问:
Show that a special sequence converge
How can I prove that $a_n = \sum_{i=1}^n \frac{1}{\sqrt{n + i}}$ converge. I know that if a sequence $(b_n)$ converge to $b$, then $(\sqrt{b_n})$ converge to $\sqrt{b}$ and $\frac{n}{\sqrt{2n}} \leq a_n \leq \frac{n}{\sqrt{n + 1}}$, but I am not sure how to finish this. Can you help?

答: You're correct in your approach. Just observe that both $\frac{n}{\sqrt{n^2 + n}}$ and $\frac{n}{\sqrt{n^2 + 1}}$ converge to $1$ (substitute $n \rightarrow \frac{1}{x}$) and use sandwich theorem.

Title: Show that a special sequence converge


问:
Riemannian metric for surface of negative Euler characteristic
So, I want to equip a surface of negative Euler characteristic with a
Riemannian metric of negative curvature.

I know from the uniformization theorem, that a metric of constant curvature exists

Now, if M is compact (for example a sphere with a finite number > 2 of puncture points):

-I know from Gauss-Bonnet, that this metric has to be of negative curvature

If M is not compact:

-the curvature can't be $> 0$ because the universal covering of M is not the sphere
(because the sphere is compact)

Is there a way to rule out the possibility that the metric is of constant curvature = $0$ (flat)

答: The Gauss-Bonnet theorem applies equally well in the case of a complete metric of finite area.  Therefore negativity of Euler characteristic implies negative sign of curvature.</p>

ContentLicense: CC BY-SA 3.0


问:
How to find if two series are analytic continuations of each other
Show that the series $a) \sum_{n=0}^{\infty} \frac{z^n}{2^{n+1}}$ and $b) \sum_{n=0}^{\infty} \frac{(z-i)^n}{(2-i)^{n+1}}$ are analytic continuations of each other.

I have performed the ratio test for both series and found that $a)$ converges for $|z|<1$ and $b)$ for $|\frac{z-i}{2-i}|<1$. I have a hint that says $b)$ also converges for $|z-i|< \sqrt{5}$.

I am not sure where the second convergence of $b)$, ($|z-i|< \sqrt{5}$), comes from.
I am also not really sure what to do next. My notes are very vague. Wouldn't the convergence of $|\frac{z-i}{2-i}|<1$ not be an analytic continuation since $1 \ngtr 1$. But the second convergence would.

I don't know if I am saying that right,  but that is what I am getting from my notes.

答:
For $|z|<2$, the series $\sum_{n=0}^\infty \frac{z^n}{2^{n+1}}$ represents the function $f(z)= \frac1{2-z}$. Then, note that we can write for $|z-i|<|2-i|=\sqrt 5$

$$\begin{align}
f(z)&=\frac{1}{2-z}\\\\
&=\frac{1}{(2-i)-(z-i)}\\\\
&=\frac1{2-i}\left(\frac1{1-\frac{z-i}{2-i}}\right)\\\\
&=\sum_{n=0}^\infty \frac{(z-i)^n}{(2-i)^{n+1}}
\end{align}$$

Hence, we see that $f(z)=\sum_{n=0}^\infty \frac{(z-i)^n}{(2-i)^{n+1}}$ for $|z-i|<\sqrt 5$ is indeed the analytic continuation of $f(z)=\sum_{n=0}^\infty \frac{z^n}{2^{n+1}}$ for $|z|<2$.

ContentLicense: CC BY-SA 4.0


问:
Determining convergence of $\frac{(x^2)}{x!}$
I'm trying to determine the convergence of the infinite series

$$\frac{x^2}{x!}$$

The factorial clues me in to using the ratio test, but I get stuck when trying to cancel.

答:
Just for your curiosity.

$$S=5\sum_{n=0}^\infty \frac{n^6}{(3n)!}x^n$$ can express in terms of hypergeometric functions (do not worry : you will learn about them quite soon).

Making $x=1$ gives the nice
$$S=\frac{5 }{2187 \sqrt{e}}\left(203 e^{3/2}-20 \sqrt{3} \sin \left(\frac{\sqrt{3}}{2}\right)+70 \cos
   \left(\frac{\sqrt{3}}{2}\right)\right)$$ which is
$$1.28786523601314644544347506207\cdots$$


问:
From the set $\mathbb{Z}_8=\mathbb{Z}/8\mathbb{Z}$ select all elements for which there exists an inverse element with respect to multiplication
I have a misunderstanding with what it means
select all elements for which there exists an inverse element with respect to multiplication
does it mean that an element from $\{0,1,2,3,4,5,6,7\}$ and for example $4\times2=8$ therefore it's the identity element? but doesn't that also apply for $4\times0$ for exapmle? or do i have to consider the elements $\{0,1,2,3,4,5,6,7\}$ as just numbers and $a^{-1}=\frac{1}{a}$?

so to be more precise i have difficulties understanding the set of elements that has an inverse element with respect to multiplication from the set $\mathbb{Z}/8$

I would appriciate any help!
Thanks in advance!

答: You want to determine the elements $x$ for which $\gcd(x,8)=1.$ For such $x$ there are integers $a,b$ for which $ax+8b=1,$ so that $a$ is an inverse for $x$ mod $8.$ More precisely, the element in $\mathbb{Z}/8\mathbb{Z}$ which is congruent to $a$ mod $8$ is the inverse you want.


问:
If we transform a set of unit vectors in $R^n$ using some $n \times n$ permutation matrix, is the centroid also transformed by the permutation matrix?
Given a set of unit vectors $V$ in $R^n$, the set's centroid $c^V$, and some $n \times n$ permutation matrix $P$, we define $V'=\{Pv~|~\forall v \in V\}$.

Is it true that $c^{V'} = Pc^V$?

If so, can we generalize $P$ to any $n \times n$ matrix $A$?

答: $P$ performs a linear transformation. The centroid is a linear combination of the points. Linear transformations preserve linear combinations, by definition. Hence the centroid also transforms via $P$.


问:
Prove the convergence of $x_{n+1}=\sqrt{\alpha +x_n}$, with $x_0=\sqrt \alpha$ using fixed point theorem
Consider the well known sequence $x_{n+1}=\sqrt{\alpha +x_n}, x_0=\sqrt \alpha, \alpha>0$.

How would one use the fixed point theorem to prove that it converges to $\sqrt{a} \ \ \ \forall a>0?$

Trying to see whether $f(x) = \sqrt{\alpha+x}$ is a contructor yields following:
$$|f(x)-f(y)|=|\frac{1}{2\sqrt{\alpha+\xi}}| |x-y| \le \frac{1}{2\sqrt{\alpha}} |x-y|,$$
which suggests that $f(x)$ is a contractor only when $\sqrt{\alpha} > \frac{1}{2} $ or $\alpha>\frac{1}{4}$.  But this is obviously false. What's wrong here?

答:
The derivative of $f(x) =\sqrt{\alpha + x}$ is $\frac{1}{2\sqrt{\alpha + x}}$.  If $x \geq 1$, then $$\frac{1}{2\sqrt{\alpha + x}} \leq \frac{1}{2\sqrt{\alpha + 1}} \leq \frac 1 2.$$

This shows that $f(x) = \sqrt{\alpha + x}$ is a contraction on $[1,\infty)$, so there is a fixed point in $[1,\infty)$.

If $x < 1$, since the sequence defined by iterating $f$ is nondecreasing, either it converges to a point below 1 or eventually it ends up in $[1,\infty)$.  In either case it is convergent.


问:
Image of a group equal to his kernel
If we have a function $f: G\rightarrow G$ , is there an example of a group $G$ that $imf=kerf$?

I try to think about the trivial group but there is another one?

Note: $f$ is Homomorphism.

答:
All you are essentially asking for is a homomorphism $f: G\rightarrow H$ such that $\operatorname{im}(f)\cong\ker(f)$. To see this, alter the codomain of $f$ to make $f$ surjective, so $f: G\rightarrow\operatorname{im}(f)$, and then compose this with the isomorphism $\phi:\operatorname{im}(f)\rightarrow\ker(f)$ to get a map $\phi f$ which, after altering the codomain again, has the required properties.

Asking for homomorphisms with isomorphism $\operatorname{im}(f)\cong\ker(f)$ rather than equality $\operatorname{im}(f)=\ker(f)$ is easier but, as the above says, equivalent.

For example, we can see that any group of order $p^2$ gives us an example: such a group $G$ has a normal subgroup $N$ of order $p$ (why?), and then the quotient group $G/N$ also has order $p$. If $f: G\rightarrow G/N$ is the associated homomorphism, we have $\operatorname{im}(f)\cong\ker(f)$ (equivalently, $N\cong G/N$), as up to isomorphism there is a single group of order $p$.

In particular, this means that cyclic groups of order $p^2$ give examples, which is a different class of examples from the other two answers.

Title: Image of a group equal to his kernel


问:
Approximate: $\sum_{\Im(\rho)>0}\frac{1}{|\rho-\frac{1}{2}|^2}$ where $\rho$ is non trivial zeros of the zeta function
I am reading Kevin Broughan's Equivalents of Riemann Hypothesis Vol. 1 (p. 38). If $\rho$ denote the non trivial zeros of Riemann zeta function in the strip $0<\Re(\rho)<1$, consider,$$S=\sum_{\Im(\rho)>0}\frac{1}{|\rho-\frac{1}{2}|^2}.$$
Question: Find the approximate value of $S$.

Attempt:
Riemann $\xi$ function can be written as
$$\xi(s)=\xi(0)\prod_{\Im(\rho)>0}(1-\frac{s}{\rho})(1-\frac{s}{\bar{\rho}}),$$
where $\bar{\rho}$ denotes complex conjugate.

Taking logarithm on both sides,

$$\log(\xi(s))=\log(\xi(0))+\sum_{\Im(\rho)>0}\log(1-\frac{s}{\rho})+\sum_{\Im(\rho)>0}\log(1-\frac{s}{\bar{\rho}})$$

and differentiating both sides w.r.t. $s$,

$$\frac{\xi'(s)}{\xi(s)}=\sum_{\Im(\rho)>0}[\frac{1}{s-\rho}+\frac{1}{s-\bar{\rho}}].$$

After this I cannot think on it.
Please find an approximate value of $$S=\sum_{\Im(\rho)>0}\frac{1}{|\rho-\frac{1}{2}|^2}.$$
Edit
$$\xi(s)=\xi(0)\prod_{\rho}(1-\frac{s}{\rho})$$
Taking log we get
$$\log(\xi(s))=\log(\xi(0))+\sum_{\rho}\log(1-\frac{s}{\rho})$$
Differentiating w.r.t. s,
$$\frac{\xi'(s)}{\xi(s)} = \sum_{\rho}\frac{1}{s-\rho}$$
Differentiating w.r.t. s again,
$$\frac{d}{ds}\frac{\xi'(s)}{\xi(s)} =- \sum_{\rho}\frac{1}{(s-\rho)^2}$$
Putting $s=1/2$ and using $\xi'(1/2)=0$,
$$\frac{\xi''(\frac{1}{2})}{\xi(\frac{1}{2})}=-\sum_{\rho} \frac{1}{(\frac{1}{2}-\rho)^2}$$

答:
A comment that got too long:

Note that $S=\sum_{\Im(\rho)>0}\frac{1}{|\rho-\frac{1}{2}|^2}=\frac{1}{2}\sum_{\rho}\frac{1}{|\rho-\frac{1}{2}|^2}$ since for conjugate roots $|\rho-1/2|=|\bar \rho -1/2|$ and the results of Broughan below involve the full sum

$2S$ is not that different from $\sum_{\Im(\rho)>0}\frac{1}{|\rho|^2}$ which is estimated in Broughan by $0.04619..$, so an estimate of $0.0463$, for example, is probably very easy (leave details to OP as I am not that fond of pointless numerical calculations - note that )

Up to some height, $H$ (see Broughan who takes $H=600,269.6771$) where all the zeroes are on the critical line the difference is just the sum of $\frac{1}{4t^2(t^2+1/4)} \le \frac{1}{4t^4}$ where those roots are $1/2+it$ and we know that $t$ increases quite fast starting at $14+$ so $1/(4t^4)$ starts at something lower than $0.00001$ and decreasing fast.

So if one needs better numerics take the table of zeroes of $\zeta$ up to the Broughan range $H$ above and compute $\sum 1/(4t^4), \rho=1/2+it, 14 < t \le H$)

Beyond that range, $S$ and the Broughan sum part differ by at most a factor of $1-\frac{1}{H}$ and that is fairly negligible not to speak that the part of either sum beyond that range is really small.


答:
Observe that $a_n = \sum_{i=1}^n \frac{1}{\sqrt{n^2+i}} \leq \sum_{i=1}^n \frac{1}{\sqrt{n^2+1}} = \frac{n}{\sqrt{n^2+1}}$. In the same way you will find that $a_n = \sum_{i=1}^n \frac{1}{\sqrt{n^2+i}} \geq \sum_{i=1}^n \frac{1}{\sqrt{n^2+n}} = \frac{n}{\sqrt{n^2+n}}$

Therefore your idea is correct and also Kaind's idea to use the sandwich theorem is good. The only thing is that both sequences converge to $1$, not to $0$. For the left one you could use the fact that $\sqrt{n^2+n} \leq n + \sqrt{n}$ and then you get
$$\frac{n}{\sqrt{n^2+n}} \geq \frac{n}{n + \sqrt{n}} = \frac{1}{1+\frac{1}{\sqrt{n}}}$$
and the last fraction obiously converges to $1$.
For the other one we find that
$$\frac{n}{\sqrt{n^2+1}} \leq \frac{n}{\sqrt{n^2}} = 1$$
So with the sandwich theorem we get $a_n \to 1$ as $n \to \infty$.

Title: Show that a special sequence converge

Body: How can I prove that $a_n = \sum_{i=1}^n \frac{1}{\sqrt{n + i}}$ converge. I know that if a sequence $(b_n)$ converge to $b$, then $(\sqrt{b_n})$ converge to $\sqrt{b}$ and $\frac{n}{\sqrt{2n}} \leq a_n \leq \frac{n}{\sqrt{n + 1}}$, but I am not sure how to finish this. Can you help?


