问:
Question about induction with summations involved
I need to prove the following but I'm struggling to do so:

$$\left(r+1\right)\sum _{j=0}^n\:\frac{\left(j+r\right)!}{j!}=\frac{\left(n+r+1\right)!}{n!\:}$$

So firstly I considered the case when $n = 0$ and I got that $(r+1)!$ was on both the LHS and RHS so it was true for $n = 0$.

I assumed it was true when $n=k$ but I'm not sure what to do for the $n=k+1$ step

答:
The left-hand side with $n=k+1$ is
$$
(r+1)\sum_{j=0}^{k+1}\frac{(j+r)!}{j!}=
(r+1)\sum_{j=0}^k\frac{(j+r)!}{j!}+(r+1)\frac{(k+1+r)!}{(k+1)!}
$$
By the induction hypothesis, you can go on with
$$
=\frac{(k+r+1)!}{k!}+(r+1)\frac{(k+1+r)!}{(k+1)!}
$$
Collect the common terms and finish up.


问:
Inequality of harmonic number $\left| \sum_{i=1}^{n}\frac{1}{i}-\log{n}-\gamma \right| \leq \frac{10}{n}$
In my Number Theory textbook, it was quoted without proof that for all positive integers $n$,
$$\left| \sum_{i=1}^{n}\frac{1}{i}-\log{n}-\gamma \right| \leq \frac{10}{n}$$
where $\gamma = 0.577...$ is the Euler–Mascheroni constant.

From my Calculus courses, I knew that
$\displaystyle\lim_{n\to\infty}\left(\sum_{i=1}^{n}\frac{1}{i}-\log{n}\right)=\gamma$

However, I am not aware of the inequality above.

I have no ideas of how to prove it, and couldn't find sources about it.

Is it a well-known result? Is it approachable? Is there a name for it? Are there resources about it?

Alternatively, a proof is also very welcomed.

Thank you very much.

答:
Let $u_n=\sum_{i=}^n \frac{1}{i}-\log n$ and $v_n=\sum_{i=1}^n\frac{1}{i}-\log(n+1)$, then $(u_n)$ is a decreasing sequence and $(v_n)$ is a nondecreasing sequence. Since they both converge towards $\gamma$, we have $v_n\leqslant\gamma\leqslant u_n$ for all $n$, which means that
$$ 0\leqslant\sum_{i=1}^n\frac{1}{i}-\log n-\gamma\leqslant u_n-v_n=\log\left(1+\frac{1}{n}\right)\leqslant\frac{1}{n} $$

ContentLicense: CC BY-SA 4.0


问:
Find the volume of the solid generated by revolving the triangular region enclosed by $y = |x|$ and $y = 1$ about the line x = −2.
Question:
Find the volume of the solid generated by revolving the triangular region enclosed by $y = |x|$ and $y = 1$ about the line $x = −2$.

My solution:

$$\pi*9*1 - \pi*1*1 - 1/3 *\pi* 1 - 1/3 *\pi* 1 = 22/3 \pi$$

Use the large cylinder minus the small cylinder and minus two cones.

But the correct answer is 4pi

Did I do anything wrong?

答: Here is an outline of a method that works: Disc/Washer Method. You need to write the equations in terms of $y$ because you rotate about a vertical line. The integral will be of the form $\int(y+2)^2dy-\int(2-y)^2dy$ where $y$ is going from $0$ to $1$. Try to understand how this integral is set up by "translating" $y=|x|$ as well as $x=-2$ two units to the right. Now you work this out. (The correct answer is indeed $4\pi$)


问:
$\xi$ be a primitive cube root of unity
Could anyone tell me first of all the below problem is wrong or okay?I am not able to figure out how to solve

$\xi$ be a primitive cube root of unity define $A=\begin{pmatrix}\xi^{-1}&0\\0&\xi\end{pmatrix}$, for a vector $v=(v_1,v_2)\in\mathbb{R}^2$ define $$|v|_A=\sqrt{|vAv^T|}$$ if $w=(1,1)$ what is the value of $|w|_A$?

答:
$$
\xi^{-1}+\xi=\frac{1+\xi^2}{\xi}
$$

Since $\xi$ is a primitive cubic root of $1$, $\xi^2+\xi+1=0$, so $1+\xi^2=-\xi$;
hence
$$
\begin{pmatrix}1 & 1\end{pmatrix}
\begin{pmatrix}\xi^{-1} & 0 \\ 0 & \xi\end{pmatrix}
\begin{pmatrix}1 \\ 1\end{pmatrix}
=\xi^{-1}+\xi
=\frac{1+\xi^2}{\xi}
=\frac{-\xi}{\xi}=-1.
$$
Thus
$$
|v|_A=\sqrt{\lvert-1\rvert}=1
$$

ContentLicense: CC BY-SA 3.0

Title: $\xi$ be a primitive cube root of unity


问:
Prove or disprove that if X is homeomorphic to Y in $\mathbb{R}^n$ then $\mathbb{R}^n\setminus X$ is homeomorphic to $\mathbb{R}^n \setminus Y$
Intuitively if I imagine homeomorphisms to be continuously transforming from one set to the other, it feels like it is true. However, I'm struggling to find an explicit homeomorphism f.</p>

答: I don't think it's true.  The whole $\Bbb R^n$ is homeomorphic to a ball.  But the complement of a ball is not homeomorphic to the empty set.


问:
Predicate Logic translation from English to formula
I am studying for an exam and I have these exercises I practiced.

Let:
Monkey(x) = "x is a monkey"
Intelligent(x) = "x is an intelligent being"
Likes(x, y) = "x likes y"

No monkey likes a tiger.

let Tigers(x) = "x is a tiger"

∀x(Tiger(x) => ∀y¬Likes(y, x)

There exists an intelligent monkey.

∃x(Monkey(x) => Intelligent(x))

Every monkey likes another monkey

∀x(Monkey(x) => ∀yLikes(y, x))

There is only one smart being who likes all other monkeys

∃!x(Monkey(x) => ∀yLikes(y, x))

Everybody likes my monkey, but my monkey doesn't like nobody else but be.

∀x(Monkey(x) ∧  ∃(y)(Monkey(y) ∧ x  ̸= y ⇒ Likes(y, x)))

I could use some feedback, please?
how would you write this?

答:
Make sure that $y$ is a Monkey. Also, it is a little unusual to your sentence starts with introducing the Tigers, rather than the Monkeys, given that the subject term of the sentence is the Monkeys. Now, as it turns out, you can make this work, because both Tigers and Monkeys are universally quantified, but you're lucky it does. I would urge you to start by introducting the Monkeys, and then the Tigers.

Use a conjunction here instead of a conditional

$y$ needs to be a Monkey too. You may also have to explicitly specific that $y$ is not the same as $x$, since it says that $x$ likes some other monkey. And finally, it is some other monkey, so you need an existential for the $y$

Yet again, $y$ needs to be a Monkey.... and different from $x$. And I suppose that for the 'smart' you should use the Intelligent predicate. Finally, the English sentence is ambiguous: is there exactly one monkey that is Intelligent .. and that one intelligent monkey happens to like all other monkeys?  Or is there exactly one monkey that has both properties of being intelligent and liking all other monkeys?

This should not be a universal at the start: it's a bout one monkey.  In fact, it's about a specific monkey ... 'my monkey' ... there is not something in the key that lets you refer to 'my monkey'?


答: $\Bbb R^n\setminus\{0\}$ is homeomorphic to $\{x\in\Bbb R^n\,:\, \lVert x\rVert>1\}$, but a closed ball is not homeomorphic to a point.

Title: Prove or disprove that if X is homeomorphic to Y in $\mathbb{R}^n$ then $\mathbb{R}^n\setminus X$ is homeomorphic to $\mathbb{R}^n \setminus Y$

Body: Intuitively if I imagine homeomorphisms to be continuously transforming from one set to the other, it feels like it is true. However, I'm struggling to find an explicit homeomorphism f.</p>


问:
How to prove a limit with a recurrence?
$s_1 = 1$ and $s_{n+1} = \dfrac{s_n + 1}{3}$ for $n \in \Bbb N$.

How do you find $\displaystyle \lim_{x\to \infty} s_n$?

Then how do you prove that the value is the limit using the definition of the limit of a sequence?  $|s_n - s| < \epsilon$ for $n$ sufficiently large.

I know the limit is $\frac{1}{2}$. I'm just having trouble with the recurrence aspect.

答:
Solve the recurrence, the limit falls out of the solution. By "generatingfuctionology", define $S(z) = \sum_{n \ge 0} s_n z^n$, multiply your recurrence by $z^n$ and add for $n \ge 0$ to get:
$$
\frac{S(z) - s_0}{z} = \frac{1}{3} S(z) + \frac{1}{3} \cdot \frac{1}{1 - z}
$$
From the recurrence "backwards" you have $s_0 = 2$, so:
$$
S(z) = \frac{1}{2} \cdot \frac{1}{1 - z} + \frac{3}{2} \cdot \frac{1}{1 - z / 3}
$$
This is two geometric series:
$$
s_n = \frac{1}{2} + \frac{3}{2} \cdot 3^{-n}
$$
Thus:
$$
\lim_{n \to \infty} s_n = \frac{1}{2}
$$</p>

Title: How to prove a limit with a recurrence?


问:
Is it possible to prove from the definition of big $O$ that $5n^3+7n+1$ is $O(n^3)$?
Is it possible to prove from the definition of big O that $5n^3+7n+1$ is $O(n^3)$? Can this be generalised to any case where you have to (and what is the procedure for  working it out?) I guess the main question is there a specific term for this kind of procedure?

from the workings I've been given in class given $T\le O(f(n))$ if $n=1$ then $5n^3+7n+1 = 5^3+7+1 (23)$. Although I'm not sure how that relates to proving the above.

Thanks,

P.L.G.

答:
All you need show to prove $f(n) = O(g(n))$ is that for some $N$ and some $C$ that

$$
|f(n)| < c|g(n)|\quad\text{for all}\quad n\ge N.
$$
You need not be at all efficient about this.  So, to show $5n^3+7n+1 = O(n^3)$, assume $n \ge N = 1$.

$$
\begin{aligned}
5n^3+7n+1 &\le5n^3+7n^3 + 1\cdot n^3\\
&= 13n^3
\end{aligned}
$$

You can reduce the constant $13$ to $5+\epsilon$ by taking $N$ large enough.  For example, find $N$ such that $$7n + 1 < n^3\quad\text{for}\quad n > N.$$


问:
Is the beta distribution the only conjugate prior distribution to the binomial distribution?

Is the beta distribution the only conjugate prior distribution to the binomial distribution?

I understand that the beta distribution is a conjugate prior however I can find very little information online which confirms if this is the only conjugate prior distribution and so I have posted this question.

Many thanks

答:
It is slightly misleading to say the beta distribution is the conjugate prior for a binomially distributed likelihood, as there is more than one beta distribution. It would be better to say that beta distributions are a conjugate family of distributions for   binomially distributed likelihoods.

The reason they are conjugate is that their densities are proportional to $\theta^\alpha (1-\theta)^\beta$ and when multiplied by a binomial likelihood you get a posterior density from the same beta distribution family though with different $\alpha$ and $\beta$.

There are other families for which this would be true:  indeed with a well-behaved  function $f(x)$ on $[0,1]$ (non-negative and integrable to a finite positive value), you could have a family of distributions with density proportional to  $f(\theta)\theta^\alpha (1-\theta)^\beta$ so that when multiplied by a binomial likelihood you get a posterior density from the same distribution family though with different $\alpha$ and $\beta$.

For example you could get strange wavy priors such as $2\cos^2(10\pi \theta)$ leading to a conjugate family with densities proportional to $\cos^2(10\pi \theta)\theta^\alpha (1-\theta)^\beta$.  These particular distributions are not well-known (I just invented them), not as easy to handle and not as useful, but they make up a conjugate family with some curious properties: if say you observed $7000$ successes and $13000$ failures, then your posterior distribution would be bimodal.


问:
Why is $\phi$ an isometry?
I want to prove this statement but honestly I'm not sure if it's true or not and don't see how I can proceed.

Let $X$ a normed vector space and $\phi$ : $X\rightarrow (X')'$ , the linear mapping given by : $(\phi(x))(f)=f(x) \;, f\in X $.

If $X$ is a Hilbert space (with its norm induced by the dot product), we can say that  $\phi$ is an isometry (bijective)  preserving the operator norm on $E'$ and $(E')'$.

Thanks in advance for your help.

答:
We need to show that $\|\phi(x)\| = \|x\|$ for every $x\in X$.

Fix $x \in X$ and notice that for every $f \in X'$ we have
$$|\phi(x)(f)| = |f(x)| \le \|f\|\|x\|$$
which means $\|\phi(x)\| \le \|x\|$.

To show the reverse inclusion, we need to find $f \in X'$ such that $\|f\|=1$ and $f(x)=\|x\|$. In this case we have
$$\|\phi(x)\| \ge \frac{|\phi(x)(f)|}{\|f\|} = |f(x)|=\|x\|.$$

If $X$ is a Hilbert space, we can define such a functional simply as $y \mapsto \left\langle y, \frac{x}{\|x\|}\right\rangle$ via the Riesz representation theorem. In general we have to use a corrolary of the Hahn-Banach theorem which gives us exactly such an $f$.

Title: Why is $\phi$ an isometry?


答:
Let $X$ be a normed TVS. Fix $x\in X.$ Now, $\|f(x)\|\le \|f\|\cdot \|x\|$ so $\|\phi(x)f\|\le \|f\|\cdot \|x\|$ so

$\tag1 \|\phi(x)\|\le \|x\|.$

On the other hand, the Hahn-Banach theorem gives us an $f\in X'$ such that $\|f\|=1$ and $\phi(x)f=\|f(x)\|=\|x\|$ so

$\tag2\|x\|=\phi(x)f\le \|\phi(x)\|\cdot \|f\|=\|\phi(x)\|$

so $\phi$ is indeed an isometry.

Title: Why is $\phi$ an isometry?

Body:
I want to prove this statement but honestly I'm not sure if it's true or not and don't see how I can proceed.

Let $X$ a normed vector space and $\phi$ : $X\rightarrow (X')'$ , the linear mapping given by : $(\phi(x))(f)=f(x) \;, f\in X $.

If $X$ is a Hilbert space (with its norm induced by the dot product), we can say that  $\phi$ is an isometry (bijective)  preserving the operator norm on $E'$ and $(E')'$.

Thanks in advance for your help.


