}(X_i = +1) = 0.45$ for all $i = 1, 2, ..., 10,000.$

$10,000$ is so large that, by the Central Limit Theorem, $W_{10,000}$ is essentially a normal random variable with mean $\mu_{10,000} = 1,000$, variance $\sigma^2_{10,000} = 10,000(0.99) = 9,900,$ and standard deviation of $\sigma_{10,000} = \sqrt{9,900} \approx 99.5$ chips.

To have run out of chips after $10,000$ bets, i.e. $W_{10,000} = 0$, would correspond to a z-score of $$z = \frac{0 - 1,000}{99.5} = -10,$$ which has an absurdly small probability, of an order of magnitude around $e^{-10^2/2} \approx 10^{-22}$, or about one sextillionth.

The probability of me being bankrupt at an earlier round than the $10,000$th is even lower than the probability of me being bankrupt at the $10,000$th round. In fact, I won't even have the opportunity to go broke until round $2,000$. So we can upper bound my losing probability by about $8,000 \times 10^{-22} \approx 10^{-18}$, or about one in a quintillion.

To put this in perspective, if I played this game once a second, it would take me a minimum of $10^{18}$ seconds $\approx 32$ billion years to see my first loss!

This explains Ionza Ieggiera's comment above:

Unfortunately, for  $T = 10,000$, the script will fail by exceeding the time and memory limits of the online Magma calculator.

Even if Magma was able to simulate this game a billion times a second, it would take decades upon decades for Magma to register anything other than a win!

Accelerating my betting disproportionately increases my risk of going broke

No strategy offers a higher probability of success than the one-chip strategy, but betting more can speed up the time it takes for me to win. Since the one-chip strategy offers such a ridiculously safe bet, this is a tradeoff we're probably willing to make. But even a relatively modest gain in time saved can come with a surprisingly steep increase in risk.

Let's say I want to go $10 \times$ as fast, so I'm betting $C_k = 10$ chips each round, and it would take me $1,000$ rounds to hit the $10,000$ chip goal. I expect to lose about $0.1 C_k = 1$ chip per round with this strategy, so after $1,000$ rounds I expect to have $$\Bbb{E}[W_{1,000}] = 2,000 - 1,000(1) = 1,000$$ chips left.

(Note that using the one-chip strategy, we would expect to have far more of our starting bankroll left at this point in the game: $\Bbb{E}[W_{1,000}] = 2,000 - 1,000(0.1) = 1,900$ chips left instead. But with the one-chip strategy, we still have $9,000$ more bets to place, whereas with the ten-chip strategy, we're already finished by round $1,000$, if we make it that far.)

The resulting analysis is very similar: my actual bankroll $W_{1,000}$ after $10,000$ rounds, using the ten-chip strategy, is given by $W_{1,000} = 2,000 + 10 \sum_{i = 1}^{1,000} X_i$, where $X_i$ are iid $\{ \pm 1 \}$-valued random variables with $\Bbb{P}(X_i = +1) = 0.45$ for all $i = 1, 2, ..., 1,000.$

$W_{1,000}$ is again basically normal per CLT, with mean $\mu_{1,000} = 1,000$.

This time, though, because we're betting $10 \times$ as much per round, our variance is $100 \times$ bigger, which corresponds to a standard deviation $10 \times$ bigger. The expected value of a single bet is $-1$ chips, so the variance of a single bet is $$(10 - (-1))^2 (0.45) + (-10 - (-1))^2 (0.55) = 99 \text{ chips,}$$ and the variance of our $1,000$ bets is $\sigma^2_{1,000} = 1,000 \times 99 = 99,000$. This gives us a standard deviation of $\sigma_{1,000} = \sqrt{99,000} \approx 314.6$ chips. So the likelihood of being broke at round $W_{1,000}$ corresponds to the probability of a $z$-score of $$z = \frac{0 - 1,000}{314.6} = -3.178,$$ which is still pretty low probability (around $0.007-0.008$) but now firmly within the realm of possibility. And that's just the probability of being broke at bet $1,000$--my probability of going broke at or before bet $1,000$ is actually slightly higher. So, my impatience has multiplied my probability of losing by a thousand million million times, and deeply cut into my margin of error. At the point where you're still playing $1,000$ rounds of this game in a row, the tradeoff doesn't seem particularly worth it. Betting $2$ or $5$ chips per round, while still keeping the probability of a win essentially certain, offers an even more modest speedup. And of course, neither the five-chip strategy, nor the two-chip strategy, offers a higher win probability than the one-chip strategy.

ContentLicense: CC BY-SA 4.0

Title: Probability Game Optimal Strategy


问:
Burnside's lemma for colourings on a triangle-square
Suppose that we have $n$ colours available and we paint the edges and vertices of an equilateral triangle (resp. square) and we want to count how many colourings there are, up to equivalence, where two colourings are equivalent if one can be transformed into the other by applying a symmetry of the triangle (resp. square). How many colourings are there?

I know how to use Burnside's lemma to count the colourings when we only paint either the edges or the vertices of a regular $n$-gon, but not both. In the above problem, I initally thought that the case of painting the vertices and edges of a triangle was exactly the same as painting just the vertices of a regular hexagon, and similarly I considered working with a regular octagon in the place of the vertices and edges of a square. It seems intuitive, but something is not convincing me: Is the group of symmetries of the vertices and edges of a regular $n$-gon isomorphic to the group of symmetries of just the edges of a regular $2n$-gon (the dihedral group of $4n$ elements). Another possibility that passed through my mind was: just multiply the orbits of the action of $D_3$ by $2$ to get the colourings when we paint the vertices and edges.

Is there any merit in my thoughts? If so, please point out my mistakes. Thanks in advance, and sorry for the long text!

答:
For the triangle case, the symmetry group will still be $D_3$ (of six elements), and not of that a hexagon. The reason is you cannot bring the vertex to an edge. With that in mind, we consider $D_3$ acting on $X$, all possible colorings of vertices and edges with $n$ colors, so $|X| = n^6$ (before considering symmetries). Applying Burnside's theorem, we have the number of orbits $|X/D_3|$ to be
$$|X/D_3| = \frac{1}{|D_3|} \sum_{g\in D_3} |Fix(g)|,$$
where $Fix(g)$ is set of colorings fixed by $g$. Now we analyze this a bit, with $D_3 = \{1,r,r^2 , f_1,f_2,f_3\}$, where $r^i$ are the rotations and $f_j$ are the flips. Then

$|Fix(1)| = n^6$;

$|Fix(r)| = |Fix(r^2)| = n^2$ (one color choice for the vertices, and another color choice for the edges);

$|Fix(f_j)|=n^4$ (one color choice for the vertex where the line of symmetry goes through, one color choice for the edge where the line of symmetry goes across, one color choice for  the pair of vertices across the line of symmetry, and one color choice for the pair of edges across the line of symmetry).

Hence $|X/D_3| = \frac{1}{6}(n^6+ 2n^2 + 3n^4)=\frac{1}{6}n^2(n^4+2+3n^2)=\frac{1}{6}n^2(n^2+2)(n^2+1)$. Here I factor it to see that indeed we get an integer as $n^2(n^2+2)(n^2+1)$ is a multiple of 6.

For the square case you similarly consider the symmetry of a square $D_4$ with eight element, and not of an octagon for the same reason that we cannot bring a vertex onto an edge.


问:
Proof that $a+b+\frac{1}{1+\sqrt{ab}}\geq\frac{5}{2}$
Kazakstan 2012

Suppose that $a, b \in\mathbb{R}$, and $a,b>0$. If $\frac{1}{a}+\frac{1}{b}=2$

prove that $a+b+\frac{1}{1+\sqrt{ab}}\geq\frac{5}{2}.$

My idea :

$a+b+\frac{1}{1+\sqrt{ab}}$ can be written as $ab(\frac{a+b}{ab}+\frac{1-\sqrt{ab}}{1-ab})=2ab+\frac{1-\sqrt{ab}}{ab-a^2b^2}$, but where  can I use the fact that $\frac{1}{a}+\frac{1}{b}=2?$

答:
Here is a solution that does not involve derivatives.

By AM-GM, we have that $a+b \geq 2\sqrt{ab}$. Hence it suffices for us to prove that $2\sqrt{ab}+\dfrac{1}{1+\sqrt{ab}} \geq \dfrac{5}{2}.$

From the given condition, $\dfrac{1}{a} + \dfrac{1}{b} =2 \Rightarrow a+b=2ab \Rightarrow 2ab \geq 2\sqrt{ab} \Rightarrow \sqrt{ab} \geq 1$. Hence, letting $x=\sqrt{ab}, x \geq 1$, we have to prove that:

\begin{align}
& 2x+\dfrac{1}{1+x} \geq \dfrac{5}{2} \\
& \iff \dfrac{2x(1+x)+1}{1+x} \geq \dfrac{5}{2} \\
& \iff \dfrac{2x^2+2x+1}{1+x} \geq \dfrac{5}{2} \\
& \iff 2(2x^2+2x+1) \geq 5+5x \\
& \iff 4x^2-x \geq  3 \\
& \iff x(4x-1) \geq 3 \\
\end{align}
But the last inequality is obvious since $x \geq 1$ and $4x-1 \geq 4-1=3$; hence we are done.


问:
Complements of sum of vector space
Let $V_1,V_2,W_1,W_2,X$ be vector spaces over the field $k$.

Assume that $V_1\oplus W_1=X$ and $V_2\oplus W_2=X$.

Is $(V_1+V_2)\oplus(W_1\cap W_2)=X$ ?

Here $\,V\oplus W=X\,$ means $\,V+W=X\,$ and $\,V\cap W=\{0\}\,$.

答:
No, this is not automatic.

Here's a counter-example in $\mathbb C^2$: Let $\{b_1,b_2\}$ be a basis.

Set $\,V_1=V_2=\langle b_1\rangle$, $W_1=\langle b_1+b_2\rangle$, and $W_2=\langle b_2\rangle\,$.

Then $\,V_1 +V_2 =V_1\,$ and $\,W_1\cap W_2=\{0\}\,$.

Title: Complements of sum of vector space


问:
sum of polynoms of given property
I have $P(x)$ a polynomial with degree $n$ ,$P(x) \ge 0$ for all $x \in$ real.

I have to prove that:

$f(x)=P(x)+P'(x)+P"(x)+......+P^{n}(x) \ge 0$ for all $x$.

I tried different methods to solve it but I got stuck.Any suggestion or advice is welcomed.

答:
you can let
$$G(x)=f(x)e^{-x},-\infty<x<\infty$$
then we have
$$G'(x)=f'(x)e^{-x}-f(x)e^{-x}=-P(x)e^{-x}\le 0$$
so this $G(x)$ is decreasing
and
$$\lim_{x\to+\infty}G(x)=\lim_{x\to+\infty}f(x)e^{-x}=0$$
$$f(x)e^{-x}=G(x)\ge \lim_{x\to+\infty}G(x)=0$$
so $$f(x)\ge 0$$</p>

ContentLicense: CC BY-SA 3.0

Title: sum of polynoms of given property


问:
When can you define addition on a set of equivalence classes?
Suppose I am given an equivalence relation on set A which is endowed with addition & closure under addition.

Then, to have addition on equivalence classes of A to be well-defined, it is to show:

$[a]+[b] = [a+b]$

Suppose I have shown that if $\forall a,b \in A, a \sim a', b\sim b'$ then $a+b \sim a'+b'$, then to my understanding, I proved that there exist an equivalence class $[a+b]$ so that $[a]+[b] \subset [a+b]$.

But then, now I need to show the other direction. I found it is easiest to define $c \in A$ so $b+c = 0$. So if $x \in [a+b]$ then $x \sim a+b$, and using what I proved, $x+c \sim a+(b+c) = a$. Letting $r = x-b$ and $s = b$ shows the other direction.

My question now is, to do this, (A,+) seems to need to have the property that all elements must have an identity, an inverse, and is associative, so what exactly must (A,+) be for all this to work, a group? abelian group? and why?

Note: Also I think I'm failing to consider a lot of things when thinking about this, so it will be very helpful if anyone could point many of those things out.

答:
If $A$ is a set, $+$ is a binary operation on $A$ and $\sim$ denotes an equivalence relation on $A$ then we can define a binary operation on $A/\sim$ by means of:$$[a]+[b]:=[a+b]\tag1$$if and only if $$\forall a,a',b,b'\in A [a\sim a'\wedge b\sim b'\implies a+b=a'+b']\tag2$$

"...then to my understanding, I proved that there exist an equivalence class $[a+b]$ so that $[a]+[b] \subset [a+b]$..."

No, $(1)$ tells us directly that $[a+b]=[a]+[b]$ and there is no such thing needed as "..showing the other direction..".

The only thing you really need is a proof that $(2)$ is valid. It is actually the proof that the operation "defined" in $(1)$ is indeed well defined.


问:
Darboux Theorem Question
Regarding:

Let $f:(0, \infty) \to \mathbb{R}$ be a differentiable function for which $f'(x)f'(\frac{1}{x})<0$ for every $0<x \ne 1$.

Then $f'(1)=0$.

--

So I am trying to prove this, and I said:

We know that $f$ is differentiable at $(0, \infty)$, so there exists a $0< \delta < 1$ such that $f$ is differentiable at $[1- \delta, 1+\delta]$ .

So there exists a $x_{0} \in [1- \delta, 1+\delta]$ such that $f'(x_{0})f'(\frac{1}{x_{0}})<0$

So by Darboux Theorem, there exists a $x_{0}<c< \frac{1}{x_{0}}$ such that $f'(c)=0$.

But now, I do not know how to link $c$ with $1$...

Any hints?