问:
Show $\mathcal{R}(A) \supseteq \mathcal{R}(B) \Leftrightarrow \mathrm{rank} A = \mathrm{rank} (A | B)$
I need help to prove:

Given two matrices $A \in \mathbb{R}^{m \times n}$ and $B \in \mathbb{R}^{m \times k}$. The following equivalence is true:
$$
\mathcal{R}(A) \supseteq \mathcal{R}(B)
\quad \Leftrightarrow \quad
\mathrm{rank} A = \mathrm{rank} (A | B).
$$
$\mathcal{R} (\cdot)$ denotes the image of a matrix and $(A | B)$ is a matrix with the columns of $A$ and $B$ next to each other.

I don't really know how to start with the proof. Could someone give me a hint?

答:
$rank(A)$ is the dimension of $\mathcal{R}(A)$, which is the span of the columns of $A$. Since $\mathcal{R}(B)\subset\mathcal{R}(A)$, the columns of $B$ are in the span of the columns of $A$, so the dimension of the column space does not increase when you attach $B$ to $A$. Hence the rank does not change.

The reasoning can be reversed. The rank is unchanged as long as you add columns which are linear combination of the previous ones.


问:
Prove that $(I_n-A)^{-1}$ exists
Let $A \in M_{nxn}(\mathbb{C})$, and assume $A^5=\mathbb{O}_{nxn}$ (A zero matrix). Prove that $(I_n-A)^{-1}$ exists and what does it equal to? So the property that $A^5=\mathbb{O}$ is useful here, and to prove that the inverse exists, should I use the determinant method (calculate that the determinant cannot be $0$?). To find what it is equal to, I have an intuition that it is the identity matrix + $A^1 + A^2$ and such. What would be the exact answer?

Title: Prove that $(I_n-A)^{-1}$ exists

Body:
\begin{align*}
(I - A^5) &= (I - A)(I + A + A^2 + A^3 + A^4) \\
\Rightarrow |(I - A^5)| &= |(I - A)||(I + A + A^2 + A^3 + A^4)| \\
\Rightarrow 1 &= |(I - A)||(I + A + A^2 + A^3 + A^4)| \\
\Rightarrow |(I - A)| &\neq 0 \\
\end{align*}
Qed.


答:
Let $A \in \mathcal{M}_{n}(\Bbb{C})$ be a $n \times n$ matrix such that $A^5 = \mathcal{O}_n$. Note that

\begin{align*}
(I_n - A)(I_n + A + A^2 + A^3 +A^4) & = I_n + A + A^2 +A^3 + A^4 - A - A^2 - A^3 -A^4 - A^5 \\
& = I_n + A^5 \\
& = I_n + \mathcal{O}_n \\
& = I_n
\end{align*}

Then, $I_n - A$ is invertible, and $A^{-1} = I_n + A + A^2 + A^3 + A^4$.

Title: Prove that $(I_n-A)^{-1}$ exists

Body: Let $A \in M_{nxn}(\mathbb{C})$, and assume $A^5=\mathbb{O}_{nxn}$ (A zero matrix). Prove that $(I_n-A)^{-1}$ exists and what does it equal to? So the property that $A^5=\mathbb{O}$ is useful here, and to prove that the inverse exists, should I use the determinant method (calculate that the determinant cannot be $0$?). To find what it is equal to, I have an intuition that it is the identity matrix + $A^1 + A^2$ and such. What would be the exact answer?


问:
List the set of points of discontinuity of piecewise function
List the set of points of discontinuity of $f:(0,\infty)\to\mathbb R$, defined by

$$f(x)=\begin{cases}x-[x]\text{ if [x] is even}\\1-x+[x]\text{ if [x] is odd}\end{cases}$$

Body: We see $$x\in(0,1), \text{floor}(x)=0 ~~~~~\text{so}~~~f(x)=x$$ $$x\in[1,2), \text{floor}(x)=1 ~~~~~\text{so}~~~f(x)=2-x$$ $$x\in[2,3), \text{floor}(x)=2 ~~~~~\text{so}~~~f(x)=x-2\\\ .\\\ .\\\ .$$ It seems the function is continuous.


问:
differentiate with respect to a function
Let's say I have this function $f(x)=x$. I want to differentiate with respect to $x^2$. So I want to calculate $\large\frac{df(x)}{dx^2}$. In general, how can I calculate the derivative of a function $f(x)$ with respect to a function $g(x)$, so $\large\frac{df(x)}{dg(x)}$? </p>

<p>(I dont know whether this is a good notation)?</p>

答:
So it's true that the correct answer is that if we want to differentiate f(x) with respect to another g(x) that we take $$\frac{f'(x)}{g'(x)}$$, but I think just calling this chain rule cancellation is not exactly correct (though it is both intuitive and also gives the right answer... lol).

To formalize things, let's say we have (1) $$f(t) = h(t,g(t))$$, which for simplicity in applying the multivariate chain rule we can write as $$f(t) = h(x(t), y(t))$$

So then (2) $$\frac{df}{dy} = \frac{dh}{dx}\frac{dx}{dy} + \frac{dh}{dy}\frac{dy}{dy}$$. We know that $$\frac{dx}{dy} = \frac{dt}{dg} = \frac{1}{\frac{dg}{dt}} = \frac{1}{g'(t)}$$, by the inverse derivative theorem, and we know $$\frac{dy}{dy} = 1$$.

So in (1) if we multiply $$\frac{dh}{dy}$$ by $$1  = \frac{g'(t)}{g'(t)}$$, we get (3) $$\frac{df}{dy} = \frac{\frac{dh}{dx}*1 + \frac{dh}{dy}\frac{dy}{dt}}{g'(t)}$$, since y=g, and then since x=t, we can write $$1 = \frac{dx}{dt}$$, which gives us

$$\frac{df}{dy} = \frac{\frac{dh}{dx}*\frac{dx}{dt} + \frac{dh}{dy}\frac{dy}{dt}}{g'(t)}$$, which is just $$\frac{f'(t)}{g'(t)}$$.


问:
Are all $*$-algebras with matrix representations projective?
If a complex unital $*$-algebra $\mathscr{A}$ has a (bounded) matrix representation (i.e. a unital $*$-homomorphism) $$\psi:\mathscr{A}\rightarrow M_d(\mathbb{C}),$$ such that $\|\psi(a)\|_{\infty}<+\infty$ for all $a\in \mathscr{A}$, does $\psi$ necessarily factor through the $C^*$-enveloping algebra of $\mathscr{A}$, and does this mean every such $\mathscr{A}$ with this property is a projective object in the category of unital $*$-algebras?

答:
This is trivial from the universal property of the $C^*$-enveloping algebra: $M_d(\mathbb{C})$ is a $C^*$-algebra, so any homomorphism to it factors through the $C^*$-enveloping algebra.

This doesn't have anything to do with being projective, though.  For a simple example, let $\mathscr{A}=M_2(\mathbb{C})\times M_3(\mathbb{C})$.  This has matrix representations, but it is not projective: there is an epimorphism $\mathscr{A}\to M_2(\mathbb{C})$ (the first projection) which does not split since there are no homomorphisms $M_2(\mathbb{C})\to M_3(\mathbb{C})$.


问:
Why $L$ is the eigenspace of $L_A$?
$A=\frac{1}{\sqrt{5}}\begin{pmatrix}  1&2\\2&-1 \end{pmatrix}$

Let $L_A$ be a reflection of $R^2$ about a line $L$ through the origin.

Then $L$ is the one dimensional eigenspace of $L_A$ corresponding to the eigenvalue 1.

I know the eigenvalues are 1,-1. But why it should be 1?

答: Because having an eigenvalue 1 in a direction means that vectors in that direction are not changed, so since your reflection has a unique fixed line, it must be in the eigendirection corresponding to 1.</p>

Title: Why $L$ is the eigenspace of $L_A$?


答:
Alternative approach.

As $n \to \infty$ the geometric mean of $(n!)$ goes to $\frac{n}{e}$.

Therefore, the denominator will go to $\left(\frac{3x}{e}\right)^{(3x)} \approx x^{(3x)}$, since $3 \approx e$.

Therefore, you are (approximately) evaluating the

$\displaystyle \lim_{x\to\infty} \frac{5x^6}{x^{(3x)}} = 0$.

The easiest way to see this is to compare the two exponents, $(6)$ and $(3x)$.

Title: Determining convergence of $\frac{(x^2)}{x!}$

Body:
I'm trying to determine the convergence of the infinite series

$$\frac{x^2}{x!}$$

The factorial clues me in to using the ratio test, but I get stuck when trying to cancel.


答:
If $A^5=0_n$ then $\mu\mid x^5$ where $\mu\in\mathbb{C}[x]$ is the minimal polynomial of $A$. Then $\mu=x^t$ for some $t\in\{1,2,3,4,5\}$. It is known that the minimal polynomial share the same roots as the characteristic polynomial of $A$. Hence $\left|xI_n-A\right|=x^n$ and $\left|I_n-A\right|=1$.

There is a result, easy to prove, garanting us that
$$\sum_{k=0}^\infty{B^k}=(I_n-B)^{-1}$$
whenever $\rho(B)<1$ where $\rho(B)$ is the spectral radius of $B$.

Using that in our case, since $\rho(A)=0$, we have the following:
$$(I_n-A)^{-1}=\sum_{k=0}^4{A^k}.$$

Title: Prove that $(I_n-A)^{-1}$ exists


问:
Epsilon delta proof questions
Hi and thanks for reading. I am not new to, but have almost no hand on experience with, epsilon delta proofs. I have an example I would like to confirm is correct, which is as follows.

Question

Prove that $\lim_{x \to 3} x^2 = 9$.

Answer

We need to show that, for every $\epsilon > 0$, there exists a $\delta > 0$ such that $|x-3|<\delta \implies |x^2 - 9|<\epsilon$. Now,

$\begin{aligned}|x-3|<\delta &\implies -\delta < x-3< \delta \\ &\implies -\delta+6<x+3<\delta+6 \\ &\implies -\delta - 6 < x+3 < \delta + 6 \\ & \implies |x+3| < \delta + 6 \\ &\implies |x-3| |x+3| < \delta (\delta +6) = \delta^2 + 6\delta \\ &\implies |x^2 - 9| < \delta^2 + 6\delta\end{aligned}$

So, if we let $\epsilon = \delta^2 + 6\delta$, we have proven the limit.

I have used Excel with different values of x and different values of delta to check this, and it seems to be true. I'd just like to check I haven't made any careless errors. I also see many people "proving" limits the wrong way round, ie assuming epsilon to be given and working backwards. This would be fine if every step was an equivalence, but if not then the proof fails. Is this not correct? I have even seen textbooks using this method and it seems very counter intuitive to me.

答:
Given $\varepsilon>0$, you are supposed to find a $\delta>0$ such that$$|x-3|<\delta\implies|x^2-9|<\varepsilon.\tag1$$You did not do that, although it follows from what you did that $\delta=\sqrt{9+\varepsilon}-3$ will work.

On the other hand, note that, if $|x-3|<1$,\begin{align}|x^2-9|&=|x-3||x+3|\\&<|x-3|\bigl(|x-3|+6\bigr)\\&<7|x-3|.\end{align}So, take $\delta=\min\left\{1,\frac\varepsilon7\right\}$, and then $(1)$ will hold.

Title: Epsilon delta proof questions


问:
Random Markov Field, Mean-Field inference and potential functions... perplexities
I'd need help in understanding some concepts presented on a paper that I've started to read recently.

The general setting is about seeing graph structures from a probabilistic persepctive, in particular:

Let $\mathcal{G} = (V,\mathcal{E})$ be a graph with a multi-variate random variable $\textbf{X}_i$ associated to each node $i \in V$, and our goal is trying to infer some latent variables $\textbf{Z}_i$ given the initial observed variables $\textbf{X}_i$. The graph is said to be defining a Markov Random Field, writing

\begin{equation}
p(\{\textbf{z}_i\}, \{\textbf{x}_i\}) \propto \prod_{i \in V}\Phi(\textbf{x}_i,\textbf{z}_i)\prod_{(i,j) \in \mathcal{E}}\Psi(\textbf{z}_i,\textbf{z}_j)
\end{equation}

where I have used the shorthand $\{\textbf{x}_i\}$ for $\{\textbf{x}_i \, , \forall i \in V\}$, and $\Phi,\,\Psi$ are so-called non-negative Potential Functions.

It is also stated that $\Phi(\textbf{x}_i, \textbf{z}_i)$ works as the 'likelihood' of a node feature vector $\textbf{x}_i$ given its latent representation $\textbf{z}_i$, while $\Psi$ controls the dependency between connected nodes.

I just wanted to ask if any one could explain me a little better the roles of these functions $\Phi, \Psi$, why the first would act as a likelihood function, and also the overall perspective lying at the bottom of the above equation that gives a factorization of the joint distribution.

Thanks!

James

答: Informally, there are two sources of information for $z$, the measurements $x$ and the "opinions" of the neighbors. Given the true value of $z$ we may have a model of the probability to see a measurement $x$. This model can be also interpreted in the other direction: given a measurement $x$, what can be said about the true value $z$. This is called the likelihood to stress that it is not a probability. The $\Phi$ function weights $z$ given $x$, so it acts like a likelihood.


问:
Find the determinant of a "piecewise" matrix
Let $n$ be a positive integer, and $A_n$ is a $n \times n$ matrix, where the $(i,j)^\text{th}$ entry is $1$, if $i \leq j$ and it is $-1$ otherwise. What is the determinant of $A_n$?

My first thought is that it's similar to a piecewise function, and I found out that generally it's a weird "upper triangular" shape, where the diagonal and the lower part of the matrix is $1$, and the upper part of the matrix is $-1$. Where should I proceed? I plan on doing some row manipulation and maybe have a matrix in RREF, which is easy since it's all just one. Should the determinant be $1$?

