问:
What is the Structure morphism of an Algebraic Space?
Definition 5.1.10 of Olsson's book "Algebraic Spaces and Stacks" gives the definition of an algebraic space as follows:

An algebraic space over a scheme $S$ is a presheaf $X:\text{Sch}/S^{op}\to \text{Set}$ such that

$X$ is a sheaf in the étale topology

$\Delta: X \to X\times_S X$ is representable by schemes

There exists an $S$-scheme $U$ and an étale surjection $U\to X$.

Later, the book states an algebraic space $X$ over $S$ is quasi-separated if the structure morphism $X\to S$ is quasi-separated. However, I don't see any "structure morphism" in the definition of an algebraic space. I understand that it should be a natural transformation $X\to \text{Hom}(-,S)$ but I don't see a natural such map.

答:
Caveat: I don't know anything about algebraic spaces. This answer just seems like the only "natural" interpretation.

$S$ (equipped with the identity map to $S$) is the terminal scheme over $S$. So $\mathrm{Hom}(-,S)$ is the terminal presheaf on $\mathrm{Sch}/S$. Why? For an $S$-scheme $Y$ with structure map $\alpha\colon Y\to S$, $\mathrm{Hom}(Y,S)$ is the singleton set $\{\alpha\}$.

So there is a unique choice of "structure map" $\eta\colon X\to \mathrm{Hom}(-,S)$. Concretely,  $\eta_Y\colon X(Y)\to \mathrm{Hom}(Y,S) = \{\alpha\}$ is the constant function with value $\alpha$.


问:
Whats the volume of parallelopiped with two of its sides eqal?
If the volume of a parallelopiped of sides $\vec{a},\vec{b},\vec{c}$ is given by $$\det\begin{pmatrix}\vec{a} & \vec{b} &\vec{c}\end{pmatrix}$$
But if two of those vectors are equal then $\det\begin{pmatrix}\vec{a} & \vec{b} &\vec{c}\end{pmatrix}=0$.

Here how can the volume of a paralleopiped with two equal sides be 0?

答: Try drawing it. A vector is an object with a length and direction. If two vectors are equal, then they have equal length and more importantly for this problem, they have the same direction. If you draw two of the three sides on top of one another, you get a parallelogram. A parallelogram has no volume.</p>

ContentLicense: CC BY-SA 4.0


问:
Probability of 4 colors sequence with repetition (8 choices of color)
Knowing that the sequence contains exactly two colors, what is the probability that one of the two colors will be repeated exactly three times?

A = The probability that one of the two colors will be repeated exactly three times

B = The probability the sequence contains exactly two colors

I know there are $8^4 = 4096$ possible sequences.

Select 2 colors among 8 $ {8 \choose 2} = 28$

choose 2 place among 4  ${4 \choose 2} = 6$

I'm not sure but I think $P(A|B) = 8/14$

Knowing that a color is repeated at least twice, what is the probability that it will be repeated exactly
three times?

A =The probability that a color is repeated at least twice

B = The probability that a color is repeated exactly three times

答:
Knowing that the sequence contains exactly two colors, what is the probability that one of the two colors will be repeated exactly three times?

It is given that there are only two colours in this particular sequence. How many possible sequences of length 4 can be built from a choice of two colours?
How many of these contain exactly three of the same colour?

Knowing that a color is repeated at least twice, what is the probability that it will be repeated exactly three times?

The sequence is of length four, and you know that two of the choices are already particular colour, call it colour $c$. So you only need to think about what the other two choices in the sequence might be. How many possibilities are there for the last two choices, and how many of them contain colour $c$ exactly once (so that the total sequence contains $c$ exactly three times)?


答:
Is $\log  x + \log y = \log (x+y)? $

No..

$\log  x + \log y = \log (x y). $

A numerical solution that finds a root is required.

Title: Solving for $\theta$ using log

Body:
0.88 = 0.95 x 0.5 x ($e ^{-0.06129 + \theta} + e^ {-0.04129 + \theta}$)

Does this equate to

log($\frac {0.88 \times 2}{0.95}$) = log($e ^{-0.06129 + \theta}) + log( e^ {-0.04129 + \theta}$)


答:
Your starting equation is $$0.88 = 0.95*0.5 * (e ^{-0.06129 + \theta} + e^ {-0.04129 + \theta})$$
which then equals $$\frac{0.88*2}{0.95} =  e ^{-0.06129 + \theta} + e^ {-0.04129 + \theta}$$.

This is the farthest we can go because $log(a+b)$ is not equal to $log(a) + log(b)$. For example, if we have $log(10+10) = log(20)$, this is not equal to $log(10) + log(10) = 1 + 1 = 2$.

We need to rewrite your solution from here on. We can factor out $e^\theta $ from this equation as doobdood noted $$\frac{0.88*2}{0.95} =  e ^{-0.06129 + \theta} + e^ {-0.04129 + \theta}$$.

to get $$\frac{0.88*2}{0.95} = e^\theta*( e ^{-0.06129} + e^ {-0.04129 })$$.
Now we can simplify this further to get
$$\frac{0.88*2}{0.95*( e ^{-0.06129} + e^ {-0.04129 })} = e^\theta$$. Now take the natural logarithm of both sides to get $\theta = \operatorname{ln}(\frac{0.88*2}{0.95*( e ^{-0.06129} + e^ {-0.04129 })})$

$\operatorname{ln}(x)$ is the logarithm with base e, aka the natural logarithm, by the way

Title: Solving for $\theta$ using log


答:
We can be a little clever here and use the fact that $e^{a+b}=e^ae^b$. Rewriting your equation yields:
$$0.88=0.95\cdot0.5\cdot\left(e^\theta e^{-0.06129}+e^\theta e^{-0.04129}\right)$$
From here, you can factor out the $e^\theta$ and solve for $\theta$.

Title: Solving for $\theta$ using log


问:
Showing $1+p$ is an element of order $p^{n-1}$ in $(\mathbb{Z}/p^n\mathbb{Z})^\times$
I'm trying to work through Dummit & Foote, but I've gotten stuck on the following question:

Let $p$ be an odd prime and let $n$ be a positive integer. Use the
  binomial theorem to show that $(1+p)^{p^{n-1}} \equiv 1\bmod{p^n}$ but
  $(1+p)^{p^{n-2}} \not \equiv 1\bmod{p^n}$. Deduce that $1+p$ is an
  element of order $p^{n-1}$ in the multiplicative group
  $(\mathbb{Z}/p^n\mathbb{Z})^\times$.

The trouble I'm having is mostly with respect to the first implication, since I'm not completely confident with what I've done so far.  I started by letting $m = p^{n-1}$ and $a_i = {m\choose i}/m$. I then ended up with
\begin{align*}
(1+p)^m = \sum_{k=0}^{m} {m\choose k}p^k &= 1 + {m \choose 1}p + {m \choose 2}p^2 + \cdots + {m \choose m-1}p^{m-1}
+ p^m\\
&= 1 + mp + a_2mp^2 + a_3mp^3 + \cdots + a_{m-1}mp^{m-1}+ p^m\\
& = 1 + p^n + a_2p^{n+1} + a_3p^{n+2} + \cdots + p^m\\
&= 1 + p^n(1+ a_2p + a_3p^2+ \cdots + p^{m-n})\\
&\equiv 1 \bmod{p^n}
\end{align*}

...but based on some numerical tests I did, the $a_i$'s aren't necessarily integers, so this doesn't work. Would someone please point out what I'm missing here?

For the second implication, I can see a clean application of Euler's theorem starts things off. To finish, I'd like to show that no power of $(1+p)$ less than $(1+p)^{p^{n-1}}$ is congruent to $1\bmod{p^n}$, but I'm not sure how knowing the first part plays into this.

Any pointers would be appreciated.

答:
I struggled with this a bit with the details, so here is my take (which piggy backs on the other solutions) just in case it helps others:

This is an answer to Dummit & Foote's "Abstract algebra", 3rd ed, Problem 21 in Section 2.3.

Also, note that ashpool's is the only other answer that explicitly uses the fact that $p$ is odd.

The key result (which Eric Auld notes above and is proved here) is that $\nu_p( \binom{p^l}{k}) = l-\nu_p(k) $ (for $k = 1,....,p^{l}$), where $\nu_p(n)$ is the largest $k$ such that $p^k \mid n$.

Another relevant result is that since $p$ is prime, $\nu_p(ab) = \nu_p(a) + \nu_p(b)$.

However, we need some estimates for $\nu_p(k)$.
We have $p^{\nu_p(k)} \le k$ (by definition) and so $\nu_p(k) \le {\log k \over \log p}$. Also, note that $p \ge 3 > e$ (the latter is used momentarily).

I claim that ${\log k \over \log p} \le k-1$ for $k \ge 1$. Note that this is true for $k=1$ and since (differentiating both sides) ${1 \over k \log p} \le 1$ for $k \ge 1$, we see that this is true for $k \ge 1$. In a similar manner, we can show that ${\log k \over \log p} \le k-2$ for $k \ge 3$.

(Presumably $n \ge 1$ for this part.)
Using the binomial theorem we get
$(1+p)^{p^{n-1}} = 1 + \sum_{k=1}^{p^{n-1}} \binom{p^{n-1}}{k} p^k$, and
$\nu_p(\binom{p^{n-1}}{k} p^k) = n-1 -\nu_p(k)+k$, and using the above, we have $\nu_p(\binom{p^{n-1}}{k} p^k) \ge  n$. Hence
$[(1+p)^{p^{n-1}}]_{p^n} = [1]_{p^n}$.

(Presumably $n \ge 2$ for this part.)
The same approach works for the next part,
$(1+p)^{p^{n-2}} = \sum_{k=0}^{2} \binom{p^{n-2}}{k} p^k + \sum_{k=3}^{p^{n-2}} \binom{p^{n-2}}{k} p^k$, and for the latter sum, note that (since $k \ge 3$) we have
$\nu_p(\binom{p^{n-2}}{k} p^k) = n-2-\nu_p(k)+k \ge n$, and so
$(1+p)^{p^{n-2}} = 1 + p^{n-1} +{p^{n-2} (p^{n-2}-1) \over 2} p^2 = 1 + p^{n-1} +{p^{n-2}-1 \over 2} p^n$. Since $p$ is odd, we see that ${p^{n-2}-1 \over 2} $ is an integer and hence $[(1+p)^{p^{n-2}}]_{p^n} = [1+p^{n-1}]_{p^n}$.

The last part of the question is to show that $1+p$ is an element of order $p^{n-1}$ in $(\mathbb{Z} / p^n \mathbb{Z})^\times$. The result follows from the above noting that $\gcd(1+p,p^n) = 1$, so $[1+p]_{p^n} \in (\mathbb{Z} / p^n \mathbb{Z})^\times$.

ContentLicense: CC BY-SA 3.0


答:
Using the definition of conditional probability, it is known that

$$\begin{split}P(repeat\_3|2\_colors)&=\frac{P(repeat\_3,2\_colors)}{P(2\_colors)}\\
&=\frac{2{4 \choose 3}{8 \choose 2}}{\left[{4\choose 1}+{4\choose 2}+{4\choose 3}\right]\cdot {8\choose 2}}\\
&=\frac{4}{7}\end{split}$$

The numerator comes from choosing 2 colors and three spots as you have done. However since the number of spots is not half the total number of spots you have to multiply by 2. The bottom is the number of ways you can get 2 colors.

Your answer for the first part looks correct. For the second part, again use the definition of conditional probability. We get

$$\begin{split}P(repeat\_thrice|repeat\_atleasttwice)&=\frac{P(repeat\_thrice,repeat\_atleasttwice)}{P(repeat\_atleasttwice)}\\
&=\frac{P(repeat\_thrice)}{P(repeat\_atleasttwice)}\\
&=\frac{\frac{{4\choose 3}}{8^4}*7}{1-\frac{{4\choose 1}7^3}{8^4}-\frac{7^4}{8^4}}\\
&=\frac{7/1024}{323/4096}\\
&=\frac{28}{323}\end{split}$$

Title: Probability of 4 colors sequence with repetition (8 choices of color)

Body:
Knowing that the sequence contains exactly two colors, what is the probability that one of the two colors will be repeated exactly three times?

A = The probability that one of the two colors will be repeated exactly three times

B = The probability the sequence contains exactly two colors

I know there are $8^4 = 4096$ possible sequences.

Select 2 colors among 8 $ {8 \choose 2} = 28$

choose 2 place among 4  ${4 \choose 2} = 6$

I'm not sure but I think $P(A|B) = 8/14$

Knowing that a color is repeated at least twice, what is the probability that it will be repeated exactly
three times?

A =The probability that a color is repeated at least twice

B = The probability that a color is repeated exactly three times


问:
How to interpret minimal sufficient statistics?
I know the definition of minimal sufficient statistics is:

A sufficient statistic T(X) is called a minimal sufficient statistic if, for any other sufficient statistic T'(X), T(X) is a function of T'(X).

I want to know how to interpret this definition from function view. T is smaller than T', if T is a function of T'. What does that mean?

答:
The notion of sufficiency is data compression.

A sufficient stat is a stat which in some sense contains all useful information about the sample $X$. (for estimating the true parameter, of course)

The definition of a sufficient stat is the following: the conditional distribution of X(sample) given T(X) does not depend on $\theta$, which is saying in more human language $T(X)$ is a summary of $X$. Immediately, we see each sufficient stat defines an equivalence relation on the sample space. Again, we're compressing data so a set of samples which yield repetitive information about the true parameter is compressed to a single equivalence class. For example, suppose we do $n$ iid Bernoulli $\theta$ trials, any set of samples such that each sample in the set has the same sum may be regarded as one sample for the purpose of estimating $\theta$.

We want maximal compression, which is a sensible way to understand the samples. All that the definition of MSufficientStat T says is if $T'(x) = T'(Y)$, then T(X) = T(Y). In order to define T as a composition of T' and a function g on the image of T', we may have to even define g in a funny way, but the point is it always exists. I guess we can think about the equivalence classes induced by a 'smaller' sufficient stat as grouping and merging the equivalence classes induced by a 'larger' sufficient stat.


问:
Find all the points where a function is not differentiable
Let $f\left( x \right)$ be a differentiable function $\mathbb{R}\to \mathbb{R}$. Find all the points where the function $\left| f\left( x \right) \right|$ is not differentiable.
Of course, I understand that $\left| f\left( x \right) \right|$ is not differentiable where its graph has a corner because of the modulus, e.g., ${{\delta }_{+}}f\left( a \right)+{{\delta }_{-}}f\left( a \right)=0$. Obviously, it is only possible where $f\left( x \right)$ changes its sign since $f\left( x \right)$ is be a differentiable function.
Is there any way to write this informal idea in a more formal manner?

答: $|f(x)|$ is not differentiable when both $f(x)=0$ and $f’(x)\neq 0$. You need the second part because of functions like $x^3$, and you don’t have to check if it changes signs because the only way for a zero of a differentiable function to not change signs is for the differential to also be 0.