Thanks! :)

答:
As you noted, we have $$f'(1-\delta_n)f'\left(\frac {1}{1-\delta_n} \right)<0$$ for any positive sequence $\{\delta_n\}$ such that $\lim\delta_n = 0$. Take $1+\varepsilon_n = \dfrac{1}{1-\delta_n}>1$, where $\lim \varepsilon_n = 0$. Then by the given condition,
$$f'(1-\delta_n)f'\left( 1+\varepsilon_n \right)<0$$
WLOG, assume that $f'(1-\delta_n)<0$ and $f'\left( 1+\varepsilon_n \right)>0$.

By sending $n$ to infinity and remembering Darboux's theorem, we see that $f'(1)=0$.

Title: Darboux Theorem Question


问:
Time complexity and big-O of double loop algorithm
I am playing around with calculating the time complexity of the following code:

```<code>    for (int i = 0; i <= n/2; i+=3){
        for (int j = i; j <= n/4; j+=2) {
            x++;
        }
    }
</code>```

I know that its big-O complexity is N^2. However, I try to extract some formula that I could deduce that based on that formula. What I have done so far.

I know since the inner loop terminates at n/4 we can actually rewrite this code to:

```<code>for (int i = 0; i <= n/4; i+=3){
    for (int j = i; j <= n/4; j+=2) {
        x++;
    }
}
</code>```