答:
Let $A$ be the $n \times n$ matrix given by
\begin{align*}
a_{ij} = \begin{cases} 1 & \text{if } i \leq j \\ -1 & \text{otherwise} \end{cases}.
\end{align*}
Then, $\det{A} = \det{B}$ where the matrix $B$ is such that
\begin{align*}
b_{ij} = \begin{cases} 1 & \text{if } i = 1 \\ 2 & \text{if } i \neq 1 \wedge i \leq j \\ 0 & \text{otherwise} \end{cases}.
\end{align*}
This is true because you can add the first row to all the other $n-1$ rows and the determinant will be the same. Next, note that $B$ is a triangular matrix and the entries of the diagonal are such that
\begin{align*}
b_{ii} = \begin{cases} 1 & \text{if } i = 1 \\ 2 &\text{otherwise} \end{cases}.
\end{align*}
Therefore,
\begin{align*}
\det{A} & = \det{B} \\
& = 1 \times \prod_{i=2}^n 2 \\
& = 2^{n-1}. & \square
\end{align*}


答:
I think it would be good to do row manipulations. If I am not wrong, the upper part of the matrix (including the diagonal) has the positive $1$ elements. Then we can add the first row to the $i$-th row for $i=2, \dots, n$. We will get a matrix of following form:
$$\begin{pmatrix}
1 & 1&1&...&1\\
0 & 2&2&...&2\\
\vdots & \ddots& \ddots &...&\vdots\\
0 & 0&0&...&2\\
\end{pmatrix}$$
These row manipulations don't change the determinant and the determinant of the resulting matrix is obviously $\det(A) = 2^{n-1}$.

Title: Find the determinant of a "piecewise" matrix

Body:
Let $n$ be a positive integer, and $A_n$ is a $n \times n$ matrix, where the $(i,j)^\text{th}$ entry is $1$, if $i \leq j$ and it is $-1$ otherwise. What is the determinant of $A_n$?

My first thought is that it's similar to a piecewise function, and I found out that generally it's a weird "upper triangular" shape, where the diagonal and the lower part of the matrix is $1$, and the upper part of the matrix is $-1$. Where should I proceed? I plan on doing some row manipulation and maybe have a matrix in RREF, which is easy since it's all just one. Should the determinant be $1$?


问:
Rooms and people expected value
I have $x$ number of rooms. I have $7x$ people.

One by one, each person randomly and independently chooses a room to enter. Multiple people can choose the same room.

What's the expected number of rooms that are never entered? What's the expected number of rooms that are entered exactly $y$ times such that $1 \leq y \leq x - 1$?

答:
The expected value of a certain room be empty is the probability of this$$
\left( 1-\frac1x\right)^{7x}.
$$
By linearity of expectation the expected number of empty rooms is
$$x\left( 1-\frac1x\right)^{7x}.
$$

Title: Rooms and people expected value


问:
How to show that $\prod_{p\text{ prime}} \frac{p}{p-1}$ diverges
I was just reading a Wikipedia article regarding the existence of infinitely many primes in certain infinite arithmetic progressions, and I read something interesting- that Euler had once discovered (175) the fact that

$$\sum_{n=1}^{\infty}\frac{1}{n}=\zeta(1)=\prod_{p \text{ prime}}\frac{p}{p-1}$$

and therefore that the latter diverges to infinity. My question is as follows:

Does there exist an elementary proof that $$\prod_{p \text{ prime}}\frac{p}{p-1}$$ diverges without using the fact that $$\sum_{n=1}^{\infty}\frac{1}{n}$$ diverges?

Expecting that this problem would be on MSE already, I searched for this problem, but did not find it. Nevertheless, I would not be surprised to find that this question already does exist here in some crevasse I failed to check. In that case, of course feel free to let me know.

答:
There are lots of elementary proofs of this fact, and indeed the stronger "Mertens theorem" that gives the exact asymptotic rate of divergence; there is also "Chebyshev's theorem" that gives a lower bound on the number of primes up to $x$ that's strong enough to imply this divergence via partial summation.

However, in most of these methods, some formula of the form $\sum_{n\le x} \frac1n = \log x + O(1)$ is bound to be used multiple times; and that formula is a stronger statement then the mere divergence of the harmonic series.

In the end, the divergence of $\sum_{n=1}^\infty \frac1n$ follows immediately from the elementary inequality $\sum_{n=1}^N \frac1n > \int_1^{N+1} \frac1x\,dx = \log(N+1)$; so I'm not sure that trying to avoid this fact is particularly natural.


问:
Is there a recursively enumerable subset of true arithmetic that is not density 0?
Let $S$ denote all sentences over the signature of $PA$ and $Th(\mathcal{N}) \subseteq S$ denote the theory of true arithmetic. Next, pick your favorite computable bijection $f : S \to \mathbb{N}$. Can there be a set $A \subseteq f(Th(\mathcal{N}))$ that is both recursively enumerable and has nonzero lower asymptotic density? Additionally, does the answer depend on $f$?

I highly suspect there can be no such $A$ for every $f$, but when choosing an appropriate $f$, one might be able to pick such a set consisting only of tautologies/otherwise "trivially true" statements.

答:
This depends wildly on $f$. For example:

Fix some coinfinite recursive $X\subseteq\mathbb{N}$ with asymptotic density $1$ (say, the non-powers of $2$). Now letting $x_i$ denote the $i$th element of $X$ in increasing order, take some recursive $f$ such that $f^{-1}(x_i)=$ "$\underline{i}=\underline{i}$" (where $\underline{n}$ is the numeral corresponding to $n$). This gets a recursive set of true sentences with asymptotic density $1$ under $f$.

Alternately, choosing a recursive $f$ so that $f^{-1}(x_i)=$ "$\neg(\underline{i}=\underline{i})$" produces the opposite effect: the set of all true sentences has asymptotic density $0$ under $f$.

The point is that we can always rearrange a bi-infinite set to get whatever density phenomenon we want, and while $Th(\mathcal{N})$ is extremely complicated there are very simple infinite pieces of it and its complement.


问:
A way to write derivatives
There are many notations when it comes to derivatives such as $f'(x)$ and $\frac{dy}{dx}$. As of the moment, I can't find a reference that uses $dy/dx$ instead of $\frac{dy}{dx}$.

Is it acceptable to write $\frac{dy}{dx}$ as $dy/dx$ or not?

答: Yes, absolutely. Writing $\frac{dy}{dx}$ as $dy/dx$ is much the same as writing $\frac{3}{5}$ as $3/5$. There is no mathematical difference between the two notations—only a typographical one.

Title: A way to write derivatives


问:
Please recommend a resource for set / relations / functions questions?
I had done a course in logic and have a basic idea about indirect reasoning, can you recommend me a textbook containing only important questions which are indirect reasoning based questions?</p>

答:
I think that if you want to become familiar with the mathematical reasoning and the basic ideas behind set theory, you can start by something more informal / intuitive.

I strongly recommend the book “Proofs and Fundamentals”, by Ethan D. Bloch. It has a really nice introduction to these topics and he really focus on the way one should write proofs.

After that, if you really want something more mature about set theory, I recommend the classic “Naive Set Theory” by Paul R. Halos.


问:
Why is the maximum number of non-attacking pairs of queens in the 8-queen problem?
In the 8-queen’s problem we want the number of attacking pairs of queens to be zero in the solution assignment of the queens.

A possible fitness function is the number of non-attacking pairs of queens that we are interested to maximize which has the maximum value $\dbinom{8}{2}$ = 28.

I'm very sorry but I can't figure out why so ... Could you help me understand it as if I was a 10 years old? I'm quity rusty on my maths (but I'm very motivated :))

答:
If we have $8$ queens, there are $\binom 82 = 28$ ways to choose two of them: that's the number of pairs of queens we have, total.

What is $\binom 82$, exactly? It is $\frac{8 \cdot 7}{2}$. We have $8$ ways to choose the first queen in the pair, and $7$ ways to choose the second queen, different from the first. But then we divide by $2$, because each pair of queens $\{X,Y\}$ was counted twice: taking $X$ as the first queen and $Y$ as the second, and taking $Y$ as the first queen and $X$ as the second.

Each pair of queens can either attack each other, or not. For our fitness function, we go through all $28$ pairs of queens, and add $1$ to the score if they do not attack each other.

Since there are only $28$ pairs, the maximum score we can get in this way is $28$.


问:
Prove a invertible square matrix has imaginary numbers
Let $C$ be an invertible square matrix. Assume that $C$ is a complex number matrix, and that at least one entry is not entirely a real number (so including the complex number $i$). Prove or disprove that the inverse of $C$, $C^{-1}$ has at least one entry that is also not entirely a real number. I did some rough work, and this seems to be true. We can calculate the inverse by using the adjunct or adjoint of the matrix. When you calculate that, you calculate for cofactors. When you calculate for cofactors, you are also calculating for determinants. This means if there's a complex number involved, it is always going to be there somewhere. Now, how do I write this as a formal proof?

答:
Suppose that all entries of an invertible matrix $A$ are real numbers. Then its determinant is real (and not $0$) and all entries of the adjugate matrix of $A$ are real numbers (its entries are determinants of real matrices). So, $A^{-1}$ is a real matrix too.

Now, apply this to $C^{-1}$: if it was a real matrix, $C$ would be a real matrix too.

By the way, this works for every field.


问:
is zero-measured set always closed?
Is zero-measured set in $\mathbb{R}^n$ always closed?
Or can you please give an example of zero-measured non-closed set in $\mathbb{R}^n$? Here, measure means Lebesgue measure.

答:
The comments already give two examples, but I'm going to chime in here so this question stops showing up as unanswered. Both comments give examples of countable sets that are not closed. Every countable set has Lebesgue measure $0$, so any non-closed countable set will do.

There's also non-closed uncountable sets of measure $0$. Let $C$ be the standard Cantor set in $[0, 1]$. It is an uncountable closed set of measure $0$. The union of two sets of measure $0$ still has measure $0$, so $\mathbb{Q} \cup C$ has measure $0$, and it's definitely not closed.

Even better, this link shows that there are plenty of Lebesgue measure $0$ sets that aren't even Borel.

Title: is zero-measured set always closed?


问:
How to find the expectation of $\log x$？
Let $X_1,...,X_n$ be a random sample. The pdf is $f(x\mid\theta)=\theta x^{\theta-1},0<x<1,\theta>0$.
I want to know $\mathbf{E}(\log x)$.

$$\int_0^1 \theta x^{\theta-1} \log x \;dx$$

I don't know how to solve this integral.

答:
I don't know how to solve this integral.

by parts

$$\underbrace{\left[x^{\theta}\log x\right]_0^1}_{=0}-\int_0^1 x^{\theta-1}dx$$

you find $E(\log x)=-\frac{1}{\theta}$


答:
Substitute $u = \log x, x = e^u, du = \frac{dx}{x}$ to get

$$ \int_{-\infty}^{0} \theta e^{\theta u} u \;du.$$

The goal is to write this in terms of the Gamma function:

$$\Gamma(s) = \int_0^\infty t^{s - 1}e^{-t} \; dt. $$

So you want to set $t = - \theta u$ above. Alternatively, you can integrate by parts (differentiating $u$ and integrating $e^{\theta u}$).

Integrating by parts is likely simpler, but the Gamma function shows up often enough in probability that it's worth keeping in mind as well.

Title: How to find the expectation of $\log x$？

Body:
Let $X_1,...,X_n$ be a random sample. The pdf is $f(x\mid\theta)=\theta x^{\theta-1},0<x<1,\theta>0$.
I want to know $\mathbf{E}(\log x)$.

$$\int_0^1 \theta x^{\theta-1} \log x \;dx$$

I don't know how to solve this integral.


答:
This is a more general proof of you statment.

First I want remark that the expression:
$$\lim_{x \to a}f(x)=L$$
Means $\forall \varepsilon >0$ If  $0<|x-a|< \delta$ implies $|f(x)-L|< \varepsilon$

Intuitively it expression say that if $x$ be near of $a$ namely $x\in(a-\delta,a+\delta)$ then $f(x)$ stay near of $L$. In other words $f(x)\in (L-\varepsilon, L+\varepsilon)$.

And the idea of make a proof is propose a $\delta$ in terms of $\varepsilon$.

Since $\varepsilon$ is given.

Now let´s to the general proof.

$$\lim_{x \to a}x^2=a^2$$
$\forall \varepsilon>0$ $\exists \delta>0$ such that $0<|x-a|<\delta$ implies $|x^2-a^2|< \varepsilon$.

Given the fact that
$|x^2-a^2|< \varepsilon$ then
$|x-a||x+a|<\varepsilon$ now we need bound $|x+a|$ since $|x-a|$ are bounded by $\delta$.

Suppose that $\delta=1$ and then $|x-a|<1$
now $|x+a|=|x-a+2a|<1+2|a|$.

Therefore
$\delta(1+2|a|)<\varepsilon$  and hence $\delta=Min(1,\frac{\varepsilon}{1+2|a|})$.

Frome here we need assume that
$0<|x-a|<Min(1,\frac{\varepsilon}{1+2|a|})$ and show that in fact $|x^2-a^2|<\varepsilon$.

Assume that  $Min(1,\frac{\varepsilon}{1+2|a|})=\frac{\varepsilon}{1+2|a|}$ then
$$|x-a|<\frac{\varepsilon}{1+2|a|}$$
$$|x-a||x+a|<\frac{\varepsilon}{1+2|a|}\cdot (1+2|a|)$$
$$ |x^2-a^2|<\varepsilon$$
and hence $$\lim_{x \to a}x^2=a^2$$ as well.
Your case is a particular case of this one.

I hope that it was helpful.

Title: Epsilon delta proof questions

Body:
Hi and thanks for reading. I am not new to, but have almost no hand on experience with, epsilon delta proofs. I have an example I would like to confirm is correct, which is as follows.

Question

Prove that $\lim_{x \to 3} x^2 = 9$.

Answer

We need to show that, for every $\epsilon > 0$, there exists a $\delta > 0$ such that $|x-3|<\delta \implies |x^2 - 9|<\epsilon$. Now,

$\begin{aligned}|x-3|<\delta &\implies -\delta < x-3< \delta \\ &\implies -\delta+6<x+3<\delta+6 \\ &\implies -\delta - 6 < x+3 < \delta + 6 \\ & \implies |x+3| < \delta + 6 \\ &\implies |x-3| |x+3| < \delta (\delta +6) = \delta^2 + 6\delta \\ &\implies |x^2 - 9| < \delta^2 + 6\delta\end{aligned}$

So, if we let $\epsilon = \delta^2 + 6\delta$, we have proven the limit.