问:
Showing that the free group of a disjoint union is isomorphic to the free product of the corresponding free groups
P. Aluffi's "Algebra: Chapter $\it 0$", exercise II.$5.8$.

Still more generally, prove that $F(A\amalg B)=F(A)*F(B)$ and that $F^{ab}(A\amalg B)=F^{ab}(A)\oplus F^{ab}(B)$ for all sets $A,B$. $($That is, the constructions $F,F^{ab}$ 'preserve coproducts'.$)$

Here $F(A)$ is the free groups on the set $A$, $A\amalg B$ is the disjoint union of $A$ and $B$, and $G*H$ is the free product of $G$ and $H$ (i.e. the coproduct in $\sf Grp$). All of those are characterized by their usual universal properties which will be used extensively for the proof.$^*$

Proof. We will show that $F(A\amalg B)$ satisfies the universal property of $F(A)*F(B)$. For startes, we will construct the (canonical) inclusion homomorphisms. Thus, consider the following diagrams

Here $\iota_A,\iota_B,\iota$ are the inclusion of $A,B,A\amalg B$ into their respective free groups. The (unique) group homomorphisms $I_A,I_B$ are induced by considering the compositions $\iota\circ i_B$ and $\iota\circ i_B$ and the universal properties of $F(A)$ and $F(B)$. Hence they are such that
$$I_A\circ\iota_A=\iota\circ i_A,~~~I_B\circ\iota_B=\iota\circ i_B$$
Now, suppose we are given group homomorphisms $g_A\colon F(A)\to G,\,g_B\colon F(B)\to G$ to some arbitrary group $G$. We can consider them in particular as set-functions and precomposing with $\iota_A$ and $\iota_B$, respectively, gives us the following

The unique map $g$ is given by the universal property of $A\amalg B$ and such that
$$g\circ i_A=g_A\circ\iota_A,~~~g\circ i_B=g_B\circ\iota_B$$
Finally, $\overline{g}$ induces a unique map $\overline{f}$ such that

by the universal property of $F(A\amalg B)$ and so $\overline{g}\circ\iota=g$. Composing gives us uniquely determined group homomorphisms $\overline{g}\circ I_A\colon F(A)\to G$ and $\overline{g}\circ I_B\colon F(B)\to G$ factoring through $F(A\amalg B)$. It remains to show the following

$$\overline{g}\circ I_A=g_A,~~~\overline{g}\circ I_B=g_B$$

But using the given commutativity relations we see that

$$(\overline{g}\circ I_A)\circ\iota_A=\overline{g}\circ(I_A\circ\iota_A)=\overline{g}\circ(\iota\circ i_A)=(\overline{g}\circ\iota)\circ i_A=g\circ i_A=g_A\circ\iota_A$$

and hence both triangles in the following diagram commute

The right triangle corresponds to the fact the by $g_A\circ\iota_A$ universally induced map is $g_A$ itself. But from the left triangles we see that $\overline{f}\circ I_A$ makes the corresponding diagram commute as well and hence $\overline{f}\circ I_A=g_A$ by the uniquesness of the induced map. The same argument, with all $A$s replaced by $B$s, yields $\overline{f}\circ I_B=g_B$. Thus, we conclude that $F(A\amalg B)$ satisfies the universal property of $F(A)*F(B)$ as every pair of group homomorphisms $g_A\colon F(A)\to G,\,g_B\colon F(B)\to G$ factors uniquely through $F(A\amalg B)$ using $I_A,I_B$ and $\overline{g}$. Hence, $F(A\amalg B)\cong F(A)*F(B)$.

The argument is precisely the same for $F^{ab}$ where we note that $G*H=G\oplus H=G\times H$ for abelian groups.   $\square$

Is the given proof correct; if so, can it be (substantially) improved? If not, where did I went wrong? I am not sure how to show actual equality instead 'mere' isomorphy (which is enough for me, to be honest) and the last part, i.e. showing that $\overline{g}\circ I_A=g_A$, is a little bit sketchy to me, even though I am quite sure the argument works.

Thanks in advance!

$^*$ I know that this preservation essentially boils down to 'left adjoints presever colimits' as the (binary) coproduct is a simple colimit and the free functor $F\colon\sf{Set}\to\sf{Grp}$ is left adjoint to the forgetful functor $\sf{Grp}\to\sf{Set}$. However, I would like to not dabble to deep into category theoretic terrain if it does not make the proof easier/more understandable. So the given proof is more or less from scratch.

答:
So, let me argue for proving the other universal property; that is, showing that $F(A)*F(B)$ has the universal property of $F(A\amalg B)$.

Let $u_A\colon A\to F(A)$ and $u_B\colon B\to F(B)$ be the canonical maps from the set to the corresponding free group. Let $\iota_A\colon F(A)\to F(A)*F(B)$ and $\iota_B\colon F(B)\to F(A)*F(B)$ be the canonical inclusions into the free product. And let $j_A\colon A\to A\amalg B$ and $j_B\colon B\to A\amalg B$ be the set-theoretic inclusions into the disjoint union/set-theoretic coproduct.

We want to show that there is a map $u_{A\amalg B}\colon A\amalg B\to F(A)*F(B)$ such that for every group $G$ and every set-theoretic map $f\colon A\amalg B\to G$, there exists a unique group homomorphism $\mathscr{F}\colon F(A)*F(B)\to G$ such that $\mathscr{F}\circ u_{A\amalg B}=f$.

So, first, the maps $\iota_A\circ u_A\colon A\to F(A)*F(B)$ and $\iota_B\circ u_B\colon B\to F(A)*F(B)$ yield a unique map $u_{A\amalg B}\to F(A)*F(B)$ such that $u_{A\amalg B}\circ j_A=\iota_A\circ u_A$ and $u_{A\amalg B}\circ j_B=\iota_B\circ u_B$.

Now let $f\colon A\amalg B\to G$. The map $f\circ j_A\colon A\to G$ induces a morphism $F_A\colon F(A)\to G$ with $F_A\circ u_A=f\circ j_A$; similarly, we have a morphism $F_B\colon F(B)\to G$ with $F_B\circ u_B=f\circ j_B$. And the maps $F_A$ and $F_B$ induce a morphism $\mathscr{F}\colon F(A)*F(B)\to G$ with $\mathscr{F}\circ \iota_A=F_A$ and $\mathscr{F}\circ \iota_B=F_B$. We want to show that this $\mathscr{F}$ satisfies $\mathscr{F}\circ u_{A\amalg B}=f$, and that it is the unique map with this property.

The universal property of $A\amalg B$ tells us that $f$ is the unique map $g\colon A\amalg B\to G$ such that $g\circ j_A=f\circ j_A$ and $g\circ j_B=f\circ j_B$. So if we can prove that $\mathscr{F}\circ u_{A\amalg B}$ also has this property, then we will have the equality with $f$.

Now,
$$\begin{align*}
(\mathscr{F}\circ u_{A\amalg B})\circ j_A&=\mathscr{F}\circ (u_{A\amalg B}\circ j_A)\\
&= \mathscr{F}\circ (i_A\circ u_A)\\
&=(\mathscr{F}\circ i_A)\circ u_A\\
&= F_A\circ u_A\\
&= f\circ j_A;\\
(\mathscr{F}\circ u_{A\amalg B})\circ j_B &= \mathscr{F}\circ (u_{A\amalg B}\circ j_B)\\
&= \mathscr{F}\circ (i_B\circ u_B)\\
&= (\mathscr{F}\circ i_B)\circ u_B\\
&= F_B\circ u_B\\
&= f\circ j_B.
\end{align*}$$
Therefore, $\mathscr{F}\circ u_{A\amalg B}=f$, as desired.

As for uniqueness, let $\mathscr{G}\colon F(A)*F(B)\to G$ be such that $\mathscr{G}\circ u_{A\amalg B} = f$. We want to show that $\mathscr{G}=\mathscr{F}$. Since $\mathscr{F}$ is the unique map such that $\mathscr{F}\circ i_A=F_A$ and $\mathscr{F}\circ i_B=F_B$, it is enough to show that $\mathscr{G}$ has this property as well.

To show that $\mathscr{G}\circ i_A=F_A$, it is enough to show that it has the universal property of $F_A$, namely, that $(\mathscr{G}\circ i_A)\circ u_A=f\circ j_A$. And, indeed,
$$\begin{align*}
(\mathscr{G}\circ i_A)\circ u_A &= \mathscr{G}\circ (i_A\circ u_A)\\
&= \mathscr{G}\circ(u_{A\amalg B}\circ j_A)\\
&= (\mathscr{G}\circ u_{A\amalg B})\circ j_A\\
&= f\circ j_A,
\end{align*}$$
as desired. Thus, $\mathscr{G}\circ i_A=F_A$, as needed. Symmetrically, $\mathscr{G}\circ i_B=F_B$. This proves that $\mathscr{G}=\mathscr{F}$, giving uniqueness, and we are done.


问:
Derive SDE for $ Z_t = \frac {X_t}{Y_t} $
Derive SDE for $ Z_t = \frac {X_t}{Y_t} $ using the 2 SDE below:

$ dX_t = rX_tdt + \sigma_XX_tdW_t $

$ dY_t = rY_tdt + \sigma_XY_td\tilde {W}_t $

I got $ dZ_t = Z_t\sigma_X(dW_t - d\tilde{W}_t) $, is this correct or should I use product rule but I'm quite uncertain? Thank you

答:
Given two stochastic processes $X_t$ and $Y_t$ and letting $Z_t = \frac{X_t}{Y_t}$, then $dZ_t$ is given by the quotient rule
$$dZ_t = \frac{Y_tdX_t - X_tdY_t - dX_tdY_t}{Y_t^2} + \frac{X_t}{Y^3_t}(dY_t)^2$$

This can be found by using Ito's multidimensional formula using $f(x, y) = \frac{x}{y}$.

Now Letting $dX_t = rX_tdt + \sigma_x X_t dW_t$ and $dY_t = rY_tdt + \sigma_x Y_t d\bar{W}_t$ we have

$$(dY_t)^2 = r^2Y_t^2(dt)^2 + 2r\sigma_x Y_t^2dtd\bar{W}_t + \sigma^2_xY_t^2(d\bar{W}_t)^2 = \sigma_x^2Y_t^2dt$$

$$dX_tdY_t = r^2X_tY_t(dt)^2 + r\sigma_x X_tY_tdtd\bar{W} + r\sigma_xX_tY_tdtdW_t + \sigma_x^2X_tY_tdW_td\bar{W_t}$$
$$= \sigma_x^2X_tY_t\rho dt = 0$$
This is assuming $dW_t$ and $d\bar{W_t}$ are independent, $\rho = 0$. Otherwise if $\rho \ne 0$ and we would still have $\sigma_x^2X_tY_t\rho dt$. Using what we have above, we get

$$dZ_t = \frac{Y_t(rX_tdt + \sigma_xX_tdW_t) - X_t(rY_tdt + \sigma_xY_td\bar{W_t})}{Y_t^2} + \frac{X_t}{Y_t^3}\sigma_x^2Y_t^2dt$$
$$= \frac{\sigma_xX_tdW_t - \sigma_xX_td\bar{W_t} + \sigma_x^2X_tdt}{Y_t}$$
$$dZ_t = \sigma_x^2Z_tdt + \sigma_xZ_t(dW_t - d\bar{W_t})$$


问:
Is projection of points onto plane considered a linear transformation
I have a set of points in $\mathbb{R}^3$ and I need to project them onto a plane with origin $p$ and normal $n$. I can project individual point $v$ like this:
$$x = v + \frac{(p-v)\cdot n}{n\cdot n}n$$
However I wonder if projection of a set of points onto line is considered a linear transformation. Such that we can use a matrix that can be calculated once? Or do we have to perform the same operation on each point individually?

答:
It will not be linear if the plane does not contain the origin (since it will send $v=0$ to a nonzero point.)

If it does contain the origin, then $x = v - \frac{v \cdot n}{n \cdot n} n$ can be written as $x = Mv$ where $M = I - \frac{1}{n \cdot n} nn^\top$.


答:
Start with the definition of $\gamma$ as

$$\gamma=\lim_{N\to\infty}\left(\sum_{i=1}^N{1\over i}-\ln N\right)$$

so

$$\begin{align}
\sum_{i=1}^n{1\over i}-\ln n-\gamma
&=\lim_{N\to\infty}\left(\ln N-\ln n-\sum_{i=n+1}^N{1\over i} \right)\\
&=\lim_{N\to\infty}\left(\int_n^N{dt\over t}-\sum_{i=n+1}^N{1\over i}\right)\\
&=\lim_{N\to\infty}\sum_{i=n+1}^N\int_{i-1}^i\left({1\over t}-{1\over i}\right)dt\\
&\lt\lim_{N\to\infty}\sum_{i=n+1}^N\left({1\over i-1}-{1\over i}\right)\\
&={1\over n}
\end{align}$$

Remark: A somewhat more careful estimate on the integral, using the concave nature of $1/t$, gives the inequality $\lt1/(2n)$, mentioned by Henry in comments.

Title: Inequality of harmonic number $\left| \sum_{i=1}^{n}\frac{1}{i}-\log{n}-\gamma \right| \leq \frac{10}{n}$

Body:
In my Number Theory textbook, it was quoted without proof that for all positive integers $n$,
$$\left| \sum_{i=1}^{n}\frac{1}{i}-\log{n}-\gamma \right| \leq \frac{10}{n}$$
where $\gamma = 0.577...$ is the Euler–Mascheroni constant.

From my Calculus courses, I knew that
$\displaystyle\lim_{n\to\infty}\left(\sum_{i=1}^{n}\frac{1}{i}-\log{n}\right)=\gamma$

However, I am not aware of the inequality above.

I have no ideas of how to prove it, and couldn't find sources about it.

Is it a well-known result? Is it approachable? Is there a name for it? Are there resources about it?

Alternatively, a proof is also very welcomed.

Thank you very much.


问:
Am I correctly using the conditional probability notation in this problem?
Problem: Two numbers are chose at random from among the numbers 1 to 10 without replacement.

Goal: Find the probability that the second number chosen is 5.

Official Solution:
$A_{i}:i \in \left\{1,2,...,10\right\}$ is event that first number chosen is $i$. $B$ is the event that second number chosen is 5. Use the total probability law to get: $$\begin{equation}P(B)=\sum_{i=1}^{10}P(B|A_{i})P(A_{i}) = 9 \, \frac{1}{9} \, \frac{1}{10}=\frac{1}{10}\end{equation}$$

My Solution:$$\begin{equation*}P(B|A_{i}) = \frac{n(B)}{n(S)} = \frac{9}{9 \cdot 10} = \frac{1}{10}\end{equation*}$$

Question: How can we at all write $P(B)$? Is probability for event $B$ not inherently conditional? You can't choose the second number without choosing the first number, so how could one write anything other than $P(B|A)$?