If we relax a bit we can say that this double loop will be approximately the same as :

```<code>for (int i = 0; i < n/4; i+=3){
    for (int j = i; j < n/4; j+=2) {
        x++;
    }
}
</code>```

just to facilitate the math. I am stuck at this point, I think we can rewrite that as :

$$\sum_{k=0}^{n/12} n/8 - 3i/2 $$  but I am not sure if it makes any sense.

答: The inner loop runs $n/8-i/2$ times, give or take a $O(1)$ correction due to the parity of $n/4-i$, provided $i\le n/4$, which happens for $n/12+O(1)$ values of $i$. Therefore, $x$ increments a number of times $\sim$-equivalent to$$\tfrac{n^2}{96}-\sum_{k=1}^{n/12}(3k)/2\sim\tfrac{n^2}{96}-\tfrac34(\tfrac{n}{12})^2=\tfrac{n^2}{192},$$where $f\sim g$ iff $f-g\in o(n^2)$.


问:
If $a^m_{1}+a^m_{2}+\cdots+a^m_{n}\in\mathbb Z$, show $a_{i}\in\mathbb Z$

Let $a_{1},a_{2},\dots,a_{n}$ be rational numbers. If for any positive integer $m$
$$a^m_{1}+a^m_{2}+\cdots+a^m_{n}\in\mathbb Z$$
show that $a_{i}\in\mathbb Z$, $i=1,2,\dots,n$.

I tried: let $S(m)=a^m_{1}+a^m_{2}+\cdots+a^m_{n}$, then  use Newton's identities, I can prove $e_{i}\in\mathbb Z$ where $e_{i}$ is the $i$-th elementary symmetric polynomial, but I can't prove every $a_{i}\in\mathbb Z$. Thanks

答:
Easily deduced from the following more general result (I'm still trying to find the source):

Let $a_1,\ldots,a_n\geqslant 1$ be distinct rational numbers. Suppose that there are nonzero real numbers $c_1,\ldots,c_n$ and a sequence $\{b_m\}$ of integers such that $$\lim_{m\to\infty}\left(b_m-\sum_{k=1}^n c_k a_k^m\right)=0.$$ Then $a_1,\ldots,a_n$ are integers.

The proof I know uses induction on $n$. Denote $r_m=b_m-\sum_{k=1}^n c_k a_k^m$, take arbitrary $1\leqslant\ell\leqslant n$, and let $a_\ell=p/q$ with $\gcd(p,q)=1$. Then $$qr_{m+1}-pr_m=(qb_{m+1}-pb_m)-\sum_{k\neq\ell}c_k(qa_k-p)a_k^m$$ tends to $0$ as $m\to\infty$. If $n>1$, this immediately gives the induction step: let $b_m'=qb_{m+1}-pb_m$ and $c_k'=c_k(qa_k-p)$ for $k\neq\ell$ (the latter are nonzero because $a_k$ are distinct); by induction, $a_k$ are integers for $k\neq\ell$, and since $\ell$ is arbitrary, we're done.

Now consider [the induction base] $n(=\ell)=1$. The above reads: $qb_{m+1}-pb_m\to 0$ as $m\to\infty$. Since these are integers, we have $qb_{m+1}-pb_m=0$ for all sufficiently large $m$, and we may also assume that $b_m\neq 0$ for these $m$ (since $a_1\geqslant 1$). Then $q^k b_{m+k}=p^k b_m$ for all $k$, hence $q^k\mid b_m$ for all $k$, which is possible only if $q=1$.


答: Note if $c\neq 1$ then $f'(c)f'(1/c)<0$ by hypothesis. This implies $f'(c)\neq 0$. Hence there is only one possible value left for $c\in (x_0,1/x_0)$, which is $c=1$.

Title: Darboux Theorem Question

Body:
Regarding:

Let $f:(0, \infty) \to \mathbb{R}$ be a differentiable function for which $f'(x)f'(\frac{1}{x})<0$ for every $0<x \ne 1$.

Then $f'(1)=0$.

--

So I am trying to prove this, and I said:

We know that $f$ is differentiable at $(0, \infty)$, so there exists a $0< \delta < 1$ such that $f$ is differentiable at $[1- \delta, 1+\delta]$ .

So there exists a $x_{0} \in [1- \delta, 1+\delta]$ such that $f'(x_{0})f'(\frac{1}{x_{0}})<0$

So by Darboux Theorem, there exists a $x_{0}<c< \frac{1}{x_{0}}$ such that $f'(c)=0$.

But now, I do not know how to link $c$ with $1$...

Any hints?

Thanks! :)


问:
Proving that Euclidean space having the infinity metric is a complete metric space (stuck)
I am trying to prove that the space ${\mathbb{R}}^k$ with the $\infty$-metric is a complete metric space.

I know that I need to show that every Cauchy sequence in the metric space ${\mathbb{R}}^k$ with the $\infty$-metric converges to a point $x\in{\mathbb{R}}^k$. Part of my work so far involved proving that the space ${\mathbb{R}}^k$ with the old Pythagorean norm is a complete metric space, but I’m not sure if I should be using that at all in this proof. Here goes:

Proof:

Suppose that $k\in{\mathbb{N}}$ and that $\left(x_{n}\right)$ is a Cauchy sequence in the space ${\mathbb{R}}^k$ with the $\infty$-metric.
Suppose that $\epsilon>0$.
Using the fact that $\left(x_{n}\right)$ is a Cauchy sequence, choose an integer $N$ such that the inequality $||x_{m}-x_{n}||_\infty<\frac{\epsilon}{2}$ holds for all integers $m$ and $n$ such that $m\ge{N}$ and $n\ge{N}$.

This is where I’m stuck. I feel like I want to use the triangle inequality from here…

Since the space ${\mathbb{R}}^k$ is a metric space, we know that the inequality $$||{x_n}-{x_m}||_{\infty}\le||{x_n}-{x}||_{\infty}+||{x_m}-{x}||_{\infty}\lt\frac{\epsilon}{2}+\frac{\epsilon}{2}=\epsilon$$ holds for all positive integers $n$ and $m$.

However, I know that this isn’t a good thing to state. I think that I’m assuming that there is some point $x$ in the space that the sequence converges to… this is exactly what I'm trying to prove!
I’m totally stuck. Any advice would be GREATLY appreciated!

EDIT: I'm including the updated version of my proof based on advice from Brian M. Scott. Here it is:

Proof:

Suppose that $k\in{\mathbb{N}}$ and that $\left(x_{n}\right)$ is a Cauchy sequence in the space ${\mathbb{R}}^k$ with the $\infty$-metric.

Suppose that $\epsilon>0$.

For each $n\in\mathbb{Z}^{+}$ we have $x_{n}= \left(x_{n,1},x_{n,2},\ldots,x_{n,k}\right)$.

Consider the real-valued sequences $\left(x_{n,j}\right)$ for each $j=1,2,\ldots,k$.

Using the fact that $\left(x_{n}\right)$ is a Cauchy sequence, choose $N\in\mathbb{Z}$ such that the inequality $||x_{m} – x_{n}||_{\infty}\lt \frac{\epsilon}{2}$ for all $n\ge N$ and $m\ge N$.

Since the inequality $||x_{m,j} – x_{n,j}||_{\infty}\le||x_{m} – x_{n}||_{\infty}$ holds for all $m$ and $n$, we have the inequality $||x_{m,j} – x_{n,j}||_{\infty}\lt\frac{\epsilon}{2}$ for all $m\ge N$ and $n\ge N$.

Thus for $j=1,2,\ldots,k$ and all $n\in\mathbb{Z}^{+}$ the sequence $\left( x_{n,j} \right)$ is a Cauchy sequence in $\mathbb{R}$.

Using the fact that $\mathbb{R}$ is complete, define $$x\left(j\right)=\lim_{n\to\infty}x_{n,j}$$ for $j=1,2,\ldots,k$.

The point $x=\left(x_{1},x_{2},\cdots,x_{k}\right)$ is the limit of $\left(x_{n}\right)$ under the Pythagorean metric.

From the inequality $$||x-x_{n}||_\infty\le||x-x_{m}||_{\infty}+||x_{m}-x_{n}||_{\infty}\lt\frac{\epsilon}{2}+\frac{\epsilon}{2}=\epsilon$$ that holds for all $n\ge N$ and $m\ge N$ we see that the sequence $\left(x_{n}\right)$ converges to the point $x\in\mathbb{R}^{k}$ and the space $\mathbb{R}^{k}$ is a complete metric space.

答:
Since $$\max_{1\leq i\leq d}|x_i|^2\leq\sum_{i=1}^d |x_i|^2\leq d\ \max_{1\leq i\leq d}|x_i|^2$$ one has
$$\|{\bf x}\|_\infty\leq\|{\bf x}\|_2\leq\sqrt{d}\>\|{\bf x}\|_\infty\qquad(x\in{\mathbb R}^d)\ .$$
Therefore the identity map
$${\rm id}:\quad\bigl({\mathbb R}^d,\ \|\cdot\|_\infty\bigr)\ \to\ \bigl({\mathbb R}^d, \ \|\cdot\|_2\bigr)$$
is Lipschitz in both directions. This means that distances between points measured one way or the other differ by a factor at most $\sqrt{d}$. It follows that a sequence $({\bf x}_n)_{n\geq0}$ in ${\mathbb R}^d$ is Cauchy with respect to $\|\cdot\|_\infty$ iff it is Cauchy with respect to $\|\cdot\|_2$.</p>


问:
If $\frac{\cos x+\cos y+\cos z}{\cos(x+y+z)}=\frac{\sin x+\sin y+\sin z}{\sin(x+y+z)}=k$, then find $\cos(x+y)+\cos(y+z)+\cos(z+x)$.

If $x,y,z \in\mathbb{R}$ and  $$\frac{\cos x+\cos y+\cos z}{\cos \left(x+y+z\right)}=\frac{\sin x+\sin y+\sin z}{\sin \left(x+y+z\right)}=k$$
Find the value of $\cos(x+y)+\cos(y+z)+\cos(z+x)$.

I was working through this problem and found an interesting solution to it. How would you solve it?

Edit: I found the same question here and here. So, this should be marked as duplicate.

答:
Rewrite the equations as
$$\cos x +\cos y+\cos z =k\cos(x+y+z) \\ \sin x+\sin y+\sin z=k\sin(x+y+z)$$

Multiply the second equation by $i$ (where $i=\sqrt{-1}$) , and add both the equations. Using Euler's formula, we are left with a single equation, that is,$$e^{ix}+e^{iy}+e^{iz}=ke^{i(x+y+z)}$$
Now, we have to find $\cos(x+y)+\cos(y+z)+\cos(z+x)=\Re\left(e^{i(x+y)}+e^{i(y+z)}+e^{i(z+x)}\right)$.

 Now, we let $\alpha=e^{ix},\beta=e^{iy},\gamma=e^{iz}$, so that we have $\alpha+\beta+\gamma=k\alpha \beta\gamma $ and we need to find $\alpha\beta+\beta\gamma+\gamma\alpha$.

 Dividing by $\alpha\beta\gamma$ in the last equation results in $(\alpha\beta)^{-1}+(\beta\gamma)^{-1}+(\gamma\alpha)^{-1}=k$.

 But notice that $\Re(\alpha\beta)=\Re\left((\alpha\beta)^{-1}\right)$ since cosine is an even function and hence $\Re\left((\alpha\beta)^{-1}+(\beta\gamma)^{-1}+(\gamma\alpha)^{-1}\right)=\Re(\alpha\beta+\beta\gamma+\gamma\alpha)=k$.

 If instead of real part, we take the imaginary parts, we get $\Im\left((\alpha\beta)^{-1}+(\beta\gamma)^{-1}+(\gamma\alpha)^{-1}\right)=-\Im\left(\alpha\beta+\beta\gamma+\gamma\alpha\right) \implies \sin(x+y)+\sin(y+z)+\sin(z+x)=0$.


答:
Suppose the result to be false and consider a counterexample with $n$  minimal. Let $b$ be the denominator of any $a_i$ and replace all the $a_i$ by $a_ib$. Since at least one $a_ib$ is an integer it can be ignored and so, by minimality of $n$ all the $a_ib$ are integers.

Similarly, we need only consider $b$ to be a prime $p$ say. Let $c_i=a_ip$, then each $c_i$ is coprime to $p$.

Now suppose $p$ is odd and consider just powers $m=(p-1)p^k$. Then each $c_i^m$ is congruent to 1 modulo $p^{k+1}$ and therefore the total sum of the $c_i^m$ is congruent to $n$ modulo $p^{k+1}$. For sufficiently large $k$ this sum cannot be divisible by $p^m$, a contradiction.

The case $p=2$ is handled the same way.

Title: If $a^m_{1}+a^m_{2}+\cdots+a^m_{n}\in\mathbb Z$, show $a_{i}\in\mathbb Z$

Body:
Let $a_{1},a_{2},\dots,a_{n}$ be rational numbers. If for any positive integer $m$
$$a^m_{1}+a^m_{2}+\cdots+a^m_{n}\in\mathbb Z$$
show that $a_{i}\in\mathbb Z$, $i=1,2,\dots,n$.

I tried: let $S(m)=a^m_{1}+a^m_{2}+\cdots+a^m_{n}$, then  use Newton's identities, I can prove $e_{i}\in\mathbb Z$ where $e_{i}$ is the $i$-th elementary symmetric polynomial, but I can't prove every $a_{i}\in\mathbb Z$. Thanks


问:
Why is it that Complex Numbers are algebraically closed?
I find it curious that Complex Numbers give enough flexibility to be algebraically closed, where the reals, rational numbers do not.  For the reals it is easy to see that they cannot be used to solve equations like $x^2 + 1 =0$.  Geometrically, one can look at the number line as see that any $x$ squared yields a positive number which when added to one cannot get you back to zero.  In the complex case, however, we are working with the plane.  In this case exponents stretch and rotate any given $x$.  It is easy to therefore see in the particular circumstance that if $x=i$ that $x^2$ rotates it to $-1$ which when added to one yields the desired result (i.e. $0$).  So because the Complex Numbers are algebraically closed, I conclude that any polynomial equation with complex coefficients my be solved by choosing one or more $x$'s in the plane and rotating them and stretching them such that they will combine using the given coefficients to produce the RHS.

Question: Why is it that we do not need a larger space than the plane to solve Complex polynomial equations?

I have tried to find a sufficient answer through Google, but was not able.  I also searched M.SE and could not find a sufficient answer.  I am not a mathematician, so I am looking for an intuitive answer if possible.

答:
Let $p(x)$ be an irreducible polynomial in $\Bbb R[x]$. It may be factored as $p(x)=\prod (x-\alpha_i)$ in $\Bbb R^a$, and hence $p'(x)=\sum_j\prod_{i\ne j}(x-\alpha_i)$. For each linear factor of $p(x)$, there is exactly a term in the sum that is not divisible by it, and so $\gcd(p,p')=0$. Thus, every irreducible polynomial in $\Bbb R[x]$ is separable, and every finite extension of $\Bbb R$ is Galois.

Let $K/\Bbb R$ be a finite extension. It is Galois. Its Galois group $G=Gal(K/\Bbb R)$ has order $[K:\Bbb R]=[K:\Bbb R(i)][\Bbb R(i):\Bbb R]$, and $[\Bbb R(i):\Bbb R]=2$, so $2\mid [K:\Bbb R]$. Therefore, $G$ has a Sylow 2-subgroup $H$. Denote its fixed field by $K^H(=\{a\in K\mid \sigma a=a\;\forall \sigma\in H\})$, then $[G:H]=[K^H:\Bbb R]$ by the Galois correspondence. This quantity is an odd integer since $H$ is a 2-subgroup with maximal order.

By the primitive element theorem, $K^H=\Bbb R(\beta)$ for some $\beta\in K^H$, and hence $[K^H:\Bbb R]=[\Bbb R(\beta):\Bbb R]$, the degree of the minimal polynomial of $\beta$ over $\Bbb R$, is odd. But every odd degree polynomial in $\Bbb R[x]$ has a root in $\Bbb R$, so the minimal polynomial, being irreducible, must have degree $1$. In other words, $K^H=\Bbb R$ and $[K:\Bbb R]=|G|=|H|=[K:K^H]$, which is a power of 2. Consequently, $[K:\Bbb R(i)]$ is also a power of 2.

Suppose that $[K:\Bbb R(i)]\ne 1$ i.e. $Gal(K/\Bbb R(i))$ is not trivial. Then it has a subgroup of index 2, which corresponds to an extension $E$ of degree 2 over $\Bbb R(i)$. Then the minimal polynomial of an element $\gamma$ in $E\setminus \Bbb R(i)$ is of the form $x^2+bx+c$ where $b,c\in \Bbb R(i)$. By completing the square, we have $(x+\dfrac b2)^2=\dfrac{b^2}4-c$. The right hand side lies in $\Bbb R(i)$ and thus has a square root in $\Bbb R(i)$, so the original polynomial actually is reducible, yielding a contradiction. Therefore, $K=\Bbb R(i)$, which completes the proof.