I have used Excel with different values of x and different values of delta to check this, and it seems to be true. I'd just like to check I haven't made any careless errors. I also see many people "proving" limits the wrong way round, ie assuming epsilon to be given and working backwards. This would be fine if every step was an equivalence, but if not then the proof fails. Is this not correct? I have even seen textbooks using this method and it seems very counter intuitive to me.


问:
Numbers of students registered for various courses
I have a probability problem that I’m struggling with. Any help would be appreciated.

Here is the problem:

A class has 28 students. In this class, each student is registered in 2 languages courses (at most) chosen among English, German or Spanish. The number of students registered in each of these 3 languages lessons is 25 for English, 18 for German and 13 for Spanish.

1)  There are many students registered for both English and Spanish. True or False ?

2)  All Students registered in German are registered in “English” ? True ? False ? We cannot tell ?

I cannot figure out how to find out this.
Can you please help on this?

Regards,

答:
Here's how to arrive at the solution:

There are 28 students in the class, 25 of which are in the English course. So how many students are not in the English course?

There are 13 students in the Spanish course. Given the answer to the previous question, how many at least have to also be in the English course? (This gives you the answer to the first question)

Given the last answer, what is the minimum number of people attending the English course that also attend the Spanish course?

Given that no one has all three courses, how many people in the English course can at most attend the German course?

How does that number compare to the number of people attending the German course? (This answers the second question)


答:
This integral can be solved by the use of integration by parts,
\begin{equation*}
\int uvdx=u\int vdx-\int \frac{du}{dx}\left(\int vdx\right) dx
\end{equation*}

\begin{gather*}
\int x^{n}\ln xdx\\
Taking\ \ln x\ as\ u\ and\ x^{n} \ as\ v,\\
\int x^{n}\ln xdx\\
=\ln x\cdotp \frac{x^{n+1}}{n+1} -\int \frac{1}{x} \cdotp \frac{x^{n+1}}{n+1}\\
=\ln x\cdotp \frac{x^{n+1}}{n+1} -\frac{1}{n+1}\int x^{n} dx\\
=\ln x\cdotp \frac{x^{n+1}}{n+1} -\frac{x^{n}}{( n+1)^{2}}\\
\int ^{1}_{0} x^{n}\ln xdx=\frac{-1}{( n+1)^{2}}
\end{gather*}
For your integral,
\begin{equation*}
n=\theta -1
\end{equation*}
Can you complete the rest?


问:
Find a closed form for this recurrence relation
For some value of $k$, $a_j=0$ for all $j\leq k$. Then, for $n\geq 1$,

$$a_{n+k}=\frac{n}{n+k}-\frac{1}{n+k}\sum_{i=k+1}^{n+k}a_i$$

What is a closed form for $a_k$?

I found that $$a_{n+k}=\frac{n}{n+k}-\frac{1}{n+k}a_{n+k}-\frac{1}{n+k}\sum_{i=k+1}^{n+k-1}a_i$$
so that
$$\frac{n+k+1}{n+k}a_{n+k}=\frac{n}{n+k}-\frac{1}{n+k}\sum_{i=k+1}^{n+k-1}a_i$$
and
$$a_{n+k}=\frac{n}{n+k+1}-\frac{1}{n+k+1}\sum_{i=k+1}^{n+k-1}a_i$$
But I don't know where to go from here. I calculated the first few terms, but I couldn't find a pattern. Can anyone help me out?

Body:
$$
\cases{
(n+k)a_{n+k}=n-\sum_{i=k+1}^{n+k}a_i\\
(n+k-1)a_{n+k-1}=n-1-\sum_{i=k+1}^{n+k-1}a_i
}
$$

after subtraction

$$
(n+k)a_{n+k}-(n+k-1)a_{n+k-1} = 1-a_{n+k}\Rightarrow (n+k+1)a_{n+k}-(n+k-1)a_{n+k-1} =1
$$

now making $m = n+k$ we have

$$
(m+1)a_m-(m-1)a_{m-1}=1 \Rightarrow a_m = \frac{c_0}{m(m+1)}+\frac 12
$$

or

$$
a_{n+k} = \frac{c_0}{(n+k)(n+k+1)}+\frac 12
$$


问:
Leibniz notation convention for a differential equation involving a function called 'd(t)'
I am composing a report on electromagnetism in nanomagnets. I need to construct a differential equation involving the differentiable portions of this function for an error analysis metric (Lyapunov exponent).

Unfortunately, the function is called d(t)
and I need to write an expression involving its derivative with respect to time.

I would much prefer not to switch to another notation such as $$\dot d$$ or $$d'(t)$$ since I have already used Leibniz notation throughout the entirety of my report and because I need the flexibility of Leibniz's notation that allows multiplying and dividing by dt.

What would be the best way to write this using Leibniz's notation without sparking confusion?

$$\frac{d}{dt}d(t)$$
$$\frac{dd}{dt}$$
$$\frac{d}{dt}(d(t))$$

答:
I would set the differential d’s upright, like so:
$$
\frac{\mathrm{d}d}{\mathrm{d}t}.
$$
You should probably do that everywhere in that case. (I like doing that anyways.)

Other than that, you can try simply pointing it out the first time. (“For conventional reasons, the function is called $d$. We apologize for the unfortunate clash with the diffential operator.”)


问:
Intuitive explanation of Buckingham Pi Theorem?
I'm currently taking a class about model building in applied math. We spent quite a bit of time on Buckingham $\Pi$ theorem. I understand it on a shallow level. I know how to solve the homework and quiz problems but I just don't understand why it works.

My next big assignment is a presentation and essay. Buckingham's theorem is a big part of it. More specifically, we are given a problem with the following variables:

\begin{array}
\hline
D    & Diameter                    \\ \hline
V   & Velocity                    \\ \hline
\rho & Density                     \\ \hline
F    & Force                       \\ \hline
n    & Revolutions/second      \\ \hline
g    & Acceleration\phantom{-}due\phantom{-}to\phantom{-}gravity \\ \hline
\mu  & Viscosity                   \\ \hline
\end{array}

We are asked to use Buckingham's theorem to derive the following equation:

$$ F = \rho D^2 v^2 \phi(\frac{nD}{V},\frac{gD}{V^2}, \frac{\mu}{\rho D V})$$

where $\phi$ is a function. I know how to get the answer but I can't tell you why it works. This was fine when it was just homework or quizzes problems but now I have to do a presentation on it. Why does Buckingham's theorem work here? From what I understand it doesn't always work - sometimes it gives you a formula that has no coherent physical meaning. We have to use theory from physics to justify the use of Buckingham's theorem but what would this justification look like? This problem is about a ship and the force of its propeller if that makes a difference. Furthermore, what exactly is the relationship between the equation above and the actual equation from the Buckingham theorem $f(\Pi_1, \Pi_2, \Pi_3, \Pi_4) = 0$?

答:
Using Buckingham $\Pi$ theorem is like working a mechanical puzzle:
 (Source: https://en.wikipedia.org/wiki/Mechanical_puzzle)

You employ all your backgrounds (intuition + ...) to solve it.


问:
How to simplify $a^n - b^n$?
How to simplify $a^n - b^n$?

If it would be $(a+b)^n$, then I could use the binomial theorem, but it's a bit different, and I have no idea how to solve it.

Thanks in advance.

答:
If you are looking for this???

$$
a^n-b^n=(a-b)\Big(\sum_{i=0}^{n-1}a^{n-1-i}b^i\Big)
$$

Title: How to simplify $a^n - b^n$?


问:
Find basis and dimension of subspace
I have finite dimensional vector space F and subspace V, such that for its vector of subspace, sum of coordinates of a vector equals zero.

How do I find basis and dimension of this subspace?

答: Suppose that $v=(x_1,\dots,x_n)\in V$ where $n=\dim F$. Thus, $x_n=-\sum\limits_{i=1}^{n-1}x_i$ and $x_1,...,x_{n-1}$ are free. This means that the dimension is $n-1$ and to find a base you can put $x_i=1$, $x_j=0$ if $i\neq j<n$ and $x_n=-1$.

Title: Find basis and dimension of subspace


问:
A clarification of partial derivatives
I have been wondering, why do textbooks say things like "partial derivative with respect to $x$", for say a function from $\mathbb{R}^4$ to $\mathbb{R}$. The function could easily have been written with variables like $a,b,c,d$. Shouldn't it really be "partial derivative with respect to the first coordinate"? Is there a textbook that doesn't make this error?

答:
Well, it is customary to define the first coordinate as x. Had they chosen to define the first coordinate as a, then it would say the partial derivative with respect to a. You'll never find a partial derivative with respect to a coordinate where that coordinate has not yet been assigned a variable. Does that make sense?
In other words, first we define the $i^{th}$ coordinate as some variable, and then we say "the partial derivative with respect to ____". The blank space being the given variable.

Title: A clarification of partial derivatives


问:
Weak continuity of K-L divergence function
If $P_n$ and $Q_n$ are two pmf's of a discrete set (say $A$) with common support and $P_n \to P$ and $Q_n \to Q$ where the convergence is pointwise here (even weak would be fine here I guess), then
$$ D(P_n\|Q_n) \to D(P\|Q)$$
follows from continuity and the fact that it is a finite sum. To recall, for finite/countable sets, KL Divergence is given by
$$D(P\|Q) = \sum_{k=1}^{|A|}P(k)\log_2\left(\frac{P(k)}{Q(k)}\right)$$

Now for uncountable sets, one would use a slightly different definition of divergence (if $A$ here were uncountable but a measurable space with a Borel Sigma algebra). For any two measures $P$ and $Q$ on $A$ where $P \ll Q$ ($P$ is absolutely continuous wrt $Q$), we have
$$D(P\|Q) = \int_A \frac{dP}{dQ}\log\left(\frac{dP}{dQ}\right)dQ$$
where $\frac{dP}{dQ}$ is the Radon Nikodym derivative of $P$ wrt $Q$.
My question is, Suppose $P_n \to^w P$ and $Q_n \to^w Q$, and $P_n \ll Q_n ~ \forall n$, does it follow that
$$ D(P_n\|Q_n) \to D(P\|Q)$$
? (Here "$\to^w$" denotes weak convergence).

My attempt: My guess was that it would not hold every time but in certain cases. To characterize them, I started with understanding $D(P_n\|Q)$. This is simply
$$D(P_n\|Q) = \int_A \frac{dP_n}{dQ}\log\left(\frac{dP_n}{dQ}\right)dQ$$.
Only now I am not sure how to show $\frac{dP_n}{dQ} \to \frac{dP}{dQ} ~ a.e$

Source and References: While one would encounter these terms in any good information theory book, I would recommend Csiszar and Korner "Coding theorems for discrete memoryless systems". There are also quite a few papers that deal with this kind of divergence. However the source of the problem is not from these books but was on a research problem I was tackling years back and am now revisiting.

Hence I appreciate ideas and tips on proceeding rather than outright answers. Feel free to request further clarification.

Update1: I had mentioned above that $P \ll Q$. If it helps we may also let $Q \ll P$.

答:
@Ashok comment resolved this completely, so I am rewriting it as an answer. Posner, "Random Coding Strategies for Minimum Entropy", IEEE Trans Info Theory, 1975 showed that KL divergence is lower-semicontinuous in the topology of weak convergence. In particular, if $P_n \to P$ and $Q_n \to Q$ weakly, then
$\lim_n D(P_n\Vert Q_n) \ge D(P\Vert Q)$. A very nice summary is provided in the excellent Y. Polyanskiy, Y. Wu, "Lecture notes on Information Theory" (see Theorem 3.6)

Note that the first claim in your question is incorrect: it is not the case that $\lim_n D(P_n \Vert Q_n) = D(P\Vert Q)$ for all countable/discrete alphabets, even if all the distributions have the same support, because the continuity property you state only holds for finite alphabets. Here is a counterexample.

Let $P(i)=\frac{1}{Z}e^{-i^2}$, where $Z=\sum_{i=1}^\infty e^{-i^2}$ is finite normalization constant, and let $P=Q=Q_n$, so that $D(P\Vert Q)=0$. Let $\alpha_n=\sum_{i=1}^n P(i)$ indicate partial sums, and define $P_n$ in the following manner:
\begin{align}
P_n(i) = \begin{cases}
(1-n^{-2}) P(i) & \text{for $i\in\{1,\dots,n-1\}$}\\
1-(1-n^{-2})\alpha_{n-1} & \text{for $i=n$}\\
0 & \text{otherwise}
\end{cases}
\end{align}
It is easy to see that $P_n$ is a valid probability distribution and that $P_n \to P$ in total variation, hence also in the topology of weak convergence.
Now write
$$D(P_n\Vert Q_n)=D(P_n\Vert P)=\alpha_{n-1}(1-n^{-2})\ln(1-n^{-2})+P_n(n)\ln\frac{P_n(n)}{P(n)}.$$
Since $\lim_{n} \alpha_{n-1}=1$ and $\lim_n (1-n^{-2})\ln(1-n^{-2})=0$,
\begin{align}
\lim_n D(P_n\Vert Q_n) &= \lim_n P_n(n)\ln\frac{P_n(n)}{P(n)}\\
&= \lim_n -P_n(n)\ln{P(n)}\\
&=\lim_n P_n(n)(n^2 + \ln Z)= \lim_n \frac{P_n(n)}{n^{-2}}
\end{align}
where we used that $\lim_n P_n(n)=0$ and the definition of $P(n)$. Finally, a simple application of L'Hospital's rule shows that  $\lim_n P_n(n)/n^{-2}=1$.
Combining gives
$$\lim_n D(P_n\Vert Q_n) =1 > 0 = D(P\Vert Q).$$


问:
$ L^2 (\mathbb{R}, \mathbb{C}) \otimes \mathbb{C} ^2 $ is the same as $L^2(\mathbb{R}, \mathbb{C}) \times L^2(\mathbb{R}, \mathbb{C})$
I want to show that $ L^2 (\mathbb{R}, \mathbb{C}) \otimes \mathbb{C} ^2 $ can be thought of as the same space as $L^2(\mathbb{R}, \mathbb{C}) \times L^2(\mathbb{R}, \mathbb{C})$ Perhaps they are isometrically isomorphic.

What is causing me trouble is the fact that we have an infinite-dimensional space here. Perhaps I could take an orthonormal Schauder basis for $ L^2 $, but the fact that these spaces are the same feel like it should be obvious.