答:
Alternative approach:

Question: How can we at all write $P(B)$? Is probability for event $B$ not inherently conditional? You can't choose the second number without choosing the first number, so how could one write anything other than $P(B|A)$?

First thought experiment:

Lottery with tickets 1 through 10.  A mediator picks two of the lottery tickets.  He then arbitrarily chooses one of the two tickets and places it in Envelope-1.  He then places the other ticket in Envelope-2.  Clearly, the chance of the ticket-5 being in Envelope-1 is the same as the chance of the ticket-5 being in Envelope-2.

Second thought experiment:

Same as the first, but instead, the mediator picks the two numbers one at a time.  The first number picked goes in Envelope-1.  The second number picked goes in Envelope-2.  The analysis is identical to the first thought experiment.  It doesn't matter whether Envelope-1 contains the first ticket (re second thought experiment) or whether it is 50-50 whether Envelope-1 contains the first ticket or the second ticket (re first thought experiment).

Therefore, the problem can be reduced to the probability of drawing a 5 on the first draw.  Therefore, without involving any conditional probabilities, $p(B) = \frac{1}{10}.$

However, the only way to utilize the above approach is to first develop your intuition to the point that you are absolutely certain that my approach is valid.

Addendum

Justifying the analysis from a different perspective.

The tickets 1 thru 10 are lined up in a row.  Then the mediator randomly assigns a number 1 thru 10 (without replacement) to each ticket.  The assigned number represents when the ticket will be drawn.

That is, if the ticket-5 is assigned the number 1, it will be drawn first.  If the ticket-5 is assigned the number 2, it will be drawn second.

The ticket-5 is equally likely to be assigned the number 1 as it is to be assigned the number 2.  Therefore, from this alternate perspective, (again) the chance of ticket-5 being drawn second is the same as the chance of ticket-5 being drawn first.


问:
Integrate $\int_{0}^{1} u^{\frac{\alpha}{\beta}-\alpha}(1-\alpha) \log \left((1-\alpha) u^{-\alpha}\right) d u $
Where $c(u, v)=u^{-\alpha}(1-\alpha) $ is the Marshall-Olkin copula density, we have the following integral:
\begin{align}
I_{1} &=\iint_{E} c(u, v) \log c(u, v) d u d v \\
&=\int_{0}^{1} \int_{0}^{u^{\frac{\alpha}{\beta}}} u^{-\alpha}(1-\alpha) \log \left(u^{-\alpha}(1-\alpha)\right) d v d u
\end{align}

They calculate this integral as follows:
$$
\begin{aligned}
I_{1} &=\int_{0}^{1} u^{\frac{\alpha}{\beta}-\alpha}(1-\alpha) \log \left((1-\alpha) u^{-\alpha}\right) d u \\
&=(1-\alpha) \log (1-\alpha) \int_{0}^{1} u^{\frac{\alpha}{\beta}-\alpha} d u-\alpha(1-\alpha) \int_{0}^{1} u^{\frac{\alpha}{\beta}-\alpha} \log u d u \\
&=\frac{(1-\alpha) \beta \log (1-\alpha)}{\alpha-\alpha \beta+\alpha}+\frac{\alpha(1-\alpha) \beta^{2}}{(\alpha-\alpha \beta+\beta)^{2}}
\end{aligned}
$$
In the last 3 lines, what are all the steps in-between? I don't understand how the first becomes the second, and the second becomes the last line.

答:
The first equality follows from standard properties of the logarithm, explicitly:
$$\log{((1-\alpha)u^{-\alpha})}=\log{(1-\alpha)}+\log{(u^{-\alpha})} = \log{(1-\alpha)}-\alpha\log{(u)}$$
The second follow by the standard integral of $x^{a}$ and integration by parts, explicitly:
$$\int_{0}^{1} u^{\frac{\alpha}{\beta}-\alpha} du=\left[\frac{u^{\frac{\alpha}{\beta}-\alpha+1}}{\frac{\alpha}{\beta}-\alpha+1}\right]_{0}^{1}=\frac{\beta}{\alpha-\alpha \beta+\beta}$$ (I think the $+\alpha$ should be a $+\beta$ in the denominator)
\begin{align}
\int_{0}^{1} u^{\frac{\alpha}{\beta}-\alpha} \log u du &=\left[\frac{u^{\frac{\alpha}{\beta}-\alpha+1}\log u}{\frac{\alpha}{\beta}-\alpha+1}\right]_0^1-\int_{0}^{1}\frac{u^{\frac{\alpha}{\beta}-\alpha}}{\frac{\alpha}{\beta}-\alpha+1}du\\
&=-\left[\frac{u^{\frac{\alpha}{\beta}-\alpha+1}}{(\frac{\alpha}{\beta}-\alpha+1)^2}\right]^1_0\\
&=-\frac{\beta^2}{(\alpha-\alpha \beta+\beta)^2}
\end{align}
Hope this helps!


问:
$\mathbb{P}(\textrm{sin}\:X>\frac{1}{2})$ if $X$ is exponentially distributed with parameter $\lambda$
Let $X$ be exponentially distributed with parameter $\lambda$. Find $\mathbb{P}(\textrm{sin}\:X>\frac{1}{2})$.

Tried a few times to integrate this, but can't get to the final result.

答: Hint: First find $\{x \ge 0: \sin(x) > 1/2\}$.  This is the union of a sequence of intervals.  Find the probability of each, and then sum a geometric  series.


问:
Show using induction (coupled linear recurrences)
Some homework help would be greatly appreciated, took a screenshot and made an image to make it easier to show and get help with.

(2) Consider the numbers defined recursively by $a_1=3$, $c_1=5$, $$a_{n+1}=3a_n+2c_n+1\text{ and }c_{n+1}=4a_n+3c_n+2\;.$$ Show using induction that for all $n\in\Bbb Z^+$, $$a_n^2+(a_n+1)^2=c_n^2\;.$$

答:
My guess is that you were too lazy to do the expansions.

Here's a way that you don't have to fully expand out everything.

$c_{n+1} ^2 = (4 a_n + 3 c_n + 2)^2 = (4a_n +2)^2 + 2 ( 4 a_n +2) (3 c_n) + 9 c_n^2$

$a_{n+1}^2 + (a_{n+1} +1)^2 = (3a_n+1) ^2 + (3a_n+2)^2 + 2 (6a_n + 3)(2c_n) + 8 c_n^2  $

We want to show that the RHS is equal to each other. This is equivalent to
$$c_n^2 = 2 a_n ^2 + 2 a_n + 1,$$

which is true by the induction hypothesis. Hence, the LHS is equal to each other and we are done.


问:
Exchanging definite integral and infinite sum in context of Fourier series
I am writing a mathematics paper on the convergence of the Fourier series for periodic functions, and in the first section (where I define the Fourier series), I also derive the standard coefficients $a_n, b_n$. The standard way of doing this is to multiply both sides of the function by $\cos{kx}$ or $\sin{kx}$ depending on which coefficient you are calculating, and integrating from $-\pi$ to $\pi$, as follows:

$$\int_{-\pi}^\pi f(x)\cos(kx) \, dx = \int_{-\pi}^\pi a_0 \cos{kx} \, dx + \int_{-\pi}^\pi \left(\sum_{n=1}^\infty a_n\cos{nx}\cos{kx} + b_n\cos{nx}\sin{kx} \right)\,dx.$$

However, from this point, the standard derivation of the coefficients involves interchanging the integral and the sum, but to me this is not inherently obvious. For example, if $a_n = 1$ for all $n$, the summation does not converge and we would be integrating something infinite, which doesn't make sense. I've looked at a few posts that talk about interchanging the order of the sum and integral, and most of them don't really seem to apply.

Monotone convergence and Vitali convergence don't seem to apply since we can't make statements about pointwise convergence of $a_n\cos{kx}$ or $b_n\sin{kx}$ without being circular. Fubini's theorem tells us that if $\int\sum_n |f_n(x)| \,dx $ is finite, then we can interchange the summation and integral. However, we can't guarantee that such an integral is finite unless $\int_{-\pi}^\pi f(x) \cos(x)\, dx$ is finite--which led me to assume that the finiteness of the sum is a sufficient condition for interchanging the integral and sum, but my professor said this was insufficient.

Another thing I am thinking about is that pointwise convergence does not preserve integrability, but uniform convergence does. Does this mean that we can only derive the coefficients $a_n, b_n$ when $f$ is uniformly convergent? In that case, why would anyone care about pointwise convergence with the Dirichlet kernel proof?

I guess put simply, my question is, what can we stipulate about $f$ so that the interchange above is valid?

答: How about the dominated convergence theorem? If you know the coefficients $a_n$ and $b_n$ decay rapidly enough so that the partial sums $|\sum_{n}^{N} a_n\cos(kx)+b_n\sin(kx)|$ are dominated by an integrable function, then the interchange is legal by dominated convergence. How fast the coefficients decay depends on the smoothness of $f$. For example if $f$ is twice continuously differentiable you can show using integration by parts that the coefficients grow like $\frac{1}{n^2}$. In this case the partial sums are indeed dominated by an $L^1[-\pi,\pi]$ function, namely a constant function.


问:
topological realizations and graphs
In the following I'm asking not for hand-waving conceptual explanations but for something a bit more formal about graphs and topological realizations.

My understanding is that these objects are functions that takes as input a specific graph and returns a topological space based on that graph, that is called "also" the topological realization of that graph. So, here comes the first question:

Q0: What is a topological realization of a graph? what is the topology of it?

Q1: Is there only  one definition? or it could be many depends on what?

Q2: What kind of properties the topological realization of a
graph and the graph itself share?

Q3: How can I prove that if the top. real. of a graph is a finite top. space, then the graph is also finite? If so, how? it seems true, but I don't know where to start with.

答:
Here's the way I would turn a graph into a topological space:

Start with a point $P_v$ for every vertex $v$.

Take an interval $I_{vw}$ for every edge $vw$, but identify one endpoint of $I_{vw}$ with $P_v$ and the other with $P_w$.

I'm sure there are other ways to do it if you try to think of them hard enough, but this seems the most natural.

With this approach, finite graphs do not give finite topological spaces (and I struggle to come up with a good way to represent any graph by a finite topological space). However, finite graphs are precisely the ones that correspond to compact topological spaces.

Title: topological realizations and graphs


问:
Hatcher Theorem 1.7 Detail
In his proof of Theorem 1.7, Hatcher uses the following covering space of $S^1$. Let $f:\mathbb{R}\to\mathbb{R}^3$ be given by $f(x)=(\cos2\pi x,\sin2\pi x,x)$. We then take the range $f[\mathbb{R}]$ of $f$, with the projection $p:\mathbb{R}^3\to\mathbb{R}^2$. Hatcher takes a loop $l$ with basepoint $(1,0)$ in $S^1$ and lifts it to a path $\tilde{l}$ in $f[\mathbb{R}]$ with endpoints $(1,0,0)$ and $(1,0,n)$ for $n\in\mathbb{Z}$. The path $\tilde{\omega}_n$ given by $\tilde{\omega}_n(x)=(\cos2\pi nx,\sin2\pi nx,nx)$ has the same endpoints, so he takes the linear homotopy $h_t=t\tilde{l}+(1-t)\tilde{\omega}_n$ between them. He then claims that $ph_t$ is a homotopy between $l$ and $\omega_n$, defined to be $p\tilde{\omega}_n$.

My problem is this: wouldn’t $h_t$ generally leave the helix $f[\mathbb{R}]$, or more importantly, the cylinder $S^1\times\mathbb{R}$ containing it? Then for any $x$ such that $h_t(x)$ is not in the cylinder, we would have $ph_t(x)\not\in S^1$, so $ph_t$ is not a homotopy in $S^1$. Wouldn’t we need to make $h_t$ something like $f(tf^{-1}\tilde{l}+(1-t)f^{-1}\tilde{\omega}_n)$?

答:
OK, here's an answer recapitulating my comment above so that this question is not left unanswered.

The part in question of the proof of Theorem 1.7 is to show that every loop $f: I \to S^1$ at the basepoint $x_0$ is homotopic to a "standard" loop $\omega_n: I \to S^1$ that winds around the circle counterclockwise $n$ times at a constant speed for some $n \in \mathbb{Z}$.

Using the theory of covering spaces, we show that $f: I \to S^1$ has a unique lift $\tilde{f}: I \to \mathbb{R}$ with $\tilde{f}(0) = 0$.  Moreover, we have $\tilde{f}(1) = n$ for some $n \in \mathbb{Z}$.  We want to show that $\tilde{f}$ is homotopic to the map $\tilde{\omega}_n: x \mapsto nx$; once we have this, composing with the projection $p: \mathbb{R} \to S^1$ gives a homotopy between $p \circ \tilde{f} = f$ and $p \circ \tilde{\omega}_n = \omega_n$.

Hatcher claims that linear interpolation between $\tilde{f}$ and $\tilde{\omega}_n$ works, using the linear structure of the codomain $\mathbb{R}$.  Of course, even though the helix described right after the statement of the theorem is topologically equivalent to $\mathbb{R}$, it is not convex in $\mathbb{R}^3$, so you are correct that if we tried to linearly interpolate in $\mathbb{R}^3$, the homotopy may "leave" the helix and is thus invalid.

Fortunately, this is not what Hatcher is proposing to do.  Instead, we linearly interpolate within $\mathbb{R}$ and not its embedding in $\mathbb{R}^3$; you can imagine this as sliding points along the helix between the endpoints of the homotopy at a uniform rate - so the homotopy never "leaves" the helix - if you want to visualize the covering projection $p: \mathbb{R} \to S^1$ literally as a projection in $\mathbb{R}^3$.

Occasionally, it may help with geometric visualization and developing intuition to imagine $\tilde{f}: I \to \mathbb{R}$ as a path on a helix in 3D, but this is not needed for, and not used in, the actual proof.

Title: Hatcher Theorem 1.7 Detail


问:
Are there two non-congruent quadrilaterals with same sets of sides and angles?

Are there two non-congruent quadrilaterals with same sets of sides and angles?

It is relatively easy to construct such pentagons. Trying to construct quadrilaterals allows for seemingly multiple degrees of freedom, yet my every attempt ends up not working out.