问:
Powers of a matrix, largest required collection for linear dependence to occur, reduction to vectors
Consider the dependence equation

$$
A^m + c_1A^{m-1}+...+c_{m-1}A + c_mI = 0
$$

where $A \in \Bbb{R}^{n \times n}$ and $I$ is the identity matrix. Asked to show that there exists an $m> 0$ such that some choice of $c_i$ yields a solution we note that the dimension of the vector  space of square matrices is $n^2$ and that therefore, at the latest, $m>n^2-1$ yields a solution.

My question relates to this: If we chose to multiply on the right by some general $x \in \Bbb{R}^n$ such that

$$
A^mx + c_1A^{m-1}x+...+c_{m-1}Ax + c_mIx = 0
$$

we have $m+1$ vectors and seek a linear dependence relation among them. Then we would say that dependence is achieved at the latest when $m>n-1$. Square matrices say something about transformations of a vector space as I understand it. Yet I can not quite grasp where the loss of information comes from. What goes wrong when we right multiply by $x$, reducing it to a question about vectors instead of matrices? Does it have something to do with only considering one x instead of a set of vectors spanning the underlying space? Thanks in advance.

答: Actually, Cayley-Hamilon implies the characteristic polynomial $\chi_A$ of $A$ annihilates $A$. Since $\deg \chi_A = n$, we conclude that the set $\{I,A,A^2, \ldots, A^n\}$ is linearly dependent so at the latest $m > n-1$ yields a solution, same as your vectors.


问:
Evaluate $y'''+3y''+3y'+y = e^{it}$
$y'''+3y''+3y'+y = e^{it}$

I've got this method of undermined coefficients but i am unable to find the particular solution. I got the complimentary solution as $$Y_c = {C_1}{e^{it}}+ {C_2}{t}{e^{it}}+{C_3}{t^{2}}{e^{it}}$$ but keep getting the wrong answer when I try to find the particular.

答: For $y'''+3y''+3y'+y = 0$ characteristic polynomial is $\lambda^3+3\lambda^2 + 3\lambda+1 = (\lambda+1)^3$, so complementary solution is $y_c = {C_1}{e^{-t}}+ {C_2}{t}{e^{-t}}+{C_3}{t^{2}}{e^{-t}}$.

Title: Evaluate $y'''+3y''+3y'+y = e^{it}$


问:
Venn diagram related question
An analysis of the survey of $320$ school pupils highlighted the following facts:

•   $50$ pupils live in New Town, travel to school by bus and have canteen lunch.

•   $110$ pupils live in New Town and have canteen lunch

•   $160$ pupils live in New Town

•   $90$ pupils travel by bus and have canteen lunch

•   $130$ pupils travel by bus

•   $30$ pupils live in New Town, travel to school by bus but do not have canteen lunch

•   $90$ pupils do not live in New Town, do not travel to school by bus and do not have canteen lunch

Depict the above information in the form of a Venn diagram.
The Principal wishes to know how many pupils have lunch at the school canteen.  Please help him find that out, and also help us understand how you arrived at the answer.

答:
Each of the pupils has three boolean properties N (lives on New Town or not), B (travels by bus or not) and C (has canteen lunch or not). There are 2^3 = 8 possible cases which can be depicted as follows:

The yellow figures were directly given in the question. The red figures can be derived. Example: 90 pupils travel by bus and have canteen lunch. Therefore, $|B C| = |N B C| + |N' B C|$ must be 90. We know from "50 pupils live in New Town, travel to school by bus and have canteen lunch" that $|N B C|$ is 50 and conclude $|N' B C|$ to be 40.

The canteen pupils sum up to $50+60+40+20 = 170$

As a Venn Diagram, this would look as follows:

Title: Venn diagram related question


问:
Curve behavior of derivative
I wonder whether there is a possibility to check how the derivative of a function behaves.

Taking the following as example:

I notice that in the first and last region, the curve is increasing but the derivative behaves differently there. For the first, it is decreasing and for the latter it is increasing.
Or is there a pattern I don't see?

答: Concavity of the graph. The derivative is decreasing as function is concave downwards and increasing as it is concave upwards.</p>

Title: Curve behavior of derivative


问:
Hard time figuring out word problem
Computers are beginning to lose power. The number of computers losing power doubles every minute. At 10 minutes half of the computers have lost power. After how many minutes will all computers be without power?

The only thing I could come up with is

$$y- (2^9)x = .5y$$

Or a summation?

$$y-x2^i \text{ from 1 to 9?}$$

I have no clue.

Is it even solvable?

答: It is 11 minutes because every minute the number doubles. So if half the computers are turned off at ten minutes and number of computers turned of doubles every minute it would be 1/2 *2 =2/2=1 in the next minute so at EXACTLY 11 minutes all computers would turn off.</p>

Title: Hard time figuring out word problem


Title: Evaluate $y'''+3y''+3y'+y = e^{it}$

Body:
$y'''+3y''+3y'+y = e^{it}$

I've got this method of undermined coefficients but i am unable to find the particular solution. I got the complimentary solution as $$Y_c = {C_1}{e^{it}}+ {C_2}{t}{e^{it}}+{C_3}{t^{2}}{e^{it}}$$ but keep getting the wrong answer when I try to find the particular.

Body:
$$y'''+3y''+3y'+y = e^{it}$$
Try for the particular solution:
$$y_p=Ae^{it}$$
$$ -iA-3A+3iA+A=1$$
$$A=\dfrac 1{2(i-1)}$$
$$A=\dfrac {i+1}{2(i-1)(i+1)}$$
$$A=\dfrac {i+1}{2(i^2-1)}$$
$$A=-\dfrac 14 (i+1)$$


问:
A-Level Normal Distribution Exam Question
When Ali competes in long jump competitions, the length of his jumps are normally distributed with mean 5.2m and standard deviation 0.7m.

Ali needs to jump 6m to qualify for the school team. What is the probability that he qualifies with one jump? What is the probability that, if allowed three jumps, that he qualifies for the team?

I was able to solve part 1 of the question (0.127 to 3sf) and, assuming that each jump is independent in part 2, I tried to use the binomial distribution, with X - B(3,0.127) and with the number of successes being 1, I got the answer of 0.290371149 (with 0.127 as p) or 0.2896389432 (with 0.1265489545 as p). However, the mark scheme says that the answer is 0.334, and I don't know where I have gone wrong.

Any advice would be greatly appreciated.

答:
The point is that you calcluated the probability that Ali makes exactly one jump over $6$m. It is possible that Ali cleared $6$m on two, or even all three, jumps. So you need to take those possibilities into account too.

The easiest way to do this: Ali fails to make the team if and only if he jumps less than $6$m on all three attempts. This occurs with probability $(1-0.127)^3$, so the probability that this doesn't happen (and hence Ali makes the team) is $$1-(1-0.127)^3\approx0.33$$


问:
Calculation of the volume of a 3D region with triple integral
I have to calculate the volume of the following region:

$$D=\{(x,y,x)\in \mathbb{R}^3: x^2+4y^2 \leq \pi^2, \quad x^2 +y^2 > \frac{\pi^2}{4} ,\quad x>0, \quad 0<z<x\cdot \cos(y)\}.$$ The projection of the region on the $xy$ plane is the region between an ellipse and a circle, so my attempt was to calculate a triple integral exploiting the fact that the domain is normal with respect to the $xy$ plane:
$$V=\iiint_D dxdydz=\int\int_A \left(\int_{z=0}^{z=x  \cdot \cos(y)}dz\right )dxdy$$

where $$A=\{(x,y)\in \mathbb{R}^2:     -\frac{\pi}{2}\leq y\leq \frac{\pi}{2},\sqrt{\frac{\pi^2}{4}-y^2}\quad \leq x\leq \sqrt{\pi^2-4y^2}  \}$$
After a bit of algebra I arrived to the conclusion that: V=$6$. Now I don't know if the result is correct.

答:
$$\int_0^{x \cos y}  \, dz=x \cos y$$
$$\int_{\frac{1}{2} \sqrt{\pi ^2-4 y^2}}^{\sqrt{\pi ^2-4 y^2}} x \cos y \, dx=\frac{3}{8} \left(\pi ^2-4 y^2\right) \cos y$$
And finally
$$\int_{-\frac{\pi }{2}}^{\frac{\pi }{2}} \frac{3}{8} \left(\pi ^2-4 y^2\right) \cos y \, dy=6$$

Last integral can be solved by parts
$$\int \frac{3}{8} \left(\pi ^2-4 y^2\right) \cos y \, dy=\frac{3}{8} \left(\pi ^2-4 y^2\right) \sin y-\int \left(-3 y \sin y \right)\, dy=$$
$$=\frac{3}{8} \left(\pi ^2-4 y^2\right) \sin y-\left(3y \cos y-\int 3\cos y\,dy\right)=$$
$$=\frac{3}{8} \left(\pi ^2-4 y^2\right) \sin y-3y\cos y+3\sin y+C=$$
$$=\frac{3}{8} \left(8-4 y^2+\pi ^2\right) \sin y-3 y \cos y+C$$
Thus the last definite integral becomes
$$\int_{-\frac{\pi }{2}}^{\frac{\pi }{2}} \frac{3}{8} \left(\pi ^2-4 y^2\right) \cos y \, dy=\left[\frac{3}{8} \left(8-4 y^2+\pi ^2\right) \sin y-3 y \cos y\right]_{-\pi/2}^{\pi/2}=3-(-3)=6$$


问:
Product of permutation matrices
I want to prove that the product of two permutation matrices is itself a permutation matrix. But I don't know how. Please help!</p>

答:
Hint:

If $P$ is a permutation matrix and $A$ any matrix then $AP$ is a matrix consisting of the columns of $A$ permuted. Similarly $PA$ consists the rows of $A$ permuted. Therefore if $A$ is also a permutation matrix ...

Title: Product of permutation matrices


问:
Why is the cyclotomic polynomial quotient ring isomorphic to a product of finite fields?
I am looking at the ring $R = \mathbb{Z}[X]/(\Phi_m(X), p)$ where $\Phi_m$ is the $m$th cyclotomic polynomial and $p$ is a prime. What I want to show is that $R$ can be split into $l$ copies of $\mathbb{F}_{p^d}$ where $d$ is the smallest integer such that $p^d \equiv 1 \mod m$ and $l\cdot d = \phi(m)$, the degree of the polynomial.

The condition in some sources I have looked at says that $p$ cannot divide $m$ and that $p$ cannot be ramified in $\mathbb{Q}[\zeta_m]$ where $\zeta_m$ is the $m$th root of unity. I cannot make the connection of why this is necessary. I don't even know if the polynomial have to be cyclotomic, or that it is sufficient that they are irreducible in $\mathbb{Z}[X]$. Any tips would be much appreciated.

答:
So $R$ is actually $\mathbb{F}_p[X]/(\Phi_m(X))$.

Assume that $R$ is a product of fields. In particular, $R$ is reduced (that is the problem for general $m$); in other words, $\Phi_m$ mod $p$ is separable. But if $p|m$, using complex roots, we can see that $\Phi_m(X)=\Phi_{m/p}(X^p)$ so is not separable mod $p$ (it’s equal to $\Phi_{m/p}(X)^p)$.

Now, assume that $p$ does not divide $m$. Then $\Phi_m$ divides $X^m-1$ which is separable mod $p$, so $\Phi_m$ is separable mod $p$, and coprime mod $p$ to any $\Phi_d$ with $d|m$ and $d < m$.

In particular, if $F$ is a field of characteristics $p$ and $\omega \in F$ is a root of $\Phi_m$, then $\omega$ has multiplicative order $m$ – and conversely.

Let $\Phi_m=P_1\ldots P_l$ be the decomposition in products of irreducible polynomials mod $p$, with the $P_i$ being monic, irreducible mod $p$ and pairwise distinct. Then $R$ is the product of the $\mathbb{F}_p[X]/(P_i)$ (by the CRT) which are actually fields generated by an element which is a root of $\Phi_m$, so is an element of multiplicative order $m$.

To conclude, we show that if $F$ is a field of degree $d$ over $\mathbb{F}_p$ generated by an element $\omega$ of multiplicative order $m$, then $d$ is the multiplicative order $u$ of $p$ mod $m$.
This shows that all the $P_i$ have the same degree.

Indeed, $1=\omega^{|F^{\times}|}=\omega^{p^d-1}$ so that $m|p^d-1$ and $u|d$. But $F’=\{z \in F,\,z^{p^u}=z\}$ is a subfield of $F$ containing $\mathbb{F}_p$ and $\omega$ so is $F$. But $F’$ is the set of roots of the polynomial $X^{p^u}-X$ so $|F|=|F’| \leq p^u$ so $d \leq u$ and $d=u$.


问:
Find QR Factorization
I'm self studying Linear Algebra and am having trouble with QR factorization. I want to find the QR factorization of Matrix A.

My attempt: I identified $v_1$, $v_2$, and $v_3$ as the 3 column vectors of $A$ respectively.

I now want to find the orthonormal basis and have gotten $u_1 = [0\; 0\; 1\; 0]$, $u_2 = [0\; -1\; 0\; 0]$, $u_3 = 0$.

I think I have made a mistake here but am not sure what to do. I think $u_3$ should not be $0$ but cannot find a mistake in my calculations.

To get $u_3$, I calculated $v_3-(u_1 \cdot v_3)u_1 - (v_3 \cdot u_2)u_2$. Then, I divided it by $||u_3||$.

答:
$$A=\begin{bmatrix}0&-3&0\\0&0&0\\2&0&0\\0&0&4\end{bmatrix}$$

According to your notation,

$$v_1=\begin{bmatrix}0\\0\\2\\0\end{bmatrix},v_2=\begin{bmatrix}-3\\0\\0\\0\end{bmatrix},v_3=\begin{bmatrix}0\\0\\0\\4\end{bmatrix}$$

We need to apply the Gram Schmidt orthonormalization process to find the $Q$ matrix.

$w_1=v_1=\begin{bmatrix}0\\0\\2\\0\end{bmatrix}, ||w_1||=2 \implies u_1=\frac{w_1}{||w_1||}=\begin{bmatrix}0\\0\\1\\0\end{bmatrix}$

$w_2=v_2-(v_2.u_1)u_1=v_2-(0)u_1=\begin{bmatrix}-3\\0\\0\\0\end{bmatrix}, ||w_2||=3 \implies u_2=\frac{w_2}{||w_2||}=\begin{bmatrix}-1\\0\\0\\0\end{bmatrix}$

$w_3=v_3-(v_3.u_1)u_1-(v_3.u_2)u_2=v_3-(0)u_1-(0)u_2=\begin{bmatrix}0\\0\\0\\4\end{bmatrix}, ||w_3||=4 \implies u_3=\frac{w_3}{||w_3||}=\begin{bmatrix}0\\0\\0\\1\end{bmatrix}$

Therefore, we can obtain our $Q$ matrix using $u_1,u_2,u_3$.

$$Q=\begin{bmatrix}0&-1&0\\0&0&0\\1&0&0\\0&0&1\end{bmatrix}$$

$R$ can be obtained using the following relation.

$$R=\begin{bmatrix}v_1.u_1&v_2.u_1&v_3.u_1\\
0&v_2.u_2&v_3.u_2\\
0&0&v_3.u_3\end{bmatrix}$$

$v_1.u_1=\begin{bmatrix}0\\0\\2\\0\end{bmatrix}.\begin{bmatrix}0\\0\\1\\0\end{bmatrix}=2, v_2.u_1=\begin{bmatrix}-3\\0\\0\\0\end{bmatrix}.\begin{bmatrix}0\\0\\1\\0\end{bmatrix}=0,