My question is, is it correct that $ L^2 (\mathbb{R}, \mathbb{C}) \otimes \mathbb{C} ^2 $ can be thought of as the same space as $L^2(\mathbb{R}, \mathbb{C}) \times L^2(\mathbb{R}, \mathbb{C})$? If so, how would one prove it?

答:
So far the answers given have directed attention towards "complicated things" like what is the tensor product for infinite-dimensional vector spaces. Let me offer another answer based on the totally opposite viewpoint of focusing on "simple things" like asking what is $\mathbb{C}^2$?

Well, $\mathbb{C}^2=L^2(\{1,2\},\mathbb{C})$, i.e., it is the space of complex-valued functions on the finite set $X=\{1,2\}$. One could also put on it the $\sigma$-algebra
$\{\varnothing,\{1\},\{2\},\{1,2\}\}$ and the counting measure and then there would be no cheating when viewing $\mathbb{C}^2$ as an honest $L^2$ space.

Now the raison d'être of tensor products is to correspond to functions on the Cartesian product when the individual vector spaces are presented to us as spaces of functions. So
$$
L^2(\mathbb{R},\mathbb{C})\otimes \mathbb{C}^2=L^2(\mathbb{R},\mathbb{C})\otimes L^2(\{1,2\},\mathbb{C})=L^2(\mathbb{R}\times\{1,2\},\mathbb{C})
$$
and elements of the last space trivially correspond to pairs of functions $f_1(x)$, $f_2(x)$ in $L^2$.


问:
Derivative using Product Rule
I have a derivative:

$$ \frac{d}{dx}[\sqrt{10x}]$$

I start by using $ ^n\sqrt{a^{x}}=a^\frac{x}{n}$ to give me:

$$\frac{d}{dx}[(10x)^\frac{1}{2}]$$

I can factor out 10 to give me:

$$\frac{d}{dx}[(10(x))^\frac{1}{2}]$$

This is the part that loses me.  I know I need to use the Product Rule to $10(x)$, but I must not be setting it up correctly.  When I use the Product Rule I get:

$$(f*g)'=f'*g+f*g'$$
$$(f*g)'=(10)'*(x)+(10)*(x)'$$
$$(f*g)'=0*(x)+(10)*1$$
$$(f*g)'=10$$

What I've been told the result should be after using the Product Rule is:
$$\frac{d}{dx}[10^\frac{1}{2}x^\frac{1}{2}]$$

Where would I be going wrong when it comes to using the Product Rule?

答:
When you are multiplying two functions and one of them is just a constant, then when you take the derivative, you can just pull the constant out.  No need for the product rule in that case.

For this problem, you don't need the chain rule either, just the power rule!
$$ \frac{d}{dx}[10x]^{1/2} = 10^{1/2} \frac{d}{dx} x^{1/2} = 10^{1/2}\cdot \frac{1}{2} \cdot x^{-1/2} = \frac{\sqrt{10}}{2 \sqrt{x}}$$

If you want to use the chain rule instead, then $10x$ is your inside function and $\sqrt{x}$ is your outside function.

A couple of points: you wrote $\frac{d}{dx} = [10x]^{1/2}$ when what you mean is $\frac{d}{dx} [10x]^{1/2}$.  That is not an equation, it is an expression, the derivative operator, acting on a function.

Also, your final result does not have calculus done, it was just the bit of algebra that says $[10x]^{1/2} = 10^{1/2} x^{1/2}$.  Maybe you understood that, but your first mistake makes it a little hard to tell.

Title: Derivative using Product Rule


答:
Since constants have a zero derivative, don't bother applying the product rule to them, just factor them out.

i.e. $(c\times f(x))' = 0\times f(x)+c\times f'(x)=c\times f'(x)$ is the same as $c\times (f(x))'$ where $c$ was factored out.

by factoring out the constant

$(\sqrt{10x})'=(\sqrt{10}\times\sqrt{x})'=\sqrt{10}\times (\sqrt{x})'=\sqrt{10}\times (x^\frac 12)'=\sqrt{10}\times \frac 12x^{-\frac 12}=\dfrac{\sqrt{10}}{2\sqrt{x}}$

by chain rule

$\require{cancel}((10x)^\frac 12)'=\frac 12(10x)^{-\frac 12}\times (10x)'=\frac 12(\cancel{10}x)^{-\frac 12}\times \underbrace{10}_{10^\frac 12\cdot\cancel{10^\frac 12}}=\frac 1210^\frac12x^{-\frac12}=\dfrac{\sqrt{10}}{2\sqrt{x}}$

And this is the same result.

Title: Derivative using Product Rule

Body:
I have a derivative:

$$ \frac{d}{dx}[\sqrt{10x}]$$

I start by using $ ^n\sqrt{a^{x}}=a^\frac{x}{n}$ to give me:

$$\frac{d}{dx}[(10x)^\frac{1}{2}]$$

I can factor out 10 to give me:

$$\frac{d}{dx}[(10(x))^\frac{1}{2}]$$

This is the part that loses me.  I know I need to use the Product Rule to $10(x)$, but I must not be setting it up correctly.  When I use the Product Rule I get:

$$(f*g)'=f'*g+f*g'$$
$$(f*g)'=(10)'*(x)+(10)*(x)'$$
$$(f*g)'=0*(x)+(10)*1$$
$$(f*g)'=10$$

What I've been told the result should be after using the Product Rule is:
$$\frac{d}{dx}[10^\frac{1}{2}x^\frac{1}{2}]$$

Where would I be going wrong when it comes to using the Product Rule?


问:
Cauchy product on exponential-looking power series
Original posting by dioxen here: Double summation including power and factorial

I am finding some trouble in computing the following sum:

$$\sum_{k=0}^\infty \frac{x^k}{k!}\;\sum_{m=0}^k\frac {y^m}{m!}$$

Could you please provide a result?

Thanks in advance

After a failed attempt at getting to the binomial expansion, I tried to play around with it a little bit. I came up with two ideas of a solution, but can't see it through. Both involve applying the Cauchy product.
\begin{equation}
\Bigg(\sum_{n=0}^\infty a_n \Bigg) \Bigg(\sum_{n=0}^{\infty} b_n \Bigg) = \sum_{n=0}^{\infty} c_n \space \text{ where } c_n = \sum_{k=0}^n a_k b_{n-k} \space \space \space \space \space \space \space \space  \space \space \space \space (1)
\end{equation}

1: In this problem, we cound treat $$ c_n= \frac{x^n}{n!}\;\sum_{k=0}^n\frac {y^k}{k!}=\frac{1}{n!}\sum_{k=0}^n\frac {(xy)^{n-k}}{(n-k)!}x^{k}=\frac{1}{n!}\sum_{k=0}^n a_k b_{ n-k}$$

So if we're thinking within the framework of the first two summations in (1), our $a_n$ and $b_n$ are as follows: $a_n=x^n$ and $ b_n = \frac {(xy)^{n}}{n!}\\$

our only problem is the we're left with the $\frac{1}{n!}$ in front of the summation. If we thought up some way to get rid of it, we'd have a nice geometric series and $e^{xy}$. But I can't think of a way...any leads here?

2: Alternatively, we could leave the $\frac{x^n}{n!}$ outside and treat it as a number (because it is, before we choose to sum it up over $n$). $$ c_n= \frac{x^n}{n!}\;\sum_{k=0}^n\frac {y^k}{k!} 1^{n-k}=\frac{x^n}{n!}\sum_{k=0}^n a_k b_{ n- k}$$ which gives us $a_n=\frac{y^n}{n!}$ and $b_n=1$. Could we do something about $\frac{x^n}{n!}$?

$$
\Bigg(\sum_{n=0}^\infty \frac{x^n}{n!} \Bigg) \Bigg(\sum_{n=0}^{\infty} \frac{x^n}{n!} \frac{y^n}{n!} \Bigg)
$$
Does this even make sense? Because otherwise I can't think of a way this sum would be computable.

答: Sadly, I believe there is no closed form for your double summation. The closest form I can get is $$\begin{eqnarray}\sum_{k=0}^\infty \frac{x^k}{k!}\;\sum_{m=0}^k\frac {y^m}{m!} &=& \sum_{k=0}^\infty \frac{x^k}{k!} \; {}_0F_1\left(\left.\begin{array}{c} - \\ k+1 \end{array} \; \right| xy \right) \\ &=& \frac 1{\pi}\sum_{k=0}^\infty \left(\frac xy \right)^{\frac k2} \; \int_0^{\pi} e^{2 \sqrt {xy} \cos(t)} \cos (kt) dt \end{eqnarray}$$</p>


问:
Heron's Formula
Heron's Formula gives the area of a triangle when the length of all three sides are known. There is no need to calculate angles or other distances in the triangle first to determine its area. The formula is given by

$Area=\sqrt{p(p-a)(p-b)(p-c)}$, where $p=\frac{a+b+c}{c}$, $a, b, c$ are sides of the triangle and $p$ is the perimeter of the triangle. The following is my concern:

If one of the sides of the triangle is greater than $p$, then the Area will not be a real number (which shouldn't be true).

Example.  Let the sides of a triangle be 175 metre, 88 metre and 84 metre, then $p=173.5$. Therefore,
$Area=\sqrt{-1991498.0625}$, which not a real value.

Therefore, the following is my question:

Why shouldn't the area be expressed as $Area=\sqrt{|p(p-a)(p-b)(p-c)|}$?

答:
Recall that, for a triangle with sides $a, b, c$, we always have $a + b > c, b + c > a, c + a > b$.

This means that $p = \frac {a + b + c} 2 > a$ and similarly $p > b$ and $p > c$.

Thus you don't have to worry about the area being non-real.

Title: Heron's Formula


答:
If $p$ is less than one of the sides then you don't have a triangle - one of the sides would have to be longer than the other two combined.

So if $p\lt a$ we have $a+b+c\lt 2a$ 0r $b+c\lt a$

Title: Heron's Formula

Body:
Heron's Formula gives the area of a triangle when the length of all three sides are known. There is no need to calculate angles or other distances in the triangle first to determine its area. The formula is given by

$Area=\sqrt{p(p-a)(p-b)(p-c)}$, where $p=\frac{a+b+c}{c}$, $a, b, c$ are sides of the triangle and $p$ is the perimeter of the triangle. The following is my concern:

If one of the sides of the triangle is greater than $p$, then the Area will not be a real number (which shouldn't be true).

Example.  Let the sides of a triangle be 175 metre, 88 metre and 84 metre, then $p=173.5$. Therefore,
$Area=\sqrt{-1991498.0625}$, which not a real value.

Therefore, the following is my question:

Why shouldn't the area be expressed as $Area=\sqrt{|p(p-a)(p-b)(p-c)|}$?


问:
How to show that a continuous function is bounded on an interval
For a continuous function $f :[0, \infty) \rightarrow [0, \infty)$, $(x_n)$ is a sequence of positive real numbers diverging to infinity. If $f(x_n) \rightarrow 0$, is $f$ bounded on $[0,\infty)$?

Are there any theorems that can help in proving/disproving whether $f$ is bounded on $[0, \infty)$ or not?

答:
No, take $f(x) = |x\sin(x)|$ and $x_n = 2\pi n$

But, if the statement is true for every sequence then $f$ is bounded with limit $0$ at infinity.


问:
How to prove that the curve $x^2+y^2-3=0$ has no rational points?
There is a question 6.20 from the Book of Proof by Hammock that reads as follows:  We say that a point $P=(x,y) \in R^2$ is rational if both x and y are rational. More precisely, P is rational if $P=(x,y) \in Q^2$.  An equation $F(x,y)=0$ is said to have a rational pointif there exists $x_0, y_0 \in Q$ such that $F(x_0,y_0)= 0$.  For example, the cruve $x^2+y^2-1=0$ has rational point $(x_0,y_0)=(1,0)$.  Show that the curve $x^2+y^2-3=0$ has no rational points.

I can only take a guess at how to start a proof.  Doing algebra on both sides of $x^2+y^2-3=0$ yields $y = \sqrt{3-x^2}$.  Now, assume y is rational -- can x be rational?

Thus, let y = $a\over b$ for some integers a and b, $b \ne  0$,$gcd(a,b)=1$.  Then let y = $c\over d$ for some integers c and d, $d \ne  0$,$gcd(c,d)=1$.  Hence, ${a \over b} = \sqrt{3 - {{c^2} \over {d^2}}}$.

I think I may want to prove some sort of contradiction in terms of rational and irrational?  This as far as I got in my scratchwork for a proof.

答:
$ y=\sqrt {3-x^2}$

For $y$ to be rational, we must have $ x^2\le3$ and $y=1\implies |x|\le\sqrt 3$ but there doesn't exist such a rational number $x$ except $|x|=1$ and for any $1<|x|<\sqrt 3, y\notin\Bbb{Q}$. Again for $x=\pm1, y=\sqrt 2\notin \Bbb{Q}$. Also $x$ cannot be $\pm\sqrt 3$.  Hence proved.


答:
Suppose $x=\frac{m}{n}$ and $y=\frac{a}{b}$, where both fractions are reduced. Then after a little algebra your equation says:

$$m^2b^2+n^2a^2=3b^2n^2$$

This is an integer equation. Two terms are divisible by $b^2$, so $b^2$ must divide the third term, $n^2a^2$. Since $a$ and $b$ are relatively prime, then actually $b^2$ divides $n^2$.

But similarly, $n^2$ divides $b^2$. So you may conclude $n=b$ and the equation reduces to

$$m^2+a^2=3n^2$$

The residues of squares mod $3$ are just $0$ and $1$, not $2$. So whne you look at that equation mod $3$:

$$m^2+a^2\equiv0$$

the only way it can work out is if $m^2\equiv a^2\equiv0$. So $3$ must divide both $m$ and $a$. But then look again at

$$m^2+a^2=3n^2$$

Now we know $9$ divides the left side. So $3$ must divide $n^2$. This is a contradiction, because now we are saying $3$ divides both $m$ and $n$, but $\frac{m}{n}$ was supposed to be a reduced fraction.

So the assumption that we could have a rational $x$ and $y$ was false.

Title: How to prove that the curve $x^2+y^2-3=0$ has no rational points?