答:
Yes.  There are two degrees of freedom in the solution space, so we can add two extra constraints to make the computation easier.  Here’s a solution with a 90° angle and a 45° angle:

The side lengths, from longest to shortest, are
$$\begin{align*}
1&, \\
α &\approx 0.838045, \\
(2 + \sqrt2) + (6 + 4\sqrt2) α + \sqrt2 α^2 - (12 + 8\sqrt2) α^3 &\approx 0.454531, \\
(3 + 3\sqrt2) + (9 + 7\sqrt2) α - (2 + 2\sqrt2) α^2 - (16 + 12\sqrt2) α^3 &\approx 0.284498,
\end{align*}$$

where $α$ is the positive root of

$$1 - \sqrt2 - 2α - 2α^2 + (4 - \sqrt2)α^3 + 4α^4 = 0.$$


问:
△ with altitude from  and angle bisector of  intersecting at $P$. Find $BP$
Let $\triangle ABC$ be a triangle with side lengths $AB = 17, BC = 28, AC = 25$. Let the altitude from $A$ to $BC$ and the angle bisector of angle $B$ meet at $P$. Given the length of $BP$ can be expressed as $\frac{a\sqrt{b}}{c}$ for positive integers $a, b, c$ where $gcd(a, c) = 1$ and $b$ is not divisible by the square of any prime, find $a + b + c$.

I drew my diagram out, and assumed that the intersection point is the midpoint of the altitude. Then i used pythag to find the length of the segment where the altitude intersects BC. I got a huge answer which is 99% incorrect so i need help on solving it.

答:
Let $\angle B=2\beta$ and $D$ be the foot of the altitude from $A$. Then
$$\cos\beta = \frac{BD}{BP}, \>\>\> \cos 2\beta = \frac{BD}{17}
$$
leading to
$$BP= \frac{17}{\cos\beta}(2\cos^2\beta-1)\tag1$$
From the cosine rule
$$\cos2\beta= \frac{ 17^2+28^2-25^2}{2\cdot 17\cdot 28}=2\cos\beta^2-1
$$
which yields $\cos^2\beta=\frac{25}{34}$. Plug into (1) to obtain
$$BP= \frac{8\sqrt{34}}5$$
Thus, $a+b+c= 47$.


问:
Topologies and manifolds
This question might seem philosophical a bit:

in a standard manifolds introductory course. when one talks about open , closed sets in $\mathbb{R}^n$ it's always the standard euclidian topology that we think about. Is there a more general framework in which one defines manifolds in different topologies in a way that generalize the standard setting that is closer to the physical world of course.

Thank you in advance for your answers

答:
Wrapping up the comments under the original question: One viewpoint of (differentiable) manifolds is that manifolds are those objects that can be approximated locally by a linear space, the "model space", so that concepts of calculus can be transported from the model space to the manifold.

The model space needs to be at least a topological vector space, a vector space that is also a topological space such that the algebraic operations are continuous, to make sense of notions of calculus. You first need to be able to define e.g. the derivative of a path in your model space before you can define the derivative of a path in the manifold.

If you take a curve in your model space $E$:

$$
c: \mathbb{R} \to E
$$

you'd need to be able to write down the definition of the derivative $c'$:

$$
c'(t) := \lim_{s \to 0} \frac{1}{s} (c(t+s) - c(t))
$$

In order to make sense of this formula you need the linear structure, a topology, and the continuity of addition and scalar multiplication.

The simplest example of a topological vector space is of course $\mathbb{R}^n$. It is a theorem that on a finite dimensional vector space there is one and only one topology that turns the vector space into a topological vector space, which in the case of $\mathbb{R}^n$ is the canoncial topology induced by, for example, the Euclidean norm.

Therefore, when one talks about differentiable manifolds modelled on $\mathbb{R}^n$, $\mathbb{R}^n$ is always equipped with this unique topology. The same applies to complex manifolds modelled on $\mathbb{C}^n$, too, of course. Things get more interesting for model spaces that are infinite dimensional, like Fréchet spaces. But as long as you are talking about differentiable manifolds modelled on finite dimensional spaces, the topology is fixed by the requirement that the model space needs to be a topological vector space.

Title: Topologies and manifolds


问:
Elliptic Deformations of Multiplication
Consider the squared absolute value function on complex numbers $a+bi \rightarrow a^2+b^2$.

It's easy to verify that the squared absolute value is a homomorphism over multiplication. That is
$$ |(a+bi)(c+di)|^2 =  (ac - bd) + (ad+bc)i = (ac-bd)^2 + (ad+bc)^2 = (ac)^2-2acbd+(bd)^2 + (ad)^2 + 2abcd + (bc)^2 = (ac)^2+(bd)^2+(ad)^2+(bc)^2= (a^2+b^2)(c^2+d^2)= |(a+bi)|^2|(c+di)|^2$$

If we define an elliptic absolute value as the map given a choice of $(u,v)$ and a complex number $a+bi \rightarrow \frac{a^2}{u^2} + \frac{b^2}{v^2}$ we can easily see that unless we have $u=v=1$ this function almost surely doesn't respect multiplication. (We can try some difference choices of $u,v$ to see this for ourselves).

So now we get to a very subjective question, what if we insist there ought to be an operation like multiplication, what we shall call a $(u,v)$-deformation of multiplication, which has the property that the $(u,v)$-elliptic absolute value respects this operation, and this operation is in some sense as "natural" as possible. We could probably say that the foci of the ellipse generated by $(u,v)$ would replace the function of 0 w.r.t this multiplication.

What would be some candidates for such a multiplication?

Some Preliminary thoughts:

If we have a complex function $\Psi$ that maps the unit circle to an ellipse centered at 0. Then we can define multiplication on the ellipse by taking two points $\rho_1, \rho_2$ and computing $\Psi(\Psi^{-1}(\rho_1)\Psi^{-1}(\rho_2))$.

This new multiplication operation at the least would "respect" our elliptic squared absolute values for the unit ellipse of a particular $(u,v)$ class.

It's meaningful then to ask, given a $(u,v)$ pair what is the space of meromorphic functions that map complex circles of radius $r$ to the ellipse $x^2/u^2 + y^2/v^2 = r^2$.

These would create can candidates for making an elliptical multiplication.

答:
My two cents comment: I find the representation of your question a bit tautological. Here is why.

My understanding of your question

Let $\tau : \mathbb{C} \rightarrow \mathbb{R_+}$ denote our ellipstic squared absolute value, that is:
$$\tau(a+ib)= \dfrac{a^2}{u^2}+\dfrac{b^2}{v^2}$$
So if I understand correctly, you want to find a binary operation $*$ from $\mathbb{C} \times \mathbb{C}$ to $\mathbb{C}$ such that:

$(\mathbb{C}, *)$ is a semigroup

$\tau( a*b) = \tau(a)\tau(b)$

My main comment
Our quest for such multiplication for $\tau$ is equivalent to a question for a multiplication that preserves the structure of the usual absolute value.

Explanation

You see, for any bijective function $\phi: \mathbb{C} \rightarrow \mathbb{C}$

I can define another binary operation $\times$ such that:
$$\phi(a)\times \phi(b) = \phi( a*b)$$
(Thank to the bijectivity of $\phi$)

Furthermore, we can see that $(\mathbb{C},\times)$ is a semigroup.
Indeed, for any complex numbers $x,y,z$, there are three complex numbers $a,b,c$ such that: $$\phi(a)=x, \phi(b)=y , \phi(c)=z$$
and therefore,
$$(x \times y ) \times z = ( \phi(a) \times \phi(b) ) \times  \phi(c) = \phi( a*b) \times \phi(c) = \phi ( (a*b)*c)$$
and  likewise,
$$x \times( y  \times z )=  \phi(a) \times (\phi(b)  \times  \phi(c) )= \phi( a) \times \phi(b*c) = \phi ( a*(b*c))$$
as $*$ is associative, $\times$ also is.

Besides, also by definition of $\phi$
$$ \tilde{\tau}( a \times b) =\tilde{\tau}( a ) \tilde{\tau}( b ) $$
where $\tilde{\tau} =\tau\circ \phi^{-1} $

Then, by choosing $\phi$ to be an appropriate deformation function in $\mathbb{C}$, say $\phi(a+ib)= a/u+ib/v$ ,  we can have
$$\tilde{tau}( a+ib)= a^2+b^2$$

In short
We are just finding something for the usual absolute value.

Title: Elliptic Deformations of Multiplication


问:
How would you go about modelling this as a Markov chain?
There is two machines which break down at different rates, µ$_A$ for machine A and µ$_B$ for machine B. When they break down, a machine can be fixed by one of two repairmen. Assume that two repairmen, X and Y, have different abilities, and they repair in exponential random times with parameters λ$_X$ and λ$_Y$ , respectively. (But the rate doesn’t depend on which machine they repair.)

How would you go about modelling this as a Markov chain?.

答:
Assume that if both machines are working when one fails, then the repairmen have equal probability of being chosen to fix the machine. The state space is
$$
S=\{(1,1), (0_X,1), (0_Y,1), (1,0_X), (1,0_Y), (0_X,0_Y), (0_Y,0_X) \}.
$$
The transition rates are given by
$$
q_{(i,j),(i',j')} = \begin{cases}
\frac{\mu_A}2,& (i,j)=(1,1)\text{ and } (i',j') \in \{(0_X,1),(0_Y,1)\}\\
\frac{\mu_B}2,& (i,j)=(1,1)\text{ and } (i',j') \in \{(1,0_X),(1,0_Y)\}\\
\lambda_X,& (i,j) \in \{(0_X,1),(1,0_X)\} \text{ and } (i',j') = (1,1)\\
\lambda_Y,& (i,j) \in \{(0_Y,1),(1,0_Y)\} \text{ and } (i',j') = (1,1)\\
\mu_A,& (i,j) = (1,0_X)\text{ and } (i',j') = (0_Y,0_X)\\
\mu_A,& (i,j) = (1,0_Y)\text{ and } (i',j') = (0_X,0_Y)\\
\mu_B,& (i,j) = (0_X,1)\text{ and } (i',j') = (0_X,0_Y)\\
\mu_B,& (i,j) = (0_Y,1)\text{ and } (i',j') = (0_Y,0_X)\\
\lambda_X,& (i,j) = (0_X,0_Y)\text{ and } (i',j') = (1,0_Y)\\
\lambda_X,& (i,j) = (0_Y,0_X)\text{ and } (i',j') = (0_Y,1)\\
\lambda_Y,& (i,j) = (0_X,0_Y)\text{ and } (i',j') = (0_X,1)\\
\lambda_Y,& (i,j) = (0_Y,0_X)\text{ and } (i',j') = (1,0_X)\\
0,& \text{otherwise.}
\end{cases}
$$
Let $Z(t)$ be the state of the system at time $t$, then $\{Z(t):t\geqslant 0\}$ is a continuous-time Markov chain with generator matrix
$$
Q = \small\left(
\begin{array}{ccccccc}
 -\left(\mu _A+\mu _B\right) & \frac{\mu _A}{2} & \frac{\mu _B}{2} & \frac{\mu _A}{2} & \frac{\mu _B}{2} & 0 & 0 \\
 \lambda _X & -\left(\mu _B+\lambda _X\right) & 0 & 0 & 0 & \mu _B & 0 \\
 \lambda _Y & 0 & -\left(\mu _B+\lambda _Y\right) & 0 & 0 & 0 & \mu _B \\
 \lambda _X & 0 & 0 & -\left(\mu _A+\lambda _X\right) & 0 & 0 & \mu _A \\
 \lambda _Y & 0 & 0 & 0 & -\left(\mu _B+\lambda _Y\right) & \mu _B & 0 \\
 0 & \lambda _Y & 0 & 0 & \lambda _X & -\left(\lambda _X+\lambda _Y\right) & 0 \\
 0 & 0 & \lambda _X & \lambda _Y & 0 & 0 & -\left(\lambda _X+\lambda _Y\right) \\
\end{array}
\right).
$$
The process has a unique stationary distribution $\pi$ which satisfies
$$
\pi_{(i,j)} = \lim_{t\to\infty} \mathbb P(Z_t = (i,j))
$$
(independent of the distribution of $Z_0$). We can find $\pi$ by computing the matrix exponential $e^{Qt}$ (which is the unique solution to the Kolmogorov backward equation $P'(t)=QP(t)$, $P'(0)=Q$) and taking any of the rows of $\lim_{t\to\infty} e^{Qt}$. More practically, $\pi$ satisfies the system of linear equations $\pi Q=0$. Note that $Q$ is singular (i.e. $\det Q=0$) as its rows all sum to zero, so we must replace one of the equations with $\sum_{(i,j)\in S} \pi_{(i,j)}=1$. However, due to the size of this matrix and the number of parameters, the closed form solution is a bit unwieldy. For example, I found that
$$
\pi_{(1,1)} = \tiny\frac{2 \lambda _X \lambda _Y \left(2 \mu _B+\lambda _X+\lambda _Y\right) \left(\mu _A+\mu _B+\lambda _X+\lambda _Y\right)}{\lambda _X^2 \left(\mu _B \left(3 \mu _A+10 \lambda _Y\right)+\left(\mu _A+2 \lambda _Y\right){}^2+6 \mu _B^2\right)+\lambda _X \left(\mu _B \left(7 \mu _A \mu _B+4 \mu _A^2+5 \mu _B^2\right)+\lambda _Y^2 \left(6 \mu _A+8 \mu _B\right)+\lambda _Y \left(\mu _A+3 \mu _B\right) \left(3 \mu _A+4 \mu _B\right)+2 \lambda _Y^3\right)+\left(\mu _B \left(3 \mu _A+4 \lambda _Y\right)+2 \lambda _Y \left(\mu _A+\lambda _Y\right)+\mu _B^2\right) \left(\mu _B \left(\mu _A+\mu _B\right)+\mu _A \lambda _Y\right)+2 \lambda _X^3 \left(\mu _B+\lambda _Y\right)}
$$
(the denominator is broken into two lines to prevent page stretching).


问:
Find volume of the solid obtained
Find volume of the solid obtained by rotating about the y-axis the region bounded by $y= 4-(x-4)^2$ and $y = 0$.