v_3.u_1=\begin{bmatrix}0\\0\\0\\4\end{bmatrix}.\begin{bmatrix}0\\0\\1\\0\end{bmatrix}=0$

$v_2.u_2=\begin{bmatrix}-3\\0\\0\\0\end{bmatrix}.\begin{bmatrix}-1\\0\\0\\0\end{bmatrix}=3,
v_3.u_2=\begin{bmatrix}0\\0\\0\\4\end{bmatrix}.\begin{bmatrix}-1\\0\\0\\0\end{bmatrix}=0$

$v_3.u_3=\begin{bmatrix}0\\0\\0\\4\end{bmatrix}.\begin{bmatrix}0\\0\\0\\1\end{bmatrix}=4$

$\therefore$ $R=\begin{bmatrix}2&0&0\\0&3&0\\0&0&4\end{bmatrix}$

$$\implies \begin{bmatrix}0&-3&0\\0&0&0\\2&0&0\\0&0&4\end{bmatrix}=\begin{bmatrix}0&-1&0\\0&0&0\\1&0&0\\0&0&1\end{bmatrix}\begin{bmatrix}2&0&0\\0&3&0\\0&0&4\end{bmatrix}$$

Title: Find QR Factorization


问:
compute $ \iint_De^{-x^2-4y^2} \ dxdy$
The question is:
$$ \iint_De^{-x^2-4y^2} \ dxdy, \quad D=\{(x,y):0\leq x\leq2y\}$$
This is how i've tried to attack this but i was getting nowhere with it :
$$ \int_{0}^{\infty}e^{-4y} \ dy \int_{0}^{2y}e^{-x^2}dx$$
Am i on the right path? how should i proceed

Any suggestion?

答: The substitution $x=r\cos\theta,\,y=\tfrac12r\sin\theta$ of Jacobian $r/2$ writes the integral as$$\int_{\pi/4}^{\pi/2}d\theta\int_0^\infty\tfrac12re^{-r^2}dr=\tfrac{\pi}{16}.$$

Title: compute $ \iint_De^{-x^2-4y^2} \ dxdy$


问:
Absolute values (definition proof)
Prove that:

a) $x \le |x|$ and

b) $|xy| = |x||y| $

Attempt:

a) If $x \ge 0: x = |x| \le |x|$.

If $ x < 0: x = |-x| = |x| \le |x|.$

b) If $xy \ge 0: |xy| = xy = |x||y|.$

If $xy < 0: |xy| = -xy = |-x||y|$ or $|x||-y| = |x||y|$.

Is there anything that is wrong?

答:
Nothing is wrong. Your "attempt" is correct.

(Note that if $xy \geq 0$, we can also have $x < 0, y < 0$, but the statement still holds.)

Title: Absolute values (definition proof)


问:
prove that e^x is not Lipschitz on R
I know that the definition of Lipschitz function is:
If A⊆R and f:A→ R is function. Then, f is said to be Lipschitz on A if there exists a constant K>0 such that,|f(x)−f(y)|≤K|x−y| for all x,y∈A.
but how to prove that e^x is not Lipschitz on R?</p>

答:
Because the derivative isn't bounded.
Convexity makes it very straighforward, though still true while being not convex.
$$f(y) - f(x) \geq f'(x) (y-x) $$

Title: prove that e^x is not Lipschitz on R


答:
If any of the variables is $0$ then the product are zero so the equality is true. Hence, we assume $x,y \ne 0$

Assume $x < 0$ and $y < 0$. Then $xy>0$ which means $$|x||y| = (-x)(-y) = xy = |xy|$$
The case $x > 0$ and $y>0$ is the same.

Assume $x>0$ and $y<0$. Then $xy < 0$ which means $$|x||y| = x(-y) = -xy = |xy|$$
The case $x<0$ and $y>0$ is the same.

Title: Absolute values (definition proof)

Body:
Prove that:

a) $x \le |x|$ and

b) $|xy| = |x||y| $

Attempt:

a) If $x \ge 0: x = |x| \le |x|$.

If $ x < 0: x = |-x| = |x| \le |x|.$

b) If $xy \ge 0: |xy| = xy = |x||y|.$

If $xy < 0: |xy| = -xy = |-x||y|$ or $|x||-y| = |x||y|$.

Is there anything that is wrong?


问:
If $a+b+c=abc$ then $\sum\limits_{cyc}\frac{1}{7a+b}\leq\frac{\sqrt3}{8}$
Let $a$, $b$ and $c$ be positive numbers such that $a+b+c=abc$. Prove that:
$$\frac{1}{7a+b}+\frac{1}{7b+c}+\frac{1}{7c+a}\leq\frac{\sqrt3}{8}$$

I tried C-S:
$$\left(\sum_{cyc}\frac{1}{7a+b}\right)^2\leq\sum_{cyc}\frac{1}{(ka+mb+c)(7a+b)^2}\sum_{cyc}(ka+mb+c)=$$
$$=\sum_{cyc}\frac{(k+m+1)(a+b+c)}{(ka+mb+c)(7a+b)^2}.$$
Thus, it remains to prove that
$$\sum_{cyc}\frac{k+m+1}{(ka+mb+c)(7a+b)^2}\leq\frac{3}{64abc},$$
but I did not find non-negative values of $k$ and $m$, for which the last inequality is true.

If we replace $7$ with $8$ so for $(a,b,c)||(28,1,5)$ this inequality would be wrong. Around this point the starting inequality is true, but we see that we can'not free use AM-GM because in AM-GM the equality occurs, when all variables are equal.

Thank you!

答:
This inequality is something hard

$  a\geq b\geq\sqrt{3}\geq c>0$  such that $abc=a+b+c$ and constraints $(C)$ and $(2)$ and $(3)$  then we have :

$$\frac{1}{7b+a}+\frac{1}{7a+c}+\frac{1}{7c+b}\leq \left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\frac{1}{1+7\left(\frac{\frac{b}{a}+\frac{a}{c}+\frac{c}{b}+\frac{a}{b}+\frac{b}{c}+\frac{c}{a}+3}{9}\right)^{\frac{1}{3}}}$$

And :

$$\frac{1}{7b+a}\leq \left(\frac{1}{b}\right)\frac{1}{1+7\left(\frac{\frac{b}{a}+\frac{a}{c}+\frac{c}{b}+\frac{a}{b}+\frac{b}{c}+\frac{c}{a}+3}{9}\right)^{\frac{1}{3}}}\quad (C)$$

$$\frac{1}{7c+b}+\frac{1}{7a+c}\leq \left(\frac{1}{c}+\frac{1}{a}\right)\frac{1}{1+7\left(\frac{\frac{b}{a}+\frac{a}{c}+\frac{c}{b}+\frac{a}{b}+\frac{b}{c}+\frac{c}{a}+3}{9}\right)^{\frac{1}{3}}}\quad (2)$$

Simplify the condition $(C)$ is easy comparing the function :

$$f(x)=\frac{1}{7x+1}$$

Or:

$$f\Big(\frac{6+\frac{a}{b}}{7}\Big)\leq f\left(\left(\frac{\frac{b}{a}+\frac{a}{c}+\frac{c}{b}+\frac{a}{b}+\frac{b}{c}+\frac{c}{a}+3}{9}\right)^{\frac{1}{3}}\right)$$

For the constraint $(2)$ we know that :

$$f(a)=\frac{1}{7c+b}+\frac{1}{7a+c}-\left(\frac{1}{c}+\frac{1}{a}\right)\frac{1}{1+7\left(\frac{\frac{b}{a}+\frac{a}{c}+\frac{c}{b}+\frac{a}{b}+\frac{b}{c}+\frac{c}{a}+3}{9}\right)^{\frac{1}{3}}}$$

is an increasing function (as the difference of a positive decreasing function and a negative increasing function)  on $(\sqrt{3},\frac{\frac{3}{2}\sqrt{3}-\frac{1}{2}c+c}{c(\frac{3}{2}\sqrt{3}-\frac{1}{2}c)-1})$ with :
$$b=\frac{a+c}{ac-1}$$

So we have to find an approximation of the roots !

We can choose $b\geq\frac{3}{2}\sqrt{3}-\frac{1}{2}c\quad (3)$

A bit of algebra and we get a polynomial with a root equal to $\sqrt{3}$ and one other root .Now the magic key is,the inequality and the constraints $(C),(2)$ are homogenous so we can introduce a coefficient and play with it keeping in mind the others constraints.

I continue later...Thanks!


答: $e^x$ is not uniformly continuous and so is not Lipschitz. Take for example $x_n=\ln n,\ $ $y_n=\ln(n+1)$ then $y_n-x_n\to 0$ but $e^{y_n}-e^{x_n}\to 1$. Roughly speaking $e^x$ diverges to $+\infty$ with big velocity, bigger than a linear function

Title: prove that e^x is not Lipschitz on R

Body:
I know that the definition of Lipschitz function is:
If A⊆R and f:A→ R is function. Then, f is said to be Lipschitz on A if there exists a constant K>0 such that,|f(x)−f(y)|≤K|x−y| for all x,y∈A.
but how to prove that e^x is not Lipschitz on R?</p>


答:
$\newcommand{\R}{\mathbb{R}}$
This kind of problem can be easily tackled using proof by contradiction.

Suppose $e^x$ is Lipschitz on $\R$. Then, there is a constant $L > 0$ such that for all $x,y \in \R$, $|e^x - e^y| \leq L|x-y|$. But according to the mean value theorem, there is $c\in (x,y)$ such that
$$
\frac{e^x - e^y}{x - y} = e^c
$$
Therefore, if we raise both $x$ and $y$, it can be arbitrarily large and of course, it can be larger than any $L > 0$, which is a contradiction.

Title: prove that e^x is not Lipschitz on R


问:
Length of a segment in rectangular parallelepiped
I am stuck on a problem from a book. I have attached the picture of that question below.

My approach was to draw a line segment between C and X and another line segment between C and Y. This should make triangle XCY. Since XCY is a right triangle (assuming angle XCY is right), I can use pythagorean theorem to find the length of segment XY( answer to the question). Because the lengths of segments CX and CY can be found by using pythagorean theorem on triangle CXD and CBY respectively. The book's solution manual however gives a different answer.

I have explained the author's solution below, in case you require:

The author's approach is to make a segment XL where XL is parallel to GH and the point L lies on CG. Another segment LM which is parallel to FG where the point M lies on segment FB. Then create a triangle XYL. First, the author finds the length of segment YL using pythagorean on right triangle LMY ( point M is to the right of point Y on FB). Length of segment XL is known to be equal to GH. So finally the author, uses pythagorean theorem  on triangle XYM to find the answer (ie. the length of segment XY).

While I agree with the author's solution. I don't understand why my solution is wrong and gives a different answer than the author. Is it because my assumption of triangle XCY being a right triangle is wrong? Or something else?

Thanks

答:
Introduce a $3$D reference system with origin in $A$.

Points have coordinates $Y=(4,3,0);\;X=(0,5,3)$

$$XY=\sqrt{(5-3)^2+(3+0)^2+(4+0)^2}=\sqrt{29}$$


问:
Shading A Venn Diagram Using A Specific Equation
The expression is: $A\triangle(B\cap C')$. The $\triangle$ refers to $-$ and the $\cap$ refers to an intersection, whilst the $\;{}'$ refers to the prime of $C$.

There are no numbers or items included in this problem. I have drawn a Venn diagram with a background (for the universal set) and circles $A$, $B$ and $C$ which equally overlap. The problem is, I am unsure how to shade it. I also must provide the logic used to solve the equation. Thanks a bunch!

答: A△(B∩C′) = A∩(B∩C')' They are equivalent! So, shade all of A that is not in B. Also shade the intersection of A, B & C (You should have shaded A, A/C AND A/B/C.).</p>


问:
Which Hecke algebra is used in representation theory?
Which Hecke algebra is used in representation theory or more specifically in the study of Langlands's conjecture ?

From here, the Hecke algebra is constructed from a locally compact topological group and its closed subgroup.

While from here, the Hecke algebra is defined in terms of Hecke operators which are coming  from Modular forms.

This makes me confused.

Are the above two definition literally and mathematically equivalent ?

If not, which one is used in the study of Local/Global Langlands's conjecture ?

Any explanation will help me.

答:
Both are used and they are basically equivalent.  What I'm going to say is technically incorrect in some details but the general idea is right.

To make life easy, assume $f$ is a modular cusp form for $\operatorname{SL}_2(\mathbb Z)$ which is a normalized eigenfunction of all Hecke operators $T_p$.  There is a way (actually several, slightly inequivalent ways) to associate $f$ to an automorphic form $\phi: \operatorname{GL}_2(\mathbb Q) \backslash \operatorname{GL}_2(\mathbb A) \rightarrow \mathbb C$.  Let $G = \operatorname{GL}_2$.  The function $\phi$ lives inside the space $L^2(G(\mathbb Q)Z_G(\mathbb A) \backslash G(\mathbb A))$ and generates an irreducible representation $\Pi$ inside there.

Let $G_p = G(\mathbb Q_p)$ and $K_p = G(\mathbb Z_p)$.  There are unique irreducible, admissible representations $\pi_p$ of $G_p$ and a representation $\pi_{\infty}$ of $G_{\infty} = G(\mathbb R)$ such that $\Pi$ contains the "infinite tensor product" representation $\otimes_{p \leq \infty} \pi_p$ as a dense subspace (some work needs to be done to make sense out of an infinite tensor product).  Assume each representation $\pi_p$ of $G_p$ has a nonzero vector fixed by $K_p$.

Let $H_p = \mathscr C_c^{\infty}(K_p \backslash G_p/K_p)$ be the convolution ring of locally constant and left and right bi-$K_p$ invariant complex valued functions on $G_p$.  This is one of the kinds of Hecke algebras you were considering.  These particular Hecke algebras turn out to be commutative rings with unity.  Let $H_{\operatorname{fin}}$ be the infinite tensor product of the rings $H_p : p < \infty$.

The function $\phi$ lies in $\otimes_{p \leq \infty} \pi_p$ and in fact is itself equal to an infinite tensor product $\phi = \otimes_{p \leq \infty} \phi_p$ with $\phi_p \in \pi_p$.  The Hecke operators $T_{p^n} : n \in \mathbb N$ scale the cusp form $f$, but if we identify $f$ with the automorphic form $\phi$, then the $T_{p^n}$ affect only the component $\phi_p$.  In fact, $T_{p^n}$ identifies with a certain element in $H_p$, and $H_p$ is generated as an algebra by the $T_{p^n}$.

In this way, the tensor product of the "local Hecke algebras" $H_p$ form the "global finite Hecke algebra" $H_{\operatorname{fin}}$, which can also be thought of as being generated by the operators $T_{p^n}$, for $p$ prime and $n\in \mathbb N$.


问:
How far could you get in Graph Theory, whilst knowing almost nothing about almost all other fields?
Apologies if the content of this post is not mathematical enough and is consequently in the wrong place. For context, I graduated with my masters in mathematics about 8/9 years ago and then went on to do other things, leaving maths behind. Recently I've got into learning Graph Theory (initially to help someone who was writing a module) and am having thoughts about taking up the torch again and doing a phd.

Now, I am able to learn the material (a 4th year graph theory module) and solve the exercises without too much difficulty and believe that Graph Theory is something I could be good at. When I have finished, I have a Springer book lined up - Graduate texts in Mathematics. However I remember next to nothing about Number Theory, Group Theory, Rings, Fields, Analysis, Topology, Probability, Set Theory, Logic etc... (basically the whole of the rest of maths). Now at the moment it seems that if one has solid foundation in mathematics, a feel for combinatorics, and the ability to abstract/visualise, this is all you need to do Graph Theory, and that all these other areas are almost entirely separate.