问:
Struggling with the Missing Side Length with One Length
In class, we have been learning about using proportions to solve for missing lengths of triangles, however, this problem has been leaving me confused. Both of the hypotenuses of the two right triangles add up to get 35. I need to find the missing length for the triangle on the left's hypotenuse. I tried YouTube and all sorts of methods, but not one that fits for this question.

答:
If $x$ is the length of the hypothenuse you need to find, then by similarity, the other hypothenuse has length $8x/12$. You'll then get an equation you can easily solve.

\begin{align*}
\frac{2x}{3}+x&=35\\
\frac{5x}{3}&=35\\
x&=21
\end{align*}


问:
What is the ratio of matrices which have full rank to total number of matrices over a finite field.
We know that
\begin{align*}
\frac{|GL_n(\mathbb{F}_q)|}{|\mathcal{M}_N(\mathbb{F}_q)|} &= \frac{\prod \limits_{k=0}^{n-1} (q^n - q^k)}{q^{n^2}} \\
&= \prod_{k=1}^n \left( 1-q^{-k} \right) \\
&= \left(\frac{1}{q}, \frac{1}{q} \right)_n,
\end{align*}
where the last expression denotes the q-Pockhammer symbol.

How can we generalize this to $m\times n$ matrices. I mean what is the ratio of matrices which have full rank to total number of matrices.

Thanks.

答:
For an $m \times n$ matrix of rank $m$ (where $m \le n$), there are $q^n-1$ possibilities for the first row, then $q^n - q^1$ for the second, etc., or
$\prod_{j=0}^{m-1} (q^n - q^j)$ in all.


问:
Evaluating a limit using the Stolz-Cesàro theorem
I have been trying to compute the following limit-

$$\lim_{n\to \infty} \dfrac {2021(1^{2020}+2^{2020}+3^{2020}....+n^{2020}) - n^{2021}}{2021(1^{2019}+2^{2019}+3^{2019}.....+n^{2019})} =L$$

By the Stolz-Cesàro theorem, letting the numerator and denominator be $a_n$ and $b_n$ respectively yields

$$L=\lim_{n\to \infty} n+1-\frac{1}{2021}\bigg((n+1)^2-n^2\left(\frac {n}{n+1}^{}\right)^{2019}\bigg)$$

Now, I don't see how binomial expansions(perhaps some other technique?) can help here. Any help would be appreciated.

Body:
By continuing your approach you can see the limit is
$$ \begin{align} L &= \lim n+1 - \dfrac{1}{2021(n+1)^{2019}} \left( (n+1)^{2021} - (n+1 -1)^{2021} \right) \\
&= \lim n+1 - \dfrac{1}{2021(n+1)^{2019}}\left( \binom{2021}{1}(n+1)^{2020} - \binom{2021}{2} (n+1)^{2019} + \cdots \right) \\
&= \boxed{1010} \end{align}$$


问:
Simplified form of $\left(6-\frac{2}{x}\right)\div\left(9-\frac{1}{x^2}\right)$.
Tried this one a couple of times but can't seem to figure it out.

I am trying to simplify the expression:

$$\left(6-\frac{2}{x}\right)\div\left(9-\frac{1}{x^2}\right)$$

So my attempt at this is:

$$=\bigg(\dfrac{6x}{x}-\dfrac{2}{x}\bigg)\div\bigg(\dfrac{9x^2}{x^2}-\dfrac{1}{x^2}\bigg)$$

$$=\bigg(\dfrac{6x-2}{x}\bigg)\div\bigg(\dfrac{9x^2-1}{x^2}\bigg)$$

$$=\dfrac{6x-2}{x}\cdot\dfrac{x^2}{9x^2-1}$$

$$=\dfrac{(6x-2)(x^2)}{(x)(9x^2-1)}$$

$$=\dfrac{6x^3-2x^2}{9x^3-x}$$

This is the part that I get stuck at. I can't decide what to factor out:
$$=\dfrac{x(6x^3-2x^2)}{x(9x^3-x)}$$

$$=\dfrac{(6x^2-2x)}{(9x^2-1)}$$

Edit, missed a difference of squares:

$$=\dfrac{2x^2(6x^3-2x^2)}{x(9x^3-x)}$$

$$=\dfrac{2x^2(3x-1)}{x(3x-1)(3x+1)}$$

Giving a final answer of:
$$=\boxed{\dfrac{2x}{3x+1}}$$

答:
$$\dfrac{(6x^3-2x^2)}{x(9x^3-x)} = \dfrac{2x^2(3x - 1)}{x(9x^2 - 1)} $$

Cancel common factor of $x$ in numerator and denominator gives us:

$$\dfrac{2x^2(3x - 1)}{x(9x^2 - 1)} = \frac{2x(3x-1)}{\left[9x^2 - 1\right]}$$

Now we have a difference of squares in the denominator, and can factor it:

$$\frac{2x(3x-1)}{\color{blue}{\bf \left[9x^2 - 1\right]}}= \frac{2x(3x-1)}{\color{blue}{\bf \left[(3x)^2 - 1\right]}} = \frac{2x(3x - 1)}{(3x - 1)(3x + 1)}$$

Now, cancel like terms: Note that  $\color{blue}{(3x - 1)}$ is a factor in the numerator and in the denominator, so we proceed to cancel:

$$ \frac{2x\color{blue}{\bf (3x - 1)}}{\color{blue}{\bf(3x - 1)}(3x + 1)} = \frac{ 2x}{3x + 1}$$


答:
I don't think it will be too hard to write down the numerator and denominator, because the matrices with full rank will have $n$ linearly independent columns, so the first column can be any nonzero vector, there are $q^2$ linear combinations of the first two columns, etc... and that gives $(q^m-1)q^m-q)\dots(q^m-q^{n-1})$.

Meanwhile the denominator is just $q^{mn}$.

Title: What is the ratio of matrices which have full rank to total number of matrices over a finite field.

Body:
We know that
\begin{align*}
\frac{|GL_n(\mathbb{F}_q)|}{|\mathcal{M}_N(\mathbb{F}_q)|} &= \frac{\prod \limits_{k=0}^{n-1} (q^n - q^k)}{q^{n^2}} \\
&= \prod_{k=1}^n \left( 1-q^{-k} \right) \\
&= \left(\frac{1}{q}, \frac{1}{q} \right)_n,
\end{align*}
where the last expression denotes the q-Pockhammer symbol.

How can we generalize this to $m\times n$ matrices. I mean what is the ratio of matrices which have full rank to total number of matrices.

Thanks.


答:
Assume colors are C-1, C-2, ..., C-8.

The easy way:

Since you are allowed to assume that exactly two colors occur in the sequence, and since the probability of exactly three out of four is unaffected by which colors are involved, assume, Without Loss of Generality that only colors C-1, C-2 are involved.

There are $2^4$ sequences possible, minus the two that only involve one color.  So the denominator will be $14$.

Assume that there are 3 of C-1, 1 of C-2.  There are 4 possible positions for C-2.  Therefore, there are 4 ways this can happen.  Multiply by 2 re 3 of C-2, 1 of C-1.

Final Answer: $\frac{8}{14}.$

The hard way : involves Bayes Theorem.

$D =$ denominator

$N =$ numerator.

First, compute the probability of there being exactly two colors in the sequence.

This will be

$$D = \frac{\binom{8}{2} \times 14}{8^4}.$$

Then compute the probability of there being exactly two colors in the sequence and one of the two colors occurs exactly 3 times.

$$N = \frac{\binom{8}{2} \times 8}{8^4}.$$

Then $$\frac{N}{D} = \frac{8}{14}.$$


问:
Distinct Mersenne numbers are relatively prime specific proof verification
As the title states, I'm supposed to prove for distinct primes $p_1,p_2 >2$ the primes dividing $2^{p_1}-1$ and $2^{p_2}-1$ are distinct.

The Wikipedia page on Theorems about Mersenne Numbers states that #6: if $m$ and $n$ are natural numbers then $m$ and $n$ are coprime if and only if $2^m-1$ and $2^n-1$ are coprime, the only one without proof. I believe I have to prove the forward $(\Longrightarrow)$ direction.

edit: this was my own work for a homework question, and I was mainly wondering if the proof was okay. there is no proof directly provided on Wikipedia, and I didn’t find the links to the proof the clearest either. this was the clearest proof to me at the time and wanted someone else’s feedback because I couldn’t believe my proof was right.

My attempt at a proof (I use $a,b$ instead of $p_1,p_2$ or $m,n$):

Suppose $(a,b)=1$ and $(2^a - 1, 2^b -1) = d$.

Since $(a,b)=1$ there exist $x,y \in \mathbb{Z}$ s.t. $ax+by=1$.

Since $a,b > 2$, $x$ and $y$ cannot both be greater than zero, otherwise $ax+by > 2+2 >1$. Neither can $x$ or $y$ be equal to zero, otherwise $x$ or $y$ would $\not\in\mathbb{Z}$, or $ax+by=0\not=1$. Neither can $x$ and $y$ be less than zero, otherwise then $ax+by <0<1$.

Thus one of $x$ or $y$ must be positive, while the other negative.

Suppose $ax>0$ then $by<0$, then $d|2^{ax}-1$ and $d|2^{-by}-1$ since

$$2^{ax}-1 = (2^a -1)((2^a)^{x-1}+(2^a)^{x-2}+\cdots+1)$$

$$2^{-by}-1 = (2^b -1)((2^b)^{-y-1}+(2^b)^{-y-2}+\cdots+1)$$

and $d|2^a -1$ and $d|2^b -1$ by definition. Then $2^{ax} - 1 = d\alpha$ and  $2^{-by} -1 = d\beta$ and then $2^{ax}-2^{-by} = 2^{-by}(2^{ax+by}-1)=2^{-by}(2^1-1)=2^{-by}  = d(\alpha -\beta)$ thus $d|2^{-by}$ and $d$ must be a power of $2$. But $d|2^a-1$ where $2^a-1$ is odd, and the only odd power of $2$ is $2^0=1$, hence $d=1$. If $(a,b)=1$ then $(2^a -1 ,2^b-1)=d=1$.

Is my proof rigorous or am I missing something?

答:
Your proof is fine, though you might also need to mention the case $ax < 0$ and $by > 0$ (though immediate).

By Wikipedia standards, most theorems are accompanied by a proof, or if not possible, a reference to the proof would be provided. In this case, the statement of this theorem is accompanied with a "[20]", which links to an archived version of Will Edgington's Mersenne Page. The various Lemmas in the section Mersenne Number Proofs supplies a proof to the claim.

Discussions on MSE that are also useful include:

Distinct Mersenne numbers are coprime

Prove that $\gcd(a^n - 1, a^m - 1) = a^{\gcd(n, m)} - 1$


问:
The Componentwise Convergence Criterion

Can someone explain to me what does it mean for a sequence of points to converge componentwise. Provide an example thanks

答:
I assume that $p_i$ is the projection onto the $i^{\text{th}}$ coordinate of $\mathbb{R}^n$.

For example, in dimension $n=2$, the sequence whose $k^{\text{th}}$ element is the point $\mathbf{u}_k = (1/k^2, (1+k)/k)$ converges to the point $(0, 1) \in \mathbb{R}^2$, since the first component converges to
$$\lim_{k\to \infty} p_1(\mathbf u_k) = \lim_{k \to \infty} \frac{1}{k^2} = 0$$
and the second component converges to
$$\lim_{k\to \infty} p_2(\mathbf u_k) = \lim_{k \to \infty} \frac{1+k}{k} = 1$$.

Title: The Componentwise Convergence Criterion


问:
Solving constrained least squares problem with modified linear constraint
could anybody give me some suggestions on solving this least-squares problem?

Find $x\in R^{n}$ that minimize $||Ax||_2$ subject to $||Bx||_2=1$, with $A\in R^{n×n} $ always full rank, $B\in R^{n×n}$ not necessary full rank.

Thanks for your help

答:
The task is to minimize $f(x) = x^\top A^\top Ax$ s.t. $g(x) = x^\top B^\top Bx - 1 = 0$. Lagrange:
$$
A^\top\!Ax = \lambda\cdot B^\top\!Bx.
$$
$\lambda = 0$ would imply that $x=0$, which is not feasible. So, $\lambda\neq 0$. Hence, $(A^\top A)^{-1}B^\top Bx = \lambda^{-1}x$ so that $\lambda^{-1}$ must be an eigenvalue of $(A^\top A)^{-1}B^\top B$ and $x$ a corresponding eigenvector. From the constraint you also get that $\|Ax\|^2 = \lambda$. So, $\lambda>0$ and $\sqrt{\lambda}$ is the optimal value. Therefore, the minimizers of this problem are the eigenvectors $x$ of $(A^\top A)^{-1}B^\top B$ with $\|Bx\|=1$ corresponding to the largest eigenvalue of $(A^\top A)^{-1}B^\top B$.

By the way, this is not really a least squares problem.


问:
Solve the differential equation ${y}''- 2{y}'- 3y= e^{3x}\left ( x^{2}+ 1 \right )$

Solve the differential equation
$${y}''- 2{y}'- 3y= e^{3x}\left ( x^{2}+ 1 \right )$$

by jp.Wolfram|Alpha that is $y= c_{1}e^{-x}+ c_{2}e^{3x}+ \frac{1}{12}e^{3x}x^{3}- \frac{1}{16}e^{3x}x^{2}+ \frac{9}{32}e^{3x}x.$ Can you explain to me why ? I realise that the result can be seperated into 2 parts actually the 1st one $c_{1}e^{-x}+ c_{2}e^{3x}$ received by solving the polynomial $a^{2}- 2a- 3$, I need to the help.

答:
Rewrite the equation as $$(e^{-3x} y''-3e^{-3x}y')+(e^{-3x}y'-3e^{-3x}y)=x^2+1$$  Now integrating $$ e^{-3x} y'+e^{-3x}y=\frac{x^3}{3}+x+C$$ $$\iff e^xy'+e^xy=e^{4x}(\frac{x^3}{3}+x+C)$$  Now again integrate both sides and use Integration by parts in RHS..

You may use the fact that $e^xy'+e^xy=(e^xy)'$


问:
Binomial Coefficient $\sum_{k=0}^n {2n \choose 2k}{2k \choose k}{2n-2k \choose n-k} $
I am trying to find the sum
$$\sum_{k=0}^n {2n \choose 2k}{2k \choose k}{2n-2k \choose n-k} $$
I know the closed-form answer and I could derive it based on the form of the answer in terms of the coefficient of some expansion but I am curious to know how to solve this problem without knowing the answer.

答:
Your summation can be written:

$$S=\sum_{k=0}^n \dfrac{(2n)!}{(2n-2k)!(2k)!}\dfrac{(2k)!}{(k)!(k)!}\dfrac{(2n-2k)!}{(n-k)!(n-k)!}$$

By cancellation and introduction of $(n)!(n)!$:

$$S=\sum_{k=0}^n \dfrac{(2n)!}{(n)!(n)!}\left(\dfrac{(n)!}{(k)!(n-k)!}\right)^2$$

$$S=\binom{2n}{n}\sum_{k=0}^n \binom{n}{k}^2$$

Finally, using using a classical result:

$$S=\binom{2n}{n}^2$$

Edit: I just found the very same question here with, in particular, an interesting combinatorial answer.