Body:
There is a question 6.20 from the Book of Proof by Hammock that reads as follows:  We say that a point $P=(x,y) \in R^2$ is rational if both x and y are rational. More precisely, P is rational if $P=(x,y) \in Q^2$.  An equation $F(x,y)=0$ is said to have a rational pointif there exists $x_0, y_0 \in Q$ such that $F(x_0,y_0)= 0$.  For example, the cruve $x^2+y^2-1=0$ has rational point $(x_0,y_0)=(1,0)$.  Show that the curve $x^2+y^2-3=0$ has no rational points.

I can only take a guess at how to start a proof.  Doing algebra on both sides of $x^2+y^2-3=0$ yields $y = \sqrt{3-x^2}$.  Now, assume y is rational -- can x be rational?

Thus, let y = $a\over b$ for some integers a and b, $b \ne  0$,$gcd(a,b)=1$.  Then let y = $c\over d$ for some integers c and d, $d \ne  0$,$gcd(c,d)=1$.  Hence, ${a \over b} = \sqrt{3 - {{c^2} \over {d^2}}}$.

I think I may want to prove some sort of contradiction in terms of rational and irrational?  This as far as I got in my scratchwork for a proof.


问:
Elegant way to solve $n\log_2(n) \le 10^6$
I'm studying Tomas Cormen Algorithms book and solve tasks listed after each chapter.

I'm curious about task 1-1. that is right after Chapter #1.

The question is: what is the best way to solve: $n\lg(n) \le 10^6$, $n \in \mathbb Z$, $\lg(n) = \log_2(n)$; ?

The simplest but longest one is substitution. Are there some elegant ways to solve this? Thank you!

Some explanations: $n$ - is what I should calculate - total quantity of input elements, $10^6$ - time in microseconds - total algorithm running time. I should figure out nmax.

答: For my money, the best way is to solve $n\log_2n=10^6$ by Newton's Method. </p>


问:
Prove that there are no prime numbers in $[n! + 2; n! + n]$
I am struggling trying to prove that there are no prime numbers in $[n! + 2; n! + n]$ gap. Could someone please help?

答:
No prime is in the gap $ [ ] $
$$(n!+2)[ n!+3,n!+4,n!+5,\dots \dots , n!+n-1](n!+n)$$
The number of terms in the sequence is $(n!+n-1)-(n!+3)+1 = n-3$
$$(n!+2)[ n!+3,n!+4,n!+5,\dots \dots , n!+n-1](n!+n)$$
Basically the sequence are all sums of $n!$, since $n!$ is the product of numbers less than it, therefore $ n! +2k$ cannot be prime, $n!+3k$ cannot also be prime,$\dots \dots$, $n!+m\cdot k$ cannot also be prime

What about the case when of $n!+ k$, when $k$ is prime, if $k$ is $\lt n$, means it's multiples are contained in $n!$ then the sum $n!+k$ is not prime

What about $n!+k$, where $k$ is prime and $k \gt n $, here the sum would be a prime number
$$(n!+2)[ n!+3,n!+4,n!+5,\dots \dots , n!+n-1](n!+n)$$
In our sequence, the largest value of $k$ is $n-1$, it happens at the last sequence $n!+\{ n-1\} $ but $n-1 \lt n$


问:
what is the variance of this? (random variables)
Can anyone here tell me how can I do this: X and Y are two random variable and W = 5X-4Y+50. Knowing that the variance of X is 99 and the variance of Y is 81 then what is the variance of W?</p>

答:
Assuming $X$ and $Y$ are independant,

$$V\left(W\right)=V\left(5X-4Y+50\right)=V\left(5X\right)+V\left(-4Y+50\right)=5^2V\left(X\right)+\left(-4\right)^2V\left(Y\right)$$


问:
Prove that for every positive real number $\epsilon$, there exists a natural number $N$ such that if $n > N$, then $\frac{1}{n^2+1} < \epsilon$.
Initially, I had:

Let $N$ be an arbitrary natural number and $n > N$. Assume $N < \sqrt{\frac{1}{\epsilon}-1}$. Since $n > N$, $n > \sqrt{\frac{1}{\epsilon}-1}$. Therefore, $ n > \sqrt{\frac{1}{\epsilon}-1}$, then $ n^2 > \frac{1}{\epsilon}-1$ , then $n^2 + 1 > \frac{1}{\epsilon}$, then $\epsilon > \frac{1}{n^2}+1$.

Obviously, this is wrong. My professor wrote comments that N should not be arbitrary, and an X next to my assumption of $N < \sqrt{\frac{1}{\epsilon}-1}$. However, I can't think of many other ways to write this?

Currenty, I have:

Let $\epsilon$ be an arbitrary positive real number. Choose $N > \sqrt{\frac{1}{\epsilon}-1}$. Assume $n > N$. Since $n > N$, $\sqrt{\frac{1}{\epsilon}-1} < N < n$. Therefore, $ n > \sqrt{\frac{1}{\epsilon}-1}$, then $ n^2 > \frac{1}{\epsilon}-1$ , then $n^2 + 1 > \frac{1}{\epsilon}$, then $\epsilon > \frac{1}{n^2+1}$.

Does this new version look right?

答:
Thing is, you are asked to choose a specific $N$. So instead of taking any $N>\sqrt{\frac{1}{\epsilon}-1}$, choose a specific one, for example $N=\lfloor \sqrt{\frac{1}{\epsilon}-1} \rfloor+1$, where $\lfloor \ \ \rfloor$ is the floor function.

Also, there is one annoying, but existing problem. Note that if $\epsilon$ is really large then $\frac{1}{\epsilon}-1$ is negative, and so its square root is not even defined. Not sure I would take off points for this (probably I would, but not much), but you should be aware of this problem. So to make it a perfect answer you should add that if $\epsilon>1$ then you can choose $N=1$ for it. If $\epsilon<1$ then just as you did.


答: All you have to do is realize that if $k< n$ then $k|n!$ so $k|n! + k$ and if $k\ne 1$ then $n! + k$ is not prime.

Title: Prove that there are no prime numbers in $[n! + 2; n! + n]$

Body: I am struggling trying to prove that there are no prime numbers in $[n! + 2; n! + n]$ gap. Could someone please help?


问:
A way to write derivatives
There are many notations when it comes to derivatives such as $f'(x)$ and $\frac{dy}{dx}$. As of the moment, I can't find a reference that uses $dy/dx$ instead of $\frac{dy}{dx}$.

Is it acceptable to write $\frac{dy}{dx}$ as $dy/dx$ or not?

答:
Writing $\frac{dy}{dx}$ or $dy/dx$ is a matter of choice, mathematical speaking they are both ratios, so I can even write as $[ dy : dx ]$

But when compiling a good math text, in textbooks or some type-settings, it's better to write as $\frac{dy}{dx}$ because it looks more like function than variable

Also consider when we want to write higher derivatives
$$\frac{d^7y}{dx^7}$$
$$d^7y/dx^7$$
vote to choose

Title: A way to write derivatives


问:
(Simple) Examples on Non-Commutative Rings
Looks like it is easier to find example of commutative rings rather than
non-commutative rings. Prabably the easiest examples of the former are
$\mathbb{Z}$ and $\mathbb{Z}_n$. We can find elaborations on these two commutative rings in various
literatures including here and here. These are quite simple and easy to comprehend.

However, the examples on simple (non-commutative kind) are not that easy.
One example is found here and it has been mentioned as "one of the simplest examples of a non-commutative ring". This is also on the easier side.

EDIT: The following two are also commutative rings.

Other examples are given in this enumeration. But as you can see, examples like Gaussian integers or Eisenstein integers are difficult for starters to comprehend.

Do you think you can give one or two simple examples on non-commutative rings, based on every day numbers?

If it is that difficult, perhaps some insight comments why this is difficult would be welcome.

答:
Here's a slightly different example. If you are comfortable with doing algebra with polynomials, then I think you will find it easy to understand.

Take any finite group $G$ (say of order $n$). Now, you can formally make a set $\{\sum_{g\in G} \alpha_g g\mid \alpha_g\in \Bbb R\}$.

Since you know how to multiply the elements of $G$, you can extend the multiplication to this set by requiring linearity to hold. So, for example, if you have $a,b,c,d\in G$ and you want to multiply $(3a+2b)(c-5d)$, you would just distribute: $3ac-15ad+2bc-10bd$. Then $ac,ad,bc,bd$ would be multiplied to be elements of $G$, and you would wind back up in the set. Addition is just done by adding "like terms."

With these operations, the set is called the group ring $\Bbb R[G]$. In fact, $G$ doesn't have to be finite, but if $G$ is infinite then you need to only use finite sums in the set I gave above. Even more, $G$ doesn't have to be a group, it just needs to have an associatve multiplication defined on it, so it could be just a monoid or semigroup.

A monoid can informally be described by thinking of a group which does not require the existence of inverses. So, if $x$ is an indeterminate, then $S=\{1,x,x^2,x^3\dots\}$ is a monoid, because $x^ix^j=x^{i+j}$ is an associative multiplication. The monoid ring $\Bbb R[S]$ for this set is something you are really familiar with: it is usualy denoted $\Bbb R [x]$ and called "the ring of polynomials over $\Bbb R$"!

For another thing, you can use any ring you want besides $\Bbb R$! You could even use $\Bbb Z$ if you so chose, or whatever other ring you wanted.

I'm confident that if you stretch your intuition for multiplying polynomials by replacing the powers of $x$ with elements of a group (or monoid), you will quickly grasp what a group ring is.

Anyhow, let's get to the point. If you choose $G$ to be a group that is not Abelian (that is, there exists $g,h\in G$ such that $gh\neq hg$) then you know for sure $\Bbb R[G]$ is not abelian: because in the ring, $gh\neq hg$.

If you would like to experiment with the smallest group which isn't commutative, you'll have to begin with the symmetries of a triangle $S_3=\{1,\sigma,\sigma^2,\tau,\sigma\tau,\sigma^2\tau\}$. To review, the multiplication obeys the relations $\sigma^3=1=\tau^2$, and $\tau\sigma=\sigma^2\tau$.

Pick two elements $p,q$ of $\Bbb Z[S_3]$. Compute $\sigma p$ and $p\sigma$. Compute $p+q$ and $p-q$ and $pq$. Have fun!


问:
Finding stationary points of $f(x,y) = x^2+y^2+\beta xy+x+2y$
I am working on a problem which can be seen below. Part i) was fine but part ii) is causing me some issues. Now the First Order Necessary Conditions state that for the interior case, then $$\nabla f(x^*) = 0$$ which I have attempted to use to find $x^*$. Taking the derivatives of the objective function I get $$\begin{bmatrix}2x + \beta y +1 \\
2y + \beta x +2 \end{bmatrix} = \begin{bmatrix}0\\ 0 \end{bmatrix}$$ which I have solved to get $(y-x) = \frac{1}{\beta - 2}$. Now I am convinced that I am wrong here because this does not seem plausible.

I would really appreciate if someone could give me some hints.

Thanks

答:
You got:
$$\begin{bmatrix}2 & \beta \\ \beta & 2 \end{bmatrix}\begin{bmatrix}x\\y \end{bmatrix} = \begin{bmatrix}-1\\ -2 \end{bmatrix}$$
Let's perform one step of row reduction:
$$\begin{bmatrix}1 & \frac{1}{2}\beta \\ 0 & 2-\frac{1}{2}\beta^2 \end{bmatrix}\begin{bmatrix}x\\y \end{bmatrix} = \begin{bmatrix}-\frac{1}{2}\\ \frac{1}{2}\beta-2 \end{bmatrix}$$
The system has no solution if $2-\frac{1}{2}\beta^2=0$ but $\frac{1}{2}\beta-2 \neq 0$, so when $\beta=2$ or $\beta=-2$. The system has infinitely many solutions if $2-\frac{1}{2}\beta^2= 0$ and $\frac{1}{2}\beta-2 = 0$, which cannot happen.

When $\beta\neq 2$ and $\beta\neq -2$ we do one more step of row reduction:
$$\begin{bmatrix}1 & 0 \\ 0 & 1 \end{bmatrix}\begin{bmatrix}x\\y \end{bmatrix} = \begin{bmatrix}-\frac{1}{2}-\frac{1}{2}\frac{\beta^2-4\beta}{4-\beta^2}\\ \frac{\beta-4}{4-\beta^2} \end{bmatrix}$$

You can now readily read the solution.


问:
Show $-u''(x) = H(x)$ and $u'(-1) = u'(1) = 0$ has infinitely many weak solutions
I want to show that the Neumann boundary value problem
$$
\begin{cases}
- u''(x) = H(x), & x \in (-1, 1)\\
u'(-1) = u'(1) = 0,
\end{cases}
$$
where $$H(x) = \begin{cases} 1, & x > 0, \\ 0, & x < 0 \end{cases}$$ is the Heaviside function, is not uniquely solvable in the weak sense, by finding infinitely many solutions.

For the test space I choose $$H^1((-1, 1); \mathbb R) := \{ u \in L^2((-1, 1); \mathbb R): u' \in L^2((-1, 1); \mathbb R) \},$$ where $u'$ is the weak derivative of $u$.
We have
$$
- \int_{-1}^{1} u''(x) v(x) dx
= \int_{-1}^{1} u'(x) v'(x) dx
$$
and
$$
\int_{0}^{1} v(x) dx
= \int_{-1}^{1} H(x) v(x) dx
$$
Thus the weak formulation is: find $u \in H^1((-1, 1); \mathbb R)$ such that
$$
\int_{-1}^{1} u'(x) v'(x) dx
= \int_{0}^{1} v(x) dx
$$
for all $v \in H^1((-1, 1); \mathbb R)$.

My idea is to write the right side as
$$
\int_{0}^{1} v(x) dx
= v(1) - \int_{-1}^{1} \max(0, x) v'(x) dx.
$$
If $v(1)$ we zero (which its not) the solutions would be
$$
u_c(x)
= \begin{cases}
c, & x < 0, \\
c - \frac{x^2}{2}, & x \ge 0
\end{cases}
$$
for any constant $c \in \mathbb R$, as the weak derivative of $u_c$ is
$$
u_c'(x) = - \max(0, x).
$$
I know that I can represent $v(1) = \int_{-1}^{1} \delta(x - 1) v(x) dx$, where $\delta$ is the Dirac distribution, but this hasn't helped. Can I maybe use the Fundamental Theorem of the Calculus of Variations (that is, $\int_{a}^{b} u(x) v'(x) dx = 0$ for all $v \in \mathbb{C}_{0}^{\infty}([a, b]; \mathbb R)$ implies the existence of a $c \in \mathbb R$ such that $u(x) = c$ almost everywhere)?