My question is, how long will this go on for? Could I, in theory, get to the frontiers of understanding in Graph Theory and come up with new ideas, with my current level of knowledge about the rest of maths? Can anyone, who is experienced in Graph Theory, provide any insight on this?

Many Thanks

答:
It really depends. Graph Theory is a vast field. You can specialize in areas of graph theory where you don't need much of the other mathematical areas, except for standard logic and very basic set theory.
If graph theory excites you I would definitely do it. Once you have found the right topic and you feel like you need more knowledge from other areas, it should be no problem to learn the things you need, especially if you already have a background in mathematics (even if you feel like that you do not remember anything which is quite normal actually). Of course, this does not cover the formalities mentioned in the comments by littleO (starting an actual PhD etc.).

However, I feel like that this alone is not a satisfying answer, so let me elaborate a little bit more information which might help you.

I'm going to list some of the topics I feel like you will need to start.

Basic mathematical logic, proof techniques (induction, proof by contradiction, ...),

Very basic set theory,

Algorithms and complexity analysis,

If you want to dive into algebraic graph theory, then of course you will need linear algebra to some extend,

If you want to dive into extremal graph theory, then you will also need probability theory,

Basic topology, this is in fact really useful if talking about planar graphs or $k$-planar graphs,

(Ramsey theory; it is actually quite useful in many situations but I wouldn't focus too much on it for now)

However, everything mentioned can become very complex very quickly. I would like to name a few examples. There is a nice concept called "Flag Algebras" by Razborov for solving problems in extremal graph theory. It is extremly useful and used quite often in research for this area. However, if you read his paper, you will quickly realize that simple mathematics is not necessarily enough. You can check it out here. Also, you can search for papers which used his method. Most of them are not necessarily using easy mathematics as well.

Proofs in structural graph theory are usually quite hefty too. Check out the famous strong perfect graph theorem and its "recent" proof by Chudnovsky et. al. You can check the paper out here. Even if it is not easy, this is actually a good example for a result which doesn't use any other mathematics than graph theory itself (of course logic and set theory...).

You can of course specialize on algorithmic aspects of graph theory as well for which you will need knowledge of computer science. A famous example is the graph colouring problem. Research for determining upper bounds for the chromatic number of certain graph families is mostly done without keeping in mind that we still need to find an efficient algorithm which does perform the actual colouring. You quickly come to the point where you want to dive into complexity theory because it is quite fascinating (polynomial reductions etc.) and you quickly realize that the problems which are researched the most in graph theory are the $\mathcal{NP}$-complete problems.

There are many  more examples, I have just named very few. In summary, however, there is nothing in my eyes that stops you from dealing with graph theory. Studying a research area is a process and during this process you will notice where your knowledge gaps are and these gaps are always closable.


问:
Are $\sum_{i=1}^{n} \xi_i / i$ and $\max\xi_n$ equally distributed?
Let the random variables $\xi_1, \xi_2, \ldots, \xi_n$ be independent, have an exponential distribution and expectation of each is 2.

Are $\sum_{i=1}^{n} \xi_i / i$ and $\max\xi_n$ equally distributed?

答:
Claim: the statement is true and may be generalized easily for  any order statistic from exponential distribution with any $\lambda$, not only $\lambda = 2$.

At first, let us note that it's sufficient to consider $\lambda = 1$ since $\frac1{\lambda}$ is a scale parameter.

See proposition 3 p.55 here:1
https://doi.org/10.1214/aoms/1177699058

So, the claim is true and may be showed by direct computation.

Moreover,
$$X_{(j)} = \sum_{i=1}^{j} \frac{\xi_i}{n-i+1}$$  in distribution.

1Chernoff, H.; Gastwirth, Joseph L.; Johns, M. V. jun., Asymptotic distribution of linear combinations of functions of order statistics with applications to estimation, Ann. Math. Stat. 38, 52-72 (1967). ZBL0157.47701.


问:
I need help with simple algebra with logarithms
Can anyone help me with this?
$$xe^{ht} = y $$
How do I eliminate "t" from the LHS of the equation? What I want to end up with is the LHS being a function of "h".

Body:
$$xe^{ht}=x(e^h)^t=y$$

$$(e^h)^t=\frac{y}{x}$$

$$e^h=\bigg(\frac{y}{x}\bigg)^{\frac{1}{t}}$$

Provided that $x,t\neq 0$.


问:
Relation between a sum of a series and the limit of a sequence
I'm stuck on this question

Let $\{a_{n}\}$ a sequence of real numbers

I need to show the series $\sum_{n=1}^{\infty}(a_{n} - a_{n-1})$ and the sequence $\{a_{n}\}$ are the same nature (convergent or divergent). Additional, I need to give a relation between the sum of the series and the limit of the sequence $\{a_{n}\}$.

Can anyone please help me?

I'm completely lost...

Thanks in advance

答: Hint: What is the $k$-th partial sum of the series? (remember that a series converges to $L$ iff the sequence of its partial sums converges to $L$, so what will happen if you found out that the partial sums of the series give back the sequence of $a_n$'s?).</p>


问:
$X_n$ ($n\ge1)$ is sequence of real numbers such that $\lim_{n \to \infty}{\frac{x_n}{n}}=0.001$ then
if $X_n$ ($n\ge1)$ is sequence of real numbers such that $\lim_{n \to \infty}{\frac{x_n}{n}}=0.001$ then,

$(A)\ \ X_n$ is a bounded sequence

$(B)\ \ X_n$ is a unbounded sequence

$(C)\ \ X_n$ is a Convvergent sequence

$(D)\ \ X_n$ is a monotonically decreasing sequence

I can only think of this particular case.

$\text{convergent sequence}\cdot \frac{1}{n}=\text{convergent sequence}\cdot \text{convergent sequence} =$finite limit

Therefore it is convergent but I think its not a correct way to solve this problem.

How do I figure out what is our sequence behavior just on the basis of limits provided?

答:
A direct proof.

If $\lim_{n \to \infty} x_n/n = c > 0$
then, for all large enough $n$,
$x_n/n > c/2$
so $x_n > cn/2$
and this is unbounded.


答:
Yes. The argument in this blog post shows that they are equal in mean, but it can be extended to show they are actually equal in distribution.

Let $a(1),\dots,a(n)$ be the (random) ordering of the indices such that $\xi_{a(1)}\leq \xi_{a(2)}\leq \cdots \leq \xi_{a(n)}$. Then
$$\max(\xi_1,\dots,\xi_n)=\xi_{a(n)}=\xi_{a(1)}+(\xi_{a(2)}-\xi_{a(1)})+\cdots+(\xi_{a(n)}-\xi_{a(n-1)}).$$
Since $\xi_{a(1)}$ is the minimum of $n$ independent $\text{Exp}(\lambda)$ random variables, its distribution is $\text{Exp}(n\lambda)$.

Since $\xi_{a(1)}$ is the smallest of the $\xi_i$, we have that $\xi_{a(2)}-\xi_{a(1)}$ is the minimum of $n-1$ independent $\text{Exp}(\lambda)$ random variables, namely
$$\xi_{a(2)}-\xi_{a(1)}=\min_{i \neq a(1)}(\xi_{i}-\xi_{a(1)}).$$
Note that $(\xi_i-\xi_{a(1)})\sim\text{Exp}(\lambda)$ for $i \neq a(1)$ by the memoryless property of the exponential distribution. Thus $\xi_{a(2)}-\xi_{a(1)}$ has the distribution $\text{Exp}((n-1)\lambda)$, and it is independent of $\xi_{a(1)}$ by the memoryless property.

Next we have $\xi_{a(3)}-\xi_{a(2)}$ is the minimum of $n-2$ independent $\text{Exp}(\lambda)$ random variables:
$$\xi_{a(3)}-\xi_{a(2)}=\min_{i \neq a(1),a(2)}(\xi_{i}-\xi_{a(2)})$$
and so has the distribution of $\text{Exp}((n-2)\lambda)$. It is independent of $\xi_{a(1)}$ and $\xi_{a(2)}-\xi_{a(1)}$ by the memoryless property.

Continuing this argument, we have that $\max(\xi_1,\dots,\xi_n)$ is the sum of $n$ independent exponential random variables with rates $n\lambda,(n-1)\lambda,\dots,\lambda$.

On the other hand, $\frac{\xi_i}{i}$ is an exponential random variable with rate $\lambda i$. Since the $\xi_i$ are independent, $\sum_{i=1}^n \frac{\xi_i}{i}$ is the sum of $n$ independent exponential random variables with rates $\lambda, 2\lambda, \dots, n\lambda$. So we conclude that $\max(\xi_1,\dots,\xi_n)$ and $\sum_{i=1}^n\frac{\xi_i}{i}$ are equal in distribution.

Body:
Let the random variables $\xi_1, \xi_2, \ldots, \xi_n$ be independent, have an exponential distribution and expectation of each is 2.

Are $\sum_{i=1}^{n} \xi_i / i$ and $\max\xi_n$ equally distributed?


问:
Two definitions of the injectivity radius
For a Riemannian manifold $M$, usually, the injectivity radius of $p \in M$ is defined by the supremum of positive numbers $r$ such that $\exp_p|_{B(0, r)}$ is a diffeomorphism onto the image. Here, $B(0,r)$ is the geodesic ball of radius $r$. However, in Klingenberg's book, the injectivity radius is defined by by the supremum of positive numbers $r$ such that $\exp_p|_{B(0, r)}$ is injective. Are these two definitions equivalent?

Considering the global rank theorem, I tried to prove the exponential map at $p$ is of constant rank, but it seems that this try does not work. In fact, Klingenberg frequently says that the exponential map is just injective instead of saying the map is a diffeomorphism. To guarantee that the exponential map at $p$ is a diffeomorphism on a geodesic ball, is it enough to show that it is injective?

Thanks!

答:
Edit: According to this answer, Klingenberg's Riemannian Geometry shows that $\exp: TM \to M$ is non-injective in an neighborhood of any critical point in Chapter 2.1, "Completeness and Cut Locus:"

The following result is from Klingenberg's Riemannian geometry book (Theorem 2.1.12):

Theorem. On a complete Riemannian manifold $(M,g)$, if $\mathrm{ker}(d\exp_p\vert_v)\neq 0$ for some $(p,v)\in TM$, then $\exp_p$ fails to be injective in every neighbourhood of $v$.

(Thus if injectivity fails infinitesimally, it also fails in every neighbourhood. This is of course false for a general smooth map.)

That being proven, Klingenberg is then free to give a weaker-looking, yet equivalent, alternative definition of injectivity radius, and apply it for the remainder of the text.


问:
measure of limsup positive for infinitely many terms
let $(X, M, \mu)$ be a measure space, and let ${E_k}$ be in $M$.  if $\mu(\cup E_k) < \infty$, and $\mu(E_k) \ge c > 0$ for infinitely many $k$'s for some $c$, then $\mu(\limsup E_k) > 0$.  Can someone explain why, I don't see where the condition  $\mu(\cup E_k) <\infty$ is used, why can this not be removed? an example might help.</p>

答:
Recall that by definition, $\limsup_k E_k=\bigcap_{l\geqslant 1}\bigcup_{j\geqslant l}A_j$. By the assumption, we have $\mu\left(\bigcup_{j\geqslant l}A_j\right)\geqslant c$ for any $l$.

Now use the following:

If $(X,\mathcal A,\mu)$ is a measure space, $(A_n,n\geqslant 1)$ is an non-increasing sequence and $\mu(A_1)$ is finite, then $\mu\left(\bigcap_nA_n\right)=\lim_n\mu(A_n)$.

The condition about finiteness of the measure of the union cannot be removed, for example when $X$ is the real line with Lebesgue measure and $A_n:=(n,+\infty)$.


问:
Differential inequality implies estimation
From the inequality
\begin{equation}
y^\prime(t) + C_1 y^\alpha(t) \le C_2, \quad y(0) = y_0
\end{equation}
for $t \in (0,T)$ and constants $C_1,C_2, \alpha > 0$,
I want to imply
\begin{equation}
y(t) \le \max\left\{y_0,\left(\frac{C_2}{C_1}\right)^{\frac1\alpha}\right\}
\end{equation}
for all $t \in (0,T)$.

I tried to do separation of variables with the transformation $z:=y^\alpha$, but I didn't get the above estimation.

Does someone know how to do this?

For $\alpha = 1$ this is contained in the proof of Lemma 4.5 in https://math.uni-paderborn.de/fileadmin/mathematik/ag-partielle-differentialgleichungen/Publikationen/Winkler/2020/winkler_2d_KS_NaSto_repulsion_homepage.pdf.

答: $y$ can not grow indefinitely, if $y'\ge 0$ then obviously $y^α\le \frac{C_2}{C_1}$, so that $y$ can not grow above $\left(\frac{C_2}{C_1}\right)^\frac1α$. On the other hand, if $y'<0$ on an initial segment, then $y$ is falling, so also $y\le y_0$ there. While a falling segment can be followed by an increasing segment, the first restriction will kick in, so that in total really $$y(x)\le\max\left(y_0,\left(\frac{C_2}{C_1}\right)^\frac1α\right).$$


问:
Determining all the possible values of $n$ under restrictions on number of factors of $n(n+2)(n+4)$

Determine all positive integers $n$ such that the number $n(n+2)(n+4)$ has at most $15$ positive divisors.

I wanted to verify my approach for the problem. It goes like this:

My Approach:

There can be two cases:

Let $n$ be even.

Now it is evident that $2|n$, $2|n+2$ and $2|n+4$

This means that $2^3|n(n+2)(n+4)$

Since we have $3$ consecutive even numbers. It is evident that either $4|(n+2)$, or $4|n$ and $4|n+4$

Suppose $4|(n+2)$, this means that there would be one more factor of $2$ than the initial number dividing the number, thus $2^4|n(n+2)(n+4)$