Title: Find volume of the solid obtained

Body:
$$V = \int\limits_{x=2}^6 2 \pi x (4 - (x-4)^2)\ dx = \frac{256 \pi }{3}$$

Mathematica code:

aa = RegionPlot3D[0 < y < 4 - (Sqrt[x^2 + z^2] - 4)^2 ,
   {x, -6, 6},
   {y, 0, 4},
   {z, -6, 6},
   PlotPoints -> 100,
   PlotStyle -> Opacity[0.5],
   BoxRatios -> {10, 4, 10},
   AxesOrigin -> {0, 0, 0},
   AxesStyle -> {{Thick, Red}, {Thick, Red}, {Thick, Red}}];
bb = Graphics3D[{Blue, Opacity[0.2],
    InfinitePlane[{0, 0, 0}, {{1, 0, 0}, {0, 0, 1}}]}];
cc = RegionPlot3D[3 < Sqrt[x^2 + z^2] < 3.2 && 0 < y < (4 - (3 - 4)^2),
   {x, -6, 6},
   {y, 0, 4},
   {z, -6, 6},
   PlotStyle -> Directive[{Opacity[0.5], Green}],
   PlotPoints -> 150];
Show[aa, bb, cc]


问:
The path of complex structure.
$J = \left( {\begin{array}{*{20}{c}}
0&{ - {I_n}}\\{{I_n}}&0\end{array}} \right)$. If $I(t)$ is a path in $\rm M(2n, \mathbb R)$ ($2n \times 2n $ real matrix) such that $I(0)=0$ and $I(t)J+JI(t)=0$, then for sufficiently small $t$ can we find a path $J(t) \in \rm M(2n, \mathbb R)$ such that (1) $J(0)=J$, (2) $J^2(t)+I^2(t)=-I_{2n}$, (3) $I(t)J(t)+J(t)I(t)=0$?</p>

答:
Define the path $J(t)\in \rm M(2n, \mathbb R)$ as a matrix whos elements are smooth functions of $t$. This condition ensures the derivative of the matrix exists w.r.t. $t$ and thus we may Taylor expand about the point $t=0$. We then write:

$J(t)=J_0+J_1t+O(t^2)$: $J_n=\dfrac{1}{n!}\left(\dfrac{d}{dt}J(t)\right)|_{t=0}$,

$I(t)=I_0+I_1t+O(t^2)$: $I_n=\dfrac{1}{n!}\left(\dfrac{d}{dt}I(t)\right)|_{t=0}$,

where we have discarded matrices multiplied by powers of $t>1$ as we are required to demonstrate the existence of $J(t)$ for "sufficiently small $t$". Now we know that $ I(0)=0$ and $J(0)=J$, so we have:

$J(t)=J+J_1t$,

$I(t)=I_1t$.

Then, because we know that:

$I(t)J+JI(t)=0$

we have that:

$I_1J+JI_1=0$.

Now we are required to satisfy:

$J^2(t)+I^2(t)=-I_{2n}$,

which, with our first order expansions first becomes:

$J^2+J_1²t²+(JJ_1+J_1J)t+I_1²t²=-I_{2n}$

and then keeping up to first order in $t$:

$J^2+(JJ_1+J_1J)t=-I_{2n}$.

But:

$J² = \left( {\begin{array}{*{20}{c}}
0&{ - {I_n}}\\{{I_n}}&0\end{array}} \right)\left( {\begin{array}{*{20}{c}}
0&{ - {I_n}}\\{{I_n}}&0\end{array}} \right)=-I_{2n}$

and so this condition is met if:

$JJ_1+J_1J=0$.

Similarly, we are required to satisfy:

$I(t)J(t)+J(t)I(t)=0$

$I_1Jt+I_1J_1t²+JI_1t+J_1I_1t²=0$

$(I_1J+JI_1)t=0$  ... but we got this for free earlier.

So, we see that a path $J(t)$ can be found that meets all the requirements if we can find a $J_1$ that anti-commutes with $J$; that is if we can satisfy:

$JJ_1+J_1J=0$.

But we allready know one matrix that anti-commutes with $J$ and so we can solve this by letting:

$J_1=I_1$.

Thus the path:

$J(t)=J+I_1t$

satisfies the requirements; which is equivalent to saying that for sufficiently small $t$ the path:

$J(t)=J+I(t)$

satisfies the requirements. Finally, as $J,I(t)\in \rm M(2n, \mathbb R)$ it follows that $J(t)\in \rm M(2n, \mathbb R)$.

Title: The path of complex structure.


问:
Prove that $10 | (n^a - n^b)$.
$n$ is a positive integer. Prove that there exists positive integers $a$ and $b$, $(a > b)$ such that $10 | (n^a - n^b)$.

I have tried to prove this by induction on $n$, but I get stuck at the induction step trying to prove it for $n = k + 1$, not knowing how to expand $((k + 1)^a - (k + 1)^b)$. Is this the wrong approach to solving this problem, or am I missing something here?

答:
As suggested in the comment by Akiva Weinberger, take $a=5$ and $b=1$.

Then $(k+1)^a-(k+1)^b=(k+1)^5-(k+1)=(k+1)[(k+1)^4-1]$

$=(k+1)(k^4+4k^3+6k^2+4k)=(k+1)k(k^3+4k^2+6k+4)$.

This is divisible by $2$ because  $(k+1)k$ is,

and modulo $5$ this is the same as $(k+1)k(k-1)(k^2+1)$, so it is divisible by $5$ too,

and therefore it is divisible by $10$.

Title: Prove that $10 | (n^a - n^b)$.


答:
You can use the following theorem: If $h(x)$ is a function such that $h(x)\in I$ (where $I$ is defined below to be the interval of convergence of the Maclaurin series for $f(x)$ and if we have a Maclaurin series expansion for a function $f(x)$ (and which converges to $f(x)$), given by $f(x)=\sum_{n\geq 0}a_n x^n$ with radius of convergence $R$ and interval of convergence $I$ then the function $f(h(x))$ will have Maclaurin series expansion $$f(h(x))=\sum_{n\geq 0}a_n (h(x))^n$$ with same radius of convergence and same interval of convergence. Furthermore this Maclaurin expansion will converge to $f(h(x))$.

Hence, $$\displaystyle sin(2z^2)=\sum_{n\geq 0} (-1)^n\frac{2^{2n+1}}{(2n+1)!}z^{4n+2}$$ has radius of convergence $R=+\infty$ and interval of convergence $(-\infty, +\infty)$.

Alternatively if you forgot this theorem, you can redo all the work and directly use the ratio test to find the radius and interval of convergence of the composed Maclaurin series. Consider $$\displaystyle  \lim_{n\to \infty} \Bigg |\frac{\frac{2^{2(n+1)+1}}{(2(n+1)+1)!}x^{4n+6}}{\frac{2^{2n+1}}{(2n+1)!}x^{4n+2}}\Bigg\vert =\lim_{n\to \infty} \frac{2^{2n+3}}{(2n+3)!}\frac{(2n+1)!}{2^{2n+1}}\bigg|\frac{x^{4n+6}}{x^{4n+2}}\bigg|$$ You're then left with: $$\lim_{n\to \infty} \frac{2^3}{(2n+2)(2n+3)}\vert x^4\vert$$

But the limit above is $0$ regardless of what $x$ can be and this shows that by the ratio test we have absolute convergence, hence convergence for all $x\in \mathbb{R}$ and the radius of convergence is $R=+\infty$

Title: Find MacLaurin Series and radius of convergence

Body:
$$\sin{2z^2}$$
$$\sin{z} = \sum_{n=0}^{\infty}(-1)^n \frac{z^{2n+1}}{(2n+1)!}$$

Let's replace z with $2z^2$ in the MacLaurin series:

$$\sin{2z^2} = \sum_{n=0}^{\infty}(-1)^n \frac{(2z^2)^{2n+1}}{(2n+1)!} = \sum_{n=0}^{\infty} (-1)^n
 \frac{2^{2n+1}}{(2n+1)!} z^{4n+2}$$

$$= z^2 - \frac{2^3}{3!}z^6 + \frac{2^5}{5!}z^{10} -+ ... = z^2 - \frac{4}{3}z^6 + \frac{4}{15}z^{10} -+...$$

But what about the radius of convergence? How do I find this?


问:
Covariant derivative question
I am watching a series of lectures about General relativity by F. Schuller.

In the Lecture 7 - Connections ( https://youtu.be/nEaiZBbCVtI?t=1245 ) he gives the properties for a covariant derivative respect to a vector field.

But instead of the Leibniz rule:

$$\nabla_v(T\otimes S)=(\nabla_v T)\otimes S + S \otimes (\nabla_vS)$$

he gives this rule:

$${ \nabla_v}(T(w,Y)) =
     ({ \nabla_v}T)(w,Y)+T({ \nabla_v}w,Y)+T(w,{ \nabla_v}y)$$

```<code>And he claims is equivalent to the Leibniz rule,
</code>```

but I cannot even derive this rule from the Leibniz rule. Some help please??

答:
His rule is better (stronger) because it tells you that the derivative commutes with taking "contractions" between tensors (e.g. $(\omega,Y)\mapsto \omega(Y)$)

You didn't specify what type of tensors your $T$ and $S$ are, but if they are say covectors, then $T\otimes S$ is a $(0,2)$-tensor (or maybe $(2,0)$, I don't know his convention), which takes in two vectors as input.

Now say $v,w$ are two vectors at a point, and choose extensions to vector fields which satisfy $\nabla v = 0,\nabla w = 0$ (you can see that derivatives on $v,w$ will cancel out if you carry them through, so the choice of extension doesn't matter).

Then, since $(T\otimes S)(v,w) = T(v)\cdot S(w)$, his rule says
$$\begin{align*}
\big(\nabla (T\otimes S)\big)(v,w)
&= \nabla\big((T\otimes S)(v,w)\big) - (T\otimes S)(\nabla v,w) - (T\otimes S)(v,\nabla w)\\
&= \nabla\big(T(v)\cdot S(w)\big)\\
&= \big[\nabla\big(T(v)\big)\big]\cdot S(w) ~+~ T(v)\cdot \big[\nabla\big(S(w)\big)\big] \\
&= \big((\nabla T)(v)\big)\cdot S(w) ~+~ T(v)\cdot \big((\nabla S)(w)\big) \\
&= \big((\nabla T)\otimes S\big)(v,w) ~+~\big(T\otimes (\nabla S)\big)(v,w)
\end{align*}
$$
so it implies your Liebniz rule in this case.

Note that he doesn't say that this "is equivalent" to what you say is the Liebniz rule, he just says that his axiom "is called the Liebniz rule".

Title: Covariant derivative question


问:
What will be the sum of this series if $a$ and $n$ is odd?
$a$ and $n$ are all odd numbers and are positive

$1+a^1...+a^{2^n-2}+a^{2^n-1}$

I need this for a larger exercise where I determine if this series can be divided by $2^n$

Thanks

答:
The finite geometric series formula states that $1+r+r^2 + r^3 +\dots + r^{k} = \frac{1-r^{(k+1)}}{1-r}$.

Note that we can just substitute in our values. We get $\frac{1-a^{2^{n}}}{1-a}$.

Im pretty sure there is no way to manipulate this to check divisibility by $2^n$ but someone could correct me if I am wrong.


问:
Embeddings of an infinite graph.
I'm asking for a formal proof or counterexample (if it's false) of the following.

If a graph (in general) has a vertex like in the picture below (of infinity degree), the
the graph cannot have a cellular embedding into any surface.

Any elaboration to help me build more intuition about this or similar questions
are very welcome.

答:
If a graph $G$ has a vertex $v$ of infinite valence then $v$ does not have a countable neighborhood basis.

To sketch a proof of this, let $E_i$ ($i \in I$) be the oriented edges of $G$ with initial vertex $G$, and let $\alpha_i : [0,1] \to E_i$ be an orientation preserving parameterization of $E_i$. For any function $R : I \to (0,1)$, the set $U(R) = \bigcup_{i \in I} \alpha_i \bigl([0,R(i))\bigr)$ is a neighborhood of $v$ (by definition of the celluar topology, also known as the CW topology). And then, for any countable set of neighborhoods $V_1,V_2,...$ of $f(v)$, one can choose a countably infinite subset $\{i_1,i_2,...\} \subset I$ and use a diagonalization construction to produce $R : I \to (0,1)$ so that $U(R)$ does not contain any of the subsets $V_j$.

But for any surface $S$ and any continuous injection $f : G \to S$, the image $f(v)$ does have a countable neighborhood basis in the subspace topology on $f(G)$, namely the intersection with $f(G)$ of any countable neighborhood basis of $f(v)$ in $S$. Therefore, $f$ is not an embedding, meaning that it does not have a continuous inverse with respect to the subspace topology on $f(G)$.

The point is that the cellular topology is a very fine topology, with many many open sets, too many for $G$ to be embeddable into a surface.

Title: Embeddings of an infinite graph.


问:
Let $X_1,X_2,\dots$ be i.i.d. and set $N_n(a,b)=\sum_{k=1}^n\mathbf1_{\{c+a/n<X_k<c+b/n\}}.$ Then $N_n(a,b)$ converges in distribution
Problem: Let $X_1,X_2,\dots$ be i.i.d. random variables with distribution function $F$ and a continuous density function $f$. Let $c\in\mathbb R$ be a number with $f(c)>0$ and consider
$$N_n(a,b)=\sum_{k=1}^n\mathbf1_{\{c+a/n<X_k<c+b/n\}}.$$
Show that $N_n(a,b)$ converges in distribution for any $a<b$ and find the limit.

My Attempt: Define the random variables
$$Y_{n,k}=\frac{\mathbf1_{\{c+a/n<X_k<c+b/n\}}-\displaystyle\int_{c+a/n}^{c+b/n}f(x)\,dx}{\sqrt n\cdot\sqrt{\displaystyle\int_{c+a/n}^{c+b/n}f(x)\,dx-\left[\displaystyle\int_{c+a/n}^{c+b/n}f(x)\,dx\right]^2}}$$
for $1\leq k\leq n$. We will check the conditions of the Lindeberg-Feller theorem. First, by construction we have $E[Y_{n,k}]=0$. Next, note that
$$Y_{n,k}^2=\frac{1}{n\cdot\left[\displaystyle\int_{c+a/n}^{c+b/n}f(x)\,dx-\left(\displaystyle\int_{c+a/n}^{c+b/n}f(x)\,dx\right)^2\right]}\left[\mathbf1_{\{c+a/n<X_k<c+b/n\}}-2\cdot\mathbf1_{\{c+a/n<X_k<c+b/n\}}\cdot\int_{c+a/n}^{c+b/n}f(x)\,dx+\left(\int_{c+a/n}^{c+b/n}f(x)\,dx\right)^2\right],$$
and taking the expectation of both sides we get that
$$E\left[Y_{n,k}^2\right]=\frac{1}n.$$
Hence,
$$\lim_{n\to\infty}\sum_{k=1}^nE\left[Y_{n,k}^2\right]=1.$$
For the third condition, we need to verify that for any $\varepsilon>0$,
$$\lim_{n\to\infty}\sum_{k=1}^n E\left[\vert Y_{n,k}\vert^2\cdot\mathbf1_{\{\vert Y_{n,k}\vert>\varepsilon\}}\right]=0.$$

I am stuck in the verification of the third condition.

I have also tried the same problem with the random variables defined as
$$Y_{n,k}=\mathbf1_{\{c+a/n<X_k<c+b/n\}}-\int_{c+a/n}^{c+b/n}f(x)\,dx,$$
and also had problems with the third condition of Lindeberg-Feller.

Does anyone have any hints on how to get started on the verification of the third condition of the Lindeberg-Feller in this example?

Thank you for your time and appreciate any feedback.

答:
So, it's about calculating the limit of a sequence of characteristic functions.

To be more precise, for any real number $\lambda $, we have:
$$\mathbb{E}( e^{ i\lambda N_n(a,b)})\underbrace{=}_{(X_i) \text{ are iids}} \left( \mathbb{E}\left( e^{ i\lambda 1_{ c+a/n < X_1< c+b/n} } \right) \right)^{n} = \left[ 1+ (e^{i\lambda}-1)\left( \int_{c+a/n}^{c+b/n}f(t)dt \right)  \right]^n$$

So you see, as $$ \lim_{n \rightarrow \infty} n \left( \int_{c+a/n}^{c+b/n}f(t)dt \right) = (b-a)f(c)$$
(Thanks to the continuity of $f$)

From which, we imly :
$$ \lim_{n \rightarrow +\infty} \mathbb{E}( e^{ i\lambda N_n(a,b)})= e^{ (b-a)f(c)(e^{i\lambda}-1) }$$

By Levy' continuity theorem, we have the conclusion.


问:
Maximum Likelihood and Sampling/Independence
If I viewed the data of a unknown weighted coin that was flipped N times and I see that there are H heads and T tails, I know that the maximum likelihood estimation for the probability of heads is just H/N and likewise for tails it is T/N. This is based off the assumption that each flip/trial was independent from each other and that the same coin was used for each trial. However, say if I only selected all the trials that resulted in heads and combined that into a different dataset and used that for maximum likelihood. Still, each trial is technically independent from each other and if so I would get that the maximum likelihood probability for heads is 100%. I know intuitively that this procedure is wrong, but I'm not sure why exactly it is. In addition, what if I selected the first half of all the trials, and then used that result to compute the maximum likelihood for the probability of the coin. Would that be an "ok" thing to do as I think it would just be computing the maximum likelihood for data with half the sample size?  Can someone explain intuitively what is going on here and clarify what is the difference between the two procedures?</p>

答:
MLE is a data driven framework. The question it asks is what probability distribution "most likely" would have led to the data observed. Independence just makes the calculations easier. Things get a lot more complicated otherwise.

So all you know about the world is the data you feed into MLE. If you have reasons to believe that the answer you get is not the correct one, then that knowledge ("prior knowledge") should be used to "adjust" the likelihood. This adjusting is the heart of Bayesian calculations.


问:
Why we are allowed to use notation $\log(x)$ instead of $\ln(x)$to denote logarithm in base $e$?
I saw a lot of answers on this site, people use $\log(f(x))$ to represent logarithm in base $e$. I have read some questions and answers about it like these:

Which is more preferable to write $\log(x)$ or $\ln(x)$

Should I assume $\log(x)$ to be $\log_e(x)$ or $\log_{10}(x)$?

From above posts, I realized in pure mathematics for advanced level it is common to use $\log(x)$ to denote logarithm in base $e$.

My question is: why we are allowed to do this? if $\log(x)$ use to denote logarithm in base $10$ too, then why we use this instead of $\ln(x)$?  As far as I know the purpose of mathematic is explaining something in most clear way and unambiguously so we should use notation proper for this purpose. but suppose the subject we are talking about has nothing to do with logarithm in base $10$ and we use $\log(x)$ to denote $\ln(x)$. but still $\log(x)$ has two meaning (logarithm in base $10$ or in base $e$)therefor it has contrast with the purpose I mentioned. why don't we avoid using this notation?

答: It really depends. Mostly in physics $\log(x)$ is assumed as base e but to be safe you should probably us $\ln(x)$. It highly depends on the context. In elementary math, $\log(x)$ is assumed as base $10$.


答:
In pure mathematics, we hardly ever need to use $\log_{10}$.

It's true that using $\ln$ would probably be more clear, but we tend to read out "$\ln$" as "log" anyway (out loud), so there is an argument to be made to just use $\log$ since it's the most natural notation which reflects how we communicate.

Title: Why we are allowed to use notation $\log(x)$ instead of $\ln(x)$to denote logarithm in base $e$?

Body:
I saw a lot of answers on this site, people use $\log(f(x))$ to represent logarithm in base $e$. I have read some questions and answers about it like these:

Which is more preferable to write $\log(x)$ or $\ln(x)$

Should I assume $\log(x)$ to be $\log_e(x)$ or $\log_{10}(x)$?

From above posts, I realized in pure mathematics for advanced level it is common to use $\log(x)$ to denote logarithm in base $e$.

My question is: why we are allowed to do this? if $\log(x)$ use to denote logarithm in base $10$ too, then why we use this instead of $\ln(x)$?  As far as I know the purpose of mathematic is explaining something in most clear way and unambiguously so we should use notation proper for this purpose. but suppose the subject we are talking about has nothing to do with logarithm in base $10$ and we use $\log(x)$ to denote $\ln(x)$. but still $\log(x)$ has two meaning (logarithm in base $10$ or in base $e$)therefor it has contrast with the purpose I mentioned. why don't we avoid using this notation?


问:
Two equalities in insurance mathematics
I'm using classical actuarial notation. $(IA)_x$ is APV of increasing insurance for $x$-years old person that pays $k$ when the person dies in year $k$. Also $(DA)_{1_{x:\bar{n|}}}$ is APV of decresing insurence for $x$-years old person that pays $n-k+1$ in case when person dies in year $k$. I need to prove these two things:

$(IA)_x=\nu q_x+\nu p_x(A_{x+1}+(IA)_{x+1});$

$(DA)_{1_{x:\bar{n|}}}=(n+1)A_{1_{x:\bar{n|}}}-(IA)_{1_{x:\bar{n|}}}$

I think I have the first one solved:

$(IA)_x=\sum_{k=0}^\infty (k+1)\nu^{k+1}\cdot {}_k p_x \cdot q_{x+k}$

$A_{x+1}=\sum_{k=0}^\infty \nu^{k+1}\cdot {}_k p_{x+1} \cdot q_{x+1+k}$

So on the right side we have:

$\nu q_x+\nu p_x(\sum_{k=0}^\infty [\nu^{k+1}\cdot {}_k p_{x+1} \cdot q_{x+1+k}]+\sum_{k=0}^\infty [(k+1)\nu^{k+1}\cdot {}_k p_{x+1} \cdot q_{x+1+k}])=$

$=\nu q_x+\nu p_x(\sum_{k=0}^\infty(k+2)\nu^{k+1}\cdot {}_k p_{x+1} \cdot q_{x+1+k})=\nu q_x+(\sum_{k=0}^\infty(k+2)\nu^{k+2}\cdot {}_{k+1} p_{x} \cdot q_{x+1+k})=$

$=\nu q_x+(\sum_{k=0}^\infty(k+1)\nu^{k+1}\cdot {}_{k} p_{x} \cdot q_{x+k})=\sum_{k=0}^\infty (k+1)\nu^{k+1}\cdot {}_k p_x \cdot q_{x+k}=(IA)_x.$

Is that correct? Could anyone give me a hint how to prove the second equality?

答:
The intuitive way to understand $$(IA)_x = v q_x + vp_x (A_{x+1} + (IA)_{x+1}) \tag{1}$$ is to condition on the survival of $(x)$ after one year.  Either $(x)$ dies in the first year with probability $q_x$, in which case the insurance pays $1$ at the end of the first year, which has present value $v$, or $(x)$ survives one year with probability $p_x$.  In this second case, we express the insurance payment as the sum of a whole life insurance of $1$, plus an increasing insurance, on $(x+1)$.  Then we discount by $v$ because one year has already passed.

Consequently, we write the RHS of Equation $(1)$ as
$$\begin{align}
vq_x + vp_x(A_{x+1} + (IA)_{x+1})
&= vq_x + vp_x \sum_{k=0}^\infty \left( 1 + (k+1) \right) v^{k+1} ({}_k p_{x+1})( q_{x+k+1}) \\
&= vq_x + \sum_{k=0}^\infty (k+2) v^{k+2} (p_x) ({}_k p_{x+1}) (q_{x+k+1}) \\
&= vq_x + \sum_{k=0}^\infty (k+2) v^{k+2} ({}_{k+1} p_x) (q_{x+k+1}) \\
&= vq_x + \sum_{k=1}^\infty (k+1) v^{k+1} ({}_k p_x) (q_{x+k} ) \\
&= \sum_{k=0}^\infty (k+1) v^{k+1} ({}_k p_x) (q_{x+k}) \\
&= (IA)_x.
\end{align}$$
This is essentially the same as your proof.

For the identity
$$\require{enclose}
(DA)_{x:\enclose{actuarial}{n}}^1 = (n+1) A_{x:\enclose{actuarial}{n}}^1 - (IA)_{x:\enclose{actuarial}{n}}^1, \tag{2}$$ it is more convenient to move the increasing insurance on the RHS to the left:
$$(DA)_{x:\enclose{actuarial}{n}}^1 + (IA)_{x:\enclose{actuarial}{n}}^1 = (n+1) A_{x:\enclose{actuarial}{n}}^1 \tag{3}$$