问:
Gaussian distribution is isotropic?
I was in a seminar today and the lecturer said that the gaussian distribution is isotropic. What does it mean for a distribution to be isotropic? It seems like he is using this property for the pseudo-independence of vectors where each entry is sampled from the normal distribution.</p>

答:
Thanks to Tomoiagă, valuable learning opportunity. Explanation for psd in his answer.

w.r.t material in course Probabilistic Deep Learning with TensorFlow 2 in coursera.

the definition for positive semi-definite:

A symmetric matrix $M \in \mathbb{R}^{d\times d}$ is positive
semi-definite if it satisfies $b^TMb \ge 0$ for all nonzero
$b\in\mathbb{R}^d$. If, in addition, we have $b^TMb = 0 \Rightarrow
> b=0$ then $M$ is positive definite.

In one word: the valid covariance matrix should be symmetry and positive (semi-)definite. However, how to check the $\Sigma$ satisfied with the requirement?

Here comes The Cholesky decomposition

For every real-valued symmetric positive-definite matrix $M$, there is a unique lower-diagonal matrix $L$ that has  positive diagonal entries for which

\begin{equation}
     LL^T = M
 \end{equation}
This is called the Cholesky decomposition of $M$

Let's build some codes.

Given a psd matrix $ \Sigma = \begin{bmatrix} 10 & 5 \\ 5 & 10 \end{bmatrix}$ and a non-psd $ \Sigma = \begin{bmatrix} 10 & 11 \\ 11 & 10 \end{bmatrix}$ as the covariance matrix

```<code>sigma = [[10., 5.],[5., 10.]]
np.linalg.cholesky(sigma)
</code>```

Output:

```<code><tf.Tensor: shape=(2, 2), dtype=float32, numpy= array([[3.1622777, 0.  ],
       [1.5811388, 2.738613 ]], dtype=float32)>
</code>```
```<code>bad_sigma = [[10., 11.], [11., 10.]]
try:
    scale_tril = tf.linalg.cholesky(bad_sigma)
except Exception as e:
    print(e)
</code>```

Output:

```<code>Cholesky decomposition was not successful. The input might not be valid.
</code>```

For convenience, a lower-triangular matrix is easier to create.

Last, the demo with isotropic Gaussian and non-isotropic ones.

```<code>import seaborn as sns
import matplotlib.pyplot as plt
%matplotlib inline

## isotropic normal
sigma = [[1., 0.],[0., 1.]]
lower_triangular = np.linalg.cholesky(sigma)
print(lower_triangular)
sigma = np.matmul(lower_triangular, np.transpose(lower_triangular))
##

bivariate_normal = np.random.multivariate_normal([0, 0], sigma, size=10000, check_valid='warn')

x1 = bivariate_normal[:, 0]
x2 = bivariate_normal[:, 1]
sns.jointplot(x1, x2, kind='kde', space=0, color='b')
</code>```

isotropic-normal

```<code># #non-isotropic normal
# sigma = [[1. , 0.6], [0.6, 1.]]
</code>```

non-isotropic-normal

Title: Gaussian distribution is isotropic?

Body:
Thanks to Tomoiagă, valuable learning opportunity. Explanation for psd in his answer.

w.r.t material in course Probabilistic Deep Learning with TensorFlow 2 in coursera.

the definition for positive semi-definite:

A symmetric matrix $M \in \mathbb{R}^{d\times d}$ is positive
semi-definite if it satisfies $b^TMb \ge 0$ for all nonzero
$b\in\mathbb{R}^d$. If, in addition, we have $b^TMb = 0 \Rightarrow
> b=0$ then $M$ is positive definite.

In one word: the valid covariance matrix should be symmetry and positive (semi-)definite. However, how to check the $\Sigma$ satisfied with the requirement?

Here comes The Cholesky decomposition

For every real-valued symmetric positive-definite matrix $M$, there is a unique lower-diagonal matrix $L$ that has  positive diagonal entries for which

\begin{equation}
     LL^T = M
 \end{equation}
This is called the Cholesky decomposition of $M$

Let's build some codes.

Given a psd matrix $ \Sigma = \begin{bmatrix} 10 & 5 \\ 5 & 10 \end{bmatrix}$ and a non-psd $ \Sigma = \begin{bmatrix} 10 & 11 \\ 11 & 10 \end{bmatrix}$ as the covariance matrix

sigma = [[10., 5.],[5., 10.]]
np.linalg.cholesky(sigma)

Output:

<tf.Tensor: shape=(2, 2), dtype=float32, numpy= array([[3.1622777, 0.  ],
       [1.5811388, 2.738613 ]], dtype=float32)>

bad_sigma = [[10., 11.], [11., 10.]]
try:
    scale_tril = tf.linalg.cholesky(bad_sigma)
except Exception as e:
    print(e)

Output:

Cholesky decomposition was not successful. The input might not be valid.

For convenience, a lower-triangular matrix is easier to create.

Last, the demo with isotropic Gaussian and non-isotropic ones.

import seaborn as sns
import matplotlib.pyplot as plt
%matplotlib inline

## isotropic normal
sigma = [[1., 0.],[0., 1.]]
lower_triangular = np.linalg.cholesky(sigma)
print(lower_triangular)
sigma = np.matmul(lower_triangular, np.transpose(lower_triangular))
##

bivariate_normal = np.random.multivariate_normal([0, 0], sigma, size=10000, check_valid='warn')

x1 = bivariate_normal[:, 0]
x2 = bivariate_normal[:, 1]
sns.jointplot(x1, x2, kind='kde', space=0, color='b')

isotropic-normal

# #non-isotropic normal
# sigma = [[1. , 0.6], [0.6, 1.]]

non-isotropic-normal


问:
If a matrix has multiple left inverse, then solution is not unique?
I just want to clarify the statement " If A has a left inverse, Ax=b has a unique solution."
This is saying that A only has at most one left inverse correct? Because if A has more than one left inverse then x would not be unique since if Y1 and Y2 and left inverse of A, x=Y1b and Y2b are both solutions.</p>

答:
If $A$ has a left inverse, then the linear transformation defined by $A$ is injective: because if $B$ is the left inverse, then from $Ax=Ay$ we can left multiply by $B$ to get $x=BAx=BAy = y$.

That means that $Ax=b$ has at most one solution; because if $x_0$ and $x_1$ are both solutions, then $b=Ax_0=Ax_1$, hence $x_0=x_1$.

However, the system $Ax=b$ could have no solutions. For example, the matrix
$$A=\left(\begin{array}{cc}
1 & 0\\
0 & 1\\
0 & 0
\end{array}\right)$$
has left inverses (infinitely many, in fact): namely, all matrices of the form
$$B=\left(\begin{array}{cc}
1 & 0 & a\\
0 & 1 & b
\end{array}\right)$$
with $a,b$ arbitrary are left inverses of $A$.

For any $b$ of the form $(r,s,0)^T$, the system $Ax=b$ has exactly one solution; for $b$ of the form $(r,s,t)^T$ with $t\neq 0$, the system $Ax=b$ has no solutions.

You do not need the matrix to have a unique left inverse. In fact, a matrix has a unique left inverse if and only if it is a square matrix and has a two-sided inverse. Matrices that have more rows than columns may have left inverses (if they are of maximum rank) and when they do, they have multiple left inverses.

In summary: if $A$ is square, and it has a left inverse, then it is invertible, it has a unique left inverse, and $Ax=b$ has a unique solution for all $b$. If $A$ has more columns than row, then it never has a left inverse. If $A$ has more rows than columns but is not of maximum rank, then it has no left inverse. And if it has more rows than columns and is of maximum rank, then it has multiple left inverses, and the system $Ax=b$ has either no solutions or exactly one solution, and both situations will occur (since $A$ will not be surjective, so some choices of $b$ will yield a system with no solutions).


问:
Show that $\prod_{k=0}^n |x-k| \le (n-1)!/2$ for $1 \le x \le n-1$.

Let $n \ge 3$, $x \in \Bbb R$ such that $1 \le x \le n-1$. Show that $\prod_{k=0}^n |x-k| \le (n-1)!/2$.

For $n=3$, $1 \le x \le 2$ we want to show $|x(x-1)(x-2)(x-3)|=x(x-1)(2-x)(3-x) \le 1$.
We see that some large bounds are easy to guess:
$$
x(x-1)(2-x)(3-x) \le 2 \times 1 \times 1 \times 2=4
$$
but not precise enough. Another try:
$$
x(x-1)(2-x)(3-x) \le x(3-x) \le 9/4
$$
the last inequality is from the study of $f(x)=x(3-x)$.
Same problem.
For $n=3$ and $1 \le x \le 2$ let $0 \le y=x-1 \le 1$.
We get using AM-GM
\begin{align}
|x(x-1)(x-2)(x-3)| &= |(y+1)(y)(y-1)(y-2)|\\
&=y(1-y^2)(2-y)\\
&\le (\frac{3-y^2}{3})^3\\
&\le  1
\end{align}

For the general case I tried to use the same approach:
let $1 \le x \le n-1$ ie. $0 \le y = (x-1)/(n-2) \le 1$. We have
\begin{align}
\prod_{k=0}^n |x-k|
&=
\prod_{k=0}^n |(n-1)y +1-k|\\
&=
\prod_{k=0}^l |(n-1)y +1-k| \prod_{k=l+1}^n |(n-1)y +1-k|\\
&=
\prod_{k=0}^l ((n-1)y +1-k) \prod_{k=l+1}^n (k-1-(n-1)y)\\
\end{align}
where $l = \lfloor (n-1)y+1\rfloor$. Applying AM-GM doesn't really help here. Suppose $y<1$ ie. $l<n$, we see that
$$
\prod_{k=0}^l ((n-1)y +1-k) \le \prod_{k=0}^l (n-k)=\frac{n!}{(n-l-1)!}
$$
and
$$
\prod_{k=l+1}^n (k-1-(n-1)y) \le \prod_{k=l}^{n-1} (k) = \frac{(n-1)!}{(l-1)!}
$$
So
$$
\prod_{k=0}^n |x-k|\le
\frac{(n-1)!}{(l-1)!}\frac{n!}{(n-l-1)!}
$$
The bound is too big...
I also tried this approach: let $f(x)= \prod_{k=0}^n (x-k)$, $f'(x)=f(x)(\sum_{k=0}^n \frac{1}{x-k})$.
I'm trying to find informations about $x$ such that $f'(x)=0$. Let $x \in \Bbb R - [|0,n|]$. So $\sum_{k=0}^n \frac{1}{x-k}=0$. But I'm stuck here.

My questions are:

Does someone have a hint or a proof?

Can someone give me some advice on how to handle such problems during an exam?

Thanks :)

答:
Consider the function $$f(x)=\prod_{k=0}^n |x-k|, \qquad x\in[1, n-1].$$ This is just a absolute value of a polynomial on a bounded closed interval, and therefore by extreme value theorem it must have a maximum  and a minimum at least once. Clearly the minimum value of $f$ is zero and achieve at each integer $1, 2, \cdots, n-1.$ Also, due to the absolute value $f(x)=f(n-x)$ and hence the graph of $f$ is symmetric about the vertical line $x=n/2.$

When $k\lt x\lt k+1$ for some $k\in\mathbb{N}$ in its domain, then by AM-GM inequality we have $$|x-k||x-(k+1)|\le\dfrac{1}{4}\left((x-k)+((k+1)-x)\right)^2=\dfrac14.$$ Or you can consider the minimum value of convex quadratic function $y=(x-k)(x-k-1)$ to get the same bound. Furthermore, we can find bounds for remaining factors by looking at distances. Hence
$$1\lt|x-(k-1)|\lt2\lt|x-(k-2)|\lt3\lt\cdots\lt k \lt|x|\lt k+1$$ and
$$1\lt|x-(k+2)|\lt2\lt|x-(k+3)|\lt3\lt\cdots\lt n-k-1 \lt|x-n|\lt n-k.$$

We can use these information to compute an upper bound for $f$ at any interval as follows.

$k=1$: $\quad f(x)\lt2.\dfrac{1}{4}. 2.3.\cdots(n-1)=\dfrac{1}{2}(n-1)!$

$k=2$: $\quad f(x)\lt3.2. \dfrac{1}{4}.2. \cdots(n-2)=\dfrac{3}{2}(n-2)!$

$k=3$: $\quad f(x)\lt4.3.2. \dfrac{1}{4}.2. \cdots(n-3)=6(n-3)!.$

Similarly

$k=r$: $\quad f(x)\lt\dfrac{1}{4}(r+1)!(n-r)!$

$k=\lfloor n/2 \rfloor$: $\quad f(x)\lt\dfrac{1}{4}(\lfloor n/2 \rfloor+1)!(n-\lfloor n/2 \rfloor)!.$

It is pretty easy to see that the sequence $g(r)=\dfrac{1}{4}(r+1)!(n-r)!$ on $r=1, 2, \cdots, \lfloor n/2 \rfloor$ is decreasing since $$\dfrac{g(r+1)}{g(r)}=\dfrac{r+2}{n-r}\le1.$$

Hence the maximum strict upper bound of $f$ occurs on $(1,2)$ and $(n-2, n-1)$ and it is exactly $$g(1)=\dfrac{1}{2}(n-1)!.$$


问:
Definition of topological group via neighborhood base -- weird difference condition
It's been over 2 years since I've seriously done point-set topology, so my apologies if this is simple. I'm working out of Liu's Algebraic Geometry and Arithmetic Curves. My question concerns a definition on page 16.

Liu defines a topological group in the standard way. Let us only consider abelian groups, and define a topological group to be a group with a topological structure so that $(x,y)\rightarrow x-y$ form $G\times G$ to $G$ is continuous.

It is not hard to see that if $U$ is an open set and $x$ is an element, then $U\rightarrow x+U$ is a homeomorphism, so we can specify the topological on a topological group $G$ by giving a neighborhood base at $0$.

Let $S$ be such a neighborhood base at $0$ of some group $G$, and let $V$ be any element in $S$. Liu claims there must exist elements $V_1,V_2\in S$, so that $V$ contains $V_1-V_2$ so that subtraction is continuous, and that the existence of such elements will make subtraction continuous. If we consider the inverse of $V$, then we would have $V_1\times V_2 \subset f^{-1}(V)$, where $f$ is subtraction. But this only gives an open neighborhood of $0$ in the inverse, and we need an open neighborhood of every point. Certainly the existence of $V_1$ and $V_2$ is necessary for subtraction to be continuous in this topology, but why is it sufficient?

Here's an attempt, using the fact that the topology is homogeneous. Let $U$ be any open set. Letting $f$ be subtraction, we must show $f^{-1}(U)$ is open. That is, given any $(a,b)\in f^{-1}(U)$, we must show an open neighborhood of $(a,b)$ is contained in $f^{-1}(U)$. Suppose $f(a,b)=c\in U$. Consider $c^{-1}U$, a neighborhood of $0$. By hypothesis, there exist $V_1,V_2$, both neighborhoods of $0$, so that $f(V_1\times V_2)\subset c^{-1}U$. Then $(V_1+a,V_2+b)$ gives the desired neighborhood of $(a,b)$ in $f^{-1}(U)$, since $f((V_1+a,V_2+b))\subset U$.