It is also evident that $3|n(n+2)(n+4)$

So now $n(n+2)(n+4)=2^4\cdot 3\cdot p_1^{a_1}\cdot p_2^{a_2}\cdots p_r^{a_r}$

This means that $\tau (n(n+2)(n+4))=5\times 2\times (a_1+1)\times (a_2+1)\times \cdots \times (a_r+1)$.

Observe that if any of the $a_i\geq 1$, then $\tau (n(n+2)(n+4))\geq 5\times 2\times 2>15$.

$\therefore $ All $a_i=0$. Thus $n(n+2)(n+4)=2^4\times 3=48=2\times 4\times 6$, which means that $\boxed{n=2}$

Suppose $4|n$ and $4|n+4$, and since we have two consecutive factors of 4, one of them must be divisible by $8$.

Thus there are $3$ additional factors of $2$ in $n(n+2)(n+4)$.

So $n(n+2)(n+4)=2^6\cdot 3\cdot p_1^{a_1}\cdot p_2^{a_2}\cdots p_r^{a_r}$

Now using the same argument as used in the previous case, we know that $a_i=0$ for all $i$.

$\therefore n(n+2)(n+4)=2^6\times 3=4\times 6\times 8$ which means that $\boxed{n=4}$.

Let $n$ be odd.

Now $3|n(n+2)(n+4)$. Also any two of them taken in pair are relatively prime.

Let $n(n+2)(n+4)=(3\times k)\cdot p_1^{a_1}\cdot p_2^{a_2}\cdots p_r^{a_r}\cdot q_1^{b_1}\cdot q_2^{b_2}\cdots q_s^{b_s}$

Now $k$ can be from the set $\{1, 3, 5,\ldots\}$.

Suppose $\tau (3k)\geq 4$, then $\tau (n(n+2)(n+4))\geq 4(a_1+1)(a_2+1)\ldots (a_r+1)(b_1+1)(b_2+1)\cdots (b_s+1)$

We need $4(a_1+1)(a_2+1)\ldots (a_r+1)(b_1+1)(b_2+1)\cdots (b_s+1)\leq 15$ or $(a_1+1)(a_2+1)\ldots (a_r+1)(b_1+1)(b_2+1)\cdots (b_s+1)\leq 3$.

But there would be at least one $a_i>1$ and $b_i>1$ because the numbers not divisible by three are greater than 1.

Thus $(a_1+1)(a_2+1)\ldots (a_r+1)(b_1+1)(b_2+1)\cdots (b_s+1)\geq 4$ which is contradictory.

$\therefore \tau (3k)\leq 3$. Now if $3\not | k$ and $k\neq 1$, then it is evident that $\tau (3k)\geq 4$ which doesn't satisfy our conditions.

$\therefore 3|k$. Since $\tau(3k)\leq 3$, the only possible values for $k$ are ${1,3}$ or $3k=3$ or $3k=9$. Which means the triplets can be $(1,3,5), (3,5,7), (5,7,9), (7,9,11)$ and $(9,11,13)$. This means that $\boxed{n=1,3,5,7,9}$

Considering both the cases together, we get $\boxed{n=1,2,3,4,5,7,9}$

Please check my approach for any mistakes. Also please provide an alternate solution less convoluted than this if possible.

THANKS

答:
Update:

As pointed out by Misha Lavrov, when $n$ is odd there's also a case where $P_n=pq$.

Denote $P_n=n(n+2)(n+4)$.

If $n$ is even then $2^4 \cdot 3 | P_n$. There are no other prime factors since there are already $10$ factors from $2^4 \cdot 3$. So the only possibilities for $P_n$ are: $2^4 \cdot 3^2, 2^4 \cdot 3, 2^5 \cdot 3, 2^6 \cdot 3$. And you can check easily only $P(2)=48$ and $P(4)=192$ work.

If $n$ is odd, then $n, n+2, n+4$ are pairwise co-prime. So the only possibilities for $P_n$ are: $P_n = pqr$ or $P_n=p^2 q r$ where $p, q, r$ are distinct prime numbers (since we can't have $4$ or more prime factors). And one of $n, n+2, n+4$ is either $3$ or $3^2$. So there are only a few cases to check. I'll leave that to you.


问:
Proof of joint Gaussian distribution in a linear equation
Let $X = AY + Z$ where $A$ is constant, $Y$ and $Z$ are independent gaussian rvs. How do you prove that $X$ and $Y$ are jointly gaussian? I know that the sum of two independent gaussians gives a gaussian distribution. But I have no idea where to start to show joint gaussian pdf.

答:
For every linear combination of $X$ and $Y$ :
$$ \mu X + \beta Y = (\beta + \mu A) Y + \mu A Z$$
And since $Y$ and $Z$ are independent, the linear combination follows a gaussian distribution. So they are jointly gaussian.


问:
I need to find a solution for $2x+\arcsin x=1$
$$2x+\arcsin x=1$$
I didn't find way to solve this problems.
Is there a difference in solving equations between $\sin$ and $\arcsin$?
For some reason I did not find the special solutions of arc.

Is there a connection to Maclaurin or Taylor?

Thanks for your help.

答:
Notice that:
$$2x-\frac \pi2\leq 2x+\arcsin(x)\leq 2x+\frac \pi2$$
So we want that:
$$2x-\frac \pi2\leq 1\leq 2x+\frac \pi2$$
And by solving this two disequations you get:
$$\frac 12-\frac \pi4 \leq x \leq \frac 12+\frac \pi4$$
But $arcsin$ is defined only in $[-1,1]$, so:
$$\frac 12-\frac \pi4 \leq x \leq 1$$
So your (eventual) zero(es) are in the interval $I=[\frac 12-\frac \pi4,1[$(you can clearly exclude $1$ verifying by yourself). Moreover the function $f(x)=2x+\arcsin(x)$ is derivable in $I$ and its derivative is:
$$2+\frac{1}{\sqrt{1-x^2}}$$
That is always $>0$ so the function is strictly increasing. Moreover since the function $f$ is continous in the interval and $f(\frac 12-\frac \pi2)-1<0$ while $f(1)-1>0$, by Bolzano theorem there must be a solution in the interval(and it's unique because the function is strictly increasing. Now that we have proved that the solution exists and is unique, to find it we have to use numerical methods.
As you said Mac-Lauren series is a possibility since $\arcsin$ is analytical,in particular the best possible choice(that would make the equation analitically solvable) would be:

$\arcsin(x)\approx x+\frac{x^3}{6}$

But to not mess with cubic equations we have to use:

$\arcsin(x) \approx x$

So:

$2x+\arcsin(x) \approx  3x$

And our equation becomes:

$3x\approx 1$

$x\approx \frac 13$

That is already pretty near(you can check). We could also continue and this time we can center the taylor series in $\frac 13$ to improve the approximation(By Peano theorem error in taylor series asimptotically decreases when we get more near to the center).

$\arcsin(x)\approx \arcsin'(1/3)x=\frac{3\sqrt{2}}{4}x$

So our equation becomes:

$(\frac{3\sqrt{2}}{4}+2)x\approx 1$

$x\approx \frac{4}{3\sqrt{2}+8}$

If you continue this way the approximation will become better and better(this is actually the Newton-Raphson method "simplifed", and in general the convergence to the root is not granted: I simplified a lot)


问:
Is the set $A = [0,1]\setminus\mathbb{Q}$ countable or not?

Is the set $A = [0,1]\setminus\mathbb{Q}$ countable or not?

What I am thinking is $A$ consist of irrational numbers in the interval $[0,1]$ hence it is subset of irrational numbers. As set of irrational numbers is uncountable so I think set $A$ is also uncountable.

答:
Yes, it is uncountable, but not for that reason. For instance, $\left\{\sqrt2+n\,\middle|\, n\in\Bbb N\right\}$ is also a set of irrational numbers, but it is countable.

However, if $[0,1]\setminus\Bbb Q$ was countable, then, since $\Bbb Q\cap[0,1]$ is countable, $[0,1]$ would be countable too, since it's the union of them.


问:
Is the order on connectivity classes total?
Suppose $X$ and $Y$ are both topological spaces. Let's say $X$ is $Y$-connected iff $\forall x_0, x_1 \in X$ $\exists y_0, y_1 \in Y$ and a continuous function $f: Y \to X$ such that $f(y_0) = x_0$ and $f(y_1) = x_1$.

Such «relative connectivity» has a following property:

If $X, Y$ and $Z$ are such topological spaces, that $X$ is $Y$-connected and $Y$ is $Z$-connected then $X$ is $Z$-connected.

Proof: Suppose $x_0, x_1 \in X$. Then $\exists y_0, y_1 \in Y$ and and a continuous function $f: Y \to X$ such that $f(y_0) = x_0$ and $f(y_1) = x_1$. There also $\exists z_0, z_1 \in Z$ and a continuous function $g: Z \to Y$ such that $f(z_0) = y_0$ and $f(z_1) = y_1$. Then $f(g(z_0)) = x_0$ and $f(g(z_1)) = x_1$, which means $X$ is $Z$-connected, Q.E.D.

Now, we can say that topological spaces $X$ and $Y$ are connectively equivalent iff $X$ is $Y$-connected and $Y$ is $X$-connected. This equivalence relation divides all topological spaces onto connectivity classes.

There exists a natural order on connectivity classes: if $C_0$ and $C_1$ are connectivity classes, we say that $C_0 \leq C_1$ iff for any topological spaces $X \in C_0$ and $Y \in C_1$ we have that $X$ is $Y$-connected.

Connectivity classes have following properties:

There exists the smallest connectivity class and it is exactly the class of $1$-element spaces

Proof:

Suppose $X := \{x\}$ is $!$-element space, and $Y$ is an arbitrary topological space. Then $f: Y \to X$ such that $\forall y \in Y f(y) = x$ is the corresponding continuous function (no matter what $y_0$ and $y_1$ we take. Thus $X$ is $Y$-connected. All we need to prove that if $|Y| > 1$ then $Y$ is not $X$-connected is to consider two distinct elements of $Y$. Thus the class of all $1$-element spaces is the smallest connectivity class.

There exists the largest connectivity class and it is exactly the class of all disconnected spaces

Proof: Suppose, $Y$ is not connected. Then it is a disjoint union of two non-empty open sets $U$ and $V$. Now, suppose $X$ is an arbitrary topological space and $x_0, x_1 \in X$. Now we take $y_0 \in U$, $y_1 \in V$ and define $f$ as

$$f(y) = \begin{cases} x_0 & \quad y \in U \\ x_1 & \quad y \in V \end{cases}$$

Those $y_0, y_1$ and $f$ satisfy our conditions. That means, $X$ is $Y$-connected.

So, we know, that there exists the largest connectivity class and it contains all disconnected spaces. To finish the proof we need to prove, that it does not contain any connected spaces:

Suppose, $X$ is not connected. Then it is a disjoint union of two non-empty open sets $U$ and $V$. Let's take $x_0 \in U$ and $x_1 \in V$. Then $Y$ will not be connected as a disjoint union of two non-empty open sets $f^{-1}(U)$ and $f^{-1}(V)$, Q.E.D.

Connectivity classes do not form a set

Proof can be found here

Now, my question is:

Is the order on connectivity classes a total order?

Or, equivalently:

Do there exist two topological spaces $X$ and $Y$ such that neither $X$ is $Y$-connected nor $Y$ is $X$-connected?

答:
It is not a total order.  Basically, if you take two connected spaces that are not path-connected for "different reasons", then they will probably be incomparable in the connectivity order.  For instance, let $X$ be the topologist's sine curve and let $Y$ be the closed long line.  Then every map $Y\to X$ is eventually constant, so $X$ is not $Y$-connected since it is not path-connected.  On the other hand, every map $X\to Y$ has image that is separable and connected, so no such map can connect the endpoints of $Y$.

(To explain the importance of avoiding path-connected spaces, the path-connected spaces are the least connectivity class if you restrict to reasonably nice (say, normal) spaces with more than one point, as a consequence of Urysohn's lemma.  See this answer for some related discussion.)

Here's a rather different sort of counterexample.  Let $X=\{a,b,c\}$ with the topology that a set is open iff it is $X$ or it does not contain $b$.  Let $Y$ have the same underlying set but the opposite topology: the open sets of $Y$ are the closed sets of $X$.  Then $X$ and $Y$ are incomparable in the connectivity order, since neither one can connect the points $a$ and $c$ in the other.  (The quick way to see this is by thinking about their specialization orders: one looks like $\wedge$ and the other looks like $\vee$.)


答:
All countable set of $R$ have Lebesgue measure equal to $0$.
So Lebesgue measure of $[0, 1] - \mathbb{Q}$ is $1$.
Eventually by contraposition, $[0, 1] - \mathbb{Q}$ is uncoutable.

Body:
Is the set $A = [0,1]\setminus\mathbb{Q}$ countable or not?

What I am thinking is $A$ consist of irrational numbers in the interval $[0,1]$ hence it is subset of irrational numbers. As set of irrational numbers is uncountable so I think set $A$ is also uncountable.


答:
What you are thinking does not work.

For example, $\{n\pi\mid n\in\mathbb N\setminus\{0\}\}$ is a subset of irrational numbers but countable.

Here's an argument that works.  If $A$ were countable, then, since $\mathbb Q$ is countable,

$A\cup \mathbb Q$ would be countable,

and therefore $[0,1]$, which is a subset of $A\cup \mathbb Q$, would be countable,

and that is a contradiction.

Title: Is the set $A = [0,1]\setminus\mathbb{Q}$ countable or not?


问:
$\sum_{i\in I}\kappa_i$ vs. $\prod_{i\in I}\kappa_i$
Let $I$ be an arbitrary set and $(\kappa_i)_{i\in I}$ a family of cardinals.

What can be said about the relationship between
$$\sum_{i\in I}\kappa_i \qquad\text{and}\qquad  \prod_{i\in I}\kappa_i\qquad?$$
I know that for $I=2$, they are equal. But what if $I$ is an arbitrary set with at least two elements, does it always hold that
$$\sum_{i\in I}\kappa_i\leq\prod_{i\in I}\kappa_i$$
or even
$$\sum_{i\in I}\kappa_i=\prod_{i\in I}\kappa_i?$$
What else can be said?

答:
Let's write $S$ for the sum $\sum_{i\in I}\kappa_i$ and $P$ for the product $\prod_{i\in I} \kappa_i$.

If $I$ is empty, then $S = 0$ and $P = 1$, so $S<P$. From now on. let's assume $I$ is non-empty.

If some $\kappa_i$ is equal to $0$, then $P = 0$, so $P<S$ unless every $\kappa_i$ is equal to $0$, in which case $P = S$. From now on, let's assume all the $\kappa_i$ are non-zero.

Let $J = \{i\in I\mid \kappa_i > 1\}$ and $K = \{i\in I\mid \kappa_i = 1\}$, and let $\lambda = \sum_{i\in K}\kappa_i = |K|$. Then $S = \left(\sum_{i\in J}\kappa_i\right) + \lambda$ and $P = \prod_{i\in J}\kappa_i$. Note that we can have $P < S$ if $\lambda$ is large enough, i.e., if many of the $\kappa_i$ are equal to $1$. But in any case, we can reduce the question to comparing $\sum_{i\in J}\kappa_i$ and $\prod_{i\in J}\kappa_i$ and $\lambda$. From now on, let's assume all the $\kappa_i$ are greater than $1$.