and now it becomes obvious what this equation means:  the sum of a $n$-year decreasing insurance on $(x)$ plus a $n$-year increasing insurance on $(x)$ equals a $n$-year insurance on $(x)$ that pays $n+1$ at any time during the term of the policy.  Think of the schedule of payments on the LHS:  given that $(x)$ dies in year $k$, for any $k \in \{1, 2, \ldots, n\}$, the decreasing insurance pays $n-k+1$, but the increasing insurance pays $k$, so the present value of the LHS is $(n+1)v^k$, which is also the present value of the RHS.  This should provide you with sufficient motivation to formulate a proof using summation notation.

Title: Two equalities in insurance mathematics


答:
We have $n^{a} - n^{b} \equiv 0\pmod 2$ since if $n \equiv 1 \pmod 2$ , then $ n^{a} - n^{b}\equiv 1-1 \equiv 0\pmod 2$ and if $n\equiv 0 \pmod 2, n^{a} - n^{b}\equiv0-0 \equiv 0\pmod 2$.

If $5|n$, then $n^{a}-n^{b}\equiv0-0\equiv 0 \pmod 5$ and so $10|n^{a}-n^{b}$.

If $5 \nmid n$, then $n^{4} \equiv 1 \pmod 5$ by Fermat's little theorem so that $n^{8} \equiv 1 \pmod 5 \implies$

$n^{8}-n^{4} \equiv 0 \pmod 5$ and altogether, in this case, $10|n^{a}-n^{b}$ as well where $a=8$ and $b=4$.

Title: Prove that $10 | (n^a - n^b)$.

Body:
$n$ is a positive integer. Prove that there exists positive integers $a$ and $b$, $(a > b)$ such that $10 | (n^a - n^b)$.

I have tried to prove this by induction on $n$, but I get stuck at the induction step trying to prove it for $n = k + 1$, not knowing how to expand $((k + 1)^a - (k + 1)^b)$. Is this the wrong approach to solving this problem, or am I missing something here?


问:
Prove: A is invertible IFF 0 is not an eigenvalue of A
I would appreciate feedback on the correctness of this proof, as well as alternative proofs.

Prove: A is invertible IFF 0 is not an eigenvalue of A.

Proof:If $A$ is invertible, letting $λ=0$ yields $Ax-λx=Ax=0$ which implies that $x=0$. So, $0$ is not an eigenvalue. If $0$ is not an eigenvalue then there is no non-zero $x$ such that $Ax=0$, and because there is only the trivial solution, A must be invertible.

答: You might want to explore the fact that $det(A) = p(0)$, where $p(x)$ is the characteristic polynomial of $A$.


问:
Positive semidefiniteness of a block matrix of positive semidefinite matrices
Given any symmetric matrix $\mathbf{M} = \begin{pmatrix}
\mathbf{A} & \mathbf{B}\\
 \mathbf{B}^\mathrm{T}& \mathbf{C}
\end{pmatrix}$, the following conditions are equivalent:

(1) $\mathbf{M}\succeq 0$ ($\mathbf{M}$ is positive semidefinite)

(2) $\mathbf{A}\succeq 0$, $\;\;(\mathbf{I}-\mathbf{A}\mathbf{A}^T )\mathbf{B}= 0$, $\;\;\mathbf{C} - \mathbf{B}^T\mathbf{A}^{\dagger }\mathbf{B} \succeq 0$.

(3) $\mathbf{C}\succeq 0$, $\;\;(\mathbf{I}-\mathbf{C}\mathbf{C}^T )\mathbf{B}= 0$, $\;\;\mathbf{A} - \mathbf{B}\mathbf{C}^{\dagger }\mathbf{B}^T \succeq 0$.

Then, if we are given that $\mathbf{A}\succeq 0$, $\mathbf{B}\succeq 0$ and $\mathbf{C}\succeq 0$, is there any way to verify (2) or (3) and therefore conclude that $\mathbf{M}$ will be positive semidefinite as well?

答: The conditions$\;\;(\mathbf{I}-\mathbf{A}\mathbf{A}^T )\mathbf{B}= 0$, $\;\;(\mathbf{I}-\mathbf{C}\mathbf{C}^T )\mathbf{B}= 0$, and $\mathbf{B} \succeq0$, do not follow from anything. W/out them the equivalence follows from the Sylvester criteria for positive definiteness and determinant formula for Schur complement. What is the source of the problem?</p>


问:
Does the sequence of functions converge uniformly?
$$f_n(x) = \begin{cases}
0,& x<0 \\
nx, &0 \leq x <1/n \\
1, &x \geq 1/n
\end{cases}$$

I suppose the sequence converges pointwise to $$f(x) = \begin{cases}
0,& x \leq0 \\
1, & x>0\\
\end{cases}$$

And the sequence will not converge uniformly on any domain with $0$ as a limit point. Is this reasoning correct?

答: Well, since the $f_n$ are continuous and the pointwise limit is not, the convergence cannot be uniform.  To do it from sratch, take $x_n=1/n^2$ and note that $|f_n(x_n)-f(x_n)|=|1/n-1|$


答: We tend to introduce $\log(x) = \log_{10}(x)$ because before you know anything about logarithms it's most convenient to think about things in terms of base $10$.  Once you learn enough about logs, you know that base $e$ is really the most convenient to work with, but at that point you've probably been introduced to the notation $\ln(x) = \log_{e}(x)$.  For many mathematicians, there comes a point when you realize the only logarithm you really care about is base $e$, so whenever possible it's convenient to just redefine $\log(x) = \log_{e}(x)$.


问:
Circle $O$ is tangential to triangle $ABC$ at $B$ and passes $C$. Express $AO$ as $\frac{a√b}{c√d}$
Let $ABC$ be a triangle with side lengths $AB = 7, BC = 8, AC = 9$. Draw a circle tangent to $AB$ at $B$ and passing through $C$. Let the center of the circle be $O$. The length of $AO$ can be expressed as $\frac{a√b}{c√d}$ for positive integers a, b, c, d where gcd(a, c) = gcd(b, d) = 1 and b, d are not divisible by the square of any prime. Find a + b + c + d.

I drew everything and cant find any patterns

答:
According to the cosine rule
$$\cos B = \frac{7^2+8^2-9^2}{2\cdot 7\cdot8}=\frac27,\>\>\>\>\>\sin B= \frac{3\sqrt5}7
$$
Then, the radius $OB$ is
$$OB = \frac{BC}{2\cos\angle OBC}=\frac4{\sin B}=\frac{28}{3\sqrt5}
$$
Per the Pythagoras' Theorem
$$AO=\sqrt{AB^2+BO^2}=\sqrt{7^2+ \frac{28^2}{45}}=\frac{7\sqrt{61}}{3\sqrt5}
$$
Thus, $a+b+c+d=76$.


问:
"Simplest" symbolic equation with "most complex" symbolic answer
The below was inspired by this question, in which the OP was surprised that the "simple" equation $x^3 + x = 1$ could have such a "complex" solution.

In rough form, the question here is a bit more general:

What is an example of the "simplest" symbolic equation whose solution is "most complex"?

That rough statement is of course inadequate to constrain or compare candidate answers.  Thus, to be specific, here are the conditions:

The equation involves a single variable, say $x$, and any number of constants.

The "complexity" of the problem equation and of the symbolic solution is the number of bytes in its expression (for instance in a MathJax representation).

The global criterion to be optimized is the ratio of the number of bytes in the solution to the number of bytes in the problem equation.

Both the problem and solution must be represented in their simplest canonical form.  For instance, one must use $\sin x$ rather than its infinite series representation.

All standard special functions and symbols (factorial !, power ^, etc.) are allowed, where the complexity is in the number of bytes in their canonical MathJax representation.

Both the problem and the solution must be symbolic, not numerical.  (Otherwise simple problem equations such as $x^2 = 2$, which have an "infinitely complex" numerical solution, would trivially optimize the criterion.)

There must be a true, single closed-form solution.  (Thus one cannot include, for example, $x^{9!} = 2$ and get a large list of solutions.)

The solution is not a simple numerical calculation, for instance $x = 9 \uparrow \uparrow \uparrow$ (using Knuth's uparrow notation) or $x = (9!)^{9!}$, etc.  As illustrated by the linked question above, it is rather the algebraic complexity of a solution.

One can get lost in nit picking (e.g., deciding which is to be preferred, \sqrt{x} versus x^{1/2}, or a/b versus \frac{a}{b}, and such).  I suspect there are some dramatic examples where such minutia are irrelevant.

答: The unique real root $x$ of $x^5-5x+12=0$ is given by $$-cx=\root5\of{(a+c)^2(b-c)}+\root5\of{(-a+c)(b-c)^2}+\root5\of{(a+c)(b+c)^2}-\root5\of{(-a+c)^2(b+c)}$$ where $c=\root4\of5$, $a=\sqrt{2\phi^{-1}}$, $b=\sqrt{2\phi}$, and $\phi=(1+\sqrt5)/2$, according to Wikipedia.


问:
Diophantic equation
What is the minimum integer $r$ such that for all integers $k\geq r$, there exist non- negative integers $x, y$ such that $k = 5x + 7y$?

答: Note how a line for target $23$ would pass through lattice points $(6,-1)$  and $(-1,4)$

Title: Diophantic equation


问:
Inverse function confusion
Im reading a book published in 1914 and there is a subject: differential of an inverse funtion. example: $y=3x$ and it says $x=\frac{y}{3}$ is an inverse function. I thought if $f$ is $y=3x$ its inverse is $x=3y$ or $y=\frac{x}{3}$. Can someone explain me please. Thanks!

答:
You're correct; it's ultimately a notational thing. For instance, your book solved for $x$ and left it as $$x = \frac y 3$$ Meanwhile, you did fundamentally the same thing, but swapped the roles of $x$ and $y$ after, to obtain $$y = \frac x 3$$ as the inverse function. What the book did, then, is write the inverse of the function as a function of $y$; you wrote it as a function of $x$. They're still representative of the same function, just notated differently.

Or, if $f(x) = 3x$, your book claims

$$f^{-1}(y) = \frac y 3$$

whereas you claim

$$f^{-1}(x) = \frac x 3$$

This sort of "equivalence up to your choice of how your name your variable" can be a bit confusing at first, but you'll get used to it over time.

Title: Inverse function confusion


问:
Radius of convergence of the Bernoulli polynomial generating function power series.
The generating function of the Bernoulli Polynomials is: $$\frac{te^{xt}}{e^t-1}=\sum_{k=0}^\infty B_k(x)\frac{t^k}{k!}.$$

Would it be right to say that the radius of convergence of this power series is $2\pi$ ? I'm not sure since the power series above is in fact a double series: $$\sum_{k=0}^\infty\left(\sum_{j=0}^k {k\choose j}B_{k-j} x^j\right)\frac{t^k}{k!}.$$

What if I were to choose a fixed value for $x$? Would the radius be $2\pi$ then, even for the double power series?

答:
I'd be inclined to construe "radius of convergence" as meaning the radius of convergence as a power series in $t$ rather than in $x$, since the $x$ is part of what it's a generating function of.

The numerator in $te^{tx}/(e^t-1)$, as a function of $t$ is an entire function, so the only singularities will be where the denominator is $0$, and then not necessarily if the numerator is also $0$ there.  The numerator is $0$ at $t=0$.   At that point the top and bottom both of a zero of multiplicity $1$, so its a removable singularity.  So it is indeed a power series (you're right-hand side is right).  The other places where $e^t-1$ is $0$ are $\pm n\pi$ for $n=1,2,3,\ldots$.  The distance from $0$ to the nearest of those is $2\pi$, so that is the radius of convergence.

If you look at the left side as a function of $x$, it's an entire function, so as a power series in $x$ you get radius $\infty$.

What happens when you view it as a function of both $x$ and $t$ is more than I will attempt to say anything about here.  (I haven't thought about that.)


答:
Let $f$ be a function of $x$, for example, $f(x) = 3x$.
What this statement means is that the value of $y$ is a function of $x$. In the example, the value of $y$ is always 3 times the value of $x$. That's why $f(x) = 3x$ and $y = 3x$ are equivalent expressions.

When we say inverse function, we don't want $y$ as function of $x$, we want $x$ as a function of $y$. This means that we want an equation which has $x$ on one side and $y$ on the other, i.e. $x = y/3$, just like your books writes. In this case, $x = y/3$ and $f^{-1}(x) = y/3$ are equivalent expressions with the exact same meaning.

Thats what an inverse function means, $x$ as a function of $y$.

Title: Inverse function confusion

Body: Im reading a book published in 1914 and there is a subject: differential of an inverse funtion. example: $y=3x$ and it says $x=\frac{y}{3}$ is an inverse function. I thought if $f$ is $y=3x$ its inverse is $x=3y$ or $y=\frac{x}{3}$. Can someone explain me please. Thanks!


问:
Is $C$ an infinite set?
Suppose $H:\mathbb{N}\times\mathbb{N}\mapsto\mathbb{N}$ is a injective map. Let $C=\{c|c=H(i,j)-H(i-1,j)$ or $H(i,j)-H(i,j-1)\}$. Is $C$ an infinite set?

答:
Suppose that $|H(i,j)-H(i-1,j)|,|H(i,j)-H(i,j-1)|\le m$ for all $i,j\in\Bbb N$. If $0\le i,j\le k\in\Bbb N$, then

$$H(0,0)-2km\le H(i,j)\le H(0,0)+2km\,.$$

(My $\Bbb N$ includes $0$.) Thus, there are only $4km+1$ possible values for $H(i,j)$ with $i,j\le k$, but there are $(k+1)^2$ pairs $\langle i,j\rangle$ with $i,j\le k$. Clearly $(k+1)^2>4km+1$ for sufficiently large $k$, so $C$ must be infinite.

Title: Is $C$ an infinite set?


答:
Will Jagy's picture suggests that $23$ cannot expressed as $7x+5y$ with non-negative integers $x$ and $y$.  Indeed, with $x=0$ we can express $0,5,10,15,20$; with $x=1$ we can express $7,12,17,22$; with $x=2$ we can express $14,19$; and with $x=3$ we can express $21$; and no other numbers less than $24$.

On the other hand, $24=7\times2+5\times2$, $25=5\times5$, $26=7\times3+5$, $27=7+5\times4$, and $28=7\times4$, and we can add multiples of $5$ to those numbers to get any higher numbers.

Title: Diophantic equation

Body: What is the minimum integer $r$ such that for all integers $k\geq r$, there exist non- negative integers $x, y$ such that $k = 5x + 7y$?


问:
Questions about the sketch of the proof : every Projective variety is complete
I cannot understand the following sketch of the proof: Every projective variety is complete using Nakayama Lemma.
The following lecture note on the 279 pages is one recently
that I have referred.

To show this, the lecturer defines the set

$$\mathcal{V}^{*}(I):=\left \{ (x^{*},y) \in \mathbb{P}^n \times Y : f(x^{*},y)=0 \right \}, $$

and since it is a closed set, then it is sufficient to show that $\pi(\mathcal{V}^{*}(I))$ is closed, or equivalently, $\overline{\pi(\mathcal{V}^{*}(I)) }=\pi(\mathcal{V}^{*}(I)) $...... (★).

To begin with, in my thoughts, (★) seems to hold: $\mathcal{V}^{*}(I) $ is closed, i.e consists of finite elements, and its projection is also finite on $Y$ whose topology is Zariski topology, so $\pi(\mathcal{V}^{*}(I)) $ is still closed.