答:
I'm confused by your "presumably ..." Use continuity of subtraction, taking the preimage $U$ of $V$, noting that $(0,0)\in U$, and so we get $V_1\times V_2\subset U$. OK, you edited, so now I have to inquire: $S$ is the basis at $0$, so what are you worrying about?

OK, given the latest version, given $(a,b)\in G\times G$, let $V+(a-b)$ be a nbhd of $a-b$ and check that subtraction maps $(a+V_1)\times (b+V_2)$ into it.


问:
I roll a fair die 4 times. Let X be the number of different outcomes that I see. Find $\mathbb{E}[X]$
I roll a fair die 4 times. Let X be the number of different outcomes that I see. Find $\mathbb{E}[X]$.

My attempt:

I know that I can write X as a sum of indicator random variables and then I can use the fact that $\mathbb{E}[1_A]=\mathbb{P}(A)$. Thus,

\begin{equation}
I_{A_1}=
\begin{cases}
1 & \text{if only one kind}\\
0 & \text{otherwise}
\end{cases}
\end{equation}
\begin{equation}
I_{A_2}=
\begin{cases}
1 & \text{if two kinds}\\
0 & \text{otherwise}
\end{cases}
\end{equation}
\begin{equation}
I_{A_3}=
\begin{cases}
1 & \text{if three kinds}\\
0 & \text{otherwise}
\end{cases}
\end{equation}
\begin{equation}
I_{A_4}=
\begin{cases}
1 & \text{all different}\\
0 & \text{otherwise}
\end{cases}
\end{equation}

Then the probabilities of event A happening for each is
$$\mathbb{P}(I_{A_1})=\frac{6\cdot1\cdot1\cdot1}{6^4}$$
$$\mathbb{P}(I_{A_2})=\frac{6\cdot5\cdot2\cdot2}{6^4}$$
$$\mathbb{P}(I_{A_3})=\frac{6\cdot5\cdot4\cdot3}{6^4}$$
$$\mathbb{P}(I_{A_4})=\frac{6\cdot5\cdot4\cdot3}{6^4}$$

The sum of these should be my desired expectation.

My question is whether or not I have found the probabilities correctly. This is my thought process for how I counted the number of choices for the numerator, using $I_{A_2}$ as an example:

There are 6 choices for choosing the first number, then we don't want to get that number again so then there are 5 choices. After that, we only want to get the first or second number again, thus there are 2 choices for the third roll and 2 choices for the fourth roll. Hence, $6\cdot5\cdot2\cdot2$.

Is this the correct way of thinking about it? Thanks

答:
This is a very brute-force-y problem, or at least I have a very brute-force-y solution. If anyone has a more elegant solution, please post it.

I want to start by saying that the sum of your probabilities should equal 1, as they comprise all the possible outcomes, yet they don't. $\mathbb P\left(A_1\right)$ and $\mathbb P\left(A_4\right)$ are correct, but the problem lies in the other two. The issue is that you fail to consider the order in which each number is rolled.

Let's start with $\mathbb P\left(A_3\right)$. Of course, the first roll can take on any of the 6 possible values. However, that does not restrict the second roll to 5 different values. If the first two rolls are the same, it's still possible to end up with 3 distinct values after 4 rolls. The same applies to the third roll. So, we should have:
$$\mathbb P\left(A_3\right)=\frac{1}{6^4}\left(6\cdot1\cdot5\cdot4+6\cdot5\cdot2\cdot4+6\cdot5\cdot4\cdot3\right)$$

There's a similar problem for $\mathbb P\left(A_2\right)$, but this one's a little more involved. We can either have 2 pairs of identical rolls or a set of 3 identical rolls and a single unique roll.
$$\mathbb P\left(A_2\right)=\frac 1{6^4}(\overbrace{6\cdot5\cdot2\cdot1+6\cdot1\cdot5\cdot1+6\cdot1\cdot1\cdot5}^\textrm{3 identical rolls}+\overbrace{6\cdot5\cdot2\cdot1+6\cdot1\cdot5\cdot1}^\textrm{2 pairs})$$

These new probabilities line all add up to 1 now, and they line up with the numerical value found with Python.


问:
¿Can a subset $\overline{S}$ of the accumulation points of a set S be greater than S itself? in $R^{n}$
Let $S \subseteq \mathbb{R}^{n}$ and $\overline{S}$ the set of all accumulation points of $S$

Then if $ S \subseteq \overline{S}$.

can $\overline{S}$ be greater than $S$?

答: Yes.  Let $S$ be the open ball of radius $354.8 \sqrt {695} \pi^3$ around the origin.  Then $S \subset \overline S$ ($\overline S$ is the closed ball of the same radius) and the inclusion is strict.


问:
Lagrange multiplier question with unit circle constraint
I'm having trouble with the following question:

Show $|ax + by| \leq (a^{2} + b^{2})^{1/2}$ if $x^{2} + y^{2} = 1$ by finding the maximum and minimum values of $f(x, y) = ax + by$ on the unit circle.

I'm learning multivariable calculus on my own, and I came across this question. I've been struggling for a couple of hours, and I'd really appreciate some help. I approach this problem using Lagrange Multipliers.

Here's what I've tried so far:

Let $f(x, y) = ax + by$ and let $g(x, y) = x^2 + y^2 - 1$. Then, by lagrange multiplier method, we have

$$a = \lambda(2x) $$

and

$$b = \lambda(2y),$$

from which we get

$$\lambda = \frac{a}{2x} = \frac{b}{2y}.$$

Then, I divided the two equations and I found $x = \frac{ay}{b}$, which I plugged into the constraint equation, but I got nowhere. Can someone please help me with this problem?

答:
You were actually well on your way to completing the proof.  Inserting your result into the constraint equation yields

$$ x \ = \frac{ay}{b} \ \ \rightarrow \ \ \left( \frac{ay}{b} \right)^2 \ + \ y^2 \ \ = \ \ 1 \ \ \Rightarrow \ \ \left( \frac{a^2 \ + \ b^2}{b^2} \right)  y^2 \ \ = \ \ 1   $$
$$   y^2 \ \ = \ \ \frac{b^2}{a^2 \ + \ b^2} \ \ , \ \  x^2 \ \ = \ \ \left( \frac{a}{b} \right)^2 · \frac{b^2}{a^2 \ + \ b^2} \ \ = \ \ \frac{a^2}{a^2 \ + \ b^2} $$
$$ \Rightarrow \ \ x \  =  \ \pm \frac{|a|}{\sqrt{a^2 \ + \ b^2}} \ \ , \ \ y \  =  \ \pm \frac{|b|}{\sqrt{a^2 \ + \ b^2}} \ \ , $$

as José Carlos Santos shows and which is presumably what you found.

The next question that arises is:  what are we to do with this result?  What you have found represents coordinates of tangent points to the unit circle for   lines represented by the function $ \ ax \ + \ by \ $ .  If we take, for instance, $ \ a > 0 \ , b > 0 \ $ (the other possible cases work similarly), then we have two lines with slope $ \ -\frac{a}{b} \ < \ 0 \ $ which contact the circle at $ \left( \frac{a}{\sqrt{a^2 \ + \ b^2}} \ , \ \frac{b}{\sqrt{a^2 \ + \ b^2}} \right) \ $ and $ \left( -\frac{a}{\sqrt{a^2 \ + \ b^2}} \ , \ -\frac{b}{\sqrt{a^2 \ + \ b^2}} \right) \ \ . $  These points lie on the lines

$$ ax \ + \ by \ \ = \ \ a·\frac{a}{\sqrt{a^2 \ + \ b^2}} \ + \ b·\frac{b}{\sqrt{a^2 \ + \ b^2}} \ \ = \ \ \frac{a^2 \ + \ b^2}{\sqrt{a^2 \ + \ b^2}} \ \ = \ \ \sqrt{a^2 \ + \ b^2} $$
and $ \  \ ax \ + \ by \ \ =  \ \ -\sqrt{a^2 \ + \ b^2} \ \ . $ [the geometric interpretation of robjohn's result]

All of the parallel lines passing through the circle are between these two lines, so we can conclude that for all points on the circle,

$$ -\sqrt{a^2 \ + \ b^2} \ \ \le \ \ ax \ + \ by \ \ \le \ \ \sqrt{a^2 \ + \ b^2} \ \ \ \text{or} \ \ \ | \ ax \ + \ by \ | \ \ \le \ \ \sqrt{a^2 \ + \ b^2} \ \ . $$


问:
Probability of 4 colors sequence with repetition (8 choices of color)
Knowing that the sequence contains exactly two colors, what is the probability that one of the two colors will be repeated exactly three times?

A = The probability that one of the two colors will be repeated exactly three times

B = The probability the sequence contains exactly two colors

I know there are $8^4 = 4096$ possible sequences.

Select 2 colors among 8 $ {8 \choose 2} = 28$

choose 2 place among 4  ${4 \choose 2} = 6$

I'm not sure but I think $P(A|B) = 8/14$

Knowing that a color is repeated at least twice, what is the probability that it will be repeated exactly
three times?

A =The probability that a color is repeated at least twice

B = The probability that a color is repeated exactly three times

答:
Your first answer is correct. It is not important which two colors are present and you can start from the point that the sequence of four has two colors.

Number of arrangements where two chosen colors are present $ = \displaystyle {2^4 - 2} = 14$. We basically take out two arrangements where only one of the two colors is present.

Arrangements where one of them is present exactly $1$ time (and so other $3$ times) $ = \displaystyle 2 \cdot 4 = 8$

So probability $ = \displaystyle \frac{8}{14} = \frac{4}{7}$.

For the second question, for a given color to repeat at least two times in sequence of four, below are the possibilities.

$(i)$ it repeats four times: $1$ arrangement

$(ii)$ it repeats three times: $4 \cdot 7 = 28$ arrangements

$(iii)$ it repeats two times: $\displaystyle {4 \choose 2} \cdot 7 + {4 \choose 2} \cdot2 \cdot{7 \choose 2} = 294 $ arrangements

Explanation -
In $(ii)$, we choose three positions out of four and for the last there is choice of $7$ colors.

In $(iii)$, we choose two positions for the given color and for the other two positions, we either have $1$ of the $7$ colors or we have $2$ of the $7$ colors and also their positions can swap.

Now probability of $(ii)$ is $\displaystyle \frac{(ii)}{(i)+(ii)+(iii)} = \frac{28}{323}$


问:
Find SVD of the given matrix

Find a decomposition $X=U \Sigma V^{T}$ of the matrix $X=\begin{bmatrix}
2 & 1 & 2\\
-2 & -1 & -2\\
4 & 2 & 4\\
2 & 1 & 2
\end{bmatrix}$  where $\Sigma$ is a rectangular diagonal matrix of

size 4x3, $U$ and $V$ are orthogonal matrices, and the upper right element of $V$ is equal to $\frac{1}{\sqrt{2}}$

I found V according to the given constraint as $\begin{bmatrix}
\frac{2}{3} & \frac{-\sqrt{2}}{6} & \frac{1}{\sqrt{2}} \\
\frac{1}{3} & \frac{2\sqrt{2}}{3} & 0\\
\frac{2}{3} & \frac{-\sqrt{2}}{6} & \frac{-1}{\sqrt{2}}
\end{bmatrix}$ and $\Sigma=\begin{bmatrix}
\sqrt{63} & 0 & 0 \\
0 & 0 & 0\\
0 & 0 & 0\\
0 & 0 & 0
\end{bmatrix}$

For finding $U$, I am using $U\Sigma=XV$. But I am stuck at this step as to how to proceed further as $\Sigma$ matrix has repeated 0 eigen values.

Can you please let me know how to find $U$ from the above equation?

答:
You already found the eigenvectors of $X^TX$. These are the rows of $V^T$.

You need to find the eigenvectors of $XX^T$. There should be one eigenvector corresponding to the eigenvalue of $\sqrt {63}$ and three corresponding to the zero eigenvalue. Arrange them as column vectors $U=\begin{pmatrix}u_1&u_2&u_3&u_4\end{pmatrix}$.

Title: Find SVD of the given matrix


问:
Struggling with the Missing Side Length with One Length
In class, we have been learning about using proportions to solve for missing lengths of triangles, however, this problem has been leaving me confused. Both of the hypotenuses of the two right triangles add up to get 35. I need to find the missing length for the triangle on the left's hypotenuse. I tried YouTube and all sorts of methods, but not one that fits for this question.

答:
Just push up the lower triangle up by $12$ units in a simple construction to remove annoying hypotenuse intersection.

$$ \sin\theta=\dfrac{12+8}{35}=\dfrac{12}{?}\to ?=21 $$


问:
Find the sum of all 5 digit numbers formed using $\{0,1,2,3,4\}$
Given solution

The sum of all the numbers is
$10 \times 24 \times 10^4 + 10 \times 18 \times 10^3 + 10 \times 18 \times 10^2 + 10 \times 18 \times 10 + 10 \times 18 \times 1. $

My work

Clearly, the number of 5 digit numbers that can be formed is
$$4\times 5\times 5 \times 5 \times 5$$
$$=2500$$

The first 5 digit number is $10000$

So sum of the numbers should be
$$S=\frac {1250}{2} (20000 + 2499)$$
$$=14061875$$

But given answer is $2599980$

Where am I going wrong?

答:
Fresh after morning coffee !

It should be clear that permuting $n$ distinct digits will yield $n!$ number strings, and that each column, by symmetry, will contain $\frac{n!}{n}$ repetitions of each digit.

With leading zeros allowed, one rep of $0,1,2,3,4$ will sum to $10$, and giving place values to each digit,  we will get

sum $= 10\cdot\dfrac{5!}5(10^4+10^3+10^2+10^1+10^0) = 10\cdot\frac{5!}5(11111)$

But we need to eliminate leading zeroes, and for this we simply subtract the sum of a $4$ digit number using the distinct digits ${1,2,3,4}$
to finally get

$\left[10\cdot\frac{5!}5(11111)\right] - \left[10\cdot\frac{4!}4(1111)\right] = \boxed{2599980}$


问:
applying polynomial on diagonalizable matrix
on page 285, of SCHAUM's "linear algebra" second edition, by Lipschutz, it says

Suppose matrix A can be diagonalized as above, say $P^{-1} A P  = D$ where $D$ is diagonal. Then $A$ has the extremely useful diagonal factorization
$$
A=PDP^{-1}
$$
Using this factorization, the algebra of $A$ reduces to the algebra of the diagonal matrix $D$.
....
and more generally, for any specific polynomial $f(t)$
$$
f(A) = f(P D P^{-1}) = P f(D) P^{-1}
$$

My question is on the polynomial part above. How do you go from $f(P D P^{-1})$ to $P f(D) P^{-1}$?