If $I$ is finite and all the $\kappa_i$ are finite, the question is about finite arithmetic. If  $|I| = 1$, then $S = P$. If $|I|>1$, then since all the $\kappa_i$ are at least $2$, we have $S < P$, except in the single case $2+2 = 2\times 2$.

If $I$ is finite and at least one of the $\kappa_i$ is infinite, then $S = P = \text{max}_{i\in I} \kappa_i$. From now on, let's assume $I$ is infinite.

Based on all the above reductions, we're left with situation that $I$ is infinite and all the $\kappa_i$ are at least $2$. I claim that in this situation, $S\leq P$.

Let $\kappa = \sup_{i\in I} \kappa_i$. Then $S = \max(\kappa,|I|)$. First, we show that $|I|\leq P$, by finding an injective function from $I$ to $\prod_{i\in I} \kappa_i$. We map $i$ to the tuple in the cartesian product which is $1$ on the $i$-coordinate and $0$ on every other coordinate. This is well-defined because $\kappa_i \geq 2$ for all $i$.

Now if $\kappa\leq |I|$, then $S = |I|\leq P$. It remains to consider the case $|I| < \kappa$. Let $I_\infty = \{i\in I\mid \kappa_i \text{ is infinite}\}$.  Since $I$ is infinite, $\kappa > \aleph_0$, so $\kappa = \sup_{i\in I} \kappa_i = \sup_{i\in I_{\infty}} \kappa_i$.  Thus $S = \kappa = \sup_{i\in I_{\infty}} \kappa_i = \sum_{i\in I_{\infty}} \kappa_i$, since $|I_\infty|\leq |I|$. And since none of the $\kappa_i$ are  $0$, we have $\prod_{i\in I_{\infty}}\kappa_i \leq P$. So we have reduced to showing that $\sum_{i\in I_{\infty}} \kappa_i \leq \prod_{i\in I_{\infty}} \kappa_i$, i.e., that $S\leq P$ under the assumption that all of the $\kappa_i$ are infinite.

For each $i$, let $s_i\colon \kappa_i\to \kappa_i$ be the successor function $\alpha\mapsto \alpha+1$ ($s_i$ is well-defined since $\kappa_i$ is infinite). Then $s_i$ is injective and does not have $0$ in its range. We define an injective function  $\bigsqcup_{i\in I} \kappa_i \to  \prod_{i\in I} \kappa_i$ by mapping $\alpha\in \kappa_i$ to the tuple which is $s_i(\alpha)$ on the $i$-coordinate and $0$ on every other coordinate. This completes the proof.

But in this situation ($I$ is infinite and all the $\kappa_i$ are at least $2$), the question of whether $S < P$ or $S = P$ can be independent of ZFC. For example, we have:
\begin{align*}
\aleph_1 + \sum_{i\in \aleph_0} 2 &= \aleph_1 + \aleph_0 = \aleph_1\\
\aleph_1 \cdot \prod_{i\in \aleph_0} 2 &= \aleph_1 \cdot 2^{\aleph_0} = 2^{\aleph_0}.
\end{align*}
In this case, the question of whether the product is strictly larger than the sum is the continuum hypothesis!

I made lots of claims without proof above. If anything is not clear to you, let me know and I'll elaborate.


问:
How do I find the matrix of a linear transformation from $\Bbb R^3 \longrightarrow \Bbb R^3$ of an orthogonal reflection on a planoe
So the questions is, " Find the matrix of the linear transformation of n orthogonal projection on the plane $2x-y+z=0$"

I don't know how I should tackle this problem. Any advice?

答:
Hint:

You know the normal vector to the plane: $\vec n=(2,-1,1)$. Hence the orthogonal projection $N=(X,Y,Z)$ of a point $M=(x,y,z)$ will lie on the line though $M$ with directing vector $\vec n$ and  satisfy the equation of the plane: $2X-Y+Z=0$

Now the parametric representation of the line yields $N=M+t\vec n=(x+2t,y-t,z+t)$.

Can you determine $t$ so that $N$ does lie in the plane, and deduce the coordinates of $N$ in function of those of $M$?


答:
First consider three basic results.

(1) It is evident that $3$ is a factor of $n(n+2)(n+4)$

(2) Let $p$ be any prime dividing two of $n,n+2,n+4$, then $p$ divides the difference between these two numbers i.e. $2$ or $4$. Then $p=2$.

(3) A number expressed as a product of prime powers, $p^aq^br^c...$, has $(a+1)(b+1)(c+1)...$ positive factors. Therefore $n(n+2)(n+4)$ has at most three prime factors. Furthermore, when there are three prime factors, we can only have $pqr$ or $p^2qr$.

$n$ odd

One of $n,n+2,n+4$ is $1,3$ or $9$ and then  $n$ is one of $1,3,5,7,9$, all of which are solutions.

$n$ even

Then $n(n+2)(n+4)$ has a factor $2^a3^br^c...$, where $a\ge 3,b\ge 1$. Since $(a+1)(b+1)\ge 8$ we must have $c=0$ i.e. only prime factors of $2$ and $3$.

Now $3$ can only be a factor of one of $n,n+2,n+4$ and therefore the other two numbers are unequal powers of $2$. If neither is less than or equal to $4$, then we would have $a\ge 1+3+4$ and then $2^83$ would give $18$ positive factors.

Therefore one of $n,n+2,n+4$ is $2$ or $4$ and then  $n$ is one of $2,4$, both of which are solutions.

Title: Determining all the possible values of $n$ under restrictions on number of factors of $n(n+2)(n+4)$

Body:
Determine all positive integers $n$ such that the number $n(n+2)(n+4)$ has at most $15$ positive divisors.

I wanted to verify my approach for the problem. It goes like this:

My Approach:

There can be two cases:

Let $n$ be even.

Now it is evident that $2|n$, $2|n+2$ and $2|n+4$

This means that $2^3|n(n+2)(n+4)$

Since we have $3$ consecutive even numbers. It is evident that either $4|(n+2)$, or $4|n$ and $4|n+4$

Suppose $4|(n+2)$, this means that there would be one more factor of $2$ than the initial number dividing the number, thus $2^4|n(n+2)(n+4)$

It is also evident that $3|n(n+2)(n+4)$

So now $n(n+2)(n+4)=2^4\cdot 3\cdot p_1^{a_1}\cdot p_2^{a_2}\cdots p_r^{a_r}$

This means that $\tau (n(n+2)(n+4))=5\times 2\times (a_1+1)\times (a_2+1)\times \cdots \times (a_r+1)$.

Observe that if any of the $a_i\geq 1$, then $\tau (n(n+2)(n+4))\geq 5\times 2\times 2>15$.

$\therefore $ All $a_i=0$. Thus $n(n+2)(n+4)=2^4\times 3=48=2\times 4\times 6$, which means that $\boxed{n=2}$

Suppose $4|n$ and $4|n+4$, and since we have two consecutive factors of 4, one of them must be divisible by $8$.

Thus there are $3$ additional factors of $2$ in $n(n+2)(n+4)$.

So $n(n+2)(n+4)=2^6\cdot 3\cdot p_1^{a_1}\cdot p_2^{a_2}\cdots p_r^{a_r}$

Now using the same argument as used in the previous case, we know that $a_i=0$ for all $i$.

$\therefore n(n+2)(n+4)=2^6\times 3=4\times 6\times 8$ which means that $\boxed{n=4}$.

Let $n$ be odd.

Now $3|n(n+2)(n+4)$. Also any two of them taken in pair are relatively prime.

Let $n(n+2)(n+4)=(3\times k)\cdot p_1^{a_1}\cdot p_2^{a_2}\cdots p_r^{a_r}\cdot q_1^{b_1}\cdot q_2^{b_2}\cdots q_s^{b_s}$

Now $k$ can be from the set $\{1, 3, 5,\ldots\}$.

Suppose $\tau (3k)\geq 4$, then $\tau (n(n+2)(n+4))\geq 4(a_1+1)(a_2+1)\ldots (a_r+1)(b_1+1)(b_2+1)\cdots (b_s+1)$

We need $4(a_1+1)(a_2+1)\ldots (a_r+1)(b_1+1)(b_2+1)\cdots (b_s+1)\leq 15$ or $(a_1+1)(a_2+1)\ldots (a_r+1)(b_1+1)(b_2+1)\cdots (b_s+1)\leq 3$.

But there would be at least one $a_i>1$ and $b_i>1$ because the numbers not divisible by three are greater than 1.

Thus $(a_1+1)(a_2+1)\ldots (a_r+1)(b_1+1)(b_2+1)\cdots (b_s+1)\geq 4$ which is contradictory.

$\therefore \tau (3k)\leq 3$. Now if $3\not | k$ and $k\neq 1$, then it is evident that $\tau (3k)\geq 4$ which doesn't satisfy our conditions.

$\therefore 3|k$. Since $\tau(3k)\leq 3$, the only possible values for $k$ are ${1,3}$ or $3k=3$ or $3k=9$. Which means the triplets can be $(1,3,5), (3,5,7), (5,7,9), (7,9,11)$ and $(9,11,13)$. This means that $\boxed{n=1,3,5,7,9}$

Considering both the cases together, we get $\boxed{n=1,2,3,4,5,7,9}$

Please check my approach for any mistakes. Also please provide an alternate solution less convoluted than this if possible.

THANKS


问:
I have question about a proof in probability
I have a question about this proof:

Let $X$ be a random variable and $n>0$. Then we have that $P(|X|\geq
n) \leq \frac{E|X|}{n}$

Proof: Let $Y: \begin{pmatrix} 0 & n\\ P(|X|<n) & P(|X| \geq n)
\end{pmatrix}$ be a random variable. Then we have that $0 \leq Y \leq
|X|$ e.g. $P(|X|-Y \geq 0) = 1$ and $E(|X|-Y) \geq 0$ so $E|X| \geq
EY$ , thus $EY = 0 \cdot P(|X| < n)+n \cdot P(|X| \geq n) = n \cdot
P(|X| \geq n) \leq E|X| = P(|X| \geq n) \leq \frac{E|X|}{n}$

And the part I'm having issues with is this here: Then we have that $0 \leq Y \leq
|X|$ e.g. $P(|X|-Y \geq 0) = 1$

If we let $P(X=0)=0.5$ and $P(X=1)=0.5$ so $n=0.5$ then we have that $P(Y=0)=0.5$ and $P(Y=0.5) = 0.5$

But can't we also work with probability $P(Y=0.5) \cdot P(X=0)=0.5 \cdot 0.5$ such that $X-Y<0$. Are we allowed to do this or not because $X$ and $Y$ might be dependent from each other?

答: Note that $Y=nI(|X|\geq n)$ where $I$ is the indicator function i.e. $I(|X|\geq n)$ equals $1$ when $|X|\geq n$ and is 0 otherwise. The claim is that $|X|\geq Y\geq 0$. It is clear that $Y\geq 0$. To see that $|X|\geq Y$ note that if $|X|\geq n$, then $|X|\geq Y=n$. If $|X|<n$, then $|X|\geq 0=Y$ as desired.


问:
Inequality with Gamma.
Consider the function $$f(x)=e^{\frac{x^2}{2}}\int^\infty_0e^{-t^2}t^{2n}\cos(2xt)dt,\quad x\in\Bbb R$$

I find a lot difficultes to prove $$\forall (x,n)\in\Bbb R^+\times\Bbb N:|f(x)|\leq C\Gamma(n+\frac{1}{2})$$
for some const $C$.
Thanks for any help.

My attempt:

Put $$f(x)=\frac{1}{\Gamma(n+\frac{1}{2})}e^{\frac{x}{2}}\int^\infty_0 e^{-t^2}t^{2n}\cos(2\sqrt{x}t)dt.$$

Let $x>0$, by the change of variable $t=\sqrt{n}s$ we get $$f(\frac{x}{n})=\frac{1}{\Gamma(n+\frac{1}{2})}n^{n}\sqrt{n}e^{\frac{x}{2n}}\int^\infty_0 e^{-n\big(s^2-2\ln(s)\big)}\cos(2\sqrt{x}s)ds. $$
Now we apply Laplace's method wich says:\
\textbf{Theorem} Consider the integral  $\int^b_a f(t)e^{-n g(t)}dt$ where $g$ is of class $\mathcal{C}^2$ on $[a,b[$ (here $a<b\leq+\infty)$ and $f$ is a continuous function on  $[a,b[$ such that

$f(a)\not=0$

$\forall t\in[a,b[:g'(t)>0$

$g'(a)=0$

$\forall t\in[a,b[:g''(t)>0$

Then $$\int^b_a f(t)e^{-n g(t)}dt\sim\sqrt{\frac{\pi}{2g''(a)}}\frac{e^{-ng(a)}f(a)}{\sqrt{n}},\quad n\to+\infty.$$

So we can write $$f(\frac{x}{n})=\frac{1}{\Gamma(n+\frac{1}{2})}\Big(n^{n}\sqrt{n}e^{\frac{x}{2n}}\int^0_{-1} e^{-n\big(s^2-2\ln(-s)\big)}\cos(2\sqrt{x}s)ds+n^{n}\sqrt{n}e^{\frac{x}{2n}}\int^\infty_1 e^{-n\big(s^2-2\ln(s)\big)}\cos(2\sqrt{x}s)ds\Big).$$
Put $$I_1=n^{n}\sqrt{n}e^{\frac{x}{2n}}\int^\infty_1 e^{-n\big(s^2-2\ln(s)\big)}\cos(2\sqrt{x}s)ds$$
with $g(s)=s^2-2\ln(s), \quad s\geq 1$ and $f(s)=\cos(2\sqrt{x}s)$, by the previous theorem we have
$$I_1\sim \sqrt{\frac{\pi}{8}}n^{n}\sqrt{n}\frac{e^{-n}}{\sqrt{n}}\cos(2\sqrt{x})$$
(we have $g'(s)=2(s-\frac{1}{s})>0$ and $g''(s)=2(1+\frac{1}{s^2})>0$)
Since $n!\sim(\frac{n}{e})^{n}\sqrt{2\pi n}$ we obtain $$I_1\sim\frac{n!}{2\sqrt{n}}\cos(2\sqrt{x})$$.

Using the same method we obtain $$I_2=n^{n}\sqrt{n}e^{\frac{x}{2n}}\int^0_{-1} e^{-n\big(s^2-2\ln(-s)\big)}\cos(2\sqrt{x}s)ds\sim\frac{n!}{2\sqrt{n}}\cos(2\sqrt{x})$$

Now by using the well-known result  $\frac{n!}{\sqrt{n}}\sim \Gamma(n+\frac{1}{2})$

we get $$\forall x>0,\qquad f(\frac{x}{n})\sim \cos(2\sqrt{x})$$

So, $\forall x>0,\qquad\exists n_0\in\Bbb N,\qquad n\geq n_0\Rightarrow  \Big|f(\frac{x}{n})\Big|\leq C$ for some constant $C>0$.

Therefore if we replace $x$ by $nx$ we obtain:\

$\forall x>0,\qquad\exists n_0\in\Bbb N,\qquad n\geq n_0\Rightarrow  \Big|f(x)\Big|\leq C$

And hence $$\sup_{(n,x)\in \{n\in\Bbb N; n\geq n_0\}\times\Bbb R^*_+}\Big|f(x)\Big|<+\infty. $$

It remains the case $(n,x)\in \{n\in\Bbb N; n\leq n_0\}\times\Bbb R^*_+$ and it is easy because if $n\leq n_0$ we have

$$f(x):=\Big(\frac{\Gamma(n+1)}{\Gamma(n+\frac{1}{2})}\Big)e^{\frac{-x}{2}}L^{\frac{-1}{2}}_{n}(x)\leq C' e^{\frac{-x}{2}}P(x) $$
for some constant $C'>0$ independent of $n$ and $P$ is polynomial with $\text{deg}(P)=n\leq n_0$.

Since $\lim_{x\to+\infty}e^{\frac{-x}{2}}P(x)=0$ we obtain $$\sup_{(n,x)\in \{n\in\Bbb N; n\leq n_0\}\times\Bbb R^*_+}\Big|f(x)\Big|<+\infty. $$

Finally: $$\sup_{(n,x)\in\Bbb N\times\Bbb R_+}\Big|f(x)\Big|<+\infty. $$

The case $x=0$ is easy because $\Psi(0)=c$.

答:
This is a possible start.

Let $t^2=y$.
$2t dt=dy$ so $dt=dy/(2\sqrt{y})$.

The integral becomes
$\int_0^{\infty} e^{-y}y^{n-1/2}\cos(2x\sqrt{y})dy/2$
which is bounded by in absolute value by
$\int_0^{\infty} e^{-y}y^{n-1/2}dy/2
=\Gamma(n+1/2)/2$.

To get rid of the $e^{x/2}$
we would have to consider how the cos
affects the integral.

Title: Inequality with Gamma.


答:
$$2x+\arcsin x=1$$
Write as $$\arcsin x=-2x+1$$
and look for intersecton of the graphs
$$
\begin{cases}
y=\arcsin x\\
y= -2x+1\\
\end{cases}
$$
In the picture below we see that there is an intersection $x_0\approx 0.3$

To find a more precise approximation we can use the Newton method to find a root of the function $$f(x)=2x+\arcsin x-1$$

Let's say we want a solution with $6$ exact decimal.
Define $$x_0=0.3;\;x_n=x_{n-1}-\frac{f(x_{n-1}}{f'(x_{n-1})}$$
We go on iterating until the difference $|x_n-x_{n-1}|<10^{-6}$

We get the following table
$$
\begin{array}{l|r|r}
 n & a_n & error\\
\hline
 0 & 0.3 &  \\
 1 & 0.331266 & 0.0312659 \\
 2 & 0.331208 & 0.0000577448 \\
 3 & 0.331208 & 2.1\times 10^{-10}\\
\end{array}
$$

The solution is then $x=0.331208$

$$...$$

Title: I need to find a solution for $2x+\arcsin x=1$

Body:
$$2x+\arcsin x=1$$
I didn't find way to solve this problems.
Is there a difference in solving equations between $\sin$ and $\arcsin$?
For some reason I did not find the special solutions of arc.

Is there a connection to Maclaurin or Taylor?

Thanks for your help.