However, the author also said "we must show for every $y \in Y$, there exists $x^{*} \in \mathbb{P}^n$ so that $(x^{*},y) \in \mathcal{V}^{*}(I) $".
In my thoughts, the author seems to verify that $\mathcal{V}^{*}(I)$ is a nonempty set, and to ascertain this, the author takes a maximal ideal $\mathfrak{m} \subset A $ that vanishes at $y$, and sets $J:=\mathfrak{m}B+I $.

Then he says "if $\mathcal{V}^{*}(J)$ is non-empty, we'll be done" (I understand this paragraph is an alternative way of showing that  $\mathcal{V}^{*}(I) \neq \phi$).
Here the author uses Nakayama's Lemma in order to verify $\mathcal{V}^{*}(J)\neq 0$.

I have some questions regarding the proof.

I am little confused on the statement that $\mathcal{V}^{*}(I)$ is closed in $\mathbb{P}^n \times Y$. I believe that it is also closed in the Zariski topology. Since the product topology is the weakest topology (or initial topology), then the product of the Zariski topology is stronger than box topology. Is this the right idea? Anyway, if we take $\mathcal{V}^{*}(I)$ in the Zariski topology, then it is closed in $\mathbb{P}^n \times Y$ because  $\mathcal{V}^{*}(I)$ is a finite set.

The author showed that $\mathcal{V}^{*}(J)\neq \phi$ (As I said, by guesswork, I firmly believe that this is an alternative proof of  $\mathcal{V}^{*}(I)\neq \phi $).
He proved this by contradiction: assuming that $\mathcal{V}^{*}(J)= \phi$, the author induces $\mathcal{V}^{*}(I)= \phi$. I can indirectly infer this by Nakayama's Lemma:  $J=\mathfrak{m}B+I $ implies $J=I$ where $I$ is a submodule of a finitely generated $A$-module.
Hence $$\phi=\mathcal{V}^{*}(J) \underset{Nakyama }{=} \mathcal{V}^{*}(I).$$
He then gives the contradiction "but it is impossible for a proper ideal $I \subset B$".
I do not understand why the author carried out the proof by applying Nakayama's Lemma, defining the set $J$ etc., because since $I$ is a proper ideal of $B$, the relation $\mathcal{V}^{*}(I)\neq \phi $ is always satisfied.
I have tried reasoning it out as follows:

It must be that $\mathcal{V}^{*}(I)\neq \phi$, since $\mathcal{V}^{*}(I)= \phi$ is
impossible for a proper ideal $I \subset B$. Hence,
$\mathcal{V}^{*}(I)\neq \phi$.

答:
Let's outline how the proof is supposed to go first - parts of your questions are based on misconceptions about the strategy. The idea is that we need to show that for any $Y$ and any nonempty closed $C\subset \Bbb P^n\times Y$, the image of $C$ in $Y$ under the projection is closed. So we want to consider the minimal possible situation: shrink $Y$ so that $Y=\overline{\pi(C)}$, therefore checking whether $\pi(C)$ is closed amounts to checking that $C\to \overline{\pi(C)}$ surjective. This is because $\overline{\pi(C)}$ is the smallest closed subset containing $\pi(C)$, so $\pi(C)$ is closed iff $\pi(C)=\overline{\pi(C)}$.

To show that $\pi(C)=\overline{\pi(C)}$, it's enough to show that there's a point of $C$ in the fiber $\pi^{-1}(y)$ for all $y\in Y=\overline{\pi(C)}$. Writing $C$ as $\mathcal{V}^*(I)$ for some choice of homogeneous ideal $I\subset B$, $\mathcal{V}^*(J)$ is the intersection of $\mathcal{V}^*(I)$ with the fiber $\pi^{-1}(y)$ - the strategy of the proof is to show that if $\mathcal{V}^*(J)$ is empty, then $\mathcal{V}^*(I)$ must be as well, which will be a contradiction because we took $C=\mathcal{V}^*(I)$ to be nonempty.

Aside from the issues addressed above, your post has some other misconceptions I believe should be corrected. Here goes:

To begin with, in my thoughts, (★) seems to hold: $\mathcal{V}^{*}(I) $ is closed, i.e consists of finite elements, and its projection is also finite on $Y$ whose topology is Zariski topology, so $\pi(\mathcal{V}^{*}(I)) $ is still closed.

Yes, $\mathcal{V}^*(I)$ is closed, but this business about "finite elements" does not make sense (what even are "finite elements"?). If you knew that the projection $\pi(\mathcal{V}^{*}(I))$ to $Y$ were closed, you would be done: every closed subset of $\Bbb P^n\times Y$ is of this form. You're getting ahead of yourself here.

I am little confused on the statement that $\mathcal{V}^{*}(I)$ is closed in $\mathbb{P}^n \times Y$. I believe that it is also closed in the Zariski topology. Since the product topology is the weakest topology (or initial topology), then the product of the Zariski topology is stronger than box topology. Is this the right idea? Anyway, if we take $\mathcal{V}^{*}(I)$ in the Zariski topology, then it is closed in $\mathbb{P}^n \times Y$ because  $\mathcal{V}^{*}(I)$ is a finite set.

When we consider the product of varieties, we put the Zariski topology on it, not the product topology. The statement that "$\mathcal{V}^{*}(I)$ is closed in $\mathbb{P}^n \times Y$" is taking $\Bbb P^n\times Y$ to have the Zariski topology. I do not understand why you think " $\mathcal{V}^{*}(I)$ is a finite set" - this is false and you should let go of this belief. Consider  $Y=\Bbb A^2$ and $n=1$, so $B=k[t_1,t_0][x_0,x_1]$ with $x_0,x_1$ homogeneous of degree one and $t_0,t_1$ homogeneous of degree zero. Then $\mathcal{V}^*((t_0))$ is a copy of $\Bbb P^1\times\Bbb A^1$ which is not finite.


答:
If $A$ is invertible, then $A$ is injective, which implies that $Null(A)=\{0\}$. Thus, the only solution to $Av=0=0\cdot v$ is $v=0$. Hence, $0$ is not an eigenvalue.

Conversely, if $0$ is not an eigenvalue, then $A=A-0I$ must be injective (otherwise, 0 is an eigenvalue. Theorem 5.6 in Axler), which implies $A$ must be invertible since $A$ is a linear operator.

Body:
I would appreciate feedback on the correctness of this proof, as well as alternative proofs.

Prove: A is invertible IFF 0 is not an eigenvalue of A.

Proof:If $A$ is invertible, letting $λ=0$ yields $Ax-λx=Ax=0$ which implies that $x=0$. So, $0$ is not an eigenvalue. If $0$ is not an eigenvalue then there is no non-zero $x$ such that $Ax=0$, and because there is only the trivial solution, A must be invertible.


问:
Atiyah-Macdonald Proposition 4.7
We have the following proposition in Atiyah-Macdonald:

Let $\mathfrak{a}$ be a decomposable ideal in a ring $A$, let $\mathfrak{a} = \bigcap_{i=1}^n\mathfrak{q}_i$ be a minimal primary decomposition, and let $r(\mathfrak{q}_i) = \mathfrak{p}_i$. Then
$$
\bigcup_{i=1}^n \mathfrak{p}_i = \{x \in A : (\mathfrak{a}:x) \neq \mathfrak{a}\}.
$$
In particular, if the zero ideal is decomposable, the set $D$ of zero-divisors of $A$ is the union of the prime ideals belonging to $0$.

I am unsure about the proof. I understand that if $\mathfrak{a}$ is decomposible, then the image (which is 0) will be decomposible in $A/\mathfrak{a}$. But I don't see how the conclusion being true in the quotient allows us to lift the result to the whole ring.

If we know the result in the quotient, namely we know that: $\cup \overline{p_i}=\{\overline{x} \in A/\mathfrak{a}: (0:\overline{x})\neq 0\}$ how can we conclude that $\cup p_i=\{x \in A/: (\mathfrak{a}:x)\neq \mathfrak{a}\}$

There is a question that is similarly titled [here] (Proposition 4.7 in Atiyah-Macdonald) but we have different things we are confused about.

答:
I guess an important point is that because $\mathfrak a\subseteq\mathfrak q_i\subseteq\mathfrak p_i$ for each $i$ we have that $\pi^{-1}(\overline{\mathfrak p_i})=\mathfrak p_i$ where $\pi$ is the projection $A\to A/\mathfrak a$ (in general you'll have that $\pi^{-1}(\overline I)=I+\mathfrak a$, so you get equality when $\mathfrak a\subseteq I$).

Now in general a pre-image of a union equals the union of the pre-images, so you get

$$\bigcup_i \mathfrak p_i=\bigcup_i\pi^{-1}(\overline{\mathfrak p_i})=\pi^{-1}(\bigcup_i\overline{\mathfrak p_i})=\pi^{-1}(\{\overline x\in A/\mathfrak a:(0:\overline x)\neq0\}),$$

and it's easy to check this last set is equal to $\{x\in A:(\mathfrak a:x)\neq\mathfrak a\}$.

Title: Atiyah-Macdonald Proposition 4.7


问:
Is it possible turn the Dirichlet ring into a Banach algebra?
The set of all arithmetic functions $f:\mathbb{Z}^{+}\to\mathbb{C}$, under pointwise addition and Dirichlet convolution, is a commutative ring, not all functions are Dirichlet invertible.

So my question is:
To what subset of the arithmetic functions and under what norms can one attach a Banach algebra structure to the Dirichlet ring? Is it even possible?

答: In fact $$\|a\| = \sum_{n=1}^\infty |a_n n^{-\sigma}|$$ is a bit more natural than my other answer.. It gives the Banach algebra of absolutely convergent Dirichlet series on $\Re(s)\ge \sigma$. There are interesting theorems about its invertible elements.


问:
the limit of $\frac{1}{\sqrt{n}}(1+\frac{2}{1+\sqrt2}+\frac{3}{1+\sqrt2+\sqrt3}+\dots+\frac{n}{1+\sqrt2+\sqrt3+\dots+\sqrt n})$ as $n\to\infty$
I need to find:

$$\lim_{n \to +\infty} \frac{1}{\sqrt{n}}(1+\frac{2}{1+\sqrt{2}} + \frac{3}{1+\sqrt{2}+\sqrt{3}} + \ldots + \frac{n}{1+\sqrt{2}+\sqrt{3}+\ldots+\sqrt{n}}), n \in \mathbb{N}$$

Looking at denominators, I see that [(...) represents any element between] : $$ 1 \le (\ldots) \le 1+\sqrt{2}+\sqrt{3}+\ldots+\sqrt{n}$$

Then I take reverses and get:
$$ 1 \ge (\ldots) \ge \frac{1}{1+\sqrt{2}+\sqrt{3}+\ldots+\sqrt{n}}$$

Then I put the other sequence on top of the former one (I see that the rightmost element is still the smallest one)
$$ 1 \ge (\ldots) \ge \frac{n}{1+\sqrt{2}+\sqrt{3}+\ldots+\sqrt{n}}$$

Then I take the sum of n elements on every end of inequality (to sum up n times the biggest element and n times the smallest element) and get:
$$ n \ge (\ldots) \ge \frac{n^2}{1+\sqrt{2}+\sqrt{3}+\ldots+\sqrt{n}}$$

Ultimately I take into consideration $\frac{1}{\sqrt{2}}$ and get:
$$ \frac{n}{\sqrt{n}} \ge (\ldots) \ge \frac{n^2}{\sqrt{n}(1+\sqrt{2}+\sqrt{3}+\ldots+\sqrt{n})}$$

Now I can use the squeeze theorem and get:

$\lim_{n \to +\infty} \frac{n}{\sqrt{n}} = \infty$

$\lim_{n \to +\infty} \frac{n^2}{\sqrt{n}(1+\sqrt{2}+\sqrt{3}+\ldots+\sqrt{n})} = \lim_{n \to +\infty} \frac{\frac{n^2}{\sqrt{n}}}{(1+\sqrt{2}+\sqrt{3}+\ldots+\sqrt{n})} \implies Stolz = \lim_{n \to +\infty} \frac{\frac{(n+1)^2}{\sqrt{n+1}}-\frac{(n)^2 }{\sqrt{n}}}{\sqrt{n+1}}$

And that is pretty disappointing - I think that the solution is wrong. Does anybody see an error in my way of thinking?

Unfortunately, I can not use integrals while doing that exercise.

答:
The Stolz–Cesàro theorem indeed suffices, with a derivative-like limit (the last one below).

Let $a_n=\sum\limits_{k=1}^n\sqrt{k}$ and $b_n=\sum\limits_{k=1}^n k/a_k$; we're computing
\begin{align*}
\lim_{n\to\infty}\frac{b_n}{\sqrt{n}}
&=\lim_{n\to\infty}\frac{b_n-b_{n-1}}{\sqrt{n}-\sqrt{n-1}}
\\&=\lim_{n\to\infty}\frac{n}{a_n}(\sqrt{n}+\sqrt{n-1})
\\&=2\lim_{n\to\infty}\frac{n^{3/2}}{a_n}
\\&=2\lim_{n\to\infty}\frac{n^{3/2}-(n-1)^{3/2}}{a_n-a_{n-1}}
\\&=2\lim_{n\to\infty}\frac{n^{3/2}-(n-1)^{3/2}}{\sqrt{n}}
\\&=\color{gray}{2\lim_{n\to\infty}\frac{1-(1-1/n)^{3/2}}{1/n}}=\mathbf{3}.
\end{align*}


答:
Notation is not nice but lets say for numbers, $\pm$ means $\{n\pm c\} = \{ n-c,n-c+1,\cdots,n+c\}$

Let me only consider the set $\{H(i,j)-H(i-1,j)\}$.

Note that, for fixed $j$, $H(i,j) \to \infty$ as $i\to \infty.$ Assume $H(i,j) - H(i-1,j) < c$.

Consider the set $\{n\pm c\}$, and note that for any $j$ there exists $n$ s.t. $H(1,j)\in \{n\pm c\}$ and further $H(i',j)\in \{n+k\pm c\}$ since they cannot make jumps larger than $c$ and it must tend to $\infty$.

Therefore, $|\{H(i,j)\in\{n \pm c \}\}|\to \infty$ as $n\to\infty$, which contradicts with injective.

Title: Is $C$ an infinite set?

Body: Suppose $H:\mathbb{N}\times\mathbb{N}\mapsto\mathbb{N}$ is a injective map. Let $C=\{c|c=H(i,j)-H(i-1,j)$ or $H(i,j)-H(i,j-1)\}$. Is $C$ an infinite set?


Title: the limit of $\frac{1}{\sqrt{n}}(1+\frac{2}{1+\sqrt2}+\frac{3}{1+\sqrt2+\sqrt3}+\dots+\frac{n}{1+\sqrt2+\sqrt3+\dots+\sqrt n})$ as $n\to\infty$

Body:
I need to find:

$$\lim_{n \to +\infty} \frac{1}{\sqrt{n}}(1+\frac{2}{1+\sqrt{2}} + \frac{3}{1+\sqrt{2}+\sqrt{3}} + \ldots + \frac{n}{1+\sqrt{2}+\sqrt{3}+\ldots+\sqrt{n}}), n \in \mathbb{N}$$

Looking at denominators, I see that [(...) represents any element between] : $$ 1 \le (\ldots) \le 1+\sqrt{2}+\sqrt{3}+\ldots+\sqrt{n}$$

Then I take reverses and get:
$$ 1 \ge (\ldots) \ge \frac{1}{1+\sqrt{2}+\sqrt{3}+\ldots+\sqrt{n}}$$

Then I put the other sequence on top of the former one (I see that the rightmost element is still the smallest one)
$$ 1 \ge (\ldots) \ge \frac{n}{1+\sqrt{2}+\sqrt{3}+\ldots+\sqrt{n}}$$

Then I take the sum of n elements on every end of inequality (to sum up n times the biggest element and n times the smallest element) and get:
$$ n \ge (\ldots) \ge \frac{n^2}{1+\sqrt{2}+\sqrt{3}+\ldots+\sqrt{n}}$$

Ultimately I take into consideration $\frac{1}{\sqrt{2}}$ and get:
$$ \frac{n}{\sqrt{n}} \ge (\ldots) \ge \frac{n^2}{\sqrt{n}(1+\sqrt{2}+\sqrt{3}+\ldots+\sqrt{n})}$$

Now I can use the squeeze theorem and get:

$\lim_{n \to +\infty} \frac{n}{\sqrt{n}} = \infty$

$\lim_{n \to +\infty} \frac{n^2}{\sqrt{n}(1+\sqrt{2}+\sqrt{3}+\ldots+\sqrt{n})} = \lim_{n \to +\infty} \frac{\frac{n^2}{\sqrt{n}}}{(1+\sqrt{2}+\sqrt{3}+\ldots+\sqrt{n})} \implies Stolz = \lim_{n \to +\infty} \frac{\frac{(n+1)^2}{\sqrt{n+1}}-\frac{(n)^2 }{\sqrt{n}}}{\sqrt{n+1}}$

And that is pretty disappointing - I think that the solution is wrong. Does anybody see an error in my way of thinking?

Unfortunately, I can not use integrals while doing that exercise.

Body:
$\sum_{k=1}^n k^{1/2}
\approx \int_0^n x^{1/2} dx
=\dfrac{n^{3/2}}{3/2}
=\dfrac23 n^{3/2}
$
and
$\sum_{k=1}^n k^{-1/2}
\approx \int_0^n x^{-1/2} dx
=\dfrac{n^{1/2}}{1/2}
=2n^{1/2}
$

so

$\begin{array}\\
s(n)
&=1+\frac{2}{1+\sqrt{2}} + \frac{3}{1+\sqrt{2}+\sqrt{3}} + \ldots + \frac{n}{1+\sqrt{2}+\sqrt{3}+\ldots+\sqrt{n}})\\
&=\sum_{k=1}^n \dfrac{k}{\sum_{j=1}^k j^{1/2}}\\
&\approx\sum_{k=1}^n \dfrac{k}{\dfrac23 k^{3/2}}\\
&=\dfrac32\sum_{k=1}^n k^{-1/2}\\
&\approx\dfrac32(2n^{1/2})\\
&=3n^{1/2}\\
\end{array}
$

so the limit is $3$.