I only knew about the relation $A^m = (P D P^{-1})^m = P D^m P^{-1}$ which is easy to show. I am not sure how it follows for any polynomial $f(t)$ as well.

答: $p(A) = \sum_{m=0}^r\alpha_mA^m$. That's all you need.


问:
Probability that position $i$ is a peak in $\sigma$, where $\sigma$ be a uniformly random permutation of $\{1,\ldots,n\}$
Let $\sigma$ be a uniformly random permutation of
$\{1,\ldots,n\}$.  That is $\sigma(1),\sigma(2),\ldots, \sigma(n)$
is a permutation and it is chosen uniformly from one of the $n!$
permutations.  Position $i$ is a peak in $\sigma$ if
$\sigma(i)$ is the maximum number amongst
$\sigma(1),\sigma(2),\ldots,\sigma(i)$. For instance if $\sigma$ is
the permutation $3,4,1,2,5$ then positions $1,2,5$ are peaks and
positions $3$ and $4$ are not. Note that position $1$ is always a
peak. Let $\sigma$ be a uniform random permutation of $\{1,2,\ldots,n\}$.

1.What is the probability that position $i$ is a peak in $\sigma$?

2.What is the expected number of peaks in $\sigma$?

Here's what I got so far, I'm not sure whether it is right or not.

So, I know $i$ is a peak in $\sigma$ if and only if $\sigma(i) > \sigma(i-1),  \sigma(i) > \sigma(i-2),  \ldots, \sigma(i) > \sigma(1) $. And thus, I set an indicator random variable $M_{ij}= \begin{cases}
                1 & , \ if \  \sigma(i) > \sigma(j)\\
                0 & , \ otherwise
            \end{cases}$

And, set $X_{i} = \sum_{j=1}^{i-1} M_{ij}$.

Thus, $P(i \ is \ a \ peak) = P(X_{i} =i-1) = P(M_{i1} =1, M_{i2}=1,\ldots, M_{i,i-1}=1)$

Since $P(M_{i1} =1) = \frac{number \ of \ elements \ j \ such \ that \ \sigma(j) < \sigma(i)}{n} = \frac{\sigma(i)-1}{n}$. After we pick the first element, there is only $\sigma(i)-2$ choices left for second element, so $P(M_{i2} =1) = \frac{\sigma(i)-2}{n-1}$, and so on, up to $i-1$ term.

Thus, $P(M_{i1} =1, M_{i2}=1,\ldots, M_{i,i-1}=1) = \frac{\sigma(i)-1}{n} \times \frac{\sigma(i)-2}{n-1} \times \ldots \times \frac{\sigma(i)-(i-1)}{n- (i-2)} = \prod_{j=1}^{i-1}  \frac{\sigma(i) -j}{n- (j-1)}$?

I was wondering is $\prod_{j=1}^{i-1}  \frac{\sigma(i) -j}{n- (j-1)}$ the correct probability or not?

And, I was wondering if the above analysis is in the right direction or does it have a better way to do it ?

答:
Let's count the number of permutations of $1,\ldots,n$ in which position $i$ is a peak. Suppose $k=\sigma(i)\in\{i,\ldots,n\}$ is the number filling the $i^{\text{th}}$ position of your permutation $\sigma$. If $\sigma(i)$ is in fact a peak, then the previous $i-1$ elements of $\sigma$ must belong to the set $\{1,\ldots,k-1\}$. There are ${k-1 \choose i-1}(i-1)!$ ways to fill in these preceding $i-1$ entries and $(n-i)!$ ways to fill in the remaining $n-i$ entries. Therefore, the probability that position $i$ is a peak equals $$\frac{
\sum_{k=i}^n{k-1 \choose i-1}(i-1)!(n-i)!}{n!}=\frac{1}{i}$$ Now let $X_i=1$ if $\sigma(i)$ is a peak and $X_i=0$ else. Set $X=X_1 + \dots + X_n$. Then $X$ counts the numbers of peaks of $\sigma$ and for each $1 \leq i \leq n$ $$\mathbb{E}(X_i)=P(X_i=1)=\frac{1}{i}$$ Finally, $$\mathbb{E}(X)=\sum_{i=1}^n\mathbb{E}(X_i)=1+\frac{1}{2}+\dots + \frac{1}{n}$$ Thank you @angryavian for the correction.


问:
Find triple integral over a tetrahedron constructed by 3 planes
The question is as follows:

Find triple integral of f(x,y,z) = xy + z by dxdydz

Over the tetrahedron D  created by the following coordinates: $(0,0,0),
(1,0,0), (0,1,0)$ and $(0,0,1)$

My answer doesn't agree with the book's answer.  I got $\frac{3}{20}$ while the books gives $\frac{1}{20}$.  I suspect that my integrands for $dxdy$ are incorrect somehow, but I still don't quite see what I did wrong >_<

My work:

(1) The region $D$ looks like the following picture

(2) So my thoughts are the following:

a. To work with $dx$, I consider the $xy$ plane (or $yx$ plane).  I note that $y = 0$ when $x = 1$ and $y =1$ when $x = 0$.  So the upper bound is the line $y = - x + 1$, or $x = 1 - y$.  The lower bound is $x = 0$.  So $x$ goes from $0$ to $1-y$.

b. With $dy$, I consider the $yz$ plane (or $zy$ plane).  Since $z = 0$ when $y = 1$ and $z=1$ when $y = 0$, the equation for upper bound of $y$ is $z = -y + 1$, or $y = 1- z$.  So $y$ goes from $0$ to $1-z$

c. With $dz$, lower bound is $0$ and upper bound is $1$

So the triple integral becomes
$$ \int_0^1   \int_0^{1-z}   \int_0^{1-y}  (xy + z) dxdydz$$

And after some lengthy works, I got $\frac{3}{20}$
But the book's answer is $\frac{1}{20}$.  So is my work wrong anywhere?
I strongly suspect that to work with $dx$, I should consider the $xz$ plane instead of $xy$ plane (I'm trying the "shooting arrow" method through the region)

Thank you in advance.

答:
The plane containing $(1,0,0)$, $(0,1,0)$, and $(0,0,1)$ is $x+y+z=1$. Therefore, your interval for the x-axis is $x=0$ to $x=1-y-z$, for the y-axis it's $y=0$ to $y=1$, and for the z-axis it's $z=0$ to $z=1$.

Therefore your integral should be:

$$\int_{0}^{1}\int_{0}^{1-z}\int_{0}^{1-y-z}(xy+z)\space{dx}\space{dy}\space{dz}$$


问:
Show spectrum of a bilateral shift is contained in a donut
Let $W$ be a bilateral shift with weight sequence $(w_n) \in \ell^{\infty}(\mathbb{Z})$, for $w_n \geq 0$ for each $n \in \mathbb{Z}$.

i.e. we have that $We_n = we_{n+1}$ for $n\in \mathbb{Z}$, where $(e_n)$ is an orthonormal basis.

Suppose that there exists $\delta >0$ such that $\delta \leq w_n \leq 1$.

Show that $\sigma(W) \subset \{ z\in \mathbb{C}: \delta \leq |z| \leq 1\}$.

Things I've done:

I showed that if $\alpha \in \sigma(W)$, then we have $\{z\in \mathbb{C}: |z|=\alpha \} \subset \sigma(W)$.

But I want to show that $\alpha \in \{ z\in \mathbb{C}: \delta \leq |z| \leq 1\}$.

Any help is appreciated!

答:
Assuming that $We_n:=w_ne_{n+1}$ and the space is $\ell^\infty(\mathbb{Z})$, then $W=LD$ where $L$ is the right-shift operator and $D$ is the multiplier operator with sequence $(w_n)$. Note that both $L$ and $D$ are invertible.

Then $\|W\|\le\|L\|\|D\|=\sup_n|w_n|\le1$; hence $\sigma(W)$ is bounded by $1$.

Now repeat for $W^{-1}$: $\|W^{-1}\|\le\|D^{-1}\|\|L^{-1}\|=\sup_n|w_n^{-1}|\le\frac{1}{\delta}$; hence $\sigma(W^{-1})$ is bounded by $1/\delta$, and thus $\sigma(W)$ is bounded below by $\delta$.


答:
Matthew Pilling's approach is correct. I just want to mention a slightly easier way to get $P(\text{$i$ is a peak})=1/i$.

Conditioned on knowing what  $\sigma(i+1),\ldots, \sigma(n)$ are, the distribution of $\sigma(1),\ldots, \sigma(i)$ is a uniform permutation of $i$ distinct numbers (namely, the $i$ numbers not equal to $\sigma(i+1),\ldots, \sigma(n)$). The largest of these $i$ numbers is equally likely to be in any of the $i$ positions, so in particular it has a $1/i$ chance of being in the $i$th position, which is the condition for making $\sigma(i)$ a peak. Since we know $P(\text{$i$ is a peak} \mid \sigma(i+1)=s_{i+1}, \ldots, \sigma(n) = s_n)=1/i$ no matter what the last $n-i$ terms are, the unconditional probability $P(\text{$i$ is a peak})$ is also $1/i$.

Title: Probability that position $i$ is a peak in $\sigma$, where $\sigma$ be a uniformly random permutation of $\{1,\ldots,n\}$

Body:
Let $\sigma$ be a uniformly random permutation of
$\{1,\ldots,n\}$.  That is $\sigma(1),\sigma(2),\ldots, \sigma(n)$
is a permutation and it is chosen uniformly from one of the $n!$
permutations.  Position $i$ is a peak in $\sigma$ if
$\sigma(i)$ is the maximum number amongst
$\sigma(1),\sigma(2),\ldots,\sigma(i)$. For instance if $\sigma$ is
the permutation $3,4,1,2,5$ then positions $1,2,5$ are peaks and
positions $3$ and $4$ are not. Note that position $1$ is always a
peak. Let $\sigma$ be a uniform random permutation of $\{1,2,\ldots,n\}$.

1.What is the probability that position $i$ is a peak in $\sigma$?

2.What is the expected number of peaks in $\sigma$?

Here's what I got so far, I'm not sure whether it is right or not.

So, I know $i$ is a peak in $\sigma$ if and only if $\sigma(i) > \sigma(i-1),  \sigma(i) > \sigma(i-2),  \ldots, \sigma(i) > \sigma(1) $. And thus, I set an indicator random variable $M_{ij}= \begin{cases}
                1 & , \ if \  \sigma(i) > \sigma(j)\\
                0 & , \ otherwise
            \end{cases}$

And, set $X_{i} = \sum_{j=1}^{i-1} M_{ij}$.

Thus, $P(i \ is \ a \ peak) = P(X_{i} =i-1) = P(M_{i1} =1, M_{i2}=1,\ldots, M_{i,i-1}=1)$

Since $P(M_{i1} =1) = \frac{number \ of \ elements \ j \ such \ that \ \sigma(j) < \sigma(i)}{n} = \frac{\sigma(i)-1}{n}$. After we pick the first element, there is only $\sigma(i)-2$ choices left for second element, so $P(M_{i2} =1) = \frac{\sigma(i)-2}{n-1}$, and so on, up to $i-1$ term.

Thus, $P(M_{i1} =1, M_{i2}=1,\ldots, M_{i,i-1}=1) = \frac{\sigma(i)-1}{n} \times \frac{\sigma(i)-2}{n-1} \times \ldots \times \frac{\sigma(i)-(i-1)}{n- (i-2)} = \prod_{j=1}^{i-1}  \frac{\sigma(i) -j}{n- (j-1)}$?

I was wondering is $\prod_{j=1}^{i-1}  \frac{\sigma(i) -j}{n- (j-1)}$ the correct probability or not?

And, I was wondering if the above analysis is in the right direction or does it have a better way to do it ?


问:
How to prove Campanato space is a Banach space
Let $p\ge1$, $\mu\ge0$ and $\Omega$ be a bounded open set in $\mathbb{R}^n$. Campanato space embraces all $u$'s which

$$[u]_{p,\mu}=[u]_{p,\mu;\Omega}=\sup_{\substack{x\in\Omega\\0<\rho<\mathrm{diam}\Omega}}\left(\rho^{-\mu}\int_{\Omega_\rho(x)}\left|u(y)-u_{x,\rho}\right|^p\,dy\right)^{\frac{1}{p}}<+\infty,$$
 where $\Omega_\rho(x)=\Omega\cap B_\rho(x)$ ($B_\rho(x)$ denotes a ball centered at $x$ with a radium $\rho$) and
$$u_{x,\rho}=\frac{1}{\left|\Omega_\rho\right|}\int_{\Omega_\rho(x)}u(y)\,dy, $$

equipped with a norm
$$\|u\|_{L^{p,\mu}}=\|u\|_{L^{p,\mu}(\Omega)}=[u]_{p,\mu;\Omega}+\|u\|_{L^p(\Omega)}.$$

Let $\{u_k\}$ be a Cauchy sequence in Campanato space, one can determine a $u$ because of the completeness of $L^p$ space. What are the next steps to prove that $u$ is also the right limit of $\{u_k\}$ in the sense of $\|\cdot\|_{L^{p,\mu}(\Omega)}$ and therefore Campanato space is complete? Thank you~

答:
Let's start with a general result.

Theorem. If $(X,\|\cdot\|)$ is a Banach space and $\rho\colon X\to [0,\infty]$ a lower semicontinuous seminorm, then $Y=\{x\in X\mid \rho(x)<\infty\}$ equipped with the norm $\|\cdot\|+\rho$ is a Banach space.

Proof: If $(x_n)$ is a Cauchy sequence in $(Y,\|\cdot\|+\rho)$, then it is also Cauchy in $(X,\|\cdot\|)$ and thus converges to a limit $x$. Since $\rho$ is lower semicontinuous, we have $\rho(x)\leq \liminf_n \rho(x_n)<\infty$. Hence $x\in Y$. Moreover,
$$
\limsup_{n\to\infty}\rho(x-x_n)\leq \limsup_{m\to\infty}\liminf_{n\to\infty}\rho(x_m-x_n)=0.
$$
Therefore $x_n\to x$ in $(Y,\|\cdot\|+\rho)$. QED

Now the completeness of the Campanato space is a simple Corollary. Indeed, to see that $[\cdot]_{p,\mu}$ is lower semicontinuous, it suffices to note that the supremum of lower semicontinuous functions is lower semicontinuous and convergence in $L^p(\Omega)$ implies convergence in $L^q(\Omega')$ for $q\leq p$ and $\Omega'\subset \Omega$.


问:
Represent in a matrix form: $\sum_{ijl}(F_{il}-F_{jl})^2W_{ij}$
How can I represent this in a matrix form:

$\sum_{ijl}(F_{il}-F_{jl})^2W_{ij}$

where all the entries are real and $W$ is  a known(constant) matrix and $F$ is a rectangular matrix. When I say matrix form, the matrix trace function can be used as well as part of the simplification. Essentially, the index $l$ is choosing a column in $F$ and then the summation is being done and so-forth.