答:
Generally, when you have a linear (inhomogeneous) equation and you want to know how many solutions there are, one thing to try is suppose $u_{1}$ and $u_{2}$ are solutions and then ask, "What equation does $u_{1} - u_{2}$ solve?"  This is true even at the level of linear algebra (not just differential equations), as you probably know.

If we do that here, if $u_{1}$ and $u_{2}$ solve your equation, then $w = u_{1} - u_{2}$ is a solution of the equation
\begin{equation*}
-w'' = 0 \quad \text{in} \, \, (-1,1), \quad w'(-1) = w'(1) = 0.
\end{equation*}
Notice that any such function must be constant due to the boundary conditions.  (Look at the equation that $f = w'$ solves; or play clever tricks with the weak formulation.)  That is, $w \equiv C$ for some $C \in \mathbb{R}$.  Hence $u_{1} \equiv u_{2} + C$.

I did not prove that there were two solutions; I just showed that any two solutions differ by a constant.  At the same time, rerunning the previous argument in reverse, we see that if $u$ is a solution, then so is $u + C$, no matter which constant $C$ I take.  That is, as soon as we have one solution, we have infinitely many.  (Again, this is consistent with linear algebra: a linear system has no solutions, a unique solution, or infinitely many.)

(The question now is: do you know that there is at least one solution?  The answer is yes at any rate.)


问:
Why such a short hack to a Partition sequence creator turns it into a prime generator?
I watched  The hardest "What comes next?" (Euler's pentagonal formula) The explanatory video by Mathologer, and I learn't a lot. Somewhere in chapter three, the presenter just throws in that a small modification to his sequence of sequences can be used to generate primes.

I created the Rosetta Code task, and Python entry that follows the video, going so far as to add the prime number generator too. But I was used to sieves for generating primes and I could not work out how this new method worked.

I still have a mental todo item to get back to this.

Another mental todo item was that I remembered from this, or probably another mathologer video where he pointed out that two sequences can lull you into thinking they are the same for several values before they then diverge. I needed to check the partition based prime generator against a sieve.

Well, this morning I added the following extra code to use Pythons sympy.primerange sieve and checked the first 100_000 primes:

```<code>if 1:
    from sympy import primerange

    n = 100_000
    print('Check that the first', n, 'partition and sieved primes are equal:')
    part_primes = list(islice(par_primes(), n))
    sieve_primes = list(primerange(2, part_primes[-1]+1))
    if part_primes == sieve_primes:
        print(f"  First {n} primes, up to {part_primes[-1]} are equal.")
    else:
        print(f"  First {n} primes are NOT equal!")
</code>```

Outputs:

```<code>Check that the first 100000 partition and sieved primes are equal:
  First 100000 primes, up to 1299709 are equal.
</code>```

So my question is: Can someone please explain what links this method and the sieve method that I am familiar with?

Thanks in advance.

(PS my level of mathematics is at around the the level shown in the video in this area).

答:
The integer partition function $p(n)$ satisfies the recurrence relation illustrated in formula (1) below where $g_n=\frac{n\,(3\, n-1)}{2}$ is the generalized pentagonal number.

$$p(n)=\sum\limits_{k\ne 0}(-1)^{k-1}\,p\left(n-g_k\right)=\sum\limits_{k>0}(-1)^{k-1}\,\left(p\left(n-g_k\right)+p\left(n-g_{-k}\right)\right)\tag{1}$$

The relationships between the integer partition function $p(n)$ and the divisor function $\sigma_1(n)$ are given in formulas (2) and (3) below.

$$\quad p(n)=\frac{1}{n}\sum\limits_{k=0}^{n-1}\,\sigma _1(n-k)\,p(k)=\frac{1}{n}\sum\limits_{k=1}^n\sigma_1(k)\,p(n-k)\,,\quad n>0\tag{2}$$

$$\quad \sigma_1(n)=\sum\limits_{k\ne 0}(-1)^{k-1}\,g_k\, p\left(n-g_k\right)=\sum\limits_{k>0}(-1)^{k-1}\,\left(g_k\, p\left(n-g_k\right)+g_{-k}\, p\left(n-g_{-k}\right)\right)\tag{3}$$

Note formula (2) above is explicitly defined as a finite sum, but formulas (1) and (3) above can also be evaluated as a finite sums since $p(n)=0\,\forall\,n<0$.

The function $\sigma_1(n)$ is related to the primes as follows:

$$\sigma_1(n)=n+1\iff n\in \mathbb{P}\tag{4}$$

If $n$ is composite then $\sigma_1(n)>n+1$ since composite values of $n$ have more divisors than just $1$ and $n$.

For more information on the partition function $p(n)$ see Weisstein, Eric W. "Partition Function P." From MathWorld--A Wolfram Web Resource, PartitionsP function, and Partition Function Identities.

Form more information on the divisor function $\sigma_1(n)$ see Weisstein, Eric W. "Divisor Function." From MathWorld--A Wolfram Web Resource and DivisorSigma Function.


问:
Constructing $g$-othogonal and symplectic basis of the vector space given a Lagrangian subspace.
Assume that for a vector space $V$ we are given a fixed symplectic form $\omega$ and a fixed inner product $g$. Given Lagrangian subspace $\Lambda$ such that $g$-orthogonal complement $\Lambda^\perp$ is also Lagrangian, construct a $g$-orthogonal and $\omega$-standard basis.

My construction: pick any orthogonal basis $(e_1,\dots , e_n)$ of $\Lambda$, and denote $\Lambda_i$ a subspace of $\Lambda$ spanned by $(e_1,\dots, e_{i-1},e_{i+1},\dots,e_n)$ (ie. span of all basis vectors except $e_i$). Then intersection of symplectic complement $\Lambda_i^{\omega}$ and $\Lambda^{\perp}$ is a one dimensional subspace, because $\Lambda_i^{\omega}$ is $n+1$ dim and contains $\Lambda$. Then by non-degeneracy there exists a vector $f_i\in \Lambda_i^{\omega}\cap \Lambda^{\perp}$ such that $\omega(e_i,f_i)=1$.

Note that $\omega(e_i,f_j)=0$ if $i \neq j$ and $\omega(f_i,f_j)=0$, hence $(e_1,\dots,e_n,f_1,\dots,f_n)$ gives a symplectic basis. Since $f_i \in \Lambda^{\perp}$ we have that $g(e_i,f_j)=0$, however I don't see how to prove that $g(f_i,f_j)=0$ for $i\neq j$. It seems to me that the choice of $f_i$ in this process is canonical so I don't see how I could change it to ensure orthogonality between $f_i$'s

答:
Regarding your method: Notice that you constructed the basis $\{f_1, \dots, f_n\}$ of $\Lambda^{\perp}$ without ever using $g|_{\Lambda^{\perp}}$, hence you can freely change $g$ on $\Lambda^{\perp}$ to produce examples for which  $\{f_1, \dots, f_n\}$  is not $g$-orthogonal. One could be tempted to apply a Gram-Schmidt process to $\{f_1, \dots, f_n\}$ to obtain a $g$-orthogonal basis, but this wouldn't give an overall $\omega$-symplectic basis.

Note that the process you described is not completely canonical, since you begin with an arbitrarily chosen $g$-orthogonal basis $\{e_1, \dots, e_n\}$ for $\Lambda$. The conclusion is thus that you should construct the bases $\{e_1, \dots, e_n\}$ and $\{f_1, \dots, f_n\}$ 'simultaneously'.

Regarding the original problem: There are undoubtly several ways to solve the problem, and you should certainly try and find one for yourself. However, for completeness, here is one possible solution.

Start with $g$-orthonormal bases $E' = \{e'_1, \dots, e'_n\}$ of $\Lambda$ and $F' = \{f'_1, \dots, f'_n\}$ of $\Lambda^{\perp}$, thereby obtaining a $g$-orthonormal basis for $V$. With respect to this last basis, $\omega$ reads as the block matrix $\left( \begin{array}{cc} 0 & M \\ -M^{T} & 0   \end{array} \right)$ where $M$ is an invertible $(n \times n)$-matrix. (In fact, $\omega$ takes this form regardless of whether $E'$ and $F'$ are orthonormal or orthogonal, and any invertible matrix $M$ can appear here.) The goal is to change $E'$ and $F'$ within $\Lambda$ and $\Lambda^{\perp}$ respectively to obtain a merely $g$-orthogonal basis for $V$ with respect to which $M$ becomes $Id$. Now, we only have the liberty to transform $E'$ into a $g$-orthogonal basis $E$ for $\Lambda$, which is possible by acting via an automorphism $P_{\Lambda} : \Lambda \to \Lambda$ which is read (with respect to $E'$) as the matrix $O_{\Lambda} D_{\Lambda}$ where $O_{\Lambda}$ is orthogonal and $D_{\Lambda}$ is invertible diagonal. Similarly, we are allowed to change $F'$ into a $g$-orthonormal basis $F$ of $\Lambda^{\perp}$ via a map $P_{\Lambda^{\perp}} = O_{\Lambda^{\perp}}D_{\Lambda^{\perp}}$. Under this process, $M$ is changed into $P_{\Lambda}^T M P_{\Lambda^{\perp}}$, which we want to equal $Id$. This means we need to find $O_{\Lambda}$, $D_{\Lambda}$, $O_{\Lambda^{\perp}}$ and $D_{\Lambda^{\perp}}$ such that $M = O_{\Lambda} D_{\Lambda}D_{\Lambda^{\perp}}^T O_{\Lambda^{\perp}}^T$. The fact that $M$ can be expressed in this form follows from polar decomposition and spectral theorem.


问:
Why does $\sin(90 + x) = \cos(x)$ and $\sin(90 - x) = \cos(x)$?
Why are both statements below true in regards of a trigonometric circumference?

$\sin(90 + x) = \cos(x)$

$\sin(90 - x) = \cos(x)$

答:
One way I have used to view the symmetry between the equations sin(x)=cos(90-x) is by visualizing two triangles:

two triangles with complementary angles

In the above image, you can find that the cosine of theta is x/hyp, while the sin of 90-theta is x/hyp.

Seeing it graphically helped he visualize how these two functions were related.


问:
Compute limit of a logarithmic term
Let $k\in\mathbb{N}$.

Is it possible to determine
$$
\lim_{n\to\infty}\frac{1}{n}\log\left(\left(\prod_{j=0}^n 4(k+j)\right)(k-1)(k+1)(k+3)\cdots(k+2n-3)\right)?
$$

Can write the log-term as sum,
\begin{align*}
&\log\left(\left(\prod_{j=0}^n 4(k+j)\right)(k-1)(k+1)(k+3)\cdots(k+2n-3)\right)\\
&=\sum_{j=k}^{n+k}\log(4)+\sum_{j=k}^{n+k}\log(j)+\log(k-1)+\sum_{i\in 2\mathbb{N}_0+1}^{2n-3}\log(k+i)
\end{align*}

The middle term tends to zero when dividing by $n$ and letting $n\to\infty$, i.e.,
$\frac{1}{n}\log(k-1)\to 0$.

The first sum is $\sum_{j=k}^{n+k}\log(4)=n\log(4)$, so that $\frac{1}{n}\sum_{j=k}^{n+k}\log(4)=\log(4)$.

答:
We have for $k\in\Bbb N\setminus\{1\}$ (since $k=1$ results in a zero inside the logarithm):
$$
L=\lim_{n\to\infty}\frac{1}{n}\log\left(\left(\prod_{j=0}^n 4(k+j)\right)(k-1)(k+1)(k+3)\cdots(k+2n-3)\right),
$$
which using the Pochhammer symbol can be simplified as
$$
\begin{aligned}
L
&=\lim_{n\to\infty}\frac{1}{n}\log\left(4^{n+1}(k+1)_n(k)_{2n-2}k(k-1)\right)\\
&=\lim_{n\to\infty}\frac{1}{n}\log\left(4^n\Gamma(n+k+1)\Gamma(2n+k-2)\frac{4k(k-1)}{k!\Gamma(k)}\right)\\
&=\lim_{n\to\infty}\frac{1}{n}\log\left(4^n\Gamma(n+k+1)\Gamma(2n+k-2)\right).
\end{aligned}
$$
Now using asymptotic expansions for the gamma function we further write
$$
\begin{aligned}
L
&=\lim_{n\to\infty}\frac{1}{n}\log\left(4^ne^{-3n}n^{n+k+1/2}(2n)^{2n+k-5/2}(2\pi)\right)\\
&=\lim_{n\to\infty}\frac{1}{n}\left(n\log4-3n+(n+k+1/2)\log n+(2n+k-5/2)\log(2n)\right)\\
&=\lim_{n\to\infty}\left(\log 4-3+\frac{n+k+1/2}{n}\log n+\frac{2n+k-5/2}{n}\log(2n)\right)\\
&=\log 4-3+\lim_{n\to\infty}\left(\log n+2\log(2n)+\underbrace{\frac{k+1/2}{n}\log n+\frac{k-5/2}{n}\log(2n)}_{\to 0}\right)\\
&=\log 4-3+\lim_{n\to\infty}\left(\log n+2\log(2n)\right)\\
&=\infty.
\end{aligned}
$$

Title: Compute limit of a logarithmic term


问:
If we localize at 2, is $S^n$ an H-space?
It is well known that the sphere $S^n$ is an H-space iff $n = 0, 1, 3, 7$. This has been proved by Adams in On the non-existence of elements of Hopf invariant one, but I have never read it and I don't know how the proof goes. However, this is a result for the "integral" spheres, but things change if we start inverting primes.

For example, if we just invert 2, then every odd sphere is an H-space. In fact, Serre proved in Groupes d'homotopie et classes de groupes abéliens that $\Omega S^{2n} \simeq S^{2n-1} \times \Omega S^{4n-1}$ away from 2 and hence $S^{2n-1}$ is an H-space since it is a retract of an H-space.