答:
Let $e_i$ denotes the $i$-th vector in the canonical basis and $u=\sum_ie_i=(1,1,\ldots,1)^\top$. Then your sum is equal to
$$
u^\top \left[\left(\sum_i e_ie_i^\top FF^\top e_iu^\top
+ \sum_j ue_j^\top FF^\top e_je_j^\top
- 2FF^\top\right) \circ W\right] u,\tag{1}
$$
where the symbol "$\circ$" denotes a Hadamard product. Since
$$
u^\top (A\circ B)u = \operatorname{trace}(B^\top A) = \operatorname{vec}(B)^\top\operatorname{vec}(A),
$$
you may also express your sum as
$$
\operatorname{trace}\left(W^\top \left(\sum_i e_ie_i^\top FF^\top e_iu^\top
+ \sum_j ue_j^\top FF^\top e_je_j^\top
- 2FF^\top\right) \right)\tag{2}
$$
or
$$
\operatorname{vec}(W)^\top
\left[\sum_i (ue_i^\top)\otimes(e_ie_i^\top) + \sum_j(e_je_j^\top)\otimes(ue_j^\top) - 2I\otimes I\right]
\operatorname{vec}(FF^\top),\tag{3}
$$
where "$\otimes$" denotes Kronecker product (tensor product). Either way, I think $\sum_{ijl}(F_{il}-F_{jl})^2W_{ij}$ is a much simpler expression.</p>


问:
Proving the inequality $\limsup (a_n+b_n)\le \limsup(a_n) +\limsup(b_n)$.
Definition: $\limsup x_n$= supremum of all subsequential limits of sequence $(x_n)=\sup E_x$, where $E_x$ is set of all subsequential limits of sequence $(x_n)$. Let $x^*=\limsup x_n$.

Given any two sequences $(a_n)$ and $(b_n)$, it is to be proven that: $\limsup (a_n+b_n)\le \limsup(a_n) +\limsup(b_n)$, provided sum on right side is not of the form $\infty -\infty$. $\tag 1$

I tried to prove it as follows:

Let $c_n=a_n+b_n$ for all $n\in \mathbb N$

Case $1: a^*, b^*, c^* \in \mathbb R$

It follows that the sequences $(a_n),(b_n),(c_n)$ are bounded. And since set of all subsequential limits of a sequence is closed, it follows that $a^*\in E_a, b^*\in E_b, c^*\in E_c\implies$ There exist subsequences $(a_{n_k}), (b_{n_l})$ and $(c_{n_m})$ such that $a_{n_k}\to a^*, b_{n_l}\to b^*, c_{n_m}\to c^* $.

For any $\epsilon \gt 0, \exists s_\epsilon\in \mathbb N$ such that $s\ge s_\epsilon \implies |a_{n_s}-a^*|\lt \epsilon/4 \space, |b_{n_s}-b^*|\lt \epsilon/4 \implies |a_{n_s}+b_{n_s}-a^*-b^*|\lt \epsilon/2$.

Now suppose on the contrary that $c^*\gt a^*+b^*$. We have: for any $\epsilon \gt 0$

$\exists M_\epsilon \in \mathbb N: m\ge M_\epsilon\implies |c_{n_m}-c^*|\lt \epsilon/2.$

Hence for $r\in \mathbb N: r\ge \sup(M_\epsilon, s_\epsilon),$ we have:

$0\lt c^*-a^*-b^*=|c^*-(a_{n_r}+b_{n_r})+ (a_{n_r}+b_{n_r}-a^*-b^*)|\le |c^*-(a_{n_r}+b_{n_r})|+ |(a_{n_r}+b_{n_r}-a^*-b^*)|\lt\epsilon \implies c^*=a^*+b^* $, which is contradiction. Therefore, $c^*\le a^*+b^*$.

Case $2$: $a^*=\infty, b^*\in \mathbb R\cup \{\infty\},$ then the result is trivial.

Case $3$: $b^*=\infty, a^*\in \mathbb R\cup \{\infty\}$, then as in Case $3$, the result is true.

Case $4$: $a^*\in \mathbb R, b^*=-\infty .$

$E_b=\{-\infty\}$ and $ (a_n)$ is bounded by some $M\gt 0$. So, $(c_n)$ is not bounded below as $(b_n)$ is not bounded below.
 So for some $M\in \mathbb N$, $n\ge M\implies c_n\lt a^*+b_n\implies c^*\le a^*+b^* $.

Case $5$: $b^*\in \mathbb R, a^*=-\infty .$

Same as case $4$.

So proved.

My question is what happens when right hand side of $(1)$ is $\infty -\infty $?

Consider for example: $(a_n)=(n), (b_n)=(-n)$, clearly $c_n=0\to \limsup c_n=0$. But can we claim that $0\gt \infty -\infty $ (I don't think so as $\infty -\infty $ is undefined.)?. If not, then $(1)$ always holds?

Please help. Thanks.

答:
Your proof is not correct. You have (possibly) distinct subsequences $(a_{n_k})$, $(b_{m_k})$ of the original sequences converging to $a^*$ and $b^*$, and these need not have any indices in common.

Also $\infty-\infty$ is undefined, and nothing can be said about $\limsup (a_n+b_n)$ if the right-hand side of $(1)$ is of that form. It can be $-\infty$, $+\infty$, or any finite real number.

Using your definition of $\limsup$ as the supremum of all subsequential  limits, I would argue as follows:

If one of $\limsup a_n$ or $\limsup b_n$ is equal to $+\infty$ then the other cannot be $-\infty$. In that case the right-hand side of $(1)$ is $+\infty$ and the inequality holds.

Therefore is suffices to consider the case that both $\limsup a_n$ and $\limsup b_n$ are a real number or $-\infty$. In particular, both sequences are bounded above.

We can also assume that $c^* = \limsup(a_n+b_n)$ is not $-\infty$.

For every $\epsilon > 0$ there is a convergent subsequence $c_{n_k}$ of $(c_n)$ with
$$
 \lim_{k \to \infty} (a_{n_k} +  b_{n_k}) =  \lim_{k \to \infty} c_{n_k} > c^* - \epsilon \, .
$$
Now $(a_{n_k})$ and $(b_{n_k})$ are both bounded sequences. Then $(a_{n_k})$ has a convergent subsequence $(a_{n_{k_l}})$ and $(b_{n_{k_l}})$ has a convergent subsequence $(b_{n_{k_{l_m}}})$. Then
$$
 c^* - \epsilon < \lim_{m \to \infty} (a_{n_{k_{l_m}}} + b_{n_{k_{l_m}}})
=  \lim_{m \to \infty} a_{n_{k_{l_m}}} +  \lim_{m \to \infty} b_{n_{k_{l_m}}} \le a^* + b^* \, .
$$
This holds for all $\epsilon > 0$, which implies that $c^*  \le  a^* + b^*$.

Remark: The difference to your proof is that we started with a convergent subsequence $(c_{n_k})$ of $(c_n)$ and then chose “sub-sub-sequences” of $(a_n)$ and $(b_n)$ which are “simultaneously” convergent.


问:
If $R$ is a finite ring, then $\exists_{n>m>0}: x^n=x^m$ for all $x\in R$
I need some help for the following proof:

If $R$ is a finite ring, then $\exists_{n>m>0}: x^n=x^m$ for all $x\in R$.

I feel there's one or more little tricks to use to see how you get to this equation. So far I've noticed that $R$ is finite, so $R^*$, the group of units, is also finite and therefore cyclic and Abelian. We can write $R^*=\{1,y,y^2\dots y^{r-1}\}$ for a generator $y$, if $|R|=r$. We see that for all $x\in R^*$ we have $x^r=1$. I thought maybe you could use this equation.

Additionally, we have that any $t\in R^*$ has $ord(t)|r$

However, I'm still getting stuck, to actually put together an equation that holds for all $x\in R$.

答:
If $R=\mathbb{Z}/8\mathbb{Z}$, then $R^*=\{\bar1,\bar3, \bar5,\bar7\}.$ Since every element is its own inverse, we have $R^*\ncong \mathbb{Z}/4\mathbb{Z}$.

If $R=Mat_n(\mathbb{F}_q)$, then $R^*=GL_n(\mathbb{F}_{q})$, where $\mathbb{F}_{q}$ is the finite field of $q$ elements.

Therefore, the group of units of a finite ring need not be cyclic nor Abelian if the ring is noncommutative. If $R$ is a finite field, then your statement about the group of units is correct.


问:
If $g(n+1) = g(n) + g(n - 1)$, and $g(1) = g(4) = 1$, then what is $g(10)$?
Let $g$ be a function such that $g(n+1) = g(n) + g(n - 1)$ for all integers $n$. If $g(1) = g(4) = 1$, then what is $g(10)$?

答:
$$g(4)=g(3)+g(2)=2g(2)+g(1)=1$$
Since $g(1)=1$, $g(2)=0$. From here, you can either solve the recursion or compute $g(10)$ directly. (In fact, this is just the Fibonacci sequence "shifted down" two terms)


问:
Change the order of integration $\int_{1}^{\frac{4}{3}}{\rm d}x\int_{\sqrt{\frac{x}{3}}}^{2- x}f\left ( x, y \right ){\rm d}y$

Change the order of integration
$$\int_{0}^{1}{\rm d}x\int_{\sqrt{\frac{x}{3}}}^{\sqrt{x}}f\left ( x, y \right ){\rm d}y+ \int_{1}^{\frac{4}{3}}{\rm d}x\int_{\sqrt{\frac{x}{3}}}^{2- x}f\left ( x, y \right ){\rm d}y$$

By Desmos_ https://www.desmos.com/calculator/npgmmkcbnn?lang=vi, we have
$$\int_{0}^{1}{\rm d}x\int_{\sqrt{\frac{x}{3}}}^{\sqrt{x}}f\left ( x, y \right ){\rm d}y= \int_{0}^{1}{\rm d}y\int_{y^{2}}^{3y^{2}}{\rm d}x$$
for the picture_ https://www.desmos.com/calculator/oy0qncqoid?lang=vi I couldn't find the inner
$$\int_{1}^{\frac{4}{3}}{\rm d}x\int_{\sqrt{\frac{x}{3}}}^{2- x}f\left ( x, y \right ){\rm d}y= \int_{\sqrt{\frac{1}{3}}}^{1}{\rm d}y\int_{a}^{b}f\left ( x, y \right ){\rm d}x$$
please help me get $a$ and $b$ that $x$ corresponds to, and how can we solve this without drawing, which as same as what I said here_ Find the relationship between $x$ and $y$ so that $y:=0\rightarrow\frac{\pi}{2}\Leftrightarrow x:=y\rightarrow\frac{\pi}{2}$ _thanks a real lot !

答:
Find the $y$ coordinates of intersection of curves $x=y^2$ and $x = 3y^2$ with line $x+y=2$.

$x=2-y = y^2 \implies y = 1$

$x=2-y = 3y^2 \implies y = \frac{2}{3}$

For $0 \leq y \leq \frac{2}{3}, y^2 \leq x \leq 3y^2$

For $\frac{2}{3} \leq y \leq 1, y^2 \leq x \leq 2-y$

So the integral becomes,

$\displaystyle \int_0^{2/3} \int_{y^2}^{3y^2} f(x,y) dx \ dy + \int_{2/3}^1 \int_{y^2}^{2-y} f(x,y) dx \ dy$


问:
Convergence in norm but not almost everywhere of $f_{t_n}=f(\cdot-t_n)$ to $f$
My question is : is there a $L^1(\mathbf R)$ function $f$ and a sequence $(t_n)_n$ of reals with limit $0$ such that $f_{t_n}=f(\cdot-t_n)$ does not converges to $f$ a.e. ?

There are several variations of this questions which are pretty much equivalent : one can drop the $L^1$ hypothesis and just assume mesurability of $f$, one could also assume that $f$ is the indicator function of a measurable subset of $[0;1]$.

What i already know :

The first thing to say is that $f_\tau$ converges to $f$ in $L^1$ norm when $\tau$ goes to zero. One way to prove that is to say that this is true for compactly suported continuous functions (because of absolute continuity) and conclude by density.

Since the $f_{t_n}$ are converging in $L^1$ norm there exists a subsequence $t_{\sigma(n)}$ such that $f_{t_{\sigma(n)}}$ converges almost everywhere to $f$. However one cannot directly use this result to conclude with the convergence a.e. of $f_{t_n}$, indeed there are many examples of sequences of functions that are converging in $L^1$ norm but not convergent a.e. In fact this show that the a.e. convergence does not corespond to any topological convergence, since the caracterisation of convergence with sub-sub-sequences doesn't apply here.

Another thing to note is that if $f$ is continuous at $x$ then $f_{t_n}(x)\to f(x)$ automatically, so if we search some $f$ such that $f_{t_n}$ does not converges a.e. to $f$ it has to be a function discontinuous on a set with positive measure. In fact for every function $g=0$ a.e. one must have that $f+g$ is discontinuous on a set with positive measure.

I believe the answer to my question is yes, but it's mostly an intuition, i don't even have euristic arguments for that. So i tried to find an example, the two kind of $f$ i have thought of are :

$f=\chi_{C}$ where $C$ is a fat cantor set or something like this kind of set and

$$f(x)=\chi_{[0;1]}\sum_{k=0}^\infty \frac{2^{-k}}{|x-q_k|^{\alpha}}$$ where $q_n$ is an enumeration of the rationals of $[0;1]$.

The problem is that it's hard to find the behavior of $|f_{t_n}(x)-f(x)|$ for general $x$ and such patological functions...

答:
Your idea of a fat Cantor set works too.  I'll repeat a construction I used
in this answer.

Let $C$ be a compact nowhere dense set of positive measure.  We can construct a sequence $t_n \to 0$ such that for every $x \in C$, there are infinitely many $n$ for which $x - t_n \notin C$.  In particular we have $\liminf_{n \to \infty} \chi_C(x-t_n) = 0$ and therefore $\chi_C(x-t_n) \not\to 1 = \chi_C(x)$.  Since $C$ has positive measure, we therefore do not have $\chi_C(\cdot-t_n) \to \chi_C$ a.e.

To construct the sequence, let $U = C^c$ and note that $U$ is open and dense.  So for each integer $m > 0$ and each $x \in C$, there exists $u \in U$ such that $|x-u| < 1/m$, which is to say that $x \in U+(x-u)$.  In other words, the collection of open sets $\{U+s : |s| < 1/m\}$ cover $C$ (indeed they cover $\mathbb{R}$).  Since $C$ is compact, there is a finite subcover $\{U+s_{m,1},\dots, U+s_{m, k_m}\}$.  That is to say, for every $x \in C$ there is some $1 \le i \le k_m$ for which $x - s_{m, i} \in U = C^c$.

Now let $(t_n)$ be the sequence
$$(s_{1,1}, \dots, s_{1, k_1}, s_{2,1}, \dots, s_{2, k_2}, \dots).$$
Since $|s_{i,m}| \le 1/m$ for every $m$, we have $t_n \to 0$, and there are infinitely many $t_n$ such that $x-t_n \notin C$ (namely, at least one of the $s_{m,i}$ for each $m$).


答:
Hint:

Set $1+xe^x=y$

Multiply numerator and denominator by $e^x$

Now $$\dfrac1{y(y-1)}=\dfrac{y-(y-1)}{y(y-1)}=?$$

Body:
Evaluate $\int {\frac{{x + 1}}{{x{{\left( {1 + x{e^x}} \right)}^2}}}dx} $

My approach is as follow

$\int {\frac{{x + 1}}{{x{{\left( {1 + x{e^x}} \right)}^2}}}dx}  \Rightarrow \int {\frac{{x + 1}}{{x\left( {{e^{2x}}} \right){{\left( {{e^{ - x}} + x} \right)}^2}}}dx}  \Rightarrow \int {\frac{{\left( {{e^{ - 2x}}} \right)\left( {1 + \frac{1}{x}} \right)}}{{{{\left( {{e^{ - x}} + x} \right)}^2}}}dx}  \Rightarrow \int {\frac{{\left( {{e^{ - 2x}}} \right)\left( {1 + \frac{1}{x}} \right)}}{{{x^2}{{\left( {\frac{{{e^{ - x}}}}{x} + 1} \right)}^2}}}dx} $

$ \Rightarrow \int {\frac{{\left( {{e^{ - 2x}}} \right)\left( {\frac{1}{{{x^2}}} + \frac{1}{{{x^3}}}} \right)}}{{\left( {\frac{{{e^{ - 2x}}}}{{{x^2}}} + \frac{{2{e^{ - x}}}}{x} + 1} \right)}}dx} $

How do I approach from here


问:
Correctness of this statement about vector subspces
I was looking at this answer and when it states:

Given $v\in V$, then it is well-known that $v=v_1+v_2$ for some $v_1\in W$ and $v_2\in W^\perp$.

And then it proceed to state:

$\beta\cup\gamma$ generates $V$.

I understand that $v=v_1+v_2$ will produce some of the elements in $V$ but that doesn't necessarily mean that it can produce all of the elements in $V$.

Is there a proof for this?

答: Yes, of course. It follows from Hilbert's projection theorem (https://en.wikipedia.org/wiki/Hilbert_projection_theorem).</p>


答:
With $$u = 1 + x e^x, \quad du = (x e^x + x) \, dx = (1 + x) e^x \, dx,$$ we have $$\int \frac{1+x}{x(1+xe^x)^2} \, dx = \int \frac{(1+x)e^x}{x e^x (1 + xe^x)^2} \, dx = \int \frac{du}{(u-1) u^2}.$$  Then partial fraction decomposition yields
$$\begin{align}
\int \frac{du}{(u-1) u^2} &= \int \frac{1}{u-1}  - \frac{1}{u} - \frac{1}{u^2} \, du \\
&= \log|u-1| - \log |u| + u^{-1} + C \\
&= \log|x| + x - \log|1 + xe^x| + \frac{1}{1 + xe^x} + C.
\end{align}$$

Title: Evaluate $\int {\frac{{x + 1}}{{x{{\left( {1 + x{e^x}} \right)}^2}}}dx} $


问:
On the density of colossally abundant numbers.
Define $\sigma(n) = \sum_{d\mid n} d$. A integer $n>1$ is said to be colossally abundant (CA) if there exists some exponent $\varepsilon > 0$ such that $$\frac{\sigma(n)}{n^{1+\varepsilon}} \geq \frac{\sigma(m)}{m^{1+\varepsilon}}$$ for every integer $m>1$. Let $p_{k} >20000$ be the largest prime factor of the CA number $n$. Is it necessarily true that $n \in [N_k, N_{k+1})$ where $N_j$ denotes the product of the first $j$ primes ?

P.S.: This result was stated without proof on page 206 of http://zakuski.utsa.edu/~jagy/Robin_1984.pdf, so either it's completely trivial or an assumption.

答:
Thank you for providing the exact reference!

My translation of the relevant passage is: "Consequently, if $N_k \le n < N_{k+1}$, with $p_k \ge 20000$, we have: [displayed equation]. If $n\le N_{k_1}$ [note the subscript is $k_1$, which I missed earlier], we use the results calculated by computer in Proposition 3 of Section 3." In the first line of the proof (same page), Robin writes: "Let $N_k = \prod_{i=1}^k p_i$ [he writes $p_k$ but surely that is a typo] and suppose that $p_k \ge 20000$ ($k\ge k_1$).", which I believe is the implicit definition of the integer $k_1$.

So I believe what is happening is:

if $n$ is at most $N_{k_1}$, which is the product of all the primes up to $20000$, then the computer results are relevant;

otherwise, there exists some integer $k\ge k_1$ such that $N_k \le n < N_{k+1}$, in which case the displayed equation is relevant.

In the OP, it is claimed that $n$ is colossally abundant and that $p_k$ is its largest prime factor; but I don't believe that either statement is necessarily true in the context of this remark. Rather, $n$ is any integer at all, and $p_k$ is only mentioned to help clarify what $N_k$ is.


问:
Range of answer: Least squares problem in a case of a rank deficient system
I am new to Algebra and I need some clarification in terms of column and row space and LS solution.

Considering the linear system of Ax=b;
In case of A is an over-determined and not full rank matrix, we want to solve this system through minimizing LS problem. If we take the x* as the minimizer, could you please explain why "x* ∈ Range (A*)" ?

I think the explanation lies within the normal equation "A* Ax* = A*b". But I am not sure how should I explain the answer. in Particular, when I try to understand the question graphically, it becomes more complicated!

答:
explain why "x* ∈ Range (A*)"

Short answer

You are probably taking this statement out of context, where the statement was followed by "without loss of generality". It is not true that any least squares solutions belongs to the row space. It is true that we can assume (for some purposes) that the least squares solution belongs to the row space, without loss of generality. Here is a step-by-step explanation for you.

Long answer

For an overdetermined full-rank problem, $\boldsymbol{A}\in\mathbb{R}^{m\times n}$, $m>n$, $\exists(\boldsymbol{A}^{*}\boldsymbol{A})^{-1}$, the least squares solution $x_+\in\mathbb{R}^n$ is in the row space $\mathrm{Range}\boldsymbol{A}^{*}$, because any vector $x\in\mathbb{R}^n$ is in the row space. The row space spans the entire $\mathbb{R}^n$, this is a consequence of $\boldsymbol{A}$ being full column rank.

Now if you have an overdetermined rank-deficient problem, $\boldsymbol{A}\in\mathbb{R}^{m\times n}$, $m>n$, $\not\exists(\boldsymbol{A}^{*}\boldsymbol{A})^{-1}$, then $\mathrm{Range}\boldsymbol{A}^{*}$ spans a proper subpace of $\mathbb{R}^n$.

A very important theorem in linear algebra says that the row space of any matrix is an orthogonal complement to the nullspace (and vice versa): $(\mathrm{Range}\boldsymbol{A}^*)^\perp=\mathrm{Null}\boldsymbol{A}$.

In our case, this means that you can add an arbitrary vector from the nullspace $x_{0}\in\mathrm{Null}\boldsymbol{A}$ to a solution $x_+$ and obtain another solution $x_1=x_++x_0$, which will be almost as good. Conversely, any solution $x_1$ is uniquely decomposed into the sum $x_1=x_++x_0$, and you can remove the nullspace component $x_0$. The remaining $x_+$ will belong to $\mathrm{Range}\boldsymbol{A}^{*}$.

Remark

One important remark pertains to what the least squares problem is precisely. Some authors define it as $\|\boldsymbol{Ax}-\boldsymbol{b}\|_{2}^{2}\to\min$, which may have multiple solutions, all equal up to the nullspace component. Other authors require, in addition, that the solution has the minimal $\ell^2$ norm: $\|x\|_2\to\min$, subject to $\|\boldsymbol{Ax}-\boldsymbol{b}\|_{2}^{2}=\min$, in which case the solution is always unique and given by $x_+:=\boldsymbol{A}^+b$, where $\boldsymbol{A}^+$ is the Moore-Penrose pseudoinverse of $\boldsymbol{A}$. To avoid ambiguity, the latter problem formulation is called the minimum-length least squares.


问:
Evaluating $\sum_{k>\frac{N}{2}}\frac {1}{N}\cdot \frac{N-k}{k}$
Assuming $N$ is even, how can I evaluate the following sum:

$$\sum_{k>\frac{N}{2}}\frac{\binom{N}{k}(k-1)!(N-k)!}{N!}\cdot\frac{N-k}{N}=
\sum_{k>\frac{N}{2}}\frac {1}{N}\cdot \frac{N-k}{k}$$

I really don't know how to do it...

Thanks!
(Not HW BTW)

答:
First thing I would do is set $N=2M$. As observed by i707107, the lhs is $0$ for $k>N$, so your summation is finite and splits as follows
$$
S_N=\sum_{k=M+1}^{2M}\frac{1}{2M}\cdot\frac{2M-k}{k}=\sum_{k=M+1}^{2M}\frac{1}{k}-\frac{1}{2M}\sum_{k=M+1}^{2M}1
$$
$$
=\sum_{k=1}^{2M}\frac{1}{k}-\sum_{k=1}^{M}\frac{1}{k}-\frac{1}{2M}\cdot M
=H_{2M}-H_M-\frac{1}{2}
$$
$$
=H_N-H_{N/2}-\frac{1}{2}
$$
where $H_N$ denotes the $N$th harmonic number. You can get asymptotics for the latter if you wish: $H_N=\log N+\gamma+o(1)$ where $\gamma\simeq 0.577\ldots$ is Euler-Mascheroni constant. It follows that
$$
S_N=\log N+\gamma -(\log(N/2)+\gamma)-\frac{1}{2}+o(1)=\log 2-\frac{1}{2}+o(1)
$$
$$
\longrightarrow\log 2-\frac{1}{2}\simeq 0.193147\ldots
$$

If you simply want to bound it from below as you say, you can use integral comparison
$$
S_N=-\frac{1}{2}+\sum_{k=M+1}^{2M}\frac{1}{k}\geq-\frac{1}{2}+\int_{M+1}^{2M+1}\frac{1}{t}dt=-\frac{1}{2}+\log\left(\frac{2M+1}{M+1}\right).
$$

Note: with the above integral comparison, completing the upper bound, you recover that the limit is $\log 2-\frac{1}{2}$ without knowing anything about Euler-Mascheroni constant.


问:
Integer-valued polynomials divisible by 2 or 3
I would like some help with the following problem:

Suppose that $p \in \operatorname{Int}(\mathbb{Z}) = \{p \in \mathbb{Q}[X] : p(\mathbb{Z}) \subseteq \mathbb{Z}\}$ is such that for all $z \in \mathbb{Z}$, either $2$ divides $p(z)$ or $3$ divides $p(z)$.
Show that the quantifiers may be reversed, i.e., that either for all $z \in \mathbb{Z}$ we have $2$ divides $p(z)$, or for all $z \in \mathbb{Z}$ we have $3$ divides $p(z)$.

I thought of using the fact that $\{\binom{X}{k} \mid k \in \mathbb{N}\}$ is a $\mathbb{Z}$-basis of $\operatorname{Int}(\mathbb{Z})$, where $$\binom{X}{k} = \frac{X (X - 1) \cdots (X - k + 1)}{k!}$$ (in particular $\binom{X}{0} = 1$).
Therefore, for a fixed $p \in \operatorname{Int}(\mathbb{Z})$, I can write $$p = \sum_{i=1}^k a_i \binom{X}{i}$$
for some $k \in \mathbb{N}$ and $a_1, \ldots, a_k \in \mathbb{Z}$.

I tried to show the contrapositive:
suppose that there is $z_2, z_3 \in \mathbb{Z}$ such that $2 \nmid p(z_2)$ and $3 \nmid p(z_3)$.
Then there is $1 \leq i, j \leq k$ such that $2 \nmid a_i$ and $3 \nmid a_j$.
But then I don't quite know how to find a $z$ with both $2 \nmid p(z)$ and $3 \nmid p(z)$,
just as I don't see how "for all $z \in \mathbb{Z}$, either $2$ divides $p(z)$ or $3$ divides $p(z)$" relates to the coefficients of $p(z)$.

Is my approach correct, and how should I proceed?

答:
Let $f$ denote the polynomial $p$ in OP, $r,s\in\Bbb N$ and $u\in\Bbb Z\setminus\{0\}$ such that $2\nmid u$, $3\nmid u$ and $g=2^r3^suf$ is a polynomial with integer coefficients.

Assume that there exists $a,b\in\Bbb Z$ such that $2\nmid f(a)$  and $3\nmid f(b)$.
By chinese remainder theorem, there exists $c\in\Bbb Z$ such that
\begin{align}
c&\equiv a\pmod{2^{r+1}}&
c&\equiv b\pmod{3^{s+1}}
\end{align}
hence
\begin{align}
g(c)&\equiv g(a)\not\equiv 0\pmod{2^{r+1}}&
g(c)&\equiv g(b)\not\equiv 0\pmod{3^{s+1}}
\end{align}
consequently, $2\nmid f(c)$ and $3\nmid f(c)$ - a contradiction.


问:
$A, B,$ and $C$ are all sets that lie in a common universal set. Prove $A = B$ given the following statements.
$(A \cup C) = (B \cup C)$ and $(A \cap C) = (B \cap C)$

I know I have to start with $A = ...$ to get to $B$. I incorporated $A = A \cap (A \cup C)$ as the first step using absorption law, but I don't know where to go from here. How do I show that it equals $(A \cup C) = (B \cup C)$? I've looked at the set identities, and think that I have to use absorption law again to show that $A \cap (A \cup C) = (A \cap (A \cup C)) \cap ((A \cap (A \cup C)) \cup (B \cup C))$. Can you guys please help me solve this problem, I'm stuck.

答:
I suggest a Venn diagram of $A, B, C$.

First solution.

Let $S_1=A$ \ $(B\cup C)$ and $S_2=B$ \ $(A\cup C)$ and $S_3=C$ \ $(A\cup B).$

Let $S_4=(A\cap B)$ \ $C$ and $S_5=(A\cap C)$ \ $B$ and $S_6=(B\cap C)$ \ $A .$

Let $S_7=A\cap B\cap C.$

Then $\{S_j:1\le j\le 7\}$ is a pairwise-disjoint family.

For brevity let $T=\cup_{j=3}^{j=7}\,S_j.$ Then $S_1\cup T=A\cup C=B\cup C=S_2\cup T.$ Therefore $$S_1=S_1\cap (S_1\cup T)=S_1\cap (S_2\cup T)=\emptyset. $$ Interchanging the subscripts $1, 2$ in the preceding line, we also have $S_2=\emptyset.$

We have $$S_5=S_5\cap ( S_5\cup S_7)=S_5\cap (A\cap C)=S_5\cap (B\cap C)=S_5\cap (S_6\cup S_7)=\emptyset.$$ And similarly we obtain $S_6=\emptyset.$

Therefore $$A=S_1\cup S_5\cup S_4\cup S_7=S_2\cup S_6\cup  S_4\cup S_7 =B.$$

Second solution.