On the other side, even if we localize at an odd prime, no even sphere can be an H-space. In fact $H(S^{2n}; R) = R\langle 1, i_{2n}\rangle$ with $\Delta i_{2n} = i_{2n} \otimes 1 + 1 \otimes i_{2n}$.  If it would be an H-space, then $0 = \Delta i_{2n}^2 = (i_{2n} \otimes 1 + 1 \otimes i_{2n})^2 = 2 \ i_{2n} \otimes i_{2n}$ and this cannot happen unless $2 = 0$ in $R$.

Hence, these arguments solve the problem for all spheres at all odd primes. But both these arguments don't work at 2.

What happens at 2? If we localize at 2, can a sphere be an H-space, other than for $n = 0, 1, 3, 7$?

I think that, at least in theory, odd spheres could have two H-space structures, one at 2 and one away from 2, that don't "match up" integrally; and even spheres could have an H-space structure at 2, because the "integral failure" already happens at odd primes.

Nevertheless, in practice, I guess that $S^n$ cannot be an H-space unless $n = 0, 1, 3, 7$, even if we localize at 2. My feeling is that the problem is exactly at 2. Maybe Adams proved exactly that this cannot happen, but since I don't know the details of the paper...

Thanks in advance!

答:
We work with the mod $2$ Steenrod algebra $\mathcal{A}$ and take $\mathbb{Z}/2$ coefficients in all cohomology groups.

Start with the Adem relation
$$Sq(2^k)Sq(2^{k+1}l)=Sq(2^{k+1}l+2^k)+\text{decomposables}.$$
By induction on degree this identity yields:

The set $\{Sq^{2^k}\mid k\geq0\}$ generates the mod $2$ Steenrod algebra $\mathcal{A}$.

On the other hand, if we take a generator $u\in H^1\mathbb{R}P^\infty$ we have $Sq^{2^k}u^{2^{k}}=u^{2^{k+1}}$ and $Sq^ju^{2^k}=0$ whenever $0<j<2^k$. Thus:

$Sq^{2^k}$ is indecomposable in $\mathcal{A}$ for each $k\geq0$.

So suppose $S^n$ admits an H-multiplication. The Hopf construction yields a map $\varphi:S^{2n+1}\rightarrow S^{n+1}$ whose mapping cone $C_\varphi=S^{n+1}\cup_\varphi e^{2n+2}$ has
$$H^*C_\varphi=\mathbb{Z}_2[x]/(x^3),\qquad |x|=n+1.$$
Thus
$$Sq^{n+1}x=x^2\neq 0.$$
From the first fact above we immediately see that $n+1$ is a power of $2$. Otherwise $Sq^{n+1}$ decomposes as a polynomial in Steenrod operations of lower degrees. Since $C_\varphi$ has no cohomology in degrees $n+1,\dots,2n-1$, this would result in $Sq^{n+1}$ annihilating $x$.

Now all of these ideas have really been 2-local. The conclusion suffers no loss if we take as input a 2-local sphere. All the cohomology groups factor through the 2-local category, and the facts about the Hopf construction hold in much greater generality.

If the 2-local sphere $S^n$ admits an H-multiplication, then $n+1$ is a power of $2$.

The difficult part is to now discount the cases $n+1=2^k$ with $k>3$. The second fact stated above shows that more intricate methods must be employed. Adams's observation is that whenever $k> 3$, then the operation $Sq^{2^k}$ can in fact be decomposed outside of $\mathcal{A}$, at least in a certain sense.

More precisely Adams constructs secondary operations $\Phi_{i,j}$ for $0\leq i\leq j$, $j\neq i+1$, of degrees $2^i+2^j-1$, and demonstrates that the formula
$$Sq^{2^{k+1}}x=\sum_{i,j<k} a_{i,j,k}\Phi_{i,j}(x)+\text{indeterminacy}$$
holds on cohomology classes $x$ satisfying $Sq^{2^r}x=0$ for $0\leq r\leq 2^k$. Here $k\geq3$ and the coefficients $a_{i,j,k}$ are certain primary cohomology operations.
As an example, if $v\in H^2\mathbb{C}P^\infty$ is a generator, then $\Phi_{0,i}$ is defined on $v^{2^j}$ for $i\leq j$ and evaluates as $\Phi_{0,j}(v^{2^j})=v^{3\cdot 2^{j-1}}$ with no indeterminacy, and $\Phi_{0,i}(v^{2^j})=0$ with no indeterminacy for $i<j$.

Returning to our putative 2-cell complex $C_\varphi$ and assuming $n+1=2^k$, $k>3$, the secondary operations $\Phi_{i,j}$ are defined on $x$. There is no trouble in computing the indeterminacy in the above formula, and it vanishes for reasons of degree. Thus Adams has succeeded in factoring the action of $Sq^{n+1}$ on this complex through trivial cohomology groups. In particular $Sq^{n+1}x=0$, which is at odds with the construction of $C_\varphi$. The conclusion is that the 2-local complex $C_\varphi$ cannot exist.

Adams: If $n\neq 0,1,3,7$, then the 2-local sphere $S^n$ cannot admit an H-space structure.


问:
Minimal and characteristic polynomials of orthogonal transformations
I am given orthogonal linear transformation $U: \mathbb{R}^4 \to \mathbb{R}^4$, represented by

$$A=\begin{bmatrix} 1/2 & 1/2 & 1/2 & -1/2 \\ 1/2 & 1/2 & -1/2 & 1/2 \\ 1/2 & -1/2 & 1/2 & 1/2 \\ 1/2 & -1/2 & -1/2 & -1/2 \end{bmatrix}$$

I want to find the characteristic and minimal polynomials of $(A+A^{-1})$ and $A$. Can I somehow use the char. and min. polynomials of the former to compute the latter? I am also required to express $A$ as a block diagonal matrix, where the blocks are $1$ and $2$ dimensional reflections and rotations, representing $U$ wrt some orthonormal basis in $\mathbb{R}^4$.

答:
Since $A$ is indeed orthogonal, as you can easily check, $A^{-1}$ coincides with the transpose of $A$, and it is then easy to compute
$$A+A^{-1}=\pmatrix{1&1&1&0\cr 1&1&-1&0\cr 1&-1&1&0\cr 0&0&0&-1}=M.$$
From the last column it is clear the minimal polynomial of $M$ has a factor $X+1$, and the image of $M+I$ is the subspace $W$ spanned by $v_1=(2,1,1,0)$, and $v_2=(1,2,-1,0)$ (its two first columns); the restriction to$~W$ of (the action of) $M$ has matrix $\binom{2~~0}{0~~2}$, since $Mv_1=2v_1$ and $Mv_2=2v_2$. So $W$ is in fact the eigenspace of $M$ for the eigenvalue $2$, and $M$ is diagonalisable with minimal polynomial $\mu=(X+1)(X-2)$, and characteristic polynomial $\chi=(X+1)^2(X-2)^2=\mu^2$, since both eigenspaces are of dimension$~2$.

For the minimal and characteristic polynomials of $A$ itself it is now easiest to consider the restrictions of its actions to the two eigenspaces of $M$ (since $A$ and $M$ clearly commute, those eigenspaces are $A$-stable). On the basis $[v_1,v_2]$ of the eigenspace of $M$ for $\lambda=2$ one obtains $\binom{1~~0}{0~~1}$ as matrix of the restriction of $A$; for the eigenspace $\ker(M+I)$ of $M$ for $\lambda=-1$ one can choose the basis $[v_3,v_4]$ where $v_3=(-1,1,1,0)$ and $v_4=(0,0,0,1)$, and the restriction of $A$ to that eigenspace on that basis is $\binom{-1/2~~~1/2}{-3/2~~~-1/2}$, whose characteristic polynomial $X^2+X+1$ being square-free (indeed irreducible in $\Bbb R[X]$) is also its minimal polynomial (the latter can also be computed by computing the relation $v_4+Mv_4+M^2v_4=0$). Then the minimal polynomial of $A$ is $(X-1)(X^2+X+1)$ and its characteristic polynomial is $(X-1)^2(X^2+X+1)$.

In passing one has found the block matrix form of $A$ after basis change to $[v_1,v_2,v_3,v_4]$, namely with the two $2\times2$ blocks mentioned above (which is in fact a $3$-block diagonal form with the two first blocks equal to $(1)$, a $1\times1$ block). For the final block other forms are possible (for appropriate bases), such as the companion matrix $\binom{0~~-1}{1~~-1}$ of $X^2+X+1$.)


问:
Papers in algebraic topology
I am an undergraduate student, I have a decent background in algebraic topology, I have studied Munkres and Armstrong. I am planning to pursue research in algebraic topology and would like a list of published papers preferably in ascending order of complexity, to go through in these summers.</p>

答:
What about "Handbook of Algebraic Topology" by I.M James?
The book description:

"Algebraic topology (also known as homotopy theory) is a flourishing branch of modern mathematics. It is very much an international subject and this is reflected in the background of the 36 leading experts who have contributed to the Handbook. Written for the reader who already has a grounding in the subject, the volume consists of 27 expository surveys covering the most active areas of research. They provide the researcher with an up-to-date overview of this exciting branch of mathematics."

Title: Papers in algebraic topology


问:
On the adjoint of a densely defined unbounded operator
Let $H,K$ be Hilbert spaces, $D(T)\subset H$ be a dense vector subspace.

Suppose $T: D(T) \rightarrow K$ and $D(T^*)=K$. What can you say about $T$?

By definition, we have $D(T^*)= \{k \in K: \exists x\in H , \langle k, Th \rangle= \langle x, h \rangle, \forall h \in D(T) \}$.

I know that since $D(T)$ is dense and such $x$ exists, we can define the adjoint of $T$ by $T^*: D(T^*) \rightarrow H$ by $T^*(h)=x$.

But not sure what I can say about $T$.

Thank you.

答:
$D(T^\ast)=K$ is equivalent to the boundedness of $T$. I assume you already know that bounded operators have an everywhere defined adjoint, so I will only prove the converse implication.

First recall that the adjoint of a densely defined operator is always closed. Since $D(T^\ast)=K$ is complete, this implies that $T^\ast$ is bounded by the closed graph theorem. Thus also its adjoint $T^{\ast\ast}$ is bounded. Moreover, if $\xi\in D(T)$ and $\eta\in D(T^\ast)$, then
$$
\langle \xi,T^\ast \eta\rangle=\langle T\xi,\eta\rangle.
$$
Thus $\xi \in D(T^{\ast\ast})$ and $T^{\ast\ast}\xi=T\xi$. Therefore $T$ is bounded as the restriction of the bounded operator $T^{\ast\ast}$.


问:
Solving $y'' - 2y' + y = \delta(t-2)$ for y(0) = 0, y'(0) = 0 by using Laplace Transforms. Need help finishing the problem.
So I'm working on this question and I've taken the Laplcace transforms for everything and separated it as $Y(s) = F(s)G(s)$, and then I've put it into the convolution equation to get $y(t)$. My problem is, I don't understand how to fully/properly integrate this when I have the $\delta(T-2)$ inside the integral. Can anyone explain what to do? I'm adding a picture of what I have until now that I am sure is correct so that there's somewhat of an explanation of my train of thought? (the u and v' at the bottom were if I was going to solve with integration by parts, which I also am not sure of how to use here).
But yeah, please help and tell me where I'm going wrong...

答:
$$y'' - 2y' + y = \delta(t-2)$$
$$\implies Y(s)=\dfrac {e^{-2s}}{(s-1)^2}$$
The inverse Laplace transform of $\dfrac 1 {(s-1)^2}$ is simply $e^tt$ then apply the formula 27:
$$L^{-1} (e^{-cs}F(s))=u(t-c) f(t-c)$$
You should get :
$$y(t)=u(t-2)e^{t-2}(t-2)$$


问:
Using the Method of Undetermined Coefficients
I have been given a question that reads:

Use the method of undetermined coefficients to find an approximation for the second derivative of the form:
$$f''(0) \approx A f(-h) + B f(0) + C f(h),$$ which is exact for all quadratic polynomials.

To be perfectly honest, I have no idea how to commence this question, nor the purpose of it, nor how it works. Any help or guidance would be much appreciated!

答:
Let's use $$f(x)=ax^2+bx+c$$
Then $$f''(0)=2a$$
Now we plug in this form of $f$ into the original equation, and we know that for quadratic polynomial is exact (so $\approx$ becomes $=$)
$$f''(0)=A\left(a(-h)^2+b(-h)+c)\right)+B(a0^2+b0+c)+C(ah^2+bh+c)$$
Now let's group terms as powers of $h$:
$$2a=h^2a(A+C)+hb(-A+C)+c(A+B+C)$$
This is valid for any value of $h$, and any values of $a,b,c$. So $$A+B+C=0\\-A+C=0\\(A+C)h^2=2$$
This means $A=C$ (in the second equation), $$2Ah^2=2\\A=\frac{1}{h^2}$$
Then finally $B=-2A$. With these:
$$f''(0)=\frac{f(-h)-2f(0)+f(h)}{h^2}$$
This is the same formula as on wikipedia, with $x=0$.


问:
A Thousand Points
A 1000 points are graphed in the co-ordinate plane. Explain why it is possible to draw a straight line in the plane so that half the points on one side of the line and half are on the other.

This question was taken from Precalculus by Stewart,Watson.

答:
Start with 2 points A  and B, draw AB's perpendicular line k to separate A and B.

Next,  add 2  points C and D. There are 3 different scenarios:

C , D on different sides of k. No action needed.

Both C,D  or one of them is on line k (assume C on line k, D  either on k below C or to the left of k).

Find a point Y slightly lower than C,  rotate k around Y CCW  by an
infinitely small angle $\delta \theta$ to move C to the right side of
k(since the size of a point is 0),  k  separates 4 points evenly.

C, D  on the same side of k (assuming left side).

Among all points on the left side, find the  point set
{$X_i$} that is closest to line k.

Move k parallelly to overlap with {$X_i$}.

Let $X_0$ be the top of all the points (for example C). Y is a point
slightly lower than $X_0$ on k.

Rotate k around Y CCW  by an infinitely small angle $\delta \theta$ to
move C to the right side of k,  then k  separates all points evenly.

Since $\delta \theta$ is an infinitely small angle, after each rotation adjustment,k is almost still  perpendicular to AB (given the adjustment time is not infinity).

Repeat these steps by adding 2x points once a time and adjust k's location and orientation accordingly  till we reach 1000 points.

Title: A Thousand Points