For any $A,B,C$ we have $$A=(A\cap C)\cup (A\setminus C)=(A\cap C)\cup ([A\cup C]\setminus C)$$ and $$B=(B\cap C)\cup (B\setminus C)=(B\cap C)\cup ([B\cup C]\setminus C)$$. Now compare the RHS of these if $A\cap C =B\cap C$ and $A\cup C=B\cup C.$


问:
How to prove the rep of SU(2) on homogeneous polynomials in 2 variables is irreducible?
The group $\mathrm{SU}(2)$ has a tautologous representation on the space $\mathbb{C}^2$ and thus a representation on the $d$th symmetric power $S^d (\mathbb{C}^2)$.  What's the easiest way to prove that this representation is irreducible?

I want to explain this in a course I'm teaching.  I want to prove this "from scratch", not using powerful tools like Schur-Weyl duality or the general classification of representations of compact simple Lie groups or simple Lie algebras.   I have an elementary method but it involves some mildly annoying calculations.  Ideally there would be some slick, quick, elementary and insightful proof.

If it helps, I'm happy to think of $S^d(\mathbb{C}^2)$ as the space of homogeneous degree-d polynomial functions on $\mathbb{C}^2$ and describe the representation in this way.  I'm happy to consider it either as a representation of $\mathrm{SU}(2)$, $\mathrm{SL}(2,\mathbb{C})$, or the Lie algebra $\mathfrak{su}(2)$ or $\mathfrak{sl}(2,\mathbb{C})$: it's pretty painless to prove irreducibility of the representation of any one of these groups or Lie algebras on $S^d(\mathbb{C}^2)$ implies irreducibility of all the others.

I'd be equally happy to prove indecomposability, since I've proved an indecomposable representation of a compact Lie group is irreducible.

答:
The action of $\mathrm{SL}(2,\mathbb{C})$ on $S^d(\mathbb{C}^2)$ preserves the image of the Veronese embedding, which is $\{ (\alpha x + \beta y)^d \mid \alpha,\beta \in \mathbb{C} \}$, and acts transitively on the nonzero elements. This set spans $S^d(\mathbb{C}^2)$.

The decomposition of a symmetric tensor into elements of this form ("rank 1" tensors) is the Waring decomposition. You can prove that this decomposition holds by using the Vandermonde determinant. Just choose any $d+1$ pairwise distinct elements of $\mathbb{C}$, say $\{\alpha_i\}$, and then $\{ (\alpha_i x + y)^d \}$ spans $S^d(\mathbb{C}^2)$. There are a few proofs of the Vandermonde determinant formula on the linked Wikipedia page, hopefully at least one of them is quick and elementary enough for your purposes.


问:
How to find the solution set for $\tan ^2 x= (\sqrt{2}-1)^2$?
The problem is as follows:

Find the set solution for the equation from below:

$\sqrt{2}-\tan (2x)=\cot \left(\frac{\pi}{4}+x\right)$

Well after doing all the necessary algebra I'm getting the following expression from below.

$\tan ^2 x= (\sqrt{2}-1)^2$

Well my intuition tells me that I have to take the inverse function for both sides of the equation not before taking the square rooth in both sides which will yield two solutions:

$\tan x= \pm \sqrt{2}-1$

But taking the inverse function will yield:

$x=\tan^{-1}\left(\sqrt{2}-1\right)$

But what it is in the brackets it is not an arc which I can remember. Could it be that I got something wrong here or what?.

Can someone help me here please?. I will really appreciate to know how exactly can I get the set solution for this.

答:
First, it should be $$\tan x = \pm \color{red}{(}\sqrt{2} - 1 \color{red}{)}.$$  Second, I think a better approach is to write instead
$$\tan^2 x = \frac{\sin^2 x}{\cos^2 x} = \frac{(1 - \cos 2x)/2}{(1 + \cos 2x)/2} = \frac{1 - \cos 2x}{1 + \cos 2x} = (\sqrt{2} - 1)^2.$$  Solving this equation for $\cos 2x$ yields $$\cos 2x = \frac{1}{\sqrt{2}},$$ hence $$2x = \frac{\pi}{4} + 2\pi k, \quad 2x = -\frac{\pi}{4} + 2\pi k, \quad k \in \mathbb Z,$$ giving $$x = \pm \frac{\pi}{8} + \pi k, \quad k \in \mathbb Z.$$  We can then verify this solution by direct substitution, noting that $\tan x$ is periodic with period $\pi$:
$$\tan^2 \left(\pm \frac{\pi}{8}\right) = 3 - 2\sqrt{2}.$$


问:
Show the operator norm of $A^T A - I_n$ can be bounded by $3\max(\delta, \delta^2)$

Let $A$ be an $m\times n$ matrix and $\delta>0.$ If all singular values of $A$ are between $1-\delta$ and $1+\delta$, $$1-\delta\leq s_n(A)\leq s_1(A) \leq 1+\delta,$$ prove $$\Vert A^TA-I_n\Vert \leq 3\max(\delta,\delta^2),$$where $s_1(A),\dots,s_n(A)$ are the singular values of $A$ and $s_1(A) \geq \cdots \geq s_n(A)$. Note that here the definition of the operator norm is $$\Vert A \Vert := \max_{x \in S^{n-1}\\ y \in S^{m-1}}\langle Ax,y\rangle=s_1(A).$$

So far, I have tried

$$\Vert A^TA-I_n\Vert = \Vert V(S^TS-I_n)V^T\Vert=s_1(A^TA-I_n).$$

But I am not sure how to connect with $3\max(\delta,\delta^2)$ . This is an extension exercise 4.1.6 from page 80 High-Dimensional Probability by Roman Vershynin. Thank you!

答:
We only need to prove for any $z \geq 0$,
\begin{equation*}
  \begin{split}
     1 -\delta \leq z \leq 1+\delta \Rightarrow |z^2 - 1| \leq 3\max(\delta,\delta^2).
  \end{split}
\end{equation*}
When $0<\delta \leq 1$, we have
\begin{equation*}
  \begin{split}
     &(1-\delta)^2 \leq z^2 \leq (1+\delta)^2 \\
     \Rightarrow & -3\delta \leq -2\delta \leq  -2\delta + \delta^2 \leq z^2 - 1 \leq 2\delta + \delta^2 \leq 3\delta \quad (\delta^2 \leq \delta)\\
     \Rightarrow & |z^2 - 1| \leq 3\delta = 3\max(\delta, \delta^2).
  \end{split}
\end{equation*}
When $\delta > 1$, we have
\begin{equation*}
  \begin{split}
     &1 -\delta < 0\leq  z \leq 1 + \delta \\
     \Rightarrow & 0 \leq z^2 \leq (1 + \delta)^2 \\
     \Rightarrow & -3\delta^2< -\delta^2 < -1 \leq z^2 - 1 \leq 2\delta + \delta^2 \leq 3\delta^2 \quad (\delta \leq \delta^2)\\
     \Rightarrow & |z^2 - 1| \leq 3\delta^2 = 3 \max(\delta,\delta^2).
  \end{split}
\end{equation*}


问:
Why does an exponential function eventually get bigger than a quadratic
I have seen the answer to this question and this one.

My $7$th grade son has this question on his homework:

How do you know an exponential expression will eventually be larger than any quadratic expression?

I can explain to him for any particular example such as $3^x$ vs. $10 x^2$ that he can just try different integer values of $x$ until he finds one, e.g. $x=6$. But, how can a $7$th grader understand that it will always be true, even $1.0001^x$ will eventually by greater than $1000 x^2$? They obviously do not know the Binomial Theorem, derivatives, Taylor series, L'Hopital's rule, Limits, etc,

Note: that is the way the problem is stated, it does not say that the base of the exponential expression has to be greater than $1$. Although for base between $0$ and $1$, it is still true that there exists some $x$ where the exponential is larger than the quadratic, the phrase "eventually" makes it sound like there is some $M$ where it is larger for all $x>M$. So, I don't like the way the question is written.

答:
This question motivates thinking about exponents and hence logarithms. But even without going into the definition of logarithms, we may consider relevant expressions of $x$-values for which the difference in rate of growth may become apparent. Consider:
$$
f(x)=a^x\\
g(x)=x^2
$$
and derive that
$$
f(2x)=a^{2x}=\left(a^x\right)^2=g(a^x)
$$
So now we can instead ask:

Does $f(2x)=g(a^x)$ eventually become larger than $g(2x)$?

So the problem now is reduced comparing the two inputs $a^x$ and $2x$ to the same function $g$ and showing that $a^x$ eventually becomes larger than $2x$ by any given factor one may wish to consider, however large.

As a corollary, this approach works for any degree polynomial, namely take $g(x)=x^n$ and consider $f(nx)=g(a^x)$ compared to $g(nx)$. Thus one compares an exponential input $a^x$ to a linear input $nx$.

Finally, in order to give the complete picture, why does $a^x$ exceed any linear term $kx$ eventually (given $a>1$)? Well, just consider:
$$
f(x+1)-f(x)=a^{x+1}-a^x=a^x(a-1)\\
g(x+1)-g(x)=k
$$
So now we only need to show that $a^x(a-1)$ eventually becomes larger than $k$ in order to see that an increase by $1$ in $x$ eventually will make $f$ grow at a faster rate than $g$. Clearly this expression $a^x(a-1)$ becomes larger an larger, and hence $f$ grows faster than a linear function with a greater slope than $g$ and will eventually catch up.


答:
Let $g(2), g(3) = b,c$ respectively.

Then

$b + c = g(4) = 1.$

$g(1) + b = 1 + b = g(2) = c.$

Solve for the two equations in two unknowns.

Title: If $g(n+1) = g(n) + g(n - 1)$, and $g(1) = g(4) = 1$, then what is $g(10)$?

Body: Let $g$ be a function such that $g(n+1) = g(n) + g(n - 1)$ for all integers $n$. If $g(1) = g(4) = 1$, then what is $g(10)$?


问:
Mirror property of angle bisector

In $\Delta ABC, A\equiv (1,2)$. If the angle bisector of $\angle B$ is $y=2x+10$ and perpendicular bisector of $AC$ is $y-x=0$, find the equation of $BC$

The solution given says that:

The image of point $A$ about the angle bisector of $C$ lies on $BC$.

The problem becomes easy after this since we now know $2$ points on the line $BC$(Another being $C$ itself)

Is the statement given in the answer a well known theorem/property? Im not exactly sure how I can prove its validity.

答:
Hint: As I said in comment the equation given for angle bisector of angle $B$ can be that of exterior, as can be seen in figure. We must use this fact that the intersection of perpendicular bisector of $AC$ and bisector of angle $B$ is on circumscribed circle of triangle. So we extend these two to intersect at $D (-10, -10)$ and draw a circle center at $O(-3, -3)$ and radius ($R=6.4$) passing $A,C$ and D. Vertex $B$ is on this circle. Vertex $C$ is the mirror of $A$ and has coordinates $C( 2, 1)$, the coordinates of $B$ is $B(-3.2, -3.4)$ so equation of $BC$ is:

$$y+3.4=\frac{1+3.4}{2+3.2}(x+3.2)$$

$$y=(\frac{4.4}{5.2}=\frac{11}{13})(x+3.2)-3.4$$

Title: Mirror property of angle bisector


问:
Are these boundary conditions consistent $y(x,0)=\cos(\pi x)$ and $y(0,t)=0$
I was asked to solve
$$\dfrac{\partial^2 y}{\partial t^2}=9\dfrac{\partial^2 y}{\partial x^2}$$
with the boundary conditions
$y(x,0)=\cos(\pi x)$,$\dfrac{\partial y}{\partial t}(x,0)$ and $y(0,t)=y(4,t)=0$

Aren't these boundary conditions inconsistent
since if $y(0,t)=0$ (for all $t$) this would imply $y(0,0)=0$
on the other hand $y(x,0)=\cos(\pi x)$ (for all $x$) this would imply that $y(0,0)=1$
Am I wrond somewhere or are these boundary condtions indeed inconsistent.

答:
These boundary conditions do not inconsistent despite the fact that the conditions of agreement in the corner are not met. This implies that your problem does not have a classic solution. Moreover it does not have continuous solution. Your solution will not have a generalized derivative everywhere (but will have derivative almost everywhere), since there will be a discontinuity of the first kind.

You can solve this problem using separation of variables (in Hilbert space), and solution will be represented as trigonometric series, which will correspond to a distribution. The derivatives of these series may not converge. But on some domains Cesàro summation can give meaning to these series. Nevertheless, on some characteristics, the Cesàro summation will not help, because there the dirac delta will appear.

Edit 1. Building a solution.

Separation of variables: $u(x, t) = X(x)T(t)$.

$X''(x) + \lambda X(x) = 0$, $X(0) = X(4) = 0$.

$T''(t) +9\lambda T(t) = 0$.

From 1) we get $X_n(x) = \sin\left(\frac{\pi n x}{4}\right)$, $\lambda_n = (\frac{\pi n}{4})^2$.
From 2) we get $T_n(t) = A_n \cos\left(\frac{3 \pi n t}{4}\right) + B_n\sin \left(\frac{3 \pi n t}{4}\right)$.

So, $u(x, t) = \sum\limits_{n=1}^\infty \left(A_n \cos\left(\frac{3 \pi n t}{4}\right) + B_n\sin\left( \frac{3 \pi n t}{4}\right)\right) \sin\left(\frac{\pi n x}{4}\right)$.

Using initial condition we obtain $A_n=\frac{1}{2}\int\limits_0^4 \cos (\pi x) \sin\left(\frac{\pi n x}{4}\right) dx$ and $B_n=0$.
We can calculate integral and get $A_n = -\frac{2 \left((-1)^n-1\right) n}{\pi
   \left(n^2-16\right)}$ for $n\neq 4$, and $A_4 = 0$.
We notice, that $A_{2m} = 0$ and $A_{2 m + 1} = \frac{8 m+4}{\pi  \left(4 m^2+4
   m-15\right)}$.

So, we have built the solution as trigonomical series $u(x, t)=\sum\limits_{m=0}^\infty \frac{8 m+4}{\pi  \left(4 m^2+4
   m-15\right)} \cos\left(\frac{3 \pi t (2 m + 1)}{4}\right) \sin\left(\frac{\pi x (2 m + 1)}{4}\right)$.


问:
Given an semi-ellipse inscribed about a square, how do I find the equation of the ellipse?
Given the following diagram:

Where:

W = (-1, 0)

X = (-1, 2)

Y = (1, 2)

Z = (1, 0)

How can I find M?

The ellipse can be assumed to be a semi-ellipse with one of the foci on $\bar{XY}$. I'm guessing that this means that one focus is at (0, 2), with the other focus at (0, -2). Now, by the definition of an ellipse, I know that the sum of the distances from those two points to any other point on the ellipse is a constant. But, having reasoned that far, I've hit a dead end. Where do I go from here?

答:
The equation of an ellipse in standard form is
$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$.

If it passes through $(1, 2)$,
then
$\frac{1}{a^2} + \frac{4}{b^2} = 1$.

However, this is one equation in two unknowns
($a$ and $b$).
To determine the unknowns,
you need another condition.

To see this visually,
imagine M getting closer to Z.
Then the upper vertex of the ellipse will move higher,
and similarly as M moves further from